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"ELECTROMAGNETICS
STEVEN W. ELLINGSON
VOLUME 2",Electromagnetics_Vol2.pdf
"ELECTROMAGNETICS
VOLUME 2",Electromagnetics_Vol2.pdf
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/T_he Open Electromagnetics Project, https://www.faculty.ece.vt.edu/swe/oem",Electromagnetics_Vol2.pdf
"ELECTROMAGNETICS
STEVEN W. ELLINGSON
VOLUME 2",Electromagnetics_Vol2.pdf
"Copyright © 2020 Steven W. Ellingson
iv
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Ellingson, Steven W., author
     Electromagnetics (V olume 2) / Steven W. Ellingson
     Pages cm
     ISBN 978-1-949373-91-2 (print)
     ISBN 978-1-949373-92-9 (ebook)
     DOI: https://doi.org/10.21061/electromagnetics-vol-2
     1.  Electromagnetism.  2. Electromagnetic theory.
     I.   Title
     QC760.E445 2020
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"Contents
Pr
eface ix
1 Preliminary Concepts 1
1.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Electromagnetic Field Theory: A Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Magnetostatics Redux 11
2.1 Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Magnetic Force on a Current-Carrying Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 T orque Induced by a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 The Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18",Electromagnetics_Vol2.pdf
"2.4 The Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.5 Force, Energy , and Potential Difference in a Magnetic Field . . . . . . . . . . . . . . . . . . . 20
3 W ave Propagation in General Media 25
3.1 Poynting’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.2 Poynting V ector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.3 W ave Equations for Lossy Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.4 Complex Permittivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.5 Loss T angent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.6 Plane W aves in Lossy Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.7 W ave Power in a Lossy Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37",Electromagnetics_Vol2.pdf
"3.7 W ave Power in a Lossy Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.8 Decibel Scale for Power Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.9 Attenuation Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.10 Poor Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.11 Good Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.12 Skin Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4 Current Flow in Imperfect Conductors 48
4.1 AC Current Flow in a Good Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.2 Impedance of a Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.3 Surface Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54",Electromagnetics_Vol2.pdf
"4.3 Surface Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
5 W ave Reflection and T ransmission 56
5.1 Plane W aves at Normal Incidence on a Planar Boundary . . . . . . . . . . . . . . . . . . . . . 56
5.2 Plane W aves at Normal Incidence on a Material Slab . . . . . . . . . . . . . . . . . . . . . . 60
5.3 T otal Transmission Through a Slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
5.4 Propagation of a Uniform Plane W ave in an Arbitrary Direction . . . . . . . . . . . . . . . . . 67
vi",Electromagnetics_Vol2.pdf
"CONTENTS vii
5.5 Decomposition of a W ave into TE and TM Components . . . . . . . . . . . . . . . . . . . . . 70
5.6 Plane W aves at Oblique Incidence on a Planar Boundary: TE Case . . . . . . . . . . . . . . . 72
5.7 Plane W aves at Oblique Incidence on a Planar Boundary: TM Case . . . . . . . . . . . . . . . 76
5.8 Angles of Reflection and Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
5.9 TE Reflection in Non-magnetic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
5.10 TM Reflection in Non-magnetic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
5.11 T otal Internal Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
5.12 Evanescent W aves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
6 W aveguides 95
6.1 Phase and Group V elocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95",Electromagnetics_Vol2.pdf
"6 W aveguides 95
6.1 Phase and Group V elocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
6.2 Parallel Plate W aveguide: Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
6.3 Parallel Plate W aveguide: TE Case, Electric Field . . . . . . . . . . . . . . . . . . . . . . . . 99
6.4 Parallel Plate W aveguide: TE Case, Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . 102
6.5 Parallel Plate W aveguide: TM Case, Electric Field . . . . . . . . . . . . . . . . . . . . . . . . 104
6.6 Parallel Plate W aveguide: The TM 0 Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.7 General Relationships for Unidirectional W aves . . . . . . . . . . . . . . . . . . . . . . . . . 108
6.8 Rectangular W aveguide: TM Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
6.9 Rectangular W aveguide: TE Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113",Electromagnetics_Vol2.pdf
"6.9 Rectangular W aveguide: TE Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6.10 Rectangular W aveguide: Propagation Characteristics . . . . . . . . . . . . . . . . . . . . . . 117
7 T ransmission Lines Redux 121
7.1 Parallel Wire Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.2 Microstrip Line Redux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
7.3 Attenuation in Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
7.4 Power Handling Capability of Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
7.5 Why 50 Ohms? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
8 Optical Fiber 138
8.1 Optical Fiber: Method of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138",Electromagnetics_Vol2.pdf
"8 Optical Fiber 138
8.1 Optical Fiber: Method of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
8.2 Acceptance Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
8.3 Dispersion in Optical Fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
9 Radiation 145
9.1 Radiation from a Current Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
9.2 Magnetic V ector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
9.3 Solution of the W ave Equation for Magnetic V ector Potential . . . . . . . . . . . . . . . . . . 150
9.4 Radiation from a Hertzian Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
9.5 Radiation from an Electrically-Short Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
9.6 Far-Field Radiation from a Thin Straight Filament of Current . . . . . . . . . . . . . . . . . . 159",Electromagnetics_Vol2.pdf
"9.6 Far-Field Radiation from a Thin Straight Filament of Current . . . . . . . . . . . . . . . . . . 159
9.7 Far-Field Radiation from a Half-W ave Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . 161
9.8 Radiation from Surface and V olume Distributions of Current . . . . . . . . . . . . . . . . . . 162
10 Antennas 166
10.1 How Antennas Radiate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
10.2 Power Radiated by an Electrically-Short Dipole . . . . . . . . . . . . . . . . . . . . . . . . . 168
10.3 Power Dissipated by an Electrically-Short Dipole . . . . . . . . . . . . . . . . . . . . . . . . 169
10.4 Reactance of the Electrically-Short Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
10.5 Equivalent Circuit Model for Transmission; Radiation Efficiency . . . . . . . . . . . . . . . . 173
10.6 Impedance of the Electrically-Short Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . 175",Electromagnetics_Vol2.pdf
"viii CONTENTS
10.7 Directivity and Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
10.8 Radiation Pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
10.9 Equivalent Circuit Model for Reception . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
10.10 Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
10.11 Potential Induced in a Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
10.12 Equivalent Circuit Model for Reception, Redux . . . . . . . . . . . . . . . . . . . . . . . . . 194
10.13 Effective Aperture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
10.14 Friis Transmission Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
A Constitutive Parameters of Some Common Materials 205",Electromagnetics_Vol2.pdf
"A Constitutive Parameters of Some Common Materials 205
A.1 Permittivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
A.2 Permeability of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
A.3 Conductivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
B Mathematical Formulas 209
B.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
B.2 V ector Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
B.3 V ector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
C Physical Constants 212
Index 213",Electromagnetics_Vol2.pdf
"Preface
About
This Book
[m0213]
Goals for this book. This book is intended to serve
as a primary textbook for the second semester of a
two-semester course in undergraduate engineering
electromagnetics. The presumed textbook for the first
semester is Electromagnetics V ol. 1, 1 which addresses
the following topics: electric and magnetic fields;
electromagnetic properties of materials;
electromagnetic waves; and devices that operate
according to associated electromagnetic principles
including resistors, capacitors, inductors,
transformers, generators, and transmission lines. The
book you are now reading – Electromagnetics V ol. 2 –
addresses the following topics:
• Chapter 1 (“Preliminary Concepts”) provides a
brief summary of conventions for units, notation,
and coordinate systems, and a synopsis of
electromagnetic field theory from V ol. 1.
• Chapter 2 (“Magnetostatics Redux”) extends the
coverage of magnetostatics in V ol. 1 to include
magnetic forces, rudimentary motors, and the",Electromagnetics_Vol2.pdf
"• Chapter 2 (“Magnetostatics Redux”) extends the
coverage of magnetostatics in V ol. 1 to include
magnetic forces, rudimentary motors, and the
Biot-Savart law .
• Chapter 3 (“W ave Propagation in General
Media”) addresses Poynting’s theorem, theory of
wave propagation in lossy media, and properties
of imperfect conductors.
• Chapter 4 (“Current Flow in Imperfect
Conductors”) addresses the frequency-dependent
distribution of current in wire conductors and
subsequently the AC impedance of wires.
1 S.W . Ellingson, Electromagnetics V ol. 1, VT Publishing, 2018.
CC BY -SA 4.0. ISBN 9780997920185.
https://doi.org/10.21061/electromagnetics- vol- 1
• Chapter 5 (“W ave Reflection and Transmission”)
addresses scattering of plane waves from planar
interfaces.
• Chapter 6 (“W aveguides”) provides an
introduction to waveguide theory via the parallel
plate and rectangular waveguides.
• Chapter 7 (“Transmission Lines Redux”) extends
the coverage of transmission lines in V ol. 1 to",Electromagnetics_Vol2.pdf
"plate and rectangular waveguides.
• Chapter 7 (“Transmission Lines Redux”) extends
the coverage of transmission lines in V ol. 1 to
include parallel wire lines, the theory of
microstrip lines, attenuation, and power-handling
capabilities. The inevitable but hard-to-answer
question “What’s so special about 50 Ω?” is
addressed at the end of this chapter.
• Chapter 8 (“Optical Fiber”) provides an
introduction to multimode fiber optics, including
the concepts of acceptance angle and modal
dispersion.
• Chapter 9 (“Radiation”) provides a derivation of
the electromagnetic fields radiated by a current
distribution, emphasizing the analysis of line
distributions using the Hertzian dipole as a
differential element.
• Chapter 10 (“ Antennas”) provides an
introduction to antennas, emphasizing equivalent
circuit models for transmission and reception
and characterization in terms of directivity and
pattern. This chapter concludes with the Friis
transmission equation.",Electromagnetics_Vol2.pdf
"and characterization in terms of directivity and
pattern. This chapter concludes with the Friis
transmission equation.
Appendices covering material properties,
mathematical formulas, and physical constants are
repeated from V ol. 1, with a few additional items.
T arget audience. This book is intended for electrical
engineering students in the third year of a bachelor of
science degree program. It is assumed that students
have successfully completed one semester of
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"x PREF ACE
engineering electromagnetics, nominally using V ol. 1.
However, the particular topics and sequence of topics
in V ol. 1 are not an essential prerequisite, and in any
event this book may be useful as a supplementary
reference when a different textbook is used. It is
assumed that readers are familiar with the
fundamentals of electric circuits and linear systems,
which are normally taught in the second year of the
degree program. It is also assumed that readers are
proficient in basic engineering mathematics, including
complex numbers, trigonometry , vectors, partial
differential equations, and multivariate calculus.
Notation, examples, and highlights. Section 1.2
summarizes the mathematical notation used in this
book. Examples are set apart from the main text as
follows:
Example 0.1. This is an example.
“Highlight boxes” are used to identify key ideas as
follows:
This is a key idea.
What
are those little numbers in square brackets?
This book is a product of the",Electromagnetics_Vol2.pdf
"follows:
This is a key idea.
What
are those little numbers in square brackets?
This book is a product of the
Open Electromagnetics Project . This project provides
a large number of sections (“modules”) which are
assembled (“remixed”) to create new and different
versions of the book. The text “ [m0213]” that you see at
the
beginning of this section uniquely identifies the
module within the larger set of modules provided by
the project. This identification is provided because
different remixes of this book may exist, each
consisting of a different subset and arrangement of
these modules. Prospective authors can use this
identification as an aid in creating their own remixes.
Why do some sections of this book seem to repeat
material presented in previous sections? In some
remixes of this book, authors might choose to
eliminate or reorder modules. For this reason, the
modules are written to “stand alone” as much as
possible. As a result, there may be some redundancy",Electromagnetics_Vol2.pdf
"modules are written to “stand alone” as much as
possible. As a result, there may be some redundancy
between sections that would not be present in a
traditional (non-remixable) textbook. While this may
seem awkward to some at first, there are clear
benefits: In particular, it never hurts to review relevant
past material before tackling a new concept. And,
since the electronic version of this book is being
offered at no cost, there is not much gained by
eliminating this useful redundancy .
Why cite Wikipedia pages as additional reading?
Many modules cite Wikipedia entries as sources of
additional information. Wikipedia represents both the
best and worst that the Internet has to offer. Most
educators would agree that citing Wikipedia pages as
primary sources is a bad idea, since quality is variable
and content is subject to change over time. On the
other hand, many Wikipedia pages are excellent, and
serve as useful sources of relevant information that is",Electromagnetics_Vol2.pdf
"other hand, many Wikipedia pages are excellent, and
serve as useful sources of relevant information that is
not strictly within the scope of the curriculum.
Furthermore, students benefit from seeing the same
material presented differently , in a broader context,
and with the additional references available as links
from Wikipedia pages. W e trust instructors and
students to realize the potential pitfalls of this type of
resource and to be alert for problems.
Acknowledgments. Here’s a list of talented and
helpful people who contributed to this book:
The staff of V irginia T ech Publishing, University
Libraries, V irginia T ech:
Acquisitions/Developmental Editor & Project
Manager: Anita W alz
Advisors: Peter Potter, Corinne Guimont
Cover, Print Production: Robert Browder
Other V irginia T ech contributors:
Accessibility: Christa Miller, Corinne Guimont,
Sarah Mease
Assessment: Anita W alz
V irginia T ech Students:
Alt text writer: Michel Comer",Electromagnetics_Vol2.pdf
"Accessibility: Christa Miller, Corinne Guimont,
Sarah Mease
Assessment: Anita W alz
V irginia T ech Students:
Alt text writer: Michel Comer
Figure designers: Kruthika Kikkeri, Sam Lally ,
Chenhao W ang
Copyediting:
Longleaf Press
External reviewers:
Randy Haupt, Colorado School of Mines
Karl W arnick, Brigham Y oung University
Anonymous faculty member, research university",Electromagnetics_Vol2.pdf
"xi
Also,
thanks are due to the students of the Fall 2019
section of ECE3106 at V irginia T ech who used the
beta version of this book and provided useful
feedback.
Finally , we acknowledge all those who have
contributed their art to Wikimedia Commons
(https://commons.wikimedia.org/) under open
licenses, allowing their work to appear as figures in
this book. These contributors are acknowledged in
figures and in the “Image Credits” section at the end
of each chapter. Thanks to each of you for your
selfless effort.
About the Open Electromagnetics
Project
[m0148]
The Open Electromagnetics Project
(https://www
.faculty .ece.vt.edu/swe/oem/) was
established at V irginia T ech in 2017 with the goal of
creating no-cost openly-licensed textbooks for
courses in undergraduate engineering
electromagnetics. While a number of very fine
traditional textbooks are available on this topic, we
feel that it has become unreasonable to insist that
students pay hundreds of dollars per book when",Electromagnetics_Vol2.pdf
"feel that it has become unreasonable to insist that
students pay hundreds of dollars per book when
effective alternatives can be provided using modern
media at little or no cost to the student. This project is
equally motivated by the desire for the freedom to
adopt, modify , and improve educational resources.
This work is distributed under a Creative Commons
BY SA license which allows – and we hope
encourages – others to adopt, modify , improve, and
expand the scope of our work.",Electromagnetics_Vol2.pdf
"xii PREF ACE
About the Author
[m0153]
Steven W . Ellingson (ellingson@vt.edu) is an
Associate
Professor at V irginia T ech in Blacksburg,
V irginia, in the United States. He received PhD and
MS degrees in Electrical Engineering from the Ohio
State University and a BS in Electrical and Computer
Engineering from Clarkson University . He was
employed by the U.S. Army , Booz-Allen & Hamilton,
Raytheon, and the Ohio State University
ElectroScience Laboratory before joining the faculty
of V irginia T ech, where he teaches courses in
electromagnetics, radio frequency electronics,
wireless communications, and signal processing. His
research includes topics in wireless communications,
radio science, and radio frequency instrumentation.
Professor Ellingson serves as a consultant to industry
and government and is the author of Radio Systems
Engineering (Cambridge University Press, 2016) and
Electromagnetics V ol. 1 (VT Publishing, 2018).",Electromagnetics_Vol2.pdf
"Chapter 1
Pr
eliminary Concepts
1.1 Units
[m0072]
The term “unit” refers to the measure used to express
a ph
ysical quantity . For example, the mean radius of
the Earth is about 6,371,000 meters; in this case, the
unit is the meter.
A number like “6,371,000” becomes a bit
cumbersome to write, so it is common to use a prefix
to modify the unit. For example, the radius of the
Earth is more commonly said to be 6371 kilometers,
where one kilometer is understood to mean
1000 meters. It is common practice to use prefixes,
such as “kilo-, ” that yield values in the range of 0.001
to 10,000. A list of standard prefixes appears in
T able 1.1.
Prefix Abbreviation Multiply by:
exa E 1018
peta
P 1015
tera T 1012
giga G 109
mega M 106
kilo k 103
milli m 10−3
micro µ 10−6
nano n 10−9
pico p 10−12
femto f 10−15
atto a 10−18
T able 1.1: Prefixes used to modify units.
Unit Abbreviation Quantifies:
ampere A electric current
coulomb
C electric charge
farad F capacitance
henry H inductance
hertz Hz frequency",Electromagnetics_Vol2.pdf
"Unit Abbreviation Quantifies:
ampere A electric current
coulomb
C electric charge
farad F capacitance
henry H inductance
hertz Hz frequency
joule J energy
meter m distance
newton N force
ohm Ω resistance
second s time
tesla T magnetic flux density
volt V electric potential
watt W power
weber Wb magnetic flux
T able 1.2: Some units that are commonly used in elec-
tromagnetics.
Writing out the names of units can also become
tedious. For this reason, it is common to use standard
abbreviations; e.g., “6731 km” as opposed to
“6371 kilometers, ” where “k” is the standard
abbreviation for the prefix “kilo” and “m” is the
standard abbreviation for “meter. ” A list of
commonly-used base units and their abbreviations are
shown in T able 1.2.
T o avoid ambiguity , it is important to always indicate
the units of a quantity; e.g., writing “6371 km” as
opposed to “6371. ” Failure to do so is a common
source of error and misunderstandings. An example is
the expression:
l= 3t",Electromagnetics_Vol2.pdf
"opposed to “6371. ” Failure to do so is a common
source of error and misunderstandings. An example is
the expression:
l= 3t
where lis length and tis time. It could be that lis in
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"2 CHAPTER 1. PRELIMINAR Y CONCEPTS
meters and tis in seconds, in which case “3” really
means “3 m/s. ” However, if it is intended that lis in
kilometers and tis in hours, then “3” really means
“3 km/h, ” and the equation is literally different. T o
patch this up, one might write “l = 3t m/s”; however,
note that this does not resolve the ambiguity we just
identified – i.e., we still don’t know the units of the
constant “3. ” Alternatively , one might write “l = 3t
where lis in meters and tis in seconds, ” which is
unambiguous but becomes quite awkward for more
complicated expressions. A better solution is to write
instead:
l= (3 m/s) t
or even better:
l= at where a= 3 m/s
since this separates the issue of units from the perhaps
more-important fact that lis proportional to tand the
constant of proportionality (a) is known.
The meter is the fundamental unit of length in the
International System of Units, known by its French
acronym “SI” and sometimes informally referred to",Electromagnetics_Vol2.pdf
"International System of Units, known by its French
acronym “SI” and sometimes informally referred to
as the “metric system. ”
In this work, we will use SI units exclusively .
Although
SI is probably the most popular for
engineering use overall, other systems remain in
common use. For example, the English system, where
the radius of the Earth might alternatively be said to
be about 3959 miles, continues to be used in various
applications and to a lesser or greater extent in
various regions of the world. An alternative system in
common use in physics and material science
applications is the CGS (“centimeter-gram-second”)
system. The CGS system is similar to SI, but with
some significant differences. For example, the base
unit of energy in the CGS system is not the “joule”
but rather the “erg, ” and the values of some physical
constants become unitless. Therefore – once again –
it is very important to include units whenever values
are stated.",Electromagnetics_Vol2.pdf
"constants become unitless. Therefore – once again –
it is very important to include units whenever values
are stated.
SI defines seven fundamental units from which all
other units can be derived. These fundamental units
are distance in meters (m), time in seconds (s),
current in amperes (A), mass in kilograms (kg),
temperature in kelvin (K), particle count in moles
(mol), and luminosity in candela (cd). SI units for
electromagnetic quantities such as coulombs (C) for
charge and volts (V) for electric potential are derived
from these fundamental units.
A frequently-overlooked feature of units is their
ability to assist in error-checking mathematical
expressions. For example, the electric field intensity
may be specified in volts per meter (V/m), so an
expression for the electric field intensity that yields
units of V/m is said to be “dimensionally correct” (but
not necessarily correct), whereas an expression that
cannot be reduced to units of V/m cannot be correct.
Additional Reading:",Electromagnetics_Vol2.pdf
"not necessarily correct), whereas an expression that
cannot be reduced to units of V/m cannot be correct.
Additional Reading:
• “International System of Units” on Wikipedia.
• “Centimetre-gram-second system of units” on
Wikipedia.",Electromagnetics_Vol2.pdf
"1.2. NOT A TION 3
1.2 Notation
[m0005]
The list below describes notation used in this book.
• V ector
s: Boldface is used to indicate a vector;
e.g., the electric field intensity vector will
typically appear as E. Quantities not in boldface
are scalars. When writing by hand, it is common
to write “ E” or “ − →E” in lieu of “E. ”
• Unit vectors: A circumflex is used to indicate a
unit vector; i.e., a vector having magnitude equal
to one. For example, the unit vector pointing in
the +xdirection will be indicated as ˆx. In
discussion, the quantity “ ˆx” is typically spoken
“x hat. ”
• Time: The symbol tis used to indicate time.
• P osition: The symbols (x,y,z ), (ρ,φ,z ), and
(r,θ,φ ) indicate positions using the Cartesian,
cylindrical, and spherical coordinate systems,
respectively . It is sometimes convenient to
express position in a manner which is
independent of a coordinate system; in this case,
we typically use the symbol r. For example,
r = ˆxx+ ˆyy+ ˆzzin the Cartesian coordinate",Electromagnetics_Vol2.pdf
"independent of a coordinate system; in this case,
we typically use the symbol r. For example,
r = ˆxx+ ˆyy+ ˆzzin the Cartesian coordinate
system.
• Phasors: A tilde is used to indicate a phasor
quantity; e.g., a voltage phasor might be
indicated as ˜V, and the phasor representation of
E will be indicated as ˜E.
• Curves, surfaces, and volumes: These
geometrical entities will usually be indicated in
script; e.g., an open surface might be indicated
as S and the curve bounding this surface might
be indicated as C. Similarly , the volume enclosed
by a closed surface S may be indicated as V.
• Integrations over curves, surfaces, and volumes
will usually be indicated using a single integral
sign with the appropriate subscript. For example:
∫
C
· · · dl is an integral over the curve C
∫
S
· · · ds is an integral over the surface S
∫
V
· · · dv is an integral over the volume V.
• Integrations over closed
curves and surfaces will
be
indicated using a circle superimposed on the",Electromagnetics_Vol2.pdf
"∫
V
· · · dv is an integral over the volume V.
• Integrations over closed
curves and surfaces will
be
indicated using a circle superimposed on the
integral sign. For example:
∮
C
· · · dl is an integral over the closed curve C
∮
S
··· ds is an integral over the closed surface S
A “closed curve” is one which forms an
unbroken loop; e.g., a circle. A “closed surface”
is one which encloses a volume with no
openings; e.g., a sphere.
• The symbol “ ∼=” means “approximately equal
to. ” This symbol is used when equality exists,
but is not being expressed with exact numerical
precision. For example, the ratio of the
circumference of a circle to its diameter is π,
where π∼= 3.14.
• The symbol “≈ ” also indicates “approximately
equal to, ” but in this case the two quantities are
unequal even if expressed with exact numerical
precision. For example, ex = 1 + x+ x2/2 + ...
as an infinite series, but ex ≈ 1 + xfor x≪ 1.
Using this approximation, e0.1 ≈ 1.1, which is",Electromagnetics_Vol2.pdf
"precision. For example, ex = 1 + x+ x2/2 + ...
as an infinite series, but ex ≈ 1 + xfor x≪ 1.
Using this approximation, e0.1 ≈ 1.1, which is
in good agreement with the actual value
e0.1 ∼= 1.1052.
• The symbol “∼ ” indicates “on the order of, ”
which is a relatively weak statement of equality
indicating that the indicated quantity is within a
factor of 10 or so of the indicated value. For
example, µ∼ 105 for a class of iron alloys, with
exact values being larger or smaller by a factor
of 5 or so.
• The symbol “≜ ” means “is defined as” or “is
equal as the result of a definition. ”
• Complex numbers: j ≜ √
−1.
• See
Appendix C for notation used to identify
commonly-used physical constants.",Electromagnetics_Vol2.pdf
"4 CHAPTER 1. PRELIMINAR Y CONCEPTS
1.3 Coordinate Systems
[m0180]
The coordinate systems most commonly used in
engineering
analysis are the Cartesian, cylindrical,
and spherical systems. These systems are illustrated
in Figures 1.1, 1.2, and 1.3, respectively . Note that the
use of variables is not universal; in particular, it is
common to encounter the use of rin lieu of ρfor the
radial coordinate in the cylindrical system, and the
use of Rin lieu of rfor the radial coordinate in the
spherical system.
Additional Reading:
• “Cylindrical coordinate system” on Wikipedia.
• “Spherical coordinate system” on Wikipedia.
y
x
z
y
x
z
c⃝ K. Kikkeri CC BY SA 4.0
Figure 1.1: Cartesian coordinate system.
y
x
ϕ
z
ρ
ϕ
z
ρ
z
c⃝ K. Kikkeri CC BY SA 4.0
Figure 1.2: Cylindrical coordinate system.
y
x
ϕz
r
θθ
ϕ
r
c⃝ K. Kikkeri CC BY SA 4.0
Figure 1.3: Spherical coordinate system.",Electromagnetics_Vol2.pdf
"1.4. ELECTROMAGNETIC FIELD THEOR Y : A REVIEW 5
1.4 Electromagnetic Field
Theory: A Review
[m0179]
This book is the second in a series of textbooks on
electromagnetics.
This section presents a summary of
electromagnetic field theory concepts presented in the
previous volume.
Electric charge and current. Charge is the ultimate
source of the electric field and has SI base units of
coulomb (C). An important source of charge is the
electron, whose charge is defined to be negative.
However, the term “charge” generally refers to a large
number of charge carriers of various types, and whose
relative net charge may be either positive or negative.
Distributions of charge may alternatively be
expressed in terms of line charge density ρl (C/m),
surface charge density ρs (C/m2), or volume charge
density ρv (C/m3). Electric current describes the net
motion of charge. Current is expressed in SI base
units of amperes (A) and may alternatively be
quantified in terms of surface current density Js",Electromagnetics_Vol2.pdf
"motion of charge. Current is expressed in SI base
units of amperes (A) and may alternatively be
quantified in terms of surface current density Js
(A/m) or volume current density J (A/m2).
Electrostatics. Electrostatics is the theory of the
electric field subject to the constraint that charge does
not accelerate. That is, charges may be motionless
(“static”) or move without acceleration (“steady
current”).
The electric field may be interpreted in terms of
energy or flux. The energy interpretation of the
electric field is referred to as electric field intensity E
(SI base units of N/C or V/m), and is related to the
energy associated with charge and forces between
charges. One finds that the electric potential (SI base
units of V) over a path C is given by
V = −
∫
C
E · dl (1.1)
The principle of independence of path means that
only the endpoints of C in Equation 1.1, and no other
details of C, matter. This leads to the finding that the
electrostatic field is conservative; i.e.,
∮
C",Electromagnetics_Vol2.pdf
"details of C, matter. This leads to the finding that the
electrostatic field is conservative; i.e.,
∮
C
E · dl = 0 (1.2)
This is referred to as Kirchoff ’s voltage law for
electrostatics. The inverse of Equation 1.1 is
E = −∇V (1.3)
That is, the electric field intensity points in the
direction in which the potential is most rapidly
decreasing, and the magnitude is equal to the rate of
change in that direction.
The flux interpretation of the electric field is referred
to as electric flux density D (SI base units of C/m 2),
and quantifies the effect of charge as a flow emanating
from the charge. Gauss’ law for electric fields states
that the electric flux through a closed surface is equal
to the enclosed charge Qencl; i.e.,
∮
S
D · ds = Qencl (1.4)
Within a material region, we find
D = ǫE (1.5)
where ǫis the permittivity (SI base units of F/m) of
the material. In free space, ǫis equal to
ǫ0 ≜ 8.854 × 10−12 F/m (1.6)
It is often convenient to quantify the permittivity of",Electromagnetics_Vol2.pdf
"the material. In free space, ǫis equal to
ǫ0 ≜ 8.854 × 10−12 F/m (1.6)
It is often convenient to quantify the permittivity of
material in terms of the unitless relative permittivity
ǫr ≜ ǫ/ǫ0.
Both E and D are useful as they lead to distinct and
independent boundary conditions at the boundary
between dissimilar material regions. Let us refer to
these regions as Regions 1 and 2, having fields
(E1,D1) and (E2,D2), respectively . Given a vector
ˆn perpendicular to the boundary and pointing into
Region 1, we find
ˆn × [E1 − E2] = 0 (1.7)
i.e., the tangential component of the electric field is
continuous across a boundary , and
ˆn · [D1 − D2] = ρs (1.8)
i.e., any discontinuity in the normal component of the
electric field must be supported by a surface charge
distribution on the boundary .",Electromagnetics_Vol2.pdf
"6 CHAPTER 1. PRELIMINAR Y CONCEPTS
Magnetostatics. Magnetostatics is the theory of the
magnetic field in response to steady current or the
intrinsic magnetization of materials. Intrinsic
magnetization is a property of some materials,
including permanent magnets and magnetizable
materials.
Like the electric field, the magnetic field may be
quantified in terms of energy or flux. The flux
interpretation of the magnetic field is referred to as
magnetic flux density B (SI base units of Wb/m 2),
and quantifies the field as a flow associated with, but
not emanating from, the source of the field. The
magnetic flux Φ (SI base units of Wb) is this flow
measured through a specified surface. Gauss’ law for
magnetic fields states that
∮
S
B · ds = 0 (1.9)
i.e., the magnetic flux through a closed surface is
zero. Comparison to Equation 1.4 leads to the
conclusion that the source of the magnetic field
cannot be localized; i.e., there is no “magnetic
charge” analogous to electric charge. Equation 1.9",Electromagnetics_Vol2.pdf
"conclusion that the source of the magnetic field
cannot be localized; i.e., there is no “magnetic
charge” analogous to electric charge. Equation 1.9
also leads to the conclusion that magnetic field lines
form closed loops.
The energy interpretation of the magnetic field is
referred to as magnetic field intensity H (SI base units
of A/m), and is related to the energy associated with
sources of the magnetic field. Ampere’s law for
magnetostatics states that
∮
C
H · dl = Iencl (1.10)
where Iencl is the current flowing past any open
surface bounded by C.
Within a homogeneous material region, we find
B = µH (1.11)
where µis the permeability (SI base units of H/m) of
the material. In free space, µis equal to
µ0 ≜ 4π× 10−7 H/m. (1.12)
It is often convenient to quantify the permeability of
material in terms of the unitless relative permeability
µr ≜ µ/µ0.
Both B and H are useful as they lead to distinct and
independent boundary conditions at the boundaries",Electromagnetics_Vol2.pdf
"µr ≜ µ/µ0.
Both B and H are useful as they lead to distinct and
independent boundary conditions at the boundaries
between dissimilar material regions. Let us refer to
these regions as Regions 1 and 2, having fields
(B1,H1) and (B2,H2), respectively . Given a vector
ˆn perpendicular to the boundary and pointing into
Region 1, we find
ˆn · [B1 − B2] = 0 (1.13)
i.e., the normal component of the magnetic field is
continuous across a boundary , and
ˆn × [H1 − H2] = Js (1.14)
i.e., any discontinuity in the tangential component of
the magnetic field must be supported by current on
the boundary .
Maxwell’s equations. Equations 1.2, 1.4, 1.9, and
1.10 are Maxwell’s equations for static fields in
integral form. As indicated in T able 1.3, these
equations may alternatively be expressed in
differential form. The principal advantage of the
differential forms is that they apply at each point in
space (as opposed to regions defined by C or S), and
subsequently can be combined with the boundary",Electromagnetics_Vol2.pdf
"space (as opposed to regions defined by C or S), and
subsequently can be combined with the boundary
conditions to solve complex problems using standard
methods from the theory of differential equations.
Conductivity . Some materials consist of an
abundance of electrons which are loosely-bound to
the atoms and molecules comprising the material.
The force exerted on these electrons by an electric
field may be sufficient to overcome the binding force,
resulting in motion of the associated charges and
subsequently current. This effect is quantified by
Ohm’s law for electromagnetics:
J = σE (1.15)
where J in this case is the conduction current
determined by the conductivity σ(SI base units of
S/m). Conductivity is a property of a material that
ranges from negligible (i.e., for “insulators”) to very
large for good conductors, which includes most
metals.
A perfect conductor is a material within which E is
essentially zero regardless of J. For such material,",Electromagnetics_Vol2.pdf
"metals.
A perfect conductor is a material within which E is
essentially zero regardless of J. For such material,
σ→ ∞. Perfect conductors are said to be
equipotential regions; that is, the potential difference",Electromagnetics_Vol2.pdf
"1.4. ELECTROMAGNETIC FIELD THEOR Y : A REVIEW 7
Electrostatics / Time-V arying
Magnetostatics (Dynamic)
Electric & magnetic independent possibly coupled
fields are...
Maxwell’s eqns.
∮
S D · ds = Qenc l
∮
S D · ds = Qencl
(integral)
∮
C E · dl = 0
∮
C E · dl = − ∂
∂t
∫
S B · ds∮
S B · ds = 0
∮
S B · d
s = 0∮
C H · dl = Iencl
∮
C H · dl = Iencl+
∫
S
∂
∂tD · ds
Maxwell’s eqns. ∇ · D = ρv ∇ · D = ρv
(differential) ∇ × E = 0 ∇ × E = − ∂
∂tB
∇ · B = 0 ∇ · B = 0
∇ × H = J ∇ × H = J+ ∂
∂tD
T able 1.3: Comparison of Maxwell’s equations for static and time-v arying electromagnetic fields. Differences
in the time-varying case relative to the static case are highlighted in blue.
between
any two points within a perfect conductor is
zero, as can be readily verified using Equation 1.1.
Time-varying fields. F araday’s law states that a
time-varying magnetic flux induces an electric
potential in a closed loop as follows:
V = − ∂
∂tΦ (1.16)
Setting
this equal to the left side of Equation 1.2 leads",Electromagnetics_Vol2.pdf
"potential in a closed loop as follows:
V = − ∂
∂tΦ (1.16)
Setting
this equal to the left side of Equation 1.2 leads
to the Maxwell-F araday equation in integral form:
∮
C
E · dl = − ∂
∂t
∫
S
B · ds (1.17)
where C is
the closed path defined by the edge of the
open surface S. Thus, we see that a time-varying
magnetic flux is able to generate an electric field. W e
also observe that electric and magnetic fields become
coupled when the magnetic flux is time-varying.
An analogous finding leads to the general form of
Ampere’s law:
∮
C
H · dl = Iencl +
∫
S
∂
∂tD · ds (1.18)
where
the new term is referred to as displacement
current. Through the displacement current, a
time-varying electric flux may be a source of the
magnetic field. In other words, we see that the electric
and magnetic fields are coupled when the electric flux
is time-varying.
Gauss’ law for electric and magnetic fields, boundary
conditions, and constitutive relationships
(Equations 1.5, 1.11, and 1.15) are the same in the",Electromagnetics_Vol2.pdf
"Gauss’ law for electric and magnetic fields, boundary
conditions, and constitutive relationships
(Equations 1.5, 1.11, and 1.15) are the same in the
time-varying case.
As indicated in T able 1.3, the time-varying version of
Maxwell’s equations may also be expressed in
differential form. The differential forms make clear
that variations in the electric field with respect to
position are associated with variations in the magnetic
field with respect to time (the Maxwell-Faraday
equation), and vice-versa (Ampere’s law).
Time-harmonic waves in source-free and lossless
media. The coupling between electric and magnetic
fields in the time-varying case leads to wave
phenomena. This is most easily analyzed for fields
which vary sinusoidally , and may thereby be
expressed as phasors. 1 Phasors, indicated in this
book by the tilde (“ ˜”), are complex-valued
quantities representing the magnitude and phase of
the associated sinusoidal waveform. Maxwell’s
equations in differential phasor form are:",Electromagnetics_Vol2.pdf
"quantities representing the magnitude and phase of
the associated sinusoidal waveform. Maxwell’s
equations in differential phasor form are:
∇ · ˜D = ˜ρv (1.19)
∇ × ˜E = −jω˜B (1.20)
∇ · ˜B = 0 (1.21)
∇ × ˜H = ˜J + jω˜D (1.22)
where ω≜ 2πf, and where f is frequency (SI base
1 Sinus oidally-varying fields are sometimes also said to be time-
harmonic.",Electromagnetics_Vol2.pdf
"8 CHAPTER 1. PRELIMINAR Y CONCEPTS
units of Hz). In regions which are free of sources (i.e.,
charges and currents) and consisting of loss-free
media (i.e., σ= 0), these equations reduce to the
following:
∇ · ˜E = 0 (1.23)
∇ × ˜E = −jωµ˜H (1.24)
∇ · ˜H = 0 (1.25)
∇ × ˜H = + jωǫ˜E (1.26)
where we have used the relationships D = ǫE and
B = µH to eliminate the flux densities D and B,
which are now redundant. Solving Equations
1.23–1.26 for E and H, we obtain the vector wave
equations:
∇2 ˜E + β2 ˜E = 0 (1.27)
∇2 ˜H + β2 ˜H = 0 (1.28)
where
β ≜ ω√
µǫ (1.29)
W
aves in source-free and lossless media are solutions
to the vector wave equations.
Uniform plane waves in source-free and lossless
media. An important subset of solutions to the vector
wave equations are uniform plane waves. Uniform
plane waves result when solutions are constrained to
exhibit constant magnitude and phase in a plane. For
example, if this plane is specified to be perpendicular",Electromagnetics_Vol2.pdf
"exhibit constant magnitude and phase in a plane. For
example, if this plane is specified to be perpendicular
to z(i.e., ∂/∂x = ∂/∂y = 0) then solutions for ˜E
have the form:
˜E = ˆx ˜Ex + ˆy ˜Ey (1.30)
where
˜Ex = E+
x0e−jβz + E−
x0e+jβz (1.31)
˜Ey = E+
y0e−jβz + E−
y0e+jβz (1.32)
and where E+
x0, E−
x0, E+
y0, and E−
y0 are constant
complex-valued coefficients which depend on sources
and boundary conditions. The first term and second
terms of Equations 1.31 and 1.32 correspond to waves
traveling in the +ˆz and −ˆz directions, respectively .
Because ˜H is a solution to the same vector wave
equation, the solution for H is identical except with
different coefficients.
The scalar components of the plane waves described
in Equations 1.31 and 1.32 exhibit the same
characteristics as other types of waves, including
sound waves and voltage and current waves in
transmission lines. In particular, the phase velocity of
waves propagating in the +ˆz and −ˆz direction is
vp = ω
β = 1√µǫ (1.33)
and",Electromagnetics_Vol2.pdf
"transmission lines. In particular, the phase velocity of
waves propagating in the +ˆz and −ˆz direction is
vp = ω
β = 1√µǫ (1.33)
and
the wavelength is
λ= 2π
β (1.34)
By
requiring solutions for ˜E and ˜H to satisfy the
Maxwell curl equations (i.e., the Maxwell-Faraday
equation and Ampere’s law), we find that ˜E, ˜H, and
the direction of propagation ˆk are mutually
perpendicular. In particular, we obtain the plane wave
relationships:
˜E = −ηˆk × ˜H (1.35)
˜H = 1
η
ˆk × ˜E (1.36)
where
η≜
√ µ
ǫ (1.37)
is
the wave impedance, also known as the intrinsic
impedance of the medium, and ˆk is in the same
direction as ˜E × ˜H.
The power density associated with a plane wave is
S =
⏐⏐
⏐˜E
⏐
⏐
⏐
2
2η (1.38)
where Shas
SI base units of W/m 2, and here it is
assumed that ˜E is in peak (as opposed to rms) units.
Commonly-assumed properties of materials.
Finally , a reminder about commonly-assumed
properties of the material constitutive parameters ǫ, µ,",Electromagnetics_Vol2.pdf
"Commonly-assumed properties of materials.
Finally , a reminder about commonly-assumed
properties of the material constitutive parameters ǫ, µ,
and σ. W e often assume these parameters exhibit the
following properties:
• Homogeneity. A material that is homogeneous is
uniform over the space it occupies; that is, the
values of its constitutive parameters are constant
at all locations within the material.",Electromagnetics_Vol2.pdf
"1.4. ELECTROMAGNETIC FIELD THEOR Y : A REVIEW 9
• Isotropy. A material that is isotropic behaves in
precisely the same way regardless of how it is
oriented with respect to sources, fields, and other
materials.
• Linearity. A material is said to be linear if its
properties do not depend on the sources and
fields applied to the material. Linear media
exhibit superposition; that is, the response to
multiple sources is equal to the sum of the
responses to the sources individually .
• Time-invariance. A material is said to be
time-invariant if its properties do not vary as a
function of time.
Additional Reading:
• “Maxwell’s Equations” on Wikipedia.
• “W ave Equation” on Wikipedia.
• “Electromagnetic W ave Equation” on Wikipedia.
• “Electromagnetic radiation” on Wikipedia.
[m0181]",Electromagnetics_Vol2.pdf
"10 CHAPTER 1. PRELIMINAR Y CONCEPTS
Image Credits
Fig. 1.1: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fCartesianBasis.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 1.2: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0096 fCylindricalCoordinates.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 1.3: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:Spherical Coordinate System.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 2
Magnetostatics
Redux
2.1 Lorentz Force
[m0015]
The Lorentz force is the force experienced by
charge in the presence of electric and magnetic
fields.
Consider a particle having charge q. The force Fe
experienced by the particle in the presence of electric
field intensity E is
Fe = qE
The force Fm experienced by the particle in the
presence of magnetic flux density B is
Fm = qv × B
where v is the velocity of the particle. The Lorentz
force experienced by the particle is simply the sum of
these forces; i.e.,
F = Fe + Fm
= q(E + v × B) (2.1)
The term “Lorentz force” is simply a concise way to
refer to the combined contributions of the electric and
magnetic fields.
A common application of the Lorentz force concept is
in analysis of the motions of charged particles in
electromagnetic fields. The Lorentz force causes
charged particles to exhibit distinct rotational
(“cyclotron”) and translational (“drift”) motions. This
is illustrated in Figures 2.1 and 2.2.
Additional Reading:",Electromagnetics_Vol2.pdf
"(“cyclotron”) and translational (“drift”) motions. This
is illustrated in Figures 2.1 and 2.2.
Additional Reading:
• “Lorentz force” on Wikipedia.
c⃝ St annered CC BY 2.5.
Figure 2.1: Motion of a particle bearing (left ) posi-
tive charge and (right ) negative charge. T op: Magnetic
field directed toward the viewer; no electric field. Bot-
tom: Magnetic field directed toward the viewer; elec-
tric field oriented as shown.
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"12 CHAPTER 2. MAGNETOST A TICS REDUX
c⃝ M. Biaek CC BY -SA 4.0.
Figure 2.2: Electrons moving in a circle in a magnetic
field (cyclotron motion). The electrons are produced
by an electron gun at bottom, consisting of a hot cath-
ode, a metal plate heated by a filament so it emits elec-
trons, and a metal anode at a high voltage with a hole
which accelerates the electrons into a beam. The elec-
trons are normally invisible, but enough air has been
left in the tube so that the air molecules glow pink
when struck by the fast-moving electrons.
2.2 Magnetic Force on a
Current-Carrying Wire
[m0017]
Consider an infinitesimally-thin and
perfectly-conducting
wire bearing a current I (SI base
units of A) in free space. Let B (r) be the impressed
magnetic flux density at each point r in the region of
space occupied by the wire. By impressed, we mean
that the field exists in the absence of the
current-carrying wire, as opposed to the field that is
induced by this current. Since current consists of",Electromagnetics_Vol2.pdf
"that the field exists in the absence of the
current-carrying wire, as opposed to the field that is
induced by this current. Since current consists of
charged particles in motion, we expect that B(r) will
exert a force on the current. Since the current is
constrained to flow on the wire, we expect this force
will also be experienced by the wire. Let us now
consider this force.
T o begin, recall that the force exerted on a particle
bearing charge qhaving velocity v is
Fm(r) = qv (r) × B (r) (2.2)
Thus, the force exerted on a differential amount of
charge dqis
dFm(r) = dqv (r) × B (r) (2.3)
Let dl (r) represent a differential-length segment of
the wire at r, pointing in the direction of current flow .
Then
dqv (r) = Idl (r) (2.4)
(If this is not clear, it might help to consider the units:
On the left, C·m/s = (C/s)·m = A·m, as on the right.)
Subsequently ,
dFm(r) = Idl (r) × B (r) (2.5)
There are three important cases of practical interest.",Electromagnetics_Vol2.pdf
"Subsequently ,
dFm(r) = Idl (r) × B (r) (2.5)
There are three important cases of practical interest.
First, consider a straight segment l forming part of a
closed loop of current in a spatially-uniform
impressed magnetic flux density B (r) = B0. In this
case, the force exerted by the magnetic field on such a
segment is given by Equation 2.5 with dl replaced by
l; i.e.:
Fm = Il × B0
(2.6)
Summarizing,
The force experienced by a straight segment of
current-carrying
wire in a spatially-uniform mag-
netic field is given by Equation 2.6.
The second case of practical interest is a rigid closed
loop of current in a spatially-uniform magnetic flux
density B0. If the loop consists of straight sides –
e.g., a rectangular loop – then the force applied to the
loop is the sum of the forces applied to each side
separately , as determined by Equation 2.6. However,
we wish to consider loops of arbitrary shape. T o
accommodate arbitrarily-shaped loops, let C be the",Electromagnetics_Vol2.pdf
"we wish to consider loops of arbitrary shape. T o
accommodate arbitrarily-shaped loops, let C be the
path through space occupied by the loop. Then the
force experienced by the loop is
F =
∫
C
dFm(r)
=
∫
C
Idl (r) × B0 (2.7)
Since I and B0 are constants, they may be extracted
from the integral:
F = I
[ ∫
C
dl (r)
]
× B0 (2.8)
Note the quantity in square brackets is zero.
Therefore:",Electromagnetics_Vol2.pdf
"2.2. MAGNETIC FORCE ON A CURRENT -CARR YING WIRE 13
The net force on a current-carrying loop of wire
in
a uniform magnetic field is zero.
Note that this does not preclude the possibility that
the rigid loop rotates; for example, the force on
opposite sides of the loop may be equal and opposite.
What we have found is merely that the force will not
lead to a translational net force on the loop; e.g.,
force that would propel the loop away from its current
position in space. The possibility of rotation without
translation leads to the most rudimentary concept for
an electric motor. Practical electric motors use
variations on essentially this same idea; see
“ Additional Reading” for more information.
The third case of practical interest is the force
experienced by two parallel infinitesimally-thin wires
in free space, as shown in Figure 2.3. Here the wires
are infinite in length (we’ll return to that in a
moment), lie in the x= 0 plane, are separated by",Electromagnetics_Vol2.pdf
"are infinite in length (we’ll return to that in a
moment), lie in the x= 0 plane, are separated by
distance d, and carry currents I1 and I2, respectively .
The current in wire 1 gives rise to a magnetic flux
density B1. The force exerted on wire 2 by B1 is:
F2 =
∫
C
[I2dl (r) × B1 (r)] (2.9)
where C is the path followed by I2, and dl (r) = ˆzdz.
A simple way to determine B1 in this situation is as
follows. First, if wire 1 had been aligned along the
x= y= 0 line, then the magnetic flux density
everywhere would be
ˆφµ0I1
2πρ
In
the present problem, wire 1 is displaced by d/2 in
the −ˆy direction. Although this would seem to make
the new expression more complicated, note that the
only positions where values of B1 (r) are required are
those corresponding to C; i.e., points on wire 2. For
these points,
B1 (r) = −ˆxµ0I1
2πd along C (2.10)
That
is, the relevant distance is d(not ρ), and the
direction of B1 (r) for points along C is −ˆx (not ˆφ).
Returning to Equation 2.9, we obtain:
F2 =
∫
C
[",Electromagnetics_Vol2.pdf
"direction of B1 (r) for points along C is −ˆx (not ˆφ).
Returning to Equation 2.9, we obtain:
F2 =
∫
C
[
I2 ˆzdz×
(
−ˆxµ0I1
2πd
)]
= −ˆy µ0I1I2
2πd
∫
C
dz (2.11)
The
remaining integral is simply the length of wire 2
that we wish to consider. Infinitely-long wires will
therefore result in infinite force. This is not a very
interesting or useful result. However, the force per
unit length of wire is finite, and is obtained simply by
dropping the integral in the previous equation. W e
obtain: F2
∆l = −ˆy µ0I1I2
2πd (2.12)
where ∆
lis the length of the section of wire 2 being
considered. Note that when the currents I1 and I2
flow in the same direction (i.e., have the same sign),
the magnetic force exerted by the current on wire 1
pulls wire 2 toward wire 1.
The same process can be used to determine the
magnetic force F1 exerted by the current in wire 1 on
wire 2. The result is
F1
∆l = + ˆy µ0I1I2
2πd (2.13)
c⃝ Y. Zhao CC BY -SA 4.0
Figure 2.3: Parallel current-carrying wires.",Electromagnetics_Vol2.pdf
"14 CHAPTER 2. MAGNETOST A TICS REDUX
When the currents I1 and I2 flow in the same
direction (i.e., when the product I1I2 is positive), then
the magnetic force exerted by the current on wire 2
pulls wire 1 toward wire 2.
W e are now able to summarize the results as follows:
If currents in parallel wires flow in the same di-
rection,
then the wires attract; whereas if the cur-
rents flow in opposite directions, then the wires
repel.
Also:
The magnitude of the associated force is
µ0I1I2/2πdfor
wires separated by distance din
non-magnetic media.
If the wires are fixed in position and not able to move,
these forces represent stored (potential) energy . It is
worth noting that this is precisely the energy which is
stored by an inductor – for example, the two wire
segments here might be interpreted as segments in
adjacent windings of a coil-shaped inductor.
Example 2.1. DC power cable.
A power cable connects a 12 V battery to a load
exhibiting an impedance of 10 Ω. The",Electromagnetics_Vol2.pdf
"Example 2.1. DC power cable.
A power cable connects a 12 V battery to a load
exhibiting an impedance of 10 Ω. The
conductors are separated by 3 mm by a plastic
insulating jacket. Estimate the force between the
conductors.
Solution. The current flowing in each conductor
is 12 V divided by 10 Ω, which is 1.2 A. In terms
of the theory developed in this section, a current
I1 = +1.2 A flows from the positive terminal of
the battery to the load on one conductor, and a
current I2 = −1.2 A returns to the battery on the
other conductor. The change in sign indicates
that the currents at any given distance from the
battery are flowing in opposite directions. Also
from the problem statement, d= 3 mm and the
insulator is presumably non-magnetic.
Assuming the conductors are approximately
straight, the force between conductors is
≈ µ0I1I2
2πd
∼= −96 .0 µN
with the negative sign indicating that the wires
repel.
Note
in the above example that this force is quite",Electromagnetics_Vol2.pdf
"≈ µ0I1I2
2πd
∼= −96 .0 µN
with the negative sign indicating that the wires
repel.
Note
in the above example that this force is quite
small, which explains why it is not always observed.
However, this force becomes significant when the
current is large or when many sets of conductors are
mechanically bound together (amounting to a larger
net current), as in a motor.
Additional Reading:
• “Electric motor” on Wikipedia.",Electromagnetics_Vol2.pdf
"2.3. TORQUE INDUCED BY A MAGNETIC FIELD 15
2.3 T orque Induced by a
Magnetic Field
[m0024]
A magnetic field exerts a force on current. This force
is
exerted in a direction perpendicular to the direction
of current flow . For this reason, current-carrying
structures in a magnetic field tend to rotate. A
convenient description of force associated with
rotational motion is torque. In this section, we define
torque and apply this concept to a closed loop of
current. These concepts apply to a wide range of
practical devices, including electric motors.
Figure 2.4 illustrates the concept of torque. T orque
depends on the following:
• A local origin r0,
• A point r which is connected to r0 by a
perfectly-rigid mechanical structure, and
• The force F applied at r.
In terms of these parameters, the torque T is:
T ≜ d × F (2.14)
where the lever arm d ≜ r − r0 gives the location of
r relative to r0. Note that T is a position-free vector
c⃝ C. W ang CC BY -SA 4.0",Electromagnetics_Vol2.pdf
"where the lever arm d ≜ r − r0 gives the location of
r relative to r0. Note that T is a position-free vector
c⃝ C. W ang CC BY -SA 4.0
Figure 2.4: T orque associated with a single lever arm.
which points in a direction perpendicular to both d
and F.
Note that T does not point in the direction of rotation.
Nevertheless, T indicates the direction of rotation
through a “right hand rule”: If you point the thumb of
your right hand in the direction of T, then the curled
fingers of your right hand will point in the direction of
torque-induced rotation.
Whether rotation actually occurs depends on the
geometry of the structure. For example, if T aligns
with the axis of a perfectly-rigid mechanical shaft,
then all of the work done by F will be applied to
rotation of the shaft on this axis. Otherwise, torque
will tend to rotate the shaft in other directions as well.
If the shaft is not free to rotate in these other
directions, then the effective torque – that is, the",Electromagnetics_Vol2.pdf
"If the shaft is not free to rotate in these other
directions, then the effective torque – that is, the
torque that contributes to rotation of the shaft – is
reduced.
The magnitude of T has SI base units of N·m and
quantifies the energy associated with the rotational
force. As you might expect, the magnitude of the
torque increases with increasing lever arm magnitude
|d|. In other words, the torque resulting from a
constant applied force increases with the length of the
lever arm.
T orque, like the translational force F, satisfies
superposition. That is, the torque resulting from
forces applied to multiple rigidly-connected lever
arms is the sum of the torques applied to the lever
arms individually .
Now consider the current loop shown in Figure 2.5.
The loop is perfectly rigid and is rigidly attached to a
non-conducting shaft. The assembly consisting of the
loop and the shaft may rotate without friction around
the axis of the shaft. The loop consists of four straight",Electromagnetics_Vol2.pdf
"loop and the shaft may rotate without friction around
the axis of the shaft. The loop consists of four straight
segments that are perfectly-conducting and
infinitesimally-thin. A spatially-uniform and static
impressed magnetic flux density B0 = ˆxB0 exists
throughout the domain of the problem. (Recall that an
impressed field is one that exists in the absence of any
other structure in the problem.) What motion, if any ,
is expected?
Recall that the net translational force on a current
loop in a spatially-uniform and static magnetic field is",Electromagnetics_Vol2.pdf
"16 CHAPTER 2. MAGNETOST A TICS REDUX
zero (Section 2.2). However, this does not preclude
the possibility of different translational forces acting
on each of the loop segments resulting in a rotation of
the shaft. Let us first calculate these forces. The force
FA on segment A is
FA = IlA × B0 (2.15)
where lA is a vector whose magnitude is equal to the
length of the segment and which points in the
direction of the current. Thus,
FA = I(ˆzL) × (ˆxB0)
= ˆyILB0 (2.16)
Similarly , the force FC on segment C is
FC = I(−ˆzL) × (ˆxB0)
= −ˆyILB0 (2.17)
The forces FB and FD on segments B and D,
respectively , are:
FB = I(−ˆxL) × (ˆxB0) = 0 (2.18)
and
FD = I(+ˆxL) × (ˆxB0) = 0 (2.19)
Thus, the force exerted on the current loop by the
impressed magnetic field will lead to rotation in the
+ ˆφdirection.
z
x
A
B
C
D
L
W
I
B0=
x̂B0
current
source
wir
e loop
non-conducting shaft
c⃝ C. W ang CC BY -SA 4.0
Figure 2.5: A rudimentary electric motor consisting
of a single current loop.",Electromagnetics_Vol2.pdf
"source
wir
e loop
non-conducting shaft
c⃝ C. W ang CC BY -SA 4.0
Figure 2.5: A rudimentary electric motor consisting
of a single current loop.
W e calculate the associated torque T as
T = TA + TB + TC + TD (2.20)
where TA, TB, TC, and TD are the torques
associated with segments A, B, C, and D, respectively .
For example, the torque associated with segment A is
TA = W
2 ˆx × FA
= ˆzLW
2 IB0 (2.21)
Similarly
,
TB = 0 since FB = 0 (2.22)
TC = ˆzLW
2 IB0 (2.23)
TD =
0 since FD = 0 (2.24)
Summing these contributions, we find
T = ˆzLWIB0 (2.25)
Note that T points in the +ˆz direction, indicating
rotational force exerted in the + ˆφdirection, as
expected. Also note that the torque is proportional to
the area LW of the loop, is proportional to the current
I, and is proportional to the magnetic field magnitude
B0.
The analysis that we just completed was static; that is,
it applies only at the instant depicted in Figure 2.5. If
the shaft is allowed to turn without friction, then the",Electromagnetics_Vol2.pdf
"it applies only at the instant depicted in Figure 2.5. If
the shaft is allowed to turn without friction, then the
loop will rotate in the + ˆφdirection. So, what will
happen to the forces and torque? First, note that FA
and FC are always in the +ˆy and −ˆy directions,
respectively , regardless of the rotation of the loop.
Once the loop rotates away from the position shown
in Figure 2.5, the forces FB and FD become
non-zero; however, they are always equal and
opposite, and so do not affect the rotation. Thus, the
loop will rotate one-quarter turn and then come to
rest, perhaps with some damped oscillation around
the rest position depending on the momentum of the
loop. At the rest position, the lever arms for
segments A and C are pointing in the same directions
as FA and FC, respectively . Therefore, the cross
product of the lever arm and translational force for
each segment is zero and subsequently
TA = TC = 0. Once stopped in this position, both",Electromagnetics_Vol2.pdf
"product of the lever arm and translational force for
each segment is zero and subsequently
TA = TC = 0. Once stopped in this position, both
the net translational force and the net torque are zero.",Electromagnetics_Vol2.pdf
"2.3. TORQUE INDUCED BY A MAGNETIC FIELD 17
c⃝ Ab normaal CC BY -SA 3.0
Figure 2.6: This DC electric motor uses brushes (here,
the motionless leads labeled “+ ” and “− ”) combined
with the motion of the shaft to periodically alternate
the direction of current between two coils, thereby cre-
ating nearly constant torque.
If such a device is to be used as a motor, it is
necessary to find a way to sustain the rotation. There
are several ways in which this might be accomplished.
First, one might make I variable in time. For
example, the direction of I could be reversed as the
loop passes the quarter-turn position. This reverses
FA and FC, propelling the loop toward the half-turn
position. The direction of I can be changed again as
the loop passes half-turn position, propelling the loop
toward the three-quarter-turn position. Continuing
this periodic reversal of the current sustains the
rotation. Alternatively , one may periodically reverse
the direction of the impressed magnetic field to the",Electromagnetics_Vol2.pdf
"rotation. Alternatively , one may periodically reverse
the direction of the impressed magnetic field to the
same effect. These methods can be combined or
augmented using multiple current loops or multiple
sets of time-varying impressed magnetic fields. Using
an appropriate combination of current loops,
magnetic fields, and waveforms for each, it is possible
to achieve sustained torque throughout the rotation.
An example is shown in Figure 2.6.
Additional Reading:
• “T orque” on Wikipedia.
• “Electric motor” on Wikipedia.",Electromagnetics_Vol2.pdf
"18 CHAPTER 2. MAGNETOST A TICS REDUX
2.4 The Biot-Savart Law
[m0066]
The Biot-Savart law (BSL) provides a method to
calculate
the magnetic field due to any distribution of
steady (DC) current. In magnetostatics, the general
solution to this problem employs Ampere’s law; i.e.,
∫
C
H · dl = Iencl (2.26)
in integral form or
∇ × H = J (2.27)
in differential form. The integral form is relatively
simple when the problem exhibits a high degree of
symmetry , facilitating a simple description in a
particular coordinate system. An example is the
magnetic field due to a straight and infinitely-long
current filament, which is easily determined by
solving the integral equation in cylindrical
coordinates. However, many problems of practical
interest do not exhibit the necessary symmetry . A
commonly-encountered example is the magnetic field
due to a single loop of current, which will be
addressed in Example 2.2. For such problems, the
differential form of Ampere’s law is needed.",Electromagnetics_Vol2.pdf
"due to a single loop of current, which will be
addressed in Example 2.2. For such problems, the
differential form of Ampere’s law is needed.
BSL is the solution to the differential form of
Ampere’s law for a differential-length current
element, illustrated in Figure 2.7. The current element
is I dl, where I is the magnitude of the current (SI
base units of A) and dl is a differential-length vector
indicating the direction of the current at the “source
point” r′. The resulting contribution to the magnetic
field intensity at the “field point” r is
dH(r) = I dl 1
4πR2 × ˆR (2.28)
where
R = ˆRR≜ r − r′ (2.29)
In
other words, R is the vector pointing from the
source point to the field point, and dH at the field
point is given by Equation 2.28. The magnetic field
due to a current-carrying wire of any shape may be
obtained by integrating over the length of the wire:
H(r) =
∫
C
dH(r) = I
4π
∫
C
dl × ˆR
R2 (2.30)
In
addition to obviating the need to solve a differential",Electromagnetics_Vol2.pdf
"H(r) =
∫
C
dH(r) = I
4π
∫
C
dl × ˆR
R2 (2.30)
In
addition to obviating the need to solve a differential
equation, BSL provides some useful insight into the
behavior of magnetic fields. In particular,
Equation 2.28 indicates that magnetic fields follow
the inverse square law – that is, the magnitude of the
magnetic field due to a differential current element
decreases in proportion to the inverse square of
distance (R−2). Also, Equation 2.28 indicates that the
direction of the magnetic field due to a differential
current element is perpendicular to both the direction
of current flow ˆl and the vector ˆR pointing from the
source point to field point. This observation is quite
useful in anticipating the direction of magnetic field
vectors in complex problems.
It may be helpful to note that BSL is analogous to
Coulomb’s law for electric fields, which is a solution
to the differential form of Gauss’ law , ∇ · D = ρv.
However, BSL applies only under magnetostatic",Electromagnetics_Vol2.pdf
"to the differential form of Gauss’ law , ∇ · D = ρv.
However, BSL applies only under magnetostatic
conditions. If the variation in currents or magnetic
fields over time is significant, then the problem
becomes significantly more complicated. See
“Jefimenko’s Equations” in “ Additional Reading” for
more information.
Example 2.2. Magnetic field along the axis of a
circular loop of current.
Consider a ring of radius ain the z= 0 plane,
centered on the origin, as shown in Figure 2.8.
As indicated in the figure, the current I flows in
R
 R
d
H(r)
 dl
@r'
I
c⃝ C. W ang CC BY -SA 4.0
Figure 2.7: Use of the Biot-Savart law to calculate the
magnetic field due to a line current.",Electromagnetics_Vol2.pdf
"2.4. THE BIOT -SA V AR T LA W 19
the ˆφdirection. Find the magnetic field intensity
along
the zaxis.
Solution. The source current position is given in
cylindrical coordinates as
r′ = ˆρa (2.31)
The position of a field point along the zaxis is
r = ˆzz (2.32)
Thus,
ˆRR≜ r − r′ = −ˆρa+ ˆzz (2.33)
and
R≜ |r − r′| =
√
a2 + z2 (2.34)
Equation
2.28 becomes:
dH(ˆzz) = I ˆφadφ
4π[a2 + z2] × ˆzz− ˆρa√
a2 + z2
= Ia
4π
ˆza− ˆρz
[a2 + z2]3/ 2 dφ (2.35)
No
w integrating over the current:
H(ˆzz) =
∫ 2π
0
Ia
4π
ˆza− ˆρz
[a2 + z2]3/ 2 dφ (2.36)
= I
a
4π[a2 + z2]3/ 2
∫ 2π
0
(ˆza− ˆρz) dφ
(2.37)
= I
a
4π[a2 + z2]3/ 2
(
ˆza
∫ 2π
0
dφ− z
∫ 2
π
0
ˆρdφ
)
(2.38)
The second integral is equal to zero. T o see this,
note that the integral is simply summing values
of ˆρfor all possible values of φ. Since
ˆρ(φ+ π) = −ˆρ(φ), the integrand for any given
value of φis equal and opposite the integrand π
radians later. (This is one example of a symmetry
argument.)
c⃝ K. Kikkeri CC BY SA 4.0 (modified)",Electromagnetics_Vol2.pdf
"value of φis equal and opposite the integrand π
radians later. (This is one example of a symmetry
argument.)
c⃝ K. Kikkeri CC BY SA 4.0 (modified)
Figure 2.8: Calculation of the magnetic field along
the zaxis due to a circular loop of current centered in
the z= 0 plane.
The first integral in the previous equation is
equal
to 2π. Thus, we obtain
H(ˆzz) = ˆz Ia2
2 [a2 + z2]3/ 2 (2.39)
Note
that the result is consistent with the
associated “right hand rule” of magnetostatics:
That is, the direction of the magnetic field is in
the direction of the curled fingers of the right
hand when the thumb of the right hand is aligned
with the location and direction of current. It is a
good exercise to confirm that this result is also
dimensionally correct.
Equation 2.28 extends straightforwardly to other
distributions of current. For example, the magnetic
field due to surface current Js (SI base units of A/m)
can be calculated using Equation 2.28 with I dl
replaced by
Js ds",Electromagnetics_Vol2.pdf
"field due to surface current Js (SI base units of A/m)
can be calculated using Equation 2.28 with I dl
replaced by
Js ds
where dsis the differential element of surface area.
This can be confirmed by dimensional analysis: I dl
has SI base units of A·m, as does JS ds. Similarly ,
the magnetic field due to volume current J (SI base
units of A/m 2) can be calculated using Equation 2.28
with I dlreplaced by
J dv
where dvis the differential element of volume. For a
single particle with charge q(SI base units of C) and",Electromagnetics_Vol2.pdf
"20 CHAPTER 2. MAGNETOST A TICS REDUX
velocity v (SI base units of m/s), the relevant quantity
is
qv
since C·m/s = (C/s)·m = A·m. In all of these cases,
Equation 2.28 applies with the appropriate
replacement for I dl.
Note that the quantities qv, I dl, JS ds, and J dv, all
having the same units of A·m, seem to be referring to
the same physical quantity . This physical quantity is
known as current moment. Thus, the “input” to BSL
can be interpreted as current moment, regardless of
whether the current of interest is distributed as a line
current, a surface current, a volumetric current, or
simply as moving charged particles. See “ Additional
Reading” at the end of this section for additional
information on the concept of “moment” in classical
physics.
Additional Reading:
• “Biot-Savart Law” on Wikipedia.
• “Jefimenko’s Equations” on Wikipedia.
• “Moment (physics)” on Wikipedia.
2.5 Force, Energy, and Potential
Difference in a Magnetic
Field
[m0059]
The force Fm experienced",Electromagnetics_Vol2.pdf
"• “Moment (physics)” on Wikipedia.
2.5 Force, Energy, and Potential
Difference in a Magnetic
Field
[m0059]
The force Fm experienced
 by a particle at location r
bearing charge qdue to a magnetic field is
Fm = qv × B(r) (2.40)
where v is the velocity (magnitude and direction) of
the particle, and B(r) is the magnetic flux density at
r. Now we must be careful: In this description, the
motion of the particle is not due to Fm. In fact the
cross product in Equation 2.40 clearly indicates that
Fm and v must be in perpendicular directions.
Instead, the reverse is true: i.e., it is the motion of the
particle that is giving rise to the force. The motion
described by v may be due to the presence of an
electric field, or it may simply be that that charge is
contained within a structure that is itself in motion.
Nevertheless, the force Fm has an associated
potential energy . Furthermore, this potential energy
may change as the particle moves. This change in",Electromagnetics_Vol2.pdf
"Nevertheless, the force Fm has an associated
potential energy . Furthermore, this potential energy
may change as the particle moves. This change in
potential energy may give rise to an electrical
potential difference (i.e., a “voltage”), as we shall
now demonstrate.
The change in potential energy can be quantified
using the concept of work, W. The incremental work
∆W done by moving the particle a short distance ∆l ,
over which we assume the change in Fm is
negligible, is
∆W ≈ Fm · ˆl∆l (2.41)
where in this case ˆl is the unit vector in the direction
of the motion; i.e., the direction of v. Note that the
purpose of the dot product in Equation 2.41 is to
ensure that only the component of Fm parallel to the
direction of motion is included in the energy tally .
Any component of v which is due to Fm (i.e.,
ultimately due to B) must be perpendicular to Fm, so
∆W for such a contribution must be, from
Equation 2.41, equal to zero. In other words: In the",Electromagnetics_Vol2.pdf
"ultimately due to B) must be perpendicular to Fm, so
∆W for such a contribution must be, from
Equation 2.41, equal to zero. In other words: In the
absence of a mechanical force or an electric field, the
potential energy of a charged particle remains
constant regardless of how it is moved by Fm. This
surprising result may be summarized as follows:",Electromagnetics_Vol2.pdf
"2.5. FORCE, ENERGY , AND POTENTIAL DIFFERENCE IN A MAGNETIC FIELD 21
The magnetic field does no work.
Instead,
the change of potential energy associated
with the magnetic field must be completely due to a
change in position resulting from other forces, such as
a mechanical force or the Coulomb force. The
presence of a magnetic field merely increases or
decreases this potential difference once the particle
has moved, and it is this change in the potential
difference that we wish to determine.
W e can make the relationship between potential
difference and the magnetic field explicit by
substituting the right side of Equation 2.40 into
Equation 2.41, yielding
∆W ≈ q[v × B(r)] · ˆl∆l (2.42)
Equation 2.42 gives the work only for a short distance
around r. Now let us try to generalize this result. If
we wish to know the work done over a larger
distance, then we must account for the possibility that
v × B varies along the path taken. T o do this, we may",Electromagnetics_Vol2.pdf
"distance, then we must account for the possibility that
v × B varies along the path taken. T o do this, we may
sum contributions from points along the path traced
out by the particle, i.e.,
W ≈
N∑
n=1
∆W (rn) (2.43)
where rn are positions defining the path. Substituting
the right side of Equation 2.42, we have
W ≈ q
N∑
n=1
[v × B(rn)] · ˆl(rn)∆l (2.44)
T aking the limit as ∆l → 0, we obtain
W = q
∫
C
[v × B(r)] · ˆl(r)dl (2.45)
where C is the path (previously , the sequence of rn’s)
followed by the particle. Now omitting the explicit
dependence on r in the integrand for clarity:
W = q
∫
C
[v × B] · dl (2.46)
where dl = ˆldlas usual. Now , we are able to
determine the change in potential energy for a
charged particle moving along any path in space,
given the magnetic field.
At this point, it is convenient to introduce the electric
potential difference V21 between the start point (1)
and end point (2) of C. V21 is defined as the work
done by traversing C, per unit of charge; i.e.,
V21 ≜ W",Electromagnetics_Vol2.pdf
"and end point (2) of C. V21 is defined as the work
done by traversing C, per unit of charge; i.e.,
V21 ≜ W
q (2.47)
This
has units of J/C, which is volts (V). Substituting
Equation 2.46, we obtain:
V21 =
∫
C
[v × B] · dl (2.48)
Equation 2.48 is electrical potential induced by
char
ge traversing a magnetic field.
Figure 2.9 shows a simple scenario that illustrates this
concept. Here, a straight perfectly-conducting wire of
length lis parallel to the yaxis and moves at speed v
in the +zdirection through a magnetic field B = ˆxB.
Thus,
v × B = ˆzv× ˆxB
= ˆyBv (2.49)
T aking endpoints 1 and 2 of the wire to be at y= y0
(2)
(1)
B
v
z
y
x
l
c⃝ C. W ang CC BY 4.0
Figure 2.9: A straight wire moving through a mag-
netic field.",Electromagnetics_Vol2.pdf
"22 CHAPTER 2. MAGNETOST A TICS REDUX
and y= y0 + l, respectively , we obtain
V21 =
∫ y0+l
y0
[ˆyBv] · ˆydy
= Bvl (2.50)
Thus, we see that endpoint 2 is at an electrical
potential of Bvl greater than that of endpoint 1. This
“voltage” exists even though the wire is
perfectly-conducting, and therefore cannot be
attributed to the electric field. This voltage exists even
though the force required for movement must be the
same on both endpoints, or could even be zero, and
therefore cannot be attributed to mechanical forces.
Instead, this change in potential is due entirely to the
magnetic field.
Because the wire does not form a closed loop, no
current flows in the wire. Therefore, this scenario has
limited application in practice. T o accomplish
something useful with this concept we must at least
form a closed loop, so that current may flow . For a
closed loop, Equation 2.48 becomes:
V =
∮
C
[v × B] · dl (2.51)
Examination of this equation indicates one additional",Electromagnetics_Vol2.pdf
"closed loop, Equation 2.48 becomes:
V =
∮
C
[v × B] · dl (2.51)
Examination of this equation indicates one additional
requirement: v × B must somehow vary over C. This
is because if v × B does not vary over C, the result
will be
[v × B] ·
∮
C
dl
which is zero because the integral is zero. The
following example demonstrates a practical
application of this idea.
Example 2.3. Potential induced in a
time-varying loop.
Figure 2.10 shows a modification to the problem
originally considered in Figure 2.9. Now , we
have created a closed loop using
perfectly-conducting and motionless wire to
form three sides of a rectangle, and assigned the
origin to the lower left corner. An
infinitesimally-small gap has been inserted in the
left (z = 0) side of the loop and closed with an
ideal resistor of value R. As before, B = ˆxB
(spatially uniform and time invariant) and
vB
zx
y
VT
+
-
R
c⃝ C. W ang CC BY -SA 4.0
Figure 2.10: A time-varying loop created by moving",Electromagnetics_Vol2.pdf
"(spatially uniform and time invariant) and
vB
zx
y
VT
+
-
R
c⃝ C. W ang CC BY -SA 4.0
Figure 2.10: A time-varying loop created by moving
a “shorting bar” along rails comprising two adjacent
sides of the loop.
v = ˆzv(constant). What is the voltage VT
across
the resistor and what is the current in the
loop?
Solution. Since the gap containing the resistor is
infinitesimally small,
VT =
∮
C
[v × B] · dl (2.52)
where C is the perimeter formed by the loop,
beginning at the “− ” terminal of VT and
returning to the “+” terminal of VT. Only the
shorting bar is in motion, so v = 0 for the other
three sides of the loop. Therefore, only the
portion of C traversing the shorting bar
contributes to VT. Thus, we find
VT =
∫ l
y=0
[ˆzv× ˆxB] · ˆydy= Bvl (2.53)
This potential gives rise to a current Bvl/R,
which flows in the counter-clockwise direction.
Note in the previous example that the magnetic field
has induced VT, not the current. The current is simply",Electromagnetics_Vol2.pdf
"Note in the previous example that the magnetic field
has induced VT, not the current. The current is simply
a response to the existence of the potential, regardless
of the source. In other words, the same potential VT
would exist even if the gap was not closed by a
resistor.",Electromagnetics_Vol2.pdf
"2.5. FORCE, ENERGY , AND POTENTIAL DIFFERENCE IN A MAGNETIC FIELD 23
Astute readers will notice that this analysis seems to
have a lot in common with Faraday’s law ,
V = − ∂
∂tΦ (2.54)
which
says the potential induced in a single closed
loop is proportional to the time rate of change of
magnetic flux Φ, where
Φ =
∫
S
B · ds (2.55)
and where S is the surface through which the flux is
calculated. From this perspective, we see that
Equation 2.51 is simply a special case of Faraday’s
law , pertaining specifically to “motional emf. ” Thus,
the preceding example can also be solved by
Faraday’s law , taking S to be the time-varying surface
bounded by C.
[m0182]",Electromagnetics_Vol2.pdf
"24 CHAPTER 2. MAGNETOST A TICS REDUX
Image Credits
Fig. 2.1: c⃝ Stannered, https://en.wikipedia.org/wiki/File:Charged-particle-drifts.svg,
CC BY 2.5 (https://creativecommons.org/licenses/by/2.5/deed.en). Modified by Maschen, author.
Fig. 2.2: c⃝ M. Biaek, https://en.wikipedia.org/wiki/File:Cyclotron motion.jpg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 2.3: c⃝ Y ahuiZ (Y . Zhao), https://commons.wikimedia.org/wiki/File:Figure2.3Y ahui.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified by author.
Fig. 2.4: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:T orque associated with a single lever arm.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 2.5: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Rudimentary electric motor.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 2.6: c⃝ Abnormaal, https://en.wikipedia.org/wiki/File:Electric motor.gif,
CC",Electromagnetics_Vol2.pdf
"CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 2.6: c⃝ Abnormaal, https://en.wikipedia.org/wiki/File:Electric motor.gif,
CC
BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Fig. 2.7: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Magnetic field due to a line current.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 2.8: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0104 fRing.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified by author.
Fig. 2.9: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Straight wire moving in a magnetic field.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 2.10: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Moving shorting bar of a circuit in mag field.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 3
W a
ve Propagation in General Media
3.1 Poynting’s Theorem
[m0073]
Despite the apparent complexity of electromagnetic
theory
, there are in fact merely four ways that
electromagnetic energy can be manipulated.
Electromagnetic energy can be:
• Transferred; i.e., conveyed by transmission lines
or in waves;
• Stored in an electric field (capacitance);
• Stored in a magnetic field (inductance); or
• Dissipated (converted to heat; i.e., resistance).
Poynting’s theorem is an expression of conservation
of energy that elegantly relates these various
possibilities. Once recognized, the theorem has
important applications in the analysis and design of
electromagnetic systems. Some of these emerge from
the derivation of the theorem, as opposed to the
unsurprising result. So, let us now derive the theorem.
W e begin with a statement of the theorem. Consider a
volume V that may contain materials and structures in
any combination. This is crudely depicted in",Electromagnetics_Vol2.pdf
"volume V that may contain materials and structures in
any combination. This is crudely depicted in
Figure 3.1. Also recall that power is the time rate of
change of energy . Then:
Pnet,in = PE + PM + PΩ (3.1)
where Pnet,in is the net power flow into V, PE is the
power associated with energy storage in electric fields
within V, PM is the power associated with energy
storage in magnetic fields within V, and PΩ is the
power dissipated (converted to heat) in V.
Some preliminary mathematical results. W e now
derive two mathematical relationships that will be
useful later. Let E be the electric field intensity (SI
base units of V/m) and let D be the electric flux
density (SI base units of C/m 2). W e require that the
constituents of V be linear and time-invariant;
therefore, D = ǫE where the permittivity ǫis
constant with respect to time, but not necessarily with
respect to position. Under these conditions, we find
∂
∂t (E · D) = 1
ǫ
∂
∂t (D · D) (3.2)
Note",Electromagnetics_Vol2.pdf
"respect to position. Under these conditions, we find
∂
∂t (E · D) = 1
ǫ
∂
∂t (D · D) (3.2)
Note
that E and D point in the same direction. Let ˆe
be the unit vector describing this direction. Then,
E = ˆeEand D = ˆeDwhere Eand Dare the scalar
c⃝ C. W ang CC BY -SA 4.0
Figure 3.1: Poynting’s theorem describes the fate of
power entering a region V consisting of materials and
structures capable of storing and dissipating energy .
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"26 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
components of E and D, respectively . Subsequently ,
1
ǫ
∂
∂t (D · D) = 1
ǫ
∂
∂tD2
= 1
ǫ2D ∂
∂tD
= 2
E · ∂
∂tD (3.3)
Summarizing:
∂
∂t (E · D) = 2E · ∂
∂tD (3.4)
which
is the expression we seek. It is worth noting
that the expressions on both sides of the equation
have the same units, namely , those of power.
Using the same reasoning, we find:
∂
∂t (H · B) = 2H · ∂
∂tB (3.5)
where H is
the magnetic field intensity (SI base units
of A/m) and B is the magnetic flux density (SI base
units of T).
Derivation of the theorem. W e begin with the
differential form of Ampere’s law:
∇ × H = J + ∂
∂tD (3.6)
T
aking the dot product with E on both sides:
E · (∇ × H) = E · J + E · ∂
∂tD (3.7)
Let’
s deal with the left side of this equation first. For
this, we will employ a vector identity (Equation B.27,
Appendix B.3). This identity states that for any two
vector fields F and G,
∇ · (F × G) = G · (∇ × F) − F · (∇ × G) (3.8)",Electromagnetics_Vol2.pdf
"Appendix B.3). This identity states that for any two
vector fields F and G,
∇ · (F × G) = G · (∇ × F) − F · (∇ × G) (3.8)
Substituting E for F and H for G and rearranging
terms, we find
E · (∇ × H) = H · (∇ × E) − ∇ · (E × H) (3.9)
Next, we invoke the Maxwell-Faraday equation:
∇ × E = − ∂
∂tB (3.10)
Using
this equation to replace the factor in the
parentheses in the second term of Equation 3.9, we
obtain:
E · (∇ × H) = H ·
(
− ∂
∂tB
)
− ∇ · (E × H) (3.11)
Substituting this expression for the left side of
Equation 3.7, we have
− H · ∂
∂tB − ∇ · (E × H) = E · J + E · ∂
∂tD (3.12)
Ne
xt we invoke the identities of Equations 3.4 and 3.5
to replace the first and last terms:
−1
2
∂
∂t (H · B) − ∇ · (E × H)
= E · J + 1
2
∂
∂t (E · D) (3.13)
No
w we move the first term to the right-hand side and
integrate both sides over the volume V:
−
∫
V
∇ · (E × H) dv
=
∫
V
E · J dv
+
∫
V
1
2
∂
∂t (E · D) dv
+
∫
V
1
2
∂
∂t (H · B) dv (3.14)
The
left side may be transformed into a surface",Electromagnetics_Vol2.pdf
"−
∫
V
∇ · (E × H) dv
=
∫
V
E · J dv
+
∫
V
1
2
∂
∂t (E · D) dv
+
∫
V
1
2
∂
∂t (H · B) dv (3.14)
The
left side may be transformed into a surface
integral using the divergence theorem:
∫
V
∇ · (E × H) dv=
∮
S
(E × H) · ds (3.15)
where S is the closed surface that bounds V and ds is
the outward-facing normal to this surface, as
indicated in Figure 3.1. Finally , we exchange the
order of time differentiation and volume integration in
the last two terms:
−
∮
S
(E × H) · ds
=
∫
V
E · J dv
+ 1
2
∂
∂t
∫
V
E · D dv
+ 1
2
∂
∂t
∫
V
H · B dv (3.16)",Electromagnetics_Vol2.pdf
"3.1. POYNTING’S THEOREM 27
Equation 3.16 is Poynting’s theorem. Each of the four
terms has the particular physical interpretation
identified in Equation 3.1, as we will now
demonstrate.
Power dissipated by ohmic loss. The first term of
the right side of Equation 3.16 is
PΩ ≜
∫
V
E · J dv
(3.17)
Equation
3.17 is Joule’s law. Joule’s law gives the
power dissipated due to finite conductivity of
material. The role of conductivity (σ , SI base units of
S/m) can be made explicit using the relationship
J = σE (Ohm’s law) in Equation 3.17, which yields:
PΩ =
∫
V
E · σE dv=
∫
V
σ|E|2 dv (3.18)
PΩ is due to the conversion of the electric field into
conduction current, and subsequently into heat. This
mechanism is commonly known as ohmic loss or
joule heating. It is worth noting that this expression
has the expected units. For example, in
Equation 3.17, E (SI base units of V/m) times J (SI
base units of A/m 2) yields a quantity with units of
W/m3; i.e., power per unit volume, which is power",Electromagnetics_Vol2.pdf
"base units of A/m 2) yields a quantity with units of
W/m3; i.e., power per unit volume, which is power
density . Then integration over volume yields units of
W; hence, power.
Energy storage in electric and magnetic fields. The
second term on the right side of Equation 3.16 is:
PE ≜ 1
2
∂
∂t
∫
V
E · D dv (3.19)
Recall D = ǫE
, where ǫis the permittivity (SI base
units of F/m) of the material. Thus, we may rewrite
the previous equation as follows:
PE = 1
2
∂
∂t
∫
V
E · ǫE dv
= 1
2
∂
∂t
∫
V
ǫ|E|2 dv
= ∂
∂t
(1
2
∫
V
ǫ|E|2 dv
)
(3.20)
The
quantity in parentheses is We, the energy stored
in the electric field within V.
The third term on the right side of Equation 3.16 is:
PM ≜ 1
2
∂
∂t
∫
V
H · B dv (3.21)
Recall B = µH
, where µis the permeability (SI base
units of H/m) of the material. Thus, we may rewrite
the previous equation as follows:
PM = 1
2
∂
∂t
∫
V
H · µH dv
= 1
2
∂
∂t
∫
V
µ|H|2 dv
= ∂
∂t
(1
2
∫
V
µ|H|2 dv
)
(3.22)
The
quantity in parentheses is Wm, the energy stored",Electromagnetics_Vol2.pdf
"PM = 1
2
∂
∂t
∫
V
H · µH dv
= 1
2
∂
∂t
∫
V
µ|H|2 dv
= ∂
∂t
(1
2
∫
V
µ|H|2 dv
)
(3.22)
The
quantity in parentheses is Wm, the energy stored
in the magnetic field within V.
Power flux through S. The left side of Equation 3.16
is
Pnet,in ≜ −
∮
S
(E × H) · ds (3.23)
The
quantity E × H is the P oynting vector, which
quantifies the spatial power density (SI base units of
W/m2) of an electromagnetic wave and the direction
in which it propagates. The reader has likely already
encountered this concept. Regardless, we’ll confirm
this interpretation of the quantity E × H in
Section 3.2. For now , observe that integration of the
Poynting vector over S as indicated in Equation 3.23
yields the total power flowing out of V through S.
The negative sign in Equation 3.23 indicates that the
combined quantity represents power flow in to V
through S. Finally , note the use of a single quantity
Pnet,in does not imply that power is entirely
inward-directed or outward-directed. Rather, Pnet,in",Electromagnetics_Vol2.pdf
"Pnet,in does not imply that power is entirely
inward-directed or outward-directed. Rather, Pnet,in
represents the net flux; i.e., the sum of the inward-
and outward-flowing power.
Summary . Equation 3.16 may now be stated
concisely as follows:
Pnet,in = PΩ + ∂
∂tWe + ∂
∂tWm (3.24)",Electromagnetics_Vol2.pdf
"28 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
Poynting’s theorem (Equation 3.24, with Equa-
tions
3.23, 3.17, 3.19, and 3.21) states that the net
electromagnetic power flowing into a region of
space may be either dissipated, or used to change
the energy stored in electric and magnetic fields
within that region.
Since we derived this result directly from the general
form of Maxwell’s equations, these three possibilities
are all the possibilities allowed by classical physics,
so this is a statement of conservation of power.
Finally , note the theorem also works in reverse; i.e.,
the net electromagnetic power flowing out of a region
of space must have originated from some active
source (i.e., the reverse of power dissipation) or
released from energy stored in electric or magnetic
fields.
Additional Reading:
• “Poynting vector” on Wikipedia.
3.2 Poynting V ector
[m0122]
In this section, we use Poynting’s theorem
(Section
3.1) to confirm the interpretation of the
Poynting vector",Electromagnetics_Vol2.pdf
"3.2 Poynting V ector
[m0122]
In this section, we use Poynting’s theorem
(Section
3.1) to confirm the interpretation of the
Poynting vector
S ≜ E × H (3.25)
as
the spatial power density (SI base units of W/m 2)
and the direction of power flow .
Figure 3.2 shows a uniform plane wave incident on a
homogeneous and lossless region V defined by the
closed right cylinder S. The axis of the cylinder is
aligned along the direction of propagation ˆk of the
plane wave. The ends of the cylinder are planar and
perpendicular to ˆk.
Now let us apply Poynting’s theorem:
Pnet,in = PΩ + ∂
∂tWe + ∂
∂tWm (3.26)
Since
the region is lossless, PΩ = 0. Presuming no
other electric or magnetic fields, We and Wm must
also be zero. Consequently , Pnet,in = 0. However,
this does not mean that there is no power flowing into
V. Instead, Pnet,in = 0 merely indicates that the net
power flowing into V is zero. Thus, we might instead
express the result for the present scenario as follows:
Pnet,in = Pin − Pout = 0 (3.27)",Electromagnetics_Vol2.pdf
"power flowing into V is zero. Thus, we might instead
express the result for the present scenario as follows:
Pnet,in = Pin − Pout = 0 (3.27)
c⃝ C. W ang CC BY -SA 4.0
Figure 3.2: A uniform plane wave incident on a region
bounded by the cylindrical surface S.",Electromagnetics_Vol2.pdf
"3.2. POYNTING VECTOR 29
where Pin and Pout indicate power flow explicitly
into and explicitly out of V as separate quantities.
Proceeding, let’s ignore what we know about power
flow in plane waves, and instead see where Poynting’s
theorem takes us. Here,
Pnet,in ≜ −
∮
S
(E × H) · ds = 0 (3.28)
The surface S consists of three sides: The two flat
ends, and the curved surface that connects them. Let
us refer to these sides as S1, S2, and S3, from left to
right as shown in Figure 3.2. Then,
−
∫
S1
(E × H) · ds (3.29)
−
∫
S2
(E × H) · ds
−
∫
S3
(E × H) · ds = 0
On the curved surface S2, ds is everywhere
perpendicular to E × H (i.e., perpendicular to ˆk).
Therefore, the second integral is equal to zero. On the
left end S1, the outward-facing differential surface
element ds = −ˆkds. On the right end S3,
ds = + ˆkds. W e are left with:
∫
S1
(E × H) · ˆkds (3.30)
−
∫
S3
(E × H) · ˆkds= 0
Compare this to Equation 3.27. Since ˆk is, by
definition, the direction in which E × H points, the",Electromagnetics_Vol2.pdf
"S1
(E × H) · ˆkds (3.30)
−
∫
S3
(E × H) · ˆkds= 0
Compare this to Equation 3.27. Since ˆk is, by
definition, the direction in which E × H points, the
first integral must be Pin and the second integral must
be Pout. Thus,
Pin =
∫
S1
(E × H) · ˆkds (3.31)
and it follows that the magnitude and direction of
E × H, which we recognize as the Poynting vector,
are spatial power density and direction of power flow ,
respectively .
The Poynting vector is named after English physicist
J.H. Poynting, one of the co-discoverers of the
concept. The fact that this vector points in the
direction of power flow , and is therefore also a
“pointing vector, ” is simply a remarkable coincidence.
Additional Reading:
• “Poynting vector” on Wikipedia.",Electromagnetics_Vol2.pdf
"30 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
3.3 W ave Equations for Lossy
Regions
[m0128]
The wave equations for electromagnetic propagation
in
lossless and source-free media, in differential
phasor form, are:
∇2 ˜E + ω2µǫ˜E = 0 (3.32)
∇2 ˜H + ω2µǫ˜H = 0 (3.33)
The constant ω2µǫis labeled β2, and βturns out to be
the phase propagation constant.
Now , we wish to upgrade these equations to account
for the possibility of loss. First, let’s be clear on what
we mean by “loss. ” Specifically , we mean the
possibility of conversion of energy from the
propagating wave into current, and subsequently to
heat. This mechanism is described by Ohm’s law:
˜J = σ˜E (3.34)
where σis conductivity and ˜J is conduction current
density (SI base units of A/m 2). In the lossless case, σ
is presumed to be zero (or at least negligible), so J is
presumed to be zero. T o obtain wave equations for
media exhibiting significant loss, we cannot assume
J = 0.",Electromagnetics_Vol2.pdf
"presumed to be zero. T o obtain wave equations for
media exhibiting significant loss, we cannot assume
J = 0.
T o obtain equations that account for the possibility of
significant conduction current, hence loss, we return
to the phasor forms of Maxwell’s equations:
∇ · ˜E = ˜ρv
ǫ (3.35)
∇
× ˜E = −jωµ˜H (3.36)
∇ · ˜H = 0 (3.37)
∇ × ˜H = ˜J + jωǫ˜E (3.38)
where ˜ρv is the charge density . If the region of
interest is source-free, then ˜ρv = 0. However, we may
not similarly suppress ˜J since σmay be non-zero. T o
make progress, let us identify the possible
contributions to ˜J as follows:
˜J = ˜Jimp + ˜Jind (3.39)
where ˜Jimp represents impressed sources of current
and ˜Jind represents current which is induced by loss.
An impressed current is one whose behavior is
independent of other features, analogous to an
independent current source in elementary circuit
theory . In the absence of such sources, Equation 3.39
becomes:
˜J = 0 + σ˜E (3.40)
Equation 3.38 may now be rewritten as:",Electromagnetics_Vol2.pdf
"theory . In the absence of such sources, Equation 3.39
becomes:
˜J = 0 + σ˜E (3.40)
Equation 3.38 may now be rewritten as:
∇ × ˜H = σ˜E + jωǫ˜E (3.41)
= (σ + jωǫ) ˜E (3.42)
= jωǫc˜E (3.43)
where we defined the new constant ǫc as follows:
ǫc ≜ ǫ− jσ
ω (3.44)
This
constant is known as complex permittivity. In
the lossless case, ǫc = ǫ; i.e., the imaginary part of
ǫc → 0 so there is no difference between the physical
permittivity ǫand ǫc. The effect of the material’s loss
is represented as a non-zero imaginary component of
the permittivity .
It is also common to express ǫc as follows:
ǫc = ǫ′ − jǫ′′ (3.45)
where the real-valued constants ǫ′ and ǫ′′ are in this
case:
ǫ′ = ǫ (3.46)
ǫ′′ = σ
ω (3.47)
This
alternative notation is useful for three reasons.
First, some authors use the symbol ǫto refer to both
physical permittivity and complex permittivity . In this
case, the “ǫ ′ − jǫ′′” notation is helpful in mitigating
confusion. Second, it is often more convenient to",Electromagnetics_Vol2.pdf
"case, the “ǫ ′ − jǫ′′” notation is helpful in mitigating
confusion. Second, it is often more convenient to
specify ǫ′′ at a frequency than it is to specify σ, which
may also be a function of frequency . In fact, in some
applications the loss of a material is most
conveniently specified using the ratio ǫ′′/ǫ′ (known as
loss tangent, for reasons explained elsewhere).
Finally , it turns out that nonlinearity of permittivity
can also be accommodated as an imaginary",Electromagnetics_Vol2.pdf
"3.3. W A VE EQUA TIONS FOR LOSSY REGIONS 31
component of the permittivity . The “ǫ ′′” notation
allows us to accommodate both effects – nonlinearity
and conductivity – using common notation. In this
section, however, we remain focused exclusively on
conductivity .
Complex permittivity ǫc (SI base units of F/m)
describes the combined effects of permittivity
and conductivity . Conductivity is represented as
an imaginary-valued component of the permittiv-
ity .
Returning to Equations 3.35–3.38, we obtain:
∇ · ˜E = 0 (3.48)
∇ × ˜E = −jωµ˜H (3.49)
∇ · ˜H = 0 (3.50)
∇ × ˜H = jωǫc˜E (3.51)
These equations are identical to the corresponding
equations for the lossless case, with the exception that
ǫhas been replaced by ǫc. Similarly , we may replace
the factor ω2µǫin Equations 3.32 and 3.33, yielding:
∇2 ˜E + ω2µǫc˜E = 0 (3.52)
∇2 ˜H + ω2µǫc˜H = 0 (3.53)
In the lossless case, ω2µǫc → ω2µǫ, which is β2 as
expected. For the general (i.e., possibly lossy) case,
we shall make an analogous definition",Electromagnetics_Vol2.pdf
"In the lossless case, ω2µǫc → ω2µǫ, which is β2 as
expected. For the general (i.e., possibly lossy) case,
we shall make an analogous definition
γ2 ≜ −ω2µǫc (3.54)
such that the wave equations may now be written as
follows:
∇2 ˜E − γ2 ˜E = 0
(3.55)
∇2 ˜H − γ2 ˜H = 0 (3.56)
Note
the minus sign in Equation 3.54, and the
associated changes of signs in Equations 3.55 and
3.56. For the moment, this choice of sign may be
viewed as arbitrary – we would have been equally
justified in choosing the opposite sign for the
definition of γ2. However, the choice we have made
is customary , and yields some notational convenience
that will become apparent later.
In the lossless case, βis the phase propagation
constant, which determines the rate at which the
phase of a wave progresses with increasing distance
along the direction of propagation. Given the
similarity of Equations 3.55 and 3.56 to
Equations 3.32 and 3.33, respectively , the constant γ
must play a similar role. So, we are motivated to find",Electromagnetics_Vol2.pdf
"similarity of Equations 3.55 and 3.56 to
Equations 3.32 and 3.33, respectively , the constant γ
must play a similar role. So, we are motivated to find
an expression for γ. At first glance, this is simple:
γ =
√
γ2. However, recall that every number has two
square
roots. When one declares β =
√
β2 = ω√µǫ,
there
is no concern since βis, by definition, positive;
therefore, one knows to take the positive-valued
square root. In contrast, γ2 is complex-valued, and so
the two possible values of
√
γ2 are both potentially
comple
x-valued. W e are left without clear guidance
on which of these values is appropriate and
physically-relevant. So we proceed with caution.
First, consider the special case that γis purely
imaginary; e.g., γ = jγ′′ where γ′′ is a real-valued
constant. In this case, γ2 = − (γ′′)2 and the wave
equations become
∇2 ˜E + (γ′′)2 ˜E = 0 (3.57)
∇2 ˜H + (γ′′)2 ˜H = 0 (3.58)
Comparing to Equations 3.32 and 3.33, we see γ′′
plays the exact same role as β; i.e., γ′′ is the phase",Electromagnetics_Vol2.pdf
"∇2 ˜H + (γ′′)2 ˜H = 0 (3.58)
Comparing to Equations 3.32 and 3.33, we see γ′′
plays the exact same role as β; i.e., γ′′ is the phase
propagation constant for whatever wave we obtain as
the solution to the above equations. Therefore, let us
make that definition formally:
β ≜ Im {γ} (3.59)
Be careful: Note that we are not claiming that γ′′ in
the possibly-lossy case is equal to ω2µǫ. Instead, we
are asserting exactly the opposite; i.e., that there is a
phase propagation constant in the general (possibly
lossy) case, and we should find that this constant
simplifies to ω2µǫin the lossless case.
Now we make the following definition for the real
component of γ:
α≜ Re {γ} (3.60)
Such that
γ = α+ jβ (3.61)
where αand βare real-valued constants.
Now , it is possible to determine γexplicitly in terms
of ωand the constitutive properties of the material.",Electromagnetics_Vol2.pdf
"32 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
First, note:
γ2 = ( α+ jβ)2
= α2 − β2 + j2αβ (3.62)
Expanding Equation 3.54 using Equations 3.45–3.47,
we obtain:
γ2 = −ω2µ
(
ǫ− jσ
ω
)
= −ω2µǫ+ j
ωµσ (3.63)
The real and imaginary parts of Equations 3.62 and
3.63 must be equal. Enforcing this equality yields the
following equations:
α2 − β2 = −ω2µǫ (3.64)
2αβ = ωµσ (3.65)
Equations 3.64 and 3.65 are independent
simultaneous equations that may be solved for αand
β. Sparing the reader the remaining steps, which are
purely mathematical, we find:
α= ω



µǫ′
2


√
1 +
(ǫ′′
ǫ′
)2
− 1





1/ 2
(3.66)
β = ω



µǫ′
2


√
1 +
(ǫ′′
ǫ′
)2
+ 1





1
/ 2
(3.67)
Equations
3.66 and 3.67 can be verified by confirming
that Equations 3.64 and 3.65 are satisfied. It is also
useful to confirm that the expected results are
obtained in the lossless case. In the lossless case,
σ= 0, so ǫ′′ = 0. Subsequently , Equation 3.66 yields
α= 0 and Equation 3.67 yields β = ω√
µǫ, as
e
xpected.",Electromagnetics_Vol2.pdf
"σ= 0, so ǫ′′ = 0. Subsequently , Equation 3.66 yields
α= 0 and Equation 3.67 yields β = ω√
µǫ, as
e
xpected.
The electromagnetic wave equations account-
ing
for the possibility of lossy media are Equa-
tions 3.55 and 3.56 with γ = α+ jβ, where α
and βare the positive real-valued constants deter-
mined by Equations 3.66 and 3.67, respectively .
W e conclude this section by pointing out a very useful
analogy to transmission line theory . In the section
“W ave Propagation on a TEM Transmission Line, ” 1
we found that the potential and current along a
transverse electromagnetic (TEM) transmission line
satisfy the same wave equations that we have
developed in this section, having a complex-valued
propagation constant γ = α+ jβ, and the same
physical interpretation of βas the phase propagation
constant. As is explained in another section, αtoo has
the same interpretation in both applications – that is,
as the attenuation constant.
Additional Reading:",Electromagnetics_Vol2.pdf
"the same interpretation in both applications – that is,
as the attenuation constant.
Additional Reading:
• “Electromagnetic W ave Equation” on Wikipedia.
1 This section may appear in a different volume, depending on
the version of this book.",Electromagnetics_Vol2.pdf
"3.4. COMPLEX PERMITTIVITY 33
3.4 Complex Permittivity
[m0134]
The relationship between electric field intensity E (SI
base
units of V/m) and electric flux density D (SI
base units of C/m 2) is:
D = ǫE (3.68)
where ǫis the permittivity (SI base units of F/m). In
simple media, ǫis a real positive value which does not
depend on the time variation of E. That is, the
response (D) to a change in E is observed
instantaneously and without delay .
In practical materials, however, the change in D in
response to a change in E may depend on the manner
in which E changes. The response may not be
instantaneous, but rather might take some time to
fully manifest. This behavior can be modeled using
the following generalization of Equation 3.68:
D = a0E+a1
∂
∂tE+a2
∂2
∂t2 E+a3
∂3
∂t3 E+... (3.69)
where a0, a1, a2,
and so on are real-valued constants,
and the number of terms is infinite. In practical
materials, the importance of the terms tends to
diminish with increasing order. Thus, it is common",Electromagnetics_Vol2.pdf
"and the number of terms is infinite. In practical
materials, the importance of the terms tends to
diminish with increasing order. Thus, it is common
that only the first few terms are significant. In many
applications involving common materials, only the
first term is significant; i.e., a0 ≈ ǫand an ≈ 0 for
n≥ 1. As we shall see in a moment, this distinction
commonly depends on frequency .
In the phasor domain, differentiation with respect to
time becomes multiplication by jω. Thus,
Equation 3.69 becomes
˜D = a0 ˜E + a1 (jω) ˜E + a2 (jω)2 ˜E + a3 (jω)3 ˜E + ...
= a0 ˜E + jωa1 ˜E − ω2a2 ˜E − jω3a3 ˜E + ...
=
(
a0 + jωa1 − ω2a2 − jω3a3 + ...
)˜E (3.70)
Note that the factor in parentheses is a
complex-valued number which depends on frequency
ωand materials parameters a0, a1, ... W e may
summarize this as follows:
˜D = ǫc˜E (3.71)
where ǫc is a complex-valued constant that depends
on frequency .
The notation “ǫ c” is used elsewhere in this book (e.g.,",Electromagnetics_Vol2.pdf
"˜D = ǫc˜E (3.71)
where ǫc is a complex-valued constant that depends
on frequency .
The notation “ǫ c” is used elsewhere in this book (e.g.,
Section 3.3) to represent a generalization of simple
permittivity that accommodates loss associated with
non-zero conductivity σ. In Section 3.3, we defined
ǫc ≜ ǫ′ − jǫ′′ (3.72)
where ǫ′ ≜ ǫ(i.e., real-valued, simple permittivity)
and ǫ′′ ≜ σ/ω(i.e., the effect of non-zero
conductivity). Similarly , ǫc in Equation 3.71 can be
expressed as
ǫc ≜ ǫ′ − jǫ′′ (3.73)
but in this case
ǫ′ ≜ a0 − ω2a2 + ... (3.74)
and
ǫ′′ ≜ −ωa1 + ω3a3 − ... (3.75)
When expressed as phasors, the temporal rela-
tionship
between electric flux density and elec-
tric field intensity can be expressed as a complex-
valued permittivity .
Thus, we now see two possible applications for the
concept of complex permittivity: Modeling the
delayed response of D to changing E, as described
above; and modeling loss associated with non-zero",Electromagnetics_Vol2.pdf
"concept of complex permittivity: Modeling the
delayed response of D to changing E, as described
above; and modeling loss associated with non-zero
conductivity . In practical work, it may not always be
clear precisely what combination of effects the
complex permittivity ǫc = ǫ′ − jǫ′′ is taking into
account. For example, if ǫc is obtained by
measurement, both delayed response and conduction
loss may be represented. Therefore, it is not
reasonable to assume a value of ǫ′′ obtained by
measurement represents only conduction loss (i.e., is
equal to σ/ω); in fact, the measurement also may
include a significant frequency-dependent
contribution associated with the delayed response
behavior identified in this section.
An example of the complex permittivity of a typical
dielectric material is shown in Figure 3.3. Note the
frequency dependence is quite simple and
slowly-varying at frequencies below 1 GHz or so, but
becomes relatively complex at higher frequencies.",Electromagnetics_Vol2.pdf
"34 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
c⃝ K.A . Mauritz (modified)
Figure 3.3: The relative contributions of the real and
imaginary components of permittivity for a typical di-
electric material (in this case, a polymer).
Additional Reading:
• “Permittivity” on Wikipedia.
3.5 Loss T angent
[m0132]
In Section 3.3, we found that the effect of loss due to
non-zero
conductivity σcould be concisely quantified
using the ratio
ǫ′′
ǫ′ = σ
ωǫ (3.76)
where ǫ′ and ǫ′′ are
the real and imaginary
components of the complex permittivity ǫc, and
ǫ′ ≜ ǫ. In this section, we explore this relationship in
greater detail.
Recall Ampere’s law in differential (but otherwise
general) form:
∇ × H = J + ∂
∂tD (3.77)
The
first term on the right is conduction current,
whereas the second term on the right is displacement
current. In the phasor domain, differentiation with
respect to time (∂/∂t) becomes multiplication by jω.
Therefore, the phasor form of Equation 3.77 is
∇ × ˜H = ˜J + jω˜D (3.78)",Electromagnetics_Vol2.pdf
"respect to time (∂/∂t) becomes multiplication by jω.
Therefore, the phasor form of Equation 3.77 is
∇ × ˜H = ˜J + jω˜D (3.78)
Also recall that ˜J = σ˜E and that ˜D = ǫ˜E. Thus,
∇ × ˜H = σ˜E + jωǫ˜E (3.79)
Interestingly , the total current is the sum of a
real-valued conduction current and an
imaginary-valued displacement current. This is
shown graphically in Figure 3.4.
Note that the angle δindicated in Figure 3.4 is given
by
tan δ≜ σ
ωǫ (3.80)
The
quantity tan δis referred to as the loss tangent.
Note that loss tangent is zero for a lossless (σ ≡ 0)
material, and increases with increasing loss. Thus,
loss tangent provides an alternative way to quantify
the effect of loss on the electromagnetic field within a
material.
Loss tangent presuming only ohmic (conduction)
loss
is given by Equation 3.80.",Electromagnetics_Vol2.pdf
"3.5. LOSS T ANGENT 35
Comparing Equation 3.80 to Equation 3.76, we see
loss tangent can equivalently be calculated as
tan δ= ǫ′′
ǫ (3.81)
and
subsequently interpreted as shown in Figure 3.5.
The discussion in this section has assumed that ǫc is
complex-valued solely due to ohmic loss. However, it
is explained in Section 3.4 that permittivity may also
be complex-valued as a way to model delay in the
response of D to changing E. Does the concept of
loss tangent apply also in this case? Since the math
does not distinguish between permittivity which is
complex due to loss and permittivity which is
complex due to delay , subsequent
mathematically-derived results apply in either case.
On the other hand, there may be potentially
significant differences in the physical manifestation of
these effects. For example, a material having large
loss tangent due to ohmic loss might become hot
when a large electric field is applied, whereas a
material having large loss tangent due to delayed",Electromagnetics_Vol2.pdf
"loss tangent due to ohmic loss might become hot
when a large electric field is applied, whereas a
material having large loss tangent due to delayed
response might not. Summarizing:
The expression for loss tangent given by Equa-
tion
3.81 and Figure 3.5 does not distinguish be-
tween ohmic loss and delayed response.
c⃝ C. W ang CC BY -SA 4.0
Figure 3.4: In the phasor domain, the total current
is the sum of a real-valued conduction current and an
imaginary-valued displacement current.
c⃝ C. W ang CC BY -SA 4.0
Figure 3.5: Loss tangent defined in terms of the real
and imaginary components of the complex permittiv-
ity ǫc.
Additional Reading:
• “Dielectric loss” on Wikipedia.",Electromagnetics_Vol2.pdf
"36 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
3.6 Plane W aves in Lossy Regions
[m0130]
The electromagnetic wave equations for source-free
re
gions consisting of possibly-lossy material are (see
Section 3.3):
∇2 ˜E − γ2 ˜E = 0 (3.82)
∇2 ˜H − γ2 ˜H = 0 (3.83)
where
γ2 ≜ −ω2µǫc (3.84)
W e now turn our attention to the question, what are
the characteristics of waves that propagate in these
conditions? As in the lossless case, these equations
permit waves having a variety of geometries including
plane waves, cylindrical waves, and spherical waves.
In this section, we will consider the important special
case of uniform plane waves.
T o obtain the general expression for the uniform
plane wave solution, we follow precisely the same
procedure described in the section “Uniform Plane
W aves: Derivation. ” 2 Although that section presumed
lossless media, the only difference in the present
situation is that the real-valued constant +β2 is
replaced with the complex-valued constant −γ2.",Electromagnetics_Vol2.pdf
"lossless media, the only difference in the present
situation is that the real-valued constant +β2 is
replaced with the complex-valued constant −γ2.
Thus, we obtain the desired solution through a simple
modification of the solution for the lossless case. For
a wave exhibiting uniform magnitude and phase in
planes of constant z, we find that the electric field is:
˜E = ˆx ˜Ex + ˆy ˜Ey (3.85)
where
˜Ex = E+
x0e−γz + E−
x0e+γz (3.86)
˜Ey = E+
y0e−γz + E−
y0e+γz (3.87)
where the complex-valued coefficients E+
x0, E−
x0,
E+
y0, and E−
y0 are determined by boundary conditions
(possibly including sources) outside the region of
interest. This result can be confirmed by verifying
that Equations 3.86 and 3.87 each satisfy
Equation 3.82. Also, it may be helpful to note that
these expressions are identical to those obtained for
2 Depending on the version of this book, this section may appear
in a different volume.
the voltage and current in lossy transmission lines, as",Electromagnetics_Vol2.pdf
"2 Depending on the version of this book, this section may appear
in a different volume.
the voltage and current in lossy transmission lines, as
described in the section “W ave Equation for a TEM
Transmission Line. ” 3
Let’s consider the special case of an ˆx-polarized plane
wave propagating in the +ˆz direction:
˜E = ˆxE+
x0e−γz (3.88)
W e established in Section 3.3 that γmay be written
explicitly in terms of its real and imaginary
components as follows:
γ = α+ jβ (3.89)
where αand βare positive real-valued constants
depending on frequency (ω ) and constitutive
properties of the medium; i.e., permittivity ,
permeability , and conductivity . Thus:
˜E = ˆxE+
x0e−(α+ jβ)z
= ˆxE+
x0e−αze−jβz (3.90)
Observe that the variation of phase with distance is
determined by βthrough the factor e−jβz; thus, βis
the phase propagation constant and plays precisely
the same role as in the lossless case. Observe also that
the variation in magnitude is determined by αthrough",Electromagnetics_Vol2.pdf
"the same role as in the lossless case. Observe also that
the variation in magnitude is determined by αthrough
the real-valued factor e−αz. Specifically , magnitude is
reduced inverse-exponentially with increasing
distance along the direction of propagation. Thus, α
is the attenuation constant and plays precisely the
same role as the attenuation constant for a lossy
transmission line.
The presence of loss in material gives rise to a
real-v
alued factor e−αz which describes the at-
tenuation of the wave with propagation (in this
case, along z) in the material.
W e may continue to exploit the similarity of the
potentially-lossy and lossless plane wave results to
quickly ascertain the characteristics of the magnetic
field. In particular, the plane wave relationships apply
exactly as they do in the lossless case. These
relationships are:
˜H = 1
η
ˆk × ˜E (3.91)
3 Depending on the version of this book, this section may appear
in a different volume.",Electromagnetics_Vol2.pdf
"3.7. W A VE POWER IN A LOSSY MEDIUM 37
˜E = −ηˆk × ˜H (3.92)
where ˆk is the direction of propagation and ηis the
wave impedance. In the lossless case, η=
√
µ/ǫ;
ho
wever, in the possibly-lossy case we must replace
ǫ= ǫ′ with ǫc = ǫ′ − jǫ′′. Thus:
η→ ηc =
√ µ
ǫc
=
√ µ
ǫ′ − jǫ′′
=
√ µ
ǫ′
√
1
1 − j(ǫ′′/ǫ′) (3.93)
Thus:
ηc =
√ µ
ǫ′ ·
[
1 − jǫ′′
ǫ′
] −1/ 2
(3.94)
Remarkably
, we find that the wave impedance for a
lossy material is equal to
√
µ/ǫ– the wave
impedance
we would calculate if we neglected loss
(i.e., assumed σ= 0) – times a correction factor that
accounts for the loss. This correction factor is
complex-valued; therefore, E and H are not in phase
when propagating through lossy material. W e now
see that in the phasor domain:
˜H = 1
ηc
ˆk × ˜E (3.95)
˜E = −ηcˆk × ˜H (3.96)
The plane wave relationships in media which are
possibly
lossy are given by Equations 3.95 and
3.96, with the complex-valued wave impedance
given by Equation 3.94.
3.7 W ave Power in a Lossy
Medium
[m0133]",Electromagnetics_Vol2.pdf
"lossy are given by Equations 3.95 and
3.96, with the complex-valued wave impedance
given by Equation 3.94.
3.7 W ave Power in a Lossy
Medium
[m0133]
In this section, we consider the power associated with
wa
ves propagating in materials which are potentially
lossy; i.e., having conductivity σsignificantly greater
than zero. This topic has previously been considered
in the section “W ave Power in a Lossless Medium”
for the case in loss is not significant. 4 A review of that
section may be useful before reading this section.
Recall that the Poynting vector
S ≜ E × H (3.97)
indicates the power density (i.e., W/m 2) of a wave and
the direction of power flow . This is “instantaneous”
power, applicable to waves regardless of the way they
vary with time. Often we are interested specifically in
waves which vary sinusoidally , and which
subsequently may be represented as phasors. In this
case, the time-average Poynting vector is
Save ≜ 1
2Re
{
˜E × ˜H∗
}
(3.98)
Further",Electromagnetics_Vol2.pdf
"subsequently may be represented as phasors. In this
case, the time-average Poynting vector is
Save ≜ 1
2Re
{
˜E × ˜H∗
}
(3.98)
Further
, we have already used this expression to find
that the time-average power density for a
sinusoidally-varying uniform plane wave in a lossless
medium is simply
Save = |E0|2
2η (lossless case) (3.99)
where |E0| is the peak (as opposed to RMS)
magnitude of the electric field intensity phasor, and η
is the wave impedance.
Let us now use Equation 3.98 to determine the
expression corresponding to Equation 3.99 in the case
of possibly-lossy media. W e may express the electric
and magnetic field intensities of a uniform plane wave
as
˜E = ˆxE0e−αze−jβz (3.100)
and
˜H = ˆy E0
ηc
e−αze−jβ
z (3.101)
4 Depending on the version of this book, this section may appear
in another volume.",Electromagnetics_Vol2.pdf
"38 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
where αand βare the attenuation constant and phase
propagation constant, respectively , and ηc is the
complex-valued wave impedance. As written, these
expressions describe a wave which is +ˆx-polarized
and propagates in the +ˆz direction. W e make these
choices for convenience only – as long as the medium
is homogeneous and isotropic, we expect our findings
to apply regardless of polarization and direction of
propagation. Applying Equation 3.98:
Save = 1
2Re
{
˜E × ˜H∗
}
= 1
2 ˆz Re
{
|
E0|2
η∗c
e−2αz
}
= ˆz|E0|2
2 Re
{ 1
η∗c
}
e−2αz (3.102)
Because ηc is
complex-valued when the material is
lossy , we must proceed with caution. First, let us
write ηc explicitly in terms of its magnitude |ηc| and
phase ψη:
η≜ |η| ejψη (3.103)
Then:
η∗
c = |ηc| e−jψη (3.104)
(η∗
c)−1 = |ηc|−1 e+jψη (3.105)
Re
{
(η∗
c)−1
}
= |ηc|−1 cos ψη (3.106)
Then Equation 3.102 may be written:
Save = ˆz|E0|2
2 |ηc| e−2αz cos ψη (3.107)",Electromagnetics_Vol2.pdf
"c)−1 = |ηc|−1 e+jψη (3.105)
Re
{
(η∗
c)−1
}
= |ηc|−1 cos ψη (3.106)
Then Equation 3.102 may be written:
Save = ˆz|E0|2
2 |ηc| e−2αz cos ψη (3.107)
The time-average power density of the plane
wa
ve described by Equation 3.100 in a possibly-
lossy material is given by Equation 3.107.
As a check, note that this expression gives the
expected result for lossless media; i.e., α= 0,
|ηc| = η, and ψη = 0. W e now see that the effect of
loss is that power density is now proportional to
(e−αz)2, so, as expected, power density is
proportional to the square of either |E| or |H|. The
result further indicates a one-time scaling of the
power density by a factor of
|η|
|ηc| cos ψη <1 (3.108)
relati
ve to a medium without loss.
The reduction in power density due to non-
zero
conductivity is proportional to a distance-
dependent factor e−2αz and an additional factor
that depends on the magnitude and phase of ηc.",Electromagnetics_Vol2.pdf
"3.8. DECIBEL SCALE FOR POWER RA TIO 39
3.8 Decibel Scale for Power Ratio
[m0154]
In many disciplines within electrical engineering, it is
common
to evaluate the ratios of powers and power
densities that differ by many orders of magnitude.
These ratios could be expressed in scientific notation,
but it is more common to use the logarithmic decibel
(dB) scale in such applications.
In the conventional (linear) scale, the ratio of power
P1 to power P0 is simply
G= P1
P0
(linear units) (3.109)
Here, “G” might be interpreted as “power gain. ” Note
that G< 1 if P1 <P0 and G> 1 if P1 >P0.
In the decibel scale, the ratio of power P1 to power P0
is
G≜ 10 log10
P1
P0
(dB) (3.110)
where
“dB” denotes a unitless quantity which is
expressed in the decibel scale. Note that G< 0 dB
(i.e., is “negative in dB”) if P1 <P0 and G> 0 dB if
P1 >P0.
The power gain P1/P0 in dB is given by Equa-
tion 3.110.
Alternatively , one might choose to interpret a power
ratio as a loss Lwith L≜ 1/Gin linear units, which",Electromagnetics_Vol2.pdf
"tion 3.110.
Alternatively , one might choose to interpret a power
ratio as a loss Lwith L≜ 1/Gin linear units, which
is L= −Gwhen expressed in dB. Most often, but not
always, engineers interpret a power ratio as “gain” if
the output power is expected to be greater than input
power (e.g., as expected for an amplifier) and as
“loss” if output power is expected to be less than
input power (e.g., as expected for a lossy transmission
line).
Power loss Lis the reciprocal of power gain G.
Therefore, L = −G when these quantities are
expressed in dB.
Example 3.1. Po wer loss from a long cable.
A 2 W signal is injected into a long cable. The
power arriving at the other end of the cable is
10 µW . What is the power loss in dB?
In linear units:
G= 10 µW
2 W = 5 × 10−6 (linear units)
In dB:
G= 10 log 10
(
5 × 10−6)∼= −53.0 dB
L= −G∼
= +53.0 dBThe decibel scale is used in precisely the same way to
relate
ratios of spatial power densities for waves. For",Electromagnetics_Vol2.pdf
"5 × 10−6)∼= −53.0 dB
L= −G∼
= +53.0 dBThe decibel scale is used in precisely the same way to
relate
ratios of spatial power densities for waves. For
example, the loss incurred when the spatial power
density is reduced from S0 (SI base units of W/m 2) to
S1 is
L= 10 log 10
S0
S1
(dB) (3.111)
This works because the common units of m −2 in the
numerator and denominator cancel, leaving a power
ratio.
A common point of confusion is the proper use of the
decibel scale to represent voltage or current ratios. T o
avoid confusion, simply refer to the definition
expressed in Equation 3.110. For example, let’s say
P1 = V2
1 /R1 where V1 is potential and R1 is the
impedance across which V1 is defined. Similarly , let
us define P0 = V2
0 /R0 where V0 is potential and R0
is the impedance across which V0 is defined.
Applying Equation 3.110:
G≜ 10 log10
P1
P0
(dB)
=
10 log10
V2
1 /R1
V2
0 /R0
(dB) (3.112)
Now , if R1 = R0, then
G= 10 log 10
V2
1
V2
0
(dB)
=
10 log10
(V1
V0
)2
(dB)
=
20 log10
V1
V0",Electromagnetics_Vol2.pdf
"P1
P0
(dB)
=
10 log10
V2
1 /R1
V2
0 /R0
(dB) (3.112)
Now , if R1 = R0, then
G= 10 log 10
V2
1
V2
0
(dB)
=
10 log10
(V1
V0
)2
(dB)
=
20 log10
V1
V0
(dB) (3.113)",Electromagnetics_Vol2.pdf
"40 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
However, note that this is not true if R1 ̸=R0.
A power ratio in dB is equal to 20 log10 of the
voltage ratio only if the associated impedances
are equal.
Adding to the potential for confusion on this point is
the concept of voltage gain Gv:
Gv ≜ 20 log10
V1
V0
(dB) (3.114)
which applies regardless of the associated
impedances. Note that Gv = Gonly if the associated
impedances are equal, and that these ratios are
different otherwise. Be careful!
The decibel scale simplifies common calculations.
Here’s an example. Let’s say a signal having power
P0 is injected into a transmission line having loss L.
Then the output power P1 = P0/Lin linear units.
However, in dB, we find:
10 log10 P1 = 10 log 10
P0
L
= 10
log10 P0 − 10 log10 L
Division has been transformed into subtraction; i.e.,
P1 = P0 − L (dB) (3.115)
This form facilitates easier calculation and
visualization, and so is typically preferred.
Finally , note that the units of P1 and P0 in",Electromagnetics_Vol2.pdf
"This form facilitates easier calculation and
visualization, and so is typically preferred.
Finally , note that the units of P1 and P0 in
Equation 3.115 are not dB per se, but rather dB with
respect to the original power units. For example, if P1
is in mW , then taking 10 log10 of this quantity results
in a quantity having units of dB relative to 1 mW . A
power expressed in dB relative to 1 mW is said to
have units of “dBm. ” For example, “0 dBm” means
0 dB relative to 1 mW , which is simply 1 mW .
Similarly +10 dBm is 10 mW , −10 dBm is 0.1 mW ,
and so on.
Additional Reading:
• “Decibel” on Wikipedia.
• “Scientific notation” on Wikipedia.
3.9 Attenuation Rate
[m0155]
Attenuation rate is
 a convenient way to quantify loss
in general media, including transmission lines, using
the decibel scale.
Consider a transmission line carrying a wave in the
+zdirection. Let P0 be the power at z= 0. Let P1 be
the power at z= l. Then the power at z= 0 relative
to the power at z= lis:
P0
P1
= e−2α·0",Electromagnetics_Vol2.pdf
"+zdirection. Let P0 be the power at z= 0. Let P1 be
the power at z= l. Then the power at z= 0 relative
to the power at z= lis:
P0
P1
= e−2α·0
e−2α·l = e2αl (linear units) (3.116)
where αis the attenuation constant; that is, the real
part of the propagation constant γ = α+ jβ.
Expressed in this manner, the power ratio is a loss;
that is, a number greater than 1 represents attenuation.
In the decibel scale, the loss is
10 log10
P0
P1
= 10 log10 e2αl
= 20αl log10 e
∼= 8.69αl dB (3.117)
Attenuation rate is defined as this quantity per unit
length. Dividing by l, we obtain:
attenuation rate ∼= 8.69α (3.118)
This has units of dB/length, where the units of length
are the same length units in which αis expressed. For
example, if αis expressed in units of m −1, then
attenuation rate has units of dB/m.
Attenuation rate ∼= 8.69 α is the loss in dB, per
unit length.
The utility of the attenuation rate concept is that it
allows us to quickly calculate loss for any distance of",Electromagnetics_Vol2.pdf
"unit length.
The utility of the attenuation rate concept is that it
allows us to quickly calculate loss for any distance of
wave travel: This loss is simply attenuation rate
(dB/m) times length (m), which yields loss in dB.
Example 3.2. Attenuation rate in a long cable.
A particular coaxial cable has an attenuation
constant α∼= 8.5 × 10−3 m−1. What is the
attenuation rate and the loss in dB for 100 m of
this cable?",Electromagnetics_Vol2.pdf
"3.10. POOR CONDUCTORS 41
The attenuation rate is
∼= 8.69
α∼
= 0.0738 dB/mThe loss in 100 m of this cable is
∼= (0
.0738 dB/m) (100 m) ∼
= 7.4 dBNote that it would be entirely appropriate, and
equi
valent, to state that the attenuation rate for
this cable is 7.4 dB/(100 m).
The concept of attenuation rate is used in precisely
the same way to relate ratios of spatial power
densities for unguided waves. This works because
spatial power density has SI base units of W/m 2, so
the common units of m −2 in the numerator and
denominator cancel in the power density ratio,
leaving a simple power ratio.
3.10 Poor Conductors
[m0156]
A poor conductor is
 a material for which conductivity
is low , yet sufficient to exhibit significant loss. T o be
clear, the loss we refer to here is the conversion of the
electric field to current through Ohm’s law .
The threshold of significance depends on the
application. For example, the dielectric spacer
separating the conductors in a coaxial cable might be",Electromagnetics_Vol2.pdf
"The threshold of significance depends on the
application. For example, the dielectric spacer
separating the conductors in a coaxial cable might be
treated as lossless (σ = 0) for short lengths at low
frequencies; whereas the loss of the cable for long
lengths and higher frequencies is typically significant,
and must be taken into account. In the latter case, the
material is said to be a poor conductor because the
loss is significant yet the material can still be treated
in most other respects as an ideal dielectric.
A quantitative but approximate criterion for
identification of a poor conductor can be obtained
from the concept of complex permittivity ǫc, which
has the form:
ǫc = ǫ′ − jǫ′′ (3.119)
Recall that ǫ′′ quantifies loss, whereas ǫ′ exists
independently of loss. In fact, ǫc = ǫ′ = ǫfor a
perfectly lossless material. Therefore, we may
quantify the lossiness of a material using the ratio
ǫ′′/ǫ′, which is sometimes referred to as loss tangent",Electromagnetics_Vol2.pdf
"quantify the lossiness of a material using the ratio
ǫ′′/ǫ′, which is sometimes referred to as loss tangent
(see Section 3.5). Using this quantity , we define a
poor conductor as a material for which ǫ′′ is very
small relative to ǫ′. Thus,
ǫ′′
ǫ′ ≪ 1 (poor conductor) (3.120)
A poor conductor is a material having loss tan-
gent
much less than 1, such that it behaves in
most respects as an ideal dielectric except that
ohmic loss may not be negligible.
An example of a poor conductor commonly
encountered in electrical engineering includes the
popular printed circuit board substrate material FR4
(fiberglass epoxy), which has ǫ′′/ǫ′ ∼ 0.008 over the
frequency range it is most commonly used. Another
example, already mentioned, is the dielectric spacer",Electromagnetics_Vol2.pdf
"42 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
material (for example, polyethylene) typically used in
coaxial cables. The loss of these materials may or
may not be significant, depending on the particulars
of the application.
The imprecise definition of Equation 3.120 is
sufficient to derive some characteristics exhibited by
all poor conductors. T o do so, first recall that the
propagation constant γis given in general as follows:
γ2 = −ω2µǫc (3.121)
Therefore:
γ =
√
−ω2µǫc (3.122)
In
general a number has two square roots, so some
caution is required here. In this case, we may proceed
as follows:
γ = jω√
µ
√
ǫ′ − jǫ′′
= j
ω
√
µǫ′
√
1 − jǫ′′
ǫ′ (3.123)
The
requirement that ǫ′′/ǫ′ ≪ 1 for a poor conductor
allows this expression to be “linearized. ” For this, we
invoke the binomial series representation:
(1 + x)n = 1 + nx+ n(n− 1)
2! x2 + ... (3.124)
where xand nare,
for our purposes, any constants;
and “... ” indicates the remaining terms in this infinite",Electromagnetics_Vol2.pdf
"2! x2 + ... (3.124)
where xand nare,
for our purposes, any constants;
and “... ” indicates the remaining terms in this infinite
series, with each term containing the factor xn with
n> 2. If x≪ 1, then all terms containing xn with
n≥ 2 will be very small relative to the first two terms
of the series. Thus,
(1 + x)n ≈ 1 + nx for x≪ 1 (3.125)
Applying this to the present problem:
(
1 − jǫ′′
ǫ′
)1/ 2
≈ 1 − jǫ′′
2ǫ′ (3.126)
where
we have used n= 1/2 and x= −jǫ′′/ǫ′.
Applying this approximation to Equation 3.123, we
obtain:
γ ≈ jω
√
µǫ′
(
1 − jǫ′′
2ǫ′
)
≈ jω
√
µǫ′ + ω
√
µǫ′ ǫ′′
2ǫ′ (3.127)
At
this point, we are able to identify an expression for
the phase propagation constant:
β ≜ Im {γ} ≈ ω
√
µǫ′ (poor conductor) (3.128)
Remarkably
, we find that βfor a poor conductor is
approximately equal to βfor an ideal dielectric.
For the attenuation constant, we find
α≜ Re {γ} ≈ ω
√
µǫ′ ǫ′′
2ǫ′ (poor conductor)
(3.129)
Alternati
vely , this expression may be written in the
following form:
α≈ 1
2βǫ′′",Electromagnetics_Vol2.pdf
"α≜ Re {γ} ≈ ω
√
µǫ′ ǫ′′
2ǫ′ (poor conductor)
(3.129)
Alternati
vely , this expression may be written in the
following form:
α≈ 1
2βǫ′′
ǫ′ (poor conductor) (3.130)
Presuming that ǫc is determined entirely by ohmic
loss, then
ǫ′′
ǫ′ = σ
ωǫ (3.131)
Under
this condition, Equation 3.129 may be
rewritten:
α≈ ω
√
µǫ′ σ
2ωǫ (poor conductor) (3.132)
Since ǫ′ = ǫunder these assumptions, the expression
simplifies to
α≈ σ
2
√ µ
ǫ = 1
2ση (poor conductor) (3.133)
where η≜
√
µ/ǫ′ is the wave impedance presuming
lossless
material. This result is remarkable for two
reasons: First, factors of ωhave been eliminated, so
there is no dependence on frequency separate from
the frequency dependence of the constitutive
parameters σ, µ, and ǫ. These parameters vary slowly
with frequency , so the value of αfor a poor conductor
also varies slowly with frequency . Second, we see α
is proportional to σand η. This makes it quite easy to
anticipate how the attenuation constant is affected by",Electromagnetics_Vol2.pdf
"is proportional to σand η. This makes it quite easy to
anticipate how the attenuation constant is affected by
changes in conductivity and wave impedance in poor
conductors.
Finally , what is the wave impedance in a poor
conductor? In contrast to η, ηc is potentially
complex-valued and may depend on σ. First, recall:
ηc =
√
µ
ǫ′ ·
[
1 − jǫ′′
ǫ′
] −1/ 2
(3.134)",Electromagnetics_Vol2.pdf
"3.11. GOOD CONDUCTORS 43
Applying the same approximation applied to γearlier,
this may be written
ηc ≈
√ µ
ǫ′ ·
[
1 − jǫ′′
2ǫ′
]
(poor conductor) (3.135)
W e see that for a poor conductor, Re{ ηc} ≈ ηand
that Im{ηc} ≪ Re {ηc}. The usual approximation in
this case is simply
ηc ≈ η (poor conductor) (3.136)
Additional
Reading:
• “Binomial series” on Wikipedia.
3.11 Good Conductors
[m0157]
A good conductor is a material which behaves in
most respects as a perfect conductor, yet exhibits
significant loss. Now , we have to be very careful: The
term “loss” applied to the concept of a “conductor”
means something quite different from the term “loss”
applied to other types of materials. Let us take a
moment to disambiguate this term.
Conductors are materials which are intended to
efficiently sustain current, which requires high
conductivity σ. In contrast, non-conductors are
materials which are intended to efficiently sustain the
electric field, which requires low σ. “Loss” for a",Electromagnetics_Vol2.pdf
"materials which are intended to efficiently sustain the
electric field, which requires low σ. “Loss” for a
non-conductor (see in particular “poor conductors, ”
Section 3.10) means the conversion of energy in the
electric field into current. In contrast, “loss” for a
conductor refers to energy already associated with
current, which is subsequently dissipated in
resistance. Summarizing: A good (“low-loss”)
conductor is a material with high conductivity , such
that power dissipated in the resistance of the material
is low .
A quantitative criterion for a good conductor can be
obtained from the concept of complex permittivity ǫc,
which has the form:
ǫc = ǫ′ − jǫ′′ (3.137)
Recall that ǫ′′ is proportional to conductivity (σ ) and
so ǫ′′ is very large for a good conductor. Therefore,
we may identify a good conductor using the ratio
ǫ′′/ǫ′, which is sometimes referred to as “loss
tangent” (see Section 3.5). Using this quantity we
define a good conductor as a material for which:
ǫ′′",Electromagnetics_Vol2.pdf
"tangent” (see Section 3.5). Using this quantity we
define a good conductor as a material for which:
ǫ′′
ǫ′ ≫ 1 (good conductor) (3.138)
This condition is met for most materials classified as
“metals, ” and especially for metals exhibiting very
high conductivity such as gold, copper, and
aluminum.
A good conductor is a material having loss tan-
gent
much greater than 1.",Electromagnetics_Vol2.pdf
"44 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
The imprecise definition of Equation 3.138 is
sufficient to derive some characteristics that are
common to materials over a wide range of
conductivity . T o derive these characteristics, first
recall that the propagation constant γis given in
general as follows:
γ2 = −ω2µǫc (3.139)
Therefore:
γ =
√
−ω2µǫc (3.140)
In
general, a number has two square roots, so some
caution is required here. In this case, we may proceed
as follows:
γ = jω√µ
√
ǫ′ − jǫ′′
= j
ω
√
µǫ′
√
1 − jǫ′′
ǫ′ (3.141)
Since ǫ′′/ǫ′ ≫ 1 for
a good conductor,
γ ≈ jω
√
µǫ′
√
−jǫ′′
ǫ′
≈ jω
√
µǫ′′
√
−j (3.142)
T
o proceed, we must determine the principal value of√−j. The answer is that √−j = (1 − j) /
√
2.5
Continuing:
γ ≈ j
ω
√
µǫ′′
2 + ω
√
µǫ′′
2 (3.143)
W
e are now able to identify expressions for the
attenuation and phase propagation constants:
α≜ Re {γ} ≈ ω
√
µǫ′′
2 (good conductor)
(3.144)
β ≜ Im {
γ} ≈ α (good conductor) (3.145)
Remarkably",Electromagnetics_Vol2.pdf
"attenuation and phase propagation constants:
α≜ Re {γ} ≈ ω
√
µǫ′′
2 (good conductor)
(3.144)
β ≜ Im {
γ} ≈ α (good conductor) (3.145)
Remarkably
, we find that α≈ βfor a good conductor,
and neither αnor βdepend on ǫ′.
In the special case that ǫc is determined entirely by
conductivity loss (i.e., σ >0) and is not accounting
5 You can confirm this simply by squaring this result. The easiest
way to derive this result is to work in polar form, in which −j is 1 at
an angle of −π/2, and the square root operation consists of taking
the square root of the magnitude and dividing the phase by 2.
for delayed polarization response (as described in
Section 3.4), then
ǫ′′ = σ
ω (3.146)
Under
this condition, Equation 3.144 may be
rewritten:
α≈ ω
√ µσ
2ω =
√ ωµσ
2 (good conductor) (3.147)
Since ω= 2πf, another possible form for this
expression is
α≈
√
πfµσ (good conductor) (3.148)
The
conductivity of most materials changes very
slowly with frequency , so this expression indicates",Electromagnetics_Vol2.pdf
"expression is
α≈
√
πfµσ (good conductor) (3.148)
The
conductivity of most materials changes very
slowly with frequency , so this expression indicates
that α(and β) increases approximately in proportion
to the square root of frequency for good conductors.
This is commonly observed in electrical engineering
applications. For example, the attenuation rate of
transmission lines increases approximately as √f.
This
is so because the principal contribution to the
attenuation is resistance in the conductors comprising
the line.
The attenuation rate for signals conveyed by
transmission
lines is approximately proportional
to the square root of frequency .
Let us now consider the wave impedance ηc in a good
conductor. Recall:
ηc =
√ µ
ǫ′ ·
[
1 − jǫ′′
ǫ′
] −1/ 2
(3.149)
Applying
the same approximation applied to γearlier
in this section, the previous expression may be written
ηc ≈
√ µ
ǫ′ ·
[
−jǫ′′
ǫ′
] −1/ 2
(good conductor)
≈
√ µ
ǫ′′ · 1√−j (3.150)
W
e’ve already established that √−j = (1 − j) /",Electromagnetics_Vol2.pdf
"ηc ≈
√ µ
ǫ′ ·
[
−jǫ′′
ǫ′
] −1/ 2
(good conductor)
≈
√ µ
ǫ′′ · 1√−j (3.150)
W
e’ve already established that √−j = (1 − j) /
√
2.
Applying
that result here:
ηc ≈
√ µ
ǫ′′ ·
√
2
1 − j (3.151)",Electromagnetics_Vol2.pdf
"3.11. GOOD CONDUCTORS 45
Now multiplying numerator and denominator by
1 + j, we obtain
ηc ≈
√ µ
2ǫ′′ · ( 1 + j) (3.152)
In the special case that ǫc is determined entirely by
conductivity loss and is not accounting for delayed
polarization response, then ǫ′′ = σ/ω, and we find:
ηc ≈
√
µω
2σ · (1 + j) (3.153)
There
are at least two other ways in which this
expression is commonly written. First, we can use
ω= 2πf to obtain:
ηc ≈
√
πfµ
σ · (1 + j) (3.154)
Second, we can use the fact that α≈ √πfµσ for
good
conductors to obtain:
ηc ≈ α
σ · (1 + j) (3.155)
In any event, we see that the magnitude of the wave
impedance bears little resemblance to the wave
impedance for a poor conductor. In particular, there is
no dependence on the physical permittivity ǫ′ = ǫ, as
we saw also for αand β. In this sense, the concept of
permittivity does not apply to good conductors, and
especially so for perfect conductors.
Note also that ψη, the phase of ηc, is always ≈ π/4",Electromagnetics_Vol2.pdf
"permittivity does not apply to good conductors, and
especially so for perfect conductors.
Note also that ψη, the phase of ηc, is always ≈ π/4
for a good conductor, in contrast to ≈ 0 for a poor
conductor. This has two implications that are useful
to know . First, since ηc is the ratio of the magnitude
of the electric field intensity to the magnitude of the
magnetic field intensity , the phase of the magnetic
field will be shifted by ≈ π/4 relative to the phase of
the electric field in a good conductor. Second, recall
from Section 3.7 that the power density for a wave is
proportional to cos ψη. Therefore, the extent to which
a good conductor is able to “extinguish” a wave
propagating through it is determined entirely by α,
and specifically is proportional to e−αl where lis
distance traveled through the material. In other words,
only a perfect conductor (σ → ∞) is able to
completely suppress wave propagation, whereas
waves are always able to penetrate some distance into",Electromagnetics_Vol2.pdf
"only a perfect conductor (σ → ∞) is able to
completely suppress wave propagation, whereas
waves are always able to penetrate some distance into
any conductor which is merely “good. ” A measure of
this distance is the skin depth of the material. The
concept of skin depth is presented in Section 3.12.
The dependence of βon conductivity leads to a
particularly surprising result for the phase velocity of
the beleaguered waves that do manage to propagate
within a good conductor. Recall that for both lossless
and low-loss (“poor conductor”) materials, the phase
velocity vp is either exactly or approximately c/√
ǫr,
where ǫr ≜ ǫ′/ǫ0,
resulting in typical phase velocities
within one order of magnitude of c. For a good
conductor, we find instead:
vp = ω
β ≈ ω√πfµσ (good conductor) (3.156)
and since ω= 2πf:
vp ≈
√
4πf
µσ (good conductor) (3.157)
Note that the phase velocity in a good conductor
increases with frequency and decreases with
conductivity . In contrast to poor conductors and",Electromagnetics_Vol2.pdf
"Note that the phase velocity in a good conductor
increases with frequency and decreases with
conductivity . In contrast to poor conductors and
non-conductors, the phase velocity in good
conductors is usually a tiny fraction of c. For
example, for a non-magnetic (µ ≈ µ0) good
conductor with typical σ∼ 106 S/m, we find
vp ∼ 100 km/s at 1 GHz – just ∼ 0.03% of the speed
of light in free space.
This result also tells us something profound about the
nature of signals that are conveyed by transmission
lines. Regardless of whether we analyze such signals
as voltage and current waves associated with the
conductors or in terms of guided waves between the
conductors, we find that the phase velocity is within
an order of magnitude or so of c. Thus, the
information conveyed by signals propagating along
transmission lines travels primarily within the space
between the conductors, and not within the
conductors. Information cannot travel primarily in the",Electromagnetics_Vol2.pdf
"between the conductors, and not within the
conductors. Information cannot travel primarily in the
conductors, as this would then result in apparent
phase velocity which is orders of magnitude less than
c, as noted previously . Remarkably , classical
transmission line theory employing the R′, G′, C′, L′
equivalent circuit model 6 gets this right, even though
that approach does not explicitly consider the
possibility of guided waves traveling between the
conductors.
6 Depending on the version of this book, this topic may appear in
another volume.",Electromagnetics_Vol2.pdf
"46 CHAPTER 3. W A VE PROP AGA TION IN GENERAL MEDIA
3.12 Skin Depth
[m0158]
The electric and magnetic fields of a wave are
diminished
as the wave propagates through lossy
media. The magnitude of these fields is proportional
to
e−αl
where α≜ Re {γ} is the attenuation constant (SI base
units of m −1), γis the propagation constant, and lis
the distance traveled. Although the rate at which
magnitude is reduced is completely described by α,
particular values of αtypically do not necessarily
provide an intuitive sense of this rate.
An alternative way to characterize attenuation is in
terms of skin depth δs, which is defined as the
distance at which the magnitude of the electric and
magnetic fields is reduced by a factor of 1/e. In other
words:
e−αδs = e−1 ∼= 0.368 (3.158)
Skin depth δs is the distance over which the mag-
nitude of the electric or magnetic field is reduced
by a factor of 1/e∼= 0.368.
Since power is proportional to the square of the field",Electromagnetics_Vol2.pdf
"nitude of the electric or magnetic field is reduced
by a factor of 1/e∼= 0.368.
Since power is proportional to the square of the field
magnitude, δs may also be interpreted as the distance
at which the power in the wave is reduced by a factor
of (1/e)2 ∼= 0.135. In yet other words: δs is the
distance at which ∼
= 86.5% of the power in the wave
is lost.
This definition for skin depth makes δs easy to
compute: From Equation 3.158, it is simply
δs = 1
α (3.159)
The
concept of skin depth is most commonly applied
to good conductors. For a good conductor,
α≈ √πfµσ (Section 3.11), so
δs ≈ 1√πfµσ (good conductors) (3.160)
Example 3.3. Skin depth of aluminum.
Aluminum, a good conductor, exhibits
σ≈ 3.7 × 107 S/m and µ≈ µ0 over a broad
range of radio frequencies. Using
Equation 3.160, we find δs ∼ 26 µm at 10 MHz.
Aluminum sheet which is 1/16-in ( ∼= 1.59 mm)
thick can also be said to have a thickness of
∼ 61δs at 10 MHz. The reduction in the power
density of an electromagnetic wave after",Electromagnetics_Vol2.pdf
"thick can also be said to have a thickness of
∼ 61δs at 10 MHz. The reduction in the power
density of an electromagnetic wave after
traveling through this sheet will be
∼
(
e−α(61δs)
)2
=
(
e−α(61/α)
)2
= e−122
which is effectively zero from a practical
engineering perspective. Therefore, 1/16-in
aluminum sheet provides excellent shielding
from electromagnetic waves at 10 MHz.
At 1 kHz, the situation is significantly different.
At this frequency , δs ∼ 2.6 mm, so 1/16-in
aluminum is only ∼ 0.6δs thick. In this case the
power density is reduced by only
∼
(
e−α(0.6δ s)
)2
=
(
e−α(0.6/α)
)2
= e−1.2 ≈ 0.3
This is a reduction of only ∼ 70% in power
density . Therefore, 1/16-in aluminum sheet
provides very little shielding at 1 kHz.
[m0183]",Electromagnetics_Vol2.pdf
"3.12. SKIN DEPTH 47
Image Credits
Fig. 3.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Poynting%E2%80%99s theorem illustration.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 3.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Uniform plane wave incident cylindrical surface.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 3.3: c⃝ K.A. Mauritz, https://commons.wikimedia.org/wiki/File:Dielectric responses.svg,
Used
with permission (see URL) and modified by author.
Fig. 3.4: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:T otal current in phasor domain.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 3.5: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Loss tangent definition.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 4
Curr
ent Flow in Imperfect Conductors
4.1 AC Current Flow in a Good
Conductor
[m0069]
In this section, we consider the distribution of current
in
a conductor which is imperfect (i.e., a “good
conductor”) and at frequencies greater than DC.
T o establish context, consider the simple DC circuit
shown in Figure 4.1. In this circuit, the current source
provides a steady current which flows through a
cylinder-shaped wire. As long as the conductivity σ
(SI base units of S/m) of the wire is uniform
throughout the wire, the current density J (SI base
units of A/m 2) is uniform throughout the wire.
Now let us consider the AC case. Whereas the electric
field intensity E is constant in the DC case, E exists
as a wave in the AC case. In a good conductor, the
c⃝ C. W ang CC BY -SA 4.0
Figure 4.1: Current flow in cylinder at DC.
magnitude of E decreases in proportion to e−αd
where αis the attenuation constant and dis distance
traversed by the wave. The attenuation constant",Electromagnetics_Vol2.pdf
"magnitude of E decreases in proportion to e−αd
where αis the attenuation constant and dis distance
traversed by the wave. The attenuation constant
increases with increasing σ, so the rate of decrease of
the magnitude of E increases with increasing σ.
In the limiting case of a perfect conductor, α→ ∞
and so E → 0 everywhere inside the material. Any
current within the wire must be the result of either an
impressed source or it must be a response to E.
Without either of these, we conclude that in the AC
case, J → 0 everywhere inside a perfect conductor. 1
But if J = 0 in the material, then how does current
pass through the wire? W e are forced to conclude that
the current must exist as a surface current ; i.e.,
entirely outside the wire, yet bound to the surface of
the wire. Thus:
In the AC case, the current passed by a perfectly-
conducting
material lies entirely on the surface of
the material.
The perfectly-conducting case is unobtainable in",Electromagnetics_Vol2.pdf
"conducting
material lies entirely on the surface of
the material.
The perfectly-conducting case is unobtainable in
practice, but the result gives us a foothold from which
we may determine what happens when σis not
infinite. If σis merely finite, then αis also finite and
subsequently wave magnitude may be non-zero over
finite distances.
Let us now consider the direction in which this
putative wave propagates. The two principal
directions in the present problem are parallel to the
1 Y ou might be tempted to invoke Ohm’s law (J = σE) to argue
against this conclusion. However, Ohm’s law provides no useful
information about the current in this case, since σ → ∞ at the same
time E → 0. What Ohm’s law is really saying in this case is that
E = J/σ → 0 because J must be finite and σ → ∞ .
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"4.1. AC CURRENT FLOW IN A GOOD CONDUCTOR 49
axis of the wire and perpendicular to the axis of the
wire. W aves propagating in any other direction may
be expressed as a linear combination of waves
traveling in the principal directions, so we need only
consider the principal directions to obtain a complete
picture.
Consider waves propagating in the perpendicular
direction first. In this case, we presume a boundary
condition in the form of non-zero surface current,
which we infer from the perfectly-conducting case
considered previously . W e also note that E deep
inside the wire must be weaker than E closer to the
surface, since a wave deep inside the wire must have
traversed a larger quantity of material than a wave
measured closer to the surface. Applying Ohm’s law
(J = σE), the current deep inside the wire must
similarly diminish for a good conductor. W e conclude
that a wave traveling in the perpendicular direction
exists and propagates toward the center of the wire,",Electromagnetics_Vol2.pdf
"that a wave traveling in the perpendicular direction
exists and propagates toward the center of the wire,
decreasing in magnitude with increasing distance
from the surface.
W e are unable to infer the presence of a wave
traveling in the other principal direction – i.e., along
the axis of the wire – since there is no apparent
boundary condition to be satisfied on either end of the
wire. Furthermore, the presence of such a wave would
mean that different cross-sections of the wire exhibit
different radial distributions of current. This is not
consistent with physical observations.
W e conclude that the only relevant wave is one which
travels from the surface of the wire inward. Since the
current density is proportional to the electric field
magnitude, we conclude:
In the AC case, the current passed by a wire com-
prised
of a good conductor is distributed with
maximum current density on the surface of the
wire, and the current density decays exponen-",Electromagnetics_Vol2.pdf
"prised
of a good conductor is distributed with
maximum current density on the surface of the
wire, and the current density decays exponen-
tially with increasing distance from the surface.
This phenomenon is known as the skin effect,
referring to the notion of current forming a skin-like
layer below the surface of the wire. The effect is
illustrated in Figure 4.2.
Since αincreases with increasing frequency , we see
that the specific distribution of current within the wire
by Biezl (modified) (public domain)
Figure
4.2: Distribution of AC current in a wire of cir-
cular cross-section. Shading indicates current density .
depends on frequency . In particular, the current will
be concentrated close to the surface at high
frequencies, uniformly distributed throughout the
wire at DC, and in an intermediate state for
intermediate frequencies.
Additional Reading:
• “Skin effect” on Wikipedia.",Electromagnetics_Vol2.pdf
"50 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS
4.2 Impedance of a Wire
[m0159]
The goal of this section is to determine the impedance
– the
ratio of potential to current – of a wire. The
answer to this question is relatively simple in the DC
(“steady current”) case: The impedance is found to be
equal to the resistance of the wire, which is given by
R= l
σA (DC) (4.1)
where lis
the length of the wire and Ais the
cross-sectional area of the wire. Also, the impedance
of a wire comprised of a perfect conductor at any
frequency is simply zero, since there is no mechanism
in the wire that can dissipate or store energy in this
case.
However, all practical wires are comprised of good –
not perfect – conductors, and of course many practical
signals are time-varying, so the two cases above do
not address a broad category of practical interest. The
more general case of non-steady currents in imperfect
conductors is complicated by the fact that the current",Electromagnetics_Vol2.pdf
"more general case of non-steady currents in imperfect
conductors is complicated by the fact that the current
in an imperfect conductor is not uniformly distributed
in the wire, but rather is concentrated near the surface
and decays exponentially with increasing distance
from the surface (this is determined in Section 4.1).
W e are now ready to consider the AC case for a wire
comprised of a good but imperfect conductor. What is
the impedance of the wire if the current source is
sinusoidally-varying? Equation 4.1 for the DC case
was determined by first obtaining expressions for the
potential (V ) and net current (I ) for a length-l section
of wire in terms of the electric field intensity E in the
wire. T o follow that same approach here, we require
an expression for I in terms of E that accounts for the
non-uniform distribution of current. This is quite
difficult to determine in the case of a cylindrical wire.
However, we may develop an approximate solution",Electromagnetics_Vol2.pdf
"difficult to determine in the case of a cylindrical wire.
However, we may develop an approximate solution
by considering a surface which is not cylindrical, but
rather planar. Here we go:
Consider the experiment described in Figure 4.3.
Here, a semi-infinite region filled with a
homogeneous good conductor meets a semi-infinite
region of free space along a planar interface at z= 0,
with increasing zcorresponding to increasing depth
into the material. A plane wave propagates in the +ˆz
direction, beginning from just inside the structure’s
surface. The justification for presuming the existence
of this wave was presented in Section 4.1. The
electric field intensity is given by
˜E = ˆxE0e−αze−jβz (4.2)
where E0 is an arbitrary complex-valued constant.
The current density is given by Ohm’s law of
electromagnetics:2
˜J = σ˜E = ˆxσE0e−αze−jβz (4.3)
Recall that α= δ−1
s where δs is skin depth (see
Section 3.12). Also, for a good conductor, we know
that β ≈ α(Section 3.11). Using these relationships,",Electromagnetics_Vol2.pdf
"s where δs is skin depth (see
Section 3.12). Also, for a good conductor, we know
that β ≈ α(Section 3.11). Using these relationships,
we may rewrite Equation 4.3 as follows:
˜J ≈ ˆxσE0e−z/δ s e−jz/δ s
= ˆxσE0e−(1+j)z/δ s (4.4)
The net current ˜I is obtained by integrating ˜J over
any cross-section S through which all the current
flows; i.e.,
˜I =
∫
S
˜J · ds (4.5)
Here, the simplest solution is obtained by choosing S
to be a rectangular surface that is perpendicular to the
direction of ˜J at x= 0. This is shown in Figure 4.4.
c⃝ C. W ang CC BY -SA 4.0
Figure 4.3: Experiment used to determine current
density and resistance in the AC case.",Electromagnetics_Vol2.pdf
"4.2. IMPEDANCE OF A WIRE 51
c⃝ Y . Zhao CC BY -SA 4.0
Figure 4.4: Choice of S for calculating net current I.
The dimensions of S are width W in the ydimension
and extending to infinity in the zdirection. Then we
have
˜I ≈
∫ W
y=0
∫ ∞
z=0
(
ˆxσE0e−(1+j)z/δ s
)
· (ˆx dydz )
= σE0W
∫ ∞
z=0
e−(1+j)z/δ s dz (4.6)
For convenience, let us define the constant
K ≜ (1 + j)/δs. Since Kis constant with respect to
z, the remaining integral is straightforward to
evaluate:
∫ ∞
0
e−Kzdz= − 1
Ke−Kz
⏐⏐⏐
⏐
∞
0
=
+ 1
K (4.7)
Incorporating
this result into Equation 4.6, we obtain:
˜I ≈ σE0W δs
1 + j (4.8)
W
e calculate ˜V for a length lof the wire as follows:
˜V = −
∫ 0
x=l
˜E · dl (4.9)
2 To be clear, this is the “point form” of Ohm’s law , as opposed
to the circuit theory form (V = IR).
where we have determined that x= 0 corresponds to
the “+” terminal and x= lcorresponds to the “− ”
terminal.3 The path of integration can be any path that
begins and ends at the proscribed terminals. The",Electromagnetics_Vol2.pdf
"terminal.3 The path of integration can be any path that
begins and ends at the proscribed terminals. The
simplest path to use is one along the surface, parallel
to the xaxis. Along this path, z= 0 and thus
˜E = ˆxE0. For this path:
˜V = −
∫ 0
x=l
(ˆxE0) · (ˆxdx) = E0l (4.10)
The impedance Zmeasured across terminals at x= 0
and x= lis now determined to be:
Z ≜
˜V
˜I
≈ 1 + j
σδs
· l
W (4.11)
The
resistance is simply the real part, so we obtain
R≈ l
σ(δsW) (A C case) (4.12)
The quantity Rin this case is referred to specifically
as the ohmic resistance, since it is due entirely to the
limited conductivity of the material as quantified by
Ohm’s law . 4
Note the resemblance to Equation 4.1 (the solution
for the DC case): In the AC case, the product δsW,
having units of area, plays the role of the physical
cross-section S. Thus, we see an interesting new
interpretation of the skin depth δs: It is the depth to
which a uniform (DC) current would need to flow in",Electromagnetics_Vol2.pdf
"interpretation of the skin depth δs: It is the depth to
which a uniform (DC) current would need to flow in
order to produce a resistance equal to the observed
(AC) resistance.
Equations 4.11 and 4.12 were obtained for a good
conductor filling an infinite half-space, having a flat
surface. How well do these results describe a
cylindrical wire? The answer depends on the radius of
the wire, a. For δs ≪ a, Equations 4.11 and 4.12 are
excellent approximations, since δs ≪ aimplies that
most of the current lies in a thin shell close to the
surface of the wire. In this case, the model used to
develop the equations is a good approximation for any
given radial slice through the wire, and we are
3 If this is not clear, recall that the electric field vector must point
away from positive charge (thus, the + terminal).
4 This is in contrast to other ways that voltage and current can be
related; for example, the non-linear V -I characteristic of a diode,
which is not governed by Ohm’s law .",Electromagnetics_Vol2.pdf
"52 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS
justified in replacing W with the circumference 2πa.
Thus, we obtain the following expressions:
Z ≈ 1 + j
σδs
· l
2πa ( δs ≪ a) (4.13)
and
so
R≈ l
σ(δs2πa) ( δs ≪ a) (4.14)
The impedance of a wire of length l and radius
a≫ δs is given by Equation 4.13. The resistance
of such a wire is given by Equation 4.14.
If, on the other hand, a<δ s or merely ∼ δs, then
current density is significant throughout the wire,
including along the axis of the wire. In this case, we
cannot assume that the current density decays
smoothly to zero with increasing distance from the
surface, and so the model leading to Equation 4.13 is
a poor approximation. The frequency required for
validity of Equation 4.13 can be determined by noting
that δs ≈ 1/√
πfµσ for a good conductor; therefore,
we
require
1√πfµσ ≪ a (4.15)
for
the derived expressions to be valid. Solving for f,
we find:
f ≫ 1
πµσa2 (4.16)
F
or commonly-encountered wires comprised of",Electromagnetics_Vol2.pdf
"1√πfµσ ≪ a (4.15)
for
the derived expressions to be valid. Solving for f,
we find:
f ≫ 1
πµσa2 (4.16)
F
or commonly-encountered wires comprised of
typical good conductors, this condition applies at
frequencies in the MHz regime and above.
These results lead us to one additional interesting
finding about the AC resistance of wires. Since
δs ≈ 1/√πfµσ for a good conductor, Equation 4.14
may
be rewritten in the following form:
R≈ 1
2
√
µf
πσ · l
a (4.17)
W e
have found that Ris approximately proportional
to √f. For example, increasing frequency by a factor
of
4 increases resistance by a factor of 2. This
frequency dependence is evident in all kinds of
practical wires and transmission lines. Summarizing:
The resistance of a wire comprised of a good but
imperfect
conductor is proportional to the square
root of frequency .
At this point, we have determined that resistance is
given approximately by Equation 4.17 for δs ≪ a,
corresponding to frequencies in the MHz regime and",Electromagnetics_Vol2.pdf
"given approximately by Equation 4.17 for δs ≪ a,
corresponding to frequencies in the MHz regime and
above, and by Equation 4.1 for δs ≫ a, typically
corresponding to frequencies in the kHz regime and
below . W e have also found that resistance changes
slowly with frequency; i.e., in proportion to √
f.
Thus,
it is often possible to roughly estimate
resistance at frequencies between these two frequency
regimes by comparing the DC resistance from
Equation 4.1 to the AC resistance from Equation 4.17.
An example follows.
Example 4.1. Resistance of the inner conductor
of RG-59.
Elsewhere we have considered RG-59 coaxial
cable (see the section “Coaxial Line, ” which
may appear in another volume depending on the
version of this book). W e noted that it was not
possible to determine the AC resistance per unit
length R′ for RG-59 from purely electrostatic
and magnetostatic considerations. W e are now
able to consider the resistance per unit length of",Electromagnetics_Vol2.pdf
"length R′ for RG-59 from purely electrostatic
and magnetostatic considerations. W e are now
able to consider the resistance per unit length of
the inner conductor, which is a solid wire of the
type considered in this section. Let us refer to
this quantity as R′
ic. Note that
R′ = R′
ic + R′
oc (4.18)
where R′
oc is the resistance per unit length of the
outer conductor. R′
oc remains a bit too
complicated to address here. However, R′
ic is
typically much greater than R′
oc, so R′ ∼ R′
ic.
That is, we get a pretty good idea of R′ for
RG-59 by considering the inner conductor alone.
The relevant parameters of the inner conductor
are µ≈ µ0, σ∼= 2.28 × 107 S/m, and",Electromagnetics_Vol2.pdf
"4.2. IMPEDANCE OF A WIRE 53
a∼= 0.292 mm. Using Equation 4.17, we find:
R′
ic ≜ Ric
l = 1
2
√
µf
πσ · 1
a
∼
=
(
227 µΩ · m−1 · Hz−1
/ 2
)√
f (4.19)
Using
Expression 4.16, we find this is valid only
for f ≫ 130 kHz. So, for example, we may be
confident that R′
ic ≈ 0.82 Ω/m at 13 MHz.
At the other extreme (f ≪ 130 kHz),
Equation 4.1 (the DC resistance) is a better
estimate. In this low frequency case, we estimate
that R′
ic ≈ 0.16 Ω/m and is approximately
constant with frequency .
W e now have a complete picture: As frequency
is increased from DC to 13 MHz, we expect that
R′
ic will increase monotonically from
≈ 0.16 Ω/m to ≈ 0.82 Ω/m, and will continue to
increase in proportion to √
f from that value.
Returning
to Equation 4.11, we see that resistance is
not the whole story here. The impedance
Z = R+ jX also has a reactive component X equal
to the resistance R; i.e.,
X ≈ R≈ l
σ(δs2πa) (4.20)
This
is unique to good conductors at AC; that is, we",Electromagnetics_Vol2.pdf
"to the resistance R; i.e.,
X ≈ R≈ l
σ(δs2πa) (4.20)
This
is unique to good conductors at AC; that is, we
see no such reactance at DC. Because this reactance is
positive, it is often referred to as an inductance.
However, this is misleading since inductance refers to
the ability of a structure to store energy in a magnetic
field, and energy storage is decidedly not what is
happening here. The similarity to inductance is
simply that this reactance is positive, as is the
reactance associated with inductance. As long as we
keep this in mind, it is reasonable to model the
reactance of the wire as an equivalent inductance:
Leq ≈ 1
2πf · l
σ(δs2πa) (4.21)
No
w substituting an expression for skin depth:
Leq ≈ 1
2πf ·
√
πfµ
σ · l
2πa
= 1
4π3/ 2
√ µ
σf · l
a (4.22)
for
a wire having a circular cross-section with
δs ≪ a. The utility of this description is that it
facilitates the modeling of wire reactance as an
inductance in an equivalent circuit. Summarizing:",Electromagnetics_Vol2.pdf
"δs ≪ a. The utility of this description is that it
facilitates the modeling of wire reactance as an
inductance in an equivalent circuit. Summarizing:
A practical wire may be modeled using an equiv-
alent
circuit consisting of an ideal resistor (Equa-
tion 4.17) in series with an ideal inductor (Equa-
tion 4.22). Whereas resistance increases with the
square root of frequency , inductance decreases
with the square root of frequency .
If the positive reactance of a wire is not due to
physical inductance, then to what physical
mechanism shall we attribute this effect? A wire has
reactance because there is a phase shift between
potential and current. This is apparent by comparing
Equation 4.6 to Equation 4.10. This is the same phase
shift that was found to exist between the electric and
magnetic fields propagating in a good conductor, as
explained in Section 3.11.
Example 4.2. Equi valent inductance of the
inner conductor of RG-59.
Elsewhere in the book we worked out that the",Electromagnetics_Vol2.pdf
"explained in Section 3.11.
Example 4.2. Equi valent inductance of the
inner conductor of RG-59.
Elsewhere in the book we worked out that the
inductance per unit length L′ of RG-59 coaxial
cable was about 370 nH/m. W e calculated this
from magnetostatic considerations, so the
reactance associated with skin effect is not
included in this estimate. Let’s see how L′ is
affected by skin effect for the inner conductor.
Using Equation 4.22 with µ= µ0,
σ∼= 2.28 × 107 S/m, and a∼
= 0.292 mm, we
find
Leq ≈
(
3.61 × 10−5 H·m−1·Hz1/ 2) l
√f
(4.23)
Per
unit length:
L′
eq ≜ Leq
l ≈ 3.61 × 10−5 H
·Hz1/ 2
√f (4.24)
This
equals the magnetostatic inductance per
unit length (≈ 370 nH/m) at f ≈ 9.52 kHz, and
decreases with increasing frequency .
Summarizing: The equivalent inductance",Electromagnetics_Vol2.pdf
"54 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS
associated with skin effect is as important as the
magnetostatic
inductance in the kHz regime, and
becomes gradually less important with
increasing frequency .
Recall that the phase velocity in a low-loss
transmission line is approximately 1/
√
L′C′. This
means
that skin effect causes the phase velocity in
such lines to decrease with decreasing frequency . In
other words:
Skin effect in the conductors comprising com-
mon
transmission lines leads to a form of disper-
sion in which higher frequencies travel faster than
lower frequencies.
This phenomenon is known as chromatic dispersion,
or simply “dispersion, ” and leads to significant
distortion for signals having large bandwidths.
Additional Reading:
• “Skin effect” on Wikipedia.
4.3 Surface Impedance
[m0160]
In Section 4.2, we derived the following expression
for
the impedance Zof a good conductor having
width W, length l, and which is infinitely deep:
Z ≈ 1 + j
σδs
· l",Electromagnetics_Vol2.pdf
"for
the impedance Zof a good conductor having
width W, length l, and which is infinitely deep:
Z ≈ 1 + j
σδs
· l
W (A C case) (4.25)
where σis conductivity (SI base units of S/m) and δs
is skin depth. Note that δs and σare constitutive
parameters of material, and do not depend on
geometry; whereas land W describe geometry . With
this in mind, we define the surface impedance ZS as
follows:
ZS ≜ 1 + j
σδs
(4.26)
so
that
Z ≈ ZS
l
W (4.27)
Unlik
e the terminal impedance Z, ZS is strictly a
materials property . In this way , it is like the intrinsic
or “wave” impedance η, which is also a materials
property . Although the units of ZS are those of
impedance (i.e., ohms), surface impedance is usually
indicated as having units of “Ω/□ ” (“ohms per
square”) to prevent confusion with the terminal
impedance. Summarizing:
Surface impedance ZS (Equation 4.26) is a ma-
terials property having units of Ω/□, and which
characterizes the AC impedance of a material in-",Electromagnetics_Vol2.pdf
"Surface impedance ZS (Equation 4.26) is a ma-
terials property having units of Ω/□, and which
characterizes the AC impedance of a material in-
dependently of the length and width of the mate-
rial.
Surface impedance is often used to specify sheet
materials used in the manufacture of electronic and
semiconductor devices, where the real part of the
surface impedance is more commonly known as the
surface resistance or sheet resistance.
Additional Reading:
• “Sheet resistance” on Wikipedia.
[m0184]",Electromagnetics_Vol2.pdf
"4.3. SURF ACE IMPEDANCE 55
Image Credits
Fig. 4.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Current flow in cylinder new .svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 4.2: Biezl, https://commons.wikimedia.org/wiki/File:Skin depth.svg, public domain. Modified from
original.
Fig.
4.3: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Experiment for current density and resistance.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 4.4: c⃝ Y ahuiZ (Y . Zhao), https://commons.wikimedia.org/wiki/File:Figure4.4.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified by author.",Electromagnetics_Vol2.pdf
"Chapter 5
W a
ve Reflection and T ransmission
5.1 Plane W aves at Normal
Incidence on a Planar
Boundary
[m0161]
When a plane wave encounters a discontinuity in
media,
reflection from the discontinuity and
transmission into the second medium is possible. In
this section, we consider the scenario illustrated in
Figure 5.1: a uniform plane wave which is normally
incident on the planar boundary between two
semi-infinite material regions. By
“normally-incident” we mean the direction of
propagation ˆk is perpendicular to the boundary . W e
shall assume that the media are “simple” and lossless
(i.e., the imaginary component of permittivity ǫ′′ is
equal to zero) and therefore the media are completely
defined by a real-valued permittivity and a real-valued
permeability .
Figure 5.1 shows the wave incident on the boundary ,
which is located at the z= 0 plane. The electric field
intensity ˜Ei of this wave is given by
˜Ei(z) = ˆxEi
0e−jβ1z , z≤ 0 (5.1)
where β1 = ω√
µ1ǫ1 is the phase propagation",Electromagnetics_Vol2.pdf
"intensity ˜Ei of this wave is given by
˜Ei(z) = ˆxEi
0e−jβ1z , z≤ 0 (5.1)
where β1 = ω√
µ1ǫ1 is the phase propagation
constant in Region 1 and Ei
0 is a complex-valued
constant. ˜Ei serves as the “stimulus” in this problem.
That is, all other contributions to the total field may
be expressed in terms of ˜Ei. In fact, all other
contributions to the total field may be expressed in
terms of Ei
0.
From the plane wave relationships, we determine that
the associated magnetic field intensity is
˜Hi(z) = ˆy Ei
0
η1
e−jβ1z , z≤ 0 (5.2)
where η1 =
√
µ1/ǫ1 is the wave impedance in
Region 1.
The possibility of a reflected plane wave propagating
in exactly the opposite direction is inferred from two
pieces of evidence: The general solution to the wave
equation, which includes terms corresponding to
waves traveling in both +ˆz or −ˆz; and the
geometrical symmetry of the problem, which
precludes waves traveling in any other directions. The
symmetry of the problem also precludes a change of",Electromagnetics_Vol2.pdf
"geometrical symmetry of the problem, which
precludes waves traveling in any other directions. The
symmetry of the problem also precludes a change of
polarization, so the reflected wave should have no ˆy
component. Therefore, we may be confident that the
reflected electric field has the form
˜Er(z) = ˆxBe+jβ1z , z≤ 0 (5.3)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.1: A uniform plane wave normally incident
on the planar boundary between two semi-infinite ma-
terial regions.
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"5.1. PLANE W A VES A T NORMAL INCIDENCE ON A PLANAR BOUNDAR Y 57
where Bis a complex-valued constant that remains to
be determined. Since the direction of propagation for
the reflected wave is −ˆz, we have from the plane
wave relationships that
˜Hr(z) = −ˆy B
η1
e+jβ1z , z≤ 0 (5.4)
Similarly
, we infer the existence of a “transmitted”
plane wave propagating on the z >0 side of the
boundary . The symmetry of the problem precludes
any direction of propagation other than +z, and with
no possibility of a wave traveling in the −ˆz direction
for z >0. Therefore, we may be confident that the
transmitted electric field has the form:
˜Et(z) = ˆxCe−jβ2z , z≥ 0 (5.5)
and an associated magnetic field having the form
˜Ht(z) = ˆy C
η2
e−jβ2z , z≥ 0 (5.6)
where β2 = ω√µ2ǫ2 and η2 =
√
µ2/ǫ2 are the phase
propagation constant and wave impedance,
respectively , in Region 2. The constant C, like B, is a
complex-valued constant that remains to be
determined.",Electromagnetics_Vol2.pdf
"respectively , in Region 2. The constant C, like B, is a
complex-valued constant that remains to be
determined.
At this point, the only unknowns in this problem are
the complex-valued constants Band C. Once these
values are known, the problem is completely solved.
These values can be determined by the application of
boundary conditions at z= 0. First, recall that the
tangential component of the total electric field
intensity must be continuous across a material
boundary . T o apply this boundary condition, let us
define ˜E1 and ˜E2 to be the total electric field
intensities in Regions 1 and 2, respectively . The total
field in Region 1 is the sum of incident and reflected
fields, so
˜E1(z) = ˜Ei(z) + ˜Er(z) (5.7)
The field in Region 2 is simply
˜E2(z) = ˜Et(z) (5.8)
Also, we note that all field components are already
tangent to the boundary . Thus, continuity of the
tangential component of the electric field across the
boundary requires ˜E1(0) = ˜E2(0), and therefore",Electromagnetics_Vol2.pdf
"tangential component of the electric field across the
boundary requires ˜E1(0) = ˜E2(0), and therefore
˜Ei(0) + ˜Er(0) = ˜Et(0) (5.9)
Now employing Equations 5.1, 5.3, and 5.5, we
obtain:
Ei
0 + B = C (5.10)
Clearly a second equation is required to determine
both Band C. This equation can be obtained by
enforcing the boundary condition on the magnetic
field. Recall that any discontinuity in the tangential
component of the total magnetic field intensity must
be supported by a current flowing on the surface.
There is no impressed current in this problem, and
there is no reason to suspect a current will arise in
response to the fields present in the problem.
Therefore, the tangential components of the magnetic
field must be continuous across the boundary . This
becomes the same boundary condition that we applied
to the total electric field intensity , so the remaining
steps are the same. W e define ˜H1 and ˜H2 to be the
total magnetic field intensities in Regions 1 and 2,",Electromagnetics_Vol2.pdf
"steps are the same. W e define ˜H1 and ˜H2 to be the
total magnetic field intensities in Regions 1 and 2,
respectively . The total field in Region 1 is
˜H1(z) = ˜Hi(z) + ˜Hr(z) (5.11)
The field in Region 2 is simply
˜H2(z) = ˜Ht(z) (5.12)
The boundary condition requires ˜H1(0) = ˜H2(0),
and therefore
˜Hi(0) + ˜Hr(0) = ˜Ht(0) (5.13)
Now employing Equations 5.2, 5.4, and 5.6, we
obtain:
Ei
0
η1
− B
η1
= C
η2
(5.14)
Equations
5.10 and 5.14 constitute a linear system of
two simultaneous equations with two unknowns. A
straightforward method of solution is to first eliminate
Cby substituting the left side of Equation 5.10 into
Equation 5.14, and then to solve for B. One obtains:
B = Γ 12Ei
0 (5.15)
where
Γ 12 ≜ η2 − η1
η2 + η1
(5.16)
Γ 12 is
known as a reflection coefficient. The subscript
“12” indicates that this coefficient applies for",Electromagnetics_Vol2.pdf
"58 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
incidence from Region 1 toward Region 2. W e may
now solve for Cby substituting Equation 5.15 into
Equation 5.10. W e find:
C = (1 + Γ 12) Ei
0 (5.17)
Now summarizing the solution:
˜Er(z) = ˆxΓ 12Ei
0e+jβ1z , z≤ 0 (5.18)
˜Et(z) = ˆx (1
+ Γ 12) Ei
0e−jβ2z , z≥ 0 (5.19)
Equations 5.18 and 5.19 are the reflected and
transmitted
fields, respectively , in response to the
incident field given in Equation 5.1 in the normal
incidence scenario shown in Figure 5.1.
Expressions for ˜Hr and ˜Ht may be obtained by
applying the plane wave relationships to the
preceding expressions.
It is useful to check this solution by examining some
special cases. First: If the material in Region 2 is
identical to the material in Region 1, then there should
be no reflection and ˜Et = ˜Ei. In this case, η2 = η1,
so Γ 12 = 0, and we obtain the expected result.
A second case of practical interest is when Region 2
is a perfect conductor. First, note that this may seem",Electromagnetics_Vol2.pdf
"A second case of practical interest is when Region 2
is a perfect conductor. First, note that this may seem
at first glance to be a violation of the “lossless”
assumption made at the beginning of this section.
While it is true that we did not explicitly account for
the possibility of a perfect conductor in Region 2,
let’s see what the present analysis has to say about
this case. If the material in Region 2 is a perfect
conductor, then there should be no transmission since
the electric field is zero in a perfect conductor. In this
case, η2 = 0 since the ratio of electric field intensity
to magnetic field intensity is zero in Region 2, and
subsequently Γ 12 = −1 and 1 + Γ 12 = 0. As
expected, ˜Et is found to be zero, and the reflected
electric field experiences a sign change as required to
enforce the boundary condition ˜Ei(0) + ˜Er(0) = 0.
Thus, we obtained the correct answer because we
were able to independently determine that η2 = 0 in a
perfect conductor.",Electromagnetics_Vol2.pdf
"Thus, we obtained the correct answer because we
were able to independently determine that η2 = 0 in a
perfect conductor.
When Region 2 is a perfect conductor, the reflec-
tion
coefficient Γ 12 = −1 and the solution de-
scribed in Equations 5.18 and 5.19 applies.
It may be helpful to note the very strong analogy
between reflection of the electric field component of a
plane wave from a planar boundary and the reflection
of a voltage wave in a transmission line from a
terminating impedance. In a transmission line, the
voltage reflection coefficient Γ is given by
Γ = ZL − Z0
ZL + Z0
(5.20)
where ZL is
the load impedance and Z0 is the
characteristic impedance of the transmission line.
Comparing this to Equation 5.16, we see η1 is
analogous to Z0 and η2 is analogous to ZL.
Furthermore, we see the special case η2 = η1,
considered previously , is analogous to a matched
load, and the special case η2 = 0, also considered
previously , is analogous to a short-circuit load.",Electromagnetics_Vol2.pdf
"load, and the special case η2 = 0, also considered
previously , is analogous to a short-circuit load.
In the case of transmission lines, we were concerned
about what fraction of the power was delivered into a
load and what fraction of the power was reflected
from a load. A similar question applies in the case of
plane wave reflection. In this case, we are concerned
about what fraction of the power density is
transmitted into Region 2 and what fraction of the
power density is reflected from the boundary . The
time-average power density Si
ave associated with the
incident wave is:
Si
ave = |Ei
0|2
2η1
(5.21)
presuming
that Ei
0 is expressed in peak (as opposed to
rms) units. Similarly , the time-average power density
Sr
ave associated with the reflected wave is:
Sr
ave =
⏐⏐Γ 12Ei
0
⏐
⏐2
2η1
= |Γ 12|2
⏐⏐Ei
0
⏐
⏐2
2η1
= |Γ 12|2 Si
av
e (5.22)
From the principle of conservation of power, the
power density St
ave transmitted into Region 2 must be",Electromagnetics_Vol2.pdf
"5.1. PLANE W A VES A T NORMAL INCIDENCE ON A PLANAR BOUNDAR Y 59
equal to the incident power density minus the
reflected power density . Thus:
St
ave = Si
ave − Sr
ave
=
(
1 − |Γ 12|2)
Si
ave (5.23)
In other words, the ratio of power density transmitted
into Region 2 to power density incident from
Region 1 is
St
ave
Siav e
= 1 − |Γ 12|2 (5.24)
Ag
ain, the analogy to transmission line theory is
evident.
The fractions of power density reflected and
transmitted
relative to power density incident are
given in terms of the reflection coefficient Γ 12 by
Equations 5.22 and 5.24, respectively .
Finally , note that a variety of alternative expressions
exist for the reflection coefficient Γ 12. Since the wave
impedance η=
√
µ/ǫin lossless media, and since
most
lossless materials are non-magnetic (i.e., exhibit
µ≈ µ0), it is possible to express Γ 12 purely in terms
of permittivity for these media. Assuming µ= µ0,
we find:
Γ 12 = η2 − η1
η2 + η1
=
√
µ0/ǫ2 −
√
µ0/ǫ1√
µ0/ǫ2 +
√
µ0/ǫ1
=
√ǫ1 − √ǫ2
√ǫ1 + √ǫ2",Electromagnetics_Vol2.pdf
"of permittivity for these media. Assuming µ= µ0,
we find:
Γ 12 = η2 − η1
η2 + η1
=
√
µ0/ǫ2 −
√
µ0/ǫ1√
µ0/ǫ2 +
√
µ0/ǫ1
=
√ǫ1 − √ǫ2
√ǫ1 + √ǫ2
(5.25)
Moreo
ver, recall that permittivity can be expressed in
terms of relative permittivity; that is, ǫ= ǫrǫ0.
Making the substitution above and eliminating all
extraneous factors of ǫ0, we find:
Γ 12 =
√ǫr1 − √ǫr2
√ǫr1 + √ǫr2
(5.26)
where ǫr1 and ǫr2 are
the relative permittivities in
Regions 1 and 2, respectively .
Example 5.1. Radio reflection from the surface
of the Moon.
At radio frequencies, the Moon can be modeled
as a low-loss dielectric with relative permittivity
of
about 3. Characterize the efficiency of
reflection of a radio wave from the Moon.
Solution. At radio frequencies, a wavelength is
many orders of magnitude less than the diameter
of the Moon, so we are justified in treating the
Moon’s surface as a planar boundary between
free space and a semi-infinite region of lossy
dielectric material. The reflection coefficient is",Electromagnetics_Vol2.pdf
"Moon’s surface as a planar boundary between
free space and a semi-infinite region of lossy
dielectric material. The reflection coefficient is
given approximately by Equation 5.26 with
ǫr1 ≈ 1 and ǫr2 ∼ 3. Thus, Γ 12 ∼ − 0.27.
Subsequently , the fraction of power reflected
from the Moon relative to power incident is
|Γ 12|2 ∼ 0.07; i.e., about 7%.
In optics, it is common to make the definition
n≜ √
ǫr where nis known as the index of refraction
or refractive index. 1 In terms of the indices of
refraction n1 and n2 for Regions 1 and 2,
respectively:
Γ 12 = n1 − n2
n1 + n2
(5.27)
Additional
Reading:
• “Refractive index” on Wikipedia.
1 A bit of a misnomer, since the definition applies even when
there is no refraction, as in the scenario considered in this section.",Electromagnetics_Vol2.pdf
"60 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.2 Plane W aves at Normal
Incidence on a Material Slab
[m0162]
In Section 5.1, we considered what happens when a
uniform
plane wave is normally incident on the planar
boundary between two semi-infinite media. In this
section, we consider the problem shown in Figure 5.2:
a uniform plane wave normally incident on a “slab”
sandwiched between two semi-infinite media. This
scenario arises in many practical engineering
problems, including the design and analysis of filters
and impedance-matching devices at RF and optical
frequencies, the analysis of RF propagation through
walls, and the design and analysis of radomes. The
findings of Section 5.1 are an important stepping
stone to the solution of this problem, so a review of
Section 5.1 is recommended before reading this
section.
For consistency of terminology , let us refer to the
problem considered in Section 5.1 as the
“single-boundary” problem and the present (slab)",Electromagnetics_Vol2.pdf
"For consistency of terminology , let us refer to the
problem considered in Section 5.1 as the
“single-boundary” problem and the present (slab)
problem as the “double-boundary” problem. Whereas
there are only two regions (“Region 1” and “Region
2”) in the single-boundary problem, in the
double-boundary problem there is a third region that
we shall refer to as “Region 3. ” W e assume that the
media comprising each slab are “simple” and lossless
(i.e., the imaginary component of permittivity ǫ′′ is
equal to zero) and therefore the media are completely
defined by a real-valued permittivity and a real-valued
c⃝ C. W ang CC BY -SA 4.0
Figure 5.2: A uniform plane wave normally incident
on a slab.
permeability . The boundary between Regions 1 and 2
is at z= −d, and the boundary between Regions 2
and 3 is at z= 0. Thus, the thickness of the slab is d.
In both problems, we presume an incident wave in
Region 1 incident on the boundary with Region 2
having electric field intensity
˜Ei(z) = ˆxEi",Electromagnetics_Vol2.pdf
"In both problems, we presume an incident wave in
Region 1 incident on the boundary with Region 2
having electric field intensity
˜Ei(z) = ˆxEi
0e−jβ1z (Region 1) (5.28)
where β1 = ω√
µ1ǫ1 is the phase propagation
constant in Region 1. ˜Ei serves as the “stimulus” in
this problem. That is, all other contributions to the
total field may be expressed in terms of ˜Ei.
From the plane wave relationships, we determine that
the associated magnetic field intensity is
˜Hi(z) = ˆy Ei
0
η1
e−jβ1z (Re gion 1) (5.29)
where η1 =
√
µ1/ǫ1 is the wave impedance in
Region 1.
The symmetry arguments of the single-boundary
problem apply in precisely the same way to the
double-boundary problem. Therefore, we presume
that the reflected electric field intensity is:
˜Er(z) = ˆxBe+jβ1z (Region 1) (5.30)
where Bis a complex-valued constant that remains to
be determined; and subsequently the reflected
magnetic field intensity is:
˜Hr(z) = −ˆy B
η1
e+jβ1z (Re gion 1) (5.31)",Electromagnetics_Vol2.pdf
"be determined; and subsequently the reflected
magnetic field intensity is:
˜Hr(z) = −ˆy B
η1
e+jβ1z (Re gion 1) (5.31)
Similarly , we infer the existence of a transmitted
plane wave propagating in the +ˆz direction in
Region 2. The electric and magnetic field intensities
of this wave are given by:
˜Et2(z) = ˆxCe−jβ2z (Region 2) (5.32)
and an associated magnetic field having the form:
˜Ht2(z) = ˆy C
η2
e−jβ2z (Re gion 2) (5.33)
where β2 = ω√µ2ǫ2 and η2 =
√
µ2/ǫ2 are the phase
propagation constant and wave impedance,
respectively , in Region 2. The constant C, like B, is a",Electromagnetics_Vol2.pdf
"5.2. PLANE W A VES A T NORMAL INCIDENCE ON A MA TERIAL SLAB 61
complex-valued constant that remains to be
determined.
Now let us consider the boundary between Regions 2
and 3. Note that ˜Et2 is incident on this boundary in
precisely the same manner as ˜Ei is incident on the
boundary between Regions 1 and 2. Therefore, we
infer a reflected electric field intensity in Region 2 as
follows:
˜Er2(z) = ˆxDe+jβ2z (Region 2) (5.34)
where Dis a complex-valued constant that remains to
be determined; and an associated magnetic field
˜Hr2(z) = −ˆy D
η2
e+jβ2z (Re gion 2) (5.35)
Subsequently , we infer a transmitted electric field
intensity in Region 3 as follows:
˜Et(z) = ˆxFe−jβ3z (Region 3) (5.36)
and an associated magnetic field having the form
˜Ht(z) = ˆy F
η3
e−jβ3z (Re gion 3) (5.37)
where β3 = ω√µ3ǫ3 and η3 =
√
µ3/ǫ3 are the phase
propagation constant and wave impedance,
respectively , in Region 3. The constant F, like D, is a
complex-valued constant that remains to be",Electromagnetics_Vol2.pdf
"propagation constant and wave impedance,
respectively , in Region 3. The constant F, like D, is a
complex-valued constant that remains to be
determined. W e infer no wave traveling in the (− ˆz) in
Region 3, just as we inferred no such wave in
Region 2 of the single-boundary problem.
For convenience, T able 5.1 shows a complete
summary of the field components we have just
identified. Note that the total field intensity in each
region is the sum of the field components in that
region. For example, the total electric field intensity
in Region 2 is ˜Et2 + ˜Er2.
Now observe that there are four unknown constants
remaining; namely B, C, D, and F. The
double-boundary problem is completely solved once
we have expressions for these constants in terms of
the “given” quantities in the problem statement.
Solutions for the unknown constants can be obtained
by enforcing boundary conditions on the electric and
magnetic fields at each of the two boundaries. As in",Electromagnetics_Vol2.pdf
"by enforcing boundary conditions on the electric and
magnetic fields at each of the two boundaries. As in
the single-boundary case, the relevant boundary
condition is that the total field should be continuous
across each boundary . Applying this condition to the
boundary at z= 0, we obtain:
˜Et2(0) + ˜Er2(0) = ˜Et(0) (5.38)
˜Ht2(0) + ˜Hr2(0) = ˜Ht(0) (5.39)
Applying this condition to the boundary at z= −d,
we obtain:
˜Ei(−d) + ˜Er(−d) = ˜Et2(−d) + ˜Er2(−d) (5.40)
˜Hi(−d) + ˜Hr(−d) = ˜Et2(−d) + ˜Er2(−d) (5.41)
Making substitutions from T able 5.1 and dividing out
common factors, Equation 5.38 becomes:
C+ D= F (5.42)
Equation 5.39 becomes:
C
η2
− D
η2
= F
η3
(5.43)
Equation
5.40 becomes:
Ei
0e+jβ1d+ Be−jβ1d = Ce+jβ2d+ De−jβ2d (5.44)
Equation 5.41 becomes:
Ei
0
η1
e+jβ1d − B
η1
e−jβ1d = C
η2
e+jβ2d − D
η2
e−jβ2d
(5.45)
Equations
5.42–5.45 are recognizable as a system of 4
simultaneous linear equations with the number of
unknowns equal to the number of equations. W e",Electromagnetics_Vol2.pdf
"Equations
5.42–5.45 are recognizable as a system of 4
simultaneous linear equations with the number of
unknowns equal to the number of equations. W e
could simply leave it at that, however, some very
useful insights are gained by solving this system of
equations in a particular manner. First, note the
resemblance between the situation at the z= 0
boundary in this problem and the situation at z= 0
boundary in the single-boundary problem. In fact, the
two are the same problem, with the following
transformation of variables (single-boundary →
double-boundary):
Ei
0 → C i.e., the independent variable (5.46)
B → D i.e., reflection (5.47)
C → F i.e., transmission (5.48)",Electromagnetics_Vol2.pdf
"62 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
Electric Field Intensity Magnetic Field Intensity Region of V alidity
Region 1 ˜Ei(z) = ˆxEi
0e−j β1z ˜Hi(z) = + ˆy
(
Ei
0/η1
)
e−jβ1z z≤ − d
˜Er(z) = ˆxBe+jβ1z ˜Hr(z) = −ˆy (B/η1) e+jβ1z
Region 2 ˜Et2(z) = ˆxC e−jβ2z ˜Ht2(z) = + ˆy (C/η2) e−jβ2z −d≤ z≤ 0
˜Er2(z) = ˆxDe+jβ2z ˜Hr2(z) = −ˆy (D/η2) e+jβ2z
Region 3 ˜Et(z) = ˆx Fe−jβ3z ˜Ht(z) = + ˆy (F/η3) e−jβ3z z≥ 0
T able 5.1: Field components in the double-boundary (slab) problem of Figure 5.2.
It immediately follows that
D= Γ 23C (5.49)
and
F = (1 + Γ 23) C (5.50)
where
Γ 23 ≜ η3 − η2
η3 + η2
(5.51)
At
this point, we could use the expressions we have
just derived to eliminate Dand F in the system of
four simultaneous equations identified earlier. This
would reduce the problem to that of two equations in
two unknowns – a dramatic simplification!
However, even this is more work than is necessary ,
and a little cleverness at this point pays big dividends",Electromagnetics_Vol2.pdf
"two unknowns – a dramatic simplification!
However, even this is more work than is necessary ,
and a little cleverness at this point pays big dividends
later. The key idea is that we usually have no interest
in the fields internal to the slab; in most problems, we
are interested merely in reflection into Region 1 from
the z= −dboundary and transmission into Region 3
through the z= 0 interface. With this in mind, let us
simply replace the two-interface problem in
Figure 5.2 with an “equivalent” single-boundary
problem shown in Figure 5.3. This problem is
“equivalent” in the following sense only: Fields in
Region 1 are identical to those in Region 1 of the
original problem. The material properties in the
region to the right of the z= −dboundary in the
equivalent problem seem unlikely to be equal to those
in Regions 2 or 3 of the original problem, so we
define a new wave impedance ηeq to represent this
new condition. If we can find an expression for ηeq,",Electromagnetics_Vol2.pdf
"in Regions 2 or 3 of the original problem, so we
define a new wave impedance ηeq to represent this
new condition. If we can find an expression for ηeq,
then we can develop a solution to the original
(two-boundary) problem that looks like a solution to
the equivalent (simpler, single-boundary) problem.
c⃝ C. W ang CC BY -SA 4.0
Figure 5.3: Representing the double-boundary prob-
lem as an equivalent single-boundary problem.
T o obtain an expression for ηeq, we invoke the
definition of wave impedance: It is simply the ratio of
electric field intensity to magnetic field intensity in
the medium. Thus:
ηeq ≜ ˆx · ˜E2(z2)
ˆy · ˜H2(z2)
=
ˆx ·
[
˜Et2(z2) + ˜Er2(
z2)
]
ˆy ·
[
˜Ht2(z2) + ˜Hr2( z2)
]
(5.52)
where z2 is any point in Region 2. For simplicity , let
us choose z2 = −d. Making substitutions, we obtain:
ηeq = Ce+jβ2d + De−jβ2d
(C/η2) e+j β2d − (D/η2) e−jβ2d (5.53)
Bringing the factor of η2 to the front and substituting
D= Γ 23C:
ηeq = η2
Ce+jβ2d + Γ 23Ce−jβ2d
Ce+jβ2d − Γ 23C e−jβ2d (5.54)",Electromagnetics_Vol2.pdf
"Bringing the factor of η2 to the front and substituting
D= Γ 23C:
ηeq = η2
Ce+jβ2d + Γ 23Ce−jβ2d
Ce+jβ2d − Γ 23C e−jβ2d (5.54)
Finally , we divide out the common factor of Cand
multiply numerator and denominator by e−jβ2d,",Electromagnetics_Vol2.pdf
"5.2. PLANE W A VES A T NORMAL INCIDENCE ON A MA TERIAL SLAB 63
yielding:
ηeq = η2
1 + Γ 23e−j2β2d
1 − Γ 23e−j2 β2d (5.55)
Equation 5.55 is the wave impedance in the re-
gion
to the right of the boundary in the equiva-
lent scenario shown in Figure 5.3. “Equivalent”
in this case means that the incident and reflected
fields in Region 1 are identical to those in the
original (slab) problem.
T wo comments on this expression before proceeding.
First: Note that if the materials in Regions 2 and 3 are
identical, then η2 = η3, so Γ 23 = 0, and thus
ηeq = η2, as expected. Second: Note that ηeq is, in
general, complex-valued. This may initially seem
troubling, since the imaginary component of the wave
impedance is normally associated with general (e.g.,
possibly lossy) material. W e specifically precluded
this possibility in the problem statement, so clearly
the complex value of the wave impedance is not
indicating loss. Instead, the non-zero phase of ηeq",Electromagnetics_Vol2.pdf
"the complex value of the wave impedance is not
indicating loss. Instead, the non-zero phase of ηeq
represents the ability of the standing wave inside the
slab to impart a phase shift between the electric and
magnetic fields. This is precisely the same effect that
one observes at the input of a transmission line: The
input impedance Zin is, in general, complex-valued
even if the line is lossless and the characteristic
impedance and load impedance are real-valued. 2 In
fact, the impedance looking into a transmission line is
given by an expression of precisely the same form as
Equation 5.55. This striking analogy between plane
waves at planar boundaries and voltage and current
waves in transmission lines applies broadly .
W e can now identify an “equivalent reflection
coefficient” Γ 1,eq for the scenario shown in
Figure 5.3:
Γ 1,eq ≜ ηeq − η1
ηeq + η1
(5.56)
The
quantity Γ 1,eq may now be used precisely in the
same way as Γ 12 was used in the single-boundary",Electromagnetics_Vol2.pdf
"Figure 5.3:
Γ 1,eq ≜ ηeq − η1
ηeq + η1
(5.56)
The
quantity Γ 1,eq may now be used precisely in the
same way as Γ 12 was used in the single-boundary
problem to find the reflected fields and reflected
power density in Region 1.
2 See the section “Input Impedance of a T erminated Lossless
Transmission Line” for a reminder. This section may appear in a
different volume depending on the version of this book.
Example 5.2. Reflection of WiFi from a glass
pane.
A WiFi (wireless LAN) signal at a center
frequency of 2.45 GHz is normally incident on a
glass pane which is 1 cm thick and is
well-characterized in this application as a
lossless dielectric with ǫr = 4. The source of the
signal is sufficiently distant from the window
that the incident signal is well-approximated as a
plane wave. Determine the fraction of power
reflected from the pane.
Solution. In this case, we identify Regions 1 and
3 as approximately free space, and Region 2 as
the pane. Thus
η1 = η3 = η0 ∼= 376.7 Ω (5.57)
and
η2 = η0
√ǫr",Electromagnetics_Vol2.pdf
"3 as approximately free space, and Region 2 as
the pane. Thus
η1 = η3 = η0 ∼= 376.7 Ω (5.57)
and
η2 = η0
√ǫr
∼
= 188
 .4 Ω (5.58)
The reflection coefficient from Region 2 to
Region 3 is
Γ 23 = η3 − η2
η3 + η2
= η0 − η2
η0 + η2
∼= 0.3333 (5.59)
Gi
ven f = 2.45 GHz, the phase propagation
constant in the glass is
β2 = 2π
λ = 2πf√ǫr
c
∼
= 102
 .6 rad/m (5.60)
Given d= 1 cm, the equivalent wave impedance
is
ηeq = η2
1 + Γ 23e−j2β2d
1 − Γ 23e−j2 β2d
∼
= 117.9 − j78.4 Ω (5.61)
Next we calculate
Γ 1,eq = ηeq − η1
ηeq + η1
∼
= −0.4859 − j0
.2354 (5.62)",Electromagnetics_Vol2.pdf
"64 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
The ratio of reflected power density to incident
po
wer density is simply the squared magnitude
of this reflection coefficient, i.e.:
Sr
ave
Siav e
= |Γ 1,eq |2 = 0.292 ∼= 29.2% (5.63)
where Sr
av
e and Si
ave are the reflected and
incident power densities, respectively .
Since B = Γ 1,eq Ei
0, we have:
˜Er(z) = ˆxΓ 1,eq Ei
0e+jβ1z , z≤ − d (5.64)
and ˜Hr(z) can be obtained from the plane wave
relationships. If desired, it is now quite simple to
obtain solutions for the electric and magnetic fields in
Regions 2 and 3. However, it is the usually the power
density transmitted into Region 3 that is of greatest
interest. This power density is easily determined from
the principle of conservation of power. If the loss in
Region 2 is negligible, then no power can be
dissipated there. In this case, all power not reflected
from the z= −dinterface must be transmitted into
Region 3. In other words:
St
ave
Siav e
= 1 − |Γ 1,eq |2 (5.65)
where St
av",Electromagnetics_Vol2.pdf
"from the z= −dinterface must be transmitted into
Region 3. In other words:
St
ave
Siav e
= 1 − |Γ 1,eq |2 (5.65)
where St
av
e is the transmitted power density .
Example 5.3. Transmission of WiFi through a
glass pane.
Continuing Example 5.2: What fraction of
incident power passes completely through the
glass pane?
Solution.
St
ave
Siav e
= 1 − |Γ 1,eq |2 ∼= 70.8% (5.66)
Additional
Reading:
• “Radome” on Wikipedia.
5.3 T otal T ransmission Through
a Slab
[m0163]
Section 5.2 details the solution of the “single-slab”
problem.
This problem is comprised of three material
regions: A semi-infinite Region 1, from which a
uniform plane wave is normally incident; Region 2,
the slab, defined by parallel planar boundaries
separated by distance d; and a semi-infinite Region 3,
through which the plane wave exits. The solution that
was developed assumes simple media with negligible
loss, so that the media in each region is characterized
entirely by permittivity and permeability .",Electromagnetics_Vol2.pdf
"loss, so that the media in each region is characterized
entirely by permittivity and permeability .
W e now focus on a particular class of applications
involving this structure. In this class of applications,
we seek total transmission through the slab. By “total
transmission” we mean 100% of power incident on
the slab is transmitted through the slab into Region 3,
and 0% of the power is reflected back into Region 1.
There are many applications for such a structure. One
application is the radome, a protective covering which
partially or completely surrounds an antenna, but
nominally does not interfere with waves being
received by or transmitted from the antenna. Another
application is RF and optical wave filtering; that is,
passing or rejecting waves falling within a narrow
range of frequencies. In this section, we will first
describe the conditions for total transmission, and
then shall provide some examples of these
applications.
W e begin with characterization of the media.",Electromagnetics_Vol2.pdf
"then shall provide some examples of these
applications.
W e begin with characterization of the media.
Region 1 is characterized by its permittivity ǫ1 and
permeability µ1, such that the wave impedance in
Region 1 is η1 =
√
µ1/ǫ1. Similarly , Region 2 is
characterized by its permittivity ǫ2 and permeability
µ2, such that the wave impedance in Region 2 is
η2 =
√
µ2/ǫ2. Region 3 is characterized by its
permittivity ǫ3 and permeability µ3, such that the
wave impedance in Region 3 is η3 =
√
µ3/ǫ3. The
analysis in this section also depends on β2, the phase
propagation constant in Region 2, which is given by
ω√µ2ǫ2.
Recall
that reflection from the slab is quantified by",Electromagnetics_Vol2.pdf
"5.3. TOT AL TRANSMISSION THROUGH A SLAB 65
the reflection coefficient
Γ 1,eq = ηeq − η1
ηeq + η1
(5.67)
where ηeq is
given by
ηeq = η2
1 + Γ 23e−j2β2d
1 − Γ 23e−j2 β2d (5.68)
and where Γ 23 is given by
Γ 23 = η3 − η2
η3 + η2
(5.69)
T
otal transmission requires that Γ 1,eq = 0. From
Equation 5.67 we see that Γ 1,eq is zero when
η1 = ηeq. Now employing Equation 5.68, we see that
total transmission requires:
η1 = η2
1 + Γ 23e−j2β2d
1 − Γ 23e−j2 β2d (5.70)
For convenience and clarity , let us define the quantity
P ≜ e−j2β2d (5.71)
Using this definition, Equation 5.70 becomes
η1 = η2
1 + Γ 23P
1 − Γ 23P (5.72)
Also,
let us substitute Equation 5.69 for Γ 23, and
multiply numerator and denominator by η3 + η2 (the
denominator of Equation 5.69). W e obtain:
η1 = η2
η3 + η2 + (η3 − η2)P
η3 + η2 − (η3 − η2)P (5.73)
Rearranging
the numerator and denominator, we
obtain:
η1 = η2
(1 + P)η3 + (1 − P)η2
(1 − P)η3 + (1 + P)η2
(5.74)
The parameters η1, η2, η3, β2, and ddefining any",Electromagnetics_Vol2.pdf
"the numerator and denominator, we
obtain:
η1 = η2
(1 + P)η3 + (1 − P)η2
(1 − P)η3 + (1 + P)η2
(5.74)
The parameters η1, η2, η3, β2, and ddefining any
single-slab structure that exhibits total transmis-
sion must satisfy Equation 5.74.
Our challenge now is to identify combinations of
parameters that satisfy this condition. There are two
general categories of solutions. These categories are
known as half-wave matching and quarter-wave
matching.
Half-wave matching applies when we have the same
material on either side of the slab; i.e., η1 = η3. Let
us refer to this common value of η1 and η3 as ηext.
Then the condition for total transmission becomes:
ηext = η2
(1 + P)ηext + (1 − P)η2
(1 − P)ηext + (1 + P)η2
(5.75)
For the above condition to be satisfied, we need the
fraction on the right side of the equation to be equal to
ηext/η2. From Equation 5.71, the magnitude of P is
always 1, so the first value of P you might think to try
is P = +1. In fact, this value satisfies Equation 5.75.",Electromagnetics_Vol2.pdf
"always 1, so the first value of P you might think to try
is P = +1. In fact, this value satisfies Equation 5.75.
Therefore, e−j2β2d = +1. This new condition is
satisfied when 2β2d= 2πm, where m= 1,2,3,...
(W e do not consider m≤ 0 to be valid solutions since
these would represent zero or negative values of d.)
Thus, we find
d= πm
β2
= λ2
2 m, where m= 1,2,3,... (5.76)
where λ2 = 2π/β2 is the wavelength inside the slab.
Summarizing:
T otal transmission through a slab embedded in
re
gions of material having equal wave impedance
(i.e., η1 = η3) may be achieved by setting the
thickness of the slab equal to an integer number
of half-wavelengths at the frequency of interest.
This is known as half-wave matching.
A remarkable feature of half-wave matching is that
there is no restriction on the permittivity or
permeability of the slab, and the only constraint on
the media in Regions 1 and 3 is that they have equal
wave impedance.
Example 5.4. Radome design by half-wave
matching.",Electromagnetics_Vol2.pdf
"the media in Regions 1 and 3 is that they have equal
wave impedance.
Example 5.4. Radome design by half-wave
matching.
The antenna for a 60 GHz radar is to be
protected from weather by a radome panel
positioned directly in front of the radar. The
panel is to be constructed from a low-loss
material having µr ≈ 1 and ǫr = 4. T o have
sufficient mechanical integrity , the panel must be
at least 3 mm thick. What thickness should be",Electromagnetics_Vol2.pdf
"66 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
used?
Solution
. This is a good application for
half-wave matching because the material on
either side of the slab is the same (presumably
free space) whereas the material used for the slab
is unspecified. The phase velocity in the slab is
vp = c
√ǫr
∼= 1.5 × 108 m/s (5.77)
so the wavelength in the slab is
λ2 = vp
f
∼
= 2.5 mm
 (5.78)
Thus, the minimum thickness of the slab that
satisfies the half-wave matching condition is
d= λ2/2 ∼= 1.25 mm. However, this does not
meet the 3 mm minimum-thickness requirement.
Neither does the next available thickness,
d= λ2 ∼= 2.5 mm. The next-thickest option,
d= 3λ2/2 ∼= 3.75 mm, does meet the
requirement. Therefore, we select d∼= 3.75 mm.
It
should be emphasized that designs employing
half-wave matching will be narrowband – that is, total
only for the design frequency . As frequency increases
or decreases from the design frequency , there will be
increasing reflection and decreasing transmission.",Electromagnetics_Vol2.pdf
"or decreases from the design frequency , there will be
increasing reflection and decreasing transmission.
Quarter-wave matching requires that the wave
impedances in each region are different and related in
a particular way . The quarter-wave solution is
obtained by requiring P = −1, so that
η1 = η2
(1 + P)η3 + (1 − P)η2
(1 − P)η3 + (1 + P)η2
= η2
η2
η3
(5.79)
Solving
this equation for wave impedance of the slab
material, we find
η2 = √η1η3 (5.80)
Note
that P = −1 is obtained when 2β2d= π+ 2πm
where m= 0,1,2,.... Thus, we find
d= π
2β2
+ π
β2
m
= λ2
4 + λ2
2 m, where m= 0,1,2,... (5.81)
Summarizing:
T otal transmission is achieved through a slab by
selecting η2 = √η1η3 and making the slab one-
quarter wavelength at the frequency of interest, or
some integer number of half-wavelengths thicker
if needed. This is known as quarter-wave match-
ing.
Example 5.5. Radome design by quarter-wave
matching.
The antenna for a 60 GHz radar is to be
protected from weather by a radome panel",Electromagnetics_Vol2.pdf
"ing.
Example 5.5. Radome design by quarter-wave
matching.
The antenna for a 60 GHz radar is to be
protected from weather by a radome panel
positioned directly in front of the radar. In this
case, however, the antenna is embedded in a
lossless material having µr ≈ 1 and ǫr = 2, and
the radome panel is to be placed between this
material and the outside, which we presume is
free space. The material in which the antenna is
embedded, and against which the radome panel
is installed, is quite rigid so there is no minimum
thickness requirement. However, the radome
panel must be made from a material which is
lossless and non-magnetic. Design the radome
panel.
Solution. The radome panel must be comprised
of a material having
η2 = √
η1η3 =
√ η0√
2 · η0 ∼= 317 Ω (5.82)
Since the radome panel is required to be
non-magnetic, the relative permittivity must be
given by
η2 = η0
√ǫr
⇒ ǫr =
(η0
η2
)2
∼= 1.41 (5.83)
The
phase velocity in the slab will be
vp = c√ǫr
∼= 3 × 108 m/s√
1.41
∼
= 2",Electromagnetics_Vol2.pdf
"given by
η2 = η0
√ǫr
⇒ ǫr =
(η0
η2
)2
∼= 1.41 (5.83)
The
phase velocity in the slab will be
vp = c√ǫr
∼= 3 × 108 m/s√
1.41
∼
= 2
 .53 × 108 m/s
(5.84)
so the wavelength in the slab is
λ2 = vp
f
∼
= 2.53 × 108 m/s
60 × 109 Hz
∼
= 4
 .20 mm (5.85)",Electromagnetics_Vol2.pdf
"5.4. PROP AGA TION OF A UNIFORM PLANE W A VE IN AN ARBITRAR Y DIRECTION 67
Thus, the minimum possible thickness of the
radome
panel is d= λ2/4 ∼= 1.05 mm, and the
relati
ve permittivity of the radome panel must be
ǫr ∼= 1.41.
Additional
Reading:
• “Radome” on Wikipedia.
5.4 Propagation of a Uniform
Plane W ave in an Arbitrary
Direction
[m0165]
An example of a uniform plane wave propagating in a
lossless
medium is shown in Figure 5.4. This wave is
expressed in the indicated coordinate system as
follows:
˜E = ˆxE0e−jβz (5.86)
This is a phasor-domain expression for the electric
field intensity , so E0 is a complex-valued number
representing the magnitude and reference phase of the
sinusoidally-varying wave. The term “reference
phase” is defined as the phase of ˜E at the origin of the
coordinate system. Since the phase propagation
constant βis real and positive, this wave is traveling
in the +ˆz direction through simple lossless media.
Note that electric field intensity vector is linearly",Electromagnetics_Vol2.pdf
"in the +ˆz direction through simple lossless media.
Note that electric field intensity vector is linearly
polarized in a direction parallel to ˆx. Depending on
the position in space (and, for the physical
time-domain waveform, time), ˜E points either in the
+ˆx direction or the −ˆx direction. Let us be a bit more
specific about the direction of the vector ˜E. T o do
this, let us define the reference polarization to be the
direction in which ˜E points when
Re
{
E0e−jβz}
≥ 0; i.e., when the phase of ˜E is
between −π/2 and +π/2 radians. Thus, the reference
polarization of ˜E in Equation 5.86 is always +ˆx.
Note that Equation 5.86 indicates a specific
combination of reference polarization and direction of
propagation. However, we may obtain any other
combination of reference polarization and direction of
c⃝ C. W ang CC BY -SA 4.0
Figure 5.4: The plane wave described by Equa-
tion 5.86.",Electromagnetics_Vol2.pdf
"68 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
c⃝ C. W ang CC BY -SA 4.0
Figure 5.5: The same plane wave described in a ro-
tated coordinate system, yielding Equation 5.87.
c⃝ C. W ang CC BY -SA 4.0
Figure 5.6: The same plane wave described in yet an-
other rotation of the coordinate system, yielding Equa-
tion 5.88.
propagation by rotation of the Cartesian coordinate
system. For example, if we rotate the +xaxis of the
coordinate system into the position originally
occupied by the +yaxis, then the very same wave is
expressed as
˜E = −ˆyE0e−jβz (5.87)
This is illustrated in Figure 5.5. At first glance, it
appears that the reference polarization has changed;
however, this is due entirely to our choice of
coordinate system. That is, the reference polarization
is precisely the same; it is only the coordinate system
used to describe the reference polarization that has
changed.
Let us now rotate the +zaxis of the coordinate
system around the xaxis into the position originally",Electromagnetics_Vol2.pdf
"changed.
Let us now rotate the +zaxis of the coordinate
system around the xaxis into the position originally
occupied by the −zaxis. Now the very same wave is
expressed as
˜E = + ˆyE0e+jβz (5.88)
This is illustrated in Figure 5.6. At first glance, it
appears that the direction of propagation has reversed;
but, again, it is only the coordinate system that has
changed, and not the direction of propagation.
Summarizing: Equations 5.86, 5.87, and 5.88 all
represent the same wave. They only appear to be
c⃝ C. W ang CC BY -SA 4.0
Figure 5.7: A plane wave described in ray-fixed coor-
dinates, yielding Equation 5.89.
different due to our choice for the orientation of the
coordinate system in each case.
Now consider the same thought experiment for the
infinite number of cases in which the wave does not
propagate in one of the three basis directions of the
Cartesian coordinate system. One situation in which
we face this complication is when the wave is",Electromagnetics_Vol2.pdf
"Cartesian coordinate system. One situation in which
we face this complication is when the wave is
obliquely incident on a surface. In this case, it is
impossible to select a single orientation of the
coordinate system in which the directions of
propagation, reference polarization, and surface
normal can all be described in terms of one basis
vector each. T o be clear: There is no fundamental
limitation imposed by this slipperiness of the
coordinate system. However, practical problems such
as the oblique-incidence scenario described above are
much easier to analyze if we are able to express
waves in a system of coordinates which always yields
the same expressions.
Fortunately , this is easily accomplished using
ray-fixed coordinates. Ray-fixed coordinates are a
unique set of coordinates that are determined from the
characteristics of the wave, as opposed to being
determined arbitrarily and separately from the
characteristics of the wave. The ray-fixed",Electromagnetics_Vol2.pdf
"characteristics of the wave, as opposed to being
determined arbitrarily and separately from the
characteristics of the wave. The ray-fixed
representation of a uniform plane wave is:
˜E(r) = ˆeE0e−jk·r
(5.89)
This
is illustrated in Figure 5.7. In this representation,
r is the position at which ˜E is evaluated, ˆe is the
reference polarization expressed as a unit vector, and
k is the unit vector ˆk in the direction of propagation,
times β; i.e.:
k ≜ ˆkβ (5.90)
Consider Equation 5.86 as an example. In this case,
ˆe = ˆx, k = ˆzβ, and (as always)
r = ˆxx+ ˆyy+ ˆzz (5.91)",Electromagnetics_Vol2.pdf
"5.4. PROP AGA TION OF A UNIFORM PLANE W A VE IN AN ARBITRAR Y DIRECTION 69
Thus, k · r = βz, as expected.
In ray-fixed coordinates, a wave can be represented
by one – and only one – expression, which is the same
expression regardless of the orientation of the
“global” coordinate system. Moreover, only two basis
directions (namely , ˆk and ˆe) must be defined. Should
a third coordinate be required, either ˆk × ˆe or ˆe × ˆk
may be selected as the additional basis direction.
Note that the first choice has the possibly-useful
feature that is the reference polarization of the
magnetic field intensity ˜H.
A general procedure for recasting the ray-fixed
representation into a “coordinate-bound”
representation is as follows. First, we represent k in
the new fixed coordinate system; e.g.:
k = kxˆx + kyˆy + kzˆz (5.92)
where
kx ≜ βˆk · ˆx (5.93)
ky ≜ βˆk · ˆy (5.94)
kz ≜ βˆk · ˆz (5.95)
Then,
k · r = kxx+ kyy+ kzz (5.96)
With this expression in hand, Equation 5.89 may be
rewritten as:",Electromagnetics_Vol2.pdf
"ky ≜ βˆk · ˆy (5.94)
kz ≜ βˆk · ˆz (5.95)
Then,
k · r = kxx+ kyy+ kzz (5.96)
With this expression in hand, Equation 5.89 may be
rewritten as:
˜E = ˆeE0e−jkxxe−jky ye−jkz z (5.97)
If desired, one can similarly decompose ˆe into its
Cartesian components as follows:
ˆe = ( ˆe · ˆx) ˆx + (ˆe · ˆy) ˆy + (ˆe · ˆz) ˆz (5.98)
Thus, we see that the ray-fixed representation of
Equation 5.89 accommodates all possible
combinations of direction of propagation and
reference polarization.
Example 5.6. Plane wave propagating away
from the z-axis.
A uniform plane wave exhibiting a reference
polarization of ˆz propagates away from the
z-axis. Develop representations of this wave in
ray-fixed and global Cartesian coordinates.
Solution. As always, the ray-fixed
representation
is given by Equation 5.89. Since
the reference polarization is ˆz, ˆe = ˆz.
Propagating away from the z-axis means
ˆk = ˆx cos φ+ ˆy sin φ (5.99)
where φindicates the specific direction of
propagation. For example, φ= 0 yields",Electromagnetics_Vol2.pdf
"Propagating away from the z-axis means
ˆk = ˆx cos φ+ ˆy sin φ (5.99)
where φindicates the specific direction of
propagation. For example, φ= 0 yields
ˆk = + ˆx, φ= π/2 yields ˆk = + ˆy, and so on.
Therefore,
k ≜ βˆk = β(ˆx cos φ+ ˆy sin φ) (5.100)
For completeness, note that the following factor
appears in the phase-determining exponent in
Equation 5.89:
k · r = β(xcos φ+ ysin φ) (5.101)
In this case, we see kx = βcos φ, ky = βsin φ,
and kz = 0. Thus, the wave may be expressed in
Cartesian coordinates as follows:
˜E = ˆzE0e−jβx cos φe−jβy sin φ (5.102)",Electromagnetics_Vol2.pdf
"70 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.5 Decomposition of a W ave into
TE and TM Components
[m0166]
A broad range of problems in electromagnetics
in
volve scattering of a plane wave by a planar
boundary between dissimilar media. Section 5.1
(“Plane W aves at Normal Incidence on a Planar
Boundary Between Lossless Media”) addressed the
special case in which the wave arrives in a direction
which is perpendicular to the boundary (i.e., “normal
incidence”). Analysis of the normal incidence case is
simplified by the fact that the directions of field
vectors associated with the reflected and transmitted
are the same (except possibly with a sign change) as
those of the incident wave. For the more general case
in which the incident wave is obliquely incident (i.e.,
not necessarily normally-incident), the directions of
the field vectors will generally be different. This
added complexity is easily handled if we take the
effort to represent the incident wave as the sum of two",Electromagnetics_Vol2.pdf
"added complexity is easily handled if we take the
effort to represent the incident wave as the sum of two
waves having particular polarizations. These
polarizations are referred to as transverse electric
(TE) and transverse magnetic (TM). This section
describes these polarizations and the method for
decomposition of a plane wave into TE and TM
components. W e will then be prepared to address the
oblique incidence case in a later section.
T o begin, we define the ray-fixed coordinate system
shown in Figure 5.8. In this figure, ˆki is a unit vector
indicating the direction in which the incident wave
propagates. The unit normal ˆn is perpendicular to the
boundary , and points into the region from which the
wave is incident. W e now make the following
definition:
The plane of incidence is the plane in which both
the
normal to the surface ( ˆn) and the direction of
propagation ( ˆki) lie.
The TE-TM decomposition consists of finding the
components of the electric and magnetic fields which",Electromagnetics_Vol2.pdf
"propagation ( ˆki) lie.
The TE-TM decomposition consists of finding the
components of the electric and magnetic fields which
are perpendicular (“transverse”) to the plane of
incidence. Of the two possible directions that are
perpendicular to the plane of incidence, we choose
ˆe⊥, defined as shown in Figure 5.8. From the figure,
we see that:
ˆe⊥ ≜
ˆki × ˆn
⏐⏐
⏐ˆki × ˆn
⏐
⏐
⏐
(5.103)
Defined
in this manner, ˆe⊥ is a unit vector which is
perpendicular to both ˆki, and so may serve as a basis
vector of a coordinate system which is attached to the
incident ray . The remaining basis vector for this
ray-fixed coordinate system is chosen as follows:
ˆei
∥ ≜ ˆe⊥ × ˆki (5.104)
Defined in this manner, ˆei
∥ is a unit vector which is
perpendicular to both ˆe⊥ and ˆki, and parallel to the
plane of incidence.
Let us now examine an incident uniform plane wave
in the new , ray-fixed coordinate system. W e begin
with the following phasor representation of the
electric field intensity:
˜Ei = ˆeiEi",Electromagnetics_Vol2.pdf
"in the new , ray-fixed coordinate system. W e begin
with the following phasor representation of the
electric field intensity:
˜Ei = ˆeiEi
0e−jki·r (5.105)
where ˆei is a unit vector indicating the reference
polarization, Ei
0 is a complex-valued scalar, and r is a
vector indicating the position at which ˜Ei is
evaluated. W e may express ˆei in the ray-fixed
coordinate system of Figure 5.8 as follows:
ˆei =
(ˆei · ˆe⊥
)ˆe⊥
+
(
ˆei · ˆei
∥
)
ˆei
∥
+
(
ˆei · ˆki
)
ˆki (5.106)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.8: Coordinate system for TE-TM decompo-
sition. Since both ˆki and ˆn lie in the plane of the page,
this is also the plane of incidence.",Electromagnetics_Vol2.pdf
"5.5. DECOMPOSITION OF A W A VE INTO TE AND TM COMPONENTS 71
The electric field vector is always perpendicular to the
direction of propagation, so ˆei · ˆki = 0. This leaves:
ˆei =
(ˆei · ˆe⊥
)ˆe⊥ +
(
ˆei · ˆei
∥
)
ˆei
∥ (5.107)
Substituting this expression into Equation 5.105, we
obtain:
˜Ei = ˆe⊥Ei
TEe−jki·r + ˆei
∥ Ei
TMe−jki·r (5.108)
where
Ei
TE ≜ Ei
0 ˆei · ˆe⊥ (5.109)
Ei
TM ≜ Ei
0 ˆei · ˆei
∥ (5.110)
The first term in Equation 5.108 is the transverse
electric (TE) component of ˜Ei, so-named because it is
the component which is perpendicular to the plane of
incidence. The second term in Equation 5.108 is the
transverse magnetic (TM) component of ˜Ei. The term
“TM” refers to the fact that the magnetic field
associated with this component of the electric field is
perpendicular to the plane of incidence. This is
apparent since the magnetic field vector is
perpendicular to both the direction of propagation and
the electric field vector.
Summarizing:",Electromagnetics_Vol2.pdf
"apparent since the magnetic field vector is
perpendicular to both the direction of propagation and
the electric field vector.
Summarizing:
The TE component is the component for which
˜Ei is
perpendicular to the plane of incidence.
The TM component is the component for which
˜Hi is
perpendicular to the plane of incidence;
i.e., the component for which ˜Ei is parallel to the
plane of incidence.
Finally , observe that the total wave is the sum of its
TE and TM components. Therefore, we may analyze
the TE and TM components separately , and know that
the result for the combined wave is simply the sum of
the results for the TE and TM components.
As stated at the beginning of this section, the utility of
the TE-TM decomposition is that it simplifies the
analysis of wave reflection. This is because the
analysis of the TE and TM cases is relatively simple,
whereas direct analysis of the scattering of
arbitrarily-polarized waves is relatively difficult.
Note that the nomenclature “TE” and “TM” is",Electromagnetics_Vol2.pdf
"whereas direct analysis of the scattering of
arbitrarily-polarized waves is relatively difficult.
Note that the nomenclature “TE” and “TM” is
commonly but not universally used. Sometimes “TE”
is referred to as “perpendicular” polarization,
indicated using the subscript “⊥ ” or “s” (short for
senkrecht, German for “perpendicular”).
Correspondingly , “TM” is sometimes referred to as
“parallel” polarization, indicated using the subscript
“∥” or “p. ”
Also, note that the TE component and TM component
are sometimes referred to as the TE mode and TM
mode, respectively . While the terms “component” and
“mode” are synonymous for the single plane wave
scenarios considered in this section, the terms are not
synonymous in general. For example, the wave inside
a waveguide may consist of multiple unique TE
modes which collectively comprise the TE component
of the field, and similarly the wave inside a waveguide
may consist of multiple unique TM modes which",Electromagnetics_Vol2.pdf
"of the field, and similarly the wave inside a waveguide
may consist of multiple unique TM modes which
collectively comprise the TM component of the field.
Finally , consider what happens when a plane wave is
normally-incident upon the boundary; i.e., when
ˆki = −ˆn. In this case, Equation 5.103 indicates that
ˆe⊥ = 0, so the TE-TM decomposition is undefined.
The situation is simply that both Ei and Hi are
already both perpendicular to the boundary , and there
is no single plane that can be uniquely identified as
the plane of incidence. W e refer to this case as
transverse electromagnetic (TEM).
A wave which is normally-incident on a planar
surf
ace is said to be transverse electromagnetic
(TEM) with respect to that boundary . There is no
unique TE-TM decomposition in this case.
The fact that a unique TE-TM decomposition does
not exist in the TEM case is of no consequence,
because the TEM case is easily handled as a separate
condition (see Section 5.1).",Electromagnetics_Vol2.pdf
"72 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.6 Plane W aves at Oblique
Incidence on a Planar
Boundary: TE Case
[m0167]
In this section, we consider the problem of reflection
and
transmission from a planar boundary between
semi-infinite media for a transverse electric (TE)
uniform plane wave. Before attempting this section, a
review of Sections 5.1 (“Plane W aves at Normal
Incidence on a Planar Boundary Between Lossless
Media”) and 5.5 (“Decomposition of a W ave into TE
and TM Components”) is recommended. Also, note
that this section has much in common with
Section 5.7 (“Plane W aves at Oblique Incidence on a
Planar Boundary: TM Case”), although it is
recommended to attempt the TE case first.
The TE case is illustrated in Figure 5.9. The boundary
between the two semi-infinite and lossless regions is
located at the z= 0 plane. The wave is incident from
Region 1. The electric field intensity ˜Ei
TE of this
wave is given by
˜Ei
TE(r) = ˆyEi
TEe−jki·r (5.111)",Electromagnetics_Vol2.pdf
"Region 1. The electric field intensity ˜Ei
TE of this
wave is given by
˜Ei
TE(r) = ˆyEi
TEe−jki·r (5.111)
In this expression, r is the position at which ˜Ei
TE is
evaluated, and
ki = ˆkiβ1 (5.112)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.9: A TE uniform plane wave obliquely inci-
dent on the planar boundary between two semi-infinite
material regions.
where ˆki is the unit vector indicating the direction of
propagation and β1 = ω√µ1ǫ1 is the phase
propagation constant in Region 1. ˜Ei
TE serves as the
“stimulus” in this problem, so all other contributions
to the total field will be expressed in terms of
quantities appearing in Equation 5.111.
The presence of reflected and transmitted uniform
plane waves is inferred from our experience with the
normal incidence scenario (Section 5.1). There, as
here, the symmetry of the problem indicates that the
reflected and transmitted components of the electric
field will have the same polarization as that of the",Electromagnetics_Vol2.pdf
"reflected and transmitted components of the electric
field will have the same polarization as that of the
incident electric field. This is because there is nothing
present in the problem that could account for a change
in polarization. Thus, the reflected and transmitted
fields will also be TE. Therefore, we postulate the
following expression for the reflected wave:
˜Er(r) = ˆyBe−jkr ·r (5.113)
where Bis an unknown, possibly complex-valued
constant to be determined and
kr = ˆkrβ1 (5.114)
indicates the direction of propagation, which is also
currently unknown.
Similarly , we postulate the following expression for
the transmitted wave:
˜Et(r) = ˆyCe−jkt·r (5.115)
where Cis an unknown, possibly complex-valued
constant to be determined;
kt = ˆktβ2 (5.116)
where ˆkt is the unit vector indicating the direction of
propagation; and β2 = ω√
µ2ǫ2 is the phase
propagation constant in Region 2.
At this point, the unknowns in this problem are the
constants Band C, as well as the directions ˆkr and",Electromagnetics_Vol2.pdf
"propagation constant in Region 2.
At this point, the unknowns in this problem are the
constants Band C, as well as the directions ˆkr and
ˆkt. W e may establish a relationship between Ei
TE, B,
and Cby application of boundary conditions at
z= 0. First, recall that the tangential component of
the total electric field intensity must be continuous
across material boundaries. T o apply this boundary
condition, let us define ˜E1 and ˜E2 to be the total
electric fields in Regions 1 and 2, respectively . The",Electromagnetics_Vol2.pdf
"5.6. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TE CASE 73
total field in Region 1 is the sum of incident and
reflected fields, so
˜E1(r) = ˜Ei
TE(r) + ˜Er(r) (5.117)
The total field in Region 2 is simply
˜E2(r) = ˜Et(r) (5.118)
Next, note that all electric field components are
already tangent to the boundary . Thus, continuity of
the tangential component of the electric field across
the boundary requires ˜E1(0) = ˜E2(0), and therefore
˜Ei
TE(r0) + ˜Er(r0) = ˜Et(r0) (5.119)
where r = r0 ≜ ˆxx+ ˆyysince z= 0 on the
boundary . Now employing Equations 5.111, 5.113,
and 5.115, we obtain:
ˆyEi
TEe−jki·r0 + ˆyBe−jkr ·r0 = ˆyCe−jkt·r0 (5.120)
Dropping the vector ( ˆy) since it is the same in each
term, we obtain:
Ei
TEe−jki·r0 + Be−jkr ·r0 = Ce−jkt·r0 (5.121)
For Equation 5.121 to be true at every point r0 on the
boundary , it must be true that
ki · r0 = kr · r0 = kt · r0 (5.122)
Essentially , we are requiring the phases of each field",Electromagnetics_Vol2.pdf
"boundary , it must be true that
ki · r0 = kr · r0 = kt · r0 (5.122)
Essentially , we are requiring the phases of each field
in Regions 1 and 2 to be matched at every point along
the boundary . Any other choice will result in a
violation of boundary conditions at some point along
the boundary . This phase matching criterion will
determine the directions of propagation of the
reflected and transmitted fields, which we shall do
later.
First, let us use the phase matching criterion to
complete the solution for the coefficients Band C.
Enforcing Equation 5.122, we observe that
Equation 5.121 reduces to:
Ei
TE + B = C (5.123)
A second equation is needed since we currently have
only one equation (Equation 5.123) and two
unknowns (B and C). The second equation is
c⃝ C. W ang CC BY -SA 4.0
Figure 5.10: Incident, reflected, and transmitted mag-
netic field components associated with the TE electric
field components shown in Figure 5.9.
obtained by applying the appropriate boundary",Electromagnetics_Vol2.pdf
"netic field components associated with the TE electric
field components shown in Figure 5.9.
obtained by applying the appropriate boundary
conditions to the magnetic field. The magnetic field
associated with each of the electric field components
is identified in Figure 5.10. Note the orientations of
the magnetic field vectors may be confirmed using the
plane wave relationships: Specifically , the cross
product of the electric and magnetic fields should
point in the direction of propagation. Expressions for
each of the magnetic field components is determined
formally below .
From the plane wave relationships, we determine that
the incident magnetic field intensity is
˜Hi(r) = 1
η1
ˆki × ˜Ei
TE (5.124)
where η1 =
√
µ1/ǫ1 is the wave impedance in
Region 1. T o make progress requires that we express
ˆki in the global fixed coordinate system. Here it is:
ˆki = ˆx sin ψi + ˆz cos ψi (5.125)
Thus:
˜Hi(r) =
(ˆz sin ψi − ˆx cos ψi)Ei
TE
η1
e−jki·r (5.126)
Similarly
, we determine that the reflected magnetic",Electromagnetics_Vol2.pdf
"Thus:
˜Hi(r) =
(ˆz sin ψi − ˆx cos ψi)Ei
TE
η1
e−jki·r (5.126)
Similarly
, we determine that the reflected magnetic
field has the form:
˜Hr(r) = 1
η1
ˆkr × ˜Er (5.127)",Electromagnetics_Vol2.pdf
"74 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
In the global coordinate system:
ˆkr = ˆx sin ψr − ˆz cos ψr (5.128)
Thus:
˜Hr(r) = ( ˆz sin ψr + ˆx cos ψr) B
η1
e−jkr ·r (5.129)
The
transmitted magnetic field has the form:
˜Ht(r) = 1
η2
ˆkt × ˜Et (5.130)
In
the global coordinate system:
ˆkt = ˆx sin ψt + ˆz cos ψt (5.131)
Thus:
˜Ht(r) =
(ˆz sin ψt − ˆx cos ψt)C
η2
e−jkt·r (5.132)
The
total magnetic field in Region 1 is the sum of
incident and reflected fields, so
˜H1(r) = ˜Hi(r) + ˜Hr(r) (5.133)
The magnetic field in Region 2 is simply
˜H2(r) = ˜Ht(r) (5.134)
Since there is no current on the boundary , the
tangential component of the total magnetic field
intensity must be continuous across the boundary .
Expressed in terms of the quantities already
established, this boundary condition requires:
ˆx · ˜Hi(r0) + ˆx · ˜Hr(r0) = ˆx · ˜Ht(r0) (5.135)
where “ ˆx·” selects the component of the magnetic
field that is tangent to the boundary . Evaluating this
expression, we obtain:
−
(
cos ψi)Ei
TE",Electromagnetics_Vol2.pdf
"where “ ˆx·” selects the component of the magnetic
field that is tangent to the boundary . Evaluating this
expression, we obtain:
−
(
cos ψi)Ei
TE
η1
e−jki·r0
+ (
cos ψr) B
η1
e−jkr ·r0
= −
(
cos ψt)C
η2
e−jkt·r0 (5.136)
No
w employing the phase matching condition
expressed in Equation 5.122, we find:
−
(
cos ψi)Ei
TE
η1
+ (
cos ψr) B
η1
= −
(
cos ψt)C
η2
(5.137)
Equations
5.123 and 5.137 comprise a linear system
of equations with unknowns Band C. This system of
equations is easily solved for Bas follows. First, use
Equation 5.123 to eliminate Cin Equation 5.137. The
result is:
−
(
cos ψi)Ei
TE
η1
+ (
cos ψr) B
η1
= −
(
cos ψt)Ei
T
E + B
η2
(5.138)
Solving
this equation for B, we obtain:
B = η2 cos ψi − η1 cos ψt
η2 cos ψr + η1 cos ψt Ei
T E (5.139)
W e can express this result as a reflection coefficient as
follows:
B = Γ TEEi
TE (5.140)
where
Γ TE ≜ η2 cos ψi − η1 cos ψt
η2 cos ψr + η1 cos ψt (5.141)
It
is worth noting that Equation 5.141 becomes the",Electromagnetics_Vol2.pdf
"follows:
B = Γ TEEi
TE (5.140)
where
Γ TE ≜ η2 cos ψi − η1 cos ψt
η2 cos ψr + η1 cos ψt (5.141)
It
is worth noting that Equation 5.141 becomes the
reflection coefficient for normal (TEM) incidence
when ψi = ψr = ψt = 0, as expected.
Returning to Equation 5.123, we now find
C = (1 + Γ TE) Ei
TE (5.142)
Let us now summarize the solution. Given the TE
electric field intensity expressed in Equation 5.111,
we find:
˜Er(r) = ˆyΓ TEEi
TEe−jkr ·r
(5.143)
˜Et(r) = ˆy (1
+ Γ TE) Ei
TEe−jkt·r (5.144)
This
solution is complete except that we have not yet
determined ˆkr, which is now completely determined
by ψr via Equation 5.128, and ˆkt, which is now
completely determined by ψt via Equation 5.131. In
other words, we have not yet determined the
directions of propagation ψr for the reflected wave
and ψt for the transmitted wave. However, ψr and ψi
can be found using Equation 5.122. Here we shall",Electromagnetics_Vol2.pdf
"5.6. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TE CASE 75
simply state the result, and in Section 5.8 we shall
perform this part of the derivation in detail and with
greater attention to the implications. One finds:
ψr = ψi (5.145)
and
ψt = arcsin
(β1
β2
sin ψi
)
(5.146)
Equation
5.145 is the unsurprising result that angle of
reflection equals angle of incidence. Equation 5.146 –
addressing angle of transmission – is a bit more
intriguing. Astute readers may notice that there is
something fishy about this equation: It seems possible
for the argument of arcsin to be greater than one.
This oddity is addressed in Section 5.8.
Finally , note that Equation 5.145 allows us to
eliminate ψr from Equation 5.141, yielding:
Γ TE = η2 cos ψi − η1 cos ψt
η2 cos ψi + η1 cos ψt (5.147)
Thus,
we obtain what is perhaps the most important
finding of this section:
The electric field reflection coefficient for oblique
TE
incidence, Γ TE, is given by Equation 5.147.",Electromagnetics_Vol2.pdf
"finding of this section:
The electric field reflection coefficient for oblique
TE
incidence, Γ TE, is given by Equation 5.147.
The following example demonstrates the utility of this
result.
Example 5.7. Po wer transmission at an
air-to-glass interface (TE case).
Figure 5.11 illustrates a TE plane wave incident
from air onto the planar boundary with glass.
The glass exhibits relative permittivity of 2.1.
Determine the power reflected and transmitted
relative to power incident on the boundary .
Solution. The power reflected relative to power
incident is |Γ TE|2 whereas the power
transmitted relative to power incident is
1 − |Γ TE|2. Γ TE may be calculated using
Equation 5.147. Calculating the quantities that
enter into this expression:
η1 ≈ η0 ∼= 376.7 Ω (air) (5.148)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.11: A TE uniform plane wave incident from
air to glass.
η2 ≈ η0√
2.1
∼= 260 .0 Ω (glass) (5.149)
ψi = 30 ◦ (5.150)
Note
β1
β2
≈ ω√µ0ǫ0
ω√µ0 · 2.1ǫ0
∼
= 0.690 (5.151)
so
ψt =
arcsin
(β1
β2",Electromagnetics_Vol2.pdf
"air to glass.
η2 ≈ η0√
2.1
∼= 260 .0 Ω (glass) (5.149)
ψi = 30 ◦ (5.150)
Note
β1
β2
≈ ω√µ0ǫ0
ω√µ0 · 2.1ǫ0
∼
= 0.690 (5.151)
so
ψt =
arcsin
(β1
β2
sin ψi
)
∼
= 20
 .2◦ (5.152)
Now substituting these values into
Equation 5.147, we obtain
Γ TE ∼= −0.2220 (5.153)
Subsequently , the fraction of power reflected
relative to power incident is |Γ TE|2 ∼= 0.049;
i.e., about 4.9% . 1 − |Γ T E|2 ∼= 95.1% of the
po
wer is transmitted into the glass.",Electromagnetics_Vol2.pdf
"76 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.7 Plane W aves at Oblique
Incidence on a Planar
Boundary: TM Case
[m0164]
In this section, we consider the problem of reflection
and
transmission from a planar boundary between
semi-infinite media for a transverse magnetic (TM)
uniform plane wave. Before attempting this section, a
review of Sections 5.1 (“Plane W aves at Normal
Incidence on a Planar Boundary Between Lossless
Media”) and 5.5 (“Decomposition of a W ave into TE
and TM Components”) is recommended. Also, note
that this section has much in common with
Section 5.6 (“Plane W aves at Oblique Incidence on a
Planar Boundary: TE Case”), and it is recommended
to attempt the TE case first.
In this section, we consider the scenario illustrated in
Figure 5.12. The boundary between the two
semi-infinite and lossless regions is located at the
z= 0 plane. The wave is incident from Region 1. The
magnetic field intensity ˜Hi
TM of this wave is given by
˜Hi
TM(r) = ˆyHi
TMe−jki·r (5.154)",Electromagnetics_Vol2.pdf
"z= 0 plane. The wave is incident from Region 1. The
magnetic field intensity ˜Hi
TM of this wave is given by
˜Hi
TM(r) = ˆyHi
TMe−jki·r (5.154)
In this expression, r is the position at which ˜Hi
TM is
c⃝ C. W ang CC BY -SA 4.0
Figure 5.12: A TM uniform plane wave obliquely
incident on the planar boundary between two semi-
infinite material regions.
evaluated. Also,
ki = ˆkiβ1 (5.155)
where ˆki is the unit vector indicating the direction of
propagation and β1 = ω√µ1ǫ1 is the phase
propagation constant in Region 1. ˜Hi
TM serves as the
“stimulus” in this problem, and all other contributions
to the total field may be expressed in terms of
parameters associated with Equation 5.154.
The presence of reflected and transmitted uniform
plane waves is inferred from our experience with the
normal incidence scenario (Section 5.1). There, as
here, the symmetry of the problem indicates that the
reflected and transmitted components of the magnetic
field will have the same polarization as that of the",Electromagnetics_Vol2.pdf
"reflected and transmitted components of the magnetic
field will have the same polarization as that of the
incident electric field. This is because there is nothing
present in the problem that could account for a
change in polarization. Thus, the reflected and
transmitted fields will also be TM. So we postulate
the following expression for the reflected wave:
˜Hr(r) = −ˆyBe−jkr ·r (5.156)
where Bis an unknown, possibly complex-valued
constant to be determined and
kr = ˆkrβ1 (5.157)
indicates the direction of propagation.
The reader may wonder why we have chosen −ˆy, as
opposed to +ˆy, as the reference polarization for ˜Hr.
In fact, either +ˆy or −ˆy could be used. However, the
choice is important because the form of the results we
obtain in this section – specifically , the reflection
coefficient – will be determined with respect to this
specific convention, and will be incorrect with respect
to the opposite convention. W e choose −ˆy because it",Electromagnetics_Vol2.pdf
"specific convention, and will be incorrect with respect
to the opposite convention. W e choose −ˆy because it
has a particular advantage which we shall point out at
the end of this section.
Continuing, we postulate the following expression for
the transmitted wave:
˜Ht(r) = ˆyCe−jkt·r (5.158)
where Cis an unknown, possibly complex-valued
constant to be determined and
kt = ˆktβ2 (5.159)",Electromagnetics_Vol2.pdf
"5.7. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TM CASE 77
where ˆkt is the unit vector indicating the direction of
propagation and β2 = ω√µ2ǫ2 is the phase
propagation constant in Region 2.
At this point, the unknowns in this problem are the
constants Band C, as well as the unknown directions
ˆkr and ˆkt. W e may establish a relationship between
Hi
TM, B, and Cby application of boundary
conditions at z= 0. First, we presume no impressed
current at the boundary . Thus, the tangential
component of the total magnetic field intensity must
be continuous across the boundary . T o apply this
boundary condition, let us define ˜H1 and ˜H2 to be the
total magnetic fields in Regions 1 and 2, respectively .
The total field in Region 1 is the sum of incident and
reflected fields, so
˜H1(r) = ˜Hi
TM(r) + ˜Hr(r) (5.160)
The field in Region 2 is simply
˜H2(r) = ˜Ht(r) (5.161)
Also, we note that all magnetic field components are
already tangent to the boundary . Thus, continuity of",Electromagnetics_Vol2.pdf
"˜H2(r) = ˜Ht(r) (5.161)
Also, we note that all magnetic field components are
already tangent to the boundary . Thus, continuity of
the tangential component of the magnetic field across
the boundary requires ˜H1(r0) = ˜H2(r0), where
r0 ≜ ˆxx+ ˆyysince z= 0 on the boundary .
Therefore,
˜Hi
TM(r0) + ˜Hr(r0) = ˜Ht(r0) (5.162)
Now employing Equations 5.154, 5.156, and 5.158,
we obtain:
ˆyHi
TMe−jki·r0 − ˆyBe−jkr ·r0 = ˆyCe−jkt·r0
(5.163)
Dropping the vector ( ˆy) since it is the same in each
term, we obtain:
Hi
TMe−jki·r0 − Be−jkr ·r0 = Ce−jkt·r0 (5.164)
For this to be true at every point r0 on the boundary , it
must be true that
ki · r0 = kr · r0 = kt · r0 (5.165)
Essentially , we are requiring the phases of each field
in Regions 1 and 2 to be matched at every point along
the boundary . Any other choice will result in a
violation of boundary conditions at some point along
the boundary . This expression allows us to solve for
the directions of propagation of the reflected and",Electromagnetics_Vol2.pdf
"the boundary . This expression allows us to solve for
the directions of propagation of the reflected and
transmitted fields, which we shall do later. Our
priority for now shall be to solve for the coefficients
Band C.
Enforcing Equation 5.165, we observe that
Equation 5.164 reduces to:
Hi
TM − B = C (5.166)
A second equation is needed since we currently have
only one equation (Equation 5.166) and two
unknowns (B and C). The second equation is
obtained by applying the appropriate boundary
conditions to the electric field. The electric field
associated with each of the magnetic field
components is identified in Figure 5.12. Note the
orientations of the electric field vectors may be
confirmed using the plane wave relationships:
Specifically , the cross product of the electric and
magnetic fields should point in the direction of
propagation. Expressions for each of the electric field
components is determined formally below .
From the plane wave relationships, we determine that",Electromagnetics_Vol2.pdf
"propagation. Expressions for each of the electric field
components is determined formally below .
From the plane wave relationships, we determine that
the incident electric field intensity is
˜Ei(r) = −η1 ˆki × ˜Hi
TM (5.167)
where η1 =
√
µ1/ǫ1 is the wave impedance in
Region 1. T o make progress requires that we express
ˆki in the global fixed coordinate system. Here it is:
ˆki = ˆx sin ψi + ˆz cos ψi (5.168)
Thus:
˜Ei(r) =
(ˆx cos ψi − ˆz sin ψi)
η1Hi
TMe−jki·r
(5.169)
Similarly we determine that the reflected electric field
has the form:
˜Er(r) = −η1 ˆkr × ˜Hr (5.170)
In the global coordinate system:
ˆkr = ˆx sin ψr − ˆz cos ψr (5.171)
Thus:
˜Er(r) = ( ˆx cos ψr + ˆz sin ψr) η1Be−jkr ·r (5.172)",Electromagnetics_Vol2.pdf
"78 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
The transmitted magnetic field has the form:
˜Et(r) = −η2 ˆkt × ˜Ht (5.173)
In the global coordinate system:
ˆkt = ˆx sin ψt + ˆz cos ψt (5.174)
Thus:
˜Et(r) =
(ˆx cos ψt − ˆz sin ψt)
η2Ce−jkt·r (5.175)
The total electric field in Region 1 is the sum of
incident and reflected fields, so
˜E1(r) = ˜Ei(r) + ˜Er(r) (5.176)
The electric field in Region 2 is simply
˜E2(r) = ˜Et(r) (5.177)
The tangential component of the total electric field
intensity must be continuous across the boundary .
Expressed in terms of the quantities already
established, this boundary condition requires:
ˆx · ˜Ei(r0) + ˆx · ˜Er(r0) = ˆx · ˜Et(r0) (5.178)
where “ ˆx·” selects the component of the electric field
that is tangent to the boundary . Evaluating this
expression, we obtain:
+
(
cos ψi)
η1Hi
TMe−jki·r0
+ (cos ψr) η1Be−jkr ·r0
= +
(
cos ψt)
η2Ce−jkt·r0 (5.179)
Now employing the “phase matching” condition
expressed in Equation 5.165, we find:
+
(
cos ψi)
η1Hi
TM",Electromagnetics_Vol2.pdf
"= +
(
cos ψt)
η2Ce−jkt·r0 (5.179)
Now employing the “phase matching” condition
expressed in Equation 5.165, we find:
+
(
cos ψi)
η1Hi
TM
+ (cos ψr) η1B
= +
(
cos ψt)
η2C (5.180)
Equations 5.166 and 5.180 comprise a linear system
of equations with unknowns Band C. This system of
equations is easily solved for Bas follows. First, use
Equation 5.166 to eliminate Cin Equation 5.180. The
result is:
+
(
cos ψi)
η1Hi
TM
+ (cos ψr) η1B
= +
(
cos ψt)
η2
(
Hi
TM − B
)
(5.181)
Solving this equation for B, we obtain:
B = −η1 cos ψi + η2 cos ψt
+η1 cos ψr + η2 cos ψt Hi
T M (5.182)
W e can express this result as follows:
B = Γ TMHi
TM (5.183)
where we have made the definition
Γ TM ≜ −η1 cos ψi + η2 cos ψt
+η1 cos ψr + η2 cos ψt (5.184)
W
e are now able to express the complete solution in
terms of the electric field intensity . First we make the
substitution Ei
TM ≜ η1Hi
TM in Equation 5.169,
yielding:
˜Ei(r) =
(ˆx cos ψi − ˆz sin ψi)
Ei
TMe−jki·r (5.185)
The factor η1Bin Equation 5.172 becomes
Γ TMEi",Electromagnetics_Vol2.pdf
"TM ≜ η1Hi
TM in Equation 5.169,
yielding:
˜Ei(r) =
(ˆx cos ψi − ˆz sin ψi)
Ei
TMe−jki·r (5.185)
The factor η1Bin Equation 5.172 becomes
Γ TMEi
TM, so we obtain:
˜Er(r) = ( ˆx cos ψr + ˆz sin ψr)
· Γ TMEi
TMe−jkr ·r (5.186)
Thus, we see Γ TM is the reflection coefficient for the
electric field intensity .
Returning to Equation 5.166, we now find
C = Hi
TM − B
= Hi
TM − Γ TMHi
TM
= (1 − Γ TM) Hi
TM
= (1 − Γ TM) Ei
TM/η1 (5.187)
Subsequently , Equation 5.175 becomes
˜Et(r) =
(ˆx cos ψt − ˆz sin ψt)
· (1 − Γ TM) η2
η1
Ei
TMe−jkt·
r (5.188)
This solution is complete except that we have not yet
determined ˆkr, which is now completely determined
by ψr via Equation 5.171, and ˆkt, which is now
completely determined by ψt via Equation 5.174. In
other words, we have not yet determined the
directions of propagation ψr for the reflected wave
and ψt for the transmitted wave. However, ψr and ψi",Electromagnetics_Vol2.pdf
"5.7. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TM CASE 79
can be found using Equation 5.165. Here we shall
simply state the result, and in Section 5.8 we shall
perform this part of the derivation in detail and with
greater attention to the implications. One finds:
ψr = ψi (5.189)
i.e., angle of reflection equals angle of incidence.
Also,
ψt = arcsin
(β1
β2
sin ψi
)
(5.190)
Astute
readers may notice that there is something
fishy about Equation 5.190. Namely , it seems possible
for the argument of arcsin to be greater than one.
This oddity is addressed in Section 5.8.
Now let us return to the following question, raised
near the beginning of this section: Why choose −ˆy, as
opposed to +ˆy, as the reference polarization for Hr,
as shown in Figure 5.12? T o answer this question,
first note that Γ TM (Equation 5.184) becomes the
reflection coefficient for normal (TEM) incidence
when ψi = ψt = 0. If we had chosen +ˆy as the
reference polarization for Hr, we would have instead",Electromagnetics_Vol2.pdf
"reflection coefficient for normal (TEM) incidence
when ψi = ψt = 0. If we had chosen +ˆy as the
reference polarization for Hr, we would have instead
obtained an expression for Γ TM that has the opposite
sign for TEM incidence. 3 There is nothing wrong
with this answer, but it is awkward to have different
values of the reflection coefficient for the same
physical scenario. By choosing −ˆy, the reflection
coefficient for the oblique incidence case computed
for ψi = 0 converges to the reflection coefficient
previously computed for the normal-incidence case. It
is important to be aware of this issue, as one
occasionally encounters work in which the opposite
(“+ ˆy”) reference polarization has been employed.
Finally , note that Equation 5.189 allows us to
eliminate ψr from Equation 5.184, yielding:
Γ TM = −η1 cos ψi + η2 cos ψt
+η1 cos ψi + η2 cos ψt (5.191)
Thus,
we obtain what is perhaps the most important
finding of this section:
The electric field reflection coefficient for oblique
TM",Electromagnetics_Vol2.pdf
"Thus,
we obtain what is perhaps the most important
finding of this section:
The electric field reflection coefficient for oblique
TM
incidence, Γ TM, is given by Equation 5.191.
3 Obtaining this result is an excellent way for the student to con-
firm their understanding of the derivation presented in this section.
The following example demonstrates the utility of this
result.
Example 5.8. Po wer transmission at an
air-to-glass interface (TM case).
Figure 5.13 illustrates a TM plane wave incident
from air onto the planar boundary with a glass
region. The glass exhibits relative permittivity of
2.1. Determine the power reflected and
transmitted relative to power incident on the
boundary .
Solution. The power reflected relative to power
incident is |Γ TM|2 whereas the power
transmitted relative to power incident is
1 − |Γ TM|2. Γ TM may be calculated using
Equation 5.191. Calculating the quantities that
enter into this expression:
η1 ≈ η0 ∼= 376.7 Ω (air) (5.192)
η2 ≈ η0
√
2.1
∼
= 260",Electromagnetics_Vol2.pdf
"Equation 5.191. Calculating the quantities that
enter into this expression:
η1 ≈ η0 ∼= 376.7 Ω (air) (5.192)
η2 ≈ η0
√
2.1
∼
= 260
 .0 Ω (glass) (5.193)
ψi = 30 ◦ (5.194)
Note
β1
β2
≈ ω√µ0ǫ0
ω√µ0 · 2.1ǫ0
∼
= 0.690 (5.195)
so
ψt =
arcsin
(β1
β2
sin ψi
)
∼
= 20
 .2◦ (5.196)
Now substituting these values into
Equation 5.191, we obtain
Γ TM ∼= −0.1442 (5.197)
(Did you get an answer closer to −0.1323? If so,
you probably did not use sufficient precision to
represent intermediate results. This is a good
example of a problem in which three significant
figures for results that are used in subsequent
calculations is not sufficient.)
The fraction of power reflected relative to power
incident is now determined to be",Electromagnetics_Vol2.pdf
"80 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
c⃝ C. W ang CC BY -SA 4.0
Figure 5.13: A TM uniform plane wave incident from
air to glass.
|Γ TM|2 ∼= 0 .021; i.e., about 2.1% .
1 − |Γ T
M|2 ∼
= 97.9%
of the power is transmitted
into
the glass.
Note that the result obtained in the preceding example
is different from the result for a TE wave incident
from the same direction (Example 5.7). In other
words:
The fraction of power reflected and transmitted
from
the planar boundary between dissimilar me-
dia depends on the polarization of the incident
wave relative to the boundary , as well as the an-
gle of incidence.
5.8 Angles of Reflection and
Refraction
[m0168]
Consider the situation shown in Figure 5.14: A
uniform
plane wave obliquely incident on the planar
boundary between two semi-infinite material regions.
Let a point on the boundary be represented as the
position vector
r0 = ˆxx+ ˆyy (5.198)
In both Sections 5.6 (“Plane W aves at Oblique",Electromagnetics_Vol2.pdf
"Let a point on the boundary be represented as the
position vector
r0 = ˆxx+ ˆyy (5.198)
In both Sections 5.6 (“Plane W aves at Oblique
Incidence on a Planar Boundary: TE Case”) and 5.7
(“Plane W aves at Oblique Incidence on a Planar
Boundary: TM Case”), it is found that
ki · r0 = kr · r0 = kt · r0 (5.199)
In this expression,
ki = β1 ˆki (5.200)
kr = β1 ˆkr (5.201)
kt = β2 ˆkt (5.202)
where ˆki, ˆkr, and ˆkt are unit vectors in the direction
of incidence, reflection, and transmission,
c⃝ C. W ang CC BY -SA 4.0
Figure 5.14: A uniform plane wave obliquely inci-
dent on the planar boundary between two semi-infinite
material regions.",Electromagnetics_Vol2.pdf
"5.8. ANGLES OF REFLECTION AND REFRACTION 81
respectively; and β1 and β2 are the phase propagation
constants in Region 1 (from which the wave is
incident) and Region 2, respectively . Equation 5.199
is essentially a boundary condition that enforces
continuity of the phase of the electric and magnetic
fields across the boundary , and is sometimes referred
to as the “phase matching” requirement. Since the
same requirement emerges independently in the TE
and TM cases, and since any plane wave may be
decomposed into TE and TM components, the
requirement must apply to any incident plane wave
regardless of polarization.
Equation 5.199 is the key to finding the direction of
reflection ψr and direction of transmission ψt. First,
observe:
ˆki = ˆx sin ψi + ˆz cos ψi (5.203)
ˆkr = ˆx sin ψr − ˆz cos ψr (5.204)
ˆkt = ˆx sin ψt + ˆz cos ψt (5.205)
Therefore, we may express Equation 5.199 in the
following form
β1
(ˆx sin ψi + ˆz cos ψi)
· (ˆxx+ ˆyy)
= β1 (ˆx sin ψr − ˆz cos ψr) · (ˆxx+ ˆyy)
= β2",Electromagnetics_Vol2.pdf
"Therefore, we may express Equation 5.199 in the
following form
β1
(ˆx sin ψi + ˆz cos ψi)
· (ˆxx+ ˆyy)
= β1 (ˆx sin ψr − ˆz cos ψr) · (ˆxx+ ˆyy)
= β2
(ˆx sin ψt + ˆz cos ψt)
· (ˆxx+ ˆyy) (5.206)
which reduces to
β1 sin ψi = β1 sin ψr = β2 sin ψt (5.207)
Examining the first and second terms of
Equation 5.207, and noting that ψi and ψr are both
limited to the range −π/2 to +π/2, we find that:
ψr = ψi
(5.208)
In
plain English:
Angle of reflection equals angle of incidence.
Examining
the first and third terms of Equation 5.207,
we find that
β1 sin ψi = β2 sin ψt (5.209)
Since β1 = ω√µ1ǫ1 and β2 = ω√µ2ǫ2,
Equation
5.209 expressed explicitly in terms of the
constitutive parameters is:
√µ1ǫ1 sin ψi = √µ2ǫ2 sin ψt (5.210)
Thus,
we see that ψt does not depend on frequency ,
except to the extent that the constitutive parameters
might. W e may also express this relationship in terms
of the relative values of constitutive parameters; i.e.,
µ1 = µr1µ0, ǫ1 = ǫr1ǫ0, µ2 = µr2µ0, and",Electromagnetics_Vol2.pdf
"of the relative values of constitutive parameters; i.e.,
µ1 = µr1µ0, ǫ1 = ǫr1ǫ0, µ2 = µr2µ0, and
ǫ2 = ǫr2ǫ0. In terms of the relative parameters:
√
µr1ǫr1 sin ψi = √µr2ǫr2 sin ψt (5.211)
This
is known as Snell’s law or the law of refraction.
Refraction is simply transmission with the result that
the direction of propagation is changed.
Snell’s law (Equation 5.211) determines the an-
gle
of refraction (transmission).
The associated formula for ψt explicitly is:
ψt = arcsin
(√ µr1ǫr1
µr2ǫr2
sin ψi
)
(5.212)
F
or the common special case of non-magnetic media,
one assumes µr1 = µr2 = 1. In this case, Snell’s law
simplifies to:
√ǫr1 sin ψi = √ǫr2 sin ψt (5.213)
In
optics, it is common to express permittivities in
terms of indices of refraction; e.g., n1 ≜ √ǫr1 and
n2 ≜ √ǫr2. Thus, Snell’s law in optics is often
expressed as:
n1 sin ψi = n2 sin ψt (5.214)
When both media are non-magnetic, Equation 5.212
simplifies to
ψt = arcsin
(√
ǫr1
ǫr2
sin ψi
)
(5.215)
When ǫr2 >",Electromagnetics_Vol2.pdf
"n1 sin ψi = n2 sin ψt (5.214)
When both media are non-magnetic, Equation 5.212
simplifies to
ψt = arcsin
(√
ǫr1
ǫr2
sin ψi
)
(5.215)
When ǫr2 >
ǫr1, we observe that ψt <ψi. In other
words, the transmitted wave travels in a direction that
is closer to the surface normal than the angle of
incidence. This scenario is demonstrated in the
following example.
Example 5.9. Refraction of light at an air-glass
boundary .
Figure 5.15 shows a narrow beam of light that is
incident from air to glass at an angle ψi = 60 ◦.",Electromagnetics_Vol2.pdf
"82 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
c⃝ Z. S´andor CC BY -SA 3.0
Figure 5.15: Angles of reflection and refraction for a
light wave incident from air onto glass.
As expected, the angle of reflection ψr is
observ
ed to be equal to ψi. The angle of
refraction ψt is observed to be 35◦. What is the
relative permittivity of the glass?
Solution. Since the permittivity of glass is
greater than that of air, we observe the expected
result ψt <ψi. Since glass is non-magnetic, the
expected relationship between these angles is
given by Equation 5.213. Solving that equation
for the relative permittivity of the glass, we
obtain:
ǫr2 = ǫr1
(sin ψi
sin ψt
)2
∼= 2.28 (5.216)
When
a wave travels in the reverse direction – i.e.,
from a non-magnetic medium of higher permittivity
to a medium of lower permittivity – one finds
ψt >ψr. In other words, the refraction is away from
the surface normal. Figure 5.16 shows an example
from common experience.
In non-magnetic media, when ǫr1 < ǫr2, ψt <",Electromagnetics_Vol2.pdf
"the surface normal. Figure 5.16 shows an example
from common experience.
In non-magnetic media, when ǫr1 < ǫr2, ψt <
ψi (refraction
toward the surface normal). When
ǫr1 > ǫr2, ψt > ψi (refraction away from the
surface normal).
c⃝ G. Saini CC BY -SA 4.0
Figure 5.16: Refraction accounts for the apparent dis-
placement of an underwater object from the perspec-
tive of an observer above the water.
Under certain conditions, the ǫr2 <ǫr1 case leads to
the following surprising observation: When
calculating ψt using, for example, Equation 5.212,
one finds that
√
ǫr1/ǫr2 s in ψi can be greater than 1.
Since the sin function yields values between −1 and
+1, the result of the arcsin function is undefined.
This odd situation is addressed in Section 5.11. For
now , we will simply note that this condition leads to
the phenomenon of total internal reflection. For now ,
all we can say is that when ǫr2 <ǫr1, ψt is able to
reach π/2 radians, which corresponds to propagation",Electromagnetics_Vol2.pdf
"all we can say is that when ǫr2 <ǫr1, ψt is able to
reach π/2 radians, which corresponds to propagation
parallel to the boundary . Beyond that threshold, we
must account for the unique physical considerations
associated with total internal reflection.
W e conclude this section with a description of the
common waveguiding device known as the prism,
shown in Figure 5.17. This particular device uses
refraction to change the direction of light waves
(similar devices can be used to manipulate radio
waves as well). Many readers are familiar with the use
of prisms to separate white light into its constituent
colors (frequencies), as shown in Figure 5.18. The
separation of colors is due to frequency dependence
of the material comprising the prism. Specifically , the
permittivity of the material is a function of frequency ,
and therefore the angle of refraction is a function of
frequency . Thus, each frequency is refracted by a
different amount. Conversely , a prism comprised of a",Electromagnetics_Vol2.pdf
"frequency . Thus, each frequency is refracted by a
different amount. Conversely , a prism comprised of a
material whose permittivity exhibits negligible
variation with frequency will not separate incident",Electromagnetics_Vol2.pdf
"5.9. TE REFLECTION IN NON-MAGNETIC MEDIA 83
c⃝ D-K uru CC BY -SA 3.0
Figure 5.17: A typical triangular prism.
white light into its constituent colors since each color
will be refracted by the same amount.
Additional Reading:
• “Prism” on Wikipedia.
• “Refraction” on Wikipedia.
• “Refractive index” on Wikipedia.
• “Snell’s law” on Wikipedia.
• “T otal internal reflection” on Wikipedia.
c⃝ Su idroot CC BY -SA 4.0
Figure 5.18: A color-separating prism.
5.9 TE Reflection in
Non-magnetic Media
[m0171]
Figure 5.19 shows a TE uniform plane wave incident
on
the planar boundary between two semi-infinite
material regions. In this case, the reflection coefficient
is given by:
Γ TE = η2 cos ψi − η1 cos ψt
η2 cos ψi + η1 cos ψt (5.217)
where ψi and ψt are
the angles of incidence and
transmission (refraction), respectively; and η1 and η2
are the wave impedances in Regions 1 and 2,
respectively . Many materials of practical interest are
non-magnetic; that is, they have permeability that is",Electromagnetics_Vol2.pdf
"respectively . Many materials of practical interest are
non-magnetic; that is, they have permeability that is
not significantly different from the permeability of
free space. In this section, we consider the behavior
of the reflection coefficient for this class of materials.
T o begin, recall the general form of Snell’s law:
sin ψt = β1
β2
sin ψi (5.218)
In
non-magnetic media, the permeabilities µ1 and µ2
are assumed equal to µ0. Thus:
β1
β2
= ω√µ1ǫ1
ω√µ2ǫ2
=
√ ǫ1
ǫ2
(5.219)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.19: A TE uniform plane wave obliquely inci-
dent on the planar boundary between two semi-infinite
regions of lossless non-magnetic material.",Electromagnetics_Vol2.pdf
"84 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
Since permittivity ǫcan be expressed as ǫ0 times the
relative permittivity ǫr, we may reduce further to:
β1
β2
=
√ ǫr1
ǫr2
(5.220)
No
w Equation 5.218 reduces to:
sin ψt =
√ ǫr1
ǫr2
sin ψi (5.221)
Ne
xt, note that for any value ψ, one may write cosine
in terms of sine as follows:
cos ψ=
√
1 − sin2 ψ (5.222)
Therefore,
cos ψt =
√
1 − ǫr1
ǫr2
sin2 ψi (5.223)
Also
we note that in non-magnetic media
η1 =
√ µ1
ǫ1
=
√ µ0
ǫr1ǫ0
= η0
√ǫr1
(5.224)
η2 =
√ µ2
ǫ2
=
√ µ0
ǫr2ǫ0
= η0
√ǫr2
(5.225)
where η0 is
the wave impedance in free space.
Making substitutions into Equation 5.217, we obtain:
Γ TE =
(
η0/√ǫr2
)
cos ψi −
(
η0/√ǫr1
)
cos ψt
(
η0/√ǫr2
)
cos ψi +
(
η0/√ǫr1
)
cos ψt
(5.226)
Multiplying
numerator and denominator by √ǫr2/η0,
we
obtain:
Γ TE = cos ψi −
√
ǫr2/ǫr1 cos ψt
cos ψi +
√
ǫr2/ǫr1 cos ψt (5.227)
Finally
, by substituting Equation 5.223, we obtain:
Γ TE =
cos ψi −
√
ǫr2/ǫr1 − s in 2 ψi
cos ψi +
√
ǫr2/ǫr1 − s in 2 ψi
(5.228)
This",Electromagnetics_Vol2.pdf
"Finally
, by substituting Equation 5.223, we obtain:
Γ TE =
cos ψi −
√
ǫr2/ǫr1 − s in 2 ψi
cos ψi +
√
ǫr2/ǫr1 − s in 2 ψi
(5.228)
This
expression has the advantage that it is now
entirely in terms of ψi, with no need to first calculate
ψt.
Using Equation 5.228, we can see how different
combinations of material affect the reflection
-1
-0.5
 0
 0.5
 1
 0  10  20  30  40  50  60  70  80  90
2
10
100
Reflection Coefficient
angle of incidence [deg]
Figure 5.20: The reflection coefficient Γ TE as a func-
tion of angle of incidence ψi for various media combi-
nations. Curves are labeled with the ratio ǫr2/ǫr1.
coefficient. First, we note that when ǫr1 = ǫr2 (i.e.,
same media on both sides of the boundary), Γ TE = 0
as expected. When ǫr1 >ǫr2 (e.g., wave traveling in
glass toward air), we see that it is possible for
ǫr2/ǫr1 − sin2 ψi to be negative, which makes Γ TE
complex-valued. This results in total internal
reflection, and is addressed elsewhere in another",Electromagnetics_Vol2.pdf
"ǫr2/ǫr1 − sin2 ψi to be negative, which makes Γ TE
complex-valued. This results in total internal
reflection, and is addressed elsewhere in another
section. When ǫr1 <ǫr2 (e.g., wave traveling in air
toward glass), we see that ǫr2/ǫr1 − sin2 ψi is always
positive, so Γ TE is always real-valued.
Let us continue with the ǫr1 <ǫr2 condition.
Figure 5.20 shows Γ TE plotted for various
combinations of media over all possible angles of
incidence from 0 (normal incidence) to π/2 (grazing
incidence). W e observe:
In non-magnetic media with ǫr1 < ǫr2, Γ T E is
real-valued, negative, and decreases to −1 as ψi
approaches grazing incidence.
Also note that at any particular angle of incidence,
Γ TE trends toward −1 as ǫr2/ǫr1 increases. Thus, we
observe that as ǫr2/ǫr1 → ∞, the result is
increasingly similar to the result we would obtain for
a perfect conductor in Region 2.",Electromagnetics_Vol2.pdf
"5.10. TM REFLECTION IN NON-MAGNETIC MEDIA 85
5.10 TM Reflection in
Non-magnetic Media
[m0172]
Figure 5.21 shows a TM uniform plane wave incident
on
the planar boundary between two semi-infinite
material regions. In this case, the reflection coefficient
is given by:
Γ TM = −η1 cos ψi + η2 cos ψt
+η1 cos ψi + η2 cos ψt (5.229)
where ψi and ψt are
the angles of incidence and
transmission (refraction), respectively; and η1 and η2
are the wave impedances in Regions 1 and 2,
respectively . Many materials of practical interest are
non-magnetic; that is, they have permeability that is
not significantly different from the permeability of
free space. In this section, we consider the behavior
of the reflection coefficient for this class of materials.
T o begin, recall the general form of Snell’s law:
sin ψt = β1
β2
sin ψi (5.230)
In
non-magnetic media, the permeabilities µ1 and µ2
are assumed equal to µ0. Thus:
β1
β2
= ω√µ1ǫ1
ω√µ2ǫ2
=
√ ǫ1
ǫ2
(5.231)
c⃝ C. W ang CC BY -SA 4.0",Electromagnetics_Vol2.pdf
"In
non-magnetic media, the permeabilities µ1 and µ2
are assumed equal to µ0. Thus:
β1
β2
= ω√µ1ǫ1
ω√µ2ǫ2
=
√ ǫ1
ǫ2
(5.231)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.21: A transverse magnetic uniform plane
wave obliquely incident on the planar boundary be-
tween two semi-infinite material regions.
Since permittivity ǫcan be expressed as ǫ0 times the
relative permittivity ǫr, we may reduce further to:
β1
β2
=
√ ǫr1
ǫr2
(5.232)
No
w Equation 5.230 reduces to:
sin ψt =
√ ǫr1
ǫr2
sin ψi (5.233)
Ne
xt, note that for any value ψ, one may write cosine
in terms of sine as follows:
cos ψ=
√
1 − sin2 ψ (5.234)
Therefore,
cos ψt =
√
1 − ǫr1
ǫr2
sin2 ψi (5.235)
Also
we note that in non-magnetic media
η1 =
√ µ1
ǫ1
=
√ µ0
ǫr1ǫ0
= η0
√ǫr1
(5.236)
η2 =
√ µ2
ǫ2
=
√ µ0
ǫr2ǫ0
= η0
√ǫr2
(5.237)
where η0 is
the wave impedance in free space.
Making substitutions into Equation 5.229, we obtain:
Γ TM = −
(
η0/√ǫr1
)
cos ψi +
(
η0/√ǫr2
)
cos ψt
+
(
η0/√ǫr1
)
cos ψi +
(
η0/√ǫr2
)
cos ψt
(5.238)
Multiplying",Electromagnetics_Vol2.pdf
"Γ TM = −
(
η0/√ǫr1
)
cos ψi +
(
η0/√ǫr2
)
cos ψt
+
(
η0/√ǫr1
)
cos ψi +
(
η0/√ǫr2
)
cos ψt
(5.238)
Multiplying
numerator and denominator by √ǫr2/η0,
we
obtain:
Γ TM = −
√
ǫr2/ǫr1 cos ψi + cos ψt
+
√
ǫr2/ǫr1 cos ψi + cos ψt (5.239)
Substituting Equation 5.235, we obtain:
Γ TM =
−
√
ǫr2/ǫr1 cos ψi +
√
1 − (ǫr1/ǫr2) sin2 ψi
+
√
ǫr2/ǫr1 cos ψi +
√
1 − (ǫr1/ǫr2) sin2 ψi
(5.240)
This expression has the advantage that it is now
entirely in terms of ψi, with no need to first calculate
ψt.
Finally , multiplying numerator and denominator by√
ǫr2/ǫr1, we obtain:
Γ TM =
− (ǫr2/ǫr1) cos ψi +
√
ǫr2/ǫr1 − s in 2 ψi
+ (ǫr2/ǫr1) cos ψi +
√
ǫr2/ǫr1 − s in 2 ψi
(5.241)",Electromagnetics_Vol2.pdf
"86 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
-1
-0.5
 0
 0.5
 1
 0  10  20  30  40  50  60  70  80  90
2
10
100
Reflection Coefficient
angle of incidence [deg]
Figure 5.22: The reflection coefficient Γ TM as a func-
tion of angle of incidence ψi for various media combi-
nations, parameterized as ǫr2/ǫr1.
Using Equation 5.241, we can easily see how
different combinations of material affect the reflection
coefficient. First, we note that when ǫr1 = ǫr2 (i.e.,
same media on both sides of the boundary), Γ TM = 0
as expected. When ǫr1 >ǫr2 (e.g., wave traveling in
glass toward air), we see that it is possible for
ǫr2/ǫr1 − sin2 ψi to be negative, which makes Γ TM
complex-valued. This results in total internal
reflection, and is addressed in another section. When
ǫr1 <ǫr2 (e.g., wave traveling in air toward glass),
we see that ǫr2/ǫr1 − sin2 ψi is always positive, so
Γ TM is always real-valued.
Let us continue with the ǫr1 <ǫr2 condition.
Figure 5.22 shows Γ TM plotted for various",Electromagnetics_Vol2.pdf
"Γ TM is always real-valued.
Let us continue with the ǫr1 <ǫr2 condition.
Figure 5.22 shows Γ TM plotted for various
combinations of media over all possible angles of
incidence from 0 (normal incidence) to π/2 (grazing
incidence). W e observe:
In non-magnetic media with ǫr1 < ǫr2, Γ T M is
real-valued and increases from a negative value
for normal incidence to +1 as ψi approaches
grazing incidence.
Note that at any particular angle of incidence, Γ TM
trends toward −1 as ǫr2/ǫr1 → ∞. In this respect,
the behavior of the TM component is similar to that
of the TE component. In other words: As
c⃝ C. W ang CC BY -SA 4.0
Figure 5.23: Reflection of a plane wave with angle of
incidence ψi equal to the polarizing angle ψi
B. Here
the media are non-magnetic with ǫr1 <ǫr2.
ǫr2/ǫr1 → ∞, the result for both the TE and TM
components are increasingly similar to the result we
would obtain for a perfect conductor in Region 2.
Also note that when ǫr1 <ǫr2, Γ TM changes sign",Electromagnetics_Vol2.pdf
"would obtain for a perfect conductor in Region 2.
Also note that when ǫr1 <ǫr2, Γ TM changes sign
from negative to positive as angle of incidence
increases from 0 to π/2. This behavior is quite
different from that of the TE component, which is
always negative for ǫr1 <ǫr2. The angle of incidence
at which Γ TM = 0 is referred to as Brewster’s angle,
which we assign the symbol ψi
B. Thus:
ψi
B ≜ ψi at which Γ TM = 0
(5.242)
In
the discussion that follows, here is the key point to
keep in mind:
Brewster’s angle ψi
B is the angle of incidence at
which Γ TM = 0.
Brewster’s angle is also referred to as the polarizing
angle. The motivation for the term “polarizing angle”
is demonstrated in Figure 5.23. In this figure, a plane
wave is incident with ψi = ψi
B. The wave may
contain TE and TM components in any combination.
Applying the principle of superposition, we may
consider these components separately . The TE
component of the incident wave will scatter as",Electromagnetics_Vol2.pdf
"Applying the principle of superposition, we may
consider these components separately . The TE
component of the incident wave will scatter as
reflected and transmitted waves which are also TE.
However, Γ TM = 0 when ψi = ψi
B, so the TM",Electromagnetics_Vol2.pdf
"5.10. TM REFLECTION IN NON-MAGNETIC MEDIA 87
component of the transmitted wave will be TM, but
the TM component of the reflected wave will be zero.
Thus, the total (TE+ TM) reflected wave will be
purely TE, regardless of the TM component of the
incident wave. This principle can be exploited to
suppress the TM component of a wave having both
TE and TM components. This method can be used to
isolate the TE and TM components of a wave.
Derivation of a formula for Brewster’s angle.
Brewster’s angle for any particular combination of
non-magnetic media may be determined as follows.
Γ TM = 0 when the numerator of Equation 5.241
equals zero, so:
− Rcos ψi
B +
√
R− sin2 ψi
B = 0 (5.243)
where we have made the substitution R≜ ǫr2/ǫr1 to
improve clarity . Moving the second term to the right
side of the equation and squaring both sides, we
obtain:
R2 cos2 ψi
B = R− sin2 ψi
B (5.244)
Now employing a trigonometric identity on the left
side of the equation, we obtain:
R2 (
1 − sin2 ψi
B
)
= R− sin2 ψi",Electromagnetics_Vol2.pdf
"B = R− sin2 ψi
B (5.244)
Now employing a trigonometric identity on the left
side of the equation, we obtain:
R2 (
1 − sin2 ψi
B
)
= R− sin2 ψi
B (5.245)
R2 − R2 sin2 ψi
B = R− sin2 ψi
B (5.246)
(
1 − R2)
sin2 ψi
B = R− R2 (5.247)
and finally
sin ψi
B =
√
R− R2
1 − R2 (5.248)
Although
this equation gets the job done, it is possible
to simplify further. Note that Equation 5.248 can be
interpreted as a description of the right triangle shown
in Figure 5.24. In the figure, we have identified the
length hof the vertical side and the length dof the
hypotenuse as:
h≜
√
R− R2 (5.249)
d≜
√
1 − R2 (5.250)
The
length bof the horizontal side is therefore
b=
√
d2 − h2 =
√
1 − R (5.251)
Subsequently
, we observe
tan ψi
B = h
b =
√
R− R2
√
1 − R =
√
R (5.252)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.24: Derivation of a simple formula for Brew-
ster’s angle for non-magnetic media.
Thus, we have found
tan ψi
B =
√ ǫr2
ǫr1
(5.253)
Example 5.10. Polarizing angle for an
air-to-glass interface.",Electromagnetics_Vol2.pdf
"ster’s angle for non-magnetic media.
Thus, we have found
tan ψi
B =
√ ǫr2
ǫr1
(5.253)
Example 5.10. Polarizing angle for an
air-to-glass interface.
A plane wave is incident from air onto the planar
boundary with a glass region. The glass exhibits
relative permittivity of 2.1. The incident wave
contains both TE and TM components. At what
angle of incidence ψi will the reflected wave be
purely TE?
Solution. Using Equation 5.253:
tan ψi
B =
√
ǫr2
ǫr1
=
√
2.1
1
∼= 1.449 (5.254)
Therefore,
Brewster’s angle is ψi
B ∼
= 55.4◦
. This
is
the angle ψi at which Γ TM = 0. Therefore,
when ψi = ψi
B, the reflected wave contains no
TM component and therefore must be purely TE.
Additional Reading:
• “Brewster’s angle” on Wikipedia.",Electromagnetics_Vol2.pdf
"88 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.11 T otal Internal Reflection
[m0169]
T otal internal reflection refers to a particular
condition resulting in the complete reflection of a
wave at the boundary between two media, with no
power transmitted into the second region. One way to
achieve complete reflection with zero transmission is
simply to require the second material to be a perfect
conductor. However, total internal reflection is a
distinct phenomenon in which neither of the two
media are perfect conductors. T otal internal reflection
has a number of practical applications; notably , it is
the enabling principle of fiber optics.
Consider the situation shown in Figure 5.25: A
uniform plane wave is obliquely incident on the
planar boundary between two semi-infinite material
regions. In Section 5.8, it is found that
ψr = ψi (5.255)
i.e., angle of reflection equals angle of incidence.
Also, from Snell’s law:
√
µr1ǫr1 sin ψi = √µr2ǫr2 sin ψt (5.256)
where",Electromagnetics_Vol2.pdf
"ψr = ψi (5.255)
i.e., angle of reflection equals angle of incidence.
Also, from Snell’s law:
√
µr1ǫr1 sin ψi = √µr2ǫr2 sin ψt (5.256)
where
“r” in the subscripts indicates the relative
(unitless) quantities. The associated formula for ψt
c⃝ C. W ang CC BY -SA 4.0 (modified)
Figure 5.25: A uniform plane wave obliquely inci-
dent on the planar boundary between two semi-infinite
material regions. Here µr1ǫr1 >µr2ǫr2, so ψt >ψi.
explicitly is:
ψt = arcsin
(√ µr1ǫr1
µr2ǫr2
sin ψi
)
(5.257)
From
Equation 5.257, it is apparent that when
µr1ǫr1 >µr2ǫr2, ψt >ψr; i.e., the transmitted wave
appears to bend away from the surface normal, as
shown in Figure 5.25. In fact, ψt can be as large as
π/2 (corresponding to propagation parallel to the
boundary) for angles of incidence which are less than
π/2. What happens if the angle of incidence is further
increased? When calculating ψt using
Equation 5.257, one finds that the argument of the
arcsine function becomes greater than 1. Since the",Electromagnetics_Vol2.pdf
"increased? When calculating ψt using
Equation 5.257, one finds that the argument of the
arcsine function becomes greater than 1. Since the
possible values of the sine function are between −1
and +1, the arcsine function is undefined. Clearly our
analysis is inadequate in this situation.
T o make sense of this, let us begin by identifying the
threshold angle of incidence ψi
c at which the trouble
begins. From the analysis in the previous paragraph,
√
µr1ǫr1
µr2ǫr2
sin ψi
c = 1 (5.258)
therefore,
ψi
c = arcsin
√ µr2ǫr2
µr1ǫr1
(5.259)
This
is known as the critical angle. When ψi <ψi
c,
our existing theory applies. When ψi ≥ ψi
c, the
situation is, at present, unclear. For non-magnetic
materials, Equation 5.259 simplifies to
ψi
c = arcsin
√ ǫr2
ǫr1
(5.260)
No
w let us examine the behavior of the reflection
coefficient. For example, the reflection coefficient for
TE component and non-magnetic materials is
Γ TE =
cos ψi −
√
ǫr2/ǫr1 − s in 2 ψi
cos ψi +
√
ǫr2/ǫr1 − s in 2 ψi
(5.261)",Electromagnetics_Vol2.pdf
"TE component and non-magnetic materials is
Γ TE =
cos ψi −
√
ǫr2/ǫr1 − s in 2 ψi
cos ψi +
√
ǫr2/ǫr1 − s in 2 ψi
(5.261)
From Equation 5.260, we see that
sin2 ψi
c = ǫr2
ǫr1
(5.262)",Electromagnetics_Vol2.pdf
"5.11. TOT AL INTERNAL REFLECTION 89
So, when ψi >ψi
c, we see that
ǫr2/ǫr1 − sin2 ψi <0 (ψ i >ψi
c) (5.263)
and therefore√
ǫr2/ǫr1 − s in 2 ψi = jB (ψi >ψi
c) (5.264)
where Bis a positive real-valued number. Now we
may write Equation 5.261 as follows:
Γ TE = A− jB
A+ jB ( ψi >ψi
c) (5.265)
where A≜ cos ψi is also a positive real-valued
number. Note that the numerator and denominator in
Equation 5.265 have equal magnitude. Therefore, the
magnitude of |Γ TE| = 1 and
Γ TE = ejζ (ψi >ψi
c) (5.266)
where ζis a real-valued number indicating the phase
of Γ TE. In other words,
When the angle of incidence ψi exceeds the criti-
cal angle ψi
c, the magnitude of the reflection coef-
ficient is 1. In this case, all power is reflected, and
no power is transmitted into the second medium.
This is total internal reflection.
Although we have obtained this result for TE
component, the identical conclusion is obtained for
TM component as well. This is left as an exercise for
the student.",Electromagnetics_Vol2.pdf
"component, the identical conclusion is obtained for
TM component as well. This is left as an exercise for
the student.
Example 5.11. T otal internal reflection in glass.
Figure 5.15 (Section 5.8) shows a demonstration
of refraction of a beam of light incident from air
onto a planar boundary with glass. Analysis of
that demonstration revealed that the relative
permittivity of the glass was ∼= 2.28. Figure 5.26
shows a modification of the demonstration: In
this case, a beam of light incident from glass
onto a planar boundary with air. Confirm that
total internal reflection is the expected result in
this demonstration.
Solution. Assuming the glass is non-magnetic,
the critical angle is given by Equation 5.260. In
the present example, ǫr1 ∼= 2.28 (glass), ǫr2 ∼
= 1
(air). Therefore, the critical angle ψi
c ∼= 41.5◦.
c⃝ Z. S´andor CC BY -SA 3.0
Figure 5.26: T otal internal reflection of a light wave
incident on a planar boundary between glass and air.
c⃝ Ti mwether CC BY -SA 3.0",Electromagnetics_Vol2.pdf
"Figure 5.26: T otal internal reflection of a light wave
incident on a planar boundary between glass and air.
c⃝ Ti mwether CC BY -SA 3.0
Figure 5.27: Laser light in a dielectric rod exhibiting
the Goos-H ¨anchen effect.
The angle of incidence in Figure 5.26 is seen to
be
about ∼= 50◦, which is greater than the critical
angle. Therefore, total internal reflection is
expected.
Note also that the phase ζof the reflection coefficient
is precisely zero unless ψi >ψi
c, at which point it is
both non-zero and varies with ψi. This is known as
the Goos-H ¨anchen effect. This leads to the startling
phenomenon shown in Figure 5.27. In this figure, the
Goos-H¨anchen phase shift is apparent as a
displacement between the point of incidence and the
point of reflection.",Electromagnetics_Vol2.pdf
"90 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
The presence of an imaginary component in the
reflection coefficient is odd for two reasons. First, we
are not accustomed to seeing a complex-valued
reflection coefficient emerge when the wave
impedances of the associated media are real-valued.
Second, the total reflection of the incident wave seems
to contradict the boundary condition that requires the
tangential components of the electric and magnetic
fields to be continuous across the boundary . That is,
how can these components of the fields be continuous
across the boundary if no power is transmitted across
the boundary? These considerations suggest that there
is a field on the opposite side of the boundary , but –
somehow – it must have zero power. This is exactly
the case, as is explained in Section 5.12.
Additional Reading:
• “Goos-H ¨anchen effect” on Wikipedia.
• “Snell’s law” on Wikipedia.
• “T otal internal reflection” on Wikipedia.
5.12 Evanescent W aves
[m0170]",Electromagnetics_Vol2.pdf
"• “Goos-H ¨anchen effect” on Wikipedia.
• “Snell’s law” on Wikipedia.
• “T otal internal reflection” on Wikipedia.
5.12 Evanescent W aves
[m0170]
Consider the situation shown in Figure 5.28: A
uniform
plane wave obliquely incident on the planar
boundary between two semi-infinite material regions,
and total internal reflection occurs because the angle
of incidence ψi is greater than the critical angle
ψi
c = arcsin
√
µr2ǫr2
µr1ǫr1
(5.267)
Therefore,
the reflection coefficient is
complex-valued with magnitude equal to 1 and phase
that depends on polarization and the constitutive
parameters of the media.
The total reflection of the incident wave seems to
contradict the boundary conditions that require the
tangential components of the electric and magnetic
fields to be continuous across the boundary . How can
these components of the fields be continuous across
the boundary if no power is transmitted across the
boundary? There must be a field on the opposite side",Electromagnetics_Vol2.pdf
"the boundary if no power is transmitted across the
boundary? There must be a field on the opposite side
of the boundary , but – somehow – it must have zero
power. T o make sense of this, let us attempt to find a
solution for the transmitted field.
c⃝ C. W ang CC BY -SA 4.0
Figure 5.28: A uniform plane wave obliquely inci-
dent on the planar boundary between two semi-infinite
material regions. Here µr1ǫr1 >µr2ǫr2 and ψi >ψi
c.",Electromagnetics_Vol2.pdf
"5.12. EV ANESCENT W A VES 91
W e begin by postulating a complex-valued angle of
transmission ψtc. Although the concept of a
complex-valued angle may seem counterintuitive,
there is mathematical support for this concept. For
example, consider the well-known trigonometric
identities:
sin θ= 1
j2
(
ejθ − e−j
θ)
(5.268)
cos θ= 1
2
(
ejθ + e−j
θ)
(5.269)
These identities allow us to compute values for sine
and cosine even when θis complex-valued. One may
conclude that the sine and cosine of a complex-valued
angle exist, although the results may also be
complex-valued.
Based on the evidence established so far, we presume
that ψtc behaves as follows:
ψtc ≜ ψt , ψi <ψi
c (5.270)
ψtc ≜ π/2 , ψi = ψi
c (5.271)
ψtc ≜ π/2 + jψ′′ , ψi >ψi
c (5.272)
In other words, ψtc is identical to ψt for ψi ≤ ψi
c, but
when total internal reflection occurs, we presume the
real part of ψtc remains fixed (parallel to the
boundary) and that an imaginary component jψ′′
emerges to satisfy the boundary conditions.",Electromagnetics_Vol2.pdf
"real part of ψtc remains fixed (parallel to the
boundary) and that an imaginary component jψ′′
emerges to satisfy the boundary conditions.
For clarity , let us assign the variable ψ′ to represent
the real part of ψtc in Equations 5.270–5.272. Then
we may refer to all three cases using a single
expression as follows:
ψtc = ψ′ + jψ′′ (5.273)
Now we use a well-known trigonometric identity as
follows:
sin ψtc = sin ( ψ′ + jψ′′) (5.274)
= sin ψ′ cos jψ′′ + cos ψ′ sin jψ′′ (5.275)
Using Equation 5.269, we find:
cos jψ′′ = 1
2
(
ej(jψ′
′) + e−j(jψ′′))
(5.276)
= 1
2
(
e−ψ′′
+ e+ψ′
′ )
(5.277)
= cosh ψ′′ (5.278)
In other words, the cosine of jψ′′ is simply the
hyperbolic cosine (“cosh ”) of ψ′′. Interestingly , cosh
of a real-valued argument is real-valued, so cos jψ′′ is
real-valued.
Using Equation 5.268, we find:
sin jψ′′ = 1
j2
(
ej(jψ′
′) − e−j(jψ′′))
(5.279)
= 1
j2
(
e−ψ′′
− e+ψ′
′ )
(5.280)
= j1
2
(
e+ψ′′
− e−ψ′
′ )
(5.281)
= jsinh ψ′′ (5.282)",Electromagnetics_Vol2.pdf
"sin jψ′′ = 1
j2
(
ej(jψ′
′) − e−j(jψ′′))
(5.279)
= 1
j2
(
e−ψ′′
− e+ψ′
′ )
(5.280)
= j1
2
(
e+ψ′′
− e−ψ′
′ )
(5.281)
= jsinh ψ′′ (5.282)
In other words, the sine of jψ′′ is jtimes hyperbolic
sine (“sinh ”) of ψ′′. Now note that sinh of a
real-valued argument is real-valued, so sin jψ′′ is
imaginary-valued.
Using these results, we find Equation 5.275 may be
written as follows:
sin ψtc = sin ψ′ cosh ψ′′ + jcos ψ′ sinh ψ′′ (5.283)
Using precisely the same approach, we find:
cos ψtc = cos ψ′ cosh ψ′′ − jsin ψ′ sinh ψ′′ (5.284)
Before proceeding, let’s make sure Equations 5.283
and 5.284 exhibit the expected behavior before the
onset of total internal reflection. For ψi <ψi
c,
ψ′ = ψt and ψ′′ = 0. In this case, sinh ψ′′ = 0,
cosh ψ′′ = 1, and Equations 5.283 and 5.284 yield
sin ψtc = sin ψt (5.285)
cos ψtc = cos ψt (5.286)
as expected.
When total internal reflection is in effect, ψi >ψi
c, so
ψ′ = π/2. In this case, Equations 5.283 and 5.284
yield
sin ψtc = cosh ψ′′ (5.287)",Electromagnetics_Vol2.pdf
"When total internal reflection is in effect, ψi >ψi
c, so
ψ′ = π/2. In this case, Equations 5.283 and 5.284
yield
sin ψtc = cosh ψ′′ (5.287)
cos ψtc = −jsinh ψ′′ (5.288)
Let us now consider what this means for the field in
Region 2. According to the formalism adopted in
previous sections, the propagation of wave",Electromagnetics_Vol2.pdf
"92 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
components in this region is described by the factor
e−jkt·r where
kt = β2 ˆkt
= β2
(ˆx sin ψtc + ˆz cos ψtc)
(5.289)
and
r = ˆxx+ ˆyy+ ˆzz (5.290)
so
kt · r =
(
β2 sin ψtc)
x+
(
β2 cos ψtc)
z
= ( β2 cosh ψ′′) x+ (−jβ2 sinh ψ′′) z (5.291)
Therefore, the wave in Region 2 propagates according
to
e−jkt·r = e−j(β2 cosh ψ′′)xe−(β2 sinh ψ′′)z (5.292)
Note that the constants β2 cosh ψ′′ and β2 sinh ψ′′ are
both real-valued and positive. Therefore,
Equation 5.292 describes a wave which propagates in
the +ˆx direction, but which is not uniform.
Specifically , the magnitude of the transmitted field
decreases exponentially with increasing z; i.e.,
maximum at the boundary , asymptotically
approaching zero with increasing distance from the
boundary . This wave is unlike the incident or
reflected waves (both uniform plane waves), and is
unlike the transmitted wave in the ψi <ψi
c case (also
a uniform plane wave). The transmitted wave that we",Electromagnetics_Vol2.pdf
"unlike the transmitted wave in the ψi <ψi
c case (also
a uniform plane wave). The transmitted wave that we
have derived in the ψi >ψi
c case gives the impression
of being somehow attached to the boundary , and so
may be described as a surface wave. However, in this
case we have a particular kind of surface wave,
known as an evanescent wave. Summarizing:
When total internal reflection occurs, the trans-
mitted
field is an evanescent wave; i.e., a surface
wave which conveys no power and whose mag-
nitude decays exponentially with increasing dis-
tance into Region 2.
At this point, we could enforce the “phase matching”
condition at the boundary , which would lead us to a
new version of Snell’s law that would allow us to
solve for ψ′′ in terms of ψi and the constitutive
properties of the media comprising Regions 1 and 2.
W e could subsequently determine values for the phase
propagation and attenuation constants for the
evanescent wave. It suffices to say that the magnitude",Electromagnetics_Vol2.pdf
"propagation and attenuation constants for the
evanescent wave. It suffices to say that the magnitude
of the evanescent field becomes negligible beyond a
few wavelengths of the boundary .
Finally , we return to the strangest characteristic of
this field: It acts like a wave, but conveys no power.
This field exists solely to enforce the electromagnetic
boundary conditions at the boundary , and does not
exist independently of the incident and reflected field.
The following thought experiment may provide some
additional insight.
In this experiment, a laser illuminates a planar
boundary between two material regions, with
conditions such that total internal reflection occurs.
Thus, all incident power is reflected, and an
evanescent wave exists on the opposite side of the
boundary . Next, the laser is turned off. All the light
incident on the boundary reflects from the boundary
and continues to propagate to infinity , even after light
is no longer incident on the boundary . In contrast, the",Electromagnetics_Vol2.pdf
"and continues to propagate to infinity , even after light
is no longer incident on the boundary . In contrast, the
evanescent wave vanishes at the moment laser light
ceases to illuminate the boundary . In other words, the
evanescent field does not continue to propagate along
the boundary to infinity . The reason for this is simply
that there is no power – hence, no energy – in the
evanescent wave.
Additional Reading:
• “Evanescent field” on Wikipedia.
• “Hyperbolic function” on Wikipedia.
• “T otal internal reflection” on Wikipedia.
[m0185]",Electromagnetics_Vol2.pdf
"5.12. EV ANESCENT W A VES 93
Image Credits
Fig. 5.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary .svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.2: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Upw incident on a slab.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.3: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Double-boundary problem equivalent.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.4: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Plane wave in basic coord.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.5: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Plane wave in rotated coord.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.6: c⃝ Sevenchw (C. W ang),",Electromagnetics_Vol2.pdf
"CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.6: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Plane wave in another rotation coord.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.7: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Plane wave in ray-fixed coord.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.8: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Coord system for TE-TM decomposition.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.9: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TE-polarized upw incident on planar boundary .svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.10: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Incident reflected and transmitted magnetic field components.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.11: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TE-polarized upw incident from air to glass.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.12: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM-polarized upw incident on planar boundary .svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.13: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM-polarized upw incident from air to glass.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.14: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Upw incident obliquely on planar boundary .svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"94 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
Fig. 5.15: c⃝ Z. S ´andor, https://commons.wikimedia.org/wiki/File:F%C3%A9nyt%C3%B6r%C3%A9s.jpg,
CC BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Fig. 5.16: c⃝ G. Saini, https://kids.kiddle.co/Image:Refractionn.jpg,
CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.17: c⃝ D-Kuru, https://commons.wikimedia.org/wiki/File:Prism-side-fs PNr%C2%B00117.jpg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Fig. 5.18: c⃝ Suidroot, https://commons.wikimedia.org/wiki/File:Prism-rainbow .svg,
CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.19: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TE-polarized upw incident on planar boundary .svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.21: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM-polarized upw incident on planar boundary .svg,
CC",Electromagnetics_Vol2.pdf
"Fig. 5.21: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM-polarized upw incident on planar boundary .svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.23: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Reflection of plane wave incidence angle equals polarizing angle.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.24: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Brewster%E2%80%99s angle for non-magnetic media.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 5.25: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary angle reflection equals incidence.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified.
Fig. 5.26: c⃝ Z. S ´andor,
https://en.wikipedia.org/wiki/File:T eljes f%C3%A9nyvisszaver%C5%91d%C3%A9s.jpg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).",Electromagnetics_Vol2.pdf
"https://en.wikipedia.org/wiki/File:T eljes f%C3%A9nyvisszaver%C5%91d%C3%A9s.jpg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Fig. 5.27: c⃝ Timwether, https://en.wikipedia.org/wiki/File:Laser in fibre.jpg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Fig. 5.28: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary total internal reflection.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 6
W a
veguides
6.1 Phase and Group V elocity
[m0176]
Phase velocity is the speed at which a point of
constant
phase travels as the wave propagates. 1 For a
sinusoidally-varying wave, this speed is easy to
quantify . T o see this, consider the wave:
Acos (ωt− βz+ ψ) (6.1)
where ω= 2πf is angular frequency , zis position,
and βis the phase propagation constant. At any given
time, the distance between points of constant phase is
one wavelength λ. Therefore, the phase velocity vp is
vp = λf (6.2)
Since β = 2π/λ, this may also be written as follows:
vp = ω
β (6.3)
Noting
that β = ω√µǫfor simple matter, we may
also
express vp in terms of the constitutive parameters
µand ǫas follows:
vp = 1√µǫ (6.4)
Since vp in
this case depends only on the constitutive
properties µand ǫ, it is reasonable to view phase
velocity also as a property of matter.
1 Formally , “velocity” is a vector which indicates both the di-
rection and rate of motion. It is common practice to use the terms",Electromagnetics_Vol2.pdf
"1 Formally , “velocity” is a vector which indicates both the di-
rection and rate of motion. It is common practice to use the terms
“phase velocity” and “group velocity” even though we are actually
referring merely to rate of motion. The direction is, of course, in the
direction of propagation.
Central to the concept of phase velocity is uniformity
over space and time. Equations 6.2–6.4 presume a
wave having the form of Equation 6.1, which exhibits
precisely the same behavior over all possible time t
from −∞ to +∞ and over all possible zfrom −∞ to
+∞. This uniformity over all space and time
precludes the use of such a wave to send information.
T o send information, the source of the wave needs to
vary at least one parameter as a function of time; for
example A(resulting in amplitude modulation), ω
(resulting in frequency modulation), or ψ(resulting in
phase modulation). In other words, information can
be transmitted only by making the wave non-uniform",Electromagnetics_Vol2.pdf
"phase modulation). In other words, information can
be transmitted only by making the wave non-uniform
in some respect. Furthermore, some materials and
structures can cause changes in ψor other
combinations of parameters which vary with position
or time. Examples include dispersion and propagation
within waveguides. Regardless of the cause, varying
the parameters ωor ψas a function of time means
that the instantaneous distance between points of
constant phase may be very different from λ. Thus,
the instantaneous frequency of variation as a function
of time and position may be very different from f. In
this case Equations 6.2–6.4 may not necessarily
provide a meaningful value for the speed of
propagation.
Some other concept is required to describe the speed
of propagation of such waves. That concept is group
velocity, vg, defined as follows:
Group velocity, vg, is the ratio of the apparent
change in frequency ωto the associated change in
the phase propagation constant β; i.e., ∆ω/∆β .",Electromagnetics_Vol2.pdf
"Group velocity, vg, is the ratio of the apparent
change in frequency ωto the associated change in
the phase propagation constant β; i.e., ∆ω/∆β .
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"96 CHAPTER 6. W A VEGUIDES
Letting ∆β become vanishingly small, we obtain
vg ≜ ∂ω
∂β (6.5)
Note
the similarity to the definition of phase velocity
in Equation 6.3. Group velocity can be interpreted as
the speed at which a disturbance in the wave
propagates. Information may be conveyed as
meaningful disturbances relative to a steady-state
condition, so group velocity is also the speed of
information in a wave.
Note Equation 6.5 yields the expected result for
waves in the form of Equation 6.1:
vg =
(∂β
∂ω
)−1
=
(∂
∂ωω√µǫ
)−1
= 1√µǫ = vp (6.6)
In
other words, the group velocity of a wave in the
form of Equation 6.1 is equal to its phase velocity .
T o observe a difference between vp and vg, βmust
somehow vary as a function of something other than
just ωand the constitutive parameters. Again,
modulation (introduced by the source of the wave)
and dispersion (frequency-dependent constitutive
parameters) are examples in which vg is not
necessarily equal to vp. Here’s an example involving",Electromagnetics_Vol2.pdf
"and dispersion (frequency-dependent constitutive
parameters) are examples in which vg is not
necessarily equal to vp. Here’s an example involving
dispersion:
Example 6.1. Phase and group velocity for a
material exhibiting square-law dispersion.
A broad class of non-magnetic dispersive media
exhibit relative permittivity ǫr that varies as the
square of frequency over a narrow range of
frequencies centered at ω0. For these media we
presume
ǫr = K
(ω
ω0
)2
(6.7)
where Kis
a real-valued positive constant. What
is the phase and group velocity for a
sinusoidally-varying wave in this material?
Solution. First, note
β = ω√µ0ǫ= ω√µ0ǫ0
√ǫr
=
√
K· ω2
ω0
√µ0ǫ0 (6.8)
The
phase velocity is:
vp = ω
β = ω0√
K· ω√µ0ǫ0
(6.9)
Whereas
the group velocity is:
vg = ∂ω
∂β =
(∂β
∂ω
)−1
(6.10)
=
(
∂
∂ω
√
K· ω2
ω0
√µ0ǫ0
)−1
(6.11)
=
(
2
√
K· ω
ω0
√µ0ǫ0
)−1
(6.12)
No
w simplifying using Equation 6.8:
vg =
(
2 β
ω
)−1
(6.13)
= 1
2
ω
β (6.14)
= 1
2vp (6.15)
Thus,
we see that in this case the group velocity",Electromagnetics_Vol2.pdf
"(6.12)
No
w simplifying using Equation 6.8:
vg =
(
2 β
ω
)−1
(6.13)
= 1
2
ω
β (6.14)
= 1
2vp (6.15)
Thus,
we see that in this case the group velocity
is always half the phase velocity .
Another commonly-encountered example for which
vg is not necessarily equal to vp is the propagation of
guided waves; e.g., waves within a waveguide. In
fact, such waves may exhibit phase velocity greater
than the speed of light in a vacuum, c. However, the
group velocity remains less than c, which means the
speed at which information may propagate in a
waveguide is less than c. No physical laws are
violated, since the universal “speed limit” capplies to
information, and not simply points of constant phase.
(See “ Additional Reading” at the end of this section
for more on this concept.)
Additional Reading:
• “Group velocity” on Wikipedia.",Electromagnetics_Vol2.pdf
"6.2. P ARALLEL PLA TE W A VEGUIDE: INTRODUCTION 97
• “Phase velocity” on Wikipedia.
• “Speed of light” on Wikipedia.
6.2 Parallel Plate W aveguide:
Introduction
[m0173]
A parallel plate waveguide is a device for guiding the
propag
ation of waves between two
perfectly-conducting plates. Our primary interest in
this structure is as a rudimentary model applicable to
a broad range of engineering problems. Examples of
such problems include analysis of the fields within
microstrip line and propagation of radio waves in the
ionosphere.
Figure 6.1 shows the geometry of interest. Here the
plates are located at z= 0 and z= a. The plates are
assumed to be infinite in extent, and therefore there is
no need to consider fields in the regions z <0 or
z >a. For this analysis, the region between the plates
is assumed to consist of an ideal (lossless) material
exhibiting real-valued permeability µand real-valued
permittivity ǫ.
Let us limit our attention to a region within the",Electromagnetics_Vol2.pdf
"exhibiting real-valued permeability µand real-valued
permittivity ǫ.
Let us limit our attention to a region within the
waveguide which is free of sources. Expressed in
phasor form, the electric field intensity is governed by
the wave equation
∇2 ˜E + β2 ˜E = 0 (6.16)
where
β = ω√
µǫ (6.17)
Equation
6.16 is a partial differential equation. This
equation, combined with boundary conditions
c⃝ C. W ang CC BY -SA 4.0 (modified)
Figure 6.1: Geometry for analysis of fields in a paral-
lel plate waveguide.",Electromagnetics_Vol2.pdf
"98 CHAPTER 6. W A VEGUIDES
imposed by the perfectly-conducting plates, is
sufficient to determine a unique solution. This is most
easily done in Cartesian coordinates, as we shall now
demonstrate. First we express ˜E in Cartesian
coordinates:
˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez (6.18)
This facilitates the decomposition of Equation 6.16
into separate equations governing the ˆx, ˆy, ˆz
components of ˜E:
∇2 ˜Ex + β2 ˜Ex = 0 (6.19)
∇2 ˜Ey + β2 ˜Ey = 0 (6.20)
∇2 ˜Ez + β2 ˜Ez = 0 (6.21)
Next we observe that the operator ∇2 may be
expressed in Cartesian coordinates as follows:
∇2 = ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2 (6.22)
so
the equations governing the Cartesian components
of ˜E may be written as follows:
∂2
∂x2
˜Ex + ∂2
∂y2
˜Ex + ∂2
∂z2
˜Ex = −β2 ˜Ex (6.23)
∂2
∂x2
˜Ey + ∂2
∂y2
˜Ey + ∂2
∂z2
˜Ey = −β2 ˜Ey (6.24)
∂2
∂x2
˜Ez + ∂2
∂y2
˜Ez + ∂2
∂z2
˜Ez = −β2 ˜Ez (6.25)
Let
us restrict our attention to scenarios that can be
completely described in two dimensions; namely x",Electromagnetics_Vol2.pdf
"∂y2
˜Ez + ∂2
∂z2
˜Ez = −β2 ˜Ez (6.25)
Let
us restrict our attention to scenarios that can be
completely described in two dimensions; namely x
and z, and for which there is no variation in y. This is
not necessarily required, however, it is representative
of a broad class of relevant problems, and allows
Equations 6.23–6.25 to be considerably simplified. If
there is no variation in y, then partial derivatives with
respect to yare zero, yielding:
∂2
∂x2
˜Ex + ∂2
∂z2
˜Ex = −β2 ˜Ex (6.26)
∂2
∂x2
˜Ey + ∂2
∂z2
˜Ey = −β2 ˜Ey (6.27)
∂2
∂x2
˜Ez + ∂2
∂z2
˜Ez = −β2 ˜Ez (6.28)
The
solution to the parallel plate waveguide problem
may now be summarized as follows: The electric field
˜E (Equation 6.18) is the solution to simultaneous
Equations 6.26–6.28 subject to the boundary
conditions that apply at the PEC surfaces at z= 0 and
z= d; namely , that the tangent component of ˜E is
zero at these surfaces.
At this point, the problem has been reduced to a",Electromagnetics_Vol2.pdf
"z= d; namely , that the tangent component of ˜E is
zero at these surfaces.
At this point, the problem has been reduced to a
routine exercise in the solution of partial differential
equations. However, a somewhat more useful
approach is to first decompose the total electric field
into transverse electric (TE) and transverse magnetic
(TM) components, 2 and determine the solutions to
these components separately . The TE component of
the electric field is parallel to the plates, and therefore
transverse (perpendicular) to the plane shown in
Figure 6.1. Thus, the TE component has
˜Ex = ˜Ez = 0, and only ˜Ey may be non-zero. The
TM component of the magnetic field intensity ( ˜H) is
parallel to the plates, and therefore transverse to the
plane shown in Figure 6.1. Thus, the TM component
has ˜Hx = ˜Hz = 0, and only ˜Hy may be non-zero. As
always, the total field is the sum of the TE and TM
components.
The TE and TM solutions are presented in
Sections 6.3 (Electric component of the TE solution),",Electromagnetics_Vol2.pdf
"components.
The TE and TM solutions are presented in
Sections 6.3 (Electric component of the TE solution),
6.4 (Magnetic component of the TE solution), and 6.5
(Electric component of the TM solution). The
magnetic component of the TM solution can be
determined via a straightforward variation of the
preceding three cases, and so is not presented here.
Presentation of the solution for the fields in a parallel
plate waveguide under the conditions described in the
section continues in Section 6.3.
2 For a refresher on TE-TM decomposition, see Section 5.5.",Electromagnetics_Vol2.pdf
"6.3. P ARALLEL PLA TE W A VEGUIDE: TE CASE, ELECTRIC FIELD 99
6.3 Parallel Plate W aveguide: TE
Case, Electric Field
[m0174]
In Section 6.2, the parallel plate waveguide was
introduced.
At the end of that section, we described
the decomposition of the problem into its TE and TM
components. In this section, we find the electric field
component of the TE field in the waveguide.
Figure 6.2 shows the problem addressed in this
section. (Additional details and assumptions are
addressed in Section 6.2.)
Since ˜Ex = ˜Ez = 0 for the TE component of the
electric field, Equations 6.26 and 6.28 are irrelevant,
leaving only:
∂2
∂x2
˜Ey + ∂2
∂z2
˜Ey = −β2 ˜Ey (6.29)
The
general solution to this partial differential
equation is:
˜Ey = e−jkz z[
Ae−jkxx + Be+jkxx]
+ e+jkz z[
Ce−jkxx + De+jkxx]
(6.30)
where A, B, C, and Dare complex-valued constants
and kx and kz are real-valued constants. W e have
assigned variable names to these constants with
advance knowledge of their physical interpretation;",Electromagnetics_Vol2.pdf
"and kx and kz are real-valued constants. W e have
assigned variable names to these constants with
advance knowledge of their physical interpretation;
however, at this moment they remain simply unknown
constants whose values must be determined by
enforcement of boundary conditions.
c⃝ C. W ang CC BY -SA 4.0
Figure 6.2: TE component of the electric field in a
parallel plate waveguide.
Note that Equation 6.30 consists of two terms. The
first term includes the factor e−jkz z, indicating a
wave propagating in the +ˆz direction, and the second
term includes the factor e+jkz z, indicating a wave
propagating in the −ˆz direction. If we impose the
restriction that sources exist only on the left (z < 0)
side of Figure 6.2, and that there be no structure
capable of wave scattering (in particular, reflection)
on the right (z > 0) side of Figure 6.2, then there can
be no wave components propagating in the −ˆz
direction. In this case, C = D= 0 and Equation 6.30
simplifies to:
˜Ey = e−jkz z[",Electromagnetics_Vol2.pdf
"be no wave components propagating in the −ˆz
direction. In this case, C = D= 0 and Equation 6.30
simplifies to:
˜Ey = e−jkz z[
Ae−jkxx + Be+jkxx]
(6.31)
Before proceeding, let’s make sure that Equation 6.31
is actually a solution to Equation 6.29. As we shall
see in a moment, performing this check will reveal
some additional useful information. First, note:
∂˜Ey
∂x = e−jkz z[
−Ae−j
kxx + Be+jkxx]
(jkx)
(6.32)
So:
∂2 ˜Ey
∂x2 = e−jkz z[
Ae−j
kxx + Be+jkxx](
−k2
x
)
(6.33)
Comparing this to Equation 6.31, we observe the
remarkable fact that
∂2 ˜Ey
∂x2 = −k2
x˜Ey (6.34)
Similarly
, note:
∂˜Ey
∂z = e−jkz z[
Ae−j
kxx + Be+jkxx]
(−jkz)
(6.35)
So:
∂2 ˜Ey
∂z2 = e−jkz z[
Ae−j
kxx + Be+jkxx](
−k2
z
)
= −k2
z ˜Ey (6.36)
Now summing these results:
∂2 ˜Ey
∂x2 + ∂2 ˜Ey
∂z2 = −
(
k2
x + k2
z
)˜Ey (6.37)
Comparing
Equation 6.37 to Equation 6.29, we
conclude that Equation 6.31 is indeed a solution to
Equation 6.29, but only if:
β2 = k2
x + k2
z (6.38)",Electromagnetics_Vol2.pdf
"100 CHAPTER 6. W A VEGUIDES
This confirms that kx and kz are in fact the
components of the propagation vector
k ≜ βˆk = ˆxkx + ˆyky + ˆzkz (6.39)
where ˆk is the unit vector pointing in the direction of
propagation, and ky = 0 in this particular problem.
The solution has now been reduced to finding the
constants A, B, and either kx or kz. This is
accomplished by enforcing the relevant boundary
conditions. In general, the component of ˜E that is
tangent to a perfectly-conducting surface is zero.
Applied to the present problem, this means ˜Ey = 0 at
x= 0 and ˜Ey = 0 at x= a. Referring to
Equation 6.31, the boundary condition at x= 0
means
e−jkz z[A· 1 + B· 1] = 0 (6.40)
The factor e−jkz z cannot be zero; therefore,
A+ B = 0. Since B = −A, we may rewrite
Equation 6.31 as follows:
˜Ey = e−jkz zB
[
e+jkxx − e−jkxx]
(6.41)
This expression is simplified using a trigonometric
identity:
1
2j
[
e+jkxx − e−j
kxx]
= sin kxx (6.42)
Let us now make the definition Ey0 ≜ j2B. Then:",Electromagnetics_Vol2.pdf
"This expression is simplified using a trigonometric
identity:
1
2j
[
e+jkxx − e−j
kxx]
= sin kxx (6.42)
Let us now make the definition Ey0 ≜ j2B. Then:
˜Ey = Ey0e−jkz zsin kxx (6.43)
Now applying the boundary condition at x= a:
Ey0e−jkz zsin kxa= 0 (6.44)
The factor e−jkz z cannot be zero, and Ey0 = 0 yields
only trivial solutions; therefore:
sin kxa= 0 (6.45)
This in turn requires that
kxa= mπ (6.46)
where mis an integer. Note that m= 0 is not of
interest since this yields kx = 0, which according to
Equation 6.43 yields the trivial solution ˜Ey = 0. Also
each integer value of mthat is less than zero is
excluded because the associated solution is different
from the solution for the corresponding positive value
of min sign only , which can be absorbed in the
arbitrary constant Ey0.
At this point we have uncovered a family of solutions
given by Equation 6.43 and Equation 6.46 with
m= 1,2,.... Each solution associated with a
particular value of mis referred to as a mode, which",Electromagnetics_Vol2.pdf
"given by Equation 6.43 and Equation 6.46 with
m= 1,2,.... Each solution associated with a
particular value of mis referred to as a mode, which
(via Equation 6.46) has a particular value of kx. The
value of kz for mode mis obtained using
Equation 6.38 as follows:
kz =
√
β2 − k2x
=
√
β2 −
(mπ
a
)2
(6.47)
Since kz is
specified to be real-valued, we require:
β2 −
(mπ
a
)2
>0 (6.48)
This
constrains β; specifically:
β >mπ
a (6.49)
Recall
that β = ω√µǫand ω= 2πf where f is
frequency . Solving for f, we find:
f > m
2a√µǫ (6.50)
Therefore,
each mode exists only above a certain
frequency , which is different for each mode. This
cutoff frequency fc for mode mis given by
f(m)
c ≜ m
2a√µǫ (6.51)
At
frequencies below the cutoff frequency for mode
m, modes 1 through m− 1 exhibit imaginary-valued
kz. The propagation constant must have a real-valued
component in order to propagate; therefore, these
modes do not propagate and may be ignored.
Let us now summarize the solution. For the scenario",Electromagnetics_Vol2.pdf
"component in order to propagate; therefore, these
modes do not propagate and may be ignored.
Let us now summarize the solution. For the scenario
depicted in Figure 6.2, the electric field component of
the TE solution is given by:
ˆy ˜Ey = ˆy
∞∑
m=1
˜E(m)
y
(6.52)",Electromagnetics_Vol2.pdf
"6.3. P ARALLEL PLA TE W A VEGUIDE: TE CASE, ELECTRIC FIELD 101
where
˜E(m)
y ≜
{
0, f <f(m)
c
E(m)
y0 e−jk(m)
z zsin k(m)
x x, f ≥ f(m)
c
(6.53)
where menumerates
modes (m = 1,2,...),
k(m)
z ≜
√
β2 −
[
k(m)
x
] 2
(6.54)
and
k(m)
x ≜ mπ
/a (6.55)
Finally
, E(m)
y0 is a complex-valued constant that
depends on sources or boundary conditions to the left
of the region of interest.
For the scenario depicted in Figure 6.2, the elec-
tric
field component of the TE solution is given
by Equation 6.52 with modal components deter-
mined as indicated by Equations 6.51–6.55. This
solution presumes all sources lie to the left of the
region of interest, and no scattering occurs to the
right of the region of interest.
T o better understand this result, let us examine the
lowest-order mode, m= 1. For this mode
f(1)
c = 1/2a√
µǫ, so this mode can exist if
f
>1/2a√µǫ. Also k( 1)
x = π/a, so
k(1)
z =
√
β2 −
(π
a
)2
(6.56)
Subsequently
,
˜E(1)
y = E(1)
y0 e−jk(1)
z zsin πx
a (6.57)
Note",Electromagnetics_Vol2.pdf
"f
>1/2a√µǫ. Also k( 1)
x = π/a, so
k(1)
z =
√
β2 −
(π
a
)2
(6.56)
Subsequently
,
˜E(1)
y = E(1)
y0 e−jk(1)
z zsin πx
a (6.57)
Note
that this mode has the form of a plane wave. The
plane wave propagates in the +ˆz direction with phase
propagation constant k(1)
z . Also, we observe that the
apparent plane wave is non-uniform, exhibiting
magnitude proportional to sin πx/awithin the
waveguide. This is shown in Figure 6.3 (left image).
In particular, we observe that the magnitude of the
wave is zero at the perfectly-conducting surfaces – as
is necessary to satisfy the boundary conditions – and
is maximum in the center of the waveguide.
Now let us examine the m= 2 mode. For this mode,
f(2)
c = 1/a√µǫ, so this mode can exist if
c⃝ C. W ang CC BY -SA 4.0
Figure 6.3: Magnitude of the two lowest-order TE
modes in a parallel plate waveguide. Left: m = 1,
Right: m= 2.
f >1/a√µǫ. This frequency is higher than f( 1)
c , so
the m= 1 mode can exist at any frequency at which",Electromagnetics_Vol2.pdf
"Right: m= 2.
f >1/a√µǫ. This frequency is higher than f( 1)
c , so
the m= 1 mode can exist at any frequency at which
the m= 2 mode exists. Also k(2)
x = 2π/a, so
k(2)
z =
√
β2 −
(2π
a
)2
(6.58)
Subsequently
,
˜E(2)
y = E(2)
y0 e−jk(2)
z zsin 2πx
a (6.59)
In
this case, the apparent plane wave propagates in the
+ˆz direction with phase propagation constant k(2)
z ,
which is less than k(1)
z . For m= 2, we find
magnitude is proportional to sin 2πx/a within the
waveguide (Figure 6.3, right image). As in the m= 1
case, we observe that the magnitude of the wave is
zero at the PEC surfaces; however, for m= 2, there
are two maxima with respect to x, and the magnitude
in the center of the waveguide is zero.
This pattern continues for higher-order modes. In
particular, each successive mode exhibits higher
cutoff frequency , smaller propagation constant, and
increasing integer number of sinusoidal half-periods
in magnitude.
Example 6.2. Single-mode TE propagation in a
parallel plate waveguide.",Electromagnetics_Vol2.pdf
"increasing integer number of sinusoidal half-periods
in magnitude.
Example 6.2. Single-mode TE propagation in a
parallel plate waveguide.
Consider an air-filled parallel plate waveguide
consisting of plates separated by 1 cm.
Determine the frequency range for which one",Electromagnetics_Vol2.pdf
"102 CHAPTER 6. W A VEGUIDES
(and only one) propagating TE mode is assured.
Solution. Single-mode
TE propagation is
assured by limiting frequency f to greater than
the cutoff frequency for m= 1, but lower than
the cutoff frequency for m= 2. (Any frequency
higher than the cutoff frequency for m= 2
allows at least 2 modes to exist.) Calculating the
applicable cutoff frequencies, we find:
f(1)
c = 1
2a√µ0ǫ0
∼= 15 .0 GHz (6.60)
f(2)
c = 2
2a√µ0ǫ0
∼= 30 .0 GHz (6.61)
Therefore, 15.0 GHz ≤ f ≤ 30.0 GHz .
Finally
, let us consider the phase velocity vp within
the waveguide. For lowest-order mode m= 1, this is
vp = ω
k(1)
z
= ω√
ω2µǫ−
(π
a
)2
(6.62)
Recall
that the speed of an electromagnetic wave in
unbounded space (i.e., not in a waveguide) is 1/√µǫ.
For
example, the speed of light in free space is
1/√µ0ǫ0 = c. However, the phase velocity indicated
by
Equation 6.62 is greater than 1/√µǫ; e.g., faster
than
light would travel in the same material",Electromagnetics_Vol2.pdf
"1/√µ0ǫ0 = c. However, the phase velocity indicated
by
Equation 6.62 is greater than 1/√µǫ; e.g., faster
than
light would travel in the same material
(presuming it were transparent). At first glance, this
may seem to be impossible. However, recall that
information travels at the group velocity vg, and not
necessarily the phase velocity . (See Section 6.1 for a
refresher.) Although we shall not demonstrate this
here, the group velocity in the parallel plate
waveguide is always less than 1/√
µǫ, so no physical
la
ws are broken, and signals travel somewhat slower
than the speed of light, as they do in any other
structure used to convey signals.
Also remarkable is that the speed of propagation is
different for each mode. In fact, we find that the phase
velocity increases and the group velocity decreases as
mincreases. This phenomenon is known as
dispersion, and sometimes specifically as mode
dispersion or modal dispersion.
6.4 Parallel Plate W aveguide: TE
Case, Magnetic Field
[m0175]",Electromagnetics_Vol2.pdf
"dispersion, and sometimes specifically as mode
dispersion or modal dispersion.
6.4 Parallel Plate W aveguide: TE
Case, Magnetic Field
[m0175]
In Section 6.2, the parallel plate waveguide was
introduced.
In Section 6.3, we determined the TE
component of the electric field. In this section, we
determine the TE component of the magnetic field.
The reader should be familiar with Section 6.3 before
attempting this section.
In Section 6.3, the TE component of the electric field
was determined to be:
ˆy ˜Ey = ˆy
∞∑
m=1
˜E(m)
y (6.63)
The TE component of the magnetic field may be
obtained from the Maxwell-Faraday equation:
∇ × ˜E = −jωµ˜H (6.64)
Thus:
˜H = j
ωµ ∇ × ˜E
= j
ωµ ∇ ×
(
ˆy ˜Ey
)
(6.65)
The
relevant form of the curl operator is
Equation B.16 (Appendix B.2). Although the
complete expression consists of 6 terms, all but 2 of
these terms are zero because the ˆx and ˆz components
of ˜E are zero. The two remaining terms are
−ˆx∂˜Ey/∂z and +ˆz∂˜Ey/∂x. Thus:
˜H = j
ωµ
(
−ˆx∂˜Ey
∂z + ˆz∂˜Ey
∂x
)",Electromagnetics_Vol2.pdf
"of ˜E are zero. The two remaining terms are
−ˆx∂˜Ey/∂z and +ˆz∂˜Ey/∂x. Thus:
˜H = j
ωµ
(
−ˆx∂˜Ey
∂z + ˆz∂˜Ey
∂x
)
(6.66)
Recall
that ˜Ey is the sum of modes, as indicated in
Equation 6.63. Since differentiation (i.e., ∂/∂z and
∂/∂x) is a linear operator, we may evaluate
Equation 6.66 for modes one at a time, and then sum
the results. Using this approach, we find:
∂˜E(m)
y
∂z = ∂
∂z E(m)
y0 e−jk(m)
z zs in k(m)
x x
=
(
E(m)
y0 e−jk(m)
z zsin k(m)
x x
)(
−jk(m)
z
)
(6.67)",Electromagnetics_Vol2.pdf
"6.4. P ARALLEL PLA TE W A VEGUIDE: TE CASE, MAGNETIC FIELD 103
and
∂˜E(m)
y
∂x = ∂
∂x E(m)
y0 e−jk(m)
z zs in k(m)
x x
=
(
E(m)
y0 e−jk(m)
z zcos k(m)
x x
)(
+k(m)
x
)
(6.68)
W e may now assemble a solution for the magnetic
field as follows:
ˆx ˜Hx + ˆz ˜Hz = ˆx
∞∑
m=1
˜H(m)
x
+ ˆz
∞∑
m=1
˜H(m)
z (6.69)
where
˜H(m)
x = −k(m)
z
ωµ E(m)
y0 e−jk(m)
z zs in k(m)
x x (6.70)
˜H(m)
z = + jk(m)
x
ωµ E(m)
y0 e−jk(m)
z zcos k(m)
x x (6.71)
and
modes may only exist at frequencies greater than
the associated cutoff frequencies. Summarizing:
The magnetic field component of the TE solu-
tion
is given by Equation 6.69 with modal com-
ponents as indicated by Equations 6.70 and 6.71.
Caveats pertaining to the cutoff frequencies and
the locations of sources and structures continue
to apply .
This result is quite complex, yet some additional
insight is possible. At the perfectly-conducting (PEC)
surface at x= 0, we see
˜H(m)
x (x= 0) = 0
˜H(m)
z (x= 0) = +j k(m)
x
ωµ E(m)
y0 e−jk(m)
z z (6.72)
Similarly",Electromagnetics_Vol2.pdf
"surface at x= 0, we see
˜H(m)
x (x= 0) = 0
˜H(m)
z (x= 0) = +j k(m)
x
ωµ E(m)
y0 e−jk(m)
z z (6.72)
Similarly
, on the PEC surface at x= a, we see
˜H(m)
x (x= a) = 0
˜H(m)
z (x= a) = −jk(m)
x
ωµ E(m)
y0 e−jk(m)
z z (6.73)
Thus,
we see the magnetic field vector at the PEC
surfaces is non-zero and parallel to the PEC surfaces.
Recall that the magnetic field is identically zero inside
a PEC material. Also recall that boundary conditions
require that discontinuity in the component of H
tangent to a surface must be supported by a surface
current. W e conclude that
Current flows on the PEC surfaces of the waveg-
uide.
If
this seems surprising, note that essentially the same
thing happens in a coaxial transmission line. That is,
signals in a coaxial transmission line can be described
equally well in terms of either potentials and currents
on the inner and outer conductors, or the
electromagnetic fields between the conductors. The
parallel plate waveguide is only slightly more",Electromagnetics_Vol2.pdf
"on the inner and outer conductors, or the
electromagnetic fields between the conductors. The
parallel plate waveguide is only slightly more
complicated because the field in a properly-designed
coaxial cable is a single transverse electromagnetic
(TEM) mode, whereas the fields in a parallel plate
waveguide are combinations of TE and TM modes.
Interestingly , we also find that the magnetic field
vector points in different directions depending on
position relative to the conducting surfaces. W e just
determined that the magnetic field is parallel to the
conducting surfaces at those surfaces. However, the
magnetic field is perpendicular to those surfaces at m
locations between x= 0 and x= a. These locations
correspond to maxima in the electric field.",Electromagnetics_Vol2.pdf
"104 CHAPTER 6. W A VEGUIDES
6.5 Parallel Plate W aveguide:
TM Case, Electric Field
[m0177]
In Section 6.2, the parallel plate waveguide shown in
Figure
6.4 was introduced. At the end of that section,
we decomposed the problem into its TE and TM
components. In this section, we find the TM
component of the fields in the waveguide.
“Transverse magnetic” means the magnetic field
vector is perpendicular to the plane of interest, and is
therefore parallel to the conducting surfaces. Thus,
˜H = ˆy ˜Hy, with no component in the ˆx or ˆz
directions. Following precisely the same reasoning
employed in Section 6.2, we find the governing
equation for the magnetic component of TM field is:
∂2
∂x2
˜Hy + ∂2
∂z2
˜Hy = −β2 ˜Hy (6.74)
The
general solution to this partial differential
equation is:
˜Hy = e−jkz z[
Ae−jkxx + Be+jkxx]
+ e+jkz z[
Ce−jkxx + De+jkxx]
(6.75)
where A, B, C, and Dare complex-valued constants;
and kx and kz are real-valued constants. W e have",Electromagnetics_Vol2.pdf
"+ e+jkz z[
Ce−jkxx + De+jkxx]
(6.75)
where A, B, C, and Dare complex-valued constants;
and kx and kz are real-valued constants. W e have
assigned variable names to these constants with
advance knowledge of their physical interpretation;
however, at this moment they remain simply unknown
constants whose values must be determined by
enforcement of boundary conditions.
c⃝ C. W ang CC BY -SA 4.0
Figure 6.4: TM component of the electric field in a
parallel plate waveguide.
Note that Equation 6.75 consists of two terms. The
first term includes the factor e−jkz z, indicating a
wave propagating in the +ˆz direction, and the second
term includes the factor e+jkz z, indicating a wave
propagating in the −ˆz direction. If we impose the
restriction that sources exist only on the left (z < 0)
side of Figure 6.4, and that there be no structure
capable of wave scattering (in particular, reflection)
on the right (z > 0) side of Figure 6.4, then there can
be no wave components propagating in the −ˆz",Electromagnetics_Vol2.pdf
"on the right (z > 0) side of Figure 6.4, then there can
be no wave components propagating in the −ˆz
direction. In this case, C = D= 0 and Equation 6.75
simplifies to:
˜Hy = e−jkz z[
Ae−jkxx + Be+jkxx]
(6.76)
Before proceeding, let’s make sure that Equation 6.76
is actually a solution to Equation 6.74. As in the TE
case, this check yields a constraint (in fact, the same
constraint) on the as-yet undetermined parameters kx
and kz. First, note:
∂˜Hy
∂x = e−jkz z[
−Ae−j
kxx + Be+jkxx]
(jkx)
(6.77)
So:
∂2 ˜Hy
∂x2 = e−jkz z[
Ae−j
kxx + Be+jkxx](
−k2
x
)
= −k2
x ˜Hy (6.78)
Next, note:
∂˜Hy
∂z = e−jkz z[
Ae−j
kxx + Be+jkxx]
(−jkz)
(6.79)
So:
∂2 ˜Hy
∂z2 = e−jkz z[
Ae−j
kxx + Be+jkxx](
−k2
z
)
= −k2
z ˜Hy (6.80)
Now summing these results:
∂2 ˜Hy
∂x2 + ∂2 ˜Hy
∂z2 = −
(
k2
x + k2
z
)˜Hy (6.81)
Comparing
Equation 6.81 to Equation 6.74, we
conclude that Equation 6.76 is a solution to
Equation 6.74 under the constraint that:
β2 = k2
x + k2
z (6.82)",Electromagnetics_Vol2.pdf
"6.5. P ARALLEL PLA TE W A VEGUIDE: TM CASE, ELECTRIC FIELD 105
This is precisely the same constraint identified in the
TE case, and confirms that kx and ky are in fact the
components of the propagation vector
k ≜ βˆk = ˆxkx + ˆyky + ˆzkz (6.83)
where ˆk is the unit vector pointing in the direction of
propagation, and ky = 0 in this particular problem.
Our objective in this section is to determine the
electric field component of the TM field. The electric
field may be obtained from the magnetic field using
Ampere’s law:
∇ × ˜H = jωǫ˜E (6.84)
Thus:
˜E = 1
jωǫ ∇ × ˜H
= 1
jωǫ ∇ ×
(
ˆy ˜Hy
)
(6.85)
The
relevant form of the curl operator is
Equation B.16 (Appendix B.2). Although the
complete expression consists of 6 terms, all but 2 of
these terms are zero because the ˆx and ˆz components
of ˜H are zero. The two remaining terms are
−ˆx∂˜Hy/∂z and +ˆz∂˜Hy/∂x. Thus:
˜E = 1
jωǫ
(
−ˆx∂˜Hy
∂z + ˆz∂˜Hy
∂x
)
(6.86)
W
e may further develop this expression using",Electromagnetics_Vol2.pdf
"−ˆx∂˜Hy/∂z and +ˆz∂˜Hy/∂x. Thus:
˜E = 1
jωǫ
(
−ˆx∂˜Hy
∂z + ˆz∂˜Hy
∂x
)
(6.86)
W
e may further develop this expression using
Equations 6.77 and 6.79. W e find the ˆx component of
˜E is:
˜Ex = kz
ωǫ e−jkz z[
Ae−j
kxx + Be+jkxx]
(6.87)
and the ˆz component of ˜E is:
˜Ez = kx
ωǫ e−jkz z[
−Ae−j
kxx + Be+jkxx]
(6.88)
The solution has now been reduced to the problem of
finding the constants A, B, and either kx or kz. This
is accomplished by enforcing the relevant boundary
conditions. In general, the component of the electric
field which is tangent to a perfectly-conducting
surface is zero. Applied to the present (TM) case, this
means ˜Ez(x= 0) = 0 and ˜Ez(x= a) = 0.
Referring to Equation 6.88, the boundary condition at
x= 0 means
kx
ωǫ e−jkz z[ −A(1) + B(1)] = 0 (6.89)
The factor e−jkz z always has unit magnitude, and so
cannot be zero. W e could require kx to be zero, but
this is unnecessarily restrictive. Instead, we require
A= Band we may rewrite Equation 6.88 as follows:
˜Ez = Bkx",Electromagnetics_Vol2.pdf
"this is unnecessarily restrictive. Instead, we require
A= Band we may rewrite Equation 6.88 as follows:
˜Ez = Bkx
ωǫ e−jkz z[
e+j
kxx − e−jkxx]
(6.90)
This expression is simplified using a trigonometric
identity:
sin kxa= 1
2j
[
e+jkxa − e−j
kxa]
(6.91)
Thus:
˜Ez = j2Bkx
ωǫ e−jkz zs in kxx (6.92)
Now following up with ˜Ex, beginning from
Equation 6.87:
˜Ex = kz
ωǫ e−jkz z[
Ae−j
kxx + Be+jkxx]
= Bkz
ωǫ e−jkz z[
e−j
kxx + e+jkxx]
= 2Bkz
ωǫ e−jkz zcos kxx (6.93)
F
or convenience we define the following
complex-valued constant:
Ex0 ≜ 2Bkz
ωǫ (6.94)
This
yields the following simpler expression:
˜Ex = Ex0 e−jkz zcos kxx (6.95)
Now let us apply the boundary condition at x= ato
˜Ez:
j2Bkz
ωǫ e−jkz zs in kxa= 0 (6.96)
Requiring B = 0 or kz = 0 yields only trivial
solutions, therefore, it must be true that
sin kxa= 0 (6.97)
This in turn requires that
kxa= mπ (6.98)",Electromagnetics_Vol2.pdf
"106 CHAPTER 6. W A VEGUIDES
where mis an integer. Note that this is precisely the
same relationship that we identified in the TE case.
There is an important difference, however. In the TE
case, m= 0 was not of interest because this yields
kx = 0, and the associated field turned out to be
identically zero. In the present (TM) case, m= 0 also
yields kx = 0, but the associated field is not
necessarily zero. That is, for m= 0, ˜Ez = 0 but ˜Ex
is not necessarily zero. Therefore, m= 0 as well as
m= 1, m= 2, and so on are of interest in the TM
case.
At this point, we have uncovered a family of solutions
with m= 0,1,2,.... Each solution is referred to as a
mode, and is associated with a particular value of kx.
In the discussion that follows, we shall find that the
consequences are identical to those identified in the
TE case, except that m= 0 is now also allowed.
Continuing: The value of kz for mode mis obtained
using Equation 6.82 as follows:
kz =
√
β2 − k2x
=
√
β2 −
(mπ
a
)2
(6.99)
Since kz is",Electromagnetics_Vol2.pdf
"Continuing: The value of kz for mode mis obtained
using Equation 6.82 as follows:
kz =
√
β2 − k2x
=
√
β2 −
(mπ
a
)2
(6.99)
Since kz is
specified to be real-valued, we require:
β2 −
(mπ
a
)2
>0 (6.100)
This
constrains β; specifically:
β >mπ
a (6.101)
Recall
that β = ω√µǫand ω= 2πf where f is
frequency . Solving for f, we find:
f > m
2a√µǫ (6.102)
Thus,
each mode exists only above a certain
frequency , which is different for each mode. This
cutoff frequency fc for mode mis given by
f(m)
c ≜ m
2a√µǫ (6.103)
At
frequencies below the cutoff frequency for mode
m, modes 0 through m− 1 exhibit imaginary-valued
kz and therefore do no propagate. Also, note that the
cutoff frequency for m= 0 is zero, and so this mode
is always able to propagate. That is, the m= 0 mode
may exist for any a> 0 and any f >0. Once again,
this is a remarkable difference from the TE case, for
which m= 0 is not available.
Let us now summarize the solution. With respect to
Figure 6.4, we find that the electric field component",Electromagnetics_Vol2.pdf
"which m= 0 is not available.
Let us now summarize the solution. With respect to
Figure 6.4, we find that the electric field component
of the TM field is given by:
˜E =
∞∑
m=0
[
ˆx ˜E(m)
x + ˆz ˜E(m)
z
]
(6.104)
where
˜E(m)
x ≜
{
0
, f <f(m)
c
E(m)
x0 e−jk(m)
z zcos k(m)
x x, f ≥ f(m)
c
(6.105)
and
˜E(m)
z ≜
{
0
, f <f(m)
c
jk(m)
x
k(m)
z
E(m)
x0 e−jk(m)
z zs in k(m)
x x, f ≥ f(m)
c
(6.106)
where menumerates
modes (m = 0,1,2,...) and
k(m)
z ≜
√
β2 −
[
k(m)
x
] 2
(6.107)
k(m)
x ≜ mπ
/a (6.108)
Finally
, the coefficients E(m)
x0 depend on sources
and/or boundary conditions to the left of the region of
interest.
For the scenario depicted in Figure 6.4, the elec-
tric
field component of the TM solution is given
by Equation 6.104 with modal components de-
termined as indicated by Equations 6.103–6.108.
This solution presumes all sources lie to the
left of the region of interest, with no additional
sources or boundary conditions to the right of the
region of interest.",Electromagnetics_Vol2.pdf
"left of the region of interest, with no additional
sources or boundary conditions to the right of the
region of interest.
The m= 0 mode, commonly referred to as the
“TM 0” mode, is of particular importance in the
analysis of microstrip transmission line, and is
addressed in Section 6.6.",Electromagnetics_Vol2.pdf
"6.6. P ARALLEL PLA TE W A VEGUIDE: THE TM 0 MODE 107
6.6 Parallel Plate W aveguide:
The TM 0 Mode
[m0220]
In Section 6.2, the parallel plate waveguide (also
sho
wn in Figure 6.5) was introduced. At the end of
that section we decomposed the problem into its
constituent TE and TM fields. In Section 6.5, we
determined the electric field component of the TM
field, which was found to consist of a set of discrete
modes. In this section, we address the lowest-order
(m= 0) mode of the TM field, which has special
relevance in a number of applications including
microstrip transmission lines. This mode is
commonly referred to as the “TM 0” mode.
The TM electric field intensity in the waveguide is
given by Equation 6.104 with modal components
determined as indicated by Equations 6.103–6.108
(Section 6.5). Recall that the m= 0 mode can only
exist only in the TM case, and does not exist in the
TE case. W e also noted that the cutoff frequency for
this mode is zero, so it may exist at any frequency ,",Electromagnetics_Vol2.pdf
"TE case. W e also noted that the cutoff frequency for
this mode is zero, so it may exist at any frequency ,
and within any non-zero plate separation a. For this
mode k(0)
x = 0, k(0)
z = β, and we find
˜E = ˆxE(0)
x0 e−jβz (TM0 mode) (6.109)
Remarkably , we find that this mode has the form of a
uniform plane wave which propagates in the +ˆz
direction; i.e., squarely between the plates. The phase
and group velocities of this wave are equal to each
other and ω/β, precisely as expected for a uniform
c⃝ C. W ang CC BY -SA 4.0 (modified)
Figure 6.5: TM 0 mode in a parallel plate waveguide.
plane wave in unbounded media; i.e., as if the plates
did not exist. This observation allows us to easily
determine the associated magnetic field: Using the
plane wave relationship,
˜H = 1
η
ˆk × ˜E (6.110)
= 1
ηˆz ×
(
ˆxE(0)
x0 e−j
βz
)
(6.111)
= ˆy E(0)
x0
η e−jβ
z (TM0 mode) (6.112)
Example 6.3. Guided waves in a printed circuit
board (PCB).
A very common form of PCB consists of a",Electromagnetics_Vol2.pdf
"(6.111)
= ˆy E(0)
x0
η e−jβ
z (TM0 mode) (6.112)
Example 6.3. Guided waves in a printed circuit
board (PCB).
A very common form of PCB consists of a
1.575 mm-thick slab of low-loss dielectric
having relative permittivity ≈ 4.5 sandwiched
between two copper planes. Characterize the
electromagnetic field in a long strip of this
material. Assume a single source exists at one
end of the strip, and operates at frequencies
below 10 GHz.
Solution. Let us assume that the copper planes
exhibit conductivity that is sufficiently high that
the inward-facing surfaces may be viewed as
perfectly-conducting. Also, let us limit scope to
the field deep within the “sandwich, ” and neglect
the region near the edges of the PCB. Under
these conditions, the PCB is well-modeled as a
parallel-plate waveguide. Thus, the
electromagnetic field consists of a combination
of TE and TM modes. The active (non-zero)
modes depend on the source (a mode must be
“stimulated” by the source in order to propagate)",Electromagnetics_Vol2.pdf
"of TE and TM modes. The active (non-zero)
modes depend on the source (a mode must be
“stimulated” by the source in order to propagate)
and modal cutoff frequencies. The cutoff
frequency for mode mis
f(m)
c = m
2a√µǫ (6.113)
In
this case, a= 1.575 mm, µ≈ µ0, and
ǫ≈ 4.5ǫ0. Therefore:
f(m)
c ≈ (44.9 GHz) m (6.114)
Since the cutoff frequency for m= 1 is much
greater than 10 GHz, we may rest assured that",Electromagnetics_Vol2.pdf
"108 CHAPTER 6. W A VEGUIDES
the only mode that can propagate inside the PCB
is
TM0. Therefore, the field deep inside the PCB
may be interpreted as a single plane wave having
the TM 0 structure shown in Figure 6.5,
propagating away from the source end of the
PCB. The phase velocity is simply that of the
apparent plane wave:
vp = 1√µǫ ≈ 1.41 × 108 m/s ≈ 0 .47c (6.115)
The scenario described in this example is essentially a
very rudimentary form of microstrip transmission
line.
6.7 General Relationships for
Unidirectional W aves
[m0222]
Analysis of electromagnetic waves in enclosed spaces
– and
in waveguides in particular – is quite difficult.
The task is dramatically simplified if it can be
assumed that the wave propagates in a single
direction; i.e., is unidirectional. This does not
necessarily entail loss of generality . For example:
Within a straight waveguide, waves can travel either
“forward” or “backward. ” The principle of
superposition allows one to consider these two",Electromagnetics_Vol2.pdf
"Within a straight waveguide, waves can travel either
“forward” or “backward. ” The principle of
superposition allows one to consider these two
unidirectional cases separately , and then to simply
sum the results.
In this section, the equations that relate the various
components of a unidirectional wave are derived. The
equations will be derived in Cartesian coordinates,
anticipating application to rectangular waveguides.
However, the underlying strategy is generally
applicable.
W e begin with Maxwell’s curl equations:
∇ × ˜E = −jωµ˜H (6.116)
∇ × ˜H = + jωǫ˜E (6.117)
Let us consider Equation 6.117 first. Solving for ˜E,
we have:
˜E = 1
jωǫ∇ × ˜H (6.118)
This
is actually three equations; that is, one each for
the ˆx, ˆy, and ˆz components of ˜E, respectively . T o
extract these equations, let us define the components
as follows:
˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez (6.119)
˜H = ˆx ˜Hx + ˆy ˜Hy + ˆz ˜Hz (6.120)
Now applying the equation for curl in Cartesian",Electromagnetics_Vol2.pdf
"as follows:
˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez (6.119)
˜H = ˆx ˜Hx + ˆy ˜Hy + ˆz ˜Hz (6.120)
Now applying the equation for curl in Cartesian
coordinates (Equation B.16 in Appendix B.2), we",Electromagnetics_Vol2.pdf
"6.7. GENERAL RELA TIONSHIPS FOR UNIDIRECTIONAL W A VES 109
find:
˜Ex = 1
jωǫ
(
∂˜Hz
∂y − ∂˜Hy
∂z
)
(6.121)
˜Ey = 1
jωǫ
(
∂˜Hx
∂z − ∂˜Hz
∂x
)
(6.122)
˜Ez = 1
jωǫ
(
∂˜Hy
∂x − ∂˜Hx
∂y
)
(6.123)
W
ithout loss of generality , we may assume that the
single direction in which the wave is traveling is in
the +ˆz direction. If this is the case, then ˜H, and
subsequently each component of ˜H, contains a factor
of e−jkz z where kz is the phase propagation constant
in the direction of travel. The remaining factors are
independent of z; i.e., depend only on xand y. With
this in mind, we further decompose components of ˜H
as follows:
˜Hx = ˜hx(x,y)e−jkz z (6.124)
˜Hy = ˜hy(x,y)e−jkz z (6.125)
˜Hz = ˜hz(x,y)e−jkz z (6.126)
where ˜hx(x,y), ˜hy(x,y), and ˜hz(x,y) represent the
remaining factors. One advantage of this
decomposition is that partial derivatives with respect
to zreduce to algebraic operations; i.e.:
∂˜Hx
∂z = −jkz˜hx( x,y)e−jkz z = −jkz ˜Hx (6.127)
∂˜Hy",Electromagnetics_Vol2.pdf
"to zreduce to algebraic operations; i.e.:
∂˜Hx
∂z = −jkz˜hx( x,y)e−jkz z = −jkz ˜Hx (6.127)
∂˜Hy
∂z = −jkz˜hy( x,y)e−jkz z = −jkz ˜Hy (6.128)
∂˜Hz
∂z = −jkz˜hz( x,y)e−jkz z = −jkz ˜Hz (6.129)
W e now substitute Equations 6.124–6.126 into
Equations 6.121–6.123 and then use
Equations 6.127–6.129 to eliminate partial derivatives
with respect to z. This yields:
˜Ex = 1
jωǫ
(
∂˜Hz
∂y + jkz ˜Hy
)
(6.130)
˜Ey = 1
jωǫ
(
−jkz ˜Hx − ∂˜Hz
∂x
)
(6.131)
˜Ez = 1
jωǫ
(
∂˜Hy
∂x − ∂˜Hx
∂y
)
(6.132)
Applying
the same procedure to the curl equation for
˜H (Equation 6.116), one obtains:
˜Hx = 1
−jωµ
(
∂˜Ez
∂y + jkz ˜Ey
)
(6.133)
˜Hy = 1
−jωµ
(
−j kz ˜Ex − ∂˜Ez
∂x
)
(6.134)
˜Hz = 1
−jωµ
(
∂˜Ey
∂x − ∂˜Ex
∂y
)
(6.135)
Equations
6.130–6.135 constitute a set of
simultaneous equations that represent Maxwell’s curl
equations in the special case of a unidirectional
(specifically , a +ˆz-traveling) wave. With just a little
bit of algebraic manipulation of these equations, it is",Electromagnetics_Vol2.pdf
"(specifically , a +ˆz-traveling) wave. With just a little
bit of algebraic manipulation of these equations, it is
possible to obtain expressions for the ˆx and ˆy
components of ˜E and ˜H which depend only on the ˆz
components of ˜E and ˜H. Here they are: 3
˜Ex = −j
k2ρ
(
+kz
∂˜Ez
∂x + ωµ∂˜Hz
∂y
)
(6.136)
˜Ey = +j
k2ρ
(
−kz
∂˜Ez
∂y + ωµ∂˜Hz
∂x
)
(6.137)
˜Hx = +j
k2ρ
(
ωǫ∂˜Ez
∂y − kz
∂˜Hz
∂x
)
(6.138)
˜Hy = −j
k2ρ
(
ωǫ∂˜Ez
∂x + kz
∂˜Hz
∂y
)
(6.139)
where
k2
ρ ≜ β2 − k2
z (6.140)
Wh
y define a parameter called “k ρ”? Note from the
definition that β2 = k2
ρ + k2
z. Further, note that this
equation is an expression of the Pythagorean theorem,
which relates the lengths of the sides of a right
triangle. Since βis the “overall” phase propagation
constant, and kz is the phase propagation constant for
propagation in the ˆz direction in the waveguide, kρ
must be associated with variation in fields in
directions perpendicular to ˆz. In the cylindrical",Electromagnetics_Vol2.pdf
"propagation in the ˆz direction in the waveguide, kρ
must be associated with variation in fields in
directions perpendicular to ˆz. In the cylindrical
coordinate system, this is the ˆρdirection, hence the
subscript “ρ. ” W e shall see later that kρ plays a special
3 Students are encouraged to derive these for themselves – only
algebra is required. Tip to get started: T o get ˜Ex, begin with Equa-
tion 6.130, eliminate ˜Hy using Equation 6.134, and then solve for
˜Ex.",Electromagnetics_Vol2.pdf
"110 CHAPTER 6. W A VEGUIDES
role in determining the structure of fields within the
waveguide, and this provides additional motivation to
identify this quantity explicitly in the field equations.
Summarizing: If you know the wave is unidirectional,
then knowledge of the components of ˜E and ˜H in the
direction of propagation is sufficient to determine
each of the remaining components of ˜E and ˜H. There
is one catch, however: For this to yield sensible
results, kρ may not be zero. So, ironically , these
expressions don’t work in the case of a uniform plane
wave, since such a wave would have ˜Ez = ˜Hz = 0
and kz = β(so kρ = 0), yielding values of zero
divided by zero for each remaining field component.
The same issue arises for any other transverse
electromagnetic (TEM) wave. For non-TEM waves –
and, in particular, unidirectional waves in waveguides
– either ˜Ez or ˜Hz must be non-zero and kρ will be
non-zero. In this case, Equations 6.136–6.139 are",Electromagnetics_Vol2.pdf
"– either ˜Ez or ˜Hz must be non-zero and kρ will be
non-zero. In this case, Equations 6.136–6.139 are
both usable and useful since they allow determination
of all field components given just the z components.
In fact, we may further exploit this simplicity by
taking one additional step: Decomposition of the
unidirectional wave into transverse electric (TE) and
transverse magnetic (TM) components. In this case,
TE means simply that ˜Ez = 0, and TM means simply
that ˜Hz = 0. For a wave which is either TE or TM,
Equations 6.136–6.139 reduce to one term each.
6.8 Rectangular W aveguide: TM
Modes
[m0223]
A rectangular waveguide is a conducting cylinder of
rectangular
cross section used to guide the
propagation of waves. Rectangular waveguide is
commonly used for the transport of radio frequency
signals at frequencies in the SHF band (3–30 GHz)
and higher. The fields in a rectangular waveguide
consist of a number of propagating modes which
depends on the electrical dimensions of the",Electromagnetics_Vol2.pdf
"and higher. The fields in a rectangular waveguide
consist of a number of propagating modes which
depends on the electrical dimensions of the
waveguide. These modes are broadly classified as
either transverse magnetic (TM) or transverse electric
(TE). In this section, we consider the TM modes.
Figure 6.6 shows the geometry of interest. Here the
walls are located at x= 0, x= a, y= 0, and y= b;
thus, the cross-sectional dimensions of the waveguide
are aand b. The interior of the waveguide is presumed
to consist of a lossless material exhibiting real-valued
permeability µand real-valued permittivity ǫ, and the
walls are assumed to be perfectly-conducting.
Let us limit our attention to a region within the
waveguide which is free of sources. Expressed in
phasor form, the electric field intensity within the
waveguide is governed by the wave equation
∇2 ˜E + β2 ˜E = 0 (6.141)
where
β = ω√
µǫ (6.142)
Figure 6.6: Geometry for analysis of fields in a rect-
angular
waveguide.",Electromagnetics_Vol2.pdf
"6.8. RECT ANGULAR W A VEGUIDE: TM MODES 111
Equation 6.141 is a partial differential equation. This
equation, combined with boundary conditions
imposed by the perfectly-conducting plates, is
sufficient to determine a unique solution. This
solution is most easily determined in Cartesian
coordinates, as we shall now demonstrate. First we
express ˜E in Cartesian coordinates:
˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez (6.143)
This facilitates the decomposition of Equation 6.141
into separate equations governing the ˆx, ˆy, and ˆz
components of ˜E:
∇2 ˜Ex + β2 ˜Ex = 0 (6.144)
∇2 ˜Ey + β2 ˜Ey = 0 (6.145)
∇2 ˜Ez + β2 ˜Ez = 0 (6.146)
Next we observe that the operator ∇2 may be
expressed in Cartesian coordinates as follows:
∇2 = ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2 (6.147)
so
the equations governing the Cartesian components
of ˜E may be written as follows:
∂2
∂x2
˜Ex + ∂2
∂y2
˜Ex + ∂2
∂z2
˜Ex + β2 ˜Ex = 0 (6.148)
∂2
∂x2
˜Ey + ∂2
∂y2
˜Ey + ∂2
∂z2
˜Ey + β2 ˜Ey = 0 (6.149)
∂2
∂x2
˜Ez + ∂2
∂y2
˜Ez + ∂2
∂z2",Electromagnetics_Vol2.pdf
"∂2
∂x2
˜Ex + ∂2
∂y2
˜Ex + ∂2
∂z2
˜Ex + β2 ˜Ex = 0 (6.148)
∂2
∂x2
˜Ey + ∂2
∂y2
˜Ey + ∂2
∂z2
˜Ey + β2 ˜Ey = 0 (6.149)
∂2
∂x2
˜Ez + ∂2
∂y2
˜Ez + ∂2
∂z2
˜Ez + β2 ˜Ez = 0 (6.150)
In
general, we expect the total field in the waveguide
to consist of unidirectional waves propagating in the
+ˆz and −ˆz directions. W e may analyze either of
these waves; then the other wave is easily derived via
symmetry , and the total field is simply a linear
combination (superposition) of these waves. With this
in mind, we limit our focus to the wave propagating
in the +ˆz direction.
In Section 6.7, it is shown that all components of the
electric and magnetic fields can be easily calculated
once ˜Ez and ˜Hz are known. The problem is further
simplified by decomposing the unidirectional wave
into TM and TE components. In this decomposition,
the TM component is defined by the property that
˜Hz = 0; i.e., is transverse (perpendicular) to the
direction of propagation. Thus, the TM component is",Electromagnetics_Vol2.pdf
"˜Hz = 0; i.e., is transverse (perpendicular) to the
direction of propagation. Thus, the TM component is
completely determined by ˜Ez.
Equations 6.136–6.139 simplify to become:
˜Ex = −jkz
k2ρ
∂˜Ez
∂x (6.151)
˜Ey = −jkz
k2ρ
∂˜Ez
∂y (6.152)
˜Hx =
+jωµ
k2ρ
∂˜Ez
∂y (6.153)
˜Hy = −jω
µ
k2ρ
∂˜Ez
∂x (6.154)
where
k2
ρ ≜ β2 − k2
z (6.155)
and kz is
the phase propagation constant; i.e., the
wave is assumed to propagate according to e−jkz z.
Now let us address the problem of finding ˜Ez, which
will then completely determine the TM field. As in
Section 6.7, we recognize that ˜Ez can be represented
as the propagation factor e−jkz z times a factor that
describes variation with respect to the remaining
spatial dimensions xand y:
˜Ez = ˜ez(x,y)e−jkz z (6.156)
Substitution of this expression into Equation 6.150
and dividing out the common factor of e−jkz z yields:
∂2
∂x2 ˜ez + ∂2
∂y2 ˜ez − k2
z˜ez + β2˜ez = 0 (6.157)
The
last two terms may be combined using
Equation 6.155, yielding:
∂2
∂x2 ˜ez + ∂2",Electromagnetics_Vol2.pdf
"∂2
∂x2 ˜ez + ∂2
∂y2 ˜ez − k2
z˜ez + β2˜ez = 0 (6.157)
The
last two terms may be combined using
Equation 6.155, yielding:
∂2
∂x2 ˜ez + ∂2
∂y2 ˜ez + k2
ρ˜ez = 0 (6.158)
This
is a partial differential equation for ˜ez in the
variables xand y. This equation may be solved using
the technique of separation of variables. In this
technique, we recognize that ˜ez(x,y) can be written
as the product of a function X(x) which depends only
on x, and a function Y(y) that depends only on y.
That is,
˜ez(x,y) = X(x)Y(y) (6.159)",Electromagnetics_Vol2.pdf
"112 CHAPTER 6. W A VEGUIDES
Substituting this expression into Equation 6.158, we
obtain:
Y ∂2
∂x2 X+ X ∂2
∂y2 Y + k2
ρXY = 0 (6.160)
Next dividing through by XY, we obtain:
1
X
∂2
∂x2 X+ 1
Y
∂2
∂y2 Y + k2
ρ = 0 (6.161)
Note
that the first term depends only on x, the second
term depends only on y, and the remaining term is a
constant. Therefore, the sum of the first and second
terms is a constant; namely −k2
ρ. Since these terms
depend on either xor y, and not both, the first term
must equal some constant, the second term must
equal some constant, and these constants must sum to
−k2
ρ. Therefore, we are justified in separating the
equation into two equations as follows:
1
X
∂2
∂x2 X+ k2
x = 0 (6.162)
1
Y
∂2
∂y2 Y + k2
y = 0 (6.163)
where
the new constants k2
x and k2
y must satisfy
k2
x + k2
y = k2
ρ (6.164)
Now multiplying Equations 6.162 and 6.163 by X
and Y, respectively , we find:
∂2
∂x2 X+ k2
xX = 0 (6.165)
∂2
∂y2 Y + k2
yY = 0 (6.166)
These
are familiar one-dimensional differential",Electromagnetics_Vol2.pdf
"and Y, respectively , we find:
∂2
∂x2 X+ k2
xX = 0 (6.165)
∂2
∂y2 Y + k2
yY = 0 (6.166)
These
are familiar one-dimensional differential
equations. The solutions are: 4
X = Acos (kxx) + Bsin (kxx) (6.167)
Y = Ccos (kyy) + Dsin (kyy) (6.168)
where A, B, C, and D– like kx and ky – are
constants to be determined. At this point, we observe
that the wave we seek can be expressed as follows:
˜Ez = ˜ez(x,y)e−jkz z
= X(x) Y(y) e−jkz z (6.169)
4 Students are encouraged to confirm that these are correct by
confirming that they are solutions to Equations 6.165 and 6.165, re-
spectively .
The solution is essentially complete except for the
values of the constants A, B, C, D, kx, and ky. The
values of these constants are determined by applying
the relevant electromagnetic boundary condition. In
this case, it is required that the component of ˜E that is
tangent to a perfectly-conducting wall must be zero.
Note that the ˆz component of ˜E is tangent to all four
walls; therefore:
˜Ez(x= 0) = 0 (6.170)",Electromagnetics_Vol2.pdf
"tangent to a perfectly-conducting wall must be zero.
Note that the ˆz component of ˜E is tangent to all four
walls; therefore:
˜Ez(x= 0) = 0 (6.170)
˜Ez(x= a) = 0 (6.171)
˜Ez(y= 0) = 0 (6.172)
˜Ez(y= b) = 0 (6.173)
Referring to Equation 6.169, these boundary
conditions in turn require:
X(x= 0) = 0 (6.174)
X(x= a) = 0 (6.175)
Y (y= 0) = 0 (6.176)
Y (y= b) = 0 (6.177)
Evaluating these conditions using Equations 6.167
and 6.168 yields:
A· 1 + B· 0 = 0 (6.178)
Acos (kxa) + Bsin (kxa) = 0 (6.179)
C· 1 + D· 0 = 0 (6.180)
Ccos (kyb) + Dsin (kyb) = 0 (6.181)
Equations 6.178 and 6.180 can be satisfied only if
A= 0 and C = 0, respectively . Subsequently ,
Equations 6.179 and 6.181 reduce to:
sin (kxa) = 0 (6.182)
sin (kyb) = 0 (6.183)
This in turn requires:
kx = mπ
a , m = 0,1,2... (6.184)
ky = nπ
b , n = 0,1,2... (6.185)
Each positive integer value of mand nleads to a
valid expression for ˜Ez known as a mode. Solutions
for which m= 0 or n= 0 yield kx = 0 or ky = 0,",Electromagnetics_Vol2.pdf
"Each positive integer value of mand nleads to a
valid expression for ˜Ez known as a mode. Solutions
for which m= 0 or n= 0 yield kx = 0 or ky = 0,
respectively . These correspond to zero-valued fields,
and therefore are not of interest. The most general",Electromagnetics_Vol2.pdf
"6.9. RECT ANGULAR W A VEGUIDE: TE MODES 113
expression for ˜Ez must account for all non-trivial
modes. Summarizing:
˜Ez =
∞∑
m=1
∞∑
n=1
˜E(m,n)
z (6.186)
where
˜E(m,n)
z ≜ E(m,n)
0 sin
(mπ
a x
)
sin
(nπ
b y
)
e−jk(m
,n )
z z
(6.187)
where E(m,n)
0 is an arbitrary constant (consolidating
the constants Band D), and, since
k2
x + k2
y = k2
ρ ≜ β2 − k2
z:
k(m,n)
z =
√
ω2µǫ−
(mπ
a
)2
−
(nπ
b
)2
(6.188)
Summarizing:
The TM ( ˜Hz = 0 ) component of the unidi-
rectional (+ ˆz-traveling) wave in a rectangular
waveguide is completely determined by Equa-
tion 6.186, and consists of modes as defined by
Equations 6.187, 6.188, 6.184, and 6.185. The
remaining non-zero field components can be de-
termined using Equations 6.151–6.154.
It is customary and convenient to refer to the TM
modes in a rectangular waveguide using the notation
“TM mn. ” For example, the mode TM 12 is given by
Equation 6.187 with m= 1 and n= 2.
Finally , note that values of k(m,n)
z obtained from",Electromagnetics_Vol2.pdf
"“TM mn. ” For example, the mode TM 12 is given by
Equation 6.187 with m= 1 and n= 2.
Finally , note that values of k(m,n)
z obtained from
Equation 6.188 are not necessarily real-valued. It is
apparent that for any given value of m, k(m,n)
z will be
imaginary-valued for all values of ngreater than some
value. Similarly , it is apparent that for any given value
of n, k(m,n)
z will be imaginary-valued for all values of
mgreater than some value. This phenomenon is
common to both TM and TE components, and so is
addressed in a separate section (Section 6.10).
Additional Reading:
• “W aveguide (radio frequency)” on Wikipedia.
• “Separation of variables” on Wikipedia.
6.9 Rectangular W aveguide: TE
Modes
[m0225]
A rectangular waveguide is a conducting cylinder of
rectangular
cross section used to guide the
propagation of waves. Rectangular waveguide is
commonly used for the transport of radio frequency
signals at frequencies in the SHF band (3–30 GHz)",Electromagnetics_Vol2.pdf
"propagation of waves. Rectangular waveguide is
commonly used for the transport of radio frequency
signals at frequencies in the SHF band (3–30 GHz)
and higher. The fields in a rectangular waveguide
consist of a number of propagating modes which
depends on the electrical dimensions of the
waveguide. These modes are broadly classified as
either transverse magnetic (TM) or transverse electric
(TE). In this section, we consider the TE modes.
Figure 6.7 shows the geometry of interest. Here the
walls are located at x= 0, x= a, y= 0, and y= b;
thus, the cross-sectional dimensions of the waveguide
are aand b. The interior of the waveguide is presumed
to consist of a lossless material exhibiting real-valued
permeability µand real-valued permittivity ǫ, and the
walls are assumed to be perfectly-conducting.
Let us limit our attention to a region within the
waveguide which is free of sources. Expressed in
phasor form, the magnetic field intensity within the",Electromagnetics_Vol2.pdf
"waveguide which is free of sources. Expressed in
phasor form, the magnetic field intensity within the
waveguide is governed by the wave equation:
∇2 ˜H + β2 ˜H = 0 (6.189)
where
β = ω√
µǫ (6.190)
Figure 6.7: Geometry for analysis of fields in a rect-
angular
waveguide.",Electromagnetics_Vol2.pdf
"114 CHAPTER 6. W A VEGUIDES
Equation 6.189 is a partial differential equation. This
equation, combined with boundary conditions
imposed by the perfectly-conducting plates, is
sufficient to determine a unique solution. This
solution is most easily determined in Cartesian
coordinates, as we shall now demonstrate. First we
express ˜H in Cartesian coordinates:
˜H = ˆx ˜Hx + ˆy ˜Hy + ˆz ˜Hz (6.191)
This facilitates the decomposition of Equation 6.189
into separate equations governing the ˆx, ˆy, and ˆz
components of ˜H:
∇2 ˜Hx + β2 ˜Hx = 0 (6.192)
∇2 ˜Hy + β2 ˜Hy = 0 (6.193)
∇2 ˜Hz + β2 ˜Hz = 0 (6.194)
Next we observe that the operator ∇2 may be
expressed in Cartesian coordinates as follows:
∇2 = ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2 (6.195)
so
the equations governing the Cartesian components
of ˜H may be written as follows:
∂2
∂x2
˜Hx + ∂2
∂y2
˜Hx + ∂2
∂z2
˜Hx + β2 ˜Hx = 0 (6.196)
∂2
∂x2
˜Hy + ∂2
∂y2
˜Hy + ∂2
∂z2
˜Hy + β2 ˜Hy = 0 (6.197)
∂2
∂x2
˜Hz + ∂2
∂y2
˜Hz + ∂2
∂z2
˜Hz + β2 ˜Hz = 0 (6.198)
In",Electromagnetics_Vol2.pdf
"∂z2
˜Hx + β2 ˜Hx = 0 (6.196)
∂2
∂x2
˜Hy + ∂2
∂y2
˜Hy + ∂2
∂z2
˜Hy + β2 ˜Hy = 0 (6.197)
∂2
∂x2
˜Hz + ∂2
∂y2
˜Hz + ∂2
∂z2
˜Hz + β2 ˜Hz = 0 (6.198)
In
general, we expect the total field in the waveguide
to consist of unidirectional waves propagating in the
+ˆz and −ˆz directions. W e may analyze either of
these waves; then the other wave is easily derived via
symmetry , and the total field is simply a linear
combination (superposition) of these waves. With this
in mind, we limit our focus to the wave propagating
in the +ˆz direction.
In Section 6.7, it is shown that all components of the
electric and magnetic fields can be easily calculated
once ˜Ez and ˜Hz are known. The problem is further
simplified by decomposing the unidirectional wave
into TM and TE components. In this decomposition,
the TE component is defined by the property that
˜Ez = 0; i.e., is transverse (perpendicular) to the
direction of propagation. Thus, the TE component is
completely determined by ˜Hz.",Electromagnetics_Vol2.pdf
"˜Ez = 0; i.e., is transverse (perpendicular) to the
direction of propagation. Thus, the TE component is
completely determined by ˜Hz.
Equations 6.136–6.139 simplify to become:
˜Ex = −jωµ
k2ρ
∂˜Hz
∂y (6.199)
˜Ey =
+jωµ
k2ρ
∂˜Hz
∂x (6.200)
˜Hx = −jkz
k2ρ
∂˜Hz
∂x (6.201)
˜Hy = −jkz
k2ρ
∂˜Hz
∂y (6.202)
where
k2
ρ ≜ β2 − k2
z (6.203)
and kz is
the phase propagation constant; i.e., the
wave is assumed to propagate according to e−jkz z.
Now let us address the problem of finding ˜Hz, which
will then completely determine the TE field. As in
Section 6.7, we recognize that ˜Hz can be represented
as the propagation factor e−jkz z times a factor that
describes variation with respect to the remaining
spatial dimensions xand y:
˜Hz = ˜hz(x,y)e−jkz z (6.204)
Substitution of this expression into Equation 6.198
and dividing out the common factor of e−jkz z yields:
∂2
∂x2
˜hz + ∂2
∂y2
˜hz − k2
z˜hz + β2˜hz = 0 (6.205)
The
last two terms may be combined using
Equation 6.203, yielding:
∂2
∂x2
˜hz + ∂2
∂y2",Electromagnetics_Vol2.pdf
"∂2
∂x2
˜hz + ∂2
∂y2
˜hz − k2
z˜hz + β2˜hz = 0 (6.205)
The
last two terms may be combined using
Equation 6.203, yielding:
∂2
∂x2
˜hz + ∂2
∂y2
˜hz + k2
ρ˜hz = 0 (6.206)
This
is a partial differential equation for ˜hz in the
variables xand y. This equation may be solved using
the technique of separation of variables. In this
technique, we recognize that ˜hz(x,y) can be written
as the product of a function X(x) which depends only
on x, and a function Y(y) that depends only on y.
That is,
˜hz(x,y) = X(x)Y(y) (6.207)",Electromagnetics_Vol2.pdf
"6.9. RECT ANGULAR W A VEGUIDE: TE MODES 115
Substituting this expression into Equation 6.206, we
obtain:
Y ∂2
∂x2 X+ X ∂2
∂y2 Y + k2
ρXY = 0 (6.208)
Next dividing through by XY, we obtain:
1
X
∂2
∂x2 X+ 1
Y
∂2
∂y2 Y + k2
ρ = 0 (6.209)
Note
that the first term depends only on x, the second
term depends only on y, and the remaining term is a
constant. Therefore, the sum of the first and second
terms is a constant; namely −k2
ρ. Since these terms
depend on either xor y, and not both, the first term
must equal some constant, the second term must
equal some constant, and these constants must sum to
−k2
ρ. Therefore, we are justified in separating the
equation into two equations as follows:
1
X
∂2
∂x2 X+ k2
x = 0 (6.210)
1
Y
∂2
∂y2 Y + k2
y = 0 (6.211)
where
the new constants k2
x and k2
y must satisfy
k2
x + k2
y = k2
ρ (6.212)
Now multiplying Equations 6.210 and 6.211 by X
and Y, respectively , we find:
∂2
∂x2 X+ k2
xX = 0 (6.213)
∂2
∂y2 Y + k2
yY = 0 (6.214)
These",Electromagnetics_Vol2.pdf
"ρ (6.212)
Now multiplying Equations 6.210 and 6.211 by X
and Y, respectively , we find:
∂2
∂x2 X+ k2
xX = 0 (6.213)
∂2
∂y2 Y + k2
yY = 0 (6.214)
These
are familiar one-dimensional differential
equations. The solutions are: 5
X = Acos (kxx) + Bsin (kxx) (6.215)
Y = Ccos (kyy) + Dsin (kyy) (6.216)
where A, B, C, and D– like kx and ky – are
constants to be determined. At this point, we observe
that the wave we seek can be expressed as follows:
˜Hz = ˜hz(x,y)e−jkz z
= X(x) Y(y) e−jkz z (6.217)
5 Students are encouraged to confirm that these are correct by
confirming that they are solutions to Equations 6.213 and 6.213, re-
spectively .
The solution is essentially complete except for the
values of the constants A, B, C, D, kx, and ky. The
values of these constants are determined by applying
the relevant electromagnetic boundary condition. In
this case, it is required that any component of ˜E that
is tangent to a perfectly-conducting wall must be
zero. Therefore:
˜Ey(x= 0) = 0 (6.218)",Electromagnetics_Vol2.pdf
"this case, it is required that any component of ˜E that
is tangent to a perfectly-conducting wall must be
zero. Therefore:
˜Ey(x= 0) = 0 (6.218)
˜Ey(x= a) = 0 (6.219)
˜Ex(y= 0) = 0 (6.220)
˜Ex(y= b) = 0 (6.221)
Referring to Equation 6.217 and employing
Equations 6.199–6.202, we obtain:
∂
∂xX(x= 0) = 0 (6.222)
∂
∂xX(x= a) = 0 (6.223)
∂
∂yY (y= 0) = 0 (6.224)
∂
∂yY (y= b) = 0 (6.225)
Evaluating the partial derivatives and dividing out
common factors of kx and ky, we find:
−Asin(kx · 0) + Bcos(kx · 0) = 0 (6.226)
−Asin(kx · a) + Bcos(kx · a) = 0 (6.227)
−Csin(ky · 0) + Dcos(ky · 0) = 0 (6.228)
−Csin(ky · b) + Dcos(ky · b) = 0 (6.229)
Evaluating:
−A· 0 + B· 1 = 0 (6.230)
−Asin (kxa) + Bcos (kxa) = 0 (6.231)
−C· 0 + D· 1 = 0 (6.232)
−Csin (kyb) + Dcos (kyb) = 0 (6.233)
Equations 6.230 and 6.232 can be satisfied only if
B = 0 and D= 0, respectively . Subsequently ,
Equations 6.231 and 6.233 reduce to:
sin (kxa) = 0 (6.234)
sin (kyb) = 0 (6.235)
This in turn requires:
kx = mπ",Electromagnetics_Vol2.pdf
"Equations 6.231 and 6.233 reduce to:
sin (kxa) = 0 (6.234)
sin (kyb) = 0 (6.235)
This in turn requires:
kx = mπ
a , m = 0,1,2... (6.236)
ky = nπ
b , n= 0,1,2... (6.237)",Electromagnetics_Vol2.pdf
"116 CHAPTER 6. W A VEGUIDES
Each positive integer value of mand nleads to a valid
expression for ˜Hz known as a mode. Summarizing:
˜Hz =
∞∑
m=0
∞∑
n=0
˜H(m,n)
z (6.238)
where
˜H(m,n)
z ≜ H(m,n)
0 cos
(mπ
a x
)
cos
(nπ
b y
)
e−jk(m
,n )
z z
(6.239)
where H(m,n)
0 is an arbitrary constant (consolidating
the constants Aand C), and, since
k2
x + k2
y = k2
ρ ≜ β2 − k2
z:
k(m,n)
z =
√
ω2µǫ−
(mπ
a
)2
−
(nπ
b
)2
(6.240)
Summarizing:
The TE ( ˜Ez = 0 ) component of the unidi-
rectional (+ ˆz-traveling) wave in a rectangular
waveguide is completely determined by Equa-
tion 6.238, and consists of modes as defined by
Equations 6.239, 6.240, 6.236, and 6.237. The
remaining non-zero field components can be de-
termined using Equations 6.199–6.202.
It is customary and convenient to refer to the TE
modes in a rectangular waveguide using the notation
“TE mn. ” For example, the mode TE 12 is given by
Equation 6.239 with m= 1 and n= 2.
Although Equation 6.238 implies the existence of a",Electromagnetics_Vol2.pdf
"“TE mn. ” For example, the mode TE 12 is given by
Equation 6.239 with m= 1 and n= 2.
Although Equation 6.238 implies the existence of a
TE00 mode, it should be noted that this wave has no
non-zero electric field components. This can be
determined mathematically by following the
procedure outlined above. However, this is also
readily confirmed as follows: ˜Ex is constant for the
TE00 mode because kρ = 0 for this mode, however,
˜Ex must be zero to meet the boundary conditions on
the walls at y= 0 and y= b. Similarly , ˜Ey is
constant for the TE 00 mode, however, ˜Ey must be
zero to meet the boundary conditions on the walls at
x= 0 and x= b.
Finally , note that values of k(m,n)
z obtained from
Equation 6.240 are not necessarily real-valued. It is
apparent that for any given value of m, k(m,n)
z will be
imaginary-valued for all values of ngreater than some
value. Similarly , it is apparent that for any given value
of n, k(m,n)
z will be imaginary-valued for all values of",Electromagnetics_Vol2.pdf
"value. Similarly , it is apparent that for any given value
of n, k(m,n)
z will be imaginary-valued for all values of
mgreater than some value. This phenomenon is
common to both TE and TM components, and so is
addressed in a separate section (Section 6.10).
Additional Reading:
• “W aveguide (radio frequency)” on Wikipedia.
• “Separation of variables” on Wikipedia.",Electromagnetics_Vol2.pdf
"6.10. RECT ANGULAR W A VEGUIDE: PROP AGA TION CHARACTERISTICS 117
6.10 Rectangular W aveguide:
Propagation Characteristics
[m0224]
In this section, we consider the propagation
characteristics
of TE and TM modes in rectangular
waveguides. Because these modes exhibit the same
phase dependence on z, findings of this section apply
equally to both sets of modes. Recall that the TM
modes in a rectangular waveguide are given by:
˜E(m,n)
z = E(m,n)
0 sin
(mπ
a x
)
sin
(nπ
b y
)
e−jk(m
,n )
z z
(6.241)
where E(m,n)
0 is an arbitrary constant (determined in
part by sources), and:
k(m,n)
z =
√
ω2µǫ−
(mπ
a
)2
−
(nπ
b
)2
(6.242)
The
TE modes in a rectangular waveguide are:
˜H(m,n)
z = H(m,n)
0 cos
(mπ
a x
)
cos
(nπ
b y
)
e−jk(m
,n )
z z
(6.243)
where H(m,n)
0 is an arbitrary constant (determined in
part by sources).
Cutoff frequency. First, observe that values of
k(m,n)
z obtained from Equation 6.242 are not
necessarily real-valued. For any given value of m,
(k(m,n)",Electromagnetics_Vol2.pdf
"k(m,n)
z obtained from Equation 6.242 are not
necessarily real-valued. For any given value of m,
(k(m,n)
z )2 will be negative for all values of ngreater
than some value. Similarly , for any given value of n,
(k(m,n)
z )2 will be negative for all values of mgreater
than some value. Should either of these conditions
occur, we find:
(
k(m,n)
z
)2
= ω2µǫ−
(mπ
a
)2
−
(nπ
b
)2
<0
= −
⏐⏐⏐
⏐ω2µǫ−
(mπ
a
)2
−
(nπ
b
)2⏐⏐⏐
⏐
= −α2 (6.244)
where αis
a positive real-valued constant. So:
k(m,n)
z = ±jα (6.245)
Subsequently:
e−jk(m,n )
z z = e−j(±jα) z
= e±αz (6.246)
The “+ ” sign option corresponds to a wave that grows
exponentially in magnitude with increasing z, which
is non-physical behavior. Therefore:
e−jk(m,n )
z z = e−αz (6.247)
Summarizing: When values of mor nare such that
(k(m,n)
z )2 <0, the magnitude of the associated wave
is no longer constant with z. Instead, the magnitude
of the wave decreases exponentially with increasing
z. Such a wave does not effectively convey power",Electromagnetics_Vol2.pdf
"of the wave decreases exponentially with increasing
z. Such a wave does not effectively convey power
through the waveguide, and is said to be cut off.
Since waveguides are normally intended for the
efficient transfer of power, it is important to know the
criteria for a mode to be cut off. Since cutoff occurs
when (k(m,n)
z )2 <0, cutoff occurs when:
ω2µǫ>
(mπ
a
)2
+
(nπ
b
)2
(6.248)
Since ω=
2πf:
f > 1
2π√µǫ
√ (mπ
a
)2
+
(nπ
b
)2
(6.249)
= 1
2√µǫ
√ (m
a
)2
+
(n
b
)2
(6.250)
Note
that 1/√µǫis the phase velocity vp for the
medium used in the waveguide. With this in mind, let
us define:
vpu ≜ 1√µǫ (6.251)
This
is the phase velocity in an unbounded medium
having the same permeability and permittivity as the
interior of the waveguide. Thus:
f >vpu
2
√ (m
a
)2
+
(n
b
)2
(6.252)
In
other words, the mode (m,n) avoids being cut off
if the frequency is high enough to meet this criterion.
Thus, it is useful to make the following definition:
fmn ≜ vpu
2
√ (m
a
)2
+
(n
b
)2
(6.253)",Electromagnetics_Vol2.pdf
"if the frequency is high enough to meet this criterion.
Thus, it is useful to make the following definition:
fmn ≜ vpu
2
√ (m
a
)2
+
(n
b
)2
(6.253)
The cutoff frequency fm n (Equation 6.253) is the
lowest frequency for which the mode (m,n) is
able to propagate (i.e., not cut off).",Electromagnetics_Vol2.pdf
"118 CHAPTER 6. W A VEGUIDES
Example 6.4. Cutof f frequencies for WR-90.
WR-90 is a popular implementation of
rectangular waveguide. WR-90 is air-filled with
dimensions a= 22.86 mm and b= 10.16 mm.
Determine cutoff frequencies and, in particular,
the lowest frequency at which WR-90 can be
used.
Solution. Since WR-90 is air-filled, µ≈ µ0,
ǫ≈ ǫ0, and vpu ≈ 1√
µ0ǫ0 ∼= 3.00 × 108 m/s.
Cutof
f frequencies are given by Equation 6.253.
Recall that there are no non-zero TE or TM
modes with m= 0 and n= 0. Since a>b, the
lowest non-zero cutoff frequency is achieved
when m= 1 and n= 0. In this case,
Equation 6.253 yields f10 = 6.557 GHz
; this is
the
lowest frequency that is able to propagate
efficiently in the waveguide. The next lowest
cutoff frequency is f20 = 13.114 GHz. The third
lowest cutoff frequency is f01 = 14.754 GHz.
The lowest-order TM mode that is non-zero and
not cut off is TM 11 (f11 = 16.145 GHz).
Phase velocity. The phase velocity for a wave",Electromagnetics_Vol2.pdf
"The lowest-order TM mode that is non-zero and
not cut off is TM 11 (f11 = 16.145 GHz).
Phase velocity. The phase velocity for a wave
propagating within a rectangular waveguide is greater
than that of electromagnetic radiation in unbounded
space. For example, the phase velocity of any
propagating mode in a vacuum-filled waveguide is
greater than c, the speed of light in free space. This is
a surprising result. Let us first derive this result and
then attempt to make sense of it.
Phase velocity vp in the rectangular waveguide is
given by
vp ≜ ω
k(m,n )
z
(6.254)
= ω√
ω2µǫ− ( mπ/a)2 − (nπ/b)2
(6.255)
Immediately we observe that phase velocity seems to
be different for different modes. Dividing the
numerator and denominator by β = ω√µǫ, we obtain:
vp = 1√µǫ
1√
1 − (ω2µǫ)−1
[
( mπ/a)2 + (nπ/b)2
]
(6.256)
Note that 1/√µǫis vpu, as defined earlier. Employing
Equation 6.253 and also noting that ω= 2πf,
Equation 6.256 may be rewritten in the following
form:
vp = vpu√
1 − (fmn/f)2
(6.257)
F",Electromagnetics_Vol2.pdf
"Equation 6.253 and also noting that ω= 2πf,
Equation 6.256 may be rewritten in the following
form:
vp = vpu√
1 − (fmn/f)2
(6.257)
F
or any propagating mode, f >fmn; subsequently ,
vp >vpu. In particular, vp >c for a vacuum-filled
waveguide.
How can this not be a violation of fundamental
physics? As noted in Section 6.1, phase velocity is
not necessarily the speed at which information
travels, but is merely the speed at which a point of
constant phase travels. T o send information, we must
create a disturbance in the otherwise sinusoidal
excitation presumed in the analysis so far. The
complex field structure creates points of constant
phase that travel faster than the disturbance is able to
convey information, so there is no violation of
physical principles.
Group velocity. As noted in Section 6.1, the speed at
which information travels is given by the group
velocity vg. In unbounded space, vg = vp, so the
speed of information is equal to the phase velocity in",Electromagnetics_Vol2.pdf
"velocity vg. In unbounded space, vg = vp, so the
speed of information is equal to the phase velocity in
that case. In a rectangular waveguide, the situation is
different. W e find:
vg =
(
∂k(m,n)
z
∂ω
)−1
(6.258)
= vpu
√
1 − (fmn/f)2 (6.259)
which
is always less than vpu for a propagating mode.
Note that group velocity in the waveguide depends on
frequency in two ways. First, because fmn takes on
different values for different modes, group velocity is
different for different modes. Specifically ,
higher-order modes propagate more slowly than
lower-order modes having the same frequency . This is
known as modal dispersion. Secondly , note that the
group velocity of any given mode depends on
frequency . This is known as chromatic dispersion.",Electromagnetics_Vol2.pdf
"6.10. RECT ANGULAR W A VEGUIDE: PROP AGA TION CHARACTERISTICS 119
The speed of a signal within a rectangular waveg-
uide
is given by the group velocity of the as-
sociated mode (Equation 6.259). This speed is
less than the speed of propagation in unbounded
media having the same permittivity and perme-
ability . Speed depends on the ratio fmn/f, and
generally decreases with increasing frequency for
any given mode.
Example 6.5. Speed of propagating in WR-90.
Revisiting WR-90 waveguide from Example 6.4:
What is the speed of propagation for a
narrowband signal at 10 GHz?
Solution. Let us assume that “narrowband” here
means that the bandwidth is negligible relative to
the center frequency , so that we need only
consider the center frequency . As previously
determined, the lowest-order propagating mode
is TE 10, for which f10 = 6.557 GHz. The
next-lowest cutoff frequency is
f20 = 13.114 GHz. Therefore, only the TE 10
mode is available for this signal. The group",Electromagnetics_Vol2.pdf
"next-lowest cutoff frequency is
f20 = 13.114 GHz. Therefore, only the TE 10
mode is available for this signal. The group
velocity for this mode at the frequency of
interest is given by Equation 6.259. Using this
equation, the speed of propagation is found to be
∼= 2.26 × 108 m/s
, which is about 75.5% of c.
[m0212]",Electromagnetics_Vol2.pdf
"120 CHAPTER 6. W A VEGUIDES
Image Credits
Fig. 6.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Geometry for analysis of fields in parallel plate waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified.
Fig. 6.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TE component of electric field in parallel plate waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 6.3: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Magnitude of the two lowest-order TE modes in parallel plate
waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 6.4: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM component of electric field in parallel plate waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 6.5: c⃝ Sevenchw (C. W ang),",Electromagnetics_Vol2.pdf
"CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 6.5: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM component of electric field in parallel plate waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified.",Electromagnetics_Vol2.pdf
"Chapter 7
T ransmission
Lines Redux
7.1 Parallel Wire T ransmission
Line
[m0188]
A parallel wire transmission line consists of wires
separated
by a dielectric spacer. Figure 7.1 shows a
common implementation, commonly known as “twin
lead. ” The wires in twin lead line are held in place by
a mechanical spacer comprised of the same low-loss
dielectric material that forms the jacket of each wire.
V ery little of the total energy associated with the
electric and magnetic fields lies inside this material,
so the jacket and spacer can usually be neglected for
the purposes of analysis and electrical design.
Parallel wire transmission line is often employed in
radio applications up to about 100 MHz as an
alternative to coaxial line. Parallel wire line has the
advantages of lower cost and lower loss than coaxial
line in this frequency range. However, parallel wire
line lacks the self-shielding property of coaxial cable;
i.e., the electromagnetic fields of coaxial line are",Electromagnetics_Vol2.pdf
"line lacks the self-shielding property of coaxial cable;
i.e., the electromagnetic fields of coaxial line are
isolated by the outer conductor, whereas those of
c⃝ Sp inningSpark, Inductiveload CC BY SA 3.0 (modified)
Figure 7.1: T win lead, a commonly-encountered form
of parallel wire transmission line.
c⃝ C. W ang CC BY SA 4.0
Figure 7.2: Parallel wire transmission line structure
and design parameters.
parallel wire line are exposed and prone to interaction
with nearby structures and devices. This prevents the
use of parallel wire line in many applications.
Another discriminator between parallel wire line and
coaxial line is that parallel wire line is differential. 1
The conductor geometry is symmetric and neither
conductor is favored as a signal datum (“ground”).
Thus, parallel wire line is commonly used in
applications where the signal sources and/or loads are
also differential; common examples are the dipole
antenna and differential amplifiers. 2",Electromagnetics_Vol2.pdf
"applications where the signal sources and/or loads are
also differential; common examples are the dipole
antenna and differential amplifiers. 2
Figure 7.2 shows a cross-section of parallel wire line.
Relevant parameters include the wire diameter, d; and
the center-to-center spacing, D.
1 The references in “ Additional Reading” at the end of this section
may be helpful if you are not familiar with this concept.
2 This is in contrast to “single-ended” line such as coaxial line,
which has conductors of different cross-sections and the outer con-
ductor is favored as the datum.
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"122 CHAPTER 7. TRANSMISSION LINES REDUX
c⃝ S. Lally CC BY SA 4.0
Figure 7.3: Structure of the electric and magnetic
fields for a cross-section of parallel wire line. In this
case, the wave is propagating away from the viewer.
The associated field structure is transverse
electromagnetic (TEM) and is therefore completely
described by a single cross-section along the line, as
shown in Figure 7.3. Expressions for these fields
exist, but are complex and not particularly useful
except as a means to calculate other parameters of
interest. One of these parameters is, of course, the
characteristic impedance since this parameter plays
an important role in the analysis and design of
systems employing transmission lines. The
characteristic impedance may be determined using
the “lumped element” transmission line model using
the following expression:
Z0 =
√
R′ + jωL′
G′ + jωC′ (7.1)
where R′, G′, C′,
and L′ are the resistance,
conductance, capacitance, and inductance per unit",Electromagnetics_Vol2.pdf
"Z0 =
√
R′ + jωL′
G′ + jωC′ (7.1)
where R′, G′, C′,
and L′ are the resistance,
conductance, capacitance, and inductance per unit
length, respectively . This analysis is considerably
simplified by neglecting loss; therefore, let us assume
the “low-loss” conditions R′ ≪ ωL′ and G′ ≪ ωC′.
Then we find:
Z0 ≈
√
L′
C′ (lo w loss) (7.2)
and the problem is reduced to determining inductance
and capacitance of the transmission line. These are
L′ = µ0
π ln
[
( D/d) +
√
(D/d)2 − 1
]
(7.3)
C′ = π
ǫ
ln
[
(D/d) +
√
(D/d)2 − 1
] (7.4)
Because
the wire separation Dis typically much
greater than the wire diameter d, D/d≫ 1 and so√
(D/d)2 − 1 ≈ D /d. This leads to the simplified
expressions
L′ ≈ µ0
π ln (2D/d) (D ≫ d) (7.5)
C′ ≈ πǫ
ln (2D/d) ( D≫ d) (7.6)
Now returning to Equation 7.2:
Z0 ≈ 1
π
√ µ0
ǫ ln (2D/d) (7.7)
Noting that ǫ= ǫrǫ0 and
√
µ0/ǫ0 ≜ η0, we obtain
Z0 ≈ 1
π
η0
√ǫr
ln (2D/d) (7.8)
The characteristic impedance of parallel wire
line,
assuming low-loss conditions and wire spac-",Electromagnetics_Vol2.pdf
"µ0/ǫ0 ≜ η0, we obtain
Z0 ≈ 1
π
η0
√ǫr
ln (2D/d) (7.8)
The characteristic impedance of parallel wire
line,
assuming low-loss conditions and wire spac-
ing much greater than wire diameter, is given by
Equation 7.8.
Observe that the characteristic impedance of parallel
wire line increases with increasing D/d. Since this
ratio is large, the characteristic impedance of parallel
wire line tends to be large relative to common values
of other kinds of TEM transmission line, such as
coaxial line and microstrip line. An example follows.
Example 7.1. 300 Ω twin-lead.
A commonly-encountered form of parallel wire
transmission line is 300 Ω twin-lead. Although
implementations vary , the wire diameter is
usually about 1 mm and and the wire spacing is
usually about 6 mm. The relative permittivity of
the medium ǫr ≈ 1 for the purposes of
calculating transmission line parameters, since
the jacket and spacer have only a small effect on
the fields. For these values, Equation 7.8 gives",Electromagnetics_Vol2.pdf
"calculating transmission line parameters, since
the jacket and spacer have only a small effect on
the fields. For these values, Equation 7.8 gives
Z0 ≈ 298 Ω, as expected.
Under the assumption that the wire jacket/spacer
material has a negligible effect on the electromagnetic",Electromagnetics_Vol2.pdf
"7.2. MICROSTRIP LINE REDUX 123
fields, and that the line is suspended in air so that
ǫr ≈ 1, the phase velocity vp for a parallel wire line is
approximately that of any electromagnetic wave in
free space; i.e., c. In practical twin-lead, the effect of
a plastic jacket/spacer material is to reduce the phase
velocity by a few percent up to about 20%, depending
on the materials and details of construction. So in
practice vp ≈ 0.8cto 0.9cfor twin-lead line.
Additional Reading:
• “T win-lead” on Wikipedia.
• “Differential signaling” on Wikipedia.
• Sec. 8.7 (“Differential Circuits”) in
S.W . Ellingson, Radio Systems Engineering,
Cambridge Univ . Press, 2016.
7.2 Microstrip Line Redux
[m0186]
A microstrip transmission line consists of a narrow
metallic
trace separated from a metallic ground plane
by a slab of dielectric material, as shown in
Figure 7.4. This is a natural way to implement a
transmission line on a printed circuit board, and so
accounts for an important and expansive range of",Electromagnetics_Vol2.pdf
"transmission line on a printed circuit board, and so
accounts for an important and expansive range of
applications. The reader should be aware that
microstrip is distinct from stripline, which is a
distinct type of transmission line; see “ Additional
Reading” at the end of this section for disambiguation
of these terms.
A microstrip line is single-ended 3 in the sense that the
conductor geometry is asymmetric and the one
conductor – namely , the ground plane – also normally
serves as ground for the source and load.
The spacer material is typically a low-loss dielectric
material having permeability approximately equal to
that of free space (µ ≈ µ0) and relative permittivity ǫr
in the range of 2 to about 10 or so.
The structure of a microstrip line is similar to that of a
parallel plate waveguide (Section 6.6), with the
obvious difference that one of the “plates” has finite
length in the direction perpendicular to the direction
of propagation. Despite this difference, the parallel",Electromagnetics_Vol2.pdf
"length in the direction perpendicular to the direction
of propagation. Despite this difference, the parallel
plate waveguide provides some useful insight into the
operation of the microstrip line. Microstrip line is
nearly always operated below the cutoff frequency of
3 The reference in “ Additional Reading” at the end of this section
may be helpful if you are not familiar with this concept.
hW
l


r
t


0
c⃝ 7h ead7metal7 CC BY SA 3.0 (modified)
Figure 7.4: Microstrip transmission line structure and
design parameters.",Electromagnetics_Vol2.pdf
"124 CHAPTER 7. TRANSMISSION LINES REDUX
Figure 7.5: Appr oximate structure of the electric and
magnetic fields within microstrip line, assuming TM 0
operation. The fields outside the line are possibly sig-
nificant, complicated, and not shown. In this case, the
wave is propagating away from the viewer.
all but the TM 0 mode, whose cutoff frequency is zero.
This guarantees that only the TM 0 mode can
propagate. The TM 0 mode has the form of a uniform
plane wave, as shown in Figure 7.5. The electric field
is oriented in the direction perpendicular to the plates,
and magnetic field is oriented in the direction parallel
to the plates. The direction of propagation E × H for
the TM 0 mode always points in the same direction;
namely , along the axis of the transmission line.
Therefore, microstrip lines nominally exhibit
transverse electromagnetic (TEM) field structure.
The limited width W of the trace results in a “fringing
field” – i.e., significant deviations from TM 0 field",Electromagnetics_Vol2.pdf
"The limited width W of the trace results in a “fringing
field” – i.e., significant deviations from TM 0 field
structure in the dielectric beyond the edges of the
trace and above the trace. The fringing fields may
play a significant role in determining the
characteristic impedance Z0. Since Z0 is an important
parameter in the analysis and design of systems using
transmission lines, we are motivated to not only
determine Z0 for the microstrip line, but also to
understand how variation in W affects Z0. Let us
address this issue by considering the following three
special cases, in order:
• W ≫ h, which we shall refer to as the “wide”
case.
• W ≪ h, which we shall refer to as the “narrow”
case.
• W ∼ h; i.e., the intermediate case in which h
and W equal to within an order of magnitude or
so.
Wide case. If W ≫ h, most of the energy associated
with propagating waves lies directly underneath the
trace, and Figure 7.5 provides a relatively accurate
impression of the fields in this region. The",Electromagnetics_Vol2.pdf
"with propagating waves lies directly underneath the
trace, and Figure 7.5 provides a relatively accurate
impression of the fields in this region. The
characteristic impedance Z0 may be determined using
the “lumped element” transmission line model using
the following expression:
Z0 =
√
R′ + jωL′
G′ + jωC′ (7.9)
where R′, G′, C′,
and L′ are the resistance,
conductance, capacitance, and inductance per unit
length, respectively . For the present analysis, nothing
is lost by assuming loss is negligible; therefore, let us
assume R′ ≪ ωL′ and G′ ≪ ωG′, yielding
Z0 ≈
√
L′
C′ (7.10)
Thus
the problem is reduced to determining
capacitance and inductance of the transmission line.
Wide microstrip line resembles a parallel-plate
capacitor whose plate spacing dis very small
compared to the plate area A. In this case, fringing
fields may be considered negligible and one finds that
the capacitance Cis given by
C ≈ ǫA
d (parallel plate capacitor) (7.11)
In terms of wide microstrip line, A= Wl where lis",Electromagnetics_Vol2.pdf
"the capacitance Cis given by
C ≈ ǫA
d (parallel plate capacitor) (7.11)
In terms of wide microstrip line, A= Wl where lis
length, and d= h. Therefore:
C′ ≜ C
l ≈ ǫW
h (W ≫ h) (7.12)
T
o determine L′, consider the view of the microstrip
line shown in Figure 7.6. Here a current source
applies a steady current I on the left, and a resistive
load closes the current loop on the right. Ampere’s
law for magnetostatics and the associated “right hand
rule” require that H is directed into the page and is
approximately uniform. The magnetic flux between
the trace and the ground plane is
Φ =
∫
S
B · ds (7.13)
where S is the surface bounded by the current loop
and B = µ0H. In the present problem, the area of S
is hland H is approximately constant over S, so:
Φ ≈ µ0Hhl (7.14)",Electromagnetics_Vol2.pdf
"7.2. MICROSTRIP LINE REDUX 125
c⃝ C. W ang CC BY SA 4.0
Figure 7.6: V iew from the side of a microstrip line,
used to determine L′.
where H is the magnitude of H. Next, recall that:
L≜ Φ
I (7.15)
So,
we may determine Lif we are able to obtain an
expression for I in terms of H. This can be done as
follows. First, note that we can express I in terms of
the current density on the trace; i.e.,
I = JsW (7.16)
where Js is the current density (SI base units of A/m)
on the trace. Next, we note that the boundary
condition for H on the trace requires that the
discontinuity in the tangent component of H going
from the trace (where the magnetic field intensity is
H) to beyond the trace (i.e., outside the transmission
line, where the magnetic field intensity is
approximately zero) be equal to the surface current.
Thus, Js ≈ H and
I = JsW ≈ HW (7.17)
Returning to Equation 7.15, we find:
L≜ Φ
I ≈ µ0Hhl
HW = µ0hl
W (7.18)
Subsequently
,
L′ ≜ L
l ≈ µ0h
W (W ≫ h) (7.19)
No",Electromagnetics_Vol2.pdf
"I = JsW ≈ HW (7.17)
Returning to Equation 7.15, we find:
L≜ Φ
I ≈ µ0Hhl
HW = µ0hl
W (7.18)
Subsequently
,
L′ ≜ L
l ≈ µ0h
W (W ≫ h) (7.19)
No
w the characteristic impedance is found to be
Z0 ≈
√
L′
C′ ≈
√
µ0h/W
ǫW/h =
√ µ0
ǫ
h
W (7.20)
The
factor
√
µ0/ǫis recognized as the wave
impedance η. It is convenient to express this in terms
c⃝ S. Lally CC BY SA 4.0
Figure 7.7: Electric and magnetic fields lines for nar-
row (W ≪ h) microstrip line.
of the free space wave impedance η0 ≜
√
µ0/ǫ0,
leading
to the expression we seek:
Z0 ≈ η0
√ǫr
h
W (W ≫ h) (7.21)
The characteristic impedance of “wide” (W ≫
h) microstrip
line is given approximately by
Equation 7.21.
It is worth noting that the characteristic impedance of
wide transmission line is proportional to h/W. The
factor h/W figures prominently in the narrow- and
intermediate-width cases as well, as we shall soon
see.
Narrow case. Figure 7.5 does not accurately depict
the fields in the case that W ≪ h. Instead, much of",Electromagnetics_Vol2.pdf
"see.
Narrow case. Figure 7.5 does not accurately depict
the fields in the case that W ≪ h. Instead, much of
the energy associated with the electric and magnetic
fields lies beyond and above the trace. Although these
fields are relatively complex, we can develop a
rudimentary model of the field structure by
considering the relevant boundary conditions. In
particular, the tangential component of E must be
zero on the surfaces of the trace and ground plane.
Also, we expect magnetic field lines to be
perpendicular to the electric field lines so that the
direction of propagation is along the axis of the line.
These insights lead to Figure 7.7.
Note that the field lines shown in Figure 7.7 are quite
similar to those for parallel wire line (Section 7.1). If
we choose diameter d= W and wire spacing
D= 2h, then the fields of the microstrip line in the
dielectric spacer must be similar to those of the
parallel wire line between the lines. This is shown in",Electromagnetics_Vol2.pdf
"126 CHAPTER 7. TRANSMISSION LINES REDUX
Figure 7.8: Noting the similarity of the fields in nar-
ro
w microstrip line to those in a parallel wire line.
Figure 7.9: Modeling the fields in narrow microstrip
line
as those of a parallel wire line, now introducing
the ground plane.
Figure 7.8. Note that the fields above the dielectric
spacer in the microstrip line will typically be
somewhat different (since the material above the trace
is typically different; e.g., air). However, it remains
true that we expect most of the energy to be contained
in the dielectric, so we shall assume that this
difference has a relatively small effect.
Next, we continue to refine the model by introducing
the ground plane, as shown in Figure 7.9: The fields
in the upper half-space are not perturbed when we do
this, since the fields in Figure 7.8 already satisfy the
boundary conditions imposed by the new ground
plane. Thus, we see that the parallel wire transmission
line provides a pretty good guide to the structure of",Electromagnetics_Vol2.pdf
"plane. Thus, we see that the parallel wire transmission
line provides a pretty good guide to the structure of
the fields for the narrow transmission line, at least in
the dielectric region of the upper half-space.
W e are now ready to estimate the characteristic
impedance Z0 ≈
√
L′/C′ of low-loss narrow
microstrip line. Neglecting the differences noted
above, we estimate that this is roughly equal to the
characteristic impedance of the analogous (d = W
and D= 2h) parallel wire line. Under this
assumption, we obtain from Equation 7.8:
Z0 ∼ 1
π
η0
√ǫr
ln (4h/W) (W ≪ h) (7.22)
This
estimate will exhibit only “order of magnitude”
accuracy (hence the “∼ ” symbol), since the
contribution from the fields in half of the relevant
space is ignored. However we expect that
Equation 7.22 will accurately capture the dependence
of Z0 on the parameters ǫr and h/W. These
properties can be useful for making fine adjustments
to a microstrip design.
The characteristic impedance of “narrow” (W ≪",Electromagnetics_Vol2.pdf
"properties can be useful for making fine adjustments
to a microstrip design.
The characteristic impedance of “narrow” (W ≪
h) microstrip
line can be roughly approximated
by Equation 7.22.
Intermediate case. An expression for Z0 in the
intermediate case – even a rough estimate – is difficult
to derive, and beyond the scope of this book.
However, we are not completely in the dark, since we
can reasonably anticipate that Z0 in the intermediate
case should be intermediate to estimates provided by
the wide and narrow cases. This is best demonstrated
by example. Figure 7.10 shows estimates of Z0 using",Electromagnetics_Vol2.pdf
"7.2. MICROSTRIP LINE REDUX 127
Figure 7.10: Z0 for FR4 as a function of h/W, as
determined by the “wide” and “narrow” approxima-
tions, along with the Wheeler 1977 formula. Note that
the vertical and horizontal axes of this plot are in log
scale.
the wide and narrow expressions (blue and green
curves, respectively) for a particular implementation
of microstrip line (see Example 7.2 for details). Since
Z0 varies smoothly with h/W, it is reasonable to
expect that simply averaging the values generated
using the wide and narrow assumptions would give a
pretty good estimate when h/W ∼ 1.
However, it is also possible to obtain an accurate
estimate directly . A widely-accepted and
broadly-applicable formula is provided in Wheeler
1977 (cited in “ Additional Reading” at the end of this
section). This expression is valid over the full range
of h/W, not merely the intermediate case of
h/W ∼ 1. Here it is:
Z0 ≈ 42.4 Ω
√ǫr + 1 ×
ln
[
1
+ 4h
W′
(
K+
√
K2 + 1 + 1/ǫr
2 π2
)]
(7.23)
where
K ≜ 14",Electromagnetics_Vol2.pdf
"h/W ∼ 1. Here it is:
Z0 ≈ 42.4 Ω
√ǫr + 1 ×
ln
[
1
+ 4h
W′
(
K+
√
K2 + 1 + 1/ǫr
2 π2
)]
(7.23)
where
K ≜ 14
+ 8/ǫr
11
(4h
W′
)
(7.24)
and W′ is W adjusted
to account for the thickness t
of the microstrip line. T ypically t≪ W and t≪ h,
for which W′ ≈ W. Although complicated, this
formula should not be completely surprising. For
example, note the parameters hand W appear in this
formula as the factor 4h/W, which we encountered in
the “narrow” approximation. Also, we see that the
formula indicates Z0 increases with increasing h/W
and decreasing ǫr, as we determined in both the
“narrow” and “wide” cases.
The following example provides a demonstration for
the very common case of microstrip implementation
in FR4 circuit board material.
Example 7.2. Characteristic impedance of
microstrip lines in FR4.
FR4 printed circuit boards consist of a substrate
having h∼= 1.575 mm and ǫr ≈ 4.5. Figure 7.10
shows Z0 as a function of h/W, as determined
by the wide and narrow approximations, along",Electromagnetics_Vol2.pdf
"having h∼= 1.575 mm and ǫr ≈ 4.5. Figure 7.10
shows Z0 as a function of h/W, as determined
by the wide and narrow approximations, along
with the value obtained using the Wheeler 1977
formula. The left side of the plot represents the
wide condition, whereas the right side of this
plot represents the narrow condition. The
Wheeler 1977 formula is an accurate estimate
over the entire horizontal span. Note that the
wide approximation is accurate only for
h/W <0.1 or so, improving with decreasing
h/W as expected. The narrow approximation
overestimates Z0 by a factor of about 1.4 (i.e.,
about 40%). This is consistent with our previous
observation that this approximation should yield
only about “order of magnitude” accuracy .
Nevertheless, the narrow approximation exhibits
approximately the same rate of increase with
h/W as does the Wheeler 1977 formula.
Also worth noting from Figure 7.10 is the
important and commonly-used result that
Z0 ≈ 50 Ω is obtained for h/W ≈ 0.5. Thus, a",Electromagnetics_Vol2.pdf
"Also worth noting from Figure 7.10 is the
important and commonly-used result that
Z0 ≈ 50 Ω is obtained for h/W ≈ 0.5. Thus, a
50 Ω microstrip line in FR4 has a trace width of
about 3 mm.
A useful “take away” from this example is that the
wide and narrow approximations serve as useful
guides for understanding how Z0 changes as a
function of h/W and ǫr. This is useful especially for
making adjustments to a microstrip line design once
an accurate value of Z0 is obtained from some other",Electromagnetics_Vol2.pdf
"128 CHAPTER 7. TRANSMISSION LINES REDUX
method; e.g., using the Wheeler 1977 formula or from
measurements.
FR4 circuit board construction is so common that the
result from the previous example deserves to be
highlighted:
In FR4 printed circuit board construction (sub-
strate
thickness 1.575 mm, relative permittivity
≈ 4.5, negligible trace thickness), Z0 ≈ 50 Ω
requires a trace width of about 3 mm. Z0 scales
roughly in proportion to h/W around this value.
Simpler approximations for Z0 are also commonly
employed in the design and analysis of microstrip
lines. These expressions are limited in the range of
h/W for which they are valid, and can usually be
shown to be special cases or approximations of
Equation 7.23. Nevertheless, they are sometimes
useful for quick “back of the envelope” calculations.
W avelength in microstrip line. An accurate general
formula for wavelength λin microstrip line is
similarly difficult to derive. A useful approximate",Electromagnetics_Vol2.pdf
"W avelength in microstrip line. An accurate general
formula for wavelength λin microstrip line is
similarly difficult to derive. A useful approximate
technique employs a result from the theory of uniform
plane waves in unbounded media. For such waves,
the phase propagation constant βis given by
β = ω√
µǫ (7.25)
It
turns out that the electromagnetic field structure for
the guided wave in a microstrip line is similar in the
sense that it exhibits TEM field structure, as does the
uniform plane wave. However, the guided wave is
different in that it exists in both the dielectric spacer
and the air above the spacer. Thus, we presume that β
for the guided wave can be approximated as that of a
uniform plane wave in unbounded media having the
same permeability µ0 but a different relative
permittivity , which we shall assign the symbol ǫr,eff
(for “effective relative permittivity”). Then:
β ≈ ω√
µ0 ǫr,ef f ǫ0 (low-loss microstrip)
= β0
√ǫr,ef f (7.26)",Electromagnetics_Vol2.pdf
"(for “effective relative permittivity”). Then:
β ≈ ω√
µ0 ǫr,ef f ǫ0 (low-loss microstrip)
= β0
√ǫr,ef f (7.26)
In other words, the phase propagation constant in a
microstrip line can be approximated as the free-space
phase propagation β0 ≜ ω√µ0ǫ0 times a correction
f
actor √ǫr,ef f.
Next, ǫr,eff is crudely approximated as the average of
the relative permittivity of the dielectric slab and the
relative permittivity of free space; i.e.,:
ǫr,eff ≈ ǫr + 1
2 (7.27)
V
arious refinements exist to improve on this
approximation; however, in practice, variations in the
value of ǫr for the dielectric due to manufacturing
processes typically make a more precise estimate
irrelevant.
Using this concept, we obtain
λ= 2π
β = 2π
β0
√ǫr,ef f
= λ0
√ǫr,ef f
(7.28)
where λ0 is the free-space wavelength c/f. Similarly
the phase velocity vp can be estimated using the
relationship
vp = ω
β = c√ǫr,ef f
(7.29)
i.e., the phase velocity in microstrip is slower than c
by a factor of √ǫr,ef f.",Electromagnetics_Vol2.pdf
"relationship
vp = ω
β = c√ǫr,ef f
(7.29)
i.e., the phase velocity in microstrip is slower than c
by a factor of √ǫr,ef f.
Example 7.3. W a velength and phase velocity in
microstrip in FR4 printed circuit boards.
FR4 is a low-loss fiberglass epoxy dielectric that
is commonly used to make printed circuit boards
(see “ Additional Reading” at the end of this
section). For FR4, ǫr ≈ 4.5. The effective
relative permittivity is therefore:
ǫr,eff ≈ (4.5 + 1)/2 = 2.75
Thus, we estimate the phase velocity for the
wave guided by this line to be about c/
√
2.75;
i.e.,
60% of c. Similarly , the wavelength of this
wave is about 60% of the free space wavelength.
In practice, these values are found to be slightly
less; typically 50%–55%. The difference is
attributable to the crude approximation of
Equation 7.27.
Additional Reading:
• “Microstrip” on Wikipedia.",Electromagnetics_Vol2.pdf
"7.3. A TTENUA TION IN COAXIAL CABLE 129
• “Printed circuit board” on Wikipedia.
• “Stripline” on Wikipedia.
• “Single-ended signaling” on Wikipedia.
• Sec. 8.7 (“Differential Circuits”) in
S.W . Ellingson, Radio Systems Engineering,
Cambridge Univ . Press, 2016.
• H.A. Wheeler, “Transmission Line Properties of
a Strip on a Dielectric Sheet on a Plane, ” IEEE
Trans. Microwave Theory & T echniques, V ol. 25,
No. 8, Aug 1977, pp. 631–47.
• “FR-4” on Wikipedia.
7.3 Attenuation in Coaxial Cable
[m0189]
In this section, we consider the issue of attenuation in
coaxial
transmission line. Recall that attenuation can
be interpreted in the context of the “lumped element”
equivalent circuit transmission line model as the
contributions of the resistance per unit length R′ and
conductance per unit length G′. In this model, R′
represents the physical resistance in the inner and
outer conductors, whereas G′ represents loss due to
current flowing directly between the conductors",Electromagnetics_Vol2.pdf
"outer conductors, whereas G′ represents loss due to
current flowing directly between the conductors
through the spacer material.
The parameters used to describe the relevant features
of coaxial cable are shown in Figure 7.11. In this
figure, aand bare the radii of the inner and outer
conductors, respectively . σic and σoc are the
conductivities (SI base units of S/m) of the inner and
outer conductors, respectively . Conductors are
assumed to be non-magnetic; i.e., having permeability
µequal to the free space value µ0. The spacer
material is assumed to be a lossy dielectric having
relative permittivity ǫr and conductivity σs.
Resistance per unit length. The resistance per unit
length is the sum of the resistances of the inner and
outer conductor per unit length. The resistance per
unit length of the inner conductor is determined by
σic and the effective cross-sectional area through
which the current flows. The latter is equal to the
circumference 2πatimes the skin depth δic of the
ϵr",Electromagnetics_Vol2.pdf
"which the current flows. The latter is equal to the
circumference 2πatimes the skin depth δic of the
ϵr
σ sσ ic
σ oc
b
a
Figure 7.11: Parameters defining the design of a coax-
ial
cable.",Electromagnetics_Vol2.pdf
"130 CHAPTER 7. TRANSMISSION LINES REDUX
inner conductor, so:
R′
ic ≈ 1
(2πa· δic) σic
for δic ≪ a (7.30)
This
expression is only valid for δic ≪ abecause
otherwise the cross-sectional area through which the
current flows is not well-modeled as a thin ring near
the surface of the conductor. Similarly , we find the
resistance per unit length of the outer conductor is
R′
oc ≈ 1
(2πb· δo c) σoc
for δoc ≪ t (7.31)
where δoc is the skin depth of the outer conductor and
tis the thickness of the outer conductor. Therefore,
the total resistance per unit length is
R′ = R′
ic + R′
oc
≈ 1
(2πa· δic) σic
+ 1
(2πb· δo c) σoc
(7.32)
Recall that skin depth depends on conductivity .
Specifically:
δic =
√
2/ωµσic (7.33)
δo
c =
√
2/ωµσo c (7.34)
Expanding Equation 7.32 to show explicitly the
dependence on conductivity , we find:
R′ ≈ 1
2π
√
2/ωµ0
[ 1
a√σic
+ 1
b√σoc
]
(7.35)
At
this point it is convenient to identify two particular
cases for the design of the cable. In the first case,",Electromagnetics_Vol2.pdf
"√
2/ωµ0
[ 1
a√σic
+ 1
b√σoc
]
(7.35)
At
this point it is convenient to identify two particular
cases for the design of the cable. In the first case,
“Case I, ” we assume σoc ≫ σic. Since b>a, we
have in this case
R′ ≈ 1
2π
√
2/ωµ0
[ 1
a√σic
]
= 1
2πδicσic
1
a (Case I) (7.36)
In the second case, “Case II, ” we assume σoc = σic.
In this case, we have
R′ ≈ 1
2π
√
2/ωµ0
[ 1
a√σic
+ 1
b√σic
]
= 1
2πδicσic
[ 1
a + 1
b
]
(Case II) (7.37)
A simpler way to deal with these two cases is to
represent them both using the single expression
R′ ≈ 1
2πδicσic
[ 1
a + C
b
]
(7.38)
where C =
0 in Case I and C = 1 in Case II.
Conductance per unit length. The conductance per
unit length of coaxial cable is simply that of the
associated coaxial structure at DC; i.e.,
G′ = 2πσs
ln (b/a) (7.39)
Unlik
e resistance, the conductance is independent of
frequency , at least to the extent that σs is independent
of frequency .
Attenuation. The attenuation of voltage and current",Electromagnetics_Vol2.pdf
"frequency , at least to the extent that σs is independent
of frequency .
Attenuation. The attenuation of voltage and current
waves as they propagate along the cable is
represented by the factor e−αz, where zis distance
traversed along the cable. It is possible to find an
expression for αin terms of the material and
geometry parameters using:
γ ≜
√
(R′ + jωL′) (G′ + jωC′) = α+ jβ (7.40)
where L′ and C′ are the inductance per unit length
and capacitance per unit length, respectively . These
are given by
L′ = µ
2πln (b/a) (7.41)
and
C′ = 2πǫ0ǫr
ln (b/a) (7.42)
In
principle we could solve Equation 7.40 for α.
However, this course of action is quite tedious, and a
simpler approximate approach facilitates some
additional insights. In this approach, we define
parameters αR associated with R′ and αG associated
with G′ such that
e−αRze−αGz = e−(αR+αG)z = e−αz (7.43)
which indicates
α= αR + αG (7.44)
Next we postulate
αR ≈ KR
R′
Z0
(7.45)",Electromagnetics_Vol2.pdf
"7.3. A TTENUA TION IN COAXIAL CABLE 131
where Z0 is the characteristic impedance
Z0 ≈ η0
2π
1√ǫr
ln b
a (lo w loss) (7.46)
and where KR is a unitless constant to be determined.
The justification for Equation 7.45 is as follows: First,
αR must increase monotonically with increasing R′.
Second, R′ must be divided by an impedance in order
to obtain the correct units of 1/m. Using similar
reasoning, we postulate
αG ≈ KGG′Z0 (7.47)
where KG is a unitless constant to be determined.
The following example demonstrates the validity of
Equations 7.45 and 7.47, and will reveal the values of
KR and KG.
Example 7.4. Attenuation constant for RG-59.
RG-59 is a popular form of coaxial cable having
the parameters a∼= 0.292 mm, b∼
= 1.855 mm,
σic ∼= 2.28 × 107 S/m, σs ∼
= 5.9 × 10−5 S/m,
and ǫr ∼= 2.25. The conductivity σoc of the outer
conductor is difficult to quantify because it
consists of a braid of thin metal strands.
However, σoc ≫ σic, so we may assume Case I;",Electromagnetics_Vol2.pdf
"conductor is difficult to quantify because it
consists of a braid of thin metal strands.
However, σoc ≫ σic, so we may assume Case I;
i.e., σoc ≫ σic, and subsequently C = 0.
Figure 7.12 shows the components αG and αR
computed for the particular choice
KR = KG = 1/2. The figure also shows
αG + αR, along with αcomputed using
Equation 7.40. W e find that the agreement
between these values is very good, which is
compelling evidence that the ansatz is valid and
KR = KG = 1/2.
Note that there is nothing to indicate that the results
demonstrated in the example are not generally true.
Thus, we come to the following conclusion:
The attenuation constant α ≈ αG + αR where
αG ≜ R′/
2Z0 and αR ≜ G′Z0/2.
Minimizing attenuation. Let us now consider if
there are design choices which minimize the
attenuation of coaxial cable. Since α= αR + αG, we
Figure 7.12: Comparison of α = Re { γ} to αR, αG,
and αR + αG for KR = KG = 1/2. The result for
α has been multiplied by 1.01; otherwise the curves",Electromagnetics_Vol2.pdf
"Figure 7.12: Comparison of α = Re { γ} to αR, αG,
and αR + αG for KR = KG = 1/2. The result for
α has been multiplied by 1.01; otherwise the curves
would be too close to tell apart.
may consider αR and αG independently . Let us first
consider αG:
αG ≜ 1
2G′Z0
≈ 1
2 · 2πσs
ln (b/a) · 1
2π
η0
√ǫr
ln (b/a)
= η0
2
σs
√ǫr
(7.48)
It
is clear from this result that αG is minimized by
minimizing σs/√ǫr. Interestingly the physical
dimensions aand bha
ve no discernible effect on αG.
Now we consider αR:
αR ≜ R′
2Z0
= 1
2
(1
/2πδicσic) [1/a+ C/b]
(1/2π)
(
η0/√ǫr
)
ln (b/a )
=
√ǫr
2η0δicσic
· [1
/a+ C/b]
ln (b/a) (7.49)
No
w making the substitution δic =
√
2/ωµ0σic in
order
to make the dependences on the constitutive
parameters explicit, we find:
αR = 1
2
√
2 · η0
√ ωµ0ǫr
σic
· [1
/a+ C/b]
ln (b/a) (7.50)",Electromagnetics_Vol2.pdf
"132 CHAPTER 7. TRANSMISSION LINES REDUX
Here we see that αR is minimized by minimizing
ǫr/σic. It’s not surprising to see that we should
maximize σic. However, it’s a little surprising that we
should minimize ǫr. Furthermore, this is in contrast to
αG, which is minimized by maximizing ǫr. Clearly
there is a tradeoff to be made here. T o determine the
parameters of this tradeoff, first note that the result
depends on frequency: Since αR dominates over αG
at sufficiently high frequency (as demonstrated in
Figure 7.12), it seems we should minimize ǫr if the
intended frequency of operation is sufficiently high;
otherwise the optimum value is frequency-dependent.
However, σs may vary as a function of ǫr, so a
general conclusion about optimum values of σs and
ǫr is not appropriate.
However, we also see that αR – unlike αG – depends
on aand b. This implies the existence of a
generally-optimum geometry . T o find this geometry ,
we minimize αR by taking the derivative with respect",Electromagnetics_Vol2.pdf
"on aand b. This implies the existence of a
generally-optimum geometry . T o find this geometry ,
we minimize αR by taking the derivative with respect
to a, setting the result equal to zero, and solving for a
and/or b. Here we go:
∂
∂aαR = 1
2
√
2 · η0
√ ωµ0ǫr
σic
· ∂
∂a
[1
/a+ C/b]
ln (b/a)
(7.51)
This
derivative is worked out in an addendum at the
end of this section. Using the result from the
addendum, the right side of Equation 7.51 can be
written as follows:
1
2
√
2 · η0
√ ωµ0ǫr
σic
·
[ −1
a2 ln (b/a ) + 1/a+ C/b
aln2 (b/a )
]
(7.52)
In order for ∂αR/∂a = 0, the factor in the square
brackets above must be equal to zero. After a few
steps of algebra, we find:
ln (b/a) = 1 + C
b/a (7.53)
In
Case I (σ oc ≫ σic), C = 0 so:
b/a= e∼= 2.72 (Case I) (7.54)
In Case II (σ oc = σic), C = 1. The resulting equation
can be solved by plotting the function, or by a few
iterations of trial and error; either way one quickly
finds
b/a∼= 3.59 (Case II) (7.55)",Electromagnetics_Vol2.pdf
"can be solved by plotting the function, or by a few
iterations of trial and error; either way one quickly
finds
b/a∼= 3.59 (Case II) (7.55)
Summarizing, we have found that αis minimized by
choosing the ratio of the outer and inner radii to be
somewhere between 2.72 and 3.59, with the precise
value depending on the relative conductivity of the
inner and outer conductors.
Substituting these values of b/ainto Equation 7.46,
we obtain:
Z0 ≈ 59.9 Ω√ǫr
to 76
.6 Ω√ǫr
(7.56)
as
the range of impedances of coaxial cable
corresponding to physical designs that minimize
attenuation.
Equation 7.56 gives the range of characteristic
impedances
that minimize attenuation for coaxial
transmission lines. The precise value within this
range depends on the ratio of the conductivity of
the outer conductor to that of the inner conductor.
Since ǫr ≥ 1, the impedance that minimizes
attenuation is less for dielectric-filled cables than it is
for air-filled cables. For example, let us once again",Electromagnetics_Vol2.pdf
"attenuation is less for dielectric-filled cables than it is
for air-filled cables. For example, let us once again
consider the RG-59 from Example 7.4. In that case,
ǫr ∼= 2.25 and C = 0, indicating Z0 ≈ 39.9 Ω is
optimum for attenuation. The actual characteristic
impedance of Z0 is about 75 Ω, so clearly RG-59 is
not optimized for attenuation. This is simply because
other considerations apply , including power handling
capability (addressed in Section 7.4) and the
convenience of standard values (addressed in
Section 7.5).
Addendum: Derivative of a2 ln(b/a). Evaluation of
Equation 7.51 requires finding the derivative of
a2 ln(b/a) with respect to a. Using the chain rule, we
find:
∂
∂a
[
a2 ln
(b
a
)]
=
[ ∂
∂aa2
]
ln
(b
a
)
+a2
[ ∂
∂a ln
(b
a
)]
(7.57)
Note
∂
∂aa2 = 2 a (7.58)",Electromagnetics_Vol2.pdf
"7.4. POWER HANDLING CAP ABILITY OF COAXIAL CABLE 133
and
∂
∂a ln
(b
a
)
= ∂
∂a [ln (b) − ln (a)]
= − ∂
∂a ln (a)
= − 1
a (7.59)
So:
∂
∂a
[
a2 ln
(b
a
)]
= [2a] ln
(b
a
)
+ a2
[
− 1
a
]
= 2
aln
(b
a
)
− a (7.60)
This
result is substituted for a2 ln(b/a) in
Equation 7.51 to obtain Equation 7.52.
7.4 Power Handling Capability of
Coaxial Cable
[m0190]
The term “power handling” refers to maximum power
that
can be safely transferred by a transmission line.
This power is limited because when the electric field
becomes too large, dielectric breakdown and arcing
may occur. This may result in damage to the line and
connected devices, and so must be avoided. Let Epk
be the maximum safe value of the electric field
intensity within the line, and let Pmax be the power
that is being transferred under this condition. This
section addresses the following question: How does
one design a coaxial cable to maximize Pmax for a
given Epk?
W e begin by finding the electric potential V within",Electromagnetics_Vol2.pdf
"one design a coaxial cable to maximize Pmax for a
given Epk?
W e begin by finding the electric potential V within
the cable. This can be done using Laplace’s equation:
∇2V = 0 (7.61)
Using the cylindrical (ρ,φ,z ) coordinate system with
the zaxis along the inner conductor, we have
∂V/∂φ = 0 due to symmetry . Also we set
∂V/∂z = 0 since the result should not depend on z.
Thus, we have:
1
ρ
∂
∂ρ
(
ρ∂V
∂ρ
)
= 0 (7.62)
Solving
for V, we have
V(ρ) = Aln ρ+ B (7.63)
where Aand Bare arbitrary constants, presumably
determined by boundary conditions. Let us assume a
voltage V0 measured from the inner conductor
(serving as the “+ ” terminal) to the outer conductor
(serving as the “− ” terminal). For this choice, we
have:
V(a) = V0 → Aln a+ B = V0 (7.64)
V(b) = 0 → Aln b+ B = 0 (7.65)
Subtracting the second equation from the first and
solving for A, we find A= −V0/ln (b/a).
Subsequently , Bis found to be V0 ln (b) /ln (b/a),
and so
V(ρ) = −V0
ln (b/a) ln ρ+ V0 ln
(b)
ln (b/a) (7.66)",Electromagnetics_Vol2.pdf
"134 CHAPTER 7. TRANSMISSION LINES REDUX
The electric field intensity is given by:
E = −∇V (7.67)
Again we have ∂V/∂φ = ∂V/∂z = 0, so
E = −ˆρ ∂
∂ρV (7.68)
= −ˆρ ∂
∂ρ
[ −V0
ln (b/a) ln ρ+ V0 ln
(b)
ln (b/a)
]
(7.69)
=
+ˆρ V0
ρln (b/a) (7.70)
Note
that the maximum electric field intensity in the
spacer occurs at ρ= a; i.e., at the surface of the inner
conductor. Therefore:
Epk = V0
aln (b/a ) (7.71)
The power transferred by the line is maximized when
the impedances of the source and load are matched to
Z0. In this case, the power transferred is V2
0 /2Z0.
Recall that the characteristic impedance Z0 is given in
the “low-loss” case as
Z0 ≈ 1
2π
η0
√ǫr
ln
(b
a
)
(7.72)
Therefore,
the maximum safe power is
Pmax = V2
0
2Z0
(7.73)
≈
E2
pk a2 ln2 (
b/a)
2 · (1/2π)
(
η0/√ǫr
)
ln (b/a ) (7.74)
=
πE2
pk
η0/√ǫr
a2 ln (b/a) (7.75)
Now let us consider if there is a value of awhich
maximizes Pmax. W e do this by seeing if
∂Pmax/∂a = 0 for some values of aand b. The",Electromagnetics_Vol2.pdf
"Now let us consider if there is a value of awhich
maximizes Pmax. W e do this by seeing if
∂Pmax/∂a = 0 for some values of aand b. The
derivative is worked out in an addendum at the end of
this section. Using the result from the addendum, we
find:
∂
∂aPmax =
π
E2
pk
η0/√ǫr
[2 aln (b/a) − a] (7.76)
For the above expression to be zero, it must be true
that 2 ln (b/a) − 1 = 0. Solving for b/a, we obtain:
b
a = √e∼= 1.65 (7.77)
for
optimum power handling. In other words, 1.65 is
the ratio of the radii of the outer and inner conductors
that maximizes the power that can be safely handled
by the cable.
Equation 7.75 suggests that ǫr should be maximized
in order to maximize power handling, and you
wouldn’t be wrong for noting that, however, there are
some other factors that may indicate otherwise. For
example, a material with higher ǫr may also have
higher σs, which means more current flowing through
the spacer and thus more ohmic heating. This",Electromagnetics_Vol2.pdf
"example, a material with higher ǫr may also have
higher σs, which means more current flowing through
the spacer and thus more ohmic heating. This
problem is so severe that cables that handle high RF
power often use air as the spacer, even though it has
the lowest possible value of ǫr. Also worth noting is
that σic and σoc do not matter according to the
analysis we’ve just done; however, to the extent that
limited conductivity results in significant ohmic
heating in the conductors – which we have also not
considered – there may be something to consider.
Suffice it to say , the actionable finding here concerns
the ratio of the radii; the other parameters have not
been suitably constrained by this analysis.
Substituting √
efor b/ain Equation 7.72, we find:
Z0 ≈ 30.0 Ω√ǫr
(7.78)
This
is the characteristic impedance of coaxial line
that optimizes power handling, subject to the caveats
identified above. For air-filled cables, we obtain 30 Ω.
Since ǫr ≥ 1, this optimum impedance is less for",Electromagnetics_Vol2.pdf
"identified above. For air-filled cables, we obtain 30 Ω.
Since ǫr ≥ 1, this optimum impedance is less for
dielectric-filled cables than it is for air-filled cables.
Summarizing:
The power handling capability of coaxial trans-
mission
line is optimized when the ratio of radii
of the outer to inner conductors b/ais about 1.65.
For the air-filled cables typically used in high-
power applications, this corresponds to a charac-
teristic impedance of about 30 Ω.
Addendum: Derivative of (1/a+ C/b) /ln(b/a).
Evaluation of Equation 7.75 requires finding the
derivative of (1/a+ C/b) /ln(b/a) with respect to a.",Electromagnetics_Vol2.pdf
"7.5. WHY 50 OHMS? 135
Using the chain rule, we find:
∂
∂a
[ 1/a+ C
/b
ln (b/a)
]
=
[ ∂
∂a
(1
a + C
b
)]
ln−1
(b
a
)
+
(1
a + C
b
)[ ∂
∂a ln−1
(b
a
)]
(7.79)
Note
∂
∂a
(1
a + C
b
)
= − 1
a2 (7.80)
T
o handle the quantity in the second set of square
brackets, first define v= ln u, where u= b/a. Then:
∂
∂av−1 =
[ ∂
∂vv−1
][ ∂
v
∂u
][ ∂
u
∂a
]
=
[
−v−2] [ 1
u
] [
−ba−2]
=
[
− ln−2
(b
a
)] [ a
b
] [
−ba−2]
= 1
aln−2
(b
a
)
(7.81)
So:
∂
∂a
[ 1/a+ C
/b
ln (b/a)
]
=
[
− 1
a2
]
ln−1
(b
a
)
+
(1
a + C
b
)[ 1
aln−2
(b
a
)]
(7.82)
This
result is substituted in Equation 7.75 to obtain
Equation 7.76.
7.5 Why 50 Ohms?
[m0191]
The quantity 50 Ω appears in a broad range of
applications across the field of electrical engineering.
In particular, it is a very popular value for the
characteristic impedance of transmission line, and is
commonly specified as the port impedance for signal
sources, amplifiers, filters, antennas, and other RF
components. So, what’s special about 50 Ω? The",Electromagnetics_Vol2.pdf
"commonly specified as the port impedance for signal
sources, amplifiers, filters, antennas, and other RF
components. So, what’s special about 50 Ω? The
short answer is “nothing. ” In fact, other standard
impedances are in common use – prominent among
these is 75 Ω. It is shown in this section that a broad
range of impedances – on the order of 10s of ohms –
emerge as useful values based on technical
considerations such as minimizing attenuation,
maximizing power handling, and compatibility with
common types of antennas. Characteristic
impedances up to 300 Ω and beyond are useful in
particular applications. However, it is not practical or
efficient to manufacture and sell products for every
possible impedance in this range. Instead, engineers
have settled on 50 Ω as a round number that lies near
the middle of this range, and have chosen a few other
values to accommodate the smaller number of
applications where there may be specific compelling
considerations.",Electromagnetics_Vol2.pdf
"values to accommodate the smaller number of
applications where there may be specific compelling
considerations.
So, the question becomes “what makes characteristic
impedances in the range of 10s of ohms particularly
useful?” One consideration is attenuation in coaxial
cable. Coaxial cable is by far the most popular type of
transmission line for connecting devices on separate
printed circuit boards or in separate enclosures. The
attenuation of coaxial cable is addressed in
Section 7.3. In that section, it is shown that
attenuation is minimized for characteristic
impedances in the range (60 Ω) /√
ǫr to ( 77 Ω) /√ǫr,
where ǫr is
the relative permittivity of the spacer
material. So, we find that Z0 in the range 60 Ω to
77 Ω is optimum for air-filled cable, but more like
40 Ω to 50 Ω for cables using a plastic spacer material
having typical ǫr ≈ 2.25. Thus, 50 Ω is clearly a
reasonable choice if a single standard value is to be
established for all such cable.",Electromagnetics_Vol2.pdf
"having typical ǫr ≈ 2.25. Thus, 50 Ω is clearly a
reasonable choice if a single standard value is to be
established for all such cable.
Coaxial cables are often required to carry high power
signals. In such applications, power handling",Electromagnetics_Vol2.pdf
"136 CHAPTER 7. TRANSMISSION LINES REDUX
capability is also important, and is addressed in
Section 7.4. In that section, we find the power
handling capability of coaxial cable is optimized
when the ratio of radii of the outer to inner conductors
b/ais about 1.65. For the air-filled cables typically
used in high-power applications, this corresponds to a
characteristic impedance of about 30 Ω. This is
significantly less than the 60 Ω to 77 Ω that
minimizes attenuation in air-filled cables. So, 50 Ω
can be viewed as a compromise between minimizing
attenuation and maximizing power handling in
air-filled coaxial cables.
Although the preceding arguments justify 50 Ω as a
standard value, one can also see how one might make
a case for 75 Ω as a secondary standard value,
especially for applications where attenuation is the
primary consideration.
V alues of 50 Ω and 75 Ω also offer some convenience
when connecting RF devices to antennas. For
example, 75 Ω is very close to the impedance of the",Electromagnetics_Vol2.pdf
"V alues of 50 Ω and 75 Ω also offer some convenience
when connecting RF devices to antennas. For
example, 75 Ω is very close to the impedance of the
commonly-encountered half-wave dipole antenna
(about 73 + j42 Ω), which may make impedance
matching to that antenna easier. Another
commonly-encountered antenna is the quarter-wave
monopole, which exhibits an impedance of about
36 + j21 Ω, which is close to 50 Ω. In fact, we see
that if we desire a single characteristic impedance that
is equally convenient for applications involving either
type of antenna, then 50 Ω is a reasonable choice.
A third commonly-encountered antenna is the folded
half-wave dipole. This type of antenna is similar to a
half-wave dipole but has better bandwidth, and is
commonly used in FM and TV systems and land
mobile radio (LMR) base stations. A folded
half-wave dipole has an impedance of about 300 Ω
and is balanced (not single-ended); thus, there is a
market for balanced transmission line having",Electromagnetics_Vol2.pdf
"half-wave dipole has an impedance of about 300 Ω
and is balanced (not single-ended); thus, there is a
market for balanced transmission line having
Z0 = 300 Ω. However, it is very easy and
inexpensive to implement a balun (a device which
converts the dipole output from balanced to
unbalanced) while simultaneously stepping down
impedance by a factor of 4; i.e., to 75 Ω. Thus, we
have an additional application for 75 Ω coaxial line.
Finally , note that it is quite simple to implement
microstrip transmission line having characteristic
impedance in the range 30 Ω to 75 Ω. For example,
50 Ω on commonly-used 1.575 mm FR4 requires a
width-to-height ratio of about 2, so the trace is about
3 mm wide. This is a very manageable size and easily
implemented in printed circuit board designs.
Additional Reading:
• “Dipole antenna” on Wikipedia.
• “Monopole antenna” on Wikipedia.
• “Balun” on Wikipedia.
[m0187]",Electromagnetics_Vol2.pdf
"7.5. WHY 50 OHMS? 137
Image Credits
Fig. 7.1: c⃝ SpinningSpark, Inductiveload,
https://commons.wikimedia.org/wiki/File:T win-lead cable dimension.svg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Minor modifications.
Fig. 7.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Parallel wire transmission line structure and design parameters.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 7.3: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Electric and Magnetic Field of Wire Cross-Section.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 7.4: c⃝ 7head7metal7, https://commons.wikimedia.org/wiki/File:Microstrip scheme.svg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Minor modifications from the original.
Fig. 7.6: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:V iew from the side of a microstrip line.svg,
CC",Electromagnetics_Vol2.pdf
"Fig. 7.6: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:V iew from the side of a microstrip line.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 7.7: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields for Microstrip.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 8
Optical
Fiber
8.1 Optical Fiber: Method of
Operation
[m0178]
In its simplest form, optical fiber consists of
concentric
regions of dielectric material as shown in
Figure 8.1. A cross-section through the fiber reveals a
circular region of transparent dielectric material
through which light propagates. This is surrounded
by a jacket of dielectric material commonly referred
to as cladding.
A characteristic of the design of any optical fiber is
that the permittivity of the fiber is greater than the
permittivity of the cladding. As explained in
Section 5.11, this creates conditions necessary for
total internal reflection. The mechanism of total
internal reflection contains the light within the fiber.
In the discipline of optics, the permittivity of a
c⃝ S. Lally CC BY -SA 4.0
Figure 8.1: Construction of the simplest form of opti-
cal fiber.
material is commonly quantified in terms of its index
of refraction. Index of refraction is the square root of",Electromagnetics_Vol2.pdf
"cal fiber.
material is commonly quantified in terms of its index
of refraction. Index of refraction is the square root of
relative permittivity , and is usually assigned the
symbol n. Thus, if we define the relative
permittivities ǫr,f ≜ ǫf/ǫ0 for the fiber and
ǫr,c ≜ ǫc/ǫ0 for the cladding, then
nf ≜ √
ǫr,f (8.1)
nc ≜ √ǫr,c (8.2)
and
nf >
nc (8.3)
Figure 8.2 illustrates total internal reflection in an
optical fiber. In this case, a ray of light in the fiber is
incident on the boundary with the cladding. W e may
treat the light ray as a uniform plane wave. T o see
why , consider that optical wavelengths range from
120 nm to 700 nm in free space. W avelength is
slightly shorter than this in fiber; specifically , by a
factor equal to the square root of the relative
permittivity . The fiber is on the order of millimeters
in diameter, which is about 4 orders of magnitude
greater than the wavelength. Thus, from the
perspective of the light ray , the fiber appears to be an",Electromagnetics_Vol2.pdf
"greater than the wavelength. Thus, from the
perspective of the light ray , the fiber appears to be an
unbounded half-space sharing a planar boundary with
the cladding, which also appears to be an unbounded
half-space.
Continuing under this presumption, the criterion for
total internal reflection is (from Section 5.11):
θi ≥ arcsin
√
ǫr,c
ǫr,f
= arcsin nc
nf
(8.4)
As
long as rays of light approach from angles that
satisfy this criterion, the associated power remains in
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"8.1. OPTICAL FIBER: METHOD OF OPERA TION 139
c⃝ S. Lally CC BY -SA 4.0
Figure 8.2: T otal internal reflection in optical fiber.
the fiber, and is reflected onward. Otherwise, power is
lost into the cladding.
Example 8.1. Critical angle for optical fiber.
T ypical values of nf and nc for an optical fiber
are 1.52 and 1.49, respectively . What internal
angle of incidence is required to maintain total
internal reflection?
Solution. Using Equation 8.4, θi must be greater
than about 78.8◦. In practice this is quite
reasonable since we desire light to be traveling
approximately parallel the axis of the fiber
(θi ≈ 90◦) anyway .
The total internal reflection criterion imposes a limit
on the radius of curvature of fiber optic cable. If fiber
optic cable is bent such that the radius of curvature is
too small, the critical angle will be exceeded at the
bend. This will occur even for light rays which are
traveling perfectly parallel to the axis of the fiber
before they arrive at the bend.",Electromagnetics_Vol2.pdf
"bend. This will occur even for light rays which are
traveling perfectly parallel to the axis of the fiber
before they arrive at the bend.
Note that the cladding serves at least two roles. First,
it determines the critical angle for total internal
reflection, and subsequently determines the minimum
radius of curvature for lossless operation of the fiber.
Second, it provides a place for the evanescent surface
waves associated with total internal reflection
(Section 5.12) to exist without interference from
objects in contact with the fiber.
Additional Reading:
• “Optical fiber” on Wikipedia.",Electromagnetics_Vol2.pdf
"140 CHAPTER 8. OPTICAL FIBER
8.2 Acceptance Angle
[m0192]
In this section, we consider the problem of injecting
light
into a fiber optic cable. The problem is
illustrated in Figure 8.3. In this figure, we see light
incident from a medium having index of refraction n0,
with angle of incidence θi. The light is transmitted
with angle of transmission θ2 into the fiber, and is
subsequently incident on the surface of the cladding
with angle of incidence θ3. For light to propagate
without loss within the cable, it is required that
sin θ3 ≥ nc
nf
(8.5)
since
this criterion must be met in order for total
internal reflection to occur.
Now consider the constraint that Equation 8.5
imposes on θi. First, we note that θ3 is related to θ2 as
follows:
θ3 = π
2 − θ2 (8.6)
therefore
s
in θ3 = sin
(π
2 − θ2
)
(8.7)
=
cos θ2 (8.8)
so
cos θ2 ≥ nc
nf
(8.9)
Squaring
both sides, we find:
cos2 θ2 ≥ n2
c
n2
f
(8.10)
c⃝ S. Lally CC BY -SA 4.0
Figure 8.3: Injecting light into a fiber optic cable.",Electromagnetics_Vol2.pdf
"nf
(8.9)
Squaring
both sides, we find:
cos2 θ2 ≥ n2
c
n2
f
(8.10)
c⃝ S. Lally CC BY -SA 4.0
Figure 8.3: Injecting light into a fiber optic cable.
Now invoking a trigonometric identity:
1 − sin2 θ2 ≥ n2
c
n2
f
(8.11)
so:
s
in2 θ2 ≤ 1 − n2
c
n2
f
(8.12)
No
w we relate the θ2 to θi using Snell’s law:
sin θ2 = n0
nf
sin θi (8.13)
so
Equation 8.12 may be written:
n2
0
n2
f
sin2 θi ≤ 1 − n2
c
n2
f
(8.14)
No
w solving for sin θi, we obtain:
sin θi ≤ 1
n0
√
n2
f − n2c (8.15)
This
result indicates the range of angles of incidence
which result in total internal reflection within the
fiber. The maximum value of θi which satisfies this
condition is known as the acceptance angle θa, so:
θa ≜ arcsin
(1
n0
√
n2
f − n2c
)
(8.16)
This
leads to the following insight:
In order to effectively launch light in the fiber, it
is
necessary for the light to arrive from within a
cone having half-angle θa with respect to the axis
of the fiber.
The associated cone of acceptance is illustrated in
Figure 8.4.",Electromagnetics_Vol2.pdf
"cone having half-angle θa with respect to the axis
of the fiber.
The associated cone of acceptance is illustrated in
Figure 8.4.
It is also common to define the quantity numerical
aperture NA as follows:
NA ≜ 1
n0
√
n2
f − n2c (8.17)
Note
that n0 is typically very close to 1
(corresponding to incidence from air), so it is
common to see NA defined as simply
√
n2
f − n2c.
This
parameter is commonly used in lieu of the
acceptance angle in datasheets for fiber optic cable.",Electromagnetics_Vol2.pdf
"8.3. DISPERSION IN OPTICAL FIBER 141
Example 8.2. Acceptance angle.
T ypical values of nf and nc for an optical fiber
are 1.52 and 1.49, respectively . What are the
numerical aperture and the acceptance angle?
Solution. Using Equation 8.17 and presuming
n0 = 1, we find NA ∼= 0.30. Since sin θa = N A,
we find θa = 17.5 ◦. Light must arrive from
within 17
.5◦ from the axis of the fiber in order to
ensure total internal reflection within the fiber.
Additional Reading:
• “Optical fiber” on Wikipedia.
• “Numerical aperture” on Wikipedia.
c⃝ S. Lally CC BY -SA 4.0
Figure 8.4: Cone of acceptance.
8.3 Dispersion in Optical Fiber
[m0193]
Light may follow a variety of paths through a fiber
optic
cable. Each of the paths has a different length,
leading to a phenomenon known as dispersion.
Dispersion distorts signals and limits the data rate of
digital signals sent over fiber optic cable. In this
section, we analyze this dispersion and its effect on
digital signals.",Electromagnetics_Vol2.pdf
"digital signals sent over fiber optic cable. In this
section, we analyze this dispersion and its effect on
digital signals.
Figure 8.5 shows the variety of paths that light may
take through a straight fiber optic cable. The nominal
path is shown in Figure 8.5(a), which is parallel to the
axis of the cable. This path has the shortest associated
propagation time. The path with the longest
associated propagation time is shown in Figure 8.5(c).
In this case, light bounces within the fiber, each time
approaching the core-clad interface at the critical
angle for total internal reflection. Any ray
approaching at a greater angle is not completely
reflected, and so would likely not survive to the end of
c⃝ S. Lally CC BY -SA 4.0
Figure 8.5: Paths that light may take through a straight
fiber optic cable: (a) Along axis, corresponding to
minimum path length; (b) Intermediate between (a)
and (c); (c) Maximum length, corresponding to thresh-
old angle for total internal reflection.",Electromagnetics_Vol2.pdf
"142 CHAPTER 8. OPTICAL FIBER
c⃝ S. Lally CC BY -SA 4.0
Figure 8.6: A digital signal that might be applied to
the input of a fiber optic cable.
the cable. Figure 8.5(b) represents the continuum of
possibilities between the extreme cases of (a) and (c),
with associated propagation times greater than that of
case (a) but less than that of case (c).
Regardless of how light is inserted into the fiber, all
possible paths depicted in Figure 8.5 are likely to
exist. This is because fiber is rarely installed in a
straight line, but rather follows a multiply-curved
path. Each curve results in new angles of incidence
upon the core-cladding boundary . The existence of
these paths leads to dispersion. T o see this, consider
the input signal shown in Figure 8.6. This signal has
period T, during which time a pulse of length
ton <T may be present. Let the minimum
propagation time through the fiber (as in
Figure 8.5(a)) be τmin. Let the maximum propagation",Electromagnetics_Vol2.pdf
"ton <T may be present. Let the minimum
propagation time through the fiber (as in
Figure 8.5(a)) be τmin. Let the maximum propagation
time through the fiber (as in Figure 8.5(c)) be τmax. If
τmax − τmin ≪ ton, we see little degradation in the
signal output from the fiber. Otherwise, we observe a
“smearing” of pulses at the output of the fiber, as
shown in Figure 8.7.
Any smearing may pose problems for whatever device
is detecting pulses at the receiving end. However, as
τmax − τmin becomes a larger fraction of ton, we see
that it becomes possible for adjacent pulses to
overlap. This presents a much more serious challenge
in detecting individual pulses, so let us examine the
c⃝ S. Lally CC BY -SA 4.0
Figure 8.7: The effect of dispersion on the signal out-
put from the fiber optic cable: T op: Arriving wave-
form corresponding to minimum path length, Middle:
Arriving waveform corresponding to maximum path
length, and Bottom: Sum of all arriving waveforms.",Electromagnetics_Vol2.pdf
"8.3. DISPERSION IN OPTICAL FIBER 143
overlap problem in greater detail. T o avoid overlap:
τmax − τmin + ton <T (8.18)
Let us define the quantity τ ≜ τmax − τmin. This is
sometimes referred to as the delay spread. 1 Thus, we
obtain the following requirement for overlap-free
transmission:
τ <T − ton (8.19)
Note that the delay spread imposes a minimum value
on T, which in turn imposes a maximum value on the
rate at which information can be transmitted on the
cable. So, we are motivated to calculate delay spread.
The minimum propagation time tmin is simply the
length lof the cable divided by the phase velocity vp
of the light within the core; i.e., tmin = l/vp. Recall
phase velocity is simply the speed of light in free
space cdivided by √
ǫr where ǫr is the relative
permittivity of the core material. Thus:
tmin = l
vp
= l
c/√ǫr
= l
c/nf
= lnf
c (8.20)
The
maximum propagation time tmax is different
because the maximum path length is different.",Electromagnetics_Vol2.pdf
"tmin = l
vp
= l
c/√ǫr
= l
c/nf
= lnf
c (8.20)
The
maximum propagation time tmax is different
because the maximum path length is different.
Specifically , the maximum path length is not l, but
rather l/cos θ2 where θ2 is the angle between the axis
and the direction of travel as light crosses the axis.
So, for example, θ2 = 0 for tmin since light in that
case travels along the axis. The value of θ2 for tmax is
determined by the threshold angle for total internal
reflection. This was determined in Section 8.2 to be
given by:
cos θ2 = nc
nf
(threshold value) (8.21)
so we obtain the following relationship:
tmax = l/cos θ2
vp
= lnf/nc
c/nf
=
ln2
f
cnc
(8.22)
and
subsequently we find:
τ = lnf
c
(nf
nc
− 1
)
(8.23)
1 Full disclosure: There are many ways to define “delay spread, ”
but this definition is not uncommon and is useful in the present anal-
ysis.
Note that τ increases linearly with l. Therefore, the
rate at which pulses can be sent without overlap",Electromagnetics_Vol2.pdf
"ysis.
Note that τ increases linearly with l. Therefore, the
rate at which pulses can be sent without overlap
decreases linearly with increasing length. In practical
applications, this means that the maximum
supportable data rate decreases as the length of the
cable increases. This is true independently of media
loss within the cable (which we have not yet even
considered!). Rather, this is a fundamental limitation
resulting from dispersion.
Example 8.3. Maximum supportable data rate
in multimode fiber optic cable.
A multimode fiber optic cable of length 1 m is
used to transmit data using the scheme shown in
Figure 8.6, with ton = T/2. V alues of nf and nc
for this fiber are 1.52 and 1.49, respectively .
What is the maximum supportable data rate?
Solution. Using Equation 8.23, we find the
delay spread τ ∼= 102 ps. T o avoid overlap,
T >2τ; therefore, T greater than about 204 ps
is required. The modulation scheme allows one
bit per period, so the maximum data rate is",Electromagnetics_Vol2.pdf
"T >2τ; therefore, T greater than about 204 ps
is required. The modulation scheme allows one
bit per period, so the maximum data rate is
1/T ∼= 4.9 × 109 bits per second; i.e., ∼
=
4.9 Gb/s
.
The
finding of 4.9 Gb/s may seem like a pretty high
data rate; however, consider what happens if the
length increases to 1 km. It is apparent from
Equation 8.23 that the delay spread will increase by a
factor of 1000, so the maximum supportable data rate
decreases by a factor of 1000 to a scant 4.9 Mb/s.
Again, this is independent of any media loss within
the fiber. T o restore the higher data rate over this
longer path, the dispersion must be reduced. One way
to do this is to divide the link into smaller links
separated by repeaters which can receive the
dispersed signal, demodulate it, regenerate the
original signal, and transmit the restored signal to the
next repeater. Alternatively , one may employ “single
mode” fiber, which has intrinsically less dispersion",Electromagnetics_Vol2.pdf
"next repeater. Alternatively , one may employ “single
mode” fiber, which has intrinsically less dispersion
than the multimode fiber presumed in our analysis.
Additional Reading:
• “Optical fiber” on Wikipedia.
[m0209]",Electromagnetics_Vol2.pdf
"144 CHAPTER 8. OPTICAL FIBER
Image Credits
Fig. 8.1: c⃝ S. Lally , https://commons.wikimedia.org/wiki/File:Figure 8.1-01.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified by author.
Fig. 8.2: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Internal Reflection in Optical Fiber.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.3: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Injecting Light into Optical Fiber.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.4: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Cone of Acceptance.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.5: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Paths of Light through Optic Cable.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.6: c⃝ Offaperry (S. Lally),",Electromagnetics_Vol2.pdf
"CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.6: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Digital Signal on Fiber Optic Cable.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.7: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Dispersion on Signal Output from Fiber Optic Cable.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 9
Radiation
9.1
Radiation from a Current
Moment
[m0194]
In this section, we begin to address the following
problem:
Given a distribution of impressed current
density J(r), what is the resulting electric field
intensity E(r)? One route to an answer is via
Maxwell’s equations. V iewing Maxwell’s equations
as a system of differential equations, a rigorous
mathematical solution is possible given the
appropriate boundary conditions. The rigorous
solution following that approach is relatively
complicated, and is presented beginning in
Section 9.2 of this book.
If we instead limit scope to a sufficiently simple
current distribution, a simple informal derivation is
possible. This section presents such a derivation. The
advantage of tackling a simple special case first is that
it will allow us to quickly assess the nature of the
solution, which will turn out to be useful once we do
eventually address the more general problem.
Furthermore, the results presented in this section will",Electromagnetics_Vol2.pdf
"eventually address the more general problem.
Furthermore, the results presented in this section will
turn out to be sufficient to tackle many
commonly-encountered applications.
The simple current distribution considered in this
section is known as a current moment. An example of
a current moment is shown in Figure 9.1 and in this
case is defined as follows:
∆J(r) = ˆz I ∆lδ (r) (9.1)
where I ∆l is the scalar component of the current
moment, having units of current times length (SI base
units of A·m); and δ(r) is the volumetric sampling
function1 defined as follows:
δ(r) ≜ 0 for r ̸= 0; and (9.2)∫
V
δ(r) dv≜ 1 (9.3)
where V is any volume which includes the origin
(r = 0). It is evident from Equation 9.3 that δ(r) has
SI base units of m −3. Subsequently , ∆J(r) has SI
base units of A/m 2, confirming that it is a volume
current density . However, it is the simplest possible
form of volume current density , since – as indicated
by Equation 9.2 – it exists only at the origin and
nowhere else.",Electromagnetics_Vol2.pdf
"form of volume current density , since – as indicated
by Equation 9.2 – it exists only at the origin and
nowhere else.
Although some current distributions approximate the
current moment, current distributions encountered in
common engineering practice generally do not exist
1 Als o a form of the Dirac delta function; see “ Additional Read-
ing” at the end of this section.
c⃝ C. W ang CC BY -SA 4.0
Figure 9.1: A +ˆz-directed current moment located at
the origin.
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"146 CHAPTER 9. RADIA TION
in precisely this form. Nevertheless, the current
moment turns out to be generally useful as a “building
block” from which practical distributions of current
can be constructed, via the principle of superposition.
Radiation from current distributions constructed in
this manner is calculated simply by summing the
radiation from each of the constituent current
moments.
Now let us consider the electric field intensity ∆E(r)
that is created by this current distribution. First, if the
current is steady (i.e., “DC”), this problem falls
within the domain of magnetostatics; i.e., the
outcome is completely described by the magnetic
field, and there can be no radiation. Therefore, let us
limit our attention to the “ AC” case, for which
radiation is possible. It will be convenient to employ
phasor representation. In phasor representation, the
current density is
∆ ˜J(r) = ˆz ˜I ∆lδ (r) (9.4)
where ˜I ∆l is simply the scalar current moment
expressed as a phasor.",Electromagnetics_Vol2.pdf
"current density is
∆ ˜J(r) = ˆz ˜I ∆lδ (r) (9.4)
where ˜I ∆l is simply the scalar current moment
expressed as a phasor.
Now we are ready to address the question “What is
∆ ˜E(r) due to ∆ ˜J(r)?” Without doing any math, we
know quite a bit about ∆ ˜E(r). For example:
• Since electric fields are proportional to the
currents that give rise to them, we expect ∆ ˜E(r)
to be proportional to
⏐⏐
⏐˜I ∆l
⏐
⏐
⏐.
• If we are sufficiently far from the origin, we
expect ∆ ˜E(r) to be approximately proportional
to 1/rwhere r≜ |r| is the distance from the
source current. This is because point sources
give rise to spherical waves, and the power
density in a spherical wave would be
proportional to 1/r2. Since time-average power
density is proportional to
⏐⏐⏐∆ ˜E(r)
⏐⏐⏐
2
, ∆ ˜E(r)
must be proportional to 1/r.
• If we are sufficiently far from the origin, and the
loss due to the medium is negligible, then we
expect the phase of ∆ ˜E(r) to change
approximately at rate βwhere βis the phase",Electromagnetics_Vol2.pdf
"loss due to the medium is negligible, then we
expect the phase of ∆ ˜E(r) to change
approximately at rate βwhere βis the phase
propagation constant 2π/λ. Since we expect
spherical phasefronts, ∆ ˜E(r) should therefore
contain the factor e−jβr.
• Ampere’s law indicates that a ˆz-directed current
at the origin should give rise to a ˆφ-directed
magnetic field in the z= 0 plane.2 At the same
time, Poynting’s theorem requires the cross
product of the electric and magnetic fields to
point in the direction of power flow . In the
present problem, this direction is away from the
source; i.e., +ˆr. Therefore, ∆ ˜E(z= 0) points in
the −ˆz direction. The same principle applies
outside of the z= 0 plane, so in general we
expect ∆ ˜E(r) to point in the ˆθdirection.
• W e expect ∆ ˜E(r) = 0 along the zaxis.
Subsequently
⏐⏐
⏐∆ ˜E(ˆr)
⏐
⏐
⏐must increase from zero
at θ= 0 and return to zero at θ= π. The
symmetry of the problem suggests
⏐
⏐⏐∆ ˜E(ˆr)
⏐⏐⏐is
maximum at θ= π/2. This magnitude must",Electromagnetics_Vol2.pdf
"at θ= 0 and return to zero at θ= π. The
symmetry of the problem suggests
⏐
⏐⏐∆ ˜E(ˆr)
⏐⏐⏐is
maximum at θ= π/2. This magnitude must
vary in the simplest possible way , leading us to
conclude that ∆ ˜E(ˆr) is proportional to sin θ.
Furthermore, the radial symmetry of the problem
means that ∆ ˜E(ˆr) should not depend at all on φ.
Putting these ideas together, we conclude that the
radiated electric field has the following form:
∆ ˜E(r) ≈ ˆθC
(
˜I ∆l
)
(sin θ) e−jβr
r (9.5)
where Cis
a constant which accounts for all of the
constants of proportionality identified in the preceding
analysis. Since the units of ∆ ˜E(r) are V/m, the units
of Cmust be Ω/m. W e have not yet accounted for the
wave impedance of the medium η, which has units of
Ω, so it would be a good bet based on the units that C
is proportional to η. However, here the informal
analysis reaches a dead end, so we shall simply state
the result from the rigorous solution: C = jηβ/4π.
The units are correct, and we finally obtain:",Electromagnetics_Vol2.pdf
"analysis reaches a dead end, so we shall simply state
the result from the rigorous solution: C = jηβ/4π.
The units are correct, and we finally obtain:
∆ ˜E(r) ≈ ˆθjηβ
4π
(
˜I ∆l
)
(s in θ) e−jβr
r (9.6)
Additional
evidence that this solution is correct
comes from the fact that it satisfies the wave equation
∇2∆ ˜E(r) + β2∆ ˜E(r) = 0. 3
2 This is sometimes described as the “right hand rule” of Am-
pere’s law .
3 Confirming this is straightforward (simply substitute and evalu-
ate) and is left as an exercise for the student.",Electromagnetics_Vol2.pdf
"9.2. MAGNETIC VECTOR POTENTIAL 147
Note that the expression we have obtained for the
radiated electric field is approximate (hence the “≈ ”).
This is due in part to our presumption of a simple
spherical wave, which may only be valid at distances
far from the source. But how far? An educated guess
would be distances much greater than a wavelength
(i.e., r≫ λ). This will do for now; in another section,
we shall show rigorously that this guess is essentially
correct.
W e conclude this section by noting that the current
distribution analyzed in this section is sometimes
referred to as a Hertzian dipole. A Hertzian dipole is
typically defined as a straight infinitesimally-thin
filament of current with length which is very small
relative to a wavelength, but not precisely zero. This
interpretation does not change the solution obtained
in this section, thus we may view the current moment
and the Hertzian dipole as effectively the same in
practical engineering applications.
Additional Reading:",Electromagnetics_Vol2.pdf
"and the Hertzian dipole as effectively the same in
practical engineering applications.
Additional Reading:
• “Dirac delta function” on Wikipedia.
• “Dipole antenna” (section entitled “Hertzian
Dipole”) on Wikipedia.
9.2 Magnetic V ector Potential
[m0195]
A common problem in electromagnetics is to
determine
the fields radiated by a specified current
distribution. This problem can be solved using
Maxwell’s equations along with the appropriate
electromagnetic boundary conditions. For
time-harmonic (sinusoidally-varying) currents, we
use phasor representation. 4 Given the specified
current distribution ˜J and the desired electromagnetic
fields ˜E and ˜H, the appropriate equations are:
∇ · ˜E = ˜ρv/ǫ (9.7)
∇ × ˜E = −jωµ˜H (9.8)
∇ · ˜H = 0 (9.9)
∇ × ˜H = ˜J + jωǫ˜E (9.10)
where ˜ρv is the volume charge density . In most
engineering problems, one is concerned with
propagation through media which are well-modeled
as homogeneous media with neutral charge, such as",Electromagnetics_Vol2.pdf
"engineering problems, one is concerned with
propagation through media which are well-modeled
as homogeneous media with neutral charge, such as
free space. 5 Therefore, in this section, we shall limit
our scope to problems in which ˜ρv = 0. Thus,
Equation 9.7 simplifies to:
∇ · ˜E = 0 (9.11)
T o solve the linear system of partial differential
Equations 9.8–9.11, it is useful to invoke the concept
of magnetic vector potential. The magnetic vector
potential is a vector field that has the useful property
that it is able to represent both the electric and
magnetic fields as a single field. This allows the
formidable system of equations identified above to be
reduced to a single equation which is simpler to solve.
Furthermore, this single equation turns out to be the
wave equation, with the slight difference that the
equation will be mathematically inhomogeneous,
with the inhomogeneous part representing the source
current.
4 Recall that there is no loss of generality in doing so, since any",Electromagnetics_Vol2.pdf
"with the inhomogeneous part representing the source
current.
4 Recall that there is no loss of generality in doing so, since any
other time-domain variation in the current distribution can be repre-
sented using sums of time-harmonic solutions via the Fourier trans-
form.
5 A counter-example would be propagation through a plasma,
which by definition consists of non-zero net charge.",Electromagnetics_Vol2.pdf
"148 CHAPTER 9. RADIA TION
The magnetic vector potential ˜A is defined by the
following relationship:
˜B ≜ ∇ × ˜A (9.12)
where ˜B = µ˜H is
the magnetic flux density . The
magnetic field appears in three of Maxwell’s
equations. For Equation 9.12 to be a reasonable
definition, ∇ × ˜A must yield reasonable results when
substituted for µ˜H in each of these equations. Let us
first check for consistency with Gauss’ law for
magnetic fields, Equation 9.9. Making the
substitution, we obtain:
∇ ·
(
∇ × ˜A
)
= 0 (9.13)
This turns out to be a mathematical identity that
applies to any vector field (see Equation B.24 in
Appendix B.3). Therefore, Equation 9.12 is
consistent with Gauss’ law for magnetic fields.
Next we check for consistency with Equation 9.8.
Making the substitution:
∇ × ˜E = −jω
(
∇ × ˜A
)
(9.14)
Gathering terms on the left, we obtain
∇ ×
(
˜E + jω˜A
)
= 0 (9.15)
Now , for reasons that will become apparent in just a
moment, we define a new scalar field ˜V and require it",Electromagnetics_Vol2.pdf
"∇ ×
(
˜E + jω˜A
)
= 0 (9.15)
Now , for reasons that will become apparent in just a
moment, we define a new scalar field ˜V and require it
to satisfy the following relationship:
− ∇ ˜V ≜ ˜E + jω˜A (9.16)
Using this definition, Equation 9.15 becomes:
∇ ×
(
−∇ ˜V
)
= 0 (9.17)
which is simply
∇ × ∇ ˜V = 0 (9.18)
Once again we have obtained a mathematical identity
that applies to any vector field (see Equation B.25 in
Appendix B.3). Therefore, ˜V can be any
mathematically-valid scalar field. Subsequently ,
Equation 9.12 is consistent with Equation 9.8
(Maxwell’s curl equation for the electric field) for any
choice of ˜V that we are inclined to make.
Astute readers might already realize what we’re up to
here. Equation 9.16 is very similar to the relationship
E = −∇V from electrostatics, 6 in which V is the
scalar electric potential field. Evidently Equation 9.16
is an enhanced version of that relationship that
accounts for the coupling with H (here, represented",Electromagnetics_Vol2.pdf
"is an enhanced version of that relationship that
accounts for the coupling with H (here, represented
by A) in the time-varying (decidedly non-static) case.
That assessment is correct, but let’s not get too far
ahead of ourselves: As demonstrated in the previous
paragraph, we are not yet compelled to make any
particular choice for ˜V, and this freedom will be
exploited later in this section.
Next we check for consistency with Equation 9.10.
Making the substitution:
∇ ×
(1
µ∇ × ˜A
)
= ˜J + j ωǫ˜E (9.19)
Multiplying both sides of the equation by µ:
∇ × ∇ × ˜A = µ˜J + jωµǫ˜E (9.20)
Next we use Equation 9.16 to eliminate ˜E, yielding:
∇ × ∇ × ˜A = µ˜J + jωµǫ
(
−∇ ˜V − jω˜A
)
(9.21)
After a bit of algebra, we obtain
∇ × ∇ × ˜A = ω2µǫ˜A − jωµǫ∇˜V + µ˜J (9.22)
Now we replace the left side of this equation using
vector identity B.29 in Appendix B.3:
∇ × ∇ × ˜A ≡ ∇
(
∇ · ˜A
)
− ∇2 ˜A (9.23)
Equation 9.22 becomes:
∇
(
∇ · ˜A
)
− ∇2 ˜A = ω2µǫ˜A − jωµǫ∇˜V + µ˜J
(9.24)",Electromagnetics_Vol2.pdf
"∇ × ∇ × ˜A ≡ ∇
(
∇ · ˜A
)
− ∇2 ˜A (9.23)
Equation 9.22 becomes:
∇
(
∇ · ˜A
)
− ∇2 ˜A = ω2µǫ˜A − jωµǫ∇˜V + µ˜J
(9.24)
Now multiplying both sides by −1 and rearranging
terms:
∇2 ˜A + ω2µǫ˜A = ∇
(
∇ · ˜A
)
+ jωµǫ∇˜V − µ˜J
(9.25)
Combining terms on the right side:
∇2 ˜A+ω2µǫ˜A = ∇
(
∇ · ˜A + jωµǫ˜V
)
−µ˜J (9.26)
Now consider the expression ∇ · ˜A + jωµǫ˜V
appearing in the parentheses on the right side of the
6 Note: No tilde in this expression.",Electromagnetics_Vol2.pdf
"9.2. MAGNETIC VECTOR POTENTIAL 149
equation. W e established earlier that ˜V can be
essentially any scalar field – from a mathematical
perspective, we are free to choose. Invoking this
freedom, we now require ˜V to satisfy the following
expression:
∇ · ˜A + jωµǫ˜V = 0 (9.27)
Clearly this is advantageous in the sense that
Equation 9.26 is now dramatically simplified. This
equation becomes:
∇2 ˜A + ω2µǫ˜A = −µ˜J
(9.28)
Note
that this expression is a wave equation. In fact
it is the same wave equation that determines ˜E and ˜H
in source-free regions, except the right-hand side is
not zero. Using mathematical terminology , we have
obtained an equation for ˜A in the form of an
inhomogeneous partial differential equation, where
the inhomogeneous part includes – no surprise here –
the source current ˜J.
Now we have what we need to find the
electromagnetic fields radiated by a current
distribution. The procedure is simply as follows:
1. Solve the partial differential Equation 9.28 for ˜A",Electromagnetics_Vol2.pdf
"electromagnetic fields radiated by a current
distribution. The procedure is simply as follows:
1. Solve the partial differential Equation 9.28 for ˜A
along with the appropriate electromagnetic
boundary conditions.
2. ˜H = (1/µ)∇ × ˜A
3. ˜E may now be determined from ˜H using
Equation 9.10.
Summarizing:
The magnetic vector potential ˜A is a vector field,
defined by Equation 9.12, that is able to represent
both the electric and magnetic fields simultane-
ously .
Also:
T o determine the electromagnetic fields radiated
by
a current distribution ˜J, one may solve Equa-
tion 9.28 for ˜A and then use Equation 9.12 to
determine ˜H and subsequently ˜E.
Specific techniques for performing this procedure – in
particular, for solving the differential equation – vary
depending on the problem, and are discussed in other
sections of this book.
W e conclude this section with a few comments about
Equation 9.27. This equation is known as the Lorenz
gauge condition. This constraint is not quite as",Electromagnetics_Vol2.pdf
"Equation 9.27. This equation is known as the Lorenz
gauge condition. This constraint is not quite as
arbitrary as the preceding derivation implies; rather,
there is some deep physics at work here. Specifically ,
the Lorenz gauge leads to the classical interpretation
of ˜V as the familiar scalar electric potential, as noted
previously in this section. (For additional information
on that idea, recommended starting points are
included in “ Additional Reading” at the end of this
section.)
At this point, it should be clear that the electric and
magnetic fields are not merely coupled quantities, but
in fact two aspects of the same field; namely , the
magnetic vector potential. In fact modern physics
(quantum mechanics) yields the magnetic vector
potential as a description of the “electromagnetic
force, ” a single entity which constitutes one of the
four fundamental forces recognized in modern
physics; the others being gravity , the strong nuclear
force, and the weak nuclear force. For more",Electromagnetics_Vol2.pdf
"four fundamental forces recognized in modern
physics; the others being gravity , the strong nuclear
force, and the weak nuclear force. For more
information on that concept, an excellent starting
point is the video “Quantum Invariance & The Origin
of The Standard Model” referenced at the end of this
section.
Additional Reading:
• “Lorenz gauge condition” on Wikipedia.
• “Magnetic potential” on Wikipedia.
• PBS Space Time video “Quantum Invariance &
The Origin of The Standard Model, ” available on
Y ouTube.",Electromagnetics_Vol2.pdf
"150 CHAPTER 9. RADIA TION
9.3 Solution of the W ave
Equation for Magnetic V ector
Potential
[m0196]
The magnetic vector potential ˜A due to a current
density ˜J is given by the following wave equation:
∇2 ˜A − γ2 ˜A = −µ˜J (9.29)
where γis the propagation constant, defined in the
usual manner 7
γ2 ≜ −ω2µǫ (9.30)
Equation 9.29 is a partial differential equation which
is inhomogeneous (in the mathematical sense) and
can be solved given appropriate boundary conditions.
In this section, we present the solution for arbitrary
distributions of current in free space.
The strategy is to first identify a solution for a
distribution of current that exists at a single point.
This distribution is the current moment. An example
of a current moment is shown in Figure 9.2. A
general expression for a current moment located at
the origin is:
˜J(r) = ˆl ˜I ∆lδ (r) (9.31)
where ˜I has units of current (SI base units of A), ∆l
has units of length (SI base units of m), ˆl is the",Electromagnetics_Vol2.pdf
"˜J(r) = ˆl ˜I ∆lδ (r) (9.31)
where ˜I has units of current (SI base units of A), ∆l
has units of length (SI base units of m), ˆl is the
7 Alternati vely , γ2 ≜ −ω2µǫc accounting for the possibility of
lossy media.
c⃝ C. W ang CC BY -SA 4.0
Figure 9.2: An example of a current moment located
at the origin. In this case, ˆl = ˆz.
direction of current flow , and δ(r) is the volumetric
sampling function 8 defined as follows:
δ(r) ≜ 0 for r ̸= 0; and (9.32)∫
V
δ(r) dv≜ 1 (9.33)
where V is any volume which includes the origin
(r = 0). It is evident from Equation 9.33 that δ(r) has
SI base units of m −3; i.e., inverse volume.
Subsequently J(r) has SI base units of A/m 2,
indicating that it is a volume current density . 9 – as
indicated by Equation 9.32 – it exists only at the
origin and nowhere else.
Substituting Equation 9.31 into Equation 9.29, we
obtain:
∇2 ˜A − γ2 ˜A = −µˆl ˜I ∆lδ (r) (9.34)
The general solution to this equation presuming
homogeneous and time-invariant media (i.e., µand ǫ",Electromagnetics_Vol2.pdf
"obtain:
∇2 ˜A − γ2 ˜A = −µˆl ˜I ∆lδ (r) (9.34)
The general solution to this equation presuming
homogeneous and time-invariant media (i.e., µand ǫ
constant with respect to space and time) is:
˜A(r) = ˆl µ ˜I ∆l e±γr
4πr (9.35)
T
o confirm that this is a solution to Equation 9.34,
substitute this expression into Equation 9.29 and
observe that the equality holds.
Note that Equation 9.35 indicates two solutions,
corresponding to the signs of the exponent in the
factor e±γr. W e can actually be a little more specific
by applying some common-sense physics. Recall that
γin general may be expressed in terms of real and
imaginary components as follows:
γ = α+ jβ (9.36)
where αis a real-valued positive constant known as
the attenuation constant and βis a real-valued
positive constant known as the phase propagation
constant. Thus, we may rewrite Equation 9.35 as
follows:
˜A(r) = ˆl µ ˜I ∆l e±αre±jβr
4πr (9.37)
8 Als o a form of the Dirac delta function; see “ Additional Read-",Electromagnetics_Vol2.pdf
"follows:
˜A(r) = ˆl µ ˜I ∆l e±αre±jβr
4πr (9.37)
8 Als o a form of the Dirac delta function; see “ Additional Read-
ing” at the end of this section.
9 Remember, the density of a volume current is with respect to
the area through which it flows, therefore the units are A/m 2.",Electromagnetics_Vol2.pdf
"9.3. SOLUTION OF THE W A VE EQUA TION FOR MAGNETIC VECTOR POTENTIAL 151
Now consider the factor e±αr/r, which determines
the dependence of magnitude on distance r. If we
choose the negative sign in the exponent, this factor
decays exponentially with increasing distance from
the origin, ultimately reaching zero at r→ ∞. This is
precisely the expected behavior, since we expect the
magnitude of a radiated field to diminish with
increasing distance from the source. If on the other
hand we choose the positive sign in the exponent, this
factor increases to infinity as r→ ∞. That outcome
can be ruled out on physical grounds, since it implies
the introduction of energy independently from the
source. The requirement that field magnitude
diminishes to zero as distance from the source
increases to infinity is known as the radiation
condition, and is essentially a boundary condition that
applies at r→ ∞. Invoking the radiation condition,
Equation 9.35 becomes:
˜A(r) = ˆl µ ˜I ∆l e−γr
4πr (9.38)
In",Electromagnetics_Vol2.pdf
"applies at r→ ∞. Invoking the radiation condition,
Equation 9.35 becomes:
˜A(r) = ˆl µ ˜I ∆l e−γr
4πr (9.38)
In
the loss-free (α = 0) case, we cannot rely on the
radiation condition to constrain the sign of γ.
However, in practical engineering work there is
always some media loss; i.e., αmight be negligible
but is not quite zero. Thus, we normally assume that
the solution for lossless (including free space)
conditions is given by Equation 9.38 with α= 0:
˜A(r) = ˆl µ ˜I ∆l e−jβr
4πr (9.39)
No
w consider the factor e−jβr in Equation 9.39. This
factor exclusively determines the dependence of the
phase of ˜A(r) with increasing distance rfrom the
source. In this case, we observe that surfaces of
constant phase correspond to spherical shells which
are concentric with the source. Thus, ˜A(r) is a
spherical wave.
Let us now consider a slightly more complicated
version of the problem in which the current moment
is no longer located at the origin, but rather is located",Electromagnetics_Vol2.pdf
"version of the problem in which the current moment
is no longer located at the origin, but rather is located
at r′. This is illustrated in Figure 9.3. This current
distribution can be expressed as follows:
˜J(r) = ˆl ˜I ∆lδ (r − r′) (9.40)
The solution in this case amounts to a straightforward
modification of the existing solution. T o see this, note
c⃝ C. W ang CC BY -SA 4.0
Figure 9.3: Current moment displaced from the ori-
gin.
that ˜A depends only on r, and not at all on θor φ. In
other words, ˜A depends only on the distance between
the “field point” r at which we observe ˜A, and the
“source point” r′ at which the current moment lies.
Thus we replace rwith this distance, which is
|r − r′|. The solution becomes:
˜A(r) = ˆl µ ˜I ∆l e−γ|r−r′|
4π|r − r′| (9.41)
Continuing
to generalize the solution, let us now
consider a scenario consisting of a filament of current
following a path C through space. This is illustrated in
Figure 9.4. Such a filament may be viewed as a",Electromagnetics_Vol2.pdf
"following a path C through space. This is illustrated in
Figure 9.4. Such a filament may be viewed as a
collection of a large number N of discrete current
moments distributed along the path. The contribution
of the nth current moment, located at rn, to the total
magnetic vector potential is:
∆ ˜A(r; rn) = ˆl(rn) µ ˜I(rn) ∆l e−γ|r−rn|
4π|r − rn| (9.42)
Note
that we are allowing both the current ˜I and
current direction ˆl to vary with position along C.
Assuming the medium is linear, superposition applies.
So we add these contributions to obtain the total",Electromagnetics_Vol2.pdf
"152 CHAPTER 9. RADIA TION
c⃝ C. W ang CC BY -SA 4.0
Figure 9.4: A filament of current lying along the path
C. This current distribution may be interpreted as a
collection of current moments lying along C.
magnetic vector potential:
˜A(r) ≈
N∑
n=1
∆ ˜A(r; rn) (9.43)
≈ µ
4π
N∑
n=1
ˆl( rn) ˜I(rn) e−γ|r−rn|
|r − rn| ∆l (9.44)
No
w letting ∆l → 0 so that we may replace ∆l with
the differential length dl:
˜A(r) = µ
4π
∫
C
ˆl(r′) ˜I(r′) e−γ|r−r′|
|r − r′| dl (9.45)
where
we have also replaced rn with the original
notation r′ since we once again have a continuum (as
opposed to a discrete set) of source locations.
The magnetic vector potential corresponding to
radiation
from a line distribution of current is
given by Equation 9.45.
Given ˜A(r), the magnetic and electric fields may be
determined using the procedure developed in
Section 9.2.
Additional Reading:
• “Magnetic potential” on Wikipedia.
• “Dirac delta function” on Wikipedia.
9.4 Radiation from a Hertzian
Dipole
[m0197]",Electromagnetics_Vol2.pdf
"Additional Reading:
• “Magnetic potential” on Wikipedia.
• “Dirac delta function” on Wikipedia.
9.4 Radiation from a Hertzian
Dipole
[m0197]
Section 9.1 presented an informal derivation of the
electromagnetic
field radiated by a Hertzian dipole
represented by a zero-length current moment. In this
section, we provide a rigorous derivation using the
concept of magnetic vector potential discussed in
Sections 9.2 and 9.3. A review of those sections is
recommended before tackling this section.
A Hertzian dipole is commonly defined as an
electrically-short and infinitesimally-thin straight
filament of current, in which the density of the current
is uniform over its length. The Hertzian dipole is
commonly used as a “building block” for constructing
physically-realizable distributions of current as
exhibited by devices such as wire antennas. The
method is to model these relatively complex
distributions of current as the sum of Hertzian dipoles,
which reduces the problem to that of summing the",Electromagnetics_Vol2.pdf
"method is to model these relatively complex
distributions of current as the sum of Hertzian dipoles,
which reduces the problem to that of summing the
contributions of the individual Hertzian dipoles, with
each Hertzian dipole having the appropriate (i.e.,
different) position, magnitude, and phase.
T o facilitate use of the Hertzian dipole as a building
block suitable for constructing physically-realizable
distributions of current, we choose to represent the
Hertzian dipole using an essentially equivalent current
distribution which is mathematically more versatile.
This description of the Hertzian dipole replaces the
notion of constant current over finite length with the
notion of a current moment located at a single point.
This is shown in Figure 9.5, and is given by:
∆ ˜J(r) = ˆl ˜I ∆lδ (r) (9.46)
where the product ˜I∆l (SI base units of A·m) is the
current moment, ˆl is the direction of current flow , and
δ(r) is the volumetric sampling function defined as
follows:",Electromagnetics_Vol2.pdf
"current moment, ˆl is the direction of current flow , and
δ(r) is the volumetric sampling function defined as
follows:
δ(r) ≜ 0 for r ̸= 0; and (9.47)∫
V
δ(r) dv≜ 1 (9.48)
where V is any volume which includes the origin
(r = 0). In this description, the Hertzian dipole is
located at the origin.",Electromagnetics_Vol2.pdf
"9.4. RADIA TION FROM A HER TZIAN DIPOLE 153
The solution for the magnetic vector potential due to a
ˆz-directed Hertzian dipole located at the origin was
presented in Section 9.3. In the present scenario, it is:
˜A(r) = ˆz µ ˜I ∆l e−γr
4πr (9.49)
where
the propagation constant γ = α+ jβ as usual.
Assuming lossless media (α = 0), we have
˜A(r) = ˆz µ ˜I ∆l e−jβr
4πr (9.50)
W
e obtain the magnetic field intensity using the
definition of magnetic vector potential:
˜H ≜ (1/µ)∇ × ˜A (9.51)
=
˜I ∆l
4π ∇ × ˆz e−j
βr
r (9.52)
T
o proceed, it is useful to convert ˆz into the spherical
coordinate system. T o do this, we find the component
of ˆz that is parallel to ˆr, ˆθ, and ˆφ; and then sum the
results:
ˆz = ˆr (ˆr · ˆz) + ˆθ
(
ˆθ· ˆz
)
+ ˆφ
(
ˆφ· ˆz
)
(9.53)
= ˆr cos θ− ˆθsin θ+ 0 (9.54)
Equation 9.52 requires computation of the following
quantity:
∇ × ˆze−jβr
r = ∇ ×
[
ˆr (cos θ) e−j
βr
r
−ˆθ(s
in θ) e−jβr
r
]
(9.55)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.5: A Hertzian dipole located at the origin,",Electromagnetics_Vol2.pdf
"∇ × ˆze−jβr
r = ∇ ×
[
ˆr (cos θ) e−j
βr
r
−ˆθ(s
in θ) e−jβr
r
]
(9.55)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.5: A Hertzian dipole located at the origin,
represented as a current moment. In this case, ˆl = ˆz.
At this point, it is convenient to make the following
definitions:
Cr ≜ (cos θ) e−jβr
r (9.56)
Cθ ≜ − (
sin θ) e−jβr
r (9.57)
These
definitions allow Equation 9.55 to be written
compactly as follows:
∇ × ˆze−jβr
r = ∇ ×
[
ˆrCr + ˆθ Cθ
]
(9.58)
The right side of Equation 9.58 is evaluated using
Equation B.18 (Appendix B.2). Although the
complete expression consists of 6 terms, only 2 terms
are non-zero. 10 This leaves:
∇ × ˆze−jβr
r = ˆφ1
r
[ ∂
∂r (rCθ) − ∂
∂θCr
]
(9.59)
= ˆφ(
sin θ) e−jβr
r
(
jβ + 1
r
)
(9.60)
Substituting
this result into Equation 9.52, we obtain:
˜H = ˆφ
˜I ∆l
4π (s in θ) e−jβr
r
(
jβ + 1
r
)
(9.61)
Let
us further limit our scope to the field far from the
antenna. Specifically , let us assume r≫ λ. Now we
use the relationship β = 2π/λ and determine the
following:",Electromagnetics_Vol2.pdf
"antenna. Specifically , let us assume r≫ λ. Now we
use the relationship β = 2π/λ and determine the
following:
jβ + 1
r = j2π
λ + 1
r (9.62)
≈ j2
π
λ = jβ (9.63)
Equation
9.61 becomes:
˜H ≈ ˆφj
˜I· β∆l
4π (s in θ) e−jβr
r (9.64)
where
the approximation holds for low-loss media
and r≫ λ. This expression is known as a far field
approximation, since it is valid only for distances
“far” (relative to a wavelength) from the source.
Now let us take a moment to interpret this result:
10 Specifically , two terms are zero because there is no ˆφ compo-
nent in the argument of the curl function; and another two terms are
zero because the argument of the curl function is independent of φ,
so partial derivatives with respect to φ are zero.",Electromagnetics_Vol2.pdf
"154 CHAPTER 9. RADIA TION
• Notice the factor β∆l has units of radians; that
is, it is electrical length. This tells us that the
magnitude of the radiated field depends on the
electrical length of the current moment.
• The factor e−jβr/rindicates that this is a
spherical wave; that is, surfaces of constant
phase correspond to concentric spheres centered
on the source, and magnitude is inversely
proportional to distance.
• The direction of the magnetic field vector is
always ˆφ, which is precisely what we expect; for
example, using the Biot-Savart law .
• Finally , note the factor sin θ. This indicates that
the field magnitude is zero along the direction in
which the source current flows, and is a
maximum in the plane perpendicular to this
direction.
Now let us determine the electric field radiated by the
Hertzian dipole. The direct method is to employ
Ampere’s law . That is,
˜E = 1
jωǫ∇ × ˜H (9.65)
where ˜H is
given by Equation 9.64. At field points far",Electromagnetics_Vol2.pdf
"Ampere’s law . That is,
˜E = 1
jωǫ∇ × ˜H (9.65)
where ˜H is
given by Equation 9.64. At field points far
from the dipole, the radius of curvature of the
spherical phasefronts is very large and so appear to be
locally planar. That is, from the perspective of an
observer far from the dipole, the arriving wave
appears to be a plane wave. In this case, we may
employ the plane wave relationships. The appropriate
relationship in this case is:
˜E = −ηˆr × ˜H (9.66)
where ηis the wave impedance. So we find:
˜E ≈ ˆθjη
˜I· β∆l
4π (s in θ) e−jβr
r (9.67)
Summarizing:
The electric and magnetic fields far (i.e., ≫ λ)
from
a ˆz-directed Hertzian dipole having con-
stant current ˜I over length ∆l , located at the ori-
gin, are given by Equations 9.67 and 9.64, respec-
tively .
Additional Reading:
• “Dipole antenna” (section entitled “Hertzian
Dipole”) on Wikipedia.",Electromagnetics_Vol2.pdf
"9.5. RADIA TION FROM AN ELECTRICALL Y -SHOR T DIPOLE 155
9.5 Radiation from an
Electrically-Short Dipole
[m0198]
The simplest distribution of radiating current that is
encountered
in common practice is the
electrically-short dipole (ESD). This current
distribution is shown in Figure 9.6. The two
characteristics that define the ESD are (1) the current
is aligned along a straight line, and (2) the length Lof
the line is much less than one-half of a wavelength;
i.e., L≪ λ/2. The latter characteristic is what we
mean by “electrically-short. ” 11
The current distribution of an ESD is approximately
triangular in magnitude, and approximately constant
in phase. How do we know this? First, note that
distributions of current cannot change in a complex or
rapid way over such distances which are much less
than a wavelength. If this is not immediately
apparent, recall the behavior of transmission lines:
11 A potential source of confusion is that the Hertzian dipole is",Electromagnetics_Vol2.pdf
"apparent, recall the behavior of transmission lines:
11 A potential source of confusion is that the Hertzian dipole is
also a “dipole” which is “electrically-short. ” The distinction is that
the current comprising a Hertzian dipole is constant over its length.
This condition is rarely and only approximately seen in practice,
whereas the triangular magnitude distribution is a relatively good ap-
proximation to a broad class of commonly-encountered electrically-
short wire antennas. Thus, the term “electrically-short dipole, ” as
used in this book, refers to the triangular distribution unless noted
otherwise.
c⃝ C. W ang CC BY -SA 4.0
Figure 9.6: Current distribution of the electrically-
short dipole (ESD).
c⃝ C. W ang CC BY -SA 4.0
Figure 9.7: Current distribution of the electrically-
short dipole (ESD) approximated as a large number of
Hertzian dipoles.
The current standing wave on a transmission line
exhibits a period of λ/2, regardless the source or",Electromagnetics_Vol2.pdf
"Hertzian dipoles.
The current standing wave on a transmission line
exhibits a period of λ/2, regardless the source or
termination. For the ESD, L≪ λ/2 and so we expect
an even simpler variation. Also, we know that the
current at the ends of the dipole must be zero, simply
because the dipole ends there. These considerations
imply that the current distribution of the ESD is
well-approximated as triangular in magnitude. 12
Expressed mathematically:
˜I(z) ≈ I0
(
1 − 2
L|z|
)
(9.68)
where I0 (SI
base units of A) is a complex-valued
constant indicating the maximum current magnitude
and phase.
There are two approaches that we might consider in
order to find the electric field radiated by an ESD. The
first approach is to calculate the magnetic vector
potential ˜A by integration over the current
distribution, calculate ˜H = (1/µ)∇ × ˜A, and finally
calculate ˜E from ˜H using Ampere’s law . W e shall
employ a simpler approach, shown in Figure 9.7.
Imagine the ESD as a collection of many shorter",Electromagnetics_Vol2.pdf
"calculate ˜E from ˜H using Ampere’s law . W e shall
employ a simpler approach, shown in Figure 9.7.
Imagine the ESD as a collection of many shorter
segments of current that radiate independently . The
total field is then the sum of these short segments.
Because these segments are very short relative to the
12 A more rigorous analysis leading to the same conclusion is pos-
sible, but is beyond the scope of this book.",Electromagnetics_Vol2.pdf
"156 CHAPTER 9. RADIA TION
length of the dipole as well as being short relative to a
wavelength, we may approximate the current over
each segment as approximately constant. In other
words, we may interpret each of these segments as
being, to a good approximation, a Hertzian dipole.
The advantage of this approach is that we already
have a solution for each of the segments. In
Section 9.4, it is shown that a ˆz-directed Hertzian
dipole at the origin radiates the electric field
˜E(r) ≈ ˆθjη
˜I· β∆l
4π (s in θ) e−jβr
r (9.69)
where ˜I and ∆
lmay be interpreted as the current and
length of the dipole, respectively . In this expression, η
is the wave impedance of medium in which the dipole
radiates (e.g., ≈ 377 Ω for free space), and we have
presumed lossless media such that the attenuation
constant α≈ 0 and the phase propagation constant
β = 2π/λ. This expression also assumes field points
far from the dipole; specifically , distances rthat are",Electromagnetics_Vol2.pdf
"β = 2π/λ. This expression also assumes field points
far from the dipole; specifically , distances rthat are
much greater than λ. Repurposing this expression for
the present problem, the segment at the origin radiates
the electric field:
˜E(r; z′ = 0) ≈ ˆθjηI0 · β∆l
4π (s in θ) e−jβr
r (9.70)
where
the notation z′ = 0 indicates the Hertzian
dipole is located at the origin. Letting the length ∆l
of this segment shrink to differential length dz′, we
may describe the contribution of this segment to the
field radiated by the ESD as follows:
d˜E(r; z′ = 0) ≈ ˆθjηI0 · βdz′
4π (s in θ) e−jβr
r (9.71)
Using
this approach, the electric field radiated by any
segment can be written:
d˜E(r; z′) ≈ ˆθ′jηβ
˜I(z′)
4π (s in θ′) e−jβ|r−ˆzz′|
|r − ˆzz′| dz′
(9.72)
Note
that θis replaced by θ′ since the ray r − ˆzz′
forms a different angle (i.e., θ′) with respect to ˆz.
Similarly , ˆθis replaced by ˆθ′, since it also varies with
z′. The electric field radiated by the ESD is obtained",Electromagnetics_Vol2.pdf
"Similarly , ˆθis replaced by ˆθ′, since it also varies with
z′. The electric field radiated by the ESD is obtained
by integration over these contributions:
˜E(r) ≈
∫ +L/ 2
−L/ 2
d˜E(ˆr; z′) (9.73)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.8: Parallel ray approximation for an ESD.
yielding:
˜E(r) ≈ jηβ
4π
∫ +L/ 2
−L/ 2
ˆθ′ ˜I( z′) (sin θ′) e−jβ|r−ˆzz′|
|r − ˆzz′| dz′
(9.74)
Gi
ven some of the assumptions we have already
made, this expression can be further simplified. For
example, note that θ′ ≈ θsince L≪ r. For the same
reason, ˆθ′ ≈ ˆθ. Since these variables are
approximately constant over the length of the dipole,
we may move them outside the integral, yielding:
˜E(r) ≈ ˆθjηβ
4π (s in θ)
∫ +L/ 2
−L/ 2
˜I(z′)e−jβ|r−ˆzz′|
|r − ˆzz′| dz′
(9.75)
It
is also possible to simplify the expression |r − ˆzz′|.
Consider Figure 9.8. Since we have already assumed
that r≫ L(i.e., the distance to field points is much
greater than the length of the dipole), the vector r is",Electromagnetics_Vol2.pdf
"that r≫ L(i.e., the distance to field points is much
greater than the length of the dipole), the vector r is
approximately parallel to the vector r − ˆzz′.
Subsequently , it must be true that
|r − ˆzz′| ≈ r− ˆr · ˆzz′ (9.76)
Note that the magnitude of r− ˆr · ˆzz′ must be
approximately equal to r, since r≫ L. So, insofar as
|r − ˆzz′| determines the magnitude of ˜E(r), we may
use the approximation:
|r − ˆzz′| ≈ r (magnitude) (9.77)
Insofar as |r − ˆzz′| determines phase, we have to be a
bit more careful. The part of the integrand of",Electromagnetics_Vol2.pdf
"9.5. RADIA TION FROM AN ELECTRICALL Y -SHOR T DIPOLE 157
Equation 9.75 that exhibits varying phase is
e−jβ|r−ˆzz′|. Using Equation 9.76, we find
e−jβ|r−ˆzz′| ≈ e−jβre+jβˆr·ˆzz′
(9.78)
The worst case in terms of phase variation within the
integral is for field points along the zaxis. For these
points, ˆr · ˆz = ±1 and subsequently |r − ˆzz′| varies
from z− L/2 to z+ L/2 where zis the location of
the field point. However, since L≪ λ(i.e., because
the dipole is electrically short), this difference in
lengths is much less than λ/2. Therefore, the phase
βˆr · ˆzz′ varies by much less than πradians, and
subsequently e−jβˆr·ˆzz′
≈ 1. W e conclude that under
these conditions,
e−jβ|r−ˆzz′| ≈ e−jβr (phase) (9.79)
Applying these simplifications for magnitude and
phase to Equation 9.75, we obtain:
˜E(r) ≈ ˆθjηβ
4π (s in θ) e−jβr
r
∫ +L/ 2
−L/ 2
˜I( z′)dz′
(9.80)
The integral in this equation is very easy to evaluate;
in fact, from inspection (Figure 9.6), we determine it",Electromagnetics_Vol2.pdf
"r
∫ +L/ 2
−L/ 2
˜I( z′)dz′
(9.80)
The integral in this equation is very easy to evaluate;
in fact, from inspection (Figure 9.6), we determine it
is equal to I0L/2. Finally , we obtain:
˜E(r) ≈ ˆθjηI0 · βL
8π (s in θ) e−jβr
r (9.81)
Summarizing:
The electric field intensity radiated by an ESD lo-
cated
at the origin and aligned along the zaxis is
given by Equation 9.81. This expression is valid
for r≫ λ.
It is worth noting that the variation in magnitude,
phase, and polarization of the ESD with field point
location is identical to that of a single Hertzian dipole
having current moment ˆzI0L/2 (Section 9.4).
However, the magnitude of the field radiated by the
ESD is exactly one-half that of the Hertzian dipole.
Why one-half? Simply because the integral over the
triangular current distribution assumed for the ESD is
one-half the integral over the uniform current
distribution that defines the Hertzian dipole. This
similarly sometimes causes confusion between",Electromagnetics_Vol2.pdf
"one-half the integral over the uniform current
distribution that defines the Hertzian dipole. This
similarly sometimes causes confusion between
Hertzian dipoles and ESDs. Remember that ESDs are
physically realizable, whereas Hertzian dipoles are
not.
It is common to eliminate the factor of βin the
magnitude using the relationship β = 2π/λ, yielding:
˜E(r) ≈ ˆθjηI0
4
L
λ (s in θ) e−jβr
r (9.82)
At
field points r≫ λ, the wave appears to be locally
planar. Therefore, we are justified using the plane
wave relationship ˜H = 1
ηˆr × ˜E to calculate ˜H. The
result is:
˜H(r) ≈ ˆφjI0
4
L
λ (s in θ) e−jβr
r (9.83)
Finally
, let us consider the spatial characteristics of
the radiated field. Figures 9.9 and 9.10 show the
result in a plane of constant φ. Figures 9.11 and 9.12
show the result in the z= 0 plane. Note that the
orientations of the electric and magnetic field vectors
indicate a Poynting vector ˜E × ˜H that is always
directed radially outward from the location of the",Electromagnetics_Vol2.pdf
"indicate a Poynting vector ˜E × ˜H that is always
directed radially outward from the location of the
dipole. This confirms that power flow is always
directed radially outward from the dipole. Due to the
symmetry of the problem, Figures 9.9–9.12 provide a
complete characterization of the relative magnitudes
and orientations of the radiated fields.",Electromagnetics_Vol2.pdf
"158 CHAPTER 9. RADIA TION
c⃝ S. Lally CC BY -SA 4.0
Figure 9.9: Magnitude of the radiated field in any
plane of constant φ.
c⃝ S. Lally CC BY -SA 4.0
Figure 9.10: Orientation of the electric and magnetic
fields in any plane of constant φ.
c⃝ S. Lally CC BY -SA 4.0
Figure 9.11: Magnitude of the radiated field in any
plane of constant z.
c⃝ S. Lally CC BY -SA 4.0
Figure 9.12: Orientation of the electric and magnetic
fields in the z= 0 plane.",Electromagnetics_Vol2.pdf
"9.6. F AR-FIELD RADIA TION FROM A THIN STRAIGHT FILAMENT OF CURRENT 159
9.6 Far-Field Radiation from a
Thin Straight Filament of
Current
[m0199]
A simple distribution of radiating current that is
encountered
in common practice is the thin straight
current filament, shown in Figure 9.13. The defining
characteristic of this distribution is that the current
filament is aligned along a straight line, and that the
maximum dimension of the cross-section of the
filament is much less than a wavelength. The latter
characteristic is what we mean by “thin” – i.e., the
filament is “electrically thin. ” Physical distributions
of current that meet this description include the
electrically-short dipole (Section 9.5) and the
half-wave dipole (Section 9.7) among others. A
gentler introduction to this distribution is via
Section 9.5 (“Radiation from an Electrically-Short
Dipole”), so students may want to review that section
first. This section presents the more general case.",Electromagnetics_Vol2.pdf
"Dipole”), so students may want to review that section
first. This section presents the more general case.
Let the magnitude and phase of current along the
filament be given by the phasor quantity ˜I(z) (SI base
units of A). In principle, the only constraint on ˜I(z)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.13: A thin straight distribution of radiating
current.
c⃝ C. W ang CC BY -SA 4.0
Figure 9.14: Current distribution approximated as a
set of Hertzian dipoles.
that applies generally is that it must be zero at the
ends of the distribution; i.e., ˜I(z) = 0 for |z| ≥ L/2.
Details beyond this constraint depend on Land
perhaps other factors. Nevertheless, the
characteristics of radiation from members of this class
of current distributions have much in common,
regardless of L. These become especially apparent if
we limit our scope to field points far from the current,
as we shall do here.
There are two approaches that we might consider in
order to find the electric field radiated by this",Electromagnetics_Vol2.pdf
"as we shall do here.
There are two approaches that we might consider in
order to find the electric field radiated by this
distribution. The first approach is to calculate the
magnetic vector potential ˜A by integration over the
current distribution (Section 9.3), calculate
˜H = (1/µ)∇ × ˜A, and finally calculate ˜E from ˜H
using the differential form of Ampere’s law . In this
section, we shall employ a simpler approach, shown
in Figure 9.14. Imagine the filament as a collection of
many shorter segments of current that radiate
independently . The total field is then the sum of these
short segments. Because these segments are very
short relative to the length of the filament as well as
being short relative to a wavelength, we may
approximate the current over each segment as
constant. In other words, we may interpret each
segment as being, to a good approximation, a
Hertzian dipole.
The advantage of this approach is that we already
have a solution for every segment. In Section 9.4, it is",Electromagnetics_Vol2.pdf
"160 CHAPTER 9. RADIA TION
shown that a ˆz-directed Hertzian dipole at the origin
radiates the electric field
˜E(r) ≈ ˆθjη
˜I(β∆l )
4π (s in θ) e−jβr
r (9.84)
where ˜I and ∆
lmay be interpreted as the current and
length of the filament, respectively . In this expression,
ηis the wave impedance of medium in which the
filament radiates (e.g., ≈ 377 Ω for free space), and
we have presumed lossless media such that the
attenuation constant α≈ 0 and the phase propagation
constant β = 2π/λ. This expression also assumes
field points far from the filament; specifically ,
distances rthat are much greater than λ. Repurposing
this expression for the present problem, we note that
the segment at the origin radiates the electric field:
˜E(r; z′ = 0) ≈ ˆθjη
˜I(0) (β∆l )
4π (s in θ) e−jβr
r(9.85)
where
the notation z′ = 0 indicates the Hertzian
dipole is located at the origin. Letting the length ∆l
of this segment shrink to differential length dz′, we
may describe the contribution of this segment to the",Electromagnetics_Vol2.pdf
"of this segment shrink to differential length dz′, we
may describe the contribution of this segment to the
field radiated by the ESD as follows:
d˜E(r; z′ = 0) ≈ ˆθjη
˜I(0) (βdz′)
4π (s in θ) e−jβr
r(9.86)
Using
this approach, the electric field radiated by any
segment can be written:
d˜E(r; z′) ≈ ˆθ′jηβ
˜I(z′)
4π (s in θ′) e−jβ|r−ˆzz′|
|r − ˆzz′| dz′
(9.87)
Note
that θis replaced by θ′ since the ray r − ˆzz′
forms a different angle (i.e., θ′) with respect to ˆz.
Subsequently , ˆθis replaced by ˆθ′, which varies
similarly with z′. The electric field radiated by the
filament is obtained by integration over these
contributions, yielding:
˜E(r) ≈
∫ +L/ 2
−L/ 2
d˜E(ˆr; z′) (9.88)
Expanding this expression:
˜E(r) ≈ jηβ
4π
∫ +L/ 2
−L/ 2
ˆθ′ ˜I( z′) (sin θ′) e−jβ|r−ˆzz′|
|r − ˆzz′| dz′
(9.89)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.15: Parallel ray approximation.
Given some of the assumptions we have already
made, this expression can be further simplified. For",Electromagnetics_Vol2.pdf
"Figure 9.15: Parallel ray approximation.
Given some of the assumptions we have already
made, this expression can be further simplified. For
example, note that θ′ ≈ θsince L≪ r. For the same
reason, ˆθ′ ≈ ˆθ. Since these variables are
approximately constant over the length of the
filament, we may move them outside the integral,
yielding:
˜E(r) ≈ ˆθjηβ
4π (s in θ)
∫ +L/ 2
−L/ 2
˜I(z′)e−jβ|r−ˆzz′|
|r − ˆzz′| dz′
(9.90)
It
is also possible to simplify the expression |r − ˆzz′|.
Consider Figure 9.15. Since we have already assumed
that r≫ L(i.e., the distance to field points is much
greater than the length of the filament), the vector r is
approximately parallel to the vector r − ˆzz′.
Subsequently , it must be true that
|r − ˆzz′| ≈ r− ˆr · ˆzz′ (9.91)
≈ r− z′ cos θ (9.92)
Note that the magnitude of r− ˆr · ˆzz′ must be
approximately equal to r, since r≫ L. So, insofar as
|r − ˆzz′| determines the magnitude of ˜E(r), we may
use the approximation:
|r − ˆzz′| ≈ r (magnitude) (9.93)",Electromagnetics_Vol2.pdf
"|r − ˆzz′| determines the magnitude of ˜E(r), we may
use the approximation:
|r − ˆzz′| ≈ r (magnitude) (9.93)
Insofar as |r − ˆzz′| determines phase, we have to be
more careful. The part of the integrand of
Equation 9.90 that exhibits varying phase is
e−jβ|r−ˆzz′|. Using Equation 9.91, we find
e−jβ|r−ˆzz′| ≈ e−jβre+jβz′ cos θ (9.94)",Electromagnetics_Vol2.pdf
"9.7. F AR-FIELD RADIA TION FROM A HALF-W A VE DIPOLE 161
These simplifications are known collectively as a far
field approximation, since they are valid only for
distances “far” from the source.
Applying these simplifications for magnitude and
phase to Equation 9.90, we obtain:
˜E(r) ≈ ˆθjηβ
4π
e−jβ
r
r (s in θ)
·
∫ +L/ 2
−L/ 2
˜I(z′)e+jβz′ cos θdz′ (9.95)
Finally , it is common to eliminate the factor of βin
the magnitude using the relationship β = 2π/λ,
yielding:
˜E(r) ≈ ˆθjη
2
e−jβ
r
r (s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
(9.96)
The electric field radiated by a thin, straight, ˆz-
directed
current filament of length L located at
the origin and aligned along the z axis is given
by Equation 9.96. This expression is valid for
r≫ Land r≫ λ.
At field points satisfying the conditions r≫ Land
r≫ λ, the wave appears to be locally planar.
Therefore, we are justified using the plane wave
relationship ˜H = (1/η)ˆr × ˜E to calculate ˜H.",Electromagnetics_Vol2.pdf
"r≫ λ, the wave appears to be locally planar.
Therefore, we are justified using the plane wave
relationship ˜H = (1/η)ˆr × ˜E to calculate ˜H.
As a check, one may readily verify that Equation 9.96
yields the expected result for the electrically-short
dipole (Section 9.5).
9.7 Far-Field Radiation from a
Half-W ave Dipole
[m0200]
A simple and important current distribution is that of
the
thin half-wave dipole (HWD), shown in
Figure 9.16. This is the distribution expected on a thin
straight wire having length L= λ/2, where λis
wavelength. This distribution is described
mathematically as follows:
˜I(z) ≈ I0 cos
(
πz
L
)
for | z| ≤ L
2 (9.97)
where I0 (SI
base units of A) is a complex-valued
constant indicating the maximum magnitude of the
current and its phase. Note that the current is zero at
the ends of the dipole; i.e., ˜I(z) = 0 for |z| = L/2.
Note also that this “cosine pulse” distribution is very
similar to the triangular distribution of the ESD, and",Electromagnetics_Vol2.pdf
"Note also that this “cosine pulse” distribution is very
similar to the triangular distribution of the ESD, and
is reminiscent of the sinusoidal variation of current in
a standing wave.
Since L= λ/2 for the HWD, Equation 9.97 may
equivalently be written:
˜I(z) ≈ I0 cos
(
2πz
λ
)
(9.98)
The
electromagnetic field radiated by this distribution
c⃝ C. W ang CC BY -SA 4.0
Figure 9.16: Current distribution of the half-wave
dipole (HWD).",Electromagnetics_Vol2.pdf
"162 CHAPTER 9. RADIA TION
of current may be calculated using the method
described in Section 9.6, in particular:
˜E(r) ≈ ˆθjη
2
e−jβ
r
r (s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
(9.99)
which is valid for field points r far from the dipole;
i.e., for r≫ Land r≫ λ. For the HWD, the quantity
in square brackets is
I0
λ
∫ +λ/ 4
−λ / 4
cos
(
2πz′
λ
)
e+jβ
z′ cos θdz′ (9.100)
The evaluation of this integral is straightforward, but
tedious. The integral reduces to
I0
π
cos
[(π/2) cos θ]
sin2 θ (9.101)
Substitution
into Equation 9.99 yields
˜E(r) ≈ ˆθjηI0
2π
cos
[(π/2) cos θ]
sin θ
e−jβ
r
r (9.102)
The
magnetic field may be determined from this
result using Ampere’s law . However, a simpler
method is to use the fact that the electric field,
magnetic field, and direction of propagation ˆr are
mutually perpendicular and related by:
˜H = 1
ηˆr × ˜E (9.103)
This
relationship indicates that the magnetic field will
be + ˆφ-directed.
The magnitude and polarization of the radiated field is",Electromagnetics_Vol2.pdf
"ηˆr × ˜E (9.103)
This
relationship indicates that the magnetic field will
be + ˆφ-directed.
The magnitude and polarization of the radiated field is
similar to that of the electrically-short dipole (ESD;
Section 9.5). A comparison of the magnitudes in any
radial plane containing the z-axis is shown in
Figure 9.17. For either current distribution, the
maximum magnitude of the fields occurs in the z= 0
plane. For a given terminal current I0, the maximum
magnitude is greater for the HWD than for the ESD.
Both current distributions yield zero magnitude along
the axis of the dipole. The polarization characteristics
of the fields of both current distributions are identical.
Additional Reading:
• “Dipole antenna” (section entitled “Half-wave
dipole”) on Wikipedia.
c⃝ C. W ang CC BY -SA 4.0
Figure 9.17: Comparison of the magnitude of the ra-
diated field of the HWD to that of an electrically-short
dipole also oriented along the z-axis. This result is for
any radial plane that includes the z-axis.",Electromagnetics_Vol2.pdf
"dipole also oriented along the z-axis. This result is for
any radial plane that includes the z-axis.
9.8 Radiation from Surface and
V olume Distributions of
Current
[m0221]
In Section 9.3, a solution was developed for radiation
from
current located at a single point r′. This current
distribution was expressed mathematically as a
current moment, as follows:
˜J(r) = ˆl ˜I ∆lδ (r − r′) (9.104)
where ˆl is the direction of current flow , ˜I has units of
current (SI base units of A), ∆l has units of length (SI
base units of m), and δ(r) is the volumetric sampling
function (Dirac “delta” function; SI base units of
m−3). In this formulation, ˜J has SI base units of
A/m2. The magnetic vector potential radiated by this
current, observed at the field point r, was found to be
˜A(r) = ˆl µ ˜I ∆l e−γ|r−r′|
4π|r − r′| (9.105)
This
solution was subsequently generalized to obtain
the radiation from any distribution of line current; i.e.,",Electromagnetics_Vol2.pdf
"9.8. RADIA TION FROM SURF ACE AND VOLUME DISTRIBUTIONS OF CURRENT 163
any distribution of current that is constrained to flow
along a single path through space, as along an
infinitesimally-thin wire.
In this section, we derive an expression for the
radiation from current that is constrained to flow
along a surface and from current which flows through
a volume. The solution in both cases can be obtained
by “recycling” the solution for line current as follows.
Letting ∆l shrink to the differential length dl,
Equation 9.104 becomes:
d˜J(r) = ˆl ˜I dlδ(r − r′) (9.106)
Subsequently , Equation 9.105 becomes:
d˜A(r; r′) = ˆl µ ˜I dle−γ|r−r′|
4π|r − r′| (9.107)
where
the notation d˜A(r; r′) is used to denote the
magnetic vector potential at the field point r due to
the differential-length current moment at the source
point r′. Next, consider that any distribution of
current can be described as a distribution of current
moments. By the principle of superposition, the",Electromagnetics_Vol2.pdf
"current can be described as a distribution of current
moments. By the principle of superposition, the
radiation from this distribution of current moments
can be calculated as the sum of the radiation from the
individual current moments. This is expressed
mathematically as follows:
˜A(r) =
∫
d˜A(r; r′) (9.108)
where the integral is over r′; i.e., summing over the
source current.
If we substitute Equation 9.107 into Equation 9.108,
we obtain the solution derived in Section 9.3, which is
specific to line distributions of current. T o obtain the
solution for surface and volume distributions of
current, we reinterpret the definition of the differential
current moment in Equation 9.106. Note that this
current is completely specified by its direction ( ˆl) and
the quantity ˜I dl, which has SI base units of A·m. W e
may describe the same current distribution
alternatively as follows:
d˜J(r) = ˆl ˜Js dsδ(r − r′) (9.109)
where ˜Js has units of surface current density (SI base",Electromagnetics_Vol2.pdf
"alternatively as follows:
d˜J(r) = ˆl ˜Js dsδ(r − r′) (9.109)
where ˜Js has units of surface current density (SI base
units of A/m) and dshas units of area (SI base units
of m 2). T o emphasize that this is precisely the same
current moment, note that ˜Js ds, like ˜I dl, has units
of A·m. Similarly ,
d˜J(r) = ˆl ˜J dvδ(r − r′) (9.110)
where ˜J has units of volume current density (SI base
units of A/m 2) and dvhas units of volume (SI base
units of m 3). Again, ˜J dvhas units of A·m.
Summarizing, we have found that
˜I dl= ˜Js ds= ˜J dv (9.111)
all describe the same differential current moment.
Thus, we may obtain solutions for surface and volume
distributions of current simply by replacing ˜I dlin
Equation 9.107, and subsequently in Equation 9.108,
with the appropriate quantity from Equation 9.111.
For a surface current distribution, we obtain:
˜A(r) = µ
4π
∫
S
˜Js(r′) e−γ|r−r′|
|r − r′| ds (9.112)
where ˜Js(
r′) ≜ ˆl(r′) ˜Js(r′) and where S is the",Electromagnetics_Vol2.pdf
"˜A(r) = µ
4π
∫
S
˜Js(r′) e−γ|r−r′|
|r − r′| ds (9.112)
where ˜Js(
r′) ≜ ˆl(r′) ˜Js(r′) and where S is the
surface over which the current flows. Similarly for a
volume current distribution, we obtain:
˜A(r) = µ
4π
∫
V
˜J(r′) e−γ|r−r′|
|r − r′| dv (9.113)
where ˜J
(r′) ≜ ˆl(r′) ˜J(r′) and where V is the volume
in which the current flows.
The magnetic vector potential corresponding to
radiation
from a surface and volume distribution
of current is given by Equations 9.112 and 9.113,
respectively .
Given ˜A(r), the magnetic and electric fields may be
determined using the procedure developed in
Section 9.2.
[m0217]",Electromagnetics_Vol2.pdf
"164 CHAPTER 9. RADIA TION
Image Credits
Fig. 9.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.3: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Current moment displaced from the origin.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.4: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A filament of current lying along the path c.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.5: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.6: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Current distribution of the esd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.7: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Esd approximated as a large number of hertzian dipoles.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.8: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Parallel ray approximation for an esd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.9: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Magnitude of the Radiated Field.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.10: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields in Plane of Constant Theta.svg,
CC",Electromagnetics_Vol2.pdf
"Fig. 9.10: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields in Plane of Constant Theta.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.11: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplaneMag.svg,
CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.12: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplanePol.svg,
CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.13: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A thin straight distribution of radiating current.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.14: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Esd approximated as a large number of hertzian dipoles.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"9.8. RADIA TION FROM SURF ACE AND VOLUME DISTRIBUTIONS OF CURRENT 165
Fig. 9.15: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Parallel ray approximation for an esd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.16: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Current distribution of the hwd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.17: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Comparison radiated field magnitude of hwd to esd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"Chapter 10
Antennas
10.1
How Antennas Radiate
[m0201]
An antenna is a tr ansducer; that is, a device which
converts signals in one form into another form. In the
case of an antenna, these two forms are (1)
conductor-bound voltage and current signals and (2)
electromagnetic waves. Traditional passive antennas
are capable of this conversion in either direction. In
this section, we consider the transmit case, in which a
conductor-bound signal is converted into a radiating
electromagnetic wave. Radiation from an antenna is
due to the time-varying current that is excited by the
bound electrical signal applied to the antenna
terminals.
Why ideal transmission lines don’t radiate. T o
describe the process that allows an antenna to
transmit, it is useful to first consider the scenario
depicted in Figure 10.1. Here a sinusoidal source is
c⃝ S. Lally CC BY -SA 4.0
Figure 10.1: T win lead transmission line terminated
into an open circuit, giving rise to a standing wave.",Electromagnetics_Vol2.pdf
"c⃝ S. Lally CC BY -SA 4.0
Figure 10.1: T win lead transmission line terminated
into an open circuit, giving rise to a standing wave.
applied to the input of an ideal twin lead transmission
line. The spacing between conductors is much less
than a wavelength, and the output of the transmission
line is terminated into an open circuit.
Without any additional information, we already know
two things about the current on this transmission line.
First, we know the current must be identically zero at
the end of the transmission line. Second, we know the
current on the two conductors comprising the
transmission line must be related as indicated in
Figure 10.1. That is, at any given position on the
transmission line, the current on each conductor is
equal in magnitude and flows in opposite directions.
Further note that Figure 10.1 depicts only the
situation over an interval of one-half of the period of
the sinusoidal source. For the other half-period, the",Electromagnetics_Vol2.pdf
"situation over an interval of one-half of the period of
the sinusoidal source. For the other half-period, the
direction of current will be in the direction opposite
that depicted in Figure 10.1. In other words: The
source is varying periodically , so the sign of the
current is changing every half-period. Generally ,
time-varying currents give rise to radiation.
Therefore, we should expect the currents depicted in
Figure 10.1 might radiate. W e can estimate the
radiation from the transmission line by interpreting
the current as a collection of Hertzian dipoles. For a
review of Hertzian dipoles, see Section 9.4; however,
all we need to know to address the present problem is
that superposition applies. That is, the total radiation
is the sum of the radiation from the individual
Hertzian dipoles. Once again looking back at
Figure 10.1, note that each Hertzian dipole
representing current on one conductor has an
associated Hertzian dipole representing the current on",Electromagnetics_Vol2.pdf
"Figure 10.1, note that each Hertzian dipole
representing current on one conductor has an
associated Hertzian dipole representing the current on
the other conductor, and is only a tiny fraction of a
wavelength distant. Furthermore, these pairs of
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"10.1. HOW ANTENNAS RADIA TE 167
Hertzian dipoles are identical in magnitude but
opposite in sign. Therefore, the radiated field from
any such pair of Hertzian dipoles is approximately
zero at distances sufficiently far from the transmission
line. Continuing to sum all such pairs of Hertzian
dipoles, the radiated field remains approximately zero
at distances sufficiently far from the transmission line.
W e come to the following remarkable conclusion:
The radiation from an ideal twin lead transmis-
sion
line with open circuit termination is negligi-
ble at distances much greater than the separation
between the conductors.
Before proceeding, let us be clear about one thing:
The situation at distances which are not large relative
to separation between the conductors is quite
different. This is because the Hertzian dipole pairs do
not appear to be quite so precisely collocated at field
points close to the transmission line. Subsequently
the cancellation of fields from Hertzian dipole pairs is",Electromagnetics_Vol2.pdf
"points close to the transmission line. Subsequently
the cancellation of fields from Hertzian dipole pairs is
less precise. The resulting sum fields are not
negligible and depend on the separation between the
conductors. For the present discussion, it suffices to
restrict our attention to the simpler “far field” case.
A simple antenna formed by modifying twin lead
transmission line. Let us now consider a
modification to the previous scenario, shown in
Figure 10.2. In the new scenario, the ends of the twin
lead are bent into right angles. This new section has
an overall length which is much less than one-half
wavelength. T o determine the current distribution on
the modified section, we first note that the current
must still be precisely zero at the ends of the
conductors, but not necessarily at the point at which
c⃝ S. Lally CC BY -SA 4.0
Figure 10.2: End of the twin lead transmission line
fashioned into an electrically-short dipole (ESD).
they are bent. Since the modified section is much",Electromagnetics_Vol2.pdf
"fashioned into an electrically-short dipole (ESD).
they are bent. Since the modified section is much
shorter than one-half wavelength, the current
distribution on this section must be very simple. In
fact, the current distribution must be that of the
electrically-short dipole (ESD), exhibiting magnitude
which is maximum at the center and decreasing
approximately linearly to zero at the ends. (See
Section 9.5 for a review of the ESD.) Finally , we note
that the current should be continuous at the junction
between the unmodified and modified sections.
Having established that the modified section exhibits
the current distribution of an ESD, and having
previously determined that the unmodified section of
transmission line does not radiate, we conclude that
this system radiates precisely as an ESD does. Note
also that we need not interpret the ESD portion of the
transmission line as a modification of the
transmission line: Instead, we may view this system",Electromagnetics_Vol2.pdf
"also that we need not interpret the ESD portion of the
transmission line as a modification of the
transmission line: Instead, we may view this system
as an unmodified transmission line attached to an
antenna, which in this case in an ESD.
The general case. Although we developed this
insight for the ESD specifically , the principle applies
generally . That is, any antenna – not just dipoles –
can be viewed as a structure that can support a current
distribution that radiates. Elaborating:
A transmitting antenna is a device that, when
dri
ven by an appropriate source, supports a time-
varying distribution of current resulting in an
electromagnetic wave that radiates away from the
device.
Although this may seem to be simply a restatement of
the definition at the beginning of this section – and it
is – we now see how this can happen, and also why
transmission lines typically aren’t also antennas.",Electromagnetics_Vol2.pdf
"168 CHAPTER 10. ANTENNAS
10.2 Power Radiated by an
Electrically-Short Dipole
[m0207]
In this section, we determine the total power radiated
by
an electrically-short dipole (ESD) antenna in
response to a sinusoidally-varying current applied to
the antenna terminals. This result is both useful on its
own and necessary as an intermediate result in
determining the impedance of the ESD. The ESD is
introduced in Section 9.5, and a review of that section
is suggested before attempting this section.
In Section 9.5, it is shown that the electric field
intensity in the far field of a ˆz-oriented ESD located at
the origin is
˜E(r) ≈ ˆθjηI0 · βL
8π (s in θ) e−jβr
r (10.1)
where r is
the field point, I0 is a complex number
representing the peak magnitude and phase of the
sinusoidally-varying terminal current, Lis the length
of the ESD, and βis the phase propagation constant
2π/λwhere λis wavelength. Note that L≪ λsince
this is an ESD. Also note that Equation 10.1 is valid",Electromagnetics_Vol2.pdf
"2π/λwhere λis wavelength. Note that L≪ λsince
this is an ESD. Also note that Equation 10.1 is valid
only for the far-field conditions r≫ Land r≫ λ,
and presumes propagation in simple (linear,
homogeneous, time-invariant, isotropic) media with
negligible loss.
Given that we have already limited scope to the far
field, it is reasonable to approximate the
electromagnetic field at each field point r as a plane
wave propagating radially away from the antenna;
i.e., in the ˆr direction. Under this assumption, the
time-average power density is
S(r) = ˆr
⏐⏐⏐˜E(r)
⏐⏐⏐
2
2η (10.2)
where ηis
the wave impedance of the medium. The
total power Prad radiated by the antenna is simply
S(r) integrated over any closed surface S that
encloses the antenna. Thus:
Prad =
∮
S
S(r) · ds (10.3)
where ds is the outward-facing differential element of
surface area. In other words, power density (W/m 2)
integrated over an area (m 2) gives power (W).
Anticipating that this problem will be addressed in",Electromagnetics_Vol2.pdf
"integrated over an area (m 2) gives power (W).
Anticipating that this problem will be addressed in
spherical coordinates, we note that
ds = ˆrr2 sin θdθdφ (10.4)
and subsequently:
Prad =
∫ π
θ=0
∫ 2π
φ=0

ˆr
⏐⏐⏐˜E(r)
⏐⏐⏐
2
2η

·
(ˆrr2 sin θ dθdφ
)
= 1
2η
∫ π
θ=0
∫ 2
π
φ=0
⏐⏐⏐˜E(r)
⏐⏐⏐
2
r2 sin θdθdφ (10.5)
Returning to Equation 10.1, we note:
⏐
⏐
⏐˜E(r)
⏐
⏐
⏐
2
≈ η2 |I0|2 (βL)2
64π2 (s in θ)2 1
r2 (10.6)
Substitution
into Equation 10.5 yields:
Prad ≈ η|I0|2 (βL)2
128π2
∫ π
θ=0
∫ 2
π
φ=0
sin3 θdθdφ (10.7)
Note that this may be factored into separate integrals
over θand φ. The integral over φis simply 2π,
leaving:
Prad ≈ η|I0|2 (βL)2
64π
∫ π
θ=0
s in3 θdθ (10.8)
The remaining integral is equal to 4/3, leaving:
Prad ≈ η|I0|2 (βL)2
48π (10.9)
This
completes the derivation, but it is useful to check
units. Recall that βhas SI base units of rad/m, so βL
has units of radians. This leaves η|I0|2, which has SI
base units of Ω · A2 = W , as expected.",Electromagnetics_Vol2.pdf
"units. Recall that βhas SI base units of rad/m, so βL
has units of radians. This leaves η|I0|2, which has SI
base units of Ω · A2 = W , as expected.
The power radiated by an ESD in response to the
current I0 applied
at the terminals is given by
Equation 10.9.
Finally , it is useful to consider how various
parameters affect the radiated power. First, note that
the radiated power is proportional to the square of the
terminal current. Second, note that the product
βL = 2πL/λ is the electrical length of the antenna;",Electromagnetics_Vol2.pdf
"10.3. POWER DISSIP A TED BY AN ELECTRICALL Y -SHOR T DIPOLE 169
that is, the length of the antenna expressed in radians,
where 2πradians is one wavelength. Thus, we see
that the power radiated by the antenna increases as the
square of electrical length.
Example 10.1. Po wer radiated by an ESD.
A dipole is 10 cm in length and is surrounded by
free space. A sinusoidal current having
frequency 30 MHz and peak magnitude 100 mA
is applied to the antenna terminals. What is the
power radiated by the antenna?
Solution. If no power is dissipated within the
antenna, then all power is radiated. The
wavelength λ= c/f ∼= 10 m, so L∼
= 0.01λ.
This certainly qualifies as electrically-short, so
we may use Equation 10.9. In the present
problem, η∼= 376.7 Ω (the free space wave
impedance), I0 = 100 mA, and
β = 2π/λ ∼
= 0.628 rad/m. Thus, we find that the
radiated power is ≈ 98.6 µW
.
10.3
Power Dissipated by an
Electrically-Short Dipole
[m0208]
The power delivered to an antenna by a source
connected",Electromagnetics_Vol2.pdf
"radiated power is ≈ 98.6 µW
.
10.3
Power Dissipated by an
Electrically-Short Dipole
[m0208]
The power delivered to an antenna by a source
connected
to the terminals is nominally radiated.
However, it is possible that some fraction of the
power delivered by the source will be dissipated
within the antenna. In this section, we consider the
dissipation due to the finite conductivity of materials
comprising the antenna. Specifically , we determine
the total power dissipated by an electrically-short
dipole (ESD) antenna in response to a
sinusoidally-varying current applied to the antenna
terminals. (The ESD is introduced in Section 9.5, and
a review of that section is suggested before attempting
this section.) The result allows us to determine
radiation efficiency and is a step toward determining
the impedance of the ESD.
Consider an ESD centered at the origin and aligned
along the zaxis. In Section 9.5, it is shown that the
current distribution is:
˜I(z) ≈ I0
(
1 − 2
L|z|
)
(10.10)",Electromagnetics_Vol2.pdf
"along the zaxis. In Section 9.5, it is shown that the
current distribution is:
˜I(z) ≈ I0
(
1 − 2
L|z|
)
(10.10)
where I0 (SI
base units of A) is a complex-valued
constant indicating the maximum current magnitude
and phase, and Lis the length of the ESD. This
current distribution may be interpreted as a set of
discrete very-short segments of constant-magnitude
current (sometimes referred to as “Hertzian dipoles”).
In this interpretation, the nth current segment is
located at z= zn and has magnitude ˜I(zn).
Now let us assume that the material comprising the
ESD is a homogeneous good conductor. Then each
segment exhibits the same resistance Rseg.
Subsequently , the power dissipated in the nth
segment is
Pseg(zn) = 1
2
⏐⏐⏐˜I(zn)
⏐⏐⏐
2
Rseg (10.11)
The
segment resistance can be determined as follows.
Let us assume that the wire comprising the ESD has
circular cross-section of radius a. Since the wire is a
good conductor, the segment resistance may be",Electromagnetics_Vol2.pdf
"170 CHAPTER 10. ANTENNAS
calculated using Equation 4.17 (Section 4.2,
“Impedance of a Wire”):
Rseg ≈ 1
2
√
µf
πσ · ∆l
a (10.12)
where µis
permeability , f is frequency , σis
conductivity , and ∆l is the length of the segment.
Substitution into Equation 10.11 yields:
Pseg(zn) ≈ 1
4a
√
µf
πσ
⏐⏐
⏐˜I(zn)
⏐
⏐
⏐
2
∆l (10.13)
No
w the total power dissipated in the antenna, Ploss,
can be expressed as the sum of the power dissipated
in each segment:
Ploss ≈
N∑
n=1
[
1
4a
√
µf
πσ
⏐⏐
⏐˜I(zn)
⏐
⏐
⏐
2
∆l
]
(10.14)
where N is
the number of segments. This may be
rewritten as follows:
Ploss ≈ 1
4a
√
µf
πσ
N∑
n=1
⏐
⏐
⏐˜I(
 zn)
⏐
⏐
⏐
2
∆l (10.15)
Reducing ∆l to the differential length dz′, we may
write this in the following integral form:
Ploss ≈ 1
4a
√
µf
πσ
∫ +L/ 2
z′=−L/ 2
⏐
⏐⏐˜I(
 z′)
⏐⏐⏐
2
dz′ (10.16)
It is worth noting that the above expression applies to
any straight wire antenna of length L. For the ESD
specifically , the current distribution is given by
Equation 10.10. Making the substitution:",Electromagnetics_Vol2.pdf
"any straight wire antenna of length L. For the ESD
specifically , the current distribution is given by
Equation 10.10. Making the substitution:
Ploss ≈ 1
4a
√
µf
πσ
∫ +L/ 2
z′=−L/ 2
⏐⏐
⏐
⏐I0
(
1 − 2
L|z′|
)⏐
⏐
⏐
⏐
2
dz′
≈ 1
4a
√
µf
πσ |I0|2
∫ +L/ 2
z′=−L/ 2
⏐
⏐⏐⏐1 − 2
L|z′|
⏐
⏐⏐⏐
2
dz′
(10.17)
The
integral is straightforward to solve, albeit a bit
tedious. The integral is found to be equal to L/3, so
we obtain:
Ploss ≈ 1
4a
√
µf
πσ |I0|2 · L
3
≈ L
12a
√
µf
πσ |I0|2 (10.18)
No
w let us return to interpretation of the ESD as a
circuit component. W e have found that applying the
current I0 to the terminals results in the dissipated
power indicated in Equation 10.18. A current source
driving the ESD does not perceive the current
distribution of the ESD nor does it perceive the
varying power over the length of the ESD. Instead,
the current source perceives only a net resistance
Rloss such that
Ploss = 1
2 |I0|2 Rlo ss (10.19)
Comparing Equations 10.18 and 10.19, we find
Rloss ≈ L
6a
√
µf",Electromagnetics_Vol2.pdf
"Rloss such that
Ploss = 1
2 |I0|2 Rlo ss (10.19)
Comparing Equations 10.18 and 10.19, we find
Rloss ≈ L
6a
√
µf
πσ (10.20)
The power dissipated within an ESD in response
to
a sinusoidal current I0 applied at the terminals
is 1
2 |I0|2 Rlo ss where Rloss (Equation 10.20) is
the resistance perceived by a source applied to
the ESD’s terminals.
Note that Rloss is not the impedance of the ESD.
Rloss is merely the contribution of internal loss to the
impedance of the ESD. The impedance of the ESD
must also account for contributions from radiation
resistance (an additional real-valued contribution) and
energy storage (perceived as a reactance). These
quantities are addressed in other sections of this book.
Example 10.2. Po wer dissipated within an
ESD.
A dipole is 10 cm in length, 1 mm in radius, and
is surrounded by free space. The antenna is
comprised of aluminum having conductivity
≈ 3.7 × 107 S/m and µ≈ µ0. A sinusoidal
current having frequency 30 MHz and peak",Electromagnetics_Vol2.pdf
"comprised of aluminum having conductivity
≈ 3.7 × 107 S/m and µ≈ µ0. A sinusoidal
current having frequency 30 MHz and peak
magnitude 100 mA is applied to the antenna
terminals. What is the power dissipated within
this antenna?
Solution. The wavelength λ= c/f ∼= 10 m, so
L= 10 cm ∼= 0.01λ. This certainly qualifies as
electrically-short, so we may use
Equation 10.20. In the present problem,
a= 1 mm and σ≈ 3.7 × 107 S/m. Thus, we",Electromagnetics_Vol2.pdf
"10.4. REACT ANCE OF THE ELECTRICALL Y -SHOR T DIPOLE 171
find that the loss resistance Rlo ss ≈ 9.49 mΩ.
Subsequently , the power dissipated within this
antenna is
Ploss = 1
2 |I0|2 Rlo ss ≈ 47.5 µW (10.21)
W e
conclude this section with one additional caveat:
Whereas this section focuses on the limited
conductivity of wire, other physical mechanisms may
contribute to the loss resistance of the ESD. In
particular, materials used to coat the antenna or to
provide mechanical support near the terminals may
potentially absorb and dissipate power that might
otherwise be radiated.
10.4 Reactance of the
Electrically-Short Dipole
[m0210]
For any given time-varying voltage appearing across
the
terminals of an electrically-short antenna, there
will be a time-varying current that flows in response.
The ratio of voltage to current is impedance. Just as a
resistor, capacitor, or inductor may be characterized
in terms of an impedance, any antenna may be
characterized in terms of this impedance. The",Electromagnetics_Vol2.pdf
"resistor, capacitor, or inductor may be characterized
in terms of an impedance, any antenna may be
characterized in terms of this impedance. The
real-valued component of this impedance accounts for
power which is radiated away from the antenna
(Section 10.2) and dissipated within the antenna
(Section 10.3). The imaginary component of this
impedance – i.e., the reactance – typically represents
energy storage within the antenna, in the same way
that the reactance of a capacitor or inductor represents
storage of electrical or magnetic energy , respectively .
In this section, we determine the reactance of the
electrically-short dipole (ESD).
Reactance of a zero-length dipole. W e begin with
what might seem initially to be an absurd notion: A
dipole antenna having zero length. However, such a
dipole is certainly electrically-short, and in fact serves
as an extreme condition from which we can deduce
some useful information.
Consider Figure 10.3, which shows a zero-length",Electromagnetics_Vol2.pdf
"as an extreme condition from which we can deduce
some useful information.
Consider Figure 10.3, which shows a zero-length
dipole being driven by a source via a transmission
line. It is immediately apparent that such a dipole is
equivalent to an open circuit. So, the impedance of a
zero-length dipole is equal to that of an open circuit.
c⃝ S. Lally CC BY -SA 4.0
Figure 10.3: Transmission line terminated into a zero-
length dipole, which is equivalent to an open circuit.",Electromagnetics_Vol2.pdf
"172 CHAPTER 10. ANTENNAS
What is the impedance of an open circuit? One might
be tempted to say “infinite, ” since the current is zero
independently of the voltage. However, we must now
be careful to properly recognize the real and
imaginary components of this infinite number, and
signs of these components. In fact, the impedance of
an open circuit is 0 − j∞. One way to confirm this is
via basic transmission line theory; i.e., the input
impedance of transmission line stubs. A more direct
justification is as follows: The real part of the
impedance is zero because there is no transfer of
power into the termination. The magnitude of the
imaginary part of the impedance must be infinite
because the current is zero. The sign of the imaginary
component is negative because the reflected current
experiences a sign change (required to make the total
current zero), whereas the reflected voltage does not
experience a sign change (and so is not canceled at
the terminals). Thus, the impedance of an open",Electromagnetics_Vol2.pdf
"experience a sign change (and so is not canceled at
the terminals). Thus, the impedance of an open
circuit, and the zero-length dipole, is 0 − j∞.
Reactance of a nearly-zero-length dipole. Let us
now consider what happens as the length of the dipole
increases from zero. Since such a dipole is short
compared to a wavelength, the variation with length
must be very simple. The reactance remains large and
negative, but can only increase (become less negative)
monotonically with increasing electrical length. This
trend continues until the dipole is no longer
electrically short. Thus, we conclude that the
reactance of an ESD is always large and negative.
The reactance of an ESD is very large and nega-
ti
ve, approaching the reactance of an open circuit
termination as length decreases to zero.
Note that this behavior is similar to that of a capacitor,
whose impedance is also negative and increases
toward zero with increasing frequency . Thus, the",Electromagnetics_Vol2.pdf
"whose impedance is also negative and increases
toward zero with increasing frequency . Thus, the
reactance of an ESD is sometimes represented in
circuit diagrams as a capacitor. However, the specific
dependence of reactance on frequency is different
from that of a capacitor (as we shall see in a moment),
so this model is generally valid only for analysis at
one frequency at time.
An approximate expression for the reactance of an
ESD. Sometimes it is useful to be able to estimate the
reactance XA of an ESD. Although a derivation is
beyond the scope of this text, a suitable expression is
(see, e.g., Johnson (1993) in “ Additional References”
at the end of this section):
XA ≈ − 120 Ω
πL/λ
[
ln
(L
2a
)
− 1
]
(10.22)
where Lis
length and a≪ Lis the radius of the wire
comprising the ESD. Note that this expression yields
the expected behavior; i.e., XA → −∞ as L→ 0,
and increases monotonically as Lincreases from zero.
Example 10.3. Reactance of an ESD.",Electromagnetics_Vol2.pdf
"the expected behavior; i.e., XA → −∞ as L→ 0,
and increases monotonically as Lincreases from zero.
Example 10.3. Reactance of an ESD.
A dipole is 10 cm in length, 1 mm in radius, and
is surrounded by free space. What is the
reactance of this antenna at 30 MHz?
Solution. The wavelength λ= c/f ∼= 10 m, so
L= 10 cm ∼= 0.01λ. This certainly qualifies as
electrically-short, so we may use
Equation 10.22. Given a= 1 mm, we find the
reactance XA ≈ − 11.1 kΩ . For what it’s worth,
this
antenna exhibits approximately the same
reactance as a 0.47 pF capacitor at the same
frequency .
The reactance of the ESD is typically orders of
magnitude larger than the real part of the impedance
of the ESD. The reactance of the ESD is also typically
very large compared to the characteristic impedance
of typical transmission lines (usually 10s to 100s of
ohms). This makes ESDs quite difficult to use in
practical transmit applications.
Additional Reading:
• R.C. Johnson (Ed.), Antenna Systems Handbook",Electromagnetics_Vol2.pdf
"ohms). This makes ESDs quite difficult to use in
practical transmit applications.
Additional Reading:
• R.C. Johnson (Ed.), Antenna Systems Handbook
(Ch. 4), McGraw-Hill, 1993.",Electromagnetics_Vol2.pdf
"10.5. EQUIV ALENT CIRCUIT MODEL FOR TRANSMISSION; RADIA TION EFFICIENCY 173
10.5 Equivalent Circuit Model
for T ransmission; Radiation
Efficiency
[m0202]
A radio transmitter consists of a source which
generates
the electrical signal intended for
transmission, and an antenna which converts this
signal into a propagating electromagnetic wave. Since
the transmitter is an electrical system, it is useful to be
able to model the antenna as an equivalent circuit.
From a circuit analysis point of view , it should be
possible to describe the antenna as a passive one-port
circuit that presents an impedance to the source.
Thus, we have the following question: What is the
equivalent circuit for an antenna which is
transmitting?
W e begin by emphasizing that the antenna is passive.
That is, the antenna does not add power. Invoking the
principle of conservation of power, there are only
three possible things that can happen to power that is
delivered to the antenna by the transmitter: 1",Electromagnetics_Vol2.pdf
"three possible things that can happen to power that is
delivered to the antenna by the transmitter: 1
• Power can be converted to a propagating
electromagnetic wave. (The desired outcome.)
• Power can be dissipated within the antenna.
• Energy can be stored by the antenna, analogous
to the storage of energy in a capacitor or
inductor.
W e also note that these outcomes can occur in any
combination. T aking this into account, we model the
antenna using the equivalent circuit shown in
Figure 10.4. Since the antenna is passive, it is
reasonable to describe it as an impedance ZA which
is (by definition) the ratio of voltage ˜VA to current ˜IA
at the terminals; i.e.,
ZA ≜
˜VA
˜IA
(10.23)
1 Note that “delivered” power means power accepted by the an-
tenna. W e are not yet considering power reflected from the antenna
due to impedance mismatch.
In the phasor domain, ZA is a complex-valued
quantity and therefore has, in general, a real-valued
component and an imaginary component. W e may",Electromagnetics_Vol2.pdf
"In the phasor domain, ZA is a complex-valued
quantity and therefore has, in general, a real-valued
component and an imaginary component. W e may
identify those components using the power
conservation argument made previously: Since the
real-valued component must represent power transfer
and the imaginary component must represent energy
storage, we infer:
ZA ≜ RA + jXA (10.24)
where RA represents power transferred to the
antenna, and XA represents energy stored by the
antenna. Note that the energy stored by the antenna is
being addressed in precisely the same manner that we
address energy storage in a capacitor or an inductor;
in all cases, as reactance. Further, we note that RA
consists of components Rrad and Rloss as follows:
ZA = Rrad + Rloss + jXA (10.25)
where Rrad represents power transferred to the
antenna and subsequently radiated, and Rloss
represents power transferred to the antenna and
subsequently dissipated.
T o confirm that this model works as expected,",Electromagnetics_Vol2.pdf
"represents power transferred to the antenna and
subsequently dissipated.
T o confirm that this model works as expected,
consider what happens when a voltage is applied
across the antenna terminals. The current ˜IA flows
and the time-average power PA transferred to the
c⃝ S. Lally CC BY -SA 4.0
Figure 10.4: Equivalent circuit for an antenna which
is transmitting.",Electromagnetics_Vol2.pdf
"174 CHAPTER 10. ANTENNAS
antenna is
PA = 1
2Re
{
˜VA˜I∗
A
}
(10.26)
where
we have assumed peak (as opposed to root
mean squared) units for voltage and current. Since
˜VA = ZA˜IA, we have:
PA = 1
2Re
{
( Rrad + Rloss + jXA) ˜IA˜I∗
A
}
(10.27)
which reduces to:
PA = 1
2
⏐⏐
⏐˜IA
⏐
⏐
⏐
2
Rrad + 1
2
⏐
⏐
⏐˜IA
⏐
⏐
⏐
2
Rlo
 ss (10.28)
As expected, the power transferred to the antenna is
the sum of
Prad ≜ 1
2
⏐⏐
⏐˜IA
⏐
⏐
⏐
2
Rrad (10.29)
representing
power transferred to the radiating
electromagnetic field, and
Ploss ≜ 1
2
⏐
⏐
⏐˜IA
⏐
⏐
⏐
2
Rlo
 ss (10.30)
representing power dissipated within the antenna.
The reactance XA will play a role in determining ˜IA
given ˜VA (and vice versa), but does not by itself
account for disposition of power. Again, this is
exactly analogous to the role played by inductors and
capacitors in a circuit.
The utility of this equivalent circuit formalism is that
it allows us to treat the antenna in the same manner as
any other component, and thereby facilitates analysis",Electromagnetics_Vol2.pdf
"it allows us to treat the antenna in the same manner as
any other component, and thereby facilitates analysis
using conventional electric circuit theory and
transmission line theory . For example: Given ZA, we
know how to specify the output impedance ZS of the
transmitter so as to minimize reflection from the
antenna: W e would choose ZS = ZA, since in this
case the voltage reflection coefficient would be
Γ = ZA − ZS
ZA + ZS
= 0 (10.31)
Alternati
vely , we might specify ZS so as to maximize
power transfer to the antenna: W e would choose
ZS = Z∗
A; i.e., conjugate matching.
In order to take full advantage of this formalism, we
require values for Rrad, Rloss, and XA. These
quantities are considered below .
Radiation resistance. Rrad is referred to as radiation
resistance. Equation 10.29 tells us that
Rrad = 2Prad
⏐⏐
⏐˜IA
⏐
⏐
⏐
−2
(10.32)
This equation suggests the following procedure: W e
apply current ˜IA to the antenna terminals, and then
determine the total power Prad radiated from the",Electromagnetics_Vol2.pdf
"apply current ˜IA to the antenna terminals, and then
determine the total power Prad radiated from the
antenna in response. For an example of this
procedure, see Section 10.2 (“T otal Power Radiated
by an Electrically-Short Dipole”). Given ˜IA and Prad,
one may then use Equation 10.32 to determine Rrad.
Loss resistance. Loss resistance represents the
dissipation of power within the antenna, which is
usually attributable to loss intrinsic to materials
comprising or surrounding the antenna. In many
cases, antennas are made from good conductors –
metals, in particular – so that Rloss is very low
compared to Rrad. For such antennas, loss is often so
low compared to Rrad that Rloss may be neglected.
In the case of the electrically-short dipole, Rloss is
typically very small but Rrad is also very small, so
both must be considered. In many other cases,
antennas contain materials with substantially greater
loss than metal. For example, a microstrip patch",Electromagnetics_Vol2.pdf
"both must be considered. In many other cases,
antennas contain materials with substantially greater
loss than metal. For example, a microstrip patch
antenna implemented on a printed circuit board
typically has non-negligible Rloss because the
dielectric material comprising the antenna exhibits
significant loss.
Antenna reactance. The reactance term jXA
accounts for energy stored by the antenna. This may
be due to reflections internal to the antenna, or due to
energy associated with non-propagating electric and
magnetic fields surrounding the antenna. The
presence of significant reactance (i.e., |XA|
comparable to or greater than |RA|) complicates
efforts to establish the desired impedance match to
the source. For an example, see Section 10.4
(“Reactance of the Electrically-Short Dipole”).
Radiation efficiency . When Rloss is non-negligible,
it is useful to characterize antennas in terms of their
radiation efficiency erad, defined as the fraction of",Electromagnetics_Vol2.pdf
"it is useful to characterize antennas in terms of their
radiation efficiency erad, defined as the fraction of
power which is radiated compared to the total power
delivered to the antenna; i.e.,
erad ≜ Prad
PA
(10.33)",Electromagnetics_Vol2.pdf
"10.6. IMPEDANCE OF THE ELECTRICALL Y -SHOR T DIPOLE 175
Using Equations 10.28–10.30, we see that this
efficiency can be expressed as follows:
erad = Rrad
Rrad + Rl oss
(10.34)
Once again, the equivalent circuit formalism proves
useful.
Example 10.4. Impedance of an antenna.
The total power radiated by an antenna is
60 mW when 20 mA (rms) is applied to the
antenna terminals. The radiation efficiency of
the antenna is known to be 70%. It is observed
that voltage and current are in-phase at the
antenna terminals. Determine (a) the radiation
resistance, (b) the loss resistance, and (c) the
impedance of the antenna.
Solution. From the problem statement,
Prad = 60 mW ,
⏐⏐
⏐˜IA
⏐
⏐
⏐= 20 mA (rms), and
erad = 0.7. Also, the fact that voltage and
current are in-phase at the antenna terminals
indicates that XA = 0. From Equation 10.32,
the radiation resistance is
Rrad ≈ 2 · (60 mW)
⏐⏐√
2 · 20 mA
⏐
⏐2 =
150 Ω (10.35)
Solving
Equation 10.34 for the loss resistance,
we find:
Rloss = 1 − erad
erad",Electromagnetics_Vol2.pdf
"Rrad ≈ 2 · (60 mW)
⏐⏐√
2 · 20 mA
⏐
⏐2 =
150 Ω (10.35)
Solving
Equation 10.34 for the loss resistance,
we find:
Rloss = 1 − erad
erad
Rr ad ∼= 64.3 Ω (10.36)
Since ZA = Rr
ad + Rloss + jXA, we find
ZA ∼
= 214.3 + j0 Ω
. This will be the ratio of
voltage
to current at the antenna terminals
regardless of the source current.
10.6 Impedance of the
Electrically-Short Dipole
[m0204]
In this section, we determine the impedance of the
electrically-short
dipole (ESD) antenna. The physical
theory of operation for the ESD is introduced in
Section 9.5, and additional relevant aspects of the
ESD antenna are addressed in Sections 10.1–10.4.
The concept of antenna impedance is addressed in
Section 10.5. Review of those sections is suggested
before attempting this section.
The impedance of any antenna may be expressed as
ZA = Rrad + Rloss + jXA (10.37)
where the real-valued quantities Rrad, Rloss, and XA
represent the radiation resistance, loss resistance, and
reactance of the antenna, respectively .",Electromagnetics_Vol2.pdf
"represent the radiation resistance, loss resistance, and
reactance of the antenna, respectively .
Radiation resistance. The radiation resistance of any
antenna can be expressed as:
Rrad = 2Prad|I0|−2 (10.38)
where |I0| is the magnitude of the current at the
antenna terminals, and Prad is the resulting total
power radiated. For an ESD (Section 10.2):
Prad ≈ η|I0|2 (βL)2
48π (10.39)
so
Rr
ad ≈ η(βL)2
24π (10.40)
It
is useful to have an alternative form of this
expression in terms of wavelength λ. This is derived
as follows. First, note:
βL = 2π
λ L= 2 πL
λ (10.41)
where L/λis
the antenna length in units of
wavelength. Substituting this expression into
Equation 10.40:
Rrad ≈ η
(
2πL
λ
)2 1
24π
≈ ηπ
6
(L
λ
)2
(10.42)",Electromagnetics_Vol2.pdf
"176 CHAPTER 10. ANTENNAS
Assuming free-space conditions, η∼= 376.7 Ω, which
is ≈ 120πΩ. Subsequently ,
Rrad ≈ 20π2
(L
λ
)2
(10.43)
This
remarkably simple expression indicates that the
radiation resistance of an ESD is very small (since
L≪ λfor an ESD), but increases as the square of the
length.
At this point, a warning is in order. The radiation
resistance of an “electrically-short dipole” is
sometimes said to be 80π2 (L/λ)2; i.e., 4 times the
right side of Equation 10.43. This higher value is not
for a physically-realizable ESD, but rather for the
Hertzian dipole (sometimes also referred to as an
“ideal dipole”). The Hertzian dipole is an
electrically-short dipole with a current distribution
that has uniform magnitude over the length of the
dipole.2 The Hertzian dipole is quite difficult to
realize in practice, and nearly all practical
electrically-short dipoles exhibit a current distribution
that is closer to that of the ESD. The ESD has a",Electromagnetics_Vol2.pdf
"electrically-short dipoles exhibit a current distribution
that is closer to that of the ESD. The ESD has a
current distribution which is maximum in the center
and goes approximately linearly to zero at the ends.
The factor-of-4 difference in the radiation resistance
of the Hertzian dipole relative to the (practical) ESD
is a straightforward consequence of the difference in
the current distributions.
Loss resistance. The loss resistance of the ESD is
derived in Section 10.3 and is found to be
Rloss ≈ L
6a
√
µf
πσ (10.44)
where a
, µ, and σare the radius, permeability , and
conductivity of the wire comprising the antenna, and
f is frequency .
Reactance. The reactance of the ESD is addressed in
Section 10.4. A suitable expression for this reactance
is (see, e.g., Johnson (1993) in “ Additional
References” at the end of this section):
XA ≈ − 120 Ω
πL/λ
[
ln
(L
2a
)
− 1
]
(10.45)
2 See Section 9.4 for additional information about the Hertzian
dipole.
where a≪ Lis assumed.",Electromagnetics_Vol2.pdf
"XA ≈ − 120 Ω
πL/λ
[
ln
(L
2a
)
− 1
]
(10.45)
2 See Section 9.4 for additional information about the Hertzian
dipole.
where a≪ Lis assumed.
Example 10.5. Impedance of an ESD.
A thin, straight dipole antenna operates at
30 MHz in free space. The length and radius of
the dipole are 1 m and 1 mm respectively . The
dipole is comprised of aluminum having
conductivity ≈ 3.7 × 107 S/m and µ≈ µ0.
What is the impedance and radiation efficiency
of this antenna?
Solution. The free-space wavelength at
f = 30 MHz is λ= c/f ∼= 10 m. Therefore,
L∼
= 0.1λ, and this is an ESD. Since this is an
ESD, we may compute the radiation resistance
using Equation 10.43, yielding Rrad ≈ 1.97 Ω.
The radius a= 1 mm and σ≈ 3.7 × 107 S/m;
thus, we find that the loss resistance
Rloss ≈ 94.9 mΩ using Equation 10.44. Using
Equation 10.45, the reactance is found to be
XA ≈ − 1991.8 Ω. The impedance of this
antenna is therefore
ZA = Rrad + Rloss + jXA
≈ 2.1 − j1991.8 Ω
(10.46)
and
the radiation efficiency of this antenna is",Electromagnetics_Vol2.pdf
"antenna is therefore
ZA = Rrad + Rloss + jXA
≈ 2.1 − j1991.8 Ω
(10.46)
and
the radiation efficiency of this antenna is
erad = Rrad
Rrad + Rl oss
≈ 95.4% (10.47)
In
the preceding example, the radiation efficiency is a
respectable 95.4%. However, the real part of the
impedance of the ESD is much less than the
characteristic impedance of typical transmission lines
(typically 10s to 100s of ohms). Another problem is
that the reactance of the ESD is very large. Some
form of impedance matching is required to efficiently
transfer power from a transmitter or transmission line
to this form of antenna. Failure to do so will result in
reflection of a large fraction of the power incident on
antenna terminals. A common solution is to insert
series inductance to reduce – nominally , cancel – the
negative reactance of the ESD. In the preceding
example, about 10 µH would be needed. The
remaining mismatch in the real-valued components of
impedance is then relatively easy to mitigate using",Electromagnetics_Vol2.pdf
"example, about 10 µH would be needed. The
remaining mismatch in the real-valued components of
impedance is then relatively easy to mitigate using
common “real-to-real” impedance matching",Electromagnetics_Vol2.pdf
"10.7. DIRECTIVITY AND GAIN 177
techniques.
Additional Reading:
• R.C. Johnson (Ed.), Antenna Systems Handbook
(Ch. 4), McGraw-Hill, 1993.
10.7 Directivity and Gain
[m0203]
A transmitting antenna does not radiate power
uniformly
in all directions. Inevitably more power is
radiated in some directions than others. Directivity
quantifies this behavior. In this section, we introduce
the concept of directivity and the related concepts of
maximum directivity and antenna gain.
Consider an antenna located at the origin. The power
radiated in a single direction (θ,φ) is formally zero.
This is because a single direction corresponds to a
solid angle of zero, which intercepts an area of zero at
any given distance from the antenna. Since the power
flowing through any surface having zero area is zero,
the power flowing in a single direction is formally
zero. Clearly we need a different metric of power in
order to develop a sensible description of the spatial
distribution of power flow .",Electromagnetics_Vol2.pdf
"zero. Clearly we need a different metric of power in
order to develop a sensible description of the spatial
distribution of power flow .
The appropriate metric is spatial power density; that
is, power per unit area, having SI base units of W/m 2.
Therefore, directivity is defined in terms of spatial
power density in a particular direction, as opposed to
power in a particular direction. Specifically ,
directivity in the direction (θ,φ) is:
D(θ,φ) ≜ S(r)
Sav e(r) (10.48)
In this expression, S(r) is the power density at
(r,θ,φ ); i.e., at a distance rin the direction (θ,φ).
Save(r) is the average power density at that distance;
that is, S(r) averaged over all possible directions at
distance r. Since directivity is a ratio of power
densities, it is unitless. Summarizing:
Directivity is ratio of power density in a specified
direction
to the power density averaged over all
directions at the same distance from the antenna.
Despite Equation 10.48, directivity does not depend",Electromagnetics_Vol2.pdf
"to the power density averaged over all
directions at the same distance from the antenna.
Despite Equation 10.48, directivity does not depend
on the distance from the antenna. T o be specific,
directivity is the same at every distance r. Even
though the numerator and denominator of
Equation 10.48 both vary with r, one finds that the
distance dependence always cancels because power
density and average power density are both",Electromagnetics_Vol2.pdf
"178 CHAPTER 10. ANTENNAS
proportional to r−2. This is a key point: Directivity is
a convenient way to characterize an antenna because
it does not change with distance from the antenna.
In general, directivity is a function of direction.
However, one is often not concerned about all
directions, but rather only the directivity in the
direction in which it is maximum. In fact it is quite
common to use the term “directivity” informally to
refer to the maximum directivity of an antenna. This
is usually what is meant when the directivity is
indicated to be a single number; in any event, the
intended meaning of the term is usually clear from
context.
Example 10.6. Directi vity of the
electrically-short dipole.
An electrically-short dipole (ESD) consists of a
straight wire having length L≪ λ/2. What is
the directivity of the ESD?
Solution. The field radiated by an ESD is
derived in Section 9.5. In that section, we find
that the electric field intensity in the far field of a",Electromagnetics_Vol2.pdf
"Solution. The field radiated by an ESD is
derived in Section 9.5. In that section, we find
that the electric field intensity in the far field of a
ˆz-oriented ESD located at the origin is:
˜E(r) ≈ ˆθjηI0 · βL
8π (s in θ) e−jβr
r (10.49)
where I0 represents
the magnitude and phase of
the current applied to the terminals, ηis the
wave impedance of the medium, and β = 2π/λ.
In Section 10.2, we find that the power density
of this field is:
S(r) ≈ η|I0|2 (βL)2
128π2 (s in θ)2 1
r2 (10.50)
and
we subsequently find that the total power
radiated is:
Prad ≈ η|I0|2 (βL)2
48π (10.51)
The
average power density Save is simply the
total power divided by the area of a sphere
centered on the ESD. Let us place this sphere at
distance r, with r≫ Land r≫ λas required
for the validity of Equations 10.49 and 10.50.
Then:
Sav
e = Prad
4πr2 ≈ η|
I0|2 (βL)2
192π2r2 (10.52)
Finally
the directivity is determined by applying
the definition:
D(θ,φ) ≜ S(r)
Sav e(r) (10.53)
≈ 1.5 (sin θ)2 (10.54)
The",Electromagnetics_Vol2.pdf
"I0|2 (βL)2
192π2r2 (10.52)
Finally
the directivity is determined by applying
the definition:
D(θ,φ) ≜ S(r)
Sav e(r) (10.53)
≈ 1.5 (sin θ)2 (10.54)
The
maximum directivity occurs in the θ= π/2
plane. Therefore, the maximum directivity is
1.5
, meaning the maximum power density is 1.5
times
greater than the power density averaged
over all directions.
Since directivity is a unitless ratio, it is common to
express it in decibels. For example, the maximum
directivity of the ESD in the preceding example is
10 log10 1.5 ∼= 1.76 dB. (Note “10 log 10” here since
directivity is the ratio of power-like quantities.)
Gain. The gain G(θ,φ) of an antenna is its directivity
modified to account for loss within the antenna.
Specifically:
G(θ,φ) ≜ S(r) for actual antenna
Sav e(r) for identical but lossless antenna
(10.55)
In this equation, the numerator is the actual power
density radiated by the antenna, which is less than the
nominal power density due to losses within the",Electromagnetics_Vol2.pdf
"density radiated by the antenna, which is less than the
nominal power density due to losses within the
antenna. The denominator is the average power
density for an antenna which is identical, but lossless.
Since the actual antenna radiates less power than an
identical but lossless version of the same antenna,
gain in any particular direction is always less than
directivity in that direction. Therefore, an equivalent
definition of antenna gain is
G(θ,φ) ≜ eradD(θ,φ) (10.56)
where erad is the radiation efficiency of the antenna
(Section 10.5).
Gain is directivity times radiation efficiency; that
is,
directivity modified to account for loss within
the antenna.",Electromagnetics_Vol2.pdf
"10.8. RADIA TION P A TTERN 179
The receive case. T o conclude this section, we make
one additional point about directivity , which applies
equally to gain. The preceding discussion has
presumed an antenna which is radiating; i.e.,
transmitting. Directivity can also be defined for the
receive case, in which it quantifies the effectiveness of
the antenna in converting power in an incident wave
to power in a load attached to the antenna. Receive
directivity is formally introduced in Section 10.13
(“Effective Aperture”). When receive directivity is
defined as specified in Section 10.13, it is equal to
transmit directivity as defined in this section. Thus, it
is commonly said that the directivity of an antenna is
the same for receive and transmit.
Additional Reading:
• “Directivity” on Wikipedia.
10.8 Radiation Pattern
[m0205]
The radiation pattern of
 a transmitting antenna
describes the magnitude and polarization of the field
radiated by the antenna as a function of angle relative",Electromagnetics_Vol2.pdf
"a transmitting antenna
describes the magnitude and polarization of the field
radiated by the antenna as a function of angle relative
to the antenna. A pattern may also be defined for a
receiving antenna, however, we defer discussion of
the receive case to a later section.
The concept of radiation pattern is closely related to
the concepts of directivity and gain (Section 10.7).
The principal distinction is the explicit consideration
of polarization. In many applications, the polarization
of the field radiated by a transmit antenna is as
important as the power density radiated by the
antenna. For example, a radio communication link
consists of an antenna which is transmitting separated
by some distance from an antenna which is receiving.
Directivity determines the power density delivered to
the receive antenna, but the receive antenna must be
co-polarized with the arriving wave in order to
capture all of this power.
The radiation pattern concept is perhaps best",Electromagnetics_Vol2.pdf
"co-polarized with the arriving wave in order to
capture all of this power.
The radiation pattern concept is perhaps best
explained by example. The simplest antenna
encountered in common practice is the
electrically-short dipole (ESD), which consists of a
straight wire of length Lthat is much less than
one-half of a wavelength. In Section 9.5, it is shown
that the field radiated by an ESD which is located at
the origin and aligned along the z-axis is:
˜E(r) ≈ ˆθjηI0 · βL
8π (s in θ) e−jβr
r (10.57)
where I0 represents
the magnitude and phase of the
current applied to the terminals, ηis the wave
impedance of the medium, and β = 2π/λ is the phase
propagation constant of the medium. Note that the
reference direction of the electric field is in the ˆθ
direction; therefore, a receiver must be ˆθ-polarized
relative to the transmitter’s coordinate system in order
to capture all available power. A receiver which is not
fully ˆθ-polarized will capture less power. In the",Electromagnetics_Vol2.pdf
"to capture all available power. A receiver which is not
fully ˆθ-polarized will capture less power. In the
extreme, a receiver which is ˆφ-polarized with respect
to the transmitter’s coordinate system will capture
zero power. Thus, for the ˆz-oriented ESD, we refer to
the ˆθ-polarization of the transmitted field as",Electromagnetics_Vol2.pdf
"180 CHAPTER 10. ANTENNAS
“co-polarized” or simply “co-pol, ” and the
ˆφ-polarization of the transmitted field as “cross-pol. ”
At this point, the reader may wonder what purpose is
served by defining cross polarization, since the
definition given above seems to suggest that cross-pol
should always be zero. In common engineering
practice, cross-pol is non-zero when co-pol is
different from the intended or nominal polarization of
the field radiated by the antenna. In the case of the
ESD, we observe that the electric field is always
ˆθ-polarized, and therefore we consider that to be the
nominal polarization. Since the actual polarization of
the ESD in the example is precisely the same as the
nominal polarization of the ESD, the cross-pol of the
ideal ESD is zero. If, on the other hand, we arbitrarily
define ˆθto be the nominal polarization and apply this
definition to a different antenna that does not produce
a uniformly ˆθ-polarized electric field, then we observe",Electromagnetics_Vol2.pdf
"definition to a different antenna that does not produce
a uniformly ˆθ-polarized electric field, then we observe
non-zero cross-pol. Cross-pol is similarly used to
quantify effects due to errors in position or
orientation, or due to undesired modification of the
field due to materials (e.g., feed or mounting
structures) near the transmit antenna. Summarizing:
Co-pol is commonly defined to be the intended
or
nominal polarization for a particular applica-
tion, which is not necessarily the actual polariza-
tion radiated by the antenna under consideration.
Cross-pol measures polarization in the orthogo-
nal plane; i.e., deviation from the presumed co-
pol.
Returning to the ESD: Since ˜E(r) depends only on θ
and not φ, the co-pol pattern is the same in any plane
that contains the zaxis. W e refer to any such plane as
the E-plane. In general:
The E-plane is any plane in which the nominal or
intended vector ˜E lies.
Since the ESD is ˆθ-polarized, the E-plane pattern of
the ESD is simply:",Electromagnetics_Vol2.pdf
"The E-plane is any plane in which the nominal or
intended vector ˜E lies.
Since the ESD is ˆθ-polarized, the E-plane pattern of
the ESD is simply:
⏐⏐⏐˜E(r)
⏐⏐⏐≈ ηI0 · βL
8π (s in θ) 1
r (10.58)
This
pattern is shown in Figure 10.5.
Equation 10.58 is referred to as an unnormalized
c⃝ S. Lally CC BY -SA 4.0
Figure 10.5: E-plane co-pol pattern for the ˆz-oriented
ESD. In the unnormalized pattern scaling, the dashed
circle indicates the maximum value of Equation 10.58.
pattern. The associated normalized pattern is
addressed later in this section.
Similarly , we define the H-plane as follows:
The H-plane is any plane in which the nominal or
intended vector ˜H lies, and so is perpendicular to
both the E-plane and the direction of propagation.
In the case of the ESD, the one and only H-plane is
the z= 0 plane. The H-plane pattern is shown in
Figure 10.6.
It is often useful to normalize the pattern, meaning
that we scale the pattern so that its maximum",Electromagnetics_Vol2.pdf
"Figure 10.6.
It is often useful to normalize the pattern, meaning
that we scale the pattern so that its maximum
magnitude corresponds to a convenient value. There
are two common scalings. One common scaling sets
the maximum value of the pattern equal to 1, and is
therefore independent of source magnitude and
distance. This is referred to as the normalized pattern.
The normalized co-pol pattern can be defined as
follows:
F(θ,φ) ≜
⏐⏐
⏐ˆe · ˜E(r)
⏐
⏐
⏐
⏐⏐⏐ˆe · ˜E(r)
⏐⏐⏐
max
(10.59)
where ˆe is
the co-pol reference direction, and the",Electromagnetics_Vol2.pdf
"10.8. RADIA TION P A TTERN 181
c⃝ S. Lally CC BY -SA 4.0
Figure 10.6: H-plane co-pol pattern for the ˆz-oriented
ESD. In the unnormalized pattern scaling, the radius
of the circle is the maximum value of Equation 10.58.
denominator is the maximum value of the electric
field at distance r.
A normalized pattern is scaled to a maximum
magnitude of 1, using the definition expressed in
Equation 10.59.
Note that the value of ris irrelevant, since numerator
and denominator both scale with rin the same way .
Thus, the normalized pattern, like directivity , does not
change with distance from the antenna.
For the ESD, ˆe = ˆθ, so the normalized co-pol pattern
is:
F(θ,φ) =
⏐⏐
⏐ˆθ· ˜E(r)
⏐
⏐
⏐
⏐
⏐⏐ˆθ· ˜E(r)
⏐⏐⏐
max
(ESD)
 (10.60)
which yields simply
F(θ,φ) ≈ sin θ (ESD) (10.61)
Thus, the E-plane normalized co-pol pattern of the
ESD is Figure 10.5 where the radius of the maximum
value circle is equal to 1, which is 0 dB. Similarly , the
H-plane normalized co-pol pattern of the ESD is",Electromagnetics_Vol2.pdf
"value circle is equal to 1, which is 0 dB. Similarly , the
H-plane normalized co-pol pattern of the ESD is
Figure 10.6 where the radius of the circle is equal to 1
(0 dB).
c⃝ T . Truckle CC BY -SA 4.0 (modified)
Figure 10.7: Co-pol pattern typical of a highly direc-
tional antenna, such as a yagi.
The other common scaling for patterns sets the
maximum value equal to maximum directivity .
Directivity is proportional to power density , which is
proportional to
⏐⏐
⏐˜E(r)
⏐
⏐
⏐
2
. Therefore, this “directivity
normalized” pattern can be expressed as:
Dmax|F(θ,φ)|2 (10.62)
where Dmax is the directivity in whichever direction
it is maximum. For the ESD considered previously ,
Dmax = 1.5 (Section 10.7). Therefore, the co-pol
pattern of the ESD using this particular scaling is
1.5 sin2 θ (ESD) (10.63)
Thus, the E-plane co-pol pattern of the ESD using this
scaling is similar to Figure 10.5, except that it is
squared in magnitude and the radius of the maximum",Electromagnetics_Vol2.pdf
"scaling is similar to Figure 10.5, except that it is
squared in magnitude and the radius of the maximum
value circle is equal to 1.5, which is 1.76 dB. The
H-plane co-pol pattern of the ESD, using this scaling,
is Figure 10.6 where the radius of the circle is equal
to 1.5 (1.76 dB).
Pattern lobes. Nearly all antennas in common use
exhibit directivity that is greater than that of the ESD.
The patterns of these antennas subsequently exhibit
more complex structure. Figure 10.7 shows an
example. The region around the direction of
maximum directivity is referred to as the main lobe.",Electromagnetics_Vol2.pdf
"182 CHAPTER 10. ANTENNAS
c⃝ S. Lally CC BY -SA 4.0
Figure 10.8: Half-power beamwidth (HPBW).
The main lobe is bounded on each side by a null,
where the magnitude reaches a local minimum,
perhaps zero. Many antennas also exhibit a lobe in
the opposite direction, known as a backlobe. (Many
other antennas exhibit a null in this direction.) Lobes
between the main lobe and the backlobe are referred
to as sidelobes. Other commonly-used pattern metrics
include first sidelobe level, which is the ratio of the
maximum magnitude of the sidelobe closest to the
main lobe to that of the main lobe; and front-to-back
ratio, which is the ratio of the maximum magnitude
of the backlobe to that of the main lobe.
When the main lobe is narrow , it is common to
characterize the pattern in terms of half-power
beamwidth (HPBW). This is shown in Figure 10.8.
HPBW is the width of the main lobe measured
between two points at which the directivity is
one-half its maximum value. As a pattern metric,",Electromagnetics_Vol2.pdf
"HPBW is the width of the main lobe measured
between two points at which the directivity is
one-half its maximum value. As a pattern metric,
HPBW depends on the plane in which it is measured.
In particular, HPBW may be different in the E- and
H-planes. For example, the E-plane HPBW of the
ESD is found to be 90◦, whereas the H-plane HPBW
is undefined since the pattern is constant in the
H-plane.
Omnidirectional and isotropic antennas. The ESD
is an example of an omnidirectional antenna. An
omnidirectional antenna is an antenna whose pattern
magnitude is nominally constant in a plane containing
the maximum directivity . For the ESD, this plane is
the H-plane, so the ESD is said to be omnidirectional
in the H-plane. The term “omnidirectional” does not
indicate constant pattern in all directions. (In fact,
note that the ESD exhibits pattern nulls in two
directions.)
An omnidirectional antenna, such as the ESD, ex-
hibits constant and nominally maximum directiv-
ity in one plane.",Electromagnetics_Vol2.pdf
"directions.)
An omnidirectional antenna, such as the ESD, ex-
hibits constant and nominally maximum directiv-
ity in one plane.
An antenna whose pattern is uniform in all directions
is said to be isotropic. No physically-realizable
antenna is isotropic; in fact, the ESD is about as close
as one can get. Nevertheless, the “isotropic antenna”
concept is useful as a standard against which other
antennas can be quantified. Since the pattern of an
isotropic antenna is constant with direction, the power
density radiated by such an antenna in any direction is
equal to the power density averaged over all
directions. Thus, the directivity of an isotropic
antenna is exactly 1. The directivity of any other
antenna may be expressed in units of “dBi, ” which is
simply dB relative to that of an isotropic antenna. For
example, the maximum directivity of the ESD is 1.5,
which is 10 log10 (1.5/1) = 1.76 dBi. Summarizing:
An isotropic antenna exhibits constant directiv-",Electromagnetics_Vol2.pdf
"which is 10 log10 (1.5/1) = 1.76 dBi. Summarizing:
An isotropic antenna exhibits constant directiv-
ity in every direction. Such an antenna is not
physically-realizable, but is nevertheless useful
as a baseline for describing other antennas.
Additional Reading:
• “Radiation pattern” on Wikipedia.",Electromagnetics_Vol2.pdf
"10.9. EQUIV ALENT CIRCUIT MODEL FOR RECEPTION 183
10.9 Equivalent Circuit Model
for Reception
[m0206]
In this section, we begin to address antennas as
de
vices that convert incident electromagnetic waves
into potentials and currents in a circuit. It is
convenient to represent this process in the form of a
Th´evenin equivalent circuit. The particular circuit
addressed in this section is shown in Figure 10.9. The
circuit consists of a voltage source ˜VOC and a series
impedance ZA. The source potential ˜VOC is the
potential at the terminals of the antenna when there is
no termination; i.e., when the antenna is
open-circuited. The series impedance ZA is the
output impedance of the circuit, and so determines the
magnitude and phase of the current at the terminals
once a load is connected. Given ˜VOC and the current
through the equivalent circuit, it is possible to
determine the power delivered to the load. Thus, this
model is quite useful, but only if we are able to",Electromagnetics_Vol2.pdf
"determine the power delivered to the load. Thus, this
model is quite useful, but only if we are able to
determine ˜VOC and ZA. This section provides an
informal derivation of these quantities that is
sufficient to productively address the subsequent
important topics of effective aperture and impedance
matching of receive antennas. 3
V ector effective length. With no derivation required,
we can deduce the following about ˜VOC:
• ˜VOC must depend on the incident electric field
3 Formal derivations of these quantities are provided in subse-
quent sections. The starting point is the section on reciprocity .
c⃝ S. Lally, CC BY -SA 4.0 (modified)
Figure 10.9: Th ´evenin equivalent circuit for an an-
tenna in the presence of an incident electromagnetic
wave.
intensity ˜Ei. Presumably the relationship is
linear, so ˜VOC is proportional to the magnitude
of ˜Ei.
• Since ˜Ei is a vector whereas ˜VOC is a scalar,
there must be some vector le for which
˜VOC = ˜Ei · le (10.64)",Electromagnetics_Vol2.pdf
"of ˜Ei.
• Since ˜Ei is a vector whereas ˜VOC is a scalar,
there must be some vector le for which
˜VOC = ˜Ei · le (10.64)
• Since ˜Ei has SI base units of V/m and ˜VOC has
SI base units of V , le must have SI base units of
m; i.e., length.
• W e expect that ˜VOC increases as the size of the
antenna increases, so the magnitude of le likely
increases with the size of the antenna.
• The direction of le must be related to the
orientation of the incident electric field relative
to that of the antenna, since this is clearly
important yet we have not already accounted for
this.
It may seem at this point that le is unambiguously
determined, and we need merely to derive its value.
However, this is not the case. There are in fact
multiple unique definitions of le that will reduce the
vector ˜Ei to the observed scalar ˜VOC via
Equation 10.64. In this section, we shall employ the
most commonly-used definition, in which le is
referred to as vector effective length. Following this",Electromagnetics_Vol2.pdf
"most commonly-used definition, in which le is
referred to as vector effective length. Following this
definition, the scalar part le of le = ˆlle is commonly
referred to as any of the following: effective length
(the term used in this book), effective height, or
antenna factor.
In this section, we shall merely define vector effective
length, and defer a formal derivation to Section 10.11.
In this definition, we arbitrarily set ˆl, the real-valued
unit vector indicating the direction of le, equal to the
direction in which the electric field transmitted from
this antenna would be polarized in the far field. For
example, consider a ˆz-oriented electrically-short
dipole (ESD) located at the origin. The electric field
transmitted from this antenna would have only a ˆθ
component, and no ˆφcomponent (and certainly no ˆr
component). Thus, ˆl = ˆθin this case.
Applying this definition, ˜Ei · ˆl yields the scalar
component of ˜Ei that is co-polarized with electric",Electromagnetics_Vol2.pdf
"component). Thus, ˆl = ˆθin this case.
Applying this definition, ˜Ei · ˆl yields the scalar
component of ˜Ei that is co-polarized with electric
field radiated by the antenna when transmitting. Now",Electromagnetics_Vol2.pdf
"184 CHAPTER 10. ANTENNAS
le is uniquely defined to be the factor that converts
this component into ˜VOC. Summarizing:
The vector effective length le = ˆlle is defined
as follows: ˆl is the real-valued unit vector cor-
responding to the polarization of the electric field
that would be transmitted from the antenna in the
far field. Subsequently , the effective length le is
le ≜
˜VOC
˜Ei · ˆl
(10.65)
where ˜VO
C is the open-circuit potential induced
at the antenna terminals in response to the inci-
dent electric field intensity ˜Ei.
While this definition yields an unambiguous value for
le, it is not yet clear what that value is. For most
antennas, effective length is quite difficult to
determine directly , and one must instead determine
effective length indirectly from the transmit
characteristics via reciprocity . This approach is
relatively easy (although still quite a bit of effort) for
thin dipoles, and is presented in Section 10.11.
T o provide an example of how effective length works",Electromagnetics_Vol2.pdf
"thin dipoles, and is presented in Section 10.11.
T o provide an example of how effective length works
right away , consider the ˆz-oriented ESD described
earlier in this section. Let the length of this ESD be
L. Let ˜Ei be a ˆθ-polarized plane wave arriving at the
ESD. The ESD is open-circuited, so the potential
induced in its terminals is ˜VOC. One observes the
following:
• When ˜Ei arrives from anywhere in the θ= π/2
plane (i.e., broadside to the ESD), ˜Ei points in
the −ˆz direction, and we find that le ≈ L/2. It
should not be surprising that le is proportional to
L; this expectation was noted earlier in this
section.
• When ˜Ei arrives from the directions θ= 0 or
θ= π– i.e., along the axis of the ESD – ˜Ei is
perpendicular to the axis of the ESD. In this
case, we find that le equals zero.
T aken together, these findings suggest that le should
contain a factor of sin θ. W e conclude that the vector
effective length for a ˆz-directed ESD of length Lis
le ≈ ˆθL
2 sin θ (ESD) (10.66)",Electromagnetics_Vol2.pdf
"contain a factor of sin θ. W e conclude that the vector
effective length for a ˆz-directed ESD of length Lis
le ≈ ˆθL
2 sin θ (ESD) (10.66)
Example 10.7. Potential induced in an ESD.
A thin straight dipole of length 10 cm is located
at the origin and aligned with the z-axis. A plane
wave is incident on the dipole from the direction
(θ= π/4,φ = π/2). The frequency of the wave
is 30 MHz. The magnitude of the incident
electric field is 10 µV/m (rms). What is the
magnitude of the induced open-circuit potential
when the electric field is (a) ˆθ-polarized and (b)
ˆφ-polarized?
Solution. The wavelength in this example is
c/f ∼= 10 m, so this dipole is electrically-short.
Using Equation 10.66:
le ≈ ˆθ10 cm
2 sin π
4
≈ ˆθ(3
.54 cm)
Thus, the effective length le = 3.54 cm. When
the electric field is ˆθ-polarized, the magnitude of
the induced open-circuit voltage is
⏐⏐⏐˜VOC
⏐⏐⏐=
⏐⏐⏐˜Ei · le
⏐⏐⏐
≈ (10 µV/m) ˆθ· ˆθ(3.54 cm)
≈ 354 nV rms (a)
When
the electric field is ˆφ-polarized:
⏐⏐⏐˜VOC",Electromagnetics_Vol2.pdf
"⏐⏐⏐˜VOC
⏐⏐⏐=
⏐⏐⏐˜Ei · le
⏐⏐⏐
≈ (10 µV/m) ˆθ· ˆθ(3.54 cm)
≈ 354 nV rms (a)
When
the electric field is ˆφ-polarized:
⏐⏐⏐˜VOC
⏐⏐⏐≈ (10 µV/m) ˆφ· ˆθ(3.54 cm)
≈ 0 (b)
This
is because the polarization of the incident
electric field is orthogonal to that of the ESD. In
fact, the answer to part (b) is zero for any angle
of incidence (θ,φ).
Output impedance. The output impedance ZA is
somewhat more difficult to determine without a
formal derivation, which is presented in
Section 10.12. For the purposes of this section, it
suffices to jump directly to the result:",Electromagnetics_Vol2.pdf
"10.9. EQUIV ALENT CIRCUIT MODEL FOR RECEPTION 185
The output impedance ZA of the equivalent cir-
cuit for an antenna in the receive case is equal to
the input impedance of the same antenna in the
transmit case.
This remarkable fact is a consequence of the
reciprocity property of antenna systems, and greatly
simplifies the analysis of receive antennas.
Now a demonstration of how the antenna equivalent
circuit can be used to determine the power delivered
by an antenna to an attached electrical circuit:
Example 10.8. Po wer captured by an ESD.
Continuing with part (a) of Example 10.7: If this
antenna is terminated into a conjugate-matched
load, then what is the power delivered to that
load? Assume the antenna is lossless.
Solution. First, we determine the impedance ZA
of the equivalent circuit of the antenna. This is
equal to the input impedance of the antenna in
transmission. Let RA and XA be the real and
imaginary parts of this impedance; i.e.,",Electromagnetics_Vol2.pdf
"equal to the input impedance of the antenna in
transmission. Let RA and XA be the real and
imaginary parts of this impedance; i.e.,
ZA = RA + jXA. Further, RA is the sum of the
radiation resistance Rrad and the loss resistance.
The loss resistance is zero because the antenna is
lossless. Since this is an ESD:
Rrad ≈ 20π2
(L
λ
)2
(10.67)
Therefore, RA = Rr
ad ≈ 4.93 mΩ. W e do not
need to calculate XA, as will become apparent
in the next step.
A conjugate-matched load has impedance Z∗
A,
so the potential ˜VL across the load is
˜VL = ˜VOC
Z∗
A
ZA + Z∗
A
= ˜VOC
Z∗
A
2RA
(10.68)
The
current ˜IL through the load is
˜IL =
˜VOC
ZA + Z∗
A
=
˜VOC
2RA
(10.69)
T
aking ˜VOC as an RMS quantity , the power PL
delivered to the load is
PL = Re {
VLI∗
L} =
⏐⏐
⏐˜VOC
⏐
⏐
⏐
2
4RA
(10.70)
In
part (a) of Example 10.7,
⏐
⏐
⏐˜VOC
⏐
⏐
⏐is found to be
≈ 354 nV rms, so PL ≈ 6.33 pW
.
Additional
Reading:
• “Th ´evenin’s Theorem” on Wikipedia.
• W .L. Stutzman & G.A. Thiele, Antenna Theory",Electromagnetics_Vol2.pdf
"≈ 354 nV rms, so PL ≈ 6.33 pW
.
Additional
Reading:
• “Th ´evenin’s Theorem” on Wikipedia.
• W .L. Stutzman & G.A. Thiele, Antenna Theory
and Design, 3rd Ed., Wiley , 2012. Sec. 4.2
(“Receiving Properties of Antennas”).",Electromagnetics_Vol2.pdf
"186 CHAPTER 10. ANTENNAS
10.10 Reciprocity
[m0214]
The term “reciprocity” refers to a class of theorems
that
relate the inputs and outputs of a linear system to
those of an identical system in which the inputs and
outputs are swapped. The importance of reciprocity in
electromagnetics is that it simplifies problems that
would otherwise be relatively difficult to solve. An
example of this is the derivation of the receiving
properties of antennas, which is addressed in other
sections using results derived in this section.
As an initial and relatively gentle introduction,
consider a well-known special case that emerges in
basic circuit theory: The two-port device shown in
Figure 10.10. The two-port is said to be reciprocal if
the voltage v2 appearing at port 2 due to a current
applied at port 1 is the same as v1 when the same
current is applied instead at port 2. This is normally
the case when the two-port consists exclusively of
linear passive devices such as ideal resistors,",Electromagnetics_Vol2.pdf
"the case when the two-port consists exclusively of
linear passive devices such as ideal resistors,
capacitors, and inductors. The fundamental
underlying requirement is linearity: That is, outputs
must be proportional to inputs, and superposition
must apply .
This result from basic circuit theory is actually a
special case of a more general theorem of
electromagnetics, which we shall now derive.
Figure 10.11 shows a scenario in which a current
distribution ˜J1 is completely contained within a
volume V. This current is expressed as a volume
current density , having SI base units of A/m 2. Also,
the current is expressed in phasor form, signaling that
we are considering a single frequency . This current
distribution gives rise to an electric field intensity ˜E1,
Inductiveload, public domain (modified)
Figure
10.10: T wo-port device.
c⃝ C. W ang CC BY -SA 4.0
Figure 10.11: An electromagnetic system consisting
of a current distribution ˜J1 radiating an electric field",Electromagnetics_Vol2.pdf
"c⃝ C. W ang CC BY -SA 4.0
Figure 10.11: An electromagnetic system consisting
of a current distribution ˜J1 radiating an electric field
˜E1 in the presence of linear matter.
c⃝ C. W ang CC BY -SA 4.0
Figure 10.12: The same electromagnetic system as
shown in Figure 10.11 (including the same distribution
of matter), but now with a different input, ˜J2, resulting
in a different output, ˜E2.",Electromagnetics_Vol2.pdf
"10.10. RECIPROCITY 187
having SI base units of V/m. The volume may consist
of any combination of linear time-invariant matter;
i.e., permittivity ǫ, permeability µ, and conductivity σ
are constants that may vary arbitrarily with position
but not with time.
Here’s a key idea: W e may interpret this scenario as a
“two-port” system in which ˜J1 is the input, ˜E1 is the
output, and the system’s behavior is completely
defined by Maxwell’s equations in differential phasor
form:
∇ × ˜E1 = −jωµ˜H1 (10.71)
∇ × ˜H1 = ˜J1 + jωǫ˜E1 (10.72)
along with the appropriate electromagnetic boundary
conditions. (For the purposes of our present analysis,
˜H1 is neither an input nor an output; it is merely a
coupled quantity that appears in this particular
formulation of the relationship between ˜E1 and ˜J1.)
Figure 10.12 shows a scenario in which a different
current distribution ˜J2 is completely contained within
the same volume V containing the same distribution",Electromagnetics_Vol2.pdf
"current distribution ˜J2 is completely contained within
the same volume V containing the same distribution
of linear matter. This new current distribution gives
rise to an electric field intensity ˜E2. The relationship
between ˜E2 and ˜J2 is governed by the same
equations:
∇ × ˜E2 = −jωµ˜H2 (10.73)
∇ × ˜H2 = ˜J2 + jωǫ˜E2 (10.74)
along with the same electromagnetic boundary
conditions. Thus, we may interpret this scenario as
the same electromagnetic system depicted in
Figure 10.11, except now with ˜J2 as the input and ˜E2
is the output.
The input current and output field in the second
scenario are, in general, completely different from the
input current and output field in the first scenario.
However, both scenarios involve precisely the same
electromagnetic system; i.e., the same governing
equations and the same distribution of matter. This
leads to the following question: Let’s say you know
nothing about the system, aside from the fact that it is",Electromagnetics_Vol2.pdf
"leads to the following question: Let’s say you know
nothing about the system, aside from the fact that it is
linear and time-invariant. However, you are able to
observe the first scenario; i.e., you know ˜E1 in
response to ˜J1. Given this very limited look into the
behavior of the system, what can you infer about ˜E2
given ˜J2, or vice-versa? At first glance, the answer
might seem to be nothing, since you lack a description
of the system. However, the answer turns out to be
that a surprising bit more information is available.
This information is provided by reciprocity . T o show
this, we must engage in some pure mathematics.
Derivation of the Lorentz reciprocity theorem.
First, let us take the dot product of ˜H2 with each side
of Equation 10.71:
˜H2 ·
(
∇ × ˜E1
)
= −jωµ˜H1 · ˜H2 (10.75)
Similarly , let us take the dot product of ˜E1 with each
side of Equation 10.74:
˜E1 ·
(
∇ × ˜H2
)
= ˜E1 · ˜J2 + jωǫ˜E1 · ˜E2 (10.76)
Next, let us subtract Equation 10.75 from",Electromagnetics_Vol2.pdf
"side of Equation 10.74:
˜E1 ·
(
∇ × ˜H2
)
= ˜E1 · ˜J2 + jωǫ˜E1 · ˜E2 (10.76)
Next, let us subtract Equation 10.75 from
Equation 10.76. The left side of the resulting equation
is
˜E1 ·
(
∇ × ˜H2
)
− ˜H2 ·
(
∇ × ˜E1
)
(10.77)
This can be simplified using the vector identity
(Appendix B.3):
∇ · (A × B) = B · (∇ × A) − A · (∇ × B) (10.78)
Yielding
˜E1 ·
(
∇ × ˜H2
)
− ˜H2 ·
(
∇ × ˜E1
)
= ∇ ·
(
˜H2 × ˜E1
)
(10.79)
So, by subtracting Equation 10.75 from
Equation 10.76, we have found:
∇·
(
˜H2 × ˜E1
)
= ˜E1 ·˜J2 +jωǫ˜E1 · ˜E2 +jωµ˜H1 · ˜H2
(10.80)
Next, we repeat the process represented by
Equations 10.75–10.80 above to generate a
complementary equation to Equation 10.80. This time
we take the dot product of ˜E2 with each side of
Equation 10.72:
˜E2 ·
(
∇ × ˜H1
)
= ˜E2 · ˜J1 + jωǫ˜E1 · ˜E2 (10.81)
Similarly , let us take the dot product of ˜H1 with each
side of Equation 10.73:
˜H1 ·
(
∇ × ˜E2
)
= −jωµ˜H1 · ˜H2 (10.82)",Electromagnetics_Vol2.pdf
"188 CHAPTER 10. ANTENNAS
Next, let us subtract Equation 10.82 from
Equation 10.81. Again using the vector identity , the
left side of the resulting equation is
˜E2 ·
(
∇ × ˜H1
)
− ˜H1 ·
(
∇ × ˜E2
)
= ∇ ·
(
˜H1 × ˜E2
)
(10.83)
So we find:
∇·
(
˜H1 × ˜E2
)
= ˜E2 ·˜J1 +jωǫ˜E1 · ˜E2 +jωµ˜H1 · ˜H2
(10.84)
Finally , subtracting Equation 10.84 from
Equation 10.80, we obtain:
∇ ·
(
˜H2 × ˜E1 − ˜H1 × ˜E2
)
= ˜E1 · ˜J2 − ˜E2 · ˜J1
(10.85)
This equation is commonly known by the name of the
theorem it represents: The Lorentz reciprocity
theorem. The theorem is a very general statement
about the relationship between fields and currents at
each point in space. An associated form of the
theorem applies to contiguous regions of space. T o
obtain this form, we simply integrate both sides of
Equation 10.85 over the volume V:
∫
V
∇ ·
(
˜H2 × ˜E1 − ˜H1 × ˜E2
)
dv
=
∫
V
(
˜E1 · ˜J2 − ˜E2 · ˜J1
)
dv (10.86)
W e now take the additional step of using the
divergence theorem (Appendix B.3) to transform the",Electromagnetics_Vol2.pdf
")
dv
=
∫
V
(
˜E1 · ˜J2 − ˜E2 · ˜J1
)
dv (10.86)
W e now take the additional step of using the
divergence theorem (Appendix B.3) to transform the
left side of the equation into a surface integral:
∮
S
(
˜H2 × ˜E1 − ˜H1 × ˜E2
)
· ds
=
∫
V
(
˜E1 · ˜J2 − ˜E2 · ˜J1
)
dv (10.87)
where S is the closed mathematical surface which
bounds V. This is also the Lorentz reciprocity
theorem, but now in integral form. This version of the
theorem relates fields on the bounding surface to
sources within the volume.
The integral form of the theorem has a particularly
useful feature. Let us confine the sources to a finite
region of space, while allowing V to grow infinitely
large, expanding to include all space. In this situation,
the closest distance between any point containing
non-zero source current and S is infinite. Because
field magnitude diminishes with distance from the
source, the fields (˜E1, ˜H1) and (˜E2, ˜H2) are all
effectively zero on S. In this case, the left side of",Electromagnetics_Vol2.pdf
"source, the fields (˜E1, ˜H1) and (˜E2, ˜H2) are all
effectively zero on S. In this case, the left side of
Equation 10.87 is zero, and we find:
∫
V
(
˜E1 · ˜J2 − ˜E2 · ˜J1
)
dv= 0
(10.88)
for
any volume V which contains all the current.
The Lorentz reciprocity theorem (Equa-
tion
10.88) describes a relationship between
one distribution of current and the resulting
fields, and a second distribution of current and
resulting fields, when both scenarios take place
in identical regions of space filled with identical
distributions of linear matter.
Why do we refer to this relationship as reciprocity?
Simply because the expression is identical when the
subscripts “1” and “2” are swapped. In other words,
the relationship does not recognize a distinction
between “inputs” and “outputs;” there are only
“ports. ”
A useful special case pertains to scenarios in which
the current distributions ˜J1 and ˜J2 are spatially
disjoint. By “spatially disjoint, ” we mean that there is",Electromagnetics_Vol2.pdf
"the current distributions ˜J1 and ˜J2 are spatially
disjoint. By “spatially disjoint, ” we mean that there is
no point in space at which both ˜J1 and ˜J2 are
non-zero; in other words, these distributions do not
overlap. (Note that the currents shown in
Figures 10.11 and 10.12 are depicted as spatially
disjoint.) T o see what happens in this case, let us first
rewrite Equation 10.88 as follows:
∫
V
˜E1 · ˜J2 dv=
∫
V
˜E2 · ˜J1 dv (10.89)
Let V1 be the contiguous volume over which ˜J1 is
non-zero, and let V2 be the contiguous volume over
which ˜J2 is non-zero. Then Equation 10.89 may be
written as follows:
∫
V2
˜E1 · ˜J2 dv=
∫
V1
˜E2 · ˜J1 dv (10.90)
The utility of this form is that we have reduced the
region of integration to just those regions where the
current exists.",Electromagnetics_Vol2.pdf
"10.10. RECIPROCITY 189
c⃝ C. W ang CC BY -SA 4.0
Figure 10.13: A two-port consisting of two dipole
antennas.
Reciprocity of two-ports consisting of antennas.
Equation 10.90 allows us to establish the reciprocity
of two-ports consisting of pairs of antennas. This is
most easily demonstrated for pairs of thin dipole
antennas, as shown in Figure 10.13. Here, port 1 is
defined by the terminal quantities ( ˜V1,˜I1) of the
antenna on the left, and port 2 is defined by the
terminal quantities ( ˜V2,˜I2) of the antenna on the
right. These quantities are defined with respect to a
small gap of length ∆l between the
perfectly-conducting arms of the dipole. Either
antenna may transmit or receive, so ( ˜V1,˜I1) depends
on ( ˜V2,˜I2), and vice-versa.
The transmit and receive cases for port 1 are
illustrated in Figure 10.14(a) and (b), respectively .
Note that ˜J1 is the current distribution on this antenna
when transmitting (i.e., when driven by the impressed
current ˜It",Electromagnetics_Vol2.pdf
"Note that ˜J1 is the current distribution on this antenna
when transmitting (i.e., when driven by the impressed
current ˜It
1) and ˜E2 is the electric field incident on the
antenna when the other antenna is transmitting. Let us
select V1 to be the cylindrical volume defined by the
exterior surface of the dipole, including the gap
between the arms of the dipole. W e are now ready to
consider the right side of Equation 10.90:
∫
V1
˜E2 · ˜J1 dv (10.91)
The electromagnetic boundary conditions applicable
to the perfectly-conducting dipole arms allow this
integral to be dramatically simplified. First, recall that
all current associated with a perfect conductor must
flow on the surface of the material, and therefore
c⃝ C. W ang CC BY -SA 4.0
Figure 10.14: The port 1 antenna (a) transmitting and
(b) receiving.
˜J1 = 0 everywhere except on the surface. Therefore,
the direction of ˜J1 is always tangent to the surface.
The tangent component of ˜E2 is zero on the surface",Electromagnetics_Vol2.pdf
"the direction of ˜J1 is always tangent to the surface.
The tangent component of ˜E2 is zero on the surface
of a perfectly-conducting material, as required by the
applicable boundary condition. Therefore,
˜E2 · ˜J1 = 0 everywhere ˜J1 is non-zero.
There is only one location where the current is
non-zero and yet there is no conductor: This location
is the gap located precisely at the terminals. Thus, we
find: ∫
V1
˜E2 · ˜J1 dv=
∫
gap
˜E2 · ˜It
1ˆl1 dl (10.92)
where the right side is a line integral crossing the gap
that defines the terminals of the antenna. W e assume
the current ˜It
1 is constant over the gap, and so may be
factored out of the integral:
∫
V1
˜E2 · ˜J1 dv= ˜It
1
∫
gap
˜E2 · ˆl1 dl (10.93)
Recall that potential between two points in space is
given by the integral of the electric field over any path
between those two points. In particular, we may
calculate the open-circuit potential ˜Vr
1 as follows:
˜Vr
1 = −
∫
gap
˜E2 · ˆl1 dl (10.94)
Remarkably , we have found:
∫
V1",Electromagnetics_Vol2.pdf
"calculate the open-circuit potential ˜Vr
1 as follows:
˜Vr
1 = −
∫
gap
˜E2 · ˆl1 dl (10.94)
Remarkably , we have found:
∫
V1
˜E2 · ˜J1 dv= −˜It
1 ˜Vr
1 (10.95)",Electromagnetics_Vol2.pdf
"190 CHAPTER 10. ANTENNAS
Applying the exact same procedure for port 2 (or by
simply exchanging subscripts), we find:
∫
V2
˜E1 · ˜J2 dv= −˜It
2 ˜Vr
2 (10.96)
Now substituting these results into Equation 10.90,
we find:
˜It
1 ˜Vr
1 = ˜It
2 ˜Vr
2
(10.97)
At
the beginning of this section, we stated that a
two-port is reciprocal if ˜V2 appearing at port 2 due to
a current applied at port 1 is the same as ˜V1 when the
same current is applied instead at port 2. If the
present two-dipole problem is reciprocal, then ˜Vr
2 due
to ˜It
1 should be the same as ˜Vr
1 when ˜It
2 = ˜It
1. Is it?
Let us set ˜It
1 equal to some particular value I0, then
the resulting value of ˜Vr
2 will be some particular
value V0. If we subsequently set ˜It
2 equal to I0, then
the resulting value of ˜Vr
1 will be, according to
Equation 10.97:
˜Vr
1 =
˜It
2 ˜Vr
2
˜It
1
= I0V0
I0
= V0 (10.98)
Therefore,
Equation 10.97 is simply a mathematical
form of the familiar definition of reciprocity from",Electromagnetics_Vol2.pdf
"1 =
˜It
2 ˜Vr
2
˜It
1
= I0V0
I0
= V0 (10.98)
Therefore,
Equation 10.97 is simply a mathematical
form of the familiar definition of reciprocity from
basic circuit theory , and we have found that our
system of antennas is reciprocal in precisely this same
sense.
The above analysis presumed pairs of straight,
perfectly conducting dipoles of arbitrary length.
However, the result is readily generalized – in fact, is
the same – for any pair of passive antennas in linear
time-invariant media. Summarizing:
The potential induced at the terminals of one an-
tenna
due to a current applied to a second antenna
is equal to the potential induced in the second an-
tenna by the same current applied to the first an-
tenna (Equation 10.97).
Additional Reading:
• “Reciprocity (electromagnetism)” on Wikipedia.
• “T wo-port network” on Wikipedia.
10.11 Potential Induced in a
Dipole
[m0215]
An electromagnetic wave incident on an antenna will
induce
a potential at the terminals of the antenna. In",Electromagnetics_Vol2.pdf
"Dipole
[m0215]
An electromagnetic wave incident on an antenna will
induce
a potential at the terminals of the antenna. In
this section, we shall derive this potential. T o simplify
the derivation, we shall consider the special case of a
straight thin dipole of arbitrary length that is
illuminated by a plane wave. However, certain aspects
of the derivation will apply to antennas generally . In
particular, the concepts of effective length (also
known as effective height ) and vector effective length
emerge naturally from this derivation, so this section
also serves as a stepping stone in the development of
an equivalent circuit model for a receiving antenna.
The derivation relies on the transmit properties of
dipoles as well as the principle of reciprocity , so
familiarity with those topics is recommended before
reading this section.
The scenario of interest is shown in Figure 10.15.
Here a thin ˆz-aligned straight dipole is located at the",Electromagnetics_Vol2.pdf
"reading this section.
The scenario of interest is shown in Figure 10.15.
Here a thin ˆz-aligned straight dipole is located at the
origin. The total length of the dipole is L. The arms
of the dipole are perfectly-conducting. The terminals
consist of a small gap of length ∆l between the arms.
The incident plane wave is described in terms of its
electric field intensity ˜Ei. The question is: What is
˜VOC, the potential at the terminals when the terminals
are open-circuited?
There are multiple approaches to solve this problem.
A direct attack is to invoke the principle that potential
is equal to the integral of the electric field intensity
over a path. In this case, the path begins at the “− ”
terminal and ends at the “+ ” terminal, crossing the
gap that defines the antenna terminals. Thus:
˜VOC = −
∫
gap
˜Egap · dl (10.99)
where ˜Egap is the electric field in the gap. The
problem with this approach is that the value of ˜Egap
is not readily available. It is not simply ˜Ei, because",Electromagnetics_Vol2.pdf
"problem with this approach is that the value of ˜Egap
is not readily available. It is not simply ˜Ei, because
the antenna structure (in particular, the
electromagnetic boundary conditions) modify the
electric field in the vicinity of the antenna. 4
4 Als o, if this were true, then the antenna itself would not matter;",Electromagnetics_Vol2.pdf
"10.11. POTENTIAL INDUCED IN A DIPOLE 191
Fortunately , we can bypass this obstacle using the
principle of reciprocity . In a reciprocity-based
strategy , we establish a relationship between two
scenarios that take place within the same
electromagnetic system. The first scenario is shown in
Figure 10.16. In this scenario, we have two dipoles.
The first dipole is precisely the dipole of interest
(Figure 10.15), except that a current ˜It
1 is applied to
the antenna terminals. This gives rise to a current
distribution ˜I(z) (SI base units of A) along the dipole,
and subsequently the dipole radiates the electric field
(Section 9.6):
˜E1(r) ≈ ˆθjη
2
e−jβ
r
r (s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
(10.100)
The second antenna is a ˆθ-aligned Hertzian dipole in
the far field, which receives ˜E1. (For a refresher on
the properties of Hertzian dipoles, see Section 9.4. A
key point is that Hertzian dipoles are vanishingly
small.) Specifically , we measure (conceptually , at",Electromagnetics_Vol2.pdf
"key point is that Hertzian dipoles are vanishingly
small.) Specifically , we measure (conceptually , at
least) the open-circuit potential ˜Vr
2 at the terminals of
the Hertzian dipole. W e select a Hertzian dipole for
this purpose because – in contrast to essentially all
only the relative spacing and orientation of the antenna terminals
w
ould matter!
c⃝ C. W ang CC BY -SA 4.0
Figure 10.15: A potential is induced at the terminals
of a thin straight dipole in response to an incident plane
wave.
c⃝ C. W ang CC BY -SA 4.0
Figure 10.16: The dipole of interest driven by current
˜It
1 radiates electric field ˜E1, resulting in open-circuit
potential ˜Vr
2 at the terminals of a Hertzian dipole in
the far field.
c⃝ C. W ang CC BY -SA 4.0
Figure 10.17: The Hertzian dipole driven by current
˜It
2 radiates electric field ˜E2, resulting in open-circuit
potential ˜Vr
1 at the terminals of the dipole of interest.",Electromagnetics_Vol2.pdf
"192 CHAPTER 10. ANTENNAS
other antennas – it is simple to determine the open
circuit potential. As explained earlier:
˜Vr
2 = −
∫
gap
˜Egap · dl (10.101)
For the Hertzian dipole, ˜Egap is simply the incident
electric field, since there is negligible structure (in
particular, a negligible amount of material) present to
modify the electric field. Thus, we have simply:
˜Vr
2 = −
∫
gap
˜E1 · dl (10.102)
Since the Hertzian dipole is very short and very far
away from the transmitting dipole, ˜E1 is essentially
constant over the gap. Also recall that we required the
Hertzian dipole to be aligned with ˜E1. Choosing to
integrate in a straight line across the gap,
Equation 10.102 reduces to:
˜Vr
2 = −˜E1(r2) · ˆθ∆l (10.103)
where ∆l is the length of the gap and r2 is the
location of the Hertzian dipole. Substituting the
expression for ˜E1 from Equation 10.100, we obtain:
˜Vr
2 ≈ − jη
2
e−jβ
r2
r2
(s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
∆l
(10.104)
where r2 = |r2|.",Electromagnetics_Vol2.pdf
"˜Vr
2 ≈ − jη
2
e−jβ
r2
r2
(s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
∆l
(10.104)
where r2 = |r2|.
The second scenario is shown in Figure 10.17. This
scenario is identical to the first scenario, with the
exception that the Hertzian dipole transmits and the
dipole of interest receives. The field radiated by the
Hertzian dipole in response to applied current ˜It
2,
evaluated at the origin, is (Section 9.4):
˜E2(r = 0) ≈ ˆθjη
˜It
2 · β∆l
4π (1) e−j
βr2
r2
(10.105)
The
“sin θ” factor in the general expression is equal to
1 in this case, since, as shown in Figure 10.17, the
origin is located broadside (i.e., at π/2 rad) relative to
the axis of the Hertzian dipole. Also note that because
the Hertzian dipole is presumed to be in the far field,
E2 may be interpreted as a plane wave in the region
of the receiving dipole of interest.
Now we ask: What is the induced potential ˜Vr
1 in the
dipole of interest? Once again, Equation 10.99 is not",Electromagnetics_Vol2.pdf
"of the receiving dipole of interest.
Now we ask: What is the induced potential ˜Vr
1 in the
dipole of interest? Once again, Equation 10.99 is not
much help, because the electric field in the gap is not
known. However, we do know that ˜Vr
1 should be
proportional to ˜E2(r = 0), since this is presumed to
be a linear system. Based on this much information
alone, there must be some vector le = ˆlle for which
˜Vr
1 = ˜E2(r = 0) · le (10.106)
This does not uniquely define either the unit vector ˆl
nor the scalar part le, since a change in the definition
of the former can be compensated by a change in the
definition of the latter and vice-versa. So at this point
we invoke the standard definition of le as the vector
effective length, introduced in Section 10.9. Thus, ˆl is
defined to be the direction in which an electric field
transmitted from the antenna would be polarized. In
the present example, ˆl = −ˆθ, where the minus sign
reflects the fact that positive terminal potential results",Electromagnetics_Vol2.pdf
"the present example, ˆl = −ˆθ, where the minus sign
reflects the fact that positive terminal potential results
in terminal current which flows in the −ˆz direction.
Thus, Equation 10.106 becomes:
˜Vr
1 = −˜E2(r = 0) · ˆθle (10.107)
W e may go a bit further and substitute the expression
for ˜E2(r = 0) from Equation 10.105:
˜Vr
1 ≈ − jη
˜It
2 · β∆l
4π
e−jβ
r2
r2
le (10.108)
No
w we invoke reciprocity . As a two-port linear
time-invariant system, it must be true that:
˜It
1 ˜Vr
1 = ˜It
2 ˜Vr
2 (10.109)
Thus:
˜Vr
1 =
˜It
2
˜It
1
˜Vr
2 (10.110)
Substituting
the expression for ˜Vr
2 from
Equation 10.104:
˜Vr
1 ≈ −
˜It
2
˜It
1
· jη
2
e−jβ
r2
r2
(s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
∆l
(10.111)
Thus, reciprocity has provided a second expression
for ˜Vr
1 . W e may solve for le by setting this expression",Electromagnetics_Vol2.pdf
"10.11. POTENTIAL INDUCED IN A DIPOLE 193
equal to the expression from Equation 10.108,
yielding
le ≈ 2π
β˜It
1
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
sin θ
(10.112)
Noting that β = 2π/λ, this simplifies to:
le ≈
[
1
˜It
1
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
sin θ
(10.113)
Thus, you can calculate le using the following
procedure:
1. Apply a current ˜It
1 to the dipole of interest.
2. Determine the resulting current distribution ˜I(z)
along the length of the dipole. (Note that
precisely this is done for the electrically-short
dipole in Section 9.5 and for the half-wave
dipole in Section 9.7.)
3. Integrate ˜I(z) over the length of the dipole as
indicated in Equation 10.113. Then divide
(“normalize”) by ˜It
1 (which is simply ˜I(0)). Note
that the result is independent of the excitation ˜It
1,
as expected since this is a linear system.
4. Multiply by sin θ.
W e have now determined that the open-circuit
terminal potential ˜VOC in response to an incident
electric field ˜Ei is",Electromagnetics_Vol2.pdf
"4. Multiply by sin θ.
W e have now determined that the open-circuit
terminal potential ˜VOC in response to an incident
electric field ˜Ei is
˜VOC = ˜Ei · le
(10.114)
where le = ˆlle is
the vector effective length defined
previously .
This result is remarkable. In plain English, we have
found that:
The potential induced in a dipole is the co-
polarized
component of the incident electric field
times a normalized integral of the transmit cur-
rent distribution over the length of the dipole,
times sine of the angle between the dipole axis
and the direction of incidence.
In other words, the reciprocity property of linear
systems allows this property of a receiving antenna to
be determined relatively easily if the transmit
characteristics of the antenna are known.
Example 10.9. Ef fective length of a thin
electrically-short dipole (ESD).
A explained in Section 9.5, the current
distribution on a thin ESD is
˜I(z) ≈ I0
(
1 − 2
L|z|
)
(10.115)
where Lis
the length of the dipole and I0 is the",Electromagnetics_Vol2.pdf
"distribution on a thin ESD is
˜I(z) ≈ I0
(
1 − 2
L|z|
)
(10.115)
where Lis
the length of the dipole and I0 is the
terminal current. Applying Equation 10.113, we
find:
le ≈
[
1
I0
∫ +L/ 2
−L/ 2
I0
(
1 − 2
L|z′|
)
e+jβ
z′ cos θdz′
]
· sin θ (10.116)
Recall β = 2π/λ, so βz′ = 2π (z′/λ). Since
this is an electrically-short dipole, z′ ≪ λover
the entire integral, and subsequently we may
assume e+jβz′ cos θ ≈ 1 over the entire integral.
Thus:
le ≈
[ ∫ +L/ 2
−L/ 2
(
1 − 2
L|z′|
)
dz′
]
s in θ
(10.117)
The integral is easily solved using standard
methods, or simply recognize that the “area
under the curve” in this case is simply one-half
“base” (L) times “height” (1). Either way , we
find
le ≈ L
2 sin θ (10.118)
Example
10.7 (Section 10.9) demonstrates how
Equation 10.114 with the vector effective length
determined in the preceding example is used to obtain
the induced potential.",Electromagnetics_Vol2.pdf
"194 CHAPTER 10. ANTENNAS
10.12 Equivalent Circuit Model
for Reception, Redux
[m0216]
Section 10.9 provides an informal derivation of an
equi
valent circuit model for a receiving antenna. This
model is shown in Figure 10.18. The derivation of
this model was informal and incomplete because the
open-circuit potential Ei · le and source impedance
ZA were not rigorously derived in that section. While
the open-circuit potential was derived in
Section 10.11 (“Potential Induced in a Dipole”), the
source impedance has not yet been addressed. In this
section, the source impedance is derived, which
completes the formal derivation of the model. Before
reading this section, a review of Section 10.10
(“Reciprocity”) is recommended.
The starting point for a formal derivation is the
two-port model shown in Figure 10.19. If the
two-port is passive, linear, and time-invariant, then
the potential v2 is a linear function of potentials and
currents present at ports 1 and 2. Furthermore, v1",Electromagnetics_Vol2.pdf
"the potential v2 is a linear function of potentials and
currents present at ports 1 and 2. Furthermore, v1
must be proportional to i1, and similarly v2 must be
proportional to i2, so any pair of “inputs” consisting
of either potentials or currents completely determines
the two remaining potentials or currents. Thus, we
may write:
v2 = Z21i1 + Z22i2 (10.119)
where Z11 and Z12 are, for the moment, simply
constants of proportionality . However, note that Z11
and Z12 have SI base units of Ω, and so we refer to
c⃝ S. Lally CC BY -SA 4.0
Figure 10.18: Th ´evenin equivalent circuit model for
an antenna in the presence of an incident electric field
Ei.
Inductiveload, public domain (modified)
Figure
10.19: T wo-port system.
these quantities as impedances. Similarly we may
write:
v1 = Z11i1 + Z12i2 (10.120)
W e can develop expressions for Z12 and Z21 as
follows. First, note that v1 = Z12i2 when i1 = 0.
Therefore, we may define Z12 as follows:
Z12 ≜ v1
i2
⏐⏐⏐
⏐
i1=0
(10.121)
The",Electromagnetics_Vol2.pdf
"follows. First, note that v1 = Z12i2 when i1 = 0.
Therefore, we may define Z12 as follows:
Z12 ≜ v1
i2
⏐⏐⏐
⏐
i1=0
(10.121)
The
simplest way to make i1 = 0 (leaving i2 as the
sole “input”) is to leave port 1 open-circuited. In
previous sections, we invoked special notation for
these circumstances. In particular, we defined ˜It
2 as i2
in phasor representation, with the superscript “t”
(“transmitter”) indicating that this is the sole “input”;
and ˜Vr
1 as v1 in phasor representation, with the
superscript “r ” (“receiver”) signaling that port 1 is
both open-circuited and the “output. ” Applying this
notation, we note:
Z12 =
˜Vr
1
˜It
2
(10.122)
Similarly:
Z21 =
˜Vr
2
˜It
1
(10.123)
In
Section 10.10 (“Reciprocity”), we established that
a pair of antennas could be represented as a passive
linear time-invariant two-port, with v1 and i1
representing the potential and current at the terminals
of one antenna (“antenna 1”), and v2 and i2
representing the potential and current at the terminals",Electromagnetics_Vol2.pdf
"of one antenna (“antenna 1”), and v2 and i2
representing the potential and current at the terminals
of another antenna (“antenna 2”). Therefore for any
pair of antennas, quantities Z11, Z12, Z22, and Z21",Electromagnetics_Vol2.pdf
"10.12. EQUIV ALENT CIRCUIT MODEL FOR RECEPTION, REDUX 195
can be identified that completely determine the
relationship between the port potentials and currents.
W e also established in Section 10.10 that:
˜It
1 ˜Vr
1 = ˜It
2 ˜Vr
2 (10.124)
Therefore,
˜Vr
1
˜It
2
=
˜Vr
2
˜It
1
(10.125)
Referring
to Equations 10.122 and 10.123, we see
that Equation 10.125 requires that:
Z12 = Z21 (10.126)
This is a key point. Even though we derived this
equality by open-circuiting ports one at a time, the
equality must hold generally since Equations 10.119
and 10.120 must apply – with the same values of Z12
and Z21 – regardless of the particular values of the
port potentials and currents.
W e are now ready to determine the Th ´evenin
equivalent circuit for a receiving antenna. Let port 1
correspond to the transmitting antenna; that is, i1 is
˜It
1. Let port 2 correspond to an open-circuited
receiving antenna; thus, i2 = 0 and v2 is ˜Vr
2 . Now
applying Equation 10.119:
v2 = Z21i1 + Z22i2
=
(
˜Vr
2 /˜It
1",Electromagnetics_Vol2.pdf
"receiving antenna; thus, i2 = 0 and v2 is ˜Vr
2 . Now
applying Equation 10.119:
v2 = Z21i1 + Z22i2
=
(
˜Vr
2 /˜It
1
)
˜It
1 + Z22 · 0
= ˜Vr
2 (10.127)
W e previously determined ˜Vr
2 from electromagnetic
considerations to be (Section 10.11):
˜Vr
2 = ˜Ei · le (10.128)
where ˜Ei is the field incident on the receiving
antenna, and le is the vector effective length as
defined in Section 10.11. Thus, the voltage source in
the Th ´evenin equivalent circuit for the receive
antenna is simply ˜Ei · le, as shown in Figure 10.18.
The other component in the Th ´evenin equivalent
circuit is the series impedance. From basic circuit
theory , this impedance is the ratio of v2 when port 2 is
open-circuited (i.e., ˜Vr
2 ) to i2 when port 2 is
short-circuited. This value of i2 can be obtained using
Equation 10.119 with v2 = 0:
0 = Z21 ˜It
1 + Z22i2 (10.129)
Therefore:
i2 = −Z21
Z22
˜It
1 (10.130)
No
w using Equation 10.123 to eliminate ˜It
1, we
obtain:
i2 = −
˜Vr
2
Z22
(10.131)
Note",Electromagnetics_Vol2.pdf
"1 + Z22i2 (10.129)
Therefore:
i2 = −Z21
Z22
˜It
1 (10.130)
No
w using Equation 10.123 to eliminate ˜It
1, we
obtain:
i2 = −
˜Vr
2
Z22
(10.131)
Note
that the reference direction for i2 as defined in
Figure 10.19 is opposite the reference direction for
short-circuit current. That is, given the polarity of v2
shown in Figure 10.19, the reference direction of
current flow through a passive load attached to this
port is from “+” to “−” through the load. Therefore,
the source impedance, calculated as the ratio of the
open-circuit potential to the short circuit current, is:
˜Vr
2
+ ˜Vr
2 /Z22
= Z22 (10.132)
W
e have found that the series impedance ZA in the
Th´evenin equivalent circuit is equal to Z22 in the
two-port model.
T o determine Z22, let us apply a current i2 = ˜It
2 to
port 2 (i.e., antenna 2). Equation 10.119 indicates that
we should see:
v2 = Z21i1 + Z22 ˜It
2 (10.133)
Solving for Z22:
Z22 = v2
˜It
2
− Z21
i1
˜It
2
(10.134)
Note
that the first term on the right is precisely the",Electromagnetics_Vol2.pdf
"v2 = Z21i1 + Z22 ˜It
2 (10.133)
Solving for Z22:
Z22 = v2
˜It
2
− Z21
i1
˜It
2
(10.134)
Note
that the first term on the right is precisely the
impedance of antenna 2 in transmission. The second
term in Equation 10.134 describes a contribution to
Z22 from antenna 1. However, our immediate interest
is in the equivalent circuit for reception of an electric
field ˜Ei in the absence of any other antenna. W e can
have it both ways by imagining that ˜Ei is generated
by antenna 1, but also that antenna 1 is far enough
away to make Z21 – the factor that determines the
effect of antenna 1 on antenna 2 – negligible. Then
we see from Equation 10.134 that Z22 is the
impedance of antenna 2 when transmitting.",Electromagnetics_Vol2.pdf
"196 CHAPTER 10. ANTENNAS
Summarizing:
The Th ´evenin equivalent circuit for an antenna
in
the presence of an incident electric field ˜Ei is
shown in Figure 10.18. The series impedance ZA
in this model is equal to the impedance of the an-
tenna in transmission.
Mutual coupling. This concludes the derivation, but
raises a follow-up question: What if antenna 2 is
present and not sufficiently far away that Z21 can be
assumed to be negligible? In this case, we refer to
antenna 1 and antenna 2 as being “coupled, ” and refer
to the effect of the presence of antenna 1 on antenna 2
as coupling. Often, this issue is referred to as mutual
coupling, since the coupling affects both antennas in a
reciprocal fashion. It is rare for coupling to be
significant between antennas on opposite ends of a
radio link. This is apparent from common experience.
For example, changes to a receive antenna are not
normally seen to affect the electric field incident at
other locations. However, coupling becomes",Electromagnetics_Vol2.pdf
"For example, changes to a receive antenna are not
normally seen to affect the electric field incident at
other locations. However, coupling becomes
important when the antenna system is a dense array;
i.e., multiple antennas separated by distances less than
a few wavelengths. It is common for coupling among
the antennas in a dense array to be significant. Such
arrays can be analyzed using a generalized version of
the theory presented in this section.
Additional Reading:
• “Th ´evenin’s Theorem” on Wikipedia.
10.13 Effective Aperture
[m0218]
When working with systems involving receiving
antennas,
it is convenient to have a single parameter
that relates incident power (as opposed to incident
electric field) to the power delivered to the receiver
electronics. This parameter is commonly known as
the effective aperture or antenna aperture.
From elementary circuit theory , the power delivered
to a load ZL is maximized when the load is conjugate
matched to the antenna impedance; i.e., when",Electromagnetics_Vol2.pdf
"to a load ZL is maximized when the load is conjugate
matched to the antenna impedance; i.e., when
ZL = Z∗
A where ZA is the impedance of the antenna.
Thus, a convenient definition for effective aperture Ae
employs the relationship:
PR,max ≜ Si
coAe (10.135)
where Si
co is the incident power density (SI base units
of W/m 2) that is co-polarized with the antenna, and
PR,max is the power delivered to a load that is
conjugate matched to the antenna impedance.
Summarizing:
Effective aperture (SI base units of m 2) is the ra-
tio of power delivered by an antenna to a conju-
gate matched load, to the incident co-polarized
power density .
As defined, a method for calculation of effective
aperture is not obvious, except perhaps through direct
measurement. In fact, there are at least three ways we
can calculate effective aperture: (a) via effective
length, (b) via thermodynamics, and (c) via
reciprocity . Each of these methods yields some
insight and are considered in turn.",Electromagnetics_Vol2.pdf
"length, (b) via thermodynamics, and (c) via
reciprocity . Each of these methods yields some
insight and are considered in turn.
Effective aperture via effective length. The
potential and current at the load of the receive antenna
can be determined using the equivalent circuit model
shown in Figure 10.20, using the Th ´evenin equivalent
circuit model for the antenna developed in
Section 10.9. In this model, the voltage source is
determined by the incident electric field intensity ˜Ei
and the vector effective length le = ˆlele of the
antenna. W e determine the associated effective
aperture as follows. Consider a co-polarized",Electromagnetics_Vol2.pdf
"10.13. EFFECTIVE APER TURE 197
sinusoidally-varying plane wave ˜Ei
co incident on the
antenna. Further, let
˜Ei
co = Eiˆe (10.136)
where ˆe is the reference direction of ˜Ei
co. The
co-polarized power density incident on the antenna is:
Si
co =
⏐⏐Ei⏐
⏐2
2η (10.137)
where ηis
the wave impedance of the medium (e.g.,
∼= 377 Ω for an antenna in free space). In response,
the time-average power delivered to the load is
PR = 1
2Re
{
˜VL˜I∗
L
}
(10.138)
where ˜VL and ˜IL are
the potential and current
phasors, respectively , at the load. W e may determine
˜VL and ˜IL from the equivalent circuit model of
Figure 10.20 using basic circuit theory . First, note
that the magnitude and phase of the voltage source is:
˜Ei
co · le = Eile
(
ˆe · ˆle
)
(10.139)
and since we have defined ˆle to be equal to ˆe,
˜Ei
co · le = Eile (10.140)
Next, note that the load impedance creates a voltage
divider with the source (antenna) impedance such that
˜VL =
(
Eile
) ZL
ZA + ZL
(10.141)",Electromagnetics_Vol2.pdf
"Next, note that the load impedance creates a voltage
divider with the source (antenna) impedance such that
˜VL =
(
Eile
) ZL
ZA + ZL
(10.141)
c⃝ S. Lally CC BY -SA 4.0 (modified)
Figure 10.20: Equivalent circuit model for an antenna
in the presence of an incident electric field ˜Ei, termi-
nated into a load impedance ZL.
Similarly , ˜IL is the voltage source output divided by
the total series resistance:
˜IL = Eile
ZA + ZL
(10.142)
Substituting
these expressions into Equation 10.138,
we obtain:
PR = 1
2
⏐⏐Eile
⏐
⏐2 RL
|ZA + ZL|2 (10.143)
Let
us identify the real and imaginary components of
ZA and ZL explicitly , as follows:
ZA = RA + jXA (10.144)
ZL = RL + jXL (10.145)
Further, let us assume that loss internal to the antenna
is negligible, so RA ≈ Rrad. If ZL is conjugate
matched for maximum power transfer, then
ZL = Z∗
A = Rrad − jXA. Therefore:
PR,max = 1
2
⏐⏐Eile
⏐
⏐2 Rrad
|ZA + Z∗
A|2
= 1
2
⏐⏐Eile
⏐
⏐2 Rrad
|2Rrad|2
=
⏐⏐Eile
⏐
⏐2
8Rrad
(10.146)
No",Electromagnetics_Vol2.pdf
"ZL = Z∗
A = Rrad − jXA. Therefore:
PR,max = 1
2
⏐⏐Eile
⏐
⏐2 Rrad
|ZA + Z∗
A|2
= 1
2
⏐⏐Eile
⏐
⏐2 Rrad
|2Rrad|2
=
⏐⏐Eile
⏐
⏐2
8Rrad
(10.146)
No
w using Equations 10.135, 10.137, and 10.146, we
find:
Ae =
⏐⏐Eile
⏐
⏐2
/8Rrad
|Ei|2 /2η
(10.147)
which
reduces to:
Ae = η|le|2
4Rrad
(10.148)
It
is not surprising that effective aperture should be
proportional to the square of the magnitude of
effective length. However, we see that the medium
and the radiation resistance of the antenna also play a
role.
Effective length and radiation resistance are easy to
calculate for wire antennas, so it is natural to use
Equation 10.148 to calculate the effective apertures of",Electromagnetics_Vol2.pdf
"198 CHAPTER 10. ANTENNAS
wire antennas. For the electrically-short dipole (ESD)
of length L, le ≈ (L/2) sin θand
Rrad ≈ 20π2 (L/λ)2. Thus, we find the effective
aperture assuming free space (i.e., η= η0) is:
Ae ≈ 0.119λ2 |sin θ|2 (lossless ESD) (10.149)
Remarkably , the effective aperture of the ESD does
not depend on its length. In fact, it depends only on
the frequency of operation.
Also worth noting is that maximum effective aperture
(i.e., the effective aperture in the θ= π/2 direction) is
Ae ≈ 0.119λ2
(lossless ESD, max.) (10.150)
Dipoles
which are not electrically-short exhibit
radiation resistance which is proportional to Lp,
where p> 2. Therefore, the effective aperture of
non-electrically-short dipoles does increase with L, as
expected. However, this increase is not dramatic
unless Lis significantly greater than λ/2. Here’s an
example:
Example 10.10. Ef fective aperture of a
half-wave dipole.
The electrically-thin half-wave dipole exhibits",Electromagnetics_Vol2.pdf
"example:
Example 10.10. Ef fective aperture of a
half-wave dipole.
The electrically-thin half-wave dipole exhibits
radiation resistance ∼= 73 Ω and effective length
λ/π. Assuming the dipole is lossless and in free
space, Equation 10.148 yields:
Ae ≈ 0.131λ2 (half-wave dipole, max.)
(10.151)
Again, this is the effective aperture for a wave
incident from broadside to the dipole.
Effective aperture via thermodynamics. A useful
insight into the concept of effective aperture can be
deduced by asking the following question: How much
power is received by an antenna when waves of equal
power density arrive from all directions
simultaneously? This question is surprisingly easy to
answer using basic principles of thermodynamics.
Thermodynamics is a field of physics that addresses
heat and its relation to radiation and electrical energy .
No previous experience with thermodynamics is
assumed in this derivation.
Consider the scenario depicted in Figure 10.21. In",Electromagnetics_Vol2.pdf
"No previous experience with thermodynamics is
assumed in this derivation.
Consider the scenario depicted in Figure 10.21. In
this scenario, the antenna is completely enclosed in a
chamber whose walls do not affect the behavior of the
antenna and which have uniform temperature T. The
load (still conjugate matched to the antenna) is
completely enclosed by a separate but identical
chamber, also at temperature T.
Consider the load chamber first. Heat causes random
acceleration of constituent charge carriers in the load,
giving rise to a randomly-varying current at the load
terminals. (This is known as Johnson-Nyquist noise.)
This current, flowing through the load, gives rise to a
potential; therefore, the load – despite being a passive
device – is actually a source of power. The power
associated with this noise is:
Pload = kTB (10.152)
where k∼= 1.38 × 10−23 J/K is Boltzmann’s constant
and Bis the bandwidth within which Pload is
measured.
Similarly , the antenna is a source of noise power",Electromagnetics_Vol2.pdf
"and Bis the bandwidth within which Pload is
measured.
Similarly , the antenna is a source of noise power
Pant. Pant can also be interpreted as captured
thermal radiation – that is, electromagnetic waves
stimulated by the random acceleration of charged
particles comprising the chamber walls. These waves
radiate from the walls and travel to the antenna. The
Rayleigh-Jeans law of thermodynamics tells us that
this radiation should have power density (i.e., SI base
units of W/m 2) equal to
2kT
λ2 B (10.153)
Chetvorno, public domain (modified)
Figure
10.21: Thermodynamic analysis of the power
exchanged between an antenna and its load.",Electromagnetics_Vol2.pdf
"10.13. EFFECTIVE APER TURE 199
per steradian of solid angle. 5 The total power
accessible to the antenna is one-half this amount,
since an antenna is sensitive to only one polarization
at a time, whereas the thermal radiation is equally
distributed among any two orthogonal polarizations.
PA is the remaining power, obtained by integrating
over all 4πsteradians. Therefore, the antenna
captures total power equal to 6
Pant =
∮
Ae(θ′,φ′)
(1
2
2kT
λ2 B
)
sin θ′dθ′dφ′
=
(k
T
λ2 B
)∮
Ae(θ′,φ′) sin θ′dθ′dφ′
(10.154)
Given the conditions of the experiment, and in
particular since the antenna chamber and the load
chamber are at the same temperature, the antenna and
load are in thermodynamic equilibrium. A
consequence of this equilibrium is that power Pant
captured by the antenna and delivered to the load
must be equal to power Pload generated by the load
and delivered to the antenna; i.e.,
Pant = Pload (10.155)
Combining Equations 10.152, 10.154, and 10.155, we
obtain:
(kT
λ2 B
)∮",Electromagnetics_Vol2.pdf
"and delivered to the antenna; i.e.,
Pant = Pload (10.155)
Combining Equations 10.152, 10.154, and 10.155, we
obtain:
(kT
λ2 B
)∮
Ae(θ′,φ′) sin θ′dθ′dφ′ = kTB
(10.156)
which reduces to:
∮
Ae(θ′,φ′) sin θ′dθ′dφ′ = λ2 (10.157)
Let ⟨Ae⟩ be the mean effective aperture of the
antenna; i.e., Ae averaged over all possible directions.
This average is simply Equation 10.157 divided by
the 4πsr of solid angle comprising all possible
directions. Thus:
⟨Ae⟩ = λ2
4π (10.158)
Recall
that a isotropic antenna is one which has the
same effective aperture in every direction. Such
5 Full disclosure: This is an approximation to the exact expres-
sion, but is very accurate at radio frequencies. See “ Additional Read-
ing” at the end of this section for additional details.
6 Recall sin θ′dθ′dφ′ is the differential element of solid angle.
antennas do not exist in practice, but the concept of an
isotropic antenna is quite useful as we shall soon see.
Since the effective aperture of an isotropic antenna",Electromagnetics_Vol2.pdf
"isotropic antenna is quite useful as we shall soon see.
Since the effective aperture of an isotropic antenna
must be the same as its mean effective aperture, we
find:
Ae = λ2
4π ≈ 0.080 λ2 (isotropic antenna)
(10.159)
Note
further that this must be the minimum possible
value of the maximum effective aperture for any
antenna.
Effective aperture via reciprocity . The fact that all
antennas have maximum effective aperture greater
than that of an isotropic antenna makes the isotropic
antenna a logical benchmark against which to
compare the effective aperture of antennas. W e can
characterize the effective aperture of any antenna as
being some unitless real-valued constant Dtimes the
effective aperture of an isotropic antenna; i.e.:
Ae ≜ Dλ2
4π (an y antenna) (10.160)
where Dmust
be greater than or equal to 1.
Astute readers might notice that Dseems a lot like
directivity. Directivity is defined in Section 10.7 as
the factor by which a transmitting antenna increases",Electromagnetics_Vol2.pdf
"directivity. Directivity is defined in Section 10.7 as
the factor by which a transmitting antenna increases
the power density of its radiation over that of an
isotropic antenna. Let us once again consider the
ESD, for which we previously determined (via the
effective length concept) that Ae ∼= 0.119λ2 in the
direction in which it is maximum. Applying
Equation 10.160 to the ESD, we find:
D≜ Ae
4π
λ2
∼= 1.50 (ESD, max.) (10.161)
Sure enough, Dis equal to the directivity of the ESD
in the transmit case.
This result turns out to be generally true; that is:
The effective aperture of any antenna is given by
Equation 10.160 where Dis the directivity of the
antenna when transmitting.
A derivation of this result for the general case is
possible using analysis similar to the thermodynamics",Electromagnetics_Vol2.pdf
"200 CHAPTER 10. ANTENNAS
presented earlier, or using the reciprocity theorem
developed in Section 10.10.
The fact that effective aperture is easily calculated
from transmit directivity is an enormously useful tool
in antenna engineering. Without this tool,
determination of effective aperture is limited to direct
measurement or calculation via effective length; and
effective length is generally difficult to calculate for
antennas which are not well-described as distributions
of current along a line. It is usually far easier to
calculate the directivity of an antenna when
transmitting, and then use Equation 10.160 to obtain
effective aperture for receive operation.
Note that this principle is itself an expression of
reciprocity . That is, one may fairly say that
“directivity” is not exclusively a transmit concept nor
a receive concept, but rather is a single quantifiable
characteristic of an antenna that applies in both the
transmit and receive case. Recall that radiation",Electromagnetics_Vol2.pdf
"characteristic of an antenna that applies in both the
transmit and receive case. Recall that radiation
patterns are used to quantify the way (transmit)
directivity varies with direction. Suddenly , we have
found that precisely the same patterns apply to the
receive case! Summarizing:
As long as the conditions required for formal
reciprocity
are satisfied, the directivity of a re-
ceiving antenna (defined via Equation 10.160) is
equal to directivity of the same antenna when
transmitting, and patterns describing receive di-
rectivity are equal to those for transmit directiv-
ity .
Finally , we note that the equivalence of transmit and
receive directivity can, again via Equation 10.160, be
used to define an effective aperture for the transmit
case. In other words, we can define the effective
aperture of a transmitting antenna to be λ2/4πtimes
the directivity of the antenna. There is no new physics
at work here; we are simply taking advantage of the",Electromagnetics_Vol2.pdf
"the directivity of the antenna. There is no new physics
at work here; we are simply taking advantage of the
fact that the concepts of effective aperture and
directivity describe essentially the same characteristic
of an antenna, and that this characteristic is the same
for both transmit and receive operation.
Additional Reading:
• “Antenna aperture” on Wikipedia.
• “Boltzmann constant” on Wikipedia.
• “Rayleigh-Jeans law” on Wikipedia.
• “Johnson-Nyquist noise” on Wikipedia.
• “Thermodynamics” on Wikipedia.",Electromagnetics_Vol2.pdf
"10.14. FRIIS TRANSMISSION EQUA TION 201
10.14 Friis T ransmission
Equation
[m0219]
A common task in radio systems applications is to
determine
the power delivered to a receiver due to a
distant transmitter. The scenario is shown in
Figure 10.22: A transmitter delivers power PT to an
antenna which has gain GT in the direction of the
receiver. The receiver’s antenna has gain GR. As
always, antenna gain is equal to directivity times
radiation efficiency , so GT and GR account for losses
internal to the antenna, but not losses due to
impedance mismatch.
A simple expression for PR can be derived as follows.
First, let us assume “free space conditions”; that is, let
us assume that the intervening terrain exhibits
negligible absorption, reflection, or other scattering of
the transmitted signal. In this case, the spatial power
density at range Rfrom the transmitter which radiates
this power through a lossless and isotropic antenna
would be:
PT
4πR2 (10.162)
that",Electromagnetics_Vol2.pdf
"density at range Rfrom the transmitter which radiates
this power through a lossless and isotropic antenna
would be:
PT
4πR2 (10.162)
that
is, total transmitted power divided by the area of
a sphere of radius Rthrough which all the power
must flow . The actual power density Si is this amount
times the gain of the transmit antenna, i.e.:
Si = PT
4πR2 GT (10.163)
The
maximum received power is the incident
co-polarized power density times the effective
Figure 10.22: Radio link accounting for antenna gains
and
spreading of the transmitted wave.
aperture Ae of the receive antenna:
PR,max = AeSi
co
= Ae
PT
4πR2 GT (10.164)
This
assumes that the receive antenna is co-polarized
with the incident electric field, and that the receiver is
conjugate-matched to the antenna. The effective
aperture can also be expressed in terms of the gain
GR of the receive antenna:
Ae = λ2
4πGR (10.165)
Thus,
Equation 10.164 may be written in the
following form:
PR,max = PTGT
( λ
4πR
)2
GR (10.166)
This",Electromagnetics_Vol2.pdf
"Ae = λ2
4πGR (10.165)
Thus,
Equation 10.164 may be written in the
following form:
PR,max = PTGT
( λ
4πR
)2
GR (10.166)
This
is the Friis transmission equation. Summarizing:
The Friis transmission equation (Equa-
tion
10.166) gives the power delivered to a
conjugate-matched receiver in response to a dis-
tant transmitter, assuming co-polarized antennas
and free space conditions.
The factor (λ/4πR)2 appearing in the Friis
transmission equation is referred to as free space path
gain. More often this is expressed as the reciprocal
quantity:
Lp ≜
( λ
4πR
)−2
(10.167)
which
is known as free space path loss. Thus,
Equation 10.166 may be expressed as follows:
PR,max = PTGTL−1
p GR (10.168)
The utility of the concept of path loss is that it may
also be determined for conditions which are different
from free space. The Friis transmission equation still
applies; one simply uses the appropriate (and
probably significantly different) value of Lp.
A common misconception is that path loss is equal to",Electromagnetics_Vol2.pdf
"applies; one simply uses the appropriate (and
probably significantly different) value of Lp.
A common misconception is that path loss is equal to
the reduction in power density due to spreading along",Electromagnetics_Vol2.pdf
"202 CHAPTER 10. ANTENNAS
the path between antennas, and therefore this
“spreading loss” increases with frequency . In fact, the
reduction in power density due to spreading between
any two distances R1 <R2 is:
PT/4πR2
1
PT/4πR2
2
=
(R1
R2
)2
(10.169)
which
is clearly independent of frequency . The path
loss Lp, in contrast, depends only on the total distance
Rand does depend on frequency . The dependence on
frequency reflects the dependence of the effective
aperture on wavelength. Thus, path loss is not loss in
the traditional sense, but rather accounts for a
combination of spreading and the λ2 dependence of
effective aperture that is common to all receiving
antennas.
Finally , note that Equation 10.168 is merely the
simplest form of the Friis transmission equation.
Commonly encountered alternative forms include
forms in which GT and/or GR are instead represented
by the associated effective apertures, and forms in
which the effects of antenna impedance mismatch",Electromagnetics_Vol2.pdf
"by the associated effective apertures, and forms in
which the effects of antenna impedance mismatch
and/or cross-polarization are taken into account.
Example 10.11. 6 GHz point-to-point link.
T errestrial telecommunications systems
commonly aggregate large numbers of
individual communications links into a single
high-bandwidth link. This is often implemented
as a radio link between dish-type antennas
having gain of about 27 dBi (that’s dB relative to
a lossless isotropic antenna) mounted on very
tall towers and operating at frequencies around 6
GHz. Assuming the minimum acceptable
receive power is −120 dBm (that’s −120 dB
relative to 1 mW; i.e., 10−15 W) and the
required range is 30 km, what is the minimum
acceptable transmit power?
Solution. From the problem statement:
GT = GR = 10 27/ 10 ∼= 501 (10.170)
λ= c
f
∼
= 3 × 108 m/s
6 × 109 Hz
∼
= 5
 .00 cm (10.171)
R= 30 km, and PR ≥ 10−15 W . W e assume
that the height and high directivity of the
antennas",Electromagnetics_Vol2.pdf
"λ= c
f
∼
= 3 × 108 m/s
6 × 109 Hz
∼
= 5
 .00 cm (10.171)
R= 30 km, and PR ≥ 10−15 W . W e assume
that the height and high directivity of the
antennas
yield conditions sufficiently close to
free space. W e further assume
conjugate-matching at the receiver, and that the
antennas are co-polarized. Under these
conditions, PR = PR,max and Equation 10.166
applies. W e find:
PT ≥ PR,max
GT (λ/4πR)2 GR
(10.172)
∼= 2
.26 × 10−7 W
∼
= 2.26 × 10−4 mW
∼
= −36.5 dBm
Additional Reading:
• “Free-space
path loss” on Wikipedia.
• “Friis transmission equation” on Wikipedia.
[m0211]",Electromagnetics_Vol2.pdf
"10.14. FRIIS TRANSMISSION EQUA TION 203
Image Credits
Fig. 10.1: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Standing W ave Creation.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.2: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Electrically-Short Dipole.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.3: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Zero-Length Dipole Open Circuit.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.4: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Transmitting Antenna Circuit.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.5: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Magnitude of the Radiated Field.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"https://commons.wikimedia.org/wiki/File:Magnitude of the Radiated Field.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.6: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplaneMag.svg,
CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.7: c⃝ T . Truckle, https://en.wikipedia.org/wiki/File:Sidelobes en.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified.
Fig. 10.8: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Half-Power Beamwidth.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.9: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified.
Fig. 10.10: Inductiveload, https://commons.wikimedia.org/wiki/File:T wo Port Circuit.svg,
public
domain. Modified.
Fig. 10.11: c⃝ Sevenchw (C. W ang),",Electromagnetics_Vol2.pdf
"Fig. 10.10: Inductiveload, https://commons.wikimedia.org/wiki/File:T wo Port Circuit.svg,
public
domain. Modified.
Fig. 10.11: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:An electromagnetic system consisting of a current distribution
radiating an electric field.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.12: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:An electromagnetic system consisting of a different current
distribution radiating an electric field.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.13: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A two-port consisting of two dipole antennas.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.14: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:The port 1 antenna a transmitting and b receiving.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).",Electromagnetics_Vol2.pdf
"204 CHAPTER 10. ANTENNAS
Fig. 10.15: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Thin straight dipole respond to incident plane wave.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.16: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Dipole of interest driven by current.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.17: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Hertzian dipole drive by current.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.18: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 10.19: Inductiveload, https://commons.wikimedia.org/wiki/File:T wo Port Circuit.svg,
public
domain. Modified.
Fig. 10.20: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg,
CC",Electromagnetics_Vol2.pdf
"public
domain. Modified.
Fig. 10.20: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified.
Fig. 10.21: Chetvorno, https://en.wikipedia.org/wiki/File:Antenna and resistor in cavity .svg,
public
domain. Modified.",Electromagnetics_Vol2.pdf
"Appendix A
Constituti
ve Parameters of Some
Common Materials
A.1 Permittivity of Some
Common Materials
[m0135]
The values below are relative permittivity ǫr ≜ ǫ/ǫ0
for
a few materials that are commonly encountered in
electrical engineering applications, and for which
permittivity emerges as a consideration. Note that
“relative permittivity” is sometimes referred to as
dielectric constant.
Here we consider only the physical (real-valued)
permittivity , which is the real part of the complex
permittivity (typically indicated as ǫ′ or ǫ′
r) for
materials exhibiting significant loss.
Permittivity varies significantly as a function of
frequency . The values below are representative of
frequencies from a few kHz to about 1 GHz. The
values given are also representative of optical
frequencies for materials such as silica that are used
in optical applications. Permittivity also varies as a
function of temperature. In applications where
precision better than about 10% is required, primary",Electromagnetics_Vol2.pdf
"function of temperature. In applications where
precision better than about 10% is required, primary
references accounting for frequency and temperature
should be consulted. The values presented here are
gathered from a variety of references, including those
indicated in “ Additional References. ”
Free Space (vacuum): ǫr ≜ 1
Solid Dielectrics:
Material ǫr Common uses
Styrofoam1 1.1
T
eflon 2 2.1
Polyethylene 2.3 coaxial cable
Polypropylene 2.3
Silica 2.4 optical fiber 3
Polystyrene 2.6
Polycarbonate 2.8
Rogers RO3003 3.0 PCB substrate
FR4 (glass epoxy laminate) 4.5 PCB substrate
1 Properly known as extruded polystyrene foam
(XPS).
2 Properly known as polytetrafluoroethylene (PTFE).
3 T ypically doped with small amounts of other
materials to slightly raise or lower the index of
refraction (= √ǫr).
Non-conducting
spacing materials used in discrete
capacitors exhibit ǫr ranging from about 5 to 50.
Semiconductors commonly appearing in electronics
– including carbon, silicon, geranium, indium",Electromagnetics_Vol2.pdf
"capacitors exhibit ǫr ranging from about 5 to 50.
Semiconductors commonly appearing in electronics
– including carbon, silicon, geranium, indium
phosphide, and so on – typically exhibit ǫr in the
range 5–15.
Glass exhibits ǫr in the range 4–10, depending on
composition.
Gasses, including air, typically exhibit ǫr ∼= 1 to
within a tiny fraction of a percent.
Liquid water typically exhibits ǫr in the range
72–81. Distilled water exhibits ǫr ≈ 81 at room
temperature, whereas sea water tends to be at the
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"206 APPENDIX A. CONSTITUTIVE P ARAMETERS OF SOME COMMON MA TERIALS
lower end of the range.
Other liquids typically exhibit ǫr in the range 10–90,
with considerable variation as a function of
temperature and frequency . Animal flesh and blood
consists primarily of liquid matter and so also exhibits
permittivity in this range.
Soil typically exhibits ǫr in the range 2.5–3.5 when
dry and higher when wet. The permittivity of soil
varies considerably depending on composition.
Additional Reading:
• CRC Handbook of Chemistry and Physics .
• “Relative permittivity” on Wikipedia.
• “Miscellaneous Dielectric Constants” on
microwaves101.com.
A.2 Permeability of Some
Common Materials
[m0136]
The values below are relative permeability
µr ≜ µ/µ0 for
a few materials that are commonly
encountered in electrical engineering applications,
and for which µr is significantly different from 1.
These materials are predominantly ferromagnetic
metals and (in the case of ferrites) materials",Electromagnetics_Vol2.pdf
"and for which µr is significantly different from 1.
These materials are predominantly ferromagnetic
metals and (in the case of ferrites) materials
containing significant ferromagnetic metal content.
Nearly all other materials exhibit µr that is not
significantly different from that of free space.
The values presented here are gathered from a variety
of references, including those indicated in “ Additional
References” at the end of this section. Be aware that
permeability may vary significantly with frequency;
values given here are applicable to the frequency
ranges for applications in which these materials are
typically used. Also be aware that materials
exhibiting high permeability are also typically
non-linear; that is, permeability depends on the
magnitude of the magnetic field. Again, values
reported here are those applicable to applications in
which these materials are typically used.
Free Space (vacuum): µr ≜ 1.
Iron (also referred to by the chemical notation “Fe”)",Electromagnetics_Vol2.pdf
"which these materials are typically used.
Free Space (vacuum): µr ≜ 1.
Iron (also referred to by the chemical notation “Fe”)
appears as a principal ingredient in many materials
and alloys employed in electrical structures and
devices. Iron exhibits µr that is very high, but which
decreases with decreasing purity . 99.95% pure iron
exhibits µr ∼ 200,000. This decreases to ∼ 5000 at
99.8% purity and is typically below 100 for purity
less than 99%.
Steel is an iron alloy that comes in many forms, with
a correspondingly broad range of permeabilities.
Electrical steel, commonly used in electrical
machinery and transformers when high permeability
is desired, exhibits µr ∼ 4000. Stainless steel,
encompassing a broad range of alloys used in
mechanical applications, exhibits µr in the range
750–1800. Carbon steel, including a broad class of
alloys commonly used in structural applications,
exhibits µr on the order of 100.",Electromagnetics_Vol2.pdf
"A.3. CONDUCTIVITY OF SOME COMMON MA TERIALS 207
Ferrites include a broad range of ceramic materials
that are combined with iron and various combinations
of other metals and are used as magnets and magnetic
devices in various electrical systems. Common
ferrites exhibit µr in the range 16–640.
Additional Reading:
• CRC Handbook of Chemistry and Physics.
• “Magnetic Materials” on microwaves101.com.
• “Permeability (electromagnetism)” on
Wikipedia.
• “Iron” on Wikipedia.
• “Electrical steel” on Wikipedia.
• “Ferrite (magnet)” on Wikipedia.
A.3 Conductivity of Some
Common Materials
[m0137]
The values below are conductivity σfor
 a few
materials that are commonly encountered in electrical
engineering applications, and for which conductivity
emerges as a consideration.
Note that materials in some applications are described
instead in terms of resistivity, which is simply the
reciprocal of conductivity .
Conductivity may vary significantly as a function of",Electromagnetics_Vol2.pdf
"instead in terms of resistivity, which is simply the
reciprocal of conductivity .
Conductivity may vary significantly as a function of
frequency . The values below are representative of
frequencies from a few kHz to a few GHz.
Conductivity also varies as a function of temperature.
In applications where precise values are required,
primary references accounting for frequency and
temperature should be consulted. The values
presented here are gathered from a variety of
references, including those indicated in “ Additional
References” at the end of this section.
Free Space (vacuum): σ≜ 0.
Commonly encountered elements:
Material σ(S/m)
Copper 5.8 × 107
Gold 4
.4 × 107
Aluminum 3.7 × 107
Iron 1.0 × 107
Platinum 0.9 × 107
Carbon 1.3 × 105
Silicon 4.4 × 10−4
W ater exhibits σranging from about 6 µS/m for
highly distilled water (thus, a very poor conductor) to
about 5 S/m for seawater (thus, a relatively good
conductor), varying also with temperature and",Electromagnetics_Vol2.pdf
"about 5 S/m for seawater (thus, a relatively good
conductor), varying also with temperature and
pressure. T ap water is typically in the range of
5–50 mS/m, depending on the level of impurities
present.
Soil typically exhibits σin the range 10−4 S/m for
dry soil to about 10−1 S/m for wet soil, varying also
due to chemical composition.",Electromagnetics_Vol2.pdf
"208 APPENDIX A. CONSTITUTIVE P ARAMETERS OF SOME COMMON MA TERIALS
Non-conductors. Most other materials that are not
well-described as conductors or semiconductors and
are dry exhibit σ <10−12 S/m. Most materials that
are considered to be insulators, including air and
common dielectrics, exhibit σ <10−15 S/m, often by
several orders of magnitude.
Additional Reading:
• CRC Handbook of Chemistry and Physics .
• “Conductivity (electrolytic)” on Wikipedia.
• “Electrical resistivity and conductivity” on
Wikipedia.
• “Soil resistivity” on Wikipedia.",Electromagnetics_Vol2.pdf
"Appendix B
Mathematical
Formulas
B.1 T rigonometry
[m0138]
ejθ = cos θ+ jsin θ (B.1)
cos θ= 1
2
(
ejθ + e−j
θ)
(B.2)
sin θ= 1
j2
(
ejθ − e−j
θ)
(B.3)
cos2 θ= 1
2 + 1
2 cos 2θ (B.4)
sin2 θ= 1
2 − 1
2 cos 2θ (B.5)
sin (a± b) = sin acos b± cos asin b (B.6)
cos (a± b) = cos acos b∓ sin asin b (B.7)
Hyperbolic trigonometric functions:
sinh θ= 1
2
(
e+θ − e−θ)
(B.8)
cos
h θ= 1
2
(
e+θ + e−θ)
(B.9)
B.2
V ector Operators
[m0139]
This section contains a summary of vector operators
expressed
in each of the three major coordinate
systems:
• Cartesian (x,y,z)
• cylindrical (ρ,φ,z)
• spherical (r,θ,φ)
Associated basis vectors are identified using a caret ( ˆ)
over the symbol. The vector operand A is expressed
in terms of components in the basis directions as
follows:
• Cartesian: A = ˆxAx + ˆyAy + ˆzAz
• cylindrical: A = ˆρAρ + ˆφAφ + ˆzAz
• spherical: A = ˆrAr + ˆθAθ + ˆφAφ
Gradient
Gradient in Cartesian coordinates:
∇f = ˆx∂f
∂x + ˆy ∂f
∂y + ˆz∂f
∂z (B.10)
Gradient
in cylindrical coordinates:",Electromagnetics_Vol2.pdf
"Gradient
Gradient in Cartesian coordinates:
∇f = ˆx∂f
∂x + ˆy ∂f
∂y + ˆz∂f
∂z (B.10)
Gradient
in cylindrical coordinates:
∇f = ˆρ∂f
∂ρ + ˆφ1
ρ
∂f
∂φ + ˆz∂f
∂z (B.11)
Electr
omagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"210 APPENDIX B. MA THEMA TICAL FORMULAS
Gradient in spherical coordinates:
∇f = ˆr∂f
∂r + ˆθ1
r
∂f
∂θ + ˆφ 1
rsin θ
∂
f
∂φ (B.12)
Di
vergence
Divergence in Cartesian coordinates:
∇ · A = ∂Ax
∂x + ∂Ay
∂y + ∂Az
∂z (B.13)
Di
vergence in cylindrical coordinates:
∇ · A = 1
ρ
∂
∂ρ (ρAρ) + 1
ρ
∂Aφ
∂φ + ∂Az
∂z (B.14)
Di
vergence in spherical coordinates:
∇ · A = 1
r2
∂
∂r
(
r2Ar
)
+ 1
rsin θ
∂
∂θ (Aθ sin θ)
+ 1
rsin θ
∂
Aφ
∂φ (B.15)
Curl
Curl
in Cartesian coordinates:
∇ × A = ˆx
(∂Az
∂y − ∂Ay
∂z
)
+ ˆy
(∂Ax
∂z − ∂Az
∂x
)
+ ˆz
(∂Ay
∂x − ∂Ax
∂y
)
(B.16)
Curl
in cylindrical coordinates:
∇ × A = ˆ ρ
(1
ρ
∂Az
∂φ − ∂Aφ
∂z
)
+ ˆφ
(∂Aρ
∂z − ∂Az
∂ρ
)
+ ˆz 1
ρ
[ ∂
∂ρ (ρAφ) − ∂
Aρ
∂φ
]
(B.17)
Curl
in spherical coordinates:
∇ × A = ˆr 1
rsin θ
[ ∂
∂θ (Aφ sin θ) − ∂
Aθ
∂φ
]
+ ˆθ1
r
[ 1
sin θ
∂Ar
∂φ − ∂
∂r (rAφ)
]
+ ˆφ1
r
[ ∂
∂r (rAθ) − ∂
Ar
∂θ
]
(B.18)
Laplacian
Laplacian
in Cartesian coordinates:
∇2f = ∂2f
∂x2 + ∂2f
∂y2 + ∂2f
∂z2 (B.19)
Laplacian
in cylindrical coordinates:
∇2f = 1
ρ
∂
∂ρ
(
ρ∂f
∂ρ
)
+ 1",Electromagnetics_Vol2.pdf
"Laplacian
in Cartesian coordinates:
∇2f = ∂2f
∂x2 + ∂2f
∂y2 + ∂2f
∂z2 (B.19)
Laplacian
in cylindrical coordinates:
∇2f = 1
ρ
∂
∂ρ
(
ρ∂f
∂ρ
)
+ 1
ρ2
∂2f
∂φ2 + ∂2f
∂z2 (B.20)
Laplacian
in spherical coordinates:
∇2f = 1
r2
∂
∂r
(
r2 ∂f
∂r
)
+ 1
r2 sin θ
∂
∂θ
(∂f
∂θ sin θ
)
+ 1
r2 sin2 θ
∂2f
∂φ2 (B.21)",Electromagnetics_Vol2.pdf
"B.3. VECTOR IDENTITIES 211
B.3 V ector Identities
[m0140]
Algebraic Identities
A · (B × C) = B · (
C × A) = C · (A × B)
(B.22)
A × (B × C) = B (A · C) − C (A · B) (B.23)
Identities Involving Differential Operators
∇ · (∇ × A) = 0 (B.24)
∇ × (∇f) = 0 (B.25)
∇ × (fA) = f(∇ × A) + ( ∇f) × A (B.26)
∇ · (A × B) = B · (∇ × A) − A · (∇ × B)
(B.27)
∇ · (∇f) = ∇2f (B.28)
∇ × ∇ × A = ∇ (∇ · A) − ∇2A (B.29)
∇2A = ∇ (∇ · A) − ∇ × (∇ × A) (B.30)
Divergence Theorem: Given a closed surface S
enclosing a contiguous volume V,
∫
V
(∇ · A) dv=
∮
S
A · ds (B.31)
where the surface normal ds is pointing out of the
volume.
Stokes’ Theorem: Given a closed curve C bounding
a contiguous surface S,
∫
S
(∇ × A) · ds =
∮
C
A · dl (B.32)
where the direction of the surface normal ds is related
to the direction of integration along C by the “right
hand rule. ”",Electromagnetics_Vol2.pdf
"Appendix C
Ph
ysical Constants
[m0141]
The speed of light in free space (c), which is the
phase velocity of any electromagnetic radiation in
free space, is ∼= 2.9979 × 108 m/s. This is commonly
rounded up to 3 × 108 m/s. This rounding incurs error
of ∼= 0.07%, which is usually much less than other
errors present in electrical engineering calculations.
The charge of an electron is ∼= −1.602 × 10−19 C.
The constant e≜ +1.602176634 × 10−19 C is known
as the “elementary charge, ” so the charge of the
electron is said to be −e.
The permittivity of free space (ǫ0) is
∼= 8.854 × 10−12 F/m.
The permeability of free space (µ0) is
4π× 10−7 H/m.
The wave impedance of free space (η0) is the ratio
of the magnitude of the electric field intensity to that
of the magnetic field intensity in free space and is√
µ0/ǫ0 ∼= 376 .7 Ω. This is also sometimes referred
to as the intrinsic impedance of free space.
Boltzmann’s constant is ∼
= 1.381 × 10−23 J/K, the
amount of energy associated with a change of one",Electromagnetics_Vol2.pdf
"to as the intrinsic impedance of free space.
Boltzmann’s constant is ∼
= 1.381 × 10−23 J/K, the
amount of energy associated with a change of one
degree of temperature. This is typically assigned the
symbol k(unfortunately , the same symbol often used
to represent wavenumber).
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2",Electromagnetics_Vol2.pdf
"Index
acceptance
angle, 140
aluminum, 43, 46, 207
Ampere’s law
general form, 7, 8, 26, 34, 105, 146, 154, 155,
159
magnetostatics, 6, 18, 124
antenna
electrically-short dipole (ESD), 155–157, 159,
161, 168–172, 175–177, 198
folded half-wave dipole, 136
half-wave dipole, 136, 159, 161–162, 198
isotropic, 199, 201
microstrip patch, 174
quarter-wave monopole, 136
yagi, 181
antenna aperture, 196
antenna factor, 183
attenuation, 129–133, 135
attenuation rate, 40–41, 44
balun, 136
binomial series, 42
Biot-Savart law , 18–20, 154
Boltzmann’s constant, 198, 212
boundary conditions, 5, 6, 8, 187, 190
Brewster’s angle, 86
carbon, 207
CGS (system of units), 2
characteristic impedance, 122, 124
charge density
line, 5
surface, 5
volume, 5
cladding, 138
conductivity
of common materials, 207–208
conductor
good, 43–45, 48, 169
perfect, 6, 43, 48, 50, 58
poor, 41–43
cone of acceptance, 140
constitutive relationships, 7
copper, 43, 207
Coulomb force, 21
Coulomb’s law , 18
coupling, 196",Electromagnetics_Vol2.pdf
"poor, 41–43
cone of acceptance, 140
constitutive relationships, 7
copper, 43, 207
Coulomb force, 21
Coulomb’s law , 18
coupling, 196
critical angle, 88
curl, 210
current, 5, 48
current density
surface, 5, 163
volume, 5, 163
current moment, 20, 150, 152, 162
cutoff frequency , 100, 106, 117, 123
cyclotron motion, 11
dB, see decibel
dBm, 40
decibel, 39–40
delay spread, 143
dielectric (material), 123
examples, 205
dielectric constant, 205
Dirac delta function, 145, 150, 162
directivity , 177–179, 199
dispersion, 54, 95, 102, 118, 141
displacement current, 7, 34
divergence, 210
divergence theorem, 26, 211
E-plane, 180
effective aperture, 196–200
effective height, 183
effective length, 183, 190
electric field intensity , 5, 11
electric flux density , 5
electrical length, 154, 168
electrically-short dipole, see antenna
electrically-short dipole (ESD), 178
electron, 212
electrostatics (description), 5
213",Electromagnetics_Vol2.pdf
"214 INDEX
English system of units, 2
evanescent waves, see waves
far field, 153, 161, 167
Faraday’s law , 7, 23
ferrite, 207
fiber optics, 88
flux
electric, 7
magnetic, 6, 7
force, 20
FR4, 41, 127, 128, 136, 205
Friis transmission equation, 201–202
gain
power, 39
voltage, 40
Gauss’ law
electric field, 5, 18
magnetic field, 6, 148
geranium, 205
glass, 205
gold, 43, 207
good conductor, 50
Goos-H¨anchen effect, 89
gradient, 209
group velocity , 95–97, 102, 118
H-plane, 180
half-power beamwidth (HPBW), 182
half-wave matching (slab), 65–66
Hertzian dipole, 147, 152–156, 159, 166, 169, 176,
191
homogeneity , 8
homogeneous (media), 8
impedance
surface, 54
wire, 50–54
independence of path, 5
index of refraction, 59, 81, 138, 205
indium phosphide, 205
inductance, 53
equivalent, 53
inverse square law , 18
iron, 206, 207
isotropic, 182
isotropic (media), 9
isotropy , 9
Jefimenko’s equations, 18
Johnson-Nyquist noise, 198
joule heating, 27
Joule’s law , 27
Kirchoff’s voltage law
electrostatics, 5",Electromagnetics_Vol2.pdf
"isotropy , 9
Jefimenko’s equations, 18
Johnson-Nyquist noise, 198
joule heating, 27
Joule’s law , 27
Kirchoff’s voltage law
electrostatics, 5
Laplace’s Equation, 133
Laplacian (operator), 210
lever arm, 15
linear (media), 9
linearity , 9
Lorentz force, 11
Lorentz reciprocity theorem, 187–188
Lorenz gauge condition, 149
loss resistance, 174
loss tangent, 30, 34–35, 41, 43
magnetic field intensity , 6
magnetic flux density , 6, 11, 148
magnetic vector potential, 147–152, 159
magnetostatics (description), 6
materials (properties), 8
materials, magnetic, 6
Maxwell’s equations, 7, 145, 147
differential phasor form, 7, 30, 108, 187
source-free lossless region, 8
static differential form, 6
static integral form, 6
Maxwell-Faraday equation, 7, 8, 26, 102
metric system, 2
microstrip, 108, 123–129, 136
mode, 100, 106, 112, 116
motional emf, 23
motor, 13, 17
mutual coupling, 196
notation, 3
numerical aperture, 140
Ohm’s law , 6, 25, 27, 30, 50
ohmic loss, 27, 34
omnidirectional, 182",Electromagnetics_Vol2.pdf
"motor, 13, 17
mutual coupling, 196
notation, 3
numerical aperture, 140
Ohm’s law , 6, 25, 27, 30, 50
ohmic loss, 27, 34
omnidirectional, 182
optical fiber, 138–143
parallel-plate waveguide, see waveguide, see
waveguide, see waveguide, see waveguide,
see waveguide
path gain, 201
path loss, 201
pattern (radiation), 179–182, 200",Electromagnetics_Vol2.pdf
"INDEX 215
normalized, 180
permeability , 6
of common materials, 206–207
relative, 6, 206
permittivity , 5
complex-valued, 30, 33–34
effective, 128
of common materials, 205–206
relative, 5, 205
phase velocity , 95–97, 118
in microstrip, 128
phasor, 7
plane of incidence, 70
plane wave relationships, 36, 73, 77, 107, 154, 157,
161
plasma, 147
platinum, 207
polarization, 179–182
polarizing angle, 86
polycarbonate, 205
polyethylene, 42, 205
polypropylene, 205
polystyrene, 205
polytetrafluoroethylene, 205
potential difference, 21
power handling, see transmission line
Poynting vector, 27–29, 37
Poynting’s theorem, 25–28, 146
printed circuit board (PCB), 123, 128, 174
material, 205
prism, 82
propagation constant, 150
attenuation, 32, 36, 40, 46, 150
phase, 30, 31, 36, 150
quarter-wave matching (slab), 66–67
radiation condition, 151
radiation efficiency , 169, 174, 178
radiation resistance, 174, 175
radome, 60, 64
ray-fixed coordinates, 68, 70
Rayleigh-Jeans law , 198
reciprocity , 186–190",Electromagnetics_Vol2.pdf
"radiation resistance, 174, 175
radome, 60, 64
ray-fixed coordinates, 68, 70
Rayleigh-Jeans law , 198
reciprocity , 186–190
rectangular waveguide, see waveguide
reference phase, 67
reference polarization, 67
reflection coefficient, 57
refraction, 81
resistivity , 207
RG-59, 52, 53, 131
right hand rule
magnetostatics, 19, 124
Stokes’ theorem, 211
sampling function, 145, 150
semiconductor (material), 205
separation of variables, 111, 114
SI (system of units), 2
silica, 205
silicon, 205, 207
skin depth, 45–46, 50, 51, 129
skin effect, 49, 53
slab, 60–67
Snell’s law , 81, 83, 85, 88, 92, 140
soil, 206, 207
speed of light, 212
standing wave, 63, 161
steel, 206
Stokes’ theorem, 211
stripline, 123
styrofoam, 205
superposition, 9, 108, 111, 114, 146, 163, 166
surface wave, see waves
T eflon, 205
thermal radiation, 198
thermodynamic equilibrium, 199
thermodynamics, 198
time-harmonic, 7, 147
time-invariance (media), 9
torque, 15
total internal reflection, 82, 84, 86, 88–90, 138
transducer, 166",Electromagnetics_Vol2.pdf
"thermodynamics, 198
time-harmonic, 7, 147
time-invariance (media), 9
torque, 15
total internal reflection, 82, 84, 86, 88–90, 138
transducer, 166
transmission line
analogy for wave propagation, 58, 63
coaxial, 52, 53, 103, 121, 129–135
differential, 121
lossy , 36
lumped-element model, 122, 124
microstrip, see microstrip
parallel wire, 121–123
power handling, 133–136
single-ended, 121, 123, 136
stripline, 123
transverse electromagnetic (TEM), 32, 124
transverse electric, 70–75, 83–84, 110, 113–116
transverse electromagnetic (TEM), 71, 103, 110, 122
transverse magnetic, 70–71, 76–80, 85–87, 110–113
twin lead, 121",Electromagnetics_Vol2.pdf
"216 INDEX
units, 1–2
vector
arithmetic, 211
identity , 211
position-free, 15
vector effective length, 183, 190, 192, 195, 196
water, 205, 207
wave equation
electromagnetic, 36, 147, 149, 150
magnetic vector potential, 149, 150
source-free lossless region, 8
source-free lossy region, 30–32
wave impedance, 8, 37, 42, 154, 212
waveguide
parallel plate, 97–108
rectangular, 110–119
wavelength
in microstrip, 128
waves
evanescent, 90–92
plane, 8, 36–37
power density , 8, 37
surface, 92
transmission line analogy , 36
unidirectional, 108–110
work, 20
yagi, see antenna",Electromagnetics_Vol2.pdf
"VOLUME 2 
Electromagnetics, volume 2, by Steven W. Ellingson is a 216-page peer-reviewed open 
textbook designed especially for electrical engineering students in the third year of a 
bachelor of science degree program. It is intended as the primary textbook for the second 
semester of a two-semester undergraduate engineering electromagnetics sequence. The 
book addresses magnetic force and the Biot-Savart law; general and lossy media; parallel 
plate and rectangular waveguides; parallel wire, microstrip, and coaxial transmission lines; 
AC current /f_low and skin depth; re/f_lection and transmission at planar boundaries; /f_ields in 
parallel plate, parallel wire, and microstrip transmission lines; optical /f_iber; and radiation 
and antennas. 
This textbook is part of the Open Electromagnetics Project led by Steven W. Ellingson 
at Virginia Tech (https://www.faculty.ece.vt.edu/swe/oem). The project goal is to",Electromagnetics_Vol2.pdf
"at Virginia Tech (https://www.faculty.ece.vt.edu/swe/oem). The project goal is to 
create publicly available, no-cost, openly licensed content for courses in engineering 
electromagnetics. The project is motivated by two things: lowering the cost of learning 
materials for students and giving faculty the freedom to adopt, modify, and improve their 
educational resources. For each book in the series, the following are freely available for all 
to use: openly licensed and accessible PDFs, problem sets and solutions, slides of /f_igures, 
editable LaTeX source /f_iles, and information for the book’s collaborator portal and listserv.
Books in this series 
Electromagnetics, volume 1, https://doi.org/10.21061/electromagnetics-vol-1
Electromagnetics, volume 2,  https://doi.org/10.21061/electromagnetics-vol-2
Advance Praise for Electromagnetics
“I commend the author for the hard work and generosity required to create a",Electromagnetics_Vol2.pdf
"Advance Praise for Electromagnetics
“I commend the author for the hard work and generosity required to create a 
book like this and to make it available for free. The overall format is pleasing and 
the material is generally rigorous and complete. Well done. ”
 — Karl Warnick, Brigham Young University
“I really liked some of the ways the author related the math to physics and 
explained why the math made sense. … The /f_igures are generally quite good. … 
The examples in the book are simple and straightforward. ”
— Randy Haupt, Colorado School of Mines
Cover Design: Robert Browder
This book is licensed with a Creative Commons 
Attribution-ShareAlike License 4.0.
Virginia Tech Publishing /uni2219 Blacksburg, Virginia",Electromagnetics_Vol2.pdf
"Algorithms and Data Structures
With Applications to Graphics and Geometry",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Algorithms and Data Structures
With Applications to Graphics and Geometry
Jurg Nievergelt
Klaus Hinrichs
Copyright © 2011 Jurg Nievergelt
Editor-In-Chief: Jurg Nievergelt
Associate Editor: Marisa Drexel Ulrich
Editorial Assistants: Jon Durden, Tessa Greenleaf, Kristyna Mauch Selph, Ernesto Serrano
For any questions about this text, please email: drexel@uga.edu
The Global Text Project is funded by the Jacobs Foundation, Zurich, Switzerland.
This book is licensed under a Creative Commons Attribution 3.0 License.
Algorithms and Data Structures 2  A Global Text",algorithms and data structures.pdf
"Table of Contents
Part I: Programming environments for motion, graphics, and geometry.................................7
1. Reducing a task to given primitives: programming motion...................................................9
A robot car, its capabilities, and the task to be performed.............................................................................. 9
Wall-following algorithm described informally............................................................................................. 10
Algorithm specified in a high-level language.................................................................................................. 11
Algorithm programmed in the robot's language............................................................................................ 12
The robot's program optimized...................................................................................................................... 12",algorithms and data structures.pdf
"2. Graphics primitives and environments.................................................................................14
Turtle graphics: a basic environment............................................................................................................. 14
QuickDraw: a graphics toolbox ...................................................................................................................... 16
A graphics frame program.............................................................................................................................. 19
3. Algorithm animation.............................................................................................................24
Computer-driven visualization: characteristics and techniques................................................................... 24
A gallery of algorithm snapshots.................................................................................................................... 27",algorithms and data structures.pdf
"Part II: Programming concepts: beyond notation....................................................................33
4. Algorithms and programs as literature: substance and form..............................................34
Programming in the large versus programming in the small........................................................................ 34
Documentation versus literature: is it meant to be read?..............................................................................35
Pascal and its dialects: lingua franca of computer science............................................................................ 40
5. Divide-and-conquer and recursion.......................................................................................45
An algorithmic principle................................................................................................................................. 45",algorithms and data structures.pdf
"Divide-and-conquer expressed as a diagram: merge sort............................................................................. 46
Recursively defined trees................................................................................................................................ 47
Recursive tree traversal.................................................................................................................................. 49
Recursion versus iteration: the Tower of Hanoi............................................................................................ 50
The flag of Alfanumerica: an algorithmic novel on iteration and recursion................................................. 52
6. Syntax.....................................................................................................................................53",algorithms and data structures.pdf
"6. Syntax.....................................................................................................................................53
Syntax and semantics..................................................................................................................................... 53
Grammars and their representation: syntax diagrams and EBNF................................................................ 54
An overly simple syntax for simple expressions............................................................................................. 57
Parenthesis-free notation for arithmetic expressions....................................................................................59
7. Syntax analysis.......................................................................................................................62",algorithms and data structures.pdf
"7. Syntax analysis.......................................................................................................................62
The role of syntax analysis.............................................................................................................................. 62
Syntax analysis of parenthesis-free expressions by counting........................................................................ 63
Analysis by recursive descent......................................................................................................................... 64
Turning syntax diagrams into a parser.......................................................................................................... 65
Part III: Objects, algorithms, programs....................................................................................67
8. Truth values, the data type 'set', and bit acrobatics.............................................................69",algorithms and data structures.pdf
"8. Truth values, the data type 'set', and bit acrobatics.............................................................69
Bits and boolean functions............................................................................................................................. 69
Swapping and crossovers: the versatile exclusive-or..................................................................................... 70
The bit sum or ""population count"".................................................................................................................. 71
9. Ordered sets...........................................................................................................................78
Sequential search............................................................................................................................................ 78",algorithms and data structures.pdf
"Binary search.................................................................................................................................................. 79
In-place permutation...................................................................................................................................... 82
10. Strings ..................................................................................................................................87
Recognizing a pattern consisting of a single string........................................................................................ 87
Recognizing a set of strings: a finite-state-machine interpreter................................................................... 88
11. Matrices and graphs: transitive closure...............................................................................93
3",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Paths in a graph.............................................................................................................................................. 93
Boolean matrix multiplication....................................................................................................................... 94
Warshall's algorithm....................................................................................................................................... 95
Minimum spanning tree in a graph................................................................................................................ 97
12. Integers...............................................................................................................................100
Operations on integers................................................................................................................................. 100",algorithms and data structures.pdf
"The Euclidean algorithm.............................................................................................................................. 102
The prime number sieve of Eratosthenes..................................................................................................... 103
Large integers............................................................................................................................................... 104
Modular number systems: the poor man's large integers............................................................................ 105
Random numbers.......................................................................................................................................... 107
13. Reals....................................................................................................................................110",algorithms and data structures.pdf
"13. Reals....................................................................................................................................110
Floating-point numbers................................................................................................................................ 110
Some dangers................................................................................................................................................. 112
Horner's method............................................................................................................................................ 113
Bisection........................................................................................................................................................ 114
Newton's method for computing the square root......................................................................................... 115",algorithms and data structures.pdf
"Newton's method for computing the square root......................................................................................... 115
14. Straight lines and circles.....................................................................................................119
Intersection.................................................................................................................................................... 119
Clipping......................................................................................................................................................... 122
Drawing digitized lines................................................................................................................................. 123
The riddle of the braiding straight lines....................................................................................................... 126",algorithms and data structures.pdf
"The riddle of the braiding straight lines....................................................................................................... 126
Digitized circles ............................................................................................................................................. 131
Part IV: Complexity of problems and algorithms...................................................................134
15. Computability and complexity...........................................................................................135
Models of computation: the ultimate RISC.................................................................................................. 135
Almost nothing is computable...................................................................................................................... 138
The halting problem is undecidable............................................................................................................. 139",algorithms and data structures.pdf
"The halting problem is undecidable............................................................................................................. 139
Computable, yet unknown............................................................................................................................ 140
Multiplication of complex numbers............................................................................................................. 142
Complexity of matrix multiplication............................................................................................................ 142
16. The mathematics of algorithm analysis.............................................................................146
Growth rates and orders of magnitude......................................................................................................... 146",algorithms and data structures.pdf
"Growth rates and orders of magnitude......................................................................................................... 146
Asymptotics................................................................................................................................................... 147
Summation formulas.................................................................................................................................... 148
Recurrence relations..................................................................................................................................... 150
Asymptotic performance of divide-and-conquer algorithms....................................................................... 153
Permutations................................................................................................................................................. 154",algorithms and data structures.pdf
"Trees.............................................................................................................................................................. 155
17. Sorting and its complexity..................................................................................................158
What is sorting? How difficult is it?............................................................................................................. 158
Types of sorting algorithms.......................................................................................................................... 160
Simple sorting algorithms that work in time Θ(n2)..................................................................................... 163
A lower bound Ω(n · log n)............................................................................................................................ 165",algorithms and data structures.pdf
"Quicksort....................................................................................................................................................... 166
Analysis for three cases: best, ""typical"", and worst...................................................................................... 169
Is it possible to sort in linear time?............................................................................................................... 174
Sorting networks........................................................................................................................................... 174
Part V: Data structures.............................................................................................................179
18. What is a data structure?...................................................................................................180",algorithms and data structures.pdf
"18. What is a data structure?...................................................................................................180
Data structures old and new......................................................................................................................... 180
Algorithms and Data Structures 4  A Global Text",algorithms and data structures.pdf
"The range of data structures studied............................................................................................................ 181
Performance criteria and measures.............................................................................................................. 182
19. Abstract data types.............................................................................................................184
Concepts: What and why?............................................................................................................................ 184
Stack.............................................................................................................................................................. 185
First-in-first-out queue................................................................................................................................. 189",algorithms and data structures.pdf
"Priority queue............................................................................................................................................... 190
Dictionary...................................................................................................................................................... 191
20. Implicit data structures.....................................................................................................196
What is an implicit data structure?.............................................................................................................. 196
Array storage................................................................................................................................................. 197
Implementation of the fixed-length fifo queue as a circular buffer............................................................ 202",algorithms and data structures.pdf
"Implementation of the fixed-length fifo queue as a circular buffer............................................................ 202
Implementation of the fixed-length priority queue as a heap..................................................................... 205
Heapsort ...................................................................................................................................................... 209
21. List structures.....................................................................................................................211
Lists, memory management, pointer variables ............................................................................................ 211
The fifo queue implemented as a one-way list ............................................................................................ 214",algorithms and data structures.pdf
"The fifo queue implemented as a one-way list ............................................................................................ 214
Tree traversal................................................................................................................................................ 214
Binary search trees....................................................................................................................................... 223
Height-balanced trees.................................................................................................................................. 228
22. Address computation ........................................................................................................239
Concepts and terminology............................................................................................................................ 239",algorithms and data structures.pdf
"The special case of small key domains ........................................................................................................ 240
The special case of perfect hashing: table contents known a priori ............................................................ 241
Conventional hash tables: collision resolution ........................................................................................... 242
Choice of hash function: randomization...................................................................................................... 246
Performance analysis .................................................................................................................................. 248
Extendible hashing ...................................................................................................................................... 249",algorithms and data structures.pdf
"A virtual radix tree: order-preserving extendible hashing.......................................................................... 251
23. Metric data structures.......................................................................................................254
Organizing the embedding space versus organizing its contents................................................................ 254
Radix trees, tries .......................................................................................................................................... 255
Quadtrees and octtrees ................................................................................................................................ 255
Spatial data structures: objectives and constraints......................................................................................257",algorithms and data structures.pdf
"Spatial data structures: objectives and constraints......................................................................................257
The grid file................................................................................................................................................... 259
Simple geometric objects and their parameter spaces................................................................................ 263
Region queries of arbitrary shape................................................................................................................ 264
Evaluating region queries with a grid file.................................................................................................... 267
Interaction between query processing and data access............................................................................... 267
Part  VI:  Interaction  between  algorithms  and  data  structures:  case  studies  in  geometric",algorithms and data structures.pdf
"Part  VI:  Interaction  between  algorithms  and  data  structures:  case  studies  in  geometric 
computation................................................................................................................................271
24. Sample problems and algorithms.....................................................................................272
Geometry and geometric computation......................................................................................................... 272
Convex hull: a multitude of algorithms........................................................................................................ 273
The uses of convexity: basic operations on polygons................................................................................... 277
Visibility in the plane: a simple algorithm whose analysis is not................................................................ 279",algorithms and data structures.pdf
"Visibility in the plane: a simple algorithm whose analysis is not................................................................ 279
25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using 
line segment intersection..........................................................................................................286
The line segment intersection test............................................................................................................... 286
The skeleton: Turning a space dimension into a time dimension.............................................................. 288
Data structures............................................................................................................................................. 288
5",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Updating the y-table and detecting an intersection.................................................................................... 289
Sweeping across intersections ..................................................................................................................... 290
Degenerate configurations, numerical errors, robustness........................................................................... 291
26. The closest pair..................................................................................................................293
The problem.................................................................................................................................................. 293
Plane-sweep applied to the closest pair problem.........................................................................................294",algorithms and data structures.pdf
"Plane-sweep applied to the closest pair problem.........................................................................................294
Implementation............................................................................................................................................ 295
Analysis......................................................................................................................................................... 297
Sweeping in three or more dimensions....................................................................................................... 298
Algorithms and Data Structures 6  A Global Text",algorithms and data structures.pdf
"Part I: Programming 
environments for motion, 
graphics, and geometry
Part I of this text book will discuss:
• simple programming environments
• program design
• informal versus formal notations
• reducing a solution to primitive operations, and programming as an activity independent of language.
The purpose of an artificial programming environment
A program can be designed with the barest of tools, paper and pencil, or in the programmer's head. In the realm 
of such informal environments, a program design may contain vague concepts expressed in an informal notation.  
Before  he  or  she  can  execute  this  program,  the  programmer  needs  a  programming  environment,  typically a 
complex  system  with many distinct  components: a computer  and  its operating system, utilities, and  program  
libraries; text and program editors; various programming languages and their processors. Such real programming  
environments force programmers to express themselves in formal notations.",algorithms and data structures.pdf
"environments force programmers to express themselves in formal notations.
Programming is the realization of a solution to a problem, expressed in terms of those operations provided by a  
given programming environment. Most programmers work in environments that provide very powerful operations  
and tools.
The more powerful a programming environment, the simpler the programming task, at least to the expert who  
has achieved mastery of this environment. Even an experienced programmer may need several months to master a  
new programming environment, and a novice may give up in frustration at the multitude of concepts and details he  
or she must understand before writing the simplest program.
The simpler a programming environment, the easier it is to write and run small programs, and the more work it  
is to write substantial, useful programs. In the early days of computing, before the proliferation of programming",algorithms and data structures.pdf
"is to write substantial, useful programs. In the early days of computing, before the proliferation of programming  
languages during the 1960s, most programmers worked in environments that were exceedingly simple by modern  
standards: Acquaintance with an assembler, a loader, and a small program library sufficed. The programs they  
wrote  were  small  compared  to  what  a  professional  programmer  writes  today.  The  simpler  a  programming 
environment is, the better suited it is for learning to program. Alas, today simple environments are hard to find!  
Even a home computer is equipped with complex software that is not easily ignored or bypassed. For the sake of  
education it is  useful to invent artificial programming environments. Their only purpose  is to illustrate  some  
important concepts in the simplest possible setting and to facilitate insight. Part I of this book introduces such a toy  
7",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
programming environment suitable for programming graphics and motion, and illustrates how it can gradually be  
enriched to approach a simple but useful graphics environment.
Textbooks on computer graphics.  The computer-driven graphics screen is a powerful new medium for  
communication.  Visualization  often  makes  it  possible  to  present  the  results  of  a  computation  in  intuitively 
appealing  ways  that  convey  insights  not  easily  gained  in  any  other  manner.  To  exploit  this  medium,  every 
programmer  must  master  basic  visualization  techniques.  We  refer  the  reader  interested  in  a  systematic 
introduction to computer graphics to such excellent textbooks as [BG 89], [FDFH 90], [NS 79], [Rog 85], [Wat 89],  
and [Wol 89].
Algorithms and Data Structures 8  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
1. Reducing a task to given 
primitives: programming 
motion
Learning objectives:
• primitives for specifying motion 
• expressing an algorithm in informal notations and in high- and low-level programming languages 
• program verification 
• program optimization
A robot car, its capabilities, and the task to be performed
Some aspects of programming can be learned  without a computer, by inventing an artificial programming  
environment as a purely mental exercise. The example of a vehicle that moves under program control in a fictitious  
landscape  is  a  microcosmos  of  programming  lore.  In  this  section  we  introduce  important  concepts  that  will 
reappear later in more elaborate settings.
The environment. Consider a two-dimensional square grid, a portion of which is enclosed by a wall made up",algorithms and data structures.pdf
"The environment. Consider a two-dimensional square grid, a portion of which is enclosed by a wall made up  
of horizontal and vertical line segments that run halfway between the grid points ( Exhibit 1.1). A robot car enclosed 
within the wall moves along this grid under computer control, one step at a time, from grid point to adjacent grid  
point. Before and after each step, the robot's state is described by a location (grid point) and a direction (north, east, 
south, or west).
Exhibit 1.1: The robot's crosshairs show its current location on the grid.
The robot is controlled by a program that uses the following commands:
left Turn 90 degrees counterclockwise.
right Turn 90 degrees clockwise.
forward Move one step, to the next grid point in front of 
you
goto # Send program control to the label #.
if touch goto # If you are touching a wall to your front, send 
program control to the label #.
Algorithms and Data Structures 9  A Global Text",algorithms and data structures.pdf
"1. Reducing a task to given primitives: programming motion
A program for the robot is a sequence of commands with distinct labels. The labels serve merely to identify the  
commands and need not be arranged either consecutively or in increasing order. Execution begins with the first  
command and proceeds to successive commands in the order in which they appear, except when flow of control is  
redirected by either of the goto commands.
Example
The following program moves the robot forward until it bumps into a wall:
1 if touch goto 4
2 forward
3 goto 1
4 { there is no command here; just a label }
In developing programs for the robot, we feel free to use any high-level language we prefer, and embed robot  
commands in it. Thus we might have expressed our wall-finding program by the simpler statement 
while not touch do forward; 
and then translated it into the robot's language.",algorithms and data structures.pdf
"while not touch do forward; 
and then translated it into the robot's language.
A program for this robot car to patrol the walls of a city consists of two parts: First, find a wall, the problem we  
just solved. Second, move along the wall forever while maintaining two conditions: 
1. Never lose touch with the wall; at all times, keep within one step of it.
2. Visit every spot along the wall in a monotonic progression.
The mental image of walking around a room with eyes closed, left arm extended, and the left hand touching the  
wall at all times will prove useful. To mirror this solution we start the robot so that it has a wall on its immediate left 
rather than in front. As the robot has no sensor on its left side, we will let it turn left at every step to sense the wall  
with its front bumper, then turn right to resume its position with the wall to its left.
Wall-following algorithm described informally",algorithms and data structures.pdf
"with its front bumper, then turn right to resume its position with the wall to its left.
Wall-following algorithm described informally
Idea of solution:  Touch the wall with your left hand; move forward, turning left or right as required to keep  
touching the wall.
Wall-following  algorithm  described  in  English: Clockwise,  starting  at  left,  look  for  the  first  direction  not 
blocked by a wall, and if found, take a step in that direction.
Let us test this algorithm on some critical configurations. The robot inside a unit square turns forever, never  
finding a direction to take a step ( Exhibit 1.2). In Exhibit 1.3 the robot negotiates a left-hand spike. After each step,  
there is a wall to its left-rear. In Exhibit 1.4 the robot enters a blind alley. At the end of the alley, it turns clockwise  
twice, then exits by the route it entered.
Exhibit 1.2: Robot in a box spins on its heels.
10",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 1.3: The robot turns around a spike.
Exhibit 1.4: Backing up in a blind alley.
Algorithm specified in a high-level language
The ideas presented informally in above section are made precise in the following elegant, concise program:
{ wall to left-rear }
loop
{ wall to left-rear }
left;
{ wall to left-front }
while touch do
 { wall to right-front }
 right;
 { wall to left-front }
endwhile;
{ wall to left-front }
forward;
{ wall to left-rear }
forever;
{ wall to left-rear }
Program verification.  The comments in braces are  program invariant s: Assertions about the state of the  
robot that are true every time the flow of control reaches the place in the program where they are written. We need  
three types of invariants to verify the wall-following program: ""wall to left-rear"", ""wall to left-front"", and ""wall to",algorithms and data structures.pdf
"three types of invariants to verify the wall-following program: ""wall to left-rear"", ""wall to left-front"", and ""wall to  
right-front"". The relationships between the robot's position and the presence of a nearby wall that must hold for  
each assertion to be true are illustrated in  Exhibit 1.5 . Shaded circles indicate points through which a wall must  
pass. Each robot command transforms its  precondition (i.e. the assertion true before the command is executed)  
into its  postcondition (i.e. the assertion true after its execution). Thus each of the commands 'left', 'right', and  
'forward' is a predicate transformer, as suggested in Exhibit 1.6.
Exhibit 1.5: Three types of invariants relate the positions of robot and wall.
Algorithms and Data Structures 11  A Global Text",algorithms and data structures.pdf
"1. Reducing a task to given primitives: programming motion
Exhibit 1.6: Robot motions as predicate transformers.
Algorithm programmed in the robot's language
A  straightforward  translation  from  the  high-level  program  into  the  robot's  low-level language  yields  the 
following seven-line wall-following program:
loop
left; 1 left
while touch do 2 if touch goto 4
3 goto 6
right; 4 right
endwhile; 5 goto 2
forward; 6 forward
forever; 7 goto 1
The robot's program optimized
In  designing a  program  it  is  best  to  follow  simple,  general  ideas,  and  to  decide  on  details  in  the  most 
straightforward  manner,  without  regard  for  the  many  alternative  ways  that are  always  available  for  handling 
details. Once a program is proven correct, and runs, then we may try to improve its efficiency, measured by time  
and memory requirements. This process of program transformation can often be done syntactically, that is merely",algorithms and data structures.pdf
"and memory requirements. This process of program transformation can often be done syntactically, that is merely  
by considering the definition of individual statements, not the algorithm as a whole. As an example, we derive a  
five-line version of the wall-following program by transforming the seven-line program in two steps.
If we have the complementary primitive  'if not touch goto #', we can simplify the flow of the program at the left  
as shown on the right side.
{ wall to left-rear } { wall to left-rear }
1 left 1 left
2 if touch goto 4 2 if not touch goto 6
3 goto 6
{ wall to right-front } { wall to right-front }
4 right 4 right
12",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
5 goto 2 5 goto 2
6 forward 6 forward
7 goto 1 7 goto 1
An optimization technique called loop rotation allows us to shorten this program by yet another instruction. It  
changes the structure of the program significantly, as we see from the way the labels have been permuted. The  
assertion ""wall to right-front"" attached to line 4 serves as an invariant of the loop  ""keep turning right while you  
can't advance"".
{ wall to right-front }
4 right
2 if touch goto 4
6 forward
1 left
7 goto 2
Programming projects
1. Design a data structure suitable for storing a wall made up of horizontal and vertical line segments in a  
square grid of bounded size. Write a ""wall-editor"", i.e. an interactive program that lets the user define and  
modify an instance of such a wall.
2. Program the wall-following algorithm and animate its execution when tracking a wall entered with the wall-",algorithms and data structures.pdf
"modify an instance of such a wall.
2. Program the wall-following algorithm and animate its execution when tracking a wall entered with the wall-
editor. Specifically, show the robot's position and orientation after each change of state.
Algorithms and Data Structures 13  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
2. Graphics primitives and 
environments
Learning objectives:
• turtle graphics
• QuickDraw: A graphics toolbox
• frame program
• interactive graphics input/output
• example: polyline input
Turtle graphics: a basic environment
Seymour Papert [Pap80] introduced the term  turtle graphics  to denote a set of primitives for line drawing.  
Originally implemented in the programming language Logo, turtle graphics primitives are now available for several  
computer systems and languages. They come in different versions, but the essential point is the same as that  
introduced in the example of the robot car: The pen (or ""turtle"") is a device that has a state (position, direction) and  
is driven by incremental operations “move” and “turn” that transform the turtle to a new state depending on its  
current state:
move(s) { take s unit steps in the direction you are facing }
turn(d) { turn counterclockwise d degrees }",algorithms and data structures.pdf
"current state:
move(s) { take s unit steps in the direction you are facing }
turn(d) { turn counterclockwise d degrees }
The turtle's initial state is set by the following operations:
moveto(x,y) { move to the position (x,y) in absolute coordinates }
turnto(d) { face d degrees from due east }
In addition, we can specify the color of the trail drawn by the moving pen:
pencolor(c) { where c = white, black, none, etc. }
Example
The following program fragment approximates a circle tangential to the x-axis at the origin by drawing a 36-
sided polygon:
moveto(0, 0); { position pen at origin }
turnto(0); { face east }
step := 7; { arbitrarily chosen step length }
do 36 times { 36 sides · 10° = 360° }
{ move(step);  turn(10) } { 10 degrees counterclockwise }
In graphics programming we are likely to use basic figures, such as circles, over and over again, each time with a  
different size and position. Thus we wish to turn a program fragment such as the circle approximation above into a",algorithms and data structures.pdf
"different size and position. Thus we wish to turn a program fragment such as the circle approximation above into a  
reusable procedure.
Algorithms and Data Structures 14  A Global Text",algorithms and data structures.pdf
"2. Graphics primitives and environments
Procedures as building blocks
A program is built from components at many different levels of complexity. At the lowest level we have the  
constructs provided by the language we use: constants, variables, operators, expressions, and simple (unstructured)  
statements. At the next higher level we have procedures: they let us refer to a program fragment of arbitrary size  
and  complexity as  a  single entity, and  build  hierarchically  nested  structures. Modern  programming languages 
provide yet another level of packaging: modules, or packages, useful for grouping related data and procedures. We  
limit our discussion to the use of procedures.
Programmers accumulate their own collection of useful program fragments. Programming languages provide  
the concept of a  procedure as the major tool for turning fragments into  reusable building blocks. A procedure  
consists of two parts with distinct purposes:",algorithms and data structures.pdf
"consists of two parts with distinct purposes:
1. The  heading specifies an important part of the procedure's external behavior through the list of  formal 
parameters: namely, what type of data moves in and out of the procedure.
2. The body implements the action performed by the procedure, processing the input data and generating the  
output data.
A program fragment that embodies a single coherent concept is best written as a procedure. This is particularly  
true if we expect to use this fragment again in a different context. The question of how general we want a procedure  
to be deserves careful thought. If the procedure is too specific, it will rarely be useful. If it is too general, it may be  
unwieldy: too large, too slow, or just too difficult to understand. The generality of a procedure depends primarily on  
the choice of formal parameters.
Example: the long road toward a procedure “circle”",algorithms and data structures.pdf
"the choice of formal parameters.
Example: the long road toward a procedure “circle”
Let us illustrate these issues by discussing design considerations for a procedure that draws a circle on the  
screen. The program fragment above for drawing a regular polygon is easily turned into
procedure ngon(n,s: integer);  { n = number of sides, s = step 
size }
var  i,j: integer;
begin
j := 360 div n;
for i := 1 to n do  { move(s);  turn(j) }
end;
But, a useful procedure to draw a circle requires additional arguments. Let us start with the following:
procedure circle(x, y, r, n: integer);
{ centered at (x, y);  r = radius;  n = number of sides }
var  a, s, i: integer;  { angle, step, counter }
begin
moveto(x, y – r);  { bottom of circle }
turnto(0);  { east }
a := 360 div n;
s := r · sin(a);  { between inscribed and circumscribed polygons }
for  i := 1  to  n  do  { move(s);  turn(a) }
end;",algorithms and data structures.pdf
"a := 360 div n;
s := r · sin(a);  { between inscribed and circumscribed polygons }
for  i := 1  to  n  do  { move(s);  turn(a) }
end;
This procedure places the burden of choosing n on the programmer. A more sophisticated, ""adaptive"" version  
might choose the number of sides on its own as a function of the radius of the circle to be drawn. We assume that  
lengths are measured in terms of pixels (picture elements) on the screen. We observe that a circle of radius r is of  
15",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
length 2 πr. We approximate it by drawing short-line segments, about 3 pixels long, thus needing about 2·r line  
segments. 
procedure circle(x, y, r: integer);  { centered at (x, y);  radius 
r}
var  a, s, i: integer;  { angle, step, counter }
begin
moveto(x, y – r);  { bottom of circle }
turnto(0);  { east }
a := 180 div r;  { 360 / (# of line segments) }
s := r · sin(a);  { between inscribed and circumscribed polygons }
for  i := 1  to  2 · r  do  { move(s);  turn(a) }
end;
This circle procedure still suffers from severe shortcomings:
1. If we discretize a circle by a set of pixels, it is an unnecessary detour to do this in two steps as done above:  
first, discretize the circle by a polygon; second, discretize the polygon by pixels. This two-step process is a  
source of unnecessary work and errors.
2. The approximation of the circle by a polygon computed from vertex to vertex leads to rounding errors that",algorithms and data structures.pdf
"source of unnecessary work and errors.
2. The approximation of the circle by a polygon computed from vertex to vertex leads to rounding errors that 
accumulate. Thus the polygon may fail to close, in particular when using integer computation with its  
inherent large rounding error.
3. The procedure attempts to draw its circle on an infinite screen. Computer screens are finite, and attempted  
drawing beyond the screen boundary may or may not cause an error. Thus the circle ought to be clipped at  
the boundaries of an arbitrarily specified rectangle.
Writing a good circle procedure is a demanding task for professionals. We started this discussion of desiderata  
and difficulties of a simple library procedure so that the reader may appreciate the thought and effort that go into  
building a useful programming environment. In chapter 14 we return to this problem and present one possible goal",algorithms and data structures.pdf
"building a useful programming environment. In chapter 14 we return to this problem and present one possible goal  
of ""the long road toward a procedure 'circle'"". We now make a huge jump from the artificially small environments  
discussed so far to one of today's realistic programming environments for graphics
QuickDraw: a graphics toolbox 
For  the  sake  of  concreteness,  the  next  few  sections  show  programs  written  for  a  specific  programming 
environment:  MacPascal using  the QuickDraw library  of graphics  routines [App 85]. It is not our purpose  to  
duplicate a manual, but only to convey the flavor of a realistic graphics package and to explain enough about  
QuickDraw for the reader to understand the few programs that follow. So our treatment is highly selective and  
biased.
Concerning the circle that we attempted to program above, QuickDraw offers five procedures for drawing circles  
and related figures:
procedure FrameOval(r: Rect);",algorithms and data structures.pdf
"and related figures:
procedure FrameOval(r: Rect);
procedure PaintOval(r: Rect);
procedure EraseOval(r: Rect);
procedure InvertOval(r: Rect);
procedure FillOval(r: Rect; pat: Pattern);
Each one inscribes an oval in an aligned rectangle r (sides parallel to the axes) so as to touch the four sides of r.  
If r is a square, the oval becomes a circle. We quote from [App 85]:
Algorithms and Data Structures 16  A Global Text",algorithms and data structures.pdf
"2. Graphics primitives and environments
FrameOval draws an outline just inside the oval that fits inside the specified rectangle, using the current  
grafPort's pen pattern, mode, and size. The outline is as wide as the pen width and as tall as the pen height.  
It's drawn with the pnPat, according to the pattern transfer mode specified by pnMode. The pen location is  
not changed by this procedure.
Right away we notice a trade-off when comparing QuickDraw to the simple turtle graphics environment we  
introduced earlier. At one stroke, “FrameOval” appears to be able to produce many different pictures, but before we  
can exploit this power, we have to learn about grafPorts, pen width, pen height, pen patterns, and pattern transfer  
modes. 'FrameOval' draws the perimeter of an oval, 'PaintOval' paints the interior as well, 'EraseOval' paints an oval 
with the current grafPort's background pattern, 'InvertOval' complements the pixels: 'white' becomes 'black', and",algorithms and data structures.pdf
"with the current grafPort's background pattern, 'InvertOval' complements the pixels: 'white' becomes 'black', and  
vice versa. 'FillOval' has an additional argument that specifies a pen pattern used for painting the interior.
We may not need to know all of this in order to use one of these procedures, but we do need to know how to  
specify a rectangle. QuickDraw has predefined a type 'Rect' that, somewhat ambiguously at the programmer's  
choice, has either of the following two interpretations:
type Rect  = record  top, left, bottom, right: integer  end;
type Rect  = record  topLeft, botRight: Point  end;
with one of the interpretations of type 'Point' being
type Point = record  v, h: integer  end;
Exhibit  2.1 illustrates  and  provides  more  information  about  these  concepts.  It  shows  a  plane  with  first 
coordinate v that runs from top to bottom, and a second coordinate h that runs from left to right. (The reason for v",algorithms and data structures.pdf
"coordinate v that runs from top to bottom, and a second coordinate h that runs from left to right. (The reason for v  
running from top to bottom, rather than vice versa as used in math books, is compatibility with text coordinates  
where lines are naturally numbered from top to bottom.) The domain of v and h are the integers from –2 15= –32768 
to 215– 1 = 32767. The points thus addressed on the screen are shown as intersections of grid lines. These lines and  
grid points are infinitely thin - they have no extension. The pixels are the unit squares between them. Each pixel is  
paired with its top left grid point. This may be enough information to let us draw a slightly fat point of radius 3  
pixels at the grid point with integer coordinates (v, h) by calling
PaintOval(v – 3, h – 3, v + 3, h + 3);
Exhibit 2.1: Screen coordinates define the location of pixels.
To understand the procedures of this section, the reader has to understand a few details about two key aspects of",algorithms and data structures.pdf
"To understand the procedures of this section, the reader has to understand a few details about two key aspects of 
interactive graphics:
• timing and synchronization of devices and program execution
• how screen pictures are controlled at the pixel level
17",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Synchronization
In interactive applications we often wish to specify a grid point by letting the user point the mouse-driven cursor  
to some spot on the screen. The 'procedure GetMouse(v, h)' returns the coordinates of the grid point where the  
cursor is located at the moment 'GetMouse' is executed. Thus we can track and paint the path of the mouse by a  
loop such as
repeat  GetMouse(v, h);  PaintOval(v – 3, h – 3, v + 3, h + 3) 
until stop;
This does not give the user any timing control over when he or she wants the computer to read the coordinates  
of the mouse cursor. Clicking the mouse button is the usual way to tell the computer ""Now!"". A predefined boolean  
function  'Button'  returns  'true'  when  the  mouse  button  is  depressed,  'false'  when  not.  We  often  synchronize 
program execution with the user's clicks by programming busy waiting loops:",algorithms and data structures.pdf
"program execution with the user's clicks by programming busy waiting loops:
repeat until Button; { waits for the button to be pressed }
while Button do; { waits for the button to be released }
The following procedure waits for the next click:
procedure waitForClick;
begin  repeat until Button;  while Button do  end;
Pixel acrobatics
The QuickDraw pen has four parameters that can be set to draw lines or paint textures of great visual variety:  
pen location 'pnLoc', pen size 'pnSize' (a rectangle of given height and width), a pen pattern 'pnPat', and a drawing  
mode 'pnMode'. The pixels affected by a motion of the pen are shown in Exhibit 2.2.
Exhibit 2.2: Footprint of the pen.
Predefined values of 'pnPat' include 'black', 'gray', and 'white'. 'pnPat' is set by calling the predefined 'procedure  
PenPat(pat: Pattern)' [e.g. 'PenPat(gray)']. As 'white' is the default background, drawing in 'white' usually serves for  
erasing.",algorithms and data structures.pdf
"PenPat(pat: Pattern)' [e.g. 'PenPat(gray)']. As 'white' is the default background, drawing in 'white' usually serves for  
erasing.
The result of drawing also depends critically on the transfer mode 'pnMode', whose values include 'patCopy',  
'patOr',  and  'patXor'.  A  transfer  mode  is  a  boolean  operation  executed  in  parallel  on  each  pair  of  pixels  in 
corresponding positions, one on the screen and one in the pen pattern.
• 'patCopy' uses the pattern pixel to overwrite the screen pixel, ignoring the latter's previous value; it is the 
default and most frequently used transfer mode.
• 'patOr' paints a black pixel if either or both the screen pixel or the pattern pixel were black; it progressively 
blackens the screen.
Algorithms and Data Structures 18  A Global Text",algorithms and data structures.pdf
"2. Graphics primitives and environments
• 'patXor' (exclusive-or, also known as ""odd parity"") sets the result to black iff exactly one of (screen pixel, 
pattern pixel) is black. A white pixel in the pen leaves the underlying screen pixel unchanged; a black pixel 
complements it. Thus a black pen inverts the screen.
'pnMode' is set by calling the predefined 'procedure PenMode(mode: integer)' [e.g. 'PenMode(patXor)'].
The meaning of the remaining predefined procedures our examples use, such as 'MoveTo' and 'LineTo', is easily  
guessed. So we terminate our peep into some key details of a powerful graphics package, and turn to examples of its  
use.
A graphics frame program
Reusable software is a time saving concept that can be practiced profitably in the small. We keep a program that  
contains nothing but a few of the most useful input/output procedures, displays samples of their results, and",algorithms and data structures.pdf
"contains nothing but a few of the most useful input/output procedures, displays samples of their results, and  
conducts a minimal dialog so that the user can step through its execution. We call this a frame program because its 
real purpose is to facilitate development and testing of new procedures by embedding them in a ready-made, tested  
environment. A simple frame program like the one below makes it very easy for a novice to write his first interactive  
graphics program.
This particular frame program contains procedures 'GetPoint', 'DrawPoint', 'ClickPoint', 'DrawLine', 'DragLine',  
'DrawCircle', and 'DragCircle' for input and display of points, lines, and circles on a screen idealized as a part of a  
Euclidean plane, disregarding the discretization due to the raster screen. Some of these procedures are so short that  
one asks why they are introduced at all. 'GetPoint', for example, only converts integer mouse coordinates v, h into a",algorithms and data structures.pdf
"one asks why they are introduced at all. 'GetPoint', for example, only converts integer mouse coordinates v, h into a  
point p with real coordinates. It enables us to refer to a point p without mentioning its coordinates explicitly. Thus,  
by bringing us closer to standard geometric notation, 'GetPoint' makes programs more readable.
The procedure 'DragLine', on the other hand, is a very useful routine for interactive input of line segments. It  
uses the rubber-band technique, which is familiar to users of graphics editors. The user presses the mouse button  
to fix the first endpoint of a line segment, and keeps it depressed while moving the mouse to the desired second  
endpoint. At all times during this motion the program keeps displaying the line segment as it would look if the  
button were released at that moment. This rubber band keeps getting drawn and erased as it moves across other",algorithms and data structures.pdf
"button were released at that moment. This rubber band keeps getting drawn and erased as it moves across other  
objects on the screen. The user should study a key detail in the procedure 'DragLine' that prevents other objects  
from  being  erased  or  modified  as  they  collide  with  the  ever-refreshed  rubber  band:  We  temporarily  set 
'PenMode(patXor)'. We encourage you to experiment by modifying this procedure in two ways:
1. Change the first call of the 'procedure DrawLine(L.p 1, L.p2, black)' to 'DrawLine(L.p 1, L.p2, white)'. You will  
have turned the procedure 'DragLine' into an artful, if somewhat random, painting brush.
2. Remove the call 'PenMode(patXor)' (thus reestablishing the default 'pnMode = patCopy'), but leave the first 
'DrawLine(L.p1, L.p 2, white)', followed by the second 'DrawLine(L.p 1, L.p 2, black)'. You now have a naive  
rubber-band routine: It alternates erasing (draw 'white') and drawing (draw 'black') the current rubber",algorithms and data structures.pdf
"rubber-band routine: It alternates erasing (draw 'white') and drawing (draw 'black') the current rubber  
band, but in so doing it modifies other objects that share pixels with the rubber band. This is our first  
example of the use of the versatile exclusive-or; others will follow later in the book.
program Frame;
  { provides mouse input and drawing of points, line segments, 
circles }
type point = record  x, y: real  end;
lineSegment = record  p1, p2: point  { endpoints }  end;
19",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
var c, p: point;
r: real;  { radius of a circle }
L: lineSegment;
procedure WaitForClick;
begin  repeat until Button;  while Button do  end;
procedure GetPoint (var p: point);
var  v, h: integer;
begin
GetMouse(v, h);
p.x := v;  p.y := h  { convert integer to real }
end;
procedure DrawPoint(p: point; pat: Pattern);
const  t = 3;  { radius of a point }
begin
PenPat(pat);
PaintOval(round(p.y) – t, round(p.x) – t, round(p.y) + t, 
round(p.x) + t)
end;
procedure ClickPoint(var p: point);
begin  WaitForClick;  GetPoint(p);  DrawPoint(p, Black)  end;
function Dist(p, q: point): real;
begin  Dist := sqrt(sqr(p.x – q.x) + sqr(p.y – q.y))  end;
procedure DrawLine(p1, p2: point; pat: Pattern);
begin
PenPat(pat);
MoveTo(round(p1.x), round(p1.y));
LineTo(round(p2.x), round(p2.y))
end;
procedure DragLine(var L: lineSegment);
begin
repeat until Button;  GetPoint(L.p1);  L.p2 := L.p1; 
PenMode(patXor);
while Button do  begin",algorithms and data structures.pdf
"end;
procedure DragLine(var L: lineSegment);
begin
repeat until Button;  GetPoint(L.p1);  L.p2 := L.p1; 
PenMode(patXor);
while Button do  begin
DrawLine(L.p1, L.p2, black);
{ replace 'black' by 'white' above to get an artistic drawing 
tool }
GetPoint(L.p2);
DrawLine(L.p1, L.p2, black)
end;
PenMode(patCopy)
end;  { DragLine }
procedure DrawCircle(c: point; r: real; pat: Pattern);
begin
PenPat(pat);
FrameOval(round(c.y – r), round(c.x – r), round(c.y + r), 
round(c.x + r))
end;
procedure DragCircle(var c: point; var r: real);
var  p: point;
begin
repeat until Button;  GetPoint(c);  r := 0.0;  PenMode(patXor);
while Button do  begin
DrawCircle(c, r, black);
GetPoint(p);
Algorithms and Data Structures 20  A Global Text",algorithms and data structures.pdf
"2. Graphics primitives and environments
r := Dist(c, p);
DrawCircle(c, r, black);
end;
PenMode(patCopy)
end;  { DragCircle }
procedure Title;
begin
ShowText;  { make sure the text window and … }
ShowDrawing;  { … the graphics window show on the screen }
WriteLn('Frame program');
WriteLn('with simple graphics and interaction routines.');
WriteLn('Click to proceed.');
WaitForClick
end;  { Title }
procedure What;
begin
WriteLn('Click a point in the drawing window.');
ClickPoint(p);
WriteLn('Drag mouse to enter a line segment.');
DragLine(L);
WriteLn('Click center of a circle and drag its radius');
DragCircle(c, r)
end;  { What }
procedure Epilog;
begin  WriteLn('Bye.')  end;
begin  { Frame }
Title;  What;  Epilog
end.  { Frame }
Example of a graphics routine: polyline input
Let us illustrate the use of the frame  program above  in developing a new graphics procedure. We choose",algorithms and data structures.pdf
"Let us illustrate the use of the frame  program above  in developing a new graphics procedure. We choose  
interactive polyline input as an example. A polyline is a chain of directed straight-line segments—the starting point  
of the next segment coincides with the endpoint of the previous one. 'Polyline' is the most useful tool for interactive  
input of most drawings made up of straight lines. The user clicks a starting point, and each subsequent click  
extends the polyline by another line segment. A double click terminates the polyline.
We  developed  'PolyLine'  starting  from  the  frame  program  above,  in  particular  the  procedure  'DragLine', 
modifying  and  adding  a  few  procedures.  Once  'Polyline'  worked,  we  simplified  the  frame  program  a  bit.  For 
example,  the  original  frame  program  uses  reals  to  represent  coordinates  of  points,  because  most  geometric",algorithms and data structures.pdf
"example,  the  original  frame  program  uses  reals  to  represent  coordinates  of  points,  because  most  geometric 
computation is done that way. A polyline on a graphics screen only needs integers, so we changed the type 'point' to  
integer coordinates. At the moment, the code for polyline input is partly in the procedure 'NextLineSegment' and in  
the procedure 'What'. In the next iteration, it would probably be combined into a single self-contained procedure,  
with all the subprocedures it needs, and the frame program would be tossed out—it has served its purpose as a  
development tool.
program PolyLine;
{ enter a chain of line segments and compute total length }
{ stop on double click }
type point = record  x, y: integer;  end;
var stop: boolean;
length: real;
21",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
p, q: point;
function EqPoints (p, q: point): boolean;
begin  EqPoints := (p.x = q.x) and (p.y = q.y)  end;
function Dist (p, q: point): real;
begin  Dist := sqrt(sqr(p.x – q.x) + sqr(p.y – q.y))  end;
procedure DrawLine (p, q: point; c: Pattern);
begin  PenPat(c);  MoveTo(p.x, p.y);  LineTo(q.x, q.y)  end;
procedure WaitForClick;
begin  repeat until Button;  while Button do  end;
procedure NextLineSegment (var stp, endp: point);
begin
endp := stp;
repeat
DrawLine(stp, endp, black);  { Try 'white' to generate artful 
pictures! }
GetMouse(endp.x, endp.y);
DrawLine(stp, endp, black)
until Button;
while Button do
end;  { NextLineSegment }
procedure Title;
begin
ShowText;  ShowDrawing;
WriteLn('Click to start a polyline.');
WriteLn('Click to end each segment.');
WriteLn('Double click to stop.')
end;  { Title }
procedure What;
begin
WaitForClick;  GetMouse(p.x, p.y);
stop := false;  length := 0.0;
PenMode(patXor);",algorithms and data structures.pdf
"end;  { Title }
procedure What;
begin
WaitForClick;  GetMouse(p.x, p.y);
stop := false;  length := 0.0;
PenMode(patXor);
while  not stop  do  begin
NextLineSegment(p, q);
stop := EqPoints(p, q);  length := length + Dist(p, q);  p := q
end
end;  { What }
procedure Epilog;
begin  WriteLn('Length of polyline = ', length);  WriteLn('Bye.') 
end;
begin  { PolyLine }
Title;  What;  Epilog
end.  { PolyLine }
Programming projects
1. Implement a simple package of turtle graphics operations on top of the graphics environment available on  
your computer.
2. Use this package to implement and test a procedure 'circle' that meets the requirements listed at the end of  
the section “Turtle graphics: a basic environment”.
Algorithms and Data Structures 22  A Global Text",algorithms and data structures.pdf
"2. Graphics primitives and environments
3. Implement your personal graphics frame program as described in “A graphics frame program”. Your effort  
will pay off in time saved later, as you will be using this program throughout the entire course.
23",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
3. Algorithm animation
I hear and I forget, I see and I remember, I do and I understand.
A picture is worth a thousand words—the art of presenting information in visual form. 
Learning objectives:
• adding animation code to a program
• examples of algorithm snapshots 
Computer-driven visualization: characteristics and techniques
The computer-driven graphics screen is a powerful new communications medium; indeed, it is the only two-way  
mass communications medium we know. Other mass communications media–the printed e.g. recorded audio and  
video—are one-way streets suitable for delivering a monolog. The unique strength of our new medium is interactive  
presentation of information. Ideally, the viewer drives the presentation, not just by pushing a start button and  
turning  a  channel  selector,  but  controls  the  presentation  at  every  step.  He  controls  the  flow  not  only  with",algorithms and data structures.pdf
"turning  a  channel  selector,  but  controls  the  presentation  at  every  step.  He  controls  the  flow  not  only  with 
commands such as ""faster"", ""slower"", ""repeat"", ""skip"", ""play this backwards"", but more important, with a barrage of  
""what if?"" questions. What if the area of this triangle becomes zero? What if we double the load on this beam? What  
if world population grows a bit faster? This powerful new medium challenges us to use it well.
When using any medium, we must ask: What can it do well, and what does it do poorly? The computer-driven  
screen is ideally suited for rapid and accurate display of information that can be deduced from large amounts of  
data by means of straightforward algorithms and lengthy computation. It can do so in response to a variety of user  
inputs as long as this variety is contained in an algorithmically tractable, narrow domain of discourse. It is not",algorithms and data structures.pdf
"inputs as long as this variety is contained in an algorithmically tractable, narrow domain of discourse. It is not  
adept at tasks that require judgment, experience, or insight. By comparison, a speaker at the blackboard is slow and  
inaccurate and can only call upon small amounts of data and tiny computations; we hope she makes up for this  
technical shortcoming by good judgment, teaching experience, and insight into the subject. By way of another  
comparison, books and films may accurately and rapidly present results based on much data and computation, but  
they lack the ability to react to a user's input.
Algorithm  animation,  the  technique  of  displaying  the  state  of  programs  in  execution,  is  ideally  suited  for 
presentation on a graphics screen. There is a need for this type of computation, and there are techniques for  
producing them. The reasons for animating programs in execution fall into two major categories, which we label  
checking and exploring.",algorithms and data structures.pdf
"producing them. The reasons for animating programs in execution fall into two major categories, which we label  
checking and exploring.
Checking 
To understand an algorithm well, it is useful to understand it from several distinct points of view. One of them is  
the static point of view on which  correctness proofs are based: Formulate invariants on the data and show that  
these are preserved under the program's operations. This abstract approach appeals to our rational mind. A second,  
equally important point of view, is dynamic: Watch the algorithm go through its paces on a variety of input data.  
This concrete approach appeals to our intuition. Whereas the static  approach relies mainly on ""thinking"", the  
dynamic approach calls mostly for ""doing"" and ""perceiving"", and thus is a prime candidate for visual human-
Algorithms and Data Structures 24  A Global Text",algorithms and data structures.pdf
"3. Algorithm animation
computer interaction. In this use of algorithm animation, the user may be checking his understanding of the  
algorithm, or may be checking the algorithm's correctness—in principle, he could reason this out, but in practice, it  
is faster and safer to have the computer animation as a double check.
Exploring
In a growing number of applications, computer visualization cannot be replaced by any other technique. This is  
the case, for example, in exploratory data analysis, where a scientist may not know a priori what she is looking for,  
and  the  only  way  to look  at a  mass  of data  is  to  generate  pictures  from  it  (see  a  special  issue  on  scientific 
visualization [Nie 89]). At times static pictures will do, but in simulations ( e.g. of the onset of turbulent flow) we  
prefer to see an animation over time.
Turning to the  techniques  of animation, computer technology is in the midst of extremely  rapid  evolution",algorithms and data structures.pdf
"prefer to see an animation over time.
Turning to the  techniques  of animation, computer technology is in the midst of extremely  rapid  evolution  
toward ever-higher-quality  interactive image generation on powerful graphics  workstations (see  [RN 91] for a  
survey of the state of the art). Fortunately, animating algorithms such as those presented in this book can be done  
adequately with the graphics tools available on low-cost workstations. These algorithms operate on discrete data  
configurations (such as matrices, trees, graphs), and use standard data structures, such as arrays and lists. For such  
limited classes of algorithms, there are software packages that help produce animations based on specifications,  
with a minimum of extra programming required. An example of an algorithm animation environment is the BALSA  
system [Bro 88, BS 85]. A more recent example is the XYZ GeoBench, which animates geometric  algorithms  
[NSDAB 91].",algorithms and data structures.pdf
"system [Bro 88, BS 85]. A more recent example is the XYZ GeoBench, which animates geometric  algorithms  
[NSDAB 91].
In our experience, the bottleneck of algorithm animation is not the extra code required, but graphic design.  
What do you want to show, and how do you display it, keeping in mind the limitations of the system you have to  
work with? The key point to consider is that data does not look like anything until we have defined a mapping from  
the data space into visual space. Defining such a mapping ranges from trivial to practically impossible.
1. For  some  kinds  of  data,  such  as  geometric  data  in  two-  and  three-dimensional  space,  or  real-valued 
functions of one or two real variables, there are natural mappings that we learned in school. These help us  
greatly in getting a feel for the data.
2. Multidimensional data (dimension ≥ 3) can be displayed on a two-dimensional screen using a number of",algorithms and data structures.pdf
"greatly in getting a feel for the data.
2. Multidimensional data (dimension ≥ 3) can be displayed on a two-dimensional screen using a number of  
straight forward techniques, such as projections into a subspace, or using color or gray level as a fourth  
dimension. But our power of perception diminishes rapidly with increasing dimensionality.
3. For discrete combinatorial data there is often no natural or accepted visual representation. As an example,  
we often draw a graph by mapping nodes into points and edges into lines. This representation is natural for  
graphs that are embedded in Euclidean space, such as a road network, and we can readily make sense of a  
map with thousands of cities and road links. When we extend it to arbitrary graphs by placing a node  
anywhere on the screen, on the other hand, we get a random crisscrossing of lines of little intuitive value.
In addition to such inherent problems of visual representation, practical difficulties of the most varied type",algorithms and data structures.pdf
"In addition to such inherent problems of visual representation, practical difficulties of the most varied type  
abound. Examples:
• Some screens are awfully small, and some data sets are awfully large for display even on the largest screens.
• An animation has to run within a narrow speed range. If it is too fast, we fail to follow, or the screen may 
flicker disturbingly; if too slow, we may lack the time to observe it.
25",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
In conclusion, we hold that it is not too difficult to animate simple algorithms as discussed here by interspersing  
drawing statements into the normal code. Independent of the algorithm to be animated, you can call on your own  
collection of display and interaction procedures that you have built up in your frame program (in the section ""A  
graphics frame program). But designing an adequate graphic representation is hard and requires a creative effort  
for each algorithm—that is where animators/programmers will spend the bulk of their effort. More on this topic in  
[NVH 86].
Example: the convex hull of points in the plane
The following program is an illustrative example for algorithm animation. 'ConvexHull' animates an  on-line 
algorithm that constructs half the convex hull (say, the upper half) of a set of points presented incrementally. It",algorithms and data structures.pdf
"algorithm that constructs half the convex hull (say, the upper half) of a set of points presented incrementally. It  
accepts one point at a time, which must lie to the right of all preceding ones, and immediately extends the convex  
hull. The algorithm is explained in detail in “sample problems and algorithms”.
program ConvexHull;  { of n ≤ 20 points in two dimensions }
const nmax = 19;  { max number of points }
r = 3; { radius of point plot }
var x, y, dx, dy: array[0 .. nmax] of integer;
b: array[0 .. nmax] of integer;  { backpointer }
n: integer;  { number of points entered so far }
px, py: integer;  { new point }
procedure PointZero;
begin
n := 0;
x[0] := 5;  y[0] := 20;  { the first point at fixed location }
dx[0] := 0;  dy[0] := 1;  { assume vertical tangent }
b[0] := 0;  { points back to itself }
PaintOval(y[0] – r, x[0] – r, y[0] + r, x[0] + r)
end;
function NextRight: boolean;
begin
if  n ≥ nmax  then
NextRight := false
else  begin
repeat until Button;",algorithms and data structures.pdf
"end;
function NextRight: boolean;
begin
if  n ≥ nmax  then
NextRight := false
else  begin
repeat until Button;
while Button do  GetMouse(px, py);
if  px ≤ x[n]  then
NextRight := false
else  begin
PaintOval(py – r, px – r, py + r, px + r);
n := n + 1;  x[n] := px;  y[n] := py;
dx[n] := x[n] – x[n – 1]; { dx > 0 } dy[n] := y[n] – y[n –1];
b[n] := n – 1;
MoveTo(px, py);  Line(–dx[n], –dy[n]);  NextRight := true
end
end
end;
procedure ComputeTangent;
var i: integer;
begin
i := b[n];
while  dy[n] · dx[i] > dy[i] · dx[n]  do  begin  { dy[n]/dx[n] > 
dy[i]/dx[i] }
Algorithms and Data Structures 26  A Global Text",algorithms and data structures.pdf
"3. Algorithm animation
i := b[i];
dx[n] := x[n] – x[i];  dy[n] := y[n] – y[i];
MoveTo(px, py);  Line(–dx[n], –dy[n]);
b[n] := i
end;
MoveTo(px, py);  PenSize(2, 2);  Line(–dx[n], –dy[n]);  PenNormal
end;
procedure Title;
begin
ShowText;  ShowDrawing;  { make sure windows lie on top }
WriteLn('The convex hull');
WriteLn('of n points in the plane sorted by x-coordinate');
WriteLn('is computed in linear time.');
Write('Click next point to the right, or Click left to quit.')
end;
begin  { ConvexHull }
Title;  PointZero;
while  NextRight  do  ComputeTangent;
Write('That's it!')
end.
A gallery of algorithm snapshots
The screen dumps shown in Exhibit 3.1 were taken from demonstration programs that we use to illustrate topics  
discussed in class. Although snapshots cannot convey the information and the impact of animations, they may give  
the reader  ideas  to try  out. We select two standard algorithm  animation  topics  (sorting  and  random  number",algorithms and data structures.pdf
"the reader  ideas  to try  out. We select two standard algorithm  animation  topics  (sorting  and  random  number 
generation), and an example showing the effect of cumulative rounding errors.
Exhibit 3.1: Initial configuration of data, …
27",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 3.1: … and snapshots from two sorting algorithms.
Visual test for randomness
Our visual system is amazingly powerful at detecting patterns of certain kinds in the midst of noise.  Random 
number generators (RNGs) are intended to simulate ""noise"" by means of simple formulas. When patterns appear in  
the visual representation of supposedly random numbers, chances are that this RNG will also fail more rigorous  
statistical tests. The eyes' pattern detection ability serves well to disqualify a faulty RNG but cannot certify one as  
adequate.  Exhibit 3.2  shows a simulation of the Galton board. In theory, the resulting density diagram should  
approximate a bellshaped Gaussian distribution. Obviously, the RNG used falls short of expectations. 
Exhibit 3.2: One look suffices to unmask a bad RNG.
Numerics of chaos, or chaos of numerical computation?",algorithms and data structures.pdf
"Exhibit 3.2: One look suffices to unmask a bad RNG.
Numerics of chaos, or chaos of numerical computation?
The following example shows the effect of rounding errors and precision in linear recurrence relations. The d-
step linear recurrence with constant coefficients in the domain of real or complex numbers,
Algorithms and Data Structures 28  A Global Text",algorithms and data structures.pdf
"3. Algorithm animation
is one of the most frequent formulas evaluated in scientific and technical computation (e.g. for the solution of  
differential equations). By proper choice of the constants c i and of initial values z 0, z 1, … , z d–1 we can generate  
sequences zk that when plotted in the plane of complex numbers form many different figures. With d= 1 and |χ 1|= 1, 
for example, we generate circles. The pictures in  Exhibit 3.3  were all generated with d = 3 and conditions that  
determine a curve that is most easily described as a circle 3 running around the perimeter of another circle 2 that  
runs around a stationary circle 1. We performed this computation with a floating-point package that lets us pick  
precision P (i.e. the number of bits in the mantissa). The resulting pictures look a bit chaotic, with a behavior we  
have come to associate with fractals—even if the mathematics of generating them is completely different, and linear",algorithms and data structures.pdf
"have come to associate with fractals—even if the mathematics of generating them is completely different, and linear  
recurrences computed without error would look much more regular. Notice that the first two images are generated  
by the same formula, with a single bit of difference in the precision used. The whim of this 1-bit difference in  
precision changes the image entirely.
29",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Algorithms and Data Structures 30  A Global Text",algorithms and data structures.pdf
"3. Algorithm animation
Exhibit 3.3: The effect of rounding errors in linear recurrence relations.
Programming projects
1. Use  your  personal  graphics  frame  program  (the  programming  project  of  “graphics  primitives  and 
environments”) to implement and animate the convex hull algorithm example.
31",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
2. Use  your  graphics  frame  program  to  implement  and  animate  the  behavior  of  recurrence  relations  as 
discussed in the section “A gallery of algorithm snapshots”.
3. Extend your graphics frame program with a set of dialog control operations sufficient to guide the user  
through the various steps of the animation of recurrence relations: in particular, to give him the options, at  
any time, to enter a new set of parameters, then execute the algorithm and animate it in either 'movie  
mode' (it runs at a predetermined speed until stopped by the user), or 'step mode' [the display changes only  
when the user enters a logical command 'next' (e.g. by clicking the mouse or hitting a specific key)].
Algorithms and Data Structures 32  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Part II: Programming 
concepts: beyond notation
Thoughts on the role of programming notations
A programming language is the main interface between a programmer and the physical machine, and a novice  
programmer will tend to identify ""programming"" with ""programming in the particular language she has learned"".  
The  realization  that there  is  much  to programming  ""beyond notation"" (i.e.  principles  that transcend  any  one 
language) is a big step forward in a programmer's development.
Part II aims to help the reader take this step forward. We present examples that are best understood by focusing  
on abstract principles of algorithm design, and only later do we grope for suitable notations to turn this principle  
into an algorithm expressed in sufficient detail to become executable. In keeping with our predilection for graphic",algorithms and data structures.pdf
"into an algorithm expressed in sufficient detail to become executable. In keeping with our predilection for graphic  
communication, the first informal expression of an algorithmic idea is often pictorial. We show by example how  
such representations, although they may be incomplete, can be turned into programs in a formal notation.
The  literature  on  programming  and  languages.  There  are  many  books  that  present  principles  of 
programming and of programming languages from a higher level of abstraction. The principles highlighted differ  
from  author  to  author,  ranging  from  intuitive  understanding  to  complete  formality.  The  following  textbooks 
provide an excellent sample from the broad spectrum of approaches: [ASS 84], [ASU 86], [Ben 82], [Ben 85], [Ben  
88], [Dij 76], [DF 88], [Gri 81], and [Mey 90].
Algorithms and Data Structures 33  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
4. Algorithms and programs 
as literature: substance and 
form
Learning objectives:
• programming in the large versus programming in the small
• large flat programs versus small deep programs
• programs as literature
• fractal pictures: snowflakes and Hilbert's space-filling curve
• recursive definition of fractals by production or rewrite rules 
• Pascal and programming notations
Programming in the large versus programming in the small
In studying and discussing the art of programming it is useful to distinguish between large programs and small  
programs, since these two types impose fundamentally different demands on the programmer.
Programming in the large
Large  programs  (e.g.  operating  systems,  database  systems,  compilers,  application  packages)  tax  our  
organizational  ability.  The  most  important  issues  to  be  dealt  with  include  requirements  analysis,  functional",algorithms and data structures.pdf
"organizational  ability.  The  most  important  issues  to  be  dealt  with  include  requirements  analysis,  functional 
specification, compatibility with other systems, how to break a large program into modules of manageable size,  
documentation, adaptability to new systems and new requirements, how to organize the team of programmers, and  
how to test the software. These issues are the staple of software engineering. When compared to the daunting  
managerial and design challenges, the task of actual coding is relatively simple. Large programs are often flat: Most 
of the listing consists of comments, interface specifications, definitions, declarations, initializations, and a lot of  
code that is executed only rarely. Although the function of any single page of source code may be rather trivial when  
considered by itself, it is difficult to understand the entire program, as you need a lot of information to understand",algorithms and data structures.pdf
"considered by itself, it is difficult to understand the entire program, as you need a lot of information to understand  
how this page relates to the whole. The classic book on programming in the large is [Bro 75].
Programming in the small
Small  programs, of  the  kind  discussed  in  this  book, challenge  our  technical  know-how  and  inventiveness. 
Algorithmic  issues dominate  the  programmer's  thinking:  Among  several  algorithms  that  all  solve  the  same 
problem, which is the most efficient under the given circumstances? How much time and space does it take? What  
data structures do we use? In contrast to large programs, small programs are usually  deep, consisting of short,  
compact code many of whose statements are executed very often. Understanding a small program may also be  
difficult, at least initially, since the chain of thought is often subtle. Once you understand it thoroughly, you can",algorithms and data structures.pdf
"difficult, at least initially, since the chain of thought is often subtle. Once you understand it thoroughly, you can  
reproduce it at any time with much less effort than was first required. Mastery of interesting small programs is the  
Algorithms and Data Structures 34  A Global Text",algorithms and data structures.pdf
"4. Algorithms and programs as literature: substance and form
best way to get started in computer science. We encourage the reader to work out all the details of the examples we  
present.
This book is concerned only with programming in the small. This decision determines our choice of  
topics to be presented, our style of presentation, and the notation we use to express programs, explanations, and  
proofs, and heavily influences our comments on techniques of programming. Our style of presentation appeals to  
the reader's intuition more than to formal rigor. We aim at highlighting the key idea of any argument that we make  
rather than belaboring the details. We take the liberty of using a free notation that suits the purpose of any specific  
argument we wish to make, trusting that the reader understands our small programs so well that he can translate  
them into the programming language of his choice. In a nut shell, we emphasize substance over form.",algorithms and data structures.pdf
"them into the programming language of his choice. In a nut shell, we emphasize substance over form.
The purpose of Part II is to help engender a fluency in using different notations. We provide yet other examples  
of unconventional notations that match the nature of the problem they are intended to describe, and we show how  
to translate  them into Pascal-like programs. Since much of the difference between programming languages is  
merely syntactic, we include two chapters that cover the basics of syntax and syntax analysis. These topics are  
important in their own right; we present them early in the hope  that they will help the student see through  
differences of notation that are merely ""syntactic sugar"".
Documentation versus literature: is it meant to be read?
It is instructive to distinguish two types of written materials, and two corresponding types of writing tasks:",algorithms and data structures.pdf
"It is instructive to distinguish two types of written materials, and two corresponding types of writing tasks:  
documents and literature.  Documents are constrained by requirements of many kinds, are read when a specific  
need arises (rarely for pleasure), and their quality is judged by criteria such as formality, conformity to a standard,  
completeness, accuracy, and consistency. Literature is a form of art free from conventions, read for education or  
entertainment, and its quality is judged by aesthetic criteria much harder to enumerate than the ones above. The  
touchstone is the question: Is it meant to be read? If the answer is ""only if necessary"", then it's a document, not  
literature.
As the name implies, the documentation of large programs is a typical document-writing chore. Much has been  
written in software engineering about documentation, a topic whose importance grows with the size and complexity",algorithms and data structures.pdf
"written in software engineering about documentation, a topic whose importance grows with the size and complexity  
of the system to be documented. We hold that small programs are not documented, they are explained. As such,  
they are literature, or ought to be. The idea of programs as literature is widely held (see, e.g. [Knu 84]). The key idea 
is that an algorithm or program is part of the text and melts into the text in the same way as a paragraph, a formula,  
or a picture does. There are also formal notations and systems designed to support a style of programming that  
integrates text and code to form a package that is both readable for humans and executable by machines [Knu 83].
Whatever notation is used for literate programming, it has to describe all phases of a program's evolution, from  
idea to specification to algorithm to program. Details of a good program cannot be understood, or at least not",algorithms and data structures.pdf
"idea to specification to algorithm to program. Details of a good program cannot be understood, or at least not  
appreciated, without an awareness of the grand design that guided the programmer. Whereas details are usually  
well expressed in some formal notation, grand designs are not. For this reason we renounce formality and attempt  
to convey ideas in whatever notation suits our purpose of insightful explanation. Let us illustrate this philosophy  
with some examples.
35",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
A snowflake
Fractal pictures are intuitively characterized by the requirement that any part of the picture, of any size, when  
sufficiently magnified, looks like the whole picture. Two pieces of information are required to define a specific  
fractal:
1. A picture primitive that serves as a building-block: Many copies of this primitive, scaled to many different  
sizes, are composed to generate the picture.
2. A recursive rule that defines the relative position of the primitives of different size.
A picture primitive is surely best defined by a drawing, and the manner of composing primitives in space again  
calls for a pictorial representation, perhaps augmented by a verbal explanation. In this style we define the fractal  
'Snowflake' by the following production rule, which we read as follows: A line segment, as shown on the left-hand",algorithms and data structures.pdf
"'Snowflake' by the following production rule, which we read as follows: A line segment, as shown on the left-hand  
side, must be replaced by a polyline, a chain of four shorter segments, as shown at the right-hand side ( Exhibit 4.1). 
We start with an initial configuration (the zero-generation) consisting of a single segment ( Exhibit 4.2). If we apply 
the production rule just once to every segment of the current generation, we obtain successively a first, second, and  
third generation, as shown in Exhibit 4.3. Further generations quickly exhaust the resolution of a graphics screen or 
the  printed  page,  so  we  stop  drawing  them.  The  curve  obtained  as  the  limit  when  this  process  is  continued 
indefinitely is a fractal. Although we cannot draw it exactly, one can study it as a mathematical object and prove  
theorems about it.
Exhibit 4.1: Production for replacing a straight-line segment by a polyline
Exhibit 4.2: The simplest initial configuration",algorithms and data structures.pdf
"theorems about it.
Exhibit 4.1: Production for replacing a straight-line segment by a polyline
Exhibit 4.2: The simplest initial configuration
Exhibit 4.3: The first three generations
The production rule drawn above is the essence of this fractal, and of the sequence of pictures that lead up to it.  
The initial configuration, on the other hand, is quite arbitrary: If we had started with a regular hexagon, rather than  
a single line segment, the pictures obtained would really have lived up to their name, snowflake. Any other initial  
configuration still generates curves with the unmistakable pattern of snowflakes, as the reader is encouraged to  
verify.
After  having familiarized  ourselves  with  the  objects  described,  let  us  turn  our  attention  to the  method  of 
description and raise three questions about the formality and executability of such notations.
1. Is our notation sufficiently formal to serve as a program for a computer to draw the family of generations of",algorithms and data structures.pdf
"1. Is our notation sufficiently formal to serve as a program for a computer to draw the family of generations of  
snowflakes? Certainly not, as we stated certain rules in colloquial language and left others completely  
unsaid, implying them only by sample drawings. As an example of the latter, consider the question: If a  
Algorithms and Data Structures 36  A Global Text",algorithms and data structures.pdf
"4. Algorithms and programs as literature: substance and form
segment is to be replaced by a ""plain with a mountain in the center"", on which side of the segment should  
the peak point? The drawings above suggest that all peaks stick out on the same side of the curve, the  
outside.
2. Could our method of description be extended and formalized to serve as a programming language for  
fractals? Of course. As an example, the production shown in  Exhibit 4.4  specifies the side on which the  
peak is to point. Every segment now has a + side and a – side. The production above states that the new  
peak is to grow over the + side of the original segment and specifies the + sides and – sides of each of the  
four  new  segments.  For  every  other  aspect  that  our  description  may  have  left  unspecified,  such  as 
placement on the screen, some notation could readily be designed to specify every detail with complete",algorithms and data structures.pdf
"placement on the screen, some notation could readily be designed to specify every detail with complete  
rigor. In “Syntax” and “Syntax analysis” we introduce some of the basic techniques for designing and using  
formal notations.
Exhibit 4.4: Refining the description to specify a ""left-right"" orientation.
3. Should we formalize this method of description and turn it into a machine-executable notation? It depends  
on the purpose for which we plan to use it. Often in this book we present just one or a few examples that  
share a common design. Our goal is for the reader to understand these few examples, not to practice the  
design of artificial programming languages. To avoid being sidetracked by a pedantic insistence on rigorous  
notation, with its inevitable overhead of introducing formalisms needed to define all details, we prefer to  
stop  when  we have  given  enough  information  for  an  attentive  reader  to grasp  the  main  idea  of  each 
example.",algorithms and data structures.pdf
"stop  when  we have  given  enough  information  for  an  attentive  reader  to grasp  the  main  idea  of  each 
example.
Hilbert's space-filling curve
Space-filling curves have been an object of mathematical curiosity since the nineteenth century, as they can be  
used to prove that the cardinality of an interval, considered as a set of points, equals the cardinality of a square (or  
any other finite two-dimensional region). The term  space-filling describes the surprising fact that such a curve  
visits every point within a square. In mathematics, space-filling curves are constructed as the limit to which an  
infinite  sequence  of curves  Ci converges.  On  a  discretized  plane, such  as  a raster-scanned  screen, no limiting 
process is needed, and typically one of the first dozen curves in the sequence already paints every pixel, so the term  
space-filling is quickly seen to be appropriate.",algorithms and data structures.pdf
"space-filling is quickly seen to be appropriate.
Let us illustrate this phenomenon using Hilbert's space-filling curve (David Hilbert, 1862–1943), whose first six  
approximations are shown in Exhibit 4.5. As the pictures suggest, Hilbert curves are best-described recursively, but  
the composition rule is more complicated than the one for snowflakes. We propose the two productions shown in  
Exhibit 4.6 to capture the essence of Hilbert (and similar) curves. This pictorial program requires explanation, but  
we hope the reader who has once understood it will find this notation useful for inventing fractals of her own. As  
always, a production is read: ""To obtain an instance of the left-handside, get instances of all the things listed on the  
right-handside"", or equivalently, ""to do the task specified by the left-hand side, do all the tasks listed on the right-
hand side"".
37",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 4.5: Six generations of the family of Hilbert curves
Exhibit 4.6: Productions for painting a square in terms of its quadrants
The left-hand side of the first production stands for the task: paint a square of given size, assuming that you  
enter at the lower left corner facing in the direction indicated by the arrow and must leave in the upper left corner,  
again facing in the direction indicated by that arrow. We assume turtle graphics primitives, where the state of the  
brush is given by a position and a direction. The hatching indicates the area to be painted. It lies to the right of the  
line that connects entry and exit corners, which we read as ""paint with your right hand"", and the hatching is in thick  
strokes. The left-hand side of the second production is similar: Paint a square ""with your left hand"" (hatching is in  
thin strokes), entering and exiting as indicated by the arrows.",algorithms and data structures.pdf
"thin strokes), entering and exiting as indicated by the arrows.
The right-hand sides of the productions are now easily explained. They say that in order to paint a square you  
must paint each of its quadrants, in the order indicated. They give explicit instructions on where to enter and exit,  
Algorithms and Data Structures 38  A Global Text",algorithms and data structures.pdf
"4. Algorithms and programs as literature: substance and form
what direction to face, and whether you are painting with your right or left hand. The last detail is to make sure that  
when the brush exits from one quadrant it gets into the correct state for entering the next. This requires the brush  
to turn by 90˚, either left or right, as the curved arrows in the pictures indicate. In the continuous plane we imagine  
the brush to ""turn on its heels"", whereas on a discrete grid it also moves to the first grid point of the adjacent  
quadrant.
These  productions  omit any rule  for  termination,  thus  simulating the limiting  process  of true  space-filling 
curves. To draw anything on the screen we need to add some termination rules that specify two things: (1) when to  
invoke the termination rule (e.g.  at some fixed depth of recursion), and (2) how to paint the square that invokes the",algorithms and data structures.pdf
"invoke the termination rule (e.g.  at some fixed depth of recursion), and (2) how to paint the square that invokes the  
termination rule (e.g. paint it all black). As was the case with snowflakes and with all fractals, the primitive pictures  
are much less important than the composition rule, so we omit it.
The  following  program  implements  a  specific  version  of  the  two  pictorial  productions  shown  above.  The 
procedure 'Walk' implements the curved arrows in the productions: the brush turns by 'halfTurn', takes a step of  
length s, and turns again by 'halfTurn'. The parameter 'halfTurn' is introduced to show the effect of cumulative  
small errors in recursive procedures. 'halfTurn = 45' causes the brush to make right-angle turns and yields Hilbert  
curves. The reader is encouraged to experiment with 'halfTurn = 43, 44, 46, 47', and other values.
program PaintAndWalk;
const pi = 3.14159;  s = 3;  { step size of walk }",algorithms and data structures.pdf
"program PaintAndWalk;
const pi = 3.14159;  s = 3;  { step size of walk }
var turtleHeading: real;  { counterclockwise, radians }
halfTurn, depth: integer;  { recursive depth of painting }
procedure TurtleTurn(angle: real);
{ turn the turtle angle degrees counterclockwise }
begin  { angle is converted to radian before adding }
turtleHeading := turtleHeading + angle · pi / 180.0
end;  { TurtleTurn }
procedure TurtleLine(dist: real);
{ draws a straight line, dist units long }
begin
Line(round(dist · cos(turtleHeading)), round(–dist·sin(turtle 
Heading)))  
end;  { TurtleLine }
procedure Walk (halfTurn: integer);
begin  TurtleTurn(halfTurn);  TurtleLine(s);  TurtleTurn(halfTurn) 
end;
procedure Qpaint (level: integer; halfTurn: integer);
begin
if  level = 0  then
TurtleTurn(2 · halfTurn)
else  begin
Qpaint(level – 1, –halfTurn);
Walk(halfTurn);
Qpaint(level – 1, halfTurn);
Walk(–halfTurn);
Qpaint(level – 1, halfTurn);
Walk(halfTurn);
Qpaint(level – 1, –halfTurn)
end
end;  { Qpaint }",algorithms and data structures.pdf
"Qpaint(level – 1, halfTurn);
Walk(–halfTurn);
Qpaint(level – 1, halfTurn);
Walk(halfTurn);
Qpaint(level – 1, –halfTurn)
end
end;  { Qpaint }
begin  { PaintAndWalk }
39",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
ShowText;  ShowDrawing;
MoveTo(100, 100);  turtleHeading := 0;  { initialize turtle 
state }
WriteLn('Enter halfTurn 0 .. 359  (45 for Hilbert curves): ');
ReadLn(halfTurn);
TurtleTurn(–halfTurn);  { init turtle turning angle }
Write('Enter depth 1 .. 6: ');  ReadLn(depth);
Qpaint(depth, halfTurn)
end.  { PaintAndWalk }
As a summary of this discourse on notation, we point to the fact that an executable program necessarily has to  
specify many details that are irrelevant from the point of view of human understanding. This book assumes that the  
reader has learned the basic steps of programming, of thinking up such details, and being able to express them  
formally in a programming language. Compare the verbosity of the one-page program above with the clarity and  
conciseness of the two pictorial productions above. The latter state the essentials of the recursive construction, and",algorithms and data structures.pdf
"conciseness of the two pictorial productions above. The latter state the essentials of the recursive construction, and  
no more, in a manner that a human can understand ""at a glance"". We aim our notation to appeal to a human mind,  
not necessarily to a computer, and choose our notation accordingly.
Pascal and its dialects: lingua franca of computer science
Lingua franca (1619):
1.  A common language that consists of Italian mixed with French, Spanish, Greek and Arabic and is spoken  
in Mediterranean ports
2. Any of various languages used as common or commercial tongues among peoples of diverse speech
3. Something resembling a common language
                                          (From Webster's Collegiate Dictionary)
Pascal as representative of today's programming languages
The definition above fits Pascal well: In the mainstream of the development of programming languages for a",algorithms and data structures.pdf
"The definition above fits Pascal well: In the mainstream of the development of programming languages for a  
couple of decades, Pascal embodies, in a simple design, some of the most important language features that became  
commonly accepted in the 1970s. This simplicity, combined with Pascal's preference for language features that are  
now  well  understood,  makes  Pascal  a  widely  understood  programming  notation.  A  few  highlights  in  the 
development of programming languages may explain how Pascal got to be a lingua franca of computer science.
Fortran  emerged  in 1954 as  the first high-level programming language to gain  acceptance and became  the 
programming language of the 1950s and early 1960s. Its appearance generated great activity in language design,  
and suddenly, around 1960, dozens of programming languages emerged. Three among these, Algol 60, COBOL, and 
Lisp, became milestones in the development of programming languages, each in its own way. Whereas COBOL",algorithms and data structures.pdf
"Lisp, became milestones in the development of programming languages, each in its own way. Whereas COBOL  
became the most widely used language of the 1960s and 1970s, and Lisp perhaps the most innovative, Algol 60  
became the most influential in several respects: it set new standards of rigor for the definition and description of a  
language, it pioneered  hierarchical block structure as the major  technique for organizing  large programs, and  
through these major technical contributions became the first of a family of mainstream programming languages  
that includes PL/1, Algol 68, Pascal, Modula-2, and Ada.
The decade of the 1960s remained one of great ferment and productivity in the field of programming languages.  
PL/1 and Algol 68, two ambitious projects that attempted to integrate many recent advances in programming  
language technology and theory, captured the lion's share of attention for several years. Pascal, a much smaller",algorithms and data structures.pdf
"language technology and theory, captured the lion's share of attention for several years. Pascal, a much smaller  
Algorithms and Data Structures 40  A Global Text",algorithms and data structures.pdf
"4. Algorithms and programs as literature: substance and form
project and language designed by Niklaus Wirth during the 1960s, ended up eclipsing both of these major efforts.  
Pascal took the best of Algol 60, in streamlined form, and added just one major extension, the then novel type  
definitions [Hoa 72]. This  lightweight edifice made it possible to implement efficient Pascal compilers on the  
microcomputers  that  mushroomed  during  the  mid  1970s  (e.g.  UCSD  Pascal),  which  opened  the  doors  to 
universities and high schools. Thus Pascal became the programming language most widely used in introductory  
computer science education, and every computer science student must be fluent in it.
Because  Pascal is so widely understood, we base  our programming notation on it but do  not adhere  to it  
slavishly. Pascal is more than 20 years old, and many of its key ideas are 30 years old. With today's insights into",algorithms and data structures.pdf
"slavishly. Pascal is more than 20 years old, and many of its key ideas are 30 years old. With today's insights into  
programming languages, many details would probably be chosen differently. Indeed, there are many ""dialects"" of  
Pascal, which typically extend the standard defined in 1969 [Wir 71] in different directions. One extension relevant  
for a publication language is that with today's hardware that supports large character sets and many different fonts  
and styles, a greater variety of symbols can be used to make the source more readable. The following examples  
introduce some of the conventions that we use often.
""Syntactic sugar"": the look of programming notations
Pascal statements lack an explicit terminator. This makes the frequent use of begin-end brackets necessary, as in  
the following program fragment, which implements the insertion sort algorithm (see chapter 17 and the section",algorithms and data structures.pdf
"the following program fragment, which implements the insertion sort algorithm (see chapter 17 and the section  
""Simple sorting algorithms that work in time""); –∞ denotes a constant ≤ any key value:
A[0] := –∞;
for i := 2 to n do  begin
j := i;
while  A[j] < A[j – 1]  do
begin  t := A[j];  A[j] := A[j – 1];  A[j – 1] := t;  j := j – 1  end;
end;
We aim at brevity and readability but wish to retain the flavor of Pascal to the extent that any new notation we  
introduce can be translated routinely into standard Pascal. Thus we write the statements above as follows:
A[0] := –∞;
for i := 2 to n do  begin
j := i;  { comments appear in italics }
while  A[j] < A[j – 1]  do  { A[j] :=: A[j – 1];  j := j – 1 }
{ braces serve as general-purpose brackets, including begin-end }
{ :=: denotes the exchange operator }
end;
Borrowing heavily from standard mathematical notation, we use conventional mathematical signs to denote",algorithms and data structures.pdf
"{ :=: denotes the exchange operator }
end;
Borrowing heavily from standard mathematical notation, we use conventional mathematical signs to denote  
operators whose Pascal designation was constrained by the small character sets typical of the early days, such as:
≠ ≤ ≥ ≠ ¬ ∧ ∨ ∈ ∉ ∩ ∪ \ |x| instead of 
<> <= >= <> not and or in not in · + – abs(x) respectively
We also use signs that may have no direct counterpart in Pascal, such 
as:
⊃ ⊇ ⊄ ⊂ ⊆ Set-theoretic relations
∞ Infinity, often used for a 
""sentinel"" (i.e. a number 
larger than all numbers to 
be processed in a given 
41",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
application)
± Plus-or-minus, used to 
define an interval [of 
uncertainty]
∑∏ Sum and product
            x Ceiling of a real number x 
(i.e. the smallest integer 
≥ x)
x Floor of a real number x (i.e. the 
largest integer ≤ x)
√ Square root
log Logarithm to the base 2
ln Natural logarithm, to the base e
iff If and only if
Although we may take a cavalier attitude toward notational differences, and readily use concise notations such  
as ∧ ∨ for the more verbose 'and', 'or', we will try to remind readers explicitly about our assumptions when there is a  
question about semantics. As an example, we assume that the boolean operators ∧ and ∨ are conditional, also called 
'cand' and 'cor': An expression containing these operators is evaluated from left to right, and the evaluation stops as  
soon as the result is known. In the expression x ∧ y, for example, x is evaluated first. If x evaluates to 'false', the",algorithms and data structures.pdf
"soon as the result is known. In the expression x ∧ y, for example, x is evaluated first. If x evaluates to 'false', the  
entire expression is 'false' without y ever being evaluated. This convention makes it possible to leave y undefined  
when x is 'false'. Only if x evaluates to 'true' do we proceed to evaluate y. An analogous convention applies to x ∨ y.
Program structure
Whereas the concise notations introduced above to denote operators can be translated almost one-to-one into a  
single line of standard Pascal, we also introduce a few extensions that may affect the program structure. In our view  
these changes make programs more elegant and easier to understand. Borrowing from many modern languages, we  
introduce a 'return()' statement to exit from procedures and functions and to return the value computed by a  
function.
Example
function gcd(u, v: integer): integer;
{ computes the greatest common divisor (gcd) of u and v }",algorithms and data structures.pdf
"function.
Example
function gcd(u, v: integer): integer;
{ computes the greatest common divisor (gcd) of u and v }
begin  if  v = 0  then  return(u)  else  return(gcd(v, u mod v)) 
end;
In this example, 'return()' merely replaces the Pascal assignments 'gcd := u' and 'gcd := gcd(v, u mod v)'. The  
latter  in  particular  illustrates  how  'return()' avoids  a  notational  blemish  in  Pascal:  On  the  left  of  the  second 
assignment, 'gcd' denotes a variable, on the right a function. 'Return()' also has the more drastic consequence that it  
causes control to exit from the surrounding procedure or function as soon as it is executed. Without entering into a  
controversy over the general advantages and disadvantages of this ""flow of control"" mechanism, let us present one  
example, typical of many search procedures, where 'return()' greatly simplifies coding. The point is that a search  
Algorithms and Data Structures 42  A Global Text",algorithms and data structures.pdf
"4. Algorithms and programs as literature: substance and form
routine terminates in one of (at least) two different ways: successfully, by having found the item in question, or  
unsuccessfully, because of a number of reasons (the item is not present, and some index is about to fall outside the  
range of a table; we cannot insert an item because the table is full, or we cannot pop a stack because it is empty,  
etc.). For the sake of efficiency as well as readability we prefer to exit from the routine as soon as a case has been  
identified and dealt with, as the following example from “Address computation:” illustrates:
function insert-into-hash-table(x: key): addr; 
var  a: addr;
begin
a := h(x);  { locate the home address of the item x to be 
inserted }
while  T[a] ≠ empty  do  begin
{ skipping over cells that are already occupied }
if  T[a] = x  then  return(a);  { x is already present; return 
its address }
a := (a + 1) mod m  { keep searching at the next address }
end;",algorithms and data structures.pdf
"if  T[a] = x  then  return(a);  { x is already present; return 
its address }
a := (a + 1) mod m  { keep searching at the next address }
end;
{ we've found an empty cell; see if there is room for x to be 
inserted }
if  n < m – 1  then  { n := n + 1;  T[a] := x }  else  err-
msg('table is full');
return(a)  { return the address where x was inserted }
end;
This  code  can  only  be  appreciated  by  comparing  it  with  alternatives  that  avoid  the  use  of  'return()'.  We 
encourage readers to try their hands at this challenge. Notice the three different ways this procedure can terminate:  
(1) no need to insert x because x is already in the table, (2) impossible to insert x because the table is full, and (3)  
the normal case when x is inserted. Standard Pascal incorporates no facilities for ""exception handling"" (e.g.  to  
cover the first two cases that should occur only rarely) and forces all three outcomes to exit the procedure at its  
textual end.",algorithms and data structures.pdf
"cover the first two cases that should occur only rarely) and forces all three outcomes to exit the procedure at its  
textual end.
Let us just mention a few other liberties that we may take. Whereas Pascal limits results of functions to certain  
simple types, we will let them be of any type: in particular, structured types, such as records and arrays. Rather  
than nesting if-then-else statements in order to discriminate among more than two mutually exclusive cases, we use 
the ""flat"" and more legible control structure:
if B1 then S1 elsif B2 then S2 elsif … else Sn ;
Our  sample  programs  do  not  return  dynamically  allocated  storage  explicitly.  They  rely  on  a  memory 
management system that retrieves free storage through ""garbage collection"". Many implementations of Pascal avoid 
garbage collection and instead provide a procedure 'dispose(…)' for the programmer to explicitly return unneeded",algorithms and data structures.pdf
"garbage collection and instead provide a procedure 'dispose(…)' for the programmer to explicitly return unneeded  
cells. If you work with such a version of Pascal and write list-processing programs that use significant amounts of  
memory, you must insert calls to 'dispose(…)' in appropriate places in your programs.
The  list  above  is  not  intended  to  be  exhaustive,  and  neither  do  we  argue  that  the  constructs  we  use  are 
necessarily superior to others commonly available. Our reason for extending the notation of Pascal (or any other  
programming language we might have chosen as a starting point) is the following: in addressing human readers, we  
believe an open-ended, somewhat informal notation is preferable to the straightjacket of any one programming  
language. The latter becomes necessary if and when we execute a program, but during the incubation period when  
43",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
our understanding slowly grows toward a firm grasp of an idea, supporting intuition is much more important than  
formality. Thus we describe data structures and algorithms with the help of figures, words, and programs as we see  
fit in any particular instance.
Programming project
1. Use your graphics frame program of “Graphics primitives and environments” to implement an editor for  
simple graphics productions such as those used to define snowflakes (e.g. 'any line segment gets replaced  
by a specified sequence of line segments'), and an interpreter that draws successive generations of the  
fractals defined by these productions.
Algorithms and Data Structures 44  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
5. Divide-and-conquer and 
recursion
Learning objectives:
• The algorithmic principle of divide-and-conquer leads directly to recursive procedures. 
• Examples: Merge sort, tree traversal. Recursion and iteration.
• My friend liked to claim ""I'm 2/3 Cherokee."" Until someone would challenge him ""Two- thirds? You mean 
1/2 , or, or maybe 3/8, how on earth can you be 2/3 of anything?"" ""It's easy,"" said Jim, ""both my parents are 
2/3.""
An algorithmic principle
Let A(D) denote the application of an algorithm A to a set of data D, producing a result R. An important class of  
algorithms, of a type called divide-and-conquer, processes data in two distinct ways, according to whether the data  
is small or large:
• If the set D is small, and/or of simple structure, we invoke a simple algorithm A 0 whose application A 0(D) 
yields R.",algorithms and data structures.pdf
"is small or large:
• If the set D is small, and/or of simple structure, we invoke a simple algorithm A 0 whose application A 0(D) 
yields R.
• If the set D is large, and/or of complex structure, we partition it into smaller subsets D 1, … , D k. For each i,  
apply A(Di) to yield a result Ri. Combine the results R1, … , Rk to yield R.
This algorithmic principle of divide-and-conquer leads  naturally to the notion of recursive procedures. The  
following example outlines  the concept in a high-level notation, highlighting the role of parameters and local  
variables.
procedure A(D: data; var R: result);
var  D1, … , Dk: data;  R1, … , Rk: result;
begin
if  simple(D) then R := A0(D)
else { D1, … , Dk := partition(D);
R1 := A(D1); … ; Rk := A(Dk);
R := combine(R1, … , Rk) }
end;
Notice how an initial data set D spawns set s D1, … , D k which, in turn, spawn children of their own. Thus the",algorithms and data structures.pdf
"R := combine(R1, … , Rk) }
end;
Notice how an initial data set D spawns set s D1, … , D k which, in turn, spawn children of their own. Thus the  
collection of all data sets generated by the partitioning scheme is a tree with root D. In order for the recursive  
procedure A(D) to terminate in all cases, the partitioning function must meet the following condition: Each branch  
of the partitioning tree, starting from the root D, eventually terminates with a data set D 0 that satisfies the predicate 
'simple(D0)', to which we can apply the algorithm.
Divide-and-conquer reduces a problem on data set D to k instances of the same problem on new sets D 1, … , D k 
that  are  ""simpler""  than  the  original  set  D.  Simpler  often  means  ""has  fewer  elements"",  but  any  measure  of 
Algorithms and Data Structures 45  A Global Text",algorithms and data structures.pdf
"5. Divide-and-conquer and recursion
""simplicity"" that monotonically heads for the predicate 'simple' will do, when algorithm A0 will finish the job. ""D is  
simple"" may mean ""D has no elements"", in which case A 0 may have to do nothing at all; or it may mean ""D has  
exactly one element"", and A0 may just mark this element as having been visited.
The following sections show examples of divide-and-conquer algorithms. As we will see, the actual workload is  
sometimes  distributed  unequally  among  different  parts  of  the  algorithm.  In  the  sorting  example,  the  step 
'R:=combine(R1, … , R k)' requires most of the work; in the ""Tower of Hanoi"" problem, the application of algorithm  
A0 takes the most effort.
Divide-and-conquer expressed as a diagram: merge sort
Suppose that we wish to sort a sequence of names alphabetically, as shown in Exhibit 5.1. We make use of the",algorithms and data structures.pdf
"Suppose that we wish to sort a sequence of names alphabetically, as shown in Exhibit 5.1. We make use of the  
divide-and-conquer strategy by partitioning a ""large"" sequence D into two subsequences D 1 and D 2, sorting each  
subsequence, and then merging them back together into sorted order. This is our algorithm A(D). If D contains at  
most one element, we do nothing at all. A0 is the identity algorithm, A0(D) = D.
Exhibit 5.1: Sorting the sequence {Z, A, S, D} by using a divide-and-conquer scheme
procedure sort(var D: sequence);
var  D1, D2: sequence;
function combine(D1, D2: sequence): sequence;
begin  { combine }
merge the two sorted sequences D1 and D2 
into a single sorted sequence D';
return(D')
end;  { combine }
begin  { sort}
if  |D| > 1  then  { split D into two sequences D1 and D2 of 
equal size;
sort(D1);  sort(D2);  D := combine(D1, D2) }
{ if |D| ≤ 1, D is trivially sorted, do nothing }
end;  { sort }
46",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
In the chapter on “sorting and its complexity”, under the section “merging and merge sorts”  we turn this divide-
and-conquer scheme into a program.
Recursively defined trees
A tree, more precisely, a rooted, ordered tree, is a data type used primarily to model any type of hierarchical  
organization.  Its  primitive  parts are  nodes and  leaves. It has  a  distinguished  node  called  the  root,  which, in 
violation of nature, is typically drawn at the top of the page, with the tree growing downward. Each node has a  
certain number of children, either leaves or nodes; leaves have no children. The exact definition of such trees can  
differ slightly with respect to details and terminology. We may define a binary tree, for example, by the condition  
that each node has either exactly, or at most, two children.",algorithms and data structures.pdf
"that each node has either exactly, or at most, two children.
The pictorial grammar shown in  Exhibit 5.2  captures this recursive definition of 'binary tree' and fixes the  
details left unspecified by the verbal description above. It uses an alphabet of three symbols: the nonterminal 'tree  
symbol', which is also the start symbol; and two terminal symbols, for 'node' and for 'leaf'.
Exhibit 5.2: The three symbols of the alphabet of a tree grammar
There are two production or rewriting rules, p1 and p2 ( Exhibit 5.3 ). The  derivation shown in  Exhibit 5.4  
illustrates the application of the production rules to generate a tree from the nonterminal start symbol.
Exhibit 5.3: Rule p1 generates a leaf, rule p2 generates a node and two new trees
Exhibit 5.4: One way to derive the tree at right
We may make the production rules more detailed by explicitly naming the coordinates associated with each",algorithms and data structures.pdf
"We may make the production rules more detailed by explicitly naming the coordinates associated with each  
symbol. On a display device such as a computer screen, the x- and y-values of a point are typically Cartesian  
coordinates with the origin in the upper-left corner. The x-values increase toward the bottom and the y-values  
increase toward the right of the display. Let (x, y) denote the screen position associated with a particular symbol,  
and let d denote the depth of a node in the tree. The root has depth 0, and the children of a node with depth d have  
depth d+1. The different levels of the tree are separated by some constant distance s. The separation between  
siblings is determined by a (rapidly decreasing) function t(d) which takes as argument the depth of the siblings and  
depends on the drawing size of the symbols and the resolution of the screen. These more detailed productions are  
shown in Exhibit 5.5.
Algorithms and Data Structures 47  A Global Text",algorithms and data structures.pdf
"5. Divide-and-conquer and recursion
Exhibit 5.5: Adding coordinate information to productions in order to control graphic layout
The translation of these two rules into high-level code is now plain:
procedure p1(x, y: coordinate);
begin
eraseTreeSymbol(x, y);
drawLeafSymbol(x, y)
end;
procedure p2(x, y: coordinate; d: level);
begin
eraseTreeSymbol(x, y);
drawNodeSymbol(x, y);
drawTreeSymbol(x + s, y – t(d + 1));
drawTreeSymbol(x + s, y + t(d + 1))
end;
If we choose t(d) = c · 2–d, these two procedures produce the display shown in Exhibit 5.6 of the tree generated  
in Exhibit 5.4.
Exhibit 5.6: Sample layout obtained by halving horizontal displacement at each successive level
Technical remark about the details of defining binary trees: Our grammar forces every node to have exactly two 
children: A child may be a node or a leaf. This lets us subsume two frequently occurring classes of binary trees  
under one common definition.",algorithms and data structures.pdf
"children: A child may be a node or a leaf. This lets us subsume two frequently occurring classes of binary trees  
under one common definition.
1. 0-2 (binary) trees.  We may identify leaves and nodes, making no distinction between them (replace the  
squares by circles in Exhibit 5.3 and Exhibit 5.4). Every node in the new tree now has either zero or two  
children, but not one. The smallest tree has a single node, the root.
2. (Arbitrary) Binary trees.  Ignore  the leaves  (drop  the squares  in  Exhibit 5.3  and  Exhibit 5.4  and  the 
branches leading into a square). Every node in the new tree now has either zero, one, or two children. The  
smallest tree (which consisted of a single leaf) now has no node at all; it is empty.
For clarity's sake, the following examples use the terminology of nodes and leaves introduced in the defining  
grammar. In some instances we point out what happens under the interpretation that leaves are dropped. 
48",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Recursive tree traversal
Recursion is a powerful tool for programming divide-and-conquer algorithms in a straightforward manner. In  
particular, when the data to be processed is defined recursively, a recursive processing algorithm that mirrors the  
structure of the data is most natural. The recursive tree traversal procedure below illustrates this point.
Traversing a tree  (in general: a graph, a data structure) means visiting every node and every leaf in an orderly  
sequence, beginning and ending at the root. What needs to be done at each node and each leaf is of no concern to  
the traversal algorithm, so we merely designate that by a call to a 'procedure visit( )'. You may think of inspecting  
the contents of all nodes and/or leaves, and writing them to a file.
Recursive tree traversals use divide-and-conquer to decompose a tree into its subtrees: At each node visited",algorithms and data structures.pdf
"Recursive tree traversals use divide-and-conquer to decompose a tree into its subtrees: At each node visited  
along the way, the two subtrees L and R to the left and right of this node must be traversed. There are three natural  
ways to sequence the node visit and the subtree traversals:
1. node; L; R  { preorder, or prefix }
2. L; node; R  { inorder or infix }
3. L; R; node  { postorder or suffix }
The following example translates this traversal algorithm into a recursive procedure:
procedure traverse(T: tree);
{ preorder, inorder, or postorder traversal of tree T with 
leaves }
begin
if  leaf(T) then visitleaf(T)
else { T is composite }
{ visit1(root(T));
traverse(leftsubtree(T));
visit2(root(T));
traverse(rightsubtree(T);
visit3(root(T)) }
end;
When leaves are ignored (i.e. a tree consisting of a single leaf is considered to be empty), the procedure body  
becomes slightly simpler:
if  not empty(T)  then  { … }",algorithms and data structures.pdf
"becomes slightly simpler:
if  not empty(T)  then  { … }
To accomplish the k-th traversal scheme (k = 1, 2, 3), 'visit k' performs the desired operation on the node, while  
the other two visits do nothing. If all three visits print out the name of the node, we obtain a sequence of node  
names  called  'triple  tree  traversal',  shown  in  Exhibit  5.7 along  with  the  three  traversal  orders  of  which  it  is 
composed. During the traversal the nodes are visited in the following sequence:
Algorithms and Data Structures 49  A Global Text",algorithms and data structures.pdf
"5. Divide-and-conquer and recursion
Exhibit 5.7: Three standard orders merged into a triple tree traversal
Recursion versus iteration: the Tower of Hanoi
The ""Tower of Hanoi"" is a stack of n disks of different sizes, held in place by a tall peg (Exhibit 5.8). The task is to 
transfer the tower from source peg S to a target peg T via an intermediate peg I, one disk at a time, without ever  
placing a larger disk on a smaller one. In this case the data set D is a tower of n disks, and the divide-and-conquer  
algorithm A partitions D asymmetrically into a small ""tower"" consisting of a single disk (the largest, at the bottom  
of the pile) and another tower D' (usually larger, but conceivably empty) consisting of the n – 1 topmost disks. The  
puzzle is solved recursively in three steps:
1. Transfer D' to the intermediate peg I.  
2. Move the largest disk to the target peg T.  
3. Transfer D' on top of the largest disk at the target peg T.",algorithms and data structures.pdf
"2. Move the largest disk to the target peg T.  
3. Transfer D' on top of the largest disk at the target peg T.
Exhibit 5.8: Initial configuration of the Tower of Hanoi.
Step 1 deserves more explanation. How do we transfer the n – 1 topmost disks from one peg to another? Notice  
that they  themselves  constitute  a tower,  to which we  may  apply  the same three-step  algorithm.  Thus we  are 
presented with successively simpler problems to solve, namely, transferring the n – 1 topmost disks from one peg to  
another, for decreasing n, until finally, for n = 0, we do nothing.
procedure Hanoi(n: integer; x, y, z: peg);
{ transfer a tower with n disks from peg x, via y, to z }
begin
if  n > 0  then  { Hanoi(n – 1, x, z, y);  move(x, z);  Hanoi(n – 
1, y, x, z) }
end;
Recursion has the advantage of intuitive clarity. Elegant and efficient as this solution may be, there is some  
complexity hidden in the bookkeeping implied by recursion.
50",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The following procedure is an equally elegant and more efficient iterative solution to this problem. It assumes  
that the pegs are cyclically ordered, and the target peg where the disks will first come to rest depends on this order  
and on the parity of n (Exhibit 5.9). For odd values of n, 'IterativeHanoi' moves the tower to peg I, for even values of 
n, to peg T.
Exhibit 5.9: Cyclic order of the pegs.
procedure IterativeHanoi(n: integer);
var  odd: boolean;  { odd represents the parity of the move }
begin
odd := true;
repeat
 case  odd  of
true: transfer smallest disk cyclically to next peg;
false: make the only legal move leaving the smallest in place
end;
odd := not odd
until  entire tower is on target peg
end;
Exercise: recursive or iterative pictures?
Chapter 4 presented some beautiful examples of recursive pictures, which would be hard to program without",algorithms and data structures.pdf
"Chapter 4 presented some beautiful examples of recursive pictures, which would be hard to program without  
recursion.  But  for  simple  recursive  pictures  iteration  is  just  as  natural.  Specify  a  convenient  set  of  graphics 
primitives  and  use  them  to  write  an  iterative  procedure  to  draw  Exhibit  5.10 to  a  nesting  depth  given  by  a 
parameter d.
Exhibit 5.10: Interleaved circles and equilateral triangles cause the radius to be exactly halved at each step.
Solution
There  are  many  choices  of  suitable  primitives  and  many  ways  to  program  these  pictures.  Specifying  an 
equilateral triangle by its center and the radius of its circumscribed circle simplifies the notation. Assume that we  
may use the procedures:
procedure circle(x, y, r: real); { coordinates of center and 
radius }
procedure equitr(x, y, r: real); { center and radius of 
circumscribed circle}
Algorithms and Data Structures 51  A Global Text",algorithms and data structures.pdf
"5. Divide-and-conquer and recursion
procedure citr(x, y, r: real; d: integer);
var vr: real;  { variable radius }
i: integer;
begin
vr := r;
for i := 1 to d do  { equitr(x, y, vr);  vr := vr/2;  circle(x, y, 
vr) }
{ show that the radius of consecutively nested circles gets
exactly halved at each step }
end;
The flag of Alfanumerica: an algorithmic novel on iteration and recursion
In the process of automating its flag industry, the United States of Alfanumerica announced a competition for  
the most elegant program to print its flag:
All solutions submitted to the prize committee fell into one of two classes, the iterative and recursive programs.  
The proponents of these two algorithm design principles could not agree on a winner, and the selection process  
sparked a civil war that split the nation into two: the Iterative States of Alfanumerica (ISA) and the Recursive States  
of Alfanumerica (RSA). Both nations fly the same flag but use entirely different production algorithms.",algorithms and data structures.pdf
"of Alfanumerica (RSA). Both nations fly the same flag but use entirely different production algorithms.
1. Write a 
procedure ISA(k: integer);
to print the ISA flag, using an iterative algorithm, of course. Assume that k is a power of 2 and k ≤ (half the  
line length of the printer).
2. Explain why the printer industry in RSA is much more innovative than the one in ISA. All modern RSA  
printers include operations for positioning the writing head anywhere within a line, and line feed works  
both forward and backward.
3. Specify  the  precise  operations  for  some  RSA  printer  of  your  design.  Using  these  operations,  write  a 
recursive 
procedure RSA(k: integer); 
to print the RSA flag.
4. Explain an unforeseen consequence of this drive to automate the flag industry of Alfanumerica: In both ISA  
and RSA, a growing number of flags can be seen fluttering in the breeze turned around by 90˚.
Exercises",algorithms and data structures.pdf
"and RSA, a growing number of flags can be seen fluttering in the breeze turned around by 90˚.
Exercises
1. Whereas divide-and-conquer algorithms usually attempt to divide the data in equal halves, the recursive  
Tower  of Hanoi  procedure  presented  in  the  section  'Recursion  versus iteration:  The  Tower of Hanoi"" 
divides the data in a very asymmetric manner: a single disk versus n – 1 disks. Why?
2. Prove by induction on n that the iterative program 'IterativeHanoi' solves the problem in 2n–1 iterations.
52
                ****************
        ********        ********
    ****    ****    ****    ****
  **  **  **  **  **  **  **  **
 * * * * * * * * * * * * * * * *
k blanks followed by k stars
twice (k/2 blanks followed by k/2 stars)
…
continue doubling and halving
down to runs length of 1.",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
6. Syntax
Learning objectives:
• syntax and semantics
• syntax diagrams and EBNF describe context-free grammars 
• terminal and nonterminal symbols 
• productions 
• definition of EBNF by itself 
• parse tree
• grammars must avoid ambiguities 
• infix, prefix, and postfix notation for arithmetic expressions
• prefix and postfix notation do not need parentheses
Syntax and semantics
Computer science has borrowed some important concepts from the study of natural languages (e.g. the notions  
of syntax and semantics).  Syntax rules prescribe how the sentences of a language are formed, independently of  
their meaning. Semantics deals with their meaning. The two sentences ""The child draws the horse"" and ""The horse  
draws the child"" are both syntactically correct according to the accepted rules of grammar. The first sentence clearly",algorithms and data structures.pdf
"draws the child"" are both syntactically correct according to the accepted rules of grammar. The first sentence clearly  
makes sense, whereas the second sentence is baffling: perhaps senseless (if ""draw"" means ""drawing a picture""),  
perhaps meaningful (if ""draw"" means ""pull""). Semantic aspects—whether a sentence is meaningful or not, and if so,  
what it means—are much more difficult to formalize and decide than syntactic issues.
However, the analogy between natural languages and programming languages does not go very far. The choice  
of  English  words  and  phrases  such  as  ""begin"",  ""end"",  ""goto"",  ""if-then-else""  lends  a  programming  language  a 
superficial similarity to natural language, but no more. The possibility of verbal encoding of mathematical formulas  
into pseudo-English has deliberately been built into COBOL; for example, ""compute velocity times time giving",algorithms and data structures.pdf
"into pseudo-English has deliberately been built into COBOL; for example, ""compute velocity times time giving  
distance"" is nothing but syntactic sugar for ""distance := velocity · time"". Much more important is the distinction  
that natural languages are not rigorously defined (neither the vocabulary, nor the syntax, and certainly not the  
semantics), whereas programming languages should be defined according to a rigorous formalism. Programming  
languages are much closer to the formal notations of mathematics than to natural languages, and programming 
notation would be a more accurate term.
The lexical part of a modern programming language [the alphabet, the set of reserved words, the construction  
rules for the identifiers (i.e. the equivalent to the vocabulary of a natural language) and the  syntax are usually  
defined formally. However, system-dependent differences are not always described precisely. The compiler often",algorithms and data structures.pdf
"defined formally. However, system-dependent differences are not always described precisely. The compiler often  
determines  in  detail  the  syntactic  correctness  of  a  program  with  respect  to  a  certain  system  (computer  and 
operating system). The semantics of a programming language could also be defined formally, but this is rarely  
done, because formal semantic definitions are extensive and difficult to read.
Algorithms and Data Structures 53  A Global Text",algorithms and data structures.pdf
"6. Syntax
The syntax of a programming language is not as important as the semantics, but good understanding of the  
syntax often helps in understanding the language. With some practice one can often guess the semantics from the  
syntax, since the syntax of a well-designed programming language is the frame that supports the semantics.
Grammars and their representation: syntax diagrams and EBNF
The syntax of modern programming languages is defined by  grammars. These are mostly of a type called  
context-free grammars , or close variants thereof, and  can be given in different notations.  Backus-Naur form  
(BNF), a milestone in the development of programming languages, was introduced in 1960 to define the syntax of  
Algol. It is the basis for other notations used today, such as EBNF (extended BNF)  and graphical representations  
such as syntax diagrams. EBNF and syntax diagrams are syntactic notations that describe exactly the context-free 
grammars of formal language theory.",algorithms and data structures.pdf
"such as syntax diagrams. EBNF and syntax diagrams are syntactic notations that describe exactly the context-free 
grammars of formal language theory.
Recursion is a central theme of all these notations: the syntactic correctness and structure of a large program  
text are  reduced  to the syntactic correctness  and  structure of its textual components. Other common notions  
include: terminal symbol, nonterminal symbol, and productions or rewriting rules that describe how nonterminal  
symbols generate strings of symbols.
The set of terminal symbols forms the alphabet of a language, the symbols from which the sentences are built. In 
EBNF a terminal symbol is enclosed in single quotation marks; in syntax diagrams a terminal symbol is represented 
by writing it in an oval:
Nonterminal symbols represent syntactic entities: statements, declarations, or expressions. Each nonterminal",algorithms and data structures.pdf
"by writing it in an oval:
Nonterminal symbols represent syntactic entities: statements, declarations, or expressions. Each nonterminal  
symbol is given a name consisting of a sequence of letters and digits, where the first character must be a letter. In  
syntax diagrams a nonterminal symbol is represented by writing its name in a rectangular box:
If a construct consists of the catenation of constructs A and B, this is expressed by
If a construct consists of either A or B, this is denoted by
If a construct may be either construct A or nothing, this is expressed by
If a construct consists of the catenation of any number of A's (including none), this is denoted by
In EBNF parentheses may be used to group entities [e.g. ( A | B )].
54",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
For each nonterminal symbol there must be at least one production that describes how this syntactic entity is  
formed from other terminal or nonterminal symbols using the composition constructs above:
The following examples show productions and the constructs they generate. A, B, C, D may denote terminal or  
nonterminal symbols.
EBNF is a formal language over a finite alphabet of symbols introduced above, built according to the rules  
explained above. Thus it is no great surprise that EBNF can be used to define itself. We use the following names for  
syntactic entities:
stmt A syntactic equation.
expr A list of alternative terms.
term A concatenation of factors.
factor A single syntactic entity or parenthesized expression.
nts Nonterminal symbol that denotes a syntactic entity. It consists of a sequence of letters and digits 
where the first character must be a letter.",algorithms and data structures.pdf
"nts Nonterminal symbol that denotes a syntactic entity. It consists of a sequence of letters and digits 
where the first character must be a letter.
ts Terminal symbol that belongs to the defined language's vocabulary. Since the vocabulary 
depends on the language to be defined there is no production for ts.
EBNF is now defined by the following productions:
EBNF = { stmt } .
Algorithms and Data Structures 55  A Global Text",algorithms and data structures.pdf
"6. Syntax
stmt = nts '=' expr '.' .
expr = term { '|' term } .
term = factor { factor } .
factor = nts | ts | '(' expr ')' | '[' expr ']' | '{' expr '}' .
nts= letter { letter | digit } .
Example: syntax of simple expressions
The following productions for the three nonterminals E(xpression), T(erm), and F(actor) can be traced back to  
Algol 60. They form the core of all grammars for arithmetic expressions. We have simplified this grammar to define  
a class of expressions that lacks, for example, a unary minus operator and many other convenient notations. These  
details  are  but not important  for  our  purpose:  namely,  understanding  how  this  grammar  assigns  the  correct 
structure to each expression. We have further simplified the grammar so that constants and variables are replaced  
by the single terminal symbol # (Exhibit 6.1):
E = T { ( '+' | '–' ) T } .
T = F { ( '·' | '/' ) F } .
F = '#' | '(' E ')' .
Exhibit 6.1: Syntax diagrams for simple arithmetic expressions.",algorithms and data structures.pdf
"E = T { ( '+' | '–' ) T } .
T = F { ( '·' | '/' ) F } .
F = '#' | '(' E ')' .
Exhibit 6.1: Syntax diagrams for simple arithmetic expressions.
From the nonterminal E we can derive different expressions. In the opposite direction we start with a sequence  
of terminal symbols and check by syntactic analysis, or parsing, whether a given sequence is a valid expression. If  
this is the case the grammar assigns to this expression a unique tree structure, the parse tree (Exhibit 6.2).
56",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 6.2: Parse tree for the expression  # · ( # ) + # / # .
Exercise: syntax diagrams for palindromes
A palindrome is a string that reads the same when read forward or backward. Examples: 0110 and 01010. 01 is  
not a palindrome, as it differs from its reverse 10.
1. What is the shortest palindrome?
2. Specify the syntax  of palindromes  over the alphabet {0,  1} in  EBNF-notation,  and  by drawing syntax  
diagrams.
Solution
1. The shortest palindrome is the null or empty string.
2. S  =  [ '0' | '1' ]  |  '0' S '0'  |  '1' S '1' (Exhibit 6.3).
Exhibit 6.3: Syntax diagram for palindromes 
An overly simple syntax for simple expressions
Why does the grammar given in previous section contain term and factor? An expression E that involves only  
binary operators (e.g. +, –, · and /) is either a primitive operand, abbreviated as #, or of the form 'E op E'. Consider",algorithms and data structures.pdf
"binary operators (e.g. +, –, · and /) is either a primitive operand, abbreviated as #, or of the form 'E op E'. Consider  
a ""simpler"" grammar for simple, parenthesis-free expressions (Exhibit 6.4):
E = '#' | E ( '+' | '–' | '·' | '/' ) E .
Algorithms and Data Structures 57  A Global Text",algorithms and data structures.pdf
"6. Syntax
Exhibit 6.4: A syntax that generates parse trees of ambiguous structure
Now the expression # · # + # can be derived from E in two different ways ( Exhibit 6.5). Such an  ambiguous 
grammar is useless since we want to derive the semantic interpretation from the syntactic structure, and the tree at  
the left contradicts the conventional operator precedence of · over +.
Exhibit 6.5: Two incompatible structures for the expression  # · # + # .
“Everything should be explained as simply as possible, but not simpler.”
(Albert Einstein)
We can salvage the idea of a grammar with a single nonterminal E by enclosing every expression of the form 'E  
op E' in parentheses, thus ensuring that every expression has a unique structure (Exhibit 6.6):
E = '#' | '(' E ( '+' | '–' | '·' | '/' ) E ')' .
Exhibit 6.6: Parentheses serve to restore unique structure.
58",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
In doing so we change the language. The more complex grammar with three nonterminals E(xpression, T(erm), 
and F(actor) lets us write expressions that are only partially parenthesized and assigns to them a unique structure 
compatible with our priority conventions: · and / have higher priority than + and –.
Exercise: the ambiguity of the dangling ""else""
The problem of the dangling ""else"" is an example of a syntax chosen to be ""too simple"" for the task it is supposed 
to handle. The syntax of several programming languages (e.g., Pascal) assigns to nested 'if-then[-else]' statements  
an ambiguous structure. It is left to the semantics of the language to disambiguate.
Let E, E 1,  E2, …  denote  Boolean  expressions,  S, S1,  S2,  … statements.  Pascal syntax  allows  two  types  of if 
statements:
if E then S
and
if E then S else S
1. Draw one syntax diagram that expresses both of these syntactic possibilities.",algorithms and data structures.pdf
"statements:
if E then S
and
if E then S else S
1. Draw one syntax diagram that expresses both of these syntactic possibilities.
2. Show all the possible syntactic structures of the statement
 if E1 then if E2 then S1 else S2
3. Propose a small modification to the Pascal language that avoids the syntactic ambiguity of the dangling  
else.  Show  that  in  your  modified  Pascal  any  arbitrarily  nested  structure  of  'if-then'  and  'if-then-else' 
statements must have a unique syntactic structure.
Parenthesis-free notation for arithmetic expressions
In the usual  infix notation for arithmetic expressions a binary operator is written between its two operands.  
Even  with  operator  precedence  conventions,  some  parentheses  are  required  to  guarantee  a  unique  syntactic 
structure.  The  selective  use  of  parentheses  complicates  the  syntax  of  infix  expressions:  Syntax  analysis, 
interpretative evaluation, and code generation all become more complicated.",algorithms and data structures.pdf
"interpretative evaluation, and code generation all become more complicated.
Parenthesis-free or Polish notation (named for the Polish logician Jan Lukasiewicz) is a simpler notation for  
arithmetic expressions. All operators are systematically written either before ( prefix notation) or after ( postfix or 
suffix notation) the operands to which they apply. We restrict our examples to the binary operators +, –, · and /.  
Operators with different arities (i.e. different numbers of arguments) are easily handled provided that the number  
of arguments used is uniquely determined by the operator symbol. To introduce the unary minus we simply need a  
different symbol than for the binary minus.
Infix a+b a+(b·c)(a+b)·c
Prefix +ab +a·bc ·+abc
Postfix ab+ abc·+ ab+c·
Postfix notation mirrors the sequence of operations performed during the evaluation of an expression. 'ab+' is",algorithms and data structures.pdf
"Postfix ab+ abc·+ ab+c·
Postfix notation mirrors the sequence of operations performed during the evaluation of an expression. 'ab+' is  
interpreted as: load a (find first operand); load b (find the second operand); add both. The syntax of arithmetic  
expressions in postfix notation is determined by the following grammar:
S = '#' | S S ( '+' | '–' | '·' | '/' )
Algorithms and Data Structures 59  A Global Text",algorithms and data structures.pdf
"6. Syntax
Exhibit 6.7: Suffix expressions have a unique structure even without the use of parentheses.
Exercises
1.  Consider the following syntax, given in EBNF:
S = A.
A = B | 'IF' A 'THEN' A 'ELSE' A.
B = C | B 'OR' C.
C = D | C 'AND' D.
D = 'x' | '(' A ')' | 'NOT' D.
(a) Determine the sets of terminal and nonterminal symbols.
(b) Give the syntax diagrams corresponding to the rules above.
(c) Which  of  the  following  expressions  is  correct  corresponding  to  the  given  syntax?  For  the  correct 
expressions show how they can be derived from the given rules:
x AND x
x NOT AND x
(x OR x) AND NOT x
IF x AND x THEN x OR x ELSE NOT x
x AND OR x
2. Extend the grammar of Section 6.3 to include the 'unary minus' (i.e. an arithmetic operator that turns any  
expression into its negative, as in –x). Do this under two different assumptions:
(a) The unary minus is denoted by a different character than the binary minus, say ¬.",algorithms and data structures.pdf
"(a) The unary minus is denoted by a different character than the binary minus, say ¬.
(b) The character – is 'overloaded' (i.e. it is used to denote both unary and binary minus). For any specific  
occurrence of –, only the context determines which operator it designates.
3. Extended Backus-Naur form and syntax diagrams
Define each of the four languages described below using both EBNF and syntax diagrams. Use the following 
conventions and notations: Uppercase letters denote nonterminal symbols. Lowercase letters and the three  
separators ',' '(' and ')' denote terminal symbols. """" stands for the empty or null string. Notice that the blank  
character does not occur in these languages, so we use it to separate distinct sentences.
60",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
L ::= a | b | … | z Letter
D ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Digit
S ::= D { D } Sequence of digits I ::= L { L | D } Identifier
(a) Real numbers (constants) in Pascal
Examples:  –3  + 3.14  10e–06  –10.0e6    but not  10e6
(b) Nonnested lists of identifiers (including the empty list)
Examples:  ()  (a)  (year, month, day)    but not  (a,(b))  and not  """"
(c) Nested lists of identifiers (including empty lists)
Examples: in addition to the examples in part (b), we have lists such as
((),())  (a, ())  (name, (first, middle, last))    but not  (a)(b)  and not  """"
(d) Parentheses expressions
Almost the same problem as part (c), except that we allow the null string, we omit identifiers and commas,  
and we allow multiple outermost pairs of parentheses.
Examples:  """"  ()  ()()  ()(())  ()(()())()()
4. Use both syntax diagrams and EBNF to define the repeated if-then-else statement:",algorithms and data structures.pdf
"Examples:  """"  ()  ()()  ()(())  ()(()())()()
4. Use both syntax diagrams and EBNF to define the repeated if-then-else statement:
if  B1  then  S1  elsif  B2  then  S2  elsif  …  else  S
Algorithms and Data Structures 61  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
7. Syntax analysis
Learning objectives:
• syntax is the frame that carries the semantics of a language
• syntax analysis
• syntax tree
• top-down parser
• syntax analysis of parenthesis-free expressions by counting
• syntax analysis by recursive descent
• recursive coroutines
The role of syntax analysis
The syntax of a language is the skeleton that carries the semantics. Therefore, we will try to get as much work as  
possible done as a side effect of syntax analysis; for example, compiling a program (i.e. translating it from one  
language into another) is a mainly semantic task. However, a good language and compiler are designed in such a  
way that syntax analysis determines where  to start with the translation process. Many processes in computer  
science are syntax-driven in this sense. Hence syntax analysis is important. In this section we derive algorithms for",algorithms and data structures.pdf
"science are syntax-driven in this sense. Hence syntax analysis is important. In this section we derive algorithms for  
syntax analysis directly from syntax diagrams. These algorithms reflect the recursive nature  of the underlying  
grammars. A program for syntax analysis is called a parser.
The composition of a sentence can be represented by a syntax tree or parse tree. The root of the tree is the start  
symbol; the leaves represent the sentence to be recognized. The tree describes how a syntactically correct sentence  
can be derived from the start symbol by applying the productions of the underlying grammar (Exhibit 7.1).
Exhibit 7.1: The unique parse tree for  # · # + # 
Top-down parsers begin with the start symbol as the goal of the analysis. In our example, ""search for an E"". The  
production for E tells us that we obtain an E if we find a sequence of T's separated by + or –. Hence we look for T's.",algorithms and data structures.pdf
"production for E tells us that we obtain an E if we find a sequence of T's separated by + or –. Hence we look for T's.  
The structure tree of an expression grows in this way as a sequence of goals from top (the root) to bottom (the  
leaves). While satisfying the goals (nonterminal symbols) the parser reads suitable symbols (terminal symbols)  
from left to right. In many practical cases a parser needs no backtrack. No backtracking is required if the current  
Algorithms and Data Structures 62  A Global Text
# á # + #
F F F
T T
E",algorithms and data structures.pdf
"7. Syntax analysis
input symbol and the nonterminal to be expanded determine uniquely the production to be applied. A recursive-
descent parser uses a set of recursive procedures to recognize its input with no backtracking.
Bottom-up  methods build the structure tree from the leaves to the root. The text is reduced until the start  
symbol is obtained.
Syntax analysis of parenthesis-free expressions by counting
Syntax analysis can be very simple. Arithmetic expressions in Polish notation are analyzed by counting. For sake  
of simplicity we assume that each operand in an arithmetic expression is denoted by the single character #. In order  
to decide whether a given string c1 c2 … cn is a correct expression in postfix notation, we form an integer sequence t 0, 
t1, … , tn according to the following rule:
t0 = 0.
ti+1 = ti + 1, if i > 0 and ci+1 is an operand.
ti+1 = ti – 1, if i > 0 and ci+1 is an operator.
Example of a correct expression:
# # # # – – + # ·
c1 c2 c3c4c5c6c7c8c9",algorithms and data structures.pdf
"ti+1 = ti – 1, if i > 0 and ci+1 is an operator.
Example of a correct expression:
# # # # – – + # ·
c1 c2 c3c4c5c6c7c8c9
t0 t1 t2t3t4t5t6t7t8t9
0 1 2 3 4 3 2 1 2 1
Example of an incorrect expression (one operator is missing):
# #   #     +    ·   #  #   /
c1 c2  c3    c4  c5  c6  c7 c8
t0 t1  t2  t3  t4   t5  t6  t7 t8
0    1     2   3  2    1   2   3  2
Theorem: The string c 1 c2 … c n over the alphabet A = { # , + , – , · , / } is a syntactically correct expression in  
postfix notation if and only if the associated integer sequence t0, t1, … , tn satisfies the following conditions:
ti > 0 for 1 ≤ i < n,  tn = 1.
Proof ⇒ : Let c1 c2 … cn be a correct arithmetic expression in postfix notation. We prove by induction on the  
length n of the string that the corresponding integer sequence satisfies the conditions.
Base of induction:  For n = 1 the only correct postfix expression is c 1 = #, and the sequence t 0 = 0, t 1 = 1 has the  
desired properties.",algorithms and data structures.pdf
"Base of induction:  For n = 1 the only correct postfix expression is c 1 = #, and the sequence t 0 = 0, t 1 = 1 has the  
desired properties.
Induction hypothesis: The theorem is correct for all expressions of length ≤ m.
Induction step: Consider a correct postfix expression S of length m + 1 > 1 over the given alphabet A. Let s = (s i) 0 ≤ i 
≤ m+1  be the integer sequence associated with S. Then S is of the form S = T U Op, where 'Op' is an operator and T  
and U are correct postfix expressions of length j ≤ m and length k ≤ m, j + k = m. Let t = (t i) 0 ≤ I ≤ j and u = (ui) 0 ≤ i ≤ k 
be the integer sequences associated with T and U. We apply the induction hypothesis to T and U. The sequence s is  
composed from t and u as follows:
63",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
s = s0 , s1 , s2 , … , sj , sj + 1 , sj + 2 , … , sm ,    sm+1
 t0 , t1 , t2 , … , tj , u1 + 1 , u2 + 1 , … , uk + 1 , 1
 0, … ,1,… ,2,1
Since t ends with 1, we add 1 to each element in u, and the subsequence therefore ends with u k + 1 = 2. Finally,  
the operator 'Op' decreases this element by 1, and s therefore ends with sm+1 = 1. Since ti > 0 for 1 ≤ i < j and ui > 0 for 
1 ≤i < k, we obtain that si > 0 for 1 ≤ i < k + 1. Hence s has the desired properties, and we have proved one direction  
of the theorem.
Proof ⇐ : We prove by induction on the length n that a string c 1 c2 … cn over A is a correct arithmetic expression  
in postfix notation if the associated integer sequence satisfies the conditions stated in the theorem.
Base of induction: For n = 1 the only sequence is t 0 = 0, t1 = 1. It follows from the definition of the sequence that  
c1 = #, which is a correct arithmetic expression in postfix notation.",algorithms and data structures.pdf
"c1 = #, which is a correct arithmetic expression in postfix notation.
Induction hypothesis: The theorem is correct for all expressions of length ≤ m.
Induction step: Let s = (si) 0 ≤ i ≤ m+1 be the integer sequence associated with a string S = c 1 c2 … cm+1 of length m + 1 
> 1 over the given alphabet A which satisfies the conditions stated in the theorem. Let j < m + 1 be the largest index  
with sj = 1. Since s1 = 1 such an index j exists. Consider the substrings T = c 1 c2 … cj and U = c j cj+1 … cm. The integer 
sequences (si) 0 ≤  i ≤ j  and (si – 1) j ≤  i ≤ m  associated with T and U both satisfy the conditions stated in the theorem.  
Hence we can apply the induction hypothesis and obtain that both T and U are correct postfix expressions. From  
the definition of the integer sequence we obtain that c m+1 is an operand 'O p'. Since T and U are correct postfix  
expressions, S = T U Op is also a correct postfix expression, and the theorem is proved.",algorithms and data structures.pdf
"expressions, S = T U Op is also a correct postfix expression, and the theorem is proved.
A similar proof  shows that the  syntactic structure  of a  postfix  expression  is  unique. The  integer sequence 
associated with a postfix expression is of practical importance: The sequence describes the depth of the stack during 
evaluation of the expression, and the largest number in the sequence is therefore the maximum number of storage  
cells needed.
Analysis by recursive descent
We return to the syntax of the simple arithmetic expressions of chapter 6 in the section “Example: syntax of  
simple expressions” ( Exhibit   7.2). Using the expression   # · (# – #) as an example, we show how these syntax  
diagrams  are  used to analyze  any expressions  by  means of a technique  called  recursive-descent parsing . The  
progress of the analysis depends on the current state and the next symbol to be read: a lookahead of exactly one",algorithms and data structures.pdf
"progress of the analysis depends on the current state and the next symbol to be read: a lookahead of exactly one  
symbol suffices to avoid backtracking. In  Exhibit 7.3  we move one step to the right after each symbol has been  
recognized, and we move vertically to step up or down in the recursion.
Algorithms and Data Structures 64  A Global Text",algorithms and data structures.pdf
"7. Syntax analysis
Exhibit  7.2: Standard syntax for simple arithmetic expressions (graphic does not match)
Exhibit 7.3: Trace of syntax analysis algorithm parsing the expression  # · ( # – # ).
Turning syntax diagrams into a parser
In a programming language that allows recursion the three syntax diagrams for simple arithmetic expressions  
can be translated directly into procedures. A nonterminal symbol corresponds to a procedure call, a loop in the  
diagram generates a while loop, and a selection is translated into an if statement. When a procedure wants to  
delegate a goal it calls another, in cyclic order: E calls T calls F calls E, and so on. Procedures implementing such a  
recursive control structure are often called recursive coroutines.
65",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The procedures that follow must be embedded into a program that provides the variable 'ch' and the procedures  
'read' and 'error'. We assume that the procedure 'error' prints an error message and terminates the program. In a  
more sophisticated implementation, 'error' would return a message to the calling procedure (e.g. 'factor'). Then this  
error message is returned up the ladder of all recursive procedure calls active at the moment.
Before the first call of the procedure 'expression', a character has to be read into 'ch'. Furthermore, we assume  
that a correct expression is terminated by a period:
…
read(ch);  expression;  if  ch ≠ '.'  then  error;
…
Exercises
1. Design recursive  algorithms to translate the simple arithmetic  expressions of chapter 6 in the section  
“Example: syntax of a simple expressions” into corresponding prefix and postfix expressions as defined in",algorithms and data structures.pdf
"“Example: syntax of a simple expressions” into corresponding prefix and postfix expressions as defined in  
chapter  6  in  the  section  “Parenthesis-free  notation  for  arithmetic  expressions”.  Same  for  the  inverse 
translations.
2. Using syntax diagrams and EBNF define a language of 'correctly nested parentheses expressions'. You have  
a bit of freedom (how much?) in defining exactly what is correctly nested and what is not, but obviously  
your definition must include expressions such as (), ((())), (()(())), and must exclude strings such as (, )(, ())
().
3. Design two parsing algorithms for your class of correctly nested parentheses expressions: one that works by  
counting, the other through recursive descent. 
Algorithms and Data Structures 66  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Part III: Objects, algorithms, 
programs
Computing with numbers and other objects
Since the introduction of computers four or five decades ago the meaning of the word  computation has kept  
expanding. Whereas ""computation"" traditionally implied ""numbers"", today we routinely compute pictures, texts,  
and many other types of objects. When classified according to the types of objects being processed, three types of  
computer  applications  stand  out  prominently  with  respect  to  the  influence  they  had  on  the  development  of 
computer science.
The first generation involved numerical computing, applied mainly to scientific and technical problems. Data to  
be processed consisted almost exclusively of numbers, or sets of numbers with a simple structure, such as vectors  
and  matrices. Programs  were  characterized  by  long execution  times  but small sets  of input and  output data.",algorithms and data structures.pdf
"and  matrices. Programs  were  characterized  by  long execution  times  but small sets  of input and  output data. 
Algorithms were more important than data structures, and many new numerical algorithms were invented. Lasting  
achievements of this first phase of computer applications include systematic study of numerical algorithms, error  
analysis, the concept of program libraries, and the first high-level programming languages, Fortran and Algol.
The second generation, hatched by the needs of commercial data processing, leads to the development of many  
new data structures. Business applications thrive on record keeping and updating, text and form processing, and  
report generation: there is not much computation in the numeric sense of the word, but a lot of reading, storing,  
moving,  and  printing  of  data.  In  other  words,  these  applications  are  data  intensive  rather  than  computation",algorithms and data structures.pdf
"moving,  and  printing  of  data.  In  other  words,  these  applications  are  data  intensive  rather  than  computation 
intensive.  By  focusing  attention  on  the  problem  of  efficient  management  of  large,  dynamically  varying  data 
collections, this phase created one of the core disciplines of computer science: data structures, and corresponding  
algorithms for managing data, such as searching and sorting.
We are  now  in  a  third  generation of computer  applications,  dominated  by  computing with  geometric  and 
pictorial objects. This change of emphasis was triggered by the advent of computers with bitmap graphics. In turn,  
this leads to the widespread use of sophisticated user interfaces that depend on graphics, and to a rapid increase in  
applications such as computer-aided design (CAD) and image processing and pattern recognition (in medicine,",algorithms and data structures.pdf
"applications such as computer-aided design (CAD) and image processing and pattern recognition (in medicine,  
cartography,  robot control).  The  young discipline  of computational geometry  has  emerged  in  response  to the 
growing  importance  of  processing  geometric  and  pictorial  objects.  It  has  created  novel  data  structures  and 
algorithms, some of which are presented in Parts V and VI.
Our selection of algorithms in Part III reflects the breadth of applications whose history we have just sketched.  
We choose the simplest types of objects from each of these different domains of computation and some of the most  
concise and elegant algorithms designed to process them. The study of typical small programs is an essential part of  
programming. A large part of computer science consists of the knowledge of how typical problems can be solved;  
and the best way to gain such knowledge is to study the main ideas that make standard programs work.",algorithms and data structures.pdf
"and the best way to gain such knowledge is to study the main ideas that make standard programs work.
Algorithms and Data Structures 67  A Global Text",algorithms and data structures.pdf
"7. Syntax analysis
Algorithms and programs
Theoretical computer science treats algorithm as a formal concept, rigorously defined in a number of ways, such  
as Turing machines or lambda calculus. But in the context of programming,  algorithm is typically used as an  
intuitive  concept  designed  to  help  people  express  solutions  to  their  problems.  The  formal  counterpart  of  an 
algorithm is a procedure or program (fragment) that expresses the algorithm in a formally defined programming  
language. The process of formalizing an algorithm as a program typically requires many decisions: some superficial  
(e.g. what type of statement is chosen to set up a loop), some of great practical consequence (e.g. for a given range  
of values of n, is the algorithm's asymptotic complexity analysis relevant or misleading?).
We present algorithms in whatever notation appears to convey the key ideas most clearly, and we have a clear",algorithms and data structures.pdf
"We present algorithms in whatever notation appears to convey the key ideas most clearly, and we have a clear  
preference for pictures. We present programs in an extended version of Pascal; readers should have little difficulty  
translating this into any programming language of their choice. Mastery of interesting small programs is the best  
way to get started in computer science. We encourage the reader to work the examples in detail.
The literature on algorithms. The development of new algorithms has been proceeding at a very rapid pace  
for several decades, and even a specialist can only stay abreast with the state of the art in some subfield, such as  
graph  algorithms,  numerical algorithms,  or geometric  algorithms. This  rapid development is sure to continue  
unabated, particularly in the increasingly important field of parallel algorithms. The cutting edge of algorithm",algorithms and data structures.pdf
"unabated, particularly in the increasingly important field of parallel algorithms. The cutting edge of algorithm  
research is published in several journals that specialize in this research topic, including the Journal of Algorithms 
and  Algorithmica.  This  literature  is  generally  accessible  only  after  a  student  has  studied  a  few  textbooks  on 
algorithms, such as [AHU 75], [Baa 88], [BB 88], [CLR 90], [GB 91], [HS 78], [Knu 73a], [Knu 81], [Knu 73b],  
[Man 89], [Meh 84a], [Meh 84b], [Meh 84c], [RND 77], [Sed 88], [Wil 86], and [Wir 86].
68",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
8. Truth values, the data 
type 'set', and bit acrobatics
Learning objectives:
• truth values, bits
• boolean variables and functions
• bit sum: four clever algorithms compared
• trade-off between time and space
Bits and boolean functions
The English mathematician George Boole (1815–1864) became one of the founders of symbolic logic when he  
endeavored to express logical arguments in mathematical form. The goal of his 1854 book The Laws Of Thought  
was ""to investigate the laws of those operations of the mind by which reasoning is performed; to give expression to  
them in the symbolic language of calculus. …""
Truth values or boolean values, named in Boole's honor, possess the smallest possible useful domain: the binary 
domain, represented by yes/no, 1/0, true/false, T/F. In the late 1940s, as the use of binary arithmetic became",algorithms and data structures.pdf
"domain, represented by yes/no, 1/0, true/false, T/F. In the late 1940s, as the use of binary arithmetic became  
standard and as information theory came to regard a two-valued quantity as the natural unit of information, the  
concise term bit was coined as an abbreviation of ""binary digit"". A bit, by any other name, is truly a primitive data  
element—at a sufficient level of detail, (almost) everything that happens in today's computers is bit manipulation.  
Just because bits are simple data quantities does not mean that processing them is necessarily simple, as we  
illustrate in this section by presenting some clever and efficient bit manipulation algorithms.
Boolean variables  range  over boolean values,  and  boolean functions  take  boolean  arguments and  produce 
boolean results. There are only four distinct boolean functions of a single boolean variable, among which 'not' is the",algorithms and data structures.pdf
"boolean results. There are only four distinct boolean functions of a single boolean variable, among which 'not' is the  
most useful: It yields the complement of its argument (i.e. turns 0 into 1, and vice versa). The other three are the  
identity and the functions that yield the constants 0 and 1. There are 16 distinct boolean functions of two boolean  
variables, of which several are frequently used, in particular: 'and', 'or'; their negations 'nand', 'nor'; the exclusive-or 
'xor'; and the implication '⊃ '. These functions are defined as follows:
a b a and b a or b a nand b a nor b a xor b a ⊃  b
0 0 0 0 1 1 0 1
0 1 0 1 1 0 1 1
1 0 0 1 1 0 1 0
1 1 1 1 0 0 0 1
Bits are the atomic data elements of today's computers, and most programming languages provide a data type  
'boolean' and built-in operators for 'and', 'or', 'not'. To avoid the necessity for boolean expressions to be fully  
Algorithms and Data Structures 69  A Global Text",algorithms and data structures.pdf
"8. Truth values, the data type 'set', and bit acrobatics
parenthesized, precedence relations are defined on these operators: 'not' takes precedence over 'and', which takes  
precedence over 'or'. Thus
x and not y or not x and y  ⇔   ((x and (not y)) or ((not x) and y)).
What can you compute with boolean variables? Theoretically everything, since large finite domains can always  
be represented by a sufficient number of boolean variables: 16-bit integers, for example, use 16 boolean variables to  
represent the integer domain –2 15 .. 2 15–1. Boolean variables are often used for program optimization in practical  
problems where efficiency is important.
Swapping and crossovers: the versatile exclusive-or
Consider the swap statement x :=: y, which we use to abbreviate the cumbersome triple:  t := x;  x := y;  y := t.  
On computers that provide bitwise boolean operations on registers, the swap operator :=: can be implemented  
efficiently without the use of a temporary variable.",algorithms and data structures.pdf
"efficiently without the use of a temporary variable.
The operator exclusive-or, often abbreviated as 'xor', is defined as
x xor y  =  x and not y or not x and y.
It yields true iff exactly one of its two arguments is true.
The bitwise boolean operation z:= x op y on n-bit registers: x[1 .. n], y[1 .. n], z[1 .. n], is defined as
for  i := 1  to  n  do  z[i] := x[i] op y[i]
With a bitwise exclusive-or, the swap x :=: y can be programmed as
x := x xor y;  y := x xor y;  x := x xor y;
It still takes three statements, but no temporary variable. Given that registers are usually in short supply, and  
that a logical operation on registers is typically just as fast as an assignment, the latter code is preferable. Exhibit 
8.1 traces the execution of this code on two 4-bit registers and shows exhaustively that the swap is performed  
correctly for all possible values of x and y.
Exhibit 8.1: Trace of registers x and y under repeated exclusive-or operations.",algorithms and data structures.pdf
"correctly for all possible values of x and y.
Exhibit 8.1: Trace of registers x and y under repeated exclusive-or operations.
Exercise: planar circuits without crossover of wires
The code above has yet another interpretation: How should we design a logical circuit that effects a logical  
crossover of two wires x and y while avoiding any physical crossover? If we had an 'xor' gate, the circuit diagram  
shown in Exhibit 8.2 would solve the problem. 'xor' gates must typically be realized as circuits built from simpler  
primitives, such as 'and', 'or', 'not'. Design a circuit consisting of 'and', 'or', 'not' gates only, which has the effect of  
crossing wires x and y while avoiding physical crossover.
Exhibit 8.2: Three exclusive-or gates in series interchange values on two wires.
70",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The bit sum or ""population count""
A computer word is a fixed-length sequence of bits, call it a bit vector. Typical word lengths are 16, 32, or 64, and 
most instructions in most computers operate on all the bits in a word at the same time, in parallel. When efficiency  
is of great importance, it is worth exploiting to the utmost the bit parallelism built into the hardware of most  
computers. Today's programming languages often fail to refer explicitly to hardware features such as registers or  
words in memory, but it is usually possible to access individual bits if one knows the representation of integers or  
other data types. In this section we take the freedom to drop the constraint of strong typing built into Pascal and  
other modern languages. We interpret the content of a register or a word in memory as it suits the need of the  
moment: a bit string, an integer, or a set.",algorithms and data structures.pdf
"moment: a bit string, an integer, or a set.
We are well aware of the dangers of such ambiguous interpretations: Programs become system and compiler  
dependent, and thus lose portability. If such ambiguity is localized in a single, small procedure, the danger may be  
kept under control, and the gain in efficiency may outweigh these drawbacks. In Pascal, for example, the type 'set' is  
especially well suited to operate at  the bit level. 'type s = set of (a, b, c)' consists of the 2 3 sets that can be formed  
from the three elements a, b, c. If the basic set M underlying the declaration of
type S = set of M
consists of n elements, then S has 2 n elements. Usually, a value of type S is internally represented by a vector of n  
contiguously allocated bits, one bit for each element of the set M. When computing with values of type S we operate  
on single bits using the boolean operators. The union of two sets of type S is obtained by applying bitwise 'or', the",algorithms and data structures.pdf
"on single bits using the boolean operators. The union of two sets of type S is obtained by applying bitwise 'or', the  
intersection by applying bitwise 'and'. The complement of a set is obtained by applying bitwise 'not'.
Example
M = {0, 1, … , 7}
Set                                 Bit vector
7       6       5       4       3       2       1        0  Elements
s1{0, 3, 4, 6} 0 1 0 1 1 0 0 1
s2{0, 1, 4, 5} 0 0 1 1 0 0 1 1
s1 ∪  s2 {0, 1, 3, 4, 5, 6} 0 1 1 1 1 0 1 1
s1 ∩  s2 {0, 4} 0 0 0 1 0 0 0 1
¬  s1 {1, 2, 5, 7} 1 0 1 0 0 1 1 0
Integers are represented on many small computers by 16 bits. We assume that a type 'w16', for ""word of length  
16"", can be defined. In Pascal, this might be
type  w16 = set of 0 .. 15;
A variable of type 'w16' is a set of at most 16 elements represented as a vector of 16 bits.
Asking for the number of elements in a set s is therefore the same as asking for the number of 1's in the bit",algorithms and data structures.pdf
"Asking for the number of elements in a set s is therefore the same as asking for the number of 1's in the bit  
pattern that represents s. The operation that counts the number of elements in a set, or the number of 1's in a word,  
is  called  the  population  count or  bit  sum. The  bit  sum  is  frequently  used  in  inner  loops  of  combinatorial 
calculations, and many a programmer has tried to make it as fast as possible. Let us look at four of these tries,  
beginning with the obvious.
Algorithms and Data Structures 71  A Global Text",algorithms and data structures.pdf
"8. Truth values, the data type 'set', and bit acrobatics
Inspect every bit
function bitsum0(w: w16): integer;
var  i, c: integer;
begin
c := 0;
for  i := 0  to  15  do  { inspect every bit }
if  i ∈  w {w[i] = 1}  then  c := c + 1;  { count the ones}
return(c)
end;
Skip the zeros
Is there a faster way? The following algorithm looks myster ious and tricky. The expression w ∩  (w – 1) contains 
both an intersection operation '∩ ', which assumes that its operands are sets, and a subtraction, which assumes that  
w is an integer:
c := 0;
while  w ≠ 0  do  { c := c + 1;  w := w ∩  (w – 1) } ;
Such mixing makes sense only if we can rely on an implicit assumption on how sets and integers are represented  
as bit vectors. With the usual binary number representation, an example shows that when the body of the loop is  
executed once, the rightmost 1 of w is replaced by 0:
w 1000100011001000
w – 1     1000100011000111
w ∩  (w – 1) 1000100011000000",algorithms and data structures.pdf
"executed once, the rightmost 1 of w is replaced by 0:
w 1000100011001000
w – 1     1000100011000111
w ∩  (w – 1) 1000100011000000
This clever code seems to look at the 1's only and skip over all the 0's: Its loop is executed only as many times as  
there are 1's in the word. This savings is worthwhile for long, sparsely populated words (few 1's and many 0's).
In the statement w := w ∩   (w – 1), w is used both as an integer (in w – 1) and as a set (as an operand in the  
intersection operation ' ∩ '). Strongly typed languages, such as Pascal, do not allow such mixing of types. In the  
following function 'bitsum 1', the  conversion routines  'w16toi' and  'itow16' are  introduced  to avoid  this  double 
interpretation of w. However, 'bitsum1' is of interest only if such a type conversion requires no extra time (i.e. if one  
knows how sets and integers are represented internally).
function bitsum1(w: w16): integer;
var  c, i: integer;  w0, w1: w16;
begin
w0 := w;  c := 0;",algorithms and data structures.pdf
"knows how sets and integers are represented internally).
function bitsum1(w: w16): integer;
var  c, i: integer;  w0, w1: w16;
begin
w0 := w;  c := 0;
while  w0 ≠ Ø  { empty set }  do  begin
i := w16toi(w0);  { w16toi converts type w16 to integer }
i := i – 1;
w1 := itow16(i);  { itow16 converts type integer to w16 }
w0 := w0 ∩  w1;  { intersection of two sets }
c := c + 1
end;
return(c)
end;
72",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Most languages provide some facility for permitting purely formal type conversions that result in no work:  
'EQUIVALENCE' statements in Fortran, 'UNSPEC' in PL/1, variant records in Pascal. Such ""conversions"" are done  
merely by interpreting the contents of a given storage location in different ways.
Logarithmic bit sum
For a computer of word length n, the following algorithm computes the bit sum of a word w running through its  
loop only  ⎡log2  n⎤ times,  as  opposed  to  n  times  for  'bitsum0'  or  up  to  n  times  for  'bitsum1'.  The  following 
description holds for arbitrary n but is understood most easily if n = 2h.
The logarithmic bit sum works on the familiar principle of divide-and-conquer. Let w denote a word consisting  
of n = 2h bits, and let S(w) be the bit sum of the bit string w. Split w into two halves and denote its left part by wL",algorithms and data structures.pdf
"of n = 2h bits, and let S(w) be the bit sum of the bit string w. Split w into two halves and denote its left part by wL  
and its right part by wR. The bit sum obviously satisfies the recursive equation S(w) = S(wL) + S(wR). Repeating  
the same argument on the substrings wL and wR, and, in turn, on the substrings they create, we arrive at a process  
to compute S(w). This process terminates when we hit substrings of length 1 [i.e. substrings consisting of a single  
bit b; in this case we have S(b) = b]. Repeated halving leads to a recursive decomposition of w, and the bit sum is  
computed by a tree of n – 1 additions as shown below for n = 4 (Exhibit 8.3).
Exhibit 8.3: Logarithmic bit sum algorithm as a result of divide-and-conquer.
This approach of treating both parts of w symmetrically and repeated halving leads to a computation of depth h  
= ⎡log2 n⎤ . To obtain a logarithmic bit sum, we apply the additional trick of performing many additions in parallel.",algorithms and data structures.pdf
"= ⎡log2 n⎤ . To obtain a logarithmic bit sum, we apply the additional trick of performing many additions in parallel.  
Notice that the total length of all operands on the same level is always n. Thus we can pack them into a single word  
and, if we arrange things cleverly, perform all the additions at the same level in one machine operation, an addition  
of two n-bit words.
Exhibit 8.4 shows how a number of the additions on short strings are carried out by a single addition on long  
strings. S(w) now denotes not only the bit sum but also its binary representation, padded with zeros to the left so as  
to have the appropriate length. Since the same algorithm is being applied to wL and wR, and since w L and wR are of 
equal length, exactly the same operations are  performed  at each stage on wL and its parts as on wR and  its  
corresponding parts. Thus if the operations of addition and shifting operate on words of length n, a single one of",algorithms and data structures.pdf
"corresponding parts. Thus if the operations of addition and shifting operate on words of length n, a single one of  
these operations can be interpreted as performing many of the same operations on the shorter parts into which w  
has been split. This logarithmic speedup works up to the word length of the computer. For n = 64, for example,  
recursive splitting generates six levels and translates into six iterations of the loop below.
Algorithms and Data Structures 73  A Global Text",algorithms and data structures.pdf
"8. Truth values, the data type 'set', and bit acrobatics
Exhibit 8.4: All processes generated by divide-and-conquer are performed in parallel 
on shared data registers.
 The algorithm is best explained with an example; we use n = 8.
w7 w6 w5 w4 w3 w2 w1 w0
w 1 1 0 1 0 0 0 1
First, extract the even-indexed bits w 6 w4 w2 w0 and place a zero to the left of each bit to obtain w even. The newly 
inserted zeros are shown in small type.
w6 w4 w2 w0
weven
0 1 0 1 0 0 0 1
Next, extract the odd-indexed bits w 7 w5 w3 w1, shift them right by one place into bit positions w 6 w4 w2 w0, and 
place a zero to the left of each bit to obtain wodd.
w7 w5 w3 w1
wodd
0 1 0 0 0 0 0 0
Then, numerically add weven and wodd, considered as integers written in base 2, to obtain w'.
w'7 w'6 w'5 w'4 w'3 w'2 w'1 w'0
weven 0 1 0 1 0 0 0 1
wodd 0 1 0 0 0 0 0 0
w' 1 0 0 1 0 0 0 1
Next, we index not bits, but pairs of bits, from right to left: (w' 1 w'0) is the zeroth pair, (w'5 w'4) is the second pair.",algorithms and data structures.pdf
"w' 1 0 0 1 0 0 0 1
Next, we index not bits, but pairs of bits, from right to left: (w' 1 w'0) is the zeroth pair, (w'5 w'4) is the second pair. 
Extract the even-indexed pairs w'5 w'4 and w'1 w'0, and place a pair of zeros to the left of each pair to obtain w'even.
74",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
w'5 w'4 w'1 w'0
w'even
0 0 0 1 0 0 0 1
Next, extract the odd-indexed pairs w' 7 w'6  and w'3 w'2 , shift them right by two places into bit positions w' 5  w'4 
and w'1 w'0 , respectively, and insert a pair of zeros to the left of each pair to obtain w'odd.
w'7 w'6 w'3 w'2
w'odd
0 0 1 0 0 0 0 0
Numerically, add w'even and w'odd to obtain w"".
w""7 w""6 w""5 w""4 w""3 w""2 w""1 w""0
w"" 0 0 1 1 0 0 0 1
Next, we index quadruples of bits, extract the quadruple w""3 w""2 w""1 w""0, and place four zeros to the left to obtain 
w""even.
w""3 w""2 w""1 w""0
w""even
0 0 0 0 0 0 0 1
Extract the quadruple w"" 7 w""6 w""5 w""4, shift it right four places into bit positions w"" 3 w""2 w""1 w""0, and place four  
zeros to the left to obtain w""odd.
w""7 w""6 w""5 w""4
w""odd
0 0 0 0 0 0 1 1
Finally, numerically add w""even and w""odd to obtain w''' = (00000100), which is the representation in base 2 of the",algorithms and data structures.pdf
"w""odd
0 0 0 0 0 0 1 1
Finally, numerically add w""even and w""odd to obtain w''' = (00000100), which is the representation in base 2 of the  
bit sum of w (4 in this example). The following function implements this algorithm.
Logarithmic bit sum implemented for a 16-bit computer:
In 'bitsum2' we apply addition and division operations directly to variables of type 'w16' without performing the  
type conversions that would be necessary in a strongly typed language such as Pascal.
function bitsum2(w: w16): integer;
const mask[0] = '0101010101010101'; 
mask[1] = '0011001100110011';
mask[2] = '0000111100001111';
mask[3] = '0000000011111111';
var  i, d: integer;  weven, wodd: w16;
begin
d := 2;
for  i := 0  to  3  do  begin
weven := w ∩  mask[i];
w := w / d;  { shift w right 2i bits }
d := d2;
wodd := w ∩  mask[i];
w := weven + wodd
end;
return(w)
end;
Algorithms and Data Structures 75  A Global Text",algorithms and data structures.pdf
"8. Truth values, the data type 'set', and bit acrobatics
Trade-off between time and space: the fastest algorithm
Are th ere still faster algorithms for computing the bit sum of a word? Is there an  optimal algorithm? The  
question of optimality of algorithms is important, but it can be answered only in special cases. To show that an  
algorithm is optimal, one must specify precisely the class of algorithms allowed and the criterion of optimality. In  
the case of bit sum algorithms, such specifications would be complicated and largely arbitrary, involving specific  
details of how computers work.
However, we can make a plausible argument that the following bit sum algorithm is the fastest possible, since it  
uses a table lookup to obtain the result in essentially one operation. The penalty for this speed is an extravagant use  
of memory space (2 n locations), thereby making the algorithm impractical except for small values of n. The choice",algorithms and data structures.pdf
"of memory space (2 n locations), thereby making the algorithm impractical except for small values of n. The choice  
of an algorithm almost always involves trade-offs among various desirable properties, and the better an algorithm is 
from one aspect, the worse it may be from another.
The algorithm is based on the idea that we can precompute the solutions to all possible questions, store the  
results, and then simply look them up when needed. As an example, for n = 3, we would store the information
Word Bit sum
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 2
1 0 0 1
1 0 1 2
1 1 0 2
1 1 1 3
What is the fastest way of looking up a word w in this table? Under assumptions similar to those used in the  
preceding algorithms, we can interpret w as an address of a memory cell that contains the bit sum of w, thus giving  
us an algorithm that requires only one memory reference.
Table lookup implemented for a 16-bit computer:
function bitsum3(w: w16): integer;",algorithms and data structures.pdf
"us an algorithm that requires only one memory reference.
Table lookup implemented for a 16-bit computer:
function bitsum3(w: w16): integer;
const  c: array[0 .. 65535] of integer = [0, 1, 1, 2, 1, 2, 2, 3, 
… , 15, 16];
begin  return(c[w])  end;
In concluding this exa mple, we notice the variety of algorithms that exist for computing the bit sum, each one  
based on entirely different principles, giving us a different trade-off between space and time. 'bitsum 0' and 'bitsum3' 
solve the problem by ""brute force"" and are simple to understand: 'bitsum 0' looks at each bit and so requires much  
time; 'bitsum3' stores the solution for each separate case and thus requires much space. The logarithmic bit sum  
algorithm  is  an  elegant  compromise:  efficient  with  respect  to  both  space  and  time,  it  merely  challenges  the 
programmer's wits.
Exercises
1. Show that there are exactly 16 distinct boolean functions of two variables.",algorithms and data structures.pdf
"programmer's wits.
Exercises
1. Show that there are exactly 16 distinct boolean functions of two variables.
2. Show that each of the boolean functions 'nand' and 'nor' is universal in the following sense: Any boolean  
function f(x, y) can be written as a nested expression involving only 'nands', and it can also be written using  
only 'nors'. Show that no other boolean function of two variables is universal.
76",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
3. Consider the logarithmic bit sum algorithm, and show that any strategy for splitting w (not just the halving  
split) requires n – 1 additions.
Algorithms and Data Structures 77  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
9. Ordered sets
Learning objectives:
• searching in ordered sets 
• sequential search. proof of program correctness
• binary search 
• in-place permutation
• nondeterministic algorithms
• cycle rotation
• cycle clipping
Sets of elements processed on a computer are always ordered according to some criterion. In the preceding  
example of the ""population count"" operation, a set is ordered arbitrarily and implicitly simply because it is mapped  
onto linear storage; a programmer using that set can ignore any order imposed by the implementation and access  
the set through functions that hide irrelevant details. In most cases, however, the order imposed on a set is not  
accidental, but is prescribed by the problem to be solved and/or the algorithm to be  used. In such cases  the  
programmer explicitly deals with issues of how to order a set and how to use any existing order to advantage.",algorithms and data structures.pdf
"programmer explicitly deals with issues of how to order a set and how to use any existing order to advantage.
Searching in ordered sets is one of the most frequent tasks performed by computers: whenever we operate on a  
data item, that item must be selected from a set of items. Searching is also an ideal ground for illustrating basic  
concepts and techniques of programming.
At times, ordered sets need to be rearranged (permuted). The chapter “Sorting and its complexity” is dedicated  
to the most frequent type of rearrangement: permuting a set of elements into ascending order. Here we discuss  
another type of rearrangement: reordering a set according to a given permutation.
Sequential search
Consider the simple case where a fixed set of n data elements is given in an array A:
const  n = … ;  { n > 0 }
type  index = 0 .. n;  elt = … ;
var  A: array[1 .. n] of elt;    or    var A: array[0 .. n] of elt;",algorithms and data structures.pdf
"const  n = … ;  { n > 0 }
type  index = 0 .. n;  elt = … ;
var  A: array[1 .. n] of elt;    or    var A: array[0 .. n] of elt;
Sequential or linear search is the simplest technique for determining whether A contains a given element x. It is  
a trivial example of an incremental algorithm, which processes a set of data one element at a time. If the search for  
x is successful, we return an index i, 1 ≤ i ≤ n, to point to x. The convention that i = 0 signals unsuccessful search is  
convenient and efficient, as it encodes all possible outcomes in a single parameter.
function find(x: elt): index; 
var  i: index;
begin
i := n;
while  (i > 0)  { can access A }  cand  (A[i] ≠ x)  { not yet 
found }  do
(1)  { (1 ≤ i ≤ n) ∧ (∀  k, i ≤ k: A[k] ≠ x) }
 i := i – 1;
Algorithms and Data Structures 78  A Global Text",algorithms and data structures.pdf
"9. Ordered sets
(2)  { (∀ k, i < k: A[k] ≠ x) ∧ ((i= 0) ∧ ((1 ≤ i ≤ n) ∧ (A[i] = x))) }
 return(i)
end;
The 'cand' operator used in the termination condition is the conditional 'and'. Evaluation proceeds from left to  
right and stops as soon as the value of the boolean expression is determined: If i > 0 yields 'false', we immediately  
terminate evaluation of the boolean expression without accessing A[i], thus avoiding an out-of-bounds error.
We have included two assertions, (1) and (2), that express the main points necessary for a formal proof of  
correctness: mainly, that each iteration of the loop extends by one element the subarray known not to contain the  
search argument x. Assertion (1) is trivially true after the initialization i := n, and remains true whenever the body  
of the while loop is about to be executed. Assertion (2) states that the loop terminates in one of two ways:
• i = 0 signals that the entire array has been scanned unsuccessfully.",algorithms and data structures.pdf
"• i = 0 signals that the entire array has been scanned unsuccessfully.
• x has been found at index i.
A formal correctness proof would have to include an argument that the loop does indeed terminate—a simple  
argument here, since i is initialized to n, decreases by 1 in each iteration, and thus will become 0 after a finite  
number of steps.
The loop is terminated by a Boolean expression composed of two terms: reaching the end of the array, i = 0, and  
testing the current array element, A[i] = x. The second term is unavoidable, but the first one can be spared by  
making sure that x is always found before the index i drops off the end of the array. This is achieved by extending  
the array by one cell A[0] and placing the search argument x in it as a sentinel. If no true element x stops the scan of 
the array, the sentinel will. Upon exit from the loop, the value of i reveals the outcome of the search, with the  
convention that 0 signals an unsuccessful search:",algorithms and data structures.pdf
"convention that 0 signals an unsuccessful search:
function find(x: elt): index; 
var  i: index;
begin
A[0] := x;  i := n;
while  A[i] ≠ x  do  i := i – 1;
return(i)
end;
How efficient is sequential search? An unsuccessful search always scans the entire array. If all n array elements  
have equal probability of being searched for, the average number of iterations of the while loop in a successful  
search is
This algorithm needs time proportional to n in the average and the worst case.
Binary search
If the data elements stored in the array A are ordered according to the order relation ≤ defined on their domain,  
that is
∀  k, 1 ≤ k < n:  A[k] ≤ A[k + 1]
the search for an element x can be made much faster because a comparison of x with any array element A[m]  
provides more information than it does in the unordered case. The result x ≠ A[m] excludes not only A[m], but also",algorithms and data structures.pdf
"provides more information than it does in the unordered case. The result x ≠ A[m] excludes not only A[m], but also  
all elements on one or the other side of A[m], depending on whether x is greater or smaller than A[m] (Exhibit 9.1).
79",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 9.1: Binary search identifies regions where the search argument is guaranteed to be absent.
The following function exploits this additional information:
const  n = … ;  { n > 0 }
type  index = 1 .. n;  elt = … ;
var  A: array[1 .. n] of elt;
function find(x: elt;  var m: index): boolean;
var  u, v: index;
begin
u := 1;  v := n;
while  u ≤ v  do  begin
(1) { (u ≤ v) ∧ (∀  k, 1 ≤ k < u: A[k] < x) ∧(∀  k, v < k ≤ n: A[k] > 
x) }
m := any value such that u ≤ m ≤ v ;
if    x < A[m]    thenv := m – 1
elsif  x > A[m]  then u := m + 1
(2) else  {x = A[m] } return(true)
end;
(3) { (u = v + 1) ∧(∀  k, 1 ≤ k < u: A[k]< x) ∧ (∀  k, v < k ≤ n: 
A[k] > x) }
return(false)
end;
u and v bound the interval of uncertainty that might contain x. Assertion (1) states that A[1], … , A[u – 1] are known  
to be smaller than x; A[v + 1], … , A[n] are known to be greater than x. Assertion (2), before exit from the function,",algorithms and data structures.pdf
"to be smaller than x; A[v + 1], … , A[n] are known to be greater than x. Assertion (2), before exit from the function,  
states that x has been found at index m. In assertion (3), u = v + 1 signals that the interval of uncertainty has shrunk  
to become empty. If there exists more than one match, this algorithm will find one of them.
This algorithm is correct independently of the choice of m but is most efficient when m is the midpoint of the  
current search interval:
m := (u + v) div 2;
With this choice of m each comparison either finds x or eliminates half of the remaining elements. Thus at most  
⎡log2 n⎤ iterations of the loop are performed in the worst case.
Exercise: binary search
The array
var A: array [1 .. n] of integer;
contains n integers in ascending order: A[1] ≤ A[2] ≤ … ≤ A[n].
(a) Write a recursive binary search
function rbs (x, u, v: integer): integer;
that returns 0 if x is not in A, and an index i such that A[i] = x if x is in A.",algorithms and data structures.pdf
"function rbs (x, u, v: integer): integer;
that returns 0 if x is not in A, and an index i such that A[i] = x if x is in A.
(b) What is the maximal depth of recursive calls of 'rbs' in terms of n?
Algorithms and Data Structures 80  A Global Text
A[m] < x
m
x < A[m]
m
If
x cannot lie in",algorithms and data structures.pdf
"9. Ordered sets
(c) Describe the advantages and disadvantages of this recursive binary search as compared to the iterative 
binary search.
Exercise: searching in a partially ordered two-dimensional array
Consider the n by m array:
var  A: array[1 .. n, 1 .. m] of integer;
and assume that the integers in each row and in each column are in ascending order; that is,
A[i, j] ≤ A[i, j + 1]for i = 1, … , n and j = 1, … , m – 1;
A[i, j] ≤ A[i + 1, j]for i = 1, … , n – 1 and j = 1, … , m.
(a) Design an algorithm that determines whether a given integer x is stored in the array A. Describe your  
algorithm in words and figures. Hint: Start by comparing x with A[1, m] (Exhibit 9.2).
Exhibit 9.2: Another example of the idea of excluded regions.
(b) Implement your algorithm by a
function IsInArray (x: integer): boolean;
(c) Show that your algorithm is correct and terminates, and determine its worst case time complexity.
Solution",algorithms and data structures.pdf
"(c) Show that your algorithm is correct and terminates, and determine its worst case time complexity.
Solution
(a) The algorithm compares x first with A[1, m]. If x is smaller than A[1, m], then x cannot be contained in the  
last column, and the search process is continued by comparing x with A[1, m – 1]. If x is greater than A[1,  
m], then x cannot be contained in the first row, and the search process is continued by comparing x with  
A[2, m]. Exhibit 9.3 shows part of a typical search process.
Exhibit 9.3: Excluded regions combine to leave only a staircase-shaped strip to examine.
(b) function IsInArray(x: integer): boolean;
var  r, c: integer;
begin
r := 1;  c := m;
while  (r ≤ n) and (c ≥ 1)  do
{1} ifx < A[r, c] then c := c – 1
elsif x > A[r, c] then r := r + 1
81",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
else { x = A[r, c] } {2} return(true);
{3}  return(false)
end;
(c) At positions {1}, {2}, and {3}, the invariant
∀  i, 1 ≤ i ≤ n,∀ j, 1 ≤ j ≤ m:
(j > c ⇒  x ≠ A[i, j]) ∧ (i < r ⇒  x ≠ A[i, j] (∗)
states that the hatched rows and columns of A do not contain x. At {2},
(1 ≤ r ≤ n) ∧ (1 ≤ c ≤ m) ∧ (x = A[r, c])
states that r and c are within index range and x has been found at (r, c). At {3},
(r = n + 1) ∨(c = 0)
states that r or c are outside the index range. This coupled with (*) implies that x is not in A: 
(r = n + 1) ∨ (c = o) ⇒  ∀  i, 1 ≤ i ,≤ n, ∀  j. 1 ≤ j ≤ m: x ≠ A[i, j].
Each iteration through the loop either decreases c by one or increases r by one. If x is not contained in the array,  
either c becomes zero or r becomes greater than n after a finite number of steps, and the algorithm terminates. In",algorithms and data structures.pdf
"either c becomes zero or r becomes greater than n after a finite number of steps, and the algorithm terminates. In  
each step, the algorithm eliminates either a row from the top or a column from the right. In the worst case it works  
its way from the upper right corner to the lower left corner in n + m – 1 steps, leading to a complexity of Θ (n + m).
In-place permutation
Representations of a permutation.  Consider an array D[1 .. n] that holds n data elements of type 'elt'.  
These are ordered by their position in the array and must be rearranged according to a specific permutation given  
in another array. Exhibit 9.4 shows an example for n = 5. Assume that a, b, c, d, e, stored in this order, are to be  
rearranged in the order c, e, d, a, b. This permutation is represented naturally by either of the two permutation  
arrays t (to) or f (from) declared as
var  t, f: array[1 .. n] of 1 .. n;",algorithms and data structures.pdf
"arrays t (to) or f (from) declared as
var  t, f: array[1 .. n] of 1 .. n;
The exhibit also shows a third representation of the same permutation: the decomposition of this permutation  
into cycles. The element in D[1] moves into D[4], the one in D[4] into D[3], the one in D[3] into D[1], closing a cycle  
that we abbreviate as (1 4 3), or (4 3 1), or (3 1 4). There is another cycle (2 5), and the entire permutation is  
represented by (1 4 3) (2 5).
Exhibit 9.4: A permutation and its representations in terms of  'to', 'from', and cycles. 
The cycle representation is intuitively most informative, as it directly reflects the decomposition of the problem  
into independent subproblems, and both the 'to' and 'from' information is easily extracted from it. But 'to' and  
'from' dispense with parentheses and lead to more concise programs.
Algorithms and Data Structures 82  A Global Text",algorithms and data structures.pdf
"9. Ordered sets
Consider the problem of executing this permutation in place: Both the given data and the result are stored in the 
same array D, and only a (small) constant amount of auxiliary storage may be used, independently of n. Let us use  
the example of in-place permutation to introduce a notation that is frequently convenient, and to illustrate how the  
choice of primitive operations affects the solution.
A multiple assignment statement will do the job, using either 'to' or 'from':
// (1 ≤ i ≤ n)  { D[t[i]] := D[i] }
or
// (1 ≤ i ≤ n)  { D[i]} := D[f[i]] }
The characteristic properties of a multiple assignment statement are:
• The left-hand side is a sequence of variables, the right-hand side is a sequence of expressions, and the two 
sequences are matched according to length and type. The value of the i-th expression on the right is 
assigned to the i-th variable on the left.",algorithms and data structures.pdf
"sequences are matched according to length and type. The value of the i-th expression on the right is 
assigned to the i-th variable on the left.
• All the expressions on the right-hand side are evaluated using the original values of all variables that occur 
in them, and the resulting values are assigned ""simultaneously"" to the variables on the left-hand side. We 
use the sign // to designate concurrent or parallel execution.
Few of today's programming languages offer multiple assignments, in particular those of variable length used  
above.  Breaking  a  multiple  assignment  into  single  assignments  usually  forces  the  programmer  to  introduce 
temporary variables. As an example, notice that the direct sequentialization:
for  i := 1  to  n  do  D[t[i]] := D[i]
or
for  i := 1  to  n  do  D[i] := D[f[i]]
is faulty, as some of the elements in D will be overwritten before they can be moved. Overwriting can be avoided at",algorithms and data structures.pdf
"is faulty, as some of the elements in D will be overwritten before they can be moved. Overwriting can be avoided at  
the cost of nearly doubling memory requirements by allocating an array A[1 .. n] of data elements for temporary  
storage:
for  i := 1  to  n  do  A[t[i]] := D[i];
for  i := 1  to  n  do  D[i] := A[i];
This,  however,  is  not  an  in-place  computation,  as  the  amount  of  auxiliary  storage  grows  with  n.  It  is 
unnecessarily inefficient: There are elegant in-place permutation algorithms based on the conventional primitive of  
the single assignment statement. They all assume that the permutation array may be destroyed as the permutation  
is being executed. If the representation of the permutation must be preserved, additional storage is required for  
bookkeeping, typically of a size proportional to n. Although this additional space may be as little as n bits, perhaps",algorithms and data structures.pdf
"bookkeeping, typically of a size proportional to n. Although this additional space may be as little as n bits, perhaps  
in order to distinguish the elements processed from those yet to be moved, such an algorithm is not technically in  
place.
Nondeterministic algorithms.  Problems of rearrangement always appear to admit many different solutions
—a phenomenon that is most apparent when one considers the multitude of sorting algorithms in the literature.  
The reason is clear: When n elements must be moved, it may not matter much which elements are moved first and  
which ones later. Thus it is useful to look for nondeterministic algorithms that refrain from specifying the precise  
sequence  of  all  actions  taken,  and  instead  merely  iterate  condition ⇒  action statements,  with  the  meaning 
""wherever  condition  applies perform the corresponding  action"". These algorithms are nondeterministic because",algorithms and data structures.pdf
"""wherever  condition  applies perform the corresponding  action"". These algorithms are nondeterministic because  
each of several distinct conditions may apply at lots of different places, and we may ""fire"" any  action that is  
83",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
currently  enabled.  Adding  sequential  control  to  a  nondeterministic  algorithm  turns  it  into  a  deterministic 
algorithm. Thus a nondeterministic algorithm corresponds to a class of deterministic ones that share common  
invariants, but differ in the order in which steps are executed. The correctness of a nondeterministic algorithm  
implies the correctness of all its sequential instances. Thus it is good algorithm design practice to develop a correct  
nondeterministic algorithm first, then turn it into a deterministic one by ordering execution of its steps with the  
goal of efficiency in mind.
Deterministic sequential algorithms come in a variety of forms depending on the choice of primitive (assignment 
or swap), data representation ('to' or 'from'), and technique. We focus on the latter and consider two techniques:",algorithms and data structures.pdf
"or swap), data representation ('to' or 'from'), and technique. We focus on the latter and consider two techniques:  
cycle rotation and cycle clipping. Cycle rotation follows naturally from the idea of decomposing a permutation into  
cycles  and  processing  one  cycle  at  a  time,  using  temporary  storage  for  a  single  element.  It  fits  the  'from' 
representation somewhat more efficiently than the 'to' representation, as the latter requires a swap of two elements  
where the former uses an assignment. Cycle clipping uses the primitive 'swap two elements' so effectively as a step  
toward executing a permutation that it needs no temporary storage for elements. Because no temporary storage is  
tied up, it is not necessary to finish processing one cycle before starting on the next one –elements can be clipped  
from their cycles in any order. Clipping works efficiently with either representation, but is easier to understand with",algorithms and data structures.pdf
"from their cycles in any order. Clipping works efficiently with either representation, but is easier to understand with  
'to'.  We  present cycle  rotation  with  'from'  and  cycle  clipping  with  'to' and  leave  the  other  two  algorithms  as 
exercises.
Cycle rotation
A search for an in-place algorithm naturally leads to the idea of processing a permutation one cycle at a time:  
every element we place at its destination bumps another one, but we avoid holding an unbounded number of  
bumped elements in temporary storage by rotating each cycle, one element at a time. This works best using the  
'from' representation. The following loop rotates the cycle that passes through an arbitrary index i:
Rotate the cycle starting at index i, updating f:
j := i;{ initialize a two-pronged fork to travel along the cycle }
p := f[j]; { p is j's predecessor in the cycle }
A := D[j]; { save a single element in an auxiliary variable A }",algorithms and data structures.pdf
"p := f[j]; { p is j's predecessor in the cycle }
A := D[j]; { save a single element in an auxiliary variable A }
while  p ≠ i  do  { D[j] := D[p];  f[j] := j;  j := p;  p := f[j]} ;
D[j] := A; { reinsert the saved element into the former cycle … }
f[j] := j; { … but now it is a fixed point }
This code works trivially for a cycle of length 1, where p = f[i] = i guards the body of the loop from ever being  
executed. The statement f[j] := j in the loop is unnecessary for rotating the cycle. Its purpose is to identify an  
element that has been placed at its final destination, so this code can be iterated for 1 ≤ i ≤ n to yield an in-place  
permutation algorithm. For the sake of efficiency we add two details: (1) We avoid unnecessary movements  A :=  
D[j];  D[j] := A  of a possibly voluminous element by guarding cycles of length 1 with the test 'i ≠ f[i]', and (2) we",algorithms and data structures.pdf
"D[j];  D[j] := A  of a possibly voluminous element by guarding cycles of length 1 with the test 'i ≠ f[i]', and (2) we  
terminate the iteration at n – 1 on the grounds that when n – 1 elements of a permutation are in their correct place,  
the n-th one is also. Using the code above, this leads to
for  i := 1  to  n – 1  do if  i ≠ f[i]  then  rotate the cycle starting at index i, updating f
Exercise
Implement  cycle  rotation  using  the  'to'  representation.  Hint: Use  the  swap  primitive  rather  than  element 
assignment.
Algorithms and Data Structures 84  A Global Text",algorithms and data structures.pdf
"9. Ordered sets
Cycle clipping
Cycle clipping is the key to elegant in-place permutation using the 'to' representation. At each step, we clip an  
arbitrary element d out of an arbitrary cycle of length > 1, thus reducing the latter's length by 1. As shown in Exhibit 
9.5, we place d at its destination, where it forms a cycle of length 1 that needs no further processing. The element it  
displaces, c, can find a (temporary) home in the cell vacated by d. It is probably out of place there, but no more so  
than it was at its previous home; its time will come to be relocated to its final destination. Since we have permuted  
elements, we must update the permutation array to reflect accurately the permutation yet to be performed. This is a  
local operation in the vicinity of the two elements that were swapped, somewhat like tightening a belt by one notch
—all but two of the elements in the clipped cycle remain unaffected. The Exhibit below shows an example. In order",algorithms and data structures.pdf
"—all but two of the elements in the clipped cycle remain unaffected. The Exhibit below shows an example. In order  
to execute the permutation (1 4 3) (2 5), we clip d from its cycle (1 4 3) by placing d at its destination D[3], thus  
bumping c into the vacant cell D[4]. This amounts to representing the cycle (1 4 3) as a product of two shorter  
cycles: the swap (3 4), which can be done right away, and the cycle (1 4) to be executed later. The cycle (2 5) remains 
unaffected. The ovals in Exhibit 9.5 indicate that corresponding entries of D and t are moved together. Exhibit 9.6 
shows what happens to a cycle clipped by a swap 
// { t[i], D[i]  :=:  t[t[i]], D[t[i]] }
Exhibit 9.5: Clipping one element out of a cycle of a permutation.
Exhibit 9.6: Effect of a swap caused by the condition i ≠ t[i].
85",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Cycles of length 1 are left alone, and the absence of cycles of length > 1 signals termination. Thus the following  
condition ⇒  action statement,  iterated  as  long  as  the  condition  i  ≠  t[i]  can  be  met,  executes  a  permutation 
represented in the array t:
∃ i:i ≠ t[i]  ⇒   // { t[i], D[i]  :=:  t[t[i]], D[t[i]] }
We use the multiple swap operator // { :=: } with the meaning: evaluate all four expressions using the original  
values of all the variables involved, then perform all four assignments simultaneously. It can be implemented using  
six single assignments and two auxiliary variables, one of type 1 .. n, the other of type 'elt'. Each swap places (at  
least) one element into its final position, say j, where it is guarded from any further swaps by virtue of j = t[j]. Thus  
the nondeterministic algorithm above executes at most n – 1 swaps: When n – 1 elements are in final position, the",algorithms and data structures.pdf
"the nondeterministic algorithm above executes at most n – 1 swaps: When n – 1 elements are in final position, the  
n-th one is also.
The conditions on i can be checked in any order, as long as they are checked exhaustively, for example:
{ (0)  (1 ≤ j < 0) ⇒  j = t[j] }
for  i := 1  to  n – 1  do
{ (1)  (1 ≤ j < i) ⇒  j = t[j] }
while  i ≠ t[i]  do  // { t[i], D[i]  :=:  t[t[i]], D[t[i]] }
{ (2)  (1 ≤ j ≤ i) ⇒  j = t[j] }
{ (3)  (1 ≤ j ≤ n – 1) ⇒  j = t[j] }
For each value of i, i is the leftmost position of the cycle that passes through i. As the while loop reduces this  
cycle to cycles of length 1, all swaps involve i and t[i] > i, as asserted by the invariant (1) (1 ≤ j < I) ⇒  j = t[j], which 
precedes the while loop. At completion of the while loop, the assertion is strengthened to include i, as stated in  
invariant (2) (1 ≤ j ≤ I) ⇒  j = t[j]. This reestablishes (1) for the next higher value of i. The vacuously true assertion",algorithms and data structures.pdf
"invariant (2) (1 ≤ j ≤ I) ⇒  j = t[j]. This reestablishes (1) for the next higher value of i. The vacuously true assertion  
(0) serves as the basis of this proof by induction. The final assertion (3) is just a restatement of assertion (2) for the  
last value of i. Since t[1] … t[n] is a permutation of 1 …n, (3) implies that n = t[n].
Exercise: cycle clipping using the 'from' representation
The nondeterministic algorithm expressed as a multiple assignment
// (1 ≤ i ≤ n)  { D[i]} := D[f[i]] }
is equally as valid for the 'from' representation as its analog 
// (1 ≤ i ≤ n)  { D[t[i]] := D[i] }
was for the 'to' representation. But in contrast to the latter, the former cannot be translated into a simple iteration  
of the condition ⇒  action statement:
∃ i: i ≠ f[i]  ⇒   // { f[i], D[i]  :=:  f[f[i]], D[f[i]] }
Why not? Can you salvage the idea of cycle clipping using the 'from' representation
Exercises",algorithms and data structures.pdf
"Why not? Can you salvage the idea of cycle clipping using the 'from' representation
Exercises
1. Write two functions that implement sequential search, one with sentinel as shown in the first section,  
""Sequential search"" the other without sentinel. Measure and compare their running time on random arrays  
of various sizes
2. Measure and compare the running times of sequential search and binary search on random arrays of size n,  
for n = 1 to n = 100. Sequential search is obviously faster for small values of n, and binary search for large n, 
but where is the crossover? Explain your observations.
Algorithms and Data Structures 86  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
10. Strings 
Learning objectives:
• searching for patterns in a string 
• finite-state machine
Most programming languages support simple operations on strings (e.g. comparison, concatenation, extraction,  
searching). Searching for a specified pattern in a string (text) is the computational kernel of most string processing  
operations. Several efficient algorithms have been developed for this potentially time-consuming operation. The  
approach presented here is very general; it allows searching for a pattern that consists not only of a single string,  
but a set of strings. The cardinality of this set influences the storage space needed, but not the time. It leads us to  
the concept of a finite-state machine (fsm).
Recognizing a pattern consisting of a single string
Problem: Given a (long) string z = z1 z2 … zn of n characters and a (usually much shorter) string p = p 1 p2 … pm of",algorithms and data structures.pdf
"Problem: Given a (long) string z = z1 z2 … zn of n characters and a (usually much shorter) string p = p 1 p2 … pm of 
m characters (the pattern), find all (nonoverlapping) occurrences of p in z. By sliding a window of length m from  
left to right along z and examining most characters z i  m times we solve the problem using m · n comparisons. By  
constructing a finite-state machine from the pattern p it suffices to examine each character zi exactly once, as shown 
in Exhibit 10.1. Each state corresponds to a prefix of the pattern, starting with the empty prefix and ending with the  
complete pattern. The input symbols are the input characters z 1, z2, … , zn of z. In the j-th step the input character z j 
leads from a state corresponding to the prefix p1 p2 … pi to
• the state with prefix p1 p2 … pi pi+1 if zj = pi+1
• a different state (often the empty prefix, λ) if zj ≠ pi+1
Example
p = barbara (Exhibit 10.1).",algorithms and data structures.pdf
"• the state with prefix p1 p2 … pi pi+1 if zj = pi+1
• a different state (often the empty prefix, λ) if zj ≠ pi+1
Example
p = barbara (Exhibit 10.1).
Exhibit 10.1: State diagram showing some of the transitions. All other state transitions lead back to the 
initial state.
Notice  that  the  pattern  'barbara',  although  it  sounds  repetitive,  cannot  overlap  with  any  part  of  itself. 
Constructing a finite-state machine for such a pattern is straightforward. But consider a self-overlapping pattern  
such as 'barbar', or 'abracadabra', or 'xx', where the first k > 0 characters are identical with the last: The text  
'barbarbar' contains two overlapping occurrences of the pattern 'barbar', and 'xxxx' contains three occurrences of  
'xx'. A finite-state machine constructed in an analogous fashion as the one used for 'barbara' always finds the first of  
Algorithms and Data Structures 87  A Global Text",algorithms and data structures.pdf
"10. Strings 
several  overlapping  occurrences  but might miss  some  of the  later  ones.  As  an  exercise, construct finite-state 
machines that detect all occurrences of self-overlapping patterns.
Recognizing a set of strings: a finite-state-machine interpreter
Finite-state machines (fsm, also called ""finite automata"") are typically used to recognize patterns that consist of  
a  set of strings. An adequate treatment of this more general problem requires introducing some concepts and  
terminology widely used in computer science.
Given a finite set A of input symbols, the  alphabet, A*  denotes the (infinite) set of all (finite) strings over A,  
including the nullstring  λ. Any subset L  ⊆  A*, finite or infinite, is called a set of strings, or a  language, over A.  
Recognizing a language L refers to the ability to examine any string z ∈ A*, one symbol at a time from left to right,  
and deciding whether or not z ∈  L.",algorithms and data structures.pdf
"and deciding whether or not z ∈  L.
A deterministic finite-state machine M is essentially given by a finite set S of states, a finite alphabet A of input 
symbols, and a transition function f: S x  A →  S. The state diagram depicts the states and the inputs, which lead  
from one state to another; thus a finite-state machine maps strings over A into sequences of states.
When treating any specific problem, it is typically useful to expand this minimal definition by specifying one or  
more of the following additional concepts. An initial state s0 S, a subset F ⊆  S of final or accepting states, a finite  
alphabet B of output symbols and an output function g: S → B, which can be used to assign certain actions to the  
states in S. We use the concepts of initial state s 0 and of accepting states F to define the notion ""recognizing a set of  
strings"":
A set L ⊆  A* of strings is recognized or accepted by the finite-state machine M = (S, A, f, s 0, F) iff all the strings",algorithms and data structures.pdf
"strings"":
A set L ⊆  A* of strings is recognized or accepted by the finite-state machine M = (S, A, f, s 0, F) iff all the strings  
in L, and no others, lead M from s0 to some state s ∈  F.
Example
Exhibit 10.3 shows the state diagram of a finite-state machine that recognizes parameter lists as defined by the  
syntax diagrams in  Exhibit 10.2. L (letter) stands for a character a .. z, D (digit) for a digit 0 .. 9.
Exhibit 10.2: Syntax diagram of simple parameter lists.
88",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 10.3: State diagram of finite-state machine to accept parameter lists. The 
starting state is '1', the single accepting state is '8'.
A straightforward implementation of a finite-state machine interpreter uses a transition matrix T to represent  
the state diagram. From the current state s the input symbol c leads to the next state T[s, c]. It is convenient to  
introduce an error state that captures all illegal transitions. The transition matrix T corresponding to Exhibit 10.3 
looks as follows:
L represents a character a .. z.
D represents a digit 0 .. 9.
! represents all characters that are not explicitly mentioned.
( ) : , ; L D !
0
1
2
3
4
5
6
7
8
0 0 0 0 0 0 0 0 0
1 2 0 0 0 0 0 0 0
2
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8
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0 0 3 0 0
3
0
6
6
0
0
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2
0
0
0
0
0
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2
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0
0
6
0
0
0
0
0
0
0
0
error state
skip blank
left parenthesis read
reading variable identifier
skip blank
colon read",algorithms and data structures.pdf
"6
0
0
2
2
0
0
0
0
0
0
2
2
0
0
3
0
0
6
0
0
0
0
0
0
0
0
error state
skip blank
left parenthesis read
reading variable identifier
skip blank
colon read
reading type identifier
skip blank
right parenthesis read
The following is a suitable environment for programming a finite-state-machine interpreter:
const  nstate = 8; { number of states, without error state }
type state = 0 .. nstate; { 0 = error state, 1 = initial state }
inchar = ' ' .. '¨'; { 64 consecutive ASCII characters }
tmatrix = array[state, inchar] of state;
var  T: tmatrix;
After initializing the transition matrix T, the procedure 'silentfsm' interprets the finite-state machine defined by  
T. It processes the sequence of input characters and jumps around in the state space, but it produces no output.
procedure silentfsm(var T: tmatrix);
var  s: state;  c: inchar;
begin
s := 1;  { initial state }
while  s ≠ 0  do  { read(c);  s := T[s, c] }
Algorithms and Data Structures 89  A Global Text",algorithms and data structures.pdf
"10. Strings 
end;
The  simple  structure  of  'silentfsm'  can  be  employed  for  a  useful  finite-state-machine  interpreter  in  which 
initialization, error condition, input processing and transitions in the state space are handled by procedures or  
functions 'initfsm', 'alive', 'processinput', and 'transition' which have to be implemented according to the desired  
behavior. The terminating procedure 'terminate' should print a message on the screen that confirms the correct  
termination of the input or shows an error condition.
procedure fsmsim(var T: tmatrix);
var  … ;
begin
initfsm;
while  alive  do  { processinput;  transition };
terminate
end;
Exercise: finite-state recognizer for multiples of 3
Consider the set of strings over the alphabet {0, 1} that represent multiples of 3 when interpreted as binary  
numbers, such as: 0, 00, 11, 00011, 110. Design two finite-state machines for recognizing this set:",algorithms and data structures.pdf
"numbers, such as: 0, 00, 11, 00011, 110. Design two finite-state machines for recognizing this set:
• Left to right: Mlr reads the strings from most significant bit to least significant.
• Right to left: Mrl reads the strings from least significant bit to most significant.
Solution
Left to right: Let rk be the number represented by the k leftmost bits, and let b be the (k + 1)-st bit, interpreted  
as an integer. Then r k+1  = 2·rk + b. The states correspond to r k mod 3 ( Exhibit 10.4). Starting state and accepting  
state: 0'.
Exhibit 10.4: Finite-state machine computes remainder modulo 3 left to right.
Right to left:  r k+1 = b·2 k + r k. Show by induction that the powers of 2 are alternatingly congruent to 1 and 2  
modulo 3 (i.e. 2k mod 3 = 1 for k even, 2 k mod 3 = 2 for k odd). Thus we need a modulo 2 counter, which appears in  
Exhibit 10.5 as two rows of three states each. Starting state: 0. Accepting states: 0 and 0'.
90",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 10.5: Finite-state machine computes remainder modulo 3 right to left.
Exercises and programming projects
1. Draw  the  state  diagram  of  several  finite-state  machines,  each  of  which  searches  a  string  z  for  all 
occurrences of an interesting pattern with repetitive parts, such as 'abaca' or 'Caracas'.
2. Draw the state diagram of finite-state machines that detect all occurrences of  a self-overlapping pattern  
such as 'abracadabra', 'barbar', or 'xx'.
3. Finite-state recognizer for various days:
Design a finite-state machine for automatic recognition of the set of nine words:
'monday','tuesday','wednesday','thursday',
'friday', 'saturday', 'sunday', 'day', 'daytime'
in a text. The underlying alphabet consists of the lowercase letters 'a' .. 'z' and the blank. Draw the state  
diagram of the finite-state machine; identify the initial state and indicate accepting states by a double circle.",algorithms and data structures.pdf
"diagram of the finite-state machine; identify the initial state and indicate accepting states by a double circle. 
It suffices to recognize membership in the set without recognizing each word individually.
4. Implementation of a pattern recognizer:
Some  useful  procedures  and  functions  require  no  parameters,  hence  most  programming  languages 
incorporate the concept of an empty parameter list. There are two reasonable syntax conventions about  
how to write the headers of parameterless procedures and functions:
(1) procedure p; function f: T;
(2) procedure p();function f(): T;
Examples: Pascal uses convention (1); Modula-2 allows both (1) and (2) for procedures, but only (2) for  
function procedures.
For each convention (1) and (2), modify the syntax diagram in Exhibit 10.2 to allow empty parameter lists,  
and draw the state diagrams of the corresponding finite-state machines.
5. Standard Pascal defines parameter lists by means of the syntax diagram shown in Exhibit 10.6.",algorithms and data structures.pdf
"5. Standard Pascal defines parameter lists by means of the syntax diagram shown in Exhibit 10.6.
Algorithms and Data Structures 91  A Global Text",algorithms and data structures.pdf
"10. Strings 
Exhibit 10.6: Syntax diagram for standard Pascal parameter lists. 
Draw a state diagram for the corresponding finite-state machine. For brevity's sake, consider the reserved words  
'function', 'var' and 'procedure' to be atomic symbols rather than strings of characters.
92",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
11. Matrices and graphs: 
transitive closure
Learning objectives:
• atomic versus structured objects
• directed versus undirected graphs
• transitive closure
• adjacency and connectivity matrix
• boolean matrix multiplication
• efficiency of an algorithm. asymptotic notation
• Warshall’s algorithm
• weighted graph
• minimum spanning tree
In any  systematic presentation of data objects, it is useful to distinguish  primitive or  atomic objects  from  
composite or structured objects. In each of the preceding chapters we have seen both types: A bit, a character, or an  
identifier is usually considered primitive; a word of bits, a string of characters, an array of identifiers is naturally  
treated as composite. Before proceeding to the most common primitive objects of computation, numbers, let us  
discuss one of the most important types of structured objects, matrices. Even when matrices are filled with the",algorithms and data structures.pdf
"discuss one of the most important types of structured objects, matrices. Even when matrices are filled with the  
simplest of primitive objects, bits, they generate interesting problems and useful algorithms.
Paths in a graph
Syntax diagrams and state diagrams are examples of a type of object that abounds in computer science: A graph 
consists of  nodes or  vertices, and of  edges or  arcs that connect a pair of nodes. Nodes and edges often have  
additional information attached to them, such as labels or numbers. If we wish to treat graphs mathematically, we  
need a definition of these objects.
Directed graph. Let N be the set of n elements {1, 2, … , n} and E a binary relation: E  N ⊆  ξ N, also denoted by 
an arrow,  → . Consider N to be the set of nodes of a directed graph G, and E the set of arcs (directed edges). A  
directed graph G may be represented by its  adjacency matrix  A ( Exhibit 11.1 ), an n  ξ  n boolean matrix whose",algorithms and data structures.pdf
"directed graph G may be represented by its  adjacency matrix  A ( Exhibit 11.1 ), an n  ξ  n boolean matrix whose  
elements A[i, j] determine the existence of an arc from i to j:
A[i, j] = true    iff    i →  j.
An arc is a path of length 1. From A we can derive all paths of any length. This leads to a relation denoted by a  
double arrow, ⇒ , called the transitive closure of E:
i ⇒  j, iff there exists a path from i to j
(i.e. a sequence of arcs i →  i1, i1 →  i2, i2 →  i3, … , i k →  j). We accept paths of length 0 (i.e. i ⇒  i for all i). This  
relation ⇒  is represented by a matrix C= A∗(Exhibit 11.1):
Algorithms and Data Structures 93  A Global Text",algorithms and data structures.pdf
"11. Matrices and graphs: transitive closure
C[i, j] = true    iff    i ⇒  j.
C stands for connectivity or reachability matrix; C = A∗ is also called transitive hull or transitive closure, since 
it is the smallest transitive relation that ""encloses"" E.
 Exhibit 11.1: Example of a directed graph with its adjacency and connectivity matrix.
(Undirected) graph. If the relation E ⊆  N ξ N is symmetric [i.e. for every ordered pair (i, j) of nodes it also  
contains the opposite pair (j, i)] we can identify the two arcs (i, j) and (j, i) with a single edge, the unordered pair (i, 
j). Books on graph theory typically start with the definition of undirected graphs (graphs, for short), but we treat  
them as a special case of directed graphs because the latter occur much more often in computer science. Whereas  
graphs are based on the concept of an edge between two nodes, directed graphs embody the concept of one-way  
arcs leading from a node to another one.
Boolean matrix multiplication",algorithms and data structures.pdf
"arcs leading from a node to another one.
Boolean matrix multiplication
Let A, B, C be n ξ n boolean matrices defined by
type nnboolean: array[1 .. n, 1 .. n] of boolean;
var  A, B, C: nnboolean;
The boolean matrix multiplication C = A · B is defined as and implemented by
and implemented by
procedure mmb(var a, b, c: nnboolean);
var  i, j, k: integer;
begin
for  i := 1  to  n  do
for  j := 1  to  n  do  begin
c[i, j] := false;
for  k := 1  to  n  do  c[i, j] := c[i, j] or (a[i, k] and 
b[k, j])  (∗∗)
end
end;
Remark:  Remember (in the section, “Pascal and its dialects: Lingua franca of computer science”) that we  
usually assume the boolean operations 'or' and 'and' to be conditional (i.e. their arguments are evaluated only as far  
as necessary to determine the value of the expression). An extension of this simple idea leads to an alternative way  
of coding boolean matrix multiplication that speeds up the innermost loop above for large values of n. Explain why",algorithms and data structures.pdf
"of coding boolean matrix multiplication that speeds up the innermost loop above for large values of n. Explain why  
the following code is equivalent to (∗∗):
k:=1;
94
1
2
3
4
5
1
2
3
4
5
1 2 3 4 5
T
T
T
T
T
T
T
T
T
T T T
T
T
T
TT
T
T
T T
C
1
2
3
4
5
1 2 3 4 5
T
T
T
T
T
T
A",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
while  not c[i, j] and (k ≤ n)  do  { c[i, j] := a[i, k] and b[k, 
j];  k := k + 1 }
Multiplication also defines powers, and this gives us a first solution to the problem of computing the transitive  
closure. If Al+1 denotes the L-th power of A, the formula 
has a clear interpretation: There exists a path of length L + 1 from i to j iff, for some node k, there exists a path of  
length L from i to k and a path of length 1 (a single arc) from k to j. Thus A 2 represents all paths of length 2; in  
general, AL represents all paths of length L, for L ≥ 1:
AL[i, j] = true  iff  there exists a path of length L from i to j.
Rather than dealing directly with the adjacency matrix A, it is more convenient to construct the matrix A' = A or  
I. The identity matrix I has the values 'true' along the diagonal, 'false' everywhere else. Thus in A' all diagonal",algorithms and data structures.pdf
"I. The identity matrix I has the values 'true' along the diagonal, 'false' everywhere else. Thus in A' all diagonal  
elements A'[i, i] = true. Then A'L describes all paths of length ≤ L (instead of exactly equal to L), for L ≥ 0.  
Therefore, the transitive closure is A∗ =  A'(n-1)
The efficiency of an algorithm is often measured by the number of ""elementary"" operations that are executed on  
a given data set. The execution time of an elementary operation [e.g. the binary boolean operators (and, or) used  
above] does not depend on the operands. To estimate the number of elementary operations performed in boolean  
matrix multiplication as a function of the matrix size n, we concentrate on the leading terms and neglect the lesser  
terms. Let us use asymptotic notation in an intuitive way; it is defined formally in Part IV.
The number of operations (and, or), executed by procedure 'mmb' when multiplying two boolean n ξ n matrices",algorithms and data structures.pdf
"The number of operations (and, or), executed by procedure 'mmb' when multiplying two boolean n ξ n matrices 
is  Θ (n3) since each of the nested  loops is iterated n times. Hence  the cost for computing A' (n–1) by repeatedly  
multiplying with A' is Θ (n4). This algorithm can be improved to Θ (n3 · log n) by repeatedly squaring: A'2, A'4, A'8 , … , 
A'k where k is the smallest power of 2 with k ≥ n – 1. It is not necessary to compute exactly A' (n–1). Instead of A'13, for 
example, it suffices to compute A' 16, the next higher power of 2, which contains all paths of length at most 16. In a  
graph with 14 nodes, this set is equal to the set of all paths of length at most 1.
Warshall's algorithm
In search of a faster algorithm we consider other ways of iterating over the set of all paths. Instead of iterating  
over paths of growing length, we iterate over an increasing number of nodes that may be used along a path from",algorithms and data structures.pdf
"over paths of growing length, we iterate over an increasing number of nodes that may be used along a path from  
node i to node j. This idea leads to an elegant algorithm due to Warshall [War 62]:
Compute a sequence of matrices B0, B1, B2, … , Bn:
B0[i, j] = A'[i, j] = true    iff    i = j  or  i →  j.
B1[i, j] = true    iff    i ⇒  j using at most node 1 along the way.
B2[i, j] = true    iff    i ⇒  j using at most nodes 1 and 2 along the way
…
Bk[i, j] = true    iff    i ⇒  j using at most nodes 1, 2, … , k along the way.
The matrices B 0, B1, … express the existence of paths that may touch an increasing number of nodes along the  
way from node i to node j; thus Bn talks about unrestricted paths and is the connectivity matrix C = Bn.
An iteration step Bk–1 → Bk is computed by the formula
Algorithms and Data Structures 95  A Global Text",algorithms and data structures.pdf
"11. Matrices and graphs: transitive closure
Bk[i, j]  =  Bk–1[i, j]  or  (Bk–1[i, k] and Bk–1[k, j]).
The cost for perfor ming one step is Θ (n2), the cost for computing the connectivity matrix  is therefore Θ (n3). A 
comparison of the formula for Warshall's algorithm wit h the formula for matrix multiplication shows that the n-ary  
'OR' has been replaced by a binary 'or'.
At first sight, the following procedure appears to execute the algorithm specified above, but a closer look reveals  
that  it  executes  something  else:  the  assignment  in  the  innermost  loop  computes  new  values  that  are  used 
immediately, instead of the old ones.
procedure warshall(var a: nnboolean);
var i, j, k: integer;
begin
for  k := 1  to  n  do
for  i := 1  to  n  do
for  j := 1  to  n  do
a[i, j] := a[i, j] or (a[i, k] and a[k, j])
{ this assignment mixes values of the old and new matrix }
end;",algorithms and data structures.pdf
"for  j := 1  to  n  do
a[i, j] := a[i, j] or (a[i, k] and a[k, j])
{ this assignment mixes values of the old and new matrix }
end;
A more thorough examination, however, shows that this ""naively"" programmed procedure computes the correct  
result in-place more efficiently than would direct application of the formulas for the matrices B k. We encourage you 
to verify that the replacement of old values by new ones leaves intact all values needed for later steps; that is, show  
that the following equalities hold:
Bk[i, k] = Bk–1[i, k]  and  Bk[k, j] = Bk–1[k, j].
Exercise: distances in a directed graph, Floyd's algorithm
Modify Warshall's algorithm so that it computes the shortest distance between any pair of nodes in a directed  
graph where each arc is assigned a length ≥ 0. We assume that the data is given in an n ξ n array of reals, where d[i, 
j] is the length of the arc between node i and node j. If no arc exists, then d[i, j] is set to ∞, a constant that is the",algorithms and data structures.pdf
"j] is the length of the arc between node i and node j. If no arc exists, then d[i, j] is set to ∞, a constant that is the  
largest real number that can be represented on the given computer. Write a procedure 'dist' that works on an array  
d of type
type  nnreal = array[1 .. n, 1 .. n] of real;
Think of the meaning of the boolean operations 'and' and 'or' in Warshall's algorithm, and find arithmetic  
operations that play an analogous role for the problem of computing distances. Explain your reasoning in words  
and pictures.
Solution
The  following  procedure  'dist'  implements  Floyd's  algorithm  [Flo  62].  We  assume  that  the  length  of  a 
nonexistent arc is ∞, that x + ∞ = ∞, and that min(x, ∞) = x for all x.
procedure dist(var d: nnreal);
var  i, j, k: integer;
begin
for  k := 1  to  n  do
for  i := 1  to  n  do
for  j := 1  to  n  do
d[i, j] := min(d[i, j], d[i, k] + d[k, j])
end;
96",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exercise: shortest paths 
In addition to the distance d[i, j] of the preceding exercise, we wish to compute a shortest path from i to j (i.e.  
one that realizes this distance). Extend the solution above and write a procedure 'shortestpath' that returns its  
result in an array 'next' of type:
type  nnn = array[1 .. n, 1 .. n] of 0 .. n;
next[i,j] contains the next node after i on a shortest path from i to 
j, or 0 if no such path exists.
Solution
procedure shortestpath(var d: nnreal; var next: nnn);
var  i, j, k: integer;
begin
for  i := 1  to  n  do
for  j := 1  to  n  do
if  d[i, j] ≠ ∞  then  next[i, j] := j  else  next[i, j] := 
0;
for  k := 1  to  n  do
for  i := 1  to  n  do
for  j := 1  to  n  do
if  d[i, k] + d[k, j] < d[i, j]  then
{ d[i, j] := d[i, k] + d[k, j];  next[i, j] := next[i, k] 
}
end;",algorithms and data structures.pdf
"for  i := 1  to  n  do
for  j := 1  to  n  do
if  d[i, k] + d[k, j] < d[i, j]  then
{ d[i, j] := d[i, k] + d[k, j];  next[i, j] := next[i, k] 
}
end;
It is easy to prove that next[i, j] = 0 at the end of the algorithm iff d[i, j] = ∞ (i.e. there is no path from i to j).
Minimum spanning tree in a graph
Consider a weighted graph G = (V, E, w), where V = {v 1, …, vn} is the set of vertices, E = {e 1, … , e m} is the set of  
edges, each edge ei is an unordered pair (vj, vk) of vertices, and w: E →  R assigns a real number to each edge, which  
we call its weight. We consider only connected graphs G, in the sense that any pair (vj, vk) of vertices is connected by 
a sequence of edges. In the following example, the edges are labeled with their weight (Exhibit 11.2).
Exhibit 11.2: Example of a minimum spanning tree.
 A tree T is a connected graph that contains no circuits: any pair (v j, vk) of vertices in T is connected by a unique",algorithms and data structures.pdf
"A tree T is a connected graph that contains no circuits: any pair (v j, vk) of vertices in T is connected by a unique  
sequence of edges. A spanning tree of a graph G is a subgraph T of G, given by its set of edges E T ⊆ E, that is a tree  
and satisfies the additional condition of being maximal, in the sense that no edge in E \ E T can be added to T  
without destroying the tree property. Observation: a connected graph G has at least one spanning tree. The weight 
Algorithms and Data Structures 97  A Global Text",algorithms and data structures.pdf
"11. Matrices and graphs: transitive closure
of a spanning tree is the sum of the weights of all its edges. A minimum spanning tree is a spanning tree of minimal 
weight. In Exhibit 11.2, the bold edges form the minimal spanning tree.
Consider the following two algorithms:
Grow:
ET := ∅ ;  { initialize to empty set } while  T is not a spanning tree  do ET := E T ∪  {a min cost edge that does  
not form a circuit when added to ET}
Shrink:
ET := E;  { initialize to set of all edges } while  T is not a spanning tree  do ET := ET \ {a max cost edge that  
leaves T connected after its removal}
Claim: The ""growing algorithm"" and ""shrinking algorithm"" determine a minimum spanning tree.
If T is a spanning tree of G and e = (v j, vk) ∉  ET, we define Ckt(e, T), ""the circuit formed by adding e to T"" as the  
set of edges in ET that form a path from v j to vk. In the example of Exhibit 11.2 with the spanning tree shown in bold 
edges we obtain Ckt((v4, v5), T) = {(v4, v1), (v1, v2), (v2, v5)}.",algorithms and data structures.pdf
"edges we obtain Ckt((v4, v5), T) = {(v4, v1), (v1, v2), (v2, v5)}.
Exercise
Show that for each edge e ∉  ET there exists exactly one such circuit. Show that for any e ∉  ET and any t ∉  Ckt(e, 
T) the graph formed by (ET \ {t}) ∪  {e} is still a spanning tree.
A local minimum spanning tree of G is a spanning tree T with the property that there exist no two edges e ∉  ET , 
t ∉  Ckt(e, T) with w(e) < w(t).
Consider the following 'exchange algorithm', which computes a local minimum spanning tree:
Exchange:
T := any spanning tree;
while  there exists e ∉  ET, t ∈  Ckt(e, T) with w(e) < w(t)  do
ET := (ET \ {t}) ∪  {e};  { exchange }
Theorem: A local minimum spanning tree for a graph G is a minimum spanning tree.
For the proof of this theorem we need:
Lemma: If T' and T"" are arbitrary spanning trees for G, T' ≠ T"", then there exist e"" ∉  ET' , e' ∉  ET"" , such that e"" 
∈  Ckt(e', T"") and e' ∈  Ckt(e"", T').",algorithms and data structures.pdf
"∈  Ckt(e', T"") and e' ∈  Ckt(e"", T').
Proof: Since T' and T"" are spanning trees for G and T' ≠ T"", there exists e"" ∈  ET"" \ ET'. Assume that Ckt(e"", T')  
⊆ T"". Then e"" and the edges in Ckt(e"", T') form a circuit in T"" that contradicts the assumption that T"" is a tree. Hence  
there must be at least one e' ∈  Ckt(e"", T') \ ET"".
Assume that for all e' ∈  Ckt(e"", T') \ ET"" we have e"" ∈  Ckt(e', T""). Then 
forms a circuit in T"" that contradicts the proposition that T"" is a tree. Hence there must be at least one e' ∈ Ckt(e"", 
T') \ ET"" with e"" ∈ Ckt(e', T"").
98",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Proof of the Theorem: Assume that T' is a local minimum spanning tree. Let T"" be a minimum spanning tree.  
If T' ≠ T"" the lemma implies the existence of e' ∈ Ckt(e"", T') \ ET"" and e"" ∈ Ckt(e', T"") \ ET'.
If w(e') < w(e""), the graph defined by the edges (E T"" \ {e""}) ∪ {e'} is a spanning tree with lower weight than T"".  
Since T"" is a minimum spanning tree, this is impossible and it follows that
w(e') ≥w (e""). (∗)
If w(e') > w(e""), the graph defined by the edges (E T' \ {e'}) ∪ {e""} is a spanning tree with lower weight than T'.  
Since T"" is a local minimum spanning tree, this is impossible and it follows that
w(e') ≤ w(e""). (∗∗)
From (∗∗) and ( ∗ ∗∗∗) it follows that w(e') = w(e"") must hold. The graph defined by the edges (E T"" \ {e""}) ∪ {e'} is  
still a spanning tree that has the same weight as T"". We replace T"" by this new minimum spanning tree and",algorithms and data structures.pdf
"still a spanning tree that has the same weight as T"". We replace T"" by this new minimum spanning tree and  
continue the replacement process. Since T' and T"" have only finitely many edges the process will terminate and T""  
will become equal to T'. This proves that T"" is a minimum spanning tree.
The theorem implies that the tree computed by 'Exchange' is a minimum spanning tree.
Exercises
1. Consider how to extend the transitive closure algorithm based on boolean matrix multiplication so that it  
computes (a) distances and (b) a shortest path.
2. Prove  that  the  algorithms  'Grow' and  'Shrink'  compute local  minimum  spanning  trees.  Thus  they  are 
minimum spanning trees by the theorem of the section entitled “Minimum spanning tree in a graph”.
Algorithms and Data Structures 99  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
12. Integers
Learning objectives:
• integers and their operations
• Euclidean algorithm
• Sieve of Eratosthenes
• large integers
• modular arithmetic
• Chinese remainder theorem
• random numbers and their generators
Operations on integers
Five basic operations account for the lion's share of integer arithmetic:
+ – · div mod
The product 'x · y', the quotient 'x div y', and the remainder 'x mod y' are related through the following div-mod 
identity:
(1)  (x div y) · y + (x mod y) = x for y ≠ 0.
Many  programming  languages  provide  these  five  operations,  but  unfortunately,  'mod'  tends  to  behave 
differently not only between different languages but also between different implementations of the same language.  
How come have we not learned in school what the remainder of a division is?
The div-mod identity, a cornerstone of number theory, defines 'mod' assuming that all the other operations are",algorithms and data structures.pdf
"The div-mod identity, a cornerstone of number theory, defines 'mod' assuming that all the other operations are  
defined. It is mostly used in the context of nonnegative integers x ≥ 0, y > 0, where everything is clear, in particular  
the convention 0 ≤ x mod y < y. One half of the domain of integers consists of negative numbers, and there are good  
reasons for extending all five basic operations to the domain of all integers (with the possible exception of y = 0),  
such as:
• Any operation with an undefined result hinders the portability and testing of programs: if the ""forbidden""  
operation does get executed by mistake, the computation may get into nonrepeatable states. Example: from  
a practical point of view it is better not to leave 'x div 0' undefined, as is customary in mathematics, but to  
define the result as '= overflow', a feature typically supported in hardware.",algorithms and data structures.pdf
"define the result as '= overflow', a feature typically supported in hardware.
• Some algorithms that we usually consider in the context of nonnegative integers have natural extensions  
into the domain of all integers (see the following sections on 'gcd' and modular number representations).
Unfortunately, the attempt to extend 'mod' to the domain of integers runs into the problem mentioned above:  
How  should  we  define  'div'  and  'mod'?  Let's  follow  the  standard  mathematical  approach  of  listing  desirable 
properties these operations might possess. In addition to the ""sacred"" div-mod identity (1) we consider:
(2) Symmetry of div: (–x) div y = x div (–y) = –(x div y).
The most plausible way to extend 'div' to negative numbers.
Algorithms and Data Structures 100  A Global Text",algorithms and data structures.pdf
"12. Integers
(3) A constraint on the possible values assumed by 'x mod y', which, for y > 0, reduces to the convention of 
nonnegative remainders:
0 ≤ x mod y < y.
This is important because a standard use of 'mod' is to partition the set of integers into y residue classes. We  
consider a weak and a strict requirement:
(3') Number of residue classes = |y|: for given y and varying x, 'x mod y' assumes exactly |y| distinct values.
(3"") In addition, we ask for nonnegative remainders: 0 ≤ x mod y < |y|.
Pondering  the consequences  of these desiderata,  we soon realize  that 'div' cannot be  extended  to negative  
arguments by means of symmetry. Even the relatively innocuous case of positive denominator y > 0 makes it  
impossible to preserve both (2) and (3""), as the following failed attempt shows:
((–3) div 2) · 2 + ((–3) mod 2) ?=? –3 Preserving (1)
(–(3 div 2)) · 2 + 1 ?=? –3 and using (2) and (3"")
(–1) · 2 + 1 ≠ –3 … fails!",algorithms and data structures.pdf
"((–3) div 2) · 2 + ((–3) mod 2) ?=? –3 Preserving (1)
(–(3 div 2)) · 2 + 1 ?=? –3 and using (2) and (3"")
(–1) · 2 + 1 ≠ –3 … fails!
Even the weak condition (3'), which we consider essential, is incompatible with (2). For y = –2, it follows from  
(1) and (2) that there are three residue classes modulo (–2): x mod (–2) yields the values 1, 0, –1; for example,
1 mod (–2) = 1, 0 mod (–2) = 0, (–1) mod (–2) = –1.
This does not go with the fact that 'x mod 2' assumes only the two values 0, 1. Since a reasonable partition into  
residue classes is more important than the superficially appealing symmetry of 'div', we have to admit that (2) was  
just wishful thinking.
Without giving any reasons, [Knu 73a] (see the chapter ""Reducing a task to given primitives; programming  
motion) defines 'mod' by means of the div-mod identity (1) as follows:
x mod y = x – y · x / y, if y ≠ 0; x mod 0 = x;",algorithms and data structures.pdf
"motion) defines 'mod' by means of the div-mod identity (1) as follows:
x mod y = x – y · x / y, if y ≠ 0; x mod 0 = x;
Thus he implicitly defines x div y = x / y, where z, the ""floor"" of z, denotes the largest integer ≤ z; the ""ceiling"" 
z denotes the smallest integer ≥ z. Knuth extends the domain of 'mod' even further by defining ""x mod 0 = x"".  
With the exception of this special case y = 0, Knuth's definition satisfies (3'): Number of residue classes = |y|. The  
definition does not satisfy (3""), but a slightly more complicated condition. For given y ≠ 0, we have 0 ≤ x mod y < y,  
if y > 0; and 0 ≥ x mod y > y, if y < 0. Knuth's definition of 'div' and 'mod' has the added advantage that it holds for  
real numbers as well, where 'mod' is a useful operation for expressing the periodic behavior of functions [e.g. tan x  
= tan (x mod π)].
Exercise: another definition of 'div' and 'mod'
1. Show that the definition",algorithms and data structures.pdf
"= tan (x mod π)].
Exercise: another definition of 'div' and 'mod'
1. Show that the definition 
in conjunction with the div-mod identity (1) meets the strict requirement (3"").
101",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Solution 
Exercise
Fill out comparable tables of values for Knuth's definition of 'div' and 'mod'.
Solution
 
The Euclidean algorithm
A famous algorithm for computing the greatest common divisor (gcd) of two natural numbers appears in Book 7  
of Euclid's Elements (ca. 300 BC). It is based on the identity gcd(u, v) = gcd(u – v, v), which can be used for u > v to  
reduce the size of the arguments, until the smaller one becomes 0.
We use these properties of the greatest common divisor of two integers u and v > 0:
gcd(u, 0) = u By convention this also holds for u = 0.
gcd(u, v) = gcd(v, u) Permutation of arguments, important for the termination of the following procedure.
gcd(u, v) = gcd(v, u – q · v) For any integer q.
The formulas above translate directly into a recursive procedure:
Algorithms and Data Structures 102  A Global Text",algorithms and data structures.pdf
"12. Integers
function gcd(u, v: integer): integer;
begin
if  v = 0  then  return(u)  else  return(gcd(v, u mod v))
end;
A test for the relative size of u and v is unnecessary. If initially u < v, the first recursive call permutes the two  
arguments, and thereafter the first argument is always larger than the second.
This simple and concise solution has a relatively high implementation cost. A stack, introduced to manage the  
recursive procedure calls, consumes space and time. In addition to the operations visible in the code (test for  
equality, assignment, and 'mod'), hidden stack maintenance operations are executed. There is an equally concise  
iterative version that requires a bit more thinking and writing, but is significantly more efficient:
function gcd(u, v: integer): integer;
var  r: integer;
begin
while  v ≠ 0  do  { r := u mod v;  u := v;  v := r };
return(u)
end;
The prime number sieve of Eratosthenes",algorithms and data structures.pdf
"var  r: integer;
begin
while  v ≠ 0  do  { r := u mod v;  u := v;  v := r };
return(u)
end;
The prime number sieve of Eratosthenes
The oldest and best-known algorithm of type sieve is named after Eratosthenes (ca. 200 BC). A set of elements is 
to be separated into two classes, the ""good"" ones and the ""bad"" ones. As is often the case in life, bad elements are  
easier to find than good ones. A sieve process successively eliminates elements that have been recognized as bad;  
each element eliminated helps in identifying further bad elements. Those elements that survive the epidemic must  
be good.
Sieve algorithms are often applicable when there is a striking asymmetry in the complexity or length of the  
proofs of the two assertions ""p is a good element"" and ""p is a bad element"". This theme occurs prominently in the  
complexity theory of problems that appear to admit only algorithms whose time requirement grows faster than",algorithms and data structures.pdf
"complexity theory of problems that appear to admit only algorithms whose time requirement grows faster than  
polynomially in the size of the input (NP completeness). Let us illustrate this asymmetry in the case of prime  
numbers, for which Eratosthenes' sieve is designed. In this analogy, ""prime"" is ""good"" and ""nonprime"" is ""bad"".
A prime is a positive integer greater than 1 that is divisible only by 1 and itself. Thus primes are defined in terms  
of their lack of an easily verified property: a prime has no factors other than the two trivial ones. To prove that 1 675  
307 419 is not prime, it suffices to exhibit a pair of factors:
1 675 307 419 = 1 234 567 · 1 357.
This verification can be done by hand. The proof that 217 – 1 is prime, on the other hand, is much more elaborate. 
In general (without knowledge of any special property this particular number might have) one has to verify, for",algorithms and data structures.pdf
"In general (without knowledge of any special property this particular number might have) one has to verify, for  
each and every number that qualifies as a candidate factor, that it is not a factor. This is obviously more time  
consuming than a mere multiplication.
Exhibiting factors through multiplication is an example of what is sometimes called a ""one-way"" or ""trapdoor""  
function: the function is easy to evaluate (just one multiplication), but its inverse is hard. In this context, the  
inverse of multiplication  is not division, but rather  factoriz ation. Muc h of modern cryptography relies on the  
difficulty of factorization.
The prime number sieve of Eratosthenes works as follow s. We mark the smallest prime, 2, and erase all of its  
multiples within the desired range 1 .. n. The smallest remaining number must be prime; we mark it and erase its  
103",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
multiples. We repeat this process for all numbers up to √n: If an integer c < n can be factored, c = a · b, then at least  
one of the factors is <√n.
{ sieve of Eratosthenes marks all the primes in 1 .. n }
const  n = … ;
var sieve: packed array [2 .. n] of boolean;
p, sqrtn, i: integer;
…
begin
for  i := 2  to  n  do  sieve[i] := true;  { initialize the 
sieve }
sqrtn := trunc(sqrt(n));
{ it suffices to consider as divisors the numbers up to √ n }
p := 2;
while  p ≤ sqrtn  do  begin
i := p · p;
while  i ≤ n  do  { sieve[i] := false;  i := i + p };
repeat  p := p + 1  until  sieve[p];
end;
end;
Large integers
The range of numbers that can  be represented directly in hardware is typically limited by the word length of the  
computer. For example, many small computers have a word length of 16 bits and thus limit integers to the range –",algorithms and data structures.pdf
"computer. For example, many small computers have a word length of 16 bits and thus limit integers to the range –
215 ≤ a < +2 15 =32768. When the built-in number system is insufficient, a variety of software techniques are used to  
extend its range. They differ greatly with respect to their properties and intended applications, but all of them come  
at an additional cost in memory and, above all, in the time required for performing arithmetic operations. Let us  
mention the most common techniques.
Double-length or double-precision integers.  Two words are used to hold an integer that squares the  
available range as compared to integers stored in one word. For a 16-bit computer we get 32-bit integers, for a 32-
bit computer we get 64-bit integers. Operations on double-precision integers are typically slower by a factor of 2 to  
4.
Variable precision integers. The idea above is extended to allocate as many words as necessary to hold a",algorithms and data structures.pdf
"4.
Variable precision integers. The idea above is extended to allocate as many words as necessary to hold a  
given  integer.  This  technique  is  used  when  the  size  of intermediate  results  that arise  during  the  course  of  a 
computation is unpredictable. It calls for list processing techniques to manage memory. The time of an operation  
depends on the size of its arguments: linearly for addition, mostly quadratically for multiplication.
Packed BCD integers. This is a compromise between double precision and variable precision that comes from 
commercial data processing. The programmer defines the maximal size of every integer variable used, typically by  
giving the maximal number of decimal digits that may be needed to express it. The compiler allocates an array of  
bytes to this variable that contains the following information: maximal length, current length, sign, and the digits.",algorithms and data structures.pdf
"bytes to this variable that contains the following information: maximal length, current length, sign, and the digits.  
The latter are stored in BCD (binary-coded decimal) representation: a decimal digit is coded in 4 bits, two of them  
are packed into a byte. Packed BCD integers are expensive in space because most of the time there is unused  
allocated  space;  and  even  more  so  in  time,  due  to  digit-by-digit  arithmetic.  They  are  unsuitable  for  lengthy 
scientific/technical computations, but OK for I/O-intensive data processing applications.
Algorithms and Data Structures 104  A Global Text",algorithms and data structures.pdf
"12. Integers
Modular number systems: the poor man's large integers
Modular arithmetic is a special-purpose technique with a narrow range of applications, but is extremely efficient  
where it applies—typically in combinatorial and number-theoretic problems. It handles addition, and particularly  
multiplication,  with  unequaled  efficiency,  but  lacks  equally  efficient  algorithms  for  division  and  comparison. 
Certain  combinatorial  problems  that  require  high  precision  can  be  solved  without  divisions  and  with  few 
comparisons; for these, modular numbers are unbeatable.
Chinese Remainder Theorem:  Let m 1, m 2, … , m k be pairwise relatively prime  positive integers, called  
moduli. Let m = m1 · m2 · … · mk be their product. Given k positive integers r1, r2, … , rk, called residues, with 0 ≤ ri < 
mi for 1 ≤ i ≤ rk, there exists exactly one integer r, 0 ≤ r < m, such that  r mod mi = ri  for 1 ≤ i ≤ k.",algorithms and data structures.pdf
"mi for 1 ≤ i ≤ rk, there exists exactly one integer r, 0 ≤ r < m, such that  r mod mi = ri  for 1 ≤ i ≤ k.
The Chinese remainder theorem is used to represent integers in the range 0 ≤ r < m uniquely as k-tuples of their  
residues modulo mi. We denote this number representation by
r ~ [r1, r2, … , rk].
The practicality of modular number systems is based on the following fact: The arithmetic operations (+ , – , ·)  
on integers r in the range 0 ≤ r< m are represented by the same operations, applied componentwise to k-tuples [r 1, 
r2, … , rk]. A modular number system replaces a single +, –, or · in a large range by k operations of the same type in  
small ranges.
If r ~ [r1, r2, … , rk], s ~ [s1, s2, … , sk], t ~ [t1, t2, … , tk], 
then:
(r + s)mod m = t⇔ (ri + si)mod mi = ti for 1 ≤ i ≤ k,
(r – s)mod m = t⇔ (ri – si)mod mi = ti for 1 ≤ i ≤ k,
(r · s)mod m = t⇔ (ri · si)mod mi = ti for 1 ≤ i ≤ k.
Example",algorithms and data structures.pdf
"(r – s)mod m = t⇔ (ri – si)mod mi = ti for 1 ≤ i ≤ k,
(r · s)mod m = t⇔ (ri · si)mod mi = ti for 1 ≤ i ≤ k.
Example
m1 = 2 and m 2 = 5, hence m = m 1 · m 2 = 2 · 5 = 10. In the following table the numbers r in the range 0 .. 9 are  
represented as pairs modulo 2 and modulo 5.
Let r = 2 and s = 3, hence r · s = 6. In modular representation: r ~ [0, 2], s ~ [1, 3], hence r · s ~ [0, 1].
A useful modular number system is formed by the moduli
m1 = 99, m2 = 100, m3 = 101, hence m = m1 · m2 · m3 = 999900.
Nearly a million integers in the range 0 ≤ r < 999900 can be represented. The conversion of a decimal number  
to its modular form is easily computed by hand by adding and subtracting pairs of digits as follows:
r mod 99: Add pairs of digits, and take the resulting sum mod 99.
r mod 100: Take the least significant pair of digits.
r mod 101: Alternatingly add and subtract pairs of digits, and take the result mod 101.",algorithms and data structures.pdf
"r mod 100: Take the least significant pair of digits.
r mod 101: Alternatingly add and subtract pairs of digits, and take the result mod 101.
The largest integer produced by operations on components is 100 2 ~ 213; it is smaller than 215 = 32768 ~ 32k and 
thus causes no overflow on a computer with 16-bit arithmetic.
105",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Example
r = 123456
r mod  99 = (56 + 34 + 12) mod 99 =   3
r mod 100 = 56
r mod 101 = (56 – 34 + 12) mod 101 = 34
r ~ [3, 56, 34]
s = 654321
s mod  99 = (21 + 43 + 65) mod 99 = 30
s mod 100 = 21
s mod 101 = (21 – 43 + 65) mod 101 = 43
s ~ [30, 21, 43]
r + s ~ [3, 56, 34] + [30, 21, 43] = [33, 77, 77]
Modular  arithmetic  has  some  shortcomings:  division,  comparison,  overflow  detection,  and  conversion  to 
decimal notation trigger intricate computations.
Exercise: Fibonacci numbers and modular arithmetic
The sequence of Fibonacci numbers
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …
is defined by 
x0 = 0, x1 = 1, xn = xn–1 + xn–2 for n ≥ 2.
Write (a) a recursive function  (b) an iterative function that computes the n-th element of this sequence. Using  
modular arithmetic, compute Fibonacci numbers up to 10 8 on a computer with 16-bit integer arithmetic, where the  
largest integer is 215 – 1 = 32767.",algorithms and data structures.pdf
"largest integer is 215 – 1 = 32767.
(c) Using moduli m1 = 999, m2 = 1000, m3 = 1001, what is the range of the integers that can be represented  
uniquely by their residues [r1, r2, r3] with respect to these moduli?
(d) Describe in words and formulas how to compute the triple [r 1, r2, r3] that uniquely represents a number 
r in this range.
(e) Modify the function in (b) to compute Fibonacci numbers in modular arithmetic with the moduli 999,  
1000, and 1001. Use the declaration
type  triple = array [1 .. 3] of integer;
and write the procedure
procedure modfib(n: integer; var r: triple);
Solution
(a) function fib(n: integer): integer;
begin
if  n ≤ 1  then  return(n)  else  return(fib(n – 1) + fib(n – 2))
end;
(b) function fib(n: integer): integer;
var  p, q, r, i: integer;
begin
if  n ≤ 1  then  return(n)
Algorithms and Data Structures 106  A Global Text",algorithms and data structures.pdf
"12. Integers
else  begin
p := 0;  q := 1;
for  i := 2  to  n  do  { r := p + q;  p := q;  q := r };
return(r)
end
end;
(c) The range is 0 .. m – 1 with m = m1 · m2 · m3 = 999 999 000.
(d) r = d1 · 1 000 000 + d2 · 1000 + d3 with 0 ≤ d1, d2, d3 ≤ 999
1 000 000 = 999 999 + 1= 1001 · 999 + 1
1000 = 999 + 1 = 1001 – 1
r1 = r mod 999 = (d1 + d2 + d3) mod 999
r2 = r mod 1000 = d3
r3 = r mod 1001 = (d1 – d2 + d3) mod 1001
(e) procedure modfib(n: integer; var r: triple);
var p, q: triple;
i, j: integer;
begin
if  n ≤ 1  then
for  j := 1  to  3  do  r[j] := n
else  begin
for  j := 1  to  3  do  { p[j] := 0;  q[j] := 1 };
for  i := 2  to  n  do  begin
for  j := 1  to  3  do  r [j] := (p[j] + q[j]) mod (998 + j);
p := q;  q := r
end
end
end;
Random numbers
The colloquial meaning of the term at random often implies ""unpredictable"". But random numbers are used in  
scientific/technical  computing  in  situations  where  unpredictability  is  neither  required  nor  desirable.  What  is",algorithms and data structures.pdf
"scientific/technical  computing  in  situations  where  unpredictability  is  neither  required  nor  desirable.  What  is 
needed in simulation, in sampling, and in the generation of test data is  not unpredictability but certain statistical  
properties. A random number generator is a program that generates a sequence of numbers that passes a number  
of specified statistical tests. Additional requirements include: it runs fast and uses little memory; it is portable to  
computers that use a different arithmetic; the sequence of random numbers generated can be reproduced (so that a  
test run can be repeated under the same conditions).
In practice, random numbers are generated by simple formulas. The most widely used class, linear congruential  
generators, given by the formula
ri+1 = (a · ri + c) mod m
are characterized by three integer constants: the multiplier a, the increment c, and the modulus m. The sequence is  
initialized with a seed r0.",algorithms and data structures.pdf
"are characterized by three integer constants: the multiplier a, the increment c, and the modulus m. The sequence is  
initialized with a seed r0.
All these  constants  must be chosen  carefully. Consider,  as  a bad  example,  a  formula  designed  to generate 
random days in the month of February:
r0 = 0,  ri+1 = (2 · ri + 1) mod 28.
It generates the sequence 0, 1, 3, 7, 15, 3, 7, 15, 3, … . Since 0 ≤ r i < m, each generator of the form above  
generates a sequence with a prefix of length < m which is followed by a period of length ≤ m. In the example, the  
107",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
prefix 0, 1 of length 2 is followed by a period 3, 7, 15 of length 3. Usually we want a long period. Results from  
number theory assert that a period of length m is obtained if the following conditions are met:
• m is chosen as a prime number.
• (a – 1) is a multiple of m.
• m does not divide c.
Example
r0 = 0,  ri+1 = (8 · ri + 1) mod 7
generates a sequence: 0, 1, 2, 3, 4, 5, 6, 0, … with a period of length 
7.
Shall we accept this as a sequence of random integers, and if not, why not? Should we prefer the sequence 4, 1, 6, 
2, 3, 0, 5, 4, … ?
For each application  of random numbers, the programmer/analyst has  to identify the important statistical  
properties required. Under normal circumstances these include:
No periodicity over the length of the sequence actually used. Example: to generate a sequence of 100 random",algorithms and data structures.pdf
"No periodicity over the length of the sequence actually used. Example: to generate a sequence of 100 random  
weekdays ∈  {Su, Mo, … , Sat}, do not pick a generator with modulus 7, which can generate a period of length at  
most 7; pick one with a period much longer than 100.
A desired distribution, most often the uniform distribution. If the range 0 .. m – 1 is partitioned into k equally  
sized intervals I1, I2, … , Ik, the numbers generated should be uniformly distributed among these intervals; this must  
be the case not only at the end of the period (this is trivially so for a generator with maximal period m), but for any  
initial part of the sequence.
Many well-known statistical tests are used to check the quality of random number generators. The run test (the 
lengths  of  monotonically  increasing  and  monotonically  decreasing  subsequences  must  occur  with  the  right",algorithms and data structures.pdf
"lengths  of  monotonically  increasing  and  monotonically  decreasing  subsequences  must  occur  with  the  right 
frequencies); the gap test (given a test interval called the ""gap"", how many consecutively generated numbers fall  
outside?); the  permutation test  (partition the sequence into subsequences of t elements; there are t! possible  
relative orderings of elements within a subsequence; each of these orderings should occur about equally often).
Exercise: visualization of random numbers
Write a program that lets its user enter t he constants a, c, m, and the seed r 0 for a linear congruential generator,  
then displays the numbers generated as dots on the screen: A pair of consecutive random numbers is interpreted as  
the (x, y)-coordinates of the dot. You will observe that most generators you enter have obvious flaws: our visual  
system is an excellent detector of regular patterns, and  most regularities correspond  to undesirable  statistical  
properties.",algorithms and data structures.pdf
"system is an excellent detector of regular patterns, and  most regularities correspond  to undesirable  statistical  
properties.
The point made above is substantiated in [PM 88].
The following simple random number generator and some of its properties are easily memorized:
r0 = 1,  ri+1 = 125 · ri mod 8192.
1. 8192 = 213, hence the remainder mod 8192 is represented by the 13 least significant bits.
2. 125 = 127 – 2 = (1111101) in binary representation.
3. Arithmetic can be done with 16-bit integers without overflow and without regard to the representation of  
negative numbers.
Algorithms and Data Structures 108  A Global Text",algorithms and data structures.pdf
"12. Integers
4. The numbers rk generated are exactly those in the range 0 ≤ r k < 8192 with rk mod 4 = 1 (i.e. the period has  
length 211 = 2048).
5. Its statistical properties are described in [Kru 69], [Knu 81] contains the most comprehensive treatment of  
the theory of random number generators.
As a conclusion of this brief introduction, remember an important rule of thumb:
Never choose a random number generator at random!
Exercises
1. Work out the details of implem enting double-precision, variable-precision, and BCD integer arithmetic, and  
estimate the time required for each operation as compared to the time of the same operation in single  
precision. For variable precision and BCD, introduce the length L of the representation as a parameter.
2. The least common multiple (lcm) of two integers u and v is the smallest integer that is a multiple of u and v.  
Design an algorithm to compute lcm(u, v).",algorithms and data structures.pdf
"Design an algorithm to compute lcm(u, v).
3. The prime decomposition of a natural number n > 0 is the (unique) multiset PD(n) = [p1, p2, … , pk] of  
primes pi whose product is n. A multiset differs from a set in that elements may occur repeatedly (e.g.  
PD(12) = [2, 2, 3]). Design an algorithm to compute PD(n) for a given n > 0. 
4. Work out the details of modular arithmetic with moduli 9, 10, 11.
5. Among the 95 linear congruential random number generators given by the formula ri+1 = a · ri mod m,  
with prime modulus m = 97 and 1 < a < 97, find out how many get disqualified ""at first sight"" by a simple  
visual test. Consider that the period of these RNGs is at most 97.
109",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
13. Reals
Learning objectives:
• floating-point numbers and their properties 
• pitfalls of numeric computation 
• Horner's method
• bisection 
• Newton's method
Floating-point numbers
Real numbers, those declared to be of type REAL in a programming language, are represented as floating-point  
numbers on most computers. A floating-point number z is represented by a (signed) mantissa m and a (signed)  
exponent e with respect to a base b: z=± m·b ±e (e.g. z=+0.11·2–1). This section presents a very brief introduction to  
floating-point arithmetic. We recommend [Gol91] as a comprehensive survey.
Floating-point numbers can only approximate real numbers, and in important ways, they behave differently.  
The major difference is due to the fact that any floating-point number system is a  finite number system , as the",algorithms and data structures.pdf
"The major difference is due to the fact that any floating-point number system is a  finite number system , as the  
mantissa m and the exponent e lie in a bounded range. Consider, as a simple example, the following number  
system:
z = ±0.b1b2 · 2±e, where b1, b2, and e may take the values 0 and 1.
The number representation is  not unique : The same real number may have many different representations,  
arranged in the following table by numerical value (lines) and constant exponent (columns).
1.5       + 0.11 · 2+1
1.0      + 0.10 · 2+1
0.75                           + 0.11 · 2±0
0.5     + 0.01 · 2+1 + 0.10 · 2±0
0.375                                               +0.11 · 2–1
0.25                           + 0.01 · 2±0 +0.10 · 2–1
0.125                                               +0.01 · 2–1
0.      +0.00 · 2+1   + 0.00 · 2±0 +0.00 · 2–1
The table is symmetric for negative numbers. Notice the cluster of representable numbers around zero. There",algorithms and data structures.pdf
"The table is symmetric for negative numbers. Notice the cluster of representable numbers around zero. There  
are only 15 different numbers, but 25= 32 different representations.
Exercise: a floating-point number system
Consider floating-point numbers represented in a 6-bit ""word"" as follows: The four bits b b 2 b1 b0 represent a  
signed mantissa, the two bits e e 0 a signed exponent to the base 2. Every number has the form x=b b 2  b1  b0·2 ee0. 
Algorithms and Data Structures 110  A Global Text",algorithms and data structures.pdf
"13. Reals
Both the exponent and the mantissa are integers represented in 2's complement form. This means that the integer  
values –2..1 are assigned to the four different representations e e0 as shown:
v e e0
      0    0  0  
1     0  1
   –2  1    0
   –1     1     1
1. Complete the following table of the values of the mantissa and their representation, and write down a  
formula to compute v from b b2 b1 b0.
v b b2 b1 b0
0 0  0  0  0
1    0  0  0   1
      …
7 0  1  1  1
   –8 1     0  0  0
      …
   –1   1     1  1  1
2. How many different number representations are there in this floating-point system?
3. How many different numbers are there in this system? Draw all of them on an axis, each number with all its 
representations.
On a byte-oriented machine, floating-point numbers are often represented by 4 bytes =32 bits: 24 bits for the  
signed mantissa, 8 bits for the signed exponent. The mantissa m is often interpreted as a fraction 0 ≤ m < 1, whose",algorithms and data structures.pdf
"signed mantissa, 8 bits for the signed exponent. The mantissa m is often interpreted as a fraction 0 ≤ m < 1, whose  
precision  is  bounded  by  23  bits;  the  8-bit  exponent  permits  scaling  within  the  range  
2–128 ≤ 2 e  ≤ 2 127. Because 32- and 64-bit floating-point number systems are so common, often coexisting on the  
same  hardware,  these  number  systems  are  often  identified  with  ""single  precision""  and  ""double  precision"", 
respectively. In recent years an IEEE standard format for-single precision floating-point numbers has emerged,  
along with standards for higher precisions: double, single extended, and double extended.
111",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The following example shows the representation of the number
+1.011110 … 0 · 2–54
in the IEEE format:
Some dangers
Floating-point computation is fraught with problems that are hard to analyze and control. Unexpected results  
abound, as the following examples show. The first two use a binary floating-point number system with a signed 2-
bit mantissa and a signed 1-bit exponent. Representable numbers lie in the range
–0.11 · 2+1 ≤ z ≤ +0.11 · 2+1.
Example: y + x = y and x ≠ 0
It suffices to choose |x| small as compared to |y|; for example,
x = 0.01 · 2–1,  y = 0.10 · 2+1.
The addition forces the mantissa of x to be shifted to the right until the exponents are equal (i.e. x is represented  
as 0.0001·2+1). Even if the sum is computed correctly as 0.1001 ·2 +1 in an accumulator of double length, storing the  
result in memory will force rounding: x + y=0.10·2+1=y.",algorithms and data structures.pdf
"result in memory will force rounding: x + y=0.10·2+1=y.
Example: Addition is not associative: (x + y) + z ≠ x + (y + z)
The following values for x, y, and z assign different values to the left and right sides.
Left side: (0.10 · 2+1 + 0.10 · 2–1) + 0.10 · 2–1 = 0.10 · 2+1
Right side: 0.10 · 2+1 + (0.10 · 2–1 + 0.10 · 2–1) = 0.11 · 2+1
A useful rule of thumb helps prevent the loss of significant digits: Add the small numbers before adding the large 
ones.
Example: ((x + y)2 – x2 – 2xy) / y2 = 1?
Let's evaluate this expression for large |x| and small |y| in a floating-point number system with five decimal  
digits.
x = 100.00, y = .01000
x + y = 100.01
(x + y)2 = 10002.0001, rounded to five digits yields 10002.
x2 = 10000.
(x + y)2 – x2 = 2.???? (four digits have been lost!)
2xy = 2.0000
(x + y)2 – x2 – 2xy = 2.???? – 2.0000 = 0.?????
Now five digits have been lost, and the result is meaningless.
Example: numerical instability",algorithms and data structures.pdf
"(x + y)2 – x2 – 2xy = 2.???? – 2.0000 = 0.?????
Now five digits have been lost, and the result is meaningless.
Example: numerical instability
Recurrence relations for sequences of numbers are prone to the phenomenon of numerical instability. Consider 
the sequence
x0 = 1.0,   x1 = 0.5,   xn+1 = 2.5 · xn – xn–1.
Algorithms and Data Structures 112  A Global Text",algorithms and data structures.pdf
"13. Reals
We first solve this linear recurrence relation in closed form by trying x i=ri for r≠0. This leads to r n+1 = 2.5 · r n– 
rn–1, and to the quadratic equation 0 = r2– 2.5 · r + 1, with the two solutions r = 2 and r = 0.5.
The general solution of the recurrence relation is a linear combination:
xi = a · 2i + b · 2–i.
The starting values x0 = 1.0 and x 1 = 0.5 determine the coefficients a=0 and b=1, and thus the sequence is given  
exactly as xi = 2–i. If the sequence x i = 2–i is computed by the recurrence relation above in a floating-point number  
system with one decimal digit, the following may happen:
x2 = 2.5 · 0.5 – 1          =0.2 (rounding the exact value 0.25),
x3 = 2.5 · 0.2 – 0.5          =0 (represented exactly with one decimal digit),
x4 = 2.5 · 0 – 0.2           =–0.2 (represented exactly with one decimal digit),
x5 = 2.5 · (–0.2)–0          =–0.5 represented exactly with one decimal digit),
x6 = 2.5 · (–0.5)–(–0.2) = –1.05 (exact) = –1.0 (rounded),",algorithms and data structures.pdf
"x5 = 2.5 · (–0.2)–0          =–0.5 represented exactly with one decimal digit),
x6 = 2.5 · (–0.5)–(–0.2) = –1.05 (exact) = –1.0 (rounded),
x7 = 2.5 · (–1) – (–0.5)    = –2.0 (represented exactly with one decimal digit),
x8 = 2.5 · (–2)–(–1)     = –4.0(represented exactly with one decimal digit).
As soon as the first rounding error has occurred, the computed sequence changes to the alternative solution x i = 
a · 2i, as can be seen from the doubling of consecutive computed values.
Exercise: floating-point number systems and calculations
(a) Consider a floating-point number system with two ternary digits t 1, t2 in the mantissa, and a ternary digit e  
in the exponent to the base 3. Every number in this system has the form x = .t 1  t2  · 3e, where t 1, t2, and e  
assume a value chosen among{0,1,2}. Draw a diagram that shows all the different numbers in this system,  
and for each number, all of its representations. How many representations are there? How many different  
numbers?",algorithms and data structures.pdf
"and for each number, all of its representations. How many representations are there? How many different  
numbers?
(b) Recall the series
which holds for |x| < 1, for example,
Use this formula to express 1/0.7 as a series of powers.
Horner's method
A polynomial of n-th degree (e.g. n = 3) is usually represented in the form
a3 · x3 + a2 · x2 + a1 · x + a0
but is better evaluated in nested form,
((a3 · x + a2) · x + a1) · x + a0.
113",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The first formula needs n multiplications of the form a i · xi and, in addition, n–1 multiplications to compute the  
powers of x. The second formula needs only n multiplications in total: The powers of x are obtained for free as a  
side effect of the coefficient multiplications.
The following procedure assumes that the (n+1) coefficients a i are stored in a sufficiently large array a of  type  
'coeff':
type  coeff = array[0 .. m] of real;
function horner(var a: coeff;  n: integer;  x: real): real;
var  i: integer;  h: real;
begin
h := a[n];
for  i := n – 1  downto  0  do  h := h · x + a[i];
return(h)
end;
Bisection
Bisection is an iterative method for solving equations of the form f(x) = 0. Assuming that the function f : R → R 
is continuous in the interval [a, b] and that f(a) · f(b) < 0, a root of the equation f(x) = 0 (a zero of f) must lie in the",algorithms and data structures.pdf
"is continuous in the interval [a, b] and that f(a) · f(b) < 0, a root of the equation f(x) = 0 (a zero of f) must lie in the  
interval [a, b] (Exhibit 13.1). Let m be the midpoint of this interval. If f(m) = 0, m is a root. If f(m) · f(a) < 0, a root  
must be contained in [a, m], and we proceed with this subinterval; if f(m) · f(b) < 0, we proceed with [m, b]. Thus at  
each iteration the interval of uncertainty  that must contain a root is half the size of the interval produced in the  
previous  iteration.  We  iterate  until  the  interval  is  smaller  than  the  tolerance  within  which  the  root  must  be 
determined.
Exhibit 13.1: As in binary search, bisection excludes half of the interval
under consideration at every step.
function bisect(function f: real;  a, b: real): real;
const  epsilon = 10–6;
var  m: real;  faneg: boolean;
begin
faneg := f(a) < 0.0;
repeat
m := (a + b) / 2.0;
if  (f(m) < 0.0) = faneg  then  a := m  else  b := m
until  |a – b| < epsilon;
return(m)",algorithms and data structures.pdf
"begin
faneg := f(a) < 0.0;
repeat
m := (a + b) / 2.0;
if  (f(m) < 0.0) = faneg  then  a := m  else  b := m
until  |a – b| < epsilon;
return(m)
Algorithms and Data Structures 114  A Global Text",algorithms and data structures.pdf
"13. Reals
end;
A sequence x1, x2, x3,… converging to x converges linearly if there exist a constant c and an index i 0 such that for 
all I > i 0: |x i+1  – x| ≤ c · |x i  – x|. An  algorithm is said to converge linearly if the sequence of approximations  
constructed by this algorithm converges linearly. In a linearly convergent algorithm each iteration adds a constant  
number of significant bits.  For example, each  iteration of bisection  halves  the interval of uncertainty  in  each  
iteration (i.e. adds one bit of precision to the result). Thus bisection converges linearly with c = 0.5. A sequence x 1, 
x2, x3,… converges quadratically if there exist a constant c and an index i 0 such that for all i > i 0: |xi+1 – x| ≤ c ·|xi – 
x|2.
Newton's method for computing the square root
Newton's method for solving equations of the form f(x) = 0 is an example of an algorithm with quadratic",algorithms and data structures.pdf
"Newton's method for solving equations of the form f(x) = 0 is an example of an algorithm with quadratic  
convergence. Let f: R → R be a continuous and differentiable function. An approximation x i+1 is obtained from xi by 
approximating f(x) in the neighborhood of x i by its tangent at the point (x i, f(xi)), and computing the intersection of  
this tangent with the x-axis (Exhibit 13.2). Hence
x ix i+1
f(x  )i
x
Exhibit 13.2: Newton's iteration approximates a curve locally by a tangent.
Newton's method is not guaranteed to converge (Exercise: construct counterexamples), but when it converges, it 
does so quadratically and therefore very fast, since each iteration doubles the number of significant bits.
To compute the square root x = √a of a real number a > 0 we consider the function f(x) = x 2 – a and solve the  
equation x2– a = 0. With f'(x)= 2 · x we obtain the iteration formula:",algorithms and data structures.pdf
"equation x2– a = 0. With f'(x)= 2 · x we obtain the iteration formula:
The  formula  that  relates  xi and  xi+1 can  be  transformed  into  an  analogous  formula  that  determines  the 
propagation of the relative error:
115",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Since
we obtain for the relative error:
Using
we get a recurrence relation for the relative error:
If we start with x0 > 0, it follows that 1+R0 > 0. Hence we obtain
R1 > R2 > R3 > … > 0.
As soon as Ri becomes small (i.e. Ri « 1), we have 1 + Ri  ≈ 1, and we obtain
Ri+1 ≈ o.5 · Ri
2
Newton's method converges quadratically as soon as x i is close enough to the true solution. With a bad initial  
guess Ri » 1 we have, on the other hand, 1 + R i ≈ Ri, and we obtain R i+1 ≈ 0.5 · R i (i.e. the computation appears to  
converge linearly until Ri « 1 and proper quadratic convergence starts).
Thus it is highly desirable to start with a good initial approximation x 0 and get quadratic convergence right from 
the beginning. We assume normalized binary floating-point numbers (i.e. a = m · 2 e with 0.5 ≤ m <1). A good  
approximation of is obtained by choosing any mantissa c with 0.5 ≤ c < 1 and halving the exponent:",algorithms and data structures.pdf
"approximation of is obtained by choosing any mantissa c with 0.5 ≤ c < 1 and halving the exponent:
In order to construct this initial approximation x 0, the programmer needs read and write access not only to a  
""real number"" but also to its components, the mantissa and exponent, for example, by procedures such as
procedure mantissa(z: real): integer;
procedure exponent(z: real): integer;
procedure buildreal(mant, exp: integer): real;
Today's programming languages often lack such facilities, and the programmer is forced to use backdoor tricks  
to construct a good initial approximation. If x 0 can be constructed by halving the exponent, we obtain the following  
upper bounds for the relative error:
R1 < 2–2, R2 < 2–5, R3 < 2–11, R4 < 2–23, R5 < 2–47,R6 < 2–95.
Algorithms and Data Structures 116  A Global Text",algorithms and data structures.pdf
"13. Reals
It is remarkable that four iterations suffice to compute an exact square root for 32-bit floating-point numbers,  
where 23 bits are used for the mantissa, one bit for the sign and eight bits for the exponent, and that six iterations  
will do for a ""number cruncher"" with a word length of 64 bits. The starting value x 0 can be further optimized by  
choosing c carefully. It can be shown that the optimal value of c for computing the square root of a real number is c  
= 1/√2 ≈ 0.707.
Exercise: square root
Consider a floating-point number system with two decimal digits in the mantissa: Every number has the form x  
= ± .d1 d2 · 10±e.
(a) How many different number representations are there in this system?
(b) How many different numbers are there in this system? Show your reasoning.
(c) Compute √50 · 102 in this number system using Newton's method with a starting value x 0 = 10. Show every 
step of the calculation. Round the result of any operation to two digits immediately.",algorithms and data structures.pdf
"step of the calculation. Round the result of any operation to two digits immediately.
Solution
(a) A number representation contains two sign bits and three decimal digits, hence there are 22 · 103 = 4000  
distinct number representations in this system.
(b) There are three sources of redundancy:
1. Multiple representations of zero
2. Exponent +0 equals exponent –0
3. Shifted mantissa: ±.d0 · 10 ±e=±.0d · 10 ±e + 1
A detailed count reveals that there are 3439 different numbers.
Zero has 2 2·10 = 40 representations, all of the form ±.00·10 ±e, with two sign bits and one decimal digit e to be  
freely chosen. Therefore, r1 = 39 must be subtracted from 4000.
If e = 0, then ±.d 1d2 · 10+0=±.d1d2 · 10–0. We assume furthermore that d 1d2 ≠ 00. The case d 1d2 = 00 has been  
covered above. Then there are 2 · 99 such pairs. Therefore, r2= 198 must be subtracted from 4000.
If d2 = 0, then ±.d10 · 10±e = ±.0d1 · 10±e+1. The case d1 = 0 has been treated above. Therefore, we assume that d1",algorithms and data structures.pdf
"If d2 = 0, then ±.d10 · 10±e = ±.0d1 · 10±e+1. The case d1 = 0 has been treated above. Therefore, we assume that d1 
≠ 0. Since ±e can assume the 18 different values –9, –8, … , –1, +0, +1, … +8, there are 2 · 9 · 18 such pairs.  
Therefore, r3 = 324 must be subtracted from 4000.
There are 4000 – r1– r2– r3 = 3439 different numbers in this system.
(c) Computing Newton's square root algorithm:
x0 = 10
x1 = .50 · (10 + 50/10) = .50 · (10 + 5) = .50 · 15 = 7.5
x2 = .50 · (7.5 + 50/7.5) = .50 · (7.5 + 6.6) = .50 · 14 = 7
x3 = .50 · = .50 · (7 + 50/7) = (7 + 7.1) = .50 · 14 = 7
117",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exercises
1. Write up all the distinct numbers in the floating-point system with number representations of the form  
z=0.b1b2 · 2 e1e2, where  b1, b 2 and e 1, e 2 may take  the values 0 and 1, and mantissa  and exponent are  
represented in 2's complement notation.
2. Provide  simple  numerical  examples  to  illustrate  floating-point  arithmetic  violations  of  mathematical 
identities.
Algorithms and Data Structures 118  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
14. Straight lines and circles
Learning objectives:
• intersection of two line segments
• degenerate configurations
• clipping
• digitized lines and circles
• Bresenham's algorithms
• braiding straight lines
Points are the simplest geometric objects; straight lines and line segments come next. Together, they make up  
the lion's share of all primitive objects used in two-dimensional geometric computation (e.g. in computer graphics).  
Using these two primitives only, we can approximate any curve and draw any picture that can be mapped onto a  
discrete raster. If we do so, most queries about complex figures get reduced to basic queries about points and line  
segments, such as: is a given point to the left, to the right, or on a given line? Do two given line segments intersect?  
As simple as these questions appear to be, they must be handled efficiently and carefully. Efficiently because these",algorithms and data structures.pdf
"As simple as these questions appear to be, they must be handled efficiently and carefully. Efficiently because these  
basic primitives of geometric computations are likely to be executed millions of times in a single program run.  
Carefully because the ubiquitous phenomenon of degenerate configurations easily traps the unwary programmer  
into overflow or meaningless results.
Intersection
The  problem  of  deciding  whether  two  line  segments  intersect  is  unexpectedly  tricky,  as  it  requires  a 
consideration  of  three  distinct  nondegenerate  cases,  as  well  as  half  a  dozen  degenerate  ones.  Starting  with 
degenerate objects, we have cases where one or both of the line segments degenerate into points. The code below  
assumes that line segments of length zero have been eliminated. We must also consider nondegenerate objects in  
degenerate configurations, as illustrated in Exhibit 14.1. Line segments A and B intersect (strictly). C and D, and E",algorithms and data structures.pdf
"degenerate configurations, as illustrated in Exhibit 14.1. Line segments A and B intersect (strictly). C and D, and E  
and F, do not intersect; the intersection point of the infinitely extended lines lies on C in the first case, but lies  
neither on E nor on F in the second case. The next three cases are degenerate: G and H intersect barely (i.e. in an  
endpoint);  I  and  J  overlap  (i.e.  they  intersect  in  infinitely  many  points);  K  and  L  do  not  intersect.  Careless 
evaluation of these last two cases is likely to generate overflow.
Exhibit 14.1: Cases to be distinguished for the segment intersection problem.
Computing the intersection point of the infinitely extended lines is a naive approach to this decision problem  
that leads to a three-step process:
Algorithms and Data Structures 119  A Global Text",algorithms and data structures.pdf
"14. Straight lines and circles
1. Check whether the two line segments are parallel (a necessary precaution before attempting to compute the  
intersection point). If so, we have a degenerate configuration that leads to one of three special cases: not  
collinear, collinear nonoverlapping, collinear overlapping
2. Compute the intersection point of the extended lines (this step is still subject to numerical problems for  
lines that are almost parallel).
3. Check whether this intersection point lies on both line segments.
If  all  we  want  is  a  yes/no  answer  to  the  intersection  question,  we  can  save  the  effort  of  computing  the 
intersection point and obtain a simpler and more robust procedure based on the following idea: two line segments  
intersect strictly iff the two endpoints of each line segment lie on opposite sides of the infinitely extended line of the  
other segment.",algorithms and data structures.pdf
"intersect strictly iff the two endpoints of each line segment lie on opposite sides of the infinitely extended line of the  
other segment.
Let L be a line given by the equation h(x, y) = a · x + b · y + c = 0, where the coefficients have been normalized  
such that a 2 + b 2 = 1. For a line L given in this Hessean normal form, and for any point p = (x, y), the function h  
evaluated at p yields the signed distance between p and L: h(p) > 0 if p lies on one side of L, h(p) < 0 if p lies on the  
other side, and h(p) = 0 if p lies on L. A line segment is usually given by its endpoints (x 1, y1) and (x 2, y2), and the  
Hessean normal form of the infinitely extended line L that passes through (x1, y1) and (x2, y2) is 
where 
is the length of the line segment, and h(x, y) is the distance of p = (x, y) from L. Two points p and q lie on opposite  
sides of L iff h(p) · h(q) < 0 ( Exhibit 14.2). h(p) = 0 or h(q) = 0 signals a degenerate configuration. Among these,",algorithms and data structures.pdf
"sides of L iff h(p) · h(q) < 0 ( Exhibit 14.2). h(p) = 0 or h(q) = 0 signals a degenerate configuration. Among these,  
h(p) = 0 and h(q) = 0 iff the segment (p, q) is collinear with L.
Exhibit 14.2: Segment s, its extended line L, and distance to points p, q as computed by function h.
type point  =  record  x, y: real  end;
segment  =  record  p1, p2: point  end;
function d(s: segment; p: point): real;
{ computes h(p) for the line L determined by s }
var dx, dy, L12: real;
120",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
begin
dx := s.p2.x – s.p1.x;  dy := s.p2.y – s.p1.y;
L12 := sqrt(dx · dx + dy · dy);
return((dy · (p.x – s.p1.x) – dx · (p.y – s.p1.y)) / L12)
end;
To optimize the intersection function, we recall the assumption L 12 > 0 and notice that we do not need the actual  
distance, only its sig n. Thus the function d used below avoids computing L 12. The function 'intersect' begins by  
checking whether the two line segments are collinear, and if so, tests them for overlap by intersecting the intervals  
obtained by projecting the line segments onto the x-axis (or onto the y-axis, if the segments are vertical). Two  
intervals [a, b] and [c, d] intersect if f min(a, b) ≤ max(c, d) and min(c, d) ≤ max(a, b). This condition could be  
simplified under the assumption that the representation of segments and intervals is ordered ""from left to right""",algorithms and data structures.pdf
"simplified under the assumption that the representation of segments and intervals is ordered ""from left to right""  
(i.e. for interval [a, b] we have a ≤ b). We do not assume this, as line segments often have a natural direction and  
cannot be ""turned around"".
function d(s: segment;  p: point): real;
begin
return((s.p2.y – s.p1.y) · (p.x – s.p1.x) – (s.p2.x – s.p1.x) · 
(p.y – s.p1.y))
end;
function overlap(a, b, c, d: real): boolean;
begin  return((min(a, b) ≤ max(c, d)) and (min(c, d) ≤ max(a, b))) 
end;
function intersect(s1, s2: segment): boolean;
var  d11, d12, d21, d22: real;
begin
d11 := d(s1, s2.p1);  d12 := d(s1, s2.p2);
if  (d11 = 0) and (d12 = 0)  then  { s1 and s2 are collinear }
if  s1.p1.x = s1.p2.x  then  { vertical }
return(overlap(s1.p1.y, s1.p2.y, s2.p1.y, s2.p2.y))
else  { not vertical }
return(overlap(s1.p1.x, s1.p2.x, s2.p1.x, s2.p2.x))
else  begin  { s1 and s2 are not collinear }
d21 := d(s2, s1.p1);  d22 := d(s2, s1.p2);",algorithms and data structures.pdf
"return(overlap(s1.p1.x, s1.p2.x, s2.p1.x, s2.p2.x))
else  begin  { s1 and s2 are not collinear }
d21 := d(s2, s1.p1);  d22 := d(s2, s1.p2);
return((d11 · d12 ≤ 0) and (d21 · d22 ≤ 0))
end
end;
In addition to the degeneracy issues we have addressed, there are numerical issues of near-degeneracy that we  
only mention. The length L12 is a condition number (i.e. an indicator of the computation's accuracy). As Exhibit 14.3 
suggests, it may be numerically impossible to tell on which side of a short line segment L a distant point p lies.
Algorithms and Data Structures 121  A Global Text",algorithms and data structures.pdf
"14. Straight lines and circles
Exhibit 14.3: A point's distance from a segment amplifies the error of the ""which side"" computation.
Conclusion: A geometric algorithm must check for degenerate configurations explicitly—the code that handles  
configurations ""in general position"" will not handle degeneracies.
Clipping
The  widespread  use  of  windows  on  graphic  screens  makes  clipping  one  of  the  most  frequently  executed 
operations: Given a rectangular window and a configuration in the plane, draw that part of the configuration which  
lies within the window. Most configurations consist of line segments, so we show how to clip a line segment given  
by its endpoints (x 1, y1) and (x 2, y2) into a window given by its four corners with coordinates {left, right}  × {top, 
bottom}.
The position of a point in relation to the window is described by four boolean variables: ll (to the left of the left",algorithms and data structures.pdf
"bottom}.
The position of a point in relation to the window is described by four boolean variables: ll (to the left of the left  
border), rr (to the right of the right border), bb (below the lower border), tt (above the upper border):
type  wcode = set of (ll, rr, bb, tt);
A point inside the window has the code ll = rr = bb = tt = false, abbreviated 0000 (Exhibit 14.4).
Exhibit 14.4: The clipping window partitions the plane into nine regions.
The procedure 'classify' determines the position of a point in relation to the window:
procedure classify(x, y: real; var c: wcode);
begin
c := Ø;  { empty set }
if  x < left  then  c := {ll}  elsif  x > right  then  c := {rr};
if  y < bottom  then  c := c ∪  {bb}  elsif  y > top  then c := c ∪  
{tt}
end;
The procedure 'clip' computes the endpoints of the clipped line segment and calls the procedure 'showline' to  
draw it:
procedure clip(x1, y1, x2, y2: real);
122
p
L",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
var  c, c1, c2: wcode;  x, y: real;  outside: boolean;
begin  { clip }
classify(x1, y1, c1);  classify(x2, y2, c2);  outside := false;
while  (c1 ≠ Ø) or (c2 ≠ Ø)  do
if  c1 ∩  c2 ≠ Ø  then
{ line segment lies completely outside the window }
{ c1 := Ø;  c2 := Ø;  outside := true }
else  begin
c := c1;
if  c = Ø  then  c := c2;
if  ll ∈  c  then  { segment intersects left }
{ y := y1 + (y2 – y1) · (left – x1) / (x2 – x1);  x := left }
elsif  rr ∈  c  then  { segment intersects right }
{ y := y1 + (y2 – y1) · (right – x1) / (x2 – x1); x := right }
elsif  bb ∈  c  then  { segment intersects bottom }
{ x := x1 + (x2 – x1) · (bottom – y1) / (y2 – y1); y := bottom }
elsif  tt ∈  c  then  { segment intersects top }
{ x := x1 + (x2 – x1) · (top – y1) / (y2 – y1);  y := top };
if  c = c1then { x1 := x;  y1 := y;  classify(x, y, c1) }
else { x2 := x;  y2 := y;  classify(x, y, c2) }
end;",algorithms and data structures.pdf
"if  c = c1then { x1 := x;  y1 := y;  classify(x, y, c1) }
else { x2 := x;  y2 := y;  classify(x, y, c2) }
end;
if  not outside  then  showline(x1, y1, x2, y2)
end;  { clip }
Drawing digitized lines
A raster graphics screen is an integer grid of pixels, each of which can be turned on or off. Euclidean geometry  
does not apply directly to such a discretized plane. Any designer using a CAD system will prefer Euclidean geometry  
to a discrete geometry as a model of the world. The problem of how to approximate the Euclidean plane by an  
integer grid turns out to be a hard question: How do we map Euclidean geometry onto a digitized space in such a  
way as to preserve the rich structure of geometry as much as possible? Let's begin with simple instances: How do  
you map a straight line onto an integer grid, and how do you draw it efficiently?  Exhibit 14.5  shows reasonable  
examples.
Exhibit 14.5: Digitized lines look like staircases.",algorithms and data structures.pdf
"examples.
Exhibit 14.5: Digitized lines look like staircases.
Consider the slope m = (y2 – y1) / (x2 – x1) of a segment with endpoints p 1 = (x1, y1) and p2 = (x2, y2). If |m| ≤ 1 we 
want one pixel blackened on each x coordinate; if |m| ≥ 1, one pixel on each y coordinate; these two requirements  
are consistent for diagonals with |m| = 1. Consider the case |m| ≤ 1. A unit step in x takes us from point (x, y) on the  
line to (x + 1, y + m). So for each x between x 1 and x 2 we paint the pixel (x, y) closest to the mathematical line  
according to the formula y = round(y 1 + m · (x – x 1)). For the case |m| > 1, we reverse the roles of x and y, taking a  
unit step in y and incrementing x by 1/m. The following procedure draws line segments with |m| ≤ 1 using unit  
steps in x.
procedure line(x1, y1, x2, y2: integer);
var  x, sx: integer;  m: real;
Algorithms and Data Structures 123  A Global Text",algorithms and data structures.pdf
"14. Straight lines and circles
begin
PaintPixel(x1, y1);
if  x1 ≠ x2  then  begin
x := x1;  sx := sgn(x2 – x1);  m := (y2 – y1) / (x2 – x1);
while  x ≠ x2  do
{ x := x + sx;  PaintPixel(x, round(y1 + m · (x – x1))) }
end
end;
This straightforward implementation has a number of disadvantages. First, it uses floating-point arithmetic to  
compute integer coordinates of pixels, a costly process. In addition, rounding errors may prevent the line from  
being reversible: reversibility means that we paint the same pixels, in reverse order, if we call the procedure with  
the two endpoints interchanged. Reversibility is desirable to avoid the following blemishes: that a line painted  
twice, from both ends, looks thicker than other lines; worse yet, that painting a line from one end and erasing it  
from the other leaves spots on the screen. A weaker constraint, which is only concerned with the result and not the  
process of painting, is easy to achieve but is less useful.",algorithms and data structures.pdf
"process of painting, is easy to achieve but is less useful.
Weak  reversibility is  most  easily  achieved  by  ordering  the  points  p1 and  p2 lexicographically  by  x  and  y 
coordinates, drawing every line from left to right, and vertical lines from bottom to top. This solution is inadequate  
for animation, where the direction of drawing is important, and the sequence in which the pixels are painted is  
determined by the application —drawing the trajectory of a falling apple from the bottom up will not do. Thus  
interactive graphics needs the stronger constraint.
Efficient  algorithms,  such  as  Bresenham's  [Bre  65],  avoid  floating-point  arithmetic  and  expensive  
multiplications through incremental computation: Starting with the current point p 1, a next point is computed as a  
function of the current point and of the line segment parameters. It turns out that only a few additions, shifts, and",algorithms and data structures.pdf
"function of the current point and of the line segment parameters. It turns out that only a few additions, shifts, and  
comparisons are required. In the following we assume that the slope m of the line satisfies |m| ≤ 1. Let
Δx = x2 – x1,    sx = sign(Δx),        Δy = y2 – y1,    sy = sign(Δy).
Assume that the pixel (x, y) is the last that has been determined to be the closest to the actual line, and we now  
want to decide whether the next pixel to be set is (x + sx, y) or (x + sx, y + sy). Exhibit 14.6 depicts the case sx = 1  
and sy = 1.
Exhibit 14.6: At the next coordinate x + sx, we identify and paint the pixel closest to the line.
Let t denote the absolute value of the difference between y and the point with abscissa x + sx on the actual line.  
Then t is given by
124",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The value of t determines the pixel to be drawn:
As the following example shows, reversibility is not an automatic consequence of the geometric fact that two  
points  determine  a  unique  line,  regardless  of  correct  rounding  or  the  order  in  which  the  two  endpoints  are 
presented. A problem arises when two grid points are equally close to the straight line (Exhibit 14.7).
Exhibit 14.7: Breaking the tie among equidistant grid points.
If the tie  is  not broken in a  consistent manner (e.g. by always  taking the  upper grid  point),  the resulting  
algorithm fails to be reversible:
All the variables introduced in this problem range over the integers, but the ratio (Δy)
(Δx)
 appears to introduce  
rational expressions. This is easily remedied by multiplying everything with Δx. We define the decision variable d as
d = |Δx| · (2 · t – 1) = sx · Δx · (2 · t – 1). (∗∗)",algorithms and data structures.pdf
"d = |Δx| · (2 · t – 1) = sx · Δx · (2 · t – 1). (∗∗)
Let di denote the decision variable which determines the pixel (x(i), y(i)) to be drawn in the i-th step. Substituting t 
and inserting x = x(i–1) and y = y(i–1) in (∗∗) we obtain
di = sx · sy · (2·Δx · y1 + 2 · (x(i–1) + sx – x1) · Δy – 2·Δx · y(i–1) – Δx · sy)
and
di+1 = sx · sy · (2·Δx · y1 + 2 · (x(i) + sx – x1) · Δy – 2·Δx · y(i) – Δx · sy).
Subtracting di from di+1, we get
di+1 – di = sx · sy · (2 · (x(i) – x(i–1)) · Δy – 2 · Δx · (y(i) – y(i–1))).
Since x(i) – x(i–1) = sx, we obtain
di+1 = di + 2 · sy · Δy – 2 · sx · Δx · sy · (y(i) – y(i–1)).
Algorithms and Data Structures 125  A Global Text",algorithms and data structures.pdf
"14. Straight lines and circles
If di < 0, or di = 0 and sy = –1, then y(i) = y(i–1), and therefore
di+1 = di + 2 · |Δy|.
If di > 0, or di = 0 and sy = 1, then y(i) = y(i–1) + sy, and therefore
di+1 = di + 2 · |Δy| – 2 · |Δx|.
This iterative computation of d i+1 from the previous d i lets us select the pixel to be drawn. The initial starting  
value for d1 is found by evaluating the formula for di, knowing that (x(0), y(0)) = (x1, y1). Then we obtain
d1 = 2 · |Δy| – |Δx|.
The arithmetic needed to evaluate these formulas is minimal: addition, subtraction and left shift (multiplication  
by 2). The following procedure implements this algorithm; it assumes that the slope of the line is between –1 and 1.
procedure BresenhamLine(x1, y1, x2, y2: integer);
var  dx, dy, sx, sy, d, x, y: integer;
begin
dx := |x2 – x1|;  sx := sgn(x2 – x1);
dy := |y2 – y1|;  sy := sgn(y2 – y1);
d := 2 · dy – dx;  x := x1;  y := y1;
PaintPixel(x, y);
while  x ≠ x2  do  begin",algorithms and data structures.pdf
"dy := |y2 – y1|;  sy := sgn(y2 – y1);
d := 2 · dy – dx;  x := x1;  y := y1;
PaintPixel(x, y);
while  x ≠ x2  do  begin
if  (d > 0) or ((d = 0) and (sy = 1))  then  { y := y + sy;– 
2·dx};
x := x + sx;  d := d + 2 · dy;
PaintPixel(x, y)
end
end;
The riddle of the braiding straight lines
Two straight lines in a plane intersect in at most one point, right? Important geometric algorithms rest on this  
well-known theorem of Euclidean geometry and would have to be reexamined if it were untrue. Is this theorem true  
for  computer lines , that is, for data objects that represent and approximate straight lines to be processed by a  
program? Perhaps yes, but mostly no.
Yes. It is possible, of course, to program geometric problems in such a way that every pair of straight lines has at  
most, or exactly, one intersection point. This is most readily achieved through symbolic computation. For example,",algorithms and data structures.pdf
"most, or exactly, one intersection point. This is most readily achieved through symbolic computation. For example,  
if the intersection of L 1 and L 2 is denoted by an expression 'Intersect(L 1, L 2)' that is ne ver evaluated but simply  
combined with other expres sions to represent a geometric construction, we are free to postulate that 'Intersect(L 1, 
L2)' is a point.
No.  For  reasons  of  efficiency,  most  computer  applications  of  geometry  require  the  immediate  numerical 
evaluation of every geometric operation. This calculation is done in a discrete, finite number system in which case  
the theorem need not be true. This fact is most easily seen if we work with a discrete plane of pixels, and we  
represent a straight line by the set of all pixels to uched by an ideal mathematical line.  Exhibit 14.8  shows three  
digitized straight lines in such a square grid model of plane geometry. Two of the lines intersect in a common",algorithms and data structures.pdf
"digitized straight lines in such a square grid model of plane geometry. Two of the lines intersect in a common  
interval of three pixels, whereas two others have no pixel in common, even though they obviously intersect.
126",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 14.8: Two intersecting lines may share none, one, or more pixels.
With floating-point arithmetic the situation is more complicated; but the fact remains that the Euclidean plane  
is replaced by a discrete set of points embedded in the plane—all those points whose coordinates are representable  
in the particular number system being used. Experience with numerical computation, and the hazards of rounding  
errors, suggests that the question ""In how many points can two straight lines intersect?"" admits the following  
answers:
• There is no intersection—the mathematically correct intersection cannot be represented in the number  
system.
• A set of points that lie close to each other: for example, an interval.
• Overflow aborts the calculation before a result is computed, even if the correct result is representable in the  
number system being used.
Exercise: two lines intersect in how many points?",algorithms and data structures.pdf
"number system being used.
Exercise: two lines intersect in how many points?
Construct  examples  to  illustrate  these  phenomena  when  using  floating-point arithmetic. Choose  a  suitable 
system G of floating-point numbers and two distinct straight lines
ai · x + bi · y + ci = 0  with  ai, bi, ci ∈  G, i=1, 2,
such that, when all operations are performed in G:
(a) There is no point whose coordinates x, y ∈  G satisfy both linear equations.
(b) There are many points whose coordinates x, y ∈  G satisfy both linear equations.
(c) There is exactly one point whose coordinates x, y ∈  G satisfy both linear equations, but the straightforward  
computation of x and y leads to overflow.
(d) As a consequence of (a) it follows that the definition ""two lines intersect  they share a common point"" is  
inappropriate for numerical computation. Formulate a numerically meaningful definition of the statement  
""two line segments intersect"".",algorithms and data structures.pdf
"inappropriate for numerical computation. Formulate a numerically meaningful definition of the statement  
""two line segments intersect"".
Exercise (b) may suggest that the points shared by two lines are neighbors. Pictorially, if the slopes of the two  
lines are almost identical, we expect to see a blurred, elongated intersection. We will show that worse things may  
happen: two straight lines may intersect in arbitrarily many points, and these points are separated by intervals in  
which the two lines alternate in lying on top of each other. Computer lines may be braided! To understand this  
Algorithms and Data Structures 127  A Global Text",algorithms and data structures.pdf
"14. Straight lines and circles
phenomenon, we need to clarify some concepts: What exactly is a straight line represented on a computer? What is  
an intersection?
There is no one answer, there are many! Consider the analogy of the mathematical concept of real numbers,  
defined by axioms. When we approximate real numbers on a computer, we have a choice of many different number  
systems  (e.g.  various  floating-point  number  systems,  rational  arithmetic  with  variable  precision,  interval 
arithmetic).  These  systems  are  typically  not  defined  by  means  of  axioms,  but  rather  in  terms  of  concrete 
representations  of  the  numbers  and  algorithms  for  executing  the  operations  on  these  numbers.  Similarly,  a 
computer line will be defined in terms of a concrete representation (e.g. two points, a point and a slope, or a linear  
expression).  All we  obtain  depends  on  the  formulas  we  use  and  on  the  basic  arithmetic  to operate  on  these",algorithms and data structures.pdf
"expression).  All we  obtain  depends  on  the  formulas  we  use  and  on  the  basic  arithmetic  to operate  on  these 
representations. The notion of a straight line can be formalized in many different ways, and although these are  
likely  to  be  mathematically  equivalent,  they  will  lead  to  data  objects  with  different  behavior  when  evaluated 
numerically. Performing an operation consists of evaluating a formula. Substituting a formula by a mathematically  
equivalent  one  may  lead  to  results  that  are  topologically  different,  because  equivalent  formulas  may  exhibit 
different sensitivities toward rounding errors.
Consider a computer that has only integer arithmetic, i.e. we use only the operations +, –, ·, div. Let Z be the set  
of integers. Two straight lines gi (i = 1, 2) are given by the following equations:
ai · x + bi · y + ci = 0  with  ai, bi, ci ∈  Z;  bi ≠ 0.",algorithms and data structures.pdf
"of integers. Two straight lines gi (i = 1, 2) are given by the following equations:
ai · x + bi · y + ci = 0  with  ai, bi, ci ∈  Z;  bi ≠ 0.
We consider the problem of whether two given straight lines intersect in a given point x 0. We use the following  
method: Solve the equations for y [i. e. y = E1(x) and y = E2(x)] and test whether E1(x0) is equal to E2(x0).
Is this method suitable? First, we need the following definitions:
x ∈  Z is a turn for the pair (E1, E2) iff
sign(E1(x) – E2(x))  ≠  sign(E1(x + 1) – E2(x + 1)).
An algorithm for the intersection problem is correct iff there are at most two turns.
The intuitive idea behind this definition is the recognition that rounding errors may force us to deal with an  
intersection interval rather than a single intersection point; but we wish to avoid separate intervals. The definition  
above partitions the x-axis into at most three disjoint intervals such that in the left interval the first line lies above",algorithms and data structures.pdf
"above partitions the x-axis into at most three disjoint intervals such that in the left interval the first line lies above  
or  below  the  second  line,  in  the  middle  interval  the  lines  ""intersect"",  and  in  the  right  interval  we  have  the 
complementary relation of the left one (Exhibit 14.9).
128",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 14.9: Desirable consistency condition for intersection of nearly parallel lines.
Consider the straight lines:
3 · x – 5 · y + 40 = 0 and 2 · x – 3 · y + 20 = 0
which lead to the evaluation formulas
Our naive approach compares the expressions
(3 · x + 40) div 5 and (2 · x + 20) div 3.
Using the definitions it is easy to calculate that the turns are
7, 8, 10, 11, 12, 14, 15, 22, 23, 25, 26, 27, 29, 30.
The straight lines have become step functions that intersect many times. They are braided (Exhibit 14.10).
Exhibit 14.10: Braiding straight lines violate the consistency condition of Exhibit 14.9.
Exercise: show that the straight lines
x – 2 · y = 0
k · x – (2 · k + 1) · y = 0 for any integer k > 0
Algorithms and Data Structures 129  A Global Text
g1
>g1 g2 =g1 g2 <g1 g2
g2
y
x
2 4 6 8 10 12 14 16 18 20 22 24 26 30 32 34 36 38
9
11
13
15
17
19
21
23
25
27
29
31
y
x
(3x + 40) div 5
(2x + 20) div 3",algorithms and data structures.pdf
"14. Straight lines and circles
have 2 · k – 1 turns in the first quadrant.
Is  braiding  due  merely  to  integer  arithmetic?  Certainly  not:  rounding  errors  also  occur  in  floating-point 
arithmetic,  and  we  can  construct  even  more  pathological  behavior.  As  an  example,  consider  a  floating-point 
arithmetic with a two-decimal-digit mantissa. We perform the evaluation operation:
and truncate intermediate results immediately to two decimal places. Consider the straight lines (Exhibit 14.11)
4.3 · x – 8.3 · y = 0,
1.4 · x – 2.7 · y = 0.
Exhibit 14.11: Example to be verified by manual computation.
These examples were constructed by intersecting straight lines with almost the same slope—a numerically ill-
conditioned problem. While working with integer arithmetic, we made the mistake of using the error-prone 'div'  
operator. The comparison of rational expressions does not require division.",algorithms and data structures.pdf
"operator. The comparison of rational expressions does not require division.
Let a1 · x + b1 · y + c1 = 0 and a2 · x + b2 · y + c2 = 0 be two straight lines. To find out whether they intersect at x 0, 
we have to check whether the equality
holds. This is equivalent to  b2 · c1 – b1 · c2 = x0 · (a2 · b1 – a1 · b2).
The last formula can be evaluated without error if sufficiently large integer arguments are allowed. Another way  
to evaluate this formula without error is to limit the size of the operands. For example, if a i, bi, ci, and x0 are n-digit 
binary numbers, it suffices to be able to represent 3n-digit binary numbers and to compute with n-digit and 2n-
digit binary numbers.
These  examples  demonstrate  that  programming  even  a  simple  geometric  problem  can  cause  unexpected 
difficulties. Numerical computation forces us to rethink and redefine elementary geometric concepts.
130
0.37
0.39
0.73 0.77 0.81 0.85 0.89 0.93
0.41
0.43
0.45
0.47
y
x
1.4 ×ξ
2.7
4.3 ×ξ
8.3",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Digitized circles 
The concepts, problems and techniques we have discussed in this chapter are not at all restricted to dealing with  
straight lines—they have their counterparts for any kind of digitized spatial object. Straight lines, defined by linear  
formulas, are the simplest nontrivial spatial objects and thus best suited to illustrate problems and solutions. In this 
section we show that the incremental drawing technique generalizes in a straightforward manner to more complex  
objects such as circles.
The basic parame ters that define a circle are the center coordinates (x c, y c) and the radius r. To simplify the  
presentation we first consider a circle with radius r centered around  the origin. Such a circle is given  by the  
equation
x2 + y2 = r2.
Efficient algorithms  for  drawing circles, such as  Bresenham's [Bre  77], avoid  floating-point arithmetic and",algorithms and data structures.pdf
"equation
x2 + y2 = r2.
Efficient algorithms  for  drawing circles, such as  Bresenham's [Bre  77], avoid  floating-point arithmetic and  
expensive multiplications through  incremental  computation: A new point is computed depending on the current  
point and on the circle parameters. Bresenham's circle algorithm was conceived for use with pen plotters and  
therefore generates all points on a circle centered at the origin by incrementing all the way around the circle. We  
present a modified version of his algorithm which takes advantage of the eight-way symmetry of a circle. If (x, y) is 
a point on the circle, we can easily determine seven other points lying on the circle (Exhibit 14.12). We consider only 
the 45˚ segment of the circle shown in the figure by incrementing from x = 0 to x = y = r / , and use eight-way  
symmetry to display points on the entire circle.
Exhibit 14.12: Eightfold symmetry of the circle.",algorithms and data structures.pdf
"symmetry to display points on the entire circle.
Exhibit 14.12: Eightfold symmetry of the circle.
Assume that the pixel p = (x, y) is the last that has been determined to be closest to the actual circle, and we now  
want to decide whether the next pixel to be set is p 1 = (x + 1, y) or p 2 = (x + 1, y – 1). Since we restrict ourselves to  
the 45˚ circle segment shown above these pixels are the only candidates. Now define
d' = (x + 1)2 + y2 – r2
d"" = (x + 1)2 + (y – 1)2 – r2
which are the differences between the squared distances from the center of the circle to p 1 (or p2) and to the actual  
circle. If |d'| ≤ |d""|, then p1 is closer (or equidistant) to the actual circle; if |d'| > |d""|, then p2 is closer. We define the 
decision variable d as
d = d' + d"". (∗∗)
We will show that the rule
If d ≤ 0 then select p1 else select p2.
Algorithms and Data Structures 131  A Global Text",algorithms and data structures.pdf
"14. Straight lines and circles
correctly selects the pixel that is closest to the actual circle. Exhibit 14.13 shows a small part of the pixel grid and  
illustrates the various possible ways [(1) to (5)] how the actual circle may intersect the vertical line at x + 1 in  
relation to the pixels p1 and p2.
Exhibit 14.13: For a given octant of the circle, if pixel p is lit, only two other pixels 
p1 and p2 need be examined.
In cases (1) and (2) p2 lies inside, p1 inside or on the circle, and we therefore obtain d' ≤ 0 and d"" < 0. Now d < 0,  
and applying the rule above will lead to the selection of p 1. Since |d'| ≤ |d""| this selection is correct. In case (3) p 1 
lies outside and p 2 inside the circle and we therefore obtain d' > 0 and d"" < 0. Applying the rule above will lead to  
the selection of p1 if d ≤ 0, and p2 if d > 0. This selection is correct since in this case d ≤ 0 is equivalent to |d'| ≤ |d""|.",algorithms and data structures.pdf
"the selection of p1 if d ≤ 0, and p2 if d > 0. This selection is correct since in this case d ≤ 0 is equivalent to |d'| ≤ |d""|.  
In cases (4) and (5) p 1 lies outside, p2 outside or on the circle and we therefore obtain d' > 0 and d"" ≥ 0. Now d > 0,  
and applying the rule above will lead to the selection of p2. Since |d'| > |d""| this selection is correct.
Let di denote the decision variable that determines the pixel (x (i), y(i)) to be drawn in the i-th step. Starting with  
(x(0), y(0)) = (0, r) we obtain
d1 = 3 – 2 · r.
If di ≤ 0, then (x(i), y(i)) = (x(i) + 1, y(i–1)), and therefore
di+1 = di + 4 · xi–1 + 6.
If di > 0, then (x(i), y(i)) = (x(i) + 1, y(i–1) – 1), and therefore
di+1 = di + 4 · (xi–1 – yi–1) + 10.
This iterative computation of d i+1 from the previous d i lets us select the correct pixel to be drawn in the ( i + 1)-th 
step.  The  arithmetic  needed  to  evaluate  these  formulas  is  minimal:  addition,  subtraction,  and  left  shift",algorithms and data structures.pdf
"step.  The  arithmetic  needed  to  evaluate  these  formulas  is  minimal:  addition,  subtraction,  and  left  shift 
(multiplication by 4). The following procedure 'BresenhamCircle' which implements this algorithm draws a circle  
with center (x c, y c) and radius r. It uses the procedure 'CirclePoints' to display points on the entire circle. In the  
cases x = y or r = 1 'CirclePoints' draws each of four pixels twice. This causes no problem on a raster display.
procedure BresenhamCircle(xc, yc, r: integer);
procedure CirclePoints(x, y: integer);
begin
PaintPixel(xc + x, yc + y); PaintPixel(xc – x, yc + y);
PaintPixel(xc + x, yc – y); PaintPixel(xc – x, yc – y);
PaintPixel(xc + y, yc + x); PaintPixel(xc – y, yc + x);
PaintPixel(xc + y, yc – x); PaintPixel(xc – y, yc – x)
end;
132",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
var  x, y, d: integer;
begin
x := 0;  y := r;  d := 3 – 2 · r;
while  x < y  do  begin
CirclePoints(x, y);
if  d < 0then d := d + 4 · x + 6
else { d := d + 4 · (x – y) + 10;  y := y – 1 };
x := x + 1
end;
if  x = y  then  CirclePoints(x, y)
end;    .i).Bresenham's algorithm:circle;
Exercises and programming projects
1. Design and implement an efficient geometric primitive which determines whether two aligned rectangles  
(i.e. rectangles with sides parallel to the coordinate axes) intersect.
2. Design and implement a geometric primitive
function inTriangle(t: triangle; p: point): …;
which takes a triangle t given by its three vertices and a point p and returns a ternary value: p is inside t, p  
is on the boundary of t, p is outside t.
3. Use the functions 'intersect' of in ""Intersection"" and 'inTriangle' above to program a
function SegmentIntersectsTriangle(s: segment; t: triangle): …;",algorithms and data structures.pdf
"function SegmentIntersectsTriangle(s: segment; t: triangle): …;
to check whether segment s and triangle t share common points. 'SegmentIntersectsTriangle' returns a  
ternary value: yes, degenerate, no. List all distinct cases of degeneracy that may occur, and show how your  
code handles them.
4. Implement Bresenham's incremental algorithms for drawing digitized straight lines and circles.
5. Two circles (x', y', r') and (x'', y'', r'') are given by the coordinates of their center and their radius. Find  
effective formulas for deciding the three-way question whether  (a)  the circles intersect as lines , (b)  the  
circles intersect as disks, or (c) neither. Avoid the square-root operation whenever possible.
Algorithms and Data Structures 133  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Part IV: Complexity of 
problems and algorithms
Fundamental issues of computation
A successful search for better and better algorithms naturally leads to the question ""Is there a best algorithm?"",  
whereas an unsuccessful search leads one to ask apprehensively: ""Is there any algorithm (of a certain type) to solve  
this problem?"" These questions turned out to be difficult and fertile. Historically, the question about the existence 
of an algorithm came first, and led to the concepts of computability and decidability in the 1930s. The question  
about a ""best"" algorithm led to the development of complexity theory in the 1960s.
The  study  of  these  fundamental  issues  of  computation  requires  a  mathematical  arsenal  that  includes 
mathematical  logic,  discrete  mathematics,  probability  theory,  and  certain  parts  of  analysis,  in  particular",algorithms and data structures.pdf
"mathematical  logic,  discrete  mathematics,  probability  theory,  and  certain  parts  of  analysis,  in  particular 
asymptotics.  We  introduce  a  few  of  these  topics,  mostly  by  example,  and  illustrate  the  use  of  mathematical 
techniques of algorithm analysis on the important problem of sorting.
Algorithms and Data Structures 134  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
15. Computability and 
complexity
Learning objectives:
• algorithm
• computability
• RISC: Reduced Instruction Set Computer
• Almost nothing is computable.
• The halting problem is undecidable.
• complexity of algorithms and problems
• Strassen's matrix multiplication
Models of computation: the ultimate RISC
Algorithm and computability are originally intuitive concepts. They can remain intuitive as long as we only want 
to show that some specific result can be computed by following a specific algorithm. Almost always an informal  
explanation  suffices  to  convince  someone  with  the  requisite  background  that  a  given  algorithm  computes  a 
specified result. We have illustrated this informal approach throughout Part III. Everything changes if we wish to  
show that a desired result is not computable. The question arises immediately: ""What tools are we allowed to use?""",algorithms and data structures.pdf
"show that a desired result is not computable. The question arises immediately: ""What tools are we allowed to use?""  
Everything is computable with the help of an oracle that knows the answers to all questions. The attempt to prove  
negative  results  about  the  nonexistence  of  certain  algorithms  forces  us  to  agree  on  a  rigorous  definition  of 
algorithm.
The question ""What can be computed by an algorithm, and what cannot?""  was studied intensively during the  
1930s  by  Emil Post  (1897–1954),  Alan  Turing  (1912–1954),  Alonzo  Church  (1903),  and  other  logicians.  They 
defined  various  formal  models  of  computation,  such  as  production  systems,  Turing  machines,  and  recursive 
functions, to capture the intuitive concept of ""computation by the application of precise rules"". All these different  
formal models of computation turned out to be equivalent. This fact greatly strengthens Church's thesis that the",algorithms and data structures.pdf
"formal models of computation turned out to be equivalent. This fact greatly strengthens Church's thesis that the  
intuitive concept of algorithm is formalized correctly by any one of these mathematical systems.
We will not define any of these standard models of computation. They all share the trait that they were designed  
to be conceptually simple: their primitive operations are chosen to be as weak as possible, as long as they retain  
their  property  of  being  universal  computing  systems  in  the  sense  that  they  can  simulate  any  computation 
performed on any other machine. It usually comes as a surprise to novices that the set of primitives of a universal  
computing machine can be so simple as long as these machines possess two essential ingredients:  unbounded 
memory and unbounded time.
Most simulations of a powerful computer on a simple one share three characteristics: it is straightforward in",algorithms and data structures.pdf
"memory and unbounded time.
Most simulations of a powerful computer on a simple one share three characteristics: it is straightforward in  
principle,  it  involves  laborious  coding  in  practice,  and  it  explodes  the  space  and  time  requirements  of  a 
Algorithms and Data Structures 135  A Global Text",algorithms and data structures.pdf
"15. Computability and complexity
computation. The weakness of the primitives, desirable from a theoretical point of view, has the consequence that  
as simple an operation as integer addition becomes an exercise in programming.
The model of computation used most often in algorithm analysis is significantly more powerful than a Turing  
machine in two respects: (1) its memory is not a tape, but an array, and (2) in one primitive operation it can deal  
with numbers of arbitrary size. This model of computation is called random access machine, abbreviated as RAM.  
A RAM is essentially a random access memory , also abbreviated as RAM, of unbounded capacity, as suggested in  
Exhibit 15.1. The memory consists of an infinite array of memory cells, addressed 0, 1, 2, … . Each cell can hold a  
number, say an integer, of arbitrary size, as the arrow pointing to the right suggests.
Exhibit 15.1: RAM - unbounded address space, unbounded cell size.",algorithms and data structures.pdf
"Exhibit 15.1: RAM - unbounded address space, unbounded cell size.
A RAM has an arithmetic unit and is driven by a program. The meaning of the word random is that any memory 
cell can be accessed in unit time (as opposed to a tape memory, say, where access time depends on distance). A  
further crucial assumption in the RAM model is that an arithmetic operation (+, –, ·, /) also takes unit time,  
regardless of the size of the numbers involved. This assumption is unrealistic in a computation where numbers may  
grow very large, but often is a useful assumption. As is the case with all models, the responsibility for using them  
properly lies with the user. To give the reader the flavor of a model of computation, we define a RAM whose  
architecture is rather similar to real computers, but is unrealistically simple.
The ultimate RISC
RISC stands for  Reduced Instruction Set Computer , a machine that has only a few types of instructions built",algorithms and data structures.pdf
"The ultimate RISC
RISC stands for  Reduced Instruction Set Computer , a machine that has only a few types of instructions built  
into the hardware. What is the minimum number of instructions a computer needs to be universal? In theory, one.
Consider a stored-program computer of the ""von Neumann type"" where data and program are stored in the  
same memory (John von Neumann, 1903–1957). Let the random access memory (RAM) be ""doubly infinite"": There  
is a countable infinity of memory cells addressed 0, 1, … , each of which can hold an integer of arbitrary size, or an  
instruction. We assume that the constant 1 is hardwired into memory cell 1; from 1 any other integer can be  
constructed.  There  is  a  single  type  of  ""three-address  instruction""  which  we  call  ""subtract,  test  and  jump"", 
abbreviated as
STJ  x, y, z
where x, y, and z are addresses. Its semantics is equivalent to
STJ  x, y, z   ⇔    x := x – y;  if  x ≤ 0  then  goto z;",algorithms and data structures.pdf
"abbreviated as
STJ  x, y, z
where x, y, and z are addresses. Its semantics is equivalent to
STJ  x, y, z   ⇔    x := x – y;  if  x ≤ 0  then  goto z;
x, y, and z refer to cells Cx, Cy, and Cz. The contents of Cx and Cy are treated as data (an integer); the contents of  
Cz, as an instruction (Exhibit 15.2).
136",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 15.2: Stored program computer: data and instructions share the memory.
Since this RISC has just one type of instruction, we waste no space on an op-code field. An instruction contains  
three addresses, each of which is an unbounded integer. In theory, fortunately, three unbounded integers can be  
packed into the same space required for a single unbounded integer. In the following exercise, this simple idea  
leads to a well-known technique introduced into mathematical logic by Kurt Gödel (1906 – 1978).
Exercise: Gödel numbering
(a) Motel Infinity has a countable infinity of rooms numbered 0, 1, 2, … . Every room is occupied, so the sign  
claims ""No Vacancy"". Convince the manager that there is room for one more person.
(b) Assume that a memory cell in our RISC stores an integer as a sign bit followed by a sequence d 0, d1, d2, … of",algorithms and data structures.pdf
"(b) Assume that a memory cell in our RISC stores an integer as a sign bit followed by a sequence d 0, d1, d2, … of 
decimal digits, least significant first. Devise a scheme for storing three addresses in one cell.
(c) Show how a sequence of positive integers i 1, i2, … , in of arbitrary length n can be encoded in a single natural  
number j: Given j, the sequence can be uniquely reconstructed. Gödel's solution:
Basic program fragments
This  computer  is  best  understood  by  considering  program  fragments  for  simple  tasks.  These  fragments 
implement simple operations, such as setting a variable to a given constant, or the assignment operator, that are  
given as primitives in most programming languages. Programming these fragments naturally leads us to introduce  
basic concepts of assembly language, in particular symbolic and relative addressing.
Set the content of cell 0 to 0:
STJ  0, 0, .+1",algorithms and data structures.pdf
"basic concepts of assembly language, in particular symbolic and relative addressing.
Set the content of cell 0 to 0:
STJ  0, 0, .+1
Whatever the current content of cell 0, subtract it from itself to obtain the integer 0. This instruction resides at  
some address in memory, which we abbreviate as '.', read as ""the current value of the program counter"". '.+1' is the  
next address, so regardless of the outcome of the test, control flows to the next instruction in memory.
a := b, where a and b are symbolic addresses. Use a temporary variable t:
STJ  t, t, .+1 { t := 0 }
STJ  t, b, .+1 { t := –b }
STJ  a, a, .+1{ a := 0 }
STJ  a, t, .+1 { a := –t, so now a = b }
Exercise: a program library
(a) Write RISC programs for a:= b + c, a := b · c, a := b div c, a := b mod c, a := |b|, a : = min(b, c), a := gcd(b,  
c).
Algorithms and Data Structures 137  A Global Text
0
1
2
13
14
.
.
.
0
1
STJ  0, 0, 14
program counter
Executing instruction 13 sets
cell 0 to 0, and increments",algorithms and data structures.pdf
"0
1
2
13
14
.
.
.
0
1
STJ  0, 0, 14
program counter
Executing instruction 13 sets
cell 0 to 0, and increments
the program counter.",algorithms and data structures.pdf
"15. Computability and complexity
(b) Show how this RISC can compute with rational numbers represented by a pair [a, b] of integers denoting  
numerator and denominator.
(c) (Advanced) Show that this RISC is universal, in the sense that it can simulate any computation done by any  
other computer.
The exercise of building up a RISC program library for elementary functions provides the same experience as the 
equivalent exercise for Turing machines, but leads to the goal much faster, since the primitive STJ is much more  
powerful than the primitives of a Turing machine.
The purpose of this section is to introduce the idea that conceptually simple models of computation are as  
powerful, in theory, as much more complex models, such as a high-level programming language. The next two  
sections demonstrate results of an opposite nature: Universal computers, in the sense we have just introduced, are",algorithms and data structures.pdf
"sections demonstrate results of an opposite nature: Universal computers, in the sense we have just introduced, are  
subject to striking limitations, even if we remove any limit on the memory and time they may use. We prove the  
existence of noncomputable functions and show that the ""halting problem"" is undecidable.
The theory of computability was developed in the 1930s, and greatly expanded in the 1950s and 1960s. Its basic  
ideas have become part of the foundation that any computer scientist is expected to know. Computability theory is  
not  directly  useful.  It  is  based  on  the  concept  ""computable  in  principle""  but offers  no  concept  of  a  ""feasible 
computation"". Feasibility, rather than ""possible in principle"", is the touchstone of computer science. Since the  
1960s, a theory of the complexity of computation is being developed, with the goal of partitioning the range of",algorithms and data structures.pdf
"1960s, a theory of the complexity of computation is being developed, with the goal of partitioning the range of  
computability  into  complexity  classes  according  to  time  and  space  requirements.  This  theory  is  still  in  full 
development and breaking new ground, in particular in the area of concurrent computation. We have used some of  
its concepts throughout Part III and continue to illustrate these ideas with simple examples and surprising results.
Almost nothing is computable
Consider  as  a  model  of  computation  any  programming  language,  with  the  fictitious  feature  that  it  is 
implemented  on  a  machine  with  infinite  memory  and  no  operational  time  limits.  Nevertheless  we  reach  the 
conclusion that ""almost nothing is computable"". This follows simply from the observation that there are fewer  
programs than problems to be solved (functions to be computed). Both the number of programs and the number of",algorithms and data structures.pdf
"programs than problems to be solved (functions to be computed). Both the number of programs and the number of  
functions are infinite, but the latter is an infinity of higher cardinality.
A programming language L is defined over an alphabet A= {a 1, a2, … , ak} of k characters. The set of programs in  
L is a subset of the set A ∗ of all strings over A. A ∗ is  countable, and so is its subset L, as it is in one-to-one  
correspondence with the natural numbers under the following mapping:
1. Generate all strings in A∗ in order of increasing length and, in case of equal length, in lexicographic order.
2. Erase all strings that do not represent a program according to the syntax rules of L.
3. Enumerate the remaining strings in the originally given order.
Among all programs in L we consider only those which compute a (partial) function from the set N = {1, 2, 3, …} 
of natural numbers into N. This can be recognized by their heading; for example,
function f(x: N): N;",algorithms and data structures.pdf
"of natural numbers into N. This can be recognized by their heading; for example,
function f(x: N): N;
As this is a subset of L, there exist only countably many such programs.
However, there are uncountably many functions f: N →  N, as Georg Cantor  (1845–1918) proved by his famous  
diagonalization argument. It starts by assuming the opposite, that the set {f | f: N →   N} is countable, then derives  
138",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
a contradiction. If there were only a countable number of such f unctions, we could enumerate all of them according  
to the following scheme:
f1(1)f1 f1(2) f1(3) f1(4)
f2(1) f2(3) f2(4)f2(2)f2
f3 f3(1) f3(3) f3(4)f3(2)
1 2 3 4
...
. . .
f4 f4(1) f4(3) f4(4)f4(2)
Construct a function g: N →  N, g(i) = f i(i) + 1, which is obtained by adding 1 to the diagonal elements in the  
scheme above. Hence g is different from each f i, at least for the argument i: g(i) ≠ f i(i). Therefore, our assumption  
that we have enumerated all functions f: N →  N is wrong. Since  there exists only a countable infinity of programs,  
but an uncountable infinity of functions, almost all functions are noncomputable.
The halting problem is undecidable
If we could predict, for any program P executed on any data set D, whether P terminates or not (i.e. whether it",algorithms and data structures.pdf
"The halting problem is undecidable
If we could predict, for any program P executed on any data set D, whether P terminates or not (i.e. whether it  
will get into an infinite loop), we would have an interesting and useful technique. If this prediction were based on  
rules that prescribe exactly how the pair (P, D) is to be tested, we could write a program H for it. A fundamental  
result of computability theory states that under reasonable assumptions about the model of computation, such a  
halting program H cannot exist.
Consider a programming language L that contains the constructs we will use: mainly recursive procedures and  
procedure parameters. Consider all procedures P in L that have no parameters, a property that can be recognized  
from the heading
procedure P;
This simplifies the problem by avoiding any data dependency of termination.
Assume that there exists a program H in L that takes as argument any parameterless procedure P in L and",algorithms and data structures.pdf
"Assume that there exists a program H in L that takes as argument any parameterless procedure P in L and  
decides whether P halts or loops (i.e. runs indefinitely):
Consider the behavior of the following parameterless procedure X:
procedure X;
begin  while  H(X)  do;  end; 
Consider the reference of X to itself; this trick corresponds to the diagonalization in the previous example.  
Consider further the loop
while  H(X)  do;
which is infinite if H(X) returns true (i.e. exactly when X should halt) and terminates if H(X) returns false (i.e.  
exactly when X should run forever). This trick corresponds to the change of the diagonal g(i) = fi(i) + 1. We obtain:
Algorithms and Data Structures 139  A Global Text",algorithms and data structures.pdf
"15. Computability and complexity
By definition of X: By construction of X:
The fiendishly crafted program X traps H in a web of contradictions. We blame the weakest link in the chain of  
reasoning  that  leads  to  this  contradiction,  namely  the  unsupported  assumption  of  the  existence  of  a  halting 
program H. This proves that the halting problem is undecidable.
Computable, yet unknown
In the preceding two sections we have illustrated the limitations of computability: clearly stated questions, such  
as the halting problem, are undecidable. This means that the halting question cannot be answered, in general, by 
any  computation  no matter  how extensive  in  time  and  space.  There  are, of course,  lots  of  individual  halting 
questions that can be answered, asserting that a particular program running on a particular data set terminates, or  
fails to do so. To illuminate this key concept of theoretical computer science further, the following examples will",algorithms and data structures.pdf
"fails to do so. To illuminate this key concept of theoretical computer science further, the following examples will  
highlight a different type of practical limitation of computability.
Computable or decidable is a concept that naturally involves one algorithm and a denumerably infinite set of  
problems, indexed by a parameter, say n. Is there a uniform procedure that will solve any one problem in the  
infinite set? For example, the ""question"" (really a denumerable infinity of questions) ""Can a given integer n > 2 be  
expressed as the sum of two primes?"" is decidable because there exists the algorithm 's2p' that will answer any  
single instance of this question:
procedure s2p(n: integer): boolean;
{ for n>2, s2p(n) returns true if n is the sum of two primes,
false otherwise }
function p(k: integer): integer;
{ for k>0, p(k) returns the k-th prime: p(1) = 2, p(2) = 3, p(3) 
= 5, … }
end;
begin
for all i, j such that p(i) < n and p(j )< n do
if  p(i) + p(j) = n  then  return(true);",algorithms and data structures.pdf
"= 5, … }
end;
begin
for all i, j such that p(i) < n and p(j )< n do
if  p(i) + p(j) = n  then  return(true);
return(false);
end;  { s2p }
So the general question ""Is any given integer the sum of two primes?"" is solved readily by a simple program. A  
single related question, however, is much harder: ""Is every even integer >2 the sum of two primes?"" Let's try:
4 = 2 + 2, 6 = 3 + 3,  8 = 5 + 3,  10 = 7 + 3 = 5 + 5,  12 = 7 + 5,
14 = 11 + 3 = 7 + 7,  16 = 13 + 3 = 11 + 5,  18 = 13 + 5 = 11 + 7,
20 = 17 + 3 = 13 + 7,  22 = 19 + 3 = 17 + 5 = 11 + 11, 
24 = 19 + 5 = 17 + 7 = 13 + 11,  26 = 23 + 3 = 21 + 5 = 19 + 7 = 13 + 13,
28 = 23 + 5 = 17 + 11,  30 = 23 + 7 = 19 + 11 = 17 + 13,
32 = 29 + 3 = 19 + 13,  34 = 31 + 3 = 29 + 5 = 23 + 11 = 17 + 17,
36 = 33 + 3 = 31 + 5 = 29 + 7 = 23 + 13 = 19 + 17.
140",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
A bit of experimentation suggests that the number of distinct representations as a sum of two primes increases  
as  the  target  integer  grows.  Christian  Goldbach  (1690–1764)  had  the  good  fortune  of  stating  the  plausible 
conjecture ""yes"" to a problem so hard that it has defied proof or counterexample for three centuries.
One might ask: Is the Goldbach conjecture decidable? The straight answer is that the concept of decidability  
does not apply to a single yes/no question such as Goldbach's conjecture. Asking for an algorithm that tells us  
whether  the  conjecture  is  true  or  false  is  meaninglessly  trivial.  Of  course,  there  is  such  an  algorithm!  If  the 
Goldbach conjecture is true, the algorithm that says 'yes' decides. If the conjecture is false, the algorithm that says  
'no' will do the job. The problem that we  don't know  which algorithm is the right one is quite compatible with",algorithms and data structures.pdf
"'no' will do the job. The problem that we  don't know  which algorithm is the right one is quite compatible with  
saying that  one of those two  is the right algorithm. If we package  two trivial algorithms into one, we get the  
following trivial algorithm for deciding Goldbach's conjecture:
function GoldbachOracle(): boolean:
begin  return(GoldbachIsRight)  end;
Notice that 'GoldbachOracle' is a function without arguments, and 'GoldbachIsRight' is a boolean constant,  
either true or false. Occasionally, the stark triviality of the argument above is concealed so cleverly under technical  
jargon as to sound profound. Watch out to see through the following plot.
Let us call an even integer > 2 that is not a sum of two primes a counterexample. None have been found as yet,  
but we can certainly reason about them, whether they exist or not. Define the
function G(k: cardinal): boolean;
as follows:",algorithms and data structures.pdf
"but we can certainly reason about them, whether they exist or not. Define the
function G(k: cardinal): boolean;
as follows:
Goldbach's conjecture is equivalent to G(0) = true. The (implausible) rival conjecture that there is exactly one  
counterexample is equivalent to G(0) = false, G(1) = true. Although we do not know the value of G(k) for any single  
k, the definition of G tells us a lot about this artificial function, namely: 
if G(i) = true for any i, then G(k) = true for all k > i.
With such a strong monotonicity property, how can G look? 
1. If Goldbach is right, then G is a constant: G(k) = true for all k.
2. If there are a finite number i of exceptions, then G is a step function: 
G(k) = false for k < i, G(k) = true for k ≥ i.
3. If there is an infinite number of exceptions, then G is again a constant: 
G(k) = false for all k.
Each of the infinitely many functions listed above is obviously computable. Hence G is computable. The value of",algorithms and data structures.pdf
"G(k) = false for all k.
Each of the infinitely many functions listed above is obviously computable. Hence G is computable. The value of  
G(0) determines truth or falsity of Goldbach's conjecture. Does that help us settle this time-honored mathematical  
puzzle? Obviously not. All we have done is to rephrase the honest statement with which we started this section,  
""The answer is yes or no, but I don't know which"" by the circuitous ""The answer can be obtained by evaluating a  
computable function, but I don't know which one"". 
Algorithms and Data Structures 141  A Global Text",algorithms and data structures.pdf
"15. Computability and complexity
Multiplication of complex numbers
Let us turn our attention from noncomputable functions and undecidable problems to very simple functions that 
are obviously computable, and ask about their complexity: How many primitive operations must be executed in  
evaluating a specific function? As an example, consider arithmetic operations on real numbers to be primitive, and  
consider the product z of two complex numbers x and y:
x = x1 + i · x2 and y = y1 + i · y2,
x · y = z = z1 + i · z2.
The complex product is defined in terms of operations on real numbers as follows:
z1 = x1 · y1 – x2 · y2,
z2 = x1 · y2 + x2 · y1.
It  appears  that  one  complex  multiplication  requires  four  real  multiplications  and  two  real  
additions/subtractions. Surprisingly, it turns out that multiplications can be traded for additions. We first compute  
three intermediate variables using one multiplication for each, and then obtain z by additions and subtractions:",algorithms and data structures.pdf
"three intermediate variables using one multiplication for each, and then obtain z by additions and subtractions:
p1 = (x1 + x2) · (y1 + y2),
p2 = x1 · y1,
p3 = x2 · y2,
z1 = p2 – p3, z2 = p1 – p2 – p3.
This evaluation of the complex product requires only 3 real multiplications, but 5 real additions / subtractions.  
This  trade  of one  multiplication for three additions may  not look like  a good deal in  practice, because  many  
computers have arithmetic chips with fast multiplication circuitry. In theory, however, the trade is clearly favorable.  
The cost of an addition grows linearly in the num ber of digits, whereas the cost of a multiplication using the  
standard method grows quadratically. The key idea behind this algorithm is that ""linear combinations of k products  
of sums can generate more than k products of simple terms"". Let us exploit this idea in a context where it makes a  
real difference.
Complexity of matrix multiplication",algorithms and data structures.pdf
"real difference.
Complexity of matrix multiplication
The complexity of an algorithm is given by  its time and space requirements. Time is usually measured by the  
number of operations executed, space by the number of variables needed at any one time (for input, intermediate  
results, and output). For a given algorithm it is often easy to count the number of operations performed in the worst  
and in the best case; it is usually difficult to determine the average number of operations performed (i.e. averaged  
over all possible input data). Practical algorithms often have time  complexities of the order O(log n), O(n 2), O(n ·  
log n), O(n2), and space complexity of the order O(n), where n measures the size of the input data.
The complexity of a problem is defined as the minimal complexity of all algorithms that solve this problem. It is  
almost always difficult to determine the complexity of a problem, since all possible algorithms must be considered,",algorithms and data structures.pdf
"almost always difficult to determine the complexity of a problem, since all possible algorithms must be considered,  
including those yet unknown. This may lead to surprising results that disprove obvious assumptions.
The complexity of an algorithm is an upper bound for the complexity of the problem solved by this algorithm.  
An algorithm is a witness for the assertion: You need at most this many operations to solve this problem. A specific  
algorithm never provides a lower bound on the complexity of a problem— it cannot extinguish the hope for a more  
efficient algorithm. Occasionally, algorithm designers engage in races lasting decades that result in (theoretically)  
faster and faster algorithms for solving a given problem. Volker Strassen started such a race with his 1969 paper  
142",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
""Gaussian  Elimination  Is  Not  Optimal""  [Str  69],  where  he  showed  that  matrix  multiplication  requires  fewer 
operations than had commonly been assumed necessary. The race has not yet ended.
The obvious way to multiply two n × n matrices uses three nested loops, each of which is iterated n times, as we  
saw in a transitive hull algorithm in the chapter, “Matrices and graphs: transitive closure”. The fact that the obvious  
algorithm  for  matrix  multiplication  is  of  time  complexity  Θ (n3),  however,  does  not  imply  that  the  matrix 
multiplication problem is of the same complexity.
Strassen's matrix multiplication
The standard algorithm for multiplying two n  × n matrices needs n 3 scalar multiplications and n 2 · (n – 1)  
additions; for the case of 2  × 2 matrices, eight multiplications and four additions. Seven scalar multiplications  
suffice if we accept 18 additions/subtractions.",algorithms and data structures.pdf
"suffice if we accept 18 additions/subtractions.
Evaluate seven expressions, each of which is a product of sums:
p1 = (a11 + a22) · (b11 + b22),
p2 = (a21 + a22) · b11
p3 = a11 · (b12 – b22)
p4 = a22 · (–b11 + b21) p5 = (a11 + a12) · b22
p6 = (–a11 + a21) · (b11 + b12)p7 = (a12 – a22) · (b21 + b22).
The elements of the product matrix are computed as follows:
r11 = p1 + p4 – p5 + p7,
r12 = p3 + p5,
r21 = p2 + p4, r22 = p1 – p2 + p3 + p6.
This algorithm does not rely on the commutativity of scalar multiplication. Hence it can be generalized to n × n 
matrices using the divide-and-conquer principle. For reasons of simplicity consider n to be a power of 2 (i.e. n = 2k); 
for other values of n, imagine padding the matrices with rows and columns of zeros up to the next power of 2. An n  
× n matrix is partitioned into four n/2 × n/2 matrices:
The product of two n  × n matrices by Strassen's method requires  seven (not eight) multiplications and 18",algorithms and data structures.pdf
"The product of two n  × n matrices by Strassen's method requires  seven (not eight) multiplications and 18  
additions/subtractions of n/2  × n/2 matrices. For large n, the work required for the 18 additions is negligible  
compared to the work required for even a single multiplication (why?); thus we have saved one multiplication out of 
eight, asymptotically at no cost.
Each n/2  × n/2 matrix is again partitioned recursively into four n/4  × n/4 matrices; after log 2 n partitioning  
steps we arrive at 1  × 1 matrices for which matrix multiplication is the primitive scalar multiplication. Let T(n)  
denote the number of scalar arithmetic operations used by Strassen's method for multiplying two n × n matrices. 
For n > 1, T(n) obeys the recursive equation
Algorithms and Data Structures 143  A Global Text",algorithms and data structures.pdf
"15. Computability and complexity
If we are only interested in the leading term of the solution, the constants 7 and 2 justify omitting the quadratic  
term, thus obtaining
Thus the number of primitive operations required to multiply two n  × n matrices using Strassen's method is  
proportional to n2.81, a statement that we abbreviate as ""Strassen's matrix multiplication takes time Θ (n2.81)"".
Does this asymptotic improvement lead to a more efficient program in practice? Probably not, as the ratio
grows too slowly to be of practical importance: For n ≈ 1000, for example, we have 5√1024 = 4 (remember: 210 = 
1024). A factor of 4 is not to be disdained, but there are many ways to win or lose a factor of 4. Trading an algorithm  
with  simple  code,  such  as  straightforward  matrix  multiplication,  for  another  that  requires  more  elaborate 
bookkeeping, such as Strassen's, can easily result in a fourfold increase of the constant factor that measures the",algorithms and data structures.pdf
"bookkeeping, such as Strassen's, can easily result in a fourfold increase of the constant factor that measures the  
time it takes to execute the body of the innermost loop.
Exercises
1. Prove that the set of all ordered pairs of integers is countably infinite.
2. A  recursive function  is  defined by a finite set of rules  that specify the function in terms of variables,  
nonnegative integer constants, increment ('+1'), the function itself, or an expression built from these by  
composition of functions. As an example, consider Ackermann's function defined as A(n) = An(n) for n ≥ 1, 
where Ak(n) is determined by
Ak(1) = 2 for k ≥ 1
A1(n) = A1(n–1) + 2 for n ≥ 2
Ak(n) = Ak–1(Ak(n–1)) for k ≥ 2
(a) Calculate A(1) , A(2) , A(3), A(4).
(b) Prove that
Ak(2) = 4 for k ≥ 1,
A1(n) = 2·n for n ≥ 1,
A2(n) = 2n for n ≥ 1,
A3(n) = 2A
3
(n–1) for n ≥ 2.
(c) Define the inverse of Ackermann's function as
α(n) = min{m: A(m) ≥ n}.",algorithms and data structures.pdf
"A1(n) = 2·n for n ≥ 1,
A2(n) = 2n for n ≥ 1,
A3(n) = 2A
3
(n–1) for n ≥ 2.
(c) Define the inverse of Ackermann's function as
α(n) = min{m: A(m) ≥ n}.
 Show that α (n) ≤ 3 for n ≤ 16, that α (n) ≤ 4 for n at most a ""tower"" of 65536 2's, and that α (n) →  ∞ as n 
→  ∞.
144",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
3. Complete Strassen's algorithm by showing how to multiply n × n matrices when n is not an exact power of  
2.
4. Assume that you can multiply 3 × 3 matrices using k multiplications. What is the largest k that will lead to  
an asymptotic improvement over Strassen's algorithm?
5. A permutation matrix P is an n × n matrix that has exactly one '1' in each row and each column; all other  
entries are '0'. A permutation matrix can be represented by an array
var a: array[1 .. n] of integer;
as follows: a[i] = j if the i-th row of P contains a '1' in the j-th column.
6. Prove that the product of two permutation matrices is again a permutation matrix.
7. Design  an  algorithm  that  multiplies  in  time  Θ (n)  two  permutation  matrices  given  in  the  array 
representation above, and stores the result in this same array representation.
Algorithms and Data Structures 145  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
16. The mathematics of 
algorithm analysis
Learning objectives:
• worst-case and average performance of an algorithm
• growth rate of a function
• asymptotics: O(), Ω (), ∴Θ ()
• asymptotic behavior of sums
• solution techniques for recurrence relations
• asymptotic performance of divide-and-conquer algorithms
• average number of inversions and average distance in a permutation
• trees and their properties
Growth rates and orders of magnitude
To understand a specific algorithm, it is useful to ask and answer the following questions, usually in this order:  
What is the problem to be solved? What is the main idea on which this algorithm is based? Why is it correct? How  
efficient is it?
The variety of problems is vast, and so is the variety of ""main ideas"" that lead one to design an algorithm and  
establish its correctness. True, there are general algorithmic principles or schemas which are problem-independent,",algorithms and data structures.pdf
"establish its correctness. True, there are general algorithmic principles or schemas which are problem-independent,  
but these rarely suffice: Interesting algorithms typically exploit specific features of a problem, so there is no unified  
approach  to understanding  the  logic  of algorithms. Remarkably, there  is a unified  approach  to the efficiency  
analysis  of  algorithms,  where  efficiency  is  measured  by  a  program's  time  and  storage  requirements.  This  is 
remarkable because there is great variety in (1) sets of input data and (2) environments (computers, operating  
systems, programming languages, coding techniques), and these differences have a great influence on the run time  
and storage consumed by a program. These two types of differences are overcome as follows.
Different sets of input data: worst-case and average performance
The most im portant characteristic of a set of data is its size, measured in terms of any unit convenient to the",algorithms and data structures.pdf
"The most im portant characteristic of a set of data is its size, measured in terms of any unit convenient to the  
problem at hand. This is typically the number of primitive objects in the data, such as bits, bytes, integers, or any  
monotonic function thereof, such as the magnitude of an integer. Examples: For sorting, the number n of elements  
is natural; for square matrices, the number n of rows and columns is convenient; it is a monotonic function (square  
root) of the actual size n 2 of the data. An algorithm may well behave very differently on different data sets of equal  
size n—among all possible configurations o f given size n some will be favorable, others less so. Both the worst-case 
data set of size n and the  average over all data sets of size n provide well-defined and important measures of  
efficiency.  Example:  When sorting data sets about whose order nothing is known, average performance is well",algorithms and data structures.pdf
"efficiency.  Example:  When sorting data sets about whose order nothing is known, average performance is well  
characterized by averaging run time over all n! permutations of the n elements.
Algorithms and Data Structures 146  A Global Text",algorithms and data structures.pdf
"16. The mathematics of algorithm analysis
Different environments: focus on growth rate and ignore constants
The work performed by an algorithm is expressed as a function of the problem size, typically measured by size n  
of the input data. B y focusing on the growth rate of this function but ignoring specific constants, we succeed in  
losing a lot of detail information that changes wildly from one computing environment to another, while retaining  
some  essential  information  that  is  remarkably  invariant  when  moving  a  computation  from  a  micro-  to  a 
supercomputer, from machine language to Pascal, from amateur to professional programmer. The definition of  
general measures for the complexity of problems and for the efficiency of algorithms is a major achievement of  
computer science. It is based on the notions of  asymptotic time and space complexity . Asymptotics  renounces",algorithms and data structures.pdf
"computer science. It is based on the notions of  asymptotic time and space complexity . Asymptotics  renounces  
exact measurement but states how the work grows as the problem size increases. This information often suffices to  
distinguish efficient algorithms from inefficient ones. The asymptotic behavior  of an algorithm is described by the  
O(),  Ω (),  Θ (), and o() notations. To determine the amount of work to be performed by an algorithm we count  
operations that take constant time (independently of n) and data objects that require constant storage space. The  
time required by an addition, comparison, or exchange of two numbers is typically independent of how many  
numbers we are processing; so is the storage requirement for a number.
Assume that the time required by four algorithms A1, A2, A3, and A4 is log2n, n, n · log2n, and n2, respectively. The 
following table shows that for sizes of data sets that frequently occur in practice, from n ≈ 10 3 to 106, the difference",algorithms and data structures.pdf
"following table shows that for sizes of data sets that frequently occur in practice, from n ≈ 10 3 to 106, the difference 
in growth rate translates into large numerical differences:
n A1 = log2n A2 = n A3 = n · log2n A4 = n2
25 = 32 5 25 = 32
5 · 25 = 160 210 ≈ 103
210 = 1024 10 210 ≈ 103 10 · 210 ≈ 104 220 ≈ 106
220 ≈ 106 20 220 ≈ 106 20 · 220 ≈ 2 · 107 240 ≈ 1012
For a specific algorithm these functions are to be multiplied by a constant factor proportional to the time it takes  
to execute the body of the innermost loop. When comparing different algorithms that solve the same problem, it  
may well happen that one innermost loop is 10 times faster or slower than another. It is rare that this difference  
approaches a factor of 100. Thus for n ≈ 1000 an algorithm with time complexity Θ (n · log n) will almost always be  
much more efficient than an algorithm with time complexity Θ (n2). For small n, say n = 32, an algorithm of time",algorithms and data structures.pdf
"much more efficient than an algorithm with time complexity Θ (n2). For small n, say n = 32, an algorithm of time  
complexity Θ (n2) may be more efficient than one of complexity Θ (n · log n) (e.g. if its constant is 10 times smaller).
When we wish to predict exactly how many seconds and bytes a program needs, asymptotic analysis is still  
useful but is only a small part of the work. We now have to go back over the formulas and keep track of all the  
constant factors discarded in cavalier fashion by the O() notation. We have to assign numbers to the time consumed  
by scores of primitive O(1) operations. It may be sufficient to estimate the time consuming primitives, such as  
floating-point operations; or it may be necessary to include those that are  hidden by a high-level programming  
language and answer questions such as: How long does an array access a[i, j] take? A procedure call? Incrementing  
the index i in a loop ""for i := 0 to n""?
Asymptotics",algorithms and data structures.pdf
"the index i in a loop ""for i := 0 to n""?
Asymptotics
Asymptotics  is  a  technique  used  to estimate  and  compare  the  growth  behavior  of  functions.  Consider  the 
function
147",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
f(x) is said to behave like x for x →  ∞ and like 1  /  x for x →  0. The motivation for such a statement is that both x and  
1  /  x are intuitively simpler, more easily understood functions than f(x). A complicated function is unlike any simpler  
one across its entire domain, but it usually behaves like a simpler one as x approaches some particular value. Thus  
all asymptotic statements include the qualifier x →  x0. For the purpose of algorithm analysis we are interested in  
the behavior of functions for large values of their argument, and all our definitions below assume x →  ∞.
The asymptotic behavior of functions is described by the O(), Ω (), Θ  (), and o() notations, as in f(x) ∈  O(g(x)). 
Each of these notations assigns to a given  function g the set of all functions  that are related to g in a well-defined",algorithms and data structures.pdf
"Each of these notations assigns to a given  function g the set of all functions  that are related to g in a well-defined  
way. Intuitively, O(), Ω (), Θ (), and o() are used to compare the growth of functions, as ≤, ≥, =, and < are used to  
compare numbers. O(g) is the set of all functions that are ≤ g in a precise technical sense that corresponds to the  
intuitive notion ""grows no faster than g"". The definition involves some technicalities signaled by the preamble ∃  ∃c > 
0,∃  >  ∃x0 ∈  X,  ∀  x ≥ x 0. It says that we ignore constant factors and initial behavior and are interested only in a  
function's behavior from some point on. N0 is the set of nonnegative integers, R0 the set of nonnegative reals. In the  
following definitions X stands for either N0 or R0. Let g: X →  X.
Definition of O(), ""big oh"":
 O(g) := {f: X →  X | ∃ c > 0, ∃ x0 ∈  X, ∀  x ≥  xo : f(x) ≤ c · g(x)} 
We say that f ∈  O(g), or that f grows at most as fast as g(x) for x →  ∞.
Definition of Ω (), ""omega"":",algorithms and data structures.pdf
"We say that f ∈  O(g), or that f grows at most as fast as g(x) for x →  ∞.
Definition of Ω (), ""omega"":
Ω (g) := {f: X →  X  ∃ c > 0  x0 ∈  X, ∀∀  x ≥ x0: f(x) ≥ c · g(x)}.
We say that f ∈  O(g), or that f grows at least as fast as g(x) for x →  ∞.
Definition of Θ (), ""theta"":
Θ (g) := O(g) ∩  Ω (g).
We say that f ∈ Θ (g), or that f has the same growth rate as g(x) for x →  ∞.
Definition of o(), ""small oh"":
We say that f ∈  o(g), or that f grows slower than g(x) for x →  ∞.
Notation: Most of th e literature uses = in place of our ∈ , such as in x = O(x 2). If you do so, just remember that  
this = has none of the standard properties of an equality relation—it is neither commutative nor transitive. Thus  
O(x2) = x is not used, and from x = O(x 2) and x2 = O(x2) it does not follow that x = x 2. The key to avoiding confusion  
is the insight that O() is not a function but a set of functions.
Summation formulas",algorithms and data structures.pdf
"is the insight that O() is not a function but a set of functions.
Summation formulas
log2 denotes the logarithm to the base 2, ln the natural logarithm to the base e.
Algorithms and Data Structures 148  A Global Text",algorithms and data structures.pdf
"16. The mathematics of algorithm analysis
The asymptotic behavior of a sum can be derived by comparing the sum to an integral that can be evaluated in  
closed form. Let f(x) be a monotonically increasing, integrable function. Then 
is bounded below and above by sums (Exhibit 16.1):
Exhibit 16.1: Bounding a definite integral by lower and upper sums.
Letting xi = i + 1, this inequality becomes 
so
149
f(x)
x
y
  x 0   x1   x 2   x n−1   ξ ν",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Example
By substituting 
with k > 0 in (∗) we obtain 
and therefore
Example
By substituting
 f x=ln x and ∫ln x dx=x⋅ln x−x
 in (∗ ∗ ) we obtain
(n+1)⋅ln(n+1)−n−ln(n+1)≤∑
i=1
n
ln i≤(n+1)cdotln(n+1)−n ,  
and therefore
∑
i=1
n
log2 i=(n+1)⋅log2(n+1)− n
ln 2
+g(n)with g (n)∈O(log n)
Example
By substituting 
in (∗ ∗ ) we obtain 
with g(n) ∈  O(n · log n).
Recurrence relations
A homogeneous linear recurrence relation with constant coefficients is of the form
xn = a1 · xn–1 + a2 · xn–2 + … + ak · xn–k
where the coefficients a i are independent of n and x 1, x 2, … , x n–1 are specified. There is a general technique for  
solving linear recurrence relations with constant coefficients - that is, for determining x n as a function of n. We will  
demonstrate this technique for the Fibonacci sequence which is defined by the recurrence
Algorithms and Data Structures 150  A Global Text",algorithms and data structures.pdf
"16. The mathematics of algorithm analysis
xn = xn–1 + xn–2,  x0 = 0,  x1 = 1.
We seek a solution of the form
xn = c · rn
with constants c and r to be determined. Substituting this into the Fibonacci recurrence relation yields
c · rn = c · rn–1 + c · rn–2
or
c · rn–2 · (r2 – r – 1) = 0.
This equation is satisfied if either c = 0 or r = 0 or r 2 – r – 1 = 0. We obtain the trivial solution x n = 0 for all n if c 
= 0 or r = 0. More interestingly, r2– r – 1 = 0 for
The sum of two solutions of a homogeneous linear recurrence relation is obviously also a solution, and it can be  
shown that any linear combination of solutions is again a solution. Therefore, the most general solution of the  
Fibonacci recurrence has the form
where c1 and c2 are determined as solutions of the linear equations derived from the initial conditions:
which yield 
the complete solution for the Fibonacci recurrence relation is therefore",algorithms and data structures.pdf
"which yield 
the complete solution for the Fibonacci recurrence relation is therefore 
Recurrence  relations  that  are  not  linear  with  constant  coefficients  have  no  general  solution  techniques 
comparable  to  the  one  discussed  above.  General  recurrence  relations  are  solved  (or  their  solutions  are 
approximated or bounded) by trial-and-error techniques. If the trial and error is guided by some general technique,  
it will yield at least a good estimate of the asymptotic behavior of the solution of most recurrence relations.
Example
Consider the recurrence relation 
151",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
with a > 0 and b > 0, which appears often in the average-case analysis of algorithms and data structures. When we  
know from the interpretation of this recurrence that its solution is monotonically nondecreasing, a systematic trial-
and-error process leads to the asymptotic behavior of the solution. The simplest possible try is a constant, x n = c. 
Substituting this into (∗) leads to 
so xn = c is not a solution. Since the left-hand side x n is smaller than an average of previous values on the right-hand  
side, the solution of this recurrence relation must grow faster than c. Next, we try a linear function xn = c · n: 
At this stage of the analysis it suffices to focus on the leading terms of each side: c · n on the left and (c + a) · n on 
the right. The assumption a > 0 makes the right side larger than the left, and we conclude that a linear function also",algorithms and data structures.pdf
"the right. The assumption a > 0 makes the right side larger than the left, and we conclude that a linear function also  
grows too slowly to be a solution of the recurrence relation. A new attempt with a function that grows yet faster, xn = 
c · n2, leads to 
Comparing the leading terms on both sides, we find that the left side is now larger than the right, and conclude  
that a quadratic function grows too fast. Having bounded the growth rate of the solution from below and above, we  
try functions whos e growth rate lies between that of a linear and a quadratic function, such as x n = c · n 1.5. A more 
sophisticated approach considers a family of functions of the form x n = c · n 1+e for any ε > 0: All of them grow too  
fast. This suggests xn = c · n · log2 n, which gives 
with g(n) ∈  O(n · log n) and h(n) ∈  O(log n). To match the linear terms on each side, we must choose c such that",algorithms and data structures.pdf
"with g(n) ∈  O(n · log n) and h(n) ∈  O(log n). To match the linear terms on each side, we must choose c such that
or c = a · ln 4 ≈ 1.386 · a. Hence we now know that the solution to the recurrence relation (∗) has the form
Algorithms and Data Structures 152  A Global Text",algorithms and data structures.pdf
"16. The mathematics of algorithm analysis
Asymptotic performance of divide-and-conquer algorithms
We  illustrate  the  power  of  the  techniques  developed  in  previous  sections  by  analyzing  the  asymptotic 
performance not of a specific algorithm, but rather, of an entire class of divide-and-conquer algorithms. In “Divide  
and conquer recursion” we presented the following schema for divide-and-conquer algorithms that partition the set  
of data into two parts:
A(D):  if  simple(D) then return(A0(D))
else 1. divide: partition D into D1 and D2;
2. conquer: R1 := A(D1);  R2 := A(D2);
3. combine: return(merge(R1, R2));
Assume further that the data set D can always be partitioned into two halves, D 1 and D 2, at every level of  
recursion. Two comments are appropriate:
1. For repeated halving to be possible it is not necessary that the size n of the data set D be a power of 2, n =",algorithms and data structures.pdf
"1. For repeated halving to be possible it is not necessary that the size n of the data set D be a power of 2, n =  
2k. It is not important that D be partitioned into two exact halves—approximate halves will do. Imagine  
padding any data set D whose size is not a power of 2 with dummy elements, up to the next power of 2.  
Dummies  can  always  be  found  that  do  not  disturb  the  real  computation:  for  example,  by  replicating 
elements  or  by  appending  sentinels.  Padding  is  usually  just  a  conceptual  trick  that  may  help  in 
understanding the process, but need not necessarily generate any additional data.
2. Whether or not the divide step is guaranteed to partition D into two approximate halves, on the other hand,  
depends critically on the problem and on the data structures used. Example: Binary search in an ordered  
array partitions D into halves by probing the element at the midpoint; the same idea is impractical in a",algorithms and data structures.pdf
"array partitions D into halves by probing the element at the midpoint; the same idea is impractical in a  
linked list because the midpoint is not directly accessible.
Under our assumption of halving, the time complexity T(n) of algorithm A applied to data D of size n satisfies  
the recurrence relation
where f(n) is the sum of the partitioning or splitting time and the ""stitching time"" required to merge two solutions  
of size n /  2 into a solution of size n. Repeated substitution yields
The term n · T(1) expresses the fact that every data item gets looked at, the second sums up the splitting and  
stitching time. Three typical cases occur:
(a) Constant time splitting and merging f(n) = c yields
153",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
T(n) = (T(1) + c) · n.
Example: Find the maximum of n numbers.
(b) Linear time splitting and merging f(n) = a · n + b yields
T(n) = a · n · log2 n + (T(1) + b) · n.
Examples: Mergesort, quicksort.
(c) Expensive splitting and merging: n ∈  o(f(n)) yields
T(n) = n · T(1) + O(f(n) · log n)
and therefore rarely leads to interesting algorithms.
Permutations
Inversions
Let (ak: 1 ≤ k ≤ n) be a permutation of the integers 1 .. n. A pair (a i, aj), 1 ≤ I < j ≤ n, is called an inversion iff ai > 
aj. What is the average number of inversions  in a permutation? Consider all permutations in pairs; that is, with any  
permutation A:
a1 = x1; a2 = x2; … ; an = xn
consider its inverse A', which contains the elements of A in inverse order:
a1 = xn; a2 = xn–1; … ; an = x1.
In one of these two permutations x i and x j are in the correct order, in the other, they form an inversion. Since",algorithms and data structures.pdf
"a1 = xn; a2 = xn–1; … ; an = x1.
In one of these two permutations x i and x j are in the correct order, in the other, they form an inversion. Since  
there are n·  (n – 1) /  2 pairs of elements (xi, xj) with 1 ≤ i < j ≤ n there are, on average, 
inversions.
Average distance
Let (ak: 1 ≤ k ≤ n) be a permutation of the natural numbers from 1 to n. The distance of the element a i from its  
correct position is |ai – i|. The total distance of all elements from their correct positions is 
Therefore, the average total distance (i.e. the average over all n! permutations) is
Algorithms and Data Structures 154  A Global Text",algorithms and data structures.pdf
"16. The mathematics of algorithm analysis
Let 1 ≤  i ≤ n and 1 ≤ j ≤ n. Consider all permutations for which a i is equal to j. Since there are (n – 1)! such  
permutations, we obtain
Therefore, 
the average distance of an element ai from its correct position is therefore
Trees
Trees are ubiquitous in disc rete mathematics and computer science, and this section summarizes some of the  
basic concepts, terminology, and results. Although trees come in different versions, in the context of algorithms and  
data structures, ""tree"" almost always means an ordered rooted tree. An ordered rooted tree  is either empty or it  
consists of a node, called a root, and a sequence of k ordered subtrees T1, T2, … , T k (Exhibit 16.2). The nodes of an 
ordered tree that have only empty subtrees are called leaves or external nodes, the other nodes are called  internal 
nodes (Exhibit 16.3). The roots of the subtrees attached to a node are its children; and this node is their parent.
155",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 16.2: Recursive definition of a rooted, ordered tree.
The level of a node is defined recursively. The root of a tree is at level 0. The children of a node at level t are at  
level t + 1. The level of a node is the length of the path from the root of the tree to this node. The height of a tree is  
defined as the maximum level of all leaves. The path length of a tree is the sum of the levels of all its nodes ( Exhibit 
16.3).
Exhibit 16.3: A tree of height = 4 and path length = 35.
A binary tree is  an ordered tree whose nodes have at most two children. A 0-2  binary tree is a tree in which  
every node has zero or two children but not one. A 0-2 tree with n leaves has exactly n  – 1 internal nodes. A binary  
tree of height h is called complete (completely balanced)  if it has 2h+1 – 1 nodes (Exhibit 16.4. A binary tree of height",algorithms and data structures.pdf
"tree of height h is called complete (completely balanced)  if it has 2h+1 – 1 nodes (Exhibit 16.4. A binary tree of height 
h is called  almost comple te if  all its leaves are on levels h – 1 and h, and all leaves on level h are as far left as  
possible (Exhibit 16.4).
Algorithms and Data Structures 156  A Global Text",algorithms and data structures.pdf
"16. The mathematics of algorithm analysis
Exhibit 16.4: Examples of well-balanced binary trees.
Exercises
1. Suppose that we are comparing implementations of two algorithms on the same machine. For inputs of size  
n, the first algorithm runs in 9 · n 2 steps, while the second algorithm runs in 81 · n · log2 n steps. Assuming 
that the steps in both algorithms take the same time, for which values of n does the first algorithm beat the  
second algorithm?
2. What is th e smallest value of n such that an algorithm whose running time is 256 · n2 runs faster than an  
algorithm whose running time is 2n on the same machine?
3. For each of the following functions f i(n), determine a function g(n) such that f i(n) ∈ Θ (g(n)). The function  
g(n) should be as simple as possible.
f1(n) = 0.001 ·  n7 + n2 + 2 ·  n
f2(n) = n ·  log n + log n + 1234 ·  n
f3(n) = 5 ·  n ·  log n + n2 ·  log n + n2
f4(n) = 5 ·  n ·  log n + n3 + n2 ·  log n
4. Prove formally that 1024 ·  n 2+ 5 ·  n ∈  Θ (n2).",algorithms and data structures.pdf
"f3(n) = 5 ·  n ·  log n + n2 ·  log n + n2
f4(n) = 5 ·  n ·  log n + n3 + n2 ·  log n
4. Prove formally that 1024 ·  n 2+ 5 ·  n ∈  Θ (n2).
5. Give an asymptotically tight bound for the following summation:
6. Find the most general solutions to the following recurrence relations.
7. Solve the recurrence T(√n) = 2·T() + log2 n. Hint: Make a change of variables m = log2 n.
8. Compute the number of inversions and the total distance for the permutation (3 1  2 4).
157",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
17. Sorting and its 
complexity
Learning objectives:
• What is sorting? 
• basic ideas and intrinsic complexity
• insertion sort
• selection sort
• merge sort
• distribution sort
• a lower bound Ω (n· log  n)
• Quicksort
• Sorting in linear time?
• sorting networks
What is sorting? How difficult is it?
The problem
Assume that S is a set of n elements x1, x2, … , xn drawn from a domain X, on which a total order ≤ is defined (i.e.  
a relation that satisfies the following axioms):
≤ is reflexive (i.e ∀ ∀ x ∈  X:  x ≤ x)
≤ is antisymmetric (i.e ∀ ∀ x, y ∈  X:  x ≤ y  ∧  y ≤ x ⇒  x = y)
≤ is transitive (i.e ∀ ∀ x, y, z ∈  X:  x ≤ y  ∧  y ≤ z  ⇒   x ≤ z)
≤ is total (i.e. ∀ ∀ x, y ∈  X  ⇒  x ≤ y  ∨  y ≤ x)
Sorting is the process of generating a sequence 
such that (i1, i2, … , in) is a permutation of the integers from 1 to n and",algorithms and data structures.pdf
"Sorting is the process of generating a sequence 
such that (i1, i2, … , in) is a permutation of the integers from 1 to n and 
holds.  Phrased  abstractly,  sorting  is  the  problem  of  finding  a  specific  permutation  (or  one  among  a  few 
permutations, when distinct  elements  may  have equal  values) out of n! possible  permutations  of the n given  
elements. Usually, the set S of elements to be sorted will be given in a data structure; in this case, the elements of S  
are ordered implicitly by this data structure, but not necessarily according to the desired order ≤. Typical sorting  
problems assume that S is given in an array or in a sequential file (magnetic tape), and the result is to be generated  
in the same structure. We characterize elements by their position in the structure (e.g. A[i] in the array A or by the  
Algorithms and Data Structures 158  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
value of a pointer in a sequential file). The access operations provided by the underlying data structure determine  
what sorting algorithms are possible.
Algorithms
Most sorting algorithms are refinements of the following idea:
while ∃(i, j):  i < j  ∧  A[i] > A[j]  do  A[i] :=: A[j];
where :=: denotes the exchange operator. Even sorting algorithms that do not explicitly exchange pairs of elements,  
or do not use an array as the underlying data structure, can usually be thought of as conforming to the schema  
above. An insertion sort, for example, takes one element at a time and inserts it in its proper place among those  
already sorted. To find the correct place of insertion, we can think of a ripple effect whereby the new element  
successively displaces (exchanges position with) all those larger than itself.
As the schema above shows, two types of operations are needed in order to sort:",algorithms and data structures.pdf
"As the schema above shows, two types of operations are needed in order to sort:
• collecting information about the order of the given elements
• ordering the elements (e.g. by exchanging a pair)
When designing an efficient algorithm we seek to economize the number of operations of both types: We try to  
avoid  collecting  redundant  information,  and  we  hope  to  move  an  element  as  few  times  as  possible.  The 
nondeterministic  algorithm  given  above  lets  us  perform  any  one  of  a  number  of  exchanges  at  a  given  time, 
regardless of their usefulness. For example, in sorting the sequence
x1 = 5, x2 = 2, x3 = 3, x4 = 4, x5 = 1
the nondeterministic algorithm permits any of seven exchanges
x1 :=: xi for 2 ≤ i ≤ 5  and  xj :=: x5 for 2 ≤ j ≤ 4.
We might have reached the state shown above by following an exotic sorting technique that sorts ""from the  
middle toward both ends"", and we might know at this time that the single exchange x 1 :=: x5 will complete the sort.",algorithms and data structures.pdf
"middle toward both ends"", and we might know at this time that the single exchange x 1 :=: x5 will complete the sort.  
The nondeterministic algorithm gives us no handle to express and use this knowledge.
The attempt to economize work forces us to depart from nondeterminacy and to impose a control structure that  
carefully sequences the operations to be performed so as to make maximal use of the information g ained so far. The 
resulting algorithms will be more complex  and difficult to understand. It is useful to remember, though, that  
sorting is basically a simple problem with a simple solution and that all the acrobatics in this chapter are due to our  
quest for efficiency.
Intrinsic complexity
There are obvious limits to how much we can economize. In the absence of any previously acquired information,  
it is clear that each element must be inspected and, in general, moved at least once. Thus we cannot hope to get",algorithms and data structures.pdf
"it is clear that each element must be inspected and, in general, moved at least once. Thus we cannot hope to get  
away with fewer than Ω (n) primitive operations. There are less obvious limits, we mention two of them here.
1. If information is collected  by asking binary questions only (any question that may receive one of two  
answers (e.g. a yes/no question, or a comparison of two elements that yields either ≤ or >), then at least n ·  
log2 n questions are necessary in general, as will be proved in the section ""A lower bound Ω n · logn"". Thus in 
this model of computation, sorting requires time Θ (n · log n).
2. In addition to collecting information, one must rearrange the elements. In the section ""Permutation"" in  
chapter 16, we have shown that in a permutation the average distance of an element from its correct  
159",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
position is approximately n/3. Therefore elements have to move an average distance of approximately n/3  
elements to end up at their destination. Depending on the access operations of the underlying storage  
structure, an element can be moved to its correct position in a single step of average length n/3, or in n/3  
steps of average length 1. If elements are rearranged by exchanging adjacent elements only, then on average  
Θ (n2) moving operations are required. Therefore, short steps are insufficient to obtain an efficient Θ (n · log 
n) sorting algorithm.
Practical aspects of sorting
Records instead of elements. We discuss sorting assuming only that the elements to be sorted are drawn  
from a totally ordered domain. In practice these elements are just the keys of records that contain additional data  
associated with the key: for example,
type recordtype = record
key: keytype;  { totally ordered by ≤ }",algorithms and data structures.pdf
"associated with the key: for example,
type recordtype = record
key: keytype;  { totally ordered by ≤ }
data: anytype
end;
We use the relational operators =, <, ≤ to compare keys, but in a given programming language, say Pascal, these  
may be undefined on values of type keytype. In general, they must be replaced by procedures: for example, when  
comparing strings with respect to the lexicographic order.
If the key field is only a small part of a large record, the exchange operation :=:, interpreted literally, becomes an  
unnecessarily costly copy operation. This can be avoided by leaving the record (or just its data field) in place, and  
only moving a small surrogate record consisting of a key and a pointer to its associated record.
Sort generators. On many systems, particularly in the world of commercial data processing, you may never  
need to write a sorting program, even though sorting is a frequently executed operation. Sorting is taken care of by",algorithms and data structures.pdf
"need to write a sorting program, even though sorting is a frequently executed operation. Sorting is taken care of by  
a sort generator, a program akin to a compiler; it selects a suitable sorting algorithm from its repertoire and tailors  
it to the problem at hand, depending on parameters such as the number of elements to be sorted, the resources  
available, the key type, or the length of the records.
Partially sorted sequences. The algorithms we discuss ignore any order that may exist in the sequence to be  
sorted. Many applications call for sorting files that are almost sorted, for example, the case where a sorted master 
file is updated with an unsorted transaction file. Some algorithms take advantage of any order present in the input  
data; their time complexity varies from O(n) for almost sorted files to O(n · log n) for randomly ordered files.
Types of sorting algorithms",algorithms and data structures.pdf
"data; their time complexity varies from O(n) for almost sorted files to O(n · log n) for randomly ordered files.
Types of sorting algorithms
Two important class es of incremental sorting algorithms create order by processing each element in turn and  
placing  it  in  its  correct  position.  These  classes,  insertion  sorts and  selection  sorts,  are  best  understood  as 
maintaining two disjoint, mutually exhaustive structures called 'sorted' and 'unsorted'.
Initialize: 'sorted' := Ø;  'unsorted' := {x1, x2, … , xn};
Loop: for i := 1 to n do
move an element from 'unsorted' to its correct place in 
'sorted';
The following illustrations show 'sorted'  and 'unsorted' sharing  an  array[1 .. n]. In this case the boundary  
between 'sorted' and 'unsorted' is represented by an index i that increases as more elements become ordered. The  
important distinction between the two types of sorting algorithms emerges from the question: In which of the two",algorithms and data structures.pdf
"important distinction between the two types of sorting algorithms emerges from the question: In which of the two  
Algorithms and Data Structures 160  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
structures  is  most of the work done? Insertion  sorts  remove  the first or  most  easily  accessible  element  from 
'unsorted' and search through 'sorted' to find its proper place. Selection sorts search through 'unsorted' to find the  
next element to be appended to 'sorted'.
Insertion sort
The i-th step inserts the i-th element into the sorted sequence of the first (i – 1) elements Exhibit 17.1).
Exhibit 17.1: Insertion sorts move an easily accessed element to its correct place.
Selection sort
The i-th step selects the smallest among the n – i + 1 elements not yet sorted, and moves it to the i-th position  
(Exhibit 17.2).
Exhibit 17.2: Selection sorts search for the correct element to move to an easily accessed place. 
Insertion and selection sorts repeatedly search through a large part of the entire data to find the proper place of",algorithms and data structures.pdf
"Insertion and selection sorts repeatedly search through a large part of the entire data to find the proper place of  
insertion  or  the  proper  element  to  be  moved.  Efficient  search  requires  random  access,  hence  these  sorting 
techniques are used primarily for internal sorting in central memory.
Merge sort
Merge sorts process (sub)sequences of elements in unidirectional order and thus are well suited for  external 
sorting on secondary storage media that provide sequential access only, such as magnetic tapes; or random access  
to large blocks of data, such as disks. Merge sorts are also efficient for internal sorting. The basic idea is to merge  
two sorted sequences of elements, called runs, into one longer sorted sequence. We read each of the input runs, and  
write the output run, starting with small elements and ending with the large ones. We keep comparing the smallest",algorithms and data structures.pdf
"write the output run, starting with small elements and ending with the large ones. We keep comparing the smallest  
of the remaining elements on each input run, and append the smaller of the two to the output run, until both input  
runs are exhausted (Exhibit 17.3).
161",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 17.3: Merge sorts exploit order already present.
The processor shown at left in  Exhibit 17.4  reads two tapes, A and B. Tape A contains runs 1 and 2; tape B  
contains runs 3 and 4. The processor merges runs 1 and 3 into the single run 1 & 3 on tape C, and runs 2 and 4 into  
the single run 2 & 4 on tape D. In a second merge step, the processor shown at the right reads tapes C and D and  
merges the two runs 1 & 3 and 2 & 4 into one run, 1  &  3  &  2  & 4.
Exhibit 17.4: Two merge steps in sequence.
Distribution sort
Distribution  sorts process  the  representation of an  element as  a value in a radix  number system and  use  
primitive arithmetic operations such as ""extract the k-th digit"". These sorts do not compare elements directly. They  
introduce  a  different  model  of computation  than  the  sorts  based  on  comparisons,  exchanges,  insertions,  and",algorithms and data structures.pdf
"introduce  a  different  model  of computation  than  the  sorts  based  on  comparisons,  exchanges,  insertions,  and 
deletions that we have considered thus far. As an example, consider numbers with at most three digits in radix 4  
representation. In a first step these numbers are distributed among four queues according to their least significant  
digit, and the queues are concatenated in increasing order. The process is repeated for the middle digit, and finally  
for the leftmost, most significant digit, as shown in Exhibit 17.5
Algorithms and Data Structures 162  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
Exhibit 17.5 Distribution sorts use the radix representation of keys to organize elements in buckets
We have now seen the basic ideas on which all sorting algorithms are built. It is more important to understand  
these ideas than to know dozens of algorithms based on them. To appreciate the intricacy of sorting, you must  
understand some algorithms in detail: we begin with simple ones that turn out to be inefficient.
Simple sorting algorithms that work in time Θ (n2)
If you invent your  own sorting technique without prior study of the literature, you will probably ""discover"" a  
well-known inefficient algorithm that works in time O(n 2), requires time Θ (n2) in the worst case, and thus is of time  
complexity Ω (n2). Your algorithm might be similar to one described below.
Consider in-place algorithms that work on an array declared as
var  A: array[1 .. n] of elt;",algorithms and data structures.pdf
"Consider in-place algorithms that work on an array declared as
var  A: array[1 .. n] of elt;
and place the elements in ascending order. Assume that the comparison operators are defined on values of type elt.  
Let cbest, caverage, and cworst denote the number of comparisons, and ebest, eaverage, and eworst the number of exchange
operations performed in the best, average, and worst case, respectively. Let inv average denote the average number of  
inversions in a permutation.
Insertion sort (Exhibit 17.6)
Let –∞ denote a constant ≤ any key value. The smallest value in the domain often serves as a sentinel –∞.
163",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 17.6: Straight insertion propagates a ripple-effect across the sorted part of the array.
A[0] := –∞;
for i := 2 to n do  begin
j := i;
while  A[j] < A[j – 1]  do  { A[j] :=: A[j – 1]; { exchange } 
j := j – 1 }
end;
This straight insertion sort is an Θ (n) algorithm in the best case and an Θ (n2) algorithm in the average and worst 
cases. In the program above, the point of insertion is found by a linear search interleaved with exchanges. A binary  
search is possible but does not improve the time complexity in the average and  worst cases, since the actual  
insertion still requires a linear-time ripple of exchanges.
Selection sort (Exhibit 17.7)
Exhibit 17.7: Straight selection scans the unsorted part of the array.
for i := 1 to n – 1 do  begin
minindex := i;  minkey := A[i];
for j := i + 1 to n do
if  A[j] < minkey  then  { minkey := A[j];  minindex := j }
A[i] :=: A[minindex]  { exchange }
end;",algorithms and data structures.pdf
"for j := i + 1 to n do
if  A[j] < minkey  then  { minkey := A[j];  minindex := j }
A[i] :=: A[minindex]  { exchange }
end;
Algorithms and Data Structures 164  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
The sum in the formula for the number of comparisons reflects the structure of the two nested for loops. The  
body of the inner loop is executed the same number of times for each of the three cases. Thus this straight selection  
sort is of time complexity Θ (n2).
A lower bound Ω (n · log n)
A straightforward counting argument yields a lower bound on the time complexity of any sorting algorithm that  
collects information about the ordering of the elements by asking only binary questions. A binary question has a  
two-valued answer: yes or no, true or false. A comparison of two elements, x ≤ y, is the most obvious example, but  
the following theorem holds for binary questions in general.
Theorem:  Any sorting algorithm that collects information by asking binary questions only executes at least 
binary questions both in the worst case, and averaged over all n! permutations. Thus the average and worst-case",algorithms and data structures.pdf
"binary questions both in the worst case, and averaged over all n! permutations. Thus the average and worst-case  
time complexity of such an algorithm is Ω (n · log n).
Proof: A sorting algorithm of the type considered here can be represented by a  binary decision tree . Each  
internal node in such a tree represents a binary question, and each leaf corresponds to a result of the decision  
process. The decision tree must distinguish each of the n! possible permutations of the input data from all the  
others; and thus must have at least n! leaves, one for each permutation.
Example: The decision tree shown in Exhibit 17.8 collects the information necessary to sort three elements, x, y  
and z, by comparisons between two elements.
Exhibit 17.8 The decision tree shows the possible n! Outcomes when sorting n elements.
165",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The average number of binary questions needed by a sorting algorithm is equal to the average depth of the  
leaves of this decision tree. The lemma following this theorem will show that in a binary tree with k leaves the  
average depth  of the leaves is at least log 2k. Therefore, the average depth of the leaves corresponding to the n!  
permutations is at least log2n!. Since 
it follows that on average at least 
n∗log2 n1− n
ln2
binary questions are needed, that is, the time complexity of each such sorting algorithm is  Ω (n · log n) in the  
average, and therefore also in the worst case.
Lemma: In a binary tree with k leaves the average depth of the leaves is at least log2k.
Proof: Suppose that the lemma is not true, and let T be the counterexample with the smallest number of nodes.  
T cannot consist of a single node because the lemma is true for such a tree. If the root r of T has only one child, the",algorithms and data structures.pdf
"T cannot consist of a single node because the lemma is true for such a tree. If the root r of T has only one child, the  
subtree T' rooted at this child would contain the k leaves of T that have an even smaller average depth in T' than in  
T. Since T was the counterexample with the smallest number of nodes, such a T' cannot exist. Therefore, the root r  
of T must have two children, and there must be kL > 0 leaves in the left subtree and kR > 0 leaves in the right subtree 
of r (kL + kR = k). Since T was chosen minimal, the kL leaves in the left subtree must have an average depth of at least 
log2 kL, and the k R leaves in the right subtree must have an average depth of at least log 2 kR. Therefore, the average  
depth of all k leaves in T must be at least 
It is easy to see that (∗) assumes its minimum value if kL = kR. Since ( ∗ ) has the value log2 k if kL = kR = k /  2 we have 
found a contradiction to our assumption that the lemma is false.
Quicksort",algorithms and data structures.pdf
"found a contradiction to our assumption that the lemma is false.
Quicksort
Quicksort  (C.  A.  R.  Hoare,  1962)  [Hoa  62]  combines  the  powerful  algorithmic  principle  of  divide-and-
conquer  with an efficient way of moving elements using few exchanges. The divide phase partitions the array into  
two disjoint parts: the ""small"" elements on the left and the ""large"" elements on the right. The conquer phase sorts 
each part separately. Thanks to the work of the divide phase, the merge phase requires no work at all to combine  
two partial solutions. Quicksort's efficiency depends crucially on the expectation that the divide phase cuts two  
sizable subarrays rather than merely slicing off an element at either end of the array (Exhibit 17.9).
Algorithms and Data Structures 166  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
Exhibit 17.9: Quicksort partitions the array into the ""small"" elements on the left and the ""large"" elements 
on the right.
We chose an arbitrary threshold value m to define ""small"" as ≤ m, and ""large"" as ≥ m, thus ensuring that any  
""small element"" ≤ any ""large element"". We partition an arbitrary subarray A[L .. R] to be sorted by executing a left-
to-right scan (incrementing an index i) ""concurrently"" with a right-to-left scan (decrementing j) ( Exhibit 17.10). The 
left-to-right scan pauses at the first element A[i] ≥ m, and the right-to-left scan pauses at the first element A[j] ≤ m.  
When both scans have paused, we exchange A[i] and A[j] and resume the scans. The partition is complete when the  
two scans have crossed over with j < i. Thereafter, quicksort is called recursively for A[L .. j] and A[i .. R], unless one 
or both of these subarrays consists of a single element and thus is trivially sorted. Example of partitioning (m = 16):",algorithms and data structures.pdf
"or both of these subarrays consists of a single element and thus is trivially sorted. Example of partitioning (m = 16):
 25 23 3 16 4 7 29 6
 i j
 6 23 3 16 4 7 29 25
 i j
 6 7 3 16 4 23 29 25
 i j
 6 7 3 4 16 23 29 25
j  i
Exhibit 17.10: Scanning the array concurrently from left to right and from right to left.
Although the threshold value m appeared arbitrary in the description above, it must meet criteria of correctness  
and efficiency. Correctness: if either the set of elements ≤ m or the set of elements ≥ m is empty, quicksort fails to  
terminate. Thus we require that min(xi) ≤ m ≤ max(xi). Efficiency requires that m be close to the median.
How do we find the median of n elements? The obvious answer is to sort the elements and pick the middle one,  
but  this  leads  to  a  chicken-and-egg  problem  when  trying  to  sort  in  the  first  place.  There  exist  sophisticated",algorithms and data structures.pdf
"but  this  leads  to  a  chicken-and-egg  problem  when  trying  to  sort  in  the  first  place.  There  exist  sophisticated 
algorithms  that  determine  the  exact  median  of  n  elements  in  time  O(n)  in  the  worst  case  [BFPRT  72].  The 
multiplicative constant might be large, but from a theoretical point of view this does not matter. The elements are  
partitioned into two equal-sized halves, and quicksort runs in time O(n · log n) even in the worst case. From a  
practical point of view, however, it is not worthwhile to spend much effort in finding the exact median when there  
are much cheaper ways of finding an acceptable approximation. The following techniques have all been used to pick  
a threshold m as a ""guess at the median"":
• An array element in a fixed position such as A[(L + R) div 2]. Warning: stay away from either end, A[L] or  
A[R], as these thresholds lead to poor performance if the elements are partially sorted.",algorithms and data structures.pdf
"A[R], as these thresholds lead to poor performance if the elements are partially sorted.
• An array element in a random position: a simple technique that yields good results.
• The median of three or five array elements in fixed or random positions.
167",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
• The average between the smallest and largest element. This requires a separate scan of the entire array in 
the beginning; thereafter, the average for each subarray can be calculated during the previous partitioning 
process.
The recursive procedure 'rqs' is a possible implementation of quicksort. The function 'guessmedian' must yield a  
threshold that lies on or between the smallest and largest of the elements to be sorted. If an array element is used as  
the threshold, the procedure 'rqs' should be changed in such a way that after finishing the partitioning process this  
element is in its final position between the left and right parts of the array.
procedure rqs (L, R: 1 .. n);  { sorts A[L], … , A[R] }
var  i, j: 0 .. n + 1;
procedure partition;
var  m: elt;
begin  { partition }
m := guessmedian (L, R);
{ min(A[L], … , A[R]) ≤ m ≤ max(A[L], … , A[R]) }
i := L;  j := R;
repeat",algorithms and data structures.pdf
"var  m: elt;
begin  { partition }
m := guessmedian (L, R);
{ min(A[L], … , A[R]) ≤ m ≤ max(A[L], … , A[R]) }
i := L;  j := R;
repeat
{ A[L], … , A[i – 1] ≤ m ≤ A[j + 1], … , A[R] }
while  A[i] < m  do  i := i + 1;
{ A[L], … , A[i – 1] ≤ m ≤ A[i] }
while  m < A[j]  do  j := j – 1;
{ A[j] ≤ m ≤ A[j + 1], … , A[R] }
if  i ≤ j  then  begin
A[i] :=: A[j];  { exchange }
{ i ≤ j ⇒  A[i] ≤ m ≤ A[j] }
i := i + 1;  j := j – 1
{ A[L], … , A[i – 1] ≤ m ≤ A[j + 1], … , A[R] }
end
else
{ i > j ⇒  i = j + 1 ⇒  exit }
end
until  i > j
end;  { partition }
begin  { rqs }
partition;
if L < j then  rqs(L, j);
if i < R then  rqs(i, R)
end;  { rqs }
An initial call 'rqs(1, n)' with n > 1 guarantees that L < R holds for each recursive call.
An iterative implementation of quicksort is given by the following procedure, 'iqs', which sorts the whole array  
A[1 .. n]. The boundaries of the subarrays to be sorted are maintained on a stack.
procedure iqs;
const  stacklength = … ;",algorithms and data structures.pdf
"A[1 .. n]. The boundaries of the subarrays to be sorted are maintained on a stack.
procedure iqs;
const  stacklength = … ;
type  stackelement = record  L, R: 1 .. n  end;
var i, j, L, R, s: 0 .. n;
stack: array[1 .. stacklength] of stackelement;
procedure partition;  { same as in rqs }
end;  { partition }
begin  { iqs }
s := 1;  stack[1].L := 1;  stack[1].R := n;
repeat
L := stack[s].L;  R := stack[s].R;  s := s – 1;
Algorithms and Data Structures 168  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
repeat
partition;
if  j – L < R – i  then  begin
if  i <  R then  { s := s + 1;  stack[s].L := i; 
stack[s].R := R };
R := j
end
else  begin
if  L < j  then  { s := s + 1;  stack[s].L := L; 
stack[s].R := j };
L := i
end
until  L ≥ R
until  s = 0
end;  { iqs }
After partitioning, 'iqs' pushes the bounds of the larger part onto the stack, thus making sure that part will be  
sorted later, and sorts the smaller part first. Thus the length of the stack is bounded by log2n.
For very small arrays, the overhead of managing a stack makes quicksort less efficient than simpler O(n 2) 
algorithms, such as an insertion sort. A practically efficient implementation of quicksort might switch to another  
sorting technique for subarrays of size up to 10 or 20. [Sed 78] is a comprehensive discussion of how to optimize  
quicksort.
Analysis for three cases: best, ""typical"", and worst",algorithms and data structures.pdf
"quicksort.
Analysis for three cases: best, ""typical"", and worst
Consider a quicksort algorithm that chooses a guessed median that differs from any of the elements to be sorted  
and thus partitions the array into two parts, one with k elements, the other with n – k elements. The work q(n)  
required to sort n elements satisfies the recurrence relation
The constant b measures the cost of calling quicksort for the array to be sorted. The term a · n covers the cost of  
partitioning, and the terms q(k) and q(n – k) correspond to the work involved in quicksorting the two subarrays.  
Most quicksort algorithms partition the array into three parts: the ""small"" left part, the single array element used to  
guess the median, and the ""large"" right part. Their work is expressed by the equation
We analyze  equation  (*); it is close enough  to the second  equation to have the same asymptotic solution.",algorithms and data structures.pdf
"We analyze  equation  (*); it is close enough  to the second  equation to have the same asymptotic solution.  
Quicksort's behavior in the best and worst cases are easy to analyze, but the average over all permutations is not.  
Therefore, we analyze another average which we call the typical case.
Quicksort's  best-case behavior  is obtained if we guess the correct median that partitions the array into two  
equal-sized  subarrays. For simplicity's sake the following calculation assumes that n is a power of 2, but this  
assumption does not affect the solution. Then (*) can be rewritten as
We use this recurrence equation to calculate
169",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
and substitute on the right-hand side to obtain 
Repeated substitution yields
The constant q(1), whic h measures quicksort's work on a trivially sorted array of length 1, and b, the cost of a  
single procedure call, do not affect the dominant term n · log 2n. The constant factor a in the dominant term can be  
estimated by analyzing the code of the procedure 'partition'. When these details do not matter, we summarize:  
Quicksort's time complexity in the best case is Θ (n · log n).
Quicksort's worst-case  behavior occurs when one of the two subarrays consists of a single element after each  
partitioning. In this case equation (∗) becomes
We use this recurrence equation to calculate 
and substitute on the right-hand side to obtain 
Repeated substitution yields 
Therefore the time complexity of quicksort in the worst case is Θ (n2).",algorithms and data structures.pdf
"Repeated substitution yields 
Therefore the time complexity of quicksort in the worst case is Θ (n2).
For the analysis of quicksort's typical behavior we make the plausible assumption that the array is equally likely  
to get partitioned between any two of its elements: For all k, 1 ≤ k < n, the probability that the array A is partitioned  
into the subarrays A[1 .. k] and A[k + 1 .. n] is 1  / (n – 1). Then the average work to be performed by quicksort is  
expressed by the recurrence relation
Algorithms and Data Structures 170  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
This recurrence relation approximates the recurrence relation discussed in chapter 16 well enough to have the  
same solution
Since ln 4 ≈ 1.386, quicksort's asymptotic behavior in the typical case is only about 40% worse than in the best  
case, and remains in Θ (n · log n). [Sed 77] is a thorough analysis of quicksort.
Merging and merge sorts
The internal sorting algorithms presented so far require direct access to each element. This is reflected in our  
analyses by treating an array access A[i], or each exchange A[i] :=: A[j], as a primitive operation whose cost is  
constant (independent of n). This assumption is not valid for elements stored on secondary storage devices such as  
magnetic tapes or disks. A better assumption that mirrors the realities of external sorting is that the elements to be  
sorted are stored as a sequential file f. The file is accessed through a file pointer which, at any given time, provides",algorithms and data structures.pdf
"sorted are stored as a sequential file f. The file is accessed through a file pointer which, at any given time, provides  
direct access to a single element. Accessing other elements requires repositioning of the file pointer. Sequential files  
may permit the pointer to advance in one direction only, as in the case of Pascal files, or to move backward and  
forward. In either case, our theoretical model assumes that the time required  for repositioning the pointer is  
proportional  to  the  distance  traveled.  This  assumption  obviously  favors  algorithms  that  process  (compare, 
exchange) pairs of adjacent elements, and penalizes algorithms such as quicksort that access elements in random  
positions.
The following external sorting algorithm is based on the merge sort principle. To make optimal use of the  
available main memory, the algorithm first creates init ial runs; a run is a sorted subsequence of elements f i, fi+1, … ,",algorithms and data structures.pdf
"available main memory, the algorithm first creates init ial runs; a run is a sorted subsequence of elements f i, fi+1, … , 
fj stored consecutively in file f, f k ≤ fk+1 for all k with i ≤ k ≤ j – 1. Assume that a buffer of capacity m elements is  
available in main memory to create initial runs of length m (perhaps less for the last run). In processing the r-th  
run, r = 0, 1, … , we read the m elements fr·m+1, fr·m+2, … , fr·m+m into memory, sort them internally, and write the sorted 
sequence to a modified file f, which may or may not reside in the same physical storage area as the original file f.  
This new file f is partially sorted into runs: fk ≤ fk+1 for all k with r · m + 1 ≤ k < r · m + m.
At this point we need two files, g and h, in addition to the file f, which contai ns the initial runs. In a copy phase 
we distribute the initial runs by copying half of them to g, the other half to h. In the subsequent merge phase each",algorithms and data structures.pdf
"we distribute the initial runs by copying half of them to g, the other half to h. In the subsequent merge phase each 
run of g is merged with exactly one run of h, and the resulting new run of double length is written onto f ( Exhibit 
17.11). After the first cycle, consisting of a copy phase followed by a merge phase, f contains half as many runs as it  
did before. After log2(n /  m) cycles f contains one single run, which is the sorted sequence of all elements.
171",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 17.11: Each copy-merge cycle halves the number of runs and doubles their lengths. 
Exercise: a merge sort in main memory
Consider the following procedure that sorts the array A:
const  n = … ;
var  A: array[1 .. n] of integer;
…
procedure sort (L, R: 1 .. n);
var  m: 1 .. n;
procedure combine;
var B: array [1 .. n] of integer;
i, j, k: 1 .. n;
begin  { combine }
i := L;  j := m + 1;
for k := L to R do
if  (i > m) cor ((j ≤ R) cand (A[j] < A[i]))  then
{ B[k] := A[j];  j := j + 1 }
else
{ B[k] := A[i];  i := i + 1 } ;
for k := L to R do  A[k] := B[k]
end;  { combine }
begin  { sort}
if  L < R  then
{ m := (L + R) div 2;  sort(L, m);  sort(m + 1, R);  combine }
end;  { sort }
The relational operators 'cand' and 'cor' are conditional! The procedure is initially called by
sort(1,n); 
(a) Draw a picture to show how 'sort' works on an array of eight elements.",algorithms and data structures.pdf
"sort(1,n); 
(a) Draw a picture to show how 'sort' works on an array of eight elements.
(b) Write down a recurrence relation to describe the work done in sorting n elements.
(c) Determine the asymptotic time complexity by solving this recurrence relation.
(d) Assume that 'sort' is called for m subarrays of equal size, not just for two. How does the asymptotic time  
complexity change?
Solution
(a) 'sort' depends on the algorithmic principle of divide and conquer. After dividing an array into a left and a  
right subarray whose numbers of elements differ by at most one, 'sort' calls itself recursively on these two  
subarrays. After these two calls are finished, the procedure 'combine' merges the two sorted subarrays  
A[L .. m] and A[m + 1 .. R] together in B. Finally, B is copied to A. An example is shown in Exhibit 17.12.
Algorithms and Data Structures 172  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
Exhibit 17.12: Sorting an array by using a divide-and-conquer scheme.
(b) The work w(n) performed while sorting n elements satisfies
The first term describes the cost of the two recursive calls of 'sort', the term a · n is the cost of merging the  
two sorted subarrays, and the constant b is the cost of calling 'sort' for the array.
(c) If
is substituted in (*∗), we obtain 
Continuing this substitution process results in 
173",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
since w(1) is constant the time complexity of 'sort' is Θ(n ·  log n).
(d) If 'sort' is called recursively for m subarrays of equal size, the cost w'(n) is
solving this recursive equation shows that the time complexity does not change [i.e. it is Θ (n · log n)].
Is it possible to sort in linear time?
The lower bound Ω (n · log n) has been derived for sorting algorithms that gather information about the ordering  
of the elements by binary questions and nothing else. This lower bound need not apply in other situations.
Example 1: sorting a permutation of the integers from 1 to n
If we know that the elements to be sorted are a permutation of the integers 1 .. n, it is possible to sort in time  
Θ (n) by storing element i in the array element with index i.
Example 2: sorting elements from a finite domain
Assume that the elements to be sorted are samples from a finite domain W = 1 .. w. Then it is possible to sort in",algorithms and data structures.pdf
"Assume that the elements to be sorted are samples from a finite domain W = 1 .. w. Then it is possible to sort in  
time Θ (n) if gaps between the elements are allowed (Exhibit 17.13). The gaps can be closed in time Θ (w).
Exhibit 17.13: Sorting elements from a finite domain in linear time.
Do these  examples contradict the lower bound  Ω (n  · log n)? No, because in these examples the information  
about the ordering of elements is obtained by asking questions more powerful than binary questions: namely, n-
valued questions in Example 1 and w-valued questions in Example 2.
A  k-valued  question  is  equivalent  to  log2k  binary  questions.  When  this  ""exchange  rate""  is  taken  into 
consideration, the theoretical time complexities of the two sorting techniques above are Θ (n · log n) and Θ (n · log  
w), respectively, thus conforming to the lower bound in the section ""A lower bound Ω (n · log n)"".",algorithms and data structures.pdf
"w), respectively, thus conforming to the lower bound in the section ""A lower bound Ω (n · log n)"".
Sorting algorithms that sort in linear time (expected linear time, but not in the worst case) are described in the  
literature under the terms bucket sort, distribution sort, and radix sort.
Sorting networks
The sorting algorithms above are designed to run on a sequential machine in which all operations, such as  
comparisons and exchanges, are performed one at a time with a single processor. If algorithms are to be efficient,  
they need to be rethought when the ground rules for their execution change: when the theoretician uses another  
model of computation, or when they are executed on a computer with a different architecture. This is particularly  
true of the many different types of multiprocessor architectures that have been built or conceived. When many  
processors are available  to share the workload, questions  of how to distribute  the work among them, how to",algorithms and data structures.pdf
"processors are available  to share the workload, questions  of how to distribute  the work among them, how to  
synchronize their operation, and how to transport data, prevail. It is not our intention to discuss sorting on general-
purpose parallel machines. We wish to illustrate the point that algorithms must be redesigned when the model of  
Algorithms and Data Structures 174  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
computation changes. For this purpose a discussion of special-purpose sorting networks suffices. The ""processors""  
in a sorting network are merely comparators: Their only function is to compare the values on two input wires and  
switch them onto two output wires such that the smaller is on top, the larger at the bottom (Exhibit 17.14).
Exhibit 17.14: Building block of sorting networks.
Comparators are arranged into a network in which n wires enter at the left and n wires exit at the right, as  
Exhibit 17.15 shows, where each vertical connection joining a pair of wires represents a comparator. The illustration  
also shows what happens to four input elements, chosen to be 4, 1, 3, 2 in this example, as they travel from left to  
right through the network.
Exhibit 17.15: A comparator network that fails to sort. The output of each 
comparator performing an exchange is shown in the ovals.",algorithms and data structures.pdf
"Exhibit 17.15: A comparator network that fails to sort. The output of each 
comparator performing an exchange is shown in the ovals.
A network of comparators is a  sorting network  if it sorts every input configuration. We consider an input  
configuration to consist of distinct elements, so that without loss of generality we may regard it as one of the n!  
permutations of the sequence (1, 2, … , n). A network that sorts a duplicate-free configuration will also sort a  
configuration containing duplicates.
The comparator network above correctly sorts many of its 4! = 24 input configurations, but it fails on the  
sequence (4, 1, 3, 2). Hence it is not a sorting network. It is evident that a network with a sufficient number of  
comparators in the right places will sort correctly, but as the example above shows, it is not immediately evident  
what number suffices or how the comparators should be placed. The network in  Exhibit 17.16  shows that five",algorithms and data structures.pdf
"what number suffices or how the comparators should be placed. The network in  Exhibit 17.16  shows that five  
comparators, arranged judiciously, suffice to sort four elements.
Exhibit 17.16: Five comparators suffice to sort four elements.
How can we tell if a given n etwork sorts successfully? Exhaustive testing is feasible for small networks such as  
the one above, where we can trace the flow of all 4! = 24 input configurations. Networks with a regular structure  
175
c1 c2 c3 c4 c5",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
usually admit a simpler correctness proof. For this example, we observe that c 1, c2, and c3 place the smallest element 
on the top wire. Similarly, c1, c2, and c4 place the largest on the bottom wire. This leaves the middle two elements on  
the middle two wires, which c5 then puts into place.
What  design  principles  might  lead  us  to  create  large  sorting  networks  guaranteed  to  be  correct?  Sorting 
algorithms  designed  for  a  sequential  machine  cannot,  in  general,  be  mapped  directly  into  network  notation, 
because the network is a more restricted model of computation: Whereas most sequential sorting algorithms make  
comparisons based on the outcome of previous comparisons, a sorting netw ork makes the same comparisons for all 
input  configurations.  The  same  fundamental  algorithm  design  principles  useful  when  designing  sequential 
algorithms also apply to parallel algorithms.",algorithms and data structures.pdf
"algorithms also apply to parallel algorithms.
Divide-and-conquer. Place two sorting networks for n wires next to each other, and combine them into a sorting  
network for 2 · n wires by appending a merge network to merge their outputs. In sequential computation merging  
is simple because we can choose the most useful comparison depending on the outcome of previous comparisons.  
The rigid structure of comparator networks makes merging networks harder to design.
Incremental algorithm.We place an n-th wire next to a sorting network with n – 1 wires, and either precede or  
follow the network by a ""ladder"" of comparators that tie the extra wire into the existing network, as shown in the  
following figures. This leads to designs that mirror the straight insertion and selection algorithms  in the section  
""Simple sorting algorithms that work in time Θ (n2)
Insertion sort. With the top n – 1 elements sorted, the element on the bottom wire trickles into its correct place.",algorithms and data structures.pdf
"Insertion sort. With the top n – 1 elements sorted, the element on the bottom wire trickles into its correct place.  
Induction yields the expanded diagram on the right in Exhibit 17.17.
Exhibit 17.17: Insertion sort leads by induction to the sorting network on the right.
Selection sort. The maximum element first trickles down to the bottom, then the remaining elements are sorted.  
The expanded diagram is on the right in Exhibit 17.18.
Exhibit 17.18: Selection sort leads by induction to the sorting network on the right.
Comparators can be shifted along their pair of wires so as to reduce the number of stages, provided that the  
topology of the network remains unchanged. This compression reduces both insertion and selection sort to the  
triangular network shown in  Exhibit 17.19 . Thus we see that the distinction between insertion and selection was  
more a distinction of sequential order of operations rather than one of data flow.
Algorithms and Data Structures 176  A Global Text",algorithms and data structures.pdf
"17. Sorting and its complexity
Exhibit 17.19: Shifting comparators reduces the number of stages.
Any numbe r of comparators that are aligned vertically require  only a single unit of time. The compressed  
triangular network has O(n 2) comparators, but its time complexity is 2 · n – 1  ∈  O(n). There are networks with  
better asymptotic behavior, but they are rather exotic [Knu 73b].
Exercises and programming projects
1. Implement insertion sort, selection sort, merge sort, and quicksort and animate the sorting process for each 
of  these  algorithms:  for  example,  as  shown  in  the  snapshots  in  “Algorithm  animation”.  Compare  the 
number  of  comparisons  and  exchange  operations  needed  by  the  algorithms  for  different  input 
configurations.
2. What is the smallest possible depth of a leaf in a decision tree for a sorting algorithm?
3. Show that 2 · n – 1 comparisons are necessary in the worst case to merge two sorted arrays containing n  
elements each.",algorithms and data structures.pdf
"3. Show that 2 · n – 1 comparisons are necessary in the worst case to merge two sorted arrays containing n  
elements each.
4. The most obvious method of systematically interchanging the out-of-order pairs of elements in an array
var  A: array[1 .. n] of elt;
is to scan adjacent pairs of elements from bottom to top (imagine that the array is drawn vertically, with  
A[1] at the top and A[n] at the bottom) repeatedly, interchanging those found out of order:
for  i := 1  to  n – 1  do
 for  j := n  downto  i + 1  do
if  A[j – 1] > A[j]  then  A[j – 1] :=: A[j];
This technique is known as bubble sort, since smaller elements ""bubble up"" to the top.
(a) Explain by words, figures, and an example how bubble sort works. Show that this algorithm sorts  
correctly.
(b) Determine the exact number of comparisons and exchange operations that are performed by bubble  
sort in the best, average, and worst case.
(c) What is the worst-case time complexity of this algorithm?",algorithms and data structures.pdf
"sort in the best, average, and worst case.
(c) What is the worst-case time complexity of this algorithm?
5. A sorting algorithm is called stable if it preserves the original order of equal elements. Which of the sorting  
algorithms discussed in this chapter is stable?
6. Assume that quicksort chooses the threshold m as the first element of the sequence to be sorted. Show that  
the running time of such a quicksort algorithm is Θ (n2) when the input array is sorted in nonincreasing or  
nondecreasing order.
7. Find a worst-case input configuration for a quicksort algorithm that chooses the threshold m as the median  
of the first, middle, and last elements of the sequence to be sorted.
8. Array A contains m and array B contains n different integers which are not necessarily ordered:
const m = … ;  { length of array A }
n = … ;  { length of array B }
var A: array[1 .. m] of integer;
B: array[1 .. n] of integer;
177",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
A duplicate is an integer that is contained in both A and B. Problem: How many duplicates are there in A  
and B?
(a) Determine the time complexity of the brute-force algorithm that compares each integer contained in  
one array to all integers in the other array.
(b) Write a more efficient
function duplicates: integer;
Your solution may rearrange the integers in the arrays.
(c) What is the worst-case time complexity of your improved algorithm?
Algorithms and Data Structures 178  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Part V: Data structures
The tools of bookkeeping
When thinking of algorithms we emphasize a dynamic sequence of actions: ""Take this and do that, then that,  
then  …  .""  In  human  experience,  ""take""  is  usually  a  straightforward  operation,  whereas  ""do""  means  work.  In 
programming, on the other hand, there are lots of interesting examples where ""do"" is nothing more complex than  
incrementing a counter or setting a bit; but ""take"" triggers a long, sophisticated search. Why do we need fancy data  
structures at all? Why can't we just spread out the data on a desk top? Everyday experience does not prepare us to  
appreciate the importance of data structure—it takes programming experience to see that algorithms are nothing  
without data structures. The algorithms presented so far were carefully chosen to require only the simplest of data",algorithms and data structures.pdf
"without data structures. The algorithms presented so far were carefully chosen to require only the simplest of data  
structures:  static  arrays.  The  geometric  algorithms  of  Part  VI,  on  the  other  hand,  and  lots  of  other  useful 
algorithms, depend on sophisticated data structures for their efficiency.
The key insight in understanding data structures is the recognition that an algorithm in execution is, at all times,  
in some state, chosen from a potentially huge state space. The state records such vital information as what steps  
have already been taken with what results, and what remains to be done. Data structures are the bookkeepers that  
record all this state information in a tidy manner so that any part can be accessed and updated efficiently. The  
remarkable fact is that there are a relatively small number of standard data structures that turn out to be useful in",algorithms and data structures.pdf
"remarkable fact is that there are a relatively small number of standard data structures that turn out to be useful in  
the most varied types of algorithms and problems, and constitute essential knowledge for any programmer.
The literature on data structures.  Whereas one can present some algorithms without emphasizing data  
structures,  as  we  did  in  Part  III,  it  appears  pointless  to  discuss  data  structures  without  some  of  the  typical 
algorithms  that  use  them;  at the  very  least,  access  and  update  algorithms  form  a  necessary  part of any  data 
structure. Accordingly, a new data structure is typically published in the context of a particular new algorithm. Only  
later, as one notices its general applicability, it may find its way into textbooks. The data structures that have  
become standard today can be found in many books, such as [AHU 83], [CLR 90], [GB 91], [HS 82], [Knu 73a],",algorithms and data structures.pdf
"become standard today can be found in many books, such as [AHU 83], [CLR 90], [GB 91], [HS 82], [Knu 73a],  
[Knu 73b], [Meh 84a], [Meh 84c], [RND 77], [Sam 90a], [Sam 90b], [Tar 83], and [Wir 86].
Algorithms and Data Structures 179  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
18. What is a data structure?
Learning objectives:
• data structures for manual use (e.g. edge-notched cards)
• general-purpose data structures
• abstract data types specify functional properties only
• data structures include access and maintenance algorithms and their implementation
• performance criteria and measures
• asymptotics
Data structures old and new
The discipline of data structures, as a systematic body of knowledge, is truly a creation of computer science. The  
question of how best to organize data was a lot simpler to answer in the days before the existence of computers: the  
organization had to be simple, because there was no automatic device that could have processed an elaborate data  
structure, and there is no human being with enough patience to do it. Consider two examples.
1. Manual files  and  catalogs, as used  in business offices and  libraries, exhibit several distinct organizing",algorithms and data structures.pdf
"1. Manual files  and  catalogs, as used  in business offices and  libraries, exhibit several distinct organizing  
principles, such as sequential and hierarchical order and cross-references. From today's point of view,  
however, manual files are not well-defined data structures. For good reasons, people did not rigorously  
define those aspects that we consider essential when characterizing a data structure: what constraints are  
imposed  on  the data, both  on the structure and  its content; what operations the data  structure must  
support;  what  constraints  these  operations  must  satisfy.  As  a  consequence,  searching  and  updating  a 
manual file is not typically a process that can be automated: It requires common sense, and perhaps even  
expert training, as is the case for a library catalog.
2. In manual computing (with pencil and paper or a nonprogrammable calculator) the algorithm is the focus",algorithms and data structures.pdf
"2. In manual computing (with pencil and paper or a nonprogrammable calculator) the algorithm is the focus  
of attention, not the data structure. Most frequently, the person computing writes data (input, intermediate  
results, output) in any convenient place within his field of vision, hoping to find them again when he needs  
them. Occasionally, to facilitate highly repetitive computations (such as income tax declarations), someone  
designs a form to prompt the user, one operation at a time, to write each data item into a specific field. Such 
a form specifies both an algorithm and a data structure with considerable formality. Compared to the  
general-purpose data structures we study in this chapter, however, such forms are highly special purpose.
Edge-notched cards are perhaps the most sophisticated data structures ever designed for manual use. Let us  
illustrate them with the example of a database of English words organized so as to help in solving crossword",algorithms and data structures.pdf
"illustrate them with the example of a database of English words organized so as to help in solving crossword  
puzzles. We write one word per card and index it according to which vowels it contains and which ones it does not  
contain. Across the top row of the card we punch 10 holes labeled A, E, I, O, U, ~A, ~E, ~I, ~O, ~U. When a word,  
say ABACA, exhibits a given vowel, such as A, we cut a notch above the hole for A; when it does not, such as E, we  
cut a notch above the hole for ~E (pronounced ""not E""). Exhibit 18.1 shows the encoding of the words BEAUTIFUL, 
EXETER, OMAHA, OMEGA. For example, we search for words that contain at least one E, but no U, by sticking  
Algorithms and Data Structures 180  A Global Text",algorithms and data structures.pdf
"18. What is a data structure?
two needles through the pack of cards at the holes E and ~U. EXETER and OMEGA will drop out. In principle it is  
easy to make this sample database more powerful by including additional attributes, such as ""A occurs exactly  
once"", ""A occurs exactly twice"", ""A occurs as the first letter in the word"", and so on. In practice, a few dozen  
attributes and thousands of cards will stretch this mechanical implementation of a multikey data structure to its  
limits of feasibility.
Exhibit 18.1: Encoding of different words in edge-notched cards.
In contrast to data structures suitable for manual processing, those developed for automatic data processing can  
be complex. Complexity is not a goal in itself, of course, but it may be an unavoidable consequence of the search for  
efficiency. Efficiency, as measured by processing time and memory space required, is the primary concern of the",algorithms and data structures.pdf
"efficiency. Efficiency, as measured by processing time and memory space required, is the primary concern of the  
discipline of data structures. Other criteria, such as simplicity of the code, play a role, but the first question to be  
asked when evaluating a data structure that supports a specified set of operations is typically: How much time and  
space does it require?
In contrast to the typical situation of manual computing (consideration of the algorithm comes first, data gets  
organized only as needed), programmed computing typically proceeds in the opposite direction: First we define the  
organization of the data rigorously, and from this the structure of the algorithm follows. Thus algorithm design is  
often driven by data structure design.
The range of data structures studied
We present generally useful data  structures along with the corresponding query, update,  and  maintenance",algorithms and data structures.pdf
"The range of data structures studied
We present generally useful data  structures along with the corresponding query, update,  and  maintenance  
algorithms; and we develop concepts and techniques designed to organize a vast body of knowledge into a coherent  
whole. Let us elaborate on both of these goals.
""Generally useful"" refers to data structures that occur naturally in many applications. They are relatively simple  
from the point of view of the operations they support —tables and queues of various types are typical examples.  
These basic data structures are the building blocks from which an applications programmer may construct more  
elaborate structures tailored to her particular application. Although our collection of specific data structures is  
rather small, it covers the great majority of techniques an applications programmer is likely to need.
We develop a unified scheme for understanding many data structures as special cases of general concepts. This  
includes:
181",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
• The separation of abstract data types, which specify only functional properties, from data structures, which  
also involve aspects of implementation
• The classification of all data structures into three major types: implicit data structures, lists, and address  
computation
• A rough assessment of the performance of data structures based on the asymptotic analysis of time and  
memory requirements
The simplest and most common assumption about the elements to be stored in a data structure is that they  
belong to a domain on which a total order ≤ is defined. Examples: integers ordered by magnitude, a character set  
with its alphabetic order, character strings of bounded length ordered lexicographically. We assume that each  
element in a domain requires as much storage as any other element in that domain; in other words, that a data",algorithms and data structures.pdf
"element in a domain requires as much storage as any other element in that domain; in other words, that a data  
structure manages memory fragments of fixed size. Data objects of greatly variable size or length, such as fragments  
of text, are typically not considered to be ""elements""; instead, they are broken into constituent pieces of fixed size,  
each of which becomes an element of the data structure.
The elements stored in a data structure are often processed according to the order ≤ defined on their domain.  
The topic of  sorting, which we surveyed in “Sorting and its complexity”, is closely related to the study of data  
structures: Indeed, several sorting algorithms appear ""for free"" in “List structures”, because every structure that  
implements the abstract data type  dictionary leads to a sorting algorithm by successive insertion of elements,  
followed by a traversal.
Performance criteria and measures",algorithms and data structures.pdf
"followed by a traversal.
Performance criteria and measures
The design of data structures is dominated by considerations of efficiency, specifically with respect to time and  
memory. But efficiency is a multifaceted quality not easily defined and measured. As a scientific discipline, the  
study of data structures is not directly concerned with the number of microseconds, machine  cycles, or bytes  
required by a specific program processing a given set of data on a particular system. It is concerned with general  
statements from which an expert practitioner can predict concrete outcomes for a specific processing task. Thus,  
measuring run times and memory usage is not the typical way to evaluate data structures. We need concepts and  
notations  for  expressing  the  performance  of  an  algorithm  independently  of  machine  speed,  memory  size, 
programming language, and operating system, and a host of other details that vary from run to run.",algorithms and data structures.pdf
"programming language, and operating system, and a host of other details that vary from run to run.
The solution to this problem emerged over the past two decades as the discipline of computational complexity  
was developed. In this theory, algorithms are ""executed"" on some ""mathematical machine"", carefully designed to be  
as simple as possible to reflect the bare essentials of a problem. The machine makes available certain  primitive 
operations, and we measure ""time"" by counting how many of those are executed. For a given algorithm and all the  
data sets it accepts as input, we analyze the number of primitive operations executed as a function of the size of the  
data. We are often interested in the worst case, that is, a data set of given size that causes the algorithm to run as  
long as possible, and the average case, the run time averaged over all data sets of a given size.",algorithms and data structures.pdf
"long as possible, and the average case, the run time averaged over all data sets of a given size.
Among the many different mathematical machines that have been defined in the theory of computation, data  
structures are evaluated almost exclusively with respect to a theoretical random access machine (RAM). A RAM is  
essentially a memory with as many locations as needed, each of which can hold a data element, such as an integer,  
or a real number; and a processing unit that can read from any one or two locations, operate on their content, and  
write the result back into a third location, all in one time unit. This model is rather close to actual sequential  
Algorithms and Data Structures 182  A Global Text",algorithms and data structures.pdf
"18. What is a data structure?
computers, except that it incorporates no bounds on the memory size—either in terms of the number of locations or  
the size of the content of this location. It implies, for example, that a multiplication of two very large numbers  
requires no more time than 2 · 3 does. This assumption is unrealistic for certain problems, but is an excellent one  
for most program runs that fit in central memory and do not require variable-precision arithmetic or variable-
length data elements. The point is that the programmer has to understand the model and its assumptions, and  
bears responsibility for applying it judiciously.
In this model, time and memory requirements are expressed as functions of input data size, and thus comparing  
the performance of two data structures is reduced to comparing functions. Asymptotics has proven to be just the  
right tool for this comparison: sharp enough to distinguish different growth rates, blunt enough to ignore constant",algorithms and data structures.pdf
"right tool for this comparison: sharp enough to distinguish different growth rates, blunt enough to ignore constant  
factors that differ from machine to machine.
As an example of the concise descriptions made possible by asymptotic operation counts, the following table  
evaluates several implementations for the abstract data type 'dictionary'. The four operations 'find', 'insert', 'delete',  
and  'next'  (with  respect  to  the  order  ≤)  exhibit  different  asymptotic  time  requirements  for  the  different 
implementations. The student should be able to explain and derive this table after studying this part of the book.
Ordered array Linear list Balanced tree Hash table
find O(log n) O(n) O(log n) O(1)a 
next O(1) O(1) O(log n) O(n) 
insert O(n) O(n) O(log n) O(1)a 
delete O(n) O(n) O(log n) O(1)b 
a On the average, but not necessarily in the worst case
b Deletions are possible but may degrade performance
Exercise",algorithms and data structures.pdf
"a On the average, but not necessarily in the worst case
b Deletions are possible but may degrade performance
Exercise
1. Describe the manual data structures that have been developed to organize libraries (e.g. catalogs that allow  
users to get access to the literature in their field of interest, or circulation records, which keep track of who  
has borrowed what book). Give examples of queries that can be answered by these data structures.
183",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
19. Abstract data types
Learning objectives:
• data abstraction
• abstract data types as a tool to describe the functional behavior of data structures
• examples of abstract data types: stack, fifo queue, priority queue, dictionary, string
Concepts: What and why?
A data structure organizes the data to be processed in such a way that the relations among the data elements are  
reflected and the operations to be performed on the data are supported. How these goals can be achieved efficiently 
is the central issue in data structures and a major concern of this book. In this chapter, however, we ask not how  
but what? In particular, we ask: what is the exact functional behavior a data structure must exhibit to be called a  
stack, a queue, or a dictionary or table?
There are several reasons for seeking a formal functional specification for common data structures. The primary",algorithms and data structures.pdf
"There are several reasons for seeking a formal functional specification for common data structures. The primary  
motivation  is  increased  generality  through  abstraction;  specifically,  to  separate  input/output  behavior  from 
implementation, so that the implementation can be changed without affecting any program that uses a particular  
data type. This goal led to the earlier introduction of the concept of type in programming languages: the type real is 
implemented differently on different machines, but usually a program using reals does not require modification  
when run on another machine. A secondary motivation is the ability to prove general theorems about all data  
structures that exhibit certain properties, thus avoiding the need to verify the theorem in each instance. This goal is  
akin to the one that sparked the development of algebra: from the axioms that define a field, we prove theorems",algorithms and data structures.pdf
"akin to the one that sparked the development of algebra: from the axioms that define a field, we prove theorems  
that hold equally true for real or complex numbers as well as quaternions.
The primary motivation can be further explained by calling on an analogy between data and programs. All  
programming languages support the concept of  procedural abstraction : operations or algorithms are isolated in  
procedures, thus making it easy to replace or change them without affecting other parts of the program. Other  
program parts do not know how a certain operation is realized; they know only how to call the corresponding  
procedure and what effect the procedure call will have. Modern programming languages increasingly support the  
analogous concept of  data abstraction or data encapsulation : the organization of data is encapsulated ( e.g. in a  
module or a  package) so that it is possible  to change  the data  structure without having to change  the whole  
program.",algorithms and data structures.pdf
"module or a  package) so that it is possible  to change  the data  structure without having to change  the whole  
program.
The secondary motivation for formal specification of data types remains an unrealized goal: although abstract  
data types are an active topic for theoretical research, it is difficult today to make the case that any theorem of use  
to programmers has been proved.
An abstract data type consists of a domain from which the data elements are drawn, and a set of operations.  
The specification of an abstract data type must identify the domain and define each of the operations. Identifying  
and describing the domain is generally straightforward. The definition of each operation consists of a syntactic and  
a semantic part. The syntactic part, which corresponds to a procedure heading, specifies the operation's name and  
Algorithms and Data Structures 184  A Global Text",algorithms and data structures.pdf
"19. Abstract data types
the type of each operand. We present the syntax of operations in mathematical function notation, specifying its  
domain and range. The semantic part attaches a meaning to each operation: what values it produces or what effect  
it has on its environment. We specify the semantics of abstract data types algebraically by axioms from which other  
properties may be deduced. This formal approach has the advantage that the operations are defined rigorously for  
any domain with the required properties. A formal description, however, does not always appeal to intuition, and  
often forces us to specify details that we might prefer to ignore. When every detail matters, on the other hand, a  
formal  specification  is  superior  to  a  precise  specification  in  natural  language;  the  latter  tends  to  become 
cumbersome and difficult to understand, as it often takes many words to avoid ambiguity.",algorithms and data structures.pdf
"cumbersome and difficult to understand, as it often takes many words to avoid ambiguity.
In this chapter we consider the abstract data types: stack, first-in-first-out queue, priority queue, and dictionary.  
For each of these data types, there is an ideal, unbounded version, and several versions that reflect the realities of  
finite machines. From a theoretical point of view we only need the ideal data types, but from a practical point of  
view,  that  doesn't  tell  the  whole  story:  in  order  to  capture  the  different  properties  a  programmer  intuitively 
associates with the vague concept ""stack"", for example, we are forced into specifying different types of stacks. In  
addition to the ideal  unbounded stack , we specify a  fixed-length stack  which mirrors the behavior of an array  
implementation,  and  a  variable-length  stack which  mirrors  the  behavior  of  a  list  implementation.  Similar",algorithms and data structures.pdf
"implementation,  and  a  variable-length  stack which  mirrors  the  behavior  of  a  list  implementation.  Similar 
distinctions apply to the other data types, but we only specify their unbounded versions.
Let X denote the domain from which the data elements are drawn. Stacks and fifo queues make no assumptions  
about X; priority queues and dictionaries require that a total order ≤ be defined on X. Let X ∗denote the set of all  
finite sequences over X.
Stack
A stack is also called a last-in-first-out queue, or lifo queue. A brief informal description of the abstract data type 
stack (more specifically, unbounded stack, in contrast to the versions introduced later) might merely state that the  
following operations are defined on it:
- create Create a new, empty stack.
- empty Return true if the stack is empty.
- push Insert a new element.
- top Return the element most recently inserted, if the stack is not 
empty.",algorithms and data structures.pdf
"- empty Return true if the stack is empty.
- push Insert a new element.
- top Return the element most recently inserted, if the stack is not 
empty.
- pop Remove the element most recently inserted, if the stack is not 
empty.
Exhibit 19.1 helps to clarify the meaning of these words.
Exhibit 19.1: Elements are inserted at and removed from the top of the stack.
A definition that uses conventional mathematical notation to capture the intention of the description above  
might define the operations by explicitly showing their effect on the contents of a stack. Let S = X ∗ be the set of  
185",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
possible states of a stack, let s = x1 x2 … xk ∈  S be an arbitrary stack state with k elements, and let λ denote the empty 
state of the stack, corresponding to the null string ∈  X*. Let 'cat' denote string concatenation. Define the functions
create: → S
empty: S → {true, false}
push: S × X → S
top: S – {λ} → X
pop: S – {λ} → S
as follows:
     ∀ s ∈  S,∀∀ x, y ∈  X:
create = λ
empty(λ) = true
s ≠ λ ⇒  empty(s) = false
push(s, y) = s cat y = x1 x2 … xk y
s ≠ λ top(s) = xk
s ≠ pop(s) = x1 x2 … xk–1
This definition refers explicitly to the contents of the stack. If we prefer to hide the contents and refer only to  
operations and their results, we are led to another style of formal definition of abstract data types that expresses the  
semantics of the operations by relating them to each other rather than to the explicitly listed contents of a data",algorithms and data structures.pdf
"semantics of the operations by relating them to each other rather than to the explicitly listed contents of a data  
structure. This is the commonly used approach to define abstract data types, and we follow it for the rest of this  
chapter.
Let S be a set and s0 ∈  S a distinguished state. s0 denotes the empty stack, and S is the set of stack states that can  
be obtained from the empty stack by performing finite sequences of 'push' and 'pop' operations. The following  
functions represent stack operations:
create: → S
empty: S → {true, false}
push: S  X → S
top: S – {s0} → X
pop: S – {s0} → S
The semantics of the stack operations is specified by the following 
axioms:
  ∀ s ∈  S, ∀ x ∈  X:
(1) create = s0
(2) empty(s0) = true
(3) empty(push(s, x)) = false
(4) top(push(s, x)) = x
(5) pop(push(s, x)) = s
These axioms can be described in natural language as follows:
(1) 'create' produces a stack in the distinguished state.
(2) The distinguished state is empty.",algorithms and data structures.pdf
"(1) 'create' produces a stack in the distinguished state.
(2) The distinguished state is empty.
(3) A stack is not empty after an element has been inserted.
(4) The element most recently inserted is on top of the stack.
(5) 'pop' is the inverse of 'push'.
Notice that 'create' plays a different role from the other stack operations: it is merely a mechanism for causing a  
stack to come into existence, and could have been omitted by postulating the existence of a stack in st ate s0. In any 
implementation, however, there is always some code that corresponds to 'create'. Technical note: we could identify 
Algorithms and Data Structures 186  A Global Text",algorithms and data structures.pdf
"19. Abstract data types
'create' with s 0, but we choose to make a distinction between the act of creating a new empty stack and the empty 
state that results from this creation; the latter may recur during normal operation of the stack.
Reduced sequences
Any s ∈  S is obtained from the empty stac k s0 by performing a finite sequence of 'push' and 'pop' operations. By  
axiom (5) this sequence can be reduced to a sequence that transforms s 0 into s and consists of 'push' operations  
only.
Example
s = pop(push(pop(push(push(s0, x), y)), z))
= pop(push(push(s0, x), z))
= push(s0, x)
An implementation of a stack may provide the following procedures:
procedure create(var s: stack);
function empty(s: stack): boolean;
procedure push(var s: stack; x: elt);
function top(s: stack): elt;
procedure pop(var s: stack);
Any program that uses this data type is restricted to calling these five procedures for creating and",algorithms and data structures.pdf
"procedure pop(var s: stack);
Any program that uses this data type is restricted to calling these five procedures for creating and  
operating on stacks; it is not allowed to use information about the underlying implementation. The  
procedures may only be called within the constraints of the specification; for example, 'top' and  
'pop' may be called only if the stack is not empty:
if  not empty(s)  then  pop(s);
The specification above assumes that a stack can grow without a bound; it defines an abstract data type called  
unbounded stack. However, any implementation imposes some bound on the size ( depth) of a stack: the size of the  
underlying array in an array imple → d reflect such →  limitations. The following  fixed-length stack  describes an  
implementation as an array of fixed size m, which limits the maximal stack depth.
Fixed-length stack
create:→  S
empty: S →  {true, false}
full: S →  {true, false}
push: {s ∈  S: not full(s)} × X →  S
top: S – {s0} →  X",algorithms and data structures.pdf
"Fixed-length stack
create:→  S
empty: S →  {true, false}
full: S →  {true, false}
push: {s ∈  S: not full(s)} × X →  S
top: S – {s0} →  X
pop: S – {s0} →  S
To specify the behavior of the function 'full' we need an internal function
depth: S →  {0, 1, 2, … , m}
that measures the stack depth, that is, the number of elements currently in the stack. The function 'depth' interacts  
with the other functions in the following axioms, which specify the stack semantics:
∀ s ∈  S, ∀ x ∈  X:
create = s0
empty(s) = true
not full(s) ⇒  empty(push(s, x)) = false
depth(s0) = 0
187",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
not empty(s) ⇒  depth(pop(s)) = depth(s) – 1
not full(s) ⇒  depth(push(s, x)) = depth(s) + 1
full(s) = (depth(s) = m)
not full(s) ⇒
top(push(s, x)) = x
pop(push(s, x)) = s
Variable-length stack
A stack implemented as a list may overflow at unpredictable moments depending on the contents of the entire  
memory, not just of the stack. We specify this behavior by postulating a function 'space-available'. It has no domain  
and thus acts as an oracle that chooses its value independently of the state of the stack (if we gave 'space-available' a  
domain, this would have to be the set of states of the entire memory).
create: →  S
empty: S →  {true, false}
space-available: → {true, false}
push: S × X →  S
top: S – {s0} →  X
pop: S – {s0} →  S
∀ s ∈  S, ∀ x ∈  X:
create = s0
empty(s0) = true
space-available ⇒
empty(push(s, x)) = false
top(push(s, x)) = x
pop(push(s, x)) = s
Implementation",algorithms and data structures.pdf
"∀ s ∈  S, ∀ x ∈  X:
create = s0
empty(s0) = true
space-available ⇒
empty(push(s, x)) = false
top(push(s, x)) = x
pop(push(s, x)) = s
Implementation
We have seen that abstract data types cannot capture our intuitive, vague concept of a stack in one single model.  
The rigor enforced by the formal definition makes us aware that there are different types of stacks with different  
behavior (quite apart from the issue of the domain type X, which specifies what type of elements are to be stored).  
This  clarity  is  an  advantage  whenever  we  attempt  to  process  abstract  data  types  automatically;  it  may  be  a 
disadvantage for human communication, because a rigorous definition may force us to (over)specify details.
The different types of stacks that we have introduced are directly related to different styles of implementation.  
The fixed-length stack, for example, describes the following implementation:
const  m = … ;  { maximum length of a stack }
type elt = … ;
stack =record",algorithms and data structures.pdf
"const  m = … ;  { maximum length of a stack }
type elt = … ;
stack =record
a: array[1 .. m] of elt;
d: 0 .. m;  { current depth of stack }
end;
procedure create(var s: stack);
begin  s.d := 0  end;
function empty(s: stack): boolean;
begin  return(s.d = 0)  end;
function full(s: stack): boolean;
begin  return(s.d = m)  end;
procedure push(var s: stack; x: elt);  { not to be called if the stack 
is full }
begin  s.d := s.d + 1;  s.a[s.d] := x  end;
Algorithms and Data Structures 188  A Global Text",algorithms and data structures.pdf
"19. Abstract data types
function top(s: stack): elt;  { not to be called if the stack is 
empty }
begin  return(s.a[s.d])  end;
procedure pop(var s: stack);  { not to be called if the stack is 
empty }
begin  s.d := s.d – 1  end;
Since the function 'depth' is not exported ( i.e. not made available to the user of this data type), it need not be  
provided as a procedure. Instead, we have implemented it as a variable d which also serves as a stack pointer.
Our implementation assumes that the user checks that the stack is not full before calling 'push', and that it is not  
empty before calling 'top' or 'pop'. We could, of course, write the procedures 'push', 'top', and 'pop' so as to ""protect  
themselves"" against illegal calls on a full or an empty stack simply by returning an error message to the calling  
program. This requires adding a further argument to each of these three procedures and leads to yet other types of",algorithms and data structures.pdf
"program. This requires adding a further argument to each of these three procedures and leads to yet other types of  
stacks which are formally different abstract data types from the ones we have discussed.
First-in-first-out queue
The  following  operations  (Exhibit  19.2)  are  defined  for  the  abstract  data  type  fifo  queue (first-in-first-out  
queue):
empty Return true if the queue is empty.
enqueue Insert a new element at the tail end of the queue.
front Return the front element of the queue.
dequeue Remove the front element.
Exhibit 19.2: Elements are inserted at the tail and removed from the head of the fifo queue. 
Let F be the set of queue states that can be obtained from the empty queue by performing finite sequences of  
'enqueue' and 'dequeue' operations. f 0 ∈  F denotes the empty queue. The following functions represent fifo queue  
operations:
create: →  F
empty: F →  {true, false}
enqueue: F × X →  F
front: F – {f0} →  X
dequeue: F – {f0} →  F",algorithms and data structures.pdf
"operations:
create: →  F
empty: F →  {true, false}
enqueue: F × X →  F
front: F – {f0} →  X
dequeue: F – {f0} →  F
The semantics of the fifo queue operations is specified by the 
following axioms:
∀ f ∈  F,∀ x ∈  X:
(1) create = f0
(2) empty(f0) = true
(3) empty(enqueue(f, x)) = false
(4) front(enqueue(f0, x)) = x
(5) not empty(f) ⇒  front(enqueue(f, x)) = front(f)
(6) dequeue(enqueue(f0, x)) = f0
(7) not empty(f) ⇒  dequeue(enqueue(f, x)) = enqueue(dequeue(f), x)
189",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Any f ∈  F is obtained from the empty fifo queue f 0 by performing a finite sequence of 'enqueue' and 'dequeue'  
operations. By axioms (6) and (7) this sequence can be reduced to a sequence consisting of 'enqueue' operations  
only which also transforms f0 into f.
Example
f = dequeue(enqueue(dequeue(enqueue(enqueue(f0, x), y)), z))
= dequeue(enqueue(enqueue(dequeue(enqueue(f0, x)), y), z))
= dequeue(enqueue(enqueue(f0, y), z))
= enqueue(dequeue(enqueue(f0, y)), z)
= enqueue(f0, z)
An implementation of a fifo queue may provide the following procedures:
procedure create(var f: fifoqueue);
function empty(f: fifoqueue): boolean;
procedure enqueue(var f: fifoqueue; x: elt);
function front(f: fifoqueue): elt;
procedure dequeue(var f: fifoqueue);
Priority queue
A priority queue orders the elements according to their value rather than their arrival time. Thus we assume that",algorithms and data structures.pdf
"Priority queue
A priority queue orders the elements according to their value rather than their arrival time. Thus we assume that 
a total order ≤ is defined on the domain X. In the following examples, X is the set of integers; a small integer means  
high priority. The following operations (Exhibit 19.3) are defined for the abstract data type priority queue:
- empty Return true if the queue is empty.
- insert Insert a new element into the queue.
- min Return the element of highest priority contained in the queue.
- delete Remove the element of highest priority from the queue.
Exhibit 19.3: An element's priority determines its position in a priority queue. 
Let P be the set of priority queue  states that can be obtained  from the empty queue  by performing finite  
sequences of 'insert' and 'delete' operations. The empty priority queue is denoted by p 0 ∈  P. The following functions 
represent priority queue operations:
create: →  P
empty: P →  {true, false}
insert: P × X →  P",algorithms and data structures.pdf
"represent priority queue operations:
create: →  P
empty: P →  {true, false}
insert: P × X →  P
min: P – {p0} →  X
delete: P – {p0} →  P
The semantics of the priority queue operations is specified by the following axioms. For x, y ∈  X, the function  
MIN(x, y) returns the smaller of the two values.
Algorithms and Data Structures 190  A Global Text",algorithms and data structures.pdf
"19. Abstract data types
∀ p ∈  P,∀ x ∈  X:
(1) create = p0
(2) empty(p0) = true
(3) empty(insert(p, x)) = false
(4) min(insert(p0, x)) = x
(5) not empty(p) ⇒  min(insert(p, x)) = MIN(x, min(p))
(6) delete(insert(p0, x)) = p0
(7) not empty(p)⇒
delete (insert(p,x))={ pifxminp
insertdeletep,xelse
Any p ∈  P is obtained from the empty queue p 0 by a finite sequence of 'insert' and 'delete' operations. By axioms  
(6) and (7) any such sequence can be reduced to a shorter one that also transforms p 0 into p and consists of 'insert'  
operations only.
Example
Assume that x < z, y < z.
p = delete(insert(delete(insert(insert(p0, x), z)), y))
= delete(insert(insert(delete(insert(p0, x)), z), y))
= delete(insert(insert(p0, z), y))
= insert(p0, z)
An implementation of a priority queue may provide the following procedures:
procedure create(var p: priorityqueue);
function empty(p: priorityqueue): boolean;
procedure insert(var p: priorityqueue; x: elt);
function min(p: priorityqueue): elt;",algorithms and data structures.pdf
"function empty(p: priorityqueue): boolean;
procedure insert(var p: priorityqueue; x: elt);
function min(p: priorityqueue): elt;
procedure delete(var p: priorityqueue);
Dictionary
Whereas stacks and fifo queues are designed to retrieve and process elements depending on their order of  
arrival, a dictionary (or table) is designed to process elements exclusively by their value (name). A priority queue is  
a hybrid: insertion is done according to value, as in a dictionary, and deletion according to position, as in a fifo  
queue.
The simplest type of dictionary supports the following operations:
- member Return true if a given element is contained in the 
dictionary.
- insert Insert a new element into the dictionary.
- delete Remove a given element from the dictionary.
Let D be the  set  of dictionary  states  that can be  obtained from the empty dictionary  by performing finite",algorithms and data structures.pdf
"Let D be the  set  of dictionary  states  that can be  obtained from the empty dictionary  by performing finite  
sequences of 'insert' and 'del ete' operations. d 0 ∈  D denotes the empty dictionary. Then the operations can be  
represented by functions as follows:
create: →  D
insert: D × X →  D
member: D × X →  {true, false}
delete: D × X →  D
The semantics of the dictionary operations is specified by the 
following axioms:
191",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
∀ d ∈  D,∀ x, y ∈  X:
(1) create = d0
(2) member(d0, x) = false
(3) member(insert(d, x), x) = true
(4) x ≠ y ⇒  member(insert(d, y), x) = member(d, x)
(5) delete(d0, x) = d0
(6) delete(insert(d, x), x) = delete(d, x)
(7) x ≠ y ⇒  delete(insert(d, x), y) = insert(delete(d, y), x)
Any d  ∈  D is obtained from the empty dictionary d 0 by a finite sequence of 'insert' and 'delete' operations. By  
axioms (6) and (7) any such sequence can be reduced to a shorter one that also transforms d 0 into d and consists of  
'insert' operations only.
Example
d = delete(insert(insert(insert(d0, x), y), z), y)
= insert(delete(insert(insert(d0, x), y), y), z)
= insert(delete(insert(d0, x), y), z)
= insert(insert(delete(d0, y), x), z)
= insert(insert(d0, x), z)
This  specification  allows  duplicates  to  be  inserted.  However,  axiom  (6)  guarantees  that  all  duplicates  are",algorithms and data structures.pdf
"= insert(insert(d0, x), z)
This  specification  allows  duplicates  to  be  inserted.  However,  axiom  (6)  guarantees  that  all  duplicates  are 
removed if a delete operation is performed. To prevent duplicates, the following axiom is added to the specification  
above:
(8) member(d, x) ⇒  insert(d, x) = d
In this case axiom (6) can be weakened to
(6') not member(d, x) ⇒  delete(insert(d, x), x) = d
An implementation of a dictionary may provide the following procedures:
procedure create(var d: dictionary);
function member(d: dictionary; x: elt): boolean;
procedure insert(var d: dictionary; x: elt);
procedure delete(var d: dictionary; x: elt);
In actual programming practice, a dictionary usually supports the additional operations 'find', 'predecessor', and  
'successor'. 'find' is similar to 'member' but in addition to a true/false answer, provides a pointer to the element",algorithms and data structures.pdf
"'successor'. 'find' is similar to 'member' but in addition to a true/false answer, provides a pointer to the element  
found. Both 'predecessor' and 'successor' take a pointer to an element e as an argument, and return a pointer to the  
element in  the  dictionary  that immediately precedes  or  follows  e,  according  to the  order  ≤.  Repeated  call  of 
'successor' thus processes the dictionary in sequential order.
Exercise: extending the abstract data type 'dictionary'
We have defined a dictionary as supporting the three operations 'member', 'insert' and 'delete'. But a dictionary,  
or table, usually supports additional operations based on a total ordering ≤ defined on its domain X. Let us add two  
operations that take an argument x ∈  X and deliver its two neighboring elements in the table:
succ(x)Return the successor of x in the table.
pred(x)Return the predecessor of x in the table.
Algorithms and Data Structures 192  A Global Text",algorithms and data structures.pdf
"19. Abstract data types
The successor of x is defined as the smallest of all the elements in the table which are larger than x, or as +∞ if  
none exists. The predecessor is defined symmetrically: the largest of all the elements in the table that are smaller  
than x, or –∞. Present a formal specification to describe the behavior of the table.
Solution
Let T be the set of states of the table , and t 0 a spec ial state that denotes the empty table. The functions and  
axioms are as follows:
member: T × X →  {true,false}
insert: T × X →  T
delete: T × X →  T
succ: T × X →  X ∪  {+∞}
pred: T × X →  X ∪  {–∞}
∀ t ∈  T,∀ x, y ∈  X:
 member(t0, x) = false
 member(insert(t, x), x) = true
 x ≠ y ⇒  member(insert(t, y), x) = member(t, x)
 delete(t0, x) = t0
 delete(insert(t, x), x) = delete(t, x)
 x ≠ y ⇒  delete(insert(t, x), y) = insert(delete(t, y), x)
–∞ < x < +∞
pred(t, x) < x < succ(t, x)
succ(t, x) ≠ +∞ ⇒  member(t, succ(t, x)) = true
pred(t, x) ≠ –∞ ⇒  member(t, pred(t, x)) = true",algorithms and data structures.pdf
"–∞ < x < +∞
pred(t, x) < x < succ(t, x)
succ(t, x) ≠ +∞ ⇒  member(t, succ(t, x)) = true
pred(t, x) ≠ –∞ ⇒  member(t, pred(t, x)) = true
x < y, member(t, y), y ≠ succ(t, x) ⇒  succ(t, x) < y
x > y, member(t, y), y ≠ pred(t, x) ⇒  y < pred(t, x)
Exercise: the abstract data type 'string'
We define the following operations for the abstract data type string:
- empty Return true if the string is empty.
- append Append a new element to the tail of the string.
- head Return the head element of the string.
- tail Remove the head element of the given string.
- length Return the length of the string.
- find Return the index of the first occurrence of a value within the 
string.
Let X = {a, b, … , z}, and S be the set of string states that can be obtained from the empty string by performing a  
finite number of 'append' and 'tail' operations. s 0 ∈  S denotes the empty string. The operations can be represented  
by functions as follows:
empty: S →  {true, false}
append: S × X →  S",algorithms and data structures.pdf
"by functions as follows:
empty: S →  {true, false}
append: S × X →  S
head: S – {s0} →  X
tail: S – {s0} →  S
length: S →  {0, 1, 2, … }
find: S × X →  {0, 1, 2, … }
Examples:
empty('abc') = false;  append('abc', 'd') = 'abcd';  head('abcd') = 
'a';
193",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
tail('abcd') = 'bcd';  length('abcd') = 4;  find('abcd', 'b') = 2.
(a) Give the axioms that specify the semantics of the abstract data type 'string'.
(b) The function hchop: S × X →  S returns the substring of a string s beginning with the first occurrence of a  
given value. Similarly, tchop: S × X →  S returns the substring of s beginning with head(s) and ending with  
the  last  occurrence  of  a  given  value.  Specify  the  behavior  of  these  operations  by  additional  axioms. 
Examples:
hchop('abcdabc','c')='cdabc'
tchop('abcdabc', 'b') = 'abcdab'
(c) The function cat: S  × S  →  S returns the concatenation of two sequences. Specify the behavior of 'cat' by  
additional axioms. Example:
cat('abcd', 'efg') = 'abcdefg'
(d) The function reverse: S →  S returns the given sequence in reverse order. Specify the behavior of reverse by  
additional axioms. Example:
reverse('abcd') = 'dcba'
Solution",algorithms and data structures.pdf
"additional axioms. Example:
reverse('abcd') = 'dcba'
Solution
(a) Axioms for the six 'string' operations:
∀ s ∈  S, ∀ x, y ∈  X:
empty(s0) = true
empty(append(s, x)) = false
head(append(s0, x)) = x
not empty(s) ⇒  head(s) = head(append(s, x))
tail(append(s0, x)) = s0
not empty(s) ⇒  tail(append(s, x)) = append(tail(s), x)
length(s0) = 0
length(append(s, x)) = length(s) + 1
find(s0, x) = 0
x ≠ y, find(s, x) = 0 ⇒  find(append(s, y), x) = 0
find(s, x) = 0 ⇒  find(append(s, x), x) = length(s) + 1
find(s, x) = d > 0 ⇒  find(append(s, y), x) = d
(b) Axioms for 'hchop' and 'tchop':
∀ s ∈  S, ∀ x, y ∈  X:
hchop(s0, x) = s0
not empty(s), head(s) = x ⇒  hchop(s, x) = s
not empty(s), head(s) ≠ x ⇒  hchop(s, x) = hchop(tail(s), x)
tchop(s0, x) = s0
tchop(append(s, x), x) = append(s, x)
x ≠ y ⇒  tchop(append(s, y), x) = tchop(s, x)
(c) Axioms for 'cat':
∀ s, s' ∈  S:
cat(s, s0) = s
not empty(s') ⇒  cat(s, s') = cat(append(s, head(s')), tail(s'))
(d) Axioms for 'reverse':
 ∀ s ∈  S:",algorithms and data structures.pdf
"∀ s, s' ∈  S:
cat(s, s0) = s
not empty(s') ⇒  cat(s, s') = cat(append(s, head(s')), tail(s'))
(d) Axioms for 'reverse':
 ∀ s ∈  S:
Algorithms and Data Structures 194  A Global Text",algorithms and data structures.pdf
"19. Abstract data types
reverse(s0) = s0
s ≠ s0 ⇒  reverse(s) = append(reverse(tail(s)), head(s))
Exercises
1. Implement two stacks i ν on ε array a[1 ..  m] in  such a  way that neit her stack overflows unless the total  
number of elements in both stacks together is m. The operations 'push', 'top', and 'pop' should run in O(1)  
time.
2. A double-ended queue (deque) can grow and shrink at both ends, left and right, using the procedures  
'enqueue-left',  'dequeue-left',  'enqueue-right',  and  'dequeue-right'.  Present  a  formal  specification  to 
describe the behavior of the abstract data type deque.
3. Extend the abstract data type priority queue by the operation next(x), which returns the element in the  
priority queue having the next lower priority than x.
195",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
20. Implicit data structures
Learning objectives:
• implicit data structures describe relationships among data elements implicitly by formulas and declarations
• array storage
• band matrices
• sparse matrices
• Buffers eliminate temporary speed differences among interacting producer and consumer processes.
• fifo queue implemented as a circular buffer
• priority queue implemented as a heap
• heapsort
What is an implicit data structure?
An  important  aspect  of  the  art  of  data  structure  design  is  the  efficient  representation  of  the  structural 
relationships among the data elements to be stored. Data is usually modeled as a graph, with nodes corresponding  
to data elements and links (directed arcs, or bidirectional edges) corresponding to relationships. Relationships  
often serve a double purpose. Primarily, they define the semantics of the data and thus allow programs to interpret",algorithms and data structures.pdf
"often serve a double purpose. Primarily, they define the semantics of the data and thus allow programs to interpret  
the data correctly. This aspect of relationships is highlighted in the database field: for example, in the entity-
relationship model. Secondarily, relationships provide a means of accessing data, by starting at some element and  
following  an  access  path that leads  to other  elements  of  interest.  In  studying  data  structures  we  are  mainly 
concerned with the use of relationships for access to data.
When the structure of the data is irregular, or when the structure is highly dynamic (extensively modified at run  
time),  there  is  no  practical  alternative  to  representing  the  relationships  explicitly.  This  is  the  domain  of  list 
structures, presented in the chapter on “List structures”. When the structure of the data is static and obeys a regular",algorithms and data structures.pdf
"structures, presented in the chapter on “List structures”. When the structure of the data is static and obeys a regular  
pattern, on the other hand, there are alternatives that compress the structural information. We can often replace  
many explicit links by a few formulas that tell us where to find the ""neighboring"" elements. When this approach  
works, it saves memory space and often leads to faster programs.
We use the term implicit to denote data structures in which the relationships among data elements are given  
implicitly by formulas and declarations in the program; no additional space is needed for these relationships in the  
data storage. The best known example is the array. If one looks at the area in which an array is stored, it is  
impossible to derive, from its contents, any relationships among the elements without the information that the  
elements belong to an array of a given type.",algorithms and data structures.pdf
"elements belong to an array of a given type.
Data structures always go hand in hand with the corresponding procedures for accessing and operating on the  
data. This is particularly true for implicit data structures: They simply do not exist independent of their accessing  
procedures. Separated from its code, an implicit data structure represents at best an unordered set of data. With the 
right code, it exhibits a rich structure, as is beautifully illustrated by the heap at the end of this chapter.
Algorithms and Data Structures 196  A Global Text",algorithms and data structures.pdf
"20. Implicit data structures
Array storage
A two-dimensional array declared as
var  A: array[1 .. m, 1 .. n] of elt;
is usually written in a rectangular shape:
A[1, 1] A[1, 2] … A[1, n]
A[2, 1] A[2, 2] … A[2, n]
… … … …
A[m, 1] A[m, 2] … A[m, n]
But it is stored in a linearly addressed memory, typically row by row (as shown below) or column by column (as  
in Fortran) in consecutive storage cells, starting at base address b. If an element fits into one cell, we have
address
A[1, 1] b
A[1, 2] b + 1
… …
A[1, n] b + n – 1
A[2, 1] b + n
A[2, 2] b + n + 1
… …
A[2, n] b + 2 · n – 1
… …
A[m, n] b + m · n – 1
If an element of type 'elt' occupies c storage cells, the address α(i, j) of A[i, j] is
This linear formula generalizes to k-dimensional arrays declared as
var  A: array[1 .. m1, 1 .. m2, … , 1 .. mk] of elt;
The address α(i1, i2, … , ik) of element A[i1, i2, … , ik] is
197",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The point is that access to an element A[i, j, …] invokes evaluation of a (linear) formula α(i, j, …) that tells us  
where to find this element. A high-level programming language hides most of the details of address computation,  
except when we wish to take advantage of any special structure our matrices may have. The following types of  
sparse matrices occur frequently in numerical linear algebra.
Band matrices. An n × n matrix M is called a band matrix of width 2 · b + 1 (b = 0, 1, …) if Mi,j = 0 for all i and 
j with |i – j| > b. In other words, all nonzero elements are located on the main diagonal and in b adjacent minor  
diagonals on both sides of the main diagonal. If n is large and b is small, much space is saved by storing M in a two-
dimensional array A with n · (2 · b + 1) cells rather than in an array with n2 cells:
type  bandm = array[1 .. n, –b .. b] of elt;
var  A: bandm;",algorithms and data structures.pdf
"dimensional array A with n · (2 · b + 1) cells rather than in an array with n2 cells:
type  bandm = array[1 .. n, –b .. b] of elt;
var  A: bandm;
Each row A[i, ·] stores the nonzero elements of the corresponding row of M, namely the diagonal element M i,i, 
the b elements to the left of the diagonal
Mi,i–b, Mi,i–b+1, … , Mi,i–1
and the b elements to the right of the diagonal
Mi,i+1, Mi,i+2, … , Mi,i+b.
The first and the last b rows of A contain empty cells corresponding to the triangles that stick out from M in  
Exhibit 20.1. The elements of M are stored in array A such that A[i, j] contains Mi,i+j (1 ≤ i ≤ n, –b ≤ j ≤ b). A total of  
b · (b + 1) cells in the upper left and lower right of A remain unused. It is not worth saving an additional b · (b + 1)  
cells by packing the band matrix M into an array of minimal size, as the mapping becomes irregular and the  
formula for calculating the indices of Mi,j becomes much more complicated.",algorithms and data structures.pdf
"formula for calculating the indices of Mi,j becomes much more complicated.
Exhibit 20.1: Extending the diagonals with dummy elements gives the 
band matrix the shape of a rectangular array.
Algorithms and Data Structures 198  A Global Text",algorithms and data structures.pdf
"20. Implicit data structures
Exercise: band matrices
(a) Write a procedure add(p, q: bandm; var r: bandm); 
which adds two band matrices stored in p and q and stores the result in r.
(b) Write a procedure bmv(p: bandm; v: … ; var w: … );
which multiplies a band matrix stored in p with a vector v of length n and stores the result in w.
Solution
(a) procedure add(p, q: bandm; var r: bandm);
var  i: 1 .. n;  j: –b .. b;
begin
for  i := 1  to  n  do
for  j := –b  to  b  do
r[i, j] := p[i, j] + q[i, j]
end;
(b)  type  vector = array[1 .. n] of real;
procedure bmv(p: bandm; v: vector; var w: vector);
var  i: 1 .. n;  j: –b .. b;
begin
for  i := 1  to  n  do  begin
w[i] := 0.0;
for  j := –b  to  b  do
if  (i + j ≥ 1) and (i + j ≤ n)  then  w[i] := w[i] + p[i, j] · 
v[i + j]
end
end;
Sparse matrices. A matrix is called sparse if it consists mostly of zeros. We have seen that sparse matrices of",algorithms and data structures.pdf
"v[i + j]
end
end;
Sparse matrices. A matrix is called sparse if it consists mostly of zeros. We have seen that sparse matrices of  
regular shape can be compressed efficiently using address computation. Irregularly shaped sparse matrices, on the  
other hand, do not yield gracefully to compression into a smaller array in such a way that access can be based on  
address computation. Instead, the nonzero elements may be stored in an unstructured set of records, where each  
record contains the pair ((i, j), A[i, j]) consisting of an index tuple (i, j) and the value A[i, j]. Any element that is  
absent from this set is assumed to be zero. As the position of a data element is stored explicitly as an index pair (i,  
j), this representation  is not an implicit data  structure. As a consequence, access  to a  random  element of an  
irregularly shaped sparse matrix typically requires searching for it, and thus is likely to be slower than the direct",algorithms and data structures.pdf
"irregularly shaped sparse matrix typically requires searching for it, and thus is likely to be slower than the direct  
access to an element of a matrix of regular shape stored in an implicit data structure. 
Exercise: triangular matrices
Let A and B be lower-triangular n × n-matrices; that is, all elements above the diagonal are zero: A i,j = Bi,j = 0 for 
i < j.
(a) Prove that the inverse (if it exists) and the matrix product of lower-triangular matrices are again  
lower-triangular.
(b) Devise a scheme for storing two lower-triangular matrices A and B in one array C of minimal size.  
Write a Pascal declaration for C and draw a picture of its contents.
(c) Write two functions
function A(i, j: 1 .. n): real;
 function B(i, j: 1 .. n): real;
199",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
(d) that access C and return the corresponding matrix elements.
(e) Write a procedure that computes A := A · B in place: The entries of A in C are replaced by the entries  
of the product A · B. You may use a (small) constant number of additional variables, independent of  
the size of A and B.
(f) Same as (d), but using A := A–1 · B.
Solution
(a) The inverse of an n × n-matrix exists iff the determinant of the matrix is non zero. Let A be a lower-
triangular matrix for which the inverse matrix B exists, that is, 
and
Let 1 ≤ j ≤ n. Then
and therefore B is a lower-triangular matrix.
Let A and B be lower-triangular, C := A · B:
If i < j, this sum is empty and therefore Ci,j = 0 (i. e. C is lower-triangular).
(b) A and B can be stored in an array C of size n · (n + 1) as follows (Exhibit 20.2):
const  n = … ;
var  C: array [0 .. n, 1 .. n] of real;
Algorithms and Data Structures 200  A Global Text",algorithms and data structures.pdf
"20. Implicit data structures
Exhibit 20.2: A staircase separates two triangular matrices
(c) stored in a rectangular array. (graphic does not match)
function A(i, j: 1 .. n): real
begin  if  i < j  then  return(0.0)  else  return(C[i, j])  end;
function B(i, j: 1 .. n): real;
begin  if  i < j  then  return(0.0)  else  return(C[n – i, n + 1 – 
j])  end;
(d) Because the new elements of the result matrix C overwrite the old elements of A, it is important to compute  
them in the right order. Specifically, within every row i of C, elements C i,j must be computed from left to  
right, that is, in increasing order of j.
procedure mult;
var  i, j, k: integer;  x: real;
begin
for  i := 1  to  n  do
for  j := 1  to  i  do  begin
x := 0.0;
for  k := j  to  i  do  x := x + A(i, k) · B(k, j);
C[i, j] := x
end
end;
(e) procedure invertA;
var  i, j, k: integer;  x: real;
begin
for  i := 1  to  n  do  begin
for  j := 1  to  i – 1  do  begin
x := 0.0;",algorithms and data structures.pdf
"end
end;
(e) procedure invertA;
var  i, j, k: integer;  x: real;
begin
for  i := 1  to  n  do  begin
for  j := 1  to  i – 1  do  begin
x := 0.0;
for  k := j  to i – 1  do  x := x – C[i, k] · C[k, j];
201",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
C[i, j] := x / C[i, i]
end;
C[i, i] := 1.0 / C[i, i]
end
end;
procedure AinvertedmultB;
begin  invertA;  mult  end; 
Implementation of the fixed-length fifo queue as a circular buffer
A fifo queue is needed in situations where two processes interact in the following way. A process called producer 
generates data for a process called consumer. The processes typically work in bursts: The producer may generate a  
lot of data while the consumer is busy with something else; thus the data has to be saved temporarily in a buffer,  
from which the consumer takes it as needed. A keyboard driver and an editor are an example of this producer-
consumer interaction. The keyboard driver transfers characters generated by key presses into the buffer, and the  
editor reads them from the buffer and interprets them (e.g. as control characters or as text to be inserted). It is",algorithms and data structures.pdf
"editor reads them from the buffer and interprets them (e.g. as control characters or as text to be inserted). It is  
worth remembering, though, that a buffer helps only if two processes work at about the same speed over the long  
run. If the producer is always faster, any buffer will overflow; if the consumer is always faster, no buffer is needed. A 
buffer can equalize only temporary differences in speeds.
With some knowledge about the statistical behavior of producer and consumer one can usually compute a buffer  
size that is sufficient to absorb producer bursts with high probability, and allocate the buffer statically in an array of  
fixed size. Among statically allocated buffers, a circular buffer is the natural implementation of a fifo queue.
A circular buffer is an array B, considered as a ring in which the first cell B[0] is the successor of the last cell B[m",algorithms and data structures.pdf
"A circular buffer is an array B, considered as a ring in which the first cell B[0] is the successor of the last cell B[m 
– 1], as shown in Exhibit 20.3. The elements are stored in the buffer in consecutive cells between the two pointers  
'in' and 'out': 'in' points to the empty cell into which the next element is to be inserted; 'out' points to the cell  
containing the next element to be removed. A new element is inserted by storing it in B[in] and advancing 'in' to the  
next cell. The element in B[out] is removed by advancing 'out' to the next cell.
Algorithms and Data Structures 202  A Global Text",algorithms and data structures.pdf
"20. Implicit data structures
Exhibit 20.3: Insertions move the pointer 'in', deletions the pointer 'out' counterclockwise around the array.
Notice that the pointers 'in' and 'out' meet both when the buffer gets full and when it gets empty. Clearly, we  
must be able to distinguish a full buffer from an empty one, so as to avoid insertion into the former and removal  
from the latter. At first sight it appears that the pointers 'in' and 'out' are insufficient to determine whether a  
circular buffer is full or empty. Thus the following implementation uses an additional variable n, which counts how  
many elements are in the buffer.
const  m = … ;  { length of buffer }
type  addr = 0 .. m – 1;  { index range }
var B: array[addr] of elt;  {storage}
in, out: addr;  { access to buffer }
n: 0 .. m;  { number of elements currently in buffer }
procedure create;
begin  in := 0;  out := 0;  n := 0  end;
function empty(): boolean;
begin  return(n = 0)  end;
function full(): boolean;",algorithms and data structures.pdf
"procedure create;
begin  in := 0;  out := 0;  n := 0  end;
function empty(): boolean;
begin  return(n = 0)  end;
function full(): boolean;
begin  return(n = m)  end;
procedure enqueue(x: elt);
{ not to be called if the queue is full }
begin  B[in] := x;  in := (in + 1) mod m;  n := n + 1  end;
function front(): elt;
{ not to be called if the queue is empty }
begin  return(B[out])  end;
procedure dequeue;
{ not to be called if the queue is empty }
begin  out := (out + 1) mod m;  n := n – 1  end;
203",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The producer uses only 'enqueue' and 'full', as it deletes no elements from the circular buffer. The consumer uses 
only 'front', 'dequeue', and 'empty', as it inserts no elements.
The state of the circular buffer is described by its contents and the values of 'in', 'out', and n. Since 'in' is changed  
only within 'enqueue', only the producer needs write-access to 'in'. Since 'out' is changed only by 'dequeue', only the  
consumer needs write-access to 'out'. The variable n, however, is changed by both processes and thus is a shared 
variable to which both processes have write-access (Exhibit 20.4 (a)).
Exhibit 20.4:
(a) Producer and consumer both have write-access to shared variable n.
(b) The producer has read/write-access to 'in' and read-only-access to 'out',
the consumer has read/write-access to 'out' and read-only-access to 'in'.",algorithms and data structures.pdf
"the consumer has read/write-access to 'out' and read-only-access to 'in'.
In a concurrent programming environment where several processes execute independently, access to shared  
variables must be  synchronized. Synchronization  is overhead  to be  avoided if possible. The shared  variable  n  
becomes superfluous (Exhibit 20.4 (b)) if we use the time-honored trick of leaving at least one cell free as a sentinel. 
This ensures that 'empty' and 'full', when expressed in terms of 'in' and 'out', can be distinguished. Specifically, we  
define  'empty'  as  in  =  out,  and  'full'  as  (in  +  1)  mod  m  =  out.  This  leads  to  an  elegant  and  more  efficient 
implementation of the fixed-length fifo queue by a circular buffer:
const  m = … ;  { length of buffer }
type addr = 0 .. m – 1;  { index range }
fifoqueue = record
B: array[addr] of elt;  { storage }
in, out: addr  { access to buffer }
end;
procedure create(var f: fifoqueue);
begin  f.in := 0;  f.out := 0  end;",algorithms and data structures.pdf
"B: array[addr] of elt;  { storage }
in, out: addr  { access to buffer }
end;
procedure create(var f: fifoqueue);
begin  f.in := 0;  f.out := 0  end;
function empty(f: fifoqueue): boolean;
begin  return(f.in = f.out)  end;
function full(f: fifoqueue): boolean;
begin  return((f.in + 1) mod m = f.out)  end;
procedure enqueue(var f: fifoqueue; x: elt);
{ not to be called if the queue is full }
begin  f.B[f.in] := x;  f.in := ( f.in + 1) mod m  end;
Algorithms and Data Structures 204  A Global Text",algorithms and data structures.pdf
"20. Implicit data structures
function front(f: fifoqueue): elt;
{ not to be called if the queue is empty }
begin  return(f.B[f.out])  end;
procedure dequeue(f: fifoqueue);
{ not to be called if the queue is empty }
begin  f.out := (f.out + 1) mod m  end;
Implementation of the fixed-length priority queue as a heap
A fixed-length priority queue  can be realized by a circular buffer, with elements stored in the cells between 'in'  
and 'out', and ordered according to their priority such that 'out' points to the element with highest priority ( Exhibit 
20.5). In this implementation, the operations 'min' and 'delete' have time complexity O(1), since 'out' points directly  
to the element with the highest priority. But insertion requires finding the correct cell corresponding to the priority  
of the element to be inserted, and shifting other elements in the buffer to make space. Binary search could achieve  
the former task in time O(log n), but the latter requires time O(n).",algorithms and data structures.pdf
"the former task in time O(log n), but the latter requires time O(n).
Exhibit 20.5: Implementing a fixed-length priority queue by a circular buffer.
Shifting elements to make space for a new element costs O(n) time.
Implementing a priority queue as a linear list, with elements ordered according to their priority, does not speed  
up insertion: Finding the correct position of insertion still requires time O(n) (Exhibit 20.6).
Exhibit 20.6: Implementing a fixed-length priority queue by a linear list. Finding the correct 
position for a new element costs O(n) time.
The heap is an elegant and efficient data structure for implementing a priority queue. It allows the operation  
'min' to be performed in time O(1) and allows both 'insert' and 'delete' to be performed in worst-case time O(log n).  
A heap is a binary tree that:
• obeys a structural property
• obeys an order property
• is embedded in an array in a certain way",algorithms and data structures.pdf
"A heap is a binary tree that:
• obeys a structural property
• obeys an order property
• is embedded in an array in a certain way
Structure:  The binary tree is as balanced as possible; all leaves are at two adjacent levels, and the nodes at the  
bottom level are located as far to the left as possible (Exhibit 20.7).
205",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 20.7: A heap has the structure of an almost complete binary tree.  
Order:  The element assigned to any node is ≤ the elements assigned to any children this node may have  
(Exhibit 20.8).
Exhibit 20.8: The order property implies that the smallest element is stored at the root.
The order property implies that the smallest element (the one with top priority) is stored in the root. The 'min'  
operation returns its value in time O(1), but the most obvious way to delete this element leaves a hole, which takes  
time to fill. How can the tree be reorganized so as to retain the structural and the order property? The structural  
condition requires the removal of the rightmost node on the lowest level. The element stored there –13 in our  
example–is used (temporarily) to fill the vacuum in the root. The root may now violate the order condition, but the",algorithms and data structures.pdf
"example–is used (temporarily) to fill the vacuum in the root. The root may now violate the order condition, but the  
latter can be restored by sifting 13 down the tree according to its weight ( Exhibit 20.9). If the order condition is  
violated at any node, the element in this node is exchanged with the smaller of the elements stored in its children;  
in our example, 13 is exchanged with 2. This sift-down process continues until the element finds its proper level, at  
the latest when it lands in a leaf.
Algorithms and Data Structures 206  A Global Text
1
2
39
19 10 8 4 13
6
5
7",algorithms and data structures.pdf
"20. Implicit data structures
Exhibit 20.9: Rebuilding the order property of the tree in Exhibit 20.8 after 1 has been 
removed and 13 has been moved to the root.
Insertion is handled analogously. The structural condition requires that a new node is created on the bottom  
level at the leftmost empty slot. The new element - 0 in our example - is temporarily stored in this node ( Exhibit 
20.10). If the parent node now violates the order condition, we restore it by floating the new element upward  
according to its weight. If the new element is smaller than the one stored in its parent node, these two elements - in  
our example 0 and 6 - are exchanged. This sift-up process continues until the element finds its proper level, at the  
latest when it surfaces at the root.
Exhibit 20.10: Rebuilding the order property of the tree in Exhibit 20.8 after 0 has 
been inserted in a new rightmost node on the lowest level.",algorithms and data structures.pdf
"Exhibit 20.10: Rebuilding the order property of the tree in Exhibit 20.8 after 0 has 
been inserted in a new rightmost node on the lowest level.
The number of steps executed during the sift-up process and the sift-down process is at most equal to the height  
of the tree. The structural condition implies that this height is [log2 n]. Thus both 'insert' and 'delete' in a heap work 
in time O(log n).
207",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
A binary tree can be implemented in many different ways, but the special class of trees that meets the structural  
condition stated above has a particularly efficient array implementation. A heap is a binary tree that satisfies the  
structural and the order condition and is embedded in a linear array in such a way that the children of a node with  
index i have indices 2 ·   i and 2 ·   i + 1 ( Exhibit 20.11). Thus the parent of a node with index j has index j div 2. Any  
subtree of a heap is also a heap, although it may not be stored contiguously. The order property for the heap implies  
that the elements stored at indices 2 ·   i and 2 ·   i + 1 are ≥ the element stored at index i. This order is called the heap 
order.
Exhibit 20.11: Embedding the tree of Exhibit 20.8 in a linear array.
The procedure 'restore' is a useful tool for managing a heap. It creates a heap out of a binary tree embedded in a",algorithms and data structures.pdf
"The procedure 'restore' is a useful tool for managing a heap. It creates a heap out of a binary tree embedded in a  
linear array h that satisfies the structural condition, provided that the two subtrees of the root node are already  
heaps. Procedure 'restore' is applied to subtrees of the entire heap whose nodes are stored between the indices L  
and R and whose tree structure is defined by the formulas 2 · i and 2 · i + 1.
const  m = … ;  { length of heap }
type  addr = 1 .. m;
var  h: array[addr] of elt;
procedure restore(L, R: addr);
var  i, j: addr;
begin
i := L;
while  i ≤ (R div 2)  do  begin
if  (2 · i < R) cand (h[2 · i + 1] < h[2 · i])  then  j := 2 · i + 
1  else  j := 2 · i;
if  h[j] < h[i]  then  { h[i] :=: h[j];  i := j }  else  i := R
end
end;
Since 'restore' operates along a single path from the root to a leaf in a tree with at most R – L nodes, it works in  
time O(log (R – L)).
Creating a heap",algorithms and data structures.pdf
"time O(log (R – L)).
Creating a heap
An array h can be turned into a heap as follows: for  i := n div 2  down to  1  do  restore(i, n);
Algorithms and Data Structures 208  A Global Text",algorithms and data structures.pdf
"20. Implicit data structures
This is more efficient than repeated insertion of a single element into an existing heap. Since the for loop is  
executed n div 2 times, and n – i ≤ n, the time complexity for creating a heap with n elements is O(n · log n). A more 
careful analysis shows that the time complexity for creating a heap is O(n).
Heap implementation of the fixed-length priority queue
const  m = … ;  { maximum length of heap }
type addr = 1 .. m;
priorityqueue = record
h: array[addr] of elt;  { heap storage }
n: 0 .. m  { current number of elements }
end;
procedure restore(var h: array[addr] of elt; L, R: addr);
begin  …  end;
procedure create(var p: priorityqueue);
begin  p.n := 0  end;
function empty(p: priorityqueue): boolean;
begin  return(p.n = 0)  end;
function full(p: priorityqueue): boolean;
begin  return(p.n = m)  end;
procedure insert(var p: priorityqueue; x: elt);
{ not to be called if the queue is full }
var  i: 1 .. m;
begin",algorithms and data structures.pdf
"begin  return(p.n = m)  end;
procedure insert(var p: priorityqueue; x: elt);
{ not to be called if the queue is full }
var  i: 1 .. m;
begin
p.n := p.n + 1;  p.h[p.n] := x;  i := p.n;
while  (i > 1) cand (p.h[i] < p.h[i div 2])  do
{ p.h[i] :=: p.h[i div 2];  i := i div 2 }
end;
function min(p: priorityqueue): elt;
{ not to be called if the queue is empty }
begin  return(p.h[1])  end;
procedure delete(var p: priorityqueue);
{ not to be called if the queue is empty }
begin  p.h[1] := p.h[p.n];  p.n := p.n – 1;  restore(p.h, 1, p.n) 
end;
Heapsort 
The heap is the core of an elegant O(n · log n) sorting algorithm. The following procedure 'heapsort' sorts n  
elements stored in the array h into decreasing order.
procedure heapsort(n: addr);  { sort elements stored in h[1 .. n] }
var  i: addr;
begin  { heap creation phase: the heap is built up }
for  i := n div 2  downto  1  do  restore(i, n);
{ shift-up phase: elements are extracted from heap in increasing 
order }",algorithms and data structures.pdf
"for  i := n div 2  downto  1  do  restore(i, n);
{ shift-up phase: elements are extracted from heap in increasing 
order }
for  i := n  downto  2  do  { h[i] :=: h[1];  restore(1, i – 1) }
end;
Each of the for loops is executed less than n times, and the time complexity of restore is O(log n). Thus heapsort  
always works in time O(n · log n).
209",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exercises and programming projects
 1. Block-diagonal matrices are composed of smaller matrices that line up along the diagonal and have 0  
elements everywhere else, as shown in Exhibit 20.12. Show how to store an arbitrary block-diagonal matrix  
in a minimal storage area, and write down the corresponding address computation formulas.
Exhibit 20.12: Structure of a block-diagonal matrix. 
 2. Let A be an antisymmetric n × n-matrix (i. e., all elements of the matrix satisfy Aij = –Aji).
(a) What values do the diagonal elements Aii of the matrix have?
(b) How can A be stored in a linear array c of minimal size? What is the size of c?
(c) Write a
function A(i, j: 1 .. n): real;
which returns the value of the corresponding matrix element.
 3. Show that the product of two n × n matrices of width 2 · b + 1 (b = 0, 1, …) is again a band matrix. What is",algorithms and data structures.pdf
"3. Show that the product of two n × n matrices of width 2 · b + 1 (b = 0, 1, …) is again a band matrix. What is  
the width of the product matrix? Write a procedure that computes the product of two band matrices both  
having the same width and stores the result as a band matrix of minimal width.
 4. Implement a double-ended queue (deque) by a circular buffer.
 5. What are the minimum and maximum numbers of elements in a heap of height h?
 6. Determine the time complexities of the following operations performed on a heap storing n elements.  (a) 
Searching any element. (b) Searching the largest element (i.e. the element with lowest priority).
 7. Implement heapsort and animate the sorting process, for example as shown in the snapshots in “Algorithm  
animation”. Compare the number of comparisons and exchange operations needed by heapsort and other  
sorting algorithms (e.g. quicksort) for different input configurations.",algorithms and data structures.pdf
"sorting algorithms (e.g. quicksort) for different input configurations.
 8. What is the running time of heapsort on an array h[1 .. n] that is already sorted in increasing order? What  
about decreasing order?
 9. In a k-ary heap, nodes have k children instead of 2 children.
(a) How would you represent a k-ary heap in an array?
(b) What is the height of a k-ary heap in terms of the number of elements n and k?
(c) Implement a priority queue by a k-ary heap. What are the time complexities of the operations 'insert'  
and 'delete' in terms of n and k?
Algorithms and Data Structures 210  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
21. List structures
Learning objectives:
• static vs dynamic data structures
• linear, circular and two-way lists
• fifo queue implemented as a linear list
• breadth-first and depth-first tree traversal
• traversing a binary tree without any auxiliary memory: triple tree traversal algorithm
• dictionary implemented as a binary search tree
• Balanced trees guarantee that dictionary operations can be performed in logarithmic time
• height-balanced trees
• multiway trees
Lists, memory management, pointer variables 
The spectrum of data structures ranges from static objects, such as a table of constants, to dynamic structures,  
such as lists. A list is designed so that not only the data values stored in it, but its size and shape can change at run  
time, due to insertions, deletions, or rearrangement of data elements. Most of the data structures discussed so far",algorithms and data structures.pdf
"time, due to insertions, deletions, or rearrangement of data elements. Most of the data structures discussed so far  
can change their size and shape to a limited extent. A circular buffer, for example, supports insertion at one end and  
deletion at the other, and  can  grow to a predeclared  maximal size. A heap supports deletion at one end  and  
insertion anywhere into an array. In a list, any local change can be done with an effort that is independent of the  
size of the list - provided that we know the memory locations of the data elements involved. The key to meeting this  
requirement is the idea of abandoning memory allocation in large contiguous chunks, and instead allocating it  
dynamically in the smallest chunk that will hold a given object. Because data elements are stored randomly in  
memory, not contiguously, an insertion or deletion into a list does not propagate a ripple effect that shifts other",algorithms and data structures.pdf
"memory, not contiguously, an insertion or deletion into a list does not propagate a ripple effect that shifts other  
elements around. An element inserted is allocated anywhere in memory where there is space and tied to other  
elements by  pointers (i.e. addresses of the memory locations where these elements happen to be stored at the  
moment). An element deleted does not leave a gap that needs to be filled as it would in an array. Instead, it leaves  
some free space that can be reclaimed later by a memory management process. The element deleted is likely to  
break some chains that tie other elements together; if so, the broken chains are relinked according to rules specific  
to the type of list used.
Pointers  are  the  language  feature  used  in  modern  programming  languages  to  capture  the  equivalent  of  a 
memory address. A pointer value is essentially an address, and a pointer variable ranges over addresses. A pointer,",algorithms and data structures.pdf
"memory address. A pointer value is essentially an address, and a pointer variable ranges over addresses. A pointer,  
however, may contain more information than merely an address. In Pascal and other strongly typed languages, for  
example, a pointer also references the type definition of the objects it can point to - a feature that enhances the  
compiler's ability to check for consistent use of pointer variables.
Let us illustrate these concepts with a simple example: a  one-way linear list  is a sequence of cells each of  
which (except the last) points to its successor. The first cell is the head of the list, the last cell is the tail. Since the  
Algorithms and Data Structures 211  A Global Text",algorithms and data structures.pdf
"21. List structures
tail has no successor, its pointer is assigned a predefined value 'nil', which differs from the address of any cell.  
Access to the list is provided by an external pointer 'head'. If the list is empty, 'head' has the value 'nil'. A cell stores  
an element xi and a pointer to the successor cell (Exhibit 21.1):
type cptr = ^cell;
cell = record  e: elt;  next: cptr  end;
Exhibit 21.1: A one-way linear list.
Local operations, such as insertion or deletion at a position given by a pointer p, are efficient. For example, the  
following statements insert a new cell containing an element y as successor of a cell being pointed at by p ( Exhibit
21.2):
new(q);  q^.e := y;  q^.next := p^.next;  p^.next := q;
Exhibit 21.2: Insertion as a local operation.
The successor of the cell pointed at by p is deleted by a single assignment statement (Exhibit 21.3):
p^.next := p^.next^.next;
Exhibit 21.3: Deletion as a local operation.",algorithms and data structures.pdf
"p^.next := p^.next^.next;
Exhibit 21.3: Deletion as a local operation.
An insertion or deletion at the head or tail of this list is a special case to be handled separately. To support  
insertion at the tail, an additional pointer variable 'tail' may be set to point to the tail element, if it exists.
A one-way linear list sometimes is handier if the tail points back to the head, making it a  circular list. In a  
circular list, the head and tail cells are replaced by a single entry cell, and any cell can be reached from any other  
without having to start at the external pointer 'entry' (Exhibit 21.4).
Exhibit 21.4: A circular list combines head and tail into a single entry point
212",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
In a  two-way (or  doubly linked )  list each cell contains two pointers, one to its successor, the other to its  
predecessor. The list can be traversed in both directions. Exhibit 21.5 shows a circular two-way list.
Exhibit 21.5: A circular two-way or doubly-linked list
Exercise: traversal of a singly linked list in both directions
Write a recursive
procedure traverse(p: cptr);
to traverse a singly linked list from the head to the tail and back again. At each visit of a node, call the
procedure visit(p: cptr);
Solve  the same  problem iteratively without using any additional storage beyond a few local pointers. Your  
traversal procedure may modify the structure of the list temporarily.
Solution
(a) procedure traverse(p: cptr);
begin  if  p ≠ nil  then  { visit(p);  traverse(p^.next); 
visit(p) }  end;
The initial call of this procedure is
traverse(head);
(b) procedure traverse(p: cptr);",algorithms and data structures.pdf
"visit(p) }  end;
The initial call of this procedure is
traverse(head);
(b) procedure traverse(p: cptr);
var  o, q: cptr;  i: integer;
begin
for  i := 1  to  2  do  { forward and back again }  begin
o := nil;
while  p ≠ nil  do  begin
visit(p);  q := p^.next;  p^.next := o;
o := p;  p := q  { the fork advances }
end;
p := o
end
end;
Traversal becomes simpler if we let the 'next' pointer of the tail 
cell point to this cell itself:
procedure traverse(p: cptr);
var o, q: cptr;
begin
o := nil;
while  p ≠ nil  do  begin
visit(p);  q := p^.next;  p^.next := o;
o := p;  p := q  { the fork advances }
end
end;
Algorithms and Data Structures 213  A Global Text",algorithms and data structures.pdf
"21. List structures
The fifo queue implemented as a one-way list 
It is natural to implement a fifo queue as a one-way linear list, where each element points to the next one ""in  
line"". The operation 'dequeue' occurs at the pointer 'head', and 'enqueue' is made fast by having an external pointer  
'tail' point to the last element in the queue. A crafty implementation of this data structure involves an empty cell,  
called a  sentinel, at the tail of the list. Its purpose is to make the list-handling procedures simpler and faster by  
making the empty queue look more like all other states of the queue. More precisely, when the queue is empty, the  
external pointers 'head' and 'tail' both point to the sentinel rather than having the value 'nil'. The sentinel allows  
insertion into the empty queue, and deletion that results in an empty queue, to be handled by the same code that",algorithms and data structures.pdf
"insertion into the empty queue, and deletion that results in an empty queue, to be handled by the same code that  
handles the general case of 'enqueue' and 'dequeue'. The reader should verify our claim that a sentinel simplifies the 
code by programming the plausible, but less efficient, procedures which assume that an empty queue is represented  
by head = tail = nil.
The queue is empty if and only if 'head' and 'tail' both point to the sentinel (i.e. if head = tail). An 'enqueue'  
operation is performed by inserting the new element into the sentinel cell and then creating a new sentinel.
type cptr = ^cell;
cell = record  e: elt;  next: cptr  end;
fifoqueue = record  head, tail: cptr  end;
procedure create(var f: fifoqueue);
begin  new(f.head);  f.tail := f.head  end;
function empty(f: fifoqueue): boolean;
begin  return(f.head = f.tail)  end;
procedure enqueue(var f: fifoqueue; x: elt);
begin  f.tail^.e := x;  new(f.tail^.next);  f.tail := f.tail^.next 
end;",algorithms and data structures.pdf
"procedure enqueue(var f: fifoqueue; x: elt);
begin  f.tail^.e := x;  new(f.tail^.next);  f.tail := f.tail^.next 
end;
function front(f: fifoqueue): elt;
{ not to be called if the queue is empty }
begin  return(f.head^.e)  end;
procedure dequeue(var f: fifoqueue);
{ not to be called if the queue is empty }
begin  f.head := f.head^.next  end;
Tree traversal
When  we  speak  of  trees  in  computer  science,  we  usually  mean  rooted,  ordered  trees: they  have  a 
distinguished node called the root, and the subtrees of any node are ordered. Rooted, ordered trees are best defined  
recursively: a tree T is either empty, or it is a tuple (N, T1, … , Tk), where N is the root of the tree, and T 1, … , Tk is a 
sequence of trees. Binary trees are the special case k = 2.
Trees are typically used to organize data or activities in a hierarchy: a top-level data set or activity is composed of",algorithms and data structures.pdf
"Trees are typically used to organize data or activities in a hierarchy: a top-level data set or activity is composed of 
a next level of data or activities, and so on. When one wishes to gather or survey all of the data or activities, it is  
necessary to traverse the tree, visiting (i.e. processing) the nodes in some systematic order. The visit at each node  
might be as simple as printing its contents or as complicated as computing a function that depends on all nodes in  
the tree. There are two major ways to traverse trees: breadth first and depth first.
Breadth-first  traversal visits  the  nodes  level  by  level.  This  is  useful  in  heuristic  search,  where  a  node 
represents  a  partial  solution  to  a  problem,  with  deeper  levels  representing  more  complete  solutions.  Before 
pursuing any one solution to a great depth, it may be advantageous to assess all the partial solutions at the present  
214",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
level, in order to pursue the most promising one. We do not discuss breadth-first traversal further, we merely  
suggest the following:
Exercise: breadth-first traversal
Decide  on  a  representation  for  trees  where  each  node  may  have  a  variable  number  of  children.  Write  a 
procedure for breadth-first traversal of such a tree. Hint: use a fifo queue to organize the traversal. The node to be  
visited is removed from the head of the queue, and its children are enqueued, in order, at the tail end.
Depth-first traversal always moves to the first unvisited node at the next deeper level, if there is one. It turns  
out that depth-first better fits the recursive definition of trees than breadth-first does and orders nodes in ways that  
are more often useful. We discuss depth-first for binary trees and leave the generalization to other trees to the",algorithms and data structures.pdf
"are more often useful. We discuss depth-first for binary trees and leave the generalization to other trees to the  
reader.  Depth-first  can  generate  three  basic  orders  for  traversing  a  binary  tree:  preorder,  inorder,  and 
postorder, defined recursively as:
preorder Visit root, traverse left subtree, traverse right subtree.
Inorder Traverse left subtree, visit root, traverse right subtree.
postorderTraverse left subtree, traverse right subtree, visit root.
For the tree in Exhibit 21.6we obtain the orders shown.
Exhibit 21.6: Standard orders defined on a binary tree
An arithmetic expression can be represented as a binary tree by assigning the operands to the leaves and the  
operators  to the  internal  nodes. The  basic  traversal  orders correspond  to different notations  for  representing 
arithmetic expressions. By traversing the expression tree ( Exhibit 21.7 ) in preorder, inorder, or postorder, we  
obtain the prefix, infix, or suffix notation, respectively.",algorithms and data structures.pdf
"obtain the prefix, infix, or suffix notation, respectively.
Exhibit 21.7: Standard traversal orders correspond to different notations for arithmetic expressions
A binary tree can be implemented as a list structure in many ways. The most common way uses an external  
pointer 'root' to access the root of the tree and represents each node by a cell that contains a field for an element to  
be stored, a pointer to the root of the left subtree, and a pointer to the root of the right subtree ( Exhibit 21.8). An 
empty left or right subtree may be represented by the pointer value 'nil', or by pointing at a sentinel, or, as we shall  
see, by a pointer that points to the node itself.
type nptr = ^node;
node = record  e: elt;  L, R: nptr  end;
var root: nptr;
Algorithms and Data Structures 215  A Global Text",algorithms and data structures.pdf
"21. List structures
Exhibit 21.8: Straightforward implementation of a binary tree 
The following procedure 'traverse' implements any or all of the three orders preorder, inorder, and postorder,  
depending on how the procedures 'visit1', 'visit2', and 'visit3' process the data in the node referenced by the pointer p. 
The root of the subtree to be traversed is passed through the formal parameter p. In the simplest case, a visit does  
nothing or simply prints the contents of the node.
procedure traverse(p: nptr);
begin
if  p ≠ nil  then  begin
visit1(p);  { preorder }
traverse(p^.L);
visit2(p);  { inorder }
traverse(p^.R);
visit3(p)  { postorder }
end
end; 
Traversing a tree involves both advancing from the root toward the leaves, and backing up from the leaves  
toward the root. Recursive invocations of the procedure 'traverse' build up a stack whose entries contain references",algorithms and data structures.pdf
"toward the root. Recursive invocations of the procedure 'traverse' build up a stack whose entries contain references  
to the nodes for which 'traverse' has been called. These entries provide a means of returning to a node after the  
traversal of one of its subtrees has been finished. The bookkeeping done by a stack or equivalent auxiliary structure  
can be avoided if the tree to be traversed may be modified temporarily.
The following triple-tree traversal algorithm provides an elegant and efficient way of traversing a binary tree  
without using any auxiliary memory (i.e. no stack is used and it is not assumed that a node contains a pointer to its  
parent node). The data structure is modified temporarily to retain the information needed to find the way back up  
the  tree  and  to restore  each  subtree  to its initial  condition  after  traversal.  The  triple-tree  traversal algorithm",algorithms and data structures.pdf
"the  tree  and  to restore  each  subtree  to its initial  condition  after  traversal.  The  triple-tree  traversal algorithm 
assumes that an empty subtree is encoded not by a 'nil' pointer, but rather by an L (left) or R (right) pointer that  
points to the node itself, as shown in Exhibit 21.9.
216",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 21.9: Coding of a leaf used in procedure TTT
procedure TTT;
var  o, p, q: nptr;
begin
o := nil;  p:= root;
while  p ≠ nil  do  begin
visit(p);
q := p^.L;
p^.L := p^.R;  { rotate left pointer }
p^.R := o;  { rotate right pointer }
o := p;
p := q
end
end;
In  this  procedure  the  pointers  p  (""present"")  and  o (""old"")  serve  as  a  two-pronged  fork.  The  tree  is  being 
traversed by the pointer p and the companion pointer o, which always lags one step behind p. The two pointers  
form a two-pronged fork that runs around the tree, starting in the initial condition with p pointing to the root of the  
tree, and o = nil. An auxiliary pointer q is needed temporarily to advance the fork. The while loop in 'TTT' is  
executed as long as p points to a node in the tree and is terminated when p assumes the value 'nil'. The initial value",algorithms and data structures.pdf
"executed as long as p points to a node in the tree and is terminated when p assumes the value 'nil'. The initial value  
of the o pointer gets saved as a temporary value. First it is assigned to the R pointer of the root, later to the L  
pointer. Finally, it gets assigned to p, the fork exits from the root of the tree, and the traversal of the tree is  
complete. The correctness of this algorithm is proved by induction on the number of nodes in the tree.
Induction hypothesis H: if at the beginning of an iteration of the while loop, the fork pointer p points to the root  
of a subtree with n > 0 nodes, and o has a value x that is different from any pointer value inside this subtree, then  
after 3 · n iterations the subtree will have been traversed in triple order (visiting each node exactly three times), all  
tree pointers in the subtree will have been restored to their original value, and the fork pointers will have been",algorithms and data structures.pdf
"tree pointers in the subtree will have been restored to their original value, and the fork pointers will have been  
reversed (i.e. p has the value x and o points to the root of the subtree).
Base of induction: H is true for n = 1.
Proof: The smallest tree we consider has exactly one node, the root alone. Before the while loop is executed for  
this subtree, the fork and the tree are in the initial state shown in Exhibit 21.10. Exhibit 21.11 shows the state of the  
fork and the tree after each iteration of the while loop. The node is visited in each iteration.
Exhibit 21.10 : Initial configuration for traversing a tree consisting of a single node
Algorithms and Data Structures 217  A Global Text",algorithms and data structures.pdf
"21. List structures
Exhibit 21.11: Tracing procedure TTT while traversing the smallest tree
Induction step: If H is true for all n, 0 < n ≤ k, H is also true for k + 1.
Proof: Consider a tree T with k + 1 nodes. T consists of a root and k nodes shared among the left and right  
subtrees of the root. Each of these subtrees has ≤ k nodes, so we apply the induction hypothesis to each of them.  
The  following is a  highly compressed account of the proof of the induction step,  illustrated  by  Exhibit 21.12 . 
Consider the tree with k + 1 nodes shown in state 1. The root is a node with three fields; the left and right subtrees  
are shown as triangles. The figure shows the typical case when both subtrees are nonempty. If one of the two  
subtrees is empty, the corresponding pointer points back to the root; these two cases can be handled similarly to the 
case n = 1. The fork starts out with p pointing at the root and o pointing at anything outside the subtree being",algorithms and data structures.pdf
"case n = 1. The fork starts out with p pointing at the root and o pointing at anything outside the subtree being  
traversed. We want to show that the initial state 1 is transformed in 3 ·  (k + 1) iterations into the final state 6. In the  
final state the subtrees are shaded to indicate that they have been correctly traversed; the fork has exited from the  
root, with p and o having exchanged values. To show that the algorithm correctly transforms state 1 into state 6, we  
consider the intermediate states 2 to 5, and what happens in each transition.
1 → 2 One iteration through the while loop advances the fork into the left subtree and rotates the pointers of the 
root.
2 → 3 H applied to the left subtree of the root says that this subtree will be correctly traversed, and the fork will 
exit from the subtree with pointers reversed.
3 → 4 This is the second iteration through the while loop that visits the root. The fork advances into the right",algorithms and data structures.pdf
"3 → 4 This is the second iteration through the while loop that visits the root. The fork advances into the right 
subtree, and the pointers of the root rotate a second time.
4 → 5 H applied to the right subtree of the root says that this subtree will be correctly traversed, and the fork will 
exit from the subtree with pointers reversed.
5→ 6 This is the third iteration through the while loop that visits the root. The fork moves out of the tree being 
traversed; the pointers of the root rotate a third time and thereby assume their original values.
218",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 21.12: Trace of procedure TTT, invoking the induction hypothesis
Exercise: binary trees
Consider a binary tree declared as follows:
type nptr = ^node;
node = record  L, R: nptr  end;
var root: nptr;
(a) If a node has no left or right subtree, the corresponding pointer has the value 'nil'. Prove that a binary tree  
with n nodes, n > 0, has n + 1 'nil' pointers.
Algorithms and Data Structures 219  A Global Text",algorithms and data structures.pdf
"21. List structures
(b) Write a function nodes(…): integer; that returns the number of nodes, and a function depth(…): integer;  
that returns the depth of a binary tree. The depth of the root is defined to be 0; the depth of any other node  
is the depth of its parent increased by 1. The depth of the tree is the maximum depth of its nodes.
Solution
(a) Each node contains two pointers, for a total of 2 · n pointers in the tree. There is exactly one pointer that  
points to each of n – 1 nodes, none points to the root. Thus 2 ·  n – (n – 1) = n + 1 pointers are 'nil'. This can  
also be proved by induction on the number of nodes in the tree.
(b) function nodes(p: nptr): integer;
begin
if  p = nil  then
return(0)
else
return(nodes(p^.L) + nodes(p^.R) + 1)
end;
function depth(p: nptr): integer;
begin
if  p = nil then return (–1)
else return(1 + max(depth(p^.L), depth(p^.R)))
end;
where 'max' is
function max(a, b: integer): integer;
begin  if  a > b  then  return(a)  else  return(b)  end;",algorithms and data structures.pdf
"end;
where 'max' is
function max(a, b: integer): integer;
begin  if  a > b  then  return(a)  else  return(b)  end;
Exercise: list copying
Effective  memory  management  sometimes  makes  it  desirable  or  necessary  to  copy  a  list.  For  example, 
performance may improve drastically if a list spread over several pages can be compressed into a single page. List  
copying involves a traversal of the original concurrently with a traversal of the copy, as the latter is being built up.
(a) Consider binary trees built from nodes of type 'node' and pointers of type 'nptr'. A tree is accessed through  
a pointer to the root, which is 'nil' for an empty tree
type nptr = ^ node;
node = record  e: elt;  L, R: nptr  end;
Write a recursive
function cptree(p: nptr): nptr;
to copy a tree given by a pointer p to its root, and return a pointer to the root of the copy.
(b) Consider arbitrary graphs built from nodes of a type similar to the nodes in (a), but they have an additional",algorithms and data structures.pdf
"(b) Consider arbitrary graphs built from nodes of a type similar to the nodes in (a), but they have an additional  
pointer field cn, intended to point to the copy of a node:
type  node = record  e: elt;  L, R: nptr;  cn: nptr  end;
A graph is accessed through a pointer to a node called the origin, and we are only concerned with nodes that can  
be reached from the origin; this access pointer is 'nil' for an empty graph. Write a recursive
function cpgraph(p: nptr): nptr;
220",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
to copy a graph given by a pointer p to its origin, and return a pointer to the origin of the copy. Use the field cn,  
assuming that its initial value is 'nil' in every node of the original graph; set it to 'nil' in every node of the copy.
Solution
(a) function cptree(p: nptr): nptr;
var  cp: nptr;
begin
if  p = nil  then
return(nil)
else  begin
new(cp);
cp^.e := p^.e;  cp^.L := cptree(p^.L);  cp^.R := cptree(p^.R);
return(cp)
end
end;
(b) function cpgraph(p: nptr): nptr;
var  cp: nptr;
begin
if  p = nil  then
return(nil)
elsif  p^.cn ≠ nil  then  { node has already been copied }
return(p^.cn)
else  begin
new(cp);  p^.cn := cp;  cp^.cn := nil;
cp^.e := p^.e;  cp^.L := cpgraph(p^.L);  cp^.R := cpgraph(p^.R);
return(cp)
end
end;
Exercise: list copying with constant auxiliary memory
Consider binary trees as in part (a) of the preceding exercise. Memory for the stack implied by the recursion can",algorithms and data structures.pdf
"Consider binary trees as in part (a) of the preceding exercise. Memory for the stack implied by the recursion can  
be saved by writing an iterative tree copying procedure that uses only a constant amount of auxiliary memory. This  
requires a trick, as any depth-first traversal must be able to back up from the leaves toward the root. In the triple-
tree traversal procedure, the return path is temporarily encoded in the tree being traversed. This idea can again be  
used here, but there is a simpler solution: The return path is temporarily encoded in the R-fields of the copy; the L-
fields of certain nodes of the copy point back to the corresponding node in the original. Work out the details of a  
tree-copying procedure that works with O(1) auxiliary memory.
Exercise: traversing a directed acyclic graph 
A directed graph consists of nodes and directed arcs, where each arc leads from one node to another. A directed",algorithms and data structures.pdf
"A directed graph consists of nodes and directed arcs, where each arc leads from one node to another. A directed  
graph is acyclic if the arcs form no cycles. One way to ensure that a graph is acyclic is to label nodes with distinct  
integers and to draw each arc from a lower number to a higher number. Consider a binary directed acyclic graph,  
where each node has two pointer fields, L and R, to represent at most two arcs that lead out of that node. An  
example is shown in Exhibit 21.13.
Exhibit 21.13: A rooted acyclic graph.
Algorithms and Data Structures 221  A Global Text",algorithms and data structures.pdf
"21. List structures
(a) Write a program to visit every node in a directed acyclic graph reachable from a pointer called 'root'. You  
are free to execute procedure 'visit' for each node as often as you like.
(b) Write a program similar to (a) where you are required to execute procedure 'visit' exactly once per node.  
Hint: Nodes may need to have additional fields.
Exercise: counting nodes on a square grid
Consider a network superimposed on a square grid: each node is connected to at most four neighbors in the  
directions east, north, west, south (Exhibit 21.14):
type nptr = ^node;
node = record  E, N, W, S: nptr;  status: boolean  end;
var origin: nptr;
Exhibit 21.14: A graph embedded in a square grid.
A 'nil' pointer indicates the absence of a neighbor. Neighboring nodes are doubly linked: if a pointer in node p  
points to node q, the reverse pointer of q points to p; (e.g., p^.W = q and q^.E = p). The pointer 'origin' is 'nil' or",algorithms and data structures.pdf
"points to node q, the reverse pointer of q points to p; (e.g., p^.W = q and q^.E = p). The pointer 'origin' is 'nil' or  
points to a node. Consider the problem of counting the number of nodes that can be reached from 'origin'. Assume  
that the status field of all nodes is initially set to false. How do you use this field? Write a function nn(p: nptr):  
integer; to count the number of nodes.
Solution
function nn(p: nptr): integer;
begin
if  p = nil cor p^.status  then
return(0)
else  begin 
p^.status:= true;
return(1 + nn(p^.E) + nn(p^.N) + nn(p^.W) + nn(p^.S))
end
end;
Exercise: counting nodes in an arbitrary network
We generalize the problem above to arbitrary directed graphs, such as that of Exhibit 21.15, where each node  
may have any number of neighbors. This graph is represented by a data structure defined by Exhibit 21.16 and the 
type definitions below. Each node is linked to an arbitrary number of other nodes.
Exhibit 21.15: An arbitrary (cyclic) directed graph.
222",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 21.16: A possible implementation as a list structure.
type nptr = ^node;  cptr = ^cell;
node = record  status: boolean;  np: nptr;  cp: cptr  end; 
cell = record  np: nptr;  cp: cptr  end;
var origin: nptr;
The pointer 'origin' has the value 'nil' or points to a node. Consider the problem of counting the number n of  
nodes that can be reached from 'origin'. The status field of all nodes is initially set to false. How do you use it? Write  
a function nn(p: nptr): integer; that returns n.
Binary search trees
A binary search tree  is a binary tree T where each node N stores a data element e(N) from a domain X on  
which a total order ≤ is defined, subject to the following order condition: For every node N in T, all elements in the  
left subtree L(N) of N are < e(N), and all elements in the right subtree R(N) of N are > e(N). Let x1, 2, … , x n be n 
elements drawn from the domain X.",algorithms and data structures.pdf
"elements drawn from the domain X.
Definition:  A binary search tree for x 1, x2, … , x n is a binary tree T with n nodes and a one-to-one mapping  
between the n given elements and the n nodes, such that
 ∀ N in T  ∀  N' ∈  L(N)  ∀  N"" ∈  R(N):  e(N') < e(N) < e(N"")
Exercise
Show that the following statement is equivalent to this order condition: The inorder traversal of the nodes of T  
coincides with the natural order < of the elements assigned to the nodes.
Remark:  The order condition can be relaxed to e(N') ≤ e(N) < e(N"") to accommodate multiple occurrences of  
the same value, with only minor modifications to the statements and algorithms presented in this section. For  
simplicity's sake we assume that all values in a tree are distinct.
The order condition permits binary search and thus guarantees a worst-case search time O(h) for a tree of height 
h.  Trees  that  are  well  balanced  (in  an  intuitive  sense;  see  the  next  section  for  a  definition),  that  have  not",algorithms and data structures.pdf
"h.  Trees  that  are  well  balanced  (in  an  intuitive  sense;  see  the  next  section  for  a  definition),  that  have  not 
degenerated into linear lists, have a height h = O(log n) and thus support search in logarithmic time.
Algorithms and Data Structures 223  A Global Text",algorithms and data structures.pdf
"21. List structures
Basic operations on binary search trees are most easily implemented as recursive procedures. Consider a tree  
represented as in the preceding section, with empty subtrees denoted by 'nil'. The following function 'find' searches
for an element x in a subtree pointed to by p. It returns a pointer to a node containing x if the search is successful,  
and 'nil' if it is not.
function find(x: elt; p: nptr): nptr;
begin
if  p = nil  then  return(nil)
elsif  x < p^.e  then  return(find(x, p^.L))
elsif  x > p^.e  then  return(find(x, p^.R))
else  { x = p^.e }  return(p)
end;
The following procedure 'insert' leaves the tree alone if the element x to be inserted is already stored in the tree.  
The parameter p initially points to the root of the subtree into which x is to be inserted.
procedure insert(x: elt; var p: nptr);
begin
if  p = nil  then  { new(p);  p^.e := x;  p^.L := nil;  p^.R := 
nil }
elsif  x < p^.e  then  insert(x, p^.L)
elsif  x > p^.e  then  insert(x, p^.R)",algorithms and data structures.pdf
"if  p = nil  then  { new(p);  p^.e := x;  p^.L := nil;  p^.R := 
nil }
elsif  x < p^.e  then  insert(x, p^.L)
elsif  x > p^.e  then  insert(x, p^.R)
end;
Initial call:
insert(x, root);
To delete an element x, we first have to find the node N that stores x. If this node is a leaf or semileaf (a node  
with only one subtree), it is easily deleted; but if it has two subtrees, it is more efficient to leave this node in place  
and to replace its element x by an element found in a leaf or semileaf node, and delete the latter ( Exhibit 21.17). 
Thus we distinguish three cases:
1. If N has no child, remove N.
2. If N has exactly one child, replace N by this child node.
3. If N has two children, replace x by the largest element y in the left subtree, or by the smallest element z in  
the right subtree of N. Either of these elements is stored in a node with at most one child, which is removed  
as in case (1) or (2).
224",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 21.17: Element x is deleted while preserving its node N. Node N is 
filled with a new value y, whose old node is easier to delete. 
A sentinel is again the key to an elegant iterative implementation of binary search trees. In a node with no left or  
right child, the corresponding pointer points to the sentinel. This sentinel is a node that contains no element; its left  
pointer points to the root and its right pointer points to itself. The root, if it exists, can only be accessed through the  
left pointer of the sentinel. The empty tree is represented by the sentinel alone ( Exhibit 21.18 ). A typical tree is  
shown in Exhibit 21.19.
Exhibit 21.18: The empty binary tree is represented by the sentinel which points to itself.
Exhibit 21.19: A binary tree implemented as a list structure with 
sentinel.",algorithms and data structures.pdf
"Exhibit 21.19: A binary tree implemented as a list structure with 
sentinel.
The following implementation of a dictionary as a binary search tree uses a sentinel accessed via the variable d:
type nptr = ^node;
node = record  e: elt;  L, R: nptr  end;
dictionary = nptr;
procedure create(var d: dictionary);
begin  {create sentinel }  new(d);  d^.L := d;  d^.R := d  end;
Algorithms and Data Structures 225  A Global Text",algorithms and data structures.pdf
"21. List structures
function member(d: dictionary; x: elt): boolean;
var  p: nptr;
begin
d^.e := x;  { initialize element in sentinel }
p := d^.L;  { point to root, if it exists }
while  x ≠ p^.e  do
if  x < p^.e  then  p := p^.L  else  { x > p^.e }  p := p^.R;
return(p ≠ d)
end;
Procedure 'find' searches for x. If found, p points to the node containing x, and q to its parent. If not found, p  
points to the sentinel and q to the parent-to-be of a new node into which x will be inserted.
procedure find(d: dictionary; x: elt; var p, q: nptr);
begin
d^.e := x;  p := d^.L;  q := d;
while  x ≠ p^.e  do  begin
q := p;
if  x < p^.e  then  p := p^.L  else  { x > p^.e }  p := p^.R
end
end;
procedure insert(var d: dictionary; x: elt);
var  p, q: nptr;
begin
find(d, x, p, q);
if  p = d  then  begin  { x is not yet in the tree }
new(p);  p^.e := x;  p^.L := d;  p^.R := d;
if  x ≤ q^.e  then  q^.L := p  else  { x > q^.e }  q^.R := p
end
end;
procedure delete(var d: dictionary; x: elt);",algorithms and data structures.pdf
"if  x ≤ q^.e  then  q^.L := p  else  { x > q^.e }  q^.R := p
end
end;
procedure delete(var d: dictionary; x: elt);
var  p, q, t: nptr;
begin
find(d, x, p, q);
if  p ≠ d  then  { x has been found }
if  (p^.L ≠ d) and (p^.R ≠ d)  then  begin
{ p has left and right children; find largest element in left 
subtree }
t := p^.L;  q:= p;
while  t^.R ≠ d  do  { q := t;  t := t^.R };
if  t^.e < q^.e  then  q^.L := t^.L  else  { t^.e > q^.e } 
q^.R := t^.L
p^.e := t^.e;
end
else  begin  { p has at most one child }
ifp^.L ≠ d  then{ left child only } p := p^.L
elsif p^.R ≠ d  then{ right child only } p := p^.R
else { p has no children }p := d;
if  x ≤ q^.e  then  q^.L := p  else  { x > q^.e }  q^.R := p
end
end;
In the best case of a completely balanced binary search tree for n elements, all leaves are on levels [log2  n] or 
[log2 n]– 1, and the search tree has the height [log2 n]. The cost for performing the 'member', 'insert', or 'delete'",algorithms and data structures.pdf
"[log2 n]– 1, and the search tree has the height [log2 n]. The cost for performing the 'member', 'insert', or 'delete'  
operation is bounded by the longest path from the root to a leaf (i.e. the height of the tree) and is therefore O(log  n). 
226",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Without any further provisions, a binary search tree can degenerate into a linear list in the worst case. Then the cost 
for each of the operations would be O(n).
What  is  the  expected  average  cost for  the  search  operation  in  a  randomly  generated binary  search  tree? 
""Randomly generated"" means that each permutation of the n elements to be stored in the binary search tree has the  
same probability of being chosen as the input sequence. Furthermore, we assume that the tree is generated by  
insertions only. Therefore, each of the n elements is equally likely to be chosen as the root of the tree. Le t pn be the 
expected path length of a randomly generated binary search tree storing n elements. Then
As shown in chapter 16 in the section “Recurrence relations”, this recurrence relation has the solution",algorithms and data structures.pdf
"As shown in chapter 16 in the section “Recurrence relations”, this recurrence relation has the solution
Since the average search time in randomly generated binary search trees, measured in terms of the number of  
nodes visited, is p n / n and ln 4 ≈ 1.386, it follows that the cost is O(log  n) and therefore only about 40  per cent 
higher than in the case of completely balanced binary search trees.
Balanced trees: general definition
If insertions and deletions occurred at random, and the assumption of the preceding section was realistic, we  
could let search trees grow and shrink as they please, incurring a modest increase of 40 per cent in search time over  
completely  balanced  trees.  But  real  data  are  not  random:  they  are  typically  clustered,  and  long  runs  of 
monotonically  increasing  or  decreasing  elements  occur,  often  as  the  result  of  a  previous  processing  step. 
Unfortunately, such deviation from randomness degrades the performance of search trees.",algorithms and data structures.pdf
"Unfortunately, such deviation from randomness degrades the performance of search trees.
To prevent search trees from degenerating into linear lists, we can monitor their shape and restructure them  
into a more  balanced shape whenever they have  become too skewed. Several classes of balanced search trees  
guarantee that each operation 'member', 'insert', and 'delete' can be performed in time O(log  n) in the worst case.  
Since the work to be done depends directly on the height of the tree, such a class B of search trees must satisfy the  
following two conditions (hT is the height of a tree T, nT is the number of nodes in T):
Balance condition:  ∃c > 0  ∀  T ∀  B:  hT ≤ c · log2 nT
Rebalancing condition:  If an 'insert' or 'delete' operation, performed on a tree T ∈  B, yields a tree T' ∉  B, it 
must be possible to rebalance T' in time O(log n) to yield a tree T"" ∈  B.
Example: almost complete trees",algorithms and data structures.pdf
"must be possible to rebalance T' in time O(log n) to yield a tree T"" ∈  B.
Example: almost complete trees
The  class  of  almost  complete  binary  search  trees  satisfies  the  balance  condition  but not  the  restructuring 
condition. In the worst case it takes time O(n) to restructure such a binary search tree ( Exhibit 21.20), and if 'insert' 
and 'delete' are defined to include any rebalancing that may be necessary, these operations cannot be guaranteed to  
run in time O(log n).
Algorithms and Data Structures 227  A Global Text",algorithms and data structures.pdf
"21. List structures
Exhibit 21.20: Restructuring: worst case
In the next two sections we present several classes of balanced trees that meet both conditions: the height-
balanced or AVL-trees (G. Adel'son-Vel'skii  and E. Landis, 1962) [AL 62] and various multiway trees, such as B-
trees [BM 72, Com 79] and their generalization, (a,b)-trees [Meh 84a].
AVL-trees, with their small nodes that hold a single data element, are used primarily for storing data in main  
memory. Multiway trees, with potentially large nodes that hold many elements, are also useful for organizing data  
on secondary storage devices, such as disks, that allow direct access to sizable physical data blocks. In this case, a  
node is typically chosen to fill a physical data block, which is read or written in one access operation.
Height-balanced trees
Definition:  A binary tree is height-balanced if, for each node, the heights of its two subtrees differ by at most",algorithms and data structures.pdf
"Height-balanced trees
Definition:  A binary tree is height-balanced if, for each node, the heights of its two subtrees differ by at most  
one. Height-balanced search trees are also called AVL-trees. Exhibit 21.21 to Exhibit 21.23 show various AVL-trees, 
and one that is not.
Exhibit 21.21: Examples of height-balanced trees
Exhibit 21.22: Example of a tree not height-balanced;the marked node violates the balance condition.
A ""most-skewed"" AVL-tree Th is an AVL-tree of height h with a minimal number of nodes. Starting with T 0 and 
T1 shown in Exhibit 21.23, Th is obtained by attaching Th–1 and Th–2 as subtrees to a new root.
228",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 21.23: Most skewed AVL trees of heights h = 0 through h = 4
The number of nodes in a most-skewed AVL-tree of height h is given by the recurrence relation
nh = nh–1 + nh–2 + 1,  n0 = 1,  n1 = 2.
In the section on recurrence relations in the chapter entitled “The mathematics of algorithm analysis”, it has  
been shown that the recurrence relation
mh = mh–1 + mh–2,  m0 = 0,  m1 = 1
has the solution
Since nh = mh+3 – 1 we obtain
Since 
it follows that 
and therefore nh behaves asymptotically as
Algorithms and Data Structures 229  A Global Text",algorithms and data structures.pdf
"21. List structures
Applying the logarithm results in
Therefore, the height of a worst-case AVL-tree with n nodes is about 1.4 4 · log2 n. Thus th e class of AVL-trees  
satisfies the balance condition, and the 'member' operation can always be performed in time O(log n).
We now show that the class of AVL-trees also satisfies the rebalancing condition. Thus AVL-trees support  
insertion and deletion in time O(log n). Each node N of an AVL-tree has one of the balance properties / (left-
leaning), \ (right-leaning), or – (horizontal), depending on the relative height of its two subtrees.
Two local tree operations, rotation and double rotation, allow the restructuring of height-balanced trees that 
have been disturbed by an insertion or deletion. They split a tree into subtrees and rebuild it in a different way.  
Exhibit 21.24  shows a node, marked black, that got out of balance, and how a local transformation builds an",algorithms and data structures.pdf
"Exhibit 21.24  shows a node, marked black, that got out of balance, and how a local transformation builds an  
equivalent tree (for the same elements, arranged in order) that is balanced. Each of these transformations has a  
mirror image that is not shown. The algorithms for insertion and deletion use these rebalancing operations as  
described below.
Exhibit 21.24: Two local rebalancing operations
Insertion
A new element is inserted as in the case of a binary search tree. The balance condition of the new node becomes  
– (horizontal). Starting at the new node, we walk toward the root of the tree, passing along the message that the  
height of the subtree rooted at the current node has increased by one. At each node encountered along this path, an  
operation determined by the following rules is performed. These rules depend on the balance condition of the node  
before the new element was inserted, and on the direction from which the node was entered (i.e. from its left or",algorithms and data structures.pdf
"before the new element was inserted, and on the direction from which the node was entered (i.e. from its left or  
right child).
230",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Rule I1:  If the current node has balance condition –, change it to / or \ depending on whether we entered from  
the node's left or from its right child. If the current node is the root, terminate; if not, continue to follow the path  
upward.
Rule I2:  If the current node has balance condition / or \ and is entered from the subtree that was previously  
shorter, change the balance condition to—and terminate (the height of the subtree rooted at the current node has  
not changed).
Rule I3:  If the current node has balance condition / or \ and is entered from the subtree that was previously  
taller, the balance condition of the current node is violated and gets restored as follows:
(a) If the last two steps were in the same direction (both from left children, or both from right children), an  
appropriate rotation restores all balances and the procedure terminates.",algorithms and data structures.pdf
"appropriate rotation restores all balances and the procedure terminates.
(b) If the last two steps were in opposite directions (one from a left child, the other from a right child), an  
appropriate double rotation restores all balances and the procedure terminates.
The initial insertion travels along a path from the root to  a leaf, and the rebalancing process travels back up  
along the same path. Thus the cost of an insertion in an AVL-tree is O(h), or O(log n) in the worst case. Not ice that 
an insertion calls for at most one rotation or double rotation, as shown in the example in Exhibit 21.25.
Example
Insert 1, 2, 5, 3, 4, 6, 7 into an initially empty AVL-tree ( Exhibit 21.25). The balance condition of a node is shown 
below it. Boldfaced nodes violate the balance condition.
Algorithms and Data Structures 231  A Global Text",algorithms and data structures.pdf
"21. List structures
Exhibit 21.25: Trace of consecutive insertions and the rebalancings they trigger
Deletion
An element is deleted as in the case of a binary search tree. Starting at the parent of the deleted node, walk  
towards the root, passing along the message that the height of the subtree rooted at the current node has decreased  
by one. At each node encountered, perform an operation according to the following rules. These rules depend on  
the balance condition of the node before the deletion and on the direction from which the current node and its child  
were entered.
Rule D1:  If the current node has balance condition –, change it to \ or / depending on whether we entered from  
the node's left or from its right child, and terminate (the height of the subtree rooted at the current node has not  
changed).
Rule D2:  If the current node has balance condition / or \ and is entered from the subtree that was previously",algorithms and data structures.pdf
"changed).
Rule D2:  If the current node has balance condition / or \ and is entered from the subtree that was previously  
taller, change the balance condition to – and continue upward, passing along the message that the subtree rooted at  
the current node has been shortened.
Rule D3:  If the current no de has balance condition / or \ and is entered from the subtree that was previously  
shorter, the balance condition is violated at the current node. We distinguish three subcases according to the  
232",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
balance  condition  of  the  other  child  of  the  current  node  (consider  also  the  mirror  images  of  the  following 
illustrations):
(a)
X Y
Z
a
b
X
Y
Z
b
a
rotation
An appropriate rotation restores the balance of the current node without changing the height of the subtree  
rooted at this node. Terminate.
(b)
X
Y Z
b
a
X Y Z
a
b
rotation
A rotation restores the balance of the current node. Continue upward, passing along the message that the  
subtree rooted at the current node has been shortened.
(c)
double rotation
W
a
b
c
a
b
c
X Y
Z W X Y Z
A double rotation restores the balance of the current node. Continue upward, passing along the message that  
the subtree rooted at the current node has been shortened. Similar transformations apply if either X or Y, but not  
both, are one level shorter than shown in this figure. If so, the balance conditions of some nodes differ from those",algorithms and data structures.pdf
"both, are one level shorter than shown in this figure. If so, the balance conditions of some nodes differ from those  
shown, but this has no influence on the total height of the subtree. In contrast to insertion, deletion may require  
more than one rotation or double rotation to restore all balances. Since the cost of a rotation or double rotation is  
constant, the worst-case cost for rebalancing the tree depends only on the height of the tree, and thus the cost of a  
deletion in an AVL-tree is O(log n) in the worst case.
Multiway trees
Nodes in a multiway tree may have a variable number of children. As we are interested in balanced trees, we add  
two restrictions. First, we insist that all leaves (the nodes without children) occur at the same depth. Second, we  
constrain the number of children of all internal nodes by a lower bound a and an upper bound b. Many varieties of",algorithms and data structures.pdf
"constrain the number of children of all internal nodes by a lower bound a and an upper bound b. Many varieties of  
multiway trees are known; they differ in details, but all are based on similar ideas. For example, (2,3)-trees are  
defined by the requirement that all internal nodes have either two or three children. We generalize this concept and  
discuss (a,b)-trees.
Definition:  Consider a domain X on which a total order ≤ is defined. Let a and b be integers with 2 ≤ a and 2 · 
a – 1 ≤ b. Let c(N) denote the number of children of node N. An (a,b)-tree is an ordered tree with the following  
properties:
Algorithms and Data Structures 233  A Global Text",algorithms and data structures.pdf
"21. List structures
• All leaves are at the same level
• 2 ≤ c(root) ≤ b
• For all internal nodes N except the root, a ≤ c(N) ≤ b
A node with k children contains k – 1 elements x 1 < x2 < … < x k–1 drawn from X; the subtrees corresponding to  
the k children are denoted by T 1, T2, … , T k. An (a,b)-tree supports ""c(N) search"" in the same way that a binary tree  
supports binary search, thanks to the following order condition:
• y ≤ xi for all elements y stored in subtrees T1, … , Ti
• xi < z for all elements z stored in subtrees Ti+1, … , Tk
Definition:  (a,b)-trees with b = 2 ·  a – 1 are known as B-trees [BM 72, Com 79].
The algorithms we discuss operate on internal nodes, shown in white in  Exhibit 21.26 , and ignore the leaves,  
shown in black. For the purpose of understanding search and update algorithms, leaves can be considered fictitious  
entities used only for counting. In practice, however, things are different. The internal nodes merely constitute a",algorithms and data structures.pdf
"entities used only for counting. In practice, however, things are different. The internal nodes merely constitute a  
directory to a file that is stored in the leaves. A leaf is typically a physical storage unit, such as a disk block, that  
holds all the records whose key values lie between two (adjacent) elements stored in internal nodes.
Exhibit 21.26: Example of a (3,5)-tree
The number n of elements stored in the internal nodes of an (a,b)-tree of height h is bounded by 
and thus 
this shows that the class of (a,b)-trees satisfies the balance condition h = O(log  n). We show that this class also  
meets the rebalancing condition, namely, that (a,b)-trees support insertion and deletion in time O(log n).
Insertion
Insertion of a new element x begins with a search for x that terminates unsuccessfully at a leaf. Let N be the  
parent node  of  this  leaf. If N contained  fewer  than  b –  1  elements  before  the  insertion,  insert x  into  N  and",algorithms and data structures.pdf
"parent node  of  this  leaf. If N contained  fewer  than  b –  1  elements  before  the  insertion,  insert x  into  N  and 
terminate. If N was full, we imagine b elements temporarily squeezed into the overflowing node N. Let m be the  
median of these b elements, and use m to split N into two: a left node N L populated by the (b – 1) / 2 elements 
smaller than m, and a right node N R populated by the (b – 1) /  2 elements larger than m. The condition 2 ·  a – 1 ≤ b 
ensures that  (b – 1) /  2 ≥ a – 1, in other words, that each of the two new nodes contains at least a – 1 elements.
The median element m is pushed upward into the parent node, where it serves as a separator between the two  
new nodes NL and NR that now take the place formerly inhabited by N. Thus the problem of insertion into a node at  
a given level is replaced by the same problem one level higher in the tree. The new separator element may be",algorithms and data structures.pdf
"a given level is replaced by the same problem one level higher in the tree. The new separator element may be  
absorbed in a nonfull parent, but if the parent overflows, the splitting process described is repeated recursively. At  
234",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
worst, the splitting process propagates to the root of the tree, where a new root that contains only the median  
element is created. (a,b)-trees grow at the root, and this is the reason for allowing the root to have as few as two  
children.
Deletion
Deletion of an element x begins by searching for it. As in the case of binary search trees, deletion is easiest at the  
bottom of the tree, at a node of maximal depth whose children are leaves. If x is found at a higher level of the tree,  
in a node that has internal nodes as children, x is the separator between two subtrees T L and T R. We replace x by  
another element z, either the largest element in T L or the smallest element in T R, both of which are stored in a node  
at the bottom of the tree. After this exchange, the problem is reduced to deleting an element z from a node N at the  
deepest level.",algorithms and data structures.pdf
"at the bottom of the tree. After this exchange, the problem is reduced to deleting an element z from a node N at the  
deepest level.
If deletion (of x or z) leaves N with at least a – 1 elements, we are done. If not, we try to restore N's occupancy  
condition by stealing an element from an adjacent sibling node M. If there is no sibling M that can spare an  
element, that is, if M is minimally occupied, M and N are merged into a single node L. L contains the a – 2 elements  
of N, the a – 1 elements of M, and the separator between M and N which was stored in their parent node, for a total  
of 2 ·  (a – 1) ≤ b – 1 elements. Since the parent (of the old nodes M and N, and of the new node L) lost an element in  
this merger, the parent may underflow. As in the case of insertion, this underflow can propagate to the root and  
may cause its deletion. Thus (a,b)-trees grow and shrink at the root.",algorithms and data structures.pdf
"may cause its deletion. Thus (a,b)-trees grow and shrink at the root.
Both insertion and deletion work along a single path from the root down to a l eaf and (possibly) back up. Thus  
their time is bounded by O(h), or equivalently, by O(log n): (a,b)-trees can be rebalanced in logarithmic time.
Amortized cost. The performance of (a,b)-trees is better than the worst-case analysis above suggests. It can be  
shown that the total cost of any sequence of s insertions and deletions  into an initially empty (a,b)-tree is linear in  
the length s of the sequence: whereas the worst-case cost of a single operation is O(log n), the amortized cost per 
operation is O(1) [Meh 84a]. Amortized cost is a complexity measure that involves both an average and a worst-case 
consideration. The average is taken over all operations in a sequence; the worst case is taken over all sequences.  
Although any one operation may take time O(log n), we are guaranteed that the total of all s operations in any",algorithms and data structures.pdf
"Although any one operation may take time O(log n), we are guaranteed that the total of all s operations in any  
sequence of length s can be done in time O(s), as if each single operation were done in time O(1).
Exhibit 21.27: A slightly skewed (3,5)-tree.
Exercise: insertion and deletion in a (3,5)-tree
Starting with the (3,5)-tree shown in  Exhibit 21.27 , perform the sequence of operations: insert 38, delete 10,  
delete 12, delete 50. Draw the tree after each operation.
Solution
Inserting 38 causes a leaf and its parent to split ( Exhibit 21.28 ). Deleting 10 causes underflow, remedied by  
borrowing an element from the left sibling ( Exhibit 21.29 ). Deleting 12 causes underflow in both a leaf and its  
parent, remedied by merging ( Exhibit 21.30 ). Deleting 50 causes merging at the leaf level and borrowing at the  
parent level (Exhibit 21.31).
Algorithms and Data Structures 235  A Global Text",algorithms and data structures.pdf
"21. List structures
Exhibit 21.28: Node splits propagate towards the root
Exhibit 21.29: A deletion is absorbed by borrowing
Exhibit 21.30: Another deletion propagates node merges towards the root
Exhibit 21.31: Node merges and borrowing combined
(2,3)-trees are the special case a = 2, b = 3: each node has two or three children. Exhibit 21.32 omits the leaves. 
Starting with the tree in state 1 we insert the value 9: the rightmost node at the bottom level overflows and splits,  
the median 8 moves up into the parent. The parent also overflows, and the median 6 generates a new root (state 2).  
The deletion of 1 is absorbed without any rebalancing (state 3). The deletion of 2 causes a node to underflow,  
remedied by stealing an element from a sibling: 2 is replaced by 3 and 3 is replaced by 4 (state 4). The deletion of 3  
236",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
triggers the merger of the nodes assigned to 3 and 5; this causes an underflow in their parent, which in turn  
propagates to the root and results in a tree of reduced height (state 5).
Exhibit 21.32: Tracing insertions and deletions in a (2,3)-tree
As  mentioned  earlier, multiway  trees  are  particularly useful for  managing  data  on  a  disk.  If each  node  is 
allocated to its own disk block, searching for a record triggers as many disk accesses as there are levels in the tree.  
The depth of the tree is minimized if the maximal fan-out b is maximized. We can pack more elements into a node  
by shrinking their size. As the records to be stored are normally much larger than their identifying keys, we store  
keys only in the internal nodes and store entire records in the leaves (which we had considered to be empty until",algorithms and data structures.pdf
"keys only in the internal nodes and store entire records in the leaves (which we had considered to be empty until  
now). Thus the internal nodes serve as an index that assigns to a key value the path to the corresponding leaf.
Exercises and programming projects
 1. Design and implement a list structure for storing a sparse matrix. Your implementation should provide  
procedures for inserting, deleting, changing, and reading matrix elements.
 2. Implement a fifo queue by a circular list  using only one external pointer f and a sentinel. f always points to  
the sentinel and provides access to the head and tail of the queue.
 3. Implement a double-ended queue (deque) by a doubly linked list.
 4. Binary search trees and sorting  A binary search tree given by the following declarations is used to manage  
a set of integers:
type nptr = ^node 
node = record  L, R: nptr;  x: integer  end; 
var root: nptr;
The empty tree is represented as root = nil.",algorithms and data structures.pdf
"a set of integers:
type nptr = ^node 
node = record  L, R: nptr;  x: integer  end; 
var root: nptr;
The empty tree is represented as root = nil.
(a) Draw the result of inserting the sequence 6, 15, 4, 2, 7, 12, 5, 18 into the empty tree.
Algorithms and Data Structures 237  A Global Text",algorithms and data structures.pdf
"21. List structures
(b) Write a procedure smallest(var x: integer); which returns the smallest number stored in the tree, and a  
procedure  remove  smallest;  which  deletes  it.  If  the  tree  is  empty  both  procedures  should  call  a 
procedure message('tree is empty');
(c) Write a procedure sort; that sorts the numbers stored in var a: array[1 .. n] of integer; by inserting the  
numbers into a binary search tree, then writing them back to the array in sorted order as it traverses  
the tree.
(d) Analyze the asymptotic time complexity of 'sort' in a typical and in the worst case.
(e) Does this approach lead to a sorting algorithm of time complexity Θ (ν •   λ ογ ν) 
 5. Extend the implementation of a dictionary as a binary search tree in the “Binary search trees” section to  
support the operations 'succ' and 'pred' as defined in chapter 19 in the section “Dictionary”.",algorithms and data structures.pdf
"support the operations 'succ' and 'pred' as defined in chapter 19 in the section “Dictionary”.
 6. Insertion and deletion in AVL-trees: Starting with an empty AVL-tree, insert 1, 2, 5, 6, 7, 8, 9, 3, 4, in this  
order. Draw the AVL-tree after each insertion. Now delete all elements in the opposite order of insertion  
(i.e. in last-in-first-out order). Does the AVL-tree go through the same states as during insertion but in  
reverse order?
 7. Implement an AVL-tree supporting the dictionary operations 'insert', 'delete', 'member', 'pred', and 'succ'.
 8. Explain how to find the smallest element in an (a,b)-tree and how to find  the predecessor of a given  
element in an (a,b)-tree.
 9. Implement a dictionary as a B-tree. 
238",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
22. Address computation 
Learning objectives:
• hashing
• perfect hashing
• collision resolution methods: separate chaining, coalesced chaining, open addressing (linear probing and 
double hashing)
• deletions degrade performance of a hash table
• Performance does not depend on the number of data elements stored but on the load factor of the hash table.
• randomization: transform unknown distribution into a uniform distribution
• Extendible hashing uses a radix tree to adapt the address range dynamically to the contents to be stored; 
deletions do not degrade performance.
• order-preserving extendible hashing
Concepts and terminology
The term address computation (also hashing, hash coding, scatter storage, or key-to-address transformations) 
refers to many search techniques that aim to assign an address of a storage cell to any key value x by means of a",algorithms and data structures.pdf
"refers to many search techniques that aim to assign an address of a storage cell to any key value x by means of a  
formula that depends on x only. Assigning an address to x independently of the presence or absence of other key  
values leads to faster access than is possible with the comparative search techniques discussed in earlier chapters.  
Although this goal cannot always be achieved, address computation does provide the fastest access possible in  
many practical situations.
We use the following concepts and terminology ( Exhibit 22.1). The home address a of x is obtained by means of  
a hash function h that maps the key domain X into the address space A [i.e. a = h(x)]. The address range is A = {0,  
1, … , m – 1}, where m is the number of storage cells available. The storage cells are represented by an array T[0 .. m  
– 1], the hash table; T[a] is the cell addressed by a  ∈  A. T[h(x)] is the cell where an element with key value x is",algorithms and data structures.pdf
"– 1], the hash table; T[a] is the cell addressed by a  ∈  A. T[h(x)] is the cell where an element with key value x is  
preferentially stored, but alas, not necessarily.
Exhibit 22.1: The hash function h maps a (typically large) key domain X into a (much smaller) 
address space A.
Algorithms and Data Structures 239  A Global Text",algorithms and data structures.pdf
"22. Address computation 
Each cell has a capacity of b > 0 elements; b stands for bucket capacity. The number n of elements to be stored  
is therefore bounded by m · b. Two cases are usefully distinguished, depending on whether the hash table resides on 
disk or in central memory:
1. Disk or other secondary storage device: Considerations of efficiency suggest that a bucket be identified with  
a physical unit of transfer, typically a disk block. Such a unit is usually large compared to the size of an  
element, and thus b > 1.
2. Main memory: Cell size is less important, but the code is simplest if a cell can hold exactly one element (i.e.  
b = 1).
For simplicity of exposition we assume that b = 1 unless otherwise stated; the generalization to arbitrary b is  
straightforward.
The key domain X is normally much larger than the number n of elements to be stored and the number m of",algorithms and data structures.pdf
"straightforward.
The key domain X is normally much larger than the number n of elements to be stored and the number m of  
available cells T[a]. For example, a table used for storing a few thousand identifiers might have as its key domain  
the set of strings of length at most 10 over t he alphabet {'a', 'b', … , 'z', '0', … , '9'}; its cardinality is close to 36 10. 
Thus in general the function h is many-to-one: Different key values map to the same address.
The content to be stored is a sample from the key domain: It is not under the programmer's control and is  
usually not even known when the hash function and table size are chosen. Thus we must expect collisions, that is,  
events where more than b elements to be stored are assigned the same address. Collision resolution methods are 
designed to handle this case by storing some of the colliding elements elsewhere. The more collisions that occur, the",algorithms and data structures.pdf
"designed to handle this case by storing some of the colliding elements elsewhere. The more collisions that occur, the 
longer the search time. Since the number of collisions is a random event, the search time is a random variable.  
Hash tables are known for excellent average performance and for terrible worst-case performance, which, one  
hopes, will never occur.
Address computation techniques support the operations 'find' and 'insert' (and to a lesser extent also 'delete') in  
expected time O(1). This is a remarkable difference from all other data structures that we have discussed so far, in  
that the average time complexity does not depend on the number n of elements stored, but on the load factor λ = 
n / (m · b), or, for the special case b = 1: λ = n / m. Note that 0 ≤ λ ≤ 1.
Before we consider the typical case of a hash table, we illustrate these concepts in two special cases where  
everything is simple; these represent ideals rarely attainable.",algorithms and data structures.pdf
"everything is simple; these represent ideals rarely attainable.
The special case of small key domains 
If the number of possible key values is less than or equal to the number of available storage cells, h can map X  
one-to-one into or onto A. Everything is simple and efficient because collisions never occur. Consider the following  
example:
X = {'a', 'b', … , 'z'},  A = {0, … , 25}
h(x) = ord(x) – ord('a');  that is,
h('a') = 0,  h('b') = 1,  h('c') = 2,  …  ,  h('z') = 25.
Since h is one-to-one, each key value x is implied by its address h(x). Thus we need not store the key values  
explicitly, as a single bit (present / absent) suffices:
var  T: array[0 .. 25] of boolean;
function member(x): boolean;
begin  return(T[h(x)])  end;
240",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
procedure insert(x);
begin  T[h(x)] := true  end;
procedure delete(x);
begin  T[h(x)] := false  end;
The idea of collision-free address computation can be extended to large key domains through a combination of  
address computation and list processing techniques, as we will see in the chapter ""Metric data structures"".
The special case of perfect hashing: table contents known a priori 
Certain common applications require storing a set of elements that never changes. The set of reserved words of a  
programming language is an example; when the lexical analyzer of a compiler extracts an identifier, the first issue  
to be determined is whether this is a reserved word such as 'begin' or 'while', or whether it is programmer defined.  
The special case where the table contents are known a priori, and no insertions or deletions occur, is handled more",algorithms and data structures.pdf
"The special case where the table contents are known a priori, and no insertions or deletions occur, is handled more  
efficiently by special-purpose data structures than by a general dictionary.
If the elements x1, x2, … , xn to be stored are known before the hash table is designed, the underlying key domain  
is not as important as the set of actually occurring key values. We can usually find a table size m, not much larger  
than the number n of elements to be stored, and an easily evaluated hash function h that assigns to  each xi a unique 
address from the address space {0, … , m – 1}. It takes some trial and error to find such a perfect hash function  h 
for a given set of elements, but the benefit of avoiding collisions is well worth the effort—the code that implements a 
collision-free hash table is simple and fast. A perfect hash function works for a static table only —a single insertion,",algorithms and data structures.pdf
"collision-free hash table is simple and fast. A perfect hash function works for a static table only —a single insertion, 
after h has been chosen, is likely to cause a collision and destroy the simplicity of the concept and efficiency of the  
implementation. Perfect hash functions should be generated automatically by a program.
The following unrealistically small example illustrates typical approaches to designing a perfect hash table. The  
task gets harder as the number m of available storage cells is reduced toward the minimum possible, that is, the  
number n of elements to be stored.
Example
In designing a perfect hash table for the elements 17, 20, 24, 38, and 51, we look for arithmetic patterns. These  
are most easily detected by considering the binary representations of the numbers to be stored:
    5     4     3     2     1     0  bit position
170 1 0 0 0 1
200 1 0 1 0 0
240 1 1 0 0 0
381 0 0 1 1 0
511 1 0 0 1 1",algorithms and data structures.pdf
"5     4     3     2     1     0  bit position
170 1 0 0 0 1
200 1 0 1 0 0
240 1 1 0 0 0
381 0 0 1 1 0
511 1 0 0 1 1
We observe that the least significant three bits identify each element uniquely. Therefore, the hash function h(x)  
= x mod 8 maps these five elements collision-free into the address space A = {0, … , 6}, with m = 7 and two empty  
cells. An attempt to further economize space leads us to observe that the bits in positions 1, 2, and 3, with weights 2,  
4, and 8 in the binary number representation, also identify each element uniquely, while ranging over the address  
space of minimal size A = {0, … , 4}. The function h(x) = (x div 2) mod 8 extracts these three bits and assigns the  
following addresses:
X:1720243851
A:0 2 4 3 1
Algorithms and Data Structures 241  A Global Text",algorithms and data structures.pdf
"22. Address computation 
A perfect hash table has to store each element explicitly, not just a bit (present/absent). In the example above,  
the elements 0, 1, 16, 17, 32, 33, … all map into address 0, but only 17 is present in the table. The access function  
'member(x)' is implemented as a single statement:
return ((h(x) ≤ 4) cand (T[h(x)] = x));
The boolean operator 'cand' used here is understood to be the  conditional and : Evaluation of the expression  
proceeds from left to right and stops as soon as its value is determined. In our example, h(x) > 4 suffices to assign  
'false' to the expression (h(x) ≤ 4) and (T[h(x)] = x). Thus the 'cand' operator guarantees that the table declared as:
var  T: array[0 .. 4] of element;
is accessed within its index bounds.
For table contents of realistic size it is impractical to construct a perfect hash function manually —we need a",algorithms and data structures.pdf
"For table contents of realistic size it is impractical to construct a perfect hash function manually —we need a  
program to search exhaustively through the large space of functions. The more slack m – n we allow, the denser is  
the population of perfect functions and the quicker we will find one. [Meh 84a] presents analytical results on the  
complexity of finding perfect hash functions.
Exercise: perfect hash tables
Design several perfect hash tables for the content {3, 13, 57, 71, 82, 93}.
Solution
Designing a perfect hash table is like answering a question of the type: What is the next element in the sequence  
1, 4, 9, … ? There are infinitely many answers, but some are more elegant than others. Consider:
h                                                       3  13 57  71 82 93 Address range
(x div 3) mod 7                                1 4 5 2 6 3 [1 .. 6]
x mod 13                                            3 0 5 6   4    2   [0 .. 6]",algorithms and data structures.pdf
"(x div 3) mod 7                                1 4 5 2 6 3 [1 .. 6]
x mod 13                                            3 0 5 6   4    2   [0 .. 6]
(x div 4) mod 8                                0 3  6 1 4 7 [0 .. 7]
if  x = 71  then  4  else  x mod 7   3 6 1 4 5 2 [1 .. 6]
Conventional hash tables: collision resolution 
In contrast to the special cases discussed, most applications of address computation present the data structure  
designer with greater uncertainties and less favorable conditions. Typically, the underlying key domain is much  
larger than the available address range, and not much is known about the elements to be stored. We may have an  
upper bound on n, and we may know the probability distribution that governs the random sample of elements to be  
stored. In setting up a customer list for a local business, for example, the number of customers may be bounded by",algorithms and data structures.pdf
"stored. In setting up a customer list for a local business, for example, the number of customers may be bounded by  
the population of the town, and the distribution of last names can be obtained from the telephone directory —many 
names will start with H and S, hardly any with Q and Y. On the basis of such information, but in ignorance of the  
actual table contents to be stored, we must choose the size m of the hash table and design the hash function h that  
maps the key domain X into the address space A= {0, … , m – 1}. We will then have to live with the consequences of  
these decisions, at least until we decide to rehash: that is, resize the table, redesign the hash function, and reinsert  
all the elements that we have stored so far.
Later sections present some pragmatic advice on the choice of h; for now, let us assume that an appropriate hash 
function is available. Regardless of how smart a hash function we have designed, collisions (more than b elements",algorithms and data structures.pdf
"function is available. Regardless of how smart a hash function we have designed, collisions (more than b elements  
share the same home address of a bucket of capacity b) are inevitable in practice. Thus hashing requires techniques  
242",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
for  handling  collisions.  We present  the  three  major collision  resolution techniques  in  use:  separate  chaining, 
coalesced chaining, and open addressing. The two techniques called chaining call upon list processing techniques to 
organize overflowing elements. Separate chaining is used when these lists live in an overflow area distinct from the  
hash table proper;  coalesced chaining  when the lists live in unused parts of the table.  Open addressing  uses  
address  computation  to  organize  overflowing  elements.  Each  of  these  three  techniques  comes  in  different 
variations; we illustrate one typical choice.
Separate chaining 
The memory allocated to the table is split into a primary and an overflow area. Any overflowing cell or bucket in 
the primary area is the head of a list, called the  overflow chain , that holds all elements that overflow from that",algorithms and data structures.pdf
"the primary area is the head of a list, called the  overflow chain , that holds all elements that overflow from that  
bucket. Exhibit 22.2 shows six elements inserted in the or der x1, x2, … . The first arrival resides at its home address;  
later ones get appended to the overflow chain.
Exhibit 22.2: Separate chaining handles collisions in a separate overflow 
area.
Separate chaining is easy to understand: insert, delete, and search operations are simple. In contrast to other  
collision  handling  techniques,  this  hybrid  between  address  computation  and  list  processing  has  two  major 
advantages: (1) deletions do not degrade the performance of the hash table, and (2) regardless of the number m of  
home addresses, the hash table will not overflow until the entire memory is exhausted. The size m of the table has a  
critical influence on the performance. If m « n, overflow chains are long and we have essentially a list processing",algorithms and data structures.pdf
"critical influence on the performance. If m « n, overflow chains are long and we have essentially a list processing  
technique that does not support direct access. If m » n, overflow chains are short but we waste space in the table.  
Even for the practical choice m ≈ n, separate chaining has some disadvantages:
• Two different accessing techniques are required.
• Pointers take up space; this may be a significant overhead for small elements.
Algorithms and Data Structures 243  A Global Text",algorithms and data structures.pdf
"22. Address computation 
• Memory is partitioned into two separate areas that do not share space: If the overflow area is full, the entire  
table is full, even if there is still space in the array of home cells. This consideration leads to the next  
technique.
Coalesced chaining 
The chains that emanate from overflowing buckets are stored in the empty space in the hash table rather than in  
a separate overflow area ( Exhibit 22.3). This has the advantage that all available space is utilized fully (except for  
the overhead of the pointers). However, managing the space shared between the two accessing techniques gets  
complicated.
Exhibit 22.3: Coalesced chaining handles collisions by building lists that share memory with the hash 
table.
The next technique has similar advantages (in addition, it incurs no overhead for pointers) and disadvantages;  
all things considered, it is probably the best collision resolution technique.
Open addressing",algorithms and data structures.pdf
"all things considered, it is probably the best collision resolution technique.
Open addressing 
Assign to each element x ∈  X a probe sequence a0 = h(x), a1, a2, … of addresses that fills the entire address range  
A. The intention is to store x preferentially at a 0, but if T[a 0] is occupied then at a 1, and so on, until the first empty  
cell is encountered along the probe sequence. The occupied cells along the probe sequence are called the collision 
path of x—note that the collision path is a prefix of the probe sequence. If we enforce the invariant:
If x is in the table at T[a] and if i precedes a in the probe sequence for x , then T[i] is occupied. The following fast  
and simple loop that travels along the collision path can be used to search for x:
a := h(x);
while  T[a] ≠ x  and  T[a] ≠ empty  do
a := (next address in probe sequence);
Let us work out the details so that this loop terminates correctly and the code is as concise and fast as we can  
make it.",algorithms and data structures.pdf
"Let us work out the details so that this loop terminates correctly and the code is as concise and fast as we can  
make it.
The probe sequence is defined by formulas in the program (an example of an implicit data structure) rather than  
by pointers in the data as is the case in coalesced chaining.
Example: linear probing
ai+1 = (ai + 1) mod m  is the simplest possible formula. Its only disadvantage is a phenomenon called  clustering. 
Clustering arises when the collision paths of many elements in the table overlap to a large extent, as is likely to  
happen in linear probing. Once elements have collided, linear probing will store them in consecutive cells. All  
elements  that  hash  into  this  block  of  contiguous  occupied  cells  travel  along  the  same  collision  path,  thus 
244",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
lengthening this block; this in turn increases the probability that future elements will hash into this block. Once this  
positive feedback loop gets started, the cluster keeps growing.
Double hashing  is a special type of open addressing designed to alleviate the clustering problem by letting  
different elements travel with steps of different size. The probe sequence is defined by the formulas
a0 = h(x),  δ = g(x) > 0,  ai+1 = (ai + δ ) mod m,   m prime
g is a second hash function that maps the key space X into [1 .. m – 1].
Two important important details must be solved:
• The probe sequence of each element must span the entire address range A. This is achieved if m is relatively 
prime to every step size δ, and the easiest way to guarantee this condition is to choose m prime.
• The termination condition of the search loop above is: T[a] = x or T[a] = empty. An unsuccessful search (x",algorithms and data structures.pdf
"• The termination condition of the search loop above is: T[a] = x or T[a] = empty. An unsuccessful search (x  
not in the table) can terminate only if an address a is generated with T[a] = empty. We have already insisted 
that each probe sequence  generates all addresses in A. In addition, we must guarantee  that the table  
contains at least one empty cell at all times—this serves as a sentinel to terminate the search loop.
The  following  declarations  and  procedures  implement  double  hashing.  We  assume  that  the  comparison 
operators = and ≠ are defined on X, and that X contains a special value 'empty', which differs from all values to be  
stored in the table. For example, a string of blanks might denote 'empty' in a table of identifiers. We choose to  
identify an unsuccessful search by simply returning the address of an empty cell.
const m = … ;  { size of hash table - must be prime! }
empty = … ;
type key = … ;  addr = 0 .. m – 1;  step = 1 .. m – 1;",algorithms and data structures.pdf
"const m = … ;  { size of hash table - must be prime! }
empty = … ;
type key = … ;  addr = 0 .. m – 1;  step = 1 .. m – 1;
var T: array[addr] of key;
n: integer;  { number of elements currently stored in T }
function h(x: key): addr;  { hash function for home address }
function g(x: key): step;  { hash function for step }
procedure init; 
var  a: addr;
begin
n := 0;
for a := 0 to m – 1 do  T[a] := empty
end;
function find(x: key): addr; 
var  a: addr;  d: step;
begin
a := h(x);  d := g(x);
while  (T[a] ≠ x) and (T[a] ≠ empty)  do  a := (a + d) mod m;
return(a)
end;
function insert(x: key): addr; 
var  a: addr;  d: step;
begin
a := h(x);  d := g(x);
while  T[a] ≠ empty  do  begin
if  T[a] = x  then  return(a);
a := (a + d) mod m
end;
if  n < m – 1  then  { n := n + 1;  T[a] := x }  else  err-
msg('table is full');
Algorithms and Data Structures 245  A Global Text",algorithms and data structures.pdf
"22. Address computation 
return(a)
end;
Deletion of elements creates problems, as is the case in many types of hash tables. An element to be deleted  
cannot simply be replaced by 'empty', or else it might break the collision paths of other elements still in the table—  
recall the basic invariant on which the correctness of open addressing is based. The idea of rearranging elements in  
the table so as to refill a cell that was emptied but needs to remain full is quickly abandoned as too complicated —if 
deletions are numerous, the programmer ought to choose a data structure that fully supports deletions, such as  
balanced trees implemented as list structures. A limited number of deletions can be accommodated in an open  
address hash table by using the following technique.
At any time, a cell is in one of three states:
• empty (was never occupied, the initial state of all cells)
• occupied (currently)
• deleted (used to be occupied but is currently free)",algorithms and data structures.pdf
"• empty (was never occupied, the initial state of all cells)
• occupied (currently)
• deleted (used to be occupied but is currently free)
A cell in state 'empty' terminates the find loop; a cell in state 'empty' or in state 'deleted' terminates the insert  
loop. The state diagram shown in Exhibit 22.4 describes the transitions possible in the lifetime of a cell. Deletions  
degrade the performance of a hash table, because a cell, once occupied, never returns to the virgin state 'empty'  
which alone terminates an unsuccessful find. Even if an equal number of insertions and deletions keeps a hash table 
at a low load factor λ, unsuccessful finds will ultimately scan the entire table, as all cells drift into one of the states  
'occupied' or 'deleted'. Before this occurs, the table ought to be rehashed; that is, the contents are inserted into a  
new, initially empty table.
Exhibit 22.4: This state diagram describes possible life cycles of a cell: Once occupied, a cell",algorithms and data structures.pdf
"new, initially empty table.
Exhibit 22.4: This state diagram describes possible life cycles of a cell: Once occupied, a cell 
will never again be as useful as an empty cell.
Exercise: hash table with deletions
Modify the program above to implement double hashing with deletions.
Choice of hash function: randomization
In conventional terminology, hashing is based on the concept of randomization. The purpose of randomizing  
is  to transform an  unknown distribution  over the key domain  X into a uniform  distribution,  and  to turn  
consecutive samples that may be dependent into independent samples. This task appears to call for magic, and  
indeed,  there  is  little  or  no  mathematics  that  applies  to  the  construction  of  hash  functions;  but  there  are 
commonsense  observations  worth  remembering.  These  observations  are  primarily  ""don'ts"".  They  stem  from",algorithms and data structures.pdf
"commonsense  observations  worth  remembering.  These  observations  are  primarily  ""don'ts"".  They  stem  from 
properties that sets of elements we wish to store frequently possess, and thus are based on  some knowledge about 
the populations to be stored. If we assumed strictly nothing about these populations, there would be little to say  
about hash functions: an  order-preserving  proportional mapping  of X  into A  would  be  as  good  as  any  other 
function. But in practice it is not, as the following examples show.
246",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
1. A Fortran compiler might use a hash table to store the set of identifiers it encounters in a program being  
compiled. The rules of the language and human habits conspire to make this set a highly biased sample  
from  the set of legal  Fortran  identifiers.  Example: Integer  variables  begin with I, J, K,  L, M, N; this  
convention  is  likely  to generate  a  cluster of identifiers  that begin with  one  of these  letters.  Example: 
Successive identifiers encountered cannot be considered independent samples: If X and Y have occurred,  
there is a higher chance for Z to follow than for WRKHG.  Example: Frequently, we see sequences of  
identifiers or statement numbers whose character codes form arithmetic progressions, such as A1, A2, A3,  
… or 10, 20, 30, … .
2. All file systems require or encourage the use of naming conventions, so that most file names begin or end",algorithms and data structures.pdf
"… or 10, 20, 30, … .
2. All file systems require or encourage the use of naming conventions, so that most file names begin or end  
with one of just a few prefixes or suffixes, such as ···.SYS, ···.BAK, ···.OBJ. An individual user, or a user  
community, is likely to generate additional conventions, so that most file names might begin, for example,  
with  the  initials  of  the  names  of  the  people  involved.  The  files  that  store  this  text,  for  example,  are 
structured according to 'part' and 'chapter', so we are currently in file P5 C22. In some directories, file  
names might be sorted alphabetically, so if they are inserted into a table in order, we process a monotonic  
sequence.
The purpose of a hash function is to break up all regularities that might be present in the set of elements to  
be stored. This is most reliably achieved by ""hashing"" the elements, a word for which the dictionary offers",algorithms and data structures.pdf
"be stored. This is most reliably achieved by ""hashing"" the elements, a word for which the dictionary offers  
the following explanations: (1) from the French  hache, ""battle-ax""; (2) to chop into small pieces; (3) to  
confuse, to muddle. Thus, to approximate the elusive goal of randomization, a hash function destroys  
patterns, including, unfortunately, the order < defined on X. Hashing typically proceeds in two steps.
1. Convert the element x into a number #(x). In most cases #(x) is an integer, occasionally, it is a real  
number 0 ≤ #(x) < 1. Whenever possible, this conversion of x into #(x) involves no action at all: The  
representation of x, whatever type x may be, is reinterpreted as the representation of the number #(x).  
When x is a variable-length item, for example a string, the representation of x is partitioned into pieces  
of suitable length that are ""folded"" on top of each other. For example, the four-letter word x = 'hash' is",algorithms and data structures.pdf
"of suitable length that are ""folded"" on top of each other. For example, the four-letter word x = 'hash' is  
encoded one letter per byte using the 7-bit ASCII code and a leading 0 as 01101000 01100001 01110011  
01101000. It may be folded to form a 16-bit integer by exclusive-or of the leading pair of bytes with the  
trailing pair of bytes:
   0110100001100001
xor  01110011011010000
  0001101100001001 which represents #(x) = 27 · 28 + 9 = 6921.
Such folding, by itself, is not hashing. Patterns in the representation of elements easily survive folding.  
For example, the leading 0 we have used to pad the 7-bit ASCII code to an 8-bit byte remains a zero  
regardless of x. If we had padded with a trailing zero, all #(x) would be even. Because #(x) often has the  
same representation as x, or a closely related one, we drop #() and use x slightly ambiguously to denote  
both the original element and its interpretation as a number.",algorithms and data structures.pdf
"both the original element and its interpretation as a number.
2. Scramble x [more precisely, #(x)] to obtain h(x). Any scrambling technique is a sensible try, as long as  
it avoids fairly obvious pitfalls. Rules of thumb:
Algorithms and Data Structures 247  A Global Text",algorithms and data structures.pdf
"22. Address computation 
▪ Each bit of an address h(x) should depend on all bits of the key value x. In particular, don't ignore  
any part of x in computing h(x). Thus h(x) = x mod 213 is suspect, as only the least significant 13 bits 
of x affect h(x).
▪ Make sure that arithmetic progressions such as Ch1, Ch2, Ch3, … get broken up rather than being  
mapped into arithmetic progressions. Thus h(x) = x mod k, where k is significantly smaller than the 
table size m, is suspect.
▪ Avoid any function that cannot produce a uniform distribution of addresses. Thus h(x) = x 2 is  
suspect; if x is uniformly distributed in [0, 1], the distribution of x2 is highly skewed.
A hash function must be fast and simple. All of the desiderata above are obtained by a hash function of the type:
h(x) = x mod m
where m is the table size and a prime number, and x is the key value interpreted as an integer.",algorithms and data structures.pdf
"h(x) = x mod m
where m is the table size and a prime number, and x is the key value interpreted as an integer.
No hash function is guaranteed to avoid the worst case of hashing, namely, that all elements to be stored collide  
on one address (this happens here if we store only multiples of the prime m). Thus a hash function must be judged  
in relation to the data it is being asked to store, and usually this is possible only after one has begun using it.  
Hashing provides a perfect example for the injunction that the programmer must think about the data, analyze its  
statistical properties, and adapt the program to the data if necessary.
Performance analysis 
We analyze open addressing without deletions assuming that each address  αi is chosen independently of all  
other addresses from a uniform distribution over A. This assumption is reasonable for double hashing and leads to",algorithms and data structures.pdf
"other addresses from a uniform distribution over A. This assumption is reasonable for double hashing and leads to  
the conclusion that the average cost for a search operation in a hash table is O(1) if we consider the load factor λ to 
be constant. We analyze the average number of probes executed as a function of  λ in two cases: U( λ) for an  
unsuccessful search, and S(λ) for a successful search.
Let pi denote the probability of using exactly i probes in an unsuccessful search. This event occurs if the first I – 
1 probes hit occupied cells, and the i-th probe hits an empty cell: pi = λi–1 · (1 – λ). Let qi denote the probability that 
at least i probes are used in an unsuccessful search; this occurs if the first i – 1 inspected cells are occupied: qi = λi–1. 
qi can also be expressed as the sum of the probabilities that we probe exactly j cells, for j running from i to m. Thus  
we obtain",algorithms and data structures.pdf
"qi can also be expressed as the sum of the probabilities that we probe exactly j cells, for j running from i to m. Thus  
we obtain 
The number of probes executed in a successful search for an element x equals the number of probes in an  
unsuccessful search for the same element x before it is inserted into the hash table. [Note: This holds only when  
elements are never relocated or deleted]. Thus the average number of probes needed to search for the i-th element  
inserted into the hash table is U((i – 1) / m), and S( λ) can be computed as the average of U( µ), for µ increasing in  
discrete steps from 0 to λ. It is a reasonable approximation to let µ vary continuously in the range from 0 to λ:
248",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 22.5 suggests that a reasonable operating range for a hash table keeps the load factor λ between 0.25 and 
0.75.  If  λ is  much smaller,  we waste  space, if it is larger than  75 per  cent, we get into a domain  where the  
performance degrades rapidly. Note: If all searches are successful, a hash table performs well even if loaded up to  
95 per cent—unsuccessful searching is the killer!
Table 22.1: The average number of probes per search grows rapidly as the load factor approaches 1.
λ 0.25 0.5 0.75 0.9 0.95 0.99
U(λ) 1.3 2.0 4.0 10.0 20.0 100.0
S(λ) 1.2 1.4 1.8 2.6 3.2 4.7
Exhibit 22.5: The average number of probes per search grows rapidly as the load factor approaches 1.
Thus the hash table designer should be able to estimate n within a factor of 2 —not an easy task. An incorrect  
guess may waste memory or cause poor performance, even table overflow followed by a crash. If the programmer",algorithms and data structures.pdf
"guess may waste memory or cause poor performance, even table overflow followed by a crash. If the programmer  
becomes aware that the load factor lies outside this range, she may rehash —change the size of the table, change the  
hash function, and reinsert all elements previously stored. 
Extendible hashing 
In contrast to standard  hashing  methods, extendible  forms  of hashing allow for the  dynamic  extension or  
shrinkage of the address range into which the hash function maps the keys. This has two major advantages: (1)  
Memory is allocated only as needed (it is unnecessary to determine the size of the address range a priori), and (2)  
deletion of elements does not degrade performance. As the address range changes, the hash function is changed in  
such a way that only a few elements are assigned a new address and need to be stored in a new bucket. The idea that",algorithms and data structures.pdf
"such a way that only a few elements are assigned a new address and need to be stored in a new bucket. The idea that 
makes this possible is to map the keys into a very large address space, of which only a portion is active at any given  
time.
Various extendible hashing methods differ in the way they represent and manage a smaller  active address  
range of variable size that is a subrange of a larger virtual address range. In the following we describe the method  
of extendible hashing that is especially well suited for storing data on secondary storage devices; in this case an  
address points to a physical block of secondary storage that can contain more than one element. An address is a bit  
string of maximum length k; however, at any time only a prefix of d bits is used. If all bit strings of length k are  
represented by a so-called radix tree  of height k, the active part of all bit strings is obtained by using only the upper",algorithms and data structures.pdf
"represented by a so-called radix tree  of height k, the active part of all bit strings is obtained by using only the upper  
d levels of the tree (i.e. by cutting the tree at level d). Exhibit 22.6 shows an example for d = 3.
Algorithms and Data Structures 249  A Global Text",algorithms and data structures.pdf
"22. Address computation 
Exhibit 22.6: Address space organized as a binary radix tree.
The radix tree shown in  Exhibit 22.6  (without the nodes that have been clipped) describes an active address  
range with addresses {00, 010, 011, 1} that are considered as bit strings or binary numbers. To each active node  
with address s there corresponds a bucket B that can store b records. If a new element has to be inserted into a full  
bucket B, then B is split: Instead of B we find two twin buckets B0 and B1 which have a one bit longer address than B, 
and the elements stored in B are distributed among B 0 and B 1 according to this bit. The new radix tree now has to  
point to the two data buckets B 0 and B1 instead of B; that is, the active address range must be extended locally (by  
moving the broken line in Exhibit 22.6). If the block with address 00 overflows, two new twin blocks with addresses",algorithms and data structures.pdf
"moving the broken line in Exhibit 22.6). If the block with address 00 overflows, two new twin blocks with addresses  
000 and 001 will be created which are represented by the corresponding nodes in the tree. If the overflowing bucket 
B has depth d, then d is incremented by 1 and the radix tree grows by one level.
In extendible hashing the clipped radix tree is represented by a directory that is implemented by an array. Let d  
be the maximum number of bits that are used in one of the bit strings for forming an address; in the example above, 
d = 3. Then the directory consists of 2 d entries. Each entry in this directory corresponds to an address and points to  
a physical data bucket which contains all elements that have been assigned this address by the hash function h. The  
directory for the radix tree in Exhibit 22.6 looks as shown in Exhibit 22.7.
Exhibit 22.7: The active address range of the tree in Exhibit 22.6 implemented as an array.",algorithms and data structures.pdf
"Exhibit 22.7: The active address range of the tree in Exhibit 22.6 implemented as an array.
The bucket with address 010 corresponds to a node on level 3 of the radix tree, and there is only one entry in the  
directory corresponding to this bucket. If this bucket overflows, the directory and data buckets are reorganized as  
shown in Exhibit 22.8. Two twin buckets that jointly contain fewer than b elements are merged into a single bucket.  
This keeps the average bucket occupancy at a high 70 per cent even in the presence of deletions, as probabilistic  
analysis predicts and simulation results confirm. Bucket merging may lead to halving the directory. A formerly  
large file that shrinks to a much smaller size will have its directory shrink in proportion. Thus extendible hashing,  
unlike conventional hashing, suffers no permanent performance degradation under deletions.
250",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 22.8: An overflowing bucket may trigger doubling of the directory.
A virtual radix tree: order-preserving extendible hashing
Hashing, in the usual sense of the word, destroys structure and thus buys uniformity at the cost of order.  
Extendible hashing, on the other hand, is practical without randomization and thus needs not accept its inevitable  
consequence,  the  destruction  of  order.  A  uniform  distribution  of  elements  is  not  nearly  as  important: 
Nonuniformity causes the directory to be deeper and thus larger than it would be for a uniform distribution, but it  
affects neither access time nor bucket occupancy. And the directory is only a small space overhead on top of the  
space required to store the data: It typically contains only one or a few pointers, say a dozen bytes, per data bucket",algorithms and data structures.pdf
"space required to store the data: It typically contains only one or a few pointers, say a dozen bytes, per data bucket  
of, say 1k bytes; it adds perhaps a few percent to the total space requirement of the table, so its growth is not  
critical. Thus extendible hashing remains feasible when the identity is used as the address computation function h,  
in which case data is accessible and can be processed sequentially in the order ≤ defined on the domain X.
When h preserves order, the word hashing seems out of place. If the directory resides in central memory and the 
data buckets on disk, what we are implementing is a virtual memory organized in the form of a radix tree of  
unbounded size. In contrast to conventional virtual memory, whose address space grows only at one end, this  
address space can grow anywhere: It is a virtual radix tree.
As an example, consider the domain X of character strings up to length 32, say, and assume that elements to be",algorithms and data structures.pdf
"As an example, consider the domain X of character strings up to length 32, say, and assume that elements to be  
stored are sampled according to the distribution of the first letter in English words. We obtain an approximate  
distribution by counting pages in a dictionary ( Exhibit 22.9). Encode the blank as 00000, 'a' as 00001, up to 'z' as  
11011, so that 'aah', for example, has the code 00001 00001 01000 00000 … (29 quintuples of zeros pad 'aah'  
to32letters). This address computation function h is almost an identity: It maps {' ', 'a', … , 'z'} 32 one-to-one into {0, 
1}160.  Such  an  order-preserving  address  computation  function  supports  many  useful  types  of  operations:  for 
example, range queries such as ""list in alphabetic order all the words stored from 'unix' to 'xinu' "".
Algorithms and Data Structures 251  A Global Text",algorithms and data structures.pdf
"22. Address computation 
Exhibit 22.9: Relative frequency of words beginning with a given letter in Webster's dictionary.
If there is one page of words starting with X for 160 pages of words starting with S, this suggests that if our  
active address space is partitioned into equally sized intervals, some intervals may be populated 160 times more  
densely than others. This translates into a directory that may be 160 times larger than necessary for a uniform  
distribution, or, since directories grow as powers of 2, may be 128 or 256 times larger. This sounds like a lot but  
may well be bearable, as the following estimates show.
Assume that we store 105 records on disk, with an avera ge occupancy of 100 records per bucket, requiring about  
1000 buckets. A uniform distribution generates a directory with one entry per bucket, for a total of 1k entries, say  
2k  or  4k  bytes.  The  nonuniform  distribution  above  requires  the  same  number  of  buckets,  about  1,000,  but",algorithms and data structures.pdf
"2k  or  4k  bytes.  The  nonuniform  distribution  above  requires  the  same  number  of  buckets,  about  1,000,  but 
generates a directory of 256k entries. If a pointer requires 2 to 4 bytes, this amounts to 0.5 to 1 Mbyte. This is less of  
a memory requirement than many applications require on today's personal computers. If the application warrants  
it (e.g. for an on-line reservation system) 1 Mbyte of memory is a small price to pay.
Thus we see that for large data sets, extendible hashing approximates the ideal characteristics of the special case  
we discussed in this chapter's section on “the special case of small key domains”. All it takes is a disk and a central  
memory of a size that is standard today but was practically infeasible a decade ago, impossible two decades ago, and 
unthought of three decades ago.
Exercises and programming projects
1. Design a perfect hash table for the elements 1, 10, 14, 20, 25, and 26.",algorithms and data structures.pdf
"unthought of three decades ago.
Exercises and programming projects
1. Design a perfect hash table for the elements 1, 10, 14, 20, 25, and 26.
2. The six names AL, FL, GA, NC, SC and VA must be distinguished from all other ordered pairs of uppercase  
letters. To solve this problem, these names are stored in the array T such that they can easily be found by  
means of a hash function h.
type addr = 0 .. 7;
pair = record c1, c2: 'A' .. 'Z' end;
var T: array [addr] of pair;
(a) Write a 
function h (name: pair): adr; 
which maps the six names onto different addresses in the range 'adr'.
252",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
(b) Write a 
procedure initTable; 
which initializes the entries of the hash table T.
(c) Write a
 function member (name: pair): boolean; 
which returns for any pair of uppercase letters whether it is stored in T.
3. Consider the hash function h(x) = x mod 9 for a table having nine entries. Collisions in this hash table are  
resolved by coalesced chaining. Demonstrate the insertion of the elements 14, 19, 10, 6, 11, 42, 21, 8, and 1.
4. Consider inserting the keys 14, 1, 19, 10, 6, 11, 42, 21, 8, and 17 into a hash table of length m = 13 using open  
addressing with the hash function h(x) = x mod m. Show the result of inserting these elements using
(a) Linear probing.
(b) Double hashing with the second hash function g(x) = 1 + x mod (m+1).
5. Implement a dictionary  supporting the operations 'insert', 'delete', and  'member' as a hash table with  
double hashing.",algorithms and data structures.pdf
"5. Implement a dictionary  supporting the operations 'insert', 'delete', and  'member' as a hash table with  
double hashing.
6. Implement a dictionary supporting the operations 'insert', 'delete', 'member', 'succ', and 'pred' by order-
preserving extendible hashing.
Algorithms and Data Structures 253  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
23. Metric data structures
Learning objectives:
• organizing the embedding space versus organizing its contents
• quadtrees and octtrees. grid file. two-disk-access principle
• simple geometric objects and their parameter spaces
• region queries of arbitrary shape
• approximation of complex objects by enclosing them in simple containers
Organizing the embedding space versus organizing its contents
Most of the data structures discussed so far organize the set of elements to be stored depending primarily, or  
even exclusively, on the relative values of these elements to each other and perhaps on their order of insertion into  
the data structure. Often, the only assumption made about these elements is that they are drawn from an ordered  
domain, and thus these structures support only comparative search techniques: the search argument is compared",algorithms and data structures.pdf
"domain, and thus these structures support only comparative search techniques: the search argument is compared  
against stored elements. The shape of data structures based on comparative search varies dynamically with the set  
of elements currently stored; it does not depend on the static domain from which these elements are samples. These 
techniques organize the particular contents to be stored rather than the embedding space.
The data structures discussed in this chapter mirror and organize the domain from which the elements are  
drawn—much of their structure is determined before the first element is ever inserted. This is typically done on the  
basis of fixed points of reference which are independent of the current contents, as inch marks on a measuring scale  
are independent of what is being measured. For this reason we call data structures that organize the embedding",algorithms and data structures.pdf
"are independent of what is being measured. For this reason we call data structures that organize the embedding  
space metric data structures . They are of increasing importance, in particular for spatial data, such as needed in  
computer-aided design or geographic data processing. Typically, these domains exhibit a much richer structure  
than a mere order: In two- or three-dimensional Euclidean space, for example, not only is order defined along any 
line (not just the coordinate axes), but also  distance between any two points. Most queries about spatial data  
involve the absolute position of elements in space, not just their relative position among each other. A typical query  
in graphics, for example, asks for the first object intercepted by a given ray of light. Computing the answer involves  
absolute position (the location of the ray) and relative order (nearest along the ray). A data structure that supports",algorithms and data structures.pdf
"absolute position (the location of the ray) and relative order (nearest along the ray). A data structure that supports  
direct access to objects according to their position in space can clearly be more efficient than one based merely on  
the relative position of elements.
The  terms  ""organizing  the  embedding  space""  and  ""organizing  its  contents""  suggest  two  extremes  along  a 
spectrum of possibilities. As we have seen in previous chapters, however, many data structures are hybrids that  
combine features from distinct types. This is particularly true of metric data structures: They always have aspects of  
address computation needed to locate elements in space, and they often use list processing techniques for efficient  
memory utilization.
Algorithms and Data Structures 254  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Radix trees, tries 
We have encountered binary radix trees, and a possible implementation, in chapter 22 in the section “Extendible 
hashing”.  Radix  trees  with a  branching  factor, or  fan-out,  greater  than  2  are  ubiquitous. The  Dewey  decimal 
classification used in libraries is a radix tree with a fan-out of 10. The hierarchical structure of many textbooks,  
including this one, can be seen as a radix tree with a fan-out determined by how many subsections at depth d + 1  
are packed into a section at depth d.
As another example, consider tries, a type of radix tree that permits the  retrieval of variable-length data. As we  
traverse the tree, we check whether or not the node we are visiting has any successors. Thus the trie can be very  
long along certain paths. As an example, consider a trie containing words in the English language. In Exhibit 23.1",algorithms and data structures.pdf
"long along certain paths. As an example, consider a trie containing words in the English language. In Exhibit 23.1 
below, the four words 'a', 'at', 'ate', and 'be' are shown explicitly. The letter 'a' is a word and is the first letter of other  
words. The field corresponding to 'a' contains the value 1, signaling that we have spelled a valid word, and there is a  
pointer to longer words beginning with 'a'. The letter 'b' is not a word, thus is marked by a 0, but it is the beginning  
of many words, all found by following its pointer. The string 'aa' is neither a word nor the beginning of a word, so its  
field contains 0 and its pointer is 'nil'.
Exhibit 23.1: A radix tree over the alphabet of letters stores (prefixes of) words.
Only a few words begin with 'ate', but among these there are some long ones, such as 'atelectasis'. It would be  
wasteful to introduce eight additional nodes, one for each of the characters in 'lectasis', just to record this word,",algorithms and data structures.pdf
"wasteful to introduce eight additional nodes, one for each of the characters in 'lectasis', just to record this word,  
without making significant use of the fan-out of 26 provided at each node. Thus tries typically use an ""overflow  
technique"" to handle long entries: The pointer field of the prefix 'ate' might point to a text field that contains  
'(ate-)lectasis' and '(ate-)lier'.
Quadtrees and octtrees 
Consider a s quare recursively partitioned into quadrants. Exhibit 23.2 23.2 shows such a square partitioned to  
the depth of 4. There are 4 quadrants at depth 1, separated by the thickest lines; 4 · 4 (sub-)quadrants separated by  
slightly thinner lines; 4 3 (sub-sub-)quadrants separated by yet thinner lines; and finally, 4 4 = 256 leaf quadrants  
separated by the thinnest lines. The partitioning structure described is a quadtree, a particular type of radix tree of",algorithms and data structures.pdf
"separated by the thinnest lines. The partitioning structure described is a quadtree, a particular type of radix tree of  
fan-out 4. The root corresponds to the entire square, its 4 children to the 4 quadrants at depth 1, and so on, as  
shown in the Exhibit 23.2.
255",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 23.2: A quarter circle digitized on a 16 · 16 grid, and its representation as a 4-level quadtree.
A quadtree is the obvious two-dimensional analog of the one-dimensional binary radix tree we have seen.  
Accordingly, quadtrees are frequently used to represent, store, and process spatial data, such as images. The figure  
shows a quarter circle, digitized on a 16 · 16 grid of pixels. This image is most easily represented by a 16 · 16 array of  
bits. The quadtree provides an alternative representation that is advantageous for images digitized to a high level of  
resolution.  Most graphic  images  in  practice  are  digitized  on rectangular grids  of anywhere  from  hundreds  to 
thousands of pixels on a side: for example, 512 · 512. In a quadtree, only the largest quadrants of constant color  
(black or white, in our example) are represented explicitly; their subquadrants are implicit.",algorithms and data structures.pdf
"(black or white, in our example) are represented explicitly; their subquadrants are implicit.
The quadtree in Exhibit 23.2 is interpreted as follows. Of the four children of the root, the northwest quadrant,  
labeled 1, is simple: entirely white. This fact is recorded in the root. The other three children, labeled 0, 2, and 3,  
contain both black and white pixels. As their description is not simple, it is contained in three quadtrees, one for  
each quadrant. Pointers to these subquadtrees emanate from the corresponding fields of the root.
The southwestern quadrant labeled 2 in turn has four quadrants at depth 2. Three of these, labeled 2.0, 2.1, and  
2.2, are entirely white; no pointers emanate from the corresponding fields in this node. Subquadrant 2.3 contains  
both black and white pixels; thus the corresponding field contains a pointer to a sub-subquadtree.
In this discussion we have introduced a notation to identify every quadrant at any depth of the quadtree. The",algorithms and data structures.pdf
"In this discussion we have introduced a notation to identify every quadrant at any depth of the quadtree. The  
root is identified by the null string; a quadrant at depth d is uniquely identified by a string of d radix-4 digits. This  
string can be  interpreted in various ways as  a number expressed  in base  4. Thus accessing and processing a  
quadtree is readily reduced to arithmetic.
Breadth-first addressing
Label the root 0, its children 1, 2, 3, 4, its grand children 5 through 20, and so on, one generation after the other.
Algorithms and Data Structures 256  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
   0
1  2     3  4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Notice that the children of any node i are 4 · i + 1, 4 · i + 2, 4 · i + 3, 4 · i + 4. The parent of node i is ( i – 1) div 4. 
This is similar to the address computation used in the heap of “Implicit data structures”, a binary tree where each  
node i has children 2 · i  and 2 · i + 1; and the parent of node i is obtained as i div 2.
Exercise
The string of radi x 4 digits along a path from the root to any node is called the  path address  of this node.  
Interpret the path address as an integer, most significant digit first. These integers label the nodes at depth d > 0  
consecutively from 0 to 4 d – 1. Devise a formula that transforms the path address into the breadth-first address.  
This formula can be used to store a quadtree as a one-dimensional array.
Data compression
The representation of an image as a quadtree is sometimes much more compact than its representation as a bit",algorithms and data structures.pdf
"Data compression
The representation of an image as a quadtree is sometimes much more compact than its representation as a bit  
map. Two conditions must hold for this to be true:
1. The image must be fairly large, typically hundreds of pixels on a side.
2. The image must have large areas of constant value (color).
The quadtree for the quarter circle above, for example, has only 14 nodes. A bit map of the same image requires  
256 bits. Which representation requires more storage? Certainly the quadtree. If we store it as a list, each node  
must be able to hold four pointers, say 4 or 8 bytes. If a pointer has value 'nil', indicating that its quadrant needs no  
refinement, we need a bit to indicate the color of this quadrant (white or black), or a total of 4 bits. If we store the  
quadtree breadth-first, no pointers are needed as the node relationships are expressed by address computation;",algorithms and data structures.pdf
"quadtree breadth-first, no pointers are needed as the node relationships are expressed by address computation;  
thus a node is reduced to four three-valued fields ('white', 'black', or 'refine'), conveniently stored in 8 bits, or 1 byte.  
This  implicit  data  structure  will  leave  many  unused  holes  in  memory.  Thus  quadtrees  do  not  achieve  data 
compression for small images.
Octtrees
Exactly  the  same  idea  for  three-dimensional  space  as  quadtrees  are  for  two-dimensional  space:  A  cube  is 
recursively partitioned into eight octants, using three orthogonal planes.
Spatial data structures: objectives and constraints
Metric data structures are used primarily for storing spatial data, such as points and simple geometric objects  
embedded in a multidimensional space. The most important objectives a spatial data structure must meet include:
1. Efficient handling of large, dynamically varying data sets in interactive applications",algorithms and data structures.pdf
"1. Efficient handling of large, dynamically varying data sets in interactive applications
2. Fast access to objects identified in a fully specified query
3. Efficient processing of proximity queries and region queries of arbitrary shape
4. A uniformly high memory utilization
Achieving these objectives is subject to many constraints, and results in trade-offs.
Managing disks. By ""large data set"" we mean one that must be stored on disk; only a small fraction of the data  
can be kept in central memory at any one time. Many data structures can be used in central memory, but the choice  
is  much  more  restricted  when  it  comes  to  managing  disks  because  of  the  well-known  ""memory  speed  gap"" 
257",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
phenomenon.  Central  memory  is  organized  in  small  physical  units  (a  byte,  a  word)  with  access  times  of 
approximately 1 microsecond, 10 –6 second. Disks are organize in large physical blocks (512 bytes to 5kilobytes) with  
access times ranging from 10 to 100 milliseconds (10–2 to 10–1 second). Compared to central memory, a disk delivers 
data blocks typically 10 3 times larger with a delay 10 4 times greater. In terms of the  data rate delivered to the  
central processing unit:
the disk is a storage device whose effectiveness is within an order of magnitude of that of central memory. The large  
size of a physical disk block is a potential source of inefficiency that can easily reduce the useful data rate of a disk a  
hundredfold or a thousandfold. Accessing a couple of bytes on disk, say a pointer needed to traverse a list, takes",algorithms and data structures.pdf
"hundredfold or a thousandfold. Accessing a couple of bytes on disk, say a pointer needed to traverse a list, takes  
about as long as accessing the entire disk block. Thus the game of managing disks is about minimizing the number  
of disk accesses.
Dynamically varying data.  The majority of computer applications today are interactive. That means that  
insertions, deletions, and modifications of data are at least as frequent as operations that merely process fixed data.  
Data structures that entail a systematic degradation of performance with continued use (such as ever-lengthening  
overflow  chains,  or  an  ever-increasing  number  of  cells  marked  ""deleted""  in  a  conventional  hash  table)  are 
unsuitable.  Only  structures  that  automatically  adapt  their  shape  to  accommodate  ever-changing  contents  can 
provide uniform response times.
Instantaneous response. Interactive use of computers sets another major challenge for data management:",algorithms and data structures.pdf
"provide uniform response times.
Instantaneous response. Interactive use of computers sets another major challenge for data management:  
the  goal  of  providing  ""instantaneous  response""  to  a  fully  specified  query.  ""Fully""  specified  means  that  every 
attribute relevant for the search has been provided, and that at most one element satisfies the query. Imagine the  
user clicking an icon on the screen, and the object represented by the icon appears instantaneously. In human  
terms, ""instantaneous"" is a well-defined physiological quantity, namely, about of a second, the limit of human time  
resolution. Ideally, an interactive system retrieves any single element fully specified in a query within 0.1 second.
Two-disk-access principle. We have already stated that in today's technology, a disk access typically takes  
from tens of milliseconds. Thus the goal of retrieving any single element in 0.1 second translates into ""retrieve any",algorithms and data structures.pdf
"from tens of milliseconds. Thus the goal of retrieving any single element in 0.1 second translates into ""retrieve any  
element in at most a few disk accesses"". Fortunately, it turns out that useful data structure can be designed that  
access data in a two-step process: (1) access the correct portion of a directory, and (2) access the correct data  
bucket. Under the assumption that both data and directory are so large that they are stored on disk, we call this the  
two-disk-access principle.
Proximity queries and region queries of arbitrary shape . The simplest example of a proximity query is  
the operation 'next', which we have often encountered in one-dimensional data structure traversals: Given a pointer  
to an element, get the next element (the successor or the predecessor) accor ding to the order defined on the  
domain.  Another  simple  example  is  an  interval  or  range  query  such  as  ""get  all  x  between  13  and  17"".  This",algorithms and data structures.pdf
"domain.  Another  simple  example  is  an  interval  or  range  query  such  as  ""get  all  x  between  13  and  17"".  This 
generalizes directly to k-dimensional orthogonal range queries   such as the two-dimensional query ""get all (x 1, x2) 
with 13 ≤ x1 < 17 and 3 ≤ x2 < 4"". In geometric computation, for example, many other instances of proximity queries  
are important, such as the ""nearest neighbor "" (in any direction), or intersection queries among objects. Region  
queries of arbitrary shape (not just rectangular) are able to express a variety of geometric conditions.
Algorithms and Data Structures 258  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Uniformly high memory utilization.  Any data structure that adapts its shape to dynamically changing  
contents  is  likely  to  leave  ""unused  holes""  in  storage  space:  space  that  is  currently  unused,  and  that  cannot 
conveniently  be used  for other purposes because it is  fragmented. We have encountered this  phenomenon  in  
multiway trees such as B-trees and in hash tables. It is practically unavoidable that dynamic data structures use  
their allocated space to less than 100%, and an average space utilization of 50% is often tolerable. The danger to  
avoid is a built-in bias that drives space utilization toward 0 when the file shrinks—elements get deleted but their  
space is not relinquished. The grid file, to be discussed next, achieves an average memory utilization of about 70%  
regardless of the mix of insertions or deletions.
The grid file",algorithms and data structures.pdf
"regardless of the mix of insertions or deletions.
The grid file
The  grid  file  is  a  metric  data  structure  designed  to  store  points  and  simple  geometric  objects  in 
multidimensional space so as to achieve the objectives stated above. This section describes its architecture, access  
and update algorithms, and properties. More details can be found in [NHS 84]  and [Hin 85].
Scales, directory, buckets
Consider as  an example a two-dimensional domain: the Cartesian product  X1  × X2, where X 1 = 0 .. 1999 is a  
subrange of the integers, and X 2 = a .. z is the ordered set of the 26 characters of the English alphabet. Pairs of the  
form (x1, x2), such as (1988, w), are elements from this domain.
The bit map is a natural data structure for storing a set S of elements from X1  × X2. It may be declared as
var  T: array[X1, X2] of boolean;
with the convention that
T[x1, x2] = true  ⇔   (x1, x2) ∈  S.",algorithms and data structures.pdf
"var  T: array[X1, X2] of boolean;
with the convention that
T[x1, x2] = true  ⇔   (x1, x2) ∈  S.
Basic set operations are performed by direct access to the array element corresponding to an element: find(x 1, 
x2)  is  simply  the  boolean  expression  T[x1,  x2];  insert(x1,  x2)  is  equivalent  to  T[x1,  x2]:=  'true',  delete(x1,  x2)  is 
equivalent to T[x 1, x 2] := 'false'. The bit map for our small domain requires an affordable 52k bits. Bit maps for  
realistic examples are rarely affordable, as the following reasoning shows. First, consider that x and y are just keys  
of records that hold additional data. If space is reserved in the array for this additional data, an array  element is not 
a bit but as many bytes as are needed, and all the absent records, for elements (x 1, x2) ∉  S, waste a lot of storage.  
Second,  most  domains  are  much  larger  than  the  example  above:  the  three-dimensional  Euclidean  space,  for",algorithms and data structures.pdf
"Second,  most  domains  are  much  larger  than  the  example  above:  the  three-dimensional  Euclidean  space,  for 
example, with  elements (x, y, z) taken as triples of 32-bit integers, or 64-bit floating-point numbers, requires bit  
maps of about 1030 and 1060 bits, respectively. For comparison's sake: a large disk has about 1010 bits.
Since large bit maps are extremely sparsely populated, they are amenable to data compression. The grid file is  
best understood as a practical data compression technique that stores huge, sparsely populated bit maps so as to  
support direct access. Returning to our example, imagine a historical database indexed by the year of birth and the  
first letter of the name of scientists: thus we find 'John von Neumann' under (1903, v). Our database is pictured as a 
cloud of points in the domain shown in Exhibit 23.3; because we have more scientists (or at least, more records) in",algorithms and data structures.pdf
"cloud of points in the domain shown in Exhibit 23.3; because we have more scientists (or at least, more records) in  
recent years, the density increases toward the right. Storing this database implies packing the records into buckets  
of fixed capacity to hold c (e.g. c = 3) records. The figure shows the domain partitioned by orthogonal hyperplanes  
into box-shaped grid cells, none of which contains more than c points.
259",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 23.3: Cells of a grid partition adapt their size so that no cell is populated by more than c points.
A grid file for this database contains the following components:
• Linear scales show how the domain is currently partitioned.
• The directory is an array whose elements are in one-to-one correspondence with the grid cells; each entry  
points to a data bucket that holds all the records of the corresponding grid cell.
Access to the record (1903, v) proceeds through three steps:
1. Scales transform key values to array indices: (1903, v) becomes (5, 4). Scales contain small amounts of  
data, which is kept in central memory; thus this step requires no disk access.
2. The index tuple (5, 4) provides direct access to the correct element of the directory. The directory may be  
large and occupy many pages on disk, but we can compute the address of the correct directory page and in",algorithms and data structures.pdf
"large and occupy many pages on disk, but we can compute the address of the correct directory page and in  
one disk access retrieve the correct directory element.
3. The directory element contains a pointer (disk address) of the correct data bucket for (1903, v), and the  
second disk access retrieves the correct record: [(1903, v), John von Neumann …].
Disk utilization
The grid file does not allocate a separate bucket to each grid cell —that would lead to an unacceptably low disk  
utilization. Exhibit 23.4 suggests, for example, that the two grid cells at the top right of the directory share the same  
bucket. How this bucket sharing comes about, and how it is maintained through splitting of overflowing buckets,  
and merging sparsely populated buckets, is shown in the following.
Algorithms and Data Structures 260  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Exhibit 23.4: The search for a record with key values (1903, v) starts with the scales and 
proceeds via the directory to the correct data bucket on disk.
The dynamics of splitting and merging
The dynamic behavior of the grid file is best explained by tracing an example: we show the effect of repeated  
insertions in a two-dimensional file. Instead of showing the grid directory, whose elements are  in one-to-one  
correspondence with the grid blocks, we draw the bucket pointers as originating directly from the grid blocks.
Initially, a single bucket A, of capacity c = 3 in our example, is assigned to the entire domain ( Exhibit 23.5 ). 
When bucket A overflows, the domain is split, a new bucket B is made available, and those records that lie in one  
half of the space are moved from the old bucket to the new one ( Exhibit 23.6). If bucket A overflows again, its grid",algorithms and data structures.pdf
"half of the space are moved from the old bucket to the new one ( Exhibit 23.6). If bucket A overflows again, its grid  
block (i.e. the left half of the space) is split according to some splitting policy: We assume the simplest splitting  
policy of alternating directions. Those records of A that lie in the lower-left grid block of Exhibit 23.7 are moved to a 
new bucket C. Notice that as bucket B did not overflow, it is left alone: Its region now consists of two grid blocks.  
For  effective  memory  utilization  it  is  essential  that  in  the  process  of  refining  the  grid  partition  we  need  not 
necessarily split a bucket when its region is split.
Exhibit 23.5: A growing grid file starts with a single bucket allocated to the entire key space.
261",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 23.6: An overflowing bucket triggers a refinement of the space partition.
Exhibit 23.7: Bucket A has been split into A and C, but the contents of B remain unchanged.
Assuming that records keep arriving in the lower-left corner of the space, bucket C will overflow. This will trigger 
a further refinement of the grid partition as shown in Exhibit 23.8, and a splitting of bucket C into C and D. The  
history of repeated splitting can be represented in the form of a binary tree, which imposes on the set of buckets  
currently in use (and hence on the set of regions of these buckets) a twin system (also called a buddy system): Each 
bucket and its region have a unique twin from which it split off. In Exhibit 23.8, C and D are twins, the pair (C, D) is 
A's twin, and the pair (A, (C, D)) is B's twin.
Exhibit 23.8: Bucket regions that span several cells ensure high disk utilization.",algorithms and data structures.pdf
"A's twin, and the pair (A, (C, D)) is B's twin.
Exhibit 23.8: Bucket regions that span several cells ensure high disk utilization.
Deletions trigger merging operations. In contrast to one-dimensional storage, where it is sufficient to merge  
buckets that split earlier, merging policies for multidimensional grid files need to be more general in order to  
maintain a high occupancy.
Algorithms and Data Structures 262  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Simple geometric objects and their parameter spaces
Consider a class of simple spatial objects, such as aligned rectangles in the plane (i.e. with sides parallel to the  
axes). Within its class, each object is defined by a small number of parameters. For example, an aligned rectangle is  
determined by its center (cx, cy) and the half-length of each side, dx and dy.
An object defined within its class by k parameters can be considered to be a point in a k-dimensional parameter  
space. For example, an aligned rectangle becomes a point in four-dimensional space. All of the geometric and  
topological properties of an object can be deduced from the class it belongs to and from the coordinates of its  
corresponding point in parameter space.
Different  choices  of  the  parameter  space  for  the  same  class  of  objects  are  appropriate,  depending  on",algorithms and data structures.pdf
"Different  choices  of  the  parameter  space  for  the  same  class  of  objects  are  appropriate,  depending  on 
characteristics of the data to be processed. Some considerations that may determine the choice of parameters are:
1. Distinction between location parameters and extension parameters.  For some classes of simple objects it  
is reasonable to distinguish location parameters, such as the center (cx, cy) of an aligned rectangle, from  
extension parameters, such as the half-sides dx and dy. This distinction is always possible for objects that  
can be described as Cartesian products of spheres of various dimensions. For example, a rectangle is the  
product  of  two  one-dimensional  spheres,  a  cylinder  the  product  of  a  one-dimensional  and  a  two-
dimensional sphere. Whenever this distinction can be made, cone-shaped search regions generated by  
proximity queries as described in the next section have a simple intuitive interpretation: The subspace of",algorithms and data structures.pdf
"proximity queries as described in the next section have a simple intuitive interpretation: The subspace of  
the location parameters acts as a ""mirror"" that reflects a query.
2. Independence of parameters, uniform distribution.  As an example, consider the class of all intervals on a  
straight line. If intervals are represented by their left and right endpoints, lx and rx, the constraint lx ≤ rx  
restricts all representations of these intervals by points (lx, rx) to the triangle above the diagonal. Any data  
structure that organizes the embedding space of the data points, as opposed to the particular set of points  
that must be stored, will pay some overhead for representing the unpopulated half of the embedding space.  
A coordinate transformation that distributes data all over the embedding space leads to more efficient  
storage. The phenomenon of nonuniform data distribution can be worse than this. In most applications, the",algorithms and data structures.pdf
"storage. The phenomenon of nonuniform data distribution can be worse than this. In most applications, the 
building blocks from which complex objects are built are much smaller than the space in which they are  
embedded, as the size of a brick is small compared to the size of a house. If so, parameters such as lx and rx  
that locate boundaries of an object are highly dependent on each other. Exhibit 23.9 shows short intervals  
on a long line clustering along the diagonal, leaving large regions of a large embedding space unpopulated;  
whereas the same set of intervals represented by a location parameter cx and an extension parameter dx  
fills a smaller embedding space in a much more uniform way. With the assumption of bounded dx, this data 
distribution is easier to handle.
263",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 23.9: A set of intervals represented in two different parameter spaces.
Region queries of arbitrary shape
Intersection is a basic component of other proximity queries, and thus deserves special attention. CAD design  
rules, for example, often require different objects to be separated by some minimal distance. This is equivalent to  
requiring that objects surrounded by a rim do not intersect. Given a subset Γ of a class of simple spatial objects with 
parameter space H, we consider two types of queries:
• point query Given a query point q, find all objects A ∈  Γ for which q ∈  A.
• point set query Given a query set Q of points, find all objects A ∈  Γ that intersect Q.
Point query. For a query point q compute the region in H that contains all points representing objects in Γ that 
overlap q.
1. Consider the class of intervals on a straight line. An interval given by its center cx and its half length dx",algorithms and data structures.pdf
"overlap q.
1. Consider the class of intervals on a straight line. An interval given by its center cx and its half length dx  
overlaps a point q with coordinate qx if and only if cx – dx ≤ qx ≤ cx + dx.
2. The class of aligned rectangles in the plane (with parameters cx, cy, dx, dy) can be treated as the Cartesian  
product of two classes of intervals, one along the x-axis, the other along the y-axis ( Exhibit 23.10 ). All  
rectangles that contain a given point q are represented by points in four-dimensional space that lie in the  
Cartesian product of two point-in-interval query regions. The region is shown by its projections onto the cx-
dx plane and the cy-dy plane.
Algorithms and Data Structures 264  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Exhibit 23.10: A set of aligned rectangles represented as a set of points in a four-dimensional 
parameter space. A point query is transformed into a cone-shaped region query.
3. Consider the class of circles in the plane. We represent a circle as a point in three-dimensional space by the  
coordinates of its center (cx, cy) and its radius r as parameters. All circles  that overlap a point q are  
represented in the corresponding three-dimensional space by points that lie in the cone with vertex q  
shown in  Exhibit 23.11 . The axis of the cone is parallel to the r-axis (the extension parameter), and its  
vertex q is considered a point in the cx-cy plane (the subspace of the location parameters).
Exhibit 23.11: Search cone for a point query for circles in the plane.
Point set query.   Given a query set Q of points, the region in H that contains all points representing objects A",algorithms and data structures.pdf
"Point set query.   Given a query set Q of points, the region in H that contains all points representing objects A  
∈  Γ that intersect Q is the union of the regions in H that results from the point queries for each point q ∈  Q. The 
union of cones is a particularly simple region in H if the query set Q is a simple spatial object.
265",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
1. Consider the class of intervals on a straight line. An interval i = (cx, dx) intersects a query interval Q = (cq,  
dq) if and only if its representing point lies in the shaded region shown in Exhibit 23.12; this region is given 
by the inequalities cx – dx ≤ cq + dq and cx + dx ≥ cq – dq.
Exhibit 23.12: An interval query, as a union of point queries, again gets transformed into a search cone.
2. The class of aligned rectangles in the plane is again treated as the Cartesian product of two classes of  
intervals, one along the x-axis, the other along the y-axis. If Q is also an aligned rectangle, all rectangles  
that intersect Q are represented by points in four-dimensional space lying in the Cartesian product of two  
interval intersection query regions.
3. Consider the class of circles in the plane. All circles that intersect a line segment L are represented by points",algorithms and data structures.pdf
"3. Consider the class of circles in the plane. All circles that intersect a line segment L are represented by points 
lying in the cone-shaped solid shown in Exhibit 23.13. This solid is obtained by embedding L in the cx-cy  
plane, the subspace of the location parameters, and moving the cone with vertex at q along L.
Algorithms and Data Structures 266  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Exhibit 23.13: Search region as a union of cones.
Evaluating region queries with a grid file
We have seen that proximity queries on spatial objects lead to search regions significantly more complex than  
orthogonal range queries. The grid file allows the evaluation of irregularly shaped search regions in such a way that  
the complexity of the region affects CPU time but not disk accesses. The latter limits the performance of a data base  
implementation. A query region Q is matched against the scales and converted into a set I of index tuples that refer  
to entries in the directory. Only after this preprocessing do we access disk to retrieve the correct pages of the  
directory and the correct data buckets whose regions intersect Q (Exhibit 23.14).
Exhibit 23.14: The cells of a grid partition that overlap an arbitrary query region Q are determined by 
merely looking up the scales. 
Interaction between query processing and data access",algorithms and data structures.pdf
"merely looking up the scales. 
Interaction between query processing and data access
The point of the two preceding sections was to show that in a metric data structure, intricate computations  
triggered by proximity queries can be preprocessed to a remarkable extent before the objects involved are retrieved. 
267",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Query preprocessing may involve a significant amount of computation based on small amounts of auxiliary data —
the scales and the query —that are kept in central memory. The final access of data from disk is highly selective —
data retrieved has a high chance of being part of the answer.
Contrast this to an approach where an object can be accessed only by its name (e.g. the part number) because  
the geometric information about its location in space is only included in the record for this object but is not part of  
the accessing mechanism. In such a database, all objects might have to be retrieved in order to determine which  
ones answer the query. Given that disk access is the bottleneck in most database applications, it pays to preprocess  
queries as much as possible in order to save disk accesses.
The integration of query processing and accessing mechanism developed in the preceding sections was made",algorithms and data structures.pdf
"The integration of query processing and accessing mechanism developed in the preceding sections was made  
possible by the assumption of simple objects, where each instance is described by a small number of parameters.  
What can we do when faced with a large number of irregularly shaped objects?
Complex, irregularly shaped spatial objects can be represented or approximated by simpler ones in a variety of  
ways,  for  example:  decomposition,  as  in  a  quad  tree  tessellation  of  a  figure  into  disjoint  raster  squares; 
representation as a  cover of overlapping simple shapes; and enclosing each object in a  container chosen from a  
class of simple shapes. The container technique allows efficient processing of proximity queries because it preserves  
the most important properties for proximity-based access to spatial objects, in particular: It does not break up the",algorithms and data structures.pdf
"the most important properties for proximity-based access to spatial objects, in particular: It does not break up the  
object into components that must be processed separately, and it eliminates many potential tests as unnecessary (if  
two containers don't intersect, the objects within won't either). As an example, consider finding all polygons that  
intersect a given query polygon, given that each of them is enclosed in a simple container such as a circle or an  
aligned  rectangle.  Testing  two  polygons  for  intersection  is  an  expensive  operation  compared  to  testing  their 
containers for intersection. The cheap container test excludes most of the polygons from an expensive, detailed  
intersection check.
Any approximation technique limits the primitive shapes that must be stored to one or a few types: for example,  
aligned rectangles or boxes. An instance of such a type is determined by a few parameters, such as coordinates of its",algorithms and data structures.pdf
"aligned rectangles or boxes. An instance of such a type is determined by a few parameters, such as coordinates of its  
center and its extension, and can be considered to be a point in a (higher-dimensional) parameter space. This  
transformation reduces object storage to point storage, increasing the dimensionality of the problem without loss of  
information. Combined with an efficient multi-dimensional data structure for  point storage it is the basis for an  
effective implementation of databases for spatial objects.
Exercises
1. Draw three quadtrees, one for each of the 4 · 8 pixel rectangles A, B and C outlined in Exhibit 23.15.
Algorithms and Data Structures 268  A Global Text",algorithms and data structures.pdf
"23. Metric data structures
Exhibit 23.15: The location of congruent objects greatly affects the complexity of a quadtree 
representation.
2. Consider a g rid file that stores points lying in a two-dimensional domain: the Cartesian product X1 × X2,  
where X1 = 0 .. 15 and X2 = 0 .. 15 are subranges of the integers. Buckets have a capacity of two points.
(a) Insert the points (2, 3), (13, 14), (3, 5), (6, 9), (10, 13), (11, 5), (14, 9), (7, 3), (15, 11), (9, 9), and (11, 10)  
into the initially empty grid file and show the state of the scales, the directory, and the buckets after  
each insert operation. Buckets are split such that their shapes remain as quadratic as possible.
(b) Delete the points (10, 13), (9, 9), (11, 10), and (14, 9) from the grid file obtained in a) and show the state  
of the scales, the directory, and the buckets after each delete operation. Assume that after deleting a",algorithms and data structures.pdf
"of the scales, the directory, and the buckets after each delete operation. Assume that after deleting a  
point in a bucket this bucket may be merged with a neighbor bucket if their joint occupancy does not  
exceed two points. Further, a boundary should be removed from its scale if there is no longer a bucket  
that is split with respect to this boundary.
(c) Without  imposing  further  restrictions  a  deadlock  situation  may  occur  after  a  sequence  of  delete 
operations: No bucket can merge with any of its neighbors, since the resulting bucket region would no  
longer  be  rectangular.  In  the  example  shown  in  Exhibit  23.16 the  shaded  ovals  represent  bucket 
regions. Devise a merging policy that prevents such deadlocks from occurring in a two-dimensional  
grid file.
269",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 23.16: This example shows bucket regions that cannot be merged pairwise.
3. Consider the class of circles in the plane represented as points in three-dimensional parameter space as  
proposed in chapter 23 in the section “Region queries of arbitrary shape”. Describe the search regions in  
the parameter space (a) for all the circles intersecting a given circle C,  (b) for all the circles contained in C,  
and (c) for all the circles enclosing .
Algorithms and Data Structures 270  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Part VI: Interaction between 
algorithms and data 
structures: case studies in 
geometric computation
Organizing and processing Euclidean space
In Part III we presented a varied sample of algorithms that use simple, mostly static, data structures. Part V was  
dedicated to dynamic data structures, and we presented the corresponding access and update algorithms. In this  
final part we illustrate the use of these dynamic data structures by presenting algorithms whose efficiency depends  
crucially  on  them,  in  particular  on  priority  queues  and  dictionaries.  We  choose  these  algorithms  from 
computational  geometry,  a  recently  developed  discipline  of  great  practical  importance  with  applications  in 
computer graphics, computer-aided design, and geographic databases.
If data structures are tools for organizing sets of data and their relationships, geometric data processing poses",algorithms and data structures.pdf
"If data structures are tools for organizing sets of data and their relationships, geometric data processing poses  
one of the most challenging tests. The ability to organize data embedded in the Euclidean space in such a way as to  
reflect the rich relationships due to location (e.g. touching or intersecting, contained in, distance) is of utmost  
importance for the efficiency of algorithms for processing spatial data. Data structures developed for traditional  
commercial data processing were often based on the concept of one primary key and several subordinate secondary  
keys. This asymmetry fails to support the equal role played by the Cartesian coordinate axes x, y, z, … of Euclidean  
space. If one spatial axis, say x, is identified as the primary key, there is a danger that queries involving the other  
axes, say y and z, become inordinately cumbersome to process, and therefore slow. For the sake of simplicity we",algorithms and data structures.pdf
"axes, say y and z, become inordinately cumbersome to process, and therefore slow. For the sake of simplicity we  
concentrate on two-dimensional geometric problems, and in particular on the highly successful class of plane-
sweep algorithms. Sweep algorithms do a remarkably good job at processing two-dimensional space efficiently  
using two distinct one-dimensional data structures, one for organizing the x-axis, the other for the y-axis.
Algorithms and Data Structures 271  A Global Text",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
24. Sample problems and 
algorithms
Learning objectives:
• The nature of geometric computation: three problems and algorithms chosen to illustrate the variety of 
issues encountered:
• Convex hull yields to simple and efficient algorithms, straightforward to implement and analyze.
• Objects with special properties, such as convexity, are often much simpler to process than are general 
objects.
• Visibility problems are surprisingly complex; even if this complexity does not show in the design of an 
algorithm, it sneaks into its analysis.
Geometry and geometric computation
Classical geometry, shaped by the ancient Greeks, is more axiomatic than constructive: It emphasizes axioms,  
theorems, and proofs, rather than algorithms. The typical statement of Euclidean geometry is an assertion about all  
geometric configurations with certain properties (e.g. the theorem of Pythagoras: ""In a right-angled triangle, the",algorithms and data structures.pdf
"geometric configurations with certain properties (e.g. the theorem of Pythagoras: ""In a right-angled triangle, the  
square on the hypotenuse c is equal to the sum of the squares on the two catheti a and b: c 2 = a 2 + b 2"") or an  
assertion of existence (e.g. the parallel axiom: ""Given a line L and a point P ∉  L, there is exactly one line parallel to  
L passing through P""). Constructive solutions to problems do occur, but the theorems about the impossibility of 
constructive solutions steal the glory: ""You cannot trisect an arbitrary angle using ruler and compass only,"" and the  
proverbial ""It is impossible to square the circle.""
Computational geometry, on the other hand, starts out with problems of construction so simple that, until the  
1970s, they were dismissed as trivial: ""Given n line segments in the plane, are they free of intersections? If not,  
compute (construct) all intersections."" This problem is only trivial with respect to the existence of a constructive",algorithms and data structures.pdf
"compute (construct) all intersections."" This problem is only trivial with respect to the existence of a constructive  
solution. As we will soon see, the question is far from trivial if interpreted as: How efficiently can we obtain the  
answer?
Computational geometry has some appealing features that make it ideal for learning about algorithms and data  
structures: (a) The problem statements are easily understood, intuitively meaningful, and mathematically rigorous;  
right away the student can try his own hand a t solving them, without having to worry about hidden subtleties or a  
lot of required background knowledge. (b) Problem statement, solution, and every step of the construction have  
natural visual representations that support abstract thinking and help in detecting errors of reasoning. (c) These  
algorithms are practical; it is easy to come up with examples where they can be applied.",algorithms and data structures.pdf
"algorithms are practical; it is easy to come up with examples where they can be applied.
Appealing as geometric computation is, writing geometric programs is a demanding task. Two traps lie hiding  
behind the obvious combinatorial intricacies that must be mastered, and they are particularly dangerous when they  
occur together: (a) degenerate configurations, and (b) the pitfalls of numerical computation due to discretization  
and  rounding errors. Degenerate configurations, such as those we discussed in “Straight lines and circles” on  
Algorithms and Data Structures 272  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
intersecting line segments, are special cases that often require special code. It is not always easy to envision all the  
kinds  of  degeneracies  that  may  occur  in  a  given  problem.  A  configuration  may  be  degenerate  for  a  specific 
algorithm, whereas it may be nondegenerate for a different algorithm solving the same problem. Rounding errors  
tend to cause more obviously disastrous consequences in geometric computation than, say, in linear algebra or  
differential equations. Whereas the traditional analysis of rounding errors focuses on bounding their cumulative  
value,  geometry  is  concerned  primarily  with  a  stringent  all-or-nothing  question:  Have  errors  impaired  the 
topological consistency of the data? (Remember the pathology of the braided straight lines.)
In this Part VI we aim to introduce the reader to some of the central ideas and techniques of computational",algorithms and data structures.pdf
"In this Part VI we aim to introduce the reader to some of the central ideas and techniques of computational  
geometry. For simplicity's sake we limit coverage to two-dimensional Euclidean geometry - most problems become  
a lot more complicated when we go from two- to three-dimensional configurations. We focus on a type of algorithm  
that is remarkably well suited for solving two-dimensional problems efficiently: sweep algorithms. To illustrate  
their generality and effectiveness, we use plane-sweep to solve several rather distinct problems. We will see that  
sweep  algorithms  for  different  problems  can  be  assembled  from  the  same  building  blocks:  a  skeleton  sweep 
program that sweeps a line across the plane based on a queue of events to be processed, and transition procedures  
that update the data structures (a dictionary or table, and perhaps other structures) at each event and maintain a",algorithms and data structures.pdf
"that update the data structures (a dictionary or table, and perhaps other structures) at each event and maintain a  
geometric invariant. Sweeps show convincingly how the dynamic data structures of Part V are essential for the  
efficiency.
The problems and algorithms we discuss deal with very simple objects: points and line segments. Applications of 
geometric computation such as CAD, on the other hand, typically deal with very complex objects made up of  
thousands of polygons. The simplicity of these algorithms does not deter from their utility. Complex objects get  
processed by being broken into their primitive parts, such as points, line segments, and triangles. The algorithms  
we present are some of the most basic subroutines of geometric computation, which play a role analogous to that of  
a square root routine for numerical computation: As they are called untold times, they must be correct and efficient.
Convex hull: a multitude of algorithms",algorithms and data structures.pdf
"Convex hull: a multitude of algorithms
The problem of computing the convex hull H(S) of a set S consisting of n points in the plane serves as an  
example to demonstrate how the techniques of computational geometry yield the concise and elegant solution that  
we presented in “Algorithm animation”. The convex hull of a set S of points in the plane is the smallest convex  
polygon that contains the points of S in its interior or on its boundary. Imagine a nail sticking out above each point  
and a tight rubber band surrounding the set of nails.
Many different algorithms solve this simple problem. Before we present in detail the algorithm that forms the  
basis of the program 'ConvexHull' of chapter 3, we briefly illustrate the main ideas behind three others. Most  
convex hull algorithms have an initialization step that uses the fact that we can easily identify two points of S that",algorithms and data structures.pdf
"convex hull algorithms have an initialization step that uses the fact that we can easily identify two points of S that  
lie  on the convex  hull H(S): for  example, two  points  Pmin and P max with  minimal  and  maximal x-coordinate, 
respectively. Algorithms that grow convex hulls over increasing subsets can use the segment as a (degenerate)  
convex hull to start with. Other algorithms use the segment to partition S into an upper and a lower subset, and  
compute the upper and the lower part of the hull H(S) separately.
1.  Jarvis's march [Jar 73] starts at a point on H(S), say P min, and 'walks around' by computing, at each point  
P, the next tangent to S, characterized by the property that all points of S lie on the same side of PQ
273",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 24.1: The ""gift-wrapping"" approach to building the convex hull.
2.  Divide-and-conquer comes to mind: Sort the points of S according to their x-coordinate, use the median x-
coordinate to partition S into a left half SL and a right half SR, apply this convex hull algorithm recursively to 
each half, and merge the two solutions H(S L) and H(SR) by computing the two common exterior tangents to  
H(SL) and H(SR) (Exhibit 24.2). Terminate the recursion when a set has at most three points.
Exhibit 24.2: Divide-and-conquer applies to many problems on spatial data.
3.  Quickhull [Byk 78], [Edd 77], [GS 79] uses divide-and-conquer in a different way. We start with two points  
on the convex hull H(S), say P min and Pmax. In general, if we know ≥ 2 points on H(S), say P, Q, R in Exhibit
24.3, these define a convex polygon contained in H(S). (Draw the appropriate picture for just two points",algorithms and data structures.pdf
"24.3, these define a convex polygon contained in H(S). (Draw the appropriate picture for just two points  
Pmin and P max on the convex hull.) There can be no points of S in the shaded sectors that extend outward  
from the vertices of the current polygon, PQR in the example. Any other points of S must lie either in the  
polygon PQR or in the regions extending outward from the sides.
Exhibit 24.3: Three points known to lie on the convex hull identify regions devoid of points.
For each side, such as PQ in Exhibit 24.4, let T be a point farthest from PQ among all those in the region  
extending outward from PQ, if there are any. T must lie on the convex hull, as is easily seen by considering  
Algorithms and Data Structures 274  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
the parallel to PQ that passes through T. Having processed the side PQ, we extend the convex polygon to  
include T, and we now must process 2 additional sides,PT and TQ. The reader will observe a formal analogy 
between quicksort (“Sorting and its complexity”) and quickhull, which has given the latter its name.
Exhibit  24.4:  The  point  T  farthest  from   identifies  a  new  region  of  exclusion 
(shaded).
4.  In an  incremental scan  or  sweep we sort the points of S according to their x-coordinates, and use the  
segment  PminPmax to partition S into an upper subset and a lower subset, as shown in  Exhibit 24.5 . For  
simplicity of presentation, we reduce the problem of computing H(S) to the two separate problems of  
computing the upper hull U(S) [i.e. the upper part of H(S)], shown in bold, and the lower hull L(S), drawn  
as a thin line. Our notation and pictures are chosen to describe U(S).",algorithms and data structures.pdf
"as a thin line. Our notation and pictures are chosen to describe U(S).
Exhibit 24.5: Separate computations for the upper hull and the lower hull.
Let P1, … , P n be the points of S sorted by x-coordinate, and let U i = U(P1, … , P i) be the upper hull of the first i  
points. U1 = P1 may serve as an initialization. For i = 2 to n we compute U i from Ui–1, as Exhibit 24.6 shows. Starting 
with the tentative tangent PiPi–1 shown as a thin dashed line, we retrace the upper hull U i–1 until we reach the actual  
tangent: in our example, the bold dashed line PiP2. The tangent is characterized by the fact that for j = 1, … , i–1, it  
minimizes the angle Ai,j between PiPj and the vertical.
275",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 24.6: Extending the partial upper hull U(P1, … , Pi–1) to the next point Pi
The program 'ConvexHull' presented in “Algorithm animation” as an example for algorithm animation is written  
as an on-line algorithm: Rather than reading all the data before starting the computation, it accepts one point at a  
time, which must lie to the right of all previous ones, and immediately extends the hull U i–1 to obtain U i. Thanks to 
the input restriction that the points are entered in sorted order, 'ConvexHull' becomes simpler and runs in linear  
time. This explains the two-line main body:
PointZero;  { sets first point and initializes all necessary 
variables }
while NextRight do  ComputeTangent;
There remain a few programming details that are best explained by relating Fig. 24.6 to the declarations:
var x, y, dx, dy: array[0 .. nmax] of integer;
b: array[0 .. nmax] of integer;  { backpointer }",algorithms and data structures.pdf
"var x, y, dx, dy: array[0 .. nmax] of integer;
b: array[0 .. nmax] of integer;  { backpointer }
n: integer;  { number of points entered so far }
px, py: integer;  { new point }
The coordinates of the points P i are stored in the arrays x and y. Rather than storing angles such as A i,j, we store 
quantities proportional to cos(Ai,j) and sin(Ai,j) in the arrays dx and dy. The array b holds back pointers for retracing  
the upper hull back toward the left: b[i] = j implies that P j is the predecessor of P i in U i. This explains the key  
procedure of the program:
procedure ComputeTangent;  { from Pn = (px, py) to Un–1 }
var i: integer;
begin
i := b[n];
while  dy[n] · dx[i] > dy[i] · dx[n]  do  begin  { dy[n]/dx[n] > 
dy[i]/dx[i] }
i := b[i];
Algorithms and Data Structures 276  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
dx[n] := x[n] – x[i];  dy[n] := y[n] – y[i];
MoveTo(px, py);  Line(–dx[n], –dy[n]);
b[n] := i
end;
MoveTo(px, py);  PenSize(2, 2);  Line(–dx[n], –dy[n]);  PenNormal
end;  { ComputeTangent }
The algorithm implemented by 'ConvexHull' is based on Graham's scan [Gra 72], where the points are ordered  
according to the angle as seen from a fixed internal point, and on [And 79].
The uses of convexity: basic operations on polygons
The convex hull of a set of points or objects (i.e. the smallest convex set that contains all objects) is a model  
problem in geometric computation, with many algorithms and applications. Why? As we stated in the introductory  
section, applications of geometric computation tend to deal with complex objects that often consist of thousands of  
primitive  parts,  such  as  points,  line  segments,  and  triangles.  It  is  often  effective  to  approximate  a  complex",algorithms and data structures.pdf
"primitive  parts,  such  as  points,  line  segments,  and  triangles.  It  is  often  effective  to  approximate  a  complex 
configuration by a simpler one, in particular, to package it in a container of simple shape. Many proximity queries  
can be answered by processing the container only. One of the most frequent queries in computer graphics, for  
example, asks what object, if any, is first struck by a given ray. If we find that the ray misses a container, we infer  
that it misses all objects in it without looking at them; only if the ray hits the container do we start the costly  
analysis of all the objects in it.
The convex hull is often a very effective container. Although not as simple as a rectangular box, say, convexity is  
such a strong geometric property that many algorithms that take time O(n) on an arbitrary polygon of n vertices  
require only time O(log n) on convex polygons. Let us list several such examples. We assume that a polygon G is",algorithms and data structures.pdf
"require only time O(log n) on convex polygons. Let us list several such examples. We assume that a polygon G is  
given as a (cyclic) sequence of n vertices and/or n edges that trace a closed path in the plane. Polygons may be self-
intersecting, whereas simple polygons may not. A simple polygon partitions the plane into two regions: the interior,  
which is simply connected, and the exterior, which has a hole.
Point-in-polygon test
Given a simple polygon G and a query point P (not on G), determine whether P lies inside or outside the  
polygon.
Two closely related algorithms that walk around the polygon solve this problem in time O(n). The first one  
computes the winding number of G around P. Imagine an observer at P looking at a vertex, say V, where the walk  
starts, and turning on her heels to keep watching the walker ( Exhibit 24.7). The observer will make a first (positive)",algorithms and data structures.pdf
"starts, and turning on her heels to keep watching the walker ( Exhibit 24.7). The observer will make a first (positive)  
turn α , followed by a (negative) turn β, followed by … , until the walker returns to the starting vertex V. The sum α  + 
β + … of all turning angles during one complete tour of G is: 2·π if P is inside G, and 0 if P is outside G.
277",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 24.7: Point-in polygon test by adding up all turning angles.
The second algorithm computes the crossing number of G with respect to P. Draw a semi-infinite ray R from P  
in any direction (Exhibit 24.8). During the walk around the polygon G from an arbitrary starting vertex V back to V,  
keep track of whether the current oriented edge intersects R, and if so, whether the edge crosses R from below (+1)  
or from above (–1). The sum of all these numbers is +1 if P is inside G, and 0 if P is outside G.
Exhibit 24.8: Point-in polygon test by adding up crossing numbers.
Point-in-convex-polygon test
For a convex polygon Q we use binary search to perform a point-in-polygon test in time O(log n). Consider the  
hierarchical  decomposition  of  Q  illustrated  by  the  convex  12-gon  shown  in  Exhibit  24.9.  We  choose  three",algorithms and data structures.pdf
"hierarchical  decomposition  of  Q  illustrated  by  the  convex  12-gon  shown  in  Exhibit  24.9.  We  choose  three 
(approximately) equidistant vertices as the vertices of an innermost core triangle, painted black. ""Equidistant"" here  
refers not to any Euclidean distance, but rather to the number of vertices to be traversed by traveling along the  
perimeter of Q. For a query point P we first ask, in time O(1), which of the seven regions defined by the extended  
edges of this triangular core contains P. These seven regions shown in Exhibit 24.10 are all ""triangles"" (albeit six of  
them extend to infinity), in the sense that each one is defined as the intersection of three half-spaces. Four of these  
regions provide a definite answer to the query ""Is P inside Q, or outside Q?"" One region (shown hatched in Exhibit
24.10) provides the answer 'In', three the answer 'Out'. The remaining three regions, labeled 'Uncertain', lead",algorithms and data structures.pdf
"24.10) provides the answer 'In', three the answer 'Out'. The remaining three regions, labeled 'Uncertain', lead  
recursively to a new point-in-convex-polygon test, for the same query point P, but a new convex polygon Q' which is  
Algorithms and Data Structures 278  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
the intersection of Q with one of the uncertain regions. As Q' has only about n  / 3 vertices, the depth of recursion is  
O(log n). Actually, after the first comparison against the innermost triangular core of Q, we have no longer a general 
point-in-convex-polygon problem, but one with additional information that makes all but the first test steps of a  
binary search.
Exhibit 24.9: Hierarchical approximation of a convex 12-gon as a 3-level tree of triangles. The root is in 
black, its children are in dark grey, grandchildren in light grey.
Exhibit 24.10: The plane partitioned into four regions of certainty and three of uncertainty. 
The latter are processed recursively.
Visibility in the plane: a simple algorithm whose analysis is not
Many computer graphics programs are dominated by visibility problems: Given a configuration of objects in",algorithms and data structures.pdf
"Many computer graphics programs are dominated by visibility problems: Given a configuration of objects in  
three-dimensional space, and given a point of view, what is visible? Dozens of algorithms for hidden-line or hidden-
surface elimination have been developed to solve this everyday problem that our visual system performs ""at a  
glance"". In contrast to the problems discussed above, visibility is surprisingly complex. We give a hint of this  
complexity by describing some of the details buried below the smooth surface of a ""simple"" version: computing the  
visibility of line segments in the plane.
Problem: Given n line segments in the plane, compute the sequence of (sub)segments seen by an observer at  
infinity (say, at y = –∞).
279",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The complexity of this problem was unexpected until discovered in 1986 [WS 88]. Fortunately, this complexity  
is revealed not by requiring complicated algorithms, but in the analysis of the inherent complexity of the geometric  
problem. The example shown in Exhibit 24.11 illustrates the input data. The endpoints (P 1, P10), (P2, P8), (P5, P12) of 
the three line segments labeled 1, 2, 3 are given; other points are computed by the algorithm. The required result is  
a list of visible segments, each segment described by its endpoints and by the identifier of the line of which it is a  
part:
(P1, P3, 1), (P3, P4, 2), (P5, P6, 3), (P6, P8, 2), (P7, P9, 3), (P9, P10, 1), (P11, P12, 3)
Exhibit 24.11: Example: Three line segments seen from below generate seven visible subsegments.
In se arch of algorithms, the reader is encouraged to work out the details of the first idea that might come to",algorithms and data structures.pdf
"In se arch of algorithms, the reader is encouraged to work out the details of the first idea that might come to  
mind: For each of the n 2 ordered pairs (L i, L j) of line segments, remove from L i the subsegment occluded by L j. 
Because L i can get cut into as many as n pieces, it must be managed as a sequence of subsegments. Finding the  
endpoints of Lj in this sequence will take time O(log n), leading to an overall algorithm of time complexity O(n 2 · log 
n).
After the reader has mastered the sweep algorithm for line intersection presented in “Plane-sweep: a general-
purpose algorithm for two-dimensional problems illustrated using line segment intersection”, he will see that its  
straightforward application to the line visibility problem requires time O((n + k) · log n), where k ∈  O(n2) is the  
number  of  intersections.  Thus  plane-sweep appears  to do  all  the work the  brute-force algorithm  above  does,",algorithms and data structures.pdf
"number  of  intersections.  Thus  plane-sweep appears  to do  all  the work the  brute-force algorithm  above  does, 
organized in a systematic left-to-right fashion. It keeps track of all intersections, most of which may be invisible. It  
has the potential to work in time O(n · log n) for many realistic data configurations characterized by k ∈  O(n), but 
not in the worst case.
Divide-and-conquer yields a simple two-dimensional visibility algorithm with a better worst-case performance.  
If n = 0 or 1, the problem is trivial. If n > 1, partition the set of n line segments into two (approximate) halves, solve  
both subproblems, and merge the results. There is no constraint on how the set is halved, so the divide step is easy.  
The conquer step is taken care of by recursion. Merging amounts to computing the minimum of two piecewise (not  
necessarily continuous) linear functions, in time linear in the number of pieces. The example with n = 4 shown in",algorithms and data structures.pdf
"necessarily continuous) linear functions, in time linear in the number of pieces. The example with n = 4 shown in  
Exhibit 24.12 illustrates the algorithm. f 12 is the visible front of segments 1 and 2, f 34 of segments 3 and 4, min(f 12, 
f34) of all four segments (Exhibit 24.13).
Algorithms and Data Structures 280  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
Exhibit 24.12: The four line segments will be partitioned into subsets {1, 2} and {3, 
4}.
Exhibit 24.13: The min operation merges the solutions of this divide-and-conquer 
algorithm.
The time complexity of this divide-and-conquer algorithm is obtained as follows. Given that at each level of  
recursion the relevant sets of line segments can be partitioned into (approximate) halves, the depth of recursion is  
O(log n). A merge step that processes v visible subsegments takes linear time O(v). Together, all the merge steps at  
281",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
a given depth process at most V subsegments, where V is the total number of visible subsegments. Thus the total  
time is bounded by O(V · log n). How large can V be?
Surprising theoretical results
Let V(n) be the number of visible subsegments in a given configuration of n lines, i.e. the size of the output of the 
visibility computation. For tiny n, the worst cases [V(2) = 4, V(3) = 8] are shown in Exhibit 24.14. An attempt to  
find worst-case configurations for general n leads to examples such as that shown in Figure 24.15, with V(n) = 5·n – 
8.
Exhibit 24.14: Configurations with the largest number of visible subsegments.
Figure 24.15: A family of configurations with 5·n – 8 visible subsegments.
You will find it difficult to come up with a class of configurations for which V(n) grows faster. It is tempting to",algorithms and data structures.pdf
"You will find it difficult to come up with a class of configurations for which V(n) grows faster. It is tempting to  
conjecture that V(n) ∈  O(n), but this conjecture is very hard to prove - for the good reason that it is false, as was  
discovered in [WS 88]. It turns out that V(n) ∈  Θ (n · α (n)), where α (n), the inverse of Ackermann's function (see  
“Computability and complexity”, Exercise 2), is a monotonically increasing function that grows so slowly that for  
practical purposes it can be treated as a constant, call it α .
Let us present some of the steps of how this surprising result was arrived at. Occasionally, simple geometric  
problems can be tied to deep results in other branches of mathematics. We transform the two-dimensional visibility  
problem into a combinatorial string problem. By numbering the given line segments, walking along the x-axis from  
Algorithms and Data Structures 282  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
left to right, and writing down the number of the line segment that is currently visible, we obtain a sequence of  
numbers (Exhibit 24.16).
Exhibit 24.16: The Davenport-Schinzel sequence associated with a configuration of 
segments.
A geometric configuration gives rise to a sequence u1, u2, … , um with the following properties:
1. 1  ≤ ui  ≤ n for 1 ≤ i ≤ m (numbers identify line segments).
2. ui ≠ ui+1 for 1 ≤ i ≤ m – 1 (no two consecutive numbers are equal).
3. There are no five indices 1 ≤ a < b < c < d < e ≤ m such that u a = uc = ue = r and u b = ud = s, r ≠ s. This  
condition captures  the geometric  properties of two intersecting straight lines: If we ever see r, s, r, s  
(possibly separated), we will never see r again, as this would imply that r and s intersect mo re than once  
(Exhibit 24.17).
Exhibit 24.17: The subsequence r, s, r, s excludes further occurrences of r.
Example
The sequence for the example above that shows m ≥ 5 n – 8 is",algorithms and data structures.pdf
"Exhibit 24.17: The subsequence r, s, r, s excludes further occurrences of r.
Example
The sequence for the example above that shows m ≥ 5 n – 8 is
1, 2, 1, 3, 1, … , 1, n–1, 1, n–1, n–2, n–3, … , 3, 2, n, 2, n, 3, n, … , n, n–2, n, n–1, n.
283",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Sequences with the properties 1 to 3, called Davenport-Schinzel sequences, have been studied in the context of  
linear differential equations. The maximal length of a Davenport-Schinzel sequence is k · n ·  α (n), where k is a  
constant and α (n) is the inverse of Ackermann's function (see “Computability and complexity”, Exercise 2) [HS 86].  
With increasing n, α (n) approaches infinity, albeit very slowly. This dampens the hope for a linear upper bound for  
the visibility problem, but does not yet disprove the conjecture. For the latter, we need an inverse: For any given  
Davenport-Schinzel sequence there exists a corresponding geometric configuration which yields this sequence. An  
explicit construction is given in [WS 88]. This establishes an isomorphism between the two-dimensional visibility  
problem and the Davenport-Schinzel sequences, and shows that the size of the output of the two-dimensional",algorithms and data structures.pdf
"problem and the Davenport-Schinzel sequences, and shows that the size of the output of the two-dimensional  
visibility problem can be superlinear - a result that challenges our geometric intuition.
Exercises
1. Given a set of points S, prove that the pair of points farthest from each other must be vertices of the convex  
hull H(S).
2. Assume a model of computation in which the operations addition, multiplication, and comparison are  
available at unit cost. Prove that in such a model Ω (n · log n) is a lower bound for computing, in order, the  
vertices of the convex hull H(S) of a set S of n points. Hint: Show that every algorithm which computes the  
convex hull of n given points can be used to sort n numbers.
3. Complete the second algorithm for the point-in-polygon test in chapter 24 in the section “The uses of  
convexity: basic operations on polygons” which computes the crossing number of the polygon G around",algorithms and data structures.pdf
"convexity: basic operations on polygons” which computes the crossing number of the polygon G around  
point P by addressing the special cases that arise when the semi-infinite ray R emanating from P intersects  
a vertex of G or overlaps an edge of G.
4. Consider an arbitrary (not necessarily simple) polygon G ( Exhibit 24.18). Provide an interpretation for the  
winding number w(G, P) of G around an arbitrary point P not on G, and prove that w(G, P) / 2·π of P is  
always equal to the crossing number of P with respect to any ray R emanating from P.
Exhibit 24.18: Winding number and crossing number of a polygon G with respect to P.
5. Design an algorithm that computes the area of an n-vertex simple, but not necessarily convex polygon in  
Θ (n) time.
6. We consider the problem of computing the intersection of two convex polygons which are given by their  
lists of vertices in cyclic order.
(a) Show that the intersection is again a convex polygon.",algorithms and data structures.pdf
"lists of vertices in cyclic order.
(a) Show that the intersection is again a convex polygon.
(b) Design an algorithm that computes the intersection. What is the time complexity of your algorithm?
Algorithms and Data Structures 284  A Global Text",algorithms and data structures.pdf
"24. Sample problems and algorithms
7. Intersection test for line L and [convex] polygon Q  If an (infinitely extended) line L intersects a polygon Q, 
it must intersect one of Q's edges. Thus a test for intersection of a given line L with a polygon can be  
reduced to repeated test of L for intersection with [some of] Q's edges.
(a) Prove that, in general, a test for line-polygon intersection must check at least n – 2 of Q's edges. Hint: 
Use an adversary argument. If two edges remain unchecked, they could be moved so as to invalidate  
the answer.
(b) Design a test that works in time O(log n) for decoding whether a line L intersects a convex polygon Q.
8. Divide-and-conquer algorithms may divide the space in which the data is embedded, rather than the set of  
data (the set of lines). Describe an algorithm for computing the sequence of visible segments that partitions  
the space recursively into vertical stripes, until each stripe is ""simple enough""; describe how you choose the",algorithms and data structures.pdf
"the space recursively into vertical stripes, until each stripe is ""simple enough""; describe how you choose the  
boundaries of the stripes; state advantages and disadvantages of this algorithm as compared to the one  
described in chapter 24 in the section “Visibility in the plane: a simple algorithm whose analysis is not”.  
Analyze the asymptotic time complexity of this algorithm.
285",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
25. Plane-sweep: a general-
purpose algorithm for two-
dimensional problems 
illustrated using line segment 
intersection
Learning objectives:
• line segment intersection test
• turning space dimensions into time dimensions
• updating a y table and detecting intersections
• sweeping across and intersection
Plane-sweep is an algorithm schema for two-dimensional geometry of great generality and effectiveness, and  
algorithm designers are well advised to try it first. It works for a surprisingly large set of problems, and when it  
works, tends to be very efficient. Plane-sweep is easiest to understand under the assumption of nondegenerate  
configurations. After explaining plane-sweep under this assumption, we remark on how degenerate cases can be  
handled with plane-sweep.
The line segment intersection test
We present a plane-sweep algorithm [SH 76] for the line segment intersection test:",algorithms and data structures.pdf
"handled with plane-sweep.
The line segment intersection test
We present a plane-sweep algorithm [SH 76] for the line segment intersection test:
Given n line segments in the plane, determine whether any two 
intersect;
and if so, compute a witness (i.e. a pair of segments that 
intersect).
Bounds on the c omplexity of this problem are easily obtained. The literature on computational geometry (e.g.  
[PS 85]) proves a lower bound Ω (n · log n). The obvious brute force approach of testing all n · (n – 1) / 2 pairs of  
line segments requires Θ (n2) time. This wide gap between n · log n and n 2 is a challenge to the algorithm designer,  
who strives for an optimal algorithm whose asymptotic running time O(n · log n) matches the lower bound.
Divide-and-conquer is often the first attempt to design an algorithm, a nd it comes in two variants illustrated in  
Fig. 25.1: (1) Divide the data, in this case the set of line segments, into two subsets of approximately equal size (i.e.",algorithms and data structures.pdf
"Fig. 25.1: (1) Divide the data, in this case the set of line segments, into two subsets of approximately equal size (i.e.  
n / 2 line segments), or (2) divide the embedding space, which is easily cut in exact halves.
Algorithms and Data Structures 286  A Global Text",algorithms and data structures.pdf
"25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using line segment
intersection
Exhibit 25.1: Two ways of applying divide-and-conquer to a set of objects embedded in the plane.
In the first case, we hope for a se paration into subsets S 1 and S 2 that permits an efficient test whether any line  
segment in S1 intersects some line segment in S 2. Exhibit 25.1 shows the ideal case where S 1 and S2 do not interact,  
but of course this cannot always be achieved in a nontrivial way; and even if S can be separated as the figure  
suggests,  finding  such  a  separating  line  looks  like  a  more  formidable  problem  than  the  original  intersection 
problem. Thus, in general, we have to test each line segment in S 1 against every line segment in S 2, a test that may  
take Θ (n2) time.
The second approach of dividing the embedding space has the unfortunate consequence of effectively increasing",algorithms and data structures.pdf
"take Θ (n2) time.
The second approach of dividing the embedding space has the unfortunate consequence of effectively increasing  
our data set. Every segment that straddles the dividing line gets ""cut"" (i.e. processed twice, once for each half  
space). The two resulting subproblems will be of size n' and n"", respectively, with n' + n"" > n, in the worst case n' +  
n"" = 2 · n. At recursion depth d we may have 2 d · n subsegments to process. No optimal algorithm is known that  
uses this technique.
The key idea in d esigning an optimal algorithm is the observation that those line segments that intersect a  
vertical line L at abscissa x are totally ordered: A segment s lies below segment t, written s < L t, if both intersect L at 
the current position x and the intersection of s with L lies below the intersection of t with L. With respect to this  
order a line segment may have an upper and a lower neighbor, and Exhibit 25.2 shows that s and t are neighbors at  
x.",algorithms and data structures.pdf
"order a line segment may have an upper and a lower neighbor, and Exhibit 25.2 shows that s and t are neighbors at  
x.
Exhibit 25.2: The sweep line L totally orders the segments that intersect L.
We describe the intersection test algorithm under the assumption that the configuration is nondegenerate (i.e.  
no three segments intersect in the same point). For simplicity's sake we also assume that no segment is vertical, so  
every segment has a left endpoint and a right endpoint. The latter assumption entails no loss of generality: For a  
vertical  segment,  we  can  arbitrarily  define  the  lower  endpoint  to  be  the  ""left  endpoint"",  thus  imposing  a 
lexicographic  (x,  y)-order  to  refine  the  x-order.  With  the  important  assumption  of  non-degeneracy,  two  line 
segments s and t ca n intersect at x 0 only if there exists an abscissa x < x 0 where s and t are neighbors. Thus it  
287",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
suffices to test all segment pairs that become neighbors at some time during a left-to-right sweep of L - a number  
that is usually significantly smaller than n · (n – 1) / 2.
As the sweep line L moves from left to right across the configuration, the order < L among the line segments  
intersecting L changes only at endpoints of a segment or at intersections of segments. As we intend to stop the  
sweep as soon as we discover an intersection, we need to perform the intersection test only at the left and right  
endpoints of segments. A segment t is tested at its left endpoint for intersection with its lower and upper neighbors.  
At the right endpoint of t we test its lower and upper neighbor for intersection (Exhibit 25.3).
The algorithm terminates as soon as we discover an intersecting pair of segments. Given n segments, each of
Exhibit 25.3: Three pairwise intersection tests charged to segment t.",algorithms and data structures.pdf
"Exhibit 25.3: Three pairwise intersection tests charged to segment t.
which may generate three intersection tests as shown in Exhibit 25.3 (two at its left, one at its right endpoint), we  
perform the O(1) pairwise segment intersection test at most 3 · n times. This linear bound on the number of pairs  
tested for intersection might raise the hope of finding a linear-time algorithm, but so far we have counted only the  
geometric primitive: ""Does a pair of segments intersect - yes or no?"" Hiding in the background we find bookkeeping  
operations such as ""Find the  upper and lower neighbor of a given segment"", and these turn out to be costlier than  
the geometric ones. We will find neighbors efficiently by maintaining the order < L in a data structure called a y-
table during the entire sweep.
The skeleton: Turning a space dimension into a time dimension
The name plane-sweep is derived from the image of sweeping the plane from left to right with a vertical line",algorithms and data structures.pdf
"The name plane-sweep is derived from the image of sweeping the plane from left to right with a vertical line  
(front, or cross section), stopping at every transition point (event) of a geometric configuration to update the cross  
section. All processing is done at this moving front, without any backtracking, with a look-ahead of only one point.  
The events are stored in the x-queue, and the current cross section is maintained by the y-table. The skeleton of a  
plane-sweep algorithm is as follows:
initX;  initY;
while  not emptyX  do  { e := nextX;  transition(e) }
The procedures 'initX' and 'initY' initialize the x-queue and the y-table. 'nextX' returns the next event in the x-
queue, 'emptyX' tells us whether the x-queue is empty. The procedure 'transition', the advancing mechanism of the  
sweep, embodies all the work to be done when a new event is encountered; it moves the front from the slice to the  
left of an event e to the slice immediately to the right of e.",algorithms and data structures.pdf
"left of an event e to the slice immediately to the right of e.
Data structures
For the line segment intersection test, the x-queue stores the left and right endpoints of the given line segments,  
ordered by their x-coo rdinate, as events to be processed when updating the vertical cross section. Each endpoint  
stores a reference to the corresponding line segment. We compare points by their x-coordinates when building the  
Algorithms and Data Structures 288  A Global Text",algorithms and data structures.pdf
"25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using line segment
intersection
x-queue. For simplicity of presentation we assume that no two endpoints of line segments have equal x- or y-
coordinates. The only operation to be performed on the x-queue is 'nextX': it returns the next event (i.e. the next left 
or right endpoint of a line segment to be processed). The cost for initializing the x-queue is O(n · log n), the cost for  
performing the 'nextX' operation is O(1).
The y-table contains those line segments that are currently intersected by the sweep line, ordered according to  
<L. In the slice between two events, this order does not change, and the y-table needs no updating ( Exhibit 25.4). 
The y-table is a dictionary that supports the operations 'insertY', 'deleteY', 'succY', and 'predY'. When entering the  
left endpoint of a line segment s we find the place where s is to be inserted in the ordering of the y-table by",algorithms and data structures.pdf
"left endpoint of a line segment s we find the place where s is to be inserted in the ordering of the y-table by  
comparing s to other line segments t already stored in the y-table. We can determine whether s < L t or t < L s by 
determining on which side of t the left endpoint of s lies. As we have seen in chapter 14 in the section “Intersection”,  
this tends to be more efficient than computing and comparing the intersection points of s and t with the sweep line.  
If we implement the dictionary as a balanced tree (e.g. an AVL tree), the operations 'insertY' and 'deleteY' are  
performed in O(log n) time, and 'succY' and 'predY' are performed in O(1) t ime if additional pointers in each node  
of the tree point to the successor and predecessor of the line segment stored in this node. Since there are 2 · n  
events in the x-queue and at most n line segments in the y-table the space complexity of this plane-sweep algorithm  
is O(n).",algorithms and data structures.pdf
"events in the x-queue and at most n line segments in the y-table the space complexity of this plane-sweep algorithm  
is O(n).
Exhibit 25.4: The y-table records the varying state of the sweep line L.
Updating the y-table and detecting an intersection
The procedure 'transition' maintains the order < L of the line segments intersecting the sweep line and performs  
intersection tests. At a left endpoint of a segment t, t is inserted into the y-table and tested for intersection with its  
lower and upper neighbors. At the right endpoint of t, t is deleted from the y-table and its two former neighbors are  
tested. The algorithm terminates when an intersect ion has been found or all events in the x-queue have been  
processed without finding an intersection:
procedure transition(e: event);
begin
s := segment(e);
if  leftPoint(e)  then  begin
insertY(s);
289",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
if  intersect(predY(s), s) or intersect (s, succY(s))  then
terminate('intersection found')
end
else  { e is right endpoint of s }  begin
if  intersect(predY(s), succY(s))  then
terminate('intersection found');
deleteY(s)
end
end;
With at most 2 · n events, and a call of 'transition' costing time O(log n), this plane-sweep algorithm needs O(n ·  
log n) time to perform the line segment intersection test.
Sweeping across intersections 
The plane-sweep algorithm for the line segment intersection test is easily adapted to the following more general  
problem [BO 79]:
Given n line segments, report all intersections.
In  addition to  the  left  and  right  endpoints,  the  x-queue  now  stores  intersection  points  as  events—any 
intersection detected is inserted into the x-queue as an event to be processed. When the sweep line reaches an",algorithms and data structures.pdf
"intersection detected is inserted into the x-queue as an event to be processed. When the sweep line reaches an  
intersection event the two participating line segments are swapped in the y-table ( Exhibit 25.5). The major increase 
in complexity as compared to the segment intersection test is that now we must process not only 2 · n events, but 2 ·  
n + k events, where k is the number of intersections discovered as we sweep the plane. A configuration with n / 2 
segments vertical and n  / 2 horizontal shows that, in the worst case, k  ∈  Θ (n2), which leads to an O(n 2 · log n)  
algorithm, certainly no improvement over the brute-force comparison of all pairs. In most realistic configurations,  
say  engineering drawings,  the  number  of intersections  is  much  less  than O(n2), and  thus it is  informative to  
introduce the parameter k in order to get an output-sensitive bound on the com plexity of this algorithm (i.e. a",algorithms and data structures.pdf
"introduce the parameter k in order to get an output-sensitive bound on the com plexity of this algorithm (i.e. a  
bound that adapts to the amount of data needed to report the result of the computation).
Exhibit 25.5: Sweeping across an intersection.
Other changes are comparatively minor. The x-queue must be a priority queue that supports the operation  
'insertX'; it can be implemented as a heap. The cost for initializing the x-queue remains O(n · log n). Without  
further analysis one might presume that the storage requirement of the x-queue is O(n + k), which implies that the  
cost for calling 'insertX' and 'nextX' remains O(log n), since k ∈  O(n2). A more detailed analysis [PS 91], however,  
shows that the size of the x-queue never exceeds O(n · (log n) 2). With a slight modification of the algorithm [Bro 81]  
Algorithms and Data Structures 290  A Global Text",algorithms and data structures.pdf
"25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using line segment
intersection
it can even be guaranteed that the size of the x-queue never exceeds O(n). The cost for exchanging two intersecting  
line segments in the y-table is O(log n), the costs for the other operations on the y-table remain the same. Since  
there are 2 · n left and right endpoints and k intersection events, the total cost for this algorithm is O((n + k) · log  
n). As most realistic applications are characterized by k ∈  O(n), reporting all intersections often remains an O(n ·  
log n) algorithm in practice. A time-optimal algorithm that finds all intersecting pairs of line segments in O(n · log n 
+ k) time using O(n + k) storage space is described in [CE 92].
Degenerate configurations, numerical errors, robustness
The discussion above is based on several assumptions of nondegeneracy, some of minor and some of major  
importance. Let us examine one of each type.",algorithms and data structures.pdf
"The discussion above is based on several assumptions of nondegeneracy, some of minor and some of major  
importance. Let us examine one of each type.
Whenever we access the x-queue ('nextX'), we used an implicit assumption that no two events (endpoints or  
intersections)  have  equal  x-coordinates.  The  order  of  processing  events  of  equal  x-coordinate  is  irrelevant. 
Assuming that no two events coincide at the same point in the plane, lexicographic (x, y)-ordering is a convenient  
systematic way to define 'nextX'.
More serious forms of degeneracy arise when events coincide in the plane, such as more than two segments  
intersecting  in  the  same  point.  This  type  of  degeneracy  is  particularly  difficult  to  handle  in  the  presence  of 
numerical errors, such as rounding errors. In the configuration shown in Exhibit 25.6 an endpoint of u lies exactly",algorithms and data structures.pdf
"numerical errors, such as rounding errors. In the configuration shown in Exhibit 25.6 an endpoint of u lies exactly  
or nearly on segment s. We may not care whether the intersection routine answers 'yes' or 'no' to the question ""Do s  
and u intersect?"" but we certainly expect a 'yes' when asking ""Do t and u intersect?"" This example shows that the  
slightest numerical inaccuracy can cause a serious error: The algorithm may fail to report the intersection of t and  
u, which it would clearly see if it bothered to look - but the algorithm looks the other way and never asks the  
question ""Do t and u intersect?""
Exhibit 25.6: A degenerate configuration may lead to inconsistent 
results.
The trace of the plane-sweep for reporting intersections may look as follows:
1. s is inserted into the y-table
2. t is inserted above s into the y-table, and s and t are tested for intersection: No intersection is found",algorithms and data structures.pdf
"1. s is inserted into the y-table
2. t is inserted above s into the y-table, and s and t are tested for intersection: No intersection is found
3. u is inserted below s in the y-table (since the evaluation of the function s(x) may conclude that the left  
endpoint of u lies below s); s and u are tested for intersection, but the intersection routine may conclude  
that s and u do not intersect: u remains below s
291",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
4. Delete u from the y-table
5. Delete s from the y-table
6. Delete t from the y-table
Notice  the  calamity  that  struck  at  the  critical  step  3.  The  evaluation  of  a  linear  expression  s(x)  and  the 
intersection routine for two segments both arrived at a result that, in isolation, is reasonable within the tolerance of  
the underlying arithmetic. The two results together are inconsistent! If the evaluation of s(x) concludes that the left  
endpoint of u lies below s, the intersection routine  must conclude that s and u intersect! If these two geometric  
primitives fail to coordinate their answers, catastrophe may strike. In our example, u and t never become neighbors  
in the y-table, so their intersection gets lost.
Exercises
1. Show that there may be Θ (n2) intersections in a set of n line segments.",algorithms and data structures.pdf
"in the y-table, so their intersection gets lost.
Exercises
1. Show that there may be Θ (n2) intersections in a set of n line segments.
2. Design a plane-sweep algorithm that determines in O(n · log n) time whether two simple polygons with a  
total of n vertices intersect.
3. Design a plane-sweep algorithm that determines in O(n · log n) time whether any two disks in a set of n  
disks intersect.
4. Design a plane-sweep algorithm that solves the line visibility problem discussed in chapter 24 in the section 
“Visibility in the plane: a simple algorithm whose analysis is not” in time O((n + k) · log n), where k ∈  O(n2) 
is the number of intersections of the line segments.
5. Give a configuration with the smallest possible number of line segments for which the first intersection  
point reported by the plane-sweep algorithm in chapter 25 in the section “Sweeping across intersections” is  
not the leftmost intersection point.",algorithms and data structures.pdf
"point reported by the plane-sweep algorithm in chapter 25 in the section “Sweeping across intersections” is  
not the leftmost intersection point.
6. Adapt the plane-sweep algorithm presented in chapter 25 in the section “Sweeping across intersections” to  
detect all intersections among a given set of n horizontal or vertical line segments. You may assume that the  
line segments do not overlap. What is the time complexity of this algorithm if the horizontal and vertical  
line segments intersect in k points?
7. Design a plane-sweep algorithm that finds all intersections among a given set of n rectangles all of whose  
sides are parallel to the coordinate axes. What is the time complexity of your algorithm?
Algorithms and Data Structures 292  A Global Text",algorithms and data structures.pdf
"26. The closest pair
26. The closest pair
Learning objectives:
• Applying, implementing and analyzing plane sweep
• Using plane sweep on three or more dimensions
Sweep algorithms solve many kinds of proximity problems efficiently. We present a simple sweep that solves the  
two-dimensional  closest  pair  problem  elegantly  in  asymptotically  optimal  time.  We  explain  why  sweeping 
generalizes easily, but not efficiently, to multidimensional closest pair problems.
The problem
We consi der the two-dimensional  closest pair problem : Given a set S of n points in the plane find a pair of  
points whose distance δ is smallest ( Exhibit 26.1). We measure distance using the metric d k, for any k ≥ 1, or d ∞, 
defined as:
Exhibit 26.1: Identify a closest pair among n points in the plane.
Special cases of interest include the ""Manhattan metric""  d 1, the ""Euclidean  metric""  d2, and  the ""maximum",algorithms and data structures.pdf
"Special cases of interest include the ""Manhattan metric""  d 1, the ""Euclidean  metric""  d2, and  the ""maximum  
metric"" d∞. Exhibit 26.2 shows the ""circles"" of radius 1 centered at a point p for some of these metrics.
Exhibit 26.2: The results of this chapter remain valid when distances are measured in various metrics.
 
The closest pair problem has a lower bound Ω (n · log n) in the algebraic decision tree model of computation [PS  
85]. Its solution can be obtained in asymptotically optimal time O(n  · log n) as a special case of more general  
problems, such as 'all-nearest-neighbors' [HNS 92]  (for each point, find a nearest neighbor), or constructing the  
Voronoi diagram [SH 75]. These general approaches call on powerful techniques that make the resulting algorithms  
harder to understand than one would expect for a simply stated problem such as ""find a closest pair"". The divide-",algorithms and data structures.pdf
"harder to understand than one would expect for a simply stated problem such as ""find a closest pair"". The divide-
and-conquer algorithm presented in [BS  76] solves the closest pair problem directly in optimal worst-case time  
complexity  Θ (n  ·  log  n)  using  the  Euclidean  metric  d2.  Whereas  the  recursive  divide-and-conquer  algorithm 
293",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
involves an intricate argument for combining the solutions of two equally sized subsets, the iterative plane-sweep  
algorithm [HNS 88] uses a simple incremental update: Starting with the empty set of points, keep adding a single  
point until the final solution for the entire set is obtained. A similar plane-sweep algorithm solves the closest pair  
problem for a set of convex objects [BH 92].
Plane-sweep applied to the closest pair problem
The skeleton of the general sweep algorithm presented in chapter 25 in the section “The skeleton: turning a  
space dimension into a time dimension”, with the data structures x-queue and y-table, is adapted to the closest pair  
problem as shown in  Exhibit 26.3 . The  x-queue stores the points of the set S, ordered by their x-coordinate, as  
events to be processed when updating the vertical cross section. Two pointers into the x-queue, 'tail' and 'current',",algorithms and data structures.pdf
"events to be processed when updating the vertical cross section. Two pointers into the x-queue, 'tail' and 'current',  
partition S into four disjoint subsets:
1. The discarded points to the left of 'tail' are not accessed any longer
2. The active points between 'tail' (inclusive) and 'current' (exclusive) are being queried
3. The current transition point, p, is being processed
4. The future points have not yet been looked at
The y-table stores the active points only, ordered by their y-coordinate.
Exhibit 26.3: Updating the invariant as the next point p is processed.
We need to compare points by their x-coordinates when building the x-queue, and by their y-coordinates while  
sweeping. For simplicity of presentation we assume that no two points have equal x- or y-coordinates. Points with  
equal x- or y-coordinates are handled by imposing an arbitrary, but consistent, total order on the set of points. We",algorithms and data structures.pdf
"equal x- or y-coordinates are handled by imposing an arbitrary, but consistent, total order on the set of points. We  
achieve this by defining two lexicographic orders: <x to be used for the x-queue, <y for the y-table:
Algorithms and Data Structures 294  A Global Text",algorithms and data structures.pdf
"26. The closest pair
The program of the following section initializes the x-queue and y-table with the two leftmost points being  
active, with δ equal to their distance, and starts the sweep with the third point.
The distinction between discarded and active points is motivated by the following argument. When a new point  
p is encountered we wish to answer the question whether this point forms a closest pair with one of the points to its  
left. We keep a pair of closest points seen so far, along with the corresponding minimal distance  δ. Therefore, all 
candidates that may form a new closest pair with the point p on the sweep line lie in a half circle centered at p, with  
radius δ.
The key question to be answered in striving for efficiency is how to retrieve quickly all the points seen so far that  
lie inside this half circle to the left of p, in order to compare their distance to p against the minimal distance δ seen",algorithms and data structures.pdf
"lie inside this half circle to the left of p, in order to compare their distance to p against the minimal distance δ seen 
so far. We m ay use any helpful data structure that organizes the points seen so far, as long as we can update this  
data structure efficiently across a transition. A circle (or half-circle) query is complex, at least when embedded in a  
plane-sweep algorithm that organizes data according to an orthogonal coordinate system. A rectangle query can be  
answered more efficiently. Thus we replace the half-circle query with a bounding rectangle query, accepting the fact  
that we might include some extraneous points, such as q.
The rectangle query in  Exhibit 26.3  is implemented in two steps. First, we cut off all the points to the left at  
distance ≥ δ from the sweep line. These points lie between 'tail' and 'current' in the x-queue and can be discarded",algorithms and data structures.pdf
"distance ≥ δ from the sweep line. These points lie between 'tail' and 'current' in the x-queue and can be discarded  
easily by advancing 'tail' and removing them from the y-table. Second, we consider only those points q in the δ-slice 
whose vertical distance from p is less than δ: |qy – py| < δ. These points can be found in the y-table by looking at  
successors and predecessors starting at the y-coordinate of p. In other words, we maintain the following invariant  
across a transition:
1. δ is the minimal distance between a pair of points seen so far (discarded or active).
2. The active points (found in the x-queue between 'tail' and 'current', and stored in the y-table ordered by y-
coordinates) are exactly those that lie in the interior of a δ-slice to the left of the sweep line.
3. Therefore, processing the transition point p involves three steps:
4. Delete all points q with qx ≤ px – δfrom the y-table. They are found by advancing 'tail' to the right.",algorithms and data structures.pdf
"4. Delete all points q with qx ≤ px – δfrom the y-table. They are found by advancing 'tail' to the right.
5. Insert p into the y-table.
6. Find all points q in the y-table with |qy – py| < δ by looking at the successors and predecessors of p. If such  
a point q is found and its distance from p is smaller than δ, update δ and the closest pair found so far.
Implementation
In the following implementation the x-queue is realized by an array that contains all the points sorted by their x-
coordinate. 'closestLeft' and 'closestRight' describe the pair of closest points found so far, n is the number of points  
under consideration, and t and c determine the positions of 'tail' and 'current':
xQueue: array[1 .. maxN] of point;
closestLeft, closestRight: point;
t, c, n: 1 .. maxN;
295",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
The x-queue is initialized by
procedure initX;
'initX' stores all the points into the x-queue, ordered by their x-
coordinates.
The empty y-table is created by
procedure initY;
A new point is inserted into the y-table by
procedure insertY(p: point);
A point is deleted from the y-table by
procedure deleteY(p: point);
The successor of a point in the y-table is returned by
function succY(p: point): point;
The predecessor of a point in the y-table is returned by
function predY(p: point): point;
The initialization part of the plane-sweep is as follows:
initX;  initY;
closestLeft := xQueue[1];  closestRight := xQueue[2];
delta := distance(closestLeft, closestRight);
insertY(closestLeft);  insertY(closestRight);
c := 3;
The events are processed by the loop:
while c ≤ n do  begin  transition;  c := c + 1;  { next event } 
end;
The procedure 'transition' encompasses all the work to be done for a new point:",algorithms and data structures.pdf
"while c ≤ n do  begin  transition;  c := c + 1;  { next event } 
end;
The procedure 'transition' encompasses all the work to be done for a new point:
procedure transition;
begin
{ step 1: remove points outside the δ-slice from the y-table }
current := xQueue[c];
while  current.x – xQueue[t].x ≥ delta  do  begin 
deleteY(xQueue[t]);  t := t + 1
end;
{ step 2: insert the new point into the y-table }
insertY(current);
{ step 3a: check the successors of the new point in the y-table }
check := current;
repeat
check := succY(check);
newDelta := distance(current, check);
if  newDelta < delta  then  begin
delta := newDelta;
closestLeft := check;  closestRight := current;
end;
until  check.y – current.y > delta;
{ step 3b: check the predecessors of the new point in the y-
table }
check := current;
repeat
check := predY(check);
newDelta := distance(current, check);
Algorithms and Data Structures 296  A Global Text",algorithms and data structures.pdf
"26. The closest pair
if  newDelta < delta  then  begin
delta := newDelta;
closestLeft := check;  closestRight := current;
end;
until  current.y – check.y > delta;
end;  { transition }
Analysis
We show that the algorithm described can be implemented so as to run in worst-case time O(n · log n) and space  
O(n).
If the y-table is implemented by a balanced binary tree (e.g. an AVL-tree or a 2-3-tree) the operations 'insertY',  
'deleteY', 'succY', and 'predY' can be performed in time O(log n). The space required is O(n).
'initX' builds the sorted x-queue in time O(n · log n) using space O(n). The procedure 'deleteY' is called at most  
once for each point and thus accumulates to O(n · log n). Every point is inserted once into the y-table, thus the calls  
of 'insertY' accumulate to O(n · log n).
There remains the problem of analyzing step 3. The loop in step 3a calls 'succY' once more than the number of",algorithms and data structures.pdf
"There remains the problem of analyzing step 3. The loop in step 3a calls 'succY' once more than the number of  
points in the upper half of the bounding box. Similarly, the loop in step 3b calls 'predY' once more than the number  
of points in the lower half of the bounding box. A standard counting technique shows that the bounding box is  
sparsely populated: For any metric dk, the box contains no more than a small, constant number c k of points, and for 
any k, ck ≤ 8. Thus 'succY' and 'predY' are called no more than 10 times, and step 3 costs time O(log n).
The key to this counting is the fact that no two points in the y-table can be closer than δ, and thus not many of  
them can be packed into the bounding box with sides  δ and 2 ·  δ. We partition this box into the eight pairwise  
disjoint, mutually exhaustive regions shown in  Exhibit 26.4 . These regions are half circles of diameter  δ in the",algorithms and data structures.pdf
"disjoint, mutually exhaustive regions shown in  Exhibit 26.4 . These regions are half circles of diameter  δ in the  
Manhattan metric d 1, and we first argue our case only when distances are measured in this metric. None of these  
half-circles can contain more than one point. If a half-circle contained two points at distance δ, they would have to  
be at opposite ends of the unique diameter of this half-circle. The se endpoints lie on the left or the right boundary  
of the bounding box, and these two boundary lines cannot contain any points, for the following reasons:
• No active point can be located on the left boundary of the bounding box; such a point would have been  
thrown out when the δ-slice was last updated.
• No active point can exist on the right boundary, as that x-coordinate is preempted by the transition point p  
being processed (remember our assumption of unequal x-coordinates).
297",algorithms and data structures.pdf
"This book is licensed under a Creative Commons Attribution 3.0 License
Exhibit 26.4: Only few points at pairwise distance ≥ δ can populate a box of size 2 · δ by δ.
We have shown that the bounding box can hold no more than eight points at pairwise distance ≥ δ when using 
the Manhattan metric d1. It is well known that for any points p, q, and for any k > 1: d1(p,q) > dk(p,q) > d∞(p,q). Thus 
the bounding box can hold no more than eight points at pairwise distance ≥ δ when using any distance dk or d∞.
Therefore, the calculation of the predecessors and successors of a transition point costs time O(log n) and  
accumulates to a total of O(n · log n) for all transitions. Summing up all costs results in O(n · log n) time and O(n)  
space complexity for this algorithm. Since Ω (n · log n) is a lower bound for the closest pair problem, we know that  
this algorithm is optimal.
Sweeping in three or more dimensions",algorithms and data structures.pdf
"this algorithm is optimal.
Sweeping in three or more dimensions
To  gain  insight  into  the  power  and  limitation  of  sweep  algorithms,  let  us  explore  whether  the  algorithm 
presented generalizes to higher-dimensional spaces. We illustrate our reasoning for three-dimensional space, but  
the same conclusion holds for any number of dimensions > 2. All of the following steps generalize easily.
Sort all the points according  to their  x-coordinate into the x-queue.  Sweep space with a y-z plane, and in  
processing the current transition point p, assume that we know the closest pair among all the points to the left of p,  
and their distance  δ. Then to determine whether p forms a new closest pair, look at all the points inside a half-
sphere of radius δ centered at p, extending to the left of p. In the hope of implementing this sphere query efficiently,  
we enclose this half sphere in a bounding box of side length 2 ·  δ in the y- and z-dimension, and  δ in the x-",algorithms and data structures.pdf
"we enclose this half sphere in a bounding box of side length 2 ·  δ in the y- and z-dimension, and  δ in the x-
dimension. Inside this box there can be at most a small, constant number c k of points at pairwise distance ≥ δ when 
using any distance dk or d∞.
We implement this box query in two steps: (1) by cutting off all the points farther to the left of p than δ, which is 
done by advancing 'tail' in the x-queue, and (2) by performing a square query among the points currently in the y-z-
table (which all lie in the δ-slice to the left of the sweep plane), as shown in Exhibit 26.5. Now we have reached the  
only  place  where  the  three-dimensional  algorithm  differs  substantially.  In  the  two-dimensional  case,  the 
corresponding  one-dimensional  interval  query  can  be  implemented  efficiently  in  time  O(log  n)  using  find, 
predecessor, and successor operations on a balanced tree, and using the knowledge that the size of the answer set is",algorithms and data structures.pdf
"predecessor, and successor operations on a balanced tree, and using the knowledge that the size of the answer set is  
Algorithms and Data Structures 298  A Global Text",algorithms and data structures.pdf
"26. The closest pair
bounded by a constant. In the three-dimensional case, the corresponding two-dimensional orthogonal range query  
cannot in general be answered in time O(log n) (per retrieved point) using any of the known data structures.  
Straightforward search requires time O(n), resulting in an overall time O(n 2) for the space sweep. This is not an  
interesting result for a problem that admits the trivial O(n2) algorithm of comparing every pair.
Exhibit 26.5: Sweeping a plane across three-dimensional space. Ideas generalize, but efficiency does not.
Sweeping reduces the dimensionality of a geometric problem by one, by replacing one space dimension by a  
""time  dimension"".  Reducing  a  two-dimensional  problem  to  a  sequence  of  one-dimensional  problems  is  often 
efficient because the total order defined in one dimension allows logarithmic search times. In contrast, reducing a",algorithms and data structures.pdf
"efficient because the total order defined in one dimension allows logarithmic search times. In contrast, reducing a  
three-dimensional problem to a sequence of two-dimensional problems rarely results in a gain in efficiency.
Exercises
1. Consider the following modification of the plane-sweep algorithm for solving the closest pair problem [BH  
92]. When encountering a transition point p do not process all points q in the y-table with |q y – py| < δ, but 
test only whether the distance of p to its successor or predecessor in the y-table is smaller than δ. When 
deleting a point q with q x ≤ px – δ from the y-table test whether the successor and predecessor of q in the y-
table are closer than δ. If a pair of points with a smaller distance than the current δ is found update δ and 
the closest pair found so far. Prove that this modified algorithm finds a closest pair  What is the time  
complexity of this algorithm?",algorithms and data structures.pdf
"the closest pair found so far. Prove that this modified algorithm finds a closest pair  What is the time  
complexity of this algorithm?
2. Design a divide-and-conquer algorithm which solves the closest pair problem. What is the time complexity  
of your algorithm? Hint: Partition the set of n points by a vertical line into two subsets of approximately n / 
2 points. Solve the closest pair problem recursively for both subsets. In the conquer step you should use the  
fact that δ is the smallest distance between any pair of points both belonging to the same subset. A point  
from the left subset can only have a distance smaller than δ to a point in the right subset if both points lie in 
a 2 · δ-slice to the left and to the right of the partitioning line. Therefore, you only have to match points  
lying in the left δ-slice against points lying in the right δ-slice.
299",algorithms and data structures.pdf
"University of Arkansas, Fayetteville University of Arkansas, Fayetteville 
ScholarWorks@UARK ScholarWorks@UARK 
Open Educational Resources 
2-8-2019 
University Physics I: Classical Mechanics University Physics I: Classical Mechanics 
Julio Gea-Banacloche 
University of Arkansas, Fayetteville 
Follow this and additional works at: https://scholarworks.uark.edu/oer 
 Part of the Atomic, Molecular and Optical Physics Commons, Elementary Particles and Fields and 
String Theory Commons, Engineering Physics Commons, and the Other Physics Commons 
Citation Citation 
Gea-Banacloche, J. (2019). University Physics I: Classical Mechanics. Open Educational Resources. 
Retrieved from https://scholarworks.uark.edu/oer/3 
This Textbook is brought to you for free and open access by ScholarWorks@UARK. It has been accepted for 
inclusion in Open Educational Resources by an authorized administrator of ScholarWorks@UARK. For more 
information, please contact scholar@uark.edu.",University Physics I Classical Mechanics.pdf
"University Physics I: Classical Mechanics
Julio Gea-Banacloche",University Physics I Classical Mechanics.pdf
"Cover image from NASA, https://www.nasa.gov/image-feature/jpl/not-really-starless-at-saturn",University Physics I Classical Mechanics.pdf
"University Physics I: Classical
Mechanics
Julio Gea-Banacloche
First revision, Fall 2019
This work is licensed under a Creative Commons Attribution-NonCommercial
4.0 International License.
Developed thanks to a grant from the University of Arkansas Libraries",University Physics I Classical Mechanics.pdf
ii,University Physics I Classical Mechanics.pdf
"Contents
Preface i
1 Reference frames, displacement, and velocity 1
1.1 Introduction ......................................... 1
1.1.1 Particles in classical mechanics .......................... 1
1.1.2 Aside: the atomic perspective ........................... 3
1.2 Position, displacement, velocity .............................. 3
1.2.1 Position ....................................... 4
1.2.2 Displacement .................................... 6
1.2.3 Velocity ....................................... 8
1.3 Reference frame changes and relative motion ...................... 1 4
1.4 In summary ......................................... 1 9
1.5 Examples .......................................... 2 1
1.5.1 Motion with (piecewise) constant velocity .................... 2 1
1.5.2 Addition of velocities, relative motion ...................... 2 4
iii",University Physics I Classical Mechanics.pdf
"iv CONTENTS
1.6 Problems .......................................... 2 7
2 Acceleration 31
2.1 The law of inertia...................................... 3 1
2.1.1 Inertial reference frames .............................. 3 3
2.2 Acceleration ......................................... 3 5
2.2.1 Average and instantaneous acceleration ..................... 3 5
2.2.2 Motion with constant acceleration ........................ 3 9
2.2.3 Acceleration as a vector .............................. 4 1
2.2.4 Acceleration in different reference frames .................... 4 1
2.3 Free fall ........................................... 4 2
2.4 In summary ......................................... 4 4
2.5 Examples .......................................... 4 6
2.5.1 Motion with piecewise constant acceleration................... 4 6
2.6 Problems .......................................... 4 8
3 Momentum and Inertia 51
3.1 Inertia ............................................ 5 1",University Physics I Classical Mechanics.pdf
"2.6 Problems .......................................... 4 8
3 Momentum and Inertia 51
3.1 Inertia ............................................ 5 1
3.1.1 Relative inertia and collisions ........................... 5 1
3.1.2 Inertial mass: definition and properties ..................... 5 5
3.2 Momentum ......................................... 5 6
3.2.1 Conservation of momentum; isolated systems .................. 5 6",University Physics I Classical Mechanics.pdf
"CONTENTS v
3.3 Extended systems and center of mass........................... 5 8
3.3.1 Center of mass motion for an isolated system.................. 5 9
3.3.2 Recoil and rocket propulsion ........................... 6 1
3.4 In summary ......................................... 6 1
3.5 Examples .......................................... 6 3
3.5.1 Reading a collision graph ............................. 6 3
3.5.2 Collision in different reference frames, center of mass,a n dr e c o i l....... 6 5
3.6 Problems .......................................... 6 7
4 Kinetic Energy 69
4.1 Kinetic Energy ....................................... 6 9
4.1.1 Kinetic energy in collisions ............................ 7 0
4.1.2 Relative velocity and coefficient of restitution.................. 7 4
4.2 “Convertible” and “translational” kinetic energy .................... 7 6
4.2.1 Kinetic energy and momentum in different reference frames .......... 7 9
4.3 In summary ......................................... 8 0",University Physics I Classical Mechanics.pdf
"4.2.1 Kinetic energy and momentum in different reference frames .......... 7 9
4.3 In summary ......................................... 8 0
4.4 Examples .......................................... 8 2
4.4.1 Collision graph revisited .............................. 8 2
4.4.2 Inelastic collision and explosive separation.................... 8 3
4.5 Problems .......................................... 8 6
5 Interactions and energy 89",University Physics I Classical Mechanics.pdf
"vi CONTENTS
5.1 Conservative interactions ................................. 8 9
5.1.1 Potential energy .................................. 9 0
5.1.2 Potential energy functions and “energy landscapes” .............. 9 4
5.2 Dissipation of energy and thermal energy ........................ 9 6
5.3 Fundamental interactions, and other forms of energy.................. 9 8
5.4 Conservation of energy ................................... 9 9
5.5 In summary .........................................1 0 1
5.6 Examples ..........................................1 0 3
5.6.1 Inelastic collision in the middle of a swing....................1 0 3
5.6.2 Kinetic energy to spring potential energy: block collides with spring .....1 0 4
5.7 Advanced Topics ......................................1 0 6
5.7.1 Two carts colliding and compressing a spring..................1 0 6
5.7.2 Getting the potential energy function from collisiondata ............1 0 7",University Physics I Classical Mechanics.pdf
"5.7.2 Getting the potential energy function from collisiondata ............1 0 7
5.8 Problems ..........................................1 0 9
6 Interactions, part 2: Forces 113
6.1 Force .............................................1 1 3
6.1.1 Forces and systems of particles ..........................1 1 5
6.2 Forces and potential energy ................................1 1 6
6.3 Forces not derived from a potential energy........................1 2 0
6.3.1 Tensions .......................................1 2 0",University Physics I Classical Mechanics.pdf
"CONTENTS vii
6.3.2 Normal forces ....................................1 2 2
6.3.3 Static and kinetic friction forces .........................1 2 3
6.3.4 Air resistance ....................................1 2 5
6.4 Free-body diagrams ....................................1 2 6
6.5 In summary .........................................1 2 7
6.6 Examples ..........................................1 2 9
6.6.1 Dropping an object on a weighing scale .....................1 2 9
6.6.2 Speeding up and slowing down ..........................1 3 0
6.7 Problems ..........................................1 3 4
7 Impulse, Work and Power 137
7.1 Introduction: work and impulse ..............................1 3 7
7.2 Work on a single particle .................................1 3 8
7.2.1 Work done by the net force, and the Work-Energy Theorem..........1 4 0
7.3 The “center of mass work”.................................1 4 1
7.4 Work done on a system by all the external forces....................1 4 2",University Physics I Classical Mechanics.pdf
"7.3 The “center of mass work”.................................1 4 1
7.4 Work done on a system by all the external forces....................1 4 2
7.4.1 The no-dissipation case ..............................1 4 3
7.4.2 The general case: systems with dissipation ...................1 4 7
7.4.3 Energy dissipated by kinetic friction .......................1 5 0
7.5 Power ............................................1 5 0
7.6 In summary .........................................1 5 1",University Physics I Classical Mechanics.pdf
"viii CONTENTS
7.7 Examples ..........................................1 5 3
7.7.1 Braking .......................................1 5 3
7.7.2 Work, energy and the choice of system: dissipative case ............1 5 4
7.7.3 Work, energy and the choice of system: non-dissipativec a s e..........1 5 5
7.7.4 Jumping .......................................1 5 6
7.8 Problems ..........................................1 5 8
8 Motion in two dimensions 161
8.1 Dealing with forces in two dimensions ..........................1 6 1
8.2 Projectile motion ......................................1 6 4
8.3 Inclined planes .......................................1 6 7
8.4 Motion on a circle (or part of a circle)..........................1 6 9
8.4.1 Centripetal acceleration and centripetal force ..................1 6 9
8.4.2 Kinematic angular variables ...........................1 7 2
8.5 In summary .........................................1 7 5
8.6 Examples ..........................................1 7 7",University Physics I Classical Mechanics.pdf
"8.5 In summary .........................................1 7 5
8.6 Examples ..........................................1 7 7
8.6.1 The penny on the turntable ............................1 7 7
8.7 Advanced Topics ......................................1 7 9
8.7.1 Staying on track ..................................1 7 9
8.7.2 Going around a banked curve ...........................1 8 1
8.7.3 Rotating frames of reference: Centrifugal force and Coriolis force .......1 8 4",University Physics I Classical Mechanics.pdf
"CONTENTS ix
8.8 Problems ..........................................1 8 6
9 Rotational dynamics 191
9.1 Rotational kinetic energy, and moment of inertia....................1 9 1
9.2 Angular momentum ....................................1 9 3
9.3 The cross product and rotational quantities.......................1 9 7
9.4 Torque ............................................2 0 1
9.5 Statics ............................................2 0 4
9.6 Rolling motion .......................................2 0 7
9.7 In summary .........................................2 1 1
9.8 Examples ..........................................2 1 3
9.8.1 Torques and forces on the wheels of an accelerating bicycle ..........2 1 3
9.8.2 Blocks connected by rope over a pulley with non-zero mass ..........2 1 4
9.9 Problems ..........................................2 1 7
10 Gravity 221
10.1 The inverse-square law...................................2 2 1",University Physics I Classical Mechanics.pdf
"9.9 Problems ..........................................2 1 7
10 Gravity 221
10.1 The inverse-square law...................................2 2 1
10.1.1 Gravitational potential energy ..........................2 2 4
10.1.2 Types of orbits under an inverse-square force..................2 2 7
10.1.3 Kepler’s laws ....................................2 3 3
10.2 Weight, acceleration, and the equivalence principle...................2 3 6
10.3 In summary .........................................2 4 1",University Physics I Classical Mechanics.pdf
"x CONTENTS
10.4 Examples ..........................................2 4 3
10.4.1 Orbital dynamics ..................................2 4 3
10.4.2 Orbital data from observations: Halley’s comet.................2 4 5
10.5 Advanced Topics ......................................2 4 7
10.5.1 Tidal Forces.....................................2 4 7
10.6 Problems ..........................................2 4 9
11 Simple harmonic motion 253
11.1 Introduction: the physics of oscillations.........................2 5 3
11.2 Simple harmonic motion..................................2 5 4
11.2.1 Energy in simple harmonic motion........................2 5 9
11.2.2 Harmonic oscillator subject to an external, constantf o r c e...........2 6 0
11.3 Pendulums .........................................2 6 2
11.3.1 The simple pendulum ...............................2 6 2
11.3.2 The “physical pendulum” .............................2 6 5
11.4 In summary .........................................2 6 6",University Physics I Classical Mechanics.pdf
"11.3.2 The “physical pendulum” .............................2 6 5
11.4 In summary .........................................2 6 6
11.5 Examples ..........................................2 6 8
11.5.1 Oscillator in a box (a basic accelerometer!)...................2 6 8
11.5.2 Meter stick as a physical pendulum........................2 7 0
11.6 Advanced Topics ......................................2 7 2
11.6.1 Mass on a spring damped by friction with a surface..............2 7 2",University Physics I Classical Mechanics.pdf
"CONTENTS xi
11.6.2 The Cavendish experiment: how to measureG with a torsion balance ....2 7 3
11.7 Problems ..........................................2 7 6
12 Waves in one dimension 279
12.1 Traveling waves.......................................2 7 9
12.1.1 The “wave shape” function: displacement and velocityo ft h em e d i u m .....2 8 1
12.1.2 Harmonic waves ..................................2 8 2
12.1.3 The wave velocity .................................2 8 5
12.1.4 Reflection and transmission of waves at a medium boundary .........2 8 6
12.2 Standing waves and resonance ..............................2 8 9
12.3 Conclusion, and further resources.............................2 9 3
12.4 In summary .........................................2 9 3
12.5 Examples ..........................................2 9 5
12.5.1 Displacement and density/pressure in a longitudinalw a v e...........2 9 5
12.5.2 Violin sounds ....................................2 9 8",University Physics I Classical Mechanics.pdf
"12.5.1 Displacement and density/pressure in a longitudinalw a v e...........2 9 5
12.5.2 Violin sounds ....................................2 9 8
12.6 Advanced Topics ......................................2 9 9
12.6.1 Chain of masses coupled with springs: dispersion, andl o n g - w a v e l e n g t hl i m i t .299
12.7 Problems ..........................................3 0 2
13 Thermodynamics 303
13.1 Introduction .........................................3 0 3
13.2 Introducing temperature ..................................3 0 4",University Physics I Classical Mechanics.pdf
"xii CONTENTS
13.2.1 Temperature and heat capacity..........................3 0 4
13.2.2 The gas thermometer ...............................3 0 6
13.2.3 The zero-th law...................................3 0 8
13.3 Heat and the first law...................................3 0 9
13.3.1 “Direct exchange of thermal energy” and early theories of heat ........3 0 9
13.3.2 The first law of thermodynamics.........................3 1 1
13.4 The second law and entropy................................3 1 2
13.4.1 Entropy .......................................3 1 3
13.4.2 The efficiency of heat engines...........................3 1 4
13.4.3 But what IS entropy, anyway?..........................3 1 7
13.5 In summary .........................................3 1 9
13.6 Examples ..........................................3 2 1
13.6.1 Calorimetry .....................................3 2 1
13.6.2 Equipartition of energy ..............................3 2 2",University Physics I Classical Mechanics.pdf
"13.6.1 Calorimetry .....................................3 2 1
13.6.2 Equipartition of energy ..............................3 2 2
13.7 Problems ..........................................3 2 3",University Physics I Classical Mechanics.pdf
"Preface
Students: if this is too long, at the very least read the last four paragraphs. Thank
you!
For many years Eric Mazur’sPrinciples and Practice of Physics was the required textbook for
University Physics I at the University of Arkansas. In writing this open-source replacement I
have tried to preserve some of its best features, while at thesame time condensing much of the
presentation, and reworking several sections that did not quite fit the needs of our curriculum:
primarily, the chapters on Thermodynamics, Waves, and Work.I h a v e a l s o s k i p p e d e n t i r e l y t h e
chapter on the “Principle of Relativity,” and instead distributed its contents among other chapters:
in particular, the Galilean reference frame transformations are now introduced at the very beginning
of the book, as are the law of inertia and the concept of inertial reference frames.
Over the past few decades, there has been a trend to increase the size of introductory physics text-",University Physics I Classical Mechanics.pdf
"Over the past few decades, there has been a trend to increase the size of introductory physics text-
books, by including more and more visual aids (pictures, diagrams, boxes... ), as well as lengthier
and more detailed explanations, perhaps in an attempt to reach as many students as possible, and
maybe even to take the place of the instructor altogether. Itseems to me that the result is rather
the opposite: a massive (and expensive) tome that no studentcould reasonably be expected to
read all the way through, at a time when “TL;DR” has become a popular acronym, and visual
learning aids (videotaped lectures, demonstrations, and computer simulations) are freely available
everywhere.
Our approach at the University of Arkansas, developed as a result of the work of Physics Education
Research experts John and Gay Stewart, is based instead on two essential facts. First, that different
students learn differently: some will learn best from a textbook, others will learn best from a lecture,",University Physics I Classical Mechanics.pdf
"students learn differently: some will learn best from a textbook, others will learn best from a lecture,
and most will only really learn from a hands-on approach, by working out the answers to questions
themselves. Second, just about everybody will benefit from repeated presentations of the material
to be learned, in different environments and even from slightly different points of view.
In keeping with this, we start by requiring the students to read the textbook material before coming
to lecture, and also take an online “reading quiz” where theycan check their understanding of what
xiii",University Physics I Classical Mechanics.pdf
"xiv CONTENTS
they have read. Then, in the lecture, they will have an opportunity to see the material presented
again, as a sort of executive summary delivered by, typically, a different instructor, who will also
be able to answer any questions they might have about the book’s presentation. Additionally, the
instructor will directly test the students’ understandingby means of conceptual questions asked of
the whole class, which are to be answered using clickers. Thisp r o m p t st h es t u d e n t st ot h i n kh a r d e r
about the material, and encourages them to discuss it on the spot with their classmates.
Immediately after each lecture, the students will have a labactivity where they will be able to
verify experimentally the concepts and principles to whichthey have just been introduced. Finally,
every week they will have an “open response” homework assignment where they get to apply the",University Physics I Classical Mechanics.pdf
"every week they will have an “open response” homework assignment where they get to apply the
principles, mathematically, to concrete problems. For botht h el a b sa n dt h eh o m e w o r k ,a d d i t i o n a l
assistance is provided by a group of dedicated teaching assistants, who are often able to explain
the material to the students in a way that better relates to their own experience.
In all this, the textbook is expected to play an important role, but certainly not to be the students’
only (nor even, necessarily, the primary) source of understanding or knowledge. Its job is to
start the learning process, and to stand by to provide a reference (among possibly several others)
afterwards. To fulfill this role, perhaps the most essentialrequirement is that it should bereadable,
and hence concise enough for every reading assignment to be ofm a n a g e a b l es i z e .T h i s( a sw e l la s",University Physics I Classical Mechanics.pdf
"and hence concise enough for every reading assignment to be ofm a n a g e a b l es i z e .T h i s( a sw e l la s
as e n s i b l eo r g a n i z a t i o n ,c l e a re x p l a n a t i o n s ,a n dam i n i m a lassortment of worked-out exercises and
end-of chapter problems) is what I have primarily tried to provide here.
As t u d e n tw h ow a n t sm o r ei n f o r m a t i o nt h a np r o v i d e di nt h i st extbook, or alternative explanations,
or more worked-out examples, can certainly get these from many other sources: first, of course,
the instructor and the teaching assistants, whose essentialr o l ea sl e a r n i n gf a c i l i t a t o r sh a st ob e
recognized from the start. Then, there is a variety of alternative textbooks available: the best
choice, probably, would still be Mazur’sPrinciples and Practice of Physics,s i n c ei tu s e st h es a m e
terminology and notation, introduces the material in almostt h es a m es e q u e n c e ,a n dh a st o n s",University Physics I Classical Mechanics.pdf
"terminology and notation, introduces the material in almostt h es a m es e q u e n c e ,a n dh a st o n s
of worked-out examples and self-quiz conceptual questions.T h a t b o o k i s a v a i l a b l e o n r e s e r v e i n
the Physics library, where it can be consulted by anybody. Ifas t u d e n tf e e l st h en e e df o ra n
alternative textbook that they can actually take home, one option is, of course, to actually buy
Mazur’s (which is what everybody had to do before); another option is to explore other open-source
textbooks available online, which have been around longer than this one and benefit from more
worked-out exercises and a more conventional presentation.O n e s u c h b o o k i sUniversity Physics I
at openstax.org (https://openstax.org/details/books/university-physics-volume-1).
Finally, there are also a large number of other online resources, although I would advise the students",University Physics I Classical Mechanics.pdf
"Finally, there are also a large number of other online resources, although I would advise the students
to approach them with caution, since not all of them may be totally rigorous, and some may end up
being more confusing than helpful. Some of my students have found the Khan Academy lectures
and/or the “Flipping Physics” lectures helpful; I personally would recommend the lectures of Walter
Lewin at MIT, if only for the wide array of cool demonstrationsy o uc a ns e et h e r e .",University Physics I Classical Mechanics.pdf
"CONTENTS xv
One last word, for the students who may have read this far, concerning the use of equations and
“proofs” in this book. It is essential to the nature of physicst ob ea b l et oc a s ti t sr e s u l t si n
mathematical terms, and to use math to explain and predict newr e s u l t s ;h e n c e ,e q u a t i o n sa n d
mathematical derivations are integral parts of any physicstextbook. I have, however, tried to keep
the math as simple as possible throughout, and I would not wantal e n g t h ym a t h e m a t i c a ld e r i v a t i o n
to get in the way of your reading assignment. If you are readingt h et e x ta n dc o m eu p o ns e v e r a l
lines of math, skim them at first to see if they make sense, but ify o ug e ts t u c kd on o ts p e n dt o o
much time on them: move on to the bottom line, and keep readingfrom there. You can always ask
your instructor or TA for help with the math later.",University Physics I Classical Mechanics.pdf
"much time on them: move on to the bottom line, and keep readingfrom there. You can always ask
your instructor or TA for help with the math later.
Iw o u l d ,h o w e v e r ,e n c o u r a g ey o ut or e t u r n ,e v e n t u a l l y ,t oa nyb i to fm a t ht h a ty o uf o u n dc h a l l e n g i n g
the first time around. Do try to go through all the algebra yourself! I have occasionally skipped
intermediate steps, just to keep the math from overwhelmingthe text: but these are typically
straightforward manipulations (multiplying or dividing both sides by something, moving something
from one side of the equation to the other, multiplying out a parenthesis or, conversely, pulling out
ac o m m o nf a c t o ro rd e n o m i n a t o r... ). If you actually work out, on your own, all the missing steps,
you will find it’s a great way to improve your algebra skills. This is something that will make it
much easier for you to deal with the homework and the exams later on this semester—and for the",University Physics I Classical Mechanics.pdf
"much easier for you to deal with the homework and the exams later on this semester—and for the
rest of your career as well.
P.S. A few possibly useful features.
The pdf version of this book is, of course searchable: you canuse command-F on the Mac or
control-F in Windows to search the text for any term (hint: trys e a r c h i n gt h et e x t b o o kb e f o r e
hitting Google!!). Hopefully, this will make up a little forthe lack of an actual index, which I just
haven’t found the time to compile yet.
The text is also pretty thoroughly hyperlinked: every equation number and figure number quoted is
al i n k .C l i c k i n go nt h ee q u a t i o nn u m b e r( f o ri n s t a n c e ,t h e(6.30), in this sentence) will take you to
where the equation was first displayed (in this case, in Chapter 6). Then you can use command-left
arrow (on the Mac) or alt-left arrow (on Windows) to go back tothe page you were reading before",University Physics I Classical Mechanics.pdf
"arrow (on the Mac) or alt-left arrow (on Windows) to go back tothe page you were reading before
you clicked on the link. Command- (or alt-) left and right arrows will let you go back and forth
between the two pages. (Note: these keyboard shortcuts onlywork in Adobe Acrobat, as far as I
know.)
The same thing works if you click on the reference to a figure number (actually that takes you to
the caption to the figure, so you may need to scroll up a bit to seet h efi g u r e ) .
Finally, the entries in the table of contents are links to therespective sections as well (except for
the preface, which for some reason does not work; oh, well... if you are reading this, at least you
made it here!).",University Physics I Classical Mechanics.pdf
"Chapter 1
Reference frames, displacement, and
velocity
1.1 Introduction
Classical mechanics is the branch of physics that deals withthe study of the motion of anything
(roughly speaking) larger than an atom or a molecule. That isal o to ft e r r i t o r y ,a n dt h em e t h o d s
and concepts of classical mechanics are at the foundation ofany branch of science or engineering
that is concerned with the motion of anything from a star to anamoeba—fluids, rocks, animals,
planets, and any and all kinds of machines. Moreover, even though the accurate description of
processes at the atomic level requires the (formally very different) methods of quantum mechanics,
at least three of the basic concepts of classical mechanics that we are going to study this semester,
namely, momentum, energy, and angular momentum, carry overinto quantum mechanics as well,
with the last two playing, in fact, an essential role.
1.1.1 Particles in classical mechanics",University Physics I Classical Mechanics.pdf
"with the last two playing, in fact, an essential role.
1.1.1 Particles in classical mechanics
In the study of motion, the most basic starting point is the concept of theposition of an object.
Clearly, if we want to describe accurately the position of a macroscopic object such as a car, we
may need a lot of information, including the precise shape ofthe car, whether it is turned this way
or that way, and so on; however, if all we want to know is how farthe car is from Fort Smith or
Fayetteville, we do not need any of that: we can just treat thecar as a dot, or mathematical point,
on the map—which is the way your GPS screen will show it, anyway. When we do this, we say
that are describing the car (or whatever the macroscopic object may be) as aparticle.
1",University Physics I Classical Mechanics.pdf
"2 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
In classical mechanics, an “ideal” particle is an object withn oa p p r e c i a b l es i z e — am a t h e m a t i c a l
point. In one dimension (that is to say, along a straight line), its position can be specified just by
giving a single number, the distance from some reference point, as we shall see in a moment (in
three dimensions, of course, three numbers are required). Int e r m so fe n e r g y( w h i c hi sp e r h a p st h e
most important concept in all of physics, and which we will introduce properly in due course), an
ideal particle has only one kind of energy, what we will latercall translational kinetic energy;i t
cannot have, for instance, rotational kinetic energy (sincei th a s“ n os h a p e ”f o rp r a c t i c a lp u r p o s e s ) ,
or any form of internal energy (elastic, thermal, etc.), since we assume it is too small to have any
internal structure in the first place.",University Physics I Classical Mechanics.pdf
"or any form of internal energy (elastic, thermal, etc.), since we assume it is too small to have any
internal structure in the first place.
The reason this is a useful concept is not just that we can oftent r e a te x t e n d e do b j e c t sa sp a r t i c l e si n
an approximate way (like the car in the example above), but also, and most importantly, that if we
want to be more precise in our calculations,we can always treat an extended object (mathematically)
as a collection of “particles.”The physical properties of the object, such as its energy, momentum,
rotational inertia, and so forth, can then be obtained by adding up the corresponding quantities for
all the particles making up the object. Not only that, but theinteractions between two extended
objects can also be calculated by adding up the interactionsbetween all the particles making up the
two objects. This is how, once we know the form of the gravitational force between two particles",University Physics I Classical Mechanics.pdf
"two objects. This is how, once we know the form of the gravitational force between two particles
(which is fairly simple, as we will see in Chapter 10), we can use that to calculate the force of
gravity between a planet and its satellites, which can be fairly complicated in detail, depending,
for instance, on the relative orientation of the planet and the satellite.
The mathematical tool we use to calculate these “sums” iscalculus—specifically, integration—and
you will see many examples of this... in your calculus courses. Calculus I is only a corequisite fort h i s
course, so we will not make a lot of use of it here, and in any casey o uw o u l dn e e dm u l t i d i m e n s i o n a l
integrals, which are an even more advanced subject, to do these kinds of calculations. But it may
be good for you to keep these ideas on the back of your mind. Calculus was, in fact, invented by
Sir Isaac Newton precisely for this purpose, and the developments of physics and mathematics have",University Physics I Classical Mechanics.pdf
"Sir Isaac Newton precisely for this purpose, and the developments of physics and mathematics have
been closely linked together ever since.
Anyway, back to particles, the plan for this semester is as follows: we will start our description of
motion by treating every object (even fairly large ones, sucha sc a r s )a sa“ p a r t i c l e , ”b e c a u s ew e
will only be concerned at first with its translational motionand the corresponding energy. Then
we will progressively make things more complex: by considering systems of two or more particles,
we will start to deal with theinternal energy of a system. Then we will move to the study ofrigid
bodies,w h i c ha r ea n o t h e ri m p o r t a n ti d e a l i z a t i o n : e x t e n d e do b j e cts whose parts all move together
as the object undergoes a translation or a rotation. This willa l l o wu st oi n t r o d u c et h ec o n c e p t
of rotational kinetic energy. Eventually we will considerwave motion, where different parts of",University Physics I Classical Mechanics.pdf
"of rotational kinetic energy. Eventually we will considerwave motion, where different parts of
an extended object (or “medium”) move relative to each other.S o , y o u s e e , t h e r e i s a l o g i c a l
progression here, with most parts of the course building on top of the previous ones, and energy as
one of the main connecting themes.",University Physics I Classical Mechanics.pdf
"1.2. POSITION, DISPLACEMENT, VELOCITY 3
1.1.2 Aside: the atomic perspective
As an aside, it should perhaps be mentioned that the buildingup of classical mechanics around
this concept of ideal particles had nothing to do, initially,w i t ha n yb e l i e fi n“ a t o m s , ”o ra na t o m i c
theory of matter. Indeed, for most 18th and 19th century physicists, matter was supposed to be a
continuous medium, and its (mental) division into particlesw a sj u s tam a t h e m a t i c a lc o n v e n i e n c e .
The atomic hypothesis became increasingly more plausible ast h e1 9 t hc e n t u r yw o r eo n ,a n db y
the 1920’s, when quantum mechanics came along, physicists had to face a surprising development:
matter, it turned out, was indeed made up of “elementary particles,” but these particles couldnot,
in fact, be themselves described by the laws of classical mechanics. One could not, for instance,",University Physics I Classical Mechanics.pdf
"in fact, be themselves described by the laws of classical mechanics. One could not, for instance,
attribute to them simultaneously well-defined positions andv e l o c i t i e s . Y e t ,i ns p i t eo ft h i s ,m o s t
of the conclusions of classical mechanics remain valid for macroscopic objects, because, most of the
time, it is OK to (formally) “break up” extended objects intochunks that are small enough to be
treated as particles, but large enough that one does not needquantum mechanics to describe their
behavior.
Quantum properties were first found to manifest themselves att h em a c r o s c o p i cl e v e lw h e nd e a l i n g
with thermal energy, because at one point it really became necessary to figure out where and how
the energy was stored at the truly microscopic (atomic) level. Thus, after centuries of successes,
classical mechanics met its first failure with the so-calledproblem of the specific heats,a n da",University Physics I Classical Mechanics.pdf
"classical mechanics met its first failure with the so-calledproblem of the specific heats,a n da
completely new physical theory—quantum mechanics—had to bed e v e l o p e di no r d e rt od e a lw i t h
the newly-discovered atomic world. But all this, as they say,i sa n o t h e rs t o r y ,a n df o ro u rv e r y
brief dealings with thermal physics—the last chapter in thisb o o k — w ew i l lj u s tt a k es p e c i fi ch e a t s
as given, that is to say, something you measure (or look up in atable), rather than something you
try to calculate from theory.
1.2 Position, displacement, velocity
Kinematics is the part of mechanics that deals with the mathematical description of motion, leaving
aside the question of what causes an object to move in a certainw a y . K i n e m a t i c s ,t h e r e f o r e ,d o e s
not include such things as forces or energy, which fall instead under the heading of dynamics. It",University Physics I Classical Mechanics.pdf
"not include such things as forces or energy, which fall instead under the heading of dynamics. It
may be said, then, that kinematics by itself is not true physics, but only applied mathematics; yet
it is still an essential part of classical mechanics, and itsmost natural starting point. This chapter
(and parts of the next one) will introduce the basic conceptsand methods of kinematics in one
dimension.",University Physics I Classical Mechanics.pdf
"4 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
1.2.1 Position
As stated in the previous section, we are initially interested only in describing the motion of a
“particle,” which can be thought of as a mathematical point ins p a c e . ( L a t e ro nw ew i l ls e et h a t ,
even for an extended object or system, it is often useful to consider the motion of a specific point
that we call the system’scenter of mass.) A point in three dimensions can be located by giving three
numbers, known as itsCartesian coordinates(or, more simply, itscoordinates). In two dimensions,
this works as shown in Figure1.1 below. As you can see, the coordinates of a point just tell us how
to find it by first moving a certain distancex,f r o map r e v i o u s l y - a g r e e do r i g i n ,a l o n gah o r i z o n t a l
(or x)a x i s ,a n dt h e nac e r t a i nd i s t a n c ey along a vertical (ory)a x i s . ( O r ,o fc o u r s e ,y o uc o u l d
equally well first move vertically and then horizontally.)",University Physics I Classical Mechanics.pdf
"equally well first move vertically and then horizontally.)
x axis
y axis
r
x
y
θ
Figure 1.1: The position vector,⃗r, of a point, and itsx and y components (the point’s coordinates).
The quantities x and y are taken to be positive or negative depending on what side ofthe origin
the point is on. Typically, we will always start by choosing apositive directionfor each axis, as the
direction along which the algebraic value of the corresponding coordinate increases. This is often
chosen to be to the right for the horizontal axis, and upwardsfor the vertical axis, but there is
nothing that says we cannot choose a different convention if itturns out to be more convenient. In
Figure 1.1,t h ea r r o w so nt h ea x e sd e n o t et h ep o s i t i v ed i r e c t i o nf o re a ch. Going by the grid, the
coordinates of the point shown arex =4u n i t s ,y =3u n i t s .
In two or three dimensions (and even, in a sense, in one dimension), the coordinates of a point can",University Physics I Classical Mechanics.pdf
"1.2. POSITION, DISPLACEMENT, VELOCITY 5
be interpreted as thecomponents of a vectorthat we call the point’sposition vector,a n dd e n o t e
by ⃗r(sometimes boldface letters are used for vectors, instead ofa na r r o wo nt o p ;i nt h a tc a s e ,t h e
position vector would be denoted byr). A vector is a mathematical object, with specific geometric
and algebraic properties, that physicists use to representaq u a n t i t yt h a th a sb o t ham a g n i t u d ea n d
ad i r e c t i o n .T h emagnitude of the position vector in Fig.1.1 is just the length of the arrow, which
is to say, 5 length units (by the Pythagorean theorem, the length of ⃗r,w h i c hw ew i l lo f t e nw r i t e
using absolute value bars as|⃗r|,i se q u a lt o
√
x2 + y2); this is just the straight-line distance of the
point to the origin. Thedirection of ⃗r,o nt h eo t h e rh a n d ,c a nb es p e c i fi e di nan u m b e ro fw a y s ;a",University Physics I Classical Mechanics.pdf
"point to the origin. Thedirection of ⃗r,o nt h eo t h e rh a n d ,c a nb es p e c i fi e di nan u m b e ro fw a y s ;a
common convention is to give the value of the angle that it makes with the positivex axis, which I
have denoted in the figure asθ (in this case, you can verify thatθ =t a n−1(y/x)=3 6.9◦). In three
dimensions, two angles would be needed to completely specifyt h ed i r e c t i o no f⃗r.
As you can see, giving the magnitude and direction of⃗ris a way to locate the point that is
completely equivalent to giving its coordinatesx and y.B y t h e s a m e t o k e n , t h e c o o r d i n a t e sx
and y are a way to specify the vector⃗rthat is completely equivalent to giving its magnitude and
direction. As I stated above, we callx and y the components (or sometimes, to be more specific,
the Cartesian components) of the vector⃗r.I nas e n s ea l lt h ev e c t o r st h a tw i l lb ei n t r o d u c e dl a t e r",University Physics I Classical Mechanics.pdf
"the Cartesian components) of the vector⃗r.I nas e n s ea l lt h ev e c t o r st h a tw i l lb ei n t r o d u c e dl a t e r
on this semester will derive their geometric and algebraic properties from the position vector⃗r,s o
once you know how to deal with one vector, you can deal with thema l l . T h eg e o m e t r i cp r o p e r t i e s
(by which I mean, how to relate a vector’s components to its magnitude and direction) I have
just covered, and will come back to later on in this chapter, and again in Chapter 8; the algebraic
properties (how to add vectors and multiply them by ordinarynumbers, which are calledscalars
in this context) I will introduce along the way.
For the first few chapters this semester, we are going to be primarily concerned with motion in
one dimension (that is to say, along a straight line, backwards or forwards), in which case all we
need to locate a point is one number, itsx (or y,o r z)c o o r d i n a t e ;w ed on o tt h e nn e e dt ow o r r y",University Physics I Classical Mechanics.pdf
"need to locate a point is one number, itsx (or y,o r z)c o o r d i n a t e ;w ed on o tt h e nn e e dt ow o r r y
particularly about vector algebra. Alternatively, we can simply say that a vector in one dimension
is essentially the same as its only component, which is just apositive or negative number (the
magnitude of the number being the magnitude of the vector, andi t ss i g ni n d i c a t i n gi t sd i r e c t i o n ) ,
and has the algebraic properties that follow naturally fromthat.
The description of the motion that we are aiming for is to find afunction of time,w h i c hw ed e n o t e
by x(t), that gives us the point’s position (that is to say, the valueo fx)f o ra n yv a l u eo ft h et i m e
parameter, t.( S e eE q .(1.10), below, for an example.) Remember thatx stands for a number that
can be positive or negative (depending on the side of the origin the point is on), and has dimensions",University Physics I Classical Mechanics.pdf
"can be positive or negative (depending on the side of the origin the point is on), and has dimensions
of length, so when giving a numerical value for it you must always include the appropriate units
(meters, centimeters, miles... ). Similarly, t stands for the time elapsed since some more or less
arbitrary “origin of time,” or time zero. Normallyt should always be positive, but in special cases
it may make sense to consider negative times (think of how we count years: “AD” would correspond
to “positive” and “BC” would correspond to negative—the difference being that there is actually no
year zero!). Anyway,t also is a number with dimensions, and must be reported with itsa p p r o p r i a t e",University Physics I Classical Mechanics.pdf
"6 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
units: seconds, minutes, hours, etc.
x (m)
t (s)
Figure 1.2: A possible position vs. time graph for an object moving in one dimension.
We will b e often interested in plotting the p osition of an ob ject as a function of time—that is to say,
the graph of the functionx(t). This may, in principle, have any shape, as you can see in Figure 1.2
above. In the lab, you will have a chance to use a position sensor that will automatically generate
graphs like that for you on the computer, for any moving objectt h a ty o ua i mt h ep o s i t i o ns e n s o r
at. It is, therefore, important that you learn how to “read” such graphs. For example, Figure
1.2 shows an object that starts, at the timet =0 ,ad i s t a n c e0.2m a w a y a n d t o t h e r i g h t o f t h e
origin (so x(0) = 0.2m ) , t h e n m o v e s i n t h e n e g a t i v e d i r e c t i o n t ox = −0.15 m, which it reaches at",University Physics I Classical Mechanics.pdf
"origin (so x(0) = 0.2m ) , t h e n m o v e s i n t h e n e g a t i v e d i r e c t i o n t ox = −0.15 m, which it reaches at
t =0 .5s ; t h e n t u r n s b a c k a n d m o v e s i n t h e o p po s i t e d i r e c t i o n u n t i lit reaches the pointx =0 .1m ,
turns again, and so on. Physically, this could be tracking thed a m p e do s c i l l a t i o n so fas y s t e ms u c h
as an object attached to a spring and sliding over a surface that exerts a friction force on it (see
Example 11.5.1).
1.2.2 Displacement
In one dimension, thedisplacement of an object over a given time interval is a quantity that
we denote as ∆x, and equals the difference between the object’s initial and final positions (in one
dimension, we will often call the “position coordinate” simply the “position,” for short):
∆x = xf − xi (1.1)
Here the subscripti denotes the object’s position at the beginning of the time interval considered,",University Physics I Classical Mechanics.pdf
"∆x = xf − xi (1.1)
Here the subscripti denotes the object’s position at the beginning of the time interval considered,
and the subscript f its position at the end of the interval. The symbol ∆ will consistently be
used throughout this book to denote achange in the quantity following the symbol, meaning the",University Physics I Classical Mechanics.pdf
"1.2. POSITION, DISPLACEMENT, VELOCITY 7
difference between its initial value and its final value. The time interval itself will be written as ∆t
and can be expressed as
∆t = tf − ti (1.2)
where againti and tf are the initial and final values of the time parameter (imagine, for instance,
that you are reading time in seconds on a digital clock, and youa r ei n t e r e s t e di nt h ec h a n g ei nt h e
object’s position between second 130 and second 132: thenti =1 3 0s ,t2 =1 3 2s ,a n d∆t =2s ) .
You can practice reading off displacements from Figure1.2.T h e d i s p l a c e m e n t b e t w e e nti =0 .5s
and tf =1 s ,f o ri n s t a n c e ,i s0.25 m (xi = −0.15 m,xf =0 .1m ) . O n t h e o t h e r h a n d , be t w e e n
ti =1sa n dtf =1 .3s , t h e d i s p l a c e m e n t i s ∆x =0 − 0.1= −0.1m .
Notice two important things about the displacement. First,it can be positive or negative. Positive
means the object moved, overall, in the positive direction;negative means it moved, overall, in",University Physics I Classical Mechanics.pdf
"means the object moved, overall, in the positive direction;negative means it moved, overall, in
the negative direction. Second, even when it is positive, thed i s p l a c e m e n td o e sn o ta l w a y se q u a l
the distance traveled by the object (distance, of course, isalways defined as a positive quantity),
because if the object “doubles back” on its tracks for some distance, that distance does not count
towards the overall displacement. For instance, looking again at Figure1.2 ,i nb e t w e e nt h et i m e s
ti =0 .5s a n dtf =1 .5s t h e o bj e c t m o v e d fi r s t 0.25 m in the positive direction, and then 0.15 m in
the negative direction, for a total distance traveled of 0.4m ; h o w e v e r , t h e t o t a l d i s p l a c e m e n t w a s
just 0.1m .
In spite of these quirks, the total displacement is, mathematically, a useful quantity, because often
we will have a way (that is to say, an equation) to calculate ∆x for a given interval, and then we",University Physics I Classical Mechanics.pdf
"we will have a way (that is to say, an equation) to calculate ∆x for a given interval, and then we
can rewrite Eq. (1.1)s ot h a ti tr e a d s
xf = xi +∆ x (1.3)
That is to say, if we know where the object started, and we haveaw a yt oc a l c u l a t e∆x,w ec a ne a s i l y
figure out where it ended up. You will see examples of this sortof calculation in the homework
later on.
Extension to two dimensions
In two dimensions, we write the displacement as the vector
∆⃗r= ⃗rf − ⃗ri (1.4)
The components of this vector are just the differences in the position coordinates of the two points
involved; that is, (∆⃗r)x (a subscriptx, y,e t c . ,i sas t a n d a r dw a yt or e p r e s e n tt h ex, y . . .component
of a vector) is equal toxf − xi,a n ds i m i l a r l y( ∆⃗r)y = yf − yi.",University Physics I Classical Mechanics.pdf
"8 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
x (m)
y (m)
ri
xi
yi
rf
Δr
xf
yf
Figure 1.3: The displacement vector for a particle that was initially at a point with position vector⃗ri and
ended up at a point with position vector⃗rf is thedifference of the position vectors.
Figure 1.3 shows how this makes sense. Thex component of ∆⃗rin the figure is ∆x =3 −7= −4m ;
the y component is ∆y =8 −4=4m . T h i sb a s i c a l l ys h o w sy o uh o wt os u b t r a c t( a n d ,b ye x t e nsion,
add, since ⃗rf = ⃗ri +∆ ⃗r)v e c t o r s : y o uj u s ts u b t r a c t( o ra d d )t h ec o r r e s p o n d i n gc o m ponents.
Note how, by the Pythagorean theorem, the length (or magnitude) of the displacement vector,
|∆⃗r| =
√
(xf − xi)2 +( yf − yi)2,e q u a l st h es t r a i g h t - l i n ed i s t a n c eb e t w e e nt h ei n i t i a lp o int and the
final point, just as in one dimension; of course, the particlecould have actually followed a very",University Physics I Classical Mechanics.pdf
"final point, just as in one dimension; of course, the particlecould have actually followed a very
different path from the initial to the final point, and therefore traveled a different distance.
1.2.3 Velocity
Average velocity
If you drive from Fayetteville to Fort Smith in 50 minutes, your average speed for the trip is
calculated by dividing the distance of 59.2m ib yt h et i m ei n t e r v a l :
average speed =distance
∆t = 59.2m i
50 min= 59.2m i
50 min× 60 min
1h r =7 1.0m p h ( 1 . 5 )
(this equation, incidentally, also shows you how to convertunits, and how you will be expected to",University Physics I Classical Mechanics.pdf
"1.2. POSITION, DISPLACEMENT, VELOCITY 9
work with significant figures this semester: the rule of thumbis, keep four significant figures in all
intermediate calculations, and report three in the final result).
The way we defineaverage velocity is similar to average speed, but with one important difference:
we use thedisplacement,i n s t e a do ft h ed i s t a n c e .S o ,t h ea v e r a g ev e l o c i t yvav of an object, moving
along a straight line, over a time interval ∆t is
vav = ∆x
∆t (1.6)
This definition has all the advantages and the quirks of the displacement itself. On the one hand,
it automatically comes with a sign (the same sign as the displacement, since ∆t will always be
positive), which tells us in what direction we have been traveling. On the other hand, it may not
be an accurate estimate of our averagespeed,i fw ed o u b l e db a c ka ta l l .I nt h em o s te x t r e m ec a s e ,",University Physics I Classical Mechanics.pdf
"be an accurate estimate of our averagespeed,i fw ed o u b l e db a c ka ta l l .I nt h em o s te x t r e m ec a s e ,
for a roundtrip (leave Fayetteville and return to Fayetteville), the average velocity would be zero,
since xf = xi and therefore ∆x =0 .
It is clear that this concept is not going to be very useful in general, if the object we are tracking
has a chance to double back in the time interval ∆t.A w a y t o p r e v e n t t h i s f r o m h a p p e n i n g , a n d
also getting a more meaningful estimate of the object’s speeda ta n yi n s t a n t ,i st om a k et h et i m e
interval very small. This leads to a new concept, that ofinstantaneous velocity.
Instantaneous velocity
We define the instantaneous velocity of an ob ject (a “particle”), at the timet = ti,a st h em a t h e -
matical limit
v =l i m
∆t→0
∆x
∆t (1.7)
The meaning of this is the following. Suppose we compute the ratio ∆x/∆t over successively smaller",University Physics I Classical Mechanics.pdf
"matical limit
v =l i m
∆t→0
∆x
∆t (1.7)
The meaning of this is the following. Suppose we compute the ratio ∆x/∆t over successively smaller
time intervals ∆t (all of them starting at the same timeti). For instance, we can start by making
tf = ti +1 s , t h e n t r ytf = ti +0 .5s ,t h e ntf = ti +0 .1s ,a n ds oo n .N a t u r a l l y ,a st h et i m ei n t e r v a l
becomes smaller, the corresponding displacement will alsobecome smaller—the particle has less
and less time to move away from its initial position,xi.T h e h o p e i s t h a t t h e s u c c e s s i v e r a t i o s
∆x/∆t will converge to a definite value: that is to say, that at some point we will start getting
very similar values, and that beyond a certain point making ∆t any smaller will not change any of
the significant digits of the result that we care about. This limit value is theinstantaneous velocity
of the object at the timeti.",University Physics I Classical Mechanics.pdf
"the significant digits of the result that we care about. This limit value is theinstantaneous velocity
of the object at the timeti.
When you think about it, there is something almost a bit self-contradictory about the concept
of instantaneous velocity. You cannot (in practice) determine the velocity of an object if all you
are given is a literal instant. You cannot even tell if the object is moving, if all you have is one
instant! Motion requires more than one instant, the passageof time. In fact, all the “instantaneous”
velocities that we can measure, with any instrument, are always really average velocities, only the",University Physics I Classical Mechanics.pdf
"10 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
average is taken over very short time intervals. Nevertheless, the fact is that for any reasonably
well-behaved position functionx(t), the limit in Eq. (1.7)i s mathematically well-defined, and it
equals what we call, in calculus, thederivative of the functionx(t):
v =l i m
∆t→0
∆x
∆t = dx
dt (1.8)
x (m)
t (s)0
5
5 10
(tf , xf)
Δx
Δt(ti , xi)
Figure 1.4: The slope of the green segment is the average velocity for the time interval ∆t shown. As ∆t
becomes smaller, this approaches the slope of the tangent at the point (ti,x i)
In fact, there is a nice geometric interpretation for this quantity: namely, it is the slope of a line
tangent to thex-vs-t curve at the point (ti,x i). As Figure1.4 shows, the average velocity ∆x/∆t
is the slope (rise over run) of a line segment drawn from the point (ti,x i)t ot h ep o i n t(tf ,x f )( t h e
green line in the figure). As we make the time interval smaller,b yb r i n g i n gtf closer to ti (and",University Physics I Classical Mechanics.pdf
"green line in the figure). As we make the time interval smaller,b yb r i n g i n gtf closer to ti (and
hence, also,xf closer toxi), the slope of this segment will approach the slope of the tangent line at
(ti,x i)( t h eb l u el i n e ) ,a n dt h i sw i l lb e ,b yt h ed e fi n i t i o n(1.7), the instantaneous velocity at that
point.
This geometric interpretation makes it easy to get a qualitative feeling, from the position-vs-time
graph, for when the particle is moving more or less fast. A large slope means a steep rise or fall,
and that is when the velocity will be largest—in magnitude. Asteep rise means a large positive
velocity, whereas a steep drop means a large negative velocity, by which I mean a velocity that is
given by a negative number which is large in absolute value. Int h ef u t u r e ,t os i m p l i f ys e n t e n c e s
like this one, I will just use the word “speed” to refer to the magnitude (that is to say, the absolute",University Physics I Classical Mechanics.pdf
"like this one, I will just use the word “speed” to refer to the magnitude (that is to say, the absolute
value) of the instantaneous velocity. Thus, speed (like distance) is always a positive number, by
definition, whereas velocity can be positive or negative; andas t e e ps l o p e( p o s i t i v eo rn e g a t i v e )
means the speed is large there.",University Physics I Classical Mechanics.pdf
"1.2. POSITION, DISPLACEMENT, VELOCITY 11
Conversely, looking at the samplex-vs-t graphs in this chapter, you may notice that there are times
when the tangent is horizontal, meaning it has zero slope, ands ot h ei n s t a n t a n e o u sv e l o c i t ya tt h o s e
times is zero (for instance, at the timet =1 .0s i n F i g u r e1.2). This makes sense when you think of
what the particle is actually doing at those special times: iti sj u s tc h a n g i n gd i r e c t i o n ,s oi t sv e l o c i t y
is going, for instance, from positive to negative. The way this happens is, it slows down, down...
the velocity gets smaller and smaller, and then, for just an instant (literally, a mathematical point
in time), it becomes zero before, the next instant, going negative.
We will b e coming back to this “reading of graphs” in the lab andt h eh o m e w o r k ,a sw e l la si nt h e
next chapter, when we introduce the concept of acceleration.
Motion with constant velocity",University Physics I Classical Mechanics.pdf
"next chapter, when we introduce the concept of acceleration.
Motion with constant velocity
If the instantaneous velocity of an object never changes, itmeans that it is always moving in the
same direction with the same speed. In that case, the instantaneous velocity and the average
velocity coincide, and that means we can writev =∆ x/∆t (where the size of the interval ∆t could
now be anything), and rewrite this equation in the form
∆x = v ∆t (1.9)
which is the same as
xf − xi = v (tf − ti)
Now suppose we keepti constant (that is, we fix the initial instant) but allow the time tf to change,
so we will just writet for an arbitrary value oftf ,a n dx for the corresponding value ofxf .W ee n d
up with the equation
x − xi = v (t − ti)
which we can also write as
x(t)= xi + v (t − ti)( 1 . 1 0 )
after some rearranging, and where the notationx(t)h a sb e e ni n t r o d u c e dt oe m p h a s i z et h a tw e",University Physics I Classical Mechanics.pdf
"x(t)= xi + v (t − ti)( 1 . 1 0 )
after some rearranging, and where the notationx(t)h a sb e e ni n t r o d u c e dt oe m p h a s i z et h a tw e
want to think ofx as a function oft.T h i s i s , n o t s u r p r i s i n g l y , t h e e q u a t i o n o f a s t r a i g h t l i n e —a
“curve” which is its own tangent and always has the same slope.
(Please make sure that you are not confused by the notation inEq. (1.10). The parentheses around
the t on the left-hand side mean that we are considering the position x as a function oft.O n
the other hand, the parentheses around the quantityt − ti on the right-hand side mean that we
are multiplying this quantity byv,w h i c hi sac o n s t a n th e r e . T h i sd i s t i n c t i o nw i l lb ep a r t i c u larly
important when we introduce the functionv(t)n e x t . )
Either one of equations (1.9)o r(1.11)c a nb eu s e dt os o l v ep r o b l e m si n v o l v i n gm o t i o nw i t hc o n s t a nt",University Physics I Classical Mechanics.pdf
"Either one of equations (1.9)o r(1.11)c a nb eu s e dt os o l v ep r o b l e m si n v o l v i n gm o t i o nw i t hc o n s t a nt
velocity, and again you will see examples of this in the homework.
Motion with changing velocity",University Physics I Classical Mechanics.pdf
"12 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
If the velocity changes with time, obtaining an expression for the position of the object as a
function of time may be a nontrivial task. In the next chapterwe will study an important special
case, namely, when the velocity changes at a constant rate (constant acceleration).
For the most general case, a graphical method that is sometimes useful is the following. Suppose
that we know the functionv(t), and we graph it, as in Figure1.5 below. Then the area under the
curve in between any two instants, sayti and tf ,i se q u a lt ot h et o t a ld i s p l a c e m e n to ft h eo b j e c t
over that time interval.
The idea involved is known in calculus asintegration,a n di tg o e sa sf o l l o w s .S u p p o s et h a tIb r e a k
v (m/s)
t (s)t1 t2 t3 ... tf
v1
v2
v(t)
Figure 1.5: How to get the displacement from the area under thev-vs-t curve.",University Physics I Classical Mechanics.pdf
"v (m/s)
t (s)t1 t2 t3 ... tf
v1
v2
v(t)
Figure 1.5: How to get the displacement from the area under thev-vs-t curve.
down the interval fromti to tf into equally spaced subintervals, beginning at the timeti (which I
am, equivalently, going to callt1,t h a ti s ,t1 ≡ ti,s oIh a v en o wt1,t 2,t 3,...t f ). Now suppose I treat
the object’s motion over each subinterval as if it were motionw i t hc o n s t a n tv e l o c i t y ,t h ev e l o c i t y
being that at the beginning of the subinterval. This, of course, is only an approximation, since
the velocity is constantly changing; but, if you look at Figure 1.5,y o uc a nc o n v i n c ey o u r s e l ft h a t
it will become a better and better approximation as I increaset h en u m b e ro fi n t e r m e d i a t ep o i n t s
and the rectangles shown in the figure become narrower and narrower. In this approximation, the
displacement during the first subinterval would be
∆x1 = v1(t2 − t1)( 1 . 1 1 )",University Physics I Classical Mechanics.pdf
"displacement during the first subinterval would be
∆x1 = v1(t2 − t1)( 1 . 1 1 )
where v1 = v(t1); similarly, ∆x2 = v2(t3 − t2), and so on.",University Physics I Classical Mechanics.pdf
"1.2. POSITION, DISPLACEMENT, VELOCITY 13
However, Eq. (1.11)i sj u s tt h ea r e ao ft h efi r s tr e c t a n g l es h o w nu n d e rt h ec u r v einF i g u r e1.5 (the
base of the rectangle has “length”t2 − t1,a n di t sh e i g h ti sv1). Similarly for the second rectangle,
and so on. So the sum ∆x1 +∆x2 +... is both an approximation to the area under thev-vs-t curve,
and an approximation to the total displacement ∆t.A s t h e s u b d i v i s i o n b e c o m e s fi n e r a n d fi n e r ,
and the rectangles narrower and narrower (and more numerous), both approximations become more
and more accurate. In the limit of “infinitely many,” infinitely narrow rectangles, you get both the
total displacement and the area under the curve exactly, andthey are both equal to each other.
Mathematically, we would write
∆x =
∫ tf
ti
v(t) dt (1.12)
where the stylized “S” (for “sum”) on the right-hand side is the symbol of the operation known",University Physics I Classical Mechanics.pdf
"∆x =
∫ tf
ti
v(t) dt (1.12)
where the stylized “S” (for “sum”) on the right-hand side is the symbol of the operation known
as integration in calculus. This is essentially the inverse of the process know as differentiation, by
which we got the velocity function from the position function, back in Eq. (1.8).
This graphical method to obtain the displacement from the velocity function is sometimes useful,
if you can estimate the area under thev-vs-t graph reliably. An important point to keep in mind
is that rectangles under the horizontal axis (correspondingt on e g a t i v ev e l o c i t i e s )h a v et ob ea d d e d
as having negative area (since the corresponding displacement is negative); see example 1.5.1 at
the end of this chapter.
Extension to two dimensions
In two (or more) dimensions, you define the average velocity vector as a vector⃗vav whose com-
ponents are vav,x =∆ x/∆t, vav,y =∆ y/∆t,a n ds oo n( w h e r e∆x, ∆y,... are the corresponding",University Physics I Classical Mechanics.pdf
"ponents are vav,x =∆ x/∆t, vav,y =∆ y/∆t,a n ds oo n( w h e r e∆x, ∆y,... are the corresponding
components of the displacement vector ∆⃗r). This can be written equivalently as the single vector
equation
⃗vav = ∆⃗r
∆t (1.13)
This tells you how to multiply (or divide) a vector by an ordinary number: you just multiply (or
divide) each component by that number. Note that, if the number in question is positive, this
operation does not change the direction of the vector at all,it just scales it up or down (which is
why ordinary numbers, in this context, are calledscalars). If the scalar is negative, the vector’s
direction is flipped as a result of the multiplication. Since∆t in the definition of velocity is always
positive, it follows that the average velocity vector alwaysp o i n t si nt h es a m ed i r e c t i o na st h e
displacement, which makes sense.
To get the instantaneous velocity, you just take the limit ofthe expression (1.13)a s∆ t → 0, for",University Physics I Classical Mechanics.pdf
"displacement, which makes sense.
To get the instantaneous velocity, you just take the limit ofthe expression (1.13)a s∆ t → 0, for
each component separately. The resulting vector⃗vhas components vx =l i m∆t→0 ∆x/∆t,e t c . ,
which can also be written asvx = dx/dt, vy = dy/dt, . . ..
All the results derived above hold for each spatial dimensiona n di t sc o r r e s p o n d i n gv e l o c i t yc o m -
ponent. For instance, the graphical method shown in Figure1.5 can always be used to get ∆x if",University Physics I Classical Mechanics.pdf
"14 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
the functionvx(t)i sk n o w n ,o re q u i v a l e n t l yt og e t∆y if you knowvy(t), and so on.
Introducing the velocity vector at this point does cause a little bit of a notational difficulty. For
quantities like x and ∆x,i ti sp r e t t yo b v i o u st h a tt h e ya r et h ex components of the vectors⃗r
and ∆⃗r,r e s p e c t i v e l y ;h o w e v e r ,t h eq u a n t i t yt h a tw eh a v es of a rb e enc a l l i n gs i m p l yv should more
properly be denoted asvx (or vy if the motion is along they axis). In fact, there is a convention
that if you use the symbol for a vector without the arrow on topor any x, y, . . .subscripts, you
must mean themagnitude of the vector. In this book, however, I have decidednot to follow that
convention, at least not until we get to Chapter 8 (and even then I will use it only for forces).
This is because we will spend most of our time dealing with motion in only one dimension, and",University Physics I Classical Mechanics.pdf
"This is because we will spend most of our time dealing with motion in only one dimension, and
it makes the notation unnecessarily cumbersome to keep having to write thex or y subscripts on
every component of every vector, when you really only have oned i m e n s i o nt ow o r r ya b o u ti nt h e
first place. So v will, throughout, refer to the relevant component of the velocity vector, to be
inferred from the context, until we get to Chapter 8 and actually need to deal with both avx and
a vy explicitly.
Finally, notice that the magnitude of the velocity vector,|⃗v| =
√
v2x + v2y + v2z,i se q u a lt ot h e
instantaneous speed,s i n c e ,a s∆t → 0, the magnitude of the displacement vector,|∆⃗r|,b e c o m e s
the actual distance traveled by the object in the time interval ∆t.
1.3 Reference frame changes and relative motion
Everything up to this point assumes that we are using a fixed, previously agreed uponreference",University Physics I Classical Mechanics.pdf
"1.3 Reference frame changes and relative motion
Everything up to this point assumes that we are using a fixed, previously agreed uponreference
frame.B a s i c a l l y ,t h i s i s j u s t a n o r i g i na n d a s e t o f a x e sa l o n g w h i cht om e a s u r eo u rc o o r d i n a t e s ,
as shown in Figure 1.
There are, however, a number of situations in physics that call for the use of different reference
frames, and, more importantly, that require us toconvert various physical quantities from one
reference frame to another. For instance, imagine you are onab o a to nar i v e r ,r o w i n gd o w n s t r e a m .
You are moving with a certain velocity relative to the water around you, but the water itself is
flowing with a different velocity relative to the shore, and youra c t u a lv e l o c i t yr e l a t i v et ot h es h o r e
is the sum of those two quantities. Ships generally have to dothis kind of calculation all the time,",University Physics I Classical Mechanics.pdf
"is the sum of those two quantities. Ships generally have to dothis kind of calculation all the time,
as do airplanes: the “airspeed” is the speed of a plane relative to the air around it, but that air is
usually moving at a substantial speed relative to the earth.
The way we deal with all these situations is by introducing twor e f e r e n c ef r a m e s ,w h i c hh e r eIa m
going to call A and B. One of them, say A, is “at rest” relative tot h ee a r t h ,a n dt h eo t h e ro n ei s
“at rest” relative to something else—which means, really, moving along with that something else.
(For instance, a reference frame at rest “relative to the river” would be a frame that’s moving along",University Physics I Classical Mechanics.pdf
"1.3. REFERENCE FRAME CHANGES AND RELATIVE MOTION 15
with the river water, like a piece of driftwood that you couldmeasure your progress relative to.)
In any case, graphically, this will look as in Figure1.6,w h i c hIh a v ed r a w nf o rt h et w o - d i m e n s i o n a l
case because I think it makes it easier to visualize what’s going on:
A
B
xAB
xBP
xAP
yBPyAP
yAB
P
rBP
rAP
rAB
Figure 1.6: Position vectors and coordinates of a pointP in two different reference frames, A and B.
In the reference frame A, the pointP has position coordinates (xAP ,y AP ). Likewise, in the reference
frame B, its coordinates are (xBP ,y BP ). As you can see, the notation chosen is such that every
coordinate in A will have an “A” as a first subscript, while thesecond subscript indicates the object
to which it refers, and similarly for coordinates in B.
The coordinates (xAB,y AB)a r es p e c i a l :t h e ya r et h ec o o r d i n a t e s ,i nt h er e f e r e n c ef r ame A, of the",University Physics I Classical Mechanics.pdf
"The coordinates (xAB,y AB)a r es p e c i a l :t h e ya r et h ec o o r d i n a t e s ,i nt h er e f e r e n c ef r ame A, of the
origin of reference frame B. This is enough to fully locate the frameBi nA ,a sl o n ga st h ef r a m e s
are not rotated relative to each other.
The thin colored lines I have drawn along the axes in Figure 1.6a r ei n t e n d e dt om a k ei tc l e a rt h a t
the following equations hold:
xAP = xAB + xBP
yAP = yAB + yBP (1.14)
Although the figure is drawn for the easy case where all these quantities are positive, you should
be able to convince yourself that Eqs. (1.14)h o l da l s ow h e no n eo rm o r eo ft h ec o o r d i n a t e sh a v e
negative values.",University Physics I Classical Mechanics.pdf
"16 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
All these coordinates are also the components of the respective position vectors, shown in the figure
and color-coded by reference frame (so, for instance,⃗rAP is the position vector ofP in the frame
A), so the equations (1.14)c a nb ew r i t t e nm o r ec o m p a c t l ya st h es i n g l ev e c t o re q u a t i o n
⃗rAP = ⃗rAB + ⃗rBP (1.15)
From all this you can see how to add vectors: algebraically, you just add their components sepa-
rately, as in Eqs. (1.14); graphically, you draw them so the tip of one vector coincides with the tail
of the other (we call this “tip-to-tail”), and then draw the sum vector from the tail of the first one
to the tip of the other one. (In general, to get two arbitrary vectors tip-to-tail you may need to
displace one of them; this is OK provided you do not change itsorientation, that is, provided you
only displace it, not rotate it. We’ll see how this works in a moment with velocities, and later on",University Physics I Classical Mechanics.pdf
"only displace it, not rotate it. We’ll see how this works in a moment with velocities, and later on
with forces.)
Of course, I showed you already how tosubtract vectors with Fig.1.3:a g a i n , a l g e b r a i c a l l y , y o u
just subtract the corresponding coordinates, whereas graphically you draw them with a common
origin, and then draw the vector from the tip of the vector youare subtracting to the tip of the
other one. If you read the previous paragraph again, you can see that Fig.1.3 can equally well be
used to show that ∆⃗r= ⃗rf −⃗ri,a st os h o wt h a t⃗rf = ⃗ri +∆ ⃗r.
In a similar way, you can see graphically from Fig.1.6 (or algebraically from Eq. (1.15)) that the
position vector ofP in the frame B is given by⃗rBP = ⃗rAP −⃗rAB.T h el a s tt e r mi nt h i se x p r e s s i o n
can be written in a different way, as follows. If I follow the convention I have introduced above, the
quantity xBA (with the order of the subscripts reversed) would be thex coordinate of the origin of",University Physics I Classical Mechanics.pdf
"quantity xBA (with the order of the subscripts reversed) would be thex coordinate of the origin of
frame A in frame B, and algebraically that would be equal to−xAB,a n ds i m i l a r l yyBA = −yAB.
Hence the vector equality⃗rAB = −⃗rBA holds. Then,
⃗rBP = ⃗rAP −⃗rAB = ⃗rAP + ⃗rBA (1.16)
This is, in a way, the “inverse” of Eq. (1.15): it tells us how to get the position ofP in the frame
Bi fw ek n o wi t sp o s i t i o ni nt h ef r a m eA .
Let me show next you how all this extends to displacements andvelocities. Suppose the pointP
indicates the position of a particle at the timet.O v e rat i m ei n t e r v a l∆t,b o t ht h ep o s i t i o no ft h e
particle and the relative position of the two reference frames may change. We can add yet another
subscript, i or f,( f o ri n i t i a la n dfi n a l )t ot h ec o o r d i n a t e s ,a n dw r i t e ,f o re xample,
xAP,i = xAB,i + xBP,i
xAP,f = xAB,f + xBP,f (1.17)
Subtracting these equations gives us the corresponding displacements:",University Physics I Classical Mechanics.pdf
"xAP,i = xAB,i + xBP,i
xAP,f = xAB,f + xBP,f (1.17)
Subtracting these equations gives us the corresponding displacements:
∆xAP =∆ xAB +∆ xBP (1.18)",University Physics I Classical Mechanics.pdf
"1.3. REFERENCE FRAME CHANGES AND RELATIVE MOTION 17
Dividing Eq. (1.18)b y∆t we get the average velocities1,a n dt h e nt a k i n gt h el i m i t∆t → 0w eg e t
the instantaneous velocities. This applies in the same way tot h ey coordinates, and the result is
the vector equation
⃗vAP = ⃗vBP + ⃗vAB (1.19)
Ih a v er e a r r a n g e dt h et e r m so nt h er i g h t - h a n ds i d et o( h o p e f ully) make it easier to visualize what’s
going on. In words: the velocity of the particleP relative to (ormeasured in)f r a m eAi se q u a lt o
the (vector) sum of the velocity of the particle as measured inf r a m eB ,p l u st h ev e l o c i t yo ff r a m e
Br e l a t i v et of r a m eA .
The result (1.19)i sj u s tw h a tw ew o u l dh a v ee x p e c t e df r o mt h ee x a m p l e sIm e n t ioned at the
beginning of this section, like rowing in a river or an airplane flying in the wind. For instance, for",University Physics I Classical Mechanics.pdf
"beginning of this section, like rowing in a river or an airplane flying in the wind. For instance, for
the airplane ⃗vBP could be its “airspeed” (only it has to be a vector, so it wouldbe more like its
“airvelocity”: that is, its velocity relative to the air around it), and⃗vAB would be the velocity of
the air relative to the earth (the wind velocity, at the location of the airplane). In other words, A
represents the earth frame of reference and B the air, or wind,f r a m eo fr e f e r e n c e . T h e n ,⃗vAP would
be the “true” velocity of the airplane relative to the earth.You can see how it would be important
to add these quantities as vectors, in general, by considering what happens when you fly in a cross
wind, or try to row across a river, as in Figure1.7 below.
vEb
vRb
vER
Figure 1.7: Rowing across a river. If you head “straight across” the river (with velocity vector⃗vRb in the",University Physics I Classical Mechanics.pdf
"vEb
vRb
vER
Figure 1.7: Rowing across a river. If you head “straight across” the river (with velocity vector⃗vRb in the
moving frame of the river, which is flowing with velocity⃗vER in the frame of the earth), your actual velocity
relative to the shore will be the vector⃗vEb. This is an instance of Eq. (1.19), with frame A being E (the
earth), frame B being R (the river), and “b” (for “boat”) standing for the point P we are tracking.
As you can see from this couple of examples, Equation (1.19)i so f t e nu s e f u la si ti sw r i t t e n ,b u t
1We have made a very natural assumption, that the time interval∆ t is the same for observers tracking the
particle’s motion in frames A and B, respectively (where eacho b s e r v e ri su n d e r s t o o dt ob em o v i n ga l o n gw i t hh i so r
her frame). This, however, turns out to benot true when any of the velocities involved is close to the speedof light,",University Physics I Classical Mechanics.pdf
"her frame). This, however, turns out to benot true when any of the velocities involved is close to the speedof light,
and so the simple addition of velocities formula (1.19)d o e sn o th o l di nE i n s t e i n ’ sr e l a t i v i t yt h e o r y .
(This is actually the first bit of real physics I have told you about in this book, so far; unfortunately, you will have
no use for it this semester!)",University Physics I Classical Mechanics.pdf
"18 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
sometimes the information we have is given to us in a different way: for instance, we could be given
the velocity of the object in frame A (⃗vAP ), and the velocity of frame B as seen in frame A (⃗vAB),
and told to calculate the velocity of the object as seen in frame B. This can be easily accomplished
if we note that the vector⃗vAB is equal to−⃗vBA;t h a ti st os a y ,t h ev e l o c i t yo ff r a m eBa ss e e nf r o m
frame A is just the opposite of the velocity of frame A as seen from frame B. Hence, Eq. (1.19)c a n
be rewritten as
⃗vAP = ⃗vBP −⃗vBA (1.20)
For most of the next few chapters we are going to be consideringo n l ym o t i o ni no n ed i m e n s i o n ,
and so we will write Eq. (1.19)( o r(1.20)) without the vector symbols, and it will be understood
that v refers to the component of the vector⃗valong the coordinate axis of interest.",University Physics I Classical Mechanics.pdf
"that v refers to the component of the vector⃗valong the coordinate axis of interest.
Aq u a n t i t yt h a tw i l lb ep a r t i c u l a r l yi m p o r t a n tl a t e ro ni st he relative velocityof two objects, which
we could label 1 and 2. The velocity of object 2 relative to object 1 is, by definition, the velocity
which an observer moving along with 1 would measure for object2 . S oi ti sj u s tas i m p l ef r a m e
change: let the earth frame be frame E and the frame moving witho b j e c t1b ef r a m e1 ,t h e nt h e
velocity we want isv12 (“velocity of object 2 in frame 1”). If we make the change A→ 1, B→ E,
and P→ 2i nE q .(1.20), we get
v12 = vE2 − vE1 (1.21)
In other words, the velocity of 2 relative to 1 is just the velocity of 2 minus the velocity of 1. This
is again a familiar effect: if you are driving down the highway at5 0m i l e sp e rh o u r ,a n dt h ec a ri n
front of you is driving at 55, then its velocity relative to youi s5m p h ,w h i c hi st h er a t ea tw h i c h",University Physics I Classical Mechanics.pdf
"front of you is driving at 55, then its velocity relative to youi s5m p h ,w h i c hi st h er a t ea tw h i c h
it is moving away from you (in the forward direction, assumedto be the positive one).
It is important to realize that all these velocities arereal velocities, each in its own reference frame.
Something may be said to be truly moving at some velocity in oner e f e r e n c ef r a m e ,a n dj u s ta s
truly moving with a different velocity in a different reference frame. I will have a lot more to say
about this in the next chapter, but in the meantime you can reflect on the fact that, if a car moving
at 55 mph collides with another one moving at 50 mph in the samedirection, the damage will be
basically the same as if the first car had been moving at 5 mph andt h es e c o n do n eh a db e e na t
rest. For practical purposes, where you are concerned, another car’s velocity relative to yoursis
that car’s “real” velocity.
Resources",University Physics I Classical Mechanics.pdf
"rest. For practical purposes, where you are concerned, another car’s velocity relative to yoursis
that car’s “real” velocity.
Resources
Ag o o da p pf o rp r a c t i c i n gh o wt oa d dv e c t o r s( a n dh o wt ob r e a kthem up into components, mag-
nitude and direction, etc.) may be found here:
https://phet.colorado.edu/en/simulation/vector-addition.
Perhaps the most dramatic demonstration of how Eq. (1.19)w o r k si nt h er e a lw o r l di si nt h i s
episode ofMythbusters: https://www.youtube.com/watch?v=BLuI118nhzc.( I ft h i sl i n kd o e sn o t
work, do a search for “Mythbusters cancel momentum.”) They shoot a ball from the bed of a truck,",University Physics I Classical Mechanics.pdf
"1.4. IN SUMMARY 19
with a velocity (relative to the truck) of 60 mph backwards, while the truck is moving forward at 60
mph. I think the result is worth watching. (Do not be distracted by their talk about momentum.
We will get there, in time.)
A very old, but also very good, educational video about different frames of reference is this one:
https://www.youtube.com/watch?v=sS17fCom0Ns.Y o u s h o u l d t r y t o w a t c h a t l e a s t p a r t o f i t .
Many things will be relevant to later parts of the course, including projectile motion, and the whole
discussion of relative motion coming up next, in Chapter 2.
1.4 In summary
1. To describe the motion of an object in one dimension we treati ta sam a t h e m a t i c a lp o i n t ,
and consider itsposition coordinate, x (often shortened to just theposition), as a function of
time: x(t).
2. Numerically, the position coordinate is the distance to achosen origin, with a positive or",University Physics I Classical Mechanics.pdf
"time: x(t).
2. Numerically, the position coordinate is the distance to achosen origin, with a positive or
negative sign depending on which side of the origin the pointis. For every problem, when we
introduce a coordinate axis we need to specify a positive direction. Starting from the origin
in that direction, the position coordinate is positive and increasing, whereas going from the
origin in the opposite direction (negative direction) it becomes increasingly negative.
3. The displacement of an object over a time interval from an initial time ti to a final timetf
is the quantity ∆x = xf −xi,w h e r exf is the position of the object at the final time (or, the
final position), andxi the position at the initial time (or initial position).
4. The average velocity of an object over the time interval from ti to tf is defined asvav =∆ x/∆t,
where ∆t = tf − ti.
5. The instantaneous velocity (often just called thevelocity)o fa no b j e c ta tt h et i m et is the",University Physics I Classical Mechanics.pdf
"where ∆t = tf − ti.
5. The instantaneous velocity (often just called thevelocity)o fa no b j e c ta tt h et i m et is the
limit value of the quantity ∆x/∆t,c a l c u l a t e df o rs u c c e s s i v e l ys h o r t e rt i m ei n t e r v a l s∆t,a l l
with the same initial timeti = t.T h i s i s , m a t h e m a t i c a l l y ,t h e d e fi n i t i o n o f t h e d e r i v a t i v e of
the functionx(t)a tt h et i m et,w h i c hw ee x p r e s sa sv = dx/dt.
6. Graphically, the instantaneous velocity of the object atthe timet is the slope of the tangent
line to thex-vs-t graph at the timet.
7. The instantaneous velocity of an object is a positive or negative quantity depending on
whether the object, at that instant, is moving in the positiveo rt h en e g a t i v ed i r e c t i o n .
8. For an object moving withconstant velocity v,t h ep o s i t i o nf u n c t i o ni sg i v e nb y[ E q .(1.10)]:
x(t)= xi + v (t − ti)",University Physics I Classical Mechanics.pdf
"20 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
where ti is an arbitrarily chosen initial time andxi the position at that time. This can also
be written in the form given by Eq. (1.9). The argument (t)o nt h el e f t - h a n ds i d eo f(1.10)i s
optional, andti is often set equal to zero, giving justx = xi + vt.T h i s , h o w e v e r ,i sn o tq u i t e
as generally applicable as the result (1.9)o r(1.10).
9. For an object moving with changing velocity, the total displacement in between timesti and
tf is equal to the total area under thev-vs-t curve in between those times; areas below the
horizontal (t)a x i sm u s tb et r e a t e da sn e g a t i v e .
10. In two or more dimensions one introduces, for every pointin space, aposition vector whose
components are just the Cartesian coordinates of that point;t h e nt h ed i s p l a c e m e n tv e c t o ri s
defined as ∆⃗r= ⃗rf − ⃗ri,t h ea v e r a g ev e l o c i t yv e c t o ri s⃗vav =∆ ⃗r /∆t,a n dt h ei n s t a n t a n e o u s",University Physics I Classical Mechanics.pdf
"defined as ∆⃗r= ⃗rf − ⃗ri,t h ea v e r a g ev e l o c i t yv e c t o ri s⃗vav =∆ ⃗r /∆t,a n dt h ei n s t a n t a n e o u s
velocity vector is the limit of this as ∆t goes to zero. Vectors are added by adding their
components separately; to multiply a vector by an ordinary number, orscalar,w ej u s tm u l t i p l y
each component by that number.
11. When tracking the motion of an object, “P”, in two differentreference frames, A and B,
the position vectors are related by⃗rAP = ⃗rAB + ⃗rBP ,a n dl i k e w i s et h ev e l o c i t yv e c t o r s :
⃗vAP = ⃗vAB + ⃗vBP .H e r e , t h e fi r s t s u b s c r i p t t e l l s y o u i n w h i c h r e f e r e n c e f r a m eyou are
measuring, and the second subscript what it is that you are looking at; ⃗rAB is the position
vector of the origin of frame B as seen in frame A, and⃗vAB its velocity.",University Physics I Classical Mechanics.pdf
"1.5. EXAMPLES 21
1.5 Examples
1.5.1 Motion with (piecewise) constant velocity
You leave your house on your bicycle to go visit a friend. At your normal speed of 9 mph, you know
it takes you 6 minutes to get there. This time, though, when youh a v et r a v e l e dh a l ft h ed i s t a n c e
you realize you forgot a book at home that you were going to return to your friend, so you turn
around and pedal at twice your normal speed, get back home, grab the book, and start off again
for your friend’s house at 18 mph (imagine you are really fit topull this off!)
(a) How far away from you does your friend live?
(b) What is the total distance you travel on this trip?
(c) How long did the whole trip take?
(d) Draw a position versus time and a velocity versus time graph for the whole trip. Use SI units for
both graphs. Neglect the time it takes you to stop and turn around, and also the time it takes you",University Physics I Classical Mechanics.pdf
"both graphs. Neglect the time it takes you to stop and turn around, and also the time it takes you
to run into your house and grab the book (in other words, assumet h o s ec h a n g e si ny o u rd i r e c t i o n
of motion happen instantly).
(e) Show explicitly, using yourv-vs-t graph, that the graphical method of Figure1.5 gives you the
total displacement for your trip.
Solution
Ia mg o i n gt ow o r ko u tt h i sp r o b l e mu s i n gb o t hm i l e sa n dS Iu n i ts, the first because it seems most
natural, and the second because we are asked to use SI units forp a r t( d ) ,s ow em i g h ta sw e l lu s e
them from the start. In general, you should use SI units whenever you can. If you are unsure of
what to do in a specific problem, ask your instructor!
(a) We are told that at 9 miles per hour it would take 6 minutes tog e tt h e r e ,s ol e tu su s e
∆x = v∆t (1.22)",University Physics I Classical Mechanics.pdf
"(a) We are told that at 9 miles per hour it would take 6 minutes tog e tt h e r e ,s ol e tu su s e
∆x = v∆t (1.22)
with v =9 m p ha n d∆t =6 m i n . W eh a v et oe i t h e rc o n v e r tt h eh o u r st om i n u t e s ,o rv i c e-versa.
Again, in this case it seems easiest to realize that 6 min equals 1/10 of an hour, so:
∆x =
(
9 miles
hr
)
× 0.1h r=0 .9m i l e s. (1.23)
In SI units, 9 mph = 4.023 m/s, and 6 min = 360 s, so ∆x =1 4 4 8m .
(b) This is just a matter of keeping track of the distance traveled in the various parts of the trip.
You start by riding half the distance to your friend’s house,which is to say, 0.45 miles, and then
you ride that again back home, so that’s 0.9m i l e s , a n d t h e n y o u ’ r e b a c k w h e r e y o u s t a r t e d , s o y o u
still have to go the 0.9m i l e s t o y o u r f r i e n d ’ s h o u s e . S o o v e r a l l , y o u r i d e f o r 1.8m i l e s , o r 2 8 9 7m .",University Physics I Classical Mechanics.pdf
"22 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
(c) The whole trip consists, as detailed above, of 0.45 miles at 9 mph, and the rest, which is
1.35 miles, at 18 mph. Applying ∆t =∆ x/v to each of these intervals, we get a total time of
∆t = 0.45 miles
9m p h + 1.35 miles
18 mph =0 .125 hours
=0 .125 × 60 min = 7.5m i n
=7 .5 × 60 s = 450 s (1.24)
(d) The graphs are shown below. Details on how to get them follow.
x (m)
v (m/s)
t (s)
t (s)
• First interval: fromt =0t o t =1 8 0s( 3m i n ,w h i c hi sw h a ti tw o u l dt a k et oc o v e rh a l ft h ed i s tance
to your friend’s house at 9 mph). The velocity is a constantv =4 .023 m/s. For the position graph,
use Eq. (1.10)w i t hxi =0 ,ti =0a n dv =4 .023 m/s.
• Second interval: fromt =1 8 0 st ot =2 7 0 s( i tt a k e sy o uh a l fo f3 m i n ,w h i c hi st os a y9 0 s ,t o
cover the same distance as above at twice the speed). The velocity is a constantv = −8.046 m/s",University Physics I Classical Mechanics.pdf
"1.5. EXAMPLES 23
(twice what it was earlier, but in the opposite direction). For the position graph, use Eq. (1.10)
with xi =7 2 4 m( t h i si sh a l fo ft h ed i s t a n c et oy o u rf r i e n d ’ sh o u s e ,a ndt h es t a r t i n gp o s i t i o nf o r
this interval),ti =1 8 0sa n dv = −8.046 m/s.
• Third interval: fromt =2 7 0st ot =4 5 0s . T h ev e l o c i t yi sac o n s t a n tv =8 .046 m/s (same speed
as just before, but in the opposite direction). For the position graph, use Eq. (1.10)w i t hxi =0m
(you start back at your house),ti =2 7 0sa n dv =8 .046 m/s.
If you are familiar with the software packageMathematica,t h ep o s i t i o ng r a p hw a sp r o d u c e du s i n g
the command
Plot[If[t<180, 4.023 t, If[t<270, 4.023*180-8.046 (t-180), 8.046 (t-270)]],{t, 0, 450}]
and the velocity graph was produced using
Plot[If[t<180, 4.023 , If[t<270, -8.046, 8.046]],{t, 0, 450}]
(and then connecting the horizontal lines by hand, which is not necessary, but helps to visualize",University Physics I Classical Mechanics.pdf
"(and then connecting the horizontal lines by hand, which is not necessary, but helps to visualize
what’s going on).
The graphs could also have been produced using the free plotting software package Gnuplot (avail-
able here: http://www.gnuplot.info/download.html)w i t ht h ef o l l o w i n gc o m m a n d s :
gnuplot> set dummy t
gnuplot> f(t) = t<180 ? 4.023*t : t<270 ? 4.023*180-8.046*(t-180) : 8.046*(t-270)
gnuplot> plot [0:450] f(t)
The first line sets the default independent variable tot (instead of x,w h i c hi sw h a tG n u p l o te x -
pects). The second line defines the piecewise function usingthe ternary operator (? :) borrowed
from the C programming language. The third line plots the function over the range indicated.
(e) For this we need to find the area under thev-vs-t graph we just plotted. Basically, we have three
rectangles: the first one has base 180 units (s) and height 4 units (m/s), so its area is 4×180 = 720",University Physics I Classical Mechanics.pdf
"rectangles: the first one has base 180 units (s) and height 4 units (m/s), so its area is 4×180 = 720
(m). The second rectangle has base 90 units and height−8( n e g a t i v e ,b e c a u s ei ti sb e l o wt h e
horizontal axis!), so its area is−720. The last one has base 180 units again (from 270 to 450) and
height 8, so its area is 8× 180 = 1440. So the total area “under” thev-vs-t curve is
720 − 720 + 1440 = 1440 meters
which is (approximately) your total displacement, that is,the 9 miles to your friend’s house. (Of
course, we would have obtained a more accurate result if we hadu s e dt h em o r ea c c u r a t ev a l u e sf o r
the “heights” of 4.023, −8.046, and 8.046, but if all we have to go by is the graph, such accuracy
is pretty much impossible.)",University Physics I Classical Mechanics.pdf
"24 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
1.5.2 Addition of velocities, relative motion
This example was inspired by the “race on a moving sidewalk” demo at
http://physics.bu.edu/~duffy/classroom.html.
Please go take a look at it!
Two girls, Ann and Becky (yes, A and B) decide to have a race while they wait for a plane at a
nearly-deserted airport. Ann will run on the moving walkway,t ot h ee n do fi t( w h i c hi s3 0 ma w a y )
and back, whereas Becky will run alongside her on the (non-moving) floor, also 30 m out and back.
The walkway moves at 1 m/s, and the girls both run at the same constant speed of 5 m/srelative
to the surface they are standing on.
(a) Relative to the (non-moving) floor, what is Ann’s velocityf o rt h efi r s tl e go fh e rr a c e ,w h e ns h e
is moving in the same direction as the walkway (take that to bethe positive direction)? What is
her velocity for the return leg?
(b) How long does it take each of the girls to complete their race?",University Physics I Classical Mechanics.pdf
"her velocity for the return leg?
(b) How long does it take each of the girls to complete their race?
(c) When both girls are running in the positive direction, what is Becky’s velocity relative to Ann?
(That is, how fast does Ann see Becky move, and in what direction?)
(d) When Ann turns around and starts running in the negative direction, but Becky is still running
in the positive direction, what is Becky’s velocity relativet oA n n ?
(e) What is the total distance Ann runsin the moving walkway’s frame of reference?
Solution
Ia mg o i n gt os o l v et h i si nt h ef o r m a tt h a ty o uw i l lb er e q u i r e dto use this semester for most of
the homework and exam problems. I will not be able to do this fore v e r ys i n g l ee x a m p l e ,b u ty o u
should! Please follow this carefully.
To begin with, you must draw a sketch of the situation described in the problem, detailed enough
to include all the relevant information you are given. Here ism i n e :
vWA= 5 m/s
vFW= 1 m/s",University Physics I Classical Mechanics.pdf
"to include all the relevant information you are given. Here ism i n e :
vWA= 5 m/s
vFW= 1 m/s
vFB= 5 m/s
30 m
going out
Alice
Becky
vWA= –5 m/s
vFW= 1 m/s
vFB= –5 m/s
coming back",University Physics I Classical Mechanics.pdf
"1.5. EXAMPLES 25
Note that I have drawn one picture for each half of the race, andt h a ta l lt h ei n f o r m a t i o ng i v e ni n
the text of the problem is there. The figure makes it clear alsothe notation I will be using for each
of the girls’ velocities, and to see at a glance what is happening.
You should next state what kind of problem this is and what basic result (theorem, principle, or
equation(s)) you are going to use to solve it. For this problem, you could say:
“This is a relative motion/reference frame transformationproblem. I will use Eq. (1.19)
⃗vAP = ⃗vBP + ⃗vAB
as well as the basic equation for motion with constant velocity:”
∆x = v∆t
After that, solve each part in turn, and make sure to show all your work!
Part (a): LetF stand for the floor frame of reference, andW the walkway frame. In the notation
of Section 1.3, we havevFW =1m / s . F o rt h efi r s tl e go fh e rr a c e ,w ea r et o l dt h a tA n n ’ sv e l ocity",University Physics I Classical Mechanics.pdf
"of Section 1.3, we havevFW =1m / s . F o rt h efi r s tl e go fh e rr a c e ,w ea r et o l dt h a tA n n ’ sv e l ocity
relative to the walkwayis 5 m/s, sovWA =5 m / s . T h e n ,b yE q .(1.19)( w i t ht h ef o l l o w i n gc h a n g e
of indices: A → F, B → W,a n dP → A),
vFA = vFW + vWA =1 m
s +5 m
s =6 m
s (1.25)
(when you see an equation like this, full of subscripts, it isag o o dp r a c t i c et or e a di to u t ,m e n t a l l y ,
to yourself: “Ann’s velocity relative to the floor equals thevelocity of the walkway relative to the
floor plus Ann’s velocity relative to the walkway.” Then takeam o m e n tt os e ei fi tm a k e ss e n s e !
Here is a place where the picture can be really helpful.)
For the return leg, use the same formula, but note that now hervelocity relative to the walkway is
negative, vWA = −5m / s , s i n c e s h e i s m o v i n g i n t h e o p po s i t e d i r e c t i o n :
vFA = vFW + vWA =1 m
s − 5 m
s = −4 m
s (1.26)",University Physics I Classical Mechanics.pdf
"negative, vWA = −5m / s , s i n c e s h e i s m o v i n g i n t h e o p po s i t e d i r e c t i o n :
vFA = vFW + vWA =1 m
s − 5 m
s = −4 m
s (1.26)
Part (b): Relative to the floor reference frame, we have just seen that Ann first covers 30 m at a
speed of 6 m/s, and then the same 30 m at a speed of 4 m/s, so her total time is
∆tA = 30m
6m/s + 30m
4m/s =5s+7 .5s =1 2.5s ( 1 . 2 7 )
whereas Becky just runs 30 m at 5 m/s both ways, so it takes her 6se i t h e rw a y ,f o rat o t a lo f1 2 s ,
which means she wins the race.",University Physics I Classical Mechanics.pdf
"26 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
Part (c): The quantity we want is written, in the notation of Section 1.3,vAB (“velocity of Becky
relative to Ann”). To calculate this, we just need to know thevelocities of both girls in some frame
of reference (the same for both!), then subtract Ann’s velocity from Becky’s (this is what Eq. (1.21)
is saying). In this case, if we just choose the floor’s reference frame, we havevFA =6 m / sa n d
vFB =5m / s ,s o
vAB = vFB − vFA =5 m
s − 6 m
s = −1 m
s (1.28)
The negative sign makes sense: Ann sees Beckyfalling behind her,s or e l a t i v et oh e rB e c k yi sm o v i n g
backwards,w h i c hi st os a y ,i nt h ed i r e c t i o nw eh a v ei d e n t i fi e da sn e g a t ive.
Part (d): Again we use the same equation, and Becky’s velocityi ss t i l lt h es a m e ,b u tn o wA n n ’ s
velocity isvFA = −4m / s ( n o t e t h e n e g a t i v e s i g n ! ) , s o
vAB = vFB − vFA =5 m
s −
(
−4 m
s
)
=9 m
s (1.29)",University Physics I Classical Mechanics.pdf
"velocity isvFA = −4m / s ( n o t e t h e n e g a t i v e s i g n ! ) , s o
vAB = vFB − vFA =5 m
s −
(
−4 m
s
)
=9 m
s (1.29)
Part (e): You may find this a bit surprising, but if you think about it the explanation for why
Ann lost the race, despite her running at the same speed as Becky relative to the surface she was
standing on,h a st ob et h a ts h ea c t u a l l yr a nal o n g e rd i s t a n c eo nt h a ts u r face! Since she was running
for a total of 12.5s a t a c o n s t a n tspeed (not velocity!) of 5 m/s in the walkway frame, then in that
frame she ran a distanced = |v|∆t =5 × 12.5=6 2.5m . T h a t i s t h e t o t a l l e n g t h o f w a l k w a y t h a t
she actually stepped on.",University Physics I Classical Mechanics.pdf
"1.6. PROBLEMS 27
1.6 Problems
Problem 1
x (m)
t (s)
The above figure is the position (in meters) versus time (in seconds) graph of an object in motion.
Only the segments betweent =1 sa n dt =2 s ,a n db e t w e e nt =4 sa n dt =5 s ,a r es t r a i g h tl i n e s .
The peak of the curve is att =3s , x =4m .
Answer the following questions, and provide a brief justification for your answer in every case.
(a) At what time(s) is the object’s velocity equal to zero?
(b) For what range(s) of times is the object moving with constant velocity?
(c) What is the object’s position coordinate att =1s ?
(d) What is the displacement of the object betweent =1sa n dt =4s ?
(e) What is thedistance traveled betweent =1sa n dt =4s ?
(f) What is the instantaneous velocity of the object att =1 .5s ?
(g) What is its average velocity betweent =1sa n dt =3s ?
Problem 2
Ap a r t i c l ei si n i t i a l l ya txi =3 m ,yi = −5m , a n d a f t e r a w h i l e i t i s f o u n d a t t h e c o o r d i n a t e s",University Physics I Classical Mechanics.pdf
"Problem 2
Ap a r t i c l ei si n i t i a l l ya txi =3 m ,yi = −5m , a n d a f t e r a w h i l e i t i s f o u n d a t t h e c o o r d i n a t e s
xf = −4m ,yf =2m .
(a) On the grid below (next page), draw the initial and final position vectors, and the displacement
vector.
(b) What are the components of the displacement vector?
(c) What are the magnitude and direction of the displacementvector? (You can specify the direction
by the angle it makes with either the positivex or the positivey axis.)",University Physics I Classical Mechanics.pdf
"28 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY
x
y
Problem 3
Marshall Dillon is riding at 30 mph after the robber of the Dodge City bank, who has a head start
of 15 minutes, but whose horse can only make 25 mph on a good day.H o w l o n g d o e s i t t a k e f o r
Dillon to catch up with the bad guy, and how far from Dodge Cityare they when this happens?
(Assume the road is straight, for simplicity.)
Problem 4
The picture below shows the velocity versus time graph of thefirst 21 seconds of a race between
two friends, “Red” and “Green.”
(a) Who is ahead att =1 0s ,a n db yh o wm u c h ?
(b) Who passes the 100 m marker first?
5 10 15 200
0
5
t (s)
v (m/s)",University Physics I Classical Mechanics.pdf
"1.6. PROBLEMS 29
Problem 5
You are trying to pass a truck on the highway. The truck is driving at 55 mph, so you speed up to
60 mph and move over to the left lane. If the truck is 17 m long, and your car is 3 m long
(a) how long does it take you to pass the truck completely?
(b) How far (along the highway) have you traveled in that time?
Note: to answer part (a) look at the problem from the perspective of the truck driver. How far are
you going relative to him, and how far would it take you to cover2 0 ma tt h a ts p e e d ?
Problem 6
Suppose the position function of a particle moving in one dimension is given by
x(t)=5+3 t +2 t2 − 0.5t3 (1.30)
where the coefficients are such that the result will be in metersi fy o ue n t e rt h et i m ei ns e c o n d s .
What is the particle’s velocity att =2s ? T h e r ea r et w ow a y sy o uc a nd ot h i s :
• If you know calculus, calculate the derivative of (1.30)a n de v a l u a t ei ta tt =2s .",University Physics I Classical Mechanics.pdf
"• If you know calculus, calculate the derivative of (1.30)a n de v a l u a t ei ta tt =2s .
• If you donot yet know how to take derivatives, calculate the limit in the definition (1.8).
That is to say, calculate ∆x/∆t with ti =2sa n d∆t equal, first, to 0.1s , t h e n t o 0.01 s, and
then to 0.001 s. You will need to keep more than the usual 4 decimals in thei n t e r m e d i a t e
calculations if you want an accurate result, but you should still report only 3 significant digits
in the final result.
Problem 7
Suppose you are rowing across a river, as in Figure1.7.Y o u rs p e e di s2m i l e sp e rh o u r r e l a t i v et o
the current, which is moving at a leisurely 1 mile per hour. Ifthe river is 10 m wide,
(a) How far downstream do you end up?
(b) To row straight across you would need to have an upstream velocity component (relative to the
current). How large would that be?
(c) If your rowing speed is still only 2 miles per hour, how longd o e si tt a k ey o ut or o wa c r o s st h e",University Physics I Classical Mechanics.pdf
"current). How large would that be?
(c) If your rowing speed is still only 2 miles per hour, how longd o e si tt a k ey o ut or o wa c r o s st h e
river now?",University Physics I Classical Mechanics.pdf
"30 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY",University Physics I Classical Mechanics.pdf
"Chapter 2
Acceleration
2.1 The law of inertia
There is something funny about motion with constant velocity: it is indistinguishable from rest.
Of course, you can usually tell whether you are movingrelative to something else.B u t i f y o u a r e
enjoying a smooth airplane ride, without looking out the window, you have no idea how fast you
are moving, or even, indeed (if the flight is exceptionally smooth) whether you are moving at all.
Ia ma c t u a l l yw r i t i n gt h i so na na i r p l a n e . T h efl i g h ts c r e e ni nforms me that I am moving at 480
mph relative to the ground, but I do not feel anything like that: just a gentle rocking up and down
and sideways that gives me no clue as to what my forward velocity is.
If I were to drop something, I know from experience that it would fall on a straight line—relative
to me, that is. If it falls from my hand it will land at my feet, just as if we were all at rest. But",University Physics I Classical Mechanics.pdf
"to me, that is. If it falls from my hand it will land at my feet, just as if we were all at rest. But
we are not at rest. In the half second or so it takes for the object to fall,t h ea i r p l a n eh a sm o v e d
forward 111 meters relative to the ground. Yet the (hypothetical) object I drop does not land 300
feet behind me! It moves forward with me as it falls, even though I am not touching it.It keeps
its initial forward velocity,even though it is no longer in contact with me or anything connected
to the airplane.
(At this point you might think that the object is still in contact with the air inside the plane, which
is moving with the plane, and conjecture that maybe it is the air inside the plane that “pushes
forward” on the object as it falls and keeps it from moving backwards. This is not necessarily a
dumb idea, but a moment’s reflection will convince you that itis impossible. We are all familiar with",University Physics I Classical Mechanics.pdf
"dumb idea, but a moment’s reflection will convince you that itis impossible. We are all familiar with
the way air pushes on things moving through it, and we know thatt h ef o r c ea no b j e c te x p e r i e n c e s
depends on its mass and its shape, so if that was what was happening, dropping objects of different
masses and shapes I would see them falling in all kind of different ways—as I would, in fact, if I
31",University Physics I Classical Mechanics.pdf
"32 CHAPTER 2. ACCELERATION
were dropping things from rest outdoors in a strong wind. Butthat is not what we experience on
an airplane at all. The air, in fact, has no effect on the forwardmotion of the falling object. It
does not push it in any way, because it is moving at the same velocity. This, in fact, reinforces our
previous conclusion: the object keeps its forward velocitywhile it is falling, in the absence of any
external influence.)
This remarkable observation is one of the most fundamental principles of physics (yes, we have
started to learn physics now!), which we callthe law of inertia.I t c a nb e s t a t e da s f o l l o w s :i n t h e
absence of any external influence (orforce)a c t i n go ni t ,a no b j e c ta tr e s tw i l ls t a ya tr e s t ,w h i l ea n
object that is already moving with some velocity will keep that same velocity (speed and direction
of motion)—at least until it is, in fact, acted upon by some force.",University Physics I Classical Mechanics.pdf
"of motion)—at least until it is, in fact, acted upon by some force.
Please let that sink in for a moment, before we start backtracking, which we have to do now on
several accounts. First, I have used repeatedly the term “force,” but I have not defined it properly.
Or have I? What if I just said that forces are precisely any “external influences” that may cause a
change in the velocity of an object? That will work, I think, until it is time to explore the concept
in more detail, a few chapters from now.
Next, I need to draw your attention to the fact that the objectI( h y p o t h e t i c a l l y )d r o p p e dd i dn o t
actually keep itstotal initial velocity: it only kept its initialforward velocity. In the downward
direction, it was speeding up from the moment it left my hand,as would any other falling object
(and as we shall see later in this chapter). But this actuallymakes sense in a certain way: there",University Physics I Classical Mechanics.pdf
"(and as we shall see later in this chapter). But this actuallymakes sense in a certain way: there
was no forward force, so the forward velocity remained constant; there was, however, a vertical
force acting all along (the force of gravity), and so the object did speed up in that direction. This
observation is, in fact, telling us something profound aboutt h ew o r l d ’ sg e o m e t r y : n a m e l y ,t h a t
forces and velocities arevectors,a n dl a w ss u c ha st h el a wo fi n e r t i aw i l lt y p i c a l l ya p p l yt ot he
vector as a whole, as well as to each component separately (that is to say, each dimension of space).
This anticipates, in fact, the way we will deal, later on, withm o t i o ni nt w oo rm o r ed i m e n s i o n s ;
but we do not need to worry about that for a few chapters still.
Finally, it is worth spending a moment reflecting on how radically the law of inertia seems to",University Physics I Classical Mechanics.pdf
"Finally, it is worth spending a moment reflecting on how radically the law of inertia seems to
contradict our intuition about the way the world works. Whatit seems to be telling us is that, if
we throw or push an object, it should continue to move foreverwith the same speed and in the
same direction with which it set out—something that we know isc e r t a i n l yn o tt r u e . B u tw h a t ’ s
happening in “real life” is that, just because we have left something alone, it doesn’t mean the world
has left it alone. After we lose contact with the object, all sorts of other forces will continue to act
on it. A ball we throw, for instance, will experience air resistance or drag (the same effect I was
worrying about in that paragraph in parenthesis in the previous page), and that will slow it down.
An object sliding on a surface will experience friction, andthat will slow it down too. Perhaps the",University Physics I Classical Mechanics.pdf
"An object sliding on a surface will experience friction, andthat will slow it down too. Perhaps the
closest thing to the law of inertia in action that you may get tos e ei sah o c k e yp u c ks l i d i n go nt h e
ice: it is remarkable (perhaps even a bit frightening) to seehow little it slows down, but even so
the ice does a exert a (very small) frictional force that wouldb r i n gt h ep u c kt oas t o pe v e n t u a l l y .",University Physics I Classical Mechanics.pdf
"2.1. THE LAW OF INERTIA 33
This is why, historically, the law of inertia was not discovered until people started developing an
appreciation for frictional forces, and the way they are constantly acting all around us to oppose
the relative motion of any objects trying to slide past each other.
This mention of relative motion, in a way, brings us full circle. Yes, relative motion is certainly
detectable, and for objects in contact it actually results int h eo c c u r r e n c eo ff o r c e so ft h ef r i c t i o n a l ,
or drag, variety. Butabsolute (that is, without reference to anything external) motion with constant
velocity is fundamentally undetectable. And in view of the law of inertia, it makes sense: if no force
is required to keep me moving with constant velocity, it follows that as long as I am moving with
constant velocity I should not be feeling any net force actingo nm e ;n o rw o u l da n yo t h e rd e t e c t i o n
apparatus I might be carrying with me.",University Physics I Classical Mechanics.pdf
"apparatus I might be carrying with me.
What we do feel in our bodies, and what we can detect with our inertial navigation systems (now
you may start to guess why they are called “inertial”), is achange in our velocity, which is to say,
our acceleration (to be defined properly in a moment). We rely, ultimately, on the law of inertia
to detect accelerations: if my plane is shaking up and down, because of turbulence (as, in fact, it
is right now!), the water in my cup may not stay put. Or, rather,t h ew a t e rm a yt r yt os t a yp u t
(really, to keep moving, at any moment, with whatever velocity it has at that moment), but if the
cup, which is connected to my hand which is connected, ultimately, to this bouncy plane, moves
suddenly out from under it, not all of the water’s parts will bea b l et oa d j u s tt h e i rv e l o c i t i e st ot h e
new velocity of the cup in time to prevent a spill.",University Physics I Classical Mechanics.pdf
"new velocity of the cup in time to prevent a spill.
This is the next very interesting fact about the physical world that we are about to discover: forces
cause accelerations, or changes in velocity, but they do so in different degrees for different objects;
and, moreover, the ultimate change in velocitytakes time.T h e fi r s t p a r t o f t h i s s t a t e m e n t h a s t o
do with the concept ofinertial mass,t ob ei n t r o d u c e di nt h en e x tc h a p t e r ;t h es e c o n dp a r tw ea r e
going to explore right now, after a brief detour to defineinertial reference frames.
2.1.1 Inertial reference frames
The example I just gave you of what happens when a plane in flighte x p e r i e n c e st u r b u l e n c ep o i n t s
to an important phenomenon, namely, that there may be times where the law of inertia may not
seem to apply in a certain reference frame. By this I mean that an object that I left at rest, like the",University Physics I Classical Mechanics.pdf
"seem to apply in a certain reference frame. By this I mean that an object that I left at rest, like the
water in my cup, may suddenly start to move—relative to the reference frame coordinates—even
though nothing and nobody is acting on it. More dramaticallystill, if a car comes to a sudden
stop, the passengers may be “projected forward”—they were initially at rest relative to the car
frame, but now they find themselves moving forward (always inthe car reference frame), to the
point that, if they are not wearing seat belts, they may end uphitting the dashboard, or the seat
in front of them.",University Physics I Classical Mechanics.pdf
"34 CHAPTER 2. ACCELERATION
Again, nobody has pushed on them, and in fact what we can see inthis case, from outside the car,
is nothing but the law of inertia at work: the passengers werejust keeping their initial velocity,
when the car suddenly slowed down under and around them. So there is nothing wrong with the
law of inertia, butthere is a problem with the reference frame:i fI w a n tt od e s c r i b e t h e m o t i o no f
objects in a reference frame like a plane being shaken up or a car that is speeding up or slowing
down, I need to allow for the fact that objects may move—alwaysr e l a t i v et ot h a tf r a m e — i na n
apparent violation of the law of inertia.
The way we deal with this in physics is by introducing the veryimportant concept of aninertial
reference frame,b yw h i c hw em e a nar e f e r e n c ef r a m ei nw h i c ha l lo b j e c t sw i l l ,at all times, be
observed to move (or not move) in a way fully consistent with the law of inertia. In other words,",University Physics I Classical Mechanics.pdf
"observed to move (or not move) in a way fully consistent with the law of inertia. In other words,
the law of inertia has to holdwhen we use that frame’s own coordinates to calculate the objects’
velocities.T h i s , o f c o u r s e , i s w h a t w e a l w a y s d o i n s t i n c t i v e l y :w h e n I amo nap l a n eIl o c a t et h e
various objects around me relative to the plane frame itself,n o tr e l a t i v et ot h ed i s t a n tg r o u n d .
To ascertain whether a frame is inertial or not, we start by checking to see if the description of
motion using that frame’s coordinates obeys the law of inertia: does an object left at rest on the
counter in the laboratory stay at rest? If set in motion, doesit move with constant velocity on a
straight line? The Earth’s surface, as it turns out, isnot quite a perfect inertial reference frame,
but itis good enough that it made it possible for us to discover the lawof inertia in the first place!",University Physics I Classical Mechanics.pdf
"but itis good enough that it made it possible for us to discover the lawof inertia in the first place!
What spoils the inertial-ness of an Earth-bound reference frame is the Earth’s rotation, which,
as we shall see later, is an example ofaccelerated motion.I n f a c t , i f y o u t h i n k a b o u t t h e g r o s s l y
non-inertial frames I have introduced above—the bouncy plane, the braking car—they all have
this in common: that their velocities are changing; they arenot moving with constant speed on a
straight line.
So, once you have found an inertial reference frame, to decidew h e t h e ra n o t h e ro n ei si n e r t i a lo r
not is simple: if it is moving with constant velocity (relative to the first, inertial frame), then it is
itself inertial; if not, it is not. I will show you how this works, formally, in a little bit (section 2.2.4,
below), after I (finally!) get around to properly introducingt h ec o n c e p to fa c c e l e r a t i o n .",University Physics I Classical Mechanics.pdf
"below), after I (finally!) get around to properly introducingt h ec o n c e p to fa c c e l e r a t i o n .
It is a fundamental principle of physics thatthe laws of physics take the same form in all inertial
reference frames.T h e l a w o f i n e r t i a i s , o f c o u r s e , a n e x a m p l e o f s u c h a l a w .S i nce all inertial frames
are moving with constant velocity relative to each other, this is another way to say that absolute
motion is undetectable, and all motion is ultimately relative. Accordingly, this principle is known
as the principle of relativity.",University Physics I Classical Mechanics.pdf
"2.2. ACCELERATION 35
2.2 Acceleration
2.2.1 Average and instantaneous acceleration
Just as we defined average velocity in the previous chapter, using the concept of displacement (or
change in position) over a time interval ∆t,w ed e fi n eaverage accelerationover the time ∆t using
the change in velocity:
aav = ∆v
∆t = vf − vi
tf − ti
(2.1)
Here, vi and vf are the initial and final velocities, respectively, that is tos a y ,t h ev e l o c i t i e sa t
the beginning and the end of the time interval ∆t.A s w a s t h e c a s e w i t h t h e a v e r a g e v e l o c i t y ,
though, the average acceleration is a concept of somewhat limited usefulness, so we might as well
proceed straight away to the definition of theinstantaneous acceleration(or just “the” acceleration,
without modifiers), through the same sort of limiting processb yw h i c hw ed e fi n e dt h ei n s t a n t a n e o u s
velocity:
a =l i m
∆t→0
∆v
∆t (2.2)",University Physics I Classical Mechanics.pdf
"velocity:
a =l i m
∆t→0
∆v
∆t (2.2)
Everything that we said in the previous chapter about the relationship between velocity and position
can now be said about the relationship between accelerationand velocity. For instance (if you know
calculus), the acceleration as a function of time is the derivative of the velocity as a function of
time, which makes it the second derivative of the position function:
a = dv
dt = d2x
dt2 (2.3)
(and if you donot know calculus yet, do not worry about the superscripts “2” onthat last expres-
sion! It is just a weird notation that you will learn someday.)
Similarly, we can “read off” the instantaneous accelerationfrom a velocity versus time graph, by
looking at the slope of the line tangent to the curve at any point. However, if what we are given is
a position versus time graph, the connection to the acceleration is morei n d i r e c t .F i g u r e2.1 (next",University Physics I Classical Mechanics.pdf
"a position versus time graph, the connection to the acceleration is morei n d i r e c t .F i g u r e2.1 (next
page) provides you with such an example. See if you can guess atw h a tp o i n t sa l o n gt h i sc u r v et h e
acceleration is positive, negative, or zero.
The way to do this “from scratch,” as it were, is to try to figureout what the velocity is doing,
first, and infer the acceleration from that. Here is how that would go:
Starting att =0 ,a n dk e e p i n ga ne y eo nt h es l o p eo ft h ex-vs-t curve, we can see that the velocity
starts at zero or near zero and increases steadily for a while,u n t i lt is a little bit more than 2 s (let
us say,t =2 .2sf o rd e fi n i t e n e s s ) . T h a tw o u l dc o r r e s p o n dt oap e r i o do fp o sitive acceleration, since
∆v would be positive for every ∆t in that range.",University Physics I Classical Mechanics.pdf
"36 CHAPTER 2. ACCELERATION
x (m)
t (s)0
5
5 10
Figure 2.1: A possible position vs. time graph for an object whose acceleration changes with time.
Between t =2 .2sa n dt =2 .5s ,a st h eo b j e c tm o v e sf r o mx =2mt o x =4m ,t h ev e l o c i t yd o e s
not appear to change very much, and the acceleration would correspondingly be zero or near zero.
Then, aroundt =2 .5s , t h e v e l o c i t y s t a r t s t o d e c r e a s e n o t i c e a b l y , be c o m i n g ( instantaneously) zero
at t =3s( x =6m ) . T h a tw o u l dc o r r e s p o n dt oan e g a t i v ea c c e l e r a t i o n . N o te, however, that the
velocity afterwards continues to decrease, becoming more and more negative until aroundt =4
s. This also corresponds to a negative acceleration: even though the object is speeding up, it is
speeding up in the negative direction, so ∆v,a n dh e n c ea,i sn e g a t i v ef o re v e r yt i m ei n t e r v a lt h e r e .",University Physics I Classical Mechanics.pdf
"speeding up in the negative direction, so ∆v,a n dh e n c ea,i sn e g a t i v ef o re v e r yt i m ei n t e r v a lt h e r e .
We conclude thata< 0f o ra l lt i m e sb e t w e e nt =2 .5sa n dt =4s .
Next, as we just look pastt =4 s ,s o m e t h i n ge l s ei n t e r e s t i n gh a p p e n s : t h eo b j e c ti ss t i llg o i n g
in the negative direction (negative velocity), but now it isslowing down. Mathematically, that
corresponds to apositive acceleration, since the algebraic value of the velocity is inf a c ti n c r e a s i n g
(a number like−3i sl a r g e rt h a nan u m b e rl i k e−5). Another way to think about it is that, if we
have less and less of a negative thing, our overall trend is positive. So the acceleration is positive
all the way fromt =4st h r o u g ht =5s( w h e r et h ev e l o c i t yi si n s t a n t a n e o u s l yz e r oa st h eo b j e ct’s
direction of motion reverses), and beyond, until aboutt =6s ,s i n c eb e t w e e nt =5sa n dt =6st h e",University Physics I Classical Mechanics.pdf
"direction of motion reverses), and beyond, until aboutt =6s ,s i n c eb e t w e e nt =5sa n dt =6st h e
velocity is positive and growing.
You can probably figure out on your own now what happens aftert =6 s ,r e a s o n i n ga sId i d
above, but you may also have noticed a pattern that makes thiskind of analysis a lot easier.
The acceleration (as those with a knowledge of calculus may have understood already), being
proportional to the second derivative of the functionx(t)w i t hr e s p e c tt ot,i sd i r e c t l yr e l a t e dt o
the curvature of the x-vs-t graph. As figure 2.2 below shows, if the graph isconcave (sometimes",University Physics I Classical Mechanics.pdf
"2.2. ACCELERATION 37
called “concave upwards”), the acceleration is positive, whereas it is negative whenever the graph
is convex (or “concave downwards”). It is (instantly) zero at those points where the curvature
changes (which you may know asinflection points), as well as over stretches of time when the
x-vs-t graph is a straight line (motion with constant velocity).
xx x
tt t
a > 0 a < 0 a = 0
Figure 2.2: What thex-vs-t curves look like for the different possible signs of the acceleration.
Figure 2.3 (in the next page) shows position, velocity, and acceleration versus time for a hypothetical
motion case. Please study it carefully until every feature ofe v e r yg r a p hm a k e ss e n s e ,r e l a t i v et o
the other two! You will see many other examples of this in the homework and the lab.
Notice that, in all these figures, the sign ofx or v at any given time has nothing to do with the sign",University Physics I Classical Mechanics.pdf
"Notice that, in all these figures, the sign ofx or v at any given time has nothing to do with the sign
of a at that same time. It is true that, for instance, a negativea,i fs u s t a i n e df o ras u ffi c i e n t l yl o n g
time, will eventually result in a negativev (as happens, for instance, in Fig.2.3 over the interval
from t =1t o t =4 s )b u tt h i sm a yt a k eal o n gt i m e ,d e p e n d i n go nt h es i z eo fa and the initial
value ofv.T h e g r a p h i c a lc l u e st of o l l o w ,i n s t e a d ,a r e :t h ea c c e l e r a tion is given by the slope of the
tangent to thev-vs-t curve, or the curvature of thex-vs-t curve, as explained in Fig.2.2;a n dt h e
velocity is given by the slope of the tangent to thex-vs-t curve.
(Note: To make the interpretation of Figure2.3 simpler, I have chosen the acceleration to be
“piecewise constant,” that is to say, constant over extendedt i m ei n t e r v a l sa n dc h a n g i n gi nv a l u e",University Physics I Classical Mechanics.pdf
"“piecewise constant,” that is to say, constant over extendedt i m ei n t e r v a l sa n dc h a n g i n gi nv a l u e
discontinuously from one interval to the next. This is physically unrealistic: in any real-life situa-
tion, the acceleration would be expected to change more or less smoothly from instant to instant.
We will see examples of that later on, when we start looking atrealistic models of collisions.)",University Physics I Classical Mechanics.pdf
"38 CHAPTER 2. ACCELERATION
Position (meters)Acceleration (m/s2) Velocity (m/s)
Time (seconds)
1234567890
0
5
10
15
20
25
30
35
1234567890
0
5
10
15
20
-5
-10
1234567890
0
5
10
15
20
-5
-10
Time (seconds)
Time (seconds)
Figure 2.3: Sample position, velocity and acceleration vs. time graphs for motionwith piecewise-constant
acceleration.",University Physics I Classical Mechanics.pdf
"2.2. ACCELERATION 39
2.2.2 Motion with constant acceleration
Ap a r t i c u l a rk i n do fm o t i o nt h a ti sb o t hr e l a t i v e l ys i m p l ea ndv e r yi m p o r t a n ti np r a c t i c ei sm o t i o n
with constant acceleration (see Figure2.3 again for examples). Ifa is constant, it means that the
velocity changes with time at a constant rate, by a fixed numbero fm / se a c hs e c o n d . ( T h e s ea r e ,
incidentally, the units of acceleration: meters per secondper second, or m/s2.) The change in
velocity over a time interval ∆t is then given by
∆v = a ∆t (2.4)
which can also be written
v = vi + a (t − ti)( 2 . 5 )
Equation (2.5)i st h ef o r mo ft h ev e l o c i t yf u n c t i o n(v as a function oft)f o rm o t i o nw i t hc o n s t a n t
acceleration. This, in turn, has to be the derivative with respect to time of the corresponding
position function. If you know simple derivatives, then, youc a nv e r i f yt h a tt h ea p p r o p r i a t ef o r m",University Physics I Classical Mechanics.pdf
"position function. If you know simple derivatives, then, youc a nv e r i f yt h a tt h ea p p r o p r i a t ef o r m
of the position function must be
x = xi + vi(t − ti)+ 1
2 a (t − ti)2 (2.6)
or in terms of intervals,
∆x = vi∆t + 1
2 a (∆t)2 (2.7)
Most often Eq. (2.6)i sw r i t t e nw i t ht h ei m p l i c i ta s s u m p t i o nt h a tt h ei n i t i a lv alue oft is zero:
x = xi + vit + 1
2 at2 (2.8)
This is simpler, but not as general as Eq. (2.6). Always make sure that you know what conditions
apply for any equation you decide to use!
As you can see from Eq. (2.5), for intervals during which the acceleration is constant,the velocity
vs. time curve should be a straight line. Figure2.3 (previous page) illustrates this. Equation (2.6),
on the other hand, shows that for those same intervals the position vs. time curve should be a
(portion of a) parabola, and again this can be seen in Figure2.3 (sometimes, if the acceleration is",University Physics I Classical Mechanics.pdf
"(portion of a) parabola, and again this can be seen in Figure2.3 (sometimes, if the acceleration is
small, the curvature of the graph may be hard to see; this happens in Figure2.3 for the interval
between t =4sa n dt =5s ) .
The observation thatv-vs-t is a straight line when the acceleration is constant providesu sw i t h
as i m p l ew a yt od e r i v eE q .(2.7), when combined with the result (from the end of the previous
chapter) that the displacement over a time interval ∆t equals the area under thev-vs-t curve for
that time interval. Indeed, consider the situation shown inFigure 2.4.T h e t o t a l a r e a u n d e r t h e
segment shown is equal to the area of a rectangle of base ∆t and height vi,p l u st h ea r e ao fa",University Physics I Classical Mechanics.pdf
"40 CHAPTER 2. ACCELERATION
triangle of base ∆t and height vf − vi.S i n c evf − vi = a∆t,s i m p l eg e o m e t r yi m m e d i a t e l yy i e l d s
Eq. (2.7), or its equivalent (2.6).
v (m/s)
t (s)ti tf
vi
vf
v(t)
Figure 2.4: Graphical way to find the displacement for motion with constant acceleration.
Lastly, consider what happens if we solve Eq. (2.4)f o r∆t and substitute the result in (2.7). We
get
∆x = vi∆v
a + (∆v)2
2a (2.9)
Letting ∆v = vf − vi,al i t t l ea l g e b r ay i e l d s
v2
f − v2
i =2 a∆x (2.10)
This is a handy little result that can also be seen to follow, more directly, from the work-energy
theorems to be introduced in Chapter 71.
1In fact, equation (2.10)t u r n so u tt ob eso handy that you will probably find yourself using it over and over this
semester, and you may even be tempted to use it for problems involving motion in two dimensions. However, unless",University Physics I Classical Mechanics.pdf
"semester, and you may even be tempted to use it for problems involving motion in two dimensions. However, unless
you really know what you are doing, you should resist the temptation, since it is very easy to use (2.10)i n c o r r e c t l y
when the acceleration and the displacement do not lie along the same line. You should use the appropriate form of
aw o r k - e n e r g yt h e o r e mi n s t e a d .",University Physics I Classical Mechanics.pdf
"2.2. ACCELERATION 41
2.2.3 Acceleration as a vector
In two (or more) dimensions we introduce the average acceleration vector
⃗aav = ∆⃗v
∆t = 1
∆t (⃗vf −⃗vi)( 2 . 1 1 )
whose components areaav,x =∆ vx/∆t,e t c . . T h ei n s t a n t a n e o u sa c c e l e r a t i o ni st h e nt h ev e c t o rg iven
by the limit of Eq. (2.11)a s∆t → 0, and its components are, therefore,ax = dvx/dt, ay = dvy/dt, . . .
Note that, since⃗vi and ⃗vf in Eq. (2.11)a r ev e c t o r s ,a n dh a v et ob es u b t r a c t e da ss u c h ,t h ea c c e l e r -
ation vector will be nonzero whenever⃗vi and ⃗vf are different, even if, for instance, their magnitudes
(which are equal to the object’s speed) are the same. In otherwords, you have accelerated motion
whenever thedirection of motion changes, even if the speed does not.
As long as we are working in one dimension, I will follow the same convention for the acceleration",University Physics I Classical Mechanics.pdf
"As long as we are working in one dimension, I will follow the same convention for the acceleration
as the one I introduced for the velocity in Chapter 1: namely,Iw i l lu s et h es y m b o la,w i t h o u t
as u b s c r i p t ,t or e f e rt ot h er e l e v a n tc o m p o n e n to ft h ea c c e l eration (ax,a y,... ), and not to the
magnitude of the vector⃗a.
2.2.4 Acceleration in different reference frames
In Chapter 1 you saw that the following relation (Eq. (1.19)) holds between the velocities of a
particle P measured in two different reference frames, A and B:
⃗vAP = ⃗vAB + ⃗vBP (2.12)
What about the acceleration? An equation like (2.12)w i l lh o l df o rt h ei n i t i a la n dfi n a lv e l o c i t i e s ,
and subtracting them we will get
∆⃗vAP =∆ ⃗vAB +∆⃗vBP (2.13)
Now suppose that reference frame B moves withconstant velocityrelative to frame A. In that case,
⃗vAB,f = ⃗vAB,i,s o∆⃗vAB =0 ,a n dt h e n ,d i v i d i n gE q .(2.13)b y∆t,a n dt a k i n gt h el i m i t∆t → 0, we
get",University Physics I Classical Mechanics.pdf
"⃗vAB,f = ⃗vAB,i,s o∆⃗vAB =0 ,a n dt h e n ,d i v i d i n gE q .(2.13)b y∆t,a n dt a k i n gt h el i m i t∆t → 0, we
get
⃗aAP = ⃗aBP (for constant⃗vAB)( 2 . 1 4 )
So, if two reference frames are moving at constant velocity relative to each other, observers in both
frames measure thesame acceleration for any object they might both be tracking.
The result (2.14)m e a n s ,i np a r t i c u l a r ,t h a ti fw eh a v ea ni n e r t i a lf r a m et h e nany frame moving
at constant velocity relative to it will be inertial too, since the respective observers’ measurements",University Physics I Classical Mechanics.pdf
"42 CHAPTER 2. ACCELERATION
will agree that an object’s velocity does not change (otherwise put, its acceleration is zero) when no
forces act on it. Conversely, an accelerated frame willnot be an inertial frame, because Eq. (2.14)
will not hold. This is consistent with the examples I mentioned in Section 2.1 (the bouncing plane,
the car coming to a stop). Another example of a non-inertial frame would be a car going around
ac u r v e ,e v e ni fi ti sg o i n ga tc o n s t a n ts p e e d ,s i n c e ,a sIj u s tpointed out above, this is also an
accelerated system. This is confirmed by the fact that objectsi ns u c hac a rt e n dt om o v e — r e l a t i v e
to the car—towards the outside of the curve, even though no actual force is acting on them.
2.3 Free fall
An important example of motion with (approximately) constant acceleration is provided byfree fall
near the surface of the Earth. We say that an object is in “freefall” when the only force acting on it",University Physics I Classical Mechanics.pdf
"near the surface of the Earth. We say that an object is in “freefall” when the only force acting on it
is the force of gravity (the word “fall” here may be a bit misleading, since the object could actually
be moving upwards some of the time, if it has been thrown straight up, for instance). The space
station is in free fall, but because it is nowhere near the surface of the earth its direction of motion
(and hence its acceleration, regarded as a two-dimensionalvector) is constantly changing. Right
next to the surface of the earth, on the other hand, the planet’s curvature is pretty much negligible
and gravity provides an approximately constant, vertical acceleration, which, in the absence of
other forces, turns out to bethe same for every object,r e g a r d l e s so fi t ss i z e ,s h a p e ,o rw e i g h t .
The above result—that, in the absence of other forces, all objects should fall to the earth at the",University Physics I Classical Mechanics.pdf
"The above result—that, in the absence of other forces, all objects should fall to the earth at the
same rate, regardless of how big or heavy they are—is so contrary to our common experience that
it took many centuries to discover it. The key, of course, as with the law of inertia, is to realize
that, under normal circumstances, frictional forces are, inf a c t ,a c t i n ga l lt h et i m e ,s oa no b j e c t
falling through the atmosphere is neverreally in “free” fall: there is always, at a minimum, and in
addition to the force of gravity, an air drag force that opposes its motion. The magnitude of this
force does depend on the object’s size and shape (basically,on how “aerodynamic” the object is);
and thus a golf ball, for instance, falls much faster than a flats h e e to fp a p e r .Y e t ,i fy o uc r u m p l e
up the sheet of paper till it has the same size and shape as the golf ball, you can see for yourself",University Physics I Classical Mechanics.pdf
"up the sheet of paper till it has the same size and shape as the golf ball, you can see for yourself
that they do fall at approximately the same rate! The equalityc a nn e v e rb ee x a c t ,h o w e v e r ,u n l e s s
you get rid completely of air drag, either by doing the experiment in an evacuated tube, or (in a
somewhat extreme way), by doing it on the surface of the moon,as the Apollo 15 astronauts did
with a hammer and a feather back in 19712.
This still leaves us with something of a mystery, however: thef o r c eo fg r a v i t yi st h eo n l yf o r c e
known to have the property that it imparts all objects thesame acceleration, regardless of their
mass or constitution. A way to put this technically is that thef o r c eo fg r a v i t yo na no b j e c ti s
2The video of this is available online:https://www.youtube.com/watch?v=oYEgdZ3iEKA.I t i s , h o w e v e r , p r e t t y",University Physics I Classical Mechanics.pdf
"2The video of this is available online:https://www.youtube.com/watch?v=oYEgdZ3iEKA.I t i s , h o w e v e r , p r e t t y
low resolution and hard to see. A very impressive modern-daydemonstration involving feathers and a bowling ball
in a completely evacuated (airless) room is available here:https://www.youtube.com/watch?v=E43-CfukEgs.",University Physics I Classical Mechanics.pdf
"2.3. FREE FALL 43
proportional to that object’sinertial mass,aq u a n t i t yt h a tw ew i l li n t r o d u c ep r o p e r l yi nt h en e x t
chapter. For the time being, we will simply record here that this acceleration, near the surface of
the earth, has a magnitude of approximately 9.8m / s2,aq u a n t i t yt h a ti sd e n o t e db yt h es y m b o lg.
Thus, if we take the upwards direction as positive (as is usually done), we get for the acceleration
of an object in free falla = −g,a n dt h ee q u a t i o n so fm o t i o nb e c o m e
∆v = −g ∆t (2.15)
∆y = vi∆t − 1
2 g (∆t)2 (2.16)
where I have usedy instead of x for the position coordinate, since that is a more common choice
for a vertical axis. Note that we could as well have chosen thedownward direction as positive, and
that may be a more natural choice in some problems. Regardless, the quantityg is always defined",University Physics I Classical Mechanics.pdf
"that may be a more natural choice in some problems. Regardless, the quantityg is always defined
to be positive:g =9 .8m / s2.T h e a c c e l e r a t i o n ,t h e n , i sg or −g,d e p e n d i n go nw h i c hd i r e c t i o nw e
take to be positive.
In practice, the value ofg changes a little from place to place around the earth, for various reasons
(it is somewhat sensitive to the density of the ground below you, and it decreases as you climb
higher away from the center of the earth). In a later chapter wew i l ls e eh o wt oc a l c u l a t et h ev a l u e
of g from the mass and radius of the earth, and also how to calculatet h ee q u i v a l e n tq u a n t i t yf o r
other planets.
In the meantime, we can use equations like (2.15)a n d(2.16)( a sw e l la s(2.10), with the appropriate
substitutions) to answer a number of interesting questionsabout objects thrown or dropped straight
up or down (always, of course, assuming that air drag is negligible). For instance, back at the",University Physics I Classical Mechanics.pdf
"up or down (always, of course, assuming that air drag is negligible). For instance, back at the
beginning of this chapter I mentioned that if I dropped an object it might take about half a second
to hit the ground. If you use Eq. (2.16)w i t hvi =0( s i n c eIa md r o p p i n gt h eo b j e c t ,n o tt h r o w i n g
it down, its initial velocity is zero), and substitute ∆t =0 .5s , y o u g e t ∆y =1 .23 m (about 4 feet).
This is a reasonable height from which to drop something.
On the other hand, you may note that half a second is not a very long time in which to make
accurate observations (especially if you do not have modernelectronic equipment), and as a result
of that there was considerable confusion for many centuriesas to the precise way in which objects
fell. Some people believed that the speed did increase in somew a ya st h eo b j e c tf e l l ,w h i l eo t h e r s
appear to have believed that an object dropped would “instantaneously” (that is, at soon as it",University Physics I Classical Mechanics.pdf
"appear to have believed that an object dropped would “instantaneously” (that is, at soon as it
left your hand) acquire some speed and keep that unchanged allt h ew a yd o w n . I nr e a l i t y ,i nt h e
presence of air drag, what happens is a combination of both: initially the speed increases at an
approximately constant rate (free, or nearly free fall), butt h ed r a gf o r c ei n c r e a s e sw i t ht h es p e e d
as well, until eventually it balances out the force of gravity, and from that point on the speed does
not increase anymore: we say that the object has reached “terminal velocity.” Some objects reach
terminal velocity almost instantly, whereas others (the more “aerodynamic” ones) may take a long
time to do so. This accounts for the confusion that prevailedbefore Galileo’s experiments in the
early 1600’s.",University Physics I Classical Mechanics.pdf
"44 CHAPTER 2. ACCELERATION
Galileo’s main insight, on the theoretical side, was the realization that it was necessary to separate
clearly the effect of gravity and the effect of the drag force. Experimentally, his big idea was to
use an inclined plane to slow down the “fall” of an object, so ast om a k ea c c u r a t em e a s u r e m e n t s
possible (and also, incidentally, reduce the air drag force!). These “inclined planes” were just
basically ramps down which he sent small balls (like marbles)r o l l i n g . B yc h a n g i n gt h es t e e p n e s s
of the ramp he could control how slowly the balls moved. He reasoned that, ultimately, the force
that made the balls go down was essentially the same force of gravity, only not the whole force,
but just a fraction of it. Today we know that, in fact, an objects l i d i n g(not rolling!) up or down
on a frictionless incline will experience an acceleration directed downwardsa l o n gt h ei n c l i n ea n d",University Physics I Classical Mechanics.pdf
"on a frictionless incline will experience an acceleration directed downwardsa l o n gt h ei n c l i n ea n d
with a magnitude equal tog sin θ,w h e r eθ is the angle that the slope makes with the horizontal:
a = g sin θ (inclined plane, takingdownwards to be positive) (2.17)
(for some reason, it seems more natural, when dealing with inclined planes, to take the downward
direction as positive!). Equation (2.17)m a k e ss e n s ei nt h et w oe x t r e m ec a s e si nw h i c ht h ep l a n ei s
completely vertical (θ =9 0◦, a = g)a n dc o m p l e t e l yh o r i z o n t a l(θ =0 ◦, a =0 ) . F o ri n t e r m e d i a t e
values, you will carry out experiments in the lab to verify this result.
We will show, in a later chapter, how Eq. (2.17)c o m e sa b o u tf r o mac a r e f u lc o n s i d e r a t i o no fa l lt h e
forces acting on the object; we will also see, later on, how itneeds to be modified for the case of a",University Physics I Classical Mechanics.pdf
"forces acting on the object; we will also see, later on, how itneeds to be modified for the case of a
rolling, rather than a sliding, object. This modification does not affect Galileo’s main conclusion,
which was, basically, that the natural falling motion in theabsence of friction or drag forces is
motion withconstant acceleration(at least, near the surface of the earth, whereg is constant to a
very good approximation).
2.4 In summary
1. The law of inertiastates that, if no external influences (forces) are acting onan object, then,
if the object is initially at rest it will stay at rest, and if iti si n i t i a l l ym o v i n gi tw i l lc o n t i n u e
to move with constant velocity (unchanging speed and direction).
2. Reference frames in which the law of inertia is seen to hold(when the velocities of objects are
calculated from their coordinates in that frame) are calledinertial.A r e f e r e n c e f r a m e t h a t i s",University Physics I Classical Mechanics.pdf
"calculated from their coordinates in that frame) are calledinertial.A r e f e r e n c e f r a m e t h a t i s
moving at constant velocity relative to an inertial frame isalso an inertial frame. Conversely,
accelerated reference frames are non-inertial.
3. Motion with constant velocity is fundamentally indistinguishable from no motion at all (i.e.,
rest). As long as the velocity (of the objects involved) doesnot change, onlyrelative motion
can be detected. This is known as theprinciple of relativity.A n o t h e r w a y t o s t a t e i t
is that the laws of physics must take the same form in all inertial reference frames (so you
cannot single out one as being in “absolute rest” or “absolutem o t i o n ” ) .",University Physics I Classical Mechanics.pdf
"2.4. IN SUMMARY 45
4. Changes in velocity are detectable, and, by (1) above, are evidence of unbalanced forces
acting on an object.
5. The rate of change of an object’s velocity is the object’sacceleration:t h e a v e r a g e a c c e l e r a t i o n
over a time interval ∆t is aav =∆ v/∆t,a n dt h ei n s t a n t a n e o u sa c c e l e r a t i o na tat i m et is the
limit of the average acceleration calculated for successively shorter time intervals ∆t,a l lw i t h
the same initial timeti = t.M a t h e m a t i c a l l y ,t h i sm e a n st h ea c c e l e r a t i o ni st h ed e r i v ative of
the velocity function,a = dv/dt.
6. In a velocity versus time graph, the acceleration can be read from the slope of the line tangent
to the curve (just like the velocity in a position versus timegraph).
7. In a position versus time graph, the regions with positiveacceleration correspond to a concave
curvature (like a parabola opening up), and those with negative acceleration correspond to a",University Physics I Classical Mechanics.pdf
"curvature (like a parabola opening up), and those with negative acceleration correspond to a
convex curvature (like a parabola opening down). Points of inflection (where the curvature
changes) and straight lines correspond to points where the acceleration is zero.
8. The basic equations used to describe motion with constantacceleration are (2.4), (2.7)a n d
(2.10)a b o v e .A l t e r n a t i v ef o r m so ft h e s ea r ea l s op r o v i d e di nt h etext.
9. In more than one dimension, a change in thedirection of the velocity vector results in a
nonzero acceleration, even if the object’s speed does not change.
10. An object is said to be infree fall when the only force acting on it is gravity. All objects
in free fall experience the same acceleration at the same point in their motion, regardless of
their mass or composition. Near the surface of the earth, thisa c c e l e r a t i o ni sa p p r o x i m a t e l y
constant and has a magnitudeg =9 .8m / s2.",University Physics I Classical Mechanics.pdf
"constant and has a magnitudeg =9 .8m / s2.
11. An object sliding on a frictionless inclined plane experiences (if air drag is negligible) an
acceleration directed downward along the incline and with amagnitude g sin θ,w h e r eθ is the
angle the incline makes with the horizontal.",University Physics I Classical Mechanics.pdf
"46 CHAPTER 2. ACCELERATION
2.5 Examples
2.5.1 Motion with piecewise constant acceleration
Construct the position vs. time, velocity vs. time, and acceleration vs. time graphs for the motion
described below. For each of the intervals (a)–(d) you’ll need to figure out the position (height)
and velocity of the rocket at the beginning and the end of the interval, and the acceleration for the
interval. In addition, for interval (b) you need to figure outthe maximum height reached by the
rocket and the time at which it occurs. For interval (d) you need to figure out its duration, that is
to say, the time at which the rocket hits the ground.
(a) A rocket is shot upwards, accelerating from rest to a finalvelocity of 20 m/s in 1 s as it
burns its fuel. (Treat the acceleration as constant during this interval.)
(b) Fromt =1st o t =4 s ,w i t ht h ef u e le x h a u s t e d ,t h er o c k e tfl i e su n d e rt h ei n fl u ence of gravity",University Physics I Classical Mechanics.pdf
"(b) Fromt =1st o t =4 s ,w i t ht h ef u e le x h a u s t e d ,t h er o c k e tfl i e su n d e rt h ei n fl u ence of gravity
alone. At some point during this time interval (you need to figure out when!) it stops climbing and
starts falling.
(c) At t =4 sap a r a c h u t eo p e n s ,s u d d e n l yc a u s i n ga nu p w a r d sa c c e l e r ation (again, treat it as
constant) lasting 1 s; at the end of this interval, the rocket’s velocity is 5 m/s downwards.
(d) The last part of the motion, with the parachute deployed,is with constant velocity of 5 m/s
downwards until the rocket hits the ground.
Solution:
(a) For this first interval (for which I will use a subscript “1”t h r o u g h o u t )w eh a v e
∆y1 = 1
2a1(∆t1)2 (2.18)
using Eq. (2.6)f o rm o t i o nw i t hc o n s t a n ta c c e l e r a t i o nw i t hz e r oi n i t i a lv elocity (I am using the",University Physics I Classical Mechanics.pdf
"2a1(∆t1)2 (2.18)
using Eq. (2.6)f o rm o t i o nw i t hc o n s t a n ta c c e l e r a t i o nw i t hz e r oi n i t i a lv elocity (I am using the
variable y,i n s t e a do fx,f o rt h ev e r t i c a lc o o r d i n a t e ;t h i si sm o r eo rl e s sc u s t o m a r y,b u t ,o fc o u r s e ,I
could have usedx just as well).
Since the acceleration is constant, it is equal to its averagev a l u e :
a1 = ∆v
∆t =2 0m
s2 .
Substituting this into (2.18)w eg e tt h eh e i g h ta tt =1 si s1 0m . T h ev e l o c i t ya tt h a tt i m e ,o f
course, isvf1 =2 0 m / s ,a sw ew e r et o l di nt h es t a t e m e n to ft h ep r o b l e m .
(b) This part is free fall with initial velocityvi2 =2 0m / s . T ofi n dh o wh i g ht h er o c k e tc l i m b s ,u s e
Eq. (2.15)i nt h ef o r mvtop − vi2 = −g(ttop − ti2), with vtop =0( a st h er o c k e tc l i m b s ,i t sv e l o c i t y
decreases, and it stops climbing when its velocity is zero).This gives usttop =3 .04 s as the time at",University Physics I Classical Mechanics.pdf
"2.5. EXAMPLES 47
which the rocket reaches the top of its trajectory, and then starts coming down. The corresponding
displacement is, by Eq. (2.16),
∆ytop = vi2(ttop − ti2) − 1
2g(ttop − ti2)2 =2 0.4m
so the maximum height it reaches is 30.4m .
At the end of the full 3-second interval, the rocket’s displacement is
∆y2 = vi2∆t2 − 1
2g(∆t2)2 =1 5.9m
(so its height is 25.9ma b o v et h eg r o u n d ) ,a n dt h efi n a lv e l o c i t yi s
vf2 = vi2 − g∆t2 = −9.43 m
s .
(c) The acceleration for this part is (vf3 −vi3)/∆t3 =( −5+9 .43)/1=4 .43 m/s2.N o t et h e p o s i t i v e
sign. The displacement is
∆y3 = −9.43 × 1+ 1
2 × 4.43 × 12 = −7.22 m
so the final height is 25.9 − 7.21 = 18.7m .
(d) This is just motion with constant speed to cover 18.7ma t5m / s .T h et i m ei tt a k e si s3.74 s.
The graphs for this motion are shown earlier in the chapter, inF i g u r e2 . 3 .",University Physics I Classical Mechanics.pdf
"48 CHAPTER 2. ACCELERATION
2.6 Problems
Problem 1
You get on your bicycle and ride it with a constant acceleration of 0.5m / s2 for 20 s. After that,
you continue riding at a constant velocity for a distance of 200 m. Finally, you slow to a stop, with
ac o n s t a n ta c c e l e r a t i o n ,o v e rad i s t a n c eo f2 0 m .
v (m/s)
t (s)
0
0
x (m)
t (s)
0
0
a (m/s2)
t (s)
0",University Physics I Classical Mechanics.pdf
"2.6. PROBLEMS 49
(a) How far did you travel while you were accelerating at 0.5m / s2,a n dw h a tw a sy o u rv e l o c i t ya t
the end of that interval?
(b) After that, how long did it take you to cover the next 200 m?
(c) What was your acceleration while you were slowing down toas t o p ,a n dh o wl o n gd i di tt a k e
you to come to a stop?
(d) Considering the whole trip, what was your average velocity?
(e) Plot the position versus time, velocity versus time, andacceleration versus time graphs for the
whole trip, in the grids provided above. Values at the beginning and end of each interval must
be exact. Slopes and curvatures must be represented accurately. Do not draw any of the curves
beyond the time the rider stops (or, if you do, make sure what you draw makes sense!).
Problem 2
You throw an ob ject straight upwards and catch it again, whenit comes down to the same initial
height, 2 s later.
(a) How high did it rise above its initial height?",University Physics I Classical Mechanics.pdf
"height, 2 s later.
(a) How high did it rise above its initial height?
(b) With what initial speed did you throw it?
(Note: for this problem you should use the fact that, if air drag is negligible, the object will return
to its initial height with the same speed it had initially.)
Problem 3
You are trying to catch up with a car that is in front of you on theh i g h w a y . I n i t i a l l yy o ua r e
both moving at 25m/s, and the distance between you is 100m. Yous t e po nt h eg a sa n ds u s t a i na
constant acceleration for a time ∆t =3 0s ,a tw h i c hp o i n ty o uh a v ep u l l e de v e nw i t ht h eo t h e rc a r .
(a) What is 25 m/s, in miles per hour?
(b) What was your acceleration over the 30 s time interval?
(c) How fast were you going when you caught up with the other car?
Problem 4
Go back to Problem 4 of Chapter 1, and use the information in thefi g u r et od r a wa na c c u r a t e
position vs. time graph for both runners.
Problem 5",University Physics I Classical Mechanics.pdf
"position vs. time graph for both runners.
Problem 5
Ac h i l do nas l e ds l i d e s( s t a r t i n gf r o mr e s t )d o w na ni c ys l o p ethat makes an angle of 15◦ with the
horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides
for 2.0s w i t h a c o n s t a n t n e g a t i v e a c c e l e r a t i o n o f−1.5m / s2 with respect to her direction of motion.
She then slides up another icy slope that makes a 20◦ angle with the horizontal.
(a) How fast was the child going when she reached the bottom ofthe first slope? How long did it
take her to get there?
(b) How long was the flat stretch at the bottom?
(c) How fast was the child going as she started up the second slope?
(d) How far up the second slope did she slide?",University Physics I Classical Mechanics.pdf
50 CHAPTER 2. ACCELERATION,University Physics I Classical Mechanics.pdf
"Chapter 3
Momentum and Inertia
3.1 Inertia
In everyday language, we speak of something or someone “having a large inertia” to mean, essen-
tially, that they are very difficult to set in motion. This usageo ft h ew o r d“ i n e r t i a ”i sc o n s i s t e n t
with the “law of inertia” we introduced in the previous chapter (which states, among other things,
that an object at rest, if left to itself, will just remain at rest), but it goes a bit beyond that by
trying to quantify just how hard it may be to get the object to move.
We do know, from exp erience, that lighter ob jects are easierto set in motion than heavier objects,
but most of us probably have an intuition that gravity (the force that pulls an object towards
the earth and hence determines its weight) is not involved inan essential way here. Imagine, for
instance, the difference between slapping a volleyball and a bowling ball. It is not hard to believe",University Physics I Classical Mechanics.pdf
"instance, the difference between slapping a volleyball and a bowling ball. It is not hard to believe
that the latter would hurt as much if we did it while floating infree fall in the space station (in a
state of effective “weightlessness”) as if we did it right hereon the surface of the earth. In other
words, it is not (necessarily) how heavy something feels, butj u s th o wmassive it is.
But just what is this “massiveness” quality that we associatei n t u i t i v e l yw i t hal a r g ei n e r t i a ? I s
there a way (other than resorting to the weight again) to assign to it a numerical value?
3.1.1 Relative inertia and collisions
One possible way to determine therelative inertias of two objects, conceptually, at least, is to try
to use one of them to set the other one in motion. Most of us are familiar with what happens
51",University Physics I Classical Mechanics.pdf
"52 CHAPTER 3. MOMENTUM AND INERTIA
when two identical objects (presumably, therefore, havingthe same inertia) collide: if the collision
is head-on (so the motion, before and after, is confined to a straight line), they basically exchange
velocities. For instance, a billiard ball hitting another one will stop dead and the second one will set
off with the same speed as the first one. The toy sometimes called“ N e w t o n ’ sb a l l s ”o r“ N e w t o n ’ s
cradle” also shows this effect. Intuitively, we understand that what it takes to stop the first ball is
exactly the same as it would take to set the second one in motionw i t ht h es a m ev e l o c i t y .
But what if the objects colliding have different inertias? We expect that the change in their
velocities as a result of the collision will be different: the velocity of the object with the largest
inertia will not change very much, and conversely, the changei nt h ev e l o c i t yo ft h eo b j e c tw i t ht h e",University Physics I Classical Mechanics.pdf
"inertia will not change very much, and conversely, the changei nt h ev e l o c i t yo ft h eo b j e c tw i t ht h e
smallest inertia will be comparatively larger. A velocity vs. time graph for the two objects might
look somewhat like the one sketched in Fig.3.1.
-0.4
-0.2
 0
 0.2
 0.4
 0.6
 0.8
 1
 1.2
 0  2  4  6  8  10
1
1
2
2
1 m/s
v (m/s)
t (ms)
Figure 3.1: An example of a velocity vs. time graph for a collision of two objects with different inertias.
In this picture, object 1, initially moving with velocityv1i =1m / s ,c o l l i d e sw i t ho b j e c t2 ,i n i t i a l l y
at rest. After the collision, which here is assumed to take a millisecond or so, object 1 actually
bounces back, so its final velocity isv1f = −1/3m / s , w h e r e a s o bj e c t 2 e n d s u p m o v i n g t o t h e r i g h t
with velocityv2f =2 /3m / s . S ot h ec h a n g ei nt h ev e l o c i t yo fo bj e c t1i s∆v1 = v1f −v1i = −4/3m / s ,
whereas for object 2 we have ∆v2 = v2f − v2i =2 /3m / s .",University Physics I Classical Mechanics.pdf
"whereas for object 2 we have ∆v2 = v2f − v2i =2 /3m / s .
It is tempting to use this ratio, ∆v1/∆v2,a sam e a s u r eo ft h erelative inertia of the two objects,
only we’d want to use it upside down and with the opposite sign:t h a t i s ,s i n c e ∆v2/∆v1 = −1/2
we would say that object 2 hastwice the inertia of object 1. But then we have to ask: is this a",University Physics I Classical Mechanics.pdf
"3.1. INERTIA 53
reliable, repeatable measure? Will it work for any kind of collision (within reason, of course: we
clearly need to stay in one dimension, and eliminate externali n fl u e n c e ss u c ha sf r i c t i o n ) ,a n df o r
any initial velocity?
To begin with, we have reason to expect that it does not matterwhether we shoot object 1 towards
object 2 or object 2 towards object 1, because we learned in thep r e v i o u sc h a p t e rt h a tonly relative
motion is detectable,a n dt h er e l a t i v em o t i o ni st h es a m ei nb o t hc a s e s . C o n s i d e r ,for instance, what
the collision in Figure3.1 appears like to a hypothetical observer moving along with object 1, at 1
m/s. To him, object 1 appears to be at rest, and it is object 2 that is coming towards him, with a
velocity of−1m / s . T o s e e w h a t t h e o u t c o m e o f t h e c o l l i s i o n l o o k s l i k e t o h im, just add the same",University Physics I Classical Mechanics.pdf
"velocity of−1m / s . T o s e e w h a t t h e o u t c o m e o f t h e c o l l i s i o n l o o k s l i k e t o h im, just add the same
−1m / s t o t h e fi n a l v e l o c i t i e s w e o b t a i n e d be f o r e : o bj e c t 1 w i l lend up moving atv1f = −4/3m / s ,
and object 2 would move atv2f = −1/3m / s , a n d w e w o u l d h a v e a s i t u a t i o n l i k e t h e o n e s h o w n i n
Figure 3.2,w h e r eb o t hc u r v e sh a v es i m p l yb e e ns h i f t e dd o w nb y1m / s :
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
 0
 0.2
 0  2  4  6  8  10
1
2
v (m/s)
t (ms)
1 2
−1 m/s
Figure 3.2: Another example (really the same collision as in Figure 1, only as seen byan observer initially
moving to the right at 1 m/s).
But then, this is exactly what we should expect to find also in our laboratory if we actually did
send the second object at 1 m/s towards the first one sitting atrest. All the individual velocities",University Physics I Classical Mechanics.pdf
"send the second object at 1 m/s towards the first one sitting atrest. All the individual velocities
have changed relative to Figure3.1,b u tt h evelocity changes,∆ v1 and ∆v2,a r ec l e a r l ys t i l lt h e
same, and therefore so is our (tentative) measure of the objects’ relative inertia.
Clearly, the same argument can be used to conclude that the same result will be obtained when
both objects are initially moving towards each other, as longa st h e i rrelative velocityis the same as",University Physics I Classical Mechanics.pdf
"54 CHAPTER 3. MOMENTUM AND INERTIA
in these examples, namely, 1 m/s. However, unless we do the experiments we cannot really predict
what will happen if we increase (or decrease) their relativevelocity. In fact, we could imagine
smashing the two objects at very high speed, so they might evenb e c o m es e r i o u s l ym a n g l e di nt h e
process. Yet, experimentally (and this is not at all an obvious result!), we would still find the same
value of −1/2f o rt h er a t i o∆v2/∆v1,a tl e a s ta sl o n ga st h ec o l l i s i o ni sn o ts ov i o l e n tt h a tt h e
objects actually break up into pieces.
Perhaps the most surprising result of our experiments wouldbe the following: imagine that the
objects have a “sticky” side (for instance, the small black rectangles shown in the pictures could
be strips of Velcro), and we turn them around so that when theycollide they will end up stuck to
each other. In this case (which, as we shall see later, is termed acompletely inelasticcollision),",University Physics I Classical Mechanics.pdf
"each other. In this case (which, as we shall see later, is termed acompletely inelasticcollision),
the v-vs-t graph might look like Figure3.3 below.
Now the two objects end up moving together to the right, fairlys l o w l y :v1f = v2f =1 /3m / s . T h e
velocity changes are ∆v1 = −2/3m / s a n d ∆v2 =1 /3 m/s, both of which are different from what
they were before, in Figs.3.1 and 3.2:y e t ,t h er a t i o∆v2/∆v1 is still equal to−1/2, just as in all
the previous cases.
 0
 0.2
 0.4
 0.6
 0.8
 1
 0  2  4  6  8  10
1 2
1 m/s
1
2
v (m/s)
t (ms)
Figure 3.3: What would happen if the objects in Figure 1 became stuck togetherwhen they collided.",University Physics I Classical Mechanics.pdf
"3.1. INERTIA 55
3.1.2 Inertial mass: definition and properties
At this point, it would seem reasonable to assume that this ratio, ∆v2/∆v1,i s ,i nf a c t ,t e l l i n gu s
something about anintrinsic property of the two objects, what we have called above their “relative
inertia.” It is easy, then, to see how one could assign a valueto the inertia of any object (at least,
conceptually): choose a “standard” object, and decide, arbitrarily, that its inertia will have the
numerical value of 1, in whichever units you choose for it (these units will turn out, in fact, to be
kilograms, as you will see in a minute). Then, to determine thei n e r t i ao fa n o t h e ro b j e c t ,w h i c h
we will label with the subscript 1, just arrange a one-dimensional collision between object 1 and
the standard, under the right conditions (basically, no netexternal forces), measure the velocity
changes ∆v1 and ∆vs,a n dt a k et h eq u a n t i t y−∆vs/∆v1 as the numerical value of the ratio of the",University Physics I Classical Mechanics.pdf
"changes ∆v1 and ∆vs,a n dt a k et h eq u a n t i t y−∆vs/∆v1 as the numerical value of the ratio of the
inertia of object 1 to the inertia of the standard object. In symbols, using the letterm to represent
an object’s inertia,
m1
ms
= −∆vs
∆v1
(3.1)
But, sincems =1b yd e fi n i t i o n ,t h i sg i v e su sd i r e c t l yt h en u m e r i c a lv a l u eof m1.
The reason we use the letterm is, as you must have guessed, because, in fact, the inertia defined
in this way turns out to be identical to what we have traditionally called “mass.” More precisely,
the quantity defined this way is an object’sinertial mass.T h e r e m a r k a b l ef a c t ,m e n t i o n e de a r l i e r ,
that the force of gravity between two objects turns out to be proportional to their inertial masses,
allows us to determine the inertial mass of an object by the more traditional procedure of simply
weighing it, rather than elaborately staging a collision between it and the standard kilogram on",University Physics I Classical Mechanics.pdf
"weighing it, rather than elaborately staging a collision between it and the standard kilogram on
an ice-hockey rink. But, in principle, we could conceive of the existence of two different quantities
that should be called “inertial mass” and “gravitational mass,” and the identity (or more precisely,
the—so far as we know—exact proportionality) of the two is a rather mysterious experimental fact1.
In any case, by the way we have constructed it, the inertial mass, defined as in Eq. (3.1), does
capture, in a quantitative way, the concept that we were trying to express at the beginning of
the chapter: namely, how difficult it may be to set an object in motion. In principle, however,
other experiments would need to be conducted to make sure thati td o e sh a v et h ep r o p e r t i e sw e
have traditionally associated with the concept of mass. Forinstance, suppose we join together two",University Physics I Classical Mechanics.pdf
"have traditionally associated with the concept of mass. Forinstance, suppose we join together two
objects of massm.I s t h e m a s s o f t h e r e s u l t i n g o b j e c t 2m?C o l l i s i o n e x p e r i m e n t s w o u l d , i n d e e d ,
show this to be the case with great accuracy in the macroscopicw o r l d( w i t hw h i c hw ea r ec o n c e r n e d
this semester), but this is a good example of how you cannot take anything for granted: at the
microscopic level, it is again a fact that the inertial mass ofa na t o m i cn u c l e u si sal i t t l eless than
the sum of the masses of all its constituent protons and neutrons2.
Probably the last thing that would need to be checked is thatthe ratio of inertias is independent
1This fact, elevated to the category of a principle by Einstein( t h eequivalence principle)i st h es t a r t i n gp o i n to f
the general theory of relativity.",University Physics I Classical Mechanics.pdf
"the general theory of relativity.
2And this is not just an unimportant bit of trivia: all of nuclear power depends on this small difference.",University Physics I Classical Mechanics.pdf
"56 CHAPTER 3. MOMENTUM AND INERTIA
of the standard.S u p p o s e t h a t w e h a v e t w o o b j e c t s , t o w h i c h w e h a v e a s s i g n e d masses m1 and m2
by arranging for each to collide with the “standard object” independently. If we now arrange for
ac o l l i s i o nb e t w e e no b j e c t s1a n d2d i r e c t l y ,w i l lw ea c t u a l lyfi n dt h a tt h er a t i oo ft h e i rv e l o c i t y
changes is given by the ratio of the separately determined masses m1 and m2?W ec e r t a i n l yw o u l d
need that to be the case, in order for the concept of inertia tobe truly useful; but again, we should
not assume anything until we have tested it! Fortunately, thet e s t sw o u l di n d e e dr e v e a lt h a t ,i n
every case, the expected relationship holds3
− ∆v2
∆v1
= m1
m2
(3.2)
At this point, we are not just in possession of a useful definition of inertia, but also of a veritable
law of nature,a sIw i l le x p l a i nn e x t .
3.2 Momentum",University Physics I Classical Mechanics.pdf
"law of nature,a sIw i l le x p l a i nn e x t .
3.2 Momentum
For an ob ject of (inertial) massm moving, in one dimension, with velocityv,w ed e fi n ei t smomen-
tum as
p = mv (3.3)
(the choice of the letterp for momentum is apparently related to the Latin word “impetus”).
We can think of momentum as a sort of extension of the concept ofi n e r t i a ,f r o ma no b j e c ta tr e s t
to an object in motion. When we speak of an object’s inertia, wet y p i c a l l yt h i n ka b o u tw h a ti t
may take to get it moving; when we speak of its momentum, we typically think of that it may take
to stop it (or perhaps deflect it). So, both the inertial massm and the velocityv are involved in
the definition.
We may also observe that what looks like inertia in some reference frame may look like momentum
in another. For instance, if you are driving in a car towing a trailer behind you, the trailer has
only a large amount of inertia, but no momentum, relative to you, because its velocity relative to",University Physics I Classical Mechanics.pdf
"only a large amount of inertia, but no momentum, relative to you, because its velocity relative to
you is zero; however, the trailer definitely has a large amounto fm o m e n t u m( b yv i r t u eo fb o t hi t s
inertial mass and its velocity) relative to somebody standing by the side of the road.
3.2.1 Conservation of momentum; isolated systems
For a system of ob jects, we treat the momentum as anadditive quantity. So, if two colliding objects,
of massesm1 and m2,h a v ei n i t i a lv e l o c i t i e sv1i and v2i,w es a yt h a tt h et o t a li n i t i a lm o m e n t u mo f
3Equation 3.2 actually is found to hold also at the microscopic (orquantum)l e v e l ,a l t h o u g ht h e r ew ep r e f e rt o
state the result by saying that conservation of momentum holds (see the following section).",University Physics I Classical Mechanics.pdf
"3.2. MOMENTUM 57
the system ispi = m1v1i + m2v2i,a n ds i m i l a r l yi ft h efi n a lv e l o c i t i e sa r ev1f and v2f ,t h et o t a lfi n a l
momentum will bepf = m1v1f + m2v2f .
We then assert thatthe total momentum of the system is not changed by the collision.M a t h e m a t -
ically, this means
pi = pf (3.4)
or
m1v1i + m2v2i = m1v1f + m2v2f (3.5)
But this last equation, in fact, follows directly from Eq. (3.2): to see this, move all the quantities
in Eq. (3.5)h a v i n gt od ow i t ho b j e c t1t oo n es i d eo ft h ee q u a ls i g n ,a n dt hose having to do with
object 2 to the other side. You then get
m1 (v1i − v1f )= m2 (v2f − v2i)
−m1∆v1 = m2∆v2 (3.6)
which is just another way to write Equation (3.2). Hence, the result (3.2)e n s u r e st h ec o n s e r v a t i o n
of the total momentum of a system of any two interacting objects (“particles”), regardless of the
form the interaction takes, as long as there are no external forces acting on them.",University Physics I Classical Mechanics.pdf
"form the interaction takes, as long as there are no external forces acting on them.
Momentum conservation is one of the most important principles in all of physics, so let me take
al i t t l et i m et oe x p l a i nh o ww eg o th e r ea n de l a b o r a t eo nt h i sresult. First, as I just mentioned,
we have been more or less implicitly assuming that the two interacting objects form anisolated
system, by which we mean that, throughout, they interact withn o t h i n go t h e rt h a ne a c ho t h e r .
(Equivalently, there are no external forces acting on them.)
It is pretty much impossible to set up a system so that it isreally isolated in this strict sense;
instead, in practice, we settle for making sure that the external forces on the two objectscancel
out.T h i s i s w h a t h a p p e n s o n t h e a i r t r a c k s w i t h w h i c h y o u w i l l b e doing experiments this semester:
gravity is acting on the carts, but that force is balanced outby the upwards push of the air from",University Physics I Classical Mechanics.pdf
"gravity is acting on the carts, but that force is balanced outby the upwards push of the air from
the track. A system on which there is nonet external force is as good as isolated for practical
purposes, and we will refer to it as such. (It is harder, of course, to completely eliminate friction
and drag forces, so we just have to settle for approximately isolated systems in practice.)
Secondly, we have assumed so far that the motion of the two objects is restricted to a straight
line—one dimension. In fact, momentum is avector quantity (just like velocity is), so in general
we should write
⃗p= m⃗ v
and conservation of momentum, in general, holds as a vector equation for any isolated system in
three dimensions:
⃗pi = ⃗pf (3.7)",University Physics I Classical Mechanics.pdf
"58 CHAPTER 3. MOMENTUM AND INERTIA
What this means, in turn, is that each separate component (x, y and z)o ft h em o m e n t u mw i l l
be separately conserved (so Eq. (3.7)i se q u i v a l e n tt ot h r e es c a l a re q u a t i o n s ,i nt h r e ed i m e n s i ons).
When we get to study the vector nature of forces, we will see aninteresting implication of this,
namely, that it is possible for one component of the momentumvector to be conserved, but not
another—depending on whether there is or there isn’t a net external force in that direction or not.
For example, anticipating things a bit, when you throw an ob ject horizontally, as long as you can
ignore air drag, there is no horizontal force acting on it, ands ot h a tc o m p o n e n to ft h em o m e n t u m
vector is conserved, but the vertical component is changingall the time because of the (vertical)
force of gravity.
Thirdly, although this may not be immediately obvious, for ani s o l a t e ds y s t e mo ft w oc o l l i d i n g",University Physics I Classical Mechanics.pdf
"force of gravity.
Thirdly, although this may not be immediately obvious, for ani s o l a t e ds y s t e mo ft w oc o l l i d i n g
objects the momentum is truly conserved throughout the wholec o l l i s i o np r o c e s s . I ti sn o tj u s ta
matter of comparing the initial and final velocities: at any oft h et i m e ss h o w ni nF i g u r e s1t h r o u g h
3, if we were to measurev1 and v2 and compute m1v1 + m2v2,w ew o u l do b t a i nt h es a m er e s u l t .
In other words, the total momentum of an isolated system isconstant:i t h a s t h e s a m e v a l u e a t a l l
times.
Finally, all these examples have involved interactions between only two particles. Can we really
generalize this to conclude that the total momentum of an isolated system of any number of particles
is constant, even when all the particles may be interacting with each other simultaneously? Here,
again, the experimental evidence is overwhelmingly in favoro ft h i sh y p o t h e s i s4,b u tm u c ho fo u r",University Physics I Classical Mechanics.pdf
"again, the experimental evidence is overwhelmingly in favoro ft h i sh y p o t h e s i s4,b u tm u c ho fo u r
confidence on its validity comes in fact from a considerationof the nature of the internal interactions
themselves. It is a mathematical fact that all of the interactions so far known to physics have the
property of conserving momentum, whether acting individually or simultaneously. No experiments
have ever suggested the existence of an interaction that doesn o th a v et h i sp r o p e r t y .
3.3 Extended systems and center of mass
Consider a collection of particles with massesm1,m 2,... ,a n dl o c a t e d ,a ts o m eg i v e ni n s t a n t ,a t
positions x1,x 2 ... .( W e a r e s t i l l , f o r t h e t i m e b e i n g , c o n s i d e r i n g o n l y m o t i o n ino n ed i m e n s i o n ,b u t
all these results generalize easily to three dimensions.) The center of massof such a system is a
mathematical point whose position coordinate is given by
xcm = m1x1 + m2x2 + ...",University Physics I Classical Mechanics.pdf
"mathematical point whose position coordinate is given by
xcm = m1x1 + m2x2 + ...
m1 + m2 + ... (3.8)
Clearly, the denominator of (3.8)i sj u s tt h et o t a lm a s so ft h es y s t e m ,w h i c hw em a yj u s td e n o t e
by M.I fa l lt h ep a r t i c l e sh a v et h es a m em a s s ,t h ec e n t e ro fm a s sw ill be somehow “in the middle”
4For an important piece of indirect evidence, just consider that any extended object is in reality a collection
of interacting particles, and the experiments establishingc o n s e r v a t i o no fm o m e n t u ma l m o s ta l w a y si n v o l v es u c h
extended objects. See the following section for further thoughts on this matter.",University Physics I Classical Mechanics.pdf
"3.3. EXTENDED SYSTEMS AND CENTER OF MASS 59
of all of them; otherwise, it will tend to be closer to the moremassive particle(s). The “particles”
in question could be spread apart, or they could literally bethe “parts” into which we choose to
subdivide, for computational purposes, a single extended object.
If the particles are in motion, the position of the center of mass, xcm,w i l li ng e n e r a lc h a n g ew i t h
time. For a solid object, where all the parts are moving together, the displacement of the center of
mass will just be the same as the displacement of any part of theo b j e c t . I nt h em o s tg e n e r a lc a s e ,
we will have (by subtractingxcmi from xcmf )
∆xcm = 1
M (m1∆x1 + m2∆x2 + ... )( 3 . 9 )
Dividing Eq. (3.9)b y∆t and taking the limit as ∆t → 0, we get the instantaneous velocity of the
center of mass:
vcm = 1
M (m1v1 + m2v2 + ... )( 3 . 1 0 )
But this is just
vcm = psys
M (3.11)
where psys = m1v1 + m2v2 + ... is the total momentum of the system.",University Physics I Classical Mechanics.pdf
"vcm = 1
M (m1v1 + m2v2 + ... )( 3 . 1 0 )
But this is just
vcm = psys
M (3.11)
where psys = m1v1 + m2v2 + ... is the total momentum of the system.
3.3.1 Center of mass motion for an isolated system
Equation (3.11)i sav e r yi n t e r e s t i n gr e s u l t . S i n c et h et o t a lm o m e n t u mo fa nisolated system is
constant, it tells us that the center of mass of an isolated system of particles moves at constant
velocity, regardless of the relative motion of the particlest h e m s e l v e so rt h e i rp o s s i b l ei n t e r a c t i o n s
with each other. As indicated above, this generalizes straightforwardly to more than one dimension,
so we can write⃗vcm = ⃗psys/M.T h u s , w e c a n s a y t h a t , f o r a n i s o l a t e d s y s t e m i n s p a c e , n o t o nly
the speed, but also the direction of motion of its center of mass does not change with time.
Clearly this result is a sort of generalization of the law of inertia. For a solid, extended object, it",University Physics I Classical Mechanics.pdf
"Clearly this result is a sort of generalization of the law of inertia. For a solid, extended object, it
does, in fact, provide us with the precise form that the law ofinertia must take: in the absence of
external forces,the center of masswill just move on a straight line with constant velocity, whereas
the object itself may be moving in any way that does not affect thec e n t e ro fm a s st r a j e c t o r y .
Specifically, the most general motion of an isolated rigid body is a straight line motion of its center
of mass at constant speed, combined with a possible rotationof the object as a whole around the
center of mass.
For a system that consists of separate parts, on the other hand, the center of mass is generally
just a point in space, which may or may not coincide at any timewith the position of any of the
parts, but which will nonetheless move at constant velocityas long as the system is isolated. This",University Physics I Classical Mechanics.pdf
"parts, but which will nonetheless move at constant velocityas long as the system is isolated. This
is illustrated by Figure3.4,w h e r et h ep o s i t i o nv s . t i m ec u r v e sh a v eb e e nd r a w nf o rt h ec olliding",University Physics I Classical Mechanics.pdf
"60 CHAPTER 3. MOMENTUM AND INERTIA
objects of Figure3.1.I h a v ea s s u m e d t h a to b j e c t 1 s t a r t so u t a tx1i = −5m m a tt =0 ,a n do b j e c t
2s t a r t sa tx2i =0a t t =0 . B e c a u s eo b j e c t2h a st w i c et h ei n e r t i ao fo b j e c t1 ,t h ep o sition of the
center of mass, as given by Eq. (3.8), will always be
xcm = x1/3+2 x2/3
that is to say, the center of mass will always be in between objects 1 and 2, and its distance from
object 2 will always be half its distance to object 1:
|xcm − x1| = 2
3|x1 − x2|
|xcm − x2| = 1
3|x1 − x2|
Figure 4 shows that this simple prescription does result in motion with constant velocity for the
center of mass (the green straight line), even though thex-vs-t curves of the two objects themselves
look relatively complicated. (Please check it out! Take a ruler to Fig.3.4 and verify that the center
of mass is, at every instant, where it is supposed to be.)
-5
-4
-3
-2
-1
 0
 1
 2
 3
 4
 0  2  4  6  8  10
x1(t)
x2(t)
xcm
x (mm)",University Physics I Classical Mechanics.pdf
"of mass is, at every instant, where it is supposed to be.)
-5
-4
-3
-2
-1
 0
 1
 2
 3
 4
 0  2  4  6  8  10
x1(t)
x2(t)
xcm
x (mm)
t (ms)
Figure 3.4: Position vs. time graph for the objects colliding in Figure 1. The greenline shows the position
of the center of mass as a function of time.
The concept of center of mass gives us an important way to simplify the description of the motion
of potentially complicated systems. We will make good use ofit in forthcoming chapters.
Av e r yn i c ed e m o n s t r a t i o no ft h em o t i o no ft h ec e n t e ro fm a s sint w o - b o d yo n e - d i m e n s i o n a lc o l l i s i o n s
can be found at
https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html
(you need to check the “center of mass” box to see it).",University Physics I Classical Mechanics.pdf
"3.4. IN SUMMARY 61
3.3.2 Recoil and rocket propulsion
As we have just seen, you cannot alter the motion of your centero fm a s sw i t h o u tr e l y i n go ns o m e
external force—which is to say, some kind of external support. This is actually something you may
have experienced when you are resting on a very slippery surface and you just cannot “get a grip”
on it. There is, however, one way to circumvent this problem which, in fact, relies on conservation
of momentum itself: if you are carrying something with you, and can throw it away from you at
high speed, you will recoil as a result of that. If you can keepthrowing things, you (with your store
of as yet unthrown things) will speed up a little more every time. This is, in essence, the principle
behind rocket propulsion.
Mathematically, consider two objects, of massesm1 and m2,i n i t i a l l ya tr e s t ,s ot h e i rt o t a lm o -",University Physics I Classical Mechanics.pdf
"behind rocket propulsion.
Mathematically, consider two objects, of massesm1 and m2,i n i t i a l l ya tr e s t ,s ot h e i rt o t a lm o -
mentum is zero. If mass 1 is thrown away from mass 2 with a speedv1f ,t h e n ,b yc o n s e r v a t i o no f
momentum (always assuming the system is isolated) we must have
0= m1v1f + m2v2f (3.12)
and thereforev2f = −m1v1f /m2.T h i s i s h o w a r o c k e t m o v e sf o r w a r d , b y c o n s t a n t l ye x p e l l i n gmass
(the hot exhaust gas) backwards at a high velocity. Note that,e v e nt h o u g hb o t ho b j e c t sm o v e ,t h e
center of mass of the whole system doesnot (in the absence of any external force), as discussed
above.
3.4 In summary
1. The inertia of an object is a measure of its tendency to resist changes in its motion. It is
quantified by theinertial mass (measured in kilograms).
2. A system of objects is calledisolated (for practical purposes) when there are nonet (or",University Physics I Classical Mechanics.pdf
"quantified by theinertial mass (measured in kilograms).
2. A system of objects is calledisolated (for practical purposes) when there are nonet (or
unbalanced)e x t e r n a lf o r c e sa c t i n go na n yo ft h eo b j e c t s( t h eo b j e c t sm ays t i l li n t e r a c tw i t h
each other).
3. When two objects forming an isolated system collide in onedimension, the changes in their
velocities are inversely proportional to their inertial masses:
∆v1
∆v2
= −m2
m1
This may be used, in principle, as a way to define the inertial mass operationally.
4. The inertial mass thus defined turns out to be exactly (as fara sw ek n o w )p r o p o r t i o n a lt ot h e
object’sgravitational mass,w h i c hd e t e r m i n e st h eg r a v i t a t i o n a lf o r c eo fa t t r a c t i o nb etween it
and any other object. For this reason, most often we measure ano b j e c t ’ si n e r t i a lm a s ss i m p l y
by weighing it.",University Physics I Classical Mechanics.pdf
"62 CHAPTER 3. MOMENTUM AND INERTIA
5. The momentum of an object of inertial massm moving with a velocity⃗vis defined as⃗p= m⃗ v.
The total momentum of a system of objects is defined as the (vector) sum of all the individual
momenta.
6. (Conservation of momentum) The momentum of an isolated system remains always con-
stant,r e g a r d l e s so fh o wt h ep a r t st h a tm a k eu pt h es y s t e mm a yi n t e ract with one another.
7. In one dimension, thecenter of massof a system of particles is a mathematical point whose
x coordinate is given by Equation (3.8)a b o v e . ( I nm o r ed i m e n s i o n s ,j u s tc h a n g et h ex’s in
Eq. (3.8)t oy and z to getycm and zcm.)
8. The center of mass of a system always moves with a velocity
⃗vcm = ⃗psys
M
where ⃗psys is the total momentum of the system, andM its total mass.
9. It follows from 8 and 6 above that for an isolated system, thec e n t e ro fm a s sm u s ta l w a y sb e",University Physics I Classical Mechanics.pdf
"9. It follows from 8 and 6 above that for an isolated system, thec e n t e ro fm a s sm u s ta l w a y sb e
at rest or moving with constant velocity. This result generalizes the law of inertia to extended
objects, or systems of particles.",University Physics I Classical Mechanics.pdf
"3.5. EXAMPLES 63
3.5 Examples
3.5.1 Reading a collision graph
The graph shows a collision between two carts (possibly equipped with magnets so that they repel
each other before they actually touch) on an air track. The inertia (mass) of cart 1 is 1 kg. Note:
this is aposition vs. time graph!
(a) What are the initial velocities of the carts?
(b) What are the final velocities of the carts?
(c) What is the mass of the second cart?
(d) Does the air track appear to be level? Why? (Hint: does thegraph show any evidence of
acceleration, for either cart, outside of the collision region?)
(e) At the collision time, is the acceleration of the first cartp o s i t i v eo rn e g a t i v e ? H o wa b o u tt h e
second cart? (Justify your answers.)
(f) For the system consisting of the two carts, what is its initial (total) momentum? What is its
final momentum?
(g) Imagine now that one of the magnets is reversed, so when thec a r t sc o l l i d et h e ys t i c kt oe a c h",University Physics I Classical Mechanics.pdf
"final momentum?
(g) Imagine now that one of the magnets is reversed, so when thec a r t sc o l l i d et h e ys t i c kt oe a c h
other. What would then be the final momentum of the system? Whatw o u l db ei t sfi n a lv e l o c i t y ?
x (m)
t (s)
cart 2
cart 1
Figure 3.5: A collision between two carts.
Solution
(a) All the velocities are to be calculated by picking an easystraight part of each curve and
calculating
v = ∆x
∆t",University Physics I Classical Mechanics.pdf
"64 CHAPTER 3. MOMENTUM AND INERTIA
for suitable intervals. In this way one gets
v1i = −1 m
s
v2i =0 .5 m
s
(b) Similarly, one gets
v1f =1 m
s
v2f = −0.5 m
s
(c) Use this equation, or equivalent (conservation of momentum is OK)
m2
m1
= −∆v1
∆v2
m2
m1
= − 1 − (−1)
−0.5 − 0.5 =2
so the mass of the second cart is 2 kg.
(d) Yes, the track appears to be level because the carts do notshow any evidence of acceleration
outside of the collision region (the position vs. time curvesa r es t r a i g h tl i n e so u t s i d eo ft h er e g i o n
approximately given by 4.5s <t< 5.5s ) .
(e) The acceleration of the first cart is positive. You can seethis either graphically (the curve is
like a parabola that opens upwards, i.e., concave), or algebraically (the cart’s velocity increases,
going from−1m / s t o 1m / s )
Similarly, the acceleration of the second cart is negative.The curve is like a parabola that opens",University Physics I Classical Mechanics.pdf
"going from−1m / s t o 1m / s )
Similarly, the acceleration of the second cart is negative.The curve is like a parabola that opens
downwards, i.e., convex; or, algebraically, the cart’s velocity decreases, going from 0.5m / s t o
−0.5m / s .
(f) The initial momentum of the system is
pi = m1v1i + m2v2i =( 1k g )×
(
−1 m
s
)
+( 2k g )×
(
0.5 m
s
)
=0
The final momentum is
pf = m1v1f + m2v2f =( 1k g )×
(
1 m
s
)
+( 2k g )×
(
−0.5 m
s
)
=0
You could also just say that the final momentum should be the same as the initial momentum,
since the system appears to be isolated.
(g) The momentum should be conserved in this case as well, sopf =0 . T h ev e l o c i t yw o u l db e
vf = pf
m1 + m2
=0",University Physics I Classical Mechanics.pdf
"3.5. EXAMPLES 65
3.5.2 Collision in different reference frames, center of mass, and recoil
An 80-kg hockey player (call him player 1), moving at 3 m/s to the right, collides with a 90-kg
player (player 2) who was moving at 2 m/s to the left. For a briefm o m e n t ,t h e ya r es t u c ks l i d i n g
together as they grab at each other.
(a) What is their joint velocity as they slide together?
(b) What was the velocity of their center of mass before and after the collision?
(c) What does the collision look like to another player that was skating initially at 1.5m / s t o
the right? Give all the initial and final velocities as seen bythis player, and show explicitly that
momentum is also conserved in this player’s frame of reference.
(d) Eventually, the 90-kg player manages to push the other oneb a c k ,i ns u c haw a yt h a tp l a y e r1
(the 80-kg player) ends up moving at 1 m/s to the leftrelative to player 2.W h a t a r e t h e i r fi n a l
velocities in the earth frame of reference?",University Physics I Classical Mechanics.pdf
"velocities in the earth frame of reference?
Solution
(a) Call the initial velocitiesv1i and v2i,t h ej o i n tfi n a lv e l o c i t yvf ,a n da s s u m et h et w op l a y e r sa r e
an isolated system for practical purposes. Then conservation of momentum reads
m1v1i + m2v2i =( m1 + m2)vf (3.13)
Solving for the final velocity, we get
vf = m1v1i + m2v2i
m1 + m2
(3.14)
Substituting the values given, we get
vf = 80 × 3 − 90 × 2
170 =0 .353 m
s (3.15)
(b) According to Eq. (3.10), the velocity of the center of mass,vcm,i sj u s tt h es a m ea sw h a tw e
just calculated (Eq. (3.14)a b o v e ) . T h i sm a k e ss e n s e :a f t e rt h ec o l l i s i o n ,i ft h ep l a y ers are moving
together, their system’s center of mass has to be moving withthem. Also, if the system is isolated,
the center of mass velocity should be the same before and aftert h ec o l l i s i o n . S ot h ea n s w e ri s
vcm = vf =0 .353 m/s",University Physics I Classical Mechanics.pdf
"the center of mass velocity should be the same before and aftert h ec o l l i s i o n . S ot h ea n s w e ri s
vcm = vf =0 .353 m/s
(c) Let me refer to this third player as “player 3,” and introduce a subscript “3” to refer to the
quantities as seen in his frame of reference. Let also the subscript “E”d e n o t et h eo r i g i n a l ,“ E a r t h ”
reference frame. From Eq. (1.19), we have then (for player 1, for instance)
v31 = v3E + vE1 = vE1 − vE3 (3.16)
because v3E,t h e“ v e l o c i t yo ft h eE a r t hi np l a y e r3 ’ sr e f e r e n c ef r a m e , ”isc l e a r l ye q u a lt o−vE3,t h e
negative of the velocity of player 3 relative to the Earth. Basically, what Eq. (3.16)i ss a y i n gi s",University Physics I Classical Mechanics.pdf
"66 CHAPTER 3. MOMENTUM AND INERTIA
that to convert all the Earth-frame velocities to the reference frame of player 3, we just need to
subtract 1.5m / s f r o m t h e m . T h i s g i v e s u s
v31,i =3 m
s − 1.5 m
s =1 .5 m
s
v32,i = −2 m
s − 1.5 m
s = −3.5 m
s
v31,f = v32,f =0 .353 m
s − 1.5 m
s = −1.147 m
s (3.17)
The total initial momentum in player’s 3 reference frame is then
psys,i = m1v31,i + m2v32,i =8 0× 1.5+9 0× (−3.5) =−195 kg· m
s (3.18)
and the final momentum is
psys,f =( m1 + m2)v31,f =1 7 0× (−1.147) =−195 kg· m
s (3.19)
So the total momentum is conserved in player 3’s reference frame. The reason for this is that this
is an inertial reference frame, because the velocity of player 3 does not change.
(d) For this part of the problem, we are back to the original reference frame (the Earth reference
frame), and we can drop the “E”s u b s c r i p t . F o rt h i sn e wp r o c e s s ,t h efi n a lv e l o c i t i e sf r o mpart",University Physics I Classical Mechanics.pdf
"frame), and we can drop the “E”s u b s c r i p t . F o rt h i sn e wp r o c e s s ,t h efi n a lv e l o c i t i e sf r o mpart
(a) become the initial velocities, so we havev1i = v2i =0 .353 m/s. [Note: alternatively, since the
system is isolated throughout, it would be OK in this case to use the velocitiesbefore the collision
to calculate its total momentum, which also needs to be conserved in this process.] We are also
told that the final velocity of player 1relative to player 2 is v21,f = v1f − v2f = −1m / s . S o w e
have two equations to solve:
(m1 + m2) ×
(
0.353 m
s
)
= m1v1f + m2v2f (conservation of momentum)
v1f − v2f = −1 m
s (final relative velocity) (3.20)
Leaving aside the units for the moment, to make the equationsmore readable (the final units will
work out, if we make sure to use SI units all along), we have:
(80 + 90)× 0.353 = 80v1f +9 0v2f
v1f = v2f − 1( 3 . 2 1 )
Now substitute the second equation, which I have “solved” already forv1f ,i nt h efi r s te q u a t i o na n d",University Physics I Classical Mechanics.pdf
"v1f = v2f − 1( 3 . 2 1 )
Now substitute the second equation, which I have “solved” already forv1f ,i nt h efi r s te q u a t i o na n d
solve forv2f .T h e r e s u l t i sv2f =0 .824 m/s, which, when substituted back in the relative velocity
equation, givesv1f = −0.176 m/s.",University Physics I Classical Mechanics.pdf
"3.6. PROBLEMS 67
3.6 Problems
Problem 1
This figure shows the position vs. time graph for two objects before and after they collide. Assume
that they form an isolated system.
(a) What are the velocities of the two objects before and aftert h ec o l l i s i o n ? ( H i n t :y o uw i l lg e ta
more accurate result if you choose the initial and final timeswhere the lines go exactly through a
point on the grid shown.)
(b) Given the result in (a), what is the ratio of the inertias oft h et w oo b j e c t s ?
1 m
3 m
2 m
1 s 2 s 3 s
object 1 obje
ct 1
object 2
object 2
Collision region
1.5 s
1.75 s
x (m)
t (s)
Problem 2
Ac a ra n dat r u c kc o l l i d eo nav e r ys l i p p e r yh i g h w a y . T h ec a r ,with a mass of 1600 kg, was initially
moving at 50 mph. The truck, with a mass of 3000 kg, hit the car from behind at 65 mph. Assume
the two vehicles form an isolated system in what follows.",University Physics I Classical Mechanics.pdf
"the two vehicles form an isolated system in what follows.
(a) If, immediately after the collision, the vehicles separate and the truck’s velocity is found to be
55 mph in the same direction it was going, how fast (in miles perh o u r )i st h ec a rm o v i n g ?
(b) If instead the vehicles end up stuck together, what will bet h e i rc o m m o nv e l o c i t yi m m e d i a t e l y
after the collision?
Problem 3
A4 - k gg u nfi r e sa0.012-kg bullet at a 3-kg block of wood that is initially at rest.T h e b u l l e t i s
embedded in the block, and they move together, immediately after the impact, with a velocity of
3.5m / s .",University Physics I Classical Mechanics.pdf
"68 CHAPTER 3. MOMENTUM AND INERTIA
(a) What was the velocity of the bullet just before impact?
(b) In order to shoot a bullet at this speed, what must have beent h er e c o i ls p e e do ft h eg u n ?
Problem 4
A2 - k go b j e c t ,m o v i n ga t1 m / s ,c o l l i d e sw i t ha1 - k go b j e c tt h ati si n i t i a l l ya tr e s t . A f t e rt h e
collision, the two objects are found to move away from each other at 1 m/s. Assume they form an
isolated system.
(a) What are their actual final velocities in the Earth reference frame?
(b) What is the velocity of the center of mass of this system? Does it change as a result of the
collision?
Problem 5
Imagine you are stranded on a frozen lake (that means no friction—no traction!), with just a bow
and a quiver of arrows. Each arrow has a mass of 0.02 kg, and with your bow can shoot them at
as p e e do f9 0 m / s( r e l a t i v et oy o u — b u ty o um i g h ta sw e l la s s u m ethat this is the arrow’s velocity",University Physics I Classical Mechanics.pdf
"as p e e do f9 0 m / s( r e l a t i v et oy o u — b u ty o um i g h ta sw e l la s s u m ethat this is the arrow’s velocity
relative to the earth, since, as you will see, your recoil velocity will end up being pretty small
anyway). So you decide to use them to propel yourself back to shore.
(a) Suppose your mass (plus the bow and arrows) is 70 kg. When you shoot an arrow, starting
from rest, with what speed do you recoil?
(b) Suppose you try to be really clever, and tie a string to thearrow, with the other end of the
string tied around your waist. The idea is to get the arrow to pull you forward. Will this work?
(Hint: remember part (a). What will happen when the string becomes taut?)
Problem 6
An object’s position function is given byx1(t)=5+1 0 t (with x1 in meters ift is in seconds). A
second object’s position function isx2(t)=5 − 6t.
(a) If the first object’s mass is 1/3t h em a s so ft h es e c o n do n e ,w h a ti st h ep o s i t i o no ft h es y s t e m’s",University Physics I Classical Mechanics.pdf
"(a) If the first object’s mass is 1/3t h em a s so ft h es e c o n do n e ,w h a ti st h ep o s i t i o no ft h es y s t e m’s
center of mass as a function of time?
(b) Under the same assumption, what is the velocity of the system’s center of mass?",University Physics I Classical Mechanics.pdf
"Chapter 4
Kinetic Energy
4.1 Kinetic Energy
For a long time in the development of classical mechanics, physicists were aware of the existence
of two different quantities that one could define for an object ofi n e r t i am and velocity v.O n e
was the momentum,mv,a n dt h eo t h e rw a ss o m e t h i n gp r o p o r t i o n a lt omv2.D e s p i t et h e i ro b v i o u s
similarities, these two quantities exhibited different properties and seemed to be capturing different
aspects of motion.
When things got finally sorted out, in the second half of the 19th century, the quantity1
2 mv2 came
to be recognized as a form ofenergy—itself perhaps the most important concept in all of physics.
Kinetic energy,a st h i sq u a n t i t yi sc a l l e d ,m a yb et h em o s to b v i o u sa n di n t u itively understandable
kind of energy, and so it is a good place to start our study of thes u b j e c t .
We will use the letterK to denote kinetic energy, and, since it is a form of energy, wewill express",University Physics I Classical Mechanics.pdf
"We will use the letterK to denote kinetic energy, and, since it is a form of energy, wewill express
it in the units especially named for this purpose, which is tosay joules (J). 1 joule is 1 kg· m2/s2.
In the definition
K = 1
2mv2 (4.1)
the letter v is meant to represent themagnitude of the velocity vector, that is to say, thespeed
of the particle. Hence, unlike momentum,kinetic energy is not a vector, but a scalar:t h e r e i s n o
sense of direction associated with it. In three dimensions,one could write
K = 1
2m
(
v2
x + v2
y + v2
z
)
(4.2)
There is, therefore, some amount of kinetic energy associated with each component of the velocity
vector, but in the end they are all added together in a lump sum.
69",University Physics I Classical Mechanics.pdf
"70 CHAPTER 4. KINETIC ENERGY
For a system of particles, we will treat kinetic energy as an additive quantity, just like we did for
momentum, so the total kinetic energy of a system will just bethe sum of the kinetic energies of
all the particles making up the system. Note that, unlike momentum, this is a scalar (not a vector)
sum, and most importantly, that kinetic energy is, by definition, always positive, so there can be no
question of a “cancellation” of one particle’s kinetic energy by another, again unlike what happened
with momentum. Two objects of equal mass moving with equal speeds in opposite directions have
at o t a lm o m e n t u mo fz e r o ,b u tt h e i rt o t a lk i n e t i ce n e r g yi sd efinitely nonzero. Basically, the kinetic
energy of a system can never be zero as long as there is any kindof motion going on in the system.
4.1.1 Kinetic energy in collisions
To gain some further insights into the concept of kinetic energy, and the ways in which it is different",University Physics I Classical Mechanics.pdf
"4.1.1 Kinetic energy in collisions
To gain some further insights into the concept of kinetic energy, and the ways in which it is different
from momentum, it is useful to look at it in the same setting inwhich we “discovered” momentum,
namely, one-dimensional collisions in an isolated system.If we look again at the collision represented
in Figure 1 of Chapter 3, reproduced below,
-0.4
-0.2
 0
 0.2
 0.4
 0.6
 0.8
 1
 1.2
 0  2  4  6  8  10
1
1
2
2
1 m/s
v (m/s)
t (ms)
Figure 4.1: Elastic collision in an isolated system. (Figure3.1.)
we can use the definition (4.1)t oc a l c u l a t et h ei n i t i a la n dfi n a lv a l u e so fK for each object, and for
the system as a whole. Remember we found that, for this particular system,m2 =2 m1,s ow ec a n
just set m1 =1 k ga n dm2 =2 k g ,f o rs i m p l i c i t y . T h ei n i t i a la n dfi n a lv e l o c i t i e sa r ev1i =1 m / s ,",University Physics I Classical Mechanics.pdf
"4.1. KINETIC ENERGY 71
v2i =0 ,v1f = −1/3m / s ,v2f =2 /3m / s , a n d s o t h e k i n e t i c e n e r g i e s a r e
K1i = 1
2 J,K 2i =0 ; K1f = 1
18 J,K 2f = 4
9 J
Note that 1/18 + 4/9=9 /18 = 1/2, and so
Ksys,i = K1i + K2i = 1
2 J= K1f + K2f = Ksys,f
In words, we find that, in this collision, the final value of thetotal kinetic energy is the same
as its initial value, and so it looks like we have “discovered” another conserved quantity (besides
momentum) for this system.
This belief may be reinforced if we look next at the collisiondepicted in Figure 2 of Chapter 3,
again reproduced below. Recall I pointed out back then that wec a nt h i n ko ft h i sa sb e i n gr e a l l y
the same collision as depicted in Figure3.1,o n l yl o o k e da tf r o ma n o t h e rf r a m eo fr e f e r e n c e( o n e
moving initially to the right at 1 m/s). We will have more to saya b o u th o wt ot r a n s f o r mq u a n t i t i e s
from a frame of reference to another by the end of the chapter.
-1.4
-1.2
-1",University Physics I Classical Mechanics.pdf
"from a frame of reference to another by the end of the chapter.
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
 0
 0.2
 0  2  4  6  8  10
1
2
v (m/s)
t (ms)
1 2
−1 m/s
Figure 4.2:Another elastic collision, equivalent to the one in Figure 1 as seen fromanother reference frame.
(Figure 3.2.)
In any case, as observed there, all we need to do is add−1m / s t o a l l t h e v e l o c i t i e s i n t h e p r e v i o u s
problem, so we havev1i =0 ,v2i = −1m / s ,v1f = −4/3m / s ,v2f = −1/3m / s . T h e c o r r e s po n d i n g
kinetic energies are, accordingly,K1i =0 ,K2i =1 J ,K1f = 8
9 J, K2f = 1
9 J. These are all different",University Physics I Classical Mechanics.pdf
"72 CHAPTER 4. KINETIC ENERGY
from the values we had in the previous example, but note that once again the total kinetic energy
after the collision equals the total kinetic energy before—namely, 1 J in this case1 .
Things are, however, very different when we consider the thirdcollision example shown in Chapter
3, namely, the one where the two objects are stuck together after the collision.
 0
 0.2
 0.4
 0.6
 0.8
 1
 0  2  4  6  8  10
1 2
1 m/s
1
2
v (m/s)
t (ms)
Figure 4.3: A totally inelastic collision. (Figure3.3.)
Their joint final velocity, consistent with conservation ofmomentum, isv1f = v2f =1 /3m / s . S i n c e
the system starts as in Figure4.1,i t sk i n e t i ce n e r g yi si n i t i a l l yKsys,i = 1
2 J, but after the collision
we have only
Ksys,f = 1
2(3 kg)
( 1
3
m
s
) 2
= 1
6 J
So kinetic energy is not conserved in this case at all.
What this shows, however, is that unlike the total momentum ofas y s t e m ,w h i c hi sc o m p l e t e l y",University Physics I Classical Mechanics.pdf
"What this shows, however, is that unlike the total momentum ofas y s t e m ,w h i c hi sc o m p l e t e l y
unaffected by internal interactions, the total kinetic energyd o e sd e p e n do nt h ed e t a i l so ft h e
interaction, and thus conveys some information about its nature. We can then refine our study
of collisions to distinguish two kinds: the ones where the initial kinetic energy is recovered after
the collision, which we will callelastic,a n dt h eo n e sw h e r ei ti sn o t ,w h i c hw ec a l linelastic.A
1This is, of course, consistent with theprinciple of relativity It o l dy o ua b o u ti nC h a p t e r2 : i ft h ep r o c e s si n
Fig. 4.2 is really the same as the one in Fig.4.1,o n l yv i e w e di nad i ff e r e n ti n e r t i a lr e f e r e n c ef r a m e ,t h e n ,ife n e r g yi s
seen to be conserved in one frame, it should also be seen to be conserved in the other. More on this below, in Section
4.2.1.",University Physics I Classical Mechanics.pdf
"4.1. KINETIC ENERGY 73
special case of inelastic collision is the one calledtotally inelastic,w h e r et h et w oo b j e c t se n du p
stuck together, as in Figure4.3.A sw es h a l l s e el a t e r ,t h e k i n e t i ce n e r g y“ d e fi c i t ”i sl a r g e sti nt h a t
case.
Ih a v es a i da b o v et h a ti na ne l a s t i cc o l l i s i o nt h ek i n e t i ce n ergy is “recovered,” and I prefer this
terminology to “conserved,” because, in fact, unlike the total momentum, the total kinetic energy of
as y s t e md o e snot remain constant throughout the interaction, not even duringa ne l a s t i cc o l l i s i o n .
The simplest example to show this would be an elastic, head-onc o l l i s i o nb e t w e e nt w oo b j e c t so f
equal mass, moving at the same speed towards each other. In thec o u r s eo ft h ec o l l i s i o n ,b o t h
objects are brought momentarily to a halt before they reversed i r e c t i o na n db o u n c eb a c k ,a n da t",University Physics I Classical Mechanics.pdf
"objects are brought momentarily to a halt before they reversed i r e c t i o na n db o u n c eb a c k ,a n da t
that instant, the total kinetic energy is zero.
You can also examine Figures4.1 and 4.2 above, and calculate, from the graphs, the value of the
total kinetic energy during the collision. You will see thatit dips to a minimum, and then comes
back to its initial value (see also Figure4.5,l a t e ri nt h i sc h a p t e r ) . C o n v e n t i o n a l l y ,w em a yt a l ko f
kinetic energy as being “conserved” in elastic collisions,but it is important to realize that we are
looking at a different kind of “conservation” than what we had with the total momentum, which
was constant before, during, and after the interaction, as long as the system remained isolated.
Elastic collisions do suggest that, whatever the ultimate nature of this thing we call “energy” might
be, it may be possible tostore it in some form (in this case, during the course of the collision),",University Physics I Classical Mechanics.pdf
"be, it may be possible tostore it in some form (in this case, during the course of the collision),
and then recover it, as kinetic energy, eventually. This paves the way for the introduction of other
kinds of “energy” besides kinetic energy, as we shall see in alater chapter, and the possibility of
interconversion to take place among these kinds. For the moment, we shall simply say that in
an elastic collision some amount of kinetic energy is temporarily stored as some kind of “internal
energy,” and after the collision this is converted back intokinetic energy; whereas, in an inelastic
collision, some amount of kinetic energy gets irrevocably converted into some “internal energy,”
and we never get it back.
Since whatever ultimately happens depends on the details andt h en a t u r eo ft h ei n t e r a c t i o n ,w e
will be led to distinguish between “conservative” interactions, where kinetic energy isreversibly",University Physics I Classical Mechanics.pdf
"will be led to distinguish between “conservative” interactions, where kinetic energy isreversibly
stored as some other form of energy somewhere, and “dissipative” interactions, where the energy
conversion is, at least in part, irreversible. Clearly, elastic collisions are associated with conservative
interactions and inelastic collisions are associated withdissipative interactions. This preliminary
classification of interactions will have to be reviewed a little more carefully, however, in the next
chapter.",University Physics I Classical Mechanics.pdf
"74 CHAPTER 4. KINETIC ENERGY
4.1.2 Relative velocity and coefficient of restitution
An interesting property of elastic collisions can be disclosed from a careful study of figures4.1 and
4.2.I n b o t h c a s e s , a s y o u c a n s e e , t h erelative velocity of the two objects colliding has the same
magnitude (but opposite sign) before and after the collision. In other words:in an elastic collision,
the objects end up moving apart at the same rate as they originally came together.
Recall that, in Chapter 1, we defined the velocity of object 2 relative to object 1 as the quantity
v12 = v2 − v1 (4.3)
(compare Eq. (1.21); and similarly the velocity of object 1 relative to object 2is v21 = v1 − v2.
With this definition you can check that, indeed, the collisions shown in Figs.4.1 and 4.2 satisfy the
equality
v12,i = −v12,f (4.4)
(note that we could equally well have usedv21 instead of v12). For instance, in Fig.4.1, v12,i =",University Physics I Classical Mechanics.pdf
"equality
v12,i = −v12,f (4.4)
(note that we could equally well have usedv21 instead of v12). For instance, in Fig.4.1, v12,i =
v2i − v1i = −1m / s , w h e r e a sv12,f =2 /3 − (−1/3) = 1 m/s. So the objects are initially moving
towards each other at a rate of 1 m per second, and they end up moving apart just as fast, at 1
mp e rs e c o n d . V i s u a l l y ,y o us h o u l dn o t i c et h a tt h ed i s t a n c ebetween the red and blue curves is
the same before and after (but not during) the collision; thefact that they cross accounts for the
difference in sign of the relative velocity, which in turns means simply that before the collision they
were coming together, and afterwards they are moving apart.
It takes only a little algebra to show that Eq. (4.4)f o l l o w sf r o mt h ej o i n tc o n d i t i o n so fc o n s e r v a t i o n
of momentum and conservation of kinetic energy. The first one(pi = pf )c l e a r l yh a st h ef o r m
m1v1i + m2v2i = m1v1f + m2v2f (4.5)",University Physics I Classical Mechanics.pdf
"of momentum and conservation of kinetic energy. The first one(pi = pf )c l e a r l yh a st h ef o r m
m1v1i + m2v2i = m1v1f + m2v2f (4.5)
whereas the second one (Ki = Kf )c a nb ew r i t t e na s
1
2m1v2
1i + 1
2m2v2
2i = 1
2m1v2
1f + 1
2m2v2
2f (4.6)
We can cancel out all the factors of 1/2i nE q .(4.6)2,t h e nr e a r r a n g ei ts ot h a tq u a n t i t i e sb e l o n g i n g
to object 1 are on one side, and quantities belonging to object2a r eo nt h eo t h e r .W eg e t
m1
(
v2
1i − v2
1f
)
= −m2
(
v2
2i − v2
2f
)
m1 (v1i − v1f )( v1i + v1f )= −m2 (v2i − v2f )( v2i + v2f )( 4 . 7 )
(using the fact thata2 − b2 =( a + b)(a − b)). Note, however, that Eq. (4.5)c a na l s ob er e w r i t t e n
as
m1 (v1i − v1f )= −m2 (v2i − v2f )
2You may be wondering, just why do we define kinetic energy withaf a c t o r1/2i nf r o n t ,a n y w a y ? T h e r ei sn o
good answer at this point. Let’s just say it will make the definition of “potential energy” simpler later, particularly",University Physics I Classical Mechanics.pdf
"good answer at this point. Let’s just say it will make the definition of “potential energy” simpler later, particularly
as regards its relationship toforce.",University Physics I Classical Mechanics.pdf
"4.1. KINETIC ENERGY 75
This immediately allows us to cancel out the corresponding factors in Eq (4.7), so we are left with
v1i + v1f = v2i + v2f ,w h i c hc a nb er e w r i t t e na s
v1f − v2f = v2i − v1i (4.8)
and this is equivalent to (4.4).
So, in an elastic collision the speed at which the two objectsmove apart is the same as the speed
at which they came together, whereas, in what is clearly the opposite extreme, in a totally inelastic
collision the final relative speed iszero—the objects do not move apart at all after they collide.
This suggests that we can quantify how inelastic a collisionis by the ratio of the final to the
initial magnitude of the relative velocity. This ratio is denoted bye and is called thecoefficient of
restitution.F o r m a l l y ,
e = −v12,f
v12,i
= −v2f − v1f
v2i − v1i
(4.9)
For an elastic collision,e =1 ,a sr e q u i r e db yE q .(4.4). For a totally inelastic collision, like the one",University Physics I Classical Mechanics.pdf
"v12,i
= −v2f − v1f
v2i − v1i
(4.9)
For an elastic collision,e =1 ,a sr e q u i r e db yE q .(4.4). For a totally inelastic collision, like the one
depicted in Fig. 3,e =0 . F o rac o l l i s i o nt h a ti si n e l a s t i c ,b u tn o tt o t a l l yi n e l a stic, e will have some
value in between these two extremes. This knowledge can be used to “design” inelastic collisions
(for homework problems, for instance!): just pick a value for e,b e t w e e n0a n d1 ,i nE q .(4.9), and
combine this equation with the conservation of momentum requirement (4.5). The two equations
then allow you to calculate the final velocities for any valueso f m1, m2,a n dt h ei n i t i a lv e l o c i t i e s .
Figure 4.4 below, for example, shows what the collision in Figure4.1 would have been like, if
the coefficient of restitution had been 0.6i n s t e a do f1 . Y o uc a nc h e c k ,b ys o l v i n g(4.5)a n d(4.9)
together, and using the initial velocities, thatv1f = −1/15 m/s =−0.0667 m/s, andv2f =8 /15 m/s",University Physics I Classical Mechanics.pdf
"together, and using the initial velocities, thatv1f = −1/15 m/s =−0.0667 m/s, andv2f =8 /15 m/s
=0 .533 m/s.
-0.4
-0.2
 0
 0.2
 0.4
 0.6
 0.8
 1
 1.2
 0  2  4  6  8  10
v (m/s)
t (ms)
1
2
Figure 4.4: An e =0 .6 collision between objects with the same inertias and initial velocities as in Figure
1.",University Physics I Classical Mechanics.pdf
"76 CHAPTER 4. KINETIC ENERGY
Although, as I just mentioned, for most “normal” collisionsthe coefficient of restitution will be
ap o s i t i v en u m b e rb e t w e e n1a n d0 ,t h e r ec a nb ee x c e p t i o n st othis. If one of the objects passes
through the other (like a bullet through a target, for instance), the value ofe will be negative
(although still between 0 and 1 in magnitude). Ande can be greater than 1 for so-called “explosive
collisions,” where some amount of extra energy is released,and converted into kinetic energy, as
the objects collide. (For instance, two hockey players colliding on the rink and pushing each other
away.) In this case, the objects may well fly apart faster thanthey came together.
An extreme example of a situation withe> 0i sa nexplosive separation,w h i c hi sw h e nt h et w o
objects are initially moving together and then fly apart. In that case, the denominator of Eq. (4.9)",University Physics I Classical Mechanics.pdf
"objects are initially moving together and then fly apart. In that case, the denominator of Eq. (4.9)
is zero, and soe is formally infinite. This suggests, what is in fact the case,namely, that although
explosive processes are certainly important, describing them through the coefficient of restitution
is rare, even when it would be formally possible. In practice,u s eo ft h ec o e ffi c i e n to fr e s t i t u t i o ni s
mostly limited to the elastic-to-completely inelastic range, that is, 0≤ e ≤ 1.
4.2 “Convertible” and “translational” kinetic energy
Figure 4.5 shows how the total kinetic energy varies with time, for the two objects shown colliding
in Figure4.1 ,d e p e n d i n go nt h ed e t a i l so ft h ec o l l i s i o n ,n a m e l y ,o nt h ev alue ofe.T h e t h r e e c u r v e s
shown cover the elastic case,e =1( F i g u r e4.1), the totally inelastic case,e =0( F i g u r e4.3), and",University Physics I Classical Mechanics.pdf
"shown cover the elastic case,e =1( F i g u r e4.1), the totally inelastic case,e =0( F i g u r e4.3), and
the inelastic case withe =0 .6o fF i g u r e4.4.R e c a l l t h a t t h e t o t a l m o m e n t u m i s c o n s e r v e d i n a l l
three cases.
 0
 0.1
 0.2
 0.3
 0.4
 0.5
 0  2  4  6  8  10
e = 1
e = 0.6
e = 0
Ksys (J)
t (ms)
Figure 4.5: The total kinetic energy as a function of time for the collisions shownin Figures 1, 3 and 4,
respectively.",University Physics I Classical Mechanics.pdf
"4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 77
Figure 4.5 shows that the greatest loss of kinetic energy happens for thet o t a l l yi n e l a s t i cc o l l i s i o n ,
which, as we will see in a moment, is, in fact, a general result.T h a t b e i n g t h e c a s e , t h e fi g u r e
also shows that it may not be always be possible to bring the total kinetic energy down to zero,
even temporarily. The reason for this is that, if momentum isconserved, the velocity of the center
of mass cannot change, so if the center of mass was moving before the collision, it must still be
moving afterwards; and, as mentioned in this chapter’s introduction, as long as there is motion in
as y s t e m ,i t st o t a lk i n e t i ce n e r g yc a n n o tb ez e r o .
All of this suggests that it should be possible to break up a system’s total kinetic energy into two
parts: one part associated with the motion of the center of mass, which cannot change in any",University Physics I Classical Mechanics.pdf
"parts: one part associated with the motion of the center of mass, which cannot change in any
momentum-conserving collision, and one part associated with the relative motion of the parts that
make up the system. This second part would vanish irreversibly in a totally inelastic collision,
whereas it would recover its original value in an elastic collision.
The way to see this mathematically, for a system of two objectsw i t hm a s s e sm1 and m2,i st o
introduce the center of mass velocityvcm [Eq. (3.10)]
vcm = m1v1 + m2v2
m1 + m2
and the relative velocityv12 = v2 − v1 (Eq. (4.3)a b o v e ) ,a n do b s e r v et h a tt h ev e l o c i t i e sv1 and v2
can be written, respectively, as
v1 = vcm − m2
m1 + m2
v12
v2 = vcm + m1
m1 + m2
v12 (4.10)
Substituting the equations (4.10)i n t ot h ee x p r e s s i o nKsys = 1
2 m1v2
1 + 1
2 m2v2
2,o n efi n d st h a tt h e
cross-terms vanish, and all that is left is
Ksys = 1
2(m1 + m2)v2
cm + 1
2
m1m2
2 + m2m2
1
(m1 + m2)2 v2
12",University Physics I Classical Mechanics.pdf
"1 + 1
2 m2v2
2,o n efi n d st h a tt h e
cross-terms vanish, and all that is left is
Ksys = 1
2(m1 + m2)v2
cm + 1
2
m1m2
2 + m2m2
1
(m1 + m2)2 v2
12
Af a c t o ro f(m1 + m2)m a yb ec a n c e l e di nt h el a s tt e r m ,a n dt h efi n a le x p r e s s i o nt a kes the form
Ksys = Kcm + Kconv (4.11)
where the center of mass kinetic energy (ortranslational energy)i sj u s tw h a to n ew o u l dh a v ei ft h e
whole system was a single particle of massM = m1 + m2 moving at the center of mass speed:
Kcm = 1
2Mv2
cm (4.12)
and the “convertible energy” Kconv is the part associated with the relative motion, which can be",University Physics I Classical Mechanics.pdf
"78 CHAPTER 4. KINETIC ENERGY
made to vanish entirely in an inelastic collision3:
Kconv = 1
2
m1m2
m1 + m2
v2
12 = 1
2µv2
12 (4.13)
The last equation implicitly defines a useful quantity that wec a l lt h ereduced massof a system of
two particles, and denote byµ:
µ = m1m2
m1 + m2
(4.14)
Equation (4.11), with the definitions (4.12)a n d(4.13), pretty much explains everything that we
see going on in Figure4.5.T h e t o t a l k i n e t i c e n e r g y i s t h e s u m o f t w o t e r m s , t h e fi r s t o f which,
Kcm,c a nn e v e rc h a n g e : i ti s ,i nf a c t ,a sc o n s t a n ta st h et o t a lm o mentum itself, since it involves
the center of mass velocity,vcm,w h i c hi sp r o p o r t i o n a lt ot h et o t a lm o m e n t u mo ft h es y s t e m( recall
equation (3.11)). The term that can, and does change, is the second one, the convertible energy. In
fact, in an ordinary collision in which the objects do not passt h r o u g he a c ho t h e r ,t h e r em u s tb ea t",University Physics I Classical Mechanics.pdf
"fact, in an ordinary collision in which the objects do not passt h r o u g he a c ho t h e r ,t h e r em u s tb ea t
least an instant in time whenKconv =0 . T h i si sb e c a u s ei ti n v o l v e st h er e l a t i v ev e l o c i t y ,a n ds ince
the relative velocity must change sign at some point (the objects are initially coming together, but
end up moving apart), it must be zero at that time.
This explains why all the curves in Fig.4.5 have the same minimum value (even though they may
reach it at different times): that value is clearlyKcm for the system (sinceKconv is zero at that
time). It is the same for all the curves because all the systemsc o n s i d e r e dh a v et h es a m et o t a lm a s s
and momentum (as determined by the initial velocities)—we just chose them that way.
Since Kcm cannot change for an isolated system, the maximum kinetic energy that can be lost in a",University Physics I Classical Mechanics.pdf
"Since Kcm cannot change for an isolated system, the maximum kinetic energy that can be lost in a
collision in such a system is the initial value ofKconv,w h i c hw ew o u l dd e n o t ea sKconv,i.T h i s i s , i n
fact, completely lost in a totally inelastic collision, since in that casev12,f =0 ,a n dE q .(4.13)t h e n
gives Kconv,f =0 . I nf a c t ,u s i n gE q .(4.9), we can relate the final value of the convertible energy
to its initial value via the coefficient of restitution:
Kconv,f = 1
2µv2
12,f = 1
2µe2v2
12,i = e2Kconv,i (4.15)
Thus, for example, in a collision withe =0 .6, the final value of the convertible energy would be
only 0.36 times its initial value: 64% of it would have been “lost.” (This is not, however, the same
as 64% of thetotal initial energy, since the latter still includesKcm,w h i c hd o e sn o tc h a n g e . ) W e
can also write Eq. (4.15)a s
∆Ksys =
(
e2 − 1
)
Kconv,i =
(
e2 − 1
) 1
2µv2
12,i (4.16)",University Physics I Classical Mechanics.pdf
"can also write Eq. (4.15)a s
∆Ksys =
(
e2 − 1
)
Kconv,i =
(
e2 − 1
) 1
2µv2
12,i (4.16)
since the only possible change inKsys must come from the convertible energy.
3Although the name “convertible energy” makes sense in this context, it is not, as far as I can tell, in general
usage. I have borrowed it from Mazur’sThe Principles and Practice of Physics,b u ty o us h o u l dp r o b a b l yn o te x p e c t
to find it in other textbooks.",University Physics I Classical Mechanics.pdf
"4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 79
Although we have derived the decomposition (4.11)f o rt h ev e r yr e s t r i c t e ds i t u a t i o no ft w oo b j e c t s
moving in one dimension, the basic result is quite general: first, everything in the derivation works
if v1 and v2 are replaced by vectors⃗v1 and ⃗v2,s ot h er e s u l t sh o l d si nt h r e ed i m e n s i o n sa sw e l l .
Second, for a system of any number of particles, one still canwrite Ksys as Kcm+a n o t h e rt e r m
that depends only on the relative motion of all the pairs of particles. This “generalized convertible
energy,” orkinetic energy of relative motionwould have the form
Krel = 1
2µ12v2
12 + 1
2µ13v2
13 + ... + 1
2µ23v2
23 + ...
(in this expression, something likeµ23 means a reduced mass like the one in Eq. (4.14), only for
masses m2 and m3,a n ds of o r t h ) .
When we get to the study of rotational motion, for instance, wew i l ls e et h a tt h et o t a lk i n e t i c",University Physics I Classical Mechanics.pdf
"masses m2 and m3,a n ds of o r t h ) .
When we get to the study of rotational motion, for instance, wew i l ls e et h a tt h et o t a lk i n e t i c
energy of an extended rigid object can be written asKcm + Krot,w h e r eKrot,t h er o t a t i o n a lk i n e t i c
energy, is just the same kind of thing as what we have called the“ c o n v e r t i b l ee n e r g y ”h e r e .
All of the above still leaves unanswered the question of whathappens to the convertible energy
that is lost in an inelastic collision. Just what is it that itgets converted into? The answer to this
question will be the subject of the following chapter.
4.2.1 Kinetic energy and momentum in different reference frames
Ih a v ep o i n t e do u tr e p e a t e d l yb e f o r et h a ta l lm o t i o ni sr e l a tive, and so, to some extent, kinetic
energy and momentum must be somewhat relative as well. A car inaf r e i g h tt r a i nh a sal o to f",University Physics I Classical Mechanics.pdf
"energy and momentum must be somewhat relative as well. A car inaf r e i g h tt r a i nh a sal o to f
momentum relative to an observer on the ground, but its momentum relative to another car on the
same train is zero, since they are not moving relative to eachother. The same could be said about
its kinetic energy.
In general, if you have a system with a total momentum⃗psys and inertiaM,i t sc e n t e ro fm a s sw i l l
have a velocity⃗vcm = ⃗psys/M.T h e n , i f y o u w e r e t o m o v e a l o n g s i d e t h e s y s t e m w i t h a v e l o c i ty
exactly equal to⃗vcm,t h et o t a lm o m e n t u mo ft h es y s t e mr e l a t i v et oy o uw o u l db ez e ro. If the system
was a solid object, it would not “hit” you if you made contact;there would be no collision. It may
help here to think, for instance, of aircraft refueling in flight: if the two planes’ velocities are exactly
matched, they can make contact without any damage, just as ifthey were at rest. A reference frame",University Physics I Classical Mechanics.pdf
"matched, they can make contact without any damage, just as ifthey were at rest. A reference frame
moving at a system’s center of mass velocity is, for this reason, called azero-momentum framefor
the system in question.
Clearly, in such a reference frame, the translational kinetic energy of the system,Kcm = 1
2 Mv2
cm,
will also be zero (since, in that frame, the center of mass is not moving at all). However, the
relative motion term,Kconv,w o u l db ecompletely unaffectedby the change in reference frame. This
is because, as you may have noticed by now, to convert velocities from one frame of reference to",University Physics I Classical Mechanics.pdf
"80 CHAPTER 4. KINETIC ENERGY
another we just add or subtract from all the velocities the relative velocity of the two frames. This
operation, however, will not change any of the relative velocities of the parts of the system, since
these are all differences to begin with. Mathematically,
(v2 + v′) − (v1 + v′)= v2 − v1
regardless of the value ofv′.
So there something we might callabsolute (as opposed to “relative”) about the convertible kinetic
energy: it is the same, it will have the same value, for any observer, regardless of how fast or in
what direction that observer may be moving relative to the system as a whole. We may think of it
as anintrinsic (meaning, observer-independent) property of the system.
4.3 In summary
1. The kinetic energy of a particle of massm moving with velocityv is defined asK = 1
2 mv2.I t
is a scalar quantity, and it is always positive. For a system ofp a r t i c l e so ra ne x t e n d e do b j e c t ,",University Physics I Classical Mechanics.pdf
"2 mv2.I t
is a scalar quantity, and it is always positive. For a system ofp a r t i c l e so ra ne x t e n d e do b j e c t ,
we defineKsys as the sum of the kinetic energies of all the particles makingup the system.
2. For any system, the total kinetic energy can be written as the sum of thetranslational (or
center of mass)k i n e t i ce n e r g y ,Kcm,a n da n o t h e rt e r mt h a ti n v o l v e st h em o t i o no ft h ep a r t s
of the system relative to each other. (See Eq. (4.11)a b o v e . ) T h et r a n s l a t i o n a lk i n e t i ce n e r g y
is constant for an isolated system, and is always given byKcm = 1
2 Mv2
cm.
3. The kinetic energy of relative motion (which, in the context of collisions, is called thecon-
vertible energy)i sg i v e n ,f o rt h es p e c i a lc a s eo fas y s t e mc o n s i s t i n go ft w oparticles (or two
non-rotating extended objects), byKconv = 1
2 µv2
12,w h e r eµ = m1m2/(m1 +m2)i st h er e d u c e d",University Physics I Classical Mechanics.pdf
"non-rotating extended objects), byKconv = 1
2 µv2
12,w h e r eµ = m1m2/(m1 +m2)i st h er e d u c e d
mass, andv12 = v2 − v1 is the relative velocity of the two objects.
4. In a one-dimensional collision between two objects that don o tp a s st h r o u g he a c ho t h e r ,t h e
convertible energy always drops to zero at some point, as a result of the interaction; that is,
it is converted entirely into some other form of energy. At thee n do ft h ei n t e r a c t i o n ,a l lt h e
convertible energy may be recovered (elastic collision), oro n l yp a r to fi t( i n e l a s t i cc o l l i s i o n ) ,
or none of it (completely inelastic collision).
5. In terms of thecoefficient of restitutione,d e fi n e da se = −v12,f /v12,i,e l a s t i cc o l l i s i o n sh a v e
e =1 ,t o t a l l yi n e l a s t i cc o l l i s i o n sh a v ee =0 ,a n di n e l a s t i cc o l l i s i o n s0<e< 1. The total
change in kinetic energy in the collision can be written as ∆Ksys =∆ Kconv =( e2 −1)Kconv,i.",University Physics I Classical Mechanics.pdf
"change in kinetic energy in the collision can be written as ∆Ksys =∆ Kconv =( e2 −1)Kconv,i.
6. Another way to say the above is that in an elastic collisionin one dimension, the two objects
move apart after the collision at the same rate (relative speed) at which they approached each
other initially. In a totally inelastic collision, conversely, the two objects do not move apart
at all after the collision—they become “stuck together.”",University Physics I Classical Mechanics.pdf
"4.3. IN SUMMARY 81
7. Besides the cases considered above, one may have collisions where the objects pass through
each other, givinge< 0, and “explosive collisions,” wheree> 1. In these latter collisions
some internal source of energy is converted into additionalkinetic energy when the objects
interact. The extreme case of this is anexplosive separation,w h i c hi st h er e v e r s eo fat o t a l l y
inelastic collision—two objects initially moving togetherfl ya p a r t ,w i t han e ti n c r e a s ei nt h e
system’s kinetic energy.
8. The translational kinetic energy of a system will, in general, have different values for observers
moving with different velocities. The convertible kinetic energy, on the other hand, is seen
by all observers to have the same value, regardless of their relative state of motion.",University Physics I Classical Mechanics.pdf
"82 CHAPTER 4. KINETIC ENERGY
4.4 Examples
4.4.1 Collision graph revisited
Look again at the collision graph from example 3.5.1 from thepoint of view of the kinetic energy
of the two carts.
(a) What is the initial kinetic energy of the system?
(b) How much of this is in the center of mass motion, and how mucho fi sc o n v e r t i b l e ?
(c) Does the convertible kinetic energy go to zero at some point during the collision? If so, when?
Is it fully recovered after the collision is over?
(d) What kind of collision is this? (Elastic, inelastic, etc.) What is the coefficient of restitution?
Solution
(a) From the solution to example 3.5.1 we know that
v1i = −1 m
s v2i =0 .5 m
s
v1f =1 m
s v2f = −0.5 m
s
and m1 =1k ga n dm2 =2k g . S ot h ei n i t i a lk i n e t i ce n e r g yi s
Ksys,i = 1
2m1v2
1i + 1
2m2v2
2i =0 .5J+0 .25 J = 0.75 J (4.17)
(b) To calculateKcm = 1
2 (m1 + m2)v2
cm,w en e e dvcm,w h i c hi nt h i sc a s ei se q u a lt o
vcm = m1v1i + m2v2i
m1 + m2
= −1+2 × 0.5
3 =0",University Physics I Classical Mechanics.pdf
"(b) To calculateKcm = 1
2 (m1 + m2)v2
cm,w en e e dvcm,w h i c hi nt h i sc a s ei se q u a lt o
vcm = m1v1i + m2v2i
m1 + m2
= −1+2 × 0.5
3 =0
so Kcm =0 ,w h i c hm e a n sa l lt h ek i n e t i ce n e r g yi sc o n v e r t i b l e . W ec a nalso calculate that directly:
Kconv,i = 1
2µv2
12,i = 1
2
( 1 × 2
1+2 kg
)
×
(
0.5 m
s − (−1) m
s
)2
= 1.52
3 J=0 .75 J (4.18)
(c) If we look at figure3.5,w ec a ns e et h a tt h ec a r t sd on o tp a s st h r o u g he a c ho t h e r ,s ot heir
relative velocity must be zero at some point, and with that, the convertible energy. In fact, the
figure makes it quite clear thatboth v1 and v2 are zero att =5 s ,s oa tt h a tp o i n ta l s ov12 =0 ,
and the convertible energyKconv =0 . ( A n ds oi st h et o t a lKsys =0a tt h a tt i m e ,s i n c eKcm =0
throughout.)",University Physics I Classical Mechanics.pdf
"4.4. EXAMPLES 83
On the other hand, it is also clear thatKconv is fully recovered after the collision is over, since the
relative velocity just changes sign:
v12,i = v2i − v1i =0 .5 m
s − (−1) m
s =1 .5 m
s
v12,f = v2f − v1f = −0.5 m
s − 1 m
s = −1.5 m
s (4.19)
Therefore
Kconv,f = 1
2µv2
12,f = 1
2µv2
12,i = Kconv,i
(d) Since the total kinetic energy (which in this case is onlyconvertible energy) is fully recovered
when the collision is over, the collision is elastic. Using equation (4.19), we can see that the
coefficient of restitution is
e = −v12,f
v12,i
= −−1.5
1.5 =1
as it should be.
4.4.2 Inelastic collision and explosive separation
Analyze example 3.5.2 from the point of view of the system’s kinetic energy. In particular, answer
the following questions:
(a) What is the total kinetic energy of the system (i)b e f o r et h ep l a y e r sc o l l i d e ,(ii)r i g h ta f t e rt h e",University Physics I Classical Mechanics.pdf
"(a) What is the total kinetic energy of the system (i)b e f o r et h ep l a y e r sc o l l i d e ,(ii)r i g h ta f t e rt h e
collision, when they are holding to one another, and (iii)a f t e rt h e ys e p a r a t e . H o wm u c ho ft h i s
energy is translational (that is, center-of-mass kinetic energy), and how much is convertible?
(b) Answer the same questions from the point of view of the player who is skating at a constant
1.5m / s t o t h e r i g h t ( p l a y e r 3 )
(To avoid needless repetition, you may use already established results, such as conservation of
momentum.)
Solution (a) Before the players collide, we have
Ksys,i = 1
2m1v2
1i + 1
2m2v2
2i = 1
2 (80 kg)×
(
3 m
s
)2
+ 1
2 (90 kg)×
(
−2 m
s
)2
=5 4 0J ( 4 . 2 0 )
While they are still holding to each other, we know from the solution to example 3.5.2 that their
joint velocity is 0.353 , and that this has to be also the velocity of their center ofm a s s ,w h i c hi s
unchanged by the collision. So, we have
Kcm = 1",University Physics I Classical Mechanics.pdf
"unchanged by the collision. So, we have
Kcm = 1
2(m1 + m2)v2
cm = 1
2 (170 kg)
(
0.353 m
s
)2
=1 0.6J ( 4 . 2 1 )",University Physics I Classical Mechanics.pdf
"84 CHAPTER 4. KINETIC ENERGY
This isKcm throughout, as well asKsys right after the collision, since the collision is totally inelastic
and that means thatKconv drops to zero. Also, subtracting this from (4.20)w i l lg i v eu st h ei n i t i a l
value of the convertible energy, without the need for a separate calculation, so
Kconv,i = Ksys,i − Kcm =5 4 0J− 10.6J =5 2 9.4J ≃ 529 J (4.22)
After the separation, the new total kinetic energy (for whichIw i l lu s et h es u b s c r i p tf)i s
Ksys,i = 1
2m1v2
1f + 1
2m2v2
2f = 1
2 (80 kg)×
(
−0.176 m
s
)2
+ 1
2 (90 kg)×
(
0.824 m
s
)2
=3 1.8J ( 4 . 2 3 )
where I have gotten the values forv1f and v2f from the solution to part (d) of Example 3.5.2.
Subtracting Kcm from this will give us the final value of the convertible energy:
Kconv,f = Ksys,f − Kcm =3 1.8J − 10.6J=2 1 .2J ( 4 . 2 4 )
To summarize, then, we have:
• Before the collision:
Ksys,i =5 4 0J,K cm =1 0.6J ,K conv,i =5 2 9.4J",University Physics I Classical Mechanics.pdf
"To summarize, then, we have:
• Before the collision:
Ksys,i =5 4 0J,K cm =1 0.6J ,K conv,i =5 2 9.4J
• Right after the collision (players still holding to each other):
Ksys = Kcm =1 0.6J ,K conv =0
• After the (explosive) separation:
Ksys,f =3 1.8J ,K cm =1 0.6J ,K conv,i =2 1.2J
So, in the collision, approximately 529 J of kinetic energy “disappeared” from the system (or, we
could say, were “converted into some form of internal energy”), whereas the players’ pushing on
each other managed to put about 21 J of kinetic energy back intot h es y s t e m ;w ew i l le x p l o r et h e s e
kinds of processes in more detail in the following chapter!
(b) We need to repeat all the above calculations with all the velocities shifted down by 1.5m / s ,
to bring them to the reference frame of player 3. Instead of putting a subscript “3” on all the
quantities, since we already have tons of subscripts to worrya b o u t ,I ’ mg o i n gt of o l l o wa na l t e r n a t i v e",University Physics I Classical Mechanics.pdf
"quantities, since we already have tons of subscripts to worrya b o u t ,I ’ mg o i n gt of o l l o wa na l t e r n a t i v e
convention and use a “prime” superscript (′)t od e n o t ea l lt h eq u a n t i t i e si nt h i sf r a m eo fr e f e r e n c e .
In brief, we have
K′
sys,i = 1
2m1(v′
1i)2 + 1
2m2(v′
2i)2 = 1
2 (80 kg)×
(
1.5 m
s
)2
+ 1
2 (90 kg)×
(
−3.5 m
s
)2
=6 4 1.3J ( 4 . 2 5 )",University Physics I Classical Mechanics.pdf
"4.4. EXAMPLES 85
K′
cm = 1
2(m1 + m2)(v′
cm)2 = 1
2 (170 kg)
(
0.353 m
s − 1.5 m
s
)2
=1 1 1.8J ( 4 . 2 6 )
K′
conv,i = K′
sys,i − K′
cm =6 4 1.3J − 111.8J =5 2 9.5J ≃ 529 J (4.27)
This shows explicitly that the convertible energy, as I pointed out earlier in this chapter, is the same
in every reference frame! (The equality is exact, if you keepenough decimals in the calculation.)
Knowing this, we can simplify the calculation of the final kinetic energy, after the explosive sepa-
ration: the convertible energy,K′
conv,f ,w i l lb et h es a m ea si nt h ee a r t hr e f e r e n c ef r a m e ,t h a ti st o
say, 21.2J , a n d t h e t o t a l k i n e t i c e n e r g y w i l l beK′
sys,f = K′
cm + K′
conv,f =1 1 1.8J + 2 1.2J =1 3 3J .
So, in this frame of reference, we have (to three significant figures):
K′
sys,i =6 4 1J,K ′
cm =1 1 2J,K ′
conv,i =5 2 9J ( b e f o r et h ec o l l i s i o n )
K′
sys = K′
cm =1 1 2J,K ′
conv =0 ( r i g h ta f t e rt h ec o l l i s i o n )
K′
sys,f =1 3 3J,K ′",University Physics I Classical Mechanics.pdf
"K′
sys = K′
cm =1 1 2J,K ′
conv =0 ( r i g h ta f t e rt h ec o l l i s i o n )
K′
sys,f =1 3 3J,K ′
cm =1 1 2J,K ′
conv,i =2 1.2J ( a f t e r t h e s e p a r a t i o n )
So, even though the total kinetic energy is different in the tworeference frames, all the (inertial)
observers will agree as to the amount of kinetic energy “lost”i nt h ec o l l i s i o n ,a sw e l la st h ea m o u n t
of kinetic energy put back into the system by the players’ pushing on each other.",University Physics I Classical Mechanics.pdf
"86 CHAPTER 4. KINETIC ENERGY
4.5 Problems
Problem 1
A7 1 - k gm a nc a nt h r o wa1 - k gb a l lw i t ham a x i m u ms p e e do f6m / sr elative to himself. Imagine
that one day he decides to try to do that on roller skates. Starting from rest, he throws the ball
as hard as he can, so it ends up moving at 6 m/s relative to him, but he himself is recoiling as a
result of the throw.
(a) Assuming conservation of momentum, find the velocities oft h em a na n dt h eb a l lr e l a t i v et ot h e
ground.
(b) What is the kinetic energy of the system right after the throw? (By the system here we mean
the man and the ball throughout.) Where did this kinetic energy come from?
(c) Is the man’s reference frame inertial throughout this process? Why or why not?
(d) Does the center of mass of the system move at all throughoutt h i sp r o c e s s ?
Problem 2
Analyze Problem 1 from Chapter 3 from the point of view of the system’s kinetic energy. In
particular, answer the following questions:",University Physics I Classical Mechanics.pdf
"Problem 2
Analyze Problem 1 from Chapter 3 from the point of view of the system’s kinetic energy. In
particular, answer the following questions:
(a) What is the total kinetic energy of the system before and after the collision? How much of this
energy is translational (that is, center-of-mass kinetic energy), and how much is convertible?
(b) What kind of collision is this? (Elastic, inelastic, etc.) What is the coefficient of restitution?
Problem 3
Analyze Problem 2 from Chapter 3 from the point of view of the system’s kinetic energy. In
particular, answer the following questions:
(a) What is the coefficient of restitution for the collision described in part (a) of the problem, and
how much kinetic energy is “lost” in that collision?
(b) What is the coefficient of restitution for the collision described in part (b) of the problem, and
how much kinetic energy is “lost” in that collision?
Problem 4",University Physics I Classical Mechanics.pdf
"how much kinetic energy is “lost” in that collision?
Problem 4
A0 .012-kg bullet, traveling at 850 m/s, hits a 2-kg block of woodthat is initially at rest, and goes
straight through it. Assume that the final velocity of the bullet relative to the blockis 400 m/s, and
that the system is isolated.
(a) What is the coefficient of restitution for this collision?
(b) How much kinetic energy is “lost” in the collision?
(c) What is the final velocity of the block?
Problem 5
A2 - k go b j e c t ,m o v i n ga t1 m / s ,c o l l i d e sw i t ha1 - k go b j e c tt h ati si n i t i a l l ya tr e s t . A s s u m et h e y
form an isolated system.",University Physics I Classical Mechanics.pdf
"4.5. PROBLEMS 87
(a) What is the initial kinetic energy of the system? How muchof this is center of mass energy,
and how much is convertible?
(b) What is the maximum amount of kinetic energy that could be“lost” (converted to other forms
of energy) in this collision?
(c) If 60% of the amount you calculated in part (b) is in fact converted into other forms of energy
in the collision, what are the final velocities of the two objects?",University Physics I Classical Mechanics.pdf
88 CHAPTER 4. KINETIC ENERGY,University Physics I Classical Mechanics.pdf
"Chapter 5
Interactions and energy
5.1 Conservative interactions
Let me summarize the physical concepts and principles we havee n c o u n t e r e ds of a ri no u rs t u d y
of classical mechanics. We have “discovered” one importantquantity, the inertia or inertial mass
of an object, and introduced two different quantities based onthat concept, the momentumm⃗ v
and the kinetic energy1
2 mv2. We found that these quantities have different but equally intriguing
properties. The total momentum of a system is insensitive tothe interactions between the parts
that make up the system, and therefore it stays constant in thea b s e n c eo fe x t e r n a li n fl u e n c e s( a
more general statement of the law of inertia, the first important principle we encountered). The
total kinetic energy, on the other hand, changes while any sort of interaction is taking place, but
in some cases it may actually return to its original value afterwards.",University Physics I Classical Mechanics.pdf
"in some cases it may actually return to its original value afterwards.
In this chapter, we will continue to explore this intriguingbehavior of the kinetic energy, and use
it to gain some important insights into the kinds of interactions we encounter in classical physics.
In the next chapter, on the other hand, we will return to the momentum perspective and use it to
formally introduce the concept of force. Hence, we can say that this chapter deals with interactions
from an energy point of view, whereas next chapter will deal with them from a force point of view.
In the previous chapter I suggested that what was going on in ane l a s t i cc o l l i s i o nc o u l db ei n t e r -
preted, or described (perhaps in a figurative way) more or lessa sf o l l o w s : a st h eo b j e c t sc o m e
together, the total kinetic energy goes down, but it is as if itw a sb e i n gt e m p o r a r i l ys t o r e da w a y",University Physics I Classical Mechanics.pdf
"together, the total kinetic energy goes down, but it is as if itw a sb e i n gt e m p o r a r i l ys t o r e da w a y
somewhere, and as the objects separate, that “stored energy”i sf u l l yr e c o v e r e da sk i n e t i ce n e r g y .
Whether this does happen or not in any particular collision (that is, whether the collision is elastic
or not) depends, as we have seen, on the kind of interaction (“bouncy” or “sticky,” for instance)
that takes place between the objects.
89",University Physics I Classical Mechanics.pdf
"90 CHAPTER 5. INTERACTIONS AND ENERGY
We are going to take the ab ove description literally, and usethe name conservative interaction
for any interaction that can “store and restore” kinetic energy in this way. The “stored energy”
itself—which is not actually kinetic energy while it remains stored, since it isnot given by the value
of 1
2 mv2 at that time—we are going to callpotential energy.T h u s , c o n s e r v a t i v e i n t e r a c t i o n s w i l l
be those that have a “potential energy” associated with them,a n dv i c e - v e r s a .
5.1.1 Potential energy
Perhaps the simplest and clearest example of the storage andrecovery of kinetic energy is what
happens when you throw an object straight upwards, as it risesa n de v e n t u a l l yf a l l sb a c kd o w n .
The object leaves your hand with some kinetic energy; as it rises it slows down, so its kinetic energy
goes down, down... all the way down to zero, eventually, as it momentarily stopsat the top of its",University Physics I Classical Mechanics.pdf
"goes down, down... all the way down to zero, eventually, as it momentarily stopsat the top of its
rise. Then it comes down, and its kinetic energy starts to increase again, until eventually, as it
comes back to your hand, it has very nearly the same kinetic energy it started out with (exactly
the same, actually, if you neglect air resistance).
The interaction responsible for this change in the object’skinetic energy is, of course, the gravi-
tational interaction between it and the Earth, so we are goingt os a yt h a tt h e“ m i s s i n g ”k i n e t i c
energy is temporarily stored asgravitational potential energyof the system formed by the Earth
and the object.
We even have a way to describ e what is going on mathematically.R e c a l l t h e e q u a t i o nv2
f − v2
i =
2a∆x for motion under constant acceleration. Let us usey instead of x,f o rt h ev e r t i c a lm o t i o n ;
let a = −g,a n dl e tvf just be the generic velocity,v,a ts o m ea r b i t r a r yh e i g h ty.W eh a v e",University Physics I Classical Mechanics.pdf
"let a = −g,a n dl e tvf just be the generic velocity,v,a ts o m ea r b i t r a r yh e i g h ty.W eh a v e
v2 − v2
i = −2g(y − yi)
Now multiply both sides of this equation by1
2 m:
1
2mv2 − 1
2mv2
i = −mg(y − yi)( 5 . 1 )
The left-hand side of (5.1)i sj u s tt h ec h a n g ei nk i n e t i ce n e r g y( f r o mi t si n i t i a lv a l u ewhen the object
was launched). We will interpret the right-hand side as the negative of the change in gravitational
potential energy. To make this clearer, rearrange Eq. (5.1)b ym o v i n ga l lt h e“ i n i t i a l ”q u a n t i t i e st o
one side: 1
2mv2 + mgy = 1
2mv2
i + mgyi (5.2)
We see, then, that the quantity1
2 mv2 + mgy stays constant (always equal to its initial value) as
the object goes up and down. Let us define thegravitational potential energyof a system formed
by the Earth and an object a heighty above the Earth’s surface as the following simple function
of y:
UG(y)= mgy (5.3)",University Physics I Classical Mechanics.pdf
"5.1. CONSERVATIVE INTERACTIONS 91
Then we see from Eq. (5.2)t h a t
K + UG =c o n s t a n t ( 5 . 4 )
This is a statement of conservation of energy under the gravitational interaction. For any interaction
that has a potential energy associated with it, the quantityK + U is called the (total)mechanical
energy.
Figure 5.1 shows how the kinetic and potential energies of an object thrown straight up change with
time. To calculateK Ih a v eu s e dt h ee q u a t i o nv = vi − gt (taking ti =0 ) ;t oc a l c u l a t eUG = mgy,
Ih a v eu s e dy = yi + vit − 1
2 gt2.Ih a v ea r b i t r a r i l ya s s u m e dt h a tt h eo b j e c th a sam a s so f1 k gand
an initial velocity of 2 m/s, and it is thrown from an initial height of 0.5m a bo v e t h e g r o u n d . N o t e
how the change in potential energy exactly mirrors the changei nk i n e t i ce n e r g y( s o∆UG = −∆K,
as indicated by Eq. (5.1)), and the total mechanical energy remains equal to its initial value of 6.9J
throughout.
 0
 1
 2
 3",University Physics I Classical Mechanics.pdf
"as indicated by Eq. (5.1)), and the total mechanical energy remains equal to its initial value of 6.9J
throughout.
 0
 1
 2
 3
 4
 5
 6
 7
 0  0.05  0.1  0.15  0.2  0.25  0.3  0.35  0.4
K
UG
energy (J)
t (s)
Figure 5.1: Potential and kinetic energy as a function of time for a system consisting of the earth and a
1-kg object sent upwards withvi = 2 m/s from a height of 0.5m .
There is something about potential energy that probably needs to be mentioned at this point.
Because I have chosen to launch the object from 0.5m a bo v e t h e g r o u n d , a n d I h a v e c h o s e n t o
measure y from the ground, I started out with a potential energy ofmgyi =4 .9J . T h i s m a k e s
sense, in a way: it tells you that if you simply dropped the object from this height, it would have
picked up an amount of kinetic energy equal to 4.9J b y t h e t i m e i t r e a c h e d t h e g r o u n d . B u t ,
actually, where I choose the vertical origin of coordinatesis arbitrary. I could start measuringy",University Physics I Classical Mechanics.pdf
"92 CHAPTER 5. INTERACTIONS AND ENERGY
from any height I wanted to—for instance, taking the initialheight of my hand to correspond to
y =0 . T h i sw o u l ds h i f tt h eb l u ec u r v ei nF i g .5.1 down by 4.9J , b u t i t w o u l d n o t c h a n g e a n y o f
the physics. The only important thing I really want the potential energy for is to calculate the
kinetic energy the object will lose or gainas it moves from one height to another,a n df o rt h a to n l y
changes in potential energy matter. I can always add or subtract any (constant) number1 to or
from U,a n di tw i l ls t i l lb et r u et h a t∆K = −∆U.
What about potential energy in the context in which we first encountered it, that of elastic collisions
in one dimension? Imagine that we have two carts collide on anair track, and one of them, let us
say cart 2, is fitted with a spring. As the carts come together,they compress the spring, and some",University Physics I Classical Mechanics.pdf
"say cart 2, is fitted with a spring. As the carts come together,they compress the spring, and some
of their kinetic energy is “stored” in it as elastic potentiale n e r g y .I np h y s i c s ,w eu s et h ef o l l o w i n g
expression for the potential energy stored in what we call anideal spring2:
Uspr(x)= 1
2k(x − x0)2 (5.5)
where k is something called the spring constant;x0 is the “equilibrium length” of the spring (when
it is neither compressed nor stretched); andx its actual length, sox>x 0 means the spring is
stretched, and x<x 0 means it is compressed. For the system of the two carts colliding, we can
take the potential energy to be given by Eq. (5.5)i ft h ed i s t a n c eb e t w e e nt h ec a r t si sl e s st h a nx0,
and 0 (corresponding to a relaxed spring) otherwise. If we putc a r t1o nt h el e f ta n dc a r t2o nt h e
right, then the distance between them isx2 − x1,a n ds ow ec a nw r i t e ,f o rt h ew h o l ei n t e r a c t i o n
U(x2 − x1)= 1",University Physics I Classical Mechanics.pdf
"right, then the distance between them isx2 − x1,a n ds ow ec a nw r i t e ,f o rt h ew h o l ei n t e r a c t i o n
U(x2 − x1)= 1
2k(x2 − x1 − x0)2 if x2 − x1 <x 0
0 otherwise (5.6)
This is enough to solve for the motion of the two carts, given the initial conditions. To see how,
look in the “Examples” section at the end of this chapter. Here, I will just give you the result.
For the calculation, shown in Fig.5.2 below, I have chosen cart 1 to have a mass of 1 kg, an initial
position (at t =0 )o fx1i = −5c m a n d a n i n i t i a l v e l o c i t y o f 1m / s , w h e r e a s c a r t 2 h a s a m a s s of
2k ga n ds t a r t sa tr e s ta tx2i =0 . Ih a v ea s s u m e dt h es p r i n gh a sal e n g t ho fx0 =2 c ma n da
spring constant k =1 0 0 0 J / m2 (which sounds like a lot but isn’t really). The collision begins at
tc =( x2i − x0 − x1i)/v1i =0 .03 s, which is the time it takes cart 1 to travel the 3 cm separating it",University Physics I Classical Mechanics.pdf
"tc =( x2i − x0 − x1i)/v1i =0 .03 s, which is the time it takes cart 1 to travel the 3 cm separating it
from the end of the spring. Prior to that point, the total kinetic energyKsys =0 .5J , a n d t h e t o t a l
potential energyU =0 .
1Of course, some choices may result in the potential energy, and even the total energy, beingnegative sometimes!
If this notion of a negative total energy bothers you a bit, wait until the chapter on gravity (Chapter 10), where we
will try to make some sense out of it...
2An “ideal spring” is basically defined, mathematically, by this expression, or by the corresponding force equation
(6.21)( w h i c hw ew i l ls t u d yi nt h en e x tc h a p t e r ,a n dw h i c hg o e sb yt hen a m eo fHooke’s law); usually, we also require
that the spring be “massless” (by which we mean that its mass should be negligible compared to all the other masses",University Physics I Classical Mechanics.pdf
"that the spring be “massless” (by which we mean that its mass should be negligible compared to all the other masses
involved in any given problem). Of course, for Eq. (5.5)t oh o l df o rx<x 0,i tm u s tb ep o s s i b l et oc o m p r e s st h es p r i n g
as well as stretch it, which is not always possible with some springs.",University Physics I Classical Mechanics.pdf
"5.1. CONSERVATIVE INTERACTIONS 93
As a result of the collision, the spring compresses and undergoes “half a cycle” of oscillation with
an “angular frequency”ω =
√
k/µ (where µ is, as in previous chapters, the “reduced mass” of the
system, µ = m1m2/(m1 + m2)). That is, the spring is compressed and then pushes out untili tg e t s
back to its equilibrium length3.T h i s l a s t s f r o mt = tc until t = tc + π/ω,d u r i n gw h i c ht i m et h e
potential and kinetic energies of the system can be written as
U(t)= 1
2µv2
12,i sin2[ω(t − tc)]
K(t)= Kcm + 1
2µv2
12,i cos2[ω(t − tc)] (5.7)
(don’t worry, all this will make a lot more sense after we get toC h a p t e r1 1o ns i m p l eh a r m o n i c
motion, I promise!). After t = tc + π/ω,t h ei n t e r a c t i o ni so v e r ,a n dK and U go back to their
initial values.
 0
 0.1
 0.2
 0.3
 0.4
 0.5
 0  0.02  0.04  0.06  0.08  0.1  0.12  0.14
K
U
energy (J)
t (s)",University Physics I Classical Mechanics.pdf
"initial values.
 0
 0.1
 0.2
 0.3
 0.4
 0.5
 0  0.02  0.04  0.06  0.08  0.1  0.12  0.14
K
U
energy (J)
t (s)
Figure 5.2: Potential and kinetic energy as a function of time for a system of two carts colliding and
compressing a spring in the process.
If you compare Figure5.2 with Figure 4.5 of Chapter 4, you’ll see that the kinetic energy curve
looks very similar, except for the time scale, which here is hundredths of a second and over there
was taken to be milliseconds. The quantity that determines the time scale here is the “half period”
of oscillation,π/ω = π
√
µ/k =0 .081 s for the values ofk and µ assumed here. We could make this
smaller by making the spring stiffer (increasingk), or the blocks lighter (reducingµ), but there’s
not much point in trying, since the collisions in Chapters 3 and 4 were all just made up in any case.
3As noted earlier, we shall always assume our springs to be “massless,” that is, that their inertia is negligible. In",University Physics I Classical Mechanics.pdf
"3As noted earlier, we shall always assume our springs to be “massless,” that is, that their inertia is negligible. In
turn, negligible inertia means that the spring does not “keepg o i n g ” : i ts t o p ss t r e t c h i n ga ss o o na si ti sb a c kt oi t s
original length.",University Physics I Classical Mechanics.pdf
"94 CHAPTER 5. INTERACTIONS AND ENERGY
The main point is that this kind of physical setup (a cart fittedw i t has p r i n g )w o u l di n d e e dg i v e
us an elastic collision, and a kinetic energy curve very muchlike the ones I used, for illustration
purposes, in Chapter 4; only now we also have a potential energy curve to go with it, and to show
where the energy is “hiding” while the collision lasts.
(You might wonder, anyway, what kind of potential energy function would actually produce the
made-up elastic collision curves in Chapters 3 and 4? The (perhaps surprising) answer is, I do
not really know, and I have no way to find out! If you are curiousabout why, again look at the
“Examples” section at the end of the chapter.)
5.1.2 Potential energy functions and “energy landscapes”
The potential energy function of a system, as illustrated inthe above examples, serves to let us know
how much energy can be stored in, or extracted from, the systemb yc h a n g i n gi t sconfiguration,",University Physics I Classical Mechanics.pdf
"how much energy can be stored in, or extracted from, the systemb yc h a n g i n gi t sconfiguration,
that is to say, the positions of its parts relative to each other. We have seen this in the case of
the gravitational force (the “configuration” in this case being the distance between the object and
the earth), and just now in the case of a spring (how stretchedor compressed it is). In all these
cases we should think of the potential energy as being a property of the system as a whole, not any
individual part; it is, very loosely speaking, something akin to a “stress” in the system that can be
turned into motion under the right conditions.
It is a consequence of the principle of conservation of momentum that, if the interaction between
two particles can be described by a potential energy function, this should be a function only of their
relative position, that is, the quantityx1 − x2 (or x2 − x1), and not of the individual coordinates,",University Physics I Classical Mechanics.pdf
"relative position, that is, the quantityx1 − x2 (or x2 − x1), and not of the individual coordinates,
x1 and x2,s e p a r a t e l y4.T h e e x a m p l eo ft h es p r i n g i nt h ep r e v i o u ss e c t i o ni l l u s t r a tes this, whereas
the gravitational potential energy example shows how this can be simplified in an important case:
in Eq. (5.3), the heighty of the object above the ground is really a measure of the distance between
the object and the earth, something that we could write, in full generality, as|⃗ro − ⃗rE| (where ⃗ro
and ⃗rE are the position vectors of the Earth and the object, respectively). However, since we do
not expect the Earth to move very much as a result of the interaction, we can take its position to
be constant, and only include the position of the object explicitly in our potential energy function,
as we did above5.
Generally speaking, then, we can identify a large class of problems where a “small” object or",University Physics I Classical Mechanics.pdf
"as we did above5.
Generally speaking, then, we can identify a large class of problems where a “small” object or
“particle” interacts with a much more massive one, and it is agood approximation to write the
potential energy of the whole system as a function of only theposition of the particle. In one
4We will see why in the next chapter! But, if you want to peek ahead, nothing’s preventing you from reading
sections 6.1 and 6.2 right now. Basically, to conserve momentum we need Eq. (6.6)t oh o l d ,a n da sy o uc a ns e ef r o m
Eq. (6.18), having the potential energy depend only onx1 − x2 ensures that.
5This will change in Chapter 10, when we get to study gravity over a planetary scale.",University Physics I Classical Mechanics.pdf
"5.1. CONSERVATIVE INTERACTIONS 95
dimension, then, we have a situation where, once the initialconditions (the particle’s initial position
and velocity) are known, the motion of the particle can be completely determined from the function
U(x), where x is the particle’s position at any given time. This can be done,u s i n gc a l c u l u s ,
essentially by the method illustrated in Example 5.6.3 at thee n do ft h i sc h a p t e r( n a m e l y ,l e t
v = ±
√
2m(E − U(x)) and solve the resulting differential equation); but it is also possible to get
some pretty valuable insights into the particle’s motion without using any calculus at all, through
am o s t l ygraphical approach that I would like to show you next.
 0
 5
 10
 15
 20
 25
 30
-4 -3 -2 -1  0  1  2  3  4
energy (J)
U(x)
E = U+K
x (m)
Figure 5.3: A hypothetical potential energy curve for a particle in one dimension. The horizontal red",University Physics I Classical Mechanics.pdf
"energy (J)
U(x)
E = U+K
x (m)
Figure 5.3: A hypothetical potential energy curve for a particle in one dimension. The horizontal red
line shows the total mechanical energy under the assumption thatthe particle starts out atx = −2 m with
Ki = 8 J. The green line assumes the particle starts instead from rest at x =1m .
In Figure 5.3 above I have assumed, as an example, that the potential energyo ft h es y s t e m ,a sa
function of the position of the particle, is given by the function U(x)= −x4/4+9 x2/2+2 x +1 ( i n
joules, if x is given in meters). Consider then what happens if the particle has a massm =4 k g
and is found initially atxi = −2m , w i t h a v e l o c i t yvi =2 m / s . ( T h i ss c e n a r i og o e sw i t ht h er e d
lines in Fig.5.3,s op l e a s ei g n o r et h eg r e e nl i n e sf o rt h et i m eb e i n g . ) I t sk i netic energy will then be
Ki =8 J ,w h e r e a st h ep o t e n t i a le n e r g yw i l lb eU(−2) = 11 J. The total mechanical energy is then",University Physics I Classical Mechanics.pdf
"Ki =8 J ,w h e r e a st h ep o t e n t i a le n e r g yw i l lb eU(−2) = 11 J. The total mechanical energy is then
E =1 9J ,a si n d i c a t e db yt h er e dh o r i z o n t a ll i n e .
Now, as the particle moves, the total energy remains constant, so as it moves to the right, its poten-
tial energy goes down at first, and consequently its kinetic energy goes up—that is, it accelerates.
At some point, however (aroundx = −0.22 m) the potential energy starts to go up, and so the
particle starts to slow down, although it keeps going, because K = E −U is still nonzero. However,
when the particle eventually reaches the pointx =2 m ,t h ep o t e n t i a le n e r g yU(2) = 19 J, and the
kinetic energy becomes zero.",University Physics I Classical Mechanics.pdf
"96 CHAPTER 5. INTERACTIONS AND ENERGY
At that point, the particle stops and turns around, just likean object thrown vertically upwards.
As it moves “down the potential energy hill,” it recovers thekinetic energy it used to have, so that
when it again reaches the starting pointx = −2m , i t s s pe e d i s a g a i n 2m / s , b u t n o w i t i s m o v i n g
in the opposite direction, so it just passes through and overthe next “hill” (since it has enough
total energy to do so), and eventually moves outside the region shown in the figure.
As another example, consider what would have happened if theparticle had been released at, say,
x =1m ,b u tw i t hz e r ov e l o c i t y . ( T h i si si l l u s t r a t e db yt h eg r e enl i n e si nF i g .5.3.) Then the total
energy would be just the potential energyU(1) = 7.25 J. The particle could not possibly move to
the right, since that would require the total energy to go up.It can only move to the left, since",University Physics I Classical Mechanics.pdf
"the right, since that would require the total energy to go up.It can only move to the left, since
in that direction U(x)d e c r e a s e s( i n i t i a l l y ,a tfi r s t ) ,a n dt h a tm e a n sK can increase (recall K is
always positive as long as the particle is in motion). So the particle speeds up to the left until,
past the pointx = −0.22 m,U(x)s t a r t st oi n c r e a s ea g a i na n dK has to go down. Eventually, as
the figure shows, we reach a point (which we can calculate to bex = −1.548 m) whereU(x)i so n c e
again equal to 7.25 J. This leaves no room for any kinetic energy, so the particle has to stop and
turn back. The resulting motion consists of the particle oscillating back and forth forever between
x = −1.548 m andx =1m .
At this point, you may have noticed that the motion I have described as following from theU(x)
function in Figure5.3 resembles very much the motion of a car on a roller-coaster having the shape",University Physics I Classical Mechanics.pdf
"function in Figure5.3 resembles very much the motion of a car on a roller-coaster having the shape
shown, or maybe a ball rolling up and down hills like the ones shown in the picture. In fact, the
correspondence can be madeexact—if we substitute sliding for rolling, since rolling motionhas
complications of its own. Given an arbitrary potential energy functionU(x)f o rap a r t i c l eo fm a s s
m,i m a g i n et h a ty o ub u i l da“ l a n d s c a p e ”o fh i l l sa n dv a l l e y sw hose heighty above the horizontal,
for a given value of the horizontal coordinatex,i sg i v e nb yt h ef u n c t i o ny(x)= U(x)/mg.( N o t e
that mg is just a constant scaling factor that does not change the shape of the curve.) Then, for
an object of massm sliding without friction over that landscape, under the influence of gravity,
the gravitational potential energy at any pointx would be UG(x)= mgy = U(x), and therefore",University Physics I Classical Mechanics.pdf
"the gravitational potential energy at any pointx would be UG(x)= mgy = U(x), and therefore
its speed at any point will be precisely the same as that of theoriginal particle, if it starts at the
same point with the same velocity.
This notion of an “energy landscape” can be extended to more than one dimension (although they
are hard to visualize in three!), or generalized to deal withconfiguration parameters other than
as i n g l ep a r t i c l e ’ sp o s i t i o n . I tc a nb ev e r yu s e f u li nan u m b ero fd i s c i p l i n e s( n o tj u s tp h y s i c s ) ,t o
predict the ways in which the configuration of a system may be likely to change.
5.2 Dissipation of energy and thermal energy
From all the foregoing, it is clear that when an interaction can be completely described by a
potential energy function we can define a quantity, which we have called the total mechanical",University Physics I Classical Mechanics.pdf
"5.2. DISSIPATION OF ENERGY AND THERMAL ENERGY 97
energy of the system,Emech = K + U,t h a ti sc o n s t a n tt h r o u g h o u tt h ei n t e r a c t i o n . H o w e v e r ,w e
already know from our study of inelastic collisions that thisi sr a r e l yt h ec a s e . E s s e n t i a lt ot h e
concept of potential energy is the idea of “storage and retrieval” of the kinetic energy of the system
during the interaction process. When kinetic energy simplydisappears from the system and does
not come back, a full description of the process in terms of a potential energy is not possible.
Processes in which some amount of mechanical energy disappears (that is, it cannot be found any-
where anymore as either macroscopic kinetic or potential energy) are calleddissipative.M y s t e r i o u s
as they may appear at first sight, there is actually a simple, intuitive explanation for them. All
macroscopic systems consist of a great number of small partsthat enjoy, at the microscopic level,",University Physics I Classical Mechanics.pdf
"macroscopic systems consist of a great number of small partsthat enjoy, at the microscopic level,
some degree of independence from each other and from the bodyto which they belong. Macro-
scopic motion of an object requires all these parts to move together as a whole, at least on average;
however, a collision with another object may very well “rattle” all these parts and leave them in a
more or less disorganized state. If the total energy is conserved, then after the collision the object’s
atoms or molecules may be, on average, vibrating faster or banging against each other more often
than before, but they will do so in random directions, so thisincreased “agitation” will not be
perceived as macroscopic motion of the object as a whole.
This kind of random agitation at the microscopic level that Ihave just introduced is what we know
today as thermal energy,a n di ti sb yf a rt h em o s tc o m m o n“ s i n k ”o rr e s e r v o i rw h e r em a croscopic",University Physics I Classical Mechanics.pdf
"today as thermal energy,a n di ti sb yf a rt h em o s tc o m m o n“ s i n k ”o rr e s e r v o i rw h e r em a croscopic
mechanical energy is “dissipated.” In our example of an inelastic collision, the energy the objects
had is not gone from the universe, in fact it is still right there inside the objects themselves; it is
just in a disorganized or incoherent state from which, as youcan imagine, it would be pretty much
impossible to retrieve it, since we would have to somehow getall the randomly-moving parts to get
back to moving in the same direction again.
We will have a lot more to say ab out thermal energy in a later chapter, but for the moment you may
want to think of it as essentiallynoise:i t i sw h a t i sl e f t ( t h e r e s i d u a l m o t i o n a l o rc o n fi g u r a t i o n a l
energy, at the microscopic level) after you remove the average, macroscopically-observable kinetic",University Physics I Classical Mechanics.pdf
"energy, at the microscopic level) after you remove the average, macroscopically-observable kinetic
or potential energy. So, for example, for a solid object moving with a velocityvcm,t h ek i n e t i c
part of its thermal energy would be the sum of the kinetic energies of all its microscopic parts,
calculated in its center of mass(or zero-momentum) reference frame;t h a tw a yy o ur e m o v ef r o m
every molecule’s velocity the quantityvcm,w h i c ht h e ya l lm u s th a v ei nc o m m o n — o na v e r a g e( s i n c e
the body as a whole is moving with that velocity)6.
In order to establish conservation of energy as a fact (whichwas one of the greatest scientific
triumphs of the 19th century) it was clearly necessary to showe x p e r i m e n t a l l yt h a tac e r t a i na m o u n t
6Note that thermal energy is not necessarily just kinetic; itmay have a configurational component to it as well. For",University Physics I Classical Mechanics.pdf
"6Note that thermal energy is not necessarily just kinetic; itmay have a configurational component to it as well. For
example, imagine a collection of vibrating diatomic molecules. You may think of each one as two atoms connected by
as p r i n g . T h el e n g t ho ft h e“ s p r i n g ”a tr e s td e t e r m i n e st h em olecule’s nominalchemical energy;t h e r m a lv i b r a t i o n s
cause this length to change, resulting in a net increase in energy that—as for two masses connected by a spring—has
both a kinetic and a configurational (or “potential”) component.",University Physics I Classical Mechanics.pdf
"98 CHAPTER 5. INTERACTIONS AND ENERGY
of mechanical energy lost always resulted in the same predictable increase in the system’s thermal
energy. Thermal energy is largely “invisible” at the macroscopic level, but we detect it indirectly
through an object’stemperature.T h e c r u c i a l e x p e r i m e n t s t o e s t a b l i s h w h a t a t t h e t i m e w a s c alled
the “mechanical equivalent of heat” were carried out by JamesP r e s c o t tJ o u l ei nt h e1 8 5 0 ’ s ,a n d
required exceedingly precise measurements of temperature(in fact, getting the experiments done
was only half the struggle; the other half was getting the scientific establishment to believe that he
could measure changes in temperature so accurately!)
5.3 Fundamental interactions, and other forms of energy
At the most fundamental (microscopic) level, physicists today believe that there are only four (or
three, depending on your perspective) basic interactions:gravity, electromagnetism, the strong nu-",University Physics I Classical Mechanics.pdf
"three, depending on your perspective) basic interactions:gravity, electromagnetism, the strong nu-
clear interaction (responsible for holding atomic nuclei together), and the weak nuclear interaction
(responsible for certain nuclear processes, such as the transmutation of a proton into a neutron7
and vice-versa). In a technical sense, at the quantum level,electromagnetism and the weak nuclear
interactions can be regarded as separate manifestations ofas i n g l e ,c o n s i s t e n tq u a n t u mfi e l dt h e o r y ,
so they are sometimes referred to as “the electroweak interaction.”
All of these interactions are conservative, in the sense thatf o ra l lo ft h e mo n ec a nd e fi n et h ee q u i v -
alent of a “potential energy function” (generalized, as necessary, to conform to the requirements
of quantum mechanics and relativity), so that for a system ofelementary particles interacting via",University Physics I Classical Mechanics.pdf
"of quantum mechanics and relativity), so that for a system ofelementary particles interacting via
any one of these interactions the total kinetic plus potential energy is a constant of the motion.
For gravity (which we do not really know how to “quantize” anyway!), this function immediately
carries over to the macroscopic domain without any changes,as we shall see in a later chapter, and
the gravitational potential energy function I introduced earlier in this chapter is an approximation
to it valid near the surface of the earth (gravity is such a weakf o r c et h a tt h eg r a v i t a t i o n a li n t e r a c -
tion between any two earth-bound objects is virtually negligible, so we only have to worry about
gravitational energy when one of the objects involved is theearth itself).
As for the strong and weak nuclear interactions, they are onlya p p r e c i a b l eo v e rt h es c a l eo fa na t o m i c",University Physics I Classical Mechanics.pdf
"As for the strong and weak nuclear interactions, they are onlya p p r e c i a b l eo v e rt h es c a l eo fa na t o m i c
nucleus, so there is no question of them directly affecting anymacroscopic mechanical processes.
They are responsible, however, for various nuclear reactions in the course of whichnuclear energyis,
most commonly, transformed into electromagnetic energy (X-o rg a m m ar a y s )a n dt h e r m a le n e r g y .
All the other forms of energy one encounters at the microscopic, and even the macroscopic, level
have their origin in electromagnetism. Some of them, like thee l e c t r o s t a t i ce n e r g yi nac a p a c i t o ro r
the magnetic interaction between two permanent magnets, ares t r a i g h t f o r w a r de n o u g hs c a l e - u p so f
their microscopic counterparts, and may allow for a potential energy description at the macroscopic
7Plus a positron and a neutrino.",University Physics I Classical Mechanics.pdf
"5.4. CONSERVATION OF ENERGY 99
level (and you will learn more about them next semester!). Many others, however, are more subtle
and involve quantum mechanical effects (such as the exclusionprinciple) in a fundamental way.
Among the most important of these ischemical energy,w h i c hi sa ne x t r e m e l yi m p o r t a n ts o u r c e
of energy for all kinds of macroscopic processes: combustion( a n de x p l o s i o n s ! ) ,t h ep r o d u c t i o n
of electrical energy in batteries, and all the biochemical processes that power our own bodies.
However, the conversion of chemical energy into macroscopicm e c h a n i c a le n e r g yi sa l m o s ta l w a y s
ad i s s i p a t i v ep r o c e s s( t h a ti s ,o n ei nw h i c hs o m eo ft h ei n i t ial chemical energy ends up irreversibly
converted into thermal energy), so it is generally impossible to describe them using a (macroscopic)
potential energy function (except, possibly, for electrochemical processes, with which we will not",University Physics I Classical Mechanics.pdf
"potential energy function (except, possibly, for electrochemical processes, with which we will not
be concerned here).
For instance, consider a chemical reaction in which some amount of chemical energy is converted
into kinetic energy of the molecules forming the reaction products. Even when care is taken to
“channel” the motion of the reaction products in a particulard i r e c t i o n( f o re x a m p l e ,t op u s ha
cylinder in a combustion engine), a lot of the individual molecules will end up flying in the “wrong”
direction, striking the sides of the container, etc. In otherw o r d s ,w ee n du pw i t hal o to ft h e
chemical energy being converted intodisorganized microscopic agitation—which is to say,thermal
energy.
Electrostatic and quantum effects are also responsible for thee l a s t i cp r o p e r t i e so fm a t e r i a l s ,w h i c h
can sometimes be described by macroscopic potential energy functions, at least to a first approxi-",University Physics I Classical Mechanics.pdf
"can sometimes be described by macroscopic potential energy functions, at least to a first approxi-
mation (like the spring we studied earlier in the chapter). They are also responsible for the adhesive
forces between surfaces that play an important role in friction, and various other kinds of what
might be called “structural energies,” most of which play only a relatively small part in the energy
balance where macroscopic objects are involved.
5.4 Conservation of energy
Today, physics is pretty much founded on the belief that the energy of a closed system (defined
as one that does not exchange energy with its surroundings—more on this in a minute) is always
conserved:t h a t i s , i n t e r n a l p r o c e s s e s a n d i n t e r a c t i o n s w i l l o n l y c a use energy to be “converted”
from one form into another, but the total, after all the formsof energy available to the system have",University Physics I Classical Mechanics.pdf
"from one form into another, but the total, after all the formsof energy available to the system have
been carefully accounted for, will not change. This belief isb a s e do nc o u n t l e s se x p e r i m e n t s ,o nt h e
one hand, and, on the other, on the fact that all the fundamental interactions that we are aware of
do conserve a system’s total energy.
Of course, recognizing whether a system is “closed” or not depends on having first a complete
catalogue of all the ways in which energy can be stored and exchanged—to make sure that there is,
in fact, no exchange of energy going on with the surroundings.N o t e ,i n c i d e n t a l l y ,t h a ta“ c l o s e d ”",University Physics I Classical Mechanics.pdf
"100 CHAPTER 5. INTERACTIONS AND ENERGY
system is not necessarily the same thing as an “isolated” system: the former relates to the total
energy, the latter to the total momentum. A parked car gettingh o t t e ri nt h es u ni sn o tac l o s e d
system (it is absorbing energy all the time) but, as far as itstotal momentum is concerned, it is
certainly fair to call it “isolated.” (And as you keep this inmind, make sure you also do not mistake
“isolated” for “insulated”!) Hopefully all these conceptswill be further clarified when we introduce
the additional auxiliary concepts of force, work, and heat (although the latter will not come until
the end of the semester).
For a closed system, we can state the principle of conservation of energy (somewhat symbolically)
in the form
K + U + Esource + Ediss =c o n s t a n t ( 5 . 8 )
where K is the total, macroscopic, kinetic energy;U the sum of all the applicable potential energies",University Physics I Classical Mechanics.pdf
"where K is the total, macroscopic, kinetic energy;U the sum of all the applicable potential energies
associated with the system’sinternal interactions; Esource is any kind of internal energy (such as
chemical energy) that isnot described by a potential energy function, but can increase the system’s
mechanical energy; andEdiss stands for the contents of the “dissipated energy reservoir”—typically
thermal energy. As with the potential energyU,t h ea b s o l u t ev a l u eo fEsource and Ediss does not
(usually) really matter: all we are interested in is how muchthey change in the course of the process
under consideration.
(a)
(c)
KEdiss Ugravity KEdiss Ugravity
KEdiss Ugravity KEdiss UgravityUelastic
Uelastic
Uelastic
Uelastic
(b)
(d)
Figure 5.4: Energy bar diagrams for a system formed by the earth and a ball thrown downwards. (a) As
the ball leaves the hand. (b) Just before it hits the ground. (c) During the collision, at the time of maximum",University Physics I Classical Mechanics.pdf
"the ball leaves the hand. (b) Just before it hits the ground. (c) During the collision, at the time of maximum
compression. (d) At the top of the first bounce. The total number of energy “units” is the same in all the
diagrams, as required by the principle of conservation of energy. From the diagrams you can tell that the
coefficient of restitutione =
√
7/9.
Figure 5.4 above is an example of this kind of “energy accounting” for a ball bouncing on the
ground. If the ball is thrown down, the system formed by the ball and the earth initially has both",University Physics I Classical Mechanics.pdf
"5.5. IN SUMMARY 101
gravitational potential energy, and kinetic energy (diagram (a)). Note that we could write the total
kinetic energy asKcm + Kconv,a sw ed i di nt h ep r e v i o u sc h a p t e r ,b u tb e c a u s eo ft h el a r g em ass
of the earth, the center of mass of the system is essentially the center of the earth, which, in our
earth-bound coordinate system, does not move at all, soKcm is, to an excellent approximation,
zero. Then, the reduced mass of the system, µ = mbME/(mb + ME)i s ,a l s ot oa ne x c e l l e n t
approximation, just equal to the mass of the ball, soKconv = 1
2 µv2
12 = 1
2 mb(vb − ve)2 = 1
2 mbv2
b
(again, because the earth does not move). So all the kinetic energy that we have is the kinetic
energy of the ball,and it is all, in principle, convertible (as you can see if you replace the ball, for
instance, with a bean bag).
As the ball falls, gravitational potential energy is being converted into kinetic energy, and the ball",University Physics I Classical Mechanics.pdf
"instance, with a bean bag).
As the ball falls, gravitational potential energy is being converted into kinetic energy, and the ball
speeds up. As it is about to hit the ground (diagram (b)), the potential energy is zero and the kinetic
energy is maximum. During the collision with the ground, allthe kinetic energy is temporarily
converted into other forms of energy, which are essentiallyelastic energy of deformation (like the
energy in a spring) and some thermal energy (diagram (c)). When it bounces back, its kinetic
energy will only be a fractione2 of what it had before the collision (wheree is the coefficient of
restitution). This kinetic energy is all converted into gravitational potential energy as the ball
reaches the top of its bounce (diagram (d)). Note there is mored i s s i p a t e de n e r g yi nd i a g r a m( d )
than in (c); this is because I have assumed that dissipation ofe n e r g yt a k e sp l a c eb o t hd u r i n gt h e",University Physics I Classical Mechanics.pdf
"than in (c); this is because I have assumed that dissipation ofe n e r g yt a k e sp l a c eb o t hd u r i n gt h e
compression and the subsequent expansion of the ball.
5.5 In summary
1. Forconservative interactionsone can define a potential energyU,s u c ht h a tt h a ti nt h ec o u r s e
of the interaction the total mechanical energyE = U + K of the system remains constant,
even asK and U separately change. The functionU is a measure of the energy stored in the
configuration of the system, that is, the relative position of all its parts.
2. The potential energy function for a system of two particlesm u s tb eaf u n c t i o no ft h e i rrelative
position only: U(x1 −x2). However, if one of the objects is very massive, so it does notm o v e
during the interaction, its position may be taken to be the origin of coordinates, andU written
as a function of the lighter object’s coordinate alone.
3. For a system formed by the earth and an object of massm at a heighty above the ground,",University Physics I Classical Mechanics.pdf
"as a function of the lighter object’s coordinate alone.
3. For a system formed by the earth and an object of massm at a heighty above the ground,
the gravitational potential energy can be written asUG = mgy (approximately, as long asy
is much smaller than the radius of the earth).
4. The elastic potential energy stored in an ideal spring of spring constantk and relaxed length
x0,w h e ns t r e t c h e do rc o m p r e s s e dt oa na c t u a ll e n g t hx,i sUspr = 1
2 k(x − x0)2.
5. For an object in one dimension, with position coordinatex,w h i c hi sp a r to fas y s t e mw i t h
potential energy U(x), the motion can be predicted from the “energy landscape” formed by",University Physics I Classical Mechanics.pdf
"102 CHAPTER 5. INTERACTIONS AND ENERGY
the graph of the functionU(x). The idea, elaborated in Section 5.1.2 above, is to imagine
the equivalent motion of an object sliding without frictionover the same landscape, under
the influence of gravity.
6. The fundamental interactions currently known in physicsare gravity, the strong nuclear in-
teraction and the electroweak interaction (which includesall electromagnetic phenomena).
These are all conservative.
7. At a macroscopic level, one finds a number of interactions and associated energies that are de-
rived from electromagnetism and quantum mechanics. Two important examples are chemical
energy, and elastic energy (which is energy associated withthe elasticity or “springiness” of
ab o d y ) .E l a s t i ce n e r g yc a no f t e nb ed e s c r i b e da p p r o x i m a t e lyb yap o t e n t i a le n e r g yf u n c t i o n ,
and as such be included in calculations of the total mechanical energy of a system.",University Physics I Classical Mechanics.pdf
"and as such be included in calculations of the total mechanical energy of a system.
8. Interactions between macroscopic objects almost alwaysinvolve the conversion of some type
of energy into another. Typically, some of the total mechanical energy is always lost in the
conversion process, because it is impossible to keep at leasts o m eo ft h ee n e r g yf r o ms p r e a d i n g
itself randomly among the microscopic parts that make up theinteracting objects. This is
an intrinsically irreversible process known asdissipation of energy.
9. Most of the time thedissipated energyends up asthermal energy,w h i c hi se n e r g ya s s o c i a t e d
with a random agitation at the atomic or molecular level.
10. A closed system is one that does not exchange energy with its surroundings. This is not
necessarily the same thing as an isolated system (which is onet h a td o e sn o te x c h a n g em o -",University Physics I Classical Mechanics.pdf
"necessarily the same thing as an isolated system (which is onet h a td o e sn o te x c h a n g em o -
mentum with its surroundings). For a closed system, the sum ofi t sm a c r o s c o p i cm e c h a n i c a l
energy (kinetic + potential) and all its other “internal” energies (chemical, thermal), must
be a constant.",University Physics I Classical Mechanics.pdf
"5.6. EXAMPLES 103
5.6 Examples
5.6.1 Inelastic collision in the middle of a swing
Tarzan swings on a vine to rescue a helpless explorer (as usual) from some attacking animal or
another. He begins his swing from a branch a height of 15 m abovet h eg r o u n d ,g r a b st h ee x p l o r e r
at the bottom of his swing, and continues the swing, upwards this time, until they both land safely
on another branch. Suppose that Tarzan weighs 90 kg and the explorer weighs 70 , and that Tarzan
doesn’t just drop from the branch, but pushes himself off so that he starts the swing with a speed
of 5 m/s. How high a branch can he and the explorer reach?
Solution
Let us break this down into parts. The first part of the swing involves the conversion of some
amount of initial gravitational potential energy into kinetic energy. Then comes the collision with
the explorer, which is completely inelastic and we can analyze using conservation of momentum",University Physics I Classical Mechanics.pdf
"the explorer, which is completely inelastic and we can analyze using conservation of momentum
(assuming Tarzan and the explorer form an isolated system fort h eb r i e ft i m et h ec o l l i s i o nl a s t s ) .
After that, the second half of their swing involves the complete conversion of their kinetic energy
into gravitational potential energy.
Let m1 be Tarzan’s mass,m2 the explorer’s mass,hi the initial height, andhf the final height.
We also have three velocities to worry ab out (or, more prop erly in this case, speeds, since their
direction is of no concern, as long as they all point the way they are supposed to): Tarzan’s initial
velocity at the beginning of the swing, which we may callvtop;h i sv e l o c i t ya tt h eb o t t o mo ft h e
swing, just before he grabs the explorer, which we may callvbot1,a n dh i sv e l o c i t yj u s ta f t e rh eg r a b s",University Physics I Classical Mechanics.pdf
"swing, just before he grabs the explorer, which we may callvbot1,a n dh i sv e l o c i t yj u s ta f t e rh eg r a b s
the explorer, which we may callvbot2.( I f y o ufi n d t h o s e s u b s c r i p t s c o n f u s i n g , I a ms o r r y , t h e ya r e
the best I could do; please feel free to make up your own.)
• First part: the downswing.We apply conservation of energy, in the form (5.8), to the first part of
the swing. The system we consider consists of Tarzan and the earth, and it has kinetic energy as
well as gravitational potential energy. We ignore the sourcee n e r g ya n dt h ed i s s i p a t e de n e r g yt e r m s ,
and consider the system closed despite the fact that Tarzan ish o l d i n go n t oav i n e( a sw es h a l ls e ei n
ac o u p l eo fc h a p t e r s ,t h ev i n ed o e sn o“ w o r k ”o nT a r z a n — m e a n ing, it does not change his energy,
only his direction of motion—because the force it exerts on Tarzan is always perpendicular to his
displacement):
Ktop + UG",University Physics I Classical Mechanics.pdf
"only his direction of motion—because the force it exerts on Tarzan is always perpendicular to his
displacement):
Ktop + UG
top = Kbot1 + UG
bot (5.9)
In terms of the quantities I introduced above, this equationbecomes:
1
2m1v2
top + m1ghi = 1
2m1v2
bot1 +0
which can be solved to give
v2
bot1 = v2
top +2 ghi (5.10)",University Physics I Classical Mechanics.pdf
"104 CHAPTER 5. INTERACTIONS AND ENERGY
(note that this is just the familiar result (2.10)f o rf r e ef a l l ! T h i si sb e c a u s e ,a sIp o i n t e do u ta b o v e ,
the vine does no work on the system.). Substituting, we get
vbot1 =
√
 (
5 m
s
)2
+2
(
9.8 m
s2
)
× (15 m) = 17.9 m
s
• Second part: the completely inelastic collision.The explorer is initially at rest (we assume he has
not seen the wild beast ready to pounce yet, or he has seen it andh ei sp a r a l y z e db yf e a r ! ) .A f t e r
Tarzan grabs him they are moving together with a speedvbot2.C o n s e r v a t i o no fm o m e n t u mg i v e s
m1vbot1 =( m1 + m2)vbot2 (5.11)
which we can solve to get
vbot2 = m1vbot1
m1 + m2
= (90 kg)× (17.9m / s )
160 kg =1 0m
s
• Third part: the upswing.Here we use again conservation of energy in the form
Kbot2 + UG
bot = Kf + UG
f (5.12)
where the subscriptf refers to the very end of the swing, when they both safely reacht h e i rn e w",University Physics I Classical Mechanics.pdf
"Kbot2 + UG
bot = Kf + UG
f (5.12)
where the subscriptf refers to the very end of the swing, when they both safely reacht h e i rn e w
branch, and all their kinetic energy has been converted to gravitational potential energy, soKf =0
(which means that is as high as they can go, unless they start climbing the vine!). This equation
can be rewritten as 1
2(m1 + m2)v2
bot2 +0=0+( m1 + m2)ghf
and solving forhf we get
hf = v2
bot2
2g = (10 m/s)2
2 × 9.8m / s2 =5 .15 m
5.6.2 Kinetic energy to spring potential energy: block collides with spring
Ab l o c ko fm a s sm is sliding on a frictionless, horizontal surface, with a velocity vi.I t h i t sa ni d e a l
spring, of spring constantk,w h i c hi sa t t a c h e dt ot h ew a l l . T h es p r i n gc o m p r e s s e su n t i lthe block
momentarily stops, and then starts expanding again, so the block ultimately bounces off.
(a) In the absence of dissipation, what is the block’s final speed?
(b) By how much is the spring compressed?
Solution",University Physics I Classical Mechanics.pdf
"(a) In the absence of dissipation, what is the block’s final speed?
(b) By how much is the spring compressed?
Solution
This is a simpler version of the problem considered in Section5 . 1 . 1 ,a n di nt h en e x te x a m p l e .T h e",University Physics I Classical Mechanics.pdf
"5.6. EXAMPLES 105
problem involves the conversion of kinetic energy into elastic potential energy, and back. In the
absence of dissipation, Eq. (5.8), specialized to this system (the spring and the block) reads:
K + Uspr =c o n s t a n t ( 5 . 1 3 )
For part (a), we consider the whole process where the spring starts relaxed and ends relaxed, so
Uspr
i = Uspr
f =0 . T h e r e f o r e ,w em u s ta l s oh a v eKf = Ki,w h i c hm e a n st h eb l o c k ’ sfi n a ls p e e di s
the same as its initial speed. As explained in the chapter, this is characteristic of a conservative
interaction.
For part (b), we take the final state to be the instant where thespring is maximally compressed and
the block is momentarily at rest, so all the energy in the system is spring (which is to say, elastic)
potential energy. If the spring is compressed a distanced (that is, x − x0 = −d in Eq. (5.5)), this
potential energy is1",University Physics I Classical Mechanics.pdf
"potential energy. If the spring is compressed a distanced (that is, x − x0 = −d in Eq. (5.5)), this
potential energy is1
2 kd2,s os e t t i n gt h a te q u a lt ot h es y s t e m ’ si n i t i a le n e r g yw eg e t :
Ki +0=0+ 1
2kd2 (5.14)
or 1
2mv2
i = 1
2kd2
which can be solved to get
d =
√
 m
k vi",University Physics I Classical Mechanics.pdf
"106 CHAPTER 5. INTERACTIONS AND ENERGY
5.7 Advanced Topics
5.7.1 Two carts colliding and compressing a spring
Unlike the example 5.6.2,w h i c hc o n s i d e r e das t a t i o n a r ys p r i n ga n da s k e do n l yq u e s t ions about
initial and final states, this example is intended to show youhow one can use “energy methods” to
solve for the actual motion of a relatively complicated system as a function of time. The system
is the two carts colliding, one of them fitted with a spring, considered in Section 5.1.1. Although
all the physics involved is straightforward, the math is at ahigher level than you will be using this
semester, so I’m presenting this here as a “curiosity” only.
First, recall the total kinetic energy for a collision problem can be written asK = Kcm + Kconv,
where (if the system is isolated)Kcm remains constant throughout. Then, the total mechanical",University Physics I Classical Mechanics.pdf
"where (if the system is isolated)Kcm remains constant throughout. Then, the total mechanical
energy E = K + U = Kcm + Kconv + U.T h i si sa l s oc o n s t a n t ,a n db e f o r et h ei n t e r a c t i o nh a p p e n s ,
when U =0 ,w eh a v eE = Kcm + Kconv,i,s os e t t i n gt h e s et w ot h i n g se q u a la n dc a n c e l i n go u tKcm
we get
Kconv = Kconv,i − U (5.15)
where the potential energy function is given by Eq. (5.6). Introducing the relative coordinate
x12 = x2 − x1,a n dt h er e l a t i v ev e l o c i t yv12,E q .(5.15)b e c o m e s
1
2µv2
12 = 1
2µv2
12,i − 1
2k(x12 − x0)2 (5.16)
an equation that must hold while the interaction is going on.We can solve this forv12,a n dt h e n
notice that bothx12 and v12 are functions of time, and the latter is the derivative with respect to
time of the former, so
v12 = ±
√
v2
12,i − (k/µ)( x12 − x0)2 (5.17)
dx12
dt = ±
√
v2
12,i − (k/µ)( x12(t) − x0)2 (5.18)",University Physics I Classical Mechanics.pdf
"time of the former, so
v12 = ±
√
v2
12,i − (k/µ)( x12 − x0)2 (5.17)
dx12
dt = ±
√
v2
12,i − (k/µ)( x12(t) − x0)2 (5.18)
(The “±”s i g nm e a n st h a tt h eq u a n t i t yo nt h er i g h t - h a n ds i d eh a st ob enegative at first, when the
carts are coming together, and positive later, when they arecoming apart. This is because I have
assumed cart 1 starts to the left of cart 2, so going in cart 2, ass e e nf r o mc a r t1 ,a p p e a r st ob e
moving to the left.)
Equation (5.18) is what is known, in calculus, as a differential equation. Theproblem is to find a
function oft, x12(t), such that when you take its derivative you get the expression on the right-hand
side. If you know how to calculate derivatives, you can checkthat the solution is in fact
x12(t)= x0 − v12,i
ω sin[ω(t − tc)] for tc ≤ t ≤ tc + π/ω (5.19)",University Physics I Classical Mechanics.pdf
"5.7. ADVANCED TOPICS 107
where the quantityω =
√
k/µ,a n dt h et i m etc is the time cart 1 first makes contact with the
spring: tc =( x2i −x0 −x1i)/v1i.T h e s o l u t i o n(5.19)i sv a l i df o ra sl o n ga st h es p r i n gi sc o m p r e s s e d ,
which is to say, for as long asx12(t) <x 0,o rs i n [ω(t − tc)] > 0, which translates to the condition
on t shown above.
Having a solution forx12,w ec o u l dn o wo b t a i ne x p l i c i tr e s u l t sf o rx1(t)a n dx2(t)s e p a r a t e l y ,u s i n g
the fact thatx1 = xcm −m2x12/(m1 +m2), andx2 = xcm +m1x12/(m1 +m2)( c o m p a r eE q s .(4.10),
in chapter 4), and finding the position of the center of mass asaf u n c t i o no ft i m ei sat r i v i a lp r o b l e m ,
since it just moves with constant velocity.
We do not, however, need to do any of this in order to generate the plots of the kinetic and potential
energy shown in Fig.5.2:t h e p o t e n t i a le n e r g y d e p e n d s o n l y o nx2 − x1,w h i c hi sg i v e ne x p l i c i t l y",University Physics I Classical Mechanics.pdf
"energy shown in Fig.5.2:t h e p o t e n t i a le n e r g y d e p e n d s o n l y o nx2 − x1,w h i c hi sg i v e ne x p l i c i t l y
by Eq. (5.19), and the kinetic energy is equal toKcm + Kconv,w h e r eKcm is constant andKconv
is given by Eq. (5.16), which can also be easily rewritten in terms of Eq. (5.19). The results are
Eqs. (5.7)i nt h et e x t .
5.7.2 Getting the potential energy function from collisiondata
Consider the collision illustrated in Figure3.4 (back in Chapter 3). Can we tell what the potential
energy function is for the interaction between the two carts?
At first sight, it may seem that all the information necessaryto “reconstruct” the functionU(x1−x2)
is available already, at least in graphical form: From Figure 3.4 you could get the value ofx2 − x1
at any timet;t h e nf r o mF i g u r e4 . 5y o uc a ng e tt h ev a l u eo fK (in the elastic-collision scenario)
for the same value oft;a n dt h e ny o uc o u l dp l o tU = E − K (where E is the total energy) as a",University Physics I Classical Mechanics.pdf
"for the same value oft;a n dt h e ny o uc o u l dp l o tU = E − K (where E is the total energy) as a
function ofx2 − x1.
But there is a catch: Figure3.4 shows that the colliding objects never get any closer thanx2 −x1 ≃
0.28 mm, so we have no way to tell whatU(x2−x1)i sf o rs m a l l e rv a l u e so fx2−x1.T h i s i s e s s e n t i a l l y
the problem faced by particle physicists when they use collisions (which they do regularly) to try
to determine the precise nature of the interactions betweenthe particles they study!
You can check this for yourself. The functions I used forx1(t)a n dx2(t)i nfi g u r e3.4 are
x1(t)= 1
3
(
(2t − 10) erf(10− 2t)+1 0e r f ( 1 0 )+t − e−4(t−5)2
√
π
)
− 5
x2(t)= 1
3
(
(5 − t)e r f ( 1 0− 2t) − 5e r f ( 1 0 )+t + e−4(t−5)2
2√
π
)
(5.20)
Here, “erf” is the so-called “error function,” which you canfind in any decent library of mathemat-",University Physics I Classical Mechanics.pdf
"108 CHAPTER 5. INTERACTIONS AND ENERGY
ical functions. This looks complicated, but it just gives yout h es h a p e sy o uw a n tf o rt h ev e l o c i t y
curves. The derivative of the above is
v1(t)= 1
3 (1 + 2 erf(10− 2t))
v2(t)= 1
3 (1 − erf(10 − 2t)) (5.21)
and you may want to try plotting these for yourself; the results h o u l db eF i g u r e3.1.
Now, assume (as I did for figure4.5)t h a tm1 =1k g ,a n dm2 =2k g ,a n du s et h e s ev a l u e sa n dt h e
results (5.21)( a s s u m e dt ob ei nm / s )t oc a l c u l a t eKsys as a function oft.T h e nU = Esys − Ksys,
with Esys =1 /2J :
U = 1
2 − 1
2m1v2
1(t) − 1
2m2v2
2(t)= 1
3
(
1 − erf2(10 − 2t)
)
(5.22)
and now do a parametric plot ofU versus x2 − x1,u s i n gt as a parameter. You will end up with a
figure like the one below:
U (J)
x2 - x1 (mm)
Figure 5.5: The potential energy function reconstructed from the information available for the collision",University Physics I Classical Mechanics.pdf
"figure like the one below:
U (J)
x2 - x1 (mm)
Figure 5.5: The potential energy function reconstructed from the information available for the collision
shown in Figs. 3.1, 3.4, 4.5. No information can be gathered from those figures (nor from theexplicit
expressions (5.20)a n d(5.21) above) on the values ofU for x2 − x1 < 0.28 mm, the distance of closest
approach of the two carts.",University Physics I Classical Mechanics.pdf
"5.8. PROBLEMS 109
5.8 Problems
Problem 1
Ap a r t i c l ei si nar e g i o nw h e r et h ep o t e n t i a le n e r g yh a st h ef orm U =5 /x (in joules, if x is in
meters).
(a) Sketch this potential energy function forx> 0.
(b) Assuming the particle starts at rest atx =0 .5m , w h i c h w a y w i l l i t g o i f r e l e a s e d ? W h y ?
(c) Under the assumption in part (b), what will be the particle’s kinetic energy after it has moved
0.1m f r o m i t s o r i g i n a l po s i t i o n ?
(d) Now assume that initially the particle is atx =1 m ,m o v i n gt o w a r d st h el e f tw i t ha ni n i t i a l
velocity vi =2 m / s . I ft h em a s so ft h ep a r t i c l ei s1 k g ,h o wc l o s et ot h eo r i gin can it get before it
stops?
Problem 2
A“ b a l l i s t i cp e n d u l u m ”i sad e v i c e( n o wl a r g e l yo b s o l e t e ,b utv e r yu s e f u li ni t sd a y )t om e a s u r et h e",University Physics I Classical Mechanics.pdf
"A“ b a l l i s t i cp e n d u l u m ”i sad e v i c e( n o wl a r g e l yo b s o l e t e ,b utv e r yu s e f u li ni t sd a y )t om e a s u r et h e
speed of a bullet as it hits a target. Let the target be a block ofw o o ds u s p e n d e df r o mas t r i n g ,a s
in the figure below. When the bullet hits, it is embedded in thewood, and together they swing,
like a pendulum, to some maximum heighth.T h e q u e s t i o ni s ,h o wd oy o ufi n d t h ei n i t i a ls p e e do f
the bullet (vi)i fy o uk n o wt h em a s so ft h eb u l l e t(m1), the mass of the block (m2), and the height
h?
h
(a) (b) 
Figure 5.6: Ballistic pendulum. (a) Before the bullet hits. (b) After the bullet hitsand is embedded in the
block, at the maximum height of the swing.
Problem 3
You drop a 0.5k g b a l l f r o m a h e i g h t o f 2m , a n d i t b o u n c e s b a c k t o a h e i g h t o f 1.5m . C o n s i d e rt h e
system formed by the ball and the Earth, so we can speak properly of its gravitational potential",University Physics I Classical Mechanics.pdf
"110 CHAPTER 5. INTERACTIONS AND ENERGY
energy.
(a) What is the kinetic energy of the ball just before it hits the ground?
(b) What is the kinetic energy of the ball just after it bouncesu p ?
(c) What is the coefficient of restitution for this collision?
(d) What kind of collision is this (elastic, inelastic, etc.)? Why?
(e) If the coefficient of restitution does not change, how highwould the ball rise on asecond bounce?
(f) On the graphs below, draw the energy bar diagrams for the system: (1) as the ball leaves your
hand; (2) just before it hits the ground (assumeh =0f o rp r a c t i c a lp u r p o s e s ) ;( 3 )j u s ta f t e ri t
leaves the ground on its way up (h =0s t i l l ) ,a n d( 4 )a tt h et o po fi t s( fi r s t )b o u n c e . M a k es u r et o
do this to scale, consistent with the values for the energiesyou have calculated above.
Esource KU Ediss
Esource KU Ediss
Esource KU Ediss
Esource KU Ediss
(1) As the ball leaves your hand (2) Just before it hits the ground",University Physics I Classical Mechanics.pdf
"Esource KU Ediss
Esource KU Ediss
Esource KU Ediss
Esource KU Ediss
(1) As the ball leaves your hand (2) Just before it hits the ground
(3) Just after it leaves the ground
 on its way up
(4) At the top of its bounce
Problem 4
A6 0 - k gs k y d i v e rj u m p sf r o ma na i r p l a n e4 0 0 0 ma b o v et h ee a r t h.A f t e r f a l l i n g 4 5 0 m , h e r e a c h e s
at e r m i n a ls p e e do f5 5 m / s( a b o u t1 2 0m p h ) . T h i sm e a n st h a ta f ter this time his speed does not
increase any more.
(a) At the moment of the jump, what is the initial (gravitational) potential energy of the system
formed by the earth and the skydiver? (TakeUG =0a tg r o u n dl e v e l . )
(b) After the skydiver has fallen 450 m, what is the (gravitational) potential energy of the system?
(Call this the “final” potential energy.)
(c) What is the final kinetic energy of the diver at that time?
(d) Assume the initial kinetic energy of the skydiver is zero.I s ∆K = −∆U for this system? If",University Physics I Classical Mechanics.pdf
"(d) Assume the initial kinetic energy of the skydiver is zero.I s ∆K = −∆U for this system? If
not, explain what happened to the “missing” energy.
(e) Can the skydiver and the earth below (excluding the atmosphere!)b e c o n s i d e r e d a c l o s e d s y s t e m",University Physics I Classical Mechanics.pdf
"5.8. PROBLEMS 111
here? Explain.
(f) After the skydiver reaches terminal speed (and before heopens his parachute), he falls for a while
at constant speed. What kind of energy conversion is taking place during this time? (Consider the
system to be the earth, the skydiver, and the air around him).
Problem 5
You shoot a 1-kg pro jectile straight up from a spring toy gun,and find that it reaches a height of
5m . ( H o w d o y o u fi g u r e o u t t h e h e i g h t ? F r o m t h e t i m e o f fl i g h t , o fcourse! See problem 2 from
Chapter 2.) You also measure that when you load the gun, the spring compresses a distance 10 cm.
What is the value of the spring constant?",University Physics I Classical Mechanics.pdf
112 CHAPTER 5. INTERACTIONS AND ENERGY,University Physics I Classical Mechanics.pdf
"Chapter 6
Interactions, part 2: Forces
6.1 Force
As we saw in the previous chapter, when an interaction can be described by a potential energy
function, it is possible to use this to get a full solution forthe motion of the objects involved, at
least in one dimension. In fact, energy-based methods (knowna st h eL a g r a n g i a na n dH a m i l t o n i a n
methods) can be also generalized to deal with problems in three dimensions, and they also provide
the most direct pathway to quantum mechanics and quantum fieldt h e o r y .I tm i g h tb ep o s s i b l et o
write an advanced textbook on classical mechanics without mentioning the concept of force at all.
On the other hand, as you may have also gathered from the example I worked out at the end of the
previous chapter (section 5.6.3), solving for the equationof motion using energy-based methods may
involve somewhat advanced math, even in just one dimension,and it only gets more complicated",University Physics I Classical Mechanics.pdf
"involve somewhat advanced math, even in just one dimension,and it only gets more complicated
in higher dimensions. There is also the question of how to dealw i t hi n t e r a c t i o n st h a ta r en o t
conservative (at the macroscopic level) and therefore cannot be described by a potential energy
function of the macroscopic coordinates. And, finally, therea r es p e c i a l i z e dp r o b l e ma r e a s( s u c ha s
the entire field of statics) where you actuallywant to know the forces acting on the various objects
involved. For all these reasons, the concept of force will beintroduced here, and the next few
chapters will illustrate how it may be used to solve a varietyof elementary problems in classical
mechanics. This does not mean, however, that we are going to forget about energy from now on:
as we will see, energy methods will continue to provide usefuls h o r t c u t si nav a r i e t yo fs i t u a t i o n s
as well.",University Physics I Classical Mechanics.pdf
"as we will see, energy methods will continue to provide usefuls h o r t c u t si nav a r i e t yo fs i t u a t i o n s
as well.
We start, as usual, by considering two ob jects that form an isolated system, so they interact with
each other and with nothing else. As we have seen, under thesecircumstances their individual
momenta change, but the total momentum remains constant. Weare going to take therate of
113",University Physics I Classical Mechanics.pdf
"114 CHAPTER 6. INTERACTIONS, PART 2: FORCES
change of each object’s momentum as a measure of theforce exerted on it by the other object.
Mathematically, this means we will write for theaverage force exerted by 1 on 2over the time
interval ∆t the expression
(F12)av = ∆p2
∆t (6.1)
Please observe the notation we are going to use: the subscripts on the symbolF are in the order
“by,on”, as in “force exerted by” (object identified by first subscript) “on” (object identified by
second subscript). (The comma is more or less optional.)
You can also see from Eq. (6.1)t h a tt h eS Iu n i t so ff o r c ea r ek g· m/s2.T h i s c o m b i n a t i o n o f u n i t s
has the special name “newton,” and it’s abbreviated by an uppercase N.
In the same way as above, we can write the average force exertedb yo b j e c t2o no b j e c t1 :
(F21)av = ∆p1
∆t (6.2)
and we know, by conservation of momentum, that we must have ∆p1 = −∆p2,s ow eg e to u rfi r s t
important result,
(F12)av = −(F21)av (6.3)",University Physics I Classical Mechanics.pdf
"∆t (6.2)
and we know, by conservation of momentum, that we must have ∆p1 = −∆p2,s ow eg e to u rfi r s t
important result,
(F12)av = −(F21)av (6.3)
That is, whenever two objects interact, they always exert equal(in magnitude) and opposite (in
direction) forces on each other.T h i s i s m o s t o f t e n c a l l e dNewton’s third lawof motion, or
informally “the law of action and reaction.”
We might as well now proceed along familiar lines and take thelimit of Eqs. (6.1)a n d(6.2)a b o v e ,
as ∆t goes to zero, in order to introduce the more general concept oft h ei n s t a n t a n e o u sf o r c e( o r
just the “force,” without any further qualifiers). We then get
F12 = dp2
dt
F21 = dp1
dt (6.4)
and, since Eq. (6.3)s h o u l dh o l df o rat i m ei n t e r v a lo fa n ys i z e ,
F12 = −F21 (6.5)
Now, under most circumstances the mass of, say, object 2 willnot change during the interaction,
so we can write
F12 = d
dt(m2v2)= m2
dv2
dt = m2a2 (6.6)",University Physics I Classical Mechanics.pdf
"so we can write
F12 = d
dt(m2v2)= m2
dv2
dt = m2a2 (6.6)
This is the result that we often refer to as “F = ma”, also known asNewton’s second law
of motion: the (net) force acting on an object is equal to the product of its inertial mass and its
acceleration.T h e f o r m u l a t i o n i n t e r m s o f t h e r a t e o f c h a n g e o f m o m e n t u m , asi nE q s .(6.4), is,",University Physics I Classical Mechanics.pdf
"6.1. FORCE 115
however, somewhat more general, so it is technically preferred, even though this semester we will
directly useF = ma throughout.
If you want an example of a physical situation whereF = dp/dt is not equivalent to F = ma,
consider a system where object 1 is a rocket, including its fuel, and “object” 2 are the gases ejected
by the rocket. In this case, the mass of both “objects” is constantly changing, as the fuel is burned
and more gases are ejected, and so the more general formF = dp/dt needs to be used to calculate
the force on the rocket (the thrust) at any given time.
At this point you may be wondering just what is Newton’sfirst law? It is just the law of inertia:
an object on which no force acts will stay at rest if it is initially at rest, or will move with constant
velocity.
6.1.1 Forces and systems of particles
What if you had, say, three objects (let us make them “particles,” for simplicity), all interacting",University Physics I Classical Mechanics.pdf
"velocity.
6.1.1 Forces and systems of particles
What if you had, say, three objects (let us make them “particles,” for simplicity), all interacting
with one another? In physics we find that all our interactionsare pairwise additive, that is, we can
write the total potential energy of the system as the sum of thep o t e n t i a le n e r g i e sa s s o c i a t e dw i t h
each pair of particles separately. As we will see in a moment,this means that the corresponding
forces are additive too, so that, for instance, the total force on particle 1 could be written as
Fall,1 = F21 + F31 = dp1
dt (6.7)
Consider now the most general case of a system that has an arbitrary number of particles, and is
not isolated; that is, there are other objects, outside the system, that exert forces on some or all of
the particles that make up the system. We will call theseexternal forces.T h e s u m o f a l l t h e f o r c e s",University Physics I Classical Mechanics.pdf
"the particles that make up the system. We will call theseexternal forces.T h e s u m o f a l l t h e f o r c e s
(both internal and external) acting on all the particles willt a k eaf o r ml i k et h i s :
Ftotal = Fext,1 + F21 + F31 + ... + Fext,2 + F12 + F32 + ... + ... = dp1
dt + dp2
dt + ... (6.8)
where Fext,1 is the sum of all the external forces acting on particle 1, andso on. But now, observe
that because of Newton’s third law, Eq. (6.5), for every term of the formFij appearing in the sum
(6.8), there is a corresponding termFji = −Fij (you can see this explicitly already in Eq. (6.8)
with F12 and F21), so all those terms (which represent all the internal forces) are going to cancel
out, and we will be left only with the sum of the external forces:
Fext,1 + Fext,2 + ... = dp1
dt + dp2
dt + ... (6.9)
The left-hand side of this equation is the sum of all the external forces; the right-hand side is the",University Physics I Classical Mechanics.pdf
"dt + dp2
dt + ... (6.9)
The left-hand side of this equation is the sum of all the external forces; the right-hand side is the
rate of change of the total momentum of the system. But the total momentum of the system is",University Physics I Classical Mechanics.pdf
"116 CHAPTER 6. INTERACTIONS, PART 2: FORCES
just equal toMvcm (compare Eq. (3.11), in the “Momentum” chapter). So we have
Fext,all = dpsys
dt = d
dt(Mvcm)( 6 . 1 0 )
This extends a previous result. We already knew that in the absence of external forces, the mo-
mentum of a system remained constant. Now we see that the system’s momentum responds to the
net external force as if the whole system was a single particleo fm a s se q u a lt ot h et o t a lm a s sM
and moving at the center of mass velocityvcm.I n f a c t ,a s s u m i n g t h a tM does not change we can
rewrite Eq. (6.10)i nt h ef o r m
Fext,all = Macm (6.11)
where acm is the acceleration of the center of mass. This is the key result that allows us to treat
extended objects as if they were particles: as far as the motion of the center of mass is concerned,
all the internal forces cancel out (as we already saw in our study of collisions), and the point",University Physics I Classical Mechanics.pdf
"all the internal forces cancel out (as we already saw in our study of collisions), and the point
representing the center of mass responds to the sum of the external forces as if it were just a
particle of massM subject to Newton’s second law,F = ma.T h e r e s u l t (6.11)a p p l i e se q u a l l yw e l l
to an extended solid object that we choose to mentally break upi n t oac o l l e c t i o no fp a r t i c l e s ,a s
to an actual collection of separate particles, or even to a collection of separate extended objects;
in the latter case, we would just have each object’s motion represented by the motion of its own
center of mass.
Finally, note that all the results above generalize to more than one dimension. In fact, forces
are vectors (just like velocity, acceleration and momentum), and all ofthe above equations, in 3
dimensions, apply separately to each vector component. In one dimension, we just need to be aware
of the sign of the forces, whenever we add several of them together.",University Physics I Classical Mechanics.pdf
"of the sign of the forces, whenever we add several of them together.
6.2 Forces and potential energy
In the last chapter I mentioned a special case that we encounter often, in which a lighter object is
interacting with a much more massive one, so that the massiveone essentially does not move at all
as a result of the interaction. Note that this does not contradict Newton’s 3rd law, Eq. (6.5): the
forces the two objects exert on each otherare the same in magnitude, but the acceleration of each
object is inversely proportional to its mass, soF12 = −F21 implies
m2a2 = −m1a1 (6.12)
and so if, for instance,m2 ≫ m1,w eg e t|a2| = |a1|m1/m2 ≪| a1|.I n w o r d s , t h e m o r e m a s s i v e
object is less responsive than the less massive one to a forceof the same magnitude. This is just
how we came up with the concept of inertial mass in the first place!
Anyway, you’ll recall that in this situation I could just write the potential energy function of the",University Physics I Classical Mechanics.pdf
"Anyway, you’ll recall that in this situation I could just write the potential energy function of the
whole system as a function of only the lighter object’s coordinate, U(x). I am going to use this",University Physics I Classical Mechanics.pdf
"6.2. FORCES AND POTENTIAL ENERGY 117
simplified setup to show you a very interesting relationshipbetween potential energies and forces.
Suppose this is a closed system in which no dissipation of energy is taking place. Then the total
mechanical energy is a constant:
Emech = 1
2mv2 + U(x)=c o n s t a n t ( 6 . 1 3 )
(Here, m is the mass of the lighter object, andv its velocity; the more massive object does not
contribute to the total kinetic energy, since it does not move!)
As the lighter object moves, bothx and v in Eq. (6.13)c h a n g ew i t ht i m e( r e c a l l ,f o ri n s t a n c e ,o u r
study of “energy landscapes” in the previous chapter, section 5.1.2). So I can take the derivative
of Eq. (6.13)w i t hr e s p e c tt ot i m e ,u s i n gt h ec h a i nr u l e ,a n dn o t i n gt h a t ,since the whole thing is a
constant, the total value of the derivative must be zero:
0= d
dt
( 1
2m
(
v(t)
)2 + U
(
x(t)
))
= mv(t) dv
dt + dU
dx
dx
dt (6.14)",University Physics I Classical Mechanics.pdf
"constant, the total value of the derivative must be zero:
0= d
dt
( 1
2m
(
v(t)
)2 + U
(
x(t)
))
= mv(t) dv
dt + dU
dx
dx
dt (6.14)
But note thatdx/dt is just the same asv(t). So I can cancel that on both terms, and then I am
left with
m dv
dt = −dU
dx (6.15)
But dv/dt is just the accelerationa,a n dF = ma.S ot h i st e l l sm et h a t
F = −dU
dx (6.16)
and this is how you can always get the force from a potential energy function.
Let us check it right away for the force of gravity: we know that UG = mgy,s o
FG = −dUG
dy = − d
dy(mgy)= −mg (6.17)
Is this right? It seems to be! Recall all objects fall with thesame acceleration, −g (assuming the
upwards direction to be positive), so ifF = ma,w em u s th a v eFG = −mg.S o t h e g r a v i t a t i o n a l
force exerted by the earth on any object (which I would denotein full byFG
E,o)i sp r o p o r t i o n a lt o
the inertial mass of the object—in fact, it is what we call theobject’sweight—but since to get the",University Physics I Classical Mechanics.pdf
"E,o)i sp r o p o r t i o n a lt o
the inertial mass of the object—in fact, it is what we call theobject’sweight—but since to get the
acceleration you have to divide the force by the inertial mass, that cancels out, anda ends up being
the same for all objects, regardless of how heavy they are.
Now that we have this result under our belt, we can move on to thes l i g h t l ym o r ec h a l l e n g i n gc a s e
of two objects of comparable masses interacting through a potential energy function that must be,
as I pointed out in the previous chapter, a function of just ther e l a t i v ec o o r d i n a t ex12 = x2 − x!.",University Physics I Classical Mechanics.pdf
"118 CHAPTER 6. INTERACTIONS, PART 2: FORCES
Ic l a i mt h a ti nt h a tc a s ey o uc a na g a i ng e tt h ef o r c eo no b j e c t1, F21,b yt a k i n gt h ed e r i v a t i v e
of U(x2 − x1)w i t hr e s p e c tt ox1 (leaving x2 alone), and reciprocally, you getF12 by taking the
derivative ofU(x2 − x1)w i t hr e s p e c tt ox2.H e r ei sh o wi tw o r k s ,a g a i nu s i n g t h ec h a i nr u l e :
F21 = − d
dx1
U(x12)= − dU
dx12
d
dx1
(x2 − x1)= dU
dx12
F12 = − d
dx2
U(x12)= − dU
dx12
d
dx2
(x2 − x1)= − dU
dx12
(6.18)
and you can see that this automatically ensures thatF21 = −F12.I n f a c t ,i t w a si no r d e r t oe n s u r e
this that I required thatU should depend only on thedifference of x1 and x2,r a t h e rt h a no ne a c h
one separately. Since we got the conditionF21 = −F12 originally from conservation of momentum,
you can see now how the two things are related1.
The only example we have seen so far of this kind of potential energy function was in last chapter’s",University Physics I Classical Mechanics.pdf
"you can see now how the two things are related1.
The only example we have seen so far of this kind of potential energy function was in last chapter’s
Section 5.1.1, for two carts interacting through an “ideal”spring. I told you there that the potential
energy of the system could be written as1
2 k(x2 −x1 −x0)2,w h e r ek was the “spring constant” and
x0 the relaxed length of the spring. If you apply Eqs. (6.18)t ot h i sf u n c t i o n ,y o uw i l lfi n dt h a tt h e
force exerted (through the spring) by cart 2 on cart 1 is
F21 = k(x2 − x1 − x0)( 6 . 1 9 )
Note that this force will be negative under the assumptions wem a d el a s tc h a p t e r ,n a m e l y ,t h a tc a r t
2i so nt h er i g h t ,c a r t1o nt h el e f t ,a n dt h es p r i n gi sc o m p r e s sed, so thatx2 − x1 <x 0.S i m i l a r l y ,
F12 = −k(x2 − x1 − x0)( 6 . 2 0 )
and this one, as it should, is positive.",University Physics I Classical Mechanics.pdf
"F12 = −k(x2 − x1 − x0)( 6 . 2 0 )
and this one, as it should, is positive.
The results (6.19)a n d(6.20)b a s i c a l l yt e l ly o uw h a tw em e a nb ya n“ i d e a ls p r i n g ”i np h y s ics: it is
as p r i n gt h a tp u l l s( i fs t r e t c h e d )o rp u s h e s( i fc o m p r e s s e d )with a force that is proportional to the
change from its equilibrium length. Thus, if you fasten one end of the spring atx =0 ,a n ds t r e t c h
it or compress it so that the other end is atx,t h es p r i n gw i l lr e s p o n db ye x e r t i n gaf o r c e
Fspr = −k(x − x0)( 6 . 2 1 )
As you can see, this is negative ifx>x 0 > 0( s p r i n gs t r e t c h e d ,p u l l i n gf o r c e )a n dp o s i t i v ei fx<x 0
(spring compressed, pushing force). In fact, the spring exerts an equal (in magnitude) and opposite
(in direction) force at the other end (the one attached to thewall), so Eq. (6.21)o n l yg i v e st h e",University Physics I Classical Mechanics.pdf
"(in direction) force at the other end (the one attached to thewall), so Eq. (6.21)o n l yg i v e st h e
correct sign of the force at the end that is denoted by the coordinate value x.E q u a t i o n s (6.19)
and (6.20)a r eab i tc l e a r e ri nt h i sr e s p e c t : E q .(6.19)g i v e st h ec o r r e c ts i g no ft h ef o r c ea tp o i n t
x1,a n dE q .(6.20)t h ec o r r e c ts i g na tp o i n tx2.
1The result (6.18)g e n e r a l i z e st om o r ed i m e n s i o n s ,b u tt od oi tp r o p e r l yy o un eed vectors and partial derivative
notation, and I’m already bending the notational rules a little bit here...",University Physics I Classical Mechanics.pdf
"6.2. FORCES AND POTENTIAL ENERGY 119
Figure 6.1 shows, in black, all the forces exerted by a spring with one fixed end, according as to
whether it is relaxed, compressed, or stretched. I have assumed that it is pushed or pulled by a
hand (not shown) at the “free” end, hence the subscript “h”, whereas the subscript “w”s t a n d sf o r
“wall.” Note that the wall and the hand, in turn, exert equal and opposite forces on the spring,
shown in red in the figure.
xx0
Fs,w
spr
Fs,w
spr
Fs,h
spr
Fs,h
spr
Fh,s
c
Fh,s
c
Fw,s
c
Fw,s
c
Figure 6.1: Forces (in black) exertedby a spring with one end attached to a wall and the other pushed
or pulled by a hand (not shown). In every case the force is proportional to the change in the length of the
spring from its equilibrium, or relaxed, value, shown here asx0. For this figure I have set the proportionality
constant k =1 . T h ef o r c e se x e r t e don the spring, by the wall and by the hand, are shown in red.",University Physics I Classical Mechanics.pdf
"constant k =1 . T h ef o r c e se x e r t e don the spring, by the wall and by the hand, are shown in red.
Equation (6.21)i sg e n e r a l l yr e f e r r e dt oa sHooke’s law,a f t e rt h eB r i t i s hs c i e n t i s tR o b e r tH o o k e( a
contemporary of Newton). Of course, it is not a “law” at all, merely a useful approximation to the
way most springs behave as long as you do not stretch them or compress them too much2.
An o t eo nt h ew a yt h ef o r c e sh a v eb e e nl a b e l e di nF i g u r e6.1.Ih a v eu s e dt h eg e n e r i cs y m b o l“c”,
which stands for “contact,” to indicate the type of force exerted by the wall and the hand on the
spring. In fact, since each pair of forces (by the hand on the spring and by the spring on the hand,
for instance), at the point of contact, arises from one and thes a m ei n t e r a c t i o n ,Is h o u l dh a v eu s e d
the same “type” notation for both, but it is widespread practice to use a superscript like “spr” to",University Physics I Classical Mechanics.pdf
"the same “type” notation for both, but it is widespread practice to use a superscript like “spr” to
denote a force whose origin is, ultimately, a spring’s elasticity. This does not change the fact that
the spring force, at the point where it is applied, is indeed acontact force.
So, next, a word on “contact” forces. Basically, what we meanby that is forces that arise where
objects “touch,” and we mean this by opposition to what are called instead “field” forces (such as
gravity, or magnetic or electrostatic forces) which “act atad i s t a n c e . ” T h ed i s t i n c t i o ni sa c t u a l l y
2Assuming that youcan compress them! Some springs, such as slinkies, actually cannot be compressed because
their coils are already in contact when they are relaxed. Nevertheless, Eq. (6.21)w i l ls t i l la p p l ya p p r o x i m a t e l yt o
such a spring when it is stretched, that is, whenx>x 0.",University Physics I Classical Mechanics.pdf
"120 CHAPTER 6. INTERACTIONS, PART 2: FORCES
only meaningful at the macroscopic level, since at the microscopic level objects neverreally touch,
and all forces are field forces, it is just that some are “long range” and some are “short range.” For
our purposes, really, the word “contact” will just be a convenient, catch-all sort of moniker that we
will use to label the force vectors when nothing more specificwill do.
6.3 Forces not derived from a potential energy
As we have seen in the previous section, for interactions thata r ea s s o c i a t e dw i t hap o t e n t i a le n e r g y ,
we are always able to determine the forces from the potentialenergy by simple differentiation. This
means that we do not have to rely exclusively on an equation ofthe typeF = ma,l i k e(6.4)o r(6.6),
to infer the value of a force from the observed acceleration; rather,we can work in reverse, and
predict the value of the acceleration (and from it all the subsequentmotion) from our knowledge
of the force.",University Physics I Classical Mechanics.pdf
"predict the value of the acceleration (and from it all the subsequentmotion) from our knowledge
of the force.
Ih a v es a i db e f o r et h a t ,o nam i c r o s c o p i cl e v e l ,a l lt h ei n t e ractions can be derived from potential
energies, yet at the macroscopic level this is not generallytrue: we have many kinds of interactions
for which the associated “stored” or converted energy cannot, in general, be written as a function
of the macroscopic position variables for the objects makingu pt h es y s t e m( b yw h i c hIm e a n ,
typically, the positions of their centers of mass). So what dow ed oi nt h o s ec a s e s ?
The forces of this type with which we shall deal this semesteractually fall into two different
categories: the ones that do not dissipate energy, and that we could,i nf a c t ,a s s o c i a t ew i t ha
potential energy if we wanted to3,a n dt h eo n e st h a td e fi n i t e l yd i s s i p a t ee n e r g ya n dn e e ds p e c ial",University Physics I Classical Mechanics.pdf
"potential energy if we wanted to3,a n dt h eo n e st h a td e fi n i t e l yd i s s i p a t ee n e r g ya n dn e e ds p e c ial
handling. The former category includes the normal force, tension, and the static friction force;
the second category includes the force of kinetic (or sliding) friction, and air resistance. A brief
description of all these forces, and the methods to deal withthem, follows.
6.3.1 Tensions
Tension is the force exerted by a stretched spring, and, similarly, byo b j e c t ss u c ha sc a b l e s ,r o p e s ,
and strings in response to a stretching force (or load) applied to them. It is ultimately an elastic
force, so, as I said above, we could in principle describe it byap o t e n t i a le n e r g y ,b u ti np r a c t i c e
cables, strings and the like are so stiff that it is often all right to neglect their change in length
altogether and assume thatno potential energy is, in fact, stored in them. The price we payfor",University Physics I Classical Mechanics.pdf
"altogether and assume thatno potential energy is, in fact, stored in them. The price we payfor
this simplification (and itis as i m p l i fi c a t i o n )i st h a tw ea r el e f tw i t h o u ta ni n d e p e n d e n tway to
determine the value of the tension in any specific case; we justh a v et oi n f e ri tf r o mt h ea c c e l e r a t i o n
3If we wanted to complicate our life, that is...",University Physics I Classical Mechanics.pdf
"6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 121
of the object on which it acts (since it is a reaction force, itcan assume any value as required to
adjust to any circumstance—up to the point where the rope snaps, anyway).
Thus, for instance, in the picture below, which shows two blocks connected by a rope over a pulley,
the tension force exerted by the rope on block 1 must equalm1a1,w h e r ea1 is the acceleration of
that block, provided there are no other horizontal forces (such as friction) acting on it. For the
hanging block, on the other hand, the net force is the sum of thet e n s i o no nt h eo t h e re n do ft h e
rope (pulling up) and gravity, pulling down. If we choose theupward direction as positive, we can
write Newton’s second law for the second block as
Ft
r,2 − m2g = m2a2 (6.22)
Two things need to be realized now. First, if the rope is inextensible, both blocks travel the same",University Physics I Classical Mechanics.pdf
"Ft
r,2 − m2g = m2a2 (6.22)
Two things need to be realized now. First, if the rope is inextensible, both blocks travel the same
distance in the same time, so their speeds are always the same,a n dh e n c et h emagnitude of their
accelerations will always be the same as well; only the sign may be different depending on which
direction we choose as positive. If we take to the right to be positive for the horizontal motion, we
will havea2 = −a1.I ’ mj u s tg o i n gt oc a l la1 = a,s ot h e na2 = −a.
1
2
Fr,1
t
Fr,2
t
FE,2
G
Fs,1
fr
a1
a2
Figure 6.2: Two blocks joined by a massless, inextensible strength threaded over a massless pulley. An
optional friction force (in red, wherefr could be eithers or k) is shown for use later, in the discussion in
subsection 3.3. In this subsection, however, it is assumed to be zero.
The second thing to note is that, if the rope’s mass is negligible, it will, like an ideal spring, pull",University Physics I Classical Mechanics.pdf
"The second thing to note is that, if the rope’s mass is negligible, it will, like an ideal spring, pull
with a force with the samemagnitude on both ends. With our specific choices (up and to the right
is positive), we then haveFt
r,2 = Ft
r,1,a n dI ’ mj u s tg o i n gt oc a l lt h i sq u a n t i t yFt.A l l t h i s y i e l d s ,",University Physics I Classical Mechanics.pdf
"122 CHAPTER 6. INTERACTIONS, PART 2: FORCES
then, the following two equations:
Ft = m1a
Ft − m2g = −m2a (6.23)
The system (6.23)c a nb ee a s i l ys o l v e dt og e t
a = m2g
m1 + m2
Ft = m1m2g
m1 + m2
(6.24)
6.3.2 Normal forces
Normal force is the reaction force with which a surface pushesb a c kw h e ni ti sb e i n gp u s h e do n .
Again, this works very much like an extremely stiff spring, this time under compression instead of
tension. And, again, we will eschew the potential energy treatment by assuming that the surface’s
actual displacement is entirely negligible, and we will justc a l c u l a t et h ev a l u eo fFn as whatever is
needed in order to make Newton’s second law work. Note that this force will always be perpendicular
to the surface, by definition (the word “normal” means “perpendicular” here); the task of dealing
with a sideways push on the surface will be delegated to the static friction force, to be covered
next.",University Physics I Classical Mechanics.pdf
"with a sideways push on the surface will be delegated to the static friction force, to be covered
next.
If I am just standing on the floor and not falling through it, then e tv e r t i c a lf o r c ea c t i n go nm e
must be zero. The force of gravity on me ismg downwards, and so the upwards normal force must
match this value, so for this situationFn = mg.B u t d o n ’ t g e t t o o a t t a c h e d t o t h e n o t i o n t h a t
the normal force must always be equal tomg,s i n c et h i sw i l lo f t e nn o tb et h ec a s e . I m a g i n e ,f o r
instance, a person standing inside an elevator at the time itis accelerating upwards. With the
upwards direction as positive, Newton’s second law for the person reads
Fn − mg = ma (6.25)
and therefore for this situation
Fn = mg + ma (6.26)
If you were weighing yourself on a bathroom scale in the elevator, this is the upwards force that the
bathroom scale would have to exert on you, and it would do thatby compressing a spring inside,",University Physics I Classical Mechanics.pdf
"bathroom scale would have to exert on you, and it would do thatby compressing a spring inside,
and it would record the “extra” compression (beyond that required by your actual weight,mg)
as extra weight. Conversely, if the elevator were accelerating downward, the scale would record
you as being lighter. In the extreme case in which the cable ofthe elevator broke and you, the
elevator and the scale ended up (briefly, before the emergencyb r a k ec a u g h to n )i nf r e ef a l l ,y o u
would all be falling with the same acceleration, you would notb ep u s h i n gd o w no nt h es c a l ea ta l l ,
and its normal force as well as your recorded weight would be zero. This is ultimately the reason",University Physics I Classical Mechanics.pdf
"6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 123
for the apparent weightlessness experienced by the astronauts in the space station, where the force
of gravity is, in fact, not very much smaller than on the surface of the earth. (We will return to
this effect after we have a good grip on two-dimensional, and inparticular circular, motion.)
6.3.3 Static and kinetic friction forces
The static friction force is a force that prevents two surfaces in contact from slipping relative to
each other. It is an extremely useful force, since we would notb ea b l et od r i v eac a r ,o rr i d ea
bicycle, or even walk, without it—as we know from experience,i fw eh a v ee v e rt r i e dt od oa n yo f
those things on a low-friction surface (such as a sheet of ice).
The science behind friction (known technically astribology)i sa c t u a l l yn o tv e r ys i m p l ea ta l l ,a n d
it is of great current interest for many reasons—whether theultimate goal is to develop ways to",University Physics I Classical Mechanics.pdf
"it is of great current interest for many reasons—whether theultimate goal is to develop ways to
reduce friction or to increase it. On an elementary level, weare all aware of the fact that even a
surface that looks smooth on a macroscopic scale will actually exhibit irregularities, such as ridges
and valleys, under a microscope. It makes sense, then, that when two such surfaces are pressed
together, the bumps on one of them will hit, and be held in placeb y ,t h eb u m p so nt h eo t h e ro n e ,
and that will prevent sliding until and unless a sufficient force is applied to temporarily “flatten”
the bumps enough to allow the thing to move4.
As long as this does not happen, that is, as long as the surfacesd onot slide relative to each other,
we say we are dealing with thestatic friction force, which is, at least approximately, an elasticf o r c e
that does not dissipate energy: the small distortion of the “bumps” on the surfaces that takes place",University Physics I Classical Mechanics.pdf
"that does not dissipate energy: the small distortion of the “bumps” on the surfaces that takes place
when you push on them typically happens slowly enough, and issmall enough, to be reversible, so
that when you stop pushing the two surfaces just go back to their initial state. This is no longer
the case once the surfaces start sliding relative to each other. At that point the character of the
friction force changes, and we have to deal with thesliding,o rkinetic friction force, as I will explain
below.
The static friction force is also, like tension and the normalf o r c e ,ar e a c t i o nf o r c et h a tw i l la d j u s t
itself, within limits, to take any value required to preventslippage in a given circumstance. Hence,
its actual value in a particular situation cannot really be ascertained until the other relevant forces—
the other forces pushing or pulling on the object—are known.",University Physics I Classical Mechanics.pdf
"the other forces pushing or pulling on the object—are known.
For instance, for the system in Figure6.2,i m a g i n et h e r ei saf o r c eo fs t a t i cf r i c t i o nb e t w e e nb l o c k
1a n dt h es u r f a c eo nw h i c hi tr e s t s ,s u ffi c i e n t l yl a r g et ok e e pitf r o ms l i d i n ga l t o g e t h e r . H o wl a r g e
4This picture based, essentially, on classical physics, leaves out an atomic-scale effect that may be important
in some cases, which is the formation of weak bonds between thea t o m so fb o t hs u r f a c e s ,r e s u l t i n gi na na c t u a l
“adhesive” force. This is, for instance, how geckos can run upv e r t i c a lw a l l s . F o ro u rp u r p o s e s ,h o w e v e r ,t h ec l a s s i c a l
picture (of small ridges and valleys bumping into each other)w i l ls u ffi c et oq u a l i t a t i v e l yu n d e r s t a n da l lt h ee x a m p l e s
we will cover this semester.",University Physics I Classical Mechanics.pdf
"124 CHAPTER 6. INTERACTIONS, PART 2: FORCES
does this have to be? If there is no acceleration (a =0 ) ,t h ee q u i v a l e n to fs y s t e m(6.23)w i l lb e
Fs
s,1 + Ft =0
Ft − m2g =0 ( 6 . 2 7 )
where Fs
s,1 is the force of static friction exerted by the surface on block1 ,a n dw ea r eg o i n gt o
let the math tell us what sign it is supposed to have. Solving the system (6.27)w ej u s tg e tt h e
condition
Fs
s,1 = −m2g (6.28)
so this is how largeFs
s,1 has to be in order to keep the whole system from moving in this case.
There is an empirical formula that tells us approximately howl a r g et h ef o r c eo fs t a t i cf r i c t i o ncan
get in a given situation. The idea behind it is that, microscopically, the surfaces are in contact only
near the top of their respective ridges. If you press them together harder, some of the ridges get
flattened and the effective contact area increases; this in turnm a k e st h es u r f a c e sm o r er e s i s t a n tt o",University Physics I Classical Mechanics.pdf
"flattened and the effective contact area increases; this in turnm a k e st h es u r f a c e sm o r er e s i s t a n tt o
slippage. A direct measure of how strongly the two surfaces press against each other is, actually,
just the normal force they exert on each other. So, in general,w ee x p e c tt h em a x i m u mf o r c et h a t
static friction will be able to exert to be proportional to the normal force between the surfaces:
⏐⏐Fs
s1,s2
⏐⏐
max = µs
⏐⏐Fn
s1,s2
⏐⏐ (6.29)
where s1a n ds2j u s tm e a n“ s u r f a c e1 ”a n d“ s u r f a c e2 , ”r e s p e c t i v e l y ,a n dt hen u m b e rµs is known
as thecoefficient of static friction:i ti sa t a b u l a t e dq u a n t i t yt h a ti sd e t e r m i n e d e x p e r i m e n t a lly, by
testing the slippage of different surfaces against each otherunder different loads.
In our example, the normal force exerted by the surface on block 1 has to be equal tom1g,s i n c e",University Physics I Classical Mechanics.pdf
"In our example, the normal force exerted by the surface on block 1 has to be equal tom1g,s i n c e
there is no vertical acceleration for that block, and so the maximum value thatFs may have in
this case isµsm1g,w h a t e v e rµs might happen to be. In fact, this setup would give us a way to
determine µs for these two surfaces: start with a small value ofm2,a n dg r a d u a l l yi n c r e a s ei tu n t i l
the system starts moving. At that point we will know thatm2g has just exceeded the maximum
possible value of |Fs
12|,n a m e l y ,µsm1g,a n ds oµs =( m2)max/m1,w h e r e(m2)max is the largest
mass we can hang before the system starts moving.
By contrast with all of the above, thekinetic friction force, which always acts so as to oppose the
relative motion of the two surfaces when theyare actually slipping, is not elastic, it is definitely
dissipative, and, most interestingly, it is alsonot much of a reactive force, meaning that its value",University Physics I Classical Mechanics.pdf
"dissipative, and, most interestingly, it is alsonot much of a reactive force, meaning that its value
can be approximately predicted for any given circumstance,and does not depend much on things
such as how fast the surfaces are actually moving relative toeach other. It does depend on how
hard the surfaces are pressing against each other, as quantified by the normal force, and on another
tabulated quantity known as thecoefficient of kinetic friction:
⏐⏐⏐Fk
s1,s2
⏐⏐⏐= µk
⏐⏐Fn
s1,s2
⏐⏐ (6.30)",University Physics I Classical Mechanics.pdf
"6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 125
Note that, unlike for static friction, this isnot the maximum possible value of|Fk|,b u ti t sactual
value; so if we knowFn (and µk)w ek n o wFk without having to solve any other equations (its sign
does depend on the direction of motion, of course). The coefficient µk is typically a little smaller
than µs,r e fl e c t i n gt h ef a c tt h a to n c ey o ug e ts o m e t h i n gy o uh a v eb e e npushing on to move, keeping
it in motion with constant velocity usually does not requirethe same amount of force.
To finish off with our example in Figure 2, suppose the systemis moving, and there is a kinetic
friction forceFk
s,1 between block 1 and the surface. The equations (6.23)t h e nh a v et ob ec h a n g e d
to
Ft − µkm1g = m1a
Ft − m2g = −m2a (6.31)
and the solution now is
a = m2 − µkm1
m1 + m2
g
Ft = m1m2(1 +µk)
m1 + m2
g (6.32)
You may ask, why does kinetic friction dissipate energy? A qualitative answer is that, as the surfaces",University Physics I Classical Mechanics.pdf
"m1 + m2
g
Ft = m1m2(1 +µk)
m1 + m2
g (6.32)
You may ask, why does kinetic friction dissipate energy? A qualitative answer is that, as the surfaces
slide past each other, their small (sometimes microscopic)ridges are constantly “bumping” into
each other; so you have lots of microscopic collisions happening all the time, and they cannot all
be perfectly elastic. So mechanical energy is being “lost.”In fact, it is primarily being converted
to thermal energy, as you can verify experimentally: this iswhy you rub your hands together to
get warm, for instance. More dramatically, this is how some people (those who really know what
they are doing!) can actually start a fire by rubbing sticks together.
6.3.4 Air resistance
Air resistance is an instance of fluid resistance ordrag,af o r c et h a to p p o s e st h em o t i o no fa no b j e c t
through a fluid. Microscopically, you can think of it as beingdue to the constant collisions of the",University Physics I Classical Mechanics.pdf
"through a fluid. Microscopically, you can think of it as beingdue to the constant collisions of the
object with the air molecules, as it cleaves its way through the air. As a result of these collisions,
some of its momentum is transferred to the air, as well as someof its kinetic energy, which ends up
as thermal energy (as in the case of kinetic friction discussed above). The very high temperatures
that air resistance can generate can be seen, in a particularly dramatic way, on the re-entry of
spacecraft into the atmosphere.
Unlike kinetic friction between solid surfaces, the fluid drag force does depend on the velocity of
the object (relative to the fluid), as well as on a number of other factors having to do with the
object’s shape and the fluid’s density and viscosity. Very roughly speaking, for low velocities the",University Physics I Classical Mechanics.pdf
"126 CHAPTER 6. INTERACTIONS, PART 2: FORCES
drag force is proportional to the object’s speed, whereas forh i g hv e l o c i t i e si ti sp r o p o r t i o n a lt ot h e
square of the speed.
In principle, one can use the appropriate drag formula together with Newton’s second law to
calculate the effect of air resistance on a simple object throwno rd r o p p e d ;i np r a c t i c e ,t h i sr e q u i r e s
as o m e w h a tm o r ea d v a n c e dm a t ht h a nw ew i l lb eu s i n gt h i sc o u r se, and the formulas themselves
are complicated, so I will not introduce them here.
One aspect of air resistance that deserves to be mentioned iswhat is known as “terminal velocity”
(which I already introduced briefly in Section 2.3). Since airr e s i s t a n c ei n c r e a s e sw i t hs p e e d ,i f
you drop an object from a sufficiently great height, the upwardsd r a gf o r c eo ni tw i l li n c r e a s ea si t",University Physics I Classical Mechanics.pdf
"you drop an object from a sufficiently great height, the upwardsd r a gf o r c eo ni tw i l li n c r e a s ea si t
accelerates, until at some point it will become as large as thed o w n w a r df o r c eo fg r a v i t y . A tt h a t
point, the net force on the object is zero, so it stops accelerating, and from that point on it continues
to fall with constant velocity. When the Greek philosopher Aristotle was trying to figure out the
motion of falling bodies, he reasoned that, since air was justa n o t h e rfl u i d ,h ec o u l ds l o wd o w nt h e
fall (in order to study it better) without changing the physics by dropping objects in liquids instead
of air. The problem with this approach, though, is that terminal velocity is reached much faster in a
liquid than in air, so Aristotle missed entirely the early stage of approximately constant acceleration,
and concluded (wrongly) that the natural way all objects fellw a sw i t hc o n s t a n tv e l o c i t y . I tt o o k",University Physics I Classical Mechanics.pdf
"and concluded (wrongly) that the natural way all objects fellw a sw i t hc o n s t a n tv e l o c i t y . I tt o o k
almost two thousand years until Galileo disproved that notion by coming up with a better method
to slow down the falling motion—namely, by using inclined planes.
6.4 Free-body diagrams
As Figure6.1 shows, trying to draw every single force acting on every single object can very quickly
become pretty messy. And anyway, this is not usually what we need: what we need is to separate
cleanly all the forces acting on any given object, one objectat a time, so we can apply Newton’s
second law,Fnet = ma,t oe a c ho b j e c ti n d i v i d u a l l y .
In order to accomplish this, we use what are known asfree-body diagrams.I n a f r e e - b o d y d i a g r a m ,
ap o t e n t i a l l yv e r yc o m p l i c a t e do b j e c ti sr e p l a c e ds y m b o l i cally by a dot or a small circle, and all",University Physics I Classical Mechanics.pdf
"ap o t e n t i a l l yv e r yc o m p l i c a t e do b j e c ti sr e p l a c e ds y m b o l i cally by a dot or a small circle, and all
the forces acting on the object are drawn (approximately to scale and properly labeled) as acting
on the dot. Regardless of whether a force is a pulling or pushing force, the convention is to always
draw itas a vector that originates at the dot.I f t h e s y s t e mi s a c c e l e r a t i n g ,i t i sa l s o a g o o d i d e a t o
indicate the acceleration’s direction also somewhere on thed i a g r a m .
The figure below (next page) shows, as an example, a free-bodydiagram for block 1 in Figure6.2,
in the presence of both a nonzero acceleration and a kinetic friction force. The diagram includes
all the forces, even gravity and the normal force, which wereleft out of the picture in Figure6.2.",University Physics I Classical Mechanics.pdf
"6.5. IN SUMMARY 127
Fr,1
t
Fs,1
n
FE,1
G
Fs,1
k
a
Figure 6.3: Free-body diagram for block 1 in Figure6.2, with the friction force adjusted so as to be
compatible with a nonzero acceleration to the right.
Note that I have drawnFn and the force of gravityFG
E,1 as having the same magnitude, since
there is no vertical acceleration for that block. If I know thev a l u eo fµk,Is h o u l da l s ot r yt od r a w
Fk = µkFn approximately to scale with the other two forces. Then, sinceIk n o wt h a tt h e r ei sa n
acceleration to the right, I need to drawFt greater thanFk,s i n c et h en e tf o r c eo nt h eb l o c km u s t
be to the right as well. And, if I were drawing a free-body diagram for block 2, I would have to
make sure that I drew its weight,FG
E,2,a sb e i n gg r e a t e ri nm a g n i t u d et h a nFt,s i n c et h en e tf o r c e
on that block needs to be downwards.
6.5 In summary
1. Whenever two objects interact, they exertforces on each other that are equal in magnitude",University Physics I Classical Mechanics.pdf
"on that block needs to be downwards.
6.5 In summary
1. Whenever two objects interact, they exertforces on each other that are equal in magnitude
and opposite in direction (Newton’s 3rd law).
2. Forces are vectors, and they are additive. The total (or net) force on an object or system is
equal to the rate of change of its total momentum (Newton’s 2nd law). If the system’s mass is
constant, this can be written asFext,all = Macm,w h e r eM is the system’s total mass andacm
is the acceleration of its center of mass. Only theexternal forces contribute to this equation;
the internal forces cancel out because of point 1 above.
3. For any interaction that can be derived from a potential energy function U(x1 − x2), the
force exerted by object 2 on object 1 is equal to−dU/dx1 (where the derivative is calculated
treating x2 as a constant), and vice-versa.
4. The force of gravity on an object near the surface of the earth is known as the object’sweight,",University Physics I Classical Mechanics.pdf
"treating x2 as a constant), and vice-versa.
4. The force of gravity on an object near the surface of the earth is known as the object’sweight,
and it is equal (in magnitude) tomg,w h e r em is the object’s inertial mass.
5. An ideal spring whose relaxed length isx0,w h e ns t r e t c h e do rc o m p r e s s e dt oal e n g t hx,e x e r t s
ap u l l i n go rp u s h i n gf o r c e ,r e s p e c t i v e l y ,a tb o t he n d s ,w i t hmagnitude k|x − x0|,w h e r ek is
called the spring constant.",University Physics I Classical Mechanics.pdf
"128 CHAPTER 6. INTERACTIONS, PART 2: FORCES
6. When dealing with macroscopic objects we introduce several “constraint” forces whose values
need to be determined from the accelerations through Newton’s second law: thetension Ft
in ropes, strings or cables; thenormal force Fn exerted by a surface in response to applied
pressure; and the static friction forceFs that prevents surfaces from slipping past each other.
7. The maximum possible value of the static friction force isµs|Fn|,w h e r eµs is the coefficient
of static friction.
8. The force of sliding or kinetic friction,Fk,a p p e a r sw h e nt w os u r f a c e sa r es l i d i n gp a s te a c h
other. Its magnitude isµk|Fn| (µk is the coefficient of static friction), and its sign is such as
to oppose the sliding motion. Unlike the forces in 6 above, itis a dissipative force.
9. A free-body diagramis a way to depictall (and only)t h ef o r c e sacting on an object. The",University Physics I Classical Mechanics.pdf
"9. A free-body diagramis a way to depictall (and only)t h ef o r c e sacting on an object. The
object should be represented as a small circle or dot. The forces should all be drawn as vectors
originating on the dot, with their directions correctly shown and their lengths approximately
to scale. The acceleration of the object should also be indicated elsewhere in the picture. The
forces should be labeled like this:Ftype
by,on.",University Physics I Classical Mechanics.pdf
"6.6. EXAMPLES 129
6.6 Examples
6.6.1 Dropping an object on a weighing scale
(Short version) Suppose you drop a 5-kg object on a spring scale from a height of 1 m. If the spring
constant isk =2 0, 000 N/m, what will the scale read?
(Long version) OK, let’s break that up into parts. Suppose that a spring scale is just a platform (of
negligible mass) sitting on top of a spring. If you put an object of massm on top of it, the spring
compresses so that (in equilibrium) it exerts an upwards force that matches that of gravity.
(a) If the spring constant isk and the object’s mass ism and the whole system is at rest, what
distance is the spring compressed?
(b) If you drop the object from a heighth,w h a ti st h e( i n s t a n t a n e o u s )maximum compression of
the spring as the object is brought to a momentary rest? (Thispart is anenergy problem! Assume
that h is much greater than the actual compression of the spring, soyou can neglect that when",University Physics I Classical Mechanics.pdf
"that h is much greater than the actual compression of the spring, soyou can neglect that when
calculating the change in gravitational potential energy.)
(c) What mass would give you that same compression, if you weret op l a c ei tg e n t l yo nt h es c a l e ,
and wait until all the oscillations died down?
(d) OK, now answer the question at the top!
Solution
(a) The forces acting on the object sitting at rest on the platform are the force of gravity,FG
E,o =
−mg,a n dt h en o r m a lf o r c ed u et ot h ep l a t f o r m ,Fn
p,o.T h i s l a s t f o r c e i s e q u a l , i n m a g n i t u d e , t o
the force exerted on the platform by the spring (it has to be, because the platform itself is being
pushed down by a forceFn
o,p = −Fn
p,o,a n dt h i sh a st ob eb a l a n c e db yt h es p r i n gf o r c e ) . T h i sm e a n s
we can, for practical purposes, pretend the platform is not there and just set the upwards force on
the object equal to the spring force,Fspr",University Physics I Classical Mechanics.pdf
"we can, for practical purposes, pretend the platform is not there and just set the upwards force on
the object equal to the spring force,Fspr
s,p = −k(x − x0). So, Newton’s second law gives
Fnet = FG
E,o + Fspr
s,p = ma =0 ( 6 . 3 3 )
For a compressed spring,x − x0 is negative, and we can just letd = x0 − x be the distance the
spring is compressed. Then Eq. (6.33)g i v e s
−mg + kd =0
so
d = mg/k (6.34)
when you just set an object on the scale and let it come to rest.
(b) This part, as the problem says, is a conservation of energys i t u a t i o n . T h es y s t e mf o r m e db y
the spring, the object and the earth starts out with some gravitational potential energy, and ends",University Physics I Classical Mechanics.pdf
"130 CHAPTER 6. INTERACTIONS, PART 2: FORCES
up, with the object momentarily at rest, with only spring potential energy:
UG
i + Uspr
i = UG
f + Uspr
f
mgyi +0= mgyf + 1
2kd2
max (6.35)
where I have used the subscript “max” on the compression distance to distinguish it from what
Ic a l c u l a t e di np a r t( a )( t h i sk i n do fm a k e ss e n s ea l s ob e c a u set h es c a l ei sg o i n gt os w i n gu pa n d
down, and we want only the maximum compression, which will give us the largest reading). The
problem said to ignore the compression when calculating thechange in UG,m e a n i n gt h a t ,i fw e
measure height from the top of the scale,yi = h and yf =0 . T h e n ,s o l v i n gE q .(6.35)f o rdmax,w e
get
dmax =
√
2mgh
k (6.36)
(c) For this part, let us rewrite Eq. (6.34)a s meq = kdmax/g,w h e r emeq is the “equivalent” mass
that you would have to place on the scale (gently) to get the same reading as in part (b). Using
then Eq. (6.36),
meq = k
g
√
2mgh
k =
√
2mkh
g (6.37)",University Physics I Classical Mechanics.pdf
"then Eq. (6.36),
meq = k
g
√
2mgh
k =
√
2mkh
g (6.37)
(d) Now we can substitute the values given:m =5 k g ,h =1 m ,k =2 0, 000 N/m. The result is
meq =1 4 3k g .
(Note: if you found the purely algebraic treatment above confusing, try substituting numerical
values in Eqs. (6.34)a n d(6.36). The first equation tells you that if you just place the 5-kg mass
on the scale it will compress a distanced =2 .45 mm. The second tells you that if you drop it it
will compress the spring a distancedmax =7 0 m m ,a b o u t2 8.6t i m e sm o r e ,w h i c hc o r r e s p o n d st o
an “equivalent mass” 28.6t i m e sg r e a t e rt h a n5 k g ,w h i c hi st os a y ,1 4 3 k g .N o t ea l s ot h at1 4 3 k gi s
an equivalent weight of 309 pounds, so if you want to try this onab a t h r o o ms c a l eI ’ da d v i s ey o u
to use smaller weights and drop them from a much smaller height!)
6.6.2 Speeding up and slowing down",University Physics I Classical Mechanics.pdf
"to use smaller weights and drop them from a much smaller height!)
6.6.2 Speeding up and slowing down
(a) A 1400-kg car, starting from rest, accelerates to a speedof 30 mph in 10 s. What is the force
on the car (assumed constant) over this period of time?
(b) Where does this force comes from? That is, what is the (external) object that exerts this force
on the car, and what is the nature of this force?
(c) Draw a free-body diagram for the car. Indicate the direction of motion, and the direction of
the acceleration. (d) Now assume that the driver, travelingat 30 mph, sees a red light ahead and",University Physics I Classical Mechanics.pdf
"6.6. EXAMPLES 131
pushes on the brake pedal. Assume that the coefficient of staticf r i c t i o nb e t w e e nt h et i r e sa n dt h e
road is µs =0 .7, and that the wheels don’t “lock”: that is to say, they continue rolling without
slipping on the road as they slow down. What is the car’s minimum stopping distance?
(e) Draw a free-body diagram of the car for the situation in (d). Again indicate the direction of
motion, and the direction of the acceleration. (f) Now assumet h a tt h ed r i v e ra g a i nw a n t st os t o p
as in part (c), but he presses on the brakes too hard, so the wheels lock, and, moreover, the road
is wet, and the coefficient of kinetic friction is onlyµk =0 .2. What is the distance the car travels
now before coming to a stop?
Solution
(a) First, let us convert 30 mph to meters per second. There are1 , 609 meters to a mile, and 3, 600
seconds to an hour, so 30 mph = 10× 1609/3600 m/s = 13.4m / s .",University Physics I Classical Mechanics.pdf
"seconds to an hour, so 30 mph = 10× 1609/3600 m/s = 13.4m / s .
Next, for constant acceleration, we can use Eq. (2.4): ∆v = a∆t.S o l v i n gf o ra,
a = ∆v
∆t = 13.4m / s
10 s =1 .34 m
s2
Finally, sinceF = ma,w eh a v e
F = ma =1 4 0 0k g× 1.34 m
s2 =1 8 8 0N
(b) The force must be provided by the road, which is the only thing external to the car that is in
contact with it. The force is, in fact, the force ofstatic friction between the car and the tires. As
explained in the chapter, this is a reaction force (the tirespush on the road, and the road pushes
back). It is static friction because the tires are not slipping relative to the road. In fact, we will
see in Chapter 9 that the point of the tire in contact with the road has an instantaneous velocity
of zero (see Figure9.8).
(c) This is the free-body diagram. Note the force
of static friction pointingforward,i nt h ed i r e c t i o n
of the acceleration. The forces have been drawn
to scale.
Fr,c
s
Fr,c
n
FE,c
G
a v",University Physics I Classical Mechanics.pdf
"132 CHAPTER 6. INTERACTIONS, PART 2: FORCES
(d) This is the opposite of part (a): the driver now relies on the force of static friction toslow down
the car. The shortest stopping distance will correspond to the largest (in magnitude) acceleration,
as per our old friend, Eq. (2.10):
v2
f − v2
i =2 a∆x (6.38)
In turn, the largest acceleration will correspond to the largest force. As explained in the chapter,
the static friction force cannot exceedµsFn (Eq. (6.29)). So, we have
Fs
max = µsFn = µsmg
since, in this case, we expect the normal force to be equal to the force of gravity. Then
|amax| = Fs
max
m = µsmg
m = µsg
We can substitute this into Eq. (6.38)w i t han e g a t i v es i g n ,s i n c et h ea c c e l e r a t i o na c t si nt h eo p posite
direction to the motion (and we are implicitly taking the direction of motion to be positive). Also
note that the final velocity we want is zero,vf =0 . W eg e t
−v2
i =2 a∆x = −2µsg∆x
From here, we can solve for ∆x:
∆x = v2
i",University Physics I Classical Mechanics.pdf
"note that the final velocity we want is zero,vf =0 . W eg e t
−v2
i =2 a∆x = −2µsg∆x
From here, we can solve for ∆x:
∆x = v2
i
2µsg = (13.4m / s )2
2 × 0.7 × 9.81 m/s2 =1 3.1m
(e) Here is the free-body diagram. The interesting
feature is that the force of static friction has re-
versed direction relative to parts (a)–(c). It is also
much larger than before. (The forces are again to
scale.)
Fr,c
s
Fr,c
n
FE,c
G
a v
(f) The math for this part is basically identical to that in part (d). The difference, physically, is
that now you are dealing with the force ofkinetic (or “sliding”) friction, and that is always given
by Fk = µkFn (this is not an upper limit, it’s just whatFk is). So we havea = −Fk/m = −µkg,
and, just as before (but withµk replacing µs),
∆x = v2
i
2µkg = (13.4m / s )2
2 × 0.2 × 9.81 m/s2 =4 5.8m",University Physics I Classical Mechanics.pdf
"6.6. EXAMPLES 133
This is a huge distance, close to half a football field! If thesen u m b e r sa r ea c c u r a t e ,y o uc a ns e e
that locking your brakes in the rain can have some pretty bad consequences.",University Physics I Classical Mechanics.pdf
"134 CHAPTER 6. INTERACTIONS, PART 2: FORCES
6.7 Problems
Problem 1
(a) Draw a free-body diagram for the skydiver in Problem 4 of Chapter 5.
(b) What is the magnitude of the air drag force on the skydiver,a f t e rh er e a c h e st e r m i n a ls p e e d ?
Problem 2
Ab o o ki ss e n ts l i d i n ga l o n gat a b l ew i t ha ni n i t i a lv e l o c i t yof2 m / s . I ts l i d e sf o r1.5m be f o r e
coming to a stop. What is the coefficient of kinetic friction between the book and the table?
Problem 3
You are pulling on a block of mass 4 kg that is attached, via a rope of negligible mass, to another
block, of mass 6 kg. The coefficient of kinetic friction betweent h eb l o c k sa n dt h es u r f a c eo nw h i c h
they are sliding isµk.Y o u fi n d t h a t w h e n y o u a p p l y a f o r c e o f 2 0 N , t h e w h o l e t h i n g m o ves at
constant velocity.
(a) Draw a free-body diagram for each of the two blocks
(b) What is the coefficient of kinetic friction between the blocks and the surface?",University Physics I Classical Mechanics.pdf
"(a) Draw a free-body diagram for each of the two blocks
(b) What is the coefficient of kinetic friction between the blocks and the surface?
(c) What is the tension in the rope?
Problem 4
Ab o xo fm a s s2 k gi ss i t t i n go nt o po fas l e do fm a s s5 k g ,w h i c hi sresting on top of a frictionless
surface (ice).
(a) What is the normal force exerted by the box on the sled? (Andb yt h es l e db a c ko nt h eb o x . )
(b) If you pull on the sled with a force of 35 N, how large does thec o e ffi c i e n to fs t a t i cf r i c t i o n ,
µs,b e t w e e nt h eb o xa n dt h es l e dh a v et ob e ,i no r d e rf o rt h eb o xt omove with the sled? Draw
free-body diagrams for the box and for the sled under this assumption (that they move together).
(c) Suppose thatµs is less than the value you got in part (b), so the box starts to slide back (relative
to the sled). If the coefficient of kinetic frictionµk between the box and the sled is 0.15, what is",University Physics I Classical Mechanics.pdf
"to the sled). If the coefficient of kinetic frictionµk between the box and the sled is 0.15, what is
the acceleration of the sled, and what is the acceleration ofthe box, while they are sliding relative
to each other (so, before the box falls off, and while you are still pulling with a 35-N force)? Draw
again the free-body diagrams appropriate to this situation.
Problem 5
You stick two ob jects together, one with a mass of 10 kg and onewith a mass of 5 kg, using a glue
that is supposed to be able to provide up to 19 N of force beforeit fails. Suppose you then pull on
the 10 kg block with a force of 30 N.
(a) What is the acceleration of the whole system?
(b) What is the force exerted on the 5 kg block, and where does itc o m ef r o m ?D o e st h eg l u eh o l d ?
(c) Now suppose you pull on the 5 kg block instead with the sameforce. Does the glue hold this
time?",University Physics I Classical Mechanics.pdf
"6.7. PROBLEMS 135
Problem 6
Draw a free-body diagram for a 70-kg person standing in an elevator carrying a 15-kg backpack (do
not consider the backpack a part of the person!). (a) if the elevator is not moving, and (b) if the
elevator is accelerating downwards at 2 m/s2.I n e a c h c a s e , w h a t i s t h e m a g n i t u d e o f t h e n o r m a l
force exerted on the person by the floor?",University Physics I Classical Mechanics.pdf
"136 CHAPTER 6. INTERACTIONS, PART 2: FORCES",University Physics I Classical Mechanics.pdf
"Chapter 7
Impulse, Work and Power
7.1 Introduction: work and impulse
In physics, “work” (or “doing work”) is what we call the process through which a force changes
the energy of an object it acts on (or the energy of a system to which the object belongs). It is,
therefore, a very technical term with a very specific meaningthat may seem counterintuitive at
times.
For instance, as it turns out, in order to change the energy ofan object on which it acts, the force
needs to be at least partly in line with the displacement of theo b j e c td u r i n gt h et i m ei ti sa c t i n g .
Af o r c et h a ti sp e r p e n d i c u l a rt ot h ed i s p l a c e m e n td o e sn ow o rk—it does not change the object’s
energy.
Imagine a satellite in a circular orbit around the earth. Theearth is constantly pulling on it with
af o r c e( g r a v i t y )d i r e c t e dt o w a r d st h ec e n t e ro ft h eo r b i ta tany given time. This force is always",University Physics I Classical Mechanics.pdf
"af o r c e( g r a v i t y )d i r e c t e dt o w a r d st h ec e n t e ro ft h eo r b i ta tany given time. This force is always
perpendicular to the displacement, which is along the orbit,a n ds oi td o e sn ow o r k : t h es a t e l l i t e
moves always at the same speed, so its kinetic energy does notchange.
The force does change the satellite’s momentum, however: it keeps bendingthe trajectory, and
therefore changing the direction (albeit not the magnitude)o ft h es a t e l l i t e ’ sm o m e n t u mv e c t o r . O f
course, it is obvious that a force must change an object’s momentum, because that is pretty much
how we defined force anyway. Recall Eq. (6.1) for the average force on an object: ⃗Fav =∆ ⃗p /∆t.
We can rearrange this to read
∆⃗p= ⃗Fav∆t (7.1)
For a constant force, the product of the force and the time overw h i c hi ti sa c t i n gi sc a l l e dt h e
137",University Physics I Classical Mechanics.pdf
"138 CHAPTER 7. IMPULSE, WORK AND POWER
impulse,u s u a l l yd e n o t e da s⃗J
⃗J = ⃗F ∆t (7.2)
Clearly, the impulse given by a force to an object is equal to the change in the object’s momentum
(by Eq. (7.1)), as long as it is the only force (or, alternatively, the netforce) acting on it. If the
force is not constant, we break up the time interval ∆t into smaller subintervals and add all the
pieces, pretty much as we did with Figure1.5 in Chapter 1 in order to calculate the displacement
for a variable velocity. Formally this results in an integral:
⃗J =
∫ tf
ti
⃗F(t) dt (7.3)
Graphically, thex component of the impulse is equal to the area under the curve of Fx versus time,
and similarly for the other components. You will get to see howi tw o r k si nal a be x p e r i m e n tt h i s
semester.
There is not a whole lot more to be said about impulse. The mainlesson to be learned from Eq. (7.1)
is that one can get a desired change in momentum—bring an object to a stop, for instance—either",University Physics I Classical Mechanics.pdf
"is that one can get a desired change in momentum—bring an object to a stop, for instance—either
by using a large force over a short time, or a smaller force overal o n g e rt i m e . I ti se a s yt os e e
how different circumstances may call for different strategies:sometimes you may want to make the
force as small as possible, if the object on which you are acting is particularly fragile; other times
you may just need to make the time as short as possible instead.
Of course, to bring something to a stop you not only need to remove its momentum, but also its
(kinetic) energy. If the former task takes time, the latter,it turns out, takesdistance.W o r k i s
am u c hr i c h e rs u b j e c tt h a ni m p u l s e ,n o to n l yb e c a u s e ,a sIh a vei n d i c a t e da b o v e ,t h ea c t u a lw o r k
done depends on the relative orientation of the force and displacement vectors, but also because",University Physics I Classical Mechanics.pdf
"done depends on the relative orientation of the force and displacement vectors, but also because
there is only one kind of momentum, but many different kinds of energy, and one of the things that
typically happens when work is done is theconversion of one type of energy into another.
So there is a lot of ground to cover, but we’ll start small, in the next section, with the simplest
kind of system, and the simplest kind of energy.
7.2 Work on a single particle
Consider a particle that undergoes a displacement ∆x while a constant forceF acts on it. In one
dimension, thework done by the force on the particle is defined by
W = F∆x (constant force) (7.4)
and it is positive if the force and the displacement have the same sign (that is, if they point in the
same direction), and negative otherwise.",University Physics I Classical Mechanics.pdf
"7.2. WORK ON A SINGLE PARTICLE 139
In three dimensions, the force will be a vector⃗F with components (Fx,F y,F z), and the displacement,
likewise, will be a vector ∆⃗rwith components (∆x, ∆y, ∆z). The work will be defined then as
W = Fx∆x + Fy∆y + Fz∆z (7.5)
This expression is an instance of what is known as thedot product (or inner product,o r scalar
product)o ft w ov e c t o r s .G i v e nt w ov e c t o r s⃗A and ⃗B,t h e i rd o tp r o d u c ti sd e fi n e d ,i nt e r m so ft h e i r
components, as
⃗A · ⃗B = AxBx + AyBy + AzBz (7.6)
This can also be expressed in terms of the vectors’ magnitudes, | ⃗A| and | ⃗B|,a n dt h ea n g l et h e y
make, in the following form:
⃗A · ⃗B = | ⃗A|| ⃗B| cos φ (7.7)
A
¶
B
Figure 7.1: Illustrating the angle φ to be used when calculating the dot product of two vectors by the
formula (7.7). One way to think of this formula is that you take the projection ofvector ⃗A onto vector ⃗B",University Physics I Classical Mechanics.pdf
"formula (7.7). One way to think of this formula is that you take the projection ofvector ⃗A onto vector ⃗B
(indicated here by the blue lines), which is equal to| ⃗A| cos φ, then multiply that by the length of⃗B (or
vice-versa, of course).
Figure 7.1 shows what I mean by the angleφ in this expression. The equality of the two definitions,
Eqs. (7.6)a n d(7.7), is proved in mathematics textbooks. The advantage of Eq. (7.7)i st h a ti ti s
independent of the choice of a system of coordinates.
Using the dot product notation, the work done by a constant force can be written as
W = ⃗F · ∆⃗r (7.8)
Equation (7.7)t h e ns h o w st h a t ,a sIm e n t i o n e di nt h ei n t r o d u c t i o n ,w h e nt hef o r c ei sp e r p e n d i c u l a r
to the displacement (φ =9 0◦)t h ew o r ki td o e si sz e r o . Y o uc a na l s os e et h i sd i r e c t l yf r o mEq. (7.5),
by choosing thex axis to point in the direction of the force (soFy = Fz =0 ) ,a n dt h ed i s p l a c e m e n t",University Physics I Classical Mechanics.pdf
"by choosing thex axis to point in the direction of the force (soFy = Fz =0 ) ,a n dt h ed i s p l a c e m e n t
to point along any of the other two axes (so ∆x =0 ) : t h er e s u l ti sW =0 .
If the force is not constant, again we follow the standard procedure of breaking up the total
displacement into pieces that are short enough that the forcem a yb et a k e nt ob ec o n s t a n to v e r",University Physics I Classical Mechanics.pdf
"140 CHAPTER 7. IMPULSE, WORK AND POWER
each of them, calculating all those (possibly very small) “pieces of work,” and adding them all
together. In one dimension, the final result can be expressedas the integral
W =
∫ xf
xi
F(x)dx (variable force) (7.9)
So the work is given by the “area” under theF-vs-x curve. In more dimensions, we have to write
ak i n do fm u l t i v a r i a b l ei n t e g r a lk n o w na saline integral.T h a t i sa d v a n c e dc a l c u l u s ,s o w e w i l ln o t
go there this semester.
7.2.1 Work done by the net force, and the Work-Energy Theorem
So much for the math and the definitions. Where does the energycome in? Let us suppose that
F is either the only force or thenet force on the particle—the sum of all the forces acting on
the particle. Again, for simplicity we will assume that it isconstant (does not change) while the
particle undergoes the displacement ∆x.H o w e v e r , n o w ∆x and Fnet are related: a constant net",University Physics I Classical Mechanics.pdf
"particle undergoes the displacement ∆x.H o w e v e r , n o w ∆x and Fnet are related: a constant net
force means a constant acceleration,a = Fnet/m,a n df o rc o n s t a n ta c c e l e r a t i o nw ek n o wt h ef o r m u l a
v2
f − v2
i =2 a∆x applies. Therefore, we can write
Wnet = Fnet∆x = ma∆x = m1
2
(
v2
f − v2
i
)
(7.10)
which is to say
Wnet =∆ K (7.11)
In words,the work done by the net force acting on a particle as it moves equals the change in the
particle’s kinetic energy in the course of its displacement.T h i s r e s u l t i s o f t e n r e f e r r e d t o a sthe
Work-Energy Theorem.
As you may have guessed from our calling it a “theorem,” the result (7.11)i sv e r yg e n e r a l . I th o l d s
in three dimensions, and it holds also when the force isn’t constant throughout the displacement—
you just have to use the correct equation to calculate the worki nt h o s ec a s e s . I tw o u l da p p l yt o",University Physics I Classical Mechanics.pdf
"you just have to use the correct equation to calculate the worki nt h o s ec a s e s . I tw o u l da p p l yt o
the work done by the net force on an extended object, also, provided it is OK to treat the extended
object as a particle—so basically, a rigid object that is moving as a whole and not doing anything
fancy such as spinning while doing so.
Another possible direction in which to generalize (7.11)m i g h tb ea sf o l l o w s . B yd e fi n i t i o n ,a
“particle” has no other kind of energy, besides (translational) kinetic energy. Also, and for the
same reason (namely, the absence of internal structure), ithas no “internal” forces—all the forces
acting on it are external. So—for this very simple system—wecould rephrase the result (7.11)b y
saying that the work done by the netexternal force acting on the system (the particle in this case)
is equal to the change in itstotal energy. It is in fact in this form that we will ultimately generalize",University Physics I Classical Mechanics.pdf
"is equal to the change in itstotal energy. It is in fact in this form that we will ultimately generalize
(7.11)t od e a lw i t ha r b i t r a r ys y s t e m s .",University Physics I Classical Mechanics.pdf
"7.3. THE “CENTER OF MASS WORK” 141
Before we go there, however, I would like to take a little detour to explore another “reasonable”
extension of the result (7.11), as well as its limitations.
7.3 The “center of mass work”
All the physics I used in order to derive the result (7.11)w a sF = ma,a n dt h ee x p r e s s i o nv2
f −v2
i =
2a∆x,w h i c ha p p l i e sw h e n e v e rw eh a v em o t i o nw i t hc o n s t a n ta c c e l eration. Now, we know that for
an arbitrary system, of total massM, Fext,net = Macm [see Eq. (6.11)]. That is enough, then, to
ensure that, ifFext,net is a constant, we will have
Fext,net∆xcm =∆ Kcm (7.12)
where Kcm,t h et r a n s l a t i o n a lk i n e t i ce n e r g y ,i s ,a su s u a l ,Kcm = 1
2 Mv2
cm,a n d∆xcm is the displace-
ment of the center of mass. The result (7.12)h o l d sf o ra na r b i t r a r ys y s t e m ,a sl o n ga sFext,net is
constant, and can be generalized by means of an integral (as inE q .(7.9)) when it is variable.",University Physics I Classical Mechanics.pdf
"constant, and can be generalized by means of an integral (as inE q .(7.9)) when it is variable.
So it seems that we could define the left-hand side of Eq. (7.12)a s“ t h ew o r kd o n eo nt h ec e n t e ro f
mass,” and take that as the natural generalization to a systemo ft h er e s u l t(7.11)f o rap a r t i c l e .
Most physicists would, in fact, be OK with that, but educatorsn o w a d a y sf r o w no nt h a ti d e a ,f o ra
couple of reasons.
First, it seems that it is essential to the notion of work thatone should multiply the force by the
displacement of the object on which it is acting.M o r e p r e c i s e l y , i n t h e d e fi n i t i o n (7.4), we want
the displacementof the point of applicationof the force1.B u t t h e r ea r e m a n ye x a m p l e so fs y s t e m s
where there is nothing at the precise location of the center ofm a s s ,a n dc e r t a i n l yn of o r c ea c t i n g
precisely there.",University Physics I Classical Mechanics.pdf
"where there is nothing at the precise location of the center ofm a s s ,a n dc e r t a i n l yn of o r c ea c t i n g
precisely there.
This is not necessarily a problem in the case of a rigid objectwhich is not doing anything funny, just
moving as a whole so that every part has the same displacement,b e c a u s et h e nt h ed i s p l a c e m e n to f
the center of mass would simply stand for the displacement ofany point at which an external force
might actually be applied. But for manydeformable systems, this would not be case. In fact, for
such systems one can usually show thatFext,net∆xcm is actuallynot the work done on the system
by the net external force. A simple example of such a system isshown below, in Figure7.2.
1As the name implies, this is the precise point at which the force is applied. For contact forces (other than friction;
see later), this is easily identified. For gravity, a sum overall the forces exerted on all the particles that make up the",University Physics I Classical Mechanics.pdf
"see later), this is easily identified. For gravity, a sum overall the forces exerted on all the particles that make up the
object may be shown to be equivalent to a single resultant force acting at a point called thecenter of gravity,w h i c h ,
for our purposes (objects in uniform or near-uniform gravitational fields) will be the same as the center of mass.",University Physics I Classical Mechanics.pdf
"142 CHAPTER 7. IMPULSE, WORK AND POWER
Fs,2
spr
Fs,1
spr
Fh,2
c
Fh,2
c
∆x1 ∆x2∆xcm
1
1
2
2
Figure 7.2: A system of two blocks connected by a spring. A constant externalf o r c e ,⃗Fc
h,2, is applied to the
block on the right. Initially the spring is relaxed, but as soon as block 2starts to move it stretches, pulling
back on block 2 and pulling forward on block 1. Because of the stretching of the spring, the displacements
∆x1,∆ xcm and ∆x2 are all different, and the work done by the external force,Fc
h,2∆x2, is different from
the “center of mass work”Fc
h,2∆xcm.
In this figure, the two blocks are connected by a spring, and thee x t e r n a lf o r c ei sa p p l i e dt ot h e
block on the right (block 2). If the blocks have the same mass,the center of mass of the system
is a point exactly halfway between them. If the spring startsin its relaxed state, it will stretch at
first, so that the center of mass will lag behind block 2, andFc
h,2∆x2,w h i c hi st h eq u a n t i t yt h a t",University Physics I Classical Mechanics.pdf
"first, so that the center of mass will lag behind block 2, andFc
h,2∆x2,w h i c hi st h eq u a n t i t yt h a t
we should properly call the “work done by the net external force” willnot be equal toFc
h,2∆xcm.
We find ourselves, therefore, with a very general and p otentially rather useful result, Eq. (7.12),
that looks a lot like it should be “the work done on the system byt h en e te x t e r n a lf o r c e ”b u t ,i n
fact, is that only sometimes. On the other hand, the result isso useful that simply referring to it
all the time by “Eq. (7.12)” will not do. I propose, therefore, to call it the “center ofmass work,”
in between quotation marks, just so we all know what we are talking about, and remember the
caveats that go with it.
We can now move to thereal theorem relating the work of the external forces on a system to
the change in its energy. What we have seen so far are really just straightforward applications",University Physics I Classical Mechanics.pdf
"the change in its energy. What we have seen so far are really just straightforward applications
of Newton’s second law. The main result coming up is deeper than that, since it involves also,
ultimately, the principle of conservation of energy.
7.4 Work done on a system by all the external forces
Consider the most general possible system, one that might contain any number of particles, with
possibly many forces, both internal and external, acting oneach of them. I will again, for simplicity,",University Physics I Classical Mechanics.pdf
"7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 143
start by considering what happens over a time interval so short that all the forces are approximately
constant (the final result will hold for arbitrarily long timei n t e r v a l s ,j u s tb ya d d i n g ,o ri n t e g r a t i n g ,
over many such short intervals). I will also work explicitlyonly the one-dimensional case, although
again that turns out to not be a real restriction.
Let then Wall,1 be the work done on particle 1 by all the forces acting on it,Wall,2 the work done
on particle 2, and so on. The total work is the sumWall,sys = Wall,1 + Wall,2 + ... .H o w e v e r ,
by the results of section 7.2, we haveWall,1 =∆ K1 (the change in kinetic energy of particle 1),
Wall,2 =∆ K2,a n ds oo n ,s oa d d i n ga l lt h e s eu pw eg e t
Wall,sys =∆ Ksys (7.13)
where ∆Ksys is the change in kinetic energy of the whole system.
So far, of course, this is nothing new. To learn something elsew en e e dt ol o o kn e x ta tt h ew o r k",University Physics I Classical Mechanics.pdf
"So far, of course, this is nothing new. To learn something elsew en e e dt ol o o kn e x ta tt h ew o r k
done by the internal forces. It is helpful here to start by considering the “no-dissipation case” in
which all the internal forces can be derived from a potentialenergy2.W e w i l l c o n s i d e r t h e c a s e
where dissipative processes happen inside the system afterwe have gained a full understanding of
the result we will obtain for this simpler case.
7.4.1 The no-dissipation case
The internal forces are, by definition, forces that arise fromt h ei n t e r a c t i o n sb e t w e e np a i r so f
particles that are both inside the system. Because of Newton’s 3rd law, the forceF12 (we will omit
the “type” superscript for now) exerted by particle 1 on particle 2 must be the negative ofF21,t h e
force exerted by particle 2 on particle 1. Hence, the work associated with this interaction for this
pair of particles can be written
W(1, 2) =F12∆x2 + F21∆x1 = F12 (∆x2 − ∆x1)( 7 . 1 4 )",University Physics I Classical Mechanics.pdf
"pair of particles can be written
W(1, 2) =F12∆x2 + F21∆x1 = F12 (∆x2 − ∆x1)( 7 . 1 4 )
Notice that ∆x2 − ∆x1 can be rewritten asx2,f − x2,i − x1,f + x1,i = x12,f − x12,i =∆ x12,w h e r e
x12 = x2 − x1 is the relative position coordinate of the two particles. Therefore,
W(1, 2) =F12∆x12 (7.15)
Now, if the interaction in question is associated with a potential energy, as I showed in section 6.2,
F12 = −dU/dx12.A s s u m et h ed i s p l a c e m e n t∆x12 is so small that we can replace the derivative by
just the ratio ∆U/∆x12 (which is consistent with our assumption that the force is approximately
constant over the time interval considered); the result willt h e nb e
W(1, 2) =F12∆x12 ≃− ∆U
∆x12
∆x12 = −∆U (7.16)
2Or else they do no work at all: the magnetic force between moving charges is an example of the latter.",University Physics I Classical Mechanics.pdf
"144 CHAPTER 7. IMPULSE, WORK AND POWER
Adding up very many such “infinitesimal” displacements willlead to the same final result, where
∆U will be the change in the potential energy over the whole process. This can also be proved
using calculus, without any approximations:
W(1, 2) =
∫ x12,f
x12,i
F12dx12 = −
∫ x12,f
x12,i
dU
dx12
dx12 = −∆U. (7.17)
We can apply this to every pair of particles and every internali n t e r a c t i o n ,a n dt h e na d du pa l lt h e
results. On one side, we will get the total work done on the system by all the internal forces; on
the other side, we will get the negative of the change in the system’s total internal energy:
Wint,sys = −∆Usys. (7.18)
In words, the work done by all the (conservative) internal forces is equal to the change in the
system’s potential energy.
Let us now put Eqs. (7.13)a n d(7.18) together: the difference between the work done by all the
forces and the work done by the internal forces is, of course,the work done by theexternal forces,",University Physics I Classical Mechanics.pdf
"forces and the work done by the internal forces is, of course,the work done by theexternal forces,
but according to Eqs. (7.13)a n d(7.18), this is equal to
Wext,sys = Wall,sys − Wint,sys =∆ Ksys +∆ Usys (7.19)
which is the change in the totalmechanical (kinetic plus potential) energy of the system. If we
further assume that the system, in the absence of the externalf o r c e s ,i sc l o s e d ,t h e nt h e r ea r en o
other processes (such as the absorption of heat) by which thetotal energy of the system might
change, and we get the simple result thatthe work done by the external forces equals the change in
the system’s total energy:
Wext,sys =∆ Esys (7.20)
As a first application of the result (7.20), consider again the blocks connected by a spring shown in
Fig. 2. You can see now why the work done by the external forceFc
h,2 has to be different, and in
fact larger, than the “center of mass work”: the latter only gives us the change in thetranslational",University Physics I Classical Mechanics.pdf
"h,2 has to be different, and in
fact larger, than the “center of mass work”: the latter only gives us the change in thetranslational
energy, but the former has to give us the change in thetotal energy—translational, convertible,
and potential:
Fc
h,2∆xcm =∆ Kcm
Fc
h,2∆x2 =∆ Kcm +∆ Kconv +∆ Uspr (7.21)
As another example, imagine you throw a ball of massm upwards (see Figure7.3,n e x tp a g e ) ,a n d
it reaches a maximum heighth above the point where your hand started to move. Let us define
the system to be the ball and the earth, so the force exerted byyour hand is an external force.
Then you do work on the system during the throw, which in the figure is the interval, from A to
B, during which your hand is on contact with the ball. The bar diagram on the side shows that
some of this work goes into increasing the system’s (gravitational) potential energy (because the",University Physics I Classical Mechanics.pdf
"7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 145
ball goes up a little while in contact with your hand), and therest, which is typically most of it,
goes into increasing the system’s kinetic energy (in this case, just the ball’s; the earth’s kinetic
energy does not change in any measurable way!).
A
B
C
rising,
from A to C
∆K ∆U Wext
falling,
from C to B
∆K ∆U Wext
the throw,
from A to B
∆K ∆U Wext
h
Figure 7.3: Tossing a ball into the air. We consider the system formed by the balland the earth. The
force exerted by the hand (which is in contact with the ball from point A to point B) is therefore an external
force. The diagrams show the system’s energy balance over threedifferent intervals.
So how much work did you actually do? If we knew the distance from A to B, and the magnitude
of the force you exerted, and if we could assume that your forcew a sc o n s t a n tt h r o u g h o u t ,w ec o u l d",University Physics I Classical Mechanics.pdf
"of the force you exerted, and if we could assume that your forcew a sc o n s t a n tt h r o u g h o u t ,w ec o u l d
calculate W from the definition (7.4). But in this case, and many others like it, it is actually
easier to find out how much total energy the system gained and just use Eq. (7.20). To find ∆E in",University Physics I Classical Mechanics.pdf
"146 CHAPTER 7. IMPULSE, WORK AND POWER
practice, all we have to do is see how high the ball rises. At theb a l l ’ sm a x i m u mh e i g h t( p o i n tC ) ,
as the second diagram shows, all the energy in the system is gravitational potential energy, and (as
long as the system stays closed), all that energy is still equal to the work you did initially, so if the
distance from A to C ish you must have done an amount of work
Wyou =∆ UG = mgh (7.22)
The third diagram in Figure7.3 shows the work-energy balance for another time interval, during
which the ball falls from C to B. Over this time, noexternal forces act on the ball (recall we have
taken the system to be the ball and the earth, so gravity is aninternal force). Then, the work done
by the external forces is zero, and the change in the total energy of the system is also zero. The
diagram just shows an increase in kinetic energy at the expense of an equal decrease in potential
energy.",University Physics I Classical Mechanics.pdf
"diagram just shows an increase in kinetic energy at the expense of an equal decrease in potential
energy.
What about the work done by theinternal forces? Eq. (7.18)t e l l su st h a tt h i sw o r ki se q u a lt o
the negative of the change in potential energy. In this case,the internal force is gravity, and the
corresponding energy is gravitational potential energy. This change in potential energy is clearly
visible in all the diagrams; however, when you add to it the change in kinetic energy, the result
is always equal to the work done by the external forceonly.P u t o t h e r w i s e , t h e i n t e r n a l f o r c e s d o
not change the system’s total energy, they just “redistribute” it among different kinds—as in, for
instance, the last diagram in Fig.7.3,w h e r ey o uc a nc l e a r l ys e et h a tg r a v i t yi sc a u s i n gt h ek i n e t ic
energy of the system to increase at the expense of the potential energy.",University Physics I Classical Mechanics.pdf
"energy of the system to increase at the expense of the potential energy.
We will use diagrams like the ones in Figure7.3 to look at the work-energy balance for different
systems. The idea is that the sum of all the columns on the left(the change in the system’s total
energy) has to equal the result on the far-right column (the work done by the net external force):
that is the content of the theorem (7.20). Note that, unlike the energy diagrams we used in Chapter
5, these columns representchanges in the energy, so they could be positive or negative.
Just as for the earlier energy diagrams, the picture we get will be different, even for the same
physical situation, depending on the choice of system. Thisis illustrated in Figure7.4 below (next
page), where I have taken the same throw shown in Fig.7.3,b u tn o wt h es y s t e mI ’ ml o o k i n ga ti s
the ball only. This means gravity is now an external force, asis the force of the hand, and the ball",University Physics I Classical Mechanics.pdf
"the ball only. This means gravity is now an external force, asis the force of the hand, and the ball
only has kinetic energy. Normally one would show the sum of thew o r kd o n eb yt h et w oe x t e r n a l
forces on a single column, but here I have chosen to break it upinto two columns for clarity.
As you can see, during the throw the hand does positive work, whereas gravity does a comparatively
small amount of negative work, and the change in kinetic energy is the sum of the two. For the
longer interval from A to C (second diagram), gravity continues to do negative work until all the
kinetic energy of the ball is gone. For the interval from C to B,t h eo n l ye x t e r n a lf o r c ei sg r a v i t y ,
which now does positive work, equal to the increase in the ball’s kinetic energy.",University Physics I Classical Mechanics.pdf
"7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 147
the throw, from A to B
∆K Wgrav Whand
rising, from A to C
∆K WhandWgrav
}
Wext
}
Wext
∆K WhandWgrav}
Wext
falling, from C to B
Figure 7.4: Work-energy balance diagrams for the same toss illustrated in Fig.7.3, but now the system is
taken to be the ball only.
Of course, the numerical value of the actual work done by any particular force does not depend on
our choice of system: in each case, gravity does the same amount of work in the processes illustrated
in Fig. 7.4 as in those illustrated in Fig.7.3. The difference, however, is that for the system in
Fig. 7.4,g r a v i t yi sa ne x t e r n a lf o r c e ,a n dn o wt h ew o r ki td o e sa c t u a lly changes the system’s total
energy, because the gravitational potential energy is nownot included in that total.
Formally, it works like this: in the case shown in Fig.7.3,w h e r et h es y s t e mi st h eb a l la n dt h e",University Physics I Classical Mechanics.pdf
"Formally, it works like this: in the case shown in Fig.7.3,w h e r et h es y s t e mi st h eb a l la n dt h e
earth, we have ∆K +∆ UG = Whand.B yt h er e s u l t (7.18), however, we have ∆UG = −Wgrav ,a n d
so this equation can be rearranged to read ∆K = Wgrav + Whand,w h i c hi sj u s tt h er e s u l t(7.20)
when the system is the ball alone.
Ultimately, the reason we emphasize the importance of the choice of system is to prevent double
counting: if you want to count the work done by gravity as contributing to the change in the system’s
total energy, it means that you are, implicitly, treating gravity as an external force, and therefore
your system must be something that does not have, by itself, gravitational potential energy (the
case of the ball in Figure7.4); conversely, if you insist on counting gravitational potential energy
as contributing to the system’s total energy, then you must treat gravity as an internal force, and",University Physics I Classical Mechanics.pdf
"as contributing to the system’s total energy, then you must treat gravity as an internal force, and
leave it out of the calculation of the work done on the system byt h ee x t e r n a lf o r c e s ,w h i c ha r et h e
only ones that can change the system’s total energy.
7.4.2 The general case: systems with dissipation
We are now ready to consider what happ ens when some of the internal interactions in a system are
not conservative. There are two key observations to keep in mind: first, of course, that energy will
always be conserved in a closed system, regardless of whethert h ei n t e r n a lf o r c e sa r e“ c o n s e r v a t i v e ”
or not: if they are not, it merely means that they will convertsome of the “organized,” mechanical
energy, into disorganized (primarily thermal) energy.",University Physics I Classical Mechanics.pdf
"148 CHAPTER 7. IMPULSE, WORK AND POWER
The second observation is that the work done by an external force on a system does not depend on
where the force comes from—that is to say, what physical arrangement we use to produce the force.
Only the value of the force at each step and the displacement oft h ep o i n to fa p p l i c a t i o na r ei n v o l v e d
in the definition (7.9). This means, in particular, that we can use a conservative interaction to do
the work for us. It turns out, then, that the generalization oft h er e s u l t(7.20)t oa p p l yt oa l ls o r t s
of interactions becomes straightforward.
To see the idea, consider, for example, the situation in Figure 7.5 below. This is essentially the same
as Figure 6.2,w h i c hw ea n a l y z e di nd e t a i lf r o mt h ep e r s p e c t i v eo ff o r c e sand accelerations in the
previous chapter. Here I have broken it up into two systems. System A, outlined in blue, consists",University Physics I Classical Mechanics.pdf
"previous chapter. Here I have broken it up into two systems. System A, outlined in blue, consists
of block 1 and the surface on which it slides, and includes a dissipative interaction—namely, kinetic
friction—between the block and the surface. The force doingwork on this system is the tension
force from the rope,⃗Ft
r,1.
1
2
Fr,1
t
Fr,2
t
FE,2
G
Fs,1
fr
a1
a2
A
B
Figure 7.5: Block sliding on a surface, with friction, being pulled by a rope attached to a block falling
under the action of gravity. The motion of this system was solved for in Section 6.3.
Because the rope is assumed to have negligible mass, this force is the same in magnitude as the",University Physics I Classical Mechanics.pdf
"7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 149
force ⃗Ft
r,2 that is doing negative work on system B. System B, outlined in magenta, consists of
block 2 and the earth and thus it includes only one internal interaction, namely gravity, which is
conservative. This means that we can immediately apply the theorem (7.20)t oi t ,a n dc o n c l u d e
that the work done on B by⃗Ft
r,2 is equal to the change in system B’s total energy:
Wr,B =∆ EB (7.23)
However, since the rope is inextensible, the two blocks movethe same distance in the same time,
and the force exerted on each by the rope is the same in magnitude, so the work done by the rope
on system A is equal in magnitude but opposite in sign to the work it does on system B:
Wr,A = −Wr,B = −∆EB (7.24)
Now consider thetotal system formed by A+B. Assuming it is a closed system, its totale n e r g y
must be constant, and so any change in the total energy of B mustb ee q u a la n do p p o s i t et h e",University Physics I Classical Mechanics.pdf
"must be constant, and so any change in the total energy of B mustb ee q u a la n do p p o s i t et h e
corresponding change in the total energy of A: ∆EB = −∆EA.T h e r e f o r e ,
Wr,A = −∆EB =∆ EA (7.25)
So we conclude that the work done by the external force on system A must be equal to the total
change in system A’s energy. In other words, Eq. (7.20)a p p l i e st os y s t e mAa sw e l l ,a si td o e st o
system B, even though the interaction between the parts thatmake up system A is dissipative.
Although I have shown this to be true just for one specific example, the argument is quite general:
if I use a conservative system B to do some work on another system A, two things happen: first, by
virtue of (7.20), the work done by B comes at the expense of its total energy, so Wext,A = −∆EB.
Second, if A and B together form a closed system, the change inA’s energy must be equal and
opposite the change in B’s energy, so ∆EA = −∆EB = Wext,A.S o t h e r e s u l t (7.20)h o l d sf o rA ,",University Physics I Classical Mechanics.pdf
"opposite the change in B’s energy, so ∆EA = −∆EB = Wext,A.S o t h e r e s u l t (7.20)h o l d sf o rA ,
regardless of whether its internal interactions are conservative or not.
What is essential in the above reasoning is that A and B together should form a closed system,
that is, one that does not exchange energy with its environment. It is very important, therefore,
if we want to apply the theorem (7.20)t oag e n e r a ls y s t e m — t h a ti s ,o n et h a ti n c l u d e sd i s s i p a t i v e
interactions—that we draw the boundary of the system in suchaw a ya st oe n s u r et h a tno dissipation
is happening at the boundary.F o r e x a m p l e , i n t h e s i t u a t i o n i l l u s t r a t e d i n F i g .7.5,i fw ew a n tt h e
result (7.25)t oa p p l yw em u s tt a k es y s t e mAt oi n c l u d eb o t hb l o c k1and the surface on which
it slides.T h e r e a s o n f o r t h i s i s t h a t t h e e n e r g y “ d i s s i p a t e d ” b y k i n e tic friction when two objects",University Physics I Classical Mechanics.pdf
"it slides.T h e r e a s o n f o r t h i s i s t h a t t h e e n e r g y “ d i s s i p a t e d ” b y k i n e tic friction when two objects
rub together goes into both objects. So, as the block slides,kinetic friction is converting some of
its kinetic energy into thermal energy, but not all this thermal energy stays inside block 1. Put
otherwise, in the presence of friction, block 1 by itself is not a closed system: it is “leaking” energy
to the surface. On the other hand, when you include (enough of)t h es u r f a c ei nt h es y s t e m ,y o u
can be sure to have “caught” all the dissipated energy, and ther e s u l t(7.20)a p p l i e s .",University Physics I Classical Mechanics.pdf
"150 CHAPTER 7. IMPULSE, WORK AND POWER
7.4.3 Energy dissipated by kinetic friction
In the situation illustrated in Fig.7.5,w em i g h tc a l c u l a t et h ee n e r g yd i s s i p a t e db yk i n e t i cf r i c t ion
by indirect means. For instance, we can use the fact that the energy of system A is of two kinds,
kinetic and “dissipated,” and therefore, by theorem (7.20), we have
∆K +∆ Ediss = Ft
r,1∆x1 (7.26)
Back in section 6.3, we used Newton’s laws to solve for the acceleration of this system and the
tension in the rope; using those results, we can calculate thed i s p l a c e m e n t∆x1 over any time
interval, and the corresponding change inK,a n dt h e nw ec a ns o l v eE q .(7.26)f o r∆Ediss.
If we do this, we will find out that, in fact, the following result holds,
∆Ediss = −Fk
s,1∆x1 (7.27)
where Fk
s,1 is the force of kinetic friction exerted by the surface on block 1, and must be understood",University Physics I Classical Mechanics.pdf
"∆Ediss = −Fk
s,1∆x1 (7.27)
where Fk
s,1 is the force of kinetic friction exerted by the surface on block 1, and must be understood
to be negative in this equation (so that ∆Ediss will come out positive, as it must be).
It is tempting to think of the productFk
s,1∆x as the work done by the force of kinetic friction on
the block, and most of the time there is nothing wrong with that, but it is important to realize
that the “point of application” of the friction force is not asingle point: rather, the force is
“distributed,” that is to say, spread over the whole contactarea between the block and the surface.
As a consequence of this, a more general expression for the energy dissipated by kinetic friction
between an objecto and a surfaces should be
∆Ediss = |Fk
s,o||∆xso| (7.28)
where I am using the Chapter 1 subscript notationxAB to refer to “the position ofB in the frame
of A”( o r“ r e l a t i v et oA”); in other words, ∆xso is the change in the position of the object relative",University Physics I Classical Mechanics.pdf
"of A”( o r“ r e l a t i v et oA”); in other words, ∆xso is the change in the position of the object relative
to the surface or, more simply,the distance that the object and the surface slip past each other
(while rubbing against each other, and hence dissipating energy). If the surfaces is at rest (relative
to the Earth), ∆xso reduces to ∆xEo,t h ed i s p l a c e m e n to ft h eo b j e c ti nt h eE a r t hr e f e r e n c ef r a m e,
and we can remove the subscriptE,a sw et y p i c a l l yd o ,f o rs i m p l i c i t y ;h o w e v e r ,i nt h er a r ec a ses
when both the surface and the objet are moving (as in part (c) ofP r o b l e m3i nC h a p t e r6 ,t h e
sled problem) what matters is how far they moverelative to each other.I n t h a t c a s e w e h a v e
|∆xso| = |∆xo − ∆xs| (with both ∆xo and ∆xs measured in the Earth reference frame).
7.5 Power
By “power” we mean the rate at which work is done, which is to say, the rate at which energy is",University Physics I Classical Mechanics.pdf
"7.5 Power
By “power” we mean the rate at which work is done, which is to say, the rate at which energy is
taken in, or given out, or converted from one form to another.The SI unit of power is the watt (W),",University Physics I Classical Mechanics.pdf
"7.6. IN SUMMARY 151
which is equal to 1 J/s. The average power going into or comingout of a system by mechanical
means, that is to say, through the action of a force applied atap o i n tu n d e r g o i n gad i s p l a c e m e n t
∆x,w i l lb e
Pav = ∆E
∆t = W
∆t = F ∆x
∆t (7.29)
assuming the force is constant. Note that in the limit when ∆t goes to zero, this gives us the
instantaneous power associated with the forceF as
P = Fv (7.30)
where v is the (instantaneous) velocity of the point of applicationof the force. This one-dimensional
result generalizes to three dimensions as
P = ⃗F · ⃗v (7.31)
using again the dot product notation.
An important goal of this chapter has been to develop a set of tools that you may use to find
out where power is spent, and how much: in any practical situation, which systems are giving
energy and which are taking it in, what forms of energy conversion are taking place, and where and",University Physics I Classical Mechanics.pdf
"energy and which are taking it in, what forms of energy conversion are taking place, and where and
through which means are the exchanges and conversions happening. These are extremely important
practical questions; the problems and exercises that you will see here will give you a feel for the
variety of situations that can already be analyzed by this “systems-based” approach, but in a way
they will still do little more than scratch the surface.
7.6 In summary
1. The change in the momentum of a system produced by a force⃗F acting over a time ∆t is
given the name of “impulse” and denoted by⃗J.F o r a c o n s t a n t f o r c e , w e h a v e⃗J =∆ ⃗p= ⃗F∆t.
2. Work, or “doing work” is the name given in physics to the process by which an applied force
brings about a change in the energy of an object, or of a systemthat contains the object on
which the force is acting.
3. The work done by a constant force⃗F acting on an object or system is given byW = ⃗F · ∆⃗r,",University Physics I Classical Mechanics.pdf
"which the force is acting.
3. The work done by a constant force⃗F acting on an object or system is given byW = ⃗F · ∆⃗r,
where the dot represents the “dot” or “scalar” product of thetwo vectors, and ∆⃗ris the
displacement undergone by the point of application of the force while the force is acting.I f
the force is perpendicular to the displacement, it does no work.
4. For a system that is otherwise closed, the net sum of the amounts of work done by all the
external forces is equal to the change in the system’s total energy, when all the types of energy
are included. Note that, for deformable systems, the displacement of the point of application
may be different for different forces.",University Physics I Classical Mechanics.pdf
"152 CHAPTER 7. IMPULSE, WORK AND POWER
5. The result in 4 above holds only provided the boundary of thes y s t e mi sn o td r a w na ta
physical surface on which dissipation occurs. Put otherwise, kinetic friction or other similar
dissipative forces (drag, air resistance) must be includedas internal,not external forces.
6. The work done by the internal forces in a closed system results only in the conversion of one
type of energy into another, always keeping the total energyconstant.
7. For a system with no internal energy, like a particle, the work done by all the external forces
equals the change in kinetic energy. This result is sometimesc a l l e dthe Work-Energy theorem
in a narrow sense.
8. For any system, if ⃗Fext,net (assumed constant) is the sum of all the external forces, the
following result holds:
⃗Fext,net · ∆⃗rcm =∆ Kcm
where Kcm is the translational (or “center of mass”) kinetic energy, and ∆⃗rcm is the displace-",University Physics I Classical Mechanics.pdf
"following result holds:
⃗Fext,net · ∆⃗rcm =∆ Kcm
where Kcm is the translational (or “center of mass”) kinetic energy, and ∆⃗rcm is the displace-
ment of the center of mass. This is only sometimes equal to thenet work done on the system
by the external forces.
9. For an objecto sliding on a surfaces,t h ee n e r g yd i s s i p a t e db yk i n e t i cf r i c t i o nc a nb ed i r e c t l y
calculated as
∆Ediss = |Fk
s,o||∆xso|
where |∆xso| = |∆xo − ∆xs| is the distance that the two surfaces in contact slip past each
other. This expression, with a negative sign, can be used to take the place of the “work done
by friction” in applications of the results 7 and 8 above to systems involving kinetic friction
forces.
10. The power of a system is the rate at which it does work, that is to say, takes in or gives up
energy: Pav =∆ E/∆t.W h e nt h i si sd o n e b ym e a n so fa na p p l i e d f o r c eF,t h ei n s t a n t a n e o u s",University Physics I Classical Mechanics.pdf
"energy: Pav =∆ E/∆t.W h e nt h i si sd o n e b ym e a n so fa na p p l i e d f o r c eF,t h ei n s t a n t a n e o u s
power can be written asP = Fv ,o r ,i nt h r e ed i m e n s i o n s ,⃗F · ⃗v.",University Physics I Classical Mechanics.pdf
"7.7. EXAMPLES 153
7.7 Examples
7.7.1 Braking
Suppose you are riding your bicycle and hit the brakes to cometo a stop. Assuming no slippage
between the tire and the road:
(a) Which force is responsible for removing your momentum? (By “you” I mean throughout “you
and the bicycle.”)
(b) Which force is responsible for removing your kinetic energy?
Solution
(a) According to what we saw in previous chapters, for example, Eq. (6.10)
∆psys
∆t = Fext,net (7.32)
the total momentum of the system can only be changed by the action of an external force, and
the only available external force is the force of static friction between the tire and the road (static,
because we assume no slippage). So it is this force that removes the forward momentum from the
system. The stopping distance, ∆xcm,a n dt h ef o r c e ,c a nb er e l a t e du s i n gE q .(7.12):
Fs
r,t∆xcm =∆ Kcm (7.33)
(b) Now, here is an interesting fact: the force of static friction, although fully responsible for",University Physics I Classical Mechanics.pdf
"Fs
r,t∆xcm =∆ Kcm (7.33)
(b) Now, here is an interesting fact: the force of static friction, although fully responsible for
stopping your center of mass motiondoes no work in this case.T h a t i sb e c a u s et h e p o i n tw h e r e i t
is applied—the point of the tire that is momentarily in contact with the road—is also momentarily
at rest relative to the road: it is, precisely,not slipping,s o∆x in the equationW = F∆x is zero.
By the time that bit of the tire has moved on, so you actually have a nonzero ∆x,y o un ol o n g e r
have anF:t h e f o r c e o f s t a t i c f r i c t i o n i s n o l o n g e r a c t i n g o n t h a t b i t oft h et i r e ,i ti sa c t i n go na
different bit—on which it will, again, do no work, for the same reason.
So, as you bring your bicycle to a halt the workWext,sys =0 ,a n di tf o l l o w sf r o mE q .(7.20)t h a t
the total energy of your system is, in fact, conserved:all your initial kinetic energy is converted to",University Physics I Classical Mechanics.pdf
"the total energy of your system is, in fact, conserved:all your initial kinetic energy is converted to
thermal energy by the brake pad rubbing on the wheel, and the internal force responsible for that
conversion is the force ofkinetic friction between the pad and the wheel.",University Physics I Classical Mechanics.pdf
"154 CHAPTER 7. IMPULSE, WORK AND POWER
7.7.2 Work, energy and the choice of system: dissipative case
Consider again the situation shown in Figure7.5.L e tm1 =1k g ,m2 =2k g ,a n dµk =0 .3. Use the
solutions provided in Section 6.3 to calculate the work doneby all the forces, and the changes in all
energies, when the system undergoes a displacement of 0.5m , a n dr e p r e s e n tt h ec h a n g e sg r a p h i c a l l y
using bar diagrams like the ones in Figure7.3 (for system A and B separately)
Solution
From Eq. (6.32), we have
a = m2 − µkm1
m1 + m2
g =5 .55 m
s2
Ft = m1m2(1 +µk)
m1 + m2
g =8 .49 N (7.34)
We can use the acceleration to calculate the change in kinetice n e r g y ,s i n c ew eh a v eE q .(2.10)f o r
motion with constant acceleration:
v2
f − v2
i =2 a∆x =2 ×
(
5.55 m
s2
)
× 0.5m=5 .55 m2
s2 (7.35)
so the change in kinetic energy of the two blocks is
∆K1 = 1
2m1
(
v2
f − v2
i
)
=2 .78 J
∆K2 = 1
2m2
(
v2
f − v2
i
)
=5 .55 J (7.36)",University Physics I Classical Mechanics.pdf
"s2 (7.35)
so the change in kinetic energy of the two blocks is
∆K1 = 1
2m1
(
v2
f − v2
i
)
=2 .78 J
∆K2 = 1
2m2
(
v2
f − v2
i
)
=5 .55 J (7.36)
We can also use the tension to calculate the work done by the external force on each system:
Wext,A = Ft
r,1∆x =( 8.49 N)× (0.5m )=4 .25 J
Wext,B = Ft
r,2∆y =( 8.49 N)× (−0.5m )= −4.25 J (7.37)
Lastly, we need the change in the gravitational potential energy of system B:
∆UG
B = m2g∆y =( 2k g )×
(
9.8 m
s2
)
× (−0.5m )= −9.8J ( 7 . 3 8 )
and the increase in dissipated energy in system A, which we cang e tf r o mE q .(7.28):
∆Ediss = −Fk
s,1∆x = µkFn
s,1∆x = µkm1g∆x =0 .3 × (1 kg)×
(
9.8 m
s2
)
× (0.5m )=1 .47 J (7.39)
We can now put all this together to show that Eq. (7.20)i n d e e dh o l d s :
Wext,A =∆ EA =∆ K1 +∆ Ediss =2 .78 J + 1.47 J = 4.25 J
Wext,B =∆ EB =∆ K2 +∆ UG
B =5 .55 J− 9.8J= −4.25 J (7.40)",University Physics I Classical Mechanics.pdf
"7.7. EXAMPLES 155
To plot all this as energy bars, if you do not have access to a very precise drawing program, you
typically have to make some approximations. In this case, wesee that ∆K2 =2 ∆K1 (exactly),
whereas ∆K1 ≃ 2∆Ediss,s ow ec a nu s eo n eb o xt or e p r e s e n tEdiss,t w ob o x e sf o r∆K1,t h r e ef o r
Wext,A,f o u rf o r∆K2,a n ds oo n . T h er e s u l ti ss h o w ni ng r e e ni nt h ep i c t u r eb e l o w ;the blue bars
have been drawn more exactly to scale, and are shown for your information only.
system A
∆K1 ∆Ediss Wext
system B
∆K2 ∆UG Wext
7.7.3 Work, energy and the choice of system: non-dissipativec a s e
Suppose you hang a spring from the ceiling, then attach a blockt ot h ee n do ft h es p r i n ga n dl e tg o .
The block starts swinging up and down on the spring. Considerthe initial time just before you let
go, and the final time when the block momentarily stops at the bottom of the swing. For each of",University Physics I Classical Mechanics.pdf
"go, and the final time when the block momentarily stops at the bottom of the swing. For each of
the choices of a system listed below, find the net energy changeo ft h es y s t e mi nt h i sp r o c e s s ,a n d
relate it explicitly to the work done on the system by an external force (or forces)
(a) System is the block and the spring.
(b) System is the block alone.
(c) System is the block and the earth.
Solution
(a) The block alone has kinetic energy, and the spring alone has (elastic) potential energy, so the
total energy of this system is the sum of these two. For the interval considered, the change in
kinetic energy is zero, because the block starts and ends (momentarily) at rest, so only the spring
energy changes. This has to be equal to the work done by gravity, which is the only external force.
So, if the spring stretches a distanced,i t sp o t e n t i a le n e r g yg o e sf r o mz e r ot o1
2 kd2,a n dt h eb l o c k",University Physics I Classical Mechanics.pdf
"So, if the spring stretches a distanced,i t sp o t e n t i a le n e r g yg o e sf r o mz e r ot o1
2 kd2,a n dt h eb l o c k
falls the same distance, so gravity does an amount of work equal tomgd,a n dw eh a v e
Wgrav = mgd =∆ Esys =∆ K +∆ Uspr =0+ 1
2kd2 (7.41)",University Physics I Classical Mechanics.pdf
"156 CHAPTER 7. IMPULSE, WORK AND POWER
(b) If the system is the block alone, the only energy it has is kinetic energy, which, as stated above,
does not see a net change in this process. This means the net work done on the block by the
external forces must be zero. The external forces in this casea r et h es p r i n gf o r c ea n dg r a v i t y ,s o
we have
Wspr + Wgrav =∆ K =0 ( 7 . 4 2 )
We have calculatedWgrav above, so from this we get that the work done by the spring on theb l o c k ,
as it stretches, is−mgd,o r( b yE q .(7.41)) −1
2kd2.N o t et h a t t h e f o r c e e x e r t e db y t h e s p r i n g i snot
constant as it stretches (or compresses) so we cannot just useE q .(7.4)t oc a l c u l a t ei t ;r a t h e r ,w e
need to calculate it as an integral, as in Eq. (7.9), or derive it in some indirect way as we have just
done here.
(c) If the system is the block and the earth, it has kinetic energy and gravitational potential energy.",University Physics I Classical Mechanics.pdf
"done here.
(c) If the system is the block and the earth, it has kinetic energy and gravitational potential energy.
The force exerted by the spring is an external force now, so wehave:
Wspr =∆ Esys =∆ K +∆ UG =0 − mgd (7.43)
so we end up again with the result thatWspr = −mgd = −1
2 kd2.N o t et h a tb o t ht h ew o r kd o n eb y
the spring and the work done by gravity are equal to the negative of the changes in their respective
potential energies, as they should be.
7.7.4 Jumping
For a standing jump, you start standing straight (A) so your body’s center of mass is at a height
h1 above the ground. You then bend your knees so your center of mass is now at a (lower) height
h2 (B). Finally, you straighten your legs, pushing hard on the ground, and take off, so your center
of mass ends up achieving a maximum height,h3,a b o v et h eg r o u n d( C ) .A n s w e rt h ef o l l o w i n g
questions in as much detail as you can.",University Physics I Classical Mechanics.pdf
"questions in as much detail as you can.
(a) Consider the system to be your body only. In going from (A)to (B), which external forces are
acting on it? How do their magnitudes compare, as a function oft i m e ?
(b) In going from (A) to (B), does any of the forces you identified in part (a) do work on your
body? If so, which one, and by how much? Does your body’s energyi n c r e a s eo rd e c r e a s ea sa
result of this? Into what kind of energy do you think this workis primarily converted?
(c) In going from (B) to (C), which external forces are actingon you? (Not all of them need to be
acting all the time.) How do their magnitudes compare, as a function of time?
(d) In going from (B) to (C), does any of the forces you identified do work on your body? If so,
which one, and by how much? Does your body’s kinetic energy seean e tc h a n g ef r o m( B )t o( C ) ?
What other energy change needs to take place in order for Eq. (7.20)( a l w a y sw i t hy o u rb o d ya s",University Physics I Classical Mechanics.pdf
"What other energy change needs to take place in order for Eq. (7.20)( a l w a y sw i t hy o u rb o d ya s
the system) to be valid for this process?
Solution",University Physics I Classical Mechanics.pdf
"7.7. EXAMPLES 157
(a) The external forces on your body are gravity, pointing down, and the normal force from the
floor, pointing up. Initially, as you start lowering your center of mass, the normal force has to be
slightly smaller than gravity, since your center of mass acquires a small downward acceleration.
However, eventuallyFn would have to exceedFG in order to stop the downward motion.
(b) The normal force does no work, because its point of application (the soles of your feet) does
not move, so ∆x in the expressionW = F∆x (Eq. (7.4)) is zero.
Gravity, on the other hand, does positive work, since you mayalways treat the center of mass as the
point of application of gravity (see Section 7.3, footnote).W e h a v eFG
y = −mg,a n d∆y = h2 −h1,
so
Wgrav = FG
y ∆y = −mg(h2 − h1)= mg(h1 − h2)
Since this is the net work done by all the external forces on mybody, and it is positive, the total
energy in my body must have increased (by the theorem (7.20): Wext,sys =∆ Esys). In this case,",University Physics I Classical Mechanics.pdf
"energy in my body must have increased (by the theorem (7.20): Wext,sys =∆ Esys). In this case,
it is clear that the main change has to be an increase in my body’selastic potential energy,a sm y
muscles tense for the jump. (An increase in thermal energy isalways possible too.)
(c) During the jump, the external forces acting on me are againg r a v i t ya n dt h en o r m a lf o r c e ,
which together determine the acceleration of my center of mass. At the beginning of the jump, the
normal force has to be much stronger than gravity, to give me alarge upwards acceleration. Since
the normal force is a reaction force, I accomplish this by pushing very hard with my feet on the
ground, as I extend my leg’s muscles: by Newton’s third law, the ground responds with an equal
and opposite force upwards.
As my legs continue to stretch, and move upwards, the force they exert on the ground decreases,",University Physics I Classical Mechanics.pdf
"and opposite force upwards.
As my legs continue to stretch, and move upwards, the force they exert on the ground decreases,
and so doesFn,w h i c he v e n t u a l l yb e c o m e sl e s st h a nFG.A tt h a tp o i n t ( p r o b a b l ye v e nb e f o r e m y
feet leave the ground) the acceleration of my center of mass becomes negative (that is, pointing
down). This ultimately causes my upwards motion to stop, andmy body to come down.
(d) The only force that does work on my body during the processdescribed in (c) is gravity, since,
again, the point of application ofFn is the point of contact between my feet and the ground, and
that point does not move up or down—it is always level with theground. So Wext,sys = Wgrav ,
which in this case is actuallynegative: Wgrav = −mg(h3 − h2).
In going from (B) to (C), there is no change in your kinetic energy, since you start at rest and end
(momentarily) with zero velocity at the top of the jump. So thef a c tt h a tt h e r ei san e tn e g a t i v e",University Physics I Classical Mechanics.pdf
"(momentarily) with zero velocity at the top of the jump. So thef a c tt h a tt h e r ei san e tn e g a t i v e
work done on you means that the energy inside your body must have gone down. Clearly, some of
this is just a decrease in elastic potential energy. However,s i n c eh3 (the final height of your center
of mass) is greater thanh1 (its initial height at (A), before crouching), there is anet loss of energy
in your body as a result of the whole process. The most obviousplace to look for this loss is in
chemical energy: you “burned” some calories in the process,primarily when pushing hard against
the ground.",University Physics I Classical Mechanics.pdf
"158 CHAPTER 7. IMPULSE, WORK AND POWER
7.8 Problems
Problem 1 In a mattress test, you drop a 7.0k g bo w l i n g b a l l f r o m a h e i g h t o f 1.5m a bo v e a
mattress, which as a result compresses 15 cm as the ball comesto a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth doon the ball as it falls, for the first
part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball?
(e) Now model the mattress as a single spring with an unknown spring constant k,a n dc o n s i d e r
the whole system formed by the ball, the earth and the mattress. By how much does the potential
energy of the mattress increase as it compresses?
(f) What is the value of the spring constantk?",University Physics I Classical Mechanics.pdf
"energy of the mattress increase as it compresses?
(f) What is the value of the spring constantk?
Problem 2 Ab l o c ko fm a s s1 k gi ss i t t i n go nt o po fac o m p r e s s e ds p r i n go fspring constant
k =3 0 0N / ma n de q u i l i b r i u ml e n g t h2 0c m . I n i t i a l l yt h es p r i n gisc o m p r e s s e d1 0 c m ,a n dt h eb l o c k
is held in place by someone pushing down on it with his hand. Att =0 ,t h eh a n di sr e m o v e d( t h i s
involves no work), the spring expands and the block flies upwards.
(a) Draw a free-body diagram for the block while the hand is still pressing down. Try to get the
forces approximately to scale. The following question should help.
(b) What must be the force (magnitude and direction) exertedby the hand on the block?
(c) How much elastic potential energy was stored in the springi n i t i a l l y ?
(d) Taking the system formed by the block and the earth, how much total work is done on it by",University Physics I Classical Mechanics.pdf
"(d) Taking the system formed by the block and the earth, how much total work is done on it by
the spring, as it expands to its equilibrium length? (You do not need to do a new calculation here,
just think of conservation of energy.)
(e) How high does the block rise above its initial position?
(f) Treating the block alone as the system, how much net work isd o n eo ni tb yt h et w oe x t e r n a l
forces (the spring and gravity) from the time just before it starts moving to the time it reaches its
maximum height? (Again, no calculation is necessary if you can justify your answer.)
Problem 3 Ac r a n ei sl i f t i n ga5 0 0 - k go b j e c ta tac o n s t a n ts p e e do f0.5m / s . W h a t i s t h e po w e r
output of the crane?
Problem 4 In a crash test, a car, initially moving at 30 m/s, hits a wall and crumples to a halt.
In the process of crumpling, the center of mass of the car movesf o r w a r dad i s t a n c eo f1 m .",University Physics I Classical Mechanics.pdf
"In the process of crumpling, the center of mass of the car movesf o r w a r dad i s t a n c eo f1 m .
(a) If the car has a mass of 1, 800 kg, what is the magnitude of the average force acting on itwhile
it stops? What, physically, is this force?
(b) Does the force you found in (a) actually do any work on the car? (Think carefully!)
(c) What is the net change in the car’s kinetic energy? Where does all that kinetic energy go?",University Physics I Classical Mechanics.pdf
"7.8. PROBLEMS 159
Problem 5 Ab l o c ko fm a s s3 k gs l i d e so nah o r i z o n t a l ,r o u g hs u r f a c et o w ards a spring with
k =5 0 0N / m . T h ek i n e t i cf r i c t i o nc o e ffi c i e n tb e t w e e nt h eb l o c kand the surface isµk =0 .6. If the
block’s speed is 5 m/s at the instant it first makes contact witht h es p r i n g ,
(a) Find the maximum compression of the spring.
(b) Draw work-energy bar diagrams for the process of the blockc o m i n gt oah a l t ,t a k i n gt h es y s t e m
to be the block and the surface only.",University Physics I Classical Mechanics.pdf
"160 CHAPTER 7. IMPULSE, WORK AND POWER",University Physics I Classical Mechanics.pdf
"Chapter 8
Motion in two dimensions
8.1 Dealing with forces in two dimensions
We have b een able to get a lot of physics from our study of (mostly) one-dimensional motion only,
but it goes without saying that the real world is a lot richer than that, and there are a number of
new and interesting phenomena that appear when one considersm o t i o ni nt w oo rt h r e ed i m e n s i o n s .
The purpose of this chapter is to introduce you to some of the simplest two-dimensional situations
of physical interest.
Ac o m m o nf e a t u r et oa l lt h e s ep r o b l e m si st h a tt h ef o r c e sa c t ing on the objects under consideration
will typically not line up with the displacements. This means, in practice, that we need to pay
more attention to the vector nature of these quantities thanwe have done so far. This section
will present a brief reminder of some basic properties of vectors, and introduce a couple of simple
principles for the analysis of the systems that will follow.",University Physics I Classical Mechanics.pdf
"principles for the analysis of the systems that will follow.
To begin with, recall that a vector is a quantity that has botham a g n i t u d ea n dad i r e c t i o n .
The magnitude of the vector just tells us how big it is: the magnitude of the velocity vector, for
instance, is the speed, that is, just how fast something is moving. When working with vectors in
one dimension, we have typically assumed that the entire vector (whether it was a velocity, an
acceleration or a force) lay along the line of motion of the system, and all we had to do to indicate
the direction was to give the vector’s magnitude an appropriate sign. For the problems that follow,
however, it will become essential to break up the vectors intot h e i rcomponents along an appropriate
set of axes. This involves very simple geometry, and followsthe example of the position vector⃗r,
whose components are just the Cartesian coordinates of the point it locates in space (as shown in",University Physics I Classical Mechanics.pdf
"whose components are just the Cartesian coordinates of the point it locates in space (as shown in
Figure 1.1). For a generic vector, for instance, a force, like the one shown in Figure8.1 below, the
components Fx and Fy may be obtained from a right triangle, as indicated there:
161",University Physics I Classical Mechanics.pdf
"162 CHAPTER 8. MOTION IN TWO DIMENSIONS
F
F
Fx
Fy
x
y
θ θ180° − θ
Fx
Figure 8.1: The components of a vector that makes an angleθ with the positivex axis. Two examples are
shown, for θ< 90◦ (in which caseFx > 0) and for 90◦ <θ< 180◦ (in which caseFx < 0). In both cases,
Fy > 0.
The triangle will always have the vector’s magnitude (| ⃗F| in this case) as the hypothenuse. The
two other sides should be parallel to the coordinate axes. Their lengths are the corresponding
components, except for a sign that depends on the orientationo ft h ev e c t o r .I fw eh a p p e nt ok n o w
the angleθ that the vector makes with the positivex axis, the following relations will always hold:
Fx = | ⃗F| cos θ
Fy = | ⃗F| sin θ
| ⃗F| =
√
F2x + F2y
θ =t a n−1 Fy
Fx
(8.1)
Note, however, that in general this angleθ may not be one of the interior angles of the triangle (as
shown on the right diagram in Fig.8.1), and that in that case it may just be simpler to calculate",University Physics I Classical Mechanics.pdf
"shown on the right diagram in Fig.8.1), and that in that case it may just be simpler to calculate
the magnitude of the components using trigonometry and an interior angle (such as 180◦ −θ in the
example), and give them the appropriate signs “by hand.” In the example on the right, the length
of the horizontal side of the triangle is equal to| ⃗F| cos(180◦ − θ), which is a positive quantity; the
correct value forFx,h o w e v e r ,i st h en e g a t i v en u m b e r| ⃗F| cos θ = −| ⃗F| cos(180◦ − θ).
In any case, it is important not to get fixated on the notion that“ t h ex component will always be
proportional to the cosine ofθ.” The symbolθ is just a convenient one to use for a generic angle.
There are four sections in this chapter, and in every one therei sa θ used with a different meaning.
When in doubt, just draw the appropriate right triangle and remember from your trigonometry
classes which side goes with the sine, and which with the cosine.",University Physics I Classical Mechanics.pdf
"classes which side goes with the sine, and which with the cosine.
For the problems that we are going to study in this chapter, thei d e ai st ob r e a ku pa l lt h ef o r c e s
involved into components along properly-chosen coordinatea x e s ,t h e na d da l lt h ec o m p o n e n t sa l o n g
any given direction, and applyFnet = ma along that direction: that is to say, we will write (and",University Physics I Classical Mechanics.pdf
"8.1. DEALING WITH FORCES IN TWO DIMENSIONS 163
eventually solve) the equations
Fnet,x = max
Fnet,y = may (8.2)
We can show that Eqs. (8.2)m u s th o l df o ra n yc h o i c eo fo r t h o g o n a lx and y axes, based on the
fact that we know⃗Fnet = m⃗ aholds along one particular direction, namely, the directionc o m m o n
to ⃗Fnet and ⃗a,a n dt h ef a c tt h a tw eh a v ed e fi n e dt h ep r o j e c t i o np r o c e d u r et obe the same for any
kind of vector. Figure8.2 shows how this works. Along the dashed line you just have the situation
that is by now familiar to us from one-dimensional problems,where ⃗alies along ⃗F (assumed here
to be the net force), and| ⃗F| = m|⃗a|.H o w e v e r , i n t h e fi g u r e I h a v e c h o s e n t h e a x e s t o m a k e a n
angle θ with this direction. Then, if you look at the projections of⃗F and ⃗aalong the x axis, you
will find
ax = |⃗a| cos θ
Fx = | ⃗F| cos θ = m|⃗a| cos θ = max (8.3)",University Physics I Classical Mechanics.pdf
"will find
ax = |⃗a| cos θ
Fx = | ⃗F| cos θ = m|⃗a| cos θ = max (8.3)
and similarly, Fy = may.I n w o r d s ,each component of the force vector is responsible for only
the corresponding component of the acceleration.A f o r c e i n t h ex direction does not cause any
acceleration in they direction, and vice-versa.
a F
ax
Fx
ay
Fy
x
y
θ
Figure 8.2:If you take the familiar, one-dimensional (see the black dashed line)form of⃗F = m⃗ a,a n dp r o j e c t
it onto orthogonal, rotated axes, you get the general two-dimensional case, showing that each orthogonal
component of the acceleration is proportional, via the massm, to only the corresponding component of the
force (Eqs. (8.2)).
In the rest of the chapter we shall see how to use Eqs. (8.2)i nan u m b e ro fe x a m p l e s .O n et h i n gI
can anticipate is that, in general, we will try to choose our axes (unlike in Fig.8.2 above) so that
one of them does coincide with the direction of the acceleration, so the motion along the other",University Physics I Classical Mechanics.pdf
"one of them does coincide with the direction of the acceleration, so the motion along the other
direction is either nonexistent (v =0 )o rt r i v i a l( c o n s t a n tv e l o c i t y ) .",University Physics I Classical Mechanics.pdf
"164 CHAPTER 8. MOTION IN TWO DIMENSIONS
8.2 Projectile motion
Projectile motion is basically just free fall, only with theunderstanding that the object we are
tracking was “projected,” or “shot,” with some initial velocity (as opposed to just dropped from
rest). Unlike in the previous cases of free fall that we have studied so far, we will now assume that
the initial velocity has a horizontal component, as a resultof which, instead of just going straight
up and/or down, the object will describe (ignoring air resistance, as usual) aparabola in a vertical
plane.
The plane in question is determined by the initial velocity (more precisely, the horizontal component
of the initial velocity) and gravity. A generic trajectory iss h o w ni nF i g u r e8.3,s h o w i n gt h ef o r c e
and acceleration vectors (constant throughout) and the velocity vector at various points along the
path.
y
x
ymax height
xmax height xrange
vi
yi vx,i
vy,i
FG
a",University Physics I Classical Mechanics.pdf
"path.
y
x
ymax height
xmax height xrange
vi
yi vx,i
vy,i
FG
a
Figure 8.3: A typical projectile trajectory. The velocity vector (in green) is shown at the initial time, the
point of maximum height, and the point where the projectile is back toits initial height.
Conceptually, the problem turns out to be extremely simple ifw ea p p l yt h eb a s i cp r i n c i p l ei n t r o -
duced in Section 8.1. The force is vertical throughout; so, after the throw, there is no horizontal
acceleration, and the vertical acceleration is just−g,j u s ta si ta l w a y sw a si no u re a r l i e r ,o n e -
dimensional free-fall problems:
ax = Fx
m =0
ay = Fy
m = −g (8.4)",University Physics I Classical Mechanics.pdf
"8.2. PROJECTILE MOTION 165
The overall motion, then, is a combination of motion with constant velocity horizontally, and
motion with constant acceleration vertically, and we can write down the corresponding equations
of motion immediately:
vx = vx,i
vy = vy,i − gt
x = xi + vx,it
y = yi + vy,it − 1
2gt2 (8.5)
where (xi,y i)a r et h ec o o r d i n a t e so ft h el a u n c h i n gp o i n t( t h e r ei su s u a l lyn or e a s o nt om a k exi
anything other than zero, so we will do that below), and (vx,i,v y,i)t h ei n i t i a lc o m p o n e n t so ft h e
velocity vector.
By eliminating t in between the last two Eqs. (8.5), we get the equation of the trajectory in the
x-y plane:
y = yi + vy,i
vx,i
x − g
2v2
x,i
x2 (8.6)
which, as indicated earlier, and as shown in Fig.8.3,i si n d e e dt h ee q u a t i o no fap a r a b o l a .
The apex of the parabola (highest point in the trajectory) isat xmax height = vx,ivy,i/g.W e c a n g e t",University Physics I Classical Mechanics.pdf
"The apex of the parabola (highest point in the trajectory) isat xmax height = vx,ivy,i/g.W e c a n g e t
this result from calculus, or from a comparison of Eq. (8.6)w i t ht h ec a n o n i c a lf o r mo fap a r a b o l a ,
or we can use some physics: the maximum height is reached, as usual, when the vertical velocity
becomes momentarily zero, so solving thevy equation (8.5)f o rtmax height and substituting in thex
equation, we get
tmax height = vy,i
g
xmax height = vx,ivy,i
g
ymax height = yi +
v2
y,i
2g (8.7)
The last of these equations should look familiar. It is, indeed a variation on our old friendv2
f −v2
i =
−2g∆y,o n l yn o wi n s t e a do ft h ef u l lv e l o c i t y⃗vwe have to use only the vertical velocity component
vy.J u s t l i k e f o r o n e - d i m e n s i o n a l m o t i o n , t h i s r e s u l t f o l l o w sagain from conservation of energy:
throughout the flight, we must haveK + UG =c o n s t a n t ,o n l yn o wt h e r ei sac o m p o n e n tt ot h e",University Physics I Classical Mechanics.pdf
"throughout the flight, we must haveK + UG =c o n s t a n t ,o n l yn o wt h e r ei sac o m p o n e n tt ot h e
kinetic energy—the part associated with the horizontal motion—which remains constant on its
own. In general, the kinetic energy of a particle will be1
2 m|⃗v|2,w h e r e|⃗v| is the magnitude of the
velocity vector—that is to say, the speed. In two dimensions,t h i sg i v e s
K = 1
2mv2
x + 1
2mv2
y (8.8)",University Physics I Classical Mechanics.pdf
"166 CHAPTER 8. MOTION IN TWO DIMENSIONS
For pro jectile motion, however,vx does not change, so any change inK will affect only the second
term in Eq. (8.8). Conservation of energy between any two instantsi and f gives
K + UG = 1
2mv2
x,i + 1
2mv2
y,i + mgyi = 1
2mv2
x,i + 1
2mv2
y,f + mgyf (8.9)
The 1
2 mv2
x,i term cancels, and therefore
v2
y,f − v2
y,i = −2g(yf − yi)( 8 . 1 0 )
Another quantity of interest is the projectile’srange,o rm a x i m u mh o r i z o n t a ld i s t a n c et r a v e l e d . W e
can calculate it from Eqs. (8.5), by settingy equal to the final height, then solving fort (which
generally requires solving a quadratic equation), and thensubstituting the result in the equation
for x.I n t h e s i m p l e c a s e w h e n t h e fi n a l h e i g h t i s t h e s a m e a s t h e i n i tial height, we can avoid the
need for calculating altogether, and just reason, from the fact that the trajectory is symmetric, that",University Physics I Classical Mechanics.pdf
"need for calculating altogether, and just reason, from the fact that the trajectory is symmetric, that
the total horizontal distance traveled will be twice the distance to the point where the maximum
height is reached, that is,xrange =2 xmax height:
xrange = 2vx,ivy,i
g (only ifyf = yi)( 8 . 1 1 )
As you can see, all these equations depend on the initial values of the components of the velocity
vector ⃗vi.I f ⃗vi makes an angleθ with the horizontal, and we simplify the notation by callingits
magnitude vi,t h e nw ec a nw r i t e
vx,i = vi cos θ
vy,i = vi sin θ (8.12)
In terms ofvi and θ,t h er a n g ee q u a t i o n(8.11)b e c o m e s
xrange = v2
i sin(2θ)
g (only ifyf = yi)( 8 . 1 3 )
since 2 sinθ cos θ =s i n ( 2θ). This tells us that for any given launch speed, the maximum range is
achieved when the launch angleθ =4 5◦ (always assuming the final height is the same as the initial
height).",University Physics I Classical Mechanics.pdf
"achieved when the launch angleθ =4 5◦ (always assuming the final height is the same as the initial
height).
In real life, of course, there will always be air resistance,and all these results will be modified
somewhat. Mathematically, things become a lot more complicated: the drag force depends on the
speed, which involvesboth components of the velocity, so the horizontal and vertical motions are
no longer decoupled: not only is there now anFx,b u ti t sv a l u ea ta n yg i v e nt i m ed e p e n d sb o t ho n
vx and vy.P h y s i c a l l y , y o u m a y t h i n k o f t h e d r a g f o r c e a s d o i n g n e g a t i vew o r ko nt h ep r o j e c t i l e ,
and hence removing kinetic energy from it. Less kinetic energy means, basically, that it will not
travel quite as far either vertically or horizontally. Surprisingly, the optimum launch angle remains
pretty close to 45◦,a tl e a s ti ft h es i m u l a t i o n sa tt h i sl i n ka r ea c c u r a t e :",University Physics I Classical Mechanics.pdf
"pretty close to 45◦,a tl e a s ti ft h es i m u l a t i o n sa tt h i sl i n ka r ea c c u r a t e :
https://phet.colorado.edu/en/simulation/projectile-motion",University Physics I Classical Mechanics.pdf
"8.3. INCLINED PLANES 167
8.3 Inclined planes
Back in Chapter 2, I stated without proof that the acceleration of an object sliding, without friction,
down an inclined plane making an angleθ with the horizontal wasg sin θ.Ic a ns h o wy o un o ww h y
this is so, and introduce friction as well.
FG
θ θ
F n
F k
a
θ
F k
FG
Eb
sb
sb
sb
F n
sb
Eb
ax
y
x
y
Figure 8.4: A block sliding down an inclined plane. The corresponding free-body diagram is shown on the
right.
Figure 8.4 above shows, on the left, a block sliding down an inclined plane and all the forces acting
on it. These are more clearly seen on the free-body diagram onthe right. I have labeled all the
forces using the⃗Ftype
by,on convention introduced back in Chapter 6 (so, for instance,⃗Fk
sb is the force of
kinetic friction exerted by the surface on the block); however, later on, for algebraic manipulations,
and especially wherex and y components need to be taken, I will drop the “by, on” subscripts, and",University Physics I Classical Mechanics.pdf
"and especially wherex and y components need to be taken, I will drop the “by, on” subscripts, and
just let the “type” superscript identify the force in question.
The diagrams also show the coordinate axes I have chosen: thex axis is along the plane, and the
y is perpendicular to it. The advantage of this choice is obvious: the motion is entirely along one
of the axes, and two of the forces (the normal force and the friction) already lie along the axes.
The only force that does not is the block’s weight (that is, thef o r c eo fg r a v i t y ) ,s ow en e e dt o
decompose it into itsx and y components. For this, we can make use of the fact, which follows
from basic geometry, that the angle of the incline,θ,i sa l s ot h ea n g l eb e t w e e nt h ev e c t o r⃗Fg and
the negative y axis. This means we have
Fg
x = Fg sin θ
Fg
y = −Fg cos θ (8.14)
Equations (8.14)a l s os h o wa n o t h e rc o n v e n t i o nIw i l la d o p tf r o mn o w ,n a m e l y ,that whenever the",University Physics I Classical Mechanics.pdf
"Fg
y = −Fg cos θ (8.14)
Equations (8.14)a l s os h o wa n o t h e rc o n v e n t i o nIw i l la d o p tf r o mn o w ,n a m e l y ,that whenever the
symbol for a vector is shownwithout an arrow on topor an x or y subscript, it will be understood
to refer to themagnitude of the vector, which is always a positive number by definition.",University Physics I Classical Mechanics.pdf
"168 CHAPTER 8. MOTION IN TWO DIMENSIONS
Newton’s second law, as given by equations (8.4)a p p l i e dt ot h i ss y s t e m ,t h e nr e a d s :
Fg
x + Fk
x = max = Fg sin θ − Fk (8.15)
for the motion along the plane, and
Fg
y + Fn
y = may = −Fg cos θ + Fn (8.16)
for the direction perpendicular to the plane. Of course, since there is no motion in this direction,
ay is zero. This gives us immediately the value of the normal force:
Fn = Fg cos θ = mg cos θ (8.17)
since Fg = mg.W e c a n a l s o u s e t h e r e s u l t (8.17), together with Eq. (6.30), to get the magnitude
of the friction force, assuming we know the coefficient of kinetic (or sliding) friction,µk:
Fk = µkFn = µkmg cos θ (8.18)
Substituting this andFg = mg in Eq. (8.15), we get
max = mg sin θ − µkmg cos θ (8.19)
We can eliminate the mass to obtain finally
ax = g (sin θ − µk cos θ)( 8 . 2 0 )
which is the desired result. In the absence of friction (µk =0 )t h i sg i v e sa = g sin θ,a sw eh a di n",University Physics I Classical Mechanics.pdf
"ax = g (sin θ − µk cos θ)( 8 . 2 0 )
which is the desired result. In the absence of friction (µk =0 )t h i sg i v e sa = g sin θ,a sw eh a di n
Chapter 2. Note that, if you reduce the tilt of the surface (that is, makeθ smaller), the cosθ term
in Eq. (8.20)g r o w sa n dt h es i nθ term gets smaller, so we must make sure that we do not use this
equation whenθ is too small or we would get the absurd result thatax < 0, that is, that the force
of kinetic friction has overcome gravity and is acceleratingt h eo b j e c tupwards!
Of course, we know from experience that what happens whenθ is very small is that the block does
not slide: it is held in place by the force of static friction. Thediagram for such a situation looks
the same as Fig.8.4,e x c e p tt h a t⃗a=0 ,t h ef o r c eo ff r i c t i o ni sFs instead of Fk,a n do fc o u r s ei t s
magnitude must match that of thex component of gravity. Equation (8.15)t h e nb e c o m e s
max =0= Fg sin θ − Fs (8.21)",University Physics I Classical Mechanics.pdf
"magnitude must match that of thex component of gravity. Equation (8.15)t h e nb e c o m e s
max =0= Fg sin θ − Fs (8.21)
Recall from Chapter 6 that the force of static friction does not have a fixed value: rather, it will
match the applied force up to a maximum value given by Eq. (6.29):
Fs
max = µsFn = µsmg cos θ (8.22)
where I have used Eq. (8.17), since clearly the equation (8.16)s t i l la p p l i e sa l o n gt h ev e r t i c a l
direction. So, on the one hand we have the requirement thatFs = mg sin θ to keep the block from",University Physics I Classical Mechanics.pdf
"8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 169
sliding, and on the other hand the constraintFs ≤ µsmg cos θ.P u t t i n gt h e s et o g e t h e rw ec o n c l u d e
that the block willnot slide as long as
mg sin θ ≤ µsmg cos θ (8.23)
or
tan θ ≤ µs (8.24)
In short, as long asθ is small enough to satisfy Eq. (8.24), the block will not move. Onceθ exceeds
the value tan−1 µs,w ec a na p p l yt h er e s u l t(8.20)f o rt h ea c c e l e r a t i o n . N o t et h a t ,s i n c ew ea l w a y s
have µs ≥ µk,t h er e s u l t(8.20)w i l la l w a y sb ep o s i t i v ei fθ> tan−1 µs,t h a ti s ,i fs i nθ>µ s cos θ.
What if we send the block slidingup the plane instead? The acceleration would still be pointing
down (since the object would be slowing down all the while), but now the force of kinetic friction
would point in the directionopposite that indicated in Figure8.4,s i n c ei ta l w a y sm u s to p p o s e",University Physics I Classical Mechanics.pdf
"would point in the directionopposite that indicated in Figure8.4,s i n c ei ta l w a y sm u s to p p o s e
the motion. If you go through the same analysis I carried out above, you will get thatax =
g (sin θ + µk cos θ)i nt h a tc a s e ,s i n c en o wf r i c t i o na n dg r a v i t ya r ew o r k i n gt o gether to slow the
motion down.
8.4 Motion on a circle (or part of a circle)
The last example of motion in two dimensions that I will consider in this chapter is motion on
ac i r c l e . T h e r ea r em a n ye x a m p l e so fc i r c u l a r( o rn e a r - c i r c ular) motion in nature, particularly in
astronomy (as we shall see in a later chapter, the orbits of most planets and many satellites are
very nearly circular). There are also many devices that we usea l lt h et i m et h a ti n v o l v er o t a t i n g
or spinning objects (wheels, gears, turntables, turbines... ). All of these can be mathematically
described as collections of particles moving in circles.",University Physics I Classical Mechanics.pdf
"described as collections of particles moving in circles.
In this section, I will first introduce the concept ofcentripetal force,w h i c hi st h ef o r c en e e d e d
to bend an object’s trajectory into a circle (or an arc of a circle), and then I will also introduce
an u m b e ro fq u a n t i t i e st h a ta r eu s e f u lf o rt h ed e s c r i p t i o no fcircular motion in general, such as
angular velocity and angular acceleration. The dynamics ofrotational motion (questions having to
do with rotational energy, and a new important quantity, angular momentum) will be the subject
of the next chapter.
8.4.1 Centripetal acceleration and centripetal force
As you know by now, the law of inertia states that, in the absence of external forces, an object will
move with constant speed on a straight line. A circle is not a straight line, so an object will not
naturally follow a circular path unless there is a force acting on it.",University Physics I Classical Mechanics.pdf
"170 CHAPTER 8. MOTION IN TWO DIMENSIONS
Another way to see this is to go back to the definition of acceleration. If an object has a velocity
vector ⃗v(t)a tt h et i m et, and a different velocity vector⃗v(t +∆ t)a tt h el a t e rt i m et +∆ t,t h e ni t s
average acceleration over the time interval ∆t is the quantity⃗vav =( ⃗v(t +∆ t) −⃗v(t))/∆t.T h i si s
nonzero even if thespeed does not change (that is, even if the two velocity vectors havet h es a m e
magnitude), as long as they have different directions, as you can see from Figure8.5 below. Thus,
motion on a circle (or an arc of a circle), even at constant speed, is accelerated motion,a n d ,b y
Newton’s second law, accelerated motion requires a force tomake it happen.
θ
R
s
v(t) v(t)v(t+Δt)
v(t+Δt)
Δv
θ
P
Q
Figure 8.5: A particle moving along an arc of a circle of radiusR. The positions and velocities at the times
t and t +∆ t are shown. The diagram on the right shows the velocity difference, ∆⃗v= ⃗v(t +∆ t) −⃗v(t).",University Physics I Classical Mechanics.pdf
"t and t +∆ t are shown. The diagram on the right shows the velocity difference, ∆⃗v= ⃗v(t +∆ t) −⃗v(t).
We can find out how large this acceleration, and the associatedf o r c e ,h a v et ob e ,b ya p p l y i n ga
little geometry and trigonometry to the situation depictedin Figure8.5.H e r eap a r t i c l ei sm o v i n g
along an arc of a circle of radiusr,s ot h a ta tt h et i m et it is at point P and at the later timet+∆t
it is at point Q. The length of the arc between P and Q (the distance it has traveled) iss = Rθ,
where the angleθ is understood to be in radians. I have assumed the speed to be constant, so the
magnitude of the velocity vector,v,i sj u s te q u a lt ot h er a t i oo ft h ed i s t a n c et r a v e l e d( a l o n gt he
circle), to the time elapsed:v = s/∆t.
Despite the speed being constant, the motion is accelerated,a sIj u s ts a i da b o v e ,b e c a u s et h e
direction of the velocity vector changes. The diagram on theright shows the velocity difference",University Physics I Classical Mechanics.pdf
"direction of the velocity vector changes. The diagram on theright shows the velocity difference
vector ∆⃗v= ⃗v(t +∆ t) − ⃗v(t). We can get its length from trigonometry: if we split the angle θ in
half, we get two right triangles, and for each of them|∆⃗v|/2= v sin(θ/2). Thus, we have
|⃗aav| = |∆⃗v|
∆t = 2v sin(θ/2)
∆t (8.25)
for the magnitude of the average acceleration vector. The instantaneous acceleration is obtained by
taking the limit of this expression as ∆t → 0. In this limit, the angleθ = s/R = v∆t/R becomes",University Physics I Classical Mechanics.pdf
"8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 171
very small, and we can use the so-called “small angle approximation,” which states that sinx ≃ x
when x is small and expressed in radians. Therefore, by Eq. (8.25),
|⃗aav| = 2v sin(θ/2)
∆t ≃ vθ
∆t = v2∆t/R
∆t (8.26)
This expression becomes exact as ∆t → 0, and then ∆t cancels out, showing the instantaneous
acceleration has magnitude
|⃗a| = ac = v2
R (8.27)
This acceleration is called thecentripetal acceleration,w h i c hi sw h yIh a v ed e n o t e di tb yt h es y m b o l
ac.T h e r e a s o n f o r t h a t n a m e i s t h a t i t i salways pointing towards the center of the circle.Y o u
can kind of see this from Figure8.5:i fy o ut a k et h ev e c t o r∆⃗vshown there, and move it (without
changing its direction, so it stays ‘parallel to itself”) tothe midpoint of the arc, halfway between
points P and Q, you will see that it does point almost straightto the center of the circle. (A more",University Physics I Classical Mechanics.pdf
"points P and Q, you will see that it does point almost straightto the center of the circle. (A more
mathematically rigorous proof of this fact, using calculus,w i l lb ep r e s e n t e di nt h en e x tc h a p t e r ,
section 9.3.)
The force ⃗Fc needed to provide this acceleration is called thecentripetal force,a n db yN e w t o n ’ s
second law it has to satisfy⃗Fc = m⃗ ac.T h u s ,t h ec e n t r i p e t a lf o r c eh a sm a g n i t u d e
Fc = mac = mv2
R (8.28)
and, like the acceleration⃗ac,i sa l w a y sd i r e c t e dt o w a r d st h ec e n t e ro ft h ec i r c l e .
Physically, the centripetal forceFc,a sg i v e nb yE q .(8.28), is what it takes to bend the trajectory so
as to keep it precisely on an arc of a circle of radiusR and with constant speedv.N o t et h a t ,s i n c e
⃗Fc is always perpendicular to the displacement (which, over anys h o r tt i m ei n t e r v a l ,i se s s e n t i a l l y",University Physics I Classical Mechanics.pdf
"⃗Fc is always perpendicular to the displacement (which, over anys h o r tt i m ei n t e r v a l ,i se s s e n t i a l l y
tangent to the circle), it doesno work on the object, and therefore (by Eq. (7.11)) its kinetic energy
does not change, sov does indeed stay constant when the centripetal force equalsthe net force.
Note also that “centripetal” is just a job description: it isnot an e wt y p eo ff o r c e . I na n yg i v e n
situation, the role of the centripetal force will be played byo n eo ft h ef o r c e sw ea r ea l r e a d yf a m i l i a r
with, such as the tension on a rope (or an appropriate component thereof) when you are swinging
an object in a horizontal circle, or gravity in the case of themoon or any other satellite.
At this point, if you have never heard about the centripetal force before, you may be feeling a little
confused, since you almost certainly have heard, instead, about a so-calledcentrifugal force that",University Physics I Classical Mechanics.pdf
"confused, since you almost certainly have heard, instead, about a so-calledcentrifugal force that
tends to push spinning things away from the center of rotation. In fact, however, this “centrifugal
force” does not really exist: the “force” that you may feel pushing you towards the outside of a curve
when you ride in a vehicle that makes a sharp turn is really nothing but your own inertia—your
body “wants” to keep moving on a straight line, but the car, bybending its trajectory, is preventing
it from doing so. The impression that you get that you would flyradially out, as opposed to along a
tangent, is also entirely due to the fact that the reference frame you are in (the car) is continuously",University Physics I Classical Mechanics.pdf
"172 CHAPTER 8. MOTION IN TWO DIMENSIONS
changing its direction of motion. You will find this effect illustrated in some detail in an example
in the “Advanced Topics” section, if you want to look at it in more depth.
On the other hand, getting a car to safely negotiate a turn is actually an important example of a
situation that requires a definitecentripetal force. On a flat surface (see the “Advanced Topics”
section for a treatment of a banked curve!), you rely entirelyo nt h ef o r c eo fs t a t i cf r i c t i o nt ok e e p
you on the track, which can typically be modeled as an arc of a circle with some radiusR.S o , i f
you are traveling at a speedv,y o un e e dFs = mv2/R.R e c a l l i n g t h a t t h e f o r c e o f s t a t i c f r i c t i o n
cannot exceed µsFn,a n dt h a to nafl a ts u r f a c ey o uw o u l dj u s th a v eFn = Fg = mg,y o us e ey o u
need to keepmv2/R smaller thanµsmg;o r ,c a n c e l i n gt h em a s s ,
v2
R <µ sg (8.29)",University Physics I Classical Mechanics.pdf
"need to keepmv2/R smaller thanµsmg;o r ,c a n c e l i n gt h em a s s ,
v2
R <µ sg (8.29)
This is the condition that has to hold in order to be able to maket h et u r ns a f e l y . T h em a x i m u m
speed is then vmax = √
µsgR,w h i c h ,a sy o uc a ns e e ,w i l ld e p e n do nt h es t a t eo ft h er o a d( f or
instance, if the road is wet the coefficientµs will be smaller). The posted, recommended speed will
typically take this into consideration and will be as low as ith a st ob et ok e e py o us a f e . N o t i c et h a t
the left-hand side of Eq. (8.29)i n c r e a s e sa st h esquare of the speed, so doubling your speed makes
that term four times larger! Do not even think of taking a turnat 60 mph if the recommended
speed is 30, and do not exceed the recommended speedat all if the road is wet or your tires are
worn.
8.4.2 Kinematic angular variables
Consider a particle moving on a circle, as in Figure8.6 below (next page). Of course, we can",University Physics I Classical Mechanics.pdf
"worn.
8.4.2 Kinematic angular variables
Consider a particle moving on a circle, as in Figure8.6 below (next page). Of course, we can
just use the regular, cartesian coordinates,x and y,t od e s c r i b ei t sm o t i o n . B u t ,i naw a y ,t h i si s
carrying around more information than we typically need, andi ti sa l s on o tv e r yt r a n s p a r e n t : a
value ofx and y does not immediately tell us how far the object has traveled along the circle itself.
Instead, the most convenient way to describe the motion of thep a r t i c l e ,i fw ek n o wt h er a d i u so f
the circle, is to give theangle θ that the position vector makes with some reference axis at any
given time, as shown in Fig.8.6.I fw ec h o o s et h ex axis as the reference, then the conversion from
ad e s c r i p t i o nb a s e do nt h er a d i u sR and the angleθ to a description in terms ofx and y is simply
x = R cos θ
y = R sin θ (8.30)",University Physics I Classical Mechanics.pdf
"ad e s c r i p t i o nb a s e do nt h er a d i u sR and the angleθ to a description in terms ofx and y is simply
x = R cos θ
y = R sin θ (8.30)
so knowing the functionθ(t)w ec a ni m m e d i a t e l yg e tx(t)a n dy(t), if we need them. (Note: in this
section I am using an uppercaseR for the magnitude of the position vector, to emphasize that it
is a constant, equal to the radius of the circle.)",University Physics I Classical Mechanics.pdf
"8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 173
+
r
v
P (x,y)
x
y
R sin θθ
R cos θ
Figure 8.6: A particle moving on a circle. The position vector has lengthR,s ot h ex and y coordinates
are R cos θ and R sin θ, respectively. The conventional positive direction of motion is indicated. The velocity
vector is always, as usual, tangent to the trajectory.
Although the angleθ itself is not a vector quantity, nor a component of a vector, iti sc o n v e n i e n t
to allow for the possibility that it might be negative. The standard convention is thatθ grows in
the counterclockwise direction from the reference axis, andd e c r e a s e si nt h ec l o c k w i s ed i r e c t i o n . O f
course, you can always get to any angle by coming from either direction, so the angle by itself does
not tell you how the particle got there. Information on the direction of motion at any given time
is best captured by the concept of theangular velocity,w h i c hw er e p r e s e n tb yt h es y m b o lω and",University Physics I Classical Mechanics.pdf
"is best captured by the concept of theangular velocity,w h i c hw er e p r e s e n tb yt h es y m b o lω and
define in a manner analogous to the way we defined the ordinary velocity: if ∆θ = θ(t +∆ t) −θ(t)
is theangular displacement over a time ∆t,t h e n
ω =l i m
∆t→0
∆θ
∆t = dθ
dt (8.31)
The standard convention is also to use radians as an angle measure in this context, so that the
units ofω will be radians per second, or rad/s. Note that the radian is adimensionless unit, so it
“disappears” from a calculation when the final result does notc a l lf o ri t( a si nE q .(8.35)b e l o w ) .
For motion with constant angular velocity, we clearly will have
θ(t)= θi + ω(t − ti)o r ∆ θ = ω∆t (constant ω)( 8 . 3 2 )",University Physics I Classical Mechanics.pdf
"174 CHAPTER 8. MOTION IN TWO DIMENSIONS
where ω is positive for counterclockwise motion, and negative for clockwise. (There is as e n s ei n
which it is useful to think ofω as a vector, but, since it is not immediately obvious how or why, I
will postpone discussion of this next chapter, after I have introduced angular momentum.)
When ω changes with time, we can introduce anangular acceleration α,d e fi n e d ,a g a i n ,i nt h e
obvious way:
α =l i m
∆t→0
∆ω
∆t = dω
dt (8.33)
Then for motion with constant angular acceleration we have the formulas
ω(t)= ωi + α(t − ti)o r ∆ ω = α∆t (constant α)
θ(t)= θi + ωi(t − ti)+ 1
2α(t − ti)2 or ∆ θ = ωi∆t + 1
2α(∆t)2 (constant α)( 8 . 3 4 )
Equations (8.32)a n d(8.34)c o m p l e t e l yp a r a l l e lt h ec o r r e s p o n d i n ge q u a t i o n sf o rm o t ion in one
dimension that we saw in Chapter 1. In fact, of course, a circlei sj u s tal i n et h a th a sb e e nb e n t",University Physics I Classical Mechanics.pdf
"dimension that we saw in Chapter 1. In fact, of course, a circlei sj u s tal i n et h a th a sb e e nb e n t
in a uniform way, so the distance traveled along the circle itself is simply proportional to the angle
swept by the position vector⃗r.A sa l r e a d yp o i n t e do u ti nc o n n e c t i o nw i t hF i g .8.5,i fw ee x p r e s s e d
θ in radians then the length of the arc corresponding to an angular displacement ∆θ would be
s = R|∆θ| (8.35)
so multiplying Eqs. (8.32)o r( 8.34)b y R directly gives the distance traveled along the circle in
each case.
Δθ
s
r(t)
r(t+∆t)
∆r
Figure 8.7: A small angular displacement. The distance traveled along the circle,s = R∆θ, is almost
identical to the straight-line distance|∆⃗r| between the initial and final positions; the two quantities become
the same in the limit ∆t → 0.
Figure 8.7 shows that, for very small angular displacements, it does notm a t t e rw h e t h e rt h ed i s t a n c e",University Physics I Classical Mechanics.pdf
"the same in the limit ∆t → 0.
Figure 8.7 shows that, for very small angular displacements, it does notm a t t e rw h e t h e rt h ed i s t a n c e
traveled is measured along the circle itself or on a straightline; that is,s ≃| ∆⃗r|.D i v i d i n gb y∆t,
using Eq. (8.35)a n dt a k i n gt h e∆t → 0l i m i tw eg e tt h ef o l l o w i n gu s e f u lr e l a t i o n s h i pb e t w e e nt h e
angular velocity and the instantaneous speedv (defined in the ordinary way as the distance traveled
per unit time, or the magnitude of the velocity vector):
|⃗v| = R|ω| (8.36)",University Physics I Classical Mechanics.pdf
"8.5. IN SUMMARY 175
As we shall see later, the productRα is also a useful quantity. It isnot,h o w e v e r ,e q u a lt ot h em a g -
nitude of the acceleration vector, but only one of its two components, thetangential acceleration:
at = Rα (8.37)
The sign convention here is that a positiveat represents a vector that is tangent to the circle and
points in the direction of increasingθ (that is, counterclockwise); the full acceleration vectoris
equal to the sum of this vector and thecentripetal accelerationvector, introduced in the previous
subsection, which always points towards the center of the circle and has magnitude
ac = v2
R = Rω2 (8.38)
(making use of Eqs. (8.29)a n d(8.36)). These results will be formally established in the next chapter,
after we introduce the vector product, although you could also verify them right now—if you are
familiar enough with derivatives at this point—by using thechain rule to take the derivatives with",University Physics I Classical Mechanics.pdf
"familiar enough with derivatives at this point—by using thechain rule to take the derivatives with
respect to time of the components of the position vector, as given in Eq. (8.30)( w i t hθ = θ(t), an
arbitrary function of time).
The main thing to remember about the radial and tangential components of the acceleration is that
the radial component (the centripetal acceleration) isalways there for circular motion, whether the
angular velocity is constant or not, whereas the tangentialacceleration is only nonzero if the angular
velocity is changing, that is to say, if the particle is slowing down or speeding up as it turns.
8.5 In summary
1. To solve problems involving motion in two dimensions, youshould break up all the forces
into their components along a suitable pair of orthogonal axes, then apply Newton’s second
law to each direction separately:Fnet,x = max, Fnet,y = may.I t i sc o n v e n i e n tt oc h o o s ey o u r
axes so that at least one of eitherax or ay will be zero.",University Physics I Classical Mechanics.pdf
"axes so that at least one of eitherax or ay will be zero.
2. An object thrown with some horizontal velocity componentand moving under the influence
of gravity alone (near the surface of the Earth) will follow aparabola in a vertical plane.
This results from horizontal motion with constant velocity,a n dv e r t i c a lm o t i o nw i t hc o n s t a n t
acceleration equal to−g,a sd e s c r i b e db ye q u a t i o n s(8.7).
3. To analyze motion up or down an inclined plane, it is convenient to choose your axes so that
the x axis lies along the surface, and they axis is perpendicular to the surface. Then, ifθ is
the angle the incline makes with the horizontal, the force ofgravity on the object will also
make an angleθ with thenegative y axis.
4. Recall that the force of kinetic friction will always pointi nad i r e c t i o no p p o s i t et h em o t i o n ,
and will have magnitudeFk = µkFn,w h e r e a st h ef o r c eo fs t a t i cf r i c t i o nw i l la l w a y st a k e",University Physics I Classical Mechanics.pdf
"176 CHAPTER 8. MOTION IN TWO DIMENSIONS
on whatever value is necessary to keep the object from moving,u pt oam a x i m u mv a l u eo f
Fs
max = µsFn.
5. An object moving in an arc of a circle of radiusR with a speedv experiences a centripetal
acceleration of magnitude ac = v2/R.“ C e n t r i p e t a l ” m e a n s t h e c o r r e s p o n d i n g v e c t o r p o i n t s
towards the center of the circle. Accordingly, to get an object of massm to move on such a
path requires acentripetal forceFc = mv2/R.
6. To describe the motion of a particle on a circle of radiusR,w eu s ea na n g u l a rp o s i t i o nv a r i a b l e
θ(t), in terms of which we define angular displacement ∆θ,a n g u l a rv e l o c i t yω = dθ/dt,a n d
angular acceleration α = dω/dt.T h e e q u a t i o n s f o r m o t i o n i n o n e d i m e n s i o n w i t h c o n s t a n t
acceleration apply to circular motion with constantα with the changesx → θ, v → ω and
a → α.",University Physics I Classical Mechanics.pdf
"acceleration apply to circular motion with constantα with the changesx → θ, v → ω and
a → α.
7. The displacement along the circle,s,c o r r e s p o n d i n gt oa na n g u l a rd i s p l a c e m e n t∆θ,i s( i n
magnitude) s = R|∆θ|.S i m i l a r l y ,t h e( l i n e a r )s p e e do ft h ep a r t i c l e( m a g n i t u d eofi t sv e l o c i t y
vector) is equal tov = R|ω|,a n dt h et a n g e n t i a lc o m p o n e n to fi t sa c c e l e r a t i o nv e c t o rh as
magnitude at = R|α|.I n a d d i t i o n t o t h i s , t h e p a r t i c l e a l w a y s h a s a r a d i a l a c c e l eration
component ar equal to the centripetal acceleration of point 5 above.",University Physics I Classical Mechanics.pdf
"8.6. EXAMPLES 177
8.6 Examples
You will work out a rather thorough example of pro jectile motion in the lab, and Section 8.3
above already has the problem of a block sliding down an inclined plane worked out for you. The
following example will show you how to use the kinematic angular variables of section8.4.2 to deal
with motion in a circle, and to calculate the centripetal acceleration in a simple situation. The
section on Advanced Topics deals with a few more challenging(but interesting) examples.
8.6.1 The penny on the turntable
Suppose that you have a penny sitting on a turntable, a distance d =1 0 c mf r o mt h ea x i so f
rotation. Assume the turntable starts moving, steadily spinning up from rest, in such a way that
after 1.3s e c o n d si th a sr e a c h e di t sfi n a lr o t a t i o nr a t eo f3 3.3r p m ( r e v o l u t i o n s pe r m i n u t e ) . A n s w e r
the following questions:",University Physics I Classical Mechanics.pdf
"the following questions:
(a) What was the turntable’s angular acceleration over the time interval fromt =0t o t =1 .3s ?
(b) How many turns (complete and fractional) did the turntable make before reaching its final
velocity?
(c) Assuming the penny has not slipped, what is its centripetal acceleration once the turntable
reaches its final velocity?
(d) How large does the static friction coefficient between thepenny and the turntable have to be
for the penny not to slip throughout this process?
Solution
(a) We are told that the turntable spins up “steadily” fromt =0t o t =1 .3s . T h e w o r d “ s t e a d i l y ”
here is a keyword that means the (angular) acceleration is constant (that is, the angular velocity
increases at a constant rate).
What is this rate? For constantα,w eh a v e ,f r o mE q .(8.34), α =∆ ω/∆t.H e r e ,t h e t i m ei n t e r v a l
∆t =1 .3, s o w e j u s t n e e d t o fi n d ∆ω.B yd e fi n i t i o n , ∆ω = ωf − ωi,a n ds i n c ew es t a r tf r o mr e s t ,",University Physics I Classical Mechanics.pdf
"∆t =1 .3, s o w e j u s t n e e d t o fi n d ∆ω.B yd e fi n i t i o n , ∆ω = ωf − ωi,a n ds i n c ew es t a r tf r o mr e s t ,
ωi =0 . S ow ej u s tn e e dωf .W ea r et o l dt h a t“ t h efi n a lr o t a t i o nr a t e ”i s3 3.3r p m ( r e v o l u t i o n s pe r
minute). What does this tell us about the angular velocity?
The angular velocity is the number of radians an object movingi nac i r c l e( s u c ha st h ep e n n yi n
this example) travels per second. A complete turn around thecircle, orrevolution,c o r r e s p o n d st o
180◦,o re q u i v a l e n t l y2π radians. So, 33.3r e v o l u t i o n s ,o rt u r n s ,p e rm i n u t em e a n s3 3.3×2π radians
per 60s, that is,
ωf = 33.3 × 2π rad
60 s =3 .49 rad
s (8.39)",University Physics I Classical Mechanics.pdf
"178 CHAPTER 8. MOTION IN TWO DIMENSIONS
The angular acceleration, therefore, is
α = ∆ω
∆t = ωf − ωi
∆t = 3.49 rad/s
1.3s =2 .68 rad
s2 (8.40)
(b) The way to answer this question is to find out the total angular displacement, ∆θ,o ft h ep e n n y
over the time interval considered (fromt =0t o t =1 .3s ) , a n d t h e n c o n v e r t t h i s t o a n u m be r o f
turns, using the relationship 2π rad = 1 turn. To get ∆θ,w es h o u l du s et h ee q u a t i o n(8.34)f o r
motion with constant angular acceleration:
∆θ = ωi∆t + 1
2α(∆t)2 (8.41)
We start from rest, soωi =0 ,W ek n o w∆t =1 .3s , a n d w e j u s t c a l c u l a t e dα =2 .68 rad/s2,s ow e
have
∆θ = 1
2 × 2.68 rad
s2 × (1.3s )2 =2 .26 rad (8.42)
This is less than 2π radians, so it takes the turntable less than one complete revolution to reach its
final angular velocity. To be precise, since 2π radians is one turn, 2.26 rad will be 2.26/(2π)t u r n s ,
which is to say, 0.36 turns—a little more than 1/3o fat u r n .",University Physics I Classical Mechanics.pdf
"which is to say, 0.36 turns—a little more than 1/3o fat u r n .
(c) For the questions above, the penny just served as a markerto keep track of the revolutions of
the turntable. Now, we turn to the dynamics of the motion of thep e n n yi t s e l f . F i r s t ,t og e ti t s
angular acceleration, we can just use Eq. (8.38), in the form
ac = Rω2 =0 .1m ×
(
3.49 rad
s
) 2
=1 .22 m
s2 (8.43)
noticing that R,t h er a d i u so ft h ec i r c l eo nw h i c ht h ep e n n ym o v e s ,i sj u s tt h edistance d to the
axis of rotation that we were given at the beginning of the problem, andω,i t sa n g u l a rv e l o c i t y ,i s
just the final angular velocity of the turntable (assuming, asw ea r et o l d ,t h a tt h ep e n n yh a sn o t
slipped relative to the turntable).
(d) Finally, how about the force needed to keep the penny fromslipping—that is to say, to keep
it moving with the turntable? This is just the centripetal force needed “bend” the trajectory of",University Physics I Classical Mechanics.pdf
"it moving with the turntable? This is just the centripetal force needed “bend” the trajectory of
the penny into a circle of radiusR,s o Fc = mac,w h e r em is the mass of the penny andac is the
centripetal acceleration we just calculated. Physically,we know that this force has to be provided by
the static (as long as the penny does not slip!) friction force between the penny and the turntable.
We know thatFs ≤ µsFn,a n dw eh a v ef o rt h en o r m a lf o r c e ,i nt h i ss i m p l es i t u a t i o n ,just Fn = mg.
Therefore, settingFs = mac we have:
mac = Fs ≤ µsFn = µsmg (8.44)
This is equivalent to the single inequalitymac ≤ µsmg,w h e r ew ec a nc a n c e lo u tt h em a s so ft h e
penny to conclude that we must haveac ≤ µsg,a n dt h e r e f o r e
µs ≥ ac
g = 1.22 m/s2
9.8m / s2 =0 .124 (8.45)",University Physics I Classical Mechanics.pdf
"8.7. ADVANCED TOPICS 179
8.7 Advanced Topics
8.7.1 Staying on track
(This example studies a situation that you could easily setupe x p e r i m e n t a l l ya th o m e( y o uc a nu s e
aw h o l es p h e r ei n s t e a do fah a l f - s p h e r e ! ) ,a l t h o u g ht og e tt hen u m b e r st ow o r ko u ty o ur e a l l yn e e d
to make sure that the friction between the surface and the object you choose is truly negligible.
Essentially the same mathematical approach could be used tostudy the problem of a skier going
over a mogul, or a car losing contact with the road if it is goingt o of a s to v e rah i l l . )
As m a l lo b j e c ti sp l a c e da tt h et o po fas m o o t h( f r i c t i o n l e s s )dome shaped like a half-sphere of
radius R,a n dg i v e nas m a l lp u s hs oi ts t a r t ss l i d i n gd o w nt h ed o m e ,i n itially moving very slowly
(vi ≃ 0), but picking up speed as it goes, until at some point it fliesoff the surface.",University Physics I Classical Mechanics.pdf
"(vi ≃ 0), but picking up speed as it goes, until at some point it fliesoff the surface.
(a) At that point, when the object loses contact with the surface, what is the angle that its position
vector (with origin at the center of the sphere) makes with thev e r t i c a l ?
(b) How far away from the sphere does the object land?
θ
θmax
F n
sb
FG
Ebθ
θ
(a) (b) 
Figure 8.8: An object (small block) sliding on a hemispherical dome. The drawing (a) shows the angle
θmax at which the object flies off (red dashed line), and a smaller, generic angle θ. The drawing (b) shows
the free-body diagram corresponding to the angleθ.
Solution
(a) As we saw in Section 8.4, in order to get an object to move along an arc of a circle, a centripetal
force of magnitude mv2/r is required. As long as our object is in contact with the surface, the
forces acting on it are the normal force (which points along the radial direction, so it makes a",University Physics I Classical Mechanics.pdf
"forces acting on it are the normal force (which points along the radial direction, so it makes a
negative contribution to the centrifugal force) and gravity, which has a componentmg cos θ along
the radius, towards the center of the circle (see Figure8.8(b), the dashed light blue line). So, the
centripetal force equation reads
mv2
R = mg cos θ − Fn (8.46)",University Physics I Classical Mechanics.pdf
"180 CHAPTER 8. MOTION IN TWO DIMENSIONS
The next thing we need to do is find the value of the speedv for a given angleθ.I f w e t r e a t
the object as a particle, its only energy is kinetic energy, and ∆K = Wnet (Eq. (7.11)), where
Wnet is the work done on the particle by the net force acting on it. The normal force is always
perpendicular to the displacement, so it does no work, whereas gravity is always vertical and does
work Wgrav = −mg∆y (taking upwards as positive, so ∆y is negative). In fact, from Figure8.8(a)
(follow the dashed blue line) you can see that for a given angle θ,t h eh e i g h to ft h eo b j e c ta b o v e
the ground isR cos θ,s ot h ev e r t i c a ld i s p l a c e m e n tf r o mi t si n i t i a lp o s i t i o ni s
∆y = −(R − R cos θ)( 8 . 4 7 )
Hence we have, for the change in kinetic energy,
1
2mv2 − 1
2mv2
i = mgR − mgR cos θ (8.48)
Assuming, as we are told in the text of the problem, thatvi ≃ 0, we getv2 ≃ 2gR −2gR cos θ,a n d
using this in Eq. (8.46)",University Physics I Classical Mechanics.pdf
"i = mgR − mgR cos θ (8.48)
Assuming, as we are told in the text of the problem, thatvi ≃ 0, we getv2 ≃ 2gR −2gR cos θ,a n d
using this in Eq. (8.46)
2mg − 2mg cos θ = mg cos θ − Fn (8.49)
or Fn =3 mg cos θ −2mg.T h i s s h o w s t h a tFn starts out (whenθ =0 )h a v i n gi t su s u a lv a l u eo fmg,
and then it becomes progressively smaller as the object slides down. The point where the object
loses contact with the surface is whenFn =0 ,a n dt h a th a p p e n sf o r
3c o sθmax =2 ( 8 . 5 0 )
or θmax =c o s−1(2/3) = 48.2◦.
Recalling that ∆y = −(R −R cos θ), we see that when cosθ =2 /3, the object has fallen a distance
R/3; put otherwise, its height above the ground at the time it flies off is 2R/3, or 2/3o ft h ei n i t i a l
height.
(b) This is just a projectile problem now. We just have to find the values of the initial conditions
(xi,y i,v x,i and vy,i)a n ds u b s t i t u t ei nt h ee q u a t i o n s(8.5). By inspecting the figure, you can see
that, at the time the object flies off,",University Physics I Classical Mechanics.pdf
"that, at the time the object flies off,
xi = R sin θmax =0 .745R
yi = R cos θmax =0 .667R (8.51)
Also, we found above thatv2 ≃ 2gR − 2gR cos θ,a n dw h e nθ = θmax this gives v2 =0 .667gR,
or v =0 .816√
gR.T h e p r o j e c t i o n a n g l e i n t h i s c a s e i s−θmax;t h a ti s ,t h ei n i t i a lv e l o c i t yo ft h e
projectile (dashed red arrow in Figure8.8(a)) is at an angle 48.2◦ below the positivex axis, so we
have:
vx,i = vi cos θmax =0 .544
√
gR
vy,i = −vi sin θmax = −0.609
√
gR (8.52)",University Physics I Classical Mechanics.pdf
"8.7. ADVANCED TOPICS 181
Now we just use these results in Eqs. (8.5). Specifically, we want to know how long it takes for the
object to reach the ground, so we use the last equation (8.5)w i t hy =0a n ds o l v ef o rt:
0= yi + vy,it − 1
2gt2 (8.53)
The result is t =0 .697
√
R/g.( Y o u d o n o t n e e d t o c a r r y t h e “g”t h r o u g h o u t ;i tw o u l db eO K
to substitute 9.8m / s2 for it. I have just kept it in symbolic form so far to make it clear that the
quantities we derive will have the right units.) Substituting this in the equation forx,w eg e t
x = xi + vx,it =0 .745R +0 .544
√
gR × 0.697
√
R/g =1 .125R (8.54)
(Note how theg cancels, so we would get the same result on any planet!) Sincethe sphere has a
radius R,t h eo b j e c tf a l l sad i s t a n c e0.125R away from the sphere.
8.7.2 Going around a banked curve
Roadway engineers often bank a curve, especially if it is a very tight turn, so the cars will not have",University Physics I Classical Mechanics.pdf
"8.7.2 Going around a banked curve
Roadway engineers often bank a curve, especially if it is a very tight turn, so the cars will not have
to rely on friction alone to provide the required centripetalf o r c e . T h ep i c t u r es h o w sac a rg o i n g
around such a curve, which we can model as an arc of a circle of radius r.I nt e r m so fr,t h eb a n k
angle θ,a n dt h ec o e ffi c i e n to fs t a t i cf r i c t i o n ,fi n dt h em a x i m u ms a f espeed around the curve.
F n
sc
FG
Ec
F s
sc
x
y
θ
F n
sc
FG
Ec
F s
sc
x
y
a
Figure 8.9: A car going around a banked curve (sketch and free-body diagram). The center of the circle is
towards the right.
The figure shows the appropriate choice of axes for this problem. The criterion is, again, to choose
the axes so that one of them will coincide with the direction oft h ea c c e l e r a t i o n . I nt h i sc a s e ,t h e
acceleration is all centripetal, that is to say, pointing, horizontally, towards the center of the circle",University Physics I Classical Mechanics.pdf
"acceleration is all centripetal, that is to say, pointing, horizontally, towards the center of the circle
on which the car is traveling.
It may seem strange to see the force of static friction pointing down the slope, but recall that for a
car turning on a flat surface it would have been pointing inwards (towards the center of the circle),",University Physics I Classical Mechanics.pdf
"182 CHAPTER 8. MOTION IN TWO DIMENSIONS
so this is the natural extension of that. In general, you should always try to imagine which way the
object would slide if friction disappeared altogether:⃗Fs must point in the directionopposite that.
Thus, for a car traveling at a reasonable speed, the directioni nw h i c hi tw o u l ds k i di su pt h es l o p e ,
and that means⃗Fs must point down the slope. But, for a car just sitting still onthe tilted road,
⃗Fs must point upwards, and we shall see in a moment that in generalt h e r ei sam i n i m u mv e l o c i t y
required for the force of static friction to point in the direction we have chosen.
Apart from this, the main difference with the flat surface case ist h a tn o wt h en o r m a lf o r c eh a sa
component along the direction of the acceleration, so it helps to keep the car moving in a circle.
On the other hand, note that we now lose (for centripetal purposes) a little bit of the friction force,",University Physics I Classical Mechanics.pdf
"On the other hand, note that we now lose (for centripetal purposes) a little bit of the friction force,
since it is pointing slightly downwards. This, however, is more than compensated for by the fact
that the normal force is greater now than it would be for a flat surface, since the car is now, so to
speak, “driving into” the road somewhat.
The dashed blue lines in the free-body diagram are meant to indicate that the angleθ of the bank
is also the angle between the normal force and the positivey axis, as well as the angle that⃗Fs
makes below the positivex axis. It follows that the components of these two forces alongt h ea x e s
shown are:
Fn
x = Fn sin θ
Fn
y = Fn cos θ (8.55)
and
Fs
x = Fs cos θ
Fn
y = −Fs sin θ (8.56)
The vertical force equation is then:
0= may = Fn
y + Fs
y − FG = Fn cos θ − Fs sin θ − mg (8.57)
This shows thatFn =( mg + Fs sin θ)/ cos θ is indeed greater than justmg for this problem, and",University Physics I Classical Mechanics.pdf
"y + Fs
y − FG = Fn cos θ − Fs sin θ − mg (8.57)
This shows thatFn =( mg + Fs sin θ)/ cos θ is indeed greater than justmg for this problem, and
must increase as the angleθ increases (since cosθ decreases with increasing θ). The horizontal
equation is:
max = Fn
x + Fs
x = Fn sin θ + Fs cos θ = mv2
r (8.58)
where I have already substituted the value of the centripetala c c e l e r a t i o nf o rax.E q u a t i o n s(8.57)
and (8.58)f o r mas y s t e mt h a tn e e d st ob es o l v e df o rt h et w ou n k n o w n sFn and Fs.T h e r e s u l ti s :
Fn = mg cos θ + mv2
r sin θ
Fs = −mg sin θ + mv2
r cos θ (8.59)",University Physics I Classical Mechanics.pdf
"8.7. ADVANCED TOPICS 183
Note that the second equation would haveFs becoming negative ifv2 <g rtan θ.T h i sm e a n st h a t
below that speed, the force of static friction must actuallypoint up the slope, as discussed above.
We can call this particular sp eed, for whichFs becomes zero, vno friction:
vno friction =
√
gr tan θ (8.60)
What this means is that it is possible to arrange the banking angle so that a car going at a specific
speed would not have to rely on friction at all in order to makethe curve: the normal force would
be just right to provide the required centripetal acceleration. A car going at that speed would not
feel either pulled down or pushed up the slope. However, a cargoing faster than that would tend
to “fly off”, and the static friction force would be required topull it in and keep it on the curve,
whereas a car moving more slowly would tend to slide down and would have to be pushed up by",University Physics I Classical Mechanics.pdf
"whereas a car moving more slowly would tend to slide down and would have to be pushed up by
the friction force. Friction, therefore, provides a range ofs a f es p e e d st od r i v ei nt h i sc a s e ,j u s ta s
it did in the flat surface case.
We can calculate the maximum safe sp eed as we did b efore, recalling that we must always have
Fs ≤ µsFn.S u b s t i t u t i n gE q s .(8.59)i nt h i se x p r e s s i o n ,a n ds o l v i n gf o rv,w eg e tt h ec o n d i t i o n
vmax = √
gr
√
µs +t a nθ
1 − µs tan θ (8.61)
This reproduces our result (8.29)f o rθ =0( afl a tr o a d ) ,a si ts h o u l d .
To put some numbers into this, suppose the curve has a radius of2 0 m ,a n dt h ec o e ffi c i e n to fs t a t i c
friction between the tires and the road isµs =0 .7. Then, for a flat surface, we getvmax =1 1.7m / s ,
or about 26 mph, whereas for a bank angle ofθ =1 0◦ (the angle chosen for the figure above) we
get vmax =1 4m / s ,o ra b o u t3 1m p h .",University Physics I Classical Mechanics.pdf
"or about 26 mph, whereas for a bank angle ofθ =1 0◦ (the angle chosen for the figure above) we
get vmax =1 4m / s ,o ra b o u t3 1m p h .
Equation (8.61)a c t u a l l yi n d i c a t e st h a tt h em a x i m u mv e l o c i t yw o u l d“ b e c o mei n fi n i t e ”f o rafi n i t e
bank angle, namely, if 1− µs tan θ =0 ,o rt a nθ =1 /µs (if µs =0 .7, this corresponds toθ =5 5◦).
This is mathematically correct, but of course we cannot takeit literally: it assumes that there is
no limit to how large a normal force the roadway may exert without sustaining damage, and also
that Fs can become arbitrarily large as long as it stays below the bound Fs ≤ µsFn.N e i t h e r o f
these assumptions would hold in real life for very large speeds. Also, the angleθ =t a n−1(1/µs)i s
much too steep: recall that, according to Eq. (8.24), the force of friction will only be able to keep
an object (initially at rest) from sliding down the slope if tan θ ≤ µs,w h i c hf o rµs =0 .7m e a n s",University Physics I Classical Mechanics.pdf
"an object (initially at rest) from sliding down the slope if tan θ ≤ µs,w h i c hf o rµs =0 .7m e a n s
θ ≤ 35◦.S o , w i t h a b a n k a n g l e o f5 5◦ you might drive on the curve, provided you were going fast
enough, but you could not park on it—the car would slide down!Bottom line, use Eq. (8.61)o n l y
for moderate values ofθ... and do not exceedθ =t a n−1 µs if you want a car to be able to drive
around the curve slowly without sliding down into the ditch.",University Physics I Classical Mechanics.pdf
"184 CHAPTER 8. MOTION IN TWO DIMENSIONS
8.7.3 Rotating frames of reference: Centrifugal force and Coriolis force
Imagine you are inside a rotating cylindrical room of radiusR.T h e r ei sam e t a lp u c ko nt h efl o o r ,
ad i s t a n c er from the axis of rotation, held in place with an electromagnet. At some time you
switch off the electromagnet and the puck is free to slide without friction. Find where the puck
strikes the wall, and show that, if it was not too far away fromthe wall to begin with, it appears
as if it had moved straight for the wall as soon as it was released.
Solution
The picture looks as shown below, to an observer in aninertial frame, looking down. The puck
starts at point A, with instantaneous velocityωr pointing straight to the left at the moment it is
released, so it just moves straight (in the inertial frame) until it hits the wall at point B. From the
cyan-colored triangle shown, we can see that it travels a distance
√
R2 − r2,w h i c ht a k e sat i m e
∆t =
√",University Physics I Classical Mechanics.pdf
"cyan-colored triangle shown, we can see that it travels a distance
√
R2 − r2,w h i c ht a k e sat i m e
∆t =
√
R2 − r2
ωr (8.62)
In this time, the room rotates counterclockwise through an angle ∆θroom = ω∆t:
∆θroom =
√
R2 − r2
r (8.63)
O
A
A‘
B
B‘ rR
Figure 8.10: The motion of the puck (cyan) and the wall (magenta) as seen by aninertial observer.
This is the angle shown in magenta in the figure. As a result of this rotation, the point A’ that was
initially on the wall straight across from the puck has moved(following the magenta dashed line)
to the position B’, so to an observer in the rotating room, looking at things from the point O, the
puck appears to head for the wall and drift a little to the rightw h i l ed o i n gs o .",University Physics I Classical Mechanics.pdf
"8.7. ADVANCED TOPICS 185
The cyan angle in the picture, which we could call ∆θpart,h a st a n g e n te q u a lt o
√
R2 − r2/r,s ow e
have
∆θroom =t a n( ∆θpart)( 8 . 6 4 )
This tells us the two angles are going to be pretty close if theya r es m a l le n o u g h ,w h i c hi sw h a t
happens if the puck starts close enough to the wall in the firstplace. The picture shows, for clarity,
the case whenr =0 .7R,w h i c hg i v e s∆θroom =1 .02 rad, and ∆θpart =t a n−1(1.02) = 0.8r a d . F o r
r =0 .9R,o nt h eo t h e rh a n d ,o n efi n d s∆θroom =0 .48 rad, and ∆θpart =t a n−1(0.48) = 0.45 rad.
In terms of pseudoforces (forces that do not, physically, exist, but may be introduced to describe
mathematically the motion of objects in non-inertial frameso fr e f e r e n c e ) ,t h en o n - i n e r t i a lo b s e r v e r
would say that the puck heads towards the wall because of acentrifugal force (that is, a force",University Physics I Classical Mechanics.pdf
"would say that the puck heads towards the wall because of acentrifugal force (that is, a force
pointing away from the center of rotation), and while doing soi td r i f t st ot h er i g h tb e c a u s eo ft h e
so-called Coriolis force.",University Physics I Classical Mechanics.pdf
"186 CHAPTER 8. MOTION IN TWO DIMENSIONS
8.8 Problems
Problem 1Ap i t c h e rt h r o w saf a s t b a l lh o r i z o n t a l l ya tas p e e do f4 2m / s .Neglecting air resistance,
(a) How long does it take for the ball to reach the batter, a distance of 18.4m a w a y ?
(b) How much does the ball drop vertically in this time?
(c) What is the vertical component of the ball’s velocity as itr e a c h e st h eb a t t e r ?
(Just to set the record straight, a real fastball would probably not drop that much, because of a
lift force, called the Magnus effect, due to the interaction ofthe air with the backspin of the ball!)
Problem 2 Two blocks are connected by a massless string threaded over amassless, frictionless
pulley, as shown in the picture. The mass of block 1 is 2 kg and the mass of block 2 is 1.5k g .T h e
angle of the incline is 30 degrees. There is friction betweenthe block and the inclined surface.
1
2
30º",University Physics I Classical Mechanics.pdf
"angle of the incline is 30 degrees. There is friction betweenthe block and the inclined surface.
1
2
30º
(a) Start by assuming that the coefficient of static friction iss t r o n ge n o u g ht ok e e pt h es y s t e mf r o m
moving, and draw free-body diagrams for the two blocks. Try tog e ta l lt h ef o r c e sa p p r o x i m a t e l y
to scale. (The following questions may be helpful.)
(b) If the system is not moving, what is the magnitude of the tension?
(c) What is the magnitude of the normal force?
(d) How large does the coefficient of static friction have to beto keep the system from moving?
(e) Now suppose the system is moving, and the coefficient of sliding (kinetic) friction is 0.2. What
is the acceleration of the system?
(f) What is the tension now?
(g) What is the rate at which energy is dissipated (instantaneous dissipated power) when the
system’s velocity is 3 m/s?",University Physics I Classical Mechanics.pdf
"(f) What is the tension now?
(g) What is the rate at which energy is dissipated (instantaneous dissipated power) when the
system’s velocity is 3 m/s?
Problem 3A6 0 - k gs k i e rs t a r t ss l i d i n gf r o mr e s tf r o mt h et o po fas l o p ethat makes an angle of 30◦
with the horizontal. Assume the bottom of the slope is 100 m below the top (measured vertically).
(a) What is the change in the gravitational potential energyof the system formed by the skier and
the Earth, as the skier slides from the top to the bottom of theslope?",University Physics I Classical Mechanics.pdf
"8.8. PROBLEMS 187
(b) What is the work done by gravity on the skier for the processd e s c r i b e da b o v e ? ( T h i n k i n gn o w
of the skier only as the system.)
(c) If you could neglect the friction between the skis and thesnow, what would be the speed of the
skier at the bottom of the slope? Why?
(d) If the speed of the skier is only 30 m/s as she reaches the bottom of the slope, how much energy
was dissipated by friction?
(e) Draw a free-body diagram of the skier as she is sliding downt h es l o p e .M a k es u r ey o ui n c l u d e
friction, and indicate the direction of the acceleration. Use your diagram to answer the next couple
of questions.
(f) What is the magnitude of the normal force exerted by the ground on the skier?
(g) Under the same assumptions as in part (d) above, what is thec o e ffi c i e n to ff r i c t i o nb e t w e e n
the skis and the snow?
(h) Again under the assumption that her final speed is 30 m/s, what is the acceleration of the skier
along the slope?",University Physics I Classical Mechanics.pdf
"the skis and the snow?
(h) Again under the assumption that her final speed is 30 m/s, what is the acceleration of the skier
along the slope?
Problem 4 Ac h i l di sd r a g g i n ga2 k gs l e dt h r o u g hafl a tp a t c ho fs n o w( c o e fficient of kinetic
friction: 0.1) with a constant force, by pulling on a rope at an angle of 25◦ to the horizontal. The
sled is moving at a constant speed.
(a) Draw a free body diagram for the sled as it is being pulled.
(b) Find the magnitude of all the forces acting on the sled.
(c) If the child were to suddenly release the rope, what wouldbe the new value of the friction force?
What would be the sled’s acceleration? (Take the initial direction of motion to be positive.)
Problem 5 Am a ni ss w i n g i n ga no b j e c t ,a t t a c h e dt oas t r i n g ,i nac i r c l eover his head (see the
sketches).
v
Top view
Side view
(a) Draw a free-body diagram for the object. Notice the stringi sn o td r a w nh o r i z o n t a li nt h e“ s i d e",University Physics I Classical Mechanics.pdf
"sketches).
v
Top view
Side view
(a) Draw a free-body diagram for the object. Notice the stringi sn o td r a w nh o r i z o n t a li nt h e“ s i d e
view” diagram above. Why is that?
(b) If the mass of the object is 1 kg, what is the vertical component of the tension?
(c) If the object makes 3 turns per second, and the radius of thec i r c l e( a ss e e ni nt h et o pv i e w )i s
0.8m ,w h a ti st h ec e n t r i p e t a la c c e l e r a t i o no ft h eo b j e c t ?",University Physics I Classical Mechanics.pdf
"188 CHAPTER 8. MOTION IN TWO DIMENSIONS
(d) Which force component in your diagram provides this centripetal acceleration?
(e) Based on your results above, what is the angle the string must make with the horizontal?
Problem 6Ag o l fb a l li sh i ti ns u c haw a yt h a ti tt r a v e l s3 0 0 mh o r i z o n t a lly and stays in the air
at o t a lo f6 s . W h a tw a si t si n i t i a lv e l o c i t y ? ( g i v eh o r i z o n t ala n dv e r t i c a lc o m p o n e n t s ,a n da l s o
magnitude and direction).
Problem 7A2 k gb l o c ki si n i t i a l l ya tr e s ta tt h et o po fa3 5◦ incline, and is then allowed to slide
down the incline. The coefficient of kinetic friction isµk =0 .25.
(a) Draw a free-body diagram for the block.
(b) Find the components of the gravitational force in the given coordinate system.
(c) Find the normal force and the force of kinetic friction.
(d) Find the magnitude of the acceleration of the block.
(e) What is the final speed of the block after sliding 0.75 m?",University Physics I Classical Mechanics.pdf
"(d) Find the magnitude of the acceleration of the block.
(e) What is the final speed of the block after sliding 0.75 m?
(f) If the coefficient of static friction isµs =0 .4, what would the maximum value of the angleθ be
for the block to not slide when it is released?
Problem 8 You throw a ball for your dog to fetch. The ball leaves your handw i t has p e e do f
2m / s , a t a n a n g l e o f 3 0◦ to the horizontal, and from a height of 1.5m a bo v e t h e g r o u n d . T h e m a s s
of the ball is 0.5k g . N e g l e c t a i r r e s i s t a n c e i n w h a t f o l l o w s .
(a) What is the acceleration of the ball while it is in flight? Report it as a vector, that is, specify
magnitude and direction (or vertical and horizontal components; if the latter, specify which direc-
tion(s) you take as positive).
(b) What is the kinetic energy of the ball as it leaves your hand?
(c) Consider the Earth as being in the system. What is the potential energy of the Earth-ball",University Physics I Classical Mechanics.pdf
"(c) Consider the Earth as being in the system. What is the potential energy of the Earth-ball
system (1) as the ball leaves your hand, (2) at its maximum height, and (3) as it finally hits the
ground?
(d) How high does the ball riseabove the ground?
(e) What is the kinetic energy of the ball as it hits the ground?
(f) Now let the system be the ball alone. How much work does theEarth do on the ball while it is
in flight? (from start to finish)
(g) What is thevelocity of the ball as it hits the ground? Report it as a vector.
(h) How far away from you (horizontally) does the ball land?
Problem 9Am a ni ss t a n d i n go nt h ep l a t f o r mo fam e r r y - g o - r o u n d ,n o th o lding on to anything.
The merry-go-round is turning at a constant rate, and makes acomplete turn every 10 s.
(a) What is the merry-go-round’s angular velocity?
(b) If the man is standing at a distance of 2 m from the center ofthe merry-go-round, what is his
centripetal acceleration?",University Physics I Classical Mechanics.pdf
"(b) If the man is standing at a distance of 2 m from the center ofthe merry-go-round, what is his
centripetal acceleration?
(c) Which actual force acting on the man is responsible for this acceleration?
(d) What is the minimum value ofµs,t h es t a t i cf r i c t i o nc o e ffi c i e n t ,b e t w e e nt h es o l e so ft h em a n’s
shoes and the platform?",University Physics I Classical Mechanics.pdf
"8.8. PROBLEMS 189
(e) The power is turned off and the platform slows down to a stopwith a constant angular accel-
eration of−0.02 rad/s2.H o wl o n gd o e si tt a k ef o ri tt os t o pc o m p l e t e l y ?
(f) What is the man’s tangential acceleration during that time? Does his centripetal acceleration
change? Why?
(g) How many turns does the platform make before coming to a complete stop?",University Physics I Classical Mechanics.pdf
190 CHAPTER 8. MOTION IN TWO DIMENSIONS,University Physics I Classical Mechanics.pdf
"Chapter 9
Rotational dynamics
Rotational motion, which involves an object spinning arounda na x i s ,o rr e v o l v i n ga r o u n dap o i n t
in space, is actually rather common in nature, so much so thatGalileo thought (mistakenly) that
circular motion, rather than motion on a straight line, was the “natural,” or “unforced” state of
motion for any body. Galileo was wrong, but there is at least one sense in which it is true that
rotational motion, once started, can go on forever in the absence of external forces. The underlying
principle is theconservation of angular momentum,w h i c hIw i l li n t r o d u c el a t e ri nt h i sc h a p t e r .
As pointed out in the previous chapter, rotational motion isalso extremely important in mechanical
devices. In every case, the rotation of an extended, rigid body can be mathematically described as
ac o l l e c t i o no fc i r c u l a rm o t i o n sb yt h ep a r t i c l e sm a k i n gu pthe body. Two very important quantities",University Physics I Classical Mechanics.pdf
"ac o l l e c t i o no fc i r c u l a rm o t i o n sb yt h ep a r t i c l e sm a k i n gu pthe body. Two very important quantities
for dealing with such collections of particles in rotation are the rotational kinetic energy, and the
angular momentum. These will both be introduced, and their properties explored, in this chapter.
9.1 Rotational kinetic energy, and moment of inertia
If a particle of massm is moving on a circle of radiusR,w i t hi n s t a n t a n e o u ss p e e dv,t h e ni t sk i n e t i c
energy is
Krot = 1
2mv2 = 1
2mR2ω2 (9.1)
using |⃗v| = R|ω|,E q .(8.36). Note that, at this stage, there is no real reason for the subscript “rot”:
equation (9.1)i sa l lo ft h ep a r t i c l e ’ sk i n e t i ce n e r g y . T h ed i s t i n c t i o nw ill only become important later
in the chapter, when we consider extended objects whose motion is a combination of translation
(of the center of mass) and rotation (around the center of mass).
191",University Physics I Classical Mechanics.pdf
"192 CHAPTER 9. ROTATIONAL DYNAMICS
Now, consider the kinetic energy of an extended object that isr o t a t i n ga r o u n ds o m ea x i s . W e
may treat the object as being made up of many “particles” (small parts) of massesm1, m2 ... .I f
the object is rigid, all the particles move together, in the sense that they all rotate through the
same angle in the same time, which means they all have the sameangular velocity. However, the
particles that are farther away from the axis of rotation areactually moving faster—they have a
larger v,a c c o r d i n gt oE q .(8.36). So the expression for the total kinetic energy in terms of all the
particles’ speeds is complicated, but in terms of the (common) angular velocity is simple:
Krot = 1
2m1v2
1 + 1
2m2v2
2 + 1
2m3v2
3 + ...
= 1
2
(
m1r2
1 + m2r2
2 + m3r2
3 + ...
)
ω2
= 1
2Iω2 (9.2)
where r1,r 2,... represent the distance of the 1st, 2nd... particle to the axis of rotation, and on the
last line I have introduced the quantity
I =
∑",University Physics I Classical Mechanics.pdf
"last line I have introduced the quantity
I =
∑
all particles
mr2 (9.3)
which is usually called themoment of inertiaof the object about the axis considered. In general,
the expression (9.3)i se v a l u a t e da sa ni n t e g r a l ,w h i c hc a nb ew r i t t e ns y m b o l i c ally asI =
∫
r2 dm;
the “mass element”dm can be expressed in terms of the local density asρdV ,w h e r eV is a volume
element. The integral is a multidimensional integral that may require somewhat sophisticated
calculus skills, so we will not be calculating any of these this semester; rather, we will rely on the
tabulated values for I for objects of different, simple, shapes. For instance, for a homogeneous
cylinder of total massM and radius R,r o t a t i n ga r o u n di t sc e n t r a la x i s ,I = 1
2 MR2;f o rah o l l o w
sphere rotating through an axis through its center,I = 2
3 MR2,a n ds oo n .
As you can see, the expression (9.2)f o rt h ek i n e t i ce n e r g yo far o t a t i n gb o d y ,1",University Physics I Classical Mechanics.pdf
"3 MR2,a n ds oo n .
As you can see, the expression (9.2)f o rt h ek i n e t i ce n e r g yo far o t a t i n gb o d y ,1
2 Iω2,p a r a l l e l st h e
expression 1
2 mv2 for a moving particle, with the replacement ofv by ω,a n dm by I.T h i ss u g g e s t s
that I is some sort of measure of a solid object’srotational inertia,b yw h i c hw em e a nt h er e s i s t a n c e
it offers to being set into rotation about the axis being considered. We will see later on, when we
introduce the torque, that this interpretation forI is indeed correct.
It should be stressed that the moment of inertia depends, in general, not just on the shape and
mass distribution of the object, but also on the axis of rotation. In general, the formula (9.3)s h o w s
that, the more mass you put farther away from the axis of rotation, the largerI will be. Thus,
for instance, a thin rod of lengthl has a moment of inertiaI = 1
12 Ml2 when rotating around a",University Physics I Classical Mechanics.pdf
"for instance, a thin rod of lengthl has a moment of inertiaI = 1
12 Ml2 when rotating around a
perpendicular axis through its midpoint, whereas it has thelarger I = 1
3 Ml2 when rotating around
ap e r p e n d i c u l a ra x i st h r o u g ho n eo fi t se n d p o i n t s .",University Physics I Classical Mechanics.pdf
"9.2. ANGULAR MOMENTUM 193
9.2 Angular momentum
Back in Chapter 3 we introduced the momentum of an object moving in one dimension asp = mv,
and found that it had the interesting property of being conserved in collisions between objects
that made up an isolated system. It seems natural to ask whether the corresponding rotational
quantity, formed by multiplying the “rotational inertia”I and the angular velocityω,h a sa n y
interesting properties as well. We are tentatively1 going to call the quantityIω angular momentum
(to distinguish it from the “ordinary,” orlinear momentum, mv), and build towards a better
understanding, and a formal definition of it, in the remaindero ft h i ss e c t i o n .
Two things are soon apparent: one is that, unlike the rotational kinetic energy, which was just
plain kinetic energy, the quantityIω really is different from the ordinary momentum, since it has",University Physics I Classical Mechanics.pdf
"plain kinetic energy, the quantityIω really is different from the ordinary momentum, since it has
different dimensions (see the “extra” factor ofR in Eq. (9.4)b e l o w ) . T h eo t h e ri st h a tt h e r ea r e ,i n
fact, systems in nature where this quantity appears to remainc o n s t a n tt oag o o da p p r o x i m a t i o n .
For instance, the Earth spinning around its axis, at a constant rate of 2π radians every 24 hours,
has, by virtue of that, a constant angular momentum (a constant I and ω).
It is best, however, to start by thinking about how we would define “angular momentum” for a
system consisting of a single particle, and then building upfrom there, as we have been doing
all semester with every new concept. For a particle moving inac i r c l e ,a c c o r d i n gt ot h ep r e v i o u s
section, the moment of inertia is justI = mR2,a n dt h e r e f o r eIω is just mR2ω.U s i n g E q . (8.36),",University Physics I Classical Mechanics.pdf
"section, the moment of inertia is justI = mR2,a n dt h e r e f o r eIω is just mR2ω.U s i n g E q . (8.36),
we can write (the letterL is the conventional symbol for angular momentum; do not mistake it for
al e n g t h ! )
L = Iω = ±mR|v| (particle moving in a circle) (9.4)
where, for consistency with our sign convention forω,w es h o u l du s et h ep o s i t i v es i g ni ft h er o t a t i o n
is counterclockwise, and the negative sign if it is clockwise.
We can again readily think of examples in nature where the quantity (9.4)i sc o n s e r v e d : f o ri n s t a n c e ,
the moon, if we treat it as a particle orbiting the Earth in an approximately circular orbit2,h a s
then an approximately constant angular momentum, as given byE q .(9.4). This example is also
interesting because it offers an inkling of an important difference between ordinary momentum,
and angular momentum: it appears that the latter can be conserved even in the presence of some",University Physics I Classical Mechanics.pdf
"and angular momentum: it appears that the latter can be conserved even in the presence of some
kinds of external forces (in this case, the force of gravity due to the Earth that keeps the moon on
its orbit).
On the other hand, it is not immediately obvious how to generalize the definition (9.4)t oo t h e r
kinds of motion. If we simply try something likeL = mvr,w h e r er is the distance to a fixed axis,
or to a fixed point, then we find that this yields a quantity thatis constantly changing, even for
the simplest possible physical system, namely, a particle moving on a straight line with constant
1As we shall see below, in generalIω is equal to only one component of the angular momentumvector.
2You can peek ahead at Figure10.1,i nt h en e x tc h a p t e r ,t os e eh o wg o o da na p p r o x i m a t i o nt h i sm ight be!",University Physics I Classical Mechanics.pdf
"194 CHAPTER 9. ROTATIONAL DYNAMICS
velocity. Yet, we would like to defineL in such a way that it will remain constant when, in fact,
nothing in the particle’s actual state of motion is changing.
The way to do this, for a particle moving on a straight line, isto define L as the product ofmv
times, not the distance of the particle to a point, but the distance of the particle’s line of motionto
the point considered. The “line of motion” is just a straightline that contains the velocity vector
at any given time. The distance between a line and a point O is,by definition, the shortest distance
from O to any point on the line; it is given by the length of a segment drawn perpendicular to the
line through the point O.
For a particle moving in a circle, the line of motion at any timei st a n g e n tt ot h ec i r c l e ,a n ds ot h e
distance between the line of motion and the center of the circle is just the radiusR,s ow er e c o v e r",University Physics I Classical Mechanics.pdf
"distance between the line of motion and the center of the circle is just the radiusR,s ow er e c o v e r
the definition Eq. (9.4). For the general case, on the other hand, we have the situation shown in
Fig. 9.1:i ft h e i n s t a n t a n e o u sv e l o c i t yo f t h e p a r t i c l e i s⃗v,a n dw ed r a wt h ep o s i t i o nv e c t o ro ft h e
particle, ⃗r,w i t ht h ep o i n tOa st h eo r i g i n ,t h e nt h ed i s t a n c eb e t w e e nOa ndt h el i n eo fm o t i o n
(sometimes also called theperpendicular distance between O and the particle) is given byr sin θ,
where θ is the angle between the vectors⃗rand ⃗v.
v
r
θ
r sin θ = r' sin θ'
v
O
r'
θ'
ϖ–θ'
Figure 9.1: For a particle moving on a straight line, the distance from the point O to the particle’s line of
motion (dashed line) is equal tor sin θ at every point in the trajectory (two possible points are shown in the
figure). The distance is the length of the blue segment. The figure shows there is some freedom in choosing",University Physics I Classical Mechanics.pdf
"figure). The distance is the length of the blue segment. The figure shows there is some freedom in choosing
how the angleθ between ⃗rand ⃗vis to be measured, since the sine ofθ is the same as the sine ofπ − θ.
We therefore try and define angular momentum relative to a p oint O as the product
L = ±mr|v| sin θ (9.5)
with the positive sign if the point O is to the left of the line ofm o t i o na st h ep a r t i c l ep a s s e sb y
(which corresponds to counterclockwise motion on the circle), and negative otherwise.
There are two very good things about the definition (9.5): the first one is that there is already
in existence a mathematical operation between vectors, called the vector,o r cross, product,
according to which|L|,a sd e fi n e db y(9.5), would just be given by|⃗r× ⃗p|,w h e r e⃗r× ⃗pis the cross
product of⃗rand ⃗p;Iw i l lh a v eal o tm o r et os a ya b o u tt h i si nt h en e x ts u b s e c t i o n.T h e s e c o n d g o o d",University Physics I Classical Mechanics.pdf
"9.2. ANGULAR MOMENTUM 195
thing is that, with this definition, the angular momentum willb ec o n s e r v e di na ni m p o r t a n tk i n d
of process, namely, a collision that converts linear to rotational motion, as illustrated in Fig.9.2
below.
v
r
θ
O
1 2
Figure 9.2: Collision between two particles, 1 and 2, of equal masses. Particle 2 istied by a massless string
to the point O. After the collision, particle 1 is at rest and particle 2 moves in the circle shown.
In the picture, particle 1 is initially moving with constantvelocity ⃗vand particle 2 is initially at
rest, tied by a massless string to the point O. If the particlesh a v et h es a m em a s s ,c o n s e r v a t i o no f
(ordinary) momentum and kinetic energy means that when theycollide they exchange velocities:
particle 1 comes to a stop and particle 2 starts to move to the right with the same speedv,b u t
immediately the string starts pulling on it and bending its path into a circle. Assuming negligible",University Physics I Classical Mechanics.pdf
"immediately the string starts pulling on it and bending its path into a circle. Assuming negligible
friction between the string and the pivoting point (and all the surfaces involved), the speed of
particle 2 will remain constant as it rotates, by conservation of kinetic energy, because the tension
in the string is always perpendicular to the particle’s displacement vector, so it does no work on it.
All of the above means that angular momentum is conserved: before the collision it was equal to
−mvr sin θ = −mvR for particle 1, and 0 for particle 2; after the collision it iszero for particle 1 and
Iω = mR2ω = −mRv for particle 2 (noteω is negative, because the rotation is clockwise). Kinetic
energy is also conserved, for the reasons argued above. On theo t h e rh a n d ,o r d i n a r ym o m e n t u mi s
only conserved until just after the collision, when the string starts pulling on particle 2, since this
represents an external force on the system.",University Physics I Classical Mechanics.pdf
"only conserved until just after the collision, when the string starts pulling on particle 2, since this
represents an external force on the system.
So we have found a situation in which linear motion is converted to circular motion while angular
momentum, as defined by Eq. (9.5), is conserved, and this despite the presence of an externalforce.
It is true that, in fact, we were able to solve for the final motion without making explicit use of
conservation of angular momentum: we only had to invoke conservation of ordinary momentum for
the short duration of the collision, and conservation of kinetic energy. But it is easy to generalize
the problem depicted in Figure9.2 to one that cannot be solved by these methods, butcan be
solved if angular momentum is constant. Suppose that we replace particle 2 and the string by a",University Physics I Classical Mechanics.pdf
"196 CHAPTER 9. ROTATIONAL DYNAMICS
thin rod of massm and length l pivoted at one end. What happens now when particle 1 strikes
the rod?
vi
r
θ
O
1 vf
Figure 9.3: Collision between a particle initially moving with velocity⃗v1 and a rod of lengthl pivoted at
an endpoint. The particle strikes the rod perpendicularly, at the other end. After the collision, the particle
is moving along the same line with velocity⃗vf , and the rod is rotating around the point O with an angular
velocity ω.
If the particle strikes the rod perpendicularly, and the collision happens very fast (that is, it is over
before the rod has time to move a significant distance), we mayassume the force exerted by the rod
on the particle (a normal force) is along the original line ofmotion, so the particle will continue to
move on that line. On the other hand, this time we cannot neglect the force exerted by the pivot
on the bar during the collision time, since the bar is one single rigid object. This means the system",University Physics I Classical Mechanics.pdf
"on the bar during the collision time, since the bar is one single rigid object. This means the system
is not isolated during the collision, and we cannot rely on ordinary momentum conservation.
Suppose, however, that the angular momentumL is conserved, as well as the total energy (the
pivot does no work on the system, since there is no displacement of that point). The initial angular
momentum is Li = −mvil.T h e fi n a l a n g u l a r m o m e n t u m i s−mvf l for the particle (note that,
if the particle bounces back,vf will be negative) andIω for the rod. The initial kinetic energy
is Ki = 1
2 mv2
i ,a n dt h efi n a lk i n e t i ce n e r g yi s1
2 mv2
f for the particle and1
2 Iω2 for the rod (recall
Eq. (9.2), so we have to solve the system
−mvil = −mvf l + Iω
1
2mv2
i = 1
2mv2
f + 1
2Iω2 (9.6)
The general solution, for arbitrary values of all the constants, is
vf = ml2 − I
ml2 + I vi
ω = − 2ml
ml2 + I vi (9.7)",University Physics I Classical Mechanics.pdf
"9.3. THE CROSS PRODUCT AND ROTATIONAL QUANTITIES 197
Note that this does reduce to our previous results for the collision of two particles if we make
I = ml2 (which one could always do, by choosing the mass of the rod,M,a p p r o p r i a t e l y ) . O nt h e
other hand, as indicated at the end of the previous subsection, for generalM we haveI = 1
3 Ml2
for the rod, so we can cancell2 almost everywhere and end up with
vf = 3m − M
3m + M vi
ω = − 6m
3m + M
vi
l (9.8)
In particular, we see that ifm = M the particle continues to move forward with 1/2o fi t si n i t i a l
velocity, and the rod spins withω = −(3/2)vi/l,w h i c hi sa c t u a l l yal a r g e ra n g u l a rv e l o c i t yt h a n
what we found for the system in Fig.9.2.
This example shows how useful conservation of angular momentum can be, but, of course, we do not
really know yet whether angular momentum is actually conserved in this problem! I will address",University Physics I Classical Mechanics.pdf
"really know yet whether angular momentum is actually conserved in this problem! I will address
this very important question—when is angular momentum conserved—in the section after next,
which is to say, after I have properly developed angular momentum as avector quantity.
9.3 The cross product and rotational quantities
The cross, or vector, product of two vectors⃗A and ⃗B is denoted by ⃗A × ⃗B.I t i s d e fi n e d a s a
vector perpendicular to both⃗A and ⃗B (that is to say, to the plane that contains them both), with
am a g n i t u d eg i v e nb y
| ⃗A × ⃗B| = AB sin θ (9.9)
where A and B are the magnitudes of⃗A and ⃗B,r e s p e c t i v e l y ,a n dθ is the angle between⃗A and ⃗B,
when they are drawn either with the same origin or tip-to-tail.
The specific direction of⃗A × ⃗B depends on the relative orientation of the two vectors. Basically,
if ⃗B is counterclockwise from⃗A,w h e nl o o k i n gd o w no nt h ep l a n ei nw h i c ht h e yl i e ,a s s u m i n gt hey",University Physics I Classical Mechanics.pdf
"if ⃗B is counterclockwise from⃗A,w h e nl o o k i n gd o w no nt h ep l a n ei nw h i c ht h e yl i e ,a s s u m i n gt hey
are drawn with a common origin, then⃗A × ⃗B points upwards from that plane; otherwise, it points
downward (into the plane). One can also use the so-calledright-hand rule,i l l u s t r a t e di nF i g u r e9.4
(next page) to figure out the direction of⃗A× ⃗B.N o t e t h a t , b y t h i s d e fi n i t i o n , t h e d i r e c t i o n o f⃗A× ⃗B
is the opposite of the direction of⃗B × ⃗A (as also illustrated in Fig.9.4). Hence, the cross-product
is non-commutative: the order of the factors makes a difference.
⃗A × ⃗B = −⃗B × ⃗A (9.10)",University Physics I Classical Mechanics.pdf
"198 CHAPTER 9. ROTATIONAL DYNAMICS
B
A × B
B × AA
A
B
Figure 9.4: The “right-hand rule” to determine the direction of the cross product. Line up the first vector
with the fingers, and the second vector with the flat of the hand, and the thumb will point in the correct
direction. In the first drawing, we are looking at the plane formed by⃗A and ⃗B from above; in the second
drawing, we are looking at the plane from below, and calculating⃗B × ⃗A.
It follows from Eq. (9.10)t h a tt h ec r o s s - p r o d u c to fa n yv e c t o rw i t hi t s e l fm u s tb ez e ro. In fact,
according to Eq. (9.9), the cross product of any two vectors that are parallel to each other is zero,
since in that caseθ =0 ,a n ds i n 0=0 . I nt h i sr e s p e c t ,t h ec r o s sp r o d u c ti st h eo p p osite of the
dot product that we introduced in Chapter 7: it is maximum whent h ev e c t o r sb e i n gm u l t i p l i e d",University Physics I Classical Mechanics.pdf
"dot product that we introduced in Chapter 7: it is maximum whent h ev e c t o r sb e i n gm u l t i p l i e d
are orthogonal, and zero when they are parallel. (And, of course, the result of⃗A × ⃗B is a vector,
whereas ⃗A · ⃗B is a scalar.)
Besides not being commutative, the cross product also does not have the associative property of
ordinary multiplication: ⃗A×( ⃗B × ⃗C) is different from (⃗A× ⃗B)× ⃗C.Y o u c a ns e e t h i s e a s i l y f r o m t h e
fact that, if ⃗A = ⃗B,t h es e c o n de x p r e s s i o nw i l lb ez e r o ,b u tt h efi r s to n eg e n e r a lly will be nonzero
(since ⃗A × ⃗C is not parallel, but rather perpendicular to⃗A).
In spite of these oddities, the cross product is extremely useful in physics. We will use it to define
the angular momentum vector⃗L of a particle, relative to a point O, as follows:
⃗L = ⃗r× ⃗p= m⃗ r×⃗v (9.11)
where ⃗ris the position vector of the particle, relative to the pointO. This definition gives us",University Physics I Classical Mechanics.pdf
"⃗L = ⃗r× ⃗p= m⃗ r×⃗v (9.11)
where ⃗ris the position vector of the particle, relative to the pointO. This definition gives us
ac o n s t a n tv e c t o rf o rap a r t i c l em o v i n go nas t r a i g h tl i n e ,a sdiscussed in the previous section:
the magnitude of ⃗L,a c c o r d i n gt oE q .(9.9)w i l lb emrv sin θ,w h i c h ,a ss h o w ni nF i g .9.1,d o e s
not change as the particle moves. As for the direction, it is always perpendicular to the plane
containing ⃗rand ⃗v(the plane of the paper, in Fig.9.1), and if you imagine moving⃗vto point O,
keeping it parallel to itself, and apply the right-hand rule,y o uw i l ls e et h a t⃗L in Fig. 9.1 should
point into the plane of the paper at all times.
To see how the definition (9.11)w o r k sf o rap a r t i c l em o v i n gi nac i r c l e ,c o n s i d e ra g a i nt h esituation
shown in Figure8.6 in the previous chapter, but now extend it to three dimensions, as in Fig.9.5,",University Physics I Classical Mechanics.pdf
"shown in Figure8.6 in the previous chapter, but now extend it to three dimensions, as in Fig.9.5,
on the next page. It is straightforward to verify that, for thed i r e c t i o no fm o t i o ns h o w n ,t h ec r o s s",University Physics I Classical Mechanics.pdf
"9.3. THE CROSS PRODUCT AND ROTATIONAL QUANTITIES 199
product ⃗r× ⃗vwill always point upwards, along the positivez axis. Furthermore, since ⃗rand ⃗v
always stay perpendicular, the magnitude of⃗L,b yE q .(9.9), will always be|⃗L| = mR|⃗v|.T a k i n g
note ofI = mR2 and of Eq. (8.36), we see we have then
|⃗L| =
(
mR2) |⃗v|
R = I|ω| (9.12)
+
r v
P (x,y)
x
y
R sin θ
θ
R cos θ
z
L
ω
Figure 9.5: A particle moving on a circle in thex-y plane. For the direction of rotation shown, the vectors
⃗L = m⃗ r×⃗vand ⃗ωlie along thez axis, in the positive direction.
This suggests that we should define theangular velocity vector, ⃗ω,a sav e c t o ro fm a g n i t u d e|ω|,
pointing along the positivez axis if the motion in thex-y plane is counterclockwise as seen from
above (and in the opposite direction otherwise). Then this will hold as a vector equation:
⃗L = I⃗ ω (9.13)
It may seem a very strange choice to have the angular velocitypoint along thez axis, when the",University Physics I Classical Mechanics.pdf
"⃗L = I⃗ ω (9.13)
It may seem a very strange choice to have the angular velocitypoint along thez axis, when the
particle is moving in thex-y plane, but in a certain way it makes sense. Suppose the particle is
moving with constant angular velocity: the directions of⃗rand ⃗vare constantly changing, but⃗ωis
pointing along the positivez direction, which does remain fixed throughout.
There are some other neat things we can do with⃗ωas defined above. Consider the cross product
⃗ω×⃗r.I n s p e c t i o n o f F i g u r e9.5 and of Eq. (8.36)s h o w st h a tt h i si sn o t h i n go t h e rt h a nt h eo r d i n a r y
velocity vector,⃗v:
⃗v= ⃗ω×⃗r (9.14)
We can also take the derivative of⃗ωto obtain the angular acceleration vector⃗α,s ot h a tE q .(8.33)
will hold as a vector equation:
⃗α=l i m
∆t→0
⃗ω(t +∆ t) − ⃗ω(t)
dt = d⃗ ω
dt (9.15)",University Physics I Classical Mechanics.pdf
"200 CHAPTER 9. ROTATIONAL DYNAMICS
For the motion depicted in Fig.9.5,t h ev e c t o r⃗αwill point along the positivez axis if the vector⃗ωis
growing (which means the particle is speeding up), and alongthe negativez axis if⃗ωis decreasing.
One important property the cross product does have is thedistributive property with respect to
the sum: (
⃗A + ⃗B
)
× ⃗C = ⃗A × ⃗C + ⃗B × ⃗C (9.16)
This, it turns out, is all that’s necessary in order to be ableto apply the product rule of differenti-
ation to calculate the derivative of a cross product; you justh a v et ob ec a r e f u ln o tt oc h a n g et h e
order of the factors in doing so. We can then take the derivative of both sides of Eq. (9.14)t og e t
an expression for the acceleration vector:
⃗a= d⃗ v
dt = d⃗ ω
dt ×⃗r+ ⃗ω× d⃗ r
dt
= ⃗α×⃗r+ ⃗ω×⃗v (9.17)
The first term on the right-hand side,⃗α× ⃗r,l i e si nt h ex-y plane, and is perpendicular to⃗r;i ti s ,",University Physics I Classical Mechanics.pdf
"dt ×⃗r+ ⃗ω× d⃗ r
dt
= ⃗α×⃗r+ ⃗ω×⃗v (9.17)
The first term on the right-hand side,⃗α× ⃗r,l i e si nt h ex-y plane, and is perpendicular to⃗r;i ti s ,
therefore, tangential to the circle. In fact, looking at its magnitude, it is clear that this is just the
tangential acceleration vector,w h i c hIi n t r o d u c e d( a sas c a l a r )i nE q .(8.37).
As for the second term in (9.17), ⃗ω× ⃗v,n o t i n gt h a t⃗ωand ⃗vare always perpendicular, it is clear
its magnitude is|ω||⃗v| = Rω2 = v2/R (making use of Eq. (8.36)a g a i n ) . T h i si sj u s tt h em a g n i t u d e
of the centripetal acceleration we studied in the previous chapter (section 8.4). Also, using the
right-hand rule in Fig.9.5,y o uc a ns e et h a t⃗ω×⃗valways points inwards, towards the center of the
circle; that is, along the direction of−⃗r.P u t t i n g a l l o f t h i s t o g e t h e r , w e c a n w r i t e t h i s v e c t o r a s",University Physics I Classical Mechanics.pdf
"circle; that is, along the direction of−⃗r.P u t t i n g a l l o f t h i s t o g e t h e r , w e c a n w r i t e t h i s v e c t o r a s
just −ω2⃗r,a n dt h ew h o l ea c c e l e r a t i o nv e c t o ra st h es u mo fat a n g e n t i a land a centripetal (radial)
component, as follows:
⃗a= ⃗at + ⃗ac
⃗at = ⃗α×⃗r
⃗ac = −ω2⃗r (9.18)
To conclude this section, let me return to the angular momentum vector, and ask the question of
whether, in general, the angular momentum of a rotating system, defined as the sum of Eq. (9.11)
over all the particles that make up the system, will or not satisfy the vector equation⃗L = I⃗ ω.W e
have seen that this indeed works for a particle moving in a circle. It will, therefore, also work for
any object that is essentially flat, and rotating about an axisp e r p e n d i c u l a rt oi t ,s i n c ei nt h a tc a s e
all its parts are just moving in circles around a common center. This was the case for the thin rod",University Physics I Classical Mechanics.pdf
"all its parts are just moving in circles around a common center. This was the case for the thin rod
we considered in connection with Figure9.3 in the previous subsection.
However, if the system is a three-dimensional object rotating about an arbitrary axis, the result
⃗L = I⃗ ωdoes not generally hold. The reason is, mathematically, thatt h em o m e n to fi n e r t i aI
is defined (Eq. (9.3)) in terms of the distances of the particles to anaxis,w h e r e a st h ea n g u l a r",University Physics I Classical Mechanics.pdf
"9.4. TORQUE 201
momentum involves the particle’s distance to apoint. For particles at different “heights” along the
axis of rotation, these quantities are different. It can be shown that, in the general case, all we can
say is thatLz = Iωz,i fw ec a l lz the axis of rotation and calculate⃗L relative to a point on that
axis.
On the other hand, if the axis of rotation is an axis of symmetryo ft h eo b j e c t ,t h e n⃗L has only a z
component, and the result⃗L = I⃗ ωholds as a vector equation. Most of the systems we will consider
this semester will be covered under this clause, or under the“essentially flat” clause mentioned
above.
In what follows we will generally assume thatI has only a z component, and we will drop the
subscript z in the equationLz = Iωz,s ot h a tL and ω will not necessarily be the magnitudes of
their respective vectors, but numbers that could be positiveo rn e g a t i v e ,d e p e n d i n go nt h ed i r e c t i o n",University Physics I Classical Mechanics.pdf
"their respective vectors, but numbers that could be positiveo rn e g a t i v e ,d e p e n d i n go nt h ed i r e c t i o n
of rotation (clockwise or counterclockwise). This is essentially the same convention we used for
vectors in one dimension, such as⃗aor ⃗p,i nt h ee a r l yc h a p t e r s ;i ti sfi n ef o ra l lt h ec a s e si nw h i c h
the (direction of the) axis of rotation does not change with time, which are the only situations we
will consider this semester.
9.4 Torque
We are finally in a p osition to answer the question, when is angular momentum conserved? To do
this, we will simply take the derivative of⃗L with respect to time, and use Newton’s laws to find
out under what circumstances it is equal to zero.
Let us start with a particle and calculate
d⃗L
dt = d
dt (m⃗ r×⃗v)= md⃗ r
dt ×⃗v+ m⃗ r× d⃗ v
dt (9.19)
The first term on the right-hand side goes as⃗v×⃗v,w h i c hi sz e r o . T h es e c o n dt e r mc a nb er e w r i t t e n",University Physics I Classical Mechanics.pdf
"dt ×⃗v+ m⃗ r× d⃗ v
dt (9.19)
The first term on the right-hand side goes as⃗v×⃗v,w h i c hi sz e r o . T h es e c o n dt e r mc a nb er e w r i t t e n
as m⃗ r×⃗a.B u t ,a c c o r d i n gt oN e w t o n ’ ss e c o n dl a w ,m⃗ a= ⃗Fnet.S o ,w ec o n c l u d et h a t
d⃗L
dt = ⃗r× ⃗Fnet (9.20)
So the angular momentum, like the ordinary momentum, will beconserved if the net force on the
particle is zero, but also, and this is an important difference,w h e nt h en e tf o r c ei sp a r a l l e l( o r
antiparallel) to the position vector. For motion on a circlewith constant speed, this is precisely
what happens: the force acting on the particle is the centripetal force, which can be written as
⃗Fc = m⃗ ac = −mω2⃗r(using Eq. (9.18)), so⃗r× ⃗Fc =0 ,a n dt h ea n g u l a rm o m e n t u mi sc o n s t a n t .",University Physics I Classical Mechanics.pdf
"202 CHAPTER 9. ROTATIONAL DYNAMICS
The quantity ⃗r× ⃗F is called the torque of a force around a point (the origin from which⃗ris
calculated, typically a pivot point or center of rotation).It is denoted with the Greek letterτ,
“tau”:
⃗τ= ⃗r× ⃗F (9.21)
For an extended ob ject or system, the rate of change of the angular momentum vector would be
given by the sum of the torques of all the forces acting on all the particles. For each torque one
needs to use the position vector of the particle on which the force is acting. As was the case when
calculating the rate of change of the ordinary momentum of anextended system (Section 6.1.1),
Newton’s third law, with a small additional assumption, leads to the cancellation of the torques
due to the internal forces3,a n ds ow ea r el e f tw i t ho n l y
⃗τext,all = d⃗Lsys
dt (9.22)
It goes without saying that all the torques and angular momenta need to be calculated relative to
the same point.",University Physics I Classical Mechanics.pdf
"⃗τext,all = d⃗Lsys
dt (9.22)
It goes without saying that all the torques and angular momenta need to be calculated relative to
the same point.
The torque of a force around a point is basically a measure of how effective the force would be at
causing a rotation around that point. Since|⃗r× ⃗F| = rF sin θ,y o uc a ns e et h a ti td e p e n d so nt h r e e
things: the magnitude of the force, the distance from the center of rotation to the point where the
force is applied, and the angle at which the force is applied.All of this can be understood pretty
well from Figure9.6 below, especially if you have ever had to use a wrench to tighten or loosen a
bolt:
AB OF1
F4F3F2
rA
rBθ
Figure 9.6: The torque around the point O of each of the forces shown is a measure of how effective it is
at causing the rod to turn around that point.
Clearly, the force ⃗F1 will not cause a rotation at all, and accordingly its torque isz e r o( s i n c ei t",University Physics I Classical Mechanics.pdf
"Clearly, the force ⃗F1 will not cause a rotation at all, and accordingly its torque isz e r o( s i n c ei t
is parallel to⃗rA). On the other hand, of all the forces shown, the most effectiveone is ⃗F3:i t i s
applied the farthest away from O, for the greatest leverage (again, think of your experiences with
3The additional assumption is that the force between any two particles lies along the line connecting the two
particles (which means it is parallel or antiparallel to thevector ⃗r1 − ⃗r2). In that case, ⃗r1 × ⃗F12 + ⃗r2 × ⃗F21 =
(⃗r1 − ⃗r2) × ⃗F12 =0 . M o s tf o r c e si nn a t u r es a t i s f yt h i sc o n d i t i o n .",University Physics I Classical Mechanics.pdf
"9.4. TORQUE 203
wrenches). It is also perpendicular to the rod, for maximum effect (sinθ =1 ) . T h ef o r c e⃗F2,b y
contrast, although also applied at the point A is at a disadvantage because of the relatively small
angle it makes with⃗rA.I fy o ui m a g i n eb r e a k i n gi tu pi n t oc o m p o n e n t s ,p a r a l l e la n dperpendicular
to the rod, only the perpendicular component (whose magnitude is F2 sin θ) would be effective at
causing a rotation; the other component, the one parallel tothe rod, would be wasted, like⃗F1.
In order to calculate torques, then, we basically need to find,f o re v e r yf o r c e ,t h ec o m p o n e n tt h a ti s
perpendicular to the position vector of its point of application. Clearly, for this purpose we can no
longer represent an extended body as a mere dot, as we did for the free-body diagrams in Chapter
6. What we need is a more careful sketch of the object, just detailed enough that we can tell how",University Physics I Classical Mechanics.pdf
"6. What we need is a more careful sketch of the object, just detailed enough that we can tell how
far from the center of rotation and at what angle each force isapplied. That kind of diagram is
called anextended free-body diagram.
Figure 9.6 could be an example of an extended free-body diagram, for an object being acted on by
four forces. Typically, though, instead of drawing the vectors ⃗rA and ⃗rB we would just indicate their
lengths on the diagram (or maybe even leave them out altogether, if we do not want to overload
the diagram with detail). I will show a couple of examples of extended free-body diagrams in the
next couple of sections.
As indicated above, to calculate the torque of each force acting on an extended object you should
use the position vector⃗rof the point where the force is applied. This is typically unambiguous for
contact forces4,b u tw h a ta b o u tg r a v i t y ? I np r i n c i p l e ,t h ef o r c eo fg r a v i t ywould act on all of the",University Physics I Classical Mechanics.pdf
"contact forces4,b u tw h a ta b o u tg r a v i t y ? I np r i n c i p l e ,t h ef o r c eo fg r a v i t ywould act on all of the
particles making up the body, and we would have to add up all thec o r r e s p o n d i n gt o r q u e s :
⃗τG = ⃗r1 × ⃗FG
E,1 + ⃗r2 × ⃗FG
E,2 + ... (9.23)
We can, however, simplify this substantially by noting that(near the surface of the Earth, at any
rate), all the forcesFG
E,1,F G
E,2,... point in the same direction (which is to say, down), and they are
all proportional to each particle’s mass. If I let the total mass of the object beM,a n dt h et o t a lf o r c e
due to gravity on the object be⃗FG
E,obj ,t h e nIh a v e⃗FG
E,1 = m1 ⃗FG
E,obj /M, ⃗FG
E,2 = m2 ⃗FG
E,obj /M, . . .,
and I can rewrite Eq. (9.23)a s
⃗τG = m1⃗r1 + m2⃗r2 + ...
M × ⃗FG
E,obj = ⃗rcm × ⃗FG
E,obj (9.24)
where ⃗rcm is the position vector of the object’s center of mass. So to findt h et o r q u ed u et og r a v i t y",University Physics I Classical Mechanics.pdf
"E,obj = ⃗rcm × ⃗FG
E,obj (9.24)
where ⃗rcm is the position vector of the object’s center of mass. So to findt h et o r q u ed u et og r a v i t y
on an extended object, just take the total force of gravity onthe object (that is to say, theweight
of the object) to be applied at its center of mass. (Obviously,t h e n ,t h et o r q u eo fg r a v i t ya r o u n d
the center of mass itself will be zero, but in some important cases an object may be pivoted at a
point other than its center of mass.)
4Actually, friction forces and normal forces may be “spread out” over a whole surface, but, if the object has enough
symmetry, it is usually OK to have them “act” at the midpoint oft h a ts u r f a c e . T h i sc a nb ep r o v e da l o n gt h el i n e s
of the derivation for gravity that follows.",University Physics I Classical Mechanics.pdf
"204 CHAPTER 9. ROTATIONAL DYNAMICS
Coming back to Eq. (9.22), the main message of this section (other, of course, than thed e fi n i t i o n
of torque itself), is that the rate of change of an object or system’s angular momentum is equal to
the net torque due to the external forces. Two special resultsf o l l o wf r o mt h i so n e . F i r s t ,i ft h en e t
external torque is zero, angular momentum will be conserved,a sw a st h ec a s e ,i np a r t i c u l a r ,f o rt h e
collision illustrated earlier, in Fig.9.3,b e t w e e nap a r t i c l ea n dar o dp i v o t e da to n ee n d .T h eo n l y
external force in that case was the force exerted on the rod, att h ep i v o tp o i n t ,b yt h ep i v o ti t s e l f ,
but the torque of that force around that point is obviously zero, since ⃗r=0 ,s oo u ra s s u m p t i o n
that the total angular momentum around that point was conserved was legitimate.",University Physics I Classical Mechanics.pdf
"that the total angular momentum around that point was conserved was legitimate.
Secondly, if⃗L = I⃗ ωholds, and the moment of inertiaI does not change with time, we can rewrite
Eq. (9.22)a s
⃗τext,all = I⃗ α (9.25)
which is basically the rotational equivalent of Newton’s, second law, ⃗F = m⃗ a.W e w i l l u s e t h i s
extensively in the remainder of this chapter.
Finally, note that situations where the moment of inertia ofas y s t e m ,I,c h a n g e sw i t ht i m ea r e
relatively easy to arrange for any deformable system. Especially interesting is the case when the
external torque is zero, soL is constant, and a change inI therefore brings about a change in
ω = L/I:t h i si sh o w ,f o ri n s t a n c e ,a ni c e - s k a t e rc a nm a k eh e r s e l fs pin faster by bringing her arms
closer to the axis of rotation (reducing herI), and, conversely, slow down her spin by stretching
out her arms. This can be done even in the absence of a contact point with the ground: high-board",University Physics I Classical Mechanics.pdf
"out her arms. This can be done even in the absence of a contact point with the ground: high-board
divers, for instance, also spin up in this way when they curl their bodies into a ball. Note that,
throughout the dive, the diver’s angular momentum around itsc e n t e ro fm a s si sc o n s t a n t ,s i n c e
the only force acting on him (gravity, neglecting air resistance) has zero torque about that point.
In all the cases just mentioned, the angular momentum is constant, but the rotational kinetic energy
changes. This is due to the work done by the internal forces (oft h ei c e - s k a t e r ’ so rt h ed i v e r ’ sb o d y ) ,
converting some internal energy (such as elastic muscular energy) into rotational kinetic energy, or
vice-versa. A convenient expression for a system’s rotational kinetic energy when⃗L = I⃗ ωholds is
Krot = 1
2Iω2 = L2
I (9.26)
which shows explicitly howK would change ifI changed andL remained constant.
9.5 Statics",University Physics I Classical Mechanics.pdf
"Krot = 1
2Iω2 = L2
I (9.26)
which shows explicitly howK would change ifI changed andL remained constant.
9.5 Statics
Statics is the branch of mechanics concerned with the forcesand stresses5 needed to keep a system at
rest, in a stable equilibrium—so that it will not move, bend orc o l l a p s e . I ti s ,o b v i o u s l y ,e x t r e m e l y
5A“ s t r e s s ”i sa“ d i s t r i b u t e d ”f o r c ei na ne x t e n d e do b j e c t ,v arying continuously from one point to another.",University Physics I Classical Mechanics.pdf
"9.5. STATICS 205
important in engineering (particularly in mechanical engineering). In an introductory physics
course, we can only deal with it at a very elementary level, byignoring altogether the deformation
of extended objects such as planks and beams (and the associated stresses), and just imposing two
simple conditions for static equilibrium:fi r s t ,t h en e t( e x t e r n a l )f o r c eo nt h es y s t e mm u s tb ez e r o ,
to make sure its center of mass stays at rest; and second, the net (external) torque on the system
must also be zero, so that it does not rotate. These conditionsc a nb es y m b o l i c a l l ye x p r e s s e da s
∑ ⃗Fext =0
∑
⃗τext =0 (9.27)
You may ask about which point one should calculate the torques. The answer is that, as long as
the first condition is satisfied (sum of the forces is zero), itdoes not matter! The proof is simple,
but you are welcome to skip it if you are not interested.",University Physics I Classical Mechanics.pdf
"but you are welcome to skip it if you are not interested.
Suppose you have two points, A and B, around which to calculatet h et o r q u e s . L e t⃗rA1,⃗ rA2,...
be the position vectors of the points of application of the forces ⃗F1, ⃗F2 ... ,r e l a t i v et op o i n tA ,a n d
⃗rB1,⃗ rB2,... ,t h es a m e ,b u tr e l a t i v et op o i n tB .I fy o ug oa l lt h ew a yb a c kt oFigure 1.6 (in Chapter
1), you can see that these vectors only differ from the first set byt h es i n g l ec o n s t a n tv e c t o r⃗rAB
that gives the position of point B relative to point A:⃗rA1 = ⃗rAB + ⃗rB1,e t c . T h e n ,f o rt h es u mo f
torques around A we have
⃗rA1 × ⃗F1 + ⃗rA2 × ⃗F2 + ... =( ⃗rAB + ⃗rB1) × ⃗F1 +( ⃗rAB + ⃗rA2) × ⃗F2 + ...
= ⃗rAB ×
(
⃗F1 + ⃗F2 ...
)
+ ⃗rB1 × ⃗F1 + ⃗rB2 × ⃗F2 + ... (9.28)
The first term on the last line is zero if the sum of all the forcesi sz e r o ,a n dw h a ti sl e f ti st h e",University Physics I Classical Mechanics.pdf
"+ ⃗rB1 × ⃗F1 + ⃗rB2 × ⃗F2 + ... (9.28)
The first term on the last line is zero if the sum of all the forcesi sz e r o ,a n dw h a ti sl e f ti st h e
sum of all the torques around B. This, indeed, for statics problems, as long as we are enforcing∑ ⃗Fext =0 ,i td o e sn o tm a t t e ra b o u tw h i c hp o i n tw ec h o o s et oc a l c u l a tet h et o r q u e . An a t u r a l
choice is the system’s center of mass, since that is typicallyap o i n to fh i g hs y m m e t r y ,b u tw em a y
also choose a point where there are many applied forces, and sog e tr i do ft h e mq u i c k l y( s i n c et h e i r
torques about that point will be zero).
The way all this works is probably best illustrated with an example. Figure 9.9 (next page) shows a
classic one, a ladder leaning against a wall. The sketch on thel e f ts h o w st h ea n g l e sa n dd i m e n s i o n s
involved, whereas the proper extended free-body diagram, showing all the forces and their points
of application, is on the right.",University Physics I Classical Mechanics.pdf
"involved, whereas the proper extended free-body diagram, showing all the forces and their points
of application, is on the right.
The minimum number of forces needed to balance the system is four: the weight of the ladder
(acting at the center of mass), a normal force from the ground,a n o t h e rn o r m a lf o r c ef r o mt h e
wall, and a force of static friction from the ground that prevents the ladder from slipping. In real
life there should also be a force of static friction from the wall, pointing upwards (also to prevent
slippage); and, of course, if there is a person on the ladder she will exert an additional force down
on it (equal to her weight), applied at whatever point she is standing. I am not going to consider",University Physics I Classical Mechanics.pdf
"206 CHAPTER 9. ROTATIONAL DYNAMICS
any of these complications, just to keep the example simple,but they could be dealt with in exactly
the same way.
θ
ϖ/2– θ
l
Fwl
n
Fgl
n
FEl
G
Fgl
s
Figure 9.7: A ladder leaning against a frictionless wall: sketch and extended free-body diagram.
With the convention that a vector quantity without an arrow ont o pr e p r e s e n t st h a tv e c t o r ’ s
magnitude, the equation for the balance of the vertical forces reads
FN
gl − mg =0 ( 9 . 2 9 )
For the horizontal forces, we have
FN
wl − Fs
gl =0 ( 9 . 3 0 )
Then, taking torques around the point where the ladder is in contact with the ground, neither of
the two forces applied at that point will contribute, and thecondition that the sum of the torques
equal zero becomes
FN
wll sin θ − mg l
2 cos θ =0 ( 9 . 3 1 )
This is because the angle made by the force of gravity with theposition vector of its point of
application is π
2 − θ,a n ds i n (π",University Physics I Classical Mechanics.pdf
"This is because the angle made by the force of gravity with theposition vector of its point of
application is π
2 − θ,a n ds i n (π
2 − θ)=c o sθ.F r o m t h e fi r s t e q u a t i o n w e g e t t h a tFN
gl = mg;f r o m
the second we get that the other normal force,FN
wl = Fs
gl.I f w e s u b s t i t u t e t h i s i n (9.31), and cancel
out l,t h el e n g t ho ft h el a d d e r ,w eg e tt h ec o n d i t i o n
Fs
gl = 1
2mg cot θ (9.32)
But the force of static friction cannot exceedµsFN
gl = µsmg,s o ,s e t t i n gt h er i g h t - h a n ds i d eo f(9.32)
to be lower than or equal toµsmg,a n dc a n c e l i n gt h ec o m m o nf a c t o rmg,w eg e tt h ec o n d i t i o n
cot θ ≤ 2µs, or tan θ ≥ 1
2µs
(9.33)
for theminimum angle θ at which we can lean the ladder before it slips and falls.",University Physics I Classical Mechanics.pdf
"9.6. ROLLING MOTION 207
9.6 Rolling motion
As a step up from a statics problem, we may consider a situationi nw h i c ht h es u mo ft h ee x t e r n a l
forces is zero, as well as the sum of the external torques, yetthe system is moving. We call
this “unforced motion.” The first condition,∑ ⃗Fext =0 ,m e a n st h a tt h ec e n t e ro fm a s so ft h e
system must be moving with constant velocity; the second condition means that the total angular
momentum must be constant. For a rigid body, this means that the most general kind of unforced
motion can be described as a translation of the center of masswith constant velocity, accompanied
by a rotation with constant angular velocity around the center of mass. For an extended, deformable
system, on the other hand, the presence of internal forces canm a k et h eg e n e r a lm o t i o nal o tm o r e
complicated. Just think, for instance, of the solar system:although everything is, loosely speaking,",University Physics I Classical Mechanics.pdf
"complicated. Just think, for instance, of the solar system:although everything is, loosely speaking,
revolving around the sun, the motions of individual planetsand (especially) moons can be fairly
complicated.
As i m p l ee x a m p l eo f( f o rp r a c t i c a lp u r p o s e s )u n f o r c e dm o t i oni sp r o v i d e db yas y m m e t r i c ,r i g i d
object (such as a ball, or a wheel) rolling on a flat surface. Then o r m a la n dg r a v i t yf o r c e sc a n c e l
each other, and since they lie along the same line their torques cancel too, so both⃗vcm and ⃗L
remain constant. In principle, you could imagine removing the ground and gravity and nothing
would change: the same motion (in the absence of air resistance) would just continue forever.
In practice, there is energy dissipation associated with rolling motion, primarily because, if the
rolling object is not perfectly rigid6, then, as it rolls, different parts of it get compressed under the",University Physics I Classical Mechanics.pdf
"rolling object is not perfectly rigid6, then, as it rolls, different parts of it get compressed under the
combined pressure of gravity and the normal force, expand again, get compressed again... This
kind of constant “squishing” ends up converting macroscopick i n e t i ce n e r g yi n t ot h e r m a le n e r g y :
you may have noticed that the tires on a car get warm as you drivea r o u n d ,a n dy o um a ya l s o
be familiar with the fact that you get a better gas mileage (less energy dissipation) when your
tires are inflated to the right pressure than when they are low(because they are more “rigid,” less
deformable, in the first case).
This conversion of mechanical energy into thermal energy canb ef o r m a l l yd e s c r i b e db yi n t r o d u c i n g
another “friction” force that we call the force ofrolling friction.E v e n t u a l l y , r o l l i n g f r i c t i o n a l o n e",University Physics I Classical Mechanics.pdf
"another “friction” force that we call the force ofrolling friction.E v e n t u a l l y , r o l l i n g f r i c t i o n a l o n e
would bring any rolling object to a stop, even in the absence ofa i rr e s i s t a n c e . I ti s ,h o w e v e r ,u s u a l l y
much weaker than sliding friction, so we will continue to ignore it from now on. You may have
noticed already that typically an object can roll on a surfacem u c hf a r t h e rt h a ni tc a ns l i d ew i t h o u t
rolling on the same surface. In fact, what happens often is that, if you try to send the object (for
instance, a billiard ball) sliding, it will lose kinetic energy rapidly to the force of kinetic friction,
but it will also start spinning under the influence of the sameforce, until a critical point is reached
6There is a simple argument, based on Einstein’s theory of relativity, that shows an infinitely rigid object cannot",University Physics I Classical Mechanics.pdf
"6There is a simple argument, based on Einstein’s theory of relativity, that shows an infinitely rigid object cannot
exist: if it did, you could send a signal instantly from one endo fi tt ot h eo t h e r ,j u s tb yp u s h i n go rp u l l i n go ny o u r
end. In practice, such motion cannot reach the other end faster than the speed of a sound wave (that is to say, a
compression wave) in the material. We will study such waves (which imply the medium is not infinitely rigid) in
Chapter 12.",University Physics I Classical Mechanics.pdf
"208 CHAPTER 9. ROTATIONAL DYNAMICS
when thecondition for rolling without slippingis satisfied:
|vcm| = R|ω| (9.34)
At this point, the object will start rolling without slipping, and losing speed at a much slower rate.
The origin of the condition (9.34)i sf a i r l ys t r a i g h t f o r w a r d . Y o uc a ni m a g i n ea no b j e c tt h a tis
rolling without slipping as “measuring the surface” as it rolls (or vice-versa, the surface measuring
the circumference of the object as its different points are pressed against it in succession). So, after it
has completed a revolution (2π radians), it should have literally “covered” a distance on the surface
equal to 2πR,t h a ti s ,a d v a n c e dad i s t a n c e2πR.B u t t h e s a m e h a s t o b e t r u e , p r o p o r t i o n a t e l y ,
for any rotation angle ∆θ other that 2π:s i n c e t h e l e n g t h o f t h e c o r r e s p o n d i n g a r c i ss = R|∆θ|,",University Physics I Classical Mechanics.pdf
"for any rotation angle ∆θ other that 2π:s i n c e t h e l e n g t h o f t h e c o r r e s p o n d i n g a r c i ss = R|∆θ|,
in a rotation over an angle|∆θ| the center of mass of the object must have advanced a distance
|∆xcm| = s = R|∆θ|.D i v i d i n gb y∆t as ∆t → 0t h e ny i e l d sE q .(9.34).
θ
vcm
Figure 9.8: Left: illustrating the rolling without slipping condition. The cyan line on the surface has the
same length as the cyan-colored arc, and will be the distance traveled by the disk when it has turned through
an angle θ. Right: velocities for four points on the edge of the disk. The pink arrows are the velocities in
the center of mass frame. In the Earth reference frame, the velocity of the center of mass,⃗vcm, in green, has
to be added to each of them. The resultant is shown in blue for two ofthem.
Note that, unlike Eq. (8.36), which it very much resembles, Eq. (9.34)i s not a“ v e c t o ri d e n t i t yi n",University Physics I Classical Mechanics.pdf
"Note that, unlike Eq. (8.36), which it very much resembles, Eq. (9.34)i s not a“ v e c t o ri d e n t i t yi n
disguise”: there is nothing like Eq. (9.14)t h a tw ec o u l ds u b s t i t u t ef o ri ti no r d e rt om a k et h es i g n s
automatically come out right. You should just treat it as a relationship between the magnitudes of
⃗vcm and ⃗ωand just pick the signs appropriately for each circumstance,b a s e do ny o u rc o n v e n t i o n
for positive directions of translation and rotation.
In fact, we could use Eq. (9.14)t ofi n dt h ev e l o c i t yo fa n yp o i n to nt h ec i r c l e ,i fw eg ot oar e ference
frame where the center is at rest—which is to say, the center ofm a s s( C M )r e f e r e n c ef r a m e ;t h e n ,t o
go back to the Earth frame, we just have to add⃗vcm (as a vector) to the vector we obtained in the
CM frame. Fig.9.8 shows the result. Note, particularly, that the point at the very bottom of the",University Physics I Classical Mechanics.pdf
"CM frame. Fig.9.8 shows the result. Note, particularly, that the point at the very bottom of the
circle has a velocity−R|ω| in the CM frame, but when we go back to the Earth frame, its velocity
is −R|ω| + vcm = −R|ω| + R|ω| =0( b yt h ec o n d i t i o n(9.34)). Thus, as long as the condition for
rolling without slipping holds, the point (or points) on therolling object that are momentarily in",University Physics I Classical Mechanics.pdf
"9.6. ROLLING MOTION 209
contact with the surface have zero instantaneous velocity.This means that, even if there was a
force acting on the object at that point (such as the force of static friction), it would do no work,
since the instantaneous powerFv for a force applied there would always be equal to zero.
We do not actually need the force of static friction to keep anobject rolling on a flat surface (as I
mentioned above, the motion could in principle go on “unforced” forever), but things are different
on an inclined plane. Fig.9.9 shows an object rolling down an inclined plane, and the corresponding
extended free-body diagram.
vcm
x
y
θ FG
Fn
Fs
Figure 9.9: An object rolling down an inclined plane, and the extended free-bodydiagram. Note that
neither gravity (applied at the CM) nor the normal force (whose lineof action passes through the CM) exert
a torque around the center of mass; only the force of static friction, ⃗Fs,d o e s .",University Physics I Classical Mechanics.pdf
"a torque around the center of mass; only the force of static friction, ⃗Fs,d o e s .
The basic equations we use to solve for the object’s motion aret h es u mo ff o r c e se q u a t i o n :
∑ ⃗Fext = M⃗ acm (9.35)
the net torque equation, with torques taken around the centero fm a s s7
∑
⃗τext = I⃗ α (9.36)
and the extension of the condition of rolling without slipping, (9.34), to the accelerations:
|acm| = R|α| (9.37)
7You may feel a little uneasy about the fact that the CM frame isnow an accelerated, and thereforenon-inertial,
frame. How do we know Eq. (9.36)e v e na p p l i e st h e r e ?T h i si s ,i n d e e d ,an o n - t r i v i a lp o i n t .However, as we shall see
in a later chapter, being in a uniformly-accelerated reference frame is equivalent to being in a uniform gravitational
field, and we have just shown that, for all torque-related purposes, one may treat such a field as a single force applied",University Physics I Classical Mechanics.pdf
"field, and we have just shown that, for all torque-related purposes, one may treat such a field as a single force applied
to the center of mass of an object. Such a force (or “pseudoforce” in this case) clearly does not contribute to the
torque about the center of mass, and so Eq. (9.36)a p p l i e si nt h eC Mf r a m e .",University Physics I Classical Mechanics.pdf
"210 CHAPTER 9. ROTATIONAL DYNAMICS
For the situation shown in Fig.9.9,i fw et a k ed o w nt h ep l a n ea st h ep o s i t i v ed i r e c t i o nf o rl i n e ar
motion, and clockwise torques as negative, we have to writeacm = −Rα.I nt h e d i r e c t i o np e r p e n -
dicular to the plane, we conclude from (9.35)t h a tFn = Mg cos θ,a ne q u a t i o nw ew i l ln o ta c t u a l l y
need8;i nt h ed i r e c t i o na l o n gt h ep l a n e ,w eh a v e
Macm = Mg sin θ − Fs (9.38)
and the torque equation just gives−FsR = Iα,w h i c hw i t hacm = −Rα becomes
FsR = I acm
R (9.39)
We can eliminateFs in between these two equations and solve foracm:
acm = g sin θ
1+ I/(MR2) (9.40)
Now you can see why, earlier in the semester, we were always careful to assume that all the objects
we sent down inclined planes weresliding,n o tr o l l i n g ! T h ea c c e l e r a t i o nf o rar o l l i n go b j e c ti snever",University Physics I Classical Mechanics.pdf
"we sent down inclined planes weresliding,n o tr o l l i n g ! T h ea c c e l e r a t i o nf o rar o l l i n go b j e c ti snever
equal to simplyg sin θ.M o s t r e m a r k a b l y , t h e c o r r e c t i o n f a c t o r d e p e n d s o n l y o n t h eshape of the
rolling object, and not on its mass or size, since the ratio ofI to MR2 is independent ofm and R
for any given geometry. Thus, for instance, for a disk,I = 1
2 MR2,s oacm = 2
3 g sin θ,w h e r e a sf o ra
hoop, I = MR2,s o acm = 1
2 g sin θ.S o a n y d i s k o r s o l i d c y l i n d e r w i l l a l w a y sr o l l d o w n t h e i n c l ine
faster thanany hoop or hollow cylinder, regardless of mass or size.
This rather surprising result may be better understood in terms of energy. First, let me show (a
result that is somewhat overdue) that for a rigid object thatis rotating around an axis passing
through its center of mass with angular velocityω we can write the total kinetic energy as
K = Kcm + Krot = 1
2Mv2
cm + 1",University Physics I Classical Mechanics.pdf
"through its center of mass with angular velocityω we can write the total kinetic energy as
K = Kcm + Krot = 1
2Mv2
cm + 1
2Iω2 (9.41)
This is because for every particle the velocity can be writtena s⃗v= ⃗vcm + ⃗v′,w h e r e⃗v′ is the velocity
relative to the center of mass (that is, in the CM frame). Sincei nt h i sf r a m et h em o t i o ni sas i m p l e
rotation, we have|v′| = ωr,w h e r er is the particle’s distance to the axis. Therefore, the kinetic
energy of that particle will be
1
2mv2 = 1
2⃗v· ⃗v= 1
2m
(
⃗vcm + ⃗v′
)
·
(
⃗vcm + ⃗v′
)
= 1
2mv2
cm + 1
2mv′2 + m⃗ vcm · ⃗v′
= 1
2mv2
cm + 1
2mr2ω2 + ⃗vcm · ⃗p′ (9.42)
8Unless we were trying to answer a question such as “how steep does the plane have to be for rolling without
slipping to become impossible?”",University Physics I Classical Mechanics.pdf
"9.7. IN SUMMARY 211
(Note how I have made use of thedot product to calculate the magnitude squared of a vector.)
On the last line, the quantity⃗p′ is the momentum of that particle in the CM frame. Adding those
momenta for all the particles should give zero, since, as we saw in an earlier chapter, the center
of mass frameis the zero momentum frame. Then, adding the contributions of all particles to the
first and second terms in9.42 gives Eq. (9.41).
Coming back to our rolling body, using Eq. (9.41)a n dt h ec o n d i t i o no fr o l l i n gw i t h o u ts l i p p i n g
(9.34), we see that the ratio of the translational to the rotationalk i n e t i ce n e r g yi s
Kcm
Krot
= mv2
cm
Iω2 = mR2
I (9.43)
The amount of energy available to accelerate the object initially is just the gravitational potential
energy of the object-earth system, and that has to be split between translational and rotational in",University Physics I Classical Mechanics.pdf
"energy of the object-earth system, and that has to be split between translational and rotational in
the proportion (9.43). An object with a proportionately largerI is one that, for a given angular
velocity, needs more rotational kinetic energy, because more of its mass is away from the rotation
axis. This leaves less energy available for its translational motion.
Resources
Unfortunately, we will not really have enough time this semester to explore further the many inter-
esting effects that follow from the vector nature of Eq. (9.20), but you are at least subconsciously
familiar with some of them if you have ever learned to ride a bicycle! A few interesting Internet
references (some of which could perhaps inspire a good Honorsp r o j e c t ! )a r et h ef o l l o w i n g :
• Walter Lewin’s lecture on gyroscopic motion (and rolling motion):
https://www.youtube.com/watch?v=N92FYHHT1qM
• A“ V e r i t a s i u m ”v i d e oo n“ a n t i g r a v i t y ” :",University Physics I Classical Mechanics.pdf
"https://www.youtube.com/watch?v=N92FYHHT1qM
• A“ V e r i t a s i u m ”v i d e oo n“ a n t i g r a v i t y ” :
https://www.youtube.com/watch?v=GeyDf4ooPdo
https://www.youtube.com/watch?v=tLMpdBjA2SU
• And the old trick of putting a gyroscope (flywheel) in a suitcase:
https://www.youtube.com/watch?v=zdN6zhZSJKw
If any of the above links are dead, try googling them. (You maywant to let me know, too!)
9.7 In summary
(Note: this summary makes extensive use of cross products, but does not include a summary of
cross product properties. Please refer to Section9.3 for that!)
1. The angular velocity and acceleration of a particle movingi nac i r c l ec a nb et r e a t e da sv e c t o r s
perpendicular to the plane of the circle,⃗ωand ⃗α,r e s p e c t i v e l y . T h ed i r e c t i o no f⃗ωis such",University Physics I Classical Mechanics.pdf
"212 CHAPTER 9. ROTATIONAL DYNAMICS
that the relation⃗v= ⃗ω×⃗ralways holds, where⃗ris the (instantaneous) position vector of the
particle on the circle.
2. The particle’s kinetic energy can be written asKrot = 1
2 Iω2,w h e r eI = mR2 is therotational
inertia or moment of inertia.F o r a n e x t e n d e d o b j e c t r o t a t i n g a b o u t a n a x i s ,Krot = 1
2 Iω2
also applies ifI is defined as the sum of the quantitiesmr2 for all the particles making up
the object, wherer is the particle’s distance to the rotation axis.
3. For a rigid object that is rotating around an axis passing through its center of mass with
angular velocityω the total kinetic energy can be written asK = Kcm+Krot = 1
2 Mv2
cm+1
2 Iω2.
This would apply also to a non-rigid system, provided all theparticles have the same angular
velocity.
4. The angular momentum, ⃗L,o fap a r t i c l ea b o u tap o i n tOi sd e fi n e da s⃗L = ⃗r× ⃗p= m⃗ r×⃗v,",University Physics I Classical Mechanics.pdf
"velocity.
4. The angular momentum, ⃗L,o fap a r t i c l ea b o u tap o i n tOi sd e fi n e da s⃗L = ⃗r× ⃗p= m⃗ r×⃗v,
where ⃗ris the position vector of the particle relative to the originO, and⃗vand ⃗pits velocity
and momentum vectors. For an extended object or system,⃗L is defined as the sum of the
quantities m⃗ r×⃗vfor all the particles making up the system.
5. For a solid object rotating around a symmetry axis,⃗L = I⃗ ω.T h i s a p p l i e s a l s o t o a n e s s e n t i a l l y
flat object rotating about a perpendicular axis, even if it isnot an axis of symmetry.
6. The torque, ⃗τ,o faf o r c ea b o u tap o i n tOi sd e fi n e da sτ = ⃗r× ⃗F,w h e r e⃗ris the position
vector of the point of application of the force relative to theo r i g i nO .I ti sam e a s u r eo ft h e
effectiveness of the force at causing a rotation around that point.
7. The rate of change of a system’s angular momentum about a point O is equal to the sum of",University Physics I Classical Mechanics.pdf
"7. The rate of change of a system’s angular momentum about a point O is equal to the sum of
the torques, about that same point, of all theexternal forces acting on the system:∑ ⃗τext =
d⃗Lsys/dt.H e n c e , a n g u l a r m o m e n t u m i s c o n s t a n t w h e n e v e r a l l t h e e x t e rnal torques vanish
(conservation of angular momentum).
8. For the cases considered in point 7. above, if the moment ofinertia I is constant, the equation∑ ⃗τext = d⃗Lsys/dt can be written in the form∑ ⃗τext = I⃗ α,w h i c hs t r o n g l yr e s e m b l e st h e
familiar ∑ ⃗Fext = m⃗ a.N o t e , h o w e v e r , t h a t d e f o r m a b l e s y s t e m s w h e r eI may change with
time as a result of internal forces are relatively common, andf o rt h o s es y s t e m st h i ss i m p l e r
equation would not apply.
9. For an object to be instatic equilibrium,w er e q u i r et h a tb o t ht h es u mo ft h ee x t e r n a lf o r c e s",University Physics I Classical Mechanics.pdf
"equation would not apply.
9. For an object to be instatic equilibrium,w er e q u i r et h a tb o t ht h es u mo ft h ee x t e r n a lf o r c e s
and of the external torques be equal to zero:∑ ⃗Fext =0a n d∑ ⃗τext =0 . N o t et h a ti ft h e
first condition applies, it does not matter about which pointwe calculate the torque, so we
are free to choose whichever is most convenient.
10. For a rigid object of radiusR rolling without slipping on some surface, the relations|vcm| =
R|ω| and |acm| = R|α| hold. The relative signs of, for instance,vcm and ω (understood here as
the relevant components of their respective vectors) need tob ec h o s e ns oa st ob ec o n s i s t e n t
with whatever convention one has adopted for the positive direction of motion and the positive
direction of rotation (typically, a counterclockwise rotation is considered positive).",University Physics I Classical Mechanics.pdf
"9.8. EXAMPLES 213
9.8 Examples
This was a long chapter, in part because it contains a number ofu s e f u lw o r k e d - o u te x a m p l e s ;s o
please make sure not to overlook them! Section 9.2.2 showed acouple of examples of problems that
can be solved using conservation of angular momentum. Section 9.3 shows you how to set up and
solve the equilibrium equations for a leaning ladder, whichis the archetype of all statics problems;
and Section 9.4 also solves for you the problem of a generic object rolling down an inclined plane.
The first couple of additional examples in this section show you have to set up and solve the
equations of motion for somewhat more complicated systems,and you should study them carefully.
The third one is slightly more sophisticated and you may treati ta so p t i o n a lr e a d i n g .
9.8.1 Torques and forces on the wheels of an accelerating bicycle
Consider an accelerating bicycle. The rider exerts a torqueon the pedals, which is transmitted to",University Physics I Classical Mechanics.pdf
"Consider an accelerating bicycle. The rider exerts a torqueon the pedals, which is transmitted to
the rear wheel by the chain (possibly amplified by the gears, etc). How does this “drive” torque on
the rear wheel (call itτd)r e l a t et ot h efi n a la c c e l e r a t i o no ft h ec e n t e ro fm a s so ft h ebicycle?
Fr,ftFr,rt
ss
Solution
We need first to figure out how many external forces, at a minimum, we have to deal with. As the
bicycle accelerates, two things happen: the wheels (both wheels) turn faster, so there must be a
net torque (clockwise in the picture, if the bicycle is accelerating to the right) oneach wheel; and
the center of mass of the system accelerates, so there must bean e te x t e r n a lf o r c eo nt h ew h o l e
system. The system is only in contact with the road, and so, aslong as no slippage happens, the
only external source of torques or forces on the wheels has tobe the force of static friction between
the tires snd the road.",University Physics I Classical Mechanics.pdf
"214 CHAPTER 9. ROTATIONAL DYNAMICS
For the front wheel, this is in fact the only external force, and the only force of any sort that exerts
at o r q u eo nt h a tw h e e l( t h e r ea r ef o r c e sa c t i n ga tt h ea x l e ,b utt h e ye x e r tn ot o r q u ea r o u n dt h e
axle). Since the torque has to be clockwise, then, the force ofs t a t i cf r i c t i o no nt h ef r o n tw h e e l ,
applied as it is at the point of contact with the road, must point backwards,t h a ti s ,o p p o s i t et h e
direction of motion. We get then one equation of motion (of thet y p e(9.25)) for that wheel:
− Fs
r,ft R = Iα (9.44)
where the subscript “ft” stands for “front tire”, and the wheel is supposed to have a radiusR and
moment of inertiaI.
For the rear wheel, we have the “drive torque”τd,e x e r t e db yt h ec h a i n ,a n da n o t h e rt o r q u ee x e r t e d
by the force of static friction,⃗Fs
r,rt,b e t w e e nt h a tt i r ea n dt h er o a d . H o w e v e r ,n o wt h ef o r c e⃗Fs
r,rt",University Physics I Classical Mechanics.pdf
"by the force of static friction,⃗Fs
r,rt,b e t w e e nt h a tt i r ea n dt h er o a d . H o w e v e r ,n o wt h ef o r c e⃗Fs
r,rt
needs to pointforward.T h i s i s b e c a u s e t h e n e t e x t e r n a l f o r c e o n t h e w h o l e b i c y c l e -rider system
is ⃗Fs
r,rt + ⃗Fs
r,ft,a n dt h a th a st op o i n tf o r w a r d ,o rt h ec e n t e ro fm a s sc o u l dn e ver accelerate in that
direction. Since we have established that⃗Fs
r,ft has to point backwards, it follows that⃗Fs
r,rt needs
to be larger, and in the forward direction. This means we get,for the center of mass acceleration,
the equation (Fnet = Macm)
Fs
r,rt − Fs
r,ft = Macm (9.45)
and for the rear wheel, the torque equation
Fs
r,rtR − τd = Iα (9.46)
Ia mf o l l o w i n gt h ec o n v e n t i o nt h a tc l o c k w i s et o r q u e sa r en e gative, and also that a force symbol
without an arrow on top represents the magnitude of the force.I fac l o c k w i s ea n g u l a ra c c e l e r a t i o n",University Physics I Classical Mechanics.pdf
"without an arrow on top represents the magnitude of the force.I fac l o c k w i s ea n g u l a ra c c e l e r a t i o n
is likewise negative, the condition of rolling without slipping [Eq. (9.37)] needs to be written as
acm = −Rα (9.47)
These are all the equations we need to relate the accelerationt oτd.W ec a ns t a r tb ys o l v i n g(9.44)
for Fs
r,ft and substituting in (9.45), then likewise solving (9.46)f o rFs
r,rt and substituting in (9.45).
The result is Iα + τd
R + Iα
R = Macm (9.48)
then use Eq (9.47)t ow r i t eα = −acm/R,a n ds o l v ef o racm:
acm = τd
MR +2 I/R (9.49)
9.8.2 Blocks connected by rope over a pulley with non-zero mass
Consider again the setup illustrated in Figure6.2,b u tn o wa s s u m et h a tt h ep u l l e yh a sam a s sM
and radius R.F o r s i m p l i c i t y , l e a v e t h e f r i c t i o n f o r c e o u t .W h a t i s n o w t hea c c e l e r a t i o no ft h e
system?",University Physics I Classical Mechanics.pdf
"9.8. EXAMPLES 215
Solution
The figure below shows the setup, plus free-body diagrams forthe two blocks (the vertical forces
on block 1 have been left out to avoid cluttering the figure, since they are not relevant here), and
an extended free-body diagram for the pulley. (You can see from the pulley diagram that there has
to be another force acting on it, to balance the two forces shown. This would be a contact force at
the axle, directed upwards and to the left. If this was a statics problem, I would have to include
it, but since it does not exert a torque around the axis of rotation, it does not contribute to the
dynamics of the system, so I have left it out as well.)
1
2
Fr,1
t
Fr,2
t
FE,2
Ga1
a2
1 2
Fr,pl
t
Fr,pd
t
The key new feature of this problem is that the tension on the string has to have different values
on either side of the pulley, because there has to be a net torque on the pulley. Hence, the leftward
force on the pulley (Ft",University Physics I Classical Mechanics.pdf
"on either side of the pulley, because there has to be a net torque on the pulley. Hence, the leftward
force on the pulley (Ft
r,pl)h a st ob es m a l l e rt h a nt h ed o w n w a r df o r c e(Ft
r,pd).
On the other hand, as long as the mass of the rope is negligible,i tw i l ls t i l lb et h ec a s et h a tt h e
horizontal part of the rope will pull with equal strength on block 1 and on the pulley, and similarly
the vertical part of the rope will pull with equal strength onthe pulley and on block 2. (To make
this point clearer, I have “color-coded” these matching forces in the figure.) This means that we
can writeFt
r,pl = Ft
r,1 and Ft
r,pd = Ft
r,2,a n dw r i t et h et o r q u ee q u a t i o n(9.25)f o rt h ep u l l e ya s
Ft
r,1R − Ft
r,2R = Iα (9.50)
We also haveF = ma for each block:
Ft
r,1 = m1a (9.51)
Ft
r,2 − m2g = −m2a (9.52)
where I have takena to be a = |⃗a1| = |⃗a2|.T h e c o n d i t i o n o f r o l l i n g w i t h o u t s l i p p i n g , E q . (9.37),",University Physics I Classical Mechanics.pdf
"where I have takena to be a = |⃗a1| = |⃗a2|.T h e c o n d i t i o n o f r o l l i n g w i t h o u t s l i p p i n g , E q . (9.37),
applied to the pulley, gives then
− Rα = a (9.53)
since, in the situation shown,α will be negative, anda has been defined as positive. Substituting
Eqs. (9.51), (9.52), and (9.53)i n t o(9.50), we get
m1aR − (m2g − m2a)R = −Ia
R (9.54)",University Physics I Classical Mechanics.pdf
"216 CHAPTER 9. ROTATIONAL DYNAMICS
which is easily solved fora:
a = m2g
m1 + m2 + I/R2 (9.55)
If you look at the structure of this equation, it all makes sense. The numerator is the force of gravity
on block 2, which is, ultimately, the force responsible for setting the whole thing in motion. The
denominator is, essentially, the inertia of the system: ordinary inertia for the blocks, and rotational
inertia for the pulley. Note further that, if we treat the pulley as a flat, homogeneous disk of mass
M,t h e nI = 1
2 MR2,a n dt h ed e n o m i n a t o ro f(9.55)b e c o m e sj u s tm1 + m2 + M/2.",University Physics I Classical Mechanics.pdf
"9.9. PROBLEMS 217
9.9 Problems
Problem 1
An ice skater has a moment of inertia equal to 1.9k g· m2 when she is rotating with her arms stretched
out, at a rate of 2 revolutions per second. She then brings herarms in, lined up with her axis of
rotation, so her moment of inertia becomes 1.1k g· m2.
(a) What is her new angular velocity? (You may use radians persecond, or revolutions per second
if you prefer.)
(b) What is the change in her kinetic energy?
(c) Where did this energy come from?
Problem 2
Two identical pucks, each of inertiam,a r ec o n n e c t e dt oar o do fl e n g t h2r and negligible inertia
that is pivoted about its center (that is, there is some sort ofp i nt h o u g hi t sc e n t e r ,a r o u n dw h i c h
it can rotate without friction). A third puck of inertiam/2s t r i k e so n eo ft h ec o n n e c t e dp u c k s
perpendicular to the rod with a speedvi.A s s u m e t h ec o l l i s i o ni se l a s t i c .",University Physics I Classical Mechanics.pdf
"perpendicular to the rod with a speedvi.A s s u m e t h ec o l l i s i o ni se l a s t i c .
(a) Draw a diagram of the situation, clearly labeling the direction of vi and what direction the
connected pucks will rotate.
(b) Is the total momentum of the system (the three pucks and ther o d )c o n s e r v e dt h r o u g h o u tt h e
interaction? Why? Is the system isolated, or can you identifya ne x t e r n a lf o r c ea c t i n go ni t ?
(c) Is the total kinetic energy of the system conserved? Why?If you found an external force in
part (b), explain why it does or does not do work on the system.
(d) Is the total angular momentum of the system conserved? Why? If you found an external force
in part (b), explain why it does or does not exert a torque on thes y s t e m .
(e) Write down an expression for the moment of inertia (rotational inertia) of the system formed
by the connected pucks.",University Physics I Classical Mechanics.pdf
"(e) Write down an expression for the moment of inertia (rotational inertia) of the system formed
by the connected pucks.
(f) Based on all of the above, write down equations expressingt h ec o n s e r v a t i o no ft h et w oq u a n t i t i e s
that are in fact conserved. These equations should involve only the given data (masses, length of
rod); the initial and final velocities,vi and vf ,o ft h ef r e ep u c k ;a n dω,t h ea n g u l a rs p e e do ft h e
connected pucks after the collision. (Assume the final velocity of the free puck lies along the same
line as its initial velocity, that is, it does not bounce off atsome random angle.)
Problem 3
Ap l a n ko fl e n g t hl =2 mi sh i n g e da to n ee n dt oaw a l l . T h eo t h e re n di sb e i n g( t e m p orarily)
supported by a worker who is holding it up with his hand, keeping the plank horizontal. The plank
has a mass of 20kg, and there is also a toolbox of mass 5 kg sitting on it, 50 cm away from the",University Physics I Classical Mechanics.pdf
"has a mass of 20kg, and there is also a toolbox of mass 5 kg sitting on it, 50 cm away from the
worker (1.5m a w a y f r o m t h e w a l l ) .
(a) Draw a free body diagram and an extended free-body diagramf o rt h ep l a n k .
(b) What are the magnitudes of (1) the upwards force exerted byt h ew o r k e ro nt h ep l a n ka n d( 2 )
the force at the hinge?",University Physics I Classical Mechanics.pdf
"218 CHAPTER 9. ROTATIONAL DYNAMICS
(c) If the worker were to let go of the plank, what would its angular acceleration be as it starts
swinging down? The moment of inertia isI = 1
3 Ml2.( N o t e :assume the toolbox stops pressing
down on the plank immediately.T h i si sag o o da p p r o x i m a t i o n ,a sy o us h a l ls e eb e l o w . )
(d) Consider a point on the plank located immediately below the toolbox. As the plank swings,
this point moves in a circle of radius 1.5m . W h a t i s i t slinear (tangential) acceleration as it starts
going down, and how does it compare to the acceleration of gravity?
Problem 4
As o l i ds p h e r eo fr a d i u s5 c ma n dm a s s0.2k g i s r o l l i n g w i t h o u t s l i p p i n g , o n l e v e l g r o u n d , a t a
constant speed of 0.5m / s . I t i s h e a d i n g t o w a r d a r a m p t h a t m a k e s a n a n g l e o f 3 0◦ with the
horizontal.
(a) What is the angular velocity of the sphere, in radians persecond?",University Physics I Classical Mechanics.pdf
"horizontal.
(a) What is the angular velocity of the sphere, in radians persecond?
(b) If the moment of inertia of a solid, homogeneous sphere isI = 2
5 mR2,w h a ti st h ei n i t i a la n g u l a r
momentum of this sphere?
(c) What is the translational kinetic energy of the sphere?
(d) What is its rotational kinetic energy?
(e) How high (vertically) will the sphere rise as it goes up ther a m p ,s t i l lr o l l i n gw i t h o u ts l i p p i n g ,
before it stops and turns around?
(f) Draw an extended free-body diagram for the sphere as it rolls up the ramp. On the diagram,
indicate the direction of rotation, and the direction of theacceleration of the center of mass.
(g) Referring to the extended free-body diagram in the previous question, which force is responsible
for the change in the angular momentum of the sphere as it rollsu pt h er a m p ? E x p l a i nb r i e fl y :
Why that force and not another one? Does the direction you havea s s u m e df o rt h i sf o r c ea g r e e",University Physics I Classical Mechanics.pdf
"Why that force and not another one? Does the direction you havea s s u m e df o rt h i sf o r c ea g r e e
with the torque it has to provide? (If not, you’d better go backa n dfi xt h a tr i g h tn o w ! )
(h) Calculate the acceleration of the sphere as it rolls up ther a m p .
Problem 5
Av e r yl i g h t ,i n e x t e n s i b l es t r i n gi sw r a p p e da r o u n dac y l i n drical spool. The end of the string is
held fixed, and the spool is released so it starts falling, as the string unwinds. Because the spool is
basically hollow, you can take its moment of inertia to beI = mR2.
(a) Draw an extended free-body diagram for the spool as it unrolls.
(b) Find the linear acceleration of the spool as it accelerates toward the ground.
(c) Let the mass of the spool be 0.1k g . W h a t i s i t s t r a n s l a t i o n a l k i n e t i c e n e r g y a f t e r i t h a s f allen
for 0.5s ?
(d) What is its rotational kinetic energy at that time?",University Physics I Classical Mechanics.pdf
"for 0.5s ?
(d) What is its rotational kinetic energy at that time?
(e) What is the tension in the string? Does this change as the spool falls? (Remember the mass of
the string is negligible.)
(f) If the radius of the spool is 3 cm, what is the magnitude of the torque (around the center of
mass) exerted by the tension?
Problem 6",University Physics I Classical Mechanics.pdf
"9.9. PROBLEMS 219
A2 0 - k gp l a n ko fl e n g t hl =4mi ss u p p o r t e da tb o t he n d sa ss h o w ni nt h efi g u r e . A6 0 - k gm a nis
standing a distancel/3f r o mt h er i g h te n do ft h ep l a n k .
1.33 m 
4 m 
(a) Draw an extended free-body diagram for the plank. Try to get the scale of the forces at least
qualitatively right.
(b) Find the upwards force on the plank exerted by each of the two supports.",University Physics I Classical Mechanics.pdf
220 CHAPTER 9. ROTATIONAL DYNAMICS,University Physics I Classical Mechanics.pdf
"Chapter 10
Gravity
10.1 The inverse-square law
Up to this point, all I have told you about gravity is that, neart h es u r f a c eo ft h eE a r t h ,t h e
gravitational force exerted by the Earth on an object of massm is FG = mg.T h i s i s , i n d e e d , a
pretty good approximation, but it does not really tell you anything about what the gravitational
force is where other objects or distances are involved.
The first comprehensive theory of gravity, formulated by Isaac Newton in the late 17th century,
postulates that any two “particles” with massesm1 and m2 will exert an attractive force (a “pull”)
on each other, whose magnitude is proportional to the producto ft h em a s s e s ,a n di n v e r s e l yp r o -
portional to the square of the distance between them. Mathematically, we write
FG
12 = Gm1m2
r2
12
(10.1)
Here, r12 is just the magnitude of the position vector of particle 2 relative to particle 1 (sor12",University Physics I Classical Mechanics.pdf
"FG
12 = Gm1m2
r2
12
(10.1)
Here, r12 is just the magnitude of the position vector of particle 2 relative to particle 1 (sor12
is, indeed, the distance between the two particles), andG is a constant, known as “Newton’s
constant” or thegravitational constant,w h i c ha tt h et i m eo fN e w t o ns t i l lh a dn o tb e e nd e t e r m i n e d
experimentally. You can see from Eq. (10.1)t h a tG is simply the magnitude, in newtons, of
the attractive force between two 1-kg masses a distance of 1 mapart. This turns out to have the
ridiculously small valueG =6 .674×10−11 Nm 2/kg2 (or, as is more commonly written, m3kg−1s−2).
It was first measured by Henry Cavendish in 1798, in what was, without a doubt, an experimental
tour de force for that time (more on that later, but you may peeka tt h e“ A d v a n c e dT o p i c s ”
section of next chapter if you are curious already). As you cans e e ,g r a v i t ya saf o r c eb e t w e e na n y",University Physics I Classical Mechanics.pdf
"section of next chapter if you are curious already). As you cans e e ,g r a v i t ya saf o r c eb e t w e e na n y
two ordinary objects is absolutely insignificant, and it takes the mass of a planet to make it into
something you can feel.
221",University Physics I Classical Mechanics.pdf
"222 CHAPTER 10. GRAVITY
Equation (10.1), as stated, applies to particles, that is to say, in practice, to any objects that
are very small compared to the distance between them. The netforce between extended masses
can be obtained using calculus, by breaking up the two objectsi n t ov e r ys m a l lp i e c e sa n da d d i n g
(vectorially!) the force exerted by every small part of one object on every small part of the other
object. This requires the use of integral calculus at a fairlya d v a n c e dl e v e l ,a n df o ri r r e g u l a r l y -
shaped objects can only be computed numerically. For spherically-symmetric objects, however, it
turns out that the result (10.1)s t i l lh o l d se x a c t l y ,p r o v i d e dt h eq u a n t i t yr12 is taken to be the
distance between the center of the spheres. The same result also holds for the force between a
finite-size sphere and a “particle,” again with the distanceto the particle measured from the center
of the sphere.",University Physics I Classical Mechanics.pdf
"finite-size sphere and a “particle,” again with the distanceto the particle measured from the center
of the sphere.
For the rest of this chapter, we will simply assume that Eq. (10.1)i sag o o da p p r o x i m a t i o nt o
be used in any of the problems we will encounter involving extended objects. You can estimate
visually how good it may be, for instance, when applied to theEarth-moon system (as we will do
in a moment), from a look at Figure10.1 below:
Figure 10.1: The moon, the earth, and the distance between them, all approximately to scale.
Based on this picture it seems that it might be OK to treat the moon as a “particle,” but that it
would not do, in general, to neglect the radius of the Earth; that is to say, we should use forr12
the distance from the moon to the center of the Earth, not justto its surface.
Before we get there, however, let us start closer to home and see what happens if we try to use",University Physics I Classical Mechanics.pdf
"Before we get there, however, let us start closer to home and see what happens if we try to use
Eq. (10.1)t oc a l c u l a t et h ef o r c ee x e r t e db yt h eE a r t ho na no b j e c ti fm ass m near its surface—say,
at a heighth above the ground. Clearly, if the radius of the Earth isRE,t h ed i s t a n c eo ft h eo b j e c t
to the center of the Earth isRE + h,a n dt h i si sw h a tw es h o u l du s ef o rr12 in Eq. (10.1). However,
noting that the radius of the Earth is about 6, 000 km (more precisely,RE =6 .37×106 m), and the
tallest mountain peak is only about 9 km above sea level, you can see that it will almost always be
av e r yg o o da p p r o x i m a t i o nt os e tr12 equal to justRE,w h i c hr e s u l t si naf o r c e
FG
E,o ≃ GMEm
R2
E
= m 6.674 × 10−11 m3kg−1s−2 × 5.97 × 1024 kg
(6.37 × 106 m)2 = m × 9.82m
s2 (10.2)
where I have used the currently known valueME =5 .97 × 1024 kg for the mass of the Earth. As",University Physics I Classical Mechanics.pdf
"(6.37 × 106 m)2 = m × 9.82m
s2 (10.2)
where I have used the currently known valueME =5 .97 × 1024 kg for the mass of the Earth. As
you can see, we recover the familiar resultFG = mg,w i t hg ≃ 9.8m / s2,w h i c hw eh a v eb e e nu s i n g
all semester. We can rewrite this result (canceling the massm)i nt h ef o r m
gE = GME
R2
E
(10.3)
Here I have put a subscriptE on g to emphasize that this is the acceleration of gravity near the
surface of the Earth, and that the same formula could be used tofi n dt h ea c c e l e r a t i o no fg r a v i t y",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 223
near the surface of any other planet or moon, just replacingME and RE by the mass and radius of
the planet or moon in question. Thus, we could speak ofgmoon, gMars,e t c . ,a n di ns o m eh o m e w o r k
problems you will be asked to calculate these quantities. Clearly, besides telling you how fast things
fall on a given planet, the quantitygplanet allows you to figure out how much something would weigh
on that planet’s surface (just multiplygplanet by the mass of the object); alternatively, the ratio of
gplanet to gE will be the ratio of the object’s weight on that planet to its weight here on Earth.
Of course, historically, this is not what Newton and his contemporaries would have done: they had
measurements of objects in free fall (or sliding on inclinedplanes) that would have given them the
value of g,a n dt h e ye v e nh a dap r e t t yg o o di d e ao ft h er a d i u so ft h eE a r t h1,b u tt h e yd i dn o t",University Physics I Classical Mechanics.pdf
"value of g,a n dt h e ye v e nh a dap r e t t yg o o di d e ao ft h er a d i u so ft h eE a r t h1,b u tt h e yd i dn o t
know eitherG or the mass of the Earth, so all they could get from Eq. (10.3)w a st h ev a l u eo ft h e
product GME.I t w a so n l yac e n t u r yl a t e r ,w h e n C a v e n d i s hm e a s u r e dG,t h a tt h e yc o u l dg e tf r o m
that the mass of the earth (as a result of which, he became knowna s“ t h em a nw h ow e i g h e dt h e
earth”!)
What Newton could do, however, with just this knowledge of thev a l u eo fGME,w a ss o m e t h i n g
that, for its time, was even more dramatic and far-reaching:namely, he could “prove” his intuition
that the same fundamental interaction—gravity—that causesa na p p l et of a l ln e a rt h es u r f a c eo f
the Earth, reaching out hundreds of thousands of miles away into space, also provides the force
needed to keep the moon on its orbit. This brought together Earth-bound science (physics) and",University Physics I Classical Mechanics.pdf
"needed to keep the moon on its orbit. This brought together Earth-bound science (physics) and
“celestial” science (astronomy) in a way that no one had everdreamed of before.
To see how this works, let us start by assuming that the moon does move in a circle, with the Earth
motionless at the center (these are all approximations, as wes h a l ls e el a t e r ,b u tt h e yg i v et h er i g h t
order of magnitude at the end, which is all that Newton could have hoped for anyway). The force
FG
E,m then has to provide the centripetal forceFc = Mmoonω2re,m,w h e r eω is the moon’s angular
velocity. We can cancel the moon’s mass from both terms and write this condition as
GME
r2e,m
= ω2re,m (10.4)
The moon revolves around the earth once about every 29 days, which is about 29× 24 × 3600 =
2.5 × 106 s. So ω is 2π radians per 2.5m i l l i o ns e c o n d s ,o rω =2 .5 × 10−6 rad/s. Substituting this
into Eq. (10.4), as well as the resultGME = gR2
E (note that, as stated above, we do not need to",University Physics I Classical Mechanics.pdf
"into Eq. (10.4), as well as the resultGME = gR2
E (note that, as stated above, we do not need to
know G and ME separately), we getre,m =3 .99 × 108 m, pretty close to today’s accepted value of
the average Earth-moon distance, which is 3.84 × 108 km. Newton would not have knownre,m to
such an accuracy, but he would still have had a pretty good ideat h a tt h i sw a s ,i n d e e d ,t h ec o r r e c t
order of magnitude2.
1The radius of the earth was known, at least approximately, since ancient times. The Greeks first figured it out
by watching the way tall objects disappear below the horizonwhen viewed from a ship at sea. Next time somebody
tries to tell you that people used to believe the earth was flat,a s kt h e mj u s tw h i c h“ p e o p l e ”t h e ya r et a l k i n ga b o u t !
2And, yes, the first estimates of the distance between the eartha n dt h em o o na l s og ob a c kt ot h ea n c i e n tG r e e k s !",University Physics I Classical Mechanics.pdf
"2And, yes, the first estimates of the distance between the eartha n dt h em o o na l s og ob a c kt ot h ea n c i e n tG r e e k s !
You can figure it out, if you know the radius of the earth, by looking at how long it takes the moon to transit across
the earth’s shadow during a lunar eclipse.",University Physics I Classical Mechanics.pdf
"224 CHAPTER 10. GRAVITY
10.1.1 Gravitational potential energy
Ever since I introduced the concept of potential energy in Chapter 5, I have been usingUG = mgy
for the gravitational potential energy of the system formedby the Earth and an object of massm a
height y above the Earth’s surface. This works well as long as the forceo fg r a v i t yi sa p p r o x i m a t e l y
constant, which is to say, as long asy is much smaller than the radius of the earth, but obviously
it must break down at some point.
Recall that, if the interaction between two objects can be described by a potential energy function
of the objects’ coordinates,U(x1,x 2), then (in one dimension) the force exerted by object 1 on
object 2 could be written asF12 = −dU/dx2.S i n c e t h e f o r c e o f g r a v i t y d o e s l i e a l o n g t h e l i n e
joining the two particles, we can cheat a bit and treat this asao n e - d i m e n s i o n a lp r o b l e m ,w i t h
(FG",University Physics I Classical Mechanics.pdf
"joining the two particles, we can cheat a bit and treat this asao n e - d i m e n s i o n a lp r o b l e m ,w i t h
(FG
12)x = −Gm1m2/(x1 − x2)2 (I’ve put a minus sign there under the assumption that particle 1
is to the left of particle 2, that is,x1 <x 2,a n dt h ef o r c eo n2i st ot h el e f t ) ,a n dfi n dap o t e n t i a l
energy function whose derivative gives that. The answer is clearly
UG(x1,x 2)= −Gm1m2
x2 − x1
+ C (10.5)
where C is an arbitrary constant. (Please take a moment to verify for yourself that, indeed,
−dUG/dx2 = −(FG
12)x,a n da l s o−dUG/dx1 = −(FG
21)x.)
Since I have assumedx2 >x 1,t h ed e n o m i n a t o ri nE q .(10.5)i sj u s tt h ed i s t a n c eb e t w e e nt h et w o
particles, and the potential energy function could be written, in three dimensions, as
UG(r12)= −Gm1m2
r12
(10.6)
where r12 = |⃗r2 − ⃗r1|,a n dIh a v es e tt h ec o n s t a n tC equal to zero. This means that the potential",University Physics I Classical Mechanics.pdf
"UG(r12)= −Gm1m2
r12
(10.6)
where r12 = |⃗r2 − ⃗r1|,a n dIh a v es e tt h ec o n s t a n tC equal to zero. This means that the potential
energy of the system is alwaysnegative,w h i c hi s ,o nt h ef a c eo fi t ,as t r a n g er e s u l t .H o w e v e r ,t h e r e
is no way to choose the constantC in (10.5)t h a tw i l lp r e v e n tt h a t : n om a t t e rh o wb i ga n dp o s i t i v e
C might be, the first term in (10.5)c a na l w a y sb e c o m el a r g e r( i nm a g n i t u d e )a n dn e g a t i v e ,i ft he
particles are very close together. So we might as well chooseC =0 ,w h i c h ,a tl e a s t ,g i v e su st h e
somewhat comforting result that the potential energy of thesystem is zero when the particles are
“infinitely” distant from each other—that is to say, so far apart that they do not feel a force from
each other any more.
But Eq. (e10.5) also makes sense in a different way: namely, it shows that the system’s potential",University Physics I Classical Mechanics.pdf
"each other any more.
But Eq. (e10.5) also makes sense in a different way: namely, it shows that the system’s potential
energy increases as the particles are moved farther apart. Indeed, we expect,physically, that if you
separate the particles by a great distance and then release them, they will pick up a lot of speed as
they approach each other; or, put differently, that the force doing work over a large distance will
give them a large amount of kinetic energy—which must come from the system’s potential energy.
But, in fact, mathematically, Eq. (10.6)a g r e e sw i t ht h i se x p e c t a t i o n : f o ra n yfi n i t ed i s t a n c e ,UG
is negative, and it gets smallerin magnitude as the distance increases, which means algebraically",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 225
it increases (since a number like, say,−0.1i s ,i nf a c t ,g r e a t e rt h a nan u m b e rl i k e−10). So as the
particles are moved farther and farther apart, the potentiale n e r g yo ft h es y s t e mdoes increase—all
the way up to a maximum value of zero!
Still, even if it makes sense mathematically, the notion of a“negative energy” is hard to wrap your
mind around. I can only offer you two possible ways to look at it.One is to simply not think of
potential energy as being anything like a “substance,” but just an accounting device that we use to
keep track of the potential that a system has to do work for us—or (more or less equivalently) to
give us kinetic energy, which is always positive and hence mayb et h o u g h to fa st h e“ r e a l ”e n e r g y .
From this point of view, whetherU is positive or negative does not matter: all that matters is the",University Physics I Classical Mechanics.pdf
"From this point of view, whetherU is positive or negative does not matter: all that matters is the
change ∆U,a n dw h e t h e rt h i sc h a n g eh a sas i g nt h a tm a k e ss e n s e .T h i s ,a tleast, is the case here,
as I have argued in the paragraph above.
The other perspective is almost opposite, and based on Einstein’s theory of relativity: in this theory,
the total energy of a system is indeed “something like a substance,” in that it is directly related
to the system’s total inertia,m,t h r o u g ht h ef a m o u se q u a t i o nE = mc2.F r o m t h i s p o i n t o f v i e w ,
the total energy of a system of two particles, interacting gravitationally, at rest, and separated
by a distancer12,w o u l db et h es u mo ft h eg r a v i t a t i o n a lp o t e n t i a le n e r g y( n e gative), and the two
particles’ “rest energies,”m1c2 and m2c2:
Etotal = m1c2 + m2c2 − Gm1m2
r12
(10.7)",University Physics I Classical Mechanics.pdf
"particles’ “rest energies,”m1c2 and m2c2:
Etotal = m1c2 + m2c2 − Gm1m2
r12
(10.7)
and this quantitywill always be positive, unless one of the “particles” is a black hole and the other
one is inside it3!
Please note that we will not use equation (10.7)t h i ss e m e s t e ra ta l l ,s i n c ew ea r ec o n c e r n e d
only with nonrelativistic mechanics here. In other words, wew i l ln o ti n c l u d et h e“ r e s te n e r g y ”
in our calculations of a system’s total energy at all. However, if we did, we would find that a
system whose rest energy is given by Eq. (10.7)d o e s ,i nf a c t ,h a v ea ni n e r t i at h a ti sless than
the sum m1 + m2.T h i s s t r o n g l y s u g g e s t s t h a t t h e n e g a t i v i t yo f t h e p o t e n t i ale n e r g yi sn o tj u s ta
mathematical convenience, but rather it reflects a fundamental physical fact.
For a system of more than two particles, the total gravitational potential energy would be obtained",University Physics I Classical Mechanics.pdf
"For a system of more than two particles, the total gravitational potential energy would be obtained
by adding expressions like (10.6)o v e ra l lt h epairs of particles. Thus, for instance, for three particles
one would have
UG(⃗r1,⃗ r2,⃗ r3)= −Gm1m2
r12
− Gm1m3
r13
− Gm2m3
r23
(10.8)
Al a r g em a s ss u c ha st h ee a r t h ,o ras t a r ,h a sa ni n t r i n s i ca m o unt of gravitational potential energy
that can be calculated by breaking it up into small parts and performing a sum like (10.8)o v e ra l l
the possible pairs of “parts.” (As usual, this sum is usuallyevaluated as an integral, by taking the
limit of an infinite number of infinitesimally small parts.) This “self-energy” does not change with
3For a justification of this statement, please see the definition of the Schwarzschild radius, later on in this chapter.",University Physics I Classical Mechanics.pdf
"226 CHAPTER 10. GRAVITY
time, and hence does not need to be included in most energy calculations involving gravitational
forces between extended objects.
One thing that you may be wondering about, regarding Eq. (10.6)f o rt h ep o t e n t i a le n e r g yo fa
pair of particles (or, for that matter, Eq. (10.1)f o rt h ef o r c e ) ,i sw h a th a p p e n sw h e nt h ed i s t a n c e
r12 goes to zero, since the mathematical expression appears to become infinite. This is technically
true, but, in practice, it would only be a problem for a pair oftrue point particles—objects that
would literally be mathematical points, with no dimensionsat all. Such things may exist in some
sense—electrons may well be an example—but they need to be described by quantum mechanics,
which is an altogether different mathematical theory.
For finite-sized ob jects, you cannot continue to use an equation like (10.6)( o r(10.1), for the force)",University Physics I Classical Mechanics.pdf
"For finite-sized ob jects, you cannot continue to use an equation like (10.6)( o r(10.1), for the force)
when you areunder the surface of the object. If you could dig a tunnel all the way down to the
center of a hypothetical “earth” that had a constant density,t h ep o t e n t i a le n e r g yo ft h es y s t e m
formed by this “earth” and a particle of massm,ad i s t a n c er from the center, would look as shown
in Fig. 10.2.N o t i c e h o wUG becomes “flat,” indicating an equilibrium position (zero force), as
r → 0. It stands to reason that the net gravitational force at thecenter of this model “earth”
should be zero, since one would be pulled equally strongly inall directions by all the mass around.
r/RE
UG/mgRE
Figure 10.2: Gravitational potential energy of a system formed by a particle ofmass m and a hypothetical
earth with uniform density, a massM, and a radiusRE, as a function of the distancer between the particle",University Physics I Classical Mechanics.pdf
"earth with uniform density, a massM, and a radiusRE, as a function of the distancer between the particle
and the center of the “earth” (solid line). The dashed line shows theresult for a system of two (point-like)
particles. The energyUG is expressed in units ofmgRE,w h e r eg = GM/R2
E.
Finally, let me show you that the result (10.6)i sf u l l yc o n s i s t e n tw i t ht h ea p p r o x i m a t i o nUG = mgy
that we have been using up till now near the surface of the earth. (If you are not interested in
mathematical derivations, feel free to skip this next bit.)Consider a particle of massm that is",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 227
initially on the surface of the earth, and then we move it to a height h above the earth. The change
in potential energy, according to (10.6), is
UG
f − UG
i = −GMEm
RE + h + GMEm
RE
(10.9)
If we write both terms with a common denominator, we get
UG
f − UG
i = GMEm
(RE + h)RE
h ≃ GMEm
R2
E
h = mgh (10.10)
The only approximation here has been to setRE + h ≃ RE in the denominator of this expression.
Since RE is of the order of thousands of kilometers, this is an excellent approximation, as long as
h is less than, say, a few hundred meters.
10.1.2 Types of orbits under an inverse-square force
Consider a system formed by two particles (or two perfect, rigid spheres) interacting only with each
other, through their gravitational attraction. Conservation of the total momentum tells us that the
center of mass of the system is either at rest or moving with constant velocity. Let us assume that",University Physics I Classical Mechanics.pdf
"center of mass of the system is either at rest or moving with constant velocity. Let us assume that
one of the objects has a much greater mass,M,t h a nt h eo t h e r ,s ot h a t ,f o rp r a c t i c a lp u r p o s e s ,i t s
center coincides with the center of mass of the whole system.This is not a bad approximation if
what we are interested in is, for instance, the orbit of a planet around the sun. The most massive
planet, Jupiter, has only about 0.001 times the mass of the sun.
Accordingly, we will assume that the more massive object doesn o tm o v ea ta l l( b yw o r k i n gi ni t s
center of mass reference frame, if necessary—note that, by our assumptions, this will be an inertial
reference frame to a good approximation), and we will be concerned only with the motion of the less
massive object under the forceF = GMm/r2,w h e r er is the distance between the centers of the
two objects. Since this force is always pulling towards the center of the more massive object (it is",University Physics I Classical Mechanics.pdf
"two objects. Since this force is always pulling towards the center of the more massive object (it is
what is often called acentral force), its torque around that point is zero, and therefore the angular
momentum, ⃗L,o ft h el e s sm a s s i v eb o d ya r o u n dt h ec e n t e ro fm a s so ft h es y s tem is constant. This
is an interesting result: it tells us, for instance, that themotion is confined to a plane,t h es a m e
plane that the vectors⃗rand ⃗vdefined initially, since their cross-product cannot change.
In spite of this simplification, the calculation of the object’s trajectory, ororbit,r e q u i r e ss o m e
fairly advanced mathematical techniques, except for the simplest case, which is that of a circular
orbit of radius R.N o t e t h a t t h i s c a s e r e q u i r e s a v e r y p r e c i s e r e l a t i o n s h i p t ohold between the
object’s velocity and the orbit’s radius, which we can get bysetting the force of gravity equal to
the centripetal force:
GMm
R2 = mv2",University Physics I Classical Mechanics.pdf
"object’s velocity and the orbit’s radius, which we can get bysetting the force of gravity equal to
the centripetal force:
GMm
R2 = mv2
R (10.11)",University Physics I Classical Mechanics.pdf
"228 CHAPTER 10. GRAVITY
So, if we want to, say, put a satellite in a circular orbit around a central body of massM and at
ad i s t a n c eR from the center of that body, we can do it, but only provided wegive the satellite
an initial velocityv =
√
GM/R in a direction perpendicular to the radius. But what if we were
to release the satellite at the same distanceR, but with a different velocity, either in magnitude
or direction? Too much speed would pull it away from the circle, so the distance to the center,
r,w o u l dt e m p o r a r i l yi n c r e a s e ;t h i sw o u l di n c r e a s et h es y s t em’s potential energy and accordingly
reduce the satellite’s velocity, so eventually it would getpulled back; then it would speed up again,
and so on.
You may experiment with this kind of thing yourself using thePhET demo at this link:
https://phet.colorado.edu/en/simulation/gravity-and-orbits
You will find that, as long as you do not give the satellite—or planet, in the simulation—too much",University Physics I Classical Mechanics.pdf
"You will find that, as long as you do not give the satellite—or planet, in the simulation—too much
speed (more on this later!) the orbit you get is, in fact, a closed curve, the kind of curve we callan
ellipse.Ih a v ed r a w no n es u c he l l i p s ef o ry o ui nF i g .10.3.
a
b
e a
r
P A
O
Figure 10.3: An elliptical orbit. The semimajor axis isa, the semiminor axis isb, and the eccentricity
e =
√
1 − b2/a2 =0 .745 in this case.. The “center of attraction” (the sun, for instance, in the case of a
planet’s or comet’s orbit) is at the point O.
As a geometrical curve, any ellipse can be characterized by acouple of numbers,a and b,w h i c h
are the lengths of thesemimajor and semiminor axes, respectively. These lengths are shown in
the figure. Alternatively, one could specifya and a parameter known as theeccentricity,d e n o t e d
by e (do not mistake this “e”f o rt h ec o e ffi c i e n to fr e s t i t u t i o no fC h a p t e r4 ! ) ,w h i c hi se qual to
e =
√",University Physics I Classical Mechanics.pdf
"by e (do not mistake this “e”f o rt h ec o e ffi c i e n to fr e s t i t u t i o no fC h a p t e r4 ! ) ,w h i c hi se qual to
e =
√
1 − b2/a2.I f a = b,o re =0 ,t h ee l l i p s eb e c o m e sac i r c l e .
The most striking feature of the elliptical orbits under theinfluence of the 1/r2 gravitational force is
that the “central object” (the sun, for instance, if we are interested in the orbit of a planet, asteroid
or comet) isnot at the geometric center of the ellipse. Rather, it is at a special point called the
focus of the ellipse (labeled “O” in the figure, since that is the origin for the position vector of the
orbiting body). There are actually two foci, symmetricallyplaced on the horizontal (major) axis,",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 229
and the distance of each focus to the center of the ellipse is given by the productea,t h a ti s ,t h e
product of the eccentricity and the semimajor axis. (This explains why the “eccentricity” is called
that: it is a measure of how “off-center” the focus is.)
For an ob ject moving in an elliptical orbit around the sun, thed i s t a n c et ot h es u ni sm i n i m a la t
ap o i n tc a l l e dt h ep e r i h e l i o n ,a n dm a x i m a la tap o i n tc a l l e dthe aphelion. Those points are shown
in the figure and labeled “P” and “A”, respectively. For an object in orbit around the earth, the
corresponding terms are perigee and apogee; for an orbit around some unspecified central body,
the terms periapsis and apoapsis are used. There is some confusion as to whether the distances are
to be measured from the surface or from the center of the central body; here I will assume they are",University Physics I Classical Mechanics.pdf
"to be measured from the surface or from the center of the central body; here I will assume they are
all measured from the center, in which case the following relationships follow directly from Figure
10.3:
rmax =( 1+ e)a
rmin =( 1− e)a
rmin + rmax =2 a
e = rmax − rmin
2a (10.12)
The ellipse I have drawn in Fig.10.3 is actually way too eccentric to represent the orbit of any
planet in the solar system (although it could well be the orbito fac o m e t ) . T h ep l a n e tw i t ht h e
most eccentric orbit is Mercury, and that is onlye =0 .21. This means thatb =0 .978a,a na l m o s t
imperceptible deviation from a circle. I have drawn the orbitt os c a l ei nF i g .10.4,a n da sy o uc a n
see the only way you can tell it is an ellipse is, precisely, because the sun is not at the center.
Figure 10.4: Orbit of Mercury, with the sun approximately to scale.",University Physics I Classical Mechanics.pdf
"230 CHAPTER 10. GRAVITY
Since an ellipse has only two parameters, and we have two constants of the motion (the total
energy, E,a n dt h ea n g u l a rm o m e n t u m ,L), we should be able to determine what the orbit will look
like based on just those two quantities. Under the assumptionw ea r em a k i n gh e r e ,t h a tt h ev e r y
massive object does not move at all, the total energy of the system is just
E = 1
2mv2 − GMm
r (10.13)
For a circular orbit, the radiusR determines the speed (as per Eq. (10.11)), and hence the total
energy, which is easily seen to beE = −GMm
2R .I t t u r n s o u t t h a t t h i s f o r m u l a h o l d s a l s o f o r e l l i p t i c a l
orbits, if one substitutes the semimajor axisa for R:
E = −GMm
2a (10.14)
Note that the total energy (10.14)i snegative.T h i sm e a n st h a tw eh a v eabound orbit, by which I
mean, a situation where the orbiting object does not have enough kinetic energy to fly arbitrarily far",University Physics I Classical Mechanics.pdf
"mean, a situation where the orbiting object does not have enough kinetic energy to fly arbitrarily far
away from the center of attraction. Indeed, sinceUG → 0a sr →∞ ,y o uc a ns e ef r o mE q .(10.13)
that if the two objects could be infinitely far apart, the totale n e r g yw o u l de v e n t u a l l yh a v et ob e
positive, for any nonzero speed of the lighter object. So, ifE< 0, we have bound orbits, which are
ellipses (of which a circle is a special case), and conversely, ifE> 0w eh a v e“ u n b o u n d ”t r a j e c t o r i e s ,
which turn out to be hyperbolas4.T h e s e t r a j e c t o r i e sj u s t p a s s n e a r t h e c e n t e r o f a t t r a c t i o nonce,
and never return.
The special borderline case whenE =0c o r r e s p o n d st oaparabolic trajectory. In this case, the
particle also never comes back: it has just enough kinetic energy to make it “to infinity,” slowing",University Physics I Classical Mechanics.pdf
"particle also never comes back: it has just enough kinetic energy to make it “to infinity,” slowing
down all the while, sov → 0a s r →∞ .T h e i n i t i a l s p e e d n e c e s s a r y t o a c c o m p l i s h t h i s , s t a r t i n g
from an initial distanceri,i su s u a l l yc a l l e dt h e“ e s c a p ev e l o c i t y ”( a l t h o u g hi tr e a l lys h o u l db e
called the escape speed), and it is found by simply setting Eq.( 10.13)e q u a lt oz e r o ,w i t hr = ri,
and solving forv:
vesc =
√
2GM
ri
(10.15)
In general, you can calculate the escape speed from any initial distance ri to the central object,
but most often it is calculated from its surface. Note thatvesc does not depend on the mass of the
lighter object (always assuming that the heavier object doesn o tm o v ea ta l l ) .T h ee s c a p ev e l o c i t y
from the surface of the earth is about 11 km/s, or 1.1×104 m/s; but this alone would not be enough",University Physics I Classical Mechanics.pdf
"from the surface of the earth is about 11 km/s, or 1.1×104 m/s; but this alone would not be enough
to let you leave the attraction of the sun behind. The escape speed from the sun starting from a
point on the earth’s orbit is 42km/s.
To summarize all of the above, suppose you are trying to put something in orbit around a much
more massive body, and you start out a distancer away from the center of that body. If you give
the object a speed smaller than the escape speed at that point,t h er e s u l tw i l lb eE< 0a n da n
4There is apparently a way to describe a hyperbola as an ellipsew i t he c c e n t r i c i t ye> 1, but I’m definitely not
going to go there.",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 231
elliptical orbit (of which a circle is a special case, if you give it the precise speedv =
√
GM/r in
the right direction). If you give it precisely the escape speed (10.15), the total energy of the system
will be zero and the trajectory of the object will be a parabola; and if you give it more speed than
vesc,t h et o t a le n e r g yw i l lb ep o s i t i v ea n dt h et r a j e c t o r yw i l lb eah y p e r b o l a .T h i si si l l u s t r a t e di n
Fig. 10.5 below.
Figure 10.5: Possible trajectories for an object that is “released” with a sideways velocity at the lowest
point in the figure, under the gravitational attraction of a large mass represented by the black circle. Each
trajectory corresponds to a different value of the object’s initialkinetic energy: ifKcirc is the kinetic energy
needed to have a circular orbit through the point of release, the figure shows the casesKi =0 .5Kcirc (small",University Physics I Classical Mechanics.pdf
"needed to have a circular orbit through the point of release, the figure shows the casesKi =0 .5Kcirc (small
ellipse), Ki = Kcirc (circle), Ki =1 .5Kcirc (large ellipse), Ki =2 Kcirc (escape velocity, parabola), and
Ki =2 .5Kcirc (hyperbola).
Note that all the trajectories shown in Figure10.5 have the same potential energy at the “point
of release” (since the distance from that point to the centerof attraction is the same for all), so
increasing the kinetic energy at that point also means increasing the total energy (10.13)( w h i c hi s
constant throughout). So the picture shows different orbits ino r d e ro fi n c r e a s i n gt o t a le n e r g y .
For a given total energy, the total angular momentum does notchange the fundamental nature of
the orbit (bound or unbound), but it can make a big difference onthe orbit’s shape. Generally
speaking, for a given energy the orbits with less angular momentum will be “narrower,” or “more",University Physics I Classical Mechanics.pdf
"speaking, for a given energy the orbits with less angular momentum will be “narrower,” or “more
squished” than the ones with more angular momentum, since a smaller initial angular momentum
at the point of insertion means a smaller sideways velocity component. In the extreme case of zero
initial angular momentum (no sideways velocity at all), thetrajectory, regardless of the total energy,
reduces to a straight line, either straight towards or straight away from the center of attraction.",University Physics I Classical Mechanics.pdf
"232 CHAPTER 10. GRAVITY
For elliptical orbits, one can prove the result
e =
√
1 − L2
aGMm2 (10.16)
which shows how the eccentricity increases asL decreases, for a given value ofa (which is to say,
for a given total energy). I should at least sketch how to obtain this result, since it is a variant of
ap r o c e d u r et h a ty o um a yh a v et ou s ef o rs o m eh o m e w o r kp r o b l e mst h i ss e m e s t e r . Y o us t a r tb y
writing the angular momentum asL = mrP vP (or mrAvA), where A and P are the special points
shown in Figure 10.2, where⃗vand ⃗rare perpendicular. Then, you note thatrP = rmin =( 1− e)a
(or, alternatively,rA = rmax =( 1+e)a), sovP = L/[m(1 −e)a]. Then substitute these expressions
for rP and vP in Eq. (10.13), set the result equal to the total energy (10.14), and solve fore.
ri
vi
vi
vi
Figure 10.6: Effect of the “angle of insertion” on the orbit.
Figure 10.6 illustrates the effect of varying the angular momentum, for a given energy. All the",University Physics I Classical Mechanics.pdf
"Figure 10.6 illustrates the effect of varying the angular momentum, for a given energy. All the
initial velocity vectors in the figure have the same magnitude, and the release point (with position
vector ⃗ri)i st h es a m ef o ra l lt h eo r b i t s ,s ot h e ya l lh a v et h es a m ee n e r gy; indeed, you can check
that the semimajor axis of the two ellipses is the same as the radius of the circle, as required by
Eq. (10.14). The difference between the orbits is their total angular momentum. The green orbit
has the maximum angular momentum possible at the given energy, since the green velocity vector
is perpendicular to⃗ri.N o t e t h a t t h i s ( m a x i m i z i n gL for a givenE< 0) always results in a circle,
in agreement with Eq. (e10.15): the eccentricity is zero when L = Lcirc ≡
√
aGMm2,w h i c hi st h e
largest value ofL allowed in Eq. (e10.15).
For the other two orbits,⃗vi and ⃗ri make angles of 45◦ and 135◦,a n ds ot h ea n g u l a rm o m e n t u m",University Physics I Classical Mechanics.pdf
"largest value ofL allowed in Eq. (e10.15).
For the other two orbits,⃗vi and ⃗ri make angles of 45◦ and 135◦,a n ds ot h ea n g u l a rm o m e n t u m
L has magnitude L = Lcirc sin 45◦ = Lcirc/
√
2. The result are the red and blue ellipses, with
eccentricities e =
√
1 − sin2(45◦)=0 .707.",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 233
10.1.3 Kepler’s laws
The first great success of Newton’s theory was to account for the results that Johannes Kepler
had extracted from astronomical data on the motion of the planets around the sun. Kepler had
managed to find a number of regularities in a mountain of data (most of which were observations
by his mentor, the Danish astronomer Tycho Brahe), and expressed them in a succinct way in
mathematical form. These results have come to be known asKepler’s laws,a n dt h e ya r ea sf o l l o w s :
1. The planets move around the sun in elliptical orbits, withthe sun at one focus of the ellipse.
2. (Law of areas) A line that connects the planet to the sun (thep l a n e t ’ sp o s i t i o nv e c t o r )s w e e p s
equal areas in equal times.
3. The square of the orbital period of any planet is proportional to the cube of the semimajor axis
of its orbit (the same proportionality constant holds for allt h ep l a n e t s ) .",University Physics I Classical Mechanics.pdf
"of its orbit (the same proportionality constant holds for allt h ep l a n e t s ) .
Ih a v ed i s c u s s e dt h efi r s t“ l a w ”a tl e n g t hi nt h ep r e v i o u ss e c tion, and also pointed out that the math
necessary to prove it is far from trivial. The second law, on the other hand, while it sounds com-
plicated, turns out to be a straightforward consequence of the conservation of angular momentum.
To see what it means, consider Figure10.7.
O
A
B
A’
B’
rA
rB
vA∆t
vB∆t
Figure 10.7: Illustrating Kepler’s law of areas. The two gray “curved triangles” have the same area, so the
particle must take the same time to go from A to A′ as it does to go from B to B′.
Suppose that, at some timetA,t h ep a r t i c l ei sa tp o i n tA ,a n dat i m e∆t later it has moved to A′.
The area “swept” by its position vector is shown in grey in thefigure, and Kepler’s second law
states that it must be the same, for the same time interval, atany point in the trajectory; so, for",University Physics I Classical Mechanics.pdf
"234 CHAPTER 10. GRAVITY
instance, if the particle starts out at B instead, then in thesame time interval ∆t it will move to a
point B′ such that the area of the “curved triangle” OBB′ equals the area of OAA′.
Qualitatively, this means that the particle needs to move more slowly when it is farther from the
center of attraction, and faster when it is closer. Quantitatively, this actually just means that its
angular momentum is constant! To see this, note that the straight distance from A toA′ is the
displacement vector ∆⃗rA,w h i c h ,f o ras u ffi c i e n t l ys h o r ti n t e r v a l∆t,w i l lb ea p p r o x i m a t e l ye q u a lt o
⃗vA∆t.A g a i n ,f o rs m a l l∆t,t h ea r e ao ft h ec u r v e dt r i a n g l ew i l lb ea p p r o x i m a t e l yt h es ame as that
of the straight triangle OAA′.I ti saw e l l - k n o w nr e s u l ti nt r i g o n o m e t r yt h a tt h ea r e ao fatriangle",University Physics I Classical Mechanics.pdf
"of the straight triangle OAA′.I ti saw e l l - k n o w nr e s u l ti nt r i g o n o m e t r yt h a tt h ea r e ao fatriangle
is equal to 1/2 the product of the lengths of any two of its sidest i m e st h es i n eo ft h ea n g l et h e y
make. So, if the two triangles in the figures have the same areas, we must have
|⃗rA||⃗vA|∆t sin θA = |⃗rB||⃗vB|∆t sin θB (10.17)
and we recognize here the condition|⃗LA| = |⃗LB|,w h i c hi st os a y ,c o n s e r v a t i o no fa n g u l a rm o m e n -
tum. (Once the result is established for infinitesimally small ∆t,w ec a ne s t a b l i s hi tf o rfi n i t e - s i z e
areas by using integral calculus, which is to say, in essence,b yb r e a k i n gu pl a r g et r i a n g l e si n t os u m s
of many small ones.)
As for Kepler’s third result, it is easy to establish for a circular orbit, and definitely not easy for an
elliptical one. Let us callT the orbital period, that is, the time it takes for the less massive object",University Physics I Classical Mechanics.pdf
"elliptical one. Let us callT the orbital period, that is, the time it takes for the less massive object
to go around the orbit once. For a circular orbit, the angularvelocity ω can be written in terms of
T as ω =2 π/T ,a n dh e n c et h er e g u l a rs p e e dv = Rω =2 πR/T .S u b s t i t u t i n g t h i s i n E q . (10.11),
we getGM/R2 =4 π2R/T2,w h i c hc a nb es i m p l i fi e df u r t h e rt or e a d
T2 = 4π2
GM R3 (10.18)
Again, this turns out to work for an elliptical orbit if we replace R by a.
Note that the proportionality constant in Eq. (10.18)d e p e n d so n l yo nt h em a s so ft h ec e n t r a lb o d y .
For the solar system, that would be the sun, of course, and thent h ef o r m u l aw o u l da p p l yt oa n y
planet, asteroid, or comet, with the same proportionality constant. This gives you a quick way to
calculate the orbital period of anything orbiting the sun, ify o uk n o wi t sd i s t a n c e( o rv i c e - v e r s a ) ,",University Physics I Classical Mechanics.pdf
"calculate the orbital period of anything orbiting the sun, ify o uk n o wi t sd i s t a n c e( o rv i c e - v e r s a ) ,
based on the fact that you know what these quantities are for the Earth.
More generally, suppose you have two planets, 1 and 2, both orbiting the same star, at distances
R1 and R2,r e s p e c t i v e l y . T h e nt h e i ro r b i t a lp e r i o d sT1 and T2 must satisfyT2
1 =( 4π2/GM)R3
1 and
T2
2 =( 4π2/GM)R3
2.D i v i d e o n e e q u a t i o n b y t h e o t h e r , a n d t h e p r o p o r t i o n a l i t y constant cancels,
so you get
( T2
T1
) 2
=
( R2
R1
) 3
(10.19)",University Physics I Classical Mechanics.pdf
"10.1. THE INVERSE-SQUARE LAW 235
From this some simple manipulation gives you
T2 = T1
( R2
R1
) 3/2
(10.20)
Note you can expressR1 and R2 in any units you like, as long as you use the same units for both,
and similarlyT1 and T2.F o ri n s t a n c e ,i fy o uu s et h eE a r t ha sy o u rr e f e r e n c e“ p l a n e t1,” then you
know that T1 =1( i ny e a r s ) ,a n dR1 =1 ,i nA U( a nA U ,o r“ a s t r o n o m i c a lu n i t , ”i st h ed i s t a n c e
from the Earth to the sun). A hypothetical planet at a distanceo f4A Uf r o mt h es u ns h o u l dt h e n
have an orbital period of 8 Earth-years, since 43/2 =
√
43 =
√
64 = 8.
Af o r m u l aj u s tl i k e(10.18), but with a different proportionality constant, would applyto the
satellites of any given planet; for instance, the myriad of artificial satellites that orbit the Earth.
Again, you could introduce a “reference satellite” labeled1, with known period and distance to the",University Physics I Classical Mechanics.pdf
"Again, you could introduce a “reference satellite” labeled1, with known period and distance to the
Earth (the moon, for instance?), and derive again the result(10.20), which would allow you to get
the period of any other satellite, if you knew how its distancet ot h ee a r t hc o m p a r e st ot h em o o n ’ s
(or, conversely, the distance at which you would need to placei ti no r d e rt og e tad e s i r e do r b i t a l
period).
For instance, suppose I want to place a satellite on a “geosynchronous” orbit, meaning that it takes
1d a yf o ri tt oo r b i tt h eE a r t h . Ik n o wt h em o o nt a k e s2 9d a y s ,s oIc a nw r i t eE q .(10.20)a s
1=2 9 (R2/R1)3/2,o r ,s o l v i n gi t ,R2/R1 =( 1/29)2/3 =0 .106, meaning the satellite would have to
be approximately 1/10 of the Earth-moon distance from (the center of) the Earth.
In hindsight, it is somewhat remarkable that Kepler’s laws are as accurate, for the solar system,",University Physics I Classical Mechanics.pdf
"In hindsight, it is somewhat remarkable that Kepler’s laws are as accurate, for the solar system,
as they turned out to be, since they can only be mathematicallyd e r i v e df r o mN e w t o n ’ st h e o r yb y
making a number of simplifying approximations: that the sundoes not move, that the gravitational
force of the other planets has no effect on each planet’s orbit,and that the planets (and the sun)
are perfect spheres, for instance. The first two of these approximations work as well as they do
because the sun is so massive; the third one works because thesizes of all the objects involved
(including the sun) are much smaller than the correspondingorbits. Nevertheless, Newton’s work
made it clear that Kepler’s laws could only be approximatelyvalid, and scientists soon set to work
on developing ways to calculate the corrections necessary tod e a lw i t h ,f o ri n s t a n c e ,t h et r a j e c t o r i e s
of comets or the orbit of the moon.",University Physics I Classical Mechanics.pdf
"of comets or the orbit of the moon.
Of the main approximations I have listed above, the easiest one to get rid of (mathematically) is
the first one, namely, that the sun does not move. Instead, whato n efi n d si st h a t ,a sl o n ga st h es u n
and the planet are still treated as an isolated system, they will both revolve around the system’s
center of mass. Of course, the sun’s motion (a slight “wobble”) is very small, but not completely
negligible. You can even see it in the simulation I mentionedearlier, at
https://phet.colorado.edu/en/simulation/gravity-and-orbits.
What is much harder to deal with, mathematically, is the factthat none of the planets in the",University Physics I Classical Mechanics.pdf
"236 CHAPTER 10. GRAVITY
solar system actually forms an isolated system with the sun,since all the planets are really pulling
gravitationally on each other all the time. Particularly, Jupiter and Saturn have a non-negligible
influence on each other’s orbits, and on the orbits of every other planet, which can only be perceived
over centuries. Basically, the orbits still look like ellipses to a very good degree, but the ellipses
rotate very, very slowly (so they fail to exactly close in on themselves). This effect, known as orbital
precession, is most dramatic for Mercury, where the ellipse’s axes rotate by more than one degree
per century.
Nevertheless, the Newtonian theory is so accurate, and the calculation techniques developed over
the centuries so sophisticated, that by the early 20th century the precession of the orbits of all
planets except Mercury had been calculated to near exact agreement with thebest observational",University Physics I Classical Mechanics.pdf
"planets except Mercury had been calculated to near exact agreement with thebest observational
data. The unexplained discrepancy for Mercury amounted onlyt o4 3s e c o n d so fa r cp e rc e n t u r y ,
out of 5600 (an error of only 0.8%). It was eventually resolved by Einstein’s general theoryo f
relativity.
10.2 Weight, acceleration, and the equivalence principle
Whether we write it asmg or as GMm/r2,t h ef o r c eo fg r a v i t yo na no b j e c to fm a s sm has the
remarkable property—not shared by any other known force—ofbeing proportional to the inertial
mass of the object. This means that, if gravity is the only force acting on a system made up of
many particles, when you divide the force on each particle bythe particle’s mass in order to find
the particle’s acceleration, you get the same value ofa for every particle (at least, assuming that
they are all at about the same distance from the object exerting the force in the first place). Thus,",University Physics I Classical Mechanics.pdf
"they are all at about the same distance from the object exerting the force in the first place). Thus,
all the parts making up the object will accelerate together,as a whole.
Suppose that you are holding an object, while in free fall (remember that “free fall” means that
gravity is the only force acting on you), and you let go of it, asi nF i g .10.8 below. Since gravity
will give you and the object the same acceleration, you’ll findt h a ti td o e sn o t“ f a l l ”r e l a t i v et o
you—that is, it will not fall any faster nor more slowly than yourself. From your own reference
frame, you will just see it hovering motionless in front of you, in the same position (relative to you)
that it occupied before you let go of it. This is exactly what you see in videos shot aboard the
International Space Station. The result is an impression ofweightlessness, or “zero gravity”—even
though gravity is very much nonzero; the space station, and everything inside it, is constantly",University Physics I Classical Mechanics.pdf
"though gravity is very much nonzero; the space station, and everything inside it, is constantly
“falling” to the earth, it just does not hit it because it has some sideways velocity (or angular
momentum) to begin with, and the earth’s pull just bends its trajectory around enough to keep it
moving in a circle. But gravity is the only force acting on it,and on everything in it (at least until
somebody pushes himself against a wall, or something like that).
So, the kind of acceleration you get from gravity is, paradoxically, such that, if you give in to it
completely, you feel like there isno gravity.",University Physics I Classical Mechanics.pdf
"10.2. WEIGHT, ACCELERATION, AND THE EQUIVALENCE PRINCIPLE 237
(a) (b)
Figure 10.8: If you are holding something while in free fall (a) and let go, since you are all accelerating at
the same rate, it stays in the same position relative to you (b), so it appears to be weightless.
The familiar sensation of weight, on the other hand, comes precisely fromnot giving in, and rather,
enlisting other forces to fight against gravity. When you do this—when you simply stand on the
surface of the earth, for instance—your feet are supported byt h eg r o u n db e l o wy o u ,b u te v e r yo t h e r
part of your body is supported by some other part of your body,immediately above or below it,
that you can think of as a sort of spring that is either somewhats t r e t c h e do rs o m e w h a tc o m p r e s s e d .
It is primarily your skeleton, and mostly your spine, that bears most of the compressive load. (See",University Physics I Classical Mechanics.pdf
"It is primarily your skeleton, and mostly your spine, that bears most of the compressive load. (See
Fig. 10.9,n e x tp a g e . ) T h es e n s a t i o no fw e i g h ti sy o u rr e s p o n s et ot h i sload. Interestingly, even
though this constant squishing may actually result in your losing a little height in the course of
ad a y( w h i c hy o ur e c o v e ra tn i g h t ,w h e ny o ul i eh o r i z o n t a l l y ),i ti sn o tab a dt h i n g ,r a t h e rt h e
contrary: your bones have evolved so that theyneed this constant pressure to grow and replace the
mass that they would otherwise lose in a “weightless” environment.
On the other hand, as shown in Fig.10.9 (c), the same compression (or extension—for instance,
for the muscles in your arms, as they hang by your side) would result from a situation in which
you were, say, standing motionless inside a rocket that is accelerating upwards witha = g,b u t",University Physics I Classical Mechanics.pdf
"you were, say, standing motionless inside a rocket that is accelerating upwards witha = g,b u t
very far away from any gravity source. In Fig.10.9 (b), the “spring” that represents your skeleton
needs to be compressed so it can exert an upward forceFspr = mug to support the weight of your
upper body (simplified here as just a single massmu). In Fig.10.9 (c), it needs to be compressed
by the same amount, so it can exert the upward forceFspr = mua needed to give your upper body
an accelerationa = g.T h e e q u a l i t yo ft h e t w oe x p r e s s i o n si sa d i r e c t c o n s e q u e n c eof the fact that
the force of gravity is proportional to an object’s inertialmass (since the second expression is just",University Physics I Classical Mechanics.pdf
"238 CHAPTER 10. GRAVITY
Newton’s second law).
(a) gravity g, a = – g (b) gravity g,  a = 0 (c)  no gravity, a =  g
Fgl
n
Fgl
n
Fsl
spr
Fsu
spr
Fsu
spr
Fsl
spr
FEu
G FEu
G
FEl
G
FEl
G
Figure 10.9: (a) In free fall, your skeleton (represented here by a relaxed spring) does not need to support
your upper body, so there is no sensation of weight. When standingon the ground motionless under the
influence of gravity, however (b), every part of your body needsto compress a little in order to support the
weight of the parts above it (as shown here by the compressed spring). The same compression, and hence
the same subjective sensation of weight, results if you are moving upwards with an accelerationa = g,b u t
in the absence of gravity (c). (The subscriptsu and l on the forces stand for “upper” and “lower” body,
respectively.)
In general, then, when your whole body is subjected to an upward accelerationa,i tf e e l sl i k ey o u r",University Physics I Classical Mechanics.pdf
"respectively.)
In general, then, when your whole body is subjected to an upward accelerationa,i tf e e l sl i k ey o u r
weight is increased by an amountma.T h e s a m e t h i n g h o l d s r e g a r d l e s s o f d i r e c t i o n — a f o r w a r d
acceleration a on a jet pilot’s body feels like a “weight”ma pushing her against her seat. This is
why these “effective forces” (or, more precisely, the accelerations that cause them) are measured
in g’s: a “force” of, say, 5g,m e a n st h a tt h ep i l o tf e e l sp u s h e da g a i n s th e rs e a tw i t ha“ f orce” equal
to 5 times her weight. What’s really happening, of course, isthe opposite—her seat is pushing
her forward,b u th e ri n t e r n a lo r g a n sa r eb e i n gc o m p r e s s e d( i no r d e rt op rovide that same forward
acceleration) the way they would under a gravity force five times stronger than at the earth’s
surface.
The parallels between being in a constantly accelerating frame of reference and being at rest under",University Physics I Classical Mechanics.pdf
"surface.
The parallels between being in a constantly accelerating frame of reference and being at rest under
the influence of a constant gravity force go beyond the subjective sensation of weight. Figure10.10",University Physics I Classical Mechanics.pdf
"10.2. WEIGHT, ACCELERATION, AND THE EQUIVALENCE PRINCIPLE 239
illustrates what happens when you drop something while traveling in the upwardly accelerating
rocket, in the absence of gravity. From an inertial observer’s point of view, the object you drop
merely keeps the upward velocity it had the moment it left yourh a n d ;b u t ,s i n c ey o ua r ei nc o n t a c t
with the rocket, your own velocity is constantly increasing,a n da sar e s u l to ft h a ty o us e et h e
object fall—relative to you.
(a) (b)
Figure 10.10: “Dropping” an object inside a constantly accelerating rocket, away from any gravity.
From a practical point of view, this suggests a couple of waysto provide an “artificial gravity” for
astronauts who might one day have to spend a long time in space,e i t h e ru n d e re x t r e m e l yw e a k
gravity (for instance, during a trip to Mars), or, what amounts to essentially the same thing, in",University Physics I Classical Mechanics.pdf
"gravity (for instance, during a trip to Mars), or, what amounts to essentially the same thing, in
free fall (as in a space station orbiting a planet). The one most often seen in movies consists in
having the space station (or spaceship) constantly spin around an axis with some angular velocity
ω.T h e n a n y o b j e c t t h a t i s m o v i n g w i t h t h e s t a t i o n , a d i s t a n c eR away from the axis, will experience
ac e n t r i p e t a la c c e l e r a t i o no fm a g n i t u d eω2R,w h i c hw i l lf e e ll i k eag r a v i t yf o r c emω2R directed in
the opposite direction, that is to say, away from the center.People would then basically “walk on
the walls” (that is to say, sideways as seen from above, with their feet away from the rotation axis
and their heads towards the rotation axis). If somebody let goo fs o m e t h i n gt h e yw e r eh o l d i n g ,",University Physics I Classical Mechanics.pdf
"240 CHAPTER 10. GRAVITY
the object would “fall towards the wall,” just like the objectc o n s i d e r e di nE x a m p l e9 . 6 . 3( p r e v i o u s
chapter). Unfortunately, while the idea might work for a space station, it would probably be
impractical for a spaceship, since one would need a fairly large R and/or a fairly large rotation rate
to get ω2R ≃ g.( O n t h e o t h e r h a n d , p r o b a b l y e v e n s o m e t h i n g l i k e1
5 g is better than nothing, so
who knows... )
On a fundamental level, the equivalence between a constantlya c c e l e r a t e dr e f e r e n c ef r a m e ,a n d
an inertial frame with a uniform gravitational field (such as,a p p r o x i m a t e l y ,t h es u r f a c eo ft h e
earth), was elevated by Einstein to a basic principle of physics, which became the foundation of
his general theory of relativity. Thisequivalence principle asserts that it is absolutely impossible",University Physics I Classical Mechanics.pdf
"his general theory of relativity. Thisequivalence principle asserts that it is absolutely impossible
to distinguish, by any kind of physics experiment, between the two situations just mentioned: a
constantly accelerated reference frame is postulated to becompletely equivalent in every way to an
inertial frame with a uniform gravitational field.
Ar e m a r k a b l ec o n s e q u e n c eo ft h ee q u i v a l e n c ep r i n c i p l ei st hat light, despite having technically “zero
rest mass,” must bend its trajectory under the influence of gravity. This can be seen as follows.
Imagine shooting a projectile horizontally inside the rocket in Figure10.10.A l t h o u g h a n i n e r t i a l
observer, looking from the outside, would see the projectilet r a v e li nas t r a i g h tl i n e ,t h eo b s e r v e r
inside the rocket would see its path bend down, just as for theprojectiles we studied back in Chapter",University Physics I Classical Mechanics.pdf
"inside the rocket would see its path bend down, just as for theprojectiles we studied back in Chapter
8. This is for the same reason he would see the object fall, relative to him, in Figure10.10:t h e
projectile has a constant velocity, so it travels the same distance in every equal time interval, but
the rocket is accelerating, so the distance it travels in equal time intervals is constantly increasing.
In basically the same way, then, a beam of light sent horizontally inside the rocket, and traveling
with constant velocity (and, therefore, in a straight line)in an inertial frame, would be seen as
bending down in the rocket’s reference frame.
However, if the equivalence principle is true, and physicalphenomena look the same in a constantly
accelerating frame as in an inertial frame with a constant gravitational field, it follows that light
must also bend its path in the latter system, in much the same way as a projectile would. (I say",University Physics I Classical Mechanics.pdf
"must also bend its path in the latter system, in much the same way as a projectile would. (I say
“much the same way” because the effect is not just as simple as giving light an “effective mass”;
there are other relativistic effects, such as space contraction and time dilation, that must also be
reckoned with.) This gravitational bending was one of the most important early predictions of
Einstein’s General Relativity theory, and certainly the most spectacular. Since one needs the light
rays to pass vary close to a large mass to get an observable effect, the way the prediction was
verified was by looking at the apparent position of the stars that can be seen close to the edge of
the sun’s disk during a solar eclipse. The slight (apparent)shift in position predicted by Einstein
was observed by Sir Arthur Eddington during the solar eclipseo f1 9 1 9( t w oe x p e d i t i o n sw e r es e n t
to remote corners of the earth for this purpose), and it was primarily responsible for Einstein’s",University Physics I Classical Mechanics.pdf
"to remote corners of the earth for this purpose), and it was primarily responsible for Einstein’s
sudden fame among the general public of his day.
Today, with modern telescopes, this so-called “gravitational lensing” effect has become an important
tool in astronomy, allowing us to interpret the pictures taken of distant galaxies, which are often",University Physics I Classical Mechanics.pdf
"10.3. IN SUMMARY 241
shifted and/or distorted by the gravity of the galaxies thatlie in between them and us.
It has even become possible to imagine an object so dense thatit would “capture” light, attracting
it so strongly that it could not leave the object’s neighborhood. Such an object has come to be
called a black hole.I f y o u s e t t h e e s c a p e v e l o c i t y o f E q . (10.15)e q u a lt ot h es p e e do fl i g h ti n
vacuum, c,a n ds o l v ef o rri,y o uo b t a i nw h a ti sc a l l e dt h eSchwarzschild radius, rs,f o rab l a c kh o l e
of mass M;t h ei d e ab e i n gt h a t ,i no r d e rt obe ab l a c kh o l e ,t h eo b j e c th a st ob es od e n s et h a ta l l
its mass M is inside a sphere of radius smaller thanrs.P h y s i c i s t s t o d a y b e l i e v e i n t h e e x i s t e n c e
(and even what one might call the ubiquity) of black holes, ofwhich the Schwarzschild solution was",University Physics I Classical Mechanics.pdf
"(and even what one might call the ubiquity) of black holes, ofwhich the Schwarzschild solution was
only the first calculated example. Note thatrs does not define the actual, physical surface of the
object; it does, however, locate what is known as the black hole’sevent horizon.N o t h i n g c a n b e
known, through observation, about anything that might happen closer to the black hole’s center
than the distancers,s i n c en oi n f o r m a t i o nc a nb et r a n s m i t t e df a s t e rt h a nl i g h t ,and no light can
escape from a distanceri <r s.
10.3 In summary
1. In Newton’s theory of gravity, two particles of inertial masses m1 and m2,s e p a r a t e db ya
distance r12,e x e r tag r a v i t a t i o n a lf o r c eo ne a c ho t h e rw h i c hi sa t t r a c t ive, along the line
joining the two particles, and has magnitudeFG
12 = FG
21 = Gm1m2/r2
12,w i t hG =6 .674 ×
10−11 m3kg−1s−2.
2. The gravitational force between two extended objects is found by adding (vectorially) the",University Physics I Classical Mechanics.pdf
"21 = Gm1m2/r2
12,w i t hG =6 .674 ×
10−11 m3kg−1s−2.
2. The gravitational force between two extended objects is found by adding (vectorially) the
forces between all the pairs of particles that make up the objects. For objects with spherical
symmetry, the result has the same form as above, withr12 being now the distance between
the centers of the spheres.
3. The gravitational potential energy of a system of two particles of masses m1 and m2 is
UG = −Gm1m2/r12.F o r s y s t e m s o f m o r e p a r t i c l e s , o n e s h o u l d j u s t a d d t h e c o r r esponding
energies for all the possible pairs. For a pair of spheres, onem a yu s et h es a m er e s u l ta sf o r
two particles, as long as one is not interested in the spheres’g r a v i t a t i o n a ls e l f - e n e r g y .
4. The expressions given above forFG and UG reduce, respectively, tomg and mgy+C (where C
is an unimportant constant) near the surface of the earth, toag o o da p p r o x i m a t i o n ,p r o v i d e d",University Physics I Classical Mechanics.pdf
"is an unimportant constant) near the surface of the earth, toag o o da p p r o x i m a t i o n ,p r o v i d e d
the distancey to the surface is much smaller than the radius of the earth,RE.I f ME is the
mass of the earth, one hasg = GME/R2
E.
5. A good first approximation to many astronomical problems iso b t a i n e db yc o n s i d e r i n gt h e
motion of a particle (or sphere) of massm under the gravitational pull of an object (also
treated as a particle or sphere) of much larger massM,w h i c hi sa s s u m e dt on o tm o v ea ta l l .
This is sometimes called theKepler problem.",University Physics I Classical Mechanics.pdf
"242 CHAPTER 10. GRAVITY
6. The solutions to the Kepler problem are of two types, depending on the system’s total energy
E:b o u n d , e l l i p t i c a l o r b i t s ( i n c l u d i n g c i r c l e s a s a s p e c i a l case), if E< 0; and unbound
hyperbolic trajectories, ifE> 0. The special trajectory obtained whenE =0i sap a r a b o l a .
7. For the elliptical orbits, one hasE = −GMm/2a,w h e r ea is the ellipse’s semimajor axes.
The large mass object is not at the center, but at one of the focio ft h ee l l i p s e . T h ed i s t a n c e
from the focus to the center is equal toea,w h e r ee is called theeccentricity of the ellipse.
8. The escape speedof an object bound gravitationally to a massM,ad i s t a n c eri away from
that mass’s center, is obtained by setting the total energy oft h es y s t e me q u a lt oz e r o . I ti s
the speed the object needs in order to be able to just escape to“infinity” and “stop there”",University Physics I Classical Mechanics.pdf
"the speed the object needs in order to be able to just escape to“infinity” and “stop there”
(mathematically, v → 0a sr →∞ ,w h i c hm a k e sE = K + UG =0 ) .
9. The angular momentum of a particle in a Kepler trajectory (circle, ellipse, parabola or hy-
perbola), relative to the point where the large massM is located, is constant. For a given
energy, orbits with less angular momentum are more eccentric.
10. A consequence of conservation of angular momentum is Kepler’s second law, or “law of areas”:
The orbiting object’s position vector (with the origin at thel o c a t i o no ft h el a r g em a s s ) ,s w e e p s
equal areas in equal times.
11. The square of the orbital period of any object in a Kepler elliptical orbit is proportional to
the cube of the semimajor axis of the ellipse. This is Kepler’st h i r dl a w . S p e c i fi c a l l y ,o n e
finds, from Newton’s theory,T2 =( 4π2/GM)a3.
12. According to Einstein’s principle of equivalence, a constant accelerationa of a reference frame",University Physics I Classical Mechanics.pdf
"finds, from Newton’s theory,T2 =( 4π2/GM)a3.
12. According to Einstein’s principle of equivalence, a constant accelerationa of a reference frame
is experienced by every object in that reference frame as an “extra weight,” or gravitational
force, equal to−ma (that is, of magnitudema and in the direction opposite the acceleration).",University Physics I Classical Mechanics.pdf
"10.4. EXAMPLES 243
10.4 Examples
10.4.1 Orbital dynamics
In the early days of space flight, astronauts sometimes mentioned the counterintuitive aspects of
orbital flight. For example, if, from a circular orbit aroundthe Earth, they wanted to move to a
lower orbit, the way to do it was to slow down their capsule (byfiring a thruster in the direction
opposite their motion). This would take them to a lower orbit,b u tt h e nt h ec a p s u l ew o u l ds t a r t
speeding up,o ni t so w n .
Use the concepts introduced in this chapter to explain what isg o i n go ni nt h i ss c e n a r i o . L e tR
be the radius of the initial orbit. For simplicity, assume thet h r u s t e ri so no n l yf o rav e r ys h o r t
time, so you can neglect the motion of the capsule during thistime. In other words, treat it as an
instantaneous reduction in velocity, and discuss:
(a) What happens to the system’s potential and kinetic energy, and angular momentum?",University Physics I Classical Mechanics.pdf
"instantaneous reduction in velocity, and discuss:
(a) What happens to the system’s potential and kinetic energy, and angular momentum?
(b) Is the new orbit circular or elliptic? How do you know? Whati st h en e wo r b i t ’ srmax (maximum
distance to the center of the Earth)?
(c) Why does the capsule speed up in its new orbit?
(d) If the new orbit is not circular, what would the astronautsn e e dt od ot om a k ei ts o ? ( W i t h o u t
getting any closer to the Earth, that is, keepingrmin the same.)
Make sure to draw a diagram of the situation. Make it as accurate as you can.
Solution
(a) Under the assumption that the capsule barely changes position during the thruster firing, the
potential energy of the system, which is equal toUG = −GMm/R,w i l ln o tc h a n g e :UG
f = UG
i .
The kinetic energy, on the other hand, will go down, since thecapsule’s speed is reduced:Kf <K i.
Hence, the total mechanical energy of the system,E = K + UG,w i l ld e c r e a s e :Ef <E i.",University Physics I Classical Mechanics.pdf
"Hence, the total mechanical energy of the system,E = K + UG,w i l ld e c r e a s e :Ef <E i.
The angular momentum will go down, sincev goes down.
(b) The new orbit has to be elliptical, since to have a circularo r b i ta tad i s t a n c eR requires a
precise velocity (given just below Eq. (10.11)b yv =
√
GM/R), and now we have changed that.
However, since the orbit must still be a closed curve, it willcontain the starting point, which is,
by our assumption, a distanceR away from the Earth. Also, if the direction of the velocity vector
does not change as a result of the thruster firing (only the magnitude ofv is supposed to change),
it follows that at this point the velocity and the position vectors are perpendicular. For a circular
orbit, this is the case everywhere. For an elliptical orbit,this is only true at the two extreme points
labeled P and A in Figure10.3 (the perigee and apogee, respectively). So, the initial position of",University Physics I Classical Mechanics.pdf
"labeled P and A in Figure10.3 (the perigee and apogee, respectively). So, the initial position of
the capsule becomes either the perigee or the apogee of the newo r b i t .W h i c hi si t ?",University Physics I Classical Mechanics.pdf
"244 CHAPTER 10. GRAVITY
To get the answer, recall that we found in (a) that the total mechanical energyE has gone down.
But, since E is a negative number, this means themagnitude of E has gone up.T h e n , i n t h e
formula (10.14),
E = −GMm
2a
the semimajor axisa must have gonedown.F o r t h e o r i g i n a l c i r c u l a r o r b i t , w e h a da = R;n o w ,
we must havea<R .T h i s m e a n s t h a t t h e s t a r t i n g p o i n t , a d i s t a n c eR away from the (center of
the) Earth, cannot be the perigee (the point of closest approach), since at that pointr = rmin,a n d
rmin is always less thana (check again Fig.10.3,o rE q s .(10.12)). Instead, the starting point has
to be the apogee of the new orbit, and therefore the distance att h a tp o i n ti sa l s ot h em a x i m u m
distance: rmax = R.
(c) The capsule speeds up in its new orbit because, as we just saw, it starts as far away from the",University Physics I Classical Mechanics.pdf
"distance: rmax = R.
(c) The capsule speeds up in its new orbit because, as we just saw, it starts as far away from the
Earth as it’s going to get; therefore, as it moves it will startg e t t i n gc l o s e rt ot h eE a r t h ,a n dw e
know from Kepler’s second law that as it gets closer it has to speed up. (You can also say that,
as it gets closer, the gravitational potential energy of thesystem will go down, and therefore its
kinetic energy must increase.)
(d) The easiest way to change the new orbit to a circular orbitwith radiusrmin would be to perform
another speed-reduction maneuver, but this time at perigee.A tp e r i g e e ,t h ed i s t a n c et ot h eE a r t h
is already rmin,w h i c hi sw h a ty o uw a n ti tt ob e ,b u tt h ec a p s u l ei sm o v i n gt o ofast to stay on a
circular orbit (put differently, the gravitational force of the Earth at that point is too weak to bend
the orbit into a circle): that is why it eventually ends up “overshooting” the Earth on the other",University Physics I Classical Mechanics.pdf
"the orbit into a circle): that is why it eventually ends up “overshooting” the Earth on the other
side. Reducing v will further reduceE and, by the same argument as above, it will result in an
orbit with a smallera,w h i c hi sw h a ty o uw a n t( s i n c e ,a tt h em o m e n t ,a>r min,a n dy o uw a n tt h e
new a to be equal tormin).
AP",University Physics I Classical Mechanics.pdf
"10.4. EXAMPLES 245
The diagram of the situation is above (previous page). The long-dash circle is the original orbit;
the solid line is the elliptical orbit resulting from the speed reduction at point A; the short-dash
circle is the circular orbit that would result from another speed reduction at the point P. Note: the
size of the orbits is greatly exaggerated compared to those int h ee a r l ys p a c efl i g h t s ,w h i c hw e r e
much closer to the Earth!
The way to draw this kind of figure is to first draw an accurate ellipse, making sure you know where
the focus is; then draw the circles centered at the focus and touching the ellipse at the right points.
An ellipse’s equation in polar form, with the origin at one focus, isr = a + ae cos φ.
10.4.2 Orbital data from observations: Halley’s comet
Halley’s comet follows an elliptical orbit around the sun. Ati t sc l o s e s ta p p r o a c h ,i ti sad i s t a n c eo f",University Physics I Classical Mechanics.pdf
"Halley’s comet follows an elliptical orbit around the sun. Ati t sc l o s e s ta p p r o a c h ,i ti sad i s t a n c eo f
0.59 AU from the sun (an astronomical unit, AU, is defined as the average distance from the earth
to the sun: 1 AU = 1.496 × 1011 m), and it is moving at 5.4 × 104 m/s. We know its period is
approximately 76 years. Ignoring the forces exerted on the comet by the other solar system objects
(a rather rough approximation):
(a) Use the appropriate Kepler law to infer the value ofa (the semimajor axis) for the comet’s
orbit.
(b) What is the eccentricity of the comet’s orbit?
(c) Using the result in (a) and conservation of angular momentum, find the speed of the comet at
aphelion (the point in its orbit when it is farthest away fromthe sun).
Solution
(a) The “appropriate Kepler law” here is the third one. For anyt w oo b j e c t so r b i t i n g ,f o ri n s t a n c e ,
the sun, the square of their orbital periods is proportionalto the cube of their orbits’ semimajor",University Physics I Classical Mechanics.pdf
"the sun, the square of their orbital periods is proportionalto the cube of their orbits’ semimajor
axes, with the same proportionality constant (4π/GMsun;s e eE q .(10.18)). We do not even need
to calculate the proportionality constant; we can divide thee q u a t i o nf o rH a l l e y ’ sc o m e tb yt h e
equation for the earth, and get
T2
Halley
T2
earth
=
a3
Halley
a3
earth
(10.21)
where T2
earth =1y r2,a n da3
earth =1A U3,s ow eg e ti m m e d i a t e l y
aHalley =
(
762)1/3
AU = 17.9A U ( 1 0 . 2 2 )
(b) We can get this one from a look at Figure10.3:t h e p r o d u c tea,p l u st h em i n i m u md i s t a n c e
between the comet and the sun (0.59 AU) is equal toa.( T h i s i s j u s t w h a t t h e s e c o n d o f t h e",University Physics I Classical Mechanics.pdf
"246 CHAPTER 10. GRAVITY
equations (10.12)s a y sa sw e l l . ) .S ow eh a v e
e = a − rmin
a =1 − rmin
a =1 − 0.59
17.9 =0 .967 (10.23)
Note that we did not even have to convert AU to kilometers. In these types of problems, particularly,
where you have to manipulate very large numbers, it really pays off to do all the calculations
symbolically and not substitute the numbers in until the verye n d ,t os e ei fs o m e t h i n gc a n c e l so u t ,
and to prevent mistakes when copying large numbers form one line to the next; and sometimes,
like here, you do not even have to convert to other units!
(c) At the point of closest approach (perihelion), the velocity and the position vector of the comet
are perpendicular, and so the magnitude of the comet’s angular momentum is just equal toL = mrv.
The same happens at the farthest point in the orbit (aphelion), and since angular momentum is
conserved for the Kepler problem, we can write
mrminvmax = mrmaxvmin (10.24)",University Physics I Classical Mechanics.pdf
"conserved for the Kepler problem, we can write
mrminvmax = mrmaxvmin (10.24)
(the reason for this choice of subscripts is that we know thatwhen r is maximum,v is minimum,
and vice-versa). Solving forvmin,t h es p e e da ta p h e l i o n ,w eg e t
vmin = rmin
rmax
vmax = rmin
2a − rmin
vmax = 0.59
2 · 17.9 − 0.59 5.4 × 104 m
s =9 0 5m
s (10.25)
Here again the equation I used to findrmax can be derived directly from Fig.10.3 (and it is also
one of the equations (10.12): rmin + rmax =2 a). Once again, I was able to use AU throughout,
since the units of distance cancel out in the fractionrmin/rmax.",University Physics I Classical Mechanics.pdf
"10.5. ADVANCED TOPICS 247
10.5 Advanced Topics
10.5.1 Tidal Forces
Throughout this chapter we have treated the objects interacting gravitationally as if they were
particles, that is to say, as if they were non-deformable andtheir shape and relative orientation
did not matter. However, these conditions are never quite realized in real life. Some planets, like
our own Earth, are particularly susceptible to deformation,b e c a u s eo ft h el a r g ea m o u n to ffl u i d
matter on their surface, and even rocky planets and moons aresensitive totidal forces,w h i c ha r e
the differences on the gravitational pull by the central attractor on different parts of the object
considered.
If you look back on Figure10.1,f o ri n s t a n c e ,i ti se a s yt os e et h a tt h em o o nm u s tb ep u l l i n gmore
strongly on the side of the Earth that is closer to it (the leftside, on that picture) than on the",University Physics I Classical Mechanics.pdf
"strongly on the side of the Earth that is closer to it (the leftside, on that picture) than on the
farther side. This is, of course, because the force of gravityd e p e n d so nt h ed i s t a n c eb e t w e e nt h e
interacting objects, and is stronger when the objects are closer. You can easily calculate that the
force by the moon on a given volume of Earth (say, a cubic meter)i sa b o u t7 %s t r o n g e ro nt h en e a r
side than on the far side. A deformable object subject to suchap a i ro ff o r c e sw i l ln a t u r a l l yb e
stretched along the direction of the pull: in the case of the Earth, this “stretching” affects primarily
the water in the oceans that cover most of the surface, resulting in two “tidal bulges” that account
for the well-known phenomenon of tides: as the earth rotatesaround its axis, each point on the
surface passes through one of the bulges once a day, resultingi nt w oh i g ht i d e se a c hd a y( a n d ,i n",University Physics I Classical Mechanics.pdf
"surface passes through one of the bulges once a day, resultingi nt w oh i g ht i d e se a c hd a y( a n d ,i n
between, a comparatively lower water level, or low tide, twice a day as well).5
For many ob jects in the solar system, this tidal stretching has, over millions of years, resulted in a
permanent deformation. The tidal forces by the Earth on the moon are weaker than those by the
moon on the Earth (since the moon is much smaller, the differenceb e t w e e nt h eE a r t h ’ sp u l lo nt h e
moon’s near and far side is less than 1.5%), but at a time when the moon was more malleable than
at present, it was enough to produce an elongation along the Earth-moon axis that is now pretty
much frozen in place.
Once a satellite (I will use the term generically to refer either to a planet orbiting the sun, or a moon
orbiting a planet) becomes permanently deformed, a new phenomenon, known astidal locking,c a n",University Physics I Classical Mechanics.pdf
"orbiting a planet) becomes permanently deformed, a new phenomenon, known astidal locking,c a n
happen. Suppose the satellite is rotating around its own axis, in addition to orbiting around the
primary body. As you can see from the figure below, if the rotation is too fast, the gravitational
5You may ask, besides being stretched, shouldn’t the net pullof the moon cause the Earth to fall towards it?
The answer is that the Earth does not fall straight towards them o o n ,f o rt h es a m er e a s o nt h em o o nd o e sn o tf a l l
straight towards the Earth: both have a sideways velocity that keeps them in a steady orbit, for which the net
gravitational pull provides the centripetal force. Both actually orbit around the center of mass of the system, which
is located about 1, 690 km under the Earth’s surface; the Earth’s orbit (and its orbital velocity) is much smaller than
the moon’s, by virtue of its much larger mass, of course.",University Physics I Classical Mechanics.pdf
"248 CHAPTER 10. GRAVITY
forces from the primary will result in a net torque on the satellite that will tend to slow down its
rotation; conversely, if it is rotating too slowly, the torque will tend to speed up the rotation. A
torque-free situation will only happen when the satellite’sp e r i o do fr o t a t i o ne x a c t l ym a t c h e si t s
orbital period, so that it always shows the same side to the primary body. This is the situation
with the Earth’s moon, and indeed for most of the major moons oft h eg i a n tp l a n e t s .
(a) (b)
τ τ
Figure 10.11: An elongated moon revolving around a planet in a clockwise orbit, and at the same time
rotating clockwise around an axis through its center. In (a), the rotation is too fast, resulting in a counter-
clockwise “tidal torque.” In (b), the rotation is too slow, and the “tidal torque” is clockwise. In both cases,
the torque is due to the moon’s misalignment, and to the gravitational force on its near side being stronger",University Physics I Classical Mechanics.pdf
"the torque is due to the moon’s misalignment, and to the gravitational force on its near side being stronger
than on its far side, as shown by the blue force vectors.
Note that, by the same argument, we would expect the tidal forces on the Earth due to the moon
to try and bring the Earth into tidal locking with the moon—that is, to try to bring the duration
of an Earth day closer to that of a lunar month. Indeed, the moon’s tidal forceshave been slowing
down the Earth’s rotation for billions of years now, and continue to do so by about 15 microseconds
every year. This process requires dissipation of energy, (which is in fact associated with the ocean
tides: think of the frictional forces caused by the waves, asthe tide comes in and out); however,
to the extent that the Earth-moon system may be treated as isolated, its total angular momentum
cannot change, and so the slowing-down of the Earth is accompanied by a very gradual increase",University Physics I Classical Mechanics.pdf
"cannot change, and so the slowing-down of the Earth is accompanied by a very gradual increase
in the radius of the moon’s orbit—about 3.8c m pe r y e a r , c u r r e n t l y — t o k e e p t h e t o t a l a n g u l a r
momentum constant.",University Physics I Classical Mechanics.pdf
"10.6. PROBLEMS 249
10.6 Problems
Problem 1
Suppose you fire a projectile straight up from the Earth’s North Pole with a speed of 10.5k m / s .
Ignore air resistance.
(a) How far from the center of the Earth does the projectile rise? How high above the surface of
the Earth is that? (The radius of the Earth isRE =6 .37 × 106 m, and the mass of the Earth is
M =5 .97 × 1024 kg.)
(b) How different is the result you got in part (a) above from whaty o uw o u l dh a v eo b t a i n e di f
you had treated the Earth’s gravitational force as a constant( i n d e p e n d e n to fh e i g h t ) ,a sw ed i di n
previous chapters?
(c) Using the correct expression for the gravitational potential energy, what is the total energy of
the projectile-Earth system, if the projectile’s mass is 1, 000 kg?
Now assume the projectile is firedhorizontally instead, with the same speed. This time, it actually
goes into orbit! (Well, it would, if you could neglect thingslike air resistance, and mountains and",University Physics I Classical Mechanics.pdf
"goes into orbit! (Well, it would, if you could neglect thingslike air resistance, and mountains and
stuff like that. Assume it does, anyway, and answer the following questions:)
(d) What is the projectile’s angular momentum around the center of the Earth?
(e) How far from the center of the Earth does it make it this time? (You will need to use conservation
of energy and angular momentum to answer this one, unless youcan think of a shortcut... )
(f) Draw a sketch of the Earth and the projectile’s trajectory.
Problem 2
You want to put a satellite in ageosynchronous orbit around the earth. (This means the asteroid
takes 1 day to complete a turn around the earth.)
(a) At what height above the surface do you need to put it?
(b) How fast is it moving?
(c) How does the answer to (b) compare to the escape speed fromthe earth, for an objectat this
height?
Problem 3
Suppose that one day astronomers discover a new asteroid thatm o v e so nav e r ye l l i p t i c a lo r b i t",University Physics I Classical Mechanics.pdf
"height?
Problem 3
Suppose that one day astronomers discover a new asteroid thatm o v e so nav e r ye l l i p t i c a lo r b i t
around the sun. At the point of closest approach (perihelion), the asteroid is 1.61 × 108 km away
from the (center of the) sun, and its speed is 38.9k m / s .
(a) What is the escape velocity from the sun at this distance?The mass of the sun is 2× 1030 kg.
(b) The astronomers estimate the mass of the asteroid as 1012 kg. What is its kinetic energy at
perihelion?
(c) What is the gravitational potential energy of the sun-asteroid system at perihelion?
(d) What is the total energy of the sun-asteroid system? Is itpositive or negative? Is this consistent
with the assumption that the orbit is an ellipse? What would apositive total energy mean?
(e) At perihelion, the asteroid’s velocity vector is perpendicular to its position vector (as drawn",University Physics I Classical Mechanics.pdf
"250 CHAPTER 10. GRAVITY
from the sun). What is then its angular momentum?
(f) Draw a sketch of an elliptical orbit. On your sketch, indicate (1) the semimajor axis, and (2)
qualitatively, where the sun might be.
(g) The point in its orbit where the asteroid is farthest awayfrom the sun is called aphelion. Use
conservation of energy and angular momentum to figure out theasteroid’s distance to the sun at
aphelion. (Hint: if solving simultaneous equations does nota p p e a lt oy o u ,t h e r ei saf o r m u l ai n
this chapter which you can use to answer this question fairlyquickly, based on something you have
calculated already.)
(h) How fast is the asteroid moving at aphelion?
Problem 4
The mass of the moon is 7.34 × 1022 kg, and its radius is about 1.74 × 106 m
(a) What is the value of “gmoon”, that is, the acceleration of gravity for a falling object near the
surface of the moon?
(b) What is the escape speed (from the moon) for an object on thes u r f a c eo ft h em o o n ?",University Physics I Classical Mechanics.pdf
"surface of the moon?
(b) What is the escape speed (from the moon) for an object on thes u r f a c eo ft h em o o n ?
(c) What is the escape speedfrom the earthfor an object that is as far from the earth as the orbit
of the moon?
(d) At some point between the earth and the moon, an object would be pulled with equal strength
towards both bodies. How far from the earth is that point?
Problem 5
On August 17, 2017, the LIGO observatory reported the detection of gravitational waves from
the merger of two neutron stars. Neutron stars are extremelydense (“a teaspoon of neutron star
material has a mass of about a billion tons”) and very small—only about 10 or 20 km in diameter.
The stars were estimated to have been separated by about 300 kmw h e nt h em e r g e rs i g n a lb e c a m e
detectable.
Let us start a little before that. Suppose the stars have the same mass, M =2 .6 × 1030 kg",University Physics I Classical Mechanics.pdf
"detectable.
Let us start a little before that. Suppose the stars have the same mass, M =2 .6 × 1030 kg
(approximately 1.3t i m e st h em a s so ft h es u n ) ,a n da r es e p a r a t e d( c e n t e r - t o - c enter distance) by
1000 km. They pull on each other gravitationally, and as a result each one moves in a circular orbit
around their common center of mass. What is then (a) their period of revolution, and (b) their
speed? (Hint: what is the centripetal force in this case?)
Problem 6
(a) Consider two possible circular orbits for a satellite around a planet, with radiiR1 and R2.I f
R1 <R 2,w h i c ho ft h et w oo r b i t sh a s(i)t h el a r g e s tt o t a le n e r g y ,a n d(ii)t h el a r g e s tt o t a la n g u l a r
momentum? Explain.
(b) For an object in a circular orbit around a planet, how doesthe orbital velocity compare to the
escape velocity from the same orbit?
Problem 7",University Physics I Classical Mechanics.pdf
"(b) For an object in a circular orbit around a planet, how doesthe orbital velocity compare to the
escape velocity from the same orbit?
Problem 7
Jupiter’s distance to the sun is 5.2a s t r o n o m i c a lu n i t s . H o wl o n gd o e si tt a k ef o rJ u p i t e rt oc o mplete",University Physics I Classical Mechanics.pdf
"10.6. PROBLEMS 251
an orbit around the sun, in earth years? (Do not look it up! Youneed to show how you can calculate
it using what you have learned in this chapter.)",University Physics I Classical Mechanics.pdf
252 CHAPTER 10. GRAVITY,University Physics I Classical Mechanics.pdf
"Chapter 11
Simple harmonic motion
11.1 Introduction: the physics of oscillations
It is probably not an exaggeration to suggest that we are all introduced to oscillatory motion from
our first moments of life. Babies, it seems, are constantly rocked to sleep, in many cases using
devices, such as cradles and rocking chairs, that exemplifythe kind of mechanical oscillator with
which this chapter is concerned. And then, of course, there are swings, which function essentially
like the pendulum depicted below.
(a) (b) (c) (d)
F t
F G
Fnet
v
F t
F G Fnet
Figure 11.1: A simple pendulum. In (a), the equilibrium position, the tension and gravity forces balance
out. In (b), they combine to produce a restoring force (in blue) pointing back towards equilibrium. In (c),
the bob is passing through equilibrium and the net force on it at that instant is again zero, but its momentum
keeps it going. At (d) we have the mirror image of (b).
253",University Physics I Classical Mechanics.pdf
"254 CHAPTER 11. SIMPLE HARMONIC MOTION
In fact, oscillatory motion is extremely common, both in natural systems and in human-made
structures. It essentially requires only two things: a stable equilibrium configuration, where the
stability is ensured by what we call arestoring force;a n di n e r t i a ,w h i c h ,o fc o u r s e ,e v e r yp h y s i c a l
system has.
The pendulum in Figure11.1 illustrates how these things combine to produce an oscillation. As the
pendulum bob is displaced from its equilibrium position, a net force on it appears (a combination
of gravity and the tension in the string), pointing back towards the vertical. When the bob is
released, it accelerates under the influence of this force, with the result that when it reaches back
the equilibrium position, its inertia (or, if you prefer, itsm o m e n t u m )c a u s e si tt oo v e r s h o o ti t .
Once this happens, the restoring force changes direction, always trying to bring the mass back",University Physics I Classical Mechanics.pdf
"Once this happens, the restoring force changes direction, always trying to bring the mass back
to equilibrium; as a result, the bob slows down, and eventually reverses course, accelerates again
towards the vertical, overshoots it again... the process will repeat itself, until all the energy we
initially put in the system (gravitational potential energy, in this case) is dissipated away (or
damped), mostly through friction at the pivot point, though air resistance plays a small part as
well.
That the motion, in the absence of dissipation, must be symmetric around the equilibrium position
follows from conservation of energy: the speed of the bob at any given height must be the same on
either side, in order for the sum of its potential and kineticenergies to be the same. In particular,
if released from rest from some height, it will stop when it reaches the same height on the other
side. In the presence of dissipation, the motion is neither exactly symmetric, nor exactly periodic",University Physics I Classical Mechanics.pdf
"side. In the presence of dissipation, the motion is neither exactly symmetric, nor exactly periodic
(that is to say, it does not repeat itself exactly—the maximumh e i g h tg e t sl o w e re v e r yt i m e ,t h e
speed as it passes through the equilibrium position gets alsos m a l l e ra n ds m a l l e r ) ,b u tw h e nt h e
dissipation is not very large one can always define an approximate period (which we will denote
with the letterT)a st h et i m ei tt a k e st oc o m p l e t eo n ef u l ls w i n g .
The inverse of the period is thefrequency, f,w h i c ht e l l su sh o wm a n yf u l ls w i n g st h ep e n d u l u m
completes per second. These two quantities,T and f,c a nb ed e fi n e df o ra n yt y p eo fp e r i o d i c( o r
approximately periodic) motion, and will always satisfy ther e l a t i o n s h i p
f = 1
T (11.1)
The units of frequency are, of course, inverse seconds (s−1). In this context, however, this unit is
called a ”hertz,” and abbreviated Hz.",University Physics I Classical Mechanics.pdf
"T (11.1)
The units of frequency are, of course, inverse seconds (s−1). In this context, however, this unit is
called a ”hertz,” and abbreviated Hz.
11.2 Simple harmonic motion
Ap a r t i c u l a r l yi m p o r t a n tk i n do fo s c i l l a t o r ym o t i o ni sc a l led simple harmonic motion.T h i s i s w h a t
happens when the restoring force islinear in the displacement from the equilibrium position: that",University Physics I Classical Mechanics.pdf
"11.2. SIMPLE HARMONIC MOTION 255
is to say, in one dimension, ifx0 is the equilibrium position, the restoring force has the form
F = −k(x − x0)( 1 1 . 2 )
We are familiar with this from Hooke’s “law” for an ideal spring (see Chapter 6). So, an object
attached to an ideal, massless spring, as in the figure below,should perform simple harmonic
motion. This kind of oscillation is distinguished by the following characteristics:
• The position as a function of time,x(t), is a sinusoidal function.
• The period of the oscillations does not depend on their amplitude (by “amplitude” we mean
the maximum displacement from the equilibrium position).
What this second property means is that, for instance, with reference to Fig.11.2,y o uc a nd i s p l a c e
the mass a distanceA,o rA/2, orA/3, or whatever you choose, and the period (and frequency) of
the resulting oscillations will be the same regardless. (This means, actually, that if you displace it",University Physics I Classical Mechanics.pdf
"the resulting oscillations will be the same regardless. (This means, actually, that if you displace it
farther it has to end up moving faster, to cover the larger distance in the same time.)
xx0
(a)
(b)
(c)
A
A
Figure 11.2: A mass attached to a spring and sliding on a frictionless surface. Figure (a) shows the spring
in its relaxed state (the “equilibrium” position for the mass, at coordinate x0). If displaced from equilibrium
a distanceA and released (b), the mass will perform simple harmonic oscillations with amplitudeA.
Since we know that “Hooke’s law” is actually just an approximation, valid only provided that
the spring is not compressed or stretched too much, we expectthat in real life the “ideal” simple
harmonic motion properties I have listed above will only holda p p r o x i m a t e l y ,a sw e l l ;s o ,i nf a c t ,
if you stretch a spring too much you will get a different period,eventually, than if you stay in",University Physics I Classical Mechanics.pdf
"if you stretch a spring too much you will get a different period,eventually, than if you stay in
the “linear,” Hooke’s law regime. This is a general characteristic of most physical systems: simple",University Physics I Classical Mechanics.pdf
"256 CHAPTER 11. SIMPLE HARMONIC MOTION
harmonic motion only happens for relatively small oscillations, but “relatively small” can still be
fairly large sometimes, and even as an approximation it is often an extremely valuable one.
The other distinctive characteristic of simple harmonic motion is that the position function is
sinusoidal, by which I mean a sine or a cosine. Thus, for example, if the mass in Fig.11.2 is
released from restat t =0 ,a n dt h ep o s i t i o nx is measured from the equilibrium positionx0 (that
is, the pointx = x0 is taken as the origin of coordinates), the functionx(t)w i l lb e
x(t)= A cos(ωt)( 1 1 . 3 )
where the quantityω,k n o w na st h eo s c i l l a t o r ’ sangular frequency,i sg i v e nb y
ω =
√
k
m (11.4)
Here, k is the spring constant, andm the mass of the object (remember the spring is assumed to be
massless). I will prove that Eq. (11.3), together with (11.4), satisfy Newton’s second law of motion",University Physics I Classical Mechanics.pdf
"massless). I will prove that Eq. (11.3), together with (11.4), satisfy Newton’s second law of motion
for this system in a moment; first, however, I need to say a couple of things aboutω.Y o u ’ l lr e c a l l
that we have used this symbol before, in Chapter 9, to represent theangular velocity of a particle
moving in a circle (or, more generally, of any rotating object). Why bring it up again now for an
apparently completely different purpose?
+
r
v
P
x
y
R sin(ωt)θ
R cos(ωt)
Figure 11.3: A particle moving on a circle with constant angular velocityω. Assuming θ =0a t t =0 ,
we have θ = ωt, and therefore the particle’sx coordinate is given by the functionx(t)= R cos(ωt). This
means the corresponding point on thex axis (the red dot) performs simple harmonic motion with angular
frequency ω as the particle rotates.",University Physics I Classical Mechanics.pdf
"11.2. SIMPLE HARMONIC MOTION 257
The answer is that there is a very close relationship betweensimple harmonic motion and circular
motion with constant speed, as Figure11.3 illustrates: as the point P rotates with constant angular
velocity ω,i t sp r o j e c t i o no n t ot h ex axis (the red dot in the figure) performs simple harmonic motion
with angular frequencyω (and amplitudeR). (Of course, there is nothing special about thex axis;
the projection onany other axis will also perform simple harmonic motion with thesame angular
frequency; for example, the blue dot on the figure.)
If the angular velocity of the particle in Fig.11.3 is constant, then its “orbital period” (the time
needed to complete one revolution) will beT =2 π/ω,a n dt h i sw i l la l s ob et h ep e r i o do ft h e
associated harmonic motion (the time it takes for the motionto repeat itself). You can see this
directly from Eq. (11.3): if you increase the timet by 2π/ω,y o ug e tt h es a m ev a l u eo fx:
x
(
t + 2π
ω
)",University Physics I Classical Mechanics.pdf
"directly from Eq. (11.3): if you increase the timet by 2π/ω,y o ug e tt h es a m ev a l u eo fx:
x
(
t + 2π
ω
)
= A cos
[
ω
(
t + 2π
ω
)]
= A cos(ωt +2 π)= A cos(ωt)= x(t)( 1 1 . 5 )
Since the frequencyf of an oscillator is equal to 1/T,t h i sg i v e su st h ef o l l o w i n gr e l a t i o n s h i pb e t w e e n
f and ω:
f = 1
T = ω
2π (11.6)
One way to tell whether one is talking about an oscillator’s frequency (f)o ri t sa n g u l a rf r e q u e n c y
(ω)—apart from the different symbols, of course—is to pay attention to the units. The frequency
f is usually given in hertz, whereas the angular frequencyω is always given in radians per second.
Apart from the factor of 2π,t h e ya r e ,o fc o u r s e ,c o m p l e t e l ye q u i v a l e n t ;s o m e t i m e so n eis just
more convenient than the other. On the other hand, the only wayt ot e l lw h e t h e rω is a harmonic
oscillator’s angular frequency or the angular velocity of something moving in a circle is from the",University Physics I Classical Mechanics.pdf
"oscillator’s angular frequency or the angular velocity of something moving in a circle is from the
context. (In this chapter, of course, it will always be the former).
Let us go back now to Eq.11.3 for our block-on-a-spring system. The derivative with respect to
time will give us the block’s velocity. This is a simple application of the chain rule of calculus:
v(t)= dx
dt = −ωA sin(ωt)( 1 1 . 7 )
Another derivative will then give us the acceleration:
a(t)= dv
dt = −ω2A cos(ωt)( 1 1 . 8 )
Note that the acceleration is always proportional to the position, only with the opposite sign. The
proportionality constant is ω2.S i n c e t h e f o r c e e x e r t e d b y t h e s p r i n g o n t h e b l o c k i sF = −kx
(because we are measuring the position from the equilibriumposition x0), Newton’s second law,
F = ma,g i v e su s
ma = −kx (11.9)
and you can check for yourself that this will be satisfied ifx is given by Eq. (11.3), a is given by
Eq. (11.8), andω is given by Eq. (11.4).",University Physics I Classical Mechanics.pdf
"258 CHAPTER 11. SIMPLE HARMONIC MOTION
The expression (11.4)f o rω is typical of what we find for many different kinds of oscillators: the
restoring force (here represented by the spring constantk)a n dt h eo b j e c t ’ si n e r t i a(m)t o g e t h e r
determine the frequency of the motion, acting in opposite directions: a larger restoring force means
ah i g h e rf r e q u e n c y( f a s t e ro s c i l l a t i o n s )w h e r e a sal a r g e rinertia means a lower frequency (slower
oscillations—a more “sluggish” response).
The position, velocity and acceleration graphs for the motion (11.3)a r es h o w ni nF i g .11.4 below.
You may want to pay attention to some of their main features. For instance, the position and the
velocity are what we call “90◦ out of phase”: one is maximum (or minimum) when the other one
is zero. The acceleration, on the other hand, is “180◦ out of phase” (that is to say, in complete",University Physics I Classical Mechanics.pdf
"is zero. The acceleration, on the other hand, is “180◦ out of phase” (that is to say, in complete
opposition) with the position. As a result of that, all combinations of signs fora and v are possible:
the object may be moving to the right with positive or negativea c c e l e r a t i o n( d e p e n d i n go nw h i c h
side of the origin it’s on), and likewise when it is moving to the left.
A
–A
0
ωA
–ωA
0
ω2A
–ω2A
0
x(t)
v(t)
a(t)
TT/2 3T/2 2T0
Figure 11.4: Position, velocity and acceleration as a function of time for an object performing simple
harmonic motion according to Eq. (11.3).
Since the time we choose ast =0i sa r b i t r a r y ,t h ef u n c t i o ni nE q .(11.3)( w h i c ha s s u m e st h a t
t =0i sw h e nt h eo b j e c t ’ sd i s p l a c e m e n ti sm a x i m u ma n dp o s i t i v e)i sc l e a r l yn o tt h em o s tg e n e r a l
formula for simple harmonic motion. Another way to see this ist or e a l i z et h a tw ec o u l dh a v e",University Physics I Classical Mechanics.pdf
"formula for simple harmonic motion. Another way to see this ist or e a l i z et h a tw ec o u l dh a v e
started the motion differently. For instance, we could have hitt h eb l o c kw i t has h a r p ,“ i m p u l s i v e ”
force, lasting only a very short time, so it would have acquired a substantial velocity before it could
have moved very far from its initial (equilibrium) position.I n s u c h a c a s e , t h e m o t i o n w o u l d b e
better described by a sine function, such asx(t)= A sin(ωt), which is zero att =0b u tw h o s e
derivative (the object’s velocity) is maximum at that time.",University Physics I Classical Mechanics.pdf
"11.2. SIMPLE HARMONIC MOTION 259
If we stick to using cosines, for definiteness, then the most general equation for the position of a
simple harmonic oscillator is as follows:
x(t)= A cos(ωt + φ)( 1 1 . 1 0 )
where φ is what we call a “phase angle,” that allows us to match the function to the initial
conditions—by which I mean, the object’s initial position and velocity. Specifically, you can see, by
setting t =0i nE q .(11.10)a n di t sd e r i v a t i v e ,t h a tt h ei n i t i a lp o s i t i o na n dv e l o c i t yof the motion
described by Eq. (11.10)a r e
xi = A cos φ
vi = −ωA sin φ (11.11)
Conversely, if you are givenxi and vi,y o uc a nu s eE q s .(11.11)t od e t e r m i n eA and φ,w h i c hi s
what you need to know in order to use Eq. (11.10)( n o t et h a tt h ea n g u l a rf r e q u e n c y ,ω,d o e snot
depend on the initial conditions—it is always the same regardless of how you choose to start the",University Physics I Classical Mechanics.pdf
"depend on the initial conditions—it is always the same regardless of how you choose to start the
motion). Specifically, you can verify that Eqs. (11.11)i m p l yt h ef o l l o w i n g :
A2 = x2
i + v2
i
ω2 (11.12)
and then, once you knowA,y o uc a ng e tφ from eitherxi = A cos φ or vi = −ωA sin φ (in fact, since
the inverse sine and inverse cosine are both multivalued functions, you should useboth equations,
to make sure you get the correct sign forφ).
11.2.1 Energy in simple harmonic motion
Equation (11.11)a b o v ea c t u a l l yf o l l o w sf r o mt h ec o n s e r v a t i o no fe n e r g yp r inciple for a harmonic
oscillator. Consider again the mass on the spring in Figure11.3.I t s k i n e t i c e n e r g y i s c l e a r l y
K = 1
2 mv2,w h e r e a st h ep o t e n t i a le n e r g yi nt h es p r i n gi s1
2 kx2.U s i n g E q . (11.10)a n di t sd e r i v a t i v e ,
we have
Uspr = 1
2kA2 cos2(ωt + φ)
K = 1
2mω2A2 sin2(ωt + φ)( 1 1 . 1 3 )",University Physics I Classical Mechanics.pdf
"2 kx2.U s i n g E q . (11.10)a n di t sd e r i v a t i v e ,
we have
Uspr = 1
2kA2 cos2(ωt + φ)
K = 1
2mω2A2 sin2(ωt + φ)( 1 1 . 1 3 )
Recalling Eq. (11.4), note thatω2 = k/m,s oi fy o us u b s t i t u t et h i si nt h es e c o n de q u a t i o na b o v e ,
you can see that the amplitude of both the potential and the kinetic energy is the same, namely,
1
2 kA2.S i n c e ,f o ra n ya n g l eθ,i ti sa l w a y st r u et h a tc o s2 θ +s i n2 θ =1 ,w efi n d
Esys = Uspr + K = 1
2kA2 = 1
2mω2A2 (11.14)",University Physics I Classical Mechanics.pdf
"260 CHAPTER 11. SIMPLE HARMONIC MOTION
so the total energy of the system is constant (independent oftime), at it should be, in the absence
of dissipation. Figure11.5 shows how the potential and kinetic energies oscillate in opposition, so
one is maximum whenever the other is minimum. It also shows that they oscillate twice as fast as
the oscillator itself: for example, the potential energy ismaximum both when the displacement is
maximum (spring maximally stretched)and when it is minimum (spring maximally compressed).
Similarly, the kinetic energy is maximum when the oscillatorp a s s e st h r o u g ht h ee q u i l i b r i u mp o s i -
tion, regardless of whether it is moving to the left or to the right.
TT/2 3T/2 2T0
Esys
Uspr
K
0
Figure 11.5: Kinetic (red), potential (blue) and total (black) energy for the oscillator shown in Fig.11.4.
11.2.2 Harmonic oscillator subject to an external, constantf o r c e",University Physics I Classical Mechanics.pdf
"11.2.2 Harmonic oscillator subject to an external, constantf o r c e
Consider a mass hanging from an ideal spring suspended from the ceiling, as in Fig.11.6 below
(next page). Supposed the relaxed length of the spring isl,s u c ht h a t ,i nt h ea b s e n c eo fg r a v i t y ,
the object’s equilibrium position would be at the heighty0 shown in figure11.6(a). In the presence
of gravity, of course, the spring needs to stretch, to balancet h eo b j e c t ’ sw e i g h t ,a n ds ot h ea c t u a l
equilibrium position for the system will bey′
0,a ss h o w ni nfi g u r e11.6(b). The upwards force from
the spring at that point will be−k(y′
0 − y0), and to balance gravity we must have
− k(y′
0 − y0) − mg =0 ( 1 1 . 1 5 )
Suppose that we now stretch the spring beyond this new equilibrium position, so the mass is now
at a heighty (figure 11.6(c)). What happens then? The net upwards force will be−k(y −y0)−mg,
but using Eq. (11.15)t h i sc a nb er e w r i t t e na s
Fnet = −k(y − y0) − mg = −k(y − y′",University Physics I Classical Mechanics.pdf
"but using Eq. (11.15)t h i sc a nb er e w r i t t e na s
Fnet = −k(y − y0) − mg = −k(y − y′
0) −− k(y′
0 − y0) − mg = −k(y − y′
0)( 1 1 . 1 6 )",University Physics I Classical Mechanics.pdf
"11.2. SIMPLE HARMONIC MOTION 261
This is a remarkable result, because the force of gravity hasdisappeared completely from the final
expression. Basically, the system behaves as if it consistedo fj u s tas p r i n go fc o n s t a n tk with
equilibrium length l′ = l + y0 − y′
0,a n dno gravity.I n o t h e r w o r d s , t h e o n l y t h i n g g r a v i t y d o e s i s
to change the equilibrium position, so that if you now displace the mass, it will oscillate around
y′
0 instead of aroundy0.T h e o s c i l l a t i o n ’ s p e r i o d a n d f r e q u e n c y a r e t h e s a m e a s i f t hes p r i n gw a s
horizontal.
y0
y0'
y
l l'
(a) (b) (c)
Figure 11.6: (a) An ideal (massless) spring hanging from the ceiling, in its relaxed position. (b) With a
mass m hanging from its end, the spring stretches to a new lengthl′,s ot h a tk(l′ −l)= mg.( c )I ft h em a s s
is now displaced from this equilibrium position (labeledy′
0 in the figure) it will perform harmonic oscillations",University Physics I Classical Mechanics.pdf
"is now displaced from this equilibrium position (labeledy′
0 in the figure) it will perform harmonic oscillations
symmetrically around the pointy′
0, with the same frequency as if the spring was horizontal.
Although I have established this here for the specific case where the oscillator involves a spring,
and the external force is gravity, this is a completely general result, valid for any simple harmonic
oscillator, since for such a system the restoring force willalways be a linear function of the displace-
ment (which is all that is required for the math to work). As long as the external force is constant,
the frequency of the oscillations will not be affected, and onlyt h ee q u i l i b r i u mp o s i t i o nw i l lc h a n g e .
In an example at the end of the chapter (under “Advanced Topics”) I will show you how you can
make use of this to calculate the effect of friction on the horizontal mass-spring combination in
Fig. 11.2.",University Physics I Classical Mechanics.pdf
"make use of this to calculate the effect of friction on the horizontal mass-spring combination in
Fig. 11.2.
One thing you need to keep in mind, however, is that when the oscillator is subjected to an external
force, as was the case here, its energy willnot,i ng e n e r a l ,r e m a i nc o n s t a n t( u n l i k ew h a tw es a w
in Section11.2.1), since the external force will be doing work on the system asit oscillates. If the",University Physics I Classical Mechanics.pdf
"262 CHAPTER 11. SIMPLE HARMONIC MOTION
external force is constant, and does not change direction, this work will be positive half the time,
and negative half the time. If it is kinetic friction, then ofcourse it will change direction every half
cycle, and the work will be negative all the time.
In the case shown in Figure11.6,t h ee x t e r n a lf o r c ei sg r a v i t y ,w h i c hw ek n o wt ob eac o n s e r v ative
force, so the energy that will be conserved will be the total energy of the system that includes the
oscillation and the Earth, and hence also the gravitationalpotential energy (for which we can use
here the familiar formUG = mgy):
Eosc+earth = Uspr + K + UG = 1
2(y − y0)2 + 1
2mv2 + mgy =c o n s t ( 1 1 . 1 7 )
The reason it is no longer possible to combine the termsUspr + K into the constant1
2 kA2,a si n
Eq. (11.14), is that now we have
y(t)= y′
0 + A cos(ωt + φ)
v(t)= −ωA sin(ωt + φ)( 1 1 . 1 8 )
so the oscillations are centered around the new equilibriumposition y′",University Physics I Classical Mechanics.pdf
"y(t)= y′
0 + A cos(ωt + φ)
v(t)= −ωA sin(ωt + φ)( 1 1 . 1 8 )
so the oscillations are centered around the new equilibriumposition y′
0,b u tt h es p r i n ge n e r g yi s
not zero at that point: it is zero aty = y0 instead. You can check for yourself, however, that
if you substitute Eqs. (11.18)i n t oE q .(11.17), and make use of the fact thatk(y′
0 − y0)= −mg
(Eq. (11.15)), you do indeed get a constant, as you should.
11.3 Pendulums
11.3.1 The simple pendulum
Besides masses on springs, pendulums are another example ofas y s t e mt h a tw i l le x h i b i ts i m p l e
harmonic motion, at least approximately, as long as the amplitude of the oscillations is small. The
simple pendulum is just a mass (or “bob”), approximated hereas a point particle, suspended from
am a s s l e s s ,i n e x t e n s i b l es t r i n g ,a si nF i g .11.7 on the next page.
We could analyze the motion of the b ob by using the general methods introduced in Chapter 8 to",University Physics I Classical Mechanics.pdf
"We could analyze the motion of the b ob by using the general methods introduced in Chapter 8 to
deal with motion in two dimensions—breaking down all the forces into components and applying
⃗Fnet = m⃗ aalong two orthogonal directions—but this turns out to be complicated by the fact
that both the direction of motion and the direction of the acceleration are constantly changing.
Although, under the assumption of small oscillations, it turns out that simply using the vertical
and horizontal directions is good enough, this is not immediately obvious, and arguably it is not
the most pedagogical way to proceed.",University Physics I Classical Mechanics.pdf
"11.3. PENDULUMS 263
F t
F G
lθ
o
Figure 11.7: A simple pendulum. The mass of the bob ism, the length of the string isl,a n dt o r q u e sa r e
calculated around the point of suspension O. The counterclockwisedirection is taken as positive.
Instead, I will take advantage of the obvious fact that the bobm o v e so na na r co fac i r c l e ,a n dt h a t
we have developed already in Chapter 9 a whole set of tools to deal with that kind of motion. Let
us, therefore, describe the position of the pendulum by the angle it makes with the vertical,θ,a n d
let α = d2θ/dt2 be the angular acceleration; we can then write the equation ofm o t i o ni nt h ef o r m
τnet = Iα,w i t ht h et o r q u e st a k e na r o u n dt h ec e n t e ro fr o t a t i o n — w h i c his to say, the point from
which the pendulum is suspended. Then the torque due to the tension in the string is zero (since
its line of action goes through the center of rotation), andτnet is just the torque due to gravity,",University Physics I Classical Mechanics.pdf
"its line of action goes through the center of rotation), andτnet is just the torque due to gravity,
which can be written
τnet = −mgl sin θ (11.19)
The minus sign is there to enforce a consistent sign convention forθ and τ:i f ,f o ri n s t a n c e ,Ic h o o s e
counterclockwise as positive for both, then I note that whenθ is positive (pendulum to the right of
the vertical),τ is clockwise, and hence negative, and vice-versa. This is characteristic of arestoring
torque,t h a ti st os a y ,o n et h a tw i l la l w a y st r yt op u s ht h es y s t e mb a ckt oi t se q u i l i b r i u mp o s i t i o n
(the vertical in this case).
As for the moment of inertia of the bob, it is justI = ml2 (if we treat it as just a point particle),
so the equationτnet = Iα takes the form
ml2 d2θ
dt2 = −mgl sin θ (11.20)
The mass and one factor ofl cancel, and we get
d2θ
dt2 = −g
l sin θ (11.21)",University Physics I Classical Mechanics.pdf
"264 CHAPTER 11. SIMPLE HARMONIC MOTION
Equation (11.21)i sa ne x a m p l eo fw h a ti sk n o w na sadifferential equation.T h e p r o b l e m i s t o fi n d a
function of time,θ(t), that satisfies this equation; that is to say, when you take its second derivative
the result is equal to−(g/l)s i n [θ(t)]. Such functions exist and are calledelliptic functions;t h e y
are included in many modern mathematical packages, but theyare still not easy to use. More
importantly, the oscillations they describe, in general, are not of the simple harmonic type.
On the other hand, if the amplitude of the oscillations is small, so that the angleθ,e x p r e s s e di n
radians, is a small number, we can make an approximation thatgreatly simplifies the problem,
namely,
sin θ ≃ θ (11.22)
This is known as thesmall angle approximation,a n dr e q u i r e sθ to be in radians. As an example,
if θ =0 .2r a d ( w h i c h c o r r e s po n d s t o a bo u t 1 1.5◦), we find sinθ =0 .199, to three-figure accuracy.",University Physics I Classical Mechanics.pdf
"if θ =0 .2r a d ( w h i c h c o r r e s po n d s t o a bo u t 1 1.5◦), we find sinθ =0 .199, to three-figure accuracy.
With this approximation, the equation to solve becomes muchsimpler:
d2θ
dt2 = −g
l θ (11.23)
We have, in fact, already solved an equation completely equivalent to this one in the previous
section: that was equation (11.9)f o rt h em a s s - o n - a - s p r i n gs y s t e m ,w h i c hc a nb er e w r i t t e na s
d2x
dt2 = − k
m x (11.24)
since a = d2x/dt2.J u s t l i k et h es o l u t i o n st o(11.24)c o u l db ew r i t t e ni nt h ef o r mx(t)= A cos(ωt +
φ), withω =
√
k/l,t h es o l u t i o n st o(11.23)c a nb ew r i t t e na s
θ(t)= A cos(ωt + φ)
ω =
√
 g
l (11.25)
This tells us that if a pendulum is not pulled too far away fromthe vertical (say, about 10◦ or less)
it will perform approximate simple harmonic oscillations,with a period of
T = 2π
ω =2 π
√
l
g (11.26)
This depends only on the length of the pendulum, and remains constant even as the oscillations",University Physics I Classical Mechanics.pdf
"T = 2π
ω =2 π
√
l
g (11.26)
This depends only on the length of the pendulum, and remains constant even as the oscillations
wind down, which is why it became the basis for time-keeping devices, beginning with the invention
of thependulum clock by Christiaan Huygens in 1656. In particular, a pendulum of length l =1m
will have a period of almost exactly 2 s, which is what gives yout h ef a m i l i a r“ t i c k - t o c k ”r h y t h mo f
a“ g r a n d f a t h e r ’ sc l o c k , ”o n c ep e rs e c o n d( t h a ti st os a y ,o n cee v e r yh a l fp e r i o d ) .",University Physics I Classical Mechanics.pdf
"11.3. PENDULUMS 265
11.3.2 The “physical pendulum”
By a “physical pendulum” one means typically any pendulum-like device for which the moment of
inertia is not given by the simple expressionI = ml2.T h i s m e a n s t h a t t h e m a s s i s n o t c o n c e n t r a t e d
into a single point-like particle a distancel away from the point of suspension; rather, for example,
the bob could have a size that is not negligible compared tol (as in Fig.11.1), or the “string” could
have a substantial mass of its own—it could, for instance, beac h a i n ,l i k ei nap l a y g r o u n ds w i n g ,
or a metal rod, as in most pendulum clocks.
F G
o
θ
d
F G
θ
(a) (b)
cm
l
Figure 11.8: The “physical pendulum.” Figure (a) shows an arbitrary distributionof mass, pivoted at
point O, with center of mass at CM, oscillating under the restoring torque provided by gravity. Figure (b)
shows the special case of a thin rod of lengthl pivoted at one end (the distanced = l/2 in this case). In",University Physics I Classical Mechanics.pdf
"shows the special case of a thin rod of lengthl pivoted at one end (the distanced = l/2 in this case). In
both cases, there is an additional force (not shown) acting at thepivot point, to balance gravity.
Regardless of the reason, having to deal with a distributed mass means also that one needs to use
the center of mass of the system as the point of application ofthe force of gravity. When this is
done, the motion of the pendulum can again be described by theangle between the vertical and a
line connecting the point of suspension and the center of mass. If the distance between these two
points is d,t h e nt h et o r q u ed u et og r a v i t yi s−mgd sin θ,a n dt h eo n l yo t h e rf o r c eo nt h es y s t e m ,
the force at the pivot point, exerts no torque around that point, so we can write the equation of
motion in the form
I d2θ
dt2 = −mgd sin θ (11.27)
Under the small-angle approximation, this will again lead tos i m p l eh a r m o n i cm o t i o n ,o n l yn o w",University Physics I Classical Mechanics.pdf
"I d2θ
dt2 = −mgd sin θ (11.27)
Under the small-angle approximation, this will again lead tos i m p l eh a r m o n i cm o t i o n ,o n l yn o w
with an angular frequency given by
ω =
√
mgd
I (11.28)",University Physics I Classical Mechanics.pdf
"266 CHAPTER 11. SIMPLE HARMONIC MOTION
As an example, consider the oscillations of a uniform, thin rod of lengthl and mass m pivoted at
one end. We then haveI = ml2/3, andd = l/2, so Eq. (11.28)g i v e s
ω =
√
3g
2l (11.29)
This is about 22% larger than the result (11.25)f o ras i m p l ep e n d u l u mo ft h es a m el e n g t h ,i m p l y i n g
ac o r r e s p o n d i n g l ys h o r t e rp e r i o d .
11.4 In summary
1. Most stable physical systems will oscillate when displaced from their equilibrium position.
The oscillations are due to a restoring force (or a restoringtorque) and the system’s own
inertia.
2. The periodT and frequencyf of an oscillatory motion are related byf =1 /T.T h eu n i t so f
frequency are s−1 or hertz (abbreviated Hz).
3. A special (but very common) kind of oscillation issimple harmonic motion.T h i s h a p p e n s
whenever the restoring force is a linear function of (that is,i ti sp r o p o r t i o n a lt o )t h es y s t e m ’ s",University Physics I Classical Mechanics.pdf
"whenever the restoring force is a linear function of (that is,i ti sp r o p o r t i o n a lt o )t h es y s t e m ’ s
displacement from the equilibrium position.
4. The angular frequency, ω,o fas i m p l eh a r m o n i co s c i l l a t o ri sr e l a t e dt ot h er e g u l a rf requency
by ω =2 πf .O n e o f t h e p r o p e r t i e s o f s i m p l e h a r m o n i c m o t i o n i s t h a t i t s frequency does
not depend on the initial conditions, that is, on the velocityo rd i s p l a c e m e n tw i t hw h i c ht h e
motion is started.
5. If the equilibrium position is chosen to correspond tox =0 ,t h em o s tg e n e r a lf o r mo ft h e
position function for a simple harmonic oscillator isx(t)= A cos(ωt+φ), where The amplitude
A and phase angleφ are determined by the initial conditions. The velocity function is then
v(t)= −ωA sin(ωt + φ), and the accelerationa(t)= −ω2A cos(ωt + φ).
6. The total energy (potential plus kinetic) in a simple harmonic oscillator is equal toEsys =
1",University Physics I Classical Mechanics.pdf
"6. The total energy (potential plus kinetic) in a simple harmonic oscillator is equal toEsys =
1
2 mω2A2.T h e k i n e t i c a n d p o t e n t i a l e n e r g i e s o s c i l l a t e ( i n o p p o s i t ion, that is to say, each being
maximum when the other is minimum) between this value and zero.
7. A mass attached to an ideal (massless) spring is an exampleof a simple harmonic oscillator.
If the spring constant isk,t h ea n g u l a rf r e q u e n c yo ft h i ss y s t e mi sω =
√
k/m.
8. An external, constant force acting on a harmonic oscillator does not change its period (or
frequency), only its equilibrium position. For the mass-on-a-spring system, a forceF will
cause a displacement of the equilibrium position equal toF/k,i nt h ed i r e c t i o no ft h ef o r c e .",University Physics I Classical Mechanics.pdf
"11.4. IN SUMMARY 267
9. A simple pendulum (a point particle of massm suspended from a massless, inextensible string
of length l)w i l lp e r f o r mh a r m o n i co s c i l l a t i o n sa r o u n dt h ev e r t i c a lp rovided the small angle,
approximation, sinθ ≃ θ,h o l d s .T h ea n g u l a rf r e q u e n c yo ft h e s eo s c i l l a t i o n si sω =
√
g/l.
10. A rigid object of massm pivoted around some point and performing oscillations undert h e
influence of gravity is sometimes called aphysical pendulum.J u s t a s f o r t h e s i m p l e p e n d u -
lum, the oscillations will be harmonic if the small-angle approximation holds. The angular
frequency will then be
√
mgd/I,w h e r eI is the moment if inertia around the pivot point, and
d the distance from the pivot point to the center of mass.",University Physics I Classical Mechanics.pdf
"268 CHAPTER 11. SIMPLE HARMONIC MOTION
11.5 Examples
11.5.1 Oscillator in a box (a basic accelerometer!)
Consider a block-spring system inside a box, as shown in the figure. The block is attached to the
spring, which is attached to the inside wall of the box. The mass of the block is 0.2k g . F o r p a r t s
(a) through (f), assume that the box does not move.
Suppose you pull the block 10 cm to the right and release it. Thea n g u l a rf r e q u e n c yo ft h eo s c i l l a -
tions is 30 rad/s. Neglect friction between the block and thebottom of the box.
(a) What is the spring constant?
(b) What will be the amplitude of the oscillations?
(c) Taking to the right to be positive, at what point in the oscillation is the velocity minimum and
what is its minimum value?
(d) At what point in the oscillation is the acceleration minimum, and what is its minimum value?
(e) What is the total energy of the spring-block system?",University Physics I Classical Mechanics.pdf
"(e) What is the total energy of the spring-block system?
(f) If you taket =0t ob et h ei n s t a n tw h e ny o ur e l e a s et h eb l o c k ,w r i t ea ne q u a tion of motion for
the oscillation,x(t)= ? ,i d e n t i f y i n gt h ev a l u e so fa l lc o n s t a n t st h a ty o uu s e .
(g) Imagine now that the box, with the spring and block in it, starts moving to the left with an
acceleration a = −4m / s2.B y h o w m u c h d o e s t h e e q u i l i b r i u m p o s i t i o n o f t h e b l o c k s h i f t(relative
to the box), and in what direction?
Solution
Most of this is really pretty straightforward, since it is just a matter of using the equations intro-
duced in this chapter properly:
(a) Since we know that for this kind of situation, the angularfrequency, the mass and the spring
constant are related by
ω =
√
k
m
we can solve this fork:
k = mω2 =0 .2k g×
(
30 rad
s
) 2
=1 8 0N
m",University Physics I Classical Mechanics.pdf
"11.5. EXAMPLES 269
(b) The amplitude will be 10 cm, since it is released at that point with no kinetic energy.
(c) The velocity is minimum (largest in magnitude, but with anegative sign) as the object passes
through the equilibrium position moving to the left.
vmin = −ωA = −
(
30 rad
s
)
× 0.1m= −3 m
s
(d) The acceleration is minimum (again, largest in magnitude, but with a negative sign) when the
spring is maximally stretched (block is farthest to the right), since this gives you the maximal force
in the negative direction:
amin = −ω2A = −
(
30 rad
s
) 2
× 0.1 m = −90 m
s2
(e) The total energy is given by the formula (either one is acceptable)
E = 1
2mω2A2 = 1
2kA2 = 1
2 (180 N/m)× (0.1m )2 =0 .9J
(You could also useE = 1
2 mv2
max.)
(f) The result is
x(t)= A cos(ωt)= A sin
(
ωt + π
2
)
with A =0 .1m a n dω =3 0 r a d / s . Y o uc o u l da l s oj u s tw r i t et h en u m b e r sd i r e c t l yi nthe formula,",University Physics I Classical Mechanics.pdf
"(
ωt + π
2
)
with A =0 .1m a n dω =3 0 r a d / s . Y o uc o u l da l s oj u s tw r i t et h en u m b e r sd i r e c t l yi nthe formula,
but in that case you need to include the units implicitly or explicitly. What I mean by “implicitly”
is to say something like: “x(t)=0 .1c o s ( 3 0t), withx in meters andt in seconds.”
(g) The equilibrium position is where the block could sit at rest relative to the box. In that case,
relative to the ground outside the box, it would be moving witha na c c e l e r a t i o na = −4m / s2,a n d
the spring force (which is the only actual force acting on theblock) would have to provide this
acceleration:
Fspr
x = −k∆x = ma
so
∆x = −ma
k = 0.2 kg × 4m / s2
180 N/m =0 .00444 m
or 4.44 mm. This is positive, so the spring stretches—the equilibrium position for the block is
shifted to the right, relative to the box’s walls.
Another way to see this is the following. As we saw in the previous chapter (section 10.2), an",University Physics I Classical Mechanics.pdf
"shifted to the right, relative to the box’s walls.
Another way to see this is the following. As we saw in the previous chapter (section 10.2), an
accelerated reference system, with accelerationa,a p p e a r s“ f r o mt h ei n s i d e ”a sa ni n e r t i a lr e f e r e n c e",University Physics I Classical Mechanics.pdf
"270 CHAPTER 11. SIMPLE HARMONIC MOTION
system subject to a gravitational interaction that pulls anyo b j e c tw i t hm a s sm with a force equal
to ma in the direction opposite the acceleration. Therefore, inside the box, which is accelerating
towards the left, the block behaves as if there was a force of gravity of magnitudema,p u l l i n gi tt o
the right. In other words, we have a situation like the one illustrated in Fig.11.6,o n l ys i d e w a y s .
As in that case, we find the equilibrium position is shifted just enough for the force of the stretched
spring to match the “force of gravity,” and in this way we get again the equationFspr
x = ma.
To get an accelerometer, we provide the box with some readoutmechanism that can tell us the
change in the oscillator’s equilibrium position. This basicp r i n c i p l ei so n eo ft h ew a y sa c c e l e r o m e t e r s —
and so-called “inertial navigation systems”—work.
11.5.2 Meter stick as a physical pendulum",University Physics I Classical Mechanics.pdf
"and so-called “inertial navigation systems”—work.
11.5.2 Meter stick as a physical pendulum
While working on the lab on torques, you notice that a meter stick suspended from the middle
behaves a little like a pendulum, in that it performs very slowo s c i l l a t i o n sw h e ny o ut i l ti ts l i g h t l y .
Intrigued, you notice that it is suspended by a simple loop ofstring tied in a knot at the top (see
figure). You measure the period of the oscillations to be about5 s ,a n dt h ew i d t ho ft h es t i c kt ob e
about 2.5c m .
(a) What does this tell you about the quantityI/M,w h e r eM is the mass of the stick, andI its
moment of inertia around a certain point?
(b) What is the “certain point” mentioned in (a)?
Solution
As the picture below shows, the stick will behave like a physical pendulum, oscillating around the
point of suspension O, which in this case is just next to the stick, where the knot is. As seen in the",University Physics I Classical Mechanics.pdf
"point of suspension O, which in this case is just next to the stick, where the knot is. As seen in the
blown-up detail, if the width of the stick isw,t h ec e n t e ro fm a s so ft h es t i c ki sl o c a t e dad i s t a n c e
d = w/2a w a yf r o mt h ep o i n to fs u s p e n s i o n :
w
cm
O",University Physics I Classical Mechanics.pdf
"11.5. EXAMPLES 271
As shown in Section11.3.2,w eh a v et h e n
ω =
√
Mgw
2I (11.30)
Squaring this, and solving forI/M,
I
M = gw
2ω2 = 9.8m / s2 × 0.025 m
2 × (2π/5s )2 =0 .0776 m2 (11.31)
The moment of inertia is to be calculated around the point O, that is to say, the point of suspension
(where the knot is in the figure). For reference, the moment ofinertia of a thin rod of lengthl
around its midpoint isMl2/12 = 0.083l2.T h e l e n g t h o f t h e m e t e r s t i c k i s , o f c o u r s e , 1 m , s o t h e
result I/M ∼ 0.08 m2 obtained above seems reasonable.",University Physics I Classical Mechanics.pdf
"272 CHAPTER 11. SIMPLE HARMONIC MOTION
11.6 Advanced Topics
11.6.1 Mass on a spring damped by friction with a surface
Consider the system depicted in Figure11.2 in the presence of friction between the block and the
surface. Let the coefficient of kinetic friction beµk and the coefficient of static friction beµs.A s
usual, we will assume thatµs ≥ µk.
As the mass oscillates, it will experience a kinetic frictionf o r c eo fm a g n i t u d eFk = µkmg,i nt h e
direction opposite the direction of motion; that is to say, aforce that changes direction every half
period. As shown in section 11.2.2, this force does not changet h ef r e q u e n c yo ft h em o t i o n ,b u ti t
displaces the equilibrium position in the direction of the force, which is to say, closer to the starting
point for each half-swing. As a result of that, the amplitudefor each half-swing is reduced from
the previous one.",University Physics I Classical Mechanics.pdf
"point for each half-swing. As a result of that, the amplitudefor each half-swing is reduced from
the previous one.
Let the original equilibrium position (in the absence of friction) be x0 =0 . S u p p o s ew ed i s p l a c e
the mass a distanceA to the right (call this position, the starting point for the first half-swing,
x1 = A), and let go. In the presence of friction, the equilibrium position for this first half-swing
becomes the pointx′
0 = Fk/k = µkmg,s ot h er e a la m p l i t u d eo ft h i sfi r s th a l f - o s c i l l a t i o nw i l lb e
A1 = x1 − x′
0 = A − x′
0,a n dt h er e s u l t i n gm o t i o nw i l lb e
x(t)= x′
0 + A1 cos(ωt)( fi r s t h a l f - p e r i o d , 0≤ t ≤ π/ω)( 1 1 . 3 2 )
The mass then stops, momentarily, att = π/ω,a tt h ep o s i t i o nx2 = x′
0 − A1,a n dt u r n sa r o u n d
for the second half-swing. However, now the external force has reversed direction, so the new
equilibrium position is at−x′
0,a n dt h en e wa m p l i t u d ei sA2 = −x′",University Physics I Classical Mechanics.pdf
"equilibrium position is at−x′
0,a n dt h en e wa m p l i t u d ei sA2 = −x′
0 − x2 = A1 − 2x′
0 = A − 3x′
0.
The motion for this next half-period is then
x(t)= −x′
0 + A2 cos(ωt)( s e c o n d h a l f - p e r i o d ,π/ω ≤ t ≤ 2π/ω)( 1 1 . 3 3 )
Continuing the process, we see thatA1 = A−x′
0, A2 = A−3x′
0, A3 = A−5x′
0 ...A n = A−(2n−1)x′
0.
Of course, this can’t go on forever, since we require the amplitude to be a positive quantity; so the
motion will consist of onlyn half-periods, where n is an integer such thatA − (2n − 1)x′
0 > 0b u t
A − (2n +1 )x′
0 < 0. (That is to say,n is equal to the integer part of (A/x′
0 +1 )/2.)
The figure shows an example of how this would go, for the following choice of parameters: period
T =1s , µk =0 .1, andA =0 .18 m. Note that, sincex′
0 depends only on the ratiom/k =1 /ω2,t h e r e
is no need to specifym and k separately. We getω =2 π/T =2 π rad/s, x′
0 = µkg/ω2 =0 .0248 m,
and (A/x′",University Physics I Classical Mechanics.pdf
"is no need to specifym and k separately. We getω =2 π/T =2 π rad/s, x′
0 = µkg/ω2 =0 .0248 m,
and (A/x′
0 +1)/2=4 .13, which means that the motion will go on for 4 half-periods before stopping.",University Physics I Classical Mechanics.pdf
"11.6. ADVANCED TOPICS 273
Figure 11.9: Damped oscillations.
Note that, in general, the oscillator does not stop at the equilibrium position. Rather, its final
position will be at the end of the last half-swing, which is either x′
0 − An (if the number n of
half-periods is odd), or−x′
0 + An,i ft h en u m b e rn is even. Either way, at that point the spring
will be exerting a force of magnitude
Fspr = k|x′
0 − An| = k|x′
0 − A +( 2n − 1)x′
0| = k|A − 2nx′
0| <k x′
0 = Fk (11.34)
Since we expect the force of static friction,Fs,t ob eg r e a t e rt h a nFk,t h i st e l l su st h a ta tt h i s
point the spring is not exerting enough force to get the mass tom o v ea g a i n .
Note: Just for the record, this isnot the way dissipation in simple harmonic motion is usually
handled. The conventional thing is to assume a damping forcethat is proportional to the oscillator’s
velocity. You will almost certainly see this more standard approach (which leads to a relatively",University Physics I Classical Mechanics.pdf
"velocity. You will almost certainly see this more standard approach (which leads to a relatively
simple differential equation) in some later course.
11.6.2 The Cavendish experiment: how to measureG with a torsion balance
Suppose that you want to try and duplicate Cavendish’s experiment to measure directly the grav-
itational force between two masses (and hence, indirectly,the value ofG). You take two relatively
small, identical objects, each of massm,a n da t t a c ht h e mt ot h ee n d so far o do fl e n g t hl (let us
say the mass of the rod is negligible, for simplicity), makingas o r to fd u m b b e l l ;t h e ny o us u s p e n d
this from the ceiling, by the midpoint, using a nylon line.",University Physics I Classical Mechanics.pdf
"274 CHAPTER 11. SIMPLE HARMONIC MOTION
d
(a) (b) (c)
Figure 11.10:(a) Torsion balance. The extremes of the oscillation are drawn in black and gray, respectively.
(b) The view from the top. The dashed line indicates the equilibrium position. (c) In the presence of two
nearby large masses, the equilibrium position is tilted very slightly; thelight blue lines in the background
show the oscillation in the absence of the masses, for reference.
You have now made atorsion balancesimilar to the one Cavendish used. You will probably find out
that it it is very hard to keep it motionless: the slightest displacement causes it to oscillate around
an equilibrium position. The way it works is that an angular displacement θ from equilibrium
puts a small twist on the line, which results in a restoring torque τ = −κθ,w h e r eκ is thetorsion
constant for your setup. If your dumbbell has moment of inertiaI,t h e nt h ee q u a t i o no fm o t i o n
τ = Iα gives you
I d2θ
dt2 = −κθ (11.35)",University Physics I Classical Mechanics.pdf
"constant for your setup. If your dumbbell has moment of inertiaI,t h e nt h ee q u a t i o no fm o t i o n
τ = Iα gives you
I d2θ
dt2 = −κθ (11.35)
If you compare this to Eq. (11.21), and follow the derivation there, you can see that the periodo f
oscillation is
T =2 π
√
I
κ (11.36)
so if you measureT you can getκ,s i n c eI =2 m(l/2)2 = ml2/2f o rt h ed u m b b e l l .
Now suppose you bring two large masses, a distanced each from each end of the dumbbell, per-
pendicular to the dumbbell axis, and one on either side, as inthe figure. The gravitational force
FG = GmM/d2 between the large and small mass results in a net “external” torque of magnitude
τext =2 FG × l
2 = FGl (11.37)
This torque will cause a very small displacement, so small that the change ind will be practically
negligible, so you can treatFG,a n dh e n c eτext,a sac o n s t a n t . T h e nt h es i t u a t i o ni sa n a l o g o u s",University Physics I Classical Mechanics.pdf
"negligible, so you can treatFG,a n dh e n c eτext,a sac o n s t a n t . T h e nt h es i t u a t i o ni sa n a l o g o u s
to that of an oscillator subjected to a constant external force (section 11.2.2): the frequency of",University Physics I Classical Mechanics.pdf
"11.6. ADVANCED TOPICS 275
the oscillations will not change, but the equilibrium position will. In Eq. (11.15)w ef o u n dt h a t
y′
0 −y0 = Fext/k for a spring of spring constantk,w h e r ey0 was the old andy′
0 the new equilibrium
position (the force was equal to−mg;t h ed i s p l a c e m e n to ft h ee q u i l i b r i u mp o s i t i o nw i l lb ei nt h e
direction of the force). For the torsion balance, the equivalent result is
θ′
0 − θ0 = τext
κ = FGl
κ (11.38)
So, if you measure the angular displacement of the equilibrium position, you can getFG.T h i s
displacement is going to be very small, but you can try to monitor the position of the dumbbell
by, for instance, reflecting a laser from it (or, one or both ofyour small masses could be a small
laser). Tracking the oscillations of the point of laser lighto nt h ew a l l ,y o um i g h tb ea b l et od e t e c t
the very small shift predicted by Eq. (11.38).",University Physics I Classical Mechanics.pdf
"276 CHAPTER 11. SIMPLE HARMONIC MOTION
11.7 Problems
Problem 1
Ab l o c ko fm a s sm is sliding on a frictionless, horizontal surface, with a velocity vi.I t h i t s a n
ideal spring, of spring constantk,w h i c hi sa t t a c h e dt ot h ew a l l . T h es p r i n gc o m p r e s s e su n t i lthe
block momentarily stops, and then starts expanding again, sot h eb l o c ku l t i m a t e l yb o u n c e so ff( s e e
Example 5.6.2).
(a) Write down an equation of motion (a functionx(t)) for the block, which is valid for as long as
it is in contact with the spring. For simplicity, assume the block is initially moving to the right,
take the time when it first makes contact with the spring to bet =0 ,a n dl e tt h ep o s i t i o no ft h e
block at that time to bex =0 . M a k es u r et h a ty o ue x p r e s sa n yc o n s t a n t si ny o u re q u a t i o n(such
as A or ω)i nt e r m so ft h eg i v e nd a t a ,n a m e l y ,m, vi,a n dk.",University Physics I Classical Mechanics.pdf
"as A or ω)i nt e r m so ft h eg i v e nd a t a ,n a m e l y ,m, vi,a n dk.
(b) Sketch the functionx(t)f o rt h er e l e v a n tt i m ei n t e r v a l .
Problem 2
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea.
Assume for simplicity that this is simple harmonic motion (int h ev e r t i c a ld i r e c t i o n )w i t ha m p l i t u d e
5c m a n d f r e q u e n c y 2H z . T h e r e i s a bo x o n t h e fl o o r w i t h m a s sm =1k g .
(a) Assuming the box remains in contact with the floor throughout, find the maximum and mini-
mum values of the normal force exerted on it by the floor over anoscillation cycle.
(b) How large would the amplitude of the oscillations have tobecome for the box to lose contact
with the floor, assuming the frequency remains constant? (Hint: what is the value of the normal
force at the moment the box loses contact with the floor?)
Problem 3",University Physics I Classical Mechanics.pdf
"force at the moment the box loses contact with the floor?)
Problem 3
Imagine a simple pendulum swinging in an elevator. If the cable holding the elevator up was to
snap, allowing the elevator to go into free fall, what would happen to the frequency of oscillation
of the pendulum? Justify your answer.
Problem 4
Consider a block of massm attached to two springs, one on the left with spring constantk1 and
one on the right with spring constantk2.E a c h s p r i n g i s a t t a c h e do n t h e o t h e r s i d e t o a w a l l ,a n d
the block slides without friction on a horizontal surface. When the block is sitting atx =0 ,b o t h
springs are relaxed.
Write Newton’s second law,F = ma, as a differential equation for an arbitrary positionx of the
block. What is the period of oscillation of this system?
Problem 5
Consider the block hanging from a spring shown in Figure11.6.S u p p o s e t h e m a s s o ft h e b l o c ki s",University Physics I Classical Mechanics.pdf
"Problem 5
Consider the block hanging from a spring shown in Figure11.6.S u p p o s e t h e m a s s o ft h e b l o c ki s
1.5k g a n d t h e s y s t e m i s a t r e s t w h e n t h e s p r i n g h a s be e n s t r e t c h ed2 c mf r o mi t so r i g i n a ll e n g t h",University Physics I Classical Mechanics.pdf
"11.7. PROBLEMS 277
(that is, with reference to the figure,y0 − y′
0 =0 .02 m).
(a) What is the value of the spring constantk?
(b) If you stretch the spring by an additional 2 cm downward from this equilibrium position, and
release it, what will be the frequency of the oscillations?
(c) Now consider the system formed by the spring, the block, and the earth. Take the “zero” of
gravitational potential energy to be at the heighty′
0 (the equilibrium point; you may also use this
as the origin for the vertical coordinate!), and calculate all the energies in the system (kinetic,
spring/elastic, and gravitational) at the highest point inthe oscillation, the equilibrium point, and
the lowest point. Verify that the sum is indeed constant.",University Physics I Classical Mechanics.pdf
278 CHAPTER 11. SIMPLE HARMONIC MOTION,University Physics I Classical Mechanics.pdf
"Chapter 12
Waves in one dimension
12.1 Traveling waves
In our study of mechanics we have so far dealt with particle-like objects (objects that have only
translational energy), and extended, rigid objects, whichmay also have rotational energy. We have,
however, implicitly assumed that all the objects we studiedhad some internal structure, or were
to some extent deformable, whenever we allowed for the possibility of their storing other forms of
energy, such as chemical or thermal.
This chapter deals with a very common type of organized (as opposed to incoherent) motion
exhibited by extended elastic objects, namely,wave motion.( O f t e n , t h e “ o b j e c t ” i n w h i c h t h e
wave motion takes place is called a “medium.”) Waves can be “traveling” or “standing,” and we
will start with the traveling kind, since they are the ones that most clearly exhibit the characteristics
typically associated with wave motion.",University Physics I Classical Mechanics.pdf
"will start with the traveling kind, since they are the ones that most clearly exhibit the characteristics
typically associated with wave motion.
A traveling wavein a medium is adisturbance of the medium that propagates through it, in a definite
direction and with a definite velocity. By a “disturbance” wetypically mean a displacement of the
parts that make up the medium, away from their rest or equilibrium position. The idea here is to
regard each part of an elastic medium as, potentially, an oscillator, which couples to the neighboring
parts by pushing or pulling on them (for an example of how to model this mathematically, see
Advanced Topic12.6.1 at the end of this chapter). When the traveling wave reaches aparticular
location in the medium, it sets that part of the medium in motion, by giving it some energy and
momentum, which it then passes on to a neighboring part, and soo nd o w nt h el i n e .",University Physics I Classical Mechanics.pdf
"momentum, which it then passes on to a neighboring part, and soo nd o w nt h el i n e .
You can see an example of how this works in a slinky. Start by stretching the slinky somewhat,
then grab a few coils, bunch them up at one end, and release them. You should see a “compression
279",University Physics I Classical Mechanics.pdf
"280 CHAPTER 12. WAVES IN ONE DIMENSION
pulse” traveling down the slinky, with very little distortion; you may even be able to see it being
reflected at the other end, and coming back, before all its energy is dissipated away.
direction of propagation of the wave (x)
direction of displacement of the medium (x) 
Figure 12.1: A longitudinal (compression) wave pulse traveling down a slinky.
The compression pulse in the slinky in Fig.12.1 is an example of what is called alongitudinal wave,
because the displacement of the parts that make up the medium(the rings, in this case) takes
place along the same spatial dimension along which the wave travels (the horizontal direction, in
the figure). The most important examples of longitudinal waves aresound waves,w h i c hw o r kab i t
like the longitudinal waves on the slinky: a region of air (orsome other medium) is compressed, and
as it expands it pushes on a neighboring region, causing it tocompress, and passing the disturbance",University Physics I Classical Mechanics.pdf
"as it expands it pushes on a neighboring region, causing it tocompress, and passing the disturbance
along. In the process, regions of rarefaction (where the density drops below its average value) are
typically produced, alongside the regions of compression (increased density).
The opposite of a longitudinal wave is atransverse wave,i nw h i c ht h ed i s p l a c e m e n to ft h em e d i u m ’ s
parts takes place in a directionperpendicular to the wave’s direction of travel. It is actually also
relatively easy to produce a transverse wave on a slinky: again, just stretch it somewhat and give
one end a vigorous shake up and down. It is, however, a little hard to draw the resulting pulse on a
long spring with all the coils, so in Figure12.2 below I have instead drawn a transverse wave pulse
on a string,w h i c hy o uc a np r o d u c ei nt h es a m ew a y . ( S t r i n g sh a v eo t h e ra dvantages: they are",University Physics I Classical Mechanics.pdf
"on a string,w h i c hy o uc a np r o d u c ei nt h es a m ew a y . ( S t r i n g sh a v eo t h e ra dvantages: they are
also easier to describe mathematically, and they are very relevant, particularly to the production
of musical sounds.)
direction of propagation of the wave (x)
x
direction of displacement
of the medium (y) y
Figure 12.2: A transverse wave pulse traveling down a string. This pulse can be generated by giving an
end of the string a strong shake, while holding the string taut. (Youcan do this on a slinky, too.)",University Physics I Classical Mechanics.pdf
"12.1. TRAVELING WAVES 281
Perhaps the most important (and remarkable) property of wavem o t i o ni st h a ti tc a nc a r r ye n e r g y
and momentum over relatively long distanceswithout an equivalent transport of matter.A g a i n ,
think of the slinky: the “pulse” can travel through the slinky’s entire length, carrying momentum
and energy with it, but each individual ring does not move veryf a ra w a yf r o mi t se q u i l i b r i u m
position. Ideally, after the pulse has passed through a particular location in the medium, the
corresponding part of the medium returns to its equilibriumposition and does not move any more:
all the energy and momentum it momentarily acquired is passedf o r w a r d . T h es a m ei s( i d e a l l y )
true for the transverse wave on the string in Fig.12.2.
Since this is meant to be a very elementary introduction to waves, I will consider only this case of
“ideal” (technically known as “linear and dispersion-free”) wave propagation, in which the speed",University Physics I Classical Mechanics.pdf
"“ideal” (technically known as “linear and dispersion-free”) wave propagation, in which the speed
of the wave does not depend on the shape or size of the disturbance. In that case, the disturbance
retains its “shape” as it travels, as I have tried to illustrate in figures12.1 and 12.2.
12.1.1 The “wave shape” function: displacement and velocityo ft h em e d i u m .
In a slinky, what I have been calling the “parts” of the mediumare very clearly seen (they are,
naturally, the individual rings); in a “homogeneous” medium( o n ew i t hn ov i s i b l ep a r t s ) ,t h ew a y
to describe the wave is to break up the medium, in your mind, into infinitely many small parts or
“particles” (as we have been doing for extended systems all semester), and write down equations
that tell us how each part moves. Physically, you should thinko fe a c ho ft h e s e“ p a r t i c l e s ”a sb e i n g
large enough to contain many molecules, but small enough thati t sp o s i t i o ni nt h em e d i u mm a yb e",University Physics I Classical Mechanics.pdf
"large enough to contain many molecules, but small enough thati t sp o s i t i o ni nt h em e d i u mm a yb e
represented by a mathematical point.
The standard way to label each “particle” of the medium is by the position vector of its equilibrium
position (the place where the particle sits at rest in the absence of a wave). In the presence of the
wave, the particle that was initially at rest at the point⃗rwill undergo a displacement that I am
going to represent by the vector⃗ξ (where ξ is the Greek letter “xi”). This displacement will in
general be a function of time, and it may also be different for different particles, so it will also be
af u n c t i o no f⃗r,t h ee q u i l i b r i u mp o s i t i o no ft h ep a r t i c l ew ea r ec o n s i d e r i ng. The particle’s position
under the influence of the wave becomes then
⃗r+ ⃗ξ(⃗r , t) (12.1)
This is very general, and it can be given a simpler form for simple cases. For instance, for a",University Physics I Classical Mechanics.pdf
"⃗r+ ⃗ξ(⃗r , t) (12.1)
This is very general, and it can be given a simpler form for simple cases. For instance, for a
transverse wave on a string, we can label each part of the string at rest by itsx coordinate, and
then take the displacement to lie along they axis; the position vector, then, could be written in
component form as (x, ξ(x, t), 0). Similarly, we can consider a “plane” sound wave as a longitudinal
wave traveling in thex direction, where the density of the medium is independent ofy and z
(that is, it is constant on planes perpendicular to the direction of propagation). In that case, the
equilibrium coordinate x can be used to refer to a whole “slice” of the medium, and the position",University Physics I Classical Mechanics.pdf
"282 CHAPTER 12. WAVES IN ONE DIMENSION
of that slice, along thex axis, at the timet will be given byx + ξ(x, t). In both of these cases, the
displacement vectorξ reduces to a single nonzero component (along they or x axis, respectively),
which can, of course, be positive or negative. I will restrictm y s e l fi m p l i c i t l yt ot h e s es i m p l ec a s e s
and treatξ as a scalar from this point on.
Under these conditions, the functionξ(x, t)( w h i c hi so f t e nc a l l e dt h ewave function)g i v e su st h e
shape of the “displacement wave,” that is to say, the displacement of every part of the medium,
labeled by its equilibriumx-coordinate, at any instant in time. Accordingly, taking thed e r i v a t i v e
of ξ gives us the velocity of the corresponding part of the medium:
vmed = dξ
dt (12.2)
This is also, in general, a vector (along the direction of motion of the wave, if the wave is longitu-",University Physics I Classical Mechanics.pdf
"vmed = dξ
dt (12.2)
This is also, in general, a vector (along the direction of motion of the wave, if the wave is longitu-
dinal, or perpendicular to it if the wave is transverse). It isa l s oaf u n c t i o no ft i m e ,a n di ng e n e r a l
will be different from the speed of the wave itself,w h i c hw eh a v et a k e nt ob ec o n s t a n t ,a n dw h i c hI
will denote byc instead.
12.1.2 Harmonic waves
An important class of waves are those for which the wave function is sinusoidal. This means that
the different parts of the medium execute simple harmonic motion, all with the same frequency,
but each (in general) with a different phase. Specifically, foras i n u s o i d a lw a v ew eh a v e
ξ(x, t)= ξ0 sin
[2πx
λ − 2πft
]
(12.3)
In Eq. (12.3), f stands for the frequency, and plays the same role it did in theprevious chapter:
it tells us how often (that is, how many times per second) the corresponding part of the medium",University Physics I Classical Mechanics.pdf
"it tells us how often (that is, how many times per second) the corresponding part of the medium
oscillates around its equilibrium position. The constantξ0 is just the amplitude of the oscillation
(what we used to callA in the previous chapter). The constantλ,o nt h eo t h e rh a n d ,i ss o m e t i m e s
known as the “spatial period,” or, most often, thewavelength of the wave: it tells you how far you
have to travel along thex axis, from a given pointx,t ofi n da n o t h e ro n et h a ti sp e r f o r m i n gt h e
same oscillation with the same amplitude and phase.
Ac o u p l eo fs n a p s h o t so fah a r m o n i cw a v ea r es h o w ni nF i g .12.3 (next page). The figure shows the
displacement ξ, at two different times, and as a function of the coordinatex used to label the parts
into which we have broken up the medium (as explained in the previous subsection). As such, the
wave it represents could equally well be longitudinal or transverse. If it is transverse, like a wave",University Physics I Classical Mechanics.pdf
"wave it represents could equally well be longitudinal or transverse. If it is transverse, like a wave
on a string, then you can think ofξ as being essentially justy,a n dt h e nt h ed i s p l a c e m e n tc u r v e
(the blue line) just gives you the shape of the string. If the wave is longitudinal, however, then it
is a bit harder to visualize what is going on just from the plotof ξ(x, t). This is what I have tried
to do with the density plots at the bottom of the figure.",University Physics I Classical Mechanics.pdf
"12.1. TRAVELING WAVES 283
 0  2  4  6  8  10
-1
-0.5
 0
 0.5
 1
 0  2  4  6  8  10
 0  2  4  6  10
t = 0
t = ∆t
λ
cΔt
x
ξ
t = 0
t = ∆t
x
x
Figure 12.3: Top: two snapshots of a traveling harmonic wave att = 0 (solid) and att =∆ t (dashed).
The quantityξ is the displacement of a typical particle of the medium at each pointx (the wave is traveling
in the positivex direction). Units for bothx and ξ are arbitrary. Bottom: The corresponding densities, for
the case of a longitudinal wave.
Imagine the wave is longitudinal, and consider thex = π point on thet =0c u r v e( t h efi r s tz e r o ,n o t
counting the origin). A particle of the medium immediately tot h el e f to ft h a tp o i n th a sap o s i t i v e
displacement, that is, it is pushed towardsx = π,w h e r e a sas l i c eo nt h er i g h th a san e g a t i v e
displacement—which means it is also pushed towards x = π.W e t h e r e f o r e e x p e c t t h e d e n s i t y o f",University Physics I Classical Mechanics.pdf
"displacement—which means it is also pushed towards x = π.W e t h e r e f o r e e x p e c t t h e d e n s i t y o f
the medium to be highest around that point, whereas aroundx =2 π the opposite occurs: particles
to the left are pushed to the left and those to the right are pushed to the right, resulting in a
low-density region. The density plot labeledt =0a t t e m p t st os h o wt h i su s i n gag r a y s c a l ew h e r e
darker and lighter correspond to regions of higher and smaller density, respectively. At the later
time t =∆ t the high and low density regions have moved a distancec∆t to the right, as shown in
the second density plot.
Regardless of whether the wave is longitudinal or transverse, if it is harmonic, the spatial pattern
will repeat itself every wavelength; you can think of the wavelength λ as the distance between
two consecutive crests (or two consecutive troughs) of the displacement function, as shown in the",University Physics I Classical Mechanics.pdf
"two consecutive crests (or two consecutive troughs) of the displacement function, as shown in the
figure. If the wave is traveling with a speedc,a no b s e r v e rs i t t i n ga tafi x e dp o i n tx would see
the disturbance pass through that point, the particles moveup and down (or back and forth), and
the motion repeat itself after the wave has traveled a distance λ,t h a ti s ,a f t e rat i m eλ/c.T h i s
means the period of the oscillation at every point isT = λ/c,a n dt h ec o r r e s p o n d i n gf r e q u e n c y
f =1 /T = c/λ:
f = c
λ (12.4)
This is the most basic equation for harmonic waves. Making useo fi t ,E q .(12.3)c a nb er e w r i t t e n",University Physics I Classical Mechanics.pdf
"284 CHAPTER 12. WAVES IN ONE DIMENSION
as
ξ(x, t)= ξ0 sin
[2π
λ (x − ct)
]
(12.5)
This suggests that if we want to have a wave moving to the left instead, all we have to do is change
the sign of the term proportional toc,w h i c hi si n d e e dt h ec a s e .
In contrast to the wave speed, which is a constant, the speed ofa n yp a r to ft h em e d i u m ,w i t h
equilibrium position x,a tt h et i m et,c a nb ec a l c u l a t e df r o mE q s .(12.2)a n d(12.3)t ob e
vmed(x, t)=2 πfξ 0 cos
[2πx
λ − 2πft
]
= ωξ0 cos
[2πx
λ − 2πft
]
(12.6)
(where I have introduced the angular frequencyω =2 πf ). Again, this is a familiar result from the
theory of simple harmonic motion: the velocity is “90 degreeso u to fp h a s e ”w i t ht h ed i s p l a c e m e n t ,
so it is maximum or minimum where the displacement is zero (that is, when the particle is passing
through its equilibrium position in one direction or the other).",University Physics I Classical Mechanics.pdf
"through its equilibrium position in one direction or the other).
Note that the result (12.6)i m p l i e st h a t ,f o ral o n g i t u d i n a lw a v e ,the “velocity wave” is in phase with
the “density wave”:t h a ti s ,t h em e d i u mv e l o c i t yi sl a r g ea n dp o s i t i v ew h e r et h edensity is largest,
and large and negative where the density is smallest (comparet h ed e n s i t yp l o t si nF i g .12.3). If we
think of the momentum of a volume element in the medium as beingp r o p o r t i o n a lt ot h ep r o d u c to f
the instantaneous density and velocity, we see that for thiswave, which is traveling in the positive
x direction, there is more “positive momentum” than “negativem o m e n t u m ”i nt h em e d i u ma ta n y
given time (of course, if the wave had been traveling in the opposite direction, the sign ofvmed in
Eq. (12.6)w o u l dh a v eb e e nn e g a t i v e ,a n dw ew o u l dh a v ef o u n dt h eo p p o s ite result). This confirms",University Physics I Classical Mechanics.pdf
"Eq. (12.6)w o u l dh a v eb e e nn e g a t i v e ,a n dw ew o u l dh a v ef o u n dt h eo p p o s ite result). This confirms
our expectation that the wave carries a net amount of momentumi nt h ed i r e c t i o no fp r o p a g a t i o n .
Ad e t a i l e dc a l c u l a t i o n( w h i c hi sb e y o n dt h es c o p eo ft h i sb o ok) shows that the time-average of the
“momentum density” (momentum per unit volume) can be writtena s
p
V = 1
2cρ0ω2ξ2
0 (12.7)
where ρ0 is the medium’s average mass density (mass per unit volume).Interestingly, this result
applies also to a transverse wave!
As mentioned in the introduction, the wave also carries energy. Equation (12.6)c o u l db eu s e dt o
calculate the kinetic energy of a small region of the medium (with volume V and density ρ0,a n d
therefore m = ρ0V ), and its time average. This turns out to be equal to the time average of the
elastic potential energy of the same part of the medium (recall that we had the same result for",University Physics I Classical Mechanics.pdf
"elastic potential energy of the same part of the medium (recall that we had the same result for
harmonic oscillators in the previous chapter). In the end, the total time-averaged energy density
(energy per unit volume) in the region of the medium occupiedby the wave is given by
E
V = 1
2ρ0ω2ξ2
0 (12.8)",University Physics I Classical Mechanics.pdf
"12.1. TRAVELING WAVES 285
Comparing (12.7)a n d(12.8), you can see that
E
V = cp
V (12.9)
This relationship between the energy and momentum densities( o n ei sj u s tc times the other) is an
extremely general result that applies to all sorts of waves,including electromagnetic waves!
12.1.3 The wave velocity
You may ask, what determines the speed of a wave in a material medium? The answer, qualitatively
speaking, is thatc always ends up being something of the form
c ∼
√
stiffness
inertia (12.10)
where “stiffness” is some measure of how rigid the material is (how hard it is to compress it or, in
the case of a transverse wave, shear it), whereas “inertia” means some sort of mass density.
For a transverse wave on a string, for instance, we find
c =
√
Ft
µ (12.11)
where Ft is the tension in the string andµ is not the “reduced mass” of anything (sorry about
the confusion!), but a common way to write the “mass per unit length” of the string. We could",University Physics I Classical Mechanics.pdf
"the confusion!), but a common way to write the “mass per unit length” of the string. We could
also just writeµ = M/L,w h e r eM is the total mass of the string andL its length. Note that the
tension is a measure of the stiffness of the string, so this is, indeed, of the general form (12.10). For
two strings under the same tension, but with different densities, the wave will travel more slowly
on the denser one.
For a sound wave in a fluid (liquid or gas), the speed of sound isusually written
c =
√
B
ρ0
(12.12)
where ρ0 is the regular density (mass per unit volume), andB is the so-calledbulk modulus,w h i c h
gives the fluid’s resistance to a change in volume when a pressure P is applied to it:B = P/(∆V/V ).
So, once again, we get something of the form (12.10). In this case, however, we find that for many
fluids the density and the stiffness are linked, so they increaset o g e t h e r ,w h i c hm e a n sw ec a n n o t",University Physics I Classical Mechanics.pdf
"fluids the density and the stiffness are linked, so they increaset o g e t h e r ,w h i c hm e a n sw ec a n n o t
simply assume that the speed of sound will be automatically smaller in a denser medium. For
gases, this does work well: the speed of sound in a lighter gas,l i k eh e l i u m ,i sg r e a t e rt h a ni na i r ,",University Physics I Classical Mechanics.pdf
"286 CHAPTER 12. WAVES IN ONE DIMENSION
whereas in a denser gas like sulfur hexafluoride the speed of sound is less than in air1.H o w e v e r ,i f
you compare the speed of sound in water to the speed of sound inair, you find it is much greater in
water, since water is much harder to compress than air: in this case, the increase in stiffness more
than makes up for the increase in density.
The same thing happens if you go from a liquid like water to a solid, where the speed of sound is
given by
c =
√
Y
ρ0
(12.13)
where Y is, again, a measure of the stiffness of the material, called the Young modulus.S i n c e a
solid is typically even harder to compress than a liquid, thespeed of sound in solids such as metals
is much greater than in water, despite their being also denser. For reference, the speed of sound in
steel would be aboutc =5 , 000 m/s; in water, about 1, 500 m/s; and in air, “only” about 340 m/s.
12.1.4 Reflection and transmission of waves at a medium boundary",University Physics I Classical Mechanics.pdf
"12.1.4 Reflection and transmission of waves at a medium boundary
Suppose that you have two different elastic media, joined in some way at a common boundary, and
you have a wave in the first medium traveling towards the boundary. Examples of media connected
this way could be two different strings tied together, or two springs with different spring constants
joined at the ends; or, for sound waves, it could just be something like water with air above it: a
compression wave in air traveling towards the water surfacewill push on the surface and set up a
sound wave there, and vice-versa.
The first thing to notice is that, if the incident wave has a frequency f,i tw i l lc a u s et h em e d i u m
boundary, when it arrives there, to oscillate at that frequency. As a result of that, the wave that
is set up in the second medium—which we call thetransmitted wave—will also have the same",University Physics I Classical Mechanics.pdf
"is set up in the second medium—which we call thetransmitted wave—will also have the same
frequency f.A g a i n , t h i n k o f t h e t w o s t r i n g s t i e d t o g e t h e r , s o t h e fi r s t s tring “drives” the second
one at the frequencyf;o rt h es o u n da tt h ea i r - w a t e rb o u n d a r y ,d r i v i n g( p u s h i n g )the water surface
at the frequencyf.
So, the incident and transmitted waves will have the same frequency, but it is clear that, if the wave
speeds in the two media are different, they cannot have the samewavelength: since the relation
(12.4)h a st oh o l d ,w ew i l lh a v eλ1 = c1/f,a n dλ2 = c2/f.T h u s , i f a p e r i o d i c w a v e g o e s f r o m a
slower to a faster medium, its wavelength will increase, andif it goes from a faster to a slower one,
the wavelength will decrease.
It is easy to see physically why this happens, and how it has tobe the case even for non-periodic",University Physics I Classical Mechanics.pdf
"the wavelength will decrease.
It is easy to see physically why this happens, and how it has tobe the case even for non-periodic
1This effect can be used to produce “funny voices,” because of the relationship f = c/λ (Eq. (12.4)), which will
be discussed in greater detail in the section on standing waves.",University Physics I Classical Mechanics.pdf
"12.1. TRAVELING WAVES 287
waves, that is, wave pulses: a pulse going into a faster mediumw i l lw i d e ni nl e n g t h( s t r e t c h ) ,
whereas a pulse going into a slower medium will become narrower (squeezed). Imagine, for example,
several people walking in line, separated by the same distance d,a l la tt h es a m ep a c e ,u n t i lt h e y
reach a line beyond which they are supposed to start running.When the first person reaches the
line, he starts running, but the second one is still walking,so by the time the second one reaches the
line the first one has increased his distance from the second.The same thing will happen between
the second and the third, and so on: the original “bunch” willbecome spread out. (If you watch
car races, chances are you have seen this kind of thing happenalready!)
Besides setting up a transmitted wave, with the properties Ihave just discussed, the incident wave",University Physics I Classical Mechanics.pdf
"Besides setting up a transmitted wave, with the properties Ihave just discussed, the incident wave
will almost always cause areflected wave to start traveling in the first medium, moving backwards
from the boundary. The reflected wave also has the same frequency as the incident one, and since it
is traveling in the same medium, it will also have the same wavelength. A non-periodic pulse, when
reflected, will therefore not be stretched or squeezed, but itw i l lb e“ t u r n e da r o u n d ”b a c k - t o - f r o n t ,
since the first part to reach the boundary also has to be the firstt ol e a v e . S e eF i g u r e12.4 (the top
part) for an example.
c2 = 2c1
c2 = c1/2
Figure 12.4: Reflection and transmission of a pulse at the boundary where two strings of different densities
are joined. (“Before” and “after” situations are shown for eachcase.) Top figure: the string on the right is
less dense, so the pulse travels faster (the tension on both strings is supposed to be the same). The reflected",University Physics I Classical Mechanics.pdf
"less dense, so the pulse travels faster (the tension on both strings is supposed to be the same). The reflected
pulse is upright but reversed left-to-right. Bottom figure: the string on the right is more dense, so the
transmitted pulse travels more slowly. The reflected pulse is reversed left-to-right and flipped upside down.
What is the physical reason for the reflected wave? Ultimately, it has to do with the energy",University Physics I Classical Mechanics.pdf
"288 CHAPTER 12. WAVES IN ONE DIMENSION
carried by the incident wave, and whether it is possible for the transmitted wave alone to handle
the incoming energy flux or not. As we saw earlier (Eq. (12.8)), the energy per unit volume in a
harmonic wave of angular frequencyω and amplitudeξ0 is E/V = 1
2 ρ0ω2ξ2
0.I f t h e w a v e i s t r a v e l i n g
at a speed c,t h e nt h ee n e r g yflux (energy transported per unit time per unit area) is equal to
(E/V )c,w h i c hi st os a y
I = 1
2cρ0ω2ξ2
0 (12.14)
This is often called theintensity of the wave. It can be written asI = 1
2 Zω2ξ2
0,w h e r eIh a v e
defined the medium’smechanical impedance(or simply theimpedance)a s
Z = cρ0 (12.15)
(for a string, the mass per unit lengthµ instead of the mass per unit volumeρ0 should be used).
You can see that if the two media have the same impedance, thenthe energy flux in medium
2w i l le x a c t l ym a t c ht h a ti nm e d i u m1 ,p r o v i d e dt h ei n c i d e n tand transmitted waves have the",University Physics I Classical Mechanics.pdf
"2w i l le x a c t l ym a t c ht h a ti nm e d i u m1 ,p r o v i d e dt h ei n c i d e n tand transmitted waves have the
same amplitudes. In that case, there will beno reflected wave: even if the two media have different
densities and wave velocities, as long as they have the same impedance, the wave will be completely
transmitted.
On the other hand, if the media have different impedances, thenit will in general be impossible to
match the energy flux with only a transmitted wave, and reflection will occur. This is not immedi-
ately obvious, since it looks like all you have to do, to compensate for the different impedances in
Eq. (12.14), is to give the transmitted wave an amplitude that is differentf r o mt h a to ft h ei n c i d e n t
wave. But the point is precisely that, mathematically, you cannot do that without introducing a
reflected wave. This is because the actual amplitude of the oscillation at the boundary has to be",University Physics I Classical Mechanics.pdf
"reflected wave. This is because the actual amplitude of the oscillation at the boundary has to be
the same onboth sides, since the two media are connected there, and oscillating together; so, if
ξ0,inc is going to be different fromξ0,trans,y o un e e dt oh a v ea n o t h e rw a v ei nm e d i u m1 ,t h er e fl e c t e d
wave, to insure thatξ0,inc + ξ0,refl = ξ0,trans.
Another way to see this is to dig in a little deeper into the physical meaning of the impedance.
This is a worthwhile detour, because impedance in various forms recurs in a number of physics and
engineering problems. For a sound wave in a solid, for instance, we can see from Eqs. (12.13)a n d
(12.15)t h a tZ = cρ0 = √
Yρ 0;s oam e d i u mc a nh a v eal a r g ei m p e d a n c ee i t h e rb yb e i n gv e r ys tiff
(large Y )o rv e r yd e n s e( l a r g eρ0)o rb o t h ;e i t h e rw a y ,o n ew o u l dh a v et ow o r kh a r d e rt os e tu pa",University Physics I Classical Mechanics.pdf
"(large Y )o rv e r yd e n s e( l a r g eρ0)o rb o t h ;e i t h e rw a y ,o n ew o u l dh a v et ow o r kh a r d e rt os e tu pa
wave in such a medium than in one with a smaller impedance. On the other hand, once the wave
is set up, all that work gets stored as energy of the wave, so a wave in a medium with largerZ will
also carry a larger amount of energy (as is also clear from Eq.(12.14))2 for a given displacement
ξ0.
So, when a wave is trying to go from a low impedance to a large impedance medium, it will find
2In this respect, it may help you to think of the impedance of anextended medium as being somewhat analog
to the inertia (mass) of a single particle. The larger the mass, the harder it is to accelerate a particle, but once you
have given it a speedv,t h el a r g e rm a s sa l s oc a r r i e sm o r ee n e r g y .",University Physics I Classical Mechanics.pdf
"12.2. STANDING WAVES AND RESONANCE 289
it hard to set up a transmitted wave: the transmitted wave amplitude will be small (compared to
that of the incident wave), and the only way to satisfy the condition ξ0,inc + ξ0,refl = ξ0,trans will be
to set up a reflected wave with anegative amplitude3—in effect, to flip the reflected wave upside
down, in addition to left-to-right. This is the case illustrated in the bottom drawing in Figure12.4.
Conversely, you might think that a wave trying to go from a highi m p e d a n c et oal o wi m p e d a n c e
medium would have no trouble setting up a transmitted wave there, and that is true—but because
of its low impedance, the transmitted wave will still not be able to carry all the energy flux by
itself. In this case,ξ0,trans will be greater thanξ0,inc,a n dt h i sw i l la l s oc a l lf o rar e fl e c t e dw a v ei n
the first medium, only now it will be “upright,” that is,ξ0,refl = ξ0,trans − ξ0,inc > 0.",University Physics I Classical Mechanics.pdf
"the first medium, only now it will be “upright,” that is,ξ0,refl = ξ0,trans − ξ0,inc > 0.
To finish up the sub ject of impedance, note that the observation we just made, that impedance
will typically go as the square root of the product of the medium’s “stiffness” times its density, is
quite general. Hence, a medium’s density will typically be agood proxy for its impedance, at least
as long as the “stiffness” factor is independent of the density(as for strings, where it is just equal
to the tension) or, even better, increases with it (as is typically the case for sound waves in most
materials). Thus, you will often hear that a reflected wave isinverted (flipped upside down) when
it is reflected from a denser medium, without any reference tothe impedance—it is just understood
that “denser” also means “larger impedance” in this case. Also note, along these lines, that a “fixed
end,” such as the end of a string that is tied down (or, for soundw a v e s ,t h ec l o s e de n do fa no r g a n",University Physics I Classical Mechanics.pdf
"end,” such as the end of a string that is tied down (or, for soundw a v e s ,t h ec l o s e de n do fa no r g a n
pipe), is essentially equivalent to a medium with “infinite”impedance, in which case there is no
transmitted wave at that end, and all the energy is reflected.
Finally, the expressionξ0,inc + ξ0,refl that I wrote earlier, for the amplitude of the wave in the first
medium, implicitly assumes a very important property of waves, which is the phenomenon known
as interference,o re q u i v a l e n t l y ,t h e“ l i n e a rs u p e r p o s i t i o np r i n c i p l e . ” According to this principle,
when two waves overlap in the same region of space, the total displacement is just equal to the
algebraic sum of the displacements produced by each wave separately. Since the displacements are
added with their signs, one may get destructive interference if the signs are different, or constructive",University Physics I Classical Mechanics.pdf
"added with their signs, one may get destructive interference if the signs are different, or constructive
interference if the signs are the same. This will play an important role in a moment, when we start
the study of standing waves.
12.2 Standing waves and resonance
Imagine you have a sinusoidal traveling wave of the form (12.5), only traveling to the left, incident
from the right on a “fixed end” atx =0 . T h ei n c i d e n tw a v ew i l lg oa sξ0 sin[2π(x + ct)/λ]; the
reflected wave should be flipped left to right and upside down,so changex to −x and put an overall
3Ab e t t e rw a yt op u tt h i sw o u l db et os a yt h a tt h ea m p l i t u d ei sp ositive as always, but the reflected wave is 180◦
out of phase with the incident wave, so the amplitude of the total wave on the medium 1 side of the boundary is
ξ0,inc − ξ0,refl .",University Physics I Classical Mechanics.pdf
"290 CHAPTER 12. WAVES IN ONE DIMENSION
minus sign on the displacement, to get−ξ0 sin[2π(−x + ct)/λ]. The sum of the two waves in the
region x> 0i st h e n
ξ(x, t)= ξ0 sin
[2π
λ (x + ct)
]
− ξ0 sin
[2π
λ (−x + ct)
]
=2 ξ0 sin
( 2πx
λ
)
cos (2πft )( 1 2 . 1 6 )
using a trigonometrical identity for sin(a + b), andf = c/λ.
The result on the right-hand side of Eq. (12.16)i sc a l l e dastanding wave.I t d o e s n o t t r a v e l
anywhere, it just oscillates “in place”: every pointx behaves like a separate oscillator with an
amplitude 2ξ0 sin(2πx/λ). This amplitude is zero at special points, where 2x/λ is equal to an
integer. These points are callednodes.
We could think of “confining” a wave of this sort to a string fixeda tb o t he n d s ,i fw em a k et h e
string have an end atx =0a n dt h eo t h e ro n ea to n eo ft h e s ep o i n t sw h e r et h ea m p l i t u d eis zero;
this means we want the lengthL of the string to satisfy
2L = nλ (12.17)",University Physics I Classical Mechanics.pdf
"this means we want the lengthL of the string to satisfy
2L = nλ (12.17)
where n =1 , 2,... .A l t e r n a t i v e l y ,w e c a nt h i n ko fL as being fixed and Eq. (12.17)a sg i v i n gu st h e
possible values ofλ that will give us standing waves:λ =2 L/n.S i n c ef = c/λ,w es e et h a ta l l
these possible standing waves, for fixedL and c, have different frequencies that we can write as
fn = nc
2L,n =1 , 2, 3,... (12.18)
Note that these are all multiples of the frequencyf1 = c/2L.W e c a l l t h i s t h efundamental frequency
of oscillation of a string fixed at both ends. The period corresponding to this fundamental frequency
is the roundtrip time of a wave pulse around the string, 2L/c.
The first three standing waves are plotted in Figure12.5 (next page). Their wave functions are
given by the right-hand side of Eq. (12.16), for 0 ≤ x ≤ L,w i t hλ =2 L/n (n =1 , 2, 3), and",University Physics I Classical Mechanics.pdf
"given by the right-hand side of Eq. (12.16), for 0 ≤ x ≤ L,w i t hλ =2 L/n (n =1 , 2, 3), and
f = fn = nc/2L.T h e a m p l i t u d ei sa r b i t r a r y ;i nt h e fi g u r e Ih a v es e ti te q u a lto1f o rc o n v e n i e n c e .
Calling the corresponding functionun(x, t)i sm o r eo rl e s sc o m m o np r a c t i c ei no t h e rc o n t e x t s :
un(x, t)=s i n
( nπx
L
)
cos (2πfnt)( 1 2 . 1 9 )
These functions are called the normal modes of vibration of the string. In Figure 12.5 Ih a v e
shown, for each of them, the displacement at the initial time, t =0 ,a sas o l i dl i n e ,a n dt h e nh a l fa
period later as a dashed line. In addition to this, notice thatt h ew a v ef u n c t i o nv a n i s h e si d e n t i c a l l y
(the string is flat) at the quarter-period intervals,t =1 /4fn and t =3 /4fn.A t t h o s e t i m e s , t h e
wave has no elastic potential energy (since the string is unstretched): as with a simple oscillator",University Physics I Classical Mechanics.pdf
"wave has no elastic potential energy (since the string is unstretched): as with a simple oscillator
passing through the equilibrium position, all its energy iskinetic. For n> 1, there are also nodes
(places where the oscillation amplitude is always zero) at points other than the ends. Including the
endpoints, the n-th normal mode hasn +1 n o d e s . T h ep l a c e s w h e r et h eo s c i l l a t i o n a m p l i t u d ei s
largest are calledantinodes.",University Physics I Classical Mechanics.pdf
"12.2. STANDING WAVES AND RESONANCE 291
-1
 0
 1
 0  0.2  0.4  0.6  0.8  1
-1
 0
 1
-1
 0
 1
x/L
ξ/ξ0
Figure 12.5: The three lowest-frequency normal modes of vibration of a stringheld down at both ends,
corresponding to, from top to bottom,n =1 , 2, 3
Animations of these standing waves can be found in many places; one I particularly like is here:
http://newt.phys.unsw.edu.au/jw/strings.html#standing.I t a l s o s h o w s g r a p h i c a l l yh o wt h e
standing wave can be considered as a superposition of two oppositely-directed traveling waves, as
in Eq. (12.16).
If we initially bent the string into one of the shapes shown inFig. 12.5,a n dt h e nr e l e a s e di t ,i t
would oscillate at the corresponding frequencyfn,k e e p i n gt h es a m es h a p e ,o n l ys c a l i n gi tu pa n d
down by a factor cos(2πfnt)a st i m ee l a p s e d .S o ,a n o t h e rw a yt ot h i n ko fs t a n d i n gw a v e sisa st h e",University Physics I Classical Mechanics.pdf
"down by a factor cos(2πfnt)a st i m ee l a p s e d .S o ,a n o t h e rw a yt ot h i n ko fs t a n d i n gw a v e sisa st h e
natural modes of vibration of an extended system—the string, in this case, although standing waves
can be produced in any medium that can carry a traveling wave.
What I mean by a “natural mode of vibration” is the following:as i n g l eo s c i l l a t o r ,s a y ,ap e n d u l u m ,
has a single “natural” frequency; if you displace it or hit it,i tj u s to s c i l l a t e sa tt h a tf r e q u e n c yw i t h
ac o n s t a n ta m p l i t u d e . A ne x t e n d e ds y s t e m ,l i k et h es t r i n g ,can be viewed as a collection of coupled
oscillators, which may in general oscillate in many differentand complicated ways; yet, there is a
specific set of frequencies—for the string with two ends fixed,t h es e q u e n c efn of Eq. (12.18)—and
associated shapes that will result in all the parts of the string performing simple harmonic motion,",University Physics I Classical Mechanics.pdf
"associated shapes that will result in all the parts of the string performing simple harmonic motion,
in synchrony, all at the same frequency.
Of course, to produce just one of these specific modes of oscillation requires some care (“driving”
the string at the right frequency is probably the easiest way;s e en e x tp a r a g r a p h ) ;h o w e v e r ,i f
you simply hit or pluck the string in any random way, a remarkable thing happens: the resulting
motion will be, mathematically, described as a sum of sinusoidal standing waves, each with one",University Physics I Classical Mechanics.pdf
"292 CHAPTER 12. WAVES IN ONE DIMENSION
of the frequencies fn, and each with a different amplitudeAn.I n a m u s i c a l i n s t r u m e n t , t h i s
will eventually generate a superposition of sound waves withf r e q u e n c i e sf1 = c/2L, f2 =2 f1,
f3 =3 f1 ... (called, in this context, thefundamental, f1,a n di t sovertones, fn = nf1). Each one of
these frequencies corresponds to a different pitch, or musicaln o t e ,a n dt h er e s u l tw i l ls o u n dal i t t l e
like a chord, although not nearly as pronounced—we will mostly hear only the fundamental, which
corresponds to the root note of the chord, but all the notes inam a j o rt r i a da r ei nf a c tp r e s e n ti n
the vibration of a single guitar or piano string4.
But wait, there’s more! Suppose that you try to get the stringto oscillate by “driving” it: that is to
say, grabbing a hold of one end and shaking it at some frequency, only with a very small amplitude,",University Physics I Classical Mechanics.pdf
"say, grabbing a hold of one end and shaking it at some frequency, only with a very small amplitude,
so the displacement at that end remains always close to zero.In that case, you will typically get
only very small amplitude oscillations, until the driving frequency hits one of the special frequencies
fn,a tw h i c hp o i n ty o uw i l lg e tal a r g eo s c i l l a t i o nw i t ht h es h a peo ft h ec o r r e s p o n d i n gs t a n d i n g
wave. This is a phenomenon known asresonance,a n dt h efn are the resonant frequencies of this
system.
Note that the effect I just described is essentially the same asyou experience when you are “pump-
ing,” or simply pushing, a swing. Unless you do it at the rightfrequency, you do not get very far;
but if you do it at the right frequency (which is the swing’s natural frequency, the one at which it
will swing on its own), you can get huge amplitude oscillations. So, the frequencies (12.18)m a yb e",University Physics I Classical Mechanics.pdf
"will swing on its own), you can get huge amplitude oscillations. So, the frequencies (12.18)m a yb e
said to be the string’s natural oscillation frequencies in the same two ways: they are the ones at
which it will oscillate if you just pluck it, and they are the ones at which you have to drive it if you
want to get large oscillations.
Pretty much everything I have just shown you above for standing waves on a string applies to sound
waves inside a tube or pipe open at both ends. In that case, however, it is not the displacement,
but the pressure (or density) wave that must have zeros at theends (since the ends are open, the
pressure there must be just the average atmospheric pressure; note that the pressure or density
waves in a sound wave do not give the absolute pressure or density, but the deviation, positive or
negative, from the average). The math, however, is identical, and one finds the same set of normal",University Physics I Classical Mechanics.pdf
"negative, from the average). The math, however, is identical, and one finds the same set of normal
modes and resonance frequencies as above. These are then thefrequencies that would be produced
when blowing in a flute or an organ pipe open at both ends. So, both from pipes and strings we get
the same “harmonic series” of frequencies (12.18)t h a th a sb e e nt h ef o u n d a t i o no fW e s t e r nm u s i c
since at least the time of Pythagoras.
4It works like this: sayf1 corresponds to a C, thenf2 is the C above that,f3 the G above that,f4 the C above
that, andf5 the E above that.",University Physics I Classical Mechanics.pdf
"12.3. CONCLUSION, AND FURTHER RESOURCES 293
12.3 Conclusion, and further resources
This chapter on one-dimensional waves has barely scratchedthe surface of the extremely rich world
of wave phenomena. I have only given you a passing glance at interference, and I have not said
anything at all about diffraction, the Doppler effect, polarization, refraction... .M a n y o f t h e s e
things you will learn about in later courses, most likely wheny o ue n c o u n t e re l e c t r o m a g n e t i cw a v e s
(which are non-mechanical, but described by the same mathematical equation).
Waves are such an intrinsically kinetic phenomenon that theya r eb e s ta p p r e c i a t e db yw a t c h -
ing them in action, or, as a second-best alternative, througha n i m a t i o n s . Aw o n d e r f u lr e p o s -
itory of such movies and animations is PHYSCLIPS at the University of New South Wales:
http://www.animations.physics.unsw.edu.au/waves-sound/oscillations/index.html",University Physics I Classical Mechanics.pdf
"http://www.animations.physics.unsw.edu.au/waves-sound/oscillations/index.html
They also have a set of pages on the “physics of music” that I have already mentioned a couple of
times. If you are interested in this topic, you should go spends o m et i m et h e r e !
http://newt.phys.unsw.edu.au/music
Finally, closer to home, the fellows at PhET (University of Colorado), have this great interactive
app to explore waves on a string:
https://phet.colorado.edu/en/simulation/wave-on-a-s tring
12.4 In summary
1. A traveling wave in an elastic medium is a collective disturbance of the particles in the medium
(a displacement, or change in pressure or density) that carries energy and momentum from
one point of the medium to another, over a distance that is typically much larger than the
displacement of the individual particles making up the wave.
2. In a longitudinal wave, the displacement of the particlesis along the line of motion of the",University Physics I Classical Mechanics.pdf
"2. In a longitudinal wave, the displacement of the particlesis along the line of motion of the
wave; in a transverse wave, it is perpendicular to the wave’smotion.
3. An important kind of waves are periodic waves, in which thedisturbance repeats itself at
each point in the medium with a periodT.S i n u s o i d a l , p e r i o d i c w a v e s a r e c a l l e d h a r m o n i c
waves. Their spatial period is called the wavelengthλ.I ft h es p e e do ft h ew a v ei sc,o n eh a s
c = fλ,w h e r ef =1 /T is the wave frequency.
4. The time-averaged energy density in a harmonic wave (sum ofk i n e t i ca n de l a s t i cp o t e n t i a l
energy per unit volume) isE/V = ρ0ω2ξ2
0/2, where ρ0 is the medium’s density, andξ0 the
amplitude of the displacement oscillations. The time average momentum density isE/cV .
The intensity of the wave (energy carried per unit time per unit area) iscE/V .",University Physics I Classical Mechanics.pdf
"294 CHAPTER 12. WAVES IN ONE DIMENSION
5. Sound is a longitudinal compression-and-rarefaction wave in an elastic medium. It can be
described in terms of displacement, pressure or density. Thep r e s s u r eo rd e n s i t yd i s t u r b a n c e
is maximal where the displacement is zero, and vice-versa.
6. The speed of sound in a solid with Young modulusY is c =
√
Y/ρ0;i nafl u i dw i t hb u l k
modulus B,i ti sc =
√
B/ρ0.I na ni d e a lg a s ,t h i sd e p e n d s o n l yo nt h e r a t i oo fs p e c i fi c h e ats,
the molar mass, and the temperature.
7. Transverse waves on a string with mass per unit lengthµ and under a tensionFt travel with
as p e e dc =
√
Ft/µ.
8. When a wave reaches the boundary between two media, it is typically partly reflected and
partly transmitted. The incident, reflected and transmittedw a v e sa l lh a v et h es a m ef r e q u e n c y .
The transmitted wave has a wavelengthc2/f,w h e r ec2 is the wave speed in the second
medium.",University Physics I Classical Mechanics.pdf
"The transmitted wave has a wavelengthc2/f,w h e r ec2 is the wave speed in the second
medium.
9. The quantity that determines how much of the energy is reflected or transmitted is the
mechanical impedance,d e fi n e df o re a c hm e d i u ma sZ = cρ0.I f Z1 = Z2 there is no reflected
wave. IfZ1 <Z 2,t h er e fl e c t e dw a v ei si n v e r t e d( fl i p p e du p s i d e - d o w n )r e l a t ive to the incident
wave. If Z1 >Z 2,i ti su p r i g h t .
10. Standing waves arise in a medium that is confined to a regiono fs p a c e ,a n da r et h en o r m a l( o r
“natural”) modes of vibration of the system. In a standing wave, each particle of the medium
oscillates with an amplitude that is a fixed function of the particle’s position (a sinusoidal
function in one dimension). This amplitude is zero at pointscalled nodes.
11. In one dimension, all the standing wave frequencies are multiples of a fundamental frequency",University Physics I Classical Mechanics.pdf
"11. In one dimension, all the standing wave frequencies are multiples of a fundamental frequency
f1 = c/2L,w h e r eL is the length of the medium (as long as the boundary conditionsa t
both ends of the medium are identical). These are theresonant frequencies of the system: if
disturbed, it will naturally oscillate in a superposition oft h e s ef r e q u e n c i e s ,a n di fd r i v e na t
one of these frequencies, one will obtain a large response.",University Physics I Classical Mechanics.pdf
"12.5. EXAMPLES 295
12.5 Examples
12.5.1 Displacement and density/pressure in a longitudinalw a v e
The picture below shows the displacement of a medium (let’s say air) as a sound pulse travels
through it. (Don’t worry about the units on the axes right now!W e a r e o n l y i n t e r e s t e d i n q u a l i t a t i v e
results here.)
ξdisplacement 
direction of propagation of the wave (x)
x
(a) Sketch the corresponding pressure (or density) pulse. Note: pressure and density are in phase,
so one is large where the other is large. In either case what isalways plotted is thedifference
between the actual pressure or density and the average pressure (for air, atmospheric pressure) or
density of the medium.
(b) If this sound pulse is incident on water, sketch the reflected pulse, both in a displacement and
in a pressure/density plot.
Solution
(a) The purpose of this example is to help refine the intuitionyou may have gotten from Figure",University Physics I Classical Mechanics.pdf
"in a pressure/density plot.
Solution
(a) The purpose of this example is to help refine the intuitionyou may have gotten from Figure
12.3 regarding the relationship between the displacement and thep r e s s u r e / d e n s i t yi nal o n g i t u d i n a l
wave. When discussing Fig.12.3,Ia r g u e dt h a tt h ed e n s i t ys h o u l db eh i g ha tap o i n tl i k ex = π
in that figure, because the particles to the left of that pointwere being pushed to the right, and
those to the right were being pushed to the left. However, a similar argument can be made to show
that the density should be higher than its equilibrium valuewhenever the displacement curve has
a negative slope, in general.
For instance, consider pointx =1i nt h efi g u r ea b o v e . P a r t i c l e sb o t ht ot h el e f ta n dt h er i g h tof
that point are being pushed to the right (positive displacement), but the displacement is larger for
the ones on the left, which will result in a bunching atx =1 .",University Physics I Classical Mechanics.pdf
"296 CHAPTER 12. WAVES IN ONE DIMENSION
Conversely, if you look at a point with positive slope, such as x =0 ,y o us e et h a tt h ep a r t i c l e so n
the right are pushed farther to the right than the particles ont h el e f t ,w h i c hm e a n st h ed e n s i t y
around x =0w i l ld r o p .
From this you may conclude that the density versus position graph will look somewhat like the
negative of the derivative of theξ-vs.-x graph: positive when ξ falls, negative when it rises, and
zero at the “turning points” (maxima or minima ofξ(x)). This is, in fact, mathematically true,
and is illustrated in the figure below.
density/pressure
ξdisplacement 
x
x
Figure 12.6: Illustrating the relationship between displacement (blue curve) andpressure/density (red) in
a longitudinal wave. The dashed lines separate the regions where the pressure (or density) is positive (higher
than in the absence of the wave) from those where it is negative.",University Physics I Classical Mechanics.pdf
"than in the absence of the wave) from those where it is negative.
(b) If this sound wave is incident from air into water, it meansi ti sg o i n gf r o mal o wi m p e d a n c e
to a high impedance medium (both the density and the speed of sound are much greater in water
than in air, giving a much largerZ = cρ0;s e eE q .(12.15)f o rt h ed e fi n i t i o no fi m p e d a n c e ) . T h i s
means the reflected displacement pulse will be flipped upsidedown, as well as left to right (see the
figure on the next page). This is just (except for the scale, which here is arbitrary) like the bottom
part of Fig.12.4.",University Physics I Classical Mechanics.pdf
"12.5. EXAMPLES 297
However, if you now try to figure out the shape of the density/pressure wave based on the displace-
ment wave, as we did in part (a), you’ll see that it is only reversed left to right, butnot flipped
upside down! This is a general property of longitudinal waves: the reflected pressure/density wave
behaves in exactly the opposite way as the displacement wave,a sf a ra st h eu p s i d e - d o w n“ fl i p ”i s
concerned: it gets flipped when going from high impedance to low impedance, and not when going
from low to high.
density/pressure
ξdisplacement 
x
x
Figure 12.7: What the wave pulse in the previous figure would look like if reflected from a high-impedance
medium. The displacement wave is reversed left to right and flipped upside down. The pressure/density
wave is only reversed left to right.
If you are curious to see how this happens mathematically, thei d e ai st h a tt h ed e n s i t yw a v ei s",University Physics I Classical Mechanics.pdf
"wave is only reversed left to right.
If you are curious to see how this happens mathematically, thei d e ai st h a tt h ed e n s i t yw a v ei s
proportional to−dξ/dx,a n dt h er e fl e c t e dd i s p l a c e m e n tw a v eg o e sl i k eξrefl = −ξinc(−x), where the
first minus sign gives the vertical flip and the second the horizontal one. Taking the derivative of
this last expression with respect tox then removes the minus sign in front.",University Physics I Classical Mechanics.pdf
"298 CHAPTER 12. WAVES IN ONE DIMENSION
12.5.2 Violin sounds
The “sounding length” of a violin string, from the bridge to the nut at the upper end of the
fingerboard, is about 32 cm.
(a) If the string is tuned so that its fundamental frequency corresponds to a concert A (440 Hz),
what is the speed of a wave on that string?
(b) If the string’s density is 0.66 g/m (note: the “g” stands for “grams”!), what is the tensiono n
the string?
(c) When the string is played, its vibration is transmitted through the bridge to the violin plates.
At what frequency will the plates vibrate?
(d) The vibration of the plates then sets up a sound wave in air.W h a t i s t h e w a v e l e n g t h o f t h i s
wave?
Solution
(a) In Section12.2 we saw that the fundamental frequency of a string fixed at bothends isf1 = c/2L
(corresponding to Eq. (12.18)w i t hn =1 ) . S e t t i n gt h i se q u a lt o4 4 0H z ,a n ds o l v i n gf o rc,
c =2 Lf1 =2 × (0.32 m)× 440 s−1 =2 8 2m
s",University Physics I Classical Mechanics.pdf
"c =2 Lf1 =2 × (0.32 m)× 440 s−1 =2 8 2m
s
(b) From Section12.1.3,w eh a v et h a tt h es p e e do faw a v eo nas t r i n gi sc =
√
Ft/µ,w h e r eFt is
the tension andµ the mass per unit length (Eq. (12.11). Solving forFt,
Ft = c2µ =
(
282 m
s
)2
× 6.6 × 10−4 kg
m =5 2.5N
(c) The plates will vibrate at the same frequency as the string, 440 Hz, since they are driven by the
motion of the string.
(d) The basic relationship to use here is Eq. (12.4), f = c/λ,w h i c hw ec a ns o l v ef o rλ if we know
c,t h es p e e do fs o u n di na i r .I nS e c t i o n12.1.3 it was stated that the speed of sound in air is about
340 m/s, so we have
λ = c
f = 340 m/s
440 s−1 =0 .77 m",University Physics I Classical Mechanics.pdf
"12.6. ADVANCED TOPICS 299
12.6 Advanced Topics
12.6.1 Chain of masses coupled with springs: dispersion, andl o n g - w a v e l e n g t h
limit.
Consider a model of an extended elastic medium in which, for simplicity, we separate the two main
medium properties, inertia and elasticity, by describing ita sac h a i no fp o i n t - l i k em a s s e s( p a r t i c l e s )
connected by massless springs, as in Figure12.8 below. I will show you here how one can get
“ideal” wave behavior in this system, provided we work in the“long-wavelength” limit, that is to
say, we consider only waves whose wavelength is much greaterthan the average distance between
neighboring masses.
......
ξn-1 ξn+1ξn
d
Figure 12.8: A chain of masses connected by massless springs. The top picture shows the equilibrium posi-
tions with the springs relaxed, the bottom one the situation where each mass has undergone a displacement
ξ.",University Physics I Classical Mechanics.pdf
"tions with the springs relaxed, the bottom one the situation where each mass has undergone a displacement
ξ.
In the figure above I have explicitly shown then-th mass and the two springs that push and/or pull
on it, both in equilibrium (top drawing) and when the chain isin motion (bottom). In the latter
case, the length of the springs depends on the relative displacements of all three masses shown.
Specifically, the length of the spring on the left isd+ξn −ξn−1,w h e r ed is the distance between the
masses in equilibrium, and the length of the spring on the right isd + ξn+1 − ξn.I ft h el e f ts p r i n g
is stretched (length greater thand)i tw i l lp u l lt ot h el e f to nt h en-th mass, and, conversely, if the
right spring is stretched (length greater thand)i tw i l lp u l lt ot h er i g h to nt h en-th mass. So, if all
the springs have the same constantk,t h ef o r c ee q u a t i o nF = ma for massn is
man = −k(ξn − ξn−1)+ k(ξn+1 − ξn)( 1 2 . 2 0 )
which we can rewrite as
md2ξn",University Physics I Classical Mechanics.pdf
"man = −k(ξn − ξn−1)+ k(ξn+1 − ξn)( 1 2 . 2 0 )
which we can rewrite as
md2ξn
dt2 = kξn−1 − 2kξn + kξn+1 (12.21)",University Physics I Classical Mechanics.pdf
"300 CHAPTER 12. WAVES IN ONE DIMENSION
Now let us try to see if we can get a sinusoidal solution to thissystem of differential equations. By
analogy with Eq. (12.3)l e t
ξn(t)= A sin
[
2π
( xn
λ − ft
)]
where xn = nd is the equilibrium position of then-th mass. Then for each of the three masses
considered, we have
ξn−1(t)= A sin[2π((n − 1)d/λ − ft)] =A sin[2π(nd/λ − ft) − 2πd/λ]
ξn(t)= A sin[2π(nd/λ − ft)]
ξn+1(t)= A sin[2π((n +1 )d/λ − ft)] =A sin[2π(nd/λ − ft)+2 πd/λ]( 1 2 . 2 2 )
We want to substitute all this in Eq. (12.21). We can use the trigonometric identity sin(a − b)+
sin(a + b)=2s i na cos b to simplifyξn−1 + ξn+1:
ξn−1 + ξn+1 =2 A sin
[
2π
( nd
λ − ft
)]
cos
( 2πd
λ
)
(12.23)
then use 1− cos x =2s i n2(x/2) to yield
kξn−1 − 2kξn + kξn+1 = −4kA sin2
( πd
λ
)
sin
[
2π
( nd
λ − ft
)]
= −4k sin2
( πd
λ
)
ξn (12.24)
It is clear now that Eq. (12.21)w i l lb es a t i s fi e dp r o v i d e dt h ef o l l o w i n gc o n d i t i o nh o l d s :
m(2πf )2 =4 k sin2
( πd
λ
)
(12.25)",University Physics I Classical Mechanics.pdf
"m(2πf )2 =4 k sin2
( πd
λ
)
(12.25)
Or, taking the square root and simplifying,
f = 1
π
√
k
m sin
( πd
λ
)
(12.26)
This is clearly a more complicated relation betweenf and λ than just Eq. (12.4). However, since
we can argue that Eq. (12.4)m u s ta l w a y sh o l df o ras i n u s o i d a lw a v e ,w h a tw eh a v ea c t u a l lyf o u n d
is that the chain of masses and springs in Fig. (12.8)w i l ls u p p o r tas i n u s o i d a lw a v ep r o v i d e dt h e
wave velocity depends on the wavelengthas required by Eqs. (12.4)a n d(12.26):
c = λf =
√
k
m
λ
π sin
( πd
λ
)
(12.27)
This is an instance of the phenomenon calleddispersion: sinusoidal waves of different frequencies
(or wavelengths) have different velocities. One thing that happens in the presence of dispersion is
that, although a single (infinite), sinusoidal wave can travel without changing its shape (provided
f and λ satisfy Eq. (12.26)), a general pulse will be distorted as it propagates throught h em e d i u m ,",University Physics I Classical Mechanics.pdf
"f and λ satisfy Eq. (12.26)), a general pulse will be distorted as it propagates throught h em e d i u m ,
often severely so.",University Physics I Classical Mechanics.pdf
"12.6. ADVANCED TOPICS 301
In the long wavelength limit, however, the dispersion in thism o d e ld i s a p p e a r s . W ec a ns e et h i s
as follows. In that limit,λ ≫ d (the wavelength is much greater than the distance between the
masses), and thereforeπd/λ ≪ 1; we can then make the small-angle approximation in Eq. (12.27),
sin(πd/λ) ≃ πd/λ,a n de n du pw i t h
c ≃ d
√
k
m (12.28)
This is of the general form
√
stiffness/inertia (as per Eq. (12.10)). Basically, in the long-wavelength
limit, the medium appears homogeneous to the wave—it cannot“tell” that it is a chain of discrete
particles. When you consider that everything that looks homogeneous on a macroscopic scale is
actually made of discrete atoms or molecules at the microscopic level, you can see that this model
is perhaps not as artificial as it might seem, and that in general you should, in fact, expect some
kind of dispersion to occur in any medium, at sufficiently smallw a v e l e n g t h s .",University Physics I Classical Mechanics.pdf
"302 CHAPTER 12. WAVES IN ONE DIMENSION
12.7 Problems
Problem 1
When plucked, the D string on a guitar vibrates with a frequency of 147 Hz.
(a) What would happen to this frequency if you were toincrease the tension in the string?
(b) The vibration of the string eventually produces a sound wave of the same frequency, traveling
through the air. If the speed of sound in air is 340 m/s, what isthe wavelength of this wave?
Problem 2
Think of a flute as basically a cylindrical tube of length 0.6m , o pe n t o t h e a t m o s p h e r e a t bo t h
ends. If the speed of sound in air is 340 m/s
(a) What is the fundamental (lowest) frequency of a sound wavei nafl u t e ?
(b) Is this a transverse or a longitudinal wave?
(c) The speed of sound in helium is about 3 times that in air. Howw o u l dt h efl u t e ’ sr e s o n a n c e
frequencies change if you filled it with helium instead of air?
Justify each of your answers briefly.
Problem 3",University Physics I Classical Mechanics.pdf
"frequencies change if you filled it with helium instead of air?
Justify each of your answers briefly.
Problem 3
The top picture shows a wave pulse on a string (string 1) traveling to the right, where the string is
attached to another one (string 2, not shown). The bottom picture shows the reflected wave some
time later.
If the tension on both strings is the same,
(a) Is string 2 more or less dense than string 1?
(b) In which string will the wave travel faster?
(c) Sketch what the reflected wave would look like if the strings’ densities were the opposite of what
you answered in part (a).
Explain each of your answers briefly.",University Physics I Classical Mechanics.pdf
"Chapter 13
Thermodynamics
13.1 Introduction
The last two lectures this semester are about thermodynamics, an extremely important branch
of physics that developed throughout the 19th century, motivated in part by the development of
the steam engines that brought about the Industrial Revolution. Physics majors will study ther-
modynamics at much greater length in University Physics IIIand subsequent courses, whereas
Engineering and Chemistry majors will encounter it also in specialized courses in their own disci-
plines.
There is really no escaping thermodynamics, but you may wonder why bring it up here (in this
course, at this time) at all? The answer is twofold:
• From the point of view of the study of energy and its transformations, which has been one
of the major themes of this course, thermodynamics providesus with the last missing pieces:
it is here that we find out what thermal energy really is, and how it is different from other",University Physics I Classical Mechanics.pdf
"it is here that we find out what thermal energy really is, and how it is different from other
forms of energy (so much so, that we say that energy has been “dissipated” or “lost” when it
becomes thermal energy). It is also here that we deal with theother way that energy can be
transferred from a system to another (other, that is, than bydoing work): this is the “direct
transfer of thermal energy,” or what is normally called anexchange of heat.
• From the point of view of the study of motion, which has been also another running theme,
thermodynamics also represents the next logical step beyondw h a tw eh a v el e a r n e ds of a r .
Recall that we started looking at the motion of extended objects as if they were simple point
particles, moving as a whole along with their center of mass,and slowly introduced tools to
deal with more complex kinds of motion: first rigid body rotations, then elastic deformations
303",University Physics I Classical Mechanics.pdf
"304 CHAPTER 13. THERMODYNAMICS
(waves) in which the constituent parts of an object move relative to each other in a way that
looks “organized,” or synchronized, from a macroscopic perspective. What is needed next is
to account for the random motion, on a microscopic scale, of the smallest parts (atoms or
molecules) that make up an extended object. This motion is constantly happening, and it is
ak e yi n g r e d i e n to ft h ec o n c e p t so ft h e r m a le n e r g ya n dt e m p e rature.
Conceptually, thermodynamics involves the introduction oft w on e wp h y s i c a lq u a n t i t i e s ,tempera-
ture and entropy.T e m p e r a t u r e w i l l b e i n t r o d u c e d i n t h i s l e c t u r e , a n d e n t r o pyi nt h en e x to n e . I t
is interesting to note from the start, however, that these are very different from all the quantities
we have introduced so far this semester, in a fundamental way.I n c l a s s i c a lp h y s i c s , a t l e a s t ,t h e r e",University Physics I Classical Mechanics.pdf
"we have introduced so far this semester, in a fundamental way.I n c l a s s i c a lp h y s i c s , a t l e a s t ,t h e r e
is no difficulty in extending all those other quantities to thestudy of the smallest parts making
up an object: we can perfectly well talk about the position, velocity or energy of a molecule. But
temperature and entropy arestatistical quantities, which are only properly defined, from a funda-
mental point of view, for a large collection of (small) subsystems: it makes no sense to speak about
the temperature or the entropy of a single molecule. This shows that there was really a profound
change in perspective and methodology in classical physicswhen statistical mechanics(the part of
physics that provides a microscopic foundation for thermodynamics) was developed.
13.2 Introducing temperature
13.2.1 Temperature and heat capacity
The change in perspective that I just mentioned also means that it is not easy to evendefine",University Physics I Classical Mechanics.pdf
"13.2.1 Temperature and heat capacity
The change in perspective that I just mentioned also means that it is not easy to evendefine
temperature, beyond our natural intuition of “hot” and “cold,” or the somewhat circular notion
that temperature is simply “what thermometers measure.” Roughly speaking, though, temperature
is a measure of the amount (or, to be somewhat more precise, the concentration)o ft h e r m a le n e r g y
in an object. When we directly put an amount of thermal energy,∆ Eth (what we will be calling
heat in a moment), in an object, we typically observe its temperature to increase in a way that is
approximately proportional to ∆Eth,a tl e a s ta sl o n ga s∆Eth is not too large:
∆T = ∆Eth
C (13.1)
The proportionality constantC is called theheat capacityof the object: according to Eq. (13.1), a
system with a large heat capacity could absorb (or give off—theequation is supposed to apply in",University Physics I Classical Mechanics.pdf
"system with a large heat capacity could absorb (or give off—theequation is supposed to apply in
either case) a large amount of thermal energy without experiencing a large change in temperature.
If the system does not do any work in the process (recall Eq. (7.20)!), then its internal energy will
increase (or decrease) by exactly the same amount of thermalenergy it has taken in (or given off)1,
1If the systemdoes do some work (or has work done on it), then Eq. (13.8)a p p l i e s ;s e eb e l o w .",University Physics I Classical Mechanics.pdf
"13.2. INTRODUCING TEMPERATURE 305
and we can use the heat capacity2 to, ultimately, relate the system’s temperature to its energy
content in a one-to-one-way.
What is found experimentally is that the heat capacity of a homogeneous object (that is, one made
of just one substance) is, in general, proportional to its mass. This is why, instead of tables of heat
capacities, what we compile are tables ofspecific heats,w h i c ha r eh e a tc a p a c i t i e sp e rk i l o g r a m( o r
sometimes per mole, or per cubic centimeter... but all these things are ultimately proportional to
the object’s mass). In terms of a specific heatc = C/m,a n da g a i na s s u m i n gn ow o r kd o n eo rb y
the system, we can rewrite Eq. (13.1)t or e a d
∆Esys = mc∆T (13.2)
or, again,
∆T = ∆Esys
mc (13.3)
which shows what I said above, that temperature really measures, not the total energy content
of an object, but itsconcentration—the thermal energy “per unit mass,” or, if you prefer (and",University Physics I Classical Mechanics.pdf
"of an object, but itsconcentration—the thermal energy “per unit mass,” or, if you prefer (and
somewhat more fundamentally) “per molecule.” An object canhave a great deal of thermal energy
just by virtue of being huge, and yet still be pretty cold (water in the ocean is a good example).
In fact, we can also rewrite Eqs. (13.1–13.3)i nt h e( s o m e w h a tc o n t r i v e d - l o o k i n g )f o r m
C = mc = m∆Esys/m
∆T (13.4)
which tells you that an object can have a large heat capacity int w ow a y s :o n ei ss i m p l yt oh a v ea
lot of mass, and the other is to have a large specific heat. The first of these ways is kind of boring
(but potentially useful, as I will discuss below); the secondi si n t e r e s t i n g ,b e c a u s ei tm e a n st h a t
ar e l a t i v e l yl a r g ec h a n g ei nt h ei n t e r n a le n e r g yp e rm o l e c u le( r o u g h l ys p e a k i n g ,t h en u m e r a t o ro f",University Physics I Classical Mechanics.pdf
"(13.4)) will only show as a relatively small change in temperature(the denominator of (13.4); a
large numerator and a small denominator make for a large fraction!).
Put differently, and somewhat fancifully,substances with a large specific heat are very good at
hiding their thermal energy from thermometers(see Fig. 13.1 for an example). This, as I said, is
an interesting observation, but it also means that measuringh e a tc a p a c i t i e s — o r ,f o rt h a tm a t t e r ,
measuring temperature itself—may not be an easy matter. Howdo we get at the object’s internal
energy if not through its temperature? Where does one start?
2Which, it must be noted, may also be a function of the system’stemperature—another complication we will
cheerfully ignore here.",University Physics I Classical Mechanics.pdf
"306 CHAPTER 13. THERMODYNAMICS
Figure 13.1: In this simple model of a gas of diatomic molecules, each molecule can store “vibrational”
potential energy (both potential and kinetic, through the oscillations of the “spring” that models the inter-
action between the atoms), plus as at least two kinds of rotationalkinetic energy (corresponding to rotations
around the axes shown), in addition to just the translational kinetic energy of its center of mass. The latter
is the only one directly measured by a gas thermometer, so a diatomicgas has many more ways of “hiding”
its thermal energy (and hence, a larger specific heat) than a monoatomic gas.
13.2.2 The gas thermometer
Ag o o ds t a r t ,a tl e a s tc o n c e p t u a l l y ,i sp r o v i d e db yl o o k i n gatas y s t e mt h a th a sn op l a c et oh i d ei t s
thermal energy—it has to show it all, have it, as it were, in full view all the time. Such a system",University Physics I Classical Mechanics.pdf
"thermal energy—it has to show it all, have it, as it were, in full view all the time. Such a system
is what has come to be known asan ideal gas—which we model, microscopically, as a collection
of molecules (or, more properly, atoms) with no dimension andn os t r u c t u r e :j u s tp o i n t l i k et h i n g s
whizzing about and continually banging into each other and against the walls of their container.
For such a systemthe only possible kind of internal energy is the sum of the molecules’ translational
kinetic energy.W e m a y e x p e c t t h i s t o b e e a s i l y d e t e c t e d b y a t h e r m o m e t e r ( o rany other energy-
sensitive probe), because as the gas molecules bang againstthe thermometer, they will indirectly
reveal the energy they carry, both by how often and how hard they collide.
As it turns out, we can be a lot more precise than that. We can analyze the theoretical model",University Physics I Classical Mechanics.pdf
"As it turns out, we can be a lot more precise than that. We can analyze the theoretical model
of an ideal gas that we have just described fairly easily, using nothing but the concepts we have
introduced earlier in the semester (plus a few simple statistical ideas) and obtain the following
result for the gas’ pressure and volume:
PV = 2
3N⟨Ktrans⟩ (13.5)
where N is the total number of molecules, and⟨Ktrans⟩ is the average translational kinetic energy
per molecule. Now, you are very likely to have seen, in high-school chemistry, theexperimentally
derived “ideal gas law,”
PV = nRT (13.6)
where n is the number of moles, andR the “ideal gas constant.” Comparing Eq. (13.5)( at h e o r e t i c a l
prediction for a mathematical model) and Eq. (13.6)( a ne m p i r i c a lr e s u l ta p p r o x i m a t e l yv a l i df o r",University Physics I Classical Mechanics.pdf
"13.2. INTRODUCING TEMPERATURE 307
many real-world gases under a wide range of pressure and temperature, where “temperature”
literally means simply “what any good thermometer would measure”) immediately tells us what
temperature is,a tl e a s tf o rt h i se x t r e m e l ys i m p l es y s t e m : i ti sj u s tam e a s ure of the average
(translational) kinetic energy per molecule.
It would be tempting to leave it at that, and immediately generalize the result to all kinds of other
systems. After all, presumably, a thermometer inserted in aliquid is fundamentally responding to
the same thing as a thermometer inserted in an ideal gas: namely, to how often, and how hard, the
liquid’s molecules bang against the thermometer’s wall. Sowe can assume that, in fact, it must
be measuring the same thing in both cases—and that would be thea v e r a g et r a n s l a t i o n a lk i n e t i c
energy per molecule. Indeed, there is a result in classical statistical mechanics that states that for",University Physics I Classical Mechanics.pdf
"energy per molecule. Indeed, there is a result in classical statistical mechanics that states that for
any system (liquid, solid, or gas) in “thermal equilibrium” (a state that I will define more precisely
later), the average translational kinetic energy per molecule must be
⟨Ktrans⟩ = 3
2kBT (13.7)
where kB is a constant calledBoltzmann’s constant(kB =1 .38×10−23 J/K), andT,a si nE q .(13.6)
is measured in degrees Kelvin.
There is nothing wrong with this way to think about temperature, except that it is too self-
limiting. To simply identify temperature with the translational kinetic energy per molecule leaves
out a lot of other possible kinds of energy that a complex system might have (a sufficiently complex
molecule may also rotate and vibrate, for instance, as shownin Fig. 13.1;t h e s ea r es o m eo ft h e
ways the molecule can “hide” its energy from the thermometer,a sIs u g g e s t e da b o v e ) . T y p i c a l l y ,",University Physics I Classical Mechanics.pdf
"ways the molecule can “hide” its energy from the thermometer,a sIs u g g e s t e da b o v e ) . T y p i c a l l y ,
all those other forms of internal energy also go up as the temperature increases, so it would be
at least a bit misleading to think of the temperature as havingt od ow i t ho n l yKtrans,E q .(13.7)
notwithstanding. Ultimately, in fact, it is thetotal internal energy of the system that we want to
relate to the temperature, which means having to deal with those pesky specific heats I introduced in
the previous section. (As an aside, the calculation of specific heats was one of the great challenges
to the theoretical physicists of the late 19th and early 20thcentury, and eventually led to the
introduction of quantum mechanics—but that is another story!)
In any case, the ideal gas not only provides us with an insightinto the microscopic picture behind the
concept of temperature, it may also serve as a thermometer itself. Equation (13.6)s h o w st h a tt h e",University Physics I Classical Mechanics.pdf
"concept of temperature, it may also serve as a thermometer itself. Equation (13.6)s h o w st h a tt h e
volume of an ideal gas held at constant pressure will increasei naw a yt h a t ’ sd i r e c t l yp r o p o r t i o n a lt o
the temperature. This is just how a conventional, old-fashioned mercury thermometer worked—as
the temperature rose, the volume of the liquid in the tube wentu p .T h ei d e a lg a st h e r m o m e t e ri s
ab i tm o r ec u m b e r s o m e( ar e l a t i v e l ys m a l lt e m p e r a t u r ec h a n gem a yc a u s eap r e t t yl a r g ec h a n g ei n
volume), but, as I stated earlier, typically works well overav e r yl a r g et e m p e r a t u r er a n g e .
By using an ideal (or nearly ideal) gas as a thermometer, basedo nE q .(13.6), we are, in fact,
implicitly defining a specific temperature scale, the Kelvinscale (indeed, you may recall that for",University Physics I Classical Mechanics.pdf
"308 CHAPTER 13. THERMODYNAMICS
Eq. (13.6)t ow o r k ,t h et e m p e r a t u r em u s tb em e a s u r e di nd e g r e e sK e l v i n). The zero point of that
scale (what we call absolute zero) is the theoretical point atw h i c ha ni d e a lg a sw o u l ds h r i n kt o
precisely zero volume. Of course, no gas stays ideal (or evengaseous!) at such low temperatures,
but the point can easily be found by extrapolation: for instance, imagine plotting experimental
values ofV vs T,a tc o n s t a n tp r e s s u r e ,f o ran e a r l yi d e a lg a s ,u s i n ga n yk i n dof thermometer scale
to measureT,o v e raw i d er a n g eo ft e m p e r a t u r e s .T h e n ,c o n n e c tt h ep o i n t sby a straight line, and
extend the line to where it crosses theT axis (soV =0 ) ;t h a tp o i n tg i v e sy o ut h ev a l u eo fa b s o l u t e
zero in the scale you were using, such as−273.15 Celsius, for instance, or−459.67 Fahrenheit.
V
T (any scale)T = 0 (Kelvin scale)",University Physics I Classical Mechanics.pdf
"zero in the scale you were using, such as−273.15 Celsius, for instance, or−459.67 Fahrenheit.
V
T (any scale)T = 0 (Kelvin scale)
Figure 13.2: Illustrating how a gas thermometer can be used to define the Kelvin,or absolute, temperature
scale.
The connection between Kelvin (orabsolute)t e m p e r a t u r ea n dm i c r o s c o p i cm o t i o ne x p r e s s e db y
equations like (13.5)t h r o u g h(13.7)i m m e d i a t e l yt e l l su st h a ta sy o ul o w e rt h et e m p e r a t u r et h e
atoms in your system will move more and more slowly, until, when you reach absolute zero, all
microscopic motion would cease. This does not quite happen,because of quantum mechanics, and
we also believe that it is impossible to really reach absolutez e r of o ro t h e rr e a s o n s ,b u ti ti st r u e
to a very good approximation, and experimentalists have recently become very good at cooling
small ensembles of atoms to temperatures extremely close toabsolute zero, where the atoms move,",University Physics I Classical Mechanics.pdf
"small ensembles of atoms to temperatures extremely close toabsolute zero, where the atoms move,
literally, slower than snails (instead of whizzing by at close to the speed of sound, as the air molecules
do at room temperature).
13.2.3 The zero-th law
Historically, thermometers became useful because they gaveu saw a yt oq u a n t i f yo u rn a t u r a l
perception of cold and hot, but the quantity they measure, temperature, would have been pretty
useless if it had not exhibited an important property, whichwe naturally take for granted, but which
is, in fact, surprisingly not trivial. This property, whichoften goes by the name ofthe zero-th law
of thermodynamics,c a nb es t a t e da sf o l l o w s :
Suppose you place two systemsA and B in contact, so they can directly exchange thermal energy
(more about this in the next section), while isolating them from the rest of the world (so their joint",University Physics I Classical Mechanics.pdf
"13.3. HEAT AND THE FIRST LAW 309
thermal energy has no other place to go). Then, eventually, they will reach a state, calledthermal
equilibrium,i nw h i c ht h e yw i l lb o t hh a v et h es a m et e m p e r a t u r e .
This is important for many reasons, not the least of which being that that is what allows us to mea-
sure temperature with a thermometer in the first place: the thermometer tells us the temperature
of the object with which we place it in contact, by first adopting itself that temperature! Of course,
ag o o dt h e r m o m e t e rh a st ob ed e s i g n e ds ot h a ti tw i l ld ot h a tw hile changing the temperature of
the system being measured as little as possible; that is to say, the thermometer has to have a much
smaller heat capacity than the system it is measuring, so thati to n l yn e e d st og i v eo rt a k eav e r y
small amount of thermal energy in order to match its temperature. But the main point here is that",University Physics I Classical Mechanics.pdf
"small amount of thermal energy in order to match its temperature. But the main point here is that
the match actually happens, and when it does, the temperaturem e a s u r e db yt h et h e r m o m e t e rw i l l
be the same for any other systems that are, in turn, in thermalequilibrium with—that is, at the
same temperature as—the first one.
The zero-th law only assures us that thermal equilibrium wille v e n t u a l l yh a p p e n ,t h a ti s ,t h et w o
systems will eventually reach one and the same temperature;it does not tell us how long this may
take, nor even, by itself, what that final temperature will be.T h e l a t t e r p o i n t , h o w e v e r , c a n b e
easily determined if you make use of conservation of energy (the first law, coming up!) and the
concept of heat capacity introduced above (think about it foram i n u t e ) .
Still, as I said above, this result is far from trivial. Just imagine, for instance, two different ideal",University Physics I Classical Mechanics.pdf
"Still, as I said above, this result is far from trivial. Just imagine, for instance, two different ideal
gases, whose molecules have different masses, that you bring toas t a t eo fj o i n tt h e r m a le q u i l i b r i u m .
Equations (13.5)t h r o u g h(13.7)t e l lu st h a ti nt h i sfi n a ls t a t et h ea v e r a g et r a n s l a t i o n a lk inetic energy
of the “A”m o l e c u l e sa n dt h e“B”m o l e c u l e sw i l lb et h es a m e .T h i sm e a n s ,i np a r t i c u l a r ,t h a tthe
more massive molecules will end up moving more slowly, on average, so mav2
a,av = mbv2
b,av.B u t
why is that? Why should it be the kinetic energies that end up matching, on average, and not, say,
the momenta, or the molecular speeds themselves? The result,t h o u g hu n d o u b t e d l yt r u e ,d e fi e d
ar i g o r o u sm a t h e m a t i c a lp r o o ff o rd e c a d e s ,i fn o tc e n t u r i e s;Ia mn o ts u r et h a tar i g o r o u sp r o o f
exists, even now.
13.3 Heat and the first law",University Physics I Classical Mechanics.pdf
"exists, even now.
13.3 Heat and the first law
13.3.1 “Direct exchange of thermal energy” and early theories of heat
In the previous section I have considered the possibility of“direct exchange of thermal energy”
between two objects. This is a phenomenon with which we are allf a m i l i a r :w h e nac o l d e ro b j e c ti s
placed in contact with a warmer one, the warmer one cools off andt h ec o l d e ro n ew a r m su p .T h i s
“warmth” that seems to flow out of one object and into the otheris conventionally called “heat.”",University Physics I Classical Mechanics.pdf
"310 CHAPTER 13. THERMODYNAMICS
Naturally, this observation was made long before the concepto f“ e n e r g y ”w a se v e nd e v e l o p e d ,a n d
so heat was thought of, for a time, as an “invisible fluid” (called, at one point, “caloric fluid”), a sort
of indestructible “substance” that literally passed from one body to another. By “indestructible” I
mean that they had a notion of this caloric fluid being conserved: it was not created or destroyed,
only exchanged from one body to another. This makes sense, inaw a y : i fi tw a sr e a l l ys o m e t h i n g
material, how could it be created or destroyed? Conservationo fm a t t e rw a sp r e t t ym u c ha c c e p t e d
scientific “dogma” already by the end of the 18th century.
This idea of conservation of the caloric fluid led to the wholefield of “calorimetry,” as essentially
aw a yt ot r yt oq u a n t i f y( t h a ti s ,m e a s u r e )t h ea m o u n to f“ c a l oric” that materials would take",University Physics I Classical Mechanics.pdf
"aw a yt ot r yt oq u a n t i f y( t h a ti s ,m e a s u r e )t h ea m o u n to f“ c a l oric” that materials would take
in or give off. The connection with temperature led directly tothe definition of heat capacities
and specific heats, just as I have introduced them above (in section 2.1); only instead of “change
in energy” you would use “change in caloric content.” This would be measured in units, called
calories, defined by the amount of caloric that led to a given temperature change in a reference
substance, such as water.
To be precise, let 1 calorie be the amount of “caloric” neededto raise the temperature of one gram
of water by one degree Celsius at a pressure of one atmosphere.T h i s m a k e s t h e s p e c i fi c h e a t o f
water, by definition, 1 calorie/◦C· gram. Now imagine you place a hot object in a container of water,
insulated from the rest of the world, and wait until thermal equilibrium is reached. Then you can",University Physics I Classical Mechanics.pdf
"insulated from the rest of the world, and wait until thermal equilibrium is reached. Then you can
calculate the “amount of caloric that flowed into the water,”from the change in its temperature,
and if you assume that all this came from the hot object then youc a nc a l c u l a t ei t sh e a tc a p a c i t y( i n
calories/◦C) from the change inits temperature. By proceeding in this fashion, scientists developed
tables of specific heats that do not need any change today—onlyt h er e c o g n i t i o nt h a t“ c a l o r i c ”i s
not really a fluid at all, but a form of energy, and can, therefore, be measured in energy units.
Clearly, conservation of caloric was a very good idea in its own way, since much of what was
established back then still works if you simply replace the word “caloric” or “heat” by “thermal
energy.” It was, however, ultimately unsatisfactory precisely because it restricted itself to what we",University Physics I Classical Mechanics.pdf
"energy.” It was, however, ultimately unsatisfactory precisely because it restricted itself to what we
would today recognize as just one kind of energy, and so it failed to recognize thermal energy as
something that could be converted into, or from, other kindsof energy.
In hindsight, it is a bit surprising that the belief in the conservation of caloric could have held for
so long. What today appear to us like obvious instances of thetransformation of (macroscopic)
mechanical energy into thermal energy, such as the warmth generated when you rub two objects
together, were explained away as instances of mechanically“squeezing” caloric fluid out of the
objects. Around the turn of the 19th century, an American expatriate, Count Rumford, observed
one of the most egregious instances of this in the enormous amount of “heat” that was generated in
the boring of cannons (which involved, basically, a huge metal tool drilling a hole in a large metal",University Physics I Classical Mechanics.pdf
"the boring of cannons (which involved, basically, a huge metal tool drilling a hole in a large metal
cylinder). He noticed that the total mass of the metal, including all the shavings, did not appear
to change in the process, and concluded that caloric had to bevirtually massless, since enormous
quantities of it could be “squeezed” out without an appreciable mass loss. He speculated that",University Physics I Classical Mechanics.pdf
"13.3. HEAT AND THE FIRST LAW 311
caloric was not a fluid at all, but rather “a form of motion,” since only something like that could
be made to increase without any apparent limit.
Rumford’s theory was not generally accepted at the time, butlater in the 19th century the direct
conversion of mechanical energy into thermal energy was established beyond a doubt by James
Prescott Joule in a series of painstaking experiments in which he used a system of weights to turn
some vanes, or paddles, that stirred water in a container andeventually caused its temperature
to rise. By measuring the mechanical energy deficit (gravitational plus kinetic) of his system of
weights and paddles, he could tell how much energy the water must have gained, and by measuring
the water’s change in temperature he could then establish thee q u i v a l e n t“ a m o u n to fc a l o r i c ”t h a t
had gone into it. He thus established what was called “the mechanical equivalent of heat,” which",University Physics I Classical Mechanics.pdf
"had gone into it. He thus established what was called “the mechanical equivalent of heat,” which
we would express today by saying that a calorie does not measure the amount of some (nonexistent)
caloric fluid, but simply an amount of energy equal to 4.18 joules (and yes, the Joule is named after
him!).
13.3.2 The first law of thermodynamics
The upshot of all this experimentation was the full development of the concept of energy as a
conserved quantity that manifested itself in different ways and could be “converted” among different
kinds. To the observation, already familiar from macroscopic mechanics, that the energy of a system
could be changed by doing work on it (or letting it do work on itse n v i r o n m e n t )w a sa d d e dt h e
observation, coming from thermal physics, thatthermal energy could also be directly exchanged
between two objects merely by placing them in contact, without any macroscopic work being",University Physics I Classical Mechanics.pdf
"between two objects merely by placing them in contact, without any macroscopic work being
involved. The two things taken together led to the principleof conservation of energy in its most
general (pre-relativistic) form:
∆E = W + Q (13.8)
which says simply that a change in the total energy of a systemmay result from work (W)o rf r o m
“heat exchange” (Q). “Heat,” in physics usage today, is simply what we call the thermal energy
that is directly transferred from one object to another, typically by contact; the convention used
for this term is the same as for the work term, that is,Q is positive if thermal energy flows into
the system and negative if thermal energy leaves the system.
Equation (13.8)i st h efirst law of thermodynamics.N o t e t h a t , i n t e r m s o fQ,t h ep r e c i s ed e fi n i t i o n
of a system’s heat capacity isC = Q/∆T,a n ds ot h i sw i l lo n l yb ee q u a lt o∆E/∆t when the system",University Physics I Classical Mechanics.pdf
"of a system’s heat capacity isC = Q/∆T,a n ds ot h i sw i l lo n l yb ee q u a lt o∆E/∆t when the system
does no work, which is why I was careful to include that condition in the derivation of Eq. (13.2).",University Physics I Classical Mechanics.pdf
"312 CHAPTER 13. THERMODYNAMICS
13.4 The second law and entropy
The second law of thermodynamics is really little more than aformal statement of the observation
that heat always flows spontaneously from a warmer to a colderobject, and never in reverse.
More precisely, consider two systems, at different temperatures, that can exchange heat with each
other but are otherwise isolated from the rest of the world. The second law states that under those
conditions the heat will only flow from the warmer to the coldero n e .
The closure of the system—its isolation from any sources of energy—is important in the above
statement. It is certainly possible to build devices that will remove heat from a relatively cold
place (like the inside of your house on a hot summer day) and exhaust it to a warmer environment.
These devices are called refrigerators or heat pumps, and them a i nt h i n ga b o u tt h e mi st h a tt h e y",University Physics I Classical Mechanics.pdf
"These devices are called refrigerators or heat pumps, and them a i nt h i n ga b o u tt h e mi st h a tt h e y
need to be plugged in to operate: that is, they require an external energy source.
If you have an energy source, then, youcan move heat from a colder to a warmer object. To avoid
unnecessary complications and loopholes (what if the energys o u r c ei sab a t t e r yt h a ti sp h y s i c a l l y
inside your “closed” system?) an alternative formulation oft h eb a s i cp r i n c i p l e ,d u et oC l a u s i u s ,
goes as follows:
No process is possible whosesole result is the transfer of heat from a cooler to a hotter body
The words “sole result” are meant to imply that in order to accomplish this “unnatural” transfer
of heat you must draw energy from some source, and so you must be, in some way,depleting that
source (the battery, for instance). On the other hand, for ther e v e r s e ,“ s p o n t a n e o u s ”p r o c e s s — t h e",University Physics I Classical Mechanics.pdf
"source (the battery, for instance). On the other hand, for ther e v e r s e ,“ s p o n t a n e o u s ”p r o c e s s — t h e
flow from hotter to cooler—no such energy source is necessary.
Am a t h e m a t i c a lw a yt of o r m u l a t et h es e c o n dl a ww o u l db ea sf o llows. Consider two systems, in
thermal equilibrium at temperaturesT1 and T2,t h a ty o up l a c ei nc o n t a c ts ot h e yc a ne x c h a n g e
heat. For simplicity, assume that exchange of heat is all thath a p p e n s ;n ow o r ki sd o n ee i t h e rb y
the systems or on them, and no heat is transferred to or from theo u t s i d ew o r l de i t h e r . T h e n ,i f
Q1 and Q2 are the amounts of heat gained by each system, we must have, bythe conservation of
energy, Q2 = −Q1,s oo n eo ft h e s ei sp o s i t i v ea n dt h eo t h e ro n ei sn e g a t i v e ,a n d,b yt h es e c o n d
law, the system with the positiveQ (the one that gains thermal energy) must be the colder one.",University Physics I Classical Mechanics.pdf
"law, the system with the positiveQ (the one that gains thermal energy) must be the colder one.
This is ensured by the following inequality:
Q1(T2 − T1) ≥ 0( 1 3 . 9 )
So, if T2 >T 1, Q1 must be positive, and ifT1 >T 2, Q1 must be negative. (The equal sign is
there to allow for the case in whichT1 = T2,i nw h i c hc a s et h et w os y s t e m sa r ei n i t i a l l yi nt h e r m a l
equilibrium already, and no heat transfer takes place.)",University Physics I Classical Mechanics.pdf
"13.4. THE SECOND LAW AND ENTROPY 313
Equation (13.9)i sv a l i dr e g a r d l e s so ft h et e m p e r a t u r es c a l e . I fw eu s et h eKelvin scale, in which all
the temperatures are positive3,w ec a nr e w r i t ei tb yd i v i d i n gb o t hs i d e sb yt h ep r o d u c tT1T2,a n d
using Q2 = −Q1,a s
Q1
T1
+ Q2
T2
≥ 0( 1 3 . 1 0 )
This more symmetric statement of the second law is a good starting point from which to introduce
the concept ofentropy,w h i c hIw i l lp r o c e e dt od on e x t .
13.4.1 Entropy
In Equations (13.9)a n d(13.10), we have takenT1 and T2 to be the initial temperatures of the two
systems, but in general, of course, these temperatures willchange during the heat transfer process.
It is useful to consider an “infinitesimal” heat transfer,dQ,s os m a l lt h a ti tl e a d st oan e g l i g i b l e
temperature change, and then define thechange in the system’s entropyby
dS = dQ
T (13.11)",University Physics I Classical Mechanics.pdf
"temperature change, and then define thechange in the system’s entropyby
dS = dQ
T (13.11)
Here, S denotes a new system variable, the entropy, which is implicitly defined by Eq. (13.11).
That is to say, suppose you take a system from one initial statet oa n o t h e rb ya d d i n go rr e m o v i n g
as e r i e so fi n fi n i t e s i m a la m o u n t so fh e a t .W et a k et h ec h a n g eine n t r o p yo v e rt h ew h o l ep r o c e s st o
be
∆S = Sf − Si =
∫ f
i
dQ
T (13.12)
Starting from an arbitrary state, we could use this to find theentropy for any other state, at least
up to a (probably) unimportant constant (a little like what happens with the energy: the absolute
value of the energy does not typically matter, it is only the energy differences that are meaningful).
This may be easier said than done, though; there is noap r i o r iguarantee that any two arbitrary
states of a system could be connected by a process for which (13.12)c o u l db ec a l c u l a t e d ,a n d",University Physics I Classical Mechanics.pdf
"states of a system could be connected by a process for which (13.12)c o u l db ec a l c u l a t e d ,a n d
conversely, it might also happen that two states could be connected by several possible processes,
and the integral in (13.12) would have different values for all those. In other words, there is no
guarantee that the entropy thus defined will be a truestate function—something that is uniquely
determined by the other variables that characterize a system’s state in thermal equilibrium.
Nevertheless, it turns out that it is possible to show that thei n t e g r a l(13.12)i si n d e e di n d e p e n d e n t
of the “path” connecting the initial and final states, at leasta sl o n ga st h ep h y s i c a lp r o c e s s e s
considered are “reversible” (a constraint that basically amounts to the requirement that heat be
exchanged, and work done, only in small increments at a time,so that the system never departs",University Physics I Classical Mechanics.pdf
"exchanged, and work done, only in small increments at a time,so that the system never departs
3For a while in the 1970’s some people were very excited by the concept of negative absolute temperatures, but
that is mostly an artificial contrivance used to describe systems that are not really in thermal equilibrium anyway.",University Physics I Classical Mechanics.pdf
"314 CHAPTER 13. THERMODYNAMICS
too far from a state of thermal equilibrium). I will not attempt the proof here, but merely note
that this provides the following, alternative formulationof the second law of thermodynamics:
For every system in thermal equilibrium, there is a state function, the entropy, with the property
that it can never decrease for a closed system.
You can see how this covers the case considered in the previouss e c t i o n ,o ft w oo b j e c t s ,1a n d2 ,i n
thermal contact with each other but isolated from the rest ofthe world. If object 1 absorbs some
heat dQ1 while at temperatureT1 its change in entropy will bedS1 = dQ1/T1,a n ds i m i l a r l yf o r
object 2. The total change in the entropy of the closed systemformed by the two objects will then
be
dStotal = dS1 + dS2 = dQ1
T1
+ dQ2
T2
(13.13)
and the requirement that this cannot be negative (that is,Stotal must not decrease) is just the same
as Eq. (13.10), in differential form.",University Physics I Classical Mechanics.pdf
"T2
(13.13)
and the requirement that this cannot be negative (that is,Stotal must not decrease) is just the same
as Eq. (13.10), in differential form.
Once again, this simply means that the hotter object gives offthe heat and the colder one absorbs
it, but when you look at it in terms of entropy it is a bit more interesting than that. You can see
that the entropy of the hotter object decreases (negativedQ), and that of the colder one increases
(positive dQ), butby a different amount:i n f a c t , i t i n c r e a s e s s o m u c h t h a t i t m a k e s t h e t o t a l c h a n g e
in entropy for the system positive. This shows that entropy is rather different from energy (which
is simply conserved in the process). You can always make it increase just by letting a process “take
its normal course”—in this case, just letting the heat flow from the warmer to the colder object
until they reach thermal equilibrium with each other (at which point, of course, the entropy will",University Physics I Classical Mechanics.pdf
"until they reach thermal equilibrium with each other (at which point, of course, the entropy will
stop increasing, since it is a function of the state and the state will no longer change).
Although not immediately obvious from the above, the absolute (or Kelvin) temperature scale plays
an essential role in the definition of the entropy, in the senset h a to n l yi ns u c has c a l e( o ra n o t h e r
scale linearly proportional to it) is the entropy, as defined by Eq. (13.12), a state variable; that is,
only when using such a temperature scale is the integral (13.12)p a t h - i n d e p e n d e n t . T h ep r o o fo f
this (which is much too complicated to even sketch here) relies essentially on the Carnot principle,
to be discussed next.
13.4.2 The efficiency of heat engines
By the beginning of the 19th century, an industrial revolution was underway in England, due
primarily to the improvements in the efficiency of steam engines that had taken place a few decades",University Physics I Classical Mechanics.pdf
"primarily to the improvements in the efficiency of steam engines that had taken place a few decades
earlier. It was natural to ask how much this efficiency could ultimately be increased, and in 1824, a
French engineer, Nicolas Sadi Carnot, wrote a monograph thatp r o v i d e da na n s w e rt ot h i sq u e s t i o n .",University Physics I Classical Mechanics.pdf
"13.4. THE SECOND LAW AND ENTROPY 315
Carnot modeled a “heat engine” as an abstract machine that worked in a cycle. In the course of
each cycle, the engine would take in an amount of heatQh from a “hot reservoir,” give off (or
“exhaust”) an amount of heat|Qc| to a “cold reservoir,” and produce an amount of work|W|.( I
am using absolute value bars here because, from the point of view of the engine,Qc and W must
be negative quantities.) At the end of the cycle, the engine should be back to its initial state,
so ∆Eengine =0 . T h eh o ta n dc o l dr e s e r v o i r sw e r es u p p o s e dt ob es y s t e m sw ith very large heat
capacities, so that the change in their temperatures as theytook in or gave off the heat from or to
the engine would be negligible.
If ∆Eengine =0 ,w em u s th a v e
∆Eengine = Qh + Qc + W = Qh −| Qc|−| W| =0 ( 1 3 . 1 4 )
that is, the work produced by the engine must be
|W| = Qh −| Qc| (13.15)",University Physics I Classical Mechanics.pdf
"∆Eengine = Qh + Qc + W = Qh −| Qc|−| W| =0 ( 1 3 . 1 4 )
that is, the work produced by the engine must be
|W| = Qh −| Qc| (13.15)
The energy input to the engine isQh,s oi ti sn a t u r a lt od e fi n et h ee ffi c i e n c ya sϵ = |W|/Qh;t h a t
is to say, the Joules of work done per Joule of heat taken in. A value of ϵ =1w o u l dm e a na n
efficiency of 100%, that is, the complete conversion of thermale n e r g yi n t om a c r o s c o p i cw o r k . B y
Eq. (13.15), we have
ϵ = |W|
Qh
= Qh −| Qc|
Qh
=1 − |Qc|
Qh
(13.16)
which shows thatϵ will always be less than 1 as long as the heat exhausted to the cold reservoir,
Qc,i sn o n z e r o . T h i si sa l w a y sn e c e s s a r i l yt h ec a s ef o rs t e a me ngines: the steam needs to be cooled
off at the end of the cycle, so a new cycle can start again.
Carnot considered a hypothetical “reversible” engine (sometimes called aCarnot machine), which",University Physics I Classical Mechanics.pdf
"Carnot considered a hypothetical “reversible” engine (sometimes called aCarnot machine), which
could be runbackwards,w h i l ei n t e r a c t i n gw i t ht h es a m et w or e s e r v o i r s . I nb a c k w a rds mode, the
machine would work as a refrigerator or heat pump. It would take in an amount of workW per cycle
(from some external source) and use that to absorb the amountof heat|Qc| from the cold reservoir
and dump the amountQh to the hot reservoir. Carnot argued thatno heat engine could have a
greater efficiency than a reversible one working between the same heat reservoirs,a n d ,c o n s e q u e n t l y ,
that all reversible engines, regardless of their composition, would have the same efficiency when
working in between the same temperatures.H i s a r g u m e n t w a s b a s e d o n t h e o b s e r v a t i o n t h a t a
hypothetical engine with a greater efficiency than the reversible one could be used to drive a",University Physics I Classical Mechanics.pdf
"hypothetical engine with a greater efficiency than the reversible one could be used to drive a
reversible one in refrigerator mode, to produce as the sole result the transfer of some net amount
of heat from the cold to the hot reservoir4,s o m e t h i n gt h a tw ea r g u e di nS e c t i o n1s h o u l db e
impossible.
4The greater efficiency engine could produce the same amount ofwork as the reversible one while absorbing less
heat from the hot reservoir and dumping less heat to the cold one. If all the work output of this engine were used
to drive the reversible one in refrigerator mode, the resultwould be, therefore, a net flow of heat out of the cold one
and a net flow of heat into the hot one.",University Physics I Classical Mechanics.pdf
"316 CHAPTER 13. THERMODYNAMICS
What makes this result more than a theoretical curiosity is the fact that an ideal gas would, in fact,
provide a suitable working substance for a Carnot machine, ifp u tt h r o u g ht h ef o l l o w i n gc y c l e( t h e
so-called “Carnot cycle”): an isothermal expansion, followed by an adiabatic expansion, then an
isothermal compression, and finally an adiabatic compression. What makes this ideally reversible
is the fact that the heat is exchanged with each reservoir onlyw h e nt h eg a si sa t( n e a r l y )t h es a m e
temperature as the reservoir itself, so by just “nudging” thet e m p e r a t u r eu po rd o w nal i t t l eb i ty o u
can get the exchange to go either way. When the ideal gas laws are used to calculate the efficiency
of such a machine, the result (theCarnot efficiency)i s
ϵC =1 − Tc
Th
(13.17)
where the temperatures must be measured in degrees Kelvin, the natural temperature scale for an
ideal gas.",University Physics I Classical Mechanics.pdf
"ϵC =1 − Tc
Th
(13.17)
where the temperatures must be measured in degrees Kelvin, the natural temperature scale for an
ideal gas.
It is actually easy to see the connection between this resultand the entropic formulation of the
second law presented above. Suppose for a moment that Carnot’s principle does not hold, that is
to say, that we can build an engine withϵ>ϵ C =1 − Tc/Th.S i n c e(13.16)m u s th o l di na n yc a s e
(because of conservation of energy), we find that this would imply
1 − |Qc|
Qh
> 1 − Tc
Th
(13.18)
and then some very simple algebra shows that
− Qh
Th
+ |Qc|
Tc
< 0( 1 3 . 1 9 )
But now consider the total entropy of the system formed by theengine and the two reservoirs. The
engine’s entropy does not change (because it works in a cycle); the entropy of the hot reservoir goes
down by an amount−Qh/Th;a n dt h ee n t r o p yo ft h ec o l dr e s e r v o i rg o e sup by an amount|Qc|/Tc.",University Physics I Classical Mechanics.pdf
"down by an amount−Qh/Th;a n dt h ee n t r o p yo ft h ec o l dr e s e r v o i rg o e sup by an amount|Qc|/Tc.
So the left-hand side of Eq. (13.19)a c t u a l l ye q u a l st h et o t a lc h a n g ei ne n t r o p y ,a n dE q .(13.19)i s
telling us that this change isnegative (the total entropy goesdown)d u r i n gt h eo p e r a t i o no ft h i s
hypothetical heat engine whose efficiency is greater than theCarnot limit (13.17). Since this is
impossible (the total entropy of a closed system can never decrease), we conclude that the Carnot
limit must always hold.
As you can see, the seemingly trivial observation with whichIs t a r t e dt h i ss e c t i o n( n a m e l y ,t h a t
heat always flows spontaneously from a hotter object to a colder object, and never in reverse) turns
out to have profound consequences. In particular, it means that the complete conversion of thermal
energy into macroscopic work is essentially impossible5,w h i c hi sw h yw et r e a tm e c h a n i c a le n e r g y",University Physics I Classical Mechanics.pdf
"energy into macroscopic work is essentially impossible5,w h i c hi sw h yw et r e a tm e c h a n i c a le n e r g y
as “lost” once it is converted to thermal energy. By Carnot’stheorem, to convert some of that
5At least it is impossible to do using a device that runs in a cycle. For a one-use-only, you might do something
like pump heat into a gas and allow it to expand, doing work as itd o e ss o ,b u te v e n t u a l l yy o uw i l lr u no u to fr o o m
to do your expanding into...",University Physics I Classical Mechanics.pdf
"13.4. THE SECOND LAW AND ENTROPY 317
thermal energy back to work we would need to introduce a colderr e s e r v o i r( a n dt a k ea d v a n t a g e ,s o
to speak, of the natural flow of heat from hotter to colder), andt h e nw ew o u l do n l yg e tar e l a t i v e l y
small conversion efficiency, unless the cold reservoir is really at a very low Kelvin temperature (and
to create such a cold reservoir would typically require refrigeration, which again consumes energy).
It is easy to see that Carnot efficiencies for reservoirs closeto room temperature are rather pitiful.
For instance, ifTh =3 0 0Ka n dTc =2 7 3K ,t h eb e s tc o n v e r s i o ne ffi c i e n c yy o uc o u l dg e tw o u l db e
0.09, or 9%.
13.4.3 But what IS entropy, anyway?
The existence of this quantity, the entropy, which can be measured or computed (up to an arbitrary
reference constant) for any system in thermal equilibrium,is one of the great discoveries of 19th",University Physics I Classical Mechanics.pdf
"reference constant) for any system in thermal equilibrium,is one of the great discoveries of 19th
century physics. There are tables of entropies that can be putt om a n yu s e s( f o ri n s t a n c e ,i n
chemistry, to figure out which reactions will happen spontaneously and which ones will not), and
one could certainly take the point of view that those tables,plus the basic insight that the total
entropy can never decrease for a closed system, are all one needs to know about it. From this
perspective, entropy is just a convenient number that we canassign to any equilibrium state of any
system, which gives us some idea of which way it is likely to goif the equilibrium is perturbed.
Nonetheless, it is natural for a physicist to ask to what, exactly, does this number correspond?
What property of the equilibrium state is actually capturedby this quantity? Especially, in the
context of a microscopic description, since that is, by and large, how physicists have always been",University Physics I Classical Mechanics.pdf
"context of a microscopic description, since that is, by and large, how physicists have always been
trying to explain things, by breaking them up into little pieces, and figuring out what the pieces
were doing. What are the molecules or atoms of a system doing inas t a t eo fh i g he n t r o p yt h a ti s
different from a state of low entropy?
The answer to this question is provided by the branch of physics known as Statistical Mechanics,
which today is mostly quantum-based (since you need quantummechanics to describe most of what
atoms or molecules do, anyway), but which started in the context of pure classical mechanics in
the mid-to-late 1800’s and, despite this handicap, was actually able to make surprising headway
for a while.
From this microscopic, but still classical, perspective (which applies, for instance, moderately well
to an ideal gas), the entropy can be seen as a measure of thespread in the velocities and positions",University Physics I Classical Mechanics.pdf
"to an ideal gas), the entropy can be seen as a measure of thespread in the velocities and positions
of the molecules that make up the system. If you think of a probability distribution, it has a mean
value and a standard deviation. In statistical mechanics, the molecules making up the system
are described statistically, by giving the probability thatt h e ym i g h th a v eac e r t a i nv e l o c i t yo rb e
at some point or another. These probability distributions may be very narrow (small standard
deviation), if you are pretty certain of the positions or thevelocities, or very broad, if you are
not very certain at all, or rather expect the actual velocities and positions to be spread over a",University Physics I Classical Mechanics.pdf
"318 CHAPTER 13. THERMODYNAMICS
considerable range of values. A state of large entropy corresponds to a broad distribution, and a
state of small entropy to a narrow one.
For an ideal gas, the temperature determines both the averagem o l e c u l a rs p e e da n dt h es p r e a do f
the velocity distribution. This is because the average velocity is zero (since it is just as likely to
be positive or negative), so the only way to make the average speed (or root-mean-square speed)
large is to have a broad velocity distribution, which makes large speeds comparatively more likely.
Then, as the temperature increases, so does the range of velocities available to the molecules, and
correspondingly the entropy. Similarly (but more simply), for a given temperature, a gas that
occupies a smaller volume will have a smaller entropy, sincethe range of positions available to the
molecules will be smaller.
These considerations may help us understand an important property of entropy, which is that",University Physics I Classical Mechanics.pdf
"molecules will be smaller.
These considerations may help us understand an important property of entropy, which is that
it increases in all irreversible processes. To begin with, note that this makes sense, since, by
definition, these are processes that do not “reverse” spontaneously. If a process involves an increase
in the total entropy of a closed system, then the reverse process will not happen, because it would
require a spontaneous decrease in entropy, which the secondlaw forbids. But, moreover, we can
see the increase in entropy directly in many of the irreversible processes we have considered this
semester, such as the ones involving friction. As I just pointed out above, in general, we may expect
that increasing the temperature of an object will increase its entropy (other things being equal),
regardless of how the increase in temperature comes about. Now, when mechanical energy is lost",University Physics I Classical Mechanics.pdf
"regardless of how the increase in temperature comes about. Now, when mechanical energy is lost
due to friction, the temperature ofboth of the objects (surfaces) involved increases, so the total
entropy will increase as well. That marks the process as irreversible.
Another example of an irreversible process might be the mixing of two gases (or of two liquids,
like cream and coffee). Start with all the “brown” molecules tothe left of a partition, and all the
“white” molecules to the right. After you remove the partition, the system will reach an equilibrium
state in which the range of positions available to both the brown and white molecules has increased
substantially—and this is, according to our microscopic picture, a state of higher entropy (other
things, such as the average molecular speeds, being equal6).
For quantum mechanical systems, where the position and velocity are not simultaneously well",University Physics I Classical Mechanics.pdf
"For quantum mechanical systems, where the position and velocity are not simultaneously well
defined variables, one uses the more abstract concept of “state” to describe what each molecule is
doing. The entropy of a system in thermal equilibrium is thendefined as a measure of the total
number of states available to its microscopic components, compatible with the constraints that
determine the macroscopic state (such as, again, total energy, number of particles, and volume).
6In the case of cream and coffee, the average molecular speeds will not be equal—the cream will be cold and the
coffee hot—but the resulting exchange of heat is just the kindof process I described at the beginning of the chapter,
and we have seen that it, too, results in an increase in the total entropy.",University Physics I Classical Mechanics.pdf
"13.5. IN SUMMARY 319
13.5 In summary
1. Temperature is a statistical quantity that provides a (typically indirect) measure of thecon-
centration of thermal energy in a system. For a system that is (approximately) well described
by classical mechanics, the temperature, as measured by a conventional thermometer, is di-
rectly proportional to the average translational kinetic energy per molecule.
2. In a process in which a system does no work, a change in the system’s temperature is related
to a change in its total internal energy (which typically includes more than just translational
kinetic energy contributions) by ∆E = C∆T,w h e r eC is the system’sheat capacity for the
process.
3. The transfer of thermal energy between two systems withoute i t h e ro n ed o i n gm a c r o s c o p i c
work on each other is generally possible. Thermal energy transferred in this way is called
heat,a n dd e n o t e db yt h es y m b o lQ.",University Physics I Classical Mechanics.pdf
"work on each other is generally possible. Thermal energy transferred in this way is called
heat,a n dd e n o t e db yt h es y m b o lQ.
4. The actual definition of a system’s heat capacity isC = Q/∆T.F o r a h o m o g e n e o u s s y s t e m
(made of just one substance),C = mc,w h e r em is the system’s mass andc the substance’s
specific heat. Specific heats typically depend on temperaturei nn o n t r i v i a lw a y s .
5. Two systems isolated from the rest of the world but allowedto exchange thermal energy with
each other will eventually reach a state ofthermal equilibrium in which their temperatures
will be the same (zero-th law of thermodynamics).
6. The work done on (or by) a system by (or on) its environment,plus the heat given to (or
taken from) the system by its environment, always equals thenet change in the system’s total
energy (conservation of energy, or first law of thermodynamics; Eq. (13.8)).",University Physics I Classical Mechanics.pdf
"energy (conservation of energy, or first law of thermodynamics; Eq. (13.8)).
7. For any system in thermal equilibrium, there exists a statev a r i a b l e ,c a l l e dentropy,w i t ht h e
property that it can never decrease for a closed system. Whenas y s t e ma tt e m p e r a t u r eT
takes in a small amount of heatdQ,i t sc h a n g ei ne n t r o p yi sg i v e nb ydS = dQ/T.
8. This principle of never-decreasing entropy is equivalentt ot h es t a t e m e n tt h a t“ N op r o c e s si s
possible whose sole result is the transfer of heat from a cooler to a hotter body.”
9. The principle 7. is also equivalent to Carnot’s theorem, which states that “it is impossible for
an engine that operates in a cycle, taking in heat from a hot reservoir at temperatureTh and
exhausting heat to a cold reservoir at temperatureTc,t od ow o r kw i t ha ne ffi c i e n c yg r e a t e r
than 1− Tc/Th.”
10. Either one of 7., 8., or 9., above, may be regarded as an equivalent statement of thesecond",University Physics I Classical Mechanics.pdf
"than 1− Tc/Th.”
10. Either one of 7., 8., or 9., above, may be regarded as an equivalent statement of thesecond
law of thermodynamics.
11. Carnot’s theorem shows the limitations inherent in the conversion of thermal energy into
macroscopic work, which is the reason why one usually regardsm e c h a n i c a le n e r g yt h a ti s
converted into thermal energy as “lost.”",University Physics I Classical Mechanics.pdf
"320 CHAPTER 13. THERMODYNAMICS
12. Microscopically, the entropy of a system is a measure of the range of distinct states available
to its microscopic components (atoms or molecules) that arecompatible with the set of
macroscopic constraints that determine its thermal equilibrium state. More entropy means a
greater range of possible “microstates.”
13. Entropy always increases in irreversible processes.",University Physics I Classical Mechanics.pdf
"13.6. EXAMPLES 321
13.6 Examples
13.6.1 Calorimetry
The specific heat of aluminum is 900 J/kg· K, and that of water is 4186 J· K. Suppose you drop a
block of aluminum of mass 1 kg at a temperature of 80◦Ci nal i t e ro fw a t e r( w h i c ha l s oh a sa
mass of 1 kg) at a temperature of 20◦C. What is the final temperature of the system, assuming no
exchange of heat with the environment takes place? How much energy does the aluminum lose/the
water gain?
Solution
Let us callTAl the initial temperature of the aluminum,Twater the initial temperature of the water,
and Tf their final common temperature. The thermal energy given off byt h ea l u m i n u me q u a l s
∆EAl = CAl(Tf − TAl)( t h i sf o l l o w sf r o mt h ed e fi n i t i o n(13.1)o fh e a tc a p a c i t y ;w ec o u l de q u a l l y
well call this quantity “the heat given off by the aluminum”).In the same way, the thermal energy
change of the water (heat absorbed by the water) equals ∆Ewater = Cwater(Tf −Twater). If the total",University Physics I Classical Mechanics.pdf
"change of the water (heat absorbed by the water) equals ∆Ewater = Cwater(Tf −Twater). If the total
system is closed, the sum of these two quantities, each with its appropriate sign, must be zero:
0=∆ EAl +∆ Ewater = CAl(Tf − TAl)+ Cwater(Tf − Twater)( 1 3 . 2 0 )
This equation forTf has the solution
Tf = CAlTAl + CwaterTwater
CAl + Cwater
(13.21)
As you can see, the result is a weighted average of the two starting temperatures, with the corre-
sponding heat capacities as the weighting factors.
The heat capacities C are equal to the given specific heats multiplied by the respective masses.
In this case, the mass of aluminum and the mass of the water arethe same, so they will cancel
in the final result. Also, we can use the temperatures in degrees Celsius, instead of Kelvin. This
is not immediately obvious from the final expression (13.21), but if you look at (13.20)y o u ’ l ls e e
it involves only temperature differences, and those have the same value in the Kelvin and Celsius
scales.",University Physics I Classical Mechanics.pdf
"it involves only temperature differences, and those have the same value in the Kelvin and Celsius
scales.
Substituting the given values in (13.21), then, we get
Tf = 900 × 80 + 4186× 20
900 + 4186 =3 0.6◦C( 1 3 . 2 2 )
This is much closer to the initial temperature of the water, ase x p e c t e d ,s i n c ei th a st h eg r e a t e r
heat capacity. The amount of heat exchanged is
Cwater(Tf − Twater)=4 1 8 6× (30.6 − 20) = 44, 440 J = 44.4k J ( 1 3 . 2 3 )",University Physics I Classical Mechanics.pdf
"322 CHAPTER 13. THERMODYNAMICS
So, 1 kg of aluminum gives off 44.4k J o f t h e r m a l e n e r g y a n d i t s t e m pe r a t u r e d r o p s a l m o s t 5 0◦C,
from 80◦Ct o3 0.6◦C, whereas 1 kg of water takes in the same amount of thermal energy and its
temperature only rises about 10.6◦C.
13.6.2 Equipartition of energy
Estimate the speed of an oxygen molecule in air at room temperature (about 300 K).
Solution
Recall that in Section 13.2.2 I mentioned that the average translational kinetic energy of a molecule
in a system at a temperatureT is 3
2 kBT (Eq. (13.7), where kB,B o l t z m a n n ’ sc o n s t a n t ,i se q u a lt o
1.38 × 10−23 J/K. So, atT =3 0 0 K ,am o l e c u l eo fo x y g e n( o ro fa n y t h i n ge l s e ,f o rt h a tm a tter)
should have, on average, a kinetic energy of
⟨Ktrans⟩ = 3
2kBT = 3
2 × 1.38 × 10−23 × 300 J = 6.21 × 10−21 J( 1 3 . 2 4 )
Since K = 1
2 mv2,w ec a nfi g u r eo u tt h ea v e r a g ev a l u eo fv2 if we know the mass of an oxygen",University Physics I Classical Mechanics.pdf
"Since K = 1
2 mv2,w ec a nfi g u r eo u tt h ea v e r a g ev a l u eo fv2 if we know the mass of an oxygen
molecule. This is something you can look up, or derive like this: One mole of oxygen atoms has
am a s so f1 6g r a m s( 1 6i st h ea t o m i cm a s sn u m b e ro fo x y g e n )a n dcontains Avogadro’s number
of atoms, 6.02 × 1023.S o a s i n g l e a t o m h a s a m a s s o f 0.016 kg/6.02 × 1023 =2 .66 × 10−26 kg. A
molecule of oxygen contains two atoms, so it has twice the mass, m =5 .32 × 10−26 kg. Then,
⟨v2⟩ = 2⟨Ktrans⟩
m = 2 × 6.21 × 10−21 J
5.32 × 10−26 kg =2 .33 × 105 m2
s2 (13.25)
The square root of this will give us what is called the “root mean square” velocity, orvrms:
vrms =
√
2.33 × 105 m2
s2 =4 8 3m
s (13.26)
This is of the same order of magnitude as (but larger than) thespeed of sound in air at room
temperature (about 340 m/s, as you may recall from Chapter 12).",University Physics I Classical Mechanics.pdf
"13.7. PROBLEMS 323
13.7 Problems
Problem 1
Consider a system of two objects in contact, one initially hotter than the other, so they may
directly exchange thermal energy, in isolation from the resto ft h ew o r l d .A c c o r d i n gt ot h el a w so f
thermodynamics, what must happen to the system’s total energy and entropy? (Do they change,
increase, decrease, stay constant... ?)
Problem 2
Consider the same two objects in Problem 1 and suppose the heatc a p a c i t yo ft h ec o l d e ro b j e c t
is much greater than the heat capacity of the hotter one. When the system reaches thermal
equilibrium, will its final temperature will be closer to theinitial temperature of the hot object,
the colder object, or exactly halfway between the two initialt e m p e r a t u r e s ?W h y ?
Problem 3
Which of the following isnot av a l i df o r m u l a t i o no ft h es e c o n dl a wo ft h e r m o d y n a m i c s ?",University Physics I Classical Mechanics.pdf
"Problem 3
Which of the following isnot av a l i df o r m u l a t i o no ft h es e c o n dl a wo ft h e r m o d y n a m i c s ?
(a) For any system in thermal equilibrium, there exists a state variable, called entropy, with the
property that it can never decrease for a closed system.
(b) No process is possible whose sole result is the transfer ofh e a tf r o mac o o l e rt oah o t t e rb o d y .
(c) It is impossible for an engine that operates in a cycle, taking in heat from a hot reservoir at
temperature Th and exhausting heat to a cold reservoir at temperatureTc,t od ow o r kw i t ha n
efficiency greater than 1− Tc/Th.
(d) The entropy of any system goes to zero asT (the absolute, or Kelvin) temperature goes to zero.
Problem 4
Which of the following statements is true?
(a) Once the entropy of a system increases, it is impossible tob r i n gi tb a c kd o w n .
(b) Once some amount of mechanical energy is converted to thermal energy, it is impossible to turn",University Physics I Classical Mechanics.pdf
"(b) Once some amount of mechanical energy is converted to thermal energy, it is impossible to turn
any of it back into mechanical energy.
(c) It is always possible to reduce the entropy of a system, fori n s t a n c e ,b yc o o l i n gi t .
(d) All of the above statements are true.
(e) None of the above statements are true.
Other questions
• Can you tell the temperature of a gas by measuring the translational kinetic energy of a single
molecule?
• Does a shuffled deck of cards have more or less entropy (in the thermodynamic sense) than
an identical, ordered set of cards? Assume they are at the samet e m p e r a t u r e .",University Physics I Classical Mechanics.pdf
"324 CHAPTER 13. THERMODYNAMICS
• Ad i a t o m i cg a sm o l e c u l e ,s u c ha sO2,c a ns t o r ek i n e t i ce n e r g yi nt h ef o r mo fv i b r a t i o n sa n d
rotations, in addition to just translation of the center of mass. By contrast, a monoatomic
gas molecule such asC has virtually no kinetic energy (at normal temperatures) other than
translational kinetic energy. Which kind of gas do you expectt oh a v eal a r g e rm o l a rh e a t
capacity (heat capacity per molecule)?",University Physics I Classical Mechanics.pdf
"Michael Biehl
The 
Shallow 
and the 
Deep
A biased 
introduction 
to neural networks 
and old school 
machine learning
The Shallow and the Deep is a collection of lecture notes that offers an 
accessible introduction to neural networks and machine learning in general. 
However, it was clear from the beginning that these notes would not be able to 
cover this rapidly changing and growing field in its entirety. The focus lies on 
classical machine learning techniques, with a bias towards classification and 
regression. Other learning paradigms and many recent developments in, for 
instance, Deep Learning are not addressed or only briefly touched upon.
Biehl argues that having a solid knowledge of the foundations of the field is 
essential, especially for anyone who wants to explore the world of machine 
learning with an ambition that goes beyond the application of some software 
package to some data set. Therefore, The Shallow and the Deep places",The+Shallow+and+the+Deep.pdf
"learning with an ambition that goes beyond the application of some software 
package to some data set. Therefore, The Shallow and the Deep places 
emphasis on fundamental concepts and theoretical background. This also 
involves delving into the history and pre-history of neural networks, where the 
foundations for most of the recent developments were laid.
These notes aim to demystify machine learning and neural networks without 
losing the appreciation for their impressive power and versatility./uni00A0
Michael Biehl is Associate Professor of Computer 
Science at the Bernoulli Institute for Mathematics, 
Computer Science and Artificial Intelligence of the 
University of Groningen, where he joined the Intelligent 
Systems group in 2003. He also holds an honorary 
Professorship of Machine Learning at the Center for 
Systems Modelling and Quantitative Biomedicine of the 
University of Birmingham, UK. 
His research focuses on the modelling and theoretical",The+Shallow+and+the+Deep.pdf
"Systems Modelling and Quantitative Biomedicine of the 
University of Birmingham, UK. 
His research focuses on the modelling and theoretical 
understanding of neural networks and machine learning in general. The 
development of efficient training algorithms for interpretable, transparent 
systems is a topic of particular interest. A variety of interdisciplinary 
collaborations concern practical applications of machine learning in the 
biomedical domain, in astronomy and other areas.
T/h.smcp/e.smcp S/h.smcp/a.smcp/l.smcp/l.smcp/o.smcp/w.smcp /a.smcp/n.smcp/d.smcp /t.smcp/h.smcp/e.smcp D/e.smcp/e.smcp/p.smcp   Michael Biehl",The+Shallow+and+the+Deep.pdf
The Shallow and the Deep,The+Shallow+and+the+Deep.pdf
"The Shallow and the Deep
A biased introduction to neural 
networks and old school machine 
learning
Michael Biehl",The+Shallow+and+the+Deep.pdf
"Published by University of Groningen Press 
Broerstraat 4 
9712 CP Groningen 
The Netherlands
First published in the Netherlands © 2023 Michael Biehl, Bernoulli Institute for Mathematics, 
Computer Science and Artificial Intelligence, Groningen
Comments, corrections and suggestions are welcome, contact: m.biehl@rug.nl 
Please cite as: Biehl, M. (2023). The Shallow and the Deep: A biased introduction to neural networks 
and old school machine learning. University of Groningen Press.
This book has been published open access thanks to the financial support of the Open Access 
Textbook Fund of  the University of Groningen.
Cover design: Bas Ekkers 
Coverphoto: Michael Biehl 
Production: LINE UP boek en media bv 
ISBN (print) 9789403430287 
ISBN (ePDF) 9789403430270 
DOI https://doi.org/ 10.21827/648c59c1a467e
This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0",The+Shallow+and+the+Deep.pdf
"DOI https://doi.org/ 10.21827/648c59c1a467e
This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 
International License. The full licence terms are available at creativecommons.org/ licenses/
by-nc-sa/4.0/legalcode",The+Shallow+and+the+Deep.pdf
"Preface
Stop calling everything AI.
— Michael I. Jordan, in [Pre21]
The subtitle of these lecture notes is “Abiased introduction to neural networks
and old school machine learning” for good reasons. Although the aim was to
give an accessible introduction to the field, it has been clear from the beginning
that it would not end up as a comprehensive, complete overview. The focus is
on classical machine learning, many recent developments cannot be covered.
Personally, I first got in touch with neural networks in my early life as a
physicist. At the time, it was sufficient to read a handful of papers and perhaps
a little later the good book [HKP91] to be up-to-date and able to contribute a
piece to the big puzzle. Am I exaggerating and somewhat nostalgic? Probably.
But the situation has definitely changed a lot. Nowadays, an overwhelming
flood of publications makes it difficult to filter out the relevant information and
keep up with the developments.",The+Shallow+and+the+Deep.pdf
"flood of publications makes it difficult to filter out the relevant information and
keep up with the developments.
The selection of topics in these notes has been determined to a large extent
by my own research interests and early experiences. This is definitely true for
the initial focus on the simple perceptron, the hydrogen atom of neural network
research as Manfred Opper put it [Opp90]. Moreover, the bulk of this text deals
with shallow systems for supervised learning, in particular classification, which
reflects my main interest in the field.
The notes may be perceived as old school, certainly by some dedicated fol-
lowers of fashion [Dav66]. Admittedly, the text does not address the most
recent developments in e.g. Deep Learning and its applications. However, in
my humble opinion it is invaluable to have a solid background knowledge of the
basics before exploring the world of machine learning with an ambition that
goes beyond the application of some software package to some data set.",The+Shallow+and+the+Deep.pdf
"basics before exploring the world of machine learning with an ambition that
goes beyond the application of some software package to some data set.
Therefore, the emphasis is on basic concepts and theoretical background,
with specific aspects selected from a personal and clearly biased viewpoint. In
a sense, the goal is to de-mystify machine learning and neural networks without
iii",The+Shallow+and+the+Deep.pdf
"iv
losing the appreciation for their fascinating power and versatility. Very often,
this involves a look into the history and pre-history of neural networks, where
the foundations for most of the recent developments were laid.
I have aimed at pointing the interested reader to many resources for further
exploration of the area. Therefore, the list of references in the bibliography,
although by no means complete, is slightly more extensive than initially envi-
sioned.
The starting point for these notes was the desire to provide more comprehen-
sive material than the presentation slides in the MSc level course Neural Net-
works (renamed Neural Networks and Computational Intelligence later) which
I have been giving at the University of Groningen. A thorough archeological
investigation of the text and figures would also reveal traces of the courses The-
orie Neuronaler Netzwerke and Unüberwachtes Lernen, that I taught way back
when in the Physics program at the University of Würzburg.",The+Shallow+and+the+Deep.pdf
"orie Neuronaler Netzwerke and Unüberwachtes Lernen, that I taught way back
when in the Physics program at the University of Würzburg.
My writing activity was greatly boosted on the occasion of the wonderful
30th Canary Islands Winter School in 2018, devoted to Big Data analysis in
Astronomy, where I had the honor to give a series of lectures on supervised
learning, see [MSK19,Bie19] for course materials and video-recorded lectures.
Last not least I would like to acknowledge constructive feedback from several
“generations” of students who followed the course and from many colleagues
and collaborators. In particular, I thank Elisa Oostwal and Janis Norden for a
critical reading of the manuscript and many suggestions for improvements.
Groningen, June 2023
The mysterious machine learning machine
© Catharina M. Gerigk and Elina L. van den Brandhof
Reproduced with kind permission of the artists.",The+Shallow+and+the+Deep.pdf
"Contents
Preface iii
1 From neurons to networks 1
1.1 Spiking neurons and synaptic interactions . . . . . . . . . . . . . 3
1.2 Firing rate models . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Neural activity and synaptic interaction . . . . . . . . . . 5
1.2.2 Sigmoidal activation functions . . . . . . . . . . . . . . . 6
1.2.3 Hebbian learning . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Network architectures . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3.1 Attractor networks and the Hopfield model . . . . . . . . 10
1.3.2 Feed-forward layered neural networks . . . . . . . . . . . 12
1.3.3 Other architectures . . . . . . . . . . . . . . . . . . . . . . 15
2 Learning from example data 17
2.1 Learning scenarios . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.1.1 Unsupervised learning . . . . . . . . . . . . . . . . . . . . 17
2.1.2 Supervised learning . . . . . . . . . . . . . . . . . . . . . 19",The+Shallow+and+the+Deep.pdf
"2.1.1 Unsupervised learning . . . . . . . . . . . . . . . . . . . . 17
2.1.2 Supervised learning . . . . . . . . . . . . . . . . . . . . . 19
2.1.3 Other learning scenarios . . . . . . . . . . . . . . . . . . . 22
2.2 Machine Learning vs. Statistical Modelling . . . . . . . . . . . . 23
2.2.1 Differences and commonalities . . . . . . . . . . . . . . . 23
2.2.2 An example case: linear regression . . . . . . . . . . . . . 24
2.2.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3 The Perceptron 31
3.1 History and literature . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2 Linearly separable functions . . . . . . . . . . . . . . . . . . . . . 33
3.3 The Rosenblatt perceptron . . . . . . . . . . . . . . . . . . . . . 36
3.3.1 The perceptron storage problem . . . . . . . . . . . . . . 36
3.3.2 Iterative Hebbian training algorithms . . . . . . . . . . . 37
3.3.3 The Rosenblatt perceptron algorithm . . . . . . . . . . . 39",The+Shallow+and+the+Deep.pdf
"3.3.2 Iterative Hebbian training algorithms . . . . . . . . . . . 37
3.3.3 The Rosenblatt perceptron algorithm . . . . . . . . . . . 39
3.3.4 The perceptron algorithm as gradient descent . . . . . . . 41
3.3.5 The Perceptron Convergence Theorem . . . . . . . . . . . 42
3.3.6 A few remarks . . . . . . . . . . . . . . . . . . . . . . . . 45
3.4 The capacity of a hyperplane . . . . . . . . . . . . . . . . . . . . 46
v",The+Shallow+and+the+Deep.pdf
"vi CONTENTS
3.4.1 The number of linearly separable dichotomies . . . . . . . 46
3.4.2 Discussion of the result . . . . . . . . . . . . . . . . . . . 52
3.4.3 Time for a pizza or some cake . . . . . . . . . . . . . . . . 53
3.5 Learning a linearly separable rule . . . . . . . . . . . . . . . . . . 55
3.5.1 Student-teacher scenario . . . . . . . . . . . . . . . . . . . 55
3.5.2 Learning in version space . . . . . . . . . . . . . . . . . . 57
3.5.3 Generalization begins where storage ends . . . . . . . . . 60
3.5.4 Optimal generalization . . . . . . . . . . . . . . . . . . . . 62
3.6 The perceptron of optimal stability . . . . . . . . . . . . . . . . . 63
3.6.1 The stability criterion . . . . . . . . . . . . . . . . . . . . 63
3.6.2 The MinOver algorithm . . . . . . . . . . . . . . . . . . . 65
3.7 Optimal stability by quadratic optimization . . . . . . . . . . . . 67
3.7.1 Optimal stability reformulated . . . . . . . . . . . . . . . 67",The+Shallow+and+the+Deep.pdf
"3.7 Optimal stability by quadratic optimization . . . . . . . . . . . . 67
3.7.1 Optimal stability reformulated . . . . . . . . . . . . . . . 67
3.7.2 The Adaptive Linear Neuron - Adaline . . . . . . . . . . . 68
3.7.3 The Adaptive Perceptron Algorithm - AdaTron . . . . . . 73
3.7.4 Support vectors . . . . . . . . . . . . . . . . . . . . . . . . 78
3.8 Inhom. lin. sep. functions revisited . . . . . . . . . . . . . . . . . 80
3.9 Some remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
4 Beyond linear separability 83
4.1 Perceptron with errors . . . . . . . . . . . . . . . . . . . . . . . . 85
4.1.1 Minimal number of errors . . . . . . . . . . . . . . . . . . 85
4.1.2 Soft margin classifier . . . . . . . . . . . . . . . . . . . . . 87
4.2 Layered networks of perceptron-like units . . . . . . . . . . . . . 90
4.2.1 Committee and parity machines . . . . . . . . . . . . . . 91
4.2.2 The parity machine: a universal classifier . . . . . . . . . 92",The+Shallow+and+the+Deep.pdf
"4.2.1 Committee and parity machines . . . . . . . . . . . . . . 91
4.2.2 The parity machine: a universal classifier . . . . . . . . . 92
4.2.3 The capacity of machines . . . . . . . . . . . . . . . . . . 95
4.3 Support Vector Machines . . . . . . . . . . . . . . . . . . . . . . 97
4.3.1 Non-linear transformation to higher dimension . . . . . . 98
4.3.2 Large margin classifier . . . . . . . . . . . . . . . . . . . . 99
4.3.3 The kernel trick . . . . . . . . . . . . . . . . . . . . . . . 100
4.3.4 A few remarks . . . . . . . . . . . . . . . . . . . . . . . . 104
5 Feed-forward networks for regression and classification 107
5.1 Feed-forward networks as non-linear function approximators . . . 107
5.1.1 Architecture and input-output relation . . . . . . . . . . . 108
5.1.2 Universal approximators . . . . . . . . . . . . . . . . . . . 109
5.2 Gradient based training of feed-forward nets . . . . . . . . . . . . 114
5.2.1 Computing the gradient: Backpropagation of Error . . . . 116",The+Shallow+and+the+Deep.pdf
"5.2 Gradient based training of feed-forward nets . . . . . . . . . . . . 114
5.2.1 Computing the gradient: Backpropagation of Error . . . . 116
5.2.2 Batch gradient descent . . . . . . . . . . . . . . . . . . . . 116
5.2.3 Stochastic gradient descent . . . . . . . . . . . . . . . . . 119
5.2.4 Practical aspects and modifications . . . . . . . . . . . . . 122
5.3 Objective functions . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.3.1 Cost functions for regression . . . . . . . . . . . . . . . . 124
5.3.2 Cost functions for classification . . . . . . . . . . . . . . . 125
5.4 Activation functions . . . . . . . . . . . . . . . . . . . . . . . . . 127",The+Shallow+and+the+Deep.pdf
"CONTENTS vii
5.4.1 Sigmoidal and related functions . . . . . . . . . . . . . . . 127
5.4.2 One-sided and unbounded activation functions . . . . . . 128
5.4.3 Exponential and normalized activations . . . . . . . . . . 130
5.4.4 Remark: universal function approximation . . . . . . . . . 131
5.5 Specific architectures . . . . . . . . . . . . . . . . . . . . . . . . . 131
5.5.1 Popular shallow networks . . . . . . . . . . . . . . . . . . 132
5.5.2 Deep and convolutional neural networks . . . . . . . . . . 135
6 Distance-based classifiers 141
6.1 Prototype-based classifiers . . . . . . . . . . . . . . . . . . . . . . 143
6.1.1 Nearest Neighbor and Nearest Prototype Classifiers . . . 143
6.1.2 Learning Vector Quantization . . . . . . . . . . . . . . . . 144
6.1.3 LVQ training algorithms . . . . . . . . . . . . . . . . . . . 145
6.2 Distance measures and relevance learning . . . . . . . . . . . . . 148
6.2.1 LVQ beyond Euclidean distance . . . . . . . . . . . . . . 148",The+Shallow+and+the+Deep.pdf
"6.2 Distance measures and relevance learning . . . . . . . . . . . . . 148
6.2.1 LVQ beyond Euclidean distance . . . . . . . . . . . . . . 148
6.2.2 Adaptive distances in relevance learning . . . . . . . . . . 149
6.3 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 153
7 Model evaluation and regularization 155
7.1 Bias and variance, over- and underfitting . . . . . . . . . . . . . 155
7.1.1 Decomposition of the error . . . . . . . . . . . . . . . . . 156
7.1.2 The bias-variance dilemma . . . . . . . . . . . . . . . . . 158
7.1.3 Beyond the classical bias-variance trade-off (?) . . . . . . 161
7.2 Controlling the network complexity . . . . . . . . . . . . . . . . . 163
7.2.1 Early stopping . . . . . . . . . . . . . . . . . . . . . . . . 163
7.2.2 Weight decay and related concepts . . . . . . . . . . . . . 164
7.2.3 Constructive algorithms . . . . . . . . . . . . . . . . . . . 167
7.2.4 Pruning . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167",The+Shallow+and+the+Deep.pdf
"7.2.3 Constructive algorithms . . . . . . . . . . . . . . . . . . . 167
7.2.4 Pruning . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
7.2.5 Weight-sharing . . . . . . . . . . . . . . . . . . . . . . . . 169
7.2.6 Dropout . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
7.3 Cross-validation and related methods . . . . . . . . . . . . . . . . 171
7.3.1 n-fold cross-validation and related schemes . . . . . . . . 172
7.3.2 Model and parameter selection . . . . . . . . . . . . . . . 175
7.4 Performance measures . . . . . . . . . . . . . . . . . . . . . . . . 175
7.4.1 Measures for regression . . . . . . . . . . . . . . . . . . . 176
7.4.2 Measures for classification . . . . . . . . . . . . . . . . . . 176
7.4.3 Receiver Operating Characteristics . . . . . . . . . . . . . 177
7.4.4 The area under the ROC curve . . . . . . . . . . . . . . . 180
7.4.5 Alternative measures for two-class problems . . . . . . . . 181",The+Shallow+and+the+Deep.pdf
"7.4.4 The area under the ROC curve . . . . . . . . . . . . . . . 180
7.4.5 Alternative measures for two-class problems . . . . . . . . 181
7.4.6 Multi-class problems . . . . . . . . . . . . . . . . . . . . . 182
7.4.7 Averages of class-wise quality measures . . . . . . . . . . 183
7.5 Interpretable systems . . . . . . . . . . . . . . . . . . . . . . . . . 185",The+Shallow+and+the+Deep.pdf
"viii CONTENTS
8 Preprocessing and unsupervised learning 187
8.1 Normalization and transformations . . . . . . . . . . . . . . . . . 188
8.1.1 Coordinate-wise transformations . . . . . . . . . . . . . . 189
8.1.2 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . 191
8.2 Dimensionality reduction . . . . . . . . . . . . . . . . . . . . . . 192
8.2.1 Low-dimensional embedding . . . . . . . . . . . . . . . . . 194
8.2.2 Multi-dimensional Scaling . . . . . . . . . . . . . . . . . . 194
8.2.3 Neighborhood Embedding . . . . . . . . . . . . . . . . . . 195
8.2.4 Feature selection . . . . . . . . . . . . . . . . . . . . . . . 196
8.3 PCA and related methods . . . . . . . . . . . . . . . . . . . . . . 197
8.3.1 Principal Component Analysis . . . . . . . . . . . . . . . 198
8.3.2 PCA by Hebbian learning . . . . . . . . . . . . . . . . . . 200
8.3.3 Independent Component Analysis . . . . . . . . . . . . . 203
8.4 Clustering and Vector Quantization . . . . . . . . . . . . . . . . 204",The+Shallow+and+the+Deep.pdf
"8.3.3 Independent Component Analysis . . . . . . . . . . . . . 203
8.4 Clustering and Vector Quantization . . . . . . . . . . . . . . . . 204
8.4.1 Basic clustering methods . . . . . . . . . . . . . . . . . . 205
8.4.2 Competitive learning for Vector Quantization . . . . . . . 206
8.4.3 Practical issues and extensions of VQ . . . . . . . . . . . 208
8.5 Density estimation . . . . . . . . . . . . . . . . . . . . . . . . . . 211
8.5.1 Parametric density estimation . . . . . . . . . . . . . . . . 211
8.5.2 Gaussian Mixture Models . . . . . . . . . . . . . . . . . . 212
8.6 Missing values and imputation techniques . . . . . . . . . . . . . 215
8.6.1 Approaches without explicit imputation . . . . . . . . . . 216
8.6.2 Imputation based on available data . . . . . . . . . . . . . 216
8.7 Over- and undersampling, augmentation . . . . . . . . . . . . . . 218
8.7.1 Weighted cost functions . . . . . . . . . . . . . . . . . . . 218",The+Shallow+and+the+Deep.pdf
"8.7 Over- and undersampling, augmentation . . . . . . . . . . . . . . 218
8.7.1 Weighted cost functions . . . . . . . . . . . . . . . . . . . 218
8.7.2 Undersampling . . . . . . . . . . . . . . . . . . . . . . . . 218
8.7.3 Oversampling . . . . . . . . . . . . . . . . . . . . . . . . . 219
8.7.4 Data augmentation . . . . . . . . . . . . . . . . . . . . . . 220
Concluding quote 223
A Optimization 225
A.1 Multi-dimensional Taylor expansion . . . . . . . . . . . . . . . . 225
A.2 Local extrema and saddle points . . . . . . . . . . . . . . . . . . 227
A.2.1 Necessary and sufficient conditions . . . . . . . . . . . . . 227
A.2.2 Example: unsolvable systems of linear equations . . . . . 228
A.3 Constrained optimization . . . . . . . . . . . . . . . . . . . . . . 230
A.3.1 Equality constraints . . . . . . . . . . . . . . . . . . . . . 230
A.3.2 Example: under-determined linear equations . . . . . . . 231
A.3.3 Inequality constraints . . . . . . . . . . . . . . . . . . . . 232",The+Shallow+and+the+Deep.pdf
"A.3.2 Example: under-determined linear equations . . . . . . . 231
A.3.3 Inequality constraints . . . . . . . . . . . . . . . . . . . . 232
A.3.4 The Wolfe Dual for convex problems . . . . . . . . . . . . 234
A.4 Gradient based optimization . . . . . . . . . . . . . . . . . . . . . 235
A.4.1 Gradient and directional derivative . . . . . . . . . . . . . 235
A.4.2 Gradient descent . . . . . . . . . . . . . . . . . . . . . . . 236
A.4.3 The gradient under coordinate transformations . . . . . . 238
A.5 Variants of gradient descent . . . . . . . . . . . . . . . . . . . . . 239",The+Shallow+and+the+Deep.pdf
"CONTENTS ix
A.5.1 Coordinate descent . . . . . . . . . . . . . . . . . . . . . . 239
A.5.2 Constrained problems and projected gradients . . . . . . 240
A.5.3 Stochastic gradient descent . . . . . . . . . . . . . . . . . 240
A.6 Example calculation of a gradient . . . . . . . . . . . . . . . . . . 243
List of figures 246
List of algorithms 247
Abbrev. and acronyms 248
Bibliography 250
1. Abandon the idea that you are ever going to finish.
— John Steinbeck, Six Writing Tips",The+Shallow+and+the+Deep.pdf
x CONTENTS,The+Shallow+and+the+Deep.pdf
"Chapter 1
From neurons to networks
Reality is overrated anyway.
— Unknown
To understand and explain the brain’s fascinating capabilities1 remains one
of the greatest scientific challenges ever. This is particularly true for its plas-
ticity, i.e. the ability to learn from experience, to adapt to and to survive in
ever-changing environments.
Ultimately, the performance of the brain must rely on its hardware (or wet-
ware [LB89]) and emerges from the cooperative behavior of its many, relatively
simple, yet highly interconnected building blocks: the neurons. The human
cortex, for instance, comprises an estimated number of 1012 neurons and each
individual cell can be connected to thousands of others.
In this introduction to Neural Networks and Computational Intelligence we
will study artificial neural networks and related systems, designed for the pur-
pose of adaptive information processing. The degree to which these systems",The+Shallow+and+the+Deep.pdf
"pose of adaptive information processing. The degree to which these systems
relate to their biological counterparts is, generally speaking, quite limited. How-
ever, their development was greatly inspired by key aspects of biological neurons
and networks. Therefore, it is useful to be aware of the conceptual connections
between artificial and biological systems, at least on a basic level.
Quite often, technical solutions are inspired by natural systems without copy-
ing all their properties in detail. Due to biological constraints, nature (i.e. evolu-
tion) might have found highly complex solutions to certain problems that could
be dealt with in a simpler fashion in a technical realization. A somewhat over-
used example in this context is the construction of efficient aircraft, which by no
means requires the use of moving wings in order to imitate bird flight faithfully.
Of course, it is unclear a priori which of the details are essential and which",The+Shallow+and+the+Deep.pdf
"Of course, it is unclear a priori which of the details are essential and which
ones can be left out in artificial systems. Obviously, this also depends on the
1(including the capability of being fascinated)
1",The+Shallow+and+the+Deep.pdf
"2 1. FROM NEURONS TO NETWORKS
specific task and context. Consequently, the interaction between the neuro-
sciences and machine learning research continues to play an important role for
the further development of both.
In this introductory text we will consider learning systems, which draw on
only the most basic mechanisms. Therefore, this chapter is only meant as a very
brief overview, which should allow to relate some of the concepts in artificial
neural computation to their biological background. The reader should be aware
that the presentation is certainly over-simplifying and probably not quite up-
to-date in all aspects.
Recommended textbooks and other sources
In most of this introductory chapter, detailed citations concerning specific topics
will not be provided. Instead, the following list points the reader to selected
textbooks, reviews or lecture notes. They range from brief and superficial to
very comprehensive and detailed reviews of the biological background. The same",The+Shallow+and+the+Deep.pdf
"very comprehensive and detailed reviews of the biological background. The same
is true for the discussion of the different conceptual levels on which biological
systems can be modelled. References point to the full citation information in
the bibliography. Note that the selection is certainly incomplete and biased by
personal preferences.
◦ K. Guerney’s (Neural Networks) gives a very basic overview and provides
a glossary of biological or biologically inspired terms [Gue97].
◦ The first sections of Neural Networks and Learning Machines by S. Haykin
cover the relevant topics in slightly greater depth [Hay09].
◦ The classical textbook Neural Networks: An Introduction to the Theory of
Neural Computation by J.A. Hertz, A. Krogh and R.G. Palmer discusses
the inspiration from biological neurons and networks in the first chapters.
It also provides a thorough analysis of the Hopfield model from a statistical
physics perspective [HKP91].",The+Shallow+and+the+Deep.pdf
"It also provides a thorough analysis of the Hopfield model from a statistical
physics perspective [HKP91].
◦ H. Horner and R. Kühn give a brief general introduction in Neural Net-
works [HK98], including a basic discussion of the biological background.
◦ Models of biological neurons, their bio-chemistry and bio-physics are in
the focus of C. Koch’s comprehensive monograph on the Biophysics of
computation [Koc98]. It discusses the different modelling approaches and
relates them to experimental data obtained from real world neurons.
◦ T. Kohonen has introduced important prototype-based learning schemes.
An entire chapter of his seminal work Self-Organizing Maps is devoted to
the Justification of Neural Modeling [Koh97].
◦ H. Ritter, T. Martinetz and K. Schulten give an overview and also discuss
some aspects of the organization of the brain in terms of maps in their
monograph Neural Computation and Self-Organizing Maps [RMS92].",The+Shallow+and+the+Deep.pdf
"some aspects of the organization of the brain in terms of maps in their
monograph Neural Computation and Self-Organizing Maps [RMS92].
◦ M. van Rossum’s lecture notes on Neural Computation provide an overview
of biological information processing and models of neural activity, synaptic
interaction and plasticity. Moreover, modelling approaches are discussed
in some detail [Ros16].",The+Shallow+and+the+Deep.pdf
"1.1. SPIKING NEURONS AND SYNAPTIC INTERACTIONS 3
1.1 Spiking neurons and synaptic interactions
The physiology and functionality of the biological systems is highly complex,
already on the single neuron level. Sophisticated modelling frameworks have
been developed that take into account the relevant electro-chemical processes
in great detail in order to represent the biology as faithfully as possible. This
includes the famous Hodgkin-Huxley model and variants thereof.
They describe the state of cell compartments in terms of an electrostatic
potential, which is due to varying ion concentrations on both sides of the cell
membrane. A number of ion channels and pumps control the concentrations
and, thus, govern the membrane potential. The original Hodgkin-Huxley model
describes its temporal evolution in terms of four coupled ordinary differential
equations, the parameters of which can be fitted to experimental data measured
in real world neurons.",The+Shallow+and+the+Deep.pdf
"equations, the parameters of which can be fitted to experimental data measured
in real world neurons.
Whenever the membrane potential reaches a threshold value, for instance
triggered by the injection of an external current, a short, localized electrical
pulse is generated. The term action potential or the more sloppy spike will be
used synonymously. The neuron is said to fire when a spike is generated.
The action potential discharges the membrane locally and propagates along
the membrane. As illustrated in Figure 1.1 (left panel), a strongly elongated
extension is attached to the soma, the so-called axon. From a purely technical
point of view, it serves as a cable along which action potentials can travel.
Of course, the actual electro-chemical processes are significantly different
from the flow of electrons in a conventional copper cable, for instance. In fact,
action potentials jump between short gaps in the myelin sheath, an insulating",The+Shallow+and+the+Deep.pdf
"action potentials jump between short gaps in the myelin sheath, an insulating
layer around the axon. By means of saltatory conduction, action potentials
spread along the axonic branches of the firing neuron and eventually reach the
points where the branches connect to the dendrites of other neurons. Such
a connection, termed synapse, is shown schematically in Fig. 1.1 (right panel).
Upon arrival of a spike, so-called neuro-transmitters are released into the synap-
tic cleft, i.e. the gap between pre-synaptic axon branch and the post-synaptic
dendrite. The transmitters are received on the post-synaptic side by substance
specific receptors. Thus, in the synapse, the action potential is not transferred
directly through a physical contact point, but chemically.2 The effect that an ar-
riving spike has on the post-synaptic neuron depends on the detailed properties
of the synapse:
◦ If the synapse is of the excitatory type, the post-synaptic membrane po-",The+Shallow+and+the+Deep.pdf
"of the synapse:
◦ If the synapse is of the excitatory type, the post-synaptic membrane po-
tential increases upon arrival of the pre-synaptic spike,
◦ When a spike arrives at an inhibitory synapse, the post-synaptic mem-
brane potential decreases.
Both excitatory and inhibitory synapses can have varying strengths, as reflected
2 Note that also so-called gap junctions exist which can function as bi-directional electrical
synapses, see e.g. [CL04] for further information and references.",The+Shallow+and+the+Deep.pdf
"4 1. FROM NEURONS TO NETWORKS
Figure 1.1: Schematic illustration of neurons (pyramidal cells) and their con-
nections. Left: Pre-synaptic and post-synaptic neurons with soma, dendritic
tree, axon, and axonic branches. Right: The synaptic cleft with vesicles releas-
ing neuro-transmitters and corresponding receptors on the post-synaptic side.
Redrawn after [Kat66].
in the magnitude of the change that a spike imposes on the post-synaptic mem-
brane potential.
Consequently, the membrane potential of a particular cell will vary over time,
depending on the actual activities of the neurons it receives spikes from through
excitatory and inhibitory synapses. When the threshold for spike generation is
reached, the neuron fires itself and, thus, influences the potential and activity of
all its post-synaptic neighbors. All in all, a set of interconnected neurons forms
a complex dynamical system of threshold units which influence each other’s",The+Shallow+and+the+Deep.pdf
"a complex dynamical system of threshold units which influence each other’s
activity through generation and synaptic transmission of action potentials.
The origin of a very successful approach to the modelling of neuronal ac-
tivity dates back to Louis Lapicque in 1907. In the framework of the so-called
Integrate-and-Fire (IaF) model, electro-chemical details accounted for in the
Hodgkin-Huxley type of models are omitted (and were probably not known at
the time). The membrane is simply represented by its conductance and ohmic
resistance. All charge transport phenomena are combined in one effective elec-
tric current, which summarizes the individual contributions of changing ion
concentrations as well as leak currents through the membrane. Similarly, the
precise form of spikes and details of their generation and transport are ignored.
Instead, the firing is modelled as an all-or-nothing threshold process, which re-",The+Shallow+and+the+Deep.pdf
"Instead, the firing is modelled as an all-or-nothing threshold process, which re-
sults in an instantaneous discharge. A spike is represented by a structureless
Dirac delta function which defines the time point of the event. Despite its sim-
plicity compared to more realistic electro-chemical models, the IaF model can
be fitted to physiological data and yields a fairly realistic description of neuronal
activity.",The+Shallow+and+the+Deep.pdf
"1.2. FIRING RATE MODELS 5
time [ms]
Figure 1.2: Left (upper): Schematic illustration of an action potential, i.e.
a short pulse on mV - and ms-scale. Left (lower): Spikes travel along the axon
through saltatory conduction via gaps in the insulating myelin sheath. Right:
Schematic illustration of how mean firing rates are derived from a temporal
spike pattern.
1.2 Firing rate models
In another step of abstraction, the description of neural activity is simplified
by taking into account only the mean firing rate, e.g. obtained as the average
number of spikes per unit time; the concept is illustrated in Fig. 1.2 (right
panel).
The implicit assumption is that most of the information in neural process-
ing is contained in the mean activity and frequency of spikes of the neurons.
Hence, the precise timing of individual action potentials is completely disre-
garded. While the role of individual spike timing appears to be the topic of",The+Shallow+and+the+Deep.pdf
"garded. While the role of individual spike timing appears to be the topic of
ongoing debate in the neurosciences3, the simplification clearly facilitates effi-
cient simulations of very large networks of neurons and can be seen as the basis
of virtually all artificial neural networks and learning systems considered in this
text.
1.2.1 Neural activity and synaptic interaction
The firing rate picture allows for a simple mathematical description of neural
activity and synaptic interaction. Consider the mean activity Si of neuron i,
which receives input from a set J of neurons with j ∕= i. Taking into account the
fact that the firing rate of a biological neuron cannot exceed a certain maximum
due to physiological and bio-chemical constraints, we can limit Si to a range of
values 0 ≤ Si where the upper limit 1 is given in arbitrary units. The resting
state Si = 0 obviously corresponds to the absence of any spike generation.",The+Shallow+and+the+Deep.pdf
"state Si = 0 obviously corresponds to the absence of any spike generation.
The activity of neuron i is given as a (non-linear) response of incoming spikes,
which are - however - also represented only by the mean activities Sj: in
Si = h(xi) with xi =
󰁛
j∈J
wij Sj. (1.1)
3See, e.g., http://romainbrette.fr/category/blog/rate-vs-timing/ for further references.",The+Shallow+and+the+Deep.pdf
"6 1. FROM NEURONS TO NETWORKS
Here, the quantities wij ∈ R represent the strength of the synapse connecting
one neuron j ∈ J with neuron i. Positive wij > 0 increase the so-called local
potential xi if neuron j is active (Sj > 0), while wij < 0 contribute negative
terms to the weighted sum. Note that real world chemical synapses are strictly
uni-directional: even if connections wij and wji exist for a given pair of neurons,
they would be physiologically separate, independent entities.
1.2.2 Sigmoidal activation functions
A variety of different activation functions h(x) have been employed in artificial
neural networks. A few specific types of functions will be introduced in a later
chapter. Here we restrict the discussion to the by now classical sigmoidal acti-
vation which arguably captures important characteristics of biological systems.
It is plausible to assume the following mathematical properties of the acti-",The+Shallow+and+the+Deep.pdf
"It is plausible to assume the following mathematical properties of the acti-
vation function h(x) of a given neuron (subscript i omitted) with local potential
x as in Eq. (1.1):
lim
x→−∞
h(x) = 0 (resting state, absence of spike generation)
h′(x) ≥ 0 (monotonic increase of the excitation)
lim
x→+∞
h(x) = 1 (maximum possible firing rate), (1.2)
which takes into account the limitations of individual neural activity discussed
in the previous section.
Various activation or transfer functions have been suggested and considered
in the literature. In the context of feed-forward neural networks, we will discuss
several options in Sec. 5.4. A very important class of plausible activations is
given by so-called sigmoidal functions, one prominent4 example being
h(x) = 1
2
󰀕
1 + tanh
󰀅
γ(x − θ)
󰀆󰀖
(1.3)
which clearly satisfies the conditions given above. The two important param-
eters are the threshold θ, which localizes the steepest increase of activity, and",The+Shallow+and+the+Deep.pdf
"eters are the threshold θ, which localizes the steepest increase of activity, and
the gain parameter γ, which quantifies the slope. It is important to note that
θ does not directly correspond to the previously discussed threshold of the all-
or-nothing generation of individual spikes: it marks the characteristic value of
h at which the activation function is centered.
4Its popularity is partly due to the fact that the relation tanh′ = 1 − tanh2 facilitates a
very efficient computation of the derivative, see also Chapter 5.",The+Shallow+and+the+Deep.pdf
"1.2. FIRING RATE MODELS 7
Si γ
θ
xi = 󰁓
j wijSj
Si
θ
xi = 󰁓
j wijSj
Figure 1.3: Schematic illustration of symmetrized activation functions. Left:
A sigmoidal transfer function with gain γ and threshold θ in the symmetrized
representation, cf. Eq. (1.6). Right: The binary McCulloch Pitts activation as
obtained in the limit γ → ∞.
Symmetrized representation of activity
We will frequently consider a symmetrized description of neural activity in terms
of modified activation functions:
lim
x→−∞
g(x) = −1 (resting state, absence of spike generation)
g′(x) ≥ 0 (monotonic increase of the excitation)
lim
x→+∞
g(x) = 1 (maximum possible firing rate). (1.4)
An example activation analogous to Eq. (1.3) is
g(x) = tanh
󰀅
γ(x − θ)
󰀄
. (1.5)
At first sight, this appears to be just an alternative assignment of a value S = −1
to the resting state.
Note that in the original description with 0 < Sj < 1, a quiescent neuron
does not influence its postsynaptic neurons explicitly. However, keeping the",The+Shallow+and+the+Deep.pdf
"Note that in the original description with 0 < Sj < 1, a quiescent neuron
does not influence its postsynaptic neurons explicitly. However, keeping the
form of the activation as
Si = g(xi) with xi =
󰁛
j∈J
wij Sj (1.6)
implies that the absence of activity (Sj = −1) in neuron j can now increase the
firing rate of neuron i if connected through an inhibitory synapse wij < 0. This
and other mathematical subtleties are clearly biologically implausible which is
due to the somewhat artificial introduction of – in a sense – negative and positive
activities which are treated in a symmetrized fashion.
However, as we do not aim at describing biological reality, the above dis-
cussed symmetrization can be justified. In fact, it simplifies the mathematical
and computational treatment, and has contributed to, for instance, the fruitful
popularization of neural networks in the statistical physics community in the
1980s and 1990s.",The+Shallow+and+the+Deep.pdf
"8 1. FROM NEURONS TO NETWORKS
McCulloch Pitts neurons
Quite frequently, an even more drastic modification is considered: for infinite
gain γ → ∞ the sigmoidal activation becomes a step function, see Fig. 1.3 (right
panel) for an illustration. Eq. (1.5) for instance yields in this limit
g(x) = sign(x − θ) =
󰀝
+1 if x ≥ θ
−1 if x < θ. (1.7)
In this symmetrized version of a binary activation function, only two possible
states are considered: either the model neuron is totally quiescent (S = −1) or
it fires at maximum frequency, which is represented by S = +1.
The extreme abstraction to binary activation states without the flexibility
of a graded response was first discussed by McCulloch and Pitts in 1943, who
originally denoted the quiescent state by S = 0. The persisting popularity of
this model is due to its simplicity as well as its similarity to Boolean concepts
in conventional computing. In the following, we will frequently resort to binary",The+Shallow+and+the+Deep.pdf
"in conventional computing. In the following, we will frequently resort to binary
model neurons in the symmetrized version (1.7). In fact, the so-called percep-
tron, as discussed in Chapter 3, can be interpreted as a single McCulloch Pitts
unit which is connected to N input neurons.
1.2.3 Hebbian learning
Probably the most intriguing property of biological neural networks is their abil-
ity to learn. Instead of realizing only pre-wired functionalities, brains adapt to
their environment or - in higher level terms - they can learn from experience.
Many potential forms of plasticity and memory representation have been dis-
cussed in the literature, including the chemical storage of information or learning
through neurogenesis, i.e. the growth of new neurons.
A very popular and plausible paradigm of learning is synaptic plasticity. A
key mechanism, Hebbian Learning, is named after psychologist Donald Hebb,
who published his work The Organization of Behavior in 1949 [Heb49]. The",The+Shallow+and+the+Deep.pdf
"key mechanism, Hebbian Learning, is named after psychologist Donald Hebb,
who published his work The Organization of Behavior in 1949 [Heb49]. The
original hypothesis was formulated in terms of a pair of neurons, which are con-
nected through an excitatory synapse:
“When an axon of cell A is near enough to excite cell B and repeatedly
or persistently takes part in firing it, some growth process or metabolic change
takes place in one or both cells such that A’s efficiency, as one of the cells firing
B, is increased.”
This is known as Hebb’s law and sometimes rephrased as “Neurons that fire
together, wire together.” Hebbian Learning results in a memory effect which
favors the simultaneous activity of neurons A and B in the future. Hence, it
constitutes a form of learning through synaptic plasticity.
The question to which extent the Hebbian paradigm reflects the biological
reality of learning is subject of on-going debate. Alternative or complementing",The+Shallow+and+the+Deep.pdf
"reality of learning is subject of on-going debate. Alternative or complementing
mechanisms have been suggested, see [SVG+18] for a recent example. In the",The+Shallow+and+the+Deep.pdf
"1.3. NETWORK ARCHITECTURES 9
context of artificial neural networks, Hebbian synaptic plasticity provides a very
plausible basis for the representation of learning in the models.
In the mathematical framework of firing rate models presented in the previ-
ous section, we can express Hebbian Learning quite elegantly, assuming that the
synaptic change is simply proportional to the pre- and post-synaptic activity:
∆wAB ∝ SASB. (1.8)
Hence, the change ∆wAB of a particular synapse wAB depends only on locally
available information: the activities of the pre-synaptic (SB) and the post-
synaptic neuron (SA). For SA, SB > 0 this is quite close to the actual Hebbian
hypothesis.
The symmetrization with −1 < SA,B < +1 adds some biologically implau-
sible aspects to the picture. For instance, an excitatory synapse connecting A
and B would also be strengthened according to Eq. (1.8) if both neurons are
quiescent at the same time, since in this case SASB > 0. Similarly, high activ-",The+Shallow+and+the+Deep.pdf
"quiescent at the same time, since in this case SASB > 0. Similarly, high activ-
ity in A and low activity in B (or vice versa) with SASB < 0 would weaken an
excitatory or strengthen an inhibitory synapse. In Hebb’s original formulation,
however, only the presence of simultaneous activity should trigger changes of
the involved synapse. Moreover, the mathematical formalism in (1.8) facilitates
the possibility that an individual excitatory synapse can become inhibitory or
vice versa, which is also questionable from the biological point of view.
Many learning paradigms in artificial neural networks and other adaptive
systems can be interpreted as Hebbian Learning in the sense of the above dis-
cussion. Examples can be found in a variety of contexts, including supervised
and unsupervised learning, see Sec. 2 for working definitions of these terms.
Note that the actual interpretation of the term Hebbian Learning varies a",The+Shallow+and+the+Deep.pdf
"Note that the actual interpretation of the term Hebbian Learning varies a
lot in the literature. Occasionally, it is employed only in the context of unsuper-
vised learning, since feedback from the environment is quite generally assumed
to constitute non-local information. Here, we follow the wide-spread, rather
relaxed use of the term for learning processes which depend on the states of the
pre- and post-synaptic units as in Eq. (1.8).
Frequently, learning can be seen as the optimization of suitable costs which
are interpreted as a function of the network parameters, i.e. the synaptic
strengths or weights. As we will see, in many cases numerical optimization
procedures, which are for instance based on gradient descent, lead to update
rules for the weights that resemble Hebbian Learning to a large extent.
1.3 Network architectures
In the previous section we have considered types of model neurons which retain
certain aspects of their biological counterparts and allow for a mathematical",The+Shallow+and+the+Deep.pdf
"certain aspects of their biological counterparts and allow for a mathematical
formulation of neural activity, synaptic interactions, and learning. This enables
us to construct networks from, for instance, sigmoidal or McCulloch Pitts neu-
rons, and model or simulate the dynamics of neurons and/or learning processes
concerning the synaptic connections.",The+Shallow+and+the+Deep.pdf
"10 1. FROM NEURONS TO NETWORKS
In the following, only the most basic and clear-cut types of network architec-
tures are introduced and discussed, namely fully connected recurrent networks
and feed-forward layered networks. The possibilities for modifications of these
networks, as well as for hybrid and intermediate types are nearly endless. Some
more specific architectures will be introduced briefly later; in Section 5.5 and
Chapter 6 various shallow and deep networks will be addressed.
1.3.1 Attractor networks and the Hopfield model
Networks with very high or unstructured connectivity form dynamical systems
of neurons which influence each other through synaptic interaction. In a network
as shown in Figure 1.4 (left panel) the activity of a particular neuron depends
on its synaptic input. Considering discrete timesteps t one obtains an update
of the form
Si(t + 1) = g
󰀳
󰁃󰁛
j∈J
wij Sj(t)
󰀴
󰁄, (1.9)
where the sum is taken over all units j ∈ J which n
euron i receives input from",The+Shallow+and+the+Deep.pdf
"of the form
Si(t + 1) = g
󰀳
󰁃󰁛
j∈J
wij Sj(t)
󰀴
󰁄, (1.9)
where the sum is taken over all units j ∈ J which n
euron i receives input from
through a synapse wij ∕= 0. Eq. (1.9) can be interpreted as an update of all
neurons in parallel. Alternatively, units could be visited in a deterministic or
randomized sequential order. We will not discuss the subtle, yet important,
differences between parallel and sequential dynamics here and refer the reader
to the literature, e.g. [HKP91].
From an initial configuration S(0) at time t = 0 which comprises the indi-
vidual activities S(0) = (S1(0), S2(0), . . . , SN (0))⊤, the dynamics generates a
sequence of states S(t) which can be considered the system’s response to the
initial stimulus. The term recurrent networks has been coined for this type of
dynamical system.
One of the most extreme, clear-cut example of a recurrent architecture is
the fully connected Hopfield or Little-Hopfield model [Hop82,Lit74,HKP91]. A",The+Shallow+and+the+Deep.pdf
"One of the most extreme, clear-cut example of a recurrent architecture is
the fully connected Hopfield or Little-Hopfield model [Hop82,Lit74,HKP91]. A
Hopfield network comprises N neurons of the McCulloch Pitts type which are
fully connected by bi-directional synapses
wij = wji ∈ R (i, j = 1, 2, . . . N) with wii = 0 for all i. (1.10)
While the exclusion of explicit, non-zero self-interactions wii appears plausible,
the assumption of symmetric, bi-directional interactions clearly constitutes yet
another serious deviation from biological reality.
The dynamics of the binary units is given by
Si(t + 1) = sign
󰀳
󰁅󰁅󰁃
N󰁛
j=1
j∕=i
wijSj(t)
󰀴
󰁆󰁆󰁄. (1.11)
John Hopfield [Hop82] realized that the corresponding random sequential update
can be seen as a zero temperature Metropolis Monte Carlo dynamics which is",The+Shallow+and+the+Deep.pdf
"1.3. NETWORK ARCHITECTURES 11
⇒
S(t = 0) S(t ≫ 1)
Figure 1.4: Recurrent neural networks. Left: A network of N = 5 neurons
with partial connectivity and uni-directional synapses. Right: Pattern retrieval
from a noisy initial configuration in a Hopfield network of 2500 units, storing
100 activity patterns. Activities Sj = ±1 are shown as black and white ’pixels’,
respectively. Initially, 40% of the Sj are flipped with respect to the pattern. The
rightmost figure displays the system shortly before perfect retrieval is achieved.
governed by an energy function of the form
H(S(t)) = −
N󰁛
i,j=1
i<j
wij Si(t) Sj(t). (1.12)
The mathematical structure is analogous to the so-called Ising model in Statisti-
cal Physics, see [HKP91,Kob97] for background and further references. There,
the degrees of freedom Si = ±1 are typically termed spins and they repre-
sent microscopic magnetic moments. Ising-like systems have been considered",The+Shallow+and+the+Deep.pdf
"the degrees of freedom Si = ±1 are typically termed spins and they repre-
sent microscopic magnetic moments. Ising-like systems have been considered
in a variety of scientific contexts ranging from the formation of binary alloys
to abstract models of segregation in the social sciences. Positive weights wij
obviously favor pairs of aligned Si = Sj which reduce the total energy of the
system. For the modelling of magnetic materials one considers specific couplings
wij as motivated by the physical interactions. For instance, constant positive
wij = 1 are assumed in the so-called Ising ferromagnet, while randomly drawn
interactions are employed to model disordered magnetic materials, so-called spin
glasses [HKP91].
In the actual Hopfield model, however, synaptic weights wij are constructed
or learned in order to facilitate a specific form of information processing [Hop82].
From a given set of uncorrelated, N-dimensional activity patterns P = {ξµ}P
µ=1
with ξµ",The+Shallow+and+the+Deep.pdf
"From a given set of uncorrelated, N-dimensional activity patterns P = {ξµ}P
µ=1
with ξµ
i ∈ {−1, +1}, a weight matrix is constructed according to
wij = wji = 1
P
P󰁛
µ=1
ξµ
i ξµ
j for i ∕= j and wii = 0 for all i, (1.13)
where the constant pre-factor 1/P follows the convention in the literature. Ac-
cording to Eq. (1.13), we can interpret the weights as empirical averages over",The+Shallow+and+the+Deep.pdf
"12 1. FROM NEURONS TO NETWORKS
the data set. Improved versions of the weight matrix for correlated patterns are
also available. In principle, all perceptron training algorithms discussed later
could be applied (per neuron) in the Hopfield network as well.
The Hopfield network can operate as an auto-associative or content ad-
dressable memory: If the system is prepared in an initial state S(t = 0) which
differs from one of the patterns ξν ∈ P only for a limited fraction of neurons with
Si(0) = −ξν
i , the dynamics can retrieve the original ξµ from corrupted or noisy
information. Ideally, the temporal evolution under the updates (1.11) restores
the pattern nearly perfectly and S(t) approaches ξν for large t. The retrieval of
a stored pattern from a noisy initial state is illustrated in Fig. 1.4 (right panel).
Note that the two-dimensional arrangement of neurons serves purely illustrative
purposes. Since every neuron is connected to every other neuron, neighborhood",The+Shallow+and+the+Deep.pdf
"purposes. Since every neuron is connected to every other neuron, neighborhood
relations do not define a low-dimensional topology in the Hopfield model.
Successful retrieval of a stored pattern is only possible if the initial deviation
of S(0) from ξν is not too large. Moreover, only a limited number of patterns
can be stored and retrieved successfully. For random patterns with zero mean
activities ξµ
j = ±1, the statistical physics based theory of the Hopfield model
(valid in the limit N → ∞) shows that P ≤ αrN must be satisfied. The value
αr ≈ 0.14 marks the so-called capacity limit of the Hopfield model5.
Note that the weight matrix construction (1.13) can also be interpreted as
Hebbian Learning: Starting from a tabula rasa state of the synaptic strengths
with zero weights, a single term of the form ξµ
i ξµ
j is added for each activity
pattern, representing the neurons that are connected by synapse wij (and wji).
Hence, Eq. (1.13) can be written as an iteration",The+Shallow+and+the+Deep.pdf
"pattern, representing the neurons that are connected by synapse wij (and wji).
Hence, Eq. (1.13) can be written as an iteration
wij(0) = 0, wij(µ) = wij(µ − 1) + 1
P ξµ
i ξµ
j , (1.14)
where the incremental change of wij depends only on locally available informa-
tion and is of the form “pre-synaptic × post-synaptic activity.”
The Hopfield model serves as a prototypical example of highly connected
neural networks. Potential applications include pattern recognition and image
processing tasks. Perhaps more importantly, the model has provided many
theoretical and conceptual insights into neural computation and continues to
do so.
More general recurrent neural networks are applied in various domains that
require some sort of temporal or sequence-based information processing. This
includes, among others, robotics, speech or handwriting recognition.
1.3.2 Feed-forward layered neural networks
Throughout this reader, we will mainly deal with another clear-cut network",The+Shallow+and+the+Deep.pdf
"1.3.2 Feed-forward layered neural networks
Throughout this reader, we will mainly deal with another clear-cut network
architecture: layered feed-forward networks. In these systems, neurons are ar-
ranged in layers and information is processed in a well-defined direction.
5This critical value is often referred to as αc in the literature, but it should not be confused
with the storage capacity of feed-forward networks in, e.g., Sec. 4.4",The+Shallow+and+the+Deep.pdf
"1.3. NETWORK ARCHITECTURES 13
The left panel of Fig. 1.5 shows a schematic illustration of a feed-forward
architecture. A specific, single layer of units (the top layer in the illustration)
represents external input to the system in terms of neural activity. In the
biological context, one might think of the photoreceptors in the retina or other
sensory neurons which can be activated by external stimuli.
The state of the neurons in all other layers of the network is determined via
synaptic interactions, and activations of the form
S(k)
i = g
󰀳
󰁃󰁛
j
w(k)
ij S(k−1)
j
󰀴
󰁄. (1.15)
Here, the activity S(k)
i of neuron i in layer k is determined from the weighted sum
of activities in the previous layer (k−1) only: information contained in the input
is processed layer by layer. Ultimately, the last layer in the structure (bottom
layer in the illustration) represents the network’s output, i.e. its response to the
input or stimulus in the first layer. The illustration displays only a single output",The+Shallow+and+the+Deep.pdf
"input or stimulus in the first layer. The illustration displays only a single output
unit, but the extension to a layer of several outputs is straightforward.
The essential property of the feed-forward network is the directed informa-
tion processing: neurons receive only input from units in the previous layer. As
a consequence, the network can be interpreted as the parameterization of an in-
put/output relation, i.e. a mathematical function that maps the vector of input
activations to a single or several output values. This interpretation still holds if
nodes receive input from several previous layers, or in other words: connections
may “skip” layers. For the sake of clarity and simplicity, we will not consider
this option in the following.
The feed-forward property and interpretation as a simple input/output re-
lation is lost as soon as any form of feedback is present: inter-layer synapses or
backward connections feeding information into previous (“higher”)layers intro-",The+Shallow+and+the+Deep.pdf
"backward connections feeding information into previous (“higher”)layers intro-
duce feedback loops, making it necessary to describe the system in terms of its
full dynamics.
Neurons that do not communicate directly with the environment, i.e. all
units that are neither input nor output nodes, are termed hidden units (nodes,
neurons) which form hidden layers in the feed-forward architecture.
The right panel of Fig. 1.5 displays a more concrete example. The net-
work comprises one layer of hidden units, here with activities σk ∈ R, and
a single output S. The response of the system to an input configuration
ξ = (ξ1, ξ2, . . . , ξN ) ∈ RN is given as
S(ξ) = g
󰀣 K󰁛
k=1
vk σk
󰀤
= g
󰀳
󰁃
K󰁛
k=1
vk g
󰀕 N󰁛
j=1
wkj ξj
󰀖󰀴
󰁄. (1.16)
Here we assume, for simplicity, that all hidden and output nodes employ the
same activation function g(. . .). Obviously, this restriction can be relaxed by
defining layer-specific or even individual activation functions. In Eq. (1.16) the",The+Shallow+and+the+Deep.pdf
"14 1. FROM NEURONS TO NETWORKS
S(ξ)
ξj ∈ R
 ξj ∈ R
wkj ∈ R
σk = g
󰀓󰁓
j wkjξj
󰀔
vk ∈ R
S(ξ) = g
󰀃󰁓
k vkσk
󰀄
Figure 1.5: Feed-forward neural networks.6 Left: A multilayered architecture
with varying layer-size and a single output unit. Right: A convergent feed-
forward network with a layer of input neurons, one hidden layer, and a single
output unit.
quantities wkj denote the weights connecting the j-th input component to the k-
th hidden unit with k = 1, 2, . . . K, where K is the total number of hidden units
(in the example K = 3). These weights can be grouped in vectors wk ∈ RN ,
while the hidden-to-output connections are denoted by vk ∈ R.
Altogether, the architecture and connectivity, the activation function and its
parameters (gain, threshold etc.), and the set of all weights determine the actual
input/output function ξ ∈ RN → S(ξ) ∈ R parameterized by the feed-forward
network. Again, the extension to several output units, i.e. multi-dimensional",The+Shallow+and+the+Deep.pdf
"network. Again, the extension to several output units, i.e. multi-dimensional
function values, is conceptually straightforward.
Without going into details yet, we note that we control the function that is
actually implemented by setting the weights and other free parameters in the
network. If their determination is guided by a set of example data representing
a target function, the term learning is used for this adaptation or fitting process.
To be more precise, this situation constitutes an example of supervised learning
as discussed in the next section.
To summarize, a feed-forward neural network represents an adaptive param-
eterization of a, in general non-linear, functional dependence. Under rather mild
conditions, feed-forward networks with suitable, continuous activation functions
are universal approximators. Loosely speaking, this means that a network can
approximate any “non-malicious”, continuous function to arbitrary precision,",The+Shallow+and+the+Deep.pdf
"approximate any “non-malicious”, continuous function to arbitrary precision,
provided the network comprises a sufficiently large (problem dependent) num-
ber of hidden units in a suitable architecture, see Chapter 5. This property
clearly motivates the use of feed-forward nets in quite general regression tasks.
If the response of the network is discretized, for instance due to a threshold-
ing operation that yields
S(ξ) ∈ {1, 2, . . . C} , (1.17)
the system performs the assignment of all possible inputs ξ to one of C categories
6Following the author’spersonal preference, layered networks are drawn from top (input)
to bottom (output). Alternative orientations can be achieved by rotating the page.",The+Shallow+and+the+Deep.pdf
"1.3. NETWORK ARCHITECTURES 15
or classes. Hence, the feed-forward network constitutes a classifier which can be
adapted to example data by choice of weights and other free parameters.
The simplest feed-forward classifier, the so-called perceptron, will serve as a
very important example system in the following. The perceptron is defined as
a linear threshold classifier with response
S(ξ) = sign
󰀳
󰁃
N󰁛
j=1
wjξj − θ
󰀴
󰁄 (1.18)
to any possible input ξ ∈ R
N , corresponding to an assignment to one of two
classes represented as S = ±1. Comparison with Eq. (1.7) shows that it can
be interpreted as a single McCulloch Pitts neuron which receives input from N
real-valued units.
The perceptron will be discussed in detail in Chapter 3 as a basic architecture
that provides valuable insights into the fundamentals of machine learning.
1.3.3 Other architectures
Apart from the clear-cut, fully connected attractor neural networks and strictly",The+Shallow+and+the+Deep.pdf
"1.3.3 Other architectures
Apart from the clear-cut, fully connected attractor neural networks and strictly
feed-forward layered nets, a large variety of network types have been considered
and designed, often with specific application domains in mind, see Chapter 5.
Combinations of feed-forward structures with, for instance, layers of highly
interconnected units are employed in the context of Reservoir Computing, see
e.g. [SVC07,LJ09] for overviews and references.
Recently, the use of feed-forward architectures has re-gained significant pop-
ularity in the context of Deep Learning [GBC16, Hue19]. Specific designs and
architectures of Deep Neural Networks including e.g. so-called convolutional or
pooling layers will be discussed very briefly in Chapter 5.
The framework of prototype-based learning is introduced in Chapter 6. Sys-
tems like Learning Vector Quantization can also be interpreted as layered net-
works with specific, distance-based activation functions in the hidden units (the",The+Shallow+and+the+Deep.pdf
"works with specific, distance-based activation functions in the hidden units (the
prototypes) and a winner-takes-all or softmax output layer for classification or
regression, respectively. Prototype-based systems in unsupervised learning will
be discussed briefly in Chapter 6.",The+Shallow+and+the+Deep.pdf
16 1. FROM NEURONS TO NETWORKS,The+Shallow+and+the+Deep.pdf
"Chapter 2
Learning from example data
You live and learn. At any rate, you live.
— Douglas Adams
Different forms of machine learning were already briefly presented in Sec. 1.
In the following section, we focus on the most clear-cut scenarios: supervised
learning and unsupervised learning. In addition, we will briefly discuss the
interesting and fruitful relations of machine learning with statistical modelling.
2.1 Learning scenarios
The main chapters of these notes deal with supervised learning, with emphasis
on classification and regression. Several of the introduced concepts and methods
can however be transferred to unsupervised settings. This is also the case for
the prototype-based methods discussed in Chapter 6.
2.1.1 Unsupervised learning
Unsupervised learning is an umbrella term comprising various methods for the
analysis of unlabeled data. Such data sets do not contain label information as-
sociated with some pre-defined target as it would be the case in classification or",The+Shallow+and+the+Deep.pdf
"sociated with some pre-defined target as it would be the case in classification or
regression. Moreover, there is no direct feedback available from the environment
or a teacher that would facilitate the evaluation of the system’s performance. A
comparison of its response with a given ground truth or approximate represen-
tation thereof is not available or possible.
For more about the background of unsupervised learning, the reader is
referred to the literature, e.g. [Hay09, Bis95a, Bis06]. An introduction, spe-
cific algorithms and applications can also be found in lecture notes by Dalya
Baron [Bar19]. Here we only briefly discuss the framework of unsupervised data
17",The+Shallow+and+the+Deep.pdf
"18 2. LEARNING FROM EXAMPLE DATA
analysis in contrast to supervised learning. We briefly revisit some key methods
of unsupervised learning in the context of preprocessing in Chapter 8.
Potential aims of unsupervised learning are quite diverse, a few examples being:
◦ Data Reduction
Frequently it makes sense to represent large amounts of data by fewer
exemplars or prototypes, which are of the same form and dimension as the
original data and capture the essential properties of the original, larger
data set. One important framework is that of Vector Quantization.
◦ Compression:
Another form of unsupervised learning aims at replacing original data
by lower-dimensional representations without reducing the actual number
of data points, see also Chapter 8. The mapping to lower dimension
should obviously preserve information to a large extent. Compression
could be done by explicitly selecting a reduced set of features, for instance.",The+Shallow+and+the+Deep.pdf
"should obviously preserve information to a large extent. Compression
could be done by explicitly selecting a reduced set of features, for instance.
Alternative techniques provide explicit projections to a lower-dimensional
space or representations that are guided by the preservation of relative
distances or neighborhood relations.
◦ Visualization
Two or three-dimensional representations of a data set can be used for the
purpose of visualizing a given data set. Hence, it can be viewed as a special
case of compression and many techniques can used in both contexts. In
addition, more specific tools have been devised for visualization tasks only.
◦ Density Estimation:
Often, an observed data set is interpreted as being generated in a stochas-
tic process according to a model density. In a training process, parameters
of the density are optimized, for instance aiming at a high likelihood as a
measure of how well the model explains the observations, see Sec. 8.5.
◦ Clustering",The+Shallow+and+the+Deep.pdf
"measure of how well the model explains the observations, see Sec. 8.5.
◦ Clustering
One important goal of unsupervised learning is the grouping of observa-
tions into clusters of similar data points which jointly display properties
from the other groups or clusters in the data set. Most frequently, cluster-
ing is formulated in terms of a specific (dis-)similarity or distance measure,
which is used to compare different feature vectors.
◦ preprocessing
The above mentioned and other unsupervised techniques can be employed
to identify representations of a data set suitable for further processing.
Consequently, unsupervised learning is frequently considered a useful pre-
processing step also for supervised learning tasks.
Note that the above list is by far not complete. Furthermore, the goals men-
tioned here can be closely related and, often, the same methods can be applied",The+Shallow+and+the+Deep.pdf
"2.1. LEARNING SCENARIOS 19
to several of them. For instance, density estimation by means of Gaussian Mix-
ture Models (GMM) could be interpreted as a probabilistic clustering method
and the obtained centers of the GMM can also serve as prototypes in the context
of Vector Quantization.
Several relevant techniques of unsupervised learning are discussed in Chapter
8, where additional references can also be found.
In a sense, in unsupervised learning there is no “right” or “wrong”. This
can be illustrated in the context of a toy clustering problem: If we sort a num-
ber of fruit according to shape and taste, we would most likely group pears
and apples and oranges in three corresponding clusters. Alternatively, we can
sort according to color only and end up with clusters of objects with like col-
ors, e.g. combining green apples with green pears vs. yellowish and red fruit.
Without further information or requirements defined by the environment, many",The+Shallow+and+the+Deep.pdf
"Without further information or requirements defined by the environment, many
clustering strategies and outcomes can be plausible. The example also illus-
trates the fact that the choice of how the data is represented and which types
of properties/features are considered important can determine the outcome of
an unsupervised learning process to the largest extent.
The important point to keep in mind is that, ultimately, the users define
the goal of the unsupervised analysis themselves. Frequently this is done by
formulating a specific cost function or objective function which reflects the task
and guides the training process. The selection or definition of a cost function
can be quite subjective and, moreover, its optimization can even completely fail
to achieve the implicitly intended goal of the analysis, see Sec. 8.4.3 for the
discussion of Vector Quantization as an example.
As a consequence, the identification of an appropriate optimization crite-",The+Shallow+and+the+Deep.pdf
"discussion of Vector Quantization as an example.
As a consequence, the identification of an appropriate optimization crite-
rion and objective function constitutes a key difficulty in unsupervised learning.
Moreover, a suitable model and mathematical framework has to be chosen that
serves the purpose in mind.
2.1.2 Supervised learning
In supervised learning, available data comprises feature vectors1 together with
target values. The data is analysed in order to tune parameters of a model,
which can be used to predict the (hopefully correct) target values for novel data
that was not contained in the training set.
Generally speaking, supervised machine learning is a promising approach if
the target task is difficult or impossible to define in terms of a set of simple
rules, while example data is available that can be analysed.
We will consider the following major tasks in supervised learning:
◦ Regression
In regression, the task is frequently to assign a real-valued quantity to",The+Shallow+and+the+Deep.pdf
"◦ Regression
In regression, the task is frequently to assign a real-valued quantity to
each observed data point. An illustrative example could be the estimation
of the weight of a cow, based on some measured features like the animal’s
height and length.
1The discussion of non-vectorial, relational or other data structures is excluded here.",The+Shallow+and+the+Deep.pdf
"20 2. LEARNING FROM EXAMPLE DATA
◦ Classification
The second important example of supervised problems is the assignment of
observations to one of several categories or classes, i.e. to a discrete target
value. A currently somewhat overstrained example is the discrimination
of cats and dogs based on photographic images.
A variety of problems can be formulated and interpreted as regression or classi-
fication tasks, including time series prediction, risk assessment in medicine, or
the pixel-wise segmentation of an image, to name only a few.
Because target values are taken into account, we can define and evaluate
clear quality criteria, e.g. the number of misclassifications for a given test set
of data or the expected mean square error (MSE) in regression. In this sense,
supervised learning appears well defined in comparison to unsupervised tasks,
generally speaking. The well-defined quality criteria suggest naturally mean-",The+Shallow+and+the+Deep.pdf
"generally speaking. The well-defined quality criteria suggest naturally mean-
ingful objective functions which can be used to guide the learning process with
respect to the given training data.
However, also in supervised learning, a number of issues have to be addressed
carefully, including the selection of a suitable model. Mismatched, too simplistic,
or overly complex systems can hinder the success of learning. This will be
discussed from a quite general perspective in Chapter 7. Similarly, details of the
training procedure may influence the performance severely. Furthermore, the
actual representation of observations and the selection of appropriate features
is essential for the success of supervised training as well.
In the following, we will mostly consider a prototypical workflow of super-
vised learning where
a) a model or hypothesis about the target rule is formulated in a training
phase by means of analysing a set of labeled examples. This could be",The+Shallow+and+the+Deep.pdf
"a) a model or hypothesis about the target rule is formulated in a training
phase by means of analysing a set of labeled examples. This could be
done, for instance, by setting the weights of a feed-forward neural network.
and
b) the learned hypothesis, e.g. the network, can be applied to novel data in
the working phase, after training.
Frequently, an intermediate validation phase is inserted after (a) in order to
estimate the expected performance of the system in phase (b) or in order to
tune model (hyper-)parameters and compare different setups. In fact, validation
constitutes a key step in supervised learning, see Chapter 7.
It is important to keep in mind that many realistic situations deviate from
this idealized scenario. Very often, the examples available for training and
validation are not truly representative of the data that the system is confronted
with in the working phase. The statistical properties and the actual target may",The+Shallow+and+the+Deep.pdf
"with in the working phase. The statistical properties and the actual target may
even change while the system is trained. This very relevant problem is addressed
in the context of so-called continual or life-long learning.
A clear-cut strategy for the supervised training of a classifier is based on
selecting only hypotheses that are consistent with the available training data
and perfectly reproduce the target labels in the training set. As we will discuss
at length in the context of the perceptron classifier, this strategy of learning",The+Shallow+and+the+Deep.pdf
"2.1. LEARNING SCENARIOS 21
in version space relies on the assumption that (a) the target can be realized
by the trained system in principle and that (b) the training data is perfectly
reliable and noise-free. Although these assumptions are hardly ever realized in
practice, the consideration of the idealized scenario provides insight into how
learning occurs by elimination of hypotheses when more and more data becomes
available.
This can be illustrated in terms of a toy example. Assume that integer
numbers have to be assigned to one of two classes denoted as “A”or “B”. Assume
furthermore that the following example assignments are provided
4 → A 13 → B 6 → A 8 → A 11 → B
as a training set. From these observations we could conclude, for instance, that
A is the class of even integers, while B comprises all odd integers. However, we
could also come to the conclusion that all integers i < 11 belong to class A and",The+Shallow+and+the+Deep.pdf
"could also come to the conclusion that all integers i < 11 belong to class A and
all others to B. Both hypotheses are perfectly consistent with the available data
and so are many others. It is in fact possible to formulate an infinite number of
consistent hypotheses based on the few examples given.
As more data becomes available, we might have to revise or extend our
analysis accordingly. An additional example 2 → B for instance, would rule
out the above mentioned concepts, while the assignment of all prime numbers
to class B would (still) constitute a consistent hypothesis now.
We will discuss learning in version space in greater detail in the context of
the perceptron and other networks with discrete output. Note that the strategy
only makes sense if the example data is reliable and noise-free; the data itself
has to be consistent with the unknown rule that we want to infer, obviously.
The simple toy example also illustrates the fact that the space of allowed",The+Shallow+and+the+Deep.pdf
"The simple toy example also illustrates the fact that the space of allowed
hypotheses has to be limited in order to facilitate learning at all. If possible
hypotheses may be arbitrarily complex, we can always construct a consistent one
by, for instance, simply taking over the given list of examples and claiming that
“all other integers belong to class A” (or just as well “...to class B”). Obviously
this approach would not infer any useful information from the data, and such a
largely arbitrary hypothesis cannot be expected to generalize to integers outside
the training set. This is a very simple example for an insight that could be
summarized as as2
Generalization begins where storage ends.
Merely storing the example set by training a very powerful system may com-
pletely miss the ultimate goal of learning, which is inference of useful information
about the underlying rule. We will study this effect more formally with respect
to neural networks for classification.",The+Shallow+and+the+Deep.pdf
"about the underlying rule. We will study this effect more formally with respect
to neural networks for classification.
The above arguments are particularly clear in the context of classification. In
regression, the concept of consistent hypotheses has to be softened, since agree-
ment with the data set is in general quantified by a continuous error measure.
2(rephrasing “Generalization begins where learning ends” — T.M. Cover)",The+Shallow+and+the+Deep.pdf
"22 2. LEARNING FROM EXAMPLE DATA
However, the main idea of supervised learning remains the same: additional
data provides evidence for some hypotheses while others become less likely.
2.1.3 Other learning scenarios
A variety of specific, relevant scenarios can be considered which deviate from
the clear-cut simple cases of supervised learning and unsupervised learning. The
following examples highlight just some tasks or practical situations that require
specific training strategies to cope with. Citations merely point to just one
selected review, edited volume or monograph for further reference.
◦ Semi-supervised Learning [CSZ06]
Frequently, only a subset of the available data is labeled. Strategies have
been developed which, in a sense, combine supervised and unsupervised
techniques in such situations.
◦ Reinforcement Learning [WO12]
In various practical contexts, feedback on the performance of a learning
system only becomes available after a sequence of decisions has been taken,",The+Shallow+and+the+Deep.pdf
"In various practical contexts, feedback on the performance of a learning
system only becomes available after a sequence of decisions has been taken,
for instance in the form of a cumulative reward. Examples would be the
reward received only after a number of steps in a game or in a pathfinding
problem in robotics.
◦ Transfer Learning [YCC18]
If the training samples are not representative for the data that the system
is confronted with in the working phase, adjustments might be necessary
in order to maintain acceptable performance. Just one example could be
the analysis of medical images which were obtained by using similar, yet
not identical technical platforms.
◦ Lifelong or Continual Learning [DRAP15]
Drift processes in non-stationary environments can play an important role
in machine learning. The statistics of the observed example data and/or
the target itself can change while the system is being trained. A system",The+Shallow+and+the+Deep.pdf
"in machine learning. The statistics of the observed example data and/or
the target itself can change while the system is being trained. A system
that learns to detect spam e-mail messages, for instance, has to be adapted
constantly to the ever-changing strategies of the senders.
◦ Causal Learning [PJS17]
Regression systems and classifiers typically reflect correlations they have
inferred from the data, which allow to make some form of prediction based
on future observations. In general, this does not explicitly take causal
relations into account. The reliable detection of causalities in a data set is
a highly non-trivial task and requires specifically designed, sophisticated
methods of analysis.
In this material, we will focus almost exclusively on well-defined problems
of supervised learning in stationary environments. In most cases, we will as-
sume that training data is representative of the problem at hand and that it is
complete and reliable to a certain extent.",The+Shallow+and+the+Deep.pdf
"2.2. MACHINE LEARNING VS. STATISTICAL MODELLING 23
2.2 Machine Learning vs. Statistical Modelling
In the sciences it happens quite frequently that the same or very similar concepts
and techniques are developed or rediscovered in different (sub-)disciplines, either
in parallel or with significant delay.
While it is – generally speaking – quite inefficient to re-invent the wheel, a
certain level of redundancy is probably inevitable in scientific research. The
same questions can occur and re-occur in very different settings, and different
communities will come up with specific approaches and answers. Moreover, it
can be beneficial to come across certain problems in different contexts and to
view them from different angles.
It is not at all surprising that this is also true for the area of machine learning,
which has been of inter-disciplinary nature right from the start, with contribu-
tions from biology, psychology, mathematics, physics and others.
2.2.1 Differences and commonalities",The+Shallow+and+the+Deep.pdf
"tions from biology, psychology, mathematics, physics and others.
2.2.1 Differences and commonalities
An area, which is often viewed as competing, complementary, or even supe-
rior to machine learning is that of inference in statistical modelling. A simple
web-search for, say, “Statistical Modelling versus Machine Learning” will yield
numerous links to discussions of their differences and commonalities. Some of
the statements that one may very likely come across are3:
– The short answer is that there is no difference
– Machine learning is just statistics, the rest is marketing
– All machine learning algorithms are black boxes
– Machine learning is the new statistics
– Statistics is only for small data sets, machine learning is for big data
– Statistical modelling has lead to irrelevant theory and questionable
conclusions
– Whatever machine learning will look like in ten years, I’m sure statisticians
will be whining that they did it earlier and better.",The+Shallow+and+the+Deep.pdf
"conclusions
– Whatever machine learning will look like in ten years, I’m sure statisticians
will be whining that they did it earlier and better.
These and similar opinions reflect a certain level of competition, which can
be counterproductive at times, to put it mildly. In the following, we will re-
frain from choosing sides in this on-going debate. Instead, the relation between
machine learning and statistical modelling will be highlighted in terms of an
illustrative example.
One of the most comprehensive, yet accessible presentations of statistical
modelling based learning is given in the excellent textbook The Elements of
Statistical Learning by T. Hastie, R. Tibshirani, and J. Friedman [HTF01]. A
view on many important methods, including density estimation and Expectation
Maximization algorithms is provided in Neural Networks for Pattern Recognition
[Bis95a] and the more recent Pattern Recognition and Machine Learning [Bis06]
by C. Bishop.",The+Shallow+and+the+Deep.pdf
"[Bis95a] and the more recent Pattern Recognition and Machine Learning [Bis06]
by C. Bishop.
3Exact references and links are not provided in the best interest of the originators.",The+Shallow+and+the+Deep.pdf
"24 2. LEARNING FROM EXAMPLE DATA
In both, machine learning and statistical modelling, the aim is to extract
information from observations or data and to formalize it. Most frequently,
this is done by generating a mathematical model of some sort and fitting its
parameters to the available data.
Quite often, machine learning and statistical models have very similar or
identical structures and, frequently, the same mathematical tools or algorithms
are used. The differences lie in the emphasis that is usually put on different
aspects of the modelling or learning:
Generally speaking, the main aim of statistical inference is to describe, but
also explain and understand the observed data in terms of models. These usually
take into account explicit assumptions about statistical properties of the obser-
vations. This includes the possible goal of confirming or falsifying hypotheses
with a desired significance or confidence.
In Machine Learning, on the contrary, the main motivation is to make pre-",The+Shallow+and+the+Deep.pdf
"with a desired significance or confidence.
In Machine Learning, on the contrary, the main motivation is to make pre-
dictions with respect to novel data, based on patterns detected in the previous
observations. Frequently, this does not rely on explicit assumptions in terms of
statistical properties of the data but employs heuristic concepts of inference.4
The goal is not so much the faithful description or interpretation of the data,
but rather, it is the application of the derived hypotheses to novel data that is in
the center of interest. The corresponding performance, for instance quantified
as an expected error in classification or regression, is the ultimate guideline.
Obviously, these goals are far from being really disjoint in a clear-cut way.
Genuine statistical methods like Bayesian classification can clearly be used with
the exclusive aim of accurate prediction in mind. Likewise, sophisticated heuris-
tic machine learning techniques like relevance learning are designed to obtain",The+Shallow+and+the+Deep.pdf
"tic machine learning techniques like relevance learning are designed to obtain
insight into mechanisms underlying the data, see Chapter 6.
Very often, both perspectives suggest very similar or even identical meth-
ods which can be used interchangeably. Frequently, it is only the underlying
philosophy and motivation that distinguishes the two approaches.
In the following section, we will have a look at a very basic, illustrative
problem: linear regression. It will be re-visited as a prototypical supervised
learning task a couple of times. Here, however, it serves as an illustration of
the relation between machine learning and statistical modelling approaches and
their underlying concepts.
2.2.2 An example case: linear regression
Linear regression constitutes one of the earliest, most important and clearest
examples of inference.
As a, by now, historical application, consider the theory of an expanding
universe according to which the velocity v of far away galaxies should be directly",The+Shallow+and+the+Deep.pdf
"universe according to which the velocity v of far away galaxies should be directly
proportional to their distance d from the observer [Hub29]:
v = Ho d. (2.1)
4However, it is very important to realize that implicit assumptions are always made, for
instance when choosing a particular machine learning framework to begin with.",The+Shallow+and+the+Deep.pdf
"2.2. MACHINE LEARNING VS. STATISTICAL MODELLING 25
v
d
Figure 2.1: Hubble dia-
gram: the velocity v of galax-
ies as a function of their dis-
tance d, taken from [Hub29].
Note that the correct units of
v should be km/s. According
to PNAS, figure and article
[Hub29] are in the public do-
main.
Here, Ho is the so-called Hubble constant which is named after Edwin Hubble,
one of the key figures in modern astronomy. Hubble fitted an assumed linear
dependence of the form (2.1) to observational data in 1929 and obtained as a
rough estimate Ho ≈ 500km/s
Mpc , see Figure 2.1. The interested reader is referred
to the astronomy literature for details, see e.g. [Huc18] for a quick start.
Two major lessons can be learnt from this example: (a) simple linear regres-
sion has been and continues to be a highly useful tool, even for very fundamental
scientific questions, and (b) the predictive power of a fit depends strongly on
the quality of the available data. The latter statement is evidenced by the fact",The+Shallow+and+the+Deep.pdf
"the quality of the available data. The latter statement is evidenced by the fact
that more recent estimates of the Hubble constant, based on more data of better
quality, correspond to much lower values Ho ≈ 73.5km/s
Mpc [Huc18].
Obviously, a result of the form (2.1) summarizes experimental or observa-
tional data in a descriptive fashion and allows us to formulate conclusions that
we have drawn from available data. At the same time, it makes it possible
to apply the underlying hypothesis on novel data. By doing so, we can test,
confirm or falsify the model and its assumptions and detect the need for correc-
tions. The topic of validating a given model will be addressed in greater detail
in Chapter 7.
Note that the following discussion is by no means intended to claim that lin-
ear regression is a genuine machine learning method. It goes back to at least Leg-
endre and Gauss, see https://en.wikipedia.org/wiki/Linear_regression#History.",The+Shallow+and+the+Deep.pdf
"endre and Gauss, see https://en.wikipedia.org/wiki/Linear_regression#History.
Here, it merely serves as a relatively simple, illustrative example problem.
A heuristic machine learning approach
Equation (2.1) represents a simple linear dependence of a target function v(d) on
a single variable d ∈ R. In the more general setting of multiple linear regression,
a target value y(ξ) is assigned to a number of arguments which are concatenated
in an N-dimensional vector ξ ∈ RN .
In the standard setting of multiple linear regression, a set of examples
D = {ξµ, yµ}P
µ=1 with ξµ ∈ RN , yµ ∈ R (2.2)",The+Shallow+and+the+Deep.pdf
"26 2. LEARNING FROM EXAMPLE DATA
is given. A hypothesis of the form
fH(ξ) =
N󰁛
i=1
wi ξi = w⊤ξ = w · ξ with w ∈ RN (2.3)
is assumed to represent or approximate the dependence y(ξ) underlying the
observed data set D. In analogy to other machine learning scenarios considered
later, we will refer to the coefficients wj also as weights and combine them in a
vector w ∈ RN . Depending on the context, any of the equivalent notations for
the scalar product in Eq. (2.3) will be used.
Note that a constant term could be incorporated formally without explicit
modification of Eq. (2.3). This can be achieved by decorating every input vector
with an additional clamped dimension ξN+1 = −1 and introducing an auxiliary
weight wN+1 = θ:
󰁨ξ = (ξ1, ξ2, ξ3, . . . , ξN , −1)⊤ , 󰁨w = (w1, w2, w3, . . . wN , θ)⊤ ∈ RN+1
⇒ 󰁨w⊤ 󰁨ξ = w⊤ξ − θ. (2.4)
Any inhomogeneous hypothesis fH(ξ) = w⊤ξ −θ including a constant term can
be written as a homogeneous function in N + 1 dimensions for an appropriately",The+Shallow+and+the+Deep.pdf
"be written as a homogeneous function in N + 1 dimensions for an appropriately
extended input space, formally. Hence, we will not consider constant contribu-
tions to the the hypothesis fH explicitly in the following. A similar argument
will be used later in the context of linearly separable classifiers.
A quite intuitive approach to the selection of the model parameters, i.e. the
weights w, is to consider the available data and to aim at a small deviation of
fH(ξµ) from the observed values yµ. Of the many possibilities to define and
quantify this goal, the quadratic deviation or Sum of Squared Error (SSE) is
probably the most frequently used one:
ESSE = 1
2
P󰁛
µ=1
󰀕
fH(ξµ) − yµ
󰀖2
where fH(ξµ) = w⊤ξµ (2.5)
and the sum is over all examples in D.
The quadratic deviation disregards whether fH(ξµ) is greater or lower than
yµ. Note that the pre-factor 1/2 conveniently cancels out when taking deriva-
tives with respect to the weights but is otherwise irrelevant. We will fre-",The+Shallow+and+the+Deep.pdf
"tives with respect to the weights but is otherwise irrelevant. We will fre-
quently replace the SSE by the Mean Squared Error (MSE) which is defined
as EMSE = ESSE/P. The constant factor 1/P is, of course, also irrelevant for
the minimization and the properties of the optima.
Necessary and sufficient conditions for the presence of a (local) minimum in
a differentiable cost function are briefly summarized and discussed in Appendix
A.2.2, where we also point to additional literature. Here, we consider only the
necessary first order condition for a weight vector w∗ that minimizes ESSE :
∇wESSE󰀏󰀏
w=w∗
!
= 0 with ∇wESSE =
P󰁛
µ=1
󰀅
w⊤ξµ − yµ󰀆
ξµ. (2.6)",The+Shallow+and+the+Deep.pdf
"2.2. MACHINE LEARNING VS. STATISTICAL MODELLING 27
Note that the SSE is also a popular objective function in the context of re-
gression in multi-layered networks, see Chapter 5 and e.g. [Bis95a,Bis06,HTF01,
HKP91,EB01]. With the convenient matrix and vector notation5
Y =
󰀃
y1, y2, . . . , yP 󰀄⊤
∈ RP , χ =
󰁫
ξ1, ξ2, . . . , ξP
󰁬⊤
∈ RP×N (2.7)
we can rewrite Eq. (2.6) and solve it formally:
χ⊤ (χw∗ − Y )
!
= 0 ⇒ w∗ =
󰀅
χ⊤χ
󰀆−1
χ⊤
󰁿 󰁾󰁽 󰂀
χ+
left
Y (2.8)
where χ+
left is the (left) Moore-Penrose pseudoinverse of the rectangular matrix
χ [PP12,BH12]. Note that the solution can be written in precisely this form only
if the (N ×N) matrix [χ⊤χ] is non-singular and, thus, [χ⊤χ]−1 exists. This can
only be the case for P > N, i.e. when the system of P equations
󰀋
w⊤ξµ = yµ󰀌
in N unknowns is over-determined and cannot be solved exactly.
For the precise definition of the Moore-Penrose and other generalized in-
verses (also in the case P ≤ N) see e.g. the Matrix Cookbook [PP12] as a",The+Shallow+and+the+Deep.pdf
"For the precise definition of the Moore-Penrose and other generalized in-
verses (also in the case P ≤ N) see e.g. the Matrix Cookbook [PP12] as a
comprehensive source of information in the context of matrix manipulations.
Heuristically, in the case of singular matrices [χ⊤χ], one can enforce the
existence of an inverse by adding a small contribution of the N-dimensional
identity matrix IN :
w∗
γ =
󰀅
χ⊤χ + γ IN
󰀆−1
χ⊤ Y. (2.9)
Since the symmetric [χ⊤χ] has only non-negative eigenvalues, the matrix on the
r.h.s. of (2.9) is guaranteed to be non-singular for any γ > 0.
We will re-visit the problem of linear regression in the context of perceptron
training and more general regression later, see also Appendix A.3.2. There,
we will also discuss the case of under-determined systems of solvable equations󰀋
w⊤ξµ = yµ󰀌P
µ=1.
In analogy to the above, it is straightforward to show that the resulting
weights w∗
λ correspond to the minimum of the modified objective function
ESSE
λ = 1
2
P󰁛
µ=1
󰀕",The+Shallow+and+the+Deep.pdf
"weights w∗
λ correspond to the minimum of the modified objective function
ESSE
λ = 1
2
P󰁛
µ=1
󰀕
fH(ξµ) − yµ
󰀖2
+ 1
2γ w2. (2.10)
Hence, we have effectively introduced a penalty term, which favors weight vectors
with smaller norm | w |2. The concept is known as weight decay, see Sec. 7.2.
Note that nearly singular matrices [χ⊤χ] would lead to large magnitude weights
according to Eq. (2.8).
This is our first encounter of regularization, i.e. the restriction of the search
space in a learning problem with the goal of improving the outcome of the
training process. In fact, weight decay is applied in a variety of problems and
5For better readability, χ is used instead of Ξ (the properly capitalized version of ξ).",The+Shallow+and+the+Deep.pdf
"28 2. LEARNING FROM EXAMPLE DATA
is by no means restricted to linear regression. Other methods of regularization
will be discussed in the context of overfitting in neural networks in Sec. 7.2.
We will revisit linear regression again in later chapters and show that it can
also be formulated as the minimization of w2 under suitable constraints. This
approach circumvents the problem of having to choose an appropriate weight
decay parameter γ in Eqs. (2.9, 2.10).
The statistical modelling perspective
In a statistical modelling approach, we aim at explaining the observed data D in
terms of an explicit model. To this end, we have to make and formalize certain
assumptions. For instance, we can assume that the labels yµ are generated
independently according to a conditional density of the form
p(yµ | ξµ, w) = N(yµ | w⊤ξ, σ2) = 1√
2πσ exp
󰀗
− 1
2σ2
󰀃
yµ − w⊤ξµ󰀄2
󰀘
. (2.11)
Hence, we assume that the observed targets essentially reflect a linear depen-
dence but are subject to Gaussian noise:",The+Shallow+and+the+Deep.pdf
"2σ2
󰀃
yµ − w⊤ξµ󰀄2
󰀘
. (2.11)
Hence, we assume that the observed targets essentially reflect a linear depen-
dence but are subject to Gaussian noise:
yµ = w⊤ξµ + σ ηµ (2.12)
with independent, random quantities ηµ with 〈ηµ〉 = 0 and 〈ηµην〉 = δµν. In
contrast to the previous, heuristic treatment, we start from an explicit assump-
tion for how and why the observed values deviate from the linear dependence.
In the following, we consider only w as parameters of our model, while σ
is fixed. Extensions that include σ as an adaptive degree of freedom are very
well possible but not essential for the comparison with the heuristic machine
learning approach.
In the simplest case we assume that example inputs ξµ are generated inde-
pendently. Then, for a given model with weights w, the likelihood of observing
a particular set of target values Y =
󰀃
y1, y2, . . . yP 󰀄⊤
factorizes:
p(Y | w, χ) =
P󰁜
µ=1
p(yµ | ξµ, w). (2.13)
The corresponding log-likelihood reads
log p(Y |χ, w) =
P󰁛
µ=1",The+Shallow+and+the+Deep.pdf
"󰀃
y1, y2, . . . yP 󰀄⊤
factorizes:
p(Y | w, χ) =
P󰁜
µ=1
p(yµ | ξµ, w). (2.13)
The corresponding log-likelihood reads
log p(Y |χ, w) =
P󰁛
µ=1
log p(yµ |ξµ, w) = −P
2 log(2πσ2)− 1
σ2
1
2
P󰁛
µ=1
󰀃
yµ − w⊤ξµ󰀄2
,
(2.14)
where we inserted the Gaussian model (2.11). Now we note that the first term
on the r.h.s. is constant with respect to w. Furthermore, the second term is
proportional to −ESSE as given in Eq. (2.5).
We conclude that the weights w∗ that explain the data with Maximum
Likelihood under the assumption of model (2.11) are exactly those that minimize
the SSE. Hence, we arrive at the same formal solution as given in (2.8).",The+Shallow+and+the+Deep.pdf
"2.2. MACHINE LEARNING VS. STATISTICAL MODELLING 29
This correspondence of the Maximum Likelihood solution in the Gaussian
model with a quadratic error measure is of course due to the specific mathemat-
ical form of the normal distribution and can be rediscovered in various other
contexts. The assumption of Gaussian noise is rarely strictly justified, but it
is very popular and appears natural in absence of more concrete knowledge.
Frequently, it yields practical methods and can be seen as the basis of popular
techniques like Principal Component Analysis, mixture models for clustering,
or Linear Discriminant Analysis [HTF01,Bis06].
Note, however, that the statistical approach is more flexible in the sense
that we could, for instance, replace the conditional model density in (2.11) by
an alternative assumption and proceed along the same lines to obtain a suitable
objective function in terms of the associated likelihood.",The+Shallow+and+the+Deep.pdf
"an alternative assumption and proceed along the same lines to obtain a suitable
objective function in terms of the associated likelihood.
Moreover, it is possible to incorporate prior knowledge, or prior beliefs, into
the formalism. If we had reason to assume that weights with low magnitude are
more likely to occur, even before any data is observed, we could express this in
terms of an appropriate prior density, for instance
po(w) ∝ exp
󰀗
− 1
2τ2o
w2
󰀘
. (2.15)
Exploiting Bayes Theorem, P(A|B)P(B) = P(B|A)P(A), we obtain from the
data likelihood P(D|w):
p(w|D) ∝ p(D|w) po(w). (2.16)
With the proper normalization this represents the posterior probability of weights
p(w|D) after having seen a data set D = {χ, Y }, and taking into account the
data independent prior po(w).
Assuming the independent generation of ξµ again and inserting the par-
ticularly convenient Gaussian prior (2.15) we can write the logarithm of the
posterior as
log [p(w|D)] ∝ −ESSE − 1
2γ w2 + const. (2.17)",The+Shallow+and+the+Deep.pdf
"ticularly convenient Gaussian prior (2.15) we can write the logarithm of the
posterior as
log [p(w|D)] ∝ −ESSE − 1
2γ w2 + const. (2.17)
with a suitable parameter γ that depends on τo and is obtained easily by working
out the logarithm of p(w|D) from Eq. (2.16).
The important observation is that maximizing the posterior probability with
respect to the set of weights w is equivalent to minimizing the objective function
given in Eq. (2.10). Hence, the Maximum A Posteriori (MAP) estimate of
the parameters w is formally identical with the MSE estimate when amended
by an appropriate weight decay term. Not surprisingly, many different names
have been coined for this form of regularization and its variants, including L2-
regularization, Tikhonov-regularization, and ridge-regression [HTF01, Bis95a,
Bis06,DHS00].
The above discussed Maximum Likelihood and MAP results are examples of
so-called point estimates: one particular set of model parameters (here: w) is",The+Shallow+and+the+Deep.pdf
"The above discussed Maximum Likelihood and MAP results are examples of
so-called point estimates: one particular set of model parameters (here: w) is
selected according to the specific criterion in use. The statistical modelling idea
allows us to go even further: In the framework of Bayesian Inference [HTF01]",The+Shallow+and+the+Deep.pdf
"30 2. LEARNING FROM EXAMPLE DATA
we can consider all possible model settings at a time, yielding the posterior
predictive probability
p(y|ξ, D) ∝
󰁝
p(y|ξ, w) p(w|D)󰁿 󰁾󰁽 󰂀
∝p(D|w) po(w)
dN w. (2.18)
Properly normalized, this defines the probability of response y(ξ) to an arbitrary
(novel) input ξ after having seen the data set D. It is obtained as an integral
over the specific model responses p(y|ξ, w) given a particular w, but integrated
over all possible models with the posterior p(w|D) as a weighting factor.
The formalism yields a probabilistic assignment of the target y, which also
makes it possible to quantify the associated uncertainty due to the data depen-
dent variability of the model parameters. On the one hand, f this constitutes
an appealing advantage over the simpler point estimates. On the other hand,
the full formalism can be quite involved in practice. Frequently, one resorts to
convenient parametric forms of model and prior densities and/or derives easy to",The+Shallow+and+the+Deep.pdf
"convenient parametric forms of model and prior densities and/or derives easy to
handle (e.g. Gaussian) approximations of the posterior predictive distribution
(2.18).
2.2.3 Conclusion
Heuristic machine learning and statistical modeling based approaches differ sig-
nificantly in terms of their conceptual foundations. However, in practice, these
differences often blur or play a minor role. Very often, seemingly purely heuris-
tic methods of machine learning can be derived from the statistical inference
perspective under suitable model assumptions. Frequently, training algorithms
and methods are very similar if not identical. Many statistics based methods
can be used for the main goals of machine learning, i.e. prediction and gen-
eralization. At the same time, sophisticated machine learning techniques can
also aim at understanding and explaining the observed data, a goal which is
frequently attributed to statistical learning.",The+Shallow+and+the+Deep.pdf
"also aim at understanding and explaining the observed data, a goal which is
frequently attributed to statistical learning.
In summary, the author’s recommendation is to acquire knowledge of both,
statistical modeling and heuristically motivated machine learning. We should
use the best of both complementary frameworks without being dogmatic or
religious in preferring one over the other. The never-ending debates and claims
of superiority by both communities are essentially useless and even counter-
productive. Instead, efforts should be combined in order to achieve a better
understanding of data analysis and learning.",The+Shallow+and+the+Deep.pdf
"Chapter 3
The Perceptron
The perceptron has shown itself worthy despite (and even because of!) its severe
limitations. It has many features to attract attention: its linearity; its intriguing
learning theorem; its clear paradigmatic simplicity as a kind of parallel compu-
tation.
— Marvin Minsky and Seymour Papert in [MP69]
3.1 History and literature
The term perceptron is used in a variety of meanings. Throughout this text,
however, it will exclusively refer to a system representing inputs ξ ∈ RN in
a layer of units, which are connected to a single binary output unit of the
McCulloch Pitts type. Its activation S(ξ) ∈ {−1, +1} is taken to represent two
possible responses.1 It corresponds to a linear threshold classifier or – in feed-
forward network jargon – to an N–1 architecture with a single binary output,
that does not comprise hidden units or layers.
Other well-known linear threshold classifiers are Linear Discriminant Analy-",The+Shallow+and+the+Deep.pdf
"that does not comprise hidden units or layers.
Other well-known linear threshold classifiers are Linear Discriminant Analy-
sis (LDA), the Naive Bayes classifier, and the popular Logistic Regression (LR).
Overviews and comparisons can be found in e.g. [DHS00,HTF01,Bis95a,Mur22].
These classical methods are based on concepts of statistical modelling and differ
mostly in the way their parameters are determined or constructed from a given
data set.
In the literature, more general architectures with several layers and/or con-
tinuous output are often referred to as (multilayer, soft, . . .) perceptrons. We
will later consider layered networks which are constructed from perceptron-like
1Of course, any binary output, e.g. S ∈ {0, 1}, could serve the same purpose.
31",The+Shallow+and+the+Deep.pdf
"32 3. THE PERCEPTRON
units, but in these lecture notes the term perceptron always refers to the single
layer, binary classifier.
Even the very simple, limited perceptron architecture is of interest for a
multitude of reasons:
◦ Pioneered by Frank Rosenblatt [Ros58, Ros61], the perceptron has been
one of the earliest, very successful machine learning concepts and devices,
and it was even realized in hardware, see Figure 3.1.
◦ Rosenblatt also suggested an algorithm for perceptron training, which is
guaranteed to converge, provided a suitable solution exists. The corre-
sponding Perceptron Convergence Theorem is one of the most fundamen-
tal results in machine learning and has contributed largely to the initial
popularity and success of the field. It will be presented and proven in Sec.
3.3.3.
◦ It serves as a prototypical model system that provides theoretical, mathe-
matical and intuitive insights into the basic mechanisms of machine learn-",The+Shallow+and+the+Deep.pdf
"matical and intuitive insights into the basic mechanisms of machine learn-
ing. At the same time it is a building block from which to construct more
powerful systems. As Manfred Opper [Opp90] put it: “Theperceptron is
the hydrogen atom of neural network research.”
◦ In its modern, conceptually extended re-incarnation, the Support Vec-
tor Machine (SVM) [SS02,CST00,STC04,Her02,DFO20], the perceptron
persists to be used successfully in a large variety of practical applications.
The precise relation of the SVM to the simple perceptron will be discussed
in great detail in Section 4.3.
◦ The history of the perceptron provides insights into how the scientific
community deals with high expectations and disillusionments leading to
the extreme over-reaction of stalling an entire field of research [Ola96].
Several original texts from the early days of the perceptron are available
in the public domain. This includes an original article from 1958 [Ros58], the",The+Shallow+and+the+Deep.pdf
"in the public domain. This includes an original article from 1958 [Ros58], the
highly interesting official Manual of the Perceptron Mark I hardware [HRM+60]
and Rosenblatt’s monograph Principles of Neurodynamics [Ros61]. An inter-
esting TV documentation is available at [You07].
The so-called Perceptron Controversy and its perception and long-lasting
impact on the machine learning community is analysed in an article entitled A
Sociological Study of the Official History of the Perceptron Controversy by M.
Ozaran [Ola96].
Clear presentations of the Rosenblatt algorithm can be found in virtually
all texts that cover the perceptron. Discussions which are quite close to these
lecture notes (apart from notation issues) are given in the monographs by J.A.
Hertz, A. Krogh and R.G. Palmer [HKP91] and by S. Haykin in [Hay09], for
instance.
The counting argument for the number of linearly separable functions, see
Sec. 3.4, is presented in e.g. [HKP91]. In this context, it should be useful to",The+Shallow+and+the+Deep.pdf
"3.2. LINEARLY SEPARABLE FUNCTIONS 33
Figure 3.1: The Mark I Perceptron. Left: Hardware realization at Cornell
Aeronautical Laboratory. Photo reproduced with kind permission from Cornell
University Library.2 The input of the Mark I was realized via a retina of 400
photosensors. Adaptive weights were represented by potentiometers that could
be tuned by electric motors. Right: Schematic outline of the Mark I architec-
ture, based on a figure taken from [HRM+60]. The triangular, shaded region
(added to the original illustration) marks a subset of units that is commonly
referred to as the perceptron in this text.
consult the original publications by R. Winder [Win61], T.M. Cover [Cov65] and
G.J. Mitchison and R.M. Durbin [MD89a]. The latter work also presents the
extension of the counting argument to two-layered networks (machines) with K
hidden units.
3.2 Linearly separable functions
The perceptron can be viewed as the simplest feed-forward neural network. It",The+Shallow+and+the+Deep.pdf
"hidden units.
3.2 Linearly separable functions
The perceptron can be viewed as the simplest feed-forward neural network. It
responds to real-valued inputs ξ ∈ RN in terms of a binary output S ∈ {−1, +1}.
The response of a perceptron with weight vector w is obtained by applying a
threshold operation to the weighted sum of inputs:
Sw,θ(ξ) = sign (w · ξ − θ) = ±1, (3.1)
where the N-dim. weight vector and the threshold θ parameterize the specific
input/output relation. Its mathematical structure suggests an immediate geo-
metrical interpretation of the perceptron which is illustrated in Fig. 3.3: The
set of points in feature space
󰁱
󰁨ξ ∈ RN
󰀏󰀏󰀏
󰀓
w · 󰁨ξ − θ
󰀔
= 0
󰁲
(3.2)
2Cornell University News Service records, #4-3-15. Division of Rare and Manuscript Col-
lections, Cornell University Library. See https://digital.library.cornell.edu/catalog/ss:550351.",The+Shallow+and+the+Deep.pdf
"34 3. THE PERCEPTRON
ξ ∈ RN (inputs)
w ∈ RN (weights)
S = sign(w · ξ − Θ) = ±1 (output)
Figure 3.2: Illustration of the single layer perceptron with N-dimensional
inputs and a binary output of the McCulloch Pitts type.
corresponds to a (hyper-)plane orthogonal to to w with an off-set θ |w| from the
origin3. Inputs with w · ξ > θ result in perceptron output +1, while vectors ξ
with w · ξ < θ yield the response −1. Hence, the perceptron realizes a linearly
separable (lin. sep. ) function: feature vectors with perceptron output +1 are
separated by the hyperplane (3.2) from those with output −1.
Two cases can be distinguished: input/output relations of the form (3.1)
with θ ∕= 0 are called inhomogeneously lin. sep., while homogeneously lin. sep.
functions can be written as
Sw(ξ) = sign (w · ξ ) . (3.3)
For the latter, the corresponding hyperplane, cf. Fig. 3.3, has no offset (θ = 0)
and includes the origin.
In the following, we will focus on homogeneously linearly separable func-",The+Shallow+and+the+Deep.pdf
"and includes the origin.
In the following, we will focus on homogeneously linearly separable func-
tions, mostly. This does not constitute an essential restriction because any
inhomogeneously lin. sep. function can be interpreted as a homogeneous one in
a higher dimensional space: Consider the function Sw,θ(ξ) = sign (w · ξ − θ)
with w, ξ ∈ RN . Now let us define the modified (N + 1)-dimensional weight
vector
󰁨w = (w1, w2, . . . , wN , θ)⊤
and augment all feature vectors by an auxiliary, “clamped” input dimension:
󰁨ξ = (ξ1, ξ2, . . . , ξN , −1) ∈ RN+1.
We observe that
󰁨w · 󰁨ξ = w · ξ − θ and, thus, sign
󰀓
󰁨w · 󰁨ξ
󰀔
= sign (w · ξ − θ) . (3.4)
As a consequence, a non-zero threshold θ in N-dimensions can always be re-
written as an additional weight in a trivially augmented feature space and,
formally, the two cases can be treated on the same grounds. Note that the ar-
gument is analogous to the formal inclusion of a constant term in multiple linear",The+Shallow+and+the+Deep.pdf
"gument is analogous to the formal inclusion of a constant term in multiple linear
regression, see Eq. (2.4). Later, we will encounter subtleties which require more
precise considerations, but for now we will restrict ourselves to homogeneous
functions and simply refer to them as linearly separable for brevity.
3The distance of the plane from the origin is exactly θ in case of normalized w with |w| = 1.",The+Shallow+and+the+Deep.pdf
"3.2. LINEARLY SEPARABLE FUNCTIONS 35
θ
0
S =−1
S =+1w/|w|
Figure 3.3: Geometrical in-
terpretation of the perceptron.
The hyperplane orthogonal to w
with off-set θ from the origin
separates feature vectors with
output S = +1 and S = −1,
respectively.
In the following, we will also call a set of P input/output pairs
D = {ξµ, Sµ
T }P
µ=1 (3.5)
(homogeneously) linearly separable, if at least one weight vector w exists with
Sµ
w = sign (w · ξµ) = Sµ
T for all µ = 1, 2, . . . , P, (3.6)
where we use the shorthand notation Sµ
w ≡ Sw(ξµ). The labels in D are denoted
by ST = ±1, with the subscript T for target or training.
A number of interesting and important questions related to linear separabil-
ity come to mind:
(Q1) Given a linearly separable data set, (how) can we find a perceptron weight
vector w that satisfies (3.6)?
(Q2) Given an arbitrary data set with binary target labels, can we determine
whether it is linearly separable without (before) attempting to train a",The+Shallow+and+the+Deep.pdf
"(Q2) Given an arbitrary data set with binary target labels, can we determine
whether it is linearly separable without (before) attempting to train a
perceptron? Are there efficient iterative algorithms which indicate (at an
early stage) that a data set is not linearly separable?
(Q3) How serious is the restriction to linear separability in the space of binary
target functions? How many linearly separable functions exist, i.e. in how
many lin. sep. ways can we label P input vectors in N dimensions?
(Q4) Can we learn an underlying linearly separable rule from the examples con-
tained in D? How does the realization or “storage” of the given labels in
D relate to the learning of the unknown rule?
(Q5) If – for a lin. sep. D – several or many vectors w satisfy the conditions
(3.6), which one is the best? What is a meaningful measure of quality and
how can the corresponding optimal weight vector be found?
(Q6) If D is not separable, can we still approximate the target classification by",The+Shallow+and+the+Deep.pdf
"how can the corresponding optimal weight vector be found?
(Q6) If D is not separable, can we still approximate the target classification by
means of a perceptron? Which alternatives or extensions exist?
Most of these questions (Q1–Q5) will be addressed and answered in the forth-
coming sections, while Chapter 4 deals with the realization or approximation of
classification schemes beyond linear separability (Q6).",The+Shallow+and+the+Deep.pdf
"36 3. THE PERCEPTRON
3.3 The Rosenblatt perceptron
To a large extent, the success of the perceptron has been due to the existence of
a training algorithm and the associated convergence theorem, both presented by
Frank Rosenblatt [Ros58, Ros61]. In the following, we first precisely define the
basic goal of the training process, outline the general form of iterative perceptron
algorithms and present Rosenblatt’s algorithm. As a key result, we reproduce
the corresponding proof of convergence, eventually.
3.3.1 The perceptron storage problem
Here we address question (Q1) of the list given in the previous section. First
we consider the task of reproducing the labels of a given data set of form (3.5)
through a perceptron. We define the so-called perceptron storage problem (PSP)
as
Perceptron Storage Problem (I) (3.7)
For a given D = {ξµ, Sµ
T }P
µ=1 with ξµ ∈ RN and Sµ
T ∈ {−1, +1} ,
find a vector w ∈ RN with sign (w · ξµ) = Sµ
T for all µ = 1, 2, . . . , P.",The+Shallow+and+the+Deep.pdf
"For a given D = {ξµ, Sµ
T }P
µ=1 with ξµ ∈ RN and Sµ
T ∈ {−1, +1} ,
find a vector w ∈ RN with sign (w · ξµ) = Sµ
T for all µ = 1, 2, . . . , P.
The term storage refers to the fact that we are not (yet) aiming at the application
of the function Sw(ξ) to vectors ξ ∕∈ D. We are only interested in reproducing
the correct assignment of labels within the data set by means of a perceptron
network. Alternatively, this aim could be achieved by storing D in a memory
table and look up the correct Sµ
T when needed.
In order to rewrite the PSP we note that sign(w · ξ) = S ⇔ w · ξ S > 0.
Defining the so-called local potentials (the term indicates a vague relation to
the membrane potentials, cf. Sec. 1.1.)
Eµ = w · ξµ Sµ
T for µ = 1, 2, . . . , P, (3.8)
we obtain an equivalent formulation of the PSP in terms of a set of inequalities:
Perceptron Storage Problem (II) (3.9)
For a given D = {ξµ, Sµ
T }P
µ=1 with ξµ ∈ RN and Sµ
T ∈ {−1, +1} ,
find a vector w ∈ RN with Eµ ≥ c > 0 for all µ = 1, 2, . . . , P.",The+Shallow+and+the+Deep.pdf
"For a given D = {ξµ, Sµ
T }P
µ=1 with ξµ ∈ RN and Sµ
T ∈ {−1, +1} ,
find a vector w ∈ RN with Eµ ≥ c > 0 for all µ = 1, 2, . . . , P.
Here, we have introduced a constant c > 0 as a margin in terms of the conditions
Eµ > 0. Note that the actual value of c is essentially irrelevant: Consider vectors
w1 and w2 = λw1 with λ > 0. Due to the linearity of the scalar product
Eµ
1 = w1 · ξµ Sµ
T ≥ c > 0 implies Eµ
2 = w2 · ξµ Sµ
T ≥ λ c > 0. (3.10)",The+Shallow+and+the+Deep.pdf
"3.3. THE ROSENBLATT PERCEPTRON 37
The existence of weights w with all Eµ ≥ c > 0 also implies that a solution for
any other positive constant can be constructed. This is a consequence of the
fact that the function Sw(ξ) = sign(w · ξ) only depends on the direction of w
in N-dim. feature space while it is invariant under changes of the norm |w|.
3.3.2 Iterative Hebbian training algorithms
In order to answer question (Q1) in Sec. 3.2, we consider iterative learning al-
gorithms which present a single
󰁱
ξν(t), Sν(t)
T
󰁲
at time step t of the training
process. Often, the sequence of examples corresponds to repeated cyclic presen-
tation, i.e.4
ν(t) = 1, 2, 3, . . . , P, 1, 2, 3, . . . (3.11)
where each loop through the examples in D is called an epoch in the literature.
A frequently used alternative is random sequential presentation, where at each
time step t one of the examples in D is selected with equal probability 1/P.",The+Shallow+and+the+Deep.pdf
"time step t one of the examples in D is selected with equal probability 1/P.
The specific form of perceptron updates we consider in the following is:
Generic iterative perceptron updates (weights)
at discrete time step t
- determine the index ν(t) of the current training example
- compute the local potential Eν(t) = w(t) · ξν(t) Sν(t)
T
- update the weight vector according to
w(t + 1) = w(t) + 1
N f(Eν(t)) ξν(t) Sν(t)
T . (3.12)
The scaling of the update with N is arbitrary and is used here to achieve con-
sistency with the literature. In order to turn (3.12) into a practical training
algorithm, the prescription has to be completed by specifying initial condition
w, and by defining a stopping criterion, obviously.
Together with the definition of the sequence ν(t), the so-called modulation
function f(. . .) determines the actual training algorithm and we assume here
that it depends only on the local potential of the actual training example. Note",The+Shallow+and+the+Deep.pdf
"that it depends only on the local potential of the actual training example. Note
that (3.12) constitutes a realization of Hebbian learning: the change of a com-
ponent wj of the weight vector is proportional to the“pre-synaptic”input ξµ
j and
the “post-synaptic” output Sµ
T .
As a consequence, the weight vector accumulates Hebbian terms ξµ Sµ
T start-
ing from the given initialization w(0). Most frequently, we will consider a so-
called tabula rasa initialization, i.e. w(0) = 0. In this case, after performing
4Formally, this can be represented by the function ν(t) = mod[(t−1), P] + 1 for t ∈ Z+",The+Shallow+and+the+Deep.pdf
"38 3. THE PERCEPTRON
updates at time steps t = 1, 2, . . . τ the weight vector is bound to have the form
w(τ) = 1
N
P󰁛
µ=1
xµ(τ) ξµ Sµ
T . (3.13)
This implies that the resulting perceptron weight vector is a linear combi-
nation of the vectors ξµ ∈ D and the so-called embedding strengths xµ(τ) ∈ R
quantify their specific contributions.
Assuming that w(0) = 0, it is also possible to rewrite the update (3.12) in
terms of the embedding strengths. At the end of the training process, the actual
weight vector can be constructed according to Eq. (3.13). Hence, the following
formulation is equivalent to (3.12):
Generic iterative perceptron updates (embedding strengths)
at discrete time step t
- determine the index ν(t) of the current training example
- compute the local potential Eν(t) = w(t) · ξν(t) Sν(t)
T
- update the embedding strength xν(t) according to
xν(t)(t + 1) = xν(t)(t) + f(Eν(t)) (3.14)
(all other embedding strengths remain unchanged at time t).",The+Shallow+and+the+Deep.pdf
"xν(t)(t + 1) = xν(t)(t) + f(Eν(t)) (3.14)
(all other embedding strengths remain unchanged at time t).
Many perceptron algorithms can be formulated directly in the weights, cf.
(3.12), or in terms of the embedding strengths as in (3.14). While in principle
equivalent, we note that the number of variables used to represent the percep-
tron during training is N in the weight vector formulation and P if embedding
strengths are updated. Thus, the computational efficiency of training and the
corresponding storage needs will depend on the ratio P/N, in practice.
Note that the simple correspondence between (3.12) and (3.14) can be lost if
the structure of the actual updates is modified. Constraints of the form xµ ≥ 0
are imposed, for instance, in the AdaTron algorithm [AB89, BAK91] for the
perceptron of optimal stability, see Sec. 3.6. This prevents a straightforward
formulation of the training scheme as an iteration in weight space. Of course,",The+Shallow+and+the+Deep.pdf
"formulation of the training scheme as an iteration in weight space. Of course,
one can always construct the weight vector via relation (3.13) if needed.",The+Shallow+and+the+Deep.pdf
"3.3. THE ROSENBLATT PERCEPTRON 39
3.3.3 The Rosenblatt perceptron algorithm
In terms of the generic algorithm (3.12, 3.14), the Rosenblatt perceptron algo-
rithm is specified by
- tabula rasa initial conditions: w(0) = 0 or, equivalently, {xµ(0) = 0}P
µ=1
- deterministic, cyclic presentation of the examples in D according to (3.11)
- and the modulation
f(Eµ) = Θ[c − Eµ] =
󰀝
0 if Eµ > c
1 if Eµ ≤ c (3.15)
with the Heaviside function Θ[x] = 1 for x ≥ 0 and Θ[x] = 0 else.
In summary, the prescriptions (3.12) and (3.14) become:
Rosenblatt perceptron algorithm (3.16)
at discrete time step t
- determine the index ν(t) of the current example according to (3.11)
- compute the local potential Eν(t) = w(t) · ξν(t) Sν(t)
T
- update the weight vector according to
w(t + 1) = w(t) + 1
N Θ[c − Eν(t)]ξν(t) Sν(t)
T (3.17)
󰀵
󰀹󰀷
or, equivalently, increment the corresponding embedding strength
xν(t)(t + 1) = xν(t)(t) + Θ[c − Eν(t)]
(all other embedding strengths remain unchanged at time t)
󰀶
󰀺󰀸
(3.18)",The+Shallow+and+the+Deep.pdf
"xν(t)(t + 1) = xν(t)(t) + Θ[c − Eν(t)]
(all other embedding strengths remain unchanged at time t)
󰀶
󰀺󰀸
(3.18)
The update (3.16) modifies w(t) only if the example input is misclassified
by the current weight vector or correctly classified with Eν(t) < c. In this case,
a Hebbian term is added. The quantity xν(t) remains unchanged or increases
by 1 in every update step (3.18). Consequently, for tabula rasa initialization,
the resulting embedding strengths are non-negative integers.
Frequently, the simple setting c = 0 is considered and the resulting scheme
is often called the (Rosenblatt) perceptron algorithm. The underlying principle
of adding a Hebbian term for misclassified examples is referred to as “learning
from mistakes”. It is the basis of several other training algorithms discussed in
forthcoming sections.
The (c = 0)-algorithm stops as soon as all examples in D are correctly
classified: the evaluation of the modulation function yields Θ(−Eµ) = 0 in all",The+Shallow+and+the+Deep.pdf
"The (c = 0)-algorithm stops as soon as all examples in D are correctly
classified: the evaluation of the modulation function yields Θ(−Eµ) = 0 in all
forthcoming update steps. The algorithm is illustrated in terms of a simple
two-dimensional feature space and a data set D comprising six labeled inputs,
see Fig. 3.4. In this specific case, the perceptron classifies all feature vectors in",The+Shallow+and+the+Deep.pdf
"40 3. THE PERCEPTRON
t=0 t=1 t=2
t=3 t=4 t=5
t=6 — — — — — — — — — — — — — — — — — — — — — –
t = 0 : w(0) = 0 all xµ = 0
t = 1 : w(1) = 1
2 ξ1 x1 → 1
t = 2 : w(2) = w(1) zero update
t = 3 : w(3) = w(2) + 1
2 ξ3 x3 → 1
t = 4 : w(4) = w(3) zero update
t = 5 : w(5) = w(4) − 1
2 ξ5 x5 → 1
t = 6 : w(6) = w(5) alg. terminates
— — — — — — — — — — — — — — — — — — — — — –
Figure 3.4: Rosenblatt perceptron algorithm.
Illustration of the training scheme with c = 0 in (3.16) in a two-dimensional
feature space (N = 2). A set of six examples is presented sequentially. Empty
circles represent feature vectors labeled with ST = −1, filled circles mark data
from class ST = +1. Initial conditions correspond to w(0) = 0 (tabula rasa).
Only one epoch of training is considered with ν(t) = t = 1, 2, . . . , 6. At each time
step, ξt is marked by a shaded circle. The current weight vector is either updated
by adding a Hebbian term if example t is misclassified (time steps t = 1, 3, 5) or",The+Shallow+and+the+Deep.pdf
"by adding a Hebbian term if example t is misclassified (time steps t = 1, 3, 5) or
it remains unchanged if the current classification is correct already (time steps
t = 2, 4, 6). We refer to the latter as zero updates. Actual non-zero updates
are given by the addition (St
T = +1) or subtraction (St
T = −1) of 1
N ξt which is
displayed as an arrow in the illustration. The resulting weight vector is shown
in the next time step. For the specific data set considered here, all examples
are correctly classified after one epoch already. In general, the data set has to
be presented several times before the Rosenblatt algorithm terminates.",The+Shallow+and+the+Deep.pdf
"3.3. THE ROSENBLATT PERCEPTRON 41
D correctly, and the algorithm stops already after one sweep through the data
set, i.e. one epoch of training.
We will show in Sec. 3.3.5 that the Rosenblatt algorithm (3.15) converges
in a finite number of steps and finds a weight vector that solves the percep-
tron storage problem (3.7,3.9), provided the given data set is indeed linearly
separable.
3.3.4 The perceptron algorithm as gradient descent
We have introduced the Rosenblatt algorithm more or less intuitively as a form of
iterative Hebbian learning. Alternatively, we can motivate it as the minimization
of the cost function
E(w) =
󰁛
µ=1
eµ(w) = 1
N
P󰁛
µ=1
Θ[c−w · ξµSµ
T ] (c−w · ξµSµ
T ) , (3.19)
which is written as a sum over contributions eµ of the given examples in the data
set. The example specific term is zero if the data point is classified correctly
with w· ξµSµ
T > c and it contributes the linear costs (c−w· ξµSµ
T ) > 0 otherwise.",The+Shallow+and+the+Deep.pdf
"with w· ξµSµ
T > c and it contributes the linear costs (c−w· ξµSµ
T ) > 0 otherwise.
Note that the objective function (3.19) is frequently referred to as the hinge loss,
see for instance [HTF01].
Strictly speaking, the function E is not differentiable wherever w· ξµ = 0 for
one or several examples. In order to cope with this difficulty one can resort to
the consideration of the sub-derivative or sub-gradient method, see [Bot04] for
a more detailed discussion. From a practical point of view, we can ignore this
subtlety here and devise a gradient based minimization as outlined in Appendix
A.4 and Sec. 5.2.3 by setting
∇w eµ = − 1
N Θ[c − w · ξµSµ
T ] ξµSµ
T (3.20)
for a single example. Here ∇w denotes the gradient with respect to the weight
vector w ∈ RN . Hence the update (3.17) can be written as
w(t + 1) = w(t) − ∇w eν(t)
󰀏󰀏󰀏
w=w(t)
(3.21)
where the currently considered example is given by ν(t) according to (3.11). An",The+Shallow+and+the+Deep.pdf
"w(t + 1) = w(t) − ∇w eν(t)
󰀏󰀏󰀏
w=w(t)
(3.21)
where the currently considered example is given by ν(t) according to (3.11). An
update step along the negative gradient tends to decrease eν(t) and, thus, the
total cost function E.
We will encounter a number of learning algorithms in quite diverse settings
which are guided by the gradient of an objective function with respect to the
entire data set or single example contributions as above, see Appendix A.4 for
a general discussion.
The Rosenblatt algorithm resembles the minimization of E(w) by stochas-
tic gradient descent (SGD), which is presented and discussed more generally in
Appendix 5.2.3 and in Chapter 5. In contrast to SGD, however, the original
prescription as suggested by Rosenblatt and discussed here presents examples",The+Shallow+and+the+Deep.pdf
"42 3. THE PERCEPTRON
in the deterministic sequential order (3.11). The formulation in terms of embed-
ding strengths has the structure of coordinate descent as discussed in Appendix
A.5.1: at every step, only one xµ is updated. However, its potential increment
(3.18) is not given by the derivative ∂/∂xµ of E in Eq. (3.19).5
Another important difference to the generic SGD is that, due to the specific
structure of the problem, it is not necessary to introduce and fine-tune a learning
rate in order to achieve convergence. In fact, one can show that for tabula rasa
initialization w(0) = 0, a constant learning rate η as in
w(t + 1) = w(t) − η
N ∇w eν(t)
󰀏󰀏󰀏
w=w(t)
= w(t) + η
N Θ[c − Eν(t)]ξν(t) Sν(t)
T
would simply rescale the resulting embedding strengths and thus also the weight
vector by a factor η as compared to (3.21). This would affect neither the con-
vergence behavior nor the resulting classification scheme. Note, however, that",The+Shallow+and+the+Deep.pdf
"vergence behavior nor the resulting classification scheme. Note, however, that
the influence of a time-dependent η(t) and its interplay with a potential nor-
malization of the weight vector is non-trivial, see for instance [BSS94].
While the interpretation as a gradient based method helps to relate the
Rosenblatt algorithm to other machine learning schemes, we will not make use
of it explicitly in the following section. For linearly separable data sets, conver-
gence can be shown explicitly without referring to gradient descent.
3.3.5 The Perceptron Convergence Theorem
The Perceptron Convergence Theorem is one of the most important, fundamen-
tal results in the field. The guaranteed convergence of the Rosenblatt perceptron
algorithm for linearly separable problems has played a key role for the popularity
of the perceptron framework:
Perceptron Convergence Theorem (PCT) (3.22)
For linearly separable data D = {ξµ, Sµ
T }P
µ=1, the Rosenblatt perceptron",The+Shallow+and+the+Deep.pdf
"of the perceptron framework:
Perceptron Convergence Theorem (PCT) (3.22)
For linearly separable data D = {ξµ, Sµ
T }P
µ=1, the Rosenblatt perceptron
algorithm stops after a finite number of update steps (3.17) or (3.18) and
yields a weight vector w with w · ξµSµ
T ≥ c > 0 for all µ = 1, 2, . . . P.
In the following we outline the proof of convergence. We consider a linearly
separable D = {ξµ, Sµ
T }P
µ=1
, which implies that at least one solution w∗ of the
Perceptron Storage Problem (3.7,3.9) exists with
󰀋
sign (w∗· ξµ) = Sµ
T
󰀌P
µ=1 or, equivalently,
󰀋
Eµ∗ = w∗ · ξµ Sµ
T ≥ c > 0
󰀌P
µ=1
(3.23)
for some positive constant c.
5It is left to the reader to work out the derivative explicitly.",The+Shallow+and+the+Deep.pdf
"3.3. THE ROSENBLATT PERCEPTRON 43
We do not have to further specify w∗ here. In fact, for a given D there could
be many solutions of the form (3.23), but here it is sufficient to assume the
existence of at least one. We will furthermore denote its squared norm as
Q∗ ≡ w∗ · w∗ = |w∗|2. (3.24)
Note that any pair of vectors w, w∗ ∈ RN satisfies
0 ≤ (w · w∗)2
|w∗|2 |w|2 = cos2 ∠{w, w∗} ≤ 1. (3.25)
As discussed above, the algorithm yields - after t time steps - a weight vector
of the form
w(t) = 1
N
P󰁛
µ=1
xµ(t) ξµSµ
T for w(0) = 0. (3.26)
In the Rosenblatt algorithm the quantity xµ(t) is an integer that counts how
often example µ has contributed a Hebbian term to the weights, cf. Sec. 3.3.3,
The total number of non-zero updates is, therefore, given by
M(t) =
P󰁛
µ=1
xµ(t). (3.27)
Now let us consider the projection R(t) = w(t) · w∗. Inserting (3.26) and ex-
ploiting the condition (3.23) we obtain the following lower bound:
R(t) = 1
N
P󰁛
µ=1
xµ(t) [w∗ · ξµSµ
T ] = 1
N
P󰁛
µ=1
xµ(t) Eµ∗
󰁿󰁾󰁽󰂀
≥c",The+Shallow+and+the+Deep.pdf
"ploiting the condition (3.23) we obtain the following lower bound:
R(t) = 1
N
P󰁛
µ=1
xµ(t) [w∗ · ξµSµ
T ] = 1
N
P󰁛
µ=1
xµ(t) Eµ∗
󰁿󰁾󰁽󰂀
≥c
≥ 1
N c M(t). (3.28)
Similarly, we consider the squared norm Q(t) = w(t)· w(t) of the trained weight
vector. At time step t with presentation of example ν(t) it changes as
Q(t + 1) =
󰀕
w(t) + 1
N Θ
󰁫
c − Eν(t)
󰁬
ξν(t) Sν(t)
T
󰀖2
(3.29)
= Q(t) + 2
N Θ
󰁫
c − Eν(t)
󰁬
Eν(t) + 1
N2 Θ2
󰁫
c − Eν(t)
󰁬󰀏󰀏󰀏ξν(t)
󰀏󰀏󰀏
2
In any finite data set D, one of the examples will have the largest norm. We
can therefore always identify the quantity
Γ ≡ 1
N max
µ
󰁱
|ξµ|2
󰁲P
µ=1
, (3.30)
where the scaling with dimension N is convenient in the following6. Next, we
observe that Θ2(x) = Θ(x) for all x. Furthermore we note that
6We could also consider the simpler, less general case of normalized inputs |ξµ|2 = ΓN.",The+Shallow+and+the+Deep.pdf
"44 3. THE PERCEPTRON
Θ[c − Eν(t)] = 0 and Eν(t) ≥ c in a zero learning step, while
Θ[c − Eν(t)] = 1 and Eν(t) < c in a non-zero learning step.
As a consequence, we can replace all Eν(t) by c in Eq. (3.29) to obtain the upper
bound
Q(t + 1) ≤ Q(t) + 2
N c Θ[c − Eν(t)] + 1
N ΓΘ[c − Eν(t)] (3.31)
Here we exploit the fact that Q changes only in non-zero updates with Θ[. . .] = 1.
Taking into account the initial value Q(0) = 0, we can conclude that
Q(t) ≤ 1
N (2 c + Γ) M(t), (3.32)
where M(t) is the number of non-zero changes of Q. In summary, we have ob-
tained the two bounds
R(t) ≥ 1
N c M(t) and Q(t) ≤ 1
N (2c + Γ) M(t), (3.33)
respectively. Exploiting Eq. (3.25) we can write
1 ≥ (w(t) · w∗)2
(|w(t)||w∗|)2 = R2(t)
Q∗ Q(t) ≥
1
N2 c2 M2(t)
Q∗ 1
N (2c + Γ)M(t) = c2
Q∗ N (2c + Γ) M(t).
(3.34)
We conclude that
M(t) ≤ M∗ = (2c + Γ) N Q∗
c2 , (3.35)
where the right hand side involves only constants: N and Γ are obtained directly",The+Shallow+and+the+Deep.pdf
"(3.34)
We conclude that
M(t) ≤ M∗ = (2c + Γ) N Q∗
c2 , (3.35)
where the right hand side involves only constants: N and Γ are obtained directly
from the given data set. The constants c and Q∗ characterize the assumed
solution w∗, which is - of course - unknown a priori. However, in any case, Eq.
(3.35) implies that the number of non-zero learning steps remains finite, if a
solution w∗ exists for the given D. In other words, after at most M∗ non-zero
learning steps, the perceptron classifies all examples correctly with Eµ ≥ c > 0.
The number of required training epochs is also upper-bounded by M∗, because
- as long as the algorithm does not stop - at least one non-zero step must occur
in every epoch.
The dependence of M∗ on the constant c deserves further attention: In the
limit c → 0, the upper bound appears to diverge (M∗ → ∞). However, if D
is linearly separable, solutions w∗ with Q∗ ∝ c2 can be found for small values",The+Shallow+and+the+Deep.pdf
"is linearly separable, solutions w∗ with Q∗ ∝ c2 can be found for small values
of c. This is due to the linear dependence of Eµ∗ on |w∗| = √Q∗, which we
already discussed in the context of Eq. (3.10). In the limit c → 0 with Q∗ ∝ c2
the upper bound becomes
lim
c→0
M∗ ≈ ΓN Q∗
c2 with Q∗
c2 = const. (3.36)
Hence, we can express the PCT (3.22) without referring to a specific value of c.",The+Shallow+and+the+Deep.pdf
"3.3. THE ROSENBLATT PERCEPTRON 45
3.3.6 A few remarks
The number of training steps
According to the PCT, the required number of training steps is finite for linearly
separable data. However, their actual number depends on the detailed proper-
ties of D and can be very large, see [Roj96] for a discussion of the computational
complexity of the Rosenblatt algorithm and further references.
More efficient alternatives to the simple Rosenblatt algorithm can be devised,
for instance based on Linear Programming methods [Fle00,PAH19] as discussed
in [Roj96].
Non-separable data
For the convergence proof, we had to assume the existence of a solution, i.e.
linear separability. The Perceptron Convergence Theorem does not provide
direct insight into the algorithm’s performance if D is not separable. We will
consider the problem of finding approximative solutions with minimum or low
number of errors in non-separable data in Sec. 4.1.
Existence of a solution",The+Shallow+and+the+Deep.pdf
"number of errors in non-separable data in Sec. 4.1.
Existence of a solution
In practice, it turns out difficult to decide whether a given D is linearly separable
or not, cf. (Q2) in Sec. 3.2. If a solution is not found by the Rosenblatt algorithm
after a number of steps, this could imply that it simply should be run for more
steps or that, indeed, a solution does not exist.
A theorem borrowed from the theory of duality in optimization, see Ap-
pendix A.3.4 and [Fle00,PAH19], provides a surprisingly clear criterion for lin-
ear separablity, which is closely related to Farkas’ Lemma [Fle00]. It is known
as Gordan’s Theorem of the Alternative in the literature [BB00, Man69]. In a
notation familiar from the previous sections it reads:
Gordan’sTheorem of the Alternative (3.37)
For a given matrix χ ∈ RN×P , exactly one of the following
statements is true:
(1) a vector w ∈ RN exists with [χ⊤w]µ > 0 for all µ = 1, 2, . . . , P
or",The+Shallow+and+the+Deep.pdf
"statements is true:
(1) a vector w ∈ RN exists with [χ⊤w]µ > 0 for all µ = 1, 2, . . . , P
or
(2) a non-zero vector 󰂓y ∈ RP exists with yµ ≥ 0 for all µ and χ 󰂓y = 0.
If the matrix χ comprises the oriented input vectors of a given data set D,
χ =
󰁫
ξ1S1
T , ξ2S2
T , . . . ξP SP
T
󰁬
,
we can interpret w as the weight vector of a perceptron and statement (1)
corresponds to (homogeneous) linear separability of the data set.",The+Shallow+and+the+Deep.pdf
"46 3. THE PERCEPTRON
On the other hand, (2) states that a linear combination of the form
P󰁛
µ=1
yµ ξµ Sµ
T = 0 with {yµ ≥ 0}P
µ=1 and 󰂓y ∕= 0
exists. This goes beyond the condition of linearly dependent columns ξµSµ
T of χ,
which would be relevant in the context of solving equations. The consideration of
inequalities requires the existence of at least one non-zero vector of non-negative
coefficients 󰂓y resulting in a zero linear combination.
Observing that χ⊤ χ 󰂓y = 0 ⇔ χ󰂓y = 0 we can re-formulate (2) also in terms of
the null-space (the space of eigenvectors 󰂓y with eigenvalue zero) of the symmetric
and positive semi-definite matrix C = [χ⊤χ]/N ∈ RP×P . We will encounter and
make use of the matrix C frequently in later sections.
In practice, checking the validity of condition (2) in Gordan’s Theorem may
be quite involved: The computation of the null-space can be costly, already.
Finding non-negative coefficients 󰂓y for zero linear combinations is also non-
trivial, in general.",The+Shallow+and+the+Deep.pdf
"Finding non-negative coefficients 󰂓y for zero linear combinations is also non-
trivial, in general.
Learning the unlearnable
In this context, several algorithms have been suggested that converge for both
separable and non-separable data sets. As one example, D. Nabutovsky and
E. Domany suggest such an iterative training procedure in [ND91]. For lin.
sep. problems, a consistent perceptron vector is found. However, unlike the
Rosenblatt algorithm, the algorithm terminates also in the presence of non-
separable data and - in this case - indicates that no solution exists. In [ND91],
references to other, similar concepts are provided and the suggested scheme is
compared with techniques based on Linear Programming.
3.4 The capacity of a hyperplane
The existence of successful training algorithms for lin. sep. data is certainly of
great value. The question remains how relevant linear separability is or, in other
words, how serious the restriction to linearly separable functions is.",The+Shallow+and+the+Deep.pdf
"words, how serious the restriction to linearly separable functions is.
Linear separability depends on details of the individual, given data set. How-
ever, surprisingly general qualitative and quantitative results can be obtained
which require only mild assumptions about the data.
3.4.1 The number of linearly separable dichotomies
In this section we address (Q3) from Sec. 3.2 and determine the number C(P, N)
of linearly separable binary class assignments or dichotomies7 of P feature vec-
tors in an N-dimensional input space. Quite surprisingly, this is possible under
rather mild conditions on the input data.
7“There are two kinds of people: Those who couch everything in dichotomies and those
who do not.” — Unknown",The+Shallow+and+the+Deep.pdf
"3.4. THE CAPACITY OF A HYPERPLANE 47
Figure 3.5: One-dimensional inputs ξ can be separated into two classes by
the origin in C(P, 1) = 2 ways, independent of P.
The derivation for general P and N has been published several times in the
literature [Win61, Cov65, MD89a] and was also reviewed in [HKP91]. Here we
follow the presentation in [MD89a] to a large extent. Interestingly, the result
is closely related to very early findings concerning high-dimensional geometry
obtained by the Swiss mathematician Ludwig Schläfli in the 19th century al-
ready [Sch01].
More recently, very similar results have been re-discovered from a different
point of view in the context of Deep Networks, see [Str19, PMB14, MPCB14].
They relate to more general theorems concerning Face-Count Formulas for
Partitions of Space by Hyperplanes, which are due to T. Zaslavsky [Zas75].
Specifically, counting the number of response regions in layered networks with",The+Shallow+and+the+Deep.pdf
"Specifically, counting the number of response regions in layered networks with
piecewise linear activation functions leads to very similar considerations and
results [PMB14].
Some results for small N or P
We first address straightforward cases involving low dimensions N and/or small
numbers P of feature vectors. The first, obvious result is that
C(1, N) = 2 for all N.
It corresponds to the fact that a single point can always be mapped to S = ±1
by a a hyperplane through the origin and setting the orientation corresponding
to the target label. Here we exclude feature vectors ξ = 0, which could always
be avoided by an over-all translation of the data set.
The second observation
C(P, 1) = 2 for all P,
reflects the insight that one-dimensional inputs ξ ∈ R can only be separated by
the origin in two ways: By setting S = sign[ξ] or by the inverted S = −sign[ξ],
see Fig. 3.5 for an illustration.
In N = 2 dimensions it is easy to see that for P = 2 typically all 2P",The+Shallow+and+the+Deep.pdf
"see Fig. 3.5 for an illustration.
In N = 2 dimensions it is easy to see that for P = 2 typically all 2P
dichotomies can be realized, as displayed in Fig. 3.6 (panel a). However, we
have to exclude a special case: If the two input vectors are collinear with the
origin, it is impossible to separate them by a line representing a homogeneously
linearly separable function.
For P = 3 we find that C(3, 2) < 23, see Fig. 3.6 (panels b,c) panel, even for
generic data sets: In panel (b), of the eight possible dichotomies, the two cases
with S1 = S2 = S3 = +1 or −1 cannot be represented by a plane through the",The+Shallow+and+the+Deep.pdf
"48 3. THE PERCEPTRON
a)
 b)
 c)
 d)
Figure 3.6: Panel (a): Two two-dimensional feature vectors (large filled
circles) in general position, i.e. not collinear with the origin (small filled circle).
Here, four linearly separable dichotomies exist: either both inputs are assigned
to the same class S1 = S2 ∈ {−1, +1} or they are separated with S1 = −S2.
Panels (b,c): Three two-dimensional feature vectors in general position. Of
the 23 = 8 dichotomies only six are linearly separable, each one represented by
one of the three planes with two possible orientations. In panel (b) the two
dichotomies with S1 = S2 = S3 cannot be realized, in panel (c) the rightmost
point cannot be separated from the other two. Panel (d): As indicated by the
dotted line, two of the three feature vectors are collinear with the origin. As a
consequence only 4 different linearly separable functions can be found.
origin. Similarly, in panel (c) one of the data points cannot be separated from",The+Shallow+and+the+Deep.pdf
"origin. Similarly, in panel (c) one of the data points cannot be separated from
the other two, which also excludes two dichotomies. This implies that
C(3, 2) = 6.
Again, the result is only valid if no subset of two data points falls onto a line
through the origin. In panel (d), a counterexample is displayed. Here, only four
distinct dichotomies can be realized by a lin. sep. function, as it is not possible
to separate the collinear points by a line through the origin. As a consequence
they can only be assigned to the same class, as for instance by the two decision
boundaries shown in the illustration.
Similar, explicit insights can still be achieved for three-dimensional data,
which is left to the reader as an exercise. Again, one has to exclude specific
configurations in which two or more feature vectors are aligned. Moreover,
subsets of three data points should not fall into a two-dimensional plane that
includes the origin.
General N and P",The+Shallow+and+the+Deep.pdf
"subsets of three data points should not fall into a two-dimensional plane that
includes the origin.
General N and P
Our intuition – certainly that of the author – usually fails in higher dimensional
spaces, i.e. for N > 3. However, it is indeed possible to obtain C(P, N) for
general N and P under rather mild assumptions. Similar to the exclusion of
collinear input vectors in N = 2 and N = 3, we formulate the condition that
the data set should be in general position [Cov65]:",The+Shallow+and+the+Deep.pdf
"3.4. THE CAPACITY OF A HYPERPLANE 49
00
Figure 3.7: Two homogeneously linearly separable dichotomies DP=9
N=2 of the
same set of two-dimensional feature vectors. In both panels, all separating
planes in the grey-shaded areas would realize the same assignment of labels.
General position condition (3.38)
A set of vectors P =
󰀋
ξµ ∈ RN 󰀌P
µ=1 is in general position, if every subset
{ξρ}K
ρ=1 ⊂ P containing K ≤ N elements is linearly independent.
For P > N, obviously, subsets of more than N elements in N dimensions are
always linearly dependent. In sloppy terms, the general position conditions
requires that the vectors in P do not form more hyperplanes than necessary.
In a sense, the general position condition corresponds to the absence of
degeneracies in the data. Real world data, however, is frequently prone to
such degeneracies. Often, it is even assumed and exploited that nominally
high-dimensional feature vectors fall into lower-dimensional manifolds due to",The+Shallow+and+the+Deep.pdf
"high-dimensional feature vectors fall into lower-dimensional manifolds due to
correlations and interdependencies.
In the modelling and simulation of learning processes, e.g. when investigating
training algorithm performances, one often resorts to randomized feature vectors
of N independent components [HKP91]. The popular example of zero mean /
unit variance Gaussian random numbers, for instance, leads to data sets that
are in general position with probability one.
For the following argument we consider a fixed set of P input vectors ξ ∈ RN
in general position. A dichotomy DP
N of these feature vectors assigns a set of
binary labels:
DP
N : ξµ ∈ RN → Sµ ∈ {−1, +1} for µ = 1, 2, . . . P.
The aim is to work out how many of the in total 2P possible dichotomies cor-
respond to linearly separable functions. Their number will be referred to as
C(P, N).
As an illustration, Fig. 3.7 displays a set of P = 9 feature vectors in N = 2",The+Shallow+and+the+Deep.pdf
"C(P, N).
As an illustration, Fig. 3.7 displays a set of P = 9 feature vectors in N = 2
dimensions. The set has been labeled in two different ways, both of which are
linearly separable, as shown in the left and right panel of the figure. All planes",The+Shallow+and+the+Deep.pdf
"50 3. THE PERCEPTRON
00
ξP+1 ξP+1
Figure 3.8: An additional feature vector ξP+1 is added to the data set of Fig.
3.7. Its label is either non-ambiguously determined by the linearly separable
function as in the left panel, or the label is ambiguous, SP+1 = +1 or −1, as
shown in the right panel.
through the origin that fall into the shaded region would separate the two classes
as represented by filled (S = +1) versus empty circles (S = −1).
Now we assume that an additional feature vector ξP+1 is added to the data
set, as displayed in Figure 3.8. We note that two essentially different situations
can occur: In the left panel of the figure, all perceptrons that separate the
already known P examples would assign ξP+1 to the same class, i.e. SP+1 = −1
in the illustration. The dichotomy is said to be non-ambiguous with respect to
the new data point.
On the contrary, in the right panel the response is ambiguous with respect to",The+Shallow+and+the+Deep.pdf
"the new data point.
On the contrary, in the right panel the response is ambiguous with respect to
the additional feature vector: Some separating planes correspond to SP+1 = −1,
others to SP+1 = +1. We will denote by8
Z(P, N) the number of ambiguous
E(P, N) the number of non-ambiguous
󰀞
dichotomies DP
N w.r.t. SP+1.
We note immediately that the total number of linearly separable dichotomies is
C(P, N) = Z(P, N) + E(P, N)
since each labelling is either ambiguous or non-ambiguous with respect to SP+1.
Obviously, each non-ambiguous dichotomy DP
N contributes exactly one lin.
sep. labeling to C(P +1, N). However, each of the Z(P, N) ambiguous labellings
contributes two lin. sep. dichotomies in the enlarged data set with P + 1 feature
vectors, as we have the choice between SP+1 = +1 and −1 while retaining
separability. Hence we have
C(P + 1, N) = E(P, N) + 2 Z(P, N) = C(P, N) + Z(P, N). (3.39)
Unfortunately this is not yet a useful recursion because Z(P, N) is not known.",The+Shallow+and+the+Deep.pdf
"C(P + 1, N) = E(P, N) + 2 Z(P, N) = C(P, N) + Z(P, N). (3.39)
Unfortunately this is not yet a useful recursion because Z(P, N) is not known.
8Here the symbols E and Z are inspired by the German terms eindeutig and zweideutig.",The+Shallow+and+the+Deep.pdf
"3.4. THE CAPACITY OF A HYPERPLANE 51
ξP+1 ξP+1
HH
Figure 3.9: The projection of data into the auxiliary subspace H ⊥ ξP+1 is
either linearly separable in H (right panel, ambiguous case) or not separable as
in the non-ambiguous case (left panel), respectively.
However the following consideration shows that Z(P, N) can be related to
the number of lin. sep. dichotomies in N−1 dimensions. In Fig. 3.9 we introduce
the auxiliary (N −1)-dimensional subspace H, i.e. the hyperplane through the
origin which is orthogonal to ξP+1. In the left panel, H falls into the grey
shaded area of separating hyperplanes, while in the right panel it does not.
Next, we project the P feature vectors into the auxiliary space H as shown
in the figure. In situations exemplified by the left panel, the projections of the P
feature vectors into the (N − 1)-dim. auxiliary space are not linearly separable
in H. In the right panel however, examples with S = ±1 fall into opposite",The+Shallow+and+the+Deep.pdf
"in H. In the right panel however, examples with S = ±1 fall into opposite
half-spaces in H. In other words, their labels correspond to a linearly separable
dichotomy of the P examples in (N − 1) dimensions.
On can show that there is indeed a one-to-one correspondence between the
Z(P, N) ambiguous lin. sep. functions and all linearly separable dichotomies in
H. Assuming that there is nothing special about H, we conclude that Z(P, N) =
C(P, N − 1). Hence we obtain the recursion
C(P + 1, N) = C(P, N) + C(P, N − 1). (3.40)
The condition is, in fact, that the data points are in general position as defined
above in (3.38). If, for instance the added data point would fall onto a line with
another data point and the origin, clearly H could not be considered a generic
subspace. For data in general position, however, we can assume that H has the
same properties as any (N − 1)-dim. subspace would.
The number of linearly separable dichotomies",The+Shallow+and+the+Deep.pdf
"same properties as any (N − 1)-dim. subspace would.
The number of linearly separable dichotomies
It is straightforward to show that the following expression satisfies the recursion
and matches the above given initial values C(P, 1) = 2 and C(1, N) = 2:
C(P, N) =
󰀻
󰁁󰁁
󰀿
󰁁󰁁
󰀽
2
P for P ≤ N
2
N−1󰁛
i=0
󰀕P − 1
i
󰀖
for P > N,
(3.41)",The+Shallow+and+the+Deep.pdf
"52 3. THE PERCEPTRON
Figure 3.10: The fraction
Pls = C(P, N)/2P of linearly
separable functions versus α =
P/N for three different values
of N (5, 20, 100).
C(P, N)/2P
α = P/N
N =5
N =100
with the familiar bionomial coefficients
󰀃m
k
󰀄
= m!
k!(m−k)! .
The expression in the second line of (3.41) would also reproduce 2P for
P < N, as
󰀃m
k
󰀄
= 0 for k > m and 2 󰁓m−1
k=0
󰀃m
k
󰀄
= 2m. The formal consideration
of the two cases in (3.41) is only provided for the sake of clarity.
The proof can be done by induction and exploits that
󰀕P
i
󰀖
=
󰀕P − 1
i − 1
󰀖
+
󰀕P − 1
i
󰀖
for arbitrary i, P and
󰀕P
i
󰀖
= 0 for i < 0.
Therefore, C(P + 1, N) according to Eq. (3.41) satisfies
2
N−1󰁛
i=0
󰀕P
i
󰀖
= 2
N−1󰁛
i=0
󰀕P −1
i − 1
󰀖
+ 2
N−1󰁛
i=0
󰀕P −1
i
󰀖
= 2
N−2󰁛
i=0
󰀕P −1
i
󰀖
󰁿 󰁾󰁽 󰂀
C(P,N−1)
+ 2
N−1󰁛
i=0
󰀕P −1
i
󰀖
󰁿 󰁾󰁽 󰂀
C(P,N)
which corresponds to the recursion relation (3.40).
Instead of the numbers C(P, N) themselves, we can also consider the fraction
of linearly separable dichotomies
Pls(P, N) = C(P, N)
2P =
󰀻",The+Shallow+and+the+Deep.pdf
"Instead of the numbers C(P, N) themselves, we can also consider the fraction
of linearly separable dichotomies
Pls(P, N) = C(P, N)
2P =
󰀻
󰁁󰁁
󰀿
󰁁󰁁
󰀽
1 for P ≤ N
2
1−P
N−1󰁛
i=0
󰀕P − 1
i
󰀖
for P > N,
(3.42)
This allows to compare and represent the results graphically for different N as
shown in Fig. 3.10. The fraction Pls can be interpreted as the probability for a
set of randomly assigned labels {Sµ = ±1}P
µ=1 to be linearly separable.
3.4.2 Discussion of the result
For the above arguments and derivation it is essential to assume that the data
set is in general position. Typically, violations of this condition will reduce the
number of linearly separable dichotomies. In this sense, the results given in Eqs.
(3.41) and (3.42) can be interpreted as an upper bound even in more realistic
settings.",The+Shallow+and+the+Deep.pdf
"3.4. THE CAPACITY OF A HYPERPLANE 53
Figure 3.10 shows the fraction of linearly separable dichotomies C(P, N)/2P
as a function of P/N. This scaling allows to conveniently display several curves
for selected values of N in one graph.
The most striking and - at first sight - counterintuitive feature of the result
(3.41) is that C(P, N) = 2P for P ≤ N. It implies that all possible dichotomies
are linearly separable, as long as the number of feature vectors does not exceed
their dimension N. We have explicitly confirmed the statement for N = 2 in
the discussion of Fig. 3.6 (panel a).
For P > N, the fraction of linearly separable dichotomies decreases and
approaches its limiting value Pls(α) → 0 for α → ∞. As exemplified in Fig. 3.10
for dimensions N = 5, 20, 100, the region with Psl ≈ 1 extends with increasing
N and at the same time the decrease of Pls with α becomes steeper. Moreover,
all curves intersect in α = P/N = 2. The behavior in the simultaneous limits",The+Shallow+and+the+Deep.pdf
"all curves intersect in α = P/N = 2. The behavior in the simultaneous limits
N → ∞, P → ∞ with finite α = P/N (3.43)
is given as
P∞
ls =
󰀝 1 for α ≤ 2
0 for α > 2.
Hence, in high dimensions, linearly separability is guaranteed (with probability
one) up to P ≤ 2N, which motivates to define the so-called
storage capacity of the perceptron: αc = 2. (3.44)
In general, the storage capacities of more powerful network architectures are
non-trivial to obtain, see for instance [EB01,WRB93].
For all N and P in the perceptron we have obtained the even stronger result
that Pls = 1 exactly for α ≤ 1, i.e. up to P = N. The latter is often referred
to as the Vapnik-Chervonenkis (VC) dimension of the perceptron, see [HTF01]
for a discussion and further references.
At first sight, the storage capacity of the perceptron or other student systems
relates only to their ability to reproduce a given data set and implement the
corresponding labels. However, as we will see in the following, storage capacity",The+Shallow+and+the+Deep.pdf
"corresponding labels. However, as we will see in the following, storage capacity
and VC-dimension play an important role with respect to the generalization
ability and learning of a rule from example data.
3.4.3 Time for a pizza or some cake
It is amusing to note that the counting of lin. sep. dichotomies is closely related
to a popular mathematical puzzle known as the lazy caterer’s problem: The max-
imum number of pieces obtained by K straight cuts through a two-dimensional
pizza (or a pancake, depending on cultural background) [Wet78,MD09] is given
by
c(K, 2) =
󰀕K
0
󰀖
+
󰀕K
1
󰀖
+
󰀕K
2
󰀖
= 2 + K + K2
2 = 1, 2, 4, 7, 11, 16 . . . (3.45)",The+Shallow+and+the+Deep.pdf
"54 3. THE PERCEPTRON
⇒
Figure 3.11: The Pizza Connnection: K planar cuts through the center (•)
of an N-dim. sphere (left panel) correspond to K − 1 arbitrary straight cuts
through the (N − 1)-dim. surface of each flattened hemisphere (right panel).
The corresponding result for three-dimensional objects is known as the cake
number [Wei99,Str19], i.e. the maximum number of pieces obtained by K planar
cuts:
c(K, 3) =
󰀕K
0
󰀖
+
󰀕K
1
󰀖
+
󰀕K
2
󰀖
+
󰀕K
3
󰀖
= 6 + 5K + K3
6 = 1, 2, 4, 8, 15, 26 . . .
(3.46)
In fact, one can show that the recursion relation
c(K + 1, N) = c(K, N) + c(K, N − 1)
holds in complete analogy to Eq. (3.40), albeit with different initial values. See
also [Str19] for the proof of a similar result.
We can directly relate the C(P, N) with the generalized cake numbers c(K, N):
The left panel of Figure 3.11 displays a sphere in N = 3 dimension, which is cut
by four equatorial planes through the center. They correspond to four feature",The+Shallow+and+the+Deep.pdf
"by four equatorial planes through the center. They correspond to four feature
vectors and divide the normalized weight space into regions representing 14 lin.
sep. dichotomies. Note that one of the planes (marked by the red circle) is
oriented such that the corresponding upper hemisphere is shown in top view. A
two-dimensional simplified representation of the situation is shown in the right
panel, which displays a schematic projection onto the selected equatorial plane.
Apparently, there is a 1:1 correspondence of K planar cuts through the cen-
ter of a spherical cake to K − 1 unrestricted cuts through a two-dimensional
pizza which represents one of the hemispheres. The same holds for the lower
hemisphere in the left panel. Quite generally, this yields the simple relation
C(P, N) = 2 c(P − 1, N − 1)
between the number of lin. sep. functions and the generalized cake numbers.",The+Shallow+and+the+Deep.pdf
"3.5. LEARNING A LINEARLY SEPARABLE RULE 55
3.5 Learning a linearly separable rule
Obviously it is not the ultimate goal of perceptron training to reproduce the
labels in a given data set, only. This could be done quite efficiently by simply
storing D in memory and look it up when needed.
In general, it is the aim of machine learning to extract information from
given data and formulate (parameterize) the acquired insight as a hypothesis,
which can be applied to novel data not contained in D in the working phase.
Addressing (Q4) from Sec. 3.2, we assume that an unknown, linearly sepa-
rable function or rule exists, which assigns any possible input vector ξ ∈ RN to
the binary output SR(ξ) = ±1, where the subscript R stands for “rule”.
Training of a perceptron from D = {ξµ, Sµ
T } should infer some information
about the unknown SR(ξ) as long as the training labels Sµ
T are correlated with
the correct Sµ
R = SR(ξµ). In the simplest and most clear-cut situation we have
Sµ
T = Sµ",The+Shallow+and+the+Deep.pdf
"T are correlated with
the correct Sµ
R = SR(ξµ). In the simplest and most clear-cut situation we have
Sµ
T = Sµ
R for all µ = 1, 2, . . . , P. (3.47)
Hence, we assume that the set of training data D = {ξµ, Sµ
R}P
µ=1 comprises
perfectly reliable examples for the application of the rule. It does, for instance,
not contain mislabeled example data and is not corrupted by any form of noise
in the input or output channel.
We will use the notation Sµ
R or SR(ξµ) for labels which are given by a
rule, explicitly. Here, this indicates that the examples in D represent a linearly
separable function, indeed. Note that more general data sets D = {ξµ, Sµ
T }P
µ=1
can also be linearly separable without direct correspondence of the Sµ
T to a
lin. sep. rule: a small set of examples for a non-separable function or examples
corrupted by noise can be very well linearly separable.
3.5.1 Student-teacher scenario
Even for data sets of reliable noise-free examples only, it is not a priori clear",The+Shallow+and+the+Deep.pdf
"3.5.1 Student-teacher scenario
Even for data sets of reliable noise-free examples only, it is not a priori clear
that a perceptron is able to reproduce the labels in D correctly, since the target
might not be linearly separable.
To further simplify our considerations, we restrict ourselves to cases, in which
the rule is indeed given by a function of the form
SR(ξ) = sign (w∗ · ξ) (3.48)
for a particular weight vector w∗. In fact, all weight vectors λw∗ with λ > 0
would define the same rule and, therefore, we will assume |w∗| = 1 implicitly,
without loss of generality.
A perceptron with weights w∗ is always correct9. It is therefore referred to as
the teacher perceptron (or teacher, for short) and can be thought of as providing
the example data from which to learn. Likewise, the trained perceptron with
adaptive weights w will be termed the student.
9The characteristic trait of many teachers, at least in their self-perception.",The+Shallow+and+the+Deep.pdf
"56 3. THE PERCEPTRON
Figure 3.12: Generalization er-
ror of the perceptron in a stu-
dent/teacher scenario. For N-
dim. random input vectors gener-
ated according to an isotropic den-
sity, the probability of disagree-
ment between student vector w and
teacher w∗ is proportional to the
red shaded area, i.e. to the angle
∠(w, w∗).
w
w∗
∕=
∕=
The choice of a specific student weight vector corresponds to a particular
hypothesis, which is represented by the linearly separable function
Sw(ξ) = sign (w · ξ) for all ξ ∈ RN . (3.49)
Student-teacher scenarios have been used extensively to model machine
learning processes, aiming at a principled understanding of the relevant phe-
nomena. They conveniently allow to control the complexity of the target rule
vs. that of the trained system in model situations, thus enabling the systematic
study of a variety of setups.
In the following sections we will consider idealized situations, in which stu-",The+Shallow+and+the+Deep.pdf
"study of a variety of setups.
In the following sections we will consider idealized situations, in which stu-
dent and teacher both represent linearly separable functions. Under this con-
dition, a plausible guideline for training the student from a set D = {ξµ, Sµ
R}
is to achieve perfect agreement with the unknown teacher in terms of the given
examples. However, the agreement should not only concern data in D as in the
storage problem; it should extend or generalize to novel data, ideally.
Frequently, the so-called generalization error serves a measure of success
of the learning process. In practical situations, it would correspond to the
performance of the student with respect to novel data, for instance in a test
set which was not used for training. In our idealized setup, we can revisit
the basic geometrical interpretation of linearly separable functions. Figure 3.12
displays a student-teacher pair of weight vectors. The illustration is, obviously,",The+Shallow+and+the+Deep.pdf
"displays a student-teacher pair of weight vectors. The illustration is, obviously,
two-dimensional. Note, however, that it can be interpreted as representing
the two-dimensional subspace spanned by N-dimensional vectors w, w∗, more
generally.
We assume that a test input ξ is generated according to an isotropic, un-
structured density anywhere in RN . The corresponding generalization error 󰂃g,
i.e. the probability for a disagreement
sign(w · ξ) ∕= sign(w∗ · ξ)
between student and teacher is directly proportional to the area of the shaded
segments, i.e. to the angle ∠(w, w∗):
󰂃g = 1
π arccos
󰀕 w · w∗
|w||w∗|
󰀖
. (3.50)",The+Shallow+and+the+Deep.pdf
"3.5. LEARNING A LINEARLY SEPARABLE RULE 57
ξ
w ∈ RN
|w|=const.•
V(3)
Figure 3.13: Dual geometrical interpretation of linear separability. Illustra-
tion in terms of labeled input vectors ξ ∈ R3 and L2-normalized weight vectors
with |w| = const. on the surface of an N-dim. hypersphere.
Left: A single, labeled input ξ1 separates all weight vectors with S1
w = +1 from
those with S1
w = −1. Right: A set of P labeled input vectors (here: P = 3)
defines the (darker) region V(3) of all weight vectors w with correct response
Sw(ξµ) = Sµ
R for µ = 1, 2, 3. For clarity, the vectors ξµ are not shown.
Orthogonal student-teacher pairs would result in 󰂃g = 1/2, which corresponds
to randomly guessing the output. Perfect agreement with 󰂃g = 0 is achieved
for w 󰀂 w∗. The latter statement holds true independent of the statistical
properties of the test data, obviously.
3.5.2 Learning in version space
In the classical setup of supervised learning, the training data comprises the",The+Shallow+and+the+Deep.pdf
"3.5.2 Learning in version space
In the classical setup of supervised learning, the training data comprises the
only available information about the rule. If the target rule is known to be
linearly separable and for reliable noise-free example data, it appears natural to
require that the hypothesis is perfectly consistent with D.
But is this a promising training strategy? In other words, can we expect
to infer meaningful information about the unknown w∗ by “just” solving the
perceptron storage problem with respect to D?
In order to obtain an intuitive insight, we re-visit and extend the geometri-
cal interpretation of linear separability. Following the so-called dual geometric
interpretation of linear separability, see Fig. 3.13 (left panel), we can interpret
weight vectors w as points in an N-dim. space. Note that every vector ξ defines
a hyperplane through the origin of this space, which separates weight vectors
w ∈ RN with positive w · ξ and Sw(ξ) = +1 from those with negative scalar",The+Shallow+and+the+Deep.pdf
"w ∈ RN with positive w · ξ and Sw(ξ) = +1 from those with negative scalar
product and Sw(ξ) = −1. Hence, given the correct target label SR(ξ), the plane
orthogonal to ξ separates correct from wrong students in N-dimensional weight
space.
Consequently, a set D of P labeled feature vectors defines a region or volume
of vectors w which reproduce Sw(ξµ) = SR(ξµ) for all µ = 1, 2, . . . , P, as",The+Shallow+and+the+Deep.pdf
"58 3. THE PERCEPTRON
V(3)
= V(4)
ξ1
ξ2
ξ3
ξ4
ξ1
ξ2
ξ3
ξ4
ξ5
Figure 3.14: Illustration of perceptron learning in version space.
Left: The hyperplane associated with the additional example ξ4 does not in-
tersect the version space V(3). Consequently, all weight vectors in V(3) already
classify ξ4 correctly and V(4) = V(3). Right: The hyperplane associated with
ξ5 cuts through V(4) and, consequently, V(5) corresponds to either the small
triangular region (green) or the remaining dark area, depending on the actual
target S5
R.
illustrated in Fig. 3.13 (right panel).
The set of all perceptron weight vectors which are consistent with the P
examples in D, i.e. which give Sµ
w = Sµ
R for all µ = 1, 2, . . . , P, is termed version
space and can be defined as
V =
󰀝
w ∈RN , w2 =1
󰀏󰀏
󰀏
󰀏sign(w · ξ
µ) = Sµ
R for all {ξµ, Sµ
R} ∈ D
󰀞
. (3.51)
In the following, we will use the notation V(P), if we want to refer to the number
of examples in D explicitly.",The+Shallow+and+the+Deep.pdf
"R for all {ξµ, Sµ
R} ∈ D
󰀞
. (3.51)
In the following, we will use the notation V(P), if we want to refer to the number
of examples in D explicitly.
In the context of the previous section, C(P, N) in Eq. (3.41) can be inter-
preted as the number of different version spaces that can be constructed for a
set of P input vectors in N-dimensional space by assigning linearly separable
target labels Sµ = ±1.
It is important to note that defining V as a set of normalized vectors with
w2 = 1 is convenient but not essential. Obviously, the normalization is irrelevant
with respect to the conditions sign(w · ξµ) = Sµ
R and could be replaced by any
other constant norm or even be omitted.
The version space is non-empty, V ∕= ∅, if and only if D is linearly separable:
If a (normalized) teacher vector w∗ defines the linearly separable rule repre-
sented by the examples in D, then w∗ ∈ V, necessarily. In words: at least the
teacher vector itself must be inside version space. However, in absence of any",The+Shallow+and+the+Deep.pdf
"teacher vector itself must be inside version space. However, in absence of any
information beyond the data set D, the unknown w∗ could be located anywhere
in V with equal likelihood.
The term learning in version space refers to the idea of admitting only hy-
potheses which are perfectly consistent with the example data. Let us assume",The+Shallow+and+the+Deep.pdf
"3.5. LEARNING A LINEARLY SEPARABLE RULE 59
that, given a set D of P reliable examples for a linearly separable rule, we have
identified some vector w ∈ V. According to the Perceptron Convergence The-
orem (3.22) this is always possible, e.g. by means of the Rosenblatt perceptron
algorithm10. In our low-dimensional illustration, Fig. 3.13 (right panel), this
means that we can always place a student vector w somewhere in the darker
region representing V(3) for P = 3 training examples.
Now, consider a fourth labeled example {ξ4, S4
R} as displayed in Fig. 3.14
(left panel). The hyperplane associated with ξ4 does not intersect the version
space V(3). Consequently all student vectors w ∈ V(3) fall into the same half-
space with respect to the new example and yield the same perceptron output
sign(w · ξ4).
This implies in turn that, if the extended data set {ξµ, Sµ
R}4
µ=1 is still linearly
separable, only one value of the target label S4
R is possible, which must be
S4",The+Shallow+and+the+Deep.pdf
"R}4
µ=1 is still linearly
separable, only one value of the target label S4
R is possible, which must be
S4
R = sign(w · ξ4) for w ∈ V(3). Consequently we have that V(4) = V(3); the
version space with respect to the extended data set remains the same as for the
previously known P = 3 examples. Hence, our strategy of learning in version
space does not require modifying the hypothesis or selecting a new student
vector w to parameterize it. In this sense, the input/output pair {ξ4, S4
R} is
un-informative in the given setting.
The situation is different in the case illustrated in the right panel of Fig.
3.14. Here, the data set is amended by {ξ5, S5
R} with the plane orthogonal to
ξ5 cutting through V(4) = V(3). Elements of V(4) on one side of the hyperplane
correspond to the perceptron response Sw(ξ5) = +1, while the others yield
Sw(ξ5) = −1.
Depending on the actual target S5
R of the additional example, the version",The+Shallow+and+the+Deep.pdf
"Sw(ξ5) = −1.
Depending on the actual target S5
R of the additional example, the version
space V(5) corresponds to either the green area or the remaining lighter region
in the illustration. In any case, the extended data set {ξµ, Sµ
R}5
µ=1 is linearly
separable and the new version space V(5) of consistent weight vectors is bound
to be smaller than the previous V(4) = V(3). The volume of consistent weight
vectors w shrinks due to the information associated with the new example data.
As we add more examples to the data set, the corresponding version space
can only remain the same or decrease in size. Indeed, one can show that V will
shrink to a point with P/N → ∞ under the rather mild condition of general
position discussed in the previous section.
Together with the fact that the teacher w∗ ∈ V(P) for any P, we can con-
clude that learning from version space will enforce w → w∗ with increasing
training set size. More precisely, we conclude that the angle ∠(w, w∗) → 0",The+Shallow+and+the+Deep.pdf
"clude that learning from version space will enforce w → w∗ with increasing
training set size. More precisely, we conclude that the angle ∠(w, w∗) → 0
for unnormalized vectors w, w∗. Hence, we can expect that learning in version
space yields hypotheses which agree with the unknown rule to a large extent,
provided the data set contains many examples. In the sense of the above dis-
cussed generalization error (3.50), it will achieve perfect generalization 󰂃g → 0
for P → ∞.
10. . . with subsequent normalization in order to match the definition (3.51) precisely.",The+Shallow+and+the+Deep.pdf
"60 3. THE PERCEPTRON
Figure 3.15: The generalization
error 󰂃g vs. α = P/N as ob-
tained from the number of ambigu-
ous / non-ambiguous linearly sep-
arable functions, cf. Eq. 3.52. The
curves correspond to N = 5, 20, 100
and to the limit N → ∞, respec-
tively (from left to right).
󰂃g
α = P/N
3.5.3 Generalization begins where storage ends
The above, elementary and illustrative considerations do not provide more con-
crete mathematical relations which, for instance, quantify the generalization
error as a function of the training set size P.
However, we can re-visit and exploit the result of Section 3.4, where we
counted the number of linearly separable functions of P feature vectors in N
dimensions under quite general assumptions [Win61,Cov65,MD89a].
Assume that the training process has exploited P examples for a linearly
separable rule and that we have identified a student weight vector in V(P) which
corresponds to one of the C(P, N) linearly separable dichotomies of the data.",The+Shallow+and+the+Deep.pdf
"corresponds to one of the C(P, N) linearly separable dichotomies of the data.
Implicitly, we assume that the true rule represents any of the C(P, N)
linearly separable functions with equal probability. Alternative approaches,
e.g. from the statistical physics perspective, weight every dichotomy with the
volume of the associated version space. We refrain from a detailed discus-
sion of this subtlety and refer the reader to the more specialized literature
[HKP91,EB01,WRB93].
Now assume that we present a novel input vector ξtest with target label Stest
to the student. In complete analogy to the considerations illustrated in Fig.
3.14, the version space could be ambiguous or non-ambiguous with respect to
the perceptron output Stest
w . In the latter case we know that any w ∈ V(P) will
provide the correct response. If the version space is ambiguous with respect to
ξtest, a fraction of the weight vectors in V(P) will be correct while the remaining",The+Shallow+and+the+Deep.pdf
"ξtest, a fraction of the weight vectors in V(P) will be correct while the remaining
ones would predict an incorrect label. Therefore, in absence of more detailed
knowledge, we expect that the response of a random w ∈ V(P) will be incorrect
with probability 1/2.
As a consequence, the generalization error 󰂃g is proportional to the fraction
of ambiguous dichotomies and we obtain
󰂃g = 1
2
Z(P, N)
C(P, N) = 1
2
C(P, N −1)
C(P, N) (3.52)
with C(P, N) from Eq. (3.41). The result is displayed in Fig. 3.15 as a function of
the scaled number of examples α = P/N. The curves represent input dimensions
N = 5, 20, 100 and the limiting case N → ∞, cf. Eq. (3.43).",The+Shallow+and+the+Deep.pdf
"3.5. LEARNING A LINEARLY SEPARABLE RULE 61
++
+−
−+
−−
+−
w∗
w
Figure 3.16: Linear separability for P < N.
Left panel: Version space corresponding to two feature vectors in N = 3
dimensions with all (four) possible combinations of labels ±1. Right panel:
Assume that, as an example, the region marked +− corresponds to the set of
correct (normalized) weight vectors. The version space is large enough to allow
for significant disagreement between student w and teacher vector w∗. Even
󰂃g = 1 is possible for extreme settings with w = −w∗.
For arbitrary N and P we find that 󰂃g = 1/2 as long as P ≤ N. For larger
N, the region of 󰂃g = 1/2 extends, up to α = αc = 2 in the limit (3.43). This
finding indicates that for relatively small data sets, the perceptron output is no
better than unbiased random guessing by, for instance, flipping a coin!
Hence, we learn from the above counting argument that storage without
generalization is possible for small P/N. Figure 3.16 illustrates the situation in",The+Shallow+and+the+Deep.pdf
"generalization is possible for small P/N. Figure 3.16 illustrates the situation in
N = 3 dimensions. The version space for P < N allows to place the student
vector very far away from the (unknown) teacher vector in V(P) without causing
disagreement on the training data. Even total misalignment with w = −w∗ is
possible in an extreme setting.
The result (3.52) indicates that for a randomly selected student inside the
version space of a randomly selected lin. sep. dichotomy, storage without learning
the rule can be observed up to the Vapnik-Chervonenkis-dimension P = N. For
large N → ∞ the region extends to the storage capacity, i.e. to P = 2N.
On the positive side, we also see that for data set sizes which exceed the
storage capacity, it is inevitable for the perceptron to perform better than ran-
dom. Learning in version space is guaranteed to yield non-trivial generalization
as soon as the number of examples exceeds the storage capacity (or the VC-
dimension for finite N, P).",The+Shallow+and+the+Deep.pdf
"as soon as the number of examples exceeds the storage capacity (or the VC-
dimension for finite N, P).
These findings confirm the intuition that learning and generalization is not
possible if the hypothesis space is too flexible. A very complex student system
can implement large sets of example data without inferring useful information
about the underlying rule.",The+Shallow+and+the+Deep.pdf
"62 3. THE PERCEPTRON
Asymptotic behavior for N → ∞
For large N → ∞ with P/N = α one can show that the asymptotic form of Eq.
(3.52) reads [Cov65]
󰂃g(α) =
󰀻
󰀿
󰀽
1/2 for α ≤ 2
1
2
1
α − 1 for α > 2.
(3.53)
As an alternative to the counting argument, methods borrowed from statistical
physics have been applied to compute the typical version space volume in high
dimensions (N → ∞) and yield the same basic dependence, i.e.
󰂃g ∝ α−1 for α → ∞
for various training algorithms in the presence of noise-free data, see for instance
[EB01,WRB93,BSS94,Opp94].
However, the statistical physics based analysis and similar considerations
also show that learning in version space typically performs better than random
even for small training sets. It turns out beneficial to select specific students
in version space. Even simple algorithms like the Rosenblatt scheme typically
yield 󰂃g(α) < 1/2 already for small α. In the following sections we will consider,",The+Shallow+and+the+Deep.pdf
"yield 󰂃g(α) < 1/2 already for small α. In the following sections we will consider,
more specifically, the perceptron of optimal stability which achieves near optimal
expected performance.
3.5.4 Optimal generalization
In the student-teacher setup discussed above, we only know that the teacher
w∗ is located somewhere in V. In absence of additional knowledge it could be
anywhere in the version space with equal probability.
Learning in version space places the student w also somewhere in V. Its
generalization error 󰂃g is, under very general circumstances, a decreasing func-
tion of the angle ∠(w, w∗), which itself is a decreasing function of the Euclidean
distance |w −w∗| for normalized weight vectors w, w∗, see Fig. 3.12. As a con-
sequence, the smallest expectation value of 󰂃g over all possible positions w∗ ∈ V
would be achieved by placing the student vector in the center of mass wcm of
the version space
wcm =
󰁝
V
w dNw. (3.54)",The+Shallow+and+the+Deep.pdf
"would be achieved by placing the student vector in the center of mass wcm of
the version space
wcm =
󰁝
V
w dNw. (3.54)
By definition, it has the smallest average (angular) distance from all other points
in the set. Note that the center of mass wcm of the normalized vectors in V
itself is not normalized, but realizes the same classification as wcm/|wcm|.
In principle, the definition (3.54) immediately suggests how to determine
wcm for a given lin. sep. data set D: We would have to determine many, random
elements w(i) ∈ V independently and compute the simple empirical estimate
[Wat93]
w(est)
cm = 1
M
M󰁛
i=1
w(i). (3.55)",The+Shallow+and+the+Deep.pdf
"3.6. THE PERCEPTRON OF OPTIMAL STABILITY 63
In practice, sampling the version space with uniform density is a non-trivial
task. The theoretical background and practical strategies for how to achieve
the optimal generalization ability when learning a linearly separable rule are
discussed in e.g. [Wat93,OH91,Ruj97].
3.6 The perceptron of optimal stability
Here we approach the question (Q5) of Sec. 3.2 of how to choose a good or near
optimal weight vector in version space from a different perspective. We avoid
the explicit computation of the center of mass of V and consider a well-defined
problem of quadratic optimization instead.
3.6.1 The stability criterion
The so-called stability of the perceptron has been established as a meaningful
optimality criterion. We first consider the stability of a particular example,
which is defined as
κµ = Eµ
|w| = w · ξµ Sµ
T
|w| . (3.56)
Due to the linearity of Eµ in w, cf. Eq. (3.8), the quantity κµ is invariant under",The+Shallow+and+the+Deep.pdf
"which is defined as
κµ = Eµ
|w| = w · ξµ Sµ
T
|w| . (3.56)
Due to the linearity of Eµ in w, cf. Eq. (3.8), the quantity κµ is invariant under
a rescaling of the form w → λw (λ > 0). In terms of the geometric interpreta-
tion of linear separability, the scalar product of ξµSµ
T and w/|w| measures the
distance of the input vector from the separating hyperplane, see Fig. 3.17. More
precisely κµ is an oriented distance: For κµ > 0, the input vector is classified
correctly by the perceptron, while for κµ < 0 we have sign(w · ξµ) = −Sµ
T and
the input is located on the wrong side of the plane.
The stability (its absolute value) quantifies how robust the perceptron re-
sponse would be against small variations of ξµ. Examples with a large distance
from the hyperplane will hardly be taken to the opposite side by noise in the
input channel.
We define the stability of the perceptron as the smallest of all κµ in D:
κ(w) = min {κµ}P
µ=1 . (3.57)",The+Shallow+and+the+Deep.pdf
"input channel.
We define the stability of the perceptron as the smallest of all κµ in D:
κ(w) = min {κµ}P
µ=1 . (3.57)
Note that if the perceptron does not separate the classes correctly, κ(w) < 0
corresponds to the negative κµ with the largest absolute value. Positive stability
κ(w) > 0 indicates that w is a solution of the PSP and separates the classes
correctly in D. In this case, κ(w) corresponds to the smallest distance of any
example from the decision boundary. It quantifies the size of the gap between
the two classes or, in other words, the classification margin of the perceptron.
In a linear separable problem it appears natural to select the perceptron
weights which maximize κ(w). In principle, the concept of stability extends to
negative κµ according to Eq. (3.56). Therefore, we can also define the perceptron
of optimal stability without requiring linear separability of the data set D =
{ξµ, Sµ
T } with more general targets ST (ξµ) = ±1.",The+Shallow+and+the+Deep.pdf
"64 3. THE PERCEPTRON
κ
κµ
ξµ
wmax
κ
Figure 3.17: Stability of the perceptron. Left: The stability κµ, defined in
Eq. (3.56), corresponds to the oriented distance of ξµ from the plane orthogonal
to w. The stability of the perceptron κ(w) is defined as the smallest κµ in the
set of examples, i.e. κ(w) = minµ{κµ}. Here, all inputs are classified correctly
with κµ > 0. Right: Restricting the student hypotheses to weight vectors with
κ(w) > κ for a given data set, selects weight vectors w in the center region of
the version space. The largest possible value of κ singles out the perceptron of
optimal stability wmax.
We define the perceptron of optimal stability (not very precisely termed the
optimal perceptron, occasionally) as the target of the following problem:
Perceptron of optimal stability (3.58)
For a given data set D = {ξµ, Sµ
T }P
µ=1, find the vector wmax ∈ RN
with wmax = argmax
w ∈ RN κ(w) for κ(w) = min
󰁱
κµ = w·ξ µSµ
T
|w|
󰁲P
µ=1
.",The+Shallow+and+the+Deep.pdf
"For a given data set D = {ξµ, Sµ
T }P
µ=1, find the vector wmax ∈ RN
with wmax = argmax
w ∈ RN κ(w) for κ(w) = min
󰁱
κµ = w·ξ µSµ
T
|w|
󰁲P
µ=1
.
For linearly separable data, the search for w could be limited to the version
space V, formally. However, this restriction is non-trivial to realize [Ruj97] and
would not constitute an advantage in practice. Moreover, for more general data
sets of unknown separability, the version space might not even exist (V = ∅) and
κ(wmax) < 0. We will discuss the usefulness of a corresponding solution wmax
with negative stability κmax < 0 later and focus on linearly separable problems
in the following.
The perceptron of optimal stability wmax does not exactly coincide with
wcm, cf. Sec. 3.5.4, in general. The “center” as defined by the “maximum
possible distance from all boundaries” is identical with the center of mass only
if V has a regular, symmetric shape. However, one can expect that wmax is in",The+Shallow+and+the+Deep.pdf
"if V has a regular, symmetric shape. However, one can expect that wmax is in
general quite close to the true center of mass and could serve as an approximative
realization.",The+Shallow+and+the+Deep.pdf
"3.6. THE PERCEPTRON OF OPTIMAL STABILITY 65
As a consequence, the perceptron of optimal stability should display favor-
able (near optimal) generalization behavior when trained from a given reliable,
lin. sep. data set. In fact, the difference appears marginal from a practical
perspective. For a theoretical comparison of optimal stability and optimal gen-
eralization in the perceptron see [Wat93].
3.6.2 The MinOver algorithm
An intuitive algorithm that can be shown to achieve optimal stability for a
given lin. sep. data set D has been suggested in [KM87]. The so-called MinOver
algorithm performs Hebbian updates for the currently least stable example in
D. We assume here tabula rasa initialization, i.e. w(0) = 0, but more general
initial states could be considered.
The prescription always aims at improving the least stable example. Note
that for a given weight vector w(t), the minimal local potential coincides with
minimum stability since κµ(t) = Eµ(t)/|w(t)|.",The+Shallow+and+the+Deep.pdf
"that for a given weight vector w(t), the minimal local potential coincides with
minimum stability since κµ(t) = Eµ(t)/|w(t)|.
The MinOver update is defined as follows:
MinOver algorithm
at discrete time step t with current w(t)
- compute the local potentials Eµ(t) = w(t) · ξµ Sµ
T for all examples in D
- determine the index 󰁥µ of the training example with minimal overlap,
i.e. with the currently lowest local potential: E󰁥µ = min {Eµ(t) }P
µ=1
- update the weight vector according to
w(t + 1) = w(t) + 1
N ξ󰁥µ S󰁥µ
T (3.59)
󰀥 or, equivalently, increment the corresponding embedding strength
x󰁥µ(t + 1) = x󰁥µ(t) + 1.
󰀦
(3.60)
A few remarks:
• According to the original presentation of the algorithm [KM87], the Heb-
bian update is only performed if the currently smallest local potential also
satisfies Eν(t) ≤ c for a given c > 0. This is reminiscent of learning from
mistakes as in the Rosenblatt algorithm (3.3.3). However, as pointed out",The+Shallow+and+the+Deep.pdf
"satisfies Eν(t) ≤ c for a given c > 0. This is reminiscent of learning from
mistakes as in the Rosenblatt algorithm (3.3.3). However, as pointed out
in [KM87], optimal stability is only achieved in the limit c → ∞, which is
equivalent to (3.59).
• In the above formulation (3.59), MinOver updates the weights (or embed-
ding strengths) even if all examples in D are classified correctly already.
As the algorithm keeps performing non-zero Hebbian updates, the tempo-
ral change of the weight vector w(t) itself does not constitute a reasonable",The+Shallow+and+the+Deep.pdf
"66 3. THE PERCEPTRON
stopping criterion. Instead, one of the following quantities could be con-
sidered:
- the angular change ∠
󰀃
w(t), w(t+T)
󰀄
= 1
π arccos
󰀕 w(t) · w(t+T)
|w(t)||w(t+T)|
󰀖
or the argument of the arccos, for simplicity.
- the total change of stabilities 󰁓P
µ=1
󰁫
κµ(t) − κµ(t+T)
󰁬2
,
for example. For these and similar criteria, reasonably large numbers of
training steps T should be performed, e.g. with T ∝ P, in order to allow
for noticeable differences. In both criteria, changes of only the norm |w(t)|
are disregarded because they do not affect the classification or its stability.
• From the definition of the MinOver algorithm (3.59) we see that it can
only yield non-negative, integer embedding strengths when the initializa-
tion is tabula rasa. This feature of wmax will be encountered again in
the following section, together with several other properties of optimal
stability.
• As proven in [KM87], the MinOver algorithm converges and yields the per-",The+Shallow+and+the+Deep.pdf
"stability.
• As proven in [KM87], the MinOver algorithm converges and yields the per-
ceptron weights of optimal stability, if D is linearly separable. The proof
of convergence is similar in spirit to that of the Perceptron Convergence
Theorem (3.22). We refrain from reproducing it here. Instead, we show
only that the perceptron of optimal stability can always be written in the
form
wmax = 1
N
P󰁛
µ=1
xµ
max ξµ Sµ
T with embedding strengths {xµ
max
∈ R}P
µ=1 .
(3.61)
The existence of embedding strengths xmax will be recovered en passant in the
next section. However, it is instructive to prove the statement explicitly here.
To this end, we consider two perceptron weight vectors: the first one is assumed
to be given as a linear combination of the familiar form
w1 =
P󰁛
µ=1
xµ
1 ξµ Sµ
T with embedding strengths {xµ
1
}P
µ=1 , (3.62)
while for the second weight vector we assume that
w2 = w1 + δ with |δ| > 0 and δ · ξµ = 0 for all µ = 1, 2, . . . , P. (3.63)",The+Shallow+and+the+Deep.pdf
"1
}P
µ=1 , (3.62)
while for the second weight vector we assume that
w2 = w1 + δ with |δ| > 0 and δ · ξµ = 0 for all µ = 1, 2, . . . , P. (3.63)
Hence, w2 cannot be written in terms of embedding strengths as it contains
contributions which are orthogonal to all input vectors in D.
If we consider the local potentials with respect to w2, we observe that
Eµ
2 = w2 · ξµSµ
T = w1 · ξµSµ
T + δ · ξµSµ
T󰁿 󰁾󰁽 󰂀
=0
= Eµ
1 . (3.64)",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 67
On the other hand we have
|w2|2 = |w1 + δ|2 = |w1|2 + 2 w1 · δ󰁿 󰁾󰁽 󰂀
=0
+ |δ|2
󰁿󰁾󰁽󰂀
>0
⇒ |w2| > |w1|, (3.65)
where the mixed term vanishes because w1 is a linear combination of the ξµ⊥δ.
As a consequence, we observe that
κµ
2 = Eµ
2
|w2| = Eµ
1
|w2| < Eµ
1
|w1| = κµ
1
for all µ, and therefore κ2 < κ1. (3.66)
We conclude that any contribution orthogonal to all ξµ inevitably reduces the
stability of a given weight vector. This implies that maximum stability is in-
deed achieved by weights of the form (3.61). The result also implies that the
framework of iterative Hebbian learning is sufficient to find the solution.
Note that a non-zero δ in Eq. (3.63) cannot exist for P > N, in general.
Obviously, if span
󰀓
{ξµ}P
µ=1
󰀔
= RN , any N-dimensional vector including wmax
can be written as a linear combination. In this case, Eq. (3.61) is trivially true.
Our simple consideration does not yield restrictions on the possible values",The+Shallow+and+the+Deep.pdf
"Our simple consideration does not yield restrictions on the possible values
that the embedding strengths can assume. However, the proven convergence
of the MinOver algorithm implies that the perceptron of optimal stability can
always be written in terms of non-negative xµ
max ≥ 0. We will recover this result
more formally in the next section.
3.7 Optimal stability by quadratic optimization
Here we will exploit the fact that optimal stability can be formulated as a prob-
lem of constrained quadratic optimization. Consequently, a wealth of theoret-
ical results and techniques from optimization theory becomes available [Fle00,
PAH19]. They provide deeper insight into the structure of the problem and
allow for the identification of efficient training algorithms, as we will exemplify
in terms of the so-called AdaTron algorithm [AB89,BAK91] Before deriving and
discussing this training scheme we re-visit another closely related, prototypical",The+Shallow+and+the+Deep.pdf
"discussing this training scheme we re-visit another closely related, prototypical
training scheme from the early days of neural network models: B. Widrow and
M.E. Hoff’s Adaptive Linear Neuron or Adaline [WH60,WL90].
3.7.1 Optimal stability reformulated
For a given lin. sep. D = {ξµ, Sµ
T }P
µ=1, the perceptron of optimal stability
corresponds to the solution of the following problem:
maximize
w ∈ RN κ(w) where κ(w) = min
󰀝
κµ = Eµ
|w| = w⊤ξµSµ
T
|w|
󰀞P
µ=1
(3.67)",The+Shallow+and+the+Deep.pdf
"68 3. THE PERCEPTRON
which is just a more compact version of (3.58).
Obviously, the stability κ can be made larger by increasing the minimal Eµ
for constant norm |w|. Analogously, κ increases with decreasing norm |w| if all
local potentials obey the constraint Eµ ≥ c > 0. As discussed previously, the
actual choice of the constant c is irrelevant because Eµ is linear in w and we can
set c = 1 without loss of generality. This allows us to re-formulate the problem
as follows:
minimize
w ∈ RN
N
2 w2 subject to inequality constraints {Eµ ≥ 1}P
µ=1 . (3.68)
Hence, we have rewritten the problem of maximal stability as the optimization
of the quadratic cost function Nw2/2 under linear inequality constraints of
the form Eµ = w⊤ξµSµ
T ≥ 1. The solution wmax then displays the (optimal)
stability κmax = 1
󰀱
|wmax|. Note that the pre-factor N/2 in (3.68) is irrelevant
for the definition of the problem but is kept for convenience and consistency of
notation.
3.7.2 The Adaptive Linear Neuron - Adaline",The+Shallow+and+the+Deep.pdf
"for the definition of the problem but is kept for convenience and consistency of
notation.
3.7.2 The Adaptive Linear Neuron - Adaline
The problem of optimal stability in the formulation (3.68) involves a system of
inequalities. Before we address its solution in the following sections, we resort
to the more familiar case of constraints given by equations, i.e. we consider the
simpler optimization problem
minimize
w ∈ RN
N
2 w2 subject to constraints {Eµ = 1}P
µ=1
. (3.69)
This, in fact, has the form of a standard optimization problem with non-linear
(here: quadratic) cost function and a set of (here: linear) constraints. A brief
discussion of this type of problem is given in Appendix A.3.1.
Historically, the problem (3.69) relates to the the so-called Adaptive Linear
Neuron or Adaline model which was introduced by B. Widrow and M.E. Hoff in
1960 [WH60]. Like the Rosenblatt Perceptron, it constitutes one of the earliest",The+Shallow+and+the+Deep.pdf
"1960 [WH60]. Like the Rosenblatt Perceptron, it constitutes one of the earliest
artificial neural network models and truly groundbreaking work in the area of
machine learning. A series of very instructive and entertaining videos about B.
Widrow’s pioneering work is available at [Wid12].
Widrow realized Adaline systems in hardware as “resistors with memory”
and introduced the term Memistor [Wid60].11 The concept was also extended
to layered Madaline (Many Adaline) networks consisting of several linear units
which were combined in a majority vote for classification [WL90,Wid12].
For our purposes, the Adaline can be interpreted as a single layer perceptron
which differs from Rosenblatt’s model only in terms of the training procedure.
11Not to be confused with the more recent concept of Memristor elements [Chu71].",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 69
The Adaline framework essentially treats the problem of binary classification as
a linear regression with subsequent thresholding of the continuous w · ξµ.
We will take a rather formal perspective based on the theory of Lagrange
multipliers [Fle00, PAH19] as outlined in Appendix A.3.1. The P linear con-
straints of the form Eµ = w⊤ξµSµ
T = 1 can be incorporated in the correspond-
ing Lagrange function
L
󰀕
w, {λµ}P
µ=1
󰀖
= N
2 w2 −
P󰁛
µ=1
λµ
󰀕
w⊤ξµSµ
T − 1
󰀖
. (3.70)
With the gradient ∇w = (∂/∂w1, ∂/∂w2, ..., ∂/∂wN )⊤ in weight space we
obtain ∇wL = Nw−󰁓
µ λµξµSµ
T . Furthermore, ∂
∂λν
󰁫󰁓
µ λµ(Eµ−1)
󰁬
= (Eν−1).
Hence, the first order stationarity conditions for a solution w∗, {λ∗µ} of problem
(3.69) read
∇wL|∗ = 0 ⇒ w∗ = 1
N
P󰁛
µ=1
λ∗µ ξµSµ
T (3.71)
and ∂L
λµ
󰀏󰀏󰀏󰀏
∗
= 0 ⇒ E∗µ = w∗⊤ξµSµ
T = 1 for all µ (3.72)
where the shorthand (. . .)|∗ stands for the evaluation of (. . .) in w = w∗ and",The+Shallow+and+the+Deep.pdf
"T (3.71)
and ∂L
λµ
󰀏󰀏󰀏󰀏
∗
= 0 ⇒ E∗µ = w∗⊤ξµSµ
T = 1 for all µ (3.72)
where the shorthand (. . .)|∗ stands for the evaluation of (. . .) in w = w∗ and
λµ = λ∗µ for all µ. Note that sufficient (second oder) conditions for a solution
of the problem are non-trivial to work out.
The second condition (3.72) merely reproduces the original equality con-
straints, which have to be satisfied by any candidate solution, obviously. The
first, more interesting stationarity condition (3.71) implies that the formally in-
troduced Lagrange parameters can be identified as the embedding strengths xµ
and can be renamed accordingly. Hence, we can eliminate the weights from L
and obtain
L
󰀃
{xµ}P
µ=1
󰀄
= 1
2N
󰁛
µ,ν
xµSµ
T ξµ⊤ξνSν
T xν −
󰁛
µ
xµ
󰀥
1
N
󰁛
ν
xνSν
T ξν
󰀦⊤
ξµSµ
T +
󰁛
µ
xµ
= − 1
2N
P󰁛
ν,µ=1
xµSµ
T ξµ⊤ ξνSν
T xν +
P󰁛
µ=1
xµ. (3.73)
It turns out useful to resort to a compact notation which again exploits that
the weights are of the form w = 1
N
󰁓
µ xµξµSµ
T . We introduce the symmetric
correlation matrix",The+Shallow+and+the+Deep.pdf
"the weights are of the form w = 1
N
󰁓
µ xµξµSµ
T . We introduce the symmetric
correlation matrix
C = C⊤ ∈ RP×P with elements Cµν = 1
N Sµ
T Sν
T ξµ⊤ξν (3.74)
and define the P-dimensional vectors 󰂓x = (x1, x2, ...xP )⊤, 󰂓E = (E1, E2, ...EP )⊤
as well as the formal 󰂓1 = (1, 1, ..., 1)⊤ ∈ RP , yielding, e.g., 󰂓x⊤󰂓1 = 󰁓
µ xµ.",The+Shallow+and+the+Deep.pdf
"70 3. THE PERCEPTRON
In addition, we use the notation 󰂓a > 󰂓b, which is popular in the optimization
related literature and indicates that aµ > bµ for all µ = 1, 2, ....P. Analogous
notations are employed for component-wise relations “<”,“≥” and “≤”.
In the convenient matrix-vector notation, we have furthermore
Eν = 1
N
P󰁛
µ=1
xµ Sµ
T Sν
T ξµ⊤ξν
󰁿 󰁾󰁽 󰂀
NCµν
= [C󰂓x]ν i.e 󰂓E = C󰂓x (3.75)
w2 = 1
N2
󰁛
µ,ν
xµxν
󰁽 󰂀󰁿 󰁾
Sµ
T Sν
T ξµ⊤ξν = 1
N 󰂓x⊤C 󰂓x = 1
N 󰂓x⊤ 󰂓E. (3.76)
In this notation, the Lagrange function (3.73) becomes L(󰂓x) = −1/2 󰂓x⊤ C 󰂓x +
󰂓x⊤󰂓1, which has to be maximized (!) with respect to 󰂓x, see the discussion of
the Wolfe Dual in [Fle00, PAH19] and Appendix A.3.4. Consequently we can
re-formulate (3.69) as the following unconstrained problem:
maximize
󰂓x ∈ RP f(󰂓x) = − 1
2 󰂓x⊤C󰂓x + 󰂓x⊤󰂓1. (3.77)
Compared to (3.69), the cost function appears slightly more complicated. In
turn, however, the constraints are eliminated. While completely equivalent with",The+Shallow+and+the+Deep.pdf
"turn, however, the constraints are eliminated. While completely equivalent with
(3.69), the re-written problem is given in terms of the embedding strengths
󰂓x ∈ IRP only.
Parallel Adaline algorithm
In absence of constraints, it is straightforward to maximize f(󰂓x) and we could
employ a variety of methods, in practice. For instance, we can resort to simple
gradient ascent, cf. Appendix A.4, with tabula rasa initialization 󰂓x(0) = 0. We
can also identify an equivalent update in terms of weights with initial w(0) = 0
by exploiting the relation w(t) = 1
N
󰁓P
µ=1 xµ(t) ξµSµ
T :
Adaline algorithm, parallel updates
󰂓x(t + 1) = 󰂓x(t) + η∇xf = 󰂓x(t) + η
󰀓
󰂓1 − 󰂓E(t)
󰀔
(3.78)
󰀥
or w(t + 1) = w(t) + η
N
P󰁛
µ=1
󰀕
1 − Eµ(t)
󰀖
ξµSµ
T
󰀦
, (3.79)
where Eµ(t) = [C󰂓x(t)]µ = w(t)⊤ξµSµ
T . The learning rate η controls the magni-
tude of the update steps.
This version of the Adaline algorithm corresponds to standard, so-called
batch gradient descent in the space of embedding strengths 󰂓x. Hence, η can",The+Shallow+and+the+Deep.pdf
"This version of the Adaline algorithm corresponds to standard, so-called
batch gradient descent in the space of embedding strengths 󰂓x. Hence, η can
be finite but has to be small enough enough to ensure convergence. The pre-
cise condition depends on the mathematical properties of the extremum, see
Appendix A.4 for a general discussion.",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 71
Sequential Adaline algorithm
As an important alternative to the above batch or parallel update, sequential
gradient-based methods can be devised, which present the example data repeat-
edly in, for instance, deterministic order and update a single embedding strength
in each step. Intuitively, we would want to increase an embedding strength of
any example with Eµ < 1 and we expect the magnitude of the updated xµ to
increase with (1 − Eµ).
The sequential Adaline algorithm corresponds to coordinate ascent in terms
of 󰂓x, a variant of gradient based optimization which is briefly discussed in Ap-
pendix A.5.1:
Adaline algorithm, sequential updates (repeated presentation of D)
– at time step t, present example µ(t) = 1, 2, 3, ..., P, 1, 2, 3, ...
– update (only) the corresponding embedding strength
xµ(t)(t + 1) = xµ(t) + 󰁨η
󰀓
1 − Eµ(t)(t)
󰀔
(3.80)
󰀥
or w(t + 1) = w(t) + 󰁨η
N
󰀕
1 − Eµ(t)(t)
󰀖
ξµ(t)Sµ(t)
T
󰀦
. (3.81)
Convergence of the Adaline",The+Shallow+and+the+Deep.pdf
"xµ(t)(t + 1) = xµ(t) + 󰁨η
󰀓
1 − Eµ(t)(t)
󰀔
(3.80)
󰀥
or w(t + 1) = w(t) + 󰁨η
N
󰀕
1 − Eµ(t)(t)
󰀖
ξµ(t)Sµ(t)
T
󰀦
. (3.81)
Convergence of the Adaline
In the following we sketch the proof of convergence for the sequential Adaline
algorithm (3.80) and obtain the corresponding condition on the learning rate 󰁨η.
For the update of the current component xµ we have
xµ → xµ + δµ with δµ = 󰁨η (1 − [C󰂓x]µ) = 󰁨η (1 − Eµ) (3.82)
while all other embedding strengths remain unchanged: 󰂓δ = (0, . . . , δµ, 0, . . . 0)⊤.
Therefore, the change of the objective function under update (3.82) is
f(󰂓x + 󰂓δ) − f(󰂓x) = −1
2
󰂓δ⊤C󰂓δ − 󰂓δ⊤ C󰂓x + 󰂓δ⊤󰂓1 = −1
2(δµ)2 Cµµ + δµ (1 − Eµ)
= 󰁨η(1 − Eµ)2
󰀗
1 − 1
2 󰁨η Cµµ
󰀘
≥ 0 for 󰁨η < 2
Cµµ . (3.83)
Unless a solution 󰂓x with Eµ = 1 for all µ has been reached, the objective function
will strictly increase in at least some of the individual steps of the sequential
procedure as long as 󰁨η < 2/Cµµ.
The diagonal elements of C are proportional to the norms of the correspond-",The+Shallow+and+the+Deep.pdf
"procedure as long as 󰁨η < 2/Cµµ.
The diagonal elements of C are proportional to the norms of the correspond-
ing feature vectors, Cµµ = (ξµ)2/N, which are of course available for any given
data set. If we want to use a single, constant value of 󰁨η and guarantee monotonic
increase of f under the iteration, the learning rate has to satisfy
0 < 󰁨η < 2
max{Cµµ}P
µ=1
. (3.84)",The+Shallow+and+the+Deep.pdf
"72 3. THE PERCEPTRON
One can also show that the objective function is bounded from above if C is
positive definite and C󰂓x = 󰂓1 has a solution: while the linear term 󰂓1⊤󰂓x in f, Eq.
(3.77), can grow in an unbounded way, the negative quadratic term will always
dominate and limit the increase.
Hence, unless a solution with C󰂓x = 󰂓E = 󰂓1 exists and has been found, the
iteration performs at least one non-zero update in each sweep through the data
set and f increases monotonically. Moreover, the cost function is bounded
from above. Therefore, the coordinate-ascent-like version (3.80) of the Adaline
converges and maximizes f under constraints 󰂓E = 󰂓1, if the learning rate satisfies
(3.84).
Relation to linear regression and the SSE
Several interesting relations to other problems and algorithms are noteworthy:
As pointed out in [DO87], the coordinate ascent (3.82) is closely related to
the well-known Gauß-Seidel and Gauß-Jacobi methods [Fle00] for the iterative",The+Shallow+and+the+Deep.pdf
"the well-known Gauß-Seidel and Gauß-Jacobi methods [Fle00] for the iterative
solution of the system of linear equations C󰂓x = 󰂓1.
For non-singular C, the problem could be solved directly by 󰂓x = C−1󰂓1,
which gives
xµ = [ C−1󰂓1 ]µ =
󰁛
ν
[C−1]µν 1 and w = 1
N
󰁛
µ,ν
[C−1]µνξµSµ
T (3.85)
in weight space. If we incorporate the outputs Sµ
T = ±1 in the modified input
vectors 󰁨ξ
µ
= Sµ
T ξµ the problem becomes equivalent to regression with the par-
ticular targets yµ = (Sµ
T )2 = 1 for all data points. Hence, the Adaline problem
has the same mathematical structure as linear regression considered in Sec. 2.2.2
and Appendix A.2.2, which immediately links Eq. (3.85) to the pseudoinverse
solutions (2.8) and (A.24), respectively.
It is interesting to note that although (3.79) is the weight space equivalent
of (3.78), it cannot be interpreted as a gradient ascent along ∇wf of the cost
function (3.77). As outlined in Appendix A.4.3, even linear transformations of",The+Shallow+and+the+Deep.pdf
"function (3.77). As outlined in Appendix A.4.3, even linear transformations of
variables do not preserve the gradient property, in general.
However, (3.79, 3.81) can be seen as gradient descent in weight space with
respect to a different, yet related cost function: the Sum of Squared Errors
(SSE), cf. Eq. (2.5), for linear regression with target values Sµ
T = ±1 :
ESSE = 1
2
P󰁛
µ=1
(1 − Eµ)2 = 1
2
P󰁛
µ=1
󰀃
Sµ
T − w⊤ξµ󰀄2
(3.86)
with ∇wESSE = −
P󰁛
µ=1
(1 − Eµ)ξµSµ
T . (3.87)
Note that writing ESSE in terms of embedding strengths would involve the
quadratic form 󰂓x⊤C⊤C󰂓x which is not present in (3.77).",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 73
This consideration also indicates what the behavior of the Adaline will be
if 󰂓E = 1 cannot be satisfied: the algorithm finds an approximate solution by
minimizing the SSE.
The sequential algorithm (3.81) in weight space is equivalent to Widrow
and Hoff’s original LMS (Least Mean Square) method for the Adaline [WH60,
Wid12]. The LMS, also referred to as the delta-rule or the Widrow-Hoff al-
gorithm in the literature, can be seen as the most important ancestor of the
prominent Backpropagation of Error for multilayered neural networks, cf. Sec.
A.4 and Chapter 5. A review of the history and conceptual relations between
these classical algorithms is given in [WL90].
3.7.3 The Adaptive Perceptron Algorithm - AdaTron
In Rosenblatt’s perceptron algorithm an update is performed whenever the pre-
sented example is misclassified. All non-zero Hebbian learning steps are per-
formed with the same magnitude, independent of the example’s distance from",The+Shallow+and+the+Deep.pdf
"formed with the same magnitude, independent of the example’s distance from
the current decision plane. On the contrary, the Adaline learning rule is adaptive
in the sense that the magnitude of the update depends on the actual deviation
of Eµ from the target value 1. While this appears to make sense and is expected
to speed up learning, it also results in negative Hebbian updates (“unlearning”)
with η(1 − Eµ) < 0 if the example is correctly classified with large Eµ > 1,
already. In fact, Adaline can yield negative embedding strengths xµ < 0 in
order enforce Eµ = 1 when the example otherwise could have Eµ > 1.
This somewhat counter-intuitive feature of the Adaline is accounted for in
the Adaptive Perceptron (AdaTron) algorithm [AB89, BAK91]. It retains the
concept of adaptivity but takes into account the inequality constraints in prob-
lem (3.68), explicitly. In essence, it performs Adaline-like updates, but prohibits",The+Shallow+and+the+Deep.pdf
"lem (3.68), explicitly. In essence, it performs Adaline-like updates, but prohibits
the embedding strengths from assuming negative values. As we will see, this rel-
atively simple modification yields the perceptron of optimal stability for linearly
separable data.
The method of Lagrange multipliers has been extended to the treatment
of inequality constraints [Fle00, PAH19], see also Appendix A. Therefore, we
can make use of several well-established results from optimization theory in the
following. We return to the problem of optimal stability in the form (3.68),
which we repeat here for the sake of clarity:
Perceptron of optimal stability (weight space)
minimize
w ∈ RN
N
2 w2 subject to inequality constraints {Eµ ≥ 1}P
µ=1 .
Formally, we have to consider the same Lagrange function as for equality
constraints, already given in Eq. (3.70):
L
󰀃
w, {λµ}P
µ=1
󰀄
= N
2 w2 −
P󰁛
µ=1
λµ
󰀕
w⊤ξµSµ
T − 1
󰀖
. (3.88)",The+Shallow+and+the+Deep.pdf
"74 3. THE PERCEPTRON
The Kuhn-Tucker Theorem of optimization theory, see [Fle00,PAH19], pro-
vides the first order necessary stationarity conditions for general, non-linear
optimization problems with inequality and/or equality conditions. They are
known as the so-called Kuhn-Tucker (KT) or Karush-Kuhn-Tucker conditions
[Fle00,PAH19], see (A.26) in the Appendix. In our specific case, as worked out
as an example in the Appendix, they read
Kuhn-Tucker conditions (optimal stability)
w∗ = 1
N
P󰁛
µ=1
λ∗µ ξµSµ
T (embedding strengths λµ) (3.89)
E∗µ = w∗⊤ξνSµ
T ≥ 1 for all µ (linear separability) (3.90)
λ∗µ ≥ 0 (not all λ∗µ = 0) (non-negative multipliers) (3.91)
λ∗µ (E∗µ − 1) = 0 for all µ (complementarity). (3.92)
The first condition is the same as for equality constraints and follows directly
from ∇wL = 0. As in the case of the Adaline, it implies that the solution of
the problem can be written in the familiar form (3.61). Moreover, the Lagrange",The+Shallow+and+the+Deep.pdf
"the problem can be written in the familiar form (3.61). Moreover, the Lagrange
multipliers λµ play the role of the embedding strengths and we can set λµ = xµ
in the following considerations. The second condition (3.90) merely reflects the
original constraint, i.e. linear separability.
Intuitively, the non-negativity of the multipliers, KT-condition (3.91), re-
flects the fact that an inequality constraint is only active on one side of the
hyperplane defined by Eµ = 1. As long as Eµ > 1, the corresponding multiplier
could be set to λµ = 0 as the constraint is satisfied in the entire half-space. Since
the solution can be written with xµ = λµ due to (3.89), we formally recover the
insight from Sec. 3.6.2 which indicates that wmax can always be represented by
non-negative embedding strengths.
After renaming the Lagrange multipliers λµ to xµ, we refer to a solution
󰂓x∗ of the problem (3.68) as a KT-point. Using our convenient matrix-vector",The+Shallow+and+the+Deep.pdf
"After renaming the Lagrange multipliers λµ to xµ, we refer to a solution
󰂓x∗ of the problem (3.68) as a KT-point. Using our convenient matrix-vector
notation, the KT conditions of the simplified problem read
Kuhn-Tucker conditions (optimal stability, simplified)
󰂓E∗ = C󰂓x∗ ≥ 󰂓1 (linear separability) (3.93)
󰂓x∗ ≥ 0 (󰂓x∗ ∕= 0) (non-negative embeddings)12 (3.94)
x∗µ (E∗µ − 1) = 0 for all µ (complementarity). (3.95)
The most interesting condition is that of complementarity (3.92) and (3.95).
It states that at optimal stability, any example with non-zero embedding will
12Obviously 󰂓x = 0, w = 0 cannot be a solution.",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 75
have E∗µ = 1 while input/output pairs with E∗µ > 1 do not contribute to the
linear combination (3.61). We will discuss this property in greater detail later.
Complementarity also implies that x∗µE∗µ = x∗µ for all µ and, therefore
󰂓x∗⊤C󰂓x∗ = 󰂓x∗⊤ 󰂓E∗ =
P󰁛
µ=1
x∗µE∗µ =
P󰁛
µ=1
x∗µ = 󰂓x∗⊤󰂓1. (3.96)
Now consider two potentially different KT-points 󰂓x1 and 󰂓x2, both satisfying the
first order stationarity conditions. By definition, the matrix C is symmetric and
positive semi-definite:
󰂓u⊤C 󰂓u ∝
󰁛
µ,ν
uµSµ
T ξµ · ξνSν
T uν =
󰀣󰁛
µ
uµξµSµ
T
󰀤2
≥ 0 for all 󰂓u ∈ RP .
Setting 󰂓u = 󰂓x1 − 󰂓x2 we obtain with (3.96):
0 ≤ (󰂓x1 − 󰂓x2)⊤C(󰂓x1 − 󰂓x2) = 󰂓x⊤
1 C󰂓x1󰁿 󰁾󰁽 󰂀
=󰂓x⊤
1 󰂓1
+ 󰂓x⊤
2 C󰂓x2󰁿 󰁾󰁽 󰂀
=󰂓x⊤
2 󰂓1
−󰂓x⊤
1 C󰂓x2 − 󰂓x2
⊤C󰂓x1
= 󰂓x⊤
1
󰀓
󰂓1 − C󰂓x2
󰀔
+ 󰂓x⊤
2
󰀓
󰂓1 − C󰂓x1
󰀔
.
All components of the KT-points 󰂓x1,2 are non-negative, while all components of
vectors (󰂓1 − C󰂓x1,2) ≤ 0 due to the linear separability condition. Therefore, we",The+Shallow+and+the+Deep.pdf
"vectors (󰂓1 − C󰂓x1,2) ≤ 0 due to the linear separability condition. Therefore, we
obtain 0 ≤ (󰂓x1 − 󰂓x2)⊤C(󰂓x1 − 󰂓x2) ≤ 0 and hence:
⇒ 0 = (󰂓x1 − 󰂓x2)⊤C(󰂓x1 − 󰂓x2) ∝
󰁛
µ,ν
(xµ
1 − xµ
2
)Sµ
T ξµ · ξνSν
T (xν
1 − xν
2
)
=
󰀥󰁛
µ
(xµ
1 − xµ
2
) ξµSµ
T
󰀦2
∝ [w1 − w2]2 ⇒ w1 = w2.
We conclude that any two KT-points define the same weight vector, which
corresponds to the perceptron of optimal stability. Even if 󰂓x1 ∕= 󰂓x2, which is
possible in spite of C󰂓x1 = C󰂓x2 for singular matrices C, the solution is unique
in terms of the weights. Moreover, this finding implies that any local solution
(KT-point) of the problem (3.68) is indeed a global solution. In contrast to
many other cost function based learning algorithms, local minima do not play
a role in optimal stability. This is a special case of a more general result which
applies to all convex optimization problems [Fle00,PAH19].
The stationarity conditions also facilitate a re-formulation of the optimiza-",The+Shallow+and+the+Deep.pdf
"applies to all convex optimization problems [Fle00,PAH19].
The stationarity conditions also facilitate a re-formulation of the optimiza-
tion problem. Similar to the simplification of the Adaline problem, it essentially
amounts to the elimination of the weights and the interpretation of Lagrange
multipliers as embedding strengths. It constitutes a special case of what is
known as the Wolfe Dual in optimization theory, see [Fle00] for the general def-
inition and proof of the corresponding Duality Theorem. A brief introduction
is given in Appendix A.3.4. Duality is frequently employed to rewrite a given
problem in terms of a modified cost function with simplified constraints.",The+Shallow+and+the+Deep.pdf
"76 3. THE PERCEPTRON
As outlined in Appendix A.3.4 the following formulation is equivalent to
problem (3.68):
Perceptron of optimal stability (embedding strengths, dual problem)
maximize
󰂓x f(󰂓x) = −1
2󰂓x⊤C󰂓x + 󰂓x⊤󰂓1 subject to 󰂓x ≥ 0. (3.97)
Hence, the resulting Wolfe Dual still comprises constraints, albeit simpler ones.
In this simplified formulation of optimal stability, see also Appendix A.3.4, the
goal is to maximize the objective function (3.77) under the constraint 󰂓x ≥ 0
(with 󰂓x ∕= 0). The search for the optimum is therefore restricted to the positive
hyperoctant of RP .
These particularly simple conditions facilitate the design of a gradient-based
active set algorithm, see Appendix A.3. Moreover, it is possible to combine this
concept with coordinate ascent as discussed in Appendix A.5.1. The algorithm
satisfies the constraint by simply clipping each xµ to zero whenever the gradient
step would lead into the excluded region of xµ < 0:",The+Shallow+and+the+Deep.pdf
"satisfies the constraint by simply clipping each xµ to zero whenever the gradient
step would lead into the excluded region of xµ < 0:
AdaTron algorithm, sequential updates (repeated presentation of D)
– at time step t, present example µ(t) = 1, 2, 3, ..., P, 1, 2, 3, ...
– update (only) the corresponding embedding strength:
xµ(t)(t+1) = max
󰀝
0, xµ(t)(t) + 󰁨η
󰀕
1 − Eµ(t)
󰀖󰀞
, (3.98)
where Eν = [C󰂓x]ν. The name AdaTron has been coined for this Adaptive
Perceptron algorithm [AB89,BAK91]. By comparison with the sequential Ada-
line algorithm (3.80) we observe that the change from equality constraints
(Eµ = 1) to inequalities (Eµ ≥ 1) leads to the restriction of embedding strengths
to non-negative values. Otherwise, the update +󰁨η (1 − Eµ) is adaptive in the
sense of the discussion of Adaline. Unlike the Rosenblatt algorithm, Adaline
and AdaTron can decrease an embedding strength if Eµ is already large.
A parallel version of the algorithm can be formulated, which performs steps",The+Shallow+and+the+Deep.pdf
"and AdaTron can decrease an embedding strength if Eµ is already large.
A parallel version of the algorithm can be formulated, which performs steps
along the direction of ∇xf, but always ensures 󰂓x > 0 by limiting the step size
where necessary. Thus, for sufficiently small values of 󰁨η, the cost function will
also decrease monotonically [AB89, BAK91]. Here, we refrain from giving a
more explicit mathematical formulation and discussion of the parallel updates.
Convergence of the sequential AdaTron algorithm
As we show in the following, the sequential AdaTron converges and yields op-
timal stability if 0 < 󰁨η < 2/Cµµ for all µ, provided the data set D is linearly",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 77
separable [AB89,BAK91].12 We consider the learning step xµ → xµ + δµ with
δµ =
󰀝
󰁨η (1 − Eµ) if − xµ ≤ 󰁨η (1 − Eµ)
−xµ < 0 if − xµ > 󰁨η (1 − Eµ) (3.99)
and see that in both cases
|δµ| ≤ 󰁨η |(1 − Eµ)| =⇒ 1 − Eµ
δµ ≥ 1
󰀱
󰁨η.
Hence, similar to Eq. (3.83) for the Adaline algorithm, we obtain
f(󰂓x+󰂓δ)−f(󰂓x) = −1
2
󰂓δ⊤C󰂓δ − 󰂓δ⊤ C󰂓x + 󰂓δ⊤󰂓1 = −1
2(δµ)2 Cµµ + δµ (1 − Eµ)
= (δµ)2
󰀗1 − Eµ
δµ − Cµµ
2
󰀘
≥ (δµ)2
󰀗1
󰁨η − Cµµ
2
󰀘
. (3.100)
The r.h.s. is obviously non-negative for 0 < 󰁨η ≤ 2/ Cµµ. We recover the same
condition (3.84) for the monotonic increase of f as in the sequential Adaline
algorithm.
Furthermore, it can be shown that f is bounded from above in the hyper-
octant x ≥ 0 if the data set is linearly separable. For general convex problems
with linear constraints it has been proven that the Wolfe Dual is bounded if
and only if the original problem has a solution [Fle00]. Here, this property",The+Shallow+and+the+Deep.pdf
"and only if the original problem has a solution [Fle00]. Here, this property
is closely related to Farkas’Lemma and Gordan’s Theorem of the Alternative
(3.37), which we discussed briefly in Sec. 3.3.6. The latter implies for matrices
C ∈ RP×P that
󰁱
󰂓x | C 󰂓x ≥ 󰂓1
󰁲
= ∅ ⇔ C 󰂓v = 0 for some 󰂓v ∕= 0 with {vµ ≥ 0}P
µ=1 .
Note that if such a vector 󰂓v exists, it is straightforward to show that f in Eq.
(3.77) cannot be bounded in 󰂓x > 0: Setting 󰂓x = γ󰂓v > 0, we obtain 󰂓x⊤C󰂓x = 0
and
lim
γ→∞
f(󰂓x) = lim
γ→∞
󰂓v⊤󰂓1 → ∞.
In summary, we have
a) For linearly separable D, the cost function f(󰂓x) is bounded from above in
the hyperoctant 󰂓x ≥ 0.
b) The function f(󰂓x) increases monotonically under non-zero, sequential Ada-
Tron updates with 0 < 󰁨η < 2/Cµµ.
c) By construction of the algorithm, any KT-point 󰂓x∗ of the problem, cf. Eq.
(3.93–3.95), is a fixed point of the AdaTron algorithm and vice versa.
In combination, (a-c) imply that the algorithm converges to a KT-point which",The+Shallow+and+the+Deep.pdf
"(3.93–3.95), is a fixed point of the AdaTron algorithm and vice versa.
In combination, (a-c) imply that the algorithm converges to a KT-point which
represents the (unique) perceptron of optimal stability.
12In [AB89, BAK91] normalized data with Cµµ = 1 was considered, yielding the condition
0 < 󰁨η < 2.",The+Shallow+and+the+Deep.pdf
"78 3. THE PERCEPTRON
Embedding strengths only
In contrast to the Rosenblatt or Adaline algorithm, it is not straightforward to
rewrite the AdaTron in terms of explicit updates in weight space. Obviously we
can compute the new weight vector as w(t + 1) = 󰁓P
µ=1 xµ(t + 1)ξµSµ
T after
each training step. However, a direct iteration of the weights w(t) cannot be
provided without involving the xµ(t), explicitly. This is due to the non-linear
clipping of embeddings at zero which has no simple equivalent in weight space.
Non-separable data
The outcome of the AdaTron training (3.98) for data that is not linearly sepa-
rable is not obvious. Modifications of the SSE (3.86) which take into account
that for Eµ > 1 the actual deviation is irrelevant have been considered in the
literature, see the discussion in Sec. 4.1.
For these, corresponding gradient descent methods in weight space can be
derived, which - for suitable choices of the cost function - resemble the Ada-",The+Shallow+and+the+Deep.pdf
"derived, which - for suitable choices of the cost function - resemble the Ada-
Tron scheme. However, they are not identical and, in particular, additional
constraints have to be imposed in order to achieve optimal stability for linearly
separable data.
The concept of optimal stability has been generalized in the so-called soft
margin classifier which tolerates misclassifications to a certain degree. A corre-
sponding extension of the AdaTron algorithm is discussed in Sec. 4.1.2.
Efficient algorithms
The MinOver and AdaTron are presented here as prototypical approaches to
solving the problem of optimal stability. While they are suitable for solving
the problem in relatively small data sets, the computational load may become
problematic when dealing with large numbers of examples and, consequently, a
large number of variables xµ.
A variety of algorithms have been devised aiming at computational effi-
ciency and scalability, mainly in the context of Support Vector Machines, cf.",The+Shallow+and+the+Deep.pdf
"A variety of algorithms have been devised aiming at computational effi-
ciency and scalability, mainly in the context of Support Vector Machines, cf.
Sec. 4.3. A prominent example is Sequential Minimal Optimization (SMO)
approach [Pla98]. It is the basis of many implementations, see e.g. [svm99]
for links and the SVM literature for further discussions and references [SS02,
CST00,STC04,Her02].
3.7.4 Support vectors
The treatment of optimal stability as a constrained, convex optimization prob-
lem has provided useful insights, from which practical algorithms such as the
AdaTron and other schemes emerged, see [Ruj93] for another example. The ab-
sence of local minima and the resulting availability of efficient optimization tools
is one of the foundations of the Support Vector Machine and has contributed
largely to its popularity [SS02,CST00,STC04,Her02,DFO20], see Sec. 4.3.
One of the most important observations with respect to optimal stability",The+Shallow+and+the+Deep.pdf
"largely to its popularity [SS02,CST00,STC04,Her02,DFO20], see Sec. 4.3.
One of the most important observations with respect to optimal stability
is a consequence of the complementarity condition (3.95). It implies that in a",The+Shallow+and+the+Deep.pdf
"3.7. OPTIMAL STABILITY BY QUADRATIC OPTIMIZATION 79
Eµ =1
∝wmax
Figure 3.18: Support Vectors in
the perceptron of optimal stability.
The arrow represents the normalized
weight vector wmax/|wmax|. Class
membership is indicated by filled
(ST = +1) and open symbols (ST =
−1), respectively. Support vectors,
marked as squares, fall into the two
hyperplanes with Eµ = 1, i.e. κµ =
κmax. All other examples (circles) dis-
play greater stability without being
embedded explicitly.
KT-point we have
either
󰀝 Eµ = 1
xµ ≥ 0
󰀞
or
󰀝 Eµ > 1
xµ = 0
󰀞
. (3.101)
In the geometrical interpretation of linear separability, the former set of feature
vectors with xµ ≥ 0 falls into one of the two hyperplanes with w · ξ = −1
or +1, respectively. Both planes are parallel to the decision boundary and
mark the maximum achievable gap between the two classes, see Figure 3.18.
Only these examples, marked as squares in the figure, contribute to the linear",The+Shallow+and+the+Deep.pdf
"Only these examples, marked as squares in the figure, contribute to the linear
combination (3.61) with non-zero embedding strength. They can be said to form
the support of the weights and, therefore, are referred to as the set of support
vectors. In a sense, they represent the hard cases in D which end up closest
to the decision boundary. The remaining training data display larger distances
from the separating hyperplane and do not explicitly contribute to the linear
combination (3.61).
For the support vectors we observe that wmax is the weight vector that
solves the system of linear equations Eµ = 1 with minimal norm |w|. All
other examples appear to be stabilized “accidentally”. Hence, if we were able
to identify them beforehand in a given D, we could solve the simpler problem
(3.77) restricted to the support vectors by applying, e.g., the Adaline algorithm.
Unfortunately, the set of support vectors is not known a priori and their deter-",The+Shallow+and+the+Deep.pdf
"Unfortunately, the set of support vectors is not known a priori and their deter-
mination is an integral part of the training process.
It is important to realize that, despite the special role of the support vectors,
all examples in the data set are relevant. Even if some of them end up with
zero embedding, the composition of the entire D implicitly determines the set
of support vectors and, thus, the actual weight vector.
Remark: Complementarity in the MinOver algorithm
The MinOver algorithm (3.59,3.60) as presented in Sec. 3.6.2 yields, in gen-
eral, non-zero integer xµ > 0 for all µ. However, the complementary condition
(3.101) is satisfied in the sense that the embedding strengths of the support vec-
tors increase in the training process, while those of the other examples remain",The+Shallow+and+the+Deep.pdf
"80 3. THE PERCEPTRON
relatively small. In the limit of infinitely many update steps, properly rescaled
xµ satisfy condition (3.92).
3.8 Inhom. lin. sep. functions revisited
So far, following the arguments leading to Eq. (3.4) in Sec. 3.2 we have treated
homogeneously and inhomogeneously linearly separable functions on the same
footing. However, with respect to the stability criterion we have to take a
subtlety into account: if we consider a perceptron with output
S(ξ) = sign [w · ξ − θ] with adaptive quantities w ∈ RN and θ ∈ R, (3.102)
the proper definition of the stabilities is
κµ
inh = (w · ξµ − θ) Sµ/ |w| (3.103)
which corresponds to the oriented distance of feature vectors ξµ from the sepa-
rating hyperplane given by (w· ξ −θ = 0). As a consequence, the corresponding
perceptron of optimal stability is given by the solution of the problem
min
w,θ
w2/2 subject to [w · ξµ − θ] Sµ ≥ 1 for all µ = 1, 2, . . . P. (3.104)",The+Shallow+and+the+Deep.pdf
"min
w,θ
w2/2 subject to [w · ξµ − θ] Sµ ≥ 1 for all µ = 1, 2, . . . P. (3.104)
Note that if for suitable θ, w2 can be minimized under the constraints, the choice
of θ will influence the result, i.e. the achievable maximum stability. Hence, the
optimization has to be done with respect to both w and θ.
If we naively exploit the equivalence of inhomogeneous separability in N
dimensions with homogeneous separability in N + 1 dimensions, c.f. Eq. (3.4),
we define the modified weight vector 󰁨w = [w, θ] and augmented feature vectors
󰁨ξ = [ξ, −1]. The corresponding naive definition of the stabilities reads
󰁨κµ = 󰁨w · 󰁨ξ
µ
Sµ
󰀱
|󰁨w|. (3.105)
Here 󰁨w2 corresponds to (w2 + θ2) and the resulting optimal stability problem
becomes
min
󰁨w
󰁨w2/2 subject to 󰁨w · 󰁨ξ
µ
Sµ ≥ 1 for all µ = 1, 2, . . . P, (3.106)
where the constraints are equivalent to those in Eq. (3.104). However, treating θ
as an additional weight in the spirit of (3.4) has altered the objective function in",The+Shallow+and+the+Deep.pdf
"as an additional weight in the spirit of (3.4) has altered the objective function in
comparison with (3.104) as it includes a contribution θ2 in the squared Euclidean
norm 󰁨w2. Consequently the solution of (3.106) would differ from the (correct)
optimal stability defined by Eq. (3.104).
For the sake of clarity, we have explicitly limited our discussion of opti-
mal stability to the case of homogenous separation with θ = 0. This also
applies to the Support Vector Machine in Sec. 4.3. The corresponding modifi-
cations of the algorithms and other results to the modified definitions (3.103)
and (3.104) for nonzero local threshold are conceptually straightforward, see
e.g. [Str19,HTF01,DFO20].",The+Shallow+and+the+Deep.pdf
"3.9. SOME REMARKS 81
3.9 Some remarks
In this chapter we have considered the simple perceptron as a prototypical
machine learning system. It serves as a framework in which to obtain insights
into the basic concepts of supervised learning. It also illustrates the importance
of optimization techniques and the related theory in machine learning.
The restriction to linearly separable functions is, of course, significant. In
the next chapter we consider a number of ways to deal with non-separable data
sets and address classification problems beyond linear separability.
It is remarkable that the perceptron still ranks among the most frequently
applied machine learning tools. This is due to the fact that the very successful
Support Vector Machine is frequently used with a linear kernel, cf. Sec. 4.3.
In this case, however, it reduces to a classifier that is equivalent to the simple
perceptron of optimal stability or its extension to the soft margin classifier, cf.
Sec. 4.1.2.",The+Shallow+and+the+Deep.pdf
82 3. THE PERCEPTRON,The+Shallow+and+the+Deep.pdf
"Chapter 4
Beyond linear separability
Non-linear means it’s hard to solve.
— Arthur Mattuck
In the previous sections we studied learning in version space as a basic train-
ing strategy in terms of the simple perceptron classifier. We obtained important
insights into the principles of learning a rule and obtained the concept of optimal
stability.
As pointed out already, learning in version space only makes sense under
a number of conditions, which – unfortunately – are rarely fulfilled in realistic
settings. The key assumptions are
(a) The data set D = {ξµ, Sµ
T }P
µ=1 is perfectly reliable, labels are correct and
noise–free.
(b) The unknown rule is linearly separable, or more generally: the student
complexity perfectly matches the target task.
In practice, a variety of effects can impair the reliability of the training data:
Some examples could be explicitly mislabeled by an unreliable expert to begin
with. Class labels could be inverted due to some form of noise in the communi-",The+Shallow+and+the+Deep.pdf
"with. Class labels could be inverted due to some form of noise in the communi-
cation between student and teacher. Alternatively (or additionally), some form
of noise or corruption may have distorted the input vectors ξµ in D, potentially
rendering some of the labels Sµ
T inconsistent.
In fact, real world training scenarios and data sets will hardly ever meet
the conditions (a) and/or (b). From the perspective of perceptron training, i.e.
restricting the hypothesis space to linearly separable functions, the following
situations are much more likely to occur:
83",The+Shallow+and+the+Deep.pdf
"84 4. BEYOND LINEAR SEPARABILITY
i) The unknown target rule is linearly separable, but D contains mislabeled
examples, for instance in the presence of noise. Depending on the number
of examples P and the degree of the corruption the following may occur:
i.1) In particular, small data sets D with few mislabeled samples can still
be linearly separable and the labels Sµ
T can be reproduced by a per-
ceptron student. While the non-empty version space V is consistent
with D, it is not perfectly representative of the target rule and the
success of training will be impaired.
i.2) The data set D is not linearly separable and, consequently, a version
space of linearly separable functions does not exist: V = ∅. In this
case, learning in version space is not even defined for the perceptron.
Hence, the data set still contains information (albeit noisy or cor-
rupted) about the rule, but it is not obvious how it can be extracted
efficiently.",The+Shallow+and+the+Deep.pdf
"Hence, the data set still contains information (albeit noisy or cor-
rupted) about the rule, but it is not obvious how it can be extracted
efficiently.
ii) The target rule itself is not linearly separable and would require learning
systems of greater complexity than the simple perceptron, ideally. Again,
the consequences for perceptron training depend on the size of the available
data set:
ii.1) Small data sets may be linearly separable and one can expect or hope
that a student w ∈ V at least approximates the unknown rule to a
certain extent, in this situation.
ii.2) Larger data sets become non-separable and V = ∅ should signal that
the target is, in fact, more complex.
Of course, superpositions of cases i) and ii) can also occur: we could have to
deal with a non-separable rule, represented by a data set which in addition is
subject to some form of corruption.
As mentioned earlier, it is a difficult task in practice to determine whether",The+Shallow+and+the+Deep.pdf
"subject to some form of corruption.
As mentioned earlier, it is a difficult task in practice to determine whether
a given data set D is linearly separable or not. Not finding a solution by use
of the Rosenblatt algorithm, for instance, could simply indicate that a larger
number of training steps is required. In [ND91], a perceptron-like algorithm
is presented which either finds a solution or terminates and establishes non-
separability [ND91], see also Sec. 3.3.6.
In the following, we discuss a number of basic ideas and strategies for coping
with non-separable data sets:
◦ A simple perceptron could be trained to implement D approximately, in
the sense that a large (possibly maximal) fraction of training labels is
reproduced by the perceptron. In Sec. 4.1 we will present and discuss
corresponding training schemes.
◦ More powerful, layered networks for classification can be constructed from
perceptron-like units. We will consider the so-called committee machine",The+Shallow+and+the+Deep.pdf
"◦ More powerful, layered networks for classification can be constructed from
perceptron-like units. We will consider the so-called committee machine
and parity machine as example two-layer systems in Sec. 4.2. We show
that the latter constitutes a universal classifier.",The+Shallow+and+the+Deep.pdf
"4.1. PERCEPTRON WITH ERRORS 85
◦ Many perceptrons (or other simple classifiers) can be combined into an
ensemble in order to take advantage of a wisdom of the crowd effect
[OM99,Urb00]. So-called Random Forests, i.e. ensembles of decision trees
[Bre01], constitute one of the most prominent examples in the literature,
currently. We refrain from a detailed discussion of ensembles and refer to
the literature [OM99,Urb00] for further references.
◦ A perceptron-like threshold operation of the form S = sign(. . .) with linear
argument can be applied after a non-linear transformation of the feature
vectors. Along these lines, the framework of the Support Vector Machines
(SVM) was developed. It has been particularly successful and continues
to do so. In Sec. 4.3 we will outline the concept and show that the SVM
can be seen as a (highly non-trivial) conceptual extension of the simple
perceptron.
◦ Perhaps the most frequently applied strategy is to treat the classification",The+Shallow+and+the+Deep.pdf
"perceptron.
◦ Perhaps the most frequently applied strategy is to treat the classification
as a regression problem. A network of continuous units (including the
output) can be trained by standard methods and, eventually, the output
is thresholded. We have seen one example already in terms of the Adaline
scheme, see Sec. 3.7. We return to this strategy in Chapter 5.
◦ Prototype-based systems [BHV16] offer another, very powerful framework
for classification beyond linear separability. In Chapter 6 we will discuss
the example of Learning Vector Quantization (LVQ), which implements –
depending on the details – piecewise linear or piecewise quadratic decision
boundaries. It is moreover a natural tool for multi-class classification.
4.1 Perceptron with errors
Assume that a given data set {ξµ, Sµ
T } with ξµ ∈ RN and Sµ
T = {−1, +1} is not
linearly separable due to one or several of the reasons discussed above.
In the following, we discuss extensions of perceptron training which toler-",The+Shallow+and+the+Deep.pdf
"linearly separable due to one or several of the reasons discussed above.
In the following, we discuss extensions of perceptron training which toler-
ate misclassification to a certain degree. Intuitively, this corresponds to the
assumption that the data set is nearly linearly separable or in other words: the
unknown rule can be approximated by a linearly separable function.
4.1.1 Minimal number of errors
Aiming at a linearly separable approximation, one might want to minimize
the number of misclassifications by choice of the weight vector w ∈ RN in
a simple perceptron student. While at a glance the idea appears plausible, we
should realize that the outcome will be very sensitive to individual, misclassified
examples in D.
Formally, we can consider the following problem:",The+Shallow+and+the+Deep.pdf
"86 4. BEYOND LINEAR SEPARABILITY
Minimal number of errors (Perceptron) (4.1)
For a given D = {ξµ, Sµ
T }P
µ=1 with ξµ ∈ RN and Sµ
T ∈ {−1, +1} ,
minimize
w ∈ RN Herr(w) =
P󰁛
µ=1
󰂃(w, ξµ, Sµ
T ) with 󰂃 =
󰀝
1 if sign (w · ξµ) = −Sµ
T
0 if sign (w · ξµ) = +Sµ
T .
Note that this cost function does not differentiate between nearly correct
examples with Eµ ≈ 0 close to the decision boundary and clear cases where the
feature vector falls deep into the incorrect half-space.
The above defined problem proves very difficult. Note that the number of
misclassified examples is obviously integer and can only display discontinuous
changes with w. On the other hand, ∇w Herr = 0 almost everywhere in feature
space. As a consequence, gradient based methods cannot be employed for the
minimization or approximate minimization of Herr.
A prescription that applies Hebbian update steps which can be seen as a
modification of the Rosenblatt algorithm (3.15) has been introduced by S.I.",The+Shallow+and+the+Deep.pdf
"A prescription that applies Hebbian update steps which can be seen as a
modification of the Rosenblatt algorithm (3.15) has been introduced by S.I.
Gallant [Gal90]. The so-called Pocket Algorithm relies on the stochastic pre-
sentation of simple examples in combination with the principle of learning from
mistake for a weight vector w. In addition to the randomized, yet otherwise
conventional, Rosenblatt updates of the vector w, the so far best (with respect
to Herr) weight vector 󰁥w is “putinto the pocket”. It is only replaced when the
on-going training process yields a vector w with a lower number of errors.
Pocket algorithm (4.2)
– initialize w(0) = 0 and 󰁥w = 0 (tabula rasa), set 󰁥H(0) = P
– at time step t, select a single feature vector ξµ with class label Sµ
T
randomly from the data set D with equal probability 1/P
– compute the local potential Eµ(t) = w(t) · ξµSµ
T
– for w(t), perform an update according to (Rosenblatt algorithm)
w(t+1) = w(t) + 1
N Θ
󰁫
−Eµ(t)
󰁬
ξµ Sµ
T .",The+Shallow+and+the+Deep.pdf
"T
– for w(t), perform an update according to (Rosenblatt algorithm)
w(t+1) = w(t) + 1
N Θ
󰁫
−Eµ(t)
󰁬
ξµ Sµ
T .
– compute H(t+1) = Herr󰀃
w(t+1)
󰀄
acc. to Eq. (4.1)
– update the pocket vector 󰁥w(t) and 󰁥H(t) according to
󰁥w(t+1) =
󰀫
w(t+1) if H(t+1) < 󰁥H(t) (improvement)
󰁥w(t) if H(t+1) ≥ 󰁥H(t) ( 󰁥w unchanged)
󰁥H(t+1) = Herr󰀃󰁥w(t+1)
󰀄
.",The+Shallow+and+the+Deep.pdf
"4.1. PERCEPTRON WITH ERRORS 87
Obviously, the number of errors 󰁥H(t) of the pocket vector 󰁥w(t) can never
increase under the updates (4.2). Moreover, one can show that the stochastic
selection of the training sample guarantees that, in principle, 󰁥w(t) approaches
the minimum of Herr with probability one [Gal90,Roj96].
However, this quite weak convergence in probability does not allow to make
statements about the expected number of updates required to achieve a solution
of a particular quality. Certainly it is not possible to provide upper bounds as
in the context of the Perceptron Convergence Theorem.
Several alternative, well-defined approaches have been considered which can
be shown to converge in a more conventional sense, at the expense of having to
accept sub-optimal Herr.
Along the lines discussed towards the end of Sec. 3.7.2, the gradient based
minimization of several cost functions similar to Eq. (3.86) has been consid-",The+Shallow+and+the+Deep.pdf
"minimization of several cost functions similar to Eq. (3.86) has been consid-
ered in the literature. For instance [GG91] discusses objective functions which
correspond to
H[k](w) = 󰁓P
µ=1 (c − Eµ)k Θ[c − Eµ] with Eµ = w · ξµSµ
T
in our notation. Note that the limiting case c = 0, k → 0 would recover the
non-differentiable Herr, Eq. (4.1). In [GG91] the above functions are, not quite
precisely, referred to as the perceptron cost function for k = 1 and the adatron
cost function for k = 2, respectively. We would like to point out that, despite
the conceptual similarity, the case c = 1, k = 2 is not strictly equivalent to the
AdaTron algorithm for non-separable data sets. Moreover, for lin. sep. data, any
w ∈ V gives H[2](w) = 0. Hence, optimal stability would have to be enforced
through an additional minimization of the norm w2.
4.1.2 Soft margin classifier
An explicit extension of the large margin concept has been suggested which",The+Shallow+and+the+Deep.pdf
"4.1.2 Soft margin classifier
An explicit extension of the large margin concept has been suggested which
allows for the controlled acceptance of misclassifications. For an introduction
in the context of the SVM, see e.g. [SS02, DFO20] and references therein. The
formalism presented there reduces to the soft margin perceptron in the case of
a linear kernel function, see Sec. 4.3 for the relation.
A corresponding, extended optimization problem similar to (3.68) reads:
Soft margin perceptron (weight space)
minimize
w, 󰂓β
N
2 w2 + γ
P󰁛
µ=1
βµ subject to {Eµ ≥ 1 − βµ}P
µ=1
and {βµ ≥ 0}P
µ=1
. (4.3)
Here, we introduce so-called slack variables βµ ∈ R with
󰀝 βµ = 0 ⇔ Eµ ≥ 1
βµ > 0 ⇔ Eµ < 1.",The+Shallow+and+the+Deep.pdf
"88 4. BEYOND LINEAR SEPARABILITY
We recover the unmodified problem of optimal stability (3.68) with βµ = 0 for
all µ = 1, 2, ..., P, which also implies that all Eµ ≥ 1. On the contrary, non-
zero βµ > 0 correspond to violations of the original constraints, i.e. examples
with Eµ < 1, which includes potential misclassifications (Eµ < 0). In (4.3),
the Lagrange multiplier γ ∈ R controls the extent to which non-zero βµ are
accepted.
In the by now familiar matrix-vector notation with 󰂓β = (β1, β2, ..., βP )⊤ we
can rewrite the problem in terms of embedding strengths:
Soft margin perceptron (embedding strengths)
minimize
󰂓x, 󰂓β
1
2 󰂓x⊤C󰂓x + γ󰂓β⊤󰂓1 subject to 󰂓E ≥ 󰂓1 − 󰂓β
and 󰂓β ≥ 0. (4.4)
Here, the derivation of the Wolfe Dual [Fle00] amounts to the elimination
of the slack variables 󰂓β. Similar to the error-free case, cf. Sec. 3.7, we obtain a
modified cost function with simpler constraints:
Soft margin perceptron (dual problem)
maximize
󰂓x − 1
2 󰂓x⊤C󰂓x + 󰂓x⊤󰂓1 subject to 󰂓0 ≤ 󰂓x ≤ γ 󰂓1. (4.5)",The+Shallow+and+the+Deep.pdf
"Soft margin perceptron (dual problem)
maximize
󰂓x − 1
2 󰂓x⊤C󰂓x + 󰂓x⊤󰂓1 subject to 󰂓0 ≤ 󰂓x ≤ γ 󰂓1. (4.5)
In comparison to the dual problem (3.68) for the error-free case, the non-
negative embedding strengths 󰂓x ≥ 0 are now also bounded from above. The
parameter γ limits the magnitudes of the xµ.
In analogy to the derivation of the AdaTron algorithm (3.98) we can devise
a similar, sequential projected-gradient descent algorithm:
AdaTron with errors, sequential updates (repeated presentation of D)
– at time step t, present the example µ = 1, 2, ..., P, 1, 2, ...
– perform the update
󰁨xµ(t + 1) = xµ(t) + 󰁥η (1 − [C󰂓x(t)]µ) gradient step
󰁥xµ(t + 1) = max {0, 󰁨xµ(t + 1)} non-negative embeddings...
xµ(t + 1) = min {γ, 󰁥xµ(t + 1)} ...with limited magnitude. (4.6)",The+Shallow+and+the+Deep.pdf
"4.1. PERCEPTRON WITH ERRORS 89
Eµ =1
∝wmax
Figure 4.1: Support vectors in the
soft margin perceptron.
The arrow represents the normalized
weight vector wmax/|wmax|. Filled
and open symbols correspond to the
classes Sµ
T = ±1, respectively. Sup-
port vectors, displayed as squares,
fall onto the two hyperplanes with
Eµ = 1, into the region between the
planes, or even deeper into the incor-
rect half-space. All other examples,
marked by circles, display Eµ > 1
without explicit embedding.
Compared to the original AdaTron for lin. sep. data, the only difference is the
additional restriction of the search to the 󰂓x ≤ γ 󰂓1. Obviously we recover the the
original algorithm in the limit γ → ∞.
As in the separable case, the algorithm follows the gradient of the cost func-
tion along (󰂓1 − 󰂓E), in principle. Hence, an individual embedding strength will
increase if the corresponding example has a local potential Eµ < 1, including",The+Shallow+and+the+Deep.pdf
"increase if the corresponding example has a local potential Eµ < 1, including
misclassifications with Eµ < 0. If some of the errors cannot be corrected because
D is not separable, the corresponding xµ would grow indefinitely. In the soft
margin version (4.6), however, updates are clipped at xµ = γ and the misclassi-
fication1 of the corresponding example is tolerated. For fixed γ, the problem is
well-defined and the AdaTron with errors (4.6) finds a solution efficiently. We
refrain from further analysis and a formal proof.
It is important to realize that the precise nature of the solution depends
strongly on the setting of the parameter γ: It controls the compromise between
the goals of – on the one hand – minimizing the norm w2 (maximizing the
margin) and – on the other hand – correcting misclassifications. More precisely,
the emphasis is not explicitly on the number of errors, but on the violations of
Eµ ≥ 1 and their severity.",The+Shallow+and+the+Deep.pdf
"the emphasis is not explicitly on the number of errors, but on the violations of
Eµ ≥ 1 and their severity.
If, for instance a mismatched (too small) value of γ is chosen, misclassifi-
cations will be accepted and favored, even in a linearly separable data set. In
practice, a suitable value can be determined by means of a validation procedure
which estimates the performance for different choices of γ.
In analogy to Sec. 3.7.4 the support vectors are characterized by xµ > 0,
as before. However, only examples with 0 < xµ < γ will lie exactly in one of
the planes with Eµ = 1. Clipped embedding strengths xµ = γ correspond to
examples which fall into the region between the planes in Fig. 4.1 or even deeper
into the incorrect half-space.
The soft margin concept for the toleration of misclassification is highly rel-
evant in the context of the Support Vector Machine, see Sec. 4.3.
1the violation of Eµ ≥ 1, to be more precise",The+Shallow+and+the+Deep.pdf
"90 4. BEYOND LINEAR SEPARABILITY
input layer ξ ∈ RN
adaptive weights wk ∈ RN
(k = 1, 2, . . . K)
hidden units σk = sign
󰀃
wk · ξ−θk󰀄
fixed hidden-to-output relation F(...)
binary output S
󰀃
ξ
󰀄
= F
󰀓󰀋
σk(ξ)
󰀌K
k=1
󰀔
Figure 4.2: The architecture of a “machine” as introduced in Sec. 4.2. A
number K of hidden units of the perceptron type are connected by adaptive
weights with the N-dim. input layer, K = 3 in the illustration. The bi-
nary response S(ξ) is determined by a pre-defined, fixed functional dependence
F
󰀃
σ1, σ2, ...σK󰀄
= ±1.
4.2 Layered networks of perceptron-like units
Perceptron-like units can be assembled in more powerful architectures as to
overcome the restriction to linearly separable functions. We will highlight this
in terms of a family of systems which are occasionally termed machines in
the literature, see [MD89a, EB01, WRB93, AMB+18, MZ95, BO91, SH93] and
references therein.
In Fig. 4.2 the architecture of a machine is illustrated. It comprises",The+Shallow+and+the+Deep.pdf
"references therein.
In Fig. 4.2 the architecture of a machine is illustrated. It comprises
◦ an input layer representing feature vectors ξ ∈ RN
◦ a single layer of K perceptron-like hidden units
σk(ξ) = sign
󰀃
wk · ξ − θk󰀄
◦ a set of adaptive input-to-hidden weight vectors wk ∈ RN and
local thresholds2 θk ∈ RN
◦ a single, binary output, determined by a fixed function F(σ1, σ2, ...σK)
The function F ultimately determines the network’s input/output relation.
However, it is assumed to be pre-wired and cannot be adapted in the training
process. Learning is restricted to the wk connecting input and hidden layer.
2Thresholds could be replaced by a formal weight from a clamped input as outlined in
(3.4).",The+Shallow+and+the+Deep.pdf
"4.2. LAYERED NETWORKS OF PERCEPTRON-LIKE UNITS 91
4.2.1 Committee and parity machines
Two specific machines have attracted particular interest:
CM: The committee machine combines the hidden unit states σk in a majority
vote [EB01,MD89a,WRB93,AMB+18]. This is realized by setting
FCM 󰀃󰀋
σk󰀌K
k=1
󰀄
= sign
󰀣 K󰁛
k=1
σk
󰀤
⇒ SCM(ξ)=sign
󰀣 K󰁛
k=1
sign
󰀅
wk · ξ−θk󰀆
󰀤
(4.7)
which is only well-defined for odd values of K which avoids ties 󰁓
k σk = 0.
The majority vote is reminiscent of an ensemble of independently trained
perceptrons. The CM, however, is meant to be trained as a whole [SH93].
PM: In the parity machine the output is computed as a product over the hidden
unit states σk [EB01,WRB93,BO91]:
FPM 󰀃󰀋
σk󰀌K
k=1
󰀄
=
K󰁜
k=1
σk ⇒ SPM (ξ) =
K󰁜
k=1
sign
󰀅
wk · ξ
󰀆
(4.8)
which results in a well-defined binary output SPM (ξ) = ±1 for any K.
Note that the product depends on whether the number of units with
σk = −1 is odd or even. In this sense FPM (. . .) is analogous to a par-
ity operation.",The+Shallow+and+the+Deep.pdf
"σk = −1 is odd or even. In this sense FPM (. . .) is analogous to a par-
ity operation.
The hidden-to-output relation of the PM, i.e. the parity operation, cannot be
represented by a single perceptron unit.3 We could realize the hidden-to-output
function by a more complex multi-layered network. But since F is considered
to be pre-wired and not adaptive, we do not have to specify a neural realization
here. On the contrary, the committee machine can be interpreted as a conven-
tional two-layered feed-forward neural network with an N −K−1 architecture
as discussed in Sec. 1.3.2 and illustrated in Fig. 1.5 (right panel). However,
compared to the general form of the output, Eq. (1.16), we have to use the acti-
vation function g(z) = sign(z) throughout the net and fix all hidden-to-output
to vk =1 in the CM.
Many theoretical results are available for CM and PM networks and more
general machines. Among other things, their storage capacity and generalization",The+Shallow+and+the+Deep.pdf
"general machines. Among other things, their storage capacity and generalization
ability have been addressed, see [MD89a, SH93, Opp94, PBG+94] for examples
and [EB01,WRB93,AMB+18] for further references.
3 The gist of Minsky and Papert’s book [MP69] is often reduced to this single insight,
which is a gross injustice to both, the book and the perceptron.",The+Shallow+and+the+Deep.pdf
"92 4. BEYOND LINEAR SEPARABILITY
+ + +
+ + −
+ − +
+ − −
− + +
−
−
+−
+
−
 + + +
+ + −
+ − −
+ − +
− + +
−
−
+−
+
−
Figure 4.3: Output function of Committee machine and Parity machine,
applied to identical sets of feature vectors ξ (filled and empty circles). The
illustration corresponds to the surface of an N-dim. hypersphere. In both ma-
chines, K = 3 oriented hyperplanes tesselate the feature space. Arrows point to
the respective half-space of σk = +1. The three hidden units states are marked
by ""+"" and ""-"" signs in the corresponding regions. The networks’ responses
S = ±1 are marked by empty and filled circles. Left panel: The majority
of σk determines the total response of the CM. Dashed lines mark redundant
pieces of the hyperplanes which do not separate outputs S = +1 from S = −1.
Right panel: In the PM, the total output is S = 󰁔
k σk. Every hyperplane
separates total outputs S = ±1 locally, everywhere in feature space.
4.2.2 The parity machine: a universal classifier",The+Shallow+and+the+Deep.pdf
"k σk. Every hyperplane
separates total outputs S = ±1 locally, everywhere in feature space.
4.2.2 The parity machine: a universal classifier
Here, we focus on a particularly interesting result. We show that a PM with
sufficiently many hidden units can implement any data set of the form D =
{ξµ, Sµ
T }P
µ=1 with binary training labels.
In the following, we outline a constructive proof which is based on the appli-
cation of a particular training strategy. So-called growth algorithms add units
to a neural network with subsequent training until the desired performance is
achieved and, for example, a given data set D has been implement. Mezard
and Parisi coined the term tiling algorithm for a particular procedure [MN89],
which adds neurons one by one. Several other similar growth schemes have been
suggested, see [GM90,Fre90] for examples and references.
A particular tiling-like algorithm for the PM was introduced and analysed",The+Shallow+and+the+Deep.pdf
"suggested, see [GM90,Fre90] for examples and references.
A particular tiling-like algorithm for the PM was introduced and analysed
in [BO91]. It was not necessarily designed as a practical training prescription
for realistic applications. Tiling-like training of the PM proceeds along the fol-
lowing lines:",The+Shallow+and+the+Deep.pdf
"4.2. LAYERED NETWORKS OF PERCEPTRON-LIKE UNITS 93
Tiling-like learning, (parity machine) (4.9)
(I) Initialization (m = 1):
Train the first unit with output S1(ξ) = σ1(ξ) = sign[w1 · ξ − θ1]
from the data set D1 = D , aiming at a large number of correctly classified
examples Q1.4
(II) After training of m units:
Given the PM with m hidden units {σ1, σ2, ...σm}, re-order the
indices µ of the examples such that the PM output is
Sm(ξµ) =
m󰁜
j=1
σj(ξµ) =
󰀝
+Sµ
T for 1 ≤ µ ≤ Qm
−Sµ
T for Qm < µ ≤ P,
where Qm is the number of correctly classified examples in D.
Define the new training set Dm+1 = {ξµ, [Sm(ξµ) Sµ
T ]}P
µ=1
with labels [Sm(ξµ)Sµ
T ] =
󰀝 +1 for 1 ≤ µ ≤ Qm (Sm was correct)
−1 for Qm < µ ≤ P (Sm was wrong )
(III) Training step:
add and train the next hidden unit σm+1(ξ) = sign
󰀅
wm+1 · ξ − θm+1
󰀆
as to achieve a low number of errors (P − Qm+1) w.r.t. data set Dm+1.
Note that if a solution with zero error, i.e. QM = P is found in step (III) for",The+Shallow+and+the+Deep.pdf
"as to achieve a low number of errors (P − Qm+1) w.r.t. data set Dm+1.
Note that if a solution with zero error, i.e. QM = P is found in step (III) for
the M-th hidden unit σM , the total output of the PM is
M󰁜
m=1
σm(ξµ) = Sµ
T for all examples in D,
i.e. the data set is perfectly reproduced by the PM of M units.
It is surprisingly straightforward to show that the number of correctly classi-
fied feature vectors can be increased by at least one (Qm+1 > Qm) when adding
the (m + 1)-th unit in step (III) of the procedure (4.9).
To this end, we consider a set of normalized input vectors in the procedure
(4.10). The normalization (4.11) could always be implemented in a preprocess-
ing step. The second relation (4.12) is trivially satisfied: in every data set δ can
be determined by computing all pair-wise scalar products.
4Any of the algorithms discussed in Sec. 4.1 could be used in this step. For the inhomo-
geneity, a clamped input as in Eq. (3.4) can be employed, for simplicity.",The+Shallow+and+the+Deep.pdf
"94 4. BEYOND LINEAR SEPARABILITY
Grandmother neuron (4.10)
Consider a set of feature vectors {ξµ}P
µ=1 with
|ξµ|2 = Γ for all µ = 1, 2, ..., P (4.11)
0 < δ < Γ − ξµ · ξν for all µ, ν (µ ∕= ν). (4.12)
Construct a perceptron weight vector and threshold as
w = −ξP and θ = δ − Γ. (4.13)
It results in the inhomogeneously linearly separable classification
Sw,θ(ξµ)=sign
󰀅
−ξP· ξµ−δ+Γ
󰀆
=
󰀻
󰁁󰁁󰁁
󰁁
󰀿
󰁁󰁁
󰁁
󰁁󰀽
sign
󰀅
−ξ
P · ξP
󰁿 󰁾󰁽 󰂀
=Γ
+Γ−δ
󰀆
=−1 for µ = P
sign
󰀅
− ξP · ξµ + Γ󰁿 󰁾󰁽 󰂀
>δ
−δ
󰀆
=+1 for µ ∕= P.
(4.14)
The corresponding perceptron separates exactly one feature vector, ξP , from all
others in the set. The term grandmother neuron has been coined for this type
of unit. It relates to the debatable concept that a single neuron in our brain is
activated specifically whenever we see our grandmother.5
For the tiling-like learning (4.9), this implies that, by use of a grandmother
unit, we can always separate ξP from all other examples in the training step",The+Shallow+and+the+Deep.pdf
"unit, we can always separate ξP from all other examples in the training step
(III). The hidden unit response for this input is σm+1(ξP ) = −1, which corrects
the misclassification as the incorrect output Sm(ξP ) = −SP
T is multiplied with
−1 yielding Sm+1(ξP ) = SP
T . All other PM outputs are left unchanged.
Hence, at least one error can be corrected by adding a unit to the growing
PM. With at most P units in the worst case, the number of errors is zero and
all labels in D are reproduced correctly.
A few remarks
◦ The grandmother unit (4.14) serves at best as a minimal solution in the
constructive proof - it is not suitable for practical purposes. The use of
O(P) perceptron units for the labelling of P examples would be highly
inefficient.
◦ Step (III) can be improved significantly as compared to the constructive
solution by using efficient training algorithms such as the soft margin
AdaTron, see Sec. 4.1.1.",The+Shallow+and+the+Deep.pdf
"solution by using efficient training algorithms such as the soft margin
AdaTron, see Sec. 4.1.1.
5An idea which is possibly not quite as unrealistic as it may seem, see for instance
[QRK+05] for a discussion of “Jennifer Aniston cells”.",The+Shallow+and+the+Deep.pdf
"4.2. LAYERED NETWORKS OF PERCEPTRON-LIKE UNITS 95
◦ Tiling-like learning imposes a strong ordering of the hidden units. Neurons
added to the system later are supposed to correct only the (hopefully)
very few misclassifications made by the first units. To some extent this
contradicts the attractive concept of neural networks as fault-tolerant and
robust distributed memories.
◦ The strength of the tiling concept is at the same time its major weak-
ness: Unlimited complexity and storage capacity can be achieved by
adding more and more units to the system, until error-free classification is
achieved. This will lead to inferior generalization behavior as the system
adapts to every little detail of the data. This suggests to apply a form
of early stopping, which limits the maximum number of units in the PM
according to validation performance.
We conclude that the parity machine is a universal classifier in the sense that
a PM with sufficiently many hidden units can implement any two-class data D:",The+Shallow+and+the+Deep.pdf
"a PM with sufficiently many hidden units can implement any two-class data D:
Universal classifier (parity machine) (4.15)
For a given data set D = {ξµ, Sµ
T }P
µ=1 with binary labels Sµ
T ∈ {−1, +1}
and normalized feature vectors with |ξµ|2 = const. for all µ = 1, 2, . . . P,
weight vectors wk ∈ RN and thresholds θk ∈ R exist (and can be found) with
SPM (ξµ) =
K󰁜
k=1
sign
󰀅
wk · ξµ − θk󰀆
= Sµ
T for all µ = 1, 2, . . . P.
Similar theorems have been derived for other “shallow” architectures with
a single layer of hidden units and a single binary output.6 These findings are
certainly of fundamental importance. In contrast to the Perceptron Convergence
Theorem, however, (4.15) and similar propositions are in general not associated
with practical, efficient training algorithms.
The result parallels the findings of Cybenko and others [Cyb89,HTF01] that
networks with a single, sufficiently large hidden layer of continuous non-linear",The+Shallow+and+the+Deep.pdf
"networks with a single, sufficiently large hidden layer of continuous non-linear
units constitute universal function approximators, see Chapter 5 for details.
It is interesting to note that also extremely deep and narrow networks can
be universal classifiers. As an example, stacks of single perceptron units with
shortcut connections to the input layer have been investigated in [Roj17,Roj03].
4.2.3 The capacity of machines
Compared to the simple perceptron, the greater complexity of the CM or PM
should lead to an increased storage capacity. In fact, following the work of
6Strictly speaking the PM does not fall into this class, as FPM cannot be realized by a
single perceptron-like unit.",The+Shallow+and+the+Deep.pdf
"96 4. BEYOND LINEAR SEPARABILITY
Mitchison and Durbin [MD89a], we can extend the arguments presented in Sec.
3.4 to machines.
The number CK(P, N) of different dichotomies that a machine with K hid-
den units can realize for P feature vectors in N dimensions is obviously bounded
from above as
CK(P, N) ≤ C(P, N)K (4.16)
with C(P, N) from Eq. (3.41) for the perceptron. If we assume that the network
can freely combine all lin. sep. functions realized by the perceptron units in the
hidden layer, we would expect an equality in (4.16). In general, correlations
between the hidden units and other redundancies or restrictions will reduce
CK(P, N) as compared to the upper bound.
With the total number of possible dichotomies 2P we obtain an upper bound
for the probability of a random labelling to be separable with K hidden units:
PK(P, N) = min
󰀝
1, C(P, N)K
2P
󰀞
(4.17)
where the minimum is applied to explicitly avoid PK(P, N) > 1 resulting from
the upper bound.",The+Shallow+and+the+Deep.pdf
"PK(P, N) = min
󰀝
1, C(P, N)K
2P
󰀞
(4.17)
where the minimum is applied to explicitly avoid PK(P, N) > 1 resulting from
the upper bound.
Further analytical treatment of Eq. (4.17) is involved, including the limit
N → ∞. However, we can obtain numerical estimates (upper bounds) of the
storage capacity by identifying, for given K and N, the value of P for which7
C(P, N)K
2P = 1
2. (4.18)
This marks the characteristic point αc(K) = P/N at which the probability
PK(P, N) drops, which is in line with the observation that C(2N, N)/2(2N) =
1/2 for the perceptron.
Example estimates obtained for N = 1000 and a few, small values of K are
αc(2)≈ 9.07(≈ 2 ×4.55), αc(6)≈ 40.53(≈ 6 ×6.76), αc(10)≈ 76.86(≈ 10 ×7.69)
which already shows that αc(K) displays a superlinear dependence on K. In
particular,
αc(K) > K × 2 = K × αc(1)
where the r.h.s. could be interpreted as a naive lower bound for combining K
perceptrons without any synergy effect.",The+Shallow+and+the+Deep.pdf
"αc(K) > K × 2 = K × αc(1)
where the r.h.s. could be interpreted as a naive lower bound for combining K
perceptrons without any synergy effect.
Figure 4.4 displays the numerical estimates for N = 100 and K ≤ 20. One
can show that for large values of K the capacity bound is given by [MD89a,
Opp94,BO91]
αc(K) ≤ K ln K/ ln 2 (4.19)
which is consistent with a faster than linear growth with the number of hidden
units. The fact that the capacity grows rapidly with K agrees with the finding
7 More precisely: the smallest integer P for which the upper bound exceeds 1/2.",The+Shallow+and+the+Deep.pdf
"4.3. SUPPORT VECTOR MACHINES 97
K
αc Figure 4.4: The storage capac-
ity αc(K) according to the esti-
mate (4.18), obtained for N =
100. The solid line corresponds
to the naive lower bound 2K for
combining K perceptrons.
that a (parity) machine with a sufficiently large number of units can implement
any given dichotomy.
In the context of Section 3.5.3, it also indicates that the number of hidden
units should be carefully selected in order to avoid the realization of large storage
capacity at the expense of poor generalization behavior.
The asymptotic K ln K-dependence of the storage capacity for large K has
been confirmed explicitly for the parity machine also by means of statistical
physics based considerations [EB01, BSS94, WRB93, Opp94]. Interestingly, for
the committee machine, the increase of αc appears to be slightly weaker as
αc ∝ K(ln K)1/2 for K → ∞ [Urb97].
This difference in the asymptotic growth of αc(K) is consistent with our",The+Shallow+and+the+Deep.pdf
"αc ∝ K(ln K)1/2 for K → ∞ [Urb97].
This difference in the asymptotic growth of αc(K) is consistent with our
intuitive insight that the committee machine is subject to redundancies, while
the parity machine makes full use of the separating hyperplanes in its hidden
layer.
The storage capacities and/or VC dimensions of a variety of network types
and architectures has been of great interest in the machine learning community.
This is due to the role that the capacity plays with respect to the ultimate goal
of learning, i.e. generalization. Examples from the statistical physics point of
view can be found in [EB01,BSS94,WRB93,Opp94], see also references therein.
4.3 Support Vector Machines
The Support Vector Machine (SVM) constitutes one of the most successful
frameworks in supervised learning for classification tasks. It combines the con-
ceptual simplicity of the large margin linear classifier, a.k.a. the perceptron of",The+Shallow+and+the+Deep.pdf
"ceptual simplicity of the large margin linear classifier, a.k.a. the perceptron of
optimal stability, with the power of general non-linear transformations to high-
dimensional spaces. In particular, it employs the so-called kernel trick which
makes it possible to realize the transformation implicitly.
For a detailed presentation of the SVM framework and further references
see, for instance, [SS02,CST00,STC04,Her02,DFO20]. A comprehensive repos-
itory of materials, including a short history of the approach is provided at
www.svms.org [svm99].
The concept of applying a kernel function in the context of pattern recog-
nition dates back to at least 1964, see Aizerman et al. [ABR64]. Probably the",The+Shallow+and+the+Deep.pdf
"98 4. BEYOND LINEAR SEPARABILITY
first practical version of Support Vector Machines, close to their current form,
was introduced by Boser, Guyon and Vapnik in 1992 [BGV92] and relates to
early algorithms developed by Vladimir Vapnik in the 1960s [VL63]. According
to Isabelle Guyon [Guy16], the MinOver algorithm [KM87] triggered their in-
terest in the concept of large margins which was then combined with the kernel
approach. Eventually, the important and practically relevant extension to soft
margin classification was introduced by Cortes and Vapnik in 1995 [CV95].
4.3.1 Non-linear transformation to higher dimension
The first important concept of the SVM framework exploits the fact that non-
separable data sets can become linearly separable by means of a non-linear
mapping of the form
ξ ∈ RN → Ψ(ξ) ∈ RM with components Ψj(ξ) (4.20)
where M can be different from N, in general. A function of the form
S(ξ) = sign [W · Ψ(ξ)] with weights W ∈ RM (4.21)",The+Shallow+and+the+Deep.pdf
"where M can be different from N, in general. A function of the form
S(ξ) = sign [W · Ψ(ξ)] with weights W ∈ RM (4.21)
is by definition linearly separable in the space of transformed vectors Ψ. So,
while retaining the basic structure of the perceptron, formally, we will be able to
realize functions beyond linearly separability by proper choice of the non-linear
transformation ξ → Ψ.
In fact, Rosenblatt already included this concept when introducing the Per-
ceptron: the threshold function sign(. . .) is applied to the weighted sum of states
in an association layer, cf. Fig. 3.1 showing the Mark I Perceptron. Its units are
referred to as masks or predicate units in the literature [Ros58, HRM+60], see
also [MP69]. In the hardware realization, for instance, 512 association units were
connected to subsets of the 400 photosensor units and performed a threshold
operation on an effectively randomized weighted sum of the incoming voltages,
see [HRM+60] for details.",The+Shallow+and+the+Deep.pdf
"operation on an effectively randomized weighted sum of the incoming voltages,
see [HRM+60] for details.
In the Support Vector Machine, the non-linear mapping is – in general – from
an N-dimensional space to a higher-dimensional space with M > N in order
to achieve linear separability of the classes. As an illustration of the concept
we discuss a simple example which was presented by Rainer Dietrich in [Die00]:
Consider a set of two-dimensional feature vectors ξ = (ξ1, ξ1)⊤, with two classes
separated by a non-linear decision boundary as displayed in Fig. 4.5 (left panel).
We apply the explicit transformation
Ψ(ξ1, ξ2) = (ξ2
1,
√
2 ξ1 ξ2, ξ2)⊤ ∈ R3
which is non-linear as it contains the square ξ2
1 and the product ξ1 ξ2. In the
example, the plane orthogonal to the weight vector W = (1, 1, −1)⊤ separates
the classes perfectly in M = 3 dimensions, see the center and right panels of
Fig. 4.5.
This is obviously only a toy example to illustrate the basic idea. In typical",The+Shallow+and+the+Deep.pdf
"Fig. 4.5.
This is obviously only a toy example to illustrate the basic idea. In typical
applications of the SVM the dimension N of the original feature space is already",The+Shallow+and+the+Deep.pdf
"4.3. SUPPORT VECTOR MACHINES 99
Figure 4.5: Illustration courtesy of Rainer Dietrich [Die00]. A two-dimensional
data set with two classes that are not linearly separable (left panel) can become
linearly separable after applying an appropriate non-linear transformation to a
higher-dimensional space (center and right panel, two different viewpoints).
quite large and frequently M has to satisfy M ≫ N in order to achieve linear
separability.
While the concept appears appealing, it is yet unclear how we should identify
a suitable transformation ξ → Ψ for a given problem and data set. Before
we return to this problem (and actually circumvent it elegantly), we discuss
the actual training, i.e. the choice of a suitable weight vector W in the M-
dimensional space.
4.3.2 Large margin classifier
Let us assume that for a given, non-separable data set DN =
󰀋
ξµ ∈ RN , Sµ
T
󰀌P
µ=1
we have found a suitable transformation such that
DM =
󰀋
Ψµ ∈ RM , Sµ
T
󰀌P
µ=1",The+Shallow+and+the+Deep.pdf
"󰀋
ξµ ∈ RN , Sµ
T
󰀌P
µ=1
we have found a suitable transformation such that
DM =
󰀋
Ψµ ∈ RM , Sµ
T
󰀌P
µ=1
is indeed linearly separable in M dimensions. Hence, we can apply conventional
perceptron training in the M-dimensional space and, by means of the Perceptron
Convergence Theorem (3.22), we are even guaranteed to find a solution.
However, in general, we do not have explicit control of (or reliable informa-
tion about) how difficult the task will be. On the one hand, we would wish to
use a powerful transformation to very high-dimensional Ψ in order to guarantee
separability and make it easy to find a suitable W. On the other hand, one
could expect inferior generalization behavior in that case. Along the lines of
the student-teacher scenarios discussed in Sec. 3.5.1, the corresponding version
space of consistent hypotheses W might be unnecessarily large.
The SVM aims at resolving this dilemma by determining the solution of",The+Shallow+and+the+Deep.pdf
"space of consistent hypotheses W might be unnecessarily large.
The SVM aims at resolving this dilemma by determining the solution of
maximum stability Wmax. Hence, the potentially very large freedom in select-
ing a weight vector W in the high-dim. version space is efficiently restricted",The+Shallow+and+the+Deep.pdf
"100 4. BEYOND LINEAR SEPARABILITY
and – following the arguments provided in Sec. 3.5.2 – we can expect good
generalization ability.
The mathematical structure of the corresponding problem is fully analogous
to the original (3.58). The M-dim. counterpart reads
Perceptron of optimal stability (M-dim. feature space) (4.22)
For a given data set DM = {Ψµ, Sµ
T }P
µ=1, find the vector Wmax ∈ RM
with Wmax = argmax
W κ(W) with κ(W) = min
󰁱
κµ = W·Ψ µSµ
T
|W|
󰁲P
µ=1
.
Obviously we can simply translate all results, concepts and algorithms from Sec.
3.6 to the transformed space.
So far we have assumed that the transformation ξ → Ψ exists and is explic-
itly known. Then we could for instance formulate and apply an M-dimensional
version of the MinOver algorithm (3.59,3.60). Moreover, we can apply the opti-
mization theoretical concepts and methods presented in Sec. 3.7 as exploited in
the next sections. Among other aspects, this implies that the resulting classifier",The+Shallow+and+the+Deep.pdf
"the next sections. Among other aspects, this implies that the resulting classifier
can be expressed in terms of support vectors, which ultimately motivates the
use of the term Support Vector Machine.
4.3.3 The kernel trick
In analogy to the original stability problem, cf. Sec. 3.7, we can introduce the
embedding strengths 󰂓X = (X1, X2, ..., XP )⊤ ∈ RP . With the shorthand Ψµ =
Ψ(ξµ) we also define the correlation matrix
Γ with elements Γµν = 1
M Sµ
T Ψµ · ΨνSν
T (4.23)
and analogous to Eqs. (3.76) we obtain
W = 1
M
P󰁛
µ=1
Xµ Ψµ Sµ and W2 = 1
M
󰂓X⊤ Γ 󰂓X. (4.24)
Eventually, we can re-formulate the problem (4.22) as
Perceptron of optimal stability (M-dim. feature space)
minimize
󰂓X
1
2
󰂓X⊤Γ 󰂓X subject to inequality constraints Γ 󰂓X ≥ 󰂓1 (4.25)
and proceed along the lines of Sec. 3.7 to derive, for instance, the corresponding
AdaTron algorithm, see below.",The+Shallow+and+the+Deep.pdf
"4.3. SUPPORT VECTOR MACHINES 101
The output of the M-dim. perceptron can be written as
S(ξ) = sign
󰁫
W · Ψ(ξ)
󰁬
= sign
󰀥
1
M
P󰁛
µ=1
Xµ SµΨµ · Ψ(ξ)
󰀦
. (4.26)
We note that this involves the scalar products of the M-dimensional, trans-
formed input vector with the transformed example training examples Ψµ. We
define a so-called kernel function
K : RN ×RN →R with K(ξ, 󰁥ξ) = 1
M Ψ(ξ) · Ψ(󰁥ξ) = 1
M
M󰁛
j=1
Ψj(ξ) Ψj(󰁥ξ)
(4.27)
which represents the scalar product in RM . We observe that
S(ξ) = sign
󰀥 P󰁛
µ=1
Xµ SµK(ξµ, ξ)
󰀦
(4.28)
does not involve the transformation Ψ(. . .) explicitly anymore. The kernel K is
defined as a function of pairs of original feature vectors. Similarly, we have
Eµ =
󰁫
Γ 󰂓X
󰁬µ
= Sµ
T
P󰁛
ν=1
Sν
T Xν K(ξµ, ξν). (4.29)
One can also formulate the AdaTron algorithm for optimal stability in the
M-dimensional space. The Kernel AdaTron was introduced and discussed in
[FCC98] and has been applied in a variety of practical problems. In analogy to
(3.98) it is given as",The+Shallow+and+the+Deep.pdf
"[FCC98] and has been applied in a variety of practical problems. In analogy to
(3.98) it is given as
Kernel AdaTron (sequential updates) (repeated presentation of D)
– at time step t, present example µ = 1, 2, 3, ..., P, 1, 2, 3, ...
– perform the update
Xµ(t + 1) = max
󰀫
0, Xµ(t) + 󰁥η
󰀣
1 − Sµ
T
P󰁛
ν=1
Sν
T Xν K(ξµ, ξν)
󰀤󰀬
. (4.30)
The training algorithm is also expressed in terms of the kernel and does not
require explicit use of the transformation Ψ, formally.
So far, the above insights suggest a strategy along the following lines:
a) For a given, non-separable DN , identify a suitable non-linear mapping
ξ → Ψ from N to M dimensions that achieves linear separability of DM .
b) Compute the kernel function for all pairs of example inputs:
Kµν = K(ξµ, ξν) = 1/M Ψµ · Ψν.",The+Shallow+and+the+Deep.pdf
"102 4. BEYOND LINEAR SEPARABILITY
c) Determine the embedding strengths 󰂓Xmax corresponding to optimal sta-
bility in the M-dim. weight space, for instance by use of the AdaTron
(4.30).
d) Classify an arbitrary ξ ∈ RN according to
S(ξ) = sign
󰀥 P󰁛
µ=1
Xµ
max SµK(ξµ, ξ)
󰀦
In practice, of course, the problem is to find and implement a suitable trans-
formation that yields separability in a given problem and data set. However, we
observe that once step (a) is performed, the transformation ξ → Ψ is not explic-
itly used anymore. Even the weight vector Wmax is not required explicitly: it
is not directly updated in in the training (c), nor is it used for the classification
in working phase (d). Instead, the representation (4.24) is used throughout.
Ultimately, this suggests to by-pass the explicit transformation in the first
place and replace step (a) by
a’) For a given, non-separable DN , identify a suitable kernel function K(ξ, 󰁥ξ)
and proceed from there as before.",The+Shallow+and+the+Deep.pdf
"place and replace step (a) by
a’) For a given, non-separable DN , identify a suitable kernel function K(ξ, 󰁥ξ)
and proceed from there as before.
This can only be mathematically sound if the selected kernel K(ξ, 󰁥ξ) func-
tion represents some meaningful transformation, implicitly. It is obvious that
for any transformation a kernel exists and we can work it out via the scalar prod-
ucts ψ(ξ) · ψ(󰁥ξ). However, the reverse is less clear: given a particular kernel,
can we guarantee that there is a valid, i.e. consistent, well-defined transforma-
tion? Fortunately, such statements can be made with respect to a large class of
functions without having to work out the underlying ξ → Ψ explicitly.
Sufficient conditions for a kernel to be valid can be provided according to
Mercer’s Theorem [Mer09], see also [SS02,CST00,STC04,Her02]. Without going
into the mathematical details and potential additional conditions, it can be
summarized for our purposes as:",The+Shallow+and+the+Deep.pdf
"into the mathematical details and potential additional conditions, it can be
summarized for our purposes as:
Mercer’scondition (sufficient condition for validity of a kernel) (4.31)
A given kernel function K can be written as K(ξ, 󰁥ξ) = 1
M Ψ(ξ) · Ψ(󰁥ξ),
with a transformation ξ ∈ RN → Ψ ∈ RM of the form (4.20), if
󰁝 󰁝
g(ξ) K(ξ, 󰁥ξ) g(󰁥ξ) dNξ dN󰁥ξ ≥ 0
holds true for all square-integrable functions g with
󰁝
g(ξ)2dN ξ < ∞.
Several families of kernel functions have been shown to satisfy Mercer’s condition
and are referred to as Mercer kernels, frequently. A few popular examples are
discussed in the following.",The+Shallow+and+the+Deep.pdf
"4.3. SUPPORT VECTOR MACHINES 103
Polynomial kernels
A polynomial kernel of degree q can be written as
K(ξµ, ξ) = (1 + ξµ · ξ )q yielding S(ξ) = sign
󰀥 P󰁛
µ=1
Xµ Sµ
T (1 + ξµ · ξ )q
󰀦
(4.32)
as the input-output relation of the classifier.
As a special case, let us consider the simplest polynomial kernel:
Linear kernel (q = 1)
K(ξµ, ξ) = (1 + ξµ · ξ ) with S(ξ) = sign
󰀥 P󰁛
µ=1
Xµ Sµ
T (1 + ξµ · ξ )
󰀦
(4.33)
= sign
󰀥 P󰁛
µ=1
Xµ Sµ
T
󰁿 󰁾󰁽 󰂀
≡M Θ
+
P󰁛
µ=1
Xµ Sµ
T ξµ
󰁿 󰁾󰁽 󰂀
≡MW
· ξ
󰀦
.
In this case, we can provide an immediate, almost trivial interpretation of the
kernel: it corresponds to the realization of a linearly separable function in the
original feature space (M = N) with
weights W 󰁥= w = 1
M
P󰁛
µ=1
Xµ Sµ
T ξµ and off-set Θ = 1
M
P󰁛
µ=1
XµSµ
T .
The SVM with linear kernel is applied very frequently in practice. There is even
an unfortunate trend to refer to it as “the SVM”. However – strictly speaking
– the SVM is not a classifier but a framework for classification: it has to be",The+Shallow+and+the+Deep.pdf
"– the SVM is not a classifier but a framework for classification: it has to be
specified by defining the kernel in use. In particular, the linear kernel reduces
the SVM to the familiar perceptron of optimal stability (with local threshold
Θ).
In order to take full advantage of the SVM concept, we have to employ more
sophisticated kernels. The first non-trivial choice beyond linearity corresponds
to q = 2:
Quadratic kernel (q = 2)
K(ξµ, ξ) = (1 + ξµ · ξ )2 = 1 + 2
N󰁛
j=1
ξµ
j [ξj] +
N󰁛
j,k=1
ξµ
j ξµ
k [ξj ξk]. (4.34)
Hence, the output S(ξ) in Eq. (4.32) with q = 2 corresponds to an inhomoge-
neously linearly separable function in terms of the N original features augmented
by N(N +1)/2 products of the form [ξjξk] which includes the squares of features
for j = k.",The+Shallow+and+the+Deep.pdf
"104 4. BEYOND LINEAR SEPARABILITY
As intuitively expected, the use of the quadratic kernel represents the non-
linear mapping from N-dim. feature space to in total M = N(N + 3)/2 trans-
formed features (original, squares and mixed products). An explicit formulation
is reminiscent of Quadratic Discriminant Analysis (QDA) [HTF01], albeit aim-
ing at a different objective function in the training.
Similarly, for the general polynomial kernel (4.32) the separating class bound-
ary becomes a general polynomial surface and the dimension M of the trans-
formed feature space grows rapidly with its degree q.
Next, we consider a somewhat extreme, yet very popular choice:
Radial Basis Functions (RBF) kernel
K(ξµ, ξ) = exp
󰀗
− 1
2 σ2
󰀓
ξµ − ξ
󰀔2󰀘
(4.35)
which involves the squared Euclidean distance and a width parameter σ.
In an attempt to interpret the popular RBF Kernel along the lines of the
discussion of polynomial kernels we could consider the Taylor series
exp[x] =
∞󰁛
k=1
1
k! xk = 1 + x + 1",The+Shallow+and+the+Deep.pdf
"discussion of polynomial kernels we could consider the Taylor series
exp[x] =
∞󰁛
k=1
1
k! xk = 1 + x + 1
2x2 + 1
6x3 + 1
24x4 + . . .
which shows that the dimension of the corresponding space would be M → ∞
as all powers and products of the original features are involved.
The RBF kernel has become one of the most popular choices in the literature.
The fact that an SVM with this extremely powerful kernel with, formally, M →
∞ can generalize at all demonstrates the importance of the restriction to optimal
stability (the large margin concept) which constitutes an efficient regularization
of the classifier.
4.3.4 A few remarks
Selection of kernels and parameter setting
In practice, the choice of the actual kernel function can influence the perfor-
mance of the corresponding classifier significantly. In addition, kernels may
contain parameters which have to be tuned to suitable values by means of vali-
dation techniques. The RBF-kernel is just one example of kernels that feature",The+Shallow+and+the+Deep.pdf
"dation techniques. The RBF-kernel is just one example of kernels that feature
a control parameter: the width σ in Eq. (4.35). The data-driven adaptation of
kernel parameters as part of the training process has also been discussed in the
literature, see [CCST99] for just one example.
Inhomogeneous separation of classes
In analogy to the perceptron of optimal stability, see Sec. 3.8, an offset from
the origin of the high-dimensional feature space can be considered in the SVM.
For the sake of simplicity we have restricted the discussion to the homogeneous
case and refer the reader to the literature w.r.t. the conceptually straightforward
extension to inhomogeneous separation, see e.g. [CST00,SS02].",The+Shallow+and+the+Deep.pdf
"4.3. SUPPORT VECTOR MACHINES 105
Soft-margin SVM
In addition to the choice of the kernel and potential parameters thereof, one of-
ten resorts to a soft margin version of the SVM [CV95], see also [SS02,CST00,
STC04, Her02, DFO20]. The considerations of Sec. 4.1.2 for the simple percep-
tron immediately carry over to the SVM formalism, once a kernel is defined.
The modified optimization problems (4.4) and (4.5) are easily generalized to
the case of the Support Vector Machine by replacing the embedding strengths
with 󰂓X and the correlation matrix by Γ from Eq. (4.24) and (4.23), respectively.
Consequently, we can immediately derive suitable training algorithms for
the soft margin SVM. For instance, the “AdaTron with errors” algorithm (4.6)
carries over to the kernel-based formulation in a straightforward fashion.
The soft margin extension introduces an additional parameter in the training
process: the parameter γ in (4.4) implicitly controls the tolerance of constraint",The+Shallow+and+the+Deep.pdf
"process: the parameter γ in (4.4) implicitly controls the tolerance of constraint
violations (or even misclassifications). Like potential parameters of the kernel,
it should be determined by means of a suitable validation procedure.
Overfitting
In the early days of the SVM, occasionally the claim was made that the strong
intrinsic regularization related to the large margin idea would eliminate the risk
of overfitting to a large extent, if not completely. However, practice shows that
the use of too complex kernels or low tolerance towards misclassification can
result in poor generalization.
Interestingly, the SVM offers a signal of overfitting which does not even
require the explicit estimation of the generalization error: the number of support
vectors ns with nonzero embedding strengths Xµ > 0. A relatively high fraction
ns/P indicates that the classifier may be overly specific to the given data set.
The fact that only very few examples in the data set are stabilized by embedding",The+Shallow+and+the+Deep.pdf
"The fact that only very few examples in the data set are stabilized by embedding
the others suggests inferior classification performance with respect to novel data
in the working phase. This indication can be exploited for the choice of a suitable
kernel to begin with, for the tuning of its parameters and for the choice of control
parameters in the optimization.
Efficient implementations
The remark at the end of Sec. 3.7.3 concerning computational efficiency and
scalability carries over to the kernel-based SVM as well.
Efficient implementations, for instance based on the concept of Sequential
Minimal Optimization (SMO) [Pla98] are available for a variety of platforms.
As just one source of information, the reader is referred to a list of links provided
at www.svms.org/software.html [svm99].",The+Shallow+and+the+Deep.pdf
106 4. BEYOND LINEAR SEPARABILITY,The+Shallow+and+the+Deep.pdf
"Chapter 5
Feed-forward networks for
regression and classification
The fishermen in the north of Spain have been using Deep Networks
for centuries. Their contribution should be recognized . . .
— Javier Movellan
Layered Neural Networks have regained significant popularity due to their
impressive successes in the context of Deep Learning applications such as image
classification. The basic designs and training techniques have been established
decades ago.
In the previous chapter we presented examples of layered networks con-
structed from perceptron-like threshold units. The by far most popular type of
networks comprise continuous units with differentiable activation functions. In
the following, we consider the use of such networks for regression and probabilis-
tic classification and discuss their ability to approximate continuous functions.
We present methods for their training by optimization techniques like gradient",The+Shallow+and+the+Deep.pdf
"We present methods for their training by optimization techniques like gradient
descent and variants thereof. Furthermore, we inspect specific example archi-
tectures and very briefly address the use of deep networks with many hidden
layers and strategies for their training.
5.1 Feed-forward networks as non-linear function
approximators
We first revisit the basic architecture and definition of strictly feed-forward,
layered networks. We show that, in principle, suitable networks can approximate
107",The+Shallow+and+the+Deep.pdf
"108 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
ξj ≡ σ(1)
j
w(1)
kj , j = 1, . . . N
S(1)
k , k = 1, . . . K(1)
S(2)
k , k = 1, . . . K(2)
w(2)
kj , j = 1, . . . K(1)
. . .
vk ≡ w(L)
1k , j = 1, . . . K(1)
σ(ξ) ≡ S(L)
1
Figure 5.1: A feed-forward neural network with N input units, L−1 hidden
layers and a single output unit.
any reasonable function from RN to R. This implies that layered networks can
serve as tools for quite general problems of non-linear regression. In section
5.1.2 we show explicitly, that suitable layered networks can approximate any
reasonable function to arbitrary precision. Cost function based training for the
learning from examples is discussed in 5.2 with emphasis on gradient based
methods.
5.1.1 Architecture and input-output relation
Figure 5.1 displays a multilayer neural network with a single output unit. The
generalization to several output units is formally straightforward. The figure",The+Shallow+and+the+Deep.pdf
"generalization to several output units is formally straightforward. The figure
suggests a convergent architecture with the number of hidden units per layer
decreasing towards the output. While in practice this is frequently the case, it
is by no means required in the following.
The network in Fig. 5.1 is strictly feed-forward, i.e. the state of a particular
hidden unit depends directly and only on the nodes in the previous layer. The
resulting hidden unit activation is
S(M)
k = g(M)
k
󰀳
󰁃
K(M−1)
󰁛
j=1
w(M)
kj S(M−1)
j − θ(M)
k
󰀴
󰁄 (5.1)
where adaptive weights w
(M)
kj connect the j-th unit in layer (M −1) to the k-th
unit of layer M and θ(M)
k denotes an adaptive local threshold. Alternatively,
we could introduce an additional, clamped unit S(M)
0 ≡ −1 in each layer and",The+Shallow+and+the+Deep.pdf
"5.1. NON-LINEAR FUNCTION APPROXIMATORS 109
represent the local threshold by a weight w(M−1)
k0 ≡ θ(M)
k . This would parallel
our representation of inhomogeneously linear separable functions in Eq. (3.4).
We can include the input layer in the the notation of Eq. (5.1) by defining
S(0)
j ≡ ξj. Similarly, we can rename the single output as S(L)
1 ≡ σ in the L-th
layer with K(L) = 1:
σ(ξ) ≡ S(L)
1 = gout
󰀳
󰁃
K(L−1)
󰁛
k=1
vk S(L−1)
k − θout
󰀴
󰁄
= g
(L)
1
󰀳
󰁃
K(L−1)
󰁛
k=1
w(L)
1k S(L−1)
k − θ(L)
1
󰀴
󰁄
(5.2)
with the alternative notations vk ≡ w(L)
1k and θout ≡ θ(L)
1 .
The output according to Eq. (5.2) together with (5.1) can be interpreted as
a function
σ : RN → R.
The precise form of the input-output relation is determined by the network
architecture including its connectivity and the activation functions. Frequently
we will assume that the same transfer function g(. . .) defines the activation of all
hidden units, while the output might result from a specific activation gout(. . .).",The+Shallow+and+the+Deep.pdf
"hidden units, while the output might result from a specific activation gout(. . .).
5.1.2 Universal approximators
Following the previous section we can use a feed-forward network with a single
output to implement a function from RN to R. It is a quite common concept to
realize or approximate functional dependencies by the superposition of specific
basis functions. A most prominent example is the representation in terms of
Fourier series which exploit properties of the trigonometric basis functions. Co-
efficients are chosen as to approximate a target function to a required precision.
Frequently, they are fitted to a set of discrete points and the resulting series is
used to interpolate or even extrapolate. Hence, the situation is reminiscent of
more general non-linear regression.
Neural networks of the form discussed in the previous section can be seen as
a particular framework for functional approximation: for a given architecture",The+Shallow+and+the+Deep.pdf
"a particular framework for functional approximation: for a given architecture
and connectivity, the function σ : RN → R is parameterized by the choice of all
adaptive quantities, i.e. weights and thresholds.
In the following we will see that the considered type of layered neural net-
works can approximate virtually any function1 to arbitrary precision.
A network for piecewise constant approximation
Here we show by construction that any reasonable function f : RN → R can
be approximated by a layered neural network comprising sigmoidal and linear
1Mathematical subtleties are ignored here to some extent.",The+Shallow+and+the+Deep.pdf
"110 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
z
g2
3(z)
sigm.
linear
z
g5
a(z) − g5
b (z)
Figure 5.2: Left panel: A sigmoidal activation of the form (5.3) with γ =
2, zo = 3 as an example. Right panel: The difference of two steep sigmoidals
(here: γ = 5, a = 2, b = 6, respectively) singles out arguments a ≤ z ≤ b. The
inset shows a graphical representation of the sigmoidals with equal γ connected
to input z with thresholds a, b, respectively, and a linear unit computing the
difference of their activations.
units. More precisely, we will consider functions which map inputs from a
compact subset of RN to a real valued output. We restrict the argument to
feature vectors from the hypercube ξ ∈ [0, 1]N , which can always be generalized
by transformations like rescaling and translation in input space.
Let us first consider sigmoidal neurons with, for instance, the activation
gγ
zo (z) = 1
1 + exp[−γ(z − zo)], (5.3)
with the argument z ∈ R, threshold zo and steepness parameter γ > 0, see",The+Shallow+and+the+Deep.pdf
"gγ
zo (z) = 1
1 + exp[−γ(z − zo)], (5.3)
with the argument z ∈ R, threshold zo and steepness parameter γ > 0, see
Figure 5.2 (left panel) for an example. Two (steep) sigmoidal units can be
combined in order to select a range of z-values, effectively:
Gγ
[a,b](z) = gγ
a(z) − gγ
b (z) = 1
1+e−γ(z−a) − 1
1+e−γ(z−b) ≈
󰀝 1 if a< z <b
0 else.
(5.4)
where the approximate identity becomes exact in the limit γ → ∞.2 This is
illustrated in the right panel of Fig. 5.2.
For N-dim. inputs ξ we can realize the selection of a specific interval [ai, bi]
for each dimension i = 1, 2, . . . N separately, defining regions of interest (ROI)
R(j) = [a(j)
1 , b(j)
1 ] × [a(j)
2 , b(j)
2 ] . . . × [a(j)
N , b(j)
N ], j = 1, 2, . . . M. (5.5)
These can be constructed to cover the volume of all possible inputs [0, 1]N ,
which implies that M grows exponentially with N.
If we add up the corresponding N activations for one region of interest we
have 󰁓N
i=1 Gγ
[ai,bi](ξi)
󰀝 ≈ N if ξ ∈ R
≤ N − 1 if ξ ∕∈ R.",The+Shallow+and+the+Deep.pdf
"If we add up the corresponding N activations for one region of interest we
have 󰁓N
i=1 Gγ
[ai,bi](ξi)
󰀝 ≈ N if ξ ∈ R
≤ N − 1 if ξ ∕∈ R.
2For the argument we can ignore the difference between open and closed intervals.",The+Shallow+and+the+Deep.pdf
"5.1. NON-LINEAR FUNCTION APPROXIMATORS 111
2󰁛
i=1
Gγ
[ai,bi](ξi) ξ2
ξ1
2
0
1
ξ2
ξ1
⇒
gγ
N−1
2
(. . .)
0
1
Figure 5.3: N threshold nodes (steep sigmoidal units) which select a specific
interval per input dimension can be combined by adding up their activation.
The left panel shows an illustration for N = 2 and ξj ∈ [−10, 10]. Only where
all threshold units are activated, the sum reaches its maximum N. For ξ ∕∈ R,
cf. Eq. 5.5, the total activation is 󰁓2
i=1 Gγ
[ai,bi](ξi) ≤ N − 1. Consequently, an
additional threshold operation can be employed to single out a region of interest
R ⊂ RN , as illustrated in the right panel.
Here we consider one particular ROI and have omitted the superscript (j) for
simplicity. Another threshold unit, i.e. a steep sigmoidal with zo = N −1/2 can
be applied to the sum to single out inputs ξ ∈ R, see Fig. 5.3 for an illustration
with N = 2.
Figure 5.4 displays the architecture of a layered net in which M such units",The+Shallow+and+the+Deep.pdf
"with N = 2.
Figure 5.4 displays the architecture of a layered net in which M such units
correspond to the ROI of Eq. (5.5). Eventually, we select one representative
value vj = f(ξ(j)) of the target function in each ROI, for instance with ξ(j) in
the center of R(j). The vj serve as weights for feeding the activations ϑj into a
simple, linear output unit. Therefore, the resulting network response
󰁓M
j=1 vjϑj(ξ) = vk for ξ ∈ R(k) (5.6)
amounts to a piecewise constant approximation of the target function in [0, 1]N .
The constructive argument shows that the network in principle constitutes
a universal approximator. However, a few remarks are in place:
◦ In order to achieve an accurate approximation of any non-trivial func-
tion, we would have to realize a rather fine-grained representation of in-
put space. Assuming that we split [0, 1] into, say, k equal size intervals
in each of the N feature dimensions, the network of Fig. 5.4 would com-",The+Shallow+and+the+Deep.pdf
"put space. Assuming that we split [0, 1] into, say, k equal size intervals
in each of the N feature dimensions, the network of Fig. 5.4 would com-
prise O(Nk) units in the second and third layer, which appears reason-
able. However, representing all possible combinations of N intervals in
the fourth layer requires on the order of O
󰀃
kN 󰀄
hidden units.
◦ The idea of training the feed-forward network by means of example data
appears somewhat obscured. The simple-minded setting of the weights
vj = f(ξ(j)) in Eq. (5.6) could be interpreted as learning from one ex-
ample per ROI. However, in view of the previous remark, the procedure",The+Shallow+and+the+Deep.pdf
"112 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
Figure 5.4: The constructed network for piecewise constant function approx-
imation: Each input unit (top layer) is connected to a set of sigmoidal units
in the second layer. Pairs of these connect to linear units in the third layer
which select specific intervals [ai, bi], [bi, ci], . . . as illustrated in Fig. 5.2. Each
unit marked as ϑj in the fourth layer performs a threshold operations on a par-
ticular sum of N activations in the third layer, corresponding to one selected
interval in each feature dimension. An activation ϑj = 1 indicates that ξ ∈ R(j)
and the resulting state of the linear output unit is given by vi = f(ξ(i)) which
corresponds to a representative ξ(i) ∈ R(i).
would require a number P of examples that grows exponentially with the
dimension N.
◦ In this sense, the construction of ROI parallels the use of grandmother
neurons when showing that the parity machine is a universal classifier in
Section 4.2.",The+Shallow+and+the+Deep.pdf
"neurons when showing that the parity machine is a universal classifier in
Section 4.2.
While the argument justifies and supports the use of feed-forward networks in
principle, it does not provide insight into how to design and how to train such
a system in practice.
Note that the number of layers required for universal approximation is lim-
ited. In the above construction scheme, four layers of processing units are
sufficient. Obviously, shallow networks are sufficient to achieve this property.
However, this does not imply that shallow architectures are necessarily suitable
for all practical applications. In fact, the recent success of deep networks appears
to suggest the contrary, in many cases.",The+Shallow+and+the+Deep.pdf
"5.1. NON-LINEAR FUNCTION APPROXIMATORS 113
input layer ξ ∈ RN
input-to-hidden weights
w(k) ∈ RN (k = 1, 2, . . . K)
hidden units Sk = g
󰀃
w(k) · ξ−θ(k)󰀄
hidden-to-output weights vk
linear output σ
󰀃
ξ
󰀄
= 󰁓K
k=1 vk Sk
Figure 5.5: A so-called Soft Committee Machine realizes a functional approx-
imation as considered in Cybenko’s Theorem. A number K of hidden units with
sigmoidal activation and adaptive thresholds θ(k) are connected through adap-
tive weight vectors w(k) with the N-dim. input layer, in the illustration K = 3.
The linear response σ(ξ) is determined as σ = 󰁓K
k=1
vk g
󰀃
w(k) · ξ−θ(k)󰀄
with
hidden-to-output weights vk ∈ R.
Variants of the Universal Approximation Theorem
In the literature various incarnations of the Universal Approximation Theorem
have been presented, differing in the degree of generality and practical relevance.
Some of them address particular classes of target functions, others focus on",The+Shallow+and+the+Deep.pdf
"Some of them address particular classes of target functions, others focus on
specific types of activation functions or network architectures, see [Hor89,Fri94,
LLPS93,SM15,MM92,CMB00] for examples and reviews.
In the following we present an early and important formulation which is due
to G. Cybenko [Cyb89]. The theorem states that a relatively simple network
with a single hidden layer of sigmoidal units and a linear output is a universal
function approximator:
Cybenko’sTheorem (5.7)
Consider inputs ξ ∈ [0, 1]N and a continuous, sigmoidal function
g : R → R with g(z) →
󰀝 1 for z → +∞
0 for z → −∞.
Let w(k) ∈ RN , vk ∈ R, θ(k) ∈ R for k = 1, 2, . . . K.
Then, finite sums of the form
σ(ξ) =
K󰁛
k=1
vk g
󰀓
w(k) · ξ − θ(k)
󰀔
are dense in the space of continuous functions C([0, 1])N .",The+Shallow+and+the+Deep.pdf
"114 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
This implies that for any continuous target function τ ∈ C([0, 1]N ) and a given
real number ε > 0, parameters
󰁱
w(k) ∈RN , vk ∈R, θ(k) ∈R
󰁲K
k=1
exist with
󰀏󰀏󰀏σ(ξ) − τ(ξ)
󰀏󰀏󰀏 < ε for all ξ ∈ [0, 1]N .
The parameters can be interpreted as the weights and thresholds of a network
with a single hidden layer and linear output which is illustrated in Figure 5.5.
The term Soft Committee Machine has been coined for this architecture, see
[Saa99] and references therein. The name refers to the network’s similarity
with the (discrete output) committee machine for classification tasks which is
discussed in Section 4.2.
In a sense, Cybenko’s Theorem provides a stronger and more useful statement
than the basic insight obtained by the construction in the previous section.
However, the problem remains that a very large number K of hidden units
might be required to exploit the approximation property in practice. Moreover,",The+Shallow+and+the+Deep.pdf
"might be required to exploit the approximation property in practice. Moreover,
the theorem itself states only the existence of suitable parameters, it does not
suggest how to find them.
The choice of appropriate network parameters based on example data is
addressed in the following sections. We focus on training schemes which are
based on the minimization of appropriate cost functions by means of gradient
descent techniques.
5.2 Gradient based training of feed-forward nets
As we have seen, feed-forward neural networks can serve as universal function
approximators. Hence it appears natural to employ them in the context of non-
linear input/output relations which correspond to a real-valued target function
σ : RN → R.
Extensions to multiple continuous outputs are obviously possible. Likewise,
classification schemes could be realized by an additional binary threshold oper-
ation on the output σ or by appropriate binning in the case of multiple classes.",The+Shallow+and+the+Deep.pdf
"ation on the output σ or by appropriate binning in the case of multiple classes.
Class memberships could also be represented by coding schemes applied to a
number of output units as discussed in a forthcoming section.
Formally, we concatenate all M adaptive parameters of a feed-forward net-
work in one vector W ∈ RM . The convenient flattened notation facilitates a
unified discussion of various network architectures in the following. In the Soft
Committee Machine displayed in Fig. 5.5 as just one example, we have
W =
󰀓
w(1)
1 , . . . . . . , w(K)
N , θ(1), . . . , θ(K), v1, . . . , vK
󰀔
∈ RM with M = KN+2K
and we can refer to any of the adaptive parameters as a component Wj.
For simplicity we focus on networks with a single, continuous output σ(ξ)
in the following. The goal of training is to implement or approximate a target
function τ : RN → R by adapting a given network architecture to a given set of
examples D = {ξµ, τ(ξµ)}P
µ=1.",The+Shallow+and+the+Deep.pdf
"5.2. GRADIENT BASED TRAINING OF FEED-FORWARD NETS 115
To this end, we define an error measure which is suitable for the comparison
of the network output σ(ξ) with the target function τ(ξ) for a given input vector.
A popular and intuitive choice is the simple quadratic deviation
e(σ, τ) = 1
2 (σ − τ)2 . (5.8)
Here and in the following, shorthands σ = σ(ξ), σµ = σ(ξµ), τ = τ(ξ) and
τµ = τ(ξµ) refer to a generic input ξ or a particular example input ξµ ∈ D,
respectively.
While many alternative measures can be considered, see the discussion in
Section 5.3, the quadratic error (5.8) remains very popular and is particularly
intuitive. It treats deviations σ > τ and σ < τ in a symmetric way and yields
e = 0 only for perfect agreement with the target.
Given a set of examples D we can define a corresponding training set specific
cost function3:
E(W) = 1
P
P󰁛
µ=1
eµ with eµ = e(σµ, τµ). (5.9)
In the context of regression, E(W) plays the role of the training error and",The+Shallow+and+the+Deep.pdf
"cost function3:
E(W) = 1
P
P󰁛
µ=1
eµ with eµ = e(σµ, τµ). (5.9)
In the context of regression, E(W) plays the role of the training error and
quantifies the network performance with respect to D. As in classification, the
expectation is that the trained network represents a hypothesis that can be
applied successfully to novel data in the working phase.
A plethora of numerical optimization methods could be used to optimize the
cost function in practice. Most of these employ local gradient information or
higher order derivatives of E(W) in order to iteratively find a (local) minimum
of the cost function. If possible, derivatives can be computed analytically or
are estimated numerically. A few prominent examples are the so-called Newton
and quasi-Newton methods, conjugate gradient descent, Levenberg-Marquardt
algorithm or line search methods. Here we point the reader to the literature,
e.g. [Fle00, PAH19, Str19, DFO20, SNW11, Bis95a, HKP91] where an overview",The+Shallow+and+the+Deep.pdf
"e.g. [Fle00, PAH19, Str19, DFO20, SNW11, Bis95a, HKP91] where an overview
can be obtained and further references are provided.
Relatively simple gradient descent techniques have been particularly popular
in the context of machine learning for decades, as a very early example we have
already discussed the Adaline algorithm [WH60] in Sec. 3.7.2.
From an optimization theoretical point of view, simple gradient descent is
certainly inferior to state-of-the art methods. However, it remains an impor-
tant, very popular tool in many machine learning frameworks, including power-
ful multi-layered architectures in Deep Learning. This is due to the conceptual
simplicity and hands-on character of the descent and the fact that the mathe-
matical structure of feed-forward networks appears particularly suitable for the
computation of the required gradients.
Moreover, although the training is generically formulated as an optimization,",The+Shallow+and+the+Deep.pdf
"computation of the required gradients.
Moreover, although the training is generically formulated as an optimization,
the actual minimization of E serves only as a proxy for the ultimate goal, which
3Other terms frequently used in this context are: objective function, loss function, or
energy.",The+Shallow+and+the+Deep.pdf
"116 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
is the successful application of the trained system to novel data. Hence, the
precision to which a minimum is determined can play a minor role and the
potential existence of many (suboptimal) local minima of E is not as problematic
as one might expect.
5.2.1 Computing the gradient: Backpropagation of Error
A key property of feed-forward layered neural networks with differentiable acti-
vation functions is that the network output itself is a differentiable function of
the inputs. Likewise, the output and the error measure e(σ, τ) are differentiable
with respect to the adaptive parameters in the network for any given input ξ:
∂e(σ, τ)
∂Wj
= (σ − τ) ∂σ
∂Wj
.
Consequently, also the data set specific cost function E = 󰁓P
µ=1 e(σµ, τµ) is a
differentiable function of all components of W.
For strictly feed-forward architectures as shown in Figure 5.1 we can ob-
tain derivatives of σ with respect to any network parameter Wj recursively by",The+Shallow+and+the+Deep.pdf
"tain derivatives of σ with respect to any network parameter Wj recursively by
applying the chain rule [Bis95a, Bis06, HKP91]. The mathematical structure
facilitates a very efficient calculation of the gradient: Weights and thresholds
serve as coefficients when computing the output for a given input in a feed-
forward network. The actual output is compared with the target and the error
is said to propagate back (towards the input) when computing the derivatives in
a layer-wise fashion, which involves the very same coefficients again. Gradients
for an example of a shallow network architecture are worked out explicitly in
Appendix A.6.
The term Backpropagation of Error (Backpropagation or Backprop for short)
was originally used for the efficient implementation of the gradient only [RM86].
Later it became synonymous with the the entire gradient based training of multi-
layered feed-forward networks and is nowadays mostly used in this sense.",The+Shallow+and+the+Deep.pdf
"layered feed-forward networks and is nowadays mostly used in this sense.
The ambiguity of the term partly complicates the on-going debates about
who invented Backpropagation or coined the term Deep Learning.4 Here we
refrain from taking part in these discussions of questionable usefulness. For
some original articles and reviews of the history of Backpropagation, see e.g.
[Wer74,LBH18,CR95,Ama93,WL90,HKP91,GBC16].
5.2.2 Batch gradient descent
In principle, the minimization of E(W) could be done by any suitable method
of non-linear optimization. In the context of layered neural networks, relatively
simple gradient based techniques continue to play a very important role. Here
we focus on the use of standard gradient descent. Gradient based techniques
are also discussed in Appendix A.4.
4For an example thread initiated by J. Schmidhuber in the connectionists mailing-list see
http://mailman.srv.cs.cmu.edu/pipermail/connectionists/2021-December/037086.html",The+Shallow+and+the+Deep.pdf
"5.2. GRADIENT BASED TRAINING OF FEED-FORWARD NETS 117
In analogy to Eq. (A.39) in the Appendix, the basic form of the updates is
given as
Batch gradient descent (basic form)
at discrete time step t perform an update step of the form
W(t + 1) = W(t) − η ∇W E|W=W(t) (5.10)
with the learning rate η and cost function E of the form (5.9).
At each time step, the gradient with respect to all adaptive quantities W is
computed as a sum over all examples in D:
∇W E = 1
P
P󰁛
µ=1
∇W eµ = 1
P
P󰁛
µ=1
󰀓
σ(ξµ) − τ(ξµ)
󰀔
∇W σ(ξµ), (5.11)
where the r.h.s. is given for the quadratic error (5.8) but can be worked out for
alternative cost functions as well.
The terms batch or offline gradient descent refer to the fact that the entire
set D of example data is used in every update. Careful changes of W in the
direction of −∇W E decrease the value of the cost function in each individual
step and, consequently, the descent approaches some local minimum W∗ of the",The+Shallow+and+the+Deep.pdf
"step and, consequently, the descent approaches some local minimum W∗ of the
cost function. If several local minima 5 exist, the actual stationary W∗ depends
on the initialization W(t=0) of the system.
The specific form of ∇W σ has to be worked out by means of the chain rule
in layered networks as explained in the previous section. It depends obviously
on the network architecture, the activation functions and the set of all adaptive
quantities in the system. In Appendix A.6 a specific example is given for a Soft
Committee Machine, c.f. Section 5.1.2.
The general discussion of gradient descent in the appendix shows that its
convergence near a local minimum W∗ of E is governed by the symmetric,
positive definite Hessian of second derivatives
H∗ = H(W∗) ∈ RM×M with elements H∗
ij =H∗
ji = ∂2 E(W)
∂Wi∂Wj
󰀏󰀏󰀏
󰀏
W=W∗
. (5.12)
In a local minimum, all eigenvalues {ρi}M
i=1 of H∗ are positive and can be sorted
by magnitude:
0 < ρ1 ≤ ρ2 ≤ . . . ≤ ρmax.",The+Shallow+and+the+Deep.pdf
"󰀏󰀏󰀏
󰀏
W=W∗
. (5.12)
In a local minimum, all eigenvalues {ρi}M
i=1 of H∗ are positive and can be sorted
by magnitude:
0 < ρ1 ≤ ρ2 ≤ . . . ≤ ρmax.
We show in App. A.4.2 that with a given, constant learning rate η > 0 the
following qualitative behavior can be expected near a local optimum W∗ :
5 The term refers to local properties of the function and possibly includes global minima.",The+Shallow+and+the+Deep.pdf
"118 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
a) η ≤ 1
ρmax
b) 1
ρmax
<η ≤ 2
ρmax
c) η > 2
ρmax
Figure 5.6: Illustration of the behavior of gradient descent near a local min-
imum W∗ (marked by the red dot in the center). The contour lines represent
the quadratic approximation of E(W) in the vicinity of a local minimum, the
black symbols correspond to the iterates W(t) in Eq. (5.10). For small step size
η (panel a) the iteration converges smoothly into the minimum, intermediate
step sizes (panel b) result in convergent yet oscillatory behavior. Failure to
converge is observed for too large step sizes (panel c).
(a) η ≤ 1/ ρmax
For small, finite learning rates, the iteration approaches the local minimum
smoothly and converges to lim
t→∞
W(t) = W∗, see Fig. 5.6 (a). Obviously,
with very small rates η ≈ 0, the approach can become unnecessarily slow.
(b) 1/ ρmax < η ≤ 2/ ρmax
In this regime, convergence is still achieved, but in at least one of the",The+Shallow+and+the+Deep.pdf
"(b) 1/ ρmax < η ≤ 2/ ρmax
In this regime, convergence is still achieved, but in at least one of the
eigendirections of H∗ an oscillatory behavior is observed. As illustrated
in panel (b) of Fig. 5.6, the alternating behavior occurs in eigendirections
with large curvature, which correspond to narrow troughs in the landscape
E(W), while the approach towards W∗ is smooth along directions of small
ρi in which E resembles a shallow basin.
(c) η > 2/ ρmax
Too large learning rates result in divergent behavior of the iterations as
displayed in Fig. 5.6 (c). Depending on η in relation to the individual
ρi, the distance from the minimum can increase in one, several, or all
eigendirections of H∗.
This insight is valid for any local minimum of E. However, two important
points should be noted. Firstly, the analysis presented in App. A.4.2 is only
valid close to a local minimum, where the Taylor expansion up to second order,",The+Shallow+and+the+Deep.pdf
"valid close to a local minimum, where the Taylor expansion up to second order,
Eq. (A.41), is a good approximation of E. Secondly, different minima can dis-
play very different properties in terms of the Hessian and its eigenvalues. In any
case, minima are not known in advance, which would render the training un-
necessary. Since the cost function E can have many local minima, the outcome",The+Shallow+and+the+Deep.pdf
"5.2. GRADIENT BASED TRAINING OF FEED-FORWARD NETS 119
of the training process may depend strongly on the initialization of the system.
Therefore, the practical relevance of the mathematical analysis is limited.
In practice, a relatively large η could be used in the initial phase of training,
assuming that the system is far away from any local minimum. Schemes have
been suggested in the literature, which monitor the iterations and adjust the
learning rate, for instance whenever a zigzagging behavior is observed. An
intuitive example of a heuristic step size adaptation in batch gradient descent
is presented in [PBB11].
The most important insight of this section is that in batch gradient descent,
non-zero finite learning rates can be used to reach a local minimum. This is in
contrast to the stochastic gradient descent strategy discussed in the next section.
There, the learning rate has to be reduced to zero in the course of training in",The+Shallow+and+the+Deep.pdf
"There, the learning rate has to be reduced to zero in the course of training in
order to enforce convergence of the network configuration.
5.2.3 Stochastic gradient descent
The cost function E(W), Eq. (5.9), is given as a sum over examples in D:
E(W) = 1
P
P󰁛
µ=1
eµ(W) with ∇W E = 1
P
P󰁛
µ=1
∇W eµ, (5.13)
where eµ quantifies the contribution of an individual example to the total costs.
Virtually all machine learning objectives mentioned and discussed in this text
can be written in such a form, with the actual function eµ(W) and the pre-
cise definition of D depending on the details, of course. This includes the log-
likelihood in maximum-likelihood problems, the SSE (2.5) in regression, the
quantization error in unsupervised Vector Quantization [HKP91,BHV16] or the
objective function of the Generalized LVQ scheme introduced in Chapter 6.
We note that costs of the form (5.13) can be interpreted as an empirical
average of eµ over the data set, corresponding to randomly drawing examples",The+Shallow+and+the+Deep.pdf
"We note that costs of the form (5.13) can be interpreted as an empirical
average of eµ over the data set, corresponding to randomly drawing examples
from D with equal probability 1/P. Accordingly, the gradient of E as in Eq.
(5.13) can also be seen as a data set average of the single example terms ∇W eµ.
As a consequence, we can approximate the gradient of E by computing a
restricted empirical mean over a random subset of D. As an extreme case, we
can inspect single, randomly selected examples:
Stochastic gradient descent (SGD)
at discrete time step t
- select a single example {ξµ(t), τµ(t)} randomply with equal probability
- perform an update step
W(t + 1) = W(t) + ∆W(t) = W(t) − 󰁥η(t) ∇W eµ(t)
󰀏󰀏
󰀏
W=W(t)
. (5.14)
with the learning rate 󰁥η(t) and error terms eµ as given in Eq. (5.13).",The+Shallow+and+the+Deep.pdf
"120 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
The learning rate is denoted as 󰁥η(t) in order to indicate a possible explicit
time-dependence and to distinguish it from η in batch gradient descent, Eq.
(5.10). Clearly, the computational costs per update are lower than in the batch
procedure which involves the sum of P gradient terms in each step.
The update of the form (5.14) is referred to as online or stochastic gradient
descent, in contrast to offline batch algorithms. It can be seen as a special case
of stochastic approximation, see [RM51,Bis95a,HTF01] and Appendix A.5.3.
The stochastic approximation of the gradient introduces noise in the training
process. As a consequence, E(W) may increase in individual update steps. The
intuitive motivation for stochastic descent is that this noise helps the system to
explore the search space more efficiently and to overcome barriers (e.g. saddle
points of E) which separate different local minima of the cost function.",The+Shallow+and+the+Deep.pdf
"explore the search space more efficiently and to overcome barriers (e.g. saddle
points of E) which separate different local minima of the cost function.
The index µ(t) of the presented example at time t in Eq. (5.14) is drawn
randomly from the set {1, 2, . . . , P} with equal probability 1/P. In general, the
resulting individual ∆W(t) will deviate from the direction of steepest descent
−∇W E. However, on average over the random selection of an example from D
the update (5.14) follows the negative gradient of the total costs E. Therefore,
in a local minimum W∗
∆W(t)
󰀏󰀏
∗ = − 1
P
P󰁛
µ=1
󰁥η(t) ∇W eµ󰀏󰀏
∗ = −󰁥η(t)∇W E
󰀏󰀏
∗ = 0
where (. . .) denotes the average over the selection of µ(t). The notation (. . .)|∗
indicates that a term is evaluated in W(t) = W∗.
Hence, the average update becomes zero in the local minimum. However,
individual updates remain non-zero, in general. This can be seen by considering
the averaged squared norm of ∆W:
|∆W|2
󰀏
󰀏󰀏
∗
= 1
P
P󰁛
µ=1
󰁥η2(t)
󰁫
∇W eµ |∗",The+Shallow+and+the+Deep.pdf
"the averaged squared norm of ∆W:
|∆W|2
󰀏
󰀏󰀏
∗
= 1
P
P󰁛
µ=1
󰁥η2(t)
󰁫
∇W eµ |∗
󰁬2 󰀝 = 0 only if ∇W eµ󰀏󰀏
∗ =0 for all µ
> 0 else.
In general, not all of the gradient contributions will vanish in W∗. One exception
would be a minimum of E in which all individual eµ are minimized. For the
quadratic costs with eµ = (σ(ξµ) − τµ)2/2 this would correspond to a global
minimum E(W∗) = 0, i.e. a perfectly solvable case where σ(ξµ) = τµ for all µ.
The generic behavior for constant learning rate 󰁥η > 0 is illustrated schemati-
cally in the left panel of Fig. 5.7. After reaching the vicinity of a local minimum,
the iteration follows a seemingly irregular trajectory corresponding to the ran-
dom sequence of individual gradient terms with | ∇W eµ|∗ |2 > 0.
We can enforce convergence near a local minimum in the sense of
lim
t→∞
W(t) = W∗ and lim
t→∞
∆W(t) = 0 (5.15)
by employing an explicitly time-dependent learning rate 󰁥η(t) which decreases",The+Shallow+and+the+Deep.pdf
"lim
t→∞
W(t) = W∗ and lim
t→∞
∆W(t) = 0 (5.15)
by employing an explicitly time-dependent learning rate 󰁥η(t) which decreases
appropriately with the number of descent steps. Conditions for suitable learning",The+Shallow+and+the+Deep.pdf
"5.2. GRADIENT BASED TRAINING OF FEED-FORWARD NETS 121
Figure 5.7: Schematic illustration of the behavior of stochastic gradient de-
scent near a local minimum W∗ as marked by the (red) dot in the center. The
contour lines represent the quadratic approximation of E(W) in the vicinity of
W∗. Left panel: Black symbols correspond to the iterates W(t) in Eq. (5.14).
With a constant learning rate 󰁥η > 0, the descent overshoots the point of sta-
tionarity in an oscillatory way. Right panel: The average of the most recent
(here: 3) positions W(t) of the stochastic descent (large filled circles) results in
a favorable estimate (large empty circle) of the local minimum.
rate schedules 󰁥η(t) → 0 can be found already in the seminal paper by Robbins
and Monro [RM51] which introduced the concept of stochastic approximation in
1951, originally in the context of finding zeros of a function. Further references
and more detailed discussions can be found in several textbooks, e.g. in [Bis95a,
HTF01].",The+Shallow+and+the+Deep.pdf
"and more detailed discussions can be found in several textbooks, e.g. in [Bis95a,
HTF01].
Robbins and Monro showed that schedules which satisfy lim
t→∞
󰁥η(t) = 0 with
(I) lim
T→∞
T󰁛
t=0
󰁥η(t)2 < ∞ and (II) lim
T→∞
T󰁛
t=0
󰁥η(t) → ∞ (5.16)
facilitate convergence. Intuitively, the first condition (I) states that 󰁥η(t) has to
decrease fast enough in order to achieve a truly stationary configuration with
limt→∞ ∆W(t) = 0. However, enforcing 󰁥η(t) → 0 too rapidly would result in
trivial stationarity at arbitrary positions in W-space. Therefore, condition (II)
implies that the decrease is slow enough so that the entire search space can be
explored efficiently.
Simple schedules which reduce the learning rate asymptotically like 󰁥η(t) ∝
1/t for large t satisfy both conditions in (5.16). This relates to the well-known
results that 󰁓∞
n=1 n−2 = π2/6 while 󰁓∞
n=1
n−1 → ∞. Just one possible (popu-
lar) realization of such a decrease is of the form
󰁥η(t) = a
b + t
with constant parameters a, b > 0.",The+Shallow+and+the+Deep.pdf
"122 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
Various schedules which satisfy the conditions (I) and (II) of Eq. (5.16)
can be considered, including power laws 󰁥η(t) ∝ t−β, logarithmic schedules like
󰁥η(t) ∝ 1/(t ln t), or other explicitly time-dependent schemes, see for instance
[DM92] and references therein.
Stochastic gradient descent is arguably the most popular basic scheme for the
training of neural networks, including systems with many layers in the context
of Deep Learning.
5.2.4 Practical aspects and modifications
Numerous alternative approaches or modifications of the gradient based schemes
have been suggested and are of great practical relevance, in particular in the
context of Deep Learning. Here, only a few can be mentioned and explained
briefly. Note that some of these concepts can also be useful in batch gradient
descent.
SGD-training in epochs: In practice, we do not have to actually draw a
random example from D independently at each time step. Most frequently,",The+Shallow+and+the+Deep.pdf
"SGD-training in epochs: In practice, we do not have to actually draw a
random example from D independently at each time step. Most frequently,
updates are organized in epochs, e.g. by generating a random permutation of
{1, 2, . . . P} and presenting the entire D in this order before moving on to the
next epoch with a novel randomized order of examples.
Mini-batch training: Instead of performing the stochastic approximation
with respect to a single example, a random subset of D can be employed replac-
ing the full gradient ∇W E by partial sums in each training step. In a sense, this
strategy retains the advantages of SGD, i.e. lower computational costs and the
introduction of noise, but yields more reliable estimates of the gradient. The
size of the mini-batches constitutes a hyperparameter which can be tuned in
practice to achieve good performance and efficiency.
Averaged SGD: While training with constant learning rate 󰁥η will not re-",The+Shallow+and+the+Deep.pdf
"practice to achieve good performance and efficiency.
Averaged SGD: While training with constant learning rate 󰁥η will not re-
sult in a converging behavior W(t) → W∗, one can expect the iterates W(t)
to approach the vicinity of a local minimum and to assume more or less ran-
dom positions centered around W∗. This can be exploited by considering a
(potentially moving) average of the form
Wav(t) = 1
k
󰁓k−1
j=0 W(t − j)
which takes into account the last k iterations of SGD. In the simplest setting
with k = t, the average is performed over all update steps up to t. As illus-
trated in Fig. 5.7 (right panel) the averaged Wav is expected to be closer to
the local minimum than the individual W(t), once the training has reached the
vicinity of the optimum. Averaging SGD for faster and smoother convergence
was suggested and studied in [PJ92,Rup88], originally.",The+Shallow+and+the+Deep.pdf
"5.3. OBJECTIVE FUNCTIONS 123
Local learning rates: For both, batch and stochastic gradient descent, it
has been suggested to use local learning rates for different layers, nodes, or
even individual weights in the network. As an early, simple rule of thumb,
Plaut et al. [PNH86] suggest to use local learning rates inversely proportional
to the fan-in of the given neuron, i.e. the number of units it receives input
from, see the discussion in [HKP91]. More sophisticated methods make use
of the local properties of the cost function in terms of second derivatives as
motivated by Newton’s method [Fle00, HKP91]. Frequently, only the diagonal
elements of the local Hessian are used to compute an individual learning rate
for the update of Wj which is, for instance, inversely proportional to ∂2E
(∂Wi)2 ,
see [BL89] or [HKP91] for further references.
Note that gradient based algorithms with local or even individual learning",The+Shallow+and+the+Deep.pdf
"(∂Wi)2 ,
see [BL89] or [HKP91] for further references.
Note that gradient based algorithms with local or even individual learning
rates do not follow the steepest descent in E anymore. However, they still realize
a descent procedure, see the discussion in the Appendix A.4.
Momentum: Already in [RHW86] a modification of simple gradient descent
was suggested, in which the update contains a memory term representing infor-
mation about recently performed updates. In its simplest form the update is a
linear combination of the (stochastic) gradient term and the previous update:
∆W(t) = −󰁥η ∇W eµ(W(t)) + α∆W(t − 1) with α > 0. (5.17)
The decay factor α controls the influence of previous update steps on the cur-
rent change of W. Obviously, momentum could also be incorporated in batch
gradient descent.
The idea is to overcome flat regions where ∇W E ≈ 0, by keeping the mo-
mentum of previous downhill moves. Furthermore, momentum should milden",The+Shallow+and+the+Deep.pdf
"The idea is to overcome flat regions where ∇W E ≈ 0, by keeping the mo-
mentum of previous downhill moves. Furthermore, momentum should milden
oscillatory behavior of the updates when E displays anisotropic curvatures.
Adaptive learning rate schedules: The design and optimization of learn-
ing rate schedules and schemes for the automated adaptation of 󰁥η in SGD or
η in batch algorithms plays a key role for efficient training prescriptions. Fre-
quently, the learning rate adaptation is combined with the concept of momen-
tum. Prominent examples are algorithms termed AdaGrad, RMSProp, Adam,
or variance-based SGD (vSGD). For a first introduction and further references,
Section 8.5 in [GBC16] can serve as a starting point. A systematic comparison
of several popular schemes can be found in [LBV17], there in the context of
gradient-based LVQ training.
5.3 Objective functions
So far we have discussed the training of layered networks based on the quadratic",The+Shallow+and+the+Deep.pdf
"gradient-based LVQ training.
5.3 Objective functions
So far we have discussed the training of layered networks based on the quadratic
deviation (5.8) which, arguably, constitutes the most prominent cost function
in the context of regression. However, insights into the actual target problem,",The+Shallow+and+the+Deep.pdf
"124 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
heuristic assumptions or concrete statistical models of the observations may mo-
tivate the use of alternative objective functions in the training process. More-
over, the use of differentiable neural networks for classification tasks motivates
the use of cost functions which are designed for this particular purpose.
In the following we briefly discuss very few important examples of cost func-
tions for regression and classification based on layered neural networks with
differentiable activation functions.
5.3.1 Cost functions for regression
The very intuitive quadratic deviation or SSE cost function appears plausible
and suitable for a variety of regression problems. A variety of alternative ob-
jective functions have been suggested in the literature that can be optimized by
gradient descent or similar procedures.
Heuristic cost functions: Numerous heuristic schemes have been suggested",The+Shallow+and+the+Deep.pdf
"gradient descent or similar procedures.
Heuristic cost functions: Numerous heuristic schemes have been suggested
which adjust an instantaneous objective function while training proceeds. The
goal could be to smooth out the costs initially by levelling out details, which
could help to avoid regions that contain unfavorable local minima. Gradually,
more and more detail are re-introduced and, eventually, the genuine objective
is optimized [HKP91].
As just one example, Makram-Ebeid et et al. suggest in [MSV89] the mod-
ified quadratic costs
E(W) =
P󰁛
µ=1
󰀝 γ (σµ − τµ)2 if σµ τµ > 0
(σµ − τµ)2 if σµ τµ ≤ 0 (5.18)
with the parameter γ ∈ [0, 1] increasing during the training process, e.g. accord-
ing to an explicit time-dependence γ(t). For γ = 0, deviations (σµ − τµ)2 do
not contribute to the costs if the output has the correct sign. Thus, training
will focus on achieving agreement in terms of sign(σµ) = sign(τµ), initially. As",The+Shallow+and+the+Deep.pdf
"will focus on achieving agreement in terms of sign(σµ) = sign(τµ), initially. As
γ → 1, the system is fine-tuned to achieve σµ ≈ τµ for all µ, eventually.
Minkowski-r errors: In the following we focus on a class of cost functions
that can be derived from a noise model which is assumed to describe the statis-
tical properties of the data at hand.
The popular quadratic deviation can be explicitly motivated by assuming
Gaussian distributed training labels. These could result from, e.g., additive
Gaussian noise corrupting the true target values in the available data, as in Eq.
(2.12). In Section 2.2.2 we have seen in the specific example of linear regression
that a corresponding Maximum Likelihood approach leads immediately to the
SSE-criterion (2.5).
Starting from different assumptions about the statistical properties leads to
specific choices of the cost function. Assume, for instance, that the training",The+Shallow+and+the+Deep.pdf
"5.3. OBJECTIVE FUNCTIONS 125
labels in a regression problem deviate from the true target function by indepen-
dent noise terms ηµ with
P(ηµ) ∝ e−β |ηµ|r
with r > 0, (5.19)
which is normalized to
󰁝
P(ηµ)dηµ = 1. Obviously we recover a Gaussian
density with a β-dependent variance for r = 2. The optimization of the corre-
sponding Maximum Likelihood criterion introduces a cost function of the form
E = 1
P
P󰁛
µ=1
|σµ(W) − τµ|r (5.20)
where terms independent of the network parameters W have been omitted.
This objective function is referred to as the Minkowski-r error in the literature
[HB87], see also [Bis95a].
We note again that for r = 2 the familiar MSE criterion is recovered, the
special case of r = 1 corresponds to the so-called Manhattan distance or city
block metric 6 |σ − τ|. Intuitively, costs with r < 2 will be less sensitive to
outliers, i.e. to examples with very large |σ − τ|, than the conventional SSE.
5.3.2 Cost functions for classification",The+Shallow+and+the+Deep.pdf
"outliers, i.e. to examples with very large |σ − τ|, than the conventional SSE.
5.3.2 Cost functions for classification
Heuristically, we can apply networks with differentiable activation functions and
output also for classification, retaining regression type cost functions like the
simple quadratic deviation or (5.18) in the training. In the simple case of two
classes we could perform one additional threshold operation on a single output
of the trained network to obtain a crisp binary classifier. Similar ideas can be
applied to multi-class problems.
A more systematic approach realizes network responses that can be inter-
preted as a probabilistic assignment of the input vector to one of the classes.
Here we follow to a large extent the presentation in [Bis95a]. First, we restrict
ourselves to the case of two classes, here represented by target values τµ ∈ {0, 1},
which correspond to crisp training labels in the simplest case. Moreover, we as-",The+Shallow+and+the+Deep.pdf
"which correspond to crisp training labels in the simplest case. Moreover, we as-
sume that also the output of the network satisfies 0 ≤ σ(ξµ) ≤ 1, as for instance
realized by a proper sigmoidal output activation.
After training, we want to interpret σ(ξ) as the class-membership probability
p(τ = 1|ξ) = σ(ξ) and p(τ = 0|ξ) = 1 − σ(ξ).
This can be written conveniently in the compact form
p(τ|ξ) = σ(ξ)τ [1 − σ(ξ)]1−τ . (5.21)
If we consider this as our model for the occurrence of a label τ in the data set
and we furthermore assume that the examples are generated independently, the
6Here, we ignore the subtle difficulty that |x|r it is not differentiable in x = 0 for r ≤ 1.",The+Shallow+and+the+Deep.pdf
"126 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
likelihood of generating a given set of labels {τµ}P
µ=1 with the network reads
P󰁜
µ=1
(σµ)τµ
[1 − σµ]1−τµ
with the shorthand σµ = σ(ξµ).
Maximizing this likelihood by choice of the network parameters W is equivalent
to minimizing the negative log-likelihood
E(W) = −
P󰁛
µ=1
󰀓
τµ ln σµ + (1 − τµ) ln(1 − σµ)
󰀔
. (5.22)
In contrast to the SSE, we omit the irrelevant pre-factor 1/P here for simplicity.
The cost function quantifies the similarity of the outputs σµ which we in-
terpret as probabilities with the targets τµ in terms of their cross entropy
[Hop87,BW88,SLF88]. Note that it is bounded from below by the entropy
Eo = −
P󰁛
µ=1
󰀓
τµ ln τµ + (1 − τµ) ln(1 − τµ)
󰀔
which can only be achieved if all σµ = τµ. Since the bound Eo does not depend
on W we can subtract it from the cross entropy and consider the equivalent
objective function
E(W) −Eo = −
P󰁛
µ=1
󰀓
τµ ln σµ
τµ + (1 −τµ) ln 1 − σµ
1 − τµ
󰀔
≥ 0 ≡ DKL(τ||σ). (5.23)",The+Shallow+and+the+Deep.pdf
"objective function
E(W) −Eo = −
P󰁛
µ=1
󰀓
τµ ln σµ
τµ + (1 −τµ) ln 1 − σµ
1 − τµ
󰀔
≥ 0 ≡ DKL(τ||σ). (5.23)
This is the so-called relative entropy or Kullback-Leibler divergence DKL(τ||σ)
between the specific probability distributions σ and τ, see for instance [Bis95a]
for a discussion in the machine learning context. 7
The cross entropy E(W), Eq. (5.22), or equivalently the Kullback-Leibler
divergence DKL constitute differentiable objective function of the adaptive pa-
rameters W and can be minimized by gradient-descent based or other opti-
mization methods for a given data set D = {ξµ, τµ}P
µ=1 . This way we achieve
a network with outputs σ(ξ) that can be interpreted as a class membership
probability.
Multi-class problems: The formalism can be extended to multiclass prob-
lems with targets τµ
k ∈ [0, 1], k ∈ {1, 2, . . . , C} which satisfy 󰁓C
k=1 τµ
k = 1. In
the simple case of crisp training labels we have that for each example exactly
one τµ",The+Shallow+and+the+Deep.pdf
"k=1 τµ
k = 1. In
the simple case of crisp training labels we have that for each example exactly
one τµ
j = 1 which indicates that ξµ is assigned to class j in the training data.
Obviously we have to consider a network architecture and activations which
can represent C assignment probabilities σk ∈ [0, 1] with 󰁓C
k=1 σk = 1. This
can be achieved, for instance, with a layer of C output units with a so-called
soft-max or normalized exponential activation, see also the following Sec. 5.4.
7Note that the KL-divergence is in general non-symmetric: DKL(τ||σ) ∕= DKL(σ||τ).",The+Shallow+and+the+Deep.pdf
"5.4. ACTIVATION FUNCTIONS 127
Now the equivalent of cost function (5.23) becomes
DKL(τ||σ) = −
P󰁛
µ=1
C󰁛
k=1
τµ
k ln
󰀕σµ
k
τµ
k
󰀖
, (5.24)
which reduces to (5.23) in the binary case with C = 2 where σ1 = 1 − σ2 and
τ1 = 1 − τ2.
Given a network structure with C outputs σk as described above, we can
determine the adaptive parameters of the system by minimization of the above
differentiable cost function. All gradient-based or more sophisticated techniques
discussed here can be applied.
5.4 Activation functions
When designing a neural network for a given task, the key step is the choice of
the network architecture and size. The choice of activation functions is equally
important, as it should reflect properties of the problem and the data. Obvi-
ously, the output unit (or units) should realize the appropriate range of possible
responses in a regression problem. In classification, properly defined outputs
should encode the crisp or probabilistic class assignments. The choice of acti-",The+Shallow+and+the+Deep.pdf
"should encode the crisp or probabilistic class assignments. The choice of acti-
vations in intermediate, hidden layers influences the complexity of the network
and can be crucial for the success of training, e.g. by gradient descent or other
techniques.
So far we have mainly considered threshold or sigmoidal activation functions,
with the notable exception of simple linear units when constructing a universal
function approximator in 5.1.2. A large variety of activation functions have been
suggested and investigated in the literature, see e.g. [HKP91, Bis95a, GBC16].
In this section, only a few important and/or popular choices are presented.
In the following we refrain from including gain parameters, local thresh-
olds or similar parameters in the description, the corresponding extensions are
straightforward. Similarly the range of activations can be trivially shifted and
scaled: for example, a sigmoidal function h(x) with 0 ≤ h(x) ≤ 1 can be",The+Shallow+and+the+Deep.pdf
"scaled: for example, a sigmoidal function h(x) with 0 ≤ h(x) ≤ 1 can be
transformed as g(x) = 2h(x) − 1 with −1 ≤ g(x) ≤ 1. It is understood that
all functions given as g(x) in the following can be modified like a g(γx) + b if
needed.
5.4.1 Sigmoidal and related functions
In Chapter 1 we motivated the use of sigmoidal activation functions as a rough
approximation of biological neuron responses in a firing rate picture. A number
of functions satisfy the conditions (1.2) or (1.4), see Fig. 5.8 for a few prominent
examples. The left panel displays erf[x] (chain line, green), tanh[x] (dashed,
blue), and the logistic function 1/(1 + exp[−x]). The latter was shifted and
scaled to realize the range g(x) ∈ [−1, 1]. In the right panel, the Heaviside step
function of the McCulloch Pitts neuron and a piecewise linear activation, which
resembles a sigmoidal function, are shown.",The+Shallow+and+the+Deep.pdf
"128 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
x
g(x)
x
g(x)
Figure 5.8: Sigmoidal and related activation functions. Left panel: three
differential sigmoidal functions. Right panel: The limiting case of the McCul-
loch Pitts activation (Heaviside step function) and a piecewise linear function.
5.4.2 One-sided and unbounded activation functions
The simplest of all activation functions, the trivial identity g(x) = x is un-
bounded. In the context of biologically inspired firing rate models this does not
make sense, as there are no limits to the frequency of spike generation. However,
in artificial networks, linear neurons are very often employed and indeed useful
for specific tasks, e.g. as trainable output units attached to an otherwise more
complex network for regression. Examples will be presented in Sec. 5.5.
Recitfied Linear Unit and variations thereof: The so-called Rectified
Linear Unit or ReLU activation [NH10], see Fig. 5.9 (left panel), has gained",The+Shallow+and+the+Deep.pdf
"Recitfied Linear Unit and variations thereof: The so-called Rectified
Linear Unit or ReLU activation [NH10], see Fig. 5.9 (left panel), has gained
significant popularity in the context of Deep Networks 8. In its original form it
corresponds to
g(x) = max{0, x} =
󰀝 0 if x < 0
x if x ≥ 0 with g′(x) =
󰀝 0 if x < 0
1 if x > 0. (5.25)
We ignore here the subtlety that the ReLU function is not continuously differ-
entiable in x = 0.
Note that the function (5.25) has been known and used for long in various
mathematical, technical and engineering fields, ranging from signal processing
and filtering to finance mathematics. Depending on the context, it is know as
the ramp function, hockey stick, or hinge function.
In the literature, several advantages are associated with the ReLU activation
when compared to, for instance, sigmoidal activations [GBC16]:
- Obviously, the ReLU is computationally cheap, and so is its derivative.",The+Shallow+and+the+Deep.pdf
"when compared to, for instance, sigmoidal activations [GBC16]:
- Obviously, the ReLU is computationally cheap, and so is its derivative.
8As stated in [GBC16] : “In modern neural networks, the default recommendation is to
use the rectified linear unit . . . ”.",The+Shallow+and+the+Deep.pdf
"5.4. ACTIVATION FUNCTIONS 129
x
g(x)
x
g(x)
Figure 5.9: Unbounded and one-sided activation functions. Left panel: Sim-
ple linear activation g(x) = x (dotted), Rectified Linear Unit ReLU (solid), Eq.
(5.25), and leaky ReLU, Eq. (5.26), with a = 0.25 (dashed). Right panel: Ex-
ponential linear unit ELU (dashed, black), cf. Eq. (5.28), Swish (solid, green),
Eq. (5.29), and Softplus (dotted, blue), Eq. (5.27).
- The ReLU is one-sided, i.e. g(x) = 0 for x < 0. As a consequence, a
considerable fraction of units will display zero activity in a given network,
typically. In this sense, a network of rectified linear units realizes sparse
activity which is considered advantageous in many cases.
- When computing derivatives via the chain rule in a network of many
layers, the multiplication of many derivatives |g′| < 1 causes the so-called
problem of vanishing gradients. Supposedly, the problem is absent in
ReLU networks where g′(x) = 1 at least for x > 0.",The+Shallow+and+the+Deep.pdf
"problem of vanishing gradients. Supposedly, the problem is absent in
ReLU networks where g′(x) = 1 at least for x > 0.
- Several empirical comparisons of networks with ReLU and other acti-
vations have been published, in which ReLU networks display favorable
training behavior and performance. Recently, theoretical studies of model
situations seem to support these claims [OSB20].
In the so-called leaky ReLU (LReLU) activation [HZRS15], the constant zero
for x < 0 is replaced by a linear dependence, usually with a slope 0 < a < 1,
see also the left panel of Fig. 5.9. Hence, it reads
g(x) =
󰀝 a x if x < 0
x if x > 0 with g′(x) =
󰀝 a if x < 0
1 if x ≥ 0. (5.26)
Differentiable, unbounded activations: Several differentiable functions
which maintain or approximate the linear behavior g(x) = x for large posi-
tive arguments x > 0 have been suggested in the literature. Fig. 5.9 (right
panel) displays three examples:
- the so-called Softplus function [GBB11]
g(x) = ln(1 + exp[x]) (5.27)",The+Shallow+and+the+Deep.pdf
"130 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
- the Exponential Linear Unit (ELU) [CUH16] with
g(x) =
󰀝
exp[x] − 1 for x < 0
x for x ≥ 0, (5.28)
- the Swish function [EYG92] with
g(x) = x
1 + exp[−x]. (5.29)
Interestingly, the Swish activation is even non-monotonic and displays a min-
imum in a negative value of the argument. Several empirical studies seem to
show favorable convergence behavior of gradient descent based training and
improved performance of Swish-networks compared to other activation func-
tions [EYG92,VRV+20].
5.4.3 Exponential and normalized activations
Frequently, units within a particular layer are coupled, e.g. through a normal-
ization, which deviates from the activation by local synaptic interaction.
Softmax function: The most prominent example is the representation of
assignment probabilities in an output layer of C units {σk}C
k=1. As outlined in
Sec. 5.3.2, the activations should obey 0 ≤ σk ≤ 1 individually. In addition,󰁓C",The+Shallow+and+the+Deep.pdf
"k=1. As outlined in
Sec. 5.3.2, the activations should obey 0 ≤ σk ≤ 1 individually. In addition,󰁓C
k=1 σk = 1 is required to justify their interpretation as probabilities.
An obvious and popular choice is to set σk = gβ
󰀃
{xk}C
k=1
󰀄
with the C pre-
activations xk and the so-called soft-max or normalized exponential activation:
gβ
󰀃
{xk}C
k=1
󰀄
= exp[βxk]󰁓C
j=1 exp[βxj]
. (5.30)
Note that the required normalization couples the units σk and their states
cannot be interpreted as independently activated by synaptic interaction: each
unit depends on all pre-activations in the layer. For β → 0 all activations will
be equal (σk = 1/C), while for β → ∞ the unit with maximum xk is singled
out with σk = 1.
Radial Basis Functions (RBF): Another popular class of activations also
deviates from the familiar concept of synaptic interactions. The RBF activation
of a given unit σ with input from neurons {sj}L
j=1 which are concatenated in
the vector 󰂓s ∈ RL is computed as",The+Shallow+and+the+Deep.pdf
"of a given unit σ with input from neurons {sj}L
j=1 which are concatenated in
the vector 󰂓s ∈ RL is computed as
σ = g (||󰂓s −󰂓c||) with 󰂓c ∈ RL. (5.31)
The activation depends on the Euclidean distance of the activation vector 󰂓s
from the adaptive center vectors 󰂓c. The term Radial Basis Function refers to
the fact that σ is isotropically centered around 󰂓c.",The+Shallow+and+the+Deep.pdf
"5.5. SPECIFIC ARCHITECTURES 131
A most prominent example is the Gaussian RBF, which is frequently referred
to as the RBF:
σ = exp
󰀅
−β (󰂓s −󰂓c)2󰀆
with parameter β > 0. (5.32)
Frequently, normalized Gaussian RBF are considered in a layer of hidden or
output units σk(k = 1, 2, . . . K):
σk(󰂓s) = exp
󰀅
−β(󰂓s −󰂓ck)2󰀆
󰁓K
j=1 exp [−β(󰂓s −󰂓cj)2]
. (5.33)
In the limit β → ∞, the normalized Gaussian RBF singles out the unit with
smallest (󰂓s −󰂓ck)2, i.e. with the closest center vector for a given 󰂓s.
A popular network architecture for regression comprises a potentially high-
dimensional input layer, a single hidden layer with K units (5.33), and a linear
output unit with adjustable weights. It is described briefly in Sec. 5.5.
5.4.4 Remark: universal function approximation
In Sec. 5.1.2 we constructed a piecewise constant function approximator using
sigmoidal and linear units. It is interesting to note that it is often straight-",The+Shallow+and+the+Deep.pdf
"sigmoidal and linear units. It is interesting to note that it is often straight-
forward to extend these considerations to other activation functions. Note for
instance, that the combination of two ReLU units, equipped with suitable local
thresholds and gain parameters, can replace a piecewise linear activation of the
type displayed in Fig. 5.8 (right panel):
max
󰀝
0, x − a
b − a
󰀞
− max
󰀝
0, x − b
b − a
󰀞
=
󰀻
󰀿
󰀽
0 for x < a
x−a
b−a for a ≤ x < b
1 for x ≥ b.
Using the resulting piecewise linear activation, we can implement the selection
of ROI, cf. Sec. 5.1.2, in analogy to the sigmoidal activations assumed there.
Hence, networks of ReLU and/or piecewise linear units also constitute universal
approximators. Similar arguments can be provided for large families of activa-
tion functions.
Similarly, units with normalized RBF activation, Eq. (5.32), can be readily
used to define ROI in input space and facilitate universal function approximation",The+Shallow+and+the+Deep.pdf
"used to define ROI in input space and facilitate universal function approximation
when combined with piecewise constant representations as in Sec. 5.1.2.
5.5 Specific architectures
In this section we consider a selection of network architectures which play a role
in practical applications and can be handled with the algorithmic approaches
that we have studied so far. In the next section, specific shallow networks are in-
troduced. In Sec. 5.5.2 we briefly discuss the design and training of multilayered
deep neural networks.",The+Shallow+and+the+Deep.pdf
"132 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
5.5.1 Popular shallow networks
So far, we have developed training prescriptions in terms of shallow, feed-forward
architectures with only one or very few hidden layers. In particular we have seen
that a single hidden layer is sufficient to provide universal function approxima-
tions. An important example that we already discussed is the parity machine
for classification, with hidden and output units of the McCulloch Pitts type9.
A soft version of the committee machine, cf. Sec. 4.2, with sigmoidal ac-
tivation can be shown to be provide universal function approximation in the
context of regression, see Sec. 5.1.2. Analogous proofs exist for similar archi-
tectures with alternative hidden activations.
Radial Basis Function networks
Radial basis functions (RBF) as activation functions have been addressed in
Sec. 5.4 already. Frequently, N −K −1 architectures with K RBF hidden units",The+Shallow+and+the+Deep.pdf
"Sec. 5.4 already. Frequently, N −K −1 architectures with K RBF hidden units
and linear output units are referred to as RBF Networks [BL88,MD89b].
In the popular case of Gaussian RBF and a single, linear output with bias
wo we have the input-output relation
σ(ξ) =
M󰁛
j=1
wjφj (ξ) + wo with φj (ξ) = exp
󰀥
−(ξ − cj)2
2σ2
j
󰀦
. (5.34)
This corresponds to Eq. (5.32) with unit-specific parameters βj = 1/(2σ2
j ). Each
unit is equipped with an adaptive vector ci ∈ RN which defines the center of
the receptive field. The response of the unit to a given input ξ ∈ R depends on
its Euclidean distance from the center vector. Here, we also include an adaptive
local bias wo ∈ R in the activation. Instead, we could introduce an additional
hidden unit with constant activation φo(ξ) = 1 for all ξ, similar to the clamped
input employed in Eqs. (2.4) and (3.4).
RBF networks of this form are universal approximators, see the general
discussion in [Bis95a] and specifically, [GP90]. Hence, we can employ networks",The+Shallow+and+the+Deep.pdf
"RBF networks of this form are universal approximators, see the general
discussion in [Bis95a] and specifically, [GP90]. Hence, we can employ networks
of the type (5.34) for general regression tasks.
The complexity of the RBF network can be increased by allowing for adaptive
(inverse) covariance matrices in the Gaussian activations:
φj (ξ) = exp
󰁫
−(ξ − cj)⊤ Σ−1
j (ξ − cj)
󰁬
. (5.35)
Nominally, this introduces N(N +1)/2 additional adaptive parameters per sym-
metric matrix Σj ∈ RN×N . In turn, fewer units might be required to achieve
the same accuracy and performance as a larger network with hidden unit ac-
tivations of the form (5.34). Modifications can be considered, such as the re-
striction to diagonal Σj or the pooling of covariances with a single adaptive
Σ = Σ1 = . . . = ΣM .
9The parity machine is strictly speaking not an (N −K−1) architecture, see Sec. 4.2.",The+Shallow+and+the+Deep.pdf
"5.5. SPECIFIC ARCHITECTURES 133
input layer ξ ∈ RN
fixed random input-to-hidden weights
wm ∈ RN (m = 1, 2, . . . M)
hidden units σm = g (wm · ξ)
hidden-to-output weights vm
linear output S
󰀃
ξ
󰀄
=
M󰁛
m=1
vm σm
Figure 5.10: Illustration of an Extreme Learning Machine (ELM). The N-
dimensional input is connected to a hidden layer with M ≫ N units by fixed
(non-adaptive) random weights. In the example, the single output unit is linear,
e.g. for the purpose of regression. Extensions to a threshold unit or a full layer
of outputs are straightforward.
The RBF architecture could be used for classification tasks by attaching
a single or multiple output classifier to the hidden layer {φj(ξ)}M
j=1 . A more
natural approach is to normalize the M activations in the hidden layer as in
(5.33) and interpret them as probabilistic class assignments:
󰁥φj(ξ) = φj(ξ)󰁓M
k=1 φk(ξ)
with φj from (5.34) or (5.35). (5.36)
These non-local activations satisfy 󰁥φj ∈ [0, 1] and 󰁓
j
󰁥φj = 1 and hence we can",The+Shallow+and+the+Deep.pdf
"󰁥φj(ξ) = φj(ξ)󰁓M
k=1 φk(ξ)
with φj from (5.34) or (5.35). (5.36)
These non-local activations satisfy 󰁥φj ∈ [0, 1] and 󰁓
j
󰁥φj = 1 and hence we can
train the system according to a classification specific cost function like (5.23)
for binary problems and (5.24) in a multi-class setting.
Remark: RBF systems for classification display a striking similarity with
prototype-based classifiers. In an LVQ system as presented in Chapter 6, the
prototypes correspond to the center vectors cj and the softmax scheme of the
classifier would be replaced by a crisp Nearest Prototype classification (NPC).
Similarly, the matrix Σ−1
j in (5.35) is the equivalent of a prototype-specific, local
relevance matrix Λj in Eq. (6.13), see Sec. 6.2.2.
Extreme Learning Machines
Frank Rosenblatt already suggested randomized connections from an input layer
to a so-called association layer, which then was classified by threshold units in
the Mark I realization of the perceptron [HRM+60], see Section 3, Fig. 3.1.",The+Shallow+and+the+Deep.pdf
"the Mark I realization of the perceptron [HRM+60], see Section 3, Fig. 3.1.
More recently, random projections have been become popular as a technique to
achieve sparse, low-dimensional representations of high-dimensional data sets,
see for instance [BM01,LHC06].
A specific feed-forward architecture, termed the Extreme Learning Machine
(ELM), was introduced in 2004 by Huang et al. [HZS06]. It is schematically",The+Shallow+and+the+Deep.pdf
"134 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
original input ξ ∈ RN
wm ∈ RN (m = 1, 2, . . . M)
encoder
decoder
latent variables ym = g (wm · ξ)
vm ∈ RN (m = 1, 2, . . . M)
ξrec =
M󰁛
m=1
ymvm
Figure 5.11: An example of a shallow auto-encoder network: The N-dim.
inputs are represented in a hidden layer with M < N units. Here, N (linear)
output units represent the target reconstruction ξrec ∈ RN , the extension to
non-linear output units is obviously possible.
represented in Fig. 5.10. The random mapping of inputs to a high-dimensional
hidden layer makes it, for instance, possible to separate classes or perform re-
gression tasks with a single linear (threshold) unit which would not be sufficient
to realize the target in terms of the original data. This is similar in spirit to the
basic idea of the Support Vector Machine, cf. 4.3. The relation of ELM and
SVM was first discussed in [FV10].
Shallow autoencoders",The+Shallow+and+the+Deep.pdf
"basic idea of the Support Vector Machine, cf. 4.3. The relation of ELM and
SVM was first discussed in [FV10].
Shallow autoencoders
A particular feed-forward type of networks can be used to find a low-dimensional
representation of high-dimensional feature vectors {ξµ}P
µ=1. To this end, we can
employ a so-called auto-encoder as illustrated in Figure 5.11. The encoder rep-
resents N-dim. input vectors ξ in a single hidden layer of M < N units. The
output ξrec is again N-dimensional and the goal is to minimize the reconstruc-
tion error in the decoder:
Erec = 1
2
P󰁛
µ=1
(ξµ
rec − ξµ)2 . (5.37)
Training amounts to the (e.g. gradient based) minimization of Erec with respect
to the network weights wm, vm ∈ RN . Obviously, the mathematical structure is
the same as in function approximation, with the special target to approximate
the identity function RN → RN . The resulting M-dimensional latent variables
{ym}P
m=1 serve as the low-dimensional representations of high-dimensional data.",The+Shallow+and+the+Deep.pdf
"{ym}P
m=1 serve as the low-dimensional representations of high-dimensional data.
In Fig. 5.11 the output units are assumed to be linear, the possible general-
ization to non-linear reconstructions is straightforward. One special case is par-
ticularly interesting: for linear activations g in the hidden layer and linear recon-
struction ξrec, the minimization of the reconstruction error (5.37) is analogous",The+Shallow+and+the+Deep.pdf
"5.5. SPECIFIC ARCHITECTURES 135
to the well known Principal Component Analysis, e.g. [Bis95a,HKP91,HTF01].
More precisely, the weight vectors wm, which minimize Erec, span the same
sub-space as the M leading principal components of the data set.
Using sigmoidal or other non-trivial hidden and/or output activations in
the auto-encoder network, generalizes the concept of PCA to non-linear low-
dimensional representations and reconstructions. In the next section we will
also briefly mention deep auto-encoders, where several hidden layers represent
and process the data internally [GBC16].
Obviously we can exploit the latent variables of an auto-encoder immediately
for the purpose of visualizing complex, high-dim. data if M = 2 or 3. Moreover,
after having trained the auto-encoder to realize a faithful internal representation,
one could attach a feed-forward classifier or regression network to the hidden
layer and apply supervised learning to realize a target function, e.g. of the type",The+Shallow+and+the+Deep.pdf
"layer and apply supervised learning to realize a target function, e.g. of the type
RM → R.
5.5.2 Deep and convolutional neural networks
At a glance, the term Deep Learning refers to the use of feed-forward neural
networks with many hidden layers. While this over-simplified definition ignores
several aspects of Deep Learning, e.g. the potential use of feedback and recurrent
systems, we will focus here on deep feed-forward architectures. As phrased by
Goodfellow, Bengio and Courville [GBC16]:
[There is no] consensus about how much depth a model requires to
qualify as ’deep’. However, deep learning can safely be regarded as
the study of models that either involve a greater amount of compo-
sition of learned functions or learned concepts than traditional ma-
chine learning does.
The enormous success of Deep Learning and its popularity after, say, 2010, can
be attributed to a number of developments, including the following:",The+Shallow+and+the+Deep.pdf
"The enormous success of Deep Learning and its popularity after, say, 2010, can
be attributed to a number of developments, including the following:
◦ The availability of large amounts of data, e.g. from image data bases or
large collections of commercial data in e-commerce
◦ The pre-training of deep networks or sub-networks from unspecific data
bases, followed by task-specific fine tuning (transfer learning)
◦ The ever-increasing computational power, made available through super-
computers or local solutions (e.g. GPU)
◦ The refinement of (mostly) gradient-based training techniques, e.g. w.r.t.
to the automatic adaptation of learning rates or efficient regularization
techniques such as drop out
◦ The exploitation and combination of concepts that had been developed
earlier for shallow networks, e.g. the ideas of weight-sharing or momentum.",The+Shallow+and+the+Deep.pdf
"136 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
Figure 5.12: A 3 × 3 ’convolutional’filter kernel is applied to a 3 × 3 image,
here zero-padded. The 3 × 3 kernel with weights denoted in the illustration
is centered on every pixel of the image to obtain the pixel values in the 3 × 3
convolved image. Note that the operation does not reduce the dimension of the
data.
◦ The use of activation functions that (supposedly) improve the efficiency
and performance of networks in training and working phase, for example
the Rectified Linear Unit (ReLU), cf. Sec. 5.4.2
◦ The consideration of particular architectures, e.g. Convolutional Neural
Networks (CNN), designed for the analysis of specific types of data, such
as images, language, time series or other data with a low-dimensional
spatial, temporal or functional structure.
Quite a few of these concepts have been known well before the rise of Deep
Learning. Ultimately their combination made the great success of Deep Net-",The+Shallow+and+the+Deep.pdf
"Quite a few of these concepts have been known well before the rise of Deep
Learning. Ultimately their combination made the great success of Deep Net-
works possible, see e.g. [Sch15,LBH18].
Convolutional and pooling layers
All convolutional neural networks (CNN) share some characteristic design fea-
tures which facilitate the processing of structured data. For instance, in images
or time series data, we expect localized information: Time series like the €/$
exchange rate will display short time correlations that decrease over time. Like-
wise, pixels in a photographic image are expected to be similar in intensity and
color if they belong to the same object, while far away pixels may be totally
unrelated. In the following we will discuss CNN in the context of images, the
transfer to other structured data is straightforward.
The localization is accounted for by connecting units in a first layer to limited
neighborhoods or patches of the input data, see Fig. 5.12 for an illustration. By",The+Shallow+and+the+Deep.pdf
"neighborhoods or patches of the input data, see Fig. 5.12 for an illustration. By
choice of the weights in such a filter kernel, nodes can implement a particular
local operation or convolution. 10 The weights can be adapted in the training
10The term is used somewhat loosely in the context of Deep Learning.",The+Shallow+and+the+Deep.pdf
"5.5. SPECIFIC ARCHITECTURES 137
Figure 5.13: The 3 × 3 convolved image from Fig. 5.12 is reduced to 2 × 2
pixels by applying a max-pooling (upper) or average pooling (lower) to all 2 ×2
patches in the 3 × 3 filtered image.
process, e.g. by using gradient based Backpropagation of Error for the entire
network. A number K of different filters is applied to all patches of the same
size in the input data. As each filter applies the same operation, the number
of adaptive weights depends only on K and on the dimension and type of the
kernels, while it is independent of the dimension of the input. This basic concept
of weight-sharing was already present in very early network models, see below
for Fukushima’s Neocognitron as an example.
The first convolutional layer in a CNN therefore represents the input data
in terms of many versions of the image obtained through a potentially large set
of adaptive filters. These may include but are not limited to (approximations",The+Shallow+and+the+Deep.pdf
"of adaptive filters. These may include but are not limited to (approximations
of) intuitive operations like the detection of edges or other local patterns.
Most frequently, after convolution, a pooling operation is applied in a sub-
sequent layer. Pooling reduces the dimensionality by combining, usually small,
patches of nodes into a super-pixel. Popular examples replace plaquettes of 2×2
or 3 × 3 nodes by the average activation (average pooling) or by the maximum
activity in the patch, see Fig. 5.13. Typically, the pooling nodes are hard wired,
i.e. not trainable, although one could for instance consider adaptive weighted
averages for pooling.
After the first convolution and pooling, the input image is represented by
a number of filtered and dimension reduced versions. Frequently, convolutional
and pooling layers are stacked in alternating fashion, yielding increasingly ab-
stract representations of decreasing dimension. Ultimately, one or several train-",The+Shallow+and+the+Deep.pdf
"stract representations of decreasing dimension. Ultimately, one or several train-
able dense layers, fully connected as in conventional feed-forward architectures,
are employed to represent the target classification or regression.
A large number of network similar architectures have been developed and
are made available in the public domain, see https://modelzoo.co for an example
repository. Many of these networks have been pre-trained on more or less generic
data sets and can be fine-tuned by the user for the specific task at hand.
Despite the simplifications through weight-sharing and similar regularization
techniques, Deep Networks often comprise a huge number of adaptive weights
and consequently can be very data hungry. In fact, it appears often surpris-",The+Shallow+and+the+Deep.pdf
"138 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
Figure 5.14: An early deep architecture (schematic) known as the Neocog-
nitron, first introduced by K. Fukushima in 1980 (1979 in Japanese) [Fuk80].
Besides input and output (recognition), several layers of so-called S-cells and
C-cells are stacked see the text for details. Illustration redrawn after [Fuk19]
with kind permission from the author.
ing that heavily over-parameterized networks can be trained successfully at all
and suffer less drastically from over-fitting effects than one might expect on
theoretical grounds.
Early examples: Neocognitron and LeNet
Inspired by an early model of human vision of Hubel and Wiesel [HW59], Kuni-
hiko Fukushima introduced the so-called Neocognitron network architecture as
early as 1979 (in Japanese) and 1980 [Fuk80,Fuk88], see preceding S-also [Fuk19]
for a more recent presentation and discussion of different versions of the basic
architecture.",The+Shallow+and+the+Deep.pdf
"for a more recent presentation and discussion of different versions of the basic
architecture.
The Neocognitron already comprises many elements of modern Convolu-
tional Neural Networks. As illustrated in Figure 5.14, the network consists of
an input and output layer, with stacked hidden layers of alternating types. Units
are typically connected to patches of nodes in the preceding layer. Feature ex-
traction layers (shaded blue in Fig. 5.14) employ local filters to these patches.
Their units correspond to the simple neurons or S-cells suggested by Hubel and
Wiesel, which are activated, for instance, by characteristic patterns like straight
lines of a particular orientation. Nodes in a subsequent layer of complex or
C-cells perform pooling operations in the sense that they are activated inde-
pendent of the precise location of the stimulation within their receptive field in
the preceding S-layer. Thanks to the averaging or pooling C-cells, the Neocogni-",The+Shallow+and+the+Deep.pdf
"the preceding S-layer. Thanks to the averaging or pooling C-cells, the Neocogni-
tron is, to a certain degree, insensitive to shifts and distortions of input patterns.",The+Shallow+and+the+Deep.pdf
"5.5. SPECIFIC ARCHITECTURES 139
Figure 5.15: A deep architecture (schematic) known as LeNet, specifically
LeNet-5, introduced by LeCun et al. in [LBD+89]. Image available under license
CC BY3.0 at https://www.researchgate.net/publication/319905492_Image_retrieval_
method_based_on_metric_learning_for_convolutional_neural_network/figures.
The sequence of S and C layers represents the input in decreasing detail and,
ultimately, the network response (e.g. a classification) is provided in the output
layer. In contrast to more recent CNN architectures, Fukushima did not train
the Neocognitron end-to-end by gradient descent techniques. Instead, the filters
realized by S-cells were either pre-wired or adapted by means of unsupervised
learning techniques.
The Neocognitron has been studied and used in the context of brain-inspired
pattern recognition, including handwritten digit recognition and similar tasks.
It constitutes a groundbreaking work that inspired many, if not all modern Deep",The+Shallow+and+the+Deep.pdf
"It constitutes a groundbreaking work that inspired many, if not all modern Deep
Networks for visual pattern recognition and similar tasks.
Another groundbreaking architecture, known as LeNet is due to LeCun and
collaborators [LBD+89]. It is an early example of a Convolutional Neural Net-
work (CNN) for pattern recognition and image processing and can be considered
the starting point for this popular type of architecture. The structure is simi-
lar to the above discussed Neocognitron. Alternating convolutional and pooling
layers process an input with increasing abstraction towards a fully connected
output layer. The LeNet can be trained from end-to-end by gradient based
Backpropagation. It was initially introduced for the task of handwritten digit
recognition (ZIP-code reading).
Groups of nodes perform the same task on different patches of the input.
Consequently, many nodes can share the same weight values which reduces",The+Shallow+and+the+Deep.pdf
"Groups of nodes perform the same task on different patches of the input.
Consequently, many nodes can share the same weight values which reduces
the effective number of adaptive quantities drastically. LeNet constitutes a
very early example of weight-sharing, which to date plays an important role
in the training of deep networks, see also Sec. 7.2.5. Further improvements of
the network performance were achieved by developing and applying a specific
pruning technique named Optimal Brain Damage [LDS90], which we introduce
in Sec. 7.2.4.
Appraisal and critique of deep learning
The success of Deep Learning, mainly in the context of image processing, has
triggered a lot of excitement in, both, academia and the general public. This
includes exaggerated claims with respect to general intelligence or applications",The+Shallow+and+the+Deep.pdf
"140 5. NETWORKS FOR REGRESSION AND CLASSIFICATION
in critical areas like clinical medicine.
Recently, several scholars have expressed criticism of Deep Learning and the
hype surrounding it, with [Mar18,Pre21,Zad19,AN20] being just a few examples.
In a sense, the situation is highly reminiscent of the strong expectations and the
later disappointments in previous waves of machine learning popularity.
In the biased opinion of the author of this text, several deplorable trends
can be observed in academia and in the general public:
◦ Deep Learning is often presented as fundamentally new and totally differ-
ent from so-called conventional, supposedly old school machine learning,
ignoring decades of research that facilitated the recent developments.
◦ The mis-identification of machine learning with image analysis only, in
particular in the non-scientific media. While image classification, scene
analysis, face recognition etc. are certainly relevant and particularly ap-",The+Shallow+and+the+Deep.pdf
"particular in the non-scientific media. While image classification, scene
analysis, face recognition etc. are certainly relevant and particularly ap-
pealing problems, machine learning should be seen from a much broader
perspective.
◦ A wide-spread tendency to use overly complex DNN to tackle even rela-
tively simple, specific applications without investing a thoughtful analysis
and often without a proper critical evaluation of the performance or com-
parison with baseline techniques.
◦ The exclusive use of ready-made programming environments while having
a limited understanding of the basic underlying principles.11 Frequently
such systems offer only limited user control or the options are not exploited
properly.
◦ A lack of theoretical insight into Deep Learning, or - much worse - a lack
of interest into a better understanding of the relevant phenomena.
The last points are particularly unfortunate in view of the many interesting",The+Shallow+and+the+Deep.pdf
"of interest into a better understanding of the relevant phenomena.
The last points are particularly unfortunate in view of the many interesting
challenges and open questions posed by Deep Learning, some of which are sum-
marized in [Sej20].
Despite these and other points of criticism, Deep Learning will play an im-
portant role in forthcoming years. It will certainly facilitate the exploration of
new and exciting application areas. At the same time Deep Learning will con-
tinue to provide highly interesting theoretical challenges that deserve significant
attention.
11Even worse, this is often combined with displaying code in illegible small but colorful fonts
on a black background.",The+Shallow+and+the+Deep.pdf
"Chapter 6
Distance-based classifiers
One can state, without exaggeration, that the observation of and the search for
similarities and differences are the basis of all human knowledge.
— Alfred Nobel
The use of distances or dissimilarities for the comparison of observations
with a set of labeled reference data points provides a simple yet powerful tool
for classification. In particular, the use of prototypes or exemplars, derived from
a given data set, is the basis for a very successful family of machine learning
approaches. Prototype-based classifiers are appealing for a number of reasons.
The extraction of information from previously observed data in terms of typi-
cal representatives, the prototypes, is particularly transparent and intuitive, in
contrast to many, more black-box like systems. The same is true for the working
phase, in which novel data are compared with the prototypes by use of a suitable
(dis-)similarity or distance measure.",The+Shallow+and+the+Deep.pdf
"phase, in which novel data are compared with the prototypes by use of a suitable
(dis-)similarity or distance measure.
Prototype systems are frequently employed for the unsupervised analysis of
complex data sets, aiming at the detection of underlying structures, such as
clusters or hierarchical relations, see also Chapter 8 and, for instance, [HTF01,
Bis95a, DHS00]. Competitive Vector Quantization, the well-known K-means
algorithm or Self-Organizing Maps are prominent examples for the use of pro-
totypes in the context of unsupervised learning [Koh97,HTF01,DHS00].
In the following the emphasis is on supervised learning in prototype-based
systems. In particular, we focus on the framework of Learning Vector Quan-
tization (LVQ) for classification. Besides the most basic concepts and train-
ing prescriptions we present extensions of the framework to unconventional
distances and to the use of adaptive measures in so-called relevance learning
schemes [BHV16].",The+Shallow+and+the+Deep.pdf
"distances and to the use of adaptive measures in so-called relevance learning
schemes [BHV16].
The aim of this chapter is far from giving a complete review of the ongoing
fundamental and application oriented research in the context of prototype-based
141",The+Shallow+and+the+Deep.pdf
"142 6. DISTANCE-BASED CLASSIFIERS
Figure 6.1: Left panel: Illustration of the Nearest Neighbor (NN) Classi-
fier for an artificial data set containing three different classes. Right panel:
A corresponding NPC scheme for the same data. Prototypes are represented
by larger symbols. Both schemes are based on Euclidean distance and yield
piecewise linear decision boundaries.
learning. It provides, at best, first insights into supervised schemes and can serve
as a starting point for the interested reader.
The emphasis will be on Teuvo Kohonen’s Learning Vector Quantization and
its extensions [Koh97]. Examples for training prescriptions are given and the use
of unconventional distance measures is discussed. As an important conceptual
extension of LVQ, Relevance Learning is introduced, with Matrix Relevance
LVQ serving as an example.
In a sense, the philosophies behind LVQ and the SVM are diametrically
opposed to each other: while support vectors represent the difficult cases in",The+Shallow+and+the+Deep.pdf
"In a sense, the philosophies behind LVQ and the SVM are diametrically
opposed to each other: while support vectors represent the difficult cases in
the data set, which are closest to the decision boundary, cf. Sec. 4.3, LVQ
represents the classes by - supposedly - typical exemplars relatively far from the
class borders.
Note that LVQ systems could be formulated and interpreted as layered neural
networks with specific, distance-based activations and a crisp output reflecting
the Winner-Takes-All principle. In fact, after years of denying the relation in
the literature, it has become popular again to point out the conceptual vicinity
to neural networks. Recent publications also discuss the embedding of LVQ
modules in deep learning approaches [VMC16,VBVS17].",The+Shallow+and+the+Deep.pdf
"6.1. PROTOTYPE-BASED CLASSIFIERS 143
6.1 Prototype-based classifiers
Among the many frameworks developed for supervised machine learning, proto-
type-based systems are particularly intuitive, flexible, and easy to implement.
Although we restrict the discussion to classification problems, many of the con-
cepts carry over to regression or, to a certain extent, also to unsupervised learn-
ing, see [BHV16].
Several prototype-based classifiers have been considered in the literature.
Some of them can be derived from well-known unsupervised schemes like the
Self-Organizing-Map or the Neural Gas [Koh97, RMS92, HSV05], which can be
extended in terms of a posterior labelling of the prototypes. Here, the focus is
on the so-called Learning Vector Quantization (LVQ), a framework which was
originally suggested by Teuvo Kohonen [Koh97]. As a starting point for the dis-
cussion, we briefly revisit the well-known k-Nearest-Neighbor (kNN) approach
to classification, see [DHS00,CH67].",The+Shallow+and+the+Deep.pdf
"cussion, we briefly revisit the well-known k-Nearest-Neighbor (kNN) approach
to classification, see [DHS00,CH67].
6.1.1 Nearest Neighbor and Nearest Prototype Classifiers
Nearest Neighbor classifiers [DHS00, CH67] constitute one of the simplest and
most popular classification schemes. In this classical approach, a number of
labeled feature vectors is stored in a reference set:
D = {ξµ, yµ = y(ξµ)}P
µ=1
with ξµ ∈ RN . In contrast to the discussion of the perceptron and similar
systems, here we do not have to restrict the presentation to binary labels. We
therefore denote the (possibly multi-class) labels by yµ ∈ {1, 2, . . . C} where C
is the number of classes.
An arbitrary novel feature vector or query ξ ∈ RN can be classified ac-
cording to its (dis-) similarities to the samples stored in the reference data.
To this end, its distance from all reference vectors ξµ ∈ D has to be computed.
Most frequently, the simple (squared) Euclidean distance is used in this context:",The+Shallow+and+the+Deep.pdf
"Most frequently, the simple (squared) Euclidean distance is used in this context:
d(ξ, ξµ) = (ξ − ξµ)2. The query ξ is then assigned to the class of its Nearest
Neighbor exemplar in D. In the more general kNN classifier, the assignment is
determined by means of a voting scheme that considers the k closest reference
vectors [CH67].
The NN or kNN classifier is obviously very easy to implement as it does
not even require a training phase. Nevertheless one can show that the kNN
approach bears the potential to realize Bayes optimal performance if the number
k of neighbors is chosen carefully [HTF01, DHS00, CH67]. Consequently, the
method serves, to date, as an important baseline algorithm and is frequently
used as a benchmark to compare performances with.
Fig. 6.1 (left panel) illustrates the NN classifier and displays how the system
implements piecewise linear class borders. Several difficulties are evident already",The+Shallow+and+the+Deep.pdf
"implements piecewise linear class borders. Several difficulties are evident already
in this simple illustration. Class borders can be overly complex, for instance if
single data points in the set have been classified incorrectly. Furthermore, the",The+Shallow+and+the+Deep.pdf
"144 6. DISTANCE-BASED CLASSIFIERS
fact that every data point contributes with equal weight can lead to overfitting
effects because the classifier over-rates the importance of individual examples.
As a consequence, it might not perform well when presented with novel, unseen
data.
Straightforward implementations of kNN compute and sort the distances of
ξ from all available examples in D. While methods for efficient sorting can
reduce the computational costs to a certain degree, the problem persists and is
definitely relevant for very large data sets.
Both drawbacks could be attenuated by reducing the number of reference
data in an intelligent way while keeping the most relevant properties of the data.
Indeed, the selection of a suitable subset of reference vectors by thinning out
D was already suggested in [Har68]. An alternative, essentially bottom-to-top
approach is considered in the following sections.
6.1.2 Learning Vector Quantization",The+Shallow+and+the+Deep.pdf
"approach is considered in the following sections.
6.1.2 Learning Vector Quantization
This successful and particularly intuitive approach to classification was intro-
duced and put forward by Teuvo Kohonen [Koh97,RMS92,BHV16,Koh95,SK99,
NE14]. The basic idea is to replace the potentially large set of labeled example
data by relatively few, representative prototype vectors.
LVQ was originally motivated as a simplifying approximation of a Bayes
classifier under the assumption that the underlying density of data corresponds
to a superposition of Gaussians [Koh97]. LVQ replaces the actual density esti-
mation by a simple and robust method of supervised Vector Quantization. Each
of the C classes is to be represented by (at least) one representative. Formally,
we consider the set of prototype vectors
󰀋
wj, cj 󰀌M
j=1 with wj ∈ RN and cj ∈ {1, 2, . . . C}. (6.1)
Here, the prototype labels cj = c(wj) indicate which class the corresponding",The+Shallow+and+the+Deep.pdf
"󰀋
wj, cj 󰀌M
j=1 with wj ∈ RN and cj ∈ {1, 2, . . . C}. (6.1)
Here, the prototype labels cj = c(wj) indicate which class the corresponding
prototype is supposed to represent. The so-called Nearest Prototype classifier
(NPC) assigns an arbitrary, e.g. novel, feature vector ξ to the class c∗ = c(w∗)
of the closest prototype
w∗(ξ) with d
󰀓
w∗(ξ), ξ
󰀔
= min
󰁱
d(wj, ξ)
󰁲M
j=1
(6.2)
where ties can be broken arbitrarily.
In the following, the closest prototype w∗(ξ) of a given input vector will be
referred to as the winner. For brevity we will frequently omit the argument of
w∗(ξ) and use the shorthand w∗ when it is obvious which input vector it refers
to.
Figure 6.1 (right panel) illustrates the NPC concept: class borders corre-
sponding to relatively few prototypes are smoother than the corresponding NN
decision boundaries shown in the left panel. Consequently, an NPC classifier",The+Shallow+and+the+Deep.pdf
"6.1. PROTOTYPE-BASED CLASSIFIERS 145
can be expected to be more robust and less prone to overfitting effects.
The performance of LVQ systems has proven to be competitive in a variety of
practical classification problems [Neu02]. In addition, their flexibility and inter-
pretability constitute important advantages of prototype-based classifiers since
prototypes are obtained and can be interpreted within the space of observed
data, directly. This feature facilitates the discussion with domain experts and
stands in contrast to many other, less transparent machine learning frameworks.
An LVQ system for nearest prototype classification can be interpreted as a
neural network with a single hidden layer. The prototypes correspond to hidden
units with a distance-based activation which feed into a winner-takes-all output
unit. This appealing analogy is elaborated in, for example, [RKSV20]. However,
it is not essential for the following.
6.1.3 LVQ training algorithms",The+Shallow+and+the+Deep.pdf
"unit. This appealing analogy is elaborated in, for example, [RKSV20]. However,
it is not essential for the following.
6.1.3 LVQ training algorithms
So far we have not addressed the question of where and how to place the pro-
totypes for a given data set. A variety of LVQ training algorithms have been
suggested in the literature [Koh97,SK99,Koh90,NE14,BGH07,Gho21,Wit10].
The first, original scheme suggested by Kohonen [Koh97] is known as LVQ1.
Essentially, it already includes all aspects of the many modifications that were
suggested later. The algorithm can be summarized in terms of the following
steps:
LVQ1 algorithm, random sequential presentation of data
– at time step t, select a single feature vector ξµ with class label yµ
randomly from the data set D with uniform probability 1/P.
– identify the winning prototype, i.e. the currently closest prototype
w∗
µ = w∗(ξµ) given by d(w∗
µ, ξµ) = min
󰀋
d(wj, ξµ)
󰀌M
j=1
(6.3)
with class label c∗
µ = c(w∗
µ).",The+Shallow+and+the+Deep.pdf
"w∗
µ = w∗(ξµ) given by d(w∗
µ, ξµ) = min
󰀋
d(wj, ξµ)
󰀌M
j=1
(6.3)
with class label c∗
µ = c(w∗
µ).
– perform a Winner-Takes-All (WTA) update: (6.4)
w∗
µ(t+1) =w∗
µ(t) + ηw Ψ(c∗
µ, yµ)
󰀃
ξµ−w∗
µ
󰀄
with Ψ(c, y)=
󰀝 +1 if c=y
−1 else.
The magnitude of the update is controlled by the learning rate ηw. The actual
update step (6.4) moves the winning prototype even closer to the presented
feature vector if w∗
µ and the example carry the same label as indicated by",The+Shallow+and+the+Deep.pdf
"146 6. DISTANCE-BASED CLASSIFIERS
Ψ(c∗
µ, yµ) = +1. On the contrary, w∗
µ is moved farther away from ξµ if the
winning prototype represents a class different from yµ, i.e. Ψ(c∗
µ
, yµ) = −1.
A popular initialization strategy is to place prototypes in the class-conditional
mean vectors in the data set, i.e.
wj(0) =
P󰁛
µ=1
δ[yµ, cj] ξµ
󰀯 P󰁛
µ=1
δ[yµ, cj]
with the Kronecker-delta δ[i, j]. If several prototypes are employed per class,
independent random variations could be added in order to avoid coinciding pro-
totypes, initially. More sophisticated initialization procedures can be realized,
for instance by applying a K-means procedure in each class separately.
After repeated presentations of the entire training set, the prototypes should
represent their respective class by assuming class-typical positions in feature
space, ideally.
Numerous modifications of this basic LVQ scheme have been considered in
the literature, see for instance [Koh90, NE14, BGH07, Gho21, Wit10] and ref-",The+Shallow+and+the+Deep.pdf
"Numerous modifications of this basic LVQ scheme have been considered in
the literature, see for instance [Koh90, NE14, BGH07, Gho21, Wit10] and ref-
erences therein. In particular, several approaches based on differentiable cost-
functions have been suggested. They allow for training in terms of gradient
descent or other optimization schemes. Note that LVQ1 and many other heuris-
tic schemes cannot be interpreted as descent algorithms in a straightforward
fashion.
One particular cost function based algorithm is the so-called Robust Soft
LVQ (RSLVQ) which has been motivated in the context of statistical mod-
elling [SO03]. The popular Generalized LVQ (GLVQ) [SY95, SY98] is guided
by an objective function that relates to the concept of large margin classifica-
tion [CGBNT03]:
EGLV Q =
P󰁛
µ=1
Φ(eµ) with eµ = d(wJ
µ, ξµ) − d(wK
µ , ξµ)
d(wJµ, ξµ) + d(wKµ , ξµ). (6.5)
Here, the vector wJ
µ denotes the closest of all prototypes which carry the same
label as the example ξµ, i.e. cJ
µ",The+Shallow+and+the+Deep.pdf
"d(wJµ, ξµ) + d(wKµ , ξµ). (6.5)
Here, the vector wJ
µ denotes the closest of all prototypes which carry the same
label as the example ξµ, i.e. cJ
µ
= yµ. Similarly, wK
µ denotes the closest proto-
type with a label different from yµ. For short, we will frequently refer to these
vectors as the correct winner wJ
µ and the incorrect winner wK
µ , respectively.
The cost function (6.5) in general comprises a non-linear and monotonically
increasing function Φ(e). A particularly simple choice is the identity Φ(e) = e,
while the authors of [SY95,SY98] suggest the use of a sigmoidal Φ(e) = 1/[1 +
exp(−γ e)], where γ > 0 controls its steepness.
Negative values eµ < 0 indicate that the corresponding training example is
correctly classified in the NPC scheme, since then d(wJ
µ, ξµ) < d(wK
µ , ξµ).1 For
large values of the steepness γ the costs approximate the number of misclassified
1Note that the argument of Φ obeys −1 ≤ eµ ≤ 1.",The+Shallow+and+the+Deep.pdf
"6.1. PROTOTYPE-BASED CLASSIFIERS 147
training data, while for small γ the minimization of EGLV Q corresponds to
maximizing the margin-like quantities eµ.
A popular and conceptually simple strategy to optimize EGLV Q is stochastic
gradient descent in which single examples are presented in randomized order
[RM51,Bot91,FP96]. In contrast to LVQ1, two prototypes are updated in each
step of the GLVQ procedure:
Generalized LVQ (GLVQ), stochastic gradient descent
– at time step t, select a single feature vector ξµ with class label yµ
randomly from the data set D with uniform probability 1/P.
– identify the correct and incorrect winners, i.e. the prototypes
wJ
µ with d(wJ
µ, ξµ) = min
󰁱
d(wj, ξµ)
󰀏󰀏 cj
µ = yµ
󰁲M
j=1
wK
µ with d(wK
µ , ξµ) = min
󰁱
d(wj, ξµ)
󰀏󰀏 cj
µ ∕= yµ
󰁲M
j=1
(6.6)
with class labels cJ
µ = yµ and cK
µ
∕= yµ, respectively.
– update both winning prototypes according to:
wJ
µ(t+1) = wJ
µ(t) − ηw
∂Φ(eµ)
∂wJµ
wK
µ (t+1) = wK
µ (t) − ηw
∂Φ(eµ)
∂wKµ
, (6.7)",The+Shallow+and+the+Deep.pdf
"– update both winning prototypes according to:
wJ
µ(t+1) = wJ
µ(t) − ηw
∂Φ(eµ)
∂wJµ
wK
µ (t+1) = wK
µ (t) − ηw
∂Φ(eµ)
∂wKµ
, (6.7)
where the gradients are evaluated in wL
µ (t) for L = J, K.
For the full form of the gradient terms we refer the reader to [SY95,SY98]. Note
that – if Euclidean distance is used – the chain rule implies that the updates
are along the gradients
∂d(wL
µ , ξµ)
∂wLµ
∝
󰀃
wL
µ − ξµ󰀄
for L = J, K. (6.8)
Moreover, the signs of the pre-factors in (6.7) are given by Ψ(cL, yµ) = ±1 as in
(6.1.3). In essence, GLVQ performs updates which move the correct (incorrect)
prototype towards (away) from the feature vector, respectively. Hence, the basic
concept of the intuitive LVQ1 is preserved in GLVQ.
In both LVQ1 and GLVQ, very often a decreasing learning rate ηw is used
to ensure convergence of the prototype positions [RM51]. Alternatively, schemes
for automated learning rate adaptation or more sophisticated optimization meth-",The+Shallow+and+the+Deep.pdf
"for automated learning rate adaptation or more sophisticated optimization meth-
ods can be applied, see e.g. [SNW11], which we will not discuss here.",The+Shallow+and+the+Deep.pdf
"148 6. DISTANCE-BASED CLASSIFIERS
6.2 Distance measures and relevance learning
So far, the discussion focussed on Euclidean distance as a standard measure for
the comparison of data points and prototypes. This choice appears natural and
it is arguably the most popular one. One has to be aware, however, that other
choices may be more suitable for real world data. Depending on the application
at hand, unconventional measures might outperform Euclidean distance by far.
Hence, the selection of a specific distance constitutes a key step in the design of
prototype-based models. In turn, the possibility to choose a distance based on
prior information and insight into the problem contributes to the flexibility of
the approach.
6.2.1 LVQ beyond Euclidean distance
As discussed above, training prescriptions based on Euclidean metrics generi-
cally yield prototype displacements along the vector (ξµ − w) as in Eq. (6.8).
Replacing the Euclidean distance by a more general, differentiable measure",The+Shallow+and+the+Deep.pdf
"Replacing the Euclidean distance by a more general, differentiable measure
δ(ξµ, w), allows for the analogous derivation of LVQ training schemes. This
is conveniently done in cost function based schemes like GLVQ, cf. Eq. (6.5),
but it is also possible for the more heuristic LVQ1, which will serve as an exam-
ple here. As a generalization of Eq. (6.4) we obtain the analogous WTA update
from example µ at time t:
w∗
µ(t + 1) = w∗
µ(t) − ηw Ψ(c∗
µ, yµ) 1
2
∂δ(w∗
µ,ξµ)
∂w∗
µ
. (6.9)
Obviously, the winner w∗
µ has to be determined by use of the same measure δ,
for the sake of consistency.
Along these lines, LVQ update rules can be derived for quite general dissim-
ilarities, provided the distance δ is differentiable with respect to the prototype
positions. Note that the formalism does not require metric properties of δ. As
a minimal condition, non-negativity δ(w, ξ) ≥ 0 should be satisfied for w ∕= ξ
and δ(ξ, ξ) = 0.
Note that cost function based approaches can also employ non-differentiable",The+Shallow+and+the+Deep.pdf
"and δ(ξ, ξ) = 0.
Note that cost function based approaches can also employ non-differentiable
measures if one resorts to alternative optimization strategies which do not re-
quire the use of gradients [SNW11]. Alternatively, differentiable approximations
of non-differentiable δ can be used, see [HV05] for a discussion thereof.
In the following, we mention just a few prominent alternatives to the stan-
dard Euclidean metrics that have been used in the context of LVQ classifiers. We
refer to, e.g., [BHV16,BHV14,HV05] for more detailed discussions and further
references.
Statistical properties of a given data set can be taken into account explicitly
by employing the well-known Mahalanobis distance [Mah36]. This classical
measure is a popular tool in the analysis of data sets. Duda et al. present a
detailed discussion and several application examples [DHS00].
Standard Minkowski distances satisfy metric properties for values of p ≥ 1",The+Shallow+and+the+Deep.pdf
"6.2. DISTANCE MEASURES AND RELEVANCE LEARNING 149
in
dp(ξ, 󰁥ξ) =
󰁫󰁓N
j=1
󰀏󰀏󰀏ξj − 󰁥ξj
󰀏󰀏󰀏
p󰁬1/p
for ξ, 󰁥ξ ∈ RN , (6.10)
which includes Euclidean distance as a special case for p = 2. Larger (smaller)
values of p put emphasis on the components ξj and 󰁥ξj with larger (smaller)
deviations |ξj − 󰁥ξj|, respectively. For instance, in the limit p → ∞ we have
d∞(ξ, 󰁥ξ) = max
j=1,...N
󰀏
󰀏󰀏ξ
j − 󰁥ξj
󰀏󰀏󰀏.
Setting p ∕= 2 has been shown to improve performance in several practical
applications, see [BBL07,GW10] for specific examples.
The squared Euclidean distance can be rewritten in terms of scalar products:
d(w, ξ)2 = (w · w − 2w · ξ + ξ · ξ) . (6.11)
So-called kernelized distances [Sch01] replace all inner products in (6.11) by a
kernel function κ:
dκ(w, ξ)2 = κ(w, w) − 2κ(w, ξ) + κ(ξ, ξ). (6.12)
As in the SVM formalism, the function κ can be associated with a non-linear
transformation from RN to a potentially higher-dimensional feature space. In",The+Shallow+and+the+Deep.pdf
"transformation from RN to a potentially higher-dimensional feature space. In
SVM training one takes advantage of the fact that data can become linearly
separable due to the transformation, as discussed in Sec. 4.3. Similarly, kernel
distances can be employed in the context of LVQ in order to achieve better
classification performance, see [VKNR12] for a particular application.
As a last example, statistical divergences can be used to quantify the dis-
similarity of densities or histogram data. For instance, image data is frequently
characterized by color or other histograms. Similarly, text can be represented
by frequency counts in a bag of words approach. In the corresponding classifi-
cation problems, the task would be to discriminate between class-characteristic
histograms. Euclidean distance is frequently insensitive to the relevant dis-
criminative properties of histograms. Hence, the classification performance can",The+Shallow+and+the+Deep.pdf
"criminative properties of histograms. Hence, the classification performance can
benefit from using specific measures, such as statistical divergences. The well-
known Kullback-Leibler divergence is just one example of many measures that
have been suggested in the literature. For further references and an example
application in the context of LVQ see [MSS+11,Mwe14]. There, it is also demon-
strated that even non-symmetric divergences can be employed properly in the
context of LVQ, as long as the measures are used in a consistent way.
6.2.2 Adaptive distances in relevance learning
In the previous subsection, a few alternative distance measures have been dis-
cussed. In practice, a particular one could be selected based on prior insights
or according to an empirical comparison in a validation procedure.",The+Shallow+and+the+Deep.pdf
"150 6. DISTANCE-BASED CLASSIFIERS
The elegant framework of relevance learning allows for a significant con-
ceptual extension of distance-based classification. It is particularly suitable for
prototype systems and was was introduced and put forward in the context of
LVQ in [HV02,SBH09,Sch10,Bun11,SBS+10,BSH+12], for instance.
Relevance learning has proven useful in a variety of applications, including
biomedical problems and image processing tasks, see for instance [Bie17].
In this very elegant approach, only the parametric form of the distance
measure is fixed in advance. Its parameters are considered adaptive quantities
which can be adjusted or optimized in the data-driven training phase. The basic
idea is very versatile and can be employed in a variety of learning tasks. We
present here only one particularly clear-cut and successful example in the context
of supervised learning: the so-called Matrix Relevance LVQ for classification
[SBH09].",The+Shallow+and+the+Deep.pdf
"of supervised learning: the so-called Matrix Relevance LVQ for classification
[SBH09].
Similar to several other schemes (e.g. [WS09, BAP+12, BCLC15]), Matrix
Relevance LVQ employs a generalized quadratic distance of the form
δΛ(w, ξ) = (w − ξ)⊤ Λ (w − ξ) =
N󰁛
i,j=1
(wi − ξi) Λij (wj − ξj). (6.13)
Heuristically, diagonal entries of Λ quantify the importance of single feature
dimensions in the distance and can also account for potentially different mag-
nitudes of the features. Pairs of features are weighted by off-diagonal elements,
which reflect the interplay of the different dimensions. Note that for Λ = IN /N,
Eq. (6.13) recovers the simple squared Euclidean distance.
In order to fulfill the minimal requirement of non-negativity, δΛ ≥ 0, a con-
venient re-parameterization is introduced in terms of an auxiliary, unrestricted
matrix Ω ∈ RN×N :
Λ = Ω⊤Ω, i.e. δΛ(w, ξ) = [Ω (w − ξ)]2 . (6.14)
Hence, δΛ can be interpreted as the conventional squared Euclidean distance",The+Shallow+and+the+Deep.pdf
"matrix Ω ∈ RN×N :
Λ = Ω⊤Ω, i.e. δΛ(w, ξ) = [Ω (w − ξ)]2 . (6.14)
Hence, δΛ can be interpreted as the conventional squared Euclidean distance
but after a linear transformation of feature space. Note that Eqs. (6.13, 6.14)
define only a pseudo-metric in RN since Λ can be singular with rank(Λ) < N
implying that δΛ(w, ξ) = 0 is possible even if w ∕= ξ.
Obviously, we could employ a fixed distance of the form (6.13) in GLVQ or
LVQ1 as outlined in the previous sections. The key idea of relevance learning,
however, is to consider the elements of the relevance matrix Λ ∈ RN×N as
adaptive quantities which can be optimized in the data-driven training process.
Numerous simplifications or extensions of the basic idea have been sug-
gested in the literature. The restriction to diagonal matrices Λ corresponds
to the original formulation of Relevance LVQ in [HV02], which assigns a single,
non-negative weighting factor to each dimension in feature space. Rectangular",The+Shallow+and+the+Deep.pdf
"non-negative weighting factor to each dimension in feature space. Rectangular
(M × N)-matrices matrices Ω with M < N can be used to parameterize a low-
rank relevance matrix [BSH+12]. The corresponding low-dimensional intrinsic
representation of data facilitates, for instance, the class-discriminative visualiza-
tion of complex data [BSH+12]. The flexibility of the LVQ classifier is enhanced",The+Shallow+and+the+Deep.pdf
"6.2. DISTANCE MEASURES AND RELEVANCE LEARNING 151
significantly when local distances are used, i.e. when separate relevance matrices
are employed per class or even per prototype [SBH09,BSH+12].
Here we restrict the discussion to the simplest case of a single, N ×N matrix
Ω corresponding to a global distance measure. The heuristic extension of the
LVQ1 prescription by means of relevance matrices is briefly discussed in [BHV16]
and its convergence behavior is analysed in [BHS+16].
Gradient based updates for the simultaneous adaptation of prototypes and
relevance matrix can be derived from a suitable cost function. We observe that
∂δΛ(w, ξ)
∂w = Ω⊤Ω(ξ−w) and ∂δΛ(w, ξ)
∂Ω = Ω (w − ξ) (w − ξ)⊤ . (6.15)
The full forms of the gradients with respect to the terms eµ in the GLVQ
cost function are presented in [SBH09], for instance. They yield the so-called
Generalized Matrix Relevance LVQ (GMLVQ) scheme, which can be formulated
as a stochastic gradient descent procedure:",The+Shallow+and+the+Deep.pdf
"Generalized Matrix Relevance LVQ (GMLVQ) scheme, which can be formulated
as a stochastic gradient descent procedure:
Generalized Matrix LVQ (GMLVQ), stochastic gradient descent
– at time step t, select a single feature vector ξµ with class label yµ
randomly from the data set D with uniform probability 1/P.
– with respect to the distance δΛ (6.13) with Λ = Ω(t)⊤Ω(t),
identify the correct and incorrect winners, i.e. the prototypes
wJ
µ with δΛ(wJ
µ, ξµ) = min
󰁱
δΛ(wj, ξµ)
󰀏󰀏 cj
µ = yµ
󰁲M
j=1
wK
µ with δΛ(wK
µ , ξµ) = min
󰁱
δΛ(wj, ξµ)
󰀏󰀏 cj
µ ∕= yµ
󰁲M
j=1
(6.16)
with class labels cJ
µ = yµ and cK
µ
∕= yµ, respectively.
– update both winning prototypes and the matrix Ω according to:
wJ
µ(t + 1) = wJ
µ(t) − ηw
∂Φ(eµ)
∂wJµ
wK
µ (t + 1) = wK
µ (t) − ηw
∂Φ(eµ)
∂wKµ
,
Ω(t + 1) = Ω(t) − ηΩ
∂Φ(eµ)
∂Ω . (6.17)
where the gradients are evaluated in Ω(t) and wL
µ (t) for L = J, K.
In both GMLVQ and Matrix LVQ1, the relevance matrix is updated in order
to decrease or increase δΛ(wL",The+Shallow+and+the+Deep.pdf
"µ (t) for L = J, K.
In both GMLVQ and Matrix LVQ1, the relevance matrix is updated in order
to decrease or increase δΛ(wL
µ , ξµ) for the winning prototype(s), depending on
the class labels in the, by now, familiar way.",The+Shallow+and+the+Deep.pdf
"152 6. DISTANCE-BASED CLASSIFIERS
Figure 6.2: Visualization of the Generalized Matrix Relevance LVQ system
as obtained from the z-score transformed Iris flower data set, see Sec. 6.2.2 for
details.
Left panel: Class prototypes are shown as bar plots with respect to the four
feature space components in the left column. The right column shows the eigen-
value spectrum of Λ, the diagonal elements of Λ, and the off-diagonal elements
in a gray-scale representation (top to bottom).
Right panel: Projection of the P = 150 feature vectors onto the two leading
eigenvectors of the relevance matrix Λ.
Frequently, the learning rate of the matrix updates is chosen to be relatively
small, ηΩ ≪ ηw, in the stochastic gradient descent procedure. This follows the
intuition that the prototypes should be enabled to follow changes in the distance
measure. The relative scaling can be different in batch gradient realizations of
GMLVQ as for instance in [VWB21].",The+Shallow+and+the+Deep.pdf
"measure. The relative scaling can be different in batch gradient realizations of
GMLVQ as for instance in [VWB21].
The matrix Ω can be initialized as the N-dim. identity or in terms of indepen-
dent random elements. In order to avoid numerical difficulties, a normalization
of the form 󰁓
i Λii = 󰁓
i,j Ω2
i,j = 1 is frequently imposed [SBH09].
In the following we illustrate Matrix Relevance LVQ in terms of a classical
benchmark data set. In the famous Iris flower data set [Fis36], four numerical
features are used to characterize 150 samples from three different classes which
correspond to particular species of Iris flowers. We obtained the data set as
provided at [Lic13] and used one prototype per class and a global relevance
matrix Λ ∈ R4×4. For the training, we employed the freely available beginner’s
toolbox for GMLVQ with default parameter setting [VWB21]. An additional
z-score transformation was applied, resulting in re-scaled features with zero",The+Shallow+and+the+Deep.pdf
"toolbox for GMLVQ with default parameter setting [VWB21]. An additional
z-score transformation was applied, resulting in re-scaled features with zero
mean and unit variance in the data set, see Sec. 8.1.1. This allows for an
immediate interpretation of the relevances, without having to take into account
the potentially different magnitudes of the features.
Figure 6.2 visualizes the obtained classifier. The resulting LVQ system
achieves almost perfect, error-free classification of the training data. It also",The+Shallow+and+the+Deep.pdf
"6.3. CONCLUDING REMARKS 153
displays very good generalization behavior with respect to validation or test set
performance not presented here.
In the left panel, the prototypes after training and the resulting relevance
matrix and its eigenvalues are displayed. As discussed above, the diagonal ele-
ments Λii can be interpreted as the relevance of features i in the classification.
Apparently, features 3 and 4 are the dominant ones in the Iris classification prob-
lem. The off-diagonal elements represent the contribution of pairs of different
features. Here, also the interplay of features 3 and 4 appears to be important.
In more realistic and challenging data sets, Relevance Matrix LVQ can pro-
vide valuable insights into the problem. GMLVQ has been exploited to identify
the most relevant or irrelevant features, e.g. in the context of medical diag-
nosis problems, see [Bie17] for a variety of applications. A recent application",The+Shallow+and+the+Deep.pdf
"nosis problems, see [Bie17] for a variety of applications. A recent application
in the context of galaxy classification based on astronomical catalogue data is
presented in [NWB18,NWB+19].
In an N-dimensional feature space, the GMLVQ relevance matrix intro-
duces O(N2) additional adaptive quantities. As a consequence, one might
expect strong overfitting effects due to the large number of free model parame-
ters. However, as observed empirically and analysed theoretically, the relevance
matrix displays a strong tendency to become singular and displays very low
rank(Λ) = O(1) ≪ N after training [BHS+16]. This effect can be interpreted as
an implicit, intrinsic mechanism of regularization, which limits the complexity
of the distance measure, effectively.
In addition, the low rank relevance matrix allows for the discriminative vi-
sualization of the data by projecting feature vectors (and prototypes) onto its
leading eigenvectors. As an illustrative example, Fig. 6.2 displays the Iris flower",The+Shallow+and+the+Deep.pdf
"leading eigenvectors. As an illustrative example, Fig. 6.2 displays the Iris flower
data set.
6.3 Concluding remarks
Prototype-based models continue to play a highly significant role in putting
forward advanced machine learning techniques. We encourage the reader to
explore recent developments in the literature. Challenging problems, such as
the analysis of functional data, non-vectorial data or relational data, to name
only very few, are currently being addressed, see [BHV16, NE14] for further
references. At the same time, exciting application areas are being explored in a
large variety of domains.
Most recently, prototype-based systems are also re-considered in the context
of Deep Learning [GBC16,Sch15,LBH18,Hue19]. The combination of multilayer
network architectures with prototype- and distance-based modules appears very
promising and is the subject of on-going research, see, for instance, [SHRV18,
VMC16] and references therein.",The+Shallow+and+the+Deep.pdf
154 6. DISTANCE-BASED CLASSIFIERS,The+Shallow+and+the+Deep.pdf
"Chapter 7
Model evaluation and
regularization
Accuracy is not enough.
— Paulo Lisboa
In supervised learning the aim is to infer relevant information from given
data, to parameterize it in terms of a model, and to apply it to novel data
successfully. It is obviously essential to know or at least have some estimate of
the performance that can be expected in the working phase.
In this chapter we discuss several aspects related to the evaluation and val-
idation of supervised learning. In Sec. 7.1, we take a rather general perspective
on overfitting and underfitting effects without necessarily addressing a partic-
ular classifier or regression framework. We present methods for controlling the
complexity of neural networks in Sec. 7.2. Cross-validation and related methods
are in the focus of Sec. 7.3. Specific quality measures beyond the simple overall
accuracy for the evaluation of classifiers and regression systems are presented",The+Shallow+and+the+Deep.pdf
"accuracy for the evaluation of classifiers and regression systems are presented
in 7.4.3. Finally, we address the importance of interpretable models in machine
learning in Sec. 7.5.
7.1 Bias and variance, over- and underfitting
Different sources of error can influence the performance of supervised learning
systems. Here we decompose the expected prediction error into two main con-
tributions, see e.g. [HTF01,Bis06]: the so-called bias corresponds to systematic
deviations of trained models from the true target, while the term variance refers
to variations of the model performance when trained from different realizations
155",The+Shallow+and+the+Deep.pdf
"156 7. MODEL EVALUATION AND REGULARIZATION
K = 1 K = 3 K = 7
DA
DB
DC
Figure 7.1: Illustration of the bias-variance dilemma in regression. In each
row of graphs, a particular set of 10 points {xi, yi}10
i=1 is approximated by least
square linear regression (K = 1), by a cubic fit (K = 3), and by fitting a
polynomial of degree seven (K = 7). Rows correspond to three randomized,
independently generated data sets DA,B,C.
of the training data1.
Frequently, a so-called irreproducible error is considered as a third, inde-
pendent contribution [HTF01]. It could, for instance, stem from intrinsic noise
in the test data which cannot be predicted even with perfect knowledge of the
target rule. As the irreproducible error is beyond our control anyway, we refrain
from including it in the discussion.
7.1.1 Decomposition of the error
For the purpose of illustration, we consider a simple one-dimensional regression
problem. The obtained insights, however, carry over to much more complex",The+Shallow+and+the+Deep.pdf
"problem. The obtained insights, however, carry over to much more complex
systems. In our example, least squares fits are based on data sets of the form
D = {xµ, yµ}P
µ=1. They contain real-valued arguments xµ ∈ R , e.g. equidistant
values in [−1, 1], and their corresponding target labels yµ ∈ R. The data sets
1Note that the terms bias and variance are used in many different scientific contexts with
area specific meanings.",The+Shallow+and+the+Deep.pdf
"7.1. BIAS AND VARIANCE, OVER- AND UNDERFITTING 157
represent a function f(x) which is of course unknown to the learning system.
We assume that the training labels are noisy versions of the true targets:
yµ = f(xµ) + rµ with 〈rµ〉 = 0 and 〈rµ rν〉 = ρ2δµν (7.1)
with the Kronecker-delta δµν. Hence, the deviation of the training labels from
the underlying target function is given by uncorrelated, zero mean random quan-
tities rµ in each data point. Further details are irrelevant for the argument, but
we could, for instance, consider independent Gaussian random numbers with
variance ρ2, i.e. rµ ∼ N(0, ρ). We perform polynomial fits of the form
fK(x) =
K󰁛
j=0
ajxj with coefficients aj ∈ R (7.2)
for powers xj with maximum degree K. Given a data set D = {xµ, yµ}P
µ=1, the
aj can be determined by minimizing the familiar quadratic deviation
ESSE = 1
2
P󰁛
µ=1
󰀓
fH(xµ) − yµ
󰀔2
. (7.3)
Fig. 7.1 displays three randomized data sets DA,B,C with P = 11 equidistant",The+Shallow+and+the+Deep.pdf
"ESSE = 1
2
P󰁛
µ=1
󰀓
fH(xµ) − yµ
󰀔2
. (7.3)
Fig. 7.1 displays three randomized data sets DA,B,C with P = 11 equidistant
xµ and noisy yµ representing the underlying target function f(x) = x3. For
each of the three slightly different data sets, polynomial least square fits were
performed with K = 1 (linear), K = 3 (cubic) and with degree K = 7. Hence,
the same data sets were analysed by using models of different complexity.
In order to obtain some insight into the interplay of model complexity and
expected performance, we consider the thought experiment of performing the
same training/fitting processes for a very large number of slightly different data
sets of the same size, which all represent the target.
We denote by 〈. . .〉D an average over many randomized realizations of the
data set or – more formally – over the probability density of the training data in
D. In this sense, the expected total quadratic deviation of a hypothesis function",The+Shallow+and+the+Deep.pdf
"D. In this sense, the expected total quadratic deviation of a hypothesis function
fH from the true target f in an arbitrary point x ∈ R is given by
󰁇󰀅
fH(x) − f(x)
󰀆2󰁈
D
, (7.4)
where the randomness of D is reflected in the outcome fH of the training. We
could also consider the integrated deviation over a range of x-values, which
would relate the SSE to the familiar generalization error. However, the following
argument would proceed in complete analogy due to the linearity of, both,
integration and averaging. Performing the square in Eq. (7.4) we obtain
󰀍
f2
H(x)
󰀎
D − 2 〈fH(x)〉D f(x) + f2(x). (7.5)
Note that the true target f(x) obviously does not depend on the data and can
be left out from the averages.",The+Shallow+and+the+Deep.pdf
"158 7. MODEL EVALUATION AND REGULARIZATION
For the sake of brevity, we omit the argument x ∈ R of the functions fH and
f in the following. Including redundant terms (*) which add up to zero we can
rewrite (7.5) as
〈fH〉2
D󰁿 󰁾󰁽 󰂀
∗
−2 〈fH(x)〉D f + f2 +
󰀍
f2
H
󰀎
D −2 〈fH〉2
D
+ 〈fH〉2
D
󰁿 󰁾󰁽 󰂀
∗
(7.6)
and obtain a decomposition of the expected quadratic deviation in x:
󰁇󰀅
fH −f
󰀆2󰁈
D
=
󰀓
f −〈fH〉D
󰀔2
󰁿 󰁾󰁽 󰂀
bias
2
+
󰀟󰀓
fH −〈fH〉D
󰀔2󰀠
D󰁿 󰁾󰁽 󰂀
variance
. (7.7)
The equality with (7.6) is straightforward to show by expanding the squares
and exploiting that 〈fH 〈fH〉D〉D = 〈fH〉2
D
=
󰁇
〈fH〉2
D
󰁈
D
.
Hence we can identify two contributions to the total expected error:
◦ Bias (squared): (f − 〈fH〉D)2
The bias term quantifies the deviation of the mean prediction from the
true target, where the average is over many randomized data sets and
corresponding training processes. A small bias indicates that there is very
little systematic deviation of the hypotheses from the unknown target rule.
◦ Variance:
󰁇󰀃",The+Shallow+and+the+Deep.pdf
"little systematic deviation of the hypotheses from the unknown target rule.
◦ Variance:
󰁇󰀃
fH − 〈fH〉D
󰀄2󰁈
D
The variance measures how much the individual predictions, obtained after
training on a given D, typically differ from the mean prediction. The
observation of a small variance implies that the outcome of the learning
is robust with respect to details of the training data.
Similar considerations apply to more general learning problems, including
classification schemes [Dom00].
7.1.2 The bias-variance dilemma
Ideally, we would like to achieve low variance and low bias at the same time, i.e.
a robust and faithful approximation of the target rule. Both goals are clearly
legitimate, but very often they constitute conflicting aims in practice, as further
illustrated in the following.
This is often referred to as the bias-variance dilemma or trade-off [HTF01,
Bis95a, Bis06, Dom00]. It is closely related to the problem of overfitting in",The+Shallow+and+the+Deep.pdf
"Bis95a, Bis06, Dom00]. It is closely related to the problem of overfitting in
unnecessarily complex systems and its counterpart, the so-called underfitting in
simplistic models.
The concept of bias and variance is illustrated in Fig. 7.2 (left panel). In
the illustration, the fits of two different models are displayed in the space of
adaptive quantities, e.g. weights in a neural network, while bias and variance
are defined in terms of the prediction error. However, we can assume that the
deviation in weight space from the target is in general associated with the error.",The+Shallow+and+the+Deep.pdf
"7.1. BIAS AND VARIANCE, OVER- AND UNDERFITTING 159
model B
target
fits of model A
model parameter 2
adaptive parameter 1
prediction error
test set
performance
← high bias high variance →
training
set performance
a measure of the model complexity
Figure 7.2: Left panel: Illustration of the bias-variance Dilemma. The true
target is represented by the filled circle in the center. Fits obtained from dif-
ferent data sets Di in model A (open circles) show low bias and large variance,
while model B (filled circles) displays large systematic bias with smaller vari-
ance. Right panel: Schematic illustration of underfitting and overfitting (af-
ter [HTF01]): expected error with respect to a test set (generalization error)
and training set performance as a function of the model complexity, e.g. K in
the polynomial fits of Fig. 7.1.
Overfitting: low bias – high variance
In terms of our polynomial regression example we can achieve low bias by em-",The+Shallow+and+the+Deep.pdf
"the polynomial fits of Fig. 7.1.
Overfitting: low bias – high variance
In terms of our polynomial regression example we can achieve low bias by em-
ploying powerful models with large degree K. In the extreme case of K = P,
for instance, we can generate models which perfectly reproduce the data points,
fK(xµ) = yµ for all µ, in each individual training process.
Because the training labels themselves are assumed to be unbiased with
〈yµ〉D = f(xµ), cf. Eq. (7.1), the averaged fit result will also be in exact
agreement with the target in the arguments xµ. In fact, since the objective
function (7.3) of the training treats positive and negative deviations symmetri-
cally as well, there is no reason to expect systematic deviations of the fits with
fK(x) > f(x) or fK(x) < f(x) for all fits in some arbitrary value of x.
However, using a very flexible model with large K will result in fits which
are very specific to the individual data set. As can be seen in Fig. 7.1, (right",The+Shallow+and+the+Deep.pdf
"are very specific to the individual data set. As can be seen in Fig. 7.1, (right
column), already for a moderate degree of K = 7, very different models emerge
from the individual training processes.
For the sample points themselves, the variance of the nearly perfect fit would
be essentially determined by the statistical variance of the training labels ρ2 in
Eq. (7.1). However, when interpolating or even extrapolating to x ∈/ {xµ}P
µ=1,
the different fits will vary a lot in their prediction fK(x). Consider, for instance,
the extrapolation to x = ±1.1 in Fig. 7.1 as a pronounced example of the effect.
Underfitting: low variance – high bias
If emphasis is put on the robustness of the model, i.e. low variance, we would
prefer simple models with low degree K in (7.2). This should prevent the fits",The+Shallow+and+the+Deep.pdf
"160 7. MODEL EVALUATION AND REGULARIZATION
from being overly specific to the individual data sets. As illustrated in the left
column of Fig. 7.1, we achieve nearly identical linear models from the different
data sets. However, a price is paid for the robustness: systematic deviations
occur in each training procedure. We observe, for instance, that the linear
fits (left column) obtained from DA,B,C are virtually identical. However, they
display quite large deviations from the sample points – which represent a non-
linear function, after all. These deviations are systematic in the sense that
they are reproduced qualitatively in each data set. For instance, for the first
sample point x1 = −1 we can see that always fK=1(x) > y1. As a consequence,
interpolation and extrapolation will also be subject to systematic errors.
Matched model complexity
In our example, fits of degree K = 3 seem to constitute an ideal compromise.
In the sample points, they achieve small deviations with no systematic ten-",The+Shallow+and+the+Deep.pdf
"In the sample points, they achieve small deviations with no systematic ten-
dency and, consequently, have relatively low bias. At the same time the fits
fK=3(x) appear also robust against variations of the data set, corresponding to
a relatively small variance.
This is of course not surprising, since polynomials of degree K = 3 perfectly
match the complexity of the underlying, true target function. It is important to
realize that this kind of information is rarely available in practical situations.
In fact, in absence of knowledge about the complexity of the target rule, it is
one of the key challenges in supervised learning to select an appropriate model
that achieves a good compromise with respect to bias and variance. In the
following sections we will consider a variety of ways to control the complexity
of a learning system with emphasis on feed-forward neural networks.
The trade-off
The above considerations suggest that there is a trade-off between the goals of",The+Shallow+and+the+Deep.pdf
"The trade-off
The above considerations suggest that there is a trade-off between the goals of
small variance and bias [HTF01, NMB+18, Dom00]. Indeed, in many machine
learning scenarios one observes such a trade-off which is illustrated in Fig. 7.2
(right panel). It shows schematically the possible dependence of the prediction
performance in the training set and the generalization error (test set perfor-
mance) as a function of the model complexity.
In our simple example, we could use the polynomial degree K as a measure
of the latter. It could be also interpreted as, for instance, the degree of a
polynomial kernel in the SVM, the number of hidden units in a two-layer neural
network, or the number of prototypes per class in an LVQ system. Similarly, the
x-axis could correspond to a continuous parameter that controls the flexibility
of the training algorithm, e.g. a weight decay parameter or the training time
itself [HKP91,Bis95a,Bis06].",The+Shallow+and+the+Deep.pdf
"of the training algorithm, e.g. a weight decay parameter or the training time
itself [HKP91,Bis95a,Bis06].
Generically, we expect the training error to be lower than the generalization
error for any model. After all, the actual optimization process is based on the
available training examples.
Simplistic models that cannot cope with the complexity of the task display,
both, poor training set and d due to large systematic bias. Increasing the",The+Shallow+and+the+Deep.pdf
"7.1. BIAS AND VARIANCE, OVER- AND UNDERFITTING 161
model’s flexibility will reduce the bias and, consequently, training and test set
error decrease with K in Fig. 7.2 (right panel). However, overly training set
specific models display overfitting: while the training error typically decreases
further with increasing K, the test set error displays a U-shaped dependence
which reflects the increase of the model variance.
It is important to realize that the extent to which the actual behavior follows
the scenario in a practical situation depends on the detailed properties of the
data and the problem at hand. While one should be aware of the possible
implications of the bias-variance dilemma, the plausibility of the above discussed
trade-off must not be over-interpreted as a mathematical proof. Note that the
decomposition (7.7) itself does not imply the existence of a trade-off, strictly
speaking.
As argued and demonstrated in e.g. [Sch93] and [NMB+18], a given practical",The+Shallow+and+the+Deep.pdf
"speaking.
As argued and demonstrated in e.g. [Sch93] and [NMB+18], a given practical
problem does not necessarily display the U-shaped dependence of the general-
ization error shown in Fig. 7.2 (right panel). There is also no general guarantee
that measures which reduce the variance in complex models will really improve
the performance of the system [Sch93]. In many practical problems, however,
the assumed bias-variance trade-off can indeed be controlled to a certain degree
and may serve as a guiding principle for the model selection process.
According to the above considerations, a reliable estimate of the expected
generalization ability would be highly desirable in any given supervised learning
scenario. It would be very useful to be able to compare and evaluate the use of
different approaches, e.g. SVM and LVQ, in a given practical problem. Similarly,
the expected performance should guide the selection of model parameters like",The+Shallow+and+the+Deep.pdf
"the expected performance should guide the selection of model parameters like
the number of hidden units in a neural network. The aim is to select the most
suitable student complexity in a situation as sketched in Fig. 7.2. The same
applies to selecting a training procedure and suitable parameter values, e.g. the
learning rate.
In Sec. 7.3 we present the basic idea of how to obtain estimates of the
generalization performance by means of cross-validation and related schemes.
7.1.3 Beyond the classical bias-variance trade-off (?)
The unreasonable effectiveness of Deep Learning in Artificial Intelligence [Sej20]
seems to raise some doubts about the validity of the bias-variance trade-off.
Deep Learning systems are frequently heavily over-parameterized with very large
numbers of layers, units and weights. For instance, the currently very popular
Large Language Models can comprise billions of adaptive parameters, see Table",The+Shallow+and+the+Deep.pdf
"Large Language Models can comprise billions of adaptive parameters, see Table
2.1 in the preprint version of [BMR+20]. According to the reasoning of the
previous sections, one would expect serious overfitting effects in such extremely
powerful systems. In practice, however, over-parameterized Deep Learning sys-
tems are trained and applied with great success.
In this context, a publication by Belkin et al. [BHMM19] has attracted a
lot of attention. The authors discuss the so-called double descent or peaking
phenomenon which is illustrated in Fig. 7.3. Beyond the classical under- and
overfitting scenario of Fig. 7.2, the test error frequently undergoes a second",The+Shallow+and+the+Deep.pdf
"162 7. MODEL EVALUATION AND REGULARIZATION
prediction error test error
training
error
number of parameters, model complexity
interpolation
threshold
“classical”
regime
“modern”
regime
Figure 7.3: Illustration of the double descent phenomenon, after [BHMM19].
descent as a function of the number of adaptive parameters. This descent occurs
in the over-parameterized regime, while the test error displays a peak at the so-
called interpolation threshold. The illustration 7.3 refers to what is sometimes
called model-wise double descent: for a given problem or data set, models of
increasing complexity are considered. Similar peaking effects can be observed
in models of fixed complexity with varying data set size in the so-called sample-
wise double descent [You21,LVM+20,Vie23].
Double descent is the subject of ongoing discussions and apparently has led
several researchers and practicioners to the somewhat hasty conclusion that the",The+Shallow+and+the+Deep.pdf
"several researchers and practicioners to the somewhat hasty conclusion that the
bias-variance trade-off generalization is wrong [You21]. Moreover, it is often
assumed that double descent is a relatively novel phenomenon that was discov-
ered specifically in Deep Learning, motivating the terms classical and modern
regime in Fig. 7.3. However, as already mentioned in [BHMM19], double descent
occurs also in much simpler settings, including shallow networks and elemen-
tary regression systems. In [LVM+20], the authors present a brief prehistory of
double descent, pointing out that it had been observed already in basic learning
problems like linear regression or perceptron training, e.g. [OKKN90].
Plausible explanations for the occurrence of double descent have been pro-
vided by several authors. It is important to note that, depending on the details
of problem and method, it is incorrect to naively identify the number of parame-",The+Shallow+and+the+Deep.pdf
"of problem and method, it is incorrect to naively identify the number of parame-
ters with the capacity or complexity of the model, as we suggestively did in Fig.
7.3. As one example, Daniela Witten presents an insightful discussion in terms
of fitting cubic splines to a number of data points in [Wit20] (a twitter thread).
There, the interpolation threshold corresponds to the situation in which the
number of parameters exactly matches the number of data points. This is anal-
ogous to other regression or interpolation schemes, such as the poynomial fits
discussed in Sec. 7.1. Right at the interpolation threshold, there is only one
possible solution for a given data set and the resulting model is very sensitive
to small variations or noise. Above the interpolation threshold, many fits are
possible. The specific selection of the solution with the minimum norm of coeffi-
cients restricts the flexibility of the system drastically, resulting in the observed",The+Shallow+and+the+Deep.pdf
"cients restricts the flexibility of the system drastically, resulting in the observed
peaking and double descent. On the contrary, fitting under other types of reg-",The+Shallow+and+the+Deep.pdf
"7.2. CONTROLLING THE NETWORK COMPLEXITY 163
ularization, cf. Sec. 7.2, would not necessarily display the peaking and double
descent [Wit20]. The influence of implicit and explicit regularization on the
emergence of double descent is also discussed in the context of ordinary least
squares regression in [KLS20].
In general, explicit regularization as well as details of the training prescrip-
tion can play an important role in determining the effective flexibility of a sys-
tem. The important conclusion is that, if the model complexity is taken into
account correctly, the bias-variance trade-off is still valid.
7.2 Controlling the network complexity
So far we have discussed the control of the student complexity in terms of the
actual model design, i.e. by choosing the degree of a polynomial fit, the number
of prototypes in LVQ, or the size of a hidden layer in a neural network. These
choices are made prior to the actual training and can be evaluated by comparing",The+Shallow+and+the+Deep.pdf
"choices are made prior to the actual training and can be evaluated by comparing
different settings after training.
In a variety of approaches the (effective) complexity of a learning system
is controlled by imposing constraints on the training process in a given archi-
tecture. Appropriate restrictions can prevent the training algorithm from fully
exploring the space of adaptive quantities. We discuss two basic methods: the
so-called early stopping strategy in Sec. 7.2.1 and the concept of weight decay
in 7.2.2, respectively. The latter is an important example for regularization by
introducing a penalty term into the objective function that guides the train-
ing. Here, we will use the term regularization more generally for all methods of
implicit or explicit complexity control.
Constructive algorithms which incorporate the addition of units or layers
into the training process are discussed in Sec. 7.2.3. In Sec. 7.2.4 we present",The+Shallow+and+the+Deep.pdf
"into the training process are discussed in Sec. 7.2.3. In Sec. 7.2.4 we present
so-called pruning procedures that remove unnecessary weights or units during
or after training. Eventually, two techniques that are particular relevant in the
context of Deep Learning, weight-sharing and Dropout, are presented in Sec.
7.2.5 and 7.2.6, respectively.
In practice, all these methods require or benefit from reliable estimates of
the performance with respect to the prediction on novel data. For now we
assume that such estimates are available, for instance by computing suitable
error measures on a large representative test set. Practical methods like the
well-known n-fold cross-validation will be discussed in Sec. 7.3.
7.2.1 Early stopping
A conceptually very simple idea is to end the training process before the system
becomes overly specific to the data set. For example, if we manually stop gradi-
ent descent updates after a suitable number tmax of epochs, the resulting weight",The+Shallow+and+the+Deep.pdf
"ent descent updates after a suitable number tmax of epochs, the resulting weight
configuration may display a relatively low value of the objective function, yet
without representing one particular local minimum too faithfully. Fig. 7.4 (left",The+Shallow+and+the+Deep.pdf
"164 7. MODEL EVALUATION AND REGULARIZATION
W(0) = 0
W(tmax)
W(0)=0
W∗W∗
Figure 7.4: Schematic illustration of early stopping and weight decay. Ellipses
correspond to contour lines of the objective function. The blue solid lines rep-
resent the unrestricted hypothetical updates by, for instance, gradient descent.
Left panel: after tmax epochs, the training is stopped, which hinders the weight
configuration from reaching the local minimum W∗ (red dot).
Right panel: weights are initialized tabula rasa and restricted to small norms.
Circles correspond to lines of equal norm |W|.
panel) shows an illustration of the effect of early stopping on the optimization
process.
Being one of the most intuitive concepts of regularization, early stopping
has been discussed very early in the context of neural networks, see for example
[BC91,SL92]. A thorough discussion and further references can be found in text
books as well, e.g. in [Bis06,GBC16].",The+Shallow+and+the+Deep.pdf
"[BC91,SL92]. A thorough discussion and further references can be found in text
books as well, e.g. in [Bis06,GBC16].
The early stopping parameter tmax plays a role that is comparable to the
degree K in the example of polynomial fits, cf.7.1. Note that tmax and other
parameters that control the training process are often referred to as hyper-
parameters in order to distinguish them from the actual adaptive quantities,
e.g. weights and thresholds in a neural network.
In order to set discrete parameter like the number of hidden units in the
network, we have to train different systems separately and compare their train-
ing and test set retrospectively. In early stopping we can monitor the system
on the fly and stop as soon as overtraining effects set in. The proper choice of
tmax based on heuristic criteria and the use of cross-validation, cf. Sec. 7.3.1 has
been addressed in the literature, see e.g. [Pre97, Pre98, STW11] for examples
and [GBC16,Bis06] for general discussions.",The+Shallow+and+the+Deep.pdf
"been addressed in the literature, see e.g. [Pre97, Pre98, STW11] for examples
and [GBC16,Bis06] for general discussions.
7.2.2 Weight decay and related concepts
We have encountered weight decay as a regularization technique already in the
discussion of simple linear regression in Sec. 2.2.2. There, a penalty term was
added to the objective function that prevents the L2-norm of the weight vec-
tor from growing arbitrarily large. Depending on the perspective, this can be
motivated heuristically in a pragmatic machine learning approach or on the",The+Shallow+and+the+Deep.pdf
"7.2. CONTROLLING THE NETWORK COMPLEXITY 165
basis of assuming prior knowledge in the context of statistical learning the-
ory [HTF01, HKP91, Bis95a, Bis06, DHS00]. In the linear regression problem,
weight decay facilitates the construction or computation of a meaningful solu-
tion of the regression problem.
Here we extend the concept to the implementation of non-linear functions in
non-linear networks. Modifications of weight decay, e.g. by considering general
Minkowski-norms or heuristically motivated penalty terms are discussed at the
end of this subsection.
Weight decay in non-linear neural networks
The concept of weight decay generalizes to a variety of classification and re-
gression problems, including the training of non-linear layered neural networks,
see [Hin86,KSV88] for early works. The influence of weight decay has also been
investigated in model scenarios from the statistical physics of learning perspec-
tive, see [Bös96,ABS99,SR98] for examples.",The+Shallow+and+the+Deep.pdf
"investigated in model scenarios from the statistical physics of learning perspec-
tive, see [Bös96,ABS99,SR98] for examples.
Compared to the case of linear regression, the formulation of weight decay
based on statistical learning theory, is less obvious for non-linear neural net-
works. However, the heuristic interpretation remains valid: restricting the mag-
nitude of weights prevents the system from exploring the search space exhaus-
tively and thus limits the effective complexity of the network. Figure 7.4 (right
panel) illustrates the effect of limiting the Euclidean L2-norm of the weights W.
The effect of weight decay can be motivated in terms of a single non-linear
unit. Assume that the activation of the unit is given by the non-linear function
g(x) ∈ R with x = w · ξ, w, ξ ∈ RN . (7.8)
For weight vectors with small norm |w| ≈ 0 we have x ≈ 0 and a Taylor
expansion implies that the activation is approximately
g(x) ≈ g(0) + g′(0) x + 1
2g′′ x2 + . . . (7.9)",The+Shallow+and+the+Deep.pdf
"expansion implies that the activation is approximately
g(x) ≈ g(0) + g′(0) x + 1
2g′′ x2 + . . . (7.9)
Thus, the activation is effectively linearized. By the same argument, the output
of a layered network of in principle non-linear units, will become nearly linear
if the magnitude of the weights are very small.
If we represent the set of all weights in a network by W and consider gra-
dient descent with respect a cost function E(W), we can limit the norm of W
heuristically by reducing the magnitude of weights in or after each update step:
W(t + 1) = W(t)
󰁫
1 − γ
󰁬
− η ∇W E|W(t) (7.10)
with the (small) weight decay parameter γ > 0. The update can also be inter-
preted as gradient descent with respect to a modified objective function:
󰁥E = E + 1
2
γ
η |W|2 with ∇W 󰁥E = ∇W E + γ
η W (7.11)",The+Shallow+and+the+Deep.pdf
"166 7. MODEL EVALUATION AND REGULARIZATION
which also leads to (7.10).2
As an alternative to the use of a penalty term, sometimes a constraint of
the type |W|2 ≤ c with constant c > 0 is imposed. This can be done explicitly
by projecting back onto the sphere in weight space |W|2 = c whenever the
constraint is violated by the updates. This so-called max-norm regularization
is considered in [SHK+14] in combination with Dropout, see Sec. 7.2.6. Weight
decay as given by Eqs. (7.10, 7.11) can be interpreted as a soft implementation
with Lagrange parameter γ/η.
Variants of weight decay
Following an argument presented in e.g. [HKP91] and [NH92b], the penalty term
∝ 󰁓
j W2
j in Eq. (7.11) favors several non-zero weights of similar magnitude
over a combination of zero weights with a few larger Wj. This can be seen by
comparing the penalty of a pair of weights {w/2, w/2} to that of {w, 0}:
󰀓w
2
󰀔2
+
󰀓w
2
󰀔2
< w2 + 02.
In the context of sparse classifiers or regression systems the aim is a system",The+Shallow+and+the+Deep.pdf
"󰀓w
2
󰀔2
+
󰀓w
2
󰀔2
< w2 + 02.
In the context of sparse classifiers or regression systems the aim is a system
with a significant fraction of zero or very small weights, which could be removed.
This can be achieved by employing modified penalty terms and update rules,
for instance the one discussed in [HKP91]:
󰁥E = E + 1
2
γ
η
󰁛
j
W2
j
1 + W2
j
with ∂ 󰁥E
Wk
= γ
η
Wk
(1 + W2
k )2 . (7.12)
Compared to (7.11) this leads to a Wk-dependent decay term in the gradient de-
scent updates which favors the further decrease of small weights. Consequently,
the modified weight decay, together with a removal of weights with |Wj| ≈ 0
after training, can serve as a method for pruning the neural network. Further
methods for the removal of unnecessary weights in a trained network are briefly
presented in Sec. 7.2.4.
A family of systematic variations of weight decay can be based on Lp-
regularization, where the penalty is given by the more general term
||W||p =
󰀓󰁛
j
Wp
j
󰀔1/p
. (7.13)",The+Shallow+and+the+Deep.pdf
"regularization, where the penalty is given by the more general term
||W||p =
󰀓󰁛
j
Wp
j
󰀔1/p
. (7.13)
Eq. (7.13) constitutes a proper norm only for p ≥ 1. Formally, extensions
to 0 < p < 1 are possible but involve mathematical subtleties. The familiar
Euclidean norm is recovered for p = 2. Another case of particular interest is
p = 1, corresponding to the so-called Manhattan norm. In linear regression,
a (non-differentiable) penalty term proportional to 󰁓
j |Wj| appears in the La-
grangian form of the so-called Least Absolut Shrinkage And Selection Operator
2Here, the scaling of γ with η merely guarantees formal equivalence with Eq. (7.10).",The+Shallow+and+the+Deep.pdf
"7.2. CONTROLLING THE NETWORK COMPLEXITY 167
(LASSO), see [HTF01] for a thorough discussion and comparison with L2-based
Ridge Regression. Similar to the above discussed heuristic penalty (7.12), L1
regularization can also be used to enforce some weights to become exactly zero.
It thus also relates to feature selection and pruning, cf. Sec. 7.2.4.
7.2.3 Constructive algorithms
In the context of classification we have discussed constructive algorithms such
as the so-called tiling algorithms, cf. Sec. 4.2.2. In these schemes, units or layers
are added to the network until the given, labeled data set can be implemented.
Similar ideas have been applied in layered networks for regression. Reviews of
suggested methods and corresponding references can be found in [KY97,SC10].
A key issue of all constructive algorithms is the need to avoid overfitting due
to the addition of too many units. Suitable stopping criteria are discussed
in [KY97].",The+Shallow+and+the+Deep.pdf
"to the addition of too many units. Suitable stopping criteria are discussed
in [KY97].
A popular constructive algorithm for regression is the so-called Cascade-
Correlation algorithm suggested by Fahlmann and Lebiere in 1997 [FL90]. Very
similar to the tiling-like algorithm (4.9), the original Cascade-Correlation scheme
adds hidden units one at a time. However, in contrast to the construction in
(4.9), the added node receives input from all previous hidden units. The new
unit is trained by maximizing the correlation of its output with the residual
error achieved so far, see [FL90] for details. The specific algorithm can lead
to an architecture with many single unit hidden layers, i.e. a deep and narrow
network as opposed to the shallow and wide architectures considered in Sec.
4.2.2 or 5.1.2.
7.2.4 Pruning
The concept of pruning or trimming a neural network is diametrically opposed to
that of constructive algorithms. The idea of pruning is to first train a relatively",The+Shallow+and+the+Deep.pdf
"that of constructive algorithms. The idea of pruning is to first train a relatively
complex system which is capable of realizing the desired task to a satisfactory
extent with respect to the given training data. In order to avoid or reduce
overfitting, the network is then simplified by removing weights and/or nodes
from the system without detoriating its performance too much. The dilution of
trained networks has been considered very early, see [HKP91] for references.
Pruning is usually done after training, and the system may be retrained
thereafter. Likewise, it can be realized in intermediate steps of the training
procedure. Pruning is considered an important ingredient of neural network
training also in recent applications of machine learning. However, as Hugo
Tessier puts it on https://towardsdatascience.com [Tes21]:
“Unfortunately, the dozens, if not hundreds of papers published each year are
revealing the hidden complexity of a supposedly straightforward idea.”",The+Shallow+and+the+Deep.pdf
"“Unfortunately, the dozens, if not hundreds of papers published each year are
revealing the hidden complexity of a supposedly straightforward idea.”
In the literature, many recently suggested pruning procedures appear to be
closely related to early works like [LDS90, HSW93], see [Ree93] for a survey.
According to reviews like [BGFG20], many authors fail to relate and compare
their work properly to early publications in the area. Consequently, we restrict",The+Shallow+and+the+Deep.pdf
"168 7. MODEL EVALUATION AND REGULARIZATION
ourself to the discussion of some early works that represent the basic ideas and
inspired later, more specific schemes. We also limit the discussion to strategies
for the removal of weights rather than entire units or layers. The latter is
frequently referred to as structural pruning, see for instance [AK13] for a review
and references.
Modified weight decay procedures can be employed for the selection of unim-
portant weights as outlined in Sec. 7.2.2. In the following we present two classical
methods which are not based on weight decay, but remove weights explicitly ac-
cording to their importance for the minimization of the cost function that guides
the training process.
Optimal Brain Damage (OBD) and Optimal Brain Surgeon(OBS)
Two classical pruning algorithms with slightly macabre names have been sug-
gested in the 1990s already: LeCun, Denker and Solla’s Optimal Brain Damage",The+Shallow+and+the+Deep.pdf
"gested in the 1990s already: LeCun, Denker and Solla’s Optimal Brain Damage
(OBD) [LDS90] and the Optimal Brain Surgeon (OBS) by Hassibi, Stork and
Wolff [HSW93]. Both schemes are based on a ranking of individual weights Wj
according to their saliency, i.e. the sensitivity of the system with respect to their
removal (setting Wj = 0).
Assume a network has been trained and the weight configuration is suffi-
ciently close to a local minimum W∗ of the cost function E(W) with E(W∗) =
E∗. A Taylor expansion for W = W∗ + U yields
E(W) ≈ E∗+ 1
2
󰁛
i,j
Ui H∗
ij Uj = E∗+ 1
2
󰁛
i
H∗
ii U2
i + 1
2
󰁛
i,j(i∕=j)
Ui H∗
ij Uj. (7.14)
Here, the linear term vanishes in the minimum and H∗
ij = ∂2E
∂Wi∂Wj
󰀏󰀏󰀏
∗
is an ele-
ment of the Hesse matrix computed in W∗. Moreover, we have simply separated
diagonal and off-diagonal contributions of the quadratic term.
In [LDS90], the authors suggest to avoid the computation of the full, po-
tentially very high-dimensional Hesse matrix and focus on the diagonal terms.",The+Shallow+and+the+Deep.pdf
"tentially very high-dimensional Hesse matrix and focus on the diagonal terms.
Assuming that H∗ is approximately diagonal, i.e. dominated by the H∗
jj, Eq.
(7.14) reduces to
E(W) ≈ E∗ + 1
2
󰁛
j
H∗
jjU2
j . (7.15)
An efficient computation of the diagonal elements of H∗ is also outlined in
[LDS90]. The so-called saliencies
sk = H∗
kkW2
k (7.16)
can be used as a guideline for the selection of weights that could be removed
from the system without increasing E significantly. The OBD procedure can be
summarized as follows (after [LDS90]):",The+Shallow+and+the+Deep.pdf
"7.2. CONTROLLING THE NETWORK COMPLEXITY 169
Optimal Brain Damage (OBD) (7.17)
1. Train a network until a local minimum W∗ of E is reached or sufficiently
well approximated
2. Compute the diagonal second derivatives H∗
kk
3. Compute the saliencies skk = H∗
kkW2
k
4. Sort the weights by saliency and set some low-saliency weights to zero
5. Potentially retrain the remaining weights
6. Go to step 1.
The alternative scheme of Optimal Brain Surgery (OBS) as suggested in
[HSW93] follows a similar line of thought. However, there the saliencies are
defined as
󰁥sk = 1
2
W2
k
[H∗−1]kk
. (7.18)
Compared to Eq. (7.18), the factor H∗
kk is replaced by 1/[H∗−1]kk with H∗−1
denoting the inverse of the Hessian H∗. Note that for diagonal matrices H∗
the definitions (7.16) and (7.18) are identical. Further differences between OBD
and OBS concern details of the procedure, see [HSW93].
Numerous modifications and extensions of OBD and OBS have been pro-",The+Shallow+and+the+Deep.pdf
"and OBS concern details of the procedure, see [HSW93].
Numerous modifications and extensions of OBD and OBS have been pro-
posed in the literature. For instance, the authors of [PHL96] suggest a scheme
which computes the saliencies with respect to the estimated generalization error
rather than based on the cost function or training error. Hence, their pruning
procedures, termed γOBS and γOBD, are more closely related to the actual
goal of training.
7.2.5 Weight-sharing
An elementary and intuitive method to reduce the flexibility of a neural network
is to consider subsets of weights that assume the same value. This so-called
weight-sharing usually relies on insights into the problem and data. Assume
that we want to apply a set of (adaptive) filters to patches of a given input
image. It appears natural to train and use only one shared set of weights per
type of filter. This reduces the effective number of weights drastically and",The+Shallow+and+the+Deep.pdf
"type of filter. This reduces the effective number of weights drastically and
consequently simplifies the training a lot. The strategy is ubiquitous in the
context of, for instance, Convolutional Neural Networks for image classification.
For defining the sets of shared weights in advance, prior knowledge is re-
quired which might not always be available. Nowlan and Hinton suggested a
soft version of weight-sharing in [NH92b,NH92a]. They propose a penalty term",The+Shallow+and+the+Deep.pdf
"170 7. MODEL EVALUATION AND REGULARIZATION
full network with all
units and connections
network after dilution
in Dropout
Figure 7.5: Illustration of regularization by Dropout (redrawn after
[SHK+14]). Left panel: the network with all nodes and weights. Right
panel: four randomly selected units (red circles) and their incoming and out-
going weights are temporarily removed.
that enforces the distribution of weights to follow a mixture of Gaussians, the
parameters of which are also subject to updates in the training process. Eventu-
ally, weights can be grouped according to their membership to the contributing
Gaussians. Within these clusters, weights can display very similar values but
are not necessarily identical.
Further discussions of hard or soft weight-sharing and references can be
found in e.g. [UMW17]. More recently, the authors of [OLLB20] conclude that
weight-sharing “is a pragmatic optimization approach” but “it is not a necessity",The+Shallow+and+the+Deep.pdf
"weight-sharing “is a pragmatic optimization approach” but “it is not a necessity
in computer vision applications.” They also argue that approximate weight-
sharing emerges in a self-organized way in unrestricted CNN, for example when
trained from images that display translational invariance.
7.2.6 Dropout
We have already discussed methods from the dilution of networks by removing
weights from a given network, either by weight decay and related methods or
by applying pruning techniques after training, see [HKP91] as well as sections
7.2.2 and 7.2.4.
In the so-called Dropout regularization [HSK+12, SHK+14, GBC16], a ran-
dom dilution is applied in the training process. More concretely: in every
individual update step, e.g. in stochastic gradient descent, individual input and
hidden units are excluded from the network. Only the remaining subnetwork is
trained as usual. DropConnect has been suggested as a variation of the basic",The+Shallow+and+the+Deep.pdf
"trained as usual. DropConnect has been suggested as a variation of the basic
idea that excludes randomly selected weights instead of units [WZZ+13].
In Dropout, at each update step a randomly determined subnetwork of lim-
ited complexity is considered. Therefore, Dropout reduces the flexibility of
the system and restricts its adaptation to the details of the training data. The
removal of nodes occurs independently with probability (1−p), the hyperparam-
eter p determines the fraction of units present in the subnetwork. According",The+Shallow+and+the+Deep.pdf
"7.3. CROSS-VALIDATION AND RELATED METHODS 171
to [SHK+14, GBC16], a value of p = 1/2 is typically used for hidden units,
while p is close to 1 for input units (e.g. p = 0.8). Note that an individual
Dropout dilution could by chance remove an entire layer or some other way cut
all connections from input to output. Such configurations have to be excluded
explicitly, but are very unlikely to occur in large networks.
In the working phase and for testing purposes, the full network is used.
Updating with Dropout will yield larger individual weights than conventional
training. This is compensated for by multiplying all weights that were included
in the Dropout by a factor p in the full network.
Dropout can be interpreted as to simulate the training of an ensemble of
simpler systems (the subnetworks). In the working phase, the complete network
yields an estimate of the corresponding ensemble average. Moreover, due to
this analogy, the potential usefulness of Dropout goes beyond the purpose of",The+Shallow+and+the+Deep.pdf
"this analogy, the potential usefulness of Dropout goes beyond the purpose of
regularization: in the working phase it can be used for uncertainty estimation,
see [GG16].
While Dropout and DropConnect were introduced and are mostly used in
the context of deep neural networks, the concept can be transferred to other
machine learning systems, see [RKSV20] for the consideration of Dropout in
Learning Vector Quantization.
7.3 Cross-validation and related methods
In supervised learning, the availability of well-defined performance measures
allows us to formulate the training process as the optimization of a suitable
objective function. However, one has to be aware that this does not necessarily
reflect the ultimate goal of the learning. The cost function can only be defined
with respect to the training set, while the generic goal of machine learning is to
apply the inferred hypothesis to novel, unseen data. Objective functions serve,
at best, as proxies for the actual aim of training.",The+Shallow+and+the+Deep.pdf
"apply the inferred hypothesis to novel, unseen data. Objective functions serve,
at best, as proxies for the actual aim of training.
As a consequence, the strict minimization of the training error, for example,
can be even counter-productive as it may lead to overtraining or overfitting ef-
fects. From a more positive perspective, this also implies that we should not
take optimization too seriously in the machine learning context. For instance,
the existence of local minima in gradient descent based learning frequently turns
out much less problematic than expected. In fact, the very success of simple-
minded techniques like stochastic gradient descent relates to the fact that strict
minimization of the cost function is usually not the primary goal of machine
learning and could be even harmful. In this sense, the use of SGD can be
interpreted as an implicit regularization which helps to avoid overfitting. Simi-",The+Shallow+and+the+Deep.pdf
"interpreted as an implicit regularization which helps to avoid overfitting. Simi-
larly, several of the explicit regularization techniques discussed in the previous
sections hinder the strict minimization of the cost function in order to achieve
better generalization behavior.
On the downside, it becomes necessary to acquire reliable information about
the expected performance on novel data, if we do not want to face unpleasant
surprises in the working phase. Clearly, the training itself and the performance",The+Shallow+and+the+Deep.pdf
"172 7. MODEL EVALUATION AND REGULARIZATION
Figure 7.6: The requirement that the training data should be representative
of the actual task at hand seems obvious, but is not always met in practice.
© Jonathan van Engelenhoven, see https://www.instagram.com/banjoofjustice
for more of his work. Cartoon reproduced from [Vie23] with kind permission of
the artist.
Remark: MNIST (see e.g. https://paperswithcode.com/dataset/mnist) is a pop-
ular benchmark database of handwritten digits.
on the actual training data does not provide us with such insights.
Validation procedures can be employed which allow us to at least estimate
the expected generalization performance [HTF01,Bis06]. The key idea is rather
obvious: split the available data into a training set and a disjoint test set of
examples. The former is then used for the adaptation of the model, which is
eventually applied to the test set. This way, we can simulate working phase
behavior while using only the available data.",The+Shallow+and+the+Deep.pdf
"eventually applied to the test set. This way, we can simulate working phase
behavior while using only the available data.
Obviously, the data is assumed to be representative for the task at hand, see
Fig. 7.6 for a tounge-in-cheek illustration. This crucial assumption was already
discussed in Sec. 2.1.2 in the context of the generic workflow of supervised
learning.
7.3.1 n-fold cross-validation and related schemes
The simple idea of splitting the available data randomly into one training and
one test set has several problems:
◦ If only relatively few examples are available, as it is very often the case
in practical problems,3 we cannot afford to disregard the information con-
3Image classification tasks have become one of the few prominent exceptions due to the
availability of very large databases.",The+Shallow+and+the+Deep.pdf
"7.3. CROSS-VALIDATION AND RELATED METHODS 173
tained in a subset of these in the training.
◦ The composition of the subsets could be lucky or unlucky in the sense
that the test set might contain only very difficult or very easy cases. As
a consequence, the test set performance might be overly pessimistic or
optimistic, respectively.
◦ Performing a single test only cannot give valid insight into the robustness
of the system with respect to details of the training set, i.e. the variance
in the sense of Sec. 7.1.1.
All these issues are addressed in a very popular standard approach known as
n-fold cross-validation, see e.g. [HTF01, Bis06, Ras18]. The idea is to split the
available data randomly into a number n of disjoint subsets of (nearly) equal
size, train on n−1 of the subsets and use the remaining subset for the validation:
n-fold cross validation (7.19)
• Generate subsets of D of (approximately) equal size:
D = ∪n
i=1Di with ∩(Di, Dj) = ∅ for i ∕= j and cardinalities |Di| = P/n.",The+Shallow+and+the+Deep.pdf
"• Generate subsets of D of (approximately) equal size:
D = ∪n
i=1Di with ∩(Di, Dj) = ∅ for i ∕= j and cardinalities |Di| = P/n.
• For i = 1 : n
◦ Train the system on Dtrain
i = D \ Di
◦ Determine a suitable performance measure on Di
• Compute average and variance of the performance measures over the n
training processes.
For each split we train the classifier or regression system on (1 − 1/n) P ex-
amples and evaluate its performance with respect to the P/n left out samples.
Eventually, we have obtained n systems trained on slightly different data sets
with n estimates of the performance, for instance in terms of the accuracies of
a classifier or the MSE in regression problems.
While the n validation sets are disjoint, we have to be aware that the training
sets strongly overlap. The obtained estimates of the generalization error and,
even more so, of the training error are definitely not statistically independent.",The+Shallow+and+the+Deep.pdf
"even more so, of the training error are definitely not statistically independent.
Hence, the mean or variance obtained over the n-fold training process should
not be over-interpreted. Nevertheless, the procedure will provide us with some
insight into the expected generalization performance and the robustness of the
system with respect to small changes in the training set.
Obviously, the parameter n in n-fold cross validation will influence the work-
load and the quality of the results:
◦ For large n, many training processes have to performed, each one based
on a large fraction of the available data. In turn, the individual validation
sets will be relatively small.",The+Shallow+and+the+Deep.pdf
"174 7. MODEL EVALUATION AND REGULARIZATION
◦ For small values of n we obtain fewer, but more reliable individual esti-
mates from larger validation sets. At the same time, the computational
workload is reduced in comparison with the use of larger n. However,
averages are performed over few individual results, only. Moreover, each
training process make use of a relatively small subset of the data and
cannot take full advantage of the available information.
In a practical situation, the choice of n will depend primarily on the number
P of available examples to begin with. For a more detailed discussion and
corresponding references see, for instance, [HTF01, Ras18]. In the literature, a
canonical value of n = 10 has become a standard choice, apparently.
Variants
Many variations of the basic idea of cross-validation have been considered in the
literature [Ras18]. The split into n disjoint subsets of data may also suffer from",The+Shallow+and+the+Deep.pdf
"literature [Ras18]. The split into n disjoint subsets of data may also suffer from
lucky/unlucky set composition. Therefore, one often resorts to a few repeti-
tions of the n-fold scheme, performed with randomized splits and an additional
average over the realizations.
In principle one could aim at realizing all possible splits of the data into
(P −p) training and p validation examples. Their number
󰀃P
p
󰀄
grows rapidly with
P and the computational costs can become unrealistically high. Alternatively,
in repeated random sub-sampling validation, also known as Monte Carlo cross
validation, one generates a number of independently generated random splits
into P − p and p examples and computes averages and variances accordingly.
This is closely related to so-called bootstrap methods [HTF01], which differ from
cross-validation by resampling subsets of data with replacement. We refrain
from a thorough comparison and discussion of the advantages and disadvantages",The+Shallow+and+the+Deep.pdf
"from a thorough comparison and discussion of the advantages and disadvantages
of these variations of cross-validation. We refer the reader to, for example,
[Efr83,ER97,Bur89,BHT23] and to textbooks like [HTF01,Bis95a,Bis06].
Leave-one-out cross-validation
As a popular extreme case, one often resorts to the so-called Leave-One-Out
validation [HTF01, Bis95a, Bis06, Ras18], in particular for very small data sets.
It follows the idea of cross-validation, but selects just one example as the smallest
possible validation set in each training run. Hence, we set n = P and run P
training processes to obtain an average of the performance measure of interest.
It is important to note that the Leave-One-Out estimate can be unreliable.
It even bears the risk of systematically yielding misleading results: In very small
data sets, leaving out one sample from a specific class can lead to a bias in the
training set towards the other class(es), which may result in overly pessimistic",The+Shallow+and+the+Deep.pdf
"training set towards the other class(es), which may result in overly pessimistic
estimates of the generalization performance [Ras18, APW+09]. A modification
that mildens the effect is known by the self-explanatory name Leave-one-out
from each class, generating validation sets that represent all classes [APW+09]
with equal weight.",The+Shallow+and+the+Deep.pdf
"7.4. PERFORMANCE MEASURES 175
7.3.2 Model and parameter selection
The above discussed validation schemes can be employed in the context of
model selection and, similarly, for the setting of parameters or hyper-parameters
[Ras18]. We can use, for instance, n-fold cross-validation to compare the ex-
pected performance of different classifiers or regression systems. We can also
employ it to determine the size of a hidden layer in a feed-forward neural net-
work, to set algorithm parameters like the learning rate in gradient descent, or
to select a particular kernel in an SVM, to name just a few examples. In the
illustration of Fig. 7.2, for instance, we would select the model complexity that
corresponds to the minimum of the U-shaped generalization error curve.
Similarly, (hyper-)parameters of the regularization schemes discussed in Sec.
7.2 can be tuned according to the performance of the system w.r.t. the validation",The+Shallow+and+the+Deep.pdf
"7.2 can be tuned according to the performance of the system w.r.t. the validation
sets. However, one has to be aware of the risk to over-interpret or even mis-use
the results of cross-validation. As an example, consider gradient based training
of a network with one hidden layer on a given data set D. Assume that, on
the basis of n-fold cross-validation, we conclude that systems with, say, K = 3
hidden units yield the best generalization ability.4
Is it justified to expect the observed, averaged performance for K = 3 when
applying the system to novel data? The problem is that we have used all of D
to determine the supposedly best parameter setting. This constitutes a data-
driven learning process and could be subject to overfitting effects in itself: The
supposedly best choice of K may be very specific to D and could fail in the
working phase.
In order to obtain a more reliable estimate of the expected performance we
would have to perform an extended validation procedure. We could split D",The+Shallow+and+the+Deep.pdf
"In order to obtain a more reliable estimate of the expected performance we
would have to perform an extended validation procedure. We could split D
into training set Dtrain and Dtest once, then apply n-fold cross-validation on
Dtrain in order to determine a suitable value of K. Eventually, a system with
the supposedly best setting can be re-trained on Dtrain and validated on Dtest
to obtain a more realistic estimate. Now, of course, we face the problem of
lucky/unlucky set compositions again, which suggests to perform a full loop
along the lines of n-fold cross-validation (or one of the discussed variants).
Strictly speaking, this has to be done separately for every independent pa-
rameter or hyperparameter in an additional layer of validation. Obviously, prac-
tical limitations apply, in particular when only small data sets are available.
7.4 Performance measures for regression
and classification
Despite the conceptual clarity of supervised learning, even the choice of an",The+Shallow+and+the+Deep.pdf
"7.4 Performance measures for regression
and classification
Despite the conceptual clarity of supervised learning, even the choice of an
appropriate measure of success can constitute a non-trivial issue in practice.
4For the difficulty to even define “thebest” see the next sections.",The+Shallow+and+the+Deep.pdf
"176 7. MODEL EVALUATION AND REGULARIZATION
7.4.1 Measures for regression
In regression, a differentiable objective or loss function typically guides the
training process. It appears natural to consider the same function also for the
evaluation of a system in validation or working phase. Most frequently, the
familiar mean squared error (MSE) is used in both contexts.
Depending on the application domain, alternative measures different from
the loss function can be employed for the evaluation of regression models. Two
popular measures that are essentially different from the MSE are:
◦ Mean Absolute Error [WM05]
MAE = 1
Q
󰁓Q
µ=1 |σµ − τµ| (7.20)
which weights deviations from the target differently than the MSE in a set
of Q test or validation samples. The MAE also satisfies 0 ≤ MAE ≤ ∞
where lower values correspond to better quality of the regression.
◦ Coefficient of Determination (CoD) [DS98]. 5
CoD = 1 −
󰁓
µ(σµ − τµ)2
󰁓
µ(τµ − 〈τ〉µ)2 (7.21)",The+Shallow+and+the+Deep.pdf
"◦ Coefficient of Determination (CoD) [DS98]. 5
CoD = 1 −
󰁓
µ(σµ − τµ)2
󰁓
µ(τµ − 〈τ〉µ)2 (7.21)
where 〈τµ〉 is the mean target value in the data set. The CoD compares
the mean squared deviation of the predictions from the targets, scaled by
the variance of the target values in the data set. The measure satisfies
−1 ≤ CoD < 1, a value of CoD = 0 indicates that all predictions are
σµ = 〈τµ〉, CoD = 1 corresponds to perfect regression with all σµ =τµ.
A variety of further evaluation criteria, e.g. based on distance metrics or cor-
relations, is available in the literature. A typology of measures is provided
in [Bot19].
7.4.2 Measures for classification
The actual goal of machine learning for classification problems is to achieve a
model that assigns data to the correct class with high probability in the working
phase. Very often, the actual objective functions used in training provide at best
a proxy of this ultimate goal. Loss functions for probabilistic classifiers based on",The+Shallow+and+the+Deep.pdf
"a proxy of this ultimate goal. Loss functions for probabilistic classifiers based on
cross-entropy like (5.23) and (5.24) could be used for training and evaluation.
Most frequently, the evaluation of crisp classifiers is guided by criteria that
directly relate to the accuracy with respect to the validation or test data. An
overview of a variety of performance measures for classification is given in [SL09].
Assume we are comparing the performances of two different classifiers, A and
B, which have been trained to perform a given binary classification. By means
5In the literature, the CoD is most frequently termed R2. The author refuses to denote a
quantity that can become negative by a square.",The+Shallow+and+the+Deep.pdf
"7.4. PERFORMANCE MEASURES 177
of cross-validation we can obtain the estimates for the generalization ability in
terms of the overall error as, say,
󰂃A
g = 0.05 and 󰂃B
g
= 0.30.
Apparently, we could conclude that A is the better classifier and should be used
in the working phase.
A closer look into the available data D, however, might reveal that it con-
sists of 95% class 1 samples, while only 5% of the data represent class 2. We
furthermore might find that classifier 1 trivially assigns all feature vectors to
class 1, resulting in 95% accuracy in D. On the other hand, model B might
have learned from the data and provides 70% correct responses in both classes
of the data set.
Clearly, this insight might make us reconsider our previous evaluation of
classifier A as the better one. If we are just after good overall accuracy and
have reason to believe that the true prevalence of class 1 data is also about 95%
in the real world, we can - of course - settle for the trivial model. If our main",The+Shallow+and+the+Deep.pdf
"in the real world, we can - of course - settle for the trivial model. If our main
goal is to detect and identify the relatively rare occurrences of class 2, classifier
B is obviously to be preferred.
This somewhat extreme example illustrates two major questions that arise
in the practical approach to classification problems:
◦ How can we cope with strongly biased data sets when evaluating the per-
formance of a classifier?
◦ Can we evaluate classifiers beyond their overall accuracies in order to
obtain better insight into the performance?
Here we do not address the question of how to take class bias into account in
the training process. Some strategies for the training from imbalanced data sets
will be discussed in Section 8.7.
7.4.3 Receiver Operating Characteristics
For two-class problems, both of the above mentioned questions can be addressed
in the framework of the so-called Receiver Operating Characteristics (ROC)",The+Shallow+and+the+Deep.pdf
"in the framework of the so-called Receiver Operating Characteristics (ROC)
[HTF01,Bis95a,Bis06,DHS00,Faw06]. The concept and terminology goes back
to signal processing tasks originally, but has become popular in the machine
learning community.
Most classifiers we have discussed obtain a binary assignment by applying
a threshold operation to a so-called discriminative function g. In terms of the
simple perceptron, for instance, we assign an input ξ ∈ RN to class S = ±1
according to
S = sign [g(ξ)] with g(ξ) =
N󰁛
j=1
wjξj, (7.22)
as discussed in Chapter 3 in great detail. Having trained the perceptron as
to implement the homogeneous lin. sep. function (7.22), we can introduce a",The+Shallow+and+the+Deep.pdf
"178 7. MODEL EVALUATION AND REGULARIZATION
threshold Θ after training and consider the modified classification
SΘ = sign [g(ξ) − Θ] . (7.23)
While this is formally identical with the consideration of an inhomogeneously
lin. sep. function, see Sec. 3.3, here the perspective is different: We assume the
threshold is introduced and varied manually after training. Furthermore, the
concept could be applied to any discriminatory function for binary classification.
Quite generally, for a large family of classifiers it is possible to realize and
control a class bias by tuning Θ in Eq. 7.23. For very large negative Θ → −∞,
all inputs will be assigned to class SΘ = −1, while large positive Θ → +∞
result in SΘ = +1 exclusively.
In a similar way, probabilistic models can be used for crisp classification by
thresholding the class membership probability which serves as the discriminative
function g in Eq. 7.23. Employing a probability threshold 0 < Θ < 1 different",The+Shallow+and+the+Deep.pdf
"function g in Eq. 7.23. Employing a probability threshold 0 < Θ < 1 different
from 1/2 imposes a bias towards one of the two classes.
Very often it is important to distinguish the two class-specific errors that
can occur: If a feature vector which is truly from class −1 is misclassified as
S = +1, we account this as a false positive or false alarm type of error. The
terminology reflects the idea that class +1 is to be detected, for instance in
a medical test which discriminates diseased (positive test result) from healthy
control patients (negative outcome). Analogously, the term false negative error
is used when the classifier misses to detect a truly positive case.6 Similarly, the
complementary true positive or true negative rates correspond to the class-wise
accuracies in the two-class problem.
The introduction of a controlled bias can be achieved in other classification
frameworks as well and is, by no means, limited to linear classifiers. For instance,",The+Shallow+and+the+Deep.pdf
"frameworks as well and is, by no means, limited to linear classifiers. For instance,
we can modify the Nearest Prototype Classification (NPC) in LVQ, Eq. (6.2).
Identifying w∗
(−1), the closest one among all prototypes representing class −1
and the closest class-(+1)-prototype w∗
(+1), we can assign an arbitrary feature
vector ξ to class +1 if
d
󰀓
w∗
(+1), ξ
󰀔
< d
󰀓
w∗
(−1), ξ
󰀔
− Θ (7.24)
and to class −1 else, thus introducing a margin Θ in the comparison of distances.
Similarly, we could consider the output unit activation in a multilayered feed-
forward neural network as the discriminative function and perform a biased
thresholding along the same lines in order to obtain a crisp class assignment.
For a given value of the threshold Θ we can obtain, e.g. from a validation or
test set, the observed absolute number of false positive classifications FP, false
negatives FN, true positives TP and true negatives TN. The corresponding
rates are defined as
fpr= FP
FP +TN , tpr= TP",The+Shallow+and+the+Deep.pdf
"negatives FN, true positives TP and true negatives TN. The corresponding
rates are defined as
fpr= FP
FP +TN , tpr= TP
FN +TP , fnr= FN
FN +TP , tnr= TN
FP +TN . (7.25)
6In the literature, other terms like type I/II errors are used frequently, but these are avoided
here for the sake of clarity.",The+Shallow+and+the+Deep.pdf
"7.4. PERFORMANCE MEASURES 179
true positive rate tpr
false positive rate fpr
Θ→+∞
Θ=0
Θ→−∞
ξ−
δΘ
Θ−
g(ξ)
Figure 7.7: Left panel: Schematic illustration of Receiver Operating Char-
acteristics. The extreme working points with Θ → ±∞ are marked by empty
circles. A filled circle corresponds to an unbiased classifier with Θ = 0, while
the dashed line represents random, biased guesses. Right panel: Illustration
of a two-class data set with discriminative function g(ξ). Feature vectors from
the negative (positive) class are displayed as green (light) and red (dark) filled
circles, respectively. A randomly selected negative example ξ− is marked by
the large filled circle and corresponds to g(ξ−) = Θ−. The variation of the
threshold by δΘ is referred to in the arguments employed in Sec. 7.4.4 to obtain
the statistical interpretation of the AUC.
Different names are used for the same quantities in the literature, depending on",The+Shallow+and+the+Deep.pdf
"the statistical interpretation of the AUC.
Different names are used for the same quantities in the literature, depending on
the actual context and discipline. In medicine, for instance, the term sensitivity
(SENS) is frequently used for the tpr, while specificity (SPEC) refers to the
tnr. An overview of the many quantities that can be derived from the four basic
quantities TP, TN, FP, FN and rates (7.25) can be found in [Wik22], which also
provides further relevant references.
The quantities in Eq. (7.25) are not independent: Obviously, they satisfy
tpr + fnr = 1 and tnr + fpr = 1.
Consequently, two of the four rates can be selected to fully characterize the
classification SΘ(ξ).
In the framework of Receiver Operating Characteristics (ROC) one deter-
mines tpr(Θ) and fpr(Θ) for a meaningful range of thresholds Θ and displays
the true positive rate as a function of the false positive rate by eliminating the
threshold parameter7.",The+Shallow+and+the+Deep.pdf
"the true positive rate as a function of the false positive rate by eliminating the
threshold parameter7.
Figure 7.7 (left panel) displays an example ROC curve for illustration. The
lower left corner, as marked by an empty circle, would correspond to the extreme
7For efficient implementation ideas see [DHS00,Faw06].",The+Shallow+and+the+Deep.pdf
"180 7. MODEL EVALUATION AND REGULARIZATION
setting Θ → ∞ with all inputs assigned to the negative class. Obviously, the
false positive rate is zero for this setting, the classifier does not give any false
alarms. On the other hand, no positive cases are detected and tpr = 0, as
well. The upper right corner in tpr = fpr = 1, also marked by an open circle,
corresponds to Θ → −∞ in (7.23) or (7.24): The classifier simply assigns every
feature vector to the positive class, thus maximizing the true positive rate at
the expense of having fpr = 1. An ideal, error-free classifier would obtain fpr =
0, tpr = 1 in the upper left corner of the ROC graph.
The performance of an unmodified classifier with Θ = 0 is marked by a filled
circle in the illustration 7.7 (left panel). It could correspond, for instance, to the
NPC in LVQ or the homogeneous, unbiased perceptron, Eq. (7.22). By selecting
a particular threshold −∞ < Θ < +∞, the user can realize any combination of",The+Shallow+and+the+Deep.pdf
"a particular threshold −∞ < Θ < +∞, the user can realize any combination of
{tpr, tpr} that is available along the ROC curve. This way, the domain expert
can adjust the actual classifier according to the specific needs in the problem at
hand. In medical diagnosis systems, for instance, high sensitivity (tpr) might
be more important than specificity (1 − fpr) or vice versa.
To a certain extent, we can also compensate for the effects of unbalanced
training data: In the illustrative example shown in Fig. 7.7 (left panel), the
classifier with Θ = 0 realizes very low fpr, which might be a consequence of an
over-representation of negative cases in the data set D. An objective function
which is related to the number of mis-classification will favor classifiers with
small fpr over those with higher tpr. In retrospect, this can be compensated
for by biasing the classifier towards the detection of positive cases and move the
working point closer to the upper left corner in the ROC.",The+Shallow+and+the+Deep.pdf
"for by biasing the classifier towards the detection of positive cases and move the
working point closer to the upper left corner in the ROC.
The hypothetical, best possible ROC is obviously given by the step function
including the ideal working point fpr = 0, tpr = 1, i.e. the complete square in
Fig. 7.7. On the other hand, a completely random guess with biased probability
tpr = fpr for assignments to class +1 would correspond to the diagonal, i.e. the
dashed line in the left panel of the illustration.
7.4.4 The area under the ROC curve
When evaluating different classifiers (or frameworks, rather) one often resorts
to the comparison of the area under the ROC curve, the so-called AUROC or
less precisely AUC [Faw06]. Intuitively the AUC with 0 ≤ AUC ≤ 1 provides
information about the degree to which the ROC deviates from the diagonal with
AUC = 1/2. Clearly, an AUC > 1/2 indicates better–than–random classifica-
tion and the AUC is often used as a single numerical quality measure for the",The+Shallow+and+the+Deep.pdf
"AUC = 1/2. Clearly, an AUC > 1/2 indicates better–than–random classifica-
tion and the AUC is often used as a single numerical quality measure for the
evaluation of classifiers. In principle, the precise shape of the ROC should be
taken into account as well, as individual ROC can differ significantly from the
idealized shape displayed in Fig. 7.7.
The AUC with respect to novel data can be estimated, for instance, in the
course of cross-validation along the lines of Sec. 7.3. It provides better insight
into the performance of the trained system than a single specific working point.
Therefore, it serves as the basis for model selection or the setting of parameters.",The+Shallow+and+the+Deep.pdf
"7.4. PERFORMANCE MEASURES 181
Moreover, the AUC can be associated with a well-defined statistical inter-
pretation. Fig. 7.7 (right panel) illustrates a two-class data set which can be
classified according to a discriminative function which, in the illustration, is
assumed to increase monotonically along the g(ξ)-axis. Note that here it is
convenient, but not necessary, to argue in terms of a linear classifier like the
perceptron, in which the weight vector w defines the discriminative direction.
In the illustration, a particular, e.g. randomly selected, negative example ξ−
is marked by a filled circle with the value g(ξ−) of the discriminative function.
In other words, in a classifier with Θ = g(ξ−) the considered example would be
located precisely at the decision boundary.
Now assume that we select a random example ξ+ from the positive class,
i.e. one of the feature vectors marked as red circles in the illustration. The",The+Shallow+and+the+Deep.pdf
"Now assume that we select a random example ξ+ from the positive class,
i.e. one of the feature vectors marked as red circles in the illustration. The
probability for such an example to satisfy g(ξ+) > Θ− = g(ξ−) is given precisely
by tpr(Θ−), which is the fraction of positive examples located on the correct
side of the decision boundary defined by g(ξ) = Θ−.
On the other hand, the local density of negative examples is given by the
derivative dfpr/dΘ in Θ−: Shifting the threshold by δΘ, as marked by the gray
shaded area, will result in correcting the output for δfpr = (dfpr/dΘ) δΘ many
samples from the negative class.
In summary, this implies that for a pair of feature vectors comprising one
randomly selected ξ− from the negative class and one randomly selected ξ+ from
the positive class, the probability that g(ξ+) > g(ξ−) is given by the integral
󰁝 +∞
−∞
tpr(ϑ) dfpr
dϑ dϑ =
󰁝 1
0
tpr dfpr = AUC.
Hence, the AUC quantifies the probability with which a randomly selected pair",The+Shallow+and+the+Deep.pdf
"󰁝 +∞
−∞
tpr(ϑ) dfpr
dϑ dϑ =
󰁝 1
0
tpr dfpr = AUC.
Hence, the AUC quantifies the probability with which a randomly selected pair
{ξ−, ξ+} is ordered according to class membership in terms of the discriminative
function g(. . .). This corresponds to the probability that a threshold value Θ
exists, at which the classifier would separate such a pair of inputs correctly.
This intuitive interpretation of the AUC in the ROC-analysis also makes it
possible to perform the training of a classifier in such a way that the expected
AUC is maximized, for details see [HR04,VKHB16].
7.4.5 Alternative measures for two-class problems
A variety of evaluation criteria for binary classification schemes have been sug-
gested in the literature. The Precision-Recall (PR) formalism [DG06] can be
considered as an alternative to the ROC. The PR evaluation is also based on
the four quantities TP, TN, FP, and FN. Precision (Prec) and Recall (Rec) are
defined as
Prec = TP
TP + FP Rec = TP
TP + FN = tpr. (7.26)",The+Shallow+and+the+Deep.pdf
"the four quantities TP, TN, FP, and FN. Precision (Prec) and Recall (Rec) are
defined as
Prec = TP
TP + FP Rec = TP
TP + FN = tpr. (7.26)
Similar to the ROC, a PR curve displaying Prec vs. Rec can be generated
by varying a bias parameter. Also, the area under the PR curve is often pro-
vided as a single quality parameter. However, unlike the AUROC it lacks an",The+Shallow+and+the+Deep.pdf
"182 7. MODEL EVALUATION AND REGULARIZATION
appealing statistical interpretation. For a discussion of supposed disadvantages
or advantages of the PR formalism over the ROC see [DG06] and references
therein.
Like other quantities, Prec and Rec can also be computed at a single, spe-
cific working point of the classifier. Various application domain specific measures
have been defined, see e.g. [Wik22,KEP17,KHV14] for overviews and references.
In fact, the large number of related quality measures can be a source of consid-
erable confusion. Here we present only two popular examples:
While the overall accuracy for a test set of in total ntot samples is computed
as (TP + TN)/ntot, the so-called balanced accuracy corresponds to an equal
weight average over classes:
BAC = tpr + tnr
2 = 1
2
󰀕 TP
TP + FN + TN
TN + FP
󰀖
. (7.27)
It is supposedly more suitable for class imbalance in training and validation sets.
Here, TP, FP and FN are the absolute numbers of true positives, false positives",The+Shallow+and+the+Deep.pdf
"Here, TP, FP and FN are the absolute numbers of true positives, false positives
and false negatives observed in the data set. Similar claims have been made for
the popular F1-measure, which is given by the harmonic mean of Prec and Rec:
F1 = TP
TP + (FP + FN)/2 = 2 Prec · Rec
Prec + Rec (7.28)
7.4.6 Multi-class problems
The evaluation of multi-class systems is more subtle. Several single valued
measures can be computed, see below, but a multi-dimensional generalization
of the ROC formalism or the PR scheme is far from obvious.
Confusion matrix
Most commonly, the so-called confusion matrix is provided in order to sum-
marize the class-specific performance of a multi-class system with respect to a
given data set. Illustration 7.8 displays the confusion matrix of a hypotheti-
cal 4-class classification problem. In the left panel, each element corresponds
to the number of feature vectors which belong to class i and are assigned to",The+Shallow+and+the+Deep.pdf
"to the number of feature vectors which belong to class i and are assigned to
class j by the classifier. In the right panel of Fig. 7.8, percentages (rounded)
are displayed. Here, diagonal elements correspond to the class-wise accuracies.
Off-diagonal elements provide insight into which classes are relatively easy or
difficult to separate.
Note that although the confusion matrix provides detailed information about
class-wise performances, it still corresponds to a single working point of the clas-
sifier. A multi-dimensional ROC- or PR-like analysis for multi-class problems is
non-trivial [Faw06,HT01]. However, many of the above mentioned single quan-
tities for a given working point can be extended to multi-class problems in a
straightforward fashion, see [KEP17]. An obvious example is the multi-class
BAC which weights every class equally by 1/C in a C-class problem.",The+Shallow+and+the+Deep.pdf
"7.4. PERFORMANCE MEASURES 183
predicted class
[abs.] 1 2 3 4
true class
1
2
3
4
sum
predicted class
[%] 1 2 3 4
true class
1
2
3
4
sum
Figure 7.8: Confusion matrix of a hypothetical imbalanced 4-class problem.
Left panel: matrix elements correspond to the absolute number of samples
from class i which are assigned to class j. Right panel: the corresponding
percentages. Diagonal entries represent the class-wise accuracies.
7.4.7 Averages of class-wise quality measures
A set of class-wise quantities can be derived from the confusion matrix or by
considering sub-classification schemes of the type class i vs. all others. This
way, all measures suitable for two-class problems, like AUROC or F1, can be
considered in C-class problems as well.
Averages of class-wise quantities can be computed in different ways, e.g. by
weighting each class equally or by taking the number of examples per class
into account. In the following, this subtlety is illustrated in terms of Precision",The+Shallow+and+the+Deep.pdf
"into account. In the following, this subtlety is illustrated in terms of Precision
(Prec) and Recall (Rec) as defined in Eq. (7.26) for two-class problems and the
F1-measure (7.28).
From the confusion matrix c(i, j) in terms of sample counts of a C-class
problem we can determine the class-wise quantities
TPi = c(i, i), FNi =
󰁛
j,(∕=i)
c(i, j), FPi =
󰁛
j(∕=i)
c(j, i) for i = 1, 2 . . . C. (7.29)
For the matrix displayed in Fig. 7.8 (left panel) we would obtain TP1 = 72, FN1 =
6 + 11 + 4 = 21, and FP1 = 3 + 13 + 7 = 23 with respect to the first class.
Macro-averages
Based on Eq. (7.29) we can also compute the class-wise Precision and Recall
values in analogy with Eq. (7.26):
Preci = TPi
TPi + FPi
, Reci = TPi
TPi + FNi
. (7.30)",The+Shallow+and+the+Deep.pdf
"184 7. MODEL EVALUATION AND REGULARIZATION
The so-called macro-averages
Precmac = 1
C
C󰁛
i=1
Preci and Recmac = 1
C
C󰁛
i=1
Reci. (7.31)
are obtained with equal weight assigned to the C classes. Alternatively, a
weighted macro-average of the form
Precw−mac = 1
ntot
C󰁛
i=1
ni Preci and Recw−mac = 1
ntot
C󰁛
i=1
niReci (7.32)
is frequently considered. Here ni denotes the number of samples in class i and
the total number is ntot = 󰁓C
i=1 ni.
Now we have at least two options for defining a macro-F1 measure:
a) as an arithmetic mean of the class-wise Fi
1 given by
Fi
1 = 2 Preci Reci
Preci + Reci
, Fmac−a
1 = 1
C
C󰁛
i=1
Fi
1 (7.33)
b) or as a harmonic mean of macro-Recall and macro-Precision, i.e.
Fmac−b
1 = 2 Precmac Recmac
Precmac + Recmac (7.34)
Unfortunately, both measures appear in the literature under the same name, of-
ten without clarifying which version was used. A recently published note [OB21]
compares Fmac−a
1 and Fmac−b
1 . The authors show that the two quantities can",The+Shallow+and+the+Deep.pdf
"compares Fmac−a
1 and Fmac−b
1 . The authors show that the two quantities can
differ significantly with Fmac−b
1 ≥ Fmac−a
1 . They demonstrate that Fmac−b
1 can
yield overly large scores in class-imbalanced problems, while Fmac−a
1 appears to
be more robust in that respect. The authors conclude that “at the very least,
researchers should indicate which formula they are using.”
Micro-averages
In contrast to the above discussed macro-averages, micro-averaging considers
all data points without taking class-specifics into account. It is obtained by
considering the averages
󰁦TP = 1
C
C󰁛
i=1
TPi, 󰁧TN = 1
C
C󰁛
i=1
TNi, 󰁦FP = 1
C
C󰁛
i=1
FPi, 󰁧FN = 1
C
C󰁛
i=1
FNi
(7.35)
and computing
Precmic =
󰁦TP
󰁦TP + 󰁦FP
and Recmic =
󰁦TP
󰁦TP + 󰁧FN
. (7.36)",The+Shallow+and+the+Deep.pdf
"7.5. INTERPRETABLE SYSTEMS 185
We note that
( 󰁦TP + 󰁦FP) = 1
C
C󰁛
i=1
(TPi + FPi) = 1
C
C󰁛
i=1
(TPi + FNi) = 󰁦TP + 󰁧FN = ntot
C ,
as both sums add up all elements of the confusion matrix. As a consequence
Precmic = Recmic = 1
ntot
C󰁛
i=1
TPi, = Fmic
1 . (7.37)
Since Precmic = Recmic, their harmonic mean Fmic
1 also equals the overall
accuracy.
In general, macro-averages put the same emphasis on each class which makes
sense if the aim is an approximately class-independent performance. The micro-
average targets the overall quality, while the weighted macro-average constitutes
a compromise between the micro- and macro-approach.
Which averaging procedure should be used to evaluate a multi-class sys-
tem depends on the actual target and the preference of the user. The class-
composition in the training data set compared to the one that is expected in
the working phase plays an important role. Assume, for instance, that a par-",The+Shallow+and+the+Deep.pdf
"the working phase plays an important role. Assume, for instance, that a par-
ticular class is represented by very few samples in the training set, but can be
expected to be more frequent in the real world data. Then, micro-averaging
bears the risk of disregarding the class in the evaluation with potential poor
performance in the working phase as a consequence.
7.5 Interpretable systems
The above discussed quality measures and validation procedures focus on the
performance of a trained system in terms of classification or regression accura-
cies. This is also true for the considered more sophisticated measures beyond
overall or class-wise accuracies. Accuracy appears to be a natural evaluation
criterion, if the goal is to, say, distinguish cats from dogs in images or perhaps
discriminate diseased patients from healthy controls in a diagnosis problem.
However, machine learning systems should be evaluated and compared to",The+Shallow+and+the+Deep.pdf
"discriminate diseased patients from healthy controls in a diagnosis problem.
However, machine learning systems should be evaluated and compared to
each other also according to complementary criteria. Potentially, some of these
cannot even be expressed in terms of simple quantitative measures. As an
illustration, we discuss an entertaining and frequently quoted example [Nik17].
It illustrates and summarizes an important issue in machine learning along the
lines of the opening quote of this chapter: “Accuracy is not enough.”
The story is that a classifier was trained to distinguish dogs from wolves
based on a labeled set of still images. It seemed to work perfectly in training and
validation, but failed completely on novel photos in the working phase. Even-
tually, a check of the database showed that all wolves had been photographed
in the snow, while dogs were shown on green grass. The classifier had “learned”
to distinguish the image backgrounds, not the actual animals.",The+Shallow+and+the+Deep.pdf
"186 7. MODEL EVALUATION AND REGULARIZATION
As usual with stories like that, it is told in many versions, see [gwe09] for an
interesting account of similar examples of supposedly mislead classifiers. Ap-
parently, the origin of the wolves vs. dogs problem is [RSG16], a publication in
which the authors purposefully trained a classifier on the misleading data. The
aim was to illustrate the problem and to test a method for explain the inner
workings of the classifier.
However, the moral of the story is definitely relevant: unnoticed biases in
data sets can result in seemingly good or even excellent performances. The
effect is frequently much more subtle and more difficult to detect than in the
wolves vs. dogs problem. As a particularly important example, medical data
sets are frequently prone to selection biasses that facilitate a seemingly successful
discrimination of diseased from healthy subjects. For example, the age or gender",The+Shallow+and+the+Deep.pdf
"discrimination of diseased from healthy subjects. For example, the age or gender
distribution in the classes could be different, while being essentially unrelated
with the actual target diagnosis. Even the more frequent occurrence of missing
values in one of the groups could be exploited by the machine learning system,
resulting in seemingly good yet useless performance. It is the nature of machine
learning systems that they are excellent artefact detectors.
Hence, the evaluation and comparison of supervised learning models in terms
of accuracy only (or similar performance oriented criteria) can be misleading
and even dangerous. Responsible use of machine learning techniques requires at
least a certain degree of insight into what is the basis of the system’s response.
Which features, for instance, appear most relevant in a classification scheme?
In the example of wolves vs. dogs: is it the properties of the animals or the
color of the background that the assignment relies upon?",The+Shallow+and+the+Deep.pdf
"In the example of wolves vs. dogs: is it the properties of the animals or the
color of the background that the assignment relies upon?
In this sense, machine learning systems should be transparent and inter-
pretable. At the very least, an effort should be made to understand how a given
classifier or regression system works. Intuitive prototype-based systems and rel-
evance learning constitute just two examples of approaches that can be useful
in this context. A qualitative or quantitative study of the importance of given
features can be performed in a variety of learning frameworks.
The motivation for favoring white-box approaches is not limited to the de-
tection of potential biases. Interpretable models also facilitate the discussion
with the domain expert and increase the user acceptance for machine learning
based support systems. Consequently, the topic of improved interpretability
has attracted considerable interest within the machine learning community and",The+Shallow+and+the+Deep.pdf
"has attracted considerable interest within the machine learning community and
continuous to do so. Partly, these efforts aim at closing the gap (if any) between
the goals of statistical modelling and machine learning discussed in Sec. 2.2.
For recent reviews and research articles we refer the reader to several spe-
cial sessions which have been organized at the European Symposium on Neural
Networks (ESANN) [VMGL12,BL13,BBVZ17]. The overview articles and ses-
sion contributions should provide a useful starting point for further explorations
of the topic. A comprehensive discussion of explainability and interpretability
with many references can also be found in [Gho21].",The+Shallow+and+the+Deep.pdf
"Chapter 8
Preprocessing and
unsupervised learning
I will let the data speak for itself when it cleans itself.
— Unknown
In most of these lecture notes we have implicitly assumed that feature vectors
and labels are provided ready-to-use for training. In practical situations, this is
rarely the case. In general, real world data sets have to be thoroughly checked
for inconsistencies, missing values or other issues. A list of useful rules for initial
data analysis is provided in [BCS+22].
Quite often, outliers are removed from the data, e.g. single data points that
appear implausible because they are very different from the bulk of the data or
display unrealistic values of individual features. Such a cleaning of the data has
to be performed with utmost care and in a controlled, reproducible way. The
criteria and goals of data set curation and cleaning are usually very specific to
the application domain. Therefore, it should rely on insights from the domain",The+Shallow+and+the+Deep.pdf
"the application domain. Therefore, it should rely on insights from the domain
experts and it should never be guided by the idea of making the data consistent
with a given research goal or hypothesis. For instance, removing suspected
outliers may seriously affect the overall quality of the data set and introduce
unwanted biases.1
Here, we focus on broadly applicable and popular preprocessing steps. The
design of a classifier or regression system usually begins with the choice of
how to represent the data to the system. This can amount to the extraction
of engineered features, e.g. derived from images. But also seemingly simple
data sets of directly observed numerical feature vectors often require careful
preprocessing.
1(or in the worst case even wanted biases that support the desired result)
187",The+Shallow+and+the+Deep.pdf
"188 8. PREPROCESSING AND UNSUPERVISED LEARNING
Depending on the type of data at hand, a multitude of preprocessing steps
can be considered. Some operations have little or no effect on the subsequent
analysis. Others may seem natural but are far from trivial and can even be
harmful. Here we will highlight only a few key techniques and selected popular
methods:
◦ Frequently, some form of normalization or coordinate transformation
is performed which then facilitates the application of a particular machine
learning framework. This can account for mismatched scaling or skewed
distributions of feature values, for instance.
◦ Dimensionality reduction plays a key role in learning problems, where
the (nominal) dimension of feature vectors is relatively high compared to
their intrinsic dimension and/or to the number of example data. The iden-
tification of specifically two- or three-dimensional representations plays an
important role in the exploration and in the human-understandable vi-",The+Shallow+and+the+Deep.pdf
"tification of specifically two- or three-dimensional representations plays an
important role in the exploration and in the human-understandable vi-
sualization of data sets or the trained systems. Feature selection can
be interpreted as a specific form of dimensionality reduction, aiming at
a the identification of a subset of available features that is suitable and
sufficient for the given task.
◦ Unsupervised density estimation, Vector Quantization and cluster-
ing can provide important insights into the structure of the data. For
example, the detection of pronounced clusters of data in a preprocessing
step can help to design a specific classifier or regression system.
◦ Often, the imputation of missing values in incomplete data sets is nec-
essary, in particular when data is scarce and incomplete feature vectors
cannot be simply discarded.
◦ Under- and oversampling techniques are applied when data sets are
imbalanced with respect to the classes in a supervised learning problem.",The+Shallow+and+the+Deep.pdf
"◦ Under- and oversampling techniques are applied when data sets are
imbalanced with respect to the classes in a supervised learning problem.
Similarly, data augmentation aims at enriching the training data in
order to achieve better robustness against noise or to impose certain in-
variances in the training process, e.g. by generating rotated or shifted
training images in object recognition.
The above goals and methods are obviously interrelated. For example, dimen-
sionality reduction by Principal Component Analysis obviously constitutes also
a coordinate transformation. Similarly, density estimation can be used for the
imputation of missing values.
8.1 Normalization and transformations
Probably the most frequently used preprocessing steps concern some form of
normalization or transformation of the feature vectors prior to the actual ma-
chine learning analysis. Clearly, such operations should never be applied blindly
without evaluating their effect on the subsequent analysis.",The+Shallow+and+the+Deep.pdf
"8.1. NORMALIZATION AND TRANSFORMATIONS 189
8.1.1 Coordinate-wise transformations
Frequently, coordinate- or feature-wise transformations are applied in order to
harmonize the range or the statistical properties of the features.
Centering and z-score transformation
Given a set of observations or measurements resulting in N-dim. features vectors
IP =
󰁱
󰁨ξ
µ
∈ RN
󰁲P
µ=1
, it can be convenient and/or useful to subtract the mean
observed in the data set. As a result, one obtains centered feature vectors
ξµ = 󰁨ξ
µ
− 󰁨m where 󰁨m = 1
P
P󰁛
µ=1
󰁨ξ
µ
with 1
P
P󰁛
µ=1
ξµ = 0. (8.1)
Here, we subtract the empirical mean 󰁨m as observed in the data set. In (rare)
cases where the true mean of the corresponding probability density is known, it
could be used to replace the empirical one.
In the N-dimensional Euclidean space, the centering (8.1) corresponds to
a simple translation of all data points, which has no effect on, for instance,",The+Shallow+and+the+Deep.pdf
"a simple translation of all data points, which has no effect on, for instance,
pair-wise Euclidean distances or angles defined by triplets of data points.
A slightly more involved transformation is frequently applied in order to
obtain zero mean, unit variance features:
zµ
j =
󰁨ξ µ
j − 󰁨mj
󰁨σj
with 󰁨σ2
j = 1
P
P󰁛
µ=1
󰀓
󰁨ξ µ
j − 󰁨mj
󰀔2
(8.2)
Here, 󰁨mj is the empirical mean of component 󰁨ξ µ
j in IP as defined in Eq. (8.1) and
󰁨σj is the corresponding standard deviation. As a result, the so-called standard
scores or z-scores zµ
j display zero mean and unit variance in the given data set.
Hence, zµ
j quantifies by how many standard deviations a feature value differs
from the empirical mean in the data set. Positive (negative) zµ
j correspond to
above (below) average values, respectively.
The z-score transformation or standardization is used to account for features
that scale differently, which can be detrimental in a supervised or unsupervised",The+Shallow+and+the+Deep.pdf
"that scale differently, which can be detrimental in a supervised or unsupervised
analysis of the data. The transformation renders the representation independent
of linear rescaling of individual features and choice of units of measure (as
in miles or centimeters for the length of an object). On a univariate level,
considering only single features, the effect of the monotonic transformation is
not critical. However, one should be aware that it can alter the relations between
features in a non-trivial way. As one example, a z-score transformation can affect
the outcome of a Vector Quantization scheme, see the discussion of Fig. 8.6 in
Sec. 8.4.3.
A variety of similar linear transformations can be motivated for specific
problems and data sets. For instance, we could employ the feature median and
interquartile range (IQR) for a scaling analogous to (8.2) in order to achieve
zero median and unit IQR data.",The+Shallow+and+the+Deep.pdf
"190 8. PREPROCESSING AND UNSUPERVISED LEARNING
n
󰁨ξ
n
log(󰁨ξ)
Figure 8.1: Left panel: skewed histogram of a of a feature 󰁨ξ that displays
many small and relatively few large values in a given data set (illustration).
Right panel: histogram of the log-transformed feature log(󰁨ξ).
Min-Max feature scaling
Sometimes it is desirable to transform features to a fixed range of values, e.g.
reflecting the specific requirements of the machine learning system in use. With
minj = min
󰁱
󰁨ξµ
j
󰁲P
µ=1
and maxj = max
󰁱
󰁨ξµ
j
󰁲P
µ=1
we can achieve that
ξj =
󰁨ξj − minj
maxj − minj
∈ [0, 1]. (8.3)
Note that the use of minimum and maximum values can be very sensitive to
the presence or absence of extreme values in a data set.
This or similar transformations are occasionally also referred to as normal-
ization in the literature. In order to avoid confusion, we will use the term only
in the context of the actual Lp-normalization of vectors as described in the
following subsection.",The+Shallow+and+the+Deep.pdf
"in the context of the actual Lp-normalization of vectors as described in the
following subsection.
Non-linear feature transformations
The effect of linear feature-wise transformations can be highly non-trivial, for
instance in the context of classification or unsupervised clustering, cf. Sec. 8.4.
Obviously, the impact of non-linear transformations can be even greater. A
large variety of such transformations can be considered, having specific goals
in mind and taking into account domain knowledge about the properties of the
data set at hand.
As just one particular example we discuss briefly the popular log-transform.
Often, features of the considered data display a skewed distribution of values.
The illustration in Figure 8.1 (left panel) shows the histogram of a non-negative
feature which frequently assumes small values and only rarely larger ones. In
such cases, a feature-wise logarithmic transformation of the form
ξj = log(󰁨ξj) or ξj = log(󰁨ξj + 󰂃) with 󰂃 > 0 (8.4)",The+Shallow+and+the+Deep.pdf
"8.1. NORMALIZATION AND TRANSFORMATIONS 191
Figure 8.2: Astrology as an
artefact of implicit normalization
by projecting stars onto the sur-
face of a sphere. The right panel
can also be viewed as an example
of (mental) overfitting.
Reproduced with kind permis-
sion from John Atkinson for
non-commercial use, see https://
wronghands1.com/ for more of his
brilliant cartoons.
reshapes the histogram to appear – loosely speaking – more Gaussian.2
The transformed data is exactly Gaussian only if the original data follows a
log-normal distribution of the form
P(󰁨ξj) ∝ exp
󰀥
−(log 󰁨ξj − 󰁨µ)2
2󰁨σ2
󰀦
. (8.5)
While this is rarely exactly the case in real world data, skewed distributions
similar to the histogram in Fig. 8.1 (left) are not uncommon in practice, e.g. in
the context of medical data relating to bio-markers, see [Bie17] for an example.
In any case, which transformations make sense in a particular problem de-
pends on available domain knowledge and insights into the data structure.",The+Shallow+and+the+Deep.pdf
"In any case, which transformations make sense in a particular problem de-
pends on available domain knowledge and insights into the data structure.
8.1.2 Normalization
One of the most frequently applied type of transformation is the normalization
of observed N-dimensional feature vectors. Based on so-called Lp-norms
||󰁨ξ||p =
󰀥 N󰁛
j=1
| 󰁨ξj |p
󰀦1/p
we can consider vectors ξ = 󰁨ξ
󰀱
||󰁨ξ||p, (8.6)
which then all display Lp-norm one.
For p = 2, corresponding to the familiar Euclidean norm, we obtain vectors
of the same length in feature space. In other words, all data points are projected
onto the unit sphere in RN . Using Euclidean distance after L2- normalization is
equivalent to measuring distances in terms of angles between the original feature
vectors. The L2-normalization appears to be a natural thing to do, for instance
in the context of the Perceptron, see Chapter 3.
In more general contexts, however, the effects of normalization can be non-",The+Shallow+and+the+Deep.pdf
"in the context of the Perceptron, see Chapter 3.
In more general contexts, however, the effects of normalization can be non-
trivial and even lead to artefacts and mis-interpretations of the data. Astrology
2The choice 󰂃 = 1 is often used for non-negative features as it maps 󰁨ξj = 0 to ξj = 0.",The+Shallow+and+the+Deep.pdf
"192 8. PREPROCESSING AND UNSUPERVISED LEARNING
is a superb example: unrelated stars appear to form meaningful clusters when
they are implicitly projected onto a sphere, see Fig. 8.2 for a tongue-in-cheek
illustration.
As another example, L1-normalization using the so-called Manhattan norm
yields transformed feature vectors with 󰁓
j |ξj| = 1. For non-negative features
e.g. representing amounts of chemical components in a sample, the normalized
data can be interpreted as concentrations. Similarly, event counts would be
transformed to normalized frequencies or probabilities.
Normalization as a preprocessing step can be helpful and greatly beneficial
in practical problems. In any case, it should be applied based on available
domain knowledge, and its effects on the subsequent analysis should be carefully
evaluated.
8.2 Dimensionality reduction
The low-dimensional representation of high-dimensional data plays an impor-",The+Shallow+and+the+Deep.pdf
"evaluated.
8.2 Dimensionality reduction
The low-dimensional representation of high-dimensional data plays an impor-
tant role in the context of data analysis and machine learning, see e.g. [Bis06,
HTF01, LV07, MPH09, BBH12] for a variety of methods and concepts. Several
interconnected, overlapping motivations for dimension reduction can be identi-
fied:
◦ Preprocessing
If high-dimensional data are presented to a machine learning system, the
number of adaptive quantities will be (at least) of the same order. This
may hinder successful training, in particular if the number of available
examples is limited. Dimension reduction can help to overcome this diffi-
culty.
◦ Exploration
Low-dimensional representations help to explore a given data set prior
to, say, supervised learning. Linear or non-linear projections can reveal
structures in the data, e.g. clusters or subspaces in which the feature
vectors are located. Such insights can help to design appropriate systems",The+Shallow+and+the+Deep.pdf
"structures in the data, e.g. clusters or subspaces in which the feature
vectors are located. Such insights can help to design appropriate systems
in the subsequent analysis by taking specific properties of the data into
account.
◦ Visualization
Closely related to the previous point, two- or three-dimensional represen-
tations provide visualizations of the data set which facilitate interaction
with the user or domain expert. Before further analysis, visualization can
give useful insight into the structure of the problem. In retrospect, e.g.
after the training of a classifier, visualization helps to evaluate its perfor-
mance and provides information about regions where classes overlap or
individual misclassified data points are located [SHH20].
It is important to realize that the intrinsic dimensionality of feature vectors
ξ ∈ RN does not necessarily coincide with the nominal dimension N. For",The+Shallow+and+the+Deep.pdf
"8.2. DIMENSIONALITY REDUCTION 193
Figure 8.3: Left panel: schematic illustration of three-dimensional data
points falling into a two-dimensional manifold. Right panel: the special case
of a linear subspace or hyperplane that contains the data points.
instance, a set of vectors {ξµ}P
µ=1 could fall into (or close to) a low-dimensional
manifold M ⊂ RN as illustrated in Fig. 8.3. A particular simple example is a
linear subspace, i.e. a hyperplane which contains the linear dependent feature
vectors, an illustration is shown in the right panel of Fig. 8.3.
Note that the popular term curse of dimensionality [Bel57] is avoided here,
because (a) it is not obvious that a nominally high dimension is necessarily detri-
mental for the performance of machine learning methods (see e.g. [HKK+10,
GT18]) and (b) superstition brings bad luck.
We can distinguish two essentially different concepts for the low-dimensional
representation of high-dim. data: in one family of approaches, each original data",The+Shallow+and+the+Deep.pdf
"representation of high-dim. data: in one family of approaches, each original data
point is represented by an individual counter-part in a low-dim. latent space
without requiring a parameterized mapping between the spaces. The positions
of the representatives are directly determined by means of optimizing a suitable
cost function which is based on the aim of (approximately) preserving neigh-
borhood relations, pair-wise distances, or the overall topology of the original
data points. These approaches do not require an explicit linear or non-linear
functional mapping from the high- to the low-dimensional space. As an impor-
tant example, we present Multi-dimensional Scaling (MDS) below and briefly
mention a few other popular methods.
In the second major framework, an explicit mapping is determined which
provides linear or non-linear projections of the original data into the latent
space. The projection is optimized according to a specific criterion which is",The+Shallow+and+the+Deep.pdf
"space. The projection is optimized according to a specific criterion which is
evaluated w.r.t. a given data set. While these methods are less flexible due
to the pre-defined form of the actual mapping, they offer the possibility to
project out-of-sample data after training. The most popular example of the
basic concept is the well-known Principal Component Analysis (PCA) which is
discussed in Sec. 8.3 also from a neural network perspective.",The+Shallow+and+the+Deep.pdf
"194 8. PREPROCESSING AND UNSUPERVISED LEARNING
ξ2
ξ3
ξ1
 y1
y2
||ξµ − ξν||
||󰂓y µ − 󰂓y ν||
Figure 8.4: Left panel: three-dimensional data points ξµ ∈ R3 which display
a relatively low variance in component ξ3 . Center: the two-dim. represen-
tations 󰂓y µ ∈ R2 as obtained by metric MDS. Right panel: scatter plot of
pair-wise Euclidean distances in N = 3 dimensions vs. distances in M = 2 after
MDS.
8.2.1 Low-dimensional embedding
Consider a given set of N-dim. feature vectors IP =
󰀋
ξµ ∈ RN 󰀌P
µ=1 with a
distance measure dN (ξ, ξ′) that quantifies the pair-wise dissimilarity of data
points in IP. As a straightforward example we can think of the Euclidean
distance in N dimensions, but the consideration of any meaningful dissimilarity
is possible. We refer to the pair-wise distances as dµν
N = dN (ξµ, ξν) in the
following.
A number of methods aim at finding an embedding of the data points in
a low-dimensional Euclidean vector space that preserves relations between the",The+Shallow+and+the+Deep.pdf
"A number of methods aim at finding an embedding of the data points in
a low-dimensional Euclidean vector space that preserves relations between the
individual feature vectors in the original space, as much as possible. The goal
could be to approximately reproduce the pair-wise distances themselves, their
rank structure, or associated probability densities.
8.2.2 Multi-dimensional Scaling
The goal of metric Multi-dimensional Scaling (MDS) 3 is to find representatives
󰂓y µ ∈ RM with M < N, which preserve pair-wise distances and their relations
as far as possible. To this end we consider coordinates yµ
j ∈ R (j = 1, . . . M
and µ = 1, . . . P) in the target space. There, we evaluate the dissimilarities of
the corresponding vectors {󰂓y µ}P
µ=1 by means of a metric dM (󰂓y, 󰂓y ′). Again, this
measure could be based on any reasonable vector norm, with Euclidean metric
being the classical and by far most popular choice. Note that, in general, the",The+Shallow+and+the+Deep.pdf
"being the classical and by far most popular choice. Note that, in general, the
measures dN and dM need not be of the same type. An example cost function
3So-called classical MDS is also known as Principal Coordinates Analysis and is not dis-
cussed here [Mea92].",The+Shallow+and+the+Deep.pdf
"8.2. DIMENSIONALITY REDUCTION 195
to optimize in MDS is the quadratic deviation
E
󰀓
{󰂓y µ}P
µ=1
󰀔
=
P󰁛
µ,ν=1
µ<ν
󰀗
dµν
N − dM (󰂓y µ, 󰂓y ν)
󰀘2
. (8.7)
All coordinates yµ
j are considered degrees of freedom that can be obtained by
minimization of E.
A (local) minimum of (8.7) corresponds to an arrangement of P points in M
dimensions which reflects the pair-wise distances dµν
N as well as possible. For
M < N it is in general not possible to obtain a perfect solution with E = 0.
Obviously, we can also not expect to find a unique minimum of E, the actual
outcome of MDS will depend on the initial configuration of 󰂓y µ and on the poten-
tial randomness in the training process. Figure 8.4 illustrates MDS in terms of a
simple example based on Euclidean distance in both spaces: three-dimensional
feature vectors ξµ (left panel) display a relatively small variance in component
ξ3. The corresponding two-dimensional representations 󰂓y µ, obtained by min-",The+Shallow+and+the+Deep.pdf
"ξ3. The corresponding two-dimensional representations 󰂓y µ, obtained by min-
imizing the quadratic deviation (8.7), is displayed in the center panel. The
scatter plot in the right panel shows that very similar pair-wise distances have
been achieved in M = 2 dimensions. Quantitative measures, e.g. a correlation
coefficient, could be obtained in order to evaluate the quality of the MDS result.
Note that the quality of the embedding is invariant under, e.g., simultaneous
translations or rotations of the 󰂓yµ.
The term MDS is frequently meant to imply the use of Euclidean distances
in RN and RM , and minimizing the quadratic deviation (8.7). Various modifica-
tions of the basic idea have been suggested and applied in practice. Obviously,
a variety of distance measures dN and dM could be considered. Moreover, spe-
cific cost functions can be chosen which put, for instance, different emphasis
on small or large distances. A good overview of MDS related methods can be",The+Shallow+and+the+Deep.pdf
"on small or large distances. A good overview of MDS related methods can be
obtained from [LV07] and [MPH09]. In particular [BBH12] discusses several
choices in a unified framework, including the so-called Isomap [TdSL00], Sam-
mon mapping [Sam69], Local Linear Embedding (LLE) [RS00], and Laplacian
Eigenmaps [BN03].
8.2.3 Neighborhood Embedding
Another popular family of methods is often used for the visual inspection of
complex data sets. In the original Stochastic Neighborhood Embedding (SNE)
as introduced by Hinton and Roweis [HR03], the data is characterized in terms
of Gaussian probabilities in the original and in the embedding space. These are
replaced by long-tailed student-t distributions in the popular t-distributed SNE
(t-SNE) that was later suggested by van der Maaten and Hinton [MH08].
Both, SNE and t-SNE, aim at the minimization of the Kullback-Leibler
divergence as a measure of (dis-)similarity between the assumed densities in",The+Shallow+and+the+Deep.pdf
"Both, SNE and t-SNE, aim at the minimization of the Kullback-Leibler
divergence as a measure of (dis-)similarity between the assumed densities in
the original and the embedding space. The optimization can be done using
gradient-based methods. Recently, an alternative known as Uniform manifold",The+Shallow+and+the+Deep.pdf
"196 8. PREPROCESSING AND UNSUPERVISED LEARNING
approximation and projection (UMAP) has become popular [MHM20]. Its key
feature is the assumption that data is distributed in (or close to) a particular
manifold along which distances are computed.
Embedding methods are very popular, mainly in the context of data ex-
ploration and visualization. For more detailed presentations of neighborhood
embedding, we refer the reader to the original literature and reviews like [LV07,
BBH12,BBK20]. One of the main drawbacks of these methods is that, generally
speaking, the embedding coordinates 󰂓y are not directly interpretable and their
connection to the original feature space is often unclear. In this sense, embed-
ding methods are not very appealing as preprocessing steps for further super-
vised learning (regression or classification). Moreover, out-of-sample extension,
i.e. the post-hoc embedding of data points that were not in IP is non-trivial.",The+Shallow+and+the+Deep.pdf
"i.e. the post-hoc embedding of data points that were not in IP is non-trivial.
One option is to add them to the set and then recompute all 󰂓y µ on the basis of
the extended data, which is obviously very costly. Alternatively, one can try to
obtain an explicit mapping function Ψ that approximately realizes Ψ : ξµ → 󰂓y µ
and can be applied to novel data points, see for instance [EHT20] for a Deep
Learning based approach.
8.2.4 Feature selection
A rather direct way of reducing the dimensionality of input data is to select
a subset of features, neglecting all others. It appears to be a natural idea to
simply try all different combinations of features and select the best possible
subset. In practice, however, the number of subsets with k features selected
from N dimensions is
󰀃N
k
󰀄
and grows very rapidly with k and N.
An overview of basic feature selection methods can be found in [GE03,CS14,
JBB15]. Three main strategies for the selection of feature subsets have been",The+Shallow+and+the+Deep.pdf
"JBB15]. Three main strategies for the selection of feature subsets have been
considered in the literature:
◦ Filter methods:
So-called filter methods aim at the identification of useful features with-
out actually training a classification or regression model. In unsupervised
approaches, features are selected or rejected based on their statistical prop-
erties, independent of the actual target problem. For instance, correlations
between different features could be exploited in order to discard redundant
features in a given data set.
Supervised filtering takes into account the label information in the data.
For instance, in the context of regression, individual features can be evalu-
ated and selected in terms of their correlations with the target. Similarly,
in classification problems, mutual information or cross entropy between
features and the target can serve as a selection criterion. Most frequently,
these criteria are applied in a univariate fashion, feature by feature. This",The+Shallow+and+the+Deep.pdf
"these criteria are applied in a univariate fashion, feature by feature. This
bears the risk of missing the relevance of combinations of features which
would be revealed in an appropriate multivariate analysis.
◦ Wrapper methods:
In the wrapper approach, the idea is to consider different combinations of",The+Shallow+and+the+Deep.pdf
"8.3. PCA AND RELATED METHODS 197
features in terms of the performance of trained regressors or classifiers. For
each candidate set of features, training and potentially validation schemes
have to be performed, which can result in considerable computational
costs. An advantage of the wrapper approach over univariate filter meth-
ods is that, in principle, favorable combinations of features can be found.
Obviously, there is no straightforward way to find the optimal set of fea-
tures for a given task. One can start with the full set of available features
and remove single ones from the set, in every step selecting the one that
has the least impact on the performance after training. Alternatively, one
could add features to the set, following a greedy strategy to improve the
quality of the classification or regression in every step.
◦ Embedded methods:
In embedded methods the selection of a subset of features is an integral",The+Shallow+and+the+Deep.pdf
"◦ Embedded methods:
In embedded methods the selection of a subset of features is an integral
part of training a specific type of model, i.e. classifier or regression system.
A popular example would be the training of decision trees in a Random
Forest [Bre01], in which a tree selects a particular feature in every branch-
ing step. Hence, the actual feature set is compiled while the system is
trained. While this can be more efficient than the wrapper approach, it is
- in a sense - limited to an essentially univariate evaluation of features.
8.3 PCA and related projection methods
A variety of methods derives an explicit mapping of the form
Ψ : RN → RM with 󰂓y = Ψ(ξ) ∈ RM (8.8)
directly from the data set IP. The mapping Ψ is parameterized and obtained
in a data driven process. Hence, unlike MDS or other embedding methods,
the mapping can be applied to out-of-sample data ξ /∈ P immediately. This is
particularly important for methods of supervised learning, where we can derive",The+Shallow+and+the+Deep.pdf
"particularly important for methods of supervised learning, where we can derive
Ψ from the training data and apply the same mapping to novel data in the
working phase.
In principle, we can obtain a meaningful projection by parameterizing the
function Ψ in a suitable way and then optimizing its parameters according to
an appropriate cost function. All criteria discussed in the context of embedding
in the previous section could be employed for the identification of an explicit
mapping as well.
We discuss here only methods which employ linear projections of the data.
Most methods, including Principal and Independent Component Analysis can
be extended in various ways. Non-linear versions can be formulated, for in-
stance, in terms of shallow or deep neural autoencoders as discussed in Sec.
5.5.1. Kernelized versions also exist, see for instance [SSM98, LG19] and refer-
ences therein.",The+Shallow+and+the+Deep.pdf
"198 8. PREPROCESSING AND UNSUPERVISED LEARNING
8.3.1 Principal Component Analysis
Due to their simplicity and intuitive nature, linear mappings are of particular
interest and practical relevance. For example, Principal Component Analysis
(PCA) is one of the most important and frequently used explicit mappings
in the context of data analysis, unsupervised learning, low-dimensional rep-
resentation and visualization. Primarily, PCA is employed to determine low-
dimensional representations. In addition it can be used to identify supposedly
non-informative contributions in a given data set. The basic idea is to disregard
linear combinations of features that hardly vary over the observed data.
Like many other standard and well-established methods, PCA can be mo-
tivated and derived from a variety of perspectives. Here we follow a mostly
heuristic line of thoughts and point out the relation to the theoretical back-
grounds in passing.",The+Shallow+and+the+Deep.pdf
"heuristic line of thoughts and point out the relation to the theoretical back-
grounds in passing.
For convenience we will assume throughout the following that the data set
used for the identification of the mapping Ψ is centered, i.e. 1
P
󰁓P
µ=1 ξµ = 0. If
this has been achieved by a centering of the form (8.1), the same transformation
using the empirical mean in IP has to be performed on novel input vectors, before
Ψ can be applied.
The direction w = u1 ∈ RN along which the centered data displays the
largest variance is of particular interest. We can formulate the search as an
optimization problem with respect to the empirical variance along w:
Evar = 1
P
P󰁛
µ=1
(yµ)2 with the projections yµ = w · ξµ. (8.9)
It is plausible to assume that the corresponding solution of largest variance, is
the direction in which most of the information about ξ is contained. In fact,
this is rigorously true under the assumption that all features follow a normal",The+Shallow+and+the+Deep.pdf
"this is rigorously true under the assumption that all features follow a normal
distribution. If we define the empirical covariance matrix 4
C ∈ RN×N with elements Cij = 1
P
P󰁛
µ=1
ξµ
i ξµ
j , (8.10)
we can rewrite the objective function, Eq. (8.9), as a quadratic form
󰀵
󰀷Evar =
N󰁛
j,k=1
wj wk
1
P
P󰁛
µ=1
ξµ
j ξµ
k
󰀶
󰀸 = w⊤ C w. (8.11)
The matrix C is positive semi-definite by definition, the quadratic form is non-
negative but otherwise unbounded. We can assume that C has ordered eigen-
values
λ1 ≥ λ2 ≥ . . . ≥ λN ≥ 0. (8.12)
4not to be confused with the complementary P ×P matrix C with elements Cµν ∝ 󰁓
j ξµ
j ξν
j
defined in (3.74) in the context of classification.",The+Shallow+and+the+Deep.pdf
"8.3. PCA AND RELATED METHODS 199
Restricting the search to normalized w with |w| = 1, it is straightforward to
show that the maximum of Evar is achieved if w is the eigenvalue of C with the
largest eigenvalue λ1:
w = u1 with Cu1 = λ1u1.
In case of degenerate leading eigenvalues, λ1 = λ2 = . . . = λm, we have to
consider the corresponding m-dimensional eigenspace. Note that the variance
associated with the normalized u1 is given directly by its eigenvalue:
1
P
P󰁛
µ=1
(u1 · ξµ)2 = u⊤
1 Cu1 = λ1(u1)2 = λ1. (8.13)
Once we have determined u1 as the direction of largest variation, we proceed
by identifying the direction u2 which displays the largest variance in the space
orthogonal to u1. In general we can determine the ordered sequence of directions
um in which the data displays the largest variance with um · uk for all k =
1, 2 . . . m − 1. Hence, we identify the M leading Principal Components of IP
with the leading eigenvectors of C
{wk}M
k=1 with projections yµ
k = wk · ξµ and variance 1",The+Shallow+and+the+Deep.pdf
"with the leading eigenvectors of C
{wk}M
k=1 with projections yµ
k = wk · ξµ and variance 1
P
P󰁛
µ=1
(yµ
k )2 = λk.
(8.14)
In the simple illustration shown in Fig. 8.3 (right panel), the vectors marked
in red correspond to the two leading eigenvectors or principal components. They
can serve as coordinate axes defining the two-dimensional subspace of largest
variation in the data set.
The M-dim. vectors 󰂓y µ = (yµ
1 , yµ
2 , . . . yµ
M )⊤ serve as M-dimensional repre-
sentations of the data set. The associated linear subspace displays the largest
variances in the data set. Under appropriate assumptions of Gaussianity, this
implies that the vectors 󰂓y µ have the maximum possible information content
about the high-dimensional ξµ.
Equivalently, one can show that, for fixed dimensionality M, PCA can be
interpreted as a linear autoencoder, cf. Sec. 5.5.1, realizing a dimensionality
reducing a linear mapping and back-transformation:
󰂓y µ
k = uk · ξµ, ξµ
est =
M󰁛
k=1
yµ
k uk, (8.15)",The+Shallow+and+the+Deep.pdf
"reducing a linear mapping and back-transformation:
󰂓y µ
k = uk · ξµ, ξµ
est =
M󰁛
k=1
yµ
k uk, (8.15)
realizes the smallest quadratic reconstruction error (5.37). Introducing the ma-
trix U = [u1, u2, . . . uM ]⊤ ∈ RM×N we can write (8.15) in the compact form
󰂓yµ = U ξµ, ξµ
est
= U⊤ 󰂓yµ. (8.16)
Generically, we will apply PCA to high-dimensional data sets, in particular
if the number of examples is lower than the nominal dimension, i.e. P < N.",The+Shallow+and+the+Deep.pdf
"200 8. PREPROCESSING AND UNSUPERVISED LEARNING
Obviously the subspace of non-zero variability in the data set is at most P-
dimensional in this case, as reflected by the rank of C. Hence, the maximum
number of meaningful principal components is M = P. Typically, fewer com-
ponents are used: under the assumption that the leading eigenvectors carry
most information, the mapping is truncated at relatively small M, neglecting
variations along minor eigendirections as noise.
Powerful tools exist that can be used to determine the leading eigenvectors
of the symmetric, semi-definite matrix C. Often, numerical methods for the
more general singular value decomposition (SVD) are preferred [Str19,DFO20].
SVD reduces to eigenvalue decomposition for diagonalizable matrices, but it is
claimed to be numerically more stable.
Whitening
PCA can also be used for a so-called Whitening transformation. The trans-
formed coordinates
󰁥yµ
k = uk · ξµ
√λk
(8.17)",The+Shallow+and+the+Deep.pdf
"Whitening
PCA can also be used for a so-called Whitening transformation. The trans-
formed coordinates
󰁥yµ
k = uk · ξµ
√λk
(8.17)
display zero mean and unit variance for all k. Hence, on the level of second
order statistics, the data appears totally isotropic with an identity covariance
matrix. Note that this does not imply that there is no structure left in the data.
Whitening can be useful when analysing properties of the data which concern
higher order statistics as in Independent Component Analysis, see Sec. 8.3.3.
8.3.2 PCA by Hebbian learning
It is instructive to study a very simple numerical procedure to compute u1, which
corresponds to the power method or von Mises iteration [MP29, Wil65]. As we
will see, it relates to Hebbian learning and can be extended to the computation
of several leading eigenvectors from a neural network perspective.
Consider an initial vector w(0) which can be expanded in terms of the eigen-
vectors ui of C:
w(0) =
N󰁛
j=1",The+Shallow+and+the+Deep.pdf
"Consider an initial vector w(0) which can be expanded in terms of the eigen-
vectors ui of C:
w(0) =
N󰁛
j=1
aj uj with coefficients aj ∈ R and a1 ∕= 0.
Repeated multiplication (from the left) with C leads to the t-th iterate
w(t) = Ctw(0) =
N󰁛
j=1
aj λt
j uj = λt
1
󰀵
󰀷a1u1 +
N󰁛
j=2
󰀕λj
λ1
󰀖t
aj uj
󰀶
󰀸
≈ λ
t
1a1u1 for t → ∞, (8.18)
where we omit contributions that vanish like O
󰀕󰀏󰀏󰀏λ2
λ1
󰀏󰀏󰀏
t󰀖
.",The+Shallow+and+the+Deep.pdf
"8.3. PCA AND RELATED METHODS 201
Hence, assuming that 0 < λ2 < λ1, the weights w(t) will be dominated by
the leading eigenvector u1 for t → ∞. In case of degeneracies, the argument
can be extended to the m-dimensional subspace of leading eigenvalues and the
role of λ2 is taken over by λm+1 accordingly.
We can obtain a modified iteration by replacing C by the shifted matrix
[IN + η C] with η > 0 where IN is the N-dim. identity. This matrix has the
same eigenvectors as C and ordered eigenvalues 1 + ηλj. The corresponding
iteration reads
w(t) = [IN + ηC] w(t−1) = w(t−1)+ η Cw(t−1) = w(t−1)+ η 1
P
P󰁛
µ=1
yµ(t) ξµ
(8.19)
where yµ(t) = w(t − 1) · ξµ and we exploit that for general w ∈ RN
[Cw]i =
N󰁛
j=1
Cijwj = 1
P
P󰁛
µ=1
ξµ
i
N󰁛
j=1
ξµ
j wj = 1
P
P󰁛
µ=1
ξµ
i yµ.
Instead of computing the sum over µ = 1, . . . P in each iteration step (8.19)
we can update the vector w in single steps of the form
w(τ) = w(τ − 1) + η yµ(τ) ξµ (8.20)",The+Shallow+and+the+Deep.pdf
"we can update the vector w in single steps of the form
w(τ) = w(τ − 1) + η yµ(τ) ξµ (8.20)
where w(0) is the initial vector and the index µ on the r.h.s follows the sequence
µ = 1, 2 . . . P, 1, 2, . . . which corresponds to the repeated presentation of all data
in a fixed sequential order 5.
Note that the update (8.20) can be interpreted as Hebbian Learning in a
linear neuron with inputs ξj and output y = w · ξ. The update can also be
derived as the gradient based maximization of cost function Evar in Eq. (8.9).
In practice, it makes sense to normalize the weight vector after each update
step in order to avoid numerical problems of diverging or vanishing |w|:
w(τ) = w(τ − 1) + η yµ(τ) ξτ
|w(τ − 1) + η yµ(τ) ξτ | (8.21)
with yµ(τ) = w(τ − 1) · ξµ. Assuming that the previous w(τ − 1) was already
normalized, we note that
|w(τ − 1) + η yµ(τ) ξτ | =
󰁳
1 + 2η[yµ(τ)]2 + O(η2).
The last term in the square root can be neglected for small learning rates η → 0.",The+Shallow+and+the+Deep.pdf
"|w(τ − 1) + η yµ(τ) ξτ | =
󰁳
1 + 2η[yµ(τ)]2 + O(η2).
The last term in the square root can be neglected for small learning rates η → 0.
In the same limit (1 + ηz)−1/2 ≈ (1 + ηz/2) and we therefore obtain
Oja’srule
w(τ) = w(τ −1) + η
󰀣
yµ(τ) ξτ − [yµ(τ)]2w(τ −1)
󰀤
. (8.22)
5The change from Eq. (8.19) to (8.20) is reminiscent of the difference between Gauss-Seidel
and Jacobi iterations for linear systems [Fle00], see Sec. 3.7.2 It also resembles the relation
between batch and stochastic gradient descent, cf. see Appendix A.48.",The+Shallow+and+the+Deep.pdf
"202 8. PREPROCESSING AND UNSUPERVISED LEARNING
Here terms O(η2) have been omitted. The iteration (8.22) is known as Oja’s
Rule after Erkki Oja [Oja82]. It combines Hebbian learning with an appropriate
weight decay that realizes an approximate normalization of w for small learning
rates.
Further principal components
In principle, one could apply the power method also for the second and all
following principal components. Theoretically, we could avoid contributions
of u1 in the iteration (8.18) by starting from a w(0) with coefficient a1 =
0. Analogously, we would set a1 = a2 = . . . am−1 = 0 when computing um.
Similarly, one could consider the modified matrices
Cm = C −
m−1󰁛
k=1
λk[uku⊤
k ]
for the iteration of the vector wm. However, both ideas are at risk to fail in
practice, as numerical inaccuracies will always introduce small but non-zero
contributions of u1 which will blow up in the iterations, if not corrected for.
It is more promising to iterate a set of vectors {wj}M",The+Shallow+and+the+Deep.pdf
"contributions of u1 which will blow up in the iterations, if not corrected for.
It is more promising to iterate a set of vectors {wj}M
j=1 in parallel and im-
pose appropriate conditions after each step, e.g. by means of a Gram-Schmidt
orthonormalization.
Two closely related extensions of Oja’s Rule are based on this concept. We
present here only the single step variants and refer the reader to the original
literature [Oja89,San89]. For details and convergence proofs, see also [HKP91]
for a more elaborate discussion.
Oja’ssubspace algorithm and Sanger’srule (8.23)
Initialize a random wm(0) with |wm(0)| = 1 for m = 1, 2, . . . M
At discrete time step τ
- determine the index µ of the current example (sequential presentation)
- update the vectors wm(τ), m = 1, 2 . . . M
according to Oja’ssubspace algorithm:
wm(τ) = wm(τ − 1) + η yµ
m(τ)
󰀥
ξµ −
M󰁛
k=1
yµ
k (τ)wk(τ − 1)
󰀦
(8.24)
or according to Sanger’srule:
wm(τ) = wm(τ − 1) + η yµ
m(τ)
󰀥
ξµ −
m󰁛
k=1
yµ
k (τ)wk(τ − 1)
󰀦
(8.25)
where yµ",The+Shallow+and+the+Deep.pdf
"ξµ −
M󰁛
k=1
yµ
k (τ)wk(τ − 1)
󰀦
(8.24)
or according to Sanger’srule:
wm(τ) = wm(τ − 1) + η yµ
m(τ)
󰀥
ξµ −
m󰁛
k=1
yµ
k (τ)wk(τ − 1)
󰀦
(8.25)
where yµ
k (τ) = wk(τ − 1) · ξµ(τ).",The+Shallow+and+the+Deep.pdf
"8.3. PCA AND RELATED METHODS 203
The first terms in the brackets [. . .] of Eqs. (8.24, 8.25) correspond to the familiar
Hebbian learning for linear units. The remaining terms can be motivated as
approximate normalizations for k = m as in (8.20), while the terms with k ∕= m
are associated with the pair-wise orthogonalization of vectors.
Note that in Oja’s subspace algorithm the sum in Eq. (8.24) is over all indices
k = 1, 2, . . . M, while in Sanger’s rule (8.25) it is truncated to k = 1, 2, . . . m
for the m-th vector. This imposes a hierarchy among the iterated weight vec-
tors. In fact, one can show that Sanger’s algorithm yields the ordered principal
components, i.e. wk(τ) → uk for large τ. On the contrary, in Oja’s subspace
algorithm the vectors {wk}M
k=1 converge to form an arbitrary orthonormal basis
of the same subspace. They do not necessarily become identical with uk and do
not display the same order with respect to the partial variances of the data.",The+Shallow+and+the+Deep.pdf
"not display the same order with respect to the partial variances of the data.
Orthogonal vectors wk ⊥ wl result in uncorrelated projections yµ
k and yµ
l .
In fact, the updates (8.25) and (8.24) can also be derived as (single example)
gradient maximization of cost function (8.9) complemented by penalty terms of
the form −[yµ
k yµ
l ]2 for k ∕= l.
8.3.3 Independent Component Analysis
As discussed above, PCA identifies directions in which uncorrelated projections
display the largest variances. In Independent Component Analysis (ICA), the
goal is to find orthogonal directions which are independent in a broader sense
[HTF01, Sto04, HO00]. ICA is, for instance, used for Blind Source Separation
to identify independent sources in a mixed signal [CJ10]. Algorithms based
on Hebbian Learning have also been suggested for Independent Component
Analysis, see e.g. [CLG08,LG19].
Typically, the problem is addressed by considering a proxy for statistical
independence. Two popular concepts are",The+Shallow+and+the+Deep.pdf
"Analysis, see e.g. [CLG08,LG19].
Typically, the problem is addressed by considering a proxy for statistical
independence. Two popular concepts are
a) Mutual information
Here, projections are identified which carry as little information about
each other as possible. For projections x = w⊤ξ and y = v = v⊤ξ the
mutual information can be determined as the relative entropy [HTF01]
Ix,y =
󰁛
x∈X
󰁛
y∈Y
px,y(x, y) log
󰀗 px,y(x, y)
px(x) py(y)
󰀘
with the joint density px,y in the domain X ×Y and the marginal densities
px and py. Minimizing Ix,y identifies independent directions w and v in
feature space.
b) Deviation from Gaussianity
Orthogonal directions in which the projections appear least Gaussian are
expected to be most interesting and potentially relevant for the task at
hand [HO00]. They would display, for instance, very skewed or multimodal
histograms of projections.",The+Shallow+and+the+Deep.pdf
"204 8. PREPROCESSING AND UNSUPERVISED LEARNING
Figure 8.5: Left panel: three unimodal histograms with kurtosis ≈ −1 (left
panel), kurtosis ≈ 0 (center panel, corresponding to a normal density), and
kurtosis ≈ 2 (right panel), respectively.
As one example strategy for (b), we discuss here the maximization (in ab-
solute value) of the kurtosis of projections yµ = w · ξµ [HO00, HTF01]. It is
defined as
kurtosis = (y − y)4
(y − y)22 − 3, (8.26)
where (. . .) denotes means of the type y = 1
P
󰁓P
µ=1 yµ. Very often, the data is
centered and whitened in a first step, Eq. (8.17), appearing isotropic with unit
variance and zero mean in any direction. In this case, Eq. (8.26) simplifies to
kurtosis = y4 − 3. (8.27)
The kurtosis as defined here is zero for normal densities.6
Projections yµ = w⊤ξ with maximum kurtosis (in absolute value) can serve
as a proxy for directions in which the data differs most from Gaussianity. Figure",The+Shallow+and+the+Deep.pdf
"as a proxy for directions in which the data differs most from Gaussianity. Figure
8.5 displays histograms of a unimodal density with kurtosis< 0 (left panel),
which appears more bumpy than a Gaussian (center panel), and a density with
kurtosis> 0, which appears pointier. Similarly, other non-Gaussian densities,
e.g. multimodal or skewed ones, also have a non-zero kurtosis.
8.4 Clustering and Vector Quantization
The identification and exploration of structures in a given data set can constitute
a very useful preprocessing step. Quite often, methods of unsupervised learning
are applied before the actual classification or regression scheme is implemented.
The purpose can be to obtain insight into the complexity of the problem and
into the difficulties that might be expected. If, for example, the data set contains
several well-defined clusters or groups of similar feature vectors, this knowledge
could be used in the design of a specific classifier. Similarly, density estimation",The+Shallow+and+the+Deep.pdf
"could be used in the design of a specific classifier. Similarly, density estimation
techniques can be applied to obtain knowledge about the general statistical
properties of the data.
6Note that the additive term −3 is sometimes omitted in the literature. Then, the kurtosis
of a Gaussian density is obviously 3.",The+Shallow+and+the+Deep.pdf
"8.4. CLUSTERING AND VECTOR QUANTIZATION 205
After a very brief discussion of elementary distance-based clustering meth-
ods, we present two prominent and related methods of unsupervised data anal-
ysis: Vector Quantization (VQ) by competitive learning and Gaussian Mixture
Models (GMM) for density estimation.
8.4.1 Basic clustering methods
A variety of specific clustering techniques exist. Many of them are based on
suitable distance measures in feature space like the familiar Euclidean metrics in
the simplest case. In methods of hierarchical clustering the distance D(Ca, Cb)
between two clusters is derived from the pairwise d(xa, xb) of their individual
elements xa ∈ Ca and xb ∈ Cb. Prominent criteria for the computation of
cluster distances are known as
◦ single linkage with
D(Ca, Cb) = min {d(xa, xb)}xa∈Ca,xb∈Cb
,
where the closest pair of feature vectors determines D(Ca, Cb).
◦ average linkage where
D(Ca, Cb) = 1
|Ca||Cb|
󰁛
xa∈Ca
󰁛
xb∈Cb
d(xa, xb)",The+Shallow+and+the+Deep.pdf
",
where the closest pair of feature vectors determines D(Ca, Cb).
◦ average linkage where
D(Ca, Cb) = 1
|Ca||Cb|
󰁛
xa∈Ca
󰁛
xb∈Cb
d(xa, xb)
is determined by the average pair-wise distance. Here, the number of
vectors contained in a cluster C is denoted as |C|.
◦ complete linkage with
D(Ca, Cb) = max {d(xa, xb)}xa∈Ca,xb∈Cb
corresponds to the largest pairwise distance of vectors in Ca and Cb.
These and similar measures can be used in so-called hierarchical clustering ap-
proaches like
◦ agglomerative clustering (bottom up):
Here, every data point is initially considered a cluster. In every step the
two clusters Ca and Cb, which are the closest according to some criterion
D(Ca, Cb), are merged into one cluster.
◦ divisive clustering (top down):
Here one starts by assigning all of the data to one cluster. In every step,
one of the current clusters is selected and split into two sub-clusters. Var-
ious criteria can be considered in the cluster selection and to guide the",The+Shallow+and+the+Deep.pdf
"ious criteria can be considered in the cluster selection and to guide the
actual division, see e.g. [DHS00].
Unlike K-means or similar methods, hierarchical clustering does not require
a predefined number of clusters. On the contrary, the procedure generates a
hierarchical tree of clusters. Deeper levels of this so-called dendrogram represent
splits of the data set into an increasing number of sub-clusters. A particular
configuration can be chosen once the dendrogram is constructed.",The+Shallow+and+the+Deep.pdf
"206 8. PREPROCESSING AND UNSUPERVISED LEARNING
8.4.2 Competitive learning for Vector Quantization
One possible aim of unsupervised learning is the representation of a potentially
large set of feature vectors by a few typical representatives. The term Vector
Quantization (VQ) has been coined for this task. In VQ, so-called prototypes
should reflect the frequency of observations in feature space. The goal could be
to reduce storage needs or to reveal structures such as clusters in the data.
In the following we present a basic scheme for unsupervised Vector Quanti-
zation by competitive learning [Koh97, HKP91, Bis95a, HTF01, BHV16]. Tech-
nically, it resembles the methods of (supervised) Learning Vector Quantization
discussed in Sec. 6.1 in that it also employs the concept of competitive learning.
However, unsupervised VQ is applied to unlabeled feature vectors and its aim is
the faithful representation of data sets, not an actual classification or regression
task.",The+Shallow+and+the+Deep.pdf
"the faithful representation of data sets, not an actual classification or regression
task.
Like most unsupervised learning techniques, VQ is also guided by a cost
function. Here, it is optimized in terms of the prototype vectors. The objective
function measures the quality of a particular prototype configuration with re-
spect to a given set of vectors P =
󰀋
xµ ∈ RN 󰀌P
µ=1. In neuroscience jargon, the
input vectors can be interpreted as stimuli which activate the neurons or units
that are characterized by N-dim. vectors
󰀋
w1, w2, . . . wK󰀌
. The prototype vec-
tors wk ∈ RN can be seen as weight vectors, exemplars or expected stimuli,
see [BHV16].
A very popular approach to VQ is based on the crisp assignment of any data
point to the closest prototype, the so-called winner in terms of a pre-defined,
fixed distance measure. We restrict the discussion to the use of simple squared
Euclidean distance d(x, y) = (x − y)2 /2 for x, y ∈ RN . Generalizations to",The+Shallow+and+the+Deep.pdf
"Euclidean distance d(x, y) = (x − y)2 /2 for x, y ∈ RN . Generalizations to
alternative measures are in principle possible along the lines discussed in 6.1.
A corresponding, suitable cost function is given by the so-called quantization
error [Bis95a,Bis06,HTF01,BHV16]:
HV Q = 󰁓P
µ=1
1
2
󰀃
wk(µ) − xµ󰀄2
. (8.28)
Here, wk(µ) ∈ RN denotes the winning prototype with the smallest Euclidean
distance from xµ ∈ RN :
d
󰀃
wk(µ), xµ󰀄
≤ d
󰀃
wj, xµ󰀄
for all j = 1, 2, . . . , K, (8.29)
where ties are broken arbitrarily. Hence, HV Q accumulates the quadratic dis-
tances of all feature vectors from their closest prototype. In this sense, the
quantization error indicates how faithfully the set of feature vectors is repre-
sented by the prototypes.
Standard competitive VQ corresponds to the stochastic gradient descent
based minimization of HV Q, see Appendix A.48. The resulting algorithm is very
intuitive and could be obtained on purely heuristic grounds: at each discrete",The+Shallow+and+the+Deep.pdf
"intuitive and could be obtained on purely heuristic grounds: at each discrete
time step, a single randomly selected feature vector xµ is drawn with equal
probability from the data set. Then, the currently closest prototype wk(µ) is
determined according to Eq. (8.29). Similar to the supervised LVQ1 algorithm",The+Shallow+and+the+Deep.pdf
"8.4. CLUSTERING AND VECTOR QUANTIZATION 207
of Sec. 6.1, only the winner is adapted. However, in unsupervised VQ the
winning prototype is always moved closer to the considered input vector; the
update does not depend on additional information such as the labels in Learning
Vector Quantization:
Competitive learning (Vector Quantization)
– at time step t, select a single feature vector xµ randomly
from the data set P with equal probability 1/P
– identify the winning prototype, i.e.
wk(µ) with d(wk(µ), xµ) = min
󰀋
d(wj, xµ)
󰀌
– update only the winning prototype according to:
wk(µ)(t+1) = wk(µ)(t) + η(t)
󰀅
xµ − wk(µ)(t)
󰀆
. (8.30)
The term competitive learning has been coined for this and several related
training schemes as prototypes compete for updates [Koh97,Bis95a,Bis06,RMS92,
BHV16]. The algorithm described above is an example of a Winner-Takes-All
(WTA) scheme, extensions to the update of several prototypes at every time step",The+Shallow+and+the+Deep.pdf
"(WTA) scheme, extensions to the update of several prototypes at every time step
have been considered also in the context of unsupervised Vector Quantization.
As in any stochastic gradient descent, convergence of the prototype vectors
has to be guaranteed by employing a time-dependent or adaptive learning rate
η(t) which slowly approaches zero in the course of training, see the discussion
of SGD in Appendix A.5.
Competitive Vector Quantization is closely related to the well-known Lloyd’s
algorithm or K-means algorithm [HTF01, Llo82, DHS00]. In contrast to the
prescription (8.30) it considers all available data in each update of the system
and alters all prototypes at a time:
K-means algorithm (Lloyd’salgorithm)
Given a data set
󰀋
xµ ∈ RN 󰀌
(µ = 1, 2 . . . P), initialize
K prototype vectors
󰀋
wj ∈ RN 󰀌
(j = 1, 2, . . . , N).
Repeat the following steps:
A) assign every data point xµ to the nearest prototype wk(µ) with
d
󰀃
wk(µ), xµ󰀄
= min
󰀋
d(wj, xµ)
󰀌
.",The+Shallow+and+the+Deep.pdf
"Repeat the following steps:
A) assign every data point xµ to the nearest prototype wk(µ) with
d
󰀃
wk(µ), xµ󰀄
= min
󰀋
d(wj, xµ)
󰀌
.
B) compute updated prototypes (centers) wj as the mean of
all data points which were assigned to wj in (A):
wj = 󰁓P
µ=1 xµδj,k(µ)
󰀱󰁓P
µ=1
δj,k(µ). (8.31)",The+Shallow+and+the+Deep.pdf
"208 8. PREPROCESSING AND UNSUPERVISED LEARNING
8.4.3 Practical issues and extensions of VQ
The quantization error HV Q can be interpreted as a quality criterion when
comparing different prototype configurations. However, this is only meaningful
for systems with the same number of prototypes. In general, HV Q will decrease
with increasing K. Obviously, placing one prototype on each individual data
point always gives the lowest possible quantization error HV Q = 0.
A key difficulty of competitive VQ and the K-means algorithm is that the
objective function HV Q can display many sub-optimal local minima. Among
other problems, this can lead to a strong dependence of the training outcome
on the initial prototype positions. As an extreme example, placing a prototype
in an empty region of feature space can prevent it from ever being identified
as the winner for any of the data points, thus leaving it unchanged in a WTA
training process. The term dead unit has been coined for such a prototype.",The+Shallow+and+the+Deep.pdf
"training process. The term dead unit has been coined for such a prototype.
For competitive learning, rank-based schemes have been suggested which
help to overcome this problem by updating not only the winner. Instead, pro-
totypes are ranked according to their proximity to the presented feature vector,
with e.g. the rank r = 1 of the winner, r = 2 of the second closest prototype etc.
Generally, the magnitude of the update of a prototype is a decreasing function
of r. A prominent example for the application of rank-based updates is the so-
called Neural Gas algorithm [MBS93,RMS92,BHV16], which employs relatively
large numbers of prototypes representing the density of data in feature space.
The idea has also been extended in the context of supervised learning [HSV05].
In Self-Organizing Maps (SOM), competitive learning is also the key ingre-
dient [Koh97]. In addition, prototypes are associated with a low-dimensional",The+Shallow+and+the+Deep.pdf
"dient [Koh97]. In addition, prototypes are associated with a low-dimensional
latent space, in which they form a grid. Updates affect the winning prototype
as defined in conventional VQ, but neighbors of the winner in the grid are also
updated. This way, the SOM can approximately preserve and visualize topolo-
gies, i.e. neighborhood relations, in the latent space. For more information on
this popular method see e.g. [Koh97,RMS92,BHV16].
Vector Quantization vs. Clustering
Frequently, Vector Quantization is confused or even identified with clustering as
indicated by the frequent use of the suggestive term K-means clustering in the
literature. This is based on the idea that prototypes or centers should always
represent pronounced clusters of similar data points. Then, the quantization
error would correspond to the average within-cluster distances.
However, it is important to note that VQ is well-defined and useful even",The+Shallow+and+the+Deep.pdf
"error would correspond to the average within-cluster distances.
However, it is important to note that VQ is well-defined and useful even
if there are no clusters present in the data. Figure 8.6 shows a few illustra-
tive situations, where prototypes represent simple, two-dimensional data with
minimal HV Q. Panel (a) displays a single, elongated cluster represented by
prototypes which characterize the variation of observed feature vectors. Here,
the minimization of HV Q does not correspond to the identification of a set of
clusters. Panel (b) shows an idealized case of approximately spherical clusters.
Each of the two clusters can be represented by one prototype. In panel (c) of",The+Shallow+and+the+Deep.pdf
"8.4. CLUSTERING AND VECTOR QUANTIZATION 209
(a)
 (b)
(c)
 (d)
Figure 8.6: Vector Quantization: representation of two-dimensional data
points by prototypes (schematic). In each panel, 200 data points are displayed
as small (red) dots, prototype positions corresponding to minimal HV Q are
marked by filled black circles. The subpanels are referred to in the text of Sec.
8.4.3.
Fig. 8.6, clusters appear to be elongated along one of the axes in feature space.
A minimal value of the quantization error can be achieved by placing the pro-
totypes in the space between the apparent clusters, where hardly any examples
are observed. Panel (c) illustrates the fact that the outcome of VQ can be
highly sensitive to coordinate transformations, even if they are linear. Panel
(d) of the Fig. 8.6 displays a situation in which a smaller, separated cluster is
not identified at all. Although these few data points contribute large distances,",The+Shallow+and+the+Deep.pdf
"not identified at all. Although these few data points contribute large distances,
their total contribution do not have a significant influence on the position of the
prototypes when minimizing HV Q.
The number of prototypes or clusters
As discussed in Sec. 2, unsupervised learning including clustering and Vector
Quantization frequently lack simple performance measures which makes it dif-
ficult to select a model of suitable complexity. The goal of Vector Quantization
can be formulated mathematically in terms of HV Q, but the cost function is
strictly speaking only suitable for comparing systems which have the same com-
plexity, i.e. the same number of prototypes. Similar restrictions apply to explicit",The+Shallow+and+the+Deep.pdf
"210 8. PREPROCESSING AND UNSUPERVISED LEARNING
Figure 8.7: Illustration of the
elbow method: the quantization
error (per sample) as obtained
from the Iris flower data set with
P = 150 feature vectors [Fis36]
as a function of K in the K-
means algorithm.
K
HV Q/P
methods of clustering and the related criteria.
In absence of additional information like domain knowledge providing a
ground truth, it is impossible to infer the correct or optimal number of clus-
ters or prototypes from the data set P alone. In this sense, clustering and
VQ are ill-defined problems, ultimately their evaluation depends on the user’s
preferences and subjective quality criteria.
Several heuristic ideas have been suggested for the determination of a suit-
able number of clusters or prototypes. Here we illustrate the popular elbow
method [Tho53] in terms of an application of the K-means algorithm. Fig. 8.7
displays the quantization error HV Q as obtained in the Iris flower data set [Fis36]",The+Shallow+and+the+Deep.pdf
"displays the quantization error HV Q as obtained in the Iris flower data set [Fis36]
by applying the algorithm with different choices of K. In this simple example,
we can conclude that K = 3 appears to be a reasonable choice. For K = 1, 2 the
quantization is much higher, while for K > 3 no further, significant reduction
of HV Q is achieved. The curve displays a pronounced elbow shape that suggests
K = 3, hence the term elbow method.7
Of course, the insight does not come as a surprise as the data set was al-
ready considered in Sec. 6.2.2 in the context of supervised learning. It contains
samples from three different classes corresponding to a more or less pronounced
cluster structure, see Fig. 6.2 for a discriminative visualization. In more gen-
eral practical situations, the shape of the corresponding curves and the elbow is
frequently much less conclusive.
Several related and alternative methods have been suggested, some of which",The+Shallow+and+the+Deep.pdf
"frequently much less conclusive.
Several related and alternative methods have been suggested, some of which
are based on more rigorous statistical arguments. As one example, R. Tibshirani
et al. suggested the so-called gap statistics as a criterion for the estimation of
the number of clusters in K-means or other methods [TWH02].
7Depending on the actual quality criterion and the visual representation, the elbow could
also be a knee.",The+Shallow+and+the+Deep.pdf
"8.5. DENSITY ESTIMATION 211
8.5 Density estimation
A rather fundamental approach to obtaining insight into the properties of a
given data set P = {ξµ}P
µ=1 is to identify a model density that could have
generated the observed data with high likelihood. Textbooks like [Bis95a,Bis06,
HTF01] provide comprehensive overviews of relevant methods and theoretical
background.
Several basic concepts can be applied in density estimation. In paramet-
ric methods, a particular statistical model is postulated and adapted. More
precisely, a particular functional form of the density is assumed. Then, the
optimization of its parameters can be formulated as an unsupervised learning
process based on the observed feature vectors in P.
So-called non-parametric methods 8 do not assume a specific functional form
a priori, but aim to infer a descriptive density directly from the data [Bis95a].
Examples are so-called histogram methods or approaches using local kernels",The+Shallow+and+the+Deep.pdf
"Examples are so-called histogram methods or approaches using local kernels
(not to be confused with kernel functions in the context of SVM and related
methods).
Here we focus on the important family of mixture models, which are some-
times referred to as semi-parametric [Bis95a]. We first discuss some basic ideas
behind parametric density estimation and then present the particular popular
example of Gaussian Mixture Models (GMM), see e.g. [Bis95a,HTF01,DFO20].
8.5.1 Parametric density estimation
In the parametric approach we make explicit assumptions about a probability
density that could explain the observed data set. Very frequently we will assume
that feature vectors ξµ ∈ RN can be interpreted as generated independently
according to a specific identical N-dim. density. We furthermore assume an
appropriate parametric form and, hence, we can write the likelihood and log-
likelihood of the observed data as
L(Θ) =
P󰁜
µ=1
p(ξµ|Θ) and ℓ(Θ) = ln(L(Θ)) =
P󰁛
µ=1
ln p(ξµ|Θ). (8.32)",The+Shallow+and+the+Deep.pdf
"likelihood of the observed data as
L(Θ) =
P󰁜
µ=1
p(ξµ|Θ) and ℓ(Θ) = ln(L(Θ)) =
P󰁛
µ=1
ln p(ξµ|Θ). (8.32)
Here, the vector Θ ∈ RK concatenates the K parameters of the model density
p(ξ|Θ). The dimension K of Θ as well as the type of parameters depend on the
actual structure of the considered model.
The optimization of ℓ(Θ) yields the corresponding maximum-likelihood model.
Similar to the discussion in Sec. 2.13 we can extend the formalism by introduc-
ing a prior density po(Θ) and derive the resulting maximum a posteriori (MAP)
model. Furthermore, Bayesian estimation techniques can be applied to obtain a
probabilistic description of the model parameters, see Sec. 2.13 for a discussion
8Despite the suggestive name, non-parametric methods may very well comprise parameters
and hyper-parameters.",The+Shallow+and+the+Deep.pdf
"212 8. PREPROCESSING AND UNSUPERVISED LEARNING
in terms of linear regression. Here, we only follow the maximum likelihood ap-
proach and consider its application to a specific type of model densities in the
next section.
8.5.2 Gaussian Mixture Models
As in any machine learning problem, model selection is a key difficulty also
in the context of density estimation. The assumed model density should be
appropriate to represent the complexity of the data. While the formalism is
very powerful, it can be difficult to define suitable models which allow for an
analytical or efficient numerical treatment. A particularly successful concept is
the consideration of adaptive models which are defined as mixtures of specific
basis functions. Gaussian densities p(ξ|Θ) constitute a particular popular and
convenient choice. Here we restrict our analysis to the particularly simple case
with
p(ξ|Θ) =
M󰁛
m=1
pm p(ξ|m, σm, wm) (8.33)
with p(ξ|m, σm, wm) =
󰀃
2πσ2
m
󰀄−N/2
exp
󰀗
− 1
2σ2m
(ξ − wm)2
󰀘",The+Shallow+and+the+Deep.pdf
"with
p(ξ|Θ) =
M󰁛
m=1
pm p(ξ|m, σm, wm) (8.33)
with p(ξ|m, σm, wm) =
󰀃
2πσ2
m
󰀄−N/2
exp
󰀗
− 1
2σ2m
(ξ − wm)2
󰀘
which also factorizes with respects to the components ξj. For simplicity we as-
sume here that the contributing densities are isotropic with covariance matrices
σM IN . Extensions to more general N-dimensional Gaussian densities are of
course possible.
The model parameters Θ in (8.33) are given by the union of the sets󰀋
σm ∈ R, wm ∈ RN , pm ∈ R
󰀌
for m = 1, 2, . . . M. Hence, Θ ∈ RK with K =
M(N + 2) in this special case. Note that the so-called mixing parameters pm
quantify the contribution of the individual Gaussians and have to satisfy the
conditions
0 ≤ pm ≤ 1 for all m and
M󰁛
m=1
pm = 1. (8.34)
Given a particular realization of the model we assign a feature vector ξ to
one of the contributing Gaussians with probability
Qm = pm p(ξ|m, σm, wm)󰁓M
k=1 pk p(ξ|k, σk, wk)
=
σ−N
m pm exp
󰁫
− 1
2σ2m
(ξ − wm)2
󰁬
󰁓M
k=1
σ−N
k pk exp
󰁫
− 1
2σ2
k
(ξ − wk)2
󰁬. (8.35)",The+Shallow+and+the+Deep.pdf
"Qm = pm p(ξ|m, σm, wm)󰁓M
k=1 pk p(ξ|k, σk, wk)
=
σ−N
m pm exp
󰁫
− 1
2σ2m
(ξ − wm)2
󰁬
󰁓M
k=1
σ−N
k pk exp
󰁫
− 1
2σ2
k
(ξ − wk)2
󰁬. (8.35)
Note that also the Qm are properly normalized with 󰁓M
m=1
Qm = 1. Analo-
gously, we define the quantities Qµ
m by (8.35) evaluated in ξ = ξµ.
For the specific model density (8.33), the log-likelihood (8.32) reads
ℓ (Θ) =
P󰁛
µ=1
ln
󰀣 M󰁛
k=1
pkσ−N
k exp
󰀗
− 1
2σ2
k
(ξµ − wm)2
󰀘󰀤
+ const. (8.36)",The+Shallow+and+the+Deep.pdf
"8.5. DENSITY ESTIMATION 213
It is straightforward to work out the necessary conditions for a maximum of
the log-likelihood [Bis95a]. They can be written in the suggestive form
wm =
P󰁛
µ=1
Qµ
m
󰀣
Qµ
m󰁓P
ν=1 Qνm
󰀤
ξµ (8.37)
σ2
m =
P󰁛
µ=1
󰀣
Qµ
m󰁓P
ν=1 Qνm
󰀤
(ξµ − wm)2 (8.38)
pm = 1
P
P󰁛
µ=1
Qµ
m (8.39)
with the assignment probabilities Qµ
m defined in Eq. (8.35). Note that for the
derivation of (8.39) the normalization 󰁓
m pm = 1 has to be taken into account
explicitly [Bis95a]. The equations can be solved self-consistently by means of the
intuitive natural iteration method, see e.g. [Kik76]. If we interpret the r.h.s. as
to define the next iterates of the quantities on the l.h.s. we obtain the following
algorithm:
Gaussian Mixture Model (isotropic Gaussian contributions)
- initialize wm, σm, pm for m = 1, 2, . . . , M
- update the model parameters according the following iterative procedure:
wm(t + 1) =
P󰁛
µ=1
Qµ
m(t)
󰀣
Qµ
m(t)󰁓P
ν=1 Qνm(t)
󰀤
ξµ (8.40)
σ2
m(t + 1) =
P󰁛
µ=1
󰀣
Qµ
m(t)󰁓P",The+Shallow+and+the+Deep.pdf
"wm(t + 1) =
P󰁛
µ=1
Qµ
m(t)
󰀣
Qµ
m(t)󰁓P
ν=1 Qνm(t)
󰀤
ξµ (8.40)
σ2
m(t + 1) =
P󰁛
µ=1
󰀣
Qµ
m(t)󰁓P
ν=1 Qνm(t)
󰀤 󰀕
ξµ − wm(t + 1)
󰀖2
(8.41)
pm(t + 1) = 1
P
P󰁛
µ=1
Qµ
m(t). (8.42)
Note that here wm(t + 1) appears on the r.h.s. of (8.41), but the terms Qµ
m(t)
are evaluated by inserting the previous wm(t) into Eq. (8.35). While this de-
tail is not essential for the iteration to work, it is very interesting to note that
this specific form can be derived as an Expectation-Maximization (EM) proce-
dure for the optimization of the (log-)likelihood [Bis95a,HTF01,DFO20]. This
methodological framework has been formalized by Dempster et al. for quite
general problems involving incomplete data, see [DLR77]. In our density esti-
mation problem we can interpret the assignment probabilities Qµ
m
as unknown
latent variables. Detailed discussions of the very versatile EM-approach can be
found in textbooks like [Bis95a] or [HTF01].
Figure 8.8 illustrates the application of the GMM algorithm in terms of a",The+Shallow+and+the+Deep.pdf
"found in textbooks like [Bis95a] or [HTF01].
Figure 8.8 illustrates the application of the GMM algorithm in terms of a
rather simple multi-modal density of two-dimensional data points (left panel).",The+Shallow+and+the+Deep.pdf
"214 8. PREPROCESSING AND UNSUPERVISED LEARNING
Figure 8.8: Illustration of density estimation by adaptation of a Gaussian
Mixture Model. Left panel: 1000 two-dimensional data points drawn from a
multimodal density. Center panel: initial density represented by a mixture of
six Gaussians. Right panel: the model density after 20 EM-steps according to
(8.40-8.42).
A mixture of six Gaussians is adapted to the data following the iteration (8.40-
8.42), the center panel of Fig. 8.8 displays the initial configuration of the system
with pk = 1/6. After as few as 20 update steps, the density is approximated very
well. Note that the model is overly complex with 6 Gaussians representing only
4 clusters in the data. However, this does not appear to constitute a problem
in this simple setting. The complexity is reduced by coinciding Gaussians with
equal means and variances or by eliminating redundant contributions by having
mixing parameters pk → 0. The actual realization of the estimated density in",The+Shallow+and+the+Deep.pdf
"mixing parameters pk → 0. The actual realization of the estimated density in
terms of the model parameters will depend on the initialization in this example.
It is very instructive to consider the algorithm (8.40–8.42) in an extremely
simplifying limit. Assume that the variances are equal and fixed in all compo-
nents of the GMM, i.e. σ2
m = σ2 which simplifies Eq. (8.35):
Qm =
pm exp
󰁫
− 1
2σ2 (ξ − wm)2
󰁬
󰁓M
k=1 pk exp
󰁫
− 1
2σ2 (ξ − wk)2
󰁬. (8.43)
If we now take the limit σ → 0, the sum is dominated by the largest summand
(the term with the smallest (ξ − wk)2), which is exponentially larger than all
other ones. Consequently we obtain
lim
σ→0
Qm =
󰀝 1 if (ξ − wm)2 ≤ (ξ − wk)2 for all k ∕= m
0 else. (8.44)
Hence, in this limit, data points are assigned in a crisp way to the closest
wm according to Euclidean distance. In the GMM algorithm (8.40–8.42), the
updates of the variances become obsolete and the pm are simply computed as",The+Shallow+and+the+Deep.pdf
"wm according to Euclidean distance. In the GMM algorithm (8.40–8.42), the
updates of the variances become obsolete and the pm are simply computed as
the fraction of crisp assignments to wm(t), while the new wm(t+1) are obtained
as the means of the data points currently assigned to wm(t).",The+Shallow+and+the+Deep.pdf
"8.6. MISSING VALUES AND IMPUTATION TECHNIQUES 215
In the considered limit, the GMM algorithm becomes identical with the
intuitive K-means procedure presented in (8.31). This is yet another example
for the observation made in Sec. 2.2 that many heuristic learning procedures
can be interpreted as special cases or limits of more general statistical modelling
schemes.
8.6 Missing values and imputation techniques
Many real world data sets are compromised by missing values, i.e. components
of feature vectors that have not been observed, registered or communicated
properly. Missing values can occur due to a variety of more or less complex
reasons, see [Gho21] (chapter 3) and [GBB+23] for a comprehensive discussion.
Quite frequently, the use of a certain spreadsheet based software tool leads to
unwanted missingness or other artefacts and makes it difficult to excel. These
are not explicitly discussed here, but always should be taken into consideration
as a potential source of error.",The+Shallow+and+the+Deep.pdf
"are not explicitly discussed here, but always should be taken into consideration
as a potential source of error.
In the literature, the following rather coarse categorization of missingness
can be found [Lit88, GLSGFV10, LR02, Gho21, GBB+23], examples are taken
from [Bla15]9
◦ MCAR: missing completely at random
Missingness is considered to be completely at random if the absence or
presence of a feature does not depend on its value or on the value and/or
missingness of other features. As an example, an accidentally damaged
blood sample in a medical study would result in missing the observation
of a particular feature.
◦ MAR: missing at random
In this case, the fact that a feature is missing might be predictable from
other information, but does not depend on the potential value of the miss-
ing feature. As an example, similar to the one presented in [Bla15], a
person may miss an IQ test because they are ill on the day that the IQ",The+Shallow+and+the+Deep.pdf
"ing feature. As an example, similar to the one presented in [Bla15], a
person may miss an IQ test because they are ill on the day that the IQ
test is given. This missingness relates to other available information about
the person, but does not depend on the potential outcome of the test, i.e.
the feature value itself.
◦ MNAR: missing not at random
If missingness is specifically related to the feature that is missing, the
somewhat vague term not at random is used. For instance, a person might
have avoided a drug test resulting in a missing observation, because they
took drugs and the outcome would be positive. Another example would be
a feature that cannot be registered whenever it exceeds an allowed range
of values, e.g. due to technical limitations of the measurement process.
The distinction of these types of missingness is not always very clear. More-
over, it can be difficult (if not impossible) to infer the underlying reason for",The+Shallow+and+the+Deep.pdf
"over, it can be difficult (if not impossible) to infer the underlying reason for
9See also https://www-users.york.ac.uk/~mb55/intro/typemiss4.htm",The+Shallow+and+the+Deep.pdf
"216 8. PREPROCESSING AND UNSUPERVISED LEARNING
missingness from the provided data alone. The MCAR and MAR types, which
are sometimes referred to as ignorable, are certainly the least difficult to handle.
More systematic forms of missingness as in MNAR, would require sophisticated
modelling techniques to take them properly into account. Consequently, the
methods discussed in the following are most appropriate for data affected by
MCAR or MAR type missingness.
8.6.1 Approaches without explicit imputation
Depending on the frequency and nature of missing features, simple strategies
can be applied which ignore or eliminate the missingness beforehand or as an
integral part of the training process.
◦ Deletion
The most straightforward idea to handle missingness is to include only complete
feature vectors in the analysis and omit all others. This is appealing since it
does not require explicit manipulations of the data. However, disregarding a",The+Shallow+and+the+Deep.pdf
"does not require explicit manipulations of the data. However, disregarding a
subset of samples might introduce biases. In any case, simple deletion is only
feasible if enough training data are available to begin with. In addition, one
would have to reject all incomplete data in validation, test or working phase. A
similar (also problematic) strategy is to omit features entirely if they appear to
be missing in a significant fraction of the available samples.
◦ Training algorithms that can handle missingness
In some machine learning frameworks it is possible to restrict their application
to the features that are present in a given observation. As an example, the
computation and ranking of pairwise distances in a subset of samples can be re-
stricted to the set of features that is present in all involved feature vectors. The
concept could be applied, for instance, in Nearest Neighbor Classification. Sim-
ilarly, in LVQ and other prototype based systems, cf. Chapter 6, the distances",The+Shallow+and+the+Deep.pdf
"ilarly, in LVQ and other prototype based systems, cf. Chapter 6, the distances
of an incomplete test or training sample from all prototypes can be computed
and compared (excluding the missing components) for the identification of the
closest prototype, see e.g. [GBV+17]. Another example of an algorithm that
can handle missingness (and noise) without explicit imputation is the so-called
Probabilistic Random Forest [RBS19].
8.6.2 Imputation based on available data
The most widely used approach to handle missing values is imputation, i.e. the
replacement of missing features by more or less sophisticated estimates based
on the available data. Several strategies and modifications of the basic idea are
explained in the following. For details and references, see e.g. [Gho21,GBB+23].
◦ Naive estimates
A straightforward but naive idea is to replace a missing value in, say, feature",The+Shallow+and+the+Deep.pdf
"8.6. MISSING VALUES AND IMPUTATION TECHNIQUES 217
ξj by the corresponding mean or median in the data set as computed from all
instances in which ξj is present. This can, however introduce or enhance a bias
in imbalanced data sets.
As a seemingly more sophisticated choice, the class-wise mean or median of ξj
could be used for imputation in a set of labeled training data. But this estimate
would be unavailable for unlabeled feature vectors in the test or working phase.
◦ Nearest-Neighbor imputation
Applying a Nearest-Neighbor (NN) (or K-NN) approach partly solves the above
mentioned problem and constitutes a popular tool for imputation. Given an
incomplete feature vector ξ with missing value ξj, one can determine the nearest
neighbor(s) of ξ in the training set e.g. in terms of the partial Euclidean distance
w.r.t. the available dimensions. Then, the value of ξj in the nearest neighbor or
an appropriate estimate based on the K nearest neighbors can be imputed.",The+Shallow+and+the+Deep.pdf
"an appropriate estimate based on the K nearest neighbors can be imputed.
◦ Regression based imputation
If a feature ξj can be expected to be correlated with other features ξk with
k ∕= j, imputation can be based on a regression scheme. From the available
data we can infer this dependency by means of linear or non-linear regression
and obtain a prediction for the missing value.
◦ Cold deck and hot deck imputation
The term cold deck imputation has been coined for situations in which a sepa-
rate data set is used for imputing missing values in the training data or when
testing/applying the trained system. On the contrary, in hot deck settings, the
available (training) data set itself is used for determining the imputed values.
◦ Generative modelling for imputation
Methods discussed in Sec. 8.4, can be employed to estimate the density of data
or a particular feature from a given data set. If the occurrence of feature ξj",The+Shallow+and+the+Deep.pdf
"or a particular feature from a given data set. If the occurrence of feature ξj
is modelled by e.g. a mixture of Gaussians, the resulting GMM can be used to
generate random values for the imputation of missing values accordingly.
◦ Multiple imputation
Frequently, more than one imputation value is generated per missing feature.
Generative approaches or randomized versions of regression based imputation
can be used to provide multiple versions of the imputed data set. Then, training
and validation can be performed on a number of versions in order to obtain
reliable performance estimates.
As an example, the so-called Multiple Imputation by Chained Equations
(MICE) has recently become popular [ASFL11, Gho21, GBB+23]. There, in
a first step all missing values are replaced by simple mean or median impu-
tation, with the exception of one selected feature dimension, say, ξk. Next, a
regression based imputation is used to replace the missing values of ξk in the",The+Shallow+and+the+Deep.pdf
"regression based imputation is used to replace the missing values of ξk in the
data set. Subsequently the missing values of a different feature ξj(j ∕= k) are",The+Shallow+and+the+Deep.pdf
"218 8. PREPROCESSING AND UNSUPERVISED LEARNING
imputed by regression and the entire procedure is repeated until all missing fea-
tures (and naive estimates) have been replaced by regression based imputation.
The imputation of all missing values can be done with different selections of the
initial feature ξk and by varying the order of features, thus obtaining several
versions of the imputed data set.
8.7 Over- and undersampling, augmentation
Here we discuss the problems that arise when a classification problem is im-
balanced in the sense that the prevalences of the individual classes are very
different. Similar difficulties occur in regression problems with skewed and/or
multimodal distributions of the target variable, but here we restrict the discus-
sion to classification. For reviews of imbalance related strategies and further
references consult, for instance [Cha09,HG09].
Class-imbalance can be handled in various ways when validating a given",The+Shallow+and+the+Deep.pdf
"references consult, for instance [Cha09,HG09].
Class-imbalance can be handled in various ways when validating a given
classifier, see the discussion in Sec. 7.4. However, it is often necessary to take the
imbalance into account in the training phase already. Strong over-representation
of certain classes in the training data can lead to very poor performance with
respect to minority classes. Note that an imbalanced training set does not
necessarily reflect the prevalences we can expect in the working phase.
8.7.1 Weighted cost functions
Assume that a training set comprises a total of P = 󰁓C
j=1 Pj examples, with
Pj of those representing class j in a C-class problem. Virtually all objective
functions we have considered for training are of the general form
E(W) = 1
P
P󰁛
µ=1
eµ (8.45)
with the contribution eµ of a single example in the data set. We can balance
the influence of the different classes by considering the weighted cost function
Ebal(W) =
C󰁛
j=1
1
Pj
Pj󰁛
µj=1
eµj (8.46)",The+Shallow+and+the+Deep.pdf
"the influence of the different classes by considering the weighted cost function
Ebal(W) =
C󰁛
j=1
1
Pj
Pj󰁛
µj=1
eµj (8.46)
where the partial sums over µj = 1, 2, . . . Pj contain only examples from class
j ∈ {1, 2, . . . C}. Note that a gradient descent optimization of Ebal leads to
updates that are equivalent to descent in the original E, Eq. (8.45), with class-
specific learning rates ηj ∝ 1/Pj.
8.7.2 Undersampling
Another straightforward way to compensate for class imbalances is to per-
form the actual training on sets with balanced class composition. This can",The+Shallow+and+the+Deep.pdf
"8.7. OVER- AND UNDERSAMPLING, AUGMENTATION 219
be achieved by randomly selecting the same number of examples from each
class. Hence, the training set size will be limited to at most min{P1, P2, . . . PC}
examples per class. Undersampling can be limited to the actual training set
while validation and test sets can remain unbalanced, but the evaluation criteria
should be robust against imbalance, like the BAC or other measures discussed
in Sec. 7.4.2.
The random selection could be done once, disregarding the remaining data.
However, this strategy would not make use of all available information and
bears the risk of selecting atypical cases. Instead, the random undersampling
should be performed repeatedly, which enables the computation of averaged
performance measures.
8.7.3 Oversampling
One can also aim at increasing the effective influence of the underrepresented
minority classes by generating additional examples for training. The two main
ideas are described in the following.",The+Shallow+and+the+Deep.pdf
"minority classes by generating additional examples for training. The two main
ideas are described in the following.
Random oversampling
In this simple approach, the training set is augmented by exact copies of ran-
domly selected examples in the underrepresented classes. It can be realized by
random selection of examples with replacement.
For loss functions of the general form (8.45), the effect of including copied
examples in the training data is very similar to weighting the classes as in Eq.
(8.46). However, in the corrected cost function every example from a given class
has the same weight.
In practice, random oversampling in the underrepresented classes and un-
dersampling of the overrepresented ones are often combined.
Generative oversampling
Instead of using exact copies of available examples, one can aim at generating
synthetic data which reflect the statistical properties of the minority classes.
The simplest idea would be to augment the training set by noisy copies of",The+Shallow+and+the+Deep.pdf
"The simplest idea would be to augment the training set by noisy copies of
randomly selected feature vectors. In more sophisticated approaches one per-
forms a suitable density estimation and uses the resulting model for generating
additional samples.
The Synthetic Minority Oversampling Technique (SMOTE) [CBHK02] fol-
lows a similar, yet simpler concept. The basic scheme is summarized in the
following:",The+Shallow+and+the+Deep.pdf
"220 8. PREPROCESSING AND UNSUPERVISED LEARNING
SMOTE (Synthetic Minority Oversampling Technique)
- randomly select an example from the minority class
- determine its k nearest neighbors in the same class, e.g. according to
Euclidean distance
- select one of the neighbors with equal probability 1/k
- generate a new data point at a random position on the line connecting
the two selected feature vectors, assign it to the same class.
Several modifications and extensions can be considered. For example, the neigh-
bor identification and construction of the synthetic data point could be based on
different distance measures, see [GBV+17] for an example of applying geodesic
SMOTE in angle-based classification.
Practical issues
The suitability of the approaches discussed in Sec. 8.7.1–8.7.3 clearly depends
on the availability of example data from all classes and on the details of the
problem at hand.
Both the weighting of classes in the cost function and undersampling are",The+Shallow+and+the+Deep.pdf
"problem at hand.
Both the weighting of classes in the cost function and undersampling are
easy to realize since they do not rely on generating synthetic data. Clearly,
undersampling is only feasible for data sets with a reasonably large number of
examples for each class to begin with. However, if that is not the case, the
application of machine learning is questionable anyway.
Generative oversampling cannot really create fundamentally novel informa-
tion, since the augmented feature vectors more or less faithfully reflect the
properties of the original data set. As a consequence, the oversampling and
subsequent training bears the risk of overfitting to the available minority class
data. In addition, generative oversampling introduces problems of model and
parameter selection, e.g. concerning the number of neighbors in SMOTE or the
choice of a model density to begin with.
Imbalanced data can constitute a significant challenge in practical applica-",The+Shallow+and+the+Deep.pdf
"choice of a model density to begin with.
Imbalanced data can constitute a significant challenge in practical applica-
tions of machine learning. Measures to compensate the imbalance should be
applied with care and they should always be evaluated in a proper validation
process.
8.7.4 Data augmentation
In practical applications, even balanced data sets are often augmented by addi-
tional, artificial training data. Data augmentation can be naively motivated as
a way to improve the training by providing more example data. In principle,
one can employ all methods discussed for oversampling in the previous section
also for the generation of additional data. The concept of data augmentation",The+Shallow+and+the+Deep.pdf
"8.7. OVER- AND UNDERSAMPLING, AUGMENTATION 221
is discussed briefly in [GBC16] with additional references given. One survey on
image data augmentation for Deep Learning is provided in [SK19].
Obviously, some of the risks mentioned in Sec. 8.7.3 are also relevant for data
augmentation. In principle, it is not possible to generate genuinely novel infor-
mation from a given data set alone. However, a few cases in which augmentation
appears justifiable can be identified:
◦ domain knowledge based models
In specific applications, generative models of the expected data in the
working phase may be available, which are based on explicit domain knowl-
edge. Such models can be used to generate surrogate data for training and
testing. Assume that the characteristics of a novel instrument, say a tele-
scope in astronomy, are known in detail. If, in addition, key properties of
the objects of interest are known, artificial data can be generated by means",The+Shallow+and+the+Deep.pdf
"the objects of interest are known, artificial data can be generated by means
of simulations of the observation process and subsequently used to (pre-)
train a classifier or regression system, see [RMBJ21] for just one recent
example. Of course, systems trained on surrogate data should eventually
be validated and tested in a real world setting.
◦ increased robustness against noise
The robustness of a classifier w.r.t. input noise can be increased by com-
plementing the training set by noisy copies of the original data. In fact,
this strategy can be viewed as a regularization technique [Bis95b,GBC16].
Obviously the noise must not be too strong and should reflect the expected
level in real data.
◦ imposing invariances
Analogous strategies can be applied in order to achieve robustness with
respect to other expected variations in feature space. This plays an impor-
tant role in image processing tasks. The original data is often subjected",The+Shallow+and+the+Deep.pdf
"respect to other expected variations in feature space. This plays an impor-
tant role in image processing tasks. The original data is often subjected
to systematic variations, such as rotation, scaling, or skewing of objects.
A properly designed training set will result in a classifier or regression
system that displays the desired invariances.
It is essential that the applied variations are suitable for the task at hand:
while e.g. galaxy classification [NWB+19] will not be affected by arbitrary
rotations in the image plane, as all orientations should occur in the data
anway. On the contrary, handwritten character recognition is insensitive
to rotations of the symbols only in a very small range of angles.
In many cases, data set augmentation appears to be a cheap but efficient way
to increase the performance of the trained system. However, systematic alterna-
tives should always be considered. In particular, invariances could be imposed",The+Shallow+and+the+Deep.pdf
"tives should always be considered. In particular, invariances could be imposed
through preprocessing, i.e. the extraction of invariant features. Even more ele-
gantly, they can be achieved by appropriate network design and modified loss
functions which realize invariant functions, see [WBJH18] for an example and
further references.",The+Shallow+and+the+Deep.pdf
222 8. PREPROCESSING AND UNSUPERVISED LEARNING,The+Shallow+and+the+Deep.pdf
"Concluding quote
Everybody right now, they look at the current technology, and they think, “OK,
that’s what artificial neural nets are.” And they don’t realize how arbitrary it
is. We just made it up! And there’s no reason why we shouldn’t make up some-
thing else.
— Geoffrey Hinton",The+Shallow+and+the+Deep.pdf
224,The+Shallow+and+the+Deep.pdf
"Appendix A
Optimization
There is nothing objective about objective functions.
— James L. McClelland
Here we summarize some essential mathematical concepts concerning real-
valued functions of multi-dimensional arguments and their optimization. In
particular, we consider local extrema and gradient-based search strategies. We
also discuss basic aspects of constrained optimization.
Note that we refrain from providing the precise (yet usually mild) math-
ematical conditions for the validity of expansions and applicability of theo-
rems. For instance, we assume implicitly that all considered functions are
continuous and differentiable, that second derivatives are symmetric, etc. For
more rigorous presentations we refer the reader to the mathematical literature,
e.g [Fle00,PP12,PAH19,Str19,DFO20].
A.1 Multi-dimensional Taylor expansion
Consider a real-valued function of the form
f : x ∈ Rd → f(x) ∈ R. (A.1)
If the value of the function f(xo) and its derivatives are known in some point xo,",The+Shallow+and+the+Deep.pdf
"f : x ∈ Rd → f(x) ∈ R. (A.1)
If the value of the function f(xo) and its derivatives are known in some point xo,
we can perform a d-dimensional Taylor expansion to obtain the approximation
f(xo + h) ≈ f(xo) + h⊤∇f(xo) + 1
2 h⊤ H(xo) h + O
󰀃
|h|3󰀄
(A.2)
in the vicinity of xo. Here, we assume that the norm of the deviation h ∈ Rd
is small: |h| ≈ 0. The first term merely corresponds to the simple estimate
f(x) ≈ f(xo) close to the reference point. The second term takes into account
225",The+Shallow+and+the+Deep.pdf
"226 A. OPTIMIZATION
Figure A.1: Real world illustration of extrema and saddle points of a func-
tion in d = 2 dimensions (elevation z(x, y)). Zero gradients can correspond to
maxima, minima, saddle points or extended flat regions. Photo taken on the
Mt. Whitney trail (https://en.wikipedia.org/wiki/Mount_Whitney_Trail).
the first (partial) derivatives of the function with respect to the coordinates x.
With the formal vector
∇ =
󰀕 ∂
∂x1
, ∂
∂x2
, . . . ∂
∂xd
󰀖⊤
(A.3)
we obtain the gradient of f in xo, i.e. the vector of partial derivatives
∇f(xo) =
󰀕 ∂f
∂x1
, ∂f
∂x2
, , . . . , ∂f
∂xd
󰀖⊤󰀏󰀏
󰀏󰀏
󰀏
x=xo
and h⊤∇f(xo) =
d󰁛
j=1
hj
∂f
∂xj
󰀏󰀏󰀏󰀏
x=xo
.
(A.4)
The third term in (A.2) involves the (d×d)-dimensional Hesse matrix or Hessian
H(xo) of second derivatives with elements
Hij(xo) = Hji(xo) = ∂2
∂xi∂xj
H
󰀏
󰀏󰀏󰀏
x=xo
and h⊤ H(xo) h =
d󰁛
i,j=1
hiHij(xo)hj.
(A.5)
Higher order terms are obviously more involved than the quadratic approxi-",The+Shallow+and+the+Deep.pdf
"∂xi∂xj
H
󰀏
󰀏󰀏󰀏
x=xo
and h⊤ H(xo) h =
d󰁛
i,j=1
hiHij(xo)hj.
(A.5)
Higher order terms are obviously more involved than the quadratic approxi-
mation (A.2). Here we assume implicitly, that a quadratic approximation is
suitable for the extrema under consideration.",The+Shallow+and+the+Deep.pdf
"A.2. LOCAL EXTREMA AND SADDLE POINTS 227
A.2 Local extrema and saddle points
First, we consider unconstrained problems of the form
minimize
x ∈ Rd f(x)
with the real-valued objective function f. Obviously, analogous results are ob-
tained for local maxima by considering the minima of −f(x). First, we assume
that x can be chosen anywhere in Rd without any restrictions. Hence, we do
not have to consider minima at the border of allowed regions
Note that the term local minimum is not synonymous with suboptimal mini-
mum, it only implies that the function locally increases whenever moving away
from the minimum. In this sense, a global minimum is also a local minimum.
A.2.1 Necessary and sufficient conditions
An obvious, necessary condition for the presence of a local minimum of f in the
point x∗ is that all first derivatives vanish:
∇f(x∗) = 0. (A.6)
This follows immediately from the Taylor expansion (A.2) up to first order: If",The+Shallow+and+the+Deep.pdf
"point x∗ is that all first derivatives vanish:
∇f(x∗) = 0. (A.6)
This follows immediately from the Taylor expansion (A.2) up to first order: If
∇f(x∗) ∕= 0, a small displacement with h = −η∇f(x∗) with η > 0 would result
in
f(x∗ + h) ≈ f(x∗) − η|∇f(x)|2 < f(x∗)
and, hence, x∗ could not be a local minimum.
Assuming that (A.6) is satisfied, we have with the shorthand H∗ = H(x∗)
f(x∗ + h) ≈ f(x∗) + 1
2h⊤ H∗ h (A.7)
∇f(x∗ + h) ≈ H∗ h. (A.8)
From Eq. (A.7) we obtain the sufficient condition
x⊤H∗ x > 0 for all x ∈ Rd (A.9)
in the presence of a local minimum in x∗. If this is the case, small steps away
from x∗ will always lead up-hill, increasing the function value. This is obviously
only correct as long as |h| is small enough and the higher-order corrections
O(|h|3) can be neglected.
The Hessian H∗ has to be positive definite, which is equivalent to the con-
dition that all its eigenvalues are positive, i.e.
H∗ ui = ρ∗
i ui with ρ∗
i
> 0 (i=1,2,. . . d) (A.10)",The+Shallow+and+the+Deep.pdf
"dition that all its eigenvalues are positive, i.e.
H∗ ui = ρ∗
i ui with ρ∗
i
> 0 (i=1,2,. . . d) (A.10)
Note that – due to the symmetry of H∗ – all eigenvalues are real and we can
construct an orthonormal set of eigenvectors {ui}d
i=1 as a basis in Rd [PP12,
Str19,DFO20].",The+Shallow+and+the+Deep.pdf
"228 A. OPTIMIZATION
Note also that an eigenvalue ρi relates to the curvature (second derivative)
of the function f in x∗ along the direction of the normalized eigenvector ui.
Defining 󰁥fi(γ) = f(x∗ + γui) and using u⊤
i ui = 1 we have
󰁥fi(γ) ≈ (x∗) + (γui)⊤H∗(γui) = 󰁥fi(0) + γ2 ρ∗
i , i.e. ∂2 󰁥fi
∂γ2
󰀏󰀏󰀏
󰀏󰀏
γ=0
= ρ∗
i
by comparison with the conventional, one-dimensional Taylor expansion with
vanishing linear term.
If the Hessian is indefinite, i.e. if it has positive and negative eigenvalues,
the point x∗ corresponds to a saddle point: in some of the eigendirections u∗
i
it displays a local maximum, while in others it behaves like a one-dimensional
local minimum, cf. Fig. A.1.
Semi-definite H∗ with ρ∗
i
≥ 0 (or ρ∗
i
≤ 0) require more careful considerations:
Eigenvalues ρ∗
i = 0 can correspond either to an extended flat region (along ui)
around a local extremum or to a saddle point in x∗. We refrain from discussing
this subtlety in detail and refer the reader to e.g. [Fle00,Str19,DFO20].",The+Shallow+and+the+Deep.pdf
"this subtlety in detail and refer the reader to e.g. [Fle00,Str19,DFO20].
A.2.2 Example: unsolvable systems of linear equations
We consider a set of P linear equations in N variables w ∈ RN of the form
w⊤ ξµ !
= yµ for µ = 1, 2, . . . , P. (A.11)
For P > N the system is overdetermined and, in general, inconsistent, i.e.
not solvable. Here, the notation is chosen to resemble the machine learning
settings that we are considering throughout, where the coefficients and r.h.s. of
the system are given by a data set of the familiar form
D =
󰀋
ξµ ∈ RN , yµ ∈ R
󰀌P
µ=1 .
Using the matrix and vector notation introduced in Sec. 2.2.2, we define
Y =
󰀃
y1, y2, . . . yP 󰀄⊤
∈ RP and χ =
󰁫
ξ1, ξ2, . . . ξP
󰁬⊤
∈ RP×N . (A.12)
Now Eq. (A.11) is conveniently written as
χ w
!
= Y. (A.13)
If χ happens to be an invertible square matrix with N = P, the solution of the
problem is obviously w∗ = χ−1 Y. However, if the set of equations is overdeter-",The+Shallow+and+the+Deep.pdf
"problem is obviously w∗ = χ−1 Y. However, if the set of equations is overdeter-
mined and cannot be satisfied exactly, we can resort to approximative solutions.
A natural and popular choice is to minimize the corresponding Sum of Squared
Error (SSE), cf. Eq. (2.5), in the sense of a linear regression
ESSE(w) = 1
2
P󰁛
µ=1
󰀃
w⊤ξµ − yµ󰀄2
= 1
2 [χw − Y ]2
= 1
2 w⊤ 󰀅
χ⊤χ
󰀆
w − w⊤χ⊤Y + 1
2 Y ⊤Y, (A.14)",The+Shallow+and+the+Deep.pdf
"A.2. LOCAL EXTREMA AND SADDLE POINTS 229
where the last term is independent of w. We proceed as in (2.2.2) by considering
the first order, necessary conditions (A.6) for a solution w∗:
∇wESSE = [χ⊤χ]w∗ − χ⊤ Y
!
= 0. (A.15)
In Sec. 2.2.2 we have already presented the formal solution of (A.15) in terms
of the left pseudoinverse [PP12,BH12]
w∗ = χ+
left Y with χ+
left
=
󰀓󰀅
χ⊤χ
󰀆−1
χ⊤
󰀔
(A.16)
under the condition that [χ⊤χ] is invertible. Here, the (N × P)-dim. matrix
χ+
left
satisfies χ+
left
χ = IN×N with the N-dim. identity.
The assumption that [χ⊤χ] ∈ RN×N is invertible is consistent with the
assumption that the equations {w∗⊤ξµ = yµ}µ=1,2,...P cannot be satisfied si-
multaneously, which - after all - was the motivation for minimizing ESSE.
In our example we obtain the Hesse matrix of second derivatives in w∗ as
H∗
ik = ∂2ESSE
∂wi∂wk
=
P󰁛
µ=1
ξµ
i ξµ
k or H∗ = χ⊤χ ∈ RN×N . (A.17)
Note that for the quadratic ESSE a Taylor expansion up to second order would",The+Shallow+and+the+Deep.pdf
"H∗
ik = ∂2ESSE
∂wi∂wk
=
P󰁛
µ=1
ξµ
i ξµ
k or H∗ = χ⊤χ ∈ RN×N . (A.17)
Note that for the quadratic ESSE a Taylor expansion up to second order would
be exact in the minimum w∗, of course.
The sufficient second order condition (A.9) corresponds to positive definite
H∗. Here, it is guaranteed that H∗ is at least positive semi-definite:
u⊤χ⊤χu = [χu]2 ≥ 0 for any u ∈ RN .
In order to have positive definite H∗, the condition P ≥ N must be fulfilled1.
For P < N, the (N × N)-dim. matrix H∗ = [χ⊤χ] cannot have full rank and is
bound to have zero eigenvalues. Correspondingly, the system (A.11) has many
solutions. We will address this case in the discussion of constrained optimization
in the next section.
Strict positive definiteness implies that H∗ = [χ⊤χ] is invertible. We con-
clude that – under this condition – the solution (A.16) exists and corresponds to
a local minimum, which is also a global minimum since unrestricted quadratic
costs cannot display other, local minima.",The+Shallow+and+the+Deep.pdf
"a local minimum, which is also a global minimum since unrestricted quadratic
costs cannot display other, local minima.
In Sec. 2.2.2 we have briefly considered the case of singular [χ⊤χ] which
cannot be inverted. There we solved the problem by introducing a non-singular
[χ⊤χ + λIN ] with λ > 0, which corresponds to a simple form of regularization.
Alternatively, we can make use of the right pseudo-inverse which is introduced
and discussed below in Sec. A.3.2.
1P ≥ N is necessary, not sufficient. Correlations in the data set can still yield singular H∗.",The+Shallow+and+the+Deep.pdf
"230 A. OPTIMIZATION
A.3 Constrained optimization
A.3.1 Equality constraints
Frequently, one encounters optimization problems of the form
minimize
x ∈ Rd f(x) subject to n equality constraints {gi(x) = 0}n
i=1 , (A.18)
where the real-valued functions gi define additional conditions under which f(x)
has to be minimized: The constraints {gi(x)}n
i=1 define the set of allowed argu-
ments x.
As a consequence, solutions of the problem (A.18) do not necessarily cor-
respond to local minima of the (unrestricted) objective function f discussed in
Sec. A.2. If it is possible to eliminate the conditions explicitly, one can transform
the problem to an unconstrained one and proceed as before.
Formally, constraints given in the form of equations as in (A.18) can be dealt
with systematically by introducing Lagrange multipliers {λi ∈ R}n
i=1. We define
the Lagrange function or Lagrangian
L
󰀕
x, {λi}n
i=1
󰀖
= f(x) −
n󰁛
i=1
λi gi(x) (A.19)",The+Shallow+and+the+Deep.pdf
"i=1. We define
the Lagrange function or Lagrangian
L
󰀕
x, {λi}n
i=1
󰀖
= f(x) −
n󰁛
i=1
λi gi(x) (A.19)
with the real-valued multipliers λi. Solutions of the constrained problem (A.18)
satisfy the first order necessary conditions
󰀝 ∂L
∂xj
󰀏󰀏󰀏󰀏
∗
= 0
󰀞d
j=1
and
󰀝 ∂L
∂λi
󰀏󰀏󰀏󰀏
∗
= 0.
󰀞n
i=1
(A.20)
where we use the shorthand (...)|∗ for the evaluation in x = x∗ and {λi = λ∗
i } .
The second set of conditions merely corresponds to the original constraints
gi(x) = 0.
Sufficient conditions for the presence of an optimum are non-trivial to formu-
late in the presence of constraints, see [Fle00,PAH19] for a detailed discussion.
Note that the extended (d + n) × (d + n)-dimensional Hessian of the Lagrange
function L is indefinite, in general.
The conditions (A.20) can often be exploited in order to eliminate some of
the variables or, in fact, the multipliers and to simplify the structure of the
optimization problem significantly.
In Sec. 3.7.2 we present an important example of the above in terms of",The+Shallow+and+the+Deep.pdf
"optimization problem significantly.
In Sec. 3.7.2 we present an important example of the above in terms of
the Adaline problem, i.e. Widrow’s Adaptive Linear Neuron [WH60, WL90].
There, we consider the minimization of the norm |w|2 under linear equality
constraints {w⊤ξµSµ
T = 1}P
µ=1. The actual perceptron weights can be eliminated
by making use of the first order conditions, while the Lagrange multipliers play
the role of the embedding strengths 󰂓x ∈ RP . The resulting optimization problem
corresponds to an unconstrained maximization of the modified cost function
(3.77).",The+Shallow+and+the+Deep.pdf
"A.3. CONSTRAINED OPTIMIZATION 231
A.3.2 Example: under-determined linear equations
We revisit the set of linear equations considered in Sec. A.2.2
χ w = Y. (A.21)
If it represents P < N equations for the N unknowns w ∈ RN , the system
can have many solutions. Then, a number io of vectors vi ∕= 0 of χ exist with
χvi = 0. Consequently, for any given solution w of (A.21) we can construct a
continuum of solutions
w +
io󰁛
i=1
ci vi with arbitrary coefficients ci ∈ R.
The solution of minimal norm w∗ is of particular interest. We can formulate
the search for w∗ as a quadratic optimization problem with linear equality
constraints:
Example: minimal norm solution of linear equations
minimize
w∈RN
1
2 w2 subject to χ w = Y. (A.22)
Introducing multipliers 󰂓λ ∈ RP , we obtain the Lagrange function
L
󰀃
w, 󰂓λ
󰀄
= 1
2Nw2 − λ⊤ (χw − Y ) (A.23)
The first order necessary conditions (A.20) become
w − χ⊤󰂓λ
!
= 0 and χw − Y
!
= 0.
While the second condition is obvious, the first one suggests to eliminate w. We",The+Shallow+and+the+Deep.pdf
"w − χ⊤󰂓λ
!
= 0 and χw − Y
!
= 0.
While the second condition is obvious, the first one suggests to eliminate w. We
obtain the modified cost function
󰁨L = −1
2
󰂓λ⊤[χχ⊤] + 󰂓λ⊤Y and the stationarity condition − χχ⊤ 󰂓λ + Y
!
= 0.
Note that here [χχ⊤] ∈ RP×P is the counterpart of the (N ×N)-dim. matrix
[χ⊤χ] that we have encountered earlier in Eqs. (A.14 ff).
Here, we assume that P < N and that [χχ⊤] is invertible. We therefore get
immediately
󰂓λ = [χχ⊤]−1 Y ⇒ w = χ⊤[χχ⊤]−1Y.
This motivates the definition of the so-called right pseudoinverse [PP12,BH12]
χ+
right = χ⊤[χχ⊤]−1 ⇒ χ χ+
right
= IP×P , (A.24)
with the P-dim. identity matrix IP×P .",The+Shallow+and+the+Deep.pdf
"232 A. OPTIMIZATION
Remark: A unified treatment
Unsolvable over-determined and solvable under-determined systems can be treated
in a unified way [BH12]. Consider the limits
lim
γ→0+
[χ⊤χ + γ IN ]−1 χ⊤ and lim
γ→0+
χ⊤ [χ χ⊤ + γ IP ]−1 (A.25)
with the P-dim. and N-dim. identity matrices IP and IN , respectively. Both
limits exist even if χ⊤χ or χχ⊤ is singular. The limits either coincide with the
left pseudoinverse (2.8) or with χ+
right defined above. Note also that Eq. (A.25,
left) corresponds to a limit of the regularization term (2.9) that we introduced
heuristically in Sec. 2.2.2.
A.3.3 Inequality constraints
The concept of Lagrange multipliers has been extended to the presence of in-
equality constraints in optimization problems of the form
minimize
x ∈ Rd f(x) subject to n constraints {gi(x) ≥ 0}n
i=1 . (A.26)
We refrain from discussing the more general combination of inequality and equal-
ity constraints, which leads to fairly obvious extensions of the following. For",The+Shallow+and+the+Deep.pdf
"ity constraints, which leads to fairly obvious extensions of the following. For
details we refer the reader to [Fle00,PAH19,Str19,DFO20]. Note that, in princi-
ple, we could introduce pairs of inequality constraints gi(x) ≥ 0 and −gi(x) ≥ 0
simultaneously in order to include equality constraints, effectively.
Formally, the corresponding Lagrange function (A.19)
L
󰀕
x, {λi}n
i=1
󰀖
= f(x) −
n󰁛
i=1
λi gi(x)
is the starting point. First order necessary conditions for a solution of (A.26) are
given by the Kuhn-Tucker (KT) Theorem of optimization theory [Fle00,PAH19].
Here, they read
Kuhn-Tucker conditions (only inequality constraints)
∇xL|∗ = 0 ⇔ ∇xf|∗ =
n󰁛
i=1
λ∗
i ∇xgi|∗ (stationarity) (A.27)
gi(x∗) ≥ 0 for i = 1, 2, . . . n (constraints) (A.28)
λ∗
i
≥ 0 for i = 1, 2, . . . n (non-neg. multipliers) (A.29)
λ∗
i
gi(x∗) = 0 for i = 1, 2, . . . n. (complementarity) (A.30)",The+Shallow+and+the+Deep.pdf
"A.3. CONSTRAINED OPTIMIZATION 233
where we use the shorthand (...)|∗ for the evaluation in x = x∗ and {λi = λ∗
i } .
The first condition (A.27) corresponds to the stationarity of the Lagrangian
with respect to the variables x. Condition (A.28) simply represents the original
constraints, while (A.29) states that all multipliers are non-negative in the op-
timum. Individual multipliers λi > 0 correspond to so-called active constraints
with gi(x) = 0, which follows from the complementarity condition (A.30). On
the contrary, if gi(x) > 0 is satisfied with λi = 0, the constraint does not have
to be enforced explicitly and is termed inactive.
More detailed interpretations and discussions of the KT conditions can be
found in the literature, see for instance [Fle00,PAH19].
Example: optimal stability in the perceptron
As an important application of the KT theorem we exploit the necessary condi-
tions (A.27–A.30) in the problem of maximum stability for the perceptron, see
Sec. 3.7.3:
minimize",The+Shallow+and+the+Deep.pdf
"tions (A.27–A.30) in the problem of maximum stability for the perceptron, see
Sec. 3.7.3:
minimize
w ∈ RN
N
2 w2 subject to inequality constraints {Eµ ≥ 1}P
µ=1 .
The KT conditions are based on the Lagrangian
L
󰀓
w,󰂓λ
󰀔
= N
2 w2 −
P󰁛
µ=1
λµ
󰀕
w⊤ξµSµ
T − 1
󰀖
.
which is the same as for the Adaline problem with equality constraints as given
in Eq. (3.70). We work out the gradient
∇wL = N w −
P󰁛
µ=1
λµ ξµSµ
T
and obtain from (A.27-A.30) the following set of necessary conditions, which
were already presented in Sec. 3.7:
Kuhn-Tucker conditions (optimal stability)
w∗ = 1
N
P󰁛
µ=1
λ∗µ ξµSµ
T (embedding strengths λµ) (A.31)
E∗µ = w∗⊤ξνSµ
T ≥ 1 for all µ (linear separability) (A.32)
λ∗µ ≥ 0 (not all λ∗µ = 0) (non-negative multipliers) (A.33)
λ∗µ (E∗µ − 1) = 0 for all µ (complementarity). (A.34)",The+Shallow+and+the+Deep.pdf
"234 A. OPTIMIZATION
As outlined in Sec. 3.7, the first condition shows that the Lagrange multipliers
play the role of embedding strengths. The weights can be interpreted as to
result from iterative Hebbian learning and can, in fact, be eliminated from the
optimization problem.
A.3.4 The Wolfe Dual for convex problems
The concept of duality plays an important role in the analysis and solution of
optimization problems. The idea is typically to derive an alternative formulation
of a given problem, which is then easier to handle numerically or even analyt-
ically. A particularly powerful framework is that of the so-called Wolfe Dual
for quite general problems [Wol61], which is discussed in some detail in [Fle00]
and [PAH19], for instance.
The Wolfe Dual simplifies significantly in the case of so-called convex prob-
lems which are of the form (A.26) with the additional requirements that
a) the set IK = {x |gi(x) ≥ 0 for i = 1, 2, . . . n} is convex",The+Shallow+and+the+Deep.pdf
"lems which are of the form (A.26) with the additional requirements that
a) the set IK = {x |gi(x) ≥ 0 for i = 1, 2, . . . n} is convex
b) the objective function f(x) is a convex function on IK.
In particular, condition (a) is satisfied if all functions gi(x) are convex or even
linear. One of the most important results for convex optimization problem states
that every local solution of the problem is also a global solution [Fle00,PAH19].
Under rather mild assumptions2 one can show that if x∗ is a solution of
(A.26), then
󰁱
x∗,󰂓λ∗
󰁲
with the vector notation 󰂓λ = {λ1, λ2, . . . λn} solves the
following problem:
Wolfe Dual of a convex problem of the form (A.26):
maximize
x∈Rd,󰂓λ∈Rn L
󰀓
x,󰂓λ
󰀔
subject to ∇xL(x,󰂓λ) = 0 and 󰂓λ ≥ 0 (A.35)
with the Lagrangian L as defined in Eq. (A.19). Moreover, the solution satisfies
f(x∗) = L(x∗,󰂓λ∗).
Note that while in the original problem (A.26) f is minimized with respect to
x, here the maximization refers to λ and x under the constraint that ∇xL = 0.",The+Shallow+and+the+Deep.pdf
"x, here the maximization refers to λ and x under the constraint that ∇xL = 0.
For a more detailed discussion of this subtlety and the Wolfe Dual of more
general problems, see [Fle00,PAH19].
Example: quadratic optimization under linear constraints
Frequently, the condition ∇xL(x,󰂓λ) = 0 can be used to eliminate the original
variables x, resulting in a simplified optimization problem. As an example
consider a quadratic problem with linear inequality constraints, which involves
the vector of variables 󰂓x ∈ Rn, i.e. d = n, a symmetric matrix C ∈ Rn×n and a
vector of constants 󰂓b ∈ Rn:
2Most importantly, the functions f and gi should be continuously differentiable. An addi-
tional so-called regularity assumption is discussed in [Fle00], sections 9.4 and 9.5.",The+Shallow+and+the+Deep.pdf
"A.4. GRADIENT BASED OPTIMIZATION 235
Example:
minimize
󰂓x∈Rn
1
2󰂓x⊤ C 󰂓x subject to C󰂓x ≥󰂓b. (A.36)
A particular property of this example problem is that the same matrix C de-
fines the quadratic form and the linear constraints. More general examples
for the application of duality are presented in, e.g., [Fle00, PAH19]. Here, the
corresponding Wolfe Dual (A.35) reads
Example: Wolfe Dual of the quadratic problem (A.36)
maximize
󰂓x∈Rn,󰂓λ ∈ Rn
1
2󰂓x⊤ C 󰂓x −󰂓λ⊤ (C󰂓x −󰂓b) (A.37)
subject to C󰂓x = C󰂓λ and 󰂓λ ≥ 0.
Now we can conveniently exploit the constraint C󰂓x = C󰂓λ in order to eliminate
󰂓x. While it does not necessarily imply 󰂓x = 󰂓λ, we can still simplify the problem
and obtain
Example: Simplification of the Wolfe Dual (A.37)
maximize
󰂓λ ∈ Rn − 1
2
󰂓λ⊤ C 󰂓λ + 󰂓λ⊤󰂓b subject to 󰂓λ ≥ 0. (A.38)
Note that this specific example has the exact same mathematical structure as
the problem of optimal stability discussed in Section 3.7, if we set n = P and 󰂓b =",The+Shallow+and+the+Deep.pdf
"the problem of optimal stability discussed in Section 3.7, if we set n = P and 󰂓b =
󰂓1 ∈ RP and moreover assume that C ∈ RP×P is defined according to Eq. (3.74).
Renaming 󰂓λ = 󰂓x we recover the Wolfe Dual (3.97) for the problem of optimal
stability, see Sec. 3.7.
A.4 Gradient based optimization
The gradient is the basis or at least an important component of many practical
optimization techniques. Gradient descent persists to be one of the most popular
and most successful method for the training of neural networks and more general
machine learning setups. We discuss the reasons for this perhaps somewhat
surprising fact in Chapter 5.
A.4.1 Gradient and directional derivative
The gradient as defined in Eq. (A.4) is closely related to so-called directional
derivative. Consider a normalized vector a ∈ Rd with |a| = 1. According to the",The+Shallow+and+the+Deep.pdf
"236 A. OPTIMIZATION
Taylor expansion (A.2), the corresponding directional derivative in xo can be
written as
lim
α→0
f(xo + αa) − f(x)
α = a⊤ ∇f(xo).
Obviously the conventional partial derivatives are recovered by setting a = ei,
i.e. by taking the directional derivative along the coordinate unit vectors ei with
ei
k = δik.
In any direction with a⊤∇f(xo) < 0 the function decreases, while a with
a⊤∇f(xo > 0 marks directions of ascent.
In a given point xo the magnitude of the directional derivative quantifies
how rapidly the function increases or decreases in the direction of a. For a
given ∇f(xo) we see immediately that we obtain the directions of ...
... steepest ascent for a ∝ +∇f(xo)
... steepest descent for a ∝ −∇f(xo)
... stationarity for a ⊥ ∇f(xo).
The gradient is always perpendicular to the level directions along which f is
locally constant.
A.4.2 Gradient descent
The properties of the gradient can be used for the numerical minimization of a",The+Shallow+and+the+Deep.pdf
"locally constant.
A.4.2 Gradient descent
The properties of the gradient can be used for the numerical minimization of a
function f(x). Starting from an initial vector x(t = 0), we consider an iteration
of the form
x(t + 1) = x(t) − η ∇f(x(t)). (A.39)
If the step size η > 0 is sufficiently small, the Taylor expansion of f in x(t) is
dominated by the linear term and we have
f(x(t + 1)) ≈ f(x(t) − η|∇f(x(t))|2 < f(x(t)) (A.40)
in every step of the sequence x(0) → x(1) . . . → x(t) → x(t + 1) . . .
Consequently f can only decrease and the gradient descent (A.39) should ap-
proach a local minimum. Analogously, we can devise gradient ascent with up-
dates along +∇f(x(t)) in order to approach a local maximum.
It is important that the step size η is small enough in order to guarantee
validity of Eq. (A.40) and ensure convergence of the iteration. We will make
this statement more precise in the following.
The role of the step size",The+Shallow+and+the+Deep.pdf
"validity of Eq. (A.40) and ensure convergence of the iteration. We will make
this statement more precise in the following.
The role of the step size
We can investigate the behavior of gradient descent in greater detail by assuming
that x(t) is already close to a local minimum x∗ with
f(x(t)) ≈ f(x∗) + 1
2 (x(t)−x∗)⊤H∗(x(t)−x∗) and ∇f(x(t)) ≈ H∗(x(t)−x∗).
(A.41)",The+Shallow+and+the+Deep.pdf
"A.4. GRADIENT BASED OPTIMIZATION 237
Introducing the shorthand δt = x(t) − x∗ for the deviation from the optimum,
the update (A.39) at step t of the descent satisfies approximately
δt ≈ δt−1 − ηH∗δt−1 ≈ [I − ηH∗] δt−1 ≈ [I − ηH∗]2 δt−2 . . . ≈ [I − ηH∗]t δ0
with the N-dim. identity matrix I. Here, we have subtracted x∗ on both sides
of Eq. (A.39) and approximated the gradient according to (A.41).
Now we can exploit the fact that an orthonormal basis of Rd can be con-
structed from the eigenvectors ui of H∗. Hence, we can expand the initial
deviation as
δ0 =
d󰁛
i=1
ci ui with coefficients ci ∈ R
in terms of the eigenvectors. We obtain immediately
δt ≈ [I − ηH∗]t δ0 =
d󰁛
i=1
ci [1 − ηρi]t ui
The corresponding squared magnitude reads
|δt|2 ≈
d󰁛
i,j=1
cicj [1 − ηρi]t [1 − ηρj]t u⊤
i uj =
d󰁛
i=1
c2
i
[1 − ηρi]2t
where we have used that u⊤
i ui = 1 and u⊤
i
uj = 0 for i ∕= j. We obtain that
lim
t→∞
|δt|2 = 0 if and only if |1 − ηρi| < 1 for all i.",The+Shallow+and+the+Deep.pdf
"[1 − ηρi]2t
where we have used that u⊤
i ui = 1 and u⊤
i
uj = 0 for i ∕= j. We obtain that
lim
t→∞
|δt|2 = 0 if and only if |1 − ηρi| < 1 for all i.
Since all eigenvalues are positive and therefore 1 −ηρi < 1 for all i, the (t → ∞)
asymptotic behavior of δt is dominated by the largest eigenvalue of the Hessian
ρmax = max {ρ1, ρ2 . . . , ρN } : The condition for convergence of the iteration to
the local minimum is that
− 1 < 1 − ηρmax ⇔ η < ηmax = 2
ρmax
. (A.42)
For a finite range of step sizes 0 < η < ηmax, convergence is guaranteed,
once the iteration is sufficiently close to the minimum. Moreover, we note that
the factor (1 − ηρmax) changes sign for η = 1/ρmax. Hence, for η < ηmax/2 the
iteration will approach the minimum smoothly, while for ηmax/2 < η < ηmax
the orientation of δt flips in every gradient step and x(t) displays an oscillatory
behavior while still approaching the minimum as t → ∞.
Finally, for η > ηmax, the iteration diverges and moves away from x∗ in at",The+Shallow+and+the+Deep.pdf
"behavior while still approaching the minimum as t → ∞.
Finally, for η > ηmax, the iteration diverges and moves away from x∗ in at
least one eigendirection of H∗. The three different regimes are illustrated and
summarized in Figure 5.6 of Chapter 5.
It is important to realize that the above result is valid only in the vicinity
of a given, local optimum. It does not enable us to make statements concerning
the behavior or the role of the step size far away from an optimum.",The+Shallow+and+the+Deep.pdf
"238 A. OPTIMIZATION
The fact that convergence can be achieved with a constant, non-zero value
of η constitutes an important insight. However, the practical usefulness of this
insight is limited: the properties of the Hessian can be very different for every
local minimum and they are obviously not available in advance. To a large
extent, the initialization x(0) of a gradient procedure will determine which of
the (potentially many) local minima will be approached.
Close to a specific minimum, the local Hessian H(x(t)) can be seen as an
estimate of H∗ and used to select a suitable step size. More sophisticated
higher order optimization techniques like Newton- or Quasi-Newton methods
use H(x(t)) or an approximation thereof explicitly in the update, see for instance
[Fle00,PAH19]. We refrain from introducing these more involved techniques here
and restrict the discussion to relatively simple gradient-based methods which
are frequently used in the context of machine learning.",The+Shallow+and+the+Deep.pdf
"and restrict the discussion to relatively simple gradient-based methods which
are frequently used in the context of machine learning.
A.4.3 The gradient under coordinate transformations
A frequently ignored property of the gradient is that the direction of steepest
descent or ascent behaves non-trivially under coordinate transformations.
In the context of the Adaline algorithm we see in Sec. 3.7.2 that gradient
descent for the cost function of Eq. (3.77) in the space of embedding strengths
󰂓x ∈ RP is not equivalent to gradient descent in terms of the weights w ∈ RN .3
Here we consider even simpler linear transformations from x ∈ Rd to z ∈ Rd :
z = Ax with components zi =
d󰁛
j=1
Aijxj and A ∈ Rd×d.
In the following we use the notation ∇x = (∂/∂x1, ∂/∂x2, . . . ∂/∂xd)⊤ and ∇z
analogously to indicate which set of variables the gradient refers to.
Interpreting the objective as a function of z we obtain with the chain rule
∂f
∂xk
=
d󰁛
i=1
∂f
∂zi
∂zi
∂xk
=
d󰁛
i=1
Aik
∂f
∂zi",The+Shallow+and+the+Deep.pdf
"Interpreting the objective as a function of z we obtain with the chain rule
∂f
∂xk
=
d󰁛
i=1
∂f
∂zi
∂zi
∂xk
=
d󰁛
i=1
Aik
∂f
∂zi
i.e. ∇x f = A⊤∇z f, (A.43)
or, assuming that A is invertible: ∇z f = A−⊤∇x f, while z = Ax.
The transformation of the gradient is different from that of “ordinary” vec-
tors, see [Tou12] for a brief discussion. The direction of steepest descent in
the transformed coordinates does not coincide with the naive projection of the
original gradient, in general.
Therefore, we should be aware that a gradient descent procedure found for
one coordinate system does not necessarily follow the steepest descent in an-
other. This seems to cast some doubt on the distinguished role of gradient
descent. We note however, reassuringly, that directions of descent need not be
the steepest in order to be useful in numerical minimization procedures.
3It can be formulated, however, as gradient descent w.r.t. a different cost function in w.",The+Shallow+and+the+Deep.pdf
"A.5. VARIANTS OF GRADIENT DESCENT 239
For a more detailed discussion and an introduction of the so-called co-variant
or natural gradient we refer to the lecture notes of M. Toussaint as a starting
point [Tou12]. The corresponding method of Natural Gradient Descent can be
related to information theoretic metrics and was pioneered by Shun-ichi Amari
in the context of machine learning, see [Ama98] for a discussion and further
references.
A.5 Variants of gradient descent
The previous section presents and discusses plain “classical” gradient descent for
the minimization of cost functions in unrestricted optimization problems.
The basic idea of gradient descent can of course be modified and adapted to
the specific properties and needs of a given problem. In the following we briefly
discuss a number of variants which are particularly relevant in machine learning
problems. This is the case, for instance, for the popular Stochastic Gradient
Descent and its modifications and extensions.",The+Shallow+and+the+Deep.pdf
"problems. This is the case, for instance, for the popular Stochastic Gradient
Descent and its modifications and extensions.
A.5.1 Coordinate descent
Quite generally we consider the minimization of a continuously differentiable
function f(x) of d-dimensional arguments x = (x1, x2, . . . xd). As discussed, the
negative gradient −∇xf|x=x(t) marks the direction of the steepest descent in a
given point x(t). We use the shorthand (. . .)|t for (. . .)|x=x(t) in the following.
From a practical point of view it is often advantageous to resort to some
direction of descent a(t) which is not necessarily the steepest and, therefore,
only has to satisfy
a(t)⊤ ∇f|t < 0. (A.44)
A corresponding iterative procedure of the form x(t+ 1) = x(t) + η a(t) can
be used to approach a (local) minimum of f, provided a suitable step size η is
chosen.
A particularly simple example is the sequential performance of steps along
one of the coordinate axes with a(t) ∝ ei(t). It is straightforward to see that we",The+Shallow+and+the+Deep.pdf
"one of the coordinate axes with a(t) ∝ ei(t). It is straightforward to see that we
can easily satisfy Eq. (A.44) by setting
a(t) = −
󰁫
e⊤
i(t) ∇f|t
󰁬
ei(t) = − ∂f
∂xi(t)
󰀏󰀏󰀏󰀏
t
ei(t) =
󰀣
0, 0, . . . −∂f
∂xi(t)
󰀏
󰀏󰀏󰀏
t
, 0, . . . 0
󰀤
.
Hence, the iteration step changes x only in one component, i(t), and it is deter-
mined by the corresponding partial derivative of f.
In deterministic-sequential coordinate descent a recursion of the form
i(0) = 1; i(t) = [i(t−1)mod d] + 1 for t = 1, 2, 3, . . .
generates a cyclic sequence of the form i(t) = 1, 2, 3, . . . , d, 1, 2, 3, . . . , d, 1, 2, . . .",The+Shallow+and+the+Deep.pdf
"240 A. OPTIMIZATION
Essentially, the structure of the algorithm is the same as for a conventional
gradient descent step when it is performed as a loop over the d coordinates. Here,
however, in each component we make use of the previously updated coordinates,
while in conventional gradient descent the values from the previous loop are
used.
The resulting, so-called coordinate-wise descent or coordinate descent for
short, is a simple, sometimes surprisingly efficient method for minimization4. A
recent review of this classical approach can be found in, e.g., [Wri15]. Obviously,
one could also consider modifications with randomized selection of the updated
coordinate, etc.
A.5.2 Constrained problems and projected gradients
In general, the solution of constrained problems by means of gradient descent
techniques requires considerable refinement.
One important approach in this context is gradient projection [Fle00]: As-
sume that the search space is restricted to a region D by equality and/or in-",The+Shallow+and+the+Deep.pdf
"sume that the search space is restricted to a region D by equality and/or in-
equality constraints. If a step according to naive, unconstrained gradient descent
yields 󰁨x(t+1) = x(t)−η∇f ∈ D, the step is accepted and x(t+1) = 󰁨x(t). Other-
wise one determines x(t+1) as a projection into D as x(t+1) = argminD ||x−󰁨x||
in an auxiliary optimization step. Depending on the precise nature of the con-
straints, the computation of the projected gradient can be costly.
In the case of simpler constraints, for instance as given by linear inequali-
ties, one can often resort to so-called active set methods [Fle00,PAH19]. There,
the iteration proceeds by unconstrained gradient descent or ascent until one or
several constraints would be violated and, therefore, become active. Temporar-
ily, active inequalities are treated as equality constraints in the following steps
which move along the corresponding planes.
A.5.3 Stochastic gradient descent",The+Shallow+and+the+Deep.pdf
"which move along the corresponding planes.
A.5.3 Stochastic gradient descent
A specific modification of gradient descent may be applied when the cost func-
tion can be written as a sum over a number of individual functions as in
f(x) =
M󰁛
m=1
hm(x) with ∇xf(x) =
M󰁛
m=1
∇xhm(x), (A.45)
where the second identify follows from the linearity of the gradient operator.
In machine learning, very often the training process is guided by a cost
function which can be written as a sum over a given set of example data. In
supervised learning, e.g. regression, it typically corresponds to an error measure
evaluated with respect to the individual examples in the training set
D =
󰀋
ξµ ∈ RN , yµ ∈ R
󰀌P
µ=1 .
4Of course, coordinate ascent can be formulated for maximization analogously.",The+Shallow+and+the+Deep.pdf
"A.5. VARIANTS OF GRADIENT DESCENT 241
A corresponding cost function can be written in the form
E(W) = 1
P
P󰁛
µ=1
eµ(W) where eµ(W) = e(ξµ, yµ| W) (A.46)
quantifies the contribution of an individual example data to the total costs.
Quite generally, the vector W is meant to concatenate all degrees of freedom in
the trained system, e.g. all adaptive weights and thresholds of a neural network.
Obviously this is of the form (A.45) with variables W and individual functions
eµ(W). The discussion in the following refers to this machine learning setup
and notation, but carries over to more general problems of the type (A.45),
Of course, the numerical minimization of E could be aimed at by applying
standard gradient descent as for any other, more general objective function. In
analogy to Eq. (A.39) the basic form of updates would be
W(t + 1) = W(t) − η ∇W E(W(t)). (A.47)
We note, however, that costs of the form (A.46) can be interpreted as an empir-
ical average (. . .) = 󰁓",The+Shallow+and+the+Deep.pdf
"W(t + 1) = W(t) − η ∇W E(W(t)). (A.47)
We note, however, that costs of the form (A.46) can be interpreted as an empir-
ical average (. . .) = 󰁓
µ(. . .)/P over the data set, which corresponds to drawing
examples from D with equal probability 1/P :
E(w) = eµ(W).
Accordingly, the gradient of E w.r.t. W can be written as a mean of individual
gradients:
∇wE(W) = 1
P
P󰁛
µ=1
∇W eµ(W) = ∇W eµ(W).
This suggests to approximate the full gradient of E by computing a restricted
empirical mean over a random subset of examples from D. As an extreme case,
we consider one randomly selected, single example in an individual training step:
µ(t) ∈ {1, 2, . . . , P} (randomly selected with equal probability 1/P)
W(t + 1) = W(t) − 󰁥η ∇W eµ(t)(W(t)) (A.48)
where we denote the learning rate by 󰁥η to potentially distinguish it from η in
batch gradient descent. Obviously, a single update step is, in general, computa-
tionally cheaper than the full gradient descent (A.47) for which a sum over all",The+Shallow+and+the+Deep.pdf
"tionally cheaper than the full gradient descent (A.47) for which a sum over all
examples has to be performed.
Individual update steps follow a rough, stochastic approximation of the true
gradient of E, hence the term stochastic gradient descent (SGD) has been coined
for the iterative procedure. In practice, we could actually draw (with replace-
ment) a random example from D independently in each step. Frequently, up-
dates are organized in epochs, e.g. by generating a random permutation of
{1, 2, . . . P} and presenting the entire D in this order before moving on to the
next epoch with a novel randomized order of the examples.",The+Shallow+and+the+Deep.pdf
"242 A. OPTIMIZATION
On average over the random selection process, an SGD update is guided by
the true negative gradient −∇W E and, therefore, we can expect that the cost
functions typically decreases over many steps for suitable choices of 󰁥η. However,
single training steps may actually increase the objective function temporarily,
as individual terms ∇W eµ can point uphill w.r.t. E with (∇W eµ) · ∇W E > 0.
Let us now study the behavior of SGD near or in a local minimum W∗ with
∇W E(W∗) = 0. Assuming that W(t) = W∗ we have that
W(t + 1) = W(t) + ∆W(t) with ∆W = −󰁥η ∇W eµ(W∗)
for a randomly selected µ ∈ {1, 2, . . . , P}. It follows that on average over the
selection process
〈∆W(t)〉D = −󰁥η E(W∗) = 0.
In this sense, the SGD training becomes stationary in a local minimum of E and
〈∆W〉D → 0 as w → W∗. However, it is important to realize that, generally,
in an individual update step, the weight vector will change even if W = W∗ is",The+Shallow+and+the+Deep.pdf
"in an individual update step, the weight vector will change even if W = W∗ is
exactly satisfied. This can be seen from the average magnitude of the update
step
󰁇
(∆W)2
󰁈
D
= 󰁥η2
󰀟󰀓
eµ(W∗)
󰀔2󰀠
D
= 󰁥η2 1
P
P󰁛
µ=1
󰀓
∇W eµ(W∗
󰀔2
≥ 0.
Note that 〈(∆W)2〉D = 0 is possible iff all individual terms ∇W eµ(W∗) = 0,
which could mean that each contribution eµ is minimized in W∗ individually.
If, for instance, a quadratic error measure of the form eµ = (σµ − τµ)2/2 with
∇W eµ = (σµ −τµ) is employed, this would imply that σµ = τµ for all examples
simultaneously, representing a perfect solution with E(W∗) = 0.
In general, however, 〈(∆W)2〉D > 0 in the local minimum W∗. This indicates
that for constant learning rate 󰁥η > 0, the adaptive quantities W(t) will persist
to fluctuate in the vicinity of a local minimum, even in the limit t → ∞.
The generic behavior for 󰁥η > 0 is illustrated in Fig. 5.7 of Chapter 5. As
discussed there, Robbins and Monro [RM51], see also [Bis95a, HTF01], have",The+Shallow+and+the+Deep.pdf
"The generic behavior for 󰁥η > 0 is illustrated in Fig. 5.7 of Chapter 5. As
discussed there, Robbins and Monro [RM51], see also [Bis95a, HTF01], have
shown that convergence can be achieved with a time dependent learning rate
schedule with lim
t→∞
󰁥η(t) = 0 which satisfies the conditions (5.16):
(I) lim
T→∞
T󰁛
t=0
󰁥η(t)2 < ∞ and (II) lim
T→∞
T󰁛
t=0
󰁥η(t) → ∞. (A.49)
Intuitively, the first condition (I) states that 󰁥η(t) has to decrease fast enough in
order to achieve a stationary configuration, eventually. Condition (II) implies
that the decrease is slow enough so that the entire search space can be explored
efficiently without stopping the iteration too early.
Simple schedules which reduce the learning rate asymptotically like 󰁥η(t) ∝
1/t for large t satisfy both conditions in (5.16). Just one possible and popular
realization of such a decrease is of the form
󰁥η(t) = a
b + t with parameters a, b > 0.",The+Shallow+and+the+Deep.pdf
"A.6. EXAMPLE CALCULATION OF A GRADIENT 243
Sophisticated schemes have been devised in which the learning rate is not ex-
plicitly time dependent, but is adapted in the course of training. The adaptation
can be based on (estimated) second order derivatives or on the observed variance
of the gradient over several update steps, for instance. Learning rate adaptation
is frequently combined with so-called momentum terms which comprise infor-
mation about previous updates. Up-to-date textbooks such as [GBC16] provide
more details and further references. For a brief discussion, see Chapter 5, where
also other modifications of SGD are introduced.
A.6 Example calculation of a gradient
In order to exemplify the computation of gradients in a feed-forward network,
we consider here a specific two-layer architecture with N-dimensional input, K
hidden units and a single output
σ(ξ) = h
󰀳
󰁃
K󰁛
j=1
vj g
󰀓
w(j) · ξ
󰀔
󰀴
󰁄. (A.50)
All input-to-hidden weight vec
tors w(j) and hidden-to-output weights vj are",The+Shallow+and+the+Deep.pdf
"σ(ξ) = h
󰀳
󰁃
K󰁛
j=1
vj g
󰀓
w(j) · ξ
󰀔
󰀴
󰁄. (A.50)
All input-to-hidden weight vec
tors w(j) and hidden-to-output weights vj are
assumed to be adaptive, while for simplicity local thresholds are not considered
here. The output activation h(. . .) is taken to be (potentially) different from
the hidden unit activations g(. . .).
For a given data set D = {ξµ, τµ}P
µ=1, we consider the familiar quadratic
deviation
E = 1
P
P󰁛
µ=1
eµ with single example terms eµ = 1
2
󰀃
σ(ξµ) − τµ󰀄2
(A.51)
Note that only σ depends on the weights, the target values τ are fixed and given
in the data set. In the following we work out derivatives for one example term
eµ only. For convenience, we omit the index µ and write σ in short for σ(ξ).
First we compute the derivative with respect to one of the hidden-to-output
weights. Only one term in the sum 󰁓K
j=1 . . . depends on vk and we obtain
∂e
∂vk
= (σ − τ) ∂σ
∂vk
= (σ − τ) h′
󰀳
󰁃
K󰁛
j=1
vjg(w(j) · ξ)
󰀴
󰁄
󰁿 󰁾󰁽 󰂀
shorthand: δ
g(w(k) · ξ). (A.52)",The+Shallow+and+the+Deep.pdf
"j=1 . . . depends on vk and we obtain
∂e
∂vk
= (σ − τ) ∂σ
∂vk
= (σ − τ) h′
󰀳
󰁃
K󰁛
j=1
vjg(w(j) · ξ)
󰀴
󰁄
󰁿 󰁾󰁽 󰂀
shorthand: δ
g(w(k) · ξ). (A.52)
Of course, h′ has to be specified in a concrete setting. For instance, if
h(x) = tanh(γx) we have h′(x) = γ
󰀃
1 − tanh2(γx)
󰀄
.
Next, we take the derivative with respect to a single input-to-hidden weight
w(m)
n , i.e. the n-th component of the m-th input-to-hidden weight vector. With
the same shorthand δ as defined above we obtain
∂e
∂w(m)
n
= (σ − τ) ∂σ
∂w(m)
n
= δ vm g′
󰀓
w(m) · ξ
󰀔
ξn. (A.53)",The+Shallow+and+the+Deep.pdf
"244 A. OPTIMIZATION
The term vm g′(. . .) corresponds to the derivative of the only term in the sum󰁓K
j=1 . . . that contains the weight vector w(m). Finally the factor ξn appears
because
w(m) · ξ =
N󰁛
j=1
w(m)
j ξj and thus ∂(w(m) · ξ)
∂w(m)
n
= ξn.
Note that the r.h.s. of Eqs. (A.52,A.53) are just numbers as they correspond to
derivatives w.r.t. single weights. We could build a single gradient vector from
the component-wise results by combining all adaptive quantities into a vector
W as introduced in the previous section. It appears more natural, however, to
write for m = 1, 2, . . . K:
∇w(m) E = δ vm g′
󰀓
w(m) · ξ
󰀔
ξ (A.54)
where the notation ∇w(m) stands for the gradient with respect to the m-th
input-to-hidden weight vector. Note that both sides of the equation obviously
correspond to N-dim. vectors. The partial derivatives w.r.t. the vk are given by
the K additional equations (A.52).
For the gradient of the full cost function we have to perform sums over",The+Shallow+and+the+Deep.pdf
"the K additional equations (A.52).
For the gradient of the full cost function we have to perform sums over
all examples. Note that the abbreviation δ is defined for a particular single
example, as well. Hence we get
∂E
∂vk
= 1
P
P󰁛
µ=1
󰀃
σ(ξµ󰀄
−τµ)h′
󰀳
󰁃
K󰁛
j=1
vjg(w(j) · ξµ)
󰀴
󰁄
󰁿 󰁾󰁽 󰂀
shorthand: δµ
g
󰀓
w(k) · ξµ
󰀔
(A.55)
∇w(m) E = 1
P
P󰁛
µ=1
δµ vm g′
󰀓
w(m) · ξµ
󰀔
ξµ. (A.56)
In a network with more layers, the chain rule has to be applied several times and
in each layer terms similar to δ from previous layers appear. While the network
response σ is determined by propagating an input towards the output, the gra-
dient is computed by propagating the deviation (σ − τ) backwards through the
network. In both operations the same weights play the role of coefficients. This
is the basic concept behind the famous Backpropagation of Error for efficient
gradient calculation in multi-layered networks [RHW86].",The+Shallow+and+the+Deep.pdf
"List of figures
1.1 Neurons and synapses . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Action potentials and firing rate . . . . . . . . . . . . . . . . . . 5
1.3 Sigmoidal activation functions . . . . . . . . . . . . . . . . . . . . 7
1.4 Recurrent neural networks . . . . . . . . . . . . . . . . . . . . . . 11
1.5 Feed-forward neural networks . . . . . . . . . . . . . . . . . . . . 14
2.1 Simple linear regression (Hubble diagram) . . . . . . . . . . . . . 25
3.1 The Mark I Perceptron . . . . . . . . . . . . . . . . . . . . . . . . 33
3.2 Single layer perceptron . . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Geometrical interpretation of the perceptron . . . . . . . . . . . 35
3.4 Rosenblatt perceptron algorithm . . . . . . . . . . . . . . . . . . 40
3.5 Linear separability in one dimension . . . . . . . . . . . . . . . . 47
3.6 The number of lin. sep. functions for N = 2 . . . . . . . . . . . . 48",The+Shallow+and+the+Deep.pdf
"3.6 The number of lin. sep. functions for N = 2 . . . . . . . . . . . . 48
3.7 Counting lin. sep. dichotomies (I) . . . . . . . . . . . . . . . . . 49
3.8 Counting lin. sep. dichotomies (II) . . . . . . . . . . . . . . . . . 50
3.9 Counting lin. sep. dichotomies (III) . . . . . . . . . . . . . . . . . 51
3.10 The fraction of lin. sep. functions . . . . . . . . . . . . . . . . . . 52
3.11 The pizza connection . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.12 Perceptron student-teacher scenario . . . . . . . . . . . . . . . . 56
3.13 Dual geometrical interpretation of the perceptron . . . . . . . . . 57
3.14 Perceptron learning in version space . . . . . . . . . . . . . . . . 58
3.15 Generalization error as a function of α = P/N . . . . . . . . . . . 60
3.16 Version space for P < N . . . . . . . . . . . . . . . . . . . . . . . 61
3.17 Stability of the perceptron . . . . . . . . . . . . . . . . . . . . . . 64",The+Shallow+and+the+Deep.pdf
"3.17 Stability of the perceptron . . . . . . . . . . . . . . . . . . . . . . 64
3.18 Support vectors (linearly separable data) . . . . . . . . . . . . . . 79
4.1 Support vectors (soft margin) . . . . . . . . . . . . . . . . . . . . 89
4.2 Architecture of “machines” . . . . . . . . . . . . . . . . . . . . . . 90
4.3 Committee and parity machine . . . . . . . . . . . . . . . . . . . 92
4.4 Storage capacity of machines . . . . . . . . . . . . . . . . . . . . 97
4.5 SVM: Illustration of the non-linear transformation . . . . . . . . 99
5.1 Generic layered network . . . . . . . . . . . . . . . . . . . . . . . 108
245",The+Shallow+and+the+Deep.pdf
"246 LIST OF FIGURES
5.2 Interval selection by sigmoidal functions . . . . . . . . . . . . . . 110
5.3 Selection of ROI in high dimensions . . . . . . . . . . . . . . . . 111
5.4 Constructed network for universal function approximation . . . . 112
5.5 Soft Committee Machine . . . . . . . . . . . . . . . . . . . . . . . 113
5.6 Gradient descent near a minimum . . . . . . . . . . . . . . . . . 118
5.7 Stochastic gradient descent near a minimum . . . . . . . . . . . . 121
5.8 Sigmoidal and related activation functions . . . . . . . . . . . . . 128
5.9 Unbounded and one-sided activation functions . . . . . . . . . . 129
5.10 Extreme Learning Machine . . . . . . . . . . . . . . . . . . . . . 133
5.11 Shallow auto-encoder . . . . . . . . . . . . . . . . . . . . . . . . . 134
5.12 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
5.13 Pooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137",The+Shallow+and+the+Deep.pdf
"5.13 Pooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
5.14 Neocognitron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
5.15 LeNet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
6.1 Nearest Neighbor and Nearest Prototype Classifiers . . . . . . . . 142
6.2 GMLVQ system and data visualization . . . . . . . . . . . . . . . 152
7.1 Bias–variance dilemma . . . . . . . . . . . . . . . . . . . . . . . . 156
7.2 Bias and variance, underfitting and overfitting . . . . . . . . . . . 159
7.3 The double descent phenomenon . . . . . . . . . . . . . . . . . . 162
7.4 Early stopping and weight decay . . . . . . . . . . . . . . . . . . 164
7.5 Dropout regularization . . . . . . . . . . . . . . . . . . . . . . . . 170
7.6 Representative training data . . . . . . . . . . . . . . . . . . . . . 172
7.7 Receiver Operating Characteristics (ROC) . . . . . . . . . . . . . 179",The+Shallow+and+the+Deep.pdf
"7.7 Receiver Operating Characteristics (ROC) . . . . . . . . . . . . . 179
7.8 Confusion matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
8.1 Log-transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 190
8.2 Big dipper and enormous kitchen . . . . . . . . . . . . . . . . . . 191
8.3 Low-dimensional manifold . . . . . . . . . . . . . . . . . . . . . . 193
8.4 Multi-dimensional scaling . . . . . . . . . . . . . . . . . . . . . . 194
8.5 Kurtosis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
8.6 Vector Quantization . . . . . . . . . . . . . . . . . . . . . . . . . 209
8.7 Elbow method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
8.8 Gaussian Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . 214
A.1 Multi-dimensional extrema and saddle points . . . . . . . . . . . 226",The+Shallow+and+the+Deep.pdf
"List of algorithms
Adaline (parallel updates), 70
Adaline (sequential updates), 71
AdaTron (sequential updates), 76
AdaTron with errors (sequential updates), 88
Batch Gradient Descent (basic form), 117
Competitive learning (Vector Quantization), 207
Cross validation (n-fold), 173
Gaussian Mixture Model, maximum likelihood, 213
Generalized Learning Vector Quantization (GLVQ), 147
Generalized Matrix Learning Vector Quantization (GMLVQ), 151
Generic iterative perceptron updates (embedding strenghts), 38
Generic iterative perceptron updates (weights), 37
Grandmother Neuron, 93
K-means algorithm, 207
Kernel AdaTron (sequential updates), 101
Learning Vector Quantization (LVQ1), 145
Lloyd’s algorithm, 207
MinOver algorithm, 65
Oja’s subspace algorithm, 202
Optimal Brain Damage (OBD), 168
Optimal Brain Surgery (OBS), 168
Pocket algorithm, 86
Rosenblatt perceptron, 39
Sanger’s rule, 202
Stochastic Gradient Descent, 119
Synthetic Minority Oversampling Technique (SMOTE), 219",The+Shallow+and+the+Deep.pdf
"Rosenblatt perceptron, 39
Sanger’s rule, 202
Stochastic Gradient Descent, 119
Synthetic Minority Oversampling Technique (SMOTE), 219
Tiling-like learning in the parity machine, 92
247",The+Shallow+and+the+Deep.pdf
"Abbrev. and acronyms
Adaline adaptive linear element, adaptive linear neuron
AdaTron adaptive perceptron (algorithm)
AUC area under the curve
AUROC area under the receiver operating characteristics curve
BAC balanced accuracy
CM committee machine
CNN convolutional neural network
CoD coefficient of determination
ELM extreme learning machine
EM expectation-maximization (algorithm)
FP, FN false positives, false negatives (counts)
fpr, fnr false positive rate, false negative rate
GLVQ generalized learning vector quantization
GMLVQ generalized matrix relevance learning vector quantization
GMM gaussian mixture model
hom. homogeneous(ly)
inh. inhomogeneous(ly)
ICA independent component analysis
IQR interquartile range
KL Kullback-Leibler (divergence)
KT Kuhn-Tucker (e.g. KT conditions, KT theorem, KT point)
l.h.s. left hand side
lin. sep. linearly separable
LMS least mean squares
LVQ learning vector quantization
MAE mean absolute error
MAR missing at random",The+Shallow+and+the+Deep.pdf
"lin. sep. linearly separable
LMS least mean squares
LVQ learning vector quantization
MAE mean absolute error
MAR missing at random
MAP maximum a posteriori (probability)
MCAR mis ing comp etely at r ndom
MNAR missing not at random
MDS multi-dimensional scaling
MICE multiple imputation by chained equations
MSE mean squared error
NPC nearest prototype classifier {classification}
OBD optimal brain damage
OBS optimal brain surgeon
ODE ordinary differential equation
PM parity machine
248",The+Shallow+and+the+Deep.pdf
"PCA principal component analysis
PCT perceptron convergence theorem
PR precision-recall
Prec precision
PSP perceptron storage problem
RBF radial basis functions
Rec recall
ReLU rectified linear unit
r.h.s. right hand side
ROC receiver operating characteristics
ROI region of interest
RSLVQ robust soft learning vector quantization
SENS sensitivity
SPEC specificity
SGD stochastic gradient descent
SMOTE synthetic minority oversampling technique
SNE stochastic neighborhood embedding
SOM self-organizing map
SSE sum of squared errors
SVM support vector machine
t-SNE t-distributed stochastic neighborhood embedding
TP, TN true positives, true negatives (counts)
tpr, tnr true positive rate, true negative rate
UMAP uniform manifold approximation and projection
VC Vapnik-Chervonenkis, e.g. in VC-dimension
VQ vector quantization
w.r.t. with respect to
WTA winner-takes-all
249",The+Shallow+and+the+Deep.pdf
"Notes on the bibliography
References are sorted alphabetically by their BIBTEX keys for ease of brows-
ing. The BIBTEX source file is available upon request from the author or at
www.cs.rug.nl/˜biehl.
Most online sources point to the publisher’s final versions, some of which might
not be publicly available. Where possible, links to accessible preprint versions
are provided as an alternative.
All of the provided online sources have been accessed in April 2023. However,
the author cannot guarantee the correctness of the links and is not liable for
potential copyright infringements on the corresponding websites.
250",The+Shallow+and+the+Deep.pdf
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"[Wit10] A.W. Witoelar. Statistical physics of Learning Vector Quantization.
PhD thesis, University of Groningen, 2010. Online: https://www.rug.
nl/research/portal/files/14692629/11complete.pdf.
[Wit20] D. Witten. The bias-variance-tradeoff & double descent, 2020. Twitter:
https://threadreaderapp.com/thread/1292293102103748609.html.
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perceptron, madaline, and backpropagation. Proc. of the IEEE,
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book/10.1007/978-3-642-27645-3.",The+Shallow+and+the+Deep.pdf
"of-the-Art. Springer, Berlin, 2012. Online: https://link.springer.com/
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"weinberger09a/weinberger09a.pdf.
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"278 BIBLIOGRAPHY
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ican Mathematical Society. AMS, 1975. Online: https://bookstore.ams.
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"Michael Biehl
The 
Shallow 
and the 
Deep
A biased 
introduction 
to neural networks 
and old school 
machine learning
The Shallow and the Deep is a collection of lecture notes that offers an 
accessible introduction to neural networks and machine learning in general. 
However, it was clear from the beginning that these notes would not be able to 
cover this rapidly changing and growing field in its entirety. The focus lies on 
classical machine learning techniques, with a bias towards classification and 
regression. Other learning paradigms and many recent developments in, for 
instance, Deep Learning are not addressed or only briefly touched upon.
Biehl argues that having a solid knowledge of the foundations of the field is 
essential, especially for anyone who wants to explore the world of machine 
learning with an ambition that goes beyond the application of some software 
package to some data set. Therefore, The Shallow and the Deep places",The+Shallow+and+the+Deep.pdf
"learning with an ambition that goes beyond the application of some software 
package to some data set. Therefore, The Shallow and the Deep places 
emphasis on fundamental concepts and theoretical background. This also 
involves delving into the history and pre-history of neural networks, where the 
foundations for most of the recent developments were laid.
These notes aim to demystify machine learning and neural networks without 
losing the appreciation for their impressive power and versatility./uni00A0
Michael Biehl is Associate Professor of Computer 
Science at the Bernoulli Institute for Mathematics, 
Computer Science and Artificial Intelligence of the 
University of Groningen, where he joined the Intelligent 
Systems group in 2003. He also holds an honorary 
Professorship of Machine Learning at the Center for 
Systems Modelling and Quantitative Biomedicine of the 
University of Birmingham, UK. 
His research focuses on the modelling and theoretical",The+Shallow+and+the+Deep.pdf
"Systems Modelling and Quantitative Biomedicine of the 
University of Birmingham, UK. 
His research focuses on the modelling and theoretical 
understanding of neural networks and machine learning in general. The 
development of efficient training algorithms for interpretable, transparent 
systems is a topic of particular interest. A variety of interdisciplinary 
collaborations concern practical applications of machine learning in the 
biomedical domain, in astronomy and other areas.
T/h.smcp/e.smcp S/h.smcp/a.smcp/l.smcp/l.smcp/o.smcp/w.smcp /a.smcp/n.smcp/d.smcp /t.smcp/h.smcp/e.smcp D/e.smcp/e.smcp/p.smcp   Michael Biehl",The+Shallow+and+the+Deep.pdf
"Elementary
Calculus
0 v2
0
g
v2
0
2g
Michael Corral",ElementaryCalculus.pdf
"Elementary Calculus
Michael Corral
Schoolcraft College",ElementaryCalculus.pdf
"About the author :
Michael Corral is an Adjunct Faculty member of the Departmen t of Mathematics at School-
craft College. He received a B.A. in Mathematics from the Uni versity of California, Berkeley ,
and received an M.A. in Mathematics and an M.S. in Industrial & Operations Engineering
from the University of Michigan.
This text was typeset in L ATEX with the KOMA-Script bundle, using the GNU Emacs
text editor on a Fedora Linux system. The graphics were creat ed using TikZ and Gnuplot.
Copyright © 2020 Michael Corral.
Permission is granted to copy , distribute and/or modify thi s document under the terms of the
GNU Free Documentation License, V ersion 1.3 or any later ver sion published by the Free
Software Foundation; with no Invariant Sections, no Front- Cover Texts, and no Back-Cover
Texts.",ElementaryCalculus.pdf
"Preface
This book covers calculus of a single variable. It is suitabl e for a year-long (or two-semester)
course, normally known as Calculus I and II in the United Stat es. The prerequisites are high
school or college algebra, geometry and trigonometry . The b ook is designed for students in
engineering, physics, mathematics, chemistry and other sc iences.
One reason for writing this text was because I had already wri tten its sequel, Vector Cal-
culus. More importantly , I was dissatisfied with the current crop o f calculus textbooks, which
I feel are bloated and keep moving further away from the subje ct’s roots in physics. In addi-
tion, many of the intuitive approaches and techniques from t he early days of calculus—which
I think often yield more insights for students—seem to have b een lost.
I agree with the views of the late Russian mathematician V .I. Arnold on teaching mathe-
matics, in particular the idea that “Mathematics is the part of physics where experiments are",ElementaryCalculus.pdf
"matics, in particular the idea that “Mathematics is the part of physics where experiments are
cheap.”1 The ties to physics are especially important in calculus, so this book tries to introduce
new concepts with physical motivations (what other motivat ions can there be?). The book con-
tains exercises and examples that I hope will adequately pre pare students who continue on in
physics and engineering. 2
Perhaps controversially , the book uses infinitesimals, mak ing it a bit of a “throwback” or
“retro” calculus text. My justification for this heretical a ct was purely pedagogical: infinitesi-
mals make learning calculus easier , and their use aligns mor e with the way students will see
calculus in their physics, chemistry and other science clas ses and textbooks (where infinitesi-
mals are employed liberally). This might ruffle some feather s among mathematical “purists,”
but they are not the main audience for this book. That said, th e book is still compatible with the",ElementaryCalculus.pdf
"but they are not the main audience for this book. That said, th e book is still compatible with the
usual limit-based approach, so an instructor could simply i gnore the parts involving infinites-
imals and teach the material as he or she normally would. I did not want to be dogmatic, so I
used infinitesimals where I thought it made sense, and used li mits where appropriate (e.g. in
discussing continuity , series). Again, pedagogy was my pri ority .
The exercises at the end of each section are divided into thre e categories: A, B and C. The
A exercises are mostly of a routine computational nature, th e B exercises are slightly more
involved, and the C exercises usually require some effort or insight to solve. A crude way of
1 AR N O L D, V .I., “On Teaching Mathematics”, Russian Math. Surveys 53 (1998), No. 1, 229-236. An HTML version
is at https://www.uni-muenster.de/Physik.TP/~munsteg/arnold.html",ElementaryCalculus.pdf
"is at https://www.uni-muenster.de/Physik.TP/~munsteg/arnold.html
2The book covers some of the types of problems and techniques f or solving them that such students will likely
encounter . F acility with using named constants (e.g. c, h, T ) is also emphasized.
iii",ElementaryCalculus.pdf
"iv P R E FA C E
describing A, B and C would be “Easy”, “Moderate” and “Challe nging”, respectively . However ,
many of the B exercises are easy and not all the C exercises are difficult. Appendix A provides
answers and hints to many of the odd-numbered and some of the e ven-numbered exercises.
A few exercises require the student to write a computer progr am to solve numerical approx-
imation problems (e.g. numerical methods for approximatin g definite integrals). Algorithms
are presented in pseudocode, with code implementations in v arious languages (primarily Java,
but also Python, Octave, Sage). I hope the code comments will help the reader figure out what
is being done, regardless of familiarity with those languag es. Students are free to implement
solutions using the language of their choice. There are no de dicated “calculator exercises,” as
those have been rendered pointless by modern computing (wit h which students need to become
acquainted).",ElementaryCalculus.pdf
"those have been rendered pointless by modern computing (wit h which students need to become
acquainted).
Stylistically I made a conscious effort to break from an unfo rtunate but all too common
mode of writing in mathematics texts, lamented in the prefac e of a physics book: “Nothing
is more repellent to normal human beings than the clinical su ccession of definitions, axioms,
and theorems generated by the labours of pure mathematician s.”3 I have been guilty of that
sin myself, but I have changed my ways and banished all traces of that sort of thing from this
book. So you won’t find Definition 1.2, Theorem 3.3, Corollary 4.6, Lemma 5.7, Axiom 1B,
etc. Instead, I tried to borrow the best of the styles from the physics and foreign languages
textbooks I enjoyed so much in college. I also deliberately a voided what the author Gore Vidal
called the “we-ness” that prevails in academic writing. The re is no good reason for the “royal",ElementaryCalculus.pdf
"called the “we-ness” that prevails in academic writing. The re is no good reason for the “royal
we” in a textbook, and it comes off as a bit pompous, so we won’t use it.
This book is released under the GNU Free Documentation Licen se (GFDL), which allows
others to not only copy and distribute the book but also to mod ify it. For more details, see
the included copy of the GFDL. So that there is no ambiguity on this matter , anyone can
make as many copies of this book as desired and distribute it a s desired, without needing my
permission. The PDF version will always be freely available to the public at no cost (go to
http://www.mecmath.net/calculus). Feel free to contact me at mcorral@schoolcraft.edu for
any questions on this or any other matter regarding the book. I welcome your feedback.
Schoolcraft College MI CHAEL CORRAL
December 2020
3 ZI MA N , J.M., Elements of Advanced Quantum Theory , Cambridge, U .K.: Cambridge University Press, 1969.",ElementaryCalculus.pdf
"Contents
Preface iii
1 The Derivative 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 The Derivative: Limit Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 The Derivative: Infinitesimal Approach . . . . . . . . . . . . . . . . . . . . . . . 15
1.4 Derivatives of Sums, Products and Quotients . . . . . . . . . . . . . . . . . . . . 21
1.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.6 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2 Derivatives of Common Functions 37
2.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.2 Trigonometric Functions and Their Inverses . . . . . . . . . . . . . . . . . . . . 41
2.3 The Exponential and Natural Logarithm Functions . . . . . . . . . . . . . . . . 45",ElementaryCalculus.pdf
"2.3 The Exponential and Natural Logarithm Functions . . . . . . . . . . . . . . . . 45
2.4 General Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . 53
3 T opics in Differential Calculus 56
3.1 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.2 Limits: Formal Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
3.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.4 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
3.5 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.6 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4 Applications of Derivatives 90
4.1 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90",ElementaryCalculus.pdf
"4 Applications of Derivatives 90
4.1 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.2 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.3 Numerical Approximation of Roots of Functions . . . . . . . . . . . . . . . . . . 109
4.4 The Mean V alue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
5 The Integral 124
5.1 The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
v",ElementaryCalculus.pdf
"vi C ONTE NTS
5.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
5.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . 140
5.4 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
5.5 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
6 Methods of Integration 159
6.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
6.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
6.3 Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
6.4 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
6.5 Miscellaneous Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 185",ElementaryCalculus.pdf
"6.5 Miscellaneous Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 185
6.6 Numerical Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
7 Analytic Geometry and Plane Curves 202
7.1 Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
7.2 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
7.3 Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
7.4 Translations and Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
7.5 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
7.6 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
7.7 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
8 Applications of Integrals 252",ElementaryCalculus.pdf
"7.7 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
8 Applications of Integrals 252
8.1 Area Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
8.2 Average V alue of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
8.3 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
8.4 Surfaces and Solids of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
8.5 Applications in Physics and Statistics . . . . . . . . . . . . . . . . . . . . . . . . 277
9 Infinite Sequences and Series 286
9.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
9.2 Tests for Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
9.3 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300",ElementaryCalculus.pdf
"9.3 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
9.4 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
9.5 Taylor’s Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308
Appendix A Answers and Hints to Selected Exercises 315
GNU Free Documentation License 321
History 329
Index 330",ElementaryCalculus.pdf
"The Greek Alphabet
Letters Name Letters Name Letters Name
A α alpha I ι iota P ρ rho
B β beta K κ kappa Σ σ sigma
Γ γ gamma Λ λ lambda T τ tau
∆ δ delta M µ mu Υ υ upsilon
E ǫ epsilon N ν nu Φ φ phi
Z ζ zeta Ξ ξ xi X χ chi
H η eta O o omicron Ψ ψ psi
Θ θ theta Π π pi Ω ω omega
Mathematical Notation
Symbol Meaning Example
⇒ if...then; implies |x| >1 ⇒ x2 > 1
⇔ if and only if; two-way implication |x| >1 ⇔ x2 > 1
iff if and only if; two-way implication |x| >1 iff x2 > 1
⇏ does not imply |x| >1 ⇏ x > 1
∃ there exists ∃ a number c > 0
∝not⌉x〉sts there does not exist ∝not⌉x〉stsx such that x2 < 0
∃! there exists a unique ∃! x such that 2 x − 1 = 3
∀ for every ∀x ≥ 0, ∝ra⌈〉⌋allowx is a real number
≡ is identically equal to f ≡ 0 ⇒ f (x) = 0 for all x
∝ is proportional to y ∝ x2 ⇒ y = k x2 for some k
⊆ is a subset of {0, 1} ⊆ {0, 1, 2}
∈ is an element of 1 ∈ {1, 2, 3}
∉ is not an element of 1 ∉ {2, 3}
∪ union of sets {0, 1} ∪ {2, 3} = {0, 1, 2, 3}
∩ intersection of sets {0, 1} ∩ {1, 2} = {1}",ElementaryCalculus.pdf
"∉ is not an element of 1 ∉ {2, 3}
∪ union of sets {0, 1} ∪ {2, 3} = {0, 1, 2, 3}
∩ intersection of sets {0, 1} ∩ {1, 2} = {1}
∅ empty set {0, 1} ∩ {2, 3} = ∅
∴ therefore ∴ n must exist",ElementaryCalculus.pdf
"CHAPTER 1
The Derivative
1.1 Introduction
Calculus can be thought of as the analysis of curved shapes.
1 Its development grew out of
attempts to solve physical problems. For example, suppose t hat an object at rest 100 ft above
the ground is dropped. Ignoring air resistance and wind, the object will fall straight down until
it hits the ground (see Figure 1.1.1(a)). As will be proved la ter , t seconds after being dropped
the object will be s = s(t) = −16t2 +100 ft above the ground. The object will thus hit the ground
after 2.5 seconds (when s = 0). While the object’s path is a straight line, the graph of its position
s above the ground as a function of time t is curved, part of a parabola (see Figure 1.1.1(b)).
100 ft
t = 0 sec
t = 2.5 sec
object falling
(a) Path of the object
0
s = −16t2 + 100
s
t
100
2.5
(b) Position s as a function of time t
Figure 1.1.1 An object dropped from 100 ft above the ground",ElementaryCalculus.pdf
"0
s = −16t2 + 100
s
t
100
2.5
(b) Position s as a function of time t
Figure 1.1.1 An object dropped from 100 ft above the ground
How fast is the object moving before it hits the ground? This i s where calculus comes in. The
solution, presented now , will motivate much of this chapter .
1It is more than that, of course, but that definition puts us in g ood company: the first European textbook on
calculus, written by the French mathematician Guillaume de l’Hôpital in 1696, was titled Analyse des Infiniment
Petits pour l’Intelligence des Lignes Courbes (which translates as Analysis of the Infinitely Small for Understanding
Curved Lines ). That book (in French) can be obtained freely in electronic form at https://archive.org
1",ElementaryCalculus.pdf
"2 Chapter 1 • The Derivative §1.1
First, the object travels 100 ft in 2.5 seconds, so its average speed in that time is
distance traveled
time elapsed = 100 ft
2.5 seconds = 40 ft/s,
and its average velocity in that time is
change in position
change in time = final position − initial position
end time − start time = 0 ft − 100 ft
2.5 sec − 0 sec = −40 ft/s.
Unlike speed, velocity takes direction into account. Thus, the object’s downward motion means
it has negative velocity . Positive velocity implies upward motion.
Using the idea of average velocity over an interval of time, t here is a natural way to define
the object’s instantaneous velocity at a particular instant of time t:
1. Find the average velocity over an interval of time.
2. Let the interval become smaller and smaller indefinitely , shrinking to a point t. If the
average velocity over that smaller and smaller interval app roaches some value, call that
value the instantaneous velocity at time t.",ElementaryCalculus.pdf
"average velocity over that smaller and smaller interval app roaches some value, call that
value the instantaneous velocity at time t.
Figure 1.1.2 below shows how to choose the interval: for any t ime t between 0 and 2.5, use
the interval [ t, t + ∆t], where ∆t (pronounced “delta t”) is a small positive number . So ∆t is the
change in time over the interval; denote by ∆s the change in the position s over that interval.
s
t
t t + ∆t 2.50
100
∆t
∆s
s(t + ∆t)
s(t)
average velocity = ∆s
∆t = s(t + ∆t) − s(t)
∆t
Figure 1.1.2 Average velocity ∆s
∆t over the interval [ t, t + ∆t]
The average velocity of the object over the interval [ t, t +∆t] is ∆s
∆t , so since s(t) = −16t2 +100:",ElementaryCalculus.pdf
"Introduction • Section 1.1 3
∆s
∆t = s(t + ∆t) − s(t)
∆t
= −16(t + ∆t)2 + 100 − (−16t2 + 100)
∆t
= −16t2 − 32t∆t − 16(∆t)2 + 100 + 16t2 − 100
∆t
= −32t∆t − 16(∆t)2
∆t = ✚✚∆t (−32t − 16∆t)
✚✚∆t
= − 32t − 16∆t ,
Now let the interval [ t, t+∆t] get smaller and smaller indefinitely—that is, let ∆t get closer and
closer to 0. Then the average velocity ∆s
∆t = −32t −16∆t gets closer and closer to −32t −0 = −32t.
Thus, the object has instantaneous velocity −32t at time t. This calculation can be interpreted
as taking the limit of ∆s
∆t as ∆t approaches 0, written as follows:
instantaneous velocity at t = limit of average velocity over [ t, t + ∆t] as ∆t approaches to 0
= lim
∆t→0
∆s
∆t
= lim
∆t→0
(−32t − 16∆t)
= − 32t − 16(0)
= − 32t
Notice that ∆t is not replaced by 0 in the ratio ∆s
∆t until after doing as much cancellation as
possible. Notice also that the instantaneous velocity of th e object varies with t, as it should",ElementaryCalculus.pdf
"∆t until after doing as much cancellation as
possible. Notice also that the instantaneous velocity of th e object varies with t, as it should
(why?). In particular , at the instant when the object hits th e ground at time t = 2.5 sec, the
instantaneous velocity is −32(2.5) = −80 ft/s.
If this makes sense so far , then you understand the crux of the idea of what a limit is and
how to calculate a limit. The instantaneous velocity v(t) = −32t is called the derivative of the
position function s(t) = −16t2 + 100. Calculating derivatives, analyzing their properties , and
using them to solve various problems are part of differential calculus .
What does this have to do with curved shapes? Instantaneous v elocity is a special case of an
instantaneous rate of change of a function; in this case the instantaneous rate of change
of the position (height above the ground) of the object. Simi lar to how the rate of change of a",ElementaryCalculus.pdf
"of the position (height above the ground) of the object. Simi lar to how the rate of change of a
line is its slope, the instantaneous rate of change of a gener al curve represents the slope of
the curve . For example, the parabola s(t) = −16t2 +100 has slope −32t for all t. Note that the
slope of this curve varies (as a function of t), unlike the slope of a straight line.",ElementaryCalculus.pdf
"4 Chapter 1 • The Derivative §1.1
Finding the area inside curved regions is another type of pro blem that calculus can solve.
The basic idea is to use simpler regions—rectangles—whose a reas are known, then use those
to approximate the area inside the curved region. One such me thod is to draw more and more
rectangles of diminishing widths inside the curved region, 2 so that the sums of their areas
approach the area of the curved region. Figure 1.1.3 shows an example with four rectangles to
approximate the area under a curve y = f (x) over an interval [ a, b] on which f (x) ≥ 0.
y = f (x)
y
x
a b
Figure 1.1.3 The area of a curved region
The limit of these sums of rectangular areas is called an integral. The study and application
of integrals are part of integral calculus . Perhaps the most remarkable result in calculus is
that there is a connection between derivatives and integral s—the Fundamental Theorem of",ElementaryCalculus.pdf
"that there is a connection between derivatives and integral s—the Fundamental Theorem of
Calculus, discovered in the 17 th century , independently , by the two men who invented calculu s
as we know it: English physicist, astronomer and mathematic ian Isaac Newton (1642-1727)
and German mathematician and philosopher Gottfried Wilhel m von Leibniz (1646-1716).
Calculus makes extensive use of infinite sequences and series . An infinite series is just a
sum of an infinite number of terms. For example, it will be show n later in the text that
π
4 = 1 − 1
3 + 1
5 − 1
7 + 1
9 − ··· , (1.1)
where the sum on the right involves an infinite number of terms . A power series is a partic-
ular type of infinite series applied to functions; it can be th ought of as a polynomial of infinite
degree. For example, the trigonometric function sin x does not appear to be a polynomial. But
it turns out that sin x has a power series representation as
sin x = x − x3
3! + x5
5! − x7
7! + x9
9! − ··· , (1.2)",ElementaryCalculus.pdf
"it turns out that sin x has a power series representation as
sin x = x − x3
3! + x5
5! − x7
7! + x9
9! − ··· , (1.2)
where again the sum continues infinitely , and the formula hol ds for all x (in radians).
The idea of replacing a function by its power series played an important role throughout the
development of calculus, and is a powerful technique in many applications.
All the functions in this text will be functions of a single re al variable—that is, the values
that the variable can take are real numbers. Below is some sta ndard notation for commonly-
used sets of numbers:
2It will be shown later (in Chapter 5) that the rectangles do no t have to be completely inside the region.",ElementaryCalculus.pdf
"Introduction • Section 1.1 5
N = the set of all natural numbers, i.e. the set of nonnegative integers: 0 , 1, 2, 3, 4, . . .
Z = the set of all integers: 0 , ±1, ±2, ±3, ±4, . . .
Q = the set of all rational numbers m
n , where m and n are integers, with n ̸=0
R = the set of all real numbers
Note that N ⊂ Z ⊂ Q ⊂ R.
The set of real numbers consists of the rational numbers toge ther with numbers that are
not rational, called irrational numbers . For example,
∝ra⌈〉⌋allow
2 is irrational. That is, 2 is not the
square of a rational number . In fact, if the square of a ration al number q were an integer ,
then q itself would have to be an integer: write q as m/n, where m and n are positive integers
with no common positive integer divisors other than 1. Since q2 = m2/n2 simply duplicates the
integer divisors of m and n, then q2 can be an integer only if n = 1, i.e. q is an integer . Clearly
2 is not the square of an integer , and thus it cannot be the squa re of a rational number . This",ElementaryCalculus.pdf
"2 is not the square of an integer , and thus it cannot be the squa re of a rational number . This
argument also shows that
∝ra⌈〉⌋allow
3,
∝ra⌈〉⌋allow
5,
∝ra⌈〉⌋allow
6,
∝ra⌈〉⌋allow
7,
∝ra⌈〉⌋allow
8,
∝ra⌈〉⌋allow
10, and so on, are irrational. 3
It turns out that there are far more irrational numbers–and h ence real numbers—than ra-
tional numbers. In fact, whereas the rational numbers can be listed in a sequence (i.e. first,
second, third, etc.), the set of real numbers cannot. 4 For example, in the closed interval [0 , 1]
there is no “next” real number after the number 0. Thus, some i nfinite sets are larger than
others— R is larger than Q. Intervals such as [0 , 1] or R itself are examples of a continuum
of objects, i.e. no gaps exist. 5 A famous unsolved problem in mathematics—the Continuum
Hypothesis—is whether an infinite set exists that is larger in size than Q but smaller than R.
Infinity is an important notion in calculus. Whether it is the idea of infinitely large or in-",ElementaryCalculus.pdf
"Infinity is an important notion in calculus. Whether it is the idea of infinitely large or in-
finitesimally small , calculus attempts to give the idea some mathematical meani ng (typically
by way of limits). 6 The mathematical use of infinity has been a subject of philoso phical debate. 7
Though several centuries old, calculus was the beginning of modern mathematics. Classical
mathematics (e.g. algebra, geometry , trigonometry)—whos e origins date back to the ancient
Babylonians, Egyptians, and Greeks—was concerned mostly w ith the study of static quantities.
Calculus produced a way to analyze dynamic (i.e. changing) quantities. The period from the
17th through the 19 th century also saw revolutionary advances in physics, chemis try , biology
and other sciences. The birth of calculus was one part of that qualitative leap.
3This argument is due to the British philosopher Bertrand Rus sell (1872-1970). For an alternative proof that
∝ra⌈〉⌋allow
2",ElementaryCalculus.pdf
"3This argument is due to the British philosopher Bertrand Rus sell (1872-1970). For an alternative proof that
∝ra⌈〉⌋allow
2
is irrational, see pp. 97-98 in G E L FA N D, I.M. A N D A. S H E N, Algebra, Boston: Birkhäuser , 1993.
4For a proof see Ch.1 in K A MK E , E., Theory of Sets , New Y ork: Dover Publications, Inc., 1950.
5For a study of the structure of the real number system, see B U R R I L L , C.W ., Foundations of Real Numbers , New
Y ork: McGraw-Hill Book Company , 1967.
6Not everyone agrees that calculus does this satisfactorily . For example, for an alternative development of ba-
sically the same material in “standard” calculus but withou t the use of limits—called infinitesimal analysis —see
KE I S L E R , H.J., Elementary Calculus: An Infinitesimal Approach , Boston: Prindle, Weber & Schmidt, 1976.
7For example, see the essays by L. E. J . Brouwer , Hermann Weyl a nd David Hilbert in H E I J E N O O R T, J. VA N, From",ElementaryCalculus.pdf
"7For example, see the essays by L. E. J . Brouwer , Hermann Weyl a nd David Hilbert in H E I J E N O O R T, J. VA N, From
Frege to Gödel: A Source Book in Mathematical Logic, 1879-19 31, Cambridge, MA: Harvard University Press, 1967.",ElementaryCalculus.pdf
"6 Chapter 1 • The Derivative §1.1
Exercises
A
For Exercises 1-4, suppose that an object moves in a straight line such that its position s after time t
is the given function s = s(t). Find the instantaneous velocity of the object at a general time t ≥ 0. Y ou
should mimic the earlier example for the instantaneous velo city when s = −16t2 + 100.
1. s = t2 2. s = 9.8t2 3. s = −16t2 + 2t 4. s = t3
5. By equation (1.1), π = 4
(
1 − 1
3 + 1
5 − 1
7 + 1
9 − ···
)
, where the nth term in the sum inside the
parentheses is (−1)n+1
2n−1 (starting at n = 1).8 So the first approximation of π using this formula is
π ≈ 4 (1) = 4.0, and the second approximation is π ≈ 4
(
1 − 1
3
)
= 8/3 ≈ 2.66667. Continue like this until
two consecutive approximations have 3 as the first digit befo re the decimal point. How many terms
in the sum did this require? Be careful with rounding off in th e approximations.
B",ElementaryCalculus.pdf
"in the sum did this require? Be careful with rounding off in th e approximations.
B
6. In elementary geometry you learned that the area inside a cir cle of radius r > 0 is πr2 (that formula
will be proved later in the text). So in particular , let C be a circle of radius 1. Then the area inside
C is π. That area can be approximated by Eudoxus’ method of exhaustion .
9 The idea is to inscribe
regular polygons inside the circle, i.e. the vertexes of the polygons touch C. Recall from geometry
that a polygon is regular if its sides are of equal length. By i ncreasing the number of sides of the
polygons, the areas inside the polygons will approach the ar ea ( π) of C. This was an early attempt
at using what is now called a limit. 10
1
C
Figure 1.1.4 Inscribed square
1
C
Figure 1.1.5 Inscribed regular hexagon
(a) Inscribe a square inside C, as in Figure 1.1.4. Show that the area inside the square is 2. This is
a poor approximation of π = 3.14159265..., obviously .",ElementaryCalculus.pdf
"a poor approximation of π = 3.14159265..., obviously .
(b) Inscribe a regular hexagon (6-sided) inside C, as in Figure 1.1.5. Show that the area inside the
hexagon is 3
∝ra⌈〉⌋allow
3
2 ≈ 2.59807621. This is a slightly better—though still poor—appr oximation of π.
(c) Inscribe a regular dodecagon (12-sided) inside C. Show that the area inside the dodecagon is 3.
It thus takes 12 sides for the approximation to get the first di git of π correct.
(d) Inscribe a regular 100-sided polygon inside C. Show that the area inside this polygon is approx-
imately 3 .13952598. This is getting closer to π.
(e) Show that the general formula for the area inside a regular n-sided polygon inscribed inside C
is n
2 sin
(360◦
n
)
. (Hint: The double-angle identity sin 2 θ = 2 sin θ cos θ might help.)
8This, by the way , is a terrible formula for calculating π; getting just the 3.14 part requires 119 terms in the sum!
9Originally due to another ancient Greek mathematician, Ant iphon (ca. 430 B.C.)",ElementaryCalculus.pdf
"9Originally due to another ancient Greek mathematician, Ant iphon (ca. 430 B.C.)
10The great ancient Greek mathematician, physicist and astro nomer Archimedes (ca. 287-212 B.C.) used this
method, together with circumscribed regular polygons, to c alculate π.",ElementaryCalculus.pdf
"Introduction • Section 1.1 7
C
7. What is the flaw in the following “proof ” that π = 4?:
d = 1
Step 1: Draw a square around a circle of diameter d = 1. The circumference
of the circle is thus πd = π, and the perimeter of the square is 4.
d = 1
Step 2: Remove corners from the square as shown in the picture on the r ight,
so that four new corners touch the circle. Notice that the per imeter of the
resulting polygon is still 4, since the lengths of the remove d corner pieces are
duplicated in the new polygon, so that the lengths of all the v ertical sides add
up to 2 while the lengths of all the horizontal sides add up to 2 .
d = 1
Step 3: Remove corners from the polygon in Step 2, as shown in the pict ure
on the right, so that eight new corners touch the circle. The p erimeter of the
resulting polygon is again still 4.
d = 1
Step 4: Continue this procedure indefinitely , with each successive polygon
still having a perimeter of 4 and becoming increasingly indi stinguishable from",ElementaryCalculus.pdf
"still having a perimeter of 4 and becoming increasingly indi stinguishable from
the circle. Since the perimeters of the polygons always equa l 4 and approach
the circle’s circumference ( π), then π must equal 4.
8. An infinite set is countable if its members can be put into a one-to-one correspondence wi th the
members of N, the set of natural numbers (0 , 1, 2, 3, 4, . . . ). Clearly N is itself countable. The set Z of
all integers is also countable, by means of the following one -to-one correspondence with N:
N 0 1 2 3 4 5 6 7 8 . . .
Z 0 1 -1 2 -2 3 -3 4 -4 . . .
Show that Q (the set of all rational numbers) is countable. (Hint: The above correspondence for Z is
an infinite list in one dimension (the horizontal direction) . For Q think two-dimensionally .)",ElementaryCalculus.pdf
"8 Chapter 1 • The Derivative §1.2
1.2 The Derivative: Limit Approach
The following definition generalizes the example from the pr evious section (concerning instan-
taneous velocity) to a general function f (x):
The derivative of a real-valued function f (x), denoted by f ′(x), is
f ′(x) = lim
∆x→0
∆f
∆x = lim
∆x→0
f (x + ∆x) − f (x)
∆x (1.3)
for x in the domain of f , provided that the limit exists. 11
For a general function f (x), the derivative f ′(x) represents the instantaneous rate of
change of f at x, i.e. the rate at which f changes at the “instant” x. For the limit part of
the definition only the intuitive idea of how to take a limit—a s in the previous section—is
needed for now . Notice that the above definition makes the der ivative f ′ itself a function of the
variable x. The function f ′ can be evaluated at specific values of x, or you can write its general
formula f ′(x).
The (instantaneous) velocity of an object as the derivative of the object’s position as a func-",ElementaryCalculus.pdf
"formula f ′(x).
The (instantaneous) velocity of an object as the derivative of the object’s position as a func-
tion of time is only one physical application of derivatives . There are many other examples:
Field Function Derivative
Physics position velocity
velocity acceleration
momentum force
work power
Field Function Derivative
Physics angular momentum torque
Engineering electric charge electric current
magnetic flux induced voltage
Economics profit marginal profit
The limit definition can be used for finding the derivatives of simple functions.
Example 1.1
Find the derivative of the function f (x) = 1.
Solution: By definition, f (x) = 1 for all x, so:
f ′(x) = lim
∆x→0
f (x + ∆x) − f (x)
∆x
= lim
∆x→0
1 − 1
∆x
= lim
∆x→0
0
∆x
= lim
∆x→0
0
f ′(x) = 0
11Recall that the domain of f is the set of all numbers x such that f (x) is defined.",ElementaryCalculus.pdf
"The Derivative: Limit Approach • Section 1.2 9
Notice in the above example that replacing ∆x by 0 was unnecessary when taking the limit,
since the ratio f (x+∆x) − f (x)
∆x simplified to 0 before taking the limit, and the limit of 0 is 0 regard-
less of what ∆x approaches. In fact, the answer—namely , f ′(x) = 0 for all x—should have been
obvious without any calculations: the function f (x) = 1 is a constant function , so its value (1)
never changes , and thus its rate of change is always 0. Hence, its derivative is 0 everywhere.
Replacing the constant 1 by any constant yields the following important result:
The derivative of any constant function is 0.
The above discussion shows that the calculation in Example 1 .1 was unnecessary . Consider
another example where no calculation is required to find the d erivative: the function f (x) = x.
The graph of this function is just the line y = x in the x y-plane, and the rate of change of a line",ElementaryCalculus.pdf
"The graph of this function is just the line y = x in the x y-plane, and the rate of change of a line
is a constant, called its slope. The line y = x has a slope of 1, so the derivative of f (x) = x is
f ′(x) = 1 for all x. The formal calculation of the derivative, though unnecess ary , verifies this:
f ′(x) = lim
∆x→0
f (x + ∆x) − f (x)
∆x = lim
∆x→0
(x + ∆x) − x
∆x = lim
∆x→0
∆x
∆x = lim
∆x→0
1 = 1
Recall that a function whose graph is a line is called a linear function . For a general
linear function f (x) = mx + b, where m is the slope of the line and b is its y-intercept, the same
argument as above for f (x) = x yields the following result:
The derivative of any linear function is the slope of the line itself:
If f (x) = mx + b then f ′(x) = m for all x.
The function f (x) = 1 from Example 1.1 is the special case where m = 0 and b = 1; its graph
is a horizontal line, so its slope (and hence its derivative) is 0 for all x. Likewise, the function",ElementaryCalculus.pdf
"is a horizontal line, so its slope (and hence its derivative) is 0 for all x. Likewise, the function
f (x) = 2x − 1 represents a line of slope m = 2, so its derivative is 2 for all x. Figure 1.2.1 shows
these and other linear functions y = f (x).
x
y
1
f (x) = 1 slope = 0
f ′(x) = 0
0
y
x
f (x) = 2x − 1
slope = 2
f ′(x) = 2
0
−1
1
2
y
x
f (x) = −x + 2
slope = −1
f ′(x) = −1
0
2
2
Figure 1.2.1 Slopes and derivatives of lines",ElementaryCalculus.pdf
"10 Chapter 1 • The Derivative §1.2
Linear functions have a constant derivative—the constant b eing the slope of the line. The
converse turns out to be true: a function with a constant deri vative must be a linear function. 12
What types of functions do not have constant derivatives? Th e previous section discussed
such a function: the parabola s(t) = −16t2 + 100, whose derivative s′(t) = −32t is clearly not a
constant function. In general, functions that represent cu rves (i.e. not straight lines) do not
change at a constant rate—that is precisely what makes them curved. So such functions do not
have a constant derivative.
Example 1.2
Find the derivative of the function f (x) = 1
x . Also, find the instantaneous rate of change of f at x = 2.
Solution: For all x ̸=0, the derivative f ′(x) is:
f ′(x) = lim
∆x→0
f (x + ∆x) − f (x)
∆x
= lim
∆x→0
1
x + ∆x − 1
x
∆x → 0
0 , so simplify the ratio before plugging in ∆x = 0,
= lim
∆x→0
x − (x + ∆x)
(x + ∆x)x",ElementaryCalculus.pdf
"∆x→0
f (x + ∆x) − f (x)
∆x
= lim
∆x→0
1
x + ∆x − 1
x
∆x → 0
0 , so simplify the ratio before plugging in ∆x = 0,
= lim
∆x→0
x − (x + ∆x)
(x + ∆x)x
∆x (after getting a common denominator)
= lim
∆x→0
−✟✟∆x
✟✟∆x(x + ∆x)x
= lim
∆x→0
−1
(x + ∆x)x = −1
(x + 0)x
f ′(x) = − 1
x2
The instantaneous rate of change of f at x = 2 is just the derivative f ′(x) evaluated at x = 2, that is,
f ′(2) = −1
22 = −1
4 .
x
y
0
2
f (x) = 1
x
Notice that the instantaneous rate of change f ′(2) = −1
4 in the
above example is a negative number . This should make sense, s ince
the function f (x) = 1
x is changing in the negative direction at x = 2;
that is, f (x) is decreasing in value at x = 2. This is plain to see from
the graph of f (x) = 1
x shown on the right. In fact, for all x ̸=0 the
function f (x) = 1
x is decreasing as x grows.13 This is reflected in the
derivative f ′(x) = −1
x2 being negative for all x ̸=0. In general, a nega-
tive derivative means that the function is decreasing, whil e a positive",ElementaryCalculus.pdf
"derivative f ′(x) = −1
x2 being negative for all x ̸=0. In general, a nega-
tive derivative means that the function is decreasing, whil e a positive
derivative means that it is increasing.
12This will be proved in Chapter 5.
13In this text, the rate of change of f (x) is always taken in the direction of increasing x, i.e. in the positive x
direction.",ElementaryCalculus.pdf
"The Derivative: Limit Approach • Section 1.2 11
The problem with using the limit definition to find the derivat ive of a curved function is that
the calculations require more work, as the above example sho ws. As the functions become more
complicated those calculations can become difficult or even impossible. And though limits have
not yet been defined formally , for now the intuitively obviou s idea of limits suffices, namely:
For a real number a and a real-valued function f (x), say that the limit of f (x) as x ap-
proaches a equals the number L, written as
limx→a f (x) = L ,
if f (x) approaches L as x approaches a.
Equivalently , this means that f (x) can be made as close as you want to L by choosing x
close enough to a. Note that x can approach a from any direction.
Below are some simple rules for limits, which will be proved l ater:
Rules for Limits: Suppose that a is a real number and that f (x) and g(x) are real-valued",ElementaryCalculus.pdf
"Rules for Limits: Suppose that a is a real number and that f (x) and g(x) are real-valued
functions such that lim x→a f (x) and lim x→a g(x) both exist. Then:
(a) limx→a ( f (x) + g(x)) =
(
limx→a f (x)
)
+
(
limx→a g(x)
)
(b) limx→a ( f (x) − g(x)) =
(
limx→a f (x)
)
−
(
limx→a g(x)
)
(c) limx→a (k · f (x)) = k ·
(
limx→a f (x)
)
for any constant k
(d) limx→a ( f (x) · g(x)) =
(
limx→a f (x)
)
·
(
limx→a g(x)
)
(e) limx→a
f (x)
g(x) =
limx→a f (x)
limx→a g(x) , if limx→a g(x) ̸=0
The above rules say that the limit of sums, differences, cons tant multiples, products, and
quotients is the sum, difference, constant multiple, produ ct, and quotient, respectively , of the
limits. This seems intuitively obvious.
These rules can be used for finding other expressions for the d erivative. The quantity ∆x rep-
resents a small number—positive or negative—that approach es 0, but it is common in mathe-
matics texts to use the letter h instead:14
f ′(x) = lim
h→0
f (x + h) − f (x)
h (1.4)",ElementaryCalculus.pdf
"matics texts to use the letter h instead:14
f ′(x) = lim
h→0
f (x + h) − f (x)
h (1.4)
14Physics texts typically prefer the delta notation, since ∆x represents a small change in some physical quantity x.",ElementaryCalculus.pdf
"12 Chapter 1 • The Derivative §1.2
Another formulation is to set h = w − x in formula (1.4), which yields
f ′(x) = lim
h→0
f (x + h) − f (x)
h = lim
w−x→0
f (x + (w − x)) − f (x)
w − x ,
so that
f ′(x) = limw→x
f (w) − f (x)
w − x (1.5)
since w − x approaches 0 if and only if w approaches x. Another formulation replaces h by −h:
f ′(x) = lim
h→0
f (x + h) − f (x)
h = lim
−h→0
f (x + −h) − f (x)
−h = lim
−h→0
−( f (x) − f (x − h))
−h ,
and thus
f ′(x) = lim
h→0
f (x) − f (x − h)
h (1.6)
since −h approaches 0 if and only if h approaches 0. The above formulations did not use the
Limit Rules, but the following result does:
Suppose that f ′(x) exists. Then
f ′(x) = lim
h→0
f (x + h) − f (x − h)
2h . (1.7)
Proof: Since f ′(x) = lim
h→0
f (x + h) − f (x)
h = lim
h→0
f (x) − f (x − h)
h by formulas (1.4) and (1.6),
then Limit Rule (c) shows that
1
2 f ′(x) = lim
h→0
f (x + h) − f (x)
2h = lim
h→0
f (x) − f (x − h)
2h .",ElementaryCalculus.pdf
"h by formulas (1.4) and (1.6),
then Limit Rule (c) shows that
1
2 f ′(x) = lim
h→0
f (x + h) − f (x)
2h = lim
h→0
f (x) − f (x − h)
2h .
Now use the idea that a − b = (a − c) + (c − b) for all a, b, and c to write:
lim
h→0
f (x + h) − f (x − h)
2h = lim
h→0
( f (x + h) − f (x)) + ( f (x) − f (x − h))
2h
= lim
h→0
f (x + h) − f (x)
2h + lim
h→0
f (x) − f (x − h)
2h (by Limit Rule (a))
= 1
2 · f ′(x) + 1
2 · f ′(x)
= f ′(x)
QED",ElementaryCalculus.pdf
"The Derivative: Limit Approach • Section 1.2 13
As an example of using these different formulations, recall that a function f is even if
f (−x) = f (x) for all x in the domain of f , and f is odd if f (−x) = −f (x) for all x in its domain.
For example, x2, x4, and cos x are even functions; x, x3, and sin x are odd functions. The
following result is often useful:
The derivative of an even function is an odd function.
The derivative of an odd function is an even function.
To prove the first statement—the second is an exercise—suppo se that f is an even function
and that f ′(x) exists for all x in its domain. Then
f ′(−x) = lim
h→0
f (−x + h) − f (−x)
h by formula (1.4) with x replaced by −x
= lim
h→0
f (−(x − h)) − f (−x)
h
= lim
h→0
f (x − h) − f (x)
h since f is even
= lim
h→0
−( f (x) − f (x − h))
h
= − lim
h→0
f (x) − f (x − h)
h by Limit Rule (c), so
f ′(−x) = −f ′(x) by formula (1.6),
which shows that f ′ is an odd function.",ElementaryCalculus.pdf
"h
= − lim
h→0
f (x) − f (x − h)
h by Limit Rule (c), so
f ′(−x) = −f ′(x) by formula (1.6),
which shows that f ′ is an odd function.
Derivatives do not always exist, as the following example sh ows.
Example 1.3
x
y
0
f (x) = |x |
y = x y= −x
Let f (x) = |x |. Show that f ′(0) does not exist.
Solution: Recall that the absolute value function f (x) = |x | is defined as
f (x) = |x | =
{
x if x ≥ 0
−x if x < 0
The graph consists of two lines meeting at the origin. For x ≥ 0 the graph
is the line y = x, which has slope 1. For x ≤ 0 the graph is the line y = −x,
which has slope -1. These lines agree in value ( y = 0) at x = 0, but their
slopes do not agree in value at x = 0. Therefore the derivative of f does not exist at x = 0, since the
derivative of a curve is just its slope. A more “formal” proof (which amounts to the same argument) is
outlined in the exercises.",ElementaryCalculus.pdf
"14 Chapter 1 • The Derivative §1.2
If the derivative f ′(x) exists then f is differentiable at x. A differentiable function is
one that is differentiable at every point in its domain. For e xample, f (x) = x is a differentiable
function, but f (x) = |x| is not differentiable at x = 0. The act of calculating a derivative is called
differentiation. For example, differentiating the function f (x) = x yields f ′(x) = 1.
Exercises
Note: For all exercises, you can use anything discussed so fa r (including previous exercises).
A
For Exercises 1-11, find the derivative of the given function f (x) for all x (unless indicated otherwise).
1. f (x) = 0 2. f (x) = 1 − 3x 3. f (x) = (x + 1)2 4. f (x) = 2x2 − 3x + 1
5. f (x) = 1
x+1 , for all x ̸= −1 6. f (x) = −1
x+1 , for all x ̸= −1 7. f (x) = 1
x2 , for all x ̸=0
8. f (x) = ∝ra⌈〉⌋allowx, for all x > 0 (Hint: Rationalize the numerator in the definition of the der ivative.)
9. f (x) =
∝ra⌈〉⌋allow
x + 1, for all x > −1 10. f (x) =
∝ra⌈〉⌋allow",ElementaryCalculus.pdf
"9. f (x) =
∝ra⌈〉⌋allow
x + 1, for all x > −1 10. f (x) =
∝ra⌈〉⌋allow
x2 + 1 11. f (x) =
∝ra⌈〉⌋allow
x2 + 3x + 4
12. In Exercise 8 the point x = 0 was excluded when calculating f ′(x), even though x = 0 is in the
domain of f (x) = ∝ra⌈〉⌋allowx. Can you explain why x = 0 was excluded?
B
13. Show that for all functions f such that f ′(x) exists, f ′(x) = limw→x
f (x) − f (w)
x − w .
14. True or false: If f and g are differentiable functions on an interval ( a, b) and f (x) < g(x) for all x in
(a, b), then f ′(x) < g′(x) for all x in ( a, b). If true, prove it; if false, give a counterexample. Would y our
answer change if the restriction of x to ( a, b) were removed and all real x were used instead?
15. Show that the derivative of an odd function is an even functio n.
C
For Exercises 16-21, assuming that f ′(x) exists, prove the given formula.
16. f ′(x) = lim
h→0
f (x + 2h) − f (x − 2h)
4h 17. f ′(x) = lim
h→0
f (x + 3h) − f (x − 3h)
6h
18. f ′(x) = lim
h→0
f (x + 2h) − f (x − 3h)",ElementaryCalculus.pdf
"16. f ′(x) = lim
h→0
f (x + 2h) − f (x − 2h)
4h 17. f ′(x) = lim
h→0
f (x + 3h) − f (x − 3h)
6h
18. f ′(x) = lim
h→0
f (x + 2h) − f (x − 3h)
5h 19. f ′(x) = lim
h→0
f (x + ah) − f (x − bh)
(a + b)h (a, b > 0)
20. limw→x
w f (x) − x f (w)
w − x = f (x) − x f ′(x) 21. limw→x
w2 f (x) − x2 f (w)
w − x = 2x f (x) − x2 f ′(x)
22. Show that f (x) = |x | is not differentiable at x = 0, using formula (1.4) for the derivative. Here you
will have to use a part of the definition which has not been used yet: as h approaches 0, h can be
either positive or negative. Consider those two cases in sho wing that the limit is not defined at x = 0.
23. Suppose that f (a + b) = f (a) f (b) for all a and b, and f ′(0) exists. Show that f ′(x) exists for all x.",ElementaryCalculus.pdf
"The Derivative: Infinitesimal Approach • Section 1.3 15
1.3 The Derivative: Infinitesimal Approach
Traditionally a function f of a variable x is written as y = f (x). The dependent variable y is
considered a function of the independent variable x. This allows taking the derivative of y
with respect to x, i.e. the derivative of y as a function of x, denoted by dy
dx . This is simply a
different way of writing f ′(x), and is just one of many ways of denoting the derivative:
Notation for the derivative of y === f (x): The following are all equivalent:
dy
dx , f ′(x) , d
dx ( f (x)) , y′ , ˙y , ˙f (x) , df
dx , D f (x)
The notation dy
dx appears to denote a fraction: a quantity dy divided by a quantity dx. It turns
out that the derivative really can be thought of in that way , as a ratio of infinitesimals . In fact,
this was the way in which derivatives were used by the founder s of calculus—Newton and, in",ElementaryCalculus.pdf
"this was the way in which derivatives were used by the founder s of calculus—Newton and, in
particular , Leibniz.15 Even today , this is often the way in which derivatives are tho ught of and
used in fields outside of mathematics, such as physics, engin eering, and chemistry , perhaps
due to its more intuitive nature.
The concept of infinitesimals used here is based on the nilsquare infinitesimal approach
developed by J .L. Bell 16, namely:
A number δ is an infinitesimal if the conditions (a)-(d) hold:
(a) δ ̸=0
(b) if δ > 0 then δ is smaller than any positive real number
(c) if δ < 0 then δ is larger than any negative real number
(d) δ2 = 0 (and hence all higher powers of δ, such as δ3 and δ4, are also 0)
Note: Any infinitesimal multiplied by a nonzero real number i s also an infinitesimal, while
0 times an infinitesimal is 0.
The above definition says that infinitesimals are numbers whi ch are closer to 0 than any",ElementaryCalculus.pdf
"0 times an infinitesimal is 0.
The above definition says that infinitesimals are numbers whi ch are closer to 0 than any
positive or negative number without being zero themselves, and raising them to powers greater
than or equal to 2 makes them 0. So infinitesimals are not real n umbers.17 This is not a
problem, since calculus deals with other numbers, such as in finity , which are not real. An
infinitesimal can be thought of as an infinitely small number a rbitrarily close to 0 but not 0.
15It was Leibniz who created the notation dy
dx . For this reason dy
dx is called the Leibniz notation for the derivative.
Newton used the dot notation ˙y, which has fallen out of favor with mathematicians but is sti ll used by many
physicists, especially when the independent variable repr esents time. Newton called derivatives fluxions . The
prime notation f ′ is due to the French mathematician and physicist Joseph Loui s Lagrange (1736-1813).",ElementaryCalculus.pdf
"prime notation f ′ is due to the French mathematician and physicist Joseph Loui s Lagrange (1736-1813).
16 BE L L , J.L., A Primer of Infinitesimal Analysis , Cambridge, U .K.: Cambridge University Press, 1998.
17In an equivalent treatment, infinitesimals are part of the hyperreal number system . See K E I S L E R , H.J., Elemen-
tary Calculus: An Infinitesimal Approach , Boston: Prindle, Weber & Schmidt, 1976.",ElementaryCalculus.pdf
"16 Chapter 1 • The Derivative §1.3
This might seem like a strange notion, but it really is not all that different from the limit
notion where, say , you let ∆x approach 0 but not necessarily let it equal 0.18
As for the square of a nonzero infinitesimal being 0, think of h ow a calculator handles the
squares of small numbers. For example, most calculators can display 10 −8 as 0.00000001, and
will even let you add 1 to that to get 1.00000001. But when you s quare 10 −8 and add 1 to it,
most calculators will display the sum as simply 1. The calcul ator treats the square of 10 −8,
namely 10 −16, as a number so small compared to 1 that it is effectively zero .19
Notice a major difference between 0 and an infinitesimal δ: 2 · 0 and 0 are the same, but 2 δ
and δ are distinct. This holds for any nonzero constant multiple, not just the number 2.
The derivative dy
dx of a function y = f (x) can now be defined in terms of infinitesimals:",ElementaryCalculus.pdf
"The derivative dy
dx of a function y = f (x) can now be defined in terms of infinitesimals:
Let dx be an infinitesimal such that f (x + dx) is defined. Then dy = f (x + dx) − f (x) is also
an infinitesimal, and the derivative of y = f (x) at x is the ratio of dy to dx:
dy
dx = f (x + dx) − f (x)
dx (1.8)
The basic idea is that dx is an infinitesimally small change in the variable x, producing an
infinitesimally small change dy in the value of y = f (x).
Example 1.4
Show that the derivative of y = f (x) = x2 is dy
dx = 2x.
Solution: For any real number x,
dy
dx = f (x + dx) − f (x)
dx
= (x + dx)2 − x2
dx
= x2 + 2x dx + (dx)2 − x2
dx
= 2x dx + 0
dx since dx is an infinitesimal ⇒ (dx)2 = 0
= 2x✚✚dx
✚✚dx
= 2x
18The infinitesimal approach was first developed in an axiomati c manner in the landmark book R O B I N S O N, A.,
Non-Standard Analysis , Amsterdam: North-Holland, 1966. Robinson showed that for all practical purposes calcu-",ElementaryCalculus.pdf
"Non-Standard Analysis , Amsterdam: North-Holland, 1966. Robinson showed that for all practical purposes calcu-
lus can be developed without resorting to limits, with equiv alent results.
19Calculators do this for display reasons—most can show only 1 0-12 digits. Try this experiment on your calculator:
Add 10 30, −
(
1030)
, and 1 in two different ways:
(
1030 + −
(
1030))
+ 1, and 10 30 +
(
−
(
1030)
+1
)
. The first way will
give you the correct answer 1, but the second way yields 0. So a ddition is not always associative on calculators!",ElementaryCalculus.pdf
"The Derivative: Infinitesimal Approach • Section 1.3 17
Y ou might have noticed that the above example did not involve limits, and that the derivative
2x represents a real number (i.e. no infinitesimals appear in th e final answer); this will always
be the case. Infinitesimals possess another useful property :
Microstraightness Property : For the graph of a differentiable function, any part of the
curve with infinitesimal length is a straight line segment.
In other words, at the infinitesimal level differentiable curves are straig ht. The idea behind
this is simple. At various points on a nonstraight different iable curve y = f (x) the distances
along the curve between the points are not quite the same as th e lengths of the line segments
joining the points. For example, in Figure 1.3.1 the distanc e s measured along the curve from
the point A to the point B is not the same as the length of the line segment AB joining A to B.
y
x
AB
A
B
y = f (x)
s
magnify sAB
A
B",ElementaryCalculus.pdf
"the point A to the point B is not the same as the length of the line segment AB joining A to B.
y
x
AB
A
B
y = f (x)
s
magnify sAB
A
B
Figure 1.3.1 Curved real-valued distance s ̸=length of AB , but infinitesimal s = length of AB
However , as the points A and B get closer to each other , the difference between that part of
the curve joining A to B and the line segment AB becomes less noticeable. That is, the curve
is almost linear when A and B are close. The Microstraightness Property simply goes one s tep
further and says that the curve actually is linear when the distance s between the points is
infinitesimal (so that s equals the length of AB at the infinitesimal level).
At first this might seem nonsensical. After all, how could any nonstraight part of a curve
be straight? Y ou have to remember that an infinitesimal is an a bstraction—it does not exist
physically . A curve y = f (x) is also an abstraction, which exists in a purely mathematic al sense,",ElementaryCalculus.pdf
"physically . A curve y = f (x) is also an abstraction, which exists in a purely mathematic al sense,
so its geometric properties at the “normal” scale do not have to match those at the infinitesimal
scale (which can be defined in any way , provided the propertie s at that scale are consistent).
y
x
x x + dx
y = f (x)
dx
dy
This abstraction finally reveals what an instantaneous rate of change
is: the average rate of change over an infinitesimal interval . Moving
an infinitesimal amount dx away from a value x produces an infinites-
imal change dy in a differentiable function y = f (x). The average rate
of change of y = f (x) over the infinitesimal interval [ x, x+dx] is thus dy
dx ,
i.e. the slope—rise over run—of the straight line segment re presented
by the curve y = f (x) over that interval, as in the figure on the right. 20
20Notice that the figure implies that the Pythagorean Theorem d oes not apply to infinitesimal triangles. This will
be discussed in Chapter 8.",ElementaryCalculus.pdf
"18 Chapter 1 • The Derivative §1.3
The Microstraightness Property can be extended to smooth curves—that is, curves without
sharp edges or cusps. For example, circles and ellipses are s mooth, but polygons are not.
The properties of infinitesimals can be applied to determine the derivatives of the sine and
cosine functions. Consider a circle of radius 1 with center O and points A and B on the circle
such that the line segment AB is a diameter . Let C be a point on the circle such that the angle
∠B AC has an infinitesimal measure dx (in radians) as in Figure 1.3.2(a).
A B
C
O
dx
1 1
(a)
A B
C
O
dx 2dx
1 1
s = /hbracewidestBC = 2dxs = /hbracewidestBC = 2dx
(b)
Figure 1.3.2 Circle O: BC = 2 sin dx, ∠BOC = 2 ∠B AC
By Thales’ Theorem from elementary geometry , the angle ∠ ACB is a right angle. Thus:
sin dx = BC
AB = BC
2 ⇒ BC = 2 sin dx
Figure 1.3.2(b) shows that ∠O AC + ∠OC A + ∠ AOC = π. Thus, 1 = OC = O A ⇒ ∠OC A =",ElementaryCalculus.pdf
"sin dx = BC
AB = BC
2 ⇒ BC = 2 sin dx
Figure 1.3.2(b) shows that ∠O AC + ∠OC A + ∠ AOC = π. Thus, 1 = OC = O A ⇒ ∠OC A =
∠O AC = dx ⇒ ∠ AOC = π− dx − dx = π− 2dx ⇒ ∠BOC = 2dx. By the arc length formula from
trigonometry , the length s of the arc /hbracewidestBC along the circle from B to C is the radius times the
central angle ∠BOC : s = /hbracewidestBC = 1 · 2dx = 2dx. But by Microstraightness, /hbracewidestBC = BC , and thus:
2 sin dx = BC = /hbracewidestBC = 2dx ⇒ sin dx = dx
Since dx is an infinitesimal, ( dx)2 = 0. So since sin 2 dx + cos2 dx = 1, then:
cos2 dx = 1 − sin2 dx = 1 − (dx)2 = 1 − 0 = 1 ⇒ cos dx = 1
The derivative of y = sin x is then:
d
dx (sin x) = dy
dx = sin ( x + dx) − sin x
dx
= (sin x cos dx + sin dx cos x) − sin x
dx by the sine addition formula
= ✘✘✘✘✘(sin x) (1) + dx cos x − ✘✘✘sin x
dx = ✚✚dx cos x
✚✚dx , and thus:
d
dx (sin x) = cos x",ElementaryCalculus.pdf
"The Derivative: Infinitesimal Approach • Section 1.3 19
A similar argument (left as an exercise) using the cosine add ition formula shows:
d
dx (cos x) = −sin x
One of the intermediate results proved here bears closer exa mination. Namely , sin dx = dx
for an infinitesimal angle dx measured in radians. At first, it might seem that this cannot b e
true. After all, an infinitesimal dx is thought of as being infinitely close to 0, and sin 0 = 0, so
you might expect that sin dx = 0. But this is not the case. The formula sin dx = dx says that in
an infinitesimal interval around 0, the function y = sin x is identical to the line y = x (not the
line y = 0). This, in turn, suggests that for real-valued x close to 0, sin x ≈ x.
This indeed turns out to be the case. The free graphing softwa re Gnuplot 21 can display the
graphs of y = sin x and y = x. Figure 1.3.3(a) below shows how those graphs compare over t he
interval [ −π, π]. Outside the interval [ −1, 1] there is a noticeable difference.
-4
-3",ElementaryCalculus.pdf
"interval [ −π, π]. Outside the interval [ −1, 1] there is a noticeable difference.
-4
-3
-2
-1
0
1
2
3
4
-3 -2 -1 0 1 2 3
y
x
sin(x)
x
(a)
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
-0.3 -0.2 -0.1 0 0.1 0.2 0.3
y
x
sin(x)
x
(b)
Figure 1.3.3 sin dx = dx: Comparing y = sin x and y = x near x = 0
Figure 1.3.3(b) shows that there is virtually no difference in the graphs even over the non-
infinitesimal interval [ −0.3, 0.3]. So sin x ≈ x is a good approximation when x is close to 0, that
is, when |x| ≪1 (the symbol ≪ means “much less than”). This approximation is used in many
applications in engineering and physics when the angle x is assumed to be small.
Notice something else suggested by the relation sin dx = dx: there is a fundamental differ-
ence at the infinitesimal level between a line of slope 1 ( y = x) and a line of slope 0 ( y = 0). In a
real interval ( −a, a) around x = 0 the difference between the two lines can be made as small as",ElementaryCalculus.pdf
"real interval ( −a, a) around x = 0 the difference between the two lines can be made as small as
desired by choosing a > 0 small enough. But in an infinitesimal interval ( −δ, δ) around x = 0
there is unbridgeable gulf between the two lines. This is the crucial difference in sin dx being
equal to dx rather than 0.
21Available at http://www.gnuplot.info.",ElementaryCalculus.pdf
"20 Chapter 1 • The Derivative §1.3
Notice also that the value of a function at an infinitesimal ma y itself be an infinitesimal (e.g.
sin dx = dx) or a real number (e.g. cos dx = 1).
For a differentiable function f (x), df
dx = f ′(x) and so multiplying both sides by dx yields the
important relation:
df = f ′(x) dx (1.9)
Note that both sides of the above equation are infinitesimals for each value of x in the domain
of f ′, since f ′(x) would then be a real number .
The notion of an infinitesimal was fairly radical at the time ( and still is). Some mathemati-
cians embraced it, e.g. the outstanding Swiss mathematicia n Leonhard Euler (1707-1783),
who produced a large amount of work using infinitesimals. But it was too radical for many
mathematicians (and philosophers 22), enough so that by the 19 th century some mathemati-
cians (notably Augustin Cauchy and Karl Weierstrass) felt t he need to put calculus on what",ElementaryCalculus.pdf
"cians (notably Augustin Cauchy and Karl Weierstrass) felt t he need to put calculus on what
they considered a more “rigorous” footing, based on limits. 23 Y et it was precisely the notion of
an infinitesimal which lent calculus its modern character , by showing the power and usefulness
of such an abstraction (especially one that did not obey the r ules of classical mathematics).
Exercises
A
For Exercises 1-9, let dx be an infinitesimal and prove the given formula.
1. (dx + 1)2 = 2dx + 1 2. (dx + 1)3 = 3dx + 1 3. (dx + 1)−1 = 1 − dx
4. tan dx = dx 5. sin 2 dx = 2dx 6. cos 2 dx = 1
7. sin 3 dx = 3dx 8. cos 3 dx = 1 9. sin 4 dx = 4dx
10. Is cot dx defined for an infinitesimal dx? If so, then find its value. If not, then explain why .
11. In the proof of the derivative formulas for sin x and cos x, the equation cos 2 dx = 1 was solved to
give cos dx = 1. Why was the other possible solution cos dx = −1 ignored?
B
12. Show that d
dx (cos x) = −sin x.
13. Show that d",ElementaryCalculus.pdf
"give cos dx = 1. Why was the other possible solution cos dx = −1 ignored?
B
12. Show that d
dx (cos x) = −sin x.
13. Show that d
dx (cos 2 x) = −2 sin 2 x. (Hint: Use Exercises 5 and 6.)
C
14. Show that d
dx (tan x) = sec2 x. (Hint: Use Exercise 4.)
22The English philosopher George Berkeley (1685-1753) famou sly derided infinitesimals as “the ghosts of departed
quantities” in his book The Analyst (1734), which had the disquieting subtitle “ A Discourse Add ressed to an Infidel
Mathematician” (directed at Newton).
23However , the limit approach turns out, ultimately , to be equ ivalent to the infinitesimal approach. In essence,
only the terminology is different.",ElementaryCalculus.pdf
"Derivatives of Sums, Products and Quotients • Section 1.4 21
1.4 Derivatives of Sums, Products and Quotients
So far the derivatives of only a few simple functions have bee n calculated. The following rules
will make it easier to calculate derivatives of more functio ns:
Rules for Derivatives: Suppose that f and g are differentiable functions of x. Then:
Sum Rule: d
dx ( f + g) = df
dx + dg
dx
Difference Rule: d
dx ( f − g) = df
dx − dg
dx
Constant Multiple Rule: d
dx (c f ) = c · df
dx for any constant c
Product Rule: d
dx ( f · g) = f · dg
dx + g · df
dx
Quotient Rule: d
dx
(f
g
)
=
g · df
dx − f · dg
dx
g2
The above rules can be written using the prime notation for de rivatives:
Sum Rule: ( f + g)′(x) = f ′(x) + g′(x)
Difference Rule: ( f − g)′(x) = f ′(x) − g′(x)
Constant Multiple Rule: (c f )′(x) = c · f ′(x) for any constant c
Product Rule: ( f · g)′(x) = f (x) · g′(x) + g(x) · f ′(x)
Quotient Rule:
(f
g
) ′
(x) = g(x) · f ′(x) − f (x) · g′(x)
( g(x))2",ElementaryCalculus.pdf
"Product Rule: ( f · g)′(x) = f (x) · g′(x) + g(x) · f ′(x)
Quotient Rule:
(f
g
) ′
(x) = g(x) · f ′(x) − f (x) · g′(x)
( g(x))2
The proof of the Sum Rule is straightforward. Since df
dx and dg
dx both exist then:
d
dx ( f + g) = ( f + g)(x + dx) − ( f + g)(x)
dx = f (x + dx) + g(x + dx) − ( f (x) + g(x))
dx
= f (x + dx) − f (x) + g(x + dx) − g(x)
dx = f (x + dx) − f (x)
dx + g(x + dx) − g(x)
dx
= df
dx + dg
dx ✓
The proofs of the Difference and Constant Multiple Rules are similar and are left as exercises.",ElementaryCalculus.pdf
"22 Chapter 1 • The Derivative §1.4
Note that by the Product Rule, in general the derivative of a p roduct is not the product of
the derivatives. That is, d( f ·g)
dx ̸=df
dx · dg
dx . This should be obvious from some earlier examples. For
instance, if f (x) = x and g(x) = 1 then ( f · g)(x) = x · 1 = x so that d( f ·g)
dx = 1, but df
dx · dg
dx = 1 · 0 = 0.
There is a proof of the Product Rule similar to the proof of the Sum Rule (see Exercise 20),
but there is a more geometric way of seeing why the formula hol ds, described below .
g(x + dx)
f(x+dx)f (x)
df
g(x) dg
Construct a rectangle whose perpendicular sides
have lengths f (x) and g(x) for some x, as in the draw-
ing on the right. Change x by some infinitesimal
amount dx, which produces infinitesimal changes df
and dg in f (x) and g(x), respectively . Assume those
changes are positive and extend the original rectan-
gle by those amounts, creating a larger rectangle with
perpendicular sides f (x + dx) and g(x + dx). Then",ElementaryCalculus.pdf
"gle by those amounts, creating a larger rectangle with
perpendicular sides f (x + dx) and g(x + dx). Then
d( f · g) = ( f · g)(x + dx) − ( f · g)(x)
= f (x + dx) · g(x + dx) − f (x) · g(x)
= (area of outer rectangle) − (area of original rectangle)
= sum of the areas of the three shaded inner rectangles
= f (x) · dg + g(x) · df + df · dg
= f (x) · dg + g(x) · df + ( f ′(x) dx) · ( g′(x) dx)
= f (x) · dg + g(x) · df + ( f ′(x) g′(x)) · (dx)2
= f (x) · dg + g(x) · df + ( f ′(x) g′(x)) · 0
d( f · g) = f (x) · dg + g(x) · df , so dividing both sides by dx yields
d( f · g)
dx = f (x) · dg
dx + g(x) · df
dx ✓
To prove the Quotient Rule, let y = f
g , so f = g · y. If y were a differentiable function of x,
then the Product Rule would give
df
dx = d( g · y)
dx = g · dy
dx + y · dg
dx = g · dy
dx + f
g · dg
dx ⇒ g · dy
dx = df
dx − f
g · dg
dx
and so dividing both sides by g and getting a common denominator gives
dy
dx = 1
g · df
dx − f
g2 · dg
dx =
g · df
dx − f · dg
dx
g2 ✓",ElementaryCalculus.pdf
"g · dg
dx
and so dividing both sides by g and getting a common denominator gives
dy
dx = 1
g · df
dx − f
g2 · dg
dx =
g · df
dx − f · dg
dx
g2 ✓
A simple mnemonic device for remembering the Quotient Rule i s: write f
g as HI
HO —so that HI
represents the “high” (numerator) part of the quotient and H O represents the “low” (denomi-
nator) part—and think of dHI and dHO as the derivatives of HI and HO, respectively . Then
d
dx
(f
g
)
= HO·dHI − HI·dHO
HO2 , pronounced as “ho-dee-hi minus hi-dee-ho over ho-ho.”",ElementaryCalculus.pdf
"Derivatives of Sums, Products and Quotients • Section 1.4 23
Example 1.5
Use the Quotient Rule to show that d
dx (tan x) = sec2 x.
Solution: Since tan x = sin x
cos x then:
d
dx (tan x) = d
dx
(sin x
cos x
)
=
(cos x) · d
dx (sin x) − (sin x) · d
dx (cos x)
cos2 x = (cos x) · (cos x) − (sin x) · (−sin x)
cos2 x
= cos2 x + sin2 x
cos2 x = 1
cos2 x = sec2 x
Example 1.6
Use the Quotient Rule to show that d
dx (sec x) = sec x tan x.
Solution: Since sec x = 1
cos x then:
d
dx (sec x) = d
dx
( 1
cos x
)
=
(cos x) · d
dx (1) − 1 · d
dx (cos x)
cos2 x = (cos x) · 0 − 1 · (−sin x)
cos2 x
= sin x
cos2 x = 1
cos x · sin x
cos x = sec x tan x
Similar to the above examples, the derivatives of cot x and csc x can be found using the
Quotient Rule (left as exercises). The derivatives of all si x trigonometric functions are:
d
dx (cos x) = −sin x d
dx (sec x) = sec x tan x
d
dx (sin x) = cos x d
dx (csc x) = −csc x cot x
d
dx (tan x) = sec2 x d
dx (cot x) = −csc2 x",ElementaryCalculus.pdf
"d
dx (cos x) = −sin x d
dx (sec x) = sec x tan x
d
dx (sin x) = cos x d
dx (csc x) = −csc x cot x
d
dx (tan x) = sec2 x d
dx (cot x) = −csc2 x
Note that the Sum and Difference Rules can be applied to sums a nd differences, respectively ,
of not just two functions but any finite (integer) number of fu nctions. For example, for three
differentiable functions f1, f2, and f3,
d
dx ( f1 + f2 + f3) = df 1
dx + d
dx ( f2 + f3) by the Sum Rule
= df 1
dx + df 2
dx + df 3
dx by the Sum Rule again.
Continuing like this for four functions, then five functions , and so forth, the Sum and Differ-
ence Rules combined with the Constant Multiple Rule yield th e following formula:",ElementaryCalculus.pdf
"24 Chapter 1 • The Derivative §1.4
For n ≥ 1 differentiable functions f1, . . ., f n and constants c1, . . ., cn:
d
dx (c1 f1 + ··· + cn f n ) = c1
df 1
dx + ··· + cn
df n
dx (1.10)
Note that the above formula includes differences, by using n egative constants. The formula
also shows that differentiation is a linear operation , which makes d
dx a linear operator . The
idea is that d
dx “operates” on differentiable functions by taking their der ivatives with respect to
the variable x. The sum c1 f1 + ··· + cn f n is called a linear combination of functions, and the
derivative of that linear combination can be taken term by te rm, with the constant multiples
taken outside the derivatives.
A special case of the above formula is replacing the function s f1, . . . , f n by nonnegative
powers of x, making the sum a polynomial in x. In previous sections the derivatives of a few
polynomials—such as x and x2—were calculated. For the derivative of a general polynomia l,",ElementaryCalculus.pdf
"polynomials—such as x and x2—were calculated. For the derivative of a general polynomia l,
the following rule is needed:
Power Rule: d
dx
(
xn )
= n x n−1 for any integer n
There are several ways to prove this formula; one such way bei ng a proof by induction ,
which in general means using the following principle: 24
Principle of Mathematical Induction
A statement P (n) about integers n ≥ k is true for all n ≥ k if:
1. P (k) is true.
2. If P (n) is true for some integer n ≥ k then P (n + 1) is true.
The idea behind mathematical induction is simple: if a state ment is true about some initial
integer k (Step 1 above) and if the statement being true for some integer implies it is true for
the next integer (Step 2 above), then the statement being true for k implies it is true for k + 1,
which in turn implies it is true for k +2, which implies it is true for k +3, and so forth, making
it true for all integers n ≥ k.",ElementaryCalculus.pdf
"which in turn implies it is true for k +2, which implies it is true for k +3, and so forth, making
it true for all integers n ≥ k.
Typically the initial integer k will be 0 or 1. To prove the Power Rule for all integers, first
use induction to prove the rule for all nonnegative integers n ≥ 0, using k = 0 for the initial
integer . For the proof by induction, let P (n) be the statement: d
dx (xn ) = n x n−1.
1. Show that P (0) is true.
That means showing that the Power Rule holds for n = 0, i.e. d
dx
(
x0)
= 0 x0−1 = 0. But
x0 = 1 is a constant, so its derivative is 0. ✓
24As Bertrand Russell noted, the name is really a misnomer: it i s actually a definition of the natural numbers
rather than a principle, and induction technically has a dif ferent meaning.",ElementaryCalculus.pdf
"Derivatives of Sums, Products and Quotients • Section 1.4 25
2. Assuming P (n) is true for some n ≥ 0, show that P (n + 1) is true.
Assuming that d
dx (xn ) = n x n−1, show that d
dx
(
xn+1)
= (n + 1) x(n+1)−1 = (n + 1) xn. It was
shown in Section 1.2 that d
dx (x) = 1, so:
d
dx
(
xn+1)
= d
dx
(
x · xn )
= x · d
dx
(
xn )
+ xn · d
dx (x) (by the Product Rule)
= x · n x n−1 + xn · 1 (by the assumption that P (n) is true)
= n x n + xn = (n + 1) xn ✓
Thus, by induction, the Power Rule is true for all nonnegativ e integers n ≥ 0.
To show that the Power Rule is true for all negative integers n < 0, write n = −m, where m
is positive (namely , m = |n |). Then:
d
dx
(
xn )
= d
dx
(
x−m )
= d
dx
( 1
xm
)
=
xm · d
dx (1) − 1 · d
dx (xm )
(xm)2 (by the Quotient Rule)
= xm · 0 − 1 · m x m−1
x2m (by the Power Rule for positive integers)
= −m x m−1−2m = −m x −m−1 = n x n−1 ✓
Thus, the Power Rule is true for all integers, which complete s the proof. QED
Example 1.7",ElementaryCalculus.pdf
"= −m x m−1−2m = −m x −m−1 = n x n−1 ✓
Thus, the Power Rule is true for all integers, which complete s the proof. QED
Example 1.7
Find the derivative of f (x) = x4 − 4x3 + 6x2 − 4x + 1.
Solution: Differentiate the polynomial term by term and use the Power R ule:
df
dx = d
dx
(
x4 − 4x3 + 6x2 − 4x + 1
)
= d
dx
(
x4 )
− 4 · d
dx
(
x3)
+ 6 · d
dx
(
x2)
− 4 · d
dx (x) + d
dx (1)
= 4x4−1 − 4 · 3x3−1 + 6 · 2x2−1 − 4 · 1 + 0
= 4x3 − 12x2 + 12x − 4
In general, the derivative of a polynomial of degree n ≥ 0 is given by:
For any constants a0, . . . , an with n ≥ 0:
d
dx
(
an xn + an−1 xn−1 ··· + a2 x2 + a1 x + a0
)
= na n xn−1 + (n−1) an−1 xn−2 + ··· +2a2 x + a1",ElementaryCalculus.pdf
"26 Chapter 1 • The Derivative §1.4
A way to remember the Power Rule is: bring the exponent down in front of the variable then
reduce the variable’s original exponent by 1. This works eve n for negative exponents.
Example 1.8
Find the derivative of f (t) = 3t100 − 2
t100 .
Solution: Differentiate term by term:
df
dt = d
dt
(
3t100 − 2
t100
)
= d
dt
(
3t100 − 2t−100 )
= 3 · 100t99 − 2 ·
(
−100t−101 )
= 300t99 + 200
t101
Exercises
A
For Exercises 1-14, use the rules from this section to find the derivative of the given function.
1. f (x) = x2 − x − 1 2. f (x) = x8 + 2x4 + 1
3. f (x) = 2x6
3 − 3
2x6 4. f (x) = sin x + cos x
4
5. f (x) = x sin x 6. f (x) = x2 tan x
7. f (x) = sin x
x 8. f (x) = sin x
x2
9. f (t) = 2t
1 + t2 10. g(t) = 1 − t2
1 + t2
11. f (x) = ax + b
cx + d (a, b, c, d are constants) 12. F (r) = −G m1 m2
r2 (G, m1 , m2 are constants)
13. A(r) = πr2 14. V (r) = 4
3 πr3
15. Show that d
dx (cot x) = −csc2 x. 16. Show that d
dx (csc x) = −csc x cot x.
B",ElementaryCalculus.pdf
"r2 (G, m1 , m2 are constants)
13. A(r) = πr2 14. V (r) = 4
3 πr3
15. Show that d
dx (cot x) = −csc2 x. 16. Show that d
dx (csc x) = −csc x cot x.
B
17. Prove the Difference Rule. 18. Prove the Constant Multiple Rule.
19. Use the Product Rule to show that for three differentiable fu nctions f , g, and h, the derivative of
their product is ( f gh )′ = f ′ gh + f g ′ h + f gh ′ .
C
20. Provide an alternative proof of the Product Rule for two diff erentiable functions f and g of x:
(a) Show that ( df )(dg) = 0.
(b) By definition, the derivative of the product f · g is
d
dx ( f · g) = f (x + dx) · g(x + dx) − f (x) · g(x)
dx .
Use that formula along with part (a) to show that d
dx ( f · g) = f · dg
dx + g · df
dx .
(Hint: Recall that df = f (x + dx) − f (x).)",ElementaryCalculus.pdf
"The Chain Rule • Section 1.5 27
1.5 The Chain Rule
From what has been discussed so far it might be tempting to thi nk that the derivative of a
function like sin 2 x is simply cos 2 x, since the derivative of sin x is cos x. It turns out that is
not correct:
d
dx (sin 2 x) = d
dx (2 sin x cos x) (by the double-angle formula for sine)
= 2 d
dx (sin x cos x) (by the Constant Multiple Rule)
= 2
(
sin x · d
dx (cos x) + cos x · d
dx (sin x)
)
(by the Product Rule)
= 2 (sin x · (−sin x) + cos x · cos x)
= 2
(
cos2 x − sin2 x
)
= 2 cos 2 x (by the double-angle formula for cosine)
So the derivative of sin 2 x is 2 cos 2 x, not cos 2 x.
In other words, you cannot simply replace x by 2 x in the derivative formula for sin x. Instead,
regard sin 2 x as a composition of two functions: the sine function and the 2 x function. That is,
let f (u) = sin u, where the variable u itself represents a function of another variable x, namely",ElementaryCalculus.pdf
"let f (u) = sin u, where the variable u itself represents a function of another variable x, namely
u(x) = 2x. So since f is a function of u, and u is a function of x, then f is a function of x, namely:
f (x) = sin 2 x. Since f is a differentiable function of u, and u is a differentiable function of x,
then df
du and du
dx both exist (with df
du = cos u and du
dx = 2), and multiplying the derivatives shows
that f is a differentiable function of x:
df
✚✚du
· ✚✚du
dx = df
dx since the infinitesimals du cancel, so
(cos u) · 2 = df
dx ⇒ df
dx = 2 cos u = 2 cos 2 x
The above argument holds in general, and is known as the Chain Rule :
Chain Rule: If f is a differentiable function of u, and u is a differentiable function of x,
then f is a differentiable function of x, and its derivative with respect to x is:
df
dx = df
du · du
dx
Notice how simple the proof is—the infinitesimals du cancel.25",ElementaryCalculus.pdf
"df
dx = df
du · du
dx
Notice how simple the proof is—the infinitesimals du cancel.25
25Some textbooks give dire warnings to not think that du is an actual quantity that can be canceled. However , you
can safely ignore those warnings, because du is just an infinitesimal and hence can be canceled!",ElementaryCalculus.pdf
"28 Chapter 1 • The Derivative §1.5
The Chain Rule should make sense intuitively . For example, i f df
du = 4 then that means f is
increasing 4 times as fast as u, and if du
dx = 3 then u is increasing 3 times as fast as x, so overall
f should be increasing 12 = 4 · 3 times as fast as x, exactly as the Chain Rule says.
Example 1.9
Find the derivative of f (x) = sin ( x2 + x + 1).
Solution: The idea is to make a substitution u = x2 + x + 1 so that f (x) = sin u. By the Chain Rule,
df
dx = df
du · d u
dx
= d
du (sin u) · d
dx (x2 + x + 1)
= (cos u) · (2x + 1)
= (2x + 1) cos ( x2 + x + 1)
after replacing u by its definition as a function of x in the last step; the final answer for the derivative
should be in terms of x, not u.
In the Chain Rule you can think of the function in question as t he composition of an “outer”
function f and an “inner” function u; first take the derivative of the “outer” function then",ElementaryCalculus.pdf
"function f and an “inner” function u; first take the derivative of the “outer” function then
multiply by the derivative of the “inner” function. Think of the “inner” function as a box into
which you can put any function of x, and the “outer” function being a function of that empty
box.
For instance, for the function f (x) = sin ( x2 + x + 1) in the previous example, think of the
“outer” function as sin ✷, where ✷ = x2 + x + 1 is the “inner” function, so that
f (x) = sin ( x2 + x + 1)
= sin ✷
df
dx = (cos ✷) · d
dx ✷
=
(
cos x2 + x + 1
)
· d
dx x2 + x + 1
= (2x + 1) cos ( x2 + x + 1)
Example 1.10
Find the derivative of f (x) = (2x4 − 3 cos x)10.
Solution: Here the “outer” function is f (✷) = ✷10 and the “inner” function is ✷ = u = 2x4 − 3 cos x:
df
dx = df
du · d u
dx = 10 ✷9 · d
dx (✷) = 10 (2 x4 − 3 cos x)9 (8x3 + 3 sin x)",ElementaryCalculus.pdf
"The Chain Rule • Section 1.5 29
Recall that the composition f ◦ g of two functions f and g is defined as ( f ◦ g)(x) = f ( g(x)).
Using prime notation the Chain Rule can be written as:
Chain Rule: If g is a differentiable function of x, and f is a differentiable function on the
range of g, then f ◦ g is a differentiable function of x, and its derivative with respect to x is:
( f ◦ g)′(x) = f ′( g(x)) · g′(x)
Using the Chain Rule, the Power Rule can be extended to includ e exponents that are rational
numbers:26
d
dx
(
xr )
= r x r−1 for any rational number r
To prove this, let r = m/n, where m and n are integers with n ̸=0. Then y = xr = xm/n = (xm )1/n,
so that yn = xm. Taking the derivative with respect to x of both sides of this equation gives
d
dx
(
yn )
= d
dx
(
xm )
, so evaluating the left side by the Chain Rule gives
n yn−1 · dy
dx = mx m−1
n
(
xm/n
) n−1
· dy
dx = mx m−1
dy
dx = mx m−1
nx m−(m/n) = m
n xm−1−(m−(m/n)) = m
n x(m/n)−1 = r x r−1 ✓
Example 1.11",ElementaryCalculus.pdf
"n yn−1 · dy
dx = mx m−1
n
(
xm/n
) n−1
· dy
dx = mx m−1
dy
dx = mx m−1
nx m−(m/n) = m
n xm−1−(m−(m/n)) = m
n x(m/n)−1 = r x r−1 ✓
Example 1.11
Find the derivative of f (x) = ∝ra⌈〉⌋allowx.
Solution: Since ∝ra⌈〉⌋allowx = x1/2 then by the Power Rule:
df
dx = d
dx
(
x1/2
)
= 1
2 x1/2−1 = 1
2 x−1/2 = 1
2 ∝ra⌈〉⌋allowx
Example 1.12
Find the derivative of f (x) = 2
3∝ra⌈〉⌋allowx .
Solution:
df
dx = d
dx
(2
3 x−1/2
)
= 2
3 · −1
2 x−3/2 = − 1
3 x3/2
26It will be shown in Chapter 2 how to define any real number as an e xponent. The Power Rule extends to that
case as well.",ElementaryCalculus.pdf
"30 Chapter 1 • The Derivative §1.5
Exercises
A
For Exercises 1-18, find the derivative of the given function .
1. f (x) = (1 − 5x)4 2. f (x) = 5 ( x3 + x − 1)4
3. f (x) =
∝ra⌈〉⌋allow
1 − 2x 4. f (x) = (1 − x2 )
3
2
5. f (x) =
∝ra⌈〉⌋allowx
x + 1 6. f (x) =
∝ra⌈〉⌋allowx + 1∝ra⌈〉⌋allowx − 1
7. f (t) =
(1 − t
1 + t
) 4
8. f (x) =
(x2 + 1
x − 1
) 6
9. f (x) = sin2 x 10. f (x) = cos
(∝ra⌈〉⌋allowx
)
11. f (x) = 3 tan (5 x) 12. f (x) = A cos (ωx + φ) ( A, ω, φ are constants)
13. f (x) = sec ( x2) 14. f (x) = sin2
( 1
1 − x
)
+ cos2
( 1
1 − x
)
15. L(β) = 1√
1 − β2
16. f (x) = 1
πs
(
1 +
(x − l
s
) 2 ) −1
(s, l are constants)
17. f (x) = cos (cos x) 18. f (x) =
√
1 + ∝ra⌈〉⌋allowx
B
19. In a certain type of electronic circuit
27 the overall gain Av is given by
Av = A o
1 − T
where the loop gain T is a function of the open-loop gain A o .
(a) Show that
dAv
dA o
= 1
1 − T − A o
(1 − T )2
d(1 − T )
dA o
.
(b) In the case where T is directly proportional to A o , use part (a) to show that
dAv
dA o",ElementaryCalculus.pdf
"dAv
dA o
= 1
1 − T − A o
(1 − T )2
d(1 − T )
dA o
.
(b) In the case where T is directly proportional to A o , use part (a) to show that
dAv
dA o
= 1
(1 − T )2 .
(Hint: First show that A o · d(1−T)
dA o
= −T .)
20. Show that the Chain Rule can be extended to 3 functions: if u is a differentiable function of x, v is
a differentiable function of u, and f is a differentiable function of v, then
df
dx = df
dv · d v
du · d u
dx
so that f is a differentiable function of x. Notice that the 3 derivatives are linked together as in a
chain (hence the name of the rule). The Chain Rule can be exten ded to any finite number of functions
by the above technique.
27This is an example of a current-differencing negative feedback amplifier . See pp.473-479 in S C H I L L I N G, D.L.
A N D C. B E L OV E , Electronic Circuits: Discrete and Integrated , 2nd ed., New Y ork: McGraw-Hill, Inc., 1979.",ElementaryCalculus.pdf
"The Chain Rule • Section 1.5 31
21. In an internal combustion engine, as a piston moves downward the connecting rod rotates the crank
in the clockwise direction, as shown in Figure 1.5.1 below . 28
connecting rod
l
A
B
O
crank
a
s
θ
piston
Figure 1.5.1
The point A can only move vertically , causing the point B to move around a circle of radius a centered
at the point O, which is directly below the point A and does not move. As the crank rotates it makes
an angle θ with the line O A . Let l = AB and s = O A as in the picture. Assume that all lengths are
measured in centimeters, and let the time variable t be measured in minutes.
(a) Show that s = a cos θ +
(
l2 − a2 sin2 θ
) 1/2
for 0 ≤ θ ≤ π.
(b) The mean piston speed is ¯S p = 2LN , where L = 2a is the piston stroke , and N is the rotational
velocity of the crank, measured in revolutions per minute (rpm). The i nstantaneous piston
velocity is S p = ds
dt . Let R = l/a. Show that for 0 ≤ θ ≤ π,
⏐
⏐
⏐
⏐
S p
¯S p
⏐
⏐
⏐
⏐ = π
2 sin θ
[",ElementaryCalculus.pdf
"velocity is S p = ds
dt . Let R = l/a. Show that for 0 ≤ θ ≤ π,
⏐
⏐
⏐
⏐
S p
¯S p
⏐
⏐
⏐
⏐ = π
2 sin θ
[
1 + cos θ
(
R2 − sin2 θ
) 1/2
]
.
28See pp.43-45 in H E Y W O O D, J.B., Internal Combustion Engine Fundamentals , New Y ork: McGraw-Hill Inc., 1988.",ElementaryCalculus.pdf
"32 Chapter 1 • The Derivative §1.6
1.6 Higher Order Derivatives
The derivative f ′(x) of a differentiable function f (x) can be thought of as a function in its
own right, and if it is differentiable then its derivative—d enoted by f ′′(x)—is the second
derivative of f (x) (the first derivative being f ′(x)). Likewise, the derivative of f ′′(x) would
be the third derivative of f (x), written as f ′′′(x). Continuing like this yields the fourth
derivative, fifth derivative, and so on. In general the n-th derivative of f (x) is obtained
by differentiating f (x) a total of n times. Derivatives beyond the first are called higher order
derivatives.
Example 1.13
For f (x) = 3x4 find f ′′(x) and f ′′′(x).
Solution: Since f ′(x) = 12x3 then the second derivative f ′′(x) is the derivative of 12 x3 , namely:
f ′′(x) = 36x2
The third derivative f ′′′(x) is then the derivative of 36 x2, namely:
f ′′′(x) = 72x
Since the prime notation for higher order derivatives can be cumbersome (e.g. writing 50",ElementaryCalculus.pdf
"f ′′′(x) = 72x
Since the prime notation for higher order derivatives can be cumbersome (e.g. writing 50
prime marks for the fiftieth derivative), other notations ha ve been created:
Notation for the second derivative of y === f (x): The following are all equivalent:
f ′′(x) , f (2)(x) , d2 y
dx2 , d2
dx2 ( f (x)) , y′′ , y(2) , ¨y , ¨f (x) , d2 f
dx2 , D2 f (x)
Notation for the n-th derivative of y === f (x): The following are all equivalent:
f (n)(x) , d n y
dxn , d n
dxn ( f (x)) , y(n) , d n f
dxn , D n f (x)
Notice that the parentheses around n in the notation f (n)(x) indicate that n is not an exponent—
it is the number of derivatives to take. The n in the Leibniz notation dn y
dxn indicates the same
thing, and in general makes working with higher order deriva tives easier:
d2 y
dx2 = d
dx
(dy
dx
)
d3 y
dx3 = d
dx
(d2 y
dx2
)
= d2
dx2
(dy
dx
)
.
.
.
d n y
dxn = d
dx
(d n−1 y
dxn−1
)
= d n−1
dxn−1
(dy
dx
)",ElementaryCalculus.pdf
"Higher Order Derivatives • Section 1.6 33
A natural question to ask is: what do higher order derivative s represent? Recall that the
first derivative f ′(x) represents the instantaneous rate of change of a function f (x) at the value
x. So the second derivative f ′′(x) represents the instantaneous rate of change of the functio n
f ′(x) at the value x. In other words, the second derivative is a rate of change of a rate of change.
s = 0
s > 0s < 0The most famous example of this is for motion in a straight lin e:
let s(t) be the position of an object at time t as the object moves along
the line. The motion can take two directions, e.g. forward/b ackward
or up/down. Take one direction to represent positive positi on and the other to represent neg-
ative direction, as in the drawing on the right. The (instant aneous) velocity v(t) of the object
at time t is s′(t), i.e. the first derivative of s(t). The acceleration a(t) of the object at time",ElementaryCalculus.pdf
"at time t is s′(t), i.e. the first derivative of s(t). The acceleration a(t) of the object at time
t is defined as v′(t), the instantaneous rate of change of the velocity . Thus, a(t) = s′′(t), i.e.
acceleration is the second derivative of position. To summa rize:
s(t) = position at time t
v(t) = velocity at time t
= ds
dt = s′(t) = ˙s(t)
a(t) = acceleration at time t
= dv
dt = v′(t) = ˙v(t)
= d
dt
(ds
dt
)
= d2 s
dt2 = s′′(t) = ¨s(t)
Example 1.14
Ignoring wind and air resistance, the position s of a ball thrown straight up with an initial velocity of
34 m/s from a starting point 2 m off the ground is given by s(t) = −4.9t2 + 34t + 2 at time t (measured in
seconds) with s measured in meters. Find the velocity and acceleration of th e ball at any time t ≥ 0.
Solution: The ball moves in a straight vertical line, first straight up t hen straight down until it hits the
ground. Its velocity v(t) is
v(t) = d s
dt = −9.8t + 34 m/s
while its acceleration a(t) is
a(t) = d2 s
dt2 = d
dt
(d s
dt
)",ElementaryCalculus.pdf
"ground. Its velocity v(t) is
v(t) = d s
dt = −9.8t + 34 m/s
while its acceleration a(t) is
a(t) = d2 s
dt2 = d
dt
(d s
dt
)
= d
dt (−9.8t + 34) = −9.8 m/s 2,
which is the acceleration due to the force of gravity on Earth .
Note that time t = 0 is the time at which the ball was thrown, so that v(0) is the initial velocity of the
ball. Indeed, v(0) = −9.8(0) + 34 = 34 m/s, as expected.",ElementaryCalculus.pdf
"34 Chapter 1 • The Derivative §1.6
v > 0
a < 0 v < 0
a < 0
v = 0
t = 3.47
s = 0
s = 2
t = 0
Notice in Example 1.14 that the acceleration of the ball is no t
only constant but also negative. To see why this makes sense,
first consider the case where the ball is moving upward. The
ball has an initial upward velocity of 34 m/s then slows down
to 0 m/s at the instant it reaches its maximum height above the
ground. So the velocity is decreasing, i.e. its rate of chang e—the
acceleration—is negative.
The ball reaches its maximum height above the ground when its
velocity is zero, that is, when v(t) = −9.8t + 34 = 0, i.e. at time t =
34/9.8 = 3.47 seconds after being thrown (see the picture above). The ba ll then starts moving
downward and its velocity is negative (e.g. at time t = 4 s the velocity is v(4) = −9.8(4) + 34 =
−5.2 m/s). Recall that negative velocity indicates downward mo tion, while positive velocity",ElementaryCalculus.pdf
"−5.2 m/s). Recall that negative velocity indicates downward mo tion, while positive velocity
means the motion is upward (away from the Earth’s center). So in the case where the ball
begins moving downward it goes from 0 m/s to a negative veloci ty , with the ball moving faster
towards the ground, which it hits with a velocity of −33.43 m/s (why?). So again the velocity is
decreasing, which again means that the acceleration is nega tive.
Common terminology involving motion might cause some confu sion with the above discus-
sion. For example, even though the ball’s acceleration is ne gative as it falls to the ground, it
is common to say that the ball is accelerating in that situation, not decelerating (as the ball is
doing while moving upward). In general, acceleration is understood to mean that the mag-
nitude (i.e. the absolute value) of the velocity is increasing. Tha t magnitude is called the
speed of the object. Deceleration means the speed is decreasing.",ElementaryCalculus.pdf
"speed of the object. Deceleration means the speed is decreasing.
The first and second derivatives of an object’s position with respect to time represent the
object’s velocity and acceleration, respectively . Do the t hird, fourth, and other higher order
derivatives have any physical meanings? It turns out they do . The third derivative of posi-
tion is called the jerk of the object. It represents the rate of change of accelerati on, and is
often used in fields such as vehicle dynamics (e.g. minimizin g jerk to provide smoother brak-
ing). The fourth, fifth, and sixth derivatives of position ar e called snap, crackle, and pop,
respectively .
29
The zero-th derivative f (0)(x) of a function f (x) is defined to be the function f (x) itself:
f (0)(x) = f (x). There is a way to define fractional derivatives , e.g. the one-half derivative
f (1/2)(x), which will be discussed in Chapter 6.
An immediate consequence of the definition of higher order de rivatives is:
d m+n
dxm+n ( f (x)) = d m
dxm",ElementaryCalculus.pdf
"An immediate consequence of the definition of higher order de rivatives is:
d m+n
dxm+n ( f (x)) = d m
dxm
( d n
dxn ( f (x))
)
for all integers m ≥ 0 and n ≥ 0.
Recall that the factorial n! of an integer n > 0 is the product of the integers from 1 to n:
n! = 1 · 2 · 3 · ··· ·n
29Y es, those really are their names, obviously inspired by a ce rtain breakfast cereal. Snap has found some uses in
flight dynamics, e.g. minimizing snap to optimize flight path s of drones.",ElementaryCalculus.pdf
"Higher Order Derivatives • Section 1.6 35
For example:
1! = 1 3! = 1 · 2 · 3 = 6
2! = 1 · 2 = 2 4! = 1 · 2 · 3 · 4 = 24
By convention 0! is defined to be 1. The following statement ca n be proved using induction:
d n
dxn (xn ) = n! for all integers n ≥ 0
Thus,
d n+1
dxn+1
(
xn )
= d
dx
( d n
dxn
(
xn ) )
= d
dx (n!) = 0
for all integers n ≥ 0, since n! is a constant. Polynomials are linear combinations of nonn egative
powers of a variable (e.g. x), so the above result combined with the Sum Rule and the Const ant
Multiple rule—which also hold for higher-order derivative s—yields this important fact:
The ( n + 1)-st derivative (“ n plus first derivative”) of a polynomial of degree n is 0:
For any polynomial p(x) = a0 + a1 x + a2 x2 + ··· + an xn of degree n, d n+1
dxn+1 ( p(x)) = 0.
This is the basis for the commonly-used statement that “any p olynomial can be differentiated
to 0” by taking a sufficient number of derivatives. For exampl e, differentiating the polynomial",ElementaryCalculus.pdf
"to 0” by taking a sufficient number of derivatives. For exampl e, differentiating the polynomial
p(x) = 100x100 +50x99 101 times would yield 0 (as would differentiating more than 1 01 times).
Exercises
A
For Exercises 1-6 find the second derivative of the given func tion.
1. f (x) = x3 + x2 + x + 1 2. f (x) = x2 sin x 3. f (x) = cos 3x
4. f (x) = sin x
x 5. f (x) = 1
x 6. F (r) = G m1 m2
r2
7. Find the first five derivatives of f (x) = sin x. Use those to find f (100)(x) and f (2014)(x).
8. Find the first five derivatives of f (x) = cos x. Use those to find f (100)(x) and f (2014)(x).
9. If an object moves along a straight line such that its positio n s(t) at time t is directly proportional to
t for all t (written as s ∝ t), then show that the object’s acceleration is always 0.
B
10. Use induction to show that d n
dxn (xn ) = n! for all integers n ≥ 1.
11. Show that for all integers n ≥ m ≥ 1, d m
dxm (xn ) = n!
(n−m)! xn−m .",ElementaryCalculus.pdf
"36 Chapter 1 • The Derivative §1.6
12. Find the general expression for the n-th derivative of f (x) = 1
a x+b for all constants a and b (a ̸=0).
13. Show that the function y = A cos (ωt + φ) + B sin (ωt + φ) satisfies the differential equation
d2 y
dt2 + ω2 y = 0
for all constants A, B, ω, and φ.
14. If s(t) represents the position at time t of an object moving along a straight line, then show that:
s′ and s′′ have the same sign ⇒ the object is accelerating
s′ and s′′ have opposite signs ⇒ the object is decelerating
15. For all twice-differentiable functions f and g, show that ( f · g)′′ = f ′′ · g + 2 f ′ · g′ + f · g′′.
C
16. Recall that taking a derivative is a way of operating on a function. That is, think of d
dx as the
differentiation operator on the collection of differentiable functions, taking a fun ction f (x) to its
derivative function df
dx :
f (x) df
dx
d
dx
Likewise, d2
dx2 is an operator on twice-differentiable functions, taking a function f (x) to its second",ElementaryCalculus.pdf
"derivative function df
dx :
f (x) df
dx
d
dx
Likewise, d2
dx2 is an operator on twice-differentiable functions, taking a function f (x) to its second
derivative function d2f
dx2 :
f (x) d2f
dx2
d2
dx2
In general, an eigenfunction of an operator A is a function φ(x) such that A(φ(x)) = λ· φ(x), that is,
φ(x) λ· φ(x)
A
for all x in the domain of φ, for some constant λ called the eigenvalue of the eigenfunction.
(a) Show for all constants k that φ(x) = cos kx is an eigenfunction of the d2
dx2 operator , and find its
eigenvalue. That is, show that d2
dx2 (φ(x)) = λ· φ(x) for some constant λ (the eigenvalue).
(b) The wave function ψ for a particle of mass m moving in a one-dimensional box of length L, given
by
ψ(x) =
√
2
L sin πx
L for 0 ≤ x ≤ L,
is a solution (assuming zero potential energy) of the time-i ndependent Schrödinger equation
− h2
8π2 m
d2ψ
dx2 = E ψ(x)
where h is Planck’s constant and E is a constant that represents the total energy of the wave",ElementaryCalculus.pdf
"− h2
8π2 m
d2ψ
dx2 = E ψ(x)
where h is Planck’s constant and E is a constant that represents the total energy of the wave
function. Find an expression for the constant E in terms of the other constants. Notice that this
makes ψ(x) an eigenfunction of the d2
dx2 operator .",ElementaryCalculus.pdf
"CHAPTER 2
Derivatives of Common Functions
2.1 Inverse Functions
The derivatives calculated in the previous chapter were mos tly for polynomials and a few
trigonometric functions. This chapter will show how to find t he derivatives of other types of
functions, beginning in this section with inverse function s. The idea here is that if a function
is differentiable and has an inverse then that inverse funct ion is also differentiable.
x
Domain
y
Range
f
y = f (x)
Figure 2.1.1
Recall that a function is a rule that assigns a single object
y from one set (the range) to each object x from another set
(the domain). That rule can be written as y = f (x), where f is
the function (see Figure 2.1.1). There is a simple vertical rule for
determining whether a rule y = f (x) is a function: f is a function
if and only if every vertical line intersects the graph of y = f (x)
in the x y-coordinate plane at most once (see Figure 2.1.2).
y
x
y = f (x)
(a) f is a function
y
x
y = f (x)",ElementaryCalculus.pdf
"in the x y-coordinate plane at most once (see Figure 2.1.2).
y
x
y = f (x)
(a) f is a function
y
x
y = f (x)
(b) f is not a function
Figure 2.1.2 V ertical rule for functions
Recall that a function f is one-to-one (often written as 1 − 1) if it assigns distinct values
of y to distinct values of x. In other words, if x1 ̸=x2 then f (x1) ̸=f (x2). Equivalently , f is
one-to-one if f (x1) = f (x2) implies x1 = x2. There is a simple horizontal rule for determining
whether a function y = f (x) is one-to-one: f is one-to-one if and only if every horizontal line
intersects the graph of y = f (x) in the x y-coordinate plane at most once (see Figure 2.1.3).
37",ElementaryCalculus.pdf
"38 Chapter 2 • Derivatives of Common Functions §2.1
y
x
y = f (x)
(a) f is one-to-one
y
x
y = f (x)
(b) f is not one-to-one
Figure 2.1.3 Horizontal rule for one-to-one functions
If a function f is one-to-one on its domain, then f has an inverse function , denoted by f −1,
such that y = f (x) if and only if f −1( y) = x. The domain of f −1 is the range of f .
The basic idea is that f −1 “undoes” what f does, and vice versa. In other words,
f −1( f (x)) = x for all x in the domain of f , and
f ( f −1( y)) = y for all y in the range of f .
Intuitively it is clear that a function is one-to-one (and he nce invertible) when it is either
strictly increasing or strictly decreasing; if the functio n, say , increases and then decreases (as
in Figure 2.1.3(b)) then the horizontal rule would be violat ed around that “turning point.” For
differentiable functions a positive derivative means the f unction is increasing, while a negative",ElementaryCalculus.pdf
"differentiable functions a positive derivative means the f unction is increasing, while a negative
derivative means the function is decreasing (this will be pr oved in Chapter 3).
However , a function can still be one-to-one even if its deriv ative is zero only at isolated points
(i.e. not identically zero over an entire interval of points ) and either positive everywhere else
or negative everywhere else. For example, the function f (x) = x3 has derivative f ′(x) = 3x2,
which is zero only at the isolated point x = 0 and positive for all other values of x. Clearly
f is one-to-one over the set of all real numbers (why?) and henc e it has an inverse function
x = f −1( y) = 3∝ra⌈〉⌋allowy defined for all real numbers y (i.e. the range of f ).
Thus, having a derivative that is either always positive or a lways negative is sufficient for a
function to be one-to-one but not necessary. Having a nonzero derivative is necessary , though,
for the inverse function to be differentiable.",ElementaryCalculus.pdf
"function to be one-to-one but not necessary. Having a nonzero derivative is necessary , though,
for the inverse function to be differentiable.
In algebra you learned that a
b = 1
b
a
for all real numbers a ̸=0 and b ̸=0 (and hence b
a ̸=0).
The same holds true for the infinitesimals dy and dx (nonzero by definition) since they can
be treated like numbers, which immediately yields a formula for the derivative of an inverse
function:
Derivative of an Inverse Function: If y = f (x) is differentiable and has an inverse
function x = f −1( y), then f −1 is differentiable and its derivative is
dx
dy = 1
dy
dx
if dy
dx ̸=0.
The inverse of a function would still exist at a point where dy
dx = 0 but it would not be differen-
tiable there, since its derivative would be the undefined qua ntity 1
0 .",ElementaryCalculus.pdf
"Inverse Functions • Section 2.1 39
Since y is a function of x, dy
dx will be in terms of x and hence 1
dy
dx
will be in terms of x. However ,
since (by invertibility) x is a function of y, dx
dy would normally be in terms of y, not x, so that
the two sides of the equation dx
dy = 1
dy
dx
are not in the same terms! One way to handle this
discrepancy is to use the formula y = f (x) to solve for x in terms of y then substitute that
expression into dy
dx , so that dx
dy = 1
dy
dx
is now in terms of y. That might not always be possible,
however (e.g. try solving for x in the formula y = x sin x).
Example 2.1
Find the inverse f −1 of the function f (x) = x3 then find the derivative of f −1 .
Solution: The function y = f (x) = x3 is one-to-one over the set of all real numbers (why?) so it has an
inverse function x = f −1( y) defined for all real numbers, namely x = f −1( y) = 3∝ra⌈〉⌋allowy.
The derivative of f −1 is
dx
dy = 1
dy
dx
= 1",ElementaryCalculus.pdf
"inverse function x = f −1( y) defined for all real numbers, namely x = f −1( y) = 3∝ra⌈〉⌋allowy.
The derivative of f −1 is
dx
dy = 1
dy
dx
= 1
3x2 , which is in terms of x, so putting it in terms of y yields
= 1
3
(
3∝ra⌈〉⌋allowy
) 2 = 1
3 y2/3
which agrees with the derivative obtained by differentiati ng x = 3∝ra⌈〉⌋allowy directly . Note that this derivative
is defined for all y except y = 0, which occurs when x =
3∝ra⌈〉⌋allow
0 = 0, i.e. at the point ( x, y) = (0, 0).
Functions are often expressed in terms of x, so it is common to see an inverse function also
expressed in terms of x: writing the inverse of f (x) = x3 as f −1(x) = 3∝ra⌈〉⌋allowx (not as f −1( y) = 3∝ra⌈〉⌋allowy)),
as confusing as that might be. In that case, the idea is to swit ch the roles of x and y in the
original function y = f (x), making it x = f ( y), and then write y = f −1(x) and use
dy
dx = 1
dx
dy
to put the derivative of f −1 in terms of x, following the same procedure mentioned earlier .
Example 2.2",ElementaryCalculus.pdf
"dy
dx = 1
dx
dy
to put the derivative of f −1 in terms of x, following the same procedure mentioned earlier .
Example 2.2
Find the inverse f −1 of the function f (x) = x3 then find the derivative of f −1 .
Solution: Rewrite y = f (x) = x3 as x = f ( y) = y3 , so that its inverse function y = f −1(x) = 3∝ra⌈〉⌋allowx has deriva-
tive
dy
dx = 1
dx
dy
= 1
3 y2 , which is in terms of y, so putting it in terms of x yields
= 1
3
(3∝ra⌈〉⌋allowx
) 2 = 1
3x2/3
which agrees with the derivative obtained by differentiati ng y = 3∝ra⌈〉⌋allowx directly .",ElementaryCalculus.pdf
"40 Chapter 2 • Derivatives of Common Functions §2.1
To obtain a formula in prime notation for the derivative of an inverse function, notice that
for all x in the domain of an invertible differentiable function f ,
f −1( f (x)) = x ⇒ d
dx
(
f −1( f (x))
)
= d
dx (x) ⇒
(
f −1) ′
( f (x)) · f ′(x) = 1
by the Chain Rule, and hence:
(
f −1) ′
( f (x)) = 1
f ′(x) if f ′(x) ̸=0
Two equivalent ways to write this are:
(
f −1) ′
(c) = 1
f ′(a) where c = f (a) and f ′(a) ̸=0
and
(
f −1) ′
(x) = 1
f ′( f −1(x)) if f ′( f −1(x)) ̸=0
Exercises
A
For Exercises 1-8, show that the given function y = f (x) is one-to-one over the given interval, then find
the formulas for the inverse function f −1 and its derivative. Use Example 2.2 as a guide, including
putting f −1 and its derivative in terms of x.
1. f (x) = x, for all x 2. f (x) = 3x, for all x
3. f (x) = x2, for all x ≥ 0 4. f (x) = ∝ra⌈〉⌋allow
x, for all x ≥ 0
5. f (x) = 1
x , for all x > 0 6. f (x) = 1
x , for all x < 0
7. f (x) = 1",ElementaryCalculus.pdf
"3. f (x) = x2, for all x ≥ 0 4. f (x) = ∝ra⌈〉⌋allow
x, for all x ≥ 0
5. f (x) = 1
x , for all x > 0 6. f (x) = 1
x , for all x < 0
7. f (x) = 1
x2 , for all x > 0 8. f (x) = x5 , for all x
9. The unit circle x2 + y2 = 1 does not define y as a single function of x, since y = ±
∝ra⌈〉⌋allow
1 − x2 defines two
separate functions. But the part of the unit circle in the firs t quadrant, i.e. for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1,
does define y = f (x) =
∝ra⌈〉⌋allow
1 − x2 as a single function of x that is one-to-one on the interval [0 , 1]. Find
the formulas for its inverse function f −1 and its derivative.
B
10. Show that if f is differentiable and invertible, and if f −1 is twice-differentiable, then
(
f −1) ′′
(x) = − f ′′( f −1(x))
(
f ′( f −1(x))
) 3 .",ElementaryCalculus.pdf
"T rigonometric Functions and Their Inverses • Section 2.2 41
2.2 T rigonometric Functions and Their Inverses
The graphs of the six trigonometric functions are shown in Fi gure 2.2.1:
x
y
0
1
−1
π
2
π 3π
2
2π
(a) y = sin x
x
y
0
1
−1
π
2
π 3π
2
2π
(b) y = cos x
x
y
0− π
2
π
2
(c) y = tan x
x
y
0
1
−1
π
2
π 3π
2 2π
(d) y = csc x
x
y
0
1
−1
π
2
π 3π
2 2π
(e) y = sec x
x
y
0
π
2
π
(f) y = cot x
Figure 2.2.1 Graphs of the six trigonometric functions
Recall that sin x, cos x, csc x, and sec x have a period of 2 π (i.e. repeat the same values every
2π radians), while tan x and cot x have a period of π.
The derivatives of the six trigonometric functions—given i n Section 1.4—are:
d
dx (sin x) = cos x d
dx (csc x) = −csc x cot x
d
dx (cos x) = −sin x d
dx (sec x) = sec x tan x
d
dx (tan x) = sec2 x d
dx (cot x) = −csc2 x
The six trigonometric functions are not one-to-one over the ir entire domains, but recall from",ElementaryCalculus.pdf
"d
dx (tan x) = sec2 x d
dx (cot x) = −csc2 x
The six trigonometric functions are not one-to-one over the ir entire domains, but recall from
trigonometry that they are one-to-one when restricted to sm aller domains, and hence have
inverse functions, called the inverse trigonometric functions .",ElementaryCalculus.pdf
"42 Chapter 2 • Derivatives of Common Functions §2.2
For example, y = sin x is one-to-one over the interval
[
−π
2 , π
2
]
, as shown in Figure 2.2.2 below:
x
y
0
−1
1
π
2
π− π
2
−π
y = sin x
Figure 2.2.2 y = sin x is one-to-one with x restricted to
[
− π
2 , π
2
]
Similarly , recall that cos x is one-to-one over [0 , π], tan x is one-to-one over ( −π/2, π/2), csc x
is one-to-one over ( −π/2, 0) ∪ (0, π/2), sec x is one-to-one over (0 , π/2) ∪ (π/2, π), and cot x is one-
to-one over (0 , π). Hence, the inverse trigonometric functions sin −1 x, cos −1 x, tan −1 x, csc −1 x,
sec−1 x and cot −1 x are defined, 1 with the following domains and ranges:
function sin−1 x cos−1 x tan−1 x csc−1 x sec−1 x cot−1 x
domain [−1, 1] [−1, 1] (−∞, ∞) |x| ≥1 |x| ≥1 (−∞, ∞)
range [−π
2 , π
2 ] [0, π]
(
−π
2 , π
2
) (
−π
2 , 0
)
∪
(
0, π
2
) (
0, π
2
)
∪
(π
2 , π
)
(0, π)
The graphs of all six inverse trigonometric functions are sh own in Figures 2.2.3 and 2.2.4
below:
x
y
0
− π
2
π
2
1−1
(a) y = sin−1 x
x
y",ElementaryCalculus.pdf
")
(0, π)
The graphs of all six inverse trigonometric functions are sh own in Figures 2.2.3 and 2.2.4
below:
x
y
0
− π
2
π
2
1−1
(a) y = sin−1 x
x
y
0
π
1−1
(b) y = cos−1 x
x
y
0
π
2
− π
2
(c) y = tan−1 x
Figure 2.2.3 Graphs of sin −1 x, cos −1 x, tan −1 x
1The arc notation arcsin x, arccos x, arctan x, arccsc x, arcsec x, arccot x is often used in place of sin −1 x, cos −1 x,
tan−1 x, csc −1 x, sec −1 x, cot −1 x, respectively .",ElementaryCalculus.pdf
"T rigonometric Functions and Their Inverses • Section 2.2 43
x
y
0
π
2
− π
2
1−1
(a) y = csc−1 x
x
y
0
π
π
2
1−1
(b) y = sec−1 x
x
y
0
π
π
2
(c) y = cot−1 x
Figure 2.2.4 Graphs of csc −1 x, sec −1 x, cot −1 x
The derivatives of the six inverse trigonometric functions are:
d
dx (sin−1 x) = 1∝ra⌈〉⌋allow
1 − x2
(for |x| <1) d
dx (csc−1 x) = − 1
|x|
∝ra⌈〉⌋allow
x2 − 1
(for |x| >1)
d
dx (cos−1 x) = − 1∝ra⌈〉⌋allow
1 − x2
(for |x| <1) d
dx (sec−1 x) = 1
|x|
∝ra⌈〉⌋allow
x2 − 1
(for |x| >1)
d
dx (tan−1 x) = 1
1 + x2
d
dx (cot−1 x) = − 1
1 + x2
For the derivative of cos −1 x, recall that y = cos−1 x is an angle between 0 and π radians, defined
for −1 ≤ x ≤ 1. Since cos y = x by the definition of y, then dx
dy = −sin y and
sin2 y = 1 − cos2 y = 1 − x2 ⇒ sin y = ±
√
1 − x2 =
√
1 − x2
since 0 ≤ y ≤ π (which means sin y must be nonnegative). Thus:
d
dx (cos−1 x) = dy
dx = 1
dx
dy
= 1
−sin y = − 1∝ra⌈〉⌋allow
1 − x2
✓",ElementaryCalculus.pdf
"√
1 − x2
since 0 ≤ y ≤ π (which means sin y must be nonnegative). Thus:
d
dx (cos−1 x) = dy
dx = 1
dx
dy
= 1
−sin y = − 1∝ra⌈〉⌋allow
1 − x2
✓
For the derivative of sec −1 x, since y = sec−1 x is defined for |x| ≥1, then 0 ≤ y < π/2 for x ≥ 1
and π/2 < y ≤ π for x ≤ −1. Recall also that sec y and tan y are both positive when 0 < y < π/2
and are both negative when π/2 < y < π. So in both cases the product sec y tan y is nonnegative,
i.e. sec y tan y = |sec y tan y|. Thus, since sec y = x and
1 + tan2 y = sec2 y ⇒ tan2 y = sec2 y − 1 ⇒ tan y = ±
√
sec2 y − 1 = ±
√
x2 − 1
then for |x| >1:
d
dx (sec−1 x) = dy
dx = 1
dx
dy
= 1
sec y tan y = 1
|sec y tan y| = 1
⏐
⏐ x
∝ra⌈〉⌋allow
x2 − 1
⏐
⏐ = 1
|x|
∝ra⌈〉⌋allow
x2 − 1
✓",ElementaryCalculus.pdf
"44 Chapter 2 • Derivatives of Common Functions §2.2
The proofs of the derivative formulas for the remaining inve rse trigonometric functions are
similar , and are left as exercises.
Example 2.3
Find the derivative of the function y = 3 tan ( π− 2x).
Solution: By the Chain Rule with u = π− 2x, the derivative of y = 3 tan ( π− 2x) = 3 tan u is:
dy
dx = dy
du · d u
dx =
(
3 sec 2 u
)
(−2) = −6 sec 2 (π− 2x)
Example 2.4
Find the derivative of the function y = sin−1 (x/4).
Solution: By the Chain Rule with u = x/4, the derivative of y = sin−1 (x/4) = sin−1 u is:
dy
dx = dy
du · d u
dx = 1∝ra⌈〉⌋allow
1 − u2
· 1
4 = 1
4
√
1 − (x2 /16)
= 1∝ra⌈〉⌋allow
16 − x2
Exercises
A
For Exercises 1-16, find the derivative of the given function y = f (x).
1. y = sec2 3x 2. y = csc(x2 + 1) 3. y = cot 3 x 4. y = cos (tan x)
5. y = tan−1(x/3) 6. y = sec−1 (x2 + 1) 7. y = cot−1 3x 8. y = cos−1 (sin x)
9. y = cot−1(1/x) 10. y = tan−1 ∝ra⌈〉⌋allow
x 11. y =
(
sin−1 3x
) 2
12. y = tan−1 1
x + tan−1 x",ElementaryCalculus.pdf
"9. y = cot−1(1/x) 10. y = tan−1 ∝ra⌈〉⌋allow
x 11. y =
(
sin−1 3x
) 2
12. y = tan−1 1
x + tan−1 x
13. y = tan−1 x−1
x+1 14. y = x sin−1 (2x + 1) 15. y = x cot−1 x 16. y = tan−1 1
x + cot−1 x
17. Find the derivative of y = sin−1 x + cos−1 x . Explain why no derivative formulas were needed.
B
For Exercises 18-21 prove the given derivative formula.
18. d
dx (sin−1 x) = 1∝ra⌈〉⌋allow
1 − x2
19. d
dx (tan−1 x) = 1
1 + x2
20. d
dx (cot−1 x) = − 1
1 + x2 21. d
dx (csc−1 x) = − 1
|x |
∝ra⌈〉⌋allow
x2 − 1
22. The Chebyshev polynomials Tn(x) = cos (n cos−1 x) are defined for all |x | ≤1 and n = 0, 1, 2, . . . .
(a) Show that the Chebyshev polynomials Tn (x) satisfy the differential equation
(1 − x2) T ′′
n (x) − x T ′
n (x) + n2 Tn (x) = 0
(b) Find polynomial expressions for T0(x), T1(x) and T2(x).
(c) Show that Tn+1(x) + Tn−1(x) = 2x T n(x) for all n ≥ 1. (Hint: Write θ = cos−1 x so that cos θ = x)",ElementaryCalculus.pdf
"The Exponential and Natural Logarithm Functions • Section 2.3 45
2.3 The Exponential and Natural Logarithm Functions
Functions of the form ax, where the exponent x varies, are called exponential functions . Unless
otherwise noted, assume that a > 0 (0 x is just 0, and ( −1)1/2 is not a real number). Y ou already
know how ax is defined when the exponent x is a rational number (i.e. x = m/n where m and n
are integers, n ̸=0). But what if x were irrational, such as
∝ra⌈〉⌋allow
2? What would 3
∝ra⌈〉⌋allow
2 mean?
The idea is that
∝ra⌈〉⌋allow
2 = 1.414213562 . . . can be approximated by rational numbers 14/10 = 1.4,
141/100 = 1.41, 1414/1000 = 1.414, 14142/10000 = 1.4142, and so on, taking larger and larger
numerators and denominators in the rational approximation s to get more and more decimal
places of
∝ra⌈〉⌋allow
2. Then 3
∝ra⌈〉⌋allow
2 would be the number that 3 raised to those rational approxima tions
approaches:
3
∝ra⌈〉⌋allow
2 = lim
m/n→
∝ra⌈〉⌋allow
2
m/n rational
3m/n",ElementaryCalculus.pdf
"2 would be the number that 3 raised to those rational approxima tions
approaches:
3
∝ra⌈〉⌋allow
2 = lim
m/n→
∝ra⌈〉⌋allow
2
m/n rational
3m/n
Each quantity 3 m/n is defined inside the above limit, and as the rational numbers m/n get
closer to the value of
∝ra⌈〉⌋allow
2 it can be shown the limit of the values of 3 m/n will exist: 2
31.4 = 3
14
10 = 4.65553672174608
31.41 = 3
141
100 = 4.70696500171657
31.414 = 3
1414
1000 = 4.72769503526854
31.4142 = 3
14142
10000 = 4.72873393017119
31.41421 = 3
141421
100000 = 4.72878588090861
31.414213 = 3
1414213
1000000 = 4.72880146624114
.
.
. .
.
.
31.414213562... = 3
∝ra⌈〉⌋allow
2 = 4.72880438783742
Of course you would never do all this by hand—you would simply use a computer or calculator ,
which use much more efficient algorithms for calculating pow ers in general. 3
All the usual rules of exponents that you learned in algebra a pply to ax when defined in
the manner described above, with a > 0 and x varying over all real numbers. Of all the pos-",ElementaryCalculus.pdf
"the manner described above, with a > 0 and x varying over all real numbers. Of all the pos-
sible values for the base a, the one that appears the most in mathematics, the sciences a nd
engineering is the base e, defined as:
e = limx→∞
(
1 + 1
x
) x
(2.1)
The approximate value of is e = 2.71828182845905 . . . (often called the Euler number ).
2For the general case see pp.61-63 in F R A N K L I N, P ., A Treatise on Advanced Calculus , New Y ork: Dover Publica-
tions, Inc., 1964.
3For example, to see how square roots and cube roots are calcul ated see Chapter 2 in F I K E , C.T ., Computer
Evaluation of Mathematical Functions , Englewood Cliffs, New Jersey: Prentice-Hall, Inc., 1968.",ElementaryCalculus.pdf
"46 Chapter 2 • Derivatives of Common Functions §2.3
The limit in this definition means that as x becomes larger—approaching infinity ( ∞)—the
values of
(
1 + 1
x
) x
approach a number , denoted by e. More decimal places for e can be obtained
by making x sufficiently large. 4 For example, when x = 5×106 the value is 2.718281555200129.
For extremely large values of x, that is, when x ≫ 1 (the symbol ≫ means “much larger than”),
e ≈
(
1 + 1
x
) x
⇒ e1/x ≈
((
1 + 1
x
) x) 1/x
= 1 + 1
x ⇒
(
e1/x − 1
)
x ≈
(1
x
)
x = 1 ,
so letting h = 1/x, and noting that h = 1/x → 0 if and only if x → ∞, yields the useful limit: 5
lim
h→0
eh − 1
h = 1 (2.2)
Using the above limit, the derivative of y = ex can be found:
d
dx
(
ex)
= ex
Proof: Using the limit definition of the derivative for f (x) = ex,
d
dx
(
ex)
= lim
h→0
f (x + h) − f (x)
h = lim
h→0
ex+h − ex
h
= lim
h→0
ex (
eh − 1
)
h = ex · lim
h→0
eh − 1
h (since ex does not depend on h)
= ex · 1 = ex
QED",ElementaryCalculus.pdf
"h→0
f (x + h) − f (x)
h = lim
h→0
ex+h − ex
h
= lim
h→0
ex (
eh − 1
)
h = ex · lim
h→0
eh − 1
h (since ex does not depend on h)
= ex · 1 = ex
QED
In general, for a differentiable function u = u(x) as the exponent the Chain Rule yields:
d
dx
(
eu )
= eu · du
dx
Example 2.5
Find the derivative of y = 4e−x2
.
Solution: dy
dx = 4e−x2
· d
dx (−x2 ) = 4e−x2
· (−2x) = −8x e−x2
4It will be shown in Chapter 9 that this limit does in fact exist .
5This is admittedly a “hand waving” argument. In Chapter 3 a mo re exact method will be discussed for proving
limits like this one.",ElementaryCalculus.pdf
"The Exponential and Natural Logarithm Functions • Section 2.3 47
The function ex is often referred to simply as the exponential function , even though there
are obviously many exponential functions. What makes the ba se e so special? Take y = A e k t to
represent the amount of some physical quantity at time t, for some constants A and k. Then
dy
dt = d
dt
(
A e k t
)
= k · A e k t = k y ,
which says that the instantaneous rate of change of the quantity is directly proportional to
the amount present at that instant . It turns out that many physical quantities exhibit that
behavior , some of which will be discussed shortly . Converse ly , in Chapter 5 it will be shown
that any solution to the differential equation dy
dt = k y must be of the form y = A e k t for some
constant A. This is what gives the exponential function its special sig nificance.
Let f (x) = ex be the exponential function. Then f (x) > 0 for all x and f ′(x) = f (x) = ex > 0 for",ElementaryCalculus.pdf
"Let f (x) = ex be the exponential function. Then f (x) > 0 for all x and f ′(x) = f (x) = ex > 0 for
all x, and so f (x) is strictly increasing. The graph is shown in Figure 2.3.1.
x
y
0
1
y = ex
Figure 2.3.1 The exponential function
x
y
0 1
y = ln x
Figure 2.3.2 Natural logarithm function
Thus, the exponential function is one-to-one over the set of all real numbers and hence has
an inverse function, called the natural logarithm function , denoted (as a function of x) as
f −1(x) = ln x. The graph is shown in Figure 2.3.2. Below is a summary of the r elationship
between ex and ln x:
domain of ln x = all x > 0 = range of ex
range of ln x = all x = domain of ex
y = ex if and only if x = ln y (2.3)
eln x = x for all x > 0 (2.4)
ln
(
ex)
= x for all x (2.5)
The reader should be aware that many—if not most—fields outsi de of mathematics use the
notation log x instead of ln x for the natural logarithm function. 6",ElementaryCalculus.pdf
"notation log x instead of ln x for the natural logarithm function. 6
6This text almost used log x as well, prevented only by the desire for compatibility with other mathematics texts.",ElementaryCalculus.pdf
"48 Chapter 2 • Derivatives of Common Functions §2.3
From algebra you should be familiar with the following prope rties of the natural logarithm,
along with their equivalent properties in terms of the expon ential function: 7
ln (ab ) = ln a + ln b e a · eb = ea+b
ln
(a
b
)
= ln a − ln b ea
eb = ea−b
ln ab = b ln a
(
ea) b = eab
ln 1 = 0 e0 = 1
To find the derivative of y = ln x, use x = e y:
dy
dx = 1
dx
dy
= 1
d
dy (e y)
= 1
e y = 1
x
Hence:
d
dx (ln x) = 1
x
In general, for a differentiable function u = u(x), the Chain Rule yields:
d
dx (ln u) = 1
u · du
dx = u′
u
Example 2.6
Find the derivative of y = ln
(
x2 + 3x − 1
)
.
Solution: dy
dx = 1
x2 + 3x − 1 · d
dx (x2 + 3x − 1) = 2x + 3
x2 + 3x − 1
Recall that |x| = −x for x < 0, in which case ln ( −x) is defined and
d
dx (ln |x|) = d
dx (ln (−x)) = 1
−x · (−1) = 1
x .
7Note that when using the formula ln
(
a
b
)
= ln a − ln b in numerical computations—especially on hand-held",ElementaryCalculus.pdf
"dx (ln (−x)) = 1
−x · (−1) = 1
x .
7Note that when using the formula ln
(
a
b
)
= ln a − ln b in numerical computations—especially on hand-held
calculators—it is preferable to use the left side of the equa tion, i.e. ln
(
a
b
)
, since the right side ln a −ln b is vulner-
able to the problem of subtractive cancellation , which can give an incorrect answer of 0 if a and b are nearly equal.
For a discussion of subtractive cancellation see § 1.3 in H E N R I C I , P ., Essentials of Numerical Analysis, with Pocket
Calculator Demonstrations , New Y ork: John Wiley & Sons, Inc., 1982.",ElementaryCalculus.pdf
"The Exponential and Natural Logarithm Functions • Section 2.3 49
Combine that result with the derivative d
dx (ln x) = 1
x for x > 0 to get:
d
dx (ln |x|) = 1
x
Logarithmic Differentiation
For some functions it is easier to differentiate the natural logarithm of the function first and
then solve for the derivative of the original function. This technique is called logarithmic
differentiation, demonstrated in the following two examples.
Example 2.7
Find the derivative of y = xx .
Solution: For this example assume x > 0 (since x is both the base and the exponent). Note that you
cannot use the Power Rule for this function since the exponen t x is a variable, not a fixed number .
Instead, take the natural logarithm of both sides of the equa tion y = xx and then take the derivative of
both sides and solve for y′:
ln y = ln
(
xx)
= x · ln x
d
dx (ln y) = d
dx (x · ln x)
y′
y = 1 · ln x + x · 1
x
y′ = y (ln x + 1) = xx (ln x + 1)
Example 2.8
Find the derivative of y = (2x+1)7(3x3−7x+6)4",ElementaryCalculus.pdf
"d
dx (ln y) = d
dx (x · ln x)
y′
y = 1 · ln x + x · 1
x
y′ = y (ln x + 1) = xx (ln x + 1)
Example 2.8
Find the derivative of y = (2x+1)7(3x3−7x+6)4
(1+sin x)5 .
Solution: Use logarithmic differentiation by taking the natural loga rithm of y and then use properties
of logarithms to simplify the differentiation before solvi ng for y′:
ln y = ln
((2x + 1)7(3x3 − 7x + 6)4
(1 + sin x)5
)
= ln
(
(2x + 1)7(3x3 − 7x + 6)4 )
− ln
(
(1 + sin x)5 )
= ln
(
(2x + 1)7)
+ ln
(
(3x3 − 7x + 6)4 )
− ln
(
(1 + sin x)5 )
d
dx (ln y) = d
dx
(
7 ln (2 x + 1) + 4 ln (3 x3 − 7x + 6) − 5 ln (1 + sin x)
)
y′
y = 7 · 2
2x + 1 + 4 · 9x2 − 7
3x3 − 7x + 6 − 5 · cos x
1 + sin x
y′ = y ·
( 14
2x + 1 + 36x2 − 28
3x3 − 7x + 6 − 5 cos x
1 + sin x
)
= (2x + 1)7 (3x3 − 7x + 6)4
(1 + sin x)5 ·
( 14
2x + 1 + 36x2 − 28
3x3 − 7x + 6 − 5 cos x
1 + sin x
)",ElementaryCalculus.pdf
"50 Chapter 2 • Derivatives of Common Functions §2.3
Radioactive Decay
A classic example of the differential equation dy
dt = k y is the case of exponential decay of a
radioactive substance, often referred to simply as radioactive decay . In this case the general
solution y = A e k t represents the amount of the substance at time t ≥ 0, and the decay constant
k is negative: dy
dt < 0 since the substance is decaying (i.e. the amount of substan ce is decreasing)
while y > 0 , so dy
dt = k y implies that k < 0.
y
t0
A0
1
2 A0
y = A0 ek t (k < 0)
tH
Figure 2.3.3
The constant A is the initial amount of the sub-
stance, i.e. the amount at time t = 0: y(0) = A e 0t =
A e 0 = A. For this reason A is sometimes denoted
by A0. The constant k turns out to be related
to the half-life of the substance, defined as the
time tH required for half the current amount of
substance to decay (see Figure 2.3.3).
Y ou might be tempted to think that the half-life
is not a constant, that it might change depending",ElementaryCalculus.pdf
"substance to decay (see Figure 2.3.3).
Y ou might be tempted to think that the half-life
is not a constant, that it might change depending
on the amount of substance present. For exam-
ple, perhaps it would take longer for 100 g of the
substance to decay to 50 g than it would for 10 g
to decay to 5 g. However , this is not so. To see
why , pick any t ≥ 0 as the current time, so that y(t) = A0 ek t is the current amount of the
substance. By definition, that amount should be halved when t he time tH has passed, that
is, y(t + tH ) = 1
2 y(t). Then tH does turn out to be independent of the initial amount A0 and
depends only on k, since
y(t + tH ) = 1
2 y(t) ⇒ A0 ek(t+tH ) = 1
2 A0 ek t ⇒ ✚✚A0✚✚ek t · ek tH = 1
2✚✚A0✚✚ek t
⇒ ek tH = 1
2 ⇒ k tH = ln
(1
2
)
= −ln 2
and so:
tH = −ln 2
k and k = −ln 2
tH
(2.6)
Example 2.9
Suppose that 5 mg of a radioactive substance decays to 3 g in 6 h ours. Find the half-life of the substance.",ElementaryCalculus.pdf
"k and k = −ln 2
tH
(2.6)
Example 2.9
Suppose that 5 mg of a radioactive substance decays to 3 g in 6 h ours. Find the half-life of the substance.
Solution: Consider A0 = 5 mg as the initial amount, so that y(t) = 5ek t is the amount at time t ≥ 0 hours.
Use the given information that y(6) = 3 mg to find k, the decay constant of the substance:
3 = y(6) = 5ek6 ⇒ 6k = ln
(3
5
)
⇒ k = 1
6 ln 0 .6",ElementaryCalculus.pdf
"The Exponential and Natural Logarithm Functions • Section 2.3 51
Then the half-life tH is:
tH = −ln 2
k = − ln 2
1
6 ln 0 .6
= 8.14 hours
Note in the above example that the given time t = 6 was used for finding the constant k and
then the half-life tH . For the converse problem—given the half-life find the time r equired for a
certain amount to decay—you would do the opposite: use the gi ven tH to find k and then solve
for the required time t from the equation y(t) = A0 ek t.
Example 2.10
Another example of the differential equation dy
dt = k y is exponential growth of cell bacteria, in which
case k > 0 since the number of cells y(t) at time t is increasing.
Example 2.11
V C
R
I (t)
Figure 2.3.4
Another example is for the current I in a simple series electric
circuit with a constant direct current (DC) source of voltag e V , a
capacitor with capacitance C, a resistor with resistance R, and a
switch, as in Figure 2.3.4. If the capacitor is initially unc harged",ElementaryCalculus.pdf
"capacitor with capacitance C, a resistor with resistance R, and a
switch, as in Figure 2.3.4. If the capacitor is initially unc harged
when the switch is open, and if the switch is closed at time t = 0,
then the current I(t) through the circuit at time t ≥ 0 satisfies (by
Kirchoff ’s Second Law) the differential equation
RC dI
dt + I = 0 ⇒ dI
dt = − I
RC
so that I(t) = I0 e−t/RC where I0 is the initial current at t = 0. Ohm’s Law says that V = I0 R, so
I(t) = V
R e−t/RC
is the current at time t ≥ 0, which decreases exponentially .
Example 2.12
In the previous examples the quantities that decayed or grew exponentially did so as functions of
time. There are other possible variables besides time, thou gh. For example, the atmospheric pres-
sure p measured as a function of height h above the surface of the Earth satisfies—assuming constant
temperature—the differential equation
dp
dh = − w0
p0
p",ElementaryCalculus.pdf
"temperature—the differential equation
dp
dh = − w0
p0
p
where p0 is the pressure at height h = 0 (i.e. ground level) and w0 is the weight of a cubic foot of air at
pressure p0 (with air pressure measured in lbs per square foot and height measured in feet). Thus,
p(h) = p0 e− w0
p0 h .
So the atmospheric pressure decreases exponentially as the height above the ground increases.",ElementaryCalculus.pdf
"52 Chapter 2 • Derivatives of Common Functions §2.3
Exercises
A
For Exercises 1-12, find the derivative of the given function .
1. y = e2x 2. y = x ex2
3. y = e−x − ex
4. y = esin x 5. y = 1 + ex
1 − ex 6. y = 1
1 + e−2x
7. y = eex
8. y = e2 ln x 9. y = ln (3 x)
10. y = ln ( x2 + 2x + 1)4 11. y =
(
ln(tan x2)
) 3 12. y = ln ( ex + e2x )
13. Show that d
dx (ln ( kx )) = 1
x for all constants k > 0.
14. Show that d
dx (ln (xn )) = n
x for all integers n ≥ 1.
For Exercises 15-18, use logarithmic differentiation to fin d dy
dx .
15. y = xx2
16. y = xln x 17. y = xsin x 18. y = (x + 2)8 (3x − 1)7
(1 − 5x)4
B
19. Suppose it takes 8 hours for 30% of a radioactive substance to decay . Find the half-life of the
substance.
20. The radioactive isotope radium-223 has a half-life of 11.43 days. How long would it take for 3 kg of
radium-223 to decay to 1 kg?
21. If a certain cell population grows exponentially—i.e. is of the form A0 ek t with k > 0—and if the",ElementaryCalculus.pdf
"radium-223 to decay to 1 kg?
21. If a certain cell population grows exponentially—i.e. is of the form A0 ek t with k > 0—and if the
population doubles in 6 hours, how long would it take for the p opulation to quadruple?
For Exercises 22-25, use induction to prove the given formul a for all n ≥ 0.
22. dn
dxn
(
ek x )
= kn ek x (any constant k ̸=0) 23. dn
dxn (x e x ) = (x + n) ex
24. dn
dxn (x e −x) = (−1)n (x − n) e−x 25. dn+1
dxn+1 (xn ln x) = n!
x
26. Show that f ′(x) = f (x) (1 − f (x)) for the sigmoid neuron function f (x) = 1
1 + e−x . This derivative
relation is used in neural network learning algorithms.
27. If y = C e−κt cos
(∝ra⌈〉⌋allow
n2 − κ2 t + γ
)
then show that
d2 y
dt2 + 2κ dy
dt + n2 y = 0
for all constants C, n, κ, γ, with 0 ≤ κ ≤ n.
C
28. Suppose that e y + ex = e y+x . Show that dy
dx = −e y−x .
29. For an infinitesimal dx show that edx = 1 + dx. (Hint: Use d
dx (ex) = ex .)
30. For an infinitesimal dx show that ln (1 + dx) = dx.",ElementaryCalculus.pdf
"General Exponential and Logarithmic Functions • Section 2.4 53
2.4 General Exponential and Logarithmic Functions
For a general exponential function y = ax, with a > 0, use logarithmic differentiation to find its
derivative:
ln y = ln
(
ax)
= x ln a
d
dx (ln y) = d
dx (x ln a) = ln a
y′
y = ln a ⇒ y′ = y · ln a
Thus, the derivative of y = ax is:
d
dx
(
ax)
= (ln a) ax
In general, for an exponent of the form u = u(x):
d
dx
(
au )
= (ln a) au · du
dx
Example 2.13
Find the derivative of y = 2cos x .
Solution: This is the case where a = 2, so:
dy
dx = (ln 2) 2 cos x · d
dx (cos x) = −(ln 2) (sin x) 2 cos x
Note that any exponential function y = ax can be expressed in terms of the exponential
function ex. Since
ax > 0 ⇒ eln (ax ) = ax ,
and since ln ( ax) = x ln a, then:
ax = ex ln a
Computers and calculators often use the above formula to cal culate ax.
The function y = ax has an inverse for any a > 0, except for a = 1 (in that case y = 1x = 1 is",ElementaryCalculus.pdf
"The function y = ax has an inverse for any a > 0, except for a = 1 (in that case y = 1x = 1 is
just a constant function). To see this, notice that since ax > 0 for all x, and ln a < 0 for 0 < a < 1,
while ln a > 0 for a > 1, then dy
dx = (ln a) ax is always negative if 0 < a < 1 and always positive if
a > 1. Thus, y = ax is a strictly decreasing function if 0 < a < 1, and it is a strictly increasing
function if a > 1. The graphs in each case are shown in Figure 2.4.1.",ElementaryCalculus.pdf
"54 Chapter 2 • Derivatives of Common Functions §2.4
x
y
0
1
a > 1
a < 1
Figure 2.4.1 y = ax
x
y
0 1
a > 1
a < 1
Figure 2.4.2 y = loga x
Hence, for any a > 0 with a ̸=1 the function f (x) = ax is one-to-one, so it has an inverse
function, called the base a logarithm and denoted by f −1(x) = loga x. It is often spoken as “log
base a of x”. The graphs for a < 1 and a > 1 are shown in Figure 2.4.2. Note that the natural
logarithm is just the base a logarithm in the special case with a = e, i.e. ln x = loge x. The base
a logarithm has properties similar to those of the natural log arithm (and the corresponding
properties of ax):
loga (b c) = loga b + loga c a b · ac = ab+c
loga
(b
c
)
= loga b − loga c ab
ac = ab−c
loga bc = c loga b
(
ab
) c
= ab c
loga 1 = 0 a0 = 1
Note that log a x can be put in terms of the natural logarithm, since
x = aloga x ⇒ ln x = ln
(
aloga x
)
= (loga x) · (ln a)
so dividing the last expression by ln a gives:
loga x = ln x
ln a",ElementaryCalculus.pdf
"x = aloga x ⇒ ln x = ln
(
aloga x
)
= (loga x) · (ln a)
so dividing the last expression by ln a gives:
loga x = ln x
ln a
The above formula is useful on calculators that do not have a l oga x key or function. Taking
the derivative of both sides yields:",ElementaryCalculus.pdf
"General Exponential and Logarithmic Functions • Section 2.4 55
d
dx
(
loga x
)
= 1
x ln a
In general, when taking the logarithm of a function u = u(x):
d
dx
(
loga u
)
= 1
u ln a · du
dx = u′
u ln a
Example 2.14
Find the derivative of y = log2(cos 4 x).
Solution: This is the case where a = 2, so:
dy
dx = 1
(cos 4 x) (ln 2) · d
dx (cos 4 x) = − 4 sin 4 x
(ln 2) (cos 4 x)
The number a is the base of both the logarithm function log a x and the exponential function
ax. Base 2 and base 10 are the most commonly used bases other than base e. Base 10 is how
numbers are normally expressed, as combinations of powers o f 10 (e.g. 2014 = 2 ·103 + 0 ·102 +
1 · 101 + 4 · 100). Base 2 is especially useful in computer science, since com puters represent all
numbers in binary format, i.e. as a sequence of zeros and ones, indicating how m any successive
powers of two to take and then sum up. 8 For example, the number 6 is represented in binary",ElementaryCalculus.pdf
"powers of two to take and then sum up. 8 For example, the number 6 is represented in binary
format as 110, since 1 · 22 + 1 · 21 + 0 · 20 = 4 + 2 + 0 = 6.
Exercises
A
For Exercises 1-9, find the derivative of the given function.
1. y = 3x + 3−x
2 2. y = 2ln 3 x 3. y = 22x
4. y = tan−1 πx 5. y = log2 (x2 + 1) 6. y = log10 ex
7. y = sin
(
log2 πx
)
8. y = log2 42x 9. y = 8log2 x
B
10. Show that for all constants k the function y = Aa
k x
ln a satisfies the differential equation dy
dx = k y.
Does this contradict the statement made in Section 2.3 that t he only solution to that differential
equation is of the form y = A ek x ? Explain your answer .
8Binary notation leads to the joke “There are 10 kinds of peopl e in the world: those who understand binary and
those who do not.”",ElementaryCalculus.pdf
"CHAPTER 3
T opics in Differential Calculus
3.1 T angent Lines
Everyone knows that the Earth is not flat, but locally, e.g. in your immediate vicinity , isn’t
the Earth effectively flat? In other words, “flat” is a fairly good approximation of t he Earth’s
surface “near” you, and it simplifies matters enough for you t o do some useful things.
This idea of approximating curved shapes by straight shapes is a frequent theme in calculus.
Recall from Chapter 1 that by the Microstraightness Propert y a differentiable curve y = f (x)
actually is a straight line over an infinitesimal interval, having slope dy
dx . The extension of that
line to all values of x is called the tangent line :
For a curve y = f (x) that is differentiable at x = a, the tangent line to the curve at the
point P = (a, f (a)) is the unique line through P with slope m = f ′(a). P is called the point
of tangency . The equation of the tangent line is thus given by:
y − f (a) = f ′(a) · (x − a) (3.1)
y
x
P
y = f (x)
tangent
line
a",ElementaryCalculus.pdf
"of tangency . The equation of the tangent line is thus given by:
y − f (a) = f ′(a) · (x − a) (3.1)
y
x
P
y = f (x)
tangent
line
a
slope = f ′(a)
Figure 3.1.1 Tangent line
Figure 3.1.1 on the right shows the tangent line to a
curve y = f (x) at a point P . If you were to look at the curve
near P with a microscope, it would look almost identical
to its tangent line through P . Why is this line—of all pos-
sible lines through P —such a good approximation of the
curve near P ? It is because at the point P the tangent line
and the curve both have the same rate of change , namely ,
f ′(a). So the curve’s values and the line’s values change
by roughly the same amount slightly away from P (where
the line and curve have the same value), making their
values nearly equal.
56",ElementaryCalculus.pdf
"T angent Lines • Section 3.1 57
Example 3.1
x
y
1
1
−1
0
(1, 1)
y = x2
y = 2x − 1
Figure 3.1.2
Find the tangent line to the curve y = x2 at x = 1.
Solution: By formula (3.1), the equation of the tangent line is
y − f (a) = f ′(a) · (x − a)
with a = 1 and f (x) = x2. So f (a) = f (1) = 12 = 1. Both the curve
y = x2 and the tangent line pass through the point (1 , f (1)) = (1, 1). The
derivative of f (x) = x2 is f ′(x) = 2x, so f ′(a) = f ′(1) = 2(1) = 2, which
is the slope of the tangent line at (1 , 1). Hence, the equation of the
tangent line is y − 1 = 2(x − 1), or (in slope-intercept form) y = 2x − 1.
The curve and tangent line are shown in Figure 3.1.2. Near the point
(1, 1) the curve and tangent line are close together , but the sepa ration
grows farther from that point, especially in the negative x direction.
In trigonometry you probably learned about tangent lines to circles, where a
tangent line is defined as the unique line that touches the cir cle at only one point,",ElementaryCalculus.pdf
"tangent line is defined as the unique line that touches the cir cle at only one point,
as in the figure on the right. In this case the tangent line is al ways on one side of
the circle, namely , the exterior of the circle; it does not cu t through the interior of
the circle. In fact, that definition is a special case of the ca lculus definition. In general, though,
the tangent line to any other type of curve will not necessari ly be on only one side of the curve,
as it was in Example 3.1.
Example 3.2
x
y
(0, 0)
y = x3
y = 0
Figure 3.1.3
Find the tangent line to the curve y = x3 at x = 0.
Solution: Use formula (3.1) with a = 0 and f (x) = x3. Then f (a) = f (0) =
03 = 0. The derivative of f (x) = x3 is f ′(x) = 3x2, so f ′(a) = f ′(0) = 3(0)2 =
0. Hence, the equation of the tangent line is y − 0 = 0(x − 0), which is
y = 0. In other words, the tangent line is the x-axis itself.
As shown in Figure 3.1.3, the tangent line cuts through the cu rve. In",ElementaryCalculus.pdf
"y = 0. In other words, the tangent line is the x-axis itself.
As shown in Figure 3.1.3, the tangent line cuts through the cu rve. In
general it is possible for a tangent line to intersect the cur ve at more
than one point, depending on the function.
Example 3.3
x
y y = sin x y= x
(0, 0)
Figure 3.1.4
Find the tangent line to the curve y = sin x at x = 0.
Solution: Use formula (3.1) with a = 0 and f (x) = sin x. Then f (a) =
f (0) = sin 0 = 0. The derivative of f (x) = sin x is f ′(x) = cos x, so
f ′(a) = f ′(0) = cos 0 = 1. Hence, the equation of the tangent line is
y − 0 = 1(x − 0), which is y = x, as in Figure 3.1.4. Near x = 0, the
tangent line y = x is close to the line y = sin x, which was shown in
Section 1.3 (namely , sin dx = dx, so that sin x ≈ x for x ≪ 1).",ElementaryCalculus.pdf
"58 Chapter 3 • T opics in Differential Calculus §3.1
There are several important things to note about tangent lin es:
• The slope of a curve’s tangent line is the slope of the curve .
Since the slope of a tangent line equals the derivative of the curve at the point of tangency ,
the slope of a curve at a particular point can be defined as the slope of its tangent line
at that point. So curves can have varying slopes, depending o n the point, unlike straight
lines, which have a constant slope. An easy way to remember al l this is to think “slope =
derivative.”
• The tangent line to a straight line is the straight line itsel f.
This follows easily from the definition of a tangent line, but is also easy to see with the “slope
= derivative” idea: a straight line’s slope (i.e. derivativ e) never changes, so its tangent line—
having the same slope—will be parallel and hence must coinci de with the straight line (since",ElementaryCalculus.pdf
"having the same slope—will be parallel and hence must coinci de with the straight line (since
they have the points of tangency in common). For example, the tangent line to the straight
line y = −3x + 2 is y = −3x + 2 at every point on the straight line.
• The tangent line can be thought of as a limit of secant lines .
A secant line to a curve is a line that passes through two points on the curve . Figure 3.1.5
shows a secant line LPQ passing through the points P = (x0, f (x0)) and Q = (w, f (w)) on the
curve y = f (x),
y
x
y = f (x) LPQ
P
Q
TP
x0
f (x0)
w
f (w)
Figure 3.1.5 Secant line LPQ approaching the tangent line TP as Q → P
As the point Q moves along the curve toward P , the line LPQ approaches the tangent line
TP at the point P , provided the curve is smooth at P (i.e. f ′(x0) exists). This is because the
slope of LPQ is ( f (w) − f (x0))/(w − x0), and so
lim
Q→P
(
slope of LPQ
)
= limw→x0
f (w) − f (x0)
w − x0
= f ′(x0) = slope of TP",ElementaryCalculus.pdf
"slope of LPQ is ( f (w) − f (x0))/(w − x0), and so
lim
Q→P
(
slope of LPQ
)
= limw→x0
f (w) − f (x0)
w − x0
= f ′(x0) = slope of TP
which means that as Q approaches P the “limit” of the secant line LPQ has the same slope
and goes through the same point P as the tangent line TP .",ElementaryCalculus.pdf
"T angent Lines • Section 3.1 59
• Smooth curves have tangent lines, nonsmooth curves do not.
For example, think of the absolute value function f (x) = |x|. Its graph has a sharp edge
at the point (0 , 0), making it nonsmooth there, as shown in Figure 3.1.6(a) be low . There
is no real way to define a tangent line at (0 , 0), because as mentioned in Section 1.2, the
derivative of f (x) does not exist at x = 0. The same holds true for curves with cusps, as in
Figure 3.1.6(b).
x
y
0
y = |x |
(a) f (x) = |x |
y = f (x)
(b) Curve with a cusp
Figure 3.1.6 Nonsmooth curves: no tangent line at the nonsmooth points
Many lines go through the point of nonsmoothness, some of whi ch are indicated by the
dashed lines in the above figures, but none of them can be the ta ngent line. Sharp edges
and cusps have to be “smoothed out” to have a tangent line.
As a point moves along a smooth curve, the corresponding tang ent lines to the curve make",ElementaryCalculus.pdf
"and cusps have to be “smoothed out” to have a tangent line.
As a point moves along a smooth curve, the corresponding tang ent lines to the curve make
varying angles with the positive x-axis—the angle is thus a function of x. Let φ = φ(x) be the
smallest angle that the tangent line L to a curve y = f (x) makes with the positive x-axis, so
that −90◦ < φ(x) < 90◦ for all x (see Figure 3.1.7).
x
x1 x2 x3
y y = f (x)
Lφ(x2) = 0◦
L
φ(x3)
L
φ(x1)
Figure 3.1.7 The angle φ(x) between the tangent line and positive x-axis
run
rise
L
φ(x)
As Figure 3.1.7 shows, −90◦ < φ(x) < 0◦ when the tangent line L has
negative slope, 0 ◦ < φ(x) < 90◦ when L has positive slope, and φ(x) = 0◦
when L is horizontal (i.e. has zero slope). The slope of a line is usu ally
defined as the rise divided by the run in a right triangle, as sh own in the
figure on the right. The figure shows as well that by definition o f the tangent of an angle,",ElementaryCalculus.pdf
"figure on the right. The figure shows as well that by definition o f the tangent of an angle,
tan φ(x) also equals the rise (opposite) over run (adjacent). Thus, since the slope of L is f ′(x),
this means that tan φ(x) = f ′(x). In other words:",ElementaryCalculus.pdf
"60 Chapter 3 • T opics in Differential Calculus §3.1
The tangent line to a curve y = f (x) makes an angle φ(x) with the positive x-axis, given by
φ(x) = tan−1 f ′(x) . (3.2)
Example 3.4
y
x
y = e2x
(
− 1
2 , 1
e
) L
φ
Find the angle φ that the tangent line to the curve y = e2x at x = −1
2 makes
with the positive x-axis, such that −90◦ < φ < 90◦.
Solution: The angle is φ = φ(−1/2) = tan−1 f ‘(−1/2), where f (x) = e2x. Since
f ′(x) = 2e2x, then
φ = φ(−1/2) = tan−1 f ‘(−1/2) = tan−1 2e−1 = tan−1 0.7358 = 36.3◦ .
The figure on the right shows the tangent line L to the curve at x = −1
2 and the angle φ.
y
x
P
y = f (x)
L
N
normal
line
a
tangent
line
Figure 3.1.8 Normal line N
Y ou learned about perpendicular lines in elementary geome-
try . Figure 3.1.8 shows the natural way to define how a line
N can be perpendicular to a curve y = f (x) at a point P on
the curve: the line is perpendicular to the tangent line of th e
curve at P . Call this line N the normal line to the curve",ElementaryCalculus.pdf
"the curve: the line is perpendicular to the tangent line of th e
curve at P . Call this line N the normal line to the curve
at P . Since N and L are perpendicular , their slopes are neg-
ative reciprocals of each other (provided neither slope is 0 ).
The equation of the normal line follows easily:
The equation of the normal line to a curve y = f (x) at a point P = (a, f (a)) is
y − f (a) = − 1
f ′(a) · (x − a) if f ′(a) ̸=0. (3.3)
If f ′(a) = 0, then the normal line is vertical and is given by x = a.
Example 3.5
Find the normal line to the curve y = x2 at x = 1. (Note: This is the curve from Example 3.1.)
Solution: The equation of the normal line is
y − f (a) = − 1
f ′(a) · (x − a)
with a = 1, f (x) = x2 , and f ′(x) = 2x. So f (a) = 1 and f ′(a) = 2. Hence, the equation of the normal line is
y − 1 = −1
2 (x − 1), or (in slope-intercept form) y = −1
2 x + 3
2 .",ElementaryCalculus.pdf
"T angent Lines • Section 3.1 61
Exercises
A
For Exercises 1-12, find the equation of the tangent line to th e curve y = f (x) at x = a.
1. f (x) = x2 + 1; at x = 2 2. f (x) = x2 − 1; at x = 2 3. f (x) = −x2 + 1; at x = 3
4. f (x) = 1; at x = −1 5. f (x) = 4x; at x = 1 6. f (x) = ex; at x = 0
7. f (x) = x2 − 3x + 7; at x = 2 8. f (x) = x + 1
x − 1 ; at x = 0 9. f (x) = (x3 + 2x − 1)3; at x = −1
10. f (x) = tan x; at x = 0 11. f (x) = sin 2 x; at x = 0 12. f (x) =
∝ra⌈〉⌋allow
1 − x2; at x = 1/
∝ra⌈〉⌋allow
2
13. Find the equations of the tangent lines to the curve y = x3 − 2x2 + 4x + 1 which are parallel to the
line y = 3x − 5.
14. Draw an example of a curve having a tangent line that intersec ts the curve at more than one point.
For Exercises 15-17, find the angle φ that the tangent line to the curve y = f (x) at x = a makes with the
positive x-axis, such that −90◦ < φ < 90◦.
15. f (x) = x2; at x = 2 16. f (x) = cos 2 x; at x = π/6 17. f (x) = x2 + 2x − 3; at x = −1",ElementaryCalculus.pdf
"positive x-axis, such that −90◦ < φ < 90◦.
15. f (x) = x2; at x = 2 16. f (x) = cos 2 x; at x = π/6 17. f (x) = x2 + 2x − 3; at x = −1
18. Show that if φ(x) is the angle that the tangent line to a curve y = f (x) makes with the positive
x-axis such that 0 ◦ ≤ φ(x) < 180◦, then
φ(x) =













cos−1
(
1
√
1 + ( f ′(x))2
)
when f ′(x) ≥ 0
cos−1
(
−1√
1 + ( f ′(x))2
)
when f ′(x) < 0.
(Hint: Draw a right triangle.)
For Exercises 19-21, find the angle φ that the tangent line to the curve y = f (x) at x = a makes with the
positive x-axis, such that 0 ◦ ≤ φ < 180◦.
19. f (x) = x2; at x = −1 20. f (x) = e−x ; at x = 1 21. f (x) = ln 2 x; at x = 10
For Exercises 22-24, find the equation of the normal line to th e curve y = f (x) at x = a.
22. f (x) = ∝ra⌈〉⌋allowx; at x = 4 23. f (x) = x2 + 1; at x = 2 24. f (x) = x2 − 7x + 4; at x = 3
25. Find the equations of the normal lines to the curve y = x3 − 2x2 − 11x + 3 which have a slope of − 1
4 .
B",ElementaryCalculus.pdf
"25. Find the equations of the normal lines to the curve y = x3 − 2x2 − 11x + 3 which have a slope of − 1
4 .
B
26. Show that the area of the triangle formed by the x-axis, the y-axis, and the tangent line to the
curve y = 1/x at any point P is constant (i.e. the area is the same for all P ).
27. For a constant a > 0, let P be a point on the curve y = ax2 , and let Q be the point where the tangent
line to the curve at P intersects the y-axis. Show that the x-axis bisects the line segment
PQ .
28. Let P be a point on the curve y = 1/x in the first quadrant, and let Q be the point where the tangent
line to the curve at P intersects the x-axis. Show that the triangle ∆P OQ is isosceles, where O is
the origin.",ElementaryCalculus.pdf
"62 Chapter 3 • T opics in Differential Calculus §3.2
3.2 Limits: Formal Definition
So far only the intuitive notion of a limit has been used, name ly:
A real number L is the limit of f (x) as x approaches a if the values of f (x) can be made
arbitrarily close to L by picking values of x sufficiently close to a.
That notion can be put in terms of a formal definition as follow s:
Let L and a be real numbers. Then L is the limit of a function f (x) as x approaches a,
written as
limx→a f (x) = L ,
if for any given number ǫ > 0, there exists a number δ > 0, such that
| f (x) − L | <ǫ whenever 0 < |x − a | <δ .
A visual way of thinking of this definition is shown in Figure 3 .2.1 below:
y
x
y = f (x)
a − δ
(
a a + δ
)
L + ǫ
L
L − ǫ
0 < |x − a| <δ
| f (x) − L | <ǫ
Figure 3.2.1 limx→a f (x) = L
Figure 3.2.1 says that for any interval around L on the y-axis, you will be able to find at least
one small interval around x = a (but excluding a) on the x-axis that the function y = f (x) maps",ElementaryCalculus.pdf
"one small interval around x = a (but excluding a) on the x-axis that the function y = f (x) maps
completely inside that interval on the y-axis. Choosing smaller intervals around L on the
y-axis could force you to find smaller intervals around a on the x-axis.
In Figure 3.2.1, f (x) is made arbitrarily close to L (within any distance ǫ > 0) by picking x
sufficiently close to a (within some distance δ > 0). Since 0 < |x − a | <δ means that x = a itself
is excluded, the solid dot at ( a, L) could even be a hollow dot. That is, f (a) does not have to
equal L, or even be defined; f (x) just needs to approach L as x approaches a. Thus–perhaps
counter-intuitively—the existence of the limit does not ac tually depend on what happens at
x = a itself.",ElementaryCalculus.pdf
"Limits: Formal Definition • Section 3.2 63
Example 3.6
Show that limx→a x = a for any real number a.
Solution: Though the limit is obvious, the following “epsilon-delta” proof shows how to use the formal
definition. The idea is to let ǫ > 0 be given, then “work backward” from the inequality | f (x) − L | <ǫ to
get an inequality of the form |x − a| <δ, where δ > 0 usually depends on ǫ. In this case the limit is L = a
and the function is f (x) = x, so since
| f (x) − a| <ǫ ⇔ | x − a| <ǫ ,
then choosing δ = ǫ means that
0 < |x − a| <δ ⇒ | x − a| <ǫ ⇒ | f (x) − a| <ǫ ,
which by definition means that limx→a x = a.
Calculating limits in this way might seem silly since—as in E xample 3.6—it requires extra
effort for a result that is obvious. The formal definition is u sed most often in proofs of general
results and theorems. For example, the rules for limits—lis ted in Section 1.2—can be proved
by using the formal definition.
Example 3.7",ElementaryCalculus.pdf
"results and theorems. For example, the rules for limits—lis ted in Section 1.2—can be proved
by using the formal definition.
Example 3.7
Suppose that limx→a f (x) and limx→a g(x) both exist. Show that
limx→a ( f (x) + g(x)) =
(
limx→a f (x)
)
+
(
limx→a g(x)
)
Solution: Let limx→a f (x) = L1 and limx→a g(x) = L2 . The goal is to show that limx→a ( f (x) + g(x)) = L1 + L2 . So let
ǫ > 0. Then ǫ/2 > 0, and so by definition there exist numbers δ1 > 0 and δ2 > 0 such that
0 < |x − a| <δ1 ⇒ | f (x) − L1 | <ǫ/2 , and
0 < |x − a| <δ2 ⇒ | g(x) − L2 | <ǫ/2 .
Now let δ = min(δ1 , δ2 ). Then δ > 0 and
0 < |x − a| <δ ⇒ 0 < |x − a| <δ1 and 0 < |x − a| <δ2
⇒ | f (x) − L1 | <ǫ/2 and | g(x) − L2 | <ǫ/2
Since | A + B | ≤ |A | + |B | for all real numbers A and B, then
| f (x) + g(x) − (L1 + L2 )| = |( f (x) − L1 ) + ( g(x) − L2 )| ≤ |f (x) − L1 | + |g(x) − L2 |
and thus
0 < |x − a| <δ ⇒ | f (x) + g(x) − (L1 + L2 )| < ǫ/2 + ǫ/2 = ǫ
which by definition means that limx→a ( f (x) + g(x)) = L1 + L2 . ✓",ElementaryCalculus.pdf
"64 Chapter 3 • T opics in Differential Calculus §3.2
The proofs of the other limit rules are similar . 1 In general, using the formal definition will
not be necessary for evaluating limits of specific functions —in many cases a simple analysis of
the function is all that is needed, often from its graph.
Example 3.8
x
y
1
1
2
0
y = f (x)
Figure 3.2.2
Evaluate lim
x→1
f (x) for the following function:
f (x) =





x if x > 1
2 if x = 1
1 if x < 1
Solution: From the graph of f (x) in Figure 3.2.2, it is clear that as x
approaches 1 from the right (i.e. for x > 1) f (x) approaches 1 along the
line y = x, whereas as x approaches 1 from the left (i.e. for x < 1) f (x)
approaches 1 along the horizontal line y = 1. Thus, lim
x→1
f (x) = 1 .
Note that the limit did not depend on the value of f (x) at x = 1.
As Example 3.8 shows, what matters for a limit is what happens to the value of f (x) as x gets",ElementaryCalculus.pdf
"As Example 3.8 shows, what matters for a limit is what happens to the value of f (x) as x gets
near a, not at x = a itself. Figure 3.2.3(a) below shows how as x approaches a, f (x) approaches
a number different from f (a). Figure 3.2.3(b) shows that x = a does not even need to be in the
domain of f (x), i.e. f (a) does not have to be defined. So it will not always be the case th at
limx→a f (x) = f (a).
y
x
y = f (x)
a
L
f (a)
(a, f (a))
(a) limx→a f (x) = L ̸=f (a)
y
x
y = f (x)
a
L
(b) limx→a f (x) = L, f (x) not defined at x = a
Figure 3.2.3 Excluding x = a from limx→a f (x)
In Example 3.8 the direction in which x approached the number 1 did not affect the limit.
But what if f (x) had approached different values depending on how x approached 1? In that
case the limit would not exist. The following definitions and notation for one-sided limits
will make situations like that simpler to state.",ElementaryCalculus.pdf
"case the limit would not exist. The following definitions and notation for one-sided limits
will make situations like that simpler to state.
1For example, see Section 2.2 in P R O T T E R , M.H. A N D C.B. M O R R E Y, A First Course in Real Analysis , New Y ork:
Springer-V erlag, 1977.",ElementaryCalculus.pdf
"Limits: Formal Definition • Section 3.2 65
Call L the right limit of a function f (x) as x approaches a, written as
limx→a+ f (x) = L ,
if f (x) approaches L as x approaches a for values of x larger than a.
Call L the left limit of a function f (x) as x approaches a, written as
limx→a− f (x) = L ,
if f (x) approaches L as x approaches a for values of x smaller than a.
The following statement follows immediately from the above definitions:
The limit of a function exists if and only if both its right lim it and left limit exist and are
equal:
limx→a f (x) = L ⇔ limx→a− f (x) = L = limx→a+ f (x)
Example 3.9
x
y
1−1
1
2
0
y = f (x)
Figure 3.2.4
Evaluate lim
x→0−
f (x), lim
x→0+
f (x), and lim
x→0
f (x) for the following function:
f (x) =
{
x2 if x < 0
2 − x if x ≥ 0
Solution: From the graph of f (x) in Figure 3.2.4, it is clear that as x
approaches 0 from the left (i.e. for x < 0) f (x) approaches 0 along the
parabola y = x2 , whereas as x approaches 0 from the right (i.e. for",ElementaryCalculus.pdf
"approaches 0 from the left (i.e. for x < 0) f (x) approaches 0 along the
parabola y = x2 , whereas as x approaches 0 from the right (i.e. for
x > 0) f (x) approaches 2 along the line y = 2 − x. Hence, lim
x→0−
f (x) = 0
and lim
x→0+
f (x) = 2 . Thus, lim
x→0
f (x) does not exist since the left and
right limits do not agree at x = 0.
Example 3.10
x
y
0
1
−1
y = sin(1/x)
Figure 3.2.5
Evaluate lim
x→0+
sin
(1
x
)
.
Solution: For x > 0 the function f (x) = sin (1/x) is defined, and its graph
is shown in Figure 3.2.5. As x approaches 0 from the right, sin (1/ x) will
be 1 for the numbers x = 2/π, 2/5 π, 2/9 π, 2/13 π, . . . (which approach 0),
and sin (1/ x) will be −1 for the numbers x = 2/3π, 2/7 π, 2/11 π, 2/15 π, . . .
(which also approach 0). So as x approaches 0 from the right, sin (1/ x)
will oscillate between 1 and −1. Thus, lim
x→0+
sin
(1
x
)
does not exist .",ElementaryCalculus.pdf
"66 Chapter 3 • T opics in Differential Calculus §3.2
So far only finite limits have been considered, that is, L = limx→a f (x) where L is a real (i.e.
finite) number . Define an infinite limit , with L = ∞or −∞, as follows:
For a real number a, the limit of a function f (x) equals infinity as x approaches a, written
as
limx→a f (x) = ∞ ,
if f (x) grows without bound as x approaches a, i.e. f (x) can be made larger than any
positive number by picking x sufficiently close to a:
For any given number M > 0, there exists a number δ > 0, such that
f (x) > M whenever 0 < |x − a | <δ .
For a real number a, the limit of a function f (x) equals negative infinity as x approaches
a, written as
limx→a f (x) = −∞ ,
if f (x) grows negatively without bound as x approaches a, i.e. f (x) can be made smaller
than any negative number by picking x sufficiently close to a:
For any given number M < 0, there exists a number δ > 0, such that
f (x) < M whenever 0 < |x − a | <δ .",ElementaryCalculus.pdf
"For any given number M < 0, there exists a number δ > 0, such that
f (x) < M whenever 0 < |x − a | <δ .
The above definitions can be modified accordingly for one-sid ed limits. If limx→a f (x) = ∞ or
limx→a f (x) = −∞, then the line x = a is a vertical asymptote of f (x), and f (x) approaches the
line x = a asymptotically. The formal definitions are rarely needed.
Example 3.11
x
y
0
y = 1
x
Figure 3.2.6
Evaluate lim
x→0
1
x .
Solution: For x ̸=0 the function f (x) = 1
x is defined, and its graph
is shown in Figure 3.2.6. As x approaches 0 from the right, 1/ x
approaches ∞, that is,
lim
x→0+
1
x = ∞ .
As x approaches 0 from the left, 1/ x approaches −∞, that is,
lim
x→0−
1
x = −∞ .
Since the right limit and the left limit are not equal, then li m
x→0
1
x does not exist .
Note that the y-axis (i.e. the line x = 0) is a vertical asymptote for f (x) = 1
x .",ElementaryCalculus.pdf
"Limits: Formal Definition • Section 3.2 67
Example 3.12
x
y
0
y = 1
x2
Figure 3.2.7
Evaluate lim
x→0
1
x2 .
Solution: For x ̸=0 the function f (x) = 1
x2 is defined, and its graph
is shown in Figure 3.2.7. As x approaches 0 from either the right
or the left, 1/ x2 approaches ∞, that is,
lim
x→0+
1
x2 = ∞ = lim
x→0−
1
x2 .
Since the right limit and the left limit both equal ∞, then lim
x→0
1
x2 = ∞ .
Note that the y-axis (i.e. the line x = 0) is a vertical asymptote for f (x) = 1
x2 .
In the limit limx→a f (x) so far only real values of a have been considered. However , a could be
either ∞ or −∞:
For a real number L, the limit of a function f (x) equals L as x approaches ∞, written as
limx→∞ f (x) = L ,
if f (x) can be made arbitrarily close to L for x sufficiently large and positive:
For any given number ǫ > 0, there exists a number N > 0, such that
| f (x) − L | < ǫ whenever x > N .
For a real number L, the limit of a function f (x) equals L as x approaches −∞, written as",ElementaryCalculus.pdf
"| f (x) − L | < ǫ whenever x > N .
For a real number L, the limit of a function f (x) equals L as x approaches −∞, written as
limx→−∞ f (x) = L ,
if f (x) can be made arbitrarily close to L for x sufficiently small and negative:
For any given number ǫ > 0, there exists a number N < 0, such that
| f (x) − L | < ǫ whenever x < N .
The above definitions can be modified accordingly for L replaced by either ∞ or −∞.
One way to interpret the statement lim x→∞
f (x) = L is: the long-term behavior of f (x) is to
approach a steady-state at L. If limx→∞
f (x) = L or limx→−∞
f (x) = L, then the line y = L is a
horizontal asymptote of f (x), and f (x) approaches the line y = L asymptotically. Again, for
most limits of specific functions, only the intuitive notion s are needed.",ElementaryCalculus.pdf
"68 Chapter 3 • T opics in Differential Calculus §3.2
Example 3.13
From Figures 3.2.6 and 3.2.7, it is clear that
limx→∞
1
x = 0 = limx→−∞
1
x and lim x→∞
1
x2 = 0 = limx→−∞
1
x2
Note that the x-axis (i.e. the line y = 0) is a horizontal asymptote for f (x) = 1
x and f (x) = 1
x2 .
Some limits are obvious, and you can use them to calculate oth er limits:
limx→∞ xn =
{
∞ for any real n > 0
0 for any real n < 0
limx→−∞ xn =
{
∞ for n = 2, 4, 6, 8, . . .
−∞ for n = 1, 3, 5, 7, . . .
limx→∞ ex = ∞ limx→−∞ ex = 0 lim x→∞ e−x = 0 lim x→−∞ e−x = ∞
limx→∞
ln x = ∞ lim
x→0+
ln x = −∞
A related notion is that of Big O notation (that is the capital letter O, not a zero):
Say that
f (x) = O( g(x)) as x → ∞,
spoken as “ f is big O of g”, if there exist positive numbers M and x0 such that
| f (x)| ≤ M | g(x)| for all x ≥ x0.
For example, obviously 2 x3 = O(x3), by picking M = 2, with x0 any positive number . In general,",ElementaryCalculus.pdf
"| f (x)| ≤ M | g(x)| for all x ≥ x0.
For example, obviously 2 x3 = O(x3), by picking M = 2, with x0 any positive number . In general,
f (x) = O( g(x)) means that f exhibits the same long-term behavior as g, up to a constant multi-
ple. Y ou can think of g as the more basic “type” of function that describes f , as far as long-term
behavior .
Example 3.14
Show that 5 x4 − 2 = O(x4).
Solution: First, recall from algebra that |a + b| ≤ |a| + |b| for all real numbers a and b. Thus,
⏐
⏐5x4 − 2
⏐
⏐ ≤
⏐
⏐5x4 ⏐
⏐ + |−2| = 5
⏐
⏐ x4 ⏐
⏐ + 2
for all x. So since
⏐
⏐ x4 ⏐
⏐ = x4 ≥ 1 for all x ≥ 1, then
⏐
⏐5x4 − 2
⏐
⏐ ≤ 5
⏐
⏐ x4 ⏐
⏐ + 2 ≤ 5
⏐
⏐ x4 ⏐
⏐ + 2
⏐
⏐ x4 ⏐
⏐ ⇒
⏐
⏐5x4 − 2
⏐
⏐ ≤ 7
⏐
⏐ x4 ⏐
⏐ for all x ≥ 1,
which shows that 5 x4 − 2 = O(x4 ), with M = 7 and x0 = 1.",ElementaryCalculus.pdf
"Limits: Formal Definition • Section 3.2 69
Some limits need algebraic manipulation before they can be e valuated.
Example 3.15
Evaluate limx→∞
(∝ra⌈〉⌋allow
x + 1 − ∝ra⌈〉⌋allowx
)
.
Solution: Note that both
∝ra⌈〉⌋allow
x + 1 and ∝ra⌈〉⌋allowx approach ∞ as x goes to ∞, resulting in a limit of the form
∞ − ∞. This is an example of an indeterminate form , which can equal anything (as will be discussed
shortly); it does not have to equal 0 (i.e. the ∞’s do not necessarily “cancel out”). The trick here is to use
the conjugate of
∝ra⌈〉⌋allow
x + 1 − ∝ra⌈〉⌋allowx, so that
limx→∞
(∝ra⌈〉⌋allow
x + 1 −
∝ra⌈〉⌋allow
x
)
= limx→∞
(∝ra⌈〉⌋allow
x + 1 −
∝ra⌈〉⌋allow
x
)
·
∝ra⌈〉⌋allow
x + 1 + ∝ra⌈〉⌋allowx∝ra⌈〉⌋allow
x + 1 + ∝ra⌈〉⌋allowx
= limx→∞
(x + 1) − x∝ra⌈〉⌋allow
x + 1 + ∝ra⌈〉⌋allowx
= limx→∞
1∝ra⌈〉⌋allow
x + 1 + ∝ra⌈〉⌋allowx
= 0
since the numerator is 1 and both terms in the sum in the denomi nator approach ∞ (i.e. 1
∞ = 0).",ElementaryCalculus.pdf
"= limx→∞
1∝ra⌈〉⌋allow
x + 1 + ∝ra⌈〉⌋allowx
= 0
since the numerator is 1 and both terms in the sum in the denomi nator approach ∞ (i.e. 1
∞ = 0).
Some other indeterminate forms are ∞/∞, 0/0 and ∞· 0. How would you handle such limits?
One way is to use L ’Hôpital’s Rule2; a simplified form is stated below:
L ’Hôpital’s Rule: If f and g are differentiable functions and
limx→a
f (x)
g(x) = ±∞
±∞ or 0
0
then
limx→a
f (x)
g(x) = limx→a
f ′(x)
g′(x) .
The number a can be real, ∞, or −∞.
Example 3.16
Evaluate limx→∞
2x − 1
ex .
Solution: This limit is of the form ∞/∞:
limx→∞
2x − 1
ex → ∞
∞
= limx→∞
2
ex by L ’Hôpital’s Rule
= 0
since the numerator is 2 and ex → ∞as x → ∞.
Note that one way of interpreting the limit being 0 is that ex grows much faster than 2 x − 1. In fact,
using L ’Hôpital’s Rule it can be shown that ex grows much faster than any polynomial, i.e. exponential
growth outstrips polynomial growth .",ElementaryCalculus.pdf
"using L ’Hôpital’s Rule it can be shown that ex grows much faster than any polynomial, i.e. exponential
growth outstrips polynomial growth .
2For a proof, see pp.89-91 in P R O T T E R , M.H. A N D C.B. M O R R E Y, A First Course in Real Analysis , New Y ork:
Springer-V erlag, 1977.",ElementaryCalculus.pdf
"70 Chapter 3 • T opics in Differential Calculus §3.2
Example 3.17
Evaluate limx→∞
x
ln x .
Solution: This limit is of the form ∞/∞:
limx→∞
x
ln x → ∞
∞
= limx→∞
1
1
x
by L ’Hôpital’s Rule
= limx→∞ x = ∞
Note that one way of interpreting the limit being ∞ is that x grows much faster than ln x. In fact,
using L ’Hôpital’s Rule it can be shown that any polynomial gr ows much faster than ln x, i.e. polynomial
growth outstrips logarithmic growth .
Example 3.18
Evaluate limx→∞ x e−2x .
Solution: This limit is of the form ∞ ·0, which can be converted to ∞/∞:
limx→∞ x e−2x = limx→∞
x
e2x → ∞
∞
= limx→∞
1
2e2x by L ’Hôpital’s Rule
= 0
Note that the limit is another consequence of exponential gr owth outstripping polynomial growth.
Example 3.19
Evaluate limx→∞
2x2 − 7x − 5
3x2 + 2x − 1 .
Solution: This limit is of the form ∞/∞:
limx→∞
2x2 − 7x − 5
3x2 + 2x − 1 → ∞
∞
= limx→∞
4x − 7
6x + 2 by L ’Hôpital’s Rule
→ ∞
∞ , so use L ’Hôpital’s Rule again
= 4
6 = 2
3",ElementaryCalculus.pdf
"limx→∞
2x2 − 7x − 5
3x2 + 2x − 1 → ∞
∞
= limx→∞
4x − 7
6x + 2 by L ’Hôpital’s Rule
→ ∞
∞ , so use L ’Hôpital’s Rule again
= 4
6 = 2
3
Note that the limit ended up being the ratio of the leading coe fficients of the polynomials in the numer-
ator and denominator of the original limit. Note also that th e lower-order terms (degree less than 2)
ended up not mattering. In general you can always discard the lower-order terms when taking the limit
of a ratio of polynomials (i.e. a rational function ).",ElementaryCalculus.pdf
"Limits: Formal Definition • Section 3.2 71
Example 3.20
Evaluate lim
x→0
1 − cos x
x .
Solution: This limit is of the form 0/0:
lim
x→0
1 − cos x
x → 0
0
= lim
x→0
sin x
1 by L ’Hôpital’s Rule
= sin 0
1 = 0
There is an intuitive justification for L ’Hôpital’s Rule: si nce the limit lim x→a
f (x)
g(x) uses a ratio
to compare how f changes relative to g as x approaches a, then it is really the rates of change
of f and g—namely f ′ and g′, respectively—that are being compared in that ratio, which is
what L ’Hôpital’s Rule says.
The following result provides another way to calculate cert ain limits:
Squeeze Theorem : Suppose that for some functions f , g and h there is a number x0 ≥ 0
such that
g(x) ≤ f (x) ≤ h(x) for all x > x0
and that limx→∞ g(x) = limx→∞ h(x) = L. Then limx→∞ f (x) = L.
Similarly , if g(x) ≤ f (x) ≤ h(x) for all x ̸=a in some interval I containing a, and if
limx→a g(x) = limx→a h(x) = L, then limx→a f (x) = L.",ElementaryCalculus.pdf
"Similarly , if g(x) ≤ f (x) ≤ h(x) for all x ̸=a in some interval I containing a, and if
limx→a g(x) = limx→a h(x) = L, then limx→a f (x) = L.
Intuitively , the Squeeze Theorem says that if one function i s “squeezed” between two func-
tions approaching the same limit, then the function in the mi ddle must also approach that
limit. The theorem also applies to one-sided limits ( x → a+ or x → a−).
Example 3.21
Evaluate limx→∞
sin x
x .
Solution: Since −1 ≤ sin x ≤ 1 for all x, then dividing all parts of those inequalities by x > 0 yields
− 1
x ≤ sin x
x ≤ 1
x for all x > 0 ⇒ limx→∞
sin x
x = 0
by the Squeeze Theorem, since limx→∞
−1
x = 0 = limx→∞
1
x .",ElementaryCalculus.pdf
"72 Chapter 3 • T opics in Differential Calculus §3.2
Exercises
A
For Exercises 1-18 evaluate the given limit.
1. lim
x→2
x2 + 3x − 10
x2 − x − 2 2. limx→∞
x2 + 3x − 10
2x2 − x − 2 3. limx→∞
x2 + 3x − 10
2x3 − x − 2 4. limx→∞
x3 + 3x − 10
2x2 − x − 2
5. lim
x→π/2
cos x
x − π/2 6. limx→∞
x2
ex 7. limx→−∞ x2 ex 8. lim
x→0+
ln x
e1/x
9. lim
x→0
tan x − x
x − sin x 10. lim
x→0
sin 3 x
sin 4 x 11. limx→∞
cos x
x 12. lim
x→0
x sin
(1
x
)
13. lim
x→0
ln (1 − x) − sin2 x
1 − cos2 x 14. lim
x→1
( 1
ln x − 1
x − 1
)
15. lim
x→π/2
(sec x − tan x)
16. lim
x→0+
e−1/x
x 17. limx→∞
(√
x2 + 4 − x
)
18. lim
x→0
cot x
csc x
B
19. The famous “twin paradox,” a result of Einstein’s special th eory of relativity , says that if one of a
pair of twins leaves the earth in a rocket traveling at a high s peed, then he will be younger than his
twin upon returning to earth.
3 This is due to the phenomenon of time dilation , which says that a",ElementaryCalculus.pdf
"twin upon returning to earth.
3 This is due to the phenomenon of time dilation , which says that a
clock moving with a speed v relative to a clock at rest in some inertial reference frame c ounts time
slower relative to the clock at rest, by a factor of
γ = 1√
1 − β2
,
called the Lorentz factor , where β = v
c is the fraction of the speed of light c at which the clock is
moving ( c ≈ 2.998 × 108 m/sec). Notice that 0 ≤ β < 1 (why?). For example, a clock moving at half
the speed of light, so that β = 0.5, would have γ = 1.1547, meaning that the clock runs about 15 .47%
slower than the clock on earth.
(a) Evaluate lim
β→1−
γ . What is the physical interpretation of this limit?
(b) Suppose an astronaut and his twin just turned 30 years old whe n the astronaut leaves earth on
a high-speed journey through space. Upon returning to earth the astronaut is 35 and his twin
is 70. At roughly what fraction of the speed of light must the a stronaut have been traveling?
20. Show that limx→∞
p(x)",ElementaryCalculus.pdf
"is 70. At roughly what fraction of the speed of light must the a stronaut have been traveling?
20. Show that limx→∞
p(x)
ex = 0 for all polynomials p(x) of degree n ≥ 1 with a positive leading coefficient.
21. Show that limx→∞
p(x)
ln x = ∞ for all polynomials p(x) of degree n ≥ 1 with a positive leading coefficient.
22. Show that 5 x3 + 6x2 − 4x + 3 = O(x3 ). 23. Show that 2x2 + 1
x + 1 = O(x). (Hint: Consider x ≥ 1)
24. Call h(x) an infinitesimal function as x → a if limx→a h(x) = 0. That is, an infinitesimal function
approaches zero near some point. Prove the following result , where a and L are real numbers:
limx→a f (x) = L ⇔ f (x) = L + h(x) for all x, where h(x) is an infinitesimal function as x → a
3See p.154-159 in F R E N C H, A.P ., Special Relativity , Surrey , U .K.: Thomas Nelson & Sons Ltd., 1968.",ElementaryCalculus.pdf
"Continuity • Section 3.3 73
3.3 Continuity
Recall from the previous section that a limit lim x→a f (x) can exist without being equal to f (a),
or with f (a) not even being defined. Many functions encountered in appli cations, however , will
meet those conditions, and they have a special name:
A function f is continuous at x = a if
limx→a f (x) = f (a) . (3.4)
A function is continuous on an interval I if it is continuous at every point in the interval.
For a closed interval I = [a, b], a function f is continuous on I if it is continuous on the
open interval ( a, b) and if lim x→a+ f (x) = f (a) (i.e. f is right continuous at x = a) and
limx→b− f (x) = f (b) (i.e. f is left continuous at x = b). A function is discontinuous at a
point if it is not continuous there. A continuous function is one that is continuous over its
entire domain.
Equation (3.4) in the above definition implies that f (a) is defined, i.e. x = a is in the domain of",ElementaryCalculus.pdf
"entire domain.
Equation (3.4) in the above definition implies that f (a) is defined, i.e. x = a is in the domain of
f . Figure 3.3.1 below shows some examples of continuity and di scontinuity:
y
x
x1 x2 x3 x4
y = f (x)
Figure 3.3.1 Continuous at x1 , discontinuous at x2 , x3 and x4
In the above figure, f is not continuous at x = x2 because lim x→x2 f (x) ̸=f (x2); f is not contin-
uous at x = x3 because lim x→x3 f (x) does not exist (the right and left limits do not agree— f is
said to have a jump discontinuity at x = x3); and f is not continuous at x = x4 because f (x4)
is not defined. However , f is continuous at x = x1.
A function is continuous if its graph is one unbroken piece ov er its entire domain. Poly-
nomials, rational functions, trigonometric functions, ex ponential functions, and logarithmic
functions are all continuous on their domains. For example, tan x is continuous over its do-",ElementaryCalculus.pdf
"functions are all continuous on their domains. For example, tan x is continuous over its do-
main, which is broken into disjoint intervals ( −π/2, π/2), ( π/2, 3π/2), (3 π/2, 5π/2), and so forth;
the graph is unbroken on each of those intervals. However , ta n x is not continuous over all of
R, since the function is not defined at all points in R.
In the language of infinitesimals, a function f is continuous at x = a if f (a + dx) − f (a) is an
infinitesimal for any infinitesimal dx. This definition is rarely used.
Physical examples of continuous functions are position, sp eed, velocity , acceleration, temper-
ature, and pressure. Some discontinuous functions do arise in applications.",ElementaryCalculus.pdf
"74 Chapter 3 • T opics in Differential Calculus §3.3
Example 3.22
The floor function ⌊x⌋ is defined as
⌊x⌋ = the largest integer less than or equal to x .
In other words, ⌊x⌋ rounds a non-integer down to the previous integer , and integ ers stay the same. For
example, ⌊0.1⌋ =0, ⌊0.9⌋ =0, ⌊0⌋ =0, and ⌊−1.3⌋ = −2. The graph of ⌊x⌋ is shown in Figure 3.3.2(a).
x
y
−2 −1 2 30 1
−1
−2
1
2
y = ⌊x⌋
(a) Floor function ⌊x⌋
x
y
−3 −2 −1 20 1
−1
−2
1
2
y = ⌈x⌉
(b) Ceiling function ⌈x⌉
Figure 3.3.2 Floor and ceiling functions
Similarly , the ceiling function ⌈x⌉ is defined as
⌈x⌉ = the smallest integer greater than or equal to x .
In other words, ⌈x⌉ rounds a non-integer up to the next integer , and integers sta y the same. For example,
⌈0.1⌉ =1, ⌈0.9⌉ =1, ⌈1⌉ =1, and ⌈−1.3⌉ = −1. The graph of ⌈x⌉ is shown in Figure 3.3.2(b).
Clearly both ⌊x⌋ and ⌈x⌉ have jump discontinuities at the integers, but both are cont inuous at all",ElementaryCalculus.pdf
"Clearly both ⌊x⌋ and ⌈x⌉ have jump discontinuities at the integers, but both are cont inuous at all
non-integer values of x. Both functions are also examples of step functions , due to the staircase
appearance of their graphs. Step functions are useful in sit uations when you want to model a quantity
that takes only a discrete set of values. For example, in a car with a 4-gear transmission, f (x) could
be the gear the transmission has shifted to while the car trav els at speed x. Up to a certain speed the
car remains in first gear ( f = 1) and then shifts to second gear ( f = 2) after attaining that speed, then
it remains in second gear until reaching another speed, upon which the car then shifts to third gear
( f = 3), and so on. In general, discrete changes in state are often modeled with step functions.
Example 3.23
For an extreme case of discontinuity , consider the function
f (x) =
{
0 if x is rational
1 if x is irrational",ElementaryCalculus.pdf
"Example 3.23
For an extreme case of discontinuity , consider the function
f (x) =
{
0 if x is rational
1 if x is irrational
This function is discontinuous at every value of x in R, since within any positive distance δ of a real
number x—no matter how small δ is—there will be an infinite number of both rational and irrat ional",ElementaryCalculus.pdf
"Continuity • Section 3.3 75
numbers. This is a property of R. So the value of f will keep jumping between 0 and 1 no matter how
close you get to x. In other words, for any number a in R, f (a) exists but it will never equal lim x→a f (x)
because that limit will not exist.
By the various rules for limits, it is straightforward to sho w that sums, differences, con-
stant multiples, products and quotients of continuous func tions are continuous. Likewise a
continuous function of a continuous function (i.e. a compos ition of continuous functions) is
also continuous. In addition, a continuous function of a fini te limit of a function can be passed
inside the limit:
If f is a continuous function and limx→a g(x) exists and is finite, then:
f
(
limx→a g(x)
)
= limx→a f ( g(x)) (3.5)
The same relation holds for one-sided limits.
The above result is useful in evaluating the indeterminate f orms 0 0, ∞0, and 1 ∞. The idea is",ElementaryCalculus.pdf
"The same relation holds for one-sided limits.
The above result is useful in evaluating the indeterminate f orms 0 0, ∞0, and 1 ∞. The idea is
to take the natural logarithm of the limit by passing the cont inuous function ln x inside the
limit and evaluate the resulting limit.
Example 3.24
Evaluate lim
x→0+
xx .
Solution: This limit is of the form 0 0 , so let y = lim
x→0+
xx and then take the natural logarithm of y:
ln y = ln
(
lim
x→0+
xx
)
= lim
x→0+
ln xx (pass the natural logarithm function inside the limit)
= lim
x→0+
x ln x → 0 · (−∞)
= lim
x→0+
ln x
1/x → −∞
∞
= lim
x→0+
1/x
−1/x2 by L ’Hôpital’s Rule
ln y = lim
x→0+
(−x) = 0
Thus, lim
x→0+
xx = y = e0 = 1.",ElementaryCalculus.pdf
"76 Chapter 3 • T opics in Differential Calculus §3.3
There is an important relationship between differentiabil ity and continuity:
Every differentiable function is continuous.
Proof: If a function f is differentiable at x = a then f ′(a) = limx→a
f (x)−f (a)
x−a exists, so
limx→a ( f (x) − f (a)) = limx→a ( f (x) − f (a)) · x − a
x − a = limx→a
f (x) − f (a)
x − a · limx→a (x − a) = f ′(a) · 0 = 0
which means that limx→a f (x) = f (a) , i.e. f is continuous at x = a . ✓
x
y
0
y = |x |
Note that the converse is not true. For example, the absolute value
function f (x) = |x| is continuous everywhere—its graph is unbroken,
as shown in the picture on the right—but recall from Example 1 .3 in
Section 1.2 that it is not differentiable at x = 0. Continuous curves
can have sharp edges and cusps, but differentiable curves ca nnot.
Two other important theorems 4 about continuous functions are:
Extreme V alue Theorem : If f is a continuous function on a closed interval [ a, b] then f",ElementaryCalculus.pdf
"Extreme V alue Theorem : If f is a continuous function on a closed interval [ a, b] then f
attains both a maximum value and a minimum value on that inter val.
Intermediate V alue Theorem : If f is a continuous function on a closed interval [ a, b]
then f attains every value between f (a) and f (b).
y
x
a b c d
y = f (x)
Figure 3.3.3 Extreme V alue Theorem
y
x
a b c d x 0
k
1
2
y = f (x)
Figure 3.3.4 Intermediate V alue Theorem
Figure 3.3.3 shows why a closed interval is required for the E xtreme V alue Theorem, as f
attains neither a maximum nor minimum on the open interval ( c, d). The Intermediate V alue
Theorem says that continuous functions cannot “skip over” i ntermediate values between two
other function values. In Figure 3.3.4 the function f skips the value k between f (c) = 1 and
f (d) = 2 because f is not continuous over all of [ c, d]. On [ a, b] the value k is attained by f at
x = x0, i.e. f (x0) = k, since f is continuous on [ a, b].",ElementaryCalculus.pdf
"x = x0, i.e. f (x0) = k, since f is continuous on [ a, b].
4The full proofs require some advanced results. See pp.97-98 and p.558 in T AY L O R, A.E. A N D W .R. M A N N,
Advanced Calculus , 2nd ed., New Y ork: John Wiley & Sons, Inc., 1972.",ElementaryCalculus.pdf
"Continuity • Section 3.3 77
Example 3.25
Show that there is a solution to the equation cos x = x.
Solution: Let f (x) = cos x − x. Since f is continuous for all x, in particular it is continuous on [0 , 1]. So
since f (0) = 1 > 0 and f (1) = −0.459698 < 0, then by the Intermediate V alue Theorem there is a number
c in the open interval (0 , 1) such that f (c) = 0, since 0 is between the values f (0) and f (1). Hence,
cos c − c = 0, which means that cos c = c. That is, x = c is a solution of cos x = x.
Note in the above example that the Intermediate V alue Theore m does not tell you how to
find the solution, just that the solution exists. To find the solut ion you can use the bisection
method: divide the interval [0 , 1] in half and apply the Intermediate V alue Theorem to each
half-interval to determine which one contains the solution ; repeat this procedure on that half-
interval, resulting in a smaller interval containing the so lution, then repeat the procedure over",ElementaryCalculus.pdf
"interval, resulting in a smaller interval containing the so lution, then repeat the procedure over
and over , until you eventually obtain an interval so small th at the midpoint of that interval
can be taken as the solution. Listing 3.1 below shows one way o f implementing the bisection
method for Example 3.25 to find the root of f (x) = cos x − x, using the Python programming
language.
Listing 3.1 Bisection method in Python
1 import math
2
3 def f(x):
4 return math.cos(x) - x
5
6 def bisect(a, b):
7 midpt = (a+b)/2.0
8 tol = 1e-15
9 if b - a > tol:
10 val = f(midpt)
11 if val*f(a) < 0:
12 bisect(a, midpt)
13 elif val*f(b) < 0:
14 bisect(midpt, b)
15 else:
16 print(""Root = %.13f"" % (midpt))
17 else:
18 print(""Root = %.13f"" % (midpt))
19
20 bisect(0, 1)
Line 8 sets the tolerance to 10 −15: the program terminates upon reaching an interval whose
length is smaller than that. The output is shown below:
Root = 0.7390851332152",ElementaryCalculus.pdf
"length is smaller than that. The output is shown below:
Root = 0.7390851332152
This is the number obtained by taking the cosine of a number (i n radians) repeatedly .",ElementaryCalculus.pdf
"78 Chapter 3 • T opics in Differential Calculus §3.3
Exercises
A
For Exercises 1-18, indicate whether the given function f (x) is continuous or discontinuous at the given
value x = a by comparing f (a) with lim x→a f (x).
1. f (x) = |x |; at x = 0 2. f (x) = |x − 1|; at x = 0 3. f (x) = ⌊x⌋; at x = 0
4. f (x) = ⌊x⌋; at x = 0.3 5. f (x) = ⌈x⌉; at x = 0 6. f (x) = ⌈x⌉; at x = 0.5
7. f (x) = x − ⌊x⌋; at x = 0 8. f (x) = x − ⌊x⌋; at x = 1.1 9. f (x) = x − |x |; at x = 0
10. f (x) =
{
0 if x ≤ 0,
1 if x > 0;
at x = 0
11. f (x) =
{
0 if x ≤ 0,
1 if x > 0;
at x = 1
12. f (x) =
{
x2 if x ≤ 0,
1 if x > 0;
at x = 0
13. f (x) =
{
x + 1 if x ≤ 0,
1 if x > 0;
at x = 1
14. f (x) =
{
sin ( x2) if x ̸=0,
0 if x = 0;
at x = 0
15. f (x) =
{
sin (1/x) if x ̸=0,
0 if x = 0;
at x = 0
16. f (x) =
{
0 if x is rational,
1 if x is irrational;
at x =
∝ra⌈〉⌋allow
3
17. f (x) =
{
0 if x is rational,
x if x is irrational;
at x = 0
18. f (x) =
{
0 if x is rational,
x if x is irrational;
at x = 1
19. Evaluate lim",ElementaryCalculus.pdf
"3
17. f (x) =
{
0 if x is rational,
x if x is irrational;
at x = 0
18. f (x) =
{
0 if x is rational,
x if x is irrational;
at x = 1
19. Evaluate lim
x→0+
xx2
. 20. Evaluate limx→∞ x1/x . 21. Evaluate lim
x→0
(1 − x)1/x .
22. If f (x) = x2 + x − 2
x − 1 for x ̸=1, how should f (1) be defined so that f (x) is continuous at x = 1 ?
23. If f (x) = 1/x for x ̸=0, is there a way to define f (0) so that f (x) is continuous for all x ?
B
24. Can a function that is not continuous over a closed interval a ttain a maximum value and a mini-
mum value in that interval? If so, then give an example; if not then explain why .
25. Show that there is a number x such that x5 − x = 3.
26. Prove that f (x) = x8 + 3x4 − 1 has at least two distinct real roots.
27. Suppose that a function f is continuous on the interval [0 , 3], f has no roots in [0 , 3], and f (1) = 1.
Prove that f (x) > 0 for all x in [0 , 3].
28. Show that an object whose average speed is vavg over the time interval a ≤ t ≤ b will move with",ElementaryCalculus.pdf
"Prove that f (x) > 0 for all x in [0 , 3].
28. Show that an object whose average speed is vavg over the time interval a ≤ t ≤ b will move with
speed vavg at some time t in [ a, b].
29. Let f (x) = 1/(x − 1). Then f (0) = −1 < 0 and f (2) = 1 > 0. Can you conclude by the Intermediate
V alue Theorem that f (x) must be 0 for some x in [0 , 2] ? Explain.
30. Show that if f ′ and f ′′ exist and are continuous at x then
f ′′(x) = lim
h→0
f (x) − 2 f (x − h) + f (x − 2h)
h2 .
31. Show that if f ′, f ′′ and f ′′′ exist and are continuous at x then
f ′′′(x) = lim
h→0
f (x) − 3 f (x − h) + 3 f (x − 2h) − f (x − 3h)
h3 .",ElementaryCalculus.pdf
"Implicit Differentiation • Section 3.4 79
3.4 Implicit Differentiation
A function y = f (x) is usually given by an explicit formula, such as y = x2. It is then straight-
forward to find dy
dx using the differentiation rules you have learned so far . But suppose instead
that you were given merely an equation involving x and y, such as
x3 y2 esin (x y) = x2 + x y + y3 .
The set of points ( x, y) satisfying this equation describes some sort of curve in th e x y-plane,
but it might not be possible to solve for y in terms of x—that is, there might not be an explicit
formula for y as a function of the variable x. So in this case does the derivative dy
dx even have
any meaning, and if so then how would you find it?
It turns out that dy
dx does make sense in such a case, because an equation involving x and y
such as the one above implicitly defines y in terms of x in the following sense: as x varies so",ElementaryCalculus.pdf
"such as the one above implicitly defines y in terms of x in the following sense: as x varies so
does y. Hence it should be possible to find the rate of change of y with respect to the variable x
(i.e. dy
dx ). To do so, take d
dx of both sides of the equation, then assume that y really is a function
of x so that you can use the Chain Rule to solve for dy
dx . The example below illustrates this
procedure, called implicit differentiation .
Example 3.26
−4 −2 0 2 4 6
−10
−5
0
5
10
x
y
Find dy
dx given the equation x3 + 3x + 2 = y2 .
Solution: The above equation implicitly defines an elliptic curve , and its
graph is shown on the right. This curve is not a function y = f (x), since it
violates the vertical line test, but y still varies with x. To find dy
dx take d
dx of
both sides of the equation then solve for dy
dx :
d
dx (x3 + 3x + 2) = d
dx ( y2 )
3x2 + 3 = 2 y · dy
dx by the Chain Rule, so
dy
dx = 3x2 + 3
2 y
At first this might seem unsatisfying—or confusing—since dy",ElementaryCalculus.pdf
"dx ( y2 )
3x2 + 3 = 2 y · dy
dx by the Chain Rule, so
dy
dx = 3x2 + 3
2 y
At first this might seem unsatisfying—or confusing—since dy
dx is given in terms of both x and y.
However , the derivative can still be evaluated at specific po ints ( x, y) on the curve, i.e. any ( x, y) satis-
fying the original equation. For example, it is easy to check that ( x, y) = (1,
∝ra⌈〉⌋allow
6) satisfies the equation
x3 + 3x + 2 = y2 , so dy
dx (1,
∝ra⌈〉⌋allow
6) = 3(1)2+3
2
∝ra⌈〉⌋allow
6 =
∝ra⌈〉⌋allow
6
2 . Note that dy
dx is not defined when y = 0.
Notice that taking the square root of both sides of the origin al equation does not result in an explicit
formula for y, since y = ±
∝ra⌈〉⌋allow
x3 + 3x + 2 defines two functions, not just one. The beauty of implicit differen-
tiation is that the derivative dy
dx = 3x2 +3
2 y calculated above gives you a single expression for the deriv ative
of both those functions.",ElementaryCalculus.pdf
"80 Chapter 3 • T opics in Differential Calculus §3.4
An algebraic curve is defined as the set of all points ( x, y) satisfying a polynomial equation
in the variables x and y, such as x2 − 3x y4 + 1 = x5 − y2 . An elliptic curve is a special case of
an algebraic curve, where the polynomial has the specific for m x3 + ax + b = y2, such as the
equation x3 + 3x + 2 = y2 from Example 3.26. Elliptic curves have certain properties that have
found applications in cryptography . 5
Example 3.27
−3 −2 −1 0 1 2 3
−3
−2
−1
0
1
2
3
x
y
Find dy
dx given the equation x + y = x3 + y3 .
Solution: The above equation implicitly defines an algebraic curve and its
graph is shown on the right. To find dy
dx take d
dx of both sides of the equation
then solve for dy
dx :
d
dx (x + y) = d
dx (x3 + y3 )
1 + dy
dx = 3x2 + 3 y2 · dy
dx by the Chain Rule, so
dy
dx = 3x2 − 1
1 − 3 y2
Notice that the curve consists of an oval shape (an ellipse, a ctually) with a line through it. In fact,",ElementaryCalculus.pdf
"dy
dx = 3x2 − 1
1 − 3 y2
Notice that the curve consists of an oval shape (an ellipse, a ctually) with a line through it. In fact,
that line is y = −x, as can be verified by replacing each instance of y in the equation x + y = x3 + y3 by
−x (resulting in the equation 0 = 0). Y ou might be wondering how dy
dx is defined at the points where that
line intersects the ellipse: is it the slope of the line y = −x (i.e. −1), or is it the slope of the tangent line
to the ellipse at those points (which would not equal −1)? This is discussed in the exercises.
The graph was created with the free open-source graphing pro gram Gnuplot 6 using the following
Gnuplot commands (which give an idea of how to plot implicit f unctions in general):
set size square
set view 0,0
set isosamples 500,500
set contour base
set cntrparam levels discrete 0
unset surface
set grid
unset key
unset ztics
set xlabel ’x’
set ylabel ’y’
f(x,y) = x + y - x**3 - y**3
splot [-3:3][-3:3] f(x,y) lw 3",ElementaryCalculus.pdf
"unset surface
set grid
unset key
unset ztics
set xlabel ’x’
set ylabel ’y’
f(x,y) = x + y - x**3 - y**3
splot [-3:3][-3:3] f(x,y) lw 3
5For example, see Section 12.2 in B U C H MA N N, J.A., Introduction to Cryptography , New Y ork: Springer-V erlag,
2001.
6See the documentation at http://www.gnuplot.info/documentation.html",ElementaryCalculus.pdf
"Implicit Differentiation • Section 3.4 81
Example 3.28
x
y
0 1
x2 + y2 = 1
(4/5, 3/5)
Find the tangent line to the curve x2 + y2 = 1 at the point (4/5 , 3/5).
Solution: This curve is the unit circle, shown in the picture on the righ t.
First use implicit differentiation to find dy
dx :
d
dx (x2 + y2 ) = d
dx (1) ⇒ 2x + 2 y · dy
dx = 0 by the Chain Rule
⇒ dy
dx = − x
y
The slope m of the tangent line to the curve at (4/5 , 3/5) is then m = dy
dx (4/5, 3/5) = −4/5
3/5 = −4/3. Thus, the
equation of the tangent line is y − 3
5 = −4
3
(
x − 4
5
)
.
Exercises
A
For Exercises 1-9, use implicit differentiation to find dy
dx .
1. x3 y − 4x y2 = y + x2 2. x y = (x + y)3 3. (x + y)3 = (x − y + 1)2
4. x2/3 + y2/3 = a2/3 5. (x2 − y2 )2 = 2x2 + y2 6. x + y
x − y = x2 + y2
7. cos ( x y) = sin ( x2 y2 ) 8. x3 − x = y2 9. x3 y2 esin ( x y) = x2 + x y + y3
10. In Example 3.28 is it possible to solve the equation x2 + y2 = 1 explicitly for y in terms of x? Explain.",ElementaryCalculus.pdf
"10. In Example 3.28 is it possible to solve the equation x2 + y2 = 1 explicitly for y in terms of x? Explain.
11. In Example 3.28 what happens to the tangent line at the point ( 1, 0)? Why does this make sense
geometrically?
12. Find the equation of the tangent line to the curve x3 + 3x2 y + y3 = 8 at the point (2 , 0).
B
13. Find d2 y
dx2 for the curve x2 + y2 = 1. Y ou may use the results from Example 3.28.
14. Show that at every point ( x0, y0 ) on the curve y2 = 4ax, the equation of the tangent line to the curve
is y y0 = 2a(x + x0 ).
15. Show that at every point ( x0, y0 ) on the ellipse x2
a2 + y2
b2 = 1, the equation of the tangent line to the
ellipse is xx0
a2 + y y0
b2 = 1.
16. Show that at every point ( x0, y0 ) on the hyperbola x2
a2 − y2
b2 = 1, the equation of the tangent line to
the hyperbola is xx0
a2 − y y0
b2 = 1.
17. Show that dy
dx is not defined at the points of intersection of the line and ell ipse described by the",ElementaryCalculus.pdf
"the hyperbola is xx0
a2 − y y0
b2 = 1.
17. Show that dy
dx is not defined at the points of intersection of the line and ell ipse described by the
curve x + y = x3 + y3 from Example 3.27. (Hint: Factor the equation x + y = x3 + y3 .)
18. Show that the points P = (2, 4) and Q = (−31/64, −337/512) are on the elliptic curve x3 + 3x + 2 = y2
from Example 3.26, and that the tangent line to the curve at P also goes through Q.",ElementaryCalculus.pdf
"82 Chapter 3 • T opics in Differential Calculus §3.5
3.5 Related Rates
If several quantities are related by an equation, then diffe rentiating both sides of that equation
with respect to a variable (usually t, representing time) produces a relation between the rates
of change of those quantities. The known rates of change are t hen used in that relation to
determine an unknown related rate.
Example 3.29
100
300
10
h
Suppose that water is being pumped into a rectangular pool at a
rate of 60,000 cubic feet per minute. If the pool is 300 ft long , 100
ft wide, and 10 ft deep, how fast is the height of the water insi de
the pool changing?
Solution: Let V be the volume of the water in the pool. Since the
volume of a rectangular solid is the product of the length, wi dth,
and height of the solid, then
V = (300)(100)h = 30000h ft3
where h is the height of the water , as in the picture on the right. Both V and h are functions of time t
(measured in minutes), and d V",ElementaryCalculus.pdf
"where h is the height of the water , as in the picture on the right. Both V and h are functions of time t
(measured in minutes), and d V
dt = 60000 ft 3 /min was given. The goal is to find dh
dt . Since
d V
dt = d
dt (30000h) = 30000 dh
dt
then
dh
dt = 1
30000
d V
dt = 1
30000 · 60000 = 2 ft/min .
Example 3.30
100
✸
x
θ
θ
Suppose that the angle of inclination from the top of a 100 ft p ole to
the sun is decreasing at a rate of 0 .05 radians per minute. How fast
is the length of the pole’s shadow on the ground increasing wh en the
angle of inclination is π/6 radians? Y ou may assume that the pole is
perpendicular to the ground.
Solution: Let θ be the angle of inclination and let x be the length of the
shadow , as in the picture on the right. Both θ and x are functions of
time t (measured in minutes), and dθ
dt = −0.05 rad/min was given (the
derivative is negative since θ is decreasing). The goal is to find dx
dt when
θ = π/6, denoted by dx
dt
⏐
⏐
⏐
⏐
θ=π/6",ElementaryCalculus.pdf
"derivative is negative since θ is decreasing). The goal is to find dx
dt when
θ = π/6, denoted by dx
dt
⏐
⏐
⏐
⏐
θ=π/6
(the vertical bar means “evaluated at” the value of the subsc ript to the right
of the bar). Since
x = 100 cot θ ⇒ dx
dt = −100 csc 2 θ · dθ
dt = −100 csc 2 θ · (−0.05) = 5 csc 2 θ
then
dx
dt
⏐
⏐
⏐
⏐
θ=π/6
= 5 csc 2 (π/6) = 5 (2) 2 = 20 ft/min .",ElementaryCalculus.pdf
"Related Rates • Section 3.5 83
Example 3.31
The radius of a right circular cylinder is decreasing at the r ate of 3 cm/min, while the height is increas-
ing at the rate of 2 cm/min. Find the rate of change of the volum e of the cylinder when the radius is 8
cm and the height is 6 cm.
Solution: Let r, h, and V be the radius, height, and volume, respectively , of the cyli nder . Then V = πr2 h.
Since dr
dt = −3 cm/min and dh
dt = 2 cm/min, then by the Product Rule:
d V
dt = d
dt (πr2 h) =
(
2πr · d r
dt
)
h + πr2 · d h
dt ⇒ d V
dt
⏐
⏐
⏐
⏐ r = 8
h = 6
= 2π(8) (−3) (6) + π(82) (2) = −160π
cm3
min
Exercises
A
1. A stone is dropped into still water . If the radius of the circu lar outer ripple increases at the rate of 4
ft/s, how fast is the area of the circle of disturbed water inc reasing when the radius is 10 ft?
2. The radius of a sphere decreases at a rate of 3 mm/hr . Determin e how fast the volume and surface
area of the sphere are changing when the radius is 5 mm.",ElementaryCalculus.pdf
"area of the sphere are changing when the radius is 5 mm.
3. A kite 80 ft above level ground moves horizontally at a rate of 4 ft/s away from the person flying it.
How fast is the string being released at the instant when 100 f t of string have been released?
4. A 10-ft ladder is leaning against a wall on level ground. If th e bottom of the ladder is dragged away
from the wall at the rate of 5 ft/s, how fast will the top of the l adder descend at the instant when it
is 8 ft from the ground?
5. A person 6 ft tall is walking at a rate of 6 ft/s away from a light which is 15 ft above the ground. At
what rate is the end of the person’s shadow moving along the gr ound away from the light?
6. An object moves along the curve y = x3 in the x y-plane. At what points on the curve are the x and y
coordinates of the object changing at the same rate?
7. The radius of a right circular cone is decreasing at the rate o f 4 cm/min, while the height is increasing",ElementaryCalculus.pdf
"7. The radius of a right circular cone is decreasing at the rate o f 4 cm/min, while the height is increasing
at the rate of 3 cm/min. Find the rate of change of the volume of the cone when the radius is 6 cm
and the height is 7 cm.
8. Two boats leave the same dock at the same time, one goes north a t 25 mph and the other goes east
at 30 mph. How fast is the distance between the boats changing when they are 100 miles apart?
9. Repeat Exercise 8 with the angle between the boats being 110 ◦.
B
10. An angle θ changes with time. For what values of θ do sin θ and tan θ change at the same rate?
11. Repeat Example 3.30 but with the ground making a 100 ◦ angle with the pole to the left of the pole.
C
12. An upright cylindrical tank full of water is tipped over at a c onstant angular speed. Assume that
the height of the tank is at least twice its radius. Show that a t the instant the tank has been tipped",ElementaryCalculus.pdf
"the height of the tank is at least twice its radius. Show that a t the instant the tank has been tipped
45◦, water is leaving the tank twice as fast as it did at the instan t the tank was first tipped. (Hint:
Think of how the water looks inside the tank as it is being tipp ed.)",ElementaryCalculus.pdf
"84 Chapter 3 • T opics in Differential Calculus §3.6
3.6 Differentials
An ideal gas satisfies the equation P V = R T , where R is a constant and P , V , and T are the
pressure, volume per mole, and temperature, respectively , of the gas. It will be proved that
dP
P + d V
V = dT
T . (3.6)
Recall that d P, d V, and d T represent infinitesimal changes in the quantities P , V , and T ,
respectively . Notice that none of the quotients in Equation (3.6) have an infinitesimal in the
denominator . For example, d P is divided not by dx or dt, as it would be in a derivative such
as d P
dx or d P
dt . Instead it is divided by P , which is not an infinitesimal. So Equation (3.6) is an
equation that relates infinitesimals themselves, i.e. infin itesimal changes, not infinitesimal
rates of change. This is, in fact, how many physical laws are stated , for reasons that will be
discussed shortly .
Though infinitesimals have been used throughout this text, m any calculus textbooks 7 do",ElementaryCalculus.pdf
"discussed shortly .
Though infinitesimals have been used throughout this text, m any calculus textbooks 7 do
not even mention them, instead preferring to call them differentials.8 For compatibility , the
definition is given here:
For a differentiable function f (x), the differential of f (x) is
df = f ′(x) dx (3.7)
where dx is an infinitesimal change in x.
Note that this is identical to Equation (1.9) in Section 1.3.
Example 3.32
Find the differential df of f (x) = x3 .
Solution: By definition,
df = f ′(x) dx = 3x2 dx
Equivalently , this can be written as
d(x3 ) = 3x2 dx ,
which is often the way it would appear in textbooks in the scie nces.
All the rules for derivatives (e.g. sum rule, product rule) a pply to differentials, and can
be proved simply by multiplying the corresponding derivati ve rule by dx on both sides of the
equation:
7More accurately , many current calculus textbooks never mention them. Calculus texts up th rough the 1930s or",ElementaryCalculus.pdf
"equation:
7More accurately , many current calculus textbooks never mention them. Calculus texts up th rough the 1930s or
so not only mentioned infinitesimals but used them extensive ly , even to the point of the texts themselves having
titles such as Introduction to Infinitesimal Calculus .
8Though often in an unclear and sometimes confusing and misle ading manner , as will be seen later in this section.",ElementaryCalculus.pdf
"Differentials • Section 3.6 85
Let f and g be differentiable functions, and let c be a constant. Then:
(a) d(c) = 0
(b) d(c f ) = c df (Constant Multiple Rule)
(c) d( f + g) = df + dg (Sum Rule)
(d) d( f − g) = df − dg (Difference Rule)
(e) d( f g ) = f dg + g df (Product Rule)
(f) d
(f
g
)
= g df − f dg
g2 (Quotient Rule)
(g) d( f n) = n f n−1 df (Power Rule)
(h) d( f ( g)) = df
dg dg (Chain Rule)
For example, to prove (e), multiply both sides of the usual Pr oduct Rule by dx so that
d( f g )
dx = f dg
dx + g df
dx ⇒ d( f g ) = ✚✚dx
(
f dg
✚✚dx + g df
✚✚dx
)
⇒ d( f g ) = f dg + g df ✓
since the dx terms all cancel. The proofs of the other rules are similar .
The differential version of the ideal gas law in Equation (3. 6)
dP
P + d V
V = dT
T
can now be proved by taking the differential of both sides of t he equation P V = R T :
d(P V ) = d(R T ) = R · dT by the Constant Multiple Rule
V dP + P d V = P V
T dT by the Product Rule and since R = P V
T
V dP
P V + P d V
P V = dT",ElementaryCalculus.pdf
"d(P V ) = d(R T ) = R · dT by the Constant Multiple Rule
V dP + P d V = P V
T dT by the Product Rule and since R = P V
T
V dP
P V + P d V
P V = dT
T after dividing both sides by P V
dP
P + d V
V = dT
T ✓
Notice that d P
P , d V
V and d T
T represent the relative infinitesimal changes in P , V , and T , respec-
tively . The differential formulation is useful for finding o ne relative infinitesimal change when
the other two are known.",ElementaryCalculus.pdf
"86 Chapter 3 • T opics in Differential Calculus §3.6
Example 3.33
Suppose that M is the total mass of a rocket and its unburnt fuel at any time t (so M is a function of
t). Over an infinitesimal time dt a mass dm of fuel is burnt and the gas byproducts are expelled out the
rear of the rocket at a velocity vE relative to the rocket. Using the law of conservation of mome ntum
over the interval dt, show that
vE dm = M dv
where m and v are the mass of burnt fuel and the velocity of the rocket, resp ectively , at the beginning
of the time dt.
Solution: Momentum is defined as mass times velocity . The momentum of th e rocket at the beginning
of the time dt is thus M v. At the end of the time dt, the momentum of the rocket consists of two parts,
namely the momentum of the rocket and its remaining unburnt f uel, which is
((mass before dt) − (increase in burnt fuel)) × ((velocity before dt) + (increase in velocity)) (3.8)
(M − dm)(v + dv) (3.9)",ElementaryCalculus.pdf
"((mass before dt) − (increase in burnt fuel)) × ((velocity before dt) + (increase in velocity)) (3.8)
(M − dm)(v + dv) (3.9)
and the momentum of the fuel that was burnt and expelled out th e rear , which is
(v − vE ) dm .
So by conservation of momentum,
M v = (M − dm)(v + dv) + (v − vE ) dm
M v = M v − v dm + M dv − (dm)(dv) + v dm − vE dm, so
vE dm = M dv − (dm)(dv) = M dv
since ( dm)(dv) = (m′(t) dt)(v′(t) dt) = m′(t)v′(t)(dt)2 = m′(t)v′(t) · 0 = 0.
Dividing both sides of vE dm = M dv by dt yields the equation
M ˙v = ˙m v E
using the dot notation—mentioned in Section 1.3—for the der ivative with respect to the time variable
t, which is still popular with physicists. Since ˙v is just acceleration a, this formulation is the classic
equation for the acceleration of a rocket. 9
Letting f be the natural logarithm function and letting g = u in the differential version of
the Chain Rule yields the following useful result:
d(ln u) = du
u",ElementaryCalculus.pdf
"the Chain Rule yields the following useful result:
d(ln u) = du
u
This is often used in a differential version of the technique of logarithmic differentiation dis-
cussed in Section 2.3.
9For other formulations see Chapter 1 in R O S S E R , J.B., R.R. N E W T O N A N D G.L. G R O S S, Mathematical Theory
of Rocket Flight , New Y ork: McGraw-Hill Book Company , Inc., 1947.",ElementaryCalculus.pdf
"Differentials • Section 3.6 87
Example 3.34
Prove the relation dP
P + d V
V = dT
T using logarithmic differentiation.
Solution: Take the natural logarithm and then the differential of both sides of the equation P V = R T :
ln (P V ) = ln ( R T ) ⇒ ln P + ln V = ln R + ln T
⇒ d(ln P + ln V ) = d(ln R + ln T )
⇒ d P
P + d V
V = 0 + d T
T = d T
T (since ln R is a constant)
Example 3.35
The derivative of the area πr2 of a circle of radius r, as a function of r, equals its circumference 2 πr. Use
the notion of a differential as an infinitesimal change to exp lain why this makes sense geometrically .
Solution: Let A = πr2 be the area of a circle of varying radius r. Then A′(r) = 2πr, which is equivalent
to saying dA = 2πr dr. To see why this makes sense geometrically , imagine increas ing the radius by
dr, as in the picture below on the left. This increases the area A of the circle to A + dA, with dA the
infinitesimal area of the shaded ring in the picture.
r
dr slice and
roll flat
dA
2πr",ElementaryCalculus.pdf
"infinitesimal area of the shaded ring in the picture.
r
dr slice and
roll flat
dA
2πr
2π(r + dr)
dr area = dA
π dr
dr area = 0
Slice that ring along the dashed line then roll it flat, yieldi ng a trapezoid with height dr, top length
2πr (from the circumference of the inner circle of the ring), and bottom length 2 π(r + dr) (from the
circumference of the outer circle of the ring), as shown in th e picture above on the right. The triangular
edges of the trapezoid contribute nothing to the area of the t rapezoid, since (by the Microstraightness
Property) the hypotenuse of each is indeed a straight line, s o each is a right triangle with height dr and
(by symmetry) base πdr, thus having area 1
2 π(dr)2 = 0. Hence the entire area dA of the trapezoid comes
from the rectangular portion of height dr and base 2 πr, which means dA = 2πr dr, as expected.
The above example answers the question of whether it is a happ y coincidence that the deriva-",ElementaryCalculus.pdf
"The above example answers the question of whether it is a happ y coincidence that the deriva-
tive of a circle’s area turns out to be the circle’s circumfer ence—no, it is not! Some other such
cases (e.g. the derivative of a sphere’s volume is its surfac e area) are left to the exercises. Note
that a similar “coincidence” does not occur for a square: if x is the length of each side then the
area is x2, but the derivative of x2 is 2 x, which is not the perimeter of the square (i.e. 4 x). Why
does this not follow the same pattern as the circle? Think abo ut a key difference in the shape
of a square in comparison to a circle, keeping differentiabi lity in mind.",ElementaryCalculus.pdf
"88 Chapter 3 • T opics in Differential Calculus §3.6
There are many benefits to using differentials—i.e. infinite simals—in calculus. 10 For exam-
ple, recall Example 3.31 in Section 3.5 on related rates, whe re the volume V of a right circular
cylinder with radius r and height h changes with time t as
d V
dt =
(
2πr · dr
dt
)
h + πr2 · dh
dt .
The above equation forces you to consider only the derivativ e with respect to the time variable
t. What if you wanted to see the rates of change with respect to a nother variable, such as r,
h, or some other quantity? In that case using the differential version of the above equation,
namely
d V = 2πr h dr + πr2 dh
provides more flexibility—you are free to divide both sides b y any differential, not just by dt.
Many related rates problems would likely benefit from this ap proach.
Present-day calculus textbooks confuse the notion of a diff erential (infinitesimal) dx with",ElementaryCalculus.pdf
"Present-day calculus textbooks confuse the notion of a diff erential (infinitesimal) dx with
the idea of a small but real value ∆x. The two are not the same. An infinitesimal is not a real
number and cannot be assigned a real value, no matter how small; ∆x can be assigned real
values. Using dx and ∆x interchangeably is a source of much confusion for students ( likewise
for dy and ∆y). This confusion rears its head in exercises involving the l inear approximation
of a curve by its tangent line near a point x0, namely f (x) ≈ f (x0) + f ′(x0)(x − x0) when x − x0
is “small” (e.g.
∝ra⌈〉⌋allow
63 ≈ 7.9375, by using f (x) = ∝ra⌈〉⌋allowx, x = 63, x0 = 64, and x − xo = ∆x = −1). Such
exercises have nothing to do with differentials, not to ment ion having dubious value nowadays.
They are remnants of a bygone era, before the advent of modern computing obviated the need
for such (generally) poor approximations.
Exercises
A",ElementaryCalculus.pdf
"They are remnants of a bygone era, before the advent of modern computing obviated the need
for such (generally) poor approximations.
Exercises
A
1. Find the differential df of f (x) = x2 − 2x + 5. 2. Find the differential df of f (x) = sin2(x2 ).
3. Show that d
(
tan−1( y/x)
)
= x dy − y dx
x2 + y2 4. Given y2 − x y + 2x2 = 3, find dy.
5. The elasticity of a function y = f (x) is E( y) = x
y · dy
dx . Show that E( y) = d(ln y)
d(ln x) .
6. Prove the differential version of the Quotient Rule:
d
(f
g
)
= g df − f dg
g2
10For an excellent overview on this subject, see D R AY, T . A N D C.A. M A N O G U E, Putting Differentials Back into
Calculus, College Math. J . 41 (2010), 90-100. Some of the material in this section is indeb ted to that paper , which
is available at http://www.math.oregonstate.edu/bridge/papers/differentials.pdf",ElementaryCalculus.pdf
"Differentials • Section 3.6 89
7. Let y = c u n , where c and n are constants. Show that
dy
y = n du
u .
8. Obviously the derivative of the constant π2 is not 2 π. But is d(π2 ) = 2πd(π) true? Explain.
B
9. The continuity relation for an ideal gas is
P M
∝ra⌈〉⌋allow
T
= constant
where P and T are the pressure and temperature, respectively , of the gas, and M is the Mach
number. Show that
d P
P + dM
M = d T
2T .
10. For an ideal gas, satisfying the equation P V = R T as before, the Gibbs energy G is defined as
G = H − T S, where H and S are the enthalpy and entropy, respectively , of the gas.
(a) Show that
d
( G
R T
)
= 1
R T dG − G
R T 2 d T .
(b) One of the fundamental property relations for an ideal gas (which you do not need to prove) is
dG = V d P − S d T .
Use this and part (a) to show that
d
( G
R T
)
= V
R T d P − H
R T 2 d T .
11. The derivative of the volume πr2 h of a right circular cylinder of radius r and height h, as a function",ElementaryCalculus.pdf
"d
( G
R T
)
= V
R T d P − H
R T 2 d T .
11. The derivative of the volume πr2 h of a right circular cylinder of radius r and height h, as a function
of r, equals its lateral surface area 2 πrh. Use the notion of a differential as an infinitesimal change
to explain why this makes sense geometrically .
12. The derivative of the volume 4π
3 r3 of a sphere of radius r, as a function of r, equals its surface area
4πr2 . Use the notion of a differential as an infinitesimal change t o explain why this makes sense
geometrically .
13. In quantum calculus the q-differential of a function f (x) is
dq f (x) = f (qx ) − f (x) ,
and the q-derivative of f (x) is
D q f (x) = dq f (x)
dq x = f (qx ) − f (x)
qx − x = f (qx ) − f (x)
(q − 1)x .
(a) Show that for all positive integers n,
D q
(
xn )
= [n] xn−1 ,
where [ n] = 1 + q + q2 + ··· +qn−1 .
(b) Use part (a) to show that for all positive integers n,
lim
q→1
D q
(
xn )
= d
dx
(
xn )
.",ElementaryCalculus.pdf
"CHAPTER 4
Applications of Derivatives
4.1 Optimization
Many physical problems involve optimization: finding either a maximum or minimum value of
some quantity . Optimization problems often have a constraint involving two variables which
allows you to rewrite the objective function —the function to optimize—as a function of a single
variable: use the constraint to solve for one variable in ter ms of another , then substitute that
expression into the objective function.
First, the intuitive notions of maximum and minimum need cla rifying.
A function f has a global maximum at x = c if f (c) ≥ f (x) for all x in the domain of f .
Similarly , f has a global minimum at x = c if f (c) ≤ f (x) for all x in the domain of f . Say
that f has a local maximum at x = c if f (c) ≥ f (x) for all x “near” c, i.e. for all x such that
|x − c | <δ for some number δ > 0. Likewise, f has a local minimum at x = c if f (c) ≤ f (x)
for all x such that |x − c | <δ for some number δ > 0.",ElementaryCalculus.pdf
"|x − c | <δ for some number δ > 0. Likewise, f has a local minimum at x = c if f (c) ≤ f (x)
for all x such that |x − c | <δ for some number δ > 0.
In other words, a global maximum is the largest value everywh ere (“globally”), whereas a local
maximum is only the largest value “locally .” Likewise for a g lobal vs local minimum. The
picture below illustrates the differences.
y
x
a c 1 c2 b
y = f (x)
In the picture, on the interval [ a, b] the function f has a global minimum at x = a, a global
maximum at x = c1, a local minimum at x = c2, and a local maximum at x = b.
90",ElementaryCalculus.pdf
"Optimization • Section 4.1 91
Every global maximum [minimum] is a local maximum [minimum] , but not vice versa. In
physical applications global maxima or minima 1 are the primary interest. The Extreme V alue
Theorem in Section 3.3 guarantees the existence of at least o ne global maximum and at least
one global minimum for continuous functions defined on closed intervals (i.e. intervals of the
form [ a, b]). All the functions under consideration here will be diffe rentiable, and hence contin-
uous. So the only issues will be how to find the global maxima or minima, and how to handle
intervals that are not closed.
Consider again the picture from the previous page, this time looking at how the derivative
f ′ changes over [ a, b]. Intuitively it is obvious that near an internal maximum (i .e. in the
open interval ( a, b)) such as at x = c1, the function should increase before that point and then
decrease after that point. That means that f ′(x) > 0 before x = c1 and f ′(x) < 0 after the",ElementaryCalculus.pdf
"decrease after that point. That means that f ′(x) > 0 before x = c1 and f ′(x) < 0 after the
“turning point” x = c1, as shown below .
y
x
a c 1 c2 b
f ′ = 0
f ′ = 0
y = f (x)
f ′ > 0 f ′ < 0
f ′ > 0
Assuming that f ′ is continuous (which will be the case for all the functions in this section),
then this means that f ′ = 0 at x = c1, that is, f ′(c1) = 0. Similarly , near the internal minimum
at x = c2, f ′(x) < 0 before x = c2 and f ′(x) > 0 after x = c2, so that f ′(c2) = 0. Points at which the
derivative is zero are called critical points (or stationary points ) of the function. So x = c1
and x = c2 are critical points of f .
Note in the picture that f ′ goes from positive to zero to negative around x = c1, so that f ′ is
decreasing around x = c1, i.e. f ′′ = ( f ′)′ < 0. Similarly , f ′ is increasing around x = c2, i.e. f ′′ > 0.
This leads to the following test for local maxima and minima: 2
Second Derivative T est : Let x = c be a critical point of f (i.e f ′(c) = 0). Then:",ElementaryCalculus.pdf
"This leads to the following test for local maxima and minima: 2
Second Derivative T est : Let x = c be a critical point of f (i.e f ′(c) = 0). Then:
(a) If f ′′(c) > 0 then f has a local minimum at x = c.
(b) If f ′′(c) < 0 then f has a local maximum at x = c.
(c) If f ′′(c) = 0 then the test fails.
To see why the test fails when f ′′(c) = 0, consider f (x) = x3: f ′(0) = 0 and f ′′(0) = 0, yet x = 0 is
neither a local minimum nor maximum in any open interval cont aining x = 0. Section 4.2 will
present an alternative for when the Second Derivative Test f ails.
1The words “maxima” and “minima” are the traditional plural f orms of maximum and minimum, respectively .
2A formal proof requires the Mean V alue Theorem, which will be presented in Section 4.4.",ElementaryCalculus.pdf
"92 Chapter 4 • Applications of Derivatives §4.1
There is a simple visual mnemonic device for remembering the Second Derivative Test, due
to a generic minimum or maximum resembling a smile or frown, r espectively:
+++ + ++
f ′′ > 0
local min.
−−− − −−
f ′′ < 0
local max.
o o
f ′′ = 0
test fails
The “eyes” in the faces represent the sign of f ′′ at a critical point, while the “mouths” indicate
the nature of that point (when f ′′ = 0 nothing is known). The procedure for finding a global
maximum or minimum can now be stated:
How to find a global maximum or minimum
Suppose that f is defined on an interval I. There are two cases:
1. The interval I is closed : The global maximum of f will occur either at an interior
local maximum or at one of the endpoints of I whichever of these points provides the
largest value of f will be where the global maximum occurs.
Similarly , the global minimum of f will occur either at an interior local minimum or at",ElementaryCalculus.pdf
"largest value of f will be where the global maximum occurs.
Similarly , the global minimum of f will occur either at an interior local minimum or at
one of the endpoints of I; whichever of these points provides the smallest value of f will
be where the global minimum occurs.
2. The interval I is not closed and has only one critical point : If the only critical
point is a local maximum then it is a global maximum. If the onl y critical point is a local
minimum then it is a global minimum.
In each case of the above procedure try to use the Second Deriv ative Test to verify that a critical
point is a local minimum or maximum, unless it is obvious from the nature of the problem that
there can be only a minimum or only a maximum.
Example 4.1
x
y
x
y
Figure 4.1.1
Show that the rectangle with the largest area for a fixed perim eter is a
square.
Solution: Let L be the perimeter of a rectangle with sides x and y. The
idea is that L is a fixed constant, but x and y can vary . Figure 4.1.1",ElementaryCalculus.pdf
"Solution: Let L be the perimeter of a rectangle with sides x and y. The
idea is that L is a fixed constant, but x and y can vary . Figure 4.1.1
shows that there are many possible shapes for the rectangle, but in all
cases L = 2x + 2 y. Let A be the area of such a rectangle. Then A = x y,
which is a function of two variables. But
L = 2x + 2 y ⇒ y = L
2 − x ,",ElementaryCalculus.pdf
"Optimization • Section 4.1 93
and hence
A = x
(L
2 − x
)
= Lx
2 − x2
is now a function of x alone, on the open interval (0 , L/2) (since the length x is positive). Now find the
critical points of A:
A′(x) = 0 ⇒ L
2 − 2x = 0
⇒ x = L
4 is the only critical point
This problem is thus the case of a function defined on an open in terval having only one critical point.
Use the Second Derivative Test to verify that the sole critic al point x = L/4 is a local maximum for A:
A′′(x) = −2 ⇒ A′′(L/4) = −2 < 0 ⇒ A has a local maximum at x = L/4
Thus, A has a global maximum at x = L/4. Also, y = L/2 − x = L/2 − L/4 = L/4, which means that x = y,
i.e. the rectangle is a square.
Note: The constraint in this example was L = 2x + 2 y and the objective function was A = x y.
Example 4.2
h
r
Figure 4.1.2
Suppose a right circular cylindrical can with top and bottom lids will be assembled
to have a fixed volume. Find the radius and height of the can tha t minimizes the
total surface area of the can.",ElementaryCalculus.pdf
"to have a fixed volume. Find the radius and height of the can tha t minimizes the
total surface area of the can.
Solution: Let V be the fixed volume of the can with radius r and height h, as in
Figure 4.1.2. The volume V is a constant, with V = πr2 h. Let S be the total surface
area of the can, including the lids. Then
S = 2πr2 + 2πrh
where the first term in the sum on the right side of the equation is the combined
area of the two circular lids and the second term is the latera l surface area of the
can. So S is a function of r and h, but h can be eliminated since
V = πr2 h ⇒ h = V
πr2
and so
S = 2πr2 + 2πr · V
πr2 = 2πr2 + 2V
r
making S a function of r alone. Now find the critical points of S (i.e. solve S′(r) = 0):
S′(r) = 0 ⇒ 4πr − 2V
r2 = 0
⇒ r3 = V
2π
⇒ r =
3
√
V
2π is the only critical point",ElementaryCalculus.pdf
"94 Chapter 4 • Applications of Derivatives §4.1
Since both r and h are lengths and have to be positive, then 0 < r < ∞. So this is another case of a
function defined on an open interval having only one critical point. Use the Second Derivative Test to
verify that this critical point r = 3
√
V
2π is a local minimum for S:
S′′(r) = 4π + 4V
r3 ⇒ S′′
(
3
√
V
2π
)
= 4π + 4V
V
2π
= 12π > 0 ⇒ S has a local minimum at r =
3
√
V
2π
Thus, S has a global minimum at r = 3
√
V
2π , and
r · r2 = r3 = V
2π ⇒ 2r = V
πr2 = h .
Hence, r = 3
√
V
2π and h = 2 3
√
V
2π will minimize the total surface area, i.e. the height should equal the
diameter .
Note that this result can be applied to soda cans, where the vo lume is V = 12 fluid ounces ≈ 21.6 cubic
inches: both a diameter and height of about 3 .8 inches will minimize the amount (and hence the cost)
of the aluminum used for the can. Y et soda cans are not that wid e and short—they are usually thinner",ElementaryCalculus.pdf
"of the aluminum used for the can. Y et soda cans are not that wid e and short—they are usually thinner
and taller . So why is a non-optimal size used in practice? Oth er factors—e.g. packing requirements, the
need for small children to hold the can in one hand—might over ride the desire to minimize the cost of
the aluminum. The lesson is that an optimal solution for one f actor (material cost) might not always be
truly optimal when all factors are considered; compromise i s often necessary .
Example 4.3
v0
L
θ
v0 v0 sin θ
v0 cos θ
θ
Figure 4.1.3
Suppose that a projectile is launched from the ground with a fi xed
initial velocity v0 at an angle θ with the ground. What value of θ
would maximize the horizontal distance traveled by the proj ectile,
assuming the ground is flat and not sloped (i.e. horizontal)?
Solution: Let x and y represent the horizontal position and vertical
position, respectively , of the projectile at time t ≥ 0. From the trian-",ElementaryCalculus.pdf
"Solution: Let x and y represent the horizontal position and vertical
position, respectively , of the projectile at time t ≥ 0. From the trian-
gle at the bottom of Figure 4.1.3, the horizontal and vertica l com-
ponents of the initial velocity are v0 cos θ and v0 sin θ, respectively .
Since distance is the product of velocity and time, then the h orizon-
tal and vertical distances traveled by the projectile by tim e t due to
the initial velocity are ( v0 cos θ)t and ( v0 sin θ)t, respectively . Ignor-
ing wind and air resistance, the only other force on the proje ctile
will be the downward force g due to gravity , so that the equations of motion for the projec tile are:
x = (v0 cos θ)t
y = −1
2 gt2 + (v0 sin θ)t
The goal is to find θ that maximizes the length L shown in Figure 4.1.3. First write y as a function of x:
x = (v0 cos θ)t ⇒ t = x
v0 cos θ ⇒ y = −1
2 g
( x
v0 cos θ
) 2
+ (v0 sin θ) · x
v0 cos θ
⇒ y = − gx2
2v2
0 cos2 θ
+ x tan θ",ElementaryCalculus.pdf
"Optimization • Section 4.1 95
Then L is the value of x > 0 that makes y = 0:
0 = − gL2
2v2
0 cos2 θ
+ L tan θ ⇒ L =
2v2
0 sin θ cos θ
g =
v2
0 sin 2 θ
g
So L is now a function of θ, with 0 < θ < π/2 (why?). So if there is a single local maximum then it must
be the global maximum. Now get the critical points of L:
L′(θ) = 0 ⇒
2v2
0 cos 2 θ
g = 0
⇒ cos 2 θ = 0
⇒ 2θ = π
2 ⇒ θ = π
4 is the only critical point
Use the Second Derivative Test to verify that L has a local maximum at θ = π/4:
L′′(θ) = −
4v2
0 sin 2 θ
g ⇒ L′′(π/4) = −
4v2
0
g < 0
⇒ L has a local maximum at θ = π
4
Thus, L has a global maximum at θ = π
4 , i.e. the projectile travels the farthest horizontally whe n
launched at a 45 ◦ angle with the ground (with L
(π
4
)
=
v2
0
g being the maximum horizontal distance).
Note that once the formula for L as a function of θ was found to be L =
v2
0 sin 2 θ
g , calculus was not actually
needed to solve this problem. Why? Since v2",ElementaryCalculus.pdf
"v2
0 sin 2 θ
g , calculus was not actually
needed to solve this problem. Why? Since v2
0 and g are positive constants (recall g = 9.8m/s2 ), L would
have its largest value when sin 2 θ has its largest value 1, which occurs when θ = π/4.
Example 4.4
Fermat’s Principle states that light always travels along the path that takes th e least amount of time.
So suppose that a ray of light is shone from a point A onto a flat horizontal reflective surface at an angle
θ1 with the surface and then reflects off the surface at an angle θ2 to a point B. Show that Fermat’s
Principle implies that θ1 = θ2 .
Solution: Let L be the horizontal distance between A and B, let d1 be the distance the light travels
from A to the point of contact C with the surface a horizontal distance x from A, let d2 be the distance
from C to B, and let y1 and y2 be the vertical distances from A and B, respectively , to the surface, as in
the picture below .
d1
d2
A
B
C
y1
y2
θ1 θ2
x L − x
L",ElementaryCalculus.pdf
"96 Chapter 4 • Applications of Derivatives §4.1
Since time is distance divided by speed, and since the speed o f light is constant, then minimizing the
total time elapsed is equivalent to minimizing the total dis tance traveled, namely D = d1 +d2 . The basic
idea here is that Fermat’s Principle implies that for the lig ht to go from A to B in the shortest time, the
unknown point C—and hence the unknown distance x—will have to be at a point that makes θ1 = θ2 .
The distances L, y1 and y2 are constants, so the goal is to write the total distance D as a function of x,
find the x that minimizes D, then show that that value of x makes θ1 = θ2 .
First, note that C has to be between A and B as in the picture, otherwise the total distance D would
be larger than if C were directly below either A or B. This ensures that θ1 and θ2 are between 0 and
π/2, and that 0 ≤ x ≤ L.
Next, by the Pythagorean Theorem and the above picture,
d1 =
√
x2 + y2
1 and d2 =
√
(L − x)2 + y2
2",ElementaryCalculus.pdf
"π/2, and that 0 ≤ x ≤ L.
Next, by the Pythagorean Theorem and the above picture,
d1 =
√
x2 + y2
1 and d2 =
√
(L − x)2 + y2
2
and so the total distance D = d1 + d2 traveled by the light is a function of x:
D(x) =
√
x2 + y2
1 +
√
(L − x)2 + y2
2
To find the critical points of D, solve the equation D′(x) = 0:
D′(x) = x√
x2 + y2
1
− L − x√
(L − x)2 + y2
2
= 0 ⇒ x
d1
= L − x
d2
⇒ sin θ1 = sin θ2 ⇒ θ1 = θ2
since the sine function is one-to-one over the interval [0 , π
2 ].
This seems to prove the result, except for one remaining issu e to resolve: verifying that the minimum
for D really does occur at the x between 0 and L where D′(x) = 0, not at the endpoints x = 0 or x = L
of the closed interval [0 , L]. Note that using the Second Derivative Test in this case doe s not matter ,
since you would have to check the value of D at the endpoints anyway and compare those values to the
values of D at the critical points. To find expressions for the critical p oints, note that
D′(x) = 0 ⇒ x√
x2 + y2
1",ElementaryCalculus.pdf
"values of D at the critical points. To find expressions for the critical p oints, note that
D′(x) = 0 ⇒ x√
x2 + y2
1
= L − x√
(L − x)2 + y2
2
⇒ x2
x2 + y2
1
= (L − x)2
(L − x)2 + y2
2
⇒ ✘✘✘✘✘(L − x)2 x2 + x2 y2
2 = ✘✘✘✘✘(L − x)2 x2 + (L − x)2 y2
1
⇒ x y2 = (L − x) y1 ⇒ x = L y1
y1 + y2
is the only critical point,
and x is between 0 and L. Now compare the values of D2 (x) at x = 0, x = L, and x = L y1
y1 +y2
:
D2(0) = L2 + y2
1 + y2
2 + 2 y1
√
L2 + y2
2
D2 (L) = L2 + y2
1 + y2
2 + 2 y2
√
L2 + y2
1
D2
(
L y1
y1 +y2
)
= L2 + y2
1 + y2
2 + 2 y1 y2
Since y2 <
√
L2 + y2
2 and y1 <
√
L2 + y2
1 , then D2
(
L y1
y1 +y2
)
is the smallest of the three values above, so that
D2 (x) has its minimum value at x = L y1
y1 +y2
, which means D(x) has its minimum value there. ✓",ElementaryCalculus.pdf
"Optimization • Section 4.1 97
Example 4.5
A man is in a boat 4 miles off a straight coast. He wants to reach a point 10 miles down the coast in the
minimum possible time. If he can row 4 mi/hr and run 5 mi/hr , wh ere should he land the boat?
∝ra⌈〉⌋allow
x2 + 164
x 10 − x
10
YSolution: Let T be the total time traveled. The goal is to minimize T . From the
picture on the right, since time is distance divided by speed , break the total time
into two parts: the time rowing in the water and the time runni ng on the coast,
so that
T = timerow + timerun = distrow
speedrow
+ distrun
speedrun
=
∝ra⌈〉⌋allow
x2 + 16
4 + 10 − x
5 ,
where 0 ≤ x ≤ 10 is the distance along the coast where the boat lands. Then
T ′(x) = x
4
∝ra⌈〉⌋allow
x2 + 16
− 1
5 = 0 ⇒ 5x = 4
√
x2 + 16 ⇒ 25x2 = 16 ( x2 + 16) ⇒ x = 16
3 ,
so x = 16
3 is the only critical point. Thus, the (global) minimum of T will occur at x = 16
3 , x = 0, or x = 10.
Since
T (0) = 3 , T (10) =
∝ra⌈〉⌋allow
29
2 ≈ 2.693 , T
(16
3
)
= 13
5 = 2.6",ElementaryCalculus.pdf
"3 , x = 0, or x = 10.
Since
T (0) = 3 , T (10) =
∝ra⌈〉⌋allow
29
2 ≈ 2.693 , T
(16
3
)
= 13
5 = 2.6
then T ( 16
3 ) < T (10) < T (0). Hence, the global minimum occurs when landing the boat x = 16
3 ≈ 5.33 miles
down the coast.
Note that this example shows the importance of checking the e ndpoints. It was quite close between
landing about 5.33 miles down the coast (2.6 hours) or simply rowing all the way to the destination
(about 2.693 hours)—the difference is only about 5.6 minute s. With just a slight change in a few of the
numbers, the minimum could have occurred at an endpoint. Mor al: always check the endpoints!
3
Example 4.6
Find the point ( x, y) on the graph of the curve y = ∝ra⌈〉⌋allowx that is closest to the point (1 , 0).
x
y
0
D
(x, y)
(1, 0)
y = ∝ra⌈〉⌋allowx
Solution: Let ( x, y) be a point on the curve y = ∝ra⌈〉⌋allowx. Then ( x, y) = (x, ∝ra⌈〉⌋allowx ),
so by the distance formula the distance D between ( x, y) and (1 , 0) is
given by",ElementaryCalculus.pdf
"so by the distance formula the distance D between ( x, y) and (1 , 0) is
given by
D2 = (x − 1)2 + ( y − 0)2 = (x − 1)2 + (∝ra⌈〉⌋allowx )2 = (x − 1)2 + x ,
which is a function of x ≥ 0 alone. Note that minimizing D is equivalent
to minimizing D2 . Since
d(D2 )
dx = 2 ( x − 1) + 1 = 2x − 1 = 0 ⇒ x = 1
2 is the only critical point,
and since d2 (D 2)
dx2 = 2 > 0 for all x, then by the Second Derivative Test x = 1/2 is a local minimum. Hence,
the global minimum for D2 must occur at the endpoint x = 0 or at x = 1/2. But D2(0) = 1 > D2 (1/2) = 3/4,
so the global minimum occurs at x = 1/2. Hence, the closest point is ( x, y) = (1/2,
∝ra⌈〉⌋allow
1/2).
3Another possible lesson is that optimal in the mathematical sense might, again, not mean optimal in a practical
sense. After all, presumably after the man is finished with wh atever he had to do at the destination 10 miles down
the coast, he then has the inconvenience of going back about 4 .67 miles to retrieve his boat. At his running speed",ElementaryCalculus.pdf
"the coast, he then has the inconvenience of going back about 4 .67 miles to retrieve his boat. At his running speed
of 10 mph this would take 28 minutes, wiping out the 5.6 minute s he gained with his “optimal” landing spot!",ElementaryCalculus.pdf
"98 Chapter 4 • Applications of Derivatives §4.1
Example 4.7
Find the width and height of the rectangle with the largest po ssible perimeter inscribed in a semicircle
of radius r.
hr
w
2
w
Solution: Let w be the width of the rectangle and let h be the height, as
in the picture. Then the perimeter is P = 2w + 2h. By symmetry and the
Pythagorean Theorem,
h2 = r2 −
(w
2
) 2
⇒ h = 1
2
√
4r2 − w2
and so P = 2w +
∝ra⌈〉⌋allow
4r2 − w2 for 0 < w < 2r. Find the critical points of P :
P ′(w) = 2 − w∝ra⌈〉⌋allow
4r2 − w2
= 0 ⇒ w = 2
√
4r2 − w2
⇒ w2 = 16r2 − 4w2
⇒ w = 4r∝ra⌈〉⌋allow
5
is the only critical point,
and since
P ′′(w) = − 4r2
(4r2 − w2)3/2 ⇒ P ′′
(
4r∝ra⌈〉⌋allow
5
)
= − 53/2
2r < 0
then P has a local maximum at w = 4r∝ra⌈〉⌋allow
5 , by the Second Derivative Test. Since P (w) is defined for w in
the open interval (0 , 2r), the local maximum is a global maximum. For the width w = 4r∝ra⌈〉⌋allow
5 the height is
h = r∝ra⌈〉⌋allow
5 , which gives the dimensions for the maximum perimeter .",ElementaryCalculus.pdf
"5 the height is
h = r∝ra⌈〉⌋allow
5 , which gives the dimensions for the maximum perimeter .
Note: If w were extended to include the cases of “degenerate” rectangl es of zero width or height, i.e.
w = 0 or w = 2r, then the maximum perimeter would still occur at w = 4r∝ra⌈〉⌋allow
5 , since P
(
4r∝ra⌈〉⌋allow
5
)
= 10r∝ra⌈〉⌋allow
5 ≈ 4.472r is
larger than P (0) = 2r and P (2r) = 4r.
Exercises
A
1. Find the point on the curve y = x2 that is closest to the point (4 , −1/2).
2. Prove that for 0 ≤ p ≤ 1, p (1 − p) ≤ 1
4 .
3. A farmer wishes to fence a field bordering a straight stream wi th 1000 yd of fencing material. It is
not necessary to fence the side bordering the stream. What is the maximum area of a rectangular
field that can be fenced in this way?
4. The power output P of a battery is given by P = V I − R I 2 , where I, V , and R are the current,
voltage, and resistance, respectively , of the battery . If V and R are constant, find the current I that
maximizes P .",ElementaryCalculus.pdf
"voltage, and resistance, respectively , of the battery . If V and R are constant, find the current I that
maximizes P .
5. An impulse turbine consists of a high speed jet of water strik ing circularly mounted blades. The
power P generated by the turbine is P = V U (V − U ), where V is the speed of the jet and U is the
speed of the turbine. If the jet speed V is constant, find the turbine speed U that maximizes P .",ElementaryCalculus.pdf
"Optimization • Section 4.1 99
6. A man is in a boat 5 miles off a straight coast. He wants to reach a point 15 miles down the coast in
the minimum possible time. If he can row 6 mi/hr and run 10 mi/h r , where should he land the boat?
7. The current I in a voltaic cell is
I = E
R + r ,
where E is the electromotive force and R and r are the external and internal resistance, respectively .
Both E and r are internal characteristics of the cell, and hence can be tr eated as constants. The
power P developed in the cell is P = R I 2 . For which value of R is the power P maximized?
8. Find the point(s) on the ellipse x2
25 + y2
16 = 1 closest to ( −1, 0).
9. Find the maximum area of a rectangle that can be inscribed in a n ellipse x2
a2 + y2
b2 = 1, where a > 0
and b > 0 are arbitrary constants. Y our answer should be in terms of a and b.
10. Find the radius and angle of the circular sector with the maxi mum area and a fixed perimeter P .",ElementaryCalculus.pdf
"10. Find the radius and angle of the circular sector with the maxi mum area and a fixed perimeter P .
11. In Example 4.3 show that the maximum height reached by the pro jectile when launched with an
initial velocity v0 at an angle 0 < θ < π
2 to the ground is
v2
0 sin2 θ
2 g .
12. The phase velocity v of a capillary wave with surface tension T and water density p is
v =
√
2πT
λp + λg
2π
where λ is the wavelength. Find the value of λ that minimizes v.
13. For an inventory model with a constant order quantity Q > 0 and a constant linear inventory de-
pletion rate D, the total unit cost T C to maintain an average inventory of Q/2 units is
T C = C + P
Q + (I +W ) Q
2D
where C is the capital investment cost, P is the cost per order , I is the per unit interest charge per
unit time, and W is the overall inventory holding cost. Find the value of Q that minimizes T C.
a
a
a
θ θ
14. The opening of a rain gutter—shown in the figure on the right—h as",ElementaryCalculus.pdf
"a
a
a
θ θ
14. The opening of a rain gutter—shown in the figure on the right—h as
a bottom and two sides each with length a. The sides make an angle
θ with the bottom. Find the value of θ that maximizes the amount of
rain the gutter can hold.
B
E r
r0 x0
15. In an electric circuit with a supplied voltage (emf) E, a resistor
with resistance r0 , and an inductor with reactance x0 , suppose
you want to add a second resistor . If r represents the resistance
of this second resistor then the power P delivered to that resistor
is given by
P = E2 r
(r + r0 )2 + x0 2
with E, r0 , and x0 treated as constants. For which value of r is the power P maximized?",ElementaryCalculus.pdf
"100 Chapter 4 • Applications of Derivatives §4.1
16. The stress τ in the x y-plane along a varying angle φ is given by
τ = τ(φ) =
σx − σy
2 sin 2 φ + τx y cos 2 φ ,
where σx, σy, and τx y are stress components that can be treated as constants. Show that the maxi-
mum stress is
τ =
√ (
σx − σy
) 2 + 4τ2x y
2 .
(Hint: Draw a right triangle with angle 2φ after finding the critical point(s).)
17. A certain jogger can run 0.16 km/min, and walks at half that sp eed. If he runs along a circular trail
with circumference 50 km and then—before completing one ful l circle—walks back straight across
to his starting point, what is the maximum time he can spend on the run/walk?
18. A rectangular poster is to contain 50 square inches of printe d material, with 4-inch top and bottom
margins and 2-inch side margins. What dimensions for the pos ter would use the least paper?
19. Find the maximum volume of a right circular cylinder that can be inscribed in a sphere of radius 3.",ElementaryCalculus.pdf
"19. Find the maximum volume of a right circular cylinder that can be inscribed in a sphere of radius 3.
20. A figure consists of a rectangle whose top side coincides with the diameter of a semicircle atop it.
If the perimeter of the figure is 20 m, find the radius and height of the semicircle and rectangle,
respectively , that maximizes the area inside the figure.
21. A thin steel pipe 25 ft long is carried down a narrow corridor 5 .4 ft wide. At the end of the corridor
is a right-angle turn into a wider corridor . How wide must thi s corridor be in order to get the pipe
around the corner? Y ou may assume that the width of the pipe ca n be ignored.
22. A rectangle is inscribed in a right triangle, with one corner of the rectangle at the
right angle of the triangle. Show that the maximum area of the rectangle occurs
when a corner of the rectangle is at the midpoint of the hypote nuse of the triangle.",ElementaryCalculus.pdf
"when a corner of the rectangle is at the midpoint of the hypote nuse of the triangle.
23. Find the relation between the radius and height of a cylindri cal can with an open top that maxi-
mizes the volume of the can, given that the surface area of the can is always the same fixed amount.
24. An isosceles triangle is circumscribed about a circle of rad ius r. Find the height of the triangle that
minimizes the perimeter of the triangle.
25. Suppose N voltaic cells are arranged in N/x rows in parallel, with each row consisting of x cells in
series, creating a current I through an external resistance R. Each cell has internal resistance r and
EMF (voltage) e. Find the x that maximizes the current I, which—due to Ohm’s Law—is given by
I = x e
(x2 r/N) + R .
26. A single-degree-of-freedom harmonically forced vibratio n system with damping factor ζ has mag-
nification factor M F = ((1 − r2 )2 + (2ζr)2 )−1/2, where r is the frequency ratio. Find the value of r that
maximizes M F .",ElementaryCalculus.pdf
"nification factor M F = ((1 − r2 )2 + (2ζr)2 )−1/2, where r is the frequency ratio. Find the value of r that
maximizes M F .
27. At a distance x ≥ 0 from the center of a uniform ring with charge q and radius a, the magnitude E
of the electric field for points on the axis of the ring is
E = qx
4πǫ0 (a2 + x2 )3/2
where ǫ0 is the permittivity of free space. Find the distance x that maximizes E.",ElementaryCalculus.pdf
"Optimization • Section 4.1 101
28. Find the equation of the tangent line to the ellipse x2
a2 + y2
b2 = 1 in the first quadrant that forms with
the coordinate axes the right triangle with minimal area.
29. A “cold” star that has exhausted its nuclear fuel—called a white dwarf —has total energy E, given
by
E = ħ2 (3π2 N q)5/3
10π2 m
(4
3 πR3
) −2/3
− 3G M 2 N2
5R
where ħ is the reduced Planck constant, N is the number of nucleons—protons and neutrons—in the
star , q is the charge of an electron, m is the mass of an electron, M is the mass of a nucleon, G is the
gravitational constant, and R is the radius of the star . Show that the radius R that minimizes E is
R =
(9π
4
) 2/3 ħ2 q5/3
G mM 2 N1/3 .
C
30. An object of mass m has orbital angular momentum l around a black hole with Schwarzchild radius
rS and mass M. The effective potential Φ of the object is
Φ = −G M
r + l2
2m2 r2 − rS l2
2m2 r3",ElementaryCalculus.pdf
"rS and mass M. The effective potential Φ of the object is
Φ = −G M
r + l2
2m2 r2 − rS l2
2m2 r3
where G is the gravitational constant and r is the object’s distance from the black hole. Show that Φ
has a local maximum and minimum at r = r1 and r = r2 , respectively , where
r1 = l2
2G M m2

1 −
√
1 − 6G M m2 rS
l2

 and r2 = l2
2G M m2

1 +
√
1 − 6G M m2 rS
l2

 .
A
B
θ1
θ2
31. Recall Fermat’s Principle from Example 4.4, which states th at light travels
along the path that takes the least amount of time. The speed o f light in a
vacuum is approximately c = 2.998 × 108 m/s, but in some other medium (e.g.
water) light is slower . Suppose that a ray of light goes from a point A in one
medium where it moves at a speed v1 and ends up at a point B in another
medium where it moves at a speed v2. Use Fermat’s Principle to prove Snell’s
Law, which says that the light is refracted through the boundary between
the two media such that
sin θ1
sin θ2
= v1
v2",ElementaryCalculus.pdf
"Law, which says that the light is refracted through the boundary between
the two media such that
sin θ1
sin θ2
= v1
v2
where θ1 and θ2 are the angles that the light makes with the normal line perpe ndicular to the
boundary of the media in the first and second medium, respecti vely , as in the picture above.
32. A sphere of radius a is inscribed in a right circular cone, with the sphere touchi ng the base of the
cone. Find the radius and height of the cone if its volume is a m inimum.
33. Find the length of the shortest line segment from the positiv e x-axis to the positive y-axis going
through a point ( a, b) in the first quadrant.
34. Find the radius r of a circle c whose center is on a fixed circle C of radius R such that the arc length
of the part of c within C is a maximum.",ElementaryCalculus.pdf
"102 Chapter 4 • Applications of Derivatives §4.2
4.2 Curve Sketching
A function can increase between two points in different ways , as shown in Figure 4.2.1.
y
x
y = f (x)
(a) f ′′ = 0: straight
y
x
y = f (x)
(b) f ′′ > 0: concave up
y
x
y = f (x)
(c) f ′′ < 0: concave down
Figure 4.2.1 Increasing function f : f ′ > 0, different signs for f ′′
In each case in the above figure the function is increasing, so that f ′(x) > 0, but the manner in
which the function increases is determined by its concavity, that is, by the sign of the second
derivative f ′′(x). The function in the graph on the far left is linear , i.e. of t he form f (x) = ax + b
for some constants a and b, so that f ′′(x) = 0 for all x. But the functions in the other two graphs
are nonlinear . In the middle graph the derivative f ′ is increasing, so that f ′′ > 0; in this case
the function is called concave up . In the graph on the far right the derivative f ′ is decreasing,",ElementaryCalculus.pdf
"the function is called concave up . In the graph on the far right the derivative f ′ is decreasing,
so that f ′′ < 0; in this case the function is called concave down . The same definitions would
hold if the function were decreasing, as shown in Figure 4.2. 2 below:
y
x
y = f (x)
(a) f ′′ = 0
y
x
y = f (x)
(b) f ′′ > 0: concave up
y
x
y = f (x)
(c) f ′′ < 0: concave down
Figure 4.2.2 Decreasing function f : f ′ < 0, different signs for f ′′
In Figures 4.2.1(b) and 4.2.2(b) the function is below the li ne joining the points at each end,
while in Figure 4.2.1(c) and 4.2.2(c) the function is above t hat line. This turns out to be true in
general, as a result of the following theorem:",ElementaryCalculus.pdf
"Curve Sketching • Section 4.2 103
Concavity Theorem : Suppose that f is a twice-differentiable function on [ a, b]. Then:
(a) If f ′′(x) > 0 on ( a, b) then f (x) is below the line l(x) joining the points ( a, f (a)) and
(b, f (b)) for all x in ( a, b).
(b) If f ′′(x) < 0 on ( a, b) then f (x) is above the line l(x) joining the points ( a, f (a)) and
(b, f (b)) for all x in ( a, b).
y
x
a b
y = f (x)
l(x)
f (a)
f (b)
Proof: Only part (a) will be proved; the proof of part (b) is
similar and left as an exercise. So assume that f ′′(x) > 0 on
(a, b), and l(x) be the line joining ( a, f (a)) and ( b, f (b)), as in
the drawing on the right. The drawing suggests that f (x) <
l(x) over ( a, b), but this is what must be proved.
The goal is to show that g(x) = f (x) − l(x) < 0 on ( a, b),
since this will show that f (x) < l(x) on ( a, b). Since f and
l are both continuous on [ a, b] then so is g. Hence g has a
global maximum somewhere in [ a, b], by the Extreme V alue",ElementaryCalculus.pdf
"l are both continuous on [ a, b] then so is g. Hence g has a
global maximum somewhere in [ a, b], by the Extreme V alue
Theorem. Suppose the global maximum occurs at an interior po int x = c, i.e. for some c in the
open interval ( a, b). Then g′(c) = 0 and g′′(c) = f ′′(c) − l′′(c) = f ′′(c) > 0, since l(x) is a line and
hence has a second derivative of 0 for all x. Then by the Second Derivative Test g has a local
minimum at x = c, which contradicts g having a global maximum at x = c. Thus, the global
maximum of g cannot occur at an interior point, so it must occur at one of th e end points x = a
or x = b. In other words, either g(x) < g(a) or g(x) < g(b) for all x in ( a, b). But f (a) = l(a)
and f (b) = l(b), so g(a) = 0 = g(b). Hence, g(x) < 0 for all x in ( a, b), i.e. f (x) < l(x) for all x in
(a, b). ✓
Points where the concavity of a function changes have a speci al name:
A function f has an inflection point at x = c if the concavity of f changes around x = c.",ElementaryCalculus.pdf
"A function f has an inflection point at x = c if the concavity of f changes around x = c.
That is, the function goes from concave up to concave down, or vice versa.
Note that to be an inflection point it does not suffice for the se cond derivative to be 0 at
that point; the second derivative must change sign around that point, either from positive to
negative or from negative to positive. For example, f (x) = x3 has an inflection point at x = 0,
since f ′′(x) = 6x < 0 for x < 0 and f ′′(x) = 6x > 0 for x > 0, i.e. f ′′(x) changes sign around x = 0
(and of course f ′′(0) = 0). But for f (x) = x4, x = 0 is not an inflection point even though f ′′(0) = 0,
since f ′′(x) = 12x2 ≥ 0 is always nonnegative. That is, f (x) = x4 is always concave up. Figure
4.2.3 below shows the difference:",ElementaryCalculus.pdf
"104 Chapter 4 • Applications of Derivatives §4.2
y
x
0
(a) f (x) = x3: inflection point at x = 0
y
x
0
(b) f (x) = x4 : non-inflection point at x = 0
Figure 4.2.3 Inflection vs non-inflection point at x = 0 with f ′′(0) = 0
Figure 4.2.3(b) shows that a point where the second derivati ve is 0 is a possible inflection point,
but you still must check that the second derivative changes s ign around that point. Using
local minima and maxima, concavity and inflection points, an d where a function increases or
decreases, you can sketch the graph of a function.
Example 4.8
Sketch the graph of f (x) = x3 − 6x2 + 9x + 1. Find all local maxima and minima, inflection points,
where the function is increasing or decreasing, and where th e function is concave up or concave down.
Solution: Since f ′(x) = 3x2 − 12x + 9 = 3 ( x − 1) ( x − 3) then x = 1 and x = 3 are the only critical points.
And since f ′′(x) = 6x − 12 then f ′′(1) = −6 < 0 and f ′′(3) = 6 > 0. So by the Second Derivative Test, f",ElementaryCalculus.pdf
"And since f ′′(x) = 6x − 12 then f ′′(1) = −6 < 0 and f ′′(3) = 6 > 0. So by the Second Derivative Test, f
has a local maximum at x = 1 and a local minimum at x = 3. Since f ′′(x) = 6x − 12 < 0 for x < 2 and
f ′′(x) = 6x −12 > 0 for x > 2, then x = 2 is an inflection point, and f is concave down for x < 2 and concave
up for x > 2. The table below shows where f is increasing and decreasing, based on the sign of f ′:
x values 3 ( x − 1) (x − 3) f ′(x) direction
x < 1 − − + f is increasing
1 < x < 3 + − − f is decreasing
x > 3 + + + f is increasing
The graph is shown below:
0
1
2
3
4
5
6
0 0 .5 1 1 .5 2 2 .5 3 3 .5 4
y
x",ElementaryCalculus.pdf
"Curve Sketching • Section 4.2 105
Example 4.9
Sketch the graph of f (x) = −x
1 + x2 . Find all local maxima and minima, inflection points, where t he func-
tion is increasing or decreasing, and where the function is c oncave up or concave down. Also indicate
any asymptotes.
Solution: Since f ′(x) = x2 −1
(1+x2 )2 then x = 1 and x = −1 are the only critical points. And since f ′′(x) = 2x (3−x2)
(1+x2)3
then f ′′(1) = 1
2 > 0 and f ′′(−1) = −1
2 < 0. So by the Second Derivative Test, f has a local minimum at
x = 1 and a local maximum at x = −1. Since f ′′(x) > 0 for x < −
∝ra⌈〉⌋allow
3, f ′′(x) < 0 for −
∝ra⌈〉⌋allow
3 < x < 0, f ′′(x) > 0 for
0 < x <
∝ra⌈〉⌋allow
3, and f ′′(x) < 0 for x >
∝ra⌈〉⌋allow
3, then x = 0, ±
∝ra⌈〉⌋allow
3 are inflection points, f is concave up for x < −
∝ra⌈〉⌋allow
3
and for 0 < x <
∝ra⌈〉⌋allow
3, and f is concave down for −
∝ra⌈〉⌋allow
3 < x < 0 and for x >
∝ra⌈〉⌋allow
3. Since f ′(x) > 0 for x < −1",ElementaryCalculus.pdf
"∝ra⌈〉⌋allow
3
and for 0 < x <
∝ra⌈〉⌋allow
3, and f is concave down for −
∝ra⌈〉⌋allow
3 < x < 0 and for x >
∝ra⌈〉⌋allow
3. Since f ′(x) > 0 for x < −1
and x > 1 then f is increasing for |x | >1. And f ′(x) < 0 for −1 < x < 1 means f is decreasing for |x | <1.
Finally , since limx→∞ f (x) = 0 and limx→−∞ f (x) = 0 then the x-axis ( y = 0) is a horizontal asymptote. There are
no vertical asymptotes (why?).
The graph is shown below:
−1
−0.5
0
0.5
1
−4 −3 −2 −1 0 1 2 3 4
y
x
If the Second Derivative Test fails then one alternative is t he following test:
First Derivative T est : For a continuous function f on an interval I, let x = c be a number
in I such that f (c) is defined, and either f ′(c) = 0 or f ′(c) does not exist. Then:
(a) If f ′(x) changes from negative to positive around x = c then f has a local minimum at
x = c.
(b) If f ′(x) changes from positive to negative around x = c then f has a local maximum at
x = c.",ElementaryCalculus.pdf
"x = c.
(b) If f ′(x) changes from positive to negative around x = c then f has a local maximum at
x = c.
This test merely states the obvious: a function decreases th en increases around a minimum,
and it increases then decreases around a maximum.",ElementaryCalculus.pdf
"106 Chapter 4 • Applications of Derivatives §4.2
Example 4.10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
−2−1.5−1−0.5 0 0 .5 1 1 .5 2
y
x
Sketch the graph of f (x) = x2/3 .
Solution: Clearly f (x) is continuous for all x, including x =
0 (since f (0) = 0), but f ′(x) = 2
3 3∝ra⌈〉⌋allowx is not defined at x = 0.
Since f ′(x) changes from negative to positive around x = 0
( f ′(x) < 0 when x < 0 and f ′(x) > 0 when x > 0), then by the
First Derivative Test f has a local minimum at x = 0. Since
f ′′(x) = − 2
9 x4/3 < 0 for all x ̸=0, then f is always concave
down. There are no vertical or horizontal asymptotes. The
graph is shown on the right.
Note that the Second Derivative Test could not be used for thi s function, since f ′(x) ̸=0 for all x (notice
also that f ′′(x) is not defined at x = 0).
A more complete alternative to the Second Derivative Test is the following: 4
Nth Derivative T est : A non-constant function f with continuous derivatives of all orders",ElementaryCalculus.pdf
"Nth Derivative T est : A non-constant function f with continuous derivatives of all orders
up to and including n > 1 at x = c has either a local minimum, local maximum or inflection
point at x = c if and only if
f (k)(c) = 0 for k = 1, 2, . . ., n − 1 and f (n)(c) ̸=0
(i.e. the nth derivative is the first nonzero derivative at x = c). If so, then:
(a) If n > 1 is even and f (n)(c) > 0 then f has a local minimum at x = c.
(b) If n > 1 is even and f (n)(c) < 0 then f has a local maximum at x = c.
(c) If n > 1 is odd then f has an inflection point at x = c.
Note that the Second Derivative Test is the special case wher e n = 2 in the N th Derivative
Test. Though this test gives necessary and sufficient condit ions for a local maximum, local
minimum, and inflection point, calculating the first n derivatives can be complicated if n is
large and the given function is not simple.
Example 4.11",ElementaryCalculus.pdf
"minimum, and inflection point, calculating the first n derivatives can be complicated if n is
large and the given function is not simple.
Example 4.11
The Second Derivative Test fails for f (x) = x4 at the critical point x = 0, since f ′′(0) = 0. But the first
4 derivatives of f (x) = x4 are f ′(x) = 4x3 , f ′′(x) = 12x2, f (3) (x) = 24x, and f (4)(x) = 24, which are all
continuous and
f (k)(0) = 0 for k = 1, 2, 3 and f (4) (0) = 24 ̸=0 .
So by the N th Derivative Test, since n = 4 is even and f (4)(0) = 24 > 0 then f (x) = x4 has a local minimum
at x = 0. Note that f (x) ≥ 0 = f (0) for all x, so x = 0 is actually a global minimum for f .
4For a proof, see pp.10-11 in K O O, D., Elements of Optimization , New Y ork: Springer-V erlag, 1977.",ElementaryCalculus.pdf
"Curve Sketching • Section 4.2 107
A common practice in many fields of science and engineering is to combine multiple named
constants (e.g. π) or variables in a function into one variable and then sketch a graph of that
function. The example below illustrates the technique.
Example 4.12
A hydrogen atom has one electron, and the probability of findi ng the electron in the ground state of the
hydrogen atom between radii r and r + dr is D(r) dr, where dr is an infinitesimal change in the radius
r (the distance from the electron to the nucleus), D(r) is the radial probability density function
D(r) = 4
a3
0
r2 e−2r/a0
and a0 ≈ 5.291772 × 10−11 m is the Bohr radius . It is useful to analyze this function in terms of r ≥ 0 in
relation to the Bohr radius a0 (e.g. r = 0.5a0 , a0 , 2 a0 , 3 a0 ). To do this, let x = r
a0
, so that
D(r) = 4
a0
( r
a0
) 2
e−2
(
r
a0
)
⇒ a0 D(x) = 4x2 e−2x
and then sketch the graph of a0 D(x), which is shown below:
0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4 5 6
a0 D
(
r
a0
)",ElementaryCalculus.pdf
"e−2
(
r
a0
)
⇒ a0 D(x) = 4x2 e−2x
and then sketch the graph of a0 D(x), which is shown below:
0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4 5 6
a0 D
(
r
a0
)
r
a0
From the graph it looks like x = 1 (i.e. r = a0 ) is a local (and global) maximum, so that the electron is
most likely to be found near r = a0 , and the probability drops off dramatically past a distance r = 3a0 .
In the exercises you will be asked to show that r = a0 is indeed a local maximum and that the inflection
points are r =
(
1 ± 1∝ra⌈〉⌋allow
2
)
a0 .
Note that the right side of the formula a0 D(x) = 4x2 e−2x does not involve a0 , which was multiplied
over to the left side. In general that is the strategy when dea ling with these sorts of functions where
variables and constants are combined. In this case the stray constant a0 can be multiplied with D since
that will not affect the location of critical and inflection p oints, nor fundamentally alter the general
shape of the graph.",ElementaryCalculus.pdf
"108 Chapter 4 • Applications of Derivatives §4.2
Example 4.13
For a single particle with two states—energy 0 and energy ǫ—in thermal contact with a reservoir at
temperature τ, the average energy U and heat capacity CV are given by
U = ǫ e−ǫ/τ
1 + e−ǫ/τ and CV = kB
(ǫ
τ
) 2 eǫ/τ
(
1 + eǫ/τ) 2
where kB ≈ 1.38065 × 10−23 J/K is the Boltzmann constant . The graph below shows both quantities as
functions of τ/ǫ (not ǫ/τ, as you might expect). See Exercise 9.
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0 1 2 3 4 5
τ/ǫ
Average Energy U /ǫ vs Heat Capacity CV /kB
U /ǫ = e−ǫ/τ
1 + e−ǫ/τ
CV /kB =
(ǫ
τ
) 2 eǫ/τ
(
1 + eǫ/τ) 2
Exercises
A
For Exercises 1-8 sketch the graph of the given function. Fin d all local maxima and minima, inflection
points, where the function is increasing or decreasing, whe re the function is concave up or concave
down, and indicate any asymptotes.
1. f (x) = x3 − 3x 2. f (x) = x3 − 3x2 + 1 3. f (x) = x e−x 4. f (x) = x2 e−x2
5. f (x) = 1
1 + x2 6. f (x) = x2",ElementaryCalculus.pdf
"down, and indicate any asymptotes.
1. f (x) = x3 − 3x 2. f (x) = x3 − 3x2 + 1 3. f (x) = x e−x 4. f (x) = x2 e−x2
5. f (x) = 1
1 + x2 6. f (x) = x2
(x − 1)2 7. f (x) = e−x − e−2x
2 8. f (x) = e−x sin x
9. Write U /ǫ and CV /kB from Example 4.13 as functions of x = τ/ǫ. Y ou do not need to sketch the graphs.
10. Show that the function D(r) = 4
a3
0
r2 e−2r/a0 from Example 4.12 has a local maximum at r = a0 and
inflection points at r =
(
1 ± 1∝ra⌈〉⌋allow
2
)
a0 .
11. Sketch the graph of Kratzer’s molecular potential V (r) = −2D
(
a
r − 1
2
a2
r2
)
as a function of x = r
a , with
a > 0 and D > 0 as constants.
12. Sketch the graph of f (K ) = 2N
∝ra⌈〉⌋allow
K e − K
kT∝ra⌈〉⌋allowπ(kT )3/2 as a function of x = K
kT , with N, k and T as positive constants.
13. Prove part (b) of the Concavity Theorem.",ElementaryCalculus.pdf
"Numerical Approximation of Roots of Functions • Section 4.3 109
4.3 Numerical Approximation of Roots of Functions
When finding critical points of a function f , you encounter the problem of solving the equation
f ′(x) = 0. The examples and exercises so far were set up carefully so t hat solutions to that
equation could be found in a simple closed form. But in practi ce this will not always be the
case—in fact it is almost never the case. For example, finding the critical points of the func tion
f (x) = sin x − x2
2 entails solving the equation f ′(x) = cos x − x = 0, for which there is no solution
in a closed-form expression.
What should you do in such a situation? 5 One possibility is to use the bisection method
mentioned in Section 3.3. In fact, in Example 3.25 the soluti on to the equation cos x = x (i.e.
cos x − x = 0) was shown to exist in the interval [0 , 1], and then a demonstration of the
bisection method was given to find that solution.",ElementaryCalculus.pdf
"cos x − x = 0) was shown to exist in the interval [0 , 1], and then a demonstration of the
bisection method was given to find that solution.
The bisection method is one of many numerical methods for finding roots of a function (i.e.
where the function is zero). Finding the critical points of a function means finding the roots
of its derivative. Though the bisection method could be used for that purpose, convergence to
each root is usually slow . A far more efficient method is Newton’s method 6, whose geometric
interpretation is shown in Figure 4.3.1 below .
y
x
0
y = f (x)
¯x x0
(x0, f (x0 ))
x1
(x1, f (x1 ))
x2
Figure 4.3.1 Newton’s method for finding a root ¯x of f (x)
The idea behind Newton’s method is simple: to find a root ¯x of a function f , choose an initial
guess x0 and then go up—or down—to the curve y = f (x) and draw the tangent line to the curve
at the point ( x0, f (x0)). Let x1 be where that tangent line intersects the x-axis, as shown above;",ElementaryCalculus.pdf
"at the point ( x0, f (x0)). Let x1 be where that tangent line intersects the x-axis, as shown above;
repeat this procedure on x1 to get the next number x2, repeat on x2 to get x3, and so on. The
resulting sequence of numbers x0, x1, x2, x3, . . . , will approach the root ¯x. Convergence under
certain conditions can be proved. 7
5Note: To “just give up”—as suggested semi-seriously by some students I have had—is not an option.
6Sometimes called the Newton-Raphson method .
7See pp.58-62 in S A AT Y, T .L. A N D J. B R A M, Nonlinear Mathematics , New Y ork: McGraw-Hill, Inc., 1964.",ElementaryCalculus.pdf
"110 Chapter 4 • Applications of Derivatives §4.3
The general formula for the number xn obtained after n ≥ 1 iterations in Newton’s method
can be determined by considering the formula for x1. First, the tangent line to y = f (x) at the
point ( x0, f (x0)) has slope f ′(x0), so the equation of the line is
y − f (x0) = f ′(x0) ( x − x0) . (4.1)
The point ( x1, 0) is (by design) also on that line, so that
0 − f (x0) = f ′(x0) ( x1 − x0) ⇒ x1 = x0 − f (x0)
f ′(x0)
provided that f ′(x0) ̸=0. The general formula for xn is given by the following algorithm:
Newton’s method : For an initial guess x0, the numbers xn for n ≥ 1 are computed itera-
tively as:
xn = xn−1 − f (xn−1)
f ′(xn−1) for n = 1, 2, 3, . . .
That is, each “next” number xn depends on the previous number xn−1. The algorithm
terminates whenever f ′(xn ) = 0, or when the desired accuracy is reached. If f ′(xn ) = 0 for
some n ≥ 0, then you could start over with a different initial guess x0.",ElementaryCalculus.pdf
"some n ≥ 0, then you could start over with a different initial guess x0.
To implement this algorithm in a programming language (for w hich Newton’s method is
well-suited), the following language-independent pseudocode can be used as a guide:
Algorithm pseudocode for Newton’s method
NEWT ON ’S METHOD
1 N ← NUMBER -OF -I TERATI ONS ✄ User supplies this value
2 x ← I NI TI AL -GUESS ✄ User supplies this value
3 for n ← 1 to N
4 do
5 if f ′(x) ̸=0
6 then
7 x ← x − f (x)
f ′(x)
8 print x
9 else
10 error “division by zero”",ElementaryCalculus.pdf
"Numerical Approximation of Roots of Functions • Section 4.3 111
Example 4.14
Use Newton’s method to find the root of f (x) = cos x − x.
Solution: Since the root is already known to be in the interval [0 , 1], choose x0 = 1 as the initial guess.
The numbers xn for n ≥ 1 can be computed with a hand-held scientific calculator , but the process is
tedious and error-prone. Using a computer is far more efficie nt and allows more flexibility .
For example, the algorithm is easily implemented in the Java programming language. Save this code
in a plain text file as newton.java:
Listing 4.1 Newton’s method in Java ( newton.java)
1 public class newton {
2 public static void main(String[] args) {
3 int N = Integer.parseInt(args[0]); //Number of iterations
4 double x = 1.0; //initial guess
5 System.out.println(""n=0: "" + x);
6 for (int i = 1; i <= N; i++) {
7 x = x - f(x)/derivf(x);
8 System.out.println(""n="" + i + "": "" + x);
9 }
10 }
11
12 //Define the function f(x)
13 public static double f(double x) {",ElementaryCalculus.pdf
"7 x = x - f(x)/derivf(x);
8 System.out.println(""n="" + i + "": "" + x);
9 }
10 }
11
12 //Define the function f(x)
13 public static double f(double x) {
14 return Math.cos(x) - x;
15 }
16
17 //Define the derivative f’(x)
18 public static double derivf(double x) {
19 return -Math.sin(x) - 1.0;
20 }
21 }
Though knowledge of Java would help, it should not be that dif ficult to figure out what the above code
is doing. The number of iterations N is passed as a command-line parameter to the program, and xn
is computed and printed for n = 0, 1, 2, . . . , N. Note that the derivative of f (x) is “hard-coded” into the
program.8 There is also no error checking for the derivative being zero at any xn . The program would
simply halt on a division by zero error .
Compile the code, then run the program with 10 iterations:
javac newton.java
java newton 10
The output is shown below:
8There are some programming language libraries for calculat ing derivatives of functions “on the fly ,” i.e. dynami-",ElementaryCalculus.pdf
"The output is shown below:
8There are some programming language libraries for calculat ing derivatives of functions “on the fly ,” i.e. dynami-
cally . For example, the GNU libmatheval C/Fortran library can perform such symbolic operations. It is available
at http://www.gnu.org/software/libmatheval/",ElementaryCalculus.pdf
"112 Chapter 4 • Applications of Derivatives §4.3
n=0: 1.0
n=1: 0.7503638678402439
n=2: 0.7391128909113617
n=3: 0.739085133385284
n=4: 0.7390851332151607
n=5: 0.7390851332151607
n=6: 0.7390851332151607
n=7: 0.7390851332151607
n=8: 0.7390851332151607
n=9: 0.7390851332151607
n=10: 0.7390851332151607
Note that the solution ¯x = 0.7390851332151607 was found after only 4 iterations; the num bers xn repeat
for n ≥ 5. This is much faster than the bisection method.
Another root-finding numerical method similar to Newton’s m ethod is the secant method ,
whose geometric interpretation is shown in Figure 4.3.2 bel ow:
y
x
0
y = f (x)
¯x x0
(x0, f (x0 ))
x1
(x1, f (x1 ))
x2
Figure 4.3.2 Secant method for finding a root ¯x of f (x)
The idea behind the secant method is simple: to find a root ¯x of a function f , choose two initial
guesses x0 and x1, then go up—or down—to the curve y = f (x) and draw the secant line through",ElementaryCalculus.pdf
"guesses x0 and x1, then go up—or down—to the curve y = f (x) and draw the secant line through
the points ( x0, f (x0)) and ( x1, f (x1)) on the curve. Let x2 be where that secant line intersects
the x-axis, as shown above; repeat this procedure on x1 and x2 to get the next number x3, and
keep repeating in this way . The resulting sequence of number s x0, x1, x2, x3, . . . , will approach
the root ¯x, under the right conditions. 9
9See pp.227-229 in D A H L Q U I S T, G. A N D Å. B J Ö R C K, Numerical Methods , Englewood Cliffs, NJ: Prentice-Hall,
Inc., 1974.",ElementaryCalculus.pdf
"Numerical Approximation of Roots of Functions • Section 4.3 113
Since the secant line through ( x0, f (x0)) and ( x1, f (x1)) has slope f (x1 )−f (x0 )
x1 −x0
, the equation of
that secant line is:
y − f (x1) = f (x1) − f (x0)
x1 − x0
(x − x1)
The point ( x2, 0) is on that line, so that
0 − f (x1) = f (x1) − f (x0)
x1 − x0
(x2 − x1) ⇒ x2 = x1 − (x1 − x0) · f (x1)
f (x1) − f (x0)
provided that x1 ̸=x0. The general formula for xn is given by the following algorithm:
Secant method : For two initial guesses x0 and x1, the numbers xn for n ≥ 2 are computed
iteratively as:
xn = xn−1 − (xn−1 − xn−2) · f (xn−1)
f (xn−1) − f (xn−2) for n = 2, 3, 4, . . . (4.2)
That is, each “next” number xn depends on the previous two numbers xn−1 and xn−2. The
algorithm terminates whenever xn = xn−1 (i.e. the numbers start repeating) or when the
desired accuracy is reached.
Algorithm pseudocode for the secant method
SECANT METHOD
1 N ← NUMBER -OF -I TERATI ONS ✄ User supplies this value",ElementaryCalculus.pdf
"desired accuracy is reached.
Algorithm pseudocode for the secant method
SECANT METHOD
1 N ← NUMBER -OF -I TERATI ONS ✄ User supplies this value
2 x0 ← FI RST-I NI TI AL -GUESS ✄ User supplies this value
3 x1 ← SECOND -I NI TI AL -GUESS ✄ User supplies this value
4 f0 ← f (x0)
5 for n ← 1 to N
6 do
7 f1 ← f (x1)
8 if f0 ̸=f1
9 then
10 x ← x1 − (x1 − x0) · f1
f1 − f0
11 print x
12 x0 ← x1
13 f0 ← f1 ✄ Re-use f1 as f0 in the next iteration
14 x1 ← x
15 else
16 error “division by zero”",ElementaryCalculus.pdf
"114 Chapter 4 • Applications of Derivatives §4.3
One difference you might have noticed between the secant met hod and Newton’s method is
that the secant method does not use derivatives. The secant m ethod replaces the derivative
in Newton’s method with the slope of a secant line which appro ximates the derivative (recall
how the tangent line is the limit of slopes of secant lines). T his might seem like a drawback,
perhaps giving a “less accurate” slope than the tangent line , but in practice it is not really a
problem. In fact, in many cases the secant method is preferab le, since computing derivatives
can often be quite complicated.
Example 4.15
Use the secant method to find the root of f (x) = cos x − x.
Solution: Since the root is already known to be in the interval [0 , 1], choose x0 = 0 and x1 = 1 as the
two initial guesses. The algorithm is easily implemented in the Java programming language. Save this
code in a plain text file as secant.java:",ElementaryCalculus.pdf
"two initial guesses. The algorithm is easily implemented in the Java programming language. Save this
code in a plain text file as secant.java:
Listing 4.2 Secant method in Java ( secant.java)
1 import java.math.*;
2 public class secant {
3 public static void main(String[] args) {
4 int N = Integer.parseInt(args[0]); //Number of iterations
5 double x0 = 0.0; //first initial guess
6 double x1 = 1.0; //second initial guess
7 double f0 = f(x0);
8 double f1;
9 double x = 0.0;
10 for (int i = 2; i <= N; i++) {
11 f1 = f(x1);
12 x = x1 - (x1 - x0)*f1/(f1 - f0);
13 x0 = x1;
14 f0 = f1; //Re-use f1 as f0 in the next iteration
15 x1 = x;
16 System.out.println(""n="" + i + "": "" + x);
17 }
18 }
19
20 //Define the function f(x)
21 public static double f(double x) {
22 return Math.cos(x) - x;
23 }
24 }
Compile the code, then run the program:
javac secant.java
java secant 10
The output is shown below:",ElementaryCalculus.pdf
"Numerical Approximation of Roots of Functions • Section 4.3 115
n=2: 0.6850733573260451
n=3: 0.736298997613654
n=4: 0.7391193619116293
n=5: 0.7390851121274639
n=6: 0.7390851332150012
n=7: 0.7390851332151607
n=8: 0.7390851332151607
n=9: NaN
n=10: NaN
Notice that the root was found after 6 iterations ( n = 7). The undefined number NaN (which stands
for “Not a Number”) was returned starting with the eighth ite ration ( n = 9) because x7 = x8 , so that
f (x8) − f (x7) = 0, causing a division by zero error in the term
x9 = x8 − (x8 − x7 ) · f (x8)
f (x8) − f (x7) .
For the function f (x) = cos x − x, the table below summarizes the results of 10 iterations of
the bisection method, Newton’s method and the secant method :
Term Bisection Newton Secant
x0 0.5 1.0 0.0
x1 0.75 0.7503638678402439 1.0
x2 0.625 0.7391128909113617 0.6850733573260451
x3 0.6875 0.739085133385284 0.736298997613654
x4 0.71875 0.7390851332151607 0.7391193619116293
x5 0.734375 0.7390851332151607 0.7390851121274639",ElementaryCalculus.pdf
"x3 0.6875 0.739085133385284 0.736298997613654
x4 0.71875 0.7390851332151607 0.7391193619116293
x5 0.734375 0.7390851332151607 0.7390851121274639
x6 0.7421875 0.7390851332151607 0.7390851332150012
x7 0.73828125 0.7390851332151607 0.7390851332151607
x8 0.740234375 0.7390851332151607 0.7390851332151607
x9 0.7392578125 0.7390851332151607 undefined
x10 0.73876953125 0.7390851332151607 undefined
Newton’s method found the root after 4 iterations, while the secant method needed 6 iterations.
After 10 iterations the bisection method had yet to find the ro ot to the same level of precision
as the other methods—it would take 52 iterations (that is, x52 ) to achieve similar accuracy to
16 decimal places.
In general Newton’s method requires fewer iterations to find a root than the secant method
does, but this does not necessarily mean that it will always b e faster . Depending on the com-
plexity of the function and its derivative, Newton’s method could involve more “expensive”",ElementaryCalculus.pdf
"plexity of the function and its derivative, Newton’s method could involve more “expensive”
operations (i.e. computing values, as opposed to assigning values) than the secant method, so
that the few more iterations possibly required by the secant method are made up for by fewer
total computations.",ElementaryCalculus.pdf
"116 Chapter 4 • Applications of Derivatives §4.3
To see this, notice that Newton’s method always requires tha t both f (xn−1) and f ′(xn−1) be
computed for the n th term xn in the sequence. The secant method needs f (xn−1) and f (xn−2)
for the n th term, but a good programmer would save the value of f (xn−1) so that it could be
re-used (and hence not re-computed) as f (xn−2) in the next iteration, resulting in potentially
fewer total computations for the secant method.
There are occasional pitfalls in using Newton’s method. For example, if f ′(xn ) = 0 for some
n ≥ 1 then Newton’s method fails, due to division by 0 in the algor ithm. The geometric reason
is clear: the tangent line to the curve at that point would be p arallel to the x-axis and hence
would not intersect it (assuming f (xn ) ̸=0). There would be no “next number” xn+1 in the
iteration! See Figure 4.3.3(a) below .
y
x
¯x xn xn−1
f ′(xn ) = 0
y = f (x)
(a) f ′(xn ) = 0
y
x
¯x x0 x1 x2 . . .
y = f (x)",ElementaryCalculus.pdf
"iteration! See Figure 4.3.3(a) below .
y
x
¯x xn xn−1
f ′(xn ) = 0
y = f (x)
(a) f ′(xn ) = 0
y
x
¯x x0 x1 x2 . . .
y = f (x)
(b) Moving away from a root
Figure 4.3.3 Newton’s method: Potential pitfalls
Another possible problem is that Newton’s method might move you away from the root, i.e.
not get closer , typically by a poor choice of x0. See Figure 4.3.3(b) above. In some extreme
cases, it is possible that Newton’s method simply loops back and forth endlessly between the
same two numbers, as in Figure 4.3.4:
y
x
¯x x0x1
y = f (x)
y
xx0 x1¯x
y = f (x)
Figure 4.3.4 Newton’s method: Infinite loop
In most cases a different choice for the initial guess x0 will fix such problems. Most textbooks
on the subject of numerical analysis discuss these issues. 10
10For example, R A L S T O N, A. A N D P . R A B I N O W I T Z, A First Course in Numerical Analysis , 2nd ed., New Y ork:
McGraw-Hill, Inc., 1978.",ElementaryCalculus.pdf
"Numerical Approximation of Roots of Functions • Section 4.3 117
There are conditions under which Newton’s method is guarant eed to work, and convergence
is fast. Newton’s method has a quadratic rate of convergence, meaning roughly that the error
terms—the differences between approximate roots and the actual r oot—are being squared in
the long term. More precisely , if the numbers xn for n ≥ 0 converge to a root ¯x, then the error
terms ǫn = xn − ¯x for n ≥ 0 satisfy the limit
limn→∞
|ǫn+1 |
|ǫn |2 = C
for some constant C. Squared error terms might sound like a bad thing, but the xn terms are
converging to the root, making the error terms closer to 0 for large n. Squaring a number ǫn
when |ǫn | <1 results in a smaller number , not a larger one.
The numerical methods that you have learned will make it poss ible to sketch the graphs of
many more functions, since finding local minima and maxima in volves finding roots of f ′, and",ElementaryCalculus.pdf
"many more functions, since finding local minima and maxima in volves finding roots of f ′, and
finding inflection points involves finding roots of f ′′. Y ou now know some methods for finding
those roots.
Finally , despite being slower , the bisection method has the nice advantage of always working.
The speed of modern computers makes the difference in algori thmic efficiency negligible in
many cases.
Exercises
A
1. Use Newton’s method to find the root of f (x) = cos x − 2x.
2. Use Newton’s method to find the positive root of f (x) = sin x − x/2.
3. Use Newton’s method to find the solution of the equation e−x = x.
4. Use Newton’s method to find the solution of the equation e−x = x2 .
5. Use Newton’s method and f (x) = x2 − 2 to approximate
∝ra⌈〉⌋allow
2 accurate to six decimal places.
6. Use Newton’s method to approximate
∝ra⌈〉⌋allow
3 accurate to six decimal places.
7. Repeat Exercise 1 with the secant method. 8. Repeat Exercise 3 with the secant method.",ElementaryCalculus.pdf
"∝ra⌈〉⌋allow
3 accurate to six decimal places.
7. Repeat Exercise 1 with the secant method. 8. Repeat Exercise 3 with the secant method.
9. Repeat Exercise 5 with the secant method. 10. Repeat Exercise 6 with the secant method.
11. Cosmic microwave background radiation is described by a function similar to f (x) = x3
−1+ex for x ≥ 0.
Use Newton’s method to find the global maximum of f accurate to four decimal places.
12. Would a different choice for x0 in either graph in Figure 4.3.4 eliminate the infinite loop? E xplain.
13. Draw a graph without any symmetry that has the infinite loop pr oblem for Newton’s method.
14. The time t( p) required for a p-way merge of a file on a single disk drive into memory is
t( p) = pa + m
ln p ,
where p > 1, a is the disk access time, and m is the time to read in one segment of the size of memory .
Find the integer closest to the value of p that minimizes t( p) when a = 24.3ms and m = 3500.3ms.",ElementaryCalculus.pdf
"118 Chapter 4 • Applications of Derivatives §4.4
4.4 The Mean Value Theorem
The difference between instantaneous and average rates of c hange has been discussed in ear-
lier sections. Recall that there is no difference between th e two for linear functions. For
nonlinear functions the average rate of change over an inter val [ a, b] of positive length (i.e.
b − a > 0) will not be the same as the instantaneous rate of change at every point in the inter-
val. However , the following theorem guarantees that they wi ll be the same at some point in
the interval:
Mean V alue Theorem: Let a and b be real numbers such that a < b, and suppose that f
is a function such that
(a) f is continuous on [ a, b], and
(b) f is differentiable on ( a, b).
Then there is at least one number c in the interval ( a, b) such that
f ′(c) = f (b) − f (a)
b − a . (4.3)
Figure 4.4.1 below shows the geometric interpretation of th e theorem:
y
x
a c b
y = f (x)
f (b)
f (a)
(b, f (b))
(a, f (a))
slope = f ′(c)",ElementaryCalculus.pdf
"Figure 4.4.1 below shows the geometric interpretation of th e theorem:
y
x
a c b
y = f (x)
f (b)
f (a)
(b, f (b))
(a, f (a))
slope = f ′(c)
slope = f (b) − f (a)
b − a
Figure 4.4.1 Mean V alue Theorem: parallel tangent line and secant line
The idea is that there is at least one point on the curve y = f (x) where the tangent line will
be parallel to the secant line joining the points ( a, f (a)) and ( b, f (b)). For each c in ( a, b) the
tangent line has slope f ′(c), while the secant line has slope f (b)−f (a)
b−a . The Mean V alue Theorem
says that these two slopes will be equal somewhere in ( a, b).
To prove the Mean V alue Theorem (sometimes called Lagrange’s Theorem ), the following
intermediate result is needed, and is important in its own ri ght:",ElementaryCalculus.pdf
"The Mean Value Theorem • Section 4.4 119
Rolle’s Theorem: Let a and b be real numbers such that a < b, and suppose that f is a
function such that
(a) f is continuous on [ a, b],
(b) f is differentiable on ( a, b), and
(c) f (a) = f (b) = 0.
Then there is at least one number c in the interval ( a, b) such that f ′(c) = 0.
y
x
a c b
y = f (x)
slope = f ′(c)
Figure 4.4.2
Figure 4.4.2 on the right shows the geometric interpre-
tation of the theorem. To prove the theorem, assume that
f is not the constant function f (x) = 0 for all x in [ a, b] (if
it were then Rolle’s Theorem would hold trivially). Then
there must be at least one x0 in ( a, b) such that either
f (x0) > 0 or f (x0) < 0. If f (x0) > 0 then by the Extreme
V alue Theorem f attains a global maximum at some x = c
in the open interval ( a, b), since f is zero at the endpoints
x = a and x = b of the closed interval [ a, b]. Then f ′(c) = 0 since f has a maximum at x = c.",ElementaryCalculus.pdf
"x = a and x = b of the closed interval [ a, b]. Then f ′(c) = 0 since f has a maximum at x = c.
Likewise if f (x0) < 0 then f attains a global minimum at some x = c in ( a, b), and thus again
f ′(c) = 0 . ✓
The Mean V alue Theorem can now be proved by applying Rolle’s T heorem to the function
F (x) = f (x) − f (a) − f (b) − f (a)
b − a (x − a)
where f satisfies the conditions of the Mean V alue Theorem. Basicall y , the function F “tilts”
the graph of f from Figure 4.4.1 to look like the graph in Figure 4.4.2. It is trivial to check
that F (a) = F (b) = 0, and F is continuous on [ a, b] and differentiable on ( a, b) since f is. Thus,
by Rolle’s Theorem, F ′(c) = 0 for some c in ( a, b). However ,
F ′(x) = f ′(x) − f (b) − f (a)
b − a
and so
F ′(c) = f ′(c) − f (b) − f (a)
b − a = 0 ⇒ f ′(c) = f (b) − f (a)
b − a ✓
Note that both the Mean V alue Theorem and Rolle’s Theorem are purely existence theorems;",ElementaryCalculus.pdf
"b − a = 0 ⇒ f ′(c) = f (b) − f (a)
b − a ✓
Note that both the Mean V alue Theorem and Rolle’s Theorem are purely existence theorems;
they tell you only that a certain number exists. The task of finding the numbers is left to
you. For the Mean V alue Theorem that task involves solving th e equation f ′(x) = f (b)−f (a)
b−a (or
f ′(x) = 0 for Rolle’s Theorem). The numerical root-finding methods f rom Section 4.3 could come
in handy , since obtaining closed-form solutions might be im possible. For that reason, the Mean
V alue Theorem is more useful for theoretical purposes. One s uch application is the following
important result:",ElementaryCalculus.pdf
"120 Chapter 4 • Applications of Derivatives §4.4
If f is a differentiable function on an interval I such that f ′(x) = 0 for all x in I, then f is a
constant function on I.
Note that I can be any interval, even the entire real line ( −∞, ∞). It is already known that
f = constant ⇒ f ′ = 0; the above result says that the converse is true. The proof i s by con-
tradiction: assume that f is not a constant function and show this contradicts the Mean V alue
Theorem. If f is not constant then there exist numbers a < b in I such that f (a) ̸=f (b). How-
ever , by the Mean V alue Theorem there must exist a number c in the interval ( a, b) such that
f ′(c) = f (b) − f (a)
b − a .
Since the derivative of f is 0 everywhere in I, then f ′(c) = 0 and so
f (b) − f (a)
b − a = 0 ⇒ f (b) − f (a) = 0 ⇒ f (a) = f (b) ,
a contradiction of f (a) ̸=f (b). Thus, f must be a constant function. ✓
Another theoretical result can be proved with the Mean V alue Theorem:",ElementaryCalculus.pdf
"a contradiction of f (a) ̸=f (b). Thus, f must be a constant function. ✓
Another theoretical result can be proved with the Mean V alue Theorem:
Let f be a differentiable function on an interval I. Then:
(a) If f ′ > 0 on I then f is increasing on I.
(b) If f ′ < 0 on I then f is decreasing on I.
To prove part (a), assume that f ′(x) > 0 for all x in I, and choose arbitrary numbers a and b
in I with a < b. To prove that f is increasing on I it suffices to show that f (a) < f (b). By the
Mean V alue Theorem there is a number c in ( a, b) (and hence in I) such that
f (b) − f (a)
b − a = f ′(c) , and so
f (b) − f (a) = (b − a) f ′(c) > 0
since b − a > 0 and f ′(c) > 0. Thus, f (b) > f (a), and so f is increasing on I. ✓
The proof of part (b) is similar and is left as an exercise. Y ou might wonder why such a proof
is necessary . After all, an intuitive explanation was provi ded in Section 1.2 for why positive",ElementaryCalculus.pdf
"is necessary . After all, an intuitive explanation was provi ded in Section 1.2 for why positive
or negative derivatives imply that a function is increasing or decreasing, respectively . That
knowledge has been assumed and used in the subsequent sectio ns. Intuitive so-called “hand-
waving” explanations, in fact, often yield more insight tha n a “formal” proof, such as the one
above. However , it is good to know that such intuition has a so lid basis and can be proved, if
needed.",ElementaryCalculus.pdf
"The Mean Value Theorem • Section 4.4 121
The Mean V alue Theorem can help in proving inequalities, oft en used in the sciences for
establishing upper or lower bounds on a quantity (e.g. worst -case scenario).
Example 4.16
Show that sin x ≤ x for all x ≥ 0.
Solution: The inequality holds trivially for x = 0, since sin 0 = 0 ≤ 0. So assume that x > 0. Then by the
Mean V alue Theorem there is a number c in (0 , x) such that for f (x) = sin x,
f (x) − f (0)
x − 0 = f ′(c) ⇒ sin x − sin 0
x − 0 = cos c
⇒ sin x = x cos c
⇒ sin x ≤ x
since cos c ≤ 1 and x > 0. Note that sin x ≤ x is a sharper inequality than sin x ≤ 1 when 0 < x < 1.
There is a useful alternative form of the Mean V alue Theorem. If a < b then let h = b − a > 0,
so that a +h = b. Then any number c in ( a, b) can be written as c = a +θh for some number θ in
(0, 1). To see this, let c be in ( a, b). Then 0 < c − a < b − a = h and so 0 < c−a
h < 1. Thus, θ = c−a
h
is in (0 , 1) and a + θh = a + (c − a) = c. Hence:",ElementaryCalculus.pdf
"h < 1. Thus, θ = c−a
h
is in (0 , 1) and a + θh = a + (c − a) = c. Hence:
Mean V alue Theorem (alternative form): Let a and h > 0 be real numbers, and suppose
that f is a function such that
(a) f is continuous on [ a, a + h], and
(b) f is differentiable on ( a, a + h).
Then there is a number θ in the interval (0 , 1) such that
f (a + h) − f (a) = h f ′(a + θh) . (4.4)
The Mean V alue Theorem is the special case of g(x) = x in the following generalization:
Extended Mean V alue Theorem: Let a and b be real numbers such that a < b, and
suppose that f and g are functions such that
(a) f and g are continuous on [ a, b],
(b) f and g are differentiable on ( a, b), and
(c) g′(x) ̸=0 for all x in ( a, b).
Then there is at least one number c in the interval ( a, b) such that
f ′(c)
g′(c) = f (b) − f (a)
g(b) − g(a) . (4.5)",ElementaryCalculus.pdf
"122 Chapter 4 • Applications of Derivatives §4.4
The Mean V alue Theorem says that the derivative of a differen tiable function will always
attain one particular value on a closed interval: the functi on’s average rate of change over the
interval. It turns out that the derivative will take on every value between its values at the
endpoints, similar to how the Intermediate V alue Theorem ap plies to continuous functions: 11
Darboux’s Theorem: If f is a differentiable function on a closed interval [ a, b] then its
derivative f ′ attains every value between f ′(a) and f ′(b).
In other words, if f ′(a) < γ < f ′(b) (or f ′(b) < γ < f ′(a)) then there is a number c in ( a, b)
such that f ′(c) = γ. If f ′ were continuous on [ a, b] then the result would follow trivially by
the Intermediate V alue Theorem for continuous functions. W hat is perhaps surprising is that
Darboux’s Theorem holds even for derivatives that are not continuous. This means that a",ElementaryCalculus.pdf
"Darboux’s Theorem holds even for derivatives that are not continuous. This means that a
discontinuous derivative cannot have the type of simple jum p discontinuities that would allow
it to “skip” over intermediate values—the points of discont inuity must be of a more complicated
type. One rough interpretation of Darboux’s Theorem is that even if a derivative is not a
continuous function, it will behave sort of as if it were.
Exercises
A
1. Does Rolle’s Theorem apply to the function f (x) = 1−| x | on the interval [ −1, 1]? If so, find the number
in ( −1, 1) that Rolle’s Theorem guarantees to exist. If not, explain why not.
2. Suppose that two horses run a race starting together and endi ng in a tie. Show that, at some time
during the race, they must have had the same speed.
3. Use the Mean V alue Theorem to show that
⏐
⏐sin A − sin B
⏐
⏐ ≤
⏐
⏐ A − B
⏐
⏐ for all A and B (in radians).
Does
⏐
⏐sin A + sin B
⏐
⏐ ≤
⏐
⏐ A + B
⏐
⏐ for all A and B? Explain.
4. Show that
⏐
⏐cos A − cos B
⏐
⏐ ≤",ElementaryCalculus.pdf
"⏐
⏐ A − B
⏐
⏐ for all A and B (in radians).
Does
⏐
⏐sin A + sin B
⏐
⏐ ≤
⏐
⏐ A + B
⏐
⏐ for all A and B? Explain.
4. Show that
⏐
⏐cos A − cos B
⏐
⏐ ≤
⏐
⏐ A − B
⏐
⏐ for all A and B (in radians). Does
⏐
⏐cos A + cos B
⏐
⏐ ≤
⏐
⏐ A + B
⏐
⏐
for all A and B? Explain.
5. Show that tan x ≥ x for all 0 ≤ x < π
2 .
6. Show that
⏐
⏐tan A − tan B
⏐
⏐ ≥
⏐
⏐ A − B
⏐
⏐ for all A and B (in radians) in
(
− π
2 , π
2
)
. Can the inequality be
extended to all A and B? Explain your answer .
B
7. Use Rolle’s Theorem to show that for all constants a and b with a > 0, f (x) = x3 − ax + b can not have
three positive roots. Also, show that it can not have three ne gative roots.
8. Use the Mean V alue Theorem to show that if f ′ < 0 on an interval I then f is decreasing on I.
11For a proof see pp.16-17 in O S T R O W S K I , A.M., Solution of Equations and Systems of Equations , 2nd ed., New
Y ork: Academic Press Inc., 1966.",ElementaryCalculus.pdf
"The Mean Value Theorem • Section 4.4 123
9. Suppose that f and g are continuous on [ a, b] and differentiable on ( a, b), and that f ′(x) > g′(x) for
all a < x < b. Show that f (b) − g(b) > f (a) − g(a).
10. Prove the Extended Mean V alue Theorem, by applying Rolle’s T heorem to the function
F (x) = f (x) − f (a) − f (b) − f (a)
g(b) − g(a) ( g(x) − g(a)) .
11. Show that ex ≥ 1 + x for all x. (Hint: Consider f (x) = ex − x.)
12. Show that ln (1 + x) < x for all x > 0. 13. Show that tan −1 x < x for all x > 0.
14. Show that for 0 < α ≤ β < π
2 ,
β− α
cos2 α ≤ tan β − tan α ≤ β− α
cos2 β .
15. Show that for 0 < a ≤ b,
b − a
b ≤ ln b
a ≤ b − a
a .
16. Show that for n > 1 and a > b,
nbn−1 (a − b) < an − bn < nan−1 (a − b) .
C
17. Show that
∝ra⌈〉⌋allow
a2 + b < a + b
2a for all positive numbers a and b.
18. Show that f (x) = cos2 x + cos2 (π
3 + x
)
− cos x cos
(π
3 + x
)
is a constant function. What is its value?",ElementaryCalculus.pdf
"2a for all positive numbers a and b.
18. Show that f (x) = cos2 x + cos2 (π
3 + x
)
− cos x cos
(π
3 + x
)
is a constant function. What is its value?
19. Suppose that f (x) is a differentiable function and that f (0) = 0 and f (1) = 1. Show that f ′(x0) = 2x0
for some x0 in the interval (0 , 1).
20. Prove the inequality ⏐
⏐
⏐
⏐
x1 + x2
1 + x1 x2
⏐
⏐
⏐
⏐ < 1 for −1 < x1 , x2 < 1
as follows:
(a) First prove the special case where x1 = x2.
(b) For the case x1 < x2 define
f (x) = x + a
1 + ax
for −1 ≤ x ≤ 1, where −1 < a < 1. Show that f is increasing on [ −1, 1], then use a = x2 and x = x1 .
Note that proving the case x2 < x1 is unnecessary (why?).
This inequality is a generalization of the same inequality f or 0 ≤ x1 , x2 < 1 in the relativistic velocity
addition law from the theory of special relativity: if object 1 has veloci ty v1 relative to a frame of
reference F , and if object 2 has a velocity v2 relative to object 1, so that x1 = v1 /c and x2 = v2/c",ElementaryCalculus.pdf
"reference F , and if object 2 has a velocity v2 relative to object 1, so that x1 = v1 /c and x2 = v2/c
represent the fractions of the speed of light c at which the objects are moving, then the fraction of
the speed of light at which object 2 is moving with respect to F is x = (x1 + x2)/(1 + x1 x2 ). So it should
be true that 0 ≤ x < 1, since nothing can move faster than the speed of light.",ElementaryCalculus.pdf
"CHAPTER 5
The Integral
5.1 The Indefinite Integral
Derivatives appear in many physical phenomena, such as the m otion of objects. Recall, for
example, that given the position function s(t) of an object moving along a straight line at time
t, you could find the velocity v(t) = s′(t) and the acceleration a(t) = v′(t) of the object at time t
by taking derivatives. Suppose the situation were reversed : given the velocity function how
would you find the position function, or given the accelerati on function how would you find the
velocity function?
s(t) v(t) a(t)
d
dt
?
d
dt
?
Figure 5.1.1 Differentiation vs antidifferentiation for motion functi ons
In this case calculating a derivative would not help, since t he reverse process is needed:
instead of differentiation you need a way of performing antidifferentiation, i.e. you would
calculate an antiderivative.
An antiderivative F (x) of a function f (x) is a function whose derivative is f (x). In other
words, F ′(x) = f (x).",ElementaryCalculus.pdf
"calculate an antiderivative.
An antiderivative F (x) of a function f (x) is a function whose derivative is f (x). In other
words, F ′(x) = f (x).
Differentiation is relatively straightforward. Y ou have l earned the derivatives of many
classes of functions (e.g. polynomials, trigonometric fun ctions, exponential and logarithmic
functions), and with the various rules for differentiation you can calculate derivatives of com-
plicated expressions involving those functions (e.g. sums , powers, products, quotients). Antid-
ifferentiation, however , is a different story .
124",ElementaryCalculus.pdf
"The Indefinite Integral • Section 5.1 125
To see some of the issues involved, consider a simple functio n like f (x) = 2x. Of course you
know that d
dx (x2) = 2x, so it seems that F (x) = x2 is the antiderivative of f (x) = 2x. But is it
the only antiderivative of f (x)? No. For example, if F (x) = x2 + 1 then F ′(x) = 2x = f (x), and so
F (x) = x2 + 1 is another antiderivative of f (x) = 2x. Likewise, so is F (x) = x2 + 2. In fact, any
function of the form F (x) = x2 + C, where C is some constant, is an antiderivative of f (x) = 2x.
Another potential issue is that functions of the form F (x) = x2 + C are just the most obvious
antiderivatives of f (x) = 2x. Could there be some other completely different function—o ne that
cannot be simplified into the form x2 +C—whose derivative also turns out to be f (x) = 2x? The
answer , luckily , is no:
Suppose that F (x) and G(x) are antiderivatives of a function f (x). Then F (x) and G(x) differ",ElementaryCalculus.pdf
"answer , luckily , is no:
Suppose that F (x) and G(x) are antiderivatives of a function f (x). Then F (x) and G(x) differ
only by a constant. That is, F (x) = G(x) + C for some constant C.
To prove this, consider the function H(x) = F (x)−G(x), defined for all x in the common domain
I of F and G. Since F ′(x) = G′(x) = f (x), then
H′(x) = F ′(x) − G′(x) = f (x) − f (x) = 0
for all x in I, so H(x) is a constant function on I, as was shown in Section 4.4 on the Mean
V alue Theorem. Thus, there is a constant C such that
H(x) = C ⇒ F (x) − G(x) = C ⇒ F (x) = G(x) + C
for all x in I. ✓
The practical consequence of the above result can be stated a s follows:
To find all antiderivatives of a function, it is necessary only to find one antiderivative and
then add a generic constant to it.
So for the function f (x) = 2x, since F (x) = x2 is one antiderivative then all antiderivatives of
f (x) are of the form F (x) = x2 + C, where C is a generic constant. Thus, functions do not have",ElementaryCalculus.pdf
"f (x) are of the form F (x) = x2 + C, where C is a generic constant. Thus, functions do not have
just one antiderivative but a whole family of antiderivatives, all differing only by a constant.
The following notation makes all this easier to express:
The indefinite integral of a function f (x) is denoted by
∫ 
f (x) dx
and represents the entire family of antiderivatives of f (x).
The large S-shaped symbol before f (x) is called an integral sign .",ElementaryCalculus.pdf
"126 Chapter 5 • The Integral §5.1
Though the indefinite integral
∫ 
f (x) dx represents all antiderivatives of f (x), the integral
can be thought of as a single object or function in its own righ t, whose derivative is f (x):
d
dx
(∫ 
f (x) dx
)
= f (x)
Y ou might be wondering what the integral sign in the indefinit e integral represents, and
why an infinitesimal dx is included. It has to do with what an infinitesimal represent s: an in-
finitesimal “piece” of a quantity . For an antiderivative F (x) of a function f (x), the infinitesimal
(or differential) dF is given by dF = F ′(x) dx = f (x) dx, and so
F (x) =
∫ 
f (x) dx =
∫ 
dF .
The integral sign thus acts as a summation symbol: it sums up t he infinitesimal “pieces” dF
of the function F (x) at each x so that they add up to the entire function F (x). Think of it as
similar to the usual summation symbol Σ used for discrete sums; the integral sign
∫ 
takes the
sum of a continuum of infinitesimal quantities instead.",ElementaryCalculus.pdf
"similar to the usual summation symbol Σ used for discrete sums; the integral sign
∫ 
takes the
sum of a continuum of infinitesimal quantities instead.
Finding (or evaluating) the indefinite integral of a function is called integrating the func-
tion, and integration is antidifferentiation.
Example 5.1
Evaluate
∫ 
0 dx.
Solution: Since the derivative of any constant function is 0, then
∫ 
0 dx = C, where C is a generic
constant.
Note: From now on C will simply be assumed to represent a generic constant, with out having to explic-
itly say so every time.
Example 5.2
Evaluate
∫ 
1 dx.
Solution: Since the derivative of F (x) = x is F ′(x) = 1, then
∫ 
1 dx = x + C.
Example 5.3
Evaluate
∫ 
x dx .
Solution: Since the derivative of F (x) = x2
2 is F ′(x) = x, then
∫ 
x dx = x2
2 + C.
Since d
dx
(
xn+1
n+1
)
= xn for any number n ̸= −1, and d
dx (ln |x|) = 1
x = x−1, then any power of x can
be integrated:",ElementaryCalculus.pdf
"The Indefinite Integral • Section 5.1 127
Power Formula:
∫ 
xn dx =







xn+1
n + 1 + C if n ̸= −1
ln |x| + C if n = −1
The following rules for indefinite integrals are immediate c onsequences of the rules for
derivatives:
Let f and g be functions and let k be a constant. Then:
1.
∫ 
k f (x) dx = k
∫ 
f (x) dx
2.
∫ 
( f (x) + g(x)) dx =
∫ 
f (x) dx +
∫ 
g(x) dx
3.
∫ 
( f (x) − g(x)) dx =
∫ 
f (x) dx −
∫ 
g(x) dx
The above rules are easily proved. For example, the first rule is a simple consequence of the
Constant Multiple Rule for derivatives: if F (x) =
∫ 
f (x) dx, then
d
dx (k F (x)) = k d
dx (F (x)) = k f (x) ⇒
∫ 
k f (x) dx = k F (x) = k
∫ 
f (x) dx . ✓
The other rules are proved similarly and are left as exercise s. Repeated use of the above rules
along with the Power Formula shows that any polynomial can be integrated term by term—in
fact any finite sum of functions can be integrated in that mann er:
For any functions f1, . . . , f n and constants k1, . . ., kn ,
∫",ElementaryCalculus.pdf
"fact any finite sum of functions can be integrated in that mann er:
For any functions f1, . . . , f n and constants k1, . . ., kn ,
∫ 
(k1 f1(x) + ··· +kn f n(x)) dx = k1
∫ 
f1(x) dx + ··· + kn
∫ 
f n(x) dx . (5.1)
Example 5.4
Evaluate
∫ 
(x7 − 3x4) dx.
Solution: Integrate term by term, pulling constant multiple outside t he integral:
∫ 
(x7 − 3x4) dx =
∫ 
x7 dx − 3
∫ 
x4 dx = x8
8 − 3x5
5 + C",ElementaryCalculus.pdf
"128 Chapter 5 • The Integral §5.1
Example 5.5
Evaluate
∫ ∝ra⌈〉⌋allowx dx .
Solution: Use the Power Formula:
∫ ∝ra⌈〉⌋allowx dx =
∫ 
x1/2 dx = x3/2
3/2 + C = 2x3/2
3 + C
Example 5.6
Evaluate
∫ ( 1
x2 + 1
x
)
dx.
Solution: Use the Power Formula and integrate term by term:
∫ ( 1
x2 + 1
x
)
dx =
∫ (
x−2 + 1
x
)
dx = x−1
−1 + ln |x | + C = −1
x + ln |x | + C
The following indefinite integrals are just re-statements o f the corresponding derivative for-
mulas for the six basic trigonometric functions:
∫ 
cos x dx = sin x + C
∫ 
sin x dx = −cos x + C
∫ 
sec2 x dx = tan x + C
∫ 
sec x tan x dx = sec x + C
∫ 
csc x cot x dx = −csc x + C
∫ 
csc2 x dx = −cot x + C
Since d
dx (ex) = ex, then:
∫ 
ex dx = ex + C",ElementaryCalculus.pdf
"The Indefinite Integral • Section 5.1 129
Example 5.7
Evaluate
∫ 
(3 sin x + 4 cos x − 5ex ) dx.
Solution: Integrate term by term:
∫ 
(3 sin x + 4 cos x − 5ex ) dx = 3
∫ 
sin x dx + 4
∫ 
cos x dx − 5
∫ 
ex dx
= −3 cos x + 4 sin x − 5ex + C
Example 5.8
Recall from Section 1.1 the example of an object dropped from a height of 100 ft. Show that the height
s(t) of the object t seconds after being dropped is s(t) = −16t2 + 100, measured in feet.
100 ft
t = 0 sec
s(t)
t > 0
Solution: When the object is dropped at time t = 0 the only force acting on it is
gravity , causing the object to accelerate downward at the kn own constant rate of
32 ft/s 2. The object’s acceleration a(t) at time t is thus a(t) = −32. If v(t) is the
object’s velocity at time t, then v′(t) = a(t), which means that
v(t) =
∫ 
a(t) dt =
∫ 
−32 dt = −32t + C
for some constant C. The constant C here is not generic—it has a specific",ElementaryCalculus.pdf
"v(t) =
∫ 
a(t) dt =
∫ 
−32 dt = −32t + C
for some constant C. The constant C here is not generic—it has a specific
value determined by the initial condition on the velocity: the object was at rest at time t = 0. That is,
v(0) = 0, which means
0 = v(0) = −32(0) + C = C ⇒ v(t) = −32t
for all t ≥ 0. Likewise, since s′(t) = v(t) then
s(t) =
∫ 
v(t) dt =
∫ 
−32t dt = −16t2 + C
for some constant C, determined by the initial condition that the object was 100 ft above the ground at
time t = 0. That is, s(0) = 100, which means
100 = s(0) = −16(0)2 + C = C ⇒ s(t) = −16t2 + 100
for all t ≥ 0. ✓
The formula for s(t) in Example 5.8 can be generalized as follows: denote the obj ect’s initial
position at time t = 0 by s0, let v0 be the object’s initial velocity (positive if thrown upward ,
negative if thrown downward), and let g represent the (positive) constant acceleration due to
gravity . By Newton’s First Law of motion the only accelerati on imparted to the object after",ElementaryCalculus.pdf
"gravity . By Newton’s First Law of motion the only accelerati on imparted to the object after
throwing it is due to gravity:
a(t) = −g ⇒ v(t) =
∫ 
a(t) dt =
∫ 
−g dt = −g t + C
for some constant C: v0 = v(0) = −g(0) + C = C. Thus, v(t) = −g t + v0 for all t ≥ 0, and so
s(t) =
∫ 
v(t) dt =
∫ 
(−g t + v0) dt = −1
2 g t2 + v0 t + C
for some constant C: s0 = s(0) = −1
2 g(0)2 + v0(0) + C = C. To summarize:",ElementaryCalculus.pdf
"130 Chapter 5 • The Integral §5.1
Free fall motion: At time t ≥ 0:
acceleration: a(t) = −g
velocity: v(t) = −g t + v0
position: s(t) = −1
2 g t2 + v0 t + s0
initial conditions: s0 = s(0) , v0 = v(0)
Note that the units are not specified—they just need to be cons istent. In metric units, g = 9.8
m/s2, while g = 32 ft/s 2 in English units.
Thinking of the indefinite integral of a function as the sum of all the infinitesimal “pieces” of
that function—for the purpose of retrieving that function— provides a handy way of integrat-
ing a differential equation to obtain the solution. The key i dea is to transform the differential
equation into an equation of differentials , which has the effect of treating functions as vari-
ables. Some examples will illustrate the technique.
Example 5.9
For any constant k, show that every solution of the differential equation dy
dt = k y is of the form y = A ek t
for some constant A. Y ou can assume that y(t) > 0 for all t.",ElementaryCalculus.pdf
"dt = k y is of the form y = A ek t
for some constant A. Y ou can assume that y(t) > 0 for all t.
Solution: Put the y terms on the left and the t terms on the right, i.e. separate the variables:
dy
y = k dt
Now integrate both sides (notice how the function y is treated as a variable):
∫ dy
y =
∫ 
k dt
ln y + C1 = kt + C2 (C1 and C2 are constants)
ln y = kt + C (combine C1 and C2 into the constant C)
y = ek t+C = ek t · eC = A ek t
where A = eC is a constant. Note that this is the formula for radioactive d ecay from Section 2.3.
Example 5.10
Recall from Section 3.6 the equation of differentials
d P
P + d V
V = d T
T
relating the pressure P , volume V and temperature T of an ideal gas. Integrate that equation to obtain
the original ideal gas law P V = R T , where R is a constant. .",ElementaryCalculus.pdf
"The Indefinite Integral • Section 5.1 131
Solution: Integrating both sides of the equation yields
∫ d P
P +
∫ d V
V =
∫ d T
T
ln P + ln V = ln T + C (C is a constant)
ln (P V ) = ln T + C
P V = eln T+C = eln T · eC = T e C = R T
where R = eC is a constant. ✓
The integration formulas in this section depended on alread y knowing the derivatives of cer-
tain functions and then “working backward” from their deriv atives to obtain the original func-
tions. Without that prior knowledge you would be reduced to g uessing, or perhaps recognizing
a pattern from some derivative you have encountered. A numbe r of integration techniques will
be presented shortly , but there are many indefinite integral s for which no simple closed form
exists (e.g.
∫ 
ex2
dx and
∫ 
sin(x2) dx).
Exercises
A
For Exercises 1-15, evaluate the given indefinite integral.
1.
∫ (
x2 + 5x − 3
)
dx 2.
∫ 
3 cos x dx 3.
∫ 
4ex dx
4.
∫ (
x5 − 8x4 − 3x3 + 1
)
dx 5.
∫ 
5 sin x dx 6.
∫ 3ex
5 dx
7.
∫ 6
x dx 8.
∫ 4
3x dx 9.
∫ (
−2",ElementaryCalculus.pdf
"∫ (
x2 + 5x − 3
)
dx 2.
∫ 
3 cos x dx 3.
∫ 
4ex dx
4.
∫ (
x5 − 8x4 − 3x3 + 1
)
dx 5.
∫ 
5 sin x dx 6.
∫ 3ex
5 dx
7.
∫ 6
x dx 8.
∫ 4
3x dx 9.
∫ (
−2
∝ra⌈〉⌋allow
x
)
dx
10.
∫ 1
3∝ra⌈〉⌋allowx dx 11.
∫ (
x + x4/3
)
dx 12.
∫ 1
3 3∝ra⌈〉⌋allowx dx
13.
∫ 
3 sec x tan x dx 14.
∫ 
5 sec2 x dx 15.
∫ 
7 csc2 x dx
16. Prove the sum and difference rules for indefinite integrals:
∫ 
( f (x) ± g(x)) dx =
∫ 
f (x) dx ±
∫ 
g(x) dx
17. Integrate both sides of the equation
d P
P + dM
M = d T
2T
to obtain the ideal gas continuity relation: P M∝ra⌈〉⌋allow
T
= constant.
18. Use the free fall motion equation for position to show that th e maximum height reached by an
object launched straight up from the ground with an initial v elocity v0 is
v2
0
2 g .",ElementaryCalculus.pdf
"132 Chapter 5 • The Integral §5.2
5.2 The Definite Integral
Recall from the last section that the integral sign in the ind efinite integral
∫ 
f (x) dx
represents a summation of the infinitesimals f (x) dx = dF for an antiderivative F (x) of f (x).
Why is the term “indefinite” used? Because the summation is in definite: the x in f (x) dx is
defined generically , meaning “ x in general,” that is, not for x in a specific range of values. The
same summation over a specific, definite range of values of x, say , over an interval [ a, b], is a
different type of integral:
The definite integral of a function f (x) over an interval [ a, b] is denoted by
∫ b
a
f (x) dx
and represents the sum of the infinitesimals f (x) dx for all x in [ a, b].
An indefinite integral yields a generic function , whereas a definite integral yields either a
number or a specific function. There are many ways to calculate the specific summa tion in a",ElementaryCalculus.pdf
"number or a specific function. There are many ways to calculate the specific summa tion in a
definite integral, one of which is motivated by a geometric in terpretation of the infinitesimal
f (x) dx as the area of a rectangle, as in Figure 5.2.1 below:
y = f (x)
f (x)
f (x + dx) − f (x)
y
x
a b x x + dx
dx
Figure 5.2.1 The infinitesimal f (x) dx as the area of a rectangle
The shaded rectangle in the above picture has height f (x) and width dx, and so its area is
f (x) dx. In fact, it appears that that area is just a little bit smalle r than the area under the
curve y = f (x) and above the x-axis between x and x + dx; there is a small gap between the
curve and the top of the rectangle, accounting for the differ ence in the area. However , the area
of that gap turns out to be zero, as shown below:",ElementaryCalculus.pdf
"The Definite Integral • Section 5.2 133
f (x)
df = f (x + dx) − f (x)
y = f (x)
x x + dx
dx
A
B
C
Figure 5.2.2 Area under the curve y = f (x) over [ x, x + dx]
By the Microstraightness Property , the curve y = f (x) shown in Figure 5.2.1 is a straight line
over the infinitesimal interval [ x, x + dx], as shown in Figure 5.2.2. 1 Thus, the part of the area
between the curve and the x-axis over the interval [ x, x + dx] consists of two parts: the area
f (x) dx of the shaded rectangle and the area of the right triangle △ABC , both of which are
shown in Figure 5.2.2. However , the area of △ABC is zero:
Area of △ABC = 1
2 (base) × (height) = 1
2 (dx)(df ) = 1
2 (dx)( f ′(x) dx) = 1
2 f ′(x)(dx)2 = 0
The function f shown in Figure 5.2.2 is increasing at x, but a similar argument could be made
if f were decreasing at x. Hence, the area between the curve y = f (x) and the x-axis comes
solely from the rectangles with area f (x) dx, as x varies from a to b. The sum of all those",ElementaryCalculus.pdf
"solely from the rectangles with area f (x) dx, as x varies from a to b. The sum of all those
rectangular areas, though, equals the definite integral of f (x) over [ a, b]. The definite integral
can thus be interpreted as an area:
For a function f (x) ≥ 0 over [ a, b], the area under the curve y = f (x) between x = a and
x = b, denoted by A, is given by
A =
∫ b
a
f (x) dx
and represents the area of the region R bounded above by y = f (x), bounded below by the
x-axis, and bounded on the sides by x = a and x = b (with a < b).
1The function f is assumed to be differentiable at x, in this case. If not then the points where f is not differentiable
can be excluded without affecting the integral.",ElementaryCalculus.pdf
"134 Chapter 5 • The Integral §5.2
y = f (x)
y
x
a b
R
Figure 5.2.3 The area A of the region R equals
∫ b
a f (x) dx
In Figure 5.2.3 the area under the curve y = f (x) between x = a and x = b is the area A of
the shaded region R, namely A =
∫ b
a f (x) dx. To calculate that area for a specific function, rect-
angles can again be used, but this time with widths that are sm all positive numbers instead
of infinitesimals. The procedure is as follows:
1. Create a partition P = {x0 < x1 < ··· <xn−1 < xn } of the interval [ a, b] into n ≥ 1 subintervals
[x0, x1], [ x1, x2], . . ., [ xn−1, xn], with x0 = a and xn = b.
2. In each subinterval [ xi−1, xi ] of P pick a number x∗
i , so that xi−1 ≤ x∗
i ≤ xi for i = 1 to n.
3. For i = 1 to n, form a rectangle whose base is the subinterval [ xi−1, xi ] of length ∆xi =
xi − xi−1 > 0 and whose height is f (x∗
i ).
4. Take the sum f (x∗
1 )∆x1 + f (x∗
2 )∆x2 + ··· +f (x∗
n)∆xn of the areas of these rectangles, called a
Riemann sum .
y = f (x)
f (x∗",ElementaryCalculus.pdf
"i ).
4. Take the sum f (x∗
1 )∆x1 + f (x∗
2 )∆x2 + ··· +f (x∗
n)∆xn of the areas of these rectangles, called a
Riemann sum .
y = f (x)
f (x∗
1 ) f (x∗
2 ) f (x∗
n )
y
x
x0a = x∗
1 x1 x∗
2 x2 . . . xn−1 x∗
n xn = b
. . .
∆x1 ∆x2 ∆xn
Figure 5.2.4 A Riemann sum of f (x) over a partition of [ a, b]
5. Take the limit of the Riemann sums as n → ∞, so that the subinterval lengths approach 0.
If the limit exists then that limit is the area A of the region R:
Area A =
∫ b
a
f (x) dx = limn→∞
n∑
i=1
f (x∗
i ) ∆xi (5.2)",ElementaryCalculus.pdf
"The Definite Integral • Section 5.2 135
The limit in Formula (5.2) should be taken over all partition s whose norm—the length of
the largest subinterval—approaches 0. In practice, howeve r , the partitions are usually chosen
so that the subintervals are of equal length, and then simply make those equal lengths smaller
and smaller by dividing the interval [ a, b] into more and more such subintervals. Note that
the points x∗
i in each subinterval can be anywhere in the subinterval—ofte n the midpoint of
the subinterval is chosen, but the left and right endpoints a re also typical choices.
In the above procedure the gaps between the rectangles and th e curve will have areas ap-
proaching 0 as the number n of subintervals grows and the subinterval lengths approach 0.
This is true if the function f is differentiable, and in fact even if f is merely continuous. 2 Thus,
the area under the curve can be defined by the above procedure.",ElementaryCalculus.pdf
"the area under the curve can be defined by the above procedure.
To calculate the area under a curve in this manner , the reader should have some familiarity
with the summation notation in Formula (5.2).
For real numbers a1, a2, . . ., an and an integer n ≥ 1,
n∑
k=1
ak = a1 + a2 + ··· + an
is the sum of a1, . . . , an. The symbol Σ is called the summation sign , which is the Greek
capital letter Sigma.
The following rules for this “Sigma notation” are intuitive ly obvious:
Let a1, a2, . . ., an, and b1, b2, . . ., bn be real numbers, and let c be a constant. Then:
(1)
n∑
k=1
(ak + bk ) =
n∑
k=1
ak +
n∑
k=1
bk
(2)
n∑
k=1
(ak − bk ) =
n∑
k=1
ak −
n∑
k=1
bk
(3)
n∑
k=1
ca k = c
n∑
k=1
ak
(4)
n∑
k=1
ak =
n∑
i=1
a i (i.e. the sum is independent of the summation index letter)
2For a proof and fuller discussion of all this, see Ch.1-2 in K N O P P, M.I., Theory of Area , Chicago: Markham",ElementaryCalculus.pdf
"2For a proof and fuller discussion of all this, see Ch.1-2 in K N O P P, M.I., Theory of Area , Chicago: Markham
Publishing Co., 1969. The book attempts to define precisely w hat an “area” actually means, including that of a
rectangle (showing agreement with the intuitive notion of w idth times height).",ElementaryCalculus.pdf
"136 Chapter 5 • The Integral §5.2
The following summation formulas can be helpful when calcul ating Riemann sums:
Let n ≥ 1 be a positive integer . Then:
(1)
n∑
k=1
1 = n
(2)
n∑
k=1
k = 1 + 2 + ··· + n = n (n + 1)
2
(3)
n∑
k=1
k2 = 12 + 22 + ··· + n2 = n (n + 1) (2n + 1)
6
(4)
n∑
k=1
k3 = 13 + 23 + ··· + n3 = n2 (n + 1)2
4
(5)
n∑
k=1
k4 = 14 + 24 + ··· + n4 = n (n + 1) (6n3 + 9n2 + n − 1)
30
Formula (1) is obvious: add the number 1 a total of n times and the sum is n.
Formula (2) can be proved by induction:
1. Show that
n∑
k=1
k = n (n + 1)
2 for n = 1:
1∑
k=1
k = 1 = 1 (1 + 1)
2 ✓
2. Assume that
n∑
k=1
k = n (n + 1)
2 for some integer n ≥ 1. Show that the formula holds for n
replaced by n + 1, that is:
n+1∑
k=1
k = (n + 1) ((n + 1) + 1)
2 = (n + 1) (n + 2)
2
To show this, note that
n+1∑
k=1
k = 1 + 2 + ··· + n + (n + 1) =
n∑
k=1
k + (n + 1)
= n (n + 1)
2 + (n + 1) = n (n + 1) + 2(n + 1)
2 = (n + 1) (n + 2)
2 ✓
3. By induction, this proves the formula for all integers n ≥ 1. QED",ElementaryCalculus.pdf
"The Definite Integral • Section 5.2 137
Formulas (3)-(5) can be proved similarly by induction (see t he exercises). The example below
shows how Formulas (2) and (3) are used in finding the limit of a Riemann sum.
Example 5.11
Use Riemann sums to calculate
∫ 2
1
x2 dx.
Solution: The definite integral is the area under the curve y = f (x) = x2 between x = 1 and x = 2, as
shown in Figure 5.2.5(a):
0
y
x
1 2
(a) Area under y = x2 over [1 , 2]
0
y
x
1 = x0 x1 x2 . . . xn−1 xn = 2
. . .
(b) Riemann sums using left endpoints: x∗
i = xi−1
Figure 5.2.5 Calculating
∫ 2
1 x2 dx
Divide the interval [1 , 2] into n subintervals of equal length ∆xi = (2 − 1)/n = 1/n for i = 1 to n, so that
the partition P is {x0 < x1 < . . . xn } where xi = 1 + i
n for i = 0, 1, . . . , n (and hence x0 = 1 and xn = 2). In
each subinterval [ xi−1 , xi ] pick the point x∗
i to be the left endpoint xi−1 , so that the rectangles appear as
in Figure 5.2.5(b). Then
∫ 2
1
x2 dx = limn→∞
n∑
i=1
f (x∗
i ) ∆xi = limn→∞
n∑
i=1",ElementaryCalculus.pdf
"i to be the left endpoint xi−1 , so that the rectangles appear as
in Figure 5.2.5(b). Then
∫ 2
1
x2 dx = limn→∞
n∑
i=1
f (x∗
i ) ∆xi = limn→∞
n∑
i=1
f (xi−1) 1
n = limn→∞
n∑
i=1
x2
i−1
1
n
= limn→∞
n∑
i=1
(
1 + i − 1
n
) 2 1
n = limn→∞
n∑
i=1
(1
n + 2
n2 (i − 1) + 1
n3 (i − 1)2
)
= limn→∞
(
n∑
i=1
1
n + 2
n2
n∑
i=1
(i − 1) + 1
n3
n∑
i=1
(i − 1)2
)
= limn→∞
(
1 + 2
n2
n−1∑
i=1
i + 1
n3
n−1∑
i=1
i2
)
= limn→∞
(
1 + 2
n2 · (n − 1)n
2 + 1
n3 · (n − 1)n(2n − 1)
6
)
(replace n by n − 1 in Formulas (2) and (3))
=
(
limn→∞ 1
)
+
(
limn→∞
n − 1
n
)
+
(
limn→∞
2n2 − 3n + 1
6n2
)
= 1 + 1
1 + 2
6 = 7
3",ElementaryCalculus.pdf
"138 Chapter 5 • The Integral §5.2
It it often simpler to use a computer to calculate approximat ions of a definite integral, by
taking the Riemann sum of a sufficiently large number of recta ngles in order to achieve the
desired accuracy . Choosing subintervals of equal length, a s in Example 5.11, makes it easier
to use an algorithm to calculate the integral.
For example, the table below summarizes the calculations of Riemann sums for the function
in Example 5.11—namely f (x) = x2 over [1 , 2]—using different values for the points x∗
i in the
subintervals (left endpoints, midpoints, and right endpoi nts):
# of rectangles Left endpoint Midpoint Right endpoint
1 1 2.25 4
2 1.625 2.3125 3.125
3 1.851851851852 2.324074074074 2.851851851852
4 1.96875 2.328125 2.71875
5 2.04 2.33 2.64
10 2.185 2.3325 2.485
100 2.31835 2.333325 2.34835
1000 2.3318335 2.33333325 2.3348335
10000 2.333183335 2.3333333325 2.333483335
100000 2.33331833335 2.333333333325 2.33334833335",ElementaryCalculus.pdf
"1000 2.3318335 2.33333325 2.3348335
10000 2.333183335 2.3333333325 2.333483335
100000 2.33331833335 2.333333333325 2.33334833335
1000000 2.333331833333 2.333333333333 2.333334833333
Due to the concavity of the curve y = x2, using the left endpoints underestimates the actual
area, whereas using the right endpoints yields an overestim ate. Using the midpoints usually
gives better results (i.e. more accuracy in fewer iteration s).
So far only definite integrals of nonnegative functions have been considered—that is, func-
tions f (x) ≥ 0 over an interval [ a, b]. If f (x) is either negative or changes sign over [ a, b], then
the definite integral can be defined as follows:
Let R be the region bounded by y = f (x) and the x-axis between x = a and x = b. If f (x) ≤ 0
over [ a, b], then ∫ b
a
f (x) dx = the negative of the area of R
If f (x) changes sign over [ a, b], then
∫ b
a
f (x) dx = the net area of R,",ElementaryCalculus.pdf
"over [ a, b], then ∫ b
a
f (x) dx = the negative of the area of R
If f (x) changes sign over [ a, b], then
∫ b
a
f (x) dx = the net area of R,
where the parts of R above the x-axis count as positive area and the parts below count as
negative area.",ElementaryCalculus.pdf
"The Definite Integral • Section 5.2 139
Note: In the definite integral
∫ b
a
f (x) dx the numbers a and b are called the limits of integra-
tion, with a being the lower limit of integration and b the upper limit of integration .
The function f (x) being integrated is called the integrand, in both definite and indefinite
integrals.
Exercises
A
1. Explain why
∫ b
a
c dx = c(b − a) for any constant c.
2. Would using left endpoints in the Riemann sums underestimat e or overestimate
∫ 2
1 ln x dx? Explain.
B
3. Use Riemann sums to calculate
∫ 1
0
x dx . 4. Use Riemann sums to calculate
∫ 1
0
x2 dx.
5. Use Riemann sums to calculate
∫ 1
0
3x2 dx. 6. Use Riemann sums to calculate
∫ 1
0
x3 dx.
7. Prove the formula
n∑
k=1
k2 = n (n + 1) (2n + 1)
6 by induction on n ≥ 1.
8. Prove the formula
n∑
k=1
k2 = n (n + 1) (2n + 1)
6 as follows:
(a) Show that
n∑
k=1
(
(k + 1)3 − k3 )
= (n + 1)3 − 1 .
(b) Show that ( k + 1)3 − k3 = 3k2 + 3k + 1 .
(c) Use the formula
n∑
k=1
k = n (n + 1)",ElementaryCalculus.pdf
"(a) Show that
n∑
k=1
(
(k + 1)3 − k3 )
= (n + 1)3 − 1 .
(b) Show that ( k + 1)3 − k3 = 3k2 + 3k + 1 .
(c) Use the formula
n∑
k=1
k = n (n + 1)
2 and parts (a) and (b) to show that
n∑
k=1
k2 = n (n + 1) (2n + 1)
6 .
9. Prove the formula
n∑
k=1
k3 = n2 (n + 1)2
4 by induction on n ≥ 1.
10. The famous quicksort algorithm in computer science is a popular method for placing objects i n
some order (e.g. numerical, alphabetical). On average the a lgorithm needs O(n log n) comparisons
to sort n objects (here log n means the natural logarithm of n). The proof of that average complexity
depends on the inequality
m−1∑
k=2
k ln k ≤
∫ m
2
x ln x dx
for all integers m > 2. Explain why that inequality is true.
C
11. Prove the formula
n∑
k=1
k4 = 14 + 24 + ··· + n4 = n (n + 1) (6n3 + 9n2 + n − 1)
30 by induction on n ≥ 1.
12. Calculate the following sum:
1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ··· + (1 + 2 + 3 + 4 + ··· +50)",ElementaryCalculus.pdf
"140 Chapter 5 • The Integral §5.3
5.3 The Fundamental Theorem of Calculus
Using Riemann sums to calculate definite integrals can be ted ious, as was seen in the previous
section. In fact the technique shown in that section depende d on the function being a low-
degree polynomial, which obviously will not always be the ca se. Luckily there is a better way ,
involving antiderivatives, given by the following theorem :
Fundamental Theorem of Calculus : Suppose that a function f is differentiable on
[a, b]. Then:
(I) The function A(x) defined on [ a, b] by
A(x) =
∫ x
a
f (t) dt
is differentiable on [ a, b], and
A′(x) = f (x)
for all x in [ a, b].
(II) If F is an antiderivative of f on [ a, b], i.e. F ′(x) = f (x) for all x in [ a, b], then
∫ b
a
f (x) dx = F (b) − F (a) .
The function A(x) in Part I of the theorem is sometimes called the area function because it
represents the area under the curve y = f (x) over the interval [ a, x], as shown in Figure 5.3.1
below .
y = f (x)
y
x
a x b
A(x)",ElementaryCalculus.pdf
"represents the area under the curve y = f (x) over the interval [ a, x], as shown in Figure 5.3.1
below .
y = f (x)
y
x
a x b
A(x)
Figure 5.3.1 The area function A(x) =
∫ x
a f (t) dt
y = f (x)
y
x
a x x + dx b
dA
Figure 5.3.2 dA = A(x + dx) − A(x)
To prove Part I, assume that f (x) ≥ 0 on [ a, b] as in Figure 5.3.1 (the proofs for f (x) either
negative or switching sign over [ a, b] are similar). The goal is to show that for any x in [ a, b]
the differential dA exists and equals f (x) dx. First, dA = A(x + dx) − A(x) is the area under the
curve y = f (x) over the interval [ x, x + dx], as shown in Figure 5.3.2 above.",ElementaryCalculus.pdf
"The Fundamental Theorem of Calculus • Section 5.3 141
By the Microstraightness Property the curve y = f (x) is a straight line over the infinitesimal
interval [ x, x + dx], so f must be either increasing, constant, or decreasing over tha t interval.
The three possibilities are shown in Figure 5.3.3:
f (x) f (x)
df
y = f (x)
x x + dx
dx
A
B
C
(a) f is increasing
f (x)
y = f (x)
x x + dx
dx
(b) f is constant
f (x)
f (x + dx)
−df
y = f (x)
x x + dx
dx
A
B
C
(c) f is decreasing
Figure 5.3.3 The three possibilities for dA
In the case where f is increasing over [ x, x + dx], the infinitesimal area dA is the sum of the
area of the rectangle of height f (x) and width dx and the area of the right triangle △ABC
shown in Figure 5.3.3(a). The area of △ABC is 1
2 (df )(dx) = 1
2 f ′(x)(dx)2 = 0, so dA = f (x) dx.
In the case where f is constant over [ x, x + dx], the infinitesimal area dA is the area of the
rectangle of height f (x) and width dx, as shown in Figure 5.3.3(b). So again, dA = f (x) dx.",ElementaryCalculus.pdf
"rectangle of height f (x) and width dx, as shown in Figure 5.3.3(b). So again, dA = f (x) dx.
In the case where f is decreasing over [ x, x + dx], the infinitesimal area dA is the sum of the
area of the rectangle of height f (x + dx) and width dx and the area of the right triangle △ABC
shown in Figure 5.3.3(c). Note that df < 0 since f is decreasing, and so the area of △ABC is
1
2 (−df )(dx) = −1
2 f ′(x)(dx)2 = 0. Thus,
dA = f (x + dx) dx = ( f (x) + df ) dx = f (x) dx + f ′(x)(dx)2 = f (x) dx + 0 = f (x) dx .
So in all three cases, d A = f (x) dx, and so A′(x) = dA
dx = f (x), which shows that A(x) is differen-
tiable and has derivative f (x). This proves Part I of the Fundamental Theorem of Calculus. ✓
To prove Part II of the theorem, let F (x) be an antiderivative of f (x) over [ a, b]. Since the are
afunction A(x) =
∫ x
a f (x) dx is also an antiderivative of f (x) over [ a, b] by Part I of the theorem,
then A(x) and F (x) differ by a constant C over [ a, b]. In other words:",ElementaryCalculus.pdf
"then A(x) and F (x) differ by a constant C over [ a, b]. In other words:
A(x) = F (x) + C for all x in [ a, b]",ElementaryCalculus.pdf
"142 Chapter 5 • The Integral §5.3
By definition A(a) = 0, since it is the area under the curve over the interval [ a, a] of zero
length. Thus,
0 = A(a) = F (a) + C ⇒ C = −F (a) ⇒ A(x) = F (x) − F (a) for all x in [ a, b]
and so ∫ b
a
f (x) dx = A(b) = F (b) − F (a)
which proves Part II of the theorem. 3 ✓
Note: In some textbooks Part I is called the First Fundamental Theorem of Calculus and Part
II is called the Second Fundamental Theorem of Calculus . The following notation provides a
shorthand way of writing F (b) − F (a):
F (x)
⏐
⏐
⏐
⏐
b
a
= F (b) − F (a)
Example 5.12
Calculate
∫ 2
1
x2 dx.
Solution: Recall from Example 5.11 in the previous section that the int egral equals 7/3. In that example
Riemann sums were used, but Part II of the Fundamental Theore m of Calculus makes the integral much
easier to calculate. Since F (x) = x3
3 is an antiderivative of f (x) = x2, then
∫ 2
1
x2 dx = x3
3
⏐
⏐
⏐
⏐
2
1
= 23
3 − 13
3 = 7
3 .",ElementaryCalculus.pdf
"easier to calculate. Since F (x) = x3
3 is an antiderivative of f (x) = x2, then
∫ 2
1
x2 dx = x3
3
⏐
⏐
⏐
⏐
2
1
= 23
3 − 13
3 = 7
3 .
Note in the above example that any antiderivative of f (x) = x2 could have been used, e.g.
F (x) = x3
3 + 5. Notice that the constant 5 cancels out when evaluating F (2) − F (1). So you do
not need to add a generic constant C to the antiderivative of f (x) in a definite integral, as you
would in an indefinite integral.
Example 5.13
Calculate
∫ π
0
sin x dx .
Solution: Since F (x) = −cos x is an antiderivative of f (x) = sin x, then
∫ π
0
sin x dx = −cos x
⏐
⏐
⏐
⏐
π
0
= −cos π − (−cos 0) = −(−1) − (−1) = 2 .
3The theorem can be proved for the weaker condition that f is merely continuous on [ a, b]. See p.173-175 in
PA R Z Y N S K I , W .R. A N D P .W . Z I P S E, Introduction to Mathematical Analysis , New Y ork: McGraw-Hill, Inc., 1982.",ElementaryCalculus.pdf
"The Fundamental Theorem of Calculus • Section 5.3 143
Example 5.14
Calculate
∫ 1
−1
x3 dx.
Solution: Since F (x) = x4
4 is an antiderivative of f (x) = x3, then
∫ 1
−1
x3 dx = x4
4
⏐
⏐
⏐
⏐
1
−1
= 14
4 − (−1)4
4 = 1
4 − 1
4 = 0 .
Example 5.14 is a special case of the following result for odd functions:
If f is an odd function, i.e. f (−x) = −f (x) for all x, then
∫ a
−a
f (x) dx = 0
for all a > 0 such that f is continuous on [ −a, a].
The idea is that since an odd function is symmetric around the origin, then the area between
the curve and the x-axis over [0 , a] will cancel out the area between the curve and the x-axis
over [ −a, 0]. Both areas are the same but one gets counted as positive an d the other negative,
as shown in Figure 5.3.4 below:
x
y
y = f (x)
a
−a +
—
Figure 5.3.4 Odd function f over [ −a, a]
x
y
y = f (x)
a−a
Figure 5.3.5 Even function f over [ −a, a]
By symmetry around the y-axis, a similar result holds for even functions (see Figure 5.3.5):",ElementaryCalculus.pdf
"y = f (x)
a−a
Figure 5.3.5 Even function f over [ −a, a]
By symmetry around the y-axis, a similar result holds for even functions (see Figure 5.3.5):
If f is an even function, i.e. f (−x) = f (x) for all x, then
∫ a
−a
f (x) dx = 2
∫ a
0
f (x) dx
for all a > 0 such that f is continuous on [ −a, a].",ElementaryCalculus.pdf
"144 Chapter 5 • The Integral §5.3
The following rules for definite integrals are a consequence of the corresponding rules for
indefinite integrals:
Let f and g be continuous functions on [ a, b] and let k be a constant. Then:
1.
∫ b
a
k f (x) dx = k
∫ b
a
f (x) dx
2.
∫ b
a
( f (x) + g(x)) dx =
∫ b
a
f (x) dx +
∫ b
a
g(x) dx
3.
∫ b
a
( f (x) − g(x)) dx =
∫ b
a
f (x) dx −
∫ b
a
g(x) dx
The following results for definite integrals are a consequen ce of the Fundamental Theorem
of Calculus:
Let f be a continuous function on [ a, b] and suppose that a < c < b. Then:
(1)
∫ a
a
f (x) dx = 0
(2)
∫ a
b
f (x) dx = −
∫ b
a
f (x) dx
(3)
∫ b
a
f (x) dx =
∫ c
a
f (x) dx +
∫ b
c
f (x) dx
For example, if F (x) is an antiderivative of f (x) on [ a, b], then
∫ c
a
f (x) dx +
∫ b
c
f (x) dx = (F (c) − F (a)) + (F (b) − F (c)) = F (b) − F (a) =
∫ b
a
f (x) dx
which proves rule (3).
The following result is a consequence of Part I of the Fundame ntal Theorem of Calculus
along with the Chain Rule:",ElementaryCalculus.pdf
"a
f (x) dx
which proves rule (3).
The following result is a consequence of Part I of the Fundame ntal Theorem of Calculus
along with the Chain Rule:
Chain Rule for integrals : Let f be a continuous function on an interval I containing
x = a, and let g(x) be a differentiable function on I. If
F (x) =
∫ g(x)
a
f (t) dt for all x in I
then F ′(x) = f ( g(x)) · g′(x) for all x in I.",ElementaryCalculus.pdf
"The Fundamental Theorem of Calculus • Section 5.3 145
Example 5.15
Let F (x) =
∫ x2
0
e−t2
dt for all x > 0. Find F ′(x).
Solution: By the Chain Rule for integrals, with f (t) = et2
and g(x) = x2 :
F ′(x) = f ( g(x)) · g′(x) = e−(x2)2
· (2x) = 2x e −x4
Exercises
A
For Exercises 1-12, evaluate the given definite integral.
1.
∫ 1
0
x2 dx 2.
∫ 1
−1
x2 dx 3.
∫ 1
0
x3 dx 4.
∫ 1
−1
(x2 + 3x − 4) dx
5.
∫ 2
1
1
x2 dx 6.
∫ 3
2
1
x3 dx 7.
∫ π/2
0
cos x dx 8.
∫ 1
0
ex dx
9.
∫ 1
−1
2ex dx 10.
∫ π
−π
sin x dx 11.
∫ 4
0
∝ra⌈〉⌋allow
x dx 12.
∫ 2
−2
x3 ex2
cos 2 x dx
13. Show that ln x =
∫ x
1
1
t dt for all x > 0. 14. Show that
∫ a
b
f (x) dx = −
∫ b
a
f (x) dx.
15. Given f (x) =
∫ x
1
t∝ra⌈〉⌋allow
t4 + 1
dt , find f ′(3) and f ′(−2).
B
16. Prove the Chain Rule for integrals.
17. Explain why for any continuous function f on [ a, b],
⏐
⏐
⏐
⏐
∫ b
a
f (x) dx
⏐
⏐
⏐
⏐ ≤
∫ b
a
| f (x)| dx .
18. Explain why if f (x) ≤ g(x) on [ a, b] then
∫ b
a
f (x) dx ≤
∫ b
a
g(x) dx .
C",ElementaryCalculus.pdf
"⏐
⏐
⏐
⏐
∫ b
a
f (x) dx
⏐
⏐
⏐
⏐ ≤
∫ b
a
| f (x)| dx .
18. Explain why if f (x) ≤ g(x) on [ a, b] then
∫ b
a
f (x) dx ≤
∫ b
a
g(x) dx .
C
19. Show that if f (x) is continuous on [ a, b] then there is a number c in ( a, b) such that
∫ b
a
f (x) dx = f (c) · (b − a) .
20. Let f (t) be a continuous function for all t ≥ 0, and for each x ≥ 0 define a function g(x) by
g(x) =
∫ x
0
(x − t) f (t) dt .
Show that g′(x) =
∫ x
0
f (t) dt for all x ≥ 0.
21. Show that for all x > 0, ∫ x
0
dt
1 + t2 +
∫ 1/x
0
dt
1 + t2
is independent of x.",ElementaryCalculus.pdf
"146 Chapter 5 • The Integral §5.4
5.4 Integration by Substitution
The integrals encountered so far—whether indefinite or defin ite—have been the simplest kind,
since the antiderivatives had known formulas. For example,
∫ 
cos x dx = sin x + C. What if the
integral were
∫ 
cos 2 x dx instead? No formula has been discussed yet for this integral , and the
answer is not sin 2 x +C, since the derivative of sin 2 x is 2 cos 2 x, not cos 2 x. But dividing sin 2 x
by 2 first and then taking the derivative would yield cos 2 x, so that
∫ 
cos 2 x dx = 1
2 sin 2 x + C.
Evaluating an integral in such a manner is often done when the function is not too compli-
cated, as the one above. Usually it will not be quite that simp le, and so a general technique
called substitution is needed. The idea behind substitution is to replace part of the function
being integrated by a new variable—typically u—so that a complicated function of x is now a
simpler function of u that you know how to integrate.",ElementaryCalculus.pdf
"being integrated by a new variable—typically u—so that a complicated function of x is now a
simpler function of u that you know how to integrate.
Example 5.16
Evaluate
∫ 
cos 2 x dx by substitution.
Solution: The 2 x in the cosine function is what makes this integral unknown, s o replace it by u: let
u = 2x. The integral is now ∫ 
cos u dx
which is a problem because the point of doing substitution is to eliminate all references to the variable
x, including in the infinitesimal dx. The entire integral needs to be in terms of u and du, but the dx is
still there. So put dx in terms of du:
u = 2x ⇒ du = 2dx ⇒ dx = 1
2 du
The integral now becomes
∫ 
cos u
(1
2 du
)
= 1
2
∫ 
cos u du = 1
2 sin u + C
by the formula already known, just with the letter u as the variable instead of x. The original integral
was in terms of x, so the final answer—for an indefinite integral—should also b e in terms of x. Thus,",ElementaryCalculus.pdf
"was in terms of x, so the final answer—for an indefinite integral—should also b e in terms of x. Thus,
the final step is to substitute back into the answer what u equals in terms of x, namely 2 x:
∫ 
cos 2 x dx = 1
2 sin u + C = 1
2 sin 2 x + C
If the procedure in the above example seems similar to making a substitution when using
the Chain Rule to take a derivative, that is because it is simi lar: you are basically doing the
same thing only in reverse. Just as for differentiation, it i s not always obvious what part of
the function is the best candidate for substitution when per forming integration. There is one
obvious rule: never make the substitution u = x, because that changes nothing.",ElementaryCalculus.pdf
"Integration by Substitution • Section 5.4 147
Example 5.17
Evaluate
∫ 
e−3x dx.
Solution: The −3x in the exponential function is what makes this integral unkn own, so make the
substitution u = −3x, which means that du = −3dx, and so dx = −1
3 du. Thus:
∫ 
e−3x dx =
∫ 
eu
(
− 1
3 du
)
= −1
3
∫ 
eu du = −1
3 eu + C = −1
3 e−3x + C
The above example can be generalized as follows:
∫ 
eax dx = 1
a eax + C for any constant a ̸=0
Example 5.17 was the special case with a = −3.
Example 5.18
Evaluate
∫ 
(1 + 4x)5 dx.
Solution: Y ou might be tempted to make the substitution u = 4x, but that would then require finding
the integral of (1 + u)5 , for which there is not yet any formula. But there is a formula for the integral of
u5 . Hence, let u = 1 + 4x, so that du = 4dx ⇒ dx = 1
4 du. Thus:
∫ 
(1 + 4x)5 dx = 1
4
∫ 
u5 du = 1
4
u6
6 + C = 1
24 (1 + 4x)6 + C
Example 5.19
Evaluate
∫ 
2x e x2
dx.
Solution: It might be unclear whether you should make the substitution u = 2x or u = x2, but the hint",ElementaryCalculus.pdf
"Example 5.19
Evaluate
∫ 
2x e x2
dx.
Solution: It might be unclear whether you should make the substitution u = 2x or u = x2, but the hint
here is that the derivative of the x2 inside the exponential function is 2 x, which appears outside the
exponential function. Indeed, you could check that letting u = 2x would result in an integral no simpler
than the current one (namely , 1
2
∫ 
u e u2 /4 du). So let u = x2, which means du = 2x dx. Thus:
∫ 
2x e x2
dx =
∫ 
eu du = eu + C = ex2
+ C
Example 5.20
Evaluate
∫ 
x
√
1 + 3x2 dx.
Solution: Note that the derivative of the 1 + 3x2 term inside the square root function is 6 x, which is
almost the function x outside the square root—all that is missing is the constant m ultiple 6. That is a
hint to let u = 1 + 3x2. Notice also that du = 6x dx ⇒ x dx = 1
6 du, and x dx is the remaining part of the
integral outside the square root. Thus:
∫ 
x
√
1 + 3x2 dx = 1
6
∫ ∝ra⌈〉⌋allowu du = 1
6
u3/2
3/2 + C = 1
9 (1 + 3x2)3/2 + C",ElementaryCalculus.pdf
"148 Chapter 5 • The Integral §5.4
Example 5.21
Evaluate
∫ x2 dx∝ra⌈〉⌋allow
x3 + 9
.
Solution: Let u = x3 + 9, so that du = 3x2 dx ⇒ x2 dx = 1
3 du. Thus:
∫ x2 dx∝ra⌈〉⌋allow
x3 + 9
=
∫ 1
3 du
∝ra⌈〉⌋allowu = 1
3
∫ 
u−1/2 du = 1
3
u1/2
1/2 + C = 2
3
√
x3 + 9 + C
Example 5.22
Evaluate
∫ 2x dx
x2 − 1 .
Solution: Notice that the numerator 2 x in the function is exactly the derivative of the denominator
x2 − 1. That is a hint to substitute on the denominator so that the i ntegral is the natural logarithm
function. Let u = x2 − 1, so that du = 2x dx. Thus:
∫ 2x dx
x2 − 1 =
∫ du
u = ln |u| + C = ln |x2 − 1| + C
Example 5.23
Evaluate
∫ 
tan x dx .
Solution: Notice that tan x = sin x
cos x and that the numerator sin x is almost the derivative of the denom-
inator cos x; all that is missing is a negative sign. That is a hint to subst itute on the denominator:
u = cos x, so that du = −sin x dx ⇒ sin x dx = −du. Thus:
∫ 
tan x dx =
∫ sin x
cos x dx
=
∫ −du",ElementaryCalculus.pdf
"u = cos x, so that du = −sin x dx ⇒ sin x dx = −du. Thus:
∫ 
tan x dx =
∫ sin x
cos x dx
=
∫ −du
u = −ln |u| + C = −ln |cos x | + C = ln |cos x |−1 + C = ln |sec x | + C
Example 5.24
Evaluate
∫ 
sec x dx .
Solution: Notice that
∫ 
sec x dx =
∫ sec x (sec x + tan x)
sec x + tan x dx =
∫ sec x tan x + sec2 x
sec x + tan x dx
and that the numerator in the last integral is the derivative of the denominator: let u = sec x + tan x,
so that du = (sec x tan x + sec2 x) dx. Thus:
∫ 
sec x dx =
∫ du
u = ln |u| + C = ln |sec x + tan x | + C
The following formulas are straightforward consequences o f substitution and the derivatives
of inverse trigonometric functions discussed in Section 2. 2:",ElementaryCalculus.pdf
"Integration by Substitution • Section 5.4 149
For any constant a > 0:
∫ dx∝ra⌈〉⌋allow
a2 − x2
= sin−1
(x
a
)
+ C (if |x| <a) (5.3)
∫ dx
a2 + x2 = 1
a tan−1
(x
a
)
+ C (5.4)
∫ dx
|x|
∝ra⌈〉⌋allow
x2 − a2
= 1
a sec−1
(x
a
)
+ C (if |x| >a) (5.5)
For example, to prove the second formula, recall that d
dx
(
tan−1 x
)
= 1
1+x2 . Make the substi-
tution u = x/a, so that x = au and dx = a du . Thus:
∫ dx
a2 + x2 =
∫ a du
a2 + a2 u2 = 1
a
∫ du
1 + u2 = 1
a tan−1 u + C = 1
a tan−1
(x
a
)
+ C ✓
Example 5.25
Evaluate
∫ dx∝ra⌈〉⌋allow
4 − 9x2
.
Solution: The 4 −9x2 inside the square root is almost of the form a2 − x2, except for the 9. The goal is to
have u2 = 9x2 , so let u = 3x, which means that dx = 1
3 du and u2 = 9x2. Thus,
∫ dx∝ra⌈〉⌋allow
4 − 9x2
= 1
3
∫ du∝ra⌈〉⌋allow
4 − u2
= 1
3 sin−1
(u
2
)
+ C = 1
3 sin−1
(3x
2
)
+ C
by Formula (5.2) with a = 2.
To use substitution with definite integrals, follow the same procedure as with indefinite",ElementaryCalculus.pdf
"2
)
+ C = 1
3 sin−1
(3x
2
)
+ C
by Formula (5.2) with a = 2.
To use substitution with definite integrals, follow the same procedure as with indefinite
integrals but add one extra step: replace the limits of integ ration x = a and x = b in the original
integral
∫ b
a f (x) dx by u = g(a) and u = g(b), respectively , in the new integral involving u, where
u = g(x) is your substitution.
Example 5.26
Evaluate
∫ 2
1
(2x + 1)3 dx.
Solution: Let u = g(x) = 2x+1, which means that dx = 1
2 du. The upper limit of integration x = 2 becomes
u = g(2) = 2(2) + 1 = 5 in the new u-based integral, while the lower limit of integration x = 1 becomes
u = g(1) = 2(1) + 1 = 3. Thus:
∫ 2
1
(2x + 1)3 dx = 1
2
∫ 5
3
u3 du = 1
8 u4
⏐
⏐
⏐
⏐
5
3
= 1
8 (54 − 34 ) = 68
Note that you could have put everything back in terms of x at the end, but there was no need to since
you would get the same numerical answer .",ElementaryCalculus.pdf
"150 Chapter 5 • The Integral §5.4
The following property of definite integrals comes in handy f or evaluating certain types of
definite integrals:
For any constant a, ∫ a
0
f (x) dx =
∫ a
0
f (a − x) dx . (5.6)
This is simple to prove, using the substitution u = a −x, so x = a −u and dx = −du, while x = 0
becomes u = a and x = a becomes u = 0 in the limits of integration:
∫ a
0
f (x) dx = −
∫ 0
a
f (a − u) du =
∫ a
0
f (a − u) du =
∫ a
0
f (a − x) dx ✓
Example 5.27
Evaluate
∫ π
0
x sin x
1 + cos2 x dx.
Solution: Let I =
∫ π
0
x sin x
1 + cos2 x dx. Then by the above property (with a = π):
I =
∫ π
0
(π− x) sin ( π− x)
1 + cos2 (π− x) dx =
∫ π
0
(π− x) sin x
1 + cos2 x dx = π
∫ π
0
sin x
1 + cos2 x dx −
∫ π
0
x sin x
1 + cos2 x dx
I = π
∫ π
0
sin x
1 + cos2 x dx − I
2I = π
∫ π
0
sin x
1 + cos2 x dx = −π tan−1 (cos x)
⏐
⏐
⏐
⏐
π
0
= −π
(
− π
4 − π
4
)
= π2
2
I = π2
4
Exercises
A
For Exercises 1-24 evaluate the given integral.
1.
∫ 
(3 cos 5 x + 4 sin 5 x) dx 2.
∫ e2x + e−2x
2 dx 3.
∫ (",ElementaryCalculus.pdf
"4 − π
4
)
= π2
2
I = π2
4
Exercises
A
For Exercises 1-24 evaluate the given integral.
1.
∫ 
(3 cos 5 x + 4 sin 5 x) dx 2.
∫ e2x + e−2x
2 dx 3.
∫ (
x e−x2
+ x2 cos x3
)
dx
4.
∫ x − 2
x2 − 4x + 9 dx 5.
∫ ex
1 + ex dx 6.
∫ 1
1 + ex dx
7.
∫ 
x
∝ra⌈〉⌋allow
x + 4 dx 8.
∫ 
cos2 x dx 9.
∫ 
tan2 x dx
10.
∫ 3∝ra⌈〉⌋allow
4 − 25x2
dx 11.
∫ 3
4 + 25x2 dx 12.
∫ 
sin2 x cos3 x dx",ElementaryCalculus.pdf
"Integration by Substitution • Section 5.4 151
13.
∫ 1
0
(2x + 1)3 dx 14.
∫ 1
0
(2x − 1)3 dx 15.
∫ 8
0
x
∝ra⌈〉⌋allow
1 + x dx
16.
∫ π/2
0
4 sin ( x/2) dx 17.
∫ π/4
0
4 sin x cos x dx 18.
∫ ∝ra⌈〉⌋allowπ
0
5x cos ( x2) dx
19.
∫ −1
−2
x
(x2 + 2)3 dx 20.
∫ ln 3
−ln 3
ex
ex + 4 dx 21.
∫ 3
1
dx∝ra⌈〉⌋allowx (x + 1)
22.
∫ 1
−1
x2 dx∝ra⌈〉⌋allow
x3 + 9
23.
∫ 2
1
dx
x2 − 6x + 9 24.
∫ 3
−3
x5 dx
ex2
25. Evaluate the indefinite integral ∫ 
sin x cos x dx
three different ways:
(a) Use the substitution u = sin x.
(b) Use the substitution u = cos x.
(c) Use the trigonometric identity 2 sin x cos x = sin 2 x.
(d) Are the three answers from parts (a)-(c) actually different ? Explain.
26. For all positive constants L, show the following:
(a)
∫ L
0
(
1 − x
L
) 2
dx = L
3
(b)
∫ L/2
−L/2
(1
2 − x
L
) 2
dx = L
3
(c)
∫ L
0
(
1 − x
L
) 3 (x
L
) 2
dx = L
60
B
27. Recall from trigonometry that sin 2 x = 1
2 (1 − cos 2 x) for all x.
(a) Use the Fundamental Theorem of Calculus to evaluate
∫ π
0
sin2 x dx .",ElementaryCalculus.pdf
"B
27. Recall from trigonometry that sin 2 x = 1
2 (1 − cos 2 x) for all x.
(a) Use the Fundamental Theorem of Calculus to evaluate
∫ π
0
sin2 x dx .
(b) Approximate the integral from part (a) by dividing the inter val [0 , π] into n = 2 subintervals of
equal length, [0 , π/2] and [ π/2, π], and finding the exact value of the sum of the areas of the
rectangles whose heights are determined at the right endpoi nts of the subintervals.
(c) Repeat part (b) with n = 3.
(d) Repeat part (b) with n = 4.
(e) Repeat part (b) with n = 6.
28. Show that
∫ 
csc x dx = −ln |csc x + cot x | + C. (Hint: See Example 5.24.)
C
29. Use the property
∫ a
0 f (x) dx =
∫ a
0 f (a − x) dx to show that
∫ π/2
0
sin2 x
sin x + cos x dx = 1∝ra⌈〉⌋allow
2
ln
(∝ra⌈〉⌋allow
2 + 1
)
.
(Hint: Use Exercise 28 and the sine addition formula.)",ElementaryCalculus.pdf
"152 Chapter 5 • The Integral §5.5
5.5 Improper Integrals
Definite integrals so far have been defined only for continuou s functions over finite closed
intervals. There are times when you will need to perform inte gration despite those conditions
not being met. For example, in quantum mechanics the Dirac delta function
4 δ is defined on R
by four properties:
x
y
0
∞
Figure 5.5.1 y = δ(x)
(1): δ(x) = 0 for all x ̸=0
(2): δ(0) = ∞
(3):
∫ ∞
−∞
δ(x) dx = 1
(4): For any continuous function f on R,
∫ ∞
−∞
f (x) δ(x) dx = f (0).
Properties (3) and (4) provide examples of one type of improper integral : an integral over
an infinite interval (in this case the entire real line R = (−∞, ∞)). Define this type of improper
integral as follows:
For a continuous function f and a real number a, define the improper integral of f over
[a, ∞) by ∫ ∞
a
f (x) dx = lim
b→∞
∫ b
a
f (x) dx ,
define the improper integral of f over ( −∞, a] by
∫ a
−∞
f (x) dx = lim
b→−∞
∫ a
b
f (x) dx ,",ElementaryCalculus.pdf
"[a, ∞) by ∫ ∞
a
f (x) dx = lim
b→∞
∫ b
a
f (x) dx ,
define the improper integral of f over ( −∞, a] by
∫ a
−∞
f (x) dx = lim
b→−∞
∫ a
b
f (x) dx ,
and define the improper integral of f over ( −∞, ∞) by
∫ ∞
−∞
f (x) dx =
∫ c
−∞
f (x) dx +
∫ ∞
c
f (x) dx ,
for any real number c (typically c = 0). If the given limit exists (i.e. is a real number) then
the improper integral is convergent; otherwise it is divergent.
4Created by the physicist P .A.M. Dirac (1902-1984), who won a Nobel Prize in physics in 1933. The function is
neither real-valued nor continuous at x = 0. The “graph” in Figure 5.5.1 is perhaps misleading, as ∞ is not an
actual point on the y-axis. One interpretation is that δ is an abstraction of an instantaneous pulse or burst of
something, preceded and followed by nothing. To learn more about this fascinating and useful function, s ee §15 in
DI R A C, P .A.M., The Principles of Quantum Mechanics , 4th ed., Oxford, UK: Oxford University Press, 1958.",ElementaryCalculus.pdf
"Improper Integrals • Section 5.5 153
The limits in the above definitions are always taken after evaluating the integral inside
the limit. Just as for “proper” definite integrals, improper integrals can be interpreted as
representing the area under a curve.
Example 5.28
Evaluate
∫ ∞
1
dx
x .
y
x
0 1
y = 1
x
Solution: For all real numbers b > 1,
∫ ∞
1
dx
x = lim
b→∞
∫ b
1
dx
x = lim
b→∞
(
ln x
⏐
⏐
⏐
⏐
b
1
)
= lim
b→∞
(ln b − ln 1) = lim
b→∞
b = ∞
and so the integral is divergent. This means that the area und er the
curve y = 1/x over the interval [1 , ∞)—as shown in the graph above—is infinite.
Example 5.29
Evaluate
∫ ∞
1
dx
x2 .
y
x
0 1
y = 1
x2
Solution: For all real numbers b > 1,
∫ ∞
1
dx
x2 = lim
b→∞
∫ b
1
dx
x2 = lim
b→∞
(
− 1
x
⏐
⏐
⏐
⏐
b
1
)
= lim
b→∞
(
− 1
b −
(
− 1
1
))
= lim
b→∞
(
1 − 1
b
)
= 1 − 0 = 1 .
This means that the area under the curve y = 1/x2 over the interval [1 , ∞)—as shown in the graph",ElementaryCalculus.pdf
"b −
(
− 1
1
))
= lim
b→∞
(
1 − 1
b
)
= 1 − 0 = 1 .
This means that the area under the curve y = 1/x2 over the interval [1 , ∞)—as shown in the graph
above—equals 1. Thus, an infinite region can have a finite area . Length and area are different and
not necessarily related concepts, as this example illustra tes. Notice that y = 1/x2 approaches the x-axis
asymptote much faster than y = 1/x does—fast enough to make the integral convergent.
Example 5.30
Evaluate
∫ 0
−∞
ex dx .
y
x
0
y = ex
Solution: For all real numbers b < 0,
∫ 0
−∞
ex dx = lim
b→−∞
∫ 0
b
ex dx = lim
b→−∞
(
ex
⏐
⏐
⏐
⏐
0
b
)
= lim
b→−∞
(1 − eb ) = 1 − 0 = 1 .
This means that the area under the curve y = ex over the interval
(−∞, 0]—as shown in the graph above—equals 1.",ElementaryCalculus.pdf
"154 Chapter 5 • The Integral §5.5
Example 5.31
Evaluate
∫ ∞
0
sin x dx .
y
x
0 2π
1
−1
y = sin x
Solution: Since
∫ ∞
0
sin x dx = lim
b→∞
∫ b
0
sin x dx
= lim
b→∞
(
−cos x
⏐
⏐
⏐
⏐
b
0
)
= lim
b→∞
(−cos b + 1)
then the integral is divergent, since lim b→∞ cos b does not exist
(cos b oscillates between 1 and -1). This means that the net area
over [0 , ∞)—counted as positive above the x-axis and negative below—is indeterminate.
Example 5.32
Evaluate
∫ ∞
−∞
dx
1 + x2 .
y
x
0
y = 1
1+x2
Solution: Split the integral at x = 0:
∫ ∞
−∞
dx
1 + x2 =
∫ 0
−∞
dx
1 + x2 +
∫ ∞
0
dx
1 + x2
=
(
lim
b→−∞
∫ 0
b
dx
1 + x2
)
+
(
lim
b→∞
∫ b
0
dx
1 + x2
)
=
(
lim
b→−∞
tan−1 x
⏐
⏐
⏐
⏐
0
b
)
+
(
lim
b→∞
tan−1 x
⏐
⏐
⏐
⏐
b
0
)
= lim
b→−∞
(tan−1 0 − tan−1 b) + lim
b→∞
(tan−1 b − tan−1 0)
= (0 − (−π/2)) + (π/2 − 0) = π
This means that the area under the curve y = 1
1+x2 over the entire real line ( −∞, ∞)—as shown in the",ElementaryCalculus.pdf
"= (0 − (−π/2)) + (π/2 − 0) = π
This means that the area under the curve y = 1
1+x2 over the entire real line ( −∞, ∞)—as shown in the
graph above—equals π. Note that if the integral were split at any number c then the answer would
be the same. Another way to evaluate the integral would have b een to use the symmetry around the
y-axis—as f (x) = 1
1+x2 is an even function—so that
∫ ∞
−∞
dx
1 + x2 = 2
∫ ∞
0
dx
1 + x2 = ··· = 2(π/2 − 0) = π .
Since the integrand is continuous over R, a common way of evaluating the integral—especially among
students—is to simply use ±∞ as actual limits of integration, thus avoiding the need to ta ke a limit:
∫ ∞
−∞
dx
1 + x2 = tan−1 x
⏐
⏐
⏐
⏐
∞
−∞
= tan−1 (∞) − tan−1 (−∞) = π
2 − −π
2 = π
This type of shortcut is fine as long as you are aware of what plu gging x = ±∞ into tan −1 x actually
means, and that there are no numbers for which the integrand i s undefined (which would yield an
improper integral of a different type, to be discussed short ly).",ElementaryCalculus.pdf
"Improper Integrals • Section 5.5 155
The second type of improper integral is of a function not cont inuous or not bounded over its
interval of integration. For example, the integral in prope rty (3) of the Dirac delta function is
of that type, since δ is discontinuous at x = 0. Define this type of improper integral as follows:
For a function f that is continuous on [ a, b) but has either a discontinuity or vertical asymp-
tote at x = b, define the improper integral of f over [ a, b) by
∫ b
a
f (x) dx = lim
c→b−
∫ c
a
f (x) dx .
Likewise, if f is continuous on ( a, b] but has either a discontinuity or vertical asymptote at
x = a, then define the improper integral of f over ( a, b] by
∫ b
a
f (x) dx = limc→a+
∫ b
c
f (x) dx .
If f is continuous on [ a, b] but has either a discontinuity or vertical asymptote at x = c for
a < c < b, then define the improper integral of f over [ a, b] by
∫ b
a
f (x) dx =
∫ c
a
f (x) dx +
∫ b
c
f (x) dx ,",ElementaryCalculus.pdf
"a < c < b, then define the improper integral of f over [ a, b] by
∫ b
a
f (x) dx =
∫ c
a
f (x) dx +
∫ b
c
f (x) dx ,
where the integrals on the right are evaluated as in the first t wo definitions. If the given
limit exists (i.e. is a real number) then the improper integr al is convergent; otherwise it
is divergent.
Adjust these definitions accordingly for infinite intervals —e.g. [ a, ∞), ( −∞, b], or
(−∞, ∞)—to be consistent with the definitions of improper integral s of that type.
Example 5.33
Evaluate
∫ 1
0
dx
x .
y
x
0 1
y = 1
x
Solution: Since x = 0 is a vertical asymptote for y = 1
x ,
∫ 1
0
dx
x = lim
c→0+
∫ 1
c
dx
x = lim
c→0+
(
ln x
⏐
⏐
⏐
⏐
1
c
)
= lim
c→0+
(ln 1 − ln c) = 0 − (−∞) = ∞
and so the integral is divergent. This means that the area und er the curve y = 1/x
over the interval (0 , 1]—as shown in the graph above—is infinite. The region is infin ite in the y direction.",ElementaryCalculus.pdf
"156 Chapter 5 • The Integral §5.5
Example 5.34
Evaluate
∫ 1
0
dx∝ra⌈〉⌋allowx .
y
x
0 1
y = 1∝ra⌈〉⌋allowx
Solution: Since x = 0 is a vertical asymptote for y = 1∝ra⌈〉⌋allowx ,
∫ 1
0
dx∝ra⌈〉⌋allowx = lim
c→0+
∫ 1
c
dx∝ra⌈〉⌋allowx = lim
c→0+
(
2
∝ra⌈〉⌋allow
x
⏐
⏐
⏐
⏐
1
c
)
= lim
c→0+
(2 − 2 ∝ra⌈〉⌋allow
c ) = 2 − 0 = 2 .
This means that the area under the curve y = 1/∝ra⌈〉⌋allowx over the interval (0 , 1]—as shown
in the graph above—equals 2. The region is infinite in the y direction.
Example 5.35
Evaluate
∫ 3
1
⌊x⌋ dx .
y
x
0 1 2 3
1
2 y = ⌊x⌋
Solution: Recall from Example 3.22 in Section 3.3 that the floor functio n
y = ⌊x⌋ has jump discontinuities at each integer value of x, as shown
in the graph on the right. The integral
∫ 3
1 ⌊x⌋ dx is thus an improper
integral over the interval [1 , 3), which needs to be split at the point of
discontinuity x = 2 within that interval:
∫ 3
1
⌊x⌋ dx =
∫ 2
1
⌊x⌋ dx +
∫ 3
2
⌊x⌋ dx
=
(
lim
b→2−
∫ b
1
⌊x⌋ dx
)
+
(
lim
c→3−
∫ c
2
⌊x⌋ dx
)
=
(
lim
b→2−
∫ b
1",ElementaryCalculus.pdf
"∫ 3
1
⌊x⌋ dx =
∫ 2
1
⌊x⌋ dx +
∫ 3
2
⌊x⌋ dx
=
(
lim
b→2−
∫ b
1
⌊x⌋ dx
)
+
(
lim
c→3−
∫ c
2
⌊x⌋ dx
)
=
(
lim
b→2−
∫ b
1
1 dx
)
+
(
lim
c→3−
∫ c
2
2 dx
)
= lim
b→2−
(
x
⏐
⏐
⏐
⏐
b
1
)
+ lim
c→3−
(
2x
⏐
⏐
⏐
⏐
c
2
)
= lim
b→2−
(b − 1) + lim
c→3−
(2c − 4) = (2 − 1) + (6 − 4) = 3
Similar to some of the above examples, the following result i s easy to prove (see the exercises):
For any real number a > 0, the improper integral
∫ ∞
a
dx
x p
is convergent if p > 1, and divergent if 0 < p ≤ 1.",ElementaryCalculus.pdf
"Improper Integrals • Section 5.5 157
The following test for convergence or divergence is sometim es helpful:
Comparison T est for Improper Integrals :
(a) If | f (x)| ≤ g(x) for all x in [ a, ∞), and if
∫ ∞
a g(x) dx is convergent, then
∫ ∞
a f (x) dx is
convergent.
(b) If f (x) ≥ g(x) ≥ 0 for all x in [ a, ∞), and if
∫ ∞
a g(x) dx is divergent, then
∫ ∞
a f (x) dx is
divergent.
The idea behind part (a) is that if −g(x) ≤ f (x) ≤ g(x) over [ a, ∞), then—thinking of improper
integrals as areas—the integral of f is “squeezed” between the two finite integrals for ±g.
There are, however , some subtle issues to prove about the lim it in the integral of f —finite
bounds might not necessarily mean the limit exists. 5
Example 5.36
Show that
∫ ∞
1
sin x
x2 dx is convergent.
y
x0
1
y = 1
x2
y = −1
x2
sin x
x2
Solution: By Example 5.29, the integral
∫ ∞
1
1
x2 dx is convergent. So
since |sin x | ≤1 for all x, then
⏐
⏐
⏐
⏐
sin x
x2
⏐
⏐
⏐
⏐ ≤ 1
x2
for all x in [1 , ∞). Thus, by the Comparison Test,",ElementaryCalculus.pdf
"∫ ∞
1
1
x2 dx is convergent. So
since |sin x | ≤1 for all x, then
⏐
⏐
⏐
⏐
sin x
x2
⏐
⏐
⏐
⏐ ≤ 1
x2
for all x in [1 , ∞). Thus, by the Comparison Test,
∫ ∞
1
sin x
x2 dx is
convergent. The graph on the right shows how the curve y = sin x
x2
is bounded between the curves y = ±1
x2 .
The rules and properties from Section 5.3 concerning definit e integrals still apply to im-
proper integrals, provided the improper integrals are conv ergent. For example, suppose a func-
tion f has a discontinuity or vertical asymptote at x = c. If both improper integrals
∫ c
a f (x) dx
and
∫ b
c f (x) dx are convergent, then the improper integral
∫ b
a f (x) dx is convergent and
∫ b
a
f (x) dx =
∫ c
a
f (x) dx +
∫ b
c
f (x) dx .
Likewise, if
∫ c
a f (x) dx and
∫ ∞
c f (x) dx are convergent, then so is
∫ ∞
a f (x) dx, with
∫ ∞
a
f (x) dx =
∫ c
a
f (x) dx +
∫ ∞
c
f (x) dx .
5See pp.140-141 in B U C K, R.C., Advanced Calculus , 2nd ed., New Y ork: McGraw-Hill Book Co., 1965.",ElementaryCalculus.pdf
"158 Chapter 5 • The Integral §5.5
Exercises
A
For Exercises 1-15, evaluate the given improper integral.
1.
∫ ∞
1
dx
x3 2.
∫ 1
0
dx
3∝ra⌈〉⌋allowx 3.
∫ ∞
0
e−x dx 4.
∫ ∞
0
e−2x dx 5.
∫ 1
−1
dx
x
6.
∫ ∞
0
x e−x2
dx 7.
∫ 0
−∞
2x dx 8.
∫ π/2
0
tan x dx 9.
∫ 1
0
ln x
x dx 10.
∫ 1
−1
dx∝ra⌈〉⌋allow
1 − x2
11.
∫ 3
0
⌈x⌉ dx 12.
∫ ∞
−∞
dx
x2 + 4 13.
∫ 1
0
dx
(x − 1)3 14.
∫ ∞
2
dx
x ln x 15.
∫ ∞
1
dx
x
∝ra⌈〉⌋allow
x2 − 1
16. In a standby system of two non-identical components, the normal operating comp onent A has a
failure rate of λA > 0 failures per unit time, while the standby component B—whic h takes over when
A fails—has a failure rate λB > 0 (with λA ̸=λB ).
(a) Find the standby system’s reliability R(t) beyond time t ≥ 0, where
R(t) =
∫ ∞
t
λA λB
λA − λB
(
e−λB x − e−λA x
)
dx .
(b) Show that the system’s mean time to failure (MTTF) m, where m =
∫ ∞
0 R(t) dt , is m = 1
λA
+ 1
λB
.
17. Show that for all a > 0,
∫ ∞
a
dx
x p is convergent if p > 1, and divergent if 0 < p ≤ 1.",ElementaryCalculus.pdf
"∫ ∞
0 R(t) dt , is m = 1
λA
+ 1
λB
.
17. Show that for all a > 0,
∫ ∞
a
dx
x p is convergent if p > 1, and divergent if 0 < p ≤ 1.
18. Show that for all a > 0,
∫ a
0
dx
x p is convergent if 0 < p < 1, and divergent if p ≥ 1.
19. Is
∫ ∞
1
dx
x + x4 convergent? Explain. 20. Is
∫ ∞
2
dx
x − ∝ra⌈〉⌋allowx convergent? Explain.
B
21. Example 5.31 showed that
∫ ∞
0 sin x dx is divergent. What is the flaw in the argument that the
integral must be 0 since each “hump” of sin x above the x-axis is canceled by one below the x-axis?
22. This exercise concerns the subtraction rule
∫ ∞
a ( f (x) − g(x)) dx =
∫ ∞
a f (x) dx −
∫ ∞
a g(x) dx.
(a) Show that 1
x(x+1) = 1
x − 1
x+1 for all x except 0 and -1
(b) Show that
∫ ∞
1
dx
x(x+1) is convergent.
(c) Show that both
∫ ∞
1
dx
x and
∫ ∞
1
dx
x+1 are divergent.
(d) Does part (c) contradict parts(a)-(b) and the subtraction r ule? Explain.
x
y
0− 1
n
1
n
n
y = Dn (x)
23. The improper integral
∫ ∞
−∞ δ(x) dx = 1 is one of the notable",ElementaryCalculus.pdf
"x
y
0− 1
n
1
n
n
y = Dn (x)
23. The improper integral
∫ ∞
−∞ δ(x) dx = 1 is one of the notable
“improprieties” of the Dirac delta function δ. One way to
think of that integral is by approximating δ by triangular
“pulse” functions Dn (for n ≥ 1), as in the picture on the right.
(a) Write a formula for each Dn (x) over all of R.
(b) Show that
∫ ∞
−∞ Dn (x) dx = 1 for all integers n ≥ 1.
(c) Show that lim n→∞ Dn (0) = ∞ =δ(0).
(d) Do the Dn functions begin to resemble δ as n → ∞?",ElementaryCalculus.pdf
"CHAPTER 6
Methods of Integration
6.1 Integration by Parts
In physics and engineering the Gamma function
1 Γ(t), defined by
Γ(t) =
∫ ∞
0
xt−1 e−x dx for all t > 0,
has found many uses. Evaluating Γ(2) entails integrating the function f (x) = x e −x. No formula
or substitution you have learned so far would be of help. Differentiating that function, on the
other hand, is easy . By the Product Rule for differentials,
d(x e −x) = x d (e−x) + d(x) e−x
= −x e −x dx + e−x dx
d(x e −x) = −x e −x dx − d(e−x)
x e −x dx = −d(x e −x) − d(e−x), so integrate both sides to get
∫ 
x e −x dx = −
∫ 
d(x e −x) −
∫ 
d(e−x) = −x e −x − e−x + C
since
∫ 
dF = F + C. Generalizing this process for functions u and v,
d(uv) = u dv + v du
u dv = d(uv) − v du
so that integrating both sides yields the integration by parts formula:
For differentiable functions u and v:
∫ 
u dv = uv −
∫ 
v du (6.1)
1Created by the Swiss mathematician, physicist and astronom er Leonhard Euler (1707-1783). The use of the",ElementaryCalculus.pdf
"∫ 
u dv = uv −
∫ 
v du (6.1)
1Created by the Swiss mathematician, physicist and astronom er Leonhard Euler (1707-1783). The use of the
Greek capital letter Γfor this function is due to the French mathematician Adrien- Marie Legendre (1752-1833).
159",ElementaryCalculus.pdf
"160 Chapter 6 • Methods of Integration §6.1
Integration by parts is just the Product Rule for derivative s in integral form, typically used
when the integral
∫ 
v du would be simpler than the original integral
∫ 
u dv .
Example 6.1
Use integration by parts to evaluate
∫ 
x e −x dx . Use the answer to evaluate Γ(2).
Solution: The original integral is always of the form
∫ 
u dv, so you must decide which parts of x e−x dx
will represent u and dv. Typically you would choose dv to be a differential that you could integrate
easily (since you will need to integrate dv to get v) and choose u to be a function whose derivative is
simpler than u (since that derivative will appear in v du, which you hope to be a simpler integral). In
this case, pick u = x and dv = e−x dx. Then du = dx and v =
∫ 
dv =
∫ 
e−x dx = −e−x (you can omit the
generic constant C for now—include it when you have evaluated
∫ 
v du). Thus,
∫ 
u dv = uv −
∫ 
v du
∫ 
x
u
e−x dx
dv
= x
u
(−e−x)
v
−
∫ 
−e−x
v
dx
du
∫",ElementaryCalculus.pdf
"∫ 
v du). Thus,
∫ 
u dv = uv −
∫ 
v du
∫ 
x
u
e−x dx
dv
= x
u
(−e−x)
v
−
∫ 
−e−x
v
dx
du
∫ 
x e −x dx = −x e −x − e−x + C
which agrees with the example at the beginning of this sectio n. Note that
∫ 
v du =
∫ 
−e−x dx, which
indeed is simpler than the original integral. The Gamma func tion value Γ(2) can now be evaluated:
Γ(2) =
∫ ∞
0
x e −x dx
= −x e −x − e−x
⏐
⏐
⏐
⏐
∞
0
= limx→∞ (−x e −x − e−x ) − (−0 e0 − e0)
= −
(
limx→∞
x
ex
)
−
(
limx→∞ e−x
)
− (0 − 1)
= 0 − 0 + 1 = 1
Example 6.2
What would happen in Example 6.1 if you let u = e−x and dv = x dx?
Solution: In this case du = −e−x dx and v =
∫ 
dv =
∫ 
x dx = 1
2 x2, so that
∫ 
x e −x dx =
∫ 
e−x
u
x dx
dv
= e−x
u
1
2 x2
v
−
∫ 1
2 x2
v
(−e−x) dx
du
= 1
2 x2 e−x + 1
2
∫ 
x2 e−x dx
which leads you in the wrong direction: a more difficult integ ral than the original.",ElementaryCalculus.pdf
"Integration by Parts • Section 6.1 161
Example 6.2 showed the importance of an appropriate choice f or u and dv. There are some
rough guidelines for that choice—as in Example 6.1—but no ru les that are guaranteed to
always work. It might not be clear when you should even attemp t integration by parts.
Example 6.3
Evaluate
∫ 
ln x dx .
Solution: Integration by parts ostensibly requires two functions in t he integral, whereas here ln x ap-
pears to be the only one. However , the choice for dv is a differential, and one exists here: dx. Choosing
dv = dx obliges you to let u = ln x. Then du = 1
x dx and v =
∫ 
dv =
∫ 
dx = x. Now integrate by parts:
∫ 
u dv = uv −
∫ 
v du
∫ 
ln x dx = (ln x) ( x) −
∫ 
x · 1
x dx
= x ln x −
∫ 
1 dx
= x ln x − x + C
Note that choosing dv = ln x dx would be pointless, as integrating dv to get v is the original problem!
Example 6.4
Evaluate
∫ 
x3 ex2
dx .
Solution: One frequently useful guideline for integration by parts is to eliminate the most complicated",ElementaryCalculus.pdf
"Example 6.4
Evaluate
∫ 
x3 ex2
dx .
Solution: One frequently useful guideline for integration by parts is to eliminate the most complicated
function in the integral by integrating it—as dv—into something simpler (which becomes v). In this
integral, ex2
is somewhat complicated but has no closed form antiderivati ve. However , x e x2
appears in
the integral and can be integrated easily (using a substitut ion as in Section 5.4). So choose dv = x e x2
dx,
which means u = x2 . Then du = 2x dx and v =
∫ 
dv =
∫ 
x e x2
dx = 1
2 ex2
. Now integrate by parts:
∫ 
u dv = uv −
∫ 
v du
∫ 
x3 ex2
dx = x2 · 1
2 ex2
−
∫ 1
2 ex2
· 2x dx
= x2
2 ex2
−
∫ 
x e x2
dx
= x2
2 ex2
− 1
2 ex2
+ C
= (x2 − 1)
2 ex2
+ C
Sometimes multiple rounds of integration by parts are neede d, as in the following example.",ElementaryCalculus.pdf
"162 Chapter 6 • Methods of Integration §6.1
Example 6.5
Evaluate
∫ 
x2 e−x dx .
Solution: This integral appears similar to the one in Example 6.1, so ch oose dv = e−x dx and u = x2 .
Then du = 2x dx and v =
∫ 
e−x dv = −e−x. Now integrate by parts:
∫ 
u dv = uv −
∫ 
v du
∫ 
x2 e−x dx = x2 · (−e−x) −
∫ 
−e−x · 2x dx
= −x2 e−x + 2
∫ 
x e −x dx (integrate by parts again)
∫ 
x2 e−x dx = −x2 e−x + 2 (−x e −x − e−x) + C (by Example 6.1)
= −x2 e−x − 2x e −x − 2 e−x + C
In the above example, notice that the u = 2x in the second integral came from the derivative
of the u = x2 in the first integral. Likewise, the dv = −e−x dx in the second integral came from
integrating the dv = e−x dx in the first integral. In general, if n rounds of integration by parts
were needed, with u i and vi representing the u and v, respectively , for round i = 1, 2, . . ., n,
then the repeated integration by parts would look like this:
∫ 
u1 dv1 = u1 v1 −
∫ 
v1 du1
= u1 v1 −
∫ 
u2 dv2
= u1 v1 −
(
u2 v2 −
∫ 
v2 du2
)",ElementaryCalculus.pdf
"then the repeated integration by parts would look like this:
∫ 
u1 dv1 = u1 v1 −
∫ 
v1 du1
= u1 v1 −
∫ 
u2 dv2
= u1 v1 −
(
u2 v2 −
∫ 
v2 du2
)
= u1 v1 − u2 v2 +
∫ 
u3 dv3
= u1 v1 − u2v2 +
(
u3 v3 −
∫ 
u4 dv4
)
= u1 v1 − u2v2 + u3 v3 −
(
u4v4 −
∫ 
u5 dv5
)
= ···
= u1 v1 − u2v2 + u3 v3 − u4 v4 + u5 v5 − ···
∫ 
un dvn
The last integral
∫ 
un dvn is one you could presumably integrate easily .",ElementaryCalculus.pdf
"Integration by Parts • Section 6.1 163
The above procedure is called the tabular method for integration by parts, since it can be
shown in a table (the arrows indicate multiplication):
u dv
u1 dv1
u2 v1 (+) + u1 v1
u3 v2 (–) − u2 v2
u3 v3 (+) + u3 v3
u4 v4 (–) − u4 v4
.
.
. .
.
. .
.
.
The idea is to differentiate down the u column and integrate down the dv column. If the u
in the original integral is a polynomial of degree n, then you know from Section 1.6 that its
(n + 1)-st derivative will be 0, at which point the tabular method terminates. The integral is
then the sum of the indicated products with alternating sign s.
For example, the tabular method on the integral from Example 6.5 looks like this:
u dv
x2 e−x dx
2x −e−x (+) + (x2) (−e−x)
2 e−x (–) − (2x) ( e−x)
0 −e−x (+) + (2) (−e−x)STOP
The integral is the sum of the products, and agrees with the re sult in Example 6.5:
∫ 
x2 e−x dx = + (x2) (−e−x) − (2x) ( e−x) + (2) (−e−x) + C = −x2 e−x − 2x e −x − 2 e−x + C
Example 6.6",ElementaryCalculus.pdf
"∫ 
x2 e−x dx = + (x2) (−e−x) − (2x) ( e−x) + (2) (−e−x) + C = −x2 e−x − 2x e −x − 2 e−x + C
Example 6.6
Evaluate
∫ 
x3 e−x dx .
Solution: Use the tabular method with u = x3 and dv = e−x dx:
u dv
x3 e−x dx
3x2 −e−x (+) + (x3) (−e−x)
6x e−x (–) − (3x2) ( e−x)
6 −e−x (+) + (6x) (−e−x)
0 e−x (–) − (6) ( e−x)STOP
∫ 
x3 e−x dx = −x3 e−x − 3x2 e−x − 6x e −x − 6 e−x + C",ElementaryCalculus.pdf
"164 Chapter 6 • Methods of Integration §6.1
Integration by parts can sometimes result in the original in tegral reappearing, allowing it
to be combined with the original integral.
Example 6.7
Evaluate
∫ 
sec3 x dx .
Solution: Let u = sec x and dv = sec2 x dx , so that du = sec x tan x dx and v =
∫ 
dv =
∫ 
sec2 x dx = tan x.
Then
∫ 
u dv = uv −
∫ 
v du
∫ 
sec3 x dx = sec x tan x −
∫ 
sec x tan2 x dx
∫ 
sec3 x dx = sec x tan x −
∫ 
sec x (sec2 x − 1) dx
∫ 
sec3 x dx = sec x tan x +
∫ 
sec x dx −
∫ 
sec3 x dx
2
∫ 
sec3 x dx = sec x tan x + ln | sec x + tan x | + C
∫ 
sec3 x dx = 1
2 (sec x tan x + ln | sec x + tan x |) + C
Example 6.8
Evaluate
∫ 
ex sin x dx .
Solution: Let u = ex and dv = sin x dx, so that du = ex dx and v =
∫ 
dv =
∫ 
sin x dx = −cos x. Then
∫ 
u dv = uv −
∫ 
v du
∫ 
ex sin x dx = −ex cos x +
∫ 
ex cos x dx
and so integration by parts is needed again, for the integral on the right: let u = ex and dv = cos x dx, so
that du = ex dx and v =
∫ 
dv =
∫",ElementaryCalculus.pdf
"and so integration by parts is needed again, for the integral on the right: let u = ex and dv = cos x dx, so
that du = ex dx and v =
∫ 
dv =
∫ 
cos x dx = sin x. Then
∫ 
ex sin x dx = −ex cos x +
(
uv −
∫ 
v du
)
∫ 
ex sin x dx = −ex cos x +
(
ex sin x −
∫ 
ex sin x dx
)
2
∫ 
ex sin x dx = −ex cos x + ex sin x
∫ 
ex sin x dx = ex
2 (sin x − cos x) + C",ElementaryCalculus.pdf
"Integration by Parts • Section 6.1 165
In Example 6.1 integration by parts was used in evaluating an improper integral. In general,
in definite or improper integrals where a and b are real numbers or ±∞,
∫ b
a
u dv = uv
⏐
⏐
⏐
⏐
b
a
−
∫ b
a
v du .
Example 6.9
Evaluate
∫ 1
0
x3
√
1 − x2 dx .
Solution: Since x3 ∝ra⌈〉⌋allow
1 − x2 = x2 · x
∝ra⌈〉⌋allow
1 − x2 , and x
∝ra⌈〉⌋allow
1 − x2 is easy to integrate (via a substitution), let
u = x2 and dv = x
∝ra⌈〉⌋allow
1 − x2 dx. Then du = 2x dx and v =
∫ 
dv =
∫ 
x
∝ra⌈〉⌋allow
1 − x2 dx = −1
3 (1 − x2 )3/2, and so:
∫ b
a
u dv = uv
⏐
⏐
⏐
⏐
b
a
−
∫ b
a
v du
∫ 1
0
x3
√
1 − x2 dx = − x2
3 (1 − x2)3/2
⏐
⏐
⏐
⏐
1
0
+
∫ 1
0
2x
3 (1 − x2)3/2 dx
= (0 − 0) +
(
−2
15 (1 − x2)5/2
⏐
⏐
⏐
⏐
1
0
)
= 0 + 2
15 = 2
15
Exercises
A
For Exercises 1-25, evaluate the given integral.
1.
∫ 
x ln x dx 2.
∫ 
x2 ex dx 3.
∫ 
x cos x dx 4.
∫ 
x 3x dx 5.
∫ 
x2 ax dx (a > 0)
6.
∫ 
ln 4 x dx 7.
∫ 
ln x2 dx 8.
∫ 
x2 sin x dx 9.
∫ 
x cos2 x dx 10.
∫ 
sin x cos 2 x dx
11.
∫",ElementaryCalculus.pdf
"∫ 
x cos x dx 4.
∫ 
x 3x dx 5.
∫ 
x2 ax dx (a > 0)
6.
∫ 
ln 4 x dx 7.
∫ 
ln x2 dx 8.
∫ 
x2 sin x dx 9.
∫ 
x cos2 x dx 10.
∫ 
sin x cos 2 x dx
11.
∫ 
sin−1 x dx 12.
∫ 
cos−1 2x dx 13.
∫ 
tan−1 3x dx 14.
∫ 
x sec2 x dx 15.
∫ 
sin x sin 3 x dx
16.
∫ ln x
x3 dx 17.
∫ 
x3 ln2 x dx 18.
∫ 
x5 ex dx 19.
∫ 2
0
x3 dx∝ra⌈〉⌋allow
4 − x2
20.
∫ 1
0
x3
√
1 + x2 dx
21.
∫ 
sin (ln x) dx 22.
∫ 
ln (1 + x2 ) dx 23.
∫ 
x tan−1 x dx 24.
∫ 
cot−1 ∝ra⌈〉⌋allow
x dx 25.
∫ 
e
∝ra⌈〉⌋allowx dx
26. Evaluate the integral
∫ 
ex sin x dx from Example 6.8 by using two rounds of the tabular method
and the formula
∫ 
u1 dv1 = u1 v1 − u2 v2 +
∫ 
v2 du2 from p.162.
B
27. Show that for all constants a and b ̸=0 :
∫ 
ea x cos bx dx = ea x (a cos bx + b sin bx )
a2 + b2 + C and
∫ 
ea x sin bx dx = ea x (a sin bx − b cos bx )
a2 + b2 + C",ElementaryCalculus.pdf
"166 Chapter 6 • Methods of Integration §6.1
28. For the Gamma function Γ(t) show the following:
(a) Γ(t + 1) = t Γ(t) for all t > 0. (Hint: Use integration by parts.)
(b) Γ(n) = (n − 1) ! for all positive integers n. (Hint: Use part (a) and induction.)
Note that by part (b) the Gamma function can be thought of as an extension of the factorial operation
to all positive real numbers. In fact, the Gamma function was created for that purpose.
29. Use Exercise 28 to prove for all integers n ≥ 1:
∫ ∞
0
rn e−r ln r dr = (n − 1) ! + n
∫ ∞
0
rn−1 e−r ln r dr
30. By the Maxwell speed distribution for gas molecules, the average speed 〈ν〉 of molecules of mass m
in a gas at temperature T is
〈ν〉 = 4π
( m
2πkT
) 3/2 ∫ ∞
0
ν3 e−mν2 /2kT dν ,
where k ≈ 1.38056 × 10−23 J/K is the Boltzmann constant. Show that
〈ν〉 =
√
8kT
πm .
31. Some physics texts write integrals in a form like this energy integral from statistical mechanics,
∫ ∞
0
ln
(
1 − αe−x2 )
d(x3 )",ElementaryCalculus.pdf
"〈ν〉 =
√
8kT
πm .
31. Some physics texts write integrals in a form like this energy integral from statistical mechanics,
∫ ∞
0
ln
(
1 − αe−x2 )
d(x3 )
which uses the differential of a function—in this case d(x3 )—instead of a variable (e.g. not just plain
dx). This often signals that integration by parts is on the way , with the added benefit of having the
v =
∫ 
dv calculation done for you—in the above integral d(x3 ) means that v = x3 , with dv = d(x3 ) not
really being needed for anything else. With that understand ing, show that for 0 < α < 1,
∫ ∞
0
ln
(
1 − αe−x2 )
d(x3 ) = −2α
∫ ∞
0
x4 e−x2
dx
1 − αe−x2 .
32. Find the flaw in the following “proof ” that 0 = 1:
Evaluating the integral
∫ dx
x using integration by parts with u = 1
x and dv = dx, so that du = −dx
x2
and v = x, shows that
∫ 
u dv = uv −
∫ 
v du
∫ dx
x =
(1
x
)
· x −
∫ 
x ·
(
− dx
x2
)
∫ dx
x = 1 +
∫ dx
x
0 = 1 ✓",ElementaryCalculus.pdf
"T rigonometric Integrals • Section 6.2 167
6.2 T rigonometric Integrals
In engineering applications you sometimes encounter integ rals of the form
∫ 
cos ( αt + φ1) cos ( βt + φ2) dt
where αt + φ1 and βt + φ2 are different angles (e.g. when the voltage and current are o ut of
phase in an AC circuit). In general, integrals involving pro ducts of sines and cosines with
“mixed” angles can be simplified with the useful product-to- sum formulas:
2
sin A cos B = 1
2 (sin ( A + B) + sin ( A − B)) (6.2)
cos A sin B = 1
2 (sin ( A + B) − sin ( A − B)) (6.3)
cos A cos B = 1
2 (cos ( A + B) + cos ( A − B)) (6.4)
sin A sin B = −1
2 (cos ( A + B) − cos ( A − B)) (6.5)
Example 6.10
Evaluate
∫ 
0.5 sin x sin 12 x dx .
Solution: Using the product-to-sum formula (6.5) with A = x and B = 12x,
sin A sin B = −1
2 (cos ( A + B) − cos ( A − B))
sin x sin 12 x = −1
2 (cos ( x + 12x) − cos ( x − 12x))
sin x sin 12 x = −1
2 (cos 13 x − cos 11 x)
since cos ( −11x) = cos 11 x. Then
∫ 
0.5 sin x sin 12 x dx = −1
4",ElementaryCalculus.pdf
"2 (cos ( x + 12x) − cos ( x − 12x))
sin x sin 12 x = −1
2 (cos 13 x − cos 11 x)
since cos ( −11x) = cos 11 x. Then
∫ 
0.5 sin x sin 12 x dx = −1
4
∫ 
(cos 13 x − cos 11 x) dx
= − 1
52 sin 13 x + 1
44 sin 11 x + C
Notice how the product-to-sum formula turned an integral of products of sines into integrals of individ-
ual cosines, which are easily integrated. The integrand is a n example of a modulated wave , commonly
used in electronic communications (e.g. radio broadcastin g). The graph is shown below:
y
x0
0.5
−0.5
y = 0.5 sin x sin 12 x
π 2π
The curves y = ±0.5 sin x (shown in dashed lines) form an amplitude envelope for the modulated wave.
2See Section 3.4 in C O R R A L, M., Trigonometry, http://mecmath.net/trig/, 2009.",ElementaryCalculus.pdf
"168 Chapter 6 • Methods of Integration §6.2
On occasion you might need to integrate trigonometric funct ions raised to powers higher
than two. For the sine function raised to odd powers of the for m 2 n + 1 (for n ≥ 1), the trick is
to replace sin 2 x by 1 − cos2 x, so that
∫ 
sin2n+1 x dx =
∫ 
(sin2 x)n sin x dx
=
∫ 
(1 − cos2 x)n sin x dx
=
∫ 
p(u) du
where p(u) is a polynomial in the variable u = cos x, and the remaining single sin x is now part
of du = −sin x dx . Y ou can then use the Power Formula to integrate that polynom ial.
Example 6.11
Evaluate
∫ 
sin3 x dx .
Solution: Let u = cos x so that du = −sin x dx :
∫ 
sin3 x dx =
∫ 
(sin2 x) sin x dx
=
∫ 
(1 − cos2 x) sin x dx
=
∫ 
(1 − u2 ) ( −du)
=
∫ 
(u2 − 1) du
= 1
3 u3 − u + C
= 1
3 cos3 x − cos x + C
In general
∫ 
sin2n+1 x dx will be a polynomial of degree 2 n + 1 in terms of cos x. Similarly , use
cos2 x = 1 − sin2 x to integrate odd powers of cos x, with the substitution u = sin x:
∫ 
cos2n+1 x dx =
∫ 
(cos2 x)n cos x dx
=",ElementaryCalculus.pdf
"cos2 x = 1 − sin2 x to integrate odd powers of cos x, with the substitution u = sin x:
∫ 
cos2n+1 x dx =
∫ 
(cos2 x)n cos x dx
=
∫ 
(1 − sin2 x)n
p(u)
cos x dx
du",ElementaryCalculus.pdf
"T rigonometric Integrals • Section 6.2 169
Integrals of the form
∫ 
sinm x cosn x dx , where either m or n is odd, can be evaluated using
the above trick for the function having the odd power .
Example 6.12
Evaluate
∫ 
sin2 x cos3 x dx .
Solution: Replace cos 2 x by 1 − sin2 x, then let u = sin x so that du = cos x dx :
∫ 
sin2 x cos3 x dx =
∫ 
sin2 x (cos2 x) cos x dx
=
∫ 
sin2 x (1 − sin2 x)
p(u)
cos x dx
du
=
∫ 
(u2 − u4 ) du
= 1
3 u3 − 1
5 u5 + C
= 1
3 sin3 x − 1
5 sin5 x + C
For even powers of sin x or cos x. Y ou would replace sin 2 x or cos 2 x with either
sin2 x = 1 − cos 2 x
2 or cos 2 x = 1 + cos 2 x
2 ,
respectively , as often as necessary , then proceed as before if odd powers occur .
Example 6.13
Evaluate
∫ 
sin4 x dx .
Solution: Replace sin 2 x by 1−cos 2 x
2 :
∫ 
sin4 x dx =
∫ (1 − cos 2 x
2
) 2
dx
= 1
4
∫ 
(1 − 2 cos 2 x + cos 2 2x) dx
= 1
4
∫ (
1 − 2 cos 2 x + 1 + cos 4 x
2
)
dx
= 1
8
∫ 
(3 − 4 cos 2 x + cos 4 x) dx
= 3x
8 − 1
4 sin 2 x + 1
32 sin 4 x + C",ElementaryCalculus.pdf
"170 Chapter 6 • Methods of Integration §6.2
Similar methods can be used for integrals of the form
∫ 
secm x tann x dx when either m is
even or n is odd. For an even power m = 2k + 2, use sec 2 x = 1 + tan2 x for all but two of the m
powers of sec x, then use the substitution u = tan x, so that du = sec2 x dx . This results in an
integral of a polynomial p(u) in terms of u = tan x:
∫ 
sec2k+2 x tann x dx =
∫ 
(sec2 x)k sec2 x tann x dx
=
∫ 
(1 + tan2 x)k tann x
p(u)
sec2 x dx
du
Likewise for an odd power n = 2k + 1, use tan 2 x = sec2 x − 1 for all but one of the n powers of
tan x, then use the substitution u = sec x, so that du = sec x tan x dx . This results in an integral
of a polynomial p(u) in terms of u = sec x:
∫ 
secm x tan2k+1 x dx =
∫ 
secm−1 x sec x (tan2 x)k tan x dx
=
∫ 
secm−1 x (sec2 x − 1)k
p(u)
sec x tan xdx
du
Mimic the above procedure for integrals of the form
∫ 
cscm x cotn x dx when either m is even",ElementaryCalculus.pdf
"=
∫ 
secm−1 x (sec2 x − 1)k
p(u)
sec x tan xdx
du
Mimic the above procedure for integrals of the form
∫ 
cscm x cotn x dx when either m is even
or n is odd, using the identity csc 2 x = 1 + cot2 x in a similar manner .
Example 6.14
Evaluate
∫ 
sec4 x tan x dx .
Solution: Use sec 2 x = 1 + tan2 x for one sec 2 x term, then substitute u = tan x, so that du = sec2 x dx :
∫ 
sec4 x tan x dx =
∫ 
sec2 x sec2 x tan x dx
=
∫ 
(1 + tan2 x) tan x sec2 x dx
=
∫ 
(1 + u2 ) u du
=
∫ 
(u + u3 ) du
= 1
2 u2 + 1
4 u4 + C
= 1
2 tan2 x + 1
4 tan4 x + C",ElementaryCalculus.pdf
"T rigonometric Integrals • Section 6.2 171
For some trigonometric integrals try putting everything in terms of sines and cosines.
Example 6.15
Evaluate
∫ cot4 x
csc5 x dx .
Solution: Put cot x and csc x in terms of sin x and cos x:
∫ cot4 x
csc5 x dx =
∫ cos4 x sin5 x
sin4 x
dx
=
∫ 
cos4 x sin x dx (now let u = cos x, du = −sin x dx )
= −
∫ 
u4 dx = −1
5 cos5 x + C
Exercises
A
For Exercises 1-12, evaluate the given integral.
1.
∫ 
sin 2 x cos 5 x dx 2.
∫ 
cos 2 x cos 5 x dx 3.
∫ 
sin 2 x sin 5 x dx 4.
∫ 
cos 2 πx sin 3 πx dx
5.
∫ 
sin3 x cos3/2 x dx 6.
∫ 
cos4 x dx 7.
∫ 
sin6 x dx 8.
∫ 
sin x sin 2 x sin 3 x dx
9.
∫ 
sec4 x dx 10.
∫ 
sec2 x tan3 x dx 11.
∫ tan3 x
sec4 x dx 12.
∫ dx
csc2 x cot x
B
13. Evaluate
∫ 
sin3 x cos3 x dx in two different ways:
(a) Use sin 3 x = (1 − cos2 x) sin x and the substitution u = cos x.
(b) Use cos 3 x = (1 − sin2 x) cos x and the substitution u = sin x.
Are the answers from parts(a) and (b) equivalent? Explain.
14. Evaluate
∫",ElementaryCalculus.pdf
"(b) Use cos 3 x = (1 − sin2 x) cos x and the substitution u = sin x.
Are the answers from parts(a) and (b) equivalent? Explain.
14. Evaluate
∫ 
sec4 x tan x dx by using sec 4 x tan x = sec3 x (sec x tan x) and the substitution u = sec x.
Is your answer equivalent to the answer in Example 6.14? Expl ain.
15. Show that
∫ 4 tan x (1 − tan2 x)
(1 + tan2 x)2 dx = −1
4 cos 4 x + C.
16. The autocorrelation function Rx (τ) of the periodic function x(t) = A cos (ωt + θ) is given by
Rx (τ) = ω
2π
∫ 2π/ω
0
x(t) x(t − τ) dt
where A, ω and θ are constants. Show that
Rx (τ) = A2
2 cos ωτ .",ElementaryCalculus.pdf
"172 Chapter 6 • Methods of Integration §6.3
6.3 T rigonometric Substitutions
One of the fundamental formulas in geometry is for the area A of a circle of radius r: A = πr2.
The calculus-based proof of that formula uses a definite inte gral evaluated by means of a
trigonometric substitution , as will now be demonstrated.
Use the circle of radius r > 0 centered at the origin (0 , 0) in the x y-plane, whose equation is
x2 + y2 = r2 (see Figure 6.3.1(a) below).
x
y
r
θ
r0
(x, y) = (r cos θ, r sin θ)x2 + y2 = r2
(a) Full circle
x
y
r−r
θ = 0θ = π
0
y =
∝ra⌈〉⌋allow
r2 − x2
(b) Upper hemisphere
Figure 6.3.1 Circle of radius r
By symmetry about the x-axis, the area A of the circle is twice the area of its upper hemi-
sphere (see Figure 6.3.1(b) above), which is the area under t he curve y =
∝ra⌈〉⌋allow
r2 − x2:
A = 2
∫ r
−r
√
r2 − x2 dx
To evaluate this integral, recall from trigonometry that an y point ( x, y) on the circle can be",ElementaryCalculus.pdf
"∝ra⌈〉⌋allow
r2 − x2:
A = 2
∫ r
−r
√
r2 − x2 dx
To evaluate this integral, recall from trigonometry that an y point ( x, y) on the circle can be
written as ( x, y) = (r cos θ, r sin θ), where 0 ≤ θ < 2π (in radians) is the angle shown in Figure
6.3.1(a). Figure 6.3.1(b) shows that as x goes from x = −r to x = r, the angle θ goes from θ = π
to θ = 0. Now substitute x = r cos θ and dx = −r sin θ dθ into the integral and change the limits
of integration from x = −r and x = r to θ = π and θ = 0, respectively:
A = 2
∫ 0
π
√
r2 − r2 cos2 θ (−r sin θ) dθ = −2
∫ π
0
√
r2 (1 − cos2 θ) ( −r sin θ) dθ
= 2
∫ π
0
r
√
sin2 θ r sin θ dθ = 2r2
∫ π
0
sin2 θ dθ = ✁2r2
∫ π
0
1 − cos 2 θ
✁2 dθ
= r2
(
θ − 1
2 sin 2 θ
) ⏐
⏐
⏐
⏐
π
0
= r2
(
π − 1
2 sin 2 π −
(
0 − 1
2 sin 0
))
= πr2 ✓",ElementaryCalculus.pdf
"T rigonometric Substitutions • Section 6.3 173
For an indefinite integral of the general form
∫ ∝ra⌈〉⌋allow
a2 − u2 du, the same calculation as above
with the substitutions u = a cos θ and du = −a sin θ dθ yields
∫ √
a2 − u2 du =
∫ √
a2 − a2 cos2 θ (−a sin θ) dθ = −a2
∫ 
sin2 θ dθ
= −a2
∫ 1 − cos 2 θ
2 dθ = −a2
2 θ + a2 sin 2 θ
4 + C ,
which is still in terms of θ. To put this back in terms of u, use θ = cos−1( u
a ), the double-angle
formula sin 2 θ = 2 sin θ cos θ, and
∝ra⌈〉⌋allow
a2 − u2 =
√
a2 sin2 θ = a sin θ. Then
∫ √
a2 − u2 du = −a2
2 cos−1
(u
a
)
+ 2a2 sin θ cos θ
4
= −a2
2 cos−1
(u
a
)
+ (a cos θ) (a sin θ)
2 + C
which results in the following formula:
∫ √
a2 − u2 du = −a2
2 cos−1
(u
a
)
+ 1
2 u
√
a2 − u2 + C (6.6)
It is left as an exercise to show that the substitution u = a sin θ gives:
∫ √
a2 − u2 du = a2
2 sin−1
(u
a
)
+ 1
2 u
√
a2 − u2 + C (6.7)
That these two seemingly different antiderivatives are equ ivalent follows immediately from
the identity sin −1 x + cos−1 x = π",ElementaryCalculus.pdf
"√
a2 − u2 + C (6.7)
That these two seemingly different antiderivatives are equ ivalent follows immediately from
the identity sin −1 x + cos−1 x = π
2 for all −1 ≤ x ≤ 1, which shows that the antiderivatives differ
by the constant πa2
4 (absorbed in the generic constant C):
a2
2 sin−1
(u
a
)
+ 1
2 u
√
a2 − u2 + C = a2
2
(π
2 − cos−1
(u
a
))
+ 1
2 u
√
a2 − u2 + C
= −a2
2 cos−1
(u
a
)
+ 1
2 u
√
a2 − u2 + C + πa2
4
C
Thus, either substitution— u = a cos θ or u = a sin θ—can be used when evaluating the integral∫ ∝ra⌈〉⌋allow
a2 − u2 du. The latter choice is sometimes preferred, to avoid the nega tive sign in du and
the resulting formula.",ElementaryCalculus.pdf
"174 Chapter 6 • Methods of Integration §6.3
Example 6.16
Evaluate
∫ √
9 − 4x2 dx .
Solution: The integrand is of the form
∝ra⌈〉⌋allow
a2 − u2 with a = 3 and u = 2x, so that du = 2dx. Then dx = 1
2 du
and so:
∫ √
9 − 4x2 dx = 1
2
∫ √
a2 − u2 du
= 1
2
(a2
2 sin−1
(u
a
)
+ 1
2 u
√
a2 − u2
)
+ C
= 9
4 sin−1
(2x
3
)
+ 1
2 x
√
9 − 4x2 + C
In general, when other methods fail, use the table below as a g uide for certain types of inte-
grals, making use of the specified substitution and trigonom etric identity:
Integral contains Substitution Identity
∝ra⌈〉⌋allow
a2 − u2 u = a sin θ 1 − sin2 θ = cos2 θ
∝ra⌈〉⌋allow
a2 + u2 u = a tan θ 1 + tan2 θ = sec2 θ
∝ra⌈〉⌋allow
u2 − a2 u = a sec θ sec2 θ − 1 = tan2 θ
For example, the substitution u = a tan θ leads to the following formula:
∫ √
a2 + u2 du = 1
2 u
√
a2 + u2 + a2
2 ln
⏐
⏐ u +
√
a2 + u2 ⏐
⏐ + C (6.8)
Similarly , the substitution u = a sec θ yields this formula:
∫ √
u2 − a2 du = 1
2 u
√
u2 − a2 − a2
2 ln
⏐
⏐ u +
√
u2 − a2 ⏐
⏐ + C (6.9)",ElementaryCalculus.pdf
"⏐ + C (6.8)
Similarly , the substitution u = a sec θ yields this formula:
∫ √
u2 − a2 du = 1
2 u
√
u2 − a2 − a2
2 ln
⏐
⏐ u +
√
u2 − a2 ⏐
⏐ + C (6.9)
The proof of each formula requires this result from Example 6 .7 in Section 6.1:
∫ 
sec3 θ dθ = 1
2 (sec θ tan θ + ln | sec θ + tan θ |) + C (6.10)
The above substitutions can be used even if no square roots ar e present.",ElementaryCalculus.pdf
"T rigonometric Substitutions • Section 6.3 175
Example 6.17
Evaluate
∫ dx
(1 + x2)2 .
Solution: Notice that this integral cannot be evaluated by using the Po wer Formula with the substi-
tution u = 1 + x2 (why?). Integration by parts does not look promising, eithe r . So try a trigonometric
substitution. The integrand contains a term of the form a2 + u2 (with a = 1 and u = x), so use the
substitution x = tan θ. Then dx = sec2 θ dθ and so
∫ dx
(1 + x2 )2 =
∫ sec2 θ dθ
(1 + tan2 θ)2
=
∫ sec2 θ dθ
(sec2 θ)2
=
∫ dθ
sec2 θ2
=
∫ 
cos2 θ dθ
=
∫ 1 + cos 2 θ
2 dθ
= θ
2 + 1
4 sin 2 θ + C
= θ
2 + 1
2 sin θ cos θ + C
by the trigonometric double-angle identity sin 2 θ = 2 sin θ cos θ.
∝ra⌈〉⌋allow1+x2
x
1
θ
The simplest way to get expressions for sin θ and cos θ in terms of x is to draw
a right triangle with an angle θ such that tan θ = x = x
1 , as in the drawing on
the right. The hypotenuse must then be
∝ra⌈〉⌋allow
1 + x2 (by the Pythagorean Theorem),",ElementaryCalculus.pdf
"1 , as in the drawing on
the right. The hypotenuse must then be
∝ra⌈〉⌋allow
1 + x2 (by the Pythagorean Theorem),
which makes it easy to read off the values of sin θ and cos θ:
sin θ = x∝ra⌈〉⌋allow
1 + x2
and cos θ = 1∝ra⌈〉⌋allow
1 + x2
Since θ = tan−1 x, putting the integral back in terms of x yields:
∫ dx
(1 + x2)2 = 1
2 tan−1 x + 1
2
x∝ra⌈〉⌋allow
1 + x2
1∝ra⌈〉⌋allow
1 + x2
+ C
= 1
2 tan−1 x + x
2 (1 + x2) + C
Note: An alternative method for getting sin θ and cos θ in terms of x would be to put tan θ = x in the
identity sec 2 θ = 1 + tan2 θ to solve for cos θ, then use the identity sin 2 θ = 1 − cos2 θ to solve for sin θ.
By completing the square, quadratic expressions in x can be put in one of the forms a2 ± u2 or
u2 − a2, enabling the use of the corresponding trigonometric subst itution.",ElementaryCalculus.pdf
"176 Chapter 6 • Methods of Integration §6.3
Example 6.18
Evaluate
∫ dx
(4x2 + 8x − 5)3/2 .
Solution: This integral cannot be evaluated by using the Power Formula , so try a trigonometric substi-
tution. Complete the square on the expression 4 x2 + 8x − 5:
4x2 + 8x − 5 = 4 ( x2 + 2x) − 5 = 4 ( x2 + 2x + 1) − 5 − 4 = 4 ( x + 1)2 − 9
This expression is now of the form u2 − a2 for u = 2 ( x + 1) and a = 3. Use the substitution u = a sec θ,
which means 2 ( x + 1) = 3 sec θ. Then 2 dx = 3 sec θ tan θ dθ and so:
∫ dx
(4x2 + 8x − 5)3/2 =
∫ dx
(4 ( x + 1)2 − 9)3/2
= 3
2
∫ sec θ tan θ dθ
(9 sec 2 θ − 9)3/2 = 3
2
∫ sec θ tan θ dθ
(9 (sec2 θ − 1))3/2
= 1
18
∫ sec θ tan θ dθ
tan3 θ
= 1
18
∫ sec θ dθ
tan2 θ
= 1
18
∫ cos θ dθ
sin2 θ
= 1
18
∫ 
csc θ cot θ dθ = − 1
18 csc θ + C
2(x+1) ∝ra⌈〉⌋allow
4x2 + 8x − 5
3
θ
To get an expression for csc θ in terms of x, draw a right triangle with
an angle θ such that sec θ = 2(x+1)
3 , as in the drawing on the right.
The side opposite θ must then be
∝ra⌈〉⌋allow",ElementaryCalculus.pdf
"an angle θ such that sec θ = 2(x+1)
3 , as in the drawing on the right.
The side opposite θ must then be
∝ra⌈〉⌋allow
4x2 + 8x − 5 (by the Pythagorean
Theorem), and hence:
csc θ = 2(x + 1)∝ra⌈〉⌋allow
4x2 + 8x − 5
Putting the integral back in terms of x yields:
∫ dx
(4x2 + 8x − 5)3/2 = − 1
18
2(x + 1)∝ra⌈〉⌋allow
4x2 + 8x − 5
+ C
= − x + 1
9
∝ra⌈〉⌋allow
4x2 + 8x − 5
+ C
Note: Trigonometric identities could have been used to obta in csc θ by knowing sec θ.
The following integrals from Section 5.4 might be helpful fo r the exercises:
∫ 
tan u du = ln |sec u | + C (6.11)
∫ 
sec u du = ln |sec u + tan u | + C (6.12)
∫ 
csc u du = −ln |csc u + cot u | + C (6.13)",ElementaryCalculus.pdf
"T rigonometric Substitutions • Section 6.3 177
Exercises
A
For Exercises 1-16, evaluate the given integral.
1.
∫ √
9 + 4x2 dx 2.
∫ √
2 − 3x2 dx 3.
∫ √
4x2 − 9 dx 4.
∫ √
x2 + 2x + 10 dx
5.
∫ ∝ra⌈〉⌋allow
1 − x2
x2 dx 6.
∫ x2 dx∝ra⌈〉⌋allow
x2 − 9
7.
∫ dx
x
∝ra⌈〉⌋allow
1 + x2
8.
∫ dx
x2
∝ra⌈〉⌋allow
a2 + x2
(a > 0)
9.
∫ x3 dx∝ra⌈〉⌋allow
x2 + 4
10.
∫ d x
(4x2 − 9)3/2 11.
∫ d x
(9 + 4x2)2 12.
∫ x2 dx∝ra⌈〉⌋allow
a2 − x2
(a > 0)
13.
∫ x3 dx∝ra⌈〉⌋allow
9 − x2
14.
∫ ∝ra⌈〉⌋allow
4 − x2
x dx 15.
∫ (x − 4) dx∝ra⌈〉⌋allow
−9x2 + 36x − 32
16.
∫ d x
(4x2 + 16x + 15)3/2
17. Prove formula (6.7) directly by using the substitution u = a sin θ.
18. Prove formula (6.8). 19. Prove formula (6.9).
20. Show that using the substitution u = a cot θ to evaluate the integral
∫ ∝ra⌈〉⌋allow
a2 + u2 dx leads to an
antiderivative equivalent to the one in formula (6.8).
21. Show that using the substitution u = a csc θ to evaluate the integral
∫ ∝ra⌈〉⌋allow
u2 − a2 dx leads to an",ElementaryCalculus.pdf
"21. Show that using the substitution u = a csc θ to evaluate the integral
∫ ∝ra⌈〉⌋allow
u2 − a2 dx leads to an
antiderivative equivalent to the one in formula (6.9).
B
22. The integrals
∫ dx
∝ra⌈〉⌋allow
x2 ± a2
can be evaluated without the use of trigonometric substitut ions, by using
differentials:
(a) For u2 = x2 ± a2 , show that
dx
u = d (x + u)
x + u .
(b) Integrate both sides of the result from part (a).
Note: In general, many integrals involving
∝ra⌈〉⌋allow
x2 ± a2 can be handled with a similar manipulation of
differentials, with varying complexity .
23. According to Newtonian physics the path of a photon grazing t he surface of the Sun should be
deflected by the Sun’s gravitational field by an angle θ, given approximately by
θ =
⏐
⏐
⏐
⏐
2 G M R
c2
∫ 0
∞
dy
(R2 + y2 )3/2
⏐
⏐
⏐
⏐
where c = 2.998×108 m/s is the speed of light, G = 6.67×10−11 N/m2 /kg2 is the gravitational constant,
M = 1.99 × 1030 kg is the mass of the Sun, and R = 6.96 × 108 m is the radius of the Sun. Show that",ElementaryCalculus.pdf
"M = 1.99 × 1030 kg is the mass of the Sun, and R = 6.96 × 108 m is the radius of the Sun. Show that
θ = 2 G M
c2 R = 4.24 × 10−6 radians = 2.43 × 10−4 degrees ≈ 0.875 seconds of arc,
where 1 second of arc = 1/3600 of 1 degree. 3
3Albert Einstein published this result in 1911, then showed i n 1915 that the true angle should be double that
amount, due to the curvature of space. Experiments verified E instein’s prediction. See pp.69-71 in S E R WAY, R.A.,
C.J. M O S E S A N D C.A. M OY E R , Modern Physics , Orlando, FL: Harcourt Brace Jovanovich Publishers, 1989.",ElementaryCalculus.pdf
"178 Chapter 6 • Methods of Integration §6.4
6.4 Partial Fractions
In the last two sections some trigonometric integrals were s implified by using various trigono-
metric identities. For integrals of rational functions—qu otients of polynomials—some alge-
braic identities (e.g. x2 − a2 = (x − a)(x + a)) will be useful in the method of partial fractions .
The idea behind this method is simple: replace a complicated rational function with simpler
ones that are easy to integrate.
For example, there is no formula for evaluating the integral
∫ dx
x2 + x
but notice that you can write
1
x2 + x = 1
x (x + 1) = 1
x − 1
x + 1 ,
called the partial fraction decomposition of 1
x2 +x , so that
∫ dx
x2 + x =
∫ (1
x − 1
x + 1
)
dx
= ln |x| − ln |x + 1| + C .
There is a systematic way to find this decomposition. First, a ssume that
1
x (x + 1) = A
x + B
x + 1
for some constants A and B. Get a common denominator on the right side:
1
x (x + 1) = A (x + 1) + B x
x (x + 1)
0x + 1
x (x + 1) = (A + B) x + A",ElementaryCalculus.pdf
"for some constants A and B. Get a common denominator on the right side:
1
x (x + 1) = A (x + 1) + B x
x (x + 1)
0x + 1
x (x + 1) = (A + B) x + A
x (x + 1)
Now equate coefficients in the numerators of both sides to sol ve for A and B:
constant term : A = 1
coefficient of x : A + B = 0 ⇒ B = −A = −1
Thus,
1
x (x + 1) = A
x + B
x + 1 = 1
x + −1
x + 1 = 1
x − 1
x + 1
as before.",ElementaryCalculus.pdf
"Partial Fractions • Section 6.4 179
The partial fraction method can be discussed in general, and its assumptions proved 4, but
only the simplest cases—linear and quadratic factors— will be considered here. In all cases it
will be assumed that the degree of the polynomial in the numer ator of the rational function is
less than the degree of the polynomial in the denominator .
Start with the most basic case, similar to the example above:
Case 1 - Distinct linear factors: A rational function p(x)
q(x) such that degree ( p(x)) <
degree ( q(x)), where q(x) is a product of n > 1 distinct linear factors
q(x) = (a1 x + b1) (a2 x + b2) ··· (an x + bn) ,
can be written as a sum of partial fractions
p(x)
q(x) = A1
a1 x + b1
+ A2
a2 x + b2
+ ··· + A n
an x + bn
for some constants A1, A2, . . . , A n. Those constants can be solved for by getting a common
denominator on the right side of the equation and then equati ng the coefficients of the
numerators of both sides.
Example 6.19
Evaluate
∫ dx",ElementaryCalculus.pdf
"denominator on the right side of the equation and then equati ng the coefficients of the
numerators of both sides.
Example 6.19
Evaluate
∫ dx
x2 − 7x + 10 .
Solution: Since x2 − 7x + 10 = (x − 2) ( x − 5), then
1
x2 − 7x + 10 = 1
(x − 2) ( x − 5) = A
x − 2 + B
x − 5
= ( A + B) x + (−5 A − 2B)
(x − 2) ( x − 5)
so that
coefficient of x : A + B = 0 ⇒ B = −A
constant term : −5 A − 2B = 1 ⇒ − 5 A + 2 A = 1 ⇒ A = −1
3 and B = 1
3
Thus,
∫ dx
x2 − 7x + 10 =
∫ (
− 1
3
x − 2 +
1
3
x − 5
)
dx
= −1
3 ln |x − 2| + 1
3 ln |x − 5| + C
4For example, see Section 5.10 in H I L L MA N , A.P ., A N D G.L. A L E X A N D E R S O N, A First Undergraduate Course in
Abstract Algebra , 3rd ed., Belmont, CA: W adsworth Publishing Co., 1983.",ElementaryCalculus.pdf
"180 Chapter 6 • Methods of Integration §6.4
If one linear factor in the denominator is repeated more than once, and all other factors are
distinct, then use the following decomposition:
Case 2 - One repeated linear factor + distinct linear factors : A rational function
p(x)
q(x) such that degree ( p(x)) < degree ( q(x)), where q(x) is a product of n > 1 distinct linear
factors and one linear factor repeated m > 1 times
q(x) = (ax + b)m (a1 x + b1) (a2 x + b2) ··· (an x + bn ) ,
can be written as a sum of partial fractions
p(x)
q(x) = A1
ax + b + A2
(ax + b)2 + ··· + A m
(ax + b)m + B1
a1 x + b1
+ ··· + Bn
an x + bn
for some constants A1, A2, . . . , A m and B1, . . . , Bn . Those constants can be solved for by
the same method as in Case 1.
Example 6.20
Evaluate
∫ x2 + x − 1
x3 + x2 dx .
Solution: Since x3 + x2 = x2 (x + 1), then
x2 + x − 1
x3 + x2 = x2 + x − 1
x2 (x + 1) = A
x + B
x2 + C
x + 1
= A x (x + 1) + B (x + 1) + C x2
x2 (x + 1)
= ( A + C) x2 + ( A + B) x + B
x2 (x + 1)
so that",ElementaryCalculus.pdf
"x3 + x2 = x2 + x − 1
x2 (x + 1) = A
x + B
x2 + C
x + 1
= A x (x + 1) + B (x + 1) + C x2
x2 (x + 1)
= ( A + C) x2 + ( A + B) x + B
x2 (x + 1)
so that
constant term : B = −1
coefficient of x : A + B = 1 ⇒ A = 1 − B = 2
coefficient of x2 : A + C = 1 ⇒ C = 1 − A = −1
Thus,
∫ x2 + x − 1
x3 + x2 dx =
∫ (2
x + −1
x2 + −1
x + 1
)
dx
= 2 ln |x | + 1
x − ln |x + 1| + C0 (C0 = generic constant)
Case 2 can be extended to more than one repeated factor—the pa rtial fraction decomposition
would then have more terms similar to those for the first repea ted factor .",ElementaryCalculus.pdf
"Partial Fractions • Section 6.4 181
Example 6.21
Evaluate
∫ dx
x2 (x + 1)2 .
Solution: Expanding Case 2 to two repeated factors,
1
x2 (x + 1)2 = A
x + B
x2 + C
x + 1 + D
(x + 1)2
= A x (x + 1)2 + B (x + 1)2 + C x2 (x + 1) + D x2
x2 (x + 1)2
= ( A + C) x3 + (2 A + B + C + D) x2 + ( A + 2B) x + B
x2 (x + 1)
so that
constant term : B = 1
coefficient of x : A + 2B = 0 ⇒ A = −2B = −2
coefficient of x3 : A + C = 0 ⇒ C = −A = 2
coefficient of x2 : 2 A + B + C + D = 0 ⇒ D = −2 A − B − C = 1
Thus,
∫ dx
x2 (x + 1)2 =
∫ (−2
x + 1
x2 + 2
x + 1 + 1
(x + 1)2
)
dx
= −2 ln |x | − 1
x + 2 ln |x + 1| − 1
x + 1 + C0 (C0 = generic constant)
The partial fraction decompositions for quadratic factors are similar to those for linear fac-
tors, except the numerators in each partial fraction can now contain linear terms. A factor of
the form ax 2 +b x+c is considered quadratic only if it cannot be factored into a p roduct of linear
terms (i.e. has no real roots) and a ̸=0.",ElementaryCalculus.pdf
"the form ax 2 +b x+c is considered quadratic only if it cannot be factored into a p roduct of linear
terms (i.e. has no real roots) and a ̸=0.
Case 3 - Distinct quadratic factors: A rational function p(x)
q(x) such that degree ( p(x)) <
degree ( q(x)), and q(x) is a product of n > 1 distinct quadratic factors
q(x) = (a1 x2 + b1 x + c1) (a2 x2 + b2 x + c2) ··· (an x2 + bn x + cn) ,
can be written as a sum of partial fractions
p(x)
q(x) = A1 x + B1
a1 x2 + b1 x + c1
+ A2 x + B2
a2 x2 + b2 x + c2
+ ··· + A n x + Bn
an x2 + bn x + cn
for some constants A1, A2, . . . , A n and B1, B2, . . . , Bn . Those constants can be solved for
by the same method as in Case 1.",ElementaryCalculus.pdf
"182 Chapter 6 • Methods of Integration §6.4
Example 6.22
Evaluate
∫ dx
(x2 + 1) ( x2 + 4) .
Solution: Neither x2 + 1 nor x2 + 4 has real roots, so by Case 3,
1
(x2 + 1) ( x2 + 4) = A x + B
x2 + 1 + C x + D
x2 + 4
= ( A x + B) ( x2 + 4) + (C x + D) ( x2 + 1)
(x2 + 1) ( x2 + 4)
= ( A + C) x3 + (B + D) x2 + (4 A + C) x + (4B + D)
(x2 + 1) ( x2 + 4)
so that
coefficient of x3 : A + C = 0 ⇒ C = −A
coefficient of x2 : B + D = 0 ⇒ D = −B
coefficient of x : 4 A + C = 0 ⇒ 4 A − A = 0 ⇒ A = 0
constant term : 4 B + D = 1 ⇒ 4B − B = 1 ⇒ B = 1
3 and D = −1
3
Thus,
∫ dx
(x2 + 1) ( x2 + 4) =
∫ ( 1
3
x2 + 1 +
− 1
3
x2 + 4
)
dx
= 1
3 tan−1 x − 1
6 tan−1
(x
2
)
+ C0
by formula (5.4) in Section 5.4.
A repeated quadratic factor is handled in the same way as a rep eated linear factor:
Case 4 - One repeated quadratic factor + distinct quadratic f actors: A rational
function p(x)
q(x) such that degree ( p(x)) < degree ( q(x)), where q(x) is a product of n > 1 distinct",ElementaryCalculus.pdf
"function p(x)
q(x) such that degree ( p(x)) < degree ( q(x)), where q(x) is a product of n > 1 distinct
quadratic factors and one quadratic factor repeated m > 1 times
q(x) = (ax 2 + b x + c)m (a1 x2 + b1 x + c1) (a2 x2 + b2 x + c2) ··· (an x2 + bn x + cn) ,
can be written as a sum of partial fractions
p(x)
q(x) = A1 x + B1
ax 2 + b x + c + ··· + A m x + Bm
(ax 2 + b x + c)m + C1 x + D1
a1 x2 + b1 x + c1
+ ··· + Cn x + D n
an x2 + bn x + cn
for some constants A1, . . . , A m, B1, . . . , Bm, C1, . . . , Cn, and D1, . . . , D n. Those constants
can be solved for by the same method as in Case 1.",ElementaryCalculus.pdf
"Partial Fractions • Section 6.4 183
Example 6.23
Evaluate
∫ dx
(x2 + 1)2 (x2 + 4) .
Solution: Neither x2 + 1 nor x2 + 4 has real roots, and x2 + 1 is repeated, so by Case 4,
1
(x2 + 1)2 (x2 + 4) = A x + B
x2 + 1 + C x + D
(x2 + 1)2 + E x + F
x2 + 4
= ( A x + B) ( x2 + 1) ( x2 + 4) + (C x + D) ( x2 + 4) + (E x + F ) ( x2 + 1)2
(x2 + 1) ( x2 + 4)
with the right side of the equation expanded as
( A + E) x5 + (B + F ) x4 + (5 A + C + 2E) x3 + (5B + D + 2F ) x2 + (4 A + 4C + E) x + (4B + 4D + F )
(x2 + 1)2 (x2 + 4)
so that equating coefficients of both sides gives
coefficient of x5 : A + E = 0 ⇒ E = −A
coefficient of x4 : B + F = 0 ⇒ F = −B
coefficient of x3 : 5 A + C + 2E = 0 ⇒ 5 A + C − 2 A = 0 ⇒ C = −3 A
coefficient of x2 : 5 B + D + 2F = 0 ⇒ 5B + D − 2F = 0 ⇒ D = −3B
coefficient of x : 4 A + 4C + E = 0 ⇒ 4 A − 12 A − A = 0 ⇒ A = 0 ⇒ C = 0 and E = 0
constant term : 4 B + 4D + F = 1 ⇒ 4B − 12B − B = 1 ⇒ B = −1
9 ⇒ D = 1
3 and F = 1
9
Thus,
∫ dx
(x2 + 1)2 (x2 + 4) =
∫ (
− 1
9
x2 + 1 +
1
3",ElementaryCalculus.pdf
"constant term : 4 B + 4D + F = 1 ⇒ 4B − 12B − B = 1 ⇒ B = −1
9 ⇒ D = 1
3 and F = 1
9
Thus,
∫ dx
(x2 + 1)2 (x2 + 4) =
∫ (
− 1
9
x2 + 1 +
1
3
(x2 + 1)2 +
1
9
x2 + 4
)
dx
= −1
9 tan−1 x + 1
3
(1
2 tan−1 x + x
2 ( x2 + 1)
)
+ 1
18 tan−1
(x
2
)
+ C0
= 1
18 tan−1 x + x
6 ( x2 + 1) + 1
18 tan−1
(x
2
)
+ C0
where the middle integral on the right is from Example 6.17 in Section 6.3.
When a rational function has a numerator with degree larger t han its denominator , dividing
the numerator by the denominator leaves the sum of a polynomi al and a new rational function
perhaps satisfying the conditions for Cases 1-4. When the nu merator and denominator have
the same degree, a trick like this might be easier .
x2 + 2
(x + 1) ( x + 2) = (x2 + 3x + 2) − 3x
(x + 1) ( x + 2) = x2 + 3x + 2
(x + 1) ( x + 2) − 3x
(x + 1) ( x + 2) = 1 − 3x
(x + 1) ( x + 2)
The last rational function on the right can be integrated usi ng partial fractions.",ElementaryCalculus.pdf
"184 Chapter 6 • Methods of Integration §6.4
Exercises
A
For Exercises 1-12, evaluate the given integral.
1.
∫ dx
x2 − x
2.
∫ x + 1
x2 − x dx 3.
∫ dx
2x2 + 3x − 2 4.
∫ dx
x2 + x − 6
5.
∫ dx
x4 − x2 6.
∫ x
(x − 2)3 dx 7.
∫ x − 1
x2 (x + 1) dx 8.
∫ x2
(x − 1)2 dx
9.
∫ x − 2
x2 (x − 1)2 dx 10.
∫ dx
x4 + x2 11.
∫ dx
x4 + 5x2 + 4 12.
∫ (x − 1)2
(x2 + 1)2 dx
B
13. For all numbers a ̸=b show that
∫ dx
(x − a) ( x − b) = 1
a − b ln
⏐
⏐
⏐
⏐
x − a
x − b
⏐
⏐
⏐
⏐ + C .
14. Let q(x) = (x − a1 ) ( x − a2 ), where a1 ̸=a2 . Show that
q′(x)
q(x) = 1
x − a1
+ 1
x − a2
.
15. For q(x) as in Exercise 14, show that
1
q(x) = 1
q′(a1 ) ( x − a1 ) + 1
q′(a2 ) ( x − a2 ) .
16. Extend Exercise 15 to three distinct linear factors: if q(x) = (x − a1 ) ( x − a2 ) ( x − a3 ) then
1
q(x) = 1
q′(a1 ) ( x − a1 ) + 1
q′(a2 ) ( x − a2 ) + 1
q′(a3 ) ( x − a3) .
This result can be extended to any n ≥ 2 distinct factors, though you do not need to prove that.
17. Find d100
dx100
( 1
x2 − 9x + 20
)
. 18. Find d2020
dx2020
(",ElementaryCalculus.pdf
"17. Find d100
dx100
( 1
x2 − 9x + 20
)
. 18. Find d2020
dx2020
(
(x2 − 1)−1 )
.
19. It is possible to use a form of partial fractions to evaluate i ntegrals that are not rational functions.
For example, evaluate the integral ∫ dx
x + x4/3
by finding constants A, B and C such that
1
x + x4/3 = 1
x (1 + x1/3) = A
x + B x −2/3 + C
1 + x1/3 .
Notice how in the second partial fraction the highest power o f x in the numerator is one less than in
the denominator , similar to a partial fraction for a quadrat ic factor in a rational function.
C
20. Find a different way to evaluate the integral in Exercise 19.",ElementaryCalculus.pdf
"Miscellaneous Integration Methods • Section 6.5 185
6.5 Miscellaneous Integration Methods
The integration methods presented so far are considered “st andard,” meaning every calculus
student should know them. This section will discuss a few add itional methods, some more
common than others. One such method is the Leibniz integral rule for “differentiation under
the integral sign.”
5 This powerful and useful method is best explained with a simp le example.
Recall from Section 5.4 that ∫ 
eαx dx = 1
α eαx + C (6.14)
for any constant α ̸=0. This antiderivative involves the variable x and the constant α. The idea
behind the Leibniz rule is to reverse those roles: view α as the variable and x as a constant.
The antiderivative is then seen as a function of α, and so its derivative can be taken with
respect to α. The big leap that this method makes is to move the differenti ation operation
inside the integral: 6
d
dα
∫ 
eαx dx =
∫ d
dα (eαx) dx =
∫ 
x e αx dx",ElementaryCalculus.pdf
"inside the integral: 6
d
dα
∫ 
eαx dx =
∫ d
dα (eαx) dx =
∫ 
x e αx dx
However , differentiating the right side of formula (6.14) s hows that
d
dα
∫ 
eαx dx = d
dα
(1
α eαx + C
)
= α (x e αx) − 1 · eαx
α2 = 1
α x e αx − 1
α2 eαx
Thus, ∫ 
x e αx dx = 1
α x e αx − 1
α2 eαx + C
which can be verified via integration by parts with the tabula r method:
u dv
x eαx dx
1 1
α eαx (+) + (x) ( 1
α eαx)
0 1
α2 eαx (–) − (1) ( 1
α2 eαx)STOP
∫ 
x e αx dx = 1
α x e αx − 1
α2 eαx + C ✓
5The renowned physicist Richard Feynman (1918-1988) famous ly lamented that the technique was no longer
being taught. See p.72 in F E Y N MA N , R.P ., Surely Y ou’re Joking, Mr . Feynman! , New Y ork: Bantam Books, 1986.
6It can be proved that this is valid when the derivative of the i ntegrand is a continuous function of α, which will
always be the case in this book. See pp.121-122 in S O K O L N I K O FF, I.S., Advanced Calculus , New Y ork: McGraw-
Hill Book Company , Inc., 1939.",ElementaryCalculus.pdf
"186 Chapter 6 • Methods of Integration §6.5
What was actually done in the above example? A known integral,
∫ 
eαx dx = 1
α eαx + C ,
was differentiated with respect to α via the Leibniz rule to produce a new integral,
∫ 
x e αx dx = 1
α x e αx − 1
α2 eαx + C ,
with the constant α treated temporarily—only during the differentiation—as a variable. In
general, that is how the Leibniz rule is used. Typically this means if you want to evaluate a
certain integral with the Leibniz rule, then you “work backw ards” to figure out which integral
you need to differentiate with respect to some constant (e.g . α) in the integrand.
Example 6.24
Use the Leibniz rule to evaluate
∫ dx
(1 + x2)2 .
Solution: By formula (5.4) in Section 5.4,
∫ dx
a2 + x2 = 1
a tan−1 (x
a
)
+ C
for any constant a > 0. So differentiate both sides with respect to a:
d
da
∫ dx
a2 + x2 = d
da
(1
a tan−1 (x
a
)
+ C
)
∫ d
da
( 1
a2 + x2
)
dx = − 1
a2 tan−1 (x
a
)
+ 1
a · 1
1 +
(x
a
) 2 · − x
a2
∫ 
− 2a
(a2 + x2 )2 dx = − 1",ElementaryCalculus.pdf
"a2 + x2 = d
da
(1
a tan−1 (x
a
)
+ C
)
∫ d
da
( 1
a2 + x2
)
dx = − 1
a2 tan−1 (x
a
)
+ 1
a · 1
1 +
(x
a
) 2 · − x
a2
∫ 
− 2a
(a2 + x2 )2 dx = − 1
a2 tan−1 (x
a
)
− x
a (a2 + x2 )
∫ dx
(a2 + x2 )2 = 1
2a3 tan−1 (x
a
)
+ x
2a2 (a2 + x2) + C
That general formula is useful in itself. In particular , for a = 1,
∫ dx
(1 + x2 )2 = 1
2 tan−1 x + x
2 (1 + x2 ) + C ,
which agrees with the result from Example 6.17 in Section 6.3 .
Notice that there was no generic constant (e.g. a or α) in the statement of the problem. When that
happens, you will need to figure out where the constant should be in order to use the Leibniz rule.
Y ou can also use differentiation under the integral sign to e valuate definite integrals.",ElementaryCalculus.pdf
"Miscellaneous Integration Methods • Section 6.5 187
Example 6.25
Show that
∫ ∞
0
e−x2
dx = 1
2
∝ra⌈〉⌋allow
π .
Solution: Let I =
∫ ∞
0 e−x2
dx. The integral is convergent, since by Exercise 11 in Section 4.4, for all x
ex2
≥ 1 + x2 ⇒ 0 ≤ e−x2
≤ 1
1 + x2
implies I is convergent by the Comparison Test, since
∫ ∞
0
1
1+x2 dx is convergent (and equals 1
2 π) by
Example 5.32 in Section 5.5. For α ≥ 0, define
φ(α) =
∫ ∞
0
α e−α2 x2
1 + x2 dx .
Then clearly φ(0) = 0, and differentiating under the integral sign shows
φ′(α) =
∫ ∞
0
−2α2 e−α2 x2
+ e−α2 x2
1 + x2 dx ⇒ φ′(0) =
∫ ∞
0
dx
1 + x2 = 1
2 π .
The substitution y = αx, so that dy = αdx, shows φ(α) can be written as
φ(α) =
∫ ∞
0
e−y2
1 +
(y
α
) 2 dy ⇒ 0 ≤ limα→∞ φ(α) ≤ I < ∞ .
Also, for α > 0,
d
dα
(1
α e−α2
φ(α)
)
= d
dα
∫ ∞
0
e−α2 (1+x2 )
1 + x2 dx =
∫ ∞
0
−2α(1 + x2 ) e−α2 (1+x2 )
1 + x2 dx
= −2α e−α2
∫ ∞
0
e−α2 x2
dx , now substitute u = αx and du = αdx to get
= −2α e−α2 1
α
∫ ∞
0
e−u2
du = −2 e−α2",ElementaryCalculus.pdf
"0
−2α(1 + x2 ) e−α2 (1+x2 )
1 + x2 dx
= −2α e−α2
∫ ∞
0
e−α2 x2
dx , now substitute u = αx and du = αdx to get
= −2α e−α2 1
α
∫ ∞
0
e−u2
du = −2 e−α2
I , and so integrating both sides yields
∫ ∞
0
d
dα
(1
α e−α2
φ(α)
)
dα = −2I
∫ ∞
0
e−α2
dα = −2I2 .
However , by the Fundamental Theorem of Calculus.
∫ ∞
0
d
dα
(1
α e−α2
φ(α)
)
dα = 1
α e−α2
φ(α)
⏐
⏐
⏐
⏐
∞
0
=
(
lim
α→∞
φ(α)
α eα2
)
−
(
lim
α→0
φ(α)
α eα2
)
= 0 −
(
lim
α→0
φ(α)
α eα2
)
→ 0
0 , so by L ’Hôpital’s Rule
= − lim
α→0
φ′(α)
eα2
+ 2α2 eα2 = −φ′(0)
1 + 0 = −1
2 π .
Thus,
−2I2 = −1
2 π ⇒ I = 1
2
∝ra⌈〉⌋allow
π
which is the desired result.",ElementaryCalculus.pdf
"188 Chapter 6 • Methods of Integration §6.5
One immediate consequence of Example 6.25 is that
∫ ∞
−∞
e−x2
dx =
∝ra⌈〉⌋allow
π
since e−x2
is an even function. The following example shows another con sequence, as well as
how useful substitutions can be in writing integrals in a dif ferent form.
Example 6.26
Show that the Gamma function Γ(t) can be written as
Γ(t) = 2
∫ ∞
0
y2t−1 e−y2
dy for all t > 0,
and that Γ
(1
2
)
= ∝ra⌈〉⌋allowπ.
Solution: Let x = y2 , so that dx = 2 y dy . Then x = 0 ⇒ y = 0 and x = ∞ ⇒ y = ∞, so
Γ(t) =
∫ ∞
0
xt−1 e−x dx =
∫ ∞
0
( y2 )t−1 e−y2
2 y dy = 2
∫ ∞
0
y2t−1 e−y2
dy .
In this form, with the help of Example 6.25 it is now easy to eva luate Γ
(1
2
)
:
Γ
(1
2
)
= 2
∫ ∞
0
y1−1 e−y2
dy = 2
∫ ∞
0
e−y2
dy = 2
(1
2
∝ra⌈〉⌋allow
π
)
=
∝ra⌈〉⌋allow
π
A function closely related to the Gamma function is the Beta function B(x, y), defined by:
B(x, y) =
∫ 1
0
tx−1 (1 − t)y−1 dt for all x > 0 and y > 0 (6.15)
It can be shown that 7
B(x, y) = Γ(x) Γ( y)",ElementaryCalculus.pdf
"B(x, y) =
∫ 1
0
tx−1 (1 − t)y−1 dt for all x > 0 and y > 0 (6.15)
It can be shown that 7
B(x, y) = Γ(x) Γ( y)
Γ(x + y) for all x > 0 and y > 0. (6.16)
Example 6.27
Show that the Beta function B(x, y) can be written as
B(x, y) =
∫ ∞
0
ux−1
(1 + u)x+y du .
Solution: Let u = t
1−t , so that t = u
1+u , 1 −t = 1
1+u , and dt = du
(1+u)2 . Then t = 0 ⇒ u = 0 and t = 1 ⇒ u = ∞,
so
B(x, y) =
∫ 1
0
tx−1 (1 − t) y−1 dt =
∫ ∞
0
( u
1 + u
) x−1 ( 1
1 + u
) y−1 du
(1 + u)2 =
∫ ∞
0
ux−1
(1 + u)x+y du .
7See p.18-19 in R A I N V I L L E , E.D., Special Functions , New Y ork: Chelsea Publishing Company , 1971.",ElementaryCalculus.pdf
"Miscellaneous Integration Methods • Section 6.5 189
Another application of substitutions in integrals is in the evaluation of fractional deriva-
tives. Recall from Section 1.6 that the zero-th derivative of a fun ction is just the function
itself, and that derivatives of order n are well-defined for integer values n ≥ 1. It turns out
that derivatives of fractional orders (e.g. n = 1
2 ) can be defined, with the Riemann-Louiville
definition being the most common:
For all 0 < α < 1, the fractional derivative of order α of a function f (x) is
dα
dxα f (x) = 1
Γ(1 − α)
d
dx
∫ x
0
f (t)
(x − t)α dt . (6.17)
Example 6.28
Calculate d1/2
dx1/2 (x) .
Solution: Here α = 1
2 and f (x) = x, so that
d1/2
dx1/2 (x) = 1
Γ(1 − 1/2)
d
dx
∫ x
0
t
(x − t)1/2 dt = 1∝ra⌈〉⌋allowπ
d
dx
∫ x
0
t dt∝ra⌈〉⌋allow
x − t
since Γ
(1
2
)
= ∝ra⌈〉⌋allowπ by Example 6.26. Use the substitution u =
∝ra⌈〉⌋allow
x − t, so that t = x − u2 and dt = −2u du.
Then t = 0 ⇒ u = ∝ra⌈〉⌋allowx and t = x ⇒ u = 0, so
d1/2",ElementaryCalculus.pdf
"∝ra⌈〉⌋allow
x − t, so that t = x − u2 and dt = −2u du.
Then t = 0 ⇒ u = ∝ra⌈〉⌋allowx and t = x ⇒ u = 0, so
d1/2
dx1/2 (x) = 1∝ra⌈〉⌋allowπ
d
dx
∫ 0
∝ra⌈〉⌋allowx
(x − u2 ) (−2u du )
u = 2∝ra⌈〉⌋allowπ
d
dx
∫ ∝ra⌈〉⌋allowx
0
(x − u2 ) du
= 2∝ra⌈〉⌋allowπ
d
dx
(
xu − 1
3 u3 ) ⏐
⏐
⏐
⏐
u=∝ra⌈〉⌋allow
x
u=0
= 2∝ra⌈〉⌋allowπ
d
dx
(
x3/2 − 1
3 x3/2
)
= 2∝ra⌈〉⌋allowπ
d
dx
(
2
3 x3/2
)
= 2∝ra⌈〉⌋allowπ
∝ra⌈〉⌋allowx
Part of the motivation for creating fractional derivatives was to find if it were possible to take
two “half ” derivatives to form a “whole” derivative:
d1/2
dx1/2
( d1/2
dx1/2 f (x)
)
= d
dx f (x)
It is left as an exercise to show that the above relation does h old for the function f (x) = x.
Derivatives with fractional order 0 < α < 1 and integer order n ≥ 1 can be combined by taking
the derivative of integer order first: 8
d n+α
dxn+α f (x) = dα
dxα
( d n
dxn f (x)
)
8For more details about fractional derivatives, as well as ex amples of their applications in physics and engineering,",ElementaryCalculus.pdf
"dxα
( d n
dxn f (x)
)
8For more details about fractional derivatives, as well as ex amples of their applications in physics and engineering,
see O L D H A M, K.B. A N D J. S PA N I E R, The Fractional Calculus , New Y ork: Academic Press, 1974.",ElementaryCalculus.pdf
"190 Chapter 6 • Methods of Integration §6.5
Recall from Section 6.3 that the trigonometric substitutio n x = r cos θ—or its sister substi-
tution x = r sin θ—was motivated by trying to find the area of a circle of radius r. To simplify
matters, let r = 1 so that points on the unit circle can be identified with the an gle θ via that
substitution, with θ as shown in Figure 6.5.1(a) below .
1
x
y
θ
1 0
(x, y) = (cos θ, sin θ)
(a) Identify points by angle
x
y
slope = 1
2
slope = − 1
2
slope = 1
slope = -1
slope = 2
slope = −2
1−1
slope = 0
0
(b) Identify points by slope
Figure 6.5.1 Points on the unit circle x2 + y2 = 1
Figure 6.5.1(b) shows a different identification of points o n the unit circle—by slope. This
will be the basis for a half-angle substitution for evaluating certain integrals.
Let A be the point ( −1, 0), then for any other point P on the unit circle draw a line from A
through P until it intersects the line x = 1, as shown in Figure 6.5.2 below:
x
y
x = 1
0 1A
P
θ/2 θ
1
t",ElementaryCalculus.pdf
"through P until it intersects the line x = 1, as shown in Figure 6.5.2 below:
x
y
x = 1
0 1A
P
θ/2 θ
1
t
∝ra⌈〉⌋allow1+t2
(a) 0 ≤ θ ≤ π
2
x
y
x = 1
1A 1
t
0
θ/2 θ
∝ra⌈〉⌋allow1+t2
(b) π
2 < θ < π
Figure 6.5.2 Half-angle substitution: t = tan
(θ
2
)
= slope of AP
From geometry you know that the inscribed angle that the line A P makes with the x-axis
is half the measure of the central angle θ. So the slope of A P is the tangent of that angle:
tan 1
2 θ = t
1 = t, which is measured along the y-axis and can take any real value. Each point on
the unit circle—except A—can be identified with that slope t.",ElementaryCalculus.pdf
"Miscellaneous Integration Methods • Section 6.5 191
Figure 6.5.2 shows only positive slopes—reflect the picture about the x-axis for negative
slopes. The figure shows that
sin 1
2 θ = t∝ra⌈〉⌋allow
1 + t2
and cos 1
2 θ = 1∝ra⌈〉⌋allow
1 + t2
so that by the double-angle identities for sine and cosine,
sin θ = 2 sin 1
2 θ cos 1
2 θ = 2 t∝ra⌈〉⌋allow
1 + t2
1∝ra⌈〉⌋allow
1 + t2
= 2t
1 + t2
and
cos θ = cos2 1
2 θ − sin2 1
2 θ = 1
1 + t2 − t2
1 + t2 = 1 − t2
1 + t2 .
Since θ = 2 tan −1 t, then
dθ = d
(
2 tan −1 t
)
= 2 dt
1 + t2 .
Below is a summary of the substitution:
Half-angle substitution: The substitution t = tan 1
2 θ yields:
sin θ = 2t
1 + t2 cos θ = 1 − t2
1 + t2 dθ = 2 dt
1 + t2
The half-angle substitution thus turns rational functions of sin θ and cos θ into rational func-
tions of t, which can be integrated using partial fractions or another method.
Example 6.29
Evaluate
∫ dθ
1 + sin θ + cos θ .
Solution: Using t = tan 1
2 θ, the denominator of the integrand is",ElementaryCalculus.pdf
"Example 6.29
Evaluate
∫ dθ
1 + sin θ + cos θ .
Solution: Using t = tan 1
2 θ, the denominator of the integrand is
1 + sin θ + cos θ = 1 + t2
1 + t2 + 2t
1 + t2 + 1 − t2
1 + t2 = 2t + 2
1 + t2
so that
∫ dθ
1 + sin θ + cos θ =
∫ 
2 dt
1+t2
2t+2
1+t2
=
∫ dt
t + 1
= ln |t + 1| + C
= ln
⏐
⏐tan 1
2 θ + 1
⏐
⏐ + C",ElementaryCalculus.pdf
"192 Chapter 6 • Methods of Integration §6.5
Example 6.30
Evaluate
∫ dθ
3 sin θ + 4 cos θ .
Solution: Using t = tan 1
2 θ, the integral becomes
∫ dθ
3 sin θ + 4 cos θ =
∫ 
2 dt
1+t2
3 2t
1+t2 + 4 1−t2
1+t2
=
∫ −1
2t2 − 3t − 2 dt
=
∫ −1
(2t + 1) ( t − 2) dt =
∫ ( A
2t + 1 + B
t − 2
)
dt
where
coefficient of t : A + 2B = 0 ⇒ A = −2B
constant term : −2 A + B = −1 ⇒ 4B + B = −1 ⇒ B = −1
5 and A = 2
5
Thus,
∫ dθ
3 sin θ + 4 cos θ =
∫ ( 2
5
2t + 1 +
− 1
5
t − 2
)
dt = 1
5 ln |2t + 1| − 1
5 ln |t − 2| + C
= 1
5 ln
⏐
⏐2 tan 1
2 θ + 1
⏐
⏐ − 1
5 ln
⏐
⏐tan 1
2 θ − 2
⏐
⏐ + C
By the half-angle substitution t = tan 1
2 θ,
sin θ
1 + cos θ =
2t
1 + t2
1 + t2
1 + t2 + 1 − t2
1 + t2
=
2t
1 + t2
2
1 + t2
= t
which yields the useful half-angle identities: 9
tan 1
2 θ = sin θ
1 + cos θ = 1 − cos θ
sin θ (6.18)
Example 6.31
Evaluate
∫ sin θ
1 + cos θ dθ .
Solution: Though you could use the half-angle substitution t = tan 1
2 θ, it is easier to use the half-angle
identity (6.18) directly , since
∫ sin θ",ElementaryCalculus.pdf
"Solution: Though you could use the half-angle substitution t = tan 1
2 θ, it is easier to use the half-angle
identity (6.18) directly , since
∫ sin θ
1 + cos θ dθ =
∫ 
tan 1
2 θ dθ = 2 ln
⏐
⏐sec 1
2 θ
⏐
⏐ + C
by formula (6.11) in Section 6.3.
9For a different derivation, see pp.79-80 in C O R R A L, M., Trigonometry, http://mecmath.net/trig/, 2009.",ElementaryCalculus.pdf
"Miscellaneous Integration Methods • Section 6.5 193
Exercises
A
For Exercises 1-12, evaluate the given integral.
1.
∫ 1 − 2 cos θ
sin θ dθ 2.
∫ dθ
3 − 5 sin θ 3.
∫ dθ
2 − sin θ 4.
∫ dθ
4 + sin θ
5.
∫ sin θ
2 − sin θ dθ 6.
∫ dθ
5 − 3 cos θ 7.
∫ dθ
1 + sin θ − cos θ 8.
∫ dθ
1 − sin θ + cos θ
9.
∫ cot θ
1 + sin θ dθ 10.
∫ 1 − cos θ
3 sin θ dθ 11.
∫ ∞
−∞
e−x2 /2 dx 12.
∫ ∞
−∞
x2 e−x6
dx
13. Consider the integral
∫ sin θ
1 + cos θ dθ from Example 6.31.
(a) Evaluate the integral using the substitution u = 1 + cos θ.
(b) Evaluate the integral using the half-angle substitution t = tan 1
2 θ.
(c) Show that the answers from parts (a) and (b) are equivalent to the result from Example 6.31.
B
φ
3
45
14. Evaluate the integral
∫ dθ
3 sin θ + 4 cos θ from Example 6.30 by noting that
∫ dθ
3 sin θ + 4 cos θ =
∫ dθ
5
(3
5 sin θ + 4
5 cos θ
)
=
∫ dθ
5
(
cos φ sin θ + sin φ cos θ
)
=
∫ dθ
5 sin ( θ + φ) = 1
5
∫ 
csc (θ + φ) dθ",ElementaryCalculus.pdf
"∫ dθ
3 sin θ + 4 cos θ =
∫ dθ
5
(3
5 sin θ + 4
5 cos θ
)
=
∫ dθ
5
(
cos φ sin θ + sin φ cos θ
)
=
∫ dθ
5 sin ( θ + φ) = 1
5
∫ 
csc (θ + φ) dθ
by the sine addition formula, where φ is the angle in the right triangle shown above. Complete the
integration and show that your answer is equivalent to the re sult from Example 6.30.
15. Show directly from the definition of the Beta function that B(x, y) = B( y, x) for all x > 0 and y > 0.
16. Show that the Beta function B(x, y) can be written as
B(x, y) =
∫ π/2
0
2 sin 2x−1(θ) cos 2 y−1(θ) dθ for all x > 0 and y > 0.
17. Use Exercise 16 and formula (6.16) to show that
∫ π/2
0
sinm θ cosn θ dθ =
Γ
(m + 1
2
)
Γ
(n + 1
2
)
2 Γ
(m + n
2 + 1
) for all m > −1 and n > −1.
18. Use Exercise 28 from Section 6.1, as well as Exercise 17 above , to show that for m = 1, 2, 3, . . . ,
∫ π/2
0
sin2m θ dθ =
∝ra⌈〉⌋allowπ Γ
(
m + 1
2
)
2 (m!) and
∫ π/2
0
sin2m+1 θ dθ =
∝ra⌈〉⌋allowπ (m!)
2 Γ
(
m + 3
2
) .",ElementaryCalculus.pdf
"194 Chapter 6 • Methods of Integration §6.5
19. Show that
∫ ∞
0
ln x
1 + x2 dx = 0. 20. Show that
∫ ∞
0
xa
ax dx = Γ(a + 1)
(ln a)a+1 for a > 1.
21. Use the result from Example 6.28 to show that
d1/2
dx1/2
( d1/2
dx1/2 (x)
)
= 1 = d
dx (x) .
22. Calculate d1/2
dx1/2 (c) for all constants c. 23. Calculate d1/3
dx1/3 (x) .
24. Show that
∫ 1
0
1∝ra⌈〉⌋allow
1 − xn dx = 1
n B
(1
n , 1
2
)
for n ≥ 1.
25. Show that the Gamma function Γ(t) can be written as
Γ(t) = pt
∫ ∞
0
ut−1 e−pu du for all t > 0 and p > 0.
26. Show that the Gamma function Γ(t) can be written as
Γ(t) =
∫ 1
0
(
ln
(1
u
)) t−1
du for all t > 0.
27. Using the result from Exercise 27 in Section 6.1 that
∫ 
ea x cos bx dx = ea x (a cos bx + b sin bx )
a2 + b2
for all constants a and b ̸=0, differentiate under the integral sign to show that for all α > 0
∫ ∞
0
x e −x sin αx dx = 2α
(1 + α2 )2 .
C
28. Use the Leibniz rule and formula (6.8) from Section 6.3 to sho w that for all a > 0,
∫ dx
∝ra⌈〉⌋allow
a2 + x2
= ln
⏐
⏐ x +
√",ElementaryCalculus.pdf
"(1 + α2 )2 .
C
28. Use the Leibniz rule and formula (6.8) from Section 6.3 to sho w that for all a > 0,
∫ dx
∝ra⌈〉⌋allow
a2 + x2
= ln
⏐
⏐ x +
√
a2 + x2 ⏐
⏐ + C .
29. Use Example 6.27 to show that the Beta function satisfies the r elation
B(x, 1 − x) =
∫ 1
0
t−x + tx−1
1 + t dt for all 0 < x < 1.
(Hint: First use a substitution to show that
∫ ∞
0
ux−1
1 + u du =
∫ ∞
0
t−x
1 + t dt.)
30. Show that for all a > −1, ∫ π/2
0
dθ
1 + a sin2 θ
= π
2
∝ra⌈〉⌋allow
1 + a
.",ElementaryCalculus.pdf
"Numerical Integration Methods • Section 6.6 195
6.6 Numerical Integration Methods
Section 5.2 showed how to obtain exact values for definite int egrals of some simple functions
(low-degree polynomials) by using areas of rectangles. For functions with no closed-form an-
tiderivative, the rectangle method typically produces an a pproximate value of the definite
integral—the more rectangles, the better the approximatio n.
y = sin ( x2 )
y
x
0 ∝ra⌈〉⌋allowπ
1
For example, suppose you wanted to evaluate
∫ ∝ra⌈〉⌋allowπ
0
sin ( x2) dx
with the rectangle method. This means finding the area of
the shaded region in the figure on the right. Computers
have eliminated the need to do these sorts of calculations
by hand. Though the rectangle method is simple to implement i n a traditional programming
language (e.g. via a looping construct), there are easier wa ys in a domain-specific language
(DSL) geared toward scientific computing. One such DSL is MAT LAB® , or its free open-source
clone Octave.
10",ElementaryCalculus.pdf
"(DSL) geared toward scientific computing. One such DSL is MAT LAB® , or its free open-source
clone Octave.
10
Implementing the rectangle method from scratch in Octave is a one-liner . For example, sup-
pose you divide the interval [0 , ∝ra⌈〉⌋allowπ] into 100,000 (10 5) subintervals of equal length, producing
100,001 (1 + 1e5 in scientific notation) equally spaced point s in [0 , ∝ra⌈〉⌋allowπ] (including 0 and ∝ra⌈〉⌋allowπ).
First use the left endpoints of the 10 5 subintervals:
octave> sum(sin(linspace(0,sqrt(pi),1+1e5)(1:end-1).^2)*sqrt(pi)/1e5)
ans = 0.8948314693913354
Now use the right endpoints:
octave> sum(sin(linspace(0,sqrt(pi),1+1e5)(2:end).^2)*sqrt(pi)/1e5)
ans = 0.8948314693913354
Finally , use the midpoints of the subintervals:
octave> sum(sin((linspace(0,sqrt(pi),1+1e5)(1:end-1)+sqrt(pi)/2e5).^2)*
sqrt(pi)/1e5)
ans = 0.8948314695305527
The true value of the integral up to 15 decimal places is 0.894 831469484145, so all three
approximations are accurate to 9 decimal places.",ElementaryCalculus.pdf
"The true value of the integral up to 15 decimal places is 0.894 831469484145, so all three
approximations are accurate to 9 decimal places.
10Octave is freely available at https://www.gnu.org/software/octave/",ElementaryCalculus.pdf
"196 Chapter 6 • Methods of Integration §6.6
The syntax in the above commands can be explained with some ex amples. The following
command creates 4 equally spaced points in the interval [1 , 7] (including x = 1 and x = 7), thus
dividing [1 , 7] into 3 subintervals each of length (7 − 1)/3 = 2:
octave> linspace(1,7,4)
ans =
1 3 5 7
Get all but the last number in the above list: 11
octave> linspace(1,7,4)(1:end-1)
ans =
1 3 5
Now square each number in that list of numbers (the dot before the exponentiation operator ^
applies the squaring operation ^2 element-wise in the list):
octave> linspace(1,7,4)(1:end-1).^2
ans =
1 9 25
Now take the sine of each of those squared numbers (measured i n radians):
octave> sin(linspace(1,7,4)(1:end-1).^2)
ans =
0.8414709848078965 0.4121184852417566 -0.132351750097 773
Now multiply each of those numbers (the heights of the rectan gles) by the width (2) of each
rectangle, then add up those areas:
octave> sum(sin(linspace(1,7,4)(1:end-1).^2)*2)",ElementaryCalculus.pdf
"rectangle, then add up those areas:
octave> sum(sin(linspace(1,7,4)(1:end-1).^2)*2)
ans = 2.24247543990376
11MATLAB would require you to use two steps:
MATLAB>> x = linspace(1,7,4);
MATLAB>> x(1:end-1)",ElementaryCalculus.pdf
"Numerical Integration Methods • Section 6.6 197
Before the advent of modern computing, the rectangle method was considered inefficient,
and so alternative methods were created. Two such methods ar e the trapezoid rule and
Simpson’s rule . The idea behind both methods is to take advantage of a nonlin ear function’s
changing slope by using nonrectangular regions. For the tra pezoid rule those regions are trape-
zoids, while Simpson’s rule uses quasi-rectangular region s whose top edges are parabolas, as
shown in Figure 6.6.1:
y
x
a b x i xi+1
y = f (x)
(a) rectangle method
y
x
a b x i xi+1
y = f (x)
(b) trapezoid rule
y
x
a b xi xi+1 xi+2
y = f (x)
(c) Simpson’s rule
Figure 6.6.1 Comparison of numerical integration methods for
∫ b
a f (x) dx
For a partition P = {a = x0 < x1 < ··· <xn−1 < xn = b} of an interval [ a, b] into n ≥ 1 subinter-
vals of equal width h = (b − a)/n, let yi = f (xi ) for i = 0, 1, . . . , n. The trapezoid rule adds up the",ElementaryCalculus.pdf
"vals of equal width h = (b − a)/n, let yi = f (xi ) for i = 0, 1, . . . , n. The trapezoid rule adds up the
areas of trapezoids on each subinterval [ xi , xi+1], with the top edge being the line segment join-
ing the points ( xi , yi ) and ( xi+1, yi+1 ). The approximation formula is straightforward to derive,
based on areas of trapezoids:
Trapezoid rule:
∫ b
a
f (x) dx ≈ h
2 ( y0 + 2 y1 + 2 y2 + ··· + 2 yn−1 + yn )
Simpson’s rule depends on pairs of neighboring subinterval s: [ x0, x1] and [ x1, x2], [ x2, x3] and
[x3, x4], . . . , [ xn−2, xn−1] and [ xn−1, xn]. Thus, n ≥ 2 must be even . The top edge of the region over
each pair [ xi , xi+1] and [ xi+1 , xi+2] is the unique parabola joining the 3 points ( xi , yi ), ( xi+1 , yi+1),
and ( xi+2, yi+2 ). The approximation formula is then: 12
Simpson’s rule:
∫ b
a
f (x) dx ≈ h
3 ( y0 + 4 y1 + 2 y2 + 4 y3 + 2 y4 + ··· + + 2 yn−2 + 4 yn−1 + yn )",ElementaryCalculus.pdf
"Simpson’s rule:
∫ b
a
f (x) dx ≈ h
3 ( y0 + 4 y1 + 2 y2 + 4 y3 + 2 y4 + ··· + + 2 yn−2 + 4 yn−1 + yn )
12For a full derivation of both formulas, see pp.144-149 in H O R N B E C K, R.W ., Numerical Methods , New Y ork:
Quantum Publishers, Inc., 1975.",ElementaryCalculus.pdf
"198 Chapter 6 • Methods of Integration §6.6
Example 6.32
Approximate the value of
∫ ∝ra⌈〉⌋allowπ
0
sin ( x2) dx by using the trapezoid rule and Simpson’s rule with n = 105
subintervals.
Solution: Since x0 = 0 and xn = ∝ra⌈〉⌋allow
π, then y0 = sin ( x2
0 ) = sin 0 = 0 and yn = sin ( x2
n ) = sin π = 0. Thus, y0
and yn contribute nothing to the summation formulas for both rules . In particular the trapezoid rule
approximation becomes
∫ ∝ra⌈〉⌋allowπ
0
sin ( x2) dx ≈ h
2 (0 + 2 y1 + 2 y2 + ··· + 2 yn−1 + 0) = h ·
(
n−1∑
k=1
yk
)
which is simple to implement in Octave:
octave> x = linspace(0,sqrt(pi),1+1e5);
octave> h = sqrt(pi)/1e5;
octave> h*sum(sin(x(2:end-1).^2))
ans = 0.8948314693913405
Likewise, the Simpson’s rule approximation becomes
∫ ∝ra⌈〉⌋allowπ
0
sin ( x2) dx ≈ h
3 (4 y1 + 2 y2 + 4 y3 + 2 y4 + ··· + + 2 yn−2 + 4 yn−1)
which can be implemented easily by using Octave’s powerful i ndexing features:
octave> (h/3)*(4*sum(sin(x(2:2:end-1).^2)) + 2*sum(sin(x(3:2:end-1).^2)))",ElementaryCalculus.pdf
"octave> (h/3)*(4*sum(sin(x(2:2:end-1).^2)) + 2*sum(sin(x(3:2:end-1).^2)))
ans = 0.8948314694841457
In the above command, the statement x(3:2:end-1) allows you to skip every other element in the list
x after position 3, by moving up the list in increments of 2 posi tions all the way to the next-to-last
position in the list ( end-1). Similarly for x(2:2:end-1), which starts at position 2 and then moves up in
increments of 2.
Note that Simpson’s rule gives essentially the true value in this case, and the value from the trapezoid
rule is virtually the same as the value produced by the built- in trapz function in Octave/MATLAB:
octave> trapz(x,sin(x.^2))
ans = 0.8948314693913402
In general you are better off using these sorts of built-in fu nctions instead of implementing your own.
Typically Simpson’s rule is slightly more efficient than the trapezoid rule, which is slightly
more efficient than the rectangle method. However , in the abo ve examples all the approxima-",ElementaryCalculus.pdf
"more efficient than the rectangle method. However , in the abo ve examples all the approxima-
tions were accurate to at least 9 decimal places (equivalent to getting the distance between
Detroit and Chicago correct within the thickness of a toothp ick). The running time of each
calculation was only a few thousandths of a second. Modern co mputing has generally made
the efficiency differences negligible.",ElementaryCalculus.pdf
"Numerical Integration Methods • Section 6.6 199
Notice that the approximations in the rectangle method, the trapezoid rule and Simpson’s
rule can all be written as linear combinations of function va lues f (a i ) multiplied by “weights”
wi : ∫ b
a
f (x) dx ≈
n∑
i=0
wi f (a i )
For example, the weights in Simpson’s rule are wi = h
3 , 2h
3 , or 4h
3 , depending on the points a i
in the interval [ a, b]. The method of Gaussian quadrature transforms an integral over any
interval [ a, b] into an integral over the specific interval [ −1, 1] and then uses a standard set of
points in [ −1, 1] and known weights for those points: 13
Gaussian quadrature: Transform the integral
∫ b
a f (x) dx into an integral over [ −1, 1] by
means of the substitution u = 1
b−a (2x − a − b), so that x = b−a
2 u + a+b
2 and dx = b−a
2 du. Then
∫ b
a
f (x) dx = b − a
2
∫ 1
−1
g(u) du ≈ b − a
2
n∑
i=1
wi g(a i )
where g(u) = f
(
b−a
2 u + a+b
2
)
, with the points a1, . . ., an and weights w1, . . . , wn given in",ElementaryCalculus.pdf
"2
∫ 1
−1
g(u) du ≈ b − a
2
n∑
i=1
wi g(a i )
where g(u) = f
(
b−a
2 u + a+b
2
)
, with the points a1, . . ., an and weights w1, . . . , wn given in
Table 6.1 below for any choice of 2 ≤ n ≤ 10 points in [ −1, 1].
T able 6.1 T able of Gaussian quadrature points and weights
n a1 , . . ., an w1 , . . . , wn
2 ±0.577350 1
3 0 8/9
±0.774597 5/9
4 ±0.339981 0.652145
±0.861136 0.347855
5 0 0.568889
±0.538469 0.478629
±0.906180 0.236927
6 ±0.238619 0.467914
±0.661209 0.360762
±0.932470 0.171324
7 0 0.417959
±0.405845 0.381830
±0.741531 0.279705
±0.949108 0.129485
n a1 , . . . , an w1 , . . . , wn
8 ±0.183435 0.362684
±0.525532 0.313707
±0.796666 0.222381
±0.960290 0.101229
9 0 0.330239
±0.324253 0.312347
±0.613371 0.260611
±0.836031 0.180648
±0.968160 0.081274
10 ±0.148874 0.295524
±0.433395 0.269267
±0.679410 0.219086
±0.865063 0.149451
±0.973907 0.066671
13The details are beyond the scope of this book. See Chapter 4 in RA L S T O N, A. A N D P . R A B I N O W I T Z, A First Course",ElementaryCalculus.pdf
"±0.973907 0.066671
13The details are beyond the scope of this book. See Chapter 4 in RA L S T O N, A. A N D P . R A B I N O W I T Z, A First Course
in Numerical Analysis , 2nd ed., New Y ork: McGraw-Hill, Inc., 1978. See also Table 1 in S T R O U D, A.H. A N D D.
SE C R E S T, Gaussian Quadrature Formulas , Englewood Cliffs, NJ: Prentice-Hall, Inc., 1966.",ElementaryCalculus.pdf
"200 Chapter 6 • Methods of Integration §6.6
Example 6.33
Approximate the value of
∫ 2
0
dx
1 + x3 by using Gaussian quadrature with n = 4 points.
Solution: For a = 0 and b = 2, use the substitution u = 1
b−a (2x−a−b) = x−1, so that x = u+1 and dx = du.
Thus, g(u) = f (u + 1) = 1
1+(u+1)3 . Using n = 4 in Table 6.1, the points ai and weights wi are
a1 = −0.339981 w1 = 0.652145
a2 = 0.339981 w2 = 0.652145
a3 = −0.861136 w3 = 0.347855
a4 = 0.861136 w4 = 0.347855
and so
∫ 2
0
dx
1 + x3 = 2 − 0
2
∫ 1
−1
g(u) du =
∫ 1
−1
du
1 + (u + 1)3 ≈
4∑
i=1
wi g(ai ) =
4∑
i=1
wi · 1
1 + (ai + 1)3
≈ 0.652145
1 + (−0.339981 + 1)3 + 0.652145
1 + (0.339981 + 1)3 + 0.347855
1 + (−0.861136 + 1)3 + 0.347855
1 + (0.861136 + 1)3
≈ 1.091621
The true value of the integral to six decimal places is 1.0900 02.
Using more points (e.g. n = 7) is easy to implement in Octave, using element-wise operat ions on arrays:
octave> a = [ 0 -0.405845 0.405845 -0.741531 0.741531 -0.949108 0.949108 ];",ElementaryCalculus.pdf
"octave> a = [ 0 -0.405845 0.405845 -0.741531 0.741531 -0.949108 0.949108 ];
octave> w = [ 0.417959 0.381830 0.381830 0.279705 0.279705 0.129485 0.129485 ];
octave> sum(w./(1 + (a+1).^3))
ans = 1.090016688064804
Gaussian quadrature can be applied to improper integrals. F or example,
∫ ∞
0
f (x) e−x dx ≈
n∑
i=1
wi f (a i )
using the points a i and weights wi in Table 6.2 14 for n = 3, 4, or 5 points in [0 , ∞):
T able 6.2 T able of Gaussian quadrature points and weights fo r
∫ ∞
0 f (x) e−x dx
n a1 , . . ., an w1 , . . ., wn
3 0.415775 0.711093
2.294280 0.278518
6.289945 0.010389
n a1 , . . . , an w1, . . . , wn
4 0.322548 0.603154
1.745761 0.357419
4.536620 0.038888
9.395071 0.000539
n a1 , . . . , an w1, . . . , wn
5 0.263560 0.521756
1.413403 0.398667
3.596426 0.075942
7.085810 0.003612
12.640801 0.000023
14See Table 6 in S T R O U D, A.H. A N D D. S E C R E S T, Gaussian Quadrature Formulas .",ElementaryCalculus.pdf
"Numerical Integration Methods • Section 6.6 201
Example 6.34
Approximate the value of
∫ ∞
0
x5 e−x dx by using Gaussian quadrature with n = 3 points in Table 6.2.
Solution: For n = 3, Table 6.2 gives a1 = 0.415775, a2 = 2.294280, a3 = 6.289945, and w1 = 0.711093,
w2 = 0.278518, w3 = 0.010389. Then for f (x) = x5 ,
∫ ∞
0
x5 e−x dx ≈
n∑
i=1
wi f (ai ) =
n∑
i=1
wi a5
i
≈ 0.711093 (0.415775)5 + 0.278518 (2.294280)5 + 0.010389 (6.289945)5
≈ 119.9974709727211
The true value is Γ(6) = 5! = 120.
Note: The points ai in Table 6.2 are the roots of the Laguerre polynomials of degree n.
Exercises
A
1. A simple pendulum of length l swings through an angle of 90 ◦ on either side of the vertical with
period P , given by
P = 4
√
l
g
∫ π/2
0
dθ√
1 − 0.5 sin 2 θ
where g = 9.8 m/s 2 is the acceleration due to gravity . Use the rectangle method (with left endpoints),
the trapezoid rule, and Simpson’s rule to write P as a constant multiple of
√
l/ g. Preferably , use a",ElementaryCalculus.pdf
"the trapezoid rule, and Simpson’s rule to write P as a constant multiple of
√
l/ g. Preferably , use a
computer and n = 105 subintervals of equal width (or n = 10 subintervals if calculating by hand).
2. Repeat Exercise 1 using Gaussian quadrature with n = 5 points.
3. Approximate the value of
∫ ∝ra⌈〉⌋allowπ
0
sin ( x2) dx by using Gaussian quadrature with n = 7 points.
4. Repeat Exercise 3 with n = 9 points. 5. Repeat Exercise 3 with n = 10 points.
6. The points ai in Table 6.1 for Gaussian quadrature are the roots of the Legendre polynomials Pn (x),
with P0 (x) = 1, P1 (x) = x, and Pn (x) defined for integers n ≥ 2 by the recursion formula
n P n (x) = (2n − 1) x P n−1(x) − (n − 1) Pn−2 (x)
(a) Write out Pn (x) explicitly in standard polynomial form for n = 2, 3, 4, 5.
(b) V erify that the roots of Pn (x) match the n points a1 , . . . , an in Table 6.1 for n = 2, 3, 4, 5.
(c) With no calculations, explain why
∫ 1
−1 P2(x) P3(x) dx =
∫ 1
−1 P3 (x) P4(x) dx =
∫ 1
−1 P4 (x) P5(x) dx = 0.",ElementaryCalculus.pdf
"(c) With no calculations, explain why
∫ 1
−1 P2(x) P3(x) dx =
∫ 1
−1 P3 (x) P4(x) dx =
∫ 1
−1 P4 (x) P5(x) dx = 0.
(d) For n = 0, 1, 2, 3, verify that ∫ 1
−1
P 2
n (x) dx = 2
2n + 1 .
7. Repeat Example 6.34 with n = 4 points.
8. Repeat Example 6.34 with n = 5 points.
9. Use Table 6.2 to approximate the value of
∫ ∞
0
ln (1 + x) e−x dx with n = 5 points.",ElementaryCalculus.pdf
"CHAPTER 7
Analytic Geometry and Plane Curves
7.1 Ellipses
If you were to ask a random person “What is a circle?” a typical response would be to kick the
can down the road: “Something that’s round.” There is a simpl e definition:
A circle is the set of all points in a plane that are a fixed distance from a fixed point in that
plane. The fixed point is the center of the circle, while the fixed distance from the center
is the circle’s radius.
Similarly , the question “What is an ellipse?” would likely b e answered with “an oval,” “some-
thing egg-shaped,” or “a squished circle.” A precise definit ion would be:
An ellipse is the set of all points in a plane such that the sum of the dista nces from each
of those points to two fixed points in the plane is the same cons tant. The two fixed points
are the foci—plural of focus—of the ellipse.
Figure 7.1.1 illustrates the above definitions, with a point P moving along each curve.
r
P
C
(a) Circle: center C, radius r = constant
P
d1 d2
F1 F2",ElementaryCalculus.pdf
"Figure 7.1.1 illustrates the above definitions, with a point P moving along each curve.
r
P
C
(a) Circle: center C, radius r = constant
P
d1 d2
F1 F2
(b) Ellipse: foci F1 and F2 , d1 + d2 = constant
Figure 7.1.1 The circle and its “squished” sibling the ellipse
Along the ellipse the sum d1 + d2 of the distances from P to the foci remains constant.
202",ElementaryCalculus.pdf
"Ellipses • Section 7.1 203
The circle’s definition makes it easy to imagine its shape, es pecially for anyone who has
drawn a circle with a compass. The definition of the ellipse, o n the other hand, might not im-
mediately suggest an “oval” shape. Its shape becomes appare nt when physically constructing
an ellipse by hand, using only the definition. Stick two pins i n a board and tie the ends of a
piece of string to the pins, with the string long enough so tha t there is some slack (see Figure
7.1.2(a)). The pins will be the foci of the ellipse.
(a) Two pins and string on a board
/C8
(b) Pull the string taut with a pencil
Figure 7.1.2 Construction of an ellipse
Pull the string taut with a pencil whose point is touching the board, then move the pencil
around as far as possible on all sides of the pins. The drawn fig ure will be an ellipse, as in
Figure 7.1.2(b). The length of the string is the constant sum d1 + d2 of distances from points",ElementaryCalculus.pdf
"Figure 7.1.2(b). The length of the string is the constant sum d1 + d2 of distances from points
on the ellipse to the foci. The symmetry of the ellipse is obvi ous.
There is some terminology connected with ellipses. The principal axis is the line containing
the foci, and the center is halfway between the foci, as in Figure 7.1.3:
principal
axis
major axis
minor
axis
F1
(focus)
F2
(focus)
C (center)V1
(vertex)
V2
(vertex)
Figure 7.1.3 The parts of an ellipse
The vertexes are the points where the ellipse intersects the principal ax is. The major axis
is the chord joining the vertexes, and the minor axis is the chord through the center that is
perpendicular to the major axis. The two semi-major axes are the halves of the major axis
joining the center to the vertexes ( CV1 and CV2 in Figure 7.1.3). Likewise the semi-minor
axes are the two halves of the minor axis. A chord through the cente r is a diameter. Notice",ElementaryCalculus.pdf
"axes are the two halves of the minor axis. A chord through the cente r is a diameter. Notice
that a circle is the special case of an ellipse with identical foci (i.e. the foci and center are all
the same point).",ElementaryCalculus.pdf
"204 Chapter 7 • Analytic Geometry and Plane Curves §7.1
Ellipses appear in nature (e.g. the orbits of planets around the Sun) and in many applica-
tions. The ancient Greeks were able to derive many propertie s of the ellipse from its purely
geometric definition. 1 Nowadays those properties are typically derived using meth ods from
analytic geometry —the study of geometric objects in the context of coordinate systems.2 Y ou
have already seen the equation of an ellipse in the x y-plane centered at the origin: x2
a2 + y2
b2 = 1,
where a > b > 0, with the x-axis as the principal axis. The equation is straightforwar d to derive
from the definition of the ellipse.
x
y
(x, y)
d1 d2
(−c, 0) (c, 0)
In the x y-plane, let the foci of an ellipse be the points
(±c, 0) for some c > 0, so that the center is the origin (0 , 0)
and the x-axis is the principal axis, as in the figure on
the right. Denote by 2 a the constant sum d1 + d2 of the
distances from points ( x, y) on the ellipse to the foci, with",ElementaryCalculus.pdf
"the right. Denote by 2 a the constant sum d1 + d2 of the
distances from points ( x, y) on the ellipse to the foci, with
a > 0. Notice that a > c, since the distance 2 c between the
foci must be less than d1 + d2 = 2a. Then by the distance
formula,
d1 + d2 = 2a
√
(x + c)2 + y2 +
√
(x − c)2 + y2 = 2a
(√
(x − c)2 + y2
) 2
=
(
2a −
√
(x + c)2 + y2
) 2
(x − c)2 + ✓✓y2 = 4a2 − 4a
√
(x + c)2 + y2 + (x + c)2 + ✓✓y2
4a
√
(x + c)2 + y2 = 4a2 + (x + c)2 − (x − c)2
✁4a
√
(x + c)2 + y2 = ✁4a2 + ✁4x c
√
(x + c)2 + y2 = a + c
a x
x2 + ✟✟2c x + c2 + y2 = a2 + ✟✟2c x + c2
a2 x2
(
1 − c2
a2
)
x2 + y2 = a2 − c2
x2
a2 + y2
a2 − c2 = 1
x2
a2 + y2
b2 = 1 where b2 = a2 − c2 (and so a > b > 0) ✓
1The word ellipse is in fact due to the Greek astronomer and geometer Apolloniu s of Perga (ca. 262-190 B.C.),
which seems an improvement over the name “thyreos” that Eucl id (ca. 360-300 B.C.) had given the shape.",ElementaryCalculus.pdf
"which seems an improvement over the name “thyreos” that Eucl id (ca. 360-300 B.C.) had given the shape.
2Pioneered by the French mathematician and philosopher René Descartes (1596-1650), for whom the Cartesian
coordinate system is named. The proposition “I think, there fore I am” ( Cogito, ergo sum ) is due to Descartes.",ElementaryCalculus.pdf
"Ellipses • Section 7.1 205
x
y
−c c−a a0
b
−b
x2
a2 + y2
b2 = 1
Figure 7.1.4
The graph of the resulting ellipse x2
a2 + y2
b2 = 1 with
a > b > 0 and foci at ( ±c, 0) is shown in Figure 7.1.4.
Since the the x-axis is principal axis then the vertexes
are found by setting y = 0: x = ±a. The vertexes are
thus ( ±a, 0), so the major axis goes from ( −a, 0) to ( a, 0)
and has length 2 a (i.e. the semi-major axis has length
a). Similarly , setting x = 0 shows the minor axis goes
from (0 , −b) to (0 , b), (i.e. the semi-minor axis has
length b). Since a > b > 0 and b2 = a2 − c2,
then c =
∝ra⌈〉⌋allow
a2 − b2. Thus, for any ellipse of the form
x2
a2 + y2
b2 = 1 with a > b > 0, the foci will be at ( ±c, 0),
where c =
∝ra⌈〉⌋allow
a2 − b2. The foci can be used to define an important geometric propert y of the
ellipse:
The eccentricity of an ellipse is the ratio of the distance between the foci to t he length of
the major axis. In the case of an ellipse x2
a2 + y2",ElementaryCalculus.pdf
"The eccentricity of an ellipse is the ratio of the distance between the foci to t he length of
the major axis. In the case of an ellipse x2
a2 + y2
b2 = 1 with a > b > 0, the eccentricity e is given
by
e = c
a =
∝ra⌈〉⌋allow
a2 − b2
a .
The eccentricity e is a measure of how “oval” an ellipse is, with 0 < e < 1. The boundary case
e = 0 is a circle, while e = 1 is a line segment; an ellipse is somewhere in between—the cl oser e
gets to 1, the more “squished” the ellipse. See Figure 7.1.5.
C
(a) Circle: e = 0
F1 F2
(b) Ellipse: e = 0.5
F1 F2
(c) Ellipse: e = 0.75
F1 F2
(d) Segment: e = 1
Figure 7.1.5 Eccentricity e
Earth’s elliptical orbit around the Sun is almost circular: the eccentricity is 0.017. Only the
orbits of V enus and Neptune (both at 0.007) have a lower eccen tricity among the nine planets
in the solar system, while Pluto’s (0.252) has the highest.
It is left as an exercise to show that the ellipse x2
a2 + y2
b2 = 1 with a > b > 0 can be written in",ElementaryCalculus.pdf
"It is left as an exercise to show that the ellipse x2
a2 + y2
b2 = 1 with a > b > 0 can be written in
terms of the eccentricity e:
y2 = (1 − e2) (a2 − x2) (7.1)",ElementaryCalculus.pdf
"206 Chapter 7 • Analytic Geometry and Plane Curves §7.1
Example 7.1
Find the area inside the ellipse x2
a2 + y2
b2 = 1.
Solution: By symmetry the area will be four times the area in the first qua drant. Solving for y in the
equation of the ellipse gives
y2 = b2 − b2 x2
a2 ⇒ y = b
√
1 − x2
a2 = b
a
√
a2 − x2
for the upper hemisphere. Thus,
Area = 4
∫ a
0
y dx = 4b
a
∫ a
0
√
a2 − x2 dx
= 4b
a
(a2
2 sin−1
(x
a
)
+ 1
2 x
√
a2 − x2
⏐
⏐
⏐
⏐
a
0
)
(by formula (6.7) in Section 6.3)
= 4b
a
(a2
2
π
2 + 0 − (0 + 0)
)
= πab
P
F1 F2
Figure 7.1.6
A remarkable feature of the ellipse is the reflection property :
light shone from one focus to any point on the ellipse will refl ect to
the other focus . Figure 7.1.6 shows light emanating from the focus
F1 and reflecting off a point P on the ellipse to the other focus F2.
Fermat’s Principle from Example 4.4 in Section 4.1 showed th at
the incoming angle θ1 (angle of incidence) of light from point A",ElementaryCalculus.pdf
"Fermat’s Principle from Example 4.4 in Section 4.1 showed th at
the incoming angle θ1 (angle of incidence) of light from point A
will equal the outgoing angle θ2 (angle of reflection) to point B
for light reflecting off a flat reflective surface at point P , as in
Figure 7.1.7(a). Fermat’s Principle also applies to curved surfaces—e.g. an ellipse—with the
angles measured relative to the tangent line to the curve at t he point of reflection, as in Figure
7.1.7(b).
A
B
P
θ1 θ2
(a) Flat surface
tangent
line
A
B
P
θ1 θ2
(b) Curved surface
Figure 7.1.7 Fermat’s Principle for reflection: θ1 = θ2",ElementaryCalculus.pdf
"Ellipses • Section 7.1 207
Notice that Fermat’s Principle is equivalent to saying that the angles α1 and α2 that the
light’s path makes with the normal line through the point of r eflection are equal, since each
angle would equal 90 ◦ − θ, as in Figure 7.1.8(a):
tangent
line
normal line
A B
P
θ θ
α1 α2
(a) α1 = α2
x
y
F1
−c
F2
c0−a a
b
N
P (x0 , y0 )
n = normal linetangent line
α1 α2
(b) Ellipse: show that α1 = α2
Figure 7.1.8 Fermat’s Principle with normal line
Thus, to prove the reflection property , it suffices to prove th at the normal line n to the ellipse
at P bisects the angle ∠F1 P F2 in Figure 7.1.8(b)—this would make α1 = α2, so that the indi-
cated path from F1 to P to F2 satisfies Fermat’s Principle. First, let P = (x0, y0) be a point on
the ellipse x2
a2 + y2
b2 = 1, with a > b > 0. Assume that P is not a vertex (i.e. y0 ̸=0), otherwise the
reflection property holds trivially . By Exercise 15 in Secti on 3.4, the equation of the tangent",ElementaryCalculus.pdf
"reflection property holds trivially . By Exercise 15 in Secti on 3.4, the equation of the tangent
line to the ellipse at P = (x0, y0) is xx0
a2 + y y0
b2 = 1 , (7.2)
so that its slope is − b2 x0
a2 y0
. Hence, the negative reciprocal a2 y0
b2 x0
is the slope of the normal line n,
whose equation—valid even when x0 = 0 (i.e. when y0 = ±b)— is then
b2 x0 ( y − y0 ) = a2 y0 (x − x0) . (7.3)
Setting y = 0 and solving for x shows the x-intercept of n is at
x = (a2 − b2) x0 y0
a2 y0
= c2
a2 x0 = e2 x0
Let N = (e2 x0, 0) be that x-intercept, as in Figure 7.1.8(b). The distance F1 N from the focus
F1 = (−c, 0) = (−ea , 0) to N is then
F1 N = e2 x0 − (−ea ) = e (a + e x0)
while the distance F2 N from the focus F2 = (c, 0) = (ea , 0) to N is
F2 N = ea − e2 x0 = e (a − e x0) .
Therefore,
F1 N
F2 N = e (a + e x0)
e (a − e x0) = a + e x0
a − e x0
.",ElementaryCalculus.pdf
"208 Chapter 7 • Analytic Geometry and Plane Curves §7.1
By the distance formula, the distance F1 P from F1 = (−ea , 0) to P = (x0, y0) is given by
(F1 P )2 = (x0 + ea )2 + y2
0
= x2
0 + 2eax 0 + e2 a2 + (1 − e2) (a2 − x2
0 ) (by formula (7.1))
(F1 P )2 = a2 + 2eax 0 + e2 x2
0 = (a + e x0)2
F1 P = a + e x0 .
Similarly , the distance F2 P from F2 = (ea , 0) to P = (x0, y0) is given by
(F2 P )2 = (x0 − ea )2 + y2
0 = x2
0 − 2eax 0 + e2a2 + (1 − e2) (a2 − x2
0)
(F2 P )2 = a2 − 2eax 0 + e2 x2
0 = (a − e x0)2
F2 P = a − e x0 .
F1 F2N
P
α1 α2
θ180◦ − θ
Thus,
F1 P
F2 P = a + e x0
a − e x0
= F1 N
F2 N ,
which means that α1 = α2:3 by the Law of Sines, and with θ =
∠F2 N P as in the figure on the right,
sin α2
F2 N = sin θ
F2 P = sin (180◦ − θ)
F2 P = sin (180◦ − θ)
F1 P · F1 P
F2 P = sin α1
F1 N · F1 N
F2 N = sin α1
F2 N
and thus sin α2 = sin α1, so that α2 = α1 (since 0 ◦ < α1, α2 < 90◦). ✓
x
y
−c
c
−b b0
a
−a
x2
b2 + y2
a2 = 1
Figure 7.1.9
An ellipse of the form
x2
b2 + y2
a2 = 1",ElementaryCalculus.pdf
"x
y
−c
c
−b b0
a
−a
x2
b2 + y2
a2 = 1
Figure 7.1.9
An ellipse of the form
x2
b2 + y2
a2 = 1
with a > b > 0 simply switches the roles of x and y in the previ-
ous examples: the principal axis is now the y-axis, the foci are at
(0, ±c), where c =
∝ra⌈〉⌋allow
a2 − b2, and the vertexes are at (0 , ±a), as in
Figure 7.1.9.
Thus, the largest denominator on the left side of an equation of
the form x2
□ 2 + y2
□ 2 = 1 tells you which axis is the principal axis. For
example, the principal axis of the ellipse x2
25 + y2
16 = 1 is the x-axis
(since 25 > 16), while the ellipse x2
4 + y2
9 = 1 has the y-axis as its
principal axis (since 9 > 4).
3This follows directly from Proposition 3 in Book VI of Euclid ’s Elements. See the purely geometric proof on
pp.125-126 in E U C L I D, Elements, (Thomas L. Heath translation), Santa Fe, NM: Green Lion Pre ss, 2002.",ElementaryCalculus.pdf
"Ellipses • Section 7.1 209
Exercises
A
1. Construct an ellipse using the procedure shown in Figure 7.1 .2. Place the two pins 7in apart and
use a 10in piece of string.
For Exercises 2-6, sketch the graph of the given ellipse, ind icate the major and minor axes and exact
locations of the foci and vertexes, and find the eccentricity e.
2. x2
25 + y2
16 = 1 3. x2
4 + y2
9 = 1 4. 4x2
25 + y2
4 = 1 5. x2 + 4 y2 = 1 6. 25x2 + 9 y2 = 225
7. Show that for a > b > 0 the ellipse x2
a2 + y2
b2 = 1 with eccentricity e can be written as y2 = (1−e2) (a2 −x2).
8. Use Example 7.1 to show the area inside the ellipse x2
a2 + y2
b2 = 1 with eccentricity e is πa2 ∝ra⌈〉⌋allow
1 − e2 .
9. For all a > b > 0, find the points of intersection of the ellipses x2
a2 + y2
b2 = 1 and x2
b2 + y2
a2 = 1.
10. Show that the vertexes are the closest and farthest points on an ellipse to either focus.
B
11. Show that any line of slope m that is tangent to the ellipse x2
a2 + y2
b2 = 1 must be of the form
y = mx ±
√",ElementaryCalculus.pdf
"B
11. Show that any line of slope m that is tangent to the ellipse x2
a2 + y2
b2 = 1 must be of the form
y = mx ±
√
a2 m2 + b2 .
12. A 10ft ladder with a mark 3ft from the top rests against a wall. If the top of the ladder slides down
the wall, with the foot of the ladder sliding away from the wal l on the ground, as in Figure 7.1.10,
then show the mark moves along part of an ellipse.
y
x
3
7
Figure 7.1.10 Exercise 12
F
P
D
G
Figure 7.1.11 Exercise 13
T1
T2
Figure 7.1.12 Exercise 14
13. Another definition of an ellipse is the set of points P for which the ratio of the distance from P to a
fixed point F (a focus) to the distance from P to a fixed line D (the directrix) is a constant e < 1 (the
eccentricity): P F
PG = e, as in Figure 7.1.11. Use this definition to show that the equa tion of an ellipse
with focus ( c, 0) can be written as x2
a2 + y2
b2 = 1 for some a > b > 0. Find the equation of the directrix.",ElementaryCalculus.pdf
"with focus ( c, 0) can be written as x2
a2 + y2
b2 = 1 for some a > b > 0. Find the equation of the directrix.
14. Show that the set of intersection points of all perpendicula r tangent lines to an ellipse form a circle,
as in Figure 7.1.12 (showing two such tangent lines T1 ⊥ T2).
15. A chord of an ellipse that passes through a focus and is perpen dicular to the major axis is a latus
rectum. Show that for the ellipse x2
a2 + y2
b2 = 1 with a > b > 0 the length of each latus rectum is 2b2
a .
16. Suppose that the normal line at one end of a latus rectum of an e llipse passes through an end of
the minor axis. Show that the eccentricity e is a root of the equation e4 + e2 − 1 = 0, then find e.
17. Show that the set of all midpoints of a family of parallel chor ds in an ellipse lie on a diameter .
(Hint: Use symmetry with chords of slope m ̸=0.)",ElementaryCalculus.pdf
"210 Chapter 7 • Analytic Geometry and Plane Curves §7.2
7.2 Parabolas
As with ellipses, you have seen parabolas (e.g. y = x2) and some of their applications (e.g.
projectile trajectories), but perhaps without knowing the ir purely geometric definition. The
alternative definition of an ellipse described in Exercise 1 3 in Section 7.1 is, in fact, similar to
the definition of the parabola:
A parabola is the set of all points in a plane that are equidistant from a fi xed point (the
focus) and a fixed line (the directrix).
D
FP
G
Figure 7.2.1 Parabola: P F = PG
Figure 7.2.1 illustrates the above definition, with a point
P moving along the parabola so that the distance from P to
the focus F equals the distance from P to the directrix D.
Note that the point halfway between the focus and directrix
must be on the parabola—that point is the vertex, which is
the point on the parabola closest to the directrix. The axis
of the parabola is the line that passes through the focus and",ElementaryCalculus.pdf
"the point on the parabola closest to the directrix. The axis
of the parabola is the line that passes through the focus and
is perpendicular to the directrix. Notice that the ratio P F
PG equals 1, whereas that ratio for
an ellipse—by the alternative definition—was the eccentric ity e < 1. The eccentricity of the
parabola, therefore, is always 1. 4
To construct a parabola from the definition, cut a piece of str ing to have the same length AB
as one side of a drafting triangle, as in Figure 7.2.2.
D
F
P
B
A
C
/C8
Figure 7.2.2 Construction of a parabola
Fasten one end of the string to the vertex A of the triangle and the other end to a pin
somewhere between A and B—the pin will be the focus F of the parabola. Hold the string taut
against the edge AB of the triangle at a point P on either side of the pin, then move the edge
BC of the triangle along the directrix D. The drawn figure will be a parabola, as the lengths",ElementaryCalculus.pdf
"BC of the triangle along the directrix D. The drawn figure will be a parabola, as the lengths
P F and P B will be equal (since the length of the string being AB = A P + P F means P F = P B).
4The eccentricity e of the parabola being 1 means there is no second vertex, unlik e the ellipse (where e < 1 forced
the existence of two vertexes in the alternative definition) .",ElementaryCalculus.pdf
"Parabolas • Section 7.2 211
x
y
(0, p)
y = −p
d1
d2
(x, y)
(x, −p)
0
−p
To derive the equation of a parabola in the x y-plane,
start with the simple case of the focus on the y-axis at
(0, p), with p > 0, and the line y = −p as the directrix,
as in the figure on the right. The vertex is then at the
origin (0 , 0). Pick a point ( x, y) whose distances d1 and
d2 from the focus (0 , p) and directrix y = −p, respectively ,
are equal. Then
d2
1 = d2
2
(x − 0)2 + ( y − p)2 = (x − x)2 + ( y + p)2
x2 +
✓✓y2 − 2 p y + p2 = ✓✓y2 + 2 p y + p2
x2 = 4 p y
In other words, y = 1
4 p x2, which is the more familiar form of a parabola. Thus, any curv e of the
form y = ax 2, with a ̸=0, is a parabola whose focus and directrix can be found by divi ding 1 by
4a: p = 1
4a , so that the focus is at
(
0, 1
4a
)
and the directrix is the line y = −1
4a . For example, the
parabola y = x2 has its focus at
(
0, 1
4
)
and its directrix is the line y = −1
4 .",ElementaryCalculus.pdf
"4a
)
and the directrix is the line y = −1
4a . For example, the
parabola y = x2 has its focus at
(
0, 1
4
)
and its directrix is the line y = −1
4 .
When p > 0 the parabola 4 p y = x2 extends upward; for p < 0 it extends downward, as in
Figure 7.2.3(a) below:
x
y
(0, p)
y = −p
0
−p
(a) 4 p y = x2: p < 0
x
y
( p, 0)
x = −p
0−p
(b) 4 px = y2: p > 0
x
y
( p, 0)
x = −p
0 −p
(c) 4 px = y2: p < 0
Figure 7.2.3 Parabolas with vertex at the origin
Switching the roles of x and y yields the parabola 4 p x = y2, with focus at ( p, 0) and directrix
x = −p. For p > 0 this parabola extends rightward, while for p < 0 it extends leftward. See
Figure 7.2.3(b) and (c).
It is left as an exercise to show that in general a curve of the f orm y = ax 2 + b x + c is a
parabola. Just like not every oval shape is an ellipse, not ev ery “cupped” or “U” shape is a
parabola (e.g. y = x4).",ElementaryCalculus.pdf
"212 Chapter 7 • Analytic Geometry and Plane Curves §7.2
The slope of the parabola 4 p y = x2 is dy
dx = 2x
4 p = x
2 p , so that the equation of the tangent line to
the parabola at a point ( x0, y0) is:
y − y0 = x0
2 p (x − x0)
2 p ( y − y0 ) = x0 x − x2
0
2 p y − 2 p y0 = x0 x − 4 p y0
2 p ( y + y0 ) = x0 x (7.4)
Likewise, switching the roles of x and y, the tangent line to the parabola 4 p x = y2 at a point
(x0, y0) is:
2 p (x + x0) = y0 y (7.5)
x
y
F = ( p, 0)
P
(x0 , y0 )
(−x0 , 0)
Q
0
β
β
Figure 7.2.4 4 px = y2
Formula (7.5) simplifies the proof of the reflection property
for parabolas: light shone from the focus to any point on the
parabola will reflect in a path parallel to the axis of the para bola.
Figure 7.2.4 shows light emanating from the focus F = ( p, 0)
and reflecting off a point P = (x0, y0) on the parabola 4 p x = y2.
If that line of reflection is parallel to the x-axis—the axis of
the parabola—then the tangent line to the parabola at ( x0, y0)",ElementaryCalculus.pdf
"If that line of reflection is parallel to the x-axis—the axis of
the parabola—then the tangent line to the parabola at ( x0, y0)
should make the same angle β with the line of reflection as it
does with the x-axis. So extend the tangent line to intersect the
x-axis and use formula (7.5) to find the x-intercept:
2 p (x + x0) = y0 y = y0 · 0 = 0 ⇒ x = −x0
Let Q = (−x0, 0), so that the distance F Q equals p + x0. The goal is to show that the angle of
incidence ∠F P Q equals the angle of reflection β. The focal radius F P has length
F P =
√
( p − x0)2 + (0 − y0 )2 =
√
p2 − 2 p x0 + x2
0 + 4 p x0 =
√
p2 + 2 p x0 + x2
0 = p + x0 .
Thus, F Q = F P in the triangle △F P Q , so that ∠F P Q = ∠F Q P = β, i.e. the light’s path does
indeed satisfy Fermat’s Principle for curved surfaces. ✓
The parabola’s reflection property shows up in some engineer ing applications, typically by
revolving part of a parabola around its axis, producing a par abolic surface in three dimen-",ElementaryCalculus.pdf
"revolving part of a parabola around its axis, producing a par abolic surface in three dimen-
sions called a paraboloid. For example, it used to be common for vehicle headlights to u se
paraboloids for their inner reflective surface, with a bulb a t the focus, so that—by the reflection
property—the light would shine straight ahead in a solid bea m. Many flashlights still operate
on that principle. The reflection property also works in the o pposite direction, which is why
satellite dishes and radio telescopes are often wide parabo loids with a signal receiver at the
focus, to maximize reception of incoming reflected signals.",ElementaryCalculus.pdf
"Parabolas • Section 7.2 213
Example 7.2
Suppose that an object is launched from the ground with an ini tial velocity v0 and at varying angles
with the ground. Show that the family of all the possible traj ectories—which are parabolic—form a
region whose boundary (called the envelope of the trajectories) is itself a parabola.
v2
0
2 g
0 v2
0
g
Solution: Recall from Example 4.3 in Section
4.1 that if the object is launched at an angle
0 < θ < π
2 with the ground, then the height y
attained by the object as a function of the hori-
zontal distance x that it travels is given by
y = − gx2
2v2
0 cos2 θ
+ x tan θ .
The curve is a parabola, with the figure on the
right showing these parabolic trajectories for 500
values of the angle θ. Clearly each parabola in-
tersects at least one other .The maximum hori-
zontal distance
v2
0
g occurs only for θ = π
4 , as was
shown in Example 4.3. The maximum vertical
height
v2
0
2 g is attained when the object is launched straight up (i.e. θ = π",ElementaryCalculus.pdf
"4 , as was
shown in Example 4.3. The maximum vertical
height
v2
0
2 g is attained when the object is launched straight up (i.e. θ = π
2 ), as was shown in Exercise 18
in Section 5.1. By symmetry only the angles 0 < θ ≤ π
2 in the same vertical plane need be considered. So
in the above figure, imagine if the trajectories for all possi ble angles were included, filling up a region
that does appear to have a parabolic boundary . This will now b e shown to be true.
First, it turns out that all the parabolas for 0 < θ < π
2 have the same directrix y =
v2
0
2 g . To see why ,
recall from Exercise 11 in Section 4.1 that the maximum heigh t reached by the object is
v2
0 sin2 θ
2 g , which
is thus the y-coordinate of the vertex of the parabola. That vertex is mid way between the focus and
the directrix. The parabola is of the form 4 p y = x2 + bx , where b is a constant that does not affect the",ElementaryCalculus.pdf
"the directrix. The parabola is of the form 4 p y = x2 + bx , where b is a constant that does not affect the
distance between the vertex and directrix 5 , and 4 p is a constant with p < 0 such that the directrix is
−p units above the vertex (since p < 0), just as in the case 4 p y = x2. The equation of the parabola then
shows that
1
4 p = − g
2v2
0 cos2 θ
⇒ p = −
v2
0 cos2 θ
2 g
so that the directrix is at
y = y-coordinate of the vertex + (−p)
=
v2
0 sin2 θ
2 g + −
(
−
v2
0 cos2 θ
2 g
)
=
v2
0
2 g (sin2 θ + cos2 θ)
y =
v2
0
2 g
It is perhaps surprising that all the parabolic trajectorie s share the same directrix y =
v2
0
2 g , which is
independent of the angle θ. Note that the heights of each vertex
(
v2
0 sin2 θ
2 g
)
and focus
(
v2
0
2 g (sin2 θ − cos2 θ)
)
do depend on θ. The common directrix is the key to the remainder of the proof .
5This will be discussed further in Section 7.4.",ElementaryCalculus.pdf
"214 Chapter 7 • Analytic Geometry and Plane Curves §7.2
Now let P be a point in the first quadrant of the x y-plane below the common directrix y =
v2
0
2 g , denoted
by D. Then P can be either inside, outside, or on the envelope, as in Figur e 7.2.5:
D
y
x
P
O
v2
0
2 g
v2
0 / g
(a) Inside the envelope
D
y
x
P
O
v2
0
2 g
v2
0 / g
(b) Outside the envelope
D
y
x
P
O
v2
0
2 g
v2
0 / g
(c) On the envelope
Figure 7.2.5 Trajectory envelope and a point P
The origin O = (0, 0) is on each trajectory , so by definition of a parabola the foc i for all the trajectories
must be a distance
v2
0
2 g from O, i.e. the distance from O to D. That is, the foci of all the trajectories must
lie on the circle C0 of radius
v2
0
2 g centered at O. If P is any other point inside the envelope, so that it lies
on at least one trajectory , then it must be a distance r > 0 below the line D. By definition of a parabola,",ElementaryCalculus.pdf
"on at least one trajectory , then it must be a distance r > 0 below the line D. By definition of a parabola,
P must be the same distance from the foci of any trajectories it belongs to. That is, the foci must be on
a circle C of radius r centered at P and touching the directrix D, as in Figure 7.2.6:
D
y
x
P
O
v2
0
2 g
v2
0 / g
CF1
F2
C0 r
(a) Two intersections F1 , F2
D
y
x
P
O
v2
0
2 g
v2
0 / g
C0
C
r
(b) No intersections
D
y
x
PF
O
v2
0
2 g
v2
0 / g
C0
C
r
(c) One intersection F
Figure 7.2.6 Circles C and C0 intersect at foci of trajectories
L
D
y
x
P
O
v2
0
2 g
v2
0
g
v2
0 / g
r
v2
0
2 g
In Figure 7.2.6(a) C and C0 intersect at two points F1 and F2 , so P
belongs to two trajectories; P must then be inside the envelope. In
Figure 7.2.6(b) C and C0 do not intersect, so P must be outside the
envelope (since it is not on a parabola with a focus on C0 ). If C and
C0 intersect at only one point F , as in Figure 7.2.6(c), then P must
be on the envelope. In that case, P is a distance r +",ElementaryCalculus.pdf
"C0 intersect at only one point F , as in Figure 7.2.6(c), then P must
be on the envelope. In that case, P is a distance r +
v2
0
2 g from O, which
is also the distance from P to the line y =
v2
0
g (denoted by L). Thus, by
definition of a parabola, P is on a parabola with focus O and directrix
L. The vertex is at
(
0,
v2
0
2 g
)
. Therefore, the envelope is a parabola: the
boundary of the shaded region in the figure on the right. ✓",ElementaryCalculus.pdf
"Parabolas • Section 7.2 215
In Example 7.2 all the trajectories were in the x y-plane only . Removing that restriction, so
that trajectories in all vertical planes through the y-axis are possible, would result in a solid
paraboloid consisting of all possible trajectories from th e origin.
Parabolas also appear in suspension bridges: the suspensio n cables
supporting a horizontal bridge (via vertical suspenders, a s in the figure
on the right) have to be parabolas if the weight of the bridge i s uniformly
distributed.6
Exercises
A
1. Construct a parabola using the procedure shown in Figure 7.2 .2.
For Exercises 2-6, sketch the graph of the given parabola and indicate the exact locations of the focus,
vertex, and directrix.
2. 8 y = x2 3. y = 8x2 4. x = y2 5. x = −3 y2 6. −1000 y = x2
7. Find the points of intersection of the parabolas 4 p y = x2 and 4 px = y2 when p > 0. What is the
equation of the line through those points?",ElementaryCalculus.pdf
"7. Find the points of intersection of the parabolas 4 p y = x2 and 4 px = y2 when p > 0. What is the
equation of the line through those points?
8. A vehicle headlight in the shape of a paraboloid is 3in deep an d has an open edge with diameter 8in.
Where should the center of the bulb be placed in order to be at t he focus, measured in inches relative
to the vertex?
9. The latus rectum of a parabola is the chord that passes through the focus and is parallel to the
directrix. Find the length of the latus rectum for the parabo la 4 p y = x2.
10. Show that the circle whose diameter is the latus rectum of a pa rabola touches the parabola’s direc-
trix at one point.
11. Find the points on the parabola 4 px = y2 such that the focal radii to those points have the same
length as the latus rectum.
12. From each end of the latus rectum of a parabola draw a line to th e point where the directrix and
axis intersect. Show that the two drawn lines are perpendicu lar .
B",ElementaryCalculus.pdf
"axis intersect. Show that the two drawn lines are perpendicu lar .
B
13. Show that any point not on a parabola is on either zero or two ta ngent lines to the parabola.
14. Show that y = mx − 2m p − m3 p is the normal line of slope m to the parabola 4 px = y2 .
15. From a point P on a parabola with vertex V let
PQ be the line segment perpendicular to the axis
at a point Q. Show that PQ 2 equals the product of QV and the length of the latus rectum.
16. Show that the curve y = ax2 + bx + c is a parabola for a ̸=0, using only the definition of a parabola.
Find the focus, vertex and directrix.
17. Show that the set of all midpoints of a family of parallel chor ds in a parabola lie on a line parallel
to the parabola’s axis.
6See pp.159-161 in S MI T H , C.E., Applied Mechanics: Statics , New Y ork: John Wiley & Sons, Inc., 1976.",ElementaryCalculus.pdf
"216 Chapter 7 • Analytic Geometry and Plane Curves §7.3
7.3 Hyperbolas
In the previous two sections you have seen curves with eccent ricity e = 0 (circles), 0 < e < 1
(ellipses) and e = 1 (parabolas). The remaining case is e > 1: the hyperbola, whose definition
is similar to the second definition of the ellipse.
A hyperbola is the set of all points in a plane such that the ratio of the dis tance from
a fixed point (a focus) to the distance from a fixed line (a directrix) is a constant e > 1,
called the eccentricity of the hyperbola.
x
y
0
y = 1
x
Figure 7.3.1
It will be shown in Section 7.4 that the curve y = 1
x is a hyperbola,
which has two branches (see Figure 7.3.1). In general a hyper bola
resembles a “wider” or less “cupped” parabola, and it has two symmet-
ric branches (and hence two foci and two directrices) as well as two
asymptotes.
The ratio of distances referred to in the definition of the hyp erbola
also appears in the second definition of the ellipse (where th e ratio is",ElementaryCalculus.pdf
"The ratio of distances referred to in the definition of the hyp erbola
also appears in the second definition of the ellipse (where th e ratio is
smaller than 1) and in the definition of the parabola (where th e ratio
equals 1). In all three cases that ratio is the eccentricity . See Figure
7.3.2(a) for the comparisons.
F e < 1
e = 1
e > 1
D
(a) Eccentricity e
Sun v < vE
v = vE
v > vE
D
(b) Orbit velocity v and escape velocity vE
Figure 7.3.2 Hyperbola, parabola and ellipse with focus F and directrix D
There is an analogue for Figure 7.3.2(a) in terms of orbits. F or example, an object approach-
ing the Sun must meet or exceed the escape velocity to overcome the Sun’s gravitational pull
and avoid returning to orbit. Figure 7.3.2(b) shows the thre e possible trajectories—hyperbola,
parabola and ellipse—in terms of the object’s velocity v and the escape velocity vE . Note the
apparent correlation between the eccentricity of the objec t’s path and its speed as a fraction of",ElementaryCalculus.pdf
"apparent correlation between the eccentricity of the objec t’s path and its speed as a fraction of
the escape velocity (i.e. v
vE
).",ElementaryCalculus.pdf
"Hyperbolas • Section 7.3 217
D
FP
G
Figure 7.3.3 Hyperbola: P F
PG = e > 1
Figure 7.3.3 illustrates the definition of a hyperbola, con-
sisting of points P whose distance P F from the focus F ex-
ceeds the distance P G to the directrix D in a way so that
the ratio P F
PG is always the same constant e > 1 (the eccen-
tricity).
x
y
(ea, 0)
x = a
e
d2
d1
0
(x, y)
Figure 7.3.4
To derive the equation of a hyperbola with eccentricity e > 1,
assume the focus is on the x-axis at ( ea , 0), with a > 0, and the
line x = a
e is the directrix, as in Figure 7.3.4. Pick a point ( x, y)
whose distances d1 and d2 from the focus ( ea , 0) and directrix
x = a
e , respectively , satisfy the condition for a hyperbola:
d1
d2
= e > 1. Then
d2
1 = e2 d2
2
(x − ea )2 + ( y − 0)2 = e2
((
x − a
e
) 2
+ ( y − y)2
)
x2 − ✘✘✘2eax + e2a2 + y2 = e2 x2 − ✘✘✘2eax + a2
(e2 − 1) x2 − y2 = (e2 − 1) a2
x2
a2 − y2
b2 = 1
where b2 = (e2 −1) a2 > 0. The hyperbola thus has two branches ( x = ±a
b
√
y2 + b2), as in Figure",ElementaryCalculus.pdf
"(e2 − 1) x2 − y2 = (e2 − 1) a2
x2
a2 − y2
b2 = 1
where b2 = (e2 −1) a2 > 0. The hyperbola thus has two branches ( x = ±a
b
√
y2 + b2), as in Figure
7.3.5. Let c = ea , so that c > a and b2 = c2 − a2. By symmetry the hyperbola has two foci
(±c, 0) and two directrices x = ±a2
c , and the lines y = ±b
a x are asymptotes. The vertexes are
the points on the hyperbola closest to the directrices.
x
y
(c, 0)
focus
vertex
(−c, 0)
focus
vertex
0 a2
c− a2
c
a−a
asymptote
y = b
a x
asymptote
y = −b
a x
conjugate
axis
transverse
axis
directrix directrix
Figure 7.3.5 Parts of the hyperbola x2
a2 − y2
b2 = 1, with b2 = c2 − a2
The center is the point midway between the foci. The transverse axis and conjugate axis
are the perpendicular lines through the foci and center , res pectively .",ElementaryCalculus.pdf
"218 Chapter 7 • Analytic Geometry and Plane Curves §7.3
In Figure 7.3.5 the vertexes are ( ±a, 0), the x-axis is the transverse axis, the center is the
origin (0 , 0), and the conjugate axis is the y-axis. Note that the existence of two foci and
directrices—when the definition of the hyperbola mentioned only a focus and a directrix—is
simply a consequence of the symmetry about both axes imposed by the equation x2
a2 − y2
b2 = 1. A
parabola, in comparison, is symmetric about only one axis.
To see why the lines y = ±b
a x are asymptotes, consider the upper half y = b
a
∝ra⌈〉⌋allow
x2 − a2 of both
branches of the hyperbola. The difference between the line y = b
a x and the upper right branch
approaches zero as x approaches infinity:
limx→∞
(b
a x − b
a
√
x2 − a2
)
= b
a limx→∞
(
x −
√
x2 − a2
)
· x +
∝ra⌈〉⌋allow
x2 − a2
x +
∝ra⌈〉⌋allow
x2 − a2
= b
a limx→∞
x2 − (x2 − a2)
x +
∝ra⌈〉⌋allow
x2 − a2
= limx→∞
ab
x +
∝ra⌈〉⌋allow
x2 − a2
= 0
Thus the line y = b",ElementaryCalculus.pdf
"x2 − a2
x +
∝ra⌈〉⌋allow
x2 − a2
= b
a limx→∞
x2 − (x2 − a2)
x +
∝ra⌈〉⌋allow
x2 − a2
= limx→∞
ab
x +
∝ra⌈〉⌋allow
x2 − a2
= 0
Thus the line y = b
a x is an (oblique) asymptote for the upper half y = b
a
∝ra⌈〉⌋allow
x2 − a2 of the right
branch in the first quadrant of the x y-plane. So by symmetry the lines y = ±b
a x are asymptotes
for both branches, i.e. for the entire hyperbola.
Conversely , given a hyperbola in the form x2
a2 − y2
b2 = 1, let c2 = a2 + b2 to get the foci ( ±c, 0)
and the directrices x = ±a2
c , while the eccentricity is e = c
a .
Example 7.3
For the hyperbola x2 − y2 = 1 find the vertexes, foci, directrices, asymptotes and eccen tricity .
Solution: Here a = b = 1, so that c2 = a2 + b2 = 2, i.e. c =
∝ra⌈〉⌋allow
2. The vertexes are thus ( ±1, 0), the
asymptotes are y = ±x, the foci are ( ±
∝ra⌈〉⌋allow
2, 0), the directrices are x = ± 1∝ra⌈〉⌋allow
2 , and the eccentricity is e =
∝ra⌈〉⌋allow
2.
x
y
c
a
y = a2
c
−c
−a
y = −a2
c
y = a
b x
y = −a
b x
Figure 7.3.6 y2
a2 − x2",ElementaryCalculus.pdf
"2 , and the eccentricity is e =
∝ra⌈〉⌋allow
2.
x
y
c
a
y = a2
c
−c
−a
y = −a2
c
y = a
b x
y = −a
b x
Figure 7.3.6 y2
a2 − x2
b2 = 1
Switching the roles of x and y produces the rotated hyper-
bola y2
a2 − x2
b2 = 1, shown in Figure 7.3.6. The transverse axis
is the y-axis, the conjugate axis is the x-axis, the vertexes
are at (0 , ±a), and the foci are at (0 , ±c), where c2 = a2 +b2.
The directrices are y = ±a2
c , the asymptotes are y = ±a
b x,
and the eccentricity is e = c
a .
For example, the hyperbola y2 − x2 = 1 has a = b = 1, so
that c =
∝ra⌈〉⌋allow
2. The vertexes are (0 , ±1), the asymptotes are
y = ±x, the foci are (0 , ±
∝ra⌈〉⌋allow
2), the directrices are y = ± 1∝ra⌈〉⌋allow
2 ,
and the eccentricity is e =
∝ra⌈〉⌋allow
2. The hyperbola y2 − x2 = 1 is
just the hyperbola x2 − y2 = 1 rotated 90 ◦.",ElementaryCalculus.pdf
"Hyperbolas • Section 7.3 219
There is another way to define a hyperbola, in terms of two foci :
A hyperbola is the set of all points in a plane such that the absolute value of the difference
of the distances from two fixed points (the foci) is a positive constant.
Figure 7.3.7 illustrates the above definition with foci F1 and F2. The difference d1 − d2 of
the distances d1 and d2 can be positive or negative depending on which branch the poi nt ( x, y)
is on, which is why the absolute value |d1 − d2 | is used.
x
y
F2F1
d1
d2
0
(x, y)
Figure 7.3.7 Hyperbola: |d1 − d2 | =constant > 0
F2F1
P
A
/C8
L
Figure 7.3.8 Construction of a hyperbola
It is left as an exercise to show that this second definition yi elds the same equation of the
hyperbola as from the first definition. The second definition i s often used as the primary def-
inition in many textbooks, perhaps because it provides a sim ple way to construct a hyperbola",ElementaryCalculus.pdf
"inition in many textbooks, perhaps because it provides a sim ple way to construct a hyperbola
by hand. Figure 7.3.8 shows the procedure for foci F1 and F2: at the focus F1 fasten one end of
a ruler of length L, and at the other end A of the ruler fasten one end of a string of length L −d
for some number 0 < d < F1 F2. Fasten the other end of the string to the focus F2 and hold the
string taut with a pencil against the ruler at a point P , while rotating the ruler about F1. The
drawn figure will be one branch of a hyperbola, since the diffe rence P F1 − P F2 will always be
the positive constant d:
P F1 − P F2 = (L − A P ) − P F2 = L − (A P + P F2) = L − (L − d) = d
Reverse the roles of F1 and F2 to draw the other branch of the hyperbola.
By Exercise 16 in Section 3.4, the tangent line to the hyperbo la x2
a2 − y2
b2 = 1 at a point ( x0, y0)
is xx0
a2 − y y0
b2 = 1 , (7.6)
so that its slope is b2 x0
a2 y0
when y0 ̸=0. Note that by the above equation, when y0 = 0 (so that",ElementaryCalculus.pdf
"is xx0
a2 − y y0
b2 = 1 , (7.6)
so that its slope is b2 x0
a2 y0
when y0 ̸=0. Note that by the above equation, when y0 = 0 (so that
x0 = ±a) the two tangent lines are the vertical lines x = ±a.",ElementaryCalculus.pdf
"220 Chapter 7 • Analytic Geometry and Plane Curves §7.3
That slope will be used in proving the reflection property for the hyperbola: Light shone
from one focus will reflect off the hyperbola in the opposite d irection from the other focus. Figure
7.3.9 shows the light’s path from focus F2 as it reflects at the point P along the line through P
and the other focus F1.
F2F1
P
Figure 7.3.9 Reflection property
x
y
F2
(c, 0)
F1
(−c, 0)
P
(x0, y0 )
α1 α2
L
θ1
θ2
θ2
θ
A
Figure 7.3.10 Hyperbola x2
a2 − y2
b2 = 1 with foci ( ±c, 0)
By Fermat’s Principle for curved surfaces, the reflection pr operty is equivalent to saying the
tangent line L to the hyperbola at the point P = (x0, y0) bisects the angle ∠F1 P F2, i.e. θ1 = θ2 as
in Figure 7.3.10. The reflection property holds trivially wh en y0 = 0 (the light reflects straight
back along the x-axis), so to show that θ1 = θ2 assume y0 ̸=0. By symmetry only x0 > 0 need be",ElementaryCalculus.pdf
"back along the x-axis), so to show that θ1 = θ2 assume y0 ̸=0. By symmetry only x0 > 0 need be
considered. For the case x0 ̸=c, let A be the x-intercept of the tangent line L. Then by Figure
7.3.10, since the sum of the angles in the triangle △F1 P A equals 180 ◦,
α1 + θ2 + (180◦ − θ) = 180◦ ⇒ θ2 = θ − α1 ⇒ tan θ2 = tan (θ − α1) .
Thus, since tan θ is the slope of the tangent line L (i.e. b2 x0
a2 y0
), and tan α1 = y0
x0 +c , then by the
subtraction formula for the tangent function
tan θ2 = tan θ − tan α1
1 + tan θ tan α1
=
b2 x0
a2 y0
− y0
x0 + c
1 + b2 x0
a2 y0
· y0
x0 + c
=
b2 x2
0 − a2 y2
0 + b2 x0 c
✭✭✭✭✭a2 y0 (x0 + c)
(a2 + b2 ) x0 y0 + a2 y0 c
✭✭✭✭✭a2 y0 (x0 + c)
= a2 b2 + b2 x0 c
c2 x0 y0 + a2 y0 c (since c2 = a2 + b2, and
x2
0
a2 −
y2
0
b2 = 1 ⇒ b2 x2
0 − a2 y2
0 = a2 b2)
= b2
✘✘✘✘✘✘(a2 + x0 c)
c y0✘✘✘✘✘✘(a2 + x0 c)
= b2
c y0
Similarly , since the sum of the angles in the triangle △F2 P A equals 180 ◦,
α2 + θ1 + θ = 180◦ ⇒ θ1 = 180◦ − (θ + α2 ) ⇒ tan θ1 = −tan (θ + α2 ) .",ElementaryCalculus.pdf
"Hyperbolas • Section 7.3 221
Thus, since tan α2 = −tan (180◦ − α2) = − y0
x−c ,
tan θ1 = − tan θ + tan α2
1 − tan θ tan α2
= −
b2 x0
a2 y0
+ −y0
x0 − c
1 − b2 x0
a2 y0
· −y0
x0 − c
= −
b2 x2
0 − a2 y2
0 − b2 x0 c
✭✭✭✭✭a2 y0 (x0 − c)
(a2 + b2 ) x0 y0 − a2 y0 c
✭✭✭✭✭a2 y0 (x0 − c)
= − a2 b2 − b2 x0 c
c2 x0 y0 − a2 y0 c = b2
✘✘✘✘✘✘(a2 − x0 c)
c y0✘✘✘✘✘✘(a2 − x0 c)
= b2
c y0
= tan θ2 ,
i.e. θ1 = θ2 (since 0 ◦ < θ1, θ2 < 90◦). ✓ Note: The case x0 = c is left as an exercise.
Ellipses, parabolas and hyperbolas are sometimes called conic sections , due to being formed
by intersections of planes with a double circular cone of unl imited extent:
(a) Ellipse (b) Parabola (c) Hyperbola
Figure 7.3.11 Conic sections
Each double cone in Figure 7.3.11 has two nappes—a cone extending upward and one ex-
tending downward. When a plane intersects only one nappe in a closed noncircular curve, as
in Figure 7.3.11(a), that curve is an ellipse. A plane that is parallel to a line on one nappe, as",ElementaryCalculus.pdf
"in Figure 7.3.11(a), that curve is an ellipse. A plane that is parallel to a line on one nappe, as
in Figure 7.3.11(b), intersects only that nappe in a parabol a. The intersection of a plane with
both nappes, as in Figure 7.3.11(c), is a hyperbola.
O
P1
P2
Q
Figure 7.3.12
To prove that the ellipse, parabola and hyperbola really are repre-
sented by the indicated conic sections, first a minor result i s needed
from three-dimensional geometry: tangent line segments to a sphere
from the same point have equal lengths, as in Figure 7.3.12. S ince the
right triangles △QOP 1 and △QOP 2 share the same hypotenuse QO
and have legs OP1 and OP2 of equal length (the radius of the sphere),
the result Q P1 = Q P2 follows by the Pythagorean Theorem.",ElementaryCalculus.pdf
"222 Chapter 7 • Analytic Geometry and Plane Curves §7.3
D
F
Q
P
G
Pc
P0
A αβ
β
C
Figure 7.3.13
For the case of a right circular double cone
(i.e. the base of each nappe is a circle in a
plane perpendicular to the axis of the cone
7),
let β be the complement of the angle that the
cone makes with its axis, as in Figure 7.3.13.
So β is the angle the cone makes with any
circular base of the cone. This constant an-
gle β, with 0 ◦ < β < 90◦, is an intrinsic prop-
erty of the cone. Let Pc be a plane that inter-
sects the lower nappe of the cone in a curve
C, such that Pc makes an angle α with any
base circle of the cone. By symmetry , only
the angles 0 ◦ < α ≤ 90◦ need be considered.
8
Inscribe a sphere in the cone so that it touches Pc at a point F , as in Figure 7.3.13 ( Pc is the
tangent plane to the sphere at F ). Let P0 be the plane through the circle where the inscribed
sphere touches the cone, and let D be the line of intersection of the planes Pc and P0. It will",ElementaryCalculus.pdf
"sphere touches the cone, and let D be the line of intersection of the planes Pc and P0. It will
be shown that F and D are the focus and directrix, respectively , of the curve C.
Let P be any point on the curve C, then let Q be the point on the plane P0 that lies on a line
through P and the cone’s vertex. Drop a perpendicular line segment fro m P to the point A in
the plane P0. From A draw a perpendicular line segment to the point G on the line D. Then
as Figure 7.3.13 shows, △Q A P and △P AG are right triangles, with
sin α = P A
P G and sin β = P A
P Q .
However , since P F and P Q are both tangent line segments to the inscribed sphere from t he
same point P , the result proved earlier shows that P Q = P F . Thus,
P G sin α = P A = P Q sin β = P F sin β ⇒ P F
P G = sin α
sin β .
Let e = P F
PG . Then e is the same constant sin α
sin β for any point P on the curve C. Thus, by
definition, e is the eccentricity of the curve C with focus F and directrix D. If 0 ◦ < α < β then",ElementaryCalculus.pdf
"sin β for any point P on the curve C. Thus, by
definition, e is the eccentricity of the curve C with focus F and directrix D. If 0 ◦ < α < β then
0◦ < sin α < sin β so that 0 < sin α
sin β < 1, which means that 0 < e < 1, and hence C is an ellipse
(by the second definition of an ellipse). Likewise, if α = β then e = 1, so that C is a parabola.
Finally , if β < α ≤ 90◦ then e > 1, so that C is a hyperbola (and will intersect both nappes of the
cone). Thus, the ellipse, parabola and hyperbola truly are c onic sections. ✓
7The proof can be extended to oblique double cones. See §364 in SA L MO N , G.S., A Treatise on Conic Sections ,
London: Longmans, Green and Co., 1929.
8The case where α = 0◦ results in a circle, which is typically not considered a coni c section.",ElementaryCalculus.pdf
"Hyperbolas • Section 7.3 223
Exercises
A
1. Construct a hyperbola using the procedure shown in Figure 7. 3.8. Place the two focus pins 7in apart
and use a 9in piece of string attached to a 12in ruler .
For Exercises 2-6, sketch the graph of the given hyperbola, i ndicate the exact locations of the foci and
vertexes, indicate each directrix and asymptote, and find th e eccentricity e.
2. x2
16 − y2
9 = 1 3. x2
8 − y2
15 = 1 4. 4x2
25 − y2
4 = 1 5. x2 − 4 y2 = 1 6. 25 y2 − 9x2 = 225
7. Find the equation of the hyperbola with foci ( ±5, 0) and vertexes ( ±3, 0).
B
8. In the second definition of the hyperbola on p.219, let 2 a be the constant absolute value of the
difference of distances from points on the hyperbola to the f oci, for some a > 0. Let the foci be at
(±c, 0) for some c > 0. Show that this second definition yields a hyperbola having an equation of the
form x2
a2 − y2
b2 = 1, with b > 0.",ElementaryCalculus.pdf
"(±c, 0) for some c > 0. Show that this second definition yields a hyperbola having an equation of the
form x2
a2 − y2
b2 = 1, with b > 0.
9. Prove the reflection property for the hyperbola (see pp.220- 221) when x0 = c for the point P = (x0, y0 )
on the hyperbola with foci at ( ±c, 0).
10. Show that a straight line parallel to an asymptote of a hyperb ola intersects the hyperbola at exactly
one point.
11. A latus rectum (plural: latus recta) of a hyperbola is a chord through eithe r focus perpendicular to
the transverse axis. Show that the latus recta of the hyperbo la x2
a2 − y2
b2 = 1 have length 2b2
a .
12. Show that the segment of an asymptote of a hyperbola between t he two directrices has the same
length as the line segment between the vertexes.
13. Show that the segment of a tangent line to a hyperbola between the hyperbola’s asymptotes has its
midpoint at the point of tangency .",ElementaryCalculus.pdf
"13. Show that the segment of a tangent line to a hyperbola between the hyperbola’s asymptotes has its
midpoint at the point of tangency .
14. The focal radii of a hyperbola are the line segments from the foci to points on the hyperbola. Show
that the lengths of the focal radii to points ( x, y) on the hyperbola x2
a2 − y2
b2 = 1 are a + ex and ex − a,
where e is the eccentricity .
15. Show that a tangent line to a hyperbola together with the hype rbola’s asymptotes bounds a triangle
of constant area (i.e. the area is independent of the point of tangency on the hyperbola).
16. Show that the product of the perpendicular distances from an y point on the hyperbola x2
a2 − y2
b2 = 1
to the asymptotes y = ±b
a x is a constant.
17. A person at a point P = (x, y) hears the crack of a rifle located at the point F1 = (−1000, 0) and the
sound of the fired bullet hitting its target located at the poi nt F2 = (1000, 0) at the same time. The",ElementaryCalculus.pdf
"sound of the fired bullet hitting its target located at the poi nt F2 = (1000, 0) at the same time. The
bullet’s speed is 2000 ft/sec and the speed of sound is 1100 ft /sec. Find an equation relating x and y.
18. Show that the set of all midpoints of a family of parallel chor ds either in one branch or between the
two branches of a hyperbola lie on a line through the center of the hyperbola.
19. Prove a second reflection property of hyperbolas: a light sho ne between the two branches and
directed toward one focus will reflect toward the other focus .",ElementaryCalculus.pdf
"224 Chapter 7 • Analytic Geometry and Plane Curves §7.4
7.4 T ranslations and Rotations
For convenience the ellipses, parabolas and hyperbolas in t he previous sections were centered
at the origin and had their foci on one of the coordinate axes. In general the center and foci of
those curves can be moved anywhere by means of coordinate transformations .
x
y
x′
y′
k
h
P = (x, y)
P ′ = (x′, y′)
x − h
y − k
O
O′
Figure 7.4.1 Translation
Y ou have probably seen in earlier courses how the graph of
a function y = f (x) can be shifted horizontally by an amount
h and vertically by an amount k by replacing x and y by x−h
and y − k, respectively:
y − k = f (x − h)
This coordinate transformation is called translation, and
can be applied to any curve in the x y-plane. The origin O =
(0, 0) is shifted to the point O′ = (h, k), which serves as the
origin of the x′ y′-plane,9 as in Figure 7.4.1. Let P = (x, y) be
a point in the x y-plane. Consider P as a point P ′ = (x′, y′) in",ElementaryCalculus.pdf
"origin of the x′ y′-plane,9 as in Figure 7.4.1. Let P = (x, y) be
a point in the x y-plane. Consider P as a point P ′ = (x′, y′) in
the x′ y′-plane, so that relative to the origin O′, the x′ and y′
coordinates of P ′ are:
x′ = x − h
y′ = y − k
Using these translation equations (i.e. the substitutions x ∝mapsto⌋〈ar→x − h and y ∝mapsto⌋〈ar→y − k), the graphs of
the conic sections can be translated to any center ( h, k):
Ellipse: For a > b > 0, an equation of the form
(x − h)2
a2 + ( y − k)2
b2 = 1
describes an ellipse with center ( h, k), vertexes ( h±a, k), and foci ( h±c, k), where c2 = a2−b2.
The eccentricity is e = c
a , and the principal axis is the line y = k.
Likewise, an equation of the form
( y − k)2
a2 + (x − h)2
b2 = 1
describes an ellipse with center ( h, k), vertexes ( h, k±a), and foci ( h, k±c), where c2 = a2−b2.
The eccentricity is e = c
a , and the principal axis is the line x = h.",ElementaryCalculus.pdf
"The eccentricity is e = c
a , and the principal axis is the line x = h.
9The prime symbol ( ′) does not indicate differentiation—it acts merely to disti nguish the new axes.",ElementaryCalculus.pdf
"T ranslations and Rotations • Section 7.4 225
Parabola: For p ̸=0, an equation of the form
(x − h)2 = 4 p( y − k)
describes a parabola with vertex ( h, k) and focus ( h, k + p). The directrix is the line y = k − p
and the axis is the line x = h.
Likewise, an equation of the form
( y − k)2 = 4 p(x − h)
describes a parabola with vertex ( h, k) and focus ( h + p, k). The directrix is the line x = h − p
and the axis is the line y = k
Hyperbola: For a ̸=0 and b ̸=0, an equation of the form
(x − h)2
a2 − ( y − k)2
b2 = 1
describes a hyperbola with center ( h, k), vertexes ( h ± a, k), and foci ( h ± c, k), where c2 =
a2 + b2. The eccentricity is e = c
a , the directrices are the lines x = h ± a2
c , the asymptotes are
the lines y − k = ±b
a (x − h), and the transverse axis is the line y = k.
Likewise, an equation of the form
( y − k)2
a2 − (x − h)2
b2 = 1
describes a hyperbola with center ( h, k), vertexes ( h, k ± a), and foci ( h, k ± c), where c2 =
a2 + b2. The eccentricity is e = c",ElementaryCalculus.pdf
"b2 = 1
describes a hyperbola with center ( h, k), vertexes ( h, k ± a), and foci ( h, k ± c), where c2 =
a2 + b2. The eccentricity is e = c
a , the directrices are the lines y = k ± a2
c , the asymptotes are
the lines y − k = ±a
b (x − h), and the transverse axis is the line x = h.
Example 7.4
x
y
y = 1
x = 2
0
2
4
(2 −
∝ra⌈〉⌋allow
3, 1) (2 +
∝ra⌈〉⌋allow
3, 1)
Figure 7.4.2 Ellipse (x−2)2
4 + ( y − 1)2 = 1
Find the vertexes and foci of the ellipse (x−2)2
4 + ( y − 1)2 = 1.
Solution: The translation coordinates are ( h, k) = (2, 1). Also,
a = 2 and b = 1. Thus, the vertexes are ( h ± a, k) = (2 ± 2, 1) =
(0, 1) and (4 , 1). Since c2 = a2 − b2 = 3 then c =
∝ra⌈〉⌋allow
3, so the foci
are ( h ± c, k) = (2 ±
∝ra⌈〉⌋allow
3, 1), as shown in Figure 7.4.2.
Note that the ellipse (x−2)2
4 +( y −1)2 = 1 in the x y-plane is
the ellipse x′2
4 + y′2 = 1 in the x′ y′-plane, for the translation
equations x′ = x − 2 and y′ = y − 1 (i.e. the x′-axis is the line
y = 1 and the y′-axis is the line x = 2).",ElementaryCalculus.pdf
"226 Chapter 7 • Analytic Geometry and Plane Curves §7.4
Example 7.5
For a ̸=0 and constants b and c, find the vertex, focus and directrix of the parabola y = ax2 + bx + c.
Solution: The idea here is to write y = ax2 + bx + c in the form ( x − h)2 = 4 p( y − k) for some h, k, and p,
by completing the square:
ax2 + bx + c = y
a
(
x2 + b
a x
)
= y − c
a
(
x2 + b
a x + b2
4a2
)
= y − c + a b2
4a2
(
x + b
2a
) 2
= 1
a
(
y + b2 − 4ac
4a
)
So h = −b
2a , k = 4a c−b2
4a , and 4 p = 1
a means p = 1
4a . Thus, the vertex is ( h, k) =
(
− b
2a , 4a c−b2
4a
)
, the focus is
(h, k + p) =
(
− b
2a , 4a c−b2 +1
4a
)
, and the directrix is the line y = k − p = 4a c−b2 −1
4a .
r
x
y
x′
y′
P = (x, y)
P ′ = (x′, y′)
θα−θ
α
O
Figure 7.4.3 Rotation
Rotation is another common coordinate transformation.
Consider the case of rotating the x y-plane about the origin
by an angle θ, as in Figure 7.4.3. The origin of the result-
ing x′ y′-plane is unchanged from the origin O = (0, 0) of the",ElementaryCalculus.pdf
"by an angle θ, as in Figure 7.4.3. The origin of the result-
ing x′ y′-plane is unchanged from the origin O = (0, 0) of the
x y-plane. To find the rotation equations for x′ and y′, let
P = (x, y) be a point in the x y-plane away from the origin.
From trigonometry you know that for r =
√
x2 + y2 ̸=0 and
the angle 0 ◦ ≤ α < 360◦ measured from the positive x-axis to
OP ,
x = r cos α and y = r sin α .
Considering P as a point P ′ = (x′, y′) in the x′ y′-plane, OP ′ makes an angle α − θ with the
positive x′-axis, so that by the sine and cosine subtraction identities ,
x′ = r cos (α− θ) = r cos α cos θ + r sin α sin θ = x cos θ + y sin θ (7.7)
and
y′ = r sin (α− θ) = r sin α cos θ − r cos α sin θ = y cos θ − x sin θ . (7.8)
Similar to the translation substitutions, the above rotati on equations allow any curve in the
x y-plane to be rotated:
To rotate a curve in the x y-plane about the origin by an angle θ, make the following sub-",ElementaryCalculus.pdf
"x y-plane to be rotated:
To rotate a curve in the x y-plane about the origin by an angle θ, make the following sub-
stitutions: x ∝mapsto⌋〈ar→x cos θ + y sin θ and y ∝mapsto⌋〈ar→ −x sin θ + y cos θ (7.9)",ElementaryCalculus.pdf
"T ranslations and Rotations • Section 7.4 227
Example 7.6
Find the equation of the ellipse x2
4 + y2 = 1 when rotated 45 ◦ counterclockwise about the origin. Simplify
the equation.
x
y
x
′
y
′
45◦
0
2
−2
2−2
Figure 7.4.4
Solution: For θ = 45◦ the substitutions are:
x ∝mapsto⌋〈ar→x cos θ + y sin θ = x cos 45 ◦ + y sin 45 ◦ = x + y∝ra⌈〉⌋allow
2
y ∝mapsto⌋〈ar→ −x sin θ + y cos θ = −x sin 45 ◦ + y cos 45 ◦ = −x + y∝ra⌈〉⌋allow
2
The equation of the rotated ellipse (shown in Figure 7.4.4) i s then:
1
4
(x + y∝ra⌈〉⌋allow
2
) 2
+
(−x + y∝ra⌈〉⌋allow
2
) 2
= 1
x2 + 2x y + y2 + 4(x2 − 2x y + y2 ) = 8
5x2 − 6x y + 5 y2 − 8 = 0
In the above example note the presence of the 6 x y term in the equation of the rotated ellipse.
In general if a conic section has a second-degree equation of the form
A x2 + B x y + C y2 + D x + E y + F = 0 (7.10)
then B ̸=0 indicates rotation, and either D ̸=0 or E ̸=0 indicates translation.
Example 7.7
Find the value of a such that rotating the hyperbola x2
a2 − y2",ElementaryCalculus.pdf
"Example 7.7
Find the value of a such that rotating the hyperbola x2
a2 − y2
a2 = 1 by 45 ◦ counterclockwise about the origin
results in the curve x y = 1.
Solution: Since θ = 45◦ then as in Example 7.6 the substitutions are again:
x ∝mapsto⌋〈ar→x + y∝ra⌈〉⌋allow
2
and y ∝mapsto⌋〈ar→−x + y∝ra⌈〉⌋allow
2
The equation of the rotated hyperbola is then:
1
a2
(x + y∝ra⌈〉⌋allow
2
) 2
− 1
a2
(−x + y∝ra⌈〉⌋allow
2
) 2
= 1
x2 + 2x y + y2 − (x2 − 2x y + y2 )
2a2 = 1
2x y
a2 = 1
Thus, when a =
∝ra⌈〉⌋allow
2 the rotated hyperbola has the equation x y = 1, which shows that the curve y = 1
x is
a hyperbola. In general any curve of the form B x y = 1 is a hyperbola for B ̸=0.",ElementaryCalculus.pdf
"228 Chapter 7 • Analytic Geometry and Plane Curves §7.4
The following result can be used for determining the type of c onic section described by a
second-degree equation: 10
The graph of A x2 +B x y +C y2 +D x +E y +F = 0 (with A, B, C not all zero) describes a curve
whose type is based on the sign of B2 − 4 AC :
(a) B2 − 4 AC < 0: an ellipse (or a circle, point, or no curve)
(b) B2 − 4 AC = 0: a parabola (or a line, two parallel lines, or no curve)
(c) B2 − 4 AC > 0: a hyperbola (or two intersecting lines)
If B ̸=0 and the curve is a conic section, then the rotation angle θ is given by:
(a) θ = 45◦ if A = C
(b) tan 2 θ = B
A−C if A ̸=C, with 0 ◦ < θ < 90◦
Recall that the rotation equations (7.7) and (7.8) for x′ and y′ in terms of x and y provided
substitutions that allowed the second-degree equation of t he rotated conic section to be found.
Conversely , to transform a second-degree equation in x and y into a “standard” conic section",ElementaryCalculus.pdf
"Conversely , to transform a second-degree equation in x and y into a “standard” conic section
equation in terms of x′ and y′ (to simplify sketching the graph), the “reverse” rotation e quations
for x and y in terms of x′ and y′ are needed:
x = x′ cos θ − y′ sin θ (7.11)
y = x′ sin θ + y′ cos θ (7.12)
Example 7.8
Determine the type of curve whose equation is 5 x2 + 4x y + 8 y2 − 36 = 0, and sketch its graph.
Solution: Since A = 5, B = 4, and C = 8, then B2 − 4 AC = −144 < 0, so the curve is an ellipse if it is a
conic section. Since B ̸=0 and A ̸=C, the rotation angle θ would be given by tan 2 θ = B
A−C = 4
−3 , with
0◦ < θ < 90◦. Then 2 θ is in the second quadrant, so that sin 2 θ = 4
5 and cos 2 θ = −3
5 . Use a half-angle
identity to find θ:
tan θ = sin 2 θ
1 + cos 2 θ =
4
5
1 + −3
5
= 2
Hence, sin θ = 2∝ra⌈〉⌋allow
5 and cos θ = 1∝ra⌈〉⌋allow
5 . Now use equations (7.11) and (7.12) to get expressions for x and y in
terms of x′ and y′:
x = x′ cos θ − y′ sin θ = x′ − 2 y′
∝ra⌈〉⌋allow
5",ElementaryCalculus.pdf
"5 . Now use equations (7.11) and (7.12) to get expressions for x and y in
terms of x′ and y′:
x = x′ cos θ − y′ sin θ = x′ − 2 y′
∝ra⌈〉⌋allow
5
y = x′ sin θ + y′ cos θ = 2x′ + y′
∝ra⌈〉⌋allow
5
10For a proof see Section 6.8 in P R O T T E R , M.H. A N D C.B. M O R R E Y, Analytic Geometry , 2nd ed., Reading, MA:
Addison-Wesley Publishing Company , Inc., 1975.",ElementaryCalculus.pdf
"T ranslations and Rotations • Section 7.4 229
x
y
x
′
y
′
64.3◦
0
3
−3
3−3
Substitute those expressions into 5 x2 + 4x y + 8 y2 − 36 = 0:
5
(x′ − 2 y′
∝ra⌈〉⌋allow
5
) 2
+ 4
(x′ − 2 y′
∝ra⌈〉⌋allow
5
) ( 2x′ + y′
∝ra⌈〉⌋allow
5
)
+ 8
(2x′ + y′
∝ra⌈〉⌋allow
5
) 2
− 36 = 0
45x′2 + 20 y′2 = 180
x′2
4 + y′2
9 = 1
So the curve’s equation in the x′ y′-plane is x′2
4 + y′2
9 = 1. This
is just the ellipse x2
4 + y2
9 = 1 rotated counterclockwise by the
angle θ = tan−1 2 ≈ 63.4◦, as shown in the figure on the right.
Exercises
A
For Exercises 1-3, sketch the graph of the given ellipse with the exact locations of the foci and vertexes.
1. (x − 3)2
25 + ( y − 2)2
16 = 1 2. (x + 1)2
9 + ( y + 3)2
4 = 1 3. 4x2 + y2 + 24x − 2 y + 21 = 0
For Exercises 4-7, sketch the graph of the given parabola wit h the exact locations of the focus, vertex
and directrix.
4. 4(x − 1)2 = 3( y + 2) 5. y = 4x2 + 24x + 21 6. 3( y + 2)2 = 5(x − 2) 7. 4x2 − 4x + 4 y − 5 = 0",ElementaryCalculus.pdf
"and directrix.
4. 4(x − 1)2 = 3( y + 2) 5. y = 4x2 + 24x + 21 6. 3( y + 2)2 = 5(x − 2) 7. 4x2 − 4x + 4 y − 5 = 0
For Exercises 8-10, sketch the graph of the given hyperbola w ith the exact locations of the foci, vertexes,
directrices, and asymptotes.
8. (x − 3)2
25 − ( y − 2)2
16 = 1 9. (x + 1)2
9 − ( y + 3)2
4 = 1 10. 4x2 − y2 + 24x − 2 y + 21 = 0
11. Find the foci, vertexes, directrices and asymptotes for the hyperbola y = 1
x in Example 7.7.
12. Find the equation of the parabola 4 y = x2 when rotated 60 ◦ counterclockwise about the origin.
13. Prove equations (7.11) and (7.12). (Hint: Use equations (7.7) and (7.8).)
B
14. Determine the type of curve whose equation is 8 x2 − 12x y + 17 y2 − 20 = 0, and sketch its graph.
15. Determine the type of curve whose equation is 7 x2 + 6x y − y2 − 32 = 0, and sketch its graph.
16. Let A′x′2 +B′x′ y′ +C′ y′2 +D′ x′ +E′ y′ +F ′ = 0 be the equation obtained by substituting the “reverse”",ElementaryCalculus.pdf
"16. Let A′x′2 +B′x′ y′ +C′ y′2 +D′ x′ +E′ y′ +F ′ = 0 be the equation obtained by substituting the “reverse”
translation equations x = x′ + h and y = y′ + k into equation (7.10). Show that A = A′, B = B′ and
C = C′ (i.e. A, B and C are invariant under translation).
17. Similar to Exercise 16, substitute the “reverse” rotation e quations (7.11) and (7.12) into equation
(7.10) to show that A + C and B2 − 4 AC are invariant under rotation.
18. Sketch the graph of the conic section whose equation is x2 − 2x y + y2 + 4x + 4 y + 10 = 0. (Hint: First
handle the rotation (since A = C use θ = 45◦ even though B2 − 4 AC = 0), then the translation (by
completing the square).)",ElementaryCalculus.pdf
"230 Chapter 7 • Analytic Geometry and Plane Curves §7.5
7.5 Hyperbolic Functions
In some textbooks you might see the sine and cosine functions called circular functions , since
any point on the unit circle x2 + y2 = 1 can be defined in terms of those functions (see Figure
7.5.1). Those definitions motivate a similar idea for the unit hyperbola x2 − y2 = 1, whose
points can be defined in terms of hyperbolic functions .
x
y
O
P = (x, y)
1
θ
x2 + y2 = 1
a
2 = sector area = 1
2 r2 θ = θ
2
x = cos θ = cos a
y = sin θ = sin a
Figure 7.5.1 Circular
y
x x
O
P = (x, y)
1
x2 − y2 = 1
a
2 = shaded area
Figure 7.5.2 Hyperbolic
For a point P = (x, y) on the unit hyperbola x2 − y2 = 1, the hyperbolic angle a is twice the
area of the shaded hyperbolic sector in Figure 7.5.2). 11 The area a
2 thus equals the area of
the right triangle with hypotenuse OP and legs of length x and y (so that the triangle’s area
is 1",ElementaryCalculus.pdf
"2 thus equals the area of
the right triangle with hypotenuse OP and legs of length x and y (so that the triangle’s area
is 1
2 x y) minus the area under the hyperbola over the interval [1 , x]. So since the upper half of
the hyperbola x2 − y2 = 1 is the function y =
∝ra⌈〉⌋allow
x2 − 1,
a
2 = (area of triangle) − (area under the hyperbola from 1 to x)
= 1
2 x y −
∫ x
1
√
u2 − 1 du
a
2 = 1
2 x
√
x2 − 1 −
(1
2 x
√
x2 − 1 − 1
2 ln
(
x +
√
x2 − 1
) )
(by formula (6.9))
a = ln
(
x +
√
x2 − 1
)
ea = x +
√
x2 − 1
(ea − x)2 =
(√
x2 − 1
) 2
e2a − 2x ea +x2 = x2 − 1
x = e2a + 1
2ea = ea + e−a
2 = cosh a
where cosh a is the hyperbolic cosine of a.
11The reason for using twice the area is merely to obtain a “clea ner” final result involving a instead of 2 a.",ElementaryCalculus.pdf
"Hyperbolic Functions • Section 7.5 231
The y-coordinate of P can then be found:
y =
√
x2 − 1 =
√ (ea + e−a
2
) 2
− 1 =
√
e2a + 2 + e−2a
4 − 4
4
=
√
e2a − 2 + e−2a
4 =
√ (ea − e−a
2
) 2
, so since a ≥ 0
y = ea − e−a
2 = sinh a
where sinh a is the hyperbolic sine of a. All six hyperbolic functions can now be defined in
general, analogous to the trigonometric (circular) functi ons:
The hyperbolic sine , hyperbolic cosine , hyperbolic tangent , hyperbolic cotan-
gent, hyperbolic secant and hyperbolic cosecant , denoted by sinh, cosh, tanh, coth,
sech and csch, respectively , are:
sinh x = ex − e−x
2 for all x cosh x = ex + e−x
2 for all x
tanh x = sinh x
cosh x for all x coth x = 1
tanh x for all x ̸=0
sech x = 1
cosh x for all x csch x = 1
sinh x for all x ̸=0
The graphs of the hyperbolic functions are shown below:
x
y
1
0
y = cosh x
y = sinh x
(a) sinh x and cosh x
x
y
1
−1
0y = tanh x
y = coth x
(b) tanh x and coth x
x
y
1
0
y = sech x
y = csch x
(c) sech x and csch x",ElementaryCalculus.pdf
"y = cosh x
y = sinh x
(a) sinh x and cosh x
x
y
1
−1
0y = tanh x
y = coth x
(b) tanh x and coth x
x
y
1
0
y = sech x
y = csch x
(c) sech x and csch x
Figure 7.5.3 Graphs of the six hyperbolic functions
The graph of y = cosh x in Figure 7.5.3(a) might look familiar: a catenary—a uniform cable
hanging from two fixed points—has the shape of a hyperbolic co sine function.",ElementaryCalculus.pdf
"232 Chapter 7 • Analytic Geometry and Plane Curves §7.5
The hyperbolic functions satisfy the following identities :
cosh2 x − sinh2 x = 1 tanh 2 x + sech2 x = 1 coth 2 x − csch2 x = 1
sinh (−x) = −sinh x cosh (−x) = cosh x tanh (−x) = −tanh x
sech (−x) = sech x csch (−x) = −csch x coth (−x) = −coth x
sinh ( u ± v) = sinh u cosh v ± cosh u sinh v sinh 2 x = 2 sinh x cosh x
cosh (u ± v) = cosh u cosh v ± sinh u sinh v cosh 2 x = cosh2 x + sinh2 x
tanh (u ± v) = tanh u ± tanh v
1 ± tanh u tanh v tanh 2 x = 2 tanh x
1 + tanh2 x
The identity cosh 2 x−sinh2 x = 1 was proved when deriving the coordinates of points on the un it
hyperbola x2 − y2 = 1 in terms of the hyperbolic angle (since such a point ( x, y) = (cosh a, sinh a)
must satisfy x2 − y2 = 1). The addition identities can be proved similarly using hy perbolic
angles (i.e. areas). 12 However , it is simpler to use the definitions of sinh and cosh i n terms of
exponential functions. For example:
sinh u cosh v + cosh u sinh v = eu − e−u",ElementaryCalculus.pdf
"exponential functions. For example:
sinh u cosh v + cosh u sinh v = eu − e−u
2 · ev + e−v
2 + eu + e−u
2 · ev − e−v
2
= eu+v + eu−v − e−u+v − e−u−v + eu+v − eu−v + e−u+v − e−u−v
4
= 2eu+v − 2e−u−v
4
= eu+v − e−(u+v)
2
= sinh ( u + v) ✓
The identity for sinh 2 x is then easy to prove by letting u = v = x in the above identity:
sinh 2 x = sinh ( x + x) = sinh x cosh x + cosh x sinh x = 2 sinh x cosh x ✓
Note that the identities cosh ( −x) = cosh x and sinh ( −x) = −sinh x mean that cosh is an even
function and sinh is an odd function. Those two functions thu s (sort of) serve as even and odd
versions of the exponential function (which is neither even nor odd). Both cosh x and sinh x
grow exponentially (the e−x term for both functions becomes negligible as x → ∞), while sinh x
decreases exponentially to −∞ as x → −∞.
12See pp.25-29 in S H E R VAT OV, V .G., Hyperbolic Functions , Boston: D.C. Heath and Company , 1963.",ElementaryCalculus.pdf
"Hyperbolic Functions • Section 7.5 233
The derivatives of the hyperbolic functions and their integ ral equivalents are:
d
dx (sinh x) = cosh x d
dx (csch x) = −csch x coth x
d
dx (cosh x) = sinh x d
dx (sech x) = −sech x tanh x
d
dx (tanh x) = sech2 x d
dx (coth x) = −csch2 x
∫ 
cosh x dx = sinh x + C
∫ 
csch x coth x dx = −csch x + C
∫ 
sinh x dx = cosh x + C
∫ 
sech x tanh x dx = −sech x + C
∫ 
sech2 x dx = tanh x + C
∫ 
csch2 x dx = −coth x + C
For example, by definition of cosh x:
d
dx (cosh x) = d
dx
(ex + e−x
2
)
= ex − e−x
2 = sinh x ✓
Example 7.9
Find the derivative of y = sinh x3.
Solution: By the Chain Rule, dy
dx = 3x2 cosh x3.
Example 7.10
Evaluate
∫ 
tanh x dx .
Solution: Use the definition of tanh x and the substitutions u = cosh x, du = sinh x dx :
∫ 
tanh x dx =
∫ sinh x
cosh x dx =
∫ du
u = ln |u| + C = ln (cosh x) + C
For any constant a > 0, both y = cosh at and y = sinh at satisfy the differential equation
y′′(t) = a2 y(t) ,",ElementaryCalculus.pdf
"∫ du
u = ln |u| + C = ln (cosh x) + C
For any constant a > 0, both y = cosh at and y = sinh at satisfy the differential equation
y′′(t) = a2 y(t) ,
which models rectilinear motion of a particle under a repuls ive force proportional to the dis-
placement. This is just one of the many reasons why hyperboli c functions appear in so many
physical applications.",ElementaryCalculus.pdf
"234 Chapter 7 • Analytic Geometry and Plane Curves §7.5
Example 7.11
In the classical theory of paramagnetism, the total number n of molecules in a gas subject to a magnetic
field of strength H is
n = 2πN
∫ π
0
e
µH
kT cos θ sin θ dθ ,
where N is the number of molecules per unit solid angle having zero po tential energy , µ is the magnetic
moment, k is Boltzmann’s constant, and T is the temperature of the gas. Show that
n = 4πN kT
µH sinh
(µH
kT
)
.
Solution: Let a = µH
kT , and let u = cos θ so that du = −sin θ dθ:
n = −2πN
∫ −1
1
eau du = 2πN
∫ 1
−1
eau du = 2πN
a eau
⏐
⏐
⏐
⏐
1
−1
= 2πN
a
(
2 · ea − e−a
2
)
= 4πN kT
µH sinh
(µH
kT
)
Since d
dx (sinh x) = cosh x = ex +e−x
2 > 0 for all x, then y = sinh x is an increasing function
and thus its inverse function x = sinh−1 y is defined. The remaining inverse hyperbolic
functions can be defined similarly , with the following domains and rang es (switching the
roles of x and y, as usual) plus their graphs:",ElementaryCalculus.pdf
"functions can be defined similarly , with the following domains and rang es (switching the
roles of x and y, as usual) plus their graphs:
function sinh−1 x cosh−1 x tanh−1 x csch−1 x sech−1 x coth−1 x
domain all x x ≥ 1 |x| <1 all x ̸=0 0 < x ≤ 1 |x| >1
range all y y ≥ 0 all y all y ̸=0 y ≥ 0 all y ̸=0
x
y
1
y = cosh−1 x
y = sinh−1 x
(a) sinh−1 x and cosh −1 x
x
y
1−1
y = coth−1 x
y = tanh−1 x
(b) tanh−1 x and coth −1
x
y
1
y = sech−1 x
y = csch−1 x
(c) sech−1 x and csch −1 x
Figure 7.5.4 Graphs of the six inverse hyperbolic functions",ElementaryCalculus.pdf
"Hyperbolic Functions • Section 7.5 235
The inverse hyperbolic functions can be expressed in terms o f the natural logarithm:
sinh−1 x = ln ( x +
√
x2 + 1 ) cosh −1 x = ln ( x +
√
x2 − 1 ) for x ≥ 1
tanh−1 x = 1
2 ln 1 + x
1 − x for |x| <1 coth −1 x = 1
2 ln x + 1
x − 1 for |x| >1
sech−1 x = ln 1 +
∝ra⌈〉⌋allow
1 − x2
x for 0 < x ≤ 1 csch −1 x = ln
(
1
x +
√
1
x2 + 1
)
for x ̸=0
Notice that the above formula for cosh −1 x was actually proved at the beginning of this section.
The formula for sinh −1 x follows from the definition of an inverse function:
y = sinh−1 x ⇒ x = sinh y = e y − e−y
2
⇒ e y − 2x − e−y = 0
⇒ e2 y − 2x e y − 1 = 0
⇒ u2 − 2xu − 1 = 0 for u = e y
⇒ u = 2x ±
√
4x2 − 4(1)(−1)
2 = x ±
√
x2 + 1
⇒ e y = u = x +
√
x2 + 1 since e y > 0
⇒ y = ln ( x +
√
x2 + 1 ) ✓
The remaining formulas can be proved similarly .
Example 7.12
y
x
O
P
(x, y) = (cosh a, sinh a)
A = (1, 0)
x2 − y2 = 1
a
2 = shaded area of O AP
Show that for the hyperbolic angle a of a point P = (x, y) on the",ElementaryCalculus.pdf
"y
x
O
P
(x, y) = (cosh a, sinh a)
A = (1, 0)
x2 − y2 = 1
a
2 = shaded area of O AP
Show that for the hyperbolic angle a of a point P = (x, y) on the
unit hyperbola x2 − y2 = 1, the area a
2 of the hyperbolic sector
O AP (the shaded region in the figure on the right) is
a
2 = 1
2 cosh−1 x .
Solution: It was shown earlier that the area of O AP is
a
2 = 1
2 ln ( x +
√
x2 − 1 )
so by the formula cosh −1 x = ln ( x +
∝ra⌈〉⌋allow
x2 − 1 ) the result follows.
This makes sense, since x = cosh a and so cosh −1 x = cosh−1(cosh a) = a, by definition of an inverse.
Y ou can use either the general formula for the derivative of a n inverse function or the above
formulas to find the derivatives of the inverse hyperbolic fu nctions:",ElementaryCalculus.pdf
"236 Chapter 7 • Analytic Geometry and Plane Curves §7.5
d
dx (sinh−1 x) = 1∝ra⌈〉⌋allow
x2 + 1
d
dx (cosh−1 x) = 1∝ra⌈〉⌋allow
x2 − 1
for x ≥ 1
d
dx (tanh−1 x) = 1
1 − x2 for |x| <1 d
dx (coth−1 x) = 1
1 − x2 for |x| >1
d
dx (sech−1 x) = −1
x
∝ra⌈〉⌋allow
1 − x2
for 0 < x ≤ 1 d
dx (csch−1 x) = −1
|x|
∝ra⌈〉⌋allow
1 + x2
for x ̸=0
For example, here is one way to find the derivative of tanh −1 x:
d
dx (tanh−1 x) = d
dx
(1
2 ln 1 + x
1 − x
)
= 1
2
d
dx (ln (1 + x) − ln (1 − x))
= 1
2
( 1
1 + x − −1
1 − x
)
= = 1 − x + (1 + x)
2(1 + x)(1 − x) = 1
1 − x2 ✓
Exercises
A
1. Prove the identities for sinh ( −x), cosh ( −x), tanh ( −x), coth ( −x), sech ( −x), and csch ( −x) on p.232.
2. Prove the identities for cosh ( u ± v), tanh ( u ± v), and tanh 2 x on p.232.
For Exercises 3-15 prove the given identity . Y our proofs can use other identities.
3. (cosh x + sinh x)r = cosh rx + sinh rx for all r 4. sinh 3 x = 3 sinh x + 4 sinh 3 x
5. sinh A cosh B = 1",ElementaryCalculus.pdf
"3. (cosh x + sinh x)r = cosh rx + sinh rx for all r 4. sinh 3 x = 3 sinh x + 4 sinh 3 x
5. sinh A cosh B = 1
2 (sinh ( A + B) + sinh ( A − B)) 6. tanh2 x + sech2 x = 1
7. sinh A sinh B = 1
2 (cosh ( A + B) − cosh ( A − B)) 8. coth2 x − csch2 x = 1
9. cosh A cosh B = 1
2 (cosh ( A + B) + cosh ( A − B)) 10. coth−1 (1
x
)
= tanh−1 x
11. cosh 2 x = cosh2 x + sinh2 x = 2 cosh 2 x − 1 = 1 + 2 sinh 2 x 12. sinh2 x
2 = cosh x − 1
2
13. cosh2 x
2 = cosh x + 1
2 14. tanh2 x
2 = cosh x − 1
cosh x + 1 15. tanh x
2 = cosh x − 1
sinh x = sinh x
cosh x + 1
16. Prove the identities for tanh −1 x, coth −1 x, sech −1 x, and csch −1 x on p.235.
17. Prove the derivative formulas for sinh x, tanh x, coth x, sech x, and csch x on p.233.
18. Prove the derivative formulas for sinh −1 x, cosh −1 x, coth −1 x, sech −1 x, and csch −1 x on p.236.
19. Show that cosh x = O(ex ) and sinh x = O(ex ). 20. Show that d
dx (tan−1(sinh x)) = sech x .
21. V erify that the curve y = tanh x has asymptotes y = ±1.",ElementaryCalculus.pdf
"dx (tan−1(sinh x)) = sech x .
21. V erify that the curve y = tanh x has asymptotes y = ±1.
22. Show that sinh dx = dx and cosh dx = 1 for any infinitesimal dx. (Hint: Use 1
1+dx = 1 − dx from
Exercise 3 in Section 1.3 along with edx = 1 + dx from Exercise 29 in Section 2.3.)",ElementaryCalculus.pdf
"Hyperbolic Functions • Section 7.5 237
23. Use Exercise 22 and the addition formula for sinh x to show that d
dx (sinh x) = cosh x.
24. Sketch the graph of f (x) = e−2x sinh x. Find all local maxima and minima, inflection points, and
vertical or horizontal asymptotes.
25. Denoting the speed of light by c, the Lorentz transformations for two inertial frames S and S′ are
x′ = x − vt√
1 − (v/c)2
and t′ = t − vx/c2
√
1 − (v/c)2
where S′ moves with speed v > 0 parallel to the x-axis of S. Let v/c = tanh ξ. Show that
x′ = −ct sinh ξ + x cosh ξ and t′ = t cosh ξ − (x/c) sinh ξ .
26. Evaluate the integral I =
∫ 3
2
dx
1−x2 in two different ways:
(a) Use d
dx (coth−1 x) = 1
1−x2 for |x | >1 to show that I = coth−1 3 − coth−1 2.
(b) Use the substitution u = 1
x in the integral, then use d
dx (tanh−1 x) = 1
1−x2 for |x | <1 to show that
I = tanh−1 1
3 − tanh−1 1
2 . Is this answer equivalent to the answer from part (a)? Expla in.
B",ElementaryCalculus.pdf
"dx (tanh−1 x) = 1
1−x2 for |x | <1 to show that
I = tanh−1 1
3 − tanh−1 1
2 . Is this answer equivalent to the answer from part (a)? Expla in.
B
27. The general solution of the differential equation y′′ = a2 y is y(t) = y1 (t) = c1 ea t + c2 e−a t, where a is
a positive constant, and c1 and c2 are arbitrary constants.
(a) V erify that y(t) = y2(t) = k1 cosh at + k2 sinh at is also a solution of y′′ = a2 y.
(b) Show that for any c1 and c2 , y1(t) = c1 ea t + c2 e−a t can be written as y1(t) = k1 cosh at +k2 sinh at
for some constants k1 and k2 in terms of c1 and c2 .
28. V erify that for positive constants β, p, l and c the function
η(x) = βx
p − βc sinh ( px/c)
p2 cosh ( pl /c)
is a solution of the differential equation (related to water displacement in a canal of length 2 l)
d2 η
dx2 − p2
c2 η = −βx p
c2 .
29. For any constant a and for s > |a| the Laplace transform L (s) of the function f (t) = sinh at is
L (s) =
∫ ∞
0
e−s t sinh at dt .
Show that L (s) = a
s2 −a2 .
C",ElementaryCalculus.pdf
"L (s) =
∫ ∞
0
e−s t sinh at dt .
Show that L (s) = a
s2 −a2 .
C
30. In quantum mechanics the scaling factor eπ/s0 of the three-particle Efimov trimer is the solution
s = s0 > 0 of the equation
s cosh
(πs
2
)
=
8 sinh
(πs
6
)
∝ra⌈〉⌋allow
3
,
Use a numerical method to approximate s0 , then calculate eπ/s0 .",ElementaryCalculus.pdf
"238 Chapter 7 • Analytic Geometry and Plane Curves §7.5
31. Continuing Example 7.11, the total magnetic moment M is defined as
M = 2πNµ
∫ π
0
e
µH
kT cos θ sin θ cos θ dθ .
(a) Use the value of n from Example 7.11 to show that M
nµ = L(a), where L(a) = coth a − 1
a is the
Langevin function and a = µH
kT .
(b) Show that for a = µH
kT , µmax > 0, and x = µmax H
kT ,
∫ µmax
0
L(a) dµ = 1
x ln
(sinh x
x
)
.
32. The age t0 of the universe (in years) is given by
t0 = τ0
∫ 1
0
dx√
2q0
x + 1 − 2q0
,
where τ0 = 2 × 1010 years is the Hubble time and q0 ≥ 0 is the deceleration parameter . For the
cosmological model with 0 < q0 < 1
2 , use the substitution x = 2q0
1−2q0
sinh θ to show that
t0 = τ0
(
1
1 − 2q0
− 2q0
(1 − 2q0 )3/2 cosh−1
(
1√
2q0
))
.
What fraction of τ0 is t0 (i.e. what is t0
τ0
) when q0 = 1
4 ? (Hint: Use Exercise 11 or 12.)
33. Circular rotations preserve the area of circular sectors (s ee Figure 7.5.5(a)). For any constant c > 0",ElementaryCalculus.pdf
"4 ? (Hint: Use Exercise 11 or 12.)
33. Circular rotations preserve the area of circular sectors (s ee Figure 7.5.5(a)). For any constant c > 0
the hyperbolic rotation φ : ( x, y) ∝mapsto⌋〈ar→(cx, y/c) moves points along the hyperbola x y = k (for k > 0), as
shown in Figure 7.5.5(b) for c > 1. This hyperbolic rotation preserves the area of hyperboli c sectors.
x
y
O 1−1
θθ
rotate OBQ by α
to OB′Q′
B
Q
B′
Q′
α
(a) Unit circle x2 + y2 = 1
y
x
0
(x, y)
(cx, y/c)
(b) Hyperbola x y = k
y
x
O
P ′
P = A′
(cosh a, sinh a)
A = (1, 0)
(c) Unit hyperbola x2 − y2 = 1
Figure 7.5.5 Circular and hyperbolic rotations
As an example of why this is true, let a > 0 and consider the unit hyperbola in Figure 7.5.5(c). Then:
(a) Let c > 0. When c ̸=1 the mapping φ : ( x, y) ∝mapsto⌋〈ar→(cx, y/c) does not move points along the unit
hyperbola. Find the formula for φ on the unit hyperbola as follows: use the rotation equations",ElementaryCalculus.pdf
"hyperbola. Find the formula for φ on the unit hyperbola as follows: use the rotation equations
from Section 7.4 to rotate the unit hyperbola 45 ◦ to a hyperbola of the form x y = k, apply φ to a
generic point on that hyperbola, then rotate that hyperbola by −45◦ back to the unit hyperbola.
(b) Use part (a) to find the value of c such that φ maps A = (1, 0) to P = (cosh a, sinh a).
(c) Let P ′ be the point that φ maps P to, and let A′ = P . Use part (b) to find the coordinates of P ′,
then show that the hyperbolic sectors O AP and O A′P ′ have the same area a/2.",ElementaryCalculus.pdf
"Parametric Equations • Section 7.6 239
7.6 Parametric Equations
Recall that Section 6.5 presented two different ways to “ide ntify” or represent points on the
unit circle—by angle and by slope, as in Figure 7.6.1:
x
y
1
θ
1 0
(x, y) = (cos θ, sin θ)
(a) Identify points by angle θ
x
y
slope = t
1−1
θ
(x, y)
0
(b) Identify points by slope t
Figure 7.6.1 Points on the unit circle x2 + y2 = 1
When identifying by the angle θ, all points ( x, y) on the unit circle can be written as
x = cos θ and y = sin θ
for any angle θ. When identifying by the slope t of lines through the point ( −1, 0), recall from
the derivation of the half-angle substitution that sin θ = 2t
1+t2 and cos θ = 1−t2
1+t2 . So all points
(x, y) on the unit circle except ( −1, 0) can be written as
x = 1 − t2
1 + t2 and y = 2t
1 + t2
for any slope t. These two distinct “identifications” are called parametrizations of the unit
circle, with parameter θ in the first case and parameter t in the second.",ElementaryCalculus.pdf
"circle, with parameter θ in the first case and parameter t in the second.
In general, one way to describe a plane curve C (i.e. a curve in the x y-plane) is to write its
x and y-coordinates as functions of a variable t:
x = x(t) and y = y(t)
These are parametric equations of C, which consists of all points ( x, y) such that x = x(t) and
y = y(t) for the parameter t in some interval I. The shorthand for this is:
C : x = x(t), y = y(t), t in I
Notice the flexibility that parametric equations provide, s ince plane curves can take any shape,
not limited to the graph of a single function y = f (x). In fact, a curve y = f (x) is the special
case where the parametric equations are x = t and y = f (t). In physical settings the parameter
t often denotes time, but it can represent anything and any sym bol can be used in its place. A
curve can have many parametrizations.",ElementaryCalculus.pdf
"240 Chapter 7 • Analytic Geometry and Plane Curves §7.6
Example 7.13
Show that for any constants ω ̸=0 and r > 0, and for t measured in radians,
x = h + r cos ωt and y = k + r sin ωt for −∞ <t < ∞
is a parametrization of the circle ( x − h)2 + ( y − k)2 = r2 with center ( h, k) and radius r.
Solution: Since ωt is similar to the angle θ in Figure 7.6.1(a), it suffices to show that ( x−h)2 +( y−k)2 = r2 :
(x − h)2 + ( y − k)2 = r2 cos2 ωt + r2 sin2 ωt = r2 ✓
The constant ω determines how fast and in which direction the circle is trac ed as the parameter t varies.
For example, for ω = 2 the circle C is traced counterclockwise at twice the speed of the paramet rization
C : x = h + r cos t, y = k + r sin t. In all cases the circle is re-traced every 2 π/ω radians. For that reason
the interval for t is often restricted to the interval [0 , 2π/ω], so that the circle is traced only once.
Example 7.14
x
y
(x, y) = (a cos t, b sin t)
−a a
b
−b",ElementaryCalculus.pdf
"Example 7.14
x
y
(x, y) = (a cos t, b sin t)
−a a
b
−b
Show that for a > 0 and b > 0 the parametric equations
x = a cos t and y = b sin t
for 0 ≤ t ≤ 2π describe an ellipse.
Solution: Since
x2
a2 + y2
b2 = a2 cos2 t
a2 + b2 sin2 t
b2 = cos2 t + sin2 t = 1
for all t, then the points ( x, y) = (a cos t, b sin t) lie on the ellipse x2
a2 + y2
b2 = 1. It should be obvious that
the entire ellipse is traced, but it is left as an exercise to s how what the parameter t represents.
Example 7.15
y
x
O
(x, y) = (cosh t, sinh t)
1
x2 − y2 = 1
t
2 = shaded area
Show that the parametric equations
x = cosh t and y = sinh t
for −∞ <t < ∞describe one branch of a hyperbola.
Solution: Since
x2 − y2 = cosh2 t − sinh2 t = 1
for all t, then the points ( x, y) = (cosh t, sinh t) lie on the unit
hyperbola x2 − y2 = 1. This was in fact shown in Section 7.5, where t is half the area of the shaded
region in the above figure for t > 0. For t < 0 the shaded region is reflected below the x-axis. Since",ElementaryCalculus.pdf
"region in the above figure for t > 0. For t < 0 the shaded region is reflected below the x-axis. Since
cosh t ≥ 1 and sinh t can take any value, then the entire right branch of the hyperb ola is traced as t
varies. Likewise the left branch has parametric equations x = −cosh t and y = sinh t. The hyperbola
is thus parametrized by area (or negative area for t < 0). In general, for a > 0 and b > 0 the hyperbola
x2
a2 − y2
b2 = 1 has parametric equations x = ±a cosh t and y = b sinh t for all t.",ElementaryCalculus.pdf
"Parametric Equations • Section 7.6 241
Example 7.16
Bézier curves 13 are used in Computer Aided Design (CAD) to join the ends of an o pen polygonal path
of noncollinear control points with a smooth curve that models the “shape” of the path. The cu rve is
created via repeated linear interpolation , illustrated in Figure 7.6.2 and described below for n = 3 points:
B0
B1
B2
A0
A1P
40%
40%
40%
(a) t = 0.4 = 40%
B0
B1
B2
B
(b) Many values of t in [0 , 1]
Figure 7.6.2 Bézier curve B with 3 control points B0, B1, B2
For three control points B0, B1, B2, pick 0 ≤ t ≤ 1. Say t = 0.4, which you can think of as a percentage:
0.4 = 40%. Let A0 and A1 be the points 40% of the way from B0 to B1 and from B1 to B2, respectively ,
as in Figure 7.6.2(a). Then the point P that is 40% of the way from A0 to A1 is on the Bézier curve B
joining B0 and B2. Do this for every t in [0 , 1] to fill out the curve B, as in Figure 7.6.2(b).",ElementaryCalculus.pdf
"joining B0 and B2. Do this for every t in [0 , 1] to fill out the curve B, as in Figure 7.6.2(b).
It can be shown via de Casteljau’s algorithm that the Bézier curve B for any three control points
B0 = (x0 , y0 ), B1 = (x1, y1 ) and B2 = (x2, y2 ) in the x y-plane has parametric equations
x = (1 − t)2 x0 + 2t (1 − t) x1 + t2 x2 and y = (1 − t)2 y0 + 2t (1 − t) y1 + t2 y2 (7.13)
for 0 ≤ t ≤ 1. Write out and simplify the explicit parametric equations for the Bézier curve B with control
points B0 = (1, 2), B1 = (2, 4) and B2 = (4, 1).
Solution: The parametric equations for B0 = (x0, y0 ) = (1, 2), B1 = (x1 , y1 ) = (2, 4), B2 = (x2, y2 ) = (4, 1) are:
x = (1 − t)2 (1) + 2t (1 − t) (2) + t2 (4) = 1 − 2t + t2 + 4t − 4t2 + 4t2 = t2 + 2t + 1
y = (1 − t)2 (2) + 2t (1 − t) (4) + t2 (1) = 2 − 4t + 2t2 + 8t − 8t2 + t2 = −5t2 + 4t + 2
y
x
B0
B1
B2
1 2 3 4
1
2
3
0
B
The Bézier curve B : x = t2 + 2t + 1, y = −5t2 + 4t + 2, 0 ≤ t ≤ 1 for",ElementaryCalculus.pdf
"y
x
B0
B1
B2
1 2 3 4
1
2
3
0
B
The Bézier curve B : x = t2 + 2t + 1, y = −5t2 + 4t + 2, 0 ≤ t ≤ 1 for
B0, B1, B2 is shown in the figure on the right. It is left as an exercise
to show that this curve is part of a parabola. In general Bézie r curves
can be created for n ≥ 3 control points in the plane, with the parametric
equations being polynomials of degree n − 1 in the parameter t. In the
exercises you will be guided in how to derive the parametric e quations
in the cases n = 3 and n = 4. Bézier curves can also be constructed
for control points in three-dimensional space. A similar co nstruct—a
Bézier surface —is used in three dimensions to model the boundary of
a polyhedron (i.e. a solid whose faces are polygons).
13Developed and popularized in the 1960s by two engineers, Pie rre Bézier and Paul de Casteljau, for vehicle body
modeling at the French automotive manufacturers Renault an d Citroën, respectively .",ElementaryCalculus.pdf
"242 Chapter 7 • Analytic Geometry and Plane Curves §7.6
A curve with parametric equations x = x(t) and y = y(t) might not be the graph of a single
function y = f (x), but the derivative dy
dx can still be found by using the differentials of x and y
as functions of t: dy = y′(t) dt and dx = x′(t) dt, so that
dy
dx = y′(t) dt
x′(t) dt = y′(t)
x′(t) =
dy
dt
dx
dt
(7.14)
when dx
dt ̸=0. The second derivative d2 y
dx2 can then be found via the Chain Rule:
d
dt
(dy
dx
)
=
( d
dx
(dy
dx
))
· dx
dt = d2 y
dx2 · dx
dt ⇒ d2 y
dx2 =
d
dt
(dy
dx
)
dx
dt
(7.15)
For t in [ a, b] with x1 = x(a) and x2 = x(b), the integral
∫ x2
x1 y dx is given by:
∫ x2
x1
y dx =
∫ b
a
y(t) x′(t) dt (7.16)
Example 7.17
A cycloid is the path of a point P on a circle rolling along a straight line. Figure 7.6.3 shows the cycloid
C traced by a circle of radius a rolling along the x-axis so that P touches the origin during the roll:
Py
a
y
x0 x a θ 2πa
t
θ
C
Figure 7.6.3 Cycloid C for a circle of radius a",ElementaryCalculus.pdf
"Py
a
y
x0 x a θ 2πa
t
θ
C
Figure 7.6.3 Cycloid C for a circle of radius a
Find parametric equations for C and find dy
dx .
Solution: For the angles t and θ—measured in radians—shown in Figure 7.6.3, t + θ + π/2 = 2π, so
t = 3π/2 − θ. The point P touches the origin as the circle rolls, so the horizontal dis tance from the
circle’s center to the y-axis is the length of the circular arc with central angle θ, namely aθ. So by the
parametrization of the circle as in Example 7.13, but with ce nter ( h, k) = (aθ, a), radius r = a, and ω = 1,
x = a θ + a cos t = a
(
θ + cos
(3π
2 − θ
))
= a (θ − sin θ)
y = a + a sin t = a
(
1 + sin
(3π
2 − θ
))
= a (1 − cos θ)
Thus, C : x = a (θ − sin θ), y = a (1 − cos θ), −∞ <θ < ∞is a parametrization of the cycloid C.",ElementaryCalculus.pdf
"Parametric Equations • Section 7.6 243
As in formula (7.14) the derivative dy
dx is given by:
dy
dx = y′(θ)
x′(θ) = a sin θ
a (1 − cos θ) = sin θ
1 − cos θ = cot 1
2 θ
Thus, dy
dx is undefined when cos θ = 1, namely , when θ = 2πk for all integers k, i.e. when x = a (θ − sin θ) =
a (2πk − sin 2 πk) = 2πka. Notice from Figure 7.6.3 that the cycloid has cusps at those values of x.
B
AA cycloid appears in the solution of the famous brachistochrone problem :14
find the plane curve joining two points A and B—where B is at a lower height
than A but not directly under it—along which an object slides frict ionless
under the force of gravity alone from A to B in the shortest time. It turns out
that the optimal path is not a straight line, but part of an inv erted (upside-
down) cycloid with a cusp at A, as in the figure on the right. 15
Exercises
A
1. For a > b > 0, in the ellipse x2
a2 + y2
b2 = 1 inscribe a circle of radius b centered at the origin, called the",ElementaryCalculus.pdf
"Exercises
A
1. For a > b > 0, in the ellipse x2
a2 + y2
b2 = 1 inscribe a circle of radius b centered at the origin, called the
minor auxiliary circle . This circle is parametrized by x = b cos t and y = b sin t for 0 ≤ t ≤ 2π, with the
angle t shown in Figure 7.6.4. From each point on that circle draw a ho rizontal line segment to the
point P on the ellipse in the same quadrant, as shown. Show that P = (a cos t, b sin t).
Note: The angle t is called the eccentric angle of the ellipse, and it is the parameter for the parametriza-
tion of the ellipse in Example 7.14.
x
y
P = (x, y)
−a a
b
−b
t
(b cos t, b sin t)
Figure 7.6.4 Exercise 1
x
y
P = (x, y)
−a a
b
−b
t
(a cos t, a sin t)
Figure 7.6.5 Exercise 2
2. Similar to Exercise 1, circumscribe a circle of radius a (called the major auxiliary circle ) around the
ellipse x2
a2 + y2
b2 = 1. This circle is parametrized by x = a cos t and y = a sin t for 0 ≤ t ≤ 2π, with the",ElementaryCalculus.pdf
"ellipse x2
a2 + y2
b2 = 1. This circle is parametrized by x = a cos t and y = a sin t for 0 ≤ t ≤ 2π, with the
eccentric angle t shown in Figure 7.6.5. From each point on that circle draw a ve rtical line segment
to the point P on the ellipse in the same quadrant, as shown. Show again that P = (a cos t, b sin t).
3. Show that the cycloid in Example 7.17 has global maxima at x = (2k + 1)πa for all integers k, and
that the cycloid is always concave down.
4. Show that the area under the cycloid in Example 7.17 over the i nterval [0 , 2πa] is 3 πa2 .
14First solved in 1696 by the Swiss physicist and mathematicia n Johann Bernoulli (1667-1748).
15See pp.60-62 in C L E G G, J.C., Calculus of V ariations , Edinburgh: Oliver & Boyd, Ltd., 1968. For Bernoulli’s proo f
see pp.644-655 in S MI T H , D.E., A Source Book in Mathematics , New Y ork: Dover Publications, Inc., 1959.",ElementaryCalculus.pdf
"244 Chapter 7 • Analytic Geometry and Plane Curves §7.6
5. The parametrization C : x = 1−t2
1+t2 , y = 2t
1+t2 , −∞ <t < ∞of the unit circle C shown earlier makes the
unit circle a rational curve , since x and y are rational functions of the parameter t. Is the ellipse
x2
a2 + y2
b2 = 1 a rational curve? Justify your answer .
6. Let P0 = (x0, y0 ) and P1 = (x1 , y1 ) be distinct points in the x y-plane. For t in [0 , 1] define
x(t) = (1 − t) x0 + t x 1 and y(t) = (1 − t) y0 + t y 1
and let P (t) = (x(t), y(t)).
(a) Show that C : x = x(t), y = y(t), 0 ≤ t ≤ 1 is a parametrization of the line segment from P0 to P1.
(b) Show that the parameter t is the proportion of the length P0 P (t) to the length P0 P1 : P0 P(t)
P0 P1
= t.
7. In an ellipsograph a rod of length a has a peg at one end and another peg a distance b from the other
end, so that the rod can slide along two thin perpendicular ra ils, with the pegs in the rails as in",ElementaryCalculus.pdf
"end, so that the rod can slide along two thin perpendicular ra ils, with the pegs in the rails as in
Figure 7.6.6. Show that a marker at the endpoint P traces an ellipse as the rod moves. (Hint: Treat
the rails as the x and y-axes. Find parametric equations, with a different angle th an in Exercise 1.)
a−b
b
P
Figure 7.6.6 Exercise 7
a
x
y
O A
B
P = (x, y)θ
C
Figure 7.6.7 Exercise 8
B
8. The end P of a thread is held taut as it is unwound from a circle of radius a starting from a point A,
as in Figure 7.6.7. Show that the path C that the end traces—called the involute of the circle—has
parametric equations x = a(cos θ + θsin θ), y = a(sin θ − θcos θ).
9. Recall that a parabola of the form y = ax2 +bx +c has a constant second derivative d2 y
dx2 = 2a. Consider
the Bézier curve B in Example 7.16.
(a) Find d2 y
dx2 for B. Is it a constant?
(b) Show that B is part of a parabola. Does this contradict part (a)? Explain . (Hint: Show the curve",ElementaryCalculus.pdf
"(a) Find d2 y
dx2 for B. Is it a constant?
(b) Show that B is part of a parabola. Does this contradict part (a)? Explain . (Hint: Show the curve
has a second-degree equation of the form (7.10).)
(c) Find the point where the curve B has a global maximum.
10. Use Exercise 6 to derive the formulas (7.13) for the parametr ic equations of the general Bézier
curve for n = 3 control points.
11. To form the Bézier curve for n = 4 control points B0 = (x0 , y0 ), B1 = (x1, y1 ), B2 = (x2, y2 ), B3 = (x3 , y3 ),
for 0 ≤ t ≤ 1 use the three points that are 100 t% of the way along the line segments B0 B1, B1 B2, B2 B3
as the control points in the n = 3 case. Show that the resulting parametrization is:
x = (1 − t)3 x0 +3t(1 − t)2 x1 +3t2 (1 − t)x2 + t3 x3 and y = (1 − t)3 y0 +3t(1 − t)2 y1 +3t2 (1 − t) y2 + t3 y3
12. Each point on a plane curve lies on some line through the origi n. Use that fact to show that",ElementaryCalculus.pdf
"12. Each point on a plane curve lies on some line through the origi n. Use that fact to show that
the equation y2 = x2 + x3 defines a rational curve (see Exercise 5). (Hint: For all real t, find the
intersections of the lines y = tx with the curve. Consider also the special case of the line x = 0.)
13. Sketch the graph of the curve C : x = 2t − 4t3 , y = t3 − 3t4 , −∞ <t < ∞.",ElementaryCalculus.pdf
"Polar Coordinates • Section 7.7 245
7.7 Polar Coordinates
Suppose that you wanted to write the equation of a spiral, lik e the one in Figure 7.7.1. The
curve is clearly not the graph of a function y = f (x) in Cartesian coordinates, as it violates the
vertical line test. However , this spiral is simple to expres s using polar coordinates .
16 Recall
that any point P distinct from the origin (denoted by O) in the x y-plane is a distance r > 0
from the origin, and the ray − − →
OP makes an angle θ with the positive x-axis, as in Figure 7.7.2.
Imagine − − →
OP swings around the “pole” at the origin.
x
y
0 1 2 3
Figure 7.7.1 Spiral
r
x
y
O
θ
P = (r, θ)
Figure 7.7.2 Polar coordinates ( r, θ)
−r
x
y
O
θ
P = (−r, θ)
Figure 7.7.3 Negative r: ( −r, θ)
The pair ( r, θ) contains the polar coordinates of P , and the positive x-axis is called the polar
axis of this coordinate system. For the angle θ measured in radians, ( r, θ) = (r, θ + 2πk) for",ElementaryCalculus.pdf
"axis of this coordinate system. For the angle θ measured in radians, ( r, θ) = (r, θ + 2πk) for
all integers k. Thus, the polar coordinates of a point are not unique. By con vention r can be
negative, by defining ( −r, θ) = (r, θ + π) for any angle θ: the ray − − →
OP is drawn in the opposite
direction from the angle θ, as in Figure 7.7.3. When r = 0, the point ( r, θ) = (0, θ) is the origin
O, regardless of the value of θ.
Example 7.18
Express the spiral in Figure 7.7.1 in polar coordinates, suc h that the distance between any two points
separated by 2 π radians is always 1.
Solution: The goal is to find some equation involving r and θ (measured in radians) that describes the
spiral. The distance between any two points separated by 2 π radians is always 1, for example:
θ = 0 ⇒ r = 1
θ = 2π ⇒ r = 2
θ = 4π ⇒ r = 3
. . .
θ = 2πk ⇒ r = 1 + k
for all integers k ≥ 0. In fact r = 1 + k when θ = 2πk for all real k ≥ 0, by the assumption about the",ElementaryCalculus.pdf
"θ = 4π ⇒ r = 3
. . .
θ = 2πk ⇒ r = 1 + k
for all integers k ≥ 0. In fact r = 1 + k when θ = 2πk for all real k ≥ 0, by the assumption about the
distance. So solving for k in terms of θ yields the polar equation r = 1 + θ
2π for all θ ≥ 0.
16Created by the Flemish mathematician Grégoire de Saint-Vin cent (1584-1667) and Italian mathematician
Bonaventura Cavalieri (1598-1647) in the 17 th century , later used by Newton in his Method of Fluxions (1671).",ElementaryCalculus.pdf
"246 Chapter 7 • Analytic Geometry and Plane Curves §7.7
Y ou might be familiar with graphing paper , for plotting poin ts or functions given in Cartesian
coordinates. Such paper consists of a rectangular grid, whe re the horizontal and vertical lines
represent where x and y, respectively , are constants, at regular intervals. Simil ar graphing
paper exists for polar coordinates, as in Figure 7.7.4.
0◦
15◦
30◦
45◦
60◦
75◦90◦
105◦
120◦
135◦
150◦
165◦
180◦
195◦
210◦
225◦
240◦
255◦
270◦ 285◦
300◦
315◦
330◦
345◦
O
Figure 7.7.4 Polar coordinate graphing paper
This polar grid is radial, not rectangular . The concentric c ircles around the origin O are
where r is constant (e.g. r = 1, r = 2), while the lines through the origin are where θ is constant,
at regular intervals for each. The angle θ shown here is in degrees, though radians are often
preferred for their “unitless” nature.",ElementaryCalculus.pdf
"Polar Coordinates • Section 7.7 247
In general, polar coordinates are useful in describing plan e curves that exhibit symmetry
about the origin (though there are other situations), which arise in many physical applications.
r
x
y
O
θ
y
x
(r, θ)(x, y)
Figure 7.7.5
Figure 7.7.5 shows how to convert between polar coordinates and Carte-
sian coordinates. Namely , for a point with polar coordinate s ( r, θ) and
Cartesian coordinates ( x, y):
Polar to Cartesian:
x = r cos θ y = r sin θ (7.17)
Cartesian to Polar:
r = ±
√
x2 + y2 tan θ = y
x if x ̸=0 (7.18)
In formula (7.18), if x = 0 then θ = π/2 or θ = 3π/2. If x ̸=0 and y ̸=0 then the two possible
solutions for θ in the equation tan θ = y
x are in opposite quadrants (for 0 ≤ θ < 2π). If the angle
θ is in the same quadrant as the point ( x, y), then r =
√
x2 + y2 (i.e. r is positive); otherwise
r = −
√
x2 + y2 (i.e. r is negative).
Example 7.19
Write the equation of the unit circle x2 + y2 = 1 in polar coordinates.",ElementaryCalculus.pdf
"r = −
√
x2 + y2 (i.e. r is negative).
Example 7.19
Write the equation of the unit circle x2 + y2 = 1 in polar coordinates.
Solution: By formula (7.18), r2 = x2 + y2 = 1, so in polar coordinates the equation is simply r = 1. In
general the circle x2 + y2 = a2 of radius a > 0 has the simpler expression r = a in polar coordinates.
Example 7.20
Write the equation x2 + ( y − 4)2 = 16 in polar coordinates.
Solution: This is the equation of a circle of radius 4 centered at the poi nt (0 , 4). Expand the equation:
x2 + ( y − 4)2 = 16
x2 + y2 − 8 y + 16 = 16
x2 + y2 = 8 y
r2 = 8 r sin θ (by formulas (7.17) and (7.18))
r = 8 sin θ
Is it valid to cancel r from both sides in the last step? Y es. The point (0 , 0) is on the circle, so canceling
r does not eliminate r = 0 as a potential solution of the equation (e.g. θ = 0 would make r = 8 sin θ = 0).
Thus, the polar equation is r = 8 sin θ.
Notice that this polar equation is actually less intuitive t han its Cartesian equivalent. From the equa-",ElementaryCalculus.pdf
"Thus, the polar equation is r = 8 sin θ.
Notice that this polar equation is actually less intuitive t han its Cartesian equivalent. From the equa-
tion x2 + ( y − 4)2 = 16 it is easy to identify the curve as a circle and read off its r adius and center; these
properties are not so obvious from the polar equation. Since the circle’s center is not the origin, there is
no symmetry about the origin, which is when polar coordinate s are often better suited.",ElementaryCalculus.pdf
"248 Chapter 7 • Analytic Geometry and Plane Curves §7.7
Derivatives in Polar Coordinates
Suppose that the polar coordinates ( r, θ) for a plane curve are related by a function: r = r(θ).
Then by formula (7.17), x = r(θ) cos θ and y = r(θ) sin θ are now parametric equations for the
curve in the parameter θ. Thus, by the Product Rule and formulas (7.14) and (7.15) fro m
Section 7.6, with dx = x′(θ) dθ and dy = y′(θ) dθ:
For a plane curve with polar equation r = r(θ),
dy
dx = r′(θ) sin θ + r(θ) cos θ
r′(θ) cos θ − r(θ) sin θ and d2 y
dx2 =
d
dθ
(dy
dx
)
r′(θ) cos θ − r(θ) sin θ . (7.19)
Example 7.21
Sketch the graph of r = 1 + cos θ.
Solution: First sketch the graph treating ( r, θ) as Cartesian coordinates, for 0 ≤ θ ≤ 2π as in Figure
7.7.6(a). Then use that graph to trace out a rough graph in pol ar coordinates, as in Figure 7.7.6(b). 17
r
θ0 π 2π
2
(a) θr-plane
y
x
2 0
1
−1
(3
2 , π
3
)
(3
2 , 5π
3
)
(1
2 , 2π
3
)
(1
2 , 4π
3
)
(b) Polar coordinates ( r, θ)",ElementaryCalculus.pdf
"r
θ0 π 2π
2
(a) θr-plane
y
x
2 0
1
−1
(3
2 , π
3
)
(3
2 , 5π
3
)
(1
2 , 2π
3
)
(1
2 , 4π
3
)
(b) Polar coordinates ( r, θ)
Figure 7.7.6 Graph of r = 1 + cos θ
To find the maxima, minima, and inflection points it is still ne cessary to find dy
dx and d2 y
dx2 . It is left as an
exercise to use formula (7.19) and double-angle identities to show that
dy
dx = −cos θ + cos 2 θ
sin θ + sin 2 θ and d2 y
dx2 = − 3 (1 + cos θ)
(sin θ + sin 2 θ)3
and that for θ in [0 , 2π], dy
dx = 0 only when θ = π
3 and 5π
3 , while dy
dx is undefined for θ = 0, 2π
3 , π, 4π
3 , and
2π. Since d2 y
dx2
⏐
⏐
⏐
θ= π
3
< 0 and d2 y
dx2
⏐
⏐
⏐
θ= 5π
3
> 0 then the curve has a local maximum when θ = π
3 and a local
minimum when θ = 5π
3 . It can also be shown that d2 y
dx2 changes sign around θ = 0, 2π
3 , π, 4π
3 , so that the
inflection points occur at those values of θ, as shown (with the correct concavity) in Figure 7.7.6(b).",ElementaryCalculus.pdf
"3 , π, 4π
3 , so that the
inflection points occur at those values of θ, as shown (with the correct concavity) in Figure 7.7.6(b).
17There are far too many interesting plane curves to cover here . For an extensive collection, see L AW R E N C E , J.D.,
A Catalog of Special Plane Curves , New Y ork: Dover Publications, Inc., 1972. See also S E G G E R N, D.H. VO N , CRC
Handbook of Mathematical Curves and Surfaces , Boca Raton, FL: CRC Press, Inc., 1990.",ElementaryCalculus.pdf
"Polar Coordinates • Section 7.7 249
Integration in Polar Coordinates
In some cases polar coordinates can simplify evaluation of a definite integral or finding an
area. To determine the polar form of a definite integral, supp ose that r is a function of θ:
r = f (θ). A polar region swept out by r = f (θ) between θ = α and θ = β would look like the
shaded region in Figure 7.7.7 with area A:
θ = α
θ = β
y
x
r = f (θ)
O
Figure 7.7.7 Area A
r
r+dr ds
y
xdθ
(r, θ)
(r + dr, θ + dθ)
Figure 7.7.8 Area dA
c a
b A C
B
Figure 7.7.9 Area = 1
2 bc sin A
A typical infinitesimal wedge of that region—shown in Figure 7.7.8—is produced by an in-
finitesimal change dθ in the angle θ, which results in an infinitesimal change dr in r. By the
Microstraightness Property , the curve r = f (θ) is a straight line with infinitesimal length ds
over that infinitesimal angle dθ. The area dA of the wedge thus equals the area of the trian-",ElementaryCalculus.pdf
"over that infinitesimal angle dθ. The area dA of the wedge thus equals the area of the trian-
gle with sides of lengths r and r + dr with included angle dθ and a side of length ds. Since
sin dθ = dθ and dr = f ′(θ) dθ, then by the formula for the area of a triangle (as shown in Fig ure
7.7.9),18
dA = 1
2 (r) (r + dr) sin dθ
= 1
2 r2 dθ + 1
2 r (dr) dθ
= 1
2 r2 dθ + 1
2 r ( f ′(θ) dθ) dθ
= 1
2 r2 dθ + 0 = 1
2 r2 dθ
since ( dθ)2 = 0. The area A of the region is the sum of these infinitesimal areas dA:
For a plane curve with polar equation r = f (θ), the area A of the region swept out between
θ = α and θ = β is:
A =
∫ θ=β
θ=α
dA =
∫ β
α
1
2 r2 dθ =
∫ β
α
1
2 ( f (θ))2 dθ (7.20)
If f is periodic then choose the angle interval [ α, β] so that the area is swept out only once.
18The formula 1
2 b c sin A for the area of a triangle △ABC is derived in most trigonometry texts. For example, see
p.54 in C O R R A L, M., Trigonometry, http://mecmath.net/trig/, 2009.",ElementaryCalculus.pdf
"250 Chapter 7 • Analytic Geometry and Plane Curves §7.7
Example 7.22
Use polar coordinates to show that the area of a circle of radi us R is πR2 .
Solution: Let the origin be the center of the circle. Then r = R is the polar equation of the circle, with
0 ≤ θ ≤ 2π sweeping out exactly one full circle. The area A inside the circle is then
A =
∫ 2π
0
1
2 r2 dθ =
∫ 2π
0
1
2 R2 dθ = 1
2 R2 θ
⏐
⏐
⏐
⏐
2π
0
= 1
2 R2 (2π− 0) = πR2 . ✓
Note the simplicity of this integral compared to the trigono metric substitution required when using
Cartesian coordinates, as in Section 6.3. Notice also that u sing a larger interval—say [0 , 4π]—for θ
would result in an incorrect area (4 πR2) even though the region inside the curve is the same.
Example 7.23
Find the area inside the curve r = 1 + cos θ.
Solution: Choose 0 ≤ θ ≤ 2π as in Example 7.21, so that the area A is:
A =
∫ 2π
0
1
2 r2 dθ =
∫ 2π
0
1
2 (1 + cos θ)2 dθ =
∫ 2π
0
(1
2 + cos θ + cos2 θ
2
)
dθ
=
∫ 2π
0
(1
2 + cos θ + 1 + cos 2 θ
4
)
dθ =
∫ 2π
0",ElementaryCalculus.pdf
"A =
∫ 2π
0
1
2 r2 dθ =
∫ 2π
0
1
2 (1 + cos θ)2 dθ =
∫ 2π
0
(1
2 + cos θ + cos2 θ
2
)
dθ
=
∫ 2π
0
(1
2 + cos θ + 1 + cos 2 θ
4
)
dθ =
∫ 2π
0
(3
4 + cos θ + cos 2 θ
4
)
dθ
= 3
4 θ + sin θ + 1
8 sin 2 θ
⏐
⏐
⏐
⏐
2π
0
= 3π
2
Exercises
A
For Exercises 1-8 write the given equation in polar coordina tes.
1. (x − 3)2 + y2 = 9 2. y = x 3. x2 − y2 = 1 4. 3x2 + 4 y2 − 6x = 9
5. y = −x 6. y = x + 1 7. y = x2 8. y = x3
9. Write the polar equation r2 = 4 cos 2 θ in Cartesian coordinates.
10. Find the tangent line to r = cos 2 θ at
(1
2 , π
6
)
. 11. Find the tangent line to r = 8 sin 2 θ at
(
2, 5π
6
)
.
12. Recall the curve r = 1 + cos θ from Example 7.21.
(a) V erify that for this curve,
dy
dx = −cos θ + cos 2 θ
sin θ + sin 2 θ and d2 y
dx2 = − 3 (1 + cos θ)
(sin θ + sin 2 θ)3 .
(b) V erify that for θ in [0 , 2π], dy
dx = 0 only when θ = π
3 and 5π
3 , and is undefined for θ = 0, 2π
3 , π, 4π
3 ,
and 2 π.
(c) V erify that d2 y
dx2
⏐
⏐
⏐
⏐
θ= π
3
< 0 and d2 y
dx2
⏐
⏐
⏐
⏐
θ= 5π
3
> 0.",ElementaryCalculus.pdf
"3 and 5π
3 , and is undefined for θ = 0, 2π
3 , π, 4π
3 ,
and 2 π.
(c) V erify that d2 y
dx2
⏐
⏐
⏐
⏐
θ= π
3
< 0 and d2 y
dx2
⏐
⏐
⏐
⏐
θ= 5π
3
> 0.
(d) V erify that d2 y
dx2 changes sign around θ = 0, 2π
3 , π, and 4π
3 .",ElementaryCalculus.pdf
"Polar Coordinates • Section 7.7 251
For Exercises 13-15, sketch the graph of the given curve and i ndicate all local maxima and minima.
13. r = 1 + sin θ 14. r = 1 − cos θ 15. r = sin 2 θ
16. Find the area inside r = 1 + sin θ. 17. Find the area inside r = sin 2 θ.
B
18. Sketch a rough graph of the meridian voltage component Eθ for a half-wavelength linear antenna:
Eθ = r(θ) =
cos ( π
2 cos θ)
sin θ
19. Show that the distance d between two points ( r1 , θ1 ) and ( r2 , θ2 ) in polar coordinates is
d =
√
r2
1 + r2
2 − 2r1 r2 cos (θ1 − θ2 ) .
20. For a point P = (r, θ) on the curve r = r(θ), let α be the angle that the tangent line through P makes
with the positive x-axis. Let ψ = α− θ be the angle between the tangent line and the line through P
and the origin. Show that tan ψ = r(θ)
r′(θ) . (Hint: Use formulas (7.19) and (3.2).)
21. For the parabola with focus at (0 , 0), vertex at (0 , −p), and directrix y = −2 p (with p > 0), show that
the polar equation is
r = 2 p
1 − sin θ .",ElementaryCalculus.pdf
"the polar equation is
r = 2 p
1 − sin θ .
22. For the ellipse with foci (0 , 0) and (2 c, 0), vertexes ( c ± a, 0), and eccentricity e = c
a (with 0 < c < a),
show that the polar equation is
r = a (1 − e2)
1 − e cos θ .
23. For the hyperbola with foci (0 , 0) and (2 c, 0), vertexes ( c±a, 0), and eccentricity e = c
a (with 0 < a < c),
show that the polar equation is
r = a (e2 − 1)
1 + e cos θ .
F1 F2
y
x−a 0 a
d1 d2
P = (x, y)
η
24. The bipolar coordinates 19 (ξ, η) of a point P = (x, y) relative to
two poles F1 = (−a, 0) and F2 = (a, 0) are given by ξ = ln d1
d2
and
η = ∠F1 P F2 , where d1 = F1 P and d2 = F2 P , as in the figure on
the right . Then −∞ <ξ < ∞except at F1 (which corresponds
to ξ = −∞) and F2 (ξ = ∞), while 0 ≤ η ≤ π for points on or
above the x-axis, and π < η < 2π below the x-axis.
(a) Show that
x = a sinh ξ
cosh ξ − cos η and y = a sin η
cosh ξ − cos η .
(Hint: Use the distance formula, the Law of Cosines, and the L aw of Sines.)",ElementaryCalculus.pdf
"x = a sinh ξ
cosh ξ − cos η and y = a sin η
cosh ξ − cos η .
(Hint: Use the distance formula, the Law of Cosines, and the L aw of Sines.)
(b) Show that for constants τ ̸=0 and σ ̸=0 or π, the curves ξ = τ and η = σ are the respective circles
(x − a coth τ)2 + y2 = a2
sinh2 τ
and x2 + ( y − a cot σ)2 = a2
sin2 σ
.
19Inspired by the lines of force and equipotential lines for an electric dipole. See pp.55-56 in S T R AT T O N , J.A.,
Electromagnetic Theory , New Y ork: McGraw-Hill Book Company , Inc., 1941.",ElementaryCalculus.pdf
"CHAPTER 8
Applications of Integrals
8.1 Area Between Curves
The “area under a curve” was defined in Chapter 5 as the area bel ow some curve y = f (x) and
above the x-axis over some interval. That was a special case of the area between curves ,
where in general one curve y = f1(x) is not necessarily always above another curve y = f2(x)
over the entire interval, as in Figure 8.1.1 for an interval [ a, b].
y
x
a b x x + dx0
y = f2 (x)
y = f1 (x)
dA
h(x) = |f1 (x) − f2(x)|
dx
Figure 8.1.1 The area A between two curves y = f1 (x) and y = f2 (x) over [ a, b]
The area A of the region between the curves in Figure 8.1.1 cannot be neg ative. Thus, a
typical infinitesimal area element dA of the region is of the form h(x) dx, where the height
function h(x) is the nonnegative difference in the y-coordinates of the curves at each x in
[a, b]: h(x) =
⏐
⏐ f1(x) − f2(x)
⏐
⏐. Hence:
The area A between two curves y = f1(x) and y = f2(x) over an interval [ a, b] is:
A =
∫ b
a
dA =
∫ b
a
⏐
⏐ f1(x) − f2 (x)
⏐",ElementaryCalculus.pdf
"⏐
⏐. Hence:
The area A between two curves y = f1(x) and y = f2(x) over an interval [ a, b] is:
A =
∫ b
a
dA =
∫ b
a
⏐
⏐ f1(x) − f2 (x)
⏐
⏐ dx (8.1)
The interval [ a, b] can be replaced by any interval—finite or infinite—over whic h the inte-
gral is defined. Neither curve is required to be above the x-axis.
252",ElementaryCalculus.pdf
"Area Between Curves • Section 8.1 253
Example 8.1
h(x)
y
x0 2
y = ex
y = e−x
Find the area between y = ex and y = e−x over [0 , 2].
Solution: Since ex ≥ e−x for x in [0 , 2], the height function h(x) for the region
between the curves over [0 , 2] is h(x) =
⏐
⏐ ex − e−x ⏐
⏐ = ex − e−x . The area A of
the region is thus
A =
∫ 2
0
(ex − e−x) dx
= ex + e−x
⏐
⏐
⏐
⏐
2
0
= e2 + e−2 − (1 + 1)
= 2 (cosh 2 − 1) .
Example 8.2
y
x0 1
y=x
y = x2
Find the area of the region bounded by y = x2 and y = x.
Solution: A “bounded” region will always mean a region of finite area, as op-
posed to unbounded regions. The curves y = x2 and y = x intersect at x = 0 and
x = 1, so the region the curves bound is the shaded region shown in the figure
on the right. Since x ≥ x2 for 0 ≤ x ≤ 1, then the height function for the region
is h(x) = |x2 − x | =x − x2 . The region’s area A is then
A =
∫ 1
0
(x − x2 ) dx
= 1
2 x − 1
3 x2
⏐
⏐
⏐
⏐
1
0
= 1
2 − 1
3 = 1
6 .
Example 8.3
y
x
0 π
4
π
3
y = sin x
y = cos x
1",ElementaryCalculus.pdf
"A =
∫ 1
0
(x − x2 ) dx
= 1
2 x − 1
3 x2
⏐
⏐
⏐
⏐
1
0
= 1
2 − 1
3 = 1
6 .
Example 8.3
y
x
0 π
4
π
3
y = sin x
y = cos x
1
Find the area of the region bounded by y = sin x and y = cos x over [0 , π/3].
Solution: As shown in the figure on the right, the curves intersect at x = π
4 ,
and cos x ≥ sin x for 0 ≤ x ≤ π
4 , while sin x ≥ cos x for π
4 ≤ x ≤ π
3 . The area A
of the region thus needs to be split into two integrals:
A =
∫ π/3
0
|sin x − cos x | dx
=
∫ π/4
0
(cos x − sin x) dx +
∫ π/3
π/4
(sin x − cos x) dx
=
(
sin x + cos x
⏐
⏐
⏐
⏐
π/4
0
)
+
(
−cos x − sin x
⏐
⏐
⏐
⏐
π/3
π/4
)
=
( 1
∝ra⌈〉⌋allow
2
+ 1∝ra⌈〉⌋allow
2
)
− (0 − 1) +
(
− 1
2 −
∝ra⌈〉⌋allow
3
2
)
−
(
− 1∝ra⌈〉⌋allow
2
− 1∝ra⌈〉⌋allow
2
)
= 4
∝ra⌈〉⌋allow
2 − 3 −
∝ra⌈〉⌋allow
3
2",ElementaryCalculus.pdf
"254 Chapter 8 • Applications of Integrals §8.1
Formula (8.1) can be extended to find the area between any numb er of curves, by splitting the
integral over subintervals with different height function s.
Example 8.4
Find the area of the region bounded by y = 6 − x2 , y = x and y = −5x above the x-axis.
y = −5x
y = x
y
x
0−1 2
y = 6 − x2
Solution: As shown in the figure on the right, above the x-axis the curve
y = 6 − x2 intersects the line y = x at x = 2 and intersects the line y = −5x
at x = −1. Since 6 −x2 ≥ −5x over [ −1, 0] and 6 −x2 ≥ x over [0 , 2], the area
A of the shaded region needs to be split into two integrals:
A =
∫ 0
−1
⏐
⏐6 − x2 − (−5x)
⏐
⏐ dx +
∫ 2
0
⏐
⏐6 − x2 − x
⏐
⏐ dx
=
∫ 0
−1
(6 − x2 + 5x) dx +
∫ 2
0
(6 − x2 − x) dx
=
(
6x − 1
3 x3 + 5
2 x2
⏐
⏐
⏐
⏐
0
−1
)
+
(
6x − 1
3 x3 − 1
2 x2
⏐
⏐
⏐
⏐
2
0
)
= 0 −
(
−6 + 1
3 + 5
2
)
+
(
12 − 8
3 − 2
)
− 0 = 21
2
For some areas between curves it might be easier to switch the roles of x and y, so that instead",ElementaryCalculus.pdf
"(
−6 + 1
3 + 5
2
)
+
(
12 − 8
3 − 2
)
− 0 = 21
2
For some areas between curves it might be easier to switch the roles of x and y, so that instead
of a vertical height function you would use a horizontal widt h function.
Example 8.5
Find the area of the region bounded by x = y2 − 2 and y = x.
y
x
0−2 −1 2
w( y) y = x
x = y2 − 2
Solution: As shown in the figure on the right, the parabola x = y2 − 2
intersects the line y = x at x = −1 and x = 2. The region has differ-
ent height functions h(x) for −2 ≤ x ≤ −1 and −1 ≤ x ≤ 2, so that two
integrals would be required for the area A. However , notice that the
width function w( y) has one definition over the entire region between
the curves x = y2 − 2 and x = y: w( y) =
⏐
⏐ y − ( y2 − 2)
⏐
⏐ = y − ( y2 − 2).
Thus, instead of integrating the vertical strips dA = h(x) dx along the
x-axis, integrate the horizontal strips dA = w( y) dy along the y-axis,
from y = −1 to y = 2:
A =
∫ 2
−1
w( y) dy =
∫ 2
−1
⏐
⏐ y − ( y2 − 2)
⏐
⏐ dy
=
∫ 2
−1",ElementaryCalculus.pdf
"from y = −1 to y = 2:
A =
∫ 2
−1
w( y) dy =
∫ 2
−1
⏐
⏐ y − ( y2 − 2)
⏐
⏐ dy
=
∫ 2
−1
( y − ( y2 − 2)) dy
= 1
2 y2 − 1
3 y3 + 2 y
⏐
⏐
⏐
⏐
2
−1
=
(
2 − 8
3 + 4
)
−
(1
2 + 1
3 − 2
)
= 9
2",ElementaryCalculus.pdf
"Area Between Curves • Section 8.1 255
θ = α
θ = β
y
x
r1
r2
O
Figure 8.1.2
The area between curves given by polar equations can be found
similarly . For example, consider curves r = r1(θ) and r = r2(θ) with
r1(θ) ≥ r2(θ) when α ≤ θ ≤ β as in Figure 8.1.2. The area A of the
region between the curves and those angles is simply the diff erence
between the “outer” and “inner” areas, each given by formula (7.20):
A =
∫ β
α
1
2 r2
1 dθ −
∫ β
α
1
2 r2
2 dθ =
∫ β
α
1
2 (r2
1 − r2
2) dθ
In general, to include cases where the “outer” and “inner” cu rves
switch positions, take the absolute value of the difference :
The area A between two polar curves r = r1(θ) and r = r2(θ) for α ≤ θ ≤ β is:
A =
∫ β
α
1
2
⏐
⏐ r2
1 − r2
2
⏐
⏐ dθ (8.2)
Example 8.6
θ=
π
6
θ=
π 3
y
x
r = 1 + cos θ
O
r = 1 − cos θ
Find the area between r = 1 + cos θ and r = 1 − cos θ for π
6 ≤ θ ≤ π
3 .
Solution: Let r1 (θ) = 1 + cos θ and r2 (θ) = 1 − cos θ. Since r1 (θ) > r2 (θ) for
π
6 ≤ θ ≤ π",ElementaryCalculus.pdf
"6 ≤ θ ≤ π
3 .
Solution: Let r1 (θ) = 1 + cos θ and r2 (θ) = 1 − cos θ. Since r1 (θ) > r2 (θ) for
π
6 ≤ θ ≤ π
3 , the area A of the region (shown in the figure on the right) is
A =
∫ π/3
π/6
1
2
⏐
⏐r2
1 − r2
2
⏐
⏐ dθ =
∫ π/3
π/6
1
2 ((1 + cos θ)2 − (1 − cos θ)2 ) dθ
=
∫ π/3
π/6
2 cos θ dθ = 2 sin θ
⏐
⏐
⏐
⏐
π/3
π/6
= 2
(∝ra⌈〉⌋allow
3
2 − 1
2
)
=
∝ra⌈〉⌋allow
3 − 1
y
x
Figure 8.1.3
Monte Carlo integration is a technique for approximating
the area of a region by taking a large number of random points
in a rectangle that encloses the region (see Figure 8.1.3). T he
idea is simple:
# of points in the region
# of points in the rectangle ≈ area of the region
area of the rectangle
For example, if 20% of the random points in the rectangle fall
inside the region, then—by randomness—you would expect the
area of the region to be about 20% of the area of the rectangle.
The more random points you take, the better the approximatio n. Since the area of the rectangle",ElementaryCalculus.pdf
"The more random points you take, the better the approximatio n. Since the area of the rectangle
is known, as well as the number of random points inside the reg ion and the rectangle, the area
of the region is easy to approximate.",ElementaryCalculus.pdf
"256 Chapter 8 • Applications of Integrals §8.1
Example 8.7
y = ex2y
1
2
3
x0 1 2 3
x2 + y2 = 9
y = 2 cos x2
Use Monte Carlo integration to approximate the area of the re gion
in the first quadrant above the curves y = ex2
and y = 2 cos x2 , and
inside the circle x2 + y2 = 9.
Solution: The region is the shaded area shown in the figure on the
right, enclosed in a rectangle of width 2 and height 3. The are a of the
rectangle is 6, and a point ( x, y) in the rectangle is inside the region
if only if the following conditions are met:
y > ex2
and y > 2 cos x2 and x2 + y2 < 9
Notice that these conditions are all in the form of inequalit ies. The
Monte Carlo integration is then simple to perform in Octave, using 10 million random points:
octave> N = 1e7;
octave> x = 2*rand(1,N);
octave> y = 3*rand(1,N);
octave> 6*(sum(y > exp(x.^2) & y > 2*cos(x.^2) & x.^2 + y.^2 < 9))/N
ans = 0.94612
The actual value—accurate to 5 decimal places—is 0.94606.",ElementaryCalculus.pdf
"octave> 6*(sum(y > exp(x.^2) & y > 2*cos(x.^2) & x.^2 + y.^2 < 9))/N
ans = 0.94612
The actual value—accurate to 5 decimal places—is 0.94606.
The rand(1,N) command returns an array of N random numbers between 0 and 1. So the statement
x = 2*rand(1,N) stores N random numbers between 0 and 2 in an array for the x-coordinates, and the
statement y = 3*rand(1,N) stores N random numbers between 0 and 3 in an array for the y-coordinates.
The statement y > exp(x.^2) returns a value of 1 if the condition y > ex2
is met, 0 otherwise. Similarly
for the statements y > 2*cos(x.^2) and x.^2 + y.^2 < 9. Joining those three statements with & sym-
bols returns a value of 1 if all three conditions are met, 0 oth erwise. The sum command thus counts how
many of the N points are inside the region. Dividing that count by N gives the ratio of points inside the
region, then multiplying by 6 (the area of the rectangle) giv es the approximate area of the region.",ElementaryCalculus.pdf
"region, then multiplying by 6 (the area of the rectangle) giv es the approximate area of the region.
Note that the size of the rectangle can affect the approximat ion—generally the larger the rectangle
the more points must be used. Note also in this example that fin ding the area by using definite integrals
would require numerical integration methods, since f (x) = ex2
and f (x) = 2 cos x2 cannot be integrated
in a closed form. In fact, even finding the points of intersect ion of the three curves—in order to split the
integrals—would require a numerical root-finding method (e .g. Newton’s method).
Exercises
A
For Exercises 1-6, find the area of the region bounded by the gi ven curves.
1. y = x2 and y = 2x + 3 2. x = −y2 + 2 y and x = 0 3. y = x2 − 1 and y = x3 − 1
4. y = x4 and y = x 5. x = y2 and x = y + 2 6. y = 4 − 4x2 and y = 1 − x2
7. Find the area between y = 4x − x2 and y = x over [0 , 4].
8. Find the area between y = cosh x and y = sinh x over [0 , ∞).",ElementaryCalculus.pdf
"7. Find the area between y = 4x − x2 and y = x over [0 , 4].
8. Find the area between y = cosh x and y = sinh x over [0 , ∞).
9. Find the area of the region defined by the inequalities 0 ≤ x ≤ y − x ≤ 1 − y ≤ 1.",ElementaryCalculus.pdf
"Area Between Curves • Section 8.1 257
10. Find the area between r = 1 + cos θ and r = 2 + 2 cos θ.
11. Find the area between r = 1 + cos θ and r = 2 + cos θ.
B
x2 − y2 = 1y
2
x1 2
x2 + y2 = 4
O
P
12. Find the area of the region in the first quadrant between the un it
hyperbola x2 − y2 = 1 and the circle x2 + y2 = 4 (i.e. the shaded region
shown in the figure on the right) in two ways:
(a) Integration using formula (8.1).
(b) Draw a line segment from the origin O to the point of intersection
P on the hyperbola, then use the areas of the resulting circula r
sector and hyperbolic sector , without resorting to integra tion.
Does your answer agree with part (a)? Explain.
13. Find the area common to the four circles of radius 5 shown in Fi gure 8.1.4. (Hint: Use symmetry .)
5
5
5
5
Figure 8.1.4 Exercise 13
y =mx+b
y = ax2A
(a) Area A
B
C
D
T
y = mx + b
y = ax2
(b) Area T
Figure 8.1.5 Exercise 14",ElementaryCalculus.pdf
"5
5
5
5
Figure 8.1.4 Exercise 13
y =mx+b
y = ax2A
(a) Area A
B
C
D
T
y = mx + b
y = ax2
(b) Area T
Figure 8.1.5 Exercise 14
14. Let A be the area of the region bounded by the parabola y = ax2 and the line y = mx + b, where a,
m, and b are positive constants (see Figure 8.1.5(a)). Let T be the area of the triangle △BC D , where
B and C are where the line intersects the parabola, and the point D on the parabola has the same
x-coordinate as the midpoint of the line segment BC (see Figure 8.1.5(b)). Show that A = 4
3 T .
15. In Example 8.7 the region was enclosed in the rectangle R = { (x, y) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 3 }. Use
Monte Carlo integration to approximate the area of the regio n again, using a different enclosing
rectangle R:
(a) R = { (x, y) : 0 ≤ x ≤ 2 , 1 ≤ y ≤ 3 }
(b) R = { (x, y) : 0 ≤ x ≤ 3 , 0 ≤ y ≤ 3 }
Are the results significantly different than before?
16. Approximate the area of the region bounded by y = x2 and y = cos x in two different ways:",ElementaryCalculus.pdf
"Are the results significantly different than before?
16. Approximate the area of the region bounded by y = x2 and y = cos x in two different ways:
(a) Use Monte Carlo integration with 10 million points.
(b) Use a numerical root-finding method to find the points of inter section of the curves, then use
those points in a numerical integration method to find the are a.
C
17. A dog is chained to a fixed point at the circular base of a cylind rical silo. The silo’s radius is 50 ft
and the chain can wrap exactly halfway around the silo. How mu ch total area can the dog roam, not
counting the area inside the silo?",ElementaryCalculus.pdf
"258 Chapter 8 • Applications of Integrals §8.2
8.2 Average Value of a Function
planet
d✸Sun
Figure 8.2.1
According to Kepler’s laws of planetary motion, a planet rev olv-
ing around the Sun follows an elliptical orbit, with the Sun a t one
focus of the ellipse, as in Figure 8.2.1. The distance d between
the planet and the Sun varies over the ellipse, reaching a min -
imum distance and a maximum distance (by the Extreme V alue
Theorem). How would you find the average distance between the
planet and the Sun over one complete orbit? The idea is to gene r-
alize the usual notion of an average of numbers.
Recall that for n numbers x1, x2, . . . , xn the average, denoted by ¯x, is simply the sum of the
numbers divided by how many numbers there are, namely
¯x = x1 + x2 + ··· + xn
n .
In statistics ¯x is called the mean of x1, x2, . . . , xn . This definition makes sense for a finite set
of numbers, but in the case of a planet revolving around the Su n, there are an uncountably",ElementaryCalculus.pdf
"of numbers, but in the case of a planet revolving around the Su n, there are an uncountably
infinite number of distances between the planet and the Sun, m aking the above definition
impossible to use. A way of taking a sum over an infinite contin uum of values is needed
instead. Such a method has already been encountered: the defi nite integral, which is merely a
sum of a continuum of infinitesimal quantities.
f (x0)
f (x1)
f (x2)
f (xn−1)
f (xn )
y
x
a = x0 x1 x2 . . . xn−1 xn = b
. . .
Figure 8.2.2 ∆xi = xi − xi−1 = (b − a)/n
To motivate the definition of the average value of
a function f over a closed interval [ a, b], denoted by
〈 f 〉, consider a partition
P = {a = x0 < x1 < x2 < ··· <xn−1 < xn = b}
that divides [ a, b] into n subintervals [ xi−1 , xi ] of
equal length ∆xi = xi − xi−1 = (b − a)/n, as in Fig-
ure 8.2.2. The n function values f (x1), f (x2), . . . ,
f (xn) constitute only a finite subset of all the func-
tion values f (x) over [ a, b], so their average would",ElementaryCalculus.pdf
"f (xn) constitute only a finite subset of all the func-
tion values f (x) over [ a, b], so their average would
be an approximation of the true function average 〈 f 〉, namely:
〈 f 〉 ≈ f (x1) + f (x2) + ··· + f (xn )
n =
n∑
i=1
f (xi )
n
By properties of summations, divide the entire sum by the con stant b − a and multiply each
term in the sum by b − a to get:",ElementaryCalculus.pdf
"Average Value of a Function • Section 8.2 259
〈 f 〉 ≈ 1
b − a
n∑
i=1
f (xi ) · b − a
n = 1
b − a
n∑
i=1
f (xi ) ∆xi
Note that the last summation on the right is just a Riemann sum for the definite integral∫ b
a f (x) dx, with the points x∗
i chosen to be the right endpoints of the intervals [ xi−1 , xi ] for i = 1
to n. Thus, taking the limit of that sum as n → ∞ (which means including more and more
function values in the average) yields the following definit ion:
The average value 〈 f 〉 of a function f over a closed interval [ a, b] is: 1
〈 f 〉 = 1
b − a
∫ b
a
f (x) dx (8.3)
Example 8.8
Find the average value of f (x) = x2 over [0 , 1].
Solution: By definition, with a = 0 and b = 1,
〈 f 〉 = 1
1 − 0
∫ 1
0
f (x) dx =
∫ 1
0
x2 dx = x3
3
⏐
⏐
⏐
⏐
1
0
= 13
3 − 03
3 = 1
3
Note that this says that if you took all the numbers between 0 a nd 1 and squared them, then the average
of those squares would be 1/3.
Example 8.9
Find the average value of f (x) = x2 over [ −1, 1].",ElementaryCalculus.pdf
"of those squares would be 1/3.
Example 8.9
Find the average value of f (x) = x2 over [ −1, 1].
Solution: By definition, with a = −1 and b = 1,
〈 f 〉 = 1
1 − (−1)
∫ 1
−1
f (x) dx = 1
2
∫ 1
−1
x2 dx = x3
6
⏐
⏐
⏐
⏐
1
−1
= 13
6 − (−1)3
6 = 1
3
Note that this is the same as the average over [0 , 1], as shown in the previous example. This should
make sense, since the function f (x) = x2 is symmetric about the y-axis, so the values of f (x) from [ −1, 0]
are the same as those from [0 , 1]. The values from [ −1, 1] just duplicate the values from [0 , 1] and hence
do not change the average.
Example 8.10
Find the average value of f (x) = sin x over [0 , π].
Solution: By definition, with a = 0 and b = π,
〈 f 〉 = 1
π− 0
∫ π
0
f (x) dx = 1
π
∫ π
0
sin x dx = − 1
π cos x
⏐
⏐
⏐
⏐
π
0
= − 1
π (cos π − cos 0) = − 1
π (−1 − 1) = 2
π
1In some statistics or mathematics texts you might see the not ation ¯f for the average value.",ElementaryCalculus.pdf
"260 Chapter 8 • Applications of Integrals §8.2
Example 8.11
x
y
0 5−5
3
−3
(x, y)
(4, 0)
d
Figure 8.2.3 x2
25 + y2
9 = 1
Find the average distance from the ellipse x2
25 + y2
9 = 1 to the point
(4, 0).
Solution: Let d represent the distance from any point ( x, y) on
the ellipse to the point (4 , 0), as in Figure 8.2.3. If ( x, y) is on
the ellipse x2
25 + y2
9 = 1 then y2 = 9(1 − x2
25 ) = 9
25 (25 − x2). So by the
distance formula, d is given by
d2 = (x − 4)2 + ( y − 0)2 = (x − 4)2 + y2
= (x − 4)2 + 9
25 (25 − x2)
= 25(x − 4)2 + 9(25 − x)2
25 = 25x2 − 200x + 400 + 225 − 9x2
25
= 16x2 − 200x + 625
25
d2 = (4x − 25)2
25 , and so taking square roots gives
d = ± 4x − 25
5 = −4x − 25
5 = 25 − 4x
5
for −5 ≤ x ≤ 5, since d = (4x − 25)/5 < 0 on [ −5, 5] and the distance cannot be negative. Note that by
symmetry of the ellipse about the x-axis, only the upper half of the ellipse is needed for the ave rage",ElementaryCalculus.pdf
"symmetry of the ellipse about the x-axis, only the upper half of the ellipse is needed for the ave rage
distance, since the lower half just duplicates the distance s. Hence, the average distance is
〈d 〉 = 1
5 − (−5)
∫ 5
−5
25 − 4x
5 dx = 1
50 (25x − 2x2)
⏐
⏐
⏐
⏐
5
−5
= 1
50 (125 − 50 − (−125 − 50)) = 5 .
Notice that the point (4 , 0) is a focus of the ellipse x2
25 + y2
9 = 1 (why?), which as it turns out makes the
calculation of the average distance fairly simple.
What if you wanted the average value of a function f that is not easily integrable? One
alternative to numerical integration techniques is the Monte Carlo method . The idea behind
it is simple: go back to the usual definition of an average, by t aking a large number N of
random numbers x1, x2, . . . , xN in [ a, b] and then using the approximation
〈 f 〉 ≈ f (x1) + f (x2) + ··· + f (xN )
N .
This might seem like taking a step backward from calculus, an d it is, but it is surprisingly",ElementaryCalculus.pdf
"〈 f 〉 ≈ f (x1) + f (x2) + ··· + f (xN )
N .
This might seem like taking a step backward from calculus, an d it is, but it is surprisingly
useful, as well as simple to implement with a computer . In add ition, it can be shown that as
the number of random points in [ a, b] increases, the approximations converge to the actual
average.",ElementaryCalculus.pdf
"Average Value of a Function • Section 8.2 261
Example 8.12
The Monte Carlo method is easy to implement in Octave/MATLAB . Typically only a “one-liner” is
needed, owing to Octave’s vectorization—i.e. the ability to perform mathematical operations on ent ire
arrays of objects all at once.
For example, recall from Example 8.8 that the average value o f f (x) = x2 over [0 , 1] is 1/3 = 0.33333 . . . .
Approximate the average value using an array of 100 million ( 108) random numbers in [0 , 1]:
octave> mean(rand(1,1e8).^2)
ans = 0.3333292094741531
The dot in the rand(1,1e8).^2 command applies the squaring operation ( ^2) to each of the 10 8 random
numbers in the array returned by the rand(1,1e8) command. The aggregate mean function then calcu-
lates the array’s mean. Trigonometric, exponential and oth er functions can be applied to arrays, with
the function evaluating each array element individually . I n general the command ( b − a).*rand(1,N)+a",ElementaryCalculus.pdf
"the function evaluating each array element individually . I n general the command ( b − a).*rand(1,N)+a
will return an array of N random numbers in the interval ( a, b).
For example, the function f (x) = sin ( x2) cannot be integrated in a closed form, but its average value
over [ π, 2π] can be approximated easily in Octave (actual average = -0.0 4154374531416104):
octave> mean(sin((pi.*rand(1,1e8)+pi).^2))
ans = -0.04153426177596753
Exercises
A
For Exercises 1-9, find the average value of the function f (x) over the given interval.
1. f (x) = 1, over [0 , 3] 2. f (x) = x, over [0 , 1] 3. f (x) = x2, over [0 , 2]
4. f (x) = x3, over [0 , 2] 5. f (x) = sin 2 x, over [0 , π/2] 6. f (x) = ex, over [ −1, 4]
7. f (x) = x3, over [ −1, 1] 8. f (x) = sin x, over [ −π/2, π/2] 9. f (x) = 1
x , over [1 , 3]
10. Electrical signals are commonly represented by a periodic waveform x(t), which is a function of",ElementaryCalculus.pdf
"x , over [1 , 3]
10. Electrical signals are commonly represented by a periodic waveform x(t), which is a function of
time t and has period T (i.e. T is the smallest positive number such that x(t + T ) = x(t) for all t). The
average power of the waveform is defined as the average value of its square ov er a single period:
⟨
x2(t)
⟩
= 1
T
∫ T
0
x2 (t) dt .
(a) Find the average power of the waveform x(t) = A cos(ωt +φ), where A > 0 and ω > 0 and φ are all
constants.
(b) The root mean square of a waveform, abbreviated as rms, is the square root of the average
power . Calculate the rms of the waveform from part (a). Write your answer in decimal form as
a percentage of the amplitude A.",ElementaryCalculus.pdf
"262 Chapter 8 • Applications of Integrals §8.2
E
R
s C11. An electric circuit with a supplied voltage (electromotive force)
E, a capacitor with capacitance C, and a resistor with resistance
R, is shown in the picture on the right. When a switch s in the
circuit is opened at time t = 0 the current I through the circuit
begins to decrease exponentially as a function of time t (mea-
sured in seconds after the switch is opened), given by
I = E
R e−t/RC
for t ≥ 0.
(a) Sketch a rough graph of I as a function of t.
(b) Note that at time t = 0 the current is I = E
R (measured in amperes), which is the familiar formula
from Ohm’s Law . That is the peak value of I. What is the current I at time t = 5RC ? Write your
answer in decimal form as a percentage of the peak current E
R (e.g. 0 .42 E
R , which would be 42%
of the peak current).
(c) Find the average current in the circuit over the time interva l [0 , 5RC ]. Write your answer in
decimal form as a percentage of the peak current.",ElementaryCalculus.pdf
"decimal form as a percentage of the peak current.
12. A spring with spring constant k and damping constant ν connects two point particles with mass m
in a gravitational wave detector . A gravitational wave pass es through the detector at time t = 0 and
induces oscillation in the spring, with a period of 2 π/Ω and energy E at time t ≥ 0 given by
E(t) = 1
4 mR 2 (
Ω2 sin2 (Ωt + φ) + ω2
0 cos2 (Ωt + φ)
)
,
where ω2
0 = 2k/m, φ = tan−1 (2νΩ/(m(ω2
0 − Ω2 )), and R is a constant.
(a) Show that the average energy 〈E 〉 over one period [0 , 2π/Ω] of oscillation is
〈E 〉 = 1
8 mR 2 (ω2
0 + Ω2 ) .
(b) Suppose a large number of identical detectors of this type ar e uniformly distributed in a planar
array at a density of σ detectors per unit area. The energy Eσ(t) imparted to each detector at
time t ≥ 0 by a gravitational wave is
Eσ(t) = νΩ2 R2 sin2 (Ωt + φ) .
Show that the average energy 〈Eσ〉 over one period [0 , 2π/Ω] of oscillation is
〈Eσ〉 = 1
2 νΩ2 R2 .
B
13. For the ellipse x2
a2 + y2",ElementaryCalculus.pdf
"Show that the average energy 〈Eσ〉 over one period [0 , 2π/Ω] of oscillation is
〈Eσ〉 = 1
2 νΩ2 R2 .
B
13. For the ellipse x2
a2 + y2
b2 = 1, with a > b > 0, the foci are the points ( c, 0) and ( −c, 0), where c =
∝ra⌈〉⌋allow
a2 − b2 .
Find the average distance from the ellipse to either of its fo ci in terms of the constants a, b, and c.
14. Write a computer program to use the Monte Carlo method with 1 m illion random points to approx-
imate the average distance from the ellipse x2
25 + y2
9 = 1 to the point (0 , 0). Use symmetry to choose
the smallest interval for the points. Could you have used for mula (8.3) instead? Explain.",ElementaryCalculus.pdf
"Arc Length and Curvature • Section 8.3 263
8.3 Arc Length and Curvature
Just like the area of a plane region can be found using calculu s, so too can the length of a plane
curve. Along the way the mystery mentioned in a footnote in Ch apter 1 will finally be solved:
what is the length of the hypotenuse of a right triangle with i nfinitesimal sides?
For a function y = f (x) denote by s the length of the piece of that curve over an interval [ a, b],
as in Figure 8.3.1 (a). Call s the curve’s arc length over [ a, b].
y
x
a b
s
y = f (x)
(a) Arc length s
dx
dyds
(b) Infinitesimal triangle
1
dy
dx
ds
dx
(c) Noninfinitesimal triangle
Figure 8.3.1 Arc length
By the Microstraightness Property , for a ≤ x < b the curve is a straight line of length ds over
the infinitesimal interval [ x, x + dx], as in Figure 8.3.1 (b), where ds > 0 is the infinitesimal
change in s over that interval. Notice that you cannot simply apply the P ythagorean Theorem
here, since that would make ds =
√
(dx)2 + (dy)2 =",ElementaryCalculus.pdf
"change in s over that interval. Notice that you cannot simply apply the P ythagorean Theorem
here, since that would make ds =
√
(dx)2 + (dy)2 =
∝ra⌈〉⌋allow
0 + 0 = 0, which is false. The trick is
to divide all sides of this infinitesimal right triangle by dx, which yields the similar—and
noninfinitesimal—right triangle shown in Figure 8.3.1(c). The Pythagorean Theorem can then
be applied to that triangle:
ds
dx =
√
12 +
(dy
dx
) 2
⇒ ds =
√
1 +
(dy
dx
) 2
dx
Summing up those infinitesimal lengths ds then yields the arc length s:
The arc length s of a curve y = f (x) over [ a, b] is:
s =
∫ b
a
ds =
∫ b
a
√
1 +
(dy
dx
) 2
dx (8.4)
That such a formula exists is of course good news, but as you ha ve probably guessed, the in-
tegral cannot be evaluated in a closed form except for a few fu nctions.2 In most cases numerical
integration methods will be required.
2This will be the last “good news/bad news” scenario in this bo ok. That’s the good news.",ElementaryCalculus.pdf
"264 Chapter 8 • Applications of Integrals §8.3
Example 8.13
y
x
0 1
y = cosh x
s1
Find the arc length of the curve y = cosh x over [0 , 1].
Solution: Since dy
dx = sinh x, then the arc length s is:
s =
∫ 1
0
√
1 +
(dy
dx
) 2
dx =
∫ 1
0
√
1 + sinh2 x dx
=
∫ 1
0
cosh x dx = sinh x
⏐
⏐
⏐
⏐
1
0
= sinh 1 − sinh 0 = sinh 1 ≈ 1.1752
Example 8.14
y
x
0−L/2 L/2
y = a cosh x
a
s
h
a
A catenary—a hanging uniform cable whose ends are fastened at the same
height h a distance L apart—has its lowest point—the apex—a distance
a > 0 above the ground. It can be shown 3 that with the apex at (0 , a), the
equation of the catenary is y = a cosh x
a . Find the arc length of the catenary .
Solution: The figure on the right shows the catenary . By symmetry the
total arc length s is twice the arc length over [0 , L/2]:
s = 2
∫ L/2
0
√
1 +
(dy
dx
) 2
dx = 2
∫ L/2
0
√
1 + sinh2 x
a dx
= 2
∫ L/2
0
cosh x
a dx = 2a sinh x
a
⏐
⏐
⏐
⏐
L/2
0
= 2a sinh L
2a
Elliptic Integrals
x
y
x2
a2 + y2
b2 = 1
−a a
b
−b",ElementaryCalculus.pdf
"∫ L/2
0
√
1 + sinh2 x
a dx
= 2
∫ L/2
0
cosh x
a dx = 2a sinh x
a
⏐
⏐
⏐
⏐
L/2
0
= 2a sinh L
2a
Elliptic Integrals
x
y
x2
a2 + y2
b2 = 1
−a a
b
−b
Suppose you tried to find the circumference s of the ellipse
x2
a2 + y2
b2 = 1, with a > b > 0, which has eccentricity e = c
a , where
c =
∝ra⌈〉⌋allow
a2 − b2. By symmetry , s is quadruple the arc length of the
upper hemisphere y = b
√
1 − x2
a2 over [0 , a]:
s = 4
∫ a
0
√
1 +
(dy
dx
) 2
dx = 4
∫ a
0
√
1 +
(
− b x
a
∝ra⌈〉⌋allow
a2 − x2
) 2
dx
= 4
∫ a
0
√
a2(a2 − x2) + b2 x2
a2(a2 − x2) dx
= 4
a
∫ a
0
√
a4 − (a2 − b2)x2
a2 − x2 dx
Now try the trigonometric substitution x = a sin θ, dx = a cos θ dθ:
3See pp.162-163 in S MI T H , C.E., Applied Mechanics: Statics , New Y ork: John Wiley & Sons, Inc., 1976.",ElementaryCalculus.pdf
"Arc Length and Curvature • Section 8.3 265
s = 4
a
∫ π/2
0
√
a4 − (a2 − b2) a2 sin2 θ
a2 − a2 sin2 θ
a cos θ dθ
= 4
∫ π/2
0
√
a2 − (a2 − b2) sin 2 θ
✘✘✘✘✘1 − sin2 θ
✘✘✘cos θ dθ
= 4a
∫ π/2
0
√
1 − a2 − b2
a2 sin2 θ dθ
s = 4a
∫ π/2
0
√
1 − e2 sin2 θ dθ (since e2 = c2
a2 = a2 − b2
a2 )
The last integral is a special case of the elliptic integral of the second kind
E(k, φ) =
∫ φ
0
√
1 − k2 sin2 θ dθ ,
with k = e and φ = π
2 . This special case is denoted by E(k) = E(k, π
2 ). Thus, the circumference s
of the ellipse is:
s = 4a E (e) = 4a E (e, π
2 )
The integral E(k) for 0 < k < 1 cannot be evaluated in a closed form. There are tables 4 for
certain values of k between 0 and 1, but a number of scientific computing applicat ions have
built-in functions to evaluate elliptic integrals.
For example, suppose you want to find the circumference s of the ellipse x2
25 + y2
9 = 1. Then
a = 5, b = 3, c =
∝ra⌈〉⌋allow
a2 − b2 = 4, and e = c
a = 0.8, so s = 4a E (e) = 20 E(0.8). In the Python-based",ElementaryCalculus.pdf
"25 + y2
9 = 1. Then
a = 5, b = 3, c =
∝ra⌈〉⌋allow
a2 − b2 = 4, and e = c
a = 0.8, so s = 4a E (e) = 20 E(0.8). In the Python-based
open-source mathematical software system Sage 5 , the elliptic integral E(k, φ) is provided by
the function elliptic_e(φ,k2). Use k = e = 0.8 and φ = π
2 :
In [1]: 20*elliptic_e(pi/2,0.8^2)
Out[1]: 25.5269988633981
The circumference is thus approximately 25.5269988633981 . In Octave/MATLAB the function
ellipke(e2) evaluates the elliptic integral E(e), with one extra step:
MATLAB>> [K,E] = ellipke(0.8^2);
MATLAB>> 20*E
ans = 25.526998863398131
4See p.609 in A B R A MO W I T Z , M. A N D I.A. S T E G U N, Handbook of Mathematical Functions , New Y ork: Dover
Publications, Inc., 1965.
5Available at https://www.sagemath.org",ElementaryCalculus.pdf
"266 Chapter 8 • Applications of Integrals §8.3
The parametric formula for arc length can be derived by divid ing all sides of the infinitesimal
right triangle in Figure 8.3.1(b) by dt, then applying the Pythagorean Theorem to the resulting
noninfinitesimal right triangle:
ds
dt =
√ (dx
dt
) 2
+
(dy
dt
) 2
⇒ ds =
√ (dx
dt
) 2
+
(dy
dt
) 2
dt
Sum up those infinitesimal lengths ds to obtain the arc length s:
The arc length s of a parametric curve x = x(t), y = y(t), a ≤ t ≤ b is:
s =
∫ b
a
ds =
∫ b
a
√ (dx
dt
) 2
+
(dy
dt
) 2
dt (8.5)
Example 8.15
y
x
1 0
1 x = cos3 t
y = sin3 t
Find the arc length of the parametric curve x = cos3 t, y = sin3 t, 0 ≤ t ≤ π/2.
Solution: Since dx
dt = −3 cos2 t sin t and dy
dt = 3 sin2 t cos t, then the arc length s is:
s =
∫ π/2
0
√ (dx
dt
) 2
+
(dy
dt
) 2
dt =
∫ π/2
0
√
(−3 cos 2 t sin t)2 + (3 sin 2 t cos t)2 dt
=
∫ π/2
0
√
9 cos 4 t sin2 t + 9 sin 4 t cos2 t dt =
∫ π/2
0
3
√
sin2 t cos2 t (cos2 t + sin2 t) dt
=
∫ π/2
0",ElementaryCalculus.pdf
"=
∫ π/2
0
√
9 cos 4 t sin2 t + 9 sin 4 t cos2 t dt =
∫ π/2
0
3
√
sin2 t cos2 t (cos2 t + sin2 t) dt
=
∫ π/2
0
3 sin t cos t dt (since cos t ≥ 0 and sin t ≥ 0 for 0 ≤ t ≤ π/2)
= 3
2 sin2 t
⏐
⏐
⏐
⏐
π/2
0
= 3
2
The polar formula for arc length can be considered a special c ase of the parametric formula.
A polar curve r = r(θ) for α ≤ θ ≤ β has Cartesian coordinates x = r(θ) cos θ and y = r(θ) sin θ,
so that
dx
dθ = dr
dθ cos θ − r sin θ and dy
dθ = dr
dθ sin θ + r cos θ .
It is left as an exercise to show that putting these derivativ es into formula (8.5)—using the
parameter θ instead of t—yields the polar arc length formula:
The arc length s of a polar curve r = r(θ) for α ≤ θ ≤ β is:
s =
∫ β
α
ds =
∫ β
α
√
r2 +
(dr
dθ
) 2
dθ (8.6)",ElementaryCalculus.pdf
"Arc Length and Curvature • Section 8.3 267
Example 8.16
Prove that the circumference of a circle of radius R is 2 πR.
Solution: Use the polar curve r = R for 0 ≤ θ ≤ 2π. Then dr
dθ = 0, so:
s =
∫ 2π
0
√
r2 +
(dr
dθ
) 2
dθ =
∫ 2π
0
√
R2 + 02 dθ =
∫ 2π
0
R d θ = R θ
⏐
⏐
⏐
⏐
2π
0
= 2πR ✓
Curvature
y
x
1 0
1
y = x2
Figure 8.3.2
In Chapter 4 you saw one simple measure of curvature: the sec-
ond derivative. From Figure 8.3.2 it is clear that the parabo la y = x2
is less curved at the point (1 , 1) than at the origin, yet d2 y
dx2 = 2 at each
point. So the second derivative—the rate of change of the slo pes dy
dx
of the tangent lines—does not fully capture the curvature of a curve;
more information is needed. It turns out that the rate of chan ge of
the angles of the tangent lines is the key to curvature.
First consider a curve with arc length s between two points A and
B on the curve. Let α be the angle between the tangent lines to the curve at A and B, as in
Figure 8.3.3(a).
α
A
B
s",ElementaryCalculus.pdf
"B on the curve. Let α be the angle between the tangent lines to the curve at A and B, as in
Figure 8.3.3(a).
α
A
B
s
(a) Small curvature
α
A
B
s
(b) Larger curvature and α
Figure 8.3.3 Curvature between A and B: same s, different α
For the same arc length s but larger angle α as in Figure 8.3.3(b), the curvature appears
greater . This suggests that curvature should measure α relative to s:
The average curvature ¯κAB of a curve between two points A and B on the curve is
¯κAB = α
s (8.7)
where s is the arc length of the curve between A and B, and α is the angle between the
tangent lines to the curve at A and B, called the angle of contingence .",ElementaryCalculus.pdf
"268 Chapter 8 • Applications of Integrals §8.3
Similar to how the instantaneous rate of change of a function at a point is the average rate
of change over an infinitesimal interval, the curvature of a c urve at a point can be defined as
the average curvature over an infinitesimal length of the cur ve:
The curvature κ of a curve at a point A on the curve is
κ = dφ
ds (8.8)
where s is the arc length function of the curve and φ is the angle that the tangent line to
the curve at A makes with the positive x-axis.
The idea is that moving an infinitesimal length ds along the curve induces an infinitesimal
difference dφ in the angles of the tangent lines. Roughly , as B moves toward A:
lim
B→A
¯κAB = lim
B→A
α
s = dφ
ds = κ
Curvature can be viewed as the instantaneous rate of change o f direction of a curve, but with
respect to arc length (i.e. distance traveled) instead of ti me.
For a curve y = f (x), recall from formula (3.2) in Section 3.1 that φ = φ(x) = tan−1 f ′(x) at",ElementaryCalculus.pdf
"For a curve y = f (x), recall from formula (3.2) in Section 3.1 that φ = φ(x) = tan−1 f ′(x) at
each point ( x, f (x)) on the curve. Thus, by the Chain Rule:
κ = dφ
ds = d (tan−1 f ′(x))
ds =
f ′′(x) dx
1 + ( f ′(x))2
ds
So since ds =
√
1 + ( f ′(x))2 dx by formula (8.4), then:
The curvature κ of a curve y = f (x) is:
κ = f ′′(x)
(1 + ( f ′(x))2 )3/2 (8.9)
The above formula makes κ a function of x. Note also that κ = 0 for a straight line, and that
the curve y = f (x) is concave up if κ > 0 and concave down if κ < 0.
Example 8.17
Find the curvature of the curve y = x2 for x = 0 and x = 1.
Solution: For f (x) = x2, f ′(x) = 2x and f ′′(x) = 2, so for any x the curvature κ = κ(x) is:
κ = f ′′(x)
(1 + ( f ′(x))2)3/2 = 2
(1 + 4x2 )3/2
In particular , κ(0) = 2 and κ(1) = 2
53/2 ≈ 0.1789. So y = x2 has more curvature at the origin than at (1 , 1).",ElementaryCalculus.pdf
"Arc Length and Curvature • Section 8.3 269
For a parametric curve x = x(t), y = y(t), the curvature κ will be a function of the parameter
t. Since dy
dx = y′(t)
x′(t) by formula (7.14) in Section 7.6, then by formula (7.15):
d2 y
dx2 =
d
dt
(dy
dx
)
dx
dt
=
d
dt
(y′(t)
x′(t)
)
x′(t) = x′(t) y′′(t) − y′(t) x′′(t)
(x′(t))3
So by formula (8.9):
κ =
d2 y
dx2
(
1 +
(dy
dx
) 2) 3/2 =
x′(t) y′′(t) − y′(t) x′′(t)
(x′(t))3
(
1 +
(y′(t)
x′(t)
) 2) 3/2 = x′(t) y′′(t) − y′(t) x′′(t)
(
(x′(t))2) 3/2
(
1 +
(y′(t)
x′(t)
) 2) 3/2
Simplify the denominator to obtain the parametric curvatur e formula:
The curvature κ of a parametric curve x = x(t), y = y(t) is:
κ = x′(t) y′′(t) − y′(t) x′′(t)
(
(x′(t))2 + ( y′(t))2) 3/2 (8.10)
The derivation of the curvature formula in polar coordinate s is left as an exercise:
The curvature κ of a polar curve r = r(θ) is:
κ = (r(θ))2 + 2 ( r′(θ))2 − r(θ) r′′(θ)
(
(r(θ))2 + (r′(θ))2) 3/2 (8.11)
Example 8.18
Find the curvature of a circle of radius R.",ElementaryCalculus.pdf
"κ = (r(θ))2 + 2 ( r′(θ))2 − r(θ) r′′(θ)
(
(r(θ))2 + (r′(θ))2) 3/2 (8.11)
Example 8.18
Find the curvature of a circle of radius R.
Solution: Use the polar curve r = r(θ) = R, so that r′(θ) = 0 = r′′(θ):
κ = (r(θ))2 + 2 (r′(θ))2 − r(θ) r′′(θ)
(
(r(θ))2 + (r′(θ))2 ) 3/2 = R2 + 2 · 02 − R · 0
(R2 + 02 )3/2 = R2
R3 = 1
R
A circle thus has constant curvature, as you would expect by s ymmetry . It turns out that any planar
curve with constant curvature is either a line or part of a cir cle.6
6See pp.62-63 in O’N E I L L , B., Elementary Differential Geometry , New Y ork: Academic Press, Inc., 1966.",ElementaryCalculus.pdf
"270 Chapter 8 • Applications of Integrals §8.3
Exercises
A
For Exercises 1-10, find the arc length of the given curve over the given interval.
1. y = x3/2 ; 1 ≤ x ≤ 4 2. y = x2 ; 0 ≤ x ≤ 1 3. y = x2/3 ; 1 ≤ x ≤ 8
4. y = x2
4 − ln x
2 ; 1 ≤ x ≤ 2 5. y = x4
4 + 1
8x2 ; 1 ≤ x ≤ 2 6. y = ln ex + 1
ex − 1 ; 1 ≤ x ≤ 2
7. x = et cos t, y = et sin t ; 0 ≤ t ≤ π 8. x = cos t + t sin t, y = sin t − t cos t ; 0 ≤ t ≤ π
9. polar curve r = 1 + cos θ ; 0 ≤ θ ≤ 2π 10. polar curve r = eθ ; 0 ≤ θ ≤ 2
11. Find the arc length of the curve in Example 8.15 for 0 ≤ t ≤ π.
12. Use formula (8.5) to find the circumference of the unit circle using two different parametrizations:
(a) x = cos t, y = sin t, 0 ≤ t ≤ 2π (b) x = 1 − t2
1 + t2 , y = 2t
1 + t2 , −∞ <t < ∞
For Exercises 13-18 find the curvature of the given curve at th e indicated points.
13. y = sin x at x = 0 and x = π
2 14. y = ln x at x = 1 15. x2
a2 + y2
b2 = 1 at ( a, 0) and (0 , b)",ElementaryCalculus.pdf
"13. y = sin x at x = 0 and x = π
2 14. y = ln x at x = 1 15. x2
a2 + y2
b2 = 1 at ( a, 0) and (0 , b)
16. y = ex at x = 0 17. x2 − y2 = 1 at (1 , 0) 18. r = 1 + cos θ at θ = 0
B
19. Prove formula (8.6). 20. Prove formula (8.11).
21. Let α and β be nonzero constants. Show that the arc length s of y = β sin x
α over the interval [0 , x0 ]
can be put in terms of the elliptic integral E(k, φ):
s =
√
α2 + β2 · E
(√
β2
α2 + β2 , x0
α
)
22. For −1 < k < 1 and −1 ≤ x ≤ 1, define u(x) and K by
u(x) =
∫ x
0
dt∝ra⌈〉⌋allow
1 − t2
∝ra⌈〉⌋allow
1 − k2 t2
and K = u(1) =
∫ 1
0
dt∝ra⌈〉⌋allow
1 − t2
∝ra⌈〉⌋allow
1 − k2 t2
so that all the square roots are defined and positive.
(a) Show that u is an increasing function of x and thus has an inverse function, call it x = sn u, with
domain [ −K , K ] and range [ −1, 1].
(b) Define cn u =
∝ra⌈〉⌋allow
1 − sn2 u and dn u =
∝ra⌈〉⌋allow
1 − k2 sn2 u. Show that:
d
du (sn u) = cn u dn u , d
du (cn u) = −sn u dn u , and d
du (dn u) = −k2 sn u cn u",ElementaryCalculus.pdf
"1 − sn2 u and dn u =
∝ra⌈〉⌋allow
1 − k2 sn2 u. Show that:
d
du (sn u) = cn u dn u , d
du (cn u) = −sn u dn u , and d
du (dn u) = −k2 sn u cn u
The functions sn u, cn u, and dn u are called the Jacobian elliptic functions .
(c) Suppose that sin φ = sn u. Show that E(k, φ) =
∫ u
0
dn2 v dv .
23. The ends of a 50 ft long catenary are fastened 40 ft apart. Use a numerical method to find how
much the apex dips below the ends. (Hint: Solve for a in Example 8.14, then use symmetry .)",ElementaryCalculus.pdf
"Surfaces and Solids of Revolution • Section 8.4 271
8.4 Surfaces and Solids of Revolution
Long before calculus was invented the ancient Greeks (e.g. A rchimedes) discovered the formu-
las for the volume and surface area of familiar three-dimens ional objects such as the sphere.
7
V olumes and surface areas of arbitrary solids and surfaces c an be found using multivariable
calculus. However , single-variable calculus can be used in the special case of the objects pos-
sessing symmetry about an axis, via methods that involve rev olving a curve or region in the
x y-plane around an axis.
For example, revolve a curve y = f (x) ≥ 0 around the x-axis, for a ≤ x ≤ b. This produces a
surface of revolution in three dimensions, as in Figure 8.4.1(a).
y
x
a b
y = f (x)
(a) Revolve y = f (x) around x-axis
y
x
a x b
y = f (x)
(b) dS over [ x, x + dx]
f (x) f (x + dx)
y
x
ds
dx
(c) Side view of frustrum
Figure 8.4.1 Surface of revolution over [ a, b] with lateral surface area S
r2
r1
l
Figure 8.4.2",ElementaryCalculus.pdf
"f (x) f (x + dx)
y
x
ds
dx
(c) Side view of frustrum
Figure 8.4.1 Surface of revolution over [ a, b] with lateral surface area S
r2
r1
l
Figure 8.4.2
To find the total lateral surface area S, pick x in [ a, b) then find the
infinitesimal surface area dS swept out over the infinitesimal interval
[x, x + dx], as in Figure 8.4.1(b). By the Microstraightness Property , the
curve y = f (x) is a straight line segment of length ds over that interval,
so that the infinitesimal surface is a frustrum—a right circular cone
with the vertex chopped off by a plane parallel to the base cir cle. From
geometry8 you might recall the formula for the lateral surface area of t he
frustrum in Figure 8.4.2: π(r1 +r2) l. Use that formula with r1 = f (x), r2 = f (x +dx) = f (x) +dy,
and l = ds =
√
1 + ( f ′(x))2 dx (by formula (8.4) in Section 8.3) as in Figure 8.4.1(c), so th at dS is
dS = π( f (x) + ( f (x) + dy))
√
1 + ( f ′(x))2 dx
= 2 π f (x)
√
1 + ( f ′(x))2 dx + π
√
1 + ( f ′(x))2 dy dx
= 2 π f (x)
√",ElementaryCalculus.pdf
"dS = π( f (x) + ( f (x) + dy))
√
1 + ( f ′(x))2 dx
= 2 π f (x)
√
1 + ( f ′(x))2 dx + π
√
1 + ( f ′(x))2 dy dx
= 2 π f (x)
√
1 + ( f ′(x))2 dx + 0
since dy dx = f ′(x) (dx)2 = 0. The surface area S is then the sum of all the areas dS:
7See Propositions 33 and 34 in On the Sphere and Cylinder , Book I , appearing in H E AT H , T .L., The Works of
Archimedes, Mineola, NY : Dover Publications, Inc., 2002. This work is a lso available at https://archive.org
8See pp.136-137 in W E L C H O N S A.M. A N D W .R. K R I C K E N B E R G E R, Solid Geometry , Boston: Ginn & Co., 1936.",ElementaryCalculus.pdf
"272 Chapter 8 • Applications of Integrals §8.4
The surface area S of the surface of revolution obtained by revolving the curve y = f (x) ≥ 0
around the x-axis for a ≤ x ≤ b is:
S =
∫ b
a
dS =
∫ b
a
2 π f (x)
√
1 + ( f ′(x))2 dx (8.12)
For a general curve y = f (x), possibly negative in [ a, b], the surface area S is:
S =
∫ b
a
dS =
∫ b
a
2 π| y| ds =
∫ b
a
2 π
⏐
⏐ f (x)
⏐
⏐
√
1 + ( f ′(x))2 dx (8.13)
y
x
a b
y =
⏐
⏐ f (x)
⏐
⏐
y = f (x)
Note that formula (8.13) holds by symmetry and formula (8.12 ). A
curve y = f (x) < 0 and the curve y =
⏐
⏐ f (x)
⏐
⏐ = −f (x) are symmet-
ric with respect to the x-axis, as in the figure on the right. Thus,
both curves sweep out the same surface of revolution when rev olved
around the x-axis. This means formula (8.13) also holds if y = f (x)
changes sign in [ a, b]: similar to the area between two curves, you
would split the integral over different subintervals depen ding on
the sign.
Example 8.19",ElementaryCalculus.pdf
"would split the integral over different subintervals depen ding on
the sign.
Example 8.19
Show that the surface area of a sphere of radius r is 4 πr2 .
y
x−r r
y =
∝ra⌈〉⌋allow
r2 − x2
Solution: Use the circle x2 + y2 = r2 . The upper half of that circle is the
curve y = f (x) =
∝ra⌈〉⌋allow
r2 − x2 over the interval [ −r, r], as in the figure on
the right. Revolving that curve around the x-axis produces a sphere
of radius r, whose surface area S is:
S =
∫ r
−r
2 π f (x)
√
1 + ( f ′(x))2 dx
=
∫ r
−r
2 π
√
r2 − x2
√
1 +
( −x∝ra⌈〉⌋allow
r2 − x2
) 2
dx
=
∫ r
−r
2 π✘✘✘✘√
r2 − x2
√
r2
✘✘✘r2 − x2 dx
= 2 πrx
⏐
⏐
⏐
⏐
r
−r
= 4 πr2 ✓
A similar derivation using a frustrum yields the surface are a S of the surface of revolution
obtained by revolving a curve y = f (x) around the y-axis, for 0 ≤ a ≤ x ≤ b:
S =
∫ b
a
dS =
∫ b
a
2 π|x| ds =
∫ b
a
2 π x
√
1 + ( f ′(x))2 dx (8.14)",ElementaryCalculus.pdf
"Surfaces and Solids of Revolution • Section 8.4 273
Now suppose you revolve the region between a curve y = f (x) ≥ 0 and the x-axis around
the x-axis, for a ≤ x ≤ b (see Figure 8.4.3(a)). This produces a solid of revolution in three
dimensions, as in Figure 8.4.3(b). Notice that this solid co nsists of the surface of revolution as
before along with its interior .
y
x
a b
y = f (x)
(a) Revolve region around x-axis
y
x
a x b
y = f (x)
(b) Solid
f (x) f (x + dx)
y
x
ds
dx
dy
(c) Infinitesimal strip
Figure 8.4.3 Solid of revolution over [ a, b] with volume area V
The goal is to find the volume V of this solid. The idea is to divide the solid into slices,
like a loaf of bread. First, the infinitesimal volume dV of the frustrum swept out by a strip
of infinitesimal width dx at x in [ a, b)—shown in Figure 8.4.3(c)—is needed. By the Micros-
traightness Property the curve y = f (x) is a straight line of length ds over the interval [ x, x+dx].",ElementaryCalculus.pdf
"traightness Property the curve y = f (x) is a straight line of length ds over the interval [ x, x+dx].
There is thus a right triangle at the top of the strip—unshade d in Figure 8.4.3(c)—whose area
A is zero: A = 1
2 (dy)(dx) = 1
2 f ′(x)(dx)2 = 0.
f (x)
dx
Figure 8.4.4
That triangle thus contributes no volume when revolved arou nd the x-axis:
the volume dV swept out by that strip all comes from the shaded rectangle of
height f (x) and width dx. That rectangle sweeps out a right circular cylinder
of radius f (x) and height dx (see Figure 8.4.4). The volume of a right circular
cylinder of radius r and height h is defined as the area of the base circle times
the height: πr2 h. Hence,
dV = π( f (x))2 dx .
The total volume V of the solid is then the sum of all those infinitesimal volumes dV:
The volume V of the solid of revolution obtained by revolving the region b etween the curve
y = f (x) and the x-axis around the x-axis for a ≤ x ≤ b is:
V =
∫ b
a
dV =
∫ b
a
π( f (x))2 dx (8.15)",ElementaryCalculus.pdf
"y = f (x) and the x-axis around the x-axis for a ≤ x ≤ b is:
V =
∫ b
a
dV =
∫ b
a
π( f (x))2 dx (8.15)
This method for finding the volume is called the disc method , since the cylinder of volume
dV resembles a disc. Think of the discs as being similar to infini tesimally thin slices of a loaf
of bread. Notice that absolute values are not needed in formu la (8.15) since f (x) is squared, so
the formula holds even when f (x) is negative.",ElementaryCalculus.pdf
"274 Chapter 8 • Applications of Integrals §8.4
Example 8.20
Show that the volume of a sphere of radius r is 4
3 πr3 .
y
x−r r
y =
∝ra⌈〉⌋allow
r2 − x2
Solution: Use the circle x2 + y2 = r2 . Revolve the region between the
upper half of the circle y = f (x) =
∝ra⌈〉⌋allow
r2 − x2 and the x-axis around the
x-axis over the interval [ −r, r], as in the figure on the right. The solid
of revolution swept out is a sphere of radius r, whose volume V is:
V =
∫ r
−r
π( f (x))2 dx =
∫ r
−r
π(r2 − x2 ) dx = πr2 x − 1
3 πx3
⏐
⏐
⏐
⏐
r
−r
=
(
πr3 − 1
3 πr3
)
−
(
−πr3 + 1
3 πr3
)
= 4
3 πr3 ✓
Instead of memorizing formula (8.15), try to remember the mo re generic approach of revolv-
ing an infinitesimal rectangular strip around an axis, which might not be the x-axis. The idea
is to find the radius r and height h—typically dx or dy—of the disc swept out by that strip, so
that the disc’s volume is dV = πr2 h. Then integrate dV over the appropriate interval to find
the volume V of the entire solid.
Example 8.21",ElementaryCalculus.pdf
"that the disc’s volume is dV = πr2 h. Then integrate dV over the appropriate interval to find
the volume V of the entire solid.
Example 8.21
Suppose the region bounded by the curve y = x2 and the x-axis for 0 ≤ x ≤ 1 is revolved around the line
x = 1. Find the volume of the resulting solid of revolution.
x = 1y
x
0 x 1
y = x2
r hx2 = y
1
Solution: The region is shaded in the figure on the right. Since the
region is revolved around a vertical axis, the disc method wi ll use
discs with height dy, not dx. At a point x in [0 , 1] go up to the curve
y = x2 and draw a horizontal rectangular strip to the line x = 1, as
shown in the figure. Let h = dy and revolve that strip around the
line x = 1, producing a disc of radius r − 1 − x and height h = dy.
Since y = x2 implies x = ∝ra⌈〉⌋allowy, the volume d V of that disc is
d V = πr2 h = π(1 − x)2 dy = π(1 − ∝ra⌈〉⌋allowy)2 dy = π(1 − 2∝ra⌈〉⌋allowy + y) dy .",ElementaryCalculus.pdf
"d V = πr2 h = π(1 − x)2 dy = π(1 − ∝ra⌈〉⌋allowy)2 dy = π(1 − 2∝ra⌈〉⌋allowy + y) dy .
The volume V of the entire solid is then the sum of those volumes d V along the y-axis for 0 ≤ y ≤ 1:
V =
∫ 1
0
d V =
∫ 1
0
π(1 − 2∝ra⌈〉⌋allowy + y) dy = π
(
y − 4
3 y3/2 + 1
2 y2
) ⏐
⏐
⏐
⏐
1
0
= π
(
1 − 4
3 + 1
2
)
= π
6
y
x
a−a 0 b−b
y = f (x)
The shell method can be used for finding the volume of a
solid with a “hole” in the middle, as in the solid of revolutio n
produced by revolving the shaded region in the figure on the
right around the y-axis. The hole in the solid between x = −a
and x = a is a result of the gap between the y-axis and the
region.",ElementaryCalculus.pdf
"Surfaces and Solids of Revolution • Section 8.4 275
To find the volume V of that solid, at a point x in [ a, b) form an infinitesimal strip of width
dx from the x-axis up to the curve y = f (x), as in Figure 8.4.5(a).
y
x
0 a x b
y = f (x)
(a) Revolve region around y-axis
f (x) f (x + dx)
y
x
ds
dx
dy
(b) Infinitesimal strip
f (x)
x dx
(c) Cylindrical shell
Figure 8.4.5 Shell method
Just like the strip in the disc method, the right triangle at t he top of this strip—as in Figure
8.4.5(b)—has zero area and thus does not contribute to the vo lume dV of the right circular
cylindrical shell swept out by the strip, shown in Figure 8.4 .5(c). The volume of that shell is
just the volume of the “outer” cylinder of radius x +dx minus the volume of the “inner” cylinder
of radius x, both with height f (x):
dV = π(x + dx)2 f (x) − π x2 f (x)
= ✘✘✘✘π x2 f (x) + 2 π x f (x) dx + π✟✟✟ ✯0
(dx)2 f (x) − ✘✘✘✘π x2 f (x)
= 2 π x f (x) dx",ElementaryCalculus.pdf
"dV = π(x + dx)2 f (x) − π x2 f (x)
= ✘✘✘✘π x2 f (x) + 2 π x f (x) dx + π✟✟✟ ✯0
(dx)2 f (x) − ✘✘✘✘π x2 f (x)
= 2 π x f (x) dx
The volume V of the entire solid is then the sum of those volumes dV, using an absolute value
to handle any sign for f (x):
The volume V of the solid of revolution obtained by revolving the region b etween the curve
y = f (x) and the x-axis around the y-axis for 0 ≤ a ≤ x ≤ b is:
V =
∫ b
a
dV =
∫ b
a
2 π x
⏐
⏐ f (x)
⏐
⏐ dx (8.16)
Example 8.22
Suppose the region bounded by the curve y = x2 and the x-axis for 0 ≤ x ≤ 1 is revolved around the
y-axis. Find the volume of the resulting solid of revolution.
y
x
0 x 1
y = x21
Solution: The region is shaded in the figure on the right. The vertical st rip at
x in [0 , 1) with infinitesimal width dx and height
⏐
⏐ f (x)
⏐
⏐ = f (x) is shown in the
figure. That strip produces the shell with volume d V in formula (8.16), so by the
shell method the volume V of the solid of revolution is:
V =
∫ 1
0
d V =
∫ 1
0
2 π x
⏐
⏐ f (x)
⏐",ElementaryCalculus.pdf
"shell method the volume V of the solid of revolution is:
V =
∫ 1
0
d V =
∫ 1
0
2 π x
⏐
⏐ f (x)
⏐
⏐ dx =
∫ 1
0
2 π x · x2 dx = π
2 x4
⏐
⏐
⏐
⏐
1
0
= π
2",ElementaryCalculus.pdf
"276 Chapter 8 • Applications of Integrals §8.4
The volume dV in formula (8.16) can be generalized to dV = 2πr hw, where r is the distance
from the axis of revolution to a generic vertical strip of infi nitesimal width w in the region, and
h is the height of the strip.
Example 8.23
Suppose the region bounded by the curves y = x2 and y = x is revolved around the y-axis. Find the
volume of the resulting solid of revolution.
y
x
0 x 1
y = x
y = x2
1
Solution: The region is shaded in the figure on the right, along with a ver tical
strip with infinitesimal width w = dx at the distance r = x from the y-axis in
the region and height h = x − x2 . That strip produces the shell with volume
d V = 2πrhw = 2πx(x − x2) dx, so that the volume V of the solid of revolution is:
V =
∫ 1
0
d V =
∫ 1
0
2 π x (x − x2) dx = 2π
3 x3 − π
2 x4
⏐
⏐
⏐
⏐
1
0
= π
6
Exercises
A
For Exercises 1-3, find the surface area of the surface of revo lution produced by revolving the given",ElementaryCalculus.pdf
"3 x3 − π
2 x4
⏐
⏐
⏐
⏐
1
0
= π
6
Exercises
A
For Exercises 1-3, find the surface area of the surface of revo lution produced by revolving the given
curve around the x-axis for the given interval.
1. y =
∝ra⌈〉⌋allow
4 − x2 for 1 ≤ x ≤ 2 2. y = cosh x for 0 ≤ x ≤ 1 3. y = x3
6 + 1
2x for 1 ≤ x ≤ 3
For Exercises 4-6, find the volume of the solid of revolution p roduced by revolving the region between
the given curve and the x-axis around the x-axis for the given interval.
4. y = x3 for 0 ≤ x ≤ 1 5. y = sin x for 0 ≤ x ≤ π 6. y = ∝ra⌈〉⌋allowx for 0 ≤ x ≤ 1
For Exercises 7-9, find the volume of the solid of revolution p roduced by revolving the region between
the given curve and the x-axis around the y-axis for the given interval.
7. y = sin ( x2) for 0 ≤ x ≤ ∝ra⌈〉⌋allowπ 8. y = sin x for 0 ≤ x ≤ π 9. y = x2 − x3 for 0 ≤ x ≤ 1
10. Revolve the region in Example 8.23 around the line x = 1 and find the volume of the resulting solid.
11. Revolving the ellipse x2
a2 + y2",ElementaryCalculus.pdf
"10. Revolve the region in Example 8.23 around the line x = 1 and find the volume of the resulting solid.
11. Revolving the ellipse x2
a2 + y2
b2 = 1 around the x-axis produces an ellipsoid, for a > b > 0. Show that
the surface area of the ellipsoid is 2 πb2 (
1 + a
eb sin−1 e
)
, where e is the eccentricity of the ellipse.
12. Show that the volume inside the ellipsoid from Exercise 11 is 4
3 πa b 2 .
13. Find the surface area and volume of a right circular cone of ra dius r and height h.
14. Formulas (8.13), (8.15) and (8.16) can be extended to includ e regions over infinite intervals—the
integrals in those formulas simply become improper integra ls. Consider the region between the
curve y = 1
x and the x-axis over the interval [1 , ∞). Revolve that region around the x-axis.
(a) Show that the surface area of the resulting surface of revolu tion is infinite.
(b) Show that the volume of the resulting solid of revolution is π.",ElementaryCalculus.pdf
"(b) Show that the volume of the resulting solid of revolution is π.
15. For 0 < a < b, revolving the region inside the circle ( x − b)2 + y2 = a2 around the y-axis produces a
donut-shaped solid of revolution called a torus. Show that the volume of the torus is 2 π2 a2 b.
16. Use formula (8.14) and symmetry to show that the torus from Ex ercise 15 has surface area 4 π2 ab.",ElementaryCalculus.pdf
"Applications in Physics and Statistics • Section 8.5 277
8.5 Applications in Physics and Statistics
This chapter concludes with a few applications showing how s ome familiar discrete sums can
be replaced by integrals, which are essentially continuous sums.
Center of Gravity
Suppose a thin uniform rod has n > 1 masses m1, . . . , mn attached, with m1 and mn at the
ends. The center of gravity of the masses is the point where—due to the Earth’s gravity—t he
rod would be balanced if a fulcrum were placed there (see Figu re 8.5.1(a)). Imagine the rod
as part of the x-axis and the weights as point masses—with each mass mk at xk —and let the
center of gravity be at ¯x, as in Figure 8.5.1(b).
m1 m2 ··· mn−1 mn
(a) Rod balanced on fulcrum
x
x1 x2 ¯x x n−1 xn
m1 m2 ··· mn−1 mn
(b) Masses and ¯x on x-axis
Figure 8.5.1 Center of gravity ¯x for masses m1 , . . . , mn
The rod is balanced if the masses do not rotate the rod, i.e. th e total torque is zero. Torque",ElementaryCalculus.pdf
"The rod is balanced if the masses do not rotate the rod, i.e. th e total torque is zero. Torque
is defined here as force times position relative to ¯x. Each mass mk applies a force mk g to the
rod—where g is the (downward) acceleration due to the Earth’s gravity—a t position ( xx − ¯x)
relative to ¯x. The total torque is thus zero if
(m1 g) ( x1 − ¯x) + (m2 g) ( x2 − ¯x) + ··· + (mn g) ( xn − ¯x) = 0
so that solving for ¯x yields:
¯x = m1 g x1 + ··· +mn g xn
m1 g + ··· +mn g = m1 x1 + ··· +mn xn
m1 + ··· +mn
=
∑ n
k=1 mk xk
∑ n
k=1 mk
(8.17)
Each quantity mk xk is called the moment of the mass mk . Thus, ¯x is the sum of the moments
divided by the total mass. This idea can be extended to region s in the x y-plane, using an
integral of a continuum of moments instead of a finite sum. The center of gravity of a planar
region is defined as the point such that any force along a line t hrough that point produces no
rotation of the region about that line.",ElementaryCalculus.pdf
"region is defined as the point such that any force along a line t hrough that point produces no
rotation of the region about that line.
9 There should thus be zero torque in both the x and y
directions, so the idea is to apply formula (8.17) in both dir ections to obtain the region’s center
of gravity ( ¯x, ¯y).
9For a proof that such a point exists, see p.206 in B R O W N, F .L., Engineering Mechanics , 2nd ed., New Y ork: John
Wiley & Sons, Inc., 1942. Some texts use the terms “center of m ass” or “centroid” instead of “center of gravity ,” and
there are differences in the meanings. However , for the situ ation presented here, where the gravitational field is
assumed to have constant magnitude and direction throughou t the region, they all mean the same thing.",ElementaryCalculus.pdf
"278 Chapter 8 • Applications of Integrals §8.5
A region can be thought of as a lamina—a thin plate with uniform density . Take the area of
the region as its mass, which makes sense given the uniform de nsity . For the region between
two curves y = f1(x) and y = f2(x) over [ a, b], with f1(x) ≥ f2(x), take a vertical slice of width
dx at some x, as in Figure 8.5.2(a). By the same arguments used in Section 8.4, all the area
from that strip comes from the rectangle of height f1(x) − f2 (x) and width dx (see the shaded
rectangle in Figure 8.5.2(b)).
y
x
0 a x b
y = f1 (x)
y = f2 (x)
(a) Region between y = f1 (x) and y = f2 (x)
f1 (x) − f2 (x)
y
x
ds
dx
dy
(x + 1
2 dx, 1
2 ( f1 (x) + f2(x)))
(b) Infinitesimal strip over [ x, x + dx]
Figure 8.5.2 Center of gravity of a region
By the assumption of uniform density , the center of gravity o f that rectangle is clearly its
geometric center , whose coordinates are
(
x + 1
2 dx, 1
2 ( f1(x) + f2(x))
)
. The entire mass of the strip",ElementaryCalculus.pdf
"geometric center , whose coordinates are
(
x + 1
2 dx, 1
2 ( f1(x) + f2(x))
)
. The entire mass of the strip
can be treated as if it is concentrated at that point. The moment mx of the strip about the
x-axis is its mass times the position of its center of gravity r elative to the x-axis (i.e. its y
coordinate):
mx = ( f1(x) − f2(x)) dx · ( 1
2 ( f1(x) + f2(x))) = 1
2 (( f1(x))2 − ( f2(x))2) dx
Similarly the moment m y of the strip about the y-axis is its mass times the x coordinate of its
center of gravity:
m y = ( f1(x) − f2(x)) dx · (x + 1
2 dx) = x ( f1(x) − f2(x)) dx + 1
2 ( f1(x) − f2 (x)) (dx)2
= x ( f1(x) − f2(x)) dx
The moments Mx and M y of the entire region about the x-axis and y-axis, respectively , are
defined as the sum of the respective moments mx and m y of all strips over [ a, b]:
Mx =
∫ b
a
mx =
∫ b
a
1
2 (( f1(x))2 − ( f2(x))2) dx and M y =
∫ b
a
m y =
∫ b
a
x ( f1(x) − f2(x)) dx",ElementaryCalculus.pdf
"Mx =
∫ b
a
mx =
∫ b
a
1
2 (( f1(x))2 − ( f2(x))2) dx and M y =
∫ b
a
m y =
∫ b
a
x ( f1(x) − f2(x)) dx
Note in formula (8.17) that the denominator is the sum of all t he masses in the system. For
the region that total mass would simply be its area M:
M =
∫ b
a
( f1(x) − f2(x)) dx
Dividing the moments Mx and M y by M yields the formula for the center of gravity:",ElementaryCalculus.pdf
"Applications in Physics and Statistics • Section 8.5 279
The center of gravity ( ¯x, ¯y) of the region between the curves y = f1(x) and y = f2(x) over
[a, b], with f1(x) ≥ f2(x), is given by:
¯x =
M y
M =
∫ b
a
x ( f1(x) − f2(x)) dx
∫ b
a
( f1(x) − f2(x)) dx
and ¯y = Mx
M =
∫ b
a
1
2 (( f1(x))2 − ( f2(x))2) dx
∫ b
a
( f1(x) − f2(x)) dx
(8.18)
Example 8.24
Find the center of gravity of the region bounded by the curve y = x2 and the x-axis for 0 ≤ x ≤ 1.
y
x
0 ¯x 1
y = x21
¯y
Solution: The region is shaded in the figure on the right. Using y = f1 (x) = x2
and y = f2 (x) = 0 in formula (8.18) yields
Mx =
∫ 1
0
1
2 ( f1 (x))2 dx =
∫ 1
0
1
2 x4 dx = 1
10 x5
⏐
⏐
⏐
⏐
1
0
= 1
10
M y =
∫ 1
0
x f 1(x) dx =
∫ 1
0
x3 dx = 1
4 x4
⏐
⏐
⏐
⏐
1
0
= 1
4
M =
∫ 1
0
f1 (x) dx =
∫ 1
0
x2 dx = 1
3 x3
⏐
⏐
⏐
⏐
1
0
= 1
3
so that the center of gravity ( ¯x, ¯y) is:
¯x =
M y
M = 1/4
1/3 = 3
4 and ¯y = Mx
M = 1/10
1/3 = 3
10
Example 8.25
y
x
0 ¯x 1
y = x
y = x2
1
¯y",ElementaryCalculus.pdf
"3
so that the center of gravity ( ¯x, ¯y) is:
¯x =
M y
M = 1/4
1/3 = 3
4 and ¯y = Mx
M = 1/10
1/3 = 3
10
Example 8.25
y
x
0 ¯x 1
y = x
y = x2
1
¯y
Find the center of gravity of the region bounded by the curves y = x and y = x2.
Solution: The region is shaded in the figure on the right. Using y = f1 (x) = x and
y = f2 (x) = x2 in formula (8.18) yields
Mx =
∫ 1
0
1
2 (( f1 (x))2 − ( f2(x))2) dx =
∫ 1
0
1
2 (x2 − x4) dx = 1
6 x3 − 1
10 x5
⏐
⏐
⏐
⏐
1
0
= 1
15
M y =
∫ 1
0
x ( f1(x) − f2 (x)) dx =
∫ 1
0
(x2 − x3) dx = 1
3 x3 − 1
4 x4
⏐
⏐
⏐
⏐
1
0
= 1
12
M =
∫ 1
0
( f1(x) − f2 (x)) dx =
∫ 1
0
(x − x2) dx = 1
2 x2 − 1
3 x3
⏐
⏐
⏐
⏐
1
0
= 1
6
so that the center of gravity ( ¯x, ¯y) is:
¯x =
M y
M = 1/12
1/6 = 1
2 and ¯y = Mx
M = 1/15
1/6 = 2
5",ElementaryCalculus.pdf
"280 Chapter 8 • Applications of Integrals §8.5
W ork
Suppose that a constant force displaces an object along a lin e in the same direction in which
the force is applied. The work done by the force is defined as the force times the displacemen t.
For example, if the constant force F moves an object from position x = a to x = b on the x-axis,
as in Figure 8.5.3(a), then the work W done by the force is:
W = force × displacement = force × (final position − initial position) = F · (b − a)
x
a b
F
(a) Constant force F
dx
F (x)
F (x + dx)
(b) V ariable force F over [ x, x + dx]
Figure 8.5.3 Work W as the effect of a force F displacing an object from x = a to x = b
Suppose now that the force F is a function of position x over [ a, b]: F = F (x). By the Micros-
traightness Property , over an infinitesimal interval [ x, x + dx] the curve y = F (x) is a straight
line, as in Figure 8.5.3(b). How should the work d W performed by F over this infinitesimal in-",ElementaryCalculus.pdf
"line, as in Figure 8.5.3(b). How should the work d W performed by F over this infinitesimal in-
terval be defined? After all, F is not constant over [ x, x+dx]—it takes every value between F (x)
and F (x + dx). It is left as an exercise to show that any value in that range can be used—they
all result in the same amount F (x) dx for the work performed. 10
For example, suppose you use the value halfway between F (x) and F (x + dx) as the value of
F : 1
2 (F (x) + F (x + dx)). Then the work d W as force times displacement is:
d W = 1
2 (F (x) + F (x + dx)) dx = 1
2 (F (x) + F (x) + F ′(x) dx) dx
= F (x) dx + 1
2 F ′(x)✟✟✟ ✯0
(dx)2
= F (x) dx
Define the total work W over [ a, b] as the sum of all the d W:
The work performed by a force F (x) in displacing an object along the x-axis from x = a to
x = b is:
W =
∫ b
a
d W =
∫ b
a
F (x) dx (8.19)
10Note how this is different than claiming that F is “essentially constant over small intervals,” as most tex tbooks",ElementaryCalculus.pdf
"a
d W =
∫ b
a
F (x) dx (8.19)
10Note how this is different than claiming that F is “essentially constant over small intervals,” as most tex tbooks
do. Instead, the additional infinitesimal force beyond F (x) contributes zero work over [ x, x + dx].",ElementaryCalculus.pdf
"Applications in Physics and Statistics • Section 8.5 281
Before continuing, some possible confusion needs to be clea red up. First, force is always a
vector—it has both a magnitude and a direction. For the forces consi dered here, which act in
a single dimension (e.g. along the x-axis), by convention the direction of the force is indicate d
by its sign: positive in the direction toward +∞, negative in the direction toward −∞. So a
force of 3 N acts in the opposite direction as a force of −3 N, but they have the same magnitude
|3| =3.
Second, work is not a vector—it is a scalar, meaning it has a magnitude but no direction.
That magnitude can have any sign, though. Work is positive if the object is displaced in the
same direction as the force, but is negative if the displacem ent is in the opposite direction of
the force. For example, if you lift an object straight up from the ground, then you did positive",ElementaryCalculus.pdf
"the force. For example, if you lift an object straight up from the ground, then you did positive
work—the object moved in the same direction as the force you u sed. However , the force of
gravity did negative work on the object as you lifted, since g ravity works downward yet the
object moved upward.
FFµ
N
−m g
Last, zero work is done by a force if no displacement in its dir ection oc-
curs. In particular , forces acting perpendicular to the lin e of displacement
perform no work. For example, consider an object of mass m on a flat hori-
zontal table top as in the figure on the right. If you push that o bject to the
right with a force F (performing positive work), then both the downward
force of gravity −m g and the upward normal force N exerted by the table
perform zero work on the object. The force of friction Fµ from the table surface does negative
work, as it opposes the force F . As an another example, no work is performed by holding a 100
lb object still and above the ground.",ElementaryCalculus.pdf
"work, as it opposes the force F . As an another example, no work is performed by holding a 100
lb object still and above the ground.
Example 8.26
x
0
compress stretchHooke’s law states that a coiled spring has an elastic restoring force
F = −kx , where x is the displacement of the end of the spring from
its equilibrium position as the spring is stretched or compr essed,
and k > 0 is the spring constant —or stiffness coefficient —specific to
the spring. This force always tries to restore the spring to i ts equi-
librium position, and the law holds only for a limited range o f x.
For a spring laid horizontally imagine it lies on the x-axis with the
equilibrium position at x = 0, as in the figure on the right.
(a) Find the spring constant k if a force of 2 N stretches the spring by 4 cm.
(b) Use part (a) to find the work performed by compressing the spri ng 3 cm.
Solution: (a) The force required to stretch the spring by an amount x is F = kx , since that force must",ElementaryCalculus.pdf
"Solution: (a) The force required to stretch the spring by an amount x is F = kx , since that force must
counter the restoring force. Thus, k = F
x = 2N
4cm = 2N
0.04m = 50 N/m.
(b) By part (a) the force required to compress the string to po sition x is F (x) = kx = 50x, since again it
must counter the restoring force. Thus, since 3 cm is 0 .03 m, the work W performed is:
W =
∫ −0.03
0
F (x) dx =
∫ −0.03
0
50x dx = 25x2
⏐
⏐
⏐
⏐
−0.03
0
= 25 (−0.03)2 − 0 = 0.0225 Nm",ElementaryCalculus.pdf
"282 Chapter 8 • Applications of Integrals §8.5
Probability
Suppose you flip two evenly balanced pennies and let X be the number of heads in the result.
Then X is a discrete random variable —discrete because it can take only a discrete set of
values (0, 1 and 2); random because its value is left to chance . The probability of a penny
being flipped heads is 50% = 1
2 , i.e. that is its theoretical likelihood since heads and tai ls are
equally likely . The sample space S of all possible outcomes is the set S = {T T , T H , H T , H H },
where H is heads and T is tails (e.g. H T means the first penny came up heads and the second
came up tails). Figure 8.5.4(a) shows a bar chart of the proba bilities—as numbers between
0 and 1—with P (X = x) denoting the probability of the event that X equals the number x.
Notice that the sum of the probabilities is 1, and P (X = x) = 0 if x is not 0, 1, or 2.
P (X = x)
x
0 1 2
1
4
1
2
probability
(a) Discrete: P (X = x)
f (x)
x
a b",ElementaryCalculus.pdf
"P (X = x)
x
0 1 2
1
4
1
2
probability
(a) Discrete: P (X = x)
f (x)
x
a b
(b) Continuous: P (a < X < b)
Figure 8.5.4 Probability: discrete vs continuous random variables
The idea behind a continuous random variable X is to fill in those gaps between the
bars in Figure 8.5.4(a), so that X would represent a continuous quantity , e.g. time, distance ,
temperature. Rather than finding P (X = x) you would find the probability that X is in a
continuum such as an interval, e.g. P (a < X < b) (see Figure 8.5.4(b)).
For a continuous random variable X define P (X = x) = 0 for all real x, and define the
probability density function for X as a function f (x) ≥ 0 such that:
(a)
∫ ∞
−∞
f (x) dx = 1
(b) P (a < X < b) =
∫ b
a
f (x) dx for all a < b (including a = −∞and b = ∞)
Notice that since P (X = a) = 0 then P (a ≤ X < b) = P (a < X < b). In general, < and ≤ are
interchangeable for events involving continuous random va riables (as well as > and ≥). In",ElementaryCalculus.pdf
"interchangeable for events involving continuous random va riables (as well as > and ≥). In
the remainder of this section it will be assumed that all rand om variables are continuous, for
which the sample space is typically all of R or some interval, finite or infinite (e.g. (0 , ∞)).",ElementaryCalculus.pdf
"Applications in Physics and Statistics • Section 8.5 283
Example 8.27
Let X be the lifetime—i.e. the time to failure—of an electronic co mponent. If the average lifetime of the
component is 700 days, then the probability density functio n f (x) for the random variable X is
f (x) =
{
λ e−λx if x ≥ 0,
0 if x < 0
(8.20)
where λ = 1
700 and x is the number of days. In this case X is said to have the exponential distribution
with parameter λ. Find the probability that the lifetime of the component is:
(a) between 600 and 800 days
(b) greater than 700 days
Solution: (a) The probability is:
P (600 < X < 800) =
∫ 800
600
f (x) dx =
∫ 800
600
1
700 e− x
700 dx = −e− x
700
⏐
⏐
⏐
⏐
800
600
= −e− 800
700 + e− 600
700 ≈ 0.1055
Thus, there is about a 10 .55% chance that the component’s lifetime will be between 600 and 800 days.
(b) The probability is:
P (X > 700) =
∫ ∞
700
f (x) dx =
∫ ∞
700
1
700 e− x
700 dx = −e− x
700
⏐
⏐
⏐
⏐
∞
700
= 0 + e−1 ≈ 0.3679
Exercises
A",ElementaryCalculus.pdf
"(b) The probability is:
P (X > 700) =
∫ ∞
700
f (x) dx =
∫ ∞
700
1
700 e− x
700 dx = −e− x
700
⏐
⏐
⏐
⏐
∞
700
= 0 + e−1 ≈ 0.3679
Exercises
A
For Exercises 1-3, find the center of gravity of the region bou nded by the given curves over the given
interval.
1. y = x3 and y = 0 ; 0 ≤ x ≤ 1 2. y = −x + 1 and y = 0 ; 0 ≤ x ≤ 1 3. y = x2 and y = x3 ; 0 ≤ x ≤ 1
4. Find the center of gravity of the region inside the circle x2 + y2 = r2 and above the x-axis.
5. Find the center of gravity of the region inside the circle x2 + y2 = r2 in the first quadrant.
6. Find the center of gravity of the region inside the ellipse x2
a2 + y2
b2 = 1 and above the x-axis.
7. Find the center of gravity of the region between the circle x2 + y2 = 4 and the ellipse x2
4 + y2 = 1 above
the x-axis.
8. Would formula (8.18) for the center of gravity change if the m ass of a region were proportional—but
not equal—to its area, say , by a constant positive proportio n δ ̸=1? Explain.",ElementaryCalculus.pdf
"not equal—to its area, say , by a constant positive proportio n δ ̸=1? Explain.
9. If a spring requires 3 N of force to be compressed 5 cm, how much work would be performed in
stretching the spring 8 cm?
10. The gravitational force F (x) exerted by the Earth on an object of mass m at a distance x from the
center of the Earth is
F (x) = − m gr 2
e
x2
where r e is the radius of the Earth. If the object is released from rest at a distance r o from the center
of the Earth, find the work performed by gravity in bringing th e object to the Earth’s surface.",ElementaryCalculus.pdf
"284 Chapter 8 • Applications of Integrals §8.5
11. Recall that the ideal gas law states that P V = R T , where R is a constant, P is the pressure, V is
the volume, and T is the temperature. It can be shown that the work W done by an ideal gas in
expanding the volume from Va to Vb is
W =
∫ Vb
Va
P d V .
Calculate W .
12. V erify that
∫ ∞
−∞
f (x) dx = 1 for the function f (x) in formula (8.20) in Example 8.27 for all λ > 0.
13. Find P (X < 300) in Example 8.27.
14. The distribution function F (x) for a random variable X is defined as F (x) = P (X ≤ x) for all x. Show
that F ′(x) = f (x), where f (x) is the probability density function for X .
B
15. Formula (8.18) can be extended to regions over an infinite int erval, provided the area is finite. Use
that fact to find the center of gravity of the region between y = e−x and the x-axis for 0 ≤ x < ∞.
16. The expected value (or mean) E[X ] of a random variable X with probability density function f (x) is
E[X ] =
∫ ∞
−∞
x f (x) dx .",ElementaryCalculus.pdf
"16. The expected value (or mean) E[X ] of a random variable X with probability density function f (x) is
E[X ] =
∫ ∞
−∞
x f (x) dx .
Show that E[X ] = 1
λ if X has the exponential distribution with parameter λ > 0.
Note: The expected value can be thought of as the weighted ave rage of all possible values of X , with
weights determined by probability . It is analogous to the id ea of a center of gravity .
17. A random variable X is said to have a normal distribution if its probability density function f (x) is
f (x) = 1
σ
∝ra⌈〉⌋allow
2π
e
(x−µ)2
2σ2 for all x
where σ > 0 and µ are constants. This is the famous “bell curve” in statistics .
(a) V erify that
∫ ∞
−∞
f (x) dx = 1. (Hint: Use Example 6.25 and a substitution.)
(b) Show that E[X ] = µ.
(c) Use numerical integration to show that P (−1 < X < 1) ≈ 0.6827 when µ = 0 and σ = 1.
18. A random variable X has the beta distribution if its probability density function f (x) is
f (x) =
{ 1
B(a,b) xa−1 (1 − x)b−1 if 0 ≤ x ≤ 1
0 elsewhere",ElementaryCalculus.pdf
"f (x) =
{ 1
B(a,b) xa−1 (1 − x)b−1 if 0 ≤ x ≤ 1
0 elsewhere
for positive constants a and b, where B(a, b) is the Beta function. Show that E[X ] = a
a+b .
19. Show that any value between F (x) and F (x +dx) for the force over [ x, x +dx] gives the same formula
d W = F (x) d x for the work performed over that interval. (Hint: Consider F (x + αdx) for 0 ≤ α ≤ 1.)
20. A drop of water of mass M is released from rest at a height sufficient for the drop to eva porate
completely , losing mass m each second (i.e. at a constant rate). Ignoring air resistan ce, show that
the work performed by gravity on the drop up to complete evapo ration is g2 M2
6m2 .",ElementaryCalculus.pdf
"Applications in Physics and Statistics • Section 8.5 285
21. This exercise is related to Einstein’s famous law E = mc2 . The relativistic momentum p of a particle
of mass m moving at a speed v along a straight line (say , the x-axis) is
p = mv√
1 − v2
c2
,
where c is the speed of light. The relativistic force on the particle along that line is
F = dp
dt ,
which is the same formula as Newton’s Second Law of motion in c lassical mechanics. Assume that
the particle starts at rest at position x1 and ends at position x2 along the x-axis. The work done by
the force F on the particle is:
W =
∫ x2
x1
F dx =
∫ x2
x1
dp
dt dx
(a) Show that
dp
dv = m
(
1 − v2
c2
) 3/2 .
(b) Use the Chain Rule formula
dp
dt = dp
dv
dv
dx
dx
dt
to show that
F dx = v dp
dv dv .
(c) Use parts (a) and (b) to show that
W =
∫ v
0
dp
dv v dv =
∫ v
0
mv
(
1 − v2
c2
) 3/2 dv .
(d) Use part (c) to show that
W = mc2
√
1 − v2
c2
− mc2 .",ElementaryCalculus.pdf
"W =
∫ v
0
dp
dv v dv =
∫ v
0
mv
(
1 − v2
c2
) 3/2 dv .
(d) Use part (c) to show that
W = mc2
√
1 − v2
c2
− mc2 .
(e) Define the relativistic kinetic energy K of the particle to be K = W , and define the total energy E
to be
E = mc2
√
1 − v2
c2
.
So by part (d), K = E − mc2 . Show that
E2 = p2 c2 + (mc2 )2 .
(Hint: Expand the right side of that equation.)
(f) What is E when the particle is at rest?
22. A median of a triangle is a line segment from a vertex to the midpoint of the opposite side, and the
three medians intersect at a common point. Show that this poi nt is a triangle’s center of gravity .",ElementaryCalculus.pdf
"CHAPTER 9
Infinite Sequences and Series
9.1 Sequences and Series
In the 5 th century B.C. the ancient Greek philosopher Zeno of Elea devi sed several paradoxes,
the most famous of which— The Dichotomy —asserts that if space is infinitely divisible then
motion is impossible. The argument goes like this: imagine a line segment of finite length, say
1 m, with a person at one end as in Figure 9.1.1.
0 . . . 1
8
1
4
1
2 1
Figure 9.1.1 Zeno’s motion paradox
Before traversing the entire distance the person would first need to travel one half the dis-
tance. Before doing that, though, he would need to travel one fourth the distance, and before
that one eighth the distance, and so on. There is thus no “first ” distance for him to traverse,
meaning his motion cannot even begin!
Obviously motion is possible, or you would not be reading thi s. Does that mean Zeno’s
reasoning is flawed? More on that later . In the meantime, noti ce a few things in Figure 9.1.1.
First, the distance markers 1
2 , 1
4 , 1",ElementaryCalculus.pdf
"reasoning is flawed? More on that later . In the meantime, noti ce a few things in Figure 9.1.1.
First, the distance markers 1
2 , 1
4 , 1
8 , . . . form an infinite sequence of numbers approaching
0. Second, the sum of the distances between successive marke rs is an infinite series which
should equal the total segment length 1:
1
2 + 1
4 + 1
8 + ··· = 1
It will be shown shortly that the sum is indeed 1, which turns o ut to have no bearing on Zeno’s
paradox. First some definitions are needed.
286",ElementaryCalculus.pdf
"Sequences and Series • Section 9.1 287
A sequence is an ordered list of objects, which in this book will always b e real numbers.
Sequences can be finite or infinite: finite if there is a last num ber in the list, infinite if every
number in the list is followed by another number (i.e. a “succ essor”). Sequences should not be
confused with sets—order matters in a sequence but not in a set, and numbers may r epeat in
a sequence but not in a set.
For example, the sequences 〈1, 2, 3〉 and 〈1, 3, 2〉 are different, since order matters. However ,
the numbers in those sequences comprise the same set {1, 2, 3}.
The simplest example of an infinite sequence is N, the set of natural numbers: 0 , 1, 2, 3, . . . .
In fact, the numbers a0, a1, a2, . . . in any infinite sequence of real numbers can be written as
the range of a function f mapping N into R:
f (n) = an (9.1)
Typically notation such as { an }∞
n=0 is used for representing an infinite sequence, or simply",ElementaryCalculus.pdf
"f (n) = an (9.1)
Typically notation such as { an }∞
n=0 is used for representing an infinite sequence, or simply
{ an } when the initial value of the index n is understood (and n is always an integer). The
intuitive notion of the limit of an infinite sequence can be stated formally:
A real number L is the limit of an infinite sequence { an }, written as
limn→∞ an = L or simply an → L ,
if for any given number ǫ > 0 there exists an integer N such that
⏐
⏐an − L
⏐
⏐ < ǫ for all n > N . (9.2)
In this case the sequence { an } is said to converge to L and is called a convergent se-
quence. A sequence that is not convergent is divergent.
In other words, a sequence { an } converges to L if the terms an can be made arbitrarily close
to L for n sufficiently large. In most cases the formal definition will n ot be needed, since by
formula (9.1) the same rules and formulas from Chapters 1 and 3 for the limit of a function",ElementaryCalculus.pdf
"formula (9.1) the same rules and formulas from Chapters 1 and 3 for the limit of a function
f (x) as x approaches ∞ apply to sequences (e.g. sums and products of limits, L ’Hôpi tal’s Rule).
All you have to do is replace x by n.
Example 9.1
For integers n ≥ 1 define an = 1
2n . Find limn→∞ an if it exists.
Solution: Since limx→∞
1
2x = 0 then replacing x by n shows that
limn→∞ an = limn→∞
1
2n = 0 .",ElementaryCalculus.pdf
"288 Chapter 9 • Infinite Sequences and Series §9.1
Example 9.2
For integers n ≥ 0 define an = 2n+1
3n+2 . Is { an } a convergent sequence? If so then find its limit.
Solution: By L ’Hôpital’s Rule, treating an integer n ≥ 0 as a real-valued variable x,
limn→∞ an = limn→∞
2n + 1
3n + 2 = limn→∞
2
3 = 2
3 .
Thus the sequence is convergent and its limit is 2
3 .
Example 9.3
For integers n ≥ 0 define an = en
3n+2 . Is { an } a convergent sequence? If so then find its limit.
Solution: By L ’Hôpital’s Rule, treating an integer n ≥ 0 as a real-valued variable x,
limn→∞ an = limn→∞
en
3n + 2 = limn→∞
en
3 = ∞
Thus the sequence is divergent.
Example 9.4
The famous Fibonacci sequence 1 { Fn } starts with the numbers 0 and 1, then each successive term is
the sum of the previous two terms: F0 = 0, F1 = 1,
Fn = Fn−1 + Fn−2 for integers n ≥ 2 (9.3)
Equation (9.3) is a recurrence relation . The first ten Fibonacci numbers are 0 , 1, 1, 2, 3, 5, 8, 13, 21, 34.",ElementaryCalculus.pdf
"Equation (9.3) is a recurrence relation . The first ten Fibonacci numbers are 0 , 1, 1, 2, 3, 5, 8, 13, 21, 34.
Clearly { Fn } is a divergent sequence, since Fn → ∞. For n ≥ 2 define an = Fn /Fn−1 . The first few values
are:
a2 = F2
F1
= 1
1 = 1 , a3 = F3
F2
= 2
1 = 2 , a4 = F4
F3
= 3
2 = 1.5 , a5 = F5
F4
= 5
3 ≈ 1.667
Show that { an } is convergent. In other words, in the Fibonacci sequence the ratios of each term to the
previous term converge to some number .
Solution: There are many ways to prove this, perhaps the simplest being to assume the sequence is
convergent and then find the limit (which would be impossible if the sequence were divergent). So
assume that an → a for some real number a. Then divide both sides of formula (9.3) by Fn−1, so that
Fn
Fn−1
= 1 + Fn−2
Fn−1
⇒ an = 1 + 1
an−1
for integers n ≥ 2
Now take the limit of both sides of the last equation as n → ∞:
limn→∞ an = limn→∞
(
1 + 1
an−1
)
⇒ a = 1 + 1
a ⇒ a2 − a − 1 = 0 ⇒ a = 1 ±
∝ra⌈〉⌋allow
5
2",ElementaryCalculus.pdf
"limn→∞ an = limn→∞
(
1 + 1
an−1
)
⇒ a = 1 + 1
a ⇒ a2 − a − 1 = 0 ⇒ a = 1 ±
∝ra⌈〉⌋allow
5
2
Since a must be positive then a = 1+
∝ra⌈〉⌋allow
5
2 . Thus, the sequence is convergent and converges to 1+
∝ra⌈〉⌋allow
5
2 ≈ 1.618.
Note: This number is the famous golden ratio , the subject of many claims regarding its appearance in
nature and supposed aesthetic appeal as a ratio of sides of a r ectangle.2
1Due to the Italian mathematician Leonardo Fibonacci (ca. 11 70-1250).
2For example, see H U N T L E Y, H.E., The Divine Proportion , New Y ork: Dover Publications, Inc., 1970.",ElementaryCalculus.pdf
"Sequences and Series • Section 9.1 289
An infinite series is the sum of an infinite sequence. If the infinite sequence is { an }∞
n=0
then the series can be written as
∞∑
n=0
an = a0 + a1 + a2 + ··· +an + ···
or simply as ∑ an when the initial value of the index n is understood. There is a natural way
to define the sum of such a series:
The sum of an infinite series
∞∑
n=0
an is
∞∑
n=0
an = limn→∞ sn (9.4)
where { sn }∞
n=0 is the sequence of partial sums of the series:
sn =
n∑
k=0
ak = a0 + a1 + a2 + ··· +an for n ≥ 0
If the partial sums { sn } converge to a real number s then the series is convergent, and
converges to s; if { sn } diverges then the series is divergent.
One important convergent series is a geometric progression :
a + ar + ar 2 + ar 3 + ··· +ar n + ··· (9.5)
with a ̸=0 and |r | <1. Multiply the n-th partial sum sn by r:
r s n = r (a + ar + ar 2 + ar 3 + ··· +ar n−1 + ar n) = ar + ar 2 + ar 3 + ar 4 + ··· +ar n + ar n+1
Now subtract r s n from sn:",ElementaryCalculus.pdf
"r s n = r (a + ar + ar 2 + ar 3 + ··· +ar n−1 + ar n) = ar + ar 2 + ar 3 + ar 4 + ··· +ar n + ar n+1
Now subtract r s n from sn:
sn − r s n = a + ar + ar 2 + ar 3 + ··· +ar n − (ar + ar 2 + ar 3 + ar 4 + ··· +ar n + ar n+1)
sn − r s n = a − ar n+1 so that
limn→∞
sn = limn→∞
a (1 − rn+1)
1 − r = a
1 − r
since rn+1 → 0 as n → ∞when |r | <1. Thus:
For a ̸=0 and |r | <1 the geometric progression
∞∑
n=0
ar n is convergent:
∞∑
n=0
ar n = a + ar + ar 2 + ar 3 + ··· +ar n + ··· = a
1 − r (9.6)",ElementaryCalculus.pdf
"290 Chapter 9 • Infinite Sequences and Series §9.1
Example 9.5
Show that
∞∑
n=1
1
2n = 1.
Solution: This is a geometric progression with a = 1
2 and r = 1
2 . So by formula (9.6) the sum is:
∞∑
n=1
1
2n =
∞∑
n=0
(1
2
)
·
(1
2
) n
= a
1 − r =
1
2
1 − 1
2
= 1 ✓
Example 9.6
Write the repeating decimal 0 .17 = 0.17171717 . . . as a rational number .
Solution: This is a geometric progression with a = 0.17 = 17
100 and r = 0.01 = 1
100 :
0.171717 . . . = 0.17 + 0.0017 + 0.000017 + ··· = 0.17 (0.01)0 + 0.17 (0.01)1 + 0.17 (0.01)2 + ···
=
∞∑
n=0
0.17 (0.01)n = a
1 − r = 0.17
1 − 0.01 = 0.17
0.99 = 17
99
Back to Zeno’s motion paradox, by Example 9.5 the sum of the in finite number of distances
between successive markers in Figure 9.1.1 is 1, as expected . This fact is often mistaken as
proof that Zeno was wrong, yet it does not actually address Ze no’s argument, since the person
is attempting to begin motion at the tail end—the “infinite en d”—of the geometric progression,",ElementaryCalculus.pdf
"is attempting to begin motion at the tail end—the “infinite en d”—of the geometric progression,
not at the beginning (i.e. at n = 0). Zeno’s point remains that there is no “first step” at that t ail
end.
In fact, even if you were to reverse the person’s position to s tart at the other end, so that
he would first move a distance 1
2 , then a distance 1
4 , and so on, as in Figure 9.1.2, then a new
problem is introduced: the person keeps moving no matter how close he is to the end point.
There is now no “last step” and motion is thus still impossibl e.
0 . . . 1
8
1
4
1
2 1
1
2
1
4
1
8
···
Figure 9.1.2 Zeno’s motion paradox in reverse
The convergence of the geometric progression has misled man y people to argue that Zeno
is wrong, by claiming it shows an infinite number of movements can be completed in a finite
amount of time. But this line of argument fails on (at least) t wo counts. First, Zeno never
argued about time—it is irrelevant to his paradox. The more f undamental flaw is that the",ElementaryCalculus.pdf
"argued about time—it is irrelevant to his paradox. The more f undamental flaw is that the
introduction of time brings along the concept of speed, typi cally taken to be constant (though it
need not be). Speed is distance over time, but it is precisely the distance part of that ratio that
Zeno rejects—that distance can never be traveled. In other w ords, it is circular—and hence
faulty—reasoning to “prove” that motion is possible by assuming it is possible.",ElementaryCalculus.pdf
"Sequences and Series • Section 9.1 291
The geometric progression also doesn’t help when consideri ng only the distances and ignor-
ing time. Even assuming each of those distances could be trav eled, the partial sums approach
1 but never actually reach it. The limit of a sequence is defined in terms of an inequality —you
can get arbitrarily close to the limit, and that is all. The eq uality in formula (9.6) is merely
a shorthand way of saying that. It is an abstraction based on p roperties of the real number
system, not necessarily based on physical reality .
Zeno’s paradox is not purely mathematical—it is about space and hence is physical, with a
tinge of philosophy . Physicists have recognized this—and n oted the flaw in the purely math-
ematical line of attack—and devised new arguments against Z eno, some more sophisticated
than others. However , they all invariably end up in some sort of circular reasoning. All this",ElementaryCalculus.pdf
"than others. However , they all invariably end up in some sort of circular reasoning. All this
calls into question the original assumption: the infinite di visibility of space. If space had some
smallest unit that could not be divided any further then ther e is no paradox—motion over a
finite distance could always be decomposed into a large but fin ite number of irreducible steps. 3
Exercises
A
For Exercises 1-8, determine if the given sequence is conver gent. If so then find its limit.
1. { n e −n }∞
n=0 2.
{ n2
3n2 + 7n − 2
} ∞
n=1
3.
{ n2
3n3 + 7n − 2
} ∞
n=1
4.
{ n3
3n2 + 7n − 2
} ∞
n=1
5.
{ n
ln n
}∞
n=2
6.
{ n !
(n + 2) !
} ∞
n=0
7.
{
sin
(nπ
2
) } ∞
n=0
8. { (−1)n cos nπ }∞
n=0
For Exercises 9-12 determine whether the given series is con vergent. If so then find its sum.
9.
∞∑
n=0
2
(2
3
) n
10.
∞∑
n=1
7−n 11.
∞∑
n=0
3n + 5n+1
6n 12.
∞∑
n=0
n
For Exercises 13-16 use a geometric progression to write the given repeating decimal as a rational
number .
13. 0.",ElementaryCalculus.pdf
"∞∑
n=0
3n + 5n+1
6n 12.
∞∑
n=0
n
For Exercises 13-16 use a geometric progression to write the given repeating decimal as a rational
number .
13. 0.
113 14. 0.9 15. 0.249 16. 0.017
B
17. In Example 9.4 define bn = Fn /Fn+1 for n ≥ 0 and show that { bn }∞
n=0 converges to
∝ra⌈〉⌋allow
5−1
2 .
18. Show that formula (4.1) for Newton’s method with f (x) = x2 − 2 and x0 = 1 yields the sequence
{ xn }∞
n=0 with
xn = 1
2 xn−1 + 1
xn−1
for n ≥ 1 ,
then assuming { xn } is convergent show that it must converge to
∝ra⌈〉⌋allow
2. (Hint: See Example 9.4.)
3As of this writing there is not yet a definitive answer as to whe ther space is continuous or discrete (quantized).
A “smallest unit” would have to be below the Planck level —around 10 −33 cm, well below current measurement
capabilities. Some recent advances in the field of loop quantum gravity suggest the possibility of quantized space.",ElementaryCalculus.pdf
"capabilities. Some recent advances in the field of loop quantum gravity suggest the possibility of quantized space.
See C H A MS E D D I N E , A.H., C O N N E S, A. & M U K H A N OV, V ., “Geometry and the quantum: basics.” J . High Energy
Phys. 2014, 98 (2014). https://doi.org/10.1007/JHEP12(2014)098",ElementaryCalculus.pdf
"292 Chapter 9 • Infinite Sequences and Series §9.1
19. In this exercise you will prove a formula for the general numb er Fn in the Fibonacci sequence.
Denote the positive and negative solutions to the equation x2 − x −1 = 0 by φ+ = 1+
∝ra⌈〉⌋allow
5
2 and φ− = 1−
∝ra⌈〉⌋allow
5
2 ,
respectively . Note that φ+ is the golden ratio mentioned in Example 9.4.
(a) Use the equation x2 − x − 1 = 0 to show that
φn+1
+ = φn
+ + φn−1
+ and φn+1
− = φn
− + φn−1
− .
(b) Use part (a) and induction to show that
Fn = φn
+ − φn
−∝ra⌈〉⌋allow
5
for n ≥ 0.
20. A ball is dropped from a height of 4 ft above the ground, and upo n each bounce off the ground the
ball bounces straight up to a height equal to 65% of its previo us height. Find the theoretical total
distance the ball could travel if it could bounce indefinitel y . Why is this physically unrealistic?
21. A continued fraction is a type of infinite sum involving a fraction with a denominat or that continues
indefinitely:
a = a0 +
1
a1 +
1
a2 +
1
a3 + ···",ElementaryCalculus.pdf
"indefinitely:
a = a0 +
1
a1 +
1
a2 +
1
a3 + ···
(a) Show that the golden ratio φ+ = 1+
∝ra⌈〉⌋allow
5
2 (a solution of x2 − x − 1 = 0) can be written as
φ+ = 1 +
1
1 +
1
1 +
1
1 + ···
.
(Hint: Look for a recurrence relation in the fraction.)
(b) Show that
∝ra⌈〉⌋allow
2 = 1 +
1
2 +
1
2 +
1
2 + ···
.
22. Show that the golden ratio φ+ = 1+
∝ra⌈〉⌋allow
5
2 can be written as φ+ =
√
1 +
√
1 +
∝ra⌈〉⌋allow
1 + ···.
23. Write a computer program to approximate the result in Exerci se 22 using 100 terms. After how
many iterations do the approximate values start repeating?
24. Would the existence of infinitesimals as a measure of space re solve Zeno’s paradox? Explain.",ElementaryCalculus.pdf
"T ests for Convergence • Section 9.2 293
9.2 Tests for Convergence
There are many ways to determine if a sequence converges—two are listed below . In all cases
changing or removing a finite number of terms in a sequence doe s not affect its convergence or
divergence:
1. Monotone Bounded T est : A sequence that is bounded and monotone—i.e. either
always increasing or always decreasing—is convergent.
2. Comparison T est : If { bn } diverges to ∞ and if { an } is a sequence such that an ≥ bn
for all n > N for some N , then { an } also diverges to ∞.
Likewise, if { cn } diverges to −∞ and an ≤ cn for all n > N for some N , then { an } also
diverges to −∞.
x
a1 a2 a3 a4 ··· → M
Figure 9.2.1
The Comparison Test makes sense intuitively , since somethi ng
larger than a quantity going to infinity must also go to infinit y .
The Monotone Bounded Test can be understood by thinking of
a bound on a sequence as a wall that the sequence can never",ElementaryCalculus.pdf
"The Monotone Bounded Test can be understood by thinking of
a bound on a sequence as a wall that the sequence can never
pass, as in Figure 9.2.1. The increasing sequence { an } in the figure moves toward M but can
never pass it. The sequence thus cannot diverge to ∞, and it cannot fluctuate back and forth
since it always increases. Thus it must converge somewhere b efore or at M.
4 Notice that the
Monotone Bounded Test tells you only that the sequence conve rges, not what it converges to.
Example 9.7
Show that the sequence { an }∞
n=1 defined for n ≥ 1 by
an = 1 · 3 · 5 ··· (2n − 1)
2 · 4 · 6 ··· (2n)
is convergent.
Solution: Notice that { an } is always decreasing, since
an+1 = 1 · 3 · 5 ··· (2n − 1) · (2n + 1)
2 · 4 · 6 ··· (2n) · (2n + 2) = an · 2n + 1
2n + 2 < an · (1) = an
for n ≥ 1. The sequence is also bounded, since 0 < an and
an = 1
2 · 3
4 · 5
6 ··· 2n − 1
2n < 1 for n ≥ 1
since each fraction in the above product is less than 1. Thus, by the Monotone Bounded Test the",ElementaryCalculus.pdf
"an = 1
2 · 3
4 · 5
6 ··· 2n − 1
2n < 1 for n ≥ 1
since each fraction in the above product is less than 1. Thus, by the Monotone Bounded Test the
sequence is convergent.
Note that for a decreasing sequence only the lower bound is ne eded for the Monotone Bounded Test, not
the upper bound. Similarly , for an increasing sequence only the upper bound matters.
4For a proof see pp.48-49 in B U C K, R.C., Advanced Calculus , 2nd ed., New Y ork: McGraw-Hill Book Co., 1965.",ElementaryCalculus.pdf
"294 Chapter 9 • Infinite Sequences and Series §9.2
Some tests for convergence of a series are listed below:
1. n-th T erm T est : If ∑ an converges then limn→∞ an = 0.
2. Ratio T est : For a series ∑ an of positive terms let
R = limn→∞
an+1
an
.
Then
(a) if R < 1 then the series converges,
(b) if R > 1 (including R = ∞) then the series diverges,
(c) if R = 1 then the test fails.
3. Integral T est : For a series
∞∑
n=1
an of positive terms let f (x) be a decreasing function on
[1, ∞) such that f (n) = an for all integers n ≥ 1. Then
∞∑
n=1
an and
∫ ∞
1
f (x) dx
either both converge or both diverge.
4. p-series T est : The series
∞∑
n=1
1
n p converges for p > 1, and diverges for p ≤ 1.
5. Comparison T est : If 0 ≤ an ≤ bn for n > N for some N , and if ∑ bn is convergent
then ∑ an is convergent. Similarly , if 0 ≤ bn ≤ an for n > N for some N , and if ∑ bn is
divergent then ∑ an is divergent.
6. Limit Comparison T est : For two series ∑ an and ∑ bn of positive terms let
L = limn→∞
an
bn
.",ElementaryCalculus.pdf
"divergent then ∑ an is divergent.
6. Limit Comparison T est : For two series ∑ an and ∑ bn of positive terms let
L = limn→∞
an
bn
.
Then
(a) if 0 < L < ∞then ∑ an and ∑ bn either both converge or both diverge,
(b) if L = 0 and ∑ bn converges then ∑ an converges,
(c) if L = ∞and ∑ bn diverges then ∑ an diverges.
7. T elescoping Series T est : Suppose ∑ an = ∑ (bn − bn+1) for some sequence { bn }. Then∑ an converges if and only if bn → L, in which case ∑ an = b1 − L.",ElementaryCalculus.pdf
"T ests for Convergence • Section 9.2 295
Most of the above tests have fairly short proofs or at least in tuitive explanations. For exam-
ple, the n-th Term Test follows from the definition of converg ence of a series: if ∑ an converges
to a number L then since each term an = sn − sn−1 is the difference of successive partial sums,
taking the limit yields
limn→∞ an = limn→∞ (sn − sn−1) = L − L = 0
by definition of the convergence of a series. ✓
Since the n-th Term Test can never be used to prove convergenc e of a series, it is often stated
in the following logically equivalent manner:
n-th T erm T est : If limn→∞ an ̸=0 then ∑ an diverges.
Example 9.8
Show that
∞∑
n=1
n
2n + 1 = 1
3 + 2
5 + 3
7 + ··· is divergent.
Solution: Since
limn→∞
n
2n + 1 = 1
2 ̸=0
then by the n-th Term Test the series diverges.
The Ratio Test takes a bit more effort to prove. 5 When the ratio R in the Ratio Test is larger",ElementaryCalculus.pdf
"2 ̸=0
then by the n-th Term Test the series diverges.
The Ratio Test takes a bit more effort to prove. 5 When the ratio R in the Ratio Test is larger
than 1 then that means the terms in the series do not approach 0 , and thus the series diverges
by the n-th Term Test. When R = 1 the test fails, meaning it is inconclusive—another test
would need to be used. When the test shows convergence it does not tell you what the series
converges to, merely that it converges.
Example 9.9
Determine if
∞∑
n=1
n
2n is convergent.
Solution: For the series general term an = n
2n ,
R = limn→∞
an+1
an
= limn→∞
n + 1
2n+1
n
2n
= limn→∞
n + 1
2n = 1
2 < 1 ,
so by the Ratio Test the series converges.
5See pp.612-613 in T AY L O R, A.E. A N D W .R. M A N N, Advanced Calculus , 2nd ed., New Y ork: John Wiley & Sons,
Inc., 1972.",ElementaryCalculus.pdf
"296 Chapter 9 • Infinite Sequences and Series §9.2
Figure 9.2.2 shows why the Integral Test works.
a1
a2 a3 a4 a5
y
x
0 1 2 3 4 5
y = f (x)
(a)
∫ ∞
1
f (x) dx >
∞∑
n=2
an
a1
a2 a3 a4
y
x
0 1 2 3 4 5
y = f (x)
(b)
∫ ∞
1
f (x) dx <
∞∑
n=1
an
Figure 9.2.2 Integral Test
In Figure 9.2.2(a) the area
∫ ∞
1 f (x) dx is greater than the total area S of all the rectangles
under the curve. Since each rectangle has height an and width 1, then S = ∑ ∞
2 an. Thus, since
removing the single term a1 from the series does not affect the convergence or divergenc e of
the series, the series converges if the improper integral co nverges, and conversely the integral
diverges if the series diverges. Similarly , in Figure 9.2.2 (b) the area
∫ ∞
1 f (x) dx is less than the
total area S = ∑ ∞
1 an of all the rectangles, so the integral converges if the serie s converges, and
the series diverges if the integral diverges. Notice how in b oth graphs the rectangles are either",ElementaryCalculus.pdf
"the series diverges if the integral diverges. Notice how in b oth graphs the rectangles are either
all below the curve or all protrude above the curve due to f (x) being a decreasing function.
Example 9.10
Show that the p-series
∞∑
n=1
1
np converges for p > 1 and diverges for p = 1.
Solution: For p ≥ 1 let f (x) = 1
x p on [1 , ∞). Then f (x) is decreasing, and f (n) = an = 1
n p for all integers
n ≥ 1. For p > 1,
∫ ∞
1
f (x) dx =
∫ ∞
1
dx
x p = −1
( p − 1) x p−1
⏐
⏐
⏐
⏐
∞
1
= 0 −
( −1
( p − 1) (1) p−1
)
= 1
p − 1 < ∞
so the integral converges. Thus, by the Integral Test, the se ries converges for p > 1. For p = 1,
∫ ∞
1
f (x) dx =
∫ ∞
1
dx
x = ln x
⏐
⏐
⏐
⏐
∞
1
= ∞
so the integral diverges. So by the Integral Test, the series diverges for p = 1. The harmonic series
∞∑
n=1
1
n = 1 + 1
2 + 1
3 + 1
4 + ···
thus diverges even though an = 1
n → 0 (which is hence not a sufficient condition for a series to con verge).",ElementaryCalculus.pdf
"∞∑
n=1
1
n = 1 + 1
2 + 1
3 + 1
4 + ···
thus diverges even though an = 1
n → 0 (which is hence not a sufficient condition for a series to con verge).
Note that this example partly proves the p-series Test. The r emaining case ( p < 1) is left as an exercise.",ElementaryCalculus.pdf
"T ests for Convergence • Section 9.2 297
The divergence part of the Comparison Test is clear enough to understand, but for the con-
vergence part with 0 ≤ an ≤ bn for all n larger than some N , ignore the (finite) number of terms
before a N and b N . Since ∑ bn converges then its partial sums must be bounded. The partial
sums for ∑ an then must also be bounded, since 0 ≤ an ≤ bn for n > N . So since an ≥ 0 means
that the partial sums for ∑ an are increasing, by the Monotone Bounded Test the partial sum s
for ∑ an must converge, i.e. ∑ an is convergent.
Example 9.11
Determine if
∞∑
n=1
1
nn is convergent.
Solution: Since nn ≥ n2 > 0 for n > 2, then
0 < 1
nn ≤ 1
n2
for n > 2. Thus, since ∑ ∞
n=1
1
n2 converges (by the p-series Test with p = 2 > 1, as in Example 9.10), the
series ∑ ∞
n=1
1
nn converges by the Comparison Test.
For the Limit Comparison Test with an
bn
→ L < ∞ and L > 0, by definition of the limit of a
sequence, an
bn",ElementaryCalculus.pdf
"n=1
1
nn converges by the Comparison Test.
For the Limit Comparison Test with an
bn
→ L < ∞ and L > 0, by definition of the limit of a
sequence, an
bn
can be made arbitrarily close to L. In particular there is an integer N such that
L
2 < an
bn
< 3L
2
for all n > N . Then
0 < an < 3L
2 bn and
∑
bn converges ⇒
∑
an converges
by the Comparison test. Likewise,
0 < L
2 bn < an and
∑
bn diverges ⇒
∑
an diverges
by the Comparison Test again. The cases L = 0 and L = ∞are handled similarly .
Example 9.12
Determine if
∞∑
n=1
n + 3
n · 2n is convergent.
Solution: Since ∑ ∞
n=1
1
2n is convergent (as part of a geometric progression) and
limn→∞
(n + 3)/(n · 2n )
1/2n = limn→∞
n + 3
n = 1
then by the Limit Comparison Test ∑ ∞
n=1
n+3
n ·2n is convergent..",ElementaryCalculus.pdf
"298 Chapter 9 • Infinite Sequences and Series §9.2
A series ∑ an is telescoping if an = bn − bn+1 for some sequence { bn }. Assume the series∑ an and sequence { bn } both start at n = 1. Then the partial sum sn for ∑ an is
sn = a1 + a2 + ··· + an = (b1 − b2) + (b2 − b3) + ··· + (bn − bn+1) = b1 − bn+1
for n ≥ 1. Thus, since b1 is a fixed number , lim n→∞ sn exists if and only if lim n→∞ bn+1 exists,
i.e. ∑ an converges if and only if { bn } converges. So if bn → L then sn → b − L, i.e. ∑ an
converges to L, which proves the Telescoping Series Test. Note that the num ber b1, as the first
number in the sequence { bn }, could be replaced by whatever the first number is, in case the
index n starts at a number different from 1.
Example 9.13
Determine if
∞∑
n=1
1
n (n + 1) is convergent. If it converges then can you find its sum?
Solution: For the sequence { bn } with bn = 1
n for n ≥ 1, each term in the series can be written as
1
n (n + 1) = 1
n − 1
n + 1 = bn − bn+1",ElementaryCalculus.pdf
"Solution: For the sequence { bn } with bn = 1
n for n ≥ 1, each term in the series can be written as
1
n (n + 1) = 1
n − 1
n + 1 = bn − bn+1
Thus, since { bn } converges to 0, by the Telescoping Series Test the series als o converges, to b1 − 0 = 1.
Convergent series have the following properties (based on s imilar properties of limits):
Let ∑ an and ∑ bn be convergent series, and let c be a number . Then:
(a) ∑ (an ± bn ) is convergent, with ∑ (an ± bn ) = ∑ an ± ∑ bn
(b) ∑ ca n is convergent, with ∑ ca n = c · ∑ an
Exercises
A
For Exercises 1-5 show that the given sequence { an }∞
n=1 is convergent.
1. an = 2 · 4 · 6 ··· (2n)
3 · 5 · 7 ··· (2n + 1) 2. an = 1 − 2n
n! 3. an = 2 · 4 · 6 ··· (2n)
1 · 3 · 5 ··· (2n − 1) · 1
2n + 2
4. an = 1
n
( 2 · 4 · 6 ··· (2n)
1 · 3 · 5 ··· (2n − 1)
) 2
5. an = 1
2 · 3
2 · 3
4 · 5
4 · 5
6 · 7
6 ··· 2n − 1
2n · 2n + 1
2n
For Exercises 6-17 determine whether the given series is con vergent.
6.
∞∑
n=0
sin
(nπ
2
)
7.
∞∑
n=1
1
n (n + 2) 8.
∞∑
n=1
1
2n 9.",ElementaryCalculus.pdf
"2n · 2n + 1
2n
For Exercises 6-17 determine whether the given series is con vergent.
6.
∞∑
n=0
sin
(nπ
2
)
7.
∞∑
n=1
1
n (n + 2) 8.
∞∑
n=1
1
2n 9.
∞∑
n=1
n
(n + 1) 2n",ElementaryCalculus.pdf
"T ests for Convergence • Section 9.2 299
10.
∞∑
n=2
1
n
∝ra⌈〉⌋allow
ln n
11.
∞∑
n=1
n !
(2n) ! 12.
∞∑
n=1
n
en 13.
∞∑
n=1
1
cosh2 n
14.
∞∑
n=1
n !
2n 15.
∞∑
n=1
1∝ra⌈〉⌋allown 16.
∞∑
n=1
1
n (2n − 1) 17.
∞∑
n=1
ln (n + 1)
n2
For Exercises 18-21 determine whether the given series is co nvergent. If convergent then find its sum.
18.
∞∑
n=1
1
(2n + 1) (2n + 3) 19.
∞∑
n=1
1
(2n + 3) (2n + 5) 20.
∞∑
n=1
2
(3n + 1) (3n + 4) 21.
∞∑
n=1
1
4n2 − 1
B
22. Continue Example 9.10 with a proof of the p-series Test for p < 1.
23. Show that { an }∞
n=1 is convergent, where
an = 1
1! + 1
2! + 1
3! + 1
4! + ··· + 1
n!
for n ≥ 1. (Hint: Use the Monotone Bounded test by using a bound on 1
n! for n > 2.)
24. Consider the series
∞∑
n=1
1
2n − 1 = 1 + 1
3 + 1
5 + 1
7 + ··· .
(a) Show that the series is divergent.
(b) The textbook Applied Mathematics for Physical Chemistry (3rd ed.), J . Barrante, provides the
following argument that the above series converges: Since
1 + 1
4 + 1
9 + 1
16 + ··· < 1 + 1
3 + 1",ElementaryCalculus.pdf
"following argument that the above series converges: Since
1 + 1
4 + 1
9 + 1
16 + ··· < 1 + 1
3 + 1
5 + 1
7 + ··· < 1 + 1
2 + 1
3 + 1
4 + ···
where the series on the left converges (by the p-series Test w ith p = 2) and the series on the right
diverges (by the p-series Test with p = 1), and since each term in the middle series is between
its corresponding terms in the left series and right series, then there must be a p-series for some
value 1 < p < 2 such that each term in the middle series is less than the corr esponding term in
that p-series. That is,
1 + 1
4 + 1
9 + 1
16 + ··· < 1 + 1
3 + 1
5 + 1
7 + ··· < 1 + 1
2p + 1
3p + 1
4p + ···
for that value of p between 1 and 2. But p > 1 for that p-series on the right, so it converges,
which means that the middle series converges! Find and expla in the flaw in this argument.
25. Wallis’ formula 6 for π is given by the infinite product
π
2 = 2
1 · 2
3 · 4
3 · 4
5 · 6
5 · 6
7 ··· 2n
2n − 1 · 2n
2n + 1 · ··· .",ElementaryCalculus.pdf
"25. Wallis’ formula 6 for π is given by the infinite product
π
2 = 2
1 · 2
3 · 4
3 · 4
5 · 6
5 · 6
7 ··· 2n
2n − 1 · 2n
2n + 1 · ··· .
Notice that this is the limit of the reciprocal of the sequenc e in Exercise 5. Write a computer program
to approximate the limit using 1 million iterations. How clo se is your approximation to π
2 ?
6Due to the English mathematician and theologian John W allis (1616-1703). For a proof of the formula see pp.738-
739 in T AY L O R, A.E. A N D W .R. M A N N, Advanced Calculus , 2nd ed., New Y ork: John Wiley & Sons, Inc., 1972.",ElementaryCalculus.pdf
"300 Chapter 9 • Infinite Sequences and Series §9.3
9.3 Alternating Series
In the last section the harmonic series
∞∑
n=1
1
n = 1 + 1
2 + 1
3 + 1
4 + 1
5 + ···
was shown to diverge. If you were to alternate the signs of suc cessive terms, as in
∞∑
n=1
(−1)n−1
n = 1 − 1
2 + 1
3 − 1
4 + 1
5 − ··· (9.7)
then it turns out that this new series—called an alternating series —converges, due to the
following test:
Alternating Series T est : If ∑ an is an alternating series —i.e. the signs of the terms an
alternate between positive and negative—such that the abso lute values of the terms are
decreasing to 0, then the series converges.
The condition for the test means that |an+1 | ≤ |an | for all n and |an | →0 as n → ∞. To see why
the test works, consider the alternating series given above by formula (9.7), with an = −1n−1
n .
The odd-numbered partial sums s1, s3, s5, . . ., can be written as
s1 = 1 , s3 = 1 −
(1
2 − 1
3
)
  
> 0
, s5 = 1 −
(1
2 − 1
3
)
  
> 0
−
(1
4 − 1
5
)
  
> 0",ElementaryCalculus.pdf
"s1 = 1 , s3 = 1 −
(1
2 − 1
3
)
  
> 0
, s5 = 1 −
(1
2 − 1
3
)
  
> 0
−
(1
4 − 1
5
)
  
> 0
, . . .
while the even-numbered partial sums s2, s4, s6, . . ., can be written as
s2 = 1 − 1
2 , s4 = 1 − 1
2 +
(1
3 − 1
4
)
  
> 0
, s6 = 1 − 1
2 +
(1
3 − 1
4
)
  
> 0
+
(1
5 − 1
6
)
  
> 0
, . . .
Thus the odd-numbered partial sums { s2n−1 } are decreasing from s1 = 1, and the even-numbered
ones { s2n } are increasing from s2 = 1 − 1
2 = 1
2 , with 1
2 < sn < 1 for all n, i.e. the partial sums are
bounded. So by the Monotone Bounded Test, both sequences { s2n−1 } and { s2n } must converge.
Since s2n − s2n−1 = a2n = −1
2n for all n ≥ 1, then
limn→∞(s2n − s2n−1) = limn→∞
−1
2n = 0 ⇒ limn→∞ s2n = limn→∞ s2n+1
Thus, the partial sums sn have a common limit, so the series converges. Notice that the key to
the convergence was having the terms decreasing in absolute value to zero.",ElementaryCalculus.pdf
"Alternating Series • Section 9.3 301
Example 9.14
Determine if
∞∑
n=2
(−1)n
ln n is convergent.
Solution: For the general term an = (−1)n
ln n , since ln ( n + 1) > ln n for n ≥ 2 and ln n → ∞as n → ∞, then
|an | decreases to 0 as n → ∞. Thus, by the Alternating Series Test the series converges.
The series ∑ ∞
n=1
(−1)n−1
n converges by the Alternating Series Test, though the series ∑ ∞
n=1
1
n
diverges. This makes ∑ ∞
n=1
(−1)n−1
n an example of a conditionally convergent series:
A series ∑ an is conditionally convergent if ∑ an converges but ∑ |an | diverges. If ∑ |an |
converges then ∑ an is absolutely convergent .
For example, ∑ ∞
n=1
(−1)n−1
n is not absolutely convergent, since ∑ ∞
n=1
1
n diverges.
Example 9.15
Is
∞∑
n=1
(−1)n−1
n2 conditionally convergent or absolutely convergent?
Solution: Since ∑ ∞
n=1
1
n2 converges (by the p-series Test) then ∑ ∞
n=1
(−1)n−1
n2 is absolutely convergent.
It turns out that absolute convergence implies ordinary con vergence:",ElementaryCalculus.pdf
"n=1
(−1)n−1
n2 is absolutely convergent.
It turns out that absolute convergence implies ordinary con vergence:
Absolute Convergence T est : If ∑ |an | converges then ∑ an converges.
The test is obvious if the terms an are all positive, so assume the series has both positive
terms (denoted by the sequence
{
apos
}
) and negative terms (denoted by
{
aneg
}
). Then the
series can be decomposed into the difference of two series:
∑
an =
∑
apos −
∑
|aneg |
Since each of the sums on the right side of the equation is part of the convergent series ∑ |an |,
then each sum itself converges (being part of a finite sum). Th us their difference, namely ∑ an,
is finite, i.e. ∑ an converges.
The test can be stated in the following logically equivalent manner:
Absolute Convergence T est : If ∑ an diverges then ∑ |an | diverges.",ElementaryCalculus.pdf
"302 Chapter 9 • Infinite Sequences and Series §9.3
One unusual feature of a conditionally convergent series is that its terms can be rearranged
to converge to any number , a result known as Riemann’s Rearrangement Theorem . For
example, the alternating harmonic series
1 − 1
2 + 1
3 − 1
4 + 1
5 − ···
consists of one divergent series of positive terms subtract ed from another series of positive
terms, namely:
1 − 1
2 + 1
3 − 1
4 + 1
5 − ··· =
(
1 + 1
3 + 1
5 + 1
7 + ···
)
−
(1
2 + 1
4 + 1
6 + ···
)
The idea is that since the first series on the right diverges th en that series has some partial
sum that could be made just larger than any positive number A. Likewise, since the second
series on the right that is being subtracted also diverges, i t has some partial sum that when
subtracted from the first partial sum results in a number just less than A. Continue like this
indefinitely , first adding a partial sum to get a number just bi gger than A then subtracting",ElementaryCalculus.pdf
"indefinitely , first adding a partial sum to get a number just bi gger than A then subtracting
another partial sum to get just less than A. Since the terms in each series approach zero, the
overall series can be made to converge to A! It also turns out that an absolutely convergent
series does not have this feature—any rearrangement of terms results in the same sum. 7
Exercises
A
For Exercises 1-5 determine whether the given alternating s eries is convergent. If convergent, then
determine if it is conditionally convergent or absolutely c onvergent.
1.
∞∑
n=1
(−1)n−1
∝ra⌈〉⌋allown 2.
∞∑
n=1
(−1)n−1 n !
2n 3.
∞∑
n=1
(−1)n−1
2n − 1 4.
∞∑
n=2
(−1)n
n ln n 5.
∞∑
n=1
(−1)n−1
n !
6. Rearrangements of a divergent alternating series can make i t appear to converge to different num-
bers. For example, find different rearrangements of the term s in the divergent series
∞∑
n=0
(−1)n = 1 − 1 + 1 − 1 + ···
so that the series appears to converge to 0, 1, -1, and 2.",ElementaryCalculus.pdf
"∞∑
n=0
(−1)n = 1 − 1 + 1 − 1 + ···
so that the series appears to converge to 0, 1, -1, and 2.
7. A guest arrives at the Aleph Null Hotel, 8 which has an infinite number of rooms, numbered Room 0,
Room 1, Room 2, and so on. The hotel manager says all the rooms a re taken, but he can still give the
guest his own room. How is that possible? Would it still be pos sible if an infinite number of guests
showed up, each wanting his own room? Explain your answers.
7For formal proofs of these statements, see pp.442-444 in K L A MBAU E R , G., Aspects of Calculus , New Y ork:
Springer-V erlag, 1986.
8The symbol ℵ0 is called aleph null and represents the cardinality of N, i.e. its size. ℵ0 is the smallest infinity .",ElementaryCalculus.pdf
"Power Series • Section 9.4 303
9.4 Power Series
A power series is an infinite series whose terms involve constants an and powers of x − c,
where x is a variable and c is a constant: ∑ an (x − c)n. In many cases c will be 0. For example,
the geometric progression
∞∑
n=0
rn = 1 + r + r2 + r3 + ··· = 1
1 − r
converges when |r | <1, i.e. for −1 < r < 1, as shown in Section 9.1. Replacing the constant r by
a variable x yields the power series
∞∑
n=0
xn = 1 + x + x2 + x3 + ··· = 1
1 − x (9.8)
that converges to 1
1−x when −1 < x < 1. Note that the series diverges for |x| ≥1, by the n-th
Term Test.
In general a power series of the form ∑ f n (x), where f n (x) = an(x − c)n is a sequence of func-
tions, has an interval of convergence defined as the set of all x such that the series con-
verges. The interval can be any combination of open or closed , as well as the extreme cases
of a single point or all real numbers. On its interval of conve rgence the power series is thus a",ElementaryCalculus.pdf
"of a single point or all real numbers. On its interval of conve rgence the power series is thus a
function of x. The radius of convergence R of a power series is defined as half the length of
the interval of convergence. In the case where the interval o f convergence is all of R you would
say R = ∞.
For example, for the above power series ∑ ∞
n=0 f n (x), where f n(x) = xn for n ≥ 0, the interval of
convergence is −1 < x < 1, so the radius of convergence is R = 1. Notice that
f (x) =
∞∑
n=0
xn for −1 < x < 1
is thus a well-defined function on the interval ( −1, 1), where it happens to equal 1
1−x . This
power series can be thought of as a polynomial of infinite degr ee.
To find the interval of convergence of a power series ∑ f n (x), you typically would use the
Ratio Test on the absolute values of the terms (since the Rati o Test requires positive terms):
r(x) = limn→∞
⏐
⏐
⏐
⏐
f n+1(x)
f n (x)
⏐
⏐
⏐
⏐ (9.9)",ElementaryCalculus.pdf
"Ratio Test on the absolute values of the terms (since the Rati o Test requires positive terms):
r(x) = limn→∞
⏐
⏐
⏐
⏐
f n+1(x)
f n (x)
⏐
⏐
⏐
⏐ (9.9)
Note that the limit r(x) in this case is a function of x. When taking the limit, though, treat x
as fixed. By the Ratio Test the power series will then converge for all x such that r(x) < 1, and
diverge when r(x) > 1. When r(x) = 1 the test is inconclusive, so you would have to check those
cases individually to see if those values of x should be added to the interval of convergence
(along with the points where r(x) < 1).",ElementaryCalculus.pdf
"304 Chapter 9 • Infinite Sequences and Series §9.4
Example 9.16
Find the interval of convergence of the power series
∞∑
n=0
xn
n ! .
Solution: For f n (x) = xn
n ! ,
r(x) = limn→∞
⏐
⏐
⏐
⏐
f n+1(x)
f n (x)
⏐
⏐
⏐
⏐ = limn→∞
⏐
⏐
⏐
⏐
xn+1/(n + 1) !
xn /n !
⏐
⏐
⏐
⏐ = |x | · limn→∞
⏐
⏐
⏐
⏐
1
n + 1
⏐
⏐
⏐
⏐ = |x | · 0 = 0
for any fixed x. Thus, r(x) = 0 < 1 for all x, so the interval of convergence is all of R, i.e. −∞ <x < ∞.
Example 9.17
Find the interval of convergence of the power series
∞∑
n=1
xn
n .
Solution: For f n (x) = xn
n ,
r(x) = limn→∞
⏐
⏐
⏐
⏐
f n+1(x)
f n (x)
⏐
⏐
⏐
⏐ = limn→∞
⏐
⏐
⏐
⏐
xn+1/(n + 1)
xn /n
⏐
⏐
⏐
⏐ = limn→∞
⏐
⏐
⏐
⏐
nx
n + 1
⏐
⏐
⏐
⏐ = |x | · limn→∞
⏐
⏐
⏐
⏐
n
n + 1
⏐
⏐
⏐
⏐ = |x | · 1 = |x |
for any fixed x. Thus, the series converges when r(x) = |x | <1 and diverges when r(x) = |x | >1.
The cases r(x) = |x | =1 need to be checked individually . For x = 1 the series is ∑ ∞
n=1
1
n , which diverges.
For x = −1 the series is ∑ ∞
n=1
(−1)n−1",ElementaryCalculus.pdf
"n=1
1
n , which diverges.
For x = −1 the series is ∑ ∞
n=1
(−1)n−1
n , which converges. Thus, the interval of convergence is −1 ≤ x < 1.
Example 9.18
Find the interval of convergence of the power series
∞∑
n=0
n ! xn .
Solution: For f n (x) = n ! xn ,
r(x) = limn→∞
⏐
⏐
⏐
⏐
f n+1(x)
f n (x)
⏐
⏐
⏐
⏐ = limn→∞
⏐
⏐
⏐
⏐
(n + 1) ! xn+1
n ! xn
⏐
⏐
⏐
⏐ = |x | · limn→∞ |n + 1| =
{
0 if x = 0,
∞ if x ̸=0.
Thus, r(x) = ∞ >1 for all x ̸=0, so the series diverges for x ̸=0. So since r(x) = 0 < 1 only for x = 0, the
interval of convergence is the single point x = 0.
It turns out that power series can be both differentiated and integrated term by term: 9
For a power series f (x) =
∞∑
n=0
an (x − c)n that converges for |x − c | <R, both
f ′(x) =
∞∑
n=1
n a n (x − c)n−1 and
∫ 
f (x) dx = C +
∞∑
n=0
an
n + 1 (x − c)n+1 (9.10)
converge for |x − c | <R.
9The proofs require uniform convergence , a stronger condition than ordinary convergence. See pp.12 9-134 in",ElementaryCalculus.pdf
"converge for |x − c | <R.
9The proofs require uniform convergence , a stronger condition than ordinary convergence. See pp.12 9-134 in
BR O MW I C H , T .J., An Introduction to the Theory of Infinite Series , 2nd ed., London: Macmillan & Co. Ltd., 1955.",ElementaryCalculus.pdf
"Power Series • Section 9.4 305
Notice that the above statement says nothing about the conve rgence of f ′(x) or
∫ 
f (x) dx at
the endpoints of the interval |x − c | <R. In each case convergence at the endpoints can be
checked individually .
Example 9.19
Write the power series form of the derivative of f (x) =
∞∑
n=0
xn and find its interval of convergence. Can
f ′(x) be written in a non-series form?.
Solution: Differentiate f (x) term by term:
f ′(x) = d
dx
(
∞∑
n=0
xn
)
= d
dx (1 + x + x2 + x3 + ···) = 0 + 1 + 2x + 3x2 + ··· =
∞∑
n=1
nx n−1
Since f (x) converges for −1 < x < 1 then so does f ′(x). Checking the endpoints, at x = 1 and x = −1 the
series for f ′(x) are ∑ ∞
n=1 n and ∑ ∞
n=1(−1)n−1 n, respectively , both of which diverge by the n-th Term Test.
Thus, the interval of convergence for f ′(x) is −1 < x < 1.
Since f (x) = 1
1−x for −1 < x < 1 then f ′(x) = 1
(1−x)2 for −1 < x < 1. Thus,
∞∑
n=1
nx n−1 = 1 + 2x + 3x2 + ··· = 1
(1 − x)2 for −1 < x < 1.
Bessel Functions",ElementaryCalculus.pdf
"1−x for −1 < x < 1 then f ′(x) = 1
(1−x)2 for −1 < x < 1. Thus,
∞∑
n=1
nx n−1 = 1 + 2x + 3x2 + ··· = 1
(1 − x)2 for −1 < x < 1.
Bessel Functions
Many applications in engineering and physics—especially t hose involving oscillations or me-
chanical vibrations—require solving differential equati ons of the form
d2 y
dx2 + 1
x
dy
dx + y = 0 , (9.11)
known as Bessel’s equation . This equation has a solution J0 (x), known as Bessel’s function
of order zero ,10 defined in terms of a power series:
J0(x) =
∞∑
n=0
(−1)n x2n
(n !)2 · 22n = 1 − x2
22 + x4
22 · 42 − x6
22 · 42 · 62 + ··· (9.12)
The Ratio Test shows that J0(x) converges for all x, since for any fixed x,
r(x) = limn→∞
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
(−1)n+1 x2n+2
((n + 1) !)2 · 22n+2
(−1)n x2n
(n !)2 · 22n
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= x2 · limn→∞
⏐
⏐
⏐
⏐
1
4 ( n + 1)2
⏐
⏐
⏐
⏐ = 0 < 1 .
10Due to the German astronomer Friedrich Wilhelm Bessel (1784 -1846), from a study of elliptic planetary motion.",ElementaryCalculus.pdf
"⏐
⏐
1
4 ( n + 1)2
⏐
⏐
⏐
⏐ = 0 < 1 .
10Due to the German astronomer Friedrich Wilhelm Bessel (1784 -1846), from a study of elliptic planetary motion.
For some historical background and examples of how Bessel’s equation arises in applications, see § I.2 and § I.8 in
MC LA C H L A N, N.W ., Bessel Functions for Engineers , London: Oxford University Press, 1934.",ElementaryCalculus.pdf
"306 Chapter 9 • Infinite Sequences and Series §9.4
The general Bessel equation of order m,
d2 y
dx2 + 1
x
dy
dx +
(
1 − m2
x2
)
y = 0 (9.13)
for m = 0, 1, 2, . . . , has a solution Jm (x), called Bessel’s function of order m:
Jm (x) =
∞∑
n=0
(−1)n 1
n ! · (n + m) !
(x
2
) 2n+m
(9.14)
For example, the Bessel function J1(x) of order 1 is
J1(x) =
∞∑
n=0
(−1)n 1
n ! · (n + 1) !
(x
2
) 2n+1
= x
2 − x3
22 · 4 + x5
22 · 42 · 6 − x7
22 · 42 · 62 · 8 + ···
Term by term differentiation shows that J′
0(x) = −J1(x):
d
dx (J0(x)) = d
dx
(
1 − x2
22 + x4
22 · 42 − x6
22 · 42 · 62 + x8
22 · 42 · 62 · 82 − ···
)
= − x
2 + x3
22 · 4 − x5
22 · 42 · 6 + x7
22 · 42 · 62 · 8 − ··· = −J1(x)
Graphs of J0(x) and J1(x) are shown in Figure 9.4.1 below . As you can see, J0(x) and J1(x)
behave as sort of “poor man’s” cosine and sine functions, res pectively .
0 5 10 15 20 25 
-1 
-0.5 
0 
0.5 
1 
x 
y 
J 0 (x) 
J 1 (x) 
Figure 9.4.1 Bessel functions J0 (x) and J1 (x)",ElementaryCalculus.pdf
"Power Series • Section 9.4 307
Exercises
A
For Exercises 1-8 find the interval of convergence of the give n power series.
1.
∞∑
n=1
nx n
(n + 1)2 2.
∞∑
n=1
nx n
2n 3.
∞∑
n=1
n2 (x − 2)n 4.
∞∑
n=0
(x + 4)n
2n
5.
∞∑
n=1
(x + 1)n
nn 6.
∞∑
n=1
nn xn 7.
∞∑
n=0
(−1)n xn 8.
∞∑
n=1
nx n
n + 1
9. Note that power series of the form ∑ ∞
n=0 an xn have an issue at x = 0 when n = 0: 0 0 is an indeter-
minate form—it can equal anything (or nothing). What value h as it implicitly been assigned so far?
What would be the technically correct way to write the series ∑ ∞
n=0 an xn so that this issue goes away?
10. Show that ∞∑
n=1
n x n = x
(1 − x)2 for −1 < x < 1.
11. Write the following infinite series as a rational number:
1
10 + 2
100 + 3
1000 + 4
10000 + ···
B
12. Differentiating term by term, verify that the Bessel functi on J0 (x) satisfies Bessel’s equation (see
equation (9.11)).
13. Show that for all m ≥ 1 the Bessel functions Jm (x) converge for all x.",ElementaryCalculus.pdf
"equation (9.11)).
13. Show that for all m ≥ 1 the Bessel functions Jm (x) converge for all x.
14. For all m ≥ 1 verify that the Bessel functions Jm (x) satisfy the general Bessel equation of order m
(see equation(9.13)).
15. For the Bessel functions J0 (x) and J1 (x) show that:
(a) d
dx (x J 1(x)) = x J 0 (x)
(b)
∫ 
J0 (x) J1 (x) dx = −1
2 J2
0 (x)
(c)
∫ 
x J 0 (x) J1 (x) dx = −1
2 x J 2
0 (x) + 1
2
∫ 
J2
0 (x) dx
(d) For all integers n ≥ 2,
∫ 
xn J0 (x) dx = xn J1 (x) + (n − 1) xn−1 J0 (x) − (n − 1)2
∫ 
xn−2 J0 (x) dx .
(Hint: Use part (a) and integration by parts twice.)
16. For all integers m ≥ 2 show that the Bessel functions Jm (x) satisfy the relations:
(a) m J m (x) + x J ′
m (x) = x J m−1 (x)
(b) Jm−1 (x) − Jm+1 (x) = 2J′
m (x)
17. Use long division to obtain the first three terms of 1
x J 2
0 (x) , then integrate term by term to show that
J0 (x)
∫ dx
x J2
0 (x)
= J0 (x) · ln x + x2
4 − 3x4
128 + ··· .",ElementaryCalculus.pdf
"x J 2
0 (x) , then integrate term by term to show that
J0 (x)
∫ dx
x J2
0 (x)
= J0 (x) · ln x + x2
4 − 3x4
128 + ··· .
This function is a Bessel function of the second kind and is another solution of Bessel’s equation.",ElementaryCalculus.pdf
"308 Chapter 9 • Infinite Sequences and Series §9.5
9.5 T aylor’s Series
In the previous section a few functions, e.g. f (x) = 1
1−x , turned out to be the sum of a power
series. This section will discuss a general method for repre senting a function as a power series,
called a T aylor’s series.11 Suppose that a function f (x) can be written as
f (x) =
∞∑
n=0
an (x − c)n
either for all x or for |x − c | <R, for some R > 0. Then f (c) = a0, and differentiating term by
term yields
f ′(x) =
∞∑
n=1
n a n (x − c)n−1 ⇒ f ′(c) = 1 · a1
f ′′(x) =
∞∑
n=2
n (n − 1) an (x − c)n−2 ⇒ f ′′(c) = 2 · 1 · a2
f ′′′(x) =
∞∑
n=3
n (n − 1) (n − 2) an (x − c)n−3 ⇒ f ′′′(c) = 3 · 2 · 1 · a3
··· ···
f (k) (x) =
∞∑
n=k
n (n − 1) (n − 2) ··· (n − k + 1) an (x − c)n−k ⇒ f (k)(c) = k! ak
so that in general (since f (0)(x) = f (x) and 0! = 1):
an = f (n)(c)
n! for n ≥ 0 (9.15)
These { an } are the T aylor’s series coefficients of f (x) at x = c. The full power series repre-
sentation of f (x) can now be stated:",ElementaryCalculus.pdf
"These { an } are the T aylor’s series coefficients of f (x) at x = c. The full power series repre-
sentation of f (x) can now be stated:
T aylor’s formula: If f (x) has a power series representation in powers of x − c, where x = c
is inside the interval of convergence, then that representa tion is unique in that interval
and is given by
f (x) =
∞∑
n=0
f (n)(c)
n! (x − c)n (9.16)
for all x in the interval. This is the T aylor’s series for f (x) about x = c.
11Named after English mathematician Brook Taylor (1685-1731 ), though such series were known to others (e.g.
James Gregory , Johann Bernoulli) before Taylor .",ElementaryCalculus.pdf
"T aylor’s Series • Section 9.5 309
Example 9.20
Find the Taylor’s series for f (x) = ex about x = 0.
Solution: Since d
dx (ex) = ex , then for all n ≥ 0,
f (n)(x) = ex ⇒ f (n)(0) = e0 = 1 .
Thus, by Taylor’s formula with c = 0:12
ex =
∞∑
n=0
f (n)(0)
n! xn =
∞∑
n=0
xn
n!
= 1 + x + x2
2! + x3
3! + x4
4! + x5
5! + ···
For what x is this Taylor’s series valid? Recall that Example 9.16 show ed the interval of convergence is
all of R. Thus the above Taylor’s series holds for all x.
Before continuing, you might be wondering why you should eve n bother with finding the
Taylor’s series—after all, in the above example why replace a simple function like ex by a far
more complicated expression? One reason is that it often hel ps simplify some computations,
especially in integrals. The idea is to use only a few terms in the series, i.e. a polynomial, as
an approximation, since polynomials are generally easier t o work with. Perhaps surprisingly ,",ElementaryCalculus.pdf
"an approximation, since polynomials are generally easier t o work with. Perhaps surprisingly ,
in many practical applications no more than two terms are nee ded, and often only one.
For example, using only the first two terms of the Taylor’s ser ies for ex in Example 9.20,
ex ≈ 1 + x is a good approximation when x is close to 0 (i.e. |x| ≪1). Using more terms does not
necessarily help—for |x| ≪1 and n > 1, xn will be effectively 0. So the added complexity would
not make the approximation significantly better .
Example 9.21
The energy density E of electromagnetic radiation at wavelength λ from a black-body at temperature
T degrees Kelvin is given by Planck’s Law of black-body radiation,
E(λ) = 8πhc
λ5 (eh c/λkT − 1)
where h is Planck’s constant, c is the speed of light, and k is Boltzmann’s constant. Show that for λ ≫ 1:
E(λ) ≈ 8πkT
λ4
Solution: Since ex ≈ 1 + x for |x | ≪1 by the Taylor series for ex , let x = hc/λkT . Then x ≪ 1 and so
E(λ) = 8πhc
λ5 (eh c/λkT − 1) ≈ 8πhc
λ5
((
1 + h c",ElementaryCalculus.pdf
"E(λ) = 8πhc
λ5 (eh c/λkT − 1) ≈ 8πhc
λ5
((
1 + h c
λkT
)
− 1
) ≈ 8πhc
λ5 h c
λkT
≈ 8πkT
λ4
12The special case of c = 0 in Taylor’s formula yields what is sometimes called the Maclaurin’s series for f (x),
though that terminology is typically not used in fields of stu dy outside mathematics.",ElementaryCalculus.pdf
"310 Chapter 9 • Infinite Sequences and Series §9.5
Example 9.22
Find the Taylor’s series for f (x) = sin x about x = 0.
Solution: The derivatives of f (x) = sin x repeat every four derivatives:
f (x) = sin x , f ′(x) = cos x , f ′′(x) = −sin x , f ′′′(x) = −cos x , f (4) (x) = sin x
So at x = 0:
f (0) = 0 , f ′(x) = 1 , f ′′(x) = 0 , f ′′′(x) = −1 , f (4) (x) = 0
So for n ≥ 0,
f (n)(0) =





0 if n is even,
1 if n = 1, 5, 9, . . . ,
−1 if n = 3, 7, 11, . . . .
Thus, by Taylor’s formula with c = 0
sin x =
∞∑
n=0
f (n)(0)
n! xn
= x − x3
3! + x5
5! − x7
7! + x9
9! − ···
=
∞∑
n=0
(−1)n x2n+1
(2n + 1)!
By the Ratio Test this series converges for all x, since for any fixed x,
r(x) = limn→∞
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
(−1)n+1 x2n+3
(2n + 3)!
(−1)n x2n+1
(2n + 1)!
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= x2 · limn→∞
⏐
⏐
⏐
⏐
1
(2n + 3) (2n + 2)
⏐
⏐
⏐
⏐ = 0 < 1 .
Example 9.23
Find the Taylor’s series for f (x) = cos x about x = 0.",ElementaryCalculus.pdf
"⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= x2 · limn→∞
⏐
⏐
⏐
⏐
1
(2n + 3) (2n + 2)
⏐
⏐
⏐
⏐ = 0 < 1 .
Example 9.23
Find the Taylor’s series for f (x) = cos x about x = 0.
Solution: The Taylor’s series can be found using the same procedure as i n Example 9.22, but it is
simpler to just differentiate the Taylor’s series for sin x term by term for all x:
cos x = d
dx (sin x) = d
dx
(
x − x3
3! + x5
5! − x7
7! + x9
9! − ···
)
= 1 − x2
2! + x4
4! − x6
6! + x8
8! − ···
=
∞∑
n=0
(−1)n x2n
(2n)!
Since the Taylor’s series for sin x converges for all x then so does its derivative. Thus the Taylor’s series
for cos x converges for all x.
Notice that the Taylor’s series for cos x has only even powers of x, while the series for sin x has only odd
powers of x. This makes sense since cos x and sin x are even and odd functions, respectively .",ElementaryCalculus.pdf
"T aylor’s Series • Section 9.5 311
Example 9.24
The function ln x is not defined at x = 0 and hence has no Taylor’s series about x = 0. Instead, find the
Taylor’s series for f (x) = ln (1 + x) about x = 0.
Solution: Take successive derivatives:
f (x) = ln (1 + x) , f ′(x) = 1
1 + x , f ′′(x) = − 1
(1 + x)2 , f ′′′(x) = 1 · 2
(1 + x)3 , f (4) (x) = −1 · 2 · 3
(1 + x)4
So f (0) = 0 and for n ≥ 1:
f (n)(x) = (−1)n−1 (n − 1)!
(1 + x)n ⇒ f (n)(0) = (−1)n−1 (n − 1)!
Thus, by Taylor’s formula,
ln (1 + x) =
∞∑
n=0
f (n)(0)
n! xn =
∞∑
n=1
(−1)n−1 (n − 1)! xn
n!
=
∞∑
n=1
(−1)n−1 xn
n
= x − x2
2 + x3
3 − x4
4 + x5
5 − ···
Use the Ratio Test to find the interval of convergence:
r(x) = limn→∞
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
(−1)n xn+1
n + 1
(−1)n−1 xn
n
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= |x | · limn→∞
⏐
⏐
⏐
⏐
n
n + 1
⏐
⏐
⏐
⏐ = |x | ·1 = |x |
So the series converges when |x | <1. Check the cases r(x) = |x | =1 individually . For x = 1 the series
is the alternating harmonic series ∑
n=1
(−1)n−1",ElementaryCalculus.pdf
"is the alternating harmonic series ∑
n=1
(−1)n−1
n , which converges. For x = −1 the series is −∑
n=1
1
n , the
negative of the harmonic series, which diverges. Thus, the s eries converges for −1 < x ≤ 1.
Example 9.25
Find the Taylor’s series for f (x) = ex2
about x = 0.
Solution: The Taylor’s series can be found using the same procedure as i n Example 9.20, but it is simpler
to just replace each occurrence of x in the Taylor’s series for ex by x2 (since the series for ex converges
for all x). In other words, make the substitution u = x2 in the Taylor’s series for eu about u = 0:
eu =
∞∑
n=0
un
n!
ex2
=
∞∑
n=0
(x2)n
n! =
∞∑
n=0
x2n
n!
= 1 + x2 + x4
2! + x6
3! + x8
4! + x10
5! + ···",ElementaryCalculus.pdf
"312 Chapter 9 • Infinite Sequences and Series §9.5
Define the n-th degree T aylor polynomial Pn (x) for a function f (x) about x = c by
Pn (x) =
n∑
k=0
f (k)(c)
k! (x − c)k
= f (c) + f ′(c)
1! (x − c) + f ′′(c)
2! (x − c)2 + ··· + f (n)(c)
n! (x − c)n
for x in the interval of convergence for the full Taylor’s series. In other words, Pn (x) is the
n-th partial sum of the Taylor’s series. Since some of the coef ficients could be zero, Pn (x) is a
polynomial of degree at most n. Thus, Pn (x) = O(xn ). For that reason Pn (x) is sometimes called
the O(xn ) approximation to f (x).
Figure 9.5.1 shows a comparison of sin x with a few of its approximations:
−8 −6 −4 −2 0 2 4 6 8
−1
−0.5
0
0.5
1
x
y
 
 
O(x7) approx.
O(x11) approx.
O(x15) approx.
sin(x)
Figure 9.5.1 sin x and Taylor’s series approximations
As you can see, the Taylor polynomials of degree 7, 11 and 15 ar e all good approximations
over the interval [ −2, 2], with the O(x15 ) approximation still being fairly good over [ −6, 6].",ElementaryCalculus.pdf
"over the interval [ −2, 2], with the O(x15 ) approximation still being fairly good over [ −6, 6].
Clearly those approximations all become poor quite quickly for |x| >6; they approach ±∞,
unlike sin x.
The following theorem shows how to measure the accuracy of th e approximations:",ElementaryCalculus.pdf
"T aylor’s Series • Section 9.5 313
Remainder Theorem :13If Pn (x) is the n-th degree Taylor polynomial about x = c for a
function f (x) in some interval containing x = c, then for all x in that interval,
f (x) = Pn (x) + Rn(x) , (9.17)
where
Rn(x) = f (n+1)(c + θ(x − c))
(n + 1)! (x − c)n+1 (9.18)
for some number θ between 0 and 1. Alternatively ,
Rn(x) = 1
n!
∫ x
c
(x − t)n f (n+1)(t) dt . (9.19)
Since the number θ is unknown in equation (9.18), usually only an upper bound on the
remainder Rn(x) can be found by that formula. For practical purposes formul a (9.19) for Rn(x)
might be easier to use (via numerical integration).
A common misconception is that hand-held calculators use Ta ylor’s series to compute values
of functions like sin x, cos x, ex, etc. However , that is typically well beyond their capabili ty ,
especially for large values of x—far too many terms would be required. Instead, many calcula -",ElementaryCalculus.pdf
"especially for large values of x—far too many terms would be required. Instead, many calcula -
tors use an algorithm called CORDIC 14 (Coordinate Rotation Digital Computer), and—perhaps
surprisingly—lookup tables. CORDIC uses the computationa lly inexpensive operation of bit-
shifting to translate large input values into a smaller rang e, then uses tables stored in memory
for values in that range, along with interpolation for numbe rs between those in the tables. 15
Exercises
A
For Exercises 1-9 write out the first three nonzero terms in th e Taylor’s series for the given function
f (x) about the given value c. Y ou may use any method you like.
1. f (x) = sin x ; c = π/2 2. f (x) = sinh x ; c = 0 3. f (x) = cosh x ; c = 0
4. f (x) = tan x ; c = 0 5. f (x) = tanh x ; c = 0 6. f (x) = sec x ; c = 0
7. f (x) = 1
1 + x2 ; c = 0 8. f (x) = 1
1 + x2 ; c = 1 9. f (x) =
∝ra⌈〉⌋allow
1 + x2 ; c = 0",ElementaryCalculus.pdf
"7. f (x) = 1
1 + x2 ; c = 0 8. f (x) = 1
1 + x2 ; c = 1 9. f (x) =
∝ra⌈〉⌋allow
1 + x2 ; c = 0
10. Use Example 9.19 from Section 9.4 to write out the first three n onzero terms in the Taylor’s series
for f (x) = 1
(1−x)3 about x = 0.
11. Use the derivative of the Taylor’s series for
∝ra⌈〉⌋allow
1 + x2 from Exercise 9 to write out the first three
nonzero terms in the Taylor’s series for f (x) = x2
∝ra⌈〉⌋allow
1+x2 about x = 0.
13For a proof see pp.171-172 in K L A MBAU E R , G., Aspects of Calculus , New Y ork: Springer-V erlag, 1986.
14See Ch.7 in S C H MI D , H., Decimal Computation , New Y ork: John Wiley & Sons, Inc., 1974.
15To see how inaccurate calculators can be, compute tan(355/2 26) in radian mode on a calculator . The true value
to 3 decimal places is −7497258.185, but few calculators produce an answer close to that.",ElementaryCalculus.pdf
"314 Chapter 9 • Infinite Sequences and Series §9.5
For Exercises 12-15 replace the function f (x) by its Taylor’s series about x = 0 to evaluate the given
indefinite integral
∫ 
f (x) dx (up to the first three nonzero terms in the series).
12.
∫ sin x
x dx 13.
∫ 
cos ( x2) dx 14.
∫ 
e−x2
dx 15.
∫ √
1 + x6 dx
16. Use d
dx (tan−1 x) = 1
1+x2 along with Exercise 7 to find the Taylor’s series for f (x) = tan−1 x about x = 0,
along with its interval of convergence.
17. Use Exercise 16 to show that
π = 4
(
1 − 1
3 + 1
5 − 1
7 + ···
)
.
18. Use the first three nonzero terms in the Taylor’s series about x = 0 for e−x2 /2 to evaluate the definite
integral ∫ 1
−1
1∝ra⌈〉⌋allow
2π
e−x2 /2 dx .
Note: The actual value (rounded to 4 decimal places) is 0.682 6.
19. Recall that the surface area S of the solid obtained by revolving the curve y = x2 around the x-axis
between x = 0 and x = 2 is given by the integral
S =
∫ 2
0
2πx2
√
1 + 4x2 dx .",ElementaryCalculus.pdf
"between x = 0 and x = 2 is given by the integral
S =
∫ 2
0
2πx2
√
1 + 4x2 dx .
The exact value of the integral rounded to 3 decimal places is S = 53.226. Use the first two nonzero
terms in the Taylor’s series for
∝ra⌈〉⌋allow
1 + 4x2 about x = 0 to approximate the integral. How close is this
approximation to the actual value? Does the approximation b ecome better if you use the first three
nonzero terms in the Taylor’s series? Justify your answer .
20. The tangential component of a space shuttle’s velocity duri ng reentry is approximately
v(t) = vc tanh
( g
vc
t + tanh−1
(v0
vc
))
where v0 is the velocity at time t = 0 and vc is the terminal velocity . If tanh −1
(
v0
vc
)
= 1
2 then show
that v(t) ≈ gt + 1
2 vc .
21. The velocity of a water wave of length L in water of depth h satisfies the equation v2 = gL
2π tanh
(
2πh
L
)
.
Show that v ≈
√
gh.
22. A disk of radius a has a charge of constant density σ. A point P lies at a distance r directly above",ElementaryCalculus.pdf
"2π tanh
(
2πh
L
)
.
Show that v ≈
√
gh.
22. A disk of radius a has a charge of constant density σ. A point P lies at a distance r directly above
the disk. The electrical potential V at point P is given by V = 2πσ(
∝ra⌈〉⌋allow
r2 + a2 − r). Show that V ≈ πa2 σ
r
for large r (i.e. r ≫ 1).
23. The fifth-degree Padé approximation uses rational functions to approximate tanh x:
tanh x ≈ x5 + 105x3 + 945x
15x4 + 420x2 + 945
Compare the values of the Padé approximation and the fifth-de gree Taylor’s series approximation
from Exercise 5, evaluated at x = 1. Which is better? The actual value of tanh(1) is 0.76159415 59558.
How do the two approximations compare at x = 2?",ElementaryCalculus.pdf
"APPENDIX A
Answers and Hints to Selected Exercises
Chapter 1
Section 1.1 (p. 6)
1. 2t 2. 19.6t 3. −32t + 2 4. 3t2
Section 1.2 (p. 14)
1. 0 3. 2x + 2 5. − 1
(x+1)2 7. − 2
x3
9. 1
2
∝ra⌈〉⌋allow
x+1 11. 2x+3
2
∝ra⌈〉⌋allow
x2 +3x+4
Section 1.3 (p. 20)
5. Hint: Use the sine double-angle formula.
7. Hint: Use Exercise 5 and the sine addition
formula.
Section 1.4 (p. 26)
1. 2x − 1 3. 4x5 + 9
x7 5. x cos x + sin x
7. x cos x−sin x
x2 9. 2−2t2
(1+t2 )2 11. ad −b c
(cx +d)2
13. 2πr
Section 1.5 (p. 30)
1. −20(1 − 5x)3 3. − 1
∝ra⌈〉⌋allow
1−2x 5. 1−x
2∝ra⌈〉⌋allowx(x+1)2
7. −8(1−t)3
(1+t)5 9. 2 sin x cos x 11. 15 sec2(5x)
13. 2x sec(x2) tan( x2) 15. β(1 − β2)−3/2
17. sin(cos x) sin x 21. Hint for part(b): Use
part(a) and the Chain Rule to find S p , then re-
call how to convert from radians per second to
revolutions per minute.
Section 1.6 (p. 35)
1. 6x + 2 3. −9 cos 3 x 5. 2
x3
Chapter 2
Section 2.1 (p. 40)
1. f −1(x) = x,
(
f −1) ′
(x) = 1
3. f −1(x) = ∝ra⌈〉⌋allow
x,
(
f −1) ′ (x) = 1
2∝ra⌈〉⌋allowx",ElementaryCalculus.pdf
"x3
Chapter 2
Section 2.1 (p. 40)
1. f −1(x) = x,
(
f −1) ′
(x) = 1
3. f −1(x) = ∝ra⌈〉⌋allow
x,
(
f −1) ′ (x) = 1
2∝ra⌈〉⌋allowx
5. f −1(x) = 1
x ,
(
f −1) ′
(x) = −1
x2
7. f −1(x) = 1∝ra⌈〉⌋allowx ,
(
f −1) ′
(x) = −1
2 x−3/2
Section 2.2 (p. 44)
1. 6 sec2 3x tan 3 x 3. −3 csc2 3x 5. 3
9+x2
7. − 3
1+9x2 9. 1
1+x2 11. 6 sin−1 3x∝ra⌈〉⌋allow
1−9x2
13. 1
1+x2 15. cot−1 x − x
1+x2
Section 2.3 (p. 52)
1. 2e2x 3. −e−x − ex 5. 2ex
(1−ex )2 7. eex
ex
9. 1
x 11. 6x(ln(tan x2 ))
2
sec2 x2
tan x2
315",ElementaryCalculus.pdf
"316 Appendix A: Answers and Hints to Selected Exercises
15. xx2
(x + 2x ln x)
17. xsin x
(
cos x ln x + sin x
x
)
19. 15.5 hours
21. 12 hours
Section 2.4 (p. 55)
1. ln 3 (3x −3−x )
2 3. (ln 2)2 22x
2x
5. 2x
(ln 2)( x2 +1) 7. cos(log2 πx)
x ln 2 9. 3x2
Chapter 3
Section 3.1 (p. 61)
1. y = 4x − 3 3. y = −6x + 10 5. y = 4x
7. y = x + 3 9. y = 240x + 176 11. y = 2x
13. y = 3x + 31
27 , y = 3x + 1 15. 75.96◦
17. 0◦ 19. 116.6◦ 21. 5.71◦
23. y = −1
4 x + 11
2
25. y = −1
4 x − 81
4 , y = −1
4 x + 1159
108
Section 3.2 (p. 72)
1. 7
3 3. 0 5. −1 7. 0 9. 2 11. 0
14. 1
2 15. 0 17. 0
Section 3.3 (p. 78)
1. continuous 3. discontinuous
5. discontinuous 7. discontinuous
9. continuous 11. continuous
13. continuous 15. discontinuous
17. continuous 19. 1 21. e−1
25. Hint: Use the Intermediate V alue Theo-
rem.
Section 3.4 (p. 81)
1. −3x2 y+4 y2 +2x
x3 −8x y−1 3. 2(x−y+1)−3(x+y)2
2(x−y+1)+3(x+y)2
5. 2x(1−(x2 −y2 ))
y(2( y2 −x2 )−1) 7. − y
x
9. −−2x−y+3x2 y2 esin(x y) +x3 y3 esin(x y) cos(x y)",ElementaryCalculus.pdf
"x3 −8x y−1 3. 2(x−y+1)−3(x+y)2
2(x−y+1)+3(x+y)2
5. 2x(1−(x2 −y2 ))
y(2( y2 −x2 )−1) 7. − y
x
9. −−2x−y+3x2 y2 esin(x y) +x3 y3 esin(x y) cos(x y)
x4 y2 esin(x y) cos(x y)+2x3 yesin(x y) −3 y2 −x
13. − x2 +y2
y3
Section 3.5 (p. 83)
1. 80π ft/s 3. 2.4 ft/s 5. 10 ft/s
7. −76π cm3/min 9. 45.14 mph
11. 155.8 ft/min
Section 3.6 (p. 88)
1. (2x − 2) dx 2. 4x sin(x2) cos( x2) dx
11. Hint: Mimic Example 3.35.
Chapter 4
Section 4.1 (p. 98)
1. (1, 1) 3. 125, 000 sq yd 5. U = V
2
7. R = r 9. 2ab 13. Q =
√
2D P
I+W
15. r =
√
r2
0 + x2
0 17. 380.62 minutes
19. 12π
∝ra⌈〉⌋allow
3 21. 12.8 ft 22. Hint: Place the
right angle of the triangle at the origin in the
x y-plane. 25. x =
√
R N
r
27. x = a∝ra⌈〉⌋allow
2 33. (a2/3 + b2/3)3/2
34. Hint: Y ou can leave your answer in terms
of R and an angle satisfying a certain equa-
tion.
Section 4.2 (p. 108)
2. local maximum at x = 0, local minimum at
x = 2, inflection pt at x = 1, increasing for x < 0
and x > 2, decreasing for 0 < x < 2, concave up",ElementaryCalculus.pdf
"2. local maximum at x = 0, local minimum at
x = 2, inflection pt at x = 1, increasing for x < 0
and x > 2, decreasing for 0 < x < 2, concave up
for x > 1, concave down for x < 1
3. local maximum at x = 1, inflection pt at
x = 2, increasing for x < 1, decreasing for x > 1,
concave up for x > 2, concave down for x < 2,
horizontal asymptote: y = 0
5. local maximum at x = 0, inflection pts at
x = ± 1
∝ra⌈〉⌋allow
3 , increasing for x < 0, decreasing for
x > 0, concave up for x < − 1∝ra⌈〉⌋allow
3 and x > 1∝ra⌈〉⌋allow
3 ,
concave down for − 1∝ra⌈〉⌋allow
3 < x < 1∝ra⌈〉⌋allow
3 , horizontal
asymptote: y = 0",ElementaryCalculus.pdf
"Appendix A: Answers and Hints to Selected Exercises 317
7. local maximum at x = ln 2, inflection pt at
x = ln 4, increasing for x < ln 2, decreasing for
x > ln 2, concave up for x > ln 4, concave down
for x < ln 4, horizontal asymptote: y = 0
Section 4.3 (p. 117)
1. x = 0.450184 3. x = 0.567143
5. x = 1.414213 11. global maximum at
x = 2.8214 14. 50
Section 4.4 (p. 122)
1. No 3. Y es 6. No 18. Hint: Calculate
f ′(x) and use the cosine addition formula.
Chapter 5
Section 5.1 (p. 131)
1. x3
3 + 5x2
2 − 3x + C 3. 4ex + C
5. −5 cos x + C 7. 6 ln|x| +C
9. −4
3 x3/2 + C 11. x2
2 + 3
7 x7/3 + C
13. 3 sec x + C 15. −7 cot x + C
Section 5.2 (p. 139)
3. 1
2 4. 1
3 5. 1 6. 1
4
Section 5.3 (p. 145)
1. 1
3 3. 1
4 5. 1
2 7. 1 9. 2e − 2e−1
11. 16
3
Section 5.4 (p. 150)
1. 3 sin 5 x −4 cos 5 x
5 + C
3. −1
2 e−x2
+ 1
3 sin x3 + C
5. ln(1 + ex ) + C
7. 2
5 (x + 4)5/2 − 8
3 (x + 4)3/2 + C
9. tan x − x + C 11. 3
10 tan−1 (5x
2
)
+ C
13. 10 15. 1192
15 17. 1 19. − 1
48
21. π
6 23. 1
2
Section 5.5 (p. 158)",ElementaryCalculus.pdf
"3 (x + 4)3/2 + C
9. tan x − x + C 11. 3
10 tan−1 (5x
2
)
+ C
13. 10 15. 1192
15 17. 1 19. − 1
48
21. π
6 23. 1
2
Section 5.5 (p. 158)
1. 1
2 3. 1 5. divergent 7. 1
ln 2
9. divergent 11. 6 13. divergent
15. π
2 19. Y es 20. No
Chapter 6
Section 6.1 (p. 165)
1. x2 ln x
2 − x2
4 + C 2. (x2 − 2x + 2)ex + C
3. x sin x + cos x + C
5. x2 ax
ln a − 2xa x
ln2 a + 2ax
ln3 a + C
7. x ln x2 − 2x + C
9. Hint: Use a double-angle identity .
11. x sin−1 x +
∝ra⌈〉⌋allow
1 − x2 + C
13. x tan−1 3x − 1
6 ln(1 + 9x2) + C
15. −3
8 sin x cos 3 x + 1
8 cos x sin 3 x + C
17. 1
4 x4 ln2 x − 1
8 x4 ln x + 1
32 x4 + C 19. 16
3
20. 2
∝ra⌈〉⌋allow
2+2
15 21. x sin(ln x)
2 − x cos(ln x)
2 + C
23. x2 tan−1 x
2 − x
2 + tan−1 x
2 + C
24. x cot−1 ∝ra⌈〉⌋allowx + ∝ra⌈〉⌋allowx + cot−1 ∝ra⌈〉⌋allowx + C
25. Hint: Try the substitution t = ∝ra⌈〉⌋allowx.
Section 6.2 (p. 171)
1. − 1
14 cos 7 x + 1
6 cos 3 x + C
3. − 1
14 sin 7 x + 1
6 sin 3 x + C
5. −2
5 cos5/2 x + 2
9 cos9/2 x + C
7. −1
4 sin 2 x + 1
48 sin2 2x + 5
16 x + 3",ElementaryCalculus.pdf
"14 cos 7 x + 1
6 cos 3 x + C
3. − 1
14 sin 7 x + 1
6 sin 3 x + C
5. −2
5 cos5/2 x + 2
9 cos9/2 x + C
7. −1
4 sin 2 x + 1
48 sin2 2x + 5
16 x + 3
64 sin 4 x + C
9. 1
3 tan3 x + tan x + C 11. 1
4 sin4 x + C
Section 6.3 (p. 177)
1. 1
2 x
∝ra⌈〉⌋allow
9 + 4x2 + 9
4 ln |2x +
∝ra⌈〉⌋allow
9 + 4x2 | +C
3. 1
2 x
∝ra⌈〉⌋allow
4x2 − 9 − 9
4 ln |2x +
∝ra⌈〉⌋allow
4x2 − 9| +C
5. −sin−1 x −
∝ra⌈〉⌋allow
1−x2
x + C
7. ln|x| −ln |1 +
∝ra⌈〉⌋allow
1 + x2 | +C
9. 1
3 (x2 + 4)3/2 − 4
∝ra⌈〉⌋allow
x2 + 4 + C
11. 1
108 tan−1 (2x
3
)
+ x
18(9+4x2) + C",ElementaryCalculus.pdf
"318 Appendix A: Answers and Hints to Selected Exercises
13. −9
∝ra⌈〉⌋allow
9 − x2 + 1
3 (9 − x2 )3/2 + C
15. −1
9
∝ra⌈〉⌋allow
−9x2 + 36x − 32 − 2
3 sin−1 (3x−6
2
)
+ C
Section 6.4 (p. 184)
1. −ln|x| +ln|x − 1| +C
3. 1
5 ln|2x − 1| −1
5 ln|x + 2| +C
5. 1
x + 1
2 ln|x − 1| −1
2 ln|x + 1| +C
7. 2 ln|x| +1
x − 2 ln|x + 1| +C
9. −3 ln|x| +2
x + 3 ln|x − 1| + 1
x−1 + C
11. 1
3 tan−1 x − 1
6 tan−1 (x
2
)
+ C
Section 6.5 (p. 193)
1. 2 ln
⏐
⏐sec 1
2 θ
⏐
⏐ − ln |sin θ| +C
3. 2
∝ra⌈〉⌋allow
3 tan−1
(
2 tan 1
2 θ−1
∝ra⌈〉⌋allow
3
)
+ C
5. 4∝ra⌈〉⌋allow
3 tan−1
(
2 tan 1
2 θ−1
∝ra⌈〉⌋allow
3
)
− θ + C
7. ln
⏐
⏐tan 1
2 θ
⏐
⏐ − ln
⏐
⏐tan 1
2 θ + 1
⏐
⏐ + C
9. ln
⏐
⏐tan 1
2 θ
⏐
⏐ − 2 ln
⏐
⏐tan 1
2 θ + 1
⏐
⏐ + C
11.
∝ra⌈〉⌋allow
2π 23. 3
2Γ( 2
3 ) x2/3
Section 6.6 (p. 201)
1. The true value is P ≈ 7.4163
√
l/ g (i.e. the
integral is ≈ 1.8541)
2. 7.416331870724302
√
l/ g
3. 0.8948311310564181
7. 119.9785845899309
9. 0.5967390281992041 (The true value is
0.5963473623231939)
Chapter 7
Section 7.1 (p. 209)",ElementaryCalculus.pdf
"√
l/ g
3. 0.8948311310564181
7. 119.9785845899309
9. 0.5967390281992041 (The true value is
0.5963473623231939)
Chapter 7
Section 7.1 (p. 209)
2. Foci: ( ±3, 0), vertexes: ( ±5, 0), e = 3
5
3. Foci: (0 , ±
∝ra⌈〉⌋allow
5 ), vertexes: (0 , ±3), e =
∝ra⌈〉⌋allow
5
3
5. Foci:
(
±
∝ra⌈〉⌋allow
3
2 , 0
)
, vertexes: ( ±1, 0), e =
∝ra⌈〉⌋allow
3
2
9.
(
± ab∝ra⌈〉⌋allow
a2 +b2 , ± ab∝ra⌈〉⌋allow
a2 +b2
)
13. Hint: Use the two
points you know for certain are on the ellipse
to find the location of the directrix.
16. Hint: Use Exercise 15 and formula (7.3).
Section 7.2 (p. 215)
2. Focus: (0 , 2), vertex: (0 , 0), directrix: y = −2
3. Focus:
(
0, 1
32
)
, vertex: (0 , 0), directrix: y =
− 1
32
4. Focus:
(1
4 , 0
)
, vertex: (0 , 0), directrix: x = −1
4
5. Focus:
(−1
12 , 0
)
, vertex: (0 , 0), directrix: x =
1
12
7. (0, 0) and (4 p, 4 p); y = x 9. |4 p |
11. (3 p, ±2
∝ra⌈〉⌋allow
3 p)
16. Focus:
(
−b
2a , 4ac −b2 +1
4a
)
, vertex:
(
−b
2a , 4ac −b2
4a
)
,
directrix: y = 4ac −b2 −1
4a
Section 7.3 (p. 223)",ElementaryCalculus.pdf
"∝ra⌈〉⌋allow
3 p)
16. Focus:
(
−b
2a , 4ac −b2 +1
4a
)
, vertex:
(
−b
2a , 4ac −b2
4a
)
,
directrix: y = 4ac −b2 −1
4a
Section 7.3 (p. 223)
2. Foci: ( ±5, 0), vertexes: ( ±4, 0), directrices:
x = ±16
5 , asymptotes: y = ±3
4 x, e = 5
4
3. Foci: ( ±
∝ra⌈〉⌋allow
23, 0), vertexes: ( ±2
∝ra⌈〉⌋allow
2, 0), direc-
trices: x = ± 8∝ra⌈〉⌋allow
23 ,
asymptotes: y = ±
∝ra⌈〉⌋allow
15
2
∝ra⌈〉⌋allow
2 x, e =
∝ra⌈〉⌋allow
23
2
∝ra⌈〉⌋allow
2
4. Foci: ( ±
∝ra⌈〉⌋allow
41
2 , 0), vertexes: ( ±5
2 , 0), directri-
ces: x = ± 25
2
∝ra⌈〉⌋allow
41 , asymptotes: y = ±4
5 x,
e =
∝ra⌈〉⌋allow
41
5
5. Foci: ( ±
∝ra⌈〉⌋allow
5
2 , 0), vertexes: ( ±1, 0), directrices:
x = ±2∝ra⌈〉⌋allow
5 , asymptotes: y = ±1
2 x, e =
∝ra⌈〉⌋allow
5
2
6. Foci: (0 , ±
∝ra⌈〉⌋allow
34), vertexes: (0 , ±3), directri-
ces: y = ± 9∝ra⌈〉⌋allow
34 , asymptotes: y = ±3
5 x, e =
∝ra⌈〉⌋allow
34
3
7. x2
9 − y2
16 = 1 17. x2
302500 − y2
697500 = 1
Section 7.4 (p. 229)
1. Foci: (0 , 2) and (6 , 2), vertexes: ( −2, 2) and
(8, 2)
3. Foci: ( −3, 1 ± 2",ElementaryCalculus.pdf
"9 − y2
16 = 1 17. x2
302500 − y2
697500 = 1
Section 7.4 (p. 229)
1. Foci: (0 , 2) and (6 , 2), vertexes: ( −2, 2) and
(8, 2)
3. Foci: ( −3, 1 ± 2
∝ra⌈〉⌋allow
3), vertexes: ( −3, −3) and
(−3, 5)",ElementaryCalculus.pdf
"Appendix A: Answers and Hints to Selected Exercises 319
5. Focus: ( −3, −239
16 ), vertex: ( −3, −15), direc-
trix: y = −241
16
7. Focus:
(1
2 , 5
4
)
, vertex:
(1
2 , 3
2
)
, directrix: y = 7
4
9. Foci: ( −1 ±
∝ra⌈〉⌋allow
13, −3), vertexes: ( −4, −3) and
(2, −3), directrices: x = −1 ± 9∝ra⌈〉⌋allow
13 , asymptotes:
y = ±2
3 (x + 1) − 3
11. Foci: (
∝ra⌈〉⌋allow
2,
∝ra⌈〉⌋allow
2) and ( −
∝ra⌈〉⌋allow
2, −
∝ra⌈〉⌋allow
2), vertexes:
(1, 1) and ( −1, −1), directices: y = −x ±
∝ra⌈〉⌋allow
2,
asymptotes: x = 0 and y = 0
15. hyperbola
Section 7.5 (p. 236)
12. Hint: Use Exercise 11.
24. local maximum at x = ln
∝ra⌈〉⌋allow
3, inflection pt
at x = ln 3, horizontal asymptote: y = 0
26. (b) Hint: See Exercise 10.
27. (b) k1 = c1 + c2, k2 = c1 − c2
30. s0 ≈ 1.006237835313385,
eπ/s0 = 22.69438187638412
33. (a) x = cx +c y
2 − y−x
2c , y = cx +c y
2 + y−x
2c
(b) c = cosh a + sinh a
(c) Hint: Use Example 7.12.
Section 7.6 (p. 243)
5. Y es 6. (a) Hint: Solve for t in terms
of x then substitute into y.",ElementaryCalculus.pdf
"2c
(b) c = cosh a + sinh a
(c) Hint: Use Example 7.12.
Section 7.6 (p. 243)
5. Y es 6. (a) Hint: Solve for t in terms
of x then substitute into y.
(b) Hint: Use the distance formula.
8. Hint: Does BP = /hbracewidest
AB ? 9. (a) − 7
2(t+1)3
(c)
(49
25 , 14
5
)
12. x = t2 − 1, y = t(t2 − 1)
Section 7.7 (p. 250)
1. r = 6 cos θ 3. r2 = sec 2 θ
5. θ = 3π
4 7. r = sec θ tan θ
9. x4 + 2x2 y2 + y4 = 4x2 − 4 y2
11. y − 1 = −3
∝ra⌈〉⌋allow
3
5 (x +
∝ra⌈〉⌋allow
3 )
13. local maxima at
(
2, π
2
)
and
(
0, 3π
2
)
,
local minima at
(1
2 , 7π
6
)
and
(1
2 , 11π
6
)
15. local maxima at
(
2
∝ra⌈〉⌋allow
2
3 , α
)
and(
−2
∝ra⌈〉⌋allow
2
3 , 2π− α
)
, local minima at
(
−2
∝ra⌈〉⌋allow
2
3 , π− α
)
and
(
2
∝ra⌈〉⌋allow
2
3 , π+ α
)
,
where α = tan−1 ∝ra⌈〉⌋allow
2 17. π
2
Chapter 8
Section 8.1 (p. 256)
1. 32
3 2. 4
3 3. 1
12 5. 9
2 7. 19
3 9. 1
12
11. 3π 13. 25
(π
3 + 1 −
∝ra⌈〉⌋allow
3
)
17. 6250π3
3 sq ft
Section 8.2 (p. 261)
1. 1 3. 4
3 5. 2
π 7. 0 9. 1
2 ln 3
Section 8.3 (p. 270)
1. 8
27 (103/2) − 1",ElementaryCalculus.pdf
"(π
3 + 1 −
∝ra⌈〉⌋allow
3
)
17. 6250π3
3 sq ft
Section 8.2 (p. 261)
1. 1 3. 4
3 5. 2
π 7. 0 9. 1
2 ln 3
Section 8.3 (p. 270)
1. 8
27 (103/2) − 1
27 (133/2) ≈ 7.634
2.
∝ra⌈〉⌋allow
5
2 + 1
4 sinh−1 2 ≈ 1.479
3. 8
27 (103/2) − 1
27 (133/2) ≈ 7.634
4. 3
4 + ln
∝ra⌈〉⌋allow
2 ≈ 1.097
7.
∝ra⌈〉⌋allow
2(eπ − 1) ≈ 31.312
9. 8 11. 3 13. κ(0) = 0, κ( π
2 ) = −1
15. κ(0) = a
b2 at ( a, 0), κ( π
2 ) = b
a2 at (0 , b)
17. −1 23. 13.27 ft
Section 8.4 (p. 276)
1. 4π 2. π
2 (2 + sinh 2) 3. 208π
9
5. π2
2 7. 2π 9. π
10 10. π
6
13. S = πr
∝ra⌈〉⌋allow
r2 + h2, V = 1
3 πr2 h
Section 8.5 (p. 283)
1.
(4
5 , 2
7
)
3.
(3
5 , 12
35
)
5.
(4r
3π , 4r
3π
)
7.
(
0, 11
4π
)
9. 0.192 Nm
11. R T log
(
Vb
Va
)
13. 0.3486 15.
(
1, 1
4
)
18. Hint: Use Exercise 28 in Section 6.1.
20. Hint: Use the equation from Section 5.1
for free fall motion to write time as a function
of height.",ElementaryCalculus.pdf
"320 Appendix A: Answers and Hints to Selected Exercises
Chapter 9
Section 9.1 (p. 291)
1. Converges to 0 2. Converges to 1
3
3. Converges to 0 5. Divergent
7. Divergent 9. 6 11. 32 13. 113
999
14. 1 15. 1
4 20. 132
7 ft 24. No
Section 9.2 (p. 298)
6. Divergent 7. Convergent
8. Divergent 9. Convergent
10. Divergent 11. Convergent
12. Convergent 13. Convergent
14. Divergent 15. Divergent
16. Convergent 17. Convergent
18. 1
6 19. 1
10 20. 1
6 21. 1
2
Section 9.3 (p. 302)
1. Conditionally convergent
3. Conditionally convergent
5. Absolutely convergent
7. Answer to second question: Y es
Section 9.4 (p. 307)
1. −1 ≤ x < 1 2. −2 < x < 2 3. 1 < x < 3
4. −6 < x < −2 5. −∞ <x < ∞ 6. x = 0
7. −1 < x < 1 10. Hint: Use Example 9.19
Section 9.5 (p. 313)
1. 1 −
(x− π
2 )2
2! +
(x− π
2 )4
4! − ···
2. x + x3
3! + x5
5! + ··· 3. 1 + x2
2! + x4
4! + ···
4. x + x3
3 + 2x5
15 + ··· 5. x − x3
3 + 2x5
15 − ···
6. 1 + x2
2 + 5x4
24 + ··· 7. 1 − x2 + x4 − ···
8. 1
2 − (x−1)
2 + (x−1)2
4 − ···
9. 1 + x2
2 − x4",ElementaryCalculus.pdf
"3 + 2x5
15 + ··· 5. x − x3
3 + 2x5
15 − ···
6. 1 + x2
2 + 5x4
24 + ··· 7. 1 − x2 + x4 − ···
8. 1
2 − (x−1)
2 + (x−1)2
4 − ···
9. 1 + x2
2 − x4
8 + ··· 11. x2 − x4
2 + 3x6
8 − ···
13. x − x5
10 + x9
216 − ··· 15. x + x7
14 − x13
104 + ···
18. 0.68485 19. 97.18; −132.605",ElementaryCalculus.pdf
"GNU Free Documentation License
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near the most prominent appearance of the work’s title, prec eding the beginning of the body of
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The “ publisher” means any person or entity that distributes copies of the Do cument to the
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If the required texts for either cover are too voluminous to fi t legibly , you should put the first",ElementaryCalculus.pdf
"verbatim copying in other respects.
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plete Transparent copy of the Document, free of added materi al. If you use the latter option,",ElementaryCalculus.pdf
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"History
This section contains the revision history of the book. For p ersons making modifications to the
book, please record the pertinent information here, follow ing the format in the first item below .
1. VERSION: 0.1
Date: 2016-01-24
Author(s): Michael Corral
Title: Elementary Calculus
Modification(s): Initial version
2. VERSION: 0.5
Date: 2020-05-27
Author(s): Michael Corral
Title: Elementary Calculus
Modification(s): Numerous corrections of typos and errors, as well as enhance ments of
some sections and more exercises.
3. VERSION: 0.5a
Date: 2020-06-03
Author(s): Michael Corral
Title: Elementary Calculus
Modification(s): Replaced the old Section 5.5 (Average V alue of a Function) wi th the new
Section 5.5 (Improper Integrals). Average V alue of a Functi on will be moved to Chapter
8 in the second half of the book.
4. VERSION: 1.0
Date: 2020-12-31
Author(s): Michael Corral
Title: Elementary Calculus
Modification(s): Full version.
329",ElementaryCalculus.pdf
"Index
Symbols
∆x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
〈 f 〉. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
˙f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
d2 y
dx2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
dn
dxn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
≫ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
≪ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
N. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4",ElementaryCalculus.pdf
"N. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
f ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
f (n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
n-th Term Test . . . . . . . . . . . . . . . . . . . . . . . . . 294
A
absolute convergence . . . . . . . . . . . . . . . . . . . 301
absolute value . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
accelerating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
algebraic curve . . . . . . . . . . . . . . . . . . . . . . . . . . 80
alternating series . . . . . . . . . . . . . . . . . . . . . . 300
angle of contingence . . . . . . . . . . . . . . . . . . . . 267",ElementaryCalculus.pdf
"alternating series . . . . . . . . . . . . . . . . . . . . . . 300
angle of contingence . . . . . . . . . . . . . . . . . . . . 267
angle of tangent line . . . . . . . . . . . . . . . . . . . . . 59
antiderivative . . . . . . . . . . . . . . . . . . . . . . . . . . 124
arc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
arc length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
parametric . . . . . . . . . . . . . . . . . . . . . . . . . 266
polar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
area between curves . . . . . . . . . . . . . . . . . . . . 252
area function . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
area under a curve . . . . . . . . . . . . . . . . . . . . . 133
asymptote . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
horizontal . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
vertical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66",ElementaryCalculus.pdf
"horizontal . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
vertical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
average value . . . . . . . . . . . . . . . . . . . . . . . . . . 258
axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . 203
B
Bézier curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
base a logarithm . . . . . . . . . . . . . . . . . . . . . . . . . 54
Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . 305
Bessel functions
order m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
order zero . . . . . . . . . . . . . . . . . . . . . . . . . . 305
Bessel’s equation . . . . . . . . . . . . . . . . . . . . . . . 305
order m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
Beta function . . . . . . . . . . . . . . . . . . . . . . . . . . . 188",ElementaryCalculus.pdf
"order m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
Beta function . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
Big O notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
bipolar coordinates . . . . . . . . . . . . . . . . . . . . . 251
bisection method . . . . . . . . . . . . . . . . . . . . . . . 109
Bohr radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
brachistochrone . . . . . . . . . . . . . . . . . . . . . . . . 243
C
catenary . . . . . . . . . . . . . . . . . . . . . . . . . . . 231, 264
ceiling function . . . . . . . . . . . . . . . . . . . . . . . . . . 74
center of gravity . . . . . . . . . . . . . . . . . . . . . . . . 277
Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Chebyshev polynomials . . . . . . . . . . . . . . . . . . 44
330",ElementaryCalculus.pdf
"Index 331
circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
definition . . . . . . . . . . . . . . . . . . . . . . . . . . 202
involute . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
circular functions . . . . . . . . . . . . . . . . . . . . . . . 230
closed interval . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
Comparison Test . . . . . . . . . . . . . . . . . . . . . . . 293
concave down . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
concave up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
conditional convergence . . . . . . . . . . . . . . . . 301
conic sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
conjugate axis . . . . . . . . . . . . . . . . . . . . . . . . . . 217
constant function . . . . . . . . . . . . . . . . . . . . . . . . . 9
Constant Multiple Rule . . . . . . . . . . . . . . . . . 21",ElementaryCalculus.pdf
"constant function . . . . . . . . . . . . . . . . . . . . . . . . . 9
Constant Multiple Rule . . . . . . . . . . . . . . . . . 21
continued fraction . . . . . . . . . . . . . . . . . . . . . . 292
continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
continuous function . . . . . . . . . . . . . . . . . . . . . 73
convergent integral . . . . . . . . . . . . . . . 152, 155
coordinate transformations . . . . . . . . . . . . 224
CORDIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
countable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
crackle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
critical point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
curvature . . . . . . . . . . . . . . . . . . . . . . . . . . 267, 268
average . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
parametric . . . . . . . . . . . . . . . . . . . . . . . . . 269",ElementaryCalculus.pdf
"average . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
parametric . . . . . . . . . . . . . . . . . . . . . . . . . 269
polar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
cycloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
D
Darboux’s Theorem . . . . . . . . . . . . . . . . . . . . . 122
decelerating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
definite integral . . . . . . . . . . . . . . . . . . . . . . . . 132
derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3, 8
n-th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Constant Multiple Rule . . . . . . . . . . . . . 21
Difference Rule . . . . . . . . . . . . . . . . . . . . . 21
exponential function . . . . . . . . . . . . . . . . 46
first . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32",ElementaryCalculus.pdf
"exponential function . . . . . . . . . . . . . . . . 46
first . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
fractional . . . . . . . . . . . . . . . . . . . . . . . . . . 189
higher order . . . . . . . . . . . . . . . . . . . . . . . . . 32
natural logarithm function . . . . . . . . . . 48
polar coordinates . . . . . . . . . . . . . . . . . . 248
Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Product Rule . . . . . . . . . . . . . . . . . . . . . . . . 21
Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . 21
second . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Sum Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
trigonometric functions . . . . . . . . . . . . . 23
with respect to x . . . . . . . . . . . . . . . . . . . . 15
zero-th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Difference Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 21",ElementaryCalculus.pdf
"zero-th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Difference Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 21
differentiable . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
differential equation . . . . . . . . . . . . . . . . . 36, 47
differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
implicit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
under the integral sign . . . . . . . . . . . . 185
Dirac delta function . . . . . . . . . . . . . . . . . . . . 152
directrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
disc method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
discontinuous . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
distribution function . . . . . . . . . . . . . . . . . . . 284
divergent integral . . . . . . . . . . . . . . . . . 152, 155",ElementaryCalculus.pdf
"distribution function . . . . . . . . . . . . . . . . . . . 284
divergent integral . . . . . . . . . . . . . . . . . 152, 155
domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8, 37
dot notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
E
e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
eccentric angle . . . . . . . . . . . . . . . . . . . . . . . . . 243
Efimov trimer . . . . . . . . . . . . . . . . . . . . . . . . . . 237
eigenfunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
eigenvalue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203",ElementaryCalculus.pdf
"area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
construction . . . . . . . . . . . . . . . . . . . . . . . . 203
definition . . . . . . . . . . . . . . . . . . . . . . . . . . 202
diameter . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
directrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
eccentric angle . . . . . . . . . . . . . . . . . . . . . 243
foci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202",ElementaryCalculus.pdf
"332 Index
latus rectum . . . . . . . . . . . . . . . . . . . . . . . 209
major auxiliary circle . . . . . . . . . . . . . . 243
major axis . . . . . . . . . . . . . . . . . . . . . . . . . . 203
minor auxiliary circle . . . . . . . . . . . . . . 243
minor axis . . . . . . . . . . . . . . . . . . . . . . . . . 203
principal axis . . . . . . . . . . . . . . . . . . . . . . 203
reflection property . . . . . . . . . . . . . . . . . 206
second definition . . . . . . . . . . . . . . . . . . . 209
semi-major axis . . . . . . . . . . . . . . . . . . . . 203
semi-minor axis . . . . . . . . . . . . . . . . . . . . 203
vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
ellipsograph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
elliptic curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
elliptic function . . . . . . . . . . . . . . . . . . . . . . . . 270",ElementaryCalculus.pdf
"elliptic curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
elliptic function . . . . . . . . . . . . . . . . . . . . . . . . 270
elliptic integral . . . . . . . . . . . . . . . . . . . . . . . . . 264
enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
escape velocity . . . . . . . . . . . . . . . . . . . . . . . . . 216
Eudoxus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Euler number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
even function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
event . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
expected value . . . . . . . . . . . . . . . . . . . . . . . . . . 284
exponential decay . . . . . . . . . . . . . . . . . . . . . . . 50
exponential distribution . . . . . . . . . . . . . . . . 283",ElementaryCalculus.pdf
"exponential decay . . . . . . . . . . . . . . . . . . . . . . . 50
exponential distribution . . . . . . . . . . . . . . . . 283
exponential functions . . . . . . . . . . . . . . . . . . . . 45
exponential growth . . . . . . . . . . . . . . . . . . . . . . 51
Extended Mean V alue Theorem . . . . . . . . 121
F
factorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Fermat’s Principle . . . . . . . . . . . . . . . . . . 95, 101
Fibonacci sequence . . . . . . . . . . . . . . . . 288, 292
First Derivative Test . . . . . . . . . . . . . . . . . . . 105
floor function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
focal radius . . . . . . . . . . . . . . . . . . . . . . . . 212, 223
fractional derivative . . . . . . . . . . . . . . . . . . . . 189
free fall formulas . . . . . . . . . . . . . . . . . . 129, 130
frustrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271",ElementaryCalculus.pdf
"free fall formulas . . . . . . . . . . . . . . . . . . 129, 130
frustrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
absolute value . . . . . . . . . . . . . . . . . . . . . . . 13
area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
average value . . . . . . . . . . . . . . . . . . . . . . 258
ceiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
circular . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
composition . . . . . . . . . . . . . . . . . . . . . . . . . 27
constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
continuous . . . . . . . . . . . . . . . . . . . . . . . . . . 73
differentiable . . . . . . . . . . . . . . . . . . . . . . . . 14
elliptic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
even . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13",ElementaryCalculus.pdf
"elliptic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
even . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
exponential . . . . . . . . . . . . . . . . . . . . . . . . . . 45
floor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
hyperbolic . . . . . . . . . . . . . . . . . . . . . . . . . . 230
infinitesimal . . . . . . . . . . . . . . . . . . . . . . . . 72
inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
inverse hyperbolic . . . . . . . . . . . . . . . . . 234
inverse trigonometric . . . . . . . . . . . . . . . 41
Langevin . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
natural logarithm . . . . . . . . . . . . . . . . . . . 47
objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13, 143",ElementaryCalculus.pdf
"objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13, 143
rational . . . . . . . . . . . . . . . . . . . . . . . . 70, 178
root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
sigmoid neuron . . . . . . . . . . . . . . . . . . . . . . 52
step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
trigonometric . . . . . . . . . . . . . . . . . . . . . . . . 41
Fundamental Theorem of Calculus . . . . 140
G
Gamma function . . . . . . . . . . . . . . . . . . 159, 188
Gaussian quadrature . . . . . . . . . . . . . . . . . . . 199
Gibbs energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
golden ratio . . . . . . . . . . . . . . . . . . . . . . . 288, 292
H
half-angle substitution . . . . . . . . . . . . . . . . . 190
half-life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50",ElementaryCalculus.pdf
"H
half-angle substitution . . . . . . . . . . . . . . . . . 190
half-life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
harmonic series . . . . . . . . . . . . . . . . . . . . . . . . 296
height function . . . . . . . . . . . . . . . . . . . . . . . . . 252
Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
horizontal asymptote . . . . . . . . . . . . . . . . . . . . 67
hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216",ElementaryCalculus.pdf
"Index 333
asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . 217
center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
conjugate axis . . . . . . . . . . . . . . . . . . . . . . 217
construction . . . . . . . . . . . . . . . . . . . . . . . . 219
definition . . . . . . . . . . . . . . . . . . . . . . . . . . 216
directrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
eccentricity . . . . . . . . . . . . . . . . . . . . . . . . 216
focal radius . . . . . . . . . . . . . . . . . . . . . . . . 223
focus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
latus rectum . . . . . . . . . . . . . . . . . . . . . . . 223
reflection property . . . . . . . . . . . . . . . . . 220
sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
transverse axis . . . . . . . . . . . . . . . . . . . . 217
vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217",ElementaryCalculus.pdf
"transverse axis . . . . . . . . . . . . . . . . . . . . 217
vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
hyperbolic angle . . . . . . . . . . . . . . . . . . . . . . . . 230
hyperbolic cosine . . . . . . . . . . . . . . . . . . . . . . . 230
hyperbolic functions . . . . . . . . . . . . . . . 230, 231
inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
hyperbolic rotation . . . . . . . . . . . . . . . . . . . . . 238
hyperbolic sector . . . . . . . . . . . . . . . . . . . . . . . 230
hyperbolic sine . . . . . . . . . . . . . . . . . . . . . . . . . 231
I
ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
implicit differentiation . . . . . . . . . . . . . . . . . . 79
improper integral . . . . . . . . . . . . . . . . . . . . . . 152
indefinite integral . . . . . . . . . . . . . . . . . . . . . . 125
indeterminate form . . . . . . . . . . . . . . . . . . 69, 75",ElementaryCalculus.pdf
"indefinite integral . . . . . . . . . . . . . . . . . . . . . . 125
indeterminate form . . . . . . . . . . . . . . . . . . 69, 75
induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
infinite series . . . . . . . . . . . . . . . . . . . . . . . . 4, 289
convergent . . . . . . . . . . . . . . . . . . . . . . . . . 289
divergent . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
partial sum . . . . . . . . . . . . . . . . . . . . . . . . 289
sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
infinitesimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
infinitesimal function . . . . . . . . . . . . . . . . . . . . 72
inflection point . . . . . . . . . . . . . . . . . . . . . . . . . 103
instantaneous rate of change . . . . . . . . . . . 3, 8
integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4",ElementaryCalculus.pdf
"integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
convergent . . . . . . . . . . . . . . . . . . . . 152, 155
definite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
divergent . . . . . . . . . . . . . . . . . . . . . 152, 155
exponential function . . . . . . . . . . . . . . . 128
improper . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
indefinite . . . . . . . . . . . . . . . . . . . . . . . . . . 125
trigonometric . . . . . . . . . . . . . . . . . 128, 167
integral sign . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
integrand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Monte Carlo . . . . . . . . . . . . . . . . . . . . . . . 255",ElementaryCalculus.pdf
"by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Monte Carlo . . . . . . . . . . . . . . . . . . . . . . . 255
numerical . . . . . . . . . . . . . . . . . . . . . . . . . . 195
tabular method . . . . . . . . . . . . . . . . . . . . 163
interval of convergence . . . . . . . . . . . . . . . . . 303
inverse function . . . . . . . . . . . . . . . . . . . . . . . . . 38
inverse hyperbolic functions . . . . . . . . . . . 234
inverse trigonometric functions . . . . . . . . . 41
involute of circle . . . . . . . . . . . . . . . . . . . . . . . . 244
irrational numbers . . . . . . . . . . . . . . . . . . . . . . . . 5
J
Jacobian elliptic functions . . . . . . . . . . . . . . 270
jerk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
jump discontinuity . . . . . . . . . . . . . . . . . . . . . . 73
L
L ’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Laguerre polynomials . . . . . . . . . . . . . . . . . . 201",ElementaryCalculus.pdf
"L
L ’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Laguerre polynomials . . . . . . . . . . . . . . . . . . 201
lamina . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
Langevin function . . . . . . . . . . . . . . . . . . . . . . 238
Laplace transform . . . . . . . . . . . . . . . . . . . . . . 237
latus rectum . . . . . . . . . . . . . . . . . . . . . . . 209, 223
of ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
of parabola . . . . . . . . . . . . . . . . . . . . . . . . . 215
left continuous . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
Legendre polynomials . . . . . . . . . . . . . . . . . . 201
Leibniz integral rule . . . . . . . . . . . . . . . . . . . 185
Leibniz notation . . . . . . . . . . . . . . . . . . . . . . . . . 15
limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3, 62
infinite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66",ElementaryCalculus.pdf
"infinite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
left . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
one-sided . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65",ElementaryCalculus.pdf
"334 Index
Limit Comparison Test . . . . . . . . . . . . . . . . . 294
limits of integration . . . . . . . . . . . . . . . . . . . . 139
line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
linear combination . . . . . . . . . . . . . . . . . . . . . . 24
linear function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
linear operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
logarithmic differentiation . . . . . . . . . . . 49, 86
Lorentz factor . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Lorentz transformation . . . . . . . . . . . . . . . . 237
M
Mach number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Maclaurin’s series . . . . . . . . . . . . . . . . . . . . . . 309
magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34",ElementaryCalculus.pdf
"Maclaurin’s series . . . . . . . . . . . . . . . . . . . . . . 309
magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
major auxiliary circle . . . . . . . . . . . . . . . . . . . 243
maximum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
Maxwell speed distribution . . . . . . . . . . . . 166
Mean V alue Theorem . . . . . . . . . . . . . . . . . . . 118
alternative form . . . . . . . . . . . . . . . . . . . 121
Extended . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
median of triangle . . . . . . . . . . . . . . . . . . . . . . 285
method of exhaustion . . . . . . . . . . . . . . . . . . . . . 6
Microstraightness Property . . . . . . . . . . . . . 17
minimum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
minor auxiliary circle . . . . . . . . . . . . . . . . . . 243
modulated wave . . . . . . . . . . . . . . . . . . . . . . . . 167
moment . . . . . . . . . . . . . . . . . . . . . . . . . . . 277, 278",ElementaryCalculus.pdf
"modulated wave . . . . . . . . . . . . . . . . . . . . . . . . 167
moment . . . . . . . . . . . . . . . . . . . . . . . . . . . 277, 278
Monotone Bounded Test . . . . . . . . . . . . . . . . 293
Monte Carlo integration . . . . . . . . . . . . . . . . 255
Monte Carlo method . . . . . . . . . . . . . . . . . . . 260
N
nappe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
natural logarithm . . . . . . . . . . . . . . . . . . . . . . . 47
natural numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Newton’s method . . . . . . . . . . . . . . . . . . . . . . . 109
norm of a partition . . . . . . . . . . . . . . . . . . . . . 135
normal distribution . . . . . . . . . . . . . . . . . . . . 284
normal line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Nth Derivative Test . . . . . . . . . . . . . . . . . . . . . 106
numerical integration . . . . . . . . . . . . . . . . . . 195
O",ElementaryCalculus.pdf
"Nth Derivative Test . . . . . . . . . . . . . . . . . . . . . 106
numerical integration . . . . . . . . . . . . . . . . . . 195
O
O notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
objective function . . . . . . . . . . . . . . . . . . . . . . . . 90
odd function . . . . . . . . . . . . . . . . . . . . . . . . 13, 143
one-to-one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
open interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
P
p-series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
p-series Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
Padé approximation . . . . . . . . . . . . . . . . . . . . 314
parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210",ElementaryCalculus.pdf
"parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
construction . . . . . . . . . . . . . . . . . . . . . . . . 210
definition . . . . . . . . . . . . . . . . . . . . . . . . . . 210
directrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
eccentricity . . . . . . . . . . . . . . . . . . . . . . . . 210
focal radius . . . . . . . . . . . . . . . . . . . . . . . . 212
focus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
latus rectum . . . . . . . . . . . . . . . . . . . . . . . 215
reflection property . . . . . . . . . . . . . . . . . 212
vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
paraboloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
parametric equations . . . . . . . . . . . . . . . . . . . 239",ElementaryCalculus.pdf
"parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
parametric equations . . . . . . . . . . . . . . . . . . . 239
parametrization of a curve . . . . . . . . . . . . . 239
partial fractions . . . . . . . . . . . . . . . . . . . . . . . . 178
partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
periodic waveform . . . . . . . . . . . . . . . . . . . . . . 261
Planck’s constant . . . . . . . . . . . . . . . . . . . . . . . . 36
plane curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
point of tangency . . . . . . . . . . . . . . . . . . . . . . . . 56
polar axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
polar coordinates . . . . . . . . . . . . . . . . . . . . . . . 245
derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 248
pop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34",ElementaryCalculus.pdf
"pop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Power Formula . . . . . . . . . . . . . . . . . . . . . . . . . 127
Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
power series . . . . . . . . . . . . . . . . . . . . . . . . . 4, 303
differentiation . . . . . . . . . . . . . . . . . . . . . 304",ElementaryCalculus.pdf
"Index 335
integration . . . . . . . . . . . . . . . . . . . . . . . . . 304
interval of convergence . . . . . . . . . . . . 303
radius of convergence . . . . . . . . . . . . . . 303
principal axis . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . 203
probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
probability density function . . . . . . . . . . . . 282
Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
product-to-sum formulas . . . . . . . . . . . . . . . 167
proof by induction . . . . . . . . . . . . . . . . . . . . . . . 24
Q
q-derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
q-differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
quantum calculus . . . . . . . . . . . . . . . . . . . . . . . 89
quicksort algorithm . . . . . . . . . . . . . . . . . . . . 139",ElementaryCalculus.pdf
"quantum calculus . . . . . . . . . . . . . . . . . . . . . . . 89
quicksort algorithm . . . . . . . . . . . . . . . . . . . . 139
Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
R
radioactive decay . . . . . . . . . . . . . . . . . . . . . . . . 50
radius of convergence . . . . . . . . . . . . . . . . . . 303
random variable . . . . . . . . . . . . . . . . . . . . . . . . 282
continuous . . . . . . . . . . . . . . . . . . . . . . . . . 282
discrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
rational curve . . . . . . . . . . . . . . . . . . . . . . . . . . 244
rational function . . . . . . . . . . . . . . . . . . . . 70, 178
rational numbers . . . . . . . . . . . . . . . . . . . . . . . . . 4
real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4",ElementaryCalculus.pdf
"rational numbers . . . . . . . . . . . . . . . . . . . . . . . . . 4
real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
recurrence relation . . . . . . . . . . . . . . . . . . . . . 288
reflection property . . . . . . . . . . . . . . . . . . . . . . 206
ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . 220
parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
regular polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
related rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
relativistic momentum . . . . . . . . . . . . . . . . . 285
relativistic velocity addition law . . . . . . . 123
reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Riemann sum . . . . . . . . . . . . . . . . . . . . . . . . . . 134
Riemann’s Rearrangement Theorem . . . 302
right continuous . . . . . . . . . . . . . . . . . . . . . . . . . 73",ElementaryCalculus.pdf
"Riemann’s Rearrangement Theorem . . . 302
right continuous . . . . . . . . . . . . . . . . . . . . . . . . . 73
Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 119
root mean square . . . . . . . . . . . . . . . . . . . . . . . 261
root of function . . . . . . . . . . . . . . . . . . . . . . . . . 109
rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
rotational velocity . . . . . . . . . . . . . . . . . . . . . . . 31
S
scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
Schrödinger equation . . . . . . . . . . . . . . . . . . . . 36
Schwarzchild radius . . . . . . . . . . . . . . . . . . . . 101
secant line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
secant method . . . . . . . . . . . . . . . . . . . . . . . . . . 112
Second Derivative Test . . . . . . . . . . . . . . . . . . 91
second-degree equation . . . . . . . . . . . . . . . . 227",ElementaryCalculus.pdf
"Second Derivative Test . . . . . . . . . . . . . . . . . . 91
second-degree equation . . . . . . . . . . . . . . . . 227
sequence . . . . . . . . . . . . . . . . . . . . . . . . 4, 286, 287
Comparison Test . . . . . . . . . . . . . . . . . . . 293
convergent . . . . . . . . . . . . . . . . . . . . . . . . . 287
divergent . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
Fibonacci . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
Monotone Bounded Test . . . . . . . . . . . 293
series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4, 286
absolutely convergent . . . . . . . . . . . . . . 301
alternating . . . . . . . . . . . . . . . . . . . . . . . . . 300
conditional convergent . . . . . . . . . . . . . 301
harmonic . . . . . . . . . . . . . . . . . . . . . . 296, 300
infinite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289",ElementaryCalculus.pdf
"harmonic . . . . . . . . . . . . . . . . . . . . . . 296, 300
infinite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
Integral Test . . . . . . . . . . . . . . . . . . . . . . . 294
Limit Comparison Test . . . . . . . . . . . . 294
n-th Term Test . . . . . . . . . . . . . . . . . . . . . 294
p-series Test . . . . . . . . . . . . . . . . . . . . . . . 294
Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . 294
rearrangement . . . . . . . . . . . . . . . . . . . . . 302
telescoping . . . . . . . . . . . . . . . . . . . . . . . . . 298
Telescoping Series Test . . . . . . . . . . . . 294
shell method . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
sigma notation . . . . . . . . . . . . . . . . . . . . . . . . . 135
sigmoid neuron function . . . . . . . . . . . . . . . . . 52
Simpson’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . 197
slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9",ElementaryCalculus.pdf
"slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
slope of a curve . . . . . . . . . . . . . . . . . . . . . . . . . . 58
smooth curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18",ElementaryCalculus.pdf
"336 Index
snap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
solid of revolution . . . . . . . . . . . . . . . . . . . . . . 273
disc method . . . . . . . . . . . . . . . . . . . . . . . . 273
speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2, 34
Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . . . 71
stationary point . . . . . . . . . . . . . . . . . . . . . . . . . 91
step function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
straight line motion . . . . . . . . . . . . . . 1, 33, 129
substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
half-angle . . . . . . . . . . . . . . . . . . . . . . . . . . 190
trigonometric . . . . . . . . . . . . . . . . . . . . . . 172
substitution method . . . . . . . . . . . . . . . . . . . . 146
Sum Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21",ElementaryCalculus.pdf
"substitution method . . . . . . . . . . . . . . . . . . . . 146
Sum Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
surface of revolution . . . . . . . . . . . . . . . . . . . 271
T
tabular integration . . . . . . . . . . . . . . . . . . . . . 163
tangent line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
angle of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Taylor polynomial . . . . . . . . . . . . . . . . . . . . . . 312
Taylor’s formula . . . . . . . . . . . . . . . . . . . . . . . . 308
Taylor’s series . . . . . . . . . . . . . . . . . . . . . . . . . . 308
Remainder Theorem . . . . . . . . . . . . . . . 313
telescoping series . . . . . . . . . . . . . . . . . . . . . . . 298
Telescoping Series Test . . . . . . . . . . . . . . . . . 294
Thales’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 18
time dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72",ElementaryCalculus.pdf
"Thales’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 18
time dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
trajectory envelope . . . . . . . . . . . . . . . . . . . . . 213
translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
transverse axis . . . . . . . . . . . . . . . . . . . . . . . . . 217
trapezoid rule . . . . . . . . . . . . . . . . . . . . . . . . . . 197
trigonometric functions . . . . . . . . . . . . . . 23, 41
inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
trigonometric integrals . . . . . . . . . . . . 128, 167
trigonometric substitution . . . . . . . . . . . . . 172
U
unit hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . 230
V
vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281",ElementaryCalculus.pdf
"V
vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
average . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
instantaneous . . . . . . . . . . . . . . . . . . . . . . . . 2
vertical asymptote . . . . . . . . . . . . . . . . . . . . . . . 66
W
W allis’ formula . . . . . . . . . . . . . . . . . . . . . . . . . 299
work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
Z
Zeno’s motion paradox . . . . . . . . . . . . 286, 290",ElementaryCalculus.pdf
"Grinstead and Snell's Introduction to Probability
The CHANCE Project1
Version dated 4 July 2006
1Copyright (C) 2006 Peter G. Doyle. This work is a version of Grinstead and Snell's
`Introduction to Probability, 2nd edition', published by the American Mathematical So-
ciety, Copyright (C) 2003 Charles M. Grinstead and J. Laurie Snell. This work is freely
redistributable under the terms of the GNU Free Documentation License.",prob.pdf
"To our wives
and in memory of
Reese T. Prosser",prob.pdf
"Contents
Preface vii
1 Discrete Probability Distributions 1
1.1 Simulation of Discrete Probabilities . . . . . . . . . . . . . . . . . . . 1
1.2 Discrete Probability Distributions . . . . . . . . . . . . . . . . . . . . 18
2 Continuous Probability Densities 41
2.1 Simulation of Continuous Probabilities . . . . . . . . . . . . . . . . . 41
2.2 Continuous Density Functions . . . . . . . . . . . . . . . . . . . . . . 55
3 Combinatorics 75
3.1 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
3.2 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
3.3 Card Shuing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
4 Conditional Probability 133
4.1 Discrete Conditional Probability . . . . . . . . . . . . . . . . . . . . 133
4.2 Continuous Conditional Probability . . . . . . . . . . . . . . . . . . . 162
4.3 Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
5 Distributions and Densities 183",prob.pdf
"4.3 Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
5 Distributions and Densities 183
5.1 Important Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 183
5.2 Important Densities . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
6 Expected Value and Variance 225
6.1 Expected Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
6.2 Variance of Discrete Random Variables . . . . . . . . . . . . . . . . . 257
6.3 Continuous Random Variables . . . . . . . . . . . . . . . . . . . . . . 268
7 Sums of Random Variables 285
7.1 Sums of Discrete Random Variables . . . . . . . . . . . . . . . . . . 285
7.2 Sums of Continuous Random Variables . . . . . . . . . . . . . . . . . 291
8 Law of Large Numbers 305
8.1 Discrete Random Variables . . . . . . . . . . . . . . . . . . . . . . . 305
8.2 Continuous Random Variables . . . . . . . . . . . . . . . . . . . . . . 316
v",prob.pdf
"vi CONTENTS
9 Central Limit Theorem 325
9.1 Bernoulli Trials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
9.2 Discrete Independent Trials . . . . . . . . . . . . . . . . . . . . . . . 340
9.3 Continuous Independent Trials . . . . . . . . . . . . . . . . . . . . . 356
10 Generating Functions 365
10.1 Discrete Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 365
10.2 Branching Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 376
10.3 Continuous Densities . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
11 Markov Chains 405
11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405
11.2 Absorbing Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . 416
11.3 Ergodic Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . 433
11.4 Fundamental Limit Theorem . . . . . . . . . . . . . . . . . . . . . . 447
11.5 Mean First Passage Time . . . . . . . . . . . . . . . . . . . . . . . . 452",prob.pdf
"11.5 Mean First Passage Time . . . . . . . . . . . . . . . . . . . . . . . . 452
12 Random Walks 471
12.1 Random Walks in Euclidean Space . . . . . . . . . . . . . . . . . . . 471
12.2 Gambler's Ruin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486
12.3 Arc Sine Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
Appendices 499
Index 503",prob.pdf
"Preface
Probability theory began in seventeenth century France when the two great French
mathematicians, Blaise Pascal and Pierre de Fermat, corresponded over two prob-
lems from games of chance. Problems like those Pascal and Fermat solved continued
to in
uence such early researchers as Huygens, Bernoulli, and DeMoivre in estab-
lishing a mathematical theory of probability. Today, probability theory is a well-
established branch of mathematics that nds applications in every area of scholarly
activity from music to physics, and in daily experience from weather prediction to
predicting the risks of new medical treatments.
This text is designed for an introductory probability course taken by sophomores,
juniors, and seniors in mathematics, the physical and social sciences, engineering,
and computer science. It presents a thorough treatment of probability ideas and
techniques necessary for a rm understanding of the subject. The text can be used",prob.pdf
"techniques necessary for a rm understanding of the subject. The text can be used
in a variety of course lengths, levels, and areas of emphasis.
For use in a standard one-term course, in which both discrete and continuous
probability is covered, students should have taken as a prerequisite two terms of
calculus, including an introduction to multiple integrals. In order to cover Chap-
ter 11, which contains material on Markov chains, some knowledge of matrix theory
is necessary.
The text can also be used in a discrete probability course. The material has been
organized in such a way that the discrete and continuous probability discussions are
presented in a separate, but parallel, manner. This organization dispels an overly
rigorous or formal view of probability and oers some strong pedagogical value
in that the discrete discussions can sometimes serve to motivate the more abstract
continuous probability discussions. For use in a discrete probability course, students",prob.pdf
"continuous probability discussions. For use in a discrete probability course, students
should have taken one term of calculus as a prerequisite.
Very little computing background is assumed or necessary in order to obtain full
benets from the use of the computing material and examples in the text. All of
the programs that are used in the text have been written in each of the languages
TrueBASIC, Maple, and Mathematica.
This book is distributed on the Web as part of the Chance Project, which is de-
voted to providing materials for beginning courses in probability and statistics. The
computer programs, solutions to the odd-numbered exercises, and current errata are
also available at this site. Instructors may obtain all of the solutions by writing to
either of the authors, at jlsnell@dartmouth.edu and cgrinst1@swarthmore.edu.
vii",prob.pdf
"viii PREFACE
FEATURES
Level of rigor and emphasis: Probability is a wonderfully intuitive and applicable
eld of mathematics. We have tried not to spoil its beauty by presenting too much
formal mathematics. Rather, we have tried to develop the key ideas in a somewhat
leisurely style, to provide a variety of interesting applications to probability, and to
show some of the nonintuitive examples that make probability such a lively subject.
Exercises: There are over 600 exercises in the text providing plenty of oppor-
tunity for practicing skills and developing a sound understanding of the ideas. In
the exercise sets are routine exercises to be done with and without the use of a
computer and more theoretical exercises to improve the understanding of basic con-
cepts. More dicult exercises are indicated by an asterisk. A solution manual for
all of the exercises is available to instructors.
Historical remarks: Introductory probability is a subject in which the funda-",prob.pdf
"all of the exercises is available to instructors.
Historical remarks: Introductory probability is a subject in which the funda-
mental ideas are still closely tied to those of the founders of the subject. For this
reason, there are numerous historical comments in the text, especially as they deal
with the development of discrete probability.
Pedagogical use of computer programs: Probability theory makes predictions
about experiments whose outcomes depend upon chance. Consequently, it lends
itself beautifully to the use of computers as a mathematical tool to simulate and
analyze chance experiments.
In the text the computer is utilized in several ways. First, it provides a labora-
tory where chance experiments can be simulated and the students can get a feeling
for the variety of such experiments. This use of the computer in probability has
been already beautifully illustrated by William Feller in the second edition of his",prob.pdf
"been already beautifully illustrated by William Feller in the second edition of his
famous text An Introduction to Probability Theory and Its Applications (New York:
Wiley, 1950). In the preface, Feller wrote about his treatment of 
uctuation in coin
tossing: \The results are so amazing and so at variance with common intuition
that even sophisticated colleagues doubted that coins actually misbehave as theory
predicts. The record of a simulated experiment is therefore included.""
In addition to providing a laboratory for the student, the computer is a powerful
aid in understanding basic results of probability theory. For example, the graphical
illustration of the approximation of the standardized binomial distributions to the
normal curve is a more convincing demonstration of the Central Limit Theorem
than many of the formal proofs of this fundamental result.
Finally, the computer allows the student to solve problems that do not lend",prob.pdf
"than many of the formal proofs of this fundamental result.
Finally, the computer allows the student to solve problems that do not lend
themselves to closed-form formulas such as waiting times in queues. Indeed, the
introduction of the computer changes the way in which we look at many problems
in probability. For example, being able to calculate exact binomial probabilities
for experiments up to 1000 trials changes the way we view the normal and Poisson
approximations.
ACKNOWLEDGMENTS
Anyone writing a probability text today owes a great debt to William Feller,
who taught us all how to make probability come alive as a subject matter. If you",prob.pdf
"PREFACE ix
nd an example, an application, or an exercise that you really like, it probably had
its origin in Feller's classic text, An Introduction to Probability Theory and Its
Applications.
We are indebted to many people for their help in this undertaking. The approach
to Markov Chains presented in the book was developed by John Kemeny and the
second author. Reese Prosser was a silent co-author for the material on continuous
probability in an earlier version of this book. Mark Kernighan contributed 40 pages
of comments on the earlier edition. Many of these comments were very thought-
provoking; in addition, they provided a student's perspective on the book. Most of
the major changes in this version of the book have their genesis in these notes.
Fuxing Hou and Lee Nave provided extensive help with the typesetting and
the gures. John Finn provided valuable pedagogical advice on the text and and
the computer programs. Karl Knaub and Jessica Sklar are responsible for the",prob.pdf
"the computer programs. Karl Knaub and Jessica Sklar are responsible for the
implementations of the computer programs in Mathematica and Maple. Jessica
and Gang Wang assisted with the solutions.
Finally, we thank the American Mathematical Society, and in particular Sergei
Gelfand and John Ewing, for their interest in this book; their help in its production;
and their willingness to make the work freely redistributable.",prob.pdf
x PREFACE,prob.pdf
"Chapter 1
Discrete Probability
Distributions
1.1 Simulation of Discrete Probabilities
Probability
In this chapter, we shall rst consider chance experiments with a nite number of
possible outcomes !1, !2, . . . , !n. For example, we roll a die and the possible
outcomes are 1, 2, 3, 4, 5, 6 corresponding to the side that turns up. We toss a coin
with possible outcomes H (heads) and T (tails).
It is frequently useful to be able to refer to an outcome of an experiment. For
example, we might want to write the mathematical expression which gives the sum
of four rolls of a die. To do this, we could let Xi, i = 1; 2; 3; 4; represent the values
of the outcomes of the four rolls, and then we could write the expression
X1 + X2 + X3 + X4
for the sum of the four rolls. The Xi's are called random variables. A random vari-
able is simply an expression whose value is the outcome of a particular experiment.
Just as in the case of other types of variables in mathematics, random variables can",prob.pdf
"Just as in the case of other types of variables in mathematics, random variables can
take on dierent values.
Let X be the random variable which represents the roll of one die. We shall
assign probabilities to the possible outcomes of this experiment. We do this by
assigning to each outcome !j a nonnegative number m(!j) in such a way that
m(!1) + m(!2) + 
 
 
 + m(!6) = 1 :
The function m(!j) is called the distribution function of the random variable X.
For the case of the roll of the die we would assign equal probabilities or probabilities
1/6 to each of the outcomes. With this assignment of probabilities, one could write
P(X  4) = 2
3
1",prob.pdf
"2 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
to mean that the probability is 2=3 that a roll of a die will have a value which does
not exceed 4.
Let Y be the random variable which represents the toss of a coin. In this case,
there are two possible outcomes, which we can label as H and T. Unless we have
reason to suspect that the coin comes up one way more often than the other way,
it is natural to assign the probability of 1/2 to each of the two outcomes.
In both of the above experiments, each outcome is assigned an equal probability.
This would certainly not be the case in general. For example, if a drug is found to
be eective 30 percent of the time it is used, we might assign a probability .3 that
the drug is eective the next time it is used and .7 that it is not eective. This last
example illustrates the intuitive frequency concept of probability. That is, if we have
a probability p that an experiment will result in outcome A, then if we repeat this",prob.pdf
"a probability p that an experiment will result in outcome A, then if we repeat this
experiment a large number of times we should expect that the fraction of times that
A will occur is about p. To check intuitive ideas like this, we shall nd it helpful to
look at some of these problems experimentally. We could, for example, toss a coin
a large number of times and see if the fraction of times heads turns up is about 1/2.
We could also simulate this experiment on a computer.
Simulation
We want to be able to perform an experiment that corresponds to a given set of
probabilities; for example, m(!1) = 1=2, m(!2) = 1=3, and m(!3) = 1=6. In this
case, one could mark three faces of a six-sided die with an !1, two faces with an !2,
and one face with an !3.
In the general case we assume that m(!1), m(!2), . . . , m(!n) are all rational
numbers, with least common denominator n. If n > 2, we can imagine a long
cylindrical die with a cross-section that is a regular n-gon. If m(!j) = nj=n, then",prob.pdf
"cylindrical die with a cross-section that is a regular n-gon. If m(!j) = nj=n, then
we can label nj of the long faces of the cylinder with an !j, and if one of the end
faces comes up, we can just roll the die again. If n = 2, a coin could be used to
perform the experiment.
We will be particularly interested in repeating a chance experiment a large num-
ber of times. Although the cylindrical die would be a convenient way to carry out
a few repetitions, it would be dicult to carry out a large number of experiments.
Since the modern computer can do a large number of operations in a very short
time, it is natural to turn to the computer for this task.
Random Numbers
We must rst nd a computer analog of rolling a die. This is done on the computer
by means of a random number generator. Depending upon the particular software
package, the computer can be asked for a real number between 0 and 1, or an integer",prob.pdf
"package, the computer can be asked for a real number between 0 and 1, or an integer
in a given set of consecutive integers. In the rst case, the real numbers are chosen
in such a way that the probability that the number lies in any particular subinterval
of this unit interval is equal to the length of the subinterval. In the second case,
each integer has the same probability of being chosen.",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 3
.203309 .762057 .151121 .623868
.932052 .415178 .716719 .967412
.069664 .670982 .352320 .049723
.750216 .784810 .089734 .966730
.946708 .380365 .027381 .900794
Table 1.1: Sample output of the program RandomNumbers.
Let X be a random variable with distribution function m(!), where ! is in the
set f!1; !2; !3g, and m(!1) = 1=2, m(!2) = 1=3, and m(!3) = 1=6. If our computer
package can return a random integer in the set f1; 2; :::; 6g, then we simply ask it
to do so, and make 1, 2, and 3 correspond to !1, 4 and 5 correspond to !2, and 6
correspond to !3. If our computer package returns a random real number r in the
interval (0; 1), then the expression
b6rc + 1
will be a random integer between 1 and 6. (The notation bxc means the greatest
integer not exceeding x, and is read \
oor of x."")
The method by which random real numbers are generated on a computer is
described in the historical discussion at the end of this section. The following",prob.pdf
"described in the historical discussion at the end of this section. The following
example gives sample output of the program RandomNumbers.
Example 1.1 (Random Number Generation) The program RandomNumbers
generates n random real numbers in the interval [0; 1], where n is chosen by the
user. When we ran the program with n = 20, we obtained the data shown in
Table 1.1. 2
Example 1.2 (Coin Tossing) As we have noted, our intuition suggests that the
probability of obtaining a head on a single toss of a coin is 1/2. To have the
computer toss a coin, we can ask it to pick a random real number in the interval
[0; 1] and test to see if this number is less than 1/2. If so, we shall call the outcome
heads; if not we call it tails. Another way to proceed would be to ask the computer
to pick a random integer from the set f0; 1g. The program CoinTosses carries
out the experiment of tossing a coin n times. Running this program, with n = 20,
resulted in:
THTTTHTTTTHTTTTTHHTT.",prob.pdf
"out the experiment of tossing a coin n times. Running this program, with n = 20,
resulted in:
THTTTHTTTTHTTTTTHHTT.
Note that in 20 tosses, we obtained 5 heads and 15 tails. Let us toss a coin n
times, where n is much larger than 20, and see if we obtain a proportion of heads
closer to our intuitive guess of 1/2. The program CoinTosses keeps track of the
number of heads. When we ran this program with n = 1000, we obtained 494 heads.
When we ran it with n = 10000, we obtained 5039 heads.",prob.pdf
"4 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
We notice that when we tossed the coin 10,000 times, the proportion of heads
was close to the \true value"" .5 for obtaining a head when a coin is tossed. A math-
ematical model for this experiment is called Bernoulli Trials (see Chapter 3). The
Law of Large Numbers, which we shall study later (see Chapter 8), will show that
in the Bernoulli Trials model, the proportion of heads should be near .5, consistent
with our intuitive idea of the frequency interpretation of probability.
Of course, our program could be easily modied to simulate coins for which the
probability of a head is p, where p is a real number between 0 and 1. 2
In the case of coin tossing, we already knew the probability of the event occurring
on each experiment. The real power of simulation comes from the ability to estimate
probabilities when they are not known ahead of time. This method has been used in",prob.pdf
"probabilities when they are not known ahead of time. This method has been used in
the recent discoveries of strategies that make the casino game of blackjack favorable
to the player. We illustrate this idea in a simple situation in which we can compute
the true probability and see how eective the simulation is.
Example 1.3 (Dice Rolling) We consider a dice game that played an important
role in the historical development of probability. The famous letters between Pas-
cal and Fermat, which many believe started a serious study of probability, were
instigated by a request for help from a French nobleman and gambler, Chevalier
de Mere. It is said that de Mere had been betting that, in four rolls of a die, at
least one six would turn up. He was winning consistently and, to get more people
to play, he changed the game to bet that, in 24 rolls of two dice, a pair of sixes
would turn up. It is claimed that de Mere lost with 24 and felt that 25 rolls were",prob.pdf
"would turn up. It is claimed that de Mere lost with 24 and felt that 25 rolls were
necessary to make the game favorable. It was un grand scandale that mathematics
was wrong.
We shall try to see if de Mere is correct by simulating his various bets. The
program DeMere1 simulates a large number of experiments, seeing, in each one,
if a six turns up in four rolls of a die. When we ran this program for 1000 plays,
a six came up in the rst four rolls 48.6 percent of the time. When we ran it for
10,000 plays this happened 51.98 percent of the time.
We note that the result of the second run suggests that de Mere was correct
in believing that his bet with one die was favorable; however, if we had based our
conclusion on the rst run, we would have decided that he was wrong. Accurate
results by simulation require a large number of experiments. 2
The program DeMere2 simulates de Mere's second bet that a pair of sixes",prob.pdf
"results by simulation require a large number of experiments. 2
The program DeMere2 simulates de Mere's second bet that a pair of sixes
will occur in n rolls of a pair of dice. The previous simulation shows that it is
important to know how many trials we should simulate in order to expect a certain
degree of accuracy in our approximation. We shall see later that in these types of
experiments, a rough rule of thumb is that, at least 95% of the time, the error does
not exceed the reciprocal of the square root of the number of trials. Fortunately,
for this dice game, it will be easy to compute the exact probabilities. We shall
show in the next section that for the rst bet the probability that de Mere wins is
1  (5=6)4 = :518.",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 5
5 10 15 20 25 30 35 40
-10
-8
-6
-4
-2
2
4
6
8
10
Figure 1.1: Peter's winnings in 40 plays of heads or tails.
One can understand this calculation as follows: The probability that no 6 turns
up on the rst toss is (5=6). The probability that no 6 turns up on either of the
rst two tosses is (5=6)2. Reasoning in the same way, the probability that no 6
turns up on any of the rst four tosses is (5=6)4. Thus, the probability of at least
one 6 in the rst four tosses is 1  (5=6)4. Similarly, for the second bet, with 24
rolls, the probability that de Mere wins is 1  (35=36)24 = :491, and for 25 rolls it
is 1  (35=36)25 = :506.
Using the rule of thumb mentioned above, it would require 27,000 rolls to have a
reasonable chance to determine these probabilities with sucient accuracy to assert
that they lie on opposite sides of .5. It is interesting to ponder whether a gambler",prob.pdf
"that they lie on opposite sides of .5. It is interesting to ponder whether a gambler
can detect such probabilities with the required accuracy from gambling experience.
Some writers on the history of probability suggest that de Mere was, in fact, just
interested in these problems as intriguing probability problems.
Example 1.4 (Heads or Tails) For our next example, we consider a problem where
the exact answer is dicult to obtain but for which simulation easily gives the
qualitative results. Peter and Paul play a game called heads or tails. In this game,
a fair coin is tossed a sequence of times|we choose 40. Each time a head comes up
Peter wins 1 penny from Paul, and each time a tail comes up Peter loses 1 penny
to Paul. For example, if the results of the 40 tosses are
THTHHHHTTHTHHTTHHTTTTHHHTHHTHHHTHHHTTTHH.
Peter's winnings may be graphed as in Figure 1.1.
Peter has won 6 pennies in this particular game. It is natural to ask for the",prob.pdf
"Peter's winnings may be graphed as in Figure 1.1.
Peter has won 6 pennies in this particular game. It is natural to ask for the
probability that he will win j pennies; here j could be any even number from 40
to 40. It is reasonable to guess that the value of j with the highest probability
is j = 0, since this occurs when the number of heads equals the number of tails.
Similarly, we would guess that the values of j with the lowest probabilities are
j = 40.",prob.pdf
"6 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
A second interesting question about this game is the following: How many times
in the 40 tosses will Peter be in the lead? Looking at the graph of his winnings
(Figure 1.1), we see that Peter is in the lead when his winnings are positive, but
we have to make some convention when his winnings are 0 if we want all tosses to
contribute to the number of times in the lead. We adopt the convention that, when
Peter's winnings are 0, he is in the lead if he was ahead at the previous toss and
not if he was behind at the previous toss. With this convention, Peter is in the lead
34 times in our example. Again, our intuition might suggest that the most likely
number of times to be in the lead is 1/2 of 40, or 20, and the least likely numbers
are the extreme cases of 40 or 0.
It is easy to settle this by simulating the game a large number of times and
keeping track of the number of times that Peter's nal winnings are j, and the",prob.pdf
"keeping track of the number of times that Peter's nal winnings are j, and the
number of times that Peter ends up being in the lead by k. The proportions over
all games then give estimates for the corresponding probabilities. The program
HTSimulation carries out this simulation. Note that when there are an even
number of tosses in the game, it is possible to be in the lead only an even number
of times. We have simulated this game 10,000 times. The results are shown in
Figures 1.2 and 1.3. These graphs, which we call spike graphs, were generated
using the program Spikegraph. The vertical line, or spike, at position x on the
horizontal axis, has a height equal to the proportion of outcomes which equal x.
Our intuition about Peter's nal winnings was quite correct, but our intuition about
the number of times Peter was in the lead was completely wrong. The simulation
suggests that the least likely number of times in the lead is 20 and the most likely",prob.pdf
"suggests that the least likely number of times in the lead is 20 and the most likely
is 0 or 40. This is indeed correct, and the explanation for it is suggested by playing
the game of heads or tails with a large number of tosses and looking at a graph of
Peter's winnings. In Figure 1.4 we show the results of a simulation of the game, for
1000 tosses and in Figure 1.5 for 10,000 tosses.
In the second example Peter was ahead most of the time. It is a remarkable
fact, however, that, if play is continued long enough, Peter's winnings will continue
to come back to 0, but there will be very long times between the times that this
happens. These and related results will be discussed in Chapter 12. 2
In all of our examples so far, we have simulated equiprobable outcomes. We
illustrate next an example where the outcomes are not equiprobable.
Example 1.5 (Horse Races) Four horses (Acorn, Balky, Chestnut, and Dolby)
have raced many times. It is estimated that Acorn wins 30 percent of the time,",prob.pdf
"have raced many times. It is estimated that Acorn wins 30 percent of the time,
Balky 40 percent of the time, Chestnut 20 percent of the time, and Dolby 10 percent
of the time.
We can have our computer carry out one race as follows: Choose a random
number x. If x < :3 then we say that Acorn won. If :3  x < :7 then Balky wins.
If :7  x < :9 then Chestnut wins. Finally, if :9  x then Dolby wins.
The program HorseRace uses this method to simulate the outcomes of n races.
Running this program for n = 10 we found that Acorn won 40 percent of the time,
Balky 20 percent of the time, Chestnut 10 percent of the time, and Dolby 30 percent",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 7
Figure 1.2: Distribution of winnings.
Figure 1.3: Distribution of number of times in the lead.",prob.pdf
"8 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
200 400 600 800 1000
1000 plays
-50
-40
-30
-20
-10
0
10
20
Figure 1.4: Peter's winnings in 1000 plays of heads or tails.
2000 4000 6000 8000 10000
10000 plays
0
50
100
150
200
Figure 1.5: Peter's winnings in 10,000 plays of heads or tails.",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 9
of the time. A larger number of races would be necessary to have better agreement
with the past experience. Therefore we ran the program to simulate 1000 races
with our four horses. Although very tired after all these races, they performed in
a manner quite consistent with our estimates of their abilities. Acorn won 29.8
percent of the time, Balky 39.4 percent, Chestnut 19.5 percent, and Dolby 11.3
percent of the time.
The program GeneralSimulation uses this method to simulate repetitions of
an arbitrary experiment with a nite number of outcomes occurring with known
probabilities. 2
Historical Remarks
Anyone who plays the same chance game over and over is really carrying out a sim-
ulation, and in this sense the process of simulation has been going on for centuries.
As we have remarked, many of the early problems of probability might well have
been suggested by gamblers' experiences.",prob.pdf
"As we have remarked, many of the early problems of probability might well have
been suggested by gamblers' experiences.
It is natural for anyone trying to understand probability theory to try simple
experiments by tossing coins, rolling dice, and so forth. The naturalist Buon tossed
a coin 4040 times, resulting in 2048 heads and 1992 tails. He also estimated the
number  by throwing needles on a ruled surface and recording how many times
the needles crossed a line (see Section 2.1). The English biologist W. F. R. Weldon1
recorded 26,306 throws of 12 dice, and the Swiss scientist Rudolf Wolf2 recorded
100,000 throws of a single die without a computer. Such experiments are very time-
consuming and may not accurately represent the chance phenomena being studied.
For example, for the dice experiments of Weldon and Wolf, further analysis of the
recorded data showed a suspected bias in the dice. The statistician Karl Pearson",prob.pdf
"recorded data showed a suspected bias in the dice. The statistician Karl Pearson
analyzed a large number of outcomes at certain roulette tables and suggested that
the wheels were biased. He wrote in 1894:
Clearly, since the Casino does not serve the valuable end of huge lab-
oratory for the preparation of probability statistics, it has no scientic
raison d'^etre. Men of science cannot have their most rened theories
disregarded in this shameless manner! The French Government must be
urged by the hierarchy of science to close the gaming-saloons; it would
be, of course, a graceful act to hand over the remaining resources of the
Casino to the Academie des Sciences for the endowment of a laboratory
of orthodox probability; in particular, of the new branch of that study,
the application of the theory of chance to the biological problems of
evolution, which is likely to occupy so much of men's thoughts in the
near future.3",prob.pdf
"evolution, which is likely to occupy so much of men's thoughts in the
near future.3
However, these early experiments were suggestive and led to important discov-
eries in probability and statistics. They led Pearson to the chi-squared test, which
1T. C. Fry, Probability and Its Engineering Uses, 2nd ed. (Princeton: Van Nostrand, 1965).
2E. Czuber, Wahrscheinlichkeitsrechnung, 3rd ed. (Berlin: Teubner, 1914).
3K. Pearson, \Science and Monte Carlo,"" Fortnightly Review, vol. 55 (1894), p. 193; cited in
S. M. Stigler, The History of Statistics (Cambridge: Harvard University Press, 1986).",prob.pdf
"10 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
is of great importance in testing whether observed data t a given probability dis-
tribution.
By the early 1900s it was clear that a better way to generate random numbers
was needed. In 1927, L. H. C. Tippett published a list of 41,600 digits obtained by
selecting numbers haphazardly from census reports. In 1955, RAND Corporation
printed a table of 1,000,000 random numbers generated from electronic noise. The
advent of the high-speed computer raised the possibility of generating random num-
bers directly on the computer, and in the late 1940s John von Neumann suggested
that this be done as follows: Suppose that you want a random sequence of four-digit
numbers. Choose any four-digit number, say 6235, to start. Square this number
to obtain 38,875,225. For the second number choose the middle four digits of this
square (i.e., 8752). Do the same process starting with 8752 to get the third number,
and so forth.",prob.pdf
"square (i.e., 8752). Do the same process starting with 8752 to get the third number,
and so forth.
More modern methods involve the concept of modular arithmetic. If a is an
integer and m is a positive integer, then by a (mod m) we mean the remainder
when a is divided by m. For example, 10 (mod 4) = 2, 8 (mod 2) = 0, and so
forth. To generate a random sequence X0; X1; X2; : : : of numbers choose a starting
number X0 and then obtain the numbers Xn+1 from Xn by the formula
Xn+1 = (aXn + c) (mod m) ;
where a, c, and m are carefully chosen constants. The sequence X0; X1; X2; : : :
is then a sequence of integers between 0 and m  1. To obtain a sequence of real
numbers in [0; 1), we divide each Xj by m. The resulting sequence consists of
rational numbers of the form j=m, where 0  j  m  1. Since m is usually a
very large integer, we think of the numbers in the sequence as being random real
numbers in [0; 1).
For both von Neumann's squaring method and the modular arithmetic technique",prob.pdf
"numbers in [0; 1).
For both von Neumann's squaring method and the modular arithmetic technique
the sequence of numbers is actually completely determined by the rst number.
Thus, there is nothing really random about these sequences. However, they produce
numbers that behave very much as theory would predict for random experiments.
To obtain dierent sequences for dierent experiments the initial number X0 is
chosen by some other procedure that might involve, for example, the time of day.4
During the Second World War, physicists at the Los Alamos Scientic Labo-
ratory needed to know, for purposes of shielding, how far neutrons travel through
various materials. This question was beyond the reach of theoretical calculations.
Daniel McCracken, writing in the Scientic American, states:
The physicists had most of the necessary data: they knew the average
distance a neutron of a given speed would travel in a given substance",prob.pdf
"The physicists had most of the necessary data: they knew the average
distance a neutron of a given speed would travel in a given substance
before it collided with an atomic nucleus, what the probabilities were
that the neutron would bounce o instead of being absorbed by the
nucleus, how much energy the neutron was likely to lose after a given
4For a detailed discussion of random numbers, see D. E. Knuth, The Art of Computer Pro-
gramming, vol. II (Reading: Addison-Wesley, 1969).",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 11
collision and so on.5
John von Neumann and Stanislas Ulam suggested that the problem be solved
by modeling the experiment by chance devices on a computer. Their work being
secret, it was necessary to give it a code name. Von Neumann chose the name
\Monte Carlo."" Since that time, this method of simulation has been called the
Monte Carlo Method.
William Feller indicated the possibilities of using computer simulations to illus-
trate basic concepts in probability in his book An Introduction to Probability Theory
and Its Applications. In discussing the problem about the number of times in the
lead in the game of \heads or tails"" Feller writes:
The results concerning 
uctuations in coin tossing show that widely
held beliefs about the law of large numbers are fallacious. These results
are so amazing and so at variance with common intuition that even
sophisticated colleagues doubted that coins actually misbehave as theory",prob.pdf
"are so amazing and so at variance with common intuition that even
sophisticated colleagues doubted that coins actually misbehave as theory
predicts. The record of a simulated experiment is therefore included.6
Feller provides a plot showing the result of 10,000 plays of heads or tails similar to
that in Figure 1.5.
The martingale betting system described in Exercise 10 has a long and interest-
ing history. Russell Barnhart pointed out to the authors that its use can be traced
back at least to 1754, when Casanova, writing in his memoirs, History of My Life,
writes
She [Casanova's mistress] made me promise to go to the casino [the
Ridotto in Venice] for money to play in partnership with her. I went
there and took all the gold I found, and, determinedly doubling my
stakes according to the system known as the martingale, I won three or
four times a day during the rest of the Carnival. I never lost the sixth
card. If I had lost it, I should have been out of funds, which amounted",prob.pdf
"four times a day during the rest of the Carnival. I never lost the sixth
card. If I had lost it, I should have been out of funds, which amounted
to two thousand zecchini.7
Even if there were no zeros on the roulette wheel so the game was perfectly fair,
the martingale system, or any other system for that matter, cannot make the game
into a favorable game. The idea that a fair game remains fair and unfair games
remain unfair under gambling systems has been exploited by mathematicians to
obtain important results in the study of probability. We will introduce the general
concept of a martingale in Chapter 6.
The word martingale itself also has an interesting history. The origin of the
word is obscure. A recent version of the Oxford English Dictionary gives examples
5D. D. McCracken, \The Monte Carlo Method,"" Scientic American, vol. 192 (May 1955),
p. 90.
6W. Feller, Introduction to Probability Theory and its Applications, vol. 1, 3rd ed. (New York:
John Wiley & Sons, 1968), p. xi.",prob.pdf
"p. 90.
6W. Feller, Introduction to Probability Theory and its Applications, vol. 1, 3rd ed. (New York:
John Wiley & Sons, 1968), p. xi.
7G. Casanova, History of My Life, vol. IV, Chap. 7, trans. W. R. Trask (New York: Harcourt-
Brace, 1968), p. 124.",prob.pdf
"12 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
of its use in the early 1600s and says that its probable origin is the reference in
Rabelais's Book One, Chapter 20:
Everything was done as planned, the only thing being that Gargantua
doubted if they would be able to nd, right away, breeches suitable to
the old fellow's legs; he was doubtful, also, as to what cut would be most
becoming to the orator|the martingale, which has a draw-bridge eect
in the seat, to permit doing one's business more easily; the sailor-style,
which aords more comfort for the kidneys; the Swiss, which is warmer
on the belly; or the codsh-tail, which is cooler on the loins.8
Dominic Lusinchi noted an earlier occurrence of the word martingale. Accord-
ing to the French dictionary Le Petit Robert, the word comes from the Provencal
word \martegalo,"" which means \from Martigues."" Martigues is a town due west of
Merseille. The dictionary gives the example of \chausses a la martinguale"" (which",prob.pdf
"Merseille. The dictionary gives the example of \chausses a la martinguale"" (which
means Martigues-style breeches) and the date 1491.
In modern uses martingale has several dierent meanings, all related to holding
down, in addition to the gambling use. For example, it is a strap on a horse's
harness used to hold down the horse's head, and also part of a sailing rig used to
hold down the bowsprit.
The Labouchere system described in Exercise 9 is named after Henry du Pre
Labouchere (1831{1912), an English journalist and member of Parliament. Labou-
chere attributed the system to Condorcet. Condorcet (1743{1794) was a political
leader during the time of the French revolution who was interested in applying prob-
ability theory to economics and politics. For example, he calculated the probability
that a jury using majority vote will give a correct decision if each juror has the
same probability of deciding correctly. His writings provided a wealth of ideas on",prob.pdf
"same probability of deciding correctly. His writings provided a wealth of ideas on
how probability might be applied to human aairs.9
Exercises
1 Modify the program CoinTosses to toss a coin n times and print out after
every 100 tosses the proportion of heads minus 1/2. Do these numbers appear
to approach 0 as n increases? Modify the program again to print out, every
100 times, both of the following quantities: the proportion of heads minus 1/2,
and the number of heads minus half the number of tosses. Do these numbers
appear to approach 0 as n increases?
2 Modify the program CoinTosses so that it tosses a coin n times and records
whether or not the proportion of heads is within .1 of .5 (i.e., between .4
and .6). Have your program repeat this experiment 100 times. About how
large must n be so that approximately 95 out of 100 times the proportion of
heads is between .4 and .6?
8Quoted in the Portable Rabelais, ed. S. Putnam (New York: Viking, 1946), p. 113.",prob.pdf
"heads is between .4 and .6?
8Quoted in the Portable Rabelais, ed. S. Putnam (New York: Viking, 1946), p. 113.
9Le Marquise de Condorcet, Essai sur l'Application de l'Analyse a la Probabilite des Decisions
Rendues a la Pluralite des Voix (Paris: Imprimerie Royale, 1785).",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 13
3 In the early 1600s, Galileo was asked to explain the fact that, although the
number of triples of integers from 1 to 6 with sum 9 is the same as the number
of such triples with sum 10, when three dice are rolled, a 9 seemed to come
up less often than a 10|supposedly in the experience of gamblers.
(a) Write a program to simulate the roll of three dice a large number of
times and keep track of the proportion of times that the sum is 9 and
the proportion of times it is 10.
(b) Can you conclude from your simulations that the gamblers were correct?
4 In raquetball, a player continues to serve as long as she is winning; a point
is scored only when a player is serving and wins the volley. The rst player
to win 21 points wins the game. Assume that you serve rst and have a
probability .6 of winning a volley when you serve and probability .5 when
your opponent serves. Estimate, by simulation, the probability that you will
win a game.",prob.pdf
"your opponent serves. Estimate, by simulation, the probability that you will
win a game.
5 Consider the bet that all three dice will turn up sixes at least once in n rolls
of three dice. Calculate f(n), the probability of at least one triple-six when
three dice are rolled n times. Determine the smallest value of n necessary for
a favorable bet that a triple-six will occur when three dice are rolled n times.
(DeMoivre would say it should be about 216 log2 = 149:7 and so would answer
150|see Exercise 1.2.17. Do you agree with him?)
6 In Las Vegas, a roulette wheel has 38 slots numbered 0, 00, 1, 2, . .. , 36. The
0 and 00 slots are green and half of the remaining 36 slots are red and half
are black. A croupier spins the wheel and throws in an ivory ball. If you bet
1 dollar on red, you win 1 dollar if the ball stops in a red slot and otherwise
you lose 1 dollar. Write a program to nd the total winnings for a player who
makes 1000 bets on red.",prob.pdf
"you lose 1 dollar. Write a program to nd the total winnings for a player who
makes 1000 bets on red.
7 Another form of bet for roulette is to bet that a specic number (say 17) will
turn up. If the ball stops on your number, you get your dollar back plus 35
dollars. If not, you lose your dollar. Write a program that will plot your
winnings when you make 500 plays of roulette at Las Vegas, rst when you
bet each time on red (see Exercise 6), and then for a second visit to Las
Vegas when you make 500 plays betting each time on the number 17. What
dierences do you see in the graphs of your winnings on these two occasions?
8 An astute student noticed that, in our simulation of the game of heads or tails
(see Example 1.4), the proportion of times the player is always in the lead is
very close to the proportion of times that the player's total winnings end up 0.
Work out these probabilities by enumeration of all cases for two tosses and",prob.pdf
"Work out these probabilities by enumeration of all cases for two tosses and
for four tosses, and see if you think that these probabilities are, in fact, the
same.
9 The Labouchere system for roulette is played as follows. Write down a list of
numbers, usually 1, 2, 3, 4. Bet the sum of the rst and last, 1 + 4 = 5, on",prob.pdf
"14 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
red. If you win, delete the rst and last numbers from your list. If you lose,
add the amount that you last bet to the end of your list. Then use the new
list and bet the sum of the rst and last numbers (if there is only one number,
bet that amount). Continue until your list becomes empty. Show that, if this
happens, you win the sum, 1 + 2 + 3 + 4 = 10, of your original list. Simulate
this system and see if you do always stop and, hence, always win. If so, why
is this not a foolproof gambling system?
10 Another well-known gambling system is the martingale doubling system. Sup-
pose that you are betting on red to turn up in roulette. Every time you win,
bet 1 dollar next time. Every time you lose, double your previous bet. Suppose
that you use this system until you have won at least 5 dollars or you have lost
more than 100 dollars. Write a program to simulate this and play it a number",prob.pdf
"more than 100 dollars. Write a program to simulate this and play it a number
of times and see how you do. In his book The Newcomes, W. M. Thack-
eray remarks \You have not played as yet? Do not do so; above all avoid a
martingale if you do.""10 Was this good advice?
11 Modify the program HTSimulation so that it keeps track of the maximum of
Peter's winnings in each game of 40 tosses. Have your program print out the
proportion of times that your total winnings take on values 0; 2; 4; : : : ; 40.
Calculate the corresponding exact probabilities for games of two tosses and
four tosses.
12 In an upcoming national election for the President of the United States, a
pollster plans to predict the winner of the popular vote by taking a random
sample of 1000 voters and declaring that the winner will be the one obtaining
the most votes in his sample. Suppose that 48 percent of the voters plan
to vote for the Republican candidate and 52 percent plan to vote for the",prob.pdf
"the most votes in his sample. Suppose that 48 percent of the voters plan
to vote for the Republican candidate and 52 percent plan to vote for the
Democratic candidate. To get some idea of how reasonable the pollster's
plan is, write a program to make this prediction by simulation. Repeat the
simulation 100 times and see how many times the pollster's prediction would
come true. Repeat your experiment, assuming now that 49 percent of the
population plan to vote for the Republican candidate; rst with a sample of
1000 and then with a sample of 3000. (The Gallup Poll uses about 3000.)
(This idea is discussed further in Chapter 9, Section 9.1.)
13 The psychologist Tversky and his colleagues11 say that about four out of ve
people will answer (a) to the following question:
A certain town is served by two hospitals. In the larger hospital about 45
babies are born each day, and in the smaller hospital 15 babies are born each",prob.pdf
"babies are born each day, and in the smaller hospital 15 babies are born each
day. Although the overall proportion of boys is about 50 percent, the actual
proportion at either hospital may be more or less than 50 percent on any day.
10W. M. Thackerey, The Newcomes (London: Bradbury and Evans, 1854{55).
11See K. McKean, \Decisions, Decisions,"" Discover, June 1985, pp. 22{31. Kevin McKean,
Discover Magazine, c
1987 Family Media, Inc. Reprinted with permission. This popular article
reports on the work of Tverksy et. al. in Judgement Under Uncertainty: Heuristics and Biases
(Cambridge: Cambridge University Press, 1982).",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 15
At the end of a year, which hospital will have the greater number of days on
which more than 60 percent of the babies born were boys?
(a) the large hospital
(b) the small hospital
(c) neither|the number of days will be about the same.
Assume that the probability that a baby is a boy is .5 (actual estimates make
this more like .513). Decide, by simulation, what the right answer is to the
question. Can you suggest why so many people go wrong?
14 You are oered the following game. A fair coin will be tossed until the rst
time it comes up heads. If this occurs on the jth toss you are paid 2j dollars.
You are sure to win at least 2 dollars so you should be willing to pay to play
this game|but how much? Few people would pay as much as 10 dollars to
play this game. See if you can decide, by simulation, a reasonable amount
that you would be willing to pay, per game, if you will be allowed to make",prob.pdf
"play this game. See if you can decide, by simulation, a reasonable amount
that you would be willing to pay, per game, if you will be allowed to make
a large number of plays of the game. Does the amount that you would be
willing to pay per game depend upon the number of plays that you will be
allowed?
15 Tversky and his colleagues12 studied the records of 48 of the Philadelphia
76ers basketball games in the 1980{81 season to see if a player had times
when he was hot and every shot went in, and other times when he was cold
and barely able to hit the backboard. The players estimated that they were
about 25 percent more likely to make a shot after a hit than after a miss.
In fact, the opposite was true|the 76ers were 6 percent more likely to score
after a miss than after a hit. Tversky reports that the number of hot and cold
streaks was about what one would expect by purely random eects. Assuming
that a player has a fty-fty chance of making a shot and makes 20 shots a",prob.pdf
"streaks was about what one would expect by purely random eects. Assuming
that a player has a fty-fty chance of making a shot and makes 20 shots a
game, estimate by simulation the proportion of the games in which the player
will have a streak of 5 or more hits.
16 Estimate, by simulation, the average number of children there would be in
a family if all people had children until they had a boy. Do the same if all
people had children until they had at least one boy and at least one girl. How
many more children would you expect to nd under the second scheme than
under the rst in 100,000 families? (Assume that boys and girls are equally
likely.)
17 Mathematicians have been known to get some of the best ideas while sitting in
a cafe, riding on a bus, or strolling in the park. In the early 1900s the famous
mathematician George Polya lived in a hotel near the woods in Zurich. He
liked to walk in the woods and think about mathematics. Polya describes the
following incident:
12ibid.",prob.pdf
"16 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
0 1 2 3-1-2-3
c.  Random walk in three dimensions.b.  Random walk in two dimensions.
a.  Random walk in one dimension.
Figure 1.6: Random walk.",prob.pdf
"1.1. SIMULATION OF DISCRETE PROBABILITIES 17
At the hotel there lived also some students with whom I usually
took my meals and had friendly relations. On a certain day one
of them expected the visit of his ancee, what (sic) I knew, but
I did not foresee that he and his ancee would also set out for a
stroll in the woods, and then suddenly I met them there. And then
I met them the same morning repeatedly, I don't remember how
many times, but certainly much too often and I felt embarrassed:
It looked as if I was snooping around which was, I assure you, not
the case.13
This set him to thinking about whether random walkers were destined to
meet.
Polya considered random walkers in one, two, and three dimensions. In one
dimension, he envisioned the walker on a very long street. At each intersec-
tion the walker 
ips a fair coin to decide which direction to walk next (see
Figure 1.6a). In two dimensions, the walker is walking on a grid of streets, and",prob.pdf
"Figure 1.6a). In two dimensions, the walker is walking on a grid of streets, and
at each intersection he chooses one of the four possible directions with equal
probability (see Figure 1.6b). In three dimensions (we might better speak of
a random climber), the walker moves on a three-dimensional grid, and at each
intersection there are now six dierent directions that the walker may choose,
each with equal probability (see Figure 1.6c).
The reader is referred to Section 12.1, where this and related problems are
discussed.
(a) Write a program to simulate a random walk in one dimension starting
at 0. Have your program print out the lengths of the times between
returns to the starting point (returns to 0). See if you can guess from
this simulation the answer to the following question: Will the walker
always return to his starting point eventually or might he drift away
forever?
(b) The paths of two walkers in two dimensions who meet after n steps can",prob.pdf
"forever?
(b) The paths of two walkers in two dimensions who meet after n steps can
be considered to be a single path that starts at (0; 0) and returns to (0; 0)
after 2n steps. This means that the probability that two random walkers
in two dimensions meet is the same as the probability that a single walker
in two dimensions ever returns to the starting point. Thus the question
of whether two walkers are sure to meet is the same as the question of
whether a single walker is sure to return to the starting point.
Write a program to simulate a random walk in two dimensions and see
if you think that the walker is sure to return to (0; 0). If so, Polya would
be sure to keep meeting his friends in the park. Perhaps by now you
have conjectured the answer to the question: Is a random walker in one
or two dimensions sure to return to the starting point? Polya answered
13G. Polya, \Two Incidents,"" Scientists at Work: Festschrift in Honour of Herman Wold, ed.",prob.pdf
"13G. Polya, \Two Incidents,"" Scientists at Work: Festschrift in Honour of Herman Wold, ed.
T. Dalenius, G. Karlsson, and S. Malmquist (Uppsala: Almquist & Wiksells Boktryckeri AB,
1970).",prob.pdf
"18 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
this question for dimensions one, two, and three. He established the
remarkable result that the answer is yes in one and two dimensions and
no in three dimensions.
(c) Write a program to simulate a random walk in three dimensions and see
whether, from this simulation and the results of (a) and (b), you could
have guessed Polya's result.
1.2 Discrete Probability Distributions
In this book we shall study many dierent experiments from a probabilistic point of
view. What is involved in this study will become evident as the theory is developed
and examples are analyzed. However, the overall idea can be described and illus-
trated as follows: to each experiment that we consider there will be associated a
random variable, which represents the outcome of any particular experiment. The
set of possible outcomes is called the sample space. In the rst part of this section,",prob.pdf
"set of possible outcomes is called the sample space. In the rst part of this section,
we will consider the case where the experiment has only nitely many possible out-
comes, i.e., the sample space is nite. We will then generalize to the case that the
sample space is either nite or countably innite. This leads us to the following
denition.
Random Variables and Sample Spaces
Denition 1.1 Suppose we have an experiment whose outcome depends on chance.
We represent the outcome of the experiment by a capital Roman letter, such as X,
called a random variable. The sample space of the experiment is the set of all
possible outcomes. If the sample space is either nite or countably innite, the
random variable is said to be discrete. 2
We generally denote a sample space by the capital Greek letter 
. As stated above,
in the correspondence between an experiment and the mathematical theory by which
it is studied, the sample space 
 corresponds to the set of possible outcomes of the",prob.pdf
"it is studied, the sample space 
 corresponds to the set of possible outcomes of the
experiment.
We now make two additional denitions. These are subsidiary to the denition
of sample space and serve to make precise some of the common terminology used
in conjunction with sample spaces. First of all, we dene the elements of a sample
space to be outcomes. Second, each subset of a sample space is dened to be an
event. Normally, we shall denote outcomes by lower case letters and events by
capital letters.
Example 1.6 A die is rolled once. We let X denote the outcome of this experiment.
Then the sample space for this experiment is the 6-element set",prob.pdf
= f1; 2; 3; 4; 5; 6g ;,prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 19
where each outcome i, for i = 1, . . . , 6, corresponds to the number of dots on the
face which turns up. The event
E = f2; 4; 6g
corresponds to the statement that the result of the roll is an even number. The
event E can also be described by saying that X is even. Unless there is reason to
believe the die is loaded, the natural assumption is that every outcome is equally
likely. Adopting this convention means that we assign a probability of 1/6 to each
of the six outcomes, i.e., m(i) = 1=6, for 1  i  6. 2
Distribution Functions
We next describe the assignment of probabilities. The denitions are motivated by
the example above, in which we assigned to each outcome of the sample space a
nonnegative number such that the sum of the numbers assigned is equal to 1.
Denition 1.2 Let X be a random variable which denotes the value of the out-
come of a certain experiment, and assume that this experiment has only nitely
many possible outcomes. Let",prob.pdf
"come of a certain experiment, and assume that this experiment has only nitely
many possible outcomes. Let 
 be the sample space of the experiment (i.e., the
set of all possible values of X, or equivalently, the set of all possible outcomes of
the experiment.) A distribution function for X is a real-valued function m whose
domain is 
 and which satises:
1. m(!)  0 ; for all ! 2 
 , and
2. P
!2
m(!) = 1 .
For any subset E of 
, we dene the probability of E to be the number P(E) given
by
P(E) =
X
!2E
m(!) :
2
Example 1.7 Consider an experiment in which a coin is tossed twice. Let X be
the random variable which corresponds to this experiment. We note that there are
several ways to record the outcomes of this experiment. We could, for example,
record the two tosses, in the order in which they occurred. In this case, we have",prob.pdf
"=fHH,HT,TH,TTg. We could also record the outcomes by simply noting the
number of heads that appeared. In this case, we have 
 =f0,1,2g. Finally, we could
record the two outcomes, without regard to the order in which they occurred. In
this case, we have 
 =fHH,HT,TTg.
We will use, for the moment, the rst of the sample spaces given above. We
will assume that all four outcomes are equally likely, and dene the distribution
function m(!) by
m(HH) = m(HT) = m(TH) = m(TT) = 1
4 :",prob.pdf
"20 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
Let E =fHH,HT,THg be the event that at least one head comes up. Then, the
probability of E can be calculated as follows:
P(E) = m(HH) + m(HT) + m(TH)
= 1
4 + 1
4 + 1
4 = 3
4 :
Similarly, if F =fHH,HTg is the event that heads comes up on the rst toss,
then we have
P(F) = m(HH) + m(HT)
= 1
4 + 1
4 = 1
2 :
2
Example 1.8 (Example 1.6 continued) The sample space for the experiment in
which the die is rolled is the 6-element set 
 = f1; 2; 3; 4; 5; 6g. We assumed that
the die was fair, and we chose the distribution function dened by
m(i) = 1
6; for i = 1; : : :; 6 :
If E is the event that the result of the roll is an even number, then E = f2; 4; 6g
and
P(E) = m(2) + m(4) + m(6)
= 1
6 + 1
6 + 1
6 = 1
2 :
2
Notice that it is an immediate consequence of the above denitions that, for
every ! 2 
,
P(f!g) = m(!) :
That is, the probability of the elementary event f!g, consisting of a single outcome",prob.pdf
"every ! 2 
,
P(f!g) = m(!) :
That is, the probability of the elementary event f!g, consisting of a single outcome
!, is equal to the value m(!) assigned to the outcome ! by the distribution function.
Example 1.9 Three people, A, B, and C, are running for the same oce, and we
assume that one and only one of them wins. The sample space may be taken as the
3-element set 
 =fA,B,Cg where each element corresponds to the outcome of that
candidate's winning. Suppose that A and B have the same chance of winning, but
that C has only 1/2 the chance of A or B. Then we assign
m(A) = m(B) = 2m(C) :
Since
m(A) + m(B) + m(C) = 1 ;",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 21
we see that
2m(C) + 2m(C) + m(C) = 1 ;
which implies that 5m(C) = 1. Hence,
m(A) = 2
5 ; m(B) = 2
5 ; m(C) = 1
5 :
Let E be the event that either A or C wins. Then E =fA,Cg, and
P(E) = m(A) + m(C) = 2
5 + 1
5 = 3
5 :
2
In many cases, events can be described in terms of other events through the use
of the standard constructions of set theory. We will brie
y review the denitions of
these constructions. The reader is referred to Figure 1.7 for Venn diagrams which
illustrate these constructions.
Let A and B be two sets. Then the union of A and B is the set
A [ B = fx j x 2 A or x 2 Bg :
The intersection of A and B is the set
A \ B = fx j x 2 A and x 2 Bg :
The dierence of A and B is the set
A  B = fx j x 2 A and x 62 Bg :
The set A is a subset of B, written A  B, if every element of A is also an element
of B. Finally, the complement of A is the set
~A = fx j x 2 
 and x 62 Ag :",prob.pdf
"of B. Finally, the complement of A is the set
~A = fx j x 2 
 and x 62 Ag :
The reason that these constructions are important is that it is typically the
case that complicated events described in English can be broken down into simpler
events using these constructions. For example, if A is the event that \it will snow
tomorrow and it will rain the next day,"" B is the event that \it will snow tomorrow,""
and C is the event that \it will rain two days from now,"" then A is the intersection
of the events B and C. Similarly, if D is the event that \it will snow tomorrow or
it will rain the next day,"" then D = B [ C. (Note that care must be taken here,
because sometimes the word \or"" in English means that exactly one of the two
alternatives will occur. The meaning is usually clear from context. In this book,
we will always use the word \or"" in the inclusive sense, i.e., A or B means that at
least one of the two events A, B is true.) The event ~B is the event that \it will not",prob.pdf
"least one of the two events A, B is true.) The event ~B is the event that \it will not
snow tomorrow."" Finally, if E is the event that \it will snow tomorrow but it will
not rain the next day,"" then E = B  C.",prob.pdf
"22 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
A   B A   B A
BA A AB
∼⊃  ⊃
A B
A   B
Figure 1.7: Basic set operations.
Properties
Theorem 1.1 The probabilities assigned to events by a distribution function on a
sample space 
 satisfy the following properties:
1. P(E)  0 for every E  
 .
2. P(
) = 1 .
3. If E  F  
, then P(E)  P(F) .
4. If A and B are disjoint subsets of 
, then P(A [ B) = P(A) + P(B) .
5. P( ~A) = 1  P(A) for every A  
 .
Proof. For any event E the probability P(E) is determined from the distribution
m by
P(E) =
X
!2E
m(!) ;
for every E  
. Since the function m is nonnegative, it follows that P(E) is also
nonnegative. Thus, Property 1 is true.
Property 2 is proved by the equations
P(
) =
X
!2
m(!) = 1 :
Suppose that E  F  
. Then every element ! that belongs to E also belongs
to F. Therefore, X
!2E
m(!) 
X
!2F
m(!) ;
since each term in the left-hand sum is in the right-hand sum, and all the terms in
both sums are non-negative. This implies that",prob.pdf
"m(!) 
X
!2F
m(!) ;
since each term in the left-hand sum is in the right-hand sum, and all the terms in
both sums are non-negative. This implies that
P(E)  P(F) ;
and Property 3 is proved.",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 23
Suppose next that A and B are disjoint subsets of 
. Then every element ! of
A [ B lies either in A and not in B or in B and not in A. It follows that
P(A [ B) = P
!2A[B m(!) = P
!2A m(!) + P
!2B m(!)
= P(A) + P(B) ;
and Property 4 is proved.
Finally, to prove Property 5, consider the disjoint union",prob.pdf
"= A [ ~A :
Since P(
) = 1, the property of disjoint additivity (Property 4) implies that
1 = P(A) + P( ~A) ;
whence P( ~A) = 1  P(A). 2
It is important to realize that Property 4 in Theorem 1.1 can be extended to
more than two sets. The general nite additivity property is given by the following
theorem.
Theorem 1.2 If A1, . .. , An are pairwise disjoint subsets of 
 (i.e., no two of the
Ai's have an element in common), then
P(A1 [ 
 
 
 [ An) =
nX
i=1
P(Ai) :
Proof. Let ! be any element in the union
A1 [ 
 
 
 [ An :
Then m(!) occurs exactly once on each side of the equality in the statement of the
theorem. 2
We shall often use the following consequence of the above theorem.
Theorem 1.3 Let A1, . . ., An be pairwise disjoint events with 
 = A1 [ 
 
 
 [ An,
and let E be any event. Then
P(E) =
nX
i=1
P(E \ Ai) :
Proof. The sets E \ A1, . .. , E \ An are pairwise disjoint, and their union is the
set E. The result now follows from Theorem 1.2. 2",prob.pdf
"24 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
Corollary 1.1 For any two events A and B,
P(A) = P(A \ B) + P(A \ ~B) :
2
Property 4 can be generalized in another way. Suppose that A and B are subsets
of 
 which are not necessarily disjoint. Then:
Theorem 1.4 If A and B are subsets of 
, then
P(A [ B) = P(A) + P(B)  P(A \ B) : (1.1)
Proof. The left side of Equation 1.1 is the sum of m(!) for ! in either A or B. We
must show that the right side of Equation 1.1 also adds m(!) for ! in A or B. If !
is in exactly one of the two sets, then it is counted in only one of the three terms
on the right side of Equation 1.1. If it is in both A and B, it is added twice from
the calculations of P(A) and P(B) and subtracted once for P(A \ B). Thus it is
counted exactly once by the right side. Of course, if A \ B = ;, then Equation 1.1
reduces to Property 4. (Equation 1.1 can also be generalized; see Theorem 3.8.) 2
Tree Diagrams",prob.pdf
"reduces to Property 4. (Equation 1.1 can also be generalized; see Theorem 3.8.) 2
Tree Diagrams
Example 1.10 Let us illustrate the properties of probabilities of events in terms
of three tosses of a coin. When we have an experiment which takes place in stages
such as this, we often nd it convenient to represent the outcomes by a tree diagram
as shown in Figure 1.8.
A path through the tree corresponds to a possible outcome of the experiment.
For the case of three tosses of a coin, we have eight paths !1, !2, . . . , !8 and,
assuming each outcome to be equally likely, we assign equal weight, 1/8, to each
path. Let E be the event \at least one head turns up."" Then ~E is the event \no
heads turn up."" This event occurs for only one outcome, namely, !8 = TTT. Thus,
~E = fTTTg and we have
P( ~E) = P(fTTTg) = m(TTT) = 1
8 :
By Property 5 of Theorem 1.1,
P(E) = 1  P( ~E) = 1  1
8 = 7
8 :
Note that we shall often nd it is easier to compute the probability that an event",prob.pdf
"By Property 5 of Theorem 1.1,
P(E) = 1  P( ~E) = 1  1
8 = 7
8 :
Note that we shall often nd it is easier to compute the probability that an event
does not happen rather than the probability that it does. We then use Property 5
to obtain the desired probability.",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 25
First toss Second toss Third toss Outcome
H
H
H
H
H
H
T
T
T
T
T
T
(Start)
ω
ω
ω
ω
ω
ω
ω
ω
1
2
3
4
5
6
7
8
H
T
Figure 1.8: Tree diagram for three tosses of a coin.
Let A be the event \the rst outcome is a head,"" and B the event \the second
outcome is a tail."" By looking at the paths in Figure 1.8, we see that
P(A) = P(B) = 1
2 :
Moreover, A\B = f!3; !4g, and so P(A\B) = 1=4: Using Theorem 1.4, we obtain
P(A [ B) = P(A) + P(B)  P(A \ B)
= 1
2 + 1
2  1
4 = 3
4 :
Since A [ B is the 6-element set,
A [ B = fHHH,HHT,HTH,HTT,TTH,TTTg ;
we see that we obtain the same result by direct enumeration. 2
In our coin tossing examples and in the die rolling example, we have assigned
an equal probability to each possible outcome of the experiment. Corresponding to
this method of assigning probabilities, we have the following denitions.
Uniform Distribution
Denition 1.3 The uniform distribution on a sample space 
 containing n ele-",prob.pdf
"Uniform Distribution
Denition 1.3 The uniform distribution on a sample space 
 containing n ele-
ments is the function m dened by
m(!) = 1
n ;
for every ! 2 
. 2",prob.pdf
"26 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
It is important to realize that when an experiment is analyzed to describe its
possible outcomes, there is no single correct choice of sample space. For the ex-
periment of tossing a coin twice in Example 1.2, we selected the 4-element set",prob.pdf
"=fHH,HT,TH,TTg as a sample space and assigned the uniform distribution func-
tion. These choices are certainly intuitively natural. On the other hand, for some
purposes it may be more useful to consider the 3-element sample space 
 = f0; 1; 2g
in which 0 is the outcome \no heads turn up,"" 1 is the outcome \exactly one head
turns up,"" and 2 is the outcome \two heads turn up."" The distribution function m
on 
 dened by the equations
m(0) = 1
4 ; m(1) = 1
2 ; m(2) = 1
4
is the one corresponding to the uniform probability density on the original sample
space 
. Notice that it is perfectly possible to choose a dierent distribution func-
tion. For example, we may consider the uniform distribution function on 
, which
is the function q dened by
q(0) = q(1) = q(2) = 1
3 :
Although q is a perfectly good distribution function, it is not consistent with ob-
served data on coin tossing.
Example 1.11 Consider the experiment that consists of rolling a pair of dice. We",prob.pdf
"served data on coin tossing.
Example 1.11 Consider the experiment that consists of rolling a pair of dice. We
take as the sample space 
 the set of all ordered pairs (i; j) of integers with 1  i  6
and 1  j  6. Thus,",prob.pdf
"= f (i; j) : 1  i; j  6 g :
(There is at least one other \reasonable"" choice for a sample space, namely the set
of all unordered pairs of integers, each between 1 and 6. For a discussion of why
we do not use this set, see Example 3.14.) To determine the size of 
, we note
that there are six choices for i, and for each choice of i there are six choices for j,
leading to 36 dierent outcomes. Let us assume that the dice are not loaded. In
mathematical terms, this means that we assume that each of the 36 outcomes is
equally likely, or equivalently, that we adopt the uniform distribution function on

 by setting
m((i; j)) = 1
36; 1  i; j  6 :
What is the probability of getting a sum of 7 on the roll of two dice|or getting a
sum of 11? The rst event, denoted by E, is the subset
E = f(1; 6); (6; 1); (2; 5); (5; 2); (3; 4); (4; 3)g :
A sum of 11 is the subset F given by
F = f(5; 6); (6; 5)g :
Consequently,
P(E) = P
!2E m(!) = 6 
 1
36 = 1
6 ;
P(F) = P
!2F m(!) = 2 
 1
36 = 1
18 :",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 27
What is the probability of getting neither snakeeyes (double ones) nor boxcars
(double sixes)? The event of getting either one of these two outcomes is the set
E = f(1; 1); (6; 6)g :
Hence, the probability of obtaining neither is given by
P( ~E) = 1  P(E) = 1  2
36 = 17
18 :
2
In the above coin tossing and the dice rolling experiments, we have assigned an
equal probability to each outcome. That is, in each example, we have chosen the
uniform distribution function. These are the natural choices provided the coin is a
fair one and the dice are not loaded. However, the decision as to which distribution
function to select to describe an experiment is not a part of the basic mathemat-
ical theory of probability. The latter begins only when the sample space and the
distribution function have already been dened.
Determination of Probabilities
It is important to consider ways in which probability distributions are determined",prob.pdf
"Determination of Probabilities
It is important to consider ways in which probability distributions are determined
in practice. One way is by symmetry. For the case of the toss of a coin, we do not
see any physical dierence between the two sides of a coin that should aect the
chance of one side or the other turning up. Similarly, with an ordinary die there
is no essential dierence between any two sides of the die, and so by symmetry we
assign the same probability for any possible outcome. In general, considerations
of symmetry often suggest the uniform distribution function. Care must be used
here. We should not always assume that, just because we do not know any reason
to suggest that one outcome is more likely than another, it is appropriate to assign
equal probabilities. For example, consider the experiment of guessing the sex of
a newborn child. It has been observed that the proportion of newborn children",prob.pdf
"a newborn child. It has been observed that the proportion of newborn children
who are boys is about .513. Thus, it is more appropriate to assign a distribution
function which assigns probability .513 to the outcome boy and probability .487 to
the outcome girl than to assign probability 1/2 to each outcome. This is an example
where we use statistical observations to determine probabilities. Note that these
probabilities may change with new studies and may vary from country to country.
Genetic engineering might even allow an individual to in
uence this probability for
a particular case.
Odds
Statistical estimates for probabilities are ne if the experiment under consideration
can be repeated a number of times under similar circumstances. However, assume
that, at the beginning of a football season, you want to assign a probability to the
event that Dartmouth will beat Harvard. You really do not have data that relates to",prob.pdf
"event that Dartmouth will beat Harvard. You really do not have data that relates to
this year's football team. However, you can determine your own personal probability",prob.pdf
"28 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
by seeing what kind of a bet you would be willing to make. For example, suppose
that you are willing to make a 1 dollar bet giving 2 to 1 odds that Dartmouth will
win. Then you are willing to pay 2 dollars if Dartmouth loses in return for receiving
1 dollar if Dartmouth wins. This means that you think the appropriate probability
for Dartmouth winning is 2/3.
Let us look more carefully at the relation between odds and probabilities. Sup-
pose that we make a bet at r to 1 odds that an event E occurs. This means that
we think that it is r times as likely that E will occur as that E will not occur. In
general, r to s odds will be taken to mean the same thing as r=s to 1, i.e., the ratio
between the two numbers is the only quantity of importance when stating odds.
Now if it is r times as likely that E will occur as that E will not occur, then the
probability that E occurs must be r=(r + 1), since we have
P(E) = r P( ~E)
and",prob.pdf
"probability that E occurs must be r=(r + 1), since we have
P(E) = r P( ~E)
and
P(E) + P( ~E) = 1 :
In general, the statement that the odds are r to s in favor of an event E occurring
is equivalent to the statement that
P(E) = r=s
(r=s) + 1
= r
r + s :
If we let P(E) = p, then the above equation can easily be solved for r=s in terms of
p; we obtain r=s = p=(1  p). We summarize the above discussion in the following
denition.
Denition 1.4 If P(E) = p, the odds in favor of the event E occurring are r : s (r
to s) where r=s = p=(1  p). If r and s are given, then p can be found by using the
equation p = r=(r + s). 2
Example 1.12 (Example 1.9 continued) In Example 1.9 we assigned probability
1/5 to the event that candidate C wins the race. Thus the odds in favor of C
winning are 1=5 : 4=5. These odds could equally well have been written as 1 : 4,
2 : 8, and so forth. A bet that C wins is fair if we receive 4 dollars if C wins and
pay 1 dollar if C loses. 2
Innite Sample Spaces",prob.pdf
"2 : 8, and so forth. A bet that C wins is fair if we receive 4 dollars if C wins and
pay 1 dollar if C loses. 2
Innite Sample Spaces
If a sample space has an innite number of points, then the way that a distribution
function is dened depends upon whether or not the sample space is countable. A
sample space is countably innite if the elements can be counted, i.e., can be put
in one-to-one correspondence with the positive integers, and uncountably innite",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 29
otherwise. Innite sample spaces require new concepts in general (see Chapter 2),
but countably innite spaces do not. If

 = f!1; !2; !3; : : :g
is a countably innite sample space, then a distribution function is dened exactly
as in Denition 1.2, except that the sum must now be a convergent innite sum.
Theorem 1.1 is still true, as are its extensions Theorems 1.2 and 1.4. One thing we
cannot do on a countably innite sample space that we could do on a nite sample
space is to dene a uniform distribution function as in Denition 1.3. You are asked
in Exercise 20 to explain why this is not possible.
Example 1.13 A coin is tossed until the rst time that a head turns up. Let the
outcome of the experiment, !, be the rst time that a head turns up. Then the
possible outcomes of our experiment are",prob.pdf
"= f1; 2; 3; : : :g :
Note that even though the coin could come up tails every time we have not allowed
for this possibility. We will explain why in a moment. The probability that heads
comes up on the rst toss is 1/2. The probability that tails comes up on the rst
toss and heads on the second is 1/4. The probability that we have two tails followed
by a head is 1/8, and so forth. This suggests assigning the distribution function
m(n) = 1=2n for n = 1, 2, 3, .. . . To see that this is a distribution function we
must show that X
!
m(!) = 1
2 + 1
4 + 1
8 + 
 
 
 = 1 :
That this is true follows from the formula for the sum of a geometric series,
1 + r + r2 + r3 + 
 
 
 = 1
1  r ;
or
r + r2 + r3 + r4 + 
 
 
 = r
1  r ; (1.2)
for 1 < r < 1.
Putting r = 1=2, we see that we have a probability of 1 that the coin eventu-
ally turns up heads. The possible outcome of tails every time has to be assigned
probability 0, so we omit it from our sample space of possible outcomes.",prob.pdf
"probability 0, so we omit it from our sample space of possible outcomes.
Let E be the event that the rst time a head turns up is after an even number
of tosses. Then
E = f2; 4; 6; 8; : : :g ;
and
P(E) = 1
4 + 1
16 + 1
64 + 
 
 
 :
Putting r = 1=4 in Equation 1.2 see that
P(E) = 1=4
1  1=4 = 1
3 :
Thus the probability that a head turns up for the rst time after an even number
of tosses is 1/3 and after an odd number of tosses is 2/3. 2",prob.pdf
"30 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
Historical Remarks
An interesting question in the history of science is: Why was probability not devel-
oped until the sixteenth century? We know that in the sixteenth century problems
in gambling and games of chance made people start to think about probability. But
gambling and games of chance are almost as old as civilization itself. In ancient
Egypt (at the time of the First Dynasty, ca. 3500 B.C.) a game now called \Hounds
and Jackals"" was played. In this game the movement of the hounds and jackals was
based on the outcome of the roll of four-sided dice made out of animal bones called
astragali. Six-sided dice made of a variety of materials date back to the sixteenth
century B.C. Gambling was widespread in ancient Greece and Rome. Indeed, in the
Roman Empire it was sometimes found necessary to invoke laws against gambling.
Why, then, were probabilities not calculated until the sixteenth century?",prob.pdf
"Why, then, were probabilities not calculated until the sixteenth century?
Several explanations have been advanced for this late development. One is that
the relevant mathematics was not developed and was not easy to develop. The
ancient mathematical notation made numerical calculation complicated, and our
familiar algebraic notation was not developed until the sixteenth century. However,
as we shall see, many of the combinatorial ideas needed to calculate probabilities
were discussed long before the sixteenth century. Since many of the chance events
of those times had to do with lotteries relating to religious aairs, it has been
suggested that there may have been religious barriers to the study of chance and
gambling. Another suggestion is that a stronger incentive, such as the development
of commerce, was necessary. However, none of these explanations seems completely
satisfactory, and people still wonder why it took so long for probability to be studied",prob.pdf
"satisfactory, and people still wonder why it took so long for probability to be studied
seriously. An interesting discussion of this problem can be found in Hacking.14
The rst person to calculate probabilities systematically was Gerolamo Cardano
(1501{1576) in his book Liber de Ludo Aleae. This was translated from the Latin
by Gould and appears in the book Cardano: The Gambling Scholar by Ore.15 Ore
provides a fascinating discussion of the life of this colorful scholar with accounts
of his interests in many dierent elds, including medicine, astrology, and mathe-
matics. You will also nd there a detailed account of Cardano's famous battle with
Tartaglia over the solution to the cubic equation.
In his book on probability Cardano dealt only with the special case that we have
called the uniform distribution function. This restriction to equiprobable outcomes
was to continue for a long time. In this case Cardano realized that the probability",prob.pdf
"was to continue for a long time. In this case Cardano realized that the probability
that an event occurs is the ratio of the number of favorable outcomes to the total
number of outcomes.
Many of Cardano's examples dealt with rolling dice. Here he realized that the
outcomes for two rolls should be taken to be the 36 ordered pairs (i; j) rather than
the 21 unordered pairs. This is a subtle point that was still causing problems much
later for other writers on probability. For example, in the eighteenth century the
famous French mathematician d'Alembert, author of several works on probability,
claimed that when a coin is tossed twice the number of heads that turn up would
14I. Hacking, The Emergence of Probability (Cambridge: Cambridge University Press, 1975).
15O. Ore, Cardano: The Gambling Scholar (Princeton: Princeton University Press, 1953).",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 31
be 0, 1, or 2, and hence we should assign equal probabilities for these three possible
outcomes.16 Cardano chose the correct sample space for his dice problems and
calculated the correct probabilities for a variety of events.
Cardano's mathematical work is interspersed with a lot of advice to the potential
gambler in short paragraphs, entitled, for example: \Who Should Play and When,""
\Why Gambling Was Condemned by Aristotle,"" \Do Those Who Teach Also Play
Well?"" and so forth. In a paragraph entitled \The Fundamental Principle of Gam-
bling,"" Cardano writes:
The most fundamental principle of all in gambling is simply equal con-
ditions, e.g., of opponents, of bystanders, of money, of situation, of the
dice box, and of the die itself. To the extent to which you depart from
that equality, if it is in your opponent's favor, you are a fool, and if in
your own, you are unjust.17",prob.pdf
"that equality, if it is in your opponent's favor, you are a fool, and if in
your own, you are unjust.17
Cardano did make mistakes, and if he realized it later he did not go back and
change his error. For example, for an event that is favorable in three out of four
cases, Cardano assigned the correct odds 3 : 1 that the event will occur. But then he
assigned odds by squaring these numbers (i.e., 9 : 1) for the event to happen twice in
a row. Later, by considering the case where the odds are 1 : 1, he realized that this
cannot be correct and was led to the correct result that when f out of n outcomes
are favorable, the odds for a favorable outcome twice in a row are f2 : n2 f2. Ore
points out that this is equivalent to the realization that if the probability that an
event happens in one experiment is p, the probability that it happens twice is p2.
Cardano proceeded to establish that for three successes the formula should be p3",prob.pdf
"Cardano proceeded to establish that for three successes the formula should be p3
and for four successes p4, making it clear that he understood that the probability
is pn for n successes in n independent repetitions of such an experiment. This will
follow from the concept of independence that we introduce in Section 4.1.
Cardano's work was a remarkable rst attempt at writing down the laws of
probability, but it was not the spark that started a systematic study of the subject.
This came from a famous series of letters between Pascal and Fermat. This corre-
spondence was initiated by Pascal to consult Fermat about problems he had been
given by Chevalier de Mere, a well-known writer, a prominent gure at the court of
Louis XIV, and an ardent gambler.
The rst problem de Mere posed was a dice problem. The story goes that he had
been betting that at least one six would turn up in four rolls of a die and winning",prob.pdf
"been betting that at least one six would turn up in four rolls of a die and winning
too often, so he then bet that a pair of sixes would turn up in 24 rolls of a pair
of dice. The probability of a six with one die is 1/6 and, by the product law for
independent experiments, the probability of two sixes when a pair of dice is thrown
is (1=6)(1=6) = 1=36. Ore18 claims that a gambling rule of the time suggested that,
since four repetitions was favorable for the occurrence of an event with probability
1/6, for an event six times as unlikely, 6 
 4 = 24 repetitions would be sucient for
16J. d'Alembert, \Croix ou Pile,"" in L'Encyclopedie, ed. Diderot, vol. 4 (Paris, 1754).
17O. Ore, op. cit., p. 189.
18O. Ore, \Pascal and the Invention of Probability Theory,"" American Mathematics Monthly,
vol. 67 (1960), pp. 409{419.",prob.pdf
"32 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
a favorable bet. Pascal showed, by exact calculation, that 25 rolls are required for
a favorable bet for a pair of sixes.
The second problem was a much harder one: it was an old problem and con-
cerned the determination of a fair division of the stakes in a tournament when the
series, for some reason, is interrupted before it is completed. This problem is now
referred to as the problem of points. The problem had been a standard problem in
mathematical texts; it appeared in Fra Luca Paccioli's book summa de Arithmetica,
Geometria, Proportioni et Proportionalita, printed in Venice in 1494,19 in the form:
A team plays ball such that a total of 60 points are required to win the
game, and each inning counts 10 points. The stakes are 10 ducats. By
some incident they cannot nish the game and one side has 50 points
and the other 20. One wants to know what share of the prize money",prob.pdf
"some incident they cannot nish the game and one side has 50 points
and the other 20. One wants to know what share of the prize money
belongs to each side. In this case I have found that opinions dier from
one to another but all seem to me insucient in their arguments, but I
shall state the truth and give the correct way.
Reasonable solutions, such as dividing the stakes according to the ratio of games
won by each player, had been proposed, but no correct solution had been found at
the time of the Pascal-Fermat correspondence. The letters deal mainly with the
attempts of Pascal and Fermat to solve this problem. Blaise Pascal (1623{1662)
was a child prodigy, having published his treatise on conic sections at age sixteen,
and having invented a calculating machine at age eighteen. At the time of the
letters, his demonstration of the weight of the atmosphere had already established
his position at the forefront of contemporary physicists. Pierre de Fermat (1601{",prob.pdf
"his position at the forefront of contemporary physicists. Pierre de Fermat (1601{
1665) was a learned jurist in Toulouse, who studied mathematics in his spare time.
He has been called by some the prince of amateurs and one of the greatest pure
mathematicians of all times.
The letters, translated by Maxine Merrington, appear in Florence David's fasci-
nating historical account of probability, Games, Gods and Gambling.20 In a letter
dated Wednesday, 29th July, 1654, Pascal writes to Fermat:
Sir,
Like you, I am equally impatient, and although I am again ill in bed,
I cannot help telling you that yesterday evening I received from M. de
Carcavi your letter on the problem of points, which I admire more than
I can possibly say. I have not the leisure to write at length, but, in a
word, you have solved the two problems of points, one with dice and the
other with sets of games with perfect justness; I am entirely satised
with it for I do not doubt that I was in the wrong, seeing the admirable",prob.pdf
"other with sets of games with perfect justness; I am entirely satised
with it for I do not doubt that I was in the wrong, seeing the admirable
agreement in which I nd myself with you now.. .
Your method is very sound and is the one which rst came to my mind
in this research; but because the labour of the combination is excessive,
I have found a short cut and indeed another method which is much
19ibid., p. 414.
20F. N. David, Games, Gods and Gambling (London: G. Grin, 1962), p. 230 .",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 33
0
1
2
3
0 1 2 3
0 0 0
8 16 32 64
20 32 48 64
6432 44 56
Number of games 
     A has won
Number of games
     B has won
Figure 1.9: Pascal's table.
quicker and neater, which I would like to tell you here in a few words:
for henceforth I would like to open my heart to you, if I may, as I am so
overjoyed with our agreement. I see that truth is the same in Toulouse
as in Paris.
Here, more or less, is what I do to show the fair value of each game,
when two opponents play, for example, in three games and each person
has staked 32 pistoles.
Let us say that the rst man had won twice and the other once; now
they play another game, in which the conditions are that, if the rst
wins, he takes all the stakes; that is 64 pistoles; if the other wins it,
then they have each won two games, and therefore, if they wish to stop
playing, they must each take back their own stake, that is, 32 pistoles
each.",prob.pdf
"then they have each won two games, and therefore, if they wish to stop
playing, they must each take back their own stake, that is, 32 pistoles
each.
Then consider, Sir, if the rst man wins, he gets 64 pistoles; if he loses
he gets 32. Thus if they do not wish to risk this last game but wish to
separate without playing it, the rst man must say: `I am certain to get
32 pistoles, even if I lost I still get them; but as for the other 32, perhaps
I will get them, perhaps you will get them, the chances are equal. Let
us then divide these 32 pistoles in half and give one half to me as well
as my 32 which are mine for sure.' He will then have 48 pistoles and the
other 16.. .
Pascal's argument produces the table illustrated in Figure 1.9 for the amount
due player A at any quitting point.
Each entry in the table is the average of the numbers just above and to the right
of the number. This fact, together with the known values when the tournament is",prob.pdf
"of the number. This fact, together with the known values when the tournament is
completed, determines all the values in this table. If player A wins the rst game,",prob.pdf
"34 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
then he needs two games to win and B needs three games to win; and so, if the
tounament is called o, A should receive 44 pistoles.
The letter in which Fermat presented his solution has been lost; but fortunately,
Pascal describes Fermat's method in a letter dated Monday, 24th August, 1654.
From Pascal's letter:21
This is your procedure when there are two players: If two players, play-
ing several games, nd themselves in that position when the rst man
needs two games and second needs three, then to nd the fair division
of stakes, you say that one must know in how many games the play will
be absolutely decided.
It is easy to calculate that this will be in four games, from which you can
conclude that it is necessary to see in how many ways four games can be
arranged between two players, and one must see how many combinations
would make the rst man win and how many the second and to share",prob.pdf
"arranged between two players, and one must see how many combinations
would make the rst man win and how many the second and to share
out the stakes in this proportion. I would have found it dicult to
understand this if I had not known it myself already; in fact you had
explained it with this idea in mind.
Fermat realized that the number of ways that the game might be nished may
not be equally likely. For example, if A needs two more games and B needs three to
win, two possible ways that the tournament might go for A to win are WLW and
LWLW. These two sequences do not have the same chance of occurring. To avoid
this diculty, Fermat extended the play, adding ctitious plays, so that all the ways
that the games might go have the same length, namely four. He was shrewd enough
to realize that this extension would not change the winner and that he now could
simply count the number of sequences favorable to each player since he had made",prob.pdf
"simply count the number of sequences favorable to each player since he had made
them all equally likely. If we list all possible ways that the extended game of four
plays might go, we obtain the following 16 possible outcomes of the play:
WWWW WLWW LWWW LLWW
WWWL WLWL LWWL LLWL
WWLW WLLW LWLW LLLW
WWLL WLLL LWLL LLLL .
Player A wins in the cases where there are at least two wins (the 11 underlined
cases), and B wins in the cases where there are at least three losses (the other
5 cases). Since A wins in 11 of the 16 possible cases Fermat argued that the
probability that A wins is 11/16. If the stakes are 64 pistoles, A should receive
44 pistoles in agreement with Pascal's result. Pascal and Fermat developed more
systematic methods for counting the number of favorable outcomes for problems
like this, and this will be one of our central problems. Such counting methods fall
under the subject of combinatorics, which is the topic of Chapter 3.
21ibid., p. 239.",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 35
We see that these two mathematicians arrived at two very dierent ways to solve
the problem of points. Pascal's method was to develop an algorithm and use it to
calculate the fair division. This method is easy to implement on a computer and easy
to generalize. Fermat's method, on the other hand, was to change the problem into
an equivalent problem for which he could use counting or combinatorial methods.
We will see in Chapter 3 that, in fact, Fermat used what has become known as
Pascal's triangle! In our study of probability today we shall nd that both the
algorithmic approach and the combinatorial approach share equal billing, just as
they did 300 years ago when probability got its start.
Exercises
1 Let 
 = fa; b; cg be a sample space. Let m(a) = 1=2, m(b) = 1=3, and
m(c) = 1=6. Find the probabilities for all eight subsets of 
.
2 Give a possible sample space 
 for each of the following experiments:",prob.pdf
"m(c) = 1=6. Find the probabilities for all eight subsets of 
.
2 Give a possible sample space 
 for each of the following experiments:
(a) An election decides between two candidates A and B.
(b) A two-sided coin is tossed.
(c) A student is asked for the month of the year and the day of the week on
which her birthday falls.
(d) A student is chosen at random from a class of ten students.
(e) You receive a grade in this course.
3 For which of the cases in Exercise 2 would it be reasonable to assign the
uniform distribution function?
4 Describe in words the events specied by the following subsets of",prob.pdf
"= fHHH; HHT; HTH; HTT; THH; THT; TTH; TTTg
(see Example 1.6).
(a) E = fHHH,HHT,HTH,HTTg.
(b) E = fHHH,TTTg.
(c) E = fHHT,HTH,THHg.
(d) E = fHHT,HTH,HTT,THH,THT,TTH,TTTg.
5 What are the probabilities of the events described in Exercise 4?
6 A die is loaded in such a way that the probability of each face turning up
is proportional to the number of dots on that face. (For example, a six is
three times as probable as a two.) What is the probability of getting an even
number in one throw?
7 Let A and B be events such that P(A \ B) = 1=4, P( ~A) = 1=3, and P(B) =
1=2. What is P(A [ B)?",prob.pdf
"36 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
8 A student must choose one of the subjects, art, geology, or psychology, as an
elective. She is equally likely to choose art or psychology and twice as likely
to choose geology. What are the respective probabilities that she chooses art,
geology, and psychology?
9 A student must choose exactly two out of three electives: art, French, and
mathematics. He chooses art with probability 5/8, French with probability
5/8, and art and French together with probability 1/4. What is the probability
that he chooses mathematics? What is the probability that he chooses either
art or French?
10 For a bill to come before the president of the United States, it must be passed
by both the House of Representatives and the Senate. Assume that, of the
bills presented to these two bodies, 60 percent pass the House, 80 percent
pass the Senate, and 90 percent pass at least one of the two. Calculate the",prob.pdf
"pass the Senate, and 90 percent pass at least one of the two. Calculate the
probability that the next bill presented to the two groups will come before the
president.
11 What odds should a person give in favor of the following events?
(a) A card chosen at random from a 52-card deck is an ace.
(b) Two heads will turn up when a coin is tossed twice.
(c) Boxcars (two sixes) will turn up when two dice are rolled.
12 You oer 3 : 1 odds that your friend Smith will be elected mayor of your city.
What probability are you assigning to the event that Smith wins?
13 In a horse race, the odds that Romance will win are listed as 2 : 3 and that
Downhill will win are 1 : 2. What odds should be given for the event that
either Romance or Downhill wins?
14 Let X be a random variable with distribution function mX(x) dened by
mX(1) = 1=5; mX(0) = 1=5; mX(1) = 2=5; mX(2) = 1=5 :
(a) Let Y be the random variable dened by the equation Y = X + 3. Find
the distribution function mY (y) of Y .",prob.pdf
"(a) Let Y be the random variable dened by the equation Y = X + 3. Find
the distribution function mY (y) of Y .
(b) Let Z be the random variable dened by the equation Z = X2. Find the
distribution function mZ(z) of Z.
*15 John and Mary are taking a mathematics course. The course has only three
grades: A, B, and C. The probability that John gets a B is .3. The probability
that Mary gets a B is .4. The probability that neither gets an A but at least
one gets a B is .1. What is the probability that at least one gets a B but
neither gets a C?
16 In a erce battle, not less than 70 percent of the soldiers lost one eye, not less
than 75 percent lost one ear, not less than 80 percent lost one hand, and not",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 37
less than 85 percent lost one leg. What is the minimal possible percentage of
those who simultaneously lost one ear, one eye, one hand, and one leg?22
*17 Assume that the probability of a \success"" on a single experiment with n
outcomes is 1=n. Let m be the number of experiments necessary to make it a
favorable bet that at least one success will occur (see Exercise 1.1.5).
(a) Show that the probability that, in m trials, there are no successes is
(1  1=n)m.
(b) (de Moivre) Show that if m = n log2 then
lim
n!1

1  1
n
m
= 1
2 :
Hint:
lim
n!1

1  1
n
n
= e1 :
Hence for large n we should choose m to be about n log 2.
(c) Would DeMoivre have been led to the correct answer for de Mere's two
bets if he had used his approximation?
18 (a) For events A1, . . . , An, prove that
P(A1 [ 
 
 
 [ An)  P(A1) + 
 
 
 + P(An) :
(b) For events A and B, prove that
P(A \ B)  P(A) + P(B)  1:
19 If A, B, and C are any three events, show that",prob.pdf
"(b) For events A and B, prove that
P(A \ B)  P(A) + P(B)  1:
19 If A, B, and C are any three events, show that
P(A [ B [ C) = P(A) + P(B) + P(C)
 P(A \ B)  P(B \ C)  P(C \ A)
+ P(A \ B \ C) :
20 Explain why it is not possible to dene a uniform distribution function (see
Denition 1.3) on a countably innite sample space. Hint: Assume m(!) = a
for all !, where 0  a  1. Does m(!) have all the properties of a distribution
function?
21 In Example 1.13 nd the probability that the coin turns up heads for the rst
time on the tenth, eleventh, or twelfth toss.
22 A die is rolled until the rst time that a six turns up. We shall see that the
probability that this occurs on the nth roll is (5=6)n1 
(1=6). Using this fact,
describe the appropriate innite sample space and distribution function for
the experiment of rolling a die until a six turns up for the rst time. Verify
that for your distribution function P
! m(!) = 1.",prob.pdf
"the experiment of rolling a die until a six turns up for the rst time. Verify
that for your distribution function P
! m(!) = 1.
22See Knot X, in Lewis Carroll, Mathematical Recreations, vol. 2 (Dover, 1958).",prob.pdf
"38 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
23 Let 
 be the sample space",prob.pdf
"= f0; 1; 2; : : :g ;
and dene a distribution function by
m(j) = (1  r)jr ;
for some xed r, 0 < r < 1, and for j = 0; 1; 2; : : :. Show that this is a
distribution function for 
.
24 Our calendar has a 400-year cycle. B. H. Brown noticed that the number of
times the thirteenth of the month falls on each of the days of the week in the
4800 months of a cycle is as follows:
Sunday 687
Monday 685
Tuesday 685
Wednesday 687
Thursday 684
Friday 688
Saturday 684
From this he deduced that the thirteenth was more likely to fall on Friday
than on any other day. Explain what he meant by this.
25 Tversky and Kahneman23 asked a group of subjects to carry out the following
task. They are told that:
Linda is 31, single, outspoken, and very bright. She majored in
philosophy in college. As a student, she was deeply concerned with
racial discrimination and other social issues, and participated in
anti-nuclear demonstrations.",prob.pdf
"racial discrimination and other social issues, and participated in
anti-nuclear demonstrations.
The subjects are then asked to rank the likelihood of various alternatives, such
as:
(1) Linda is active in the feminist movement.
(2) Linda is a bank teller.
(3) Linda is a bank teller and active in the feminist movement.
Tversky and Kahneman found that between 85 and 90 percent of the subjects
rated alternative (1) most likely, but alternative (3) more likely than alterna-
tive (2). Is it? They call this phenomenon the conjunction fallacy, and note
that it appears to be unaected by prior training in probability or statistics.
Is this phenomenon a fallacy? If so, why? Can you give a possible explanation
for the subjects' choices?
23K. McKean, \Decisions, Decisions,"" pp. 22{31.",prob.pdf
"1.2. DISCRETE PROBABILITY DISTRIBUTIONS 39
26 Two cards are drawn successively from a deck of 52 cards. Find the probability
that the second card is higher in rank than the rst card. Hint: Show that 1 =
P(higher) + P(lower) + P(same) and use the fact that P(higher) = P(lower).
27 A life table is a table that lists for a given number of births the estimated
number of people who will live to a given age. In Appendix C we give a life
table based upon 100,000 births for ages from 0 to 85, both for women and for
men. Show how from this table you can estimate the probability m(x) that a
person born in 1981 would live to age x. Write a program to plot m(x) both
for men and for women, and comment on the dierences that you see in the
two cases.
*28 Here is an attempt to get around the fact that we cannot choose a \random
integer.""
(a) What, intuitively, is the probability that a \randomly chosen"" positive
integer is a multiple of 3?",prob.pdf
"integer.""
(a) What, intuitively, is the probability that a \randomly chosen"" positive
integer is a multiple of 3?
(b) Let P3(N) be the probability that an integer, chosen at random between
1 and N, is a multiple of 3 (since the sample space is nite, this is a
legitimate probability). Show that the limit
P3 = lim
N!1
P3(N)
exists and equals 1/3. This formalizes the intuition in (a), and gives us
a way to assign \probabilities"" to certain events that are innite subsets
of the positive integers.
(c) If A is any set of positive integers, let A(N) mean the number of elements
of A which are less than or equal to N. Then dene the \probability"" of
A as
P(A) = lim
N!1
A(N)=N ;
provided this limit exists. Show that this denition would assign prob-
ability 0 to any nite set and probability 1 to the set of all positive
integers. Thus, the probability of the set of all integers is not the sum of
the probabilities of the individual integers in this set. This means that",prob.pdf
"the probabilities of the individual integers in this set. This means that
the denition of probability given here is not a completely satisfactory
denition.
(d) Let A be the set of all positive integers with an odd number of dig-
its. Show that P(A) does not exist. This shows that under the above
denition of probability, not all sets have probabilities.
29 (from Sholander24) In a standard clover-leaf interchange, there are four ramps
for making right-hand turns, and inside these four ramps, there are four more
ramps for making left-hand turns. Your car approaches the interchange from
the south. A mechanism has been installed so that at each point where there
exists a choice of directions, the car turns to the right with xed probability r.
24M. Sholander, Problem #1034, Mathematics Magazine, vol. 52, no. 3 (May 1979), p. 183.",prob.pdf
"40 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS
(a) If r = 1=2, what is your chance of emerging from the interchange going
west?
(b) Find the value of r that maximizes your chance of a westward departure
from the interchange.
30 (from Benkoski25) Consider a \pure"" cloverleaf interchange in which there
are no ramps for right-hand turns, but only the two intersecting straight
highways with cloverleaves for left-hand turns. (Thus, to turn right in such
an interchange, one must make three left-hand turns.) As in the preceding
problem, your car approaches the interchange from the south. What is the
value of r that maximizes your chances of an eastward departure from the
interchange?
31 (from vos Savant26) A reader of Marilyn vos Savant's column wrote in with
the following question:
My dad heard this story on the radio. At Duke University, two
students had received A's in chemistry all semester. But on the
night before the nal exam, they were partying in another state",prob.pdf
"students had received A's in chemistry all semester. But on the
night before the nal exam, they were partying in another state
and didn't get back to Duke until it was over. Their excuse to the
professor was that they had a 
at tire, and they asked if they could
take a make-up test. The professor agreed, wrote out a test and sent
the two to separate rooms to take it. The rst question (on one side
of the paper) was worth 5 points, and they answered it easily. Then
they 
ipped the paper over and found the second question, worth
95 points: `Which tire was it?' What was the probability that both
students would say the same thing? My dad and I think it's 1 in
16. Is that right?""
(a) Is the answer 1/16?
(b) The following question was asked of a class of students. \I was driving
to school today, and one of my tires went 
at. Which tire do you think
it was?"" The responses were as follows: right front, 58%, left front, 11%,",prob.pdf
"to school today, and one of my tires went 
at. Which tire do you think
it was?"" The responses were as follows: right front, 58%, left front, 11%,
right rear, 18%, left rear, 13%. Suppose that this distribution holds in
the general population, and assume that the two test-takers are randomly
chosen from the general population. What is the probability that they
will give the same answer to the second question?
25S. Benkoski, Comment on Problem #1034, Mathematics Magazine, vol. 52, no. 3 (May 1979),
pp. 183-184.
26M. vos Savant, Parade Magazine, 3 March 1996, p. 14.",prob.pdf
"Chapter 2
Continuous Probability
Densities
2.1 Simulation of Continuous Probabilities
In this section we shall show how we can use computer simulations for experiments
that have a whole continuum of possible outcomes.
Probabilities
Example 2.1 We begin by constructing a spinner, which consists of a circle of unit
circumference and a pointer as shown in Figure 2.1. We pick a point on the circle
and label it 0, and then label every other point on the circle with the distance, say
x, from 0 to that point, measured counterclockwise. The experiment consists of
spinning the pointer and recording the label of the point at the tip of the pointer.
We let the random variable X denote the value of this outcome. The sample space
is clearly the interval [0; 1). We would like to construct a probability model in which
each outcome is equally likely to occur.
If we proceed as we did in Chapter 1 for experiments with a nite number of",prob.pdf
"each outcome is equally likely to occur.
If we proceed as we did in Chapter 1 for experiments with a nite number of
possible outcomes, then we must assign the probability 0 to each outcome, since
otherwise, the sum of the probabilities, over all of the possible outcomes, would
not equal 1. (In fact, summing an uncountable number of real numbers is a tricky
business; in particular, in order for such a sum to have any meaning, at most
countably many of the summands can be dierent than 0.) However, if all of the
assigned probabilities are 0, then the sum is 0, not 1, as it should be.
In the next section, we will show how to construct a probability model in this
situation. At present, we will assume that such a model can be constructed. We
will also assume that in this model, if E is an arc of the circle, and E is of length
p, then the model will assign the probability p to E. This means that if the pointer",prob.pdf
"p, then the model will assign the probability p to E. This means that if the pointer
is spun, the probability that it ends up pointing to a point in E equals p, which is
certainly a reasonable thing to expect.
41",prob.pdf
"42 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0
x
Figure 2.1: A spinner.
To simulate this experiment on a computer is an easy matter. Many computer
software packages have a function which returns a random real number in the in-
terval [0; 1]. Actually, the returned value is always a rational number, and the
values are determined by an algorithm, so a sequence of such values is not truly
random. Nevertheless, the sequences produced by such algorithms behave much
like theoretically random sequences, so we can use such sequences in the simulation
of experiments. On occasion, we will need to refer to such a function. We will call
this function rnd. 2
Monte Carlo Procedure and Areas
It is sometimes desirable to estimate quantities whose exact values are dicult or
impossible to calculate exactly. In some of these cases, a procedure involving chance,
called a Monte Carlo procedure, can be used to provide such an estimate.",prob.pdf
"called a Monte Carlo procedure, can be used to provide such an estimate.
Example 2.2 In this example we show how simulation can be used to estimate
areas of plane gures. Suppose that we program our computer to provide a pair
(x; y) or numbers, each chosen independently at random from the interval [0; 1].
Then we can interpret this pair (x; y) as the coordinates of a point chosen at random
from the unit square. Events are subsets of the unit square. Our experience with
Example 2.1 suggests that the point is equally likely to fall in subsets of equal area.
Since the total area of the square is 1, the probability of the point falling in a specic
subset E of the unit square should be equal to its area. Thus, we can estimate the
area of any subset of the unit square by estimating the probability that a point
chosen at random from this square falls in the subset.
We can use this method to estimate the area of the region E under the curve",prob.pdf
"chosen at random from this square falls in the subset.
We can use this method to estimate the area of the region E under the curve
y = x2 in the unit square (see Figure 2.2). We choose a large number of points (x; y)
at random and record what fraction of them fall in the region E = f (x; y) : y  x2 g.
The program MonteCarlo will carry out this experiment for us. Running this
program for 10,000 experiments gives an estimate of .325 (see Figure 2.3).
From these experiments we would estimate the area to be about 1/3. Of course,",prob.pdf
"2.1. SIMULATION OF CONTINUOUS PROBABILITIES 43
1
x
1
y
y = x2
E
Figure 2.2: Area under y = x2:
for this simple region we can nd the exact area by calculus. In fact,
Area of E =
Z 1
0
x2 dx = 1
3 :
We have remarked in Chapter 1 that, when we simulate an experiment of this type
n times to estimate a probability, we can expect the answer to be in error by at
most 1=pn at least 95 percent of the time. For 10,000 experiments we can expect
an accuracy of 0.01, and our simulation did achieve this accuracy.
This same argument works for any region E of the unit square. For example,
suppose E is the circle with center (1=2; 1=2) and radius 1/2. Then the probability
that our random point (x; y) lies inside the circle is equal to the area of the circle,
that is,
P(E) = 
1
2
2
= 
4 :
If we did not know the value of , we could estimate the value by performing this
experiment a large number of times! 2
The above example is not the only way of estimating the value of  by a chance",prob.pdf
"experiment a large number of times! 2
The above example is not the only way of estimating the value of  by a chance
experiment. Here is another way, discovered by Buon.1
1G. L. Buon, in \Essai d'Arithmetique Morale,"" Oeuvres Completes de Buon avec Supple-
ments, tome iv, ed. Dumenil (Paris, 1836).",prob.pdf
"44 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
1
1
1000 trials
Estimate of area is .325
y = x2
E
Figure 2.3: Computing the area by simulation.
Buon's Needle
Example 2.3 Suppose that we take a card table and draw across the top surface
a set of parallel lines a unit distance apart. We then drop a common needle of
unit length at random on this surface and observe whether or not the needle lies
across one of the lines. We can describe the possible outcomes of this experiment
by coordinates as follows: Let d be the distance from the center of the needle to the
nearest line. Next, let L be the line determined by the needle, and dene  as the
acute angle that the line L makes with the set of parallel lines. (The reader should
certainly be wary of this description of the sample space. We are attempting to
coordinatize a set of line segments. To see why one must be careful in the choice
of coordinates, see Example 2.6.) Using this description, we have 0  d  1=2, and",prob.pdf
"of coordinates, see Example 2.6.) Using this description, we have 0  d  1=2, and
0    =2. Moreover, we see that the needle lies across the nearest line if and
only if the hypotenuse of the triangle (see Figure 2.4) is less than half the length of
the needle, that is,
d
sin  < 1
2 :
Now we assume that when the needle drops, the pair (; d) is chosen at random
from the rectangle 0    =2, 0  d  1=2. We observe whether the needle lies
across the nearest line (i.e., whether d  (1=2) sin). The probability of this event
E is the fraction of the area of the rectangle which lies inside E (see Figure 2.5).",prob.pdf
"2.1. SIMULATION OF CONTINUOUS PROBABILITIES 45
d
1/2
θ
Figure 2.4: Buon's experiment.
θ0
1/2
0
d
π/2
E
Figure 2.5: Set E of pairs (; d) with d < 1
2 sin .
Now the area of the rectangle is =4, while the area of E is
Area =
Z =2
0
1
2 sin  d = 1
2 :
Hence, we get
P(E) = 1=2
=4 = 2
 :
The program BuonsNeedle simulates this experiment. In Figure 2.6, we show
the position of every 100th needle in a run of the program in which 10,000 needles
were \dropped."" Our nal estimate for  is 3.139. While this was within 0.003 of
the true value for  we had no right to expect such accuracy. The reason for this
is that our simulation estimates P(E). While we can expect this estimate to be in
error by at most 0.001, a small error in P(E) gets magnied when we use this to
compute  = 2=P(E). Perlman and Wichura, in their article \Sharpening Buon's",prob.pdf
"46 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0.00
5.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
5.00
10000
3.139
Figure 2.6: Simulation of Buon's needle experiment.
Needle,""2 show that we can expect to have an error of not more than 5=pn about
95 percent of the time. Here n is the number of needles dropped. Thus for 10,000
needles we should expect an error of no more than 0.05, and that was the case here.
We see that a large number of experiments is necessary to get a decent estimate for
. 2
In each of our examples so far, events of the same size are equally likely. Here
is an example where they are not. We will see many other such examples later.
Example 2.4 Suppose that we choose two random real numbers in [0; 1] and add
them together. Let X be the sum. How is X distributed?
To help understand the answer to this question, we can use the program Are-
abargraph. This program produces a bar graph with the property that on each",prob.pdf
"abargraph. This program produces a bar graph with the property that on each
interval, the area, rather than the height, of the bar is equal to the fraction of out-
comes that fell in the corresponding interval. We have carried out this experiment
1000 times; the data is shown in Figure 2.7. It appears that the function dened
by
f(x) =

x; if 0  x  1,
2  x; if 1 < x  2
ts the data very well. (It is shown in the gure.) In the next section, we will
see that this function is the \right"" function. By this we mean that if a and b are
any two real numbers between 0 and 2, with a  b, then we can use this function
to calculate the probability that a  X  b. To understand how this calculation
might be performed, we again consider Figure 2.7. Because of the way the bars
were constructed, the sum of the areas of the bars corresponding to the interval
2M. D. Perlman and M. J. Wichura, \Sharpening Buon's Needle,"" The American Statistician,
vol. 29, no. 4 (1975), pp. 157{163.",prob.pdf
"2.1. SIMULATION OF CONTINUOUS PROBABILITIES 47
0 0.5 1 1.5 2
0
0.2
0.4
0.6
0.8
1
Figure 2.7: Sum of two random numbers.
[a; b] approximates the probability that a  X  b. But the sum of the areas of
these bars also approximates the integral
Z b
a
f(x) dx :
This suggests that for an experiment with a continuum of possible outcomes, if we
nd a function with the above property, then we will be able to use it to calculate
probabilities. In the next section, we will show how to determine the function
f(x). 2
Example 2.5 Suppose that we choose 100 random numbers in [0; 1], and let X
represent their sum. How is X distributed? We have carried out this experiment
10000 times; the results are shown in Figure 2.8. It is not so clear what function
ts the bars in this case. It turns out that the type of function which does the job
is called a normal density function. This type of function is sometimes referred to
as a \bell-shaped"" curve. It is among the most important functions in the subject",prob.pdf
"as a \bell-shaped"" curve. It is among the most important functions in the subject
of probability, and will be formally dened in Section 5.2 of Chapter 4.3. 2
Our last example explores the fundamental question of how probabilities are
assigned.
Bertrand's Paradox
Example 2.6 A chord of a circle is a line segment both of whose endpoints lie on
the circle. Suppose that a chord is drawn at random in a unit circle. What is the
probability that its length exceeds
p
3?
Our answer will depend on what we mean by random, which will depend, in turn,
on what we choose for coordinates. The sample space 
 is the set of all possible
chords in the circle. To nd coordinates for these chords, we rst introduce a",prob.pdf
"48 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
40 45 50 55 60
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
Figure 2.8: Sum of 100 random numbers.
x
y
A
B
M
θ
β
α
Figure 2.9: Random chord.
rectangular coordinate system with origin at the center of the circle (see Figure 2.9).
We note that a chord of a circle is perpendicular to the radial line containing the
midpoint of the chord. We can describe each chord by giving:
1. The rectangular coordinates (x; y) of the midpoint M, or
2. The polar coordinates (r; ) of the midpoint M, or
3. The polar coordinates (1; ) and (1; ) of the endpoints A and B.
In each case we shall interpret at random to mean: choose these coordinates at
random.
We can easily estimate this probability by computer simulation. In programming
this simulation, it is convenient to include certain simplications, which we describe
in turn:",prob.pdf
"2.1. SIMULATION OF CONTINUOUS PROBABILITIES 49
1. To simulate this case, we choose values for x and y from [1; 1] at random.
Then we check whether x2 + y2  1. If not, the point M = (x; y) lies outside
the circle and cannot be the midpoint of any chord, and we ignore it. Oth-
erwise, M lies inside the circle and is the midpoint of a unique chord, whose
length L is given by the formula:
L = 2
p
1  (x2 + y2) :
2. To simulate this case, we take account of the fact that any rotation of the
circle does not change the length of the chord, so we might as well assume in
advance that the chord is horizontal. Then we choose r from [1; 1] at random,
and compute the length of the resulting chord with midpoint (r; =2) by the
formula:
L = 2
p
1  r2 :
3. To simulate this case, we assume that one endpoint, say B, lies at (1; 0) (i.e.,
that  = 0). Then we choose a value for  from [0; 2] at random and compute
the length of the resulting chord, using the Law of Cosines, by the formula:
L =
p",prob.pdf
"the length of the resulting chord, using the Law of Cosines, by the formula:
L =
p
2  2 cos :
The program BertrandsParadox carries out this simulation. Running this
program produces the results shown in Figure 2.10. In the rst circle in this gure,
a smaller circle has been drawn. Those chords which intersect this smaller circle
have length at least
p
3. In the second circle in the gure, the vertical line intersects
all chords of length at least
p
3. In the third circle, again the vertical line intersects
all chords of length at least
p
3.
In each case we run the experiment a large number of times and record the
fraction of these lengths that exceed
p
3. We have printed the results of every
100th trial up to 10,000 trials.
It is interesting to observe that these fractions are not the same in the three cases;
they depend on our choice of coordinates. This phenomenon was rst observed by
Bertrand, and is now known as Bertrand's paradox.3 It is actually not a paradox at",prob.pdf
"Bertrand, and is now known as Bertrand's paradox.3 It is actually not a paradox at
all; it is merely a re
ection of the fact that dierent choices of coordinates will lead
to dierent assignments of probabilities. Which assignment is \correct"" depends on
what application or interpretation of the model one has in mind.
One can imagine a real experiment involving throwing long straws at a circle
drawn on a card table. A \correct"" assignment of coordinates should not depend
on where the circle lies on the card table, or where the card table sits in the room.
Jaynes4 has shown that the only assignment which meets this requirement is (2).
In this sense, the assignment (2) is the natural, or \correct"" one (see Exercise 11).
We can easily see in each case what the true probabilities are if we note thatp
3 is the length of the side of an inscribed equilateral triangle. Hence, a chord has
3J. Bertrand, Calcul des Probabilites (Paris: Gauthier-Villars, 1889).",prob.pdf
"3J. Bertrand, Calcul des Probabilites (Paris: Gauthier-Villars, 1889).
4E. T. Jaynes, \The Well-Posed Problem,"" in Papers on Probability, Statistics and Statistical
Physics, R. D. Rosencrantz, ed. (Dordrecht: D. Reidel, 1983), pp. 133{148.",prob.pdf
"50 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
 .0
1.0
 .2
 .4
 .6
 .8
1.0
.488
.227
 .0
1.0
 .2
 .4
 .6
 .8
1.0
 .0
1.0
 .2
 .4
 .6
 .8
1.0
.332
10000 10000 10000
Figure 2.10: Bertrand's paradox.
length L >
p
3 if its midpoint has distance d < 1=2 from the origin (see Figure 2.9).
The following calculations determine the probability that L >
p
3 in each of the
three cases.
1. L >
p
3 if(x; y) lies inside a circle of radius 1/2, which occurs with probability
p = (1=2)2
(1)2 = 1
4 :
2. L >
p
3 if jrj < 1=2, which occurs with probability
1=2  (1=2)
1  (1) = 1
2 :
3. L >
p
3 if 2=3 <  < 4=3, which occurs with probability
4=3  2=3
2  0 = 1
3 :
We see that our simulations agree quite well with these theoretical values. 2
Historical Remarks
G. L. Buon (1707{1788) was a natural scientist in the eighteenth century who
applied probability to a number of his investigations. His work is found in his
monumental 44-volume Histoire Naturelle and its supplements.5 For example, he",prob.pdf
"monumental 44-volume Histoire Naturelle and its supplements.5 For example, he
5G. L. Buon, Histoire Naturelle, Generali et Particular avec le Description du Cabinet du
Roy, 44 vols. (Paris: L`Imprimerie Royale, 1749{1803).",prob.pdf
"2.1. SIMULATION OF CONTINUOUS PROBABILITIES 51
Length of Number of Number of Estimate
Experimenter needle casts crossings for 
Wolf, 1850 .8 5000 2532 3.1596
Smith, 1855 .6 3204 1218.5 3.1553
De Morgan, c.1860 1.0 600 382.5 3.137
Fox, 1864 .75 1030 489 3.1595
Lazzerini, 1901 .83 3408 1808 3.1415929
Reina, 1925 .5419 2520 869 3.1795
Table 2.1: Buon needle experiments to estimate .
presented a number of mortality tables and used them to compute, for each age
group, the expected remaining lifetime. From his table he observed: the expected
remaining lifetime of an infant of one year is 33 years, while that of a man of 21
years is also approximately 33 years. Thus, a father who is not yet 21 can hope to
live longer than his one year old son, but if the father is 40, the odds are already 3
to 2 that his son will outlive him.6
Buon wanted to show that not all probability calculations rely only on algebra,
but that some rely on geometrical calculations. One such problem was his famous",prob.pdf
"but that some rely on geometrical calculations. One such problem was his famous
\needle problem"" as discussed in this chapter.7 In his original formulation, Buon
describes a game in which two gamblers drop a loaf of French bread on a wide-board

oor and bet on whether or not the loaf falls across a crack in the 
oor. Buon
asked: what length L should the bread loaf be, relative to the width W of the

oorboards, so that the game is fair. He found the correct answer (L = (=4)W)
using essentially the methods described in this chapter. He also considered the case
of a checkerboard 
oor, but gave the wrong answer in this case. The correct answer
was given later by Laplace.
The literature contains descriptions of a number of experiments that were actu-
ally carried out to estimate  by this method of dropping needles. N. T. Gridgeman8
discusses the experiments shown in Table 2.1. (The halves for the number of cross-",prob.pdf
"discusses the experiments shown in Table 2.1. (The halves for the number of cross-
ing comes from a compromise when it could not be decided if a crossing had actually
occurred.) He observes, as we have, that 10,000 casts could do no more than estab-
lish the rst decimal place of  with reasonable condence. Gridgeman points out
that, although none of the experiments used even 10,000 casts, they are surprisingly
good, and in some cases, too good. The fact that the number of casts is not always
a round number would suggest that the authors might have resorted to clever stop-
ping to get a good answer. Gridgeman comments that Lazzerini's estimate turned
out to agree with a well-known approximation to , 355=113 = 3:1415929, discov-
ered by the fth-century Chinese mathematician, Tsu Ch'ungchih. Gridgeman says
that he did not have Lazzerini's original report, and while waiting for it (knowing
6G. L. Buon, \Essai d'Arithmetique Morale,"" p. 301.
7ibid., pp. 277{278.",prob.pdf
"6G. L. Buon, \Essai d'Arithmetique Morale,"" p. 301.
7ibid., pp. 277{278.
8N. T. Gridgeman, \Geometric Probability and the Number "" Scripta Mathematika, vol. 25,
no. 3, (1960), pp. 183{195.",prob.pdf
"52 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
only the needle crossed a line 1808 times in 3408 casts) deduced that the length of
the needle must have been 5/6. He calculated this from Buon's formula, assuming
 = 355=113:
L = P(E)
2 = 1
2
355
113
1808
3408

= 5
6 = :8333 :
Even with careful planning one would have to be extremely lucky to be able to stop
so cleverly.
The second author likes to trace his interest in probability theory to the Chicago
World's Fair of 1933 where he observed a mechanical device dropping needles and
displaying the ever-changing estimates for the value of . (The rst author likes to
trace his interest in probability theory to the second author.)
Exercises
*1 In the spinner problem (see Example 2.1) divide the unit circumference into
three arcs of length 1/2, 1/3, and 1/6. Write a program to simulate the
spinner experiment 1000 times and print out what fraction of the outcomes",prob.pdf
"three arcs of length 1/2, 1/3, and 1/6. Write a program to simulate the
spinner experiment 1000 times and print out what fraction of the outcomes
fall in each of the three arcs. Now plot a bar graph whose bars have width 1/2,
1/3, and 1/6, and areas equal to the corresponding fractions as determined
by your simulation. Show that the heights of the bars are all nearly the same.
2 Do the same as in Exercise 1, but divide the unit circumference into ve arcs
of length 1/3, 1/4, 1/5, 1/6, and 1/20.
3 Alter the program MonteCarlo to estimate the area of the circle of radius
1/2 with center at (1=2; 1=2) inside the unit square by choosing 1000 points
at random. Compare your results with the true value of =4. Use your results
to estimate the value of . How accurate is your estimate?
4 Alter the program MonteCarlo to estimate the area under the graph of
y = sin x inside the unit square by choosing 10,000 points at random. Now",prob.pdf
"4 Alter the program MonteCarlo to estimate the area under the graph of
y = sin x inside the unit square by choosing 10,000 points at random. Now
calculate the true value of this area and use your results to estimate the value
of . How accurate is your estimate?
5 Alter the program MonteCarlo to estimate the area under the graph of
y = 1=(x + 1) in the unit square in the same way as in Exercise 4. Calculate
the true value of this area and use your simulation results to estimate the
value of log 2. How accurate is your estimate?
6 To simulate the Buon's needle problem we choose independently the dis-
tance d and the angle  at random, with 0  d  1=2 and 0    =2,
and check whether d  (1=2) sin. Doing this a large number of times, we
estimate  as 2=a, where a is the fraction of the times that d  (1=2) sin.
Write a program to estimate  by this method. Run your program several
times for each of 100, 1000, and 10,000 experiments. Does the accuracy of",prob.pdf
"Write a program to estimate  by this method. Run your program several
times for each of 100, 1000, and 10,000 experiments. Does the accuracy of
the experimental approximation for  improve as the number of experiments
increases?",prob.pdf
"2.1. SIMULATION OF CONTINUOUS PROBABILITIES 53
7 For Buon's needle problem, Laplace9 considered a grid with horizontal and
vertical lines one unit apart. He showed that the probability that a needle of
length L  1 crosses at least one line is
p = 4L  L2
 :
To simulate this experiment we choose at random an angle  between 0 and
=2 and independently two numbers d1 and d2 between 0 and L=2. (The two
numbers represent the distance from the center of the needle to the nearest
horizontal and vertical line.) The needle crosses a line if either d1  (L=2) sin
or d2  (L=2) cos. We do this a large number of times and estimate  as
 = 4L  L2
a ;
where a is the proportion of times that the needle crosses at least one line.
Write a program to estimate  by this method, run your program for 100,
1000, and 10,000 experiments, and compare your results with Buon's method
described in Exercise 6. (Take L = 1.)
8 A long needle of length L much bigger than 1 is dropped on a grid with",prob.pdf
"described in Exercise 6. (Take L = 1.)
8 A long needle of length L much bigger than 1 is dropped on a grid with
horizontal and vertical lines one unit apart. We will see (in Exercise 6.3.28)
that the average number a of lines crossed is approximately
a = 4L
 :
To estimate  by simulation, pick an angle  at random between 0 and =2 and
compute L sin + L cos. This may be used for the number of lines crossed.
Repeat this many times and estimate  by
 = 4L
a ;
where a is the average number of lines crossed per experiment. Write a pro-
gram to simulate this experiment and run your program for the number of
experiments equal to 100, 1000, and 10,000. Compare your results with the
methods of Laplace or Buon for the same number of experiments. (Use
L = 100.)
The following exercises involve experiments in which not all outcomes are
equally likely. We shall consider such experiments in detail in the next section,
but we invite you to explore a few simple cases here.",prob.pdf
"equally likely. We shall consider such experiments in detail in the next section,
but we invite you to explore a few simple cases here.
9 A large number of waiting time problems have an exponential distribution of
outcomes. We shall see (in Section 5.2) that such outcomes are simulated by
computing (1=) log(rnd), where  > 0. For waiting times produced in this
way, the average waiting time is 1=. For example, the times spent waiting for
9P. S. Laplace, Theorie Analytique des Probabilites (Paris: Courcier, 1812).",prob.pdf
"54 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
a car to pass on a highway, or the times between emissions of particles from a
radioactive source, are simulated by a sequence of random numbers, each of
which is chosen by computing (1=) log(rnd), where 1= is the average time
between cars or emissions. Write a program to simulate the times between
cars when the average time between cars is 30 seconds. Have your program
compute an area bar graph for these times by breaking the time interval from
0 to 120 into 24 subintervals. On the same pair of axes, plot the function
f(x) = (1=30)e(1=30)x. Does the function t the bar graph well?
10 In Exercise 9, the distribution came \out of a hat."" In this problem, we will
again consider an experiment whose outcomes are not equally likely. We will
determine a function f(x) which can be used to determine the probability of
certain events. Let T be the right triangle in the plane with vertices at the",prob.pdf
"certain events. Let T be the right triangle in the plane with vertices at the
points (0; 0); (1; 0); and (0; 1). The experiment consists of picking a point
at random in the interior of T, and recording only the x-coordinate of the
point. Thus, the sample space is the set [0; 1], but the outcomes do not seem
to be equally likely. We can simulate this experiment by asking a computer to
return two random real numbers in [0; 1], and recording the rst of these two
numbers if their sum is less than 1. Write this program and run it for 10,000
trials. Then make a bar graph of the result, breaking the interval [0; 1] into
10 intervals. Compare the bar graph with the function f(x) = 2  2x. Now
show that there is a constant c such that the height of T at the x-coordinate
value x is c times f(x) for every x in [0; 1]. Finally, show that
Z 1
0
f(x) dx = 1 :
How might one use the function f(x) to determine the probability that the
outcome is between :2 and :5?",prob.pdf
"Z 1
0
f(x) dx = 1 :
How might one use the function f(x) to determine the probability that the
outcome is between :2 and :5?
11 Here is another way to pick a chord at random on the circle of unit radius.
Imagine that we have a card table whose sides are of length 100. We place
coordinate axes on the table in such a way that each side of the table is parallel
to one of the axes, and so that the center of the table is the origin. We now
place a circle of unit radius on the table so that the center of the circle is the
origin. Now pick out a point (x0; y0) at random in the square, and an angle 
at random in the interval (=2; =2). Let m = tan. Then the equation of
the line passing through (x0; y0) with slope m is
y = y0 + m(x  x0) ;
and the distance of this line from the center of the circle (i.e., the origin) is
d =




y0  mx0
p
m2 + 1



 :
We can use this distance formula to check whether the line intersects the circle",prob.pdf
"d =




y0  mx0
p
m2 + 1



 :
We can use this distance formula to check whether the line intersects the circle
(i.e., whether d < 1). If so, we consider the resulting chord a random chord.",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 55
This describes an experiment of dropping a long straw at random on a table
on which a circle is drawn.
Write a program to simulate this experiment 10000 times and estimate the
probability that the length of the chord is greater than
p
3. How does your
estimate compare with the results of Example 2.6?
2.2 Continuous Density Functions
In the previous section we have seen how to simulate experiments with a whole
continuum of possible outcomes and have gained some experience in thinking about
such experiments. Now we turn to the general problem of assigning probabilities to
the outcomes and events in such experiments. We shall restrict our attention here
to those experiments whose sample space can be taken as a suitably chosen subset
of the line, the plane, or some other Euclidean space. We begin with some simple
examples.
Spinners
Example 2.7 The spinner experiment described in Example 2.1 has the interval",prob.pdf
"examples.
Spinners
Example 2.7 The spinner experiment described in Example 2.1 has the interval
[0; 1) as the set of possible outcomes. We would like to construct a probability
model in which each outcome is equally likely to occur. We saw that in such a
model, it is necessary to assign the probability 0 to each outcome. This does not at
all mean that the probability of every event must be zero. On the contrary, if we
let the random variable X denote the outcome, then the probability
P( 0  X  1)
that the head of the spinner comes to rest somewhere in the circle, should be equal
to 1. Also, the probability that it comes to rest in the upper half of the circle should
be the same as for the lower half, so that
P

0  X < 1
2

= P
1
2  X < 1

= 1
2 :
More generally, in our model, we would like the equation
P(c  X < d) = d  c
to be true for every choice of c and d.
If we let E = [c; d], then we can write the above formula in the form
P(E) =
Z
E
f(x) dx ;",prob.pdf
"to be true for every choice of c and d.
If we let E = [c; d], then we can write the above formula in the form
P(E) =
Z
E
f(x) dx ;
where f(x) is the constant function with value 1. This should remind the reader of
the corresponding formula in the discrete case for the probability of an event:
P(E) =
X
!2E
m(!) :",prob.pdf
"56 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
Figure 2.11: Spinner experiment.
The dierence is that in the continuous case, the quantity being integrated, f(x),
is not the probability of the outcome x. (However, if one uses innitesimals, one
can consider f(x) dx as the probability of the outcome x.)
In the continuous case, we will use the following convention. If the set of out-
comes is a set of real numbers, then the individual outcomes will be referred to
by small Roman letters such as x. If the set of outcomes is a subset of R2, then
the individual outcomes will be denoted by (x; y). In either case, it may be more
convenient to refer to an individual outcome by using !, as in Chapter 1.
Figure 2.11 shows the results of 1000 spins of the spinner. The function f(x)
is also shown in the gure. The reader will note that the area under f(x) and
above a given interval is approximately equal to the fraction of outcomes that fell",prob.pdf
"above a given interval is approximately equal to the fraction of outcomes that fell
in that interval. The function f(x) is called the density function of the random
variable X. The fact that the area under f(x) and above an interval corresponds
to a probability is the dening property of density functions. A precise denition
of density functions will be given shortly. 2
Darts
Example 2.8 A game of darts involves throwing a dart at a circular target of unit
radius. Suppose we throw a dart once so that it hits the target, and we observe
where it lands.
To describe the possible outcomes of this experiment, it is natural to take as our
sample space the set 
 of all the points in the target. It is convenient to describe
these points by their rectangular coordinates, relative to a coordinate system with
origin at the center of the target, so that each pair (x; y) of coordinates with x2+y2 
1 describes a possible outcome of the experiment. Then 
 = f (x; y) : x2 + y2  1 g",prob.pdf
"1 describes a possible outcome of the experiment. Then 
 = f (x; y) : x2 + y2  1 g
is a subset of the Euclidean plane, and the event E = f (x; y) : y > 0 g, for example,
corresponds to the statement that the dart lands in the upper half of the target,
and so forth. Unless there is reason to believe otherwise (and with experts at the",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 57
game there may well be!), it is natural to assume that the coordinates are chosen
at random. (When doing this with a computer, each coordinate is chosen uniformly
from the interval [1; 1]. If the resulting point does not lie inside the unit circle,
the point is not counted.) Then the arguments used in the preceding example show
that the probability of any elementary event, consisting of a single outcome, must
be zero, and suggest that the probability of the event that the dart lands in any
subset E of the target should be determined by what fraction of the target area lies
in E. Thus,
P(E) = area of E
area of target = area of E
 :
This can be written in the form
P(E) =
Z
E
f(x) dx ;
where f(x) is the constant function with value 1=. In particular, if E = f (x; y) :
x2 + y2  a2 g is the event that the dart lands within distance a < 1 of the center
of the target, then
P(E) = a2
 = a2 :",prob.pdf
"x2 + y2  a2 g is the event that the dart lands within distance a < 1 of the center
of the target, then
P(E) = a2
 = a2 :
For example, the probability that the dart lies within a distance 1/2 of the center
is 1/4. 2
Example 2.9 In the dart game considered above, suppose that, instead of observ-
ing where the dart lands, we observe how far it lands from the center of the target.
In this case, we take as our sample space the set 
 of all circles with centers at
the center of the target. It is convenient to describe these circles by their radii, so
that each circle is identied by its radius r, 0  r  1. In this way, we may regard",prob.pdf
"as the subset [0; 1] of the real line.
What probabilities should we assign to the events E of 
? If
E = f r : 0  r  a g ;
then E occurs if the dart lands within a distance a of the center, that is, within the
circle of radius a, and we saw in the previous example that under our assumptions
the probability of this event is given by
P([0; a]) = a2 :
More generally, if
E = f r : a  r  b g ;
then by our basic assumptions,
P(E) = P([a; b]) = P([0; b])  P([0; a])
= b2  a2
= (b  a)(b + a)
= 2(b  a)(b + a)
2 :",prob.pdf
"58 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0 0.2 0.4 0.6 0.8 1
0
0.5
1
1.5
2
0         0.2       0.4        0.6        0.8         1 
2
1.5
1
0.5
0
Figure 2.12: Distribution of dart distances in 400 throws.
Thus, P(E) =2(length of E)(midpoint of E). Here we see that the probability
assigned to the interval E depends not only on its length but also on its midpoint
(i.e., not only on how long it is, but also on where it is). Roughly speaking, in this
experiment, events of the form E = [a; b] are more likely if they are near the rim
of the target and less likely if they are near the center. (A common experience for
beginners! The conclusion might well be dierent if the beginner is replaced by an
expert.)
Again we can simulate this by computer. We divide the target area into ten
concentric regions of equal thickness.
The computer program Darts throws n darts and records what fraction of the
total falls in each of these concentric regions. The program Areabargraph then",prob.pdf
"total falls in each of these concentric regions. The program Areabargraph then
plots a bar graph with the area of the ith bar equal to the fraction of the total
falling in the ith region. Running the program for 1000 darts resulted in the bar
graph of Figure 2.12.
Note that here the heights of the bars are not all equal, but grow approximately
linearly with r. In fact, the linear function y = 2r appears to t our bar graph quite
well. This suggests that the probability that the dart falls within a distance a of the
center should be given by the area under the graph of the function y = 2r between
0 and a. This area is a2, which agrees with the probability we have assigned above
to this event. 2
Sample Space Coordinates
These examples suggest that for continuous experiments of this sort we should assign
probabilities for the outcomes to fall in a given interval by means of the area under
a suitable function.
More generally, we suppose that suitable coordinates can be introduced into the",prob.pdf
"a suitable function.
More generally, we suppose that suitable coordinates can be introduced into the
sample space 
, so that we can regard 
 as a subset of Rn. We call such a sample
space a continuous sample space. We let X be a random variable which represents
the outcome of the experiment. Such a random variable is called a continuous
random variable. We then dene a density function for X as follows.",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 59
Density Functions of Continuous Random Variables
Denition 2.1 Let X be a continuous real-valued random variable. A density
function for X is a real-valued function f which satises
P(a  X  b) =
Z b
a
f(x) dx
for all a; b 2 R. 2
We note that it is not the case that all continuous real-valued random variables
possess density functions. However, in this book, we will only consider continuous
random variables for which density functions exist.
In terms of the density f(x), if E is a subset of R, then
P(X 2 E) =
Z
E
f(x) dx :
The notation here assumes that E is a subset of R for which
R
E f(x) dx makes
sense.
Example 2.10 (Example 2.7 continued) In the spinner experiment, we choose for
our set of outcomes the interval 0  x < 1, and for our density function
f(x) =
 1; if 0  x < 1,
0; otherwise.
If E is the event that the head of the spinner falls in the upper half of the circle,
then E = f x : 0  x  1=2 g, and so
P(E) =
Z 1=2
0
1 dx = 1
2 :",prob.pdf
"then E = f x : 0  x  1=2 g, and so
P(E) =
Z 1=2
0
1 dx = 1
2 :
More generally, if E is the event that the head falls in the interval [a; b], then
P(E) =
Z b
a
1 dx = b  a :
2
Example 2.11 (Example 2.8 continued) In the rst dart game experiment, we
choose for our sample space a disc of unit radius in the plane and for our density
function the function
f(x; y) =
 1=; if x2 + y2  1,
0; otherwise.
The probability that the dart lands inside the subset E is then given by
P(E) =
Z Z
E
1
 dx dy
= 1
 
 (area of E) :
2",prob.pdf
"60 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
In these two examples, the density function is constant and does not depend
on the particular outcome. It is often the case that experiments in which the
coordinates are chosen at random can be described by constant density functions,
and, as in Section 1.2, we call such density functions uniform or equiprobable. Not
all experiments are of this type, however.
Example 2.12 (Example 2.9 continued) In the second dart game experiment, we
choose for our sample space the unit interval on the real line and for our density
the function
f(r) =

2r; if 0 < r < 1,
0; otherwise.
Then the probability that the dart lands at distance r, a  r  b, from the center
of the target is given by
P([a; b]) =
Z b
a
2r dr
= b2  a2 :
Here again, since the density is small when r is near 0 and large when r is near 1, we
see that in this experiment the dart is more likely to land near the rim of the target",prob.pdf
"see that in this experiment the dart is more likely to land near the rim of the target
than near the center. In terms of the bar graph of Example 2.9, the heights of the
bars approximate the density function, while the areas of the bars approximate the
probabilities of the subintervals (see Figure 2.12). 2
We see in this example that, unlike the case of discrete sample spaces, the
value f(x) of the density function for the outcome x is not the probability of x
occurring (we have seen that this probability is always 0) and in general f(x) is not
a probability at all. In this example, if we take  = 2 then f(3=4) = 3=2, which
being bigger than 1, cannot be a probability.
Nevertheless, the density function f does contain all the probability information
about the experiment, since the probabilities of all events can be derived from it.
In particular, the probability that the outcome of the experiment falls in an interval
[a; b] is given by
P([a; b]) =
Z b
a
f(x) dx ;",prob.pdf
"In particular, the probability that the outcome of the experiment falls in an interval
[a; b] is given by
P([a; b]) =
Z b
a
f(x) dx ;
that is, by the area under the graph of the density function in the interval [a; b].
Thus, there is a close connection here between probabilities and areas. We have
been guided by this close connection in making up our bar graphs; each bar is chosen
so that its area, and not its height, represents the relative frequency of occurrence,
and hence estimates the probability of the outcome falling in the associated interval.
In the language of the calculus, we can say that the probability of occurrence of
an event of the form [x; x + dx], where dx is small, is approximately given by
P([x; x + dx])  f(x)dx ;
that is, by the area of the rectangle under the graph of f. Note that as dx ! 0,
this probability ! 0, so that the probability P(fxg) of a single point is again 0, as
in Example 2.7.",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 61
A glance at the graph of a density function tells us immediately which events of
an experiment are more likely. Roughly speaking, we can say that where the density
is large the events are more likely, and where it is small the events are less likely.
In Example 2.4 the density function is largest at 1. Thus, given the two intervals
[0; a] and [1; 1 + a], where a is a small positive real number, we see that X is more
likely to take on a value in the second interval than in the rst.
Cumulative Distribution Functions of Continuous Random
Variables
We have seen that density functions are useful when considering continuous ran-
dom variables. There is another kind of function, closely related to these density
functions, which is also of great importance. These functions are called cumulative
distribution functions.
Denition 2.2 Let X be a continuous real-valued random variable. Then the
cumulative distribution function of X is dened by the equation",prob.pdf
"Denition 2.2 Let X be a continuous real-valued random variable. Then the
cumulative distribution function of X is dened by the equation
FX(x) = P(X  x) :
2
If X is a continuous real-valued random variable which possesses a density function,
then it also has a cumulative distribution function, and the following theorem shows
that the two functions are related in a very nice way.
Theorem 2.1 Let X be a continuous real-valued random variable with density
function f(x). Then the function dened by
F(x) =
Z x
1
f(t) dt
is the cumulative distribution function of X. Furthermore, we have
d
dxF(x) = f(x) :
Proof. By denition,
F(x) = P(X  x) :
Let E = (1; x]. Then
P(X  x) = P(X 2 E) ;
which equals Z x
1
f(t) dt :
Applying the Fundamental Theorem of Calculus to the rst equation in the
statement of the theorem yields the second statement. 2",prob.pdf
"62 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
-1 -0.5 0 0.5 1 1.5 2
0.25
0.5
0.75
1
1.25
1.5
1.75
2
f  (x)
F  (x)
X
X
Figure 2.13: Distribution and density for X = U2.
In many experiments, the density function of the relevant random variable is easy
to write down. However, it is quite often the case that the cumulative distribution
function is easier to obtain than the density function. (Of course, once we have
the cumulative distribution function, the density function can easily be obtained by
dierentiation, as the above theorem shows.) We now give some examples which
exhibit this phenomenon.
Example 2.13 A real number is chosen at random from [0; 1] with uniform prob-
ability, and then this number is squared. Let X represent the result. What is the
cumulative distribution function of X? What is the density of X?
We begin by letting U represent the chosen real number. Then X = U2. If
0  x  1, then we have
FX (x) = P(X  x)
= P(U2  x)
= P(U  px)
= px :",prob.pdf
"We begin by letting U represent the chosen real number. Then X = U2. If
0  x  1, then we have
FX (x) = P(X  x)
= P(U2  x)
= P(U  px)
= px :
It is clear that X always takes on a value between 0 and 1, so the cumulative
distribution function of X is given by
FX (x) =
8
<
:
0; if x  0;p
x; if 0  x  1;
1; if x  1:
From this we easily calculate that the density function of X is
fX(x) =
8
<
:
0; if x  0;
1=(2p
x); if 0  x  1;
0; if x > 1:
Note that FX(x) is continuous, but fX(x) is not. (See Figure 2.13.) 2",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 63
0.2 0.4 0.6 0.8 1
0.2
0.4
0.6
0.8
1
E.8
Figure 2.14: Calculation of distribution function for Example 2.14.
When referring to a continuous random variable X (say with a uniform density
function), it is customary to say that \X is uniformly distributed on the interval
[a; b]."" It is also customary to refer to the cumulative distribution function of X as
the distribution function of X. Thus, the word \distribution"" is being used in sev-
eral dierent ways in the subject of probability. (Recall that it also has a meaning
when discussing discrete random variables.) When referring to the cumulative dis-
tribution function of a continuous random variable X, we will always use the word
\cumulative"" as a modier, unless the use of another modier, such as \normal"" or
\exponential,"" makes it clear. Since the phrase \uniformly densitied on the interval
[a; b]"" is not acceptable English, we will have to say \uniformly distributed"" instead.",prob.pdf
"[a; b]"" is not acceptable English, we will have to say \uniformly distributed"" instead.
Example 2.14 In Example 2.4, we considered a random variable, dened to be
the sum of two random real numbers chosen uniformly from [0; 1]. Let the random
variables X and Y denote the two chosen real numbers. Dene Z = X + Y . We
will now derive expressions for the cumulative distribution function and the density
function of Z.
Here we take for our sample space 
 the unit square in R2 with uniform density.
A point ! 2 
 then consists of a pair (x; y) of numbers chosen at random. Then
0  Z  2. Let Ez denote the event that Z  z. In Figure 2.14, we show the set
E:8. The event Ez, for any z between 0 and 1, looks very similar to the shaded set
in the gure. For 1 < z  2, the set Ez looks like the unit square with a triangle
removed from the upper right-hand corner. We can now calculate the probability
distribution FZ of Z; it is given by
FZ(z) = P(Z  z)
= Area of Ez",prob.pdf
"64 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
-1 1 2 3
0.2
0.4
0.6
0.8
1
-1 1 2 3
0.2
0.4
0.6
0.8
1
FZ (z)
f   (z)Z
Figure 2.15: Distribution and density functions for Example 2.14.
1
E
Z
Figure 2.16: Calculation of Fz for Example 2.15.
=
8
>
>
<
>
>
:
0; if z < 0;
(1=2)z2; if 0  z  1;
1  (1=2)(2  z)2; if 1  z  2;
1; if 2 < z:
The density function is obtained by dierentiating this function:
fZ(z) =
8
>
>
<
>
>
:
0; if z < 0;
z; if 0  z  1;
2  z; if 1  z  2;
0; if 2 < z:
The reader is referred to Figure 2.15 for the graphs of these functions. 2
Example 2.15 In the dart game described in Example 2.8, what is the distribution
of the distance of the dart from the center of the target? What is its density?
Here, as before, our sample space 
 is the unit disk in R2, with coordinates
(X; Y ). Let Z =
p
X2 + Y 2 represent the distance from the center of the target. Let",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 65
-1 -0.5 0.5 1 1.5 2
0.2
0.4
0.6
0.8
1
-1 -0.5 0 0.5 1 1.5 2
0.25
0.5
0.75
1
1.25
1.5
1.75
2
F  (z)Z
f   (z)Z   
Figure 2.17: Distribution and density for Z =
p
X2 + Y 2.
E be the event fZ  zg. Then the distribution function FZ of Z (see Figure 2.16)
is given by
FZ(z) = P(Z  z)
= Area of E
Area of target :
Thus, we easily compute that
FZ(z) =
8
<
:
0; if z  0;
z2; if 0  z  1;
1; if z > 1:
The density fZ(z) is given again by the derivative of FZ(z):
fZ(z) =
8
<
:
0; if z  0;
2z; if 0  z  1;
0; if z > 1:
The reader is referred to Figure 2.17 for the graphs of these functions.
We can verify this result by simulation, as follows: We choose values for X and
Y at random from [0; 1] with uniform distribution, calculate Z =
p
X2 + Y 2, check
whether 0  Z  1, and present the results in a bar graph (see Figure 2.18). 2
Example 2.16 Suppose Mr. and Mrs. Lockhorn agree to meet at the Hanover Inn",prob.pdf
"Example 2.16 Suppose Mr. and Mrs. Lockhorn agree to meet at the Hanover Inn
between 5:00 and 6:00 P.M. on Tuesday. Suppose each arrives at a time between
5:00 and 6:00 chosen at random with uniform probability. What is the distribution
function for the length of time that the rst to arrive has to wait for the other?
What is the density function?
Here again we can take the unit square to represent the sample space, and (X; Y )
as the arrival times (after 5:00 P.M.) for the Lockhorns. Let Z = jX  Y j. Then we
have FX (x) = x and FY (y) = y. Moreover (see Figure 2.19),
FZ(z) = P(Z  z)
= P(jX  Y j  z)
= Area of E :",prob.pdf
"66 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0 0.2 0.4 0.6 0.8 1
0
0.5
1
1.5
2
Figure 2.18: Simulation results for Example 2.15.
Thus, we have
FZ(z) =
8
<
:
0; if z  0;
1  (1  z)2; if 0  z  1;
1; if z > 1:
The density fZ(z) is again obtained by dierentiation:
fZ(z) =
8
<
:
0; if z  0;
2(1  z); if 0  z  1;
0; if z > 1:
2
Example 2.17 There are many occasions where we observe a sequence of occur-
rences which occur at \random"" times. For example, we might be observing emis-
sions of a radioactive isotope, or cars passing a milepost on a highway, or light bulbs
burning out. In such cases, we might dene a random variable X to denote the time
between successive occurrences. Clearly, X is a continuous random variable whose
range consists of the non-negative real numbers. It is often the case that we can
model X by using the exponential density. This density is given by the formula
f(t) =

et; if t  0;
0; if t < 0:",prob.pdf
"model X by using the exponential density. This density is given by the formula
f(t) =

et; if t  0;
0; if t < 0:
The number  is a non-negative real number, and represents the reciprocal of the
average value of X. (This will be shown in Chapter 6.) Thus, if the average time
between occurrences is 30 minutes, then  = 1=30. A graph of this density function
with  = 1=30 is shown in Figure 2.20. One can see from the gure that even
though the average value is 30, occasionally much larger values are taken on by X.
Suppose that we have bought a computer that contains a Warp 9 hard drive.
The salesperson says that the average time between breakdowns of this type of hard
drive is 30 months. It is often assumed that the length of time between breakdowns",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 67
E
1 - z
1 - z
1 - z
1 - z
E
Figure 2.19: Calculation of FZ.
20 40 60 80 100 120
0.005
0.01
0.015
0.02
0.025
0.03
f (t) = (1/30) e 
- (1/30) t
Figure 2.20: Exponential density with  = 1=30.",prob.pdf
"68 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0 20 40 60 80 100
0
0.005
0.01
0.015
0.02
0.025
0.03
Figure 2.21: Residual lifespan of a hard drive.
is distributed according to the exponential density. We will assume that this model
applies here, with  = 1=30.
Now suppose that we have been operating our computer for 15 months. We
assume that the original hard drive is still running. We ask how long we should
expect the hard drive to continue to run. One could reasonably expect that the
hard drive will run, on the average, another 15 months. (One might also guess
that it will run more than 15 months, since the fact that it has already run for 15
months implies that we don't have a lemon.) The time which we have to wait is
a new random variable, which we will call Y . Obviously, Y = X  15. We can
write a computer program to produce a sequence of simulated Y -values. To do this,
we rst produce a sequence of X's, and discard those values which are less than",prob.pdf
"we rst produce a sequence of X's, and discard those values which are less than
or equal to 15 (these values correspond to the cases where the hard drive has quit
running before 15 months). To simulate a value of X, we compute the value of the
expression

1


log(rnd) ;
where rnd represents a random real number between 0 and 1. (That this expression
has the exponential density will be shown in Chapter 4.3.) Figure 2.21 shows an
area bar graph of 10,000 simulated Y -values.
The average value of Y in this simulation is 29.74, which is closer to the original
average life span of 30 months than to the value of 15 months which was guessed
above. Also, the distribution of Y is seen to be close to the distribution of X.
It is in fact the case that X and Y have the same distribution. This property is
called the memoryless property, because the amount of time that we have to wait
for an occurrence does not depend on how long we have already waited. The only",prob.pdf
"for an occurrence does not depend on how long we have already waited. The only
continuous density function with this property is the exponential density. 2",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 69
Assignment of Probabilities
A fundamental question in practice is: How shall we choose the probability density
function in describing any given experiment? The answer depends to a great extent
on the amount and kind of information available to us about the experiment. In
some cases, we can see that the outcomes are equally likely. In some cases, we can
see that the experiment resembles another already described by a known density.
In some cases, we can run the experiment a large number of times and make a
reasonable guess at the density on the basis of the observed distribution of outcomes,
as we did in Chapter 1. In general, the problem of choosing the right density function
for a given experiment is a central problem for the experimenter and is not always
easy to solve (see Example 2.6). We shall not examine this question in detail here
but instead shall assume that the right density is already known for each of the
experiments under study.",prob.pdf
"but instead shall assume that the right density is already known for each of the
experiments under study.
The introduction of suitable coordinates to describe a continuous sample space,
and a suitable density to describe its probabilities, is not always so obvious, as our
nal example shows.
Innite Tree
Example 2.18 Consider an experiment in which a fair coin is tossed repeatedly,
without stopping. We have seen in Example 1.6 that, for a coin tossed n times, the
natural sample space is a binary tree with n stages. On this evidence we expect
that for a coin tossed repeatedly, the natural sample space is a binary tree with an
innite number of stages, as indicated in Figure 2.22.
It is surprising to learn that, although the n-stage tree is obviously a nite sample
space, the unlimited tree can be described as a continuous sample space. To see how
this comes about, let us agree that a typical outcome of the unlimited coin tossing",prob.pdf
"this comes about, let us agree that a typical outcome of the unlimited coin tossing
experiment can be described by a sequence of the form ! = fH H T H T T H : : :g.
If we write 1 for H and 0 for T, then ! = f1 1 0 1 0 0 1 : : :g. In this way, each
outcome is described by a sequence of 0's and 1's.
Now suppose we think of this sequence of 0's and 1's as the binary expansion
of some real number x = :1101001 
 
 
 lying between 0 and 1. (A binary expansion
is like a decimal expansion but based on 2 instead of 10.) Then each outcome is
described by a value of x, and in this way x becomes a coordinate for the sample
space, taking on all real values between 0 and 1. (We note that it is possible for
two dierent sequences to correspond to the same real number; for example, the
sequences fT H H H H H : : :g and fH T T T T T: : :g both correspond to the real
number 1=2. We will not concern ourselves with this apparent problem here.)",prob.pdf
"number 1=2. We will not concern ourselves with this apparent problem here.)
What probabilities should be assigned to the events of this sample space? Con-
sider, for example, the event E consisting of all outcomes for which the rst toss
comes up heads and the second tails. Every such outcome has the form :10 
 
 
,
where  can be either 0 or 1. Now if x is our real-valued coordinate, then the value
of x for every such outcome must lie between 1=2 = :10000 
 
 
 and 3=4 = :11000 
 
 
,
and moreover, every value of x between 1/2 and 3/4 has a binary expansion of the",prob.pdf
"70 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
0
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0(start)
1
1
1
1
1
1
1
0
0
0
0 0
0
0
Figure 2.22: Tree for innite number of tosses of a coin.
form :10   
 
 
. This means that ! 2 E if and only if 1=2  x < 3=4, and in this
way we see that we can describe E by the interval [1=2; 3=4). More generally, every
event consisting of outcomes for which the results of the rst n tosses are prescribed
is described by a binary interval of the form [k=2n; (k + 1)=2n).
We have already seen in Section 1.2 that in the experiment involving n tosses,
the probability of any one outcome must be exactly 1=2n. It follows that in the
unlimited toss experiment, the probability of any event consisting of outcomes for
which the results of the rst n tosses are prescribed must also be 1=2n. But 1=2n is
exactly the length of the interval of x-values describing E! Thus we see that, just as
with the spinner experiment, the probability of an event E is determined by what",prob.pdf
"with the spinner experiment, the probability of an event E is determined by what
fraction of the unit interval lies in E.
Consider again the statement: The probability is 1/2 that a fair coin will turn up
heads when tossed. We have suggested that one interpretation of this statement is
that if we toss the coin indenitely the proportion of heads will approach 1/2. That
is, in our correspondence with binary sequences we expect to get a binary sequence
with the proportion of 1's tending to 1/2. The event E of binary sequences for which
this is true is a proper subset of the set of all possible binary sequences. It does
not contain, for example, the sequence 011011011 : : : (i.e., (011) repeated again and
again). The event E is actually a very complicated subset of the binary sequences,
but its probability can be determined as a limit of probabilities for events with a
nite number of outcomes whose probabilities are given by nite tree measures.",prob.pdf
"nite number of outcomes whose probabilities are given by nite tree measures.
When the probability of E is computed in this way, its value is found to be 1.
This remarkable result is known as the Strong Law of Large Numbers (or Law of
Averages) and is one justication for our frequency concept of probability. We shall
prove a weak form of this theorem in Chapter 8. 2",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 71
Exercises
1 Suppose you choose at random a real number X from the interval [2; 10].
(a) Find the density function f(x) and the probability of an event E for this
experiment, where E is a subinterval [a; b] of [2; 10].
(b) From (a), nd the probability that X > 5, that 5 < X < 7, and that
X2  12X + 35 > 0.
2 Suppose you choose a real number X from the interval [2; 10] with a density
function of the form
f(x) = Cx ;
where C is a constant.
(a) Find C.
(b) Find P(E), where E = [a; b] is a subinterval of [2; 10].
(c) Find P(X > 5), P(X < 7), and P(X2  12X + 35 > 0).
3 Same as Exercise 2, but suppose
f(x) = C
x :
4 Suppose you throw a dart at a circular target of radius 10 inches. Assuming
that you hit the target and that the coordinates of the outcomes are chosen
at random, nd the probability that the dart falls
(a) within 2 inches of the center.
(b) within 2 inches of the rim.
(c) within the rst quadrant of the target.",prob.pdf
"(a) within 2 inches of the center.
(b) within 2 inches of the rim.
(c) within the rst quadrant of the target.
(d) within the rst quadrant and within 2 inches of the rim.
5 Suppose you are watching a radioactive source that emits particles at a rate
described by the exponential density
f(t) = et ;
where  = 1, so that the probability P(0; T) that a particle will appear in
the next T seconds is P([0; T]) =
RT
0 et dt. Find the probability that a
particle (not necessarily the rst) will appear
(a) within the next second.
(b) within the next 3 seconds.
(c) between 3 and 4 seconds from now.
(d) after 4 seconds from now.",prob.pdf
"72 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
6 Assume that a new light bulb will burn out after t hours, where t is chosen
from [0; 1) with an exponential density
f(t) = et :
In this context,  is often called the failure rate of the bulb.
(a) Assume that  = 0:01, and nd the probability that the bulb will not
burn out before T hours. This probability is often called the reliability
of the bulb.
(b) For what T is the reliability of the bulb = 1=2?
7 Choose a number B at random from the interval [0; 1] with uniform density.
Find the probability that
(a) 1=3 < B < 2=3.
(b) jB  1=2j  1=4.
(c) B < 1=4 or 1  B < 1=4.
(d) 3B2 < B.
8 Choose independently two numbers B and C at random from the interval [0; 1]
with uniform density. Note that the point (B; C) is then chosen at random in
the unit square. Find the probability that
(a) B + C < 1=2.
(b) BC < 1=2.
(c) jB  Cj < 1=2.
(d) maxfB; Cg < 1=2.
(e) minfB; Cg < 1=2.
(f) B < 1=2 and 1  C < 1=2.
(g) conditions (c) and (f) both hold.",prob.pdf
"(b) BC < 1=2.
(c) jB  Cj < 1=2.
(d) maxfB; Cg < 1=2.
(e) minfB; Cg < 1=2.
(f) B < 1=2 and 1  C < 1=2.
(g) conditions (c) and (f) both hold.
(h) B2 + C2  1=2.
(i) (B  1=2)2 + (C  1=2)2 < 1=4.
9 Suppose that we have a sequence of occurrences. We assume that the time
X between occurrences is exponentially distributed with  = 1=10, so on the
average, there is one occurrence every 10 minutes (see Example 2.17). You
come upon this system at time 100, and wait until the next occurrence. Make
a conjecture concerning how long, on the average, you will have to wait. Write
a program to see if your conjecture is right.
10 As in Exercise 9, assume that we have a sequence of occurrences, but now
assume that the time X between occurrences is uniformly distributed between
5 and 15. As before, you come upon this system at time 100, and wait until
the next occurrence. Make a conjecture concerning how long, on the average,
you will have to wait. Write a program to see if your conjecture is right.",prob.pdf
"2.2. CONTINUOUS DENSITY FUNCTIONS 73
11 For examples such as those in Exercises 9 and 10, it might seem that at least
you should not have to wait on average more than 10 minutes if the average
time between occurrences is 10 minutes. Alas, even this is not true. To see
why, consider the following assumption about the times between occurrences.
Assume that the time between occurrences is 3 minutes with probability .9
and 73 minutes with probability .1. Show by simulation that the average time
between occurrences is 10 minutes, but that if you come upon this system at
time 100, your average waiting time is more than 10 minutes.
12 Take a stick of unit length and break it into three pieces, choosing the break
points at random. (The break points are assumed to be chosen simultane-
ously.) What is the probability that the three pieces can be used to form a
triangle? Hint: The sum of the lengths of any two pieces must exceed the",prob.pdf
"ously.) What is the probability that the three pieces can be used to form a
triangle? Hint: The sum of the lengths of any two pieces must exceed the
length of the third, so each piece must have length < 1=2. Now use Exer-
cise 8(g).
13 Take a stick of unit length and break it into two pieces, choosing the break
point at random. Now break the longer of the two pieces at a random point.
What is the probability that the three pieces can be used to form a triangle?
14 Choose independently two numbers B and C at random from the interval
[1; 1] with uniform distribution, and consider the quadratic equation
x2 + Bx + C = 0 :
Find the probability that the roots of this equation
(a) are both real.
(b) are both positive.
Hints: (a) requires 0  B2  4C, (b) requires 0  B2  4C, B  0, 0  C.
15 At the Tunbridge World's Fair, a coin toss game works as follows. Quarters
are tossed onto a checkerboard. The management keeps all the quarters, but",prob.pdf
"are tossed onto a checkerboard. The management keeps all the quarters, but
for each quarter landing entirely within one square of the checkerboard the
management pays a dollar. Assume that the edge of each square is twice the
diameter of a quarter, and that the outcomes are described by coordinates
chosen at random. Is this a fair game?
16 Three points are chosen at random on a circle of unit circumference. What is
the probability that the triangle dened by these points as vertices has three
acute angles? Hint: One of the angles is obtuse if and only if all three points
lie in the same semicircle. Take the circumference as the interval [0; 1]. Take
one point at 0 and the others at B and C.
17 Write a program to choose a random number X in the interval [2; 10] 1000
times and record what fraction of the outcomes satisfy X > 5, what fraction
satisfy 5 < X < 7, and what fraction satisfy x2 12x + 35 > 0. How do these
results compare with Exercise 1?",prob.pdf
"74 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES
18 Write a program to choose a point (X; Y ) at random in a square of side 20
inches, doing this 10,000 times, and recording what fraction of the outcomes
fall within 19 inches of the center; of these, what fraction fall between 8 and 10
inches of the center; and, of these, what fraction fall within the rst quadrant
of the square. How do these results compare with those of Exercise 4?
19 Write a program to simulate the problem describe in Exercise 7 (see Exer-
cise 17). How do the simulation results compare with the results of Exercise 7?
20 Write a program to simulate the problem described in Exercise 12.
21 Write a program to simulate the problem described in Exercise 16.
22 Write a program to carry out the following experiment. A coin is tossed 100
times and the number of heads that turn up is recorded. This experiment
is then repeated 1000 times. Have your program plot a bar graph for the",prob.pdf
"times and the number of heads that turn up is recorded. This experiment
is then repeated 1000 times. Have your program plot a bar graph for the
proportion of the 1000 experiments in which the number of heads is n, for
each n in the interval [35; 65]. Does the bar graph look as though it can be t
with a normal curve?
23 Write a program that picks a random number between 0 and 1 and computes
the negative of its logarithm. Repeat this process a large number of times and
plot a bar graph to give the number of times that the outcome falls in each
interval of length 0.1 in [0; 10]. On this bar graph plot a graph of the density
f(x) = ex. How well does this density t your graph?",prob.pdf
"Chapter 3
Combinatorics
3.1 Permutations
Many problems in probability theory require that we count the number of ways
that a particular event can occur. For this, we study the topics of permutations and
combinations. We consider permutations in this section and combinations in the
next section.
Before discussing permutations, it is useful to introduce a general counting tech-
nique that will enable us to solve a variety of counting problems, including the
problem of counting the number of possible permutations of n objects.
Counting Problems
Consider an experiment that takes place in several stages and is such that the
number of outcomes m at the nth stage is independent of the outcomes of the
previous stages. The number m may be dierent for dierent stages. We want to
count the number of ways that the entire experiment can be carried out.
Example 3.1 You are eating at Emile's restaurant and the waiter informs you",prob.pdf
"count the number of ways that the entire experiment can be carried out.
Example 3.1 You are eating at Emile's restaurant and the waiter informs you
that you have (a) two choices for appetizers: soup or juice; (b) three for the main
course: a meat, sh, or vegetable dish; and (c) two for dessert: ice cream or cake.
How many possible choices do you have for your complete meal? We illustrate the
possible meals by a tree diagram shown in Figure 3.1. Your menu is decided in three
stages|at each stage the number of possible choices does not depend on what is
chosen in the previous stages: two choices at the rst stage, three at the second,
and two at the third. From the tree diagram we see that the total number of choices
is the product of the number of choices at each stage. In this examples we have
2 
 3 
 2 = 12 possible menus. Our menu example is an example of the following
general counting technique. 2
75",prob.pdf
"76 CHAPTER 3. COMBINATORICS
ice cream
cake
ice cream
cake
ice cream
cake
ice cream
cake
ice cream
cake
ice cream
cake
(start)
soup
meat
fish
vegetable
juice
meat
fish
vegetable
Figure 3.1: Tree for your menu.
A Counting Technique
A task is to be carried out in a sequence of r stages. There are n1 ways to carry
out the rst stage; for each of these n1 ways, there are n2 ways to carry out the
second stage; for each of these n2 ways, there are n3 ways to carry out the third
stage, and so forth. Then the total number of ways in which the entire task can be
accomplished is given by the product N = n1 
 n2 
 : : : 
 nr.
Tree Diagrams
It will often be useful to use a tree diagram when studying probabilities of events
relating to experiments that take place in stages and for which we are given the
probabilities for the outcomes at each stage. For example, assume that the owner
of Emile's restaurant has observed that 80 percent of his customers choose the soup",prob.pdf
"of Emile's restaurant has observed that 80 percent of his customers choose the soup
for an appetizer and 20 percent choose juice. Of those who choose soup, 50 percent
choose meat, 30 percent choose sh, and 20 percent choose the vegetable dish. Of
those who choose juice for an appetizer, 30 percent choose meat, 40 percent choose
sh, and 30 percent choose the vegetable dish. We can use this to estimate the
probabilities at the rst two stages as indicated on the tree diagram of Figure 3.2.
We choose for our sample space the set 
 of all possible paths ! = !1, !2,
. . . , !6 through the tree. How should we assign our probability distribution? For
example, what probability should we assign to the customer choosing soup and then
the meat? If 8/10 of the customers choose soup and then 1/2 of these choose meat,
a proportion 8=10 
 1=2 = 4=10 of the customers choose soup and then meat. This
suggests choosing our probability distribution for each path through the tree to be",prob.pdf
"suggests choosing our probability distribution for each path through the tree to be
the product of the probabilities at each of the stages along the path. This results in
the probability distribution for the sample points ! indicated in Figure 3.2. (Note
that m(!1) + 
 
 
 + m(!6) = 1.) From this we see, for example, that the probability",prob.pdf
"3.1. PERMUTATIONS 77
(start)
soup
meat
fish
vegetable
juice
.8
.2
.2
.3
.3
.4
.5
.3
meat
fish
vegetable
ω (ω)
ω 
ω 
ω 
ω 
ω 
ω 
.4
.24
.16
.06
.08
.06
m
1
2
3
4
5
6
Figure 3.2: Two-stage probability assignment.
that a customer chooses meat is m(!1) + m(!4) = :46.
We shall say more about these tree measures when we discuss the concept of
conditional probability in Chapter 4. We return now to more counting problems.
Example 3.2 We can show that there are at least two people in Columbus, Ohio,
who have the same three initials. Assuming that each person has three initials,
there are 26 possibilities for a person's rst initial, 26 for the second, and 26 for the
third. Therefore, there are 263 = 17;576 possible sets of initials. This number is
smaller than the number of people living in Columbus, Ohio; hence, there must be
at least two people with the same three initials. 2
We consider next the celebrated birthday problem|often used to show that",prob.pdf
"at least two people with the same three initials. 2
We consider next the celebrated birthday problem|often used to show that
naive intuition cannot always be trusted in probability.
Birthday Problem
Example 3.3 How many people do we need to have in a room to make it a favorable
bet (probability of success greater than 1/2) that two people in the room will have
the same birthday?
Since there are 365 possible birthdays, it is tempting to guess that we would
need about 1/2 this number, or 183. You would surely win this bet. In fact, the
number required for a favorable bet is only 23. To show this, we nd the probability
pr that, in a room with r people, there is no duplication of birthdays; we will have
a favorable bet if this probability is less than one half.",prob.pdf
"78 CHAPTER 3. COMBINATORICS
Number of people Probability that all birthdays are dierent
20 .5885616
21 .5563117
22 .5243047
23 .4927028
24 .4616557
25 .4313003
Table 3.1: Birthday problem.
Assume that there are 365 possible birthdays for each person (we ignore leap
years). Order the people from 1 to r. For a sample point !, we choose a possible
sequence of length r of birthdays each chosen as one of the 365 possible dates.
There are 365 possibilities for the rst element of the sequence, and for each of
these choices there are 365 for the second, and so forth, making 365r possible
sequences of birthdays. We must nd the number of these sequences that have no
duplication of birthdays. For such a sequence, we can choose any of the 365 days
for the rst element, then any of the remaining 364 for the second, 363 for the third,
and so forth, until we make r choices. For the rth choice, there will be 365  r + 1
possibilities. Hence, the total number of sequences with no duplications is",prob.pdf
"possibilities. Hence, the total number of sequences with no duplications is
365 
 364 
 363 
 : : : 
 (365  r + 1) :
Thus, assuming that each sequence is equally likely,
pr = 365 
 364 
 : : : 
 (365  r + 1)
365r :
We denote the product
(n)(n  1) 
 
 
(n  r + 1)
by (n)r (read \n down r,"" or \n lower r""). Thus,
pr = (365)r
(365)r :
The program Birthday carries out this computation and prints the probabilities
for r = 20 to 25. Running this program, we get the results shown in Table 3.1. As
we asserted above, the probability for no duplication changes from greater than one
half to less than one half as we move from 22 to 23 people. To see how unlikely it is
that we would lose our bet for larger numbers of people, we have run the program
again, printing out values from r = 10 to r = 100 in steps of 10. We see that in
a room of 40 people the odds already heavily favor a duplication, and in a room
of 100 the odds are overwhelmingly in favor of a duplication. We have assumed",prob.pdf
"of 100 the odds are overwhelmingly in favor of a duplication. We have assumed
that birthdays are equally likely to fall on any particular day. Statistical evidence
suggests that this is not true. However, it is intuitively clear (but not easy to prove)
that this makes it even more likely to have a duplication with a group of 23 people.
(See Exercise 19 to nd out what happens on planets with more or fewer than 365
days per year.) 2",prob.pdf
"3.1. PERMUTATIONS 79
Number of people Probability that all birthdays are dierent
10 .8830518
20 .5885616
30 .2936838
40 .1087682
50 .0296264
60 .0058773
70 .0008404
80 .0000857
90 .0000062
100 .0000003
Table 3.2: Birthday problem.
We now turn to the topic of permutations.
Permutations
Denition 3.1 Let A be any nite set. A permutation of A is a one-to-one mapping
of A onto itself. 2
To specify a particular permutation we list the elements of A and, under them,
show where each element is sent by the one-to-one mapping. For example, if A =
fa; b; cg a possible permutation  would be
 =
a b c
b c a

:
By the permutation , a is sent to b, b is sent to c, and c is sent to a. The
condition that the mapping be one-to-one means that no two elements of A are
sent, by the mapping, into the same element of A.
We can put the elements of our set in some order and rename them 1, 2, .. . , n.
Then, a typical permutation of the set A = fa1; a2; a3; a4g can be written in the
form
 =
1 2 3 4",prob.pdf
"Then, a typical permutation of the set A = fa1; a2; a3; a4g can be written in the
form
 =
1 2 3 4
2 1 4 3

;
indicating that a1 went to a2, a2 to a1, a3 to a4, and a4 to a3.
If we always choose the top row to be 1 2 3 4 then, to prescribe the permutation,
we need only give the bottom row, with the understanding that this tells us where 1
goes, 2 goes, and so forth, under the mapping. When this is done, the permutation
is often called a rearrangement of the n objects 1, 2, 3, . . . , n. For example, all
possible permutations, or rearrangements, of the numbers A = f1; 2; 3g are:
123; 132; 213; 231; 312; 321 :
It is an easy matter to count the number of possible permutations of n objects.
By our general counting principle, there are n ways to assign the rst element, for",prob.pdf
"80 CHAPTER 3. COMBINATORICS
n n!
0 1
1 1
2 2
3 6
4 24
5 120
6 720
7 5040
8 40320
9 362880
10 3628800
Table 3.3: Values of the factorial function.
each of these we have n  1 ways to assign the second object, n  2 for the third,
and so forth. This proves the following theorem.
Theorem 3.1 The total number of permutations of a set A of n elements is given
by n 
 (n  1) 
 (n  2) 
 : : : 
 1. 2
It is sometimes helpful to consider orderings of subsets of a given set. This
prompts the following denition.
Denition 3.2 Let A be an n-element set, and let k be an integer between 0 and
n. Then a k-permutation of A is an ordered listing of a subset of A of size k. 2
Using the same techniques as in the last theorem, the following result is easily
proved.
Theorem 3.2 The total number of k-permutations of a set A of n elements is given
by n 
 (n  1) 
 (n  2) 
 : : : 
 (n  k + 1). 2
Factorials
The number given in Theorem 3.1 is called n factorial, and is denoted by n!. The",prob.pdf
"by n 
 (n  1) 
 (n  2) 
 : : : 
 (n  k + 1). 2
Factorials
The number given in Theorem 3.1 is called n factorial, and is denoted by n!. The
expression 0! is dened to be 1 to make certain formulas come out simpler. The
rst few values of this function are shown in Table 3.3. The reader will note that
this function grows very rapidly.
The expression n! will enter into many of our calculations, and we shall need to
have some estimate of its magnitude when n is large. It is clearly not practical to
make exact calculations in this case. We shall instead use a result called Stirling's
formula. Before stating this formula we need a denition.",prob.pdf
"3.1. PERMUTATIONS 81
n n! Approximation Ratio
1 1 .922 1.084
2 2 1.919 1.042
3 6 5.836 1.028
4 24 23.506 1.021
5 120 118.019 1.016
6 720 710.078 1.013
7 5040 4980.396 1.011
8 40320 39902.395 1.010
9 362880 359536.873 1.009
10 3628800 3598696.619 1.008
Table 3.4: Stirling approximations to the factorial function.
Denition 3.3 Let an and bn be two sequences of numbers. We say that an is
asymptotically equal to bn, and write an  bn, if
lim
n!1
an
bn
= 1 :
2
Example 3.4 If an = n + pn and bn = n then, since an=bn = 1 + 1=pn and this
ratio tends to 1 as n tends to innity, we have an  bn. 2
Theorem 3.3 (Stirling's Formula) The sequence n! is asymptotically equal to
nnenp
2n :
2
The proof of Stirling's formula may be found in most analysis texts. Let us
verify this approximation by using the computer. The program StirlingApprox-
imations prints n!, the Stirling approximation, and, nally, the ratio of these two",prob.pdf
"imations prints n!, the Stirling approximation, and, nally, the ratio of these two
numbers. Sample output of this program is shown in Table 3.4. Note that, while
the ratio of the numbers is getting closer to 1, the dierence between the exact
value and the approximation is increasing, and indeed, this dierence will tend to
innity as n tends to innity, even though the ratio tends to 1. (This was also true
in our Example 3.4 where n + pn  n, but the dierence is pn.)
Generating Random Permutations
We now consider the question of generating a random permutation of the integers
between 1 and n. Consider the following experiment. We start with a deck of n
cards, labelled 1 through n. We choose a random card out of the deck, note its label,
and put the card aside. We repeat this process until all n cards have been chosen.
It is clear that each permutation of the integers from 1 to n can occur as a sequence",prob.pdf
"82 CHAPTER 3. COMBINATORICS
Number of xed points Fraction of permutations
n = 10 n = 20 n = 30
0 .362 .370 .358
1 .368 .396 .358
2 .202 .164 .192
3 .052 .060 .070
4 .012 .008 .020
5 .004 .002 .002
Average number of xed points .996 .948 1.042
Table 3.5: Fixed point distributions.
of labels in this experiment, and that each sequence of labels is equally likely to
occur. In our implementations of the computer algorithms, the above procedure is
called RandomPermutation.
Fixed Points
There are many interesting problems that relate to properties of a permutation
chosen at random from the set of all permutations of a given nite set. For example,
since a permutation is a one-to-one mapping of the set onto itself, it is interesting to
ask how many points are mapped onto themselves. We call such points xed points
of the mapping.
Let pk(n) be the probability that a random permutation of the set f1; 2; : : :; ng
has exactly k xed points. We will attempt to learn something about these prob-",prob.pdf
"has exactly k xed points. We will attempt to learn something about these prob-
abilities using simulation. The program FixedPoints uses the procedure Ran-
domPermutation to generate random permutations and count xed points. The
program prints the proportion of times that there are k xed points as well as the
average number of xed points. The results of this program for 500 simulations for
the cases n = 10, 20, and 30 are shown in Table 3.5. Notice the rather surprising
fact that our estimates for the probabilities do not seem to depend very heavily on
the number of elements in the permutation. For example, the probability that there
are no xed points, when n = 10; 20; or 30 is estimated to be between .35 and .37.
We shall see later (see Example 3.12) that for n  10 the exact probabilities pn(0)
are, to six decimal place accuracy, equal to 1=e  :367879. Thus, for all practi-
cal purposes, after n = 10 the probability that a random permutation of the set",prob.pdf
"cal purposes, after n = 10 the probability that a random permutation of the set
f1; 2; : : :; ng has no xed points does not depend upon n. These simulations also
suggest that the average number of xed points is close to 1. It can be shown (see
Example 6.8) that the average is exactly equal to 1 for all n.
More picturesque versions of the xed-point problem are: You have arranged
the books on your book shelf in alphabetical order by author and they get returned
to your shelf at random; what is the probability that exactly k of the books end up
in their correct position? (The library problem.) In a restaurant n hats are checked
and they are hopelessly scrambled; what is the probability that no one gets his own
hat back? (The hat check problem.) In the Historical Remarks at the end of this
section, we give one method for solving the hat check problem exactly. Another",prob.pdf
"3.1. PERMUTATIONS 83
Date Snowfall in inches
1974 75
1975 88
1976 72
1977 110
1978 85
1979 30
1980 55
1981 86
1982 51
1983 64
Table 3.6: Snowfall in Hanover.
Year 1 2 3 4 5 6 7 8 9 10
Ranking 6 9 5 10 7 1 3 8 2 4
Table 3.7: Ranking of total snowfall.
method is given in Example 3.12.
Records
Here is another interesting probability problem that involves permutations. Esti-
mates for the amount of measured snow in inches in Hanover, New Hampshire, in
the ten years from 1974 to 1983 are shown in Table 3.6. Suppose we have started
keeping records in 1974. Then our rst year's snowfall could be considered a record
snowfall starting from this year. A new record was established in 1975; the next
record was established in 1977, and there were no new records established after
this year. Thus, in this ten-year period, there were three records established: 1974,
1975, and 1977. The question that we ask is: How many records should we expect",prob.pdf
"1975, and 1977. The question that we ask is: How many records should we expect
to be established in such a ten-year period? We can count the number of records
in terms of a permutation as follows: We number the years from 1 to 10. The
actual amounts of snowfall are not important but their relative sizes are. We can,
therefore, change the numbers measuring snowfalls to numbers 1 to 10 by replacing
the smallest number by 1, the next smallest by 2, and so forth. (We assume that
there are no ties.) For our example, we obtain the data shown in Table 3.7.
This gives us a permutation of the numbers from 1 to 10 and, from this per-
mutation, we can read o the records; they are in years 1, 2, and 4. Thus we can
dene records for a permutation as follows:
Denition 3.4 Let  be a permutation of the set f1; 2; : : :; ng. Then i is a record
of  if either i = 1 or (j) < (i) for every j = 1; : : :; i  1. 2
Now if we regard all rankings of snowfalls over an n-year period to be equally",prob.pdf
"of  if either i = 1 or (j) < (i) for every j = 1; : : :; i  1. 2
Now if we regard all rankings of snowfalls over an n-year period to be equally
likely (and allow no ties), we can estimate the probability that there will be k
records in n years as well as the average number of records by simulation.",prob.pdf
"84 CHAPTER 3. COMBINATORICS
We have written a program Records that counts the number of records in ran-
domly chosen permutations. We have run this program for the cases n = 10, 20, 30.
For n = 10 the average number of records is 2.968, for 20 it is 3.656, and for 30
it is 3.960. We see now that the averages increase, but very slowly. We shall see
later (see Example 6.11) that the average number is approximately log n. Since
log 10 = 2:3, log 20 = 3, and log 30 = 3:4, this is consistent with the results of our
simulations.
As remarked earlier, we shall be able to obtain formulas for exact results of
certain problems of the above type. However, only minor changes in the problem
make this impossible. The power of simulation is that minor changes in a problem
do not make the simulation much more dicult. (See Exercise 20 for an interesting
variation of the hat check problem.)
List of Permutations
Another method to solve problems that is not sensitive to small changes in the",prob.pdf
"variation of the hat check problem.)
List of Permutations
Another method to solve problems that is not sensitive to small changes in the
problem is to have the computer simply list all possible permutations and count the
fraction that have the desired property. The program AllPermutations produces
a list of all of the permutations of n. When we try running this program, we run
into a limitation on the use of the computer. The number of permutations of n
increases so rapidly that even to list all permutations of 20 objects is impractical.
Historical Remarks
Our basic counting principle stated that if you can do one thing in r ways and for
each of these another thing in s ways, then you can do the pair in rs ways. This
is such a self-evident result that you might expect that it occurred very early in
mathematics. N. L. Biggs suggests that we might trace an example of this principle
as follows: First, he relates a popular nursery rhyme dating back to at least 1730:",prob.pdf
"as follows: First, he relates a popular nursery rhyme dating back to at least 1730:
As I was going to St. Ives,
I met a man with seven wives,
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits.
Kits, cats, sacks and wives,
How many were going to St. Ives?
(You need our principle only if you are not clever enough to realize that you are
supposed to answer one, since only the narrator is going to St. Ives; the others are
going in the other direction!)
He also gives a problem appearing on one of the oldest surviving mathematical
manuscripts of about 1650 B.C., roughly translated as:",prob.pdf
"3.1. PERMUTATIONS 85
Houses 7
Cats 49
Mice 343
Wheat 2401
Hekat 16807
19607
The following interpretation has been suggested: there are seven houses, each
with seven cats; each cat kills seven mice; each mouse would have eaten seven heads
of wheat, each of which would have produced seven hekat measures of grain. With
this interpretation, the table answers the question of how many hekat measures
were saved by the cats' actions. It is not clear why the writer of the table wanted
to add the numbers together.1
One of the earliest uses of factorials occurred in Euclid's proof that there are
innitely many prime numbers. Euclid argued that there must be a prime number
between n and n! + 1 as follows: n! and n! + 1 cannot have common factors. Either
n!+1 is prime or it has a proper factor. In the latter case, this factor cannot divide
n! and hence must be between n and n! + 1. If this factor is not prime, then it
has a factor that, by the same argument, must be bigger than n. In this way, we",prob.pdf
"has a factor that, by the same argument, must be bigger than n. In this way, we
eventually reach a prime bigger than n, and this holds for all n.
The \n!"" rule for the number of permutations seems to have occurred rst in
India. Examples have been found as early as 300 B.C., and by the eleventh century
the general formula seems to have been well known in India and then in the Arab
countries.
The hat check problem is found in an early probability book written by de Mont-
mort and rst printed in 1708.2 It appears in the form of a game called Treize. In
a simplied version of this game considered by de Montmort one turns over cards
numbered 1 to 13, calling out 1, 2, . . . , 13 as the cards are examined. De Montmort
asked for the probability that no card that is turned up agrees with the number
called out.
This probability is the same as the probability that a random permutation of
13 elements has no xed point. De Montmort solved this problem by the use of a",prob.pdf
"13 elements has no xed point. De Montmort solved this problem by the use of a
recursion relation as follows: let wn be the number of permutations of n elements
with no xed point (such permutations are called derangements). Then w1 = 0 and
w2 = 1.
Now assume that n  3 and choose a derangement of the integers between 1 and
n. Let k be the integer in the rst position in this derangement. By the denition of
derangement, we have k 6= 1. There are two possibilities of interest concerning the
position of 1 in the derangement: either 1 is in the kth position or it is elsewhere. In
the rst case, the n  2 remaining integers can be positioned in wn2 ways without
resulting in any xed points. In the second case, we consider the set of integers
f1; 2; : : :; k  1; k + 1; : : :; ng. The numbers in this set must occupy the positions
f2; 3; : : :; ng so that none of the numbers other than 1 in this set are xed, and",prob.pdf
"f2; 3; : : :; ng so that none of the numbers other than 1 in this set are xed, and
1N. L. Biggs, \The Roots of Combinatorics,"" Historia Mathematica, vol. 6 (1979), pp. 109{136.
2P. R. de Montmort, Essay d'Analyse sur des Jeux de Hazard, 2d ed. (Paris: Quillau, 1713).",prob.pdf
"86 CHAPTER 3. COMBINATORICS
also so that 1 is not in position k. The number of ways of achieving this kind of
arrangement is just wn1. Since there are n  1 possible values of k, we see that
wn = (n  1)wn1 + (n  1)wn2
for n  3. One might conjecture from this last equation that the sequence fwng
grows like the sequence fn!g.
In fact, it is easy to prove by induction that
wn = nwn1 + (1)n :
Then pi = wi=i! satises
pi  pi1 = (1)i
i! :
If we sum from i = 2 to n, and use the fact that p1 = 0, we obtain
pn = 1
2!  1
3! + 
 
 
 + (1)n
n! :
This agrees with the rst n + 1 terms of the expansion for ex for x = 1 and hence
for large n is approximately e1  :368. David remarks that this was possibly
the rst use of the exponential function in probability.3 We shall see another way
to derive de Montmort's result in the next section, using a method known as the
Inclusion-Exclusion method.
Recently, a related problem appeared in a column of Marilyn vos Savant.4",prob.pdf
"Inclusion-Exclusion method.
Recently, a related problem appeared in a column of Marilyn vos Savant.4
Charles Price wrote to ask about his experience playing a certain form of solitaire,
sometimes called \frustration solitaire."" In this particular game, a deck of cards
is shued, and then dealt out, one card at a time. As the cards are being dealt,
the player counts from 1 to 13, and then starts again at 1. (Thus, each number is
counted four times.) If a number that is being counted coincides with the rank of
the card that is being turned up, then the player loses the game. Price found that
he rarely won and wondered how often he should win. Vos Savant remarked that
the expected number of matches is 4 so it should be dicult to win the game.
Finding the chance of winning is a harder problem than the one that de Mont-
mort solved because, when one goes through the entire deck, there are dierent
patterns for the matches that might occur. For example matches may occur for two",prob.pdf
"patterns for the matches that might occur. For example matches may occur for two
cards of the same rank, say two aces, or for two dierent ranks, say a two and a
three.
A discussion of this problem can be found in Riordan.5 In this book, it is shown
that as n ! 1, the probability of no matches tends to 1=e4.
The original game of Treize is more dicult to analyze than frustration solitaire.
The game of Treize is played as follows. One person is chosen as dealer and the
others are players. Each player, other than the dealer, puts up a stake. The dealer
shues the cards and turns them up one at a time calling out, \Ace, two, three,...,
3F. N. David, Games, Gods and Gambling (London: Grin, 1962), p. 146.
4M. vos Savant, Ask Marilyn, Parade Magazine, Boston Globe, 21 August 1994.
5J. Riordan, An Introduction to Combinatorial Analysis, (New York: John Wiley & Sons,
1958).",prob.pdf
"3.1. PERMUTATIONS 87
king,"" just as in frustration solitaire. If the dealer goes through the 13 cards without
a match he pays the players an amount equal to their stake, and the deal passes to
someone else. If there is a match the dealer collects the players' stakes; the players
put up new stakes, and the dealer continues through the deck, calling out, \Ace,
two, three, ...."" If the dealer runs out of cards he reshues and continues the count
where he left o. He continues until there is a run of 13 without a match and then
a new dealer is chosen.
The question at this point is how much money can the dealer expect to win from
each player. De Montmort found that if each player puts up a stake of 1, say, then
the dealer will win approximately .801 from each player.
Peter Doyle calculated the exact amount that the dealer can expect to win. The
answer is:
26516072156010218582227607912734182784642120482136091446715371962089931",prob.pdf
"answer is:
26516072156010218582227607912734182784642120482136091446715371962089931
52311343541724554334912870541440299239251607694113500080775917818512013
82176876653563173852874555859367254632009477403727395572807459384342747
87664965076063990538261189388143513547366316017004945507201764278828306
60117107953633142734382477922709835281753299035988581413688367655833113
24476153310720627474169719301806649152698704084383914217907906954976036
28528211590140316202120601549126920880824913325553882692055427830810368
57818861208758248800680978640438118582834877542560955550662878927123048
26997601700116233592793308297533642193505074540268925683193887821301442
70519791882/
33036929133582592220117220713156071114975101149831063364072138969878007
99647204708825303387525892236581323015628005621143427290625658974433971
65719454122908007086289841306087561302818991167357863623756067184986491
35353553622197448890223267101158801016285931351979294387223277033396967",prob.pdf
"65719454122908007086289841306087561302818991167357863623756067184986491
35353553622197448890223267101158801016285931351979294387223277033396967
79797069933475802423676949873661605184031477561560393380257070970711959
69641268242455013319879747054693517809383750593488858698672364846950539
88868628582609905586271001318150621134407056983214740221851567706672080
94586589378459432799868706334161812988630496327287254818458879353024498
00322425586446741048147720934108061350613503856973048971213063937040515
59533731591.
This is .803 to 3 decimal places. A description of the algorithm used to nd this
answer can be found on his Web page.6 A discussion of this problem and other
problems can be found in Doyle et al.7
The birthday problem does not seem to have a very old history. Problems of
this type were rst discussed by von Mises.8 It was made popular in the 1950s by
Feller's book.9
6P. Doyle, \Solution to Montmort's Probleme du Treize,"" http://math.ucsd.edu/~doyle/.",prob.pdf
"Feller's book.9
6P. Doyle, \Solution to Montmort's Probleme du Treize,"" http://math.ucsd.edu/~doyle/.
7P. Doyle, C. Grinstead, and J. Snell, \Frustration Solitaire,"" UMAP Journal, vol. 16, no. 2
(1995), pp. 137-145.
8R. von Mises, \Uber Aufteilungs- und Besetzungs-Wahrscheinlichkeiten,"" Revue de la Faculte
des Sciences de l'Universite d'Istanbul, N. S. vol. 4 (1938-39), pp. 145-163.
9W. Feller, Introduction to Probability Theory and Its Applications, vol. 1, 3rd ed. (New York:",prob.pdf
"88 CHAPTER 3. COMBINATORICS
Stirling presented his formula
n! 
p
2n
n
e
n
in his work Methodus Dierentialis published in 1730.10 This approximation was
used by de Moivre in establishing his celebrated central limit theorem that we
will study in Chapter 9. De Moivre himself had independently established this
approximation, but without identifying the constant . Having established the
approximation
2Bpn
for the central term of the binomial distribution, where the constant B was deter-
mined by an innite series, de Moivre writes:
.. . my worthy and learned Friend, Mr. James Stirling, who had applied
himself after me to that inquiry, found that the Quantity B did denote
the Square-root of the Circumference of a Circle whose Radius is Unity,
so that if that Circumference be called c the Ratio of the middle Term
to the Sum of all Terms will be expressed by 2=pnc. . . .11
Exercises
1 Four people are to be arranged in a row to have their picture taken. In how
many ways can this be done?",prob.pdf
"Exercises
1 Four people are to be arranged in a row to have their picture taken. In how
many ways can this be done?
2 An automobile manufacturer has four colors available for automobile exteri-
ors and three for interiors. How many dierent color combinations can he
produce?
3 In a digital computer, a bit is one of the integers f0,1g, and a word is any
string of 32 bits. How many dierent words are possible?
4 What is the probability that at least 2 of the presidents of the United States
have died on the same day of the year? If you bet this has happened, would
you win your bet?
5 There are three dierent routes connecting city A to city B. How many ways
can a round trip be made from A to B and back? How many ways if it is
desired to take a dierent route on the way back?
6 In arranging people around a circular table, we take into account their seats
relative to each other, not the actual position of any one person. Show that",prob.pdf
"relative to each other, not the actual position of any one person. Show that
n people can be arranged around a circular table in (n  1)! ways.
John Wiley & Sons, 1968).
10J. Stirling, Methodus Dierentialis, (London: Bowyer, 1730).
11A. de Moivre, The Doctrine of Chances, 3rd ed. (London: Millar, 1756).",prob.pdf
"3.1. PERMUTATIONS 89
7 Five people get on an elevator that stops at ve 
oors. Assuming that each
has an equal probability of going to any one 
oor, nd the probability that
they all get o at dierent 
oors.
8 A nite set 
 has n elements. Show that if we count the empty set and 
 as
subsets, there are 2n subsets of 
.
9 A more rened inequality for approximating n! is given by
p
2n
n
e
n
e1=(12n+1) < n! <
p
2n
n
e
n
e1=(12n) :
Write a computer program to illustrate this inequality for n = 1 to 9.
10 A deck of ordinary cards is shued and 13 cards are dealt. What is the
probability that the last card dealt is an ace?
11 There are n applicants for the director of computing. The applicants are inter-
viewed independently by each member of the three-person search committee
and ranked from 1 to n. A candidate will be hired if he or she is ranked rst
by at least two of the three interviewers. Find the probability that a candidate",prob.pdf
"by at least two of the three interviewers. Find the probability that a candidate
will be accepted if the members of the committee really have no ability at all
to judge the candidates and just rank the candidates randomly. In particular,
compare this probability for the case of three candidates and the case of ten
candidates.
12 A symphony orchestra has in its repertoire 30 Haydn symphonies, 15 modern
works, and 9 Beethoven symphonies. Its program always consists of a Haydn
symphony followed by a modern work, and then a Beethoven symphony.
(a) How many dierent programs can it play?
(b) How many dierent programs are there if the three pieces can be played
in any order?
(c) How many dierent three-piece programs are there if more than one
piece from the same category can be played and they can be played in
any order?
13 A certain state has license plates showing three numbers and three letters.
How many dierent license plates are possible",prob.pdf
"any order?
13 A certain state has license plates showing three numbers and three letters.
How many dierent license plates are possible
(a) if the numbers must come before the letters?
(b) if there is no restriction on where the letters and numbers appear?
14 The door on the computer center has a lock which has ve buttons numbered
from 1 to 5. The combination of numbers that opens the lock is a sequence
of ve numbers and is reset every week.
(a) How many combinations are possible if every button must be used once?",prob.pdf
"90 CHAPTER 3. COMBINATORICS
(b) Assume that the lock can also have combinations that require you to
push two buttons simultaneously and then the other three one at a time.
How many more combinations does this permit?
15 A computing center has 3 processors that receive n jobs, with the jobs assigned
to the processors purely at random so that all of the 3n possible assignments
are equally likely. Find the probability that exactly one processor has no jobs.
16 Prove that at least two people in Atlanta, Georgia, have the same initials,
assuming no one has more than four initials.
17 Find a formula for the probability that among a set of n people, at least two
have their birthdays in the same month of the year (assuming the months are
equally likely for birthdays).
18 Consider the problem of nding the probability of more than one coincidence
of birthdays in a group of n people. These include, for example, three people",prob.pdf
"of birthdays in a group of n people. These include, for example, three people
with the same birthday, or two pairs of people with the same birthday, or
larger coincidences. Show how you could compute this probability, and write
a computer program to carry out this computation. Use your program to nd
the smallest number of people for which it would be a favorable bet that there
would be more than one coincidence of birthdays.
*19 Suppose that on planet Zorg a year has n days, and that the lifeforms there
are equally likely to have hatched on any day of the year. We would like
to estimate d, which is the minimum number of lifeforms needed so that the
probability of at least two sharing a birthday exceeds 1/2.
(a) In Example 3.3, it was shown that in a set of d lifeforms, the probability
that no two life forms share a birthday is
(n)d
nd ;
where (n)d = (n)(n  1) 
 
 
 (n  d + 1). Thus, we would like to set this
equal to 1/2 and solve for d.
(b) Using Stirling's Formula, show that
(n)d",prob.pdf
"where (n)d = (n)(n  1) 
 
 
 (n  d + 1). Thus, we would like to set this
equal to 1/2 and solve for d.
(b) Using Stirling's Formula, show that
(n)d
nd 

1 + d
n  d
nd+1=2
ed :
(c) Now take the logarithm of the right-hand expression, and use the fact
that for small values of x, we have
log(1 + x)  x  x2
2 :
(We are implicitly using the fact that d is of smaller order of magnitude
than n. We will also use this fact in part (d).)",prob.pdf
"3.1. PERMUTATIONS 91
(d) Set the expression found in part (c) equal to  log(2), and solve for d as
a function of n, thereby showing that
d 
p
2(log 2) n :
Hint: If all three summands in the expression found in part (b) are used,
one obtains a cubic equation in d. If the smallest of the three terms is
thrown away, one obtains a quadratic equation in d.
(e) Use a computer to calculate the exact values of d for various values of
n. Compare these values with the approximate values obtained by using
the answer to part d).
20 At a mathematical conference, ten participants are randomly seated around
a circular table for meals. Using simulation, estimate the probability that no
two people sit next to each other at both lunch and dinner. Can you make an
intelligent conjecture for the case of n participants when n is large?
21 Modify the program AllPermutations to count the number of permutations
of n objects that have exactly j xed points for j = 0, 1, 2, . .. , n. Run",prob.pdf
"21 Modify the program AllPermutations to count the number of permutations
of n objects that have exactly j xed points for j = 0, 1, 2, . .. , n. Run
your program for n = 2 to 6. Make a conjecture for the relation between the
number that have 0 xed points and the number that have exactly 1 xed
point. A proof of the correct conjecture can be found in Wilf.12
22 Mr. Wimply Dimple, one of London's most prestigious watch makers, has
come to Sherlock Holmes in a panic, having discovered that someone has
been producing and selling crude counterfeits of his best selling watch. The 16
counterfeits so far discovered bear stamped numbers, all of which fall between
1 and 56, and Dimple is anxious to know the extent of the forger's work. All
present agree that it seems reasonable to assume that the counterfeits thus
far produced bear consecutive numbers from 1 to whatever the total number
is.
\Chin up, Dimple,"" opines Dr. Watson. \I shouldn't worry overly much if",prob.pdf
"far produced bear consecutive numbers from 1 to whatever the total number
is.
\Chin up, Dimple,"" opines Dr. Watson. \I shouldn't worry overly much if
I were you; the Maximum Likelihood Principle, which estimates the total
number as precisely that which gives the highest probability for the series
of numbers found, suggests that we guess 56 itself as the total. Thus, your
forgers are not a big operation, and we shall have them safely behind bars
before your business suers signicantly.""
\Stu, nonsense, and bother your fancy principles, Watson,"" counters Holmes.
\Anyone can see that, of course, there must be quite a few more than 56
watches|why the odds of our having discovered precisely the highest num-
bered watch made are laughably negligible. A much better guess would be
twice 56.""
(a) Show that Watson is correct that the Maximum Likelihood Principle
gives 56.
12H. S. Wilf, \A Bijection in the Theory of Derangements,"" Mathematics Magazine, vol. 57,
no. 1 (1984), pp. 37{40.",prob.pdf
"92 CHAPTER 3. COMBINATORICS
(b) Write a computer program to compare Holmes's and Watson's guessing
strategies as follows: x a total N and choose 16 integers randomly
between 1 and N. Let m denote the largest of these. Then Watson's
guess for N is m, while Holmes's is 2m. See which of these is closer to
N. Repeat this experiment (with N still xed) a hundred or more times,
and determine the proportion of times that each comes closer. Whose
seems to be the better strategy?
23 Barbara Smith is interviewing candidates to be her secretary. As she inter-
views the candidates, she can determine the relative rank of the candidates
but not the true rank. Thus, if there are six candidates and their true rank is
6, 1, 4, 2, 3, 5, (where 1 is best) then after she had interviewed the rst three
candidates she would rank them 3, 1, 2. As she interviews each candidate,
she must either accept or reject the candidate. If she does not accept the",prob.pdf
"candidates she would rank them 3, 1, 2. As she interviews each candidate,
she must either accept or reject the candidate. If she does not accept the
candidate after the interview, the candidate is lost to her. She wants to de-
cide on a strategy for deciding when to stop and accept a candidate that will
maximize the probability of getting the best candidate. Assume that there
are n candidates and they arrive in a random rank order.
(a) What is the probability that Barbara gets the best candidate if she inter-
views all of the candidates? What is it if she chooses the rst candidate?
(b) Assume that Barbara decides to interview the rst half of the candidates
and then continue interviewing until getting a candidate better than any
candidate seen so far. Show that she has a better than 25 percent chance
of ending up with the best candidate.
24 For the task described in Exercise 23, it can be shown13 that the best strategy",prob.pdf
"of ending up with the best candidate.
24 For the task described in Exercise 23, it can be shown13 that the best strategy
is to pass over the rst k  1 candidates where k is the smallest integer for
which 1
k + 1
k + 1 + 
 
 
 + 1
n  1  1 :
Using this strategy the probability of getting the best candidate is approxi-
mately 1=e = :368. Write a program to simulate Barbara Smith's interviewing
if she uses this optimal strategy, using n = 10, and see if you can verify that
the probability of success is approximately 1=e.
3.2 Combinations
Having mastered permutations, we now consider combinations. Let U be a set with
n elements; we want to count the number of distinct subsets of the set U that have
exactly j elements. The empty set and the set U are considered to be subsets of U.
The empty set is usually denoted by .
13E. B. Dynkin and A. A. Yushkevich, Markov Processes: Theorems and Problems, trans. J. S.
Wood (New York: Plenum, 1969).",prob.pdf
"3.2. COMBINATIONS 93
Example 3.5 Let U = fa; b; cg. The subsets of U are
; fag; fbg; fcg; fa; bg; fa; cg; fb; cg; fa; b; cg :
2
Binomial Coecients
The number of distinct subsets with j elements that can be chosen from a set with
n elements is denoted by
n
j


, and is pronounced \n choose j."" The number
n
j


is
called a binomial coecient. This terminology comes from an application to algebra
which will be discussed later in this section.
In the above example, there is one subset with no elements, three subsets with
exactly 1 element, three subsets with exactly 2 elements, and one subset with exactly
3 elements. Thus,
3
0


= 1,
3
1


= 3,
3
2


= 3, and
3
3


= 1. Note that there are
23 = 8 subsets in all. (We have already seen that a set with n elements has 2n
subsets; see Exercise 3.1.8.) It follows that
3
0

+
3
1

+
3
2

+
3
3

= 23 = 8 ;
n
0

=
n
n

= 1 :
Assume that n > 0. Then, since there is only one way to choose a set with no",prob.pdf
"3
0

+
3
1

+
3
2

+
3
3

= 23 = 8 ;
n
0

=
n
n

= 1 :
Assume that n > 0. Then, since there is only one way to choose a set with no
elements and only one way to choose a set with n elements, the remaining values
of
n
j


are determined by the following recurrence relation:
Theorem 3.4 For integers n and j, with 0 < j < n, the binomial coecients
satisfy: n
j

=
n  1
j

+
n  1
j  1

: (3.1)
Proof. We wish to choose a subset of j elements. Choose an element u of U.
Assume rst that we do not want u in the subset. Then we must choose the j
elements from a set of n1 elements; this can be done in
n1
j


ways. On the other
hand, assume that we do want u in the subset. Then we must choose the other
j  1 elements from the remaining n  1 elements of U; this can be done in
n1
j1


ways. Since u is either in our subset or not, the number of ways that we can choose
a subset of j elements is the sum of the number of subsets of j elements which have",prob.pdf
"a subset of j elements is the sum of the number of subsets of j elements which have
u as a member and the number which do not|this is what Equation 3.1 states. 2
The binomial coecient
n
j


is dened to be 0, if j < 0 or if j > n. With this
denition, the restrictions on j in Theorem 3.4 are unnecessary.",prob.pdf
"94 CHAPTER 3. COMBINATORICS
n = 0            1
10            1        10        45      120      210      252      210      120        45        10           1     
 9            1          9        36        84      126      126        84        36          9          1 
8            1          8        28        56        70        56        28          8          1
7            1          7        21        35        35        21          7          1
6            1          6        15        20        15          6          1
5            1          5        10        10          5          1
4            1          4          6          4          1
3            1          3          3          1
2            1          2          1
1            1          1
j = 0          1          2          3          4          5          6          7          8          9          10
Figure 3.3: Pascal's triangle.
Pascal's Triangle
The relation 3.1, together with the knowledge that
n
0

=",prob.pdf
"Figure 3.3: Pascal's triangle.
Pascal's Triangle
The relation 3.1, together with the knowledge that
n
0

=
n
n

= 1 ;
determines completely the numbers
n
j


. We can use these relations to determine
the famous triangle of Pascal, which exhibits all these numbers in matrix form (see
Figure 3.3).
The nth row of this triangle has the entries
n
0


,
n
1


,. .. ,
n
n


. We know that the
rst and last of these numbers are 1. The remaining numbers are determined by
the recurrence relation Equation 3.1; that is, the entry
n
j


for 0 < j < n in the
nth row of Pascal's triangle is the sum of the entry immediately above and the one
immediately to its left in the (n  1)st row. For example,
5
2


= 6 + 4 = 10.
This algorithm for constructing Pascal's triangle can be used to write a computer
program to compute the binomial coecients. You are asked to do this in Exercise 4.
While Pascal's triangle provides a way to construct recursively the binomial",prob.pdf
"While Pascal's triangle provides a way to construct recursively the binomial
coecients, it is also possible to give a formula for
n
j


.
Theorem 3.5 The binomial coecients are given by the formula
n
j

= (n)j
j! : (3.2)
Proof. Each subset of size j of a set of size n can be ordered in j! ways. Each of
these orderings is a j-permutation of the set of size n. The number of j-permutations
is (n)j, so the number of subsets of size j is
(n)j
j! :
This completes the proof. 2",prob.pdf
"3.2. COMBINATIONS 95
The above formula can be rewritten in the form
n
j

= n!
j!(n  j)! :
This immediately shows that
n
j

=
 n
n  j

:
When using Equation 3.2 in the calculation of
n
j


, if one alternates the multi-
plications and divisions, then all of the intermediate values in the calculation are
integers. Furthermore, none of these intermediate values exceed the nal value.
(See Exercise 40.)
Another point that should be made concerning Equation 3.2 is that if it is used
to dene the binomial coecients, then it is no longer necessary to require n to be
a positive integer. The variable j must still be a non-negative integer under this
denition. This idea is useful when extending the Binomial Theorem to general
exponents. (The Binomial Theorem for non-negative integer exponents is given
below as Theorem 3.7.)
Poker Hands
Example 3.6 Poker players sometimes wonder why a four of a kind beats a full",prob.pdf
"below as Theorem 3.7.)
Poker Hands
Example 3.6 Poker players sometimes wonder why a four of a kind beats a full
house. A poker hand is a random subset of 5 elements from a deck of 52 cards.
A hand has four of a kind if it has four cards with the same value|for example,
four sixes or four kings. It is a full house if it has three of one value and two of a
second|for example, three twos and two queens. Let us see which hand is more
likely. How many hands have four of a kind? There are 13 ways that we can specify
the value for the four cards. For each of these, there are 48 possibilities for the fth
card. Thus, the number of four-of-a-kind hands is 13 
 48 = 624. Since the total
number of possible hands is
52
5


= 2598960, the probability of a hand with four of
a kind is 624=2598960 = :00024.
Now consider the case of a full house; how many such hands are there? There
are 13 choices for the value which occurs three times; for each of these there are4
3

",prob.pdf
"are 13 choices for the value which occurs three times; for each of these there are4
3


= 4 choices for the particular three cards of this value that are in the hand.
Having picked these three cards, there are 12 possibilities for the value which occurs
twice; for each of these there are
4
2


= 6 possibilities for the particular pair of this
value. Thus, the number of full houses is 13 
 4 
 12 
 6 = 3744, and the probability
of obtaining a hand with a full house is 3744=2598960 = :0014. Thus, while both
types of hands are unlikely, you are six times more likely to obtain a full house than
four of a kind. 2",prob.pdf
"96 CHAPTER 3. COMBINATORICS
(start)


S
F
F
F
F
S
S
S
S
S
S
F
F
F
p
q
p 
p   q
p q  
q p  
q   p
q 
q
q
q
q
q
p
p
p
p
p
p
q
q   p
p   q
m (ω)ω
ω 
ω 
ω 
ω 
ω 
ω 
ω 
ω 
2
3
3
2
2
2
2
2
1
2
3
4
5
6
7
8
Figure 3.4: Tree diagram of three Bernoulli trials.
Bernoulli Trials
Our principal use of the binomial coecients will occur in the study of one of the
important chance processes called Bernoulli trials.
Denition 3.5 A Bernoulli trials process is a sequence of n chance experiments
such that
1. Each experiment has two possible outcomes, which we may call success and
failure.
2. The probability p of success on each experiment is the same for each ex-
periment, and this probability is not aected by any knowledge of previous
outcomes. The probability q of failure is given by q = 1  p.
2
Example 3.7 The following are Bernoulli trials processes:
1. A coin is tossed ten times. The two possible outcomes are heads and tails.
The probability of heads on any one toss is 1/2.",prob.pdf
"1. A coin is tossed ten times. The two possible outcomes are heads and tails.
The probability of heads on any one toss is 1/2.
2. An opinion poll is carried out by asking 1000 people, randomly chosen from
the population, if they favor the Equal Rights Amendment|the two outcomes
being yes and no. The probability p of a yes answer (i.e., a success) indicates
the proportion of people in the entire population that favor this amendment.
3. A gambler makes a sequence of 1-dollar bets, betting each time on black at
roulette at Las Vegas. Here a success is winning 1 dollar and a failure is losing",prob.pdf
"3.2. COMBINATIONS 97
1 dollar. Since in American roulette the gambler wins if the ball stops on one
of 18 out of 38 positions and loses otherwise, the probability of winning is
p = 18=38 = :474.
2
To analyze a Bernoulli trials process, we choose as our sample space a binary
tree and assign a probability distribution to the paths in this tree. Suppose, for
example, that we have three Bernoulli trials. The possible outcomes are indicated
in the tree diagram shown in Figure 3.4. We dene X to be the random variable
which represents the outcome of the process, i.e., an ordered triple of S's and F's.
The probabilities assigned to the branches of the tree represent the probability for
each individual trial. Let the outcome of the ith trial be denoted by the random
variable Xi, with distribution function mi. Since we have assumed that outcomes
on any one trial do not aect those on another, we assign the same probabilities",prob.pdf
"on any one trial do not aect those on another, we assign the same probabilities
at each level of the tree. An outcome ! for the entire experiment will be a path
through the tree. For example, !3 represents the outcomes SFS. Our frequency
interpretation of probability would lead us to expect a fraction p of successes on
the rst experiment; of these, a fraction q of failures on the second; and, of these, a
fraction p of successes on the third experiment. This suggests assigning probability
pqp to the outcome !3. More generally, we assign a distribution function m(!) for
paths ! by dening m(!) to be the product of the branch probabilities along the
path !. Thus, the probability that the three events S on the rst trial, F on the
second trial, and S on the third trial occur is the product of the probabilities for
the individual events. We shall see in the next chapter that this means that the
events involved are independent in the sense that the knowledge of one event does",prob.pdf
"events involved are independent in the sense that the knowledge of one event does
not aect our prediction for the occurrences of the other events.
Binomial Probabilities
We shall be particularly interested in the probability that in n Bernoulli trials there
are exactly j successes. We denote this probability by b(n; p; j). Let us calculate the
particular value b(3; p; 2) from our tree measure. We see that there are three paths
which have exactly two successes and one failure, namely !2, !3, and !5. Each of
these paths has the same probability p2q. Thus b(3; p; 2) = 3p2q. Considering all
possible numbers of successes we have
b(3; p; 0) = q3 ;
b(3; p; 1) = 3pq2 ;
b(3; p; 2) = 3p2q ;
b(3; p; 3) = p3 :
We can, in the same manner, carry out a tree measure for n experiments and
determine b(n; p; j) for the general case of n Bernoulli trials.",prob.pdf
"98 CHAPTER 3. COMBINATORICS
Theorem 3.6 Given n Bernoulli trials with probability p of success on each exper-
iment, the probability of exactly j successes is
b(n; p; j) =
n
j

pjqnj
where q = 1  p.
Proof. We construct a tree measure as described above. We want to nd the sum
of the probabilities for all paths which have exactly j successes and n  j failures.
Each such path is assigned a probability pjqnj. How many such paths are there?
To specify a path, we have to pick, from the n possible trials, a subset of j to be
successes, with the remaining n  j outcomes being failures. We can do this in
n
j


ways. Thus the sum of the probabilities is
b(n; p; j) =
n
j

pjqnj :
2
Example 3.8 A fair coin is tossed six times. What is the probability that exactly
three heads turn up? The answer is
b(6; :5; 3) =
6
3
1
2
3 1
2
3
= 20 
 1
64 = :3125 :
2
Example 3.9 A die is rolled four times. What is the probability that we obtain",prob.pdf
"b(6; :5; 3) =
6
3
1
2
3 1
2
3
= 20 
 1
64 = :3125 :
2
Example 3.9 A die is rolled four times. What is the probability that we obtain
exactly one 6? We treat this as Bernoulli trials with success = \rolling a 6"" and
failure = \rolling some number other than a 6."" Then p = 1=6, and the probability
of exactly one success in four trials is
b(4; 1=6; 1) =
4
1
1
6
1 5
6
3
= :386 :
2
To compute binomial probabilities using the computer, multiply the function
choose(n; k) by pkqnk. The program BinomialProbabilities prints out the bi-
nomial probabilities b(n; p; k) for k between kmin and kmax, and the sum of these
probabilities. We have run this program for n = 100, p = 1=2, kmin = 45, and
kmax = 55; the output is shown in Table 3.8. Note that the individual probabilities
are quite small. The probability of exactly 50 heads in 100 tosses of a coin is about
.08. Our intuition tells us that this is the most likely outcome, which is correct;",prob.pdf
".08. Our intuition tells us that this is the most likely outcome, which is correct;
but, all the same, it is not a very likely outcome.",prob.pdf
"3.2. COMBINATIONS 99
k b(n; p; k)
45 .0485
46 .0580
47 .0666
48 .0735
49 .0780
50 .0796
51 .0780
52 .0735
53 .0666
54 .0580
55 .0485
Table 3.8: Binomial probabilities for n = 100; p = 1=2.
Binomial Distributions
Denition 3.6 Let n be a positive integer, and let p be a real number between 0
and 1. Let B be the random variable which counts the number of successes in a
Bernoulli trials process with parameters n and p. Then the distribution b(n; p; k)
of B is called the binomial distribution. 2
We can get a better idea about the binomial distribution by graphing this dis-
tribution for dierent values of n and p (see Figure 3.5). The plots in this gure
were generated using the program BinomialPlot.
We have run this program for p = :5 and p = :3. Note that even for p = :3 the
graphs are quite symmetric. We shall have an explanation for this in Chapter 9. We
also note that the highest probability occurs around the value np, but that these",prob.pdf
"also note that the highest probability occurs around the value np, but that these
highest probabilities get smaller as n increases. We shall see in Chapter 6 that np
is the mean or expected value of the binomial distribution b(n; p; k).
The following example gives a nice way to see the binomial distribution, when
p = 1=2.
Example 3.10 A Galton board is a board in which a large number of BB-shots are
dropped from a chute at the top of the board and de
ected o a number of pins on
their way down to the bottom of the board. The nal position of each slot is the
result of a number of random de
ections either to the left or the right. We have
written a program GaltonBoard to simulate this experiment.
We have run the program for the case of 20 rows of pins and 10,000 shots being
dropped. We show the result of this simulation in Figure 3.6.
Note that if we write 0 every time the shot is de
ected to the left, and 1 every",prob.pdf
"dropped. We show the result of this simulation in Figure 3.6.
Note that if we write 0 every time the shot is de
ected to the left, and 1 every
time it is de
ected to the right, then the path of the shot can be described by a
sequence of 0's and 1's of length n, just as for the n-fold coin toss.
The distribution shown in Figure 3.6 is an example of an empirical distribution,
in the sense that it comes about by means of a sequence of experiments. As expected,",prob.pdf
"100 CHAPTER 3. COMBINATORICS
0 20 40 60 80 100 120
0
0.025
0.05
0.075
0.1
0.125
0.15
0 20 40 60 80 100
0.02
0.04
0.06
0.08
0.1
0.12 p = .5
n = 40
n = 80
n = 160
n = 30
n = 120
n = 270
p = .3
0
Figure 3.5: Binomial distributions.",prob.pdf
"3.2. COMBINATIONS 101
Figure 3.6: Simulation of the Galton board.
this empirical distribution resembles the corresponding binomial distribution with
parameters n = 20 and p = 1=2. 2
Hypothesis Testing
Example 3.11 Suppose that ordinary aspirin has been found eective against
headaches 60 percent of the time, and that a drug company claims that its new
aspirin with a special headache additive is more eective. We can test this claim
as follows: we call their claim the alternate hypothesis, and its negation, that the
additive has no appreciable eect, the null hypothesis. Thus the null hypothesis is
that p = :6, and the alternate hypothesis is that p > :6, where p is the probability
that the new aspirin is eective.
We give the aspirin to n people to take when they have a headache. We want to
nd a number m, called the critical value for our experiment, such that we reject
the null hypothesis if at least m people are cured, and otherwise we accept it. How",prob.pdf
"the null hypothesis if at least m people are cured, and otherwise we accept it. How
should we determine this critical value?
First note that we can make two kinds of errors. The rst, often called a type 1
error in statistics, is to reject the null hypothesis when in fact it is true. The second,
called a type 2 error, is to accept the null hypothesis when it is false. To determine
the probability of both these types of errors we introduce a function (p), dened
to be the probability that we reject the null hypothesis, where this probability is
calculated under the assumption that the null hypothesis is true. In the present
case, we have
(p) =
X
mkn
b(n; p; k) :",prob.pdf
"102 CHAPTER 3. COMBINATORICS
Note that (:6) is the probability of a type 1 error, since this is the probability
of a high number of successes for an ineective additive. So for a given n we want
to choose m so as to make (:6) quite small, to reduce the likelihood of a type 1
error. But as m increases above the most probable value np = :6n, (:6), being
the upper tail of a binomial distribution, approaches 0. Thus increasing m makes
a type 1 error less likely.
Now suppose that the additive really is eective, so that p is appreciably greater
than .6; say p = :8. (This alternative value of p is chosen arbitrarily; the following
calculations depend on this choice.) Then choosing m well below np = :8n will
increase (:8), since now (:8) is all but the lower tail of a binomial distribution.
Indeed, if we put (:8) = 1  (:8), then (:8) gives us the probability of a type 2
error, and so decreasing m makes a type 2 error less likely.",prob.pdf
"Indeed, if we put (:8) = 1  (:8), then (:8) gives us the probability of a type 2
error, and so decreasing m makes a type 2 error less likely.
The manufacturer would like to guard against a type 2 error, since if such an
error is made, then the test does not show that the new drug is better, when in
fact it is. If the alternative value of p is chosen closer to the value of p given in
the null hypothesis (in this case p = :6), then for a given test population, the
value of  will increase. So, if the manufacturer's statistician chooses an alternative
value for p which is close to the value in the null hypothesis, then it will be an
expensive proposition (i.e., the test population will have to be large) to reject the
null hypothesis with a small value of .
What we hope to do then, for a given test population n, is to choose a value
of m, if possible, which makes both these probabilities small. If we make a type 1",prob.pdf
"of m, if possible, which makes both these probabilities small. If we make a type 1
error we end up buying a lot of essentially ordinary aspirin at an in
ated price; a
type 2 error means we miss a bargain on a superior medication. Let us say that
we want our critical number m to make each of these undesirable cases less than 5
percent probable.
We write a program PowerCurve to plot, for n = 100 and selected values of m,
the function (p), for p ranging from .4 to 1. The result is shown in Figure 3.7. We
include in our graph a box (in dotted lines) from .6 to .8, with bottom and top at
heights .05 and .95. Then a value for m satises our requirements if and only if the
graph of  enters the box from the bottom, and leaves from the top (why?|which
is the type 1 and which is the type 2 criterion?). As m increases, the graph of 
moves to the right. A few experiments have shown us that m = 69 is the smallest",prob.pdf
"moves to the right. A few experiments have shown us that m = 69 is the smallest
value for m that thwarts a type 1 error, while m = 73 is the largest which thwarts a
type 2. So we may choose our critical value between 69 and 73. If we're more intent
on avoiding a type 1 error we favor 73, and similarly we favor 69 if we regard a
type 2 error as worse. Of course, the drug company may not be happy with having
as much as a 5 percent chance of an error. They might insist on having a 1 percent
chance of an error. For this we would have to increase the number n of trials (see
Exercise 28). 2
Binomial Expansion
We next remind the reader of an application of the binomial coecients to algebra.
This is the binomial expansion, from which we get the term binomial coecient.",prob.pdf
"3.2. COMBINATIONS 103
                                                            
.4 1.5 .6 .7 .8 .9 1
 .0
1.0
 .1
 .2
 .3
 .4
 .5
 .6
 .7
 .8
 .9
1.0
                                                            
.4 1.5 .6 .7 .8 .9 1
 .0
1.0
 .1
 .2
 .3
 .4
 .5
 .6
 .7
 .8
 .9
1.0
Figure 3.7: The power curve.
Theorem 3.7 (Binomial Theorem) The quantity (a + b)n can be expressed in
the form
(a + b)n =
nX
j=0
n
j

ajbnj :
Proof. To see that this expansion is correct, write
(a + b)n = (a + b)(a + b) 
 
 
(a + b) :
When we multiply this out we will have a sum of terms each of which results from
a choice of an a or b for each of n factors. When we choose j a's and (n  j) b's,
we obtain a term of the form ajbnj. To determine such a term, we have to specify
j of the n terms in the product from which we choose the a. This can be done inn
j


ways. Thus, collecting these terms in the sum contributes a term
n
j


ajbnj. 2
For example, we have
(a + b)0 = 1
(a + b)1 = a + b",prob.pdf
"j


ways. Thus, collecting these terms in the sum contributes a term
n
j


ajbnj. 2
For example, we have
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3 :
We see here that the coecients of successive powers do indeed yield Pascal's tri-
angle.
Corollary 3.1 The sum of the elements in the nth row of Pascal's triangle is 2n.
If the elements in the nth row of Pascal's triangle are added with alternating signs,
the sum is 0.",prob.pdf
"104 CHAPTER 3. COMBINATORICS
Proof. The rst statement in the corollary follows from the fact that
2n = (1 + 1)n =
n
0

+
n
1

+
n
2

+ 
 
 
 +
n
n

;
and the second from the fact that
0 = (1  1)n =
n
0


n
1

+
n
2

 
 
 
 + (1)n
n
n

:
2
The rst statement of the corollary tells us that the number of subsets of a set
of n elements is 2n. We shall use the second statement in our next application of
the binomial theorem.
We have seen that, when A and B are any two events (cf. Section 1.2),
P(A [ B) = P(A) + P(B)  P(A \ B):
We now extend this theorem to a more general version, which will enable us to nd
the probability that at least one of a number of events occurs.
Inclusion-Exclusion Principle
Theorem 3.8 Let P be a probability distribution on a sample space 
, and let
fA1; A2; : : :; Ang be a nite set of events. Then
P(A1 [ A2 [ 
 
 
 [ An) =
nX
i=1
P(Ai) 
X
1i<jn
P(Ai \ Aj)
+
X
1i<j<kn
P(Ai \ Aj \ Ak)  
 
 
 : (3.3)",prob.pdf
"P(A1 [ A2 [ 
 
 
 [ An) =
nX
i=1
P(Ai) 
X
1i<jn
P(Ai \ Aj)
+
X
1i<j<kn
P(Ai \ Aj \ Ak)  
 
 
 : (3.3)
That is, to nd the probability that at least one of n events Ai occurs, rst add
the probability of each event, then subtract the probabilities of all possible two-way
intersections, add the probability of all three-way intersections, and so forth.
Proof. If the outcome ! occurs in at least one of the events Ai, its probability is
added exactly once by the left side of Equation 3.3. We must show that it is added
exactly once by the right side of Equation 3.3. Assume that ! is in exactly k of the
sets. Then its probability is added k times in the rst term, subtracted
k
2


times in
the second, added
k
3


times in the third term, and so forth. Thus, the total number
of times that it is added is
k
1


k
2

+
k
3

 
 
 
 (1)k1
k
k

:
But
0 = (1  1)k =
kX
j=0
k
j

(1)j =
k
0


kX
j=1
k
j

(1)j1 :",prob.pdf
"3.2. COMBINATIONS 105
Hence,
1 =
k
0

=
kX
j=1
k
j

(1)j1 :
If the outcome ! is not in any of the events Ai, then it is not counted on either side
of the equation. 2
Hat Check Problem
Example 3.12 We return to the hat check problem discussed in Section 3.1, that
is, the problem of nding the probability that a random permutation contains at
least one xed point. Recall that a permutation is a one-to-one map of a set
A = fa1; a2; : : :; ang onto itself. Let Ai be the event that the ith element ai remains
xed under this map. If we require that ai is xed, then the map of the remaining
n1 elements provides an arbitrary permutation of (n1) objects. Since there are
(n  1)! such permutations, P(Ai) = (n  1)!=n! = 1=n. Since there are n choices
for ai, the rst term of Equation 3.3 is 1. In the same way, to have a particular
pair (ai; aj) xed, we can choose any permutation of the remaining n  2 elements;
there are (n  2)! such choices and thus
P(Ai \ Aj) = (n  2)!
n! = 1",prob.pdf
"there are (n  2)! such choices and thus
P(Ai \ Aj) = (n  2)!
n! = 1
n(n  1) :
The number of terms of this form in the right side of Equation 3.3 is
n
2

= n(n  1)
2! :
Hence, the second term of Equation 3.3 is
n(n  1)
2! 
 1
n(n  1) =  1
2! :
Similarly, for any specic three events Ai, Aj, Ak,
P(Ai \ Aj \ Ak) = (n  3)!
n! = 1
n(n  1)(n  2) ;
and the number of such terms is
n
3

= n(n  1)(n  2)
3! ;
making the third term of Equation 3.3 equal to 1/3!. Continuing in this way, we
obtain
P(at least one xed point) = 1  1
2! + 1
3!  
 
 
 (1)n1 1
n!
and
P(no xed point) = 1
2!  1
3! + 
 
 
 (1)n 1
n! :",prob.pdf
"106 CHAPTER 3. COMBINATORICS
Probability that no one
n gets his own hat back
3 .333333
4 .375
5 .366667
6 .368056
7 .367857
8 .367882
9 .367879
10 .367879
Table 3.9: Hat check problem.
From calculus we learn that
ex = 1 + x + 1
2!x2 + 1
3!x3 + 
 
 
 + 1
n!xn + 
 
 
 :
Thus, if x = 1, we have
e1 = 1
2!  1
3! + 
 
 
 + (1)n
n! + 
 
 

= :3678794 :
Therefore, the probability that there is no xed point, i.e., that none of the n people
gets his own hat back, is equal to the sum of the rst n terms in the expression for
e1. This series converges very fast. Calculating the partial sums for n = 3 to 10
gives the data in Table 3.9.
After n = 9 the probabilities are essentially the same to six signicant gures.
Interestingly, the probability of no xed point alternately increases and decreases
as n increases. Finally, we note that our exact results are in good agreement with
our simulations reported in the previous section. 2
Choosing a Sample Space",prob.pdf
"our simulations reported in the previous section. 2
Choosing a Sample Space
We now have some of the tools needed to accurately describe sample spaces and
to assign probability functions to those sample spaces. Nevertheless, in some cases,
the description and assignment process is somewhat arbitrary. Of course, it is to
be hoped that the description of the sample space and the subsequent assignment
of a probability function will yield a model which accurately predicts what would
happen if the experiment were actually carried out. As the following examples show,
there are situations in which \reasonable"" descriptions of the sample space do not
produce a model which ts the data.
In Feller's book,14 a pair of models is given which describe arrangements of
certain kinds of elementary particles, such as photons and protons. It turns out that
experiments have shown that certain types of elementary particles exhibit behavior",prob.pdf
"experiments have shown that certain types of elementary particles exhibit behavior
14W. Feller, Introduction to Probability Theory and Its Applications vol. 1, 3rd ed. (New York:
John Wiley and Sons, 1968), p. 41",prob.pdf
"3.2. COMBINATIONS 107
which is accurately described by one model, called \Bose-Einstein statistics,"" while
other types of elementary particles can be modelled using \Fermi-Dirac statistics.""
Feller says:
We have here an instructive example of the impossibility of selecting or
justifying probability models by a priori arguments. In fact, no pure
reasoning could tell that photons and protons would not obey the same
probability laws.
We now give some examples of this description and assignment process.
Example 3.13 In the quantum mechanical model of the helium atom, various
parameters can be used to classify the energy states of the atom. In the triplet
spin state (S = 1) with orbital angular momentum 1 (L = 1), there are three
possibilities, 0, 1, or 2, for the total angular momentum (J). (It is not assumed that
the reader knows what any of this means; in fact, the example is more illustrative
if the reader does not know anything about quantum mechanics.) We would like",prob.pdf
"if the reader does not know anything about quantum mechanics.) We would like
to assign probabilities to the three possibilities for J. The reader is undoubtedly
resisting the idea of assigning the probability of 1=3 to each of these outcomes. She
should now ask herself why she is resisting this assignment. The answer is probably
because she does not have any \intuition"" (i.e., experience) about the way in which
helium atoms behave. In fact, in this example, the probabilities 1=9; 3=9; and
5=9 are assigned by the theory. The theory gives these assignments because these
frequencies were observed in experiments and further parameters were developed in
the theory to allow these frequencies to be predicted. 2
Example 3.14 Suppose two pennies are 
ipped once each. There are several \rea-
sonable"" ways to describe the sample space. One way is to count the number of
heads in the outcome; in this case, the sample space can be written f0; 1; 2g. An-",prob.pdf
"heads in the outcome; in this case, the sample space can be written f0; 1; 2g. An-
other description of the sample space is the set of all ordered pairs of H's and T's,
i.e.,
f(H; H); (H; T); (T; H); (T; T)g:
Both of these descriptions are accurate ones, but it is easy to see that (at most) one
of these, if assigned a constant probability function, can claim to accurately model
reality. In this case, as opposed to the preceding example, the reader will probably
say that the second description, with each outcome being assigned a probability of
1=4, is the \right"" description. This conviction is due to experience; there is no
proof that this is the way reality works. 2
The reader is also referred to Exercise 26 for another example of this process.
Historical Remarks
The binomial coecients have a long and colorful history leading up to Pascal's
Treatise on the Arithmetical Triangle,15 where Pascal developed many important",prob.pdf
"Treatise on the Arithmetical Triangle,15 where Pascal developed many important
15B. Pascal, Traite du Triangle Arithmetique (Paris: Desprez, 1665).",prob.pdf
"108 CHAPTER 3. COMBINATORICS
1 1 1 1 1 1 1 1 1 1
1 2 3 4 5 6 7 8 9
1 3 6 10 15 21 28 36
1 4 10 20 35 56 84
1 5 15 35 70 126
1 6 21 56 126
1 7 28 84
1 8 36
1 9
1
Table 3.10: Pascal's triangle.
natural numbers 1 2 3 4 5 6 7 8 9
triangular numbers 1 3 6 10 15 21 28 36 45
tetrahedral numbers 1 4 10 20 35 56 84 120 165
Table 3.11: Figurate numbers.
properties of these numbers. This history is set forth in the book Pascal's Arith-
metical Triangle by A. W. F. Edwards.16 Pascal wrote his triangle in the form
shown in Table 3.10.
Edwards traces three dierent ways that the binomial coecients arose. He
refers to these as the gurate numbers, the combinatorial numbers, and the binomial
numbers. They are all names for the same thing (which we have called binomial
coecients) but that they are all the same was not appreciated until the sixteenth
century.
The gurate numbers date back to the Pythagorean interest in number pat-",prob.pdf
"century.
The gurate numbers date back to the Pythagorean interest in number pat-
terns around 540 BC. The Pythagoreans considered, for example, triangular patterns
shown in Figure 3.8. The sequence of numbers
1; 3; 6; 10; : : :
obtained as the number of points in each triangle are called triangular numbers.
From the triangles it is clear that the nth triangular number is simply the sum of
the rst n integers. The tetrahedral numbers are the sums of the triangular numbers
and were obtained by the Greek mathematicians Theon and Nicomachus at the
beginning of the second century BC. The tetrahedral number 10, for example, has
the geometric representation shown in Figure 3.9. The rst three types of gurate
numbers can be represented in tabular form as shown in Table 3.11.
These numbers provide the rst four rows of Pascal's triangle, but the table was
not to be completed in the West until the sixteenth century.
In the East, Hindu mathematicians began to encounter the binomial coecients",prob.pdf
"not to be completed in the West until the sixteenth century.
In the East, Hindu mathematicians began to encounter the binomial coecients
in combinatorial problems. Bhaskara in his Lilavati of 1150 gave a rule to nd the
16A. W. F. Edwards, Pascal's Arithmetical Triangle (London: Grin, 1987).",prob.pdf
"3.2. COMBINATIONS 109
1 3 6 10
Figure 3.8: Pythagorean triangular patterns.
Figure 3.9: Geometric representation of the tetrahedral number 10.",prob.pdf
"110 CHAPTER 3. COMBINATORICS
11
12 22
13 23 33
14 24 34 44
15 25 35 45 55
16 26 36 46 56 66
Table 3.12: Outcomes for the roll of two dice.
number of medicinal preparations using 1, 2, 3, 4, 5, or 6 possible ingredients.17 His
rule is equivalent to our formula
n
r

= (n)r
r! :
The binomial numbers as coecients of (a+b)n appeared in the works of math-
ematicians in China around 1100. There are references about this time to \the
tabulation system for unlocking binomial coecients."" The triangle to provide the
coecients up to the eighth power is given by Chu Shih-chieh in a book written
around 1303 (see Figure 3.10).18 The original manuscript of Chu's book has been
lost, but copies have survived. Edwards notes that there is an error in this copy of
Chu's triangle. Can you nd it? (Hint: Two numbers which should be equal are
not.) Other copies do not show this error.
The rst appearance of Pascal's triangle in the West seems to have come from",prob.pdf
"not.) Other copies do not show this error.
The rst appearance of Pascal's triangle in the West seems to have come from
calculations of Tartaglia in calculating the number of possible ways that n dice
might turn up.19 For one die the answer is clearly 6. For two dice the possibilities
may be displayed as shown in Table 3.12.
Displaying them this way suggests the sixth triangular number 1 + 2 + 3 + 4 +
5 + 6 = 21 for the throw of 2 dice. Tartaglia \on the rst day of Lent, 1523, in
Verona, having thought about the problem all night,""20 realized that the extension
of the gurate table gave the answers for n dice. The problem had suggested itself
to Tartaglia from watching people casting their own horoscopes by means of a Book
of Fortune, selecting verses by a process which included noting the numbers on the
faces of three dice. The 56 ways that three dice can fall were set out on each page.
The way the numbers were written in the book did not suggest the connection with",prob.pdf
"The way the numbers were written in the book did not suggest the connection with
gurate numbers, but a method of enumeration similar to the one we used for 2
dice does. Tartaglia's table was not published until 1556.
A table for the binomial coecients was published in 1554 by the German mathe-
matician Stifel.21 Pascal's triangle appears also in Cardano's Opus novum of 1570.22
17ibid., p. 27.
18J. Needham, Science and Civilization in China, vol. 3 (New York: Cambridge University
Press, 1959), p. 135.
19N. Tartaglia, General Trattato di Numeri et Misure (Vinegia, 1556).
20Quoted in Edwards, op. cit., p. 37.
21M. Stifel, Arithmetica Integra (Norimburgae, 1544).
22G. Cardano, Opus Novum de Proportionibus Numerorum (Basilea, 1570).",prob.pdf
"3.2. COMBINATIONS 111
Figure 3.10: Chu Shih-chieh's triangle. [From J. Needham, Science and Civilization
in China, vol. 3 (New York: Cambridge University Press, 1959), p. 135. Reprinted
with permission.]",prob.pdf
"112 CHAPTER 3. COMBINATORICS
Cardano was interested in the problem of nding the number of ways to choose r
objects out of n. Thus by the time of Pascal's work, his triangle had appeared as
a result of looking at the gurate numbers, the combinatorial numbers, and the
binomial numbers, and the fact that all three were the same was presumably pretty
well understood.
Pascal's interest in the binomial numbers came from his letters with Fermat
concerning a problem known as the problem of points. This problem, and the
correspondence between Pascal and Fermat, were discussed in Chapter 1. The
reader will recall that this problem can be described as follows: Two players A and
B are playing a sequence of games and the rst player to win n games wins the
match. It is desired to nd the probability that A wins the match at a time when
A has won a games and B has won b games. (See Exercises 4.1.40-4.1.42.)
Pascal solved the problem by backward induction, much the way we would do",prob.pdf
"A has won a games and B has won b games. (See Exercises 4.1.40-4.1.42.)
Pascal solved the problem by backward induction, much the way we would do
today in writing a computer program for its solution. He referred to the combina-
torial method of Fermat which proceeds as follows: If A needs c games and B needs
d games to win, we require that the players continue to play until they have played
c + d  1 games. The winner in this extended series will be the same as the winner
in the original series. The probability that A wins in the extended series and hence
in the original series is
c+d1X
r=c
1
2c+d1
c + d  1
r

:
Even at the time of the letters Pascal seemed to understand this formula.
Suppose that the rst player to win n games wins the match, and suppose that
each player has put up a stake of x. Pascal studied the value of winning a particular
game. By this he meant the increase in the expected winnings of the winner of the",prob.pdf
"game. By this he meant the increase in the expected winnings of the winner of the
particular game under consideration. He showed that the value of the rst game is
1 
 3 
 5 
 : : : 
 (2n  1)
2 
 4 
 6 
 : : : 
 (2n) x :
His proof of this seems to use Fermat's formula and the fact that the above ratio of
products of odd to products of even numbers is equal to the probability of exactly
n heads in 2n tosses of a coin. (See Exercise 39.)
Pascal presented Fermat with the table shown in Table 3.13. He states:
You will see as always, that the value of the rst game is equal to that
of the second which is easily shown by combinations. You will see, in
the same way, that the numbers in the rst line are always increasing;
so also are those in the second; and those in the third. But those in the
fourth line are decreasing, and those in the fth, etc. This seems odd.23
The student can pursue this question further using the computer and Pascal's",prob.pdf
"The student can pursue this question further using the computer and Pascal's
backward iteration method for computing the expected payo at any point in the
series.
23F. N. David, op. cit., p. 235.",prob.pdf
"3.2. COMBINATIONS 113
if each one staken 256 in
From my opponent's 256 6 5 4 3 2 1
positions I get, for the games games games games games games
1st game 63 70 80 96 128 256
2nd game 63 70 80 96 128
3rd game 56 60 64 64
4th game 42 40 32
5th game 24 16
6th game 8
Table 3.13: Pascal's solution for the problem of points.
In his treatise, Pascal gave a formal proof of Fermat's combinatorial formula as
well as proofs of many other basic properties of binomial numbers. Many of his
proofs involved induction and represent some of the rst proofs by this method.
His book brought together all the dierent aspects of the numbers in the Pascal
triangle as known in 1654, and, as Edwards states, \That the Arithmetical Triangle
should bear Pascal's name cannot be disputed.""24
The rst serious study of the binomial distribution was undertaken by James
Bernoulli in his Ars Conjectandi published in 1713.25 We shall return to this work
in the historical remarks in Chapter 8.
Exercises",prob.pdf
"Bernoulli in his Ars Conjectandi published in 1713.25 We shall return to this work
in the historical remarks in Chapter 8.
Exercises
1 Compute the following:
(a)
6
3


(b) b(5; :2; 4)
(c)
7
2


(d)
26
26


(e) b(4; :2; 3)
(f)
6
2


(g)
10
9


(h) b(8; :3; 5)
2 In how many ways can we choose ve people from a group of ten to form a
committee?
3 How many seven-element subsets are there in a set of nine elements?
4 Using the relation Equation 3.1 write a program to compute Pascal's triangle,
putting the results in a matrix. Have your program print the triangle for
n = 10.
24A. W. F. Edwards, op. cit., p. ix.
25J. Bernoulli, Ars Conjectandi (Basil: Thurnisiorum, 1713).",prob.pdf
"114 CHAPTER 3. COMBINATORICS
5 Use the program BinomialProbabilities to nd the probability that, in 100
tosses of a fair coin, the number of heads that turns up lies between 35 and
65, between 40 and 60, and between 45 and 55.
6 Charles claims that he can distinguish between beer and ale 75 percent of the
time. Ruth bets that he cannot and, in fact, just guesses. To settle this, a bet
is made: Charles is to be given ten small glasses, each having been lled with
beer or ale, chosen by tossing a fair coin. He wins the bet if he gets seven or
more correct. Find the probability that Charles wins if he has the ability that
he claims. Find the probability that Ruth wins if Charles is guessing.
7 Show that
b(n; p; j) = p
q
n  j + 1
j

b(n; p; j  1) ;
for j  1. Use this fact to determine the value or values of j which give
b(n; p; j) its greatest value. Hint: Consider the successive ratios as j increases.
8 A die is rolled 30 times. What is the probability that a 6 turns up exactly 5",prob.pdf
"8 A die is rolled 30 times. What is the probability that a 6 turns up exactly 5
times? What is the most probable number of times that a 6 will turn up?
9 Find integers n and r such that the following equation is true:
13
5

+ 2
13
6

+
13
7

=
n
r

:
10 In a ten-question true-false exam, nd the probability that a student gets a
grade of 70 percent or better by guessing. Answer the same question if the
test has 30 questions, and if the test has 50 questions.
11 A restaurant oers apple and blueberry pies and stocks an equal number of
each kind of pie. Each day ten customers request pie. They choose, with
equal probabilities, one of the two kinds of pie. How many pieces of each kind
of pie should the owner provide so that the probability is about .95 that each
customer gets the pie of his or her own choice?
12 A poker hand is a set of 5 cards randomly chosen from a deck of 52 cards.
Find the probability of a
(a) royal 
ush (ten, jack, queen, king, ace in a single suit).",prob.pdf
"Find the probability of a
(a) royal 
ush (ten, jack, queen, king, ace in a single suit).
(b) straight 
ush (ve in a sequence in a single suit, but not a royal 
ush).
(c) four of a kind (four cards of the same face value).
(d) full house (one pair and one triple, each of the same face value).
(e) 
ush (ve cards in a single suit but not a straight or royal 
ush).
(f) straight (ve cards in a sequence, not all the same suit). (Note that in
straights, an ace counts high or low.)
13 If a set has 2n elements, show that it has more subsets with n elements than
with any other number of elements.",prob.pdf
"3.2. COMBINATIONS 115
14 Let b(2n; :5; n) be the probability that in 2n tosses of a fair coin exactly n heads
turn up. Using Stirling's formula (Theorem 3.3), show that b(2n; :5; n) 
1=pn. Use the program BinomialProbabilities to compare this with the
exact value for n = 10 to 25.
15 A baseball player, Smith, has a batting average of :300 and in a typical game
comes to bat three times. Assume that Smith's hits in a game can be consid-
ered to be a Bernoulli trials process with probability .3 for success. Find the
probability that Smith gets 0, 1, 2, and 3 hits.
16 The Siwash University football team plays eight games in a season, winning
three, losing three, and ending two in a tie. Show that the number of ways
that this can happen is 8
3
5
3

= 8!
3! 3! 2! :
17 Using the technique of Exercise 16, show that the number of ways that one
can put n dierent objects into three boxes with a in the rst, b in the second,
and c in the third is n!=(a! b! c!).",prob.pdf
"can put n dierent objects into three boxes with a in the rst, b in the second,
and c in the third is n!=(a! b! c!).
18 Baumgartner, Prosser, and Crowell are grading a calculus exam. There is a
true-false question with ten parts. Baumgartner notices that one student has
only two out of the ten correct and remarks, \The student was not even bright
enough to have 
ipped a coin to determine his answers."" \Not so clear,"" says
Prosser. \With 340 students I bet that if they all 
ipped coins to determine
their answers there would be at least one exam with two or fewer answers
correct."" Crowell says, \I'm with Prosser. In fact, I bet that we should expect
at least one exam in which no answer is correct if everyone is just guessing.""
Who is right in all of this?
19 A gin hand consists of 10 cards from a deck of 52 cards. Find the probability
that a gin hand has
(a) all 10 cards of the same suit.
(b) exactly 4 cards in one suit and 3 in two other suits.
(c) a 4, 3, 2, 1, distribution of suits.",prob.pdf
"(a) all 10 cards of the same suit.
(b) exactly 4 cards in one suit and 3 in two other suits.
(c) a 4, 3, 2, 1, distribution of suits.
20 A six-card hand is dealt from an ordinary deck of cards. Find the probability
that:
(a) All six cards are hearts.
(b) There are three aces, two kings, and one queen.
(c) There are three cards of one suit and three of another suit.
21 A lady wishes to color her ngernails on one hand using at most two of the
colors red, yellow, and blue. How many ways can she do this?",prob.pdf
"116 CHAPTER 3. COMBINATORICS
22 How many ways can six indistinguishable letters be put in three mail boxes?
Hint: One representation of this is given by a sequence jLLjLjLLLj where the
j's represent the partitions for the boxes and the L's the letters. Any possible
way can be so described. Note that we need two bars at the ends and the
remaining two bars and the six L's can be put in any order.
23 Using the method for the hint in Exercise 22, show that r indistinguishable
objects can be put in n boxes in
n + r  1
n  1

=
n + r  1
r

dierent ways.
24 A travel bureau estimates that when 20 tourists go to a resort with ten hotels
they distribute themselves as if the bureau were putting 20 indistinguishable
objects into ten distinguishable boxes. Assuming this model is correct, nd
the probability that no hotel is left vacant when the rst group of 20 tourists
arrives.
25 An elevator takes on six passengers and stops at ten 
oors. We can assign",prob.pdf
"arrives.
25 An elevator takes on six passengers and stops at ten 
oors. We can assign
two dierent equiprobable measures for the ways that the passengers are dis-
charged: (a) we consider the passengers to be distinguishable or (b) we con-
sider them to be indistinguishable (see Exercise 23 for this case). For each
case, calculate the probability that all the passengers get o at dierent 
oors.
26 You are playing heads or tails with Prosser but you suspect that his coin is
unfair. Von Neumann suggested that you proceed as follows: Toss Prosser's
coin twice. If the outcome is HT call the result win. if it is TH call the result
lose. If it is TT or HH ignore the outcome and toss Prosser's coin twice again.
Keep going until you get either an HT or a TH and call the result win or lose
in a single play. Repeat this procedure for each play. Assume that Prosser's
coin turns up heads with probability p.
(a) Find the probability of HT, TH, HH, TT with two tosses of Prosser's
coin.",prob.pdf
"coin turns up heads with probability p.
(a) Find the probability of HT, TH, HH, TT with two tosses of Prosser's
coin.
(b) Using part (a), show that the probability of a win on any one play is 1/2,
no matter what p is.
27 John claims that he has extrasensory powers and can tell which of two symbols
is on a card turned face down (see Example 3.11). To test his ability he is
asked to do this for a sequence of trials. Let the null hypothesis be that he is
just guessing, so that the probability is 1/2 of his getting it right each time,
and let the alternative hypothesis be that he can name the symbol correctly
more than half the time. Devise a test with the property that the probability
of a type 1 error is less than .05 and the probability of a type 2 error is less
than .05 if John can name the symbol correctly 75 percent of the time.",prob.pdf
"3.2. COMBINATIONS 117
28 In Example 3.11 assume the alternative hypothesis is that p = :8 and that it
is desired to have the probability of each type of error less than .01. Use the
program PowerCurve to determine values of n and m that will achieve this.
Choose n as small as possible.
29 A drug is assumed to be eective with an unknown probability p. To estimate
p the drug is given to n patients. It is found to be eective for m patients.
The method of maximum likelihood for estimating p states that we should
choose the value for p that gives the highest probability of getting what we
got on the experiment. Assuming that the experiment can be considered as a
Bernoulli trials process with probability p for success, show that the maximum
likelihood estimate for p is the proportion m=n of successes.
30 Recall that in the World Series the rst team to win four games wins the
series. The series can go at most seven games. Assume that the Red Sox",prob.pdf
"30 Recall that in the World Series the rst team to win four games wins the
series. The series can go at most seven games. Assume that the Red Sox
and the Mets are playing the series. Assume that the Mets win each game
with probability p. Fermat observed that even though the series might not go
seven games, the probability that the Mets win the series is the same as the
probability that they win four or more game in a series that was forced to go
seven games no matter who wins the individual games.
(a) Using the program PowerCurve of Example 3.11 nd the probability
that the Mets win the series for the cases p = :5, p = :6, p = :7.
(b) Assume that the Mets have probability .6 of winning each game. Use
the program PowerCurve to nd a value of n so that, if the series goes
to the rst team to win more than half the games, the Mets will have a
95 percent chance of winning the series. Choose n as small as possible.",prob.pdf
"to the rst team to win more than half the games, the Mets will have a
95 percent chance of winning the series. Choose n as small as possible.
31 Each of the four engines on an airplane functions correctly on a given 
ight
with probability .99, and the engines function independently of each other.
Assume that the plane can make a safe landing if at least two of its engines
are functioning correctly. What is the probability that the engines will allow
for a safe landing?
32 A small boy is lost coming down Mount Washington. The leader of the search
team estimates that there is a probability p that he came down on the east
side and a probability 1  p that he came down on the west side. He has n
people in his search team who will search independently and, if the boy is
on the side being searched, each member will nd the boy with probability
u. Determine how he should divide the n people into two groups to search",prob.pdf
"on the side being searched, each member will nd the boy with probability
u. Determine how he should divide the n people into two groups to search
the two sides of the mountain so that he will have the highest probability of
nding the boy. How does this depend on u?
*33 2n balls are chosen at random from a total of 2n red balls and 2n blue balls.
Find a combinatorial expression for the probability that the chosen balls are
equally divided in color. Use Stirling's formula to estimate this probability.",prob.pdf
"118 CHAPTER 3. COMBINATORICS
Using BinomialProbabilities, compare the exact value with Stirling's ap-
proximation for n = 20.
34 Assume that every time you buy a box of Wheaties, you receive one of the
pictures of the n players on the New York Yankees. Over a period of time,
you buy m  n boxes of Wheaties.
(a) Use Theorem 3.8 to show that the probability that you get all n pictures
is
1 
n
1
n  1
n
m
+
n
2
n  2
n
m
 
 
 

+ (1)n1
 n
n  1
1
n
m
:
Hint: Let Ek be the event that you do not get the kth player's picture.
(b) Write a computer program to compute this probability. Use this program
to nd, for given n, the smallest value of m which will give probability
 :5 of getting all n pictures. Consider n = 50, 100, and 150 and show
that m = n log n + n log 2 is a good estimate for the number of boxes
needed. (For a derivation of this estimate, see Feller.26)
*35 Prove the following binomial identity
2n
n

=
nX
j=0
n
j
2
:",prob.pdf
"needed. (For a derivation of this estimate, see Feller.26)
*35 Prove the following binomial identity
2n
n

=
nX
j=0
n
j
2
:
Hint: Consider an urn with n red balls and n blue balls inside. Show that
each side of the equation equals the number of ways to choose n balls from
the urn.
36 Let j and n be positive integers, with j  n. An experiment consists of
choosing, at random, a j-tuple of positive integers whose sum is at most n.
(a) Find the size of the sample space. Hint: Consider n indistinguishable
balls placed in a row. Place j markers between consecutive pairs of balls,
with no two markers between the same pair of balls. (We also allow one
of the n markers to be placed at the end of the row of balls.) Show that
there is a 1-1 correspondence between the set of possible positions for
the markers and the set of j-tuples whose size we are trying to count.
(b) Find the probability that the j-tuple selected contains at least one 1.",prob.pdf
"the markers and the set of j-tuples whose size we are trying to count.
(b) Find the probability that the j-tuple selected contains at least one 1.
37 Let n (mod m) denote the remainder when the integer n is divided by the
integer m. Write a computer program to compute the numbers
n
j


(mod m)
where
n
j


is a binomial coecient and m is an integer. You can do this by
using the recursion relations for generating binomial coecients, doing all the
26W. Feller, Introduction to Probability Theory and its Applications, vol. I, 3rd ed. (New York:
John Wiley & Sons, 1968), p. 106.",prob.pdf
"3.2. COMBINATIONS 119
arithmetic using the basic function mod(n; m). Try to write your program to
make as large a table as possible. Run your program for the cases m = 2 to 7.
Do you see any patterns? In particular, for the case m = 2 and n a power
of 2, verify that all the entries in the (n  1)st row are 1. (The corresponding
binomial numbers are odd.) Use your pictures to explain why this is true.
38 Lucas27 proved the following general result relating to Exercise 37. If p is
any prime number, then
n
j


(mod p) can be found as follows: Expand n
and j in base p as n = s0 + s1p + s2p2 + 
 
 
 + skpk and j = r0 + r1p +
r2p2 + 
 
 
 +rkpk, respectively. (Here k is chosen large enough to represent all
numbers from 0 to n in base p using k digits.) Let s = (s0; s1; s2; : : : ; sk) and
r = (r0; r1; r2; : : : ; rk). Then
n
j

(mod p) =
kY
i=0
si
ri

(mod p) :
For example, if p = 7, n = 12, and j = 9, then
12 = 5 
 70 + 1 
 71 ;
9 = 2 
 70 + 1 
 71 ;
so that
s = (5; 1) ;
r = (2; 1) ;",prob.pdf
"i=0
si
ri

(mod p) :
For example, if p = 7, n = 12, and j = 9, then
12 = 5 
 70 + 1 
 71 ;
9 = 2 
 70 + 1 
 71 ;
so that
s = (5; 1) ;
r = (2; 1) ;
and this result states that
12
9

(mod p) =
5
2
1
1

(mod 7) :
Since
12
9


= 220 = 3 (mod 7), and
5
2


= 10 = 3 (mod 7), we see that the
result is correct for this example.
Show that this result implies that, for p = 2, the (pk 1)st row of your triangle
in Exercise 37 has no zeros.
39 Prove that the probability of exactly n heads in 2n tosses of a fair coin is
given by the product of the odd numbers up to 2n1 divided by the product
of the even numbers up to 2n.
40 Let n be a positive integer, and assume that j is a positive integer not exceed-
ing n=2. Show that in Theorem 3.5, if one alternates the multiplications and
divisions, then all of the intermediate values in the calculation are integers.
Show also that none of these intermediate values exceed the nal value.",prob.pdf
"Show also that none of these intermediate values exceed the nal value.
27E. Lucas, \Theorie des Functions Numeriques Simplement Periodiques,"" American J. Math.,
vol. 1 (1878), pp. 184-240, 289-321.",prob.pdf
"120 CHAPTER 3. COMBINATORICS
3.3 Card Shuing
Much of this section is based upon an article by Brad Mann,28 which is an exposition
of an article by David Bayer and Persi Diaconis.29
Rie Shues
Given a deck of n cards, how many times must we shue it to make it \random""?
Of course, the answer depends upon the method of shuing which is used and what
we mean by \random."" We shall begin the study of this question by considering a
standard model for the rie shue.
We begin with a deck of n cards, which we will assume are labelled in increasing
order with the integers from 1 to n. A rie shue consists of a cut of the deck into
two stacks and an interleaving of the two stacks. For example, if n = 6, the initial
ordering is (1; 2; 3; 4; 5; 6), and a cut might occur between cards 2 and 3. This gives
rise to two stacks, namely (1; 2) and (3; 4; 5; 6). These are interleaved to form a
new ordering of the deck. For example, these two stacks might form the ordering",prob.pdf
"new ordering of the deck. For example, these two stacks might form the ordering
(1; 3; 4; 2; 5; 6). In order to discuss such shues, we need to assign a probability
distribution to the set of all possible shues. There are several reasonable ways in
which this can be done. We will give several dierent assignment strategies, and
show that they are equivalent. (This does not mean that this assignment is the
only reasonable one.) First, we assign the binomial probability b(n; 1=2; k) to the
event that the cut occurs after the kth card. Next, we assume that all possible
interleavings, given a cut, are equally likely. Thus, to complete the assignment
of probabilities, we need to determine the number of possible interleavings of two
stacks of cards, with k and n  k cards, respectively.
We begin by writing the second stack in a line, with spaces in between each
pair of consecutive cards, and with spaces at the beginning and end (so there are",prob.pdf
"pair of consecutive cards, and with spaces at the beginning and end (so there are
n  k + 1 spaces). We choose, with replacement, k of these spaces, and place the
cards from the rst stack in the chosen spaces. This can be done in
n
k

ways. Thus, the probability of a given interleaving should be
1n
k

 :
Next, we note that if the new ordering is not the identity ordering, it is the
result of a unique cut-interleaving pair. If the new ordering is the identity, it is the
result of any one of n + 1 cut-interleaving pairs.
We dene a rising sequence in an ordering to be a maximal subsequence of
consecutive integers in increasing order. For example, in the ordering
(2; 3; 5; 1; 4; 7; 6) ;
28B. Mann, \How Many Times Should You Shue a Deck of Cards?"", UMAP Journal, vol. 15,
no. 4 (1994), pp. 303{331.
29D. Bayer and P. Diaconis, \Trailing the Dovetail Shue to its Lair,"" Annals of Applied Prob-
ability, vol. 2, no. 2 (1992), pp. 294{313.",prob.pdf
"3.3. CARD SHUFFLING 121
there are 4 rising sequences; they are (1), (2; 3; 4), (5; 6), and (7). It is easy to see
that an ordering is the result of a rie shue applied to the identity ordering if
and only if it has no more than two rising sequences. (If the ordering has two rising
sequences, then these rising sequences correspond to the two stacks induced by the
cut, and if the ordering has one rising sequence, then it is the identity ordering.)
Thus, the sample space of orderings obtained by applying a rie shue to the
identity ordering is naturally described as the set of all orderings with at most two
rising sequences.
It is now easy to assign a probability distribution to this sample space. Each
ordering with two rising sequences is assigned the value
b(n; 1=2; k)n
k

 = 1
2n ;
and the identity ordering is assigned the value
n + 1
2n :
There is another way to view a rie shue. We can imagine starting with a",prob.pdf
"k

 = 1
2n ;
and the identity ordering is assigned the value
n + 1
2n :
There is another way to view a rie shue. We can imagine starting with a
deck cut into two stacks as before, with the same probabilities assignment as before
i.e., the binomial distribution. Once we have the two stacks, we take cards, one by
one, o of the bottom of the two stacks, and place them onto one stack. If there
are k1 and k2 cards, respectively, in the two stacks at some point in this process,
then we make the assumption that the probabilities that the next card to be taken
comes from a given stack is proportional to the current stack size. This implies that
the probability that we take the next card from the rst stack equals
k1
k1 + k2
;
and the corresponding probability for the second stack is
k2
k1 + k2
:
We shall now show that this process assigns the uniform probability to each of the
possible interleavings of the two stacks.",prob.pdf
"k2
k1 + k2
:
We shall now show that this process assigns the uniform probability to each of the
possible interleavings of the two stacks.
Suppose, for example, that an interleaving came about as the result of choosing
cards from the two stacks in some order. The probability that this result occurred
is the product of the probabilities at each point in the process, since the choice
of card at each point is assumed to be independent of the previous choices. Each
factor of this product is of the form
ki
k1 + k2
;
where i = 1 or 2, and the denominator of each factor equals the number of cards left
to be chosen. Thus, the denominator of the probability is just n!. At the moment
when a card is chosen from a stack that has i cards in it, the numerator of the",prob.pdf
"122 CHAPTER 3. COMBINATORICS
corresponding factor in the probability is i, and the number of cards in this stack
decreases by 1. Thus, the numerator is seen to be k!(n  k)!, since all cards in both
stacks are eventually chosen. Therefore, this process assigns the probability
1n
k


to each possible interleaving.
We now turn to the question of what happens when we rie shue s times. It
should be clear that if we start with the identity ordering, we obtain an ordering
with at most 2s rising sequences, since a rie shue creates at most two rising
sequences from every rising sequence in the starting ordering. In fact, it is not hard
to see that each such ordering is the result of s rie shues. The question becomes,
then, in how many ways can an ordering with r rising sequences come about by
applying s rie shues to the identity ordering? In order to answer this question,
we turn to the idea of an a-shue.
a-Shues",prob.pdf
"applying s rie shues to the identity ordering? In order to answer this question,
we turn to the idea of an a-shue.
a-Shues
There are several ways to visualize an a-shue. One way is to imagine a creature
with a hands who is given a deck of cards to rie shue. The creature naturally
cuts the deck into a stacks, and then ries them together. (Imagine that!) Thus,
the ordinary rie shue is a 2-shue. As in the case of the ordinary 2-shue, we
allow some of the stacks to have 0 cards. Another way to visualize an a-shue is
to think about its inverse, called an a-unshue. This idea is described in the proof
of the next theorem.
We will now show that an a-shue followed by a b-shue is equivalent to an ab-
shue. This means, in particular, that s rie shues in succession are equivalent
to one 2s-shue. This equivalence is made precise by the following theorem.
Theorem 3.9 Let a and b be two positive integers. Let Sa;b be the set of all ordered",prob.pdf
"Theorem 3.9 Let a and b be two positive integers. Let Sa;b be the set of all ordered
pairs in which the rst entry is an a-shue and the second entry is a b-shue. Let
Sab be the set of all ab-shues. Then there is a 1-1 correspondence between Sa;b
and Sab with the following property. Suppose that (T1; T2) corresponds to T3. If
T1 is applied to the identity ordering, and T2 is applied to the resulting ordering,
then the nal ordering is the same as the ordering that is obtained by applying T3
to the identity ordering.
Proof. The easiest way to describe the required correspondence is through the idea
of an unshue. An a-unshue begins with a deck of n cards. One by one, cards are
taken from the top of the deck and placed, with equal probability, on the bottom
of any one of a stacks, where the stacks are labelled from 0 to a1. After all of the
cards have been distributed, we combine the stacks to form one stack by placing",prob.pdf
"cards have been distributed, we combine the stacks to form one stack by placing
stack i on top of stack i+1, for 0  i  a1. It is easy to see that if one starts with
a deck, there is exactly one way to cut the deck to obtain the a stacks generated by
the a-unshue, and with these a stacks, there is exactly one way to interleave them",prob.pdf
"3.3. CARD SHUFFLING 123
to obtain the deck in the order that it was in before the unshue was performed.
Thus, this a-unshue corresponds to a unique a-shue, and this a-shue is the
inverse of the original a-unshue.
If we apply an ab-unshue U3 to a deck, we obtain a set of ab stacks, which
are then combined, in order, to form one stack. We label these stacks with ordered
pairs of integers, where the rst coordinate is between 0 and a  1, and the second
coordinate is between 0 and b  1. Then we label each card with the label of its
stack. The number of possible labels is ab, as required. Using this labelling, we
can describe how to nd a b-unshue and an a-unshue, such that if these two
unshues are applied in this order to the deck, we obtain the same set of ab stacks
as were obtained by the ab-unshue.
To obtain the b-unshue U2, we sort the deck into b stacks, with the ith stack
containing all of the cards with second coordinate i, for 0  i  b  1. Then these",prob.pdf
"containing all of the cards with second coordinate i, for 0  i  b  1. Then these
stacks are combined to form one stack. The a-unshue U1 proceeds in the same
manner, except that the rst coordinates of the labels are used. The resulting a
stacks are then combined to form one stack.
The above description shows that the cards ending up on top are all those
labelled (0; 0). These are followed by those labelled (0; 1); (0; 2); : : : ; (0; b 
1); (1; 0); (1; 1); : : : ; (a  1; b  1). Furthermore, the relative order of any pair
of cards with the same labels is never altered. But this is exactly the same as an
ab-unshue, if, at the beginning of such an unshue, we label each of the cards
with one of the labels (0; 0); (0; 1); : : : ; (0; b  1); (1; 0); (1; 1); : : : ; (a  1; b  1).
This completes the proof. 2
In Figure 3.11, we show the labels for a 2-unshue of a deck with 10 cards.
There are 4 cards with the label 0 and 6 cards with the label 1, so if the 2-unshue",prob.pdf
"There are 4 cards with the label 0 and 6 cards with the label 1, so if the 2-unshue
is performed, the rst stack will have 4 cards and the second stack will have 6 cards.
When this unshue is performed, the deck ends up in the identity ordering.
In Figure 3.12, we show the labels for a 4-unshue of the same deck (because
there are four labels being used). This gure can also be regarded as an example of
a pair of 2-unshues, as described in the proof above. The rst 2-unshue will use
the second coordinate of the labels to determine the stacks. In this case, the two
stacks contain the cards whose values are
f5; 1; 6; 2; 7g and f8; 9; 3; 4; 10g :
After this 2-unshue has been performed, the deck is in the order shown in Fig-
ure 3.11, as the reader should check. If we wish to perform a 4-unshue on the
deck, using the labels shown, we sort the cards lexicographically, obtaining the four
stacks
f1; 2g; f3; 4g; f5; 6; 7g; and f8; 9; 10g :",prob.pdf
"deck, using the labels shown, we sort the cards lexicographically, obtaining the four
stacks
f1; 2g; f3; 4g; f5; 6; 7g; and f8; 9; 10g :
When these stacks are combined, we once again obtain the identity ordering of the
deck. The point of the above theorem is that both sorting procedures always lead
to the same initial ordering.",prob.pdf
"124 CHAPTER 3. COMBINATORICS
Figure 3.11: Before a 2-unshue.
Figure 3.12: Before a 4-unshue.",prob.pdf
"3.3. CARD SHUFFLING 125
Theorem 3.10 If D is any ordering that is the result of applying an a-shue and
then a b-shue to the identity ordering, then the probability assigned to D by this
pair of operations is the same as the probability assigned to D by the process of
applying an ab-shue to the identity ordering.
Proof. Call the sample space of a-shues Sa. If we label the stacks by the integers
from 0 to a  1, then each cut-interleaving pair, i.e., shue, corresponds to exactly
one n-digit base a integer, where the ith digit in the integer is the stack of which
the ith card is a member. Thus, the number of cut-interleaving pairs is equal to
the number of n-digit base a integers, which is an. Of course, not all of these
pairs leads to dierent orderings. The number of pairs leading to a given ordering
will be discussed later. For our purposes it is enough to point out that it is the
cut-interleaving pairs that determine the probability assignment.",prob.pdf
"will be discussed later. For our purposes it is enough to point out that it is the
cut-interleaving pairs that determine the probability assignment.
The previous theorem shows that there is a 1-1 correspondence between Sa;b and
Sab. Furthermore, corresponding elements give the same ordering when applied to
the identity ordering. Given any ordering D, let m1 be the number of elements
of Sa;b which, when applied to the identity ordering, result in D. Let m2 be the
number of elements of Sab which, when applied to the identity ordering, result in D.
The previous theorem implies that m1 = m2. Thus, both sets assign the probability
m1
(ab)n
to D. This completes the proof. 2
Connection with the Birthday Problem
There is another point that can be made concerning the labels given to the cards
by the successive unshues. Suppose that we 2-unshue an n-card deck until the
labels on the cards are all dierent. It is easy to see that this process produces",prob.pdf
"labels on the cards are all dierent. It is easy to see that this process produces
each permutation with the same probability, i.e., this is a random process. To see
this, note that if the labels become distinct on the sth 2-unshue, then one can
think of this sequence of 2-unshues as one 2s-unshue, in which all of the stacks
determined by the unshue have at most one card in them (remember, the stacks
correspond to the labels). If each stack has at most one card in it, then given any
two cards in the deck, it is equally likely that the rst card has a lower or a higher
label than the second card. Thus, each possible ordering is equally likely to result
from this 2s-unshue.
Let T be the random variable that counts the number of 2-unshues until all
labels are distinct. One can think of T as giving a measure of how long it takes in
the unshuing process until randomness is reached. Since shuing and unshuing",prob.pdf
"the unshuing process until randomness is reached. Since shuing and unshuing
are inverse processes, T also measures the number of shues necessary to achieve
randomness. Suppose that we have an n-card deck, and we ask for P(T  s). This
equals 1  P(T > s). But T > s if and only if it is the case that not all of the
labels after s 2-unshues are distinct. This is just the birthday problem; we are
asking for the probability that at least two people have the same birthday, given",prob.pdf
"126 CHAPTER 3. COMBINATORICS
that we have n people and there are 2s possible birthdays. Using our formula from
Example 3.3, we nd that
P(T > s) = 1 
2s
n
n!
2sn : (3.4)
In Chapter 6, we will dene the average value of a random variable. Using this
idea, and the above equation, one can calculate the average value of the random
variable T (see Exercise 6.1.41). For example, if n = 52, then the average value of
T is about 11.7. This means that, on the average, about 12 rie shues are needed
for the process to be considered random.
Cut-Interleaving Pairs and Orderings
As was noted in the proof of Theorem 3.10, not all of the cut-interleaving pairs lead
to dierent orderings. However, there is an easy formula which gives the number of
such pairs that lead to a given ordering.
Theorem 3.11 If an ordering of length n has r rising sequences, then the number
of cut-interleaving pairs under an a-shue of the identity ordering which lead to
the ordering is n + a  r
n

:",prob.pdf
"of cut-interleaving pairs under an a-shue of the identity ordering which lead to
the ordering is n + a  r
n

:
Proof. To see why this is true, we need to count the number of ways in which the
cut in an a-shue can be performed which will lead to a given ordering with r rising
sequences. We can disregard the interleavings, since once a cut has been made, at
most one interleaving will lead to a given ordering. Since the given ordering has
r rising sequences, r  1 of the division points in the cut are determined. The
remaining a  1  (r  1) = a  r division points can be placed anywhere. The
number of places to put these remaining division points is n + 1 (which is the
number of spaces between the consecutive pairs of cards, including the positions at
the beginning and the end of the deck). These places are chosen with repetition
allowed, so the number of ways to make these choices is
n + a  r
a  r

=
n + a  r
n

:",prob.pdf
"allowed, so the number of ways to make these choices is
n + a  r
a  r

=
n + a  r
n

:
In particular, this means that if D is an ordering that is the result of applying
an a-shue to the identity ordering, and if D has r rising sequences, then the
probability assigned to D by this process is
n+ar
n


an :
This completes the proof. 2",prob.pdf
"3.3. CARD SHUFFLING 127
The above theorem shows that the essential information about the probability
assigned to an ordering under an a-shue is just the number of rising sequences in
the ordering. Thus, if we determine the number of orderings which contain exactly
r rising sequences, for each r between 1 and n, then we will have determined the
distribution function of the random variable which consists of applying a random
a-shue to the identity ordering.
The number of orderings of f1; 2; : : :; ng with r rising sequences is denoted by
A(n; r), and is called an Eulerian number. There are many ways to calculate the
values of these numbers; the following theorem gives one recursive method which
follows immediately from what we already know about a-shues.
Theorem 3.12 Let a and n be positive integers. Then
an =
aX
r=1
n + a  r
n

A(n; r) : (3.5)
Thus,
A(n; a) = an 
a1X
r=1
n + a  r
n

A(n; r) :
In addition,
A(n; 1) = 1 :",prob.pdf
"an =
aX
r=1
n + a  r
n

A(n; r) : (3.5)
Thus,
A(n; a) = an 
a1X
r=1
n + a  r
n

A(n; r) :
In addition,
A(n; 1) = 1 :
Proof. The second equation can be used to calculate the values of the Eulerian
numbers, and follows immediately from the Equation 3.5. The last equation is
a consequence of the fact that the only ordering of f1; 2; : : :; ng with one rising
sequence is the identity ordering. Thus, it remains to prove Equation 3.5. We will
count the set of a-shues of a deck with n cards in two ways. First, we know that
there are an such shues (this was noted in the proof of Theorem 3.10). But there
are A(n; r) orderings of f1; 2; : : :; ng with r rising sequences, and Theorem 3.11
states that for each such ordering, there are exactly
n + a  r
n

cut-interleaving pairs that lead to the ordering. Therefore, the right-hand side of
Equation 3.5 counts the set of a-shues of an n-card deck. This completes the
proof. 2
Random Orderings and Random Processes",prob.pdf
"Equation 3.5 counts the set of a-shues of an n-card deck. This completes the
proof. 2
Random Orderings and Random Processes
We now turn to the second question that was asked at the beginning of this section:
What do we mean by a \random"" ordering? It is somewhat misleading to think
about a given ordering as being random or not random. If we want to choose a
random ordering from the set of all orderings of f1; 2; : : :; ng, we mean that we
want every ordering to be chosen with the same probability, i.e., any ordering is as
\random"" as any other.",prob.pdf
"128 CHAPTER 3. COMBINATORICS
The word \random"" should really be used to describe a process. We will say that
a process that produces an object from a (nite) set of objects is a random process
if each object in the set is produced with the same probability by the process. In
the present situation, the objects are the orderings, and the process which produces
these objects is the shuing process. It is easy to see that no a-shue is really a
random process, since if T1 and T2 are two orderings with a dierent number of
rising sequences, then they are produced by an a-shue, applied to the identity
ordering, with dierent probabilities.
Variation Distance
Instead of requiring that a sequence of shues yield a process which is random, we
will dene a measure that describes how far away a given process is from a random
process. Let X be any process which produces an ordering of f1; 2; : : :; ng. Dene
fX() be the probability that X produces the ordering . (Thus, X can be thought",prob.pdf
"fX() be the probability that X produces the ordering . (Thus, X can be thought
of as a random variable with distribution function f.) Let 
n be the set of all
orderings of f1; 2; : : :; ng. Finally, let u() = 1=j
nj for all  2 
n. The function
u is the distribution function of a process which produces orderings and which is
random. For each ordering  2 
n, the quantity
jfX()  u()j
is the dierence between the actual and desired probabilities that X produces . If
we sum this over all orderings  and call this sum S, we see that S = 0 if and only
if X is random, and otherwise S is positive. It is easy to show that the maximum
value of S is 2, so we will multiply the sum by 1=2 so that the value falls in the
interval [0; 1]. Thus, we obtain the following sum as the formula for the variation
distance between the two processes:
k fX  u k= 1
2
X
2
n
jfX()  u()j :
Now we apply this idea to the case of shuing. We let X be the process of s",prob.pdf
"k fX  u k= 1
2
X
2
n
jfX()  u()j :
Now we apply this idea to the case of shuing. We let X be the process of s
successive rie shues applied to the identity ordering. We know that it is also
possible to think of X as one 2s-shue. We also know that fX is constant on the
set of all orderings with r rising sequences, where r is any positive integer. Finally,
we know the value of fX on an ordering with r rising sequences, and we know how
many such orderings there are. Thus, in this specic case, we have
k fX  u k= 1
2
nX
r=1
A(n; r)




2s + n  r
n

=2ns  1
n!



:
Since this sum has only n summands, it is easy to compute this for moderate sized
values of n. For n = 52, we obtain the list of values given in Table 3.14.
To help in understanding these data, they are shown in graphical form in Fig-
ure 3.13. The program VariationList produces the data shown in both Table 3.14
and Figure 3.13. One sees that until 5 shues have occurred, the output of X is",prob.pdf
"3.3. CARD SHUFFLING 129
Number of Rie Shues Variation Distance
1 1
2 1
3 1
4 0.9999995334
5 0.9237329294
6 0.6135495966
7 0.3340609995
8 0.1671586419
9 0.0854201934
10 0.0429455489
11 0.0215023760
12 0.0107548935
13 0.0053779101
14 0.0026890130
Table 3.14: Distance to the random process.
5 10 15 20
0.2
0.4
0.6
0.8
1
Figure 3.13: Distance to the random process.",prob.pdf
"130 CHAPTER 3. COMBINATORICS
very far from random. After 5 shues, the distance from the random process is
essentially halved each time a shue occurs.
Given the distribution functions fX() and u() as above, there is another
way to view the variation distance k fX  u k. Given any event T (which is a
subset of Sn), we can calculate its probability under the process X and under the
uniform process. For example, we can imagine that T represents the set of all
permutations in which the rst player in a 7-player poker game is dealt a straight

ush (ve consecutive cards in the same suit). It is interesting to consider how
much the probability of this event after a certain number of shues diers from the
probability of this event if all permutations are equally likely. This dierence can
be thought of as describing how close the process X is to the random process with
respect to the event T.
Now consider the event T such that the absolute value of the dierence between",prob.pdf
"respect to the event T.
Now consider the event T such that the absolute value of the dierence between
these two probabilities is as large as possible. It can be shown that this absolute
value is the variation distance between the process X and the uniform process. (The
reader is asked to prove this fact in Exercise 4.)
We have just seen that, for a deck of 52 cards, the variation distance between
the 7-rie shue process and the random process is about :334. It is of interest
to nd an event T such that the dierence between the probabilities that the two
processes produce T is close to :334. An event with this property can be described
in terms of the game called New-Age Solitaire.
New-Age Solitaire
This game was invented by Peter Doyle. It is played with a standard 52-card deck.
We deal the cards face up, one at a time, onto a discard pile. If an ace is encountered,
say the ace of Hearts, we use it to start a Heart pile. Each suit pile must be built",prob.pdf
"say the ace of Hearts, we use it to start a Heart pile. Each suit pile must be built
up in order, from ace to king, using only subsequently dealt cards. Once we have
dealt all of the cards, we pick up the discard pile and continue. We dene the Yin
suits to be Hearts and Clubs, and the Yang suits to be Diamonds and Spades. The
game ends when either both Yin suit piles have been completed, or both Yang suit
piles have been completed. It is clear that if the ordering of the deck is produced
by the random process, then the probability that the Yin suit piles are completed
rst is exactly 1/2.
Now suppose that we buy a new deck of cards, break the seal on the package,
and rie shue the deck 7 times. If one tries this, one nds that the Yin suits win
about 75% of the time. This is 25% more than we would get if the deck were in
truly random order. This deviation is reasonably close to the theoretical maximum
of 33:4% obtained above.",prob.pdf
"truly random order. This deviation is reasonably close to the theoretical maximum
of 33:4% obtained above.
Why do the Yin suits win so often? In a brand new deck of cards, the suits are
in the following order, from top to bottom: ace through king of Hearts, ace through
king of Clubs, king through ace of Diamonds, and king through ace of Spades. Note
that if the cards were not shued at all, then the Yin suit piles would be completed
on the rst pass, before any Yang suit cards are even seen. If we were to continue
playing the game until the Yang suit piles are completed, it would take 13 passes",prob.pdf
"3.3. CARD SHUFFLING 131
through the deck to do this. Thus, one can see that in a new deck, the Yin suits are
in the most advantageous order and the Yang suits are in the least advantageous
order. Under 7 rie shues, the relative advantage of the Yin suits over the Yang
suits is preserved to a certain extent.
Exercises
1 Given any ordering  of f1; 2; : : :; ng, we can dene 1, the inverse ordering
of , to be the ordering in which the ith element is the position occupied by
i in . For example, if  = (1; 3; 5; 2; 4; 7; 6), then 1 = (1; 4; 2; 5; 3; 7; 6). (If
one thinks of these orderings as permutations, then 1 is the inverse of .)
A fall occurs between two positions in an ordering if the left position is occu-
pied by a larger number than the right position. It will be convenient to say
that every ordering has a fall after the last position. In the above example,
1 has four falls. They occur after the second, fourth, sixth, and seventh",prob.pdf
"1 has four falls. They occur after the second, fourth, sixth, and seventh
positions. Prove that the number of rising sequences in an ordering  equals
the number of falls in 1.
2 Show that if we start with the identity ordering of f1; 2; : : :; ng, then the prob-
ability that an a-shue leads to an ordering with exactly r rising sequences
equals n+ar
n


an A(n; r) ;
for 1  r  a.
3 Let D be a deck of n cards. We have seen that there are an a-shues of D.
A coding of the set of a-unshues was given in the proof of Theorem 3.9. We
will now give a coding of the a-shues which corresponds to the coding of
the a-unshues. Let S be the set of all n-tuples of integers, each between 0
and a  1. Let M = (m1; m2; : : :; mn) be any element of S. Let ni be the
number of i's in M, for 0  i  a  1. Suppose that we start with the deck
in increasing order (i.e., the cards are numbered from 1 to n). We label the
rst n0 cards with a 0, the next n1 cards with a 1, etc. Then the a-shue",prob.pdf
"rst n0 cards with a 0, the next n1 cards with a 1, etc. Then the a-shue
corresponding to M is the shue which results in the ordering in which the
cards labelled i are placed in the positions in M containing the label i. The
cards with the same label are placed in these positions in increasing order of
their numbers. For example, if n = 6 and a = 3, let M = (1; 0; 2; 2; 0; 2).
Then n0 = 2; n1 = 1; and n2 = 3. So we label cards 1 and 2 with a 0, card
3 with a 1, and cards 4, 5, and 6 with a 2. Then cards 1 and 2 are placed
in positions 2 and 5, card 3 is placed in position 1, and cards 4, 5, and 6 are
placed in positions 3, 4, and 6, resulting in the ordering (3; 1; 4; 5; 2; 6).
(a) Using this coding, show that the probability that in an a-shue, the
rst card (i.e., card number 1) moves to the ith position, is given by the
following expression:
(a  1)i1ani + (a  2)i1(a  1)ni + 
 
 
 + 1i12ni
an :",prob.pdf
"132 CHAPTER 3. COMBINATORICS
(b) Give an accurate estimate for the probability that in three rie shues
of a 52-card deck, the rst card ends up in one of the rst 26 positions.
Using a computer, accurately estimate the probability of the same event
after seven rie shues.
4 Let X denote a particular process that produces elements of Sn, and let U
denote the uniform process. Let the distribution functions of these processes
be denoted by fX and u, respectively. Show that the variation distance
k fX  u k is equal to
max
TSn
X
2T

fX()  u()

:
Hint: Write the permutations in Sn in decreasing order of the dierence
fX()  u().
5 Consider the process described in the text in which an n-card deck is re-
peatedly labelled and 2-unshued, in the manner described in the proof of
Theorem 3.9. (See Figures 3.10 and 3.13.) The process continues until the
labels are all dierent. Show that the process never terminates until at least
dlog2(n)e unshues have been done.",prob.pdf
"Chapter 4
Conditional Probability
4.1 Discrete Conditional Probability
Conditional Probability
In this section we ask and answer the following question. Suppose we assign a
distribution function to a sample space and then learn that an event E has occurred.
How should we change the probabilities of the remaining events? We shall call the
new probability for an event F the conditional probability of F given E and denote
it by P(FjE).
Example 4.1 An experiment consists of rolling a die once. Let X be the outcome.
Let F be the event fX = 6g, and let E be the event fX > 4g. We assign the
distribution function m(!) = 1=6 for ! = 1; 2; : : :; 6. Thus, P(F) = 1=6. Now
suppose that the die is rolled and we are told that the event E has occurred. This
leaves only two possible outcomes: 5 and 6. In the absence of any other information,
we would still regard these outcomes to be equally likely, so the probability of F
becomes 1/2, making P(FjE) = 1=2. 2",prob.pdf
"we would still regard these outcomes to be equally likely, so the probability of F
becomes 1/2, making P(FjE) = 1=2. 2
Example 4.2 In the Life Table (see Appendix C), one nds that in a population
of 100,000 females, 89.835% can expect to live to age 60, while 57.062% can expect
to live to age 80. Given that a woman is 60, what is the probability that she lives
to age 80?
This is an example of a conditional probability. In this case, the original sample
space can be thought of as a set of 100,000 females. The events E and F are the
subsets of the sample space consisting of all women who live at least 60 years, and
at least 80 years, respectively. We consider E to be the new sample space, and note
that F is a subset of E. Thus, the size of E is 89,835, and the size of F is 57,062.
So, the probability in question equals 57;062=89;835 = :6352. Thus, a woman who
is 60 has a 63.52% chance of living to age 80. 2
133",prob.pdf
"134 CHAPTER 4. CONDITIONAL PROBABILITY
Example 4.3 Consider our voting example from Section 1.2: three candidates A,
B, and C are running for oce. We decided that A and B have an equal chance of
winning and C is only 1/2 as likely to win as A. Let A be the event \A wins,"" B
that \B wins,"" and C that \C wins."" Hence, we assigned probabilities P(A) = 2=5,
P(B) = 2=5, and P(C) = 1=5.
Suppose that before the election is held, A drops out of the race. As in Exam-
ple 4.1, it would be natural to assign new probabilities to the events B and C which
are proportional to the original probabilities. Thus, we would have P(Bj A) = 2=3,
and P(Cj A) = 1=3. It is important to note that any time we assign probabilities
to real-life events, the resulting distribution is only useful if we take into account
all relevant information. In this example, we may have knowledge that most voters
who favor A will vote for C if A is no longer in the race. This will clearly make the",prob.pdf
"who favor A will vote for C if A is no longer in the race. This will clearly make the
probability that C wins greater than the value of 1/3 that was assigned above. 2
In these examples we assigned a distribution function and then were given new
information that determined a new sample space, consisting of the outcomes that
are still possible, and caused us to assign a new distribution function to this space.
We want to make formal the procedure carried out in these examples. Let",prob.pdf
"= f!1; !2; : : : ; !rg be the original sample space with distribution function m(!j)
assigned. Suppose we learn that the event E has occurred. We want to assign a new
distribution function m(!jjE) to 
 to re
ect this fact. Clearly, if a sample point !j
is not in E, we want m(!jjE) = 0. Moreover, in the absence of information to the
contrary, it is reasonable to assume that the probabilities for !k in E should have
the same relative magnitudes that they had before we learned that E had occurred.
For this we require that
m(!kjE) = cm(!k)
for all !k in E, with c some positive constant. But we must also have
X
E
m(!kjE) = c
X
E
m(!k) = 1 :
Thus,
c = 1P
E m(!k) = 1
P(E) :
(Note that this requires us to assume that P(E) > 0.) Thus, we will dene
m(!kjE) = m(!k)
P(E)
for !k in E. We will call this new distribution the conditional distribution given E.
For a general event F, this gives
P(FjE) =
X
F\E
m(!kjE) =
X
F\E
m(!k)
P(E) = P(F \ E)
P(E) :",prob.pdf
"For a general event F, this gives
P(FjE) =
X
F\E
m(!kjE) =
X
F\E
m(!k)
P(E) = P(F \ E)
P(E) :
We call P(FjE) the conditional probability of F occurring given that E occurs,
and compute it using the formula
P(FjE) = P(F \ E)
P(E) :",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 135
(start)

p (ω)ω
ω 
ω 
ω 
ω 
1/2
1/2
l
ll
2/5
3/5
1/2
1/2 b
w
w
b 1/5
3/10
1/4
1/4
Urn Color of ball
1
2
3
4
Figure 4.1: Tree diagram.
Example 4.4 (Example 4.1 continued) Let us return to the example of rolling a
die. Recall that F is the event X = 6, and E is the event X > 4. Note that E \ F
is the event F. So, the above formula gives
P(FjE) = P(F \ E)
P(E)
= 1=6
1=3
= 1
2 ;
in agreement with the calculations performed earlier. 2
Example 4.5 We have two urns, I and II. Urn I contains 2 black balls and 3 white
balls. Urn II contains 1 black ball and 1 white ball. An urn is drawn at random
and a ball is chosen at random from it. We can represent the sample space of this
experiment as the paths through a tree as shown in Figure 4.1. The probabilities
assigned to the paths are also shown.
Let B be the event \a black ball is drawn,"" and I the event \urn I is chosen.""
Then the branch weight 2/5, which is shown on one branch in the gure, can now",prob.pdf
"Then the branch weight 2/5, which is shown on one branch in the gure, can now
be interpreted as the conditional probability P(BjI).
Suppose we wish to calculate P(IjB). Using the formula, we obtain
P(IjB) = P(I \ B)
P(B)
= P(I \ B)
P(B \ I) + P(B \ II)
= 1=5
1=5 + 1=4 = 4
9 :
2",prob.pdf
"136 CHAPTER 4. CONDITIONAL PROBABILITY
(start)

p (ω)ω
ω 
ω 
ω 
ω 
9/20
11/20
b
w
4/9
5/9
5/11
6/11 I
II
II
I 1/5
3/10
1/4
1/4
UrnColor of ball
1
3
2
4
Figure 4.2: Reverse tree diagram.
Bayes Probabilities
Our original tree measure gave us the probabilities for drawing a ball of a given
color, given the urn chosen. We have just calculated the inverse probability that a
particular urn was chosen, given the color of the ball. Such an inverse probability is
called a Bayes probability and may be obtained by a formula that we shall develop
later. Bayes probabilities can also be obtained by simply constructing the tree
measure for the two-stage experiment carried out in reverse order. We show this
tree in Figure 4.2.
The paths through the reverse tree are in one-to-one correspondence with those
in the forward tree, since they correspond to individual outcomes of the experiment,
and so they are assigned the same probabilities. From the forward tree, we nd that",prob.pdf
"and so they are assigned the same probabilities. From the forward tree, we nd that
the probability of a black ball is
1
2 
 2
5 + 1
2 
 1
2 = 9
20 :
The probabilities for the branches at the second level are found by simple divi-
sion. For example, if x is the probability to be assigned to the top branch at the
second level, we must have
9
20 
 x = 1
5
or x = 4=9. Thus, P(IjB) = 4=9, in agreement with our previous calculations. The
reverse tree then displays all of the inverse, or Bayes, probabilities.
Example 4.6 We consider now a problem called the Monty Hall problem. This
has long been a favorite problem but was revived by a letter from Craig Whitaker
to Marilyn vos Savant for consideration in her column in Parade Magazine.1 Craig
wrote:
1Marilyn vos Savant, Ask Marilyn, Parade Magazine, 9 September; 2 December; 17 February
1990, reprinted in Marilyn vos Savant, Ask Marilyn, St. Martins, New York, 1992.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 137
Suppose you're on Monty Hall's Let's Make a Deal! You are given the
choice of three doors, behind one door is a car, the others, goats. You
pick a door, say 1, Monty opens another door, say 3, which has a goat.
Monty says to you \Do you want to pick door 2?"" Is it to your advantage
to switch your choice of doors?
Marilyn gave a solution concluding that you should switch, and if you do, your
probability of winning is 2/3. Several irate readers, some of whom identied them-
selves as having a PhD in mathematics, said that this is absurd since after Monty
has ruled out one door there are only two possible doors and they should still each
have the same probability 1/2 so there is no advantage to switching. Marilyn stuck
to her solution and encouraged her readers to simulate the game and draw their own
conclusions from this. We also encourage the reader to do this (see Exercise 11).",prob.pdf
"conclusions from this. We also encourage the reader to do this (see Exercise 11).
Other readers complained that Marilyn had not described the problem com-
pletely. In particular, the way in which certain decisions were made during a play
of the game were not specied. This aspect of the problem will be discussed in Sec-
tion 4.3. We will assume that the car was put behind a door by rolling a three-sided
die which made all three choices equally likely. Monty knows where the car is, and
always opens a door with a goat behind it. Finally, we assume that if Monty has
a choice of doors (i.e., the contestant has picked the door with the car behind it),
he chooses each door with probability 1/2. Marilyn clearly expected her readers to
assume that the game was played in this manner.
As is the case with most apparent paradoxes, this one can be resolved through
careful analysis. We begin by describing a simpler, related question. We say that",prob.pdf
"careful analysis. We begin by describing a simpler, related question. We say that
a contestant is using the \stay"" strategy if he picks a door, and, if oered a chance
to switch to another door, declines to do so (i.e., he stays with his original choice).
Similarly, we say that the contestant is using the \switch"" strategy if he picks a door,
and, if oered a chance to switch to another door, takes the oer. Now suppose
that a contestant decides in advance to play the \stay"" strategy. His only action
in this case is to pick a door (and decline an invitation to switch, if one is oered).
What is the probability that he wins a car? The same question can be asked about
the \switch"" strategy.
Using the \stay"" strategy, a contestant will win the car with probability 1/3,
since 1/3 of the time the door he picks will have the car behind it. On the other
hand, if a contestant plays the \switch"" strategy, then he will win whenever the",prob.pdf
"hand, if a contestant plays the \switch"" strategy, then he will win whenever the
door he originally picked does not have the car behind it, which happens 2/3 of the
time.
This very simple analysis, though correct, does not quite solve the problem
that Craig posed. Craig asked for the conditional probability that you win if you
switch, given that you have chosen door 1 and that Monty has chosen door 3. To
solve this problem, we set up the problem before getting this information and then
compute the conditional probability given this information. This is a process that
takes place in several stages; the car is put behind a door, the contestant picks a
door, and nally Monty opens a door. Thus it is natural to analyze this using a
tree measure. Here we make an additional assumption that if Monty has a choice",prob.pdf
"138 CHAPTER 4. CONDITIONAL PROBABILITY
1/3
1/3
1/3
1/3
1/3
1/3
1/3
1/3
1/3
1/3
1
1/2
1/2
1/2
1/2
1/2
1
1/2
1
1
1
1
Door opened 
by Monty
Door chosen 
by contestant
Path 
probabilities
Placement 
of car
1
2
3
1
2
3
1
1
2
2
3
3
2
3
3
2
3
3
1
1
2
1
1
2
1/18
1/18
1/18
1/18
1/18
1/9
1/9
1/9
1/18
1/9
1/9
1/9
1/3
1/3
Figure 4.3: The Monty Hall problem.
of doors (i.e., the contestant has picked the door with the car behind it) then he
picks each door with probability 1/2. The assumptions we have made determine the
branch probabilities and these in turn determine the tree measure. The resulting
tree and tree measure are shown in Figure 4.3. It is tempting to reduce the tree's
size by making certain assumptions such as: \Without loss of generality, we will
assume that the contestant always picks door 1."" We have chosen not to make any
such assumptions, in the interest of clarity.
Now the given information, namely that the contestant chose door 1 and Monty",prob.pdf
"such assumptions, in the interest of clarity.
Now the given information, namely that the contestant chose door 1 and Monty
chose door 3, means only two paths through the tree are possible (see Figure 4.4).
For one of these paths, the car is behind door 1 and for the other it is behind door
2. The path with the car behind door 2 is twice as likely as the one with the car
behind door 1. Thus the conditional probability is 2/3 that the car is behind door 2
and 1/3 that it is behind door 1, so if you switch you have a 2/3 chance of winning
the car, as Marilyn claimed.
At this point, the reader may think that the two problems above are the same,
since they have the same answers. Recall that we assumed in the original problem",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 139
1/3
1/3
1/3 1/2
1
Door opened 
by Monty
Door chosen 
by contestant
Unconditional 
probability
Placement 
of car
1
2
1
1 3
3 1/18
1/9
1/3
Conditional
probability
1/3
2/3
Figure 4.4: Conditional probabilities for the Monty Hall problem.
if the contestant chooses the door with the car, so that Monty has a choice of two
doors, he chooses each of them with probability 1/2. Now suppose instead that
in the case that he has a choice, he chooses the door with the larger number with
probability 3/4. In the \switch"" vs. \stay"" problem, the probability of winning
with the \switch"" strategy is still 2/3. However, in the original problem, if the
contestant switches, he wins with probability 4/7. The reader can check this by
noting that the same two paths as before are the only two possible paths in the
tree. The path leading to a win, if the contestant switches, has probability 1/3,",prob.pdf
"tree. The path leading to a win, if the contestant switches, has probability 1/3,
while the path which leads to a loss, if the contestant switches, has probability 1/4.
2
Independent Events
It often happens that the knowledge that a certain event E has occurred has no eect
on the probability that some other event F has occurred, that is, that P(FjE) =
P(F). One would expect that in this case, the equation P(EjF) = P(E) would
also be true. In fact (see Exercise 1), each equation implies the other. If these
equations are true, we might say the F is independent of E. For example, you
would not expect the knowledge of the outcome of the rst toss of a coin to change
the probability that you would assign to the possible outcomes of the second toss,
that is, you would not expect that the second toss depends on the rst. This idea
is formalized in the following denition of independent events.
Denition 4.1 Let E and F be two events. We say that they are independent if",prob.pdf
"is formalized in the following denition of independent events.
Denition 4.1 Let E and F be two events. We say that they are independent if
either 1) both events have positive probability and
P(EjF) = P(E) and P(FjE) = P(F) ;
or 2) at least one of the events has probability 0. 2",prob.pdf
"140 CHAPTER 4. CONDITIONAL PROBABILITY
As noted above, if both P(E) and P(F) are positive, then each of the above
equations imply the other, so that to see whether two events are independent, only
one of these equations must be checked (see Exercise 1).
The following theorem provides another way to check for independence.
Theorem 4.1 Two events E and F are independent if and only if
P(E \ F) = P(E)P(F) :
Proof. If either event has probability 0, then the two events are independent and
the above equation is true, so the theorem is true in this case. Thus, we may assume
that both events have positive probability in what follows. Assume that E and F
are independent. Then P(EjF) = P(E), and so
P(E \ F) = P(EjF)P(F)
= P(E)P(F) :
Assume next that P(E \ F) = P(E)P(F). Then
P(EjF) = P(E \ F)
P(F) = P(E) :
Also,
P(FjE) = P(F \ E)
P(E) = P(F) :
Therefore, E and F are independent. 2
Example 4.7 Suppose that we have a coin which comes up heads with probability",prob.pdf
"P(FjE) = P(F \ E)
P(E) = P(F) :
Therefore, E and F are independent. 2
Example 4.7 Suppose that we have a coin which comes up heads with probability
p, and tails with probability q. Now suppose that this coin is tossed twice. Using
a frequency interpretation of probability, it is reasonable to assign to the outcome
(H; H) the probability p2, to the outcome (H; T) the probability pq, and so on. Let
E be the event that heads turns up on the rst toss and F the event that tails
turns up on the second toss. We will now check that with the above probability
assignments, these two events are independent, as expected. We have P(E) =
p2 + pq = p, P(F) = pq + q2 = q. Finally P(E \ F) = pq, so P(E \ F) =
P(E)P(F). 2
Example 4.8 It is often, but not always, intuitively clear when two events are
independent. In Example 4.7, let A be the event \the rst toss is a head"" and B
the event \the two outcomes are the same."" Then
P(BjA) = P(B \ A)
P(A) = PfHHg
PfHH,HTg = 1=4
1=2 = 1
2 = P(B):",prob.pdf
"the event \the two outcomes are the same."" Then
P(BjA) = P(B \ A)
P(A) = PfHHg
PfHH,HTg = 1=4
1=2 = 1
2 = P(B):
Therefore, A and B are independent, but the result was not so obvious. 2",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 141
Example 4.9 Finally, let us give an example of two events that are not indepen-
dent. In Example 4.7, let I be the event \heads on the rst toss"" and J the event
\two heads turn up."" Then P(I) = 1=2 and P(J) = 1=4. The event I\J is the event
\heads on both tosses"" and has probability 1=4. Thus, I and J are not independent
since P(I)P(J) = 1=8 6= P(I \ J). 2
We can extend the concept of independence to any nite set of events A1, A2,
. . . , An.
Denition 4.2 A set of events fA1; A2; : : : ; Ang is said to be mutually indepen-
dent if for any subset fAi; Aj; : : : ; Amg of these events we have
P(Ai \ Aj \ 
 
 
 \ Am) = P(Ai)P(Aj) 
 
 
P(Am);
or equivalently, if for any sequence A1, A2, . .. , An with Aj = Aj or ~Aj,
P( A1 \ A2 \ 
 
 
 \ An) = P( A1)P( A2) 
 
 
 P( An):
(For a proof of the equivalence in the case n = 3, see Exercise 33.) 2
Using this terminology, it is a fact that any sequence (S; S; F; F; S; : : : ; S) of possible",prob.pdf
"Using this terminology, it is a fact that any sequence (S; S; F; F; S; : : : ; S) of possible
outcomes of a Bernoulli trials process forms a sequence of mutually independent
events.
It is natural to ask: If all pairs of a set of events are independent, is the whole
set mutually independent? The answer is not necessarily, and an example is given
in Exercise 7.
It is important to note that the statement
P(A1 \ A2 \ 
 
 
 \ An) = P(A1)P(A2) 
 
 
 P(An)
does not imply that the events A1, A2, . . . , An are mutually independent (see
Exercise 8).
Joint Distribution Functions and Independence of Random
Variables
It is frequently the case that when an experiment is performed, several dierent
quantities concerning the outcomes are investigated.
Example 4.10 Suppose we toss a coin three times. The basic random variable
X corresponding to this experiment has eight possible outcomes, which are the
ordered triples consisting of H's and T's. We can also dene the random variable",prob.pdf
"ordered triples consisting of H's and T's. We can also dene the random variable
Xi, for i = 1; 2; 3, to be the outcome of the ith toss. If the coin is fair, then we
should assign the probability 1/8 to each of the eight possible outcomes. Thus, the
distribution functions of X1, X2, and X3 are identical; in each case they are dened
by m(H) = m(T) = 1=2. 2",prob.pdf
"142 CHAPTER 4. CONDITIONAL PROBABILITY
If we have several random variables X1; X2; : : : ; Xn which correspond to a given
experiment, then we can consider the joint random variable X = (X1; X2; : : : ; Xn)
dened by taking an outcome ! of the experiment, and writing, as an n-tuple, the
corresponding n outcomes for the random variables X1; X2; : : :; Xn. Thus, if the
random variable Xi has, as its set of possible outcomes the set Ri, then the set of
possible outcomes of the joint random variable X is the Cartesian product of the
Ri's, i.e., the set of all n-tuples of possible outcomes of the Xi's.
Example 4.11 (Example 4.10 continued) In the coin-tossing example above, let
Xi denote the outcome of the ith toss. Then the joint random variable X =
(X1; X2; X3) has eight possible outcomes.
Suppose that we now dene Yi, for i = 1; 2; 3, as the number of heads which
occur in the rst i tosses. Then Yi has f0; 1; : : :; ig as possible outcomes, so at rst",prob.pdf
"occur in the rst i tosses. Then Yi has f0; 1; : : :; ig as possible outcomes, so at rst
glance, the set of possible outcomes of the joint random variable Y = (Y1; Y2; Y3)
should be the set
f(a1; a2; a3) : 0  a1  1; 0  a2  2; 0  a3  3g :
However, the outcome (1; 0; 1) cannot occur, since we must have a1  a2  a3. The
solution to this problem is to dene the probability of the outcome (1; 0; 1) to be 0.
In addition, we must have ai+1  ai  1 for i = 1; 2.
We now illustrate the assignment of probabilities to the various outcomes for
the joint random variables X and Y . In the rst case, each of the eight outcomes
should be assigned the probability 1/8, since we are assuming that we have a fair
coin. In the second case, since Yi has i + 1 possible outcomes, the set of possible
outcomes has size 24. Only eight of these 24 outcomes can actually occur, namely
the ones satisfying a1  a2  a3. Each of these outcomes corresponds to exactly",prob.pdf
"the ones satisfying a1  a2  a3. Each of these outcomes corresponds to exactly
one of the outcomes of the random variable X, so it is natural to assign probability
1/8 to each of these. We assign probability 0 to the other 16 outcomes. In each
case, the probability function is called a joint distribution function. 2
We collect the above ideas in a denition.
Denition 4.3 Let X1; X2; : : : ; Xn be random variables associated with an exper-
iment. Suppose that the sample space (i.e., the set of possible outcomes) of Xi is
the set Ri. Then the joint random variable X = (X1; X2; : : :; Xn) is dened to be
the random variable whose outcomes consist of ordered n-tuples of outcomes, with
the ith coordinate lying in the set Ri. The sample space 
 of X is the Cartesian
product of the Ri's:",prob.pdf
"= R1  R2  
 
 
  Rn :
The joint distribution function of X is the function which gives the probability of
each of the outcomes of X. 2
Example 4.12 (Example 4.10 continued) We now consider the assignment of prob-
abilities in the above example. In the case of the random variable X, the probabil-
ity of any outcome (a1; a2; a3) is just the product of the probabilities P(Xi = ai),",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 143
Not smoke Smoke Total
Not cancer 40 10 50
Cancer 7 3 10
Totals 47 13 60
Table 4.1: Smoking and cancer.
S
0 1
0 40/60 10/60
C
1 7/60 3/60
Table 4.2: Joint distribution.
for i = 1; 2; 3. However, in the case of Y , the probability assigned to the outcome
(1; 1; 0) is not the product of the probabilities P(Y1 = 1), P(Y2 = 1), and P(Y3 = 0).
The dierence between these two situations is that the value of Xi does not aect
the value of Xj, if i 6= j, while the values of Yi and Yj aect one another. For
example, if Y1 = 1, then Y2 cannot equal 0. This prompts the next denition. 2
Denition 4.4 The random variables X1, X2, . .. , Xn are mutually independent
if
P(X1 = r1; X2 = r2; : : :; Xn = rn)
= P(X1 = r1)P(X2 = r2) 
 
 
 P(Xn = rn)
for any choice of r1; r2; : : : ; rn. Thus, if X1; X2; : : : ; Xn are mutually independent,
then the joint distribution function of the random variable
X = (X1; X2; : : : ; Xn)",prob.pdf
"then the joint distribution function of the random variable
X = (X1; X2; : : : ; Xn)
is just the product of the individual distribution functions. When two random
variables are mutually independent, we shall say more brie
y that they are indepen-
dent. 2
Example 4.13 In a group of 60 people, the numbers who do or do not smoke and
do or do not have cancer are reported as shown in Table 4.1. Let 
 be the sample
space consisting of these 60 people. A person is chosen at random from the group.
Let C(!) = 1 if this person has cancer and 0 if not, and S(!) = 1 if this person
smokes and 0 if not. Then the joint distribution of fC; Sg is given in Table 4.2. For
example P(C = 0; S = 0) = 40=60, P(C = 0; S = 1) = 10=60, and so forth. The
distributions of the individual random variables are called marginal distributions.
The marginal distributions of C and S are:
pC =
 0 1
50=60 10=60

;",prob.pdf
"144 CHAPTER 4. CONDITIONAL PROBABILITY
pS =
 0 1
47=60 13=60

:
The random variables S and C are not independent, since
P(C = 1; S = 1) = 3
60 = :05 ;
P(C = 1)P(S = 1) = 10
60 
 13
60 = :036 :
Note that we would also see this from the fact that
P(C = 1jS = 1) = 3
13 = :23 ;
P(C = 1) = 1
6 = :167 :
2
Independent Trials Processes
The study of random variables proceeds by considering special classes of random
variables. One such class that we shall study is the class of independent trials.
Denition 4.5 A sequence of random variables X1, X2, . . . , Xn that are mutually
independent and that have the same distribution is called a sequence of independent
trials or an independent trials process.
Independent trials processes arise naturally in the following way. We have a
single experiment with sample space R = fr1; r2; : : : ; rsg and a distribution function
mX =
r1 r2 
 
 
 rs
p1 p2 
 
 
 ps

:
We repeat this experiment n times. To describe this total experiment, we choose",prob.pdf
"mX =
r1 r2 
 
 
 rs
p1 p2 
 
 
 ps

:
We repeat this experiment n times. To describe this total experiment, we choose
as sample space the space",prob.pdf
"= R  R  
 
 
  R;
consisting of all possible sequences ! = (!1; !2; : : : ; !n) where the value of each !j
is chosen from R. We assign a distribution function to be the product distribution
m(!) = m(!1) 
 : : : 
 m(!n) ;
with m(!j) = pk when !j = rk. Then we let Xj denote the jth coordinate of the
outcome (r1; r2; : : : ; rn). The random variables X1, . . . , Xn form an independent
trials process. 2
Example 4.14 An experiment consists of rolling a die three times. Let Xi repre-
sent the outcome of the ith roll, for i = 1; 2; 3. The common distribution function
is
mi =
 1 2 3 4 5 6
1=6 1=6 1=6 1=6 1=6 1=6

:",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 145
The sample space is R3 = R  R  R with R = f1; 2; 3; 4; 5; 6g. If ! = (1; 3; 6),
then X1(!) = 1, X2(!) = 3, and X3(!) = 6 indicating that the rst roll was a 1,
the second was a 3, and the third was a 6. The probability assigned to any sample
point is
m(!) = 1
6 
 1
6 
 1
6 = 1
216 :
2
Example 4.15 Consider next a Bernoulli trials process with probability p for suc-
cess on each experiment. Let Xj(!) = 1 if the jth outcome is success and Xj(!) = 0
if it is a failure. Then X1, X2, . . . , Xn is an independent trials process. Each Xj
has the same distribution function
mj =
0 1
q p

;
where q = 1  p.
If Sn = X1 + X2 + 
 
 
 + Xn, then
P(Sn = j) =
n
j

pjqnj ;
and Sn has, as distribution, the binomial distribution b(n; p; j). 2
Bayes' Formula
In our examples, we have considered conditional probabilities of the following form:
Given the outcome of the second stage of a two-stage experiment, nd the proba-",prob.pdf
"Given the outcome of the second stage of a two-stage experiment, nd the proba-
bility for an outcome at the rst stage. We have remarked that these probabilities
are called Bayes probabilities.
We return now to the calculation of more general Bayes probabilities. Suppose
we have a set of events H1; H2, . . . , Hm that are pairwise disjoint and such that
the sample space 
 satises the equation",prob.pdf
"= H1 [ H2 [ 
 
 
 [ Hm :
We call these events hypotheses. We also have an event E that gives us some
information about which hypothesis is correct. We call this event evidence.
Before we receive the evidence, then, we have a set of prior probabilities P(H1),
P(H2), . . . , P(Hm) for the hypotheses. If we know the correct hypothesis, we know
the probability for the evidence. That is, we know P(EjHi) for all i. We want to
nd the probabilities for the hypotheses given the evidence. That is, we want to nd
the conditional probabilities P(HijE). These probabilities are called the posterior
probabilities.
To nd these probabilities, we write them in the form
P(HijE) = P(Hi \ E)
P(E) : (4.1)",prob.pdf
"146 CHAPTER 4. CONDITIONAL PROBABILITY
Number having The results
Disease this disease + + + { { + { {
d1 3215 2110 301 704 100
d2 2125 396 132 1187 410
d3 4660 510 3568 73 509
Total 10000
Table 4.3: Diseases data.
We can calculate the numerator from our given information by
P(Hi \ E) = P(Hi)P(EjHi) : (4.2)
Since one and only one of the events H1, H2, . .. , Hm can occur, we can write the
probability of E as
P(E) = P(H1 \ E) + P(H2 \ E) + 
 
 
 + P(Hm \ E) :
Using Equation 4.2, the above expression can be seen to equal
P(H1)P(EjH1) + P(H2)P(EjH2) + 
 
 
 + P(Hm)P(EjHm) : (4.3)
Using (4.1), (4.2), and (4.3) yields Bayes' formula:
P(HijE) = P(Hi)P(EjHi)Pm
k=1 P(Hk)P(EjHk) :
Although this is a very famous formula, we will rarely use it. If the number of
hypotheses is small, a simple tree measure calculation is easily carried out, as we
have done in our examples. If the number of hypotheses is large, then we should
use a computer.",prob.pdf
"have done in our examples. If the number of hypotheses is large, then we should
use a computer.
Bayes probabilities are particularly appropriate for medical diagnosis. A doctor
is anxious to know which of several diseases a patient might have. She collects
evidence in the form of the outcomes of certain tests. From statistical studies the
doctor can nd the prior probabilities of the various diseases before the tests, and
the probabilities for specic test outcomes, given a particular disease. What the
doctor wants to know is the posterior probability for the particular disease, given
the outcomes of the tests.
Example 4.16 A doctor is trying to decide if a patient has one of three diseases
d1, d2, or d3. Two tests are to be carried out, each of which results in a positive
(+) or a negative () outcome. There are four possible test patterns ++, +,
+, and . National records have indicated that, for 10,000 people having one of",prob.pdf
"+, and . National records have indicated that, for 10,000 people having one of
these three diseases, the distribution of diseases and test results are as in Table 4.3.
From this data, we can estimate the prior probabilities for each of the diseases
and, given a particular disease, the probability of a particular test outcome. For
example, the prior probability of disease d1 may be estimated to be 3215=10;000 =
:3215. The probability of the test result +, given disease d1, may be estimated to
be 301=3215 = :094.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 147
d1 d2 d3
+ + .700 .131 .169
+ { .075 .033 .892
{ + .358 .604 .038
{ { .098 .403 .499
Table 4.4: Posterior probabilities.
We can now use Bayes' formula to compute various posterior probabilities. The
computer program Bayes computes these posterior probabilities. The results for
this example are shown in Table 4.4.
We note from the outcomes that, when the test result is ++, the disease d1 has
a signicantly higher probability than the other two. When the outcome is +,
this is true for disease d3. When the outcome is +, this is true for disease d2.
Note that these statements might have been guessed by looking at the data. If the
outcome is , the most probable cause is d3, but the probability that a patient
has d2 is only slightly smaller. If one looks at the data in this case, one can see that
it might be hard to guess which of the two diseases d2 and d3 is more likely. 2",prob.pdf
"it might be hard to guess which of the two diseases d2 and d3 is more likely. 2
Our nal example shows that one has to be careful when the prior probabilities
are small.
Example 4.17 A doctor gives a patient a test for a particular cancer. Before the
results of the test, the only evidence the doctor has to go on is that 1 woman
in 1000 has this cancer. Experience has shown that, in 99 percent of the cases in
which cancer is present, the test is positive; and in 95 percent of the cases in which
it is not present, it is negative. If the test turns out to be positive, what probability
should the doctor assign to the event that cancer is present? An alternative form
of this question is to ask for the relative frequencies of false positives and cancers.
We are given that prior(cancer) = :001 and prior(not cancer) = :999. We
know also that P(+jcancer) = :99, P(jcancer) = :01, P(+jnot cancer) = :05,
and P(jnot cancer) = :95. Using this data gives the result shown in Figure 4.5.",prob.pdf
"and P(jnot cancer) = :95. Using this data gives the result shown in Figure 4.5.
We see now that the probability of cancer given a positive test has only increased
from .001 to .019. While this is nearly a twenty-fold increase, the probability that
the patient has the cancer is still small. Stated in another way, among the positive
results, 98.1 percent are false positives, and 1.9 percent are cancers. When a group
of second-year medical students was asked this question, over half of the students
incorrectly guessed the probability to be greater than .5. 2
Historical Remarks
Conditional probability was used long before it was formally dened. Pascal and
Fermat considered the problem of points: given that team A has won m games and
team B has won n games, what is the probability that A will win the series? (See
Exercises 40{42.) This is clearly a conditional probability problem.
In his book, Huygens gave a number of problems, one of which was:",prob.pdf
"148 CHAPTER 4. CONDITIONAL PROBABILITY
.001
can
not
.01
.95
.05 +
-
.001
0
.05
.949
+
-
.051
.949
+
-
.981
1
0 can
not
.001
.05
0
.949
can
not
.019
Original Tree Reverse Tree
.99
.999
Figure 4.5: Forward and reverse tree diagrams.
Three gamblers, A, B and C, take 12 balls of which 4 are white and 8
black. They play with the rules that the drawer is blindfolded, A is to
draw rst, then B and then C, the winner to be the one who rst draws
a white ball. What is the ratio of their chances?2
From his answer it is clear that Huygens meant that each ball is replaced after
drawing. However, John Hudde, the mayor of Amsterdam, assumed that he meant
to sample without replacement and corresponded with Huygens about the dierence
in their answers. Hacking remarks that \Neither party can understand what the
other is doing.""3
By the time of de Moivre's book, The Doctrine of Chances, these distinctions
were well understood. De Moivre dened independence and dependence as follows:",prob.pdf
"were well understood. De Moivre dened independence and dependence as follows:
Two Events are independent, when they have no connexion one with
the other, and that the happening of one neither forwards nor obstructs
the happening of the other.
Two Events are dependent, when they are so connected together as that
the Probability of either's happening is altered by the happening of the
other.4
De Moivre used sampling with and without replacement to illustrate that the
probability that two independent events both happen is the product of their prob-
abilities, and for dependent events that:
2Quoted in F. N. David, Games, Gods and Gambling (London: Grin, 1962), p. 119.
3I. Hacking, The Emergence of Probability (Cambridge: Cambridge University Press, 1975),
p. 99.
4A. de Moivre, The Doctrine of Chances, 3rd ed. (New York: Chelsea, 1967), p. 6.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 149
The Probability of the happening of two Events dependent, is the prod-
uct of the Probability of the happening of one of them, by the Probability
which the other will have of happening, when the rst is considered as
having happened; and the same Rule will extend to the happening of as
many Events as may be assigned.5
The formula that we call Bayes' formula, and the idea of computing the proba-
bility of a hypothesis given evidence, originated in a famous essay of Thomas Bayes.
Bayes was an ordained minister in Tunbridge Wells near London. His mathemat-
ical interests led him to be elected to the Royal Society in 1742, but none of his
results were published within his lifetime. The work upon which his fame rests,
\An Essay Toward Solving a Problem in the Doctrine of Chances,"" was published
in 1763, three years after his death.6 Bayes reviewed some of the basic concepts of",prob.pdf
"in 1763, three years after his death.6 Bayes reviewed some of the basic concepts of
probability and then considered a new kind of inverse probability problem requiring
the use of conditional probability.
Bernoulli, in his study of processes that we now call Bernoulli trials, had proven
his famous law of large numbers which we will study in Chapter 8. This theorem
assured the experimenter that if he knew the probability p for success, he could
predict that the proportion of successes would approach this value as he increased
the number of experiments. Bernoulli himself realized that in most interesting cases
you do not know the value of p and saw his theorem as an important step in showing
that you could determine p by experimentation.
To study this problem further, Bayes started by assuming that the probability p
for success is itself determined by a random experiment. He assumed in fact that this
experiment was such that this value for p is equally likely to be any value between",prob.pdf
"experiment was such that this value for p is equally likely to be any value between
0 and 1. Without knowing this value we carry out n experiments and observe m
successes. Bayes proposed the problem of nding the conditional probability that
the unknown probability p lies between a and b. He obtained the answer:
P(a  p < bjm successes in n trials) =
Rb
a xm(1  x)nm dx
R1
0 xm(1  x)nm dx
:
We shall see in the next section how this result is obtained. Bayes clearly wanted
to show that the conditional distribution function, given the outcomes of more and
more experiments, becomes concentrated around the true value of p. Thus, Bayes
was trying to solve an inverse problem. The computation of the integrals was too
dicult for exact solution except for small values of j and n, and so Bayes tried
approximate methods. His methods were not very satisfactory and it has been
suggested that this discouraged him from publishing his results.",prob.pdf
"approximate methods. His methods were not very satisfactory and it has been
suggested that this discouraged him from publishing his results.
However, his paper was the rst in a series of important studies carried out by
Laplace, Gauss, and other great mathematicians to solve inverse problems. They
studied this problem in terms of errors in measurements in astronomy. If an as-
tronomer were to know the true value of a distance and the nature of the random
5ibid, p. 7.
6T. Bayes, \An Essay Toward Solving a Problem in the Doctrine of Chances,"" Phil. Trans.
Royal Soc. London, vol. 53 (1763), pp. 370{418.",prob.pdf
"150 CHAPTER 4. CONDITIONAL PROBABILITY
errors caused by his measuring device he could predict the probabilistic nature of
his measurements. In fact, however, he is presented with the inverse problem of
knowing the nature of the random errors, and the values of the measurements, and
wanting to make inferences about the unknown true value.
As Maistrov remarks, the formula that we have called Bayes' formula does not
appear in his essay. Laplace gave it this name when he studied these inverse prob-
lems.7 The computation of inverse probabilities is fundamental to statistics and
has led to an important branch of statistics called Bayesian analysis, assuring Bayes
eternal fame for his brief essay.
Exercises
1 Assume that E and F are two events with positive probabilities. Show that
if P(EjF) = P(E), then P(FjE) = P(F).
2 A coin is tossed three times. What is the probability that exactly two heads
occur, given that
(a) the rst outcome was a head?
(b) the rst outcome was a tail?",prob.pdf
"occur, given that
(a) the rst outcome was a head?
(b) the rst outcome was a tail?
(c) the rst two outcomes were heads?
(d) the rst two outcomes were tails?
(e) the rst outcome was a head and the third outcome was a head?
3 A die is rolled twice. What is the probability that the sum of the faces is
greater than 7, given that
(a) the rst outcome was a 4?
(b) the rst outcome was greater than 3?
(c) the rst outcome was a 1?
(d) the rst outcome was less than 5?
4 A card is drawn at random from a deck of cards. What is the probability that
(a) it is a heart, given that it is red?
(b) it is higher than a 10, given that it is a heart? (Interpret J, Q, K, A as
11, 12, 13, 14.)
(c) it is a jack, given that it is red?
5 A coin is tossed three times. Consider the following events
A: Heads on the rst toss.
B: Tails on the second.
C: Heads on the third toss.
D: All three outcomes the same (HHH or TTT).
E: Exactly one head turns up.",prob.pdf
"B: Tails on the second.
C: Heads on the third toss.
D: All three outcomes the same (HHH or TTT).
E: Exactly one head turns up.
7L. E. Maistrov, Probability Theory: A Historical Sketch, trans. and ed. Samual Kotz (New
York: Academic Press, 1974), p. 100.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 151
(a) Which of the following pairs of these events are independent?
(1) A, B
(2) A, D
(3) A, E
(4) D, E
(b) Which of the following triples of these events are independent?
(1) A, B, C
(2) A, B, D
(3) C, D, E
6 From a deck of ve cards numbered 2, 4, 6, 8, and 10, respectively, a card
is drawn at random and replaced. This is done three times. What is the
probability that the card numbered 2 was drawn exactly two times, given
that the sum of the numbers on the three draws is 12?
7 A coin is tossed twice. Consider the following events.
A: Heads on the rst toss.
B: Heads on the second toss.
C: The two tosses come out the same.
(a) Show that A, B, C are pairwise independent but not independent.
(b) Show that C is independent of A and B but not of A \ B.
8 Let 
 = fa; b; c; d; e; fg. Assume that m(a) = m(b) = 1=8 and m(c) =
m(d) = m(e) = m(f) = 3=16. Let A, B, and C be the events A = fd; e; ag,",prob.pdf
"8 Let 
 = fa; b; c; d; e; fg. Assume that m(a) = m(b) = 1=8 and m(c) =
m(d) = m(e) = m(f) = 3=16. Let A, B, and C be the events A = fd; e; ag,
B = fc; e; ag, C = fc; d; ag. Show that P(A \ B \ C) = P(A)P(B)P(C) but
no two of these events are independent.
9 What is the probability that a family of two children has
(a) two boys given that it has at least one boy?
(b) two boys given that the rst child is a boy?
10 In Example 4.2, we used the Life Table (see Appendix C) to compute a con-
ditional probability. The number 93,753 in the table, corresponding to 40-
year-old males, means that of all the males born in the United States in 1950,
93.753% were alive in 1990. Is it reasonable to use this as an estimate for the
probability of a male, born this year, surviving to age 40?
11 Simulate the Monty Hall problem. Carefully state any assumptions that you
have made when writing the program. Which version of the problem do you
think that you are simulating?",prob.pdf
"have made when writing the program. Which version of the problem do you
think that you are simulating?
12 In Example 4.17, how large must the prior probability of cancer be to give a
posterior probability of .5 for cancer given a positive test?
13 Two cards are drawn from a bridge deck. What is the probability that the
second card drawn is red?",prob.pdf
"152 CHAPTER 4. CONDITIONAL PROBABILITY
14 If P( ~B) = 1=4 and P(AjB) = 1=2, what is P(A \ B)?
15 (a) What is the probability that your bridge partner has exactly two aces,
given that she has at least one ace?
(b) What is the probability that your bridge partner has exactly two aces,
given that she has the ace of spades?
16 Prove that for any three events A, B, C, each having positive probability, and
with the property that P(A \ B) > 0,
P(A \ B \ C) = P(A)P(BjA)P(CjA \ B) :
17 Prove that if A and B are independent so are
(a) A and ~B.
(b) ~A and ~B.
18 A doctor assumes that a patient has one of three diseases d1, d2, or d3. Before
any test, he assumes an equal probability for each disease. He carries out a
test that will be positive with probability .8 if the patient has d1, .6 if he has
disease d2, and .4 if he has disease d3. Given that the outcome of the test was
positive, what probabilities should the doctor now assign to the three possible
diseases?",prob.pdf
"positive, what probabilities should the doctor now assign to the three possible
diseases?
19 In a poker hand, John has a very strong hand and bets 5 dollars. The prob-
ability that Mary has a better hand is .04. If Mary had a better hand she
would raise with probability .9, but with a poorer hand she would only raise
with probability .1. If Mary raises, what is the probability that she has a
better hand than John does?
20 The Polya urn model for contagion is as follows: We start with an urn which
contains one white ball and one black ball. At each second we choose a ball
at random from the urn and replace this ball and add one more of the color
chosen. Write a program to simulate this model, and see if you can make
any predictions about the proportion of white balls in the urn after a large
number of draws. Is there a tendency to have a large fraction of balls of the
same color in the long run?
21 It is desired to nd the probability that in a bridge deal each player receives an",prob.pdf
"same color in the long run?
21 It is desired to nd the probability that in a bridge deal each player receives an
ace. A student argues as follows. It does not matter where the rst ace goes.
The second ace must go to one of the other three players and this occurs with
probability 3/4. Then the next must go to one of two, an event of probability
1/2, and nally the last ace must go to the player who does not have an ace.
This occurs with probability 1/4. The probability that all these events occur
is the product (3=4)(1=2)(1=4) = 3=32. Is this argument correct?
22 One coin in a collection of 65 has two heads. The rest are fair. If a coin,
chosen at random from the lot and then tossed, turns up heads 6 times in a
row, what is the probability that it is the two-headed coin?",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 153
23 You are given two urns and fty balls. Half of the balls are white and half
are black. You are asked to distribute the balls in the urns with no restriction
placed on the number of either type in an urn. How should you distribute
the balls in the urns to maximize the probability of obtaining a white ball if
an urn is chosen at random and a ball drawn out at random? Justify your
answer.
24 A fair coin is thrown n times. Show that the conditional probability of a head
on any specied trial, given a total of k heads over the n trials, is k=n (k > 0).
25 (Johnsonbough8) A coin with probability p for heads is tossed n times. Let E
be the event \a head is obtained on the rst toss' and Fk the event `exactly k
heads are obtained."" For which pairs (n; k) are E and Fk independent?
26 Suppose that A and B are events such that P(AjB) = P(BjA) and P(A[B) =
1 and P(A \ B) > 0. Prove that P(A) > 1=2.",prob.pdf
"26 Suppose that A and B are events such that P(AjB) = P(BjA) and P(A[B) =
1 and P(A \ B) > 0. Prove that P(A) > 1=2.
27 (Chung9) In London, half of the days have some rain. The weather forecaster
is correct 2/3 of the time, i.e., the probability that it rains, given that she has
predicted rain, and the probability that it does not rain, given that she has
predicted that it won't rain, are both equal to 2/3. When rain is forecast,
Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with
probability 1/3. Find
(a) the probability that Pickwick has no umbrella, given that it rains.
(b) the probability that he brings his umbrella, given that it doesn't rain.
28 Probability theory was used in a famous court case: People v. Collins.10 In
this case a purse was snatched from an elderly person in a Los Angeles suburb.
A couple seen running from the scene were described as a black man with a
beard and a mustache and a blond girl with hair in a ponytail. Witnesses said",prob.pdf
"beard and a mustache and a blond girl with hair in a ponytail. Witnesses said
they drove o in a partly yellow car. Malcolm and Janet Collins were arrested.
He was black and though clean shaven when arrested had evidence of recently
having had a beard and a mustache. She was blond and usually wore her hair
in a ponytail. They drove a partly yellow Lincoln. The prosecution called a
professor of mathematics as a witness who suggested that a conservative set of
probabilities for the characteristics noted by the witnesses would be as shown
in Table 4.5.
The prosecution then argued that the probability that all of these character-
istics are met by a randomly chosen couple is the product of the probabilities
or 1/12,000,000, which is very small. He claimed this was proof beyond a rea-
sonable doubt that the defendants were guilty. The jury agreed and handed
down a verdict of guilty of second-degree robbery.",prob.pdf
"sonable doubt that the defendants were guilty. The jury agreed and handed
down a verdict of guilty of second-degree robbery.
8R. Johnsonbough, \Problem #103,"" Two Year College Math Journal, vol. 8 (1977), p. 292.
9K. L. Chung, Elementary Probability Theory With Stochastic Processes, 3rd ed. (New York:
Springer-Verlag, 1979), p. 152.
10M. W. Gray, \Statistics and the Law,"" Mathematics Magazine, vol. 56 (1983), pp. 67{81.",prob.pdf
"154 CHAPTER 4. CONDITIONAL PROBABILITY
man with mustache 1/4
girl with blond hair 1/3
girl with ponytail 1/10
black man with beard 1/10
interracial couple in a car 1/1000
partly yellow car 1/10
Table 4.5: Collins case probabilities.
If you were the lawyer for the Collins couple how would you have countered
the above argument? (The appeal of this case is discussed in Exercise 5.1.34.)
29 A student is applying to Harvard and Dartmouth. He estimates that he has
a probability of .5 of being accepted at Dartmouth and .3 of being accepted
at Harvard. He further estimates the probability that he will be accepted by
both is .2. What is the probability that he is accepted by Dartmouth if he is
accepted by Harvard? Is the event \accepted at Harvard"" independent of the
event \accepted at Dartmouth""?
30 Luxco, a wholesale lightbulb manufacturer, has two factories. Factory A sells
bulbs in lots that consists of 1000 regular and 2000 softglow bulbs each. Ran-",prob.pdf
"bulbs in lots that consists of 1000 regular and 2000 softglow bulbs each. Ran-
dom sampling has shown that on the average there tend to be about 2 bad
regular bulbs and 11 bad softglow bulbs per lot. At factory B the lot size is
reversed|there are 2000 regular and 1000 softglow per lot|and there tend
to be 5 bad regular and 6 bad softglow bulbs per lot.
The manager of factory A asserts, \We're obviously the better producer; our
bad bulb rates are .2 percent and .55 percent compared to B's .25 percent and
.6 percent. We're better at both regular and softglow bulbs by half of a tenth
of a percent each.""
\Au contraire,"" counters the manager of B, \each of our 3000 bulb lots con-
tains only 11 bad bulbs, while A's 3000 bulb lots contain 13. So our .37
percent bad bulb rate beats their .43 percent.""
Who is right?
31 Using the Life Table for 1981 given in Appendix C, nd the probability that a
male of age 60 in 1981 lives to age 80. Find the same probability for a female.",prob.pdf
"male of age 60 in 1981 lives to age 80. Find the same probability for a female.
32 (a) There has been a blizzard and Helen is trying to drive from Woodstock
to Tunbridge, which are connected like the top graph in Figure 4.6. Here
p and q are the probabilities that the two roads are passable. What is
the probability that Helen can get from Woodstock to Tunbridge?
(b) Now suppose that Woodstock and Tunbridge are connected like the mid-
dle graph in Figure 4.6. What now is the probability that she can get
from W to T? Note that if we think of the roads as being components
of a system, then in (a) and (b) we have computed the reliability of a
system whose components are (a) in series and (b) in parallel.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 155
Woodstock Tunbridge
p q
C
D
TW
.8.9
.9.8
.95
W T
p
q
(a)
(b)
(c)
Figure 4.6: From Woodstock to Tunbridge.
(c) Now suppose W and T are connected like the bottom graph in Figure 4.6.
Find the probability of Helen's getting from W to T. Hint: If the road
from C to D is impassable, it might as well not be there at all; if it is
passable, then gure out how to use part (b) twice.
33 Let A1, A2, and A3 be events, and let Bi represent either Ai or its complement
~Ai. Then there are eight possible choices for the triple (B1; B2; B3). Prove
that the events A1, A2, A3 are independent if and only if
P(B1 \ B2 \ B3) = P(B1)P(B2)P(B3) ;
for all eight of the possible choices for the triple (B1; B2; B3).
34 Four women, A, B, C, and D, check their hats, and the hats are returned in a
random manner. Let 
 be the set of all possible permutations of A, B, C, D.
Let Xj = 1 if the jth woman gets her own hat back and 0 otherwise. What",prob.pdf
"random manner. Let 
 be the set of all possible permutations of A, B, C, D.
Let Xj = 1 if the jth woman gets her own hat back and 0 otherwise. What
is the distribution of Xj? Are the Xi's mutually independent?
35 A box has numbers from 1 to 10. A number is drawn at random. Let X1 be
the number drawn. This number is replaced, and the ten numbers mixed. A
second number X2 is drawn. Find the distributions of X1 and X2. Are X1
and X2 independent? Answer the same questions if the rst number is not
replaced before the second is drawn.",prob.pdf
"156 CHAPTER 4. CONDITIONAL PROBABILITY
Y
-1 0 1 2
X -1 0 1/36 1/6 1/12
0 1/18 0 1/18 0
1 0 1/36 1/6 1/12
2 1/12 0 1/12 1/6
Table 4.6: Joint distribution.
36 A die is thrown twice. Let X1 and X2 denote the outcomes. Dene X =
min(X1; X2). Find the distribution of X.
*37 Given that P(X = a) = r, P(max(X; Y ) = a) = s, and P(min(X; Y ) = a) =
t, show that you can determine u = P(Y = a) in terms of r, s, and t.
38 A fair coin is tossed three times. Let X be the number of heads that turn up
on the rst two tosses and Y the number of heads that turn up on the third
toss. Give the distribution of
(a) the random variables X and Y .
(b) the random variable Z = X + Y .
(c) the random variable W = X  Y .
39 Assume that the random variables X and Y have the joint distribution given
in Table 4.6.
(a) What is P(X  1 and Y  0)?
(b) What is the conditional probability that Y  0 given that X = 2?
(c) Are X and Y independent?
(d) What is the distribution of Z = XY ?",prob.pdf
"(b) What is the conditional probability that Y  0 given that X = 2?
(c) Are X and Y independent?
(d) What is the distribution of Z = XY ?
40 In the problem of points, discussed in the historical remarks in Section 3.2, two
players, A and B, play a series of points in a game with player A winning each
point with probability p and player B winning each point with probability
q = 1  p. The rst player to win N points wins the game. Assume that
N = 3. Let X be a random variable that has the value 1 if player A wins the
series and 0 otherwise. Let Y be a random variable with value the number
of points played in a game. Find the distribution of X and Y when p = 1=2.
Are X and Y independent in this case? Answer the same questions for the
case p = 2=3.
41 The letters between Pascal and Fermat, which are often credited with having
started probability theory, dealt mostly with the problem of points described
in Exercise 40. Pascal and Fermat considered the problem of nding a fair",prob.pdf
"in Exercise 40. Pascal and Fermat considered the problem of nding a fair
division of stakes if the game must be called o when the rst player has won
r games and the second player has won s games, with r < N and s < N. Let
P(r; s) be the probability that player A wins the game if he has already won
r points and player B has won s points. Then",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 157
(a) P(r; N) = 0 if r < N,
(b) P(N; s) = 1 if s < N,
(c) P(r; s) = pP(r + 1; s) + qP(r; s + 1) if r < N and s < N;
and (1), (2), and (3) determine P(r; s) for r  N and s  N. Pascal used
these facts to nd P(r; s) by working backward: He rst obtained P(N 1; j)
for j = N  1, N  2, .. . , 0; then, from these values, he obtained P(N  2; j)
for j = N  1, N  2, . . ., 0 and, continuing backward, obtained all the
values P(r; s). Write a program to compute P(r; s) for given N, a, b, and p.
Warning: Follow Pascal and you will be able to run N = 100; use recursion
and you will not be able to run N = 20.
42 Fermat solved the problem of points (see Exercise 40) as follows: He realized
that the problem was dicult because the possible ways the play might go are
not equally likely. For example, when the rst player needs two more games
and the second needs three to win, two possible ways the series might go for",prob.pdf
"and the second needs three to win, two possible ways the series might go for
the rst player are WLW and LWLW. These sequences are not equally likely.
To avoid this diculty, Fermat extended the play, adding ctitious plays so
that the series went the maximum number of games needed (four in this case).
He obtained equally likely outcomes and used, in eect, the Pascal triangle to
calculate P(r; s). Show that this leads to a formula for P(r; s) even for the
case p 6= 1=2.
43 The Yankees are playing the Dodgers in a world series. The Yankees win each
game with probability .6. What is the probability that the Yankees win the
series? (The series is won by the rst team to win four games.)
44 C. L. Anderson11 has used Fermat's argument for the problem of points to
prove the following result due to J. G. Kingston. You are playing the game
of points (see Exercise 40) but, at each point, when you serve you win with
probability p, and when your opponent serves you win with probability p.",prob.pdf
"of points (see Exercise 40) but, at each point, when you serve you win with
probability p, and when your opponent serves you win with probability p.
You will serve rst, but you can choose one of the following two conventions
for serving: for the rst convention you alternate service (tennis), and for the
second the person serving continues to serve until he loses a point and then
the other player serves (racquetball). The rst player to win N points wins
the game. The problem is to show that the probability of winning the game
is the same under either convention.
(a) Show that, under either convention, you will serve at most N points and
your opponent at most N  1 points.
(b) Extend the number of points to 2N  1 so that you serve N points and
your opponent serves N  1. For example, you serve any additional
points necessary to make N serves and then your opponent serves any
additional points necessary to make him serve N 1 points. The winner",prob.pdf
"points necessary to make N serves and then your opponent serves any
additional points necessary to make him serve N 1 points. The winner
11C. L. Anderson, \Note on the Advantage of First Serve,"" Journal of Combinatorial Theory,
Series A, vol. 23 (1977), p. 363.",prob.pdf
"158 CHAPTER 4. CONDITIONAL PROBABILITY
is now the person, in the extended game, who wins the most points.
Show that playing these additional points has not changed the winner.
(c) Show that (a) and (b) prove that you have the same probability of win-
ning the game under either convention.
45 In the previous problem, assume that p = 1  p.
(a) Show that under either service convention, the rst player will win more
often than the second player if and only if p > :5.
(b) In volleyball, a team can only win a point while it is serving. Thus, any
individual \play"" either ends with a point being awarded to the serving
team or with the service changing to the other team. The rst team to
win N points wins the game. (We ignore here the additional restriction
that the winning team must be ahead by at least two points at the end of
the game.) Assume that each team has the same probability of winning
the play when it is serving, i.e., that p = 1  p. Show that in this case,",prob.pdf
"the game.) Assume that each team has the same probability of winning
the play when it is serving, i.e., that p = 1  p. Show that in this case,
the team that serves rst will win more than half the time, as long as
p > 0. (If p = 0, then the game never ends.) Hint: Dene p0 to be the
probability that a team wins the next point, given that it is serving. If
we write q = 1  p, then one can show that
p0 = p
1  q2 :
If one now considers this game in a slightly dierent way, one can see
that the second service convention in the preceding problem can be used,
with p replaced by p0.
46 A poker hand consists of 5 cards dealt from a deck of 52 cards. Let X and
Y be, respectively, the number of aces and kings in a poker hand. Find the
joint distribution of X and Y .
47 Let X1 and X2 be independent random variables and let Y1 = 1(X1) and
Y2 = 2(X2).
(a) Show that
P(Y1 = r; Y2 = s) =
X
1(a)=r
2 (b)=s
P(X1 = a; X2 = b) :",prob.pdf
"Y2 = 2(X2).
(a) Show that
P(Y1 = r; Y2 = s) =
X
1(a)=r
2 (b)=s
P(X1 = a; X2 = b) :
(b) Using (a), show that P(Y1 = r; Y2 = s) = P(Y1 = r)P(Y2 = s) so that
Y1 and Y2 are independent.
48 Let 
 be the sample space of an experiment. Let E be an event with P(E) > 0
and dene mE(!) by mE(!) = m(!jE). Prove that mE(!) is a distribution
function on E, that is, that mE(!)  0 and that P
!2
 mE(!) = 1. The
function mE is called the conditional distribution given E.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 159
49 You are given two urns each containing two biased coins. The coins in urn I
come up heads with probability p1, and the coins in urn II come up heads
with probability p2 6= p1. You are given a choice of (a) choosing an urn at
random and tossing the two coins in this urn or (b) choosing one coin from
each urn and tossing these two coins. You win a prize if both coins turn up
heads. Show that you are better o selecting choice (a).
50 Prove that, if A1, A2, . .. , An are independent events dened on a sample
space 
 and if 0 < P(Aj) < 1 for all j, then 
 must have at least 2n points.
51 Prove that if
P(AjC)  P(BjC) and P(Aj ~C)  P(Bj ~C) ;
then P(A)  P(B).
52 A coin is in one of n boxes. The probability that it is in the ith box is pi.
If you search in the ith box and it is there, you nd it with probability ai.
Show that the probability p that the coin is in the jth box, given that you
have looked in the ith box and not found it, is
p =",prob.pdf
"Show that the probability p that the coin is in the jth box, given that you
have looked in the ith box and not found it, is
p =
 pj=(1  aipi); if j 6= i;
(1  ai)pi=(1  aipi); if j = i:
53 George Wolford has suggested the following variation on the Linda problem
(see Exercise 1.2.25). The registrar is carrying John and Mary's registration
cards and drops them in a puddle. When he pickes them up he cannot read the
names but on the rst card he picked up he can make out Mathematics 23 and
Government 35, and on the second card he can make out only Mathematics
23. He asks you if you can help him decide which card belongs to Mary. You
know that Mary likes government but does not like mathematics. You know
nothing about John and assume that he is just a typical Dartmouth student.
From this you estimate:
P(Mary takes Government 35) = :5 ;
P(Mary takes Mathematics 23) = :1 ;
P(John takes Government 35) = :3 ;
P(John takes Mathematics 23) = :2 :",prob.pdf
"P(Mary takes Government 35) = :5 ;
P(Mary takes Mathematics 23) = :1 ;
P(John takes Government 35) = :3 ;
P(John takes Mathematics 23) = :2 :
Assume that their choices for courses are independent events. Show that
the card with Mathematics 23 and Government 35 showing is more likely
to be Mary's than John's. The conjunction fallacy referred to in the Linda
problem would be to assume that the event \Mary takes Mathematics 23 and
Government 35"" is more likely than the event \Mary takes Mathematics 23.""
Why are we not making this fallacy here?",prob.pdf
"160 CHAPTER 4. CONDITIONAL PROBABILITY
54 (Suggested by Eisenberg and Ghosh12) A deck of playing cards can be de-
scribed as a Cartesian product
Deck = Suit  Rank ;
where Suit = f|; }; ~; g and Rank = f2; 3; : : :; 10; J; Q; K; Ag. This just
means that every card may be thought of as an ordered pair like (}; 2). By
a suit event we mean any event A contained in Deck which is described in
terms of Suit alone. For instance, if A is \the suit is red,"" then
A = f}; ~g  Rank ;
so that A consists of all cards of the form (}; r) or (~; r) where r is any rank.
Similarly, a rank event is any event described in terms of rank alone.
(a) Show that if A is any suit event and B any rank event, then A and B are
independent. (We can express this brie
y by saying that suit and rank
are independent.)
(b) Throw away the ace of spades. Show that now no nontrivial (i.e., neither
empty nor the whole space) suit event A is independent of any nontrivial
rank event B. Hint: Here independence comes down to",prob.pdf
"empty nor the whole space) suit event A is independent of any nontrivial
rank event B. Hint: Here independence comes down to
c=51 = (a=51) 
 (b=51) ;
where a, b, c are the respective sizes of A, B and A \ B. It follows that
51 must divide ab, hence that 3 must divide one of a and b, and 17 the
other. But the possible sizes for suit and rank events preclude this.
(c) Show that the deck in (b) nevertheless does have pairs A, B of nontrivial
independent events. Hint: Find 2 events A and B of sizes 3 and 17,
respectively, which intersect in a single point.
(d) Add a joker to a full deck. Show that now there is no pair A, B of
nontrivial independent events. Hint: See the hint in (b); 53 is prime.
The following problems are suggested by Stanley Gudder in his article \Do
Good Hands Attract?""13 He says that event A attracts event B if P(BjA) >
P(B) and repels B if P(BjA) < P(B).
55 Let Ri be the event that the ith player in a poker game has a royal 
ush.",prob.pdf
"P(B) and repels B if P(BjA) < P(B).
55 Let Ri be the event that the ith player in a poker game has a royal 
ush.
Show that a royal 
ush (A,K,Q,J,10 of one suit) attracts another royal 
ush,
that is P(R2jR1) > P(R2). Show that a royal 
ush repels full houses.
56 Prove that A attracts B if and only if B attracts A. Hence we can say that
A and B are mutually attractive if A attracts B.
12B. Eisenberg and B. K. Ghosh, \Independent Events in a Discrete Uniform Probability Space,""
The American Statistician, vol. 41, no. 1 (1987), pp. 52{56.
13S. Gudder, \Do Good Hands Attract?"" Mathematics Magazine, vol. 54, no. 1 (1981), pp. 13{
16.",prob.pdf
"4.1. DISCRETE CONDITIONAL PROBABILITY 161
57 Prove that A neither attracts nor repels B if and only if A and B are inde-
pendent.
58 Prove that A and B are mutually attractive if and only if P(BjA) > P(Bj ~A).
59 Prove that if A attracts B, then A repels ~B.
60 Prove that if A attracts both B and C, and A repels B \ C, then A attracts
B [ C. Is there any example in which A attracts both B and C and repels
B [ C?
61 Prove that if B1, B2, . .. , Bn are mutually disjoint and collectively exhaustive,
and if A attracts some Bi, then A must repel some Bj.
62 (a) Suppose that you are looking in your desk for a letter from some time
ago. Your desk has eight drawers, and you assess the probability that it
is in any particular drawer is 10% (so there is a 20% chance that it is not
in the desk at all). Suppose now that you start searching systematically
through your desk, one drawer at a time. In addition, suppose that
you have not found the letter in the rst i drawers, where 0  i  7.",prob.pdf
"through your desk, one drawer at a time. In addition, suppose that
you have not found the letter in the rst i drawers, where 0  i  7.
Let pi denote the probability that the letter will be found in the next
drawer, and let qi denote the probability that the letter will be found
in some subsequent drawer (both pi and qi are conditional probabilities,
since they are based upon the assumption that the letter is not in the
rst i drawers). Show that the pi's increase and the qi's decrease. (This
problem is from Falk et al.14)
(b) The following data appeared in an article in the Wall Street Journal.15
For the ages 20, 30, 40, 50, and 60, the probability of a woman in the
U.S. developing cancer in the next ten years is 0.5%, 1.2%, 3.2%, 6.4%,
and 10.8%, respectively. At the same set of ages, the probability of a
woman in the U.S. eventually developing cancer is 39.6%, 39.5%, 39.1%,
37.5%, and 34.2%, respectively. Do you think that the problem in part",prob.pdf
"woman in the U.S. eventually developing cancer is 39.6%, 39.5%, 39.1%,
37.5%, and 34.2%, respectively. Do you think that the problem in part
(a) gives an explanation for these data?
63 Here are two variations of the Monty Hall problem that are discussed by
Granberg.16
(a) Suppose that everything is the same except that Monty forgot to nd
out in advance which door has the car behind it. In the spirit of \the
show must go on,"" he makes a guess at which of the two doors to open
and gets lucky, opening a door behind which stands a goat. Now should
the contestant switch?
14R. Falk, A. Lipson, and C. Konold, \The ups and downs of the hope function in a fruitless
search,"" in Subjective Probability, G. Wright and P. Ayton, (eds.) (Chichester: Wiley, 1994), pgs.
353-377.
15C. Crossen, \Fright by the numbers: Alarming disease data are frequently 
awed,"" Wall Street
Journal, 11 April 1996, p. B1.
16D. Granberg, \To switch or not to switch,"" in The power of logical thinking, M. vos Savant,",prob.pdf
"Journal, 11 April 1996, p. B1.
16D. Granberg, \To switch or not to switch,"" in The power of logical thinking, M. vos Savant,
(New York: St. Martin's 1996).",prob.pdf
"162 CHAPTER 4. CONDITIONAL PROBABILITY
(b) You have observed the show for a long time and found that the car is
put behind door A 45% of the time, behind door B 40% of the time and
behind door C 15% of the time. Assume that everything else about the
show is the same. Again you pick door A. Monty opens a door with a
goat and oers to let you switch. Should you? Suppose you knew in
advance that Monty was going to give you a chance to switch. Should
you have initially chosen door A?
4.2 Continuous Conditional Probability
In situations where the sample space is continuous we will follow the same procedure
as in the previous section. Thus, for example, if X is a continuous random variable
with density function f(x), and if E is an event with positive probability, we dene
a conditional density function by the formula
f(xjE) =
f(x)=P(E); if x 2 E;
0; if x 62 E:
Then for any event F, we have
P(FjE) =
Z
F
f(xjE) dx :",prob.pdf
"a conditional density function by the formula
f(xjE) =
f(x)=P(E); if x 2 E;
0; if x 62 E:
Then for any event F, we have
P(FjE) =
Z
F
f(xjE) dx :
The expression P(FjE) is called the conditional probability of F given E. As in the
previous section, it is easy to obtain an alternative expression for this probability:
P(FjE) =
Z
F
f(xjE) dx =
Z
E\F
f(x)
P(E) dx = P(E \ F)
P(E) :
We can think of the conditional density function as being 0 except on E, and
normalized to have integral 1 over E. Note that if the original density is a uniform
density corresponding to an experiment in which all events of equal size are equally
likely, then the same will be true for the conditional density.
Example 4.18 In the spinner experiment (cf. Example 2.1), suppose we know that
the spinner has stopped with head in the upper half of the circle, 0  x  1=2. What
is the probability that 1=6  x  1=3?
Here E = [0; 1=2], F = [1=6; 1=3], and F \ E = F. Hence
P(FjE) = P(F \ E)
P(E)
= 1=6
1=2
= 1
3 ;",prob.pdf
"is the probability that 1=6  x  1=3?
Here E = [0; 1=2], F = [1=6; 1=3], and F \ E = F. Hence
P(FjE) = P(F \ E)
P(E)
= 1=6
1=2
= 1
3 ;
which is reasonable, since F is 1/3 the size of E. The conditional density function
here is given by",prob.pdf
"4.2. CONTINUOUS CONDITIONAL PROBABILITY 163
f(xjE) =
2; if 0  x < 1=2;
0; if 1=2  x < 1:
Thus the conditional density function is nonzero only on [0; 1=2], and is uniform
there. 2
Example 4.19 In the dart game (cf. Example 2.8), suppose we know that the dart
lands in the upper half of the target. What is the probability that its distance from
the center is less than 1/2?
Here E = f (x; y) : y  0 g, and F = f (x; y) : x2 + y2 < (1=2)2 g. Hence,
P(FjE) = P(F \ E)
P(E) = (1=)[(1=2)(=4)]
(1=)(=2)
= 1=4 :
Here again, the size of F \ E is 1/4 the size of E. The conditional density function
is
f((x; y)jE) =
f(x; y)=P(E) = 2=; if (x; y) 2 E;
0; if (x; y) 62 E:
2
Example 4.20 We return to the exponential density (cf. Example 2.17). We sup-
pose that we are observing a lump of plutonium-239. Our experiment consists of
waiting for an emission, then starting a clock, and recording the length of time X
that passes until the next emission. Experience has shown that X has an expo-",prob.pdf
"that passes until the next emission. Experience has shown that X has an expo-
nential density with some parameter , which depends upon the size of the lump.
Suppose that when we perform this experiment, we notice that the clock reads r
seconds, and is still running. What is the probability that there is no emission in a
further s seconds?
Let G(t) be the probability that the next particle is emitted after time t. Then
G(t) =
Z 1
t
ex dx
= ex
1
t = et :
Let E be the event \the next particle is emitted after time r"" and F the event
\the next particle is emitted after time r + s."" Then
P(FjE) = P(F \ E)
P(E)
= G(r + s)
G(r)
= e(r+s)
er
= es :",prob.pdf
"164 CHAPTER 4. CONDITIONAL PROBABILITY
This tells us the rather surprising fact that the probability that we have to wait
s seconds more for an emission, given that there has been no emission in r seconds,
is independent of the time r. This property (called the memoryless property)
was introduced in Example 2.17. When trying to model various phenomena, this
property is helpful in deciding whether the exponential density is appropriate.
The fact that the exponential density is memoryless means that it is reasonable
to assume if one comes upon a lump of a radioactive isotope at some random time,
then the amount of time until the next emission has an exponential density with
the same parameter as the time between emissions. A well-known example, known
as the \bus paradox,"" replaces the emissions by buses. The apparent paradox arises
from the following two facts: 1) If you know that, on the average, the buses come",prob.pdf
"from the following two facts: 1) If you know that, on the average, the buses come
by every 30 minutes, then if you come to the bus stop at a random time, you should
only have to wait, on the average, for 15 minutes for a bus, and 2) Since the buses
arrival times are being modelled by the exponential density, then no matter when
you arrive, you will have to wait, on the average, for 30 minutes for a bus.
The reader can now see that in Exercises 2.2.9, 2.2.10, and 2.2.11, we were
asking for simulations of conditional probabilities, under various assumptions on
the distribution of the interarrival times. If one makes a reasonable assumption
about this distribution, such as the one in Exercise 2.2.10, then the average waiting
time is more nearly one-half the average interarrival time. 2
Independent Events
If E and F are two events with positive probability in a continuous sample space,
then, as in the case of discrete sample spaces, we dene E and F to be independent",prob.pdf
"then, as in the case of discrete sample spaces, we dene E and F to be independent
if P(EjF) = P(E) and P(FjE) = P(F). As before, each of the above equations
imply the other, so that to see whether two events are independent, only one of these
equations must be checked. It is also the case that, if E and F are independent,
then P(E \ F) = P(E)P(F).
Example 4.21 (Example 4.18 continued) In the dart game (see Example 4.18), let
E be the event that the dart lands in the upper half of the target (y  0) and F the
event that the dart lands in the right half of the target (x  0). Then P(E \ F) is
the probability that the dart lies in the rst quadrant of the target, and
P(E \ F) = 1

Z
E\F
1 dxdy
= Area(E \ F)
= Area(E) Area (F)
=
1

Z
E
1 dxdy
1

Z
F
1 dxdy

= P(E)P(F)
so that E and F are independent. What makes this work is that the events E and
F are described by restricting dierent coordinates. This idea is made more precise
below. 2",prob.pdf
"4.2. CONTINUOUS CONDITIONAL PROBABILITY 165
Joint Density and Cumulative Distribution Functions
In a manner analogous with discrete random variables, we can dene joint density
functions and cumulative distribution functions for multi-dimensional continuous
random variables.
Denition 4.6 Let X1; X2; : : : ; Xn be continuous random variables associated
with an experiment, and let X = (X1; X2; : : : ; Xn). Then the joint cumulative
distribution function of X is dened by
F(x1; x2; : : :; xn) = P(X1  x1; X2  x2; : : :; Xn  xn) :
The joint density function of X satises the following equation:
F(x1; x2; : : : ; xn) =
Z x1
1
Z x2
1

 
 

Z xn
1
f(t1; t2; : : :tn) dtndtn1 : : : dt1 :
2
It is straightforward to show that, in the above notation,
f(x1; x2; : : : ; xn) = @nF(x1; x2; : : :; xn)
@x1@x2 
 
 
 @xn
: (4.4)
Independent Random Variables
As with discrete random variables, we can dene mutual independence of continuous
random variables.",prob.pdf
": (4.4)
Independent Random Variables
As with discrete random variables, we can dene mutual independence of continuous
random variables.
Denition 4.7 Let X1, X2, . . . , Xn be continuous random variables with cumula-
tive distribution functions F1(x); F2(x); : : : ; Fn(x). Then these random variables
are mutually independent if
F(x1; x2; : : :; xn) = F1(x1)F2(x2) 
 
 
 Fn(xn)
for any choice of x1; x2; : : : ; xn. Thus, if X1; X2; : : :; Xn are mutually inde-
pendent, then the joint cumulative distribution function of the random variable
X = (X1; X2; : : : ; Xn) is just the product of the individual cumulative distribution
functions. When two random variables are mutually independent, we shall say more
brie
y that they are independent. 2
Using Equation 4.4, the following theorem can easily be shown to hold for mu-
tually independent continuous random variables.
Theorem 4.2 Let X1, X2, . . . , Xn be continuous random variables with density",prob.pdf
"tually independent continuous random variables.
Theorem 4.2 Let X1, X2, . . . , Xn be continuous random variables with density
functions f1(x); f2(x); : : : ; fn(x). Then these random variables are mutually in-
dependent if and only if
f(x1; x2; : : : ; xn) = f1(x1)f2(x2) 
 
 
 fn(xn)
for any choice of x1; x2; : : : ; xn. 2",prob.pdf
"166 CHAPTER 4. CONDITIONAL PROBABILITY
1
1
r
r0
ω
ω
E
2
1
1
2
1
Figure 4.7: X1 and X2 are independent.
Let's look at some examples.
Example 4.22 In this example, we dene three random variables, X1; X2, and
X3. We will show that X1 and X2 are independent, and that X1 and X3 are not
independent. Choose a point ! = (!1; !2) at random from the unit square. Set
X1 = !2
1, X2 = !2
2, and X3 = !1 + !2. Find the joint distributions F12(r1; r2) and
F23(r2; r3).
We have already seen (see Example 2.13) that
F1(r1) = P(1 < X1  r1)
= pr1; if 0  r1  1 ;
and similarly,
F2(r2) = pr2 ;
if 0  r2  1. Now we have (see Figure 4.7)
F12(r1; r2) = P(X1  r1 and X2  r2)
= P(!1  pr1 and !2  pr2)
= Area(E1)
= pr1
pr2
= F1(r1)F2(r2) :
In this case F12(r1; r2) = F1(r1)F2(r2) so that X1 and X2 are independent. On the
other hand, if r1 = 1=4 and r3 = 1, then (see Figure 4.8)
F13(1=4; 1) = P(X1  1=4; X3  1)",prob.pdf
"4.2. CONTINUOUS CONDITIONAL PROBABILITY 167
1
1
0
ω
ω
ω  + ω  = 1
1/2
1 2
1
2
Ε 2
Figure 4.8: X1 and X3 are not independent.
= P(!1  1=2; !1 + !2  1)
= Area(E2)
= 1
2  1
8 = 3
8 :
Now recalling that
F3(r3) =
8
>
>
<
>
>
:
0; if r3 < 0;
(1=2)r2
3; if 0  r3  1;
1  (1=2)(2  r3)2; if 1  r3  2;
1; if 2 < r3;
(see Example 2.14), we have F1(1=4)F3(1) = (1=2)(1=2) = 1=4. Hence, X1 and X3
are not independent random variables. A similar calculation shows that X2 and X3
are not independent either. 2
Although we shall not prove it here, the following theorem is a useful one. The
statement also holds for mutually independent discrete random variables. A proof
may be found in Renyi.17
Theorem 4.3 Let X1; X2; : : : ; Xn be mutually independent continuous random
variables and let 1(x); 2(x); : : : ; n(x) be continuous functions. Then 1(X1);
2(X2); : : : ; n(Xn) are mutually independent. 2
Independent Trials
Using the notion of independence, we can now formulate for continuous sample",prob.pdf
"2(X2); : : : ; n(Xn) are mutually independent. 2
Independent Trials
Using the notion of independence, we can now formulate for continuous sample
spaces the notion of independent trials (see Denition 4.5).
17A. Renyi, Probability Theory (Budapest: Akademiai Kiado, 1970), p. 183.",prob.pdf
"168 CHAPTER 4. CONDITIONAL PROBABILITY
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2
2.5
3
α = β =.5
α = β =1
α = β = 2
0
Figure 4.9: Beta density for  =  = :5; 1; 2:
Denition 4.8 A sequence X1, X2, . . . , Xn of random variables Xi that are
mutually independent and have the same density is called an independent trials
process. 2
As in the case of discrete random variables, these independent trials processes
arise naturally in situations where an experiment described by a single random
variable is repeated n times.
Beta Density
We consider next an example which involves a sample space with both discrete
and continuous coordinates. For this example we shall need a new density function
called the beta density. This density has two parameters ,  and is dened by
B(; ; x) =
(1=B(; ))x1(1  x)1; if 0  x  1;
0; otherwise:
Here  and  are any positive numbers, and the beta function B(; ) is given by
the area under the graph of x1(1  x)1 between 0 and 1:
B(; ) =
Z 1
0",prob.pdf
"the area under the graph of x1(1  x)1 between 0 and 1:
B(; ) =
Z 1
0
x1(1  x)1 dx :
Note that when  =  = 1 the beta density if the uniform density. When  and
 are greater than 1 the density is bell-shaped, but when they are less than 1 it is
U-shaped as suggested by the examples in Figure 4.9.
We shall need the values of the beta function only for integer values of  and ,
and in this case
B(; ) = (  1)! (  1)!
( +   1)! :
Example 4.23 In medical problems it is often assumed that a drug is eective with
a probability x each time it is used and the various trials are independent, so that",prob.pdf
"4.2. CONTINUOUS CONDITIONAL PROBABILITY 169
one is, in eect, tossing a biased coin with probability x for heads. Before further
experimentation, you do not know the value x but past experience might give some
information about its possible values. It is natural to represent this information
by sketching a density function to determine a distribution for x. Thus, we are
considering x to be a continuous random variable, which takes on values between
0 and 1. If you have no knowledge at all, you would sketch the uniform density.
If past experience suggests that x is very likely to be near 2/3 you would sketch
a density with maximum at 2/3 and a spread re
ecting your uncertainly in the
estimate of 2/3. You would then want to nd a density function that reasonably
ts your sketch. The beta densities provide a class of densities that can be t to
most sketches you might make. For example, for  > 1 and  > 1 it is bell-shaped
with the parameters  and  determining its peak and its spread.",prob.pdf
"most sketches you might make. For example, for  > 1 and  > 1 it is bell-shaped
with the parameters  and  determining its peak and its spread.
Assume that the experimenter has chosen a beta density to describe the state of
his knowledge about x before the experiment. Then he gives the drug to n subjects
and records the number i of successes. The number i is a discrete random variable,
so we may conveniently describe the set of possible outcomes of this experiment by
referring to the ordered pair (x; i).
We let m(ijx) denote the probability that we observe i successes given the value
of x. By our assumptions, m(ijx) is the binomial distribution with probability x
for success:
m(ijx) = b(n; x; i) =
n
i

xi(1  x)j ;
where j = n  i.
If x is chosen at random from [0; 1] with a beta density B(; ; x), then the
density function for the outcome of the pair (x; i) is
f(x; i) = m(ijx)B(; ; x)
=
n
i

xi(1  x)j 1
B(; )x1(1  x)1
=
n
i
 1
B(; )x+i1(1  x)+j1 :",prob.pdf
"f(x; i) = m(ijx)B(; ; x)
=
n
i

xi(1  x)j 1
B(; )x1(1  x)1
=
n
i
 1
B(; )x+i1(1  x)+j1 :
Now let m(i) be the probability that we observe i successes not knowing the value
of x. Then
m(i) =
Z 1
0
m(ijx)B(; ; x) dx
=
n
i
 1
B(; )
Z 1
0
x+i1(1  x)+j1 dx
=
n
i
B( + i;  + j)
B(; ) :
Hence, the probability density f(xji) for x, given that i successes were observed, is
f(xji) = f(x; i)
m(i)",prob.pdf
"170 CHAPTER 4. CONDITIONAL PROBABILITY
= x+i1(1  x)+j1
B( + i;  + j) ; (4.5)
that is, f(xji) is another beta density. This says that if we observe i successes and
j failures in n subjects, then the new density for the probability that the drug is
eective is again a beta density but with parameters  + i,  + j.
Now we assume that before the experiment we choose a beta density with pa-
rameters  and , and that in the experiment we obtain i successes in n trials.
We have just seen that in this case, the new density for x is a beta density with
parameters  + i and  + j.
Now we wish to calculate the probability that the drug is eective on the next
subject. For any particular real number t between 0 and 1, the probability that x
has the value t is given by the expression in Equation 4.5. Given that x has the
value t, the probability that the drug is eective on the next subject is just t. Thus,",prob.pdf
"value t, the probability that the drug is eective on the next subject is just t. Thus,
to obtain the probability that the drug is eective on the next subject, we integrate
the product of the expression in Equation 4.5 and t over all possible values of t. We
obtain:
1
B( + i;  + j)
Z 1
0
t 
 t+i1(1  t)+j1 dt
= B( + i + 1;  + j)
B( + i;  + j)
= ( + i)! ( + j  1)!
( +  + i + j)! 
 ( +  + i + j  1)!
( + i  1)! ( + j  1)!
=  + i
 +  + n :
If n is large, then our estimate for the probability of success after the experiment
is approximately the proportion of successes observed in the experiment, which is
certainly a reasonable conclusion. 2
The next example is another in which the true probabilities are unknown and
must be estimated based upon experimental data.
Example 4.24 (Two-armed bandit problem) You are in a casino and confronted by
two slot machines. Each machine pays o either 1 dollar or nothing. The probability",prob.pdf
"two slot machines. Each machine pays o either 1 dollar or nothing. The probability
that the rst machine pays o a dollar is x and that the second machine pays o
a dollar is y. We assume that x and y are random numbers chosen independently
from the interval [0; 1] and unknown to you. You are permitted to make a series of
ten plays, each time choosing one machine or the other. How should you choose to
maximize the number of times that you win?
One strategy that sounds reasonable is to calculate, at every stage, the prob-
ability that each machine will pay o and choose the machine with the higher
probability. Let win(i), for i = 1 or 2, be the number of times that you have won
on the ith machine. Similarly, let lose(i) be the number of times you have lost on
the ith machine. Then, from Example 4.23, the probability p(i) that you win if you",prob.pdf
"4.2. CONTINUOUS CONDITIONAL PROBABILITY 171
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2
2.5  Machine     Result  
     1           W
     1           L
     2           L
     1           L
     1           W
     1           L
     1           L
     1           L
     2           W
     2           L
0
0
Figure 4.10: Play the best machine.
choose the ith machine is
p(i) = win(i) + 1
win(i) + lose(i) + 2 :
Thus, if p(1) > p(2) you would play machine 1 and otherwise you would play
machine 2. We have written a program TwoArm to simulate this experiment. In
the program, the user species the initial values for x and y (but these are unknown
to the experimenter). The program calculates at each stage the two conditional
densities for x and y, given the outcomes of the previous trials, and then computes
p(i), for i = 1, 2. It then chooses the machine with the highest value for the
probability of winning for the next play. The program prints the machine chosen",prob.pdf
"probability of winning for the next play. The program prints the machine chosen
on each play and the outcome of this play. It also plots the new densities for x
(solid line) and y (dotted line), showing only the current densities. We have run
the program for ten plays for the case x = :6 and y = :7. The result is shown in
Figure 4.10.
The run of the program shows the weakness of this strategy. Our initial proba-
bility for winning on the better of the two machines is .7. We start with the poorer
machine and our outcomes are such that we always have a probability greater than
.6 of winning and so we just keep playing this machine even though the other ma-
chine is better. If we had lost on the rst play we would have switched machines.
Our nal density for y is the same as our initial density, namely, the uniform den-
sity. Our nal density for x is dierent and re
ects a much more accurate knowledge
about x. The computer did pretty well with this strategy, winning seven out of the",prob.pdf
"about x. The computer did pretty well with this strategy, winning seven out of the
ten trials, but ten trials are not enough to judge whether this is a good strategy in
the long run.
Another popular strategy is the play-the-winner strategy. As the name suggests,
for this strategy we choose the same machine when we win and switch machines
when we lose. The program TwoArm will simulate this strategy as well. In
Figure 4.11, we show the results of running this program with the play-the-winner
strategy and the same true probabilities of .6 and .7 for the two machines. After
ten plays our densities for the unknown probabilities of winning suggest to us that
the second machine is indeed the better of the two. We again won seven out of the
ten trials.",prob.pdf
"172 CHAPTER 4. CONDITIONAL PROBABILITY
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2
  Machine      Result  
     1           W
     1           W
     1           L
     2           L
     1           W
     1           W
     1           L
     2           L
     1           L
     2           W
Figure 4.11: Play the winner.
Neither of the strategies that we simulated is the best one in terms of maximizing
our average winnings. This best strategy is very complicated but is reasonably ap-
proximated by the play-the-winner strategy. Variations on this example have played
an important role in the problem of clinical tests of drugs where experimenters face
a similar situation. 2
Exercises
1 Pick a point x at random (with uniform density) in the interval [0; 1]. Find
the probability that x > 1=2, given that
(a) x > 1=4.
(b) x < 3=4.
(c) jx  1=2j < 1=4.
(d) x2  x + 2=9 < 0.
2 A radioactive material emits -particles at a rate described by the density
function
f(t) = :1e:1t :",prob.pdf
"(c) jx  1=2j < 1=4.
(d) x2  x + 2=9 < 0.
2 A radioactive material emits -particles at a rate described by the density
function
f(t) = :1e:1t :
Find the probability that a particle is emitted in the rst 10 seconds, given
that
(a) no particle is emitted in the rst second.
(b) no particle is emitted in the rst 5 seconds.
(c) a particle is emitted in the rst 3 seconds.
(d) a particle is emitted in the rst 20 seconds.
3 The Acme Super light bulb is known to have a useful life described by the
density function
f(t) = :01e:01t ;
where time t is measured in hours.",prob.pdf
"4.2. CONTINUOUS CONDITIONAL PROBABILITY 173
(a) Find the failure rate of this bulb (see Exercise 2.2.6).
(b) Find the reliability of this bulb after 20 hours.
(c) Given that it lasts 20 hours, nd the probability that the bulb lasts
another 20 hours.
(d) Find the probability that the bulb burns out in the forty-rst hour, given
that it lasts 40 hours.
4 Suppose you toss a dart at a circular target of radius 10 inches. Given that
the dart lands in the upper half of the target, nd the probability that
(a) it lands in the right half of the target.
(b) its distance from the center is less than 5 inches.
(c) its distance from the center is greater than 5 inches.
(d) it lands within 5 inches of the point (0; 5).
5 Suppose you choose two numbers x and y, independently at random from
the interval [0; 1]. Given that their sum lies in the interval [0; 1], nd the
probability that
(a) jx  yj < 1.
(b) xy < 1=2.
(c) maxfx; yg < 1=2.
(d) x2 + y2 < 1=4.
(e) x > y.",prob.pdf
"probability that
(a) jx  yj < 1.
(b) xy < 1=2.
(c) maxfx; yg < 1=2.
(d) x2 + y2 < 1=4.
(e) x > y.
6 Find the conditional density functions for the following experiments.
(a) A number x is chosen at random in the interval [0; 1], given that x > 1=4.
(b) A number t is chosen at random in the interval [0; 1) with exponential
density et, given that 1 < t < 10.
(c) A dart is thrown at a circular target of radius 10 inches, given that it
falls in the upper half of the target.
(d) Two numbers x and y are chosen at random in the interval [0; 1], given
that x > y.
7 Let x and y be chosen at random from the interval [0; 1]. Show that the events
x > 1=3 and y > 2=3 are independent events.
8 Let x and y be chosen at random from the interval [0; 1]. Which pairs of the
following events are independent?
(a) x > 1=3.
(b) y > 2=3.
(c) x > y.",prob.pdf
"174 CHAPTER 4. CONDITIONAL PROBABILITY
(d) x + y < 1.
9 Suppose that X and Y are continuous random variables with density functions
fX(x) and fY (y), respectively. Let f(x; y) denote the joint density function
of (X; Y ). Show that Z 1
1
f(x; y) dy = fX(x) ;
and Z 1
1
f(x; y) dx = fY (y) :
*10 In Exercise 2.2.12 you proved the following: If you take a stick of unit length
and break it into three pieces, choosing the breaks at random (i.e., choosing
two real numbers independently and uniformly from [0, 1]), then the prob-
ability that the three pieces form a triangle is 1/4. Consider now a similar
experiment: First break the stick at random, then break the longer piece
at random. Show that the two experiments are actually quite dierent, as
follows:
(a) Write a program which simulates both cases for a run of 1000 trials, prints
out the proportion of successes for each run, and repeats this process ten
times. (Call a trial a success if the three pieces do form a triangle.) Have",prob.pdf
"times. (Call a trial a success if the three pieces do form a triangle.) Have
your program pick (x; y) at random in the unit square, and in each case
use x and y to nd the two breaks. For each experiment, have it plot
(x; y) if (x; y) gives a success.
(b) Show that in the second experiment the theoretical probability of success
is actually 2 log2  1.
11 A coin has an unknown bias p that is assumed to be uniformly distributed
between 0 and 1. The coin is tossed n times and heads turns up j times and
tails turns up k times. We have seen that the probability that heads turns up
next time is
j + 1
n + 2 :
Show that this is the same as the probability that the next ball is black for
the Polya urn model of Exercise 4.1.20. Use this result to explain why, in the
Polya urn model, the proportion of black balls does not tend to 0 or 1 as one
might expect but rather to a uniform distribution on the interval [0; 1].
12 Previous experience with a drug suggests that the probability p that the drug",prob.pdf
"12 Previous experience with a drug suggests that the probability p that the drug
is eective is a random quantity having a beta density with parameters  = 2
and  = 3. The drug is used on ten subjects and found to be successful
in four out of the ten patients. What density should we now assign to the
probability p? What is the probability that the drug will be successful the
next time it is used?",prob.pdf
"4.3. PARADOXES 175
13 Write a program to allow you to compare the strategies play-the-winner and
play-the-best-machine for the two-armed bandit problem of Example 4.24.
Have your program determine the initial payo probabilities for each machine
by choosing a pair of random numbers between 0 and 1. Have your program
carry out 20 plays and keep track of the number of wins for each of the two
strategies. Finally, have your program make 1000 repetitions of the 20 plays
and compute the average winning per 20 plays. Which strategy seems to
be the best? Repeat these simulations with 20 replaced by 100. Does your
answer to the above question change?
14 Consider the two-armed bandit problem of Example 4.24. Bruce Barnes pro-
posed the following strategy, which is a variation on the play-the-best-machine
strategy. The machine with the greatest probability of winning is played un-
less the following two conditions hold: (a) the dierence in the probabilities",prob.pdf
"less the following two conditions hold: (a) the dierence in the probabilities
for winning is less than .08, and (b) the ratio of the number of times played
on the more often played machine to the number of times played on the less
often played machine is greater than 1.4. If the above two conditions hold,
then the machine with the smaller probability of winning is played. Write a
program to simulate this strategy. Have your program choose the initial payo
probabilities at random from the unit interval [0; 1], make 20 plays, and keep
track of the number of wins. Repeat this experiment 1000 times and obtain
the average number of wins per 20 plays. Implement a second strategy|for
example, play-the-best-machine or one of your own choice, and see how this
second strategy compares with Bruce's on average wins.
4.3 Paradoxes
Much of this section is based on an article by Snell and Vanderbei.18
One must be very careful in dealing with problems involving conditional prob-",prob.pdf
"Much of this section is based on an article by Snell and Vanderbei.18
One must be very careful in dealing with problems involving conditional prob-
ability. The reader will recall that in the Monty Hall problem (Example 4.6), if
the contestant chooses the door with the car behind it, then Monty has a choice of
doors to open. We made an assumption that in this case, he will choose each door
with probability 1/2. We then noted that if this assumption is changed, the answer
to the original question changes. In this section, we will study other examples of
the same phenomenon.
Example 4.25 Consider a family with two children. Given that one of the children
is a boy, what is the probability that both children are boys?
One way to approach this problem is to say that the other child is equally likely
to be a boy or a girl, so the probability that both children are boys is 1/2. The \text-
book"" solution would be to draw the tree diagram and then form the conditional",prob.pdf
"book"" solution would be to draw the tree diagram and then form the conditional
tree by deleting paths to leave only those paths that are consistent with the given
18J. L. Snell and R. Vanderbei, \Three Bewitching Paradoxes,"" in Topics in Contemporary
Probability and Its Applications, CRC Press, Boca Raton, 1995.",prob.pdf
"176 CHAPTER 4. CONDITIONAL PROBABILITY
First
child
Second
child
Conditional
probability
First
child
Second
child
Unconditional
probability
b
g
b
g
b
g
b
1/4
1/4
1/4
1/4
1/3
1/3
1/31/4
1/4
1/41/2
1/2
1/2
1/2
1/2
b
g
1/2
1/2
1/2
g
b
Unconditional
probability
1/2
1/2
1/2
Figure 4.12: Tree for Example 4.25.
information. The result is shown in Figure 4.12. We see that the probability of two
boys given a boy in the family is not 1/2 but rather 1/3. 2
This problem and others like it are discussed in Bar-Hillel and Falk.19 These
authors stress that the answer to conditional probabilities of this kind can change
depending upon how the information given was actually obtained. For example,
they show that 1/2 is the correct answer for the following scenario.
Example 4.26 Mr. Smith is the father of two. We meet him walking along the
street with a young boy whom he proudly introduces as his son. What is the
probability that Mr. Smith's other child is also a boy?",prob.pdf
"street with a young boy whom he proudly introduces as his son. What is the
probability that Mr. Smith's other child is also a boy?
As usual we have to make some additional assumptions. For example, we will
assume that if Mr. Smith has a boy and a girl, he is equally likely to choose either
one to accompany him on his walk. In Figure 4.13 we show the tree analysis of this
problem and we see that 1/2 is, indeed, the correct answer. 2
Example 4.27 It is not so easy to think of reasonable scenarios that would lead to
the classical 1/3 answer. An attempt was made by Stephen Geller in proposing this
problem to Marilyn vos Savant.20 Geller's problem is as follows: A shopkeeper says
she has two new baby beagles to show you, but she doesn't know whether they're
both male, both female, or one of each sex. You tell her that you want only a male,
and she telephones the fellow who's giving them a bath. \Is at least one a male?""",prob.pdf
"and she telephones the fellow who's giving them a bath. \Is at least one a male?""
19M. Bar-Hillel and R. Falk, \Some teasers concerning conditional probabilities,"" Cognition,
vol. 11 (1982), pgs. 109-122.
20M. vos Savant, \Ask Marilyn,"" Parade Magazine, 9 September; 2 December; 17 February
1990, reprinted in Marilyn vos Savant, Ask Marilyn, St. Martins, New York, 1992.",prob.pdf
"4.3. PARADOXES 177
Mr.Smith's
children
Walking with
Mr.Smith
Unconditional
probability
Mr.Smith's
children
Walking with
Mr. Smith
Unconditional
probability
b
bb b
g
g
g
b
1/4
1/8
1/8
1/8
1/8
1/4
1/4
1/4
1/4
1/4
1
1/2
1/2
1/2
1/2
1
bg
gb
gg
b
bb b
b
1/4
1/8
1/8
1/4
1/4
1/4
1
1/2
1/2
bg
gb
Conditional
probability
1/2
1/4
1/4
Figure 4.13: Tree for Example 4.26.",prob.pdf
"178 CHAPTER 4. CONDITIONAL PROBABILITY
she asks. \Yes,"" she informs you with a smile. What is the probability that the
other one is male?
The reader is asked to decide whether the model which gives an answer of 1/3
is a reasonable one to use in this case. 2
In the preceding examples, the apparent paradoxes could easily be resolved by
clearly stating the model that is being used and the assumptions that are being
made. We now turn to some examples in which the paradoxes are not so easily
resolved.
Example 4.28 Two envelopes each contain a certain amount of money. One en-
velope is given to Ali and the other to Baba and they are told that one envelope
contains twice as much money as the other. However, neither knows who has the
larger prize. Before anyone has opened their envelope, Ali is asked if she would like
to trade her envelope with Baba. She reasons as follows: Assume that the amount
in my envelope is x. If I switch, I will end up with x=2 with probability 1/2, and",prob.pdf
"in my envelope is x. If I switch, I will end up with x=2 with probability 1/2, and
2x with probability 1/2. If I were given the opportunity to play this game many
times, and if I were to switch each time, I would, on average, get
1
2
x
2 + 1
22x = 5
4x :
This is greater than my average winnings if I didn't switch.
Of course, Baba is presented with the same opportunity and reasons in the same
way to conclude that he too would like to switch. So they switch and each thinks
that his/her net worth just went up by 25%.
Since neither has yet opened any envelope, this process can be repeated and so
again they switch. Now they are back with their original envelopes and yet they
think that their fortune has increased 25% twice. By this reasoning, they could
convince themselves that by repeatedly switching the envelopes, they could become
arbitrarily wealthy. Clearly, something is wrong with the above reasoning, but
where is the mistake?",prob.pdf
"arbitrarily wealthy. Clearly, something is wrong with the above reasoning, but
where is the mistake?
One of the tricks of making paradoxes is to make them slightly more dicult than
is necessary to further befuddle us. As John Finn has suggested, in this paradox we
could just have well started with a simpler problem. Suppose Ali and Baba know
that I am going to give then either an envelope with $5 or one with $10 and I am
going to toss a coin to decide which to give to Ali, and then give the other to Baba.
Then Ali can argue that Baba has 2x with probability 1=2 and x=2 with probability
1=2. This leads Ali to the same conclusion as before. But now it is clear that this
is nonsense, since if Ali has the envelope containing $5, Baba cannot possibly have
half of this, namely $2.50, since that was not even one of the choices. Similarly, if
Ali has $10, Baba cannot have twice as much, namely $20. In fact, in this simpler",prob.pdf
"Ali has $10, Baba cannot have twice as much, namely $20. In fact, in this simpler
problem the possibly outcomes are given by the tree diagram in Figure 4.14. From
the diagram, it is clear that neither is made better o by switching. 2
In the above example, Ali's reasoning is incorrect because he infers that if the
amount in his envelope is x, then the probability that his envelope contains the",prob.pdf
"4.3. PARADOXES 179
$5$10
$10$5
1/2
1/21/2
1
11/2
In Ali©s
envelope

In Baba©s
envelope
Figure 4.14: John Finn's version of Example 4.28.
smaller amount is 1/2, and the probability that her envelope contains the larger
amount is also 1/2. In fact, these conditional probabilities depend upon the distri-
bution of the amounts that are placed in the envelopes.
For deniteness, let X denote the positive integer-valued random variable which
represents the smaller of the two amounts in the envelopes. Suppose, in addition,
that we are given the distribution of X, i.e., for each positive integer x, we are given
the value of
px = P(X = x) :
(In Finn's example, p5 = 1, and pn = 0 for all other values of n.) Then it is easy to
calculate the conditional probability that an envelope contains the smaller amount,
given that it contains x dollars. The two possible sample points are (x; x=2) and
(x; 2x). If x is odd, then the rst sample point has probability 0, since x=2 is not",prob.pdf
"(x; 2x). If x is odd, then the rst sample point has probability 0, since x=2 is not
an integer, so the desired conditional probability is 1 that x is the smaller amount.
If x is even, then the two sample points have probabilities px=2 and px, respectively,
so the conditional probability that x is the smaller amount is
px
px=2 + px
;
which is not necessarily equal to 1/2.
Steven Brams and D. Marc Kilgour21 study the problem, for dierent distri-
butions, of whether or not one should switch envelopes, if one's objective is to
maximize the long-term average winnings. Let x be the amount in your envelope.
They show that for any distribution of X, there is at least one value of x such
that you should switch. They give an example of a distribution for which there is
exactly one value of x such that you should switch (see Exercise 5). Perhaps the
most interesting case is a distribution in which you should always switch. We now
give this example.",prob.pdf
"most interesting case is a distribution in which you should always switch. We now
give this example.
Example 4.29 Suppose that we have two envelopes in front of us, and that one
envelope contains twice the amount of money as the other (both amounts are pos-
itive integers). We are given one of the envelopes, and asked if we would like to
switch.
21S. J. Brams and D. M. Kilgour, \The Box Problem: To Switch or Not to Switch,"" Mathematics
Magazine, vol. 68, no. 1 (1995), p. 29.",prob.pdf
"180 CHAPTER 4. CONDITIONAL PROBABILITY
As above, we let X denote the smaller of the two amounts in the envelopes, and
let
px = P(X = x) :
We are now in a position where we can calculate the long-term average winnings, if
we switch. (This long-term average is an example of a probabilistic concept known
as expectation, and will be discussed in Chapter 6.) Given that one of the two
sample points has occurred, the probability that it is the point (x; x=2) is
px=2
px=2 + px
;
and the probability that it is the point (x; 2x) is
px
px=2 + px
:
Thus, if we switch, our long-term average winnings are
px=2
px=2 + px
x
2 + px
px=2 + px
2x :
If this is greater than x, then it pays in the long run for us to switch. Some routine
algebra shows that the above expression is greater than x if and only if
px=2
px=2 + px
< 2
3 : (4.6)
It is interesting to consider whether there is a distribution on the positive integers
such that the inequality 4.6 is true for all even values of x. Brams and Kilgour22",prob.pdf
"such that the inequality 4.6 is true for all even values of x. Brams and Kilgour22
give the following example.
We dene px as follows:
px =
(
1
3

2
3
k1
; if x = 2k;
0; otherwise.
It is easy to calculate (see Exercise 4) that for all relevant values of x, we have
px=2
px=2 + px
= 3
5 ;
which means that the inequality 4.6 is always true. 2
So far, we have been able to resolve paradoxes by clearly stating the assumptions
being made and by precisely stating the models being used. We end this section by
describing a paradox which we cannot resolve.
Example 4.30 Suppose that we have two envelopes in front of us, and we are
told that the envelopes contain X and Y dollars, respectively, where X and Y are
dierent positive integers. We randomly choose one of the envelopes, and we open
22ibid.",prob.pdf
"4.3. PARADOXES 181
it, revealing X, say. Is it possible to determine, with probability greater than 1/2,
whether X is the smaller of the two dollar amounts?
Even if we have no knowledge of the joint distribution of X and Y , the surprising
answer is yes! Here's how to do it. Toss a fair coin until the rst time that heads
turns up. Let Z denote the number of tosses required plus 1/2. If Z > X, then we
say that X is the smaller of the two amounts, and if Z < X, then we say that X is
the larger of the two amounts.
First, if Z lies between X and Y , then we are sure to be correct. Since X and
Y are unequal, Z lies between them with positive probability. Second, if Z is not
between X and Y , then Z is either greater than both X and Y , or is less than both
X and Y . In either case, X is the smaller of the two amounts with probability 1/2,
by symmetry considerations (remember, we chose the envelope at random). Thus,
the probability that we are correct is greater than 1/2. 2
Exercises",prob.pdf
"by symmetry considerations (remember, we chose the envelope at random). Thus,
the probability that we are correct is greater than 1/2. 2
Exercises
1 One of the rst conditional probability paradoxes was provided by Bertrand.23
It is called the Box Paradox. A cabinet has three drawers. In the rst drawer
there are two gold balls, in the second drawer there are two silver balls, and
in the third drawer there is one silver and one gold ball. A drawer is picked at
random and a ball chosen at random from the two balls in the drawer. Given
that a gold ball was drawn, what is the probability that the drawer with the
two gold balls was chosen?
2 The following problem is called the two aces problem. This problem, dat-
ing back to 1936, has been attributed to the English mathematician J. H.
C. Whitehead (see Gridgeman24). This problem was also submitted to Mar-
ilyn vos Savant by the master of mathematical puzzles Martin Gardner, who
remarks that it is one of his favorites.",prob.pdf
"ilyn vos Savant by the master of mathematical puzzles Martin Gardner, who
remarks that it is one of his favorites.
A bridge hand has been dealt, i. e. thirteen cards are dealt to each player.
Given that your partner has at least one ace, what is the probability that he
has at least two aces? Given that your partner has the ace of hearts, what
is the probability that he has at least two aces? Answer these questions for
a version of bridge in which there are eight cards, namely four aces and four
kings, and each player is dealt two cards. (The reader may wish to solve the
problem with a 52-card deck.)
3 In the preceding exercise, it is natural to ask \How do we get the information
that the given hand has an ace?"" Gridgeman considers two dierent ways
that we might get this information. (Again, assume the deck consists of eight
cards.)
(a) Assume that the person holding the hand is asked to \Name an ace in
your hand"" and answers \The ace of hearts."" What is the probability",prob.pdf
"cards.)
(a) Assume that the person holding the hand is asked to \Name an ace in
your hand"" and answers \The ace of hearts."" What is the probability
that he has a second ace?
23J. Bertrand, Calcul des Probabilites, Gauthier-Uillars, 1888.
24N. T. Gridgeman, Letter, American Statistician, 21 (1967), pgs. 38-39.",prob.pdf
"182 CHAPTER 4. CONDITIONAL PROBABILITY
(b) Suppose the person holding the hand is asked the more direct question
\Do you have the ace of hearts?"" and the answer is yes. What is the
probability that he has a second ace?
4 Using the notation introduced in Example 4.29, show that in the example of
Brams and Kilgour, if x is a positive power of 2, then
px=2
px=2 + px
= 3
5 :
5 Using the notation introduced in Example 4.29, let
px =
(
2
3

1
3
k
; if x = 2k;
0; otherwise.
Show that there is exactly one value of x such that if your envelope contains
x, then you should switch.
*6 (For bridge players only. From Sutherland.25) Suppose that we are the de-
clarer in a hand of bridge, and we have the king, 9, 8, 7, and 2 of a certain
suit, while the dummy has the ace, 10, 5, and 4 of the same suit. Suppose
that we want to play this suit in such a way as to maximize the probability
of having no losers in the suit. We begin by leading the 2 to the ace, and we",prob.pdf
"of having no losers in the suit. We begin by leading the 2 to the ace, and we
note that the queen drops on our left. We then lead the 10 from the dummy,
and our right-hand opponent plays the six (after playing the three on the rst
round). Should we nesse or play for the drop?
25E. Sutherland, \Restricted Choice | Fact or Fiction?"", Canadian Master Point, November
1, 1993.",prob.pdf
"Chapter 5
Important Distributions and
Densities
5.1 Important Distributions
In this chapter, we describe the discrete probability distributions and the continuous
probability densities that occur most often in the analysis of experiments. We will
also show how one simulates these distributions and densities on a computer.
Discrete Uniform Distribution
In Chapter 1, we saw that in many cases, we assume that all outcomes of an exper-
iment are equally likely. If X is a random variable which represents the outcome
of an experiment of this type, then we say that X is uniformly distributed. If the
sample space S is of size n, where 0 < n < 1, then the distribution function m(!)
is dened to be 1=n for all ! 2 S. As is the case with all of the discrete probabil-
ity distributions discussed in this chapter, this experiment can be simulated on a
computer using the program GeneralSimulation. However, in this case, a faster",prob.pdf
"computer using the program GeneralSimulation. However, in this case, a faster
algorithm can be used instead. (This algorithm was described in Chapter 1; we
repeat the description here for completeness.) The expression
1 + bn (rnd)c
takes on as a value each integer between 1 and n with probability 1=n (the notation
bxc denotes the greatest integer not exceeding x). Thus, if the possible outcomes
of the experiment are labelled !1 !2; : : :; !n, then we use the above expression to
represent the subscript of the output of the experiment.
If the sample space is a countably innite set, such as the set of positive integers,
then it is not possible to have an experiment which is uniform on this set (see
Exercise 3). If the sample space is an uncountable set, with positive, nite length,
such as the interval [0; 1], then we use continuous density functions (see Section 5.2).
183",prob.pdf
"184 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Binomial Distribution
The binomial distribution with parameters n, p, and k was dened in Chapter 3. It
is the distribution of the random variable which counts the number of heads which
occur when a coin is tossed n times, assuming that on any one toss, the probability
that a head occurs is p. The distribution function is given by the formula
b(n; p; k) =
n
k

pkqnk ;
where q = 1  p.
One straightforward way to simulate a binomial random variable X is to compute
the sum of n independent 01 random variables, each of which take on the value 1
with probability p. This method requires n calls to a random number generator to
obtain one value of the random variable. When n is relatively large (say at least 30),
the Central Limit Theorem (see Chapter 9) implies that the binomial distribution is
well-approximated by the corresponding normal density function (which is dened
in Section 5.2) with parameters  = np and  = pnpq. Thus, in this case we",prob.pdf
"in Section 5.2) with parameters  = np and  = pnpq. Thus, in this case we
can compute a value Y of a normal random variable with these parameters, and if
1=2  Y < n + 1=2, we can use the value
bY + 1=2c
to represent the random variable X. If Y < 1=2 or Y > n + 1=2, we reject Y and
compute another value. We will see in the next section how we can quickly simulate
normal random variables.
Geometric Distribution
Consider a Bernoulli trials process continued for an innite number of trials; for
example, a coin tossed an innite sequence of times. We showed in Section 2.2 how
to assign a probability distribution to the innite tree. Thus, we can determine
the distribution for any random variable X relating to the experiment provided
P(X = a) can be computed in terms of a nite number of trials. For example, let
T be the number of trials up to and including the rst success. Then
P(T = 1) = p ;
P(T = 2) = qp ;
P(T = 3) = q2p ;
and in general,
P(T = n) = qn1p :",prob.pdf
"P(T = 1) = p ;
P(T = 2) = qp ;
P(T = 3) = q2p ;
and in general,
P(T = n) = qn1p :
To show that this is a distribution, we must show that
p + qp + q2p + 
 
 
 = 1 :",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 185
5 10 15 20
0
0.2
0.4 p= .5
5 10 15 20
0
0.05
0.1
0.15
0.2
p= .2
Figure 5.1: Geometric distributions.
The left-hand expression is just a geometric series with rst term p and common
ratio q, so its sum is p
1  q
which equals 1.
In Figure 5.1 we have plotted this distribution using the program Geometric-
Plot for the cases p = :5 and p = :2. We see that as p decreases we are more likely
to get large values for T, as would be expected. In both cases, the most probable
value for T is 1. This will always be true since
P(T = j + 1)
P(T = j) = q < 1 :
In general, if 0 < p < 1, and q = 1  p, then we say that the random variable T
has a geometric distribution if
P(T = j) = qj1p ;
for j = 1; 2; 3; : : : .
To simulate the geometric distribution with parameter p, we can simply compute
a sequence of random numbers in [0; 1), stopping when an entry does not exceed p.
However, for small values of p, this is time-consuming (taking, on the average, 1=p",prob.pdf
"However, for small values of p, this is time-consuming (taking, on the average, 1=p
steps). We now describe a method whose running time does not depend upon the
size of p. Dene Y to be the smallest integer satisfying the inequality
1  qY  rnd : (5.1)
Then we have
P(Y = j) = P

1  qj  rnd > 1  qj1

= qj1  qj
= qj1(1  q)
= qj1p :",prob.pdf
"186 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Thus, Y is geometrically distributed with parameter p. To generate Y , all we have
to do is solve Equation 5.1 for Y . We obtain
Y =
&
log(1  rnd)
log q
'
;
where the notation dxe means the least integer which is greater than or equal to x.
Since log(1rnd) and log(rnd) are identically distributed, Y can also be generated
using the equation
Y =
&
log rnd
log q
'
:
Example 5.1 The geometric distribution plays an important role in the theory of
queues, or waiting lines. For example, suppose a line of customers waits for service
at a counter. It is often assumed that, in each small time unit, either 0 or 1 new
customers arrive at the counter. The probability that a customer arrives is p and
that no customer arrives is q = 1  p. Then the time T until the next arrival has
a geometric distribution. It is natural to ask for the probability that no customer
arrives in the next k time units, that is, for P(T > k). This is given by
P(T > k) =
1X
j=k+1",prob.pdf
"arrives in the next k time units, that is, for P(T > k). This is given by
P(T > k) =
1X
j=k+1
qj1p = qk(p + qp + q2p + 
 
 
)
= qk :
This probability can also be found by noting that we are asking for no successes
(i.e., arrivals) in a sequence of k consecutive time units, where the probability of a
success in any one time unit is p. Thus, the probability is just qk, since arrivals in
any two time units are independent events.
It is often assumed that the length of time required to service a customer also
has a geometric distribution but with a dierent value for p. This implies a rather
special property of the service time. To see this, let us compute the conditional
probability
P(T > r + s j T > r) = P(T > r + s)
P(T > r) = qr+s
qr = qs :
Thus, the probability that the customer's service takes s more time units is inde-
pendent of the length of time r that the customer has already been served. Because",prob.pdf
"pendent of the length of time r that the customer has already been served. Because
of this interpretation, this property is called the \memoryless"" property, and is also
obeyed by the exponential distribution. (Fortunately, not too many service stations
have this property.) 2
Negative Binomial Distribution
Suppose we are given a coin which has probability p of coming up heads when it is
tossed. We x a positive integer k, and toss the coin until the kth head appears. We",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 187
let X represent the number of tosses. When k = 1, X is geometrically distributed.
For a general k, we say that X has a negative binomial distribution. We now
calculate the probability distribution of X. If X = x, then it must be true that
there were exactly k  1 heads thrown in the rst x  1 tosses, and a head must
have been thrown on the xth toss. There are
x  1
k  1

sequences of length x with these properties, and each of them is assigned the same
probability, namely
pk1qxk :
Therefore, if we dene
u(x; k; p) = P(X = x) ;
then
u(x; k; p) =
x  1
k  1

pkqxk :
One can simulate this on a computer by simulating the tossing of a coin. The
following algorithm is, in general, much faster. We note that X can be understood
as the sum of k outcomes of a geometrically distributed experiment with parameter
p. Thus, we can use the following sum as a means of generating X:
kX
j=1
&
log rndj
log q
'
:",prob.pdf
"p. Thus, we can use the following sum as a means of generating X:
kX
j=1
&
log rndj
log q
'
:
Example 5.2 A fair coin is tossed until the second time a head turns up. The
distribution for the number of tosses is u(x; 2; p). Thus the probability that x tosses
are needed to obtain two heads is found by letting k = 2 in the above formula. We
obtain
u(x; 2; 1=2) =
x  1
1
1
2x ;
for x = 2; 3; : : : .
In Figure 5.2 we give a graph of the distribution for k = 2 and p = :25. Note
that the distribution is quite asymmetric, with a long tail re
ecting the fact that
large values of x are possible. 2
Poisson Distribution
The Poisson distribution arises in many situations. It is safe to say that it is one of
the three most important discrete probability distributions (the other two being the
uniform and the binomial distributions). The Poisson distribution can be viewed
as arising from the binomial distribution or from the exponential density. We shall",prob.pdf
"as arising from the binomial distribution or from the exponential density. We shall
now explain its connection with the former; its connection with the latter will be
explained in the next section.",prob.pdf
"188 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
5 10 15 20 25 30
0
0.02
0.04
0.06
0.08
0.1
Figure 5.2: Negative binomial distribution with k = 2 and p = :25.
Suppose that we have a situation in which a certain kind of occurrence happens
at random over a period of time. For example, the occurrences that we are interested
in might be incoming telephone calls to a police station in a large city. We want
to model this situation so that we can consider the probabilities of events such
as more than 10 phone calls occurring in a 5-minute time interval. Presumably,
in our example, there would be more incoming calls between 6:00 and 7:00 P.M.
than between 4:00 and 5:00 A.M., and this fact would certainly aect the above
probability. Thus, to have a hope of computing such probabilities, we must assume
that the average rate, i.e., the average number of occurrences per minute, is a
constant. This rate we will denote by . (Thus, in a given 5-minute time interval,",prob.pdf
"constant. This rate we will denote by . (Thus, in a given 5-minute time interval,
we would expect about 5 occurrences.) This means that if we were to apply our
model to the two time periods given above, we would simply use dierent rates
for the two time periods, thereby obtaining two dierent probabilities for the given
event.
Our next assumption is that the number of occurrences in two non-overlapping
time intervals are independent. In our example, this means that the events that
there are j calls between 5:00 and 5:15 P.M. and k calls between 6:00 and 6:15 P.M.
on the same day are independent.
We can use the binomial distribution to model this situation. We imagine that
a given time interval is broken up into n subintervals of equal length. If the subin-
tervals are suciently short, we can assume that two or more occurrences happen
in one subinterval with a probability which is negligible in comparison with the",prob.pdf
"in one subinterval with a probability which is negligible in comparison with the
probability of at most one occurrence. Thus, in each subinterval, we are assuming
that there is either 0 or 1 occurrence. This means that the sequence of subintervals
can be thought of as a sequence of Bernoulli trials, with a success corresponding to
an occurrence in the subinterval.",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 189
To decide upon the proper value of p, the probability of an occurrence in a given
subinterval, we reason as follows. On the average, there are t occurrences in a
time interval of length t. If this time interval is divided into n subintervals, then
we would expect, using the Bernoulli trials interpretation, that there should be np
occurrences. Thus, we want
t = np ;
so
p = t
n :
We now wish to consider the random variable X, which counts the number of
occurrences in a given time interval. We want to calculate the distribution of X.
For ease of calculation, we will assume that the time interval is of length 1; for time
intervals of arbitrary length t, see Exercise 11. We know that
P(X = 0) = b(n; p; 0) = (1  p)n =

1  
n
n
:
For large n, this is approximately e. It is easy to calculate that for any xed k,
we have
b(n; p; k)
b(n; p; k  1) =   (k  1)p
kq
which, for large n (and therefore small p) is approximately =k. Thus, we have",prob.pdf
"we have
b(n; p; k)
b(n; p; k  1) =   (k  1)p
kq
which, for large n (and therefore small p) is approximately =k. Thus, we have
P(X = 1)  e ;
and in general,
P(X = k)  k
k! e : (5.2)
The above distribution is the Poisson distribution. We note that it must be checked
that the distribution given in Equation 5.2 really is a distribution, i.e., that its
values are non-negative and sum to 1. (See Exercise 12.)
The Poisson distribution is used as an approximation to the binomial distribu-
tion when the parameters n and p are large and small, respectively (see Examples 5.3
and 5.4). However, the Poisson distribution also arises in situations where it may
not be easy to interpret or measure the parameters n and p (see Example 5.5).
Example 5.3 A typesetter makes, on the average, one mistake per 1000 words.
Assume that he is setting a book with 100 words to a page. Let S100 be the number
of mistakes that he makes on a single page. Then the exact probability distribution",prob.pdf
"of mistakes that he makes on a single page. Then the exact probability distribution
for S100 would be obtained by considering S100 as a result of 100 Bernoulli trials
with p = 1=1000. The expected value of S100 is  = 100(1=1000) = :1. The exact
probability that S100 = j is b(100; 1=1000; j), and the Poisson approximation is
e:1(:1)j
j! :
In Table 5.1 we give, for various values of n and p, the exact values computed by
the binomial distribution and the Poisson approximation. 2",prob.pdf
"190 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Poisson Binomial Poisson Binomial Poisson Binomial
n = 100 n = 100 n = 1000
j  = :1 p = :001  = 1 p = :01  = 10 p = :01
0 .9048 .9048 .3679 .3660 .0000 .0000
1 .0905 .0905 .3679 .3697 .0005 .0004
2 .0045 .0045 .1839 .1849 .0023 .0022
3 .0002 .0002 .0613 .0610 .0076 .0074
4 .0000 .0000 .0153 .0149 .0189 .0186
5 .0031 .0029 .0378 .0374
6 .0005 .0005 .0631 .0627
7 .0001 .0001 .0901 .0900
8 .0000 .0000 .1126 .1128
9 .1251 .1256
10 .1251 .1257
11 .1137 .1143
12 .0948 .0952
13 .0729 .0731
14 .0521 .0520
15 .0347 .0345
16 .0217 .0215
17 .0128 .0126
18 .0071 .0069
19 .0037 .0036
20 .0019 .0018
21 .0009 .0009
22 .0004 .0004
23 .0002 .0002
24 .0001 .0001
25 .0000 .0000
Table 5.1: Poisson approximation to the binomial distribution.",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 191
Example 5.4 In his book,1 Feller discusses the statistics of 
ying bomb hits in the
south of London during the Second World War.
Assume that you live in a district of size 10 blocks by 10 blocks so that the total
district is divided into 100 small squares. How likely is it that the square in which
you live will receive no hits if the total area is hit by 400 bombs?
We assume that a particular bomb will hit your square with probability 1/100.
Since there are 400 bombs, we can regard the number of hits that your square
receives as the number of successes in a Bernoulli trials process with n = 400 and
p = 1=100. Thus we can use the Poisson distribution with  = 400 
 1=100 = 4 to
approximate the probability that your square will receive j hits. This probability
is p(j) = e44j=j!. The expected number of squares that receive exactly j hits
is then 100 
 p(j). It is easy to write a program LondonBombs to simulate this",prob.pdf
"is then 100 
 p(j). It is easy to write a program LondonBombs to simulate this
situation and compare the expected number of squares with j hits with the observed
number. In Exercise 26 you are asked to compare the actual observed data with
that predicted by the Poisson distribution.
In Figure 5.3, we have shown the simulated hits, together with a spike graph
showing both the observed and predicted frequencies. The observed frequencies are
shown as squares, and the predicted frequencies are shown as dots. 2
If the reader would rather not consider 
ying bombs, he is invited to instead consider
an analogous situation involving cookies and raisins. We assume that we have made
enough cookie dough for 500 cookies. We put 600 raisins in the dough, and mix it
thoroughly. One way to look at this situation is that we have 500 cookies, and after
placing the cookies in a grid on the table, we throw 600 raisins at the cookies. (See
Exercise 22.)",prob.pdf
"placing the cookies in a grid on the table, we throw 600 raisins at the cookies. (See
Exercise 22.)
Example 5.5 Suppose that in a certain xed amount A of blood, the average
human has 40 white blood cells. Let X be the random variable which gives the
number of white blood cells in a random sample of size A from a random individual.
We can think of X as binomially distributed with each white blood cell in the body
representing a trial. If a given white blood cell turns up in the sample, then the
trial corresponding to that blood cell was a success. Then p should be taken as
the ratio of A to the total amount of blood in the individual, and n will be the
number of white blood cells in the individual. Of course, in practice, neither of
these parameters is very easy to measure accurately, but presumably the number
40 is easy to measure. But for the average human, we then have 40 = np, so we
can think of X as being Poisson distributed, with parameter  = 40. In this case,",prob.pdf
"can think of X as being Poisson distributed, with parameter  = 40. In this case,
it is easier to model the situation using the Poisson distribution than the binomial
distribution. 2
To simulate a Poisson random variable on a computer, a good way is to take
advantage of the relationship between the Poisson distribution and the exponential
density. This relationship and the resulting simulation algorithm will be described
in the next section.
1ibid., p. 161.",prob.pdf
"192 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
0 2 4 6 8 10
0
0.05
0.1
0.15
0.2
Figure 5.3: Flying bomb hits.",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 193
Hypergeometric Distribution
Suppose that we have a set of N balls, of which k are red and N  k are blue. We
choose n of these balls, without replacement, and dene X to be the number of red
balls in our sample. The distribution of X is called the hypergeometric distribution.
We note that this distribution depends upon three parameters, namely N, k, and
n. There does not seem to be a standard notation for this distribution; we will use
the notation h(N; k; n; x) to denote P(X = x). This probability can be found by
noting that there are N
n

dierent samples of size n, and the number of such samples with exactly x red balls
is obtained by multiplying the number of ways of choosing x red balls from the set
of k red balls and the number of ways of choosing n  x blue balls from the set of
N  k blue balls. Hence, we have
h(N; k; n; x) =
k
x

Nk
nx


N
n

 :
This distribution can be generalized to the case where there are more than two",prob.pdf
"h(N; k; n; x) =
k
x

Nk
nx


N
n

 :
This distribution can be generalized to the case where there are more than two
types of objects. (See Exercise 40.)
If we let N and k tend to 1, in such a way that the ratio k=N remains xed, then
the hypergeometric distribution tends to the binomial distribution with parameters
n and p = k=N. This is reasonable because if N and k are much larger than n, then
whether we choose our sample with or without replacement should not aect the
probabilities very much, and the experiment consisting of choosing with replacement
yields a binomially distributed random variable (see Exercise 44).
An example of how this distribution might be used is given in Exercises 36 and
37. We now give another example involving the hypergeometric distribution. It
illustrates a statistical test called Fisher's Exact Test.
Example 5.6 It is often of interest to consider two traits, such as eye color and",prob.pdf
"illustrates a statistical test called Fisher's Exact Test.
Example 5.6 It is often of interest to consider two traits, such as eye color and
hair color, and to ask whether there is an association between the two traits. Two
traits are associated if knowing the value of one of the traits for a given person
allows us to predict the value of the other trait for that person. The stronger the
association, the more accurate the predictions become. If there is no association
between the traits, then we say that the traits are independent. In this example, we
will use the traits of gender and political party, and we will assume that there are
only two possible genders, female and male, and only two possible political parties,
Democratic and Republican.
Suppose that we have collected data concerning these traits. To test whether
there is an association between the traits, we rst assume that there is no association",prob.pdf
"there is an association between the traits, we rst assume that there is no association
between the two traits. This gives rise to an \expected"" data set, in which knowledge
of the value of one trait is of no help in predicting the value of the other trait. Our
collected data set usually diers from this expected data set. If it diers by quite a
bit, then we would tend to reject the assumption of independence of the traits. To",prob.pdf
"194 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Democrat Republican
Female 24 4 28
Male 8 14 22
32 18 50
Table 5.2: Observed data.
Democrat Republican
Female s11 s12 t11
Male s21 s22 t12
t21 t22 n
Table 5.3: General data table.
nail down what is meant by \quite a bit,"" we decide which possible data sets dier
from the expected data set by at least as much as ours does, and then we compute
the probability that any of these data sets would occur under the assumption of
independence of traits. If this probability is small, then it is unlikely that the
dierence between our collected data set and the expected data set is due entirely
to chance.
Suppose that we have collected the data shown in Table 5.2. The row and column
sums are called marginal totals, or marginals. In what follows, we will denote the
row sums by t11 and t12, and the column sums by t21 and t22. The ijth entry in
the table will be denoted by sij. Finally, the size of the data set will be denoted",prob.pdf
"the table will be denoted by sij. Finally, the size of the data set will be denoted
by n. Thus, a general data table will look as shown in Table 5.3. We now explain
the model which will be used to construct the \expected"" data set. In the model,
we assume that the two traits are independent. We then put t21 yellow balls and
t22 green balls, corresponding to the Democratic and Republican marginals, into
an urn. We draw t11 balls, without replacement, from the urn, and call these balls
females. The t12 balls remaining in the urn are called males. In the specic case
under consideration, the probability of getting the actual data under this model is
given by the expression
32
24

18
4


50
28

 ;
i.e., a value of the hypergeometric distribution.
We are now ready to construct the expected data set. If we choose 28 balls
out of 50, we should expect to see, on the average, the same percentage of yellow
balls in our sample as in the urn. Thus, we should expect to see, on the average,",prob.pdf
"balls in our sample as in the urn. Thus, we should expect to see, on the average,
28(32=50) = 17:92  18 yellow balls in our sample. (See Exercise 36.) The other
expected values are computed in exactly the same way. Thus, the expected data
set is shown in Table 5.4. We note that the value of s11 determines the other
three values in the table, since the marginals are all xed. Thus, in considering
the possible data sets that could appear in this model, it is enough to consider the
various possible values of s11. In the specic case at hand, what is the probability",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 195
Democrat Republican
Female 18 10 28
Male 14 8 22
32 18 50
Table 5.4: Expected data.
of drawing exactly a yellow balls, i.e., what is the probability that s11 = a? It is
32
a

 18
28a


50
28

 : (5.3)
We are now ready to decide whether our actual data diers from the expected
data set by an amount which is greater than could be reasonably attributed to
chance alone. We note that the expected number of female Democrats is 18, but
the actual number in our data is 24. The other data sets which dier from the
expected data set by more than ours correspond to those where the number of
female Democrats equals 25, 26, 27, or 28. Thus, to obtain the required probability,
we sum the expression in (5.3) from a = 24 to a = 28. We obtain a value of :000395.
Thus, we should reject the hypothesis that the two traits are independent. 2
Finally, we turn to the question of how to simulate a hypergeometric random",prob.pdf
"Finally, we turn to the question of how to simulate a hypergeometric random
variable X. Let us assume that the parameters for X are N, k, and n. We imagine
that we have a set of N balls, labelled from 1 to N. We decree that the rst k of
these balls are red, and the rest are blue. Suppose that we have chosen m balls,
and that j of them are red. Then there are k  j red balls left, and N  m balls
left. Thus, our next choice will be red with probability
k  j
N  m :
So at this stage, we choose a random number in [0; 1], and report that a red ball has
been chosen if and only if the random number does not exceed the above expression.
Then we update the values of m and j, and continue until n balls have been chosen.
Benford Distribution
Our next example of a distribution comes from the study of leading digits in data
sets. It turns out that many data sets that occur \in real life"" have the property that",prob.pdf
"sets. It turns out that many data sets that occur \in real life"" have the property that
the rst digits of the data are not uniformly distributed over the set f1; 2; : : :; 9g.
Rather, it appears that the digit 1 is most likely to occur, and that the distribution
is monotonically decreasing on the set of possible digits. The Benford distribution
appears, in many cases, to t such data. Many explanations have been given for the
occurrence of this distribution. Possibly the most convincing explanation is that
this distribution is the only one that is invariant under a change of scale. If one
thinks of certain data sets as somehow \naturally occurring,"" then the distribution
should be unaected by which units are chosen in which to represent the data, i.e.,
the distribution should be invariant under change of scale.",prob.pdf
"196 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
2 4 6 8
0
0.05
0.1
0.15
0.2
0.25
0.3
Figure 5.4: Leading digits in President Clinton's tax returns.
Theodore Hill2 gives a general description of the Benford distribution, when one
considers the rst d digits of integers in a data set. We will restrict our attention
to the rst digit. In this case, the Benford distribution has distribution function
f(k) = log10(k + 1)  log10(k) ;
for 1  k  9.
Mark Nigrini3 has advocated the use of the Benford distribution as a means
of testing suspicious nancial records such as bookkeeping entries, checks, and tax
returns. His idea is that if someone were to \make up"" numbers in these cases,
the person would probably produce numbers that are fairly uniformly distributed,
while if one were to use the actual numbers, the leading digits would roughly follow
the Benford distribution. As an example, Nigrini analyzed President Clinton's tax",prob.pdf
"the Benford distribution. As an example, Nigrini analyzed President Clinton's tax
returns for a 13-year period. In Figure 5.4, the Benford distribution values are
shown as squares, and the President's tax return data are shown as circles. One
sees that in this example, the Benford distribution ts the data very well.
This distribution was discovered by the astronomer Simon Newcomb who stated
the following in his paper on the subject: \That the ten digits do not occur with
equal frequency must be evident to anyone making use of logarithm tables, and
noticing how much faster the rst pages wear out than the last ones. The rst
signicant gure is oftener 1 than any other digit, and the frequency diminishes up
to 9.""4
2T. P. Hill, \The Signicant Digit Phenomenon,"" American Mathematical Monthly, vol. 102,
no. 4 (April 1995), pgs. 322-327.
3M. Nigrini, \Detecting Biases and Irregularities in Tabulated Data,"" working paper",prob.pdf
"no. 4 (April 1995), pgs. 322-327.
3M. Nigrini, \Detecting Biases and Irregularities in Tabulated Data,"" working paper
4S. Newcomb, \Note on the frequency of use of the dierent digits in natural numbers,"" Amer-
ican Journal of Mathematics, vol. 4 (1881), pgs. 39-40.",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 197
Exercises
1 For which of the following random variables would it be appropriate to assign
a uniform distribution?
(a) Let X represent the roll of one die.
(b) Let X represent the number of heads obtained in three tosses of a coin.
(c) A roulette wheel has 38 possible outcomes: 0, 00, and 1 through 36. Let
X represent the outcome when a roulette wheel is spun.
(d) Let X represent the birthday of a randomly chosen person.
(e) Let X represent the number of tosses of a coin necessary to achieve a
head for the rst time.
2 Let n be a positive integer. Let S be the set of integers between 1 and n.
Consider the following process: We remove a number from S at random and
write it down. We repeat this until S is empty. The result is a permutation
of the integers from 1 to n. Let X denote this permutation. Is X uniformly
distributed?
3 Let X be a random variable which can take on countably many values. Show
that X cannot be uniformly distributed.",prob.pdf
"distributed?
3 Let X be a random variable which can take on countably many values. Show
that X cannot be uniformly distributed.
4 Suppose we are attending a college which has 3000 students. We wish to
choose a subset of size 100 from the student body. Let X represent the subset,
chosen using the following possible strategies. For which strategies would it
be appropriate to assign the uniform distribution to X? If it is appropriate,
what probability should we assign to each outcome?
(a) Take the rst 100 students who enter the cafeteria to eat lunch.
(b) Ask the Registrar to sort the students by their Social Security number,
and then take the rst 100 in the resulting list.
(c) Ask the Registrar for a set of cards, with each card containing the name
of exactly one student, and with each student appearing on exactly one
card. Throw the cards out of a third-story window, then walk outside
and pick up the rst 100 cards that you nd.",prob.pdf
"card. Throw the cards out of a third-story window, then walk outside
and pick up the rst 100 cards that you nd.
5 Under the same conditions as in the preceding exercise, can you describe
a procedure which, if used, would produce each possible outcome with the
same probability? Can you describe such a procedure that does not rely on a
computer or a calculator?
6 Let X1; X2; : : : ; Xn be n mutually independent random variables, each of
which is uniformly distributed on the integers from 1 to k. Let Y denote the
minimum of the Xi's. Find the distribution of Y .
7 A die is rolled until the rst time T that a six turns up.
(a) What is the probability distribution for T?",prob.pdf
"198 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
(b) Find P(T > 3).
(c) Find P(T > 6jT > 3).
8 If a coin is tossed a sequence of times, what is the probability that the rst
head will occur after the fth toss, given that it has not occurred in the rst
two tosses?
9 A worker for the Department of Fish and Game is assigned the job of esti-
mating the number of trout in a certain lake of modest size. She proceeds as
follows: She catches 100 trout, tags each of them, and puts them back in the
lake. One month later, she catches 100 more trout, and notes that 10 of them
have tags.
(a) Without doing any fancy calculations, give a rough estimate of the num-
ber of trout in the lake.
(b) Let N be the number of trout in the lake. Find an expression, in terms
of N, for the probability that the worker would catch 10 tagged trout
out of the 100 trout that she caught the second time.
(c) Find the value of N which maximizes the expression in part (b). This",prob.pdf
"out of the 100 trout that she caught the second time.
(c) Find the value of N which maximizes the expression in part (b). This
value is called the maximum likelihood estimate for the unknown quantity
N. Hint: Consider the ratio of the expressions for successive values of
N.
10 A census in the United States is an attempt to count everyone in the country.
It is inevitable that many people are not counted. The U. S. Census Bureau
proposed a way to estimate the number of people who were not counted by
the latest census. Their proposal was as follows: In a given locality, let N
denote the actual number of people who live there. Assume that the census
counted n1 people living in this area. Now, another census was taken in the
locality, and n2 people were counted. In addition, n12 people were counted
both times.
(a) Given N, n1, and n2, let X denote the number of people counted both
times. Find the probability that X = k, where k is a xed positive
integer between 0 and n2.",prob.pdf
"times. Find the probability that X = k, where k is a xed positive
integer between 0 and n2.
(b) Now assume that X = n12. Find the value of N which maximizes the
expression in part (a). Hint: Consider the ratio of the expressions for
successive values of N.
11 Suppose that X is a random variable which represents the number of calls
coming in to a police station in a one-minute interval. In the text, we showed
that X could be modelled using a Poisson distribution with parameter ,
where this parameter represents the average number of incoming calls per
minute. Now suppose that Y is a random variable which represents the num-
ber of incoming calls in an interval of length t. Show that the distribution of
Y is given by
P(Y = k) = et (t)k
k! ;",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 199
i.e., Y is Poisson with parameter t. Hint: Suppose a Martian were to observe
the police station. Let us also assume that the basic time interval used on
Mars is exactly t Earth minutes. Finally, we will assume that the Martian
understands the derivation of the Poisson distribution in the text. What
would she write down for the distribution of Y ?
12 Show that the values of the Poisson distribution given in Equation 5.2 sum to
1.
13 The Poisson distribution with parameter  = :3 has been assigned for the
outcome of an experiment. Let X be the outcome function. Find P(X = 0),
P(X = 1), and P(X > 1).
14 On the average, only 1 person in 1000 has a particular rare blood type.
(a) Find the probability that, in a city of 10,000 people, no one has this
blood type.
(b) How many people would have to be tested to give a probability greater
than 1/2 of nding at least one person with this blood type?",prob.pdf
"blood type.
(b) How many people would have to be tested to give a probability greater
than 1/2 of nding at least one person with this blood type?
15 Write a program for the user to input n, p, j and have the program print out
the exact value of b(n; p; k) and the Poisson approximation to this value.
16 Assume that, during each second, a Dartmouth switchboard receives one call
with probability .01 and no calls with probability .99. Use the Poisson ap-
proximation to estimate the probability that the operator will miss at most
one call if she takes a 5-minute coee break.
17 The probability of a royal 
ush in a poker hand is p = 1=649;740. How large
must n be to render the probability of having no royal 
ush in n hands smaller
than 1=e?
18 A baker blends 600 raisins and 400 chocolate chips into a dough mix and,
from this, makes 500 cookies.
(a) Find the probability that a randomly picked cookie will have no raisins.",prob.pdf
"from this, makes 500 cookies.
(a) Find the probability that a randomly picked cookie will have no raisins.
(b) Find the probability that a randomly picked cookie will have exactly two
chocolate chips.
(c) Find the probability that a randomly chosen cookie will have at least
two bits (raisins or chips) in it.
19 The probability that, in a bridge deal, one of the four hands has all hearts
is approximately 6:3  1012. In a city with about 50,000 bridge players the
resident probability expert is called on the average once a year (usually late at
night) and told that the caller has just been dealt a hand of all hearts. Should
she suspect that some of these callers are the victims of practical jokes?",prob.pdf
"200 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
20 An advertiser drops 10,000 lea
ets on a city which has 2000 blocks. Assume
that each lea
et has an equal chance of landing on each block. What is the
probability that a particular block will receive no lea
ets?
21 In a class of 80 students, the professor calls on 1 student chosen at random
for a recitation in each class period. There are 32 class periods in a term.
(a) Write a formula for the exact probability that a given student is called
upon j times during the term.
(b) Write a formula for the Poisson approximation for this probability. Using
your formula estimate the probability that a given student is called upon
more than twice.
22 Assume that we are making raisin cookies. We put a box of 600 raisins into
our dough mix, mix up the dough, then make from the dough 500 cookies.
We then ask for the probability that a randomly chosen cookie will have
0, 1, 2, . . . raisins. Consider the cookies as trials in an experiment, and",prob.pdf
"We then ask for the probability that a randomly chosen cookie will have
0, 1, 2, . . . raisins. Consider the cookies as trials in an experiment, and
let X be the random variable which gives the number of raisins in a given
cookie. Then we can regard the number of raisins in a cookie as the result
of n = 600 independent trials with probability p = 1=500 for success on each
trial. Since n is large and p is small, we can use the Poisson approximation
with  = 600(1=500) = 1:2. Determine the probability that a given cookie
will have at least ve raisins.
23 For a certain experiment, the Poisson distribution with parameter  = m has
been assigned. Show that a most probable outcome for the experiment is
the integer value k such that m  1  k  m. Under what conditions will
there be two most probable values? Hint: Consider the ratio of successive
probabilities.
24 When John Kemeny was chair of the Mathematics Department at Dartmouth",prob.pdf
"probabilities.
24 When John Kemeny was chair of the Mathematics Department at Dartmouth
College, he received an average of ten letters each day. On a certain weekday
he received no mail and wondered if it was a holiday. To decide this he
computed the probability that, in ten years, he would have at least 1 day
without any mail. He assumed that the number of letters he received on a
given day has a Poisson distribution. What probability did he nd? Hint:
Apply the Poisson distribution twice. First, to nd the probability that, in
3000 days, he will have at least 1 day without mail, assuming each year has
about 300 days on which mail is delivered.
25 Reese Prosser never puts money in a 10-cent parking meter in Hanover. He
assumes that there is a probability of .05 that he will be caught. The rst
oense costs nothing, the second costs 2 dollars, and subsequent oenses cost
5 dollars each. Under his assumptions, how does the expected cost of parking",prob.pdf
"5 dollars each. Under his assumptions, how does the expected cost of parking
100 times without paying the meter compare with the cost of paying the meter
each time?",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 201
Number of deaths Number of corps with x deaths in a given year
0 144
1 91
2 32
3 11
4 2
Table 5.5: Mule kicks.
26 Feller5 discusses the statistics of 
ying bomb hits in an area in the south of
London during the Second World War. The area in question was divided into
24  24 = 576 small areas. The total number of hits was 537. There were
229 squares with 0 hits, 211 with 1 hit, 93 with 2 hits, 35 with 3 hits, 7 with
4 hits, and 1 with 5 or more. Assuming the hits were purely random, use the
Poisson approximation to nd the probability that a particular square would
have exactly k hits. Compute the expected number of squares that would
have 0, 1, 2, 3, 4, and 5 or more hits and compare this with the observed
results.
27 Assume that the probability that there is a signicant accident in a nuclear
power plant during one year's time is .001. If a country has 100 nuclear plants,
estimate the probability that there is at least one such accident during a given",prob.pdf
"estimate the probability that there is at least one such accident during a given
year.
28 An airline nds that 4 percent of the passengers that make reservations on
a particular 
ight will not show up. Consequently, their policy is to sell 100
reserved seats on a plane that has only 98 seats. Find the probability that
every person who shows up for the 
ight will nd a seat available.
29 The king's coinmaster boxes his coins 500 to a box and puts 1 counterfeit coin
in each box. The king is suspicious, but, instead of testing all the coins in
1 box, he tests 1 coin chosen at random out of each of 500 boxes. What is the
probability that he nds at least one fake? What is it if the king tests 2 coins
from each of 250 boxes?
30 (From Kemeny6) Show that, if you make 100 bets on the number 17 at
roulette at Monte Carlo (see Example 6.13), you will have a probability greater
than 1/2 of coming out ahead. What is your expected winning?",prob.pdf
"roulette at Monte Carlo (see Example 6.13), you will have a probability greater
than 1/2 of coming out ahead. What is your expected winning?
31 In one of the rst studies of the Poisson distribution, von Bortkiewicz7 con-
sidered the frequency of deaths from kicks in the Prussian army corps. From
the study of 14 corps over a 20-year period, he obtained the data shown in
Table 5.5. Fit a Poisson distribution to this data and see if you think that
the Poisson distribution is appropriate.
5ibid., p. 161.
6Private communication.
7L. von Bortkiewicz, Das Gesetz der Kleinen Zahlen (Leipzig: Teubner, 1898), p. 24.",prob.pdf
"202 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
32 It is often assumed that the auto trac that arrives at the intersection during
a unit time period has a Poisson distribution with expected value m. Assume
that the number of cars X that arrive at an intersection from the north in unit
time has a Poisson distribution with parameter  = m and the number Y that
arrive from the west in unit time has a Poisson distribution with parameter
 = m. If X and Y are independent, show that the total number X + Y
that arrive at the intersection in unit time has a Poisson distribution with
parameter  = m + m.
33 Cars coming along Magnolia Street come to a fork in the road and have to
choose either Willow Street or Main Street to continue. Assume that the
number of cars that arrive at the fork in unit time has a Poisson distribution
with parameter  = 4. A car arriving at the fork chooses Main Street with
probability 3/4 and Willow Street with probability 1/4. Let X be the random",prob.pdf
"with parameter  = 4. A car arriving at the fork chooses Main Street with
probability 3/4 and Willow Street with probability 1/4. Let X be the random
variable which counts the number of cars that, in a given unit of time, pass
by Joe's Barber Shop on Main Street. What is the distribution of X?
34 In the appeal of the People v. Collins case (see Exercise 4.1.28), the counsel
for the defense argued as follows: Suppose, for example, there are 5,000,000
couples in the Los Angeles area and the probability that a randomly chosen
couple ts the witnesses' description is 1/12,000,000. Then the probability
that there are two such couples given that there is at least one is not at all
small. Find this probability. (The California Supreme Court overturned the
initial guilty verdict.)
35 A manufactured lot of brass turnbuckles has S items of which D are defective.
A sample of s items is drawn without replacement. Let X be a random variable",prob.pdf
"A sample of s items is drawn without replacement. Let X be a random variable
that gives the number of defective items in the sample. Let p(d) = P(X = d).
(a) Show that
p(d) =
D
d

SD
sd


S
s

 :
Thus, X is hypergeometric.
(b) Prove the following identity, known as Euler's formula:
min(D;s)X
d=0
D
d
S  D
s  d

=
S
s

:
36 A bin of 1000 turnbuckles has an unknown number D of defectives. A sample
of 100 turnbuckles has 2 defectives. The maximum likelihood estimate for D
is the number of defectives which gives the highest probability for obtaining
the number of defectives observed in the sample. Guess this number D and
then write a computer program to verify your guess.
37 There are an unknown number of moose on Isle Royale (a National Park in
Lake Superior). To estimate the number of moose, 50 moose are captured and",prob.pdf
"5.1. IMPORTANT DISTRIBUTIONS 203
tagged. Six months later 200 moose are captured and it is found that 8 of
these were tagged. Estimate the number of moose on Isle Royale from these
data, and then verify your guess by computer program (see Exercise 36).
38 A manufactured lot of buggy whips has 20 items, of which 5 are defective. A
random sample of 5 items is chosen to be inspected. Find the probability that
the sample contains exactly one defective item
(a) if the sampling is done with replacement.
(b) if the sampling is done without replacement.
39 Suppose that N and k tend to 1 in such a way that k=N remains xed. Show
that
h(N; k; n; x) ! b(n; k=N; x) :
40 A bridge deck has 52 cards with 13 cards in each of four suits: spades, hearts,
diamonds, and clubs. A hand of 13 cards is dealt from a shued deck. Find
the probability that the hand has
(a) a distribution of suits 4, 4, 3, 2 (for example, four spades, four hearts,
three diamonds, two clubs).",prob.pdf
"the probability that the hand has
(a) a distribution of suits 4, 4, 3, 2 (for example, four spades, four hearts,
three diamonds, two clubs).
(b) a distribution of suits 5, 3, 3, 2.
41 Write a computer algorithm that simulates a hypergeometric random variable
with parameters N, k, and n.
42 You are presented with four dierent dice. The rst one has two sides marked 0
and four sides marked 4. The second one has a 3 on every side. The third one
has a 2 on four sides and a 6 on two sides, and the fourth one has a 1 on three
sides and a 5 on three sides. You allow your friend to pick any of the four
dice he wishes. Then you pick one of the remaining three and you each roll
your die. The person with the largest number showing wins a dollar. Show
that you can choose your die so that you have probability 2/3 of winning no
matter which die your friend picks. (See Tenney and Foster.8)
43 The students in a certain class were classied by hair color and eye color. The",prob.pdf
"matter which die your friend picks. (See Tenney and Foster.8)
43 The students in a certain class were classied by hair color and eye color. The
conventions used were: Brown and black hair were considered dark, and red
and blonde hair were considered light; black and brown eyes were considered
dark, and blue and green eyes were considered light. They collected the data
shown in Table 5.6. Are these traits independent? (See Example 5.6.)
44 Suppose that in the hypergeometric distribution, we let N and k tend to 1 in
such a way that the ratio k=N approaches a real number p between 0 and 1.
Show that the hypergeometric distribution tends to the binomial distribution
with parameters n and p.
8R. L. Tenney and C. C. Foster, Non-transitive Dominance, Math. Mag. 49 (1976) no. 3, pgs.
115-120.",prob.pdf
"204 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Dark Eyes Light Eyes
Dark Hair 28 15 43
Light Hair 9 23 32
37 38 75
Table 5.6: Observed data.
0 10 20 30 40
0
500
1000
1500
2000
2500
3000
3500
Figure 5.5: Distribution of choices in the Powerball lottery.
45 (a) Compute the leading digits of the rst 100 powers of 2, and see how well
these data t the Benford distribution.
(b) Multiply each number in the data set of part (a) by 3, and compare the
distribution of the leading digits with the Benford distribution.
46 In the Powerball lottery, contestants pick 5 dierent integers between 1 and 45,
and in addition, pick a bonus integer from the same range (the bonus integer
can equal one of the rst ve integers chosen). Some contestants choose the
numbers themselves, and others let the computer choose the numbers. The
data shown in Table 5.7 are the contestant-chosen numbers in a certain state
on May 3, 1996. A spike graph of the data is shown in Figure 5.5.",prob.pdf
"data shown in Table 5.7 are the contestant-chosen numbers in a certain state
on May 3, 1996. A spike graph of the data is shown in Figure 5.5.
The goal of this problem is to check the hypothesis that the chosen numbers
are uniformly distributed. To do this, compute the value v of the random
variable 2 given in Example 5.6. In the present case, this random variable has
44 degrees of freedom. One can nd, in a 2 table, the value v0 = 59:43 , which
represents a number with the property that a 2-distributed random variable
takes on values that exceed v0 only 5% of the time. Does your computed value
of v exceed v0? If so, you should reject the hypothesis that the contestants'
choices are uniformly distributed.",prob.pdf
"5.2. IMPORTANT DENSITIES 205
Integer Times Integer Times Integer Times
Chosen Chosen Chosen
1 2646 2 2934 3 3352
4 3000 5 3357 6 2892
7 3657 8 3025 9 3362
10 2985 11 3138 12 3043
13 2690 14 2423 15 2556
16 2456 17 2479 18 2276
19 2304 20 1971 21 2543
22 2678 23 2729 24 2414
25 2616 26 2426 27 2381
28 2059 29 2039 30 2298
31 2081 32 1508 33 1887
34 1463 35 1594 36 1354
37 1049 38 1165 39 1248
40 1493 41 1322 42 1423
43 1207 44 1259 45 1224
Table 5.7: Numbers chosen by contestants in the Powerball lottery.
5.2 Important Densities
In this section, we will introduce some important probability density functions and
give some examples of their use. We will also consider the question of how one
simulates a given density using a computer.
Continuous Uniform Density
The simplest density function corresponds to the random variable U whose value
represents the outcome of the experiment consisting of choosing a real number at
random from the interval [a; b].
f(!) =
1=(b  a); if a  !  b;",prob.pdf
"represents the outcome of the experiment consisting of choosing a real number at
random from the interval [a; b].
f(!) =
1=(b  a); if a  !  b;
0; otherwise.
It is easy to simulate this density on a computer. We simply calculate the
expression
(b  a)rnd + a :
Exponential and Gamma Densities
The exponential density function is dened by
f(x) =
ex; if 0  x < 1;
0; otherwise:
Here  is any positive constant, depending on the experiment. The reader has seen
this density in Example 2.17. In Figure 5.6 we show graphs of several exponen-
tial densities for dierent choices of . The exponential density is often used to",prob.pdf
"206 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
0 2 4 6 8 10
λ=1
λ=2
λ=1/2
Figure 5.6: Exponential densities.
describe experiments involving a question of the form: How long until something
happens? For example, the exponential density is often used to study the time
between emissions of particles from a radioactive source.
The cumulative distribution function of the exponential density is easy to com-
pute. Let T be an exponentially distributed random variable with parameter . If
x  0, then we have
F(x) = P(T  x)
=
Z x
0
et dt
= 1  ex :
Both the exponential density and the geometric distribution share a property
known as the \memoryless"" property. This property was introduced in Example 5.1;
it says that
P(T > r + s j T > r) = P(T > s) :
This can be demonstrated to hold for the exponential density by computing both
sides of this equation. The right-hand side is just
1  F(s) = es ;
while the left-hand side is
P(T > r + s)
P(T > r) = 1  F(r + s)
1  F(r)",prob.pdf
"5.2. IMPORTANT DENSITIES 207
= e(r+s)
er
= es :
There is a very important relationship between the exponential density and
the Poisson distribution. We begin by dening X1; X2; : : : to be a sequence of
independent exponentially distributed random variables with parameter . We
might think of Xi as denoting the amount of time between the ith and (i + 1)st
emissions of a particle by a radioactive source. (As we shall see in Chapter 6, we
can think of the parameter  as representing the reciprocal of the average length of
time between emissions. This parameter is a quantity that might be measured in
an actual experiment of this type.)
We now consider a time interval of length t, and we let Y denote the random
variable which counts the number of emissions that occur in the time interval. We
would like to calculate the distribution function of Y (clearly, Y is a discrete random
variable). If we let Sn denote the sum X1 + X2 + 
 
 
 + Xn, then it is easy to see
that",prob.pdf
"variable). If we let Sn denote the sum X1 + X2 + 
 
 
 + Xn, then it is easy to see
that
P(Y = n) = P(Sn  t and Sn+1 > t) :
Since the event Sn+1  t is a subset of the event Sn  t, the above probability is
seen to be equal to
P(Sn  t)  P(Sn+1  t) : (5.4)
We will show in Chapter 7 that the density of Sn is given by the following formula:
gn(x) =
(
(x)n1
(n1)! ex; if x > 0,
0; otherwise.
This density is an example of a gamma density with parameters  and n. The
general gamma density allows n to be any positive real number. We shall not
discuss this general density.
It is easy to show by induction on n that the cumulative distribution function
of Sn is given by:
Gn(x) =
8
<
:
1  ex

1 + x
1! + 
 
 
 + (x)n1
(n1)!

; if x > 0;
0; otherwise.
Using this expression, the quantity in (5.4) is easy to compute; we obtain
et (t)n
n! ;
which the reader will recognize as the probability that a Poisson-distributed random
variable, with parameter t, takes on the value n.",prob.pdf
"n! ;
which the reader will recognize as the probability that a Poisson-distributed random
variable, with parameter t, takes on the value n.
The above relationship will allow us to simulate a Poisson distribution, once
we have found a way to simulate an exponential density. The following random
variable does the job:
Y = 1
 log(rnd) : (5.5)",prob.pdf
"208 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Using Corollary 5.2 (below), one can derive the above expression (see Exercise 3).
We content ourselves for now with a short calculation that should convince the
reader that the random variable Y has the required property. We have
P(Y  y) = P

1
 log(rnd)  y

= P(log(rnd)  y)
= P(rnd  ey)
= 1  ey :
This last expression is seen to be the cumulative distribution function of an expo-
nentially distributed random variable with parameter .
To simulate a Poisson random variable W with parameter , we simply generate
a sequence of values of an exponentially distributed random variable with the same
parameter, and keep track of the subtotals Sk of these values. We stop generating
the sequence when the subtotal rst exceeds . Assume that we nd that
Sn   < Sn+1 :
Then the value n is returned as a simulated value for W.
Example 5.7 (Queues) Suppose that customers arrive at random times at a service",prob.pdf
"Then the value n is returned as a simulated value for W.
Example 5.7 (Queues) Suppose that customers arrive at random times at a service
station with one server, and suppose that each customer is served immediately if
no one is ahead of him, but must wait his turn in line otherwise. How long should
each customer expect to wait? (We dene the waiting time of a customer to be the
length of time between the time that he arrives and the time that he begins to be
served.)
Let us assume that the interarrival times between successive customers are given
by random variables X1, X2, . . . , Xn that are mutually independent and identically
distributed with an exponential cumulative distribution function given by
FX(t) = 1  et:
Let us assume, too, that the service times for successive customers are given by
random variables Y1, Y2, . . . , Yn that again are mutually independent and identically
distributed with another exponential cumulative distribution function given by
FY (t) = 1  et:",prob.pdf
"distributed with another exponential cumulative distribution function given by
FY (t) = 1  et:
The parameters  and  represent, respectively, the reciprocals of the average
time between arrivals of customers and the average service time of the customers.
Thus, for example, the larger the value of , the smaller the average time between
arrivals of customers. We can guess that the length of time a customer will spend
in the queue depends on the relative sizes of the average interarrival time and the
average service time.
It is easy to verify this conjecture by simulation. The program Queue simulates
this queueing process. Let N(t) be the number of customers in the queue at time t.",prob.pdf
"5.2. IMPORTANT DENSITIES 209
2000 4000 6000 8000 10000
10
20
30
40
50
60
2000 4000 6000 8000 10000
200
400
600
800
1000
1200 λ = 1 λ = 1
µ = .9 µ = 1.1
Figure 5.7: Queue sizes.
0 10 20 30 40 50
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
Figure 5.8: Waiting times.
Then we plot N(t) as a function of t for dierent choices of the parameters  and
 (see Figure 5.7).
We note that when  < , then 1= > 1=, so the average interarrival time is
greater than the average service time, i.e., customers are served more quickly, on
average, than new ones arrive. Thus, in this case, it is reasonable to expect that
N(t) remains small. However, if  >  then customers arrive more quickly than
they are served, and, as expected, N(t) appears to grow without limit.
We can now ask: How long will a customer have to wait in the queue for service?
To examine this question, we let Wi be the length of time that the ith customer has
to remain in the system (waiting in line and being served). Then we can present",prob.pdf
"to remain in the system (waiting in line and being served). Then we can present
these data in a bar graph, using the program Queue, to give some idea of how the
Wi are distributed (see Figure 5.8). (Here  = 1 and  = 1:1.)
We see that these waiting times appear to be distributed exponentially. This is
always the case when  < . The proof of this fact is too complicated to give here,
but we can verify it by simulation for dierent choices of  and , as above. 2",prob.pdf
"210 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Functions of a Random Variable
Before continuing our list of important densities, we pause to consider random
variables which are functions of other random variables. We will prove a general
theorem that will allow us to derive expressions such as Equation 5.5.
Theorem 5.1 Let X be a continuous random variable, and suppose that (x) is a
strictly increasing function on the range of X. Dene Y = (X). Suppose that X
and Y have cumulative distribution functions FX and FY respectively. Then these
functions are related by
FY (y) = FX (1(y)):
If (x) is strictly decreasing on the range of X, then
FY (y) = 1  FX (1(y)) :
Proof. Since  is a strictly increasing function on the range of X, the events
(X  1(y)) and ((X)  y) are equal. Thus, we have
FY (y) = P(Y  y)
= P((X)  y)
= P(X  1(y))
= FX(1(y)) :
If (x) is strictly decreasing on the range of X, then we have
FY (y) = P(Y  y)
= P((X)  y)
= P(X  1(y))
= 1  P(X < 1(y))",prob.pdf
"= FX(1(y)) :
If (x) is strictly decreasing on the range of X, then we have
FY (y) = P(Y  y)
= P((X)  y)
= P(X  1(y))
= 1  P(X < 1(y))
= 1  FX (1(y)) :
This completes the proof. 2
Corollary 5.1 Let X be a continuous random variable, and suppose that (x) is a
strictly increasing function on the range of X. Dene Y = (X). Suppose that the
density functions of X and Y are fX and fY , respectively. Then these functions
are related by
fY (y) = fX(1(y)) d
dy 1(y) :
If (x) is strictly decreasing on the range of X, then
fY (y) = fX(1(y)) d
dy1(y) :",prob.pdf
"5.2. IMPORTANT DENSITIES 211
Proof. This result follows from Theorem 5.1 by using the Chain Rule. 2
If the function  is neither strictly increasing nor strictly decreasing, then the
situation is somewhat more complicated but can be treated by the same methods.
For example, suppose that Y = X2, Then (x) = x2, and
FY (y) = P(Y  y)
= P(py  X  +py)
= P(X  +py)  P(X  py)
= FX (py)  FX(py) :
Moreover,
fY (y) = d
dyFY (y)
= d
dy(FX (py)  FX(py))
=

fX(py) + fX(py)
 1
2py :
We see that in order to express FY in terms of FX when Y = (X), we have to
express P(Y  y) in terms of P(X  x), and this process will depend in general
upon the structure of .
Simulation
Theorem 5.1 tells us, among other things, how to simulate on the computer a random
variable Y with a prescribed cumulative distribution function F. We assume that
F(y) is strictly increasing for those values of y where 0 < F(y) < 1. For this",prob.pdf
"F(y) is strictly increasing for those values of y where 0 < F(y) < 1. For this
purpose, let U be a random variable which is uniformly distributed on [0; 1]. Then
U has cumulative distribution function FU (u) = u. Now, if F is the prescribed
cumulative distribution function for Y , then to write Y in terms of U we rst solve
the equation
F(y) = u
for y in terms of u. We obtain y = F1(u). Note that since F is an increasing
function this equation always has a unique solution (see Figure 5.9). Then we set
Z = F1(U) and obtain, by Theorem 5.1,
FZ(y) = FU (F(y)) = F(y) ;
since FU (u) = u. Therefore, Z and Y have the same cumulative distribution func-
tion. Summarizing, we have the following.",prob.pdf
"212 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
y = φ(x)
x = FY (y)
Y (y)Graph of F
x
y
1
0
Figure 5.9: Converting a uniform distribution FU into a prescribed distribution FY .
Corollary 5.2 If F(y) is a given cumulative distribution function that is strictly
increasing when 0 < F(y) < 1 and if U is a random variable with uniform distribu-
tion on [0; 1], then
Y = F1(U)
has the cumulative distribution F(y). 2
Thus, to simulate a random variable with a given cumulative distribution F we
need only set Y = F1(rnd).
Normal Density
We now come to the most important density function, the normal density function.
We have seen in Chapter 3 that the binomial distribution functions are bell-shaped,
even for moderate size values of n. We recall that a binomially-distributed random
variable with parameters n and p can be considered to be the sum of n mutually
independent 0-1 random variables. A very important theorem in probability theory,",prob.pdf
"independent 0-1 random variables. A very important theorem in probability theory,
called the Central Limit Theorem, states that under very general conditions, if we
sum a large number of mutually independent random variables, then the distribution
of the sum can be closely approximated by a certain specic continuous density,
called the normal density. This theorem will be discussed in Chapter 9.
The normal density function with parameters  and  is dened as follows:
fX(x) = 1p
2e(x)2=22
:
The parameter  represents the \center"" of the density (and in Chapter 6, we will
show that it is the average, or expected, value of the density). The parameter 
is a measure of the \spread"" of the density, and thus it is assumed to be positive.
(In Chapter 6, we will show that  is the standard deviation of the density.) We
note that it is not at all obvious that the above function is a density, i.e., that its",prob.pdf
"5.2. IMPORTANT DENSITIES 213
-4 -2 2 4
0.1
0.2
0.3
0.4
σ = 1
σ = 2
Figure 5.10: Normal density for two sets of parameter values.
integral over the real line equals 1. The cumulative distribution function is given
by the formula
FX(x) =
Z x
1
1p
2e(u)2 =22
du :
In Figure 5.10 we have included for comparison a plot of the normal density for
the cases  = 0 and  = 1, and  = 0 and  = 2.
One cannot write FX in terms of simple functions. This leads to several prob-
lems. First of all, values of FX must be computed using numerical integration.
Extensive tables exist containing values of this function (see Appendix A). Sec-
ondly, we cannot write F1
X in closed form, so we cannot use Corollary 5.2 to help
us simulate a normal random variable. For this reason, special methods have been
developed for simulating a normal distribution. One such method relies on the fact
that if U and V are independent random variables with uniform densities on [0; 1],
then the random variables
X =
p",prob.pdf
"that if U and V are independent random variables with uniform densities on [0; 1],
then the random variables
X =
p
2 logU cos2V
and
Y =
p
2 logU sin 2V
are independent, and have normal density functions with parameters  = 0 and
 = 1. (This is not obvious, nor shall we prove it here. See Box and Muller.9)
Let Z be a normal random variable with parameters  = 0 and  = 1. A
normal random variable with these parameters is said to be a standard normal
random variable. It is an important and useful fact that if we write
X = Z +  ;
then X is a normal random variable with parameters  and . To show this, we
will use Theorem 5.1. We have (z) = z + , 1(x) = (x  )=, and
FX(x) = FZ
x  


;
9G. E. P. Box and M. E. Muller, A Note on the Generation of Random Normal Deviates, Ann.
of Math. Stat. 29 (1958), pgs. 610-611.",prob.pdf
"214 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
fX(x) = fZ
x  



 1

= 1p
2e(x)2=22
:
The reader will note that this last expression is the density function with parameters
 and , as claimed.
We have seen above that it is possible to simulate a standard normal random
variable Z. If we wish to simulate a normal random variable X with parameters 
and , then we need only transform the simulated values for Z using the equation
X = Z + .
Suppose that we wish to calculate the value of a cumulative distribution function
for the normal random variable X, with parameters  and . We can reduce this
calculation to one concerning the standard normal random variable Z as follows:
FX(x) = P(X  x)
= P

Z  x  


= FZ
x  


:
This last expression can be found in a table of values of the cumulative distribution
function for a standard normal random variable. Thus, we see that it is unnecessary
to make tables of normal distribution functions with arbitrary  and .",prob.pdf
"to make tables of normal distribution functions with arbitrary  and .
The process of changing a normal random variable to a standard normal ran-
dom variable is known as standardization. If X has a normal distribution with
parameters  and  and if
Z = X  
 ;
then Z is said to be the standardized version of X.
The following example shows how we use the standardized version of a normal
random variable X to compute specic probabilities relating to X.
Example 5.8 Suppose that X is a normally distributed random variable with pa-
rameters  = 10 and  = 3. Find the probability that X is between 4 and 16.
To solve this problem, we note that Z = (X  10)=3 is the standardized version
of X. So, we have
P(4  X  16) = P(X  16)  P(X  4)
= FX (16)  FX(4)
= FZ
16  10
3

 FZ
4  10
3

= FZ(2)  FZ(2) :",prob.pdf
"5.2. IMPORTANT DENSITIES 215
0 1 2 3 4 5
0
0.1
0.2
0.3
0.4
0.5
0.6
Figure 5.11: Distribution of dart distances in 1000 drops.
This last expression can be evaluated by using tabulated values of the standard
normal distribution function (see 11.5); when we use this table, we nd that FZ(2) =
:9772 and FZ(2) = :0228. Thus, the answer is .9544.
In Chapter 6, we will see that the parameter  is the mean, or average value, of
the random variable X. The parameter  is a measure of the spread of the random
variable, and is called the standard deviation. Thus, the question asked in this
example is of a typical type, namely, what is the probability that a random variable
has a value within two standard deviations of its average value. 2
Maxwell and Rayleigh Densities
Example 5.9 Suppose that we drop a dart on a large table top, which we consider
as the xy-plane, and suppose that the x and y coordinates of the dart point are",prob.pdf
"as the xy-plane, and suppose that the x and y coordinates of the dart point are
independent and have a normal distribution with parameters  = 0 and  = 1.
How is the distance of the point from the origin distributed?
This problem arises in physics when it is assumed that a moving particle in
Rn has components of the velocity that are mutually independent and normally
distributed and it is desired to nd the density of the speed of the particle. The
density in the case n = 3 is called the Maxwell density.
The density in the case n = 2 (i.e. the dart board experiment described above)
is called the Rayleigh density. We can simulate this case by picking independently a
pair of coordinates (x; y), each from a normal distribution with  = 0 and  = 1 on
(1; 1), calculating the distance r =
p
x2 + y2 of the point (x; y) from the origin,
repeating this process a large number of times, and then presenting the results in a
bar graph. The results are shown in Figure 5.11.",prob.pdf
"216 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Female Male
A 37 56 93
B 63 60 123
C 47 43 90
Below C 5 8 13
152 167 319
Table 5.8: Calculus class data.
Female Male
A 44.3 48.7 93
B 58.6 64.4 123
C 42.9 47.1 90
Below C 6.2 6.8 13
152 167 319
Table 5.9: Expected data.
We have also plotted the theoretical density
f(r) = rer2=2 :
This will be derived in Chapter 7; see Example 7.7. 2
Chi-Squared Density
We return to the problem of independence of traits discussed in Example 5.6. It
is frequently the case that we have two traits, each of which have several dierent
values. As was seen in the example, quite a lot of calculation was needed even
in the case of two values for each trait. We now give another method for testing
independence of traits, which involves much less calculation.
Example 5.10 Suppose that we have the data shown in Table 5.8 concerning
grades and gender of students in a Calculus class. We can use the same sort of",prob.pdf
"grades and gender of students in a Calculus class. We can use the same sort of
model in this situation as was used in Example 5.6. We imagine that we have an
urn with 319 balls of two colors, say blue and red, corresponding to females and
males, respectively. We now draw 93 balls, without replacement, from the urn.
These balls correspond to the grade of A. We continue by drawing 123 balls, which
correspond to the grade of B. When we nish, we have four sets of balls, with each
ball belonging to exactly one set. (We could have stipulated that the balls were
of four colors, corresponding to the four possible grades. In this case, we would
draw a subset of size 152, which would correspond to the females. The balls re-
maining in the urn would correspond to the males. The choice does not aect the
nal determination of whether we should reject the hypothesis of independence of
traits.)
The expected data set can be determined in exactly the same way as in Exam-",prob.pdf
"traits.)
The expected data set can be determined in exactly the same way as in Exam-
ple 5.6. If we do this, we obtain the expected values shown in Table 5.9. Even if",prob.pdf
"5.2. IMPORTANT DENSITIES 217
the traits are independent, we would still expect to see some dierences between
the numbers in corresponding boxes in the two tables. However, if the dierences
are large, then we might suspect that the two traits are not independent. In Ex-
ample 5.6, we used the probability distribution of the various possible data sets to
compute the probability of nding a data set that diers from the expected data
set by at least as much as the actual data set does. We could do the same in this
case, but the amount of computation is enormous.
Instead, we will describe a single number which does a good job of measuring
how far a given data set is from the expected one. To quantify how far apart the two
sets of numbers are, we could sum the squares of the dierences of the corresponding
numbers. (We could also sum the absolute values of the dierences, but we would
not want to sum the dierences.) Suppose that we have data in which we expect",prob.pdf
"not want to sum the dierences.) Suppose that we have data in which we expect
to see 10 objects of a certain type, but instead we see 18, while in another case we
expect to see 50 objects of a certain type, but instead we see 58. Even though the
two dierences are about the same, the rst dierence is more surprising than the
second, since the expected number of outcomes in the second case is quite a bit
larger than the expected number in the rst case. One way to correct for this is
to divide the individual squares of the dierences by the expected number for that
box. Thus, if we label the values in the eight boxes in the rst table by Oi (for
observed values) and the values in the eight boxes in the second table by Ei (for
expected values), then the following expression might be a reasonable one to use to
measure how far the observed data is from what is expected:
8X
i=1
(Oi  Ei)2
Ei
:
This expression is a random variable, which is usually denoted by the symbol 2,",prob.pdf
"8X
i=1
(Oi  Ei)2
Ei
:
This expression is a random variable, which is usually denoted by the symbol 2,
pronounced \ki-squared."" It is called this because, under the assumption of inde-
pendence of the two traits, the density of this random variable can be computed and
is approximately equal to a density called the chi-squared density. We choose not
to give the explicit expression for this density, since it involves the gamma function,
which we have not discussed. The chi-squared density is, in fact, a special case of
the general gamma density.
In applying the chi-squared density, tables of values of this density are used, as
in the case of the normal density. The chi-squared density has one parameter n,
which is called the number of degrees of freedom. The number n is usually easy to
determine from the problem at hand. For example, if we are checking two traits for
independence, and the two traits have a and b values, respectively, then the number",prob.pdf
"independence, and the two traits have a and b values, respectively, then the number
of degrees of freedom of the random variable 2 is (a1)(b1). So, in the example
at hand, the number of degrees of freedom is 3.
We recall that in this example, we are trying to test for independence of the
two traits of gender and grades. If we assume these traits are independent, then
the ball-and-urn model given above gives us a way to simulate the experiment.
Using a computer, we have performed 1000 experiments, and for each one, we have
calculated a value of the random variable 2. The results are shown in Figure 5.12,
together with the chi-squared density function with three degrees of freedom.",prob.pdf
"218 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
0 2 4 6 8 10 12
0
0.05
0.1
0.15
0.2
Figure 5.12: Chi-squared density with three degrees of freedom.
As we stated above, if the value of the random variable 2 is large, then we
would tend not to believe that the two traits are independent. But how large is
large? The actual value of this random variable for the data above is 4.13. In
Figure 5.12, we have shown the chi-squared density with 3 degrees of freedom. It
can be seen that the value 4.13 is larger than most of the values taken on by this
random variable.
Typically, a statistician will compute the value v of the random variable 2,
just as we have done. Then, by looking in a table of values of the chi-squared
density, a value v0 is determined which is only exceeded 5% of the time. If v  v0,
the statistician rejects the hypothesis that the two traits are independent. In the
present case, v0 = 7:815, so we would not reject the hypothesis that the two traits
are independent. 2",prob.pdf
"present case, v0 = 7:815, so we would not reject the hypothesis that the two traits
are independent. 2
Cauchy Density
The following example is from Feller.10
Example 5.11 Suppose that a mirror is mounted on a vertical axis, and is free
to revolve about that axis. The axis of the mirror is 1 foot from a straight wall
of innite length. A pulse of light is shown onto the mirror, and the re
ected ray
hits the wall. Let  be the angle between the re
ected ray and the line that is
perpendicular to the wall and that runs through the axis of the mirror. We assume
that  is uniformly distributed between =2 and =2. Let X represent the distance
between the point on the wall that is hit by the re
ected ray and the point on the
wall that is closest to the axis of the mirror. We now determine the density of X.
Let B be a xed positive quantity. Then X  B if and only if tan()  B,
which happens if and only if   arctan(B). This happens with probability
=2  arctan(B)
 :",prob.pdf
"which happens if and only if   arctan(B). This happens with probability
=2  arctan(B)
 :
10W. Feller, An Introduction to Probability Theory and Its Applications,, vol. 2, (New York:
Wiley, 1966)",prob.pdf
"5.2. IMPORTANT DENSITIES 219
Thus, for positive B, the cumulative distribution function of X is
F(B) = 1  =2  arctan(B)
 :
Therefore, the density function for positive B is
f(B) = 1
(1 + B2) :
Since the physical situation is symmetric with respect to  = 0, it is easy to see
that the above expression for the density is correct for negative values of B as well.
The Law of Large Numbers, which we will discuss in Chapter 8, states that
in many cases, if we take the average of independent values of a random variable,
then the average approaches a specic number as the number of values increases.
It turns out that if one does this with a Cauchy-distributed random variable, the
average does not approach any specic number. 2
Exercises
1 Choose a number U from the unit interval [0; 1] with uniform distribution.
Find the cumulative distribution and density for the random variables
(a) Y = U + 2.
(b) Y = U3.
2 Choose a number U from the interval [0; 1] with uniform distribution. Find",prob.pdf
"(a) Y = U + 2.
(b) Y = U3.
2 Choose a number U from the interval [0; 1] with uniform distribution. Find
the cumulative distribution and density for the random variables
(a) Y = 1=(U + 1).
(b) Y = log(U + 1).
3 Use Corollary 5.2 to derive the expression for the random variable given in
Equation 5.5. Hint: The random variables 1  rnd and rnd are identically
distributed.
4 Suppose we know a random variable Y as a function of the uniform random
variable U: Y = (U), and suppose we have calculated the cumulative dis-
tribution function FY (y) and thence the density fY (y). How can we check
whether our answer is correct? An easy simulation provides the answer: Make
a bar graph of Y = (rnd) and compare the result with the graph of fY (y).
These graphs should look similar. Check your answers to Exercises 1 and 2
by this method.
5 Choose a number U from the interval [0; 1] with uniform distribution. Find
the cumulative distribution and density for the random variables
(a) Y = jU  1=2j.",prob.pdf
"the cumulative distribution and density for the random variables
(a) Y = jU  1=2j.
(b) Y = (U  1=2)2.",prob.pdf
"220 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
6 Check your results for Exercise 5 by simulation as described in Exercise 4.
7 Explain how you can generate a random variable whose cumulative distribu-
tion function is
F(x) =
8
<
:
0; if x < 0;
x2; if 0  x  1;
1; if x > 1:
8 Write a program to generate a sample of 1000 random outcomes each of which
is chosen from the distribution given in Exercise 7. Plot a bar graph of your
results and compare this empirical density with the density for the cumulative
distribution given in Exercise 7.
9 Let U, V be random numbers chosen independently from the interval [0; 1]
with uniform distribution. Find the cumulative distribution and density of
each of the variables
(a) Y = U + V .
(b) Y = jU  V j.
10 Let U, V be random numbers chosen independently from the interval [0; 1].
Find the cumulative distribution and density for the random variables
(a) Y = max(U; V ).
(b) Y = min(U; V ).",prob.pdf
"Find the cumulative distribution and density for the random variables
(a) Y = max(U; V ).
(b) Y = min(U; V ).
11 Write a program to simulate the random variables of Exercises 9 and 10 and
plot a bar graph of the results. Compare the resulting empirical density with
the density found in Exercises 9 and 10.
12 A number U is chosen at random in the interval [0; 1]. Find the probability
that
(a) R = U2 < 1=4.
(b) S = U(1  U) < 1=4.
(c) T = U=(1  U) < 1=4.
13 Find the cumulative distribution function F and the density function f for
each of the random variables R, S, and T in Exercise 12.
14 A point P in the unit square has coordinates X and Y chosen at random in
the interval [0; 1]. Let D be the distance from P to the nearest edge of the
square, and E the distance to the nearest corner. What is the probability
that
(a) D < 1=4?
(b) E < 1=4?
15 In Exercise 14 nd the cumulative distribution F and density f for the random
variable D.",prob.pdf
"5.2. IMPORTANT DENSITIES 221
16 Let X be a random variable with density function
fX(x) =
 cx(1  x); if 0 < x < 1;
0; otherwise.
(a) What is the value of c?
(b) What is the cumulative distribution function FX for X?
(c) What is the probability that X < 1=4?
17 Let X be a random variable with cumulative distribution function
F(x) =
8
<
:
0; if x < 0;
sin2(x=2); if 0  x  1;
1; if 1 < x:
(a) What is the density function fX for X?
(b) What is the probability that X < 1=4?
18 Let X be a random variable with cumulative distribution function FX , and
let Y = X + b, Z = aX, and W = aX + b, where a and b are any constants.
Find the cumulative distribution functions FY , FZ, and FW . Hint: The cases
a > 0, a = 0, and a < 0 require dierent arguments.
19 Let X be a random variable with density function fX, and let Y = X + b,
Z = aX, and W = aX + b, where a 6= 0. Find the density functions fY , fZ,
and fW . (See Exercise 18.)",prob.pdf
"Z = aX, and W = aX + b, where a 6= 0. Find the density functions fY , fZ,
and fW . (See Exercise 18.)
20 Let X be a random variable uniformly distributed over [c; d], and let Y =
aX + b. For what choice of a and b is Y uniformly distributed over [0; 1]?
21 Let X be a random variable with cumulative distribution function F strictly
increasing on the range of X. Let Y = F(X). Show that Y is uniformly
distributed in the interval [0; 1]. (The formula X = F1(Y ) then tells us how
to construct X from a uniform random variable Y .)
22 Let X be a random variable with cumulative distribution function F. The
median of X is the value m for which F(m) = 1=2. Then X < m with
probability 1/2 and X > m with probability 1/2. Find m if X is
(a) uniformly distributed over the interval [a; b].
(b) normally distributed with parameters  and .
(c) exponentially distributed with parameter .
23 Let X be a random variable with density function fX. The mean of X is
the value  =
R",prob.pdf
"(c) exponentially distributed with parameter .
23 Let X be a random variable with density function fX. The mean of X is
the value  =
R
xfx(x) dx. Then  gives an average value for X (see Sec-
tion 6.3). Find  if X is distributed uniformly, normally, or exponentially, as
in Exercise 22.",prob.pdf
"222 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
Test Score Letter grade
 +  < x A
 < x <  +  B
   < x <  C
  2 < x <    D
x <   2 F
Table 5.10: Grading on the curve.
24 Let X be a random variable with density function fX. The mode of X is the
value M for which f(M) is maximum. Then values of X near M are most
likely to occur. Find M if X is distributed normally or exponentially, as in
Exercise 22. What happens if X is distributed uniformly?
25 Let X be a random variable normally distributed with parameters  = 70,
 = 10. Estimate
(a) P(X > 50).
(b) P(X < 60).
(c) P(X > 90).
(d) P(60 < X < 80).
26 Bridies' Bearing Works manufactures bearing shafts whose diameters are nor-
mally distributed with parameters  = 1,  = :002. The buyer's specications
require these diameters to be 1:000  :003 cm. What fraction of the manu-
facturer's shafts are likely to be rejected? If the manufacturer improves her
quality control, she can reduce the value of . What value of  will ensure",prob.pdf
"quality control, she can reduce the value of . What value of  will ensure
that no more than 1 percent of her shafts are likely to be rejected?
27 A nal examination at Podunk University is constructed so that the test
scores are approximately normally distributed, with parameters  and . The
instructor assigns letter grades to the test scores as shown in Table 5.10 (this
is the process of \grading on the curve"").
What fraction of the class gets A, B, C, D, F?
28 (Ross11) An expert witness in a paternity suit testies that the length (in
days) of a pregnancy, from conception to delivery, is approximately normally
distributed, with parameters  = 270,  = 10. The defendant in the suit is
able to prove that he was out of the country during the period from 290 to 240
days before the birth of the child. What is the probability that the defendant
was in the country when the child was conceived?
29 Suppose that the time (in hours) required to repair a car is an exponentially",prob.pdf
"was in the country when the child was conceived?
29 Suppose that the time (in hours) required to repair a car is an exponentially
distributed random variable with parameter  = 1=2. What is the probabil-
ity that the repair time exceeds 4 hours? If it exceeds 4 hours what is the
probability that it exceeds 8 hours?
11S. Ross, A First Course in Probability Theory, 2d ed. (New York: Macmillan, 1984).",prob.pdf
"5.2. IMPORTANT DENSITIES 223
30 Suppose that the number of years a car will run is exponentially distributed
with parameter  = 1=4. If Prosser buys a used car today, what is the
probability that it will still run after 4 years?
31 Let U be a uniformly distributed random variable on [0; 1]. What is the
probability that the equation
x2 + 4Ux + 1 = 0
has two distinct real roots x1 and x2?
32 Write a program to simulate the random variables whose densities are given
by the following, making a suitable bar graph of each and comparing the exact
density with the bar graph.
(a) fX(x) = ex on [0; 1) (but just do it on [0; 10]):
(b) fX(x) = 2x on [0; 1]:
(c) fX(x) = 3x2 on [0; 1]:
(d) fX(x) = 4jx  1=2j on [0; 1]:
33 Suppose we are observing a process such that the time between occurrences
is exponentially distributed with  = 1=30 (i.e., the average time between
occurrences is 30 minutes). Suppose that the process starts at a certain time",prob.pdf
"occurrences is 30 minutes). Suppose that the process starts at a certain time
and we start observing the process 3 hours later. Write a program to simulate
this process. Let T denote the length of time that we have to wait, after we
start our observation, for an occurrence. Have your program keep track of T.
What is an estimate for the average value of T?
34 Jones puts in two new lightbulbs: a 60 watt bulb and a 100 watt bulb. It
is claimed that the lifetime of the 60 watt bulb has an exponential density
with average lifetime 200 hours ( = 1=200). The 100 watt bulb also has an
exponential density but with average lifetime of only 100 hours ( = 1=100).
Jones wonders what is the probability that the 100 watt bulb will outlast the
60 watt bulb.
If X and Y are two independent random variables with exponential densities
f(x) = ex and g(x) = ex, respectively, then the probability that X is
less than Y is given by
P(X < Y ) =
Z 1
0
f(x)(1  G(x)) dx;",prob.pdf
"f(x) = ex and g(x) = ex, respectively, then the probability that X is
less than Y is given by
P(X < Y ) =
Z 1
0
f(x)(1  G(x)) dx;
where G(x) is the cumulative distribution function for g(x). Explain why this
is the case. Use this to show that
P(X < Y ) = 
 + 
and to answer Jones's question.",prob.pdf
"224 CHAPTER 5. DISTRIBUTIONS AND DENSITIES
35 Consider the simple queueing process of Example 5.7. Suppose that you watch
the size of the queue. If there are j people in the queue the next time the
queue size changes it will either decrease to j  1 or increase to j + 1. Use
the result of Exercise 34 to show that the probability that the queue size
decreases to j  1 is =( + ) and the probability that it increases to j + 1
is =( + ). When the queue size is 0 it can only increase to 1. Write a
program to simulate the queue size. Use this simulation to help formulate a
conjecture containing conditions on  and  that will ensure that the queue
will have times when it is empty.
36 Let X be a random variable having an exponential density with parameter .
Find the density for the random variable Y = rX, where r is a positive real
number.
37 Let X be a random variable having a normal density and consider the random",prob.pdf
"number.
37 Let X be a random variable having a normal density and consider the random
variable Y = eX. Then Y has a log normal density. Find this density of Y .
38 Let X1 and X2 be independent random variables and for i = 1; 2, let Yi =
i(Xi), where i is strictly increasing on the range of Xi. Show that Y1 and
Y2 are independent. Note that the same result is true without the assumption
that the i's are strictly increasing, but the proof is more dicult.",prob.pdf
"Chapter 6
Expected Value and Variance
6.1 Expected Value of Discrete Random Variables
When a large collection of numbers is assembled, as in a census, we are usually
interested not in the individual numbers, but rather in certain descriptive quantities
such as the average or the median. In general, the same is true for the probability
distribution of a numerically-valued random variable. In this and in the next section,
we shall discuss two such descriptive quantities: the expected value and the variance.
Both of these quantities apply only to numerically-valued random variables, and so
we assume, in these sections, that all random variables have numerical values. To
give some intuitive justication for our denition, we consider the following game.
Average Value
A die is rolled. If an odd number turns up, we win an amount equal to this number;
if an even number turns up, we lose an amount equal to this number. For example,",prob.pdf
"if an even number turns up, we lose an amount equal to this number. For example,
if a two turns up we lose 2, and if a three comes up we win 3. We want to decide if
this is a reasonable game to play. We rst try simulation. The program Die carries
out this simulation.
The program prints the frequency and the relative frequency with which each
outcome occurs. It also calculates the average winnings. We have run the program
twice. The results are shown in Table 6.1.
In the rst run we have played the game 100 times. In this run our average gain
is :57. It looks as if the game is unfavorable, and we wonder how unfavorable it
really is. To get a better idea, we have played the game 10,000 times. In this case
our average gain is :4949.
We note that the relative frequency of each of the six possible outcomes is quite
close to the probability 1/6 for this outcome. This corresponds to our frequency
interpretation of probability. It also suggests that for very large numbers of plays,",prob.pdf
"interpretation of probability. It also suggests that for very large numbers of plays,
our average gain should be
 = 1
1
6

 2
1
6

+ 3
1
6

 4
1
6

+ 5
1
6

 6
1
6

225",prob.pdf
"226 CHAPTER 6. EXPECTED VALUE AND VARIANCE
n = 100 n = 10000
Winning Frequency Relative Frequency Relative
Frequency Frequency
1 17 .17 1681 .1681
-2 17 .17 1678 .1678
3 16 .16 1626 .1626
-4 18 .18 1696 .1696
5 16 .16 1686 .1686
-6 16 .16 1633 .1633
Table 6.1: Frequencies for dice game.
= 9
6  12
6 = 3
6 = :5 :
This agrees quite well with our average gain for 10,000 plays.
We note that the value we have chosen for the average gain is obtained by taking
the possible outcomes, multiplying by the probability, and adding the results. This
suggests the following denition for the expected outcome of an experiment.
Expected Value
Denition 6.1 Let X be a numerically-valued discrete random variable with sam-
ple space 
 and distribution function m(x). The expected value E(X) is dened
by
E(X) =
X
x2
xm(x) ;
provided this sum converges absolutely. We often refer to the expected value as
the mean, and denote E(X) by  for short. If the above sum does not converge",prob.pdf
"the mean, and denote E(X) by  for short. If the above sum does not converge
absolutely, then we say that X does not have an expected value. 2
Example 6.1 Let an experiment consist of tossing a fair coin three times. Let
X denote the number of heads which appear. Then the possible values of X are
0; 1; 2 and 3. The corresponding probabilities are 1=8; 3=8; 3=8; and 1=8. Thus, the
expected value of X equals
0
1
8

+ 1
3
8

+ 2
3
8

+ 3
1
8

= 3
2 :
Later in this section we shall see a quicker way to compute this expected value,
based on the fact that X can be written as a sum of simpler random variables. 2
Example 6.2 Suppose that we toss a fair coin until a head rst comes up, and let
X represent the number of tosses which were made. Then the possible values of X
are 1; 2; : : :, and the distribution function of X is dened by
m(i) = 1
2i :",prob.pdf
"6.1. EXPECTED VALUE 227
(This is just the geometric distribution with parameter 1=2.) Thus, we have
E(X) =
1X
i=1
i 1
2i
=
1X
i=1
1
2i +
1X
i=2
1
2i + 
 
 

= 1 + 1
2 + 1
22 + 
 
 

= 2 :
2
Example 6.3 (Example 6.2 continued) Suppose that we 
ip a coin until a head
rst appears, and if the number of tosses equals n, then we are paid 2n dollars.
What is the expected value of the payment?
We let Y represent the payment. Then,
P(Y = 2n) = 1
2n ;
for n  1. Thus,
E(Y ) =
1X
n=1
2n 1
2n ;
which is a divergent sum. Thus, Y has no expectation. This example is called
the St. Petersburg Paradox. The fact that the above sum is innite suggests that
a player should be willing to pay any xed amount per game for the privilege of
playing this game. The reader is asked to consider how much he or she would be
willing to pay for this privilege. It is unlikely that the reader's answer is more than
10 dollars; therein lies the paradox.",prob.pdf
"willing to pay for this privilege. It is unlikely that the reader's answer is more than
10 dollars; therein lies the paradox.
In the early history of probability, various mathematicians gave ways to resolve
this paradox. One idea (due to G. Cramer) consists of assuming that the amount
of money in the world is nite. He thus assumes that there is some xed value of
n such that if the number of tosses equals or exceeds n, the payment is 2n dollars.
The reader is asked to show in Exercise 20 that the expected value of the payment
is now nite.
Daniel Bernoulli and Cramer also considered another way to assign value to
the payment. Their idea was that the value of a payment is some function of the
payment; such a function is now called a utility function. Examples of reasonable
utility functions might include the square-root function or the logarithm function.
In both cases, the value of 2n dollars is less than twice the value of n dollars. It",prob.pdf
"In both cases, the value of 2n dollars is less than twice the value of n dollars. It
can easily be shown that in both cases, the expected utility of the payment is nite
(see Exercise 20). 2",prob.pdf
"228 CHAPTER 6. EXPECTED VALUE AND VARIANCE
Example 6.4 Let T be the time for the rst success in a Bernoulli trials process.
Then we take as sample space 
 the integers 1; 2; : : : and assign the geometric
distribution
m(j) = P(T = j) = qj1p :
Thus,
E(T) = 1 
 p + 2qp + 3q2p + 
 
 

= p(1 + 2q + 3q2 + 
 
 
) :
Now if jxj < 1, then
1 + x + x2 + x3 + 
 
 
 = 1
1  x :
Dierentiating this formula, we get
1 + 2x + 3x2 + 
 
 
 = 1
(1  x)2 ;
so
E(T) = p
(1  q)2 = p
p2 = 1
p :
In particular, we see that if we toss a fair coin a sequence of times, the expected
time until the rst heads is 1/(1/2) = 2. If we roll a die a sequence of times, the
expected number of rolls until the rst six is 1/(1/6) = 6. 2
Interpretation of Expected Value
In statistics, one is frequently concerned with the average value of a set of data.
The following example shows that the ideas of average value and expected value are
very closely related.",prob.pdf
"The following example shows that the ideas of average value and expected value are
very closely related.
Example 6.5 The heights, in inches, of the women on the Swarthmore basketball
team are 5' 9"", 5' 9"", 5' 6"", 5' 8"", 5' 11"", 5' 5"", 5' 7"", 5' 6"", 5' 6"", 5' 7"", 5' 10"", and
6' 0"".
A statistician would compute the average height (in inches) as follows:
69 + 69 + 66 + 68 + 71 + 65 + 67 + 66 + 66 + 67 + 70 + 72
12 = 67:9 :
One can also interpret this number as the expected value of a random variable. To
see this, let an experiment consist of choosing one of the women at random, and let
X denote her height. Then the expected value of X equals 67.9. 2
Of course, just as with the frequency interpretation of probability, to interpret
expected value as an average outcome requires further justication. We know that
for any nite experiment the average of the outcomes is not predictable. However,
we shall eventually prove that the average will usually be close to E(X) if we repeat",prob.pdf
"we shall eventually prove that the average will usually be close to E(X) if we repeat
the experiment a large number of times. We rst need to develop some properties of
the expected value. Using these properties, and those of the concept of the variance",prob.pdf
"6.1. EXPECTED VALUE 229
X Y
HHH 1
HHT 2
HTH 3
HTT 2
THH 2
THT 3
TTH 2
TTT 1
Table 6.2: Tossing a coin three times.
to be introduced in the next section, we shall be able to prove the Law of Large
Numbers. This theorem will justify mathematically both our frequency concept
of probability and the interpretation of expected value as the average value to be
expected in a large number of experiments.
Expectation of a Function of a Random Variable
Suppose that X is a discrete random variable with sample space 
, and (x) is
a real-valued function with domain 
. Then (X) is a real-valued random vari-
able. One way to determine the expected value of (X) is to rst determine the
distribution function of this random variable, and then use the denition of expec-
tation. However, there is a better way to compute the expected value of (X), as
demonstrated in the next example.
Example 6.6 Suppose a coin is tossed 9 times, with the result
HHHTTTTHT :",prob.pdf
"demonstrated in the next example.
Example 6.6 Suppose a coin is tossed 9 times, with the result
HHHTTTTHT :
The rst set of three heads is called a run. There are three more runs in this
sequence, namely the next four tails, the next head, and the next tail. We do not
consider the rst two tosses to constitute a run, since the third toss has the same
value as the rst two.
Now suppose an experiment consists of tossing a fair coin three times. Find the
expected number of runs. It will be helpful to think of two random variables, X
and Y , associated with this experiment. We let X denote the sequence of heads and
tails that results when the experiment is performed, and Y denote the number of
runs in the outcome X. The possible outcomes of X and the corresponding values
of Y are shown in Table 6.2.
To calculate E(Y ) using the denition of expectation, we rst must nd the
distribution function m(y) of Y i.e., we group together those values of X with a",prob.pdf
"distribution function m(y) of Y i.e., we group together those values of X with a
common value of Y and add their probabilities. In this case, we calculate that the
distribution function of Y is: m(1) = 1=4; m(2) = 1=2; and m(3) = 1=4. One easily
nds that E(Y ) = 2.",prob.pdf
"230 CHAPTER 6. EXPECTED VALUE AND VARIANCE
Now suppose we didn't group the values of X with a common Y -value, but
instead, for each X-value x, we multiply the probability of x and the corresponding
value of Y , and add the results. We obtain
1
1
8

+ 2
1
8

+ 3
1
8

+ 2
1
8

+ 2
1
8

+ 3
1
8

+ 2
1
8

+ 1
1
8

;
which equals 2.
This illustrates the following general principle. If X and Y are two random
variables, and Y can be written as a function of X, then one can compute the
expected value of Y using the distribution function of X. 2
Theorem 6.1 If X is a discrete random variable with sample space 
 and distri-
bution function m(x), and if  : 
 ! R is a function, then
E((X)) =
X
x2
(x)m(x) ;
provided the series converges absolutely. 2
The proof of this theorem is straightforward, involving nothing more than group-
ing values of X with a common Y -value, as in Example 6.6.
The Sum of Two Random Variables",prob.pdf
"ing values of X with a common Y -value, as in Example 6.6.
The Sum of Two Random Variables
Many important results in probability theory concern sums of random variables.
We rst consider what it means to add two random variables.
Example 6.7 We 
ip a coin and let X have the value 1 if the coin comes up heads
and 0 if the coin comes up tails. Then, we roll a die and let Y denote the face that
comes up. What does X + Y mean, and what is its distribution? This question
is easily answered in this case, by considering, as we did in Chapter 4, the joint
random variable Z = (X; Y ), whose outcomes are ordered pairs of the form (x; y),
where 0  x  1 and 1  y  6. The description of the experiment makes it
reasonable to assume that X and Y are independent, so the distribution function
of Z is uniform, with 1=12 assigned to each outcome. Now it is an easy matter to
nd the set of outcomes of X + Y , and its distribution function. 2",prob.pdf
"of Z is uniform, with 1=12 assigned to each outcome. Now it is an easy matter to
nd the set of outcomes of X + Y , and its distribution function. 2
In Example 6.1, the random variable X denoted the number of heads which
occur when a fair coin is tossed three times. It is natural to think of X as the
sum of the random variables X1; X2; X3, where Xi is dened to be 1 if the ith toss
comes up heads, and 0 if the ith toss comes up tails. The expected values of the
Xi's are extremely easy to compute. It turns out that the expected value of X can
be obtained by simply adding the expected values of the Xi's. This fact is stated
in the following theorem.",prob.pdf
"6.1. EXPECTED VALUE 231
Theorem 6.2 Let X and Y be random variables with nite expected values. Then
E(X + Y ) = E(X) + E(Y ) ;
and if c is any constant, then
E(cX) = cE(X) :
Proof. Let the sample spaces of X and Y be denoted by 
X and 
Y , and suppose
that

X = fx1; x2; : : :g
and

Y = fy1; y2; : : :g :
Then we can consider the random variable X + Y to be the result of applying the
function (x; y) = x+y to the joint random variable (X; Y ). Then, by Theorem 6.1,
we have
E(X + Y ) =
X
j
X
k
(xj + yk)P(X = xj; Y = yk)
=
X
j
X
k
xjP(X = xj; Y = yk) +
X
j
X
k
ykP(X = xj; Y = yk)
=
X
j
xjP(X = xj) +
X
k
ykP(Y = yk) :
The last equality follows from the fact that
X
k
P(X = xj; Y = yk) = P(X = xj)
and X
j
P(X = xj; Y = yk) = P(Y = yk) :
Thus,
E(X + Y ) = E(X) + E(Y ) :
If c is any constant,
E(cX) =
X
j
cxjP(X = xj)
= c
X
j
xjP(X = xj)
= cE(X) :
2",prob.pdf
"232 CHAPTER 6. EXPECTED VALUE AND VARIANCE
X Y
a b c 3
a c b 1
b a c 1
b c a 0
c a b 0
c b a 1
Table 6.3: Number of xed points.
It is easy to prove by mathematical induction that the expected value of the sum
of any nite number of random variables is the sum of the expected values of the
individual random variables.
It is important to note that mutual independence of the summands was not
needed as a hypothesis in the Theorem 6.2 and its generalization. The fact that
expectations add, whether or not the summands are mutually independent, is some-
times referred to as the First Fundamental Mystery of Probability.
Example 6.8 Let Y be the number of xed points in a random permutation of
the set fa; b; cg. To nd the expected value of Y , it is helpful to consider the basic
random variable associated with this experiment, namely the random variable X
which represents the random permutation. There are six possible outcomes of X,",prob.pdf
"which represents the random permutation. There are six possible outcomes of X,
and we assign to each of them the probability 1=6 see Table 6.3. Then we can
calculate E(Y ) using Theorem 6.1, as
3
1
6

+ 1
1
6

+ 1
1
6

+ 0
1
6

+ 0
1
6

+ 1
1
6

= 1 :
We now give a very quick way to calculate the average number of xed points
in a random permutation of the set f1; 2; 3; : : :; ng. Let Z denote the random
permutation. For each i, 1  i  n, let Xi equal 1 if Z xes i, and 0 otherwise. So
if we let F denote the number of xed points in Z, then
F = X1 + X2 + 
 
 
 + Xn :
Therefore, Theorem 6.2 implies that
E(F) = E(X1) + E(X2) + 
 
 
 + E(Xn) :
But it is easy to see that for each i,
E(Xi) = 1
n ;
so
E(F) = 1 :
This method of calculation of the expected value is frequently very useful. It applies
whenever the random variable in question can be written as a sum of simpler random
variables. We emphasize again that it is not necessary that the summands be
mutually independent. 2",prob.pdf
"6.1. EXPECTED VALUE 233
Bernoulli Trials
Theorem 6.3 Let Sn be the number of successes in n Bernoulli trials with prob-
ability p for success on each trial. Then the expected number of successes is np.
That is,
E(Sn) = np :
Proof. Let Xj be a random variable which has the value 1 if the jth outcome is a
success and 0 if it is a failure. Then, for each Xj,
E(Xj) = 0 
 (1  p) + 1 
 p = p :
Since
Sn = X1 + X2 + 
 
 
 + Xn ;
and the expected value of the sum is the sum of the expected values, we have
E(Sn) = E(X1) + E(X2) + 
 
 
 + E(Xn)
= np :
2
Poisson Distribution
Recall that the Poisson distribution with parameter  was obtained as a limit of
binomial distributions with parameters n and p, where it was assumed that np = ,
and n ! 1. Since for each n, the corresponding binomial distribution has expected
value , it is reasonable to guess that the expected value of a Poisson distribution
with parameter  also has expectation equal to . This is in fact the case, and the",prob.pdf
"with parameter  also has expectation equal to . This is in fact the case, and the
reader is invited to show this (see Exercise 21).
Independence
If X and Y are two random variables, it is not true in general that E(X 
 Y ) =
E(X)E(Y ). However, this is true if X and Y are independent.
Theorem 6.4 If X and Y are independent random variables, then
E(X 
 Y ) = E(X)E(Y ) :
Proof. Suppose that",prob.pdf
"X = fx1; x2; : : :g
and

Y = fy1; y2; : : :g",prob.pdf
"234 CHAPTER 6. EXPECTED VALUE AND VARIANCE
are the sample spaces of X and Y , respectively. Using Theorem 6.1, we have
E(X 
 Y ) =
X
j
X
k
xjykP(X = xj; Y = yk) :
But if X and Y are independent,
P(X = xj; Y = yk) = P(X = xj)P(Y = yk) :
Thus,
E(X 
 Y ) =
X
j
X
k
xjykP(X = xj)P(Y = yk)
=
0
@X
j
xjP(X = xj)
1
A
 X
k
ykP(Y = yk)
!
= E(X)E(Y ) :
2
Example 6.9 A coin is tossed twice. Xi = 1 if the ith toss is heads and 0 otherwise.
We know that X1 and X2 are independent. They each have expected value 1/2.
Thus E(X1 
 X2) = E(X1)E(X2) = (1=2)(1=2) = 1=4. 2
We next give a simple example to show that the expected values need not mul-
tiply if the random variables are not independent.
Example 6.10 Consider a single toss of a coin. We dene the random variable X
to be 1 if heads turns up and 0 if tails turns up, and we set Y = 1  X. Then
E(X) = E(Y ) = 1=2. But X 
 Y = 0 for either outcome. Hence, E(X 
 Y ) = 0 6=
E(X)E(Y ). 2",prob.pdf
"E(X) = E(Y ) = 1=2. But X 
 Y = 0 for either outcome. Hence, E(X 
 Y ) = 0 6=
E(X)E(Y ). 2
We return to our records example of Section 3.1 for another application of the
result that the expected value of the sum of random variables is the sum of the
expected values of the individual random variables.
Records
Example 6.11 We start keeping snowfall records this year and want to nd the
expected number of records that will occur in the next n years. The rst year is
necessarily a record. The second year will be a record if the snowfall in the second
year is greater than that in the rst year. By symmetry, this probability is 1/2.
More generally, let Xj be 1 if the jth year is a record and 0 otherwise. To nd
E(Xj), we need only nd the probability that the jth year is a record. But the
record snowfall for the rst j years is equally likely to fall in any one of these years,",prob.pdf
"6.1. EXPECTED VALUE 235
so E(Xj) = 1=j. Therefore, if Sn is the total number of records observed in the
rst n years,
E(Sn) = 1 + 1
2 + 1
3 + 
 
 
 + 1
n :
This is the famous divergent harmonic series. It is easy to show that
E(Sn)  log n
as n ! 1. A more accurate approximation to E(Sn) is given by the expression
log n + 
 + 1
2n ;
where 
 denotes Euler's constant, and is approximately equal to .5772.
Therefore, in ten years the expected number of records is approximately 2:9298;
the exact value is the sum of the rst ten terms of the harmonic series which is
2.9290. 2
Craps
Example 6.12 In the game of craps, the player makes a bet and rolls a pair of
dice. If the sum of the numbers is 7 or 11 the player wins, if it is 2, 3, or 12 the
player loses. If any other number results, say r, then r becomes the player's point
and he continues to roll until either r or 7 occurs. If r comes up rst he wins, and
if 7 comes up rst he loses. The program Craps simulates playing this game a",prob.pdf
"if 7 comes up rst he loses. The program Craps simulates playing this game a
number of times.
We have run the program for 1000 plays in which the player bets 1 dollar each
time. The player's average winnings were :006. The game of craps would seem
to be only slightly unfavorable. Let us calculate the expected winnings on a single
play and see if this is the case. We construct a two-stage tree measure as shown in
Figure 6.1.
The rst stage represents the possible sums for his rst roll. The second stage
represents the possible outcomes for the game if it has not ended on the rst roll. In
this stage we are representing the possible outcomes of a sequence of rolls required
to determine the nal outcome. The branch probabilities for the rst stage are
computed in the usual way assuming all 36 possibilites for outcomes for the pair of
dice are equally likely. For the second stage we assume that the game will eventually",prob.pdf
"dice are equally likely. For the second stage we assume that the game will eventually
end, and we compute the conditional probabilities for obtaining either the point or
a 7. For example, assume that the player's point is 6. Then the game will end when
one of the eleven pairs, (1; 5), (2; 4), (3; 3), (4; 2), (5; 1), (1; 6), (2; 5), (3; 4), (4; 3),
(5; 2), (6; 1), occurs. We assume that each of these possible pairs has the same
probability. Then the player wins in the rst ve cases and loses in the last six.
Thus the probability of winning is 5/11 and the probability of losing is 6/11. From
the path probabilities, we can nd the probability that the player wins 1 dollar; it
is 244/495. The probability of losing is then 251/495. Thus if X is his winning for",prob.pdf
"236 CHAPTER 6. EXPECTED VALUE AND VARIANCE
W
L
W
L
W
L
W
L
W
L
W
L
(2,3,12) L
10
9
8
6
5
4
(7,11) W
1/3
2/3
2/5
3/5
5/11
6/11
5/11
6/11
2/5
3/5
1/3
2/3
2/9
1/12
1/9
5/36
5/36
1/9
1/12
1/9
1/36
2/36
2/45
3/45
25/396
30/396
25/396
30/396
2/45
3/45
1/36
2/36
Figure 6.1: Tree measure for craps.",prob.pdf
"6.1. EXPECTED VALUE 237
a dollar bet,
E(X) = 1
244
495

+ (1)
251
495

=  7
495  :0141 :
The game is unfavorable, but only slightly. The player's expected gain in n plays is
n(:0141). If n is not large, this is a small expected loss for the player. The casino
makes a large number of plays and so can aord a small average gain per play and
still expect a large prot. 2
Roulette
Example 6.13 In Las Vegas, a roulette wheel has 38 slots numbered 0, 00, 1, 2,
. . . , 36. The 0 and 00 slots are green, and half of the remaining 36 slots are red
and half are black. A croupier spins the wheel and throws an ivory ball. If you bet
1 dollar on red, you win 1 dollar if the ball stops in a red slot, and otherwise you
lose a dollar. We wish to calculate the expected value of your winnings, if you bet
1 dollar on red.
Let X be the random variable which denotes your winnings in a 1 dollar bet on
red in Las Vegas roulette. Then the distribution of X is given by
mX =
 1 1
20=38 18=38

;",prob.pdf
"red in Las Vegas roulette. Then the distribution of X is given by
mX =
 1 1
20=38 18=38

;
and one can easily calculate (see Exercise 5) that
E(X)  :0526 :
We now consider the roulette game in Monte Carlo, and follow the treatment
of Sagan.1 In the roulette game in Monte Carlo there is only one 0. If you bet 1
franc on red and a 0 turns up, then, depending upon the casino, one or more of the
following options may be oered:
(a) You get 1/2 of your bet back, and the casino gets the other half of your bet.
(b) Your bet is put \in prison,"" which we will denote by P1. If red comes up on
the next turn, you get your bet back (but you don't win any money). If black or 0
comes up, you lose your bet.
(c) Your bet is put in prison P1, as before. If red comes up on the next turn, you
get your bet back, and if black comes up on the next turn, then you lose your bet.
If a 0 comes up on the next turn, then your bet is put into double prison, which we",prob.pdf
"If a 0 comes up on the next turn, then your bet is put into double prison, which we
will denote by P2. If your bet is in double prison, and if red comes up on the next
turn, then your bet is moved back to prison P1 and the game proceeds as before.
If your bet is in double prison, and if black or 0 come up on the next turn, then
you lose your bet. We refer the reader to Figure 6.2, where a tree for this option is
shown. In this gure, S is the starting position, W means that you win your bet,
L means that you lose your bet, and E means that you break even.
1H. Sagan, Markov Chains in Monte Carlo, Math. Mag., vol. 54, no. 1 (1981), pp. 3-10.",prob.pdf
"238 CHAPTER 6. EXPECTED VALUE AND VARIANCE
S
W
L
E
LL
L
L
L
L
E
P1 P1 P1
P2P2 P2
Figure 6.2: Tree for 2-prison Monte Carlo roulette.
It is interesting to compare the expected winnings of a 1 franc bet on red, under
each of these three options. We leave the rst two calculations as an exercise (see
Exercise 37). Suppose that you choose to play alternative (c). The calculation for
this case illustrates the way that the early French probabilists worked problems like
this.
Suppose you bet on red, you choose alternative (c), and a 0 comes up. Your
possible future outcomes are shown in the tree diagram in Figure 6.3. Assume that
your money is in the rst prison and let x be the probability that you lose your
franc. From the tree diagram we see that
x = 18
37 + 1
37P(you lose your franc j your franc is in P2) :
Also,
P(you lose your franc j your franc is in P2) = 19
37 + 18
37x :
So, we have
x = 18
37 + 1
37
19
37 + 18
37x

:",prob.pdf
"Also,
P(you lose your franc j your franc is in P2) = 19
37 + 18
37x :
So, we have
x = 18
37 + 1
37
19
37 + 18
37x

:
Solving for x, we obtain x = 685=1351. Thus, starting at S, the probability that
you lose your bet equals
18
37 + 1
37x = 25003
49987 :
To nd the probability that you win when you bet on red, note that you can
only win if red comes up on the rst turn, and this happens with probability 18/37.
Thus your expected winnings are
1 
 18
37  1 
 25003
49987 =  687
49987  :0137 :",prob.pdf
"6.1. EXPECTED VALUE 239
P W
L
P
P
L
18/37
18/37
1/37
19/37
18/37
1
1
2
Figure 6.3: Your money is put in prison.
It is interesting to note that the more romantic option (c) is less favorable than
option (a) (see Exercise 37).
If you bet 1 dollar on the number 17, then the distribution function for your
winnings X is
PX =
 1 35
36=37 1=37

;
and the expected winnings are
1 
 36
37 + 35 
 1
37 =  1
37  :027 :
Thus, at Monte Carlo dierent bets have dierent expected values. In Las Vegas
almost all bets have the same expected value of 2=38 = :0526 (see Exercises 4
and 5). 2
Conditional Expectation
Denition 6.2 If F is any event and X is a random variable with sample space",prob.pdf
"= fx1; x2; : : :g, then the conditional expectation given F is dened by
E(XjF) =
X
j
xjP(X = xjjF) :
Conditional expectation is used most often in the form provided by the following
theorem. 2
Theorem 6.5 Let X be a random variable with sample space 
. If F1, F2, . . ., Fr
are events such that Fi \ Fj = ; for i 6= j and 
 = [jFj, then
E(X) =
X
j
E(XjFj)P(Fj) :",prob.pdf
"240 CHAPTER 6. EXPECTED VALUE AND VARIANCE
Proof. We have
X
j
E(XjFj)P(Fj) =
X
j
X
k
xkP(X = xkjFj)P(Fj)
=
X
j
X
k
xkP(X = xk and Fj occurs)
=
X
k
X
j
xkP(X = xk and Fj occurs)
=
X
k
xkP(X = xk)
= E(X) :
2
Example 6.14 (Example 6.12 continued) Let T be the number of rolls in a single
play of craps. We can think of a single play as a two-stage process. The rst stage
consists of a single roll of a pair of dice. The play is over if this roll is a 2, 3, 7,
11, or 12. Otherwise, the player's point is established, and the second stage begins.
This second stage consists of a sequence of rolls which ends when either the player's
point or a 7 is rolled. We record the outcomes of this two-stage experiment using
the random variables X and S, where X denotes the rst roll, and S denotes the
number of rolls in the second stage of the experiment (of course, S is sometimes
equal to 0). Note that T = S + 1. Then by Theorem 6.5
E(T) =
12X
j=2
E(TjX = j)P(X = j) :",prob.pdf
"equal to 0). Note that T = S + 1. Then by Theorem 6.5
E(T) =
12X
j=2
E(TjX = j)P(X = j) :
If j = 7, 11 or 2, 3, 12, then E(TjX = j) = 1. If j = 4; 5; 6; 8; 9; or 10, we can
use Example 6.4 to calculate the expected value of S. In each of these cases, we
continue rolling until we get either a j or a 7. Thus, S is geometrically distributed
with parameter p, which depends upon j. If j = 4, for example, the value of p is
3=36 + 6=36 = 1=4. Thus, in this case, the expected number of additional rolls is
1=p = 4, so E(TjX = 4) = 1 + 4 = 5. Carrying out the corresponding calculations
for the other possible values of j and using Theorem 6.5 gives
E(T) = 1
12
36

+

1 + 36
3 + 6
3
36

+

1 + 36
4 + 6
4
36

+

1 + 36
5 + 6
5
36

+

1 + 36
5 + 6
5
36

+

1 + 36
4 + 6
4
36

+

1 + 36
3 + 6
3
36

= 557
165
 3:375 : : : :
2",prob.pdf
"6.1. EXPECTED VALUE 241
Martingales
We can extend the notion of fairness to a player playing a sequence of games by
using the concept of conditional expectation.
Example 6.15 Let S1, S2, . . . , Sn be Peter's accumulated fortune in playing heads
or tails (see Example 1.4). Then
E(SnjSn1 = a; : : : ; S1 = r) = 1
2(a + 1) + 1
2(a  1) = a :
We note that Peter's expected fortune after the next play is equal to his present
fortune. When this occurs, we say the game is fair. A fair game is also called a
martingale. If the coin is biased and comes up heads with probability p and tails
with probability q = 1  p, then
E(SnjSn1 = a; : : :; S1 = r) = p(a + 1) + q(a  1) = a + p  q :
Thus, if p < q, this game is unfavorable, and if p > q, it is favorable. 2
If you are in a casino, you will see players adopting elaborate systems of play
to try to make unfavorable games favorable. Two such systems, the martingale
doubling system and the more conservative Labouchere system, were described in",prob.pdf
"doubling system and the more conservative Labouchere system, were described in
Exercises 1.1.9 and 1.1.10. Unfortunately, such systems cannot change even a fair
game into a favorable game.
Even so, it is a favorite pastime of many people to develop systems of play for
gambling games and for other games such as the stock market. We close this section
with a simple illustration of such a system.
Stock Prices
Example 6.16 Let us assume that a stock increases or decreases in value each
day by 1 dollar, each with probability 1/2. Then we can identify this simplied
model with our familiar game of heads or tails. We assume that a buyer, Mr. Ace,
adopts the following strategy. He buys the stock on the rst day at its price V .
He then waits until the price of the stock increases by one to V + 1 and sells. He
then continues to watch the stock until its price falls back to V . He buys again and
waits until it goes up to V +1 and sells. Thus he holds the stock in intervals during",prob.pdf
"waits until it goes up to V +1 and sells. Thus he holds the stock in intervals during
which it increases by 1 dollar. In each such interval, he makes a prot of 1 dollar.
However, we assume that he can do this only for a nite number of trading days.
Thus he can lose if, in the last interval that he holds the stock, it does not get back
up to V + 1; and this is the only way he can lose. In Figure 6.4 we illustrate a
typical history if Mr. Ace must stop in twenty days. Mr. Ace holds the stock under
his system during the days indicated by broken lines. We note that for the history
shown in Figure 6.4, his system nets him a gain of 4 dollars.
We have written a program StockSystem to simulate the fortune of Mr. Ace
if he uses his sytem over an n-day period. If one runs this program a large number",prob.pdf
"242 CHAPTER 6. EXPECTED VALUE AND VARIANCE
5 10 15 20
-1
-0.5
0.5
1
1.5
2
Figure 6.4: Mr. Ace's system.
of times, for n = 20, say, one nds that his expected winnings are very close to 0,
but the probability that he is ahead after 20 days is signicantly greater than 1/2.
For small values of n, the exact distribution of winnings can be calculated. The
distribution for the case n = 20 is shown in Figure 6.5. Using this distribution,
it is easy to calculate that the expected value of his winnings is exactly 0. This
is another instance of the fact that a fair game (a martingale) remains fair under
quite general systems of play.
Although the expected value of his winnings is 0, the probability that Mr. Ace is
ahead after 20 days is about .610. Thus, he would be able to tell his friends that his
system gives him a better chance of being ahead than that of someone who simply
buys the stock and holds it, if our simple random model is correct. There have been",prob.pdf
"buys the stock and holds it, if our simple random model is correct. There have been
a number of studies to determine how random the stock market is. 2
Historical Remarks
With the Law of Large Numbers to bolster the frequency interpretation of proba-
bility, we nd it natural to justify the denition of expected value in terms of the
average outcome over a large number of repetitions of the experiment. The concept
of expected value was used before it was formally dened; and when it was used,
it was considered not as an average value but rather as the appropriate value for
a gamble. For example recall, from the Historical Remarks section of Chapter 1,
Section 1.2, Pascal's way of nding the value of a three-game series that had to be
called o before it is nished.
Pascal rst observed that if each player has only one game to win, then the
stake of 64 pistoles should be divided evenly. Then he considered the case where
one player has won two games and the other one.",prob.pdf
"stake of 64 pistoles should be divided evenly. Then he considered the case where
one player has won two games and the other one.
Then consider, Sir, if the rst man wins, he gets 64 pistoles, if he loses
he gets 32. Thus if they do not wish to risk this last game, but wish",prob.pdf
"6.1. EXPECTED VALUE 243
-20 -15 -10 -5 0 5 10
0
0.05
0.1
0.15
0.2
Figure 6.5: Winnings distribution for n = 20.
to separate without playing it, the rst man must say: \I am certain
to get 32 pistoles, even if I lose I still get them; but as for the other
32 pistoles, perhaps I will get them, perhaps you will get them, the
chances are equal. Let us then divide these 32 pistoles in half and give
one half to me as well as my 32 which are mine for sure."" He will then
have 48 pistoles and the other 16.2
Note that Pascal reduced the problem to a symmetric bet in which each player
gets the same amount and takes it as obvious that in this case the stakes should be
divided equally.
The rst systematic study of expected value appears in Huygens' book. Like
Pascal, Huygens nd the value of a gamble by assuming that the answer is obvious
for certain symmetric situations and uses this to deduce the expected for the general
situation. He does this in steps. His rst proposition is",prob.pdf
"for certain symmetric situations and uses this to deduce the expected for the general
situation. He does this in steps. His rst proposition is
Prop. I. If I expect a or b, either of which, with equal probability, may
fall to me, then my Expectation is worth (a+b)=2, that is, the half Sum
of a and b.3
Huygens proved this as follows: Assume that two player A and B play a game in
which each player puts up a stake of (a + b)=2 with an equal chance of winning the
total stake. Then the value of the game to each player is (a + b)=2. For example, if
the game had to be called o clearly each player should just get back his original
stake. Now, by symmetry, this value is not changed if we add the condition that
the winner of the game has to pay the loser an amount b as a consolation prize.
Then for player A the value is still (a + b)=2. But what are his possible outcomes
2Quoted in F. N. David, Games, Gods and Gambling (London: Grin, 1962), p. 231.",prob.pdf
"2Quoted in F. N. David, Games, Gods and Gambling (London: Grin, 1962), p. 231.
3C. Huygens, Calculating in Games of Chance, translation attributed to John Arbuthnot (Lon-
don, 1692), p. 34.",prob.pdf
"244 CHAPTER 6. EXPECTED VALUE AND VARIANCE
for the modied game? If he wins he gets the total stake a + b and must pay B an
amount b so ends up with a. If he loses he gets an amount b from player B. Thus
player A wins a or b with equal chances and the value to him is (a + b)=2.
Huygens illustrated this proof in terms of an example. If you are oered a game
in which you have an equal chance of winning 2 or 8, the expected value is 5, since
this game is equivalent to the game in which each player stakes 5 and agrees to pay
the loser 3 | a game in which the value is obviously 5.
Huygens' second proposition is
Prop. II. If I expect a, b, or c, either of which, with equal facility, may
happen, then the Value of my Expectation is (a + b + c)=3, or the third
of the Sum of a, b, and c.4
His argument here is similar. Three players, A, B, and C, each stake
(a + b + c)=3
in a game they have an equal chance of winning. The value of this game to player",prob.pdf
"(a + b + c)=3
in a game they have an equal chance of winning. The value of this game to player
A is clearly the amount he has staked. Further, this value is not changed if A enters
into an agreement with B that if one of them wins he pays the other a consolation
prize of b and with C that if one of them wins he pays the other a consolation prize
of c. By symmetry these agreements do not change the value of the game. In this
modied game, if A wins he wins the total stake a + b + c minus the consolation
prizes b + c giving him a nal winning of a. If B wins, A wins b and if C wins, A
wins c. Thus A nds himself in a game with value (a + b + c)=3 and with outcomes
a, b, and c occurring with equal chance. This proves Proposition II.
More generally, this reasoning shows that if there are n outcomes
a1; a2; : : : ; an ;
all occurring with the same probability, the expected value is
a1 + a2 + 
 
 
 + an
n :
In his third proposition Huygens considered the case where you win a or b but",prob.pdf
"a1 + a2 + 
 
 
 + an
n :
In his third proposition Huygens considered the case where you win a or b but
with unequal probabilities. He assumed there are p chances of winning a, and q
chances of winning b, all having the same probability. He then showed that the
expected value is
E = p
p + q 
 a + q
p + q 
 b :
This follows by considering an equivalent gamble with p + q outcomes all occurring
with the same probability and with a payo of a in p of the outcomes and b in q of
the outcomes. This allowed Huygens to compute the expected value for experiments
with unequal probabilities, at least when these probablities are rational numbers.
Thus, instead of dening the expected value as a weighted average, Huygens
assumed that the expected value of certain symmetric gambles are known and de-
duced the other values from these. Although this requires a good deal of clever
4ibid., p. 35.",prob.pdf
"6.1. EXPECTED VALUE 245
manipulation, Huygens ended up with values that agree with those given by our
modern denition of expected value. One advantage of this method is that it gives
a justication for the expected value in cases where it is not reasonable to assume
that you can repeat the experiment a large number of times, as for example, in
betting that at least two presidents died on the same day of the year. (In fact,
three did; all were signers of the Declaration of Independence, and all three died on
July 4.)
In his book, Huygens calculated the expected value of games using techniques
similar to those which we used in computing the expected value for roulette at
Monte Carlo. For example, his proposition XIV is:
Prop. XIV. If I were playing with another by turns, with two Dice, on
this Condition, that if I throw 7 I gain, and if he throws 6 he gains
allowing him the rst Throw: To nd the proportion of my Hazard to
his.5",prob.pdf
"this Condition, that if I throw 7 I gain, and if he throws 6 he gains
allowing him the rst Throw: To nd the proportion of my Hazard to
his.5
A modern description of this game is as follows. Huygens and his opponent take
turns rolling a die. The game is over if Huygens rolls a 7 or his opponent rolls a 6.
His opponent rolls rst. What is the probability that Huygens wins the game?
To solve this problem Huygens let x be his chance of winning when his opponent
threw rst and y his chance of winning when he threw rst. Then on the rst roll
his opponent wins on 5 out of the 36 possibilities. Thus,
x = 31
36 
 y :
But when Huygens rolls he wins on 6 out of the 36 possible outcomes, and in the
other 30, he is led back to where his chances are x. Thus
y = 6
36 + 30
36 
 x :
From these two equations Huygens found that x = 31=61.
Another early use of expected value appeared in Pascal's argument to show that
a rational person should believe in the existence of God.6 Pascal said that we have",prob.pdf
"a rational person should believe in the existence of God.6 Pascal said that we have
to make a wager whether to believe or not to believe. Let p denote the probability
that God does not exist. His discussion suggests that we are playing a game with
two strategies, believe and not believe, with payos as shown in Table 6.4.
Here u represents the cost to you of passing up some worldly pleasures as
a consequence of believing that God exists. If you do not believe, and God is a
vengeful God, you will lose x. If God exists and you do believe you will gain v.
Now to determine which strategy is best you should compare the two expected
values
p(u) + (1  p)v and p0 + (1  p)(x);
5ibid., p. 47.
6Quoted in I. Hacking, The Emergence of Probability (Cambridge: Cambridge Univ. Press,
1975).",prob.pdf
"246 CHAPTER 6. EXPECTED VALUE AND VARIANCE
God does not exist God exists
p 1  p
believe u v
not believe 0 x
Table 6.4: Payos.
Age Survivors
0 100
6 64
16 40
26 25
36 16
46 10
56 6
66 3
76 1
Table 6.5: Graunt's mortality data.
and choose the larger of the two. In general, the choice will depend upon the value of
p. But Pascal assumed that the value of v is innite and so the strategy of believing
is best no matter what probability you assign for the existence of God. This example
is considered by some to be the beginning of decision theory. Decision analyses of
this kind appear today in many elds, and, in particular, are an important part of
medical diagnostics and corporate business decisions.
Another early use of expected value was to decide the price of annuities. The
study of statistics has its origins in the use of the bills of mortality kept in the
parishes in London from 1603. These records kept a weekly tally of christenings",prob.pdf
"parishes in London from 1603. These records kept a weekly tally of christenings
and burials. From these John Graunt made estimates for the population of London
and also provided the rst mortality data,7 shown in Table 6.5.
As Hacking observes, Graunt apparently constructed this table by assuming
that after the age of 6 there is a constant probability of about 5/8 of surviving
for another decade.8 For example, of the 64 people who survive to age 6, 5/8 of
64 or 40 survive to 16, 5/8 of these 40 or 25 survive to 26, and so forth. Of course,
he rounded o his gures to the nearest whole person.
Clearly, a constant mortality rate cannot be correct throughout the whole range,
and later tables provided by Halley were more realistic in this respect.9
7ibid., p. 108.
8ibid., p. 109.
9E. Halley, \An Estimate of The Degrees of Mortality of Mankind,"" Phil. Trans. Royal. Soc.,",prob.pdf
"6.1. EXPECTED VALUE 247
A terminal annuity provides a xed amount of money during a period of n years.
To determine the price of a terminal annuity one needs only to know the appropriate
interest rate. A life annuity provides a xed amount during each year of the buyer's
life. The appropriate price for a life annuity is the expected value of the terminal
annuity evaluated for the random lifetime of the buyer. Thus, the work of Huygens
in introducing expected value and the work of Graunt and Halley in determining
mortality tables led to a more rational method for pricing annuities. This was one
of the rst serious uses of probability theory outside the gambling houses.
Although expected value plays a role now in every branch of science, it retains
its importance in the casino. In 1962, Edward Thorp's book Beat the Dealer10
provided the reader with a strategy for playing the popular casino game of blackjack
that would assure the player a positive expected winning. This book forevermore",prob.pdf
"that would assure the player a positive expected winning. This book forevermore
changed the belief of the casinos that they could not be beat.
Exercises
1 A card is drawn at random from a deck consisting of cards numbered 2
through 10. A player wins 1 dollar if the number on the card is odd and
loses 1 dollar if the number if even. What is the expected value of his win-
nings?
2 A card is drawn at random from a deck of playing cards. If it is red, the player
wins 1 dollar; if it is black, the player loses 2 dollars. Find the expected value
of the game.
3 In a class there are 20 students: 3 are 5' 6"", 5 are 5'8"", 4 are 5'10"", 4 are
6', and 4 are 6' 2"". A student is chosen at random. What is the student's
expected height?
4 In Las Vegas the roulette wheel has a 0 and a 00 and then the numbers 1 to 36
marked on equal slots; the wheel is spun and a ball stops randomly in one
slot. When a player bets 1 dollar on a number, he receives 36 dollars if the",prob.pdf
"slot. When a player bets 1 dollar on a number, he receives 36 dollars if the
ball stops on this number, for a net gain of 35 dollars; otherwise, he loses his
dollar bet. Find the expected value for his winnings.
5 In a second version of roulette in Las Vegas, a player bets on red or black.
Half of the numbers from 1 to 36 are red, and half are black. If a player bets
a dollar on black, and if the ball stops on a black number, he gets his dollar
back and another dollar. If the ball stops on a red number or on 0 or 00 he
loses his dollar. Find the expected winnings for this bet.
6 A die is rolled twice. Let X denote the sum of the two numbers that turn up,
and Y the dierence of the numbers (specically, the number on the rst roll
minus the number on the second). Show that E(XY ) = E(X)E(Y ). Are X
and Y independent?
vol. 17 (1693), pp. 596{610; 654{656.
10E. Thorp, Beat the Dealer (New York: Random House, 1962).",prob.pdf
"248 CHAPTER 6. EXPECTED VALUE AND VARIANCE
*7 Show that, if X and Y are random variables taking on only two values each,
and if E(XY ) = E(X)E(Y ), then X and Y are independent.
8 A royal family has children until it has a boy or until it has three children,
whichever comes rst. Assume that each child is a boy with probability 1/2.
Find the expected number of boys in this royal family and the expected num-
ber of girls.
9 If the rst roll in a game of craps is neither a natural nor craps, the player
can make an additional bet, equal to his original one, that he will make his
point before a seven turns up. If his point is four or ten he is paid o at 2 : 1
odds; if it is a ve or nine he is paid o at odds 3 : 2; and if it is a six or eight
he is paid o at odds 6 : 5. Find the player's expected winnings if he makes
this additional bet when he has the opportunity.
10 In Example 6.16 assume that Mr. Ace decides to buy the stock and hold it",prob.pdf
"this additional bet when he has the opportunity.
10 In Example 6.16 assume that Mr. Ace decides to buy the stock and hold it
until it goes up 1 dollar and then sell and not buy again. Modify the program
StockSystem to nd the distribution of his prot under this system after
a twenty-day period. Find the expected prot and the probability that he
comes out ahead.
11 On September 26, 1980, the New York Times reported that a mysterious
stranger strode into a Las Vegas casino, placed a single bet of 777,000 dollars
on the \don't pass"" line at the crap table, and walked away with more than
1.5 million dollars. In the \don't pass"" bet, the bettor is essentially betting
with the house. An exception occurs if the roller rolls a 12 on the rst roll.
In this case, the roller loses and the \don't pass"" better just gets back the
money bet instead of winning. Show that the \don't pass"" bettor has a more
favorable bet than the roller.",prob.pdf
"money bet instead of winning. Show that the \don't pass"" bettor has a more
favorable bet than the roller.
12 Recall that in the martingale doubling system (see Exercise 1.1.10), the player
doubles his bet each time he loses. Suppose that you are playing roulette in
a fair casino where there are no 0's, and you bet on red each time. You then
win with probability 1/2 each time. Assume that you enter the casino with
100 dollars, start with a 1-dollar bet and employ the martingale system. You
stop as soon as you have won one bet, or in the unlikely event that black
turns up six times in a row so that you are down 63 dollars and cannot make
the required 64-dollar bet. Find your expected winnings under this system of
play.
13 You have 80 dollars and play the following game. An urn contains two white
balls and two black balls. You draw the balls out one at a time without
replacement until all the balls are gone. On each draw, you bet half of your",prob.pdf
"balls and two black balls. You draw the balls out one at a time without
replacement until all the balls are gone. On each draw, you bet half of your
present fortune that you will draw a white ball. What is your expected nal
fortune?
14 In the hat check problem (see Example 3.12), it was assumed that N people
check their hats and the hats are handed back at random. Let Xj = 1 if the",prob.pdf
"6.1. EXPECTED VALUE 249
jth person gets his or her hat and 0 otherwise. Find E(Xj) and E(Xj 
 Xk)
for j not equal to k. Are Xj and Xk independent?
15 A box contains two gold balls and three silver balls. You are allowed to choose
successively balls from the box at random. You win 1 dollar each time you
draw a gold ball and lose 1 dollar each time you draw a silver ball. After a
draw, the ball is not replaced. Show that, if you draw until you are ahead by
1 dollar or until there are no more gold balls, this is a favorable game.
16 Gerolamo Cardano in his book, The Gambling Scholar, written in the early
1500s, considers the following carnival game. There are six dice. Each of the
dice has ve blank sides. The sixth side has a number between 1 and 6|a
dierent number on each die. The six dice are rolled and the player wins a
prize depending on the total of the numbers which turn up.
(a) Find, as Cardano did, the expected total without nding its distribution.",prob.pdf
"prize depending on the total of the numbers which turn up.
(a) Find, as Cardano did, the expected total without nding its distribution.
(b) Large prizes were given for large totals with a modest fee to play the
game. Explain why this could be done.
17 Let X be the rst time that a failure occurs in an innite sequence of Bernoulli
trials with probability p for success. Let pk = P(X = k) for k = 1, 2, . . . .
Show that pk = pk1q where q = 1  p. Show that P
k pk = 1. Show that
E(X) = 1=q. What is the expected number of tosses of a coin required to
obtain the rst tail?
18 Exactly one of six similar keys opens a certain door. If you try the keys, one
after another, what is the expected number of keys that you will have to try
before success?
19 A multiple choice exam is given. A problem has four possible answers, and
exactly one answer is correct. The student is allowed to choose a subset of
the four possible answers as his answer. If his chosen subset contains the",prob.pdf
"the four possible answers as his answer. If his chosen subset contains the
correct answer, the student receives three points, but he loses one point for
each wrong answer in his chosen subset. Show that if he just guesses a subset
uniformly and randomly his expected score is zero.
20 You are oered the following game to play: a fair coin is tossed until heads
turns up for the rst time (see Example 6.3). If this occurs on the rst toss
you receive 2 dollars, if it occurs on the second toss you receive 22 = 4 dollars
and, in general, if heads turns up for the rst time on the nth toss you receive
2n dollars.
(a) Show that the expected value of your winnings does not exist (i.e., is
given by a divergent sum) for this game. Does this mean that this game
is favorable no matter how much you pay to play it?
(b) Assume that you only receive 210 dollars if any number greater than or
equal to ten tosses are required to obtain the rst head. Show that your",prob.pdf
"(b) Assume that you only receive 210 dollars if any number greater than or
equal to ten tosses are required to obtain the rst head. Show that your
expected value for this modied game is nite and nd its value.",prob.pdf
"250 CHAPTER 6. EXPECTED VALUE AND VARIANCE
(c) Assume that you pay 10 dollars for each play of the original game. Write
a program to simulate 100 plays of the game and see how you do.
(d) Now assume that the utility of n dollars is pn. Write an expression for
the expected utility of the payment, and show that this expression has a
nite value. Estimate this value. Repeat this exercise for the case that
the utility function is log(n).
21 Let X be a random variable which is Poisson distributed with parameter .
Show that E(X) = . Hint: Recall that
ex = 1 + x + x2
2! + x3
3! + 
 
 
 :
22 Recall that in Exercise 1.1.14, we considered a town with two hospitals. In
the large hospital about 45 babies are born each day, and in the smaller
hospital about 15 babies are born each day. We were interested in guessing
which hospital would have on the average the largest number of days with
the property that more than 60 percent of the children born on that day are",prob.pdf
"which hospital would have on the average the largest number of days with
the property that more than 60 percent of the children born on that day are
boys. For each hospital nd the expected number of days in a year that have
the property that more than 60 percent of the children born on that day were
boys.
23 An insurance company has 1,000 policies on men of age 50. The company
estimates that the probability that a man of age 50 dies within a year is .01.
Estimate the number of claims that the company can expect from beneciaries
of these men within a year.
24 Using the life table for 1981 in Appendix C, write a program to compute the
expected lifetime for males and females of each possible age from 1 to 85.
Compare the results for males and females. Comment on whether life insur-
ance should be priced dierently for males and females.
*25 A deck of ESP cards consists of 20 cards each of two types: say ten stars,",prob.pdf
"ance should be priced dierently for males and females.
*25 A deck of ESP cards consists of 20 cards each of two types: say ten stars,
ten circles (normally there are ve types). The deck is shued and the cards
turned up one at a time. You, the alleged percipient, are to name the symbol
on each card before it is turned up.
Suppose that you are really just guessing at the cards. If you do not get to
see each card after you have made your guess, then it is easy to calculate the
expected number of correct guesses, namely ten.
If, on the other hand, you are guessing with information, that is, if you see
each card after your guess, then, of course, you might expect to get a higher
score. This is indeed the case, but calculating the correct expectation is no
longer easy.
But it is easy to do a computer simulation of this guessing with information,
so we can get a good idea of the expectation by simulation. (This is similar to",prob.pdf
"so we can get a good idea of the expectation by simulation. (This is similar to
the way that skilled blackjack players make blackjack into a favorable game
by observing the cards that have already been played. See Exercise 29.)",prob.pdf
"6.1. EXPECTED VALUE 251
(a) First, do a simulation of guessing without information, repeating the
experiment at least 1000 times. Estimate the expected number of correct
answers and compare your result with the theoretical expectation.
(b) What is the best strategy for guessing with information?
(c) Do a simulation of guessing with information, using the strategy in (b).
Repeat the experiment at least 1000 times, and estimate the expectation
in this case.
(d) Let S be the number of stars and C the number of circles in the deck. Let
h(S; C) be the expected winnings using the optimal guessing strategy in
(b). Show that h(S; C) satises the recursion relation
h(S; C) = S
S + C h(S  1; C) + C
S + C h(S; C  1) + max(S; C)
S + C ;
and h(0; 0) = h(1; 0) = h(0; 1) = 0. Using this relation, write a
program to compute h(S; C) and nd h(10; 10). Compare the computed
value of h(10; 10) with the result of your simulation in (c). For more",prob.pdf
"program to compute h(S; C) and nd h(10; 10). Compare the computed
value of h(10; 10) with the result of your simulation in (c). For more
about this exercise and Exercise 26 see Diaconis and Graham.11
*26 Consider the ESP problem as described in Exercise 25. You are again guessing
with information, and you are using the optimal guessing strategy of guessing
star if the remaining deck has more stars, circle if more circles, and tossing a
coin if the number of stars and circles are equal. Assume that S  C, where
S is the number of stars and C the number of circles.
We can plot the results of a typical game on a graph, where the horizontal axis
represents the number of steps and the vertical axis represents the dierence
between the number of stars and the number of circles that have been turned
up. A typical game is shown in Figure 6.6. In this particular game, the order
in which the cards were turned up is (C; S; S; S; S; C; C; S; S; C). Thus, in this",prob.pdf
"in which the cards were turned up is (C; S; S; S; S; C; C; S; S; C). Thus, in this
particular game, there were six stars and four circles in the deck. This means,
in particular, that every game played with this deck would have a graph which
ends at the point (10; 2). We dene the line L to be the horizontal line which
goes through the ending point on the graph (so its vertical coordinate is just
the dierence between the number of stars and circles in the deck).
(a) Show that, when the random walk is below the line L, the player guesses
right when the graph goes up (star is turned up) and, when the walk is
above the line, the player guesses right when the walk goes down (circle
turned up). Show from this property that the subject is sure to have at
least S correct guesses.
(b) When the walk is at a point (x; x) on the line L the number of stars and
circles remaining is the same, and so the subject tosses a coin. Show that",prob.pdf
"circles remaining is the same, and so the subject tosses a coin. Show that
11P. Diaconis and R. Graham, \The Analysis of Sequential Experiments with Feedback to Sub-
jects,"" Annals of Statistics, vol. 9 (1981), pp. 3{23.",prob.pdf
"252 CHAPTER 6. EXPECTED VALUE AND VARIANCE
2
1
1 2 3 4 5 6 7 8 9 10
(10,2)L
Figure 6.6: Random walk for ESP.
the probability that the walk reaches (x; x) is
S
x

C
x


S+C
2x

 :
Hint: The outcomes of 2x cards is a hypergeometric distribution (see
Section 5.1).
(c) Using the results of (a) and (b) show that the expected number of correct
guesses under intelligent guessing is
S +
CX
x=1
1
2
S
x

C
x


S+C
2x

 :
27 It has been said12 that a Dr. B. Muriel Bristol declined a cup of tea stating
that she preferred a cup into which milk had been poured rst. The famous
statistician R. A. Fisher carried out a test to see if she could tell whether milk
was put in before or after the tea. Assume that for the test Dr. Bristol was
given eight cups of tea|four in which the milk was put in before the tea and
four in which the milk was put in after the tea.
(a) What is the expected number of correct guesses the lady would make if
she had no information after each test and was just guessing?",prob.pdf
"(a) What is the expected number of correct guesses the lady would make if
she had no information after each test and was just guessing?
(b) Using the result of Exercise 26 nd the expected number of correct
guesses if she was told the result of each guess and used an optimal
guessing strategy.
28 In a popular computer game the computer picks an integer from 1 to n at
random. The player is given k chances to guess the number. After each guess
the computer responds \correct,"" \too small,"" or \too big.""
12J. F. Box, R. A. Fisher, The Life of a Scientist (New York: John Wiley and Sons, 1978).",prob.pdf
"6.1. EXPECTED VALUE 253
(a) Show that if n  2k 1, then there is a strategy that guarantees you will
correctly guess the number in k tries.
(b) Show that if n  2k 1, there is a strategy that assures you of identifying
one of 2k  1 numbers and hence gives a probability of (2k  1)=n of
winning. Why is this an optimal strategy? Illustrate your result in
terms of the case n = 9 and k = 3.
29 In the casino game of blackjack the dealer is dealt two cards, one face up and
one face down, and each player is dealt two cards, both face down. If the
dealer is showing an ace the player can look at his down cards and then make
a bet called an insurance bet. (Expert players will recognize why it is called
insurance.) If you make this bet you will win the bet if the dealer's second
card is a ten card: namely, a ten, jack, queen, or king. If you win, you are
paid twice your insurance bet; otherwise you lose this bet. Show that, if the",prob.pdf
"paid twice your insurance bet; otherwise you lose this bet. Show that, if the
only cards you can see are the dealer's ace and your two cards and if your
cards are not ten cards, then the insurance bet is an unfavorable bet. Show,
however, that if you are playing two hands simultaneously, and you have no
ten cards, then it is a favorable bet. (Thorp13 has shown that the game of
blackjack is favorable to the player if he or she can keep good enough track
of the cards that have been played.)
30 Assume that, every time you buy a box of Wheaties, you receive a picture of
one of the n players for the New York Yankees (see Exercise 3.2.34). Let Xk
be the number of additional boxes you have to buy, after you have obtained
k1 dierent pictures, in order to obtain the next new picture. Thus X1 = 1,
X2 is the number of boxes bought after this to obtain a picture dierent from
the rst pictured obtained, and so forth.
(a) Show that Xk has a geometric distribution with p = (n  k + 1)=n.",prob.pdf
"the rst pictured obtained, and so forth.
(a) Show that Xk has a geometric distribution with p = (n  k + 1)=n.
(b) Simulate the experiment for a team with 26 players (25 would be more
accurate but we want an even number). Carry out a number of simula-
tions and estimate the expected time required to get the rst 13 players
and the expected time to get the second 13. How do these expectations
compare?
(c) Show that, if there are 2n players, the expected time to get the rst half
of the players is
2n
 1
2n + 1
2n  1 + 
 
 
 + 1
n + 1

;
and the expected time to get the second half is
2n
1
n + 1
n  1 + 
 
 
 + 1

:
13E. Thorp, Beat the Dealer (New York: Random House, 1962).",prob.pdf
"254 CHAPTER 6. EXPECTED VALUE AND VARIANCE
(d) In Example 6.11 we stated that
1 + 1
2 + 1
3 + 
 
 
 + 1
n  log n + :5772 + 1
2n :
Use this to estimate the expression in (c). Compare these estimates with
the exact values and also with your estimates obtained by simulation for
the case n = 26.
*31 (Feller14) A large number, N, of people are subjected to a blood test. This
can be administered in two ways: (1) Each person can be tested separately,
in this case N test are required, (2) the blood samples of k persons can be
pooled and analyzed together. If this test is negative, this one test suces
for the k people. If the test is positive, each of the k persons must be tested
separately, and in all, k + 1 tests are required for the k people. Assume that
the probability p that a test is positive is the same for all people and that
these events are independent.
(a) Find the probability that the test for a pooled sample of k people will
be positive.",prob.pdf
"these events are independent.
(a) Find the probability that the test for a pooled sample of k people will
be positive.
(b) What is the expected value of the number X of tests necessary under
plan (2)? (Assume that N is divisible by k.)
(c) For small p, show that the value of k which will minimize the expected
number of tests under the second plan is approximately 1=pp.
32 Write a program to add random numbers chosen from [0; 1] until the rst
time the sum is greater than one. Have your program repeat this experiment
a number of times to estimate the expected number of selections necessary
in order that the sum of the chosen numbers rst exceeds 1. On the basis of
your experiments, what is your estimate for this number?
*33 The following related discrete problem also gives a good clue for the answer
to Exercise 32. Randomly select with replacement t1, t2, . . ., tr from the set
(1=n; 2=n; : : :; n=n). Let X be the smallest value of r satisfying
t1 + t2 + 
 
 
 + tr > 1 :",prob.pdf
"(1=n; 2=n; : : :; n=n). Let X be the smallest value of r satisfying
t1 + t2 + 
 
 
 + tr > 1 :
Then E(X) = (1 + 1=n)n. To prove this, we can just as well choose t1, t2,
.. . , tr randomly with replacement from the set (1; 2; : : :; n) and let X be the
smallest value of r for which
t1 + t2 + 
 
 
 + tr > n :
(a) Use Exercise 3.2.36 to show that
P(X  j + 1) =
n
j
1
n
j
:
14W. Feller, Introduction to Probability Theory and Its Applications, 3rd ed., vol. 1 (New York:
John Wiley and Sons, 1968), p. 240.",prob.pdf
"6.1. EXPECTED VALUE 255
(b) Show that
E(X) =
nX
j=0
P(X  j + 1) :
(c) From these two facts, nd an expression for E(X). This proof is due to
Harris Schultz.15
*34 (Banach's Matchbox16) A man carries in each of his two front pockets a box
of matches originally containing N matches. Whenever he needs a match,
he chooses a pocket at random and removes one from that box. One day he
reaches into a pocket and nds the box empty.
(a) Let pr denote the probability that the other pocket contains r matches.
Dene a sequence of counter random variables as follows: Let Xi = 1 if
the ith draw is from the left pocket, and 0 if it is from the right pocket.
Interpret pr in terms of Sn = X1 + X2 + 
 
 
 + Xn. Find a binomial
expression for pr.
(b) Write a computer program to compute the pr, as well as the probability
that the other pocket contains at least r matches, for N = 100 and r
from 0 to 50.
(c) Show that (N  r)pr = (1=2)(2N + 1)pr+1  (1=2)(r + 1)pr+1 .
(d) Evaluate P
r pr.",prob.pdf
"from 0 to 50.
(c) Show that (N  r)pr = (1=2)(2N + 1)pr+1  (1=2)(r + 1)pr+1 .
(d) Evaluate P
r pr.
(e) Use (c) and (d) to determine the expectation E of the distribution fprg.
(f) Use Stirling's formula to obtain an approximation for E. How many
matches must each box contain to ensure a value of about 13 for the
expectation E? (Take  = 22=7.)
35 A coin is tossed until the rst time a head turns up. If this occurs on the nth
toss and n is odd you win 2n=n, but if n is even then you lose 2n=n. Then if
your expected winnings exist they are given by the convergent series
1  1
2 + 1
3  1
4 + 
 
 

called the alternating harmonic series. It is tempting to say that this should
be the expected value of the experiment. Show that if we were to do this, the
expected value of an experiment would depend upon the order in which the
outcomes are listed.
36 Suppose we have an urn containing c yellow balls and d green balls. We draw",prob.pdf
"outcomes are listed.
36 Suppose we have an urn containing c yellow balls and d green balls. We draw
k balls, without replacement, from the urn. Find the expected number of
yellow balls drawn. Hint: Write the number of yellow balls drawn as the sum
of c random variables.
15H. Schultz, \An Expected Value Problem,"" Two-Year Mathematics Journal, vol. 10, no. 4
(1979), pp. 277{78.
16W. Feller, Introduction to Probability Theory, vol. 1, p. 166.",prob.pdf
"256 CHAPTER 6. EXPECTED VALUE AND VARIANCE
37 The reader is referred to Example 6.13 for an explanation of the various op-
tions available in Monte Carlo roulette.
(a) Compute the expected winnings of a 1 franc bet on red under option (a).
(b) Repeat part (a) for option (b).
(c) Compare the expected winnings for all three options.
*38 (from Pittel17) Telephone books, n in number, are kept in a stack. The
probability that the book numbered i (where 1  i  n) is consulted for a
given phone call is pi > 0, where the pi's sum to 1. After a book is used,
it is placed at the top of the stack. Assume that the calls are independent
and evenly spaced, and that the system has been employed indenitely far
into the past. Let di be the average depth of book i in the stack. Show that
di  dj whenever pi  pj. Thus, on the average, the more popular books
have a tendency to be closer to the top of the stack. Hint: Let pij denote the",prob.pdf
"di  dj whenever pi  pj. Thus, on the average, the more popular books
have a tendency to be closer to the top of the stack. Hint: Let pij denote the
probability that book i is above book j. Show that pij = pij(1  pj) + pjipi.
*39 (from Propp18) In the previous problem, let P be the probability that at the
present time, each book is in its proper place, i.e., book i is ith from the top.
Find a formula for P in terms of the pi's. In addition, nd the least upper
bound on P, if the pi's are allowed to vary. Hint: First nd the probability
that book 1 is in the right place. Then nd the probability that book 2 is in
the right place, given that book 1 is in the right place. Continue.
*40 (from H. Shultz and B. Leonard19) A sequence of random numbers in [0; 1)
is generated until the sequence is no longer monotone increasing. The num-
bers are chosen according to the uniform distribution. What is the expected
length of the sequence? (In calculating the length, the term that destroys",prob.pdf
"length of the sequence? (In calculating the length, the term that destroys
monotonicity is included.) Hint: Let a1; a2; : : : be the sequence and let X
denote the length of the sequence. Then
P(X > k) = P(a1 < a2 < 
 
 
 < ak) ;
and the probability on the right-hand side is easy to calculate. Furthermore,
one can show that
E(X) = 1 + P(X > 1) + P(X > 2) + 
 
 
 :
41 Let T be the random variable that counts the number of 2-unshues per-
formed on an n-card deck until all of the labels on the cards are distinct. This
random variable was discussed in Section 3.3. Using Equation 3.4 in that
section, together with the formula
E(T) =
1X
s=0
P(T > s)
17B. Pittel, Problem #1195, Mathematics Magazine, vol. 58, no. 3 (May 1985), pg. 183.
18J. Propp, Problem #1159, Mathematics Magazine vol. 57, no. 1 (Feb. 1984), pg. 50.
19H. Shultz and B. Leonard, \Unexpected Occurrences of the Number e,"" Mathematics Magazine
vol. 62, no. 4 (October, 1989), pp. 269-271.",prob.pdf
"6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 257
that was proved in Exercise 33, show that
E(T) =
1X
s=0

1 
2s
n
 n!
2sn

:
Show that for n = 52, this expression is approximately equal to 11.7. (As was
stated in Chapter 3, this means that on the average, almost 12 rie shues of
a 52-card deck are required in order for the process to be considered random.)
6.2 Variance of Discrete Random Variables
The usefulness of the expected value as a prediction for the outcome of an ex-
periment is increased when the outcome is not likely to deviate too much from the
expected value. In this section we shall introduce a measure of this deviation, called
the variance.
Variance
Denition 6.3 Let X be a numerically valued random variable with expected value
 = E(X). Then the variance of X, denoted by V (X), is
V (X) = E((X  )2) :
2
Note that, by Theorem 6.1, V (X) is given by
V (X) =
X
x
(x  )2m(x) ; (6.1)
where m is the distribution function of X.
Standard Deviation",prob.pdf
"2
Note that, by Theorem 6.1, V (X) is given by
V (X) =
X
x
(x  )2m(x) ; (6.1)
where m is the distribution function of X.
Standard Deviation
The standard deviation of X, denoted by D(X), is D(X) =
p
V (X). We often
write  for D(X) and 2 for V (X).
Example 6.17 Consider one roll of a die. Let X be the number that turns up. To
nd V (X), we must rst nd the expected value of X. This is
 = E(X) = 1
1
6

+ 2
1
6

+ 3
1
6

+ 4
1
6

+ 5
1
6

+ 6
1
6

= 7
2 :
To nd the variance of X, we form the new random variable (X  )2 and
compute its expectation. We can easily do this using the following table.",prob.pdf
"258 CHAPTER 6. EXPECTED VALUE AND VARIANCE
x m(x) (x  7=2)2
1 1/6 25/4
2 1/6 9/4
3 1/6 1/4
4 1/6 1/4
5 1/6 9/4
6 1/6 25/4
Table 6.6: Variance calculation.
From this table we nd E((X  )2) is
V (X) = 1
6
25
4 + 9
4 + 1
4 + 1
4 + 9
4 + 25
4

= 35
12 ;
and the standard deviation D(X) =
p
35=12  1:707. 2
Calculation of Variance
We next prove a theorem that gives us a useful alternative form for computing the
variance.
Theorem 6.6 If X is any random variable with E(X) = , then
V (X) = E(X2)  2 :
Proof. We have
V (X) = E((X  )2) = E(X2  2X + 2)
= E(X2)  2E(X) + 2 = E(X2)  2 :
2
Using Theorem 6.6, we can compute the variance of the outcome of a roll of a
die by rst computing
E(X2) = 1
1
6

+ 4
1
6

+ 9
1
6

+ 16
1
6

+ 25
1
6

+ 36
1
6

= 91
6 ;
and,
V (X) = E(X2)  2 = 91
6 
7
2
2
= 35
12 ;
in agreement with the value obtained directly from the denition of V (X).",prob.pdf
"6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 259
Properties of Variance
The variance has properties very dierent from those of the expectation. If c is any
constant, E(cX) = cE(X) and E(X + c) = E(X) + c. These two statements imply
that the expectation is a linear function. However, the variance is not linear, as
seen in the next theorem.
Theorem 6.7 If X is any random variable and c is any constant, then
V (cX) = c2V (X)
and
V (X + c) = V (X) :
Proof. Let  = E(X). Then E(cX) = c, and
V (cX) = E((cX  c)2) = E(c2(X  )2)
= c2E((X  )2) = c2V (X) :
To prove the second assertion, we note that, to compute V (X + c), we would
replace x by x+c and  by +c in Equation 6.1. Then the c's would cancel, leaving
V (X). 2
We turn now to some general properties of the variance. Recall that if X and Y
are any two random variables, E(X+Y ) = E(X)+E(Y ). This is not always true for
the case of the variance. For example, let X be a random variable with V (X) 6= 0,",prob.pdf
"the case of the variance. For example, let X be a random variable with V (X) 6= 0,
and dene Y = X. Then V (X) = V (Y ), so that V (X) + V (Y ) = 2V (X). But
X + Y is always 0 and hence has variance 0. Thus V (X + Y ) 6= V (X) + V (Y ).
In the important case of mutually independent random variables, however, the
variance of the sum is the sum of the variances.
Theorem 6.8 Let X and Y be two independent random variables. Then
V (X + Y ) = V (X) + V (Y ) :
Proof. Let E(X) = a and E(Y ) = b. Then
V (X + Y ) = E((X + Y )2)  (a + b)2
= E(X2) + 2E(XY ) + E(Y 2)  a2  2ab  b2 :
Since X and Y are independent, E(XY ) = E(X)E(Y ) = ab. Thus,
V (X + Y ) = E(X2)  a2 + E(Y 2)  b2 = V (X) + V (Y ) :
2",prob.pdf
"260 CHAPTER 6. EXPECTED VALUE AND VARIANCE
It is easy to extend this proof, by mathematical induction, to show that the
variance of the sum of any number of mutually independent random variables is the
sum of the individual variances. Thus we have the following theorem.
Theorem 6.9 Let X1, X2, . . . , Xn be an independent trials process with E(Xj) =
 and V (Xj) = 2. Let
Sn = X1 + X2 + 
 
 
 + Xn
be the sum, and
An = Sn
n
be the average. Then
E(Sn) = n ;
V (Sn) = n2 ;
(Sn) = pn ;
E(An) =  ;
V (An) = 2
;
(An) = pn :
Proof. Since all the random variables Xj have the same expected value, we have
E(Sn) = E(X1) + 
 
 
 + E(Xn) = n ;
V (Sn) = V (X1) + 
 
 
 + V (Xn) = n2 ;
and
(Sn) = pn :
We have seen that, if we multiply a random variable X with mean  and variance
2 by a constant c, the new random variable has expected value c and variance
c22. Thus,
E(An) = E
Sn
n

= n
n =  ;
and
V (An) = V
Sn
n

= V (Sn)
n2 = n2
n2 = 2
n :",prob.pdf
"c22. Thus,
E(An) = E
Sn
n

= n
n =  ;
and
V (An) = V
Sn
n

= V (Sn)
n2 = n2
n2 = 2
n :
Finally, the standard deviation of An is given by
(An) = pn :
2",prob.pdf
"6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 261
1 2 3 4 5 6
0
0.1
0.2
0.3
0.4
0.5
0.6
2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5
2n = 10 n = 100
Figure 6.7: Empirical distribution of An.
The last equation in the above theorem implies that in an independent trials
process, if the individual summands have nite variance, then the standard devi-
ation of the average goes to 0 as n ! 1. Since the standard deviation tells us
something about the spread of the distribution around the mean, we see that for
large values of n, the value of An is usually very close to the mean of An, which
equals , as shown above. This statement is made precise in Chapter 8, where it
is called the Law of Large Numbers. For example, let X represent the roll of a fair
die. In Figure 6.7, we show the distribution of a random variable An corresponding
to X, for n = 10 and n = 100.
Example 6.18 Consider n rolls of a die. We have seen that, if Xj is the outcome",prob.pdf
"to X, for n = 10 and n = 100.
Example 6.18 Consider n rolls of a die. We have seen that, if Xj is the outcome
if the jth roll, then E(Xj) = 7=2 and V (Xj) = 35=12. Thus, if Sn is the sum of the
outcomes, and An = Sn=n is the average of the outcomes, we have E(An) = 7=2 and
V (An) = (35=12)=n. Therefore, as n increases, the expected value of the average
remains constant, but the variance tends to 0. If the variance is a measure of the
expected deviation from the mean this would indicate that, for large n, we can
expect the average to be very near the expected value. This is in fact the case, and
we shall justify it in Chapter 8. 2
Bernoulli Trials
Consider next the general Bernoulli trials process. As usual, we let Xj = 1 if the
jth outcome is a success and 0 if it is a failure. If p is the probability of a success,
and q = 1  p, then
E(Xj) = 0q + 1p = p ;
E(X2
j ) = 02q + 12p = p ;
and
V (Xj) = E(X2
j )  (E(Xj))2 = p  p2 = pq :",prob.pdf
"and q = 1  p, then
E(Xj) = 0q + 1p = p ;
E(X2
j ) = 02q + 12p = p ;
and
V (Xj) = E(X2
j )  (E(Xj))2 = p  p2 = pq :
Thus, for Bernoulli trials, if Sn = X1 +X2 +
 
 
+Xn is the number of successes,
then E(Sn) = np, V (Sn) = npq, and D(Sn) = pnpq: If An = Sn=n is the average
number of successes, then E(An) = p, V (An) = pq=n, and D(An) =
p
pq=n. We
see that the expected proportion of successes remains p and the variance tends to 0.",prob.pdf
"262 CHAPTER 6. EXPECTED VALUE AND VARIANCE
This suggests that the frequency interpretation of probability is a correct one. We
shall make this more precise in Chapter 8.
Example 6.19 Let T denote the number of trials until the rst success in a
Bernoulli trials process. Then T is geometrically distributed. What is the vari-
ance of T? In Example 4.15, we saw that
mT =
1 2 3 
 
 

p qp q2p 
 
 


:
In Example 6.4, we showed that
E(T) = 1=p :
Thus,
V (T) = E(T2)  1=p2 ;
so we need only nd
E(T2) = 1p + 4qp + 9q2p + 
 
 

= p(1 + 4q + 9q2 + 
 
 
) :
To evaluate this sum, we start again with
1 + x + x2 + 
 
 
 = 1
1  x :
Dierentiating, we obtain
1 + 2x + 3x2 + 
 
 
 = 1
(1  x)2 :
Multiplying by x,
x + 2x2 + 3x3 + 
 
 
 = x
(1  x)2 :
Dierentiating again gives
1 + 4x + 9x2 + 
 
 
 = 1 + x
(1  x)3 :
Thus,
E(T2) = p 1 + q
(1  q)3 = 1 + q
p2
and
V (T) = E(T2)  (E(T))2
= 1 + q
p2  1
p2 = q
p2 :
For example, the variance for the number of tosses of a coin until the rst",prob.pdf
"p2
and
V (T) = E(T2)  (E(T))2
= 1 + q
p2  1
p2 = q
p2 :
For example, the variance for the number of tosses of a coin until the rst
head turns up is (1=2)=(1=2)2 = 2. The variance for the number of rolls of a
die until the rst six turns up is (5=6)=(1=6)2 = 30. Note that, as p decreases, the
variance increases rapidly. This corresponds to the increased spread of the geometric
distribution as p decreases (noted in Figure 5.1). 2",prob.pdf
"6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 263
Poisson Distribution
Just as in the case of expected values, it is easy to guess the variance of the Poisson
distribution with parameter . We recall that the variance of a binomial distribution
with parameters n and p equals npq. We also recall that the Poisson distribution
could be obtained as a limit of binomial distributions, if n goes to 1 and p goes
to 0 in such a way that their product is kept xed at the value . In this case,
npq = q approaches , since q goes to 1. So, given a Poisson distribution with
parameter , we should guess that its variance is . The reader is asked to show
this in Exercise 29.
Exercises
1 A number is chosen at random from the set S = f1; 0; 1g. Let X be the
number chosen. Find the expected value, variance, and standard deviation of
X.
2 A random variable X has the distribution
pX =
 0 1 2 4
1=3 1=3 1=6 1=6

:
Find the expected value, variance, and standard deviation of X.",prob.pdf
"X.
2 A random variable X has the distribution
pX =
 0 1 2 4
1=3 1=3 1=6 1=6

:
Find the expected value, variance, and standard deviation of X.
3 You place a 1-dollar bet on the number 17 at Las Vegas, and your friend
places a 1-dollar bet on black (see Exercises 1.1.6 and 1.1.7). Let X be your
winnings and Y be her winnings. Compare E(X), E(Y ), and V (X), V (Y ).
What do these computations tell you about the nature of your winnings if
you and your friend make a sequence of bets, with you betting each time on
a number and your friend betting on a color?
4 X is a random variable with E(X) = 100 and V (X) = 15. Find
(a) E(X2).
(b) E(3X + 10).
(c) E(X).
(d) V (X).
(e) D(X).
5 In a certain manufacturing process, the (Fahrenheit) temperature never varies
by more than 2 from 62. The temperature is, in fact, a random variable F
with distribution
PF =
 60 61 62 63 64
1=10 2=10 4=10 2=10 1=10

:
(a) Find E(F) and V (F).",prob.pdf
"with distribution
PF =
 60 61 62 63 64
1=10 2=10 4=10 2=10 1=10

:
(a) Find E(F) and V (F).
(b) Dene T = F  62. Find E(T) and V (T), and compare these answers
with those in part (a).",prob.pdf
"264 CHAPTER 6. EXPECTED VALUE AND VARIANCE
(c) It is decided to report the temperature readings on a Celsius scale, that
is, C = (5=9)(F  32). What is the expected value and variance for the
readings now?
6 Write a computer program to calculate the mean and variance of a distribution
which you specify as data. Use the program to compare the variances for the
following densities, both having expected value 0:
pX =
 2 1 0 1 2
3=11 2=11 1=11 2=11 3=11

;
pY =
 2 1 0 1 2
1=11 2=11 5=11 2=11 1=11

:
7 A coin is tossed three times. Let X be the number of heads that turn up.
Find V (X) and D(X).
8 A random sample of 2400 people are asked if they favor a government pro-
posal to develop new nuclear power plants. If 40 percent of the people in the
country are in favor of this proposal, nd the expected value and the stan-
dard deviation for the number S2400 of people in the sample who favored the
proposal.
9 A die is loaded so that the probability of a face coming up is proportional to",prob.pdf
"proposal.
9 A die is loaded so that the probability of a face coming up is proportional to
the number on that face. The die is rolled with outcome X. Find V (X) and
D(X).
10 Prove the following facts about the standard deviation.
(a) D(X + c) = D(X).
(b) D(cX) = jcjD(X).
11 A number is chosen at random from the integers 1, 2, 3, . . ., n. Let X be the
number chosen. Show that E(X) = (n + 1)=2 and V (X) = (n  1)(n + 1)=12.
Hint: The following identity may be useful:
12 + 22 + 
 
 
 + n2 = (n)(n + 1)(2n + 1)
6 :
12 Let X be a random variable with  = E(X) and 2 = V (X). Dene X =
(X)=. The random variable X is called the standardized random variable
associated with X. Show that this standardized random variable has expected
value 0 and variance 1.
13 Peter and Paul play Heads or Tails (see Example 1.4). Let Wn be Peter's
winnings after n matches. Show that E(Wn) = 0 and V (Wn) = n.
14 Find the expected value and the variance for the number of boys and the",prob.pdf
"winnings after n matches. Show that E(Wn) = 0 and V (Wn) = n.
14 Find the expected value and the variance for the number of boys and the
number of girls in a royal family that has children until there is a boy or until
there are three children, whichever comes rst.",prob.pdf
"6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 265
15 Suppose that n people have their hats returned at random. Let Xi = 1 if the
ith person gets his or her own hat back and 0 otherwise. Let Sn = Pn
i=1 Xi.
Then Sn is the total number of people who get their own hats back. Show
that
(a) E(X2
i ) = 1=n.
(b) E(Xi 
 Xj) = 1=n(n  1) for i 6= j.
(c) E(S2
n) = 2 (using (a) and (b)).
(d) V (Sn) = 1.
16 Let Sn be the number of successes in n independent trials. Use the program
BinomialProbabilities (Section 3.2) to compute, for given n, p, and j, the
probability
P(jp
npq < Sn  np < jpnpq) :
(a) Let p = :5, and compute this probability for j = 1, 2, 3 and n = 10, 30, 50.
Do the same for p = :2.
(b) Show that the standardized random variable S
n = (Sn  np)=pnpq has
expected value 0 and variance 1. What do your results from (a) tell you
about this standardized quantity S
n?
17 Let X be the outcome of a chance experiment with E(X) =  and V (X) =",prob.pdf
"about this standardized quantity S
n?
17 Let X be the outcome of a chance experiment with E(X) =  and V (X) =
2. When  and 2 are unknown, the statistician often estimates them by
repeating the experiment n times with outcomes x1, x2, . . . , xn, estimating
 by the sample mean
x = 1
n
nX
i=1
xi ;
and 2 by the sample variance
s2 = 1
n
nX
i=1
(xi  x)2 :
Then s is the sample standard deviation. These formulas should remind the
reader of the denitions of the theoretical mean and variance. (Many statisti-
cians dene the sample variance with the coecient 1=n replaced by 1=(n1).
If this alternative denition is used, the expected value of s2 is equal to 2.
See Exercise 18, part (d).)
Write a computer program that will roll a die n times and compute the sample
mean and sample variance. Repeat this experiment several times for n = 10
and n = 1000. How well do the sample mean and sample variance estimate
the true mean 7/2 and variance 35/12?",prob.pdf
"and n = 1000. How well do the sample mean and sample variance estimate
the true mean 7/2 and variance 35/12?
18 Show that, for the sample mean x and sample variance s2 as dened in Exer-
cise 17,
(a) E(x) = .",prob.pdf
"266 CHAPTER 6. EXPECTED VALUE AND VARIANCE
(b) E

(x  )2

= 2=n.
(c) E(s2) = n1
n 2. Hint: For (c) write
nX
i=1
(xi  x)2 =
nX
i=1

(xi  )  (x  )

2
=
nX
i=1
(xi  )2  2(x  )
nX
i=1
(xi  ) + n(x  )2
=
nX
i=1
(xi  )2  n(x  )2;
and take expectations of both sides, using part (b) when necessary.
(d) Show that if, in the denition of s2 in Exercise 17, we replace the coe-
cient 1=n by the coecient 1=(n1), then E(s2) = 2. (This shows why
many statisticians use the coecient 1=(n  1). The number s2 is used
to estimate the unknown quantity 2. If an estimator has an average
value which equals the quantity being estimated, then the estimator is
said to be unbiased. Thus, the statement E(s2) = 2 says that s2 is an
unbiased estimator of 2.)
19 Let X be a random variable taking on values a1, a2, . .. , ar with probabilities
p1, p2, . . . , pr and with E(X) = . Dene the spread of X as follows:
 =
rX
i=1
jai  jpi :",prob.pdf
"p1, p2, . . . , pr and with E(X) = . Dene the spread of X as follows:
 =
rX
i=1
jai  jpi :
This, like the standard deviation, is a way to quantify the amount that a
random variable is spread out around its mean. Recall that the variance of a
sum of mutually independent random variables is the sum of the individual
variances. The square of the spread corresponds to the variance in a manner
similar to the correspondence between the spread and the standard deviation.
Show by an example that it is not necessarily true that the square of the
spread of the sum of two independent random variables is the sum of the
squares of the individual spreads.
20 We have two instruments that measure the distance between two points. The
measurements given by the two instruments are random variables X1 and
X2 that are independent with E(X1) = E(X2) = , where  is the true
distance. From experience with these instruments, we know the values of the
variances 2
1 and 2",prob.pdf
"distance. From experience with these instruments, we know the values of the
variances 2
1 and 2
2. These variances are not necessarily the same. From two
measurements, we estimate  by the weighted average  = wX1 + (1  w)X2.
Here w is chosen in [0; 1] to minimize the variance of .
(a) What is E()?
(b) How should w be chosen in [0; 1] to minimize the variance of ?",prob.pdf
"6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 267
21 Let X be a random variable with E(X) =  and V (X) = 2. Show that the
function f(x) dened by
f(x) =
X
!
(X(!)  x)2p(!)
has its minimum value when x = .
22 Let X and Y be two random variables dened on the nite sample space 
.
Assume that X, Y , X + Y , and X  Y all have the same distribution. Prove
that P(X = Y = 0) = 1.
23 If X and Y are any two random variables, then the covariance of X and Y is
dened by Cov(X; Y ) = E((X E(X))(Y  E(Y ))). Note that Cov(X; X) =
V (X). Show that, if X and Y are independent, then Cov(X; Y ) = 0; and
show, by an example, that we can have Cov(X; Y ) = 0 and X and Y not
independent.
*24 A professor wishes to make up a true-false exam with n questions. She assumes
that she can design the problems in such a way that a student will answer
the jth problem correctly with probability pj, and that the answers to the
various problems may be considered independent experiments. Let Sn be the",prob.pdf
"the jth problem correctly with probability pj, and that the answers to the
various problems may be considered independent experiments. Let Sn be the
number of problems that a student will get correct. The professor wishes to
choose pj so that E(Sn) = :7n and so that the variance of Sn is as large as
possible. Show that, to achieve this, she should choose pj = :7 for all j; that
is, she should make all the problems have the same diculty.
25 (Lamperti20) An urn contains exactly 5000 balls, of which an unknown number
X are white and the rest red, where X is a random variable with a probability
distribution on the integers 0, 1, 2, . .. , 5000.
(a) Suppose we know that E(X) = . Show that this is enough to allow us
to calculate the probability that a ball drawn at random from the urn
will be white. What is this probability?
(b) We draw a ball from the urn, examine its color, replace it, and then
draw another. Under what conditions, if any, are the results of the two",prob.pdf
"(b) We draw a ball from the urn, examine its color, replace it, and then
draw another. Under what conditions, if any, are the results of the two
drawings independent; that is, does
P(white; white) = P(white)2 ?
(c) Suppose the variance of X is 2. What is the probability of drawing two
white balls in part (b)?
26 For a sequence of Bernoulli trials, let X1 be the number of trials until the rst
success. For j  2, let Xj be the number of trials after the (j  1)st success
until the jth success. It can be shown that X1, X2, . . . is an independent trials
process.
20Private communication.",prob.pdf
"268 CHAPTER 6. EXPECTED VALUE AND VARIANCE
(a) What is the common distribution, expected value, and variance for Xj?
(b) Let Tn = X1 +X2 +
 
 
+Xn. Then Tn is the time until the nth success.
Find E(Tn) and V (Tn).
(c) Use the results of (b) to nd the expected value and variance for the
number of tosses of a coin until the nth occurrence of a head.
27 Referring to Exercise 6.1.30, nd the variance for the number of boxes of
Wheaties bought before getting half of the players' pictures and the variance
for the number of additional boxes needed to get the second half of the players'
pictures.
28 In Example 5.3, assume that the book in question has 1000 pages. Let X be
the number of pages with no mistakes. Show that E(X) = 905 and V (X) =
86. Using these results, show that the probability is  :05 that there will be
more than 924 pages without errors or fewer than 866 pages without errors.
29 Let X be Poisson distributed with parameter . Show that V (X) = .",prob.pdf
"more than 924 pages without errors or fewer than 866 pages without errors.
29 Let X be Poisson distributed with parameter . Show that V (X) = .
6.3 Continuous Random Variables
In this section we consider the properties of the expected value and the variance
of a continuous random variable. These quantities are dened just as for discrete
random variables and share the same properties.
Expected Value
Denition 6.4 Let X be a real-valued random variable with density function f(x).
The expected value  = E(X) is dened by
 = E(X) =
Z +1
1
xf(x) dx ;
provided the integral Z +1
1
jxjf(x) dx
is nite. 2
The reader should compare this denition with the corresponding one for discrete
random variables in Section 6.1. Intuitively, we can interpret E(X), as we did in
the previous sections, as the value that we should expect to obtain if we perform a
large number of independent experiments and average the resulting values of X.",prob.pdf
"large number of independent experiments and average the resulting values of X.
We can summarize the properties of E(X) as follows (cf. Theorem 6.2).",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 269
Theorem 6.10 If X and Y are real-valued random variables and c is any constant,
then
E(X + Y ) = E(X) + E(Y ) ;
E(cX) = cE(X) :
The proof is very similar to the proof of Theorem 6.2, and we omit it. 2
More generally, if X1, X2, . . . , Xn are n real-valued random variables, and c1, c2,
. . . , cn are n constants, then
E(c1X1 + c2X2 + 
 
 
 + cnXn) = c1E(X1) + c2E(X2) + 
 
 
 + cnE(Xn) :
Example 6.20 Let X be uniformly distributed on the interval [0; 1]. Then
E(X) =
Z 1
0
x dx = 1=2 :
It follows that if we choose a large number N of random numbers from [0; 1] and take
the average, then we can expect that this average should be close to the expected
value of 1/2. 2
Example 6.21 Let Z = (x; y) denote a point chosen uniformly and randomly from
the unit disk, as in the dart game in Example 2.8 and let X = (x2 + y2)1=2 be the
distance from Z to the center of the disk. The density function of X can easily be",prob.pdf
"distance from Z to the center of the disk. The density function of X can easily be
shown to equal f(x) = 2x, so by the denition of expected value,
E(X) =
Z 1
0
xf(x) dx
=
Z 1
0
x(2x) dx
= 2
3 :
2
Example 6.22 In the example of the couple meeting at the Inn (Example 2.16),
each person arrives at a time which is uniformly distributed between 5:00 and 6:00
PM. The random variable Z under consideration is the length of time the rst
person has to wait until the second one arrives. It was shown that
fZ(z) = 2(1  z) ;
for 0  z  1. Hence,
E(Z) =
Z 1
0
zfZ(z) dz",prob.pdf
"270 CHAPTER 6. EXPECTED VALUE AND VARIANCE
=
Z 1
0
2z(1  z) dz
=
h
z2  2
3z3
i1
0
= 1
3 :
2
Expectation of a Function of a Random Variable
Suppose that X is a real-valued random variable and (x) is a continuous function
from R to R. The following theorem is the continuous analogue of Theorem 6.1.
Theorem 6.11 If X is a real-valued random variable and if  : R ! R is a
continuous real-valued function with domain [a; b], then
E((X)) =
Z +1
1
(x)fX (x) dx ;
provided the integral exists. 2
For a proof of this theorem, see Ross.21
Expectation of the Product of Two Random Variables
In general, it is not true that E(XY ) = E(X)E(Y ), since the integral of a product is
not the product of integrals. But if X and Y are independent, then the expectations
multiply.
Theorem 6.12 Let X and Y be independent real-valued continuous random vari-
ables with nite expected values. Then we have
E(XY ) = E(X)E(Y ) :
Proof. We will prove this only in the case that the ranges of X and Y are contained",prob.pdf
"E(XY ) = E(X)E(Y ) :
Proof. We will prove this only in the case that the ranges of X and Y are contained
in the intervals [a; b] and [c; d], respectively. Let the density functions of X and Y
be denoted by fX(x) and fY (y), respectively. Since X and Y are independent, the
joint density function of X and Y is the product of the individual density functions.
Hence
E(XY ) =
Z b
a
Z d
c
xyfX(x)fY (y) dy dx
=
Z b
a
xfX(x) dx
Z d
c
yfY (y) dy
= E(X)E(Y ) :
The proof in the general case involves using sequences of bounded random vari-
ables that approach X and Y , and is somewhat technical, so we will omit it. 2
21S. Ross, A First Course in Probability, (New York: Macmillan, 1984), pgs. 241-245.",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 271
In the same way, one can show that if X1, X2, . . ., Xn are n mutually indepen-
dent real-valued random variables, then
E(X1X2 
 
 
 Xn) = E(X1) E(X2) 
 
 
 E(Xn) :
Example 6.23 Let Z = (X; Y ) be a point chosen at random in the unit square.
Let A = X2 and B = Y 2. Then Theorem 4.3 implies that A and B are independent.
Using Theorem 6.11, the expectations of A and B are easy to calculate:
E(A) = E(B) =
Z 1
0
x2 dx
= 1
3 :
Using Theorem 6.12, the expectation of AB is just the product of E(A) and E(B),
or 1/9. The usefulness of this theorem is demonstrated by noting that it is quite a
bit more dicult to calculate E(AB) from the denition of expectation. One nds
that the density function of AB is
fAB(t) =  log(t)
4
p
t ;
so
E(AB) =
Z 1
0
tfAB(t) dt
= 1
9 :
2
Example 6.24 Again let Z = (X; Y ) be a point chosen at random in the unit
square, and let W = X + Y . Then Y and W are not independent, and we have
E(Y ) = 1
2 ;
E(W) = 1 ;",prob.pdf
"square, and let W = X + Y . Then Y and W are not independent, and we have
E(Y ) = 1
2 ;
E(W) = 1 ;
E(Y W) = E(XY + Y 2) = E(X)E(Y ) + 1
3 = 7
12 6= E(Y )E(W) :
2
We turn now to the variance.
Variance
Denition 6.5 Let X be a real-valued random variable with density function f(x).
The variance 2 = V (X) is dened by
2 = V (X) = E((X  )2) :
2",prob.pdf
"272 CHAPTER 6. EXPECTED VALUE AND VARIANCE
The next result follows easily from Theorem 6.1. There is another way to calculate
the variance of a continuous random variable, which is usually slightly easier. It is
given in Theorem 6.15.
Theorem 6.13 If X is a real-valued random variable with E(X) = , then
2 =
Z 1
1
(x  )2f(x) dx :
2
The properties listed in the next three theorems are all proved in exactly the
same way that the corresponding theorems for discrete random variables were
proved in Section 6.2.
Theorem 6.14 If X is a real-valued random variable dened on 
 and c is any
constant, then (cf. Theorem 6.7)
V (cX) = c2V (X) ;
V (X + c) = V (X) :
2
Theorem 6.15 If X is a real-valued random variable with E(X) = , then (cf.
Theorem 6.6)
V (X) = E(X2)  2 :
2
Theorem 6.16 If X and Y are independent real-valued random variables on 
,
then (cf. Theorem 6.8)
V (X + Y ) = V (X) + V (Y ) :
2
Example 6.25 (continuation of Example 6.20) If X is uniformly distributed on",prob.pdf
",
then (cf. Theorem 6.8)
V (X + Y ) = V (X) + V (Y ) :
2
Example 6.25 (continuation of Example 6.20) If X is uniformly distributed on
[0; 1], then, using Theorem 6.15, we have
V (X) =
Z 1
0

x  1
2
2
dx = 1
12 :
2",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 273
Example 6.26 Let X be an exponentially distributed random variable with pa-
rameter . Then the density function of X is
fX(x) = ex :
From the denition of expectation and integration by parts, we have
E(X) =
Z 1
0
xfX(x) dx
= 
Z 1
0
xex dx
= xex




1
0
+
Z 1
0
ex dx
= 0 + ex





1
0
= 1
 :
Similarly, using Theorems 6.11 and 6.15, we have
V (X) =
Z 1
0
x2fX(x) dx  1
2
= 
Z 1
0
x2ex dx  1
2
= x2ex




1
0
+ 2
Z 1
0
xex dx  1
2
= x2ex




1
0
 2xex





1
0
 2
2 ex




1
0
 1
2 = 2
2  1
2 = 1
2 :
In this case, both E(X) and V (X) are nite if  > 0. 2
Example 6.27 Let Z be a standard normal random variable with density function
fZ(x) = 1p
2ex2=2 :
Since this density function is symmetric with respect to the y-axis, then it is easy
to show that Z 1
1
xfZ(x) dx
has value 0. The reader should recall however, that the expectation is dened to be",prob.pdf
"to show that Z 1
1
xfZ(x) dx
has value 0. The reader should recall however, that the expectation is dened to be
the above integral only if the integral
Z 1
1
jxjfZ(x) dx
is nite. This integral equals
2
Z 1
0
xfZ(x) dx ;",prob.pdf
"274 CHAPTER 6. EXPECTED VALUE AND VARIANCE
which one can easily show is nite. Thus, the expected value of Z is 0.
To calculate the variance of Z, we begin by applying Theorem 6.15:
V (Z) =
Z +1
1
x2fZ(x) dx  2 :
If we write x2 as x 
 x, and integrate by parts, we obtain
1p
2(xex2=2)




+1
1
+ 1
p
2
Z +1
1
ex2=2 dx :
The rst summand above can be shown to equal 0, since as x ! 1, ex2=2 gets
small more quickly than x gets large. The second summand is just the standard
normal density integrated over its domain, so the value of this summand is 1.
Therefore, the variance of the standard normal density equals 1.
Now let X be a (not necessarily standard) normal random variable with param-
eters  and . Then the density function of X is
fX(x) = 1p
2e(x)2=22
:
We can write X = Z + , where Z is a standard normal random variable. Since
E(Z) = 0 and V (Z) = 1 by the calculation above, Theorems 6.10 and 6.14 imply
that
E(X) = E(Z + ) =  ;
V (X) = V (Z + ) = 2 :
2",prob.pdf
"E(Z) = 0 and V (Z) = 1 by the calculation above, Theorems 6.10 and 6.14 imply
that
E(X) = E(Z + ) =  ;
V (X) = V (Z + ) = 2 :
2
Example 6.28 Let X be a continuous random variable with the Cauchy density
function
fX(x) = a

1
a2 + x2 :
Then the expectation of X does not exist, because the integral
a

Z +1
1
jxj dx
a2 + x2
diverges. Thus the variance of X also fails to exist. Densities whose variance is not
dened, like the Cauchy density, behave quite dierently in a number of important
respects from those whose variance is nite. We shall see one instance of this
dierence in Section 8.2. 2
Independent Trials",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 275
Corollary 6.1 If X1, X2, . .. , Xn is an independent trials process of real-valued
random variables, with E(Xi) =  and V (Xi) = 2, and if
Sn = X1 + X2 + 
 
 
 + Xn ;
An = Sn
n ;
then
E(Sn) = n ;
E(An) =  ;
V (Sn) = n2 ;
V (An) = 2
n :
It follows that if we set
S
n = Sn  np
n2 ;
then
E(S
n) = 0 ;
V (S
n) = 1 :
We say that S
n is a standardized version of Sn (see Exercise 12 in Section 6.2). 2
Queues
Example 6.29 Let us consider again the queueing problem, that is, the problem of
the customers waiting in a queue for service (see Example 5.7). We suppose again
that customers join the queue in such a way that the time between arrivals is an
exponentially distributed random variable X with density function
fX(t) = et :
Then the expected value of the time between arrivals is simply 1= (see Exam-
ple 6.26), as was stated in Example 5.7. The reciprocal  of this expected value",prob.pdf
"ple 6.26), as was stated in Example 5.7. The reciprocal  of this expected value
is often referred to as the arrival rate. The service time of an individual who is
rst in line is dened to be the amount of time that the person stays at the head
of the line before leaving. We suppose that the customers are served in such a way
that the service time is another exponentially distributed random variable Y with
density function
fX(t) = et :
Then the expected value of the service time is
E(X) =
Z 1
0
tfX(t) dt = 1
 :
The reciprocal  if this expected value is often referred to as the service rate.",prob.pdf
"276 CHAPTER 6. EXPECTED VALUE AND VARIANCE
We expect on grounds of our everyday experience with queues that if the service
rate is greater than the arrival rate, then the average queue size will tend to stabilize,
but if the service rate is less than the arrival rate, then the queue will tend to increase
in length without limit (see Figure 5.7). The simulations in Example 5.7 tend to
bear out our everyday experience. We can make this conclusion more precise if we
introduce the trac intensity as the product
 = (arrival rate)(average service time) = 
 = 1=
1= :
The trac intensity is also the ratio of the average service time to the average
time between arrivals. If the trac intensity is less than 1 the queue will perform
reasonably, but if it is greater than 1 the queue will grow indenitely large. In the
critical case of  = 1, it can be shown that the queue will become large but there
will always be times at which the queue is empty.22",prob.pdf
"critical case of  = 1, it can be shown that the queue will become large but there
will always be times at which the queue is empty.22
In the case that the trac intensity is less than 1 we can consider the length of
the queue as a random variable Z whose expected value is nite,
E(Z) = N :
The time spent in the queue by a single customer can be considered as a random
variable W whose expected value is nite,
E(W) = T :
Then we can argue that, when a customer joins the queue, he expects to nd N
people ahead of him, and when he leaves the queue, he expects to nd T people
behind him. Since, in equilibrium, these should be the same, we would expect to
nd that
N = T :
This last relationship is called Little's law for queues.23 We will not prove it here.
A proof may be found in Ross.24 Note that in this case we are counting the waiting
time of all customers, even those that do not have to wait at all. In our simulation
in Section 4.2, we did not consider these customers.",prob.pdf
"time of all customers, even those that do not have to wait at all. In our simulation
in Section 4.2, we did not consider these customers.
If we knew the expected queue length then we could use Little's law to obtain
the expected waiting time, since
T = N
 :
The queue length is a random variable with a discrete distribution. We can estimate
this distribution by simulation, keeping track of the queue lengths at the times at
which a customer arrives. We show the result of this simulation (using the program
Queue) in Figure 6.8.
22L. Kleinrock, Queueing Systems, vol. 2 (New York: John Wiley and Sons, 1975).
23ibid., p. 17.
24S. M. Ross, Applied Probability Models with Optimization Applications, (San Francisco:
Holden-Day, 1970)",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 277
0 10 20 30 40 50
0
0.02
0.04
0.06
0.08
Figure 6.8: Distribution of queue lengths.
We note that the distribution appears to be a geometric distribution. In the
study of queueing theory it is shown that the distribution for the queue length in
equilibrium is indeed a geometric distribution with
sj = (1  )j for j = 0; 1; 2; : : : ;
if  < 1. The expected value of a random variable with this distribution is
N = 
(1  )
(see Example 6.4). Thus by Little's result the expected waiting time is
T = 
(1  ) = 1
   ;
where  is the service rate,  the arrival rate, and  the trac intensity.
In our simulation, the arrival rate is 1 and the service rate is 1.1. Thus, the
trac intensity is 1=1:1 = 10=11, the expected queue size is
10=11
(1  10=11) = 10 ;
and the expected waiting time is
1
1:1  1 = 10 :
In our simulation the average queue size was 8.19 and the average waiting time was",prob.pdf
"and the expected waiting time is
1
1:1  1 = 10 :
In our simulation the average queue size was 8.19 and the average waiting time was
7.37. In Figure 6.9, we show the histogram for the waiting times. This histogram
suggests that the density for the waiting times is exponential with parameter ,
and this is the case. 2",prob.pdf
"278 CHAPTER 6. EXPECTED VALUE AND VARIANCE
0 10 20 30 40 50
0
0.02
0.04
0.06
0.08
Figure 6.9: Distribution of queue waiting times.
Exercises
1 Let X be a random variable with range [1; 1] and let fX(x) be the density
function of X. Find (X) and 2(X) if, for jxj < 1,
(a) fX(x) = 1=2.
(b) fX(x) = jxj.
(c) fX(x) = 1  jxj.
(d) fX(x) = (3=2)x2.
2 Let X be a random variable with range [1; 1] and fX its density function.
Find (X) and 2(X) if, for jxj > 1, fX(x) = 0, and for jxj < 1,
(a) fX(x) = (3=4)(1  x2).
(b) fX(x) = (=4) cos(x=2).
(c) fX(x) = (x + 1)=2.
(d) fX(x) = (3=8)(x + 1)2.
3 The lifetime, measure in hours, of the ACME super light bulb is a random
variable T with density function fT (t) = 2tet, where  = :05. What is the
expected lifetime of this light bulb? What is its variance?
4 Let X be a random variable with range [1; 1] and density function fX(x) =
ax + b if jxj < 1.
(a) Show that if R+1
1 fX(x) dx = 1, then b = 1=2.",prob.pdf
"4 Let X be a random variable with range [1; 1] and density function fX(x) =
ax + b if jxj < 1.
(a) Show that if R+1
1 fX(x) dx = 1, then b = 1=2.
(b) Show that if fX(x)  0, then 1=2  a  1=2.
(c) Show that  = (2=3)a, and hence that 1=3    1=3.",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 279
(d) Show that 2(X) = (2=3)b  (4=9)a2 = 1=3  (4=9)a2.
5 Let X be a random variable with range [1; 1] and density function fX(x) =
ax2 + bx + c if jxj < 1 and 0 otherwise.
(a) Show that 2a=3 + 2c = 1 (see Exercise 4).
(b) Show that 2b=3 = (X).
(c) Show that 2a=5 + 2c=3 = 2(X).
(d) Find a, b, and c if (X) = 0, 2(X) = 1=15, and sketch the graph of fX.
(e) Find a, b, and c if (X) = 0, 2(X) = 1=2, and sketch the graph of fX.
6 Let T be a random variable with range [0; 1] and fT its density function.
Find (T) and 2(T) if, for t < 0, fT (t) = 0, and for t > 0,
(a) fT (t) = 3e3t.
(b) fT (t) = 9te3t.
(c) fT (t) = 3=(1 + t)4.
7 Let X be a random variable with density function fX. Show, using elementary
calculus, that the function
(a) = E((X  a)2)
takes its minimum value when a = (X), and in that case (a) = 2(X).
8 Let X be a random variable with mean  and variance 2. Let Y = aX2 +
bX + c. Find the expected value of Y .",prob.pdf
"8 Let X be a random variable with mean  and variance 2. Let Y = aX2 +
bX + c. Find the expected value of Y .
9 Let X, Y , and Z be independent random variables, each with mean  and
variance 2.
(a) Find the expected value and variance of S = X + Y + Z.
(b) Find the expected value and variance of A = (1=3)(X + Y + Z).
(c) Find the expected value of S2 and A2.
10 Let X and Y be independent random variables with uniform density functions
on [0; 1]. Find
(a) E(jX  Y j).
(b) E(max(X; Y )).
(c) E(min(X; Y )).
(d) E(X2 + Y 2).
(e) E((X + Y )2).",prob.pdf
"280 CHAPTER 6. EXPECTED VALUE AND VARIANCE
11 The Pilsdor Beer Company runs a 
eet of trucks along the 100 mile road
from Hangtown to Dry Gulch. The trucks are old, and are apt to break
down at any point along the road with equal probability. Where should the
company locate a garage so as to minimize the expected distance from a
typical breakdown to the garage? In other words, if X is a random variable
giving the location of the breakdown, measured, say, from Hangtown, and b
gives the location of the garage, what choice of b minimizes E(jX  bj)? Now
suppose X is not distributed uniformly over [0; 100], but instead has density
function fX(x) = 2x=10;000. Then what choice of b minimizes E(jX  bj)?
12 Find E(XY ), where X and Y are independent random variables which are
uniform on [0; 1]. Then verify your answer by simulation.
13 Let X be a random variable that takes on nonnegative values and has distri-
bution function F(x). Show that
E(X) =
Z 1
0
(1  F(x)) dx :",prob.pdf
"13 Let X be a random variable that takes on nonnegative values and has distri-
bution function F(x). Show that
E(X) =
Z 1
0
(1  F(x)) dx :
Hint: Integrate by parts.
Illustrate this result by calculating E(X) by this method if X has an expo-
nential distribution F(x) = 1  ex for x  0, and F(x) = 0 otherwise.
14 Let X be a continuous random variable with density function fX(x). Show
that if Z +1
1
x2fX(x) dx < 1 ;
then Z +1
1
jxjfX(x) dx < 1 :
Hint: Except on the interval [1; 1], the rst integrand is greater than the
second integrand.
15 Let X be a random variable distributed uniformly over [0; 20]. Dene a new
random variable Y by Y = bXc (the greatest integer in X). Find the expected
value of Y . Do the same for Z = bX + :5c. Compute E

jX  Y j


and
E

jX  Zj


. (Note that Y is the value of X rounded o to the nearest
smallest integer, while Z is the value of X rounded o to the nearest integer.
Which method of rounding o is better? Why?)",prob.pdf
"smallest integer, while Z is the value of X rounded o to the nearest integer.
Which method of rounding o is better? Why?)
16 Assume that the lifetime of a diesel engine part is a random variable X with
density fX. When the part wears out, it is replaced by another with the same
density. Let N(t) be the number of parts that are used in time t. We want
to study the random variable N(t)=t. Since parts are replaced on the average
every E(X) time units, we expect about t=E(X) parts to be used in time t.
That is, we expect that
lim
t!1
E
N(t)
t

= 1
E(X) :",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 281
This result is correct but quite dicult to prove. Write a program that will
allow you to specify the density fX, and the time t, and simulate this experi-
ment to nd N(t)=t. Have your program repeat the experiment 500 times and
plot a bar graph for the random outcomes of N(t)=t. From this data, estimate
E(N(t)=t) and compare this with 1=E(X). In particular, do this for t = 100
with the following two densities:
(a) fX = et.
(b) fX = tet.
17 Let X and Y be random variables. The covariance Cov(X; Y) is dened by
(see Exercise 6.2.23)
cov(X; Y) = E((X  (X))(Y  (Y))) :
(a) Show that cov(X; Y) = E(XY)  E(X)E(Y).
(b) Using (a), show that cov(X; Y ) = 0, if X and Y are independent. (Cau-
tion: the converse is not always true.)
(c) Show that V (X + Y ) = V (X) + V (Y ) + 2cov(X; Y ).
18 Let X and Y be random variables with positive variance. The correlation of
X and Y is dened as
(X; Y ) = cov(X; Y )p
V (X)V (Y )
:",prob.pdf
"18 Let X and Y be random variables with positive variance. The correlation of
X and Y is dened as
(X; Y ) = cov(X; Y )p
V (X)V (Y )
:
(a) Using Exercise 17(c), show that
0  V
 X
(X) + Y
(Y )

= 2(1 + (X; Y )) :
(b) Now show that
0  V
 X
(X)  Y
(Y )

= 2(1  (X; Y )) :
(c) Using (a) and (b), show that
1  (X; Y )  1 :
19 Let X and Y be independent random variables with uniform densities in [0; 1].
Let Z = X + Y and W = X  Y . Find
(a) (X; Y ) (see Exercise 18).
(b) (X; Z).
(c) (Y; W).
(d) (Z; W).",prob.pdf
"282 CHAPTER 6. EXPECTED VALUE AND VARIANCE
*20 When studying certain physiological data, such as heights of fathers and sons,
it is often natural to assume that these data (e.g., the heights of the fathers
and the heights of the sons) are described by random variables with normal
densities. These random variables, however, are not independent but rather
are correlated. For example, a two-dimensional standard normal density for
correlated random variables has the form
fX;Y (x; y) = 1
2
p
1  2 
 e(x22xy+y2)=2(12 ) :
(a) Show that X and Y each have standard normal densities.
(b) Show that the correlation of X and Y (see Exercise 18) is .
*21 For correlated random variables X and Y it is natural to ask for the expected
value for X given Y . For example, Galton calculated the expected value of
the height of a son given the height of the father. He used this to show
that tall men can be expected to have sons who are less tall on the average.",prob.pdf
"the height of a son given the height of the father. He used this to show
that tall men can be expected to have sons who are less tall on the average.
Similarly, students who do very well on one exam can be expected to do less
well on the next exam, and so forth. This is called regression on the mean.
To dene this conditional expected value, we rst dene a conditional density
of X given Y = y by
fXjY (xjy) = fX;Y (x; y)
fY (y) ;
where fX;Y (x; y) is the joint density of X and Y , and fY is the density for Y .
Then the conditional expected value of X given Y is
E(XjY = y) =
Z b
a
xfXjY (xjy) dx :
For the normal density in Exercise 20, show that the conditional density of
fXjY (xjy) is normal with mean y and variance 1  2. From this we see that
if X and Y are positively correlated (0 <  < 1), and if y > E(Y ), then the
expected value for X given Y = y will be less than y (i.e., we have regression
on the mean).",prob.pdf
"expected value for X given Y = y will be less than y (i.e., we have regression
on the mean).
22 A point Y is chosen at random from [0; 1]. A second point X is then chosen
from the interval [0; Y ]. Find the density for X. Hint: Calculate fXjY as in
Exercise 21 and then use
fX(x) =
Z 1
x
fXjY (xjy)fY (y) dy :
Can you also derive your result geometrically?
*23 Let X and V be two standard normal random variables. Let  be a real
number between -1 and 1.
(a) Let Y = X +
p
1  2V . Show that E(Y ) = 0 and V ar(Y ) = 1. We
shall see later (see Example 7.5 and Example 10.17), that the sum of two
independent normal random variables is again normal. Thus, assuming
this fact, we have shown that Y is standard normal.",prob.pdf
"6.3. CONTINUOUS RANDOM VARIABLES 283
(b) Using Exercises 17 and 18, show that the correlation of X and Y is .
(c) In Exercise 20, the joint density function fX;Y (x; y) for the random vari-
able (X; Y ) is given. Now suppose that we want to know the set of
points (x; y) in the xy-plane such that fX;Y (x; y) = C for some constant
C. This set of points is called a set of constant density. Roughly speak-
ing, a set of constant density is a set of points where the outcomes (X; Y )
are equally likely to fall. Show that for a given C, the set of points of
constant density is a curve whose equation is
x2  2xy + y2 = D ;
where D is a constant which depends upon C. (This curve is an ellipse.)
(d) One can plot the ellipse in part (c) by using the parametric equations
x = r cosp
2(1  )
+ r sin p
2(1 + )
;
y = r cosp
2(1  )
 r sin p
2(1 + )
:
Write a program to plot 1000 pairs (X; Y ) for  = 1=2; 0; 1=2. For each",prob.pdf
"2(1  )
+ r sin p
2(1 + )
;
y = r cosp
2(1  )
 r sin p
2(1 + )
:
Write a program to plot 1000 pairs (X; Y ) for  = 1=2; 0; 1=2. For each
plot, have your program plot the above parametric curves for r = 1; 2; 3.
*24 Following Galton, let us assume that the fathers and sons have heights that
are dependent normal random variables. Assume that the average height is
68 inches, standard deviation is 2.7 inches, and the correlation coecient is .5
(see Exercises 20 and 21). That is, assume that the heights of the fathers
and sons have the form 2:7X + 68 and 2:7Y + 68, respectively, where X
and Y are correlated standardized normal random variables, with correlation
coecient .5.
(a) What is the expected height for the son of a father whose height is
72 inches?
(b) Plot a scatter diagram of the heights of 1000 father and son pairs. Hint:
You can choose standardized pairs as in Exercise 23 and then plot (2:7X+
68; 2:7Y + 68).",prob.pdf
"You can choose standardized pairs as in Exercise 23 and then plot (2:7X+
68; 2:7Y + 68).
*25 When we have pairs of data (xi; yi) that are outcomes of the pairs of dependent
random variables X, Y we can estimate the coorelation coecient  by
r =
P
i(xi  x)(yi  y)
(n  1)sXsY
;
where x and y are the sample means for X and Y , respectively, and sX and sY
are the sample standard deviations for X and Y (see Exercise 6.2.17). Write
a program to compute the sample means, variances, and correlation for such
dependent data. Use your program to compute these quantities for Galton's
data on heights of parents and children given in Appendix B.",prob.pdf
"284 CHAPTER 6. EXPECTED VALUE AND VARIANCE
Plot the equal density ellipses as dened in Exercise 23 for r = 4, 6, and 8, and
on the same graph print the values that appear in the table at the appropriate
points. For example, print 12 at the point (70:5; 68:2), indicating that there
were 12 cases where the parent's height was 70.5 and the child's was 68.12.
See if Galton's data is consistent with the equal density ellipses.
26 (from Hamming25) Suppose you are standing on the bank of a straight river.
(a) Choose, at random, a direction which will keep you on dry land, and
walk 1 km in that direction. Let P denote your position. What is the
expected distance from P to the river?
(b) Now suppose you proceed as in part (a), but when you get to P, you pick
a random direction (from among all directions) and walk 1 km. What
is the probability that you will reach the river before the second walk is
completed?
27 (from Hamming26) A game is played as follows: A random number X is chosen",prob.pdf
"completed?
27 (from Hamming26) A game is played as follows: A random number X is chosen
uniformly from [0; 1]. Then a sequence Y1; Y2; : : : of random numbers is chosen
independently and uniformly from [0; 1]. The game ends the rst time that
Yi > X. You are then paid (i  1) dollars. What is a fair entrance fee for
this game?
28 A long needle of length L much bigger than 1 is dropped on a grid with
horizontal and vertical lines one unit apart. Show that the average number a
of lines crossed is approximately
a = 4L
 :
25R. W. Hamming, The Art of Probability for Scientists and Engineers (Redwood City:
Addison-Wesley, 1991), p. 192.
26ibid., pg. 205.",prob.pdf
"Chapter 7
Sums of Independent
Random Variables
7.1 Sums of Discrete Random Variables
In this chapter we turn to the important question of determining the distribution of
a sum of independent random variables in terms of the distributions of the individual
constituents. In this section we consider only sums of discrete random variables,
reserving the case of continuous random variables for the next section.
We consider here only random variables whose values are integers. Their distri-
bution functions are then dened on these integers. We shall nd it convenient to
assume here that these distribution functions are dened for all integers, by dening
them to be 0 where they are not otherwise dened.
Convolutions
Suppose X and Y are two independent discrete random variables with distribution
functions m1(x) and m2(x). Let Z = X + Y . We would like to determine the dis-
tribution function m3(x) of Z. To do this, it is enough to determine the probability",prob.pdf
"tribution function m3(x) of Z. To do this, it is enough to determine the probability
that Z takes on the value z, where z is an arbitrary integer. Suppose that X = k,
where k is some integer. Then Z = z if and only if Y = z  k. So the event Z = z
is the union of the pairwise disjoint events
(X = k) and (Y = z  k) ;
where k runs over the integers. Since these events are pairwise disjoint, we have
P(Z = z) =
1X
k=1
P(X = k) 
 P(Y = z  k) :
Thus, we have found the distribution function of the random variable Z. This leads
to the following denition.
285",prob.pdf
"286 CHAPTER 7. SUMS OF RANDOM VARIABLES
Denition 7.1 Let X and Y be two independent integer-valued random variables,
with distribution functions m1(x) and m2(x) respectively. Then the convolution of
m1(x) and m2(x) is the distribution function m3 = m1  m2 given by
m3(j) =
X
k
m1(k) 
 m2(j  k) ;
for j = : : :; 2; 1; 0; 1; 2; : : :. The function m3(x) is the distribution function
of the random variable Z = X + Y . 2
It is easy to see that the convolution operation is commutative, and it is straight-
forward to show that it is also associative.
Now let Sn = X1 +X2 +
 
 
+Xn be the sum of n independent random variables
of an independent trials process with common distribution function m dened on
the integers. Then the distribution function of S1 is m. We can write
Sn = Sn1 + Xn :
Thus, since we know the distribution function of Xn is m, we can nd the distribu-
tion function of Sn by induction.
Example 7.1 A die is rolled twice. Let X1 and X2 be the outcomes, and let",prob.pdf
"tion function of Sn by induction.
Example 7.1 A die is rolled twice. Let X1 and X2 be the outcomes, and let
S2 = X1 + X2 be the sum of these outcomes. Then X1 and X2 have the common
distribution function:
m =
 1 2 3 4 5 6
1=6 1=6 1=6 1=6 1=6 1=6

:
The distribution function of S2 is then the convolution of this distribution with
itself. Thus,
P(S2 = 2) = m(1)m(1)
= 1
6 
 1
6 = 1
36 ;
P(S2 = 3) = m(1)m(2) + m(2)m(1)
= 1
6 
 1
6 + 1
6 
 1
6 = 2
36 ;
P(S2 = 4) = m(1)m(3) + m(2)m(2) + m(3)m(1)
= 1
6 
 1
6 + 1
6 
 1
6 + 1
6 
 1
6 = 3
36 :
Continuing in this way we would nd P(S2 = 5) = 4=36, P(S2 = 6) = 5=36,
P(S2 = 7) = 6=36, P(S2 = 8) = 5=36, P(S2 = 9) = 4=36, P(S2 = 10) = 3=36,
P(S2 = 11) = 2=36, and P(S2 = 12) = 1=36.
The distribution for S3 would then be the convolution of the distribution for S2
with the distribution for X3. Thus
P(S3 = 3) = P(S2 = 2)P(X3 = 1)",prob.pdf
"7.1. SUMS OF DISCRETE RANDOM VARIABLES 287
= 1
36 
 1
6 = 1
216 ;
P(S3 = 4) = P(S2 = 3)P(X3 = 1) + P(S2 = 2)P(X3 = 2)
= 2
36 
 1
6 + 1
36 
 1
6 = 3
216 ;
and so forth.
This is clearly a tedious job, and a program should be written to carry out this
calculation. To do this we rst write a program to form the convolution of two
densities p and q and return the density r. We can then write a program to nd the
density for the sum Sn of n independent random variables with a common density
p, at least in the case that the random variables have a nite number of possible
values.
Running this program for the example of rolling a die n times for n = 10; 20; 30
results in the distributions shown in Figure 7.1. We see that, as in the case of
Bernoulli trials, the distributions become bell-shaped. We shall discuss in Chapter 9
a very general theorem called the Central Limit Theorem that will explain this
phenomenon. 2
Example 7.2 A well-known method for evaluating a bridge hand is: an ace is",prob.pdf
"phenomenon. 2
Example 7.2 A well-known method for evaluating a bridge hand is: an ace is
assigned a value of 4, a king 3, a queen 2, and a jack 1. All other cards are assigned
a value of 0. The point count of the hand is then the sum of the values of the
cards in the hand. (It is actually more complicated than this, taking into account
voids in suits, and so forth, but we consider here this simplied form of the point
count.) If a card is dealt at random to a player, then the point count for this card
has distribution
pX =
 0 1 2 3 4
36=52 4=52 4=52 4=52 4=52

:
Let us regard the total hand of 13 cards as 13 independent trials with this
common distribution. (Again this is not quite correct because we assume here that
we are always choosing a card from a full deck.) Then the distribution for the point
count C for the hand can be found from the program NFoldConvolution by using
the distribution for a single card and choosing n = 13. A player with a point count",prob.pdf
"the distribution for a single card and choosing n = 13. A player with a point count
of 13 or more is said to have an opening bid. The probability of having an opening
bid is then
P(C  13) :
Since we have the distribution of C, it is easy to compute this probability. Doing
this we nd that
P(C  13) = :2845 ;
so that about one in four hands should be an opening bid according to this simplied
model. A more realistic discussion of this problem can be found in Epstein, The
Theory of Gambling and Statistical Logic.1 2
1R. A. Epstein, The Theory of Gambling and Statistical Logic, rev. ed. (New York: Academic
Press, 1977).",prob.pdf
"288 CHAPTER 7. SUMS OF RANDOM VARIABLES
20 40 60 80 100 120 1400
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
20 40 60 80 100 120 1400
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
20 40 60 80 100 120 1400
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
n = 10
n = 20

n = 30
Figure 7.1: Density of Sn for rolling a die n times.",prob.pdf
"7.1. SUMS OF DISCRETE RANDOM VARIABLES 289
For certain special distributions it is possible to nd an expression for the dis-
tribution that results from convoluting the distribution with itself n times.
The convolution of two binomial distributions, one with parameters m and p
and the other with parameters n and p, is a binomial distribution with parameters
(m+n) and p. This fact follows easily from a consideration of the experiment which
consists of rst tossing a coin m times, and then tossing it n more times.
The convolution of k geometric distributions with common parameter p is a
negative binomial distribution with parameters p and k. This can be seen by con-
sidering the experiment which consists of tossing a coin until the kth head appears.
Exercises
1 A die is rolled three times. Find the probability that the sum of the outcomes
is
(a) greater than 9.
(b) an odd number.
2 The price of a stock on a given trading day changes according to the distri-
bution
pX =
1 0 1 2",prob.pdf
"is
(a) greater than 9.
(b) an odd number.
2 The price of a stock on a given trading day changes according to the distri-
bution
pX =
1 0 1 2
1=4 1=2 1=8 1=8

:
Find the distribution for the change in stock price after two (independent)
trading days.
3 Let X1 and X2 be independent random variables with common distribution
pX =
 0 1 2
1=8 3=8 1=2

:
Find the distribution of the sum X1 + X2.
4 In one play of a certain game you win an amount X with distribution
pX =
 1 2 3
1=4 1=4 1=2

:
Using the program NFoldConvolution nd the distribution for your total
winnings after ten (independent) plays. Plot this distribution.
5 Consider the following two experiments: the rst has outcome X taking on
the values 0, 1, and 2 with equal probabilities; the second results in an (in-
dependent) outcome Y taking on the value 3 with probability 1/4 and 4 with
probability 3/4. Find the distribution of
(a) Y + X.
(b) Y  X.",prob.pdf
"290 CHAPTER 7. SUMS OF RANDOM VARIABLES
6 People arrive at a queue according to the following scheme: During each
minute of time either 0 or 1 person arrives. The probability that 1 person
arrives is p and that no person arrives is q = 1  p. Let Cr be the number of
customers arriving in the rst r minutes. Consider a Bernoulli trials process
with a success if a person arrives in a unit time and failure if no person arrives
in a unit time. Let Tr be the number of failures before the rth success.
(a) What is the distribution for Tr?
(b) What is the distribution for Cr?
(c) Find the mean and variance for the number of customers arriving in the
rst r minutes.
7 (a) A die is rolled three times with outcomes X1, X2, and X3. Let Y3 be the
maximum of the values obtained. Show that
P(Y3  j) = P(X1  j)3 :
Use this to nd the distribution of Y3. Does Y3 have a bell-shaped dis-
tribution?
(b) Now let Yn be the maximum value when n dice are rolled. Find the",prob.pdf
"tribution?
(b) Now let Yn be the maximum value when n dice are rolled. Find the
distribution of Yn. Is this distribution bell-shaped for large values of n?
8 A baseball player is to play in the World Series. Based upon his season play,
you estimate that if he comes to bat four times in a game the number of hits
he will get has a distribution
pX =
0 1 2 3 4
:4 :2 :2 :1 :1

:
Assume that the player comes to bat four times in each game of the series.
(a) Let X denote the number of hits that he gets in a series. Using the
program NFoldConvolution, nd the distribution of X for each of the
possible series lengths: four-game, ve-game, six-game, seven-game.
(b) Using one of the distribution found in part (a), nd the probability that
his batting average exceeds .400 in a four-game series. (The batting
average is the number of hits divided by the number of times at bat.)
(c) Given the distribution pX, what is his long-term batting average?",prob.pdf
"average is the number of hits divided by the number of times at bat.)
(c) Given the distribution pX, what is his long-term batting average?
9 Prove that you cannot load two dice in such a way that the probabilities for
any sum from 2 to 12 are the same. (Be sure to consider the case where one
or more sides turn up with probability zero.)
10 (Levy2) Assume that n is an integer, not prime. Show that you can nd two
distributions a and b on the nonnegative integers such that the convolution of
2See M. Krasner and B. Ranulae, \Sur une Propriete des Polynomes de la Division du Circle"";
and the following note by J. Hadamard, in C. R. Acad. Sci., vol. 204 (1937), pp. 397{399.",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 291
a and b is the equiprobable distribution on the set 0, 1, 2, . .. , n  1. If n is
prime this is not possible, but the proof is not so easy. (Assume that neither
a nor b is concentrated at 0.)
11 Assume that you are playing craps with dice that are loaded in the following
way: faces two, three, four, and ve all come up with the same probability
(1=6) + r. Faces one and six come up with probability (1=6)  2r, with 0 <
r < :02. Write a computer program to nd the probability of winning at craps
with these dice, and using your program nd which values of r make craps a
favorable game for the player with these dice.
7.2 Sums of Continuous Random Variables
In this section we consider the continuous version of the problem posed in the
previous section: How are sums of independent random variables distributed?
Convolutions
Denition 7.2 Let X and Y be two continuous random variables with density",prob.pdf
"Convolutions
Denition 7.2 Let X and Y be two continuous random variables with density
functions f(x) and g(y), respectively. Assume that both f(x) and g(y) are dened
for all real numbers. Then the convolution f  g of f and g is the function given by
(f  g)(z) =
Z +1
1
f(z  y)g(y) dy
=
Z +1
1
g(z  x)f(x) dx :
2
This denition is analogous to the denition, given in Section 7.1, of the con-
volution of two distribution functions. Thus it should not be surprising that if X
and Y are independent, then the density of their sum is the convolution of their
densities. This fact is stated as a theorem below, and its proof is left as an exercise
(see Exercise 1).
Theorem 7.1 Let X and Y be two independent random variables with density
functions fX(x) and fY (y) dened for all x. Then the sum Z = X + Y is a random
variable with density function fZ(z), where fZ is the convolution of fX and fY . 2
To get a better understanding of this important result, we will look at some
examples.",prob.pdf
"292 CHAPTER 7. SUMS OF RANDOM VARIABLES
Sum of Two Independent Uniform Random Variables
Example 7.3 Suppose we choose independently two numbers at random from the
interval [0; 1] with uniform probability density. What is the density of their sum?
Let X and Y be random variables describing our choices and Z = X + Y their
sum. Then we have
fX(x) = fY (x) =
 1 if 0  x  1,
0 otherwise;
and the density function for the sum is given by
fZ(z) =
Z +1
1
fX(z  y)fY (y) dy :
Since fY (y) = 1 if 0  y  1 and 0 otherwise, this becomes
fZ(z) =
Z 1
0
fX(z  y) dy :
Now the integrand is 0 unless 0  z  y  1 (i.e., unless z  1  y  z) and then it
is 1. So if 0  z  1, we have
fZ(z) =
Z z
0
dy = z ;
while if 1 < z  2, we have
fZ(z) =
Z 1
z1
dy = 2  z ;
and if z < 0 or z > 2 we have fZ(z) = 0 (see Figure 7.2). Hence,
fZ(z) =
8
<
:
z; if 0  z  1;
2  z; if 1 < z  2;
0; otherwise.
Note that this result agrees with that of Example 2.4. 2
Sum of Two Independent Exponential Random Variables",prob.pdf
"2  z; if 1 < z  2;
0; otherwise.
Note that this result agrees with that of Example 2.4. 2
Sum of Two Independent Exponential Random Variables
Example 7.4 Suppose we choose two numbers at random from the interval [0; 1)
with an exponential density with parameter . What is the density of their sum?
Let X, Y , and Z = X + Y denote the relevant random variables, and fX, fY ,
and fZ their densities. Then
fX(x) = fY (x) =

ex; if x  0;
0; otherwise;",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 293
0.5 1 1.5 2
0.2
0.4
0.6
0.8
1
Figure 7.2: Convolution of two uniform densities.
1 2 3 4 5 6
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Figure 7.3: Convolution of two exponential densities with  = 1.
and so, if z > 0,
fZ(z) =
Z +1
1
fX(z  y)fY (y) dy
=
Z z
0
e(zy)ey dy
=
Z z
0
2ez dy
= 2zez;
while if z < 0, fZ(z) = 0 (see Figure 7.3). Hence,
fZ(z) =

2zez; if z  0;
0; otherwise.
2",prob.pdf
"294 CHAPTER 7. SUMS OF RANDOM VARIABLES
Sum of Two Independent Normal Random Variables
Example 7.5 It is an interesting and important fact that the convolution of two
normal densities with means 1 and 2 and variances 1 and 2 is again a normal
density, with mean 1 + 2 and variance 2
1 + 2
2. We will show this in the special
case that both random variables are standard normal. The general case can be done
in the same way, but the calculation is messier. Another way to show the general
result is given in Example 10.17.
Suppose X and Y are two independent random variables, each with the standard
normal density (see Example 5.8). We have
fX(x) = fY (y) = 1p
2
ex2=2 ;
and so
fZ(z) = fX  fY (z)
= 1
2
Z +1
1
e(zy)2=2ey2=2 dy
= 1
2ez2=4
Z +1
1
e(yz=2)2
dy
= 1
2ez2=4p
 1p
Z 1
1
e(yz=2)2
dy

:
The expression in the brackets equals 1, since it is the integral of the normal density
function with  = 0 and  =
p
2. So, we have
fZ(z) = 1p
4ez2=4 :
2",prob.pdf
"function with  = 0 and  =
p
2. So, we have
fZ(z) = 1p
4ez2=4 :
2
Sum of Two Independent Cauchy Random Variables
Example 7.6 Choose two numbers at random from the interval (1; +1) with
the Cauchy density with parameter a = 1 (see Example 5.10). Then
fX(x) = fY (x) = 1
(1 + x2) ;
and Z = X + Y has density
fZ(z) = 1
2
Z +1
1
1
1 + (z  y)2
1
1 + y2 dy :",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 295
This integral requires some eort, and we give here only the result (see Section 10.3,
or Dwass3):
fZ(z) = 2
(4 + z2) :
Now, suppose that we ask for the density function of the average
A = (1=2)(X + Y )
of X and Y . Then A = (1=2)Z. Exercise 5.2.19 shows that if U and V are two
continuous random variables with density functions fU (x) and fV (x), respectively,
and if V = aU, then
fV (x) =
1
a

fU
x
a

:
Thus, we have
fA(z) = 2fZ(2z) = 1
(1 + z2) :
Hence, the density function for the average of two random variables, each having a
Cauchy density, is again a random variable with a Cauchy density; this remarkable
property is a peculiarity of the Cauchy density. One consequence of this is if the
error in a certain measurement process had a Cauchy density and you averaged
a number of measurements, the average could not be expected to be any more
accurate than any one of your individual measurements! 2
Rayleigh Density",prob.pdf
"a number of measurements, the average could not be expected to be any more
accurate than any one of your individual measurements! 2
Rayleigh Density
Example 7.7 Suppose X and Y are two independent standard normal random
variables. Now suppose we locate a point P in the xy-plane with coordinates (X; Y )
and ask: What is the density of the square of the distance of P from the origin?
(We have already simulated this problem in Example 5.9.) Here, with the preceding
notation, we have
fX(x) = fY (x) = 1p
2ex2=2 :
Moreover, if X2 denotes the square of X, then (see Theorem 5.1 and the discussion
following)
fX2 (r) =
 1
2pr (fX(pr) + fX(pr)) if r > 0;
0 otherwise.
=
 1p
2r (er=2) if r > 0;
0 otherwise.
3M. Dwass, \On the Convolution of Cauchy Distributions,"" American Mathematical Monthly,
vol. 92, no. 1, (1985), pp. 55{57; see also R. Nelson, letters to the Editor, ibid., p. 679.",prob.pdf
"296 CHAPTER 7. SUMS OF RANDOM VARIABLES
This is a gamma density with  = 1=2,  = 1=2 (see Example 7.4). Now let
R2 = X2 + Y 2. Then
fR2 (r) =
Z +1
1
fX2 (r  s)fY 2 (s) ds
= 1
4
Z +1
1
e(rs)=2 r  s
2
1=2
es s
2
1=2
ds ;
=
 1
2er2=2; if r  0;
0; otherwise.
Hence, R2 has a gamma density with  = 1=2,  = 1. We can interpret this result
as giving the density for the square of the distance of P from the center of a target
if its coordinates are normally distributed.
The density of the random variable R is obtained from that of R2 in the usual
way (see Theorem 5.1), and we nd
fR(r) =
 1
2er2=2 
 2r = rer2=2; if r  0;
0; otherwise.
Physicists will recognize this as a Rayleigh density. Our result here agrees with
our simulation in Example 5.9. 2
Chi-Squared Density
More generally, the same method shows that the sum of the squares of n independent
normally distributed random variables with mean 0 and standard deviation 1 has",prob.pdf
"normally distributed random variables with mean 0 and standard deviation 1 has
a gamma density with  = 1=2 and  = n=2. Such a density is called a chi-squared
density with n degrees of freedom. This density was introduced in Chapter 4.3.
In Example 5.10, we used this density to test the hypothesis that two traits were
independent.
Another important use of the chi-squared density is in comparing experimental
data with a theoretical discrete distribution, to see whether the data supports the
theoretical model. More specically, suppose that we have an experiment with a
nite set of outcomes. If the set of outcomes is countable, we group them into nitely
many sets of outcomes. We propose a theoretical distribution which we think will
model the experiment well. We obtain some data by repeating the experiment a
number of times. Now we wish to check how well the theoretical distribution ts
the data.
Let X be the random variable which represents a theoretical outcome in the",prob.pdf
"the data.
Let X be the random variable which represents a theoretical outcome in the
model of the experiment, and let m(x) be the distribution function of X. In a
manner similar to what was done in Example 5.10, we calculate the value of the
expression
V =
X
x
(ox  n 
 m(x))2
n 
 m(x) ;
where the sum runs over all possible outcomes x, n is the number of data points,
and ox denotes the number of outcomes of type x observed in the data. Then",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 297
Outcome Observed Frequency
1 15
2 8
3 7
4 5
5 7
6 18
Table 7.1: Observed data.
for moderate or large values of n, the quantity V is approximately chi-squared
distributed, with 1 degrees of freedom, where  represents the number of possible
outcomes. The proof of this is beyond the scope of this book, but we will illustrate
the reasonableness of this statement in the next example. If the value of V is very
large, when compared with the appropriate chi-squared density function, then we
would tend to reject the hypothesis that the model is an appropriate one for the
experiment at hand. We now give an example of this procedure.
Example 7.8 Suppose we are given a single die. We wish to test the hypothesis
that the die is fair. Thus, our theoretical distribution is the uniform distribution on
the integers between 1 and 6. So, if we roll the die n times, the expected number",prob.pdf
"the integers between 1 and 6. So, if we roll the die n times, the expected number
of data points of each type is n=6. Thus, if oi denotes the actual number of data
points of type i, for 1  i  6, then the expression
V =
6X
i=1
(oi  n=6)2
n=6
is approximately chi-squared distributed with 5 degrees of freedom.
Now suppose that we actually roll the die 60 times and obtain the data in
Table 7.1. If we calculate V for this data, we obtain the value 13.6. The graph of
the chi-squared density with 5 degrees of freedom is shown in Figure 7.4. One sees
that values as large as 13.6 are rarely taken on by V if the die is fair, so we would
reject the hypothesis that the die is fair. (When using this test, a statistician will
reject the hypothesis if the data gives a value of V which is larger than 95% of the
values one would expect to obtain if the hypothesis is true.)
In Figure 7.5, we show the results of rolling a die 60 times, then calculating V ,",prob.pdf
"values one would expect to obtain if the hypothesis is true.)
In Figure 7.5, we show the results of rolling a die 60 times, then calculating V ,
and then repeating this experiment 1000 times. The program that performs these
calculations is called DieTest. We have superimposed the chi-squared density with
5 degrees of freedom; one can see that the data values t the curve fairly well, which
supports the statement that the chi-squared density is the correct one to use. 2
So far we have looked at several important special cases for which the convolution
integral can be evaluated explicitly. In general, the convolution of two continuous
densities cannot be evaluated explicitly, and we must resort to numerical methods.
Fortunately, these prove to be remarkably eective, at least for bounded densities.",prob.pdf
"298 CHAPTER 7. SUMS OF RANDOM VARIABLES
5 10 15 20
0.025
0.05
0.075
0.1
0.125
0.15
Figure 7.4: Chi-squared density with 5 degrees of freedom.
0 5 10 15 20 25 30
0
0.025
0.05
0.075
0.1
0.125
0.15 1000 experiments
60 rolls per experiment
Figure 7.5: Rolling a fair die.",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 299
1 2 3 4 5 6 7 80
0.2
0.4
0.6
0.8
1 n = 2
n = 4
n = 6
n = 8
n = 10
Figure 7.6: Convolution of n uniform densities.
Independent Trials
We now consider brie
y the distribution of the sum of n independent random vari-
ables, all having the same density function. If X1, X2, . . . , Xn are these random
variables and Sn = X1 + X2 + 
 
 
 + Xn is their sum, then we will have
fSn (x) = (fX1  fX2  
 
 
  fXn ) (x) ;
where the right-hand side is an n-fold convolution. It is possible to calculate this
density for general values of n in certain simple cases.
Example 7.9 Suppose the Xi are uniformly distributed on the interval [0; 1]. Then
fXi (x) =
 1; if 0  x  1;
0; otherwise,
and fSn (x) is given by the formula4
fSn (x) =
 1
(n1)!
P
0jx(1)jn
j


(x  j)n1; if 0 < x < n;
0; otherwise.
The density fSn (x) for n = 2, 4, 6, 8, 10 is shown in Figure 7.6.
If the Xi are distributed normally, with mean 0 and variance 1, then (cf. Exam-
ple 7.5)",prob.pdf
"If the Xi are distributed normally, with mean 0 and variance 1, then (cf. Exam-
ple 7.5)
fXi (x) = 1p
2ex2=2 ;
4J. B. Uspensky, Introduction to Mathematical Probability (New York: McGraw-Hill, 1937),
p. 277.",prob.pdf
"300 CHAPTER 7. SUMS OF RANDOM VARIABLES
-15 -10 -5 5 10 15
0.025
0.05
0.075
0.1
0.125
0.15
0.175 n = 5
n = 10
n = 15
n = 20
n = 25
Figure 7.7: Convolution of n standard normal densities.
and
fSn (x) = 1p
2nex2=2n :
Here the density fSn for n = 5, 10, 15, 20, 25 is shown in Figure 7.7.
If the Xi are all exponentially distributed, with mean 1=, then
fXi (x) = ex ;
and
fSn (x) = ex(x)n1
(n  1)! :
In this case the density fSn for n = 2, 4, 6, 8, 10 is shown in Figure 7.8. 2
Exercises
1 Let X and Y be independent real-valued random variables with density func-
tions fX(x) and fY (y), respectively. Show that the density function of the
sum X +Y is the convolution of the functions fX(x) and fY (y). Hint: Let X
be the joint random variable (X; Y ). Then the joint density function of X is
fX(x)fY (y), since X and Y are independent. Now compute the probability
that X +Y  z, by integrating the joint density function over the appropriate",prob.pdf
"fX(x)fY (y), since X and Y are independent. Now compute the probability
that X +Y  z, by integrating the joint density function over the appropriate
region in the plane. This gives the cumulative distribution function of Z. Now
dierentiate this function with respect to z to obtain the density function of
z.
2 Let X and Y be independent random variables dened on the space 
, with
density functions fX and fY , respectively. Suppose that Z = X + Y . Find
the density fZ of Z if",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 301
5 10 15 20
0.05
0.1
0.15
0.2
0.25
0.3
0.35 n = 2
n = 4
n = 6
n = 8
n = 10
Figure 7.8: Convolution of n exponential densities with  = 1.
(a)
fX(x) = fY (x) =

1=2; if 1  x  +1;
0; otherwise.
(b)
fX(x) = fY (x) =
 1=2; if 3  x  5;
0; otherwise.
(c)
fX(x) =
 1=2; if 1  x  1;
0; otherwise.
fY (x) =

1=2; if 3  x  5;
0; otherwise.
(d) What can you say about the set E = f z : fZ(z) > 0 g in each case?
3 Suppose again that Z = X + Y . Find fZ if
(a)
fX(x) = fY (x) =
 x=2; if 0 < x < 2;
0; otherwise:
(b)
fX(x) = fY (x) =
 (1=2)(x  3); if 3 < x < 5;
0; otherwise:
(c)
fX(x) =

1=2; if 0 < x < 2;
0; otherwise;",prob.pdf
"302 CHAPTER 7. SUMS OF RANDOM VARIABLES
fY (x) =
 x=2; if 0 < x < 2;
0; otherwise:
(d) What can you say about the set E = f z : fZ(z) > 0 g in each case?
4 Let X, Y , and Z be independent random variables with
fX(x) = fY (x) = fZ(x) =
 1; if 0 < x < 1;
0; otherwise.
Suppose that W = X + Y + Z. Find fW directly, and compare your answer
with that given by the formula in Example 7.9. Hint: See Example 7.3.
5 Suppose that X and Y are independent and Z = X + Y . Find fZ if
(a)
fX(x) =

ex; if x > 0;
0; otherwise.
fY (x) =
 ex; if x > 0;
0; otherwise.
(b)
fX(x) =

ex; if x > 0;
0; otherwise.
fY (x) =

1; if 0 < x < 1;
0; otherwise.
6 Suppose again that Z = X + Y . Find fZ if
fX(x) = 1p
21
e(x1)2=22
1
fY (x) = 1p
22
e(x2)2=22
2 :
*7 Suppose that R2 = X2 + Y 2. Find fR2 and fR if
fX(x) = 1p
21
e(x1)2=22
1
fY (x) = 1p
22
e(x2)2=22
2 :
8 Suppose that R2 = X2 + Y 2. Find fR2 and fR if
fX(x) = fY (x) =
 1=2; if 1  x  1;
0; otherwise.",prob.pdf
"1
fY (x) = 1p
22
e(x2)2=22
2 :
8 Suppose that R2 = X2 + Y 2. Find fR2 and fR if
fX(x) = fY (x) =
 1=2; if 1  x  1;
0; otherwise.
9 Assume that the service time for a customer at a bank is exponentially dis-
tributed with mean service time 2 minutes. Let X be the total service time
for 10 customers. Estimate the probability that X > 22 minutes.",prob.pdf
"7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 303
10 Let X1, X2, . . . , Xn be n independent random variables each of which has
an exponential density with mean . Let M be the minimum value of the
Xj. Show that the density for M is exponential with mean =n. Hint: Use
cumulative distribution functions.
11 A company buys 100 lightbulbs, each of which has an exponential lifetime of
1000 hours. What is the expected time for the rst of these bulbs to burn
out? (See Exercise 10.)
12 An insurance company assumes that the time between claims from each of its
homeowners' policies is exponentially distributed with mean . It would like
to estimate  by averaging the times for a number of policies, but this is not
very practical since the time between claims is about 30 years. At Galambos'5
suggestion the company puts its customers in groups of 50 and observes the
time of the rst claim within each group. Show that this provides a practical
way to estimate the value of .",prob.pdf
"time of the rst claim within each group. Show that this provides a practical
way to estimate the value of .
13 Particles are subject to collisions that cause them to split into two parts with
each part a fraction of the parent. Suppose that this fraction is uniformly
distributed between 0 and 1. Following a single particle through several split-
tings we obtain a fraction of the original particle Zn = X1 
X2 
: : :
Xn where
each Xj is uniformly distributed between 0 and 1. Show that the density for
the random variable Zn is
fn(z) = 1
(n  1)!( log z)n1:
Hint: Show that Yk =  logXk is exponentially distributed. Use this to nd
the density function for Sn = Y1 +Y2 +
 
 
+Yn, and from this the cumulative
distribution and density of Zn = eSn .
14 Assume that X1 and X2 are independent random variables, each having an
exponential density with parameter . Show that Z = X1  X2 has density
fZ(z) = (1=2)ejzj :",prob.pdf
"exponential density with parameter . Show that Z = X1  X2 has density
fZ(z) = (1=2)ejzj :
15 Suppose we want to test a coin for fairness. We 
ip the coin n times and
record the number of times X0 that the coin turns up tails and the number
of times X1 = n  X0 that the coin turns up heads. Now we set
Z =
1X
i=0
(Xi  n=2)2
n=2 :
Then for a fair coin Z has approximately a chi-squared distribution with
2  1 = 1 degree of freedom. Verify this by computer simulation rst for a
fair coin (p = 1=2) and then for a biased coin (p = 1=3).
5J. Galambos, Introductory Probability Theory (New York: Marcel Dekker, 1984), p. 159.",prob.pdf
"304 CHAPTER 7. SUMS OF RANDOM VARIABLES
16 Verify your answers in Exercise 2(a) by computer simulation: Choose X and
Y from [1; 1] with uniform density and calculate Z = X + Y . Repeat this
experiment 500 times, recording the outcomes in a bar graph on [2; 2] with
40 bars. Does the density fZ calculated in Exercise 2(a) describe the shape
of your bar graph? Try this for Exercises 2(b) and Exercise 2(c), too.
17 Verify your answers to Exercise 3 by computer simulation.
18 Verify your answer to Exercise 4 by computer simulation.
19 The support of a function f(x) is dened to be the set
fx : f(x) > 0g :
Suppose that X and Y are two continuous random variables with density
functions fX(x) and fY (y), respectively, and suppose that the supports of
these density functions are the intervals [a; b] and [c; d], respectively. Find the
support of the density function of the random variable X + Y .
20 Let X1, X2, .. . , Xn be a sequence of independent random variables, all having",prob.pdf
"support of the density function of the random variable X + Y .
20 Let X1, X2, .. . , Xn be a sequence of independent random variables, all having
a common density function fX with support [a; b] (see Exercise 19). Let
Sn = X1 + X2 + 
 
 
 + Xn, with density function fSn . Show that the support
of fSn is the interval [na; nb]. Hint: Write fSn = fSn1  fX. Now use
Exercise 19 to establish the desired result by induction.
21 Let X1, X2, .. . , Xn be a sequence of independent random variables, all having
a common density function fX. Let A = Sn=n be their average. Find fA if
(a) fX(x) = (1=
p
2)ex2=2 (normal density).
(b) fX(x) = ex (exponential density).
Hint: Write fA(x) in terms of fSn (x).",prob.pdf
"Chapter 8
Law of Large Numbers
8.1 Law of Large Numbers for Discrete Random
Variables
We are now in a position to prove our rst fundamental theorem of probability.
We have seen that an intuitive way to view the probability of a certain outcome
is as the frequency with which that outcome occurs in the long run, when the ex-
periment is repeated a large number of times. We have also dened probability
mathematically as a value of a distribution function for the random variable rep-
resenting the experiment. The Law of Large Numbers, which is a theorem proved
about the mathematical model of probability, shows that this model is consistent
with the frequency interpretation of probability. This theorem is sometimes called
the law of averages. To nd out what would happen if this law were not true, see
the article by Robert M. Coates.1
Chebyshev Inequality
To discuss the Law of Large Numbers, we rst need an important inequality called
the Chebyshev Inequality.",prob.pdf
"Chebyshev Inequality
To discuss the Law of Large Numbers, we rst need an important inequality called
the Chebyshev Inequality.
Theorem 8.1 (Chebyshev Inequality) Let X be a discrete random variable
with expected value  = E(X), and let  > 0 be any positive real number. Then
P(jX  j  )  V (X)
2 :
Proof. Let m(x) denote the distribution function of X. Then the probability that
X diers from  by at least  is given by
P(jX  j  ) =
X
jxj
m(x) :
1R. M. Coates, \The Law,"" The World of Mathematics, ed. James R. Newman (New York:
Simon and Schuster, 1956.
305",prob.pdf
"306 CHAPTER 8. LAW OF LARGE NUMBERS
We know that
V (X) =
X
x
(x  )2m(x) ;
and this is clearly at least as large as
X
jxj
(x  )2m(x) ;
since all the summands are positive and we have restricted the range of summation
in the second sum. But this last sum is at least
X
jxj
2m(x) = 2 X
jxj
m(x)
= 2P(jX  j  ) :
So,
P(jX  j  )  V (X)
2 :
2
Note that X in the above theorem can be any discrete random variable, and  any
positive number.
Example 8.1 Let X by any random variable with E(X) =  and V (X) = 2.
Then, if  = k, Chebyshev's Inequality states that
P(jX  j  k)  2
k22 = 1
k2 :
Thus, for any random variable, the probability of a deviation from the mean of
more than k standard deviations is  1=k2. If, for example, k = 5, 1=k2 = :04. 2
Chebyshev's Inequality is the best possible inequality in the sense that, for any
 > 0, it is possible to give an example of a random variable for which Chebyshev's",prob.pdf
" > 0, it is possible to give an example of a random variable for which Chebyshev's
Inequality is in fact an equality. To see this, given  > 0, choose X with distribution
pX =
 +
1=2 1=2

:
Then E(X) = 0, V (X) = 2, and
P(jX  j  ) = V (X)
2 = 1 :
We are now prepared to state and prove the Law of Large Numbers.",prob.pdf
"8.1. DISCRETE RANDOM VARIABLES 307
Law of Large Numbers
Theorem 8.2 (Law of Large Numbers) Let X1, X2, . . . , Xn be an independent
trials process, with nite expected value  = E(Xj) and nite variance 2 = V (Xj).
Let Sn = X1 + X2 + 
 
 
 + Xn. Then for any  > 0,
P




Sn
n  



 

! 0
as n ! 1. Equivalently,
P




Sn
n  



< 

! 1
as n ! 1.
Proof. Since X1, X2, . . . , Xn are independent and have the same distributions, we
can apply Theorem 6.9. We obtain
V (Sn) = n2 ;
and
V (Sn
n ) = 2
n :
Also we know that
E(Sn
n ) =  :
By Chebyshev's Inequality, for any  > 0,
P




Sn
n  



 

 2
n2 :
Thus, for xed ,
P




Sn
n  



 

! 0
as n ! 1, or equivalently,
P




Sn
n  



< 

! 1
as n ! 1. 2
Law of Averages
Note that Sn=n is an average of the individual outcomes, and one often calls the Law
of Large Numbers the \law of averages."" It is a striking fact that we can start with",prob.pdf
"of Large Numbers the \law of averages."" It is a striking fact that we can start with
a random experiment about which little can be predicted and, by taking averages,
obtain an experiment in which the outcome can be predicted with a high degree
of certainty. The Law of Large Numbers, as we have stated it, is often called the
\Weak Law of Large Numbers"" to distinguish it from the \Strong Law of Large
Numbers"" described in Exercise 15.",prob.pdf
"308 CHAPTER 8. LAW OF LARGE NUMBERS
Consider the important special case of Bernoulli trials with probability p for
success. Let Xj = 1 if the jth outcome is a success and 0 if it is a failure. Then
Sn = X1 +X2 +
 
 
+Xn is the number of successes in n trials and  = E(X1) = p.
The Law of Large Numbers states that for any  > 0
P




Sn
n  p



< 

! 1
as n ! 1. The above statement says that, in a large number of repetitions of a
Bernoulli experiment, we can expect the proportion of times the event will occur to
be near p. This shows that our mathematical model of probability agrees with our
frequency interpretation of probability.
Coin Tossing
Let us consider the special case of tossing a coin n times with Sn the number of
heads that turn up. Then the random variable Sn=n represents the fraction of times
heads turns up and will have values between 0 and 1. The Law of Large Numbers
predicts that the outcomes for this random variable will, for large n, be near 1/2.",prob.pdf
"predicts that the outcomes for this random variable will, for large n, be near 1/2.
In Figure 8.1, we have plotted the distribution for this example for increasing
values of n. We have marked the outcomes between .45 and .55 by dots at the top
of the spikes. We see that as n increases the distribution gets more and more con-
centrated around .5 and a larger and larger percentage of the total area is contained
within the interval (:45; :55), as predicted by the Law of Large Numbers.
Die Rolling
Example 8.2 Consider n rolls of a die. Let Xj be the outcome of the jth roll.
Then Sn = X1 +X2 +
 
 
+Xn is the sum of the rst n rolls. This is an independent
trials process with E(Xj) = 7=2. Thus, by the Law of Large Numbers, for any  > 0
P




Sn
n  7
2



 

! 0
as n ! 1. An equivalent way to state this is that, for any  > 0,
P




Sn
n  7
2



< 

! 1
as n ! 1. 2
Numerical Comparisons
It should be emphasized that, although Chebyshev's Inequality proves the Law of",prob.pdf
"Sn
n  7
2



< 

! 1
as n ! 1. 2
Numerical Comparisons
It should be emphasized that, although Chebyshev's Inequality proves the Law of
Large Numbers, it is actually a very crude inequality for the probabilities involved.
However, its strength lies in the fact that it is true for any random variable at all,
and it allows us to prove a very powerful theorem.
In the following example, we compare the estimates given by Chebyshev's In-
equality with the actual values.",prob.pdf
"8.1. DISCRETE RANDOM VARIABLES 309
0 0.2 0.4 0.6 0.8 1
0
0.02
0.04
0.06
0.08
0.1
0 0.2 0.4 0.6 0.8 1
0
0.02
0.04
0.06
0.08
0 0.2 0.4 0.6 0.8 1
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 0.2 0.4 0.6 0.8 1
0
0.02
0.04
0.06
0.08
0.1
0.12
0 0.2 0.4 0.6 0.8 1
0
0.05
0.1
0.15
0.2
0.25
0 0.2 0.4 0.6 0.8 1
0
0.025
0.05
0.075
0.1
0.125
0.15
0.175
n=10 n=20
n=40n=30
n=60 n=100
Figure 8.1: Bernoulli trials distributions.",prob.pdf
"310 CHAPTER 8. LAW OF LARGE NUMBERS
Example 8.3 Let X1, X2, . . . , Xn be a Bernoulli trials process with probability .3
for success and .7 for failure. Let Xj = 1 if the jth outcome is a success and 0
otherwise. Then, E(Xj) = :3 and V (Xj) = (:3)(:7) = :21. If
An = Sn
n = X1 + X2 + 
 
 
 + Xn
n
is the average of the Xi, then E(An) = :3 and V (An) = V (Sn)=n2 = :21=n.
Chebyshev's Inequality states that if, for example,  = :1,
P(jAn  :3j  :1)  :21
n(:1)2 = 21
n :
Thus, if n = 100,
P(jA100  :3j  :1)  :21 ;
or if n = 1000,
P(jA1000  :3j  :1)  :021 :
These can be rewritten as
P(:2 < A100 < :4)  :79 ;
P(:2 < A1000 < :4)  :979 :
These values should be compared with the actual values, which are (to six decimal
places)
P(:2 < A100 < :4)  :962549
P(:2 < A1000 < :4)  1 :
The program Law can be used to carry out the above calculations in a systematic
way. 2
Historical Remarks
The Law of Large Numbers was rst proved by the Swiss mathematician James",prob.pdf
"way. 2
Historical Remarks
The Law of Large Numbers was rst proved by the Swiss mathematician James
Bernoulli in the fourth part of his work Ars Conjectandi published posthumously
in 1713.2 As often happens with a rst proof, Bernoulli's proof was much more
dicult than the proof we have presented using Chebyshev's inequality. Cheby-
shev developed his inequality to prove a general form of the Law of Large Numbers
(see Exercise 12). The inequality itself appeared much earlier in a work by Bien-
ayme, and in discussing its history Maistrov remarks that it was referred to as the
Bienayme-Chebyshev Inequality for a long time.3
In Ars Conjectandi Bernoulli provides his reader with a long discussion of the
meaning of his theorem with lots of examples. In modern notation he has an event
2J. Bernoulli, The Art of Conjecturing IV, trans. Bing Sung, Technical Report No. 2, Dept. of
Statistics, Harvard Univ., 1966",prob.pdf
"2J. Bernoulli, The Art of Conjecturing IV, trans. Bing Sung, Technical Report No. 2, Dept. of
Statistics, Harvard Univ., 1966
3L. E. Maistrov, Probability Theory: A Historical Approach, trans. and ed. Samual Kotz, (New
York: Academic Press, 1974), p. 202",prob.pdf
"8.1. DISCRETE RANDOM VARIABLES 311
that occurs with probability p but he does not know p. He wants to estimate p
by the fraction p of the times the event occurs when the experiment is repeated a
number of times. He discusses in detail the problem of estimating, by this method,
the proportion of white balls in an urn that contains an unknown number of white
and black balls. He would do this by drawing a sequence of balls from the urn,
replacing the ball drawn after each draw, and estimating the unknown proportion
of white balls in the urn by the proportion of the balls drawn that are white. He
shows that, by choosing n large enough he can obtain any desired accuracy and
reliability for the estimate. He also provides a lively discussion of the applicability
of his theorem to estimating the probability of dying of a particular disease, of
dierent kinds of weather occurring, and so forth.
In speaking of the number of trials necessary for making a judgement, Bernoulli",prob.pdf
"dierent kinds of weather occurring, and so forth.
In speaking of the number of trials necessary for making a judgement, Bernoulli
observes that the \man on the street"" believes the \law of averages.""
Further, it cannot escape anyone that for judging in this way about any
event at all, it is not enough to use one or two trials, but rather a great
number of trials is required. And sometimes the stupidest man|by
some instinct of nature per se and by no previous instruction (this is
truly amazing)| knows for sure that the more observations of this sort
that are taken, the less the danger will be of straying from the mark.4
But he goes on to say that he must contemplate another possibility.
Something futher must be contemplated here which perhaps no one has
thought about till now. It certainly remains to be inquired whether
after the number of observations has been increased, the probability is
increased of attaining the true ratio between the number of cases in",prob.pdf
"after the number of observations has been increased, the probability is
increased of attaining the true ratio between the number of cases in
which some event can happen and in which it cannot happen, so that
this probability nally exceeds any given degree of certainty; or whether
the problem has, so to speak, its own asymptote|that is, whether some
degree of certainty is given which one can never exceed.5
Bernoulli recognized the importance of this theorem, writing:
Therefore, this is the problem which I now set forth and make known
after I have already pondered over it for twenty years. Both its novelty
and its very great usefullness, coupled with its just as great diculty,
can exceed in weight and value all the remaining chapters of this thesis.6
Bernoulli concludes his long proof with the remark:
Whence, nally, this one thing seems to follow: that if observations of
all events were to be continued throughout all eternity, (and hence the",prob.pdf
"Whence, nally, this one thing seems to follow: that if observations of
all events were to be continued throughout all eternity, (and hence the
ultimate probability would tend toward perfect certainty), everything in
4Bernoulli, op. cit., p. 38.
5ibid., p. 39.
6ibid., p. 42.",prob.pdf
"312 CHAPTER 8. LAW OF LARGE NUMBERS
the world would be perceived to happen in xed ratios and according to
a constant law of alternation, so that even in the most accidental and
fortuitous occurrences we would be bound to recognize, as it were, a
certain necessity and, so to speak, a certain fate.
I do now know whether Plato wished to aim at this in his doctrine of
the universal return of things, according to which he predicted that all
things will return to their original state after countless ages have past.7
Exercises
1 A fair coin is tossed 100 times. The expected number of heads is 50, and the
standard deviation for the number of heads is (100 
 1=2 
 1=2)1=2 = 5. What
does Chebyshev's Inequality tell you about the probability that the number
of heads that turn up deviates from the expected number 50 by three or more
standard deviations (i.e., by at least 15)?
2 Write a program that uses the function binomial(n; p; x) to compute the exact",prob.pdf
"standard deviations (i.e., by at least 15)?
2 Write a program that uses the function binomial(n; p; x) to compute the exact
probability that you estimated in Exercise 1. Compare the two results.
3 Write a program to toss a coin 10,000 times. Let Sn be the number of heads
in the rst n tosses. Have your program print out, after every 1000 tosses,
Sn  n=2. On the basis of this simulation, is it correct to say that you can
expect heads about half of the time when you toss a coin a large number of
times?
4 A 1-dollar bet on craps has an expected winning of :0141. What does the
Law of Large Numbers say about your winnings if you make a large number
of 1-dollar bets at the craps table? Does it assure you that your losses will be
small? Does it assure you that if n is very large you will lose?
5 Let X be a random variable with E(X) = 0 and V (X) = 1. What integer
value k will assure us that P(jXj  k)  :01?
6 Let Sn be the number of successes in n Bernoulli trials with probability p for",prob.pdf
"value k will assure us that P(jXj  k)  :01?
6 Let Sn be the number of successes in n Bernoulli trials with probability p for
success on each trial. Show, using Chebyshev's Inequality, that for any  > 0
P




Sn
n  p



 

 p(1  p)
n2 :
7 Find the maximum possible value for p(1  p) if 0 < p < 1. Using this result
and Exercise 6, show that the estimate
P




Sn
n  p



 

 1
4n2
is valid for any p.
7ibid., pp. 65{66.",prob.pdf
"8.1. DISCRETE RANDOM VARIABLES 313
8 A fair coin is tossed a large number of times. Does the Law of Large Numbers
assure us that, if n is large enough, with probability > :99 the number of
heads that turn up will not deviate from n=2 by more than 100?
9 In Exercise 6.2.15, you showed that, for the hat check problem, the number
Sn of people who get their own hats back has E(Sn) = V (Sn) = 1. Using
Chebyshev's Inequality, show that P(Sn  11)  :01 for any n  11.
10 Let X by any random variable which takes on values 0, 1, 2, . . . , n and has
E(X) = V (X) = 1. Show that, for any positive integer k,
P(X  k + 1)  1
k2 :
11 We have two coins: one is a fair coin and the other is a coin that produces
heads with probability 3/4. One of the two coins is picked at random, and this
coin is tossed n times. Let Sn be the number of heads that turns up in these
n tosses. Does the Law of Large Numbers allow us to predict the proportion",prob.pdf
"n tosses. Does the Law of Large Numbers allow us to predict the proportion
of heads that will turn up in the long run? After we have observed a large
number of tosses, can we tell which coin was chosen? How many tosses suce
to make us 95 percent sure?
12 (Chebyshev8) Assume that X1, X2, . . . , Xn are independent random variables
with possibly dierent distributions and let Sn be their sum. Let mk = E(Xk),
2
k = V (Xk), and Mn = m1 + m2 + 
 
 
 + mn. Assume that 2
k < R for all k.
Prove that, for any  > 0,
P




Sn
n  Mn
n



< 

! 1
as n ! 1.
13 A fair coin is tossed repeatedly. Before each toss, you are allowed to decide
whether to bet on the outcome. Can you describe a betting system with
innitely many bets which will enable you, in the long run, to win more
than half of your bets? (Note that we are disallowing a betting system that
says to bet until you are ahead, then quit.) Write a computer program that",prob.pdf
"says to bet until you are ahead, then quit.) Write a computer program that
implements this betting system. As stated above, your program must decide
whether to bet on a particular outcome before that outcome is determined.
For example, you might select only outcomes that come after there have been
three tails in a row. See if you can get more than 50% heads by your \system.""
*14 Prove the following analogue of Chebyshev's Inequality:
P(jX  E(X)j  )  1
E(jX  E(X)j) :
8P. L. Chebyshev, \On Mean Values,"" J. Math. Pure. Appl., vol. 12 (1867), pp. 177{184.",prob.pdf
"314 CHAPTER 8. LAW OF LARGE NUMBERS
*15 We have proved a theorem often called the \Weak Law of Large Numbers.""
Most people's intuition and our computer simulations suggest that, if we toss
a coin a sequence of times, the proportion of heads will really approach 1/2;
that is, if Sn is the number of heads in n times, then we will have
An = Sn
n ! 1
2
as n ! 1. Of course, we cannot be sure of this since we are not able to toss
the coin an innite number of times, and, if we could, the coin could come up
heads every time. However, the \Strong Law of Large Numbers,"" proved in
more advanced courses, states that
P
Sn
n ! 1
2

= 1 :
Describe a sample space 
 that would make it possible for us to talk about
the event
E =

! : Sn
n ! 1
2

:
Could we assign the equiprobable measure to this space? (See Example 2.18.)
*16 In this exercise, we shall construct an example of a sequence of random vari-
ables that satises the weak law of large numbers, but not the strong law.",prob.pdf
"ables that satises the weak law of large numbers, but not the strong law.
The distribution of Xi will have to depend on i, because otherwise both laws
would be satised. (This problem was communicated to us by David Maslen.)
Suppose we have an innite sequence of mutually independent events A1; A2; : : :.
Let ai = P(Ai), and let r be a positive integer.
(a) Find an expression of the probability that none of the Ai with i > r
occur.
(b) Use the fact that x  1  ex to show that
P(No Ai with i > r occurs)  e
P1
i=r ai
(c) (The rst Borel-Cantelli lemma) Prove that if P1
i=1 ai diverges, then
P(innitely many Ai occur) = 1:
Now, let Xi be a sequence of mutually independent random variables
such that for each positive integer i  2,
P(Xi = i) = 1
2i logi; P(Xi = i) = 1
2i logi; P(Xi = 0) = 1 1
i logi:
When i = 1 we let Xi = 0 with probability 1. As usual we let Sn =
X1 + 
 
 
 + Xn. Note that the mean of each Xi is 0.",prob.pdf
"8.1. DISCRETE RANDOM VARIABLES 315
(d) Find the variance of Sn.
(e) Show that the sequence hXii satises the Weak Law of Large Numbers,
i.e. prove that for any  > 0
P




Sn
n



 

! 0 ;
as n tends to innity.
We now show that fXig does not satisfy the Strong Law of Large Num-
bers. Suppose that Sn=n ! 0. Then because
Xn
n = Sn
n  n  1
n
Sn1
n  1 ;
we know that Xn=n ! 0. From the denition of limits, we conclude that
the inequality jXij  1
2i can only be true for nitely many i.
(f) Let Ai be the event jXij  1
2i. Find P(Ai). Show that P1
i=1 P(Ai)
diverges (use the Integral Test).
(g) Prove that Ai occurs for innitely many i.
(h) Prove that
P
Sn
n ! 0

= 0;
and hence that the Strong Law of Large Numbers fails for the sequence
fXig.
*17 Let us toss a biased coin that comes up heads with probability p and assume
the validity of the Strong Law of Large Numbers as described in Exercise 15.
Then, with probability 1,
Sn
n ! p",prob.pdf
"the validity of the Strong Law of Large Numbers as described in Exercise 15.
Then, with probability 1,
Sn
n ! p
as n ! 1. If f(x) is a continuous function on the unit interval, then we also
have
f
Sn
n

! f(p) :
Finally, we could hope that
E

f
Sn
n

! E(f(p)) = f(p) :
Show that, if all this is correct, as in fact it is, we would have proven that
any continuous function on the unit interval is a limit of polynomial func-
tions. This is a sketch of a probabilistic proof of an important theorem in
mathematics called the Weierstrass approximation theorem.",prob.pdf
"316 CHAPTER 8. LAW OF LARGE NUMBERS
8.2 Law of Large Numbers for Continuous Ran-
dom Variables
In the previous section we discussed in some detail the Law of Large Numbers for
discrete probability distributions. This law has a natural analogue for continuous
probability distributions, which we consider somewhat more brie
y here.
Chebyshev Inequality
Just as in the discrete case, we begin our discussion with the Chebyshev Inequality.
Theorem 8.3 (Chebyshev Inequality) Let X be a continuous random variable
with density function f(x). Suppose X has a nite expected value  = E(X) and
nite variance 2 = V (X). Then for any positive number  > 0 we have
P(jX  j  )  2
2 :
2
The proof is completely analogous to the proof in the discrete case, and we omit
it.
Note that this theorem says nothing if 2 = V (X) is innite.
Example 8.4 Let X be any continuous random variable with E(X) =  and
V (X) = 2. Then, if  = k = k standard deviations for some integer k, then
P(jX  j  k)  2",prob.pdf
"V (X) = 2. Then, if  = k = k standard deviations for some integer k, then
P(jX  j  k)  2
k22 = 1
k2 ;
just as in the discrete case. 2
Law of Large Numbers
With the Chebyshev Inequality we can now state and prove the Law of Large
Numbers for the continuous case.
Theorem 8.4 (Law of Large Numbers) Let X1, X2, . . . , Xn be an independent
trials process with a continuous density function f, nite expected value , and nite
variance 2. Let Sn = X1 + X2 + 
 
 
 + Xn be the sum of the Xi. Then for any real
number  > 0 we have
lim
n!1
P




Sn
n  



 

= 0 ;
or equivalently,
lim
n!1
P




Sn
n  



< 

= 1 :
2",prob.pdf
"8.2. CONTINUOUS RANDOM VARIABLES 317
Note that this theorem is not necessarily true if 2 is innite (see Example 8.8).
As in the discrete case, the Law of Large Numbers says that the average value
of n independent trials tends to the expected value as n ! 1, in the precise sense
that, given  > 0, the probability that the average value and the expected value
dier by more than  tends to 0 as n ! 1.
Once again, we suppress the proof, as it is identical to the proof in the discrete
case.
Uniform Case
Example 8.5 Suppose we choose at random n numbers from the interval [0; 1]
with uniform distribution. Then if Xi describes the ith choice, we have
 = E(Xi) =
Z 1
0
x dx = 1
2 ;
2 = V (Xi) =
Z 1
0
x2 dx  2
= 1
3  1
4 = 1
12 :
Hence,
E
Sn
n

= 1
2 ;
V
Sn
n

= 1
12n ;
and for any  > 0,
P




Sn
n  1
2



 

 1
12n2 :
This says that if we choose n numbers at random from [0; 1], then the chances",prob.pdf
"12n ;
and for any  > 0,
P




Sn
n  1
2



 

 1
12n2 :
This says that if we choose n numbers at random from [0; 1], then the chances
are better than 1  1=(12n2) that the dierence jSn=n  1=2j is less than . Note
that  plays the role of the amount of error we are willing to tolerate: If we choose
 = 0:1, say, then the chances that jSn=n  1=2j is less than 0.1 are better than
1  100=(12n). For n = 100, this is about .92, but if n = 1000, this is better than
.99 and if n = 10;000, this is better than .999.
We can illustrate what the Law of Large Numbers says for this example graph-
ically. The density for An = Sn=n is determined by
fAn (x) = nfSn (nx) :
We have seen in Section 7.2, that we can compute the density fSn (x) for the
sum of n uniform random variables. In Figure 8.2 we have used this to plot the
density for An for various values of n. We have shaded in the area for which An
would lie between .45 and .55. We see that as we increase n, we obtain more and",prob.pdf
"would lie between .45 and .55. We see that as we increase n, we obtain more and
more of the total area inside the shaded region. The Law of Large Numbers tells us
that we can obtain as much of the total area as we please inside the shaded region
by choosing n large enough (see also Figure 8.1). 2",prob.pdf
"318 CHAPTER 8. LAW OF LARGE NUMBERS
n=2 n=5 n=10
n=20 n=30 n=50
Figure 8.2: Illustration of Law of Large Numbers | uniform case.
Normal Case
Example 8.6 Suppose we choose n real numbers at random, using a normal dis-
tribution with mean 0 and variance 1. Then
 = E(Xi) = 0 ;
2 = V (Xi) = 1 :
Hence,
E
Sn
n

= 0 ;
V
Sn
n

= 1
n ;
and, for any  > 0,
P




Sn
n  0



 

 1
n2 :
In this case it is possible to compare the Chebyshev estimate for P(jSn=n j  )
in the Law of Large Numbers with exact values, since we know the density function
for Sn=n exactly (see Example 7.9). The comparison is shown in Table 8.1, for
 = :1. The data in this table was produced by the program LawContinuous. We
see here that the Chebyshev estimates are in general not very accurate. 2",prob.pdf
"8.2. CONTINUOUS RANDOM VARIABLES 319
n P(jSn=nj  :1) Chebyshev
100 .31731 1.00000
200 .15730 .50000
300 .08326 .33333
400 .04550 .25000
500 .02535 .20000
600 .01431 .16667
700 .00815 .14286
800 .00468 .12500
900 .00270 .11111
1000 .00157 .10000
Table 8.1: Chebyshev estimates.
Monte Carlo Method
Here is a somewhat more interesting example.
Example 8.7 Let g(x) be a continuous function dened for x 2 [0; 1] with values
in [0; 1]. In Section 2.1, we showed how to estimate the area of the region under
the graph of g(x) by the Monte Carlo method, that is, by choosing a large number
of random values for x and y with uniform distribution and seeing what fraction of
the points P(x; y) fell inside the region under the graph (see Example 2.2).
Here is a better way to estimate the same area (see Figure 8.3). Let us choose a
large number of independent values Xn at random from [0; 1] with uniform density,
set Yn = g(Xn), and nd the average value of the Yn. Then this average is our",prob.pdf
"set Yn = g(Xn), and nd the average value of the Yn. Then this average is our
estimate for the area. To see this, note that if the density function for Xn is
uniform,
 = E(Yn) =
Z 1
0
g(x)f(x) dx
=
Z 1
0
g(x) dx
= average value of g(x) ;
while the variance is
2 = E((Yn  )2) =
Z 1
0
(g(x)  )2 dx < 1 ;
since for all x in [0; 1], g(x) is in [0; 1], hence  is in [0; 1], and so jg(x)  j  1.
Now let An = (1=n)(Y1 + Y2 + 
 
 
 + Yn). Then by Chebyshev's Inequality, we have
P(jAn  j  )  2
n2 < 1
n2 :
This says that to get within  of the true value for  =
R1
0 g(x) dx with probability
at least p, we should choose n so that 1=n2  1  p (i.e., so that n  1=2(1  p)).
Note that this method tells us how large to take n to get a desired accuracy. 2",prob.pdf
"320 CHAPTER 8. LAW OF LARGE NUMBERS
Y
X
Y = g (x)
0
1
1
Figure 8.3: Area problem.
The Law of Large Numbers requires that the variance 2 of the original under-
lying density be nite: 2 < 1. In cases where this fails to hold, the Law of Large
Numbers may fail, too. An example follows.
Cauchy Case
Example 8.8 Suppose we choose n numbers from (1; +1) with a Cauchy den-
sity with parameter a = 1. We know that for the Cauchy density the expected value
and variance are undened (see Example 6.28). In this case, the density function
for
An = Sn
n
is given by (see Example 7.6)
fAn (x) = 1
(1 + x2) ;
that is, the density function for An is the same for all n. In this case, as n increases,
the density function does not change at all, and the Law of Large Numbers does
not hold. 2
Exercises
1 Let X be a continuous random variable with mean  = 10 and variance
2 = 100=3. Using Chebyshev's Inequality, nd an upper bound for the
following probabilities.",prob.pdf
"8.2. CONTINUOUS RANDOM VARIABLES 321
(a) P(jX  10j  2).
(b) P(jX  10j  5).
(c) P(jX  10j  9).
(d) P(jX  10j  20).
2 Let X be a continuous random variable with values unformly distributed over
the interval [0; 20].
(a) Find the mean and variance of X.
(b) Calculate P(jX  10j  2), P(jX  10j  5), P(jX  10j  9), and
P(jX  10j  20) exactly. How do your answers compare with those of
Exercise 1? How good is Chebyshev's Inequality in this case?
3 Let X be the random variable of Exercise 2.
(a) Calculate the function f(x) = P(jX  10j  x).
(b) Now graph the function f(x), and on the same axes, graph the Chebyshev
function g(x) = 100=(3x2). Show that f(x)  g(x) for all x > 0, but
that g(x) is not a very good approximation for f(x).
4 Let X be a continuous random variable with values exponentially distributed
over [0; 1) with parameter  = 0:1.
(a) Find the mean and variance of X.
(b) Using Chebyshev's Inequality, nd an upper bound for the following",prob.pdf
"over [0; 1) with parameter  = 0:1.
(a) Find the mean and variance of X.
(b) Using Chebyshev's Inequality, nd an upper bound for the following
probabilities: P(jX  10j  2), P(jX  10j  5), P(jX  10j  9), and
P(jX  10j  20).
(c) Calculate these probabilities exactly, and compare with the bounds in
(b).
5 Let X be a continuous random variable with values normally distributed over
(1; +1) with mean  = 0 and variance 2 = 1.
(a) Using Chebyshev's Inequality, nd upper bounds for the following prob-
abilities: P(jXj  1), P(jXj  2), and P(jXj  3).
(b) The area under the normal curve between 1 and 1 is .6827, between
2 and 2 is .9545, and between 3 and 3 it is .9973 (see the table in
Appendix A). Compare your bounds in (a) with these exact values. How
good is Chebyshev's Inequality in this case?
6 If X is normally distributed, with mean  and variance 2, nd an upper
bound for the following probabilities, using Chebyshev's Inequality.
(a) P(jX  j  ).
(b) P(jX  j  2).",prob.pdf
"bound for the following probabilities, using Chebyshev's Inequality.
(a) P(jX  j  ).
(b) P(jX  j  2).
(c) P(jX  j  3).",prob.pdf
"322 CHAPTER 8. LAW OF LARGE NUMBERS
(d) P(jX  j  4).
Now nd the exact value using the program NormalArea or the normal table
in Appendix A, and compare.
7 If X is a random variable with mean  6= 0 and variance 2, dene the relative
deviation D of X from its mean by
D =




X  




 :
(a) Show that P(D  a)  2=(2a2).
(b) If X is the random variable of Exercise 1, nd an upper bound for P(D 
:2), P(D  :5), P(D  :9), and P(D  2).
8 Let X be a continuous random variable and dene the standardized version
X of X by:
X = X  
 :
(a) Show that P(jXj  a)  1=a2.
(b) If X is the random variable of Exercise 1, nd bounds for P(jXj  2),
P(jXj  5), and P(jXj  9).
9 (a) Suppose a number X is chosen at random from [0; 20] with uniform
probability. Find a lower bound for the probability that X lies between
8 and 12, using Chebyshev's Inequality.
(b) Now suppose 20 real numbers are chosen independently from [0; 20] with",prob.pdf
"8 and 12, using Chebyshev's Inequality.
(b) Now suppose 20 real numbers are chosen independently from [0; 20] with
uniform probability. Find a lower bound for the probability that their
average lies between 8 and 12.
(c) Now suppose 100 real numbers are chosen independently from [0; 20].
Find a lower bound for the probability that their average lies between
8 and 12.
10 A student's score on a particular calculus nal is a random variable with values
of [0; 100], mean 70, and variance 25.
(a) Find a lower bound for the probability that the student's score will fall
between 65 and 75.
(b) If 100 students take the nal, nd a lower bound for the probability that
the class average will fall between 65 and 75.
11 The Pilsdor beer company runs a 
eet of trucks along the 100 mile road from
Hangtown to Dry Gulch, and maintains a garage halfway in between. Each
of the trucks is apt to break down at a point X miles from Hangtown, where",prob.pdf
"Hangtown to Dry Gulch, and maintains a garage halfway in between. Each
of the trucks is apt to break down at a point X miles from Hangtown, where
X is a random variable uniformly distributed over [0; 100].
(a) Find a lower bound for the probability P(jX  50j  10).",prob.pdf
"8.2. CONTINUOUS RANDOM VARIABLES 323
(b) Suppose that in one bad week, 20 trucks break down. Find a lower bound
for the probability P(jA20  50j  10), where A20 is the average of the
distances from Hangtown at the time of breakdown.
12 A share of common stock in the Pilsdor beer company has a price Yn on
the nth business day of the year. Finn observes that the price change Xn =
Yn+1  Yn appears to be a random variable with mean  = 0 and variance
2 = 1=4. If Y1 = 30, nd a lower bound for the following probabilities, under
the assumption that the Xn's are mutually independent.
(a) P(25  Y2  35).
(b) P(25  Y11  35).
(c) P(25  Y101  35).
13 Suppose one hundred numbers X1, X2, . . . , X100 are chosen independently at
random from [0; 20]. Let S = X1 + X2 + 
 
 
 + X100 be the sum, A = S=100
the average, and S = (S  1000)=(10=
p
3) the standardized sum. Find lower
bounds for the probabilities
(a) P(jS  1000j  100).
(b) P(jA  10j  1).
(c) P(jSj 
p
3).",prob.pdf
"p
3) the standardized sum. Find lower
bounds for the probabilities
(a) P(jS  1000j  100).
(b) P(jA  10j  1).
(c) P(jSj 
p
3).
14 Let X be a continuous random variable normally distributed on (1; +1)
with mean 0 and variance 1. Using the normal table provided in Appendix A,
or the program NormalArea, nd values for the function f(x) = P(jXj  x)
as x increases from 0 to 4.0 in steps of .25. Note that for x  0 the table gives
NA(0; x) = P(0  X  x) and thus P(jXj  x) = 2(:5  NA(0; x). Plot by
hand the graph of f(x) using these values, and the graph of the Chebyshev
function g(x) = 1=x2, and compare (see Exercise 3).
15 Repeat Exercise 14, but this time with mean 10 and variance 3. Note that
the table in Appendix A presents values for a standard normal variable. Find
the standardized version X for X, nd values for f(x) = P(jXj  x) as in
Exercise 14, and then rescale these values for f(x) = P(jX 10j  x). Graph",prob.pdf
"Exercise 14, and then rescale these values for f(x) = P(jX 10j  x). Graph
and compare this function with the Chebyshev function g(x) = 3=x2.
16 Let Z = X=Y where X and Y have normal densities with mean 0 and standard
deviation 1. Then it can be shown that Z has a Cauchy density.
(a) Write a program to illustrate this result by plotting a bar graph of 1000
samples obtained by forming the ratio of two standard normal outcomes.
Compare your bar graph with the graph of the Cauchy density. Depend-
ing upon which computer language you use, you may or may not need to
tell the computer how to simulate a normal random variable. A method
for doing this was described in Section 5.2.",prob.pdf
"324 CHAPTER 8. LAW OF LARGE NUMBERS
(b) We have seen that the Law of Large Numbers does not apply to the
Cauchy density (see Example 8.8). Simulate a large number of experi-
ments with Cauchy density and compute the average of your results. Do
these averages seem to be approaching a limit? If so can you explain
why this might be?
17 Show that, if X  0, then P(X  a)  E(X)=a.
18 (Lamperti9) Let X be a non-negative random variable. What is the best
upper bound you can give for P(X  a) if you know
(a) E(X) = 20.
(b) E(X) = 20 and V (X) = 25.
(c) E(X) = 20, V (X) = 25, and X is symmetric about its mean.
9Private communication.",prob.pdf
"Chapter 9
Central Limit Theorem
9.1 Central Limit Theorem for Bernoulli Trials
The second fundamental theorem of probability is the Central Limit Theorem. This
theorem says that if Sn is the sum of n mutually independent random variables, then
the distribution function of Sn is well-approximated by a certain type of continuous
function known as a normal density function, which is given by the formula
f;(x) = 1p
2e(x)2=(22 ) ;
as we have seen in Chapter 4.3. In this section, we will deal only with the case that
 = 0 and  = 1. We will call this particular normal density function the standard
normal density, and we will denote it by (x):
(x) = 1p
2
ex2=2 :
A graph of this function is given in Figure 9.1. It can be shown that the area under
any normal density equals 1.
The Central Limit Theorem tells us, quite generally, what happens when we
have the sum of a large number of independent random variables each of which con-",prob.pdf
"have the sum of a large number of independent random variables each of which con-
tributes a small amount to the total. In this section we shall discuss this theorem
as it applies to the Bernoulli trials and in Section 9.2 we shall consider more general
processes. We will discuss the theorem in the case that the individual random vari-
ables are identically distributed, but the theorem is true, under certain conditions,
even if the individual random variables have dierent distributions.
Bernoulli Trials
Consider a Bernoulli trials process with probability p for success on each trial.
Let Xi = 1 or 0 according as the ith outcome is a success or failure, and let
Sn = X1 +X2 +
 
 
+Xn. Then Sn is the number of successes in n trials. We know
that Sn has as its distribution the binomial probabilities b(n; p; j). In Section 3.2,
325",prob.pdf
"326 CHAPTER 9. CENTRAL LIMIT THEOREM
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
Figure 9.1: Standard normal density.
we plotted these distributions for p = :3 and p = :5 for various values of n (see
Figure 3.5).
We note that the maximum values of the distributions appeared near the ex-
pected value np, which causes their spike graphs to drift o to the right as n in-
creased. Moreover, these maximum values approach 0 as n increased, which causes
the spike graphs to 
atten out.
Standardized Sums
We can prevent the drifting of these spike graphs by subtracting the expected num-
ber of successes np from Sn, obtaining the new random variable Sn  np. Now the
maximum values of the distributions will always be near 0.
To prevent the spreading of these spike graphs, we can normalize Sn np to have
variance 1 by dividing by its standard deviation pnpq (see Exercise 6.2.12 and Ex-
ercise 6.2.16).
Denition 9.1 The standardized sum of Sn is given by
S
n = Sn  nppnpq :
S",prob.pdf
"ercise 6.2.16).
Denition 9.1 The standardized sum of Sn is given by
S
n = Sn  nppnpq :
S
n always has expected value 0 and variance 1. 2
Suppose we plot a spike graph with the spikes placed at the possible values of S
n:
x0, x1, . . . , xn, where
xj = j  nppnpq : (9.1)
We make the height of the spike at xj equal to the distribution value b(n; p; j). An
example of this standardized spike graph, with n = 270 and p = :3, is shown in
Figure 9.2. This graph is beautifully bell-shaped. We would like to t a normal
density to this spike graph. The obvious choice to try is the standard normal density,
since it is centered at 0, just as the standardized spike graph is. In this gure, we",prob.pdf
"9.1. BERNOULLI TRIALS 327
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
Figure 9.2: Normalized binomial distribution and standard normal density.
have drawn this standard normal density. The reader will note that a horrible thing
has occurred: Even though the shapes of the two graphs are the same, the heights
are quite dierent.
If we want the two graphs to t each other, we must modify one of them; we
choose to modify the spike graph. Since the shapes of the two graphs look fairly
close, we will attempt to modify the spike graph without changing its shape. The
reason for the diering heights is that the sum of the heights of the spikes equals
1, while the area under the standard normal density equals 1. If we were to draw a
continuous curve through the top of the spikes, and nd the area under this curve,
we see that we would obtain, approximately, the sum of the heights of the spikes
multiplied by the distance between consecutive spikes, which we will call . Since",prob.pdf
"multiplied by the distance between consecutive spikes, which we will call . Since
the sum of the heights of the spikes equals one, the area under this curve would be
approximately . Thus, to change the spike graph so that the area under this curve
has value 1, we need only multiply the heights of the spikes by 1=. It is easy to see
from Equation 9.1 that
 = 1pnpq :
In Figure 9.3 we show the standardized sum S
n for n = 270 and p = :3, after
correcting the heights, together with the standard normal density. (This gure was
produced with the program CLTBernoulliPlot.) The reader will note that the
standard normal ts the height-corrected spike graph extremely well. In fact, one
version of the Central Limit Theorem (see Theorem 9.1) says that as n increases,
the standard normal density will do an increasingly better job of approximating
the height-corrected spike graphs corresponding to a Bernoulli trials process with
n summands.",prob.pdf
"the height-corrected spike graphs corresponding to a Bernoulli trials process with
n summands.
Let us x a value x on the x-axis and let n be a xed positive integer. Then,
using Equation 9.1, the point xj that is closest to x has a subscript j given by the",prob.pdf
"328 CHAPTER 9. CENTRAL LIMIT THEOREM
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
Figure 9.3: Corrected spike graph with standard normal density.
formula
j = hnp + xpnpqi ;
where hai means the integer nearest to a. Thus the height of the spike above xj
will be pnpq b(n; p; j) = pnpq b(n; p; hnp + xj
pnpqi) :
For large n, we have seen that the height of the spike is very close to the height of
the normal density at x. This suggests the following theorem.
Theorem 9.1 (Central Limit Theorem for Binomial Distributions) For the
binomial distribution b(n; p; j) we have
lim
n!1
pnpq b(n; p; hnp + xpnpqi) = (x) ;
where (x) is the standard normal density.
The proof of this theorem can be carried out using Stirling's approximation from
Section 3.1. We indicate this method of proof by considering the case x = 0. In
this case, the theorem states that
lim
n!1
pnpq b(n; p; hnpi) = 1p
2 = :3989 : : : :
In order to simplify the calculation, we assume that np is an integer, so that hnpi =
np. Then",prob.pdf
"lim
n!1
pnpq b(n; p; hnpi) = 1p
2 = :3989 : : : :
In order to simplify the calculation, we assume that np is an integer, so that hnpi =
np. Then
pnpq b(n; p; np) = pnpq pnpqnq n!
(np)! (nq)! :
Recall that Stirling's formula (see Theorem 3.3) states that
n! 
p
2n nnen as n ! 1 :",prob.pdf
"9.1. BERNOULLI TRIALS 329
Using this, we have
pnpq b(n; p; np) 
pnpq pnpqnqp
2n nnen
p2npp2nq (np)np(nq)nqenpenq ;
which simplies to 1=
p
2. 2
Approximating Binomial Distributions
We can use Theorem 9.1 to nd approximations for the values of binomial distri-
bution functions. If we wish to nd an approximation for b(n; p; j), we set
j = np + xpnpq
and solve for x, obtaining
x = j  nppnpq :
Theorem 9.1 then says that pnpq b(n; p; j)
is approximately equal to (x), so
b(n; p; j)  (x)pnpq
= 1pnpq
j  nppnpq

:
Example 9.1 Let us estimate the probability of exactly 55 heads in 100 tosses of
a coin. For this case np = 100 
 1=2 = 50 and pnpq =
p
100 
 1=2 
 1=2 = 5. Thus
x55 = (55  50)=5 = 1 and
P(S100 = 55)  (1)
5 = 1
5
 1p
2e1=2

= :0484 :
To four decimal places, the actual value is .0485, and so the approximation is
very good. 2
The program CLTBernoulliLocal illustrates this approximation for any choice",prob.pdf
"very good. 2
The program CLTBernoulliLocal illustrates this approximation for any choice
of n, p, and j. We have run this program for two examples. The rst is the
probability of exactly 50 heads in 100 tosses of a coin; the estimate is .0798, while the
actual value, to four decimal places, is .0796. The second example is the probability
of exactly eight sixes in 36 rolls of a die; here the estimate is .1093, while the actual
value, to four decimal places, is .1196.",prob.pdf
"330 CHAPTER 9. CENTRAL LIMIT THEOREM
The individual binomial probabilities tend to 0 as n tends to innity. In most
applications we are not interested in the probability that a specic outcome occurs,
but rather in the probability that the outcome lies in a given interval, say the interval
[a; b]. In order to nd this probability, we add the heights of the spike graphs for
values of j between a and b. This is the same as asking for the probability that the
standardized sum S
n lies between a and b, where a and b are the standardized
values of a and b. But as n tends to innity the sum of these areas could be expected
to approach the area under the standard normal density between a and b. The
Central Limit Theorem states that this does indeed happen.
Theorem 9.2 (Central Limit Theorem for Bernoulli Trials) Let Sn be the
number of successes in n Bernoulli trials with probability p for success, and let a
and b be two xed real numbers. Then
lim
n!1
P

a  Sn  nppnpq  b

=
Z b
a",prob.pdf
"and b be two xed real numbers. Then
lim
n!1
P

a  Sn  nppnpq  b

=
Z b
a
(x) dx :
2
This theorem can be proved by adding together the approximations to b(n; p; k)
given in Theorem 9.1.It is also a special case of the more general Central Limit
Theorem (see Section 10.3).
We know from calculus that the integral on the right side of this equation is
equal to the area under the graph of the standard normal density (x) between
a and b. We denote this area by NA(a; b). Unfortunately, there is no simple way
to integrate the function ex2=2, and so we must either use a table of values or else
a numerical integration program. (See Figure 9.4 for values of NA(0; z). A more
extensive table is given in Appendix A.)
It is clear from the symmetry of the standard normal density that areas such as
that between 2 and 3 can be found from this table by adding the area from 0 to 2
(same as that from 2 to 0) to the area from 0 to 3.
Approximation of Binomial Probabilities",prob.pdf
"(same as that from 2 to 0) to the area from 0 to 3.
Approximation of Binomial Probabilities
Suppose that Sn is binomially distributed with parameters n and p. We have seen
that the above theorem shows how to estimate a probability of the form
P(i  Sn  j) ; (9.2)
where i and j are integers between 0 and n. As we have seen, the binomial distri-
bution can be represented as a spike graph, with spikes at the integers between 0
and n, and with the height of the kth spike given by b(n; p; k). For moderate-sized
values of n, if we standardize this spike graph, and change the heights of its spikes,
in the manner described above, the sum of the heights of the spikes is approximated
by the area under the standard normal density between i and j. It turns out that
a slightly more accurate approximation is aorded by the area under the standard",prob.pdf
"9.1. BERNOULLI TRIALS 331
NA (0,z) = area of 
shaded region 
0 z
z       NA(z)           z       NA(z)            z        NA(z)            z       NA(z)   
.0    .0000            1.0    .3413             2.0    .4772             3.0   .4987 
.1    .0398            1.1    .3643             2.1    .4821             3.1   .4990 
.2    .0793            1.2    .3849             2.2    .4861             3.2   .4993     
.3    .1179            1.3    .4032             2.3    .4893             3.3   .4995
.4    .1554            1.4    .4192             2.4    .4918             3.4   .4997
.5    .1915            1.5    .4332             2.5    .4938             3.5   .4998
.6    .2257            1.6    .4452             2.6    .4953             3.6   .4998
.7    .2580            1.7    .4554             2.7    .4965             3.7   .4999
.8    .2881            1.8    .4641             2.8    .4974             3.8   .4999",prob.pdf
".8    .2881            1.8    .4641             2.8    .4974             3.8   .4999
.9    .3159            1.9    .4713             2.9    .4981             3.9   .5000  
Figure 9.4: Table of values of NA(0; z), the normal area from 0 to z.",prob.pdf
"332 CHAPTER 9. CENTRAL LIMIT THEOREM
normal density between the standardized values corresponding to (i  1=2) and
(j + 1=2); these values are
i = i  1=2  nppnpq
and
j = j + 1=2  nppnpq :
Thus,
P(i  Sn  j)  NA
 
i  1
2  nppnpq ; j + 1
2  nppnpq
!
:
It should be stressed that the approximations obtained by using the Central Limit
Theorem are only approximations, and sometimes they are not very close to the
actual values (see Exercise 12).
We now illustrate this idea with some examples.
Example 9.2 A coin is tossed 100 times. Estimate the probability that the number
of heads lies between 40 and 60 (the word \between"" in mathematics means inclusive
of the endpoints). The expected number of heads is 100
1=2 = 50, and the standard
deviation for the number of heads is
p
100 
 1=2 
 1=2 = 5. Thus, since n = 100 is
reasonably large, we have
P(40  Sn  60)  P
39:5  50
5  S
n  60:5  50
5

= P(2:1  S
n  2:1)
 NA(2:1; 2:1)
= 2NA(0; 2:1)
 :9642 :",prob.pdf
"reasonably large, we have
P(40  Sn  60)  P
39:5  50
5  S
n  60:5  50
5

= P(2:1  S
n  2:1)
 NA(2:1; 2:1)
= 2NA(0; 2:1)
 :9642 :
The actual value is .96480, to ve decimal places.
Note that in this case we are asking for the probability that the outcome will
not deviate by more than two standard deviations from the expected value. Had
we asked for the probability that the number of successes is between 35 and 65, this
would have represented three standard deviations from the mean, and, using our
1/2 correction, our estimate would be the area under the standard normal curve
between 3:1 and 3.1, or 2NA(0; 3:1) = :9980. The actual answer in this case, to
ve places, is .99821. 2
It is important to work a few problems by hand to understand the conversion
from a given inequality to an inequality relating to the standardized variable. After
this, one can then use a computer program that carries out this conversion, including",prob.pdf
"this, one can then use a computer program that carries out this conversion, including
the 1/2 correction. The program CLTBernoulliGlobal is such a program for
estimating probabilities of the form P(a  Sn  b).
Example 9.3 Dartmouth College would like to have 1050 freshmen. This college
cannot accommodate more than 1060. Assume that each applicant accepts with",prob.pdf
"9.1. BERNOULLI TRIALS 333
probability .6 and that the acceptances can be modeled by Bernoulli trials. If the
college accepts 1700, what is the probability that it will have too many acceptances?
If it accepts 1700 students, the expected number of students who matricu-
late is :6 
 1700 = 1020. The standard deviation for the number that accept isp
1700 
 :6 
 :4  20. Thus we want to estimate the probability
P(S1700 > 1060) = P(S1700  1061)
= P

S
1700  1060:5  1020
20

= P(S
1700  2:025) :
From Table 9.4, if we interpolate, we would estimate this probability to be
:5  :4784 = :0216. Thus, the college is fairly safe using this admission policy. 2
Applications to Statistics
There are many important questions in the eld of statistics that can be answered
using the Central Limit Theorem for independent trials processes. The following
example is one that is encountered quite frequently in the news. Another example",prob.pdf
"example is one that is encountered quite frequently in the news. Another example
of an application of the Central Limit Theorem to statistics is given in Section 9.2.
Example 9.4 One frequently reads that a poll has been taken to estimate the pro-
portion of people in a certain population who favor one candidate over another in
a race with two candidates. (This model also applies to races with more than two
candidates A and B, and two ballot propositions.) Clearly, it is not possible for
pollsters to ask everyone for their preference. What is done instead is to pick a
subset of the population, called a sample, and ask everyone in the sample for their
preference. Let p be the actual proportion of people in the population who are in
favor of candidate A and let q = 1p. If we choose a sample of size n from the pop-
ulation, the preferences of the people in the sample can be represented by random
variables X1; X2; : : : ; Xn, where Xi = 1 if person i is in favor of candidate A, and",prob.pdf
"variables X1; X2; : : : ; Xn, where Xi = 1 if person i is in favor of candidate A, and
Xi = 0 if person i is in favor of candidate B. Let Sn = X1 + X2 + 
 
 
 + Xn. If each
subset of size n is chosen with the same probability, then Sn is hypergeometrically
distributed. If n is small relative to the size of the population (which is typically
true in practice), then Sn is approximately binomially distributed, with parameters
n and p.
The pollster wants to estimate the value p. An estimate for p is provided by the
value p = Sn=n, which is the proportion of people in the sample who favor candidate
B. The Central Limit Theorem says that the random variable p is approximately
normally distributed. (In fact, our version of the Central Limit Theorem says that
the distribution function of the random variable
S
n = Sn  nppnpq
is approximated by the standard normal density.) But we have
p = Sn  nppnpq
rpq
n + p ;",prob.pdf
"334 CHAPTER 9. CENTRAL LIMIT THEOREM
i.e., p is just a linear function of S
n. Since the distribution of S
n is approximated
by the standard normal density, the distribution of the random variable p must also
be bell-shaped. We also know how to write the mean and standard deviation of p
in terms of p and n. The mean of p is just p, and the standard deviation is
rpq
n :
Thus, it is easy to write down the standardized version of p; it is
p = p  pp
pq=n
:
Since the distribution of the standardized version of p is approximated by the
standard normal density, we know, for example, that 95% of its values will lie within
two standard deviations of its mean, and the same is true of p. So we have
P

p  2
rpq
n < p < p + 2
rpq
n

 :954 :
Now the pollster does not know p or q, but he can use p and q = 1  p in their
place without too much danger. With this idea in mind, the above statement is
equivalent to the statement
P
 
p  2
r
pq
n < p < p + 2
r
pq
n
!
 :954 :",prob.pdf
"equivalent to the statement
P
 
p  2
r
pq
n < p < p + 2
r
pq
n
!
 :954 :
The resulting interval 
p  2ppqpn ; p + 2ppqpn

is called the 95 percent condence interval for the unknown value of p. The name
is suggested by the fact that if we use this method to estimate p in a large number
of samples we should expect that in about 95 percent of the samples the true value
of p is contained in the condence interval obtained from the sample. In Exercise 11
you are asked to write a program to illustrate that this does indeed happen.
The pollster has control over the value of n. Thus, if he wants to create a 95%
condence interval with length 6%, then he should choose a value of n so that
2ppqpn  :03 :
Using the fact that pq  1=4, no matter what the value of p is, it is easy to show
that if he chooses a value of n so that
1pn  :03 ;
he will be safe. This is equivalent to choosing
n  1111 :",prob.pdf
"9.1. BERNOULLI TRIALS 335
0.48 0.5 0.52 0.54 0.56 0.58 0.6
0
5
10
15
20
25
Figure 9.5: Polling simulation.
So if the pollster chooses n to be 1200, say, and calculates p using his sample of size
1200, then 19 times out of 20 (i.e., 95% of the time), his condence interval, which
is of length 6%, will contain the true value of p. This type of condence interval
is typically reported in the news as follows: this survey has a 3% margin of error.
In fact, most of the surveys that one sees reported in the paper will have sample
sizes around 1000. A somewhat surprising fact is that the size of the population has
apparently no eect on the sample size needed to obtain a 95% condence interval
for p with a given margin of error. To see this, note that the value of n that was
needed depended only on the number .03, which is the margin of error. In other
words, whether the population is of size 100,000 or 100,000,000, the pollster needs",prob.pdf
"words, whether the population is of size 100,000 or 100,000,000, the pollster needs
only to choose a sample of size 1200 or so to get the same accuracy of estimate of
p. (We did use the fact that the sample size was small relative to the population
size in the statement that Sn is approximately binomially distributed.)
In Figure 9.5, we show the results of simulating the polling process. The popula-
tion is of size 100,000, and for the population, p = :54. The sample size was chosen
to be 1200. The spike graph shows the distribution of p for 10,000 randomly chosen
samples. For this simulation, the program kept track of the number of samples for
which p was within 3% of .54. This number was 9648, which is close to 95% of the
number of samples used.
Another way to see what the idea of condence intervals means is shown in
Figure 9.6. In this gure, we show 100 condence intervals, obtained by computing
p for 100 dierent samples of size 1200 from the same population as before. The",prob.pdf
"p for 100 dierent samples of size 1200 from the same population as before. The
reader can see that most of these condence intervals (96, to be exact) contain the
true value of p.
The Gallup Poll has used these polling techniques in every Presidential election
since 1936 (and in innumerable other elections as well). Table 9.11 shows the results
1The Gallup Poll Monthly, November 1992, No. 326, p. 33. Supplemented with the help of",prob.pdf
"336 CHAPTER 9. CENTRAL LIMIT THEOREM
0.48 0.5 0.52 0.54 0.56 0.58 0.6
Figure 9.6: Condence interval simulation.
of their eorts. The reader will note that most of the approximations to p are within
3% of the actual value of p. The sample sizes for these polls were typically around
1500. (In the table, both the predicted and actual percentages for the winning
candidate refer to the percentage of the vote among the \major"" political parties.
In most elections, there were two major parties, but in several elections, there were
three.)
This technique also plays an important role in the evaluation of the eectiveness
of drugs in the medical profession. For example, it is sometimes desired to know
what proportion of patients will be helped by a new drug. This proportion can
be estimated by giving the drug to a subset of the patients, and determining the
proportion of this sample who are helped by the drug. 2
Historical Remarks",prob.pdf
"proportion of this sample who are helped by the drug. 2
Historical Remarks
The Central Limit Theorem for Bernoulli trials was rst proved by Abraham
de Moivre and appeared in his book, The Doctrine of Chances, rst published
in 1718.2
De Moivre spent his years from age 18 to 21 in prison in France because of his
Protestant background. When he was released he left France for England, where
he worked as a tutor to the sons of noblemen. Newton had presented a copy of
his Principia Mathematica to the Earl of Devonshire. The story goes that, while
de Moivre was tutoring at the Earl's house, he came upon Newton's work and found
that it was beyond him. It is said that he then bought a copy of his own and tore
Lydia K. Saab, The Gallup Organization.
2A. de Moivre, The Doctrine of Chances, 3d ed. (London: Millar, 1756).",prob.pdf
"9.1. BERNOULLI TRIALS 337
Year Winning Gallup Final Election Deviation
Candidate Survey Result
1936 Roosevelt 55.7% 62.5% 6.8%
1940 Roosevelt 52.0% 55.0% 3.0%
1944 Roosevelt 51.5% 53.3% 1.8%
1948 Truman 44.5% 49.9% 5.4%
1952 Eisenhower 51.0% 55.4% 4.4%
1956 Eisenhower 59.5% 57.8% 1.7%
1960 Kennedy 51.0% 50.1% 0.9%
1964 Johnson 64.0% 61.3% 2.7%
1968 Nixon 43.0% 43.5% 0.5%
1972 Nixon 62.0% 61.8% 0.2%
1976 Carter 48.0% 50.0% 2.0%
1980 Reagan 47.0% 50.8% 3.8%
1984 Reagan 59.0% 59.1% 0.1%
1988 Bush 56.0% 53.9% 2.1%
1992 Clinton 49.0% 43.2% 5.8%
1996 Clinton 52.0% 50.1% 1.9%
Table 9.1: Gallup Poll accuracy record.
it into separate pages, learning it page by page as he walked around London to his
tutoring jobs. De Moivre frequented the coeehouses in London, where he started
his probability work by calculating odds for gamblers. He also met Newton at such a
coeehouse and they became fast friends. De Moivre dedicated his book to Newton.",prob.pdf
"coeehouse and they became fast friends. De Moivre dedicated his book to Newton.
The Doctrine of Chances provides the techniques for solving a wide variety of
gambling problems. In the midst of these gambling problems de Moivre rather
modestly introduces his proof of the Central Limit Theorem, writing
A Method of approximating the Sum of the Terms of the Binomial
(a + b)n expanded into a Series, from whence are deduced some prac-
tical Rules to estimate the Degree of Assent which is to be given to
Experiments.3
De Moivre's proof used the approximation to factorials that we now call Stirling's
formula. De Moivre states that he had obtained this formula before Stirling but
without determining the exact value of the constant
p
2. While he says it is not
really necessary to know this exact value, he concedes that knowing it \has spread
a singular Elegancy on the Solution.""
The complete proof and an interesting discussion of the life of de Moivre can be",prob.pdf
"a singular Elegancy on the Solution.""
The complete proof and an interesting discussion of the life of de Moivre can be
found in the book Games, Gods and Gambling by F. N. David.4
3ibid., p. 243.
4F. N. David, Games, Gods and Gambling (London: Grin, 1962).",prob.pdf
"338 CHAPTER 9. CENTRAL LIMIT THEOREM
Exercises
1 Let S100 be the number of heads that turn up in 100 tosses of a fair coin. Use
the Central Limit Theorem to estimate
(a) P(S100  45).
(b) P(45 < S100 < 55).
(c) P(S100 > 63).
(d) P(S100 < 57).
2 Let S200 be the number of heads that turn up in 200 tosses of a fair coin.
Estimate
(a) P(S200 = 100).
(b) P(S200 = 90).
(c) P(S200 = 80).
3 A true-false examination has 48 questions. June has probability 3/4 of an-
swering a question correctly. April just guesses on each question. A passing
score is 30 or more correct answers. Compare the probability that June passes
the exam with the probability that April passes it.
4 Let S be the number of heads in 1,000,000 tosses of a fair coin. Use (a) Cheby-
shev's inequality, and (b) the Central Limit Theorem, to estimate the prob-
ability that S lies between 499,500 and 500,500. Use the same two methods
to estimate the probability that S lies between 499,000 and 501,000, and the",prob.pdf
"to estimate the probability that S lies between 499,000 and 501,000, and the
probability that S lies between 498,500 and 501,500.
5 A rookie is brought to a baseball club on the assumption that he will have a
.300 batting average. (Batting average is the ratio of the number of hits to the
number of times at bat.) In the rst year, he comes to bat 300 times and his
batting average is .267. Assume that his at bats can be considered Bernoulli
trials with probability .3 for success. Could such a low average be considered
just bad luck or should he be sent back to the minor leagues? Comment on
the assumption of Bernoulli trials in this situation.
6 Once upon a time, there were two railway trains competing for the passenger
trac of 1000 people leaving from Chicago at the same hour and going to Los
Angeles. Assume that passengers are equally likely to choose each train. How
many seats must a train have to assure a probability of .99 or better of having
a seat for each passenger?",prob.pdf
"many seats must a train have to assure a probability of .99 or better of having
a seat for each passenger?
7 Assume that, as in Example 9.3, Dartmouth admits 1750 students. What is
the probability of too many acceptances?
8 A club serves dinner to members only. They are seated at 12-seat tables. The
manager observes over a long period of time that 95 percent of the time there
are between six and nine full tables of members, and the remainder of the",prob.pdf
"9.1. BERNOULLI TRIALS 339
time the numbers are equally likely to fall above or below this range. Assume
that each member decides to come with a given probability p, and that the
decisions are independent. How many members are there? What is p?
9 Let Sn be the number of successes in n Bernoulli trials with probability .8 for
success on each trial. Let An = Sn=n be the average number of successes. In
each case give the value for the limit, and give a reason for your answer.
(a) limn!1 P(An = :8).
(b) limn!1 P(:7n < Sn < :9n).
(c) limn!1 P(Sn < :8n + :8pn).
(d) limn!1 P(:79 < An < :81).
10 Find the probability that among 10,000 random digits the digit 3 appears not
more than 931 times.
11 Write a computer program to simulate 10,000 Bernoulli trials with probabil-
ity .3 for success on each trial. Have the program compute the 95 percent
condence interval for the probability of success based on the proportion of
successes. Repeat the experiment 100 times and see how many times the true",prob.pdf
"successes. Repeat the experiment 100 times and see how many times the true
value of .3 is included within the condence limits.
12 A balanced coin is 
ipped 400 times. Determine the number x such that
the probability that the number of heads is between 200  x and 200 + x is
approximately .80.
13 A noodle machine in Spumoni's spaghetti factory makes about 5 percent de-
fective noodles even when properly adjusted. The noodles are then packed
in crates containing 1900 noodles each. A crate is examined and found to
contain 115 defective noodles. What is the approximate probability of nding
at least this many defective noodles if the machine is properly adjusted?
14 A restaurant feeds 400 customers per day. On the average 20 percent of the
customers order apple pie.
(a) Give a range (called a 95 percent condence interval) for the number of
pieces of apple pie ordered on a given day such that you can be 95 percent
sure that the actual number will fall in this range.",prob.pdf
"pieces of apple pie ordered on a given day such that you can be 95 percent
sure that the actual number will fall in this range.
(b) How many customers must the restaurant have, on the average, to be at
least 95 percent sure that the number of customers ordering pie on that
day falls in the 19 to 21 percent range?
15 Recall that if X is a random variable, the cumulative distribution function
of X is the function F(x) dened by
F(x) = P(X  x) :
(a) Let Sn be the number of successes in n Bernoulli trials with probability p
for success. Write a program to plot the cumulative distribution for Sn.",prob.pdf
"340 CHAPTER 9. CENTRAL LIMIT THEOREM
(b) Modify your program in (a) to plot the cumulative distribution F 
n(x) of
the standardized random variable
S
n = Sn  nppnpq :
(c) Dene the normal distribution N(x) to be the area under the normal
curve up to the value x. Modify your program in (b) to plot the normal
distribution as well, and compare it with the cumulative distribution
of S
n. Do this for n = 10; 50, and 100.
16 In Example 3.11, we were interested in testing the hypothesis that a new form
of aspirin is eective 80 percent of the time rather than the 60 percent of the
time as reported for standard aspirin. The new aspirin is given to n people.
If it is eective in m or more cases, we accept the claim that the new drug
is eective 80 percent of the time and if not we reject the claim. Using the
Central Limit Theorem, show that you can choose the number of trials n and
the critical value m so that the probability that we reject the hypothesis when",prob.pdf
"the critical value m so that the probability that we reject the hypothesis when
it is true is less than .01 and the probability that we accept it when it is false
is also less than .01. Find the smallest value of n that will suce for this.
17 In an opinion poll it is assumed that an unknown proportion p of the people
are in favor of a proposed new law and a proportion 1  p are against it.
A sample of n people is taken to obtain their opinion. The proportion p in
favor in the sample is taken as an estimate of p. Using the Central Limit
Theorem, determine how large a sample will ensure that the estimate will,
with probability .95, be correct to within .01.
18 A description of a poll in a certain newspaper says that one can be 95%
condent that error due to sampling will be no more than plus or minus 3
percentage points. A poll in the New York Times taken in Iowa says that
\according to statistical theory, in 19 out of 20 cases the results based on such",prob.pdf
"\according to statistical theory, in 19 out of 20 cases the results based on such
samples will dier by no more than 3 percentage points in either direction
from what would have been obtained by interviewing all adult Iowans."" These
are both attempts to explain the concept of condence intervals. Do both
statements say the same thing? If not, which do you think is the more accurate
description?
9.2 Central Limit Theorem for Discrete Indepen-
dent Trials
We have illustrated the Central Limit Theorem in the case of Bernoulli trials, but
this theorem applies to a much more general class of chance processes. In particular,
it applies to any independent trials process such that the individual trials have nite
variance. For such a process, both the normal approximation for individual terms
and the Central Limit Theorem are valid.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 341
Let Sn = X1 + X2 + 
 
 
 + Xn be the sum of n independent discrete random
variables of an independent trials process with common distribution function m(x)
dened on the integers, with mean  and variance 2. We have seen in Section 7.2
that the distributions for such independent sums have shapes resembling the nor-
mal curve, but the largest values drift to the right and the curves 
atten out (see
Figure 7.6). We can prevent this just as we did for Bernoulli trials.
Standardized Sums
Consider the standardized random variable
S
n = Sn  np
n2 :
This standardizes Sn to have expected value 0 and variance 1. If Sn = j, then
S
n has the value xj with
xj = j  np
n2 :
We can construct a spike graph just as we did for Bernoulli trials. Each spike is
centered at some xj. The distance between successive spikes is
b = 1p
n2 ;
and the height of the spike is
h =
p
n2P(Sn = j) :",prob.pdf
"centered at some xj. The distance between successive spikes is
b = 1p
n2 ;
and the height of the spike is
h =
p
n2P(Sn = j) :
The case of Bernoulli trials is the special case for which Xj = 1 if the jth
outcome is a success and 0 otherwise; then  = p and 2 = ppq.
We now illustrate this process for two dierent discrete distributions. The rst
is the distribution m, given by
m =
1 2 3 4 5
:2 :2 :2 :2 :2

:
In Figure 9.7 we show the standardized sums for this distribution for the cases
n = 2 and n = 10. Even for n = 2 the approximation is surprisingly good.
For our second discrete distribution, we choose
m =
1 2 3 4 5
:4 :3 :1 :1 :1

:
This distribution is quite asymmetric and the approximation is not very good
for n = 3, but by n = 10 we again have an excellent approximation (see Figure 9.8).
Figures 9.7 and 9.8 were produced by the program CLTIndTrialsPlot.",prob.pdf
"342 CHAPTER 9. CENTRAL LIMIT THEOREM
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
n = 2 n = 10
Figure 9.7: Distribution of standardized sums.
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
n = 3 n = 10
Figure 9.8: Distribution of standardized sums.
Approximation Theorem
As in the case of Bernoulli trials, these graphs suggest the following approximation
theorem for the individual probabilities.
Theorem 9.3 Let X1, X2, . . . , Xn be an independent trials process and let Sn =
X1 + X2 + 
 
 
 + Xn. Assume that the greatest common divisor of the dierences of
all the values that the Xj can take on is 1. Let E(Xj) =  and V (Xj) = 2. Then
for n large,
P(Sn = j)  (xj)p
n2 ;
where xj = (j  n)=
p
n2, and (x) is the standard normal density. 2
The program CLTIndTrialsLocal implements this approximation. When we
run this program for 6 rolls of a die, and ask for the probability that the sum of the",prob.pdf
"run this program for 6 rolls of a die, and ask for the probability that the sum of the
rolls equals 21, we obtain an actual value of .09285, and a normal approximation
value of .09537. If we run this program for 24 rolls of a die, and ask for the
probability that the sum of the rolls is 72, we obtain an actual value of .01724
and a normal approximation value of .01705. These results show that the normal
approximations are quite good.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 343
Central Limit Theorem for a Discrete Independent Trials Pro-
cess
The Central Limit Theorem for a discrete independent trials process is as follows.
Theorem 9.4 (Central Limit Theorem) Let Sn = X1 + X2 + 
 
 
 + Xn be the
sum of n discrete independent random variables with common distribution having
expected value  and variance 2. Then, for a < b,
lim
n!1
P

a < Sn  n
p
n2 < b

= 1p
2
Z b
a
ex2=2 dx :
2
We will give the proofs of Theorems 9.3 and Theorem 9.4 in Section 10.3. Here
we consider several examples.
Examples
Example 9.5 A die is rolled 420 times. What is the probability that the sum of
the rolls lies between 1400 and 1550?
The sum is a random variable
S420 = X1 + X2 + 
 
 
 + X420 ;
where each Xj has distribution
mX =
 1 2 3 4 5 6
1=6 1=6 1=6 1=6 1=6 1=6

We have seen that  = E(X) = 7=2 and 2 = V (X) = 35=12. Thus, E(S420) =
420 
 7=2 = 1470, 2(S420) = 420 
 35=12 = 1225, and (S420) = 35. Therefore,
P(1400  S420  1550)  P",prob.pdf
"420 
 7=2 = 1470, 2(S420) = 420 
 35=12 = 1225, and (S420) = 35. Therefore,
P(1400  S420  1550)  P
1399:5  1470
35  S
420  1550:5  1470
35

= P(2:01  S
420  2:30)
 NA(2:01; 2:30) = :9670 :
We note that the program CLTIndTrialsGlobal could be used to calculate these
probabilities. 2
Example 9.6 A student's grade point average is the average of his grades in 30
courses. The grades are based on 100 possible points and are recorded as integers.
Assume that, in each course, the instructor makes an error in grading of k with
probability jp=kj, where k = 1, 2, 3, 4, 5. The probability of no error is
then 1  (137=30)p. (The parameter p represents the inaccuracy of the instructor's
grading.) Thus, in each course, there are two grades for the student, namely the",prob.pdf
"344 CHAPTER 9. CENTRAL LIMIT THEOREM
\correct"" grade and the recorded grade. So there are two average grades for the
student, namely the average of the correct grades and the average of the recorded
grades.
We wish to estimate the probability that these two average grades dier by less
than .05 for a given student. We now assume that p = 1=20. We also assume
that the total error is the sum S30 of 30 independent random variables each with
distribution
mX :
 5 4 3 2 1 0 1 2 3 4 5
1
100
1
80
1
60
1
40
1
20
463
600
1
20
1
40
1
60
1
80
1
100

:
One can easily calculate that E(X) = 0 and 2(X) = 1:5. Then we have
P

:05  S30
30  :05


= P(1:5  S30  1:5)
= P

1:5p
30
1:5  S
30  1:5p
30
1:5

= P(:224  S
30  :224)
 NA(:224; :224) = :1772 :
This means that there is only a 17.7% chance that a given student's grade point
average is accurate to within .05. (Thus, for example, if two candidates for valedic-",prob.pdf
"average is accurate to within .05. (Thus, for example, if two candidates for valedic-
torian have recorded averages of 97.1 and 97.2, there is an appreciable probability
that their correct averages are in the reverse order.) For a further discussion of this
example, see the article by R. M. Kozelka.5 2
A More General Central Limit Theorem
In Theorem 9.4, the discrete random variables that were being summed were as-
sumed to be independent and identically distributed. It turns out that the assump-
tion of identical distributions can be substantially weakened. Much work has been
done in this area, with an important contribution being made by J. W. Lindeberg.
Lindeberg found a condition on the sequence fXng which guarantees that the dis-
tribution of the sum Sn is asymptotically normally distributed. Feller showed that
Lindeberg's condition is necessary as well, in the sense that if the condition does
not hold, then the sum Sn is not asymptotically normally distributed. For a pre-",prob.pdf
"not hold, then the sum Sn is not asymptotically normally distributed. For a pre-
cise statement of Lindeberg's Theorem, we refer the reader to Feller.6 A sucient
condition that is stronger (but easier to state) than Lindeberg's condition, and is
weaker than the condition in Theorem 9.4, is given in the following theorem.
5R. M. Kozelka, \Grade-Point Averages and the Central Limit Theorem,"" American Math.
Monthly, vol. 86 (Nov 1979), pp. 773-777.
6W. Feller, Introduction to Probability Theory and its Applications, vol. 1, 3rd ed. (New York:
John Wiley & Sons, 1968), p. 254.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 345
Theorem 9.5 (Central Limit Theorem) Let X1; X2; : : : ; Xn ; : : : be a se-
quence of independent discrete random variables, and let Sn = X1 +X2 +
 
 
+Xn.
For each n, denote the mean and variance of Xn by n and 2
n, respectively. De-
ne the mean and variance of Sn to be mn and s2
n, respectively, and assume that
sn ! 1. If there exists a constant A, such that jXnj  A for all n, then for a < b,
lim
n!1
P

a < Sn  mn
sn
< b

= 1p
2
Z b
a
ex2=2 dx :
2
The condition that jXnj  A for all n is sometimes described by saying that the
sequence fXng is uniformly bounded. The condition that sn ! 1 is necessary (see
Exercise 15).
We illustrate this theorem by generating a sequence of n random distributions on
the interval [a; b]. We then convolute these distributions to nd the distribution of
the sum of n independent experiments governed by these distributions. Finally, we",prob.pdf
"the sum of n independent experiments governed by these distributions. Finally, we
standardize the distribution for the sum to have mean 0 and standard deviation 1
and compare it with the normal density. The program CLTGeneral carries out
this procedure.
In Figure 9.9 we show the result of running this program for [a; b] = [2; 4], and
n = 1; 4; and 10. We see that our rst random distribution is quite asymmetric.
By the time we choose the sum of ten such experiments we have a very good t to
the normal curve.
The above theorem essentially says that anything that can be thought of as being
made up as the sum of many small independent pieces is approximately normally
distributed. This brings us to one of the most important questions that was asked
about genetics in the 1800's.
The Normal Distribution and Genetics
When one looks at the distribution of heights of adults of one sex in a given pop-
ulation, one cannot help but notice that this distribution looks like the normal",prob.pdf
"ulation, one cannot help but notice that this distribution looks like the normal
distribution. An example of this is shown in Figure 9.10. This gure shows the
distribution of heights of 9593 women between the ages of 21 and 74. These data
come from the Health and Nutrition Examination Survey I (HANES I). For this
survey, a sample of the U.S. civilian population was chosen. The survey was carried
out between 1971 and 1974.
A natural question to ask is \How does this come about?"". Francis Galton,
an English scientist in the 19th century, studied this question, and other related
questions, and constructed probability models that were of great importance in
explaining the genetic eects on such attributes as height. In fact, one of the most
important ideas in statistics, the idea of regression to the mean, was invented by
Galton in his attempts to understand these genetic eects.
Galton was faced with an apparent contradiction. On the one hand, he knew",prob.pdf
"Galton in his attempts to understand these genetic eects.
Galton was faced with an apparent contradiction. On the one hand, he knew
that the normal distribution arises in situations in which many small independent
eects are being summed. On the other hand, he also knew that many quantitative",prob.pdf
"346 CHAPTER 9. CENTRAL LIMIT THEOREM
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
0.5
0.6
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
-4 -2 0 2 4
0
0.1
0.2
0.3
0.4
Figure 9.9: Sums of randomly chosen random variables.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 347
50 55 60 65 70 75 80
0
0.025
0.05
0.075
0.1
0.125
0.15
Figure 9.10: Distribution of heights of adult women.
attributes, such as height, are strongly in
uenced by genetic factors: tall parents
tend to have tall ospring. Thus in this case, there seem to be two large eects,
namely the parents. Galton was certainly aware of the fact that non-genetic factors
played a role in determining the height of an individual. Nevertheless, unless these
non-genetic factors overwhelm the genetic ones, thereby refuting the hypothesis
that heredity is important in determining height, it did not seem possible for sets of
parents of given heights to have ospring whose heights were normally distributed.
One can express the above problem symbolically as follows. Suppose that we
choose two specic positive real numbers x and y, and then nd all pairs of parents
one of whom is x units tall and the other of whom is y units tall. We then look",prob.pdf
"one of whom is x units tall and the other of whom is y units tall. We then look
at all of the ospring of these pairs of parents. One can postulate the existence of
a function f(x; y) which denotes the genetic eect of the parents' heights on the
heights of the ospring. One can then let W denote the eects of the non-genetic
factors on the heights of the ospring. Then, for a given set of heights fx; yg, the
random variable which represents the heights of the ospring is given by
H = f(x; y) + W ;
where f is a deterministic function, i.e., it gives one output for a pair of inputs
fx; yg. If we assume that the eect of f is large in comparison with the eect of
W, then the variance of W is small. But since f is deterministic, the variance of H
equals the variance of W, so the variance of H is small. However, Galton observed
from his data that the variance of the heights of the ospring of a given pair of
parent heights is not small. This would seem to imply that inheritance plays a",prob.pdf
"parent heights is not small. This would seem to imply that inheritance plays a
small role in the determination of the height of an individual. Later in this section,
we will describe the way in which Galton got around this problem.
We will now consider the modern explanation of why certain traits, such as
heights, are approximately normally distributed. In order to do so, we need to
introduce some terminology from the eld of genetics. The cells in a living organism
that are not directly involved in the transmission of genetic material to ospring
are called somatic cells, and the remaining cells are called germ cells. Organisms of",prob.pdf
"348 CHAPTER 9. CENTRAL LIMIT THEOREM
a given species have their genetic information encoded in sets of physical entities,
called chromosomes. The chromosomes are paired in each somatic cell. For example,
human beings have 23 pairs of chromosomes in each somatic cell. The sex cells
contain one chromosome from each pair. In sexual reproduction, two sex cells, one
from each parent, contribute their chromosomes to create the set of chromosomes
for the ospring.
Chromosomes contain many subunits, called genes. Genes consist of molecules
of DNA, and one gene has, encoded in its DNA, information that leads to the reg-
ulation of proteins. In the present context, we will consider those genes containing
information that has an eect on some physical trait, such as height, of the organ-
ism. The pairing of the chromosomes gives rise to a pairing of the genes on the
chromosomes.
In a given species, each gene can be any one of several forms. These various",prob.pdf
"chromosomes.
In a given species, each gene can be any one of several forms. These various
forms are called alleles. One should think of the dierent alleles as potentially
producing dierent eects on the physical trait in question. Of the two alleles that
are found in a given gene pair in an organism, one of the alleles came from one
parent and the other allele came from the other parent. The possible types of pairs
of alleles (without regard to order) are called genotypes.
If we assume that the height of a human being is largely controlled by a specic
gene, then we are faced with the same diculty that Galton was. We are assuming
that each parent has a pair of alleles which largely controls their heights. Since
each parent contributes one allele of this gene pair to each of its ospring, there are
four possible allele pairs for the ospring at this gene location. The assumption is
that these pairs of alleles largely control the height of the ospring, and we are also",prob.pdf
"that these pairs of alleles largely control the height of the ospring, and we are also
assuming that genetic factors outweigh non-genetic factors. It follows that among
the ospring we should see several modes in the height distribution of the ospring,
one mode corresponding to each possible pair of alleles. This distribution does not
correspond to the observed distribution of heights.
An alternative hypothesis, which does explain the observation of normally dis-
tributed heights in ospring of a given sex, is the multiple-gene hypothesis. Under
this hypothesis, we assume that there are many genes that aect the height of an
individual. These genes may dier in the amount of their eects. Thus, we can
represent each gene pair by a random variable Xi, where the value of the random
variable is the allele pair's eect on the height of the individual. Thus, for example,
if each parent has two dierent alleles in the gene pair under consideration, then",prob.pdf
"if each parent has two dierent alleles in the gene pair under consideration, then
the ospring has one of four possible pairs of alleles at this gene location. Now the
height of the ospring is a random variable, which can be expressed as
H = X1 + X2 + 
 
 
 + Xn + W ;
if there are n genes that aect height. (Here, as before, the random variable W de-
notes non-genetic eects.) Although n is xed, if it is fairly large, then Theorem 9.5
implies that the sum X1 + X2 + 
 
 
 + Xn is approximately normally distributed.
Now, if we assume that the Xi's have a signicantly larger cumulative eect than
W does, then H is approximately normally distributed.
Another observed feature of the distribution of heights of adults of one sex in",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 349
a population is that the variance does not seem to increase or decrease from one
generation to the next. This was known at the time of Galton, and his attempts
to explain this led him to the idea of regression to the mean. This idea will be
discussed further in the historical remarks at the end of the section. (The reason
that we only consider one sex is that human heights are clearly sex-linked, and in
general, if we have two populations that are each normally distributed, then their
union need not be normally distributed.)
Using the multiple-gene hypothesis, it is easy to explain why the variance should
be constant from generation to generation. We begin by assuming that for a specic
gene location, there are k alleles, which we will denote by A1; A2; : : : ; Ak. We
assume that the ospring are produced by random mating. By this we mean that
given any ospring, it is equally likely that it came from any pair of parents in the",prob.pdf
"given any ospring, it is equally likely that it came from any pair of parents in the
preceding generation. There is another way to look at random mating that makes
the calculations easier. We consider the set S of all of the alleles (at the given gene
location) in all of the germ cells of all of the individuals in the parent generation.
In terms of the set S, by random mating we mean that each pair of alleles in S is
equally likely to reside in any particular ospring. (The reader might object to this
way of thinking about random mating, as it allows two alleles from the same parent
to end up in an ospring; but if the number of individuals in the parent population
is large, then whether or not we allow this event does not aect the probabilities
very much.)
For 1  i  k, we let pi denote the proportion of alleles in the parent population
that are of type Ai. It is clear that this is the same as the proportion of alleles in the",prob.pdf
"that are of type Ai. It is clear that this is the same as the proportion of alleles in the
germ cells of the parent population, assuming that each parent produces roughly
the same number of germs cells. Consider the distribution of alleles in the ospring.
Since each germ cell is equally likely to be chosen for any particular ospring, the
distribution of alleles in the ospring is the same as in the parents.
We next consider the distribution of genotypes in the two generations. We will
prove the following fact: the distribution of genotypes in the ospring generation
depends only upon the distribution of alleles in the parent generation (in particular,
it does not depend upon the distribution of genotypes in the parent generation).
Consider the possible genotypes; there are k(k + 1)=2 of them. Under our assump-
tions, the genotype AiAi will occur with frequency p2
i , and the genotype AiAj,
with i 6= j, will occur with frequency 2pipj. Thus, the frequencies of the genotypes",prob.pdf
"i , and the genotype AiAj,
with i 6= j, will occur with frequency 2pipj. Thus, the frequencies of the genotypes
depend only upon the allele frequencies in the parent generation, as claimed.
This means that if we start with a certain generation, and a certain distribution
of alleles, then in all generations after the one we started with, both the allele
distribution and the genotype distribution will be xed. This last statement is
known as the Hardy-Weinberg Law.
We can describe the consequences of this law for the distribution of heights
among adults of one sex in a population. We recall that the height of an ospring
was given by a random variable H, where
H = X1 + X2 + 
 
 
 + Xn + W ;
with the Xi's corresponding to the genes that aect height, and the random variable",prob.pdf
"350 CHAPTER 9. CENTRAL LIMIT THEOREM
W denoting non-genetic eects. The Hardy-Weinberg Law states that for each Xi,
the distribution in the ospring generation is the same as the distribution in the
parent generation. Thus, if we assume that the distribution of W is roughly the
same from generation to generation (or if we assume that its eects are small), then
the distribution of H is the same from generation to generation. (In fact, dietary
eects are part of W, and it is clear that in many human populations, diets have
changed quite a bit from one generation to the next in recent times. This change is
thought to be one of the reasons that humans, on the average, are getting taller. It
is also the case that the eects of W are thought to be small relative to the genetic
eects of the parents.)
Discussion
Generally speaking, the Central Limit Theorem contains more information than
the Law of Large Numbers, because it gives us detailed information about the",prob.pdf
"the Law of Large Numbers, because it gives us detailed information about the
shape of the distribution of S
n; for large n the shape is approximately the same
as the shape of the standard normal density. More specically, the Central Limit
Theorem says that if we standardize and height-correct the distribution of Sn, then
the normal density function is a very good approximation to this distribution when
n is large. Thus, we have a computable approximation for the distribution for Sn,
which provides us with a powerful technique for generating answers for all sorts
of questions about sums of independent random variables, even if the individual
random variables have dierent distributions.
Historical Remarks
In the mid-1800's, the Belgian mathematician Quetelet7 had shown empirically that
the normal distribution occurred in real data, and had also given a method for tting
the normal curve to a given data set. Laplace8 had shown much earlier that the",prob.pdf
"the normal curve to a given data set. Laplace8 had shown much earlier that the
sum of many independent identically distributed random variables is approximately
normal. Galton knew that certain physical traits in a population appeared to be
approximately normally distributed, but he did not consider Laplace's result to be
a good explanation of how this distribution comes about. We give a quote from
Galton that appears in the fascinating book by S. Stigler9 on the history of statistics:
First, let me point out a fact which Quetelet and all writers who have
followed in his paths have unaccountably overlooked, and which has an
intimate bearing on our work to-night. It is that, although characteris-
tics of plants and animals conform to the law, the reason of their doing
so is as yet totally unexplained. The essence of the law is that dierences
should be wholly due to the collective actions of a host of independent
petty in
uences in various combinations...Now the processes of hered-",prob.pdf
"should be wholly due to the collective actions of a host of independent
petty in
uences in various combinations...Now the processes of hered-
ity...are not petty in
uences, but very important ones...The conclusion
7S. Stigler, The History of Statistics, (Cambridge: Harvard University Press, 1986), p. 203.
8ibid., p. 136
9ibid., p. 281.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 351
Figure 9.11: Two-stage version of the quincunx.
is...that the processes of heredity must work harmoniously with the law
of deviation, and be themselves in some sense conformable to it.
Galton invented a device known as a quincunx (now commonly called a Galton
board), which we used in Example 3.10 to show how to physically obtain a binomial
distribution. Of course, the Central Limit Theorem says that for large values of
the parameter n, the binomial distribution is approximately normal. Galton used
the quincunx to explain how inheritance aects the distribution of a trait among
ospring.
We consider, as Galton did, what happens if we interrupt, at some intermediate
height, the progress of the shot that is falling in the quincunx. The reader is referred
to Figure 9.11. This gure is a drawing of Karl Pearson,10 based upon Galton's
notes. In this gure, the shot is being temporarily segregated into compartments at",prob.pdf
"notes. In this gure, the shot is being temporarily segregated into compartments at
the line AB. (The line A0B0 forms a platform on which the shot can rest.) If the line
AB is not too close to the top of the quincunx, then the shot will be approximately
normally distributed at this line. Now suppose that one compartment is opened, as
shown in the gure. The shot from that compartment will fall, forming a normal
distribution at the bottom of the quincunx. If now all of the compartments are
10Karl Pearson, The Life, Letters and Labours of Francis Galton, vol. IIIB, (Cambridge at the
University Press 1930.) p. 466. Reprinted with permission.",prob.pdf
"352 CHAPTER 9. CENTRAL LIMIT THEOREM
opened, all of the shot will fall, producing the same distribution as would occur if
the shot were not temporarily stopped at the line AB. But the action of stopping
the shot at the line AB, and then releasing the compartments one at a time, is
just the same as convoluting two normal distributions. The normal distributions at
the bottom, corresponding to each compartment at the line AB, are being mixed,
with their weights being the number of shot in each compartment. On the other
hand, it is already known that if the shot are unimpeded, the nal distribution is
approximately normal. Thus, this device shows that the convolution of two normal
distributions is again normal.
Galton also considered the quincunx from another perspective. He segregated
into seven groups, by weight, a set of 490 sweet pea seeds. He gave 10 seeds from
each of the seven group to each of seven friends, who grew the plants from the",prob.pdf
"each of the seven group to each of seven friends, who grew the plants from the
seeds. Galton found that each group produced seeds whose weights were normally
distributed. (The sweet pea reproduces by self-pollination, so he did not need to
consider the possibility of interaction between dierent groups.) In addition, he
found that the variances of the weights of the ospring were the same for each
group. This segregation into groups corresponds to the compartments at the line
AB in the quincunx. Thus, the sweet peas were acting as though they were being
governed by a convolution of normal distributions.
He now was faced with a problem. We have shown in Chapter 7, and Galton
knew, that the convolution of two normal distributions produces a normal distribu-
tion with a larger variance than either of the original distributions. But his data on
the sweet pea seeds showed that the variance of the ospring population was the",prob.pdf
"the sweet pea seeds showed that the variance of the ospring population was the
same as the variance of the parent population. His answer to this problem was to
postulate a mechanism that he called reversion, and is now called regression to the
mean. As Stigler puts it:11
The seven groups of progeny were normally distributed, but not about
their parents' weight. Rather they were in every case distributed about
a value that was closer to the average population weight than was that of
the parent. Furthermore, this reversion followed \the simplest possible
law,"" that is, it was linear. The average deviation of the progeny from
the population average was in the same direction as that of the parent,
but only a third as great. The mean progeny reverted to type, and
the increased variation was just sucient to maintain the population
variability.
Galton illustrated reversion with the illustration shown in Figure 9.12.12 The",prob.pdf
"variability.
Galton illustrated reversion with the illustration shown in Figure 9.12.12 The
parent population is shown at the top of the gure, and the slanted lines are meant
to correspond to the reversion eect. The ospring population is shown at the
bottom of the gure.
11ibid., p. 282.
12Karl Pearson, The Life, Letters and Labours of Francis Galton, vol. IIIA, (Cambridge at the
University Press 1930.) p. 9. Reprinted with permission.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 353
Figure 9.12: Galton's explanation of reversion.",prob.pdf
"354 CHAPTER 9. CENTRAL LIMIT THEOREM
Exercises
1 A die is rolled 24 times. Use the Central Limit Theorem to estimate the
probability that
(a) the sum is greater than 84.
(b) the sum is equal to 84.
2 A random walker starts at 0 on the x-axis and at each time unit moves 1
step to the right or 1 step to the left with probability 1/2. Estimate the
probability that, after 100 steps, the walker is more than 10 steps from the
starting position.
3 A piece of rope is made up of 100 strands. Assume that the breaking strength
of the rope is the sum of the breaking strengths of the individual strands.
Assume further that this sum may be considered to be the sum of an inde-
pendent trials process with 100 experiments each having expected value of 10
pounds and standard deviation 1. Find the approximate probability that the
rope will support a weight
(a) of 1000 pounds.
(b) of 970 pounds.
4 Write a program to nd the average of 1000 random digits 0, 1, 2, 3, 4, 5, 6, 7,",prob.pdf
"rope will support a weight
(a) of 1000 pounds.
(b) of 970 pounds.
4 Write a program to nd the average of 1000 random digits 0, 1, 2, 3, 4, 5, 6, 7,
8, or 9. Have the program test to see if the average lies within three standard
deviations of the expected value of 4.5. Modify the program so that it repeats
this simulation 1000 times and keeps track of the number of times the test is
passed. Does your outcome agree with the Central Limit Theorem?
5 A die is thrown until the rst time the total sum of the face values of the die
is 700 or greater. Estimate the probability that, for this to happen,
(a) more than 210 tosses are required.
(b) less than 190 tosses are required.
(c) between 180 and 210 tosses, inclusive, are required.
6 A bank accepts rolls of pennies and gives 50 cents credit to a customer without
counting the contents. Assume that a roll contains 49 pennies 30 percent of
the time, 50 pennies 60 percent of the time, and 51 pennies 10 percent of the
time.",prob.pdf
"the time, 50 pennies 60 percent of the time, and 51 pennies 10 percent of the
time.
(a) Find the expected value and the variance for the amount that the bank
loses on a typical roll.
(b) Estimate the probability that the bank will lose more than 25 cents in
100 rolls.
(c) Estimate the probability that the bank will lose exactly 25 cents in 100
rolls.",prob.pdf
"9.2. DISCRETE INDEPENDENT TRIALS 355
(d) Estimate the probability that the bank will lose any money in 100 rolls.
(e) How many rolls does the bank need to collect to have a 99 percent chance
of a net loss?
7 A surveying instrument makes an error of 2, 1, 0, 1, or 2 feet with equal
probabilities when measuring the height of a 200-foot tower.
(a) Find the expected value and the variance for the height obtained using
this instrument once.
(b) Estimate the probability that in 18 independent measurements of this
tower, the average of the measurements is between 199 and 201, inclusive.
8 For Example 9.6 estimate P(S30 = 0). That is, estimate the probability that
the errors cancel out and the student's grade point average is correct.
9 Prove the Law of Large Numbers using the Central Limit Theorem.
10 Peter and Paul match pennies 10,000 times. Describe brie
y what each of the
following theorems tells you about Peter's fortune.
(a) The Law of Large Numbers.
(b) The Central Limit Theorem.",prob.pdf
"following theorems tells you about Peter's fortune.
(a) The Law of Large Numbers.
(b) The Central Limit Theorem.
11 A tourist in Las Vegas was attracted by a certain gambling game in which
the customer stakes 1 dollar on each play; a win then pays the customer
2 dollars plus the return of her stake, although a loss costs her only her stake.
Las Vegas insiders, and alert students of probability theory, know that the
probability of winning at this game is 1/4. When driven from the tables by
hunger, the tourist had played this game 240 times. Assuming that no near
miracles happened, about how much poorer was the tourist upon leaving the
casino? What is the probability that she lost no money?
12 We have seen that, in playing roulette at Monte Carlo (Example 6.13), betting
1 dollar on red or 1 dollar on 17 amounts to choosing between the distributions
mX =
 1 1=2 1
18=37 1=37 18=37

or
mX =
 1 35
36=37 1=37

You plan to choose one of these methods and use it to make 100 1-dollar",prob.pdf
"mX =
 1 1=2 1
18=37 1=37 18=37

or
mX =
 1 35
36=37 1=37

You plan to choose one of these methods and use it to make 100 1-dollar
bets using the method chosen. Using the Central Limit Theorem, estimate
the probability of winning any money for each of the two games. Compare
your estimates with the actual probabilities, which can be shown, from exact
calculations, to equal .437 and .509 to three decimal places.
13 In Example 9.6 nd the largest value of p that gives probability .954 that the
rst decimal place is correct.",prob.pdf
"356 CHAPTER 9. CENTRAL LIMIT THEOREM
14 It has been suggested that Example 9.6 is unrealistic, in the sense that the
probabilities of errors are too low. Make up your own (reasonable) estimate
for the distribution m(x), and determine the probability that a student's grade
point average is accurate to within .05. Also determine the probability that
it is accurate to within .5.
15 Find a sequence of uniformly bounded discrete independent random variables
fXng such that the variance of their sum does not tend to 1 as n ! 1, and
such that their sum is not asymptotically normally distributed.
9.3 Central Limit Theorem for Continuous Inde-
pendent Trials
We have seen in Section 9.2 that the distribution function for the sum of a large
number n of independent discrete random variables with mean  and variance 2
tends to look like a normal density with mean n and variance n2. What is
remarkable about this result is that it holds for any distribution with nite mean",prob.pdf
"remarkable about this result is that it holds for any distribution with nite mean
and variance. We shall see in this section that the same result also holds true for
continuous random variables having a common density function.
Let us begin by looking at some examples to see whether such a result is even
plausible.
Standardized Sums
Example 9.7 Suppose we choose n random numbers from the interval [0; 1] with
uniform density. Let X1, X2, . . . , Xn denote these choices, and Sn = X1 + X2 +

 
 
 + Xn their sum.
We saw in Example 7.9 that the density function for Sn tends to have a normal
shape, but is centered at n=2 and is 
attened out. In order to compare the shapes
of these density functions for dierent values of n, we proceed as in the previous
section: we standardize Sn by dening
S
n = Sn  n
pn :
Then we see that for all n we have
E(S
n) = 0 ;
V (S
n) = 1 :
The density function for S
n is just a standardized version of the density function
for Sn (see Figure 9.13). 2",prob.pdf
"E(S
n) = 0 ;
V (S
n) = 1 :
The density function for S
n is just a standardized version of the density function
for Sn (see Figure 9.13). 2
Example 9.8 Let us do the same thing, but now choose numbers from the interval
[0; +1) with an exponential density with parameter . Then (see Example 6.26)",prob.pdf
"9.3. CONTINUOUS INDEPENDENT TRIALS 357
-3 -2 -1 1 2 3
0.1
0.2
0.3
0.4
n =   2
n =   3
n =   4
n = 10
Figure 9.13: Density function for S
n (uniform case, n = 2; 3; 4; 10).
 = E(Xi) = 1
 ;
2 = V (Xj) = 1
2 :
Here we know the density function for Sn explicitly (see Section 7.2). We can
use Corollary 5.1 to calculate the density function for S
n. We obtain
fSn (x) = ex(x)n1
(n  1)! ;
fSn (x) =
pn
 fSn
pnx + n


:
The graph of the density function for S
n is shown in Figure 9.14. 2
These examples make it seem plausible that the density function for the nor-
malized random variable S
n for large n will look very much like the normal density
with mean 0 and variance 1 in the continuous case as well as in the discrete case.
The Central Limit Theorem makes this statement precise.
Central Limit Theorem
Theorem 9.6 (Central Limit Theorem) Let Sn = X1 + X2 + 
 
 
 + Xn be the
sum of n independent continuous random variables with common density function p",prob.pdf
"sum of n independent continuous random variables with common density function p
having expected value  and variance 2. Let S
n = (Sn n)=pn. Then we have,",prob.pdf
"358 CHAPTER 9. CENTRAL LIMIT THEOREM
-4 -2 2
0.1
0.2
0.3
0.4
0.5n =   2
n =   3
n = 10
n = 30
Figure 9.14: Density function for S
n (exponential case, n = 2; 3; 10; 30,  = 1).
for all a < b,
lim
n!1
P(a < S
n < b) = 1p
2
Z b
a
ex2=2 dx :
2
We shall give a proof of this theorem in Section 10.3. We will now look at some
examples.
Example 9.9 Suppose a surveyor wants to measure a known distance, say of 1 mile,
using a transit and some method of triangulation. He knows that because of possible
motion of the transit, atmospheric distortions, and human error, any one measure-
ment is apt to be slightly in error. He plans to make several measurements and take
an average. He assumes that his measurements are independent random variables
with a common distribution of mean  = 1 and standard deviation  = :0002 (so,
if the errors are approximately normally distributed, then his measurements are
within 1 foot of the correct distance about 65% of the time). What can he say
about the average?",prob.pdf
"within 1 foot of the correct distance about 65% of the time). What can he say
about the average?
He can say that if n is large, the average Sn=n has a density function that is
approximately normal, with mean  = 1 mile, and standard deviation  = :0002=p
n
miles.
How many measurements should he make to be reasonably sure that his average
lies within .0001 of the true value? The Chebyshev inequality says
P




Sn
n  



 :0001

 (:0002)2
n(108) = 4
n ;
so that we must have n  80 before the probability that his error is less than .0001
exceeds .95.",prob.pdf
"9.3. CONTINUOUS INDEPENDENT TRIALS 359
We have already noticed that the estimate in the Chebyshev inequality is not
always a good one, and here is a case in point. If we assume that n is large enough
so that the density for Sn is approximately normal, then we have
P




Sn
n  



< :0001

= P

:5p
n < S
n < +:5pn


 1p
2
Z +:5pn
:5pn
ex2=2 dx ;
and this last expression is greater than .95 if :5pn  2: This says that it suces
to take n = 16 measurements for the same results. This second calculation is
stronger, but depends on the assumption that n = 16 is large enough to establish
the normal density as a good approximation to S
n, and hence to Sn. The Central
Limit Theorem here says nothing about how large n has to be. In most cases
involving sums of independent random variables, a good rule of thumb is that for
n  30, the approximation is a good one. In the present case, if we assume that the",prob.pdf
"n  30, the approximation is a good one. In the present case, if we assume that the
errors are approximately normally distributed, then the approximation is probably
fairly good even for n = 16. 2
Estimating the Mean
Example 9.10 (Continuation of Example 9.9) Now suppose our surveyor is mea-
suring an unknown distance with the same instruments under the same conditions.
He takes 36 measurements and averages them. How sure can he be that his mea-
surement lies within .0002 of the true value?
Again using the normal approximation, we get
P




Sn
n  



< :0002

= P

jS
nj < :5p
n


 2p
2
Z 3
3
ex2=2 dx
 :997 :
This means that the surveyor can be 99.7 percent sure that his average is within
.0002 of the true value. To improve his condence, he can take more measurements,
or require less accuracy, or improve the quality of his measurements (i.e., reduce
the variance 2). In each case, the Central Limit Theorem gives quantitative infor-",prob.pdf
"the variance 2). In each case, the Central Limit Theorem gives quantitative infor-
mation about the condence of a measurement process, assuming always that the
normal approximation is valid.
Now suppose the surveyor does not know the mean or standard deviation of his
measurements, but assumes that they are independent. How should he proceed?
Again, he makes several measurements of a known distance and averages them.
As before, the average error is approximately normally distributed, but now with
unknown mean and variance. 2",prob.pdf
"360 CHAPTER 9. CENTRAL LIMIT THEOREM
Sample Mean
If he knows the variance 2 of the error distribution is .0002, then he can estimate
the mean  by taking the average, or sample mean of, say, 36 measurements:
 = x1 + x2 + 
 
 
 + xn
n ;
where n = 36. Then, as before, E() = . Moreover, the preceding argument shows
that
P(j  j < :0002)  :997 :
The interval ( :0002; + :0002) is called the 99.7% condence interval for  (see
Example 9.4).
Sample Variance
If he does not know the variance 2 of the error distribution, then he can estimate
2 by the sample variance:
2 = (x1  )2 + (x2  )2 + 
 
 
 + (xn  )2
n ;
where n = 36. The Law of Large Numbers, applied to the random variables (Xi 
)2, says that for large n, the sample variance 2 lies close to the variance 2, so
that the surveyor can use 2 in place of 2 in the argument above.
Experience has shown that, in most practical problems of this type, the sample",prob.pdf
"that the surveyor can use 2 in place of 2 in the argument above.
Experience has shown that, in most practical problems of this type, the sample
variance is a good estimate for the variance, and can be used in place of the variance
to determine condence levels for the sample mean. This means that we can rely
on the Law of Large Numbers for estimating the variance, and the Central Limit
Theorem for estimating the mean.
We can check this in some special cases. Suppose we know that the error distri-
bution is normal, with unknown mean and variance. Then we can take a sample of
n measurements, nd the sample mean  and sample variance 2, and form
T
n = Sn  npn ;
where n = 36. We expect T
n to be a good approximation for S
n for large n.
t-Density
The statistician W. S. Gosset13 has shown that in this case T
n has a density function
that is not normal but rather a t-density with n degrees of freedom. (The number",prob.pdf
"n has a density function
that is not normal but rather a t-density with n degrees of freedom. (The number
n of degrees of freedom is simply a parameter which tells us which t-density to use.)
In this case we can use the t-density in place of the normal density to determine
condence levels for . As n increases, the t-density approaches the normal density.
Indeed, even for n = 8 the t-density and normal density are practically the same
(see Figure 9.15).
13W. S. Gosset discovered the distribution we now call the t-distribution while working for the
Guinness Brewery in Dublin. He wrote under the pseudonym \Student."" The results discussed here
rst appeared in Student, \The Probable Error of a Mean,"" Biometrika, vol. 6 (1908), pp. 1-24.",prob.pdf
"9.3. CONTINUOUS INDEPENDENT TRIALS 361
-6 -4 -2 2 4 6
0.1
0.2
0.3
0.4
Figure 9.15: Graph of tdensity for n = 1; 3; 8 and the normal density with  =
0;  = 1.
Exercises
Notes on computer problems:
(a) Simulation: Recall (see Corollary 5.2) that
X = F1(rnd)
will simulate a random variable with density f(x) and distribution
F(X) =
Z x
1
f(t) dt :
In the case that f(x) is a normal density function with mean  and standard
deviation , where neither F nor F1 can be expressed in closed form, use
instead
X = 
p
2 log(rnd) cos2(rnd) +  :
(b) Bar graphs: you should aim for about 20 to 30 bars (of equal width) in your
graph. You can achieve this by a good choice of the range [xmin; xmin] and
the number of bars (for instance, [  3;  + 3] with 30 bars will work in
many cases). Experiment!
1 Let X be a continuous random variable with mean (X) and variance 2(X),
and let X = (X  )= be its standardized version. Verify directly that
(X) = 0 and 2(X) = 1.",prob.pdf
"362 CHAPTER 9. CENTRAL LIMIT THEOREM
2 Let fXkg, 1  k  n, be a sequence of independent random variables, all with
mean 0 and variance 1, and let Sn, S
n, and An be their sum, standardized
sum, and average, respectively. Verify directly that S
n = Sn=pn = pnAn.
3 Let fXkg, 1  k  n, be a sequence of random variables, all with mean  and
variance 2, and Yk = X
k be their standardized versions. Let Sn and Tn be
the sum of the Xk and Yk, and S
n and T
n their standardized version. Show
that S
n = T
n = Tn=pn.
4 Suppose we choose independently 25 numbers at random (uniform density)
from the interval [0; 20]. Write the normal densities that approximate the
densities of their sum S25, their standardized sum S
25, and their average A25.
5 Write a program to choose independently 25 numbers at random from [0; 20],
compute their sum S25, and repeat this experiment 1000 times. Make a bar
graph for the density of S25 and compare it with the normal approximation",prob.pdf
"compute their sum S25, and repeat this experiment 1000 times. Make a bar
graph for the density of S25 and compare it with the normal approximation
of Exercise 4. How good is the t? Now do the same for the standardized
sum S
25 and the average A25.
6 In general, the Central Limit Theorem gives a better estimate than Cheby-
shev's inequality for the average of a sum. To see this, let A25 be the
average calculated in Exercise 5, and let N be the normal approximation
for A25. Modify your program in Exercise 5 to provide a table of the function
F(x) = P(jA25  10j  x) = fraction of the total of 1000 trials for which
jA25  10j  x. Do the same for the function f(x) = P(jN  10j  x). (You
can use the normal table, Table 9.4, or the procedure NormalArea for this.)
Now plot on the same axes the graphs of F(x), f(x), and the Chebyshev
function g(x) = 4=(3x2). How do f(x) and g(x) compare as estimates for
F(x)?
7 The Central Limit Theorem says the sums of independent random variables",prob.pdf
"F(x)?
7 The Central Limit Theorem says the sums of independent random variables
tend to look normal, no matter what crazy distribution the individual variables
have. Let us test this by a computer simulation. Choose independently 25
numbers from the interval [0; 1] with the probability density f(x) given below,
and compute their sum S25. Repeat this experiment 1000 times, and make up
a bar graph of the results. Now plot on the same graph the density (x) =
normal (x; (S25); (S25)). How well does the normal density t your bar
graph in each case?
(a) f(x) = 1.
(b) f(x) = 2x.
(c) f(x) = 3x2.
(d) f(x) = 4jx  1=2j.
(e) f(x) = 2  4jx  1=2j.
8 Repeat the experiment described in Exercise 7 but now choose the 25 numbers
from [0; 1), using f(x) = ex.",prob.pdf
"9.3. CONTINUOUS INDEPENDENT TRIALS 363
9 How large must n be before Sn = X1+X2+
 
 
+Xn is approximately normal?
This number is often surprisingly small. Let us explore this question with a
computer simulation. Choose n numbers from [0; 1] with probability density
f(x), where n = 3, 6, 12, 20, and f(x) is each of the densities in Exercise 7.
Compute their sum Sn, repeat this experiment 1000 times, and make up a
bar graph of 20 bars of the results. How large must n be before you get a
good t?
10 A surveyor is measuring the height of a cli known to be about 1000 feet.
He assumes his instrument is properly calibrated and that his measurement
errors are independent, with mean  = 0 and variance 2 = 10. He plans to
take n measurements and form the average. Estimate, using (a) Chebyshev's
inequality and (b) the normal approximation, how large n should be if he
wants to be 95 percent sure that his average falls within 1 foot of the true",prob.pdf
"inequality and (b) the normal approximation, how large n should be if he
wants to be 95 percent sure that his average falls within 1 foot of the true
value. Now estimate, using (a) and (b), what value should 2 have if he wants
to make only 10 measurements with the same condence?
11 The price of one share of stock in the Pilsdor Beer Company (see Exer-
cise 8.2.12) is given by Yn on the nth day of the year. Finn observes that
the dierences Xn = Yn+1  Yn appear to be independent random variables
with a common distribution having mean  = 0 and variance 2 = 1=4. If
Y1 = 100, estimate the probability that Y365 is
(a)  100.
(b)  110.
(c)  120.
12 Test your conclusions in Exercise 11 by computer simulation. First choose
364 numbers Xi with density f(x) = normal(x; 0; 1=4). Now form the sum
Y365 = 100 + X1 + X2 + 
 
 
 + X364, and repeat this experiment 200 times.
Make up a bar graph on [50; 150] of the results, superimposing the graph of the",prob.pdf
"Make up a bar graph on [50; 150] of the results, superimposing the graph of the
approximating normal density. What does this graph say about your answers
in Exercise 11?
13 Physicists say that particles in a long tube are constantly moving back and
forth along the tube, each with a velocity Vk (in cm/sec) at any given moment
that is normally distributed, with mean  = 0 and variance 2 = 1. Suppose
there are 1020 particles in the tube.
(a) Find the mean and variance of the average velocity of the particles.
(b) What is the probability that the average velocity is  109 cm/sec?
14 An astronomer makes n measurements of the distance between Jupiter and
a particular one of its moons. Experience with the instruments used leads
her to believe that for the proper units the measurements will be normally",prob.pdf
"364 CHAPTER 9. CENTRAL LIMIT THEOREM
distributed with mean d, the true distance, and variance 16. She performs a
series of n measurements. Let
An = X1 + X2 + 
 
 
 + Xn
n
be the average of these measurements.
(a) Show that
P

An  8pn  d  An + 8pn

 :95:
(b) When nine measurements were taken, the average of the distances turned
out to be 23.2 units. Putting the observed values in (a) gives the 95 per-
cent condence interval for the unknown distance d. Compute this in-
terval.
(c) Why not say in (b) more simply that the probability is .95 that the value
of d lies in the computed condence interval?
(d) What changes would you make in the above procedure if you wanted to
compute a 99 percent condence interval?
15 Plot a bar graph similar to that in Figure 9.10 for the heights of the mid-
parents in Galton's data as given in Appendix B and compare this bar graph
to the appropriate normal curve.",prob.pdf
"Chapter 10
Generating Functions
10.1 Generating Functions for Discrete Distribu-
tions
So far we have considered in detail only the two most important attributes of a
random variable, namely, the mean and the variance. We have seen how these
attributes enter into the fundamental limit theorems of probability, as well as into
all sorts of practical calculations. We have seen that the mean and variance of
a random variable contain important information about the random variable, or,
more precisely, about the distribution function of that variable. Now we shall see
that the mean and variance do not contain all the available information about the
density function of a random variable. To begin with, it is easy to give examples of
dierent distribution functions which have the same mean and the same variance.
For instance, suppose X and Y are random variables, with distributions
pX =
1 2 3 4 5 6
0 1=4 1=2 0 0 1=4

;
pY =
 1 2 3 4 5 6
1=4 0 0 1=2 1=4 0

:",prob.pdf
"pX =
1 2 3 4 5 6
0 1=4 1=2 0 0 1=4

;
pY =
 1 2 3 4 5 6
1=4 0 0 1=2 1=4 0

:
Then with these choices, we have E(X) = E(Y ) = 7=2 and V (X) = V (Y ) = 9=4,
and yet certainly pX and pY are quite dierent density functions.
This raises a question: If X is a random variable with range fx1; x2; : : :g of at
most countable size, and distribution function p = pX, and if we know its mean
 = E(X) and its variance 2 = V (X), then what else do we need to know to
determine p completely?
Moments
A nice answer to this question, at least in the case that X has nite range, can be
given in terms of the moments of X, which are numbers dened as follows:
365",prob.pdf
"366 CHAPTER 10. GENERATING FUNCTIONS
k = kth moment of X
= E(Xk)
=
1X
j=1
(xj)kp(xj) ;
provided the sum converges. Here p(xj) = P(X = xj).
In terms of these moments, the mean  and variance 2 of X are given simply
by
 = 1;
2 = 2  2
1 ;
so that a knowledge of the rst two moments of X gives us its mean and variance.
But a knowledge of all the moments of X determines its distribution function p
completely.
Moment Generating Functions
To see how this comes about, we introduce a new variable t, and dene a function
g(t) as follows:
g(t) = E(etX)
=
1X
k=0
ktk
k!
= E
 1X
k=0
Xktk
k!
!
=
1X
j=1
etxj p(xj) :
We call g(t) the moment generating function for X, and think of it as a convenient
bookkeeping device for describing the moments of X. Indeed, if we dierentiate
g(t) n times and then set t = 0, we get n:
dn
dtn g(t)




t=0
= g(n)(0)
=
1X
k=n
k! ktkn
(k  n)! k!





t=0
= n :
It is easy to calculate the moment generating function for simple examples.",prob.pdf
"10.1. DISCRETE DISTRIBUTIONS 367
Examples
Example 10.1 Suppose X has range f1; 2; 3; : : :; ng and pX(j) = 1=n for 1  j  n
(uniform distribution). Then
g(t) =
nX
j=1
1
netj
= 1
n(et + e2t + 
 
 
 + ent)
= et(ent  1)
n(et  1) :
If we use the expression on the right-hand side of the second line above, then it is
easy to see that
1 = g0(0) = 1
n(1 + 2 + 3 + 
 
 
 + n) = n + 1
2 ;
2 = g00(0) = 1
n(1 + 4 + 9 + 
 
 
 + n2) = (n + 1)(2n + 1)
6 ;
and that  = 1 = (n + 1)=2 and 2 = 2  2
1 = (n2  1)=12. 2
Example 10.2 Suppose now that X has range f0; 1; 2; 3; : : :; ng and pX(j) =n
j


pjqnj for 0  j  n (binomial distribution). Then
g(t) =
nX
j=0
etj
n
j

pjqnj
=
nX
j=0
n
j

(pet)jqnj
= (pet + q)n :
Note that
1 = g0(0) = n(pet + q)n1pet

t=0 = np ;
2 = g00(0) = n(n  1)p2 + np ;
so that  = 1 = np, and 2 = 2  2
1 = np(1  p), as expected. 2
Example 10.3 Suppose X has range f1; 2; 3; : : :g and pX(j) = qj1p for all j
(geometric distribution). Then
g(t) =
1X
j=1",prob.pdf
"Example 10.3 Suppose X has range f1; 2; 3; : : :g and pX(j) = qj1p for all j
(geometric distribution). Then
g(t) =
1X
j=1
etjqj1p
= pet
1  qet :",prob.pdf
"368 CHAPTER 10. GENERATING FUNCTIONS
Here
1 = g0(0) = pet
(1  qet)2




t=0
= 1
p ;
2 = g00(0) = pet + pqe2t
(1  qet)3




t=0
= 1 + q
p2 ;
 = 1 = 1=p, and 2 = 2  2
1 = q=p2, as computed in Example 6.26. 2
Example 10.4 Let X have range f0; 1; 2; 3; : : :g and let pX(j) = ej=j! for all j
(Poisson distribution with mean ). Then
g(t) =
1X
j=0
etj ej
j!
= e
1X
j=0
(et)j
j!
= eeet
= e(et1) :
Then
1 = g0(0) = e(et1)et



t=0
=  ;
2 = g00(0) = e(et1)(2e2t + et)



t=0
= 2 +  ;
 = 1 = , and 2 = 2  2
1 = .
The variance of the Poisson distribution is easier to obtain in this way than
directly from the denition (as was done in Exercise 6.2.29). 2
Moment Problem
Using the moment generating function, we can now show, at least in the case of
a discrete random variable with nite range, that its distribution function is com-
pletely determined by its moments.
Theorem 10.1 Let X be a discrete random variable with nite range fx1; x2; : : : ;",prob.pdf
"pletely determined by its moments.
Theorem 10.1 Let X be a discrete random variable with nite range fx1; x2; : : : ;
xng, distribution function p, and moment generating function g. Then g is uniquely
determined by p, and conversely.
Proof. We know that p determines g, since
g(t) =
nX
j=1
etxj p(xj) :
Conversely, assume that g(t) is known. We wish to determine the values of xj and
p(xj), for 1  j  n. We assume, without loss of generality, that p(xj) > 0 for
1  j  n, and that
x1 < x2 < : : : < xn :",prob.pdf
"10.1. DISCRETE DISTRIBUTIONS 369
We note that g(t) is dierentiable for all t, since it is a nite linear combination of
exponential functions. If we compute g0(t)=g(t), we obtain
x1p(x1)etx1 + : : : + xnp(xn)etxn
p(x1)etx1 + : : : + p(xn)etxn
:
Dividing both top and bottom by etxn , we obtain the expression
x1p(x1)et(x1xn) + : : : + xnp(xn)
p(x1)et(x1xn) + : : : + p(xn) :
Since xn is the largest of the xj's, this expression approaches xn as t goes to 1. So
we have shown that
xn = lim
t!1
g0(t)
g(t) :
To nd p(xn), we simply divide g(t) by etxn and let t go to 1. Once xn and
p(xn) have been determined, we can subtract p(xn)etxn from g(t), and repeat the
above procedure with the resulting function, obtaining, in turn, xn1; : : :; x1 and
p(xn1); : : : ; p(x1). 2
If we delete the hypothesis that X have nite range in the above theorem, then
the conclusion is no longer necessarily true.
Ordinary Generating Functions",prob.pdf
"the conclusion is no longer necessarily true.
Ordinary Generating Functions
In the special but important case where the xj are all nonnegative integers, xj = j,
we can prove this theorem in a simpler way.
In this case, we have
g(t) =
nX
j=0
etjp(j) ;
and we see that g(t) is a polynomial in et. If we write z = et, and dene the function
h by
h(z) =
nX
j=0
zjp(j) ;
then h(z) is a polynomial in z containing the same information as g(t), and in fact
h(z) = g(log z) ;
g(t) = h(et) :
The function h(z) is often called the ordinary generating function for X. Note that
h(1) = g(0) = 1, h0(1) = g0(0) = 1, and h00(1) = g00(0)g0(0) = 2 1. It follows
from all this that if we know g(t), then we know h(z), and if we know h(z), then
we can nd the p(j) by Taylor's formula:
p(j) = coecient of zj in h(z)
= h(j)(0)
j! :",prob.pdf
"370 CHAPTER 10. GENERATING FUNCTIONS
For example, suppose we know that the moments of a certain discrete random
variable X are given by
0 = 1 ;
k = 1
2 + 2k
4 ; for k  1 :
Then the moment generating function g of X is
g(t) =
1X
k=0
ktk
k!
= 1 + 1
2
1X
k=1
tk
k! + 1
4
1X
k=1
(2t)k
k!
= 1
4 + 1
2et + 1
4e2t :
This is a polynomial in z = et, and
h(z) = 1
4 + 1
2z + 1
4z2 :
Hence, X must have range f0; 1; 2g, and p must have values f1=4; 1=2; 1=4g.
Properties
Both the moment generating function g and the ordinary generating function h have
many properties useful in the study of random variables, of which we can consider
only a few here. In particular, if X is any discrete random variable and Y = X +a,
then
gY (t) = E(etY )
= E(et(X+a))
= etaE(etX)
= etagX(t) ;
while if Y = bX, then
gY (t) = E(etY )
= E(etbX)
= gX(bt) :
In particular, if
X = X  
 ;
then (see Exercise 11)
gx(t) = et=gX
t


:",prob.pdf
"10.1. DISCRETE DISTRIBUTIONS 371
If X and Y are independent random variables and Z = X + Y is their sum,
with pX, pY , and pZ the associated distribution functions, then we have seen in
Chapter 7 that pZ is the convolution of pX and pY , and we know that convolution
involves a rather complicated calculation. But for the generating functions we have
instead the simple relations
gZ(t) = gX(t)gY (t) ;
hZ(z) = hX(z)hY (z) ;
that is, gZ is simply the product of gX and gY , and similarly for hZ.
To see this, rst note that if X and Y are independent, then etX and etY are
independent (see Exercise 5.2.38), and hence
E(etXetY ) = E(etX)E(etY ) :
It follows that
gZ(t) = E(etZ) = E(et(X+Y ))
= E(etX)E(etY )
= gX(t)gY (t) ;
and, replacing t by log z, we also get
hZ(z) = hX(z)hY (z) :
Example 10.5 If X and Y are independent discrete random variables with range
f0; 1; 2; : : :; ng and binomial distribution
pX(j) = pY (j) =
n
j

pjqnj ;",prob.pdf
"f0; 1; 2; : : :; ng and binomial distribution
pX(j) = pY (j) =
n
j

pjqnj ;
and if Z = X + Y , then we know (cf. Section 7.1) that the range of X is
f0; 1; 2; : : :; 2ng
and X has binomial distribution
pZ(j) = (pX  pY )(j) =
2n
j

pjq2nj :
Here we can easily verify this result by using generating functions. We know that
gX(t) = gY (t) =
nX
j=0
etj
n
j

pjqnj
= (pet + q)n ;
and
hX(z) = hY (z) = (pz + q)n :",prob.pdf
"372 CHAPTER 10. GENERATING FUNCTIONS
Hence, we have
gZ(t) = gX(t)gY (t) = (pet + q)2n ;
or, what is the same,
hZ(z) = hX(z)hY (z) = (pz + q)2n
=
2nX
j=0
2n
j

(pz)jq2nj ;
from which we can see that the coecient of zj is just pZ(j) =
2n
j


pjq2nj. 2
Example 10.6 If X and Y are independent discrete random variables with the
non-negative integers f0; 1; 2; 3; : : :g as range, and with geometric distribution func-
tion
pX(j) = pY (j) = qjp ;
then
gX(t) = gY (t) = p
1  qet ;
and if Z = X + Y , then
gZ(t) = gX(t)gY (t)
= p2
1  2qet + q2e2t :
If we replace et by z, we get
hZ(z) = p2
(1  qz)2
= p2
1X
k=0
(k + 1)qkzk ;
and we can read o the values of pZ(j) as the coecient of zj in this expansion
for h(z), even though h(z) is not a polynomial in this case. The distribution pZ is
a negative binomial distribution (see Section 5.1). 2
Here is a more interesting example of the power and scope of the method of
generating functions.
Heads or Tails",prob.pdf
"Here is a more interesting example of the power and scope of the method of
generating functions.
Heads or Tails
Example 10.7 In the coin-tossing game discussed in Example 1.4, we now consider
the question \When is Peter rst in the lead?""
Let Xk describe the outcome of the kth trial in the game
Xk =
+1; if kth toss is heads;
1; if kth toss is tails.",prob.pdf
"10.1. DISCRETE DISTRIBUTIONS 373
Then the Xk are independent random variables describing a Bernoulli process. Let
S0 = 0, and, for n  1, let
Sn = X1 + X2 + 
 
 
 + Xn :
Then Sn describes Peter's fortune after n trials, and Peter is rst in the lead after
n trials if Sk  0 for 1  k < n and Sn = 1.
Now this can happen when n = 1, in which case S1 = X1 = 1, or when n > 1,
in which case S1 = X1 = 1. In the latter case, Sk = 0 for k = n  1, and perhaps
for other k between 1 and n. Let m be the least such value of k; then Sm = 0 and
Sk < 0 for 1  k < m. In this case Peter loses on the rst trial, regains his initial
position in the next m  1 trials, and gains the lead in the next n  m trials.
Let p be the probability that the coin comes up heads, and let q = 1  p. Let
rn be the probability that Peter is rst in the lead after n trials. Then from the
discussion above, we see that
rn = 0 ; if n even;
r1 = p (= probability of heads in a single toss);",prob.pdf
"discussion above, we see that
rn = 0 ; if n even;
r1 = p (= probability of heads in a single toss);
rn = q(r1rn2 + r3rn4 + 
 
 
 + rn2r1) ; if n > 1; n odd:
Now let T describe the time (that is, the number of trials) required for Peter to
take the lead. Then T is a random variable, and since P(T = n) = rn, r is the
distribution function for T.
We introduce the generating function hT (z) for T:
hT (z) =
1X
n=0
rnzn :
Then, by using the relations above, we can verify the relation
hT (z) = pz + qz(hT (z))2 :
If we solve this quadratic equation for hT (z), we get
hT (z) = 1 
p
1  4pqz2
2qz = 2pz
1 
p
1  4pqz2 :
Of these two solutions, we want the one that has a convergent power series in z
(i.e., that is nite for z = 0). Hence we choose
hT (z) = 1 
p
1  4pqz2
2qz = 2pz
1 +
p
1  4pqz2 :
Now we can ask: What is the probability that Peter is ever in the lead? This
probability is given by (see Exercise 10)
1X
n=0
rn = hT (1) = 1 
p
1  4pq
2q
= 1  jp  qj
2q
=

p=q; if p < q;",prob.pdf
"probability is given by (see Exercise 10)
1X
n=0
rn = hT (1) = 1 
p
1  4pq
2q
= 1  jp  qj
2q
=

p=q; if p < q;
1; if p  q;",prob.pdf
"374 CHAPTER 10. GENERATING FUNCTIONS
so that Peter is sure to be in the lead eventually if p  q.
How long will it take? That is, what is the expected value of T? This value is
given by
E(T) = h0
T (1) =
1=(p  q); if p > q;
1; if p = q:
This says that if p > q, then Peter can expect to be in the lead by about 1=(p  q)
trials, but if p = q, he can expect to wait a long time.
A related problem, known as the Gambler's Ruin problem, is studied in Exer-
cise 23 and in Section 12.2. 2
Exercises
1 Find the generating functions, both ordinary h(z) and moment g(t), for the
following discrete probability distributions.
(a) The distribution describing a fair coin.
(b) The distribution describing a fair die.
(c) The distribution describing a die that always comes up 3.
(d) The uniform distribution on the set fn; n + 1; n + 2; : : :; n + kg.
(e) The binomial distribution on fn; n + 1; n + 2; : : : ; n + kg.
(f) The geometric distribution on f0; 1; 2; : : :; g with p(j) = 2=3j+1.",prob.pdf
"(e) The binomial distribution on fn; n + 1; n + 2; : : : ; n + kg.
(f) The geometric distribution on f0; 1; 2; : : :; g with p(j) = 2=3j+1.
2 For each of the distributions (a) through (d) of Exercise 1 calculate the rst
and second moments, 1 and 2, directly from their denition, and verify that
h(1) = 1, h0(1) = 1, and h00(1) = 2  1.
3 Let p be a probability distribution on f0; 1; 2g with moments 1 = 1, 2 = 3=2.
(a) Find its ordinary generating function h(z).
(b) Using (a), nd its moment generating function.
(c) Using (b), nd its rst six moments.
(d) Using (a), nd p0, p1, and p2.
4 In Exercise 3, the probability distribution is completely determined by its rst
two moments. Show that this is always true for any probability distribution
on f0; 1; 2g. Hint: Given 1 and 2, nd h(z) as in Exercise 3 and use h(z)
to determine p.
5 Let p and p0 be the two distributions
p =
 1 2 3 4 5
1=3 0 0 2=3 0

;
p0 =
1 2 3 4 5
0 2=3 0 0 1=3

:",prob.pdf
"10.1. DISCRETE DISTRIBUTIONS 375
(a) Show that p and p0 have the same rst and second moments, but not the
same third and fourth moments.
(b) Find the ordinary and moment generating functions for p and p0.
6 Let p be the probability distribution
p =
0 1 2
0 1=3 2=3

;
and let pn = p  p  
 
 
  p be the n-fold convolution of p with itself.
(a) Find p2 by direct calculation (see Denition 7.1).
(b) Find the ordinary generating functions h(z) and h2(z) for p and p2, and
verify that h2(z) = (h(z))2.
(c) Find hn(z) from h(z).
(d) Find the rst two moments, and hence the mean and variance, of pn
from hn(z). Verify that the mean of pn is n times the mean of p.
(e) Find those integers j for which pn(j) > 0 from hn(z).
7 Let X be a discrete random variable with values in f0; 1; 2; : : :; ng and moment
generating function g(t). Find, in terms of g(t), the generating functions for
(a) X.
(b) X + 1.
(c) 3X.
(d) aX + b.",prob.pdf
"generating function g(t). Find, in terms of g(t), the generating functions for
(a) X.
(b) X + 1.
(c) 3X.
(d) aX + b.
8 Let X1, X2, . . . , Xn be an independent trials process, with values in f0; 1g
and mean  = 1=3. Find the ordinary and moment generating functions for
the distribution of
(a) S1 = X1. Hint: First nd X1 explicitly.
(b) S2 = X1 + X2.
(c) Sn = X1 + X2 + 
 
 
 + Xn.
9 Let X and Y be random variables with values in f1; 2; 3; 4; 5; 6g with distri-
bution functions pX and pY given by
pX(j) = aj ;
pY (j) = bj :
(a) Find the ordinary generating functions hX(z) and hY (z) for these distri-
butions.
(b) Find the ordinary generating function hZ(z) for the distribution Z =
X + Y .",prob.pdf
"376 CHAPTER 10. GENERATING FUNCTIONS
(c) Show that hZ(z) cannot ever have the form
hZ(z) = z2 + z3 + 
 
 
 + z12
11 :
Hint: hX and hY must have at least one nonzero root, but hZ(z) in the form
given has no nonzero real roots.
It follows from this observation that there is no way to load two dice so that
the probability that a given sum will turn up when they are tossed is the same
for all sums (i.e., that all outcomes are equally likely).
10 Show that if
h(z) = 1 
p
1  4pqz2
2qz ;
then
h(1) =

p=q; if p  q;
1; if p  q;
and
h0(1) =
 1=(p  q); if p > q;
1; if p = q:
11 Show that if X is a random variable with mean  and variance 2, and if
X = (X  )= is the standardized version of X, then
gX(t) = et=gX
t


:
10.2 Branching Processes
Historical Background
In this section we apply the theory of generating functions to the study of an
important chance process called a branching process.
Until recently it was thought that the theory of branching processes originated",prob.pdf
"important chance process called a branching process.
Until recently it was thought that the theory of branching processes originated
with the following problem posed by Francis Galton in the Educational Times in
1873.1
Problem 4001: A large nation, of whom we will only concern ourselves
with the adult males, N in number, and who each bear separate sur-
names, colonise a district. Their law of population is such that, in each
generation, a0 per cent of the adult males have no male children who
reach adult life; a1 have one such male child; a2 have two; and so on up
to a5 who have ve.
Find (1) what proportion of the surnames will have become extinct
after r generations; and (2) how many instances there will be of the
same surname being held by m persons.
1D. G. Kendall, \Branching Processes Since 1873,"" Journal of London Mathematics Society,
vol. 41 (1966), p. 386.",prob.pdf
"10.2. BRANCHING PROCESSES 377
The rst attempt at a solution was given by Reverend H. W. Watson. Because
of a mistake in algebra, he incorrectly concluded that a family name would always
die out with probability 1. However, the methods that he employed to solve the
problems were, and still are, the basis for obtaining the correct solution.
Heyde and Seneta discovered an earlier communication by Bienayme (1845) that
anticipated Galton and Watson by 28 years. Bienayme showed, in fact, that he was
aware of the correct solution to Galton's problem. Heyde and Seneta in their book
I. J. Bienayme: Statistical Theory Anticipated,2 give the following translation from
Bienayme's paper:
If . . . the mean of the number of male children who replace the number
of males of the preceding generation were less than unity, it would be
easily realized that families are dying out due to the disappearance of
the members of which they are composed. However, the analysis shows",prob.pdf
"easily realized that families are dying out due to the disappearance of
the members of which they are composed. However, the analysis shows
further that when this mean is equal to unity families tend to disappear,
although less rapidly .. . .
The analysis also shows clearly that if the mean ratio is greater than
unity, the probability of the extinction of families with the passing of
time no longer reduces to certainty. It only approaches a nite limit,
which is fairly simple to calculate and which has the singular charac-
teristic of being given by one of the roots of the equation (in which
the number of generations is made innite) which is not relevant to the
question when the mean ratio is less than unity.3
Although Bienayme does not give his reasoning for these results, he did indicate
that he intended to publish a special paper on the problem. The paper was never
written, or at least has never been found. In his communication Bienayme indicated",prob.pdf
"written, or at least has never been found. In his communication Bienayme indicated
that he was motivated by the same problem that occurred to Galton. The opening
paragraph of his paper as translated by Heyde and Seneta says,
A great deal of consideration has been given to the possible multipli-
cation of the numbers of mankind; and recently various very curious
observations have been published on the fate which allegedly hangs over
the aristocrary and middle classes; the families of famous men, etc. This
fate, it is alleged, will inevitably bring about the disappearance of the
so-called families fermees.4
A much more extensive discussion of the history of branching processes may be
found in two papers by David G. Kendall.5
2C. C. Heyde and E. Seneta, I. J. Bienayme: Statistical Theory Anticipated (New York:
Springer Verlag, 1977).
3ibid., pp. 117{118.
4ibid., p. 118.
5D. G. Kendall, \Branching Processes Since 1873,"" pp. 385{406; and \The Genealogy of Ge-",prob.pdf
"Springer Verlag, 1977).
3ibid., pp. 117{118.
4ibid., p. 118.
5D. G. Kendall, \Branching Processes Since 1873,"" pp. 385{406; and \The Genealogy of Ge-
nealogy: Branching Processes Before (and After) 1873,"" Bulletin London Mathematics Society,
vol. 7 (1975), pp. 225{253.",prob.pdf
"378 CHAPTER 10. GENERATING FUNCTIONS
2
1
0
1/4
1/4
1/4
1/4
1/4
1/4
1/2
1/16
1/8
5/16
1/2
4
3
2
1
0
0
1
2
1/64
1/32
5/64
1/8
1/16
1/16
1/16
1/16
1/2
Figure 10.1: Tree diagram for Example 10.8.
Branching processes have served not only as crude models for population growth
but also as models for certain physical processes such as chemical and nuclear chain
reactions.
Problem of Extinction
We turn now to the rst problem posed by Galton (i.e., the problem of nding the
probability of extinction for a branching process). We start in the 0th generation
with 1 male parent. In the rst generation we shall have 0, 1, 2, 3, . . . male
ospring with probabilities p0, p1, p2, p3, . . . . If in the rst generation there are k
ospring, then in the second generation there will be X1 + X2 + 
 
 
 + Xk ospring,
where X1, X2, . . . , Xk are independent random variables, each with the common
distribution p0, p1, p2, . . . . This description enables us to construct a tree, and a",prob.pdf
"distribution p0, p1, p2, . . . . This description enables us to construct a tree, and a
tree measure, for any number of generations.
Examples
Example 10.8 Assume that p0 = 1=2, p1 = 1=4, and p2 = 1=4. Then the tree
measure for the rst two generations is shown in Figure 10.1.
Note that we use the theory of sums of independent random variables to assign
branch probabilities. For example, if there are two ospring in the rst generation,
the probability that there will be two in the second generation is
P(X1 + X2 = 2) = p0p2 + p1p1 + p2p0
= 1
2 
 1
4 + 1
4 
 1
4 + 1
4 
 1
2 = 5
16 :
We now study the probability that our process dies out (i.e., that at some
generation there are no ospring).",prob.pdf
"10.2. BRANCHING PROCESSES 379
Let dm be the probability that the process dies out by the mth generation. Of
course, d0 = 0. In our example, d1 = 1=2 and d2 = 1=2 + 1=8 + 1=16 = 11=16 (see
Figure 10.1). Note that we must add the probabilities for all paths that lead to 0
by the mth generation. It is clear from the denition that
0 = d0  d1  d2  
 
 
  1 :
Hence, dm converges to a limit d, 0  d  1, and d is the probability that the
process will ultimately die out. It is this value that we wish to determine. We
begin by expressing the value dm in terms of all possible outcomes on the rst
generation. If there are j ospring in the rst generation, then to die out by the
mth generation, each of these lines must die out in m  1 generations. Since they
proceed independently, this probability is (dm1)j. Therefore
dm = p0 + p1dm1 + p2(dm1)2 + p3(dm1)3 + 
 
 
 : (10.1)
Let h(z) be the ordinary generating function for the pi:
h(z) = p0 + p1z + p2z2 + 
 
 
 :",prob.pdf
"dm = p0 + p1dm1 + p2(dm1)2 + p3(dm1)3 + 
 
 
 : (10.1)
Let h(z) be the ordinary generating function for the pi:
h(z) = p0 + p1z + p2z2 + 
 
 
 :
Using this generating function, we can rewrite Equation 10.1 in the form
dm = h(dm1) : (10.2)
Since dm ! d, by Equation 10.2 we see that the value d that we are looking for
satises the equation
d = h(d) : (10.3)
One solution of this equation is always d = 1, since
1 = p0 + p1 + p2 + 
 
 
 :
This is where Watson made his mistake. He assumed that 1 was the only solution to
Equation 10.3. To examine this question more carefully, we rst note that solutions
to Equation 10.3 represent intersections of the graphs of
y = z
and
y = h(z) = p0 + p1z + p2z2 + 
 
 
 :
Thus we need to study the graph of y = h(z). We note that h(0) = p0. Also,
h0(z) = p1 + 2p2z + 3p3z2 + 
 
 
 ; (10.4)
and
h00(z) = 2p2 + 3 
 2p3z + 4 
 3p4z2 + 
 
 
 :
From this we see that for z  0, h0(z)  0 and h00(z)  0. Thus for nonnegative",prob.pdf
"and
h00(z) = 2p2 + 3 
 2p3z + 4 
 3p4z2 + 
 
 
 :
From this we see that for z  0, h0(z)  0 and h00(z)  0. Thus for nonnegative
z, h(z) is an increasing function and is concave upward. Therefore the graph of",prob.pdf
"380 CHAPTER 10. GENERATING FUNCTIONS
1 1 1
1
1
1
0
0
0
0
0
y
z
d > 1d < 1 d = 1 0
y = z
y y
z z
y = h (z)
1 1
(a) (c)(b)
Figure 10.2: Graphs of y = z and y = h(z).
y = h(z) can intersect the line y = z in at most two points. Since we know it must
intersect the line y = z at (1; 1), we know that there are just three possibilities, as
shown in Figure 10.2.
In case (a) the equation d = h(d) has roots fd; 1g with 0  d < 1. In the second
case (b) it has only the one root d = 1. In case (c) it has two roots f1; dg where
1 < d. Since we are looking for a solution 0  d  1, we see in cases (b) and (c)
that our only solution is 1. In these cases we can conclude that the process will die
out with probability 1. However in case (a) we are in doubt. We must study this
case more carefully.
From Equation 10.4 we see that
h0(1) = p1 + 2p2 + 3p3 + 
 
 
 = m ;
where m is the expected number of ospring produced by a single parent. In case (a)",prob.pdf
"h0(1) = p1 + 2p2 + 3p3 + 
 
 
 = m ;
where m is the expected number of ospring produced by a single parent. In case (a)
we have h0(1) > 1, in (b) h0(1) = 1, and in (c) h0(1) < 1. Thus our three cases
correspond to m > 1, m = 1, and m < 1. We assume now that m > 1. Recall that
d0 = 0, d1 = h(d0) = p0, d2 = h(d1), . .. , and dn = h(dn1). We can construct
these values geometrically, as shown in Figure 10.3.
We can see geometrically, as indicated for d0, d1, d2, and d3 in Figure 10.3, that
the points (di; h(di)) will always lie above the line y = z. Hence, they must converge
to the rst intersection of the curves y = z and y = h(z) (i.e., to the root d < 1).
This leads us to the following theorem. 2
Theorem 10.2 Consider a branching process with generating function h(z) for the
number of ospring of a given parent. Let d be the smallest root of the equation
z = h(z). If the mean number m of ospring produced by a single parent is  1,",prob.pdf
"z = h(z). If the mean number m of ospring produced by a single parent is  1,
then d = 1 and the process dies out with probability 1. If m > 1 then d < 1 and
the process dies out with probability d. 2
We shall often want to know the probability that a branching process dies out
by a particular generation, as well as the limit of these probabilities. Let dn be",prob.pdf
"10.2. BRANCHING PROCESSES 381
y = z
y = h(z)
y
z
1
p0
0 d = 0 1d d d d 1 2 3
Figure 10.3: Geometric determination of d.
the probability of dying out by the nth generation. Then we know that d1 = p0.
We know further that dn = h(dn1) where h(z) is the generating function for the
number of ospring produced by a single parent. This makes it easy to compute
these probabilities.
The program Branch calculates the values of dn. We have run this program
for 12 generations for the case that a parent can produce at most two ospring and
the probabilities for the number produced are p0 = :2, p1 = :5, and p2 = :3. The
results are given in Table 10.1.
We see that the probability of dying out by 12 generations is about .6. We shall
see in the next example that the probability of eventually dying out is 2/3, so that
even 12 generations is not enough to give an accurate estimate for this probability.
We now assume that at most two ospring can be produced. Then
h(z) = p0 + p1z + p2z2 :",prob.pdf
"We now assume that at most two ospring can be produced. Then
h(z) = p0 + p1z + p2z2 :
In this simple case the condition z = h(z) yields the equation
d = p0 + p1d + p2d2 ;
which is satised by d = 1 and d = p0=p2. Thus, in addition to the root d = 1 we
have the second root d = p0=p2. The mean number m of ospring produced by a
single parent is
m = p1 + 2p2 = 1  p0  p2 + 2p2 = 1  p0 + p2 :
Thus, if p0 > p2, m < 1 and the second root is > 1. If p0 = p2, we have a double
root d = 1. If p0 < p2, m > 1 and the second root d is less than 1 and represents
the probability that the process will die out.",prob.pdf
"382 CHAPTER 10. GENERATING FUNCTIONS
Generation Probability of dying out
1 .2
2 .312
3 .385203
4 .437116
5 .475879
6 .505878
7 .529713
8 .549035
9 .564949
10 .578225
11 .589416
12 .598931
Table 10.1: Probability of dying out.
p0 = :2092
p1 = :2584
p2 = :2360
p3 = :1593
p4 = :0828
p5 = :0357
p6 = :0133
p7 = :0042
p8 = :0011
p9 = :0002
p10 = :0000
Table 10.2: Distribution of number of female children.
Example 10.9 Keytz6 compiled and analyzed data on the continuation of the
female family line among Japanese women. His estimates at the basic probability
distribution for the number of female children born to Japanese women of ages
45{49 in 1960 are given in Table 10.2.
The expected number of girls in a family is then 1.837 so the probability d of
extinction is less than 1. If we run the program Branch, we can estimate that d is
in fact only about .324. 2
Distribution of Ospring
So far we have considered only the rst of the two problems raised by Galton,",prob.pdf
"in fact only about .324. 2
Distribution of Ospring
So far we have considered only the rst of the two problems raised by Galton,
namely the probability of extinction. We now consider the second problem, that
is, the distribution of the number Zn of ospring in the nth generation. The exact
form of the distribution is not known except in very special cases. We shall see,
6N. Keytz, Introduction to the Mathematics of Population, rev. ed. (Reading, PA: Addison
Wesley, 1977).",prob.pdf
"10.2. BRANCHING PROCESSES 383
however, that we can describe the limiting behavior of Zn as n ! 1.
We rst show that the generating function hn(z) of the distribution of Zn can
be obtained from h(z) for any branching process.
We recall that the value of the generating function at the value z for any random
variable X can be written as
h(z) = E(zX) = p0 + p1z + p2z2 + 
 
 
 :
That is, h(z) is the expected value of an experiment which has outcome zj with
probability pj.
Let Sn = X1 + X2 + 
 
 
 + Xn where each Xj has the same integer-valued
distribution (pj) with generating function k(z) = p0 + p1z + p2z2 + 
 
 
 : Let kn(z)
be the generating function of Sn. Then using one of the properties of ordinary
generating functions discussed in Section 10.1, we have
kn(z) = (k(z))n ;
since the Xj's are independent and all have the same distribution.
Consider now the branching process Zn. Let hn(z) be the generating function
of Zn. Then
hn+1(z) = E(zZn+1 )
=
X
k
E(zZn+1 jZn = k)P(Zn = k) :",prob.pdf
"Consider now the branching process Zn. Let hn(z) be the generating function
of Zn. Then
hn+1(z) = E(zZn+1 )
=
X
k
E(zZn+1 jZn = k)P(Zn = k) :
If Zn = k, then Zn+1 = X1 +X2 +
 
 
+Xk where X1, X2, .. . , Xk are independent
random variables with common generating function h(z). Thus
E(zZn+1 jZn = k) = E(zX1+X2 +


+Xk ) = (h(z))k ;
and
hn+1(z) =
X
k
(h(z))kP(Zn = k) :
But
hn(z) =
X
k
P(Zn = k)zk :
Thus,
hn+1(z) = hn(h(z)) : (10.5)
If we dierentiate Equation 10.5 and use the chain rule we have
h0
n+1(z) = h0
n(h(z))h0(z):
Putting z = 1 and using the fact that h(1) = 1, h0(1) = m, and h0
n(1) = mn = the
mean number of ospring in the n'th generation, we have
mn+1 = mn 
 m :
Thus, m2 = m 
 m = m2, m3 = m2 
 m = m3, and in general
mn = mn :
Thus, for a branching process with m > 1, the mean number of ospring grows
exponentially at a rate m.",prob.pdf
"384 CHAPTER 10. GENERATING FUNCTIONS
Examples
Example 10.10 For the branching process of Example 10.8 we have
h(z) = 1=2 + (1=4)z + (1=4)z2 ;
h2(z) = h(h(z)) = 1=2 + (1=4)[1=2 + (1=4)z + (1=4)z2]
= +(1=4)[1=2 + (1=4)z + (1=4)z2]2
= 11=16 + (1=8)z + (9=64)z2 + (1=32)z3 + (1=64)z4 :
The probabilities for the number of ospring in the second generation agree with
those obtained directly from the tree measure (see Figure 1). 2
It is clear that even in the simple case of at most two ospring, we cannot easily
carry out the calculation of hn(z) by this method. However, there is one special
case in which this can be done.
Example 10.11 Assume that the probabilities p1, p2, . . . form a geometric series:
pk = bck1, k = 1, 2, . . . , with 0 < b  1  c and 0 < c < 1. Then we have
p0 = 1  p1  p2  
 
 

= 1  b  bc  bc2  
 
 

= 1  b
1  c :
The generating function h(z) for this distribution is
h(z) = p0 + p1z + p2z2 + 
 
 

= 1  b
1  c + bz + bcz2 + bc2z3 + 
 
 

= 1  b
1  c + bz",prob.pdf
"The generating function h(z) for this distribution is
h(z) = p0 + p1z + p2z2 + 
 
 

= 1  b
1  c + bz + bcz2 + bc2z3 + 
 
 

= 1  b
1  c + bz
1  cz :
From this we nd
h0(z) = bcz
(1  cz)2 + b
1  cz = b
(1  cz)2
and
m = h0(1) = b
(1  c)2 :
We know that if m  1 the process will surely die out and d = 1. To nd the
probability d when m > 1 we must nd a root d < 1 of the equation
z = h(z) ;
or
z = 1  b
1  c + bz
1  cz:",prob.pdf
"10.2. BRANCHING PROCESSES 385
This leads us to a quadratic equation. We know that z = 1 is one solution. The
other is found to be
d = 1  b  c
c(1  c) :
It is easy to verify that d < 1 just when m > 1.
It is possible in this case to nd the distribution of Zn. This is done by rst
nding the generating function hn(z).7 The result for m 6= 1 is:
hn(z) = 1  mn
 1  d
mn  d

+
mn
h
1d
mnd
i2
z
1 
h
mn1
mnd
i
z
:
The coecients of the powers of z give the distribution for Zn:
P(Zn = 0) = 1  mn 1  d
mn  d = d(mn  1)
mn  d
and
P(Zn = j) = mn
 1  d
mn  d
2


mn  1
mn  d
j1
;
for j  1. 2
Example 10.12 Let us re-examine the Keytz data to see if a distribution of the
type considered in Example 10.11 could reasonably be used as a model for this
population. We would have to estimate from the data the parameters b and c for
the formula pk = bck1. Recall that
m = b
(1  c)2 (10.6)
and the probability d that the process dies out is
d = 1  b  c
c(1  c) : (10.7)",prob.pdf
"the formula pk = bck1. Recall that
m = b
(1  c)2 (10.6)
and the probability d that the process dies out is
d = 1  b  c
c(1  c) : (10.7)
Solving Equation 10.6 and 10.7 for b and c gives
c = m  1
m  d
and
b = m
1  d
m  d
2
:
We shall use the value 1.837 for m and .324 for d that we found in the Keytz
example. Using these values, we obtain b = :3666 and c = :5533. Note that
(1  c)2 < b < 1  c, as required. In Table 10.3 we give for comparison the
probabilities p0 through p8 as calculated by the geometric distribution versus the
empirical values.
7T. E. Harris, The Theory of Branching Processes (Berlin: Springer, 1963), p. 9.",prob.pdf
"386 CHAPTER 10. GENERATING FUNCTIONS
Geometric
pj Data Model
0 .2092 .1816
1 .2584 .3666
2 .2360 .2028
3 .1593 .1122
4 .0828 .0621
5 .0357 .0344
6 .0133 .0190
7 .0042 .0105
8 .0011 .0058
9 .0002 .0032
10 .0000 .0018
Table 10.3: Comparison of observed and expected frequencies.
The geometric model tends to favor the larger numbers of ospring but is similar
enough to show that this modied geometric distribution might be appropriate to
use for studies of this kind.
Recall that if Sn = X1 + X2 + 
 
 
 + Xn is the sum of independent random
variables with the same distribution then the Law of Large Numbers states that
Sn=n converges to a constant, namely E(X1). It is natural to ask if there is a
similar limiting theorem for branching processes.
Consider a branching process with Zn representing the number of ospring after
n generations. Then we have seen that the expected value of Zn is mn. Thus we can
scale the random variable Zn to have expected value 1 by considering the random
variable",prob.pdf
"scale the random variable Zn to have expected value 1 by considering the random
variable
Wn = Zn
mn :
In the theory of branching processes it is proved that this random variable Wn
will tend to a limit as n tends to innity. However, unlike the case of the Law of
Large Numbers where this limit is a constant, for a branching process the limiting
value of the random variables Wn is itself a random variable.
Although we cannot prove this theorem here we can illustrate it by simulation.
This requires a little care. When a branching process survives, the number of
ospring is apt to get very large. If in a given generation there are 1000 ospring,
the ospring of the next generation are the result of 1000 chance events, and it will
take a while to simulate these 1000 experiments. However, since the nal result is
the sum of 1000 independent experiments we can use the Central Limit Theorem to
replace these 1000 experiments by a single experiment with normal density having",prob.pdf
"replace these 1000 experiments by a single experiment with normal density having
the appropriate mean and variance. The program BranchingSimulation carries
out this process.
We have run this program for the Keytz example, carrying out 10 simulations
and graphing the results in Figure 10.4.
The expected number of female ospring per female is 1.837, so that we are
graphing the outcome for the random variables Wn = Zn=(1:837)n. For three of",prob.pdf
"10.2. BRANCHING PROCESSES 387
5 10 15 20 25
0.5
1
1.5
2
2.5
3
Figure 10.4: Simulation of Zn=mn for the Keytz example.
the simulations the process died out, which is consistent with the value d = :3 that
we found for this example. For the other seven simulations the value of Wn tends
to a limiting value which is dierent for each simulation. 2
Example 10.13 We now examine the random variable Zn more closely for the
case m < 1 (see Example 10.11). Fix a value t > 0; let [tmn] be the integer part of
tmn. Then
P(Zn = [tmn]) = mn( 1  d
mn  d)2(mn  1
mn  d)[tmn]1
= 1
mn ( 1  d
1  d=mn )2(1  1=mn
1  d=mn )tmn+a ;
where jaj  2. Thus, as n ! 1,
mnP(Zn = [tmn]) ! (1  d)2 et
etd = (1  d)2et(1d) :
For t = 0,
P(Zn = 0) ! d :
We can compare this result with the Central Limit Theorem for sums Sn of integer-
valued independent random variables (see Theorem 9.3), which states that if t is an
integer and u = (t  n)=
p
2n, then as n ! 1,
p
2nP(Sn = u
p
2n + n) ! 1p
2eu2=2 :",prob.pdf
"integer and u = (t  n)=
p
2n, then as n ! 1,
p
2nP(Sn = u
p
2n + n) ! 1p
2eu2=2 :
We see that the form of these statements are quite similar. It is possible to prove
a limit theorem for a general class of branching processes that states that under",prob.pdf
"388 CHAPTER 10. GENERATING FUNCTIONS
suitable hypotheses, as n ! 1,
mnP(Zn = [tmn]) ! k(t) ;
for t > 0, and
P(Zn = 0) ! d :
However, unlike the Central Limit Theorem for sums of independent random vari-
ables, the function k(t) will depend upon the basic distribution that determines the
process. Its form is known for only a very few examples similar to the one we have
considered here. 2
Chain Letter Problem
Example 10.14 An interesting example of a branching process was suggested by
Free Huizinga.8 In 1978, a chain letter called the \Circle of Gold,"" believed to have
started in California, found its way across the country to the theater district of New
York. The chain required a participant to buy a letter containing a list of 12 names
for 100 dollars. The buyer gives 50 dollars to the person from whom the letter was
purchased and then sends 50 dollars to the person whose name is at the top of the
list. The buyer then crosses o the name at the top of the list and adds her own",prob.pdf
"list. The buyer then crosses o the name at the top of the list and adds her own
name at the bottom in each letter before it is sold again.
Let us rst assume that the buyer may sell the letter only to a single person.
If you buy the letter you will want to compute your expected winnings. (We are
ignoring here the fact that the passing on of chain letters through the mail is a
federal oense with certain obvious resulting penalties.) Assume that each person
involved has a probability p of selling the letter. Then you will receive 50 dollars
with probability p and another 50 dollars if the letter is sold to 12 people, since then
your name would have risen to the top of the list. This occurs with probability p12,
and so your expected winnings are 100 + 50p + 50p12. Thus the chain in this
situation is a highly unfavorable game.
It would be more reasonable to allow each person involved to make a copy of
the list and try to sell the letter to at least 2 other people. Then you would have",prob.pdf
"the list and try to sell the letter to at least 2 other people. Then you would have
a chance of recovering your 100 dollars on these sales, and if any of the letters is
sold 12 times you will receive a bonus of 50 dollars for each of these cases. We can
consider this as a branching process with 12 generations. The members of the rst
generation are the letters you sell. The second generation consists of the letters sold
by members of the rst generation, and so forth.
Let us assume that the probabilities that each individual sells letters to 0, 1,
or 2 others are p0, p1, and p2, respectively. Let Z1, Z2, . . . , Z12 be the number of
letters in the rst 12 generations of this branching process. Then your expected
winnings are
50(E(Z1) + E(Z12)) = 50m + 50m12 ;
8Private communication.",prob.pdf
"10.2. BRANCHING PROCESSES 389
where m = p1 +2p2 is the expected number of letters you sold. Thus to be favorable
we just have
50m + 50m12 > 100 ;
or
m + m12 > 2 :
But this will be true if and only if m > 1. We have seen that this will occur in
the quadratic case if and only if p2 > p0. Let us assume for example that p0 = :2,
p1 = :5, and p2 = :3. Then m = 1:1 and the chain would be a favorable game. Your
expected prot would be
50(1:1 + 1:112)  100  112 :
The probability that you receive at least one payment from the 12th generation is
1d12. We nd from our program Branch that d12 = :599. Thus, 1d12 = :401 is
the probability that you receive some bonus. The maximum that you could receive
from the chain would be 50(2 + 212) = 204;900 if everyone were to successfully sell
two letters. Of course you can not always expect to be so lucky. (What is the
probability of this happening?)
To simulate this game, we need only simulate a branching process for 12 gen-",prob.pdf
"probability of this happening?)
To simulate this game, we need only simulate a branching process for 12 gen-
erations. Using a slightly modied version of our program BranchingSimulation
we carried out twenty such simulations, giving the results shown in Table 10.4.
Note that we were quite lucky on a few runs, but we came out ahead only a
little less than half the time. The process died out by the twelfth generation in 12
out of the 20 experiments, in good agreement with the probability d12 = :599 that
we calculated using the program Branch.
Let us modify the assumptions about our chain letter to let the buyer sell the
letter to as many people as she can instead of to a maximum of two. We shall
assume, in fact, that a person has a large number N of acquaintances and a small
probability p of persuading any one of them to buy the letter. Then the distribution
for the number of letters that she sells will be a binomial distribution with mean",prob.pdf
"for the number of letters that she sells will be a binomial distribution with mean
m = Np. Since N is large and p is small, we can assume that the probability pj
that an individual sells the letter to j people is given by the Poisson distribution
pj = emmj
j! :",prob.pdf
"390 CHAPTER 10. GENERATING FUNCTIONS
Z1 Z2 Z3 Z4 Z5 Z6 Z7 Z8 Z9 Z10 Z11 Z12 Prot
1 0 0 0 0 0 0 0 0 0 0 0 -50
1 1 2 3 2 3 2 1 2 3 3 6 250
0 0 0 0 0 0 0 0 0 0 0 0 -100
2 4 4 2 3 4 4 3 2 2 1 1 50
1 2 3 5 4 3 3 3 5 8 6 6 250
0 0 0 0 0 0 0 0 0 0 0 0 -100
2 3 2 2 2 1 2 3 3 3 4 6 300
1 2 1 1 1 1 2 1 0 0 0 0 -50
0 0 0 0 0 0 0 0 0 0 0 0 -100
1 0 0 0 0 0 0 0 0 0 0 0 -50
2 3 2 3 3 3 5 9 12 12 13 15 750
1 1 1 0 0 0 0 0 0 0 0 0 -50
1 2 2 3 3 0 0 0 0 0 0 0 -50
1 1 1 1 2 2 3 4 4 6 4 5 200
1 1 0 0 0 0 0 0 0 0 0 0 -50
1 0 0 0 0 0 0 0 0 0 0 0 -50
1 0 0 0 0 0 0 0 0 0 0 0 -50
1 1 2 3 3 4 2 3 3 3 3 2 50
1 2 4 6 6 9 10 13 16 17 15 18 850
1 0 0 0 0 0 0 0 0 0 0 0 -50
Table 10.4: Simulation of chain letter (nite distribution case).",prob.pdf
"10.2. BRANCHING PROCESSES 391
Z1 Z2 Z3 Z4 Z5 Z6 Z7 Z8 Z9 Z10 Z11 Z12 Prot
1 2 6 7 7 8 11 9 7 6 6 5 200
1 0 0 0 0 0 0 0 0 0 0 0 -50
1 0 0 0 0 0 0 0 0 0 0 0 -50
1 1 1 0 0 0 0 0 0 0 0 0 -50
0 0 0 0 0 0 0 0 0 0 0 0 -100
1 1 1 1 1 1 2 4 9 7 9 7 300
2 3 3 4 2 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 -50
2 1 0 0 0 0 0 0 0 0 0 0 0
3 3 4 7 11 17 14 11 11 10 16 25 1300
0 0 0 0 0 0 0 0 0 0 0 0 -100
1 2 2 1 1 3 1 0 0 0 0 0 -50
0 0 0 0 0 0 0 0 0 0 0 0 -100
2 3 1 0 0 0 0 0 0 0 0 0 0
3 1 0 0 0 0 0 0 0 0 0 0 50
1 0 0 0 0 0 0 0 0 0 0 0 -50
3 4 4 7 10 11 9 11 12 14 13 10 550
1 3 3 4 9 5 7 9 8 8 6 3 100
1 0 4 6 6 9 10 13 0 0 0 0 -50
1 0 0 0 0 0 0 0 0 0 0 0 -50
Table 10.5: Simulation of chain letter (Poisson case).
The generating function for the Poisson distribution is
h(z) =
1X
j=0
emmjzj
j!
= em
1X
j=0
mjzj
j!
= ememz = em(z1) :
The expected number of letters that an individual passes on is m, and again to
be favorable we must have m > 1. Let us assume again that m = 1:1. Then we",prob.pdf
"be favorable we must have m > 1. Let us assume again that m = 1:1. Then we
can nd again the probability 1  d12 of a bonus from Branch. The result is .232.
Although the expected winnings are the same, the variance is larger in this case,
and the buyer has a better chance for a reasonably large prot. We again carried
out 20 simulations using the Poisson distribution with mean 1.1. The results are
shown in Table 10.5.
We note that, as before, we came out ahead less than half the time, but we also
had one large prot. In only 6 of the 20 cases did we receive any prot. This is
again in reasonable agreement with our calculation of a probability .232 for this
happening. 2",prob.pdf
"392 CHAPTER 10. GENERATING FUNCTIONS
Exercises
1 Let Z1, Z2, . . ., ZN describe a branching process in which each parent has
j ospring with probability pj. Find the probability d that the process even-
tually dies out if
(a) p0 = 1=2, p1 = 1=4, and p2 = 1=4.
(b) p0 = 1=3, p1 = 1=3, and p2 = 1=3.
(c) p0 = 1=3, p1 = 0, and p2 = 2=3.
(d) pj = 1=2j+1, for j = 0, 1, 2, . . . .
(e) pj = (1=3)(2=3)j, for j = 0, 1, 2, . . . .
(f) pj = e22j=j!, for j = 0, 1, 2, . .. (estimate d numerically).
2 Let Z1, Z2, . . ., ZN describe a branching process in which each parent has
j ospring with probability pj. Find the probability d that the process dies
out if
(a) p0 = 1=2, p1 = p2 = 0, and p3 = 1=2.
(b) p0 = p1 = p2 = p3 = 1=4.
(c) p0 = t, p1 = 1  2t, p2 = 0, and p3 = t, where t  1=2.
3 In the chain letter problem (see Example 10.14) nd your expected prot if
(a) p0 = 1=2, p1 = 0, and p2 = 1=2.
(b) p0 = 1=6, p1 = 1=2, and p2 = 1=3.
Show that if p0 > 1=2, you cannot expect to make a prot.",prob.pdf
"(a) p0 = 1=2, p1 = 0, and p2 = 1=2.
(b) p0 = 1=6, p1 = 1=2, and p2 = 1=3.
Show that if p0 > 1=2, you cannot expect to make a prot.
4 Let SN = X1 + X2 + 
 
 
 + XN , where the Xi's are independent random
variables with common distribution having generating function f(z). Assume
that N is an integer valued random variable independent of all of the Xj and
having generating function g(z). Show that the generating function for SN is
h(z) = g(f(z)). Hint: Use the fact that
h(z) = E(zSN ) =
X
k
E(zSN jN = k)P(N = k) :
5 We have seen that if the generating function for the ospring of a single
parent is f(z), then the generating function for the number of ospring after
two generations is given by h(z) = f(f(z)). Explain how this follows from the
result of Exercise 4.
6 Consider a queueing process such that in each minute either 1 or 0 customers
arrive with probabilities p or q = 1  p, respectively. (The number p is called",prob.pdf
"arrive with probabilities p or q = 1  p, respectively. (The number p is called
the arrival rate.) When a customer starts service she nishes in the next
minute with probability r. The number r is called the service rate.) Thus
when a customer begins being served she will nish being served in j minutes
with probability (1  r)j1r, for j = 1, 2, 3, . . . .",prob.pdf
"10.3. CONTINUOUS DENSITIES 393
(a) Find the generating function f(z) for the number of customers who arrive
in one minute and the generating function g(z) for the length of time that
a person spends in service once she begins service.
(b) Consider a customer branching process by considering the ospring of a
customer to be the customers who arrive while she is being served. Using
Exercise 4, show that the generating function for our customer branching
process is h(z) = g(f(z)).
(c) If we start the branching process with the arrival of the rst customer,
then the length of time until the branching process dies out will be the
busy period for the server. Find a condition in terms of the arrival rate
and service rate that will assure that the server will ultimately have a
time when he is not busy.
7 Let N be the expected total number of ospring in a branching process. Let
m be the mean number of ospring of a single parent. Show that
N = 1 +
X
pk 
 k

N = 1 + mN",prob.pdf
"m be the mean number of ospring of a single parent. Show that
N = 1 +
X
pk 
 k

N = 1 + mN
and hence that N is nite if and only if m < 1 and in that case N = 1=(1m).
8 Consider a branching process such that the number of ospring of a parent is
j with probability 1=2j+1 for j = 0, 1, 2, .. . .
(a) Using the results of Example 10.11 show that the probability that there
are j ospring in the nth generation is
p(n)
j =
 1
n(n+1)( n
n+1)j; if j  1;
n
n+1; if j = 0:
(b) Show that the probability that the process dies out exactly at the nth
generation is 1=n(n + 1).
(c) Show that the expected lifetime is innite even though d = 1.
10.3 Generating Functions for Continuous Densi-
ties
In the previous section, we introduced the concepts of moments and moment gen-
erating functions for discrete random variables. These concepts have natural ana-
logues for continuous random variables, provided some care is taken in arguments
involving convergence.
Moments",prob.pdf
"logues for continuous random variables, provided some care is taken in arguments
involving convergence.
Moments
If X is a continuous random variable dened on the probability space 
, with
density function fX, then we dene the nth moment of X by the formula
n = E(Xn) =
Z +1
1
xnfX(x) dx ;",prob.pdf
"394 CHAPTER 10. GENERATING FUNCTIONS
provided the integral
n = E(Xn) =
Z +1
1
jxjnfX(x) dx ;
is nite. Then, just as in the discrete case, we see that 0 = 1, 1 = , and
2  2
1 = 2.
Moment Generating Functions
Now we dene the moment generating function g(t) for X by the formula
g(t) =
1X
k=0
ktk
k! =
1X
k=0
E(Xk)tk
k!
= E(etX ) =
Z +1
1
etxfX(x) dx ;
provided this series converges. Then, as before, we have
n = g(n)(0) :
Examples
Example 10.15 Let X be a continuous random variable with range [0; 1] and
density function fX(x) = 1 for 0  x  1 (uniform density). Then
n =
Z 1
0
xn dx = 1
n + 1 ;
and
g(t) =
1X
k=0
tk
(k + 1)!
= et  1
t :
Here the series converges for all t. Alternatively, we have
g(t) =
Z +1
1
etxfX(x) dx
=
Z 1
0
etx dx = et  1
t :
Then (by L'H^opital's rule)
0 = g(0) = lim
t!0
et  1
t = 1 ;
1 = g0(0) = lim
t!0
tet  et + 1
t2 = 1
2 ;
2 = g00(0) = lim
t!0
t3et  2t2et + 2tet  2t
t4 = 1
3 :",prob.pdf
"10.3. CONTINUOUS DENSITIES 395
In particular, we verify that  = g0(0) = 1=2 and
2 = g00(0)  (g0(0))2 = 1
3  1
4 = 1
12
as before (see Example 6.25). 2
Example 10.16 Let X have range [ 0; 1) and density function fX(x) = ex
(exponential density with parameter ). In this case
n =
Z 1
0
xnex dx = (1)n dn
dn
Z 1
0
ex dx
= (1)n dn
dn [1
] = n!
n ;
and
g(t) =
1X
k=0
ktk
k!
=
1X
k=0
[ t
]k = 
  t :
Here the series converges only for jtj < . Alternatively, we have
g(t) =
Z 1
0
etxex dx
= e(t)x
t  




1
0
= 
  t :
Now we can verify directly that
n = g(n)(0) = n!
(  t)n+1




t=0
= n!
n :
2
Example 10.17 Let X have range (1; +1) and density function
fX(x) = 1p
2ex2=2
(normal density). In this case we have
n = 1p
2
Z +1
1
xnex2=2 dx
=
 (2m)!
2mm!; if n = 2m,
0; if n = 2m + 1.",prob.pdf
"396 CHAPTER 10. GENERATING FUNCTIONS
(These moments are calculated by integrating once by parts to show that n =
(n  1)n2, and observing that 0 = 1 and 1 = 0.) Hence,
g(t) =
1X
n=0
ntn
n!
=
1X
m=0
t2m
2mm! = et2=2 :
This series converges for all values of t. Again we can verify that g(n)(0) = n.
Let X be a normal random variable with parameters  and . It is easy to show
that the moment generating function of X is given by
et+(2 =2)t2
:
Now suppose that X and Y are two independent normal random variables with
parameters 1, 1, and 2, 2, respectively. Then, the product of the moment
generating functions of X and Y is
et(1 +2 )+((2
1 +2
2 )=2)t2
:
This is the moment generating function for a normal random variable with mean
1 + 2 and variance 2
1 + 2
2. Thus, the sum of two independent normal random
variables is again normal. (This was proved for the special case that both summands
are standard normal in Example 7.5.) 2",prob.pdf
"variables is again normal. (This was proved for the special case that both summands
are standard normal in Example 7.5.) 2
In general, the series dening g(t) will not converge for all t. But in the important
special case where X is bounded (i.e., where the range of X is contained in a nite
interval), we can show that the series does converge for all t.
Theorem 10.3 Suppose X is a continuous random variable with range contained
in the interval [M; M]. Then the series
g(t) =
1X
k=0
ktk
k!
converges for all t to an innitely dierentiable function g(t), and g(n)(0) = n.
Proof. We have
k =
Z +M
M
xkfX(x) dx ;
so
jkj 
Z +M
M
jxjkfX(x) dx
 Mk
Z +M
M
fX(x) dx = Mk :",prob.pdf
"10.3. CONTINUOUS DENSITIES 397
Hence, for all N we have
NX
k=0




ktk
k!




NX
k=0
(Mjtj)k
k!  eMjtj ;
which shows that the power series converges for all t. We know that the sum of a
convergent power series is always dierentiable. 2
Moment Problem
Theorem 10.4 If X is a bounded random variable, then the moment generating
function gX(t) of x determines the density function fX(x) uniquely.
Sketch of the Proof. We know that
gX(t) =
1X
k=0
ktk
k!
=
Z +1
1
etxf(x) dx :
If we replace t by i, where  is real and i = p1, then the series converges for
all , and we can dene the function
kX() = gX(i) =
Z +1
1
eixfX(x) dx :
The function kX () is called the characteristic function of X, and is dened by
the above equation even when the series for gX does not converge. This equation
says that kX is the Fourier transform of fX. It is known that the Fourier transform
has an inverse, given by the formula
fX(x) = 1
2
Z +1
1
eixkX() d ;",prob.pdf
"has an inverse, given by the formula
fX(x) = 1
2
Z +1
1
eixkX() d ;
suitably interpreted.9 Here we see that the characteristic function kX , and hence
the moment generating function gX, determines the density function fX uniquely
under our hypotheses. 2
Sketch of the Proof of the Central Limit Theorem
With the above result in mind, we can now sketch a proof of the Central Limit
Theorem for bounded continuous random variables (see Theorem 9.6). To this end,
let X be a continuous random variable with density function fX, mean  = 0 and
variance 2 = 1, and moment generating function g(t) dened by its series for all t.
9H. Dym and H. P. McKean, Fourier Series and Integrals (New York: Academic Press, 1972).",prob.pdf
"398 CHAPTER 10. GENERATING FUNCTIONS
Let X1, X2, . . . , Xn be an independent trials process with each Xi having density
fX, and let Sn = X1 + X2 + 
 
 
 + Xn, and S
n = (Sn  n)=
p
n2 = Sn=pn. Then
each Xi has moment generating function g(t), and since the Xi are independent,
the sum Sn, just as in the discrete case (see Section 10.1), has moment generating
function
gn(t) = (g(t))n ;
and the standardized sum S
n has moment generating function
g
n(t) =

g
 tpn
n
:
We now show that, as n ! 1, g
n(t) ! et2=2, where et2=2 is the moment gener-
ating function of the normal density n(x) = (1=
p
2)ex2=2 (see Example 10.17).
To show this, we set u(t) = log g(t), and
u
n(t) = log g
n(t)
= n log g
 t
pn

= nu
 tpn

;
and show that u
n(t) ! t2=2 as n ! 1. First we note that
u(0) = log gn(0) = 0 ;
u0(0) = g0(0)
g(0) = 1
1 = 0 ;
u00(0) = g00(0)g(0)  (g0(0))2
(g(0))2
= 2  2
1
1 = 2 = 1 :
Now by using L'H^opital's rule twice, we get
lim
n!1
u
n(t) = lim
s!1
u(t=p
s)
s1",prob.pdf
"u00(0) = g00(0)g(0)  (g0(0))2
(g(0))2
= 2  2
1
1 = 2 = 1 :
Now by using L'H^opital's rule twice, we get
lim
n!1
u
n(t) = lim
s!1
u(t=p
s)
s1
= lim
s!1
u0(t=ps)t
2s1=2
= lim
s!1
u00
 tps
t2
2 = 2 t2
2 = t2
2 :
Hence, g
n(t) ! et2=2 as n ! 1. Now to complete the proof of the Central Limit
Theorem, we must show that if g
n(t) ! et2=2, then under our hypotheses the
distribution functions F
n(x) of the S
n must converge to the distribution function
F
N (x) of the normal variable N; that is, that
F
n(a) = P(S
n  a) ! 1p
2
Z a
1
ex2=2 dx ;
and furthermore, that the density functions f
n(x) of the S
n must converge to the
density function for N; that is, that
f
n(x) ! 1p
2ex2=2 ;",prob.pdf
"10.3. CONTINUOUS DENSITIES 399
as n ! 1.
Since the densities, and hence the distributions, of the S
n are uniquely deter-
mined by their moment generating functions under our hypotheses, these conclu-
sions are certainly plausible, but their proofs involve a detailed examination of
characteristic functions and Fourier transforms, and we shall not attempt them
here.
In the same way, we can prove the Central Limit Theorem for bounded discrete
random variables with integer values (see Theorem 9.4). Let X be a discrete random
variable with density function p(j), mean  = 0, variance 2 = 1, and moment
generating function g(t), and let X1, X2, . . . , Xn form an independent trials process
with common density p. Let Sn = X1 + X2 + 
 
 
 + Xn and S
n = Sn=pn, with
densities pn and p
n, and moment generating functions gn(t) and g
n(t) =

g( t
pn )
n
:
Then we have
g
n(t) ! et2=2 ;
just as in the continuous case, and this implies in the same way that the distribution
functions F",prob.pdf
"
g( t
pn )
n
:
Then we have
g
n(t) ! et2=2 ;
just as in the continuous case, and this implies in the same way that the distribution
functions F
n(x) converge to the normal distribution; that is, that
F
n(a) = P(S
n  a) ! 1p
2
Z a
1
ex2=2 dx ;
as n ! 1.
The corresponding statement about the distribution functions p
n, however, re-
quires a little extra care (see Theorem 9.3). The trouble arises because the dis-
tribution p(x) is not dened for all x, but only for integer x. It follows that the
distribution p
n(x) is dened only for x of the form j=p
n, and these values change
as n changes.
We can x this, however, by introducing the function p(x), dened by the for-
mula
p(x) =
 p(j); if j  1=2  x < j + 1=2,
0 ; otherwise:
Then p(x) is dened for all x, p(j) = p(j), and the graph of p(x) is the step function
for the distribution p(j) (see Figure 3 of Section 9.1).
In the same way we introduce the step function pn(x) and p
n(x) associated with",prob.pdf
"for the distribution p(j) (see Figure 3 of Section 9.1).
In the same way we introduce the step function pn(x) and p
n(x) associated with
the distributions pn and p
n, and their moment generating functions gn(t) and g
n(t).
If we can show that g
n(t) ! et2=2, then we can conclude that
p
n(x) ! 1
p
2et2 =2 ;
as n ! 1, for all x, a conclusion strongly suggested by Figure 9.3.
Now g(t) is given by
g(t) =
Z +1
1
etx p(x) dx
=
+NX
j=N
Z j+1=2
j1=2
etxp(j) dx",prob.pdf
"400 CHAPTER 10. GENERATING FUNCTIONS
=
+NX
j=N
p(j)etj et=2  et=2
2t=2
= g(t)sinh(t=2)
t=2 ;
where we have put
sinh(t=2) = et=2  et=2
2 :
In the same way, we nd that
gn(t) = gn(t)sinh(t=2)
t=2 ;
g
n(t) = g
n(t)sinh(t=2pn)
t=2pn :
Now, as n ! 1, we know that g
n(t) ! et2=2, and, by L'H^opital's rule,
lim
n!1
sinh(t=2pn)
t=2pn = 1 :
It follows that
g
n(t) ! et2=2 ;
and hence that
p
n(x) ! 1
p
2
ex2=2 ;
as n ! 1. The astute reader will note that in this sketch of the proof of Theo-
rem 9.3, we never made use of the hypothesis that the greatest common divisor of
the dierences of all the values that the Xi can take on is 1. This is a technical
point that we choose to ignore. A complete proof may be found in Gnedenko and
Kolmogorov.10
Cauchy Density
The characteristic function of a continuous density is a useful tool even in cases when
the moment series does not converge, or even in cases when the moments themselves",prob.pdf
"the moment series does not converge, or even in cases when the moments themselves
are not nite. As an example, consider the Cauchy density with parameter a = 1
(see Example 5.10)
f(x) = 1
(1 + x2) :
If X and Y are independent random variables with Cauchy density f(x), then the
average Z = (X + Y )=2 also has Cauchy density f(x), that is,
fZ(x) = f(x) :
10B. V. Gnedenko and A. N. Kolomogorov, Limit Distributions for Sums of Independent Random
Variables (Reading: Addison-Wesley, 1968), p. 233.",prob.pdf
"10.3. CONTINUOUS DENSITIES 401
This is hard to check directly, but easy to check by using characteristic functions.
Note rst that
2 = E(X2) =
Z +1
1
x2
(1 + x2) dx = 1
so that 2 is innite. Nevertheless, we can dene the characteristic function kX ()
of x by the formula
kX() =
Z +1
1
eix 1
(1 + x2) dx :
This integral is easy to do by contour methods, and gives us
kX () = kY () = ejj :
Hence,
kX+Y () = (ejj)2 = e2jj ;
and since
kZ() = kX+Y (=2) ;
we have
kZ() = e2j=2j = ejj :
This shows that kZ = kX = kY , and leads to the conclusions that fZ = fX = fY .
It follows from this that if X1, X2, . . . , Xn is an independent trials process with
common Cauchy density, and if
An = X1 + X2 + 
 
 
 + Xn
n
is the average of the Xi, then An has the same density as do the Xi. This means
that the Law of Large Numbers fails for this process; the distribution of the average
An is exactly the same as for the individual terms. Our proof of the Law of Large",prob.pdf
"An is exactly the same as for the individual terms. Our proof of the Law of Large
Numbers fails in this case because the variance of Xi is not nite.
Exercises
1 Let X be a continuous random variable with values in [ 0; 2] and density fX.
Find the moment generating function g(t) for X if
(a) fX(x) = 1=2.
(b) fX(x) = (1=2)x.
(c) fX(x) = 1  (1=2)x.
(d) fX(x) = j1  xj.
(e) fX(x) = (3=8)x2.
Hint: Use the integral denition, as in Examples 10.15 and 10.16.
2 For each of the densities in Exercise 1 calculate the rst and second moments,
1 and 2, directly from their denition and verify that g(0) = 1, g0(0) = 1,
and g00(0) = 2.",prob.pdf
"402 CHAPTER 10. GENERATING FUNCTIONS
3 Let X be a continuous random variable with values in [ 0; 1) and density fX.
Find the moment generating functions for X if
(a) fX(x) = 2e2x.
(b) fX(x) = e2x + (1=2)ex.
(c) fX(x) = 4xe2x.
(d) fX(x) = (x)n1ex=(n  1)!.
4 For each of the densities in Exercise 3, calculate the rst and second moments,
1 and 2, directly from their denition and verify that g(0) = 1, g0(0) = 1,
and g00(0) = 2.
5 Find the characteristic function kX() for each of the random variables X of
Exercise 1.
6 Let X be a continuous random variable whose characteristic function kX ()
is
kX () = ejj; 1 <  < +1 :
Show directly that the density fX of X is
fX(x) = 1
(1 + x2) :
7 Let X be a continuous random variable with values in [ 0; 1], uniform density
function fX(x)  1 and moment generating function g(t) = (et  1)=t. Find
in terms of g(t) the moment generating function for
(a) X.
(b) 1 + X.
(c) 3X.
(d) aX + b.",prob.pdf
"in terms of g(t) the moment generating function for
(a) X.
(b) 1 + X.
(c) 3X.
(d) aX + b.
8 Let X1, X2, . . . , Xn be an independent trials process with uniform density.
Find the moment generating function for
(a) X1.
(b) S2 = X1 + X2.
(c) Sn = X1 + X2 + 
 
 
 + Xn.
(d) An = Sn=n.
(e) S
n = (Sn  n)=
p
n2.
9 Let X1, X2, . . . , Xn be an independent trials process with normal density of
mean 1 and variance 2. Find the moment generating function for
(a) X1.
(b) S2 = X1 + X2.",prob.pdf
"10.3. CONTINUOUS DENSITIES 403
(c) Sn = X1 + X2 + 
 
 
 + Xn.
(d) An = Sn=n.
(e) S
n = (Sn  n)=
p
n2.
10 Let X1, X2, . . . , Xn be an independent trials process with density
f(x) = 1
2ejxj; 1 < x < +1 :
(a) Find the mean and variance of f(x).
(b) Find the moment generating function for X1, Sn, An, and S
n.
(c) What can you say about the moment generating function of S
n as n !
1?
(d) What can you say about the moment generating function of An as n !
1?",prob.pdf
404 CHAPTER 10. GENERATING FUNCTIONS,prob.pdf
"Chapter 11
Markov Chains
11.1 Introduction
Most of our study of probability has dealt with independent trials processes. These
processes are the basis of classical probability theory and much of statistics. We
have discussed two of the principal theorems for these processes: the Law of Large
Numbers and the Central Limit Theorem.
We have seen that when a sequence of chance experiments forms an indepen-
dent trials process, the possible outcomes for each experiment are the same and
occur with the same probability. Further, knowledge of the outcomes of the pre-
vious experiments does not in
uence our predictions for the outcomes of the next
experiment. The distribution for the outcomes of a single experiment is sucient
to construct a tree and a tree measure for a sequence of n experiments, and we
can answer any probability question about these experiments by using this tree
measure.
Modern probability theory studies chance processes for which the knowledge",prob.pdf
"measure.
Modern probability theory studies chance processes for which the knowledge
of previous outcomes in
uences predictions for future experiments. In principle,
when we observe a sequence of chance experiments, all of the past outcomes could
in
uence our predictions for the next experiment. For example, this should be the
case in predicting a student's grades on a sequence of exams in a course. But to
allow this much generality would make it very dicult to prove general results.
In 1907, A. A. Markov began the study of an important new type of chance
process. In this process, the outcome of a given experiment can aect the outcome
of the next experiment. This type of process is called a Markov chain.
Specifying a Markov Chain
We describe a Markov chain as follows: We have a set of states, S = fs1; s2; : : : ; srg.
The process starts in one of these states and moves successively from one state to
another. Each move is called a step. If the chain is currently in state si, then",prob.pdf
"another. Each move is called a step. If the chain is currently in state si, then
it moves to state sj at the next step with a probability denoted by pij, and this
probability does not depend upon which states the chain was in before the current
405",prob.pdf
"406 CHAPTER 11. MARKOV CHAINS
state.
The probabilities pij are called transition probabilities. The process can remain
in the state it is in, and this occurs with probability pii. An initial probability
distribution, dened on S, species the starting state. Usually this is done by
specifying a particular state as the starting state.
R. A. Howard1 provides us with a picturesque description of a Markov chain as
a frog jumping on a set of lily pads. The frog starts on one of the pads and then
jumps from lily pad to lily pad with the appropriate transition probabilities.
Example 11.1 According to Kemeny, Snell, and Thompson,2 the Land of Oz is
blessed by many things, but not by good weather. They never have two nice days
in a row. If they have a nice day, they are just as likely to have snow as rain the
next day. If they have snow or rain, they have an even chance of having the same
the next day. If there is change from snow or rain, only half of the time is this a",prob.pdf
"the next day. If there is change from snow or rain, only half of the time is this a
change to a nice day. With this information we form a Markov chain as follows.
We take as states the kinds of weather R, N, and S. From the above information
we determine the transition probabilities. These are most conveniently represented
in a square array as
P =
0
@
R N S
R 1=2 1=4 1=4
N 1=2 0 1=2
S 1=4 1=4 1=2
1
A:
2
Transition Matrix
The entries in the rst row of the matrix P in Example 11.1 represent the proba-
bilities for the various kinds of weather following a rainy day. Similarly, the entries
in the second and third rows represent the probabilities for the various kinds of
weather following nice and snowy days, respectively. Such a square array is called
the matrix of transition probabilities, or the transition matrix.
We consider the question of determining the probability that, given the chain is
in state i today, it will be in state j two days from now. We denote this probability
by p(2)",prob.pdf
"in state i today, it will be in state j two days from now. We denote this probability
by p(2)
ij . In Example 11.1, we see that if it is rainy today then the event that it
is snowy two days from now is the disjoint union of the following three events: 1)
it is rainy tomorrow and snowy two days from now, 2) it is nice tomorrow and
snowy two days from now, and 3) it is snowy tomorrow and snowy two days from
now. The probability of the rst of these events is the product of the conditional
probability that it is rainy tomorrow, given that it is rainy today, and the conditional
probability that it is snowy two days from now, given that it is rainy tomorrow.
Using the transition matrix P, we can write this product as p11p13. The other two
1R. A. Howard, Dynamic Probabilistic Systems, vol. 1 (New York: John Wiley and Sons, 1971).
2J. G. Kemeny, J. L. Snell, G. L. Thompson, Introduction to Finite Mathematics, 3rd ed.
(Englewood Clis, NJ: Prentice-Hall, 1974).",prob.pdf
"11.1. INTRODUCTION 407
events also have probabilities that can be written as products of entries of P. Thus,
we have
p(2)
13 = p11p13 + p12p23 + p13p33 :
This equation should remind the reader of a dot product of two vectors; we are
dotting the rst row of P with the third column of P. This is just what is done
in obtaining the 1; 3-entry of the product of P with itself. In general, if a Markov
chain has r states, then
p(2)
ij =
rX
k=1
pikpkj :
The following general theorem is easy to prove by using the above observation and
induction.
Theorem 11.1 Let P be the transition matrix of a Markov chain. The ijth en-
try p(n)
ij of the matrix Pn gives the probability that the Markov chain, starting in
state si, will be in state sj after n steps.
Proof. The proof of this theorem is left as an exercise (Exercise 17). 2
Example 11.2 (Example 11.1 continued) Consider again the weather in the Land
of Oz. We know that the powers of the transition matrix give us interesting in-",prob.pdf
"of Oz. We know that the powers of the transition matrix give us interesting in-
formation about the process as it evolves. We shall be particularly interested in
the state of the chain after a large number of steps. The program MatrixPowers
computes the powers of P.
We have run the program MatrixPowers for the Land of Oz example to com-
pute the successive powers of P from 1 to 6. The results are shown in Table 11.1.
We note that after six days our weather predictions are, to three-decimal-place ac-
curacy, independent of today's weather. The probabilities for the three types of
weather, R, N, and S, are .4, .2, and .4 no matter where the chain started. This
is an example of a type of Markov chain called a regular Markov chain. For this
type of chain, it is true that long-range predictions are independent of the starting
state. Not all chains are regular, but this is an important class of chains that we
shall study in detail later. 2",prob.pdf
"state. Not all chains are regular, but this is an important class of chains that we
shall study in detail later. 2
We now consider the long-term behavior of a Markov chain when it starts in a
state chosen by a probability distribution on the set of states, which we will call a
probability vector. A probability vector with r components is a row vector whose
entries are non-negative and sum to 1. If u is a probability vector which represents
the initial state of a Markov chain, then we think of the ith component of u as
representing the probability that the chain starts in state si.
With this interpretation of random starting states, it is easy to prove the fol-
lowing theorem.",prob.pdf
"408 CHAPTER 11. MARKOV CHAINS
P1 =
0
@
Rain Nice Snow
Rain :500 :250 :250
Nice :500 :000 :500
Snow :250 :250 :500
1
A
P2 =
0
@
Rain Nice Snow
Rain :438 :188 :375
Nice :375 :250 :375
Snow :375 :188 :438
1
A
P3 =
0
@
Rain Nice Snow
Rain :406 :203 :391
Nice :406 :188 :406
Snow :391 :203 :406
1
A
P4 =
0
@
Rain Nice Snow
Rain :402 :199 :398
Nice :398 :203 :398
Snow :398 :199 :402
1
A
P5 =
0
@
Rain Nice Snow
Rain :400 :200 :399
Nice :400 :199 :400
Snow :399 :200 :400
1
A
P6 =
0
@
Rain Nice Snow
Rain :400 :200 :400
Nice :400 :200 :400
Snow :400 :200 :400
1
A
Table 11.1: Powers of the Land of Oz transition matrix.",prob.pdf
"11.1. INTRODUCTION 409
Theorem 11.2 Let P be the transition matrix of a Markov chain, and let u be the
probability vector which represents the starting distribution. Then the probability
that the chain is in state si after n steps is the ith entry in the vector
u(n) = uPn :
Proof. The proof of this theorem is left as an exercise (Exercise 18). 2
We note that if we want to examine the behavior of the chain under the assump-
tion that it starts in a certain state si, we simply choose u to be the probability
vector with ith entry equal to 1 and all other entries equal to 0.
Example 11.3 In the Land of Oz example (Example 11.1) let the initial probability
vector u equal (1=3; 1=3; 1=3). Then we can calculate the distribution of the states
after three days using Theorem 11.2 and our previous calculation of P3. We obtain
u(3) = uP3 = ( 1=3; 1=3; 1=3)
0
@
:406 :203 :391
:406 :188 :406
:391 :203 :406
1
A
= ( :401; :198; :401) :
2
Examples",prob.pdf
"u(3) = uP3 = ( 1=3; 1=3; 1=3)
0
@
:406 :203 :391
:406 :188 :406
:391 :203 :406
1
A
= ( :401; :198; :401) :
2
Examples
The following examples of Markov chains will be used throughout the chapter for
exercises.
Example 11.4 The President of the United States tells person A his or her in-
tention to run or not to run in the next election. Then A relays the news to B,
who in turn relays the message to C, and so forth, always to some new person. We
assume that there is a probability a that a person will change the answer from yes
to no when transmitting it to the next person and a probability b that he or she
will change it from no to yes. We choose as states the message, either yes or no.
The transition matrix is then
P =
 yes no
yes 1  a a
no b 1  b

:
The initial state represents the President's choice. 2
Example 11.5 Each time a certain horse runs in a three-horse race, he has proba-
bility 1/2 of winning, 1/4 of coming in second, and 1/4 of coming in third, indepen-",prob.pdf
"bility 1/2 of winning, 1/4 of coming in second, and 1/4 of coming in third, indepen-
dent of the outcome of any previous race. We have an independent trials process,",prob.pdf
"410 CHAPTER 11. MARKOV CHAINS
but it can also be considered from the point of view of Markov chain theory. The
transition matrix is
P =
0
@
W P S
W :5 :25 :25
P :5 :25 :25
S :5 :25 :25
1
A:
2
Example 11.6 In the Dark Ages, Harvard, Dartmouth, and Yale admitted only
male students. Assume that, at that time, 80 percent of the sons of Harvard men
went to Harvard and the rest went to Yale, 40 percent of the sons of Yale men went
to Yale, and the rest split evenly between Harvard and Dartmouth; and of the sons
of Dartmouth men, 70 percent went to Dartmouth, 20 percent to Harvard, and
10 percent to Yale. We form a Markov chain with transition matrix
P =
0
@
H Y D
H :8 :2 0
Y :3 :4 :3
D :2 :1 :7
1
A:
2
Example 11.7 Modify Example 11.6 by assuming that the son of a Harvard man
always went to Harvard. The transition matrix is now
P =
0
@
H Y D
H 1 0 0
Y :3 :4 :3
D :2 :1 :7
1
A:
2
Example 11.8 (Ehrenfest Model) The following is a special case of a model, called",prob.pdf
"P =
0
@
H Y D
H 1 0 0
Y :3 :4 :3
D :2 :1 :7
1
A:
2
Example 11.8 (Ehrenfest Model) The following is a special case of a model, called
the Ehrenfest model,3 that has been used to explain diusion of gases. The general
model will be discussed in detail in Section 11.5. We have two urns that, between
them, contain four balls. At each step, one of the four balls is chosen at random
and moved from the urn that it is in into the other urn. We choose, as states, the
number of balls in the rst urn. The transition matrix is then
P =
0
B
B
B
B
@
0 1 2 3 4
0 0 1 0 0 0
1 1=4 0 3=4 0 0
2 0 1=2 0 1=2 0
3 0 0 3=4 0 1=4
4 0 0 0 1 0
1
C
C
C
C
A
:
2
3P. and T. Ehrenfest, \Uber zwei bekannte Einwande gegen das Boltzmannsche H-Theorem,""
Physikalishce Zeitschrift, vol. 8 (1907), pp. 311-314.",prob.pdf
"11.1. INTRODUCTION 411
Example 11.9 (Gene Model) The simplest type of inheritance of traits in animals
occurs when a trait is governed by a pair of genes, each of which may be of two types,
say G and g. An individual may have a GG combination or Gg (which is genetically
the same as gG) or gg. Very often the GG and Gg types are indistinguishable in
appearance, and then we say that the G gene dominates the g gene. An individual
is called dominant if he or she has GG genes, recessive if he or she has gg, and
hybrid with a Gg mixture.
In the mating of two animals, the ospring inherits one gene of the pair from
each parent, and the basic assumption of genetics is that these genes are selected at
random, independently of each other. This assumption determines the probability
of occurrence of each type of ospring. The ospring of two purely dominant parents
must be dominant, of two recessive parents must be recessive, and of one dominant
and one recessive parent must be hybrid.",prob.pdf
"must be dominant, of two recessive parents must be recessive, and of one dominant
and one recessive parent must be hybrid.
In the mating of a dominant and a hybrid animal, each ospring must get a
G gene from the former and has an equal chance of getting G or g from the latter.
Hence there is an equal probability for getting a dominant or a hybrid ospring.
Again, in the mating of a recessive and a hybrid, there is an even chance for getting
either a recessive or a hybrid. In the mating of two hybrids, the ospring has an
equal chance of getting G or g from each parent. Hence the probabilities are 1/4
for GG, 1/2 for Gg, and 1/4 for gg.
Consider a process of continued matings. We start with an individual of known
genetic character and mate it with a hybrid. We assume that there is at least one
ospring. An ospring is chosen at random and is mated with a hybrid and this
process repeated through a number of generations. The genetic type of the chosen",prob.pdf
"process repeated through a number of generations. The genetic type of the chosen
ospring in successive generations can be represented by a Markov chain. The states
are dominant, hybrid, and recessive, and indicated by GG, Gg, and gg respectively.
The transition probabilities are
P =
0
@
GG Gg gg
GG :5 :5 0
Gg :25 :5 :25
gg 0 :5 :5
1
A:
2
Example 11.10 Modify Example 11.9 as follows: Instead of mating the oldest
ospring with a hybrid, we mate it with a dominant individual. The transition
matrix is
P =
0
@
GG Gg gg
GG 1 0 0
Gg :5 :5 0
gg 0 1 0
1
A:
2",prob.pdf
"412 CHAPTER 11. MARKOV CHAINS
Example 11.11 We start with two animals of opposite sex, mate them, select two
of their ospring of opposite sex, and mate those, and so forth. To simplify the
example, we will assume that the trait under consideration is independent of sex.
Here a state is determined by a pair of animals. Hence, the states of our process
will be: s1 = (GG; GG), s2 = (GG; Gg), s3 = (GG; gg), s4 = (Gg; Gg), s5 =
(Gg; gg), and s6 = (gg; gg).
We illustrate the calculation of transition probabilities in terms of the state s2.
When the process is in this state, one parent has GG genes, the other Gg. Hence,
the probability of a dominant ospring is 1/2. Then the probability of transition
to s1 (selection of two dominants) is 1/4, transition to s2 is 1/2, and to s4 is 1/4.
The other states are treated the same way. The transition matrix of this chain is:
P1 =
0
B
B
B
B
B
B
B
@
GG,GG GG,Gg GG,gg Gg,Gg Gg,gg gg,gg
GG,GG 1:000 :000 :000 :000 :000 :000",prob.pdf
"P1 =
0
B
B
B
B
B
B
B
@
GG,GG GG,Gg GG,gg Gg,Gg Gg,gg gg,gg
GG,GG 1:000 :000 :000 :000 :000 :000
GG,Gg :250 :500 :000 :250 :000 :000
GG,gg :000 :000 :000 1:000 :000 :000
Gg,Gg :062 :250 :125 :250 :250 :062
Gg,gg :000 :000 :000 :250 :500 :250
gg,gg :000 :000 :000 :000 :000 1:000
1
C
C
C
C
C
C
C
A
:
2
Example 11.12 (Stepping Stone Model) Our nal example is another example
that has been used in the study of genetics. It is called the stepping stone model.4
In this model we have an n-by-n array of squares, and each square is initially any
one of k dierent colors. For each step, a square is chosen at random. This square
then chooses one of its eight neighbors at random and assumes the color of that
neighbor. To avoid boundary problems, we assume that if a square S is on the
left-hand boundary, say, but not at a corner, it is adjacent to the square T on the
right-hand boundary in the same row as S, and S is also adjacent to the squares",prob.pdf
"right-hand boundary in the same row as S, and S is also adjacent to the squares
just above and below T. A similar assumption is made about squares on the upper
and lower boundaries. The top left-hand corner square is adjacent to three obvious
neighbors, namely the squares below it, to its right, and diagonally below and to
the right. It has ve other neighbors, which are as follows: the other three corner
squares, the square below the upper right-hand corner, and the square to the right
of the bottom left-hand corner. The other three corners also have, in a similar way,
eight neighbors. (These adjacencies are much easier to understand if one imagines
making the array into a cylinder by gluing the top and bottom edge together, and
then making the cylinder into a doughnut by gluing the two circular boundaries
together.) With these adjacencies, each square in the array is adjacent to exactly
eight other squares.",prob.pdf
"together.) With these adjacencies, each square in the array is adjacent to exactly
eight other squares.
A state in this Markov chain is a description of the color of each square. For this
Markov chain the number of states is kn2
, which for even a small array of squares
4S. Sawyer, \Results for The Stepping Stone Model for Migration in Population Genetics,""
Annals of Probability, vol. 4 (1979), pp. 699{728.",prob.pdf
"11.1. INTRODUCTION 413
Figure 11.1: Initial state of the stepping stone model.
Figure 11.2: State of the stepping stone model after 10,000 steps.
is enormous. This is an example of a Markov chain that is easy to simulate but
dicult to analyze in terms of its transition matrix. The program SteppingStone
simulates this chain. We have started with a random initial conguration of two
colors with n = 20 and show the result after the process has run for some time in
Figure 11.2.
This is an example of an absorbing Markov chain. This type of chain will be
studied in Section 11.2. One of the theorems proved in that section, applied to
the present example, implies that with probability 1, the stones will eventually all
be the same color. By watching the program run, you can see that territories are
established and a battle develops to see which color survives. At any time the
probability that a particular color will win out is equal to the proportion of the",prob.pdf
"probability that a particular color will win out is equal to the proportion of the
array of this color. You are asked to prove this in Exercise 11.2.32. 2
Exercises
1 It is raining in the Land of Oz. Determine a tree and a tree measure for the
next three days' weather. Find w(1); w(2); and w(3) and compare with the
results obtained from P; P2; and P3.",prob.pdf
"414 CHAPTER 11. MARKOV CHAINS
2 In Example 11.4, let a = 0 and b = 1=2. Find P; P2; and P3: What would
Pn be? What happens to Pn as n tends to innity? Interpret this result.
3 In Example 11.5, nd P, P2; and P3: What is Pn?
4 For Example 11.6, nd the probability that the grandson of a man from Har-
vard went to Harvard.
5 In Example 11.7, nd the probability that the grandson of a man from Harvard
went to Harvard.
6 In Example 11.9, assume that we start with a hybrid bred to a hybrid. Find
u(1); u(2); and u(3): What would u(n) be?
7 Find the matrices P2; P3; P4; and Pn for the Markov chain determined by
the transition matrix P =
1 0
0 1

. Do the same for the transition matrix
P =
0 1
1 0

. Interpret what happens in each of these processes.
8 A certain calculating machine uses only the digits 0 and 1. It is supposed to
transmit one of these digits through several stages. However, at every stage,
there is a probability p that the digit that enters this stage will be changed",prob.pdf
"there is a probability p that the digit that enters this stage will be changed
when it leaves and a probability q = 1p that it won't. Form a Markov chain
to represent the process of transmission by taking as states the digits 0 and 1.
What is the matrix of transition probabilities?
9 For the Markov chain in Exercise 8, draw a tree and assign a tree measure
assuming that the process begins in state 0 and moves through two stages
of transmission. What is the probability that the machine, after two stages,
produces the digit 0 (i.e., the correct digit)? What is the probability that the
machine never changed the digit from 0? Now let p = :1. Using the program
MatrixPowers, compute the 100th power of the transition matrix. Interpret
the entries of this matrix. Repeat this with p = :2. Why do the 100th powers
appear to be the same?
10 Modify the program MatrixPowers so that it prints out the average An of
the powers Pn, for n = 1 to N. Try your program on the Land of Oz example",prob.pdf
"10 Modify the program MatrixPowers so that it prints out the average An of
the powers Pn, for n = 1 to N. Try your program on the Land of Oz example
and compare An and Pn:
11 Assume that a man's profession can be classied as professional, skilled la-
borer, or unskilled laborer. Assume that, of the sons of professional men,
80 percent are professional, 10 percent are skilled laborers, and 10 percent are
unskilled laborers. In the case of sons of skilled laborers, 60 percent are skilled
laborers, 20 percent are professional, and 20 percent are unskilled. Finally, in
the case of unskilled laborers, 50 percent of the sons are unskilled laborers,
and 25 percent each are in the other two categories. Assume that every man
has at least one son, and form a Markov chain by following the profession of
a randomly chosen son of a given family through several generations. Set up",prob.pdf
"11.1. INTRODUCTION 415
the matrix of transition probabilities. Find the probability that a randomly
chosen grandson of an unskilled laborer is a professional man.
12 In Exercise 11, we assumed that every man has a son. Assume instead that
the probability that a man has at least one son is .8. Form a Markov chain
with four states. If a man has a son, the probability that this son is in a
particular profession is the same as in Exercise 11. If there is no son, the
process moves to state four which represents families whose male line has died
out. Find the matrix of transition probabilities and nd the probability that
a randomly chosen grandson of an unskilled laborer is a professional man.
13 Write a program to compute u(n) given u and P. Use this program to
compute u(10) for the Land of Oz example, with u = (0; 1; 0), and with
u = (1=3; 1=3; 1=3).
14 Using the program MatrixPowers, nd P1 through P6 for Examples 11.9",prob.pdf
"u = (1=3; 1=3; 1=3).
14 Using the program MatrixPowers, nd P1 through P6 for Examples 11.9
and 11.10. See if you can predict the long-range probability of nding the
process in each of the states for these examples.
15 Write a program to simulate the outcomes of a Markov chain after n steps,
given the initial starting state and the transition matrix P as data (see Ex-
ample 11.12). Keep this program for use in later problems.
16 Modify the program of Exercise 15 so that it keeps track of the proportion of
times in each state in n steps. Run the modied program for dierent starting
states for Example 11.1 and Example 11.8. Does the initial state aect the
proportion of time spent in each of the states if n is large?
17 Prove Theorem 11.1.
18 Prove Theorem 11.2.
19 Consider the following process. We have two coins, one of which is fair, and the
other of which has heads on both sides. We give these two coins to our friend,",prob.pdf
"other of which has heads on both sides. We give these two coins to our friend,
who chooses one of them at random (each with probability 1/2). During the
rest of the process, she uses only the coin that she chose. She now proceeds
to toss the coin many times, reporting the results. We consider this process
to consist solely of what she reports to us.
(a) Given that she reports a head on the nth toss, what is the probability
that a head is thrown on the (n + 1)st toss?
(b) Consider this process as having two states, heads and tails. By computing
the other three transition probabilities analogous to the one in part (a),
write down a \transition matrix"" for this process.
(c) Now assume that the process is in state \heads"" on both the (n  1)st
and the nth toss. Find the probability that a head comes up on the
(n + 1)st toss.
(d) Is this process a Markov chain?",prob.pdf
"416 CHAPTER 11. MARKOV CHAINS
11.2 Absorbing Markov Chains
The subject of Markov chains is best studied by considering special types of Markov
chains. The rst type that we shall study is called an absorbing Markov chain.
Denition 11.1 A state si of a Markov chain is called absorbing if it is impossible
to leave it (i.e., pii = 1). A Markov chain is absorbing if it has at least one absorbing
state, and if from every state it is possible to go to an absorbing state (not necessarily
in one step). 2
Denition 11.2 In an absorbing Markov chain, a state which is not absorbing is
called transient. 2
Drunkard's Walk
Example 11.13 A man walks along a four-block stretch of Park Avenue (see Fig-
ure 11.3). If he is at corner 1, 2, or 3, then he walks to the left or right with equal
probability. He continues until he reaches corner 4, which is a bar, or corner 0,
which is his home. If he reaches either home or the bar, he stays there.",prob.pdf
"which is his home. If he reaches either home or the bar, he stays there.
We form a Markov chain with states 0, 1, 2, 3, and 4. States 0 and 4 are
absorbing states. The transition matrix is then
P =
0
B
B
B
B
@
0 1 2 3 4
0 1 0 0 0 0
1 1=2 0 1=2 0 0
2 0 1=2 0 1=2 0
3 0 0 1=2 0 1=2
4 0 0 0 0 1
1
C
C
C
C
A
:
The states 1, 2, and 3 are transient states, and from any of these it is possible to
reach the absorbing states 0 and 4. Hence the chain is an absorbing chain. When
a process reaches an absorbing state, we shall say that it is absorbed. 2
The most obvious question that can be asked about such a chain is: What is
the probability that the process will eventually reach an absorbing state? Other
interesting questions include: (a) What is the probability that the process will end
up in a given absorbing state? (b) On the average, how long will it take for the
process to be absorbed? (c) On the average, how many times will the process be in",prob.pdf
"process to be absorbed? (c) On the average, how many times will the process be in
each transient state? The answers to all these questions depend, in general, on the
state from which the process starts as well as the transition probabilities.
Canonical Form
Consider an arbitrary absorbing Markov chain. Renumber the states so that the
transient states come rst. If there are r absorbing states and t transient states,
the transition matrix will have the following canonical form",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 417
1 2 30 4
1 11/2 1/2 1/2
1/2 1/2 1/2
Figure 11.3: Drunkard's walk.
P =
0
B
@
TR. ABS.
TR. Q
R
ABS. 0 I
1
C
A
Here I is an r-by-r indentity matrix, 0 is an r-by-t zero matrix, R is a nonzero
t-by-r matrix, and Q is an t-by-t matrix. The rst t states are transient and the
last r states are absorbing.
In Section 11.1, we saw that the entry p(n)
ij of the matrix Pn is the probability of
being in the state sj after n steps, when the chain is started in state si. A standard
matrix algebra argument shows that Pn is of the form
Pn =
0
B
@
TR. ABS.
TR. Qn

ABS. 0 I
1
C
A
where the asterisk  stands for the t-by-r matrix in the upper right-hand corner
of Pn: (This submatrix can be written in terms of Q and R, but the expression
is complicated and is not needed at this time.) The form of Pn shows that the
entries of Qn give the probabilities for being in each of the transient states after n",prob.pdf
"entries of Qn give the probabilities for being in each of the transient states after n
steps for each possible transient starting state. For our rst theorem we prove that
the probability of being in the transient states after n steps approaches zero. Thus
every entry of Qn must approach zero as n approaches innity (i.e, Qn ! 0).
Probability of Absorption
Theorem 11.3 In an absorbing Markov chain, the probability that the process
will be absorbed is 1 (i.e., Qn ! 0 as n ! 1).
Proof. From each nonabsorbing state sj it is possible to reach an absorbing state.
Let mj be the minimum number of steps required to reach an absorbing state,
starting from sj. Let pj be the probability that, starting from sj, the process will
not reach an absorbing state in mj steps. Then pj < 1. Let m be the largest of the",prob.pdf
"418 CHAPTER 11. MARKOV CHAINS
mj and let p be the largest of pj. The probability of not being absorbed in m steps
is less than or equal to p, in 2m steps less than or equal to p2, etc. Since p < 1
these probabilities tend to 0. Since the probability of not being absorbed in n steps
is monotone decreasing, these probabilities also tend to 0, hence limn!1 Qn = 0:
2
The Fundamental Matrix
Theorem 11.4 For an absorbing Markov chain the matrix I  Q has an inverse
N and N = I + Q + Q2 + 
 
 
 . The ij-entry nij of the matrix N is the expected
number of times the chain is in state sj, given that it starts in state si. The initial
state is counted if i = j.
Proof. Let (I  Q)x = 0; that is x = Qx: Then, iterating this we see that
x = Qnx: Since Qn ! 0, we have Qnx ! 0, so x = 0. Thus (I  Q)1 = N
exists. Note next that
(I  Q)(I + Q + Q2 + 
 
 
 + Qn) = I  Qn+1 :
Thus multiplying both sides by N gives
I + Q + Q2 + 
 
 
 + Qn = N(I  Qn+1) :
Letting n tend to innity we have",prob.pdf
"Thus multiplying both sides by N gives
I + Q + Q2 + 
 
 
 + Qn = N(I  Qn+1) :
Letting n tend to innity we have
N = I + Q + Q2 + 
 
 
 :
Let si and sj be two transient states, and assume throughout the remainder of
the proof that i and j are xed. Let X(k) be a random variable which equals 1
if the chain is in state sj after k steps, and equals 0 otherwise. For each k, this
random variable depends upon both i and j; we choose not to explicitly show this
dependence in the interest of clarity. We have
P(X(k) = 1) = q(k)
ij ;
and
P(X(k) = 0) = 1  q(k)
ij ;
where q(k)
ij is the ijth entry of Qk. These equations hold for k = 0 since Q0 = I.
Therefore, since X(k) is a 0-1 random variable, E(X(k)) = q(k)
ij .
The expected number of times the chain is in state sj in the rst n steps, given
that it starts in state si, is clearly
E

X(0) + X(1) + 
 
 
 + X(n)

= q(0)
ij + q(1)
ij + 
 
 
 + q(n)
ij :
Letting n tend to innity we have
E

X(0) + X(1) + 
 
 


= q(0)
ij + q(1)",prob.pdf
"E

X(0) + X(1) + 
 
 
 + X(n)

= q(0)
ij + q(1)
ij + 
 
 
 + q(n)
ij :
Letting n tend to innity we have
E

X(0) + X(1) + 
 
 


= q(0)
ij + q(1)
ij + 
 
 
 = nij :
2",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 419
Denition 11.3 For an absorbing Markov chain P, the matrix N = (I  Q)1 is
called the fundamental matrix for P. The entry nij of N gives the expected number
of times that the process is in the transient state sj if it is started in the transient
state si. 2
Example 11.14 (Example 11.13 continued) In the Drunkard's Walk example, the
transition matrix in canonical form is
P =
0
B
B
B
@
1 2 3 0 4
1 0 1=2 0
1=2 0
2 1=2 0 1=2 0 0
3 0 1=2 0 0 1=2
0 0 0 0 1 0
4 0 0 0 0 1
1
C
C
C
A:
From this we see that the matrix Q is
Q =
0
@
0 1=2 0
1=2 0 1=2
0 1=2 0
1
A ;
and
I  Q =
0
@
1 1=2 0
1=2 1 1=2
0 1=2 1
1
A :
Computing (I  Q)1, we nd
N = (I  Q)1 =
0
@
1 2 3
1 3=2 1 1=2
2 1 2 1
3 1=2 1 3=2
1
A:
From the middle row of N, we see that if we start in state 2, then the expected
number of times in states 1, 2, and 3 before being absorbed are 1, 2, and 1. 2
Time to Absorption
We now consider the question: Given that the chain starts in state si, what is the",prob.pdf
"Time to Absorption
We now consider the question: Given that the chain starts in state si, what is the
expected number of steps before the chain is absorbed? The answer is given in the
next theorem.
Theorem 11.5 Let ti be the expected number of steps before the chain is absorbed,
given that the chain starts in state si, and let t be the column vector whose ith
entry is ti. Then
t = Nc ;
where c is a column vector all of whose entries are 1.",prob.pdf
"420 CHAPTER 11. MARKOV CHAINS
Proof. If we add all the entries in the ith row of N, we will have the expected
number of times in any of the transient states for a given starting state si, that
is, the expected time required before being absorbed. Thus, ti is the sum of the
entries in the ith row of N. If we write this statement in matrix form, we obtain
the theorem. 2
Absorption Probabilities
Theorem 11.6 Let bij be the probability that an absorbing chain will be absorbed
in the absorbing state sj if it starts in the transient state si. Let B be the matrix
with entries bij. Then B is an t-by-r matrix, and
B = NR ;
where N is the fundamental matrix and R is as in the canonical form.
Proof. We have
Bij =
X
n
X
k
q(n)
ik rkj
=
X
k
X
n
q(n)
ik rkj
=
X
k
nikrkj
= (NR)ij :
This completes the proof. 2
Another proof of this is given in Exercise 34.
Example 11.15 (Example 11.14 continued) In the Drunkard's Walk example, we
found that
N =
0
@
1 2 3
1 3=2 1 1=2
2 1 2 1
3 1=2 1 3=2
1
A:
Hence,",prob.pdf
"Example 11.15 (Example 11.14 continued) In the Drunkard's Walk example, we
found that
N =
0
@
1 2 3
1 3=2 1 1=2
2 1 2 1
3 1=2 1 3=2
1
A:
Hence,
t = Nc =
0
@
3=2 1 1=2
1 2 1
1=2 1 3=2
1
A
0
@
1
1
1
1
A
=
0
@
3
4
3
1
A :",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 421
Thus, starting in states 1, 2, and 3, the expected times to absorption are 3, 4, and
3, respectively.
From the canonical form,
R =
0
@
0 4
1 1=2 0
2 0 0
3 0 1=2
1
A:
Hence,
B = NR =
0
@
3=2 1 1=2
1 2 1
1=2 1 3=2
1
A

0
@
1=2 0
0 0
0 1=2
1
A
=
0
@
0 4
1 3=4 1=4
2 1=2 1=2
3 1=4 3=4
1
A:
Here the rst row tells us that, starting from state 1, there is probability 3/4 of
absorption in state 0 and 1/4 of absorption in state 4. 2
Computation
The fact that we have been able to obtain these three descriptive quantities in
matrix form makes it very easy to write a computer program that determines these
quantities for a given absorbing chain matrix.
The program AbsorbingChain calculates the basic descriptive quantities of an
absorbing Markov chain.
We have run the program AbsorbingChain for the example of the drunkard's
walk (Example 11.13) with 5 blocks. The results are as follows:
Q =
0
B
B
@
1 2 3 4
1 :00 :50 :00 :00
2 :50 :00 :50 :00
3 :00 :50 :00 :50",prob.pdf
"walk (Example 11.13) with 5 blocks. The results are as follows:
Q =
0
B
B
@
1 2 3 4
1 :00 :50 :00 :00
2 :50 :00 :50 :00
3 :00 :50 :00 :50
4 :00 :00 :50 :00
1
C
C
A;
R =
0
B
B
@
0 5
1 :50 :00
2 :00 :00
3 :00 :00
4 :00 :50
1
C
C
A;",prob.pdf
"422 CHAPTER 11. MARKOV CHAINS
N =
0
B
B
@
1 2 3 4
1 1:60 1:20 :80 :40
2 1:20 2:40 1:60 :80
3 :80 1:60 2:40 1:20
4 :40 :80 1:20 1:60
1
C
C
A;
t =
0
B
B
@
1 4:00
2 6:00
3 6:00
4 4:00
1
C
C
A;
B =
0
B
B
@
0 5
1 :80 :20
2 :60 :40
3 :40 :60
4 :20 :80
1
C
C
A:
Note that the probability of reaching the bar before reaching home, starting
at x, is x=5 (i.e., proportional to the distance of home from the starting point).
(See Exercise 24.)
Exercises
1 In Example 11.4, for what values of a and b do we obtain an absorbing Markov
chain?
2 Show that Example 11.7 is an absorbing Markov chain.
3 Which of the genetics examples (Examples 11.9, 11.10, and 11.11) are ab-
sorbing?
4 Find the fundamental matrix N for Example 11.10.
5 For Example 11.11, verify that the following matrix is the inverse of I  Q
and hence is the fundamental matrix N.
N =
0
B
B
@
8=3 1=6 4=3 2=3
4=3 4=3 8=3 4=3
4=3 1=3 8=3 4=3
2=3 1=6 4=3 8=3
1
C
C
A :
Find Nc and NR. Interpret the results.",prob.pdf
"N =
0
B
B
@
8=3 1=6 4=3 2=3
4=3 4=3 8=3 4=3
4=3 1=3 8=3 4=3
2=3 1=6 4=3 8=3
1
C
C
A :
Find Nc and NR. Interpret the results.
6 In the Land of Oz example (Example 11.1), change the transition matrix by
making R an absorbing state. This gives
P =
0
@
R N S
R 1 0 0
N 1=2 0 1=2
S 1=4 1=4 1=2
1
A:",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 423
Find the fundamental matrix N, and also Nc and NR. Interpret the results.
7 In Example 11.8, make states 0 and 4 into absorbing states. Find the fun-
damental matrix N, and also Nc and NR, for the resulting absorbing chain.
Interpret the results.
8 In Example 11.13 (Drunkard's Walk) of this section, assume that the proba-
bility of a step to the right is 2/3, and a step to the left is 1/3. Find N; Nc,
and NR. Compare these with the results of Example 11.15.
9 A process moves on the integers 1, 2, 3, 4, and 5. It starts at 1 and, on each
successive step, moves to an integer greater than its present position, moving
with equal probability to each of the remaining larger integers. State ve is
an absorbing state. Find the expected number of steps to reach state ve.
10 Using the result of Exercise 9, make a conjecture for the form of the funda-
mental matrix if the process moves as in that exercise, except that it now",prob.pdf
"mental matrix if the process moves as in that exercise, except that it now
moves on the integers from 1 to n. Test your conjecture for several dierent
values of n. Can you conjecture an estimate for the expected number of steps
to reach state n, for large n? (See Exercise 11 for a method of determining
this expected number of steps.)
*11 Let bk denote the expected number of steps to reach n from n  k, in the
process described in Exercise 9.
(a) Dene b0 = 0. Show that for k > 0, we have
bk = 1 + 1
k

bk1 + bk2 + 
 
 
 + b0


:
(b) Let
f(x) = b0 + b1x + b2x2 + 
 
 
 :
Using the recursion in part (a), show that f(x) satises the dierential
equation
(1  x)2y0  (1  x)y  1 = 0 :
(c) Show that the general solution of the dierential equation in part (b) is
y =  log(1  x)
1  x + c
1  x ;
where c is a constant.
(d) Use part (c) to show that
bk = 1 + 1
2 + 1
3 + 
 
 
 + 1
k :
12 Three tanks ght a three-way duel. Tank A has probability 1/2 of destroying",prob.pdf
"(d) Use part (c) to show that
bk = 1 + 1
2 + 1
3 + 
 
 
 + 1
k :
12 Three tanks ght a three-way duel. Tank A has probability 1/2 of destroying
the tank at which it res, tank B has probability 1/3 of destroying the tank at
which it res, and tank C has probability 1/6 of destroying the tank at which",prob.pdf
"424 CHAPTER 11. MARKOV CHAINS
it res. The tanks re together and each tank res at the strongest opponent
not yet destroyed. Form a Markov chain by taking as states the subsets of the
set of tanks. Find N; Nc, and NR, and interpret your results. Hint: Take
as states ABC, AC, BC, A, B, C, and none, indicating the tanks that could
survive starting in state ABC. You can omit AB because this state cannot be
reached from ABC.
13 Smith is in jail and has 3 dollars; he can get out on bail if he has 8 dollars.
A guard agrees to make a series of bets with him. If Smith bets A dollars,
he wins A dollars with probability .4 and loses A dollars with probability .6.
Find the probability that he wins 8 dollars before losing all of his money if
(a) he bets 1 dollar each time (timid strategy).
(b) he bets, each time, as much as possible but not more than necessary to
bring his fortune up to 8 dollars (bold strategy).
(c) Which strategy gives Smith the better chance of getting out of jail?",prob.pdf
"bring his fortune up to 8 dollars (bold strategy).
(c) Which strategy gives Smith the better chance of getting out of jail?
14 With the situation in Exercise 13, consider the strategy such that for i < 4,
Smith bets min(i; 4i), and for i  4, he bets according to the bold strategy,
where i is his current fortune. Find the probability that he gets out of jail
using this strategy. How does this probability compare with that obtained for
the bold strategy?
15 Consider the game of tennis when deuce is reached. If a player wins the next
point, he has advantage. On the following point, he either wins the game or the
game returns to deuce. Assume that for any point, player A has probability
.6 of winning the point and player B has probability .4 of winning the point.
(a) Set this up as a Markov chain with state 1: A wins; 2: B wins; 3:
advantage A; 4: deuce; 5: advantage B.
(b) Find the absorption probabilities.
(c) At deuce, nd the expected duration of the game and the probability",prob.pdf
"advantage A; 4: deuce; 5: advantage B.
(b) Find the absorption probabilities.
(c) At deuce, nd the expected duration of the game and the probability
that B will win.
Exercises 16 and 17 concern the inheritance of color-blindness, which is a sex-
linked characteristic. There is a pair of genes, g and G, of which the former
tends to produce color-blindness, the latter normal vision. The G gene is
dominant. But a man has only one gene, and if this is g, he is color-blind. A
man inherits one of his mother's two genes, while a woman inherits one gene
from each parent. Thus a man may be of type G or g, while a woman may be
type GG or Gg or gg. We will study a process of inbreeding similar to that
of Example 11.11 by constructing a Markov chain.
16 List the states of the chain. Hint: There are six. Compute the transition
probabilities. Find the fundamental matrix N, Nc, and NR.",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 425
17 Show that in both Example 11.11 and the example just given, the probability
of absorption in a state having genes of a particular type is equal to the
proportion of genes of that type in the starting state. Show that this can
be explained by the fact that a game in which your fortune is the number of
genes of a particular type in the state of the Markov chain is a fair game.5
18 Assume that a student going to a certain four-year medical school in northern
New England has, each year, a probability q of 
unking out, a probability r
of having to repeat the year, and a probability p of moving on to the next
year (in the fourth year, moving on means graduating).
(a) Form a transition matrix for this process taking as states F, 1, 2, 3, 4,
and G where F stands for 
unking out and G for graduating, and the
other states represent the year of study.
(b) For the case q = :1, r = :2, and p = :7 nd the time a beginning student",prob.pdf
"other states represent the year of study.
(b) For the case q = :1, r = :2, and p = :7 nd the time a beginning student
can expect to be in the second year. How long should this student expect
to be in medical school?
(c) Find the probability that this beginning student will graduate.
19 (E. Brown6) Mary and John are playing the following game: They have a
three-card deck marked with the numbers 1, 2, and 3 and a spinner with the
numbers 1, 2, and 3 on it. The game begins by dealing the cards out so that
the dealer gets one card and the other person gets two. A move in the game
consists of a spin of the spinner. The person having the card with the number
that comes up on the spinner hands that card to the other person. The game
ends when someone has all the cards.
(a) Set up the transition matrix for this absorbing Markov chain, where the
states correspond to the number of cards that Mary has.
(b) Find the fundamental matrix.
(c) On the average, how many moves will the game last?",prob.pdf
"states correspond to the number of cards that Mary has.
(b) Find the fundamental matrix.
(c) On the average, how many moves will the game last?
(d) If Mary deals, what is the probability that John will win the game?
20 Assume that an experiment has m equally probable outcomes. Show that the
expected number of independent trials before the rst occurrence of k consec-
utive occurrences of one of these outcomes is (mk  1)=(m  1). Hint: Form
an absorbing Markov chain with states 1, 2, . . . , k with state i representing
the length of the current run. The expected time until a run of k is 1 more
than the expected time until absorption for the chain started in state 1. It has
been found that, in the decimal expansion of pi, starting with the 24,658,601st
digit, there is a run of nine 7's. What would your result say about the ex-
pected number of digits necessary to nd such a run if the digits are produced
randomly?",prob.pdf
"pected number of digits necessary to nd such a run if the digits are produced
randomly?
5H. Gonshor, \An Application of Random Walk to a Problem in Population Genetics,"" Amer-
ican Math Monthly, vol. 94 (1987), pp. 668{671
6Private communication.",prob.pdf
"426 CHAPTER 11. MARKOV CHAINS
21 (Roberts7) A city is divided into 3 areas 1, 2, and 3. It is estimated that
amounts u1, u2, and u3 of pollution are emitted each day from these three
areas. A fraction qij of the pollution from region i ends up the next day at
region j. A fraction qi = 1P
j qij > 0 goes into the atmosphere and escapes.
Let w(n)
i be the amount of pollution in area i after n days.
(a) Show that w(n) = u + uQ + 
 
 
 + uQn1.
(b) Show that w(n) ! w, and show how to compute w from u.
(c) The government wants to limit pollution levels to a prescribed level by
prescribing w: Show how to determine the levels of pollution u which
would result in a prescribed limiting value w.
22 In the Leontief economic model,8 there are n industries 1, 2, . . . , n. The
ith industry requires an amount 0  qij  1 of goods (in dollar value) from
company j to produce 1 dollar's worth of goods. The outside demand on the",prob.pdf
"ith industry requires an amount 0  qij  1 of goods (in dollar value) from
company j to produce 1 dollar's worth of goods. The outside demand on the
industries, in dollar value, is given by the vector d = (d1; d2; : : :; dn). Let Q
be the matrix with entries qij.
(a) Show that if the industries produce total amounts given by the vector
x = (x1; x2; : : : ; xn) then the amounts of goods of each type that the
industries will need just to meet their internal demands is given by the
vector xQ.
(b) Show that in order to meet the outside demand d and the internal de-
mands the industries must produce total amounts given by a vector
x = (x1; x2; : : :; xn) which satises the equation x = xQ + d.
(c) Show that if Q is the Q-matrix for an absorbing Markov chain, then it
is possible to meet any outside demand d.
(d) Assume that the row sums of Q are less than or equal to 1. Give an
economic interpretation of this condition. Form a Markov chain by taking",prob.pdf
"(d) Assume that the row sums of Q are less than or equal to 1. Give an
economic interpretation of this condition. Form a Markov chain by taking
the states to be the industries and the transition probabilites to be the qij.
Add one absorbing state 0. Dene
qi0 = 1 
X
j
qij :
Show that this chain will be absorbing if every company is either making
a prot or ultimately depends upon a prot-making company.
(e) Dene xc to be the gross national product. Find an expression for the
gross national product in terms of the demand vector d and the vector
t giving the expected time to absorption.
23 A gambler plays a game in which on each play he wins one dollar with prob-
ability p and loses one dollar with probability q = 1  p. The Gambler's Ruin
7F. Roberts, Discrete Mathematical Models (Englewood Clis, NJ: Prentice Hall, 1976).
8W. W. Leontief, Input-Output Economics (Oxford: Oxford University Press, 1966).",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 427
problem is the problem of nding the probability wx of winning an amount T
before losing everything, starting with state x. Show that this problem may
be considered to be an absorbing Markov chain with states 0, 1, 2, . . . , T with
0 and T absorbing states. Suppose that a gambler has probability p = :48
of winning on each play. Suppose, in addition, that the gambler starts with
50 dollars and that T = 100 dollars. Simulate this game 100 times and see
how often the gambler is ruined. This estimates w50.
24 Show that wx of Exercise 23 satises the following conditions:
(a) wx = pwx+1 + qwx1 for x = 1, 2, . . . , T  1.
(b) w0 = 0.
(c) wT = 1.
Show that these conditions determine wx. Show that, if p = q = 1=2, then
wx = x
T
satises (a), (b), and (c) and hence is the solution. If p 6= q, show that
wx = (q=p)x  1
(q=p)T  1
satises these conditions and hence gives the probability of the gambler win-
ning.",prob.pdf
"wx = (q=p)x  1
(q=p)T  1
satises these conditions and hence gives the probability of the gambler win-
ning.
25 Write a program to compute the probability wx of Exercise 24 for given values
of x, p, and T. Study the probability that the gambler will ruin the bank in a
game that is only slightly unfavorable, say p = :49, if the bank has signicantly
more money than the gambler.
*26 We considered the two examples of the Drunkard's Walk corresponding to the
cases n = 4 and n = 5 blocks (see Example 11.13). Verify that in these two
examples the expected time to absorption, starting at x, is equal to x(n  x).
See if you can prove that this is true in general. Hint: Show that if f(x) is
the expected time to absorption then f(0) = f(n) = 0 and
f(x) = (1=2)f(x  1) + (1=2)f(x + 1) + 1
for 0 < x < n. Show that if f1(x) and f2(x) are two solutions, then their
dierence g(x) is a solution of the equation
g(x) = (1=2)g(x  1) + (1=2)g(x + 1) :",prob.pdf
"dierence g(x) is a solution of the equation
g(x) = (1=2)g(x  1) + (1=2)g(x + 1) :
Also, g(0) = g(n) = 0. Show that it is not possible for g(x) to have a strict
maximum or a strict minimum at the point i, where 1  i  n  1. Use this
to show that g(i) = 0 for all i. This shows that there is at most one solution.
Then verify that the function f(x) = x(n  x) is a solution.",prob.pdf
"428 CHAPTER 11. MARKOV CHAINS
27 Consider an absorbing Markov chain with state space S. Let f be a function
dened on S with the property that
f(i) =
X
j2S
pijf(j) ;
or in vector form
f = Pf :
Then f is called a harmonic function for P. If you imagine a game in which
your fortune is f(i) when you are in state i, then the harmonic condition
means that the game is fair in the sense that your expected fortune after one
step is the same as it was before the step.
(a) Show that for f harmonic
f = Pnf
for all n.
(b) Show, using (a), that for f harmonic
f = P1f ;
where
P1 = lim
n!1
Pn =
 0
B
0 I

:
(c) Using (b), prove that when you start in a transient state i your expected
nal fortune X
k
bikf(k)
is equal to your starting fortune f(i). In other words, a fair game on
a nite state space remains fair to the end. (Fair games in general are
called martingales. Fair games on innite state spaces need not remain
fair with an unlimited number of plays allowed. For example, consider",prob.pdf
"called martingales. Fair games on innite state spaces need not remain
fair with an unlimited number of plays allowed. For example, consider
the game of Heads or Tails (see Example 1.4). Let Peter start with
1 penny and play until he has 2. Then Peter will be sure to end up
1 penny ahead.)
28 A coin is tossed repeatedly. We are interested in nding the expected number
of tosses until a particular pattern, say B = HTH, occurs for the rst time.
If, for example, the outcomes of the tosses are HHTTHTH we say that the
pattern B has occurred for the rst time after 7 tosses. Let T B be the time
to obtain pattern B for the rst time. Li9 gives the following method for
determining E(TB).
We are in a casino and, before each toss of the coin, a gambler enters, pays
1 dollar to play, and bets that the pattern B = HTH will occur on the next
9S-Y. R. Li, \A Martingale Approach to the Study of Occurrence of Sequence Patterns in",prob.pdf
"9S-Y. R. Li, \A Martingale Approach to the Study of Occurrence of Sequence Patterns in
Repeated Experiments,"" Annals of Probability, vol. 8 (1980), pp. 1171{1176.",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 429
three tosses. If H occurs, he wins 2 dollars and bets this amount that the next
outcome will be T. If he wins, he wins 4 dollars and bets this amount that
H will come up next time. If he wins, he wins 8 dollars and the pattern has
occurred. If at any time he loses, he leaves with no winnings.
Let A and B be two patterns. Let AB be the amount the gamblers win who
arrive while the pattern A occurs and bet that B will occur. For example, if
A = HT and B = HTH then AB = 2 + 4 = 6 since the rst gambler bet on
H and won 2 dollars and then bet on T and won 4 dollars more. The second
gambler bet on H and lost. If A = HH and B = HTH, then AB = 2 since the
rst gambler bet on H and won but then bet on T and lost and the second
gambler bet on H and won. If A = B = HTH then AB = BB = 8 + 2 = 10.
Now for each gambler coming in, the casino takes in 1 dollar. Thus the casino
takes in TB dollars. How much does it pay out? The only gamblers who go",prob.pdf
"Now for each gambler coming in, the casino takes in 1 dollar. Thus the casino
takes in TB dollars. How much does it pay out? The only gamblers who go
o with any money are those who arrive during the time the pattern B occurs
and they win the amount BB. But since all the bets made are perfectly fair
bets, it seems quite intuitive that the expected amount the casino takes in
should equal the expected amount that it pays out. That is, E(T B) = BB.
Since we have seen that for B = HTH, BB = 10, the expected time to reach
the pattern HTH for the rst time is 10. If we had been trying to get the
pattern B = HHH, then BB = 8 + 4 +2 = 14 since all the last three gamblers
are paid o in this case. Thus the expected time to get the pattern HHH is 14.
To justify this argument, Li used a theorem from the theory of martingales
(fair games).
We can obtain these expectations by considering a Markov chain whose states
are the possible initial segments of the sequence HTH; these states are HTH,",prob.pdf
"are the possible initial segments of the sequence HTH; these states are HTH,
HT, H, and ;, where ; is the empty set. Then, for this example, the transition
matrix is
0
B
B
@
HTH HT H ;
HTH 1 0 0 0
HT :5 0 0 :5
H 0 :5 :5 0
; 0 0 :5 :5
1
C
C
A;
and if B = HTH, E(TB) is the expected time to absorption for this chain
started in state ;.
Show, using the associated Markov chain, that the values E(T B) = 10 and
E(TB) = 14 are correct for the expected time to reach the patterns HTH and
HHH, respectively.
29 We can use the gambling interpretation given in Exercise 28 to nd the ex-
pected number of tosses required to reach pattern B when we start with pat-
tern A. To be a meaningful problem, we assume that pattern A does not have
pattern B as a subpattern. Let EA(TB) be the expected time to reach pattern
B starting with pattern A. We use our gambling scheme and assume that the
rst k coin tosses produced the pattern A. During this time, the gamblers",prob.pdf
"430 CHAPTER 11. MARKOV CHAINS
made an amount AB. The total amount the gamblers will have made when
the pattern B occurs is BB. Thus, the amount that the gamblers made after
the pattern A has occurred is BB - AB. Again by the fair game argument,
EA(TB) = BB-AB.
For example, suppose that we start with pattern A = HT and are trying to
get the pattern B = HTH. Then we saw in Exercise 28 that AB = 4 and BB
= 10 so EA(TB) = BB-AB= 6.
Verify that this gambling interpretation leads to the correct answer for all
starting states in the examples that you worked in Exercise 28.
30 Here is an elegant method due to Guibas and Odlyzko10 to obtain the expected
time to reach a pattern, say HTH, for the rst time. Let f(n) be the number
of sequences of length n which do not have the pattern HTH. Let fp(n) be the
number of sequences that have the pattern for the rst time after n tosses.
To each element of f(n), add the pattern HTH. Then divide the resulting",prob.pdf
"number of sequences that have the pattern for the rst time after n tosses.
To each element of f(n), add the pattern HTH. Then divide the resulting
sequences into three subsets: the set where HTH occurs for the rst time at
time n + 1 (for this, the original sequence must have ended with HT); the set
where HTH occurs for the rst time at time n + 2 (cannot happen for this
pattern); and the set where the sequence HTH occurs for the rst time at time
n + 3 (the original sequence ended with anything except HT). Doing this, we
have
f(n) = fp(n + 1) + fp(n + 3) :
Thus,
f(n)
2n = 2fp(n + 1)
2n+1 + 23fp(n + 3)
2n+3 :
If T is the time that the pattern occurs for the rst time, this equality states
that
P(T > n) = 2P(T = n + 1) + 8P(T = n + 3) :
Show that if you sum this equality over all n you obtain
1X
n=0
P(T > n) = 2 + 8 = 10 :
Show that for any integer-valued random variable
E(T) =
1X
n=0
P(T > n) ;
and conclude that E(T) = 10. Note that this method of proof makes very",prob.pdf
"Show that for any integer-valued random variable
E(T) =
1X
n=0
P(T > n) ;
and conclude that E(T) = 10. Note that this method of proof makes very
clear that E(T) is, in general, equal to the expected amount the casino pays
out and avoids the martingale system theorem used by Li.
10L. J. Guibas and A. M. Odlyzko, \String Overlaps, Pattern Matching, and Non-transitive
Games,"" Journal of Combinatorial Theory, Series A, vol. 30 (1981), pp. 183{208.",prob.pdf
"11.2. ABSORBING MARKOV CHAINS 431
31 In Example 11.11, dene f(i) to be the proportion of G genes in state i. Show
that f is a harmonic function (see Exercise 27). Why does this show that the
probability of being absorbed in state (GG; GG) is equal to the proportion of
G genes in the starting state? (See Exercise 17.)
32 Show that the stepping stone model (Example 11.12) is an absorbing Markov
chain. Assume that you are playing a game with red and green squares, in
which your fortune at any time is equal to the proportion of red squares at
that time. Give an argument to show that this is a fair game in the sense that
your expected winning after each step is just what it was before this step.Hint:
Show that for every possible outcome in which your fortune will decrease by
one there is another outcome of exactly the same probability where it will
increase by one.
Use this fact and the results of Exercise 27 to show that the probability that a",prob.pdf
"increase by one.
Use this fact and the results of Exercise 27 to show that the probability that a
particular color wins out is equal to the proportion of squares that are initially
of this color.
33 Consider a random walker who moves on the integers 0, 1, . . . , N, moving one
step to the right with probability p and one step to the left with probability
q = 1  p. If the walker ever reaches 0 or N he stays there. (This is the
Gambler's Ruin problem of Exercise 23.) If p = q show that the function
f(i) = i
is a harmonic function (see Exercise 27), and if p 6= q then
f(i) =
q
p
i
is a harmonic function. Use this and the result of Exercise 27 to show that
the probability biN of being absorbed in state N starting in state i is
biN =
( i
N ; if p = q;
(q
p )i1
(q
p )N 1; if p 6= q:
For an alternative derivation of these results see Exercise 24.
34 Complete the following alternate proof of Theorem 11.6. Let si be a tran-",prob.pdf
"For an alternative derivation of these results see Exercise 24.
34 Complete the following alternate proof of Theorem 11.6. Let si be a tran-
sient state and sj be an absorbing state. If we compute bij in terms of the
possibilities on the outcome of the rst step, then we have the equation
bij = pij +
X
k
pikbkj ;
where the summation is carried out over all transient states sk. Write this in
matrix form, and derive from this equation the statement
B = NR :",prob.pdf
"432 CHAPTER 11. MARKOV CHAINS
35 In Monte Carlo roulette (see Example 6.6), under option (c), there are six
states (S, W, L, E, P1, and P2). The reader is referred to Figure 6.2, which
contains a tree for this option. Form a Markov chain for this option, and use
the program AbsorbingChain to nd the probabilities that you win, lose, or
break even for a 1 franc bet on red. Using these probabilities, nd the expected
winnings for this bet. For a more general discussion of Markov chains applied
to roulette, see the article of H. Sagan referred to in Example 6.13.
36 We consider next a game called Penney-ante by its inventor W. Penney.11
There are two players; the rst player picks a pattern A of H's and T's, and
then the second player, knowing the choice of the rst player, picks a dierent
pattern B. We assume that neither pattern is a subpattern of the other pattern.
A coin is tossed a sequence of times, and the player whose pattern comes up",prob.pdf
"A coin is tossed a sequence of times, and the player whose pattern comes up
rst is the winner. To analyze the game, we need to nd the probability pA
that pattern A will occur before pattern B and the probability pB = 1  pA
that pattern B occurs before pattern A. To determine these probabilities we
use the results of Exercises 28 and 29. Here you were asked to show that, the
expected time to reach a pattern B for the rst time is,
E(TB) = BB ;
and, starting with pattern A, the expected time to reach pattern B is
EA(TB) = BB  AB :
(a) Show that the odds that the rst player will win are given by John
Conway's formula12:
pA
1  pA
= pA
pB
= BB  BA
AA  AB :
Hint: Explain why
E(TB) = E(TA or B) + pAEA(TB)
and thus
BB = E(TA or B) + pA(BB  AB) :
Interchange A and B to nd a similar equation involving the pB. Finally,
note that
pA + pB = 1 :
Use these equations to solve for pA and pB.
(b) Assume that both players choose a pattern of the same length k. Show",prob.pdf
"note that
pA + pB = 1 :
Use these equations to solve for pA and pB.
(b) Assume that both players choose a pattern of the same length k. Show
that, if k = 2, this is a fair game, but, if k = 3, the second player has
an advantage no matter what choice the rst player makes. (It has been
shown that, for k  3, if the rst player chooses a1, a2, . .. , ak, then
the optimal strategy for the second player is of the form b, a1, . .. , ak1
where b is the better of the two choices H or T.13)
11W. Penney, \Problem: Penney-Ante,"" Journal of Recreational Math, vol. 2 (1969), p. 241.
12M. Gardner, \Mathematical Games,"" Scientic American, vol. 10 (1974), pp. 120{125.
13Guibas and Odlyzko, op. cit.",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 433
11.3 Ergodic Markov Chains
A second important kind of Markov chain we shall study in detail is an ergodic
Markov chain, dened as follows.
Denition 11.4 A Markov chain is called an ergodic chain if it is possible to go
from every state to every state (not necessarily in one move). 2
In many books, ergodic Markov chains are called irreducible.
Denition 11.5 A Markov chain is called a regular chain if some power of the
transition matrix has only positive elements. 2
In other words, for some n, it is possible to go from any state to any state in
exactly n steps. It is clear from this denition that every regular chain is ergodic.
On the other hand, an ergodic chain is not necessarily regular, as the following
examples show.
Example 11.16 Let the transition matrix of a Markov chain be dened by
P =
1 2
1 0 1
2 1 0

:
Then is clear that it is possible to move from any state to any state, so the chain is",prob.pdf
"P =
1 2
1 0 1
2 1 0

:
Then is clear that it is possible to move from any state to any state, so the chain is
ergodic. However, if n is odd, then it is not possible to move from state 0 to state
0 in n steps, and if n is even, then it is not possible to move from state 0 to state 1
in n steps, so the chain is not regular. 2
A more interesting example of an ergodic, non-regular Markov chain is provided by
the Ehrenfest urn model.
Example 11.17 Recall the Ehrenfest urn model (Example 11.8). The transition
matrix for this example is
P =
0
B
B
B
B
@
0 1 2 3 4
0 0 1 0 0 0
1 1=4 0 3=4 0 0
2 0 1=2 0 1=2 0
3 0 0 3=4 0 1=4
4 0 0 0 1 0
1
C
C
C
C
A
:
In this example, if we start in state 0 we will, after any even number of steps, be in
either state 0, 2 or 4, and after any odd number of steps, be in states 1 or 3. Thus
this chain is ergodic but not regular. 2",prob.pdf
"434 CHAPTER 11. MARKOV CHAINS
Regular Markov Chains
Any transition matrix that has no zeros determines a regular Markov chain. How-
ever, it is possible for a regular Markov chain to have a transition matrix that has
zeros. The transition matrix of the Land of Oz example of Section 11.1 has pNN = 0
but the second power P2 has no zeros, so this is a regular Markov chain.
An example of a nonregular Markov chain is an absorbing chain. For example,
let
P =
 1 0
1=2 1=2

be the transition matrix of a Markov chain. Then all powers of P will have a 0 in
the upper right-hand corner.
We shall now discuss two important theorems relating to regular chains.
Theorem 11.7 Let P be the transition matrix for a regular chain. Then, as n !
1, the powers Pn approach a limiting matrix W with all rows the same vector w.
The vector w is a strictly positive probability vector (i.e., the components are all
positive and they sum to one). 2",prob.pdf
"The vector w is a strictly positive probability vector (i.e., the components are all
positive and they sum to one). 2
In the next section we give two proofs of this fundamental theorem. We give
here the basic idea of the rst proof.
We want to show that the powers Pn of a regular transition matrix tend to a
matrix with all rows the same. This is the same as showing that Pn converges to
a matrix with constant columns. Now the jth column of Pn is Pny where y is a
column vector with 1 in the jth entry and 0 in the other entries. Thus we need only
prove that for any column vector y; Pny approaches a constant vector as n tend to
innity.
Since each row of P is a probability vector, Py replaces y by averages of its
components. Here is an example:
0
@
1=2 1=4 1=4
1=3 1=3 1=3
1=2 1=2 0
1
A
0
@
1
2
3
1
A=
0
@
1=2 
 1 + 1=4 
 2 + 1=4 
 3
1=3 
 1 + 1=3 
 2 + 1=3 
 3
1=2 
 1 + 1=2 
 2 + 0 
 3
1
A=
0
@
7=4
2
3=2
1
A :
The result of the averaging process is to make the components of Py more similar",prob.pdf
"1=2 
 1 + 1=2 
 2 + 0 
 3
1
A=
0
@
7=4
2
3=2
1
A :
The result of the averaging process is to make the components of Py more similar
than those of y. In particular, the maximum component decreases (from 3 to 2)
and the minimum component increases (from 1 to 3/2). Our proof will show that
as we do more and more of this averaging to get Pny, the dierence between the
maximum and minimum component will tend to 0 as n ! 1. This means Pny
tends to a constant vector. The ijth entry of Pn, p(n)
ij , is the probability that the
process will be in state sj after n steps if it starts in state si. If we denote the
common row of W by w, then Theorem 11.7 states that the probability of being
in sj in the long run is approximately wj, the jth entry of w, and is independent
of the starting state.",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 435
Example 11.18 Recall that for the Land of Oz example of Section 11.1, the sixth
power of the transition matrix P is, to three decimal places,
P6 =
0
@
R N S
R :4 :2 :4
N :4 :2 :4
S :4 :2 :4
1
A:
Thus, to this degree of accuracy, the probability of rain six days after a rainy day
is the same as the probability of rain six days after a nice day, or six days after
a snowy day. Theorem 11.7 predicts that, for large n, the rows of P approach a
common vector. It is interesting that this occurs so soon in our example. 2
Theorem 11.8 Let P be a regular transition matrix, let
W = lim
n!1
Pn ;
let w be the common row of W, and let c be the column vector all of whose
components are 1. Then
(a) wP = w, and any row vector v such that vP = v is a constant multiple of w.
(b) Pc = c, and any column vector x such that Px = x is a multiple of c.
Proof. To prove part (a), we note that from Theorem 11.7,
Pn ! W :
Thus,
Pn+1 = Pn 
 P ! WP :",prob.pdf
"Proof. To prove part (a), we note that from Theorem 11.7,
Pn ! W :
Thus,
Pn+1 = Pn 
 P ! WP :
But Pn+1 ! W, and so W = WP, and w = wP.
Let v be any vector with vP = v. Then v = vPn, and passing to the limit,
v = vW. Let r be the sum of the components of v. Then it is easily checked that
vW = rw. So, v = rw.
To prove part (b), assume that x = Px. Then x = Pnx, and again passing to
the limit, x = Wx. Since all rows of W are the same, the components of Wx are
all equal, so x is a multiple of c. 2
Note that an immediate consequence of Theorem 11.8 is the fact that there is
only one probability vector v such that vP = v.
Fixed Vectors
Denition 11.6 A row vector w with the property wP = w is called a xed row
vector for P. Similarly, a column vector x such that Px = x is called a xed column
vector for P. 2",prob.pdf
"436 CHAPTER 11. MARKOV CHAINS
Thus, the common row of W is the unique vector w which is both a xed row
vector for P and a probability vector. Theorem 11.8 shows that any xed row vector
for P is a multiple of w and any xed column vector for P is a constant vector.
One can also state Denition 11.6 in terms of eigenvalues and eigenvectors. A
xed row vector is a left eigenvector of the matrix P corresponding to the eigenvalue
1. A similar statement can be made about xed column vectors.
We will now give several dierent methods for calculating the xed row vector
w for a regular Markov chain.
Example 11.19 By Theorem 11.7 we can nd the limiting vector w for the Land
of Oz from the fact that
w1 + w2 + w3 = 1
and
( w1 w2 w3 )
0
@
1=2 1=4 1=4
1=2 0 1=2
1=4 1=4 1=2
1
A= ( w1 w2 w3 ) :
These relations lead to the following four equations in three unknowns:
w1 + w2 + w3 = 1 ;
(1=2)w1 + (1=2)w2 + (1=4)w3 = w1 ;
(1=4)w1 + (1=4)w3 = w2 ;
(1=4)w1 + (1=2)w2 + (1=2)w3 = w3 :",prob.pdf
"w1 + w2 + w3 = 1 ;
(1=2)w1 + (1=2)w2 + (1=4)w3 = w1 ;
(1=4)w1 + (1=4)w3 = w2 ;
(1=4)w1 + (1=2)w2 + (1=2)w3 = w3 :
Our theorem guarantees that these equations have a unique solution. If the
equations are solved, we obtain the solution
w = ( :4 :2 :4) ;
in agreement with that predicted from P6, given in Example 11.2. 2
To calculate the xed vector, we can assume that the value at a particular state,
say state one, is 1, and then use all but one of the linear equations from wP = w.
This set of equations will have a unique solution and we can obtain w from this
solution by dividing each of its entries by their sum to give the probability vector w.
We will now illustrate this idea for the above example.
Example 11.20 (Example 11.19 continued) We set w1 = 1, and then solve the
rst and second linear equations from wP = w. We have
(1=2) + (1=2)w2 + (1=4)w3 = 1 ;
(1=4) + (1=4)w3 = w2 :
If we solve these, we obtain
( w1 w2 w3 ) = ( 1 1=2 1) :",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 437
Now we divide this vector by the sum of the components, to obtain the nal answer:
w = ( :4 :2 :4) :
This method can be easily programmed to run on a computer. 2
As mentioned above, we can also think of the xed row vector w as a left
eigenvector of the transition matrix P. Thus, if we write I to denote the identity
matrix, then w satises the matrix equation
wP = wI ;
or equivalently,
w(P  I) = 0 :
Thus, w is in the left nullspace of the matrix P  I. Furthermore, Theorem 11.8
states that this left nullspace has dimension 1. Certain computer programming
languages can nd nullspaces of matrices. In such languages, one can nd the xed
row probability vector for a matrix P by computing the left nullspace and then
normalizing a vector in the nullspace so the sum of its components is 1.
The program FixedVector uses one of the above methods (depending upon
the language in which it is written) to calculate the xed row probability vector for",prob.pdf
"the language in which it is written) to calculate the xed row probability vector for
regular Markov chains.
So far we have always assumed that we started in a specic state. The following
theorem generalizes Theorem 11.7 to the case where the starting state is itself
determined by a probability vector.
Theorem 11.9 Let P be the transition matrix for a regular chain and v an arbi-
trary probability vector. Then
lim
n!1
vPn = w ;
where w is the unique xed probability vector for P.
Proof. By Theorem 11.7,
lim
n!1
Pn = W :
Hence,
lim
n!1
vPn = vW :
But the entries in v sum to 1, and each row of W equals w. From these statements,
it is easy to check that
vW = w :
2
If we start a Markov chain with initial probabilities given by v, then the proba-
bility vector vPn gives the probabilities of being in the various states after n steps.
Theorem 11.9 then establishes the fact that, even in this more general class of
processes, the probability of being in sj approaches wj.",prob.pdf
"438 CHAPTER 11. MARKOV CHAINS
Equilibrium
We also obtain a new interpretation for w. Suppose that our starting vector picks
state si as a starting state with probability wi, for all i. Then the probability of
being in the various states after n steps is given by wPn = w, and is the same on all
steps. This method of starting provides us with a process that is called \stationary.""
The fact that w is the only probability vector for which wP = w shows that we
must have a starting probability vector of exactly the kind described to obtain a
stationary process.
Many interesting results concerning regular Markov chains depend only on the
fact that the chain has a unique xed probability vector which is positive. This
property holds for all ergodic Markov chains.
Theorem 11.10 For an ergodic Markov chain, there is a unique probability vec-
tor w such that wP = w and w is strictly positive. Any row vector such that
vP = v is a multiple of w. Any column vector x such that Px = x is a constant",prob.pdf
"vP = v is a multiple of w. Any column vector x such that Px = x is a constant
vector.
Proof. This theorem states that Theorem 11.8 is true for ergodic chains. The
result follows easily from the fact that, if P is an ergodic transition matrix, then
P = (1=2)I+ (1=2)P is a regular transition matrix with the same xed vectors (see
Exercises 25{28). 2
For ergodic chains, the xed probability vector has a slightly dierent inter-
pretation. The following two theorems, which we will not prove here, furnish an
interpretation for this xed vector.
Theorem 11.11 Let P be the transition matrix for an ergodic chain. Let An be
the matrix dened by
An = I + P + P2 + 
 
 
 + Pn
n + 1 :
Then An ! W, where W is a matrix all of whose rows are equal to the unique
xed probability vector w for P. 2
If P is the transition matrix of an ergodic chain, then Theorem 11.8 states
that there is only one xed row probability vector for P. Thus, we can use the",prob.pdf
"that there is only one xed row probability vector for P. Thus, we can use the
same techniques that were used for regular chains to solve for this xed vector. In
particular, the program FixedVector works for ergodic chains.
To interpret Theorem 11.11, let us assume that we have an ergodic chain that
starts in state si. Let X(m) = 1 if the mth step is to state sj and 0 otherwise. Then
the average number of times in state sj in the rst n steps is given by
H(n) = X(0) + X(1) + X(2) + 
 
 
 + X(n)
n + 1 :
But X(m) takes on the value 1 with probability p(m)
ij and 0 otherwise. Thus
E(X(m)) = p(m)
ij , and the ijth entry of An gives the expected value of H(n), that",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 439
is, the expected proportion of times in state sj in the rst n steps if the chain starts
in state si.
If we call being in state sj success and any other state failure, we could ask if
a theorem analogous to the law of large numbers for independent trials holds. The
answer is yes and is given by the following theorem.
Theorem 11.12 (Law of Large Numbers for Ergodic Markov Chains) Let
H(n)
j be the proportion of times in n steps that an ergodic chain is in state sj. Then
for any  > 0,
P

jH(n)
j  wjj > 

! 0 ;
independent of the starting state si. 2
We have observed that every regular Markov chain is also an ergodic chain.
Hence, Theorems 11.11 and 11.12 apply also for regular chains. For example, this
gives us a new interpretation for the xed vector w = (:4; :2; :4) in the Land of Oz
example. Theorem 11.11 predicts that, in the long run, it will rain 40 percent of
the time in the Land of Oz, be nice 20 percent of the time, and snow 40 percent of",prob.pdf
"the time in the Land of Oz, be nice 20 percent of the time, and snow 40 percent of
the time.
Simulation
We illustrate Theorem 11.12 by writing a program to simulate the behavior of a
Markov chain. SimulateChain is such a program.
Example 11.21 In the Land of Oz, there are 525 days in a year. We have simulated
the weather for one year in the Land of Oz, using the program SimulateChain.
The results are shown in Table 11.2.
SSRNRNSSSSSSNRSNSSRNSRNSSSNSRRRNSSSNRRSSSSNRSSNSRRRRRRNSSS
SSRRRSNSNRRRRSRSRNSNSRRNRRNRSSNSRNRNSSRRSRNSSSNRSRRSSNRSNR
RNSSSSNSSNSRSRRNSSNSSRNSSRRNRRRSRNRRRNSSSNRNSRNSNRNRSSSRSS
NRSSSNSSSSSSNSSSNSNSRRNRNRRRRSRRRSSSSNRRSSSSRSRRRNRRRSSSSR
RNRRRSRSSRRRRSSRNRRRRRRNSSRNRSSSNRNSNRRRRNRRRNRSNRRNSRRSNR
RRRSSSRNRRRNSNSSSSSRRRRSRNRSSRRRRSSSRRRNRNRRRSRSRNSNSSRRRR
RNSNRNSNRRNRRRRRRSSSNRSSRSNRSSSNSNRNSNSSSNRRSRRRNRRRRNRNRS
SSNSRSNRNRRSNRRNSRSSSRNSRRSSNSRRRNRRSNRRNSSSSSNRNSSSSSSSNR
NSRRRNSSRRRNSSSNRRSRNSSRRNRRNRSNRRRRRRRRRNSNRRRRRNSRRSSSSN
SNS
State Times Fraction
R 217 .413",prob.pdf
"NSRRRNSSRRRNSSSNRRSRNSSRRNRRNRSNRRRRRRRRRNSNRRRRRNSRRSSSSN
SNS
State Times Fraction
R 217 .413
N 109 .208
S 199 .379
Table 11.2: Weather in the Land of Oz.",prob.pdf
"440 CHAPTER 11. MARKOV CHAINS
We note that the simulation gives a proportion of times in each of the states not
too dierent from the long run predictions of .4, .2, and .4 assured by Theorem 11.7.
To get better results we have to simulate our chain for a longer time. We do this
for 10,000 days without printing out each day's weather. The results are shown in
Table 11.3. We see that the results are now quite close to the theoretical values of
.4, .2, and .4.
State Times Fraction
R 4010 .401
N 1902 .19
S 4088 .409
Table 11.3: Comparison of observed and predicted frequencies for the Land of Oz.
2
Examples of Ergodic Chains
The computation of the xed vector w may be dicult if the transition matrix
is very large. It is sometimes useful to guess the xed vector on purely intuitive
grounds. Here is a simple example to illustrate this kind of situation.
Example 11.22 A white rat is put into the maze of Figure 11.4. There are nine",prob.pdf
"grounds. Here is a simple example to illustrate this kind of situation.
Example 11.22 A white rat is put into the maze of Figure 11.4. There are nine
compartments with connections between the compartments as indicated. The rat
moves through the compartments at random. That is, if there are k ways to leave
a compartment, it chooses each of these with equal probability. We can represent
the travels of the rat by a Markov chain process with transition matrix given by
P =
0
B
B
B
B
B
B
B
B
B
B
B
B
B
@
1 2 3 4 5 6 7 8 9
1 0 1=2 0 0 0 1=2 0 0 0
2 1=3 0 1=3 0 1=3 0 0 0 0
3 0 1=2 0 1=2 0 0 0 0 0
4 0 0 1=3 0 1=3 0 0 0 1=3
5 0 1=4 0 1=4 0 1=4 0 1=4 0
6 1=3 0 0 0 1=3 0 1=3 0 0
7 0 0 0 0 0 1=2 0 1=2 0
8 0 0 0 0 1=3 0 1=3 0 1=3
9 0 0 0 1=2 0 0 0 1=2 0
1
C
C
C
C
C
C
C
C
C
C
C
C
C
A
:
That this chain is not regular can be seen as follows: From an odd-numbered
state the process can go only to an even-numbered state, and from an even-numbered",prob.pdf
"state the process can go only to an even-numbered state, and from an even-numbered
state it can go only to an odd number. Hence, starting in state i the process will
be alternately in even-numbered and odd-numbered states. Therefore, odd powers
of P will have 0's for the odd-numbered entries in row 1. On the other hand, a
glance at the maze shows that it is possible to go from every state to every other
state, so that the chain is ergodic.",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 441
1 2 3
456
7 8 9
Figure 11.4: The maze problem.
To nd the xed probability vector for this matrix, we would have to solve ten
equations in nine unknowns. However, it would seem reasonable that the times
spent in each compartment should, in the long run, be proportional to the number
of entries to each compartment. Thus, we try the vector whose jth component is
the number of entries to the jth compartment:
x = ( 2 3 2 3 4 3 2 3 2) :
It is easy to check that this vector is indeed a xed vector so that the unique
probability vector is this vector normalized to have sum 1:
w = ( 1
12
1
8
1
12
1
8
1
6
1
8
1
12
1
8
1
12 ) :
2
Example 11.23 (Example 11.8 continued) We recall the Ehrenfest urn model of
Example 11.8. The transition matrix for this chain is as follows:
P =
0
B
B
B
B
@
0 1 2 3 4
0 :000 1:000 :000 :000 :000
1 :250 :000 :750 :000 :000
2 :000 :500 :000 :500 :000
3 :000 :000 :750 :000 :250
4 :000 :000 :000 1:000 :000
1
C
C
C
C
A
:",prob.pdf
"1 :250 :000 :750 :000 :000
2 :000 :500 :000 :500 :000
3 :000 :000 :750 :000 :250
4 :000 :000 :000 1:000 :000
1
C
C
C
C
A
:
If we run the program FixedVector for this chain, we obtain the vector
w =
 0 1 2 3 4
:0625 :2500 :3750 :2500 :0625


:
By Theorem 11.12, we can interpret these values for wi as the proportion of times
the process is in each of the states in the long run. For example, the proportion of",prob.pdf
"442 CHAPTER 11. MARKOV CHAINS
times in state 0 is .0625 and the proportion of times in state 1 is .375. The astute
reader will note that these numbers are the binomial distribution 1/16, 4/16, 6/16,
4/16, 1/16. We could have guessed this answer as follows: If we consider a particular
ball, it simply moves randomly back and forth between the two urns. This suggests
that the equilibrium state should be just as if we randomly distributed the four
balls in the two urns. If we did this, the probability that there would be exactly
j balls in one urn would be given by the binomial distribution b(n; p; j) with n = 4
and p = 1=2. 2
Exercises
1 Which of the following matrices are transition matrices for regular Markov
chains?
(a) P =
:5 :5
:5 :5

.
(b) P =
:5 :5
1 0

.
(c) P =
0
@
1=3 0 2=3
0 1 0
0 1=5 4=5
1
A.
(d) P =
0 1
1 0

.
(e) P =
0
@
1=2 1=2 0
0 1=2 1=2
1=3 1=3 1=3
1
A.
2 Consider the Markov chain with transition matrix
P =
0
@
1=2 1=3 1=6
3=4 0 1=4
0 1 0
1
A :",prob.pdf
"1 0

.
(e) P =
0
@
1=2 1=2 0
0 1=2 1=2
1=3 1=3 1=3
1
A.
2 Consider the Markov chain with transition matrix
P =
0
@
1=2 1=3 1=6
3=4 0 1=4
0 1 0
1
A :
(a) Show that this is a regular Markov chain.
(b) The process is started in state 1; nd the probability that it is in state 3
after two steps.
(c) Find the limiting probability vector w.
3 Consider the Markov chain with general 2  2 transition matrix
P =
1  a a
b 1  b

:
(a) Under what conditions is P absorbing?
(b) Under what conditions is P ergodic but not regular?
(c) Under what conditions is P regular?",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 443
4 Find the xed probability vector w for the matrices in Exercise 3 that are
ergodic.
5 Find the xed probability vector w for each of the following regular matrices.
(a) P =
:75 :25
:5 :5

.
(b) P =
:9 :1
:1 :9

.
(c) P =
0
@
3=4 1=4 0
0 2=3 1=3
1=4 1=4 1=2
1
A.
6 Consider the Markov chain with transition matrix in Exercise 3, with a = b =
1. Show that this chain is ergodic but not regular. Find the xed probability
vector and interpret it. Show that Pn does not tend to a limit, but that
An = I + P + P2 + 
 
 
 + Pn
n + 1
does.
7 Consider the Markov chain with transition matrix of Exercise 3, with a = 0
and b = 1=2. Compute directly the unique xed probability vector, and use
your result to prove that the chain is not ergodic.
8 Show that the matrix
P =
0
@
1 0 0
1=4 1=2 1=4
0 0 1
1
A
has more than one xed probability vector. Find the matrix that Pn ap-
proaches as n ! 1, and verify that it is not a matrix all of whose rows are
the same.",prob.pdf
"proaches as n ! 1, and verify that it is not a matrix all of whose rows are
the same.
9 Prove that, if a 3-by-3 transition matrix has the property that its column sums
are 1, then (1=3; 1=3; 1=3) is a xed probability vector. State a similar result
for n-by-n transition matrices. Interpret these results for ergodic chains.
10 Is the Markov chain in Example 11.10 ergodic?
11 Is the Markov chain in Example 11.11 ergodic?
12 Consider Example 11.13 (Drunkard's Walk). Assume that if the walker reaches
state 0, he turns around and returns to state 1 on the next step and, simi-
larly, if he reaches 4 he returns on the next step to state 3. Is this new chain
ergodic? Is it regular?
13 For Example 11.4 when P is ergodic, what is the proportion of people who
are told that the President will run? Interpret the fact that this proportion
is independent of the starting state.",prob.pdf
"444 CHAPTER 11. MARKOV CHAINS
14 Consider an independent trials process to be a Markov chain whose states are
the possible outcomes of the individual trials. What is its xed probability
vector? Is the chain always regular? Illustrate this for Example 11.5.
15 Show that Example 11.8 is an ergodic chain, but not a regular chain. Show
that its xed probability vector w is a binomial distribution.
16 Show that Example 11.9 is regular and nd the limiting vector.
17 Toss a fair die repeatedly. Let Sn denote the total of the outcomes through
the nth toss. Show that there is a limiting value for the proportion of the rst
n values of Sn that are divisible by 7, and compute the value for this limit.
Hint: The desired limit is an equilibrium probability vector for an appropriate
seven state Markov chain.
18 Let P be the transition matrix of a regular Markov chain. Assume that there
are r states and let N(r) be the smallest integer n such that P is regular if",prob.pdf
"are r states and let N(r) be the smallest integer n such that P is regular if
and only if PN(r) has no zero entries. Find a nite upper bound for N(r).
See if you can determine N(3) exactly.
*19 Dene f(r) to be the smallest integer n such that for all regular Markov chains
with r states, the nth power of the transition matrix has all entries positive.
It has been shown,14 that f(r) = r2  2r + 2.
(a) Dene the transition matrix of an r-state Markov chain as follows: For
states si, with i = 1, 2, . . . , r2, P(i; i+1) = 1, P(r1; r) = P(r1; 1) =
1=2, and P(r; 1) = 1. Show that this is a regular Markov chain.
(b) For r = 3, verify that the fth power is the rst power that has no zeros.
(c) Show that, for general r, the smallest n such that Pn has all entries
positive is n = f(r).
20 A discrete time queueing system of capacity n consists of the person being
served and those waiting to be served. The queue length x is observed each",prob.pdf
"served and those waiting to be served. The queue length x is observed each
second. If 0 < x < n, then with probability p, the queue size is increased by
one by an arrival and, inependently, with probability r, it is decreased by one
because the person being served nishes service. If x = 0, only an arrival (with
probability p) is possible. If x = n, an arrival will depart without waiting for
service, and so only the departure (with probability r) of the person being
served is possible. Form a Markov chain with states given by the number of
customers in the queue. Modify the program FixedVector so that you can
input n, p, and r, and the program will construct the transition matrix and
compute the xed vector. The quantity s = p=r is called the trac intensity.
Describe the dierences in the xed vectors according as s < 1, s = 1, or
s > 1.
14E. Seneta, Non-Negative Matrices: An Introduction to Theory and Applications, Wiley, New
York, 1973, pp. 52-54.",prob.pdf
"11.3. ERGODIC MARKOV CHAINS 445
21 Write a computer program to simulate the queue in Exercise 20. Have your
program keep track of the proportion of the time that the queue length is j for
j = 0, 1, . . . , n and the average queue length. Show that the behavior of the
queue length is very dierent depending upon whether the trac intensity s
has the property s < 1, s = 1, or s > 1.
22 In the queueing problem of Exercise 20, let S be the total service time required
by a customer and T the time between arrivals of the customers.
(a) Show that P(S = j) = (1  r)j1r and P(T = j) = (1  p)j1p, for
j > 0.
(b) Show that E(S) = 1=r and E(T) = 1=p.
(c) Interpret the conditions s < 1, s = 1 and s > 1 in terms of these expected
values.
23 In Exercise 20 the service time S has a geometric distribution with E(S) =
1=r. Assume that the service time is, instead, a constant time of t seconds.
Modify your computer program of Exercise 21 so that it simulates a constant",prob.pdf
"Modify your computer program of Exercise 21 so that it simulates a constant
time service distribution. Compare the average queue length for the two
types of distributions when they have the same expected service time (i.e.,
take t = 1=r). Which distribution leads to the longer queues on the average?
24 A certain experiment is believed to be described by a two-state Markov chain
with the transition matrix P, where
P =
:5 :5
p 1  p

and the parameter p is not known. When the experiment is performed many
times, the chain ends in state one approximately 20 percent of the time and in
state two approximately 80 percent of the time. Compute a sensible estimate
for the unknown parameter p and explain how you found it.
25 Prove that, in an r-state ergodic chain, it is possible to go from any state to
any other state in at most r  1 steps.
26 Let P be the transition matrix of an r-state ergodic chain. Prove that, if the
diagonal entries pii are positive, then the chain is regular.",prob.pdf
"26 Let P be the transition matrix of an r-state ergodic chain. Prove that, if the
diagonal entries pii are positive, then the chain is regular.
27 Prove that if P is the transition matrix of an ergodic chain, then (1=2)(I+P)
is the transition matrix of a regular chain. Hint: Use Exercise 26.
28 Prove that P and (1=2)(I + P) have the same xed vectors.
29 In his book, Wahrscheinlichkeitsrechnung und Statistik,15 A. Engle proposes
an algorithm for nding the xed vector for an ergodic Markov chain when
the transition probabilities are rational numbers. Here is his algorithm: For
15A. Engle, Wahrscheinlichkeitsrechnung und Statistik, vol. 2 (Stuttgart: Klett Verlag, 1976).",prob.pdf
"446 CHAPTER 11. MARKOV CHAINS
(4 2 4)
(5 2 3)
(8 2 4)
(7 3 4)
(8 4 4)
(8 3 5)
(8 4 8)
(10 4 6)
(12 4 8)
(12 5 7)
(12 6 8)
(13 5 8)
(16 6 8)
(15 6 9)
(16 6 12)
(17 7 10)
(20 8 12)
(20 8 12) :
Table 11.4: Distribution of chips.
each state i, let ai be the least common multiple of the denominators of the
non-zero entries in the ith row. Engle describes his algorithm in terms of mov-
ing chips around on the states|indeed, for small examples, he recommends
implementing the algorithm this way. Start by putting ai chips on state i for
all i. Then, at each state, redistribute the ai chips, sending aipij to state j.
The number of chips at state i after this redistribution need not be a multiple
of ai. For each state i, add just enough chips to bring the number of chips at
state i up to a multiple of ai. Then redistribute the chips in the same manner.
This process will eventually reach a point where the number of chips at each",prob.pdf
"This process will eventually reach a point where the number of chips at each
state, after the redistribution, is the same as before redistribution. At this
point, we have found a xed vector. Here is an example:
P =
0
@
1 2 3
1 1=2 1=4 1=4
2 1=2 0 1=2
3 1=2 1=4 1=4
1
A:
We start with a = (4; 2; 4). The chips after successive redistributions are
shown in Table 11.4.
We nd that a = (20; 8; 12) is a xed vector.
(a) Write a computer program to implement this algorithm.
(b) Prove that the algorithm will stop. Hint: Let b be a vector with integer
components that is a xed vector for P and such that each coordinate of",prob.pdf
"11.4. FUNDAMENTAL LIMIT THEOREM 447
the starting vector a is less than or equal to the corresponding component
of b. Show that, in the iteration, the components of the vectors are
always increasing, and always less than or equal to the corresponding
component of b.
30 (Coman, Kaduta, and Shepp16) A computing center keeps information on a
tape in positions of unit length. During each time unit there is one request to
occupy a unit of tape. When this arrives the rst free unit is used. Also, during
each second, each of the units that are occupied is vacated with probability p.
Simulate this process, starting with an empty tape. Estimate the expected
number of sites occupied for a given value of p. If p is small, can you choose the
tape long enough so that there is a small probability that a new job will have
to be turned away (i.e., that all the sites are occupied)? Form a Markov chain
with states the number of sites occupied. Modify the program FixedVector",prob.pdf
"with states the number of sites occupied. Modify the program FixedVector
to compute the xed vector. Use this to check your conjecture by simulation.
*31 (Alternate proof of Theorem 11.8) Let P be the transition matrix of an ergodic
Markov chain. Let x be any column vector such that Px = x. Let M be the
maximum value of the components of x. Assume that xi = M. Show that if
pij > 0 then xj = M. Use this to prove that x must be a constant vector.
32 Let P be the transition matrix of an ergodic Markov chain. Let w be a xed
probability vector (i.e., w is a row vector with wP = w). Show that if wi = 0
and pji > 0 then wj = 0. Use this to show that the xed probability vector
for an ergodic chain cannot have any 0 entries.
33 Find a Markov chain that is neither absorbing or ergodic.
11.4 Fundamental Limit Theorem for Regular
Chains
The fundamental limit theorem for regular Markov chains states that if P is a
regular transition matrix then
lim
n!1
Pn = W ;",prob.pdf
"Chains
The fundamental limit theorem for regular Markov chains states that if P is a
regular transition matrix then
lim
n!1
Pn = W ;
where W is a matrix with each row equal to the unique xed probability row vector
w for P. In this section we shall give two very dierent proofs of this theorem.
Our rst proof is carried out by showing that, for any column vector y, Pny
tends to a constant vector. As indicated in Section 11.3, this will show that Pn
converges to a matrix with constant columns or, equivalently, to a matrix with all
rows the same.
The following lemma says that if an r-by-r transition matrix has no zero entries,
and y is any column vector with r entries, then the vector Py has entries which are
\closer together"" than the entries are in y.
16E. G. Coman, J. T. Kaduta, and L. A. Shepp, \On the Asymptotic Optimality of First-
Storage Allocation,"" IEEE Trans. Software Engineering, vol. II (1985), pp. 235-239.",prob.pdf
"448 CHAPTER 11. MARKOV CHAINS
Lemma 11.1 Let P be an r-by-r transition matrix with no zero entries. Let d be
the smallest entry of the matrix. Let y be a column vector with r components, the
largest of which is M0 and the smallest m0. Let M1 and m1 be the largest and
smallest component, respectively, of the vector Py. Then
M1  m1  (1  2d)(M0  m0) :
Proof. In the discussion following Theorem11.7, it was noted that each entry in the
vector Py is a weighted average of the entries in y. The largest weighted average
that could be obtained in the present case would occur if all but one of the entries
of y have value M0 and one entry has value m0, and this one small entry is weighted
by the smallest possible weight, namely d. In this case, the weighted average would
equal
dm0 + (1  d)M0 :
Similarly, the smallest possible weighted average equals
dM0 + (1  d)m0 :
Thus,
M1  m1 

dm0 + (1  d)M0



dM0 + (1  d)m0

= (1  2d)(M0  m0) :
This completes the proof of the lemma. 2",prob.pdf
"dM0 + (1  d)m0 :
Thus,
M1  m1 

dm0 + (1  d)M0



dM0 + (1  d)m0

= (1  2d)(M0  m0) :
This completes the proof of the lemma. 2
We turn now to the proof of the fundamental limit theorem for regular Markov
chains.
Theorem 11.13 (Fundamental Limit Theorem for Regular Chains) If P
is the transition matrix for a regular Markov chain, then
lim
n!1
Pn = W ;
where W is matrix with all rows equal. Furthermore, all entries in W are strictly
positive.
Proof. We prove this theorem for the special case that P has no 0 entries. The
extension to the general case is indicated in Exercise 5. Let y be any r-component
column vector, where r is the number of states of the chain. We assume that
r > 1, since otherwise the theorem is trivial. Let Mn and mn be, respectively,
the maximum and minimum components of the vector Pn y. The vector Pny is
obtained from the vector Pn1y by multiplying on the left by the matrix P. Hence
each component of Pny is an average of the components of Pn1y. Thus",prob.pdf
"obtained from the vector Pn1y by multiplying on the left by the matrix P. Hence
each component of Pny is an average of the components of Pn1y. Thus
M0  M1  M2  
 
 
",prob.pdf
"11.4. FUNDAMENTAL LIMIT THEOREM 449
and
m0  m1  m2  
 
 
 :
Each sequence is monotone and bounded:
m0  mn  Mn  M0 :
Hence, each of these sequences will have a limit as n tends to innity.
Let M be the limit of Mn and m the limit of mn. We know that m  M. We
shall prove that M  m = 0. This will be the case if Mn  mn tends to 0. Let d
be the smallest element of P. Since all entries of P are strictly positive, we have
d > 0. By our lemma
Mn  mn  (1  2d)(Mn1  mn1) :
From this we see that
Mn  mn  (1  2d)n(M0  m0) :
Since r  2, we must have d  1=2, so 0  1  2d < 1, so the dierence Mn  mn
tends to 0 as n tends to innity. Since every component of Pny lies between
mn and Mn, each component must approach the same number u = M = m. This
shows that
lim
n!1
Pny = u ;
where u is a column vector all of whose components equal u.
Now let y be the vector with jth component equal to 1 and all other components",prob.pdf
"Pny = u ;
where u is a column vector all of whose components equal u.
Now let y be the vector with jth component equal to 1 and all other components
equal to 0. Then Pny is the jth column of Pn. Doing this for each j proves that the
columns of Pn approach constant column vectors. That is, the rows of Pn approach
a common row vector w, or,
lim
n!1
Pn = W :
It remains to show that all entries in W are strictly positive. As before, let y
be the vector with jth component equal to 1 and all other components equal to 0.
Then Py is the jth column of P, and this column has all entries strictly positive.
The minimum component of the vector Py was dened to be m1, hence m1 > 0.
Since m1  m, we have m > 0. Note nally that this value of m is just the jth
component of w, so all components of w are strictly positive. 2
Doeblin's Proof
We give now a very dierent proof of the main part of the fundamental limit theorem",prob.pdf
"Doeblin's Proof
We give now a very dierent proof of the main part of the fundamental limit theorem
for regular Markov chains. This proof was rst given by Doeblin,17 a brilliant young
mathematician who was killed in his twenties in the Second World War.
17W. Doeblin, \Expose de la Theorie des Chaines Simple Constantes de Markov a un Nombre
Fini d'Etats,"" Rev. Mach. de l'Union Interbalkanique, vol. 2 (1937), pp. 77{105.",prob.pdf
"450 CHAPTER 11. MARKOV CHAINS
Theorem 11.14 Let P be the transition matrix for a regular Markov chain with
xed vector w. Then for any initial probability vector u, uPn ! w as n ! 1:
Proof. Let X0; X1; : : : be a Markov chain with transition matrix P started in
state si. Let Y0; Y1; : : : be a Markov chain with transition probability P started
with initial probabilities given by w. The X and Y processes are run independently
of each other.
We consider also a third Markov chain P which consists of watching both the
X and Y processes. The states for P are pairs (si; sj). The transition probabilities
are given by
P[(i; j); (k; l)] = P(i; k) 
 P(j; l) :
Since P is regular there is an N such that PN (i; j) > 0 for all i and j. Thus for the
P chain it is also possible to go from any state (si; sj) to any other state (sk; sl)
in at most N steps. That is P is also a regular Markov chain.
We know that a regular Markov chain will reach any state in a nite time. Let T",prob.pdf
"in at most N steps. That is P is also a regular Markov chain.
We know that a regular Markov chain will reach any state in a nite time. Let T
be the rst time the the chain P is in a state of the form (sk; sk). In other words,
T is the rst time that the X and the Y processes are in the same state. Then we
have shown that
P[T > n] ! 0 as n ! 1 :
If we watch the X and Y processes after the rst time they are in the same state
we would not predict any dierence in their long range behavior. Since this will
happen no matter how we started these two processes, it seems clear that the long
range behaviour should not depend upon the starting state. We now show that this
is true.
We rst note that if n  T, then since X and Y are both in the same state at
time T,
P(Xn = j j n  T) = P(Yn = j j n  T) :
If we multiply both sides of this equation by P(n  T), we obtain
P(Xn = j; n  T) = P(Yn = j; n  T) : (11.1)
We know that for all n,
P(Yn = j) = wj :
But",prob.pdf
"P(Xn = j; n  T) = P(Yn = j; n  T) : (11.1)
We know that for all n,
P(Yn = j) = wj :
But
P(Yn = j) = P(Yn = j; n  T) + P(Yn = j; n < T) ;
and the second summand on the right-hand side of this equation goes to 0 as n goes
to 1, since P(n < T) goes to 0 as n goes to 1. So,
P(Yn = j; n  T) ! wj ;
as n goes to 1. From Equation 11.1, we see that
P(Xn = j; n  T) ! wj ;",prob.pdf
"11.4. FUNDAMENTAL LIMIT THEOREM 451
as n goes to 1. But by similar reasoning to that used above, the dierence between
this last expression and P(Xn = j) goes to 0 as n goes to 1. Therefore,
P(Xn = j) ! wj ;
as n goes to 1. This completes the proof. 2
In the above proof, we have said nothing about the rate at which the distributions
of the Xn's approach the xed distribution w. In fact, it can be shown that18
rX
j=1
j P(Xn = j)  wj j 2P(T > n) :
The left-hand side of this inequality can be viewed as the distance between the
distribution of the Markov chain after n steps, starting in state si, and the limiting
distribution w.
Exercises
1 Dene P and y by
P =
 :5 :5
:25 :75

; y =
1
0

:
Compute Py, P2y, and P4y and show that the results are approaching a
constant vector. What is this vector?
2 Let P be a regular r  r transition matrix and y any r-component column
vector. Show that the value of the limiting constant vector for Pny is wy.
3 Let
P =
0
@
1 0 0
:25 0 :75
0 0 1
1
A",prob.pdf
"vector. Show that the value of the limiting constant vector for Pny is wy.
3 Let
P =
0
@
1 0 0
:25 0 :75
0 0 1
1
A
be a transition matrix of a Markov chain. Find two xed vectors of P that are
linearly independent. Does this show that the Markov chain is not regular?
4 Describe the set of all xed column vectors for the chain given in Exercise 3.
5 The theorem that Pn ! W was proved only for the case that P has no zero
entries. Fill in the details of the following extension to the case that P is
regular. Since P is regular, for some N; PN has no zeros. Thus, the proof
given shows that MnN  mnN approaches 0 as n tends to innity. However,
the dierence Mn  mn can never increase. (Why?) Hence, if we know that
the dierences obtained by looking at every Nth time tend to 0, then the
entire sequence must also tend to 0.
6 Let P be a regular transition matrix and let w be the unique non-zero xed
vector of P. Show that no entry of w is 0.",prob.pdf
"6 Let P be a regular transition matrix and let w be the unique non-zero xed
vector of P. Show that no entry of w is 0.
18T. Lindvall, Lectures on the Coupling Method (New York: Wiley 1992).",prob.pdf
"452 CHAPTER 11. MARKOV CHAINS
7 Here is a trick to try on your friends. Shue a deck of cards and deal them
out one at a time. Count the face cards each as ten. Ask your friend to look
at one of the rst ten cards; if this card is a six, she is to look at the card that
turns up six cards later; if this card is a three, she is to look at the card that
turns up three cards later, and so forth. Eventually she will reach a point
where she is to look at a card that turns up x cards later but there are not
x cards left. You then tell her the last card that she looked at even though
you did not know her starting point. You tell her you do this by watching
her, and she cannot disguise the times that she looks at the cards. In fact you
just do the same procedure and, even though you do not start at the same
point as she does, you will most likely end at the same point. Why?
8 Write a program to play the game in Exercise 7.",prob.pdf
"point as she does, you will most likely end at the same point. Why?
8 Write a program to play the game in Exercise 7.
9 (Suggested by Peter Doyle) In the proof of Theorem 11.14, we assumed the
existence of a xed vector w. To avoid this assumption, beef up the coupling
argument to show (without assuming the existence of a stationary distribution
w) that for appropriate constants C and r < 1, the distance between P n
and Pn is at most Crn for any starting distributions  and . Apply this
in the case where  = P to conclude that the sequence Pn is a Cauchy
sequence, and that its limit is a matrix W whose rows are all equal to a
probability vector w with wP = w. Note that the distance between P n and
w is at most Crn, so in freeing ourselves from the assumption about having
a xed vector we've proved that the convergence to equilibrium takes place
exponentially fast.
11.5 Mean First Passage Time for Ergodic Chains",prob.pdf
"a xed vector we've proved that the convergence to equilibrium takes place
exponentially fast.
11.5 Mean First Passage Time for Ergodic Chains
In this section we consider two closely related descriptive quantities of interest for
ergodic chains: the mean time to return to a state and the mean time to go from
one state to another state.
Let P be the transition matrix of an ergodic chain with states s1, s2, . . . , sr. Let
w = (w1; w2; : : : ; wr) be the unique probability vector such that wP = w. Then,
by the Law of Large Numbers for Markov chains, in the long run the process will
spend a fraction wj of the time in state sj. Thus, if we start in any state, the chain
will eventually reach state sj; in fact, it will be in state sj innitely often.
Another way to see this is the following: Form a new Markov chain by making
sj an absorbing state, that is, dene pjj = 1. If we start at any state other than sj,",prob.pdf
"sj an absorbing state, that is, dene pjj = 1. If we start at any state other than sj,
this new process will behave exactly like the original chain up to the rst time that
state sj is reached. Since the original chain was an ergodic chain, it was possible
to reach sj from any other state. Thus the new chain is an absorbing chain with a
single absorbing state sj that will eventually be reached. So if we start the original
chain at a state si with i 6= j, we will eventually reach the state sj.
Let N be the fundamental matrix for the new chain. The entries of N give the
expected number of times in each state before absorption. In terms of the original",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 453
1 2 3
456
7 8 9
Figure 11.5: The maze problem.
chain, these quantities give the expected number of times in each of the states before
reaching state sj for the rst time. The ith component of the vector Nc gives the
expected number of steps before absorption in the new chain, starting in state si.
In terms of the old chain, this is the expected number of steps required to reach
state sj for the rst time starting at state si.
Mean First Passage Time
Denition 11.7 If an ergodic Markov chain is started in state si, the expected
number of steps to reach state sj for the rst time is called the mean rst passage
time from si to sj. It is denoted by mij. By convention mii = 0. 2
Example 11.24 Let us return to the maze example (Example 11.22). We shall
make this ergodic chain into an absorbing chain by making state 5 an absorbing
state. For example, we might assume that food is placed in the center of the maze",prob.pdf
"state. For example, we might assume that food is placed in the center of the maze
and once the rat nds the food, he stays to enjoy it (see Figure 11.5).
The new transition matrix in canonical form is
P =
0
B
B
B
B
B
B
B
B
B
B
B
B
B
B
@
1 2 3 4 6 7 8 9 5
1 0 1=2 0 0 1=2 0 0 0
0
2 1=3 0 1=3 0 0 0 0 0 1=3
3 0 1=2 0 1=2 0 0 0 0 0
4 0 0 1=3 0 0 1=3 0 1=3 1=3
6 1=3 0 0 0 0 0 0 0 1=3
7 0 0 0 0 1=2 0 1=2 0 0
8 0 0 0 0 0 1=3 0 1=3 1=3
9 0 0 0 1=2 0 0 1=2 0 0
5 0 0 0 0 0 0 0 0 1
1
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A
:",prob.pdf
"454 CHAPTER 11. MARKOV CHAINS
If we compute the fundamental matrix N, we obtain
N = 1
8
0
B
B
B
B
B
B
B
B
B
B
@
14 9 4 3 9 4 3 2
6 14 6 4 4 2 2 2
4 9 14 9 3 2 3 4
2 4 6 14 2 2 4 6
6 4 2 2 14 6 4 2
4 3 2 3 9 14 9 4
2 2 2 4 4 6 14 6
2 3 4 9 3 4 9 14
1
C
C
C
C
C
C
C
C
C
C
A
:
The expected time to absorption for dierent starting states is given by the vec-
tor Nc, where
Nc =
0
B
B
B
B
B
B
B
B
B
B
@
6
5
6
5
5
6
5
6
1
C
C
C
C
C
C
C
C
C
C
A
:
We see that, starting from compartment 1, it will take on the average six steps
to reach food. It is clear from symmetry that we should get the same answer for
starting at state 3, 7, or 9. It is also clear that it should take one more step,
starting at one of these states, than it would starting at 2, 4, 6, or 8. Some of the
results obtained from N are not so obvious. For instance, we note that the expected
number of times in the starting state is 14/8 regardless of the state in which we
start. 2
Mean Recurrence Time",prob.pdf
"number of times in the starting state is 14/8 regardless of the state in which we
start. 2
Mean Recurrence Time
A quantity that is closely related to the mean rst passage time is the mean recur-
rence time, dened as follows. Assume that we start in state si; consider the length
of time before we return to si for the rst time. It is clear that we must return,
since we either stay at si the rst step or go to some other state sj, and from any
other state sj, we will eventually reach si because the chain is ergodic.
Denition 11.8 If an ergodic Markov chain is started in state si, the expected
number of steps to return to si for the rst time is the mean recurrence time for si.
It is denoted by ri. 2
We need to develop some basic properties of the mean rst passage time. Con-
sider the mean rst passage time from si to sj; assume that i 6= j. This may be
computed as follows: take the expected number of steps required given the outcome",prob.pdf
"computed as follows: take the expected number of steps required given the outcome
of the rst step, multiply by the probability that this outcome occurs, and add. If
the rst step is to sj, the expected number of steps required is 1; if it is to some",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 455
other state sk, the expected number of steps required is mkj plus 1 for the step
already taken. Thus,
mij = pij +
X
k6=j
pik(mkj + 1) ;
or, since P
k pik = 1,
mij = 1 +
X
k6=j
pikmkj : (11.2)
Similarly, starting in si, it must take at least one step to return. Considering
all possible rst steps gives us
ri =
X
k
pik(mki + 1) (11.3)
= 1 +
X
k
pikmki : (11.4)
Mean First Passage Matrix and Mean Recurrence Matrix
Let us now dene two matrices M and D. The ijth entry mij of M is the mean rst
passage time to go from si to sj if i 6= j; the diagonal entries are 0. The matrix M
is called the mean rst passage matrix. The matrix D is the matrix with all entries
0 except the diagonal entries dii = ri. The matrix D is called the mean recurrence
matrix. Let C be an r  r matrix with all entries 1. Using Equation 11.2 for the
case i 6= j and Equation 11.4 for the case i = j, we obtain the matrix equation
M = PM + C  D ; (11.5)
or
(I  P)M = C  D : (11.6)",prob.pdf
"case i 6= j and Equation 11.4 for the case i = j, we obtain the matrix equation
M = PM + C  D ; (11.5)
or
(I  P)M = C  D : (11.6)
Equation 11.6 with mii = 0 implies Equations 11.2 and 11.4. We are now in a
position to prove our rst basic theorem.
Theorem 11.15 For an ergodic Markov chain, the mean recurrence time for state
si is ri = 1=wi, where wi is the ith component of the xed probability vector for
the transition matrix.
Proof. Multiplying both sides of Equation 11.6 by w and using the fact that
w(I  P) = 0
gives
wC  wD = 0 :
Here wC is a row vector with all entries 1 and wD is a row vector with ith entry
wiri. Thus
(1; 1; : : :; 1) = (w1r1; w2r2; : : :; wnrn)
and
ri = 1=wi ;
as was to be proved. 2",prob.pdf
"456 CHAPTER 11. MARKOV CHAINS
Corollary 11.1 For an ergodic Markov chain, the components of the xed proba-
bility vector w are strictly positive.
Proof. We know that the values of ri are nite and so wi = 1=ri cannot be 0. 2
Example 11.25 In Example 11.22 we found the xed probability vector for the
maze example to be
w = ( 1
12
1
8
1
12
1
8
1
6
1
8
1
12
1
8
1
12 ) :
Hence, the mean recurrence times are given by the reciprocals of these probabilities.
That is,
r = ( 12 8 12 8 6 8 12 8 12) :
2
Returning to the Land of Oz, we found that the weather in the Land of Oz could
be represented by a Markov chain with states rain, nice, and snow. In Section 11.3
we found that the limiting vector was w = (2=5; 1=5; 2=5). From this we see that
the mean number of days between rainy days is 5/2, between nice days is 5, and
between snowy days is 5/2.
Fundamental Matrix
We shall now develop a fundamental matrix for ergodic chains that will play a role",prob.pdf
"between snowy days is 5/2.
Fundamental Matrix
We shall now develop a fundamental matrix for ergodic chains that will play a role
similar to that of the fundamental matrix N = (I  Q)1 for absorbing chains. As
was the case with absorbing chains, the fundamental matrix can be used to nd
a number of interesting quantities involving ergodic chains. Using this matrix, we
will give a method for calculating the mean rst passage times for ergodic chains
that is easier to use than the method given above. In addition, we will state (but
not prove) the Central Limit Theorem for Markov Chains, the statement of which
uses the fundamental matrix.
We begin by considering the case that P is the transition matrix of a regular
Markov chain. Since there are no absorbing states, we might be tempted to try
Z = (I  P)1 for a fundamental matrix. But I  P does not have an inverse. To
see this, recall that a matrix R has an inverse if and only if Rx = 0 implies x = 0.",prob.pdf
"see this, recall that a matrix R has an inverse if and only if Rx = 0 implies x = 0.
But since Pc = c we have (I  P)c = 0, and so I  P does not have an inverse.
We recall that if we have an absorbing Markov chain, and Q is the restriction
of the transition matrix to the set of transient states, then the fundamental matrix
N could be written as
N = I + Q + Q2 + 
 
 
 :
The reason that this power series converges is that Qn ! 0, so this series acts like
a convergent geometric series.
This idea might prompt one to try to nd a similar series for regular chains.
Since we know that Pn ! W, we might consider the series
I + (P  W) + (P2  W) + 
 
 
 : (11.7)",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 457
We now use special properties of P and W to rewrite this series. The special
properties are: 1) PW = W, and 2) Wk = W for all positive integers k. These
facts are easy to verify, and are left as an exercise (see Exercise 22). Using these
facts, we see that
(P  W)n =
nX
i=0
(1)i
n
i

PniWi
= Pn +
nX
i=1
(1)i
n
i

Wi
= Pn +
nX
i=1
(1)i
n
i

W
= Pn +
 nX
i=1
(1)i
n
i
!
W :
If we expand the expression (1  1)n, using the Binomial Theorem, we obtain the
expression in parenthesis above, except that we have an extra term (which equals
1). Since (1  1)n = 0, we see that the above expression equals -1. So we have
(P  W)n = Pn  W ;
for all n  1.
We can now rewrite the series in 11.7 as
I + (P  W) + (P  W)2 + 
 
 
 :
Since the nth term in this series is equal to Pn  W, the nth term goes to 0 as n
goes to innity. This is sucient to show that this series converges, and sums to",prob.pdf
"goes to innity. This is sucient to show that this series converges, and sums to
the inverse of the matrix I  P + W. We call this inverse the fundamental matrix
associated with the chain, and we denote it by Z.
In the case that the chain is ergodic, but not regular, it is not true that Pn ! W
as n ! 1. Nevertheless, the matrix I  P + W still has an inverse, as we will now
show.
Proposition 11.1 Let P be the transition matrix of an ergodic chain, and let W
be the matrix all of whose rows are the xed probability row vector for P. Then
the matrix
I  P + W
has an inverse.
Proof. Let x be a column vector such that
(I  P + W)x = 0 :
To prove the proposition, it is sucient to show that x must be the zero vector.
Multiplying this equation by w and using the fact that w(I P) = 0 and wW = w,
we have
w(I  P + W)x = wx = 0 :",prob.pdf
"458 CHAPTER 11. MARKOV CHAINS
Therefore,
(I  P)x = 0 :
But this means that x = Px is a xed column vector for P. By Theorem 11.10,
this can only happen if x is a constant vector. Since wx = 0, and w has strictly
positive entries, we see that x = 0. This completes the proof. 2
As in the regular case, we will call the inverse of the matrix I  P + W the
fundamental matrix for the ergodic chain with transition matrix P, and we will use
Z to denote this fundamental matrix.
Example 11.26 Let P be the transition matrix for the weather in the Land of Oz.
Then
I  P + W =
0
@
1 0 0
0 1 0
0 0 1
1
A
0
@
1=2 1=4 1=4
1=2 0 1=2
1=4 1=4 1=2
1
A+
0
@
2=5 1=5 2=5
2=5 1=5 2=5
2=5 1=5 2=5
1
A
=
0
@
9=10 1=20 3=20
1=10 6=5 1=10
3=20 1=20 9=10
1
A ;
so
Z = (I  P + W)1 =
0
@
86=75 1=25 14=75
2=25 21=25 2=25
14=75 1=25 86=75
1
A :
2
Using the Fundamental Matrix to Calculate the Mean First
Passage Matrix
We shall show how one can obtain the mean rst passage matrix M from the",prob.pdf
"2
Using the Fundamental Matrix to Calculate the Mean First
Passage Matrix
We shall show how one can obtain the mean rst passage matrix M from the
fundamental matrix Z for an ergodic Markov chain. Before stating the theorem
which gives the rst passage times, we need a few facts about Z.
Lemma 11.2 Let Z = (I  P + W)1, and let c be a column vector of all 1's.
Then
Zc = c ;
wZ = w ;
and
Z(I  P) = I  W :
Proof. Since Pc = c and Wc = c,
c = (I  P + W)c :
If we multiply both sides of this equation on the left by Z, we obtain
Zc = c :",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 459
Similarly, since wP = w and wW = w,
w = w(I  P + W) :
If we multiply both sides of this equation on the right by Z, we obtain
wZ = w :
Finally, we have
(I  P + W)(I  W) = I  W  P + W + W  W
= I  P :
Multiplying on the left by Z, we obtain
I  W = Z(I  P) :
This completes the proof. 2
The following theorem shows how one can obtain the mean rst passage times
from the fundamental matrix.
Theorem 11.16 The mean rst passage matrix M for an ergodic chain is deter-
mined from the fundamental matrix Z and the xed row probability vector w by
mij = zjj  zij
wj
:
Proof. We showed in Equation 11.6 that
(I  P)M = C  D :
Thus,
Z(I  P)M = ZC  ZD ;
and from Lemma 11.2,
Z(I  P)M = C  ZD :
Again using Lemma 11.2, we have
M  WM = C  ZD
or
M = C  ZD + WM :
From this equation, we see that
mij = 1  zijrj + (wM)j : (11.8)
But mjj = 0, and so
0 = 1  zjjrj + (wM)j ;",prob.pdf
"460 CHAPTER 11. MARKOV CHAINS
or
(wM)j = zjjrj  1 : (11.9)
From Equations 11.8 and 11.9, we have
mij = (zjj  zij) 
 rj :
Since rj = 1=wj,
mij = zjj  zij
wj
:
2
Example 11.27 (Example 11.26 continued) In the Land of Oz example, we nd
that
Z = (I  P + W)1 =
0
@
86=75 1=25 14=75
2=25 21=25 2=25
14=75 1=25 86=75
1
A :
We have also seen that w = (2=5; 1=5; 2=5). So, for example,
m12 = z22  z12
w2
= 21=25  1=25
1=5
= 4 ;
by Theorem 11.16. Carrying out the calculations for the other entries of M, we
obtain
M =
0
@
0 4 10=3
8=3 0 8=3
10=3 4 0
1
A :
2
Computation
The program ErgodicChain calculates the fundamental matrix, the xed vector,
the mean recurrence matrix D, and the mean rst passage matrix M. We have run
the program for the Ehrenfest urn model (Example 11.8). We obtain:
P =
0
B
B
B
B
@
0 1 2 3 4
0 :0000 1:0000 :0000 :0000 :0000
1 :2500 :0000 :7500 :0000 :0000
2 :0000 :5000 :0000 :5000 :0000
3 :0000 :0000 :7500 :0000 :2500
4 :0000 :0000 :0000 1:0000 :0000
1
C
C
C
C
A
;
w =",prob.pdf
"1 :2500 :0000 :7500 :0000 :0000
2 :0000 :5000 :0000 :5000 :0000
3 :0000 :0000 :7500 :0000 :2500
4 :0000 :0000 :0000 1:0000 :0000
1
C
C
C
C
A
;
w =
 0 1 2 3 4
:0625 :2500 :3750 :2500 :0625


;",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 461
r =
 0 1 2 3 4
16:0000 4:0000 2:6667 4:0000 16:0000


;
M =
0
B
B
B
B
@
0 1 2 3 4
0 :0000 1:0000 2:6667 6:3333 21:3333
1 15:0000 :0000 1:6667 5:3333 20:3333
2 18:6667 3:6667 :0000 3:6667 18:6667
3 20:3333 5:3333 1:6667 :0000 15:0000
4 21:3333 6:3333 2:6667 1:0000 :0000
1
C
C
C
C
A
:
From the mean rst passage matrix, we see that the mean time to go from 0 balls
in urn 1 to 2 balls in urn 1 is 2.6667 steps while the mean time to go from 2 balls in
urn 1 to 0 balls in urn 1 is 18.6667. This re
ects the fact that the model exhibits a
central tendency. Of course, the physicist is interested in the case of a large number
of molecules, or balls, and so we should consider this example for n so large that
we cannot compute it even with a computer.
Ehrenfest Model
Example 11.28 (Example 11.23 continued) Let us consider the Ehrenfest model
(see Example 11.8) for gas diusion for the general case of 2n balls. Every second,",prob.pdf
"(see Example 11.8) for gas diusion for the general case of 2n balls. Every second,
one of the 2n balls is chosen at random and moved from the urn it was in to the
other urn. If there are i balls in the rst urn, then with probability i=2n we take
one of them out and put it in the second urn, and with probability (2n  i)=2n we
take a ball from the second urn and put it in the rst urn. At each second we let
the number i of balls in the rst urn be the state of the system. Then from state i
we can pass only to state i  1 and i + 1, and the transition probabilities are given
by
pij =
8
<
:
i
2n ; if j = i  1;
1  i
2n ; if j = i + 1;
0 ; otherwise.
This denes the transition matrix of an ergodic, non-regular Markov chain (see
Exercise 15). Here the physicist is interested in long-term predictions about the
state occupied. In Example 11.23, we gave an intuitive reason for expecting that
the xed vector w is the binomial distribution with parameters 2n and 1=2. It is",prob.pdf
"the xed vector w is the binomial distribution with parameters 2n and 1=2. It is
easy to check that this is correct. So,
wi =
2n
i


22n :
Thus the mean recurrence time for state i is
ri = 22n
2n
i

 :",prob.pdf
"462 CHAPTER 11. MARKOV CHAINS
0 200 400 600 800 1000
40
45
50
55
60
65
0 200 400 600 800 1000
40
45
50
55
60
65
Time forward
Time reversed
Figure 11.6: Ehrenfest simulation.
Consider in particular the central term i = n. We have seen that this term is
approximately 1=pn. Thus we may approximate rn by pn.
This model was used to explain the concept of reversibility in physical systems.
Assume that we let our system run until it is in equilibrium. At this point, a movie
is made, showing the system's progress. The movie is then shown to you, and you
are asked to tell if the movie was shown in the forward or the reverse direction.
It would seem that there should always be a tendency to move toward an equal
proportion of balls so that the correct order of time should be the one with the
most transitions from i to i  1 if i > n and i to i + 1 if i < n.
In Figure 11.6 we show the results of simulating the Ehrenfest urn model for",prob.pdf
"most transitions from i to i  1 if i > n and i to i + 1 if i < n.
In Figure 11.6 we show the results of simulating the Ehrenfest urn model for
the case of n = 50 and 1000 time units, using the program EhrenfestUrn. The
top graph shows these results graphed in the order in which they occurred and the
bottom graph shows the same results but with time reversed. There is no apparent
dierence.",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 463
We note that if we had not started in equilibrium, the two graphs would typically
look quite dierent. 2
Reversibility
If the Ehrenfest model is started in equilibrium, then the process has no apparent
time direction. The reason for this is that this process has a property called re-
versibility. Dene Xn to be the number of balls in the left urn at step n. We can
calculate, for a general ergodic chain, the reverse transition probability:
P(Xn1 = jjXn = i) = P(Xn1 = j; Xn = i)
P(Xn = i)
= P(Xn1 = j)P(Xn = ijXn1 = j)
P(Xn = i)
= P(Xn1 = j)pji
P(Xn = i) :
In general, this will depend upon n, since P(Xn = j) and also P(Xn1 = j)
change with n. However, if we start with the vector w or wait until equilibrium is
reached, this will not be the case. Then we can dene
p
ij = wjpji
wi
as a transition matrix for the process watched with time reversed.
Let us calculate a typical transition probability for the reverse chain P = fp
ijg",prob.pdf
"Let us calculate a typical transition probability for the reverse chain P = fp
ijg
in the Ehrenfest model. For example,
p
i;i1 = wi1pi1;i
wi
=
2n
i1


22n  2n  i + 1
2n  22n
2n
i


= (2n)!
(i  1)! (2n  i + 1)!  (2n  i + 1)i! (2n  i)!
2n(2n)!
= i
2n = pi;i1 :
Similar calculations for the other transition probabilities show that P = P.
When this occurs the process is called reversible. Clearly, an ergodic chain is re-
versible if, and only if, for every pair of states si and sj, wipij = wjpji. In particular,
for the Ehrenfest model this means that wipi;i1 = wi1pi1;i. Thus, in equilib-
rium, the pairs (i; i  1) and (i  1; i) should occur with the same frequency. While
many of the Markov chains that occur in applications are reversible, this is a very
strong condition. In Exercise 12 you are asked to nd an example of a Markov chain
which is not reversible.
The Central Limit Theorem for Markov Chains",prob.pdf
"which is not reversible.
The Central Limit Theorem for Markov Chains
Suppose that we have an ergodic Markov chain with states s1; s2; : : : ; sk. It is
natural to consider the distribution of the random variables S(n)
j , which denotes",prob.pdf
"464 CHAPTER 11. MARKOV CHAINS
the number of times that the chain is in state sj in the rst n steps. The jth
component wj of the xed probability row vector w is the proportion of times that
the chain is in state sj in the long run. Hence, it is reasonable to conjecture that
the expected value of the random variable S(n)
j , as n ! 1, is asymptotic to nwj,
and it is easy to show that this is the case (see Exercise 23).
It is also natural to ask whether there is a limiting distribution of the random
variables S(n)
j . The answer is yes, and in fact, this limiting distribution is the normal
distribution. As in the case of independent trials, one must normalize these random
variables. Thus, we must subtract from S(n)
j its expected value, and then divide by
its standard deviation. In both cases, we will use the asymptotic values of these
quantities, rather than the values themselves. Thus, in the rst case, we will use",prob.pdf
"quantities, rather than the values themselves. Thus, in the rst case, we will use
the value nwj. It is not so clear what we should use in the second case. It turns
out that the quantity
2
j = 2wjzjj  wj  w2
j (11.10)
represents the asymptotic variance. Armed with these ideas, we can state the
following theorem.
Theorem 11.17 (Central Limit Theorem for Markov Chains) For an er-
godic chain, for any real numbers r < s, we have
P
 
r <
S(n)
j  nwj
q
n2
j
< s
!
! 1p
2
Z s
r
ex2=2 dx ;
as n ! 1, for any choice of starting state, where 2
j is the quantity dened in
Equation 11.10. 2
Historical Remarks
Markov chains were introduced by Andrei Andreevich Markov (1856{1922) and
were named in his honor. He was a talented undergraduate who received a gold
medal for his undergraduate thesis at St. Petersburg University. Besides being
an active research mathematician and teacher, he was also active in politics and",prob.pdf
"an active research mathematician and teacher, he was also active in politics and
patricipated in the liberal movement in Russia at the beginning of the twentieth
century. In 1913, when the government celebrated the 300th anniversary of the
House of Romanov family, Markov organized a counter-celebration of the 200th
anniversary of Bernoulli's discovery of the Law of Large Numbers.
Markov was led to develop Markov chains as a natural extension of sequences
of independent random variables. In his rst paper, in 1906, he proved that for a
Markov chain with positive transition probabilities and numerical states the average
of the outcomes converges to the expected value of the limiting distribution (the
xed vector). In a later paper he proved the central limit theorem for such chains.
Writing about Markov, A. P. Youschkevitch remarks:
Markov arrived at his chains starting from the internal needs of prob-
ability theory, and he never wrote about their applications to physical",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 465
science. For him the only real examples of the chains were literary texts,
where the two states denoted the vowels and consonants.19
In a paper written in 1913,20 Markov chose a sequence of 20,000 letters from
Pushkin's Eugene Onegin to see if this sequence can be approximately considered
a simple chain. He obtained the Markov chain with transition matrix
vowel consonant
vowel :128 :872
consonant :663 :337

:
The xed vector for this chain is (:432; :568), indicating that we should expect
about 43.2 percent vowels and 56.8 percent consonants in the novel, which was
borne out by the actual count.
Claude Shannon considered an interesting extension of this idea in his book The
Mathematical Theory of Communication,21 in which he developed the information-
theoretic concept of entropy. Shannon considers a series of Markov chain approxi-
mations to English prose. He does this rst by chains in which the states are letters",prob.pdf
"mations to English prose. He does this rst by chains in which the states are letters
and then by chains in which the states are words. For example, for the case of
words he presents rst a simulation where the words are chosen independently but
with appropriate frequencies.
REPRESENTING AND SPEEDILY IS AN GOOD APT OR COME
CAN DIFFERENT NATURAL HERE HE THE A IN CAME THE TO
OF TO EXPERT GRAY COME TO FURNISHES THE LINE MES-
SAGE HAD BE THESE.
He then notes the increased resemblence to ordinary English text when the words
are chosen as a Markov chain, in which case he obtains
THE HEAD AND IN FRONTAL ATTACK ON AN ENGLISH WRI-
TER THAT THE CHARACTER OF THIS POINT IS THEREFORE
ANOTHER METHOD FOR THE LETTERS THAT THE TIME OF
WHO EVER TOLD THE PROBLEM FOR AN UNEXPECTED.
A simulation like the last one is carried out by opening a book and choosing the
rst word, say it is the. Then the book is read until the word the appears again",prob.pdf
"rst word, say it is the. Then the book is read until the word the appears again
and the word after this is chosen as the second word, which turned out to be head.
The book is then read until the word head appears again and the next word, and,
is chosen, and so on.
Other early examples of the use of Markov chains occurred in Galton's study of
the problem of survival of family names in 1889 and in the Markov chain introduced
19See Dictionary of Scientic Biography, ed. C. C. Gillespie (New York: Scribner's Sons, 1970),
pp. 124{130.
20A. A. Markov, \An Example of Statistical Analysis of the Text of Eugene Onegin Illustrat-
ing the Association of Trials into a Chain,"" Bulletin de l'Acadamie Imperiale des Sciences de
St. Petersburg, ser. 6, vol. 7 (1913), pp. 153{162.
21C. E. Shannon and W. Weaver, The Mathematical Theory of Communication (Urbana: Univ.
of Illinois Press, 1964).",prob.pdf
"466 CHAPTER 11. MARKOV CHAINS
by P. and T. Ehrenfest in 1907 for diusion. Poincare in 1912 dicussed card shuing
in terms of an ergodic Markov chain dened on a permutation group. Brownian
motion, a continuous time version of random walk, was introducted in 1900{1901
by L. Bachelier in his study of the stock market, and in 1905{1907 in the works of
A. Einstein and M. Smoluchowsky in their study of physical processes.
One of the rst systematic studies of nite Markov chains was carried out by
M. Frechet.22 The treatment of Markov chains in terms of the two fundamental
matrices that we have used was developed by Kemeny and Snell 23 to avoid the use of
eigenvalues that one of these authors found too complex. The fundamental matrix N
occurred also in the work of J. L. Doob and others in studying the connection
between Markov processes and classical potential theory. The fundamental matrix Z
for ergodic chains appeared rst in the work of Frechet, who used it to nd the",prob.pdf
"for ergodic chains appeared rst in the work of Frechet, who used it to nd the
limiting variance for the central limit theorem for Markov chains.
Exercises
1 Consider the Markov chain with transition matrix
P =
1=2 1=2
1=4 3=4

:
Find the fundamental matrix Z for this chain. Compute the mean rst passage
matrix using Z.
2 A study of the strengths of Ivy League football teams shows that if a school
has a strong team one year it is equally likely to have a strong team or average
team next year; if it has an average team, half the time it is average next year,
and if it changes it is just as likely to become strong as weak; if it is weak it
has 2/3 probability of remaining so and 1/3 of becoming average.
(a) A school has a strong team. On the average, how long will it be before
it has another strong team?
(b) A school has a weak team; how long (on the average) must the alumni
wait for a strong team?
3 Consider Example 11.4 with a = :5 and b = :75. Assume that the President",prob.pdf
"wait for a strong team?
3 Consider Example 11.4 with a = :5 and b = :75. Assume that the President
says that he or she will run. Find the expected length of time before the rst
time the answer is passed on incorrectly.
4 Find the mean recurrence time for each state of Example 11.4 for a = :5 and
b = :75. Do the same for general a and b.
5 A die is rolled repeatedly. Show by the results of this section that the mean
time between occurrences of a given number is 6.
22M. Frechet, \Theorie des evenements en chaine dans le cas d'un nombre ni d'etats possible,""
in Recherches theoriques Modernes sur le calcul des probabilites, vol. 2 (Paris, 1938).
23J. G. Kemeny and J. L. Snell, Finite Markov Chains.",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 467
2 43
65
1
Figure 11.7: Maze for Exercise 7.
6 For the Land of Oz example (Example 11.1), make rain into an absorbing
state and nd the fundamental matrix N. Interpret the results obtained from
this chain in terms of the original chain.
7 A rat runs through the maze shown in Figure 11.7. At each step it leaves the
room it is in by choosing at random one of the doors out of the room.
(a) Give the transition matrix P for this Markov chain.
(b) Show that it is an ergodic chain but not a regular chain.
(c) Find the xed vector.
(d) Find the expected number of steps before reaching Room 5 for the rst
time, starting in Room 1.
8 Modify the program ErgodicChain so that you can compute the basic quan-
tities for the queueing example of Exercise 11.3.20. Interpret the mean recur-
rence time for state 0.
9 Consider a random walk on a circle of circumference n. The walker takes
one unit step clockwise with probability p and one unit counterclockwise with",prob.pdf
"one unit step clockwise with probability p and one unit counterclockwise with
probability q = 1  p. Modify the program ErgodicChain to allow you to
input n and p and compute the basic quantities for this chain.
(a) For which values of n is this chain regular? ergodic?
(b) What is the limiting vector w?
(c) Find the mean rst passage matrix for n = 5 and p = :5. Verify that
mij = d(n  d), where d is the clockwise distance from i to j.
10 Two players match pennies and have between them a total of 5 pennies. If at
any time one player has all of the pennies, to keep the game going, he gives
one back to the other player and the game will continue. Show that this game
can be formulated as an ergodic chain. Study this chain using the program
ErgodicChain.",prob.pdf
"468 CHAPTER 11. MARKOV CHAINS
11 Calculate the reverse transition matrix for the Land of Oz example (Exam-
ple 11.1). Is this chain reversible?
12 Give an example of a three-state ergodic Markov chain that is not reversible.
13 Let P be the transition matrix of an ergodic Markov chain and P the reverse
transition matrix. Show that they have the same xed probability vector w.
14 If P is a reversible Markov chain, is it necessarily true that the mean time
to go from state i to state j is equal to the mean time to go from state j to
state i? Hint: Try the Land of Oz example (Example 11.1).
15 Show that any ergodic Markov chain with a symmetric transition matrix (i.e.,
pij = pji) is reversible.
16 (Crowell24) Let P be the transition matrix of an ergodic Markov chain. Show
that
(I + P + 
 
 
 + Pn1)(I  P + W) = I  Pn + nW ;
and from this show that
I + P + 
 
 
 + Pn1
n ! W ;
as n ! 1.
17 An ergodic Markov chain is started in equilibrium (i.e., with initial probability",prob.pdf
"and from this show that
I + P + 
 
 
 + Pn1
n ! W ;
as n ! 1.
17 An ergodic Markov chain is started in equilibrium (i.e., with initial probability
vector w). The mean time until the next occurrence of state si is mi =P
k wkmki + wiri. Show that mi = zii=wi, by using the facts that wZ = w
and mki = (zii  zki)=wi.
18 A perpetual craps game goes on at Charley's. Jones comes into Charley's on
an evening when there have already been 100 plays. He plans to play until the
next time that snake eyes (a pair of ones) are rolled. Jones wonders how many
times he will play. On the one hand he realizes that the average time between
snake eyes is 36 so he should play about 18 times as he is equally likely to
have come in on either side of the halfway point between occurrences of snake
eyes. On the other hand, the dice have no memory, and so it would seem
that he would have to play for 36 more times no matter what the previous",prob.pdf
"eyes. On the other hand, the dice have no memory, and so it would seem
that he would have to play for 36 more times no matter what the previous
outcomes have been. Which, if either, of Jones's arguments do you believe?
Using the result of Exercise 17, calculate the expected to reach snake eyes, in
equilibrium, and see if this resolves the apparent paradox. If you are still in
doubt, simulate the experiment to decide which argument is correct. Can you
give an intuitive argument which explains this result?
19 Show that, for an ergodic Markov chain (see Theorem 11.16),
X
j
mijwj =
X
j
zjj  1 = K :
24Private communication.",prob.pdf
"11.5. MEAN FIRST PASSAGE TIME 469
- 5
 B
20
 C
- 30
 A
 15
GO
Figure 11.8: Simplied Monopoly.
The second expression above shows that the number K is independent of
i. The number K is called Kemeny's constant. A prize was oered to the
rst person to give an intuitively plausible reason for the above sum to be
independent of i. (See also Exercise 24.)
20 Consider a game played as follows: You are given a regular Markov chain
with transition matrix P, xed probability vector w, and a payo function f
which assigns to each state si an amount fi which may be positive or negative.
Assume that wf = 0. You watch this Markov chain as it evolves, and every
time you are in state si you receive an amount fi. Show that your expected
winning after n steps can be represented by a column vector g(n), with
g(n) = (I + P + P2 + 
 
 
 + Pn)f:
Show that as n ! 1, g(n) ! g with g = Zf.
21 A highly simplied game of \Monopoly"" is played on a board with four squares",prob.pdf
"Show that as n ! 1, g(n) ! g with g = Zf.
21 A highly simplied game of \Monopoly"" is played on a board with four squares
as shown in Figure 11.8. You start at GO. You roll a die and move clockwise
around the board a number of squares equal to the number that turns up on
the die. You collect or pay an amount indicated on the square on which you
land. You then roll the die again and move around the board in the same
manner from your last position. Using the result of Exercise 20, estimate
the amount you should expect to win in the long run playing this version of
Monopoly.
22 Show that if P is the transition matrix of a regular Markov chain, and W is
the matrix each of whose rows is the xed probability vector corresponding
to P, then PW = W, and Wk = W for all positive integers k.
23 Assume that an ergodic Markov chain has states s1; s2; : : : ; sk. Let S(n)
j denote
the number of times that the chain is in state sj in the rst n steps. Let w",prob.pdf
"j denote
the number of times that the chain is in state sj in the rst n steps. Let w
denote the xed probability row vector for this chain. Show that, regardless
of the starting state, the expected value of S(n)
j , divided by n, tends to wj as
n ! 1. Hint: If the chain starts in state si, then the expected value of S(n)
j
is given by the expression
nX
h=0
p(h)
ij :",prob.pdf
"470 CHAPTER 11. MARKOV CHAINS
24 In the course of a walk with Snell along Minnehaha Avenue in Minneapolis
in the fall of 1983, Peter Doyle25 suggested the following explanation for the
constancy of Kemeny's constant (see Exercise 19). Choose a target state
according to the xed vector w. Start from state i and wait until the time T
that the target state occurs for the rst time. Let Ki be the expected value
of T. Observe that
Ki + wi 
 1=wi =
X
j
PijKj + 1 ;
and hence
Ki =
X
j
PijKj :
By the maximum principle, Ki is a constant. Should Peter have been given
the prize?
25Private communication.",prob.pdf
"Chapter 12
Random Walks
12.1 Random Walks in Euclidean Space
In the last several chapters, we have studied sums of random variables with the goal
being to describe the distribution and density functions of the sum. In this chapter,
we shall look at sums of discrete random variables from a dierent perspective. We
shall be concerned with properties which can be associated with the sequence of
partial sums, such as the number of sign changes of this sequence, the number of
terms in the sequence which equal 0, and the expected size of the maximum term
in the sequence.
We begin with the following denition.
Denition 12.1 Let fXkg1
k=1 be a sequence of independent, identically distributed
discrete random variables. For each positive integer n, we let Sn denote the sum
X1 +X2 +
 
 
+Xn. The sequence fSng1
n=1 is called a random walk. If the common
range of the Xk's is Rm, then we say that fSng is a random walk in Rm. 2",prob.pdf
"n=1 is called a random walk. If the common
range of the Xk's is Rm, then we say that fSng is a random walk in Rm. 2
We view the sequence of Xk's as being the outcomes of independent experiments.
Since the Xk's are independent, the probability of any particular (nite) sequence
of outcomes can be obtained by multiplying the probabilities that each Xk takes
on the specied value in the sequence. Of course, these individual probabilities are
given by the common distribution of the Xk's. We will typically be interested in
nding probabilities for events involving the related sequence of Sn's. Such events
can be described in terms of the Xk's, so their probabilities can be calculated using
the above idea.
There are several ways to visualize a random walk. One can imagine that a
particle is placed at the origin in Rm at time n = 0. The sum Sn represents the
position of the particle at the end of n seconds. Thus, in the time interval [n1; n],",prob.pdf
"position of the particle at the end of n seconds. Thus, in the time interval [n1; n],
the particle moves (or jumps) from position Sn1 to Sn. The vector representing
this motion is just Sn Sn1, which equals Xn. This means that in a random walk,
the jumps are independent and identically distributed. If m = 1, for example, then
one can imagine a particle on the real line that starts at the origin, and at the
end of each second, jumps one unit to the right or the left, with probabilities given
471",prob.pdf
"472 CHAPTER 12. RANDOM WALKS
by the distribution of the Xk's. If m = 2, one can visualize the process as taking
place in a city in which the streets form square city blocks. A person starts at one
corner (i.e., at an intersection of two streets) and goes in one of the four possible
directions according to the distribution of the Xk's. If m = 3, one might imagine
being in a jungle gym, where one is free to move in any one of six directions (left,
right, forward, backward, up, and down). Once again, the probabilities of these
movements are given by the distribution of the Xk's.
Another model of a random walk (used mostly in the case where the range is
R1) is a game, involving two people, which consists of a sequence of independent,
identically distributed moves. The sum Sn represents the score of the rst person,
say, after n moves, with the assumption that the score of the second person is
Sn. For example, two people might be 
ipping coins, with a match or non-match",prob.pdf
"Sn. For example, two people might be 
ipping coins, with a match or non-match
representing +1 or 1, respectively, for the rst player. Or, perhaps one coin is
being 
ipped, with a head or tail representing +1 or 1, respectively, for the rst
player.
Random Walks on the Real Line
We shall rst consider the simplest non-trivial case of a random walk in R1, namely
the case where the common distribution function of the random variables Xn is
given by
fX(x) =
 1=2; if x = 1;
0; otherwise.
This situation corresponds to a fair coin being 
ipped, with Sn representing the
number of heads minus the number of tails which occur in the rst n 
ips. We note
that in this situation, all paths of length n have the same probability, namely 2n.
It is sometimes instructive to represent a random walk as a polygonal line, or
path, in the plane, where the horizontal axis represents time and the vertical axis
represents the value of Sn. Given a sequence fSng of partial sums, we rst plot the",prob.pdf
"represents the value of Sn. Given a sequence fSng of partial sums, we rst plot the
points (n; Sn), and then for each k < n, we connect (k; Sk) and (k +1; Sk+1) with a
straight line segment. The length of a path is just the dierence in the time values
of the beginning and ending points on the path. The reader is referred to Figure
12.1. This gure, and the process it illustrates, are identical with the example,
given in Chapter 1, of two people playing heads or tails.
Returns and First Returns
We say that an equalization has occurred, or there is a return to the origin at time
n, if Sn = 0. We note that this can only occur if n is an even integer. To calculate
the probability of an equalization at time 2m, we need only count the number of
paths of length 2m which begin and end at the origin. The number of such paths
is clearly 2m
m

:
Since each path has probability 22m, we have the following theorem.",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 473
5 10 15 20 25 30 35 40
-10
-8
-6
-4
-2
2
4
6
8
10
Figure 12.1: A random walk of length 40.
Theorem 12.1 The probability of a return to the origin at time 2m is given by
u2m =
2m
m

22m :
The probability of a return to the origin at an odd time is 0. 2
A random walk is said to have a rst return to the origin at time 2m if m > 0, and
S2k 6= 0 for all k < m. In Figure 12.1, the rst return occurs at time 2. We dene
f2m to be the probability of this event. (We also dene f0 = 0.) One can think
of the expression f2m22m as the number of paths of length 2m between the points
(0; 0) and (2m; 0) that do not touch the horizontal axis except at the endpoints.
Using this idea, it is easy to prove the following theorem.
Theorem 12.2 For n  1, the probabilities fu2kg and ff2kg are related by the
equation
u2n = f0u2n + f2u2n2 + 
 
 
 + f2nu0 :
Proof. There are u2n22n paths of length 2n which have endpoints (0; 0) and (2n; 0).",prob.pdf
"equation
u2n = f0u2n + f2u2n2 + 
 
 
 + f2nu0 :
Proof. There are u2n22n paths of length 2n which have endpoints (0; 0) and (2n; 0).
The collection of such paths can be partitioned into n sets, depending upon the time
of the rst return to the origin. A path in this collection which has a rst return to
the origin at time 2k consists of an initial segment from (0; 0) to (2k; 0), in which
no interior points are on the horizontal axis, and a terminal segment from (2k; 0)
to (2n; 0), with no further restrictions on this segment. Thus, the number of paths
in the collection which have a rst return to the origin at time 2k is given by
f2k22ku2n2k22n2k = f2ku2n2k22n :
If we sum over k, we obtain the equation
u2n22n = f0u2n22n + f2u2n222n + 
 
 
 + f2nu022n :
Dividing both sides of this equation by 22n completes the proof. 2",prob.pdf
"474 CHAPTER 12. RANDOM WALKS
The expression in the right-hand side of the above theorem should remind the reader
of a sum that appeared in Denition 7.1 of the convolution of two distributions. The
convolution of two sequences is dened in a similar manner. The above theorem
says that the sequence fu2ng is the convolution of itself and the sequence ff2ng.
Thus, if we represent each of these sequences by an ordinary generating function,
then we can use the above relationship to determine the value f2n.
Theorem 12.3 For m  1, the probability of a rst return to the origin at time
2m is given by
f2m = u2m
2m  1 =
2m
m


(2m  1)22m :
Proof. We begin by dening the generating functions
U(x) =
1X
m=0
u2mxm
and
F(x) =
1X
m=0
f2mxm :
Theorem 12.2 says that
U(x) = 1 + U(x)F(x) : (12.1)
(The presence of the 1 on the right-hand side is due to the fact that u0 is dened
to be 1, but Theorem 12.2 only holds for m  1.) We note that both generating",prob.pdf
"to be 1, but Theorem 12.2 only holds for m  1.) We note that both generating
functions certainly converge on the interval (1; 1), since all of the coecients are at
most 1 in absolute value. Thus, we can solve the above equation for F(x), obtaining
F(x) = U(x)  1
U(x) :
Now, if we can nd a closed-form expression for the function U(x), we will also have
a closed-form expression for F(x). From Theorem 12.1, we have
U(x) =
1X
m=0
2m
m

22mxm :
In Wilf,1 we nd that
1p1  4x =
1X
m=0
2m
m

xm :
The reader is asked to prove this statement in Exercise 1. If we replace x by x=4
in the last equation, we see that
U(x) = 1p1  x :
1H. S. Wilf, Generatingfunctionology, (Boston: Academic Press, 1990), p. 50.",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 475
Therefore, we have
F(x) = U(x)  1
U(x)
= (1  x)1=2  1
(1  x)1=2
= 1  (1  x)1=2 :
Although it is possible to compute the value of f2m using the Binomial Theorem,
it is easier to note that F0(x) = U(x)=2, so that the coecients f2m can be found
by integrating the series for U(x). We obtain, for m  1,
f2m = u2m2
2m
=
2m2
m1


m22m1
=
2m
m


(2m  1)22m
= u2m
2m  1 ;
since 2m  2
m  1

= m
2(2m  1)
2m
m

:
This completes the proof of the theorem. 2
Probability of Eventual Return
In the symmetric random walk process in Rm, what is the probability that the
particle eventually returns to the origin? We rst examine this question in the case
that m = 1, and then we consider the general case. The results in the next two
examples are due to Polya.2
Example 12.1 (Eventual Return in R1) One has to approach the idea of eventual
return with some care, since the sample space seems to be the set of all walks of",prob.pdf
"return with some care, since the sample space seems to be the set of all walks of
innite length, and this set is non-denumerable. To avoid diculties, we will dene
wn to be the probability that a rst return has occurred no later than time n. Thus,
wn concerns the sample space of all walks of length n, which is a nite set. In terms
of the wn's, it is reasonable to dene the probability that the particle eventually
returns to the origin to be
w = lim
n!1
wn :
This limit clearly exists and is at most one, since the sequence fwng1
n=1 is an
increasing sequence, and all of its terms are at most one.
2G. Polya, \Uber eine Aufgabe der Wahrscheinlichkeitsrechnung betreend die Irrfahrt im
Strassennetz,"" Math. Ann., vol. 84 (1921), pp. 149-160.",prob.pdf
"476 CHAPTER 12. RANDOM WALKS
In terms of the fn probabilities, we see that
w2n =
nX
i=1
f2i :
Thus,
w =
1X
i=1
f2i :
In the proof of Theorem 12.3, the generating function
F(x) =
1X
m=0
f2mxm
was introduced. There it was noted that this series converges for x 2 (1; 1). In
fact, it is possible to show that this series also converges for x = 1 by using
Exercise 4, together with the fact that
f2m = u2m
2m  1 :
(This fact was proved in the proof of Theorem 12.3.) Since we also know that
F(x) = 1  (1  x)1=2 ;
we see that
w = F(1) = 1 :
Thus, with probability one, the particle returns to the origin.
An alternative proof of the fact that w = 1 can be obtained by using the results
in Exercise 2. 2
Example 12.2 (Eventual Return in Rm) We now turn our attention to the case
that the random walk takes place in more than one dimension. We dene f(m)
2n to
be the probability that the rst return to the origin in Rm occurs at time 2n. The
quantity u(m)",prob.pdf
"2n to
be the probability that the rst return to the origin in Rm occurs at time 2n. The
quantity u(m)
2n is dened in a similar manner. Thus, f(1)
2n and u(1)
2n equal f2n and u2n,
which were dened earlier. If, in addition, we dene u(m)
0 = 1 and f(m)
0 = 0, then
one can mimic the proof of Theorem 12.2, and show that for all m  1,
u(m)
2n = f(m)
0 u(m)
2n + f(m)
2 u(m)
2n2 + 
 
 
 + f(m)
2n u(m)
0 : (12.2)
We continue to generalize previous work by dening
U(m)(x) =
1X
n=0
u(m)
2n xn
and
F(m)(x) =
1X
n=0
f(m)
2n xn :",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 477
Then, by using Equation 12.2, we see that
U(m)(x) = 1 + U(m)(x)F(m)(x) ;
as before. These functions will always converge in the interval (1; 1), since all of
their coecients are at most one in magnitude. In fact, since
w(m)
 =
1X
n=0
f(m)
2n  1
for all m, the series for F(m)(x) converges at x = 1 as well, and F(m)(x) is left-
continuous at x = 1, i.e.,
lim
x""1
F(m)(x) = F(m)(1) :
Thus, we have
w(m)
 = lim
x""1
F(m)(x) = lim
x""1
U(m)(x)  1
U(m)(x) ; (12.3)
so to determine w(m)
 , it suces to determine
lim
x""1
U(m)(x) :
We let u(m) denote this limit.
We claim that
u(m) =
1X
n=0
u(m)
2n :
(This claim is reasonable; it says that to nd out what happens to the function
U(m)(x) at x = 1, just let x = 1 in the power series for U(m)(x).) To prove the
claim, we note that the coecients u(m)
2n are non-negative, so U(m)(x) increases
monotonically on the interval [0; 1). Thus, for each K, we have
KX
n=0
u(m)
2n  lim
x""1
U(m)(x) = u(m) 
1X
n=0",prob.pdf
"monotonically on the interval [0; 1). Thus, for each K, we have
KX
n=0
u(m)
2n  lim
x""1
U(m)(x) = u(m) 
1X
n=0
u(m)
2n :
By letting K ! 1, we see that
u(m) =
1X
2n
u(m)
2n :
This establishes the claim.
From Equation 12.3, we see that if u(m) < 1, then the probability of an eventual
return is
u(m)  1
u(m) ;
while if u(m) = 1, then the probability of eventual return is 1.
To complete the example, we must estimate the sum
1X
n=0
u(m)
2n :",prob.pdf
"478 CHAPTER 12. RANDOM WALKS
In Exercise 12, the reader is asked to show that
u(2)
2n = 1
42n
2n
n
2
:
Using Stirling's Formula, it is easy to show that (see Exercise 13)
2n
n

 22n
pn ;
so
u(2)
2n  1
n :
From this it follows easily that
1X
n=0
u(2)
2n
diverges, so w(2)
 = 1, i.e., in R2, the probability of an eventual return is 1.
When m = 3, Exercise 12 shows that
u(3)
2n = 1
22n
2n
n
X
j;k
1
3n
n!
j!k!(n  j  k)!
2
:
Let M denote the largest value of
1
3n
n!
j!k!(n  j  k)! ;
over all non-negative values of j and k with j + k  n. It is easy, using Stirling's
Formula, to show that
M  c
n ;
for some constant c. Thus, we have
u(3)
2n  1
22n
2n
n
X
j;k
M
3n
n!
j!k!(n  j  k)!

:
Using Exercise 14, one can show that the right-hand expression is at most
c0
n3=2 ;
where c0 is a constant. Thus,
1X
n=0
u(3)
2n
converges, so w(3)
 is strictly less than one. This means that in R3, the probability of",prob.pdf
"c0
n3=2 ;
where c0 is a constant. Thus,
1X
n=0
u(3)
2n
converges, so w(3)
 is strictly less than one. This means that in R3, the probability of
an eventual return to the origin is strictly less than one (in fact, it is approximately
.34).
One may summarize these results by stating that one should not get drunk in
more than two dimensions. 2",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 479
Expected Number of Equalizations
We now give another example of the use of generating functions to nd a general
formula for terms in a sequence, where the sequence is related by recursion relations
to other sequences. Exercise 9 gives still another example.
Example 12.3 (Expected Number of Equalizations) In this example, we will de-
rive a formula for the expected number of equalizations in a random walk of length
2m. As in the proof of Theorem 12.3, the method has four main parts. First, a
recursion is found which relates the mth term in the unknown sequence to earlier
terms in the same sequence and to terms in other (known) sequences. An exam-
ple of such a recursion is given in Theorem 12.2. Second, the recursion is used
to derive a functional equation involving the generating functions of the unknown
sequence and one or more known sequences. Equation 12.1 is an example of such",prob.pdf
"sequence and one or more known sequences. Equation 12.1 is an example of such
a functional equation. Third, the functional equation is solved for the unknown
generating function. Last, using a device such as the Binomial Theorem, integra-
tion, or dierentiation, a formula for the mth coecient of the unknown generating
function is found.
We begin by dening g2m to be the number of equalizations among all of the
random walks of length 2m. (For each random walk, we disregard the equalization
at time 0.) We dene g0 = 0. Since the number of walks of length 2m equals 22m,
the expected number of equalizations among all such random walks is g2m=22m.
Next, we dene the generating function G(x):
G(x) =
1X
k=0
g2kxk :
Now we need to nd a recursion which relates the sequence fg2kg to one or both of
the known sequences ff2kg and fu2kg. We consider m to be a xed positive integer,
and consider the set of all paths of length 2m as the disjoint union
E2 [ E4 [ 
 
 
 [ E2m [ H ;",prob.pdf
"and consider the set of all paths of length 2m as the disjoint union
E2 [ E4 [ 
 
 
 [ E2m [ H ;
where E2k is the set of all paths of length 2m with rst equalization at time 2k,
and H is the set of all paths of length 2m with no equalization. It is easy to show
(see Exercise 3) that
jE2kj = f2k22m :
We claim that the number of equalizations among all paths belonging to the set
E2k is equal to
jE2kj + 22kf2kg2m2k : (12.4)
Each path in E2k has one equalization at time 2k, so the total number of such
equalizations is just jE2kj. This is the rst summand in expression Equation 12.4.
There are 22kf2k dierent initial segments of length 2k among the paths in E2k.
Each of these initial segments can be augmented to a path of length 2m in 22m2k
ways, by adjoining all possible paths of length 2m2k. The number of equalizations
obtained by adjoining all of these paths to any one initial segment is g2m2k, by",prob.pdf
"480 CHAPTER 12. RANDOM WALKS
denition. This gives the second summand in Equation 12.4. Since k can range
from 1 to m, we obtain the recursion
g2m =
mX
k=1

jE2kj + 22kf2kg2m2k

: (12.5)
The second summand in the typical term above should remind the reader of a
convolution. In fact, if we multiply the generating function G(x) by the generating
function
F(4x) =
1X
k=0
22kf2kxk ;
the coecient of xm equals
mX
k=0
22kf2kg2m2k :
Thus, the product G(x)F(4x) is part of the functional equation that we are seeking.
The rst summand in the typical term in Equation 12.5 gives rise to the sum
22m
mX
k=1
f2k :
From Exercise 2, we see that this sum is just (1u2m)22m. Thus, we need to create
a generating function whose mth coecient is this term; this generating function is
1X
m=0
(1  u2m)22mxm ;
or 1X
m=0
22mxm 
1X
m=0
u2m22mxm :
The rst sum is just (1  4x)1, and the second sum is U(4x). So, the functional
equation which we have been seeking is
G(x) = F(4x)G(x) + 1
1  4x  U(4x) :",prob.pdf
"equation which we have been seeking is
G(x) = F(4x)G(x) + 1
1  4x  U(4x) :
If we solve this recursion for G(x), and simplify, we obtain
G(x) = 1
(1  4x)3=2  1
(1  4x) : (12.6)
We now need to nd a formula for the coecient of xm. The rst summand in
Equation 12.6 is (1=2)U0(4x), so the coecient of xm in this function is
u2m+222m+1(m + 1) :
The second summand in Equation 12.6 is the sum of a geometric series with common
ratio 4x, so the coecient of xm is 22m. Thus, we obtain",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 481
g2m = u2m+222m+1(m + 1)  22m
= 1
2
2m + 2
m + 1

(m + 1)  22m :
We recall that the quotient g2m=22m is the expected number of equalizations
among all paths of length 2m. Using Exercise 4, it is easy to show that
g2m
22m 
r
2

p
2m :
In particular, this means that the average number of equalizations among all paths
of length 4m is not twice the average number of equalizations among all paths of
length 2m. In order for the average number of equalizations to double, one must
quadruple the lengths of the random walks. 2
It is interesting to note that if we dene
Mn = max
0kn
Sk ;
then we have
E(Mn) 
r
2

pn :
This means that the expected number of equalizations and the expected maximum
value for random walks of length n are asymptotically equal as n ! 1. (In fact,
it can be shown that the two expected values dier by at most 1=2 for all positive
integers n. See Exercise 9.)
Exercises
1 Using the Binomial Theorem, show that
1p1  4x =
1X",prob.pdf
"integers n. See Exercise 9.)
Exercises
1 Using the Binomial Theorem, show that
1p1  4x =
1X
m=0
2m
m

xm :
What is the interval of convergence of this power series?
2 (a) Show that for m  1,
f2m = u2m2  u2m :
(b) Using part (a), nd a closed-form expression for the sum
f2 + f4 + 
 
 
 + f2m :
(c) Using part (b), show that
1X
m=1
f2m = 1 :
(One can also obtain this statement from the fact that
F(x) = 1  (1  x)1=2 :)",prob.pdf
"482 CHAPTER 12. RANDOM WALKS
(d) Using parts (a) and (b), show that the probability of no equalization in
the rst 2m outcomes equals the probability of an equalization at time
2m.
3 Using the notation of Example 12.3, show that
jE2kj = f2k22m :
4 Using Stirling's Formula, show that
u2m  1pm :
5 A lead change in a random walk occurs at time 2k if S2k1 and S2k+1 are of
opposite sign.
(a) Give a rigorous argument which proves that among all walks of length
2m that have an equalization at time 2k, exactly half have a lead change
at time 2k.
(b) Deduce that the total number of lead changes among all walks of length
2m equals
1
2(g2m  u2m) :
(c) Find an asymptotic expression for the average number of lead changes
in a random walk of length 2m.
6 (a) Show that the probability that a random walk of length 2m has a last
return to the origin at time 2k, where 0  k  m, equals
2k
k

2m2k
mk


22m = u2ku2m2k :
(The case k = 0 consists of all paths that do not return to the origin at",prob.pdf
"2k
k

2m2k
mk


22m = u2ku2m2k :
(The case k = 0 consists of all paths that do not return to the origin at
any positive time.) Hint: A path whose last return to the origin occurs
at time 2k consists of two paths glued together, one path of which is of
length 2k and which begins and ends at the origin, and the other path
of which is of length 2m  2k and which begins at the origin but never
returns to the origin. Both types of paths can be counted using quantities
which appear in this section.
(b) Using part (a), show that if m is odd, the probability that a walk of
length 2m has no equalization in the last m outcomes is equal to 1=2,
regardless of the value of m. Hint: The answer to part a) is symmetric
in k and m  k.
7 Show that the probability of no equalization in a walk of length 2m equals
u2m.",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 483
*8 Show that
P(S1  0; S2  0; : : : ; S2m  0) = u2m :
Hint: First explain why
P(S1 > 0; S2 > 0; : : : ; S2m > 0)
= 1
2P(S1 6= 0; S2 6= 0; : : : ; S2m 6= 0) :
Then use Exercise 7, together with the observation that if no equalization
occurs in the rst 2m outcomes, then the path goes through the point (1; 1)
and remains on or above the horizontal line x = 1.
*9 In Feller,3 one nds the following theorem: Let Mn be the random variable
which gives the maximum value of Sk, for 1  k  n. Dene
pn;r =
 n
n+r
2

2n :
If r  0, then
P(Mn = r) =
 pn;r ; if r  n (mod 2);
pn;r+1 ; if r 6 n (mod 2):
(a) Using this theorem, show that
E(M2m) = 1
22m
mX
k=1
(4k  1)
 2m
m + k

;
and if n = 2m + 1, then
E(M2m+1) = 1
22m+1
mX
k=0
(4k + 1)
 2m + 1
m + k + 1

:
(b) For m  1, dene
rm =
mX
k=1
k
 2m
m + k

and
sm =
mX
k=1
k
 2m + 1
m + k + 1

:
By using the identity
n
k

=
n  1
k  1

+
n  1
k

;
show that
sm = 2rm  1
2

22m 
2m
m
",prob.pdf
"
and
sm =
mX
k=1
k
 2m + 1
m + k + 1

:
By using the identity
n
k

=
n  1
k  1

+
n  1
k

;
show that
sm = 2rm  1
2

22m 
2m
m

3W. Feller, Introduction to Probability Theory and its Applications, vol. I, 3rd ed. (New York:
John Wiley & Sons, 1968).",prob.pdf
"484 CHAPTER 12. RANDOM WALKS
and
rm = 2sm1 + 1
222m1 ;
if m  2.
(c) Dene the generating functions
R(x) =
1X
k=1
rkxk
and
S(x) =
1X
k=1
skxk :
Show that
S(x) = 2R(x)  1
2
 1
1  4x

+ 1
2
p
1  4x

and
R(x) = 2xS(x) + x
 1
1  4x

:
(d) Show that
R(x) = x
(1  4x)3=2 ;
and
S(x) = 1
2
 1
(1  4x)3=2

 1
2
 1
1  4x

:
(e) Show that
rm = m
2m  1
m  1

;
and
sm = 1
2(m + 1)
2m + 1
m

 1
2(22m) :
(f) Show that
E(M2m) = m
22m1
2m
m

+ 1
22m+1
2m
m

 1
2 ;
and
E(M2m+1) = m + 1
22m+1
2m + 2
m + 1

 1
2 :
The reader should compare these formulas with the expression for
g2m=2(2m) in Example 12.3.",prob.pdf
"12.1. RANDOM WALKS IN EUCLIDEAN SPACE 485
*10 (from K. Levasseur4) A parent and his child play the following game. A deck
of 2n cards, n red and n black, is shued. The cards are turned up one at a
time. Before each card is turned up, the parent and the child guess whether
it will be red or black. Whoever makes more correct guesses wins the game.
The child is assumed to guess each color with the same probability, so she
will have a score of n, on average. The parent keeps track of how many cards
of each color have already been turned up. If more black cards, say, than
red cards remain in the deck, then the parent will guess black, while if an
equal number of each color remain, then the parent guesses each color with
probability 1/2. What is the expected number of correct guesses that will be
made by the parent? Hint: Each of the
2n
n


possible orderings of red and
black cards corresponds to a random walk of length 2n that returns to the",prob.pdf
"made by the parent? Hint: Each of the
2n
n


possible orderings of red and
black cards corresponds to a random walk of length 2n that returns to the
origin at time 2n. Show that between each pair of successive equalizations,
the parent will be right exactly once more than he will be wrong. Explain
why this means that the average number of correct guesses by the parent is
greater than n by exactly one-half the average number of equalizations. Now
dene the random variable Xi to be 1 if there is an equalization at time 2i,
and 0 otherwise. Then, among all relevant paths, we have
E(Xi) = P(Xi = 1) =
2n2i
ni

2i
i


2n
n

 :
Thus, the expected number of equalizations equals
E
 nX
i=1
Xi

= 12n
n


nX
i=1
2n  2i
n  i
2i
i

:
One can now use generating functions to nd the value of the sum.
It should be noted that in a game such as this, a more interesting question
than the one asked above is what is the probability that the parent wins the",prob.pdf
"than the one asked above is what is the probability that the parent wins the
game? For this game, this question was answered by D. Zagier.5 He showed
that the probability of winning is asymptotic (for large n) to the quantity
1
2 + 1
2
p
2 :
*11 Prove that
u(2)
2n = 1
42n
nX
k=0
(2n)!
k!k!(n  k)!(n  k)! ;
and
u(3)
2n = 1
62n
X
j;k
(2n)!
j!j!k!k!(n  j  k)!(n  j  k)! ;
4K. Levasseur, \How to Beat Your Kids at Their Own Game,"" Mathematics Magazine vol. 61,
no. 5 (December, 1988), pp. 301-305.
5D. Zagier, \How Often Should You Beat Your Kids?"" Mathematics Magazine vol. 63, no. 2
(April 1990), pp. 89-92.",prob.pdf
"486 CHAPTER 12. RANDOM WALKS
where the last sum extends over all non-negative j and k with j +k  n. Also
show that this last expression may be rewritten as
1
22n
2n
n
X
j;k
 1
3n
n!
j!k!(n  j  k)!
2
:
*12 Prove that if n  0, then
nX
k=0
n
k
2
=
2n
n

:
Hint: Write the sum as nX
k=0
n
k
 n
n  k

and explain why this is a coecient in the product
(1 + x)n(1 + x)n :
Use this, together with Exercise 11, to show that
u(2)
2n = 1
42n
2n
n
 nX
k=0
n
k
2
= 1
42n
2n
n
2
:
*13 Using Stirling's Formula, prove that
2n
n

 22n
pn :
*14 Prove that X
j;k
 1
3n
n!
j!k!(n  j  k)!

= 1 ;
where the sum extends over all non-negative j and k such that j + k  n.
Hint: Count how many ways one can place n labelled balls in 3 labelled urns.
*15 Using the result proved for the random walk in R3 in Example 12.2, explain
why the probability of an eventual return in Rn is strictly less than one, for
all n  3. Hint: Consider a random walk in Rn and disregard all but the rst",prob.pdf
"all n  3. Hint: Consider a random walk in Rn and disregard all but the rst
three coordinates of the particle's position.
12.2 Gambler's Ruin
In the last section, the simplest kind of symmetric random walk in R1 was studied.
In this section, we remove the assumption that the random walk is symmetric.
Instead, we assume that p and q are non-negative real numbers with p +q = 1, and
that the common distribution function of the jumps of the random walk is
fX(x) =

p; if x = 1;
q; if x = 1:",prob.pdf
"12.2. GAMBLER'S RUIN 487
One can imagine the random walk as representing a sequence of tosses of a weighted
coin, with a head appearing with probability p and a tail appearing with probability
q. An alternative formulation of this situation is that of a gambler playing a sequence
of games against an adversary (sometimes thought of as another person, sometimes
called \the house"") where, in each game, the gambler has probability p of winning.
The Gambler's Ruin Problem
The above formulation of this type of random walk leads to a problem known as the
Gambler's Ruin problem. This problem was introduced in Exercise 23, but we will
give the description of the problem again. A gambler starts with a \stake"" of size s.
She plays until her capital reaches the value M or the value 0. In the language of
Markov chains, these two values correspond to absorbing states. We are interested
in studying the probability of occurrence of each of these two outcomes.",prob.pdf
"in studying the probability of occurrence of each of these two outcomes.
One can also assume that the gambler is playing against an \innitely rich""
adversary. In this case, we would say that there is only one absorbing state, namely
when the gambler's stake is 0. Under this assumption, one can ask for the proba-
bility that the gambler is eventually ruined.
We begin by dening qk to be the probability that the gambler's stake reaches 0,
i.e., she is ruined, before it reaches M, given that the initial stake is k. We note that
q0 = 1 and qM = 0. The fundamental relationship among the qk's is the following:
qk = pqk+1 + qqk1 ;
where 1  k  M  1. This holds because if her stake equals k, and she plays one
game, then her stake becomes k + 1 with probability p and k  1 with probability
q. In the rst case, the probability of eventual ruin is qk+1 and in the second case,
it is qk1. We note that since p + q = 1, we can write the above equation as
p(qk+1  qk) = q(qk  qk1) ;
or",prob.pdf
"it is qk1. We note that since p + q = 1, we can write the above equation as
p(qk+1  qk) = q(qk  qk1) ;
or
qk+1  qk = q
p(qk  qk1) :
From this equation, it is easy to see that
qk+1  qk =
q
p
k
(q1  q0) : (12.7)
We now use telescoping sums to obtain an equation in which the only unknown is
q1:
1 = qM  q0
=
M1X
k=0
(qk+1  qk) ;",prob.pdf
"488 CHAPTER 12. RANDOM WALKS
so
1 =
M1X
k=0
q
p
k
(q1  q0)
= (q1  q0)
M1X
k=0
q
p
k
:
If p 6= q, then the above expression equals
(q1  q0)(q=p)M  1
(q=p)  1 ;
while if p = q = 1=2, then we obtain the equation
1 = (q1  q0)M :
For the moment we shall assume that p 6= q. Then we have
q1  q0 =  (q=p)  1
(q=p)M  1 :
Now, for any z with 1  z  M, we have
qz  q0 =
z1X
k=0
(qk+1  qk)
= (q1  q0)
z1X
k=0
q
p
k
= (q1  q0)(q=p)z  1
(q=p)  1
=  (q=p)z  1
(q=p)M  1 :
Therefore,
qz = 1  (q=p)z  1
(q=p)M  1
= (q=p)M  (q=p)z
(q=p)M  1 :
Finally, if p = q = 1=2, it is easy to show that (see Exercise 10)
qz = M  z
M :
We note that both of these formulas hold if z = 0.
We dene, for 0  z  M, the quantity pz to be the probability that the
gambler's stake reaches M without ever having reached 0. Since the game might",prob.pdf
"12.2. GAMBLER'S RUIN 489
continue indenitely, it is not obvious that pz + qz = 1 for all z. However, one can
use the same method as above to show that if p 6= q, then
qz = (q=p)z  1
(q=p)M  1 ;
and if p = q = 1=2, then
qz = z
M :
Thus, for all z, it is the case that pz + qz = 1, so the game ends with probability 1.
Innitely Rich Adversaries
We now turn to the problem of nding the probability of eventual ruin if the gambler
is playing against an innitely rich adversary. This probability can be obtained by
letting M go to 1 in the expression for qz calculated above. If q < p, then the
expression approaches (q=p)z, and if q > p, the expression approaches 1. In the
case p = q = 1=2, we recall that qz = 1  z=M. Thus, if M ! 1, we see that the
probability of eventual ruin tends to 1.
Historical Remarks
In 1711, De Moivre, in his book De Mesura Sortis, gave an ingenious derivation
of the probability of ruin. The following description of his argument is taken from",prob.pdf
"of the probability of ruin. The following description of his argument is taken from
David.6 The notation used is as follows: We imagine that there are two players, A
and B, and the probabilities that they win a game are p and q, respectively. The
players start with a and b counters, respectively.
Imagine that each player starts with his counters before him in a pile,
and that nominal values are assigned to the counters in the following
manner. A's bottom counter is given the nominal value q=p; the next is
given the nominal value (q=p)2, and so on until his top counter which
has the nominal value (q=p)a. B's top counter is valued (q=p)a+1, and
so on downwards until his bottom counter which is valued (q=p)a+b.
After each game the loser's top counter is transferred to the top of the
winner's pile, and it is always the top counter which is staked for the
next game. Then in terms of the nominal values B's stake is always",prob.pdf
"winner's pile, and it is always the top counter which is staked for the
next game. Then in terms of the nominal values B's stake is always
q=p times A's, so that at every game each player's nominal expectation
is nil. This remains true throughout the play; therefore A's chance of
winning all B's counters, multiplied by his nominal gain if he does so,
must equal B's chance multiplied by B's nominal gain. Thus,
Pa
q
p
a+1
+ 
 
 
 +
q
p
a+b
= Pb
q
p

+ 
 
 
 +
q
p
a
: (12.8)
6F. N. David, Games, Gods and Gambling (London: Grin, 1962).",prob.pdf
"490 CHAPTER 12. RANDOM WALKS
Using this equation, together with the fact that
Pa + Pb = 1 ;
it can easily be shown that
Pa = (q=p)a  1
(q=p)a+b  1 ;
if p 6= q, and
Pa = a
a + b ;
if p = q = 1=2.
In terms of modern probability theory, de Moivre is changing the values of the
counters to make an unfair game into a fair game, which is called a martingale.
With the new values, the expected fortune of player A (that is, the sum of the
nominal values of his counters) after each play equals his fortune before the play
(and similarly for player B). (For a simpler martingale argument, see Exercise 9.) De
Moivre then uses the fact that when the game ends, it is still fair, thus Equation 12.8
must be true. This fact requires proof, and is one of the central theorems in the
area of martingale theory.
Exercises
1 In the gambler's ruin problem, assume that the gambler initial stake is 1
dollar, and assume that her probability of success on any one game is p. Let",prob.pdf
"dollar, and assume that her probability of success on any one game is p. Let
T be the number of games until 0 is reached (the gambler is ruined). Show
that the generating function for T is
h(z) = 1 
p
1  4pqz2
2pz ;
and that
h(1) =
 q=p; if q  p;
1; if q  p;
and
h0(1) =
 1=(q  p); if q > p;
1; if q = p:
Interpret your results in terms of the time T to reach 0. (See also Exam-
ple 10.7.)
2 Show that the Taylor series expansion for p1  x is
p
1  x =
1X
n=0
1=2
n

xn ;
where the binomial coecient
1=2
n


is
1=2
n

= (1=2)(1=2  1) 
 
 
(1=2  n + 1)
n! :",prob.pdf
"12.2. GAMBLER'S RUIN 491
Using this and the result of Exercise 1, show that the probability that the
gambler is ruined on the nth step is
pT (n) =
(
(1)k1
2p
1=2
k


(4pq)k; if n = 2k  1,
0; if n = 2k.
3 For the gambler's ruin problem, assume that the gambler starts with k dollars.
Let Tk be the time to reach 0 for the rst time.
(a) Show that the generating function hk(t) for Tk is the kth power of the
generating function for the time T to ruin starting at 1. Hint: Let
Tk = U1 + U2 + 
 
 
 + Uk, where Uj is the time for the walk starting at j
to reach j  1 for the rst time.
(b) Find hk(1) and h0
k(1) and interpret your results.
4 (The next three problems come from Feller.7) As in the text, assume that M
is a xed positive integer.
(a) Show that if a gambler starts with an stake of 0 (and is allowed to have a
negative amount of money), then the probability that her stake reaches
the value of M before it returns to 0 equals p(1  q1).",prob.pdf
"negative amount of money), then the probability that her stake reaches
the value of M before it returns to 0 equals p(1  q1).
(b) Show that if the gambler starts with a stake of M then the probability
that her stake reaches 0 before it returns to M equals qqM1.
5 Suppose that a gambler starts with a stake of 0 dollars.
(a) Show that the probability that her stake never reaches M before return-
ing to 0 equals 1  p(1  q1).
(b) Show that the probability that her stake reaches the value M exactly
k times before returning to 0 equals p(1  q1)(1  qqM1)k1(qqM1).
Hint: Use Exercise 4.
6 In the text, it was shown that if q < p, there is a positive probability that
a gambler, starting with a stake of 0 dollars, will never return to the origin.
Thus, we will now assume that q  p. Using Exercise 5, show that if a
gambler starts with a stake of 0 dollars, then the expected number of times
her stake equals M before returning to 0 equals (p=q)M , if q > p and 1, if",prob.pdf
"her stake equals M before returning to 0 equals (p=q)M , if q > p and 1, if
q = p. (We quote from Feller: \The truly amazing implications of this result
appear best in the language of fair games. A perfect coin is tossed until
the rst equalization of the accumulated numbers of heads and tails. The
gambler receives one penny for every time that the accumulated number of
heads exceeds the accumulated number of tails by m. The `fair entrance fee'
equals 1 independent of m."")
7W. Feller, op. cit., pg. 367.",prob.pdf
"492 CHAPTER 12. RANDOM WALKS
7 In the game in Exercise 6, let p = q = 1=2 and M = 10. What is the
probability that the gambler's stake equals M at least 20 times before it
returns to 0?
8 Write a computer program which simulates the game in Exercise 6 for the
case p = q = 1=2, and M = 10.
9 In de Moivre's description of the game, we can modify the denition of player
A's fortune in such a way that the game is still a martingale (and the calcula-
tions are simpler). We do this by assigning nominal values to the counters in
the same way as de Moivre, but each player's current fortune is dened to be
just the value of the counter which is being wagered on the next game. So, if
player A has a counters, then his current fortune is (q=p)a (we stipulate this
to be true even if a = 0). Show that under this denition, player A's expected
fortune after one play equals his fortune before the play, if p 6= q. Then, as
de Moivre does, write an equation which expresses the fact that player A's",prob.pdf
"de Moivre does, write an equation which expresses the fact that player A's
expected nal fortune equals his initial fortune. Use this equation to nd the
probability of ruin of player A.
10 Assume in the gambler's ruin problem that p = q = 1=2.
(a) Using Equation 12.7, together with the facts that q0 = 1 and qM = 0,
show that for 0  z  M,
qz = M  z
M :
(b) In Equation 12.8, let p ! 1=2 (and since q = 1  p, q ! 1=2 as well).
Show that in the limit,
qz = M  z
M :
Hint: Replace q by 1  p, and use L'Hopital's rule.
11 In American casinos, the roulette wheels have the integers between 1 and 36,
together with 0 and 00. Half of the non-zero numbers are red, the other half
are black, and 0 and 00 are green. A common bet in this game is to bet a
dollar on red. If a red number comes up, the bettor gets her dollar back, and
also gets another dollar. If a black or green number comes up, she loses her
dollar.
(a) Suppose that someone starts with 40 dollars, and continues to bet on red",prob.pdf
"dollar.
(a) Suppose that someone starts with 40 dollars, and continues to bet on red
until either her fortune reaches 50 or 0. Find the probability that her
fortune reaches 50 dollars.
(b) How much money would she have to start with, in order for her to have
a 95% chance of winning 10 dollars before going broke?
(c) A casino owner was once heard to remark that \If we took 0 and 00 o
of the roulette wheel, we would still make lots of money, because people
would continue to come in and play until they lost all of their money.""
Do you think that such a casino would stay in business?",prob.pdf
"12.3. ARC SINE LAWS 493
12.3 Arc Sine Laws
In Exercise 12.1.6, the distribution of the time of the last equalization in the sym-
metric random walk was determined. If we let 2k;2m denote the probability that
a random walk of length 2m has its last equalization at time 2k, then we have
2k;2m = u2ku2m2k :
We shall now show how one can approximate the distribution of the 's with a
simple function. We recall that
u2k  1p
k
:
Therefore, as both k and m go to 1, we have
2k;2m  1

p
k(m  k)
:
This last expression can be written as
1
m
p
(k=m)(1  k=m)
:
Thus, if we dene
f(x) = 1

p
x(1  x)
;
for 0 < x < 1, then we have
2k;2m  1
mf
k
m

:
The reason for the  sign is that we no longer require that k get large. This means
that we can replace the discrete 2k;2m distribution by the continuous density f(x)
on the interval [0; 1] and obtain a good approximation. In particular, if x is a xed
real number between 0 and 1, then we have
X
k<xm
2k;2m 
Z x
0
f(t) dt :",prob.pdf
"real number between 0 and 1, then we have
X
k<xm
2k;2m 
Z x
0
f(t) dt :
It turns out that f(x) has a nice antiderivative, so we can write
X
k<xm
2k;2m  2
 arcsinpx :
One can see from the graph of this last function that it has a minimum at x = 1=2
and is symmetric about that point. As noted in the exercise, this implies that half
of the walks of length 2m have no equalizations after time m, a fact which probably
would not be guessed.
It turns out that the arc sine density comes up in the answers to many other
questions concerning random walks on the line. Recall that in Section 12.1, a",prob.pdf
"494 CHAPTER 12. RANDOM WALKS
random walk could be viewed as a polygonal line connecting (0; 0) with (m; Sm).
Under this interpretation, we dene b2k;2m to be the probability that a random walk
of length 2m has exactly 2k of its 2m polygonal line segments above the t-axis.
The probability b2k;2m is frequently interpreted in terms of a two-player game.
(The reader will recall the game Heads or Tails, in Example 1.4.) Player A is said
to be in the lead at time n if the random walk is above the t-axis at that time, or
if the random walk is on the t-axis at time n but above the t-axis at time n  1.
(At time 0, neither player is in the lead.) One can ask what is the most probable
number of times that player A is in the lead, in a game of length 2m. Most people
will say that the answer to this question is m. However, the following theorem says
that m is the least likely number of times that player A is in the lead, and the most
likely number of times in the lead is 0 or 2m.",prob.pdf
"that m is the least likely number of times that player A is in the lead, and the most
likely number of times in the lead is 0 or 2m.
Theorem 12.4 If Peter and Paul play a game of Heads or Tails of length 2m, the
probability that Peter will be in the lead exactly 2k times is equal to
2k;2m :
Proof. To prove the theorem, we need to show that
b2k;2m = 2k;2m : (12.9)
Exercise 12.1.7 shows that b2m;2m = u2m and b0;2m = u2m, so we only need to prove
that Equation 12.9 holds for 1  k  m1. We can obtain a recursion involving the
b's and the f's (dened in Section 12.1) by counting the number of paths of length
2m that have exactly 2k of their segments above the t-axis, where 1  k  m  1.
To count this collection of paths, we assume that the rst return occurs at time 2j,
where 1  j  m  1. There are two cases to consider. Either during the rst 2j
outcomes the path is above the t-axis or below the t-axis. In the rst case, it must",prob.pdf
"outcomes the path is above the t-axis or below the t-axis. In the rst case, it must
be true that the path has exactly (2k  2j) line segments above the t-axis, between
t = 2j and t = 2m. In the second case, it must be true that the path has exactly
2k line segments above the t-axis, between t = 2j and t = 2m.
We now count the number of paths of the various types described above. The
number of paths of length 2j all of whose line segments lie above the t-axis and
which return to the origin for the rst time at time 2j equals (1=2)22jf2j. This
also equals the number of paths of length 2j all of whose line segments lie below
the t-axis and which return to the origin for the rst time at time 2j. The number
of paths of length (2m  2j) which have exactly (2k  2j) line segments above the
t-axis is b2k2j;2m2j. Finally, the number of paths of length (2m  2j) which have
exactly 2k line segments above the t-axis is b2k;2m2j. Therefore, we have
b2k;2m = 1
2
kX
j=1
f2jb2k2j;2m2j + 1
2",prob.pdf
"exactly 2k line segments above the t-axis is b2k;2m2j. Therefore, we have
b2k;2m = 1
2
kX
j=1
f2jb2k2j;2m2j + 1
2
mkX
j=1
f2jb2k;2m2j :
We now assume that Equation 12.9 is true for m < n. Then we have",prob.pdf
"12.3. ARC SINE LAWS 495
0 10 20 30 40
0
0.02
0.04
0.06
0.08
0.1
0.12
Figure 12.2: Times in the lead.
b2k;2n = 1
2
kX
j=1
f2j2k2j;2m2j + 1
2
mkX
j=1
f2j2k;2m2j
= 1
2
kX
j=1
f2ju2k2ju2m2k + 1
2
mkX
j=1
f2ju2ku2m2j2k
= 1
2u2m2k
kX
j=1
f2ju2k2j + 1
2u2k
mkX
j=1
f2ju2m2j2k
= 1
2u2m2ku2k + 1
2u2ku2m2k ;
where the last equality follows from Theorem 12.2. Thus, we have
b2k;2n = 2k;2n ;
which completes the proof. 2
We illustrate the above theorem by simulating 10,000 games of Heads or Tails, with
each game consisting of 40 tosses. The distribution of the number of times that
Peter is in the lead is given in Figure 12.2, together with the arc sine density.
We end this section by stating two other results in which the arc sine density
appears. Proofs of these results may be found in Feller.8
Theorem 12.5 Let J be the random variable which, for a given random walk of
length 2m, gives the smallest subscript j such that Sj = S2m. (Such a subscript j",prob.pdf
"length 2m, gives the smallest subscript j such that Sj = S2m. (Such a subscript j
must be even, by parity considerations.) Let 
2k;2m be the probability that J = 2k.
Then we have

2k;2m = 2k;2m :
2
8W. Feller, op. cit., pp. 93{94.",prob.pdf
"496 CHAPTER 12. RANDOM WALKS
The next theorem says that the arc sine density is applicable to a wide range
of situations. A continuous distribution function F(x) is said to be symmetric
if F(x) = 1  F(x). (If X is a continuous random variable with a symmetric
distribution function, then for any real x, we have P(X  x) = P(X  x).) We
imagine that we have a random walk of length n in which each summand has the
distribution F(x), where F is continuous and symmetric. The subscript of the rst
maximum of such a walk is the unique subscript k such that
Sk > S0; : : : ; Sk > Sk1; Sk  Sk+1; : : : ; Sk  Sn :
We dene the random variable Kn to be the subscript of the rst maximum. We
can now state the following theorem concerning the random variable Kn.
Theorem 12.6 Let F be a symmetric continuous distribution function, and let 
be a xed real number strictly between 0 and 1. Then as n ! 1, we have
P(Kn < n) ! 2
 arcsinp :
2",prob.pdf
"be a xed real number strictly between 0 and 1. Then as n ! 1, we have
P(Kn < n) ! 2
 arcsinp :
2
A version of this theorem that holds for a symmetric random walk can also be
found in Feller.
Exercises
1 For a random walk of length 2m, dene k to equal 1 if Sk > 0, or if Sk1 = 1
and Sk = 0. Dene k to equal -1 in all other cases. Thus, k gives the side
of the t-axis that the random walk is on during the time interval [k  1; k]. A
\law of large numbers"" for the sequence fkg would say that for any  > 0,
we would have
P

 < 1 + 2 + 
 
 
 + n
n < 

! 1
as n ! 1. Even though the 's are not independent, the above assertion
certainly appears reasonable. Using Theorem 12.4, show that if 1  x  1,
then
lim
n!1
P
1 + 2 + 
 
 
 + n
n < x

= 2
 arcsin
r
1 + x
2 :
2 Given a random walk W of length m, with summands
fX1; X2; : : :; Xmg ;
dene the reversed random walk to be the walk W with summands
fXm; Xm1; : : : ; X1g :
(a) Show that the kth partial sum S",prob.pdf
"fX1; X2; : : :; Xmg ;
dene the reversed random walk to be the walk W with summands
fXm; Xm1; : : : ; X1g :
(a) Show that the kth partial sum S
k satises the equation
S
k = Sm  Snk ;
where Sk is the kth partial sum for the random walk W.",prob.pdf
"12.3. ARC SINE LAWS 497
(b) Explain the geometric relationship between the graphs of a random walk
and its reversal. (It is not in general true that one graph is obtained
from the other by re
ecting in a vertical line.)
(c) Use parts (a) and (b) to prove Theorem 12.5.",prob.pdf
498 CHAPTER 12. RANDOM WALKS,prob.pdf
"Appendices
499",prob.pdf
"500 APPENDICES
NA (0,d) = area of 
shaded region 
0 d
          .00        .01        .02        .03        .04        .05        .06        .07        .08        .09          
0.0    .0000    .0040    .0080    .0120    .0160     .0199   .0239    .0279    .0319    .0359
0.1    .0398    .0438    .0478    .0517    .0557     .0596   .0636    .0675    .0714    .0753           
0.2    .0793    .0832    .0871    .0910    .0948     .0987   .1026    .1064    .1103    .1141
0.3    .1179    .1217    .1255    .1293    .1331     .1368   .1406    .1443    .1480    .1517            
0.4    .1554    .1591    .1628    .1664    .1700     .1736   .1772    .1808    .1844    .1879            
0.5    .1915    .1950    .1985    .2019    .2054     .2088   .2123    .2157    .2190    .2224            
0.6    .2257    .2291    .2324    .2357    .2389     .2422   .2454    .2486    .2517    .2549",prob.pdf
"0.6    .2257    .2291    .2324    .2357    .2389     .2422   .2454    .2486    .2517    .2549             
0.7    .2580    .2611    .2642    .2673    .2704     .2734   .2764    .2794    .2823    .2852             
0.8    .2881    .2910    .2939    .2967    .2995     .3023   .3051    .3078    .3106    .3133             
0.9    .3159    .3186    .3212    .3238    .3264     .3289   .3315    .3340    .3365    .3389             
1.0    .3413    .3438    .3461    .3485    .3508     .3531   .3554    .3577    .3599    .3621             
1.1    .3643    .3665    .3686    .3708    .3729     .3749   .3770    .3790    .3810    .3830               
1.2    .3849    .3869    .3888    .3907    .3925     .3944   .3962    .3980    .3997    .4015 
1.3    .4032    .4049    .4066    .4082    .4099     .4115   .4131    .4147    .4162    .4177 
1.4    .4192    .4207    .4222    .4236    .4251     .4265   .4279    .4292    .4306    .4319",prob.pdf
"1.4    .4192    .4207    .4222    .4236    .4251     .4265   .4279    .4292    .4306    .4319
1.5    .4332    .4345    .4357    .4370    .4382     .4394   .4406    .4418    .4429    .4441
1.6    .4452    .4463    .4474    .4484    .4495     .4505   .4515    .4525    .4535    .4545
1.7    .4554    .4564    .4573    .4582    .4591     .4599   .4608    .4616    .4625    .4633
1.8    .4641    .4649    .4656    .4664    .4671     .4678   .4686    .4693    .4699    .4706 
1.9    .4713    .4719    .4726    .4732    .4738     .4744   .4750    .4756    .4761    .4767
2.0    .4772    .4778    .4783    .4788    .4793     .4798   .4803    .4808    .4812    .4817
2.1    .4821    .4826    .4830    .4834    .4838     .4842   .4846    .4850    .4854    .4857 
2.2    .4861    .4864    .4868    .4871    .4875     .4878   .4881    .4884    .4887    .4890 
2.3    .4893    .4896    .4898    .4901    .4904     .4906   .4909    .4911    .4913    .4916",prob.pdf
"2.3    .4893    .4896    .4898    .4901    .4904     .4906   .4909    .4911    .4913    .4916
2.4    .4918    .4920    .4922    .4925    .4927     .4929   .4931    .4932    .4934    .4936
2.5    .4938    .4940    .4941    .4943    .4945     .4946   .4948    .4949    .4951    .4952
2.6    .4953    .4955    .4956    .4957    .4959     .4960   .4961    .4962    .4963    .4964  
2.7    .4965    .4966    .4967    .4968    .4969     .4970   .4971    .4972    .4973    .4974 
2.8    .4974    .4975    .4976    .4977    .4977     .4978   .4979    .4979    .4980    .4981
2.9    .4981    .4982    .4982    .4983    .4984     .4984   .4985    .4985    .4986    .4986
3.0    .4987    .4987    .4987    .4988    .4988     .4989   .4989    .4989    .4990    .4990
3.1    .4990    .4991    .4991    .4991    .4992     .4992   .4992    .4992    .4993    .4993
3.2    .4993    .4993    .4994    .4994    .4994     .4994   .4994    .4995    .4995    .4995",prob.pdf
"3.2    .4993    .4993    .4994    .4994    .4994     .4994   .4994    .4995    .4995    .4995 
3.3    .4995    .4995    .4995    .4996    .4996     .4996   .4996    .4996    .4996    .4997
3.4    .4997    .4997    .4997    .4997    .4997     .4997   .4997    .4997    .4997    .4998
3.5    .4998    .4998    .4998    .4998    .4998     .4998   .4998    .4998    .4998    .4998
3.6    .4998    .4998    .4999    .4999    .4999     .4999   .4999    .4999    .4999    .4999
3.7    .4999    .4999    .4999    .4999    .4999     .4999   .4999    .4999    .4999    .4999 
3.8    .4999    .4999    .4999    .4999    .4999     .4999   .4999    .4999    .4999    .4999
3.9    .5000    .5000    .5000    .5000    .5000     .5000   .5000    .5000    .5000    .5000    
Appendix A
Normal distribution table",prob.pdf
"APPENDICES 501
Above       . .          . .       . .       . .       . .       . .       . .  
     . .       . .       . .       . .       . .          1        3        . . 
           4               5             . .
    72.5                   . .       . .       . .       . .       . .       . .
       . .         1        2        1        2         7        2         4    
       19              6           72.2
    71.5                   . .       . .       . .       . .         1        3 
       4        3        5
      10        4         9        2         2           43             11      
    69.9
    70.5                     1      . .         1      . .         1        1   
     3      12      18      14        7         4        3         3           6
8             22          69.5
    69.5                   . .       . .         1      16        4      17     
 27      20      33      25      20       11        4         5         183",prob.pdf
"27      20      33      25      20       11        4         5         183     
        41          68.9  
    68.5                     
1      . .         7      11      16      25      31      34      48      21    
  18         4        3        . .         219             49          68.2
    67.5                   . .         3        5      14      15      36      3
8      28      38      19      11         4       . .        . .         211    
         33          67.6
    66.5                   . .         3        3        5        2      17     
 17      14      13        4       . .        . .       . .        . .          
 78             20          67.2
    65.5                     1      . .         9        5        7      11     
 11        7        7        5        2         1       . .        . .          
 66             12          66.7
    64.5                     1        1        4        4        1        5",prob.pdf
"66             12          66.7
    64.5                     1        1        4        4        1        5     
   5      . .         2      . .        . .        . .       . .        . .     
      23               5          65.8 
Below       . .            1       . .        2        4        1        2      
  2        1        1      . .        . .        . .       . .        . .        
   14               1            . .
 Totals       . .           5         7      32      59      48    117    138    
120    167      99      64       41      17       14         928           205  
          . .
Medians   . .           . .       . .    66.3   67.8   67.9   67.7   67.9   68.3
   68.5    69.0   69.0    70.0     . .        . .            . .              . 
.            . . 
Heights of the Mid-parents in inches.             Below 62.2   63.2   64.2   65.2   66.2   67.2   68.2   69
.2   70.2   71.2   72.2   73.2   Above     children.  
    Heights of the adult children.",prob.pdf
".2   70.2   71.2   72.2   73.2   Above     children.  
    Heights of the adult children.
Total number ofAdult
Mid- parents.
Medians
Number of adult children of various statures born of 205 mid-parents of various 
statures.
(All female heights have been multiplied by 1.08) Note.    In calculating the Medians, the entries have been taken as referring to
 the middle of the squares in which they stand.  The reason 
why the headings run 62.2, 63.2, &c., instead of 62.5, 63.5, &c., is that the ob
servations are unequally distributed between 62 and 63, 63 
and 64, &c., there being a strong bias in favour of integral inches.   After car
eful consideration, I concluded that the h
eadings, as adopted, 
best satisfied the conditions.  This inequality was not apparent in the case of 
the Mid-parents.  Source: F. Galton, ""Regression towards 
Mediocrity in Hereditary Stature"",  
Royal Anthropological Institute of Great Britain and Ireland,
  vol.15 (1885), p.248.
Appendix B",prob.pdf
"502 APPENDICES
Appendix C
Life Table
Number of survivors at single years of Age, out of 100,000 Born Alive, by  
Race and Sex: United States, 1990.
  0      100000      100000    100000                         43       94707         92840      96626 
  1        99073        98969      99183                         44       94453         92505      96455  
  2        99008        98894      99128                         45       94179         92147      96266 
  3        98959        98840      99085                         46       93882         91764      96057 
  4        98921        98799      99051                         47       93560         91352      95827 
  5        98890        98765      99023                         48       93211         90908      95573 
  6        98863        98735      99000                         49       92832         90429      95294  
  7        98839        98707      98980                         50       92420         89912      94987",prob.pdf
"7        98839        98707      98980                         50       92420         89912      94987 
  8        98817        98680      98962                         51       91971         89352      94650  
  9        98797        98657      98946                         52       91483         88745      94281 
10        98780        98638      98931                         53       90950         88084      93877 
11        98765        98623      98917                         54       90369         87363      93436 
12        98750        98608      98902                         55       89735         86576      92955  
13        98730        98586      98884                         56       89045         85719      92432 
14        98699        98547      98862                         57       88296         84788      91864 
15        98653        98485      98833                         58       87482         83777      91246",prob.pdf
"15        98653        98485      98833                         58       87482         83777      91246  
16        98590        98397      98797                         59       86596         82678      90571  
17        98512        98285      98753                         60       85634         81485      89835 
18        98421        98154      98704                         61       84590         80194      89033 
19        98323        98011      98654                         62       83462         78803      88162 
20        98223        97863      98604                         63       82252         77314      87223 
21        98120        97710      98555                         64       80961         75729      86216 
22        98015        97551      98506                         65       79590         74051      85141 
23        97907        97388      98456                         66       78139         72280      83995",prob.pdf
"23        97907        97388      98456                         66       78139         72280      83995 
24        97797        97221      98405                         67       76603         70414      82772 
25        97684        97052      98351                         68       74975         68445      81465 
26        97569        96881      98294                         69       73244         66364      80064 
27        97452        96707      98235                         70       71404         64164      78562 
28        97332        96530      98173                         71       69453         61847      76953 
29        97207        96348      98107                         72       67392         59419      75234 
30        97077        96159      98038                         73       65221         56885      73400 
31        96941        95962      97965                         74       62942         54249      71499",prob.pdf
"31        96941        95962      97965                         74       62942         54249      71499 
32        96800        95785      97887                         75       60557         51519      69376 
33        96652        95545      97804                         76       58069         48704      67178 
34        96497        95322      97717                         77       55482         45816      64851 
35        96334        95089      97624                         78       52799         42867      62391 
36        96161        94843      97525                         79       50026         39872      59796 
37        95978        94585      97419                         80       47168         36848      57062 
38        95787        94316      97306                         81       44232         33811      54186 
39        95588        94038      97187                         82       41227         30782      51167",prob.pdf
"39        95588        94038      97187                         82       41227         30782      51167  
40        95382        93753      97061                         83       38161         27782      48002 
41        95168        93460      96926                         84       35046         24834      44690 
42        94944        93157      96782                         85       31892         21962      41230 
Age  Both sexes  Male    Female                     Age  Both sexes   Male    Female
All races                                                           All races",prob.pdf
"Index
, estimation of, 43{46
n!, 80
absorbing Markov chain, 416
absorbing state, 416
AbsorbingChain (program), 421
absorption probabilities, 420
Ace, Mr., 241
Ali, 178
alleles, 348
AllPermutations (program), 84
ANDERSON, C. L., 157
annuity, 246
life, 247
terminal, 247
arc sine laws, 493
area, estimation of, 42
Areabargraph (program), 46
asymptotically equal, 81
Baba, 178
babies, 14, 250
Banach's Matchbox, 255
BAR-HILLEL, M., 176
BARNES, B., 175
BARNHART, R., 11
BAYER, D., 120
Bayes (program), 147
Bayes probability, 136
Bayes' formula, 146
BAYES, T., 149
beard, 153
bell-shaped, 47
Benford distribution, 195
BENKOSKI, S., 40
Bernoulli trials process, 96
BERNOULLI, D., 227
BERNOULLI, J., 113, 149, 310{312
Bertrand's paradox, 47{50
BERTRAND, J., 49, 181
BertrandsParadox (program), 49
beta density, 168
BIENAYME, I., 310, 377
BIGGS, N. L., 85
binary expansion, 69
binomial coecient, 93
binomial distribution, 99, 184
approximating a, 329
Binomial Theorem, 103
BinomialPlot (program), 99",prob.pdf
"binary expansion, 69
binomial coecient, 93
binomial distribution, 99, 184
approximating a, 329
Binomial Theorem, 103
BinomialPlot (program), 99
BinomialProbabilities (program), 98
Birthday (program), 78
birthday problem, 77
blackjack, 247, 253
blood test, 254
Bose-Einstein statistics, 107
Box paradox, 181
BOX, G. E. P., 213
boxcars, 27
BRAMS, S., 179, 182
Branch (program), 381
branching process, 376
customer, 393
BranchingSimulation (program), 386
bridge, 181, 182, 199, 203, 287
BROWN, B. H., 38
BROWN, E., 425
Buon's needle, 44{46, 51{53
BUFFON, G. L., 9, 44, 50{51
BuonsNeedle (program), 45
bus paradox, 164
calendar, 38
cancer, 147
503",prob.pdf
"504 INDEX
canonical form of an absorbing
Markov chain, 416
car, 137
CARDANO, G., 30{31, 110, 249
cars on a highway, 66
CASANOVA, G., 11
Cauchy density, 218, 400
cells, 347
Central Limit Theorem, 325
for Bernoulli Trials, 330
for Binomial Distributions, 328
for continuous independent trials
process, 357
for discrete independent random
variables, 345
for discrete independent trials
process, 343
for Markov Chains, 464
proof of, 397
chain letter, 388
characteristic function, 397
Chebyshev Inequality, 305, 316
CHEBYSHEV, P. L., 313
chi-squared density, 216, 296
Chicago World's Fair, 52
chord, random, 47, 54
chromosomes, 348
CHU, S.-C., 110
CHUNG, K. L., 153
Circle of Gold, 388
Clinton, Bill, 196
clover-leaf interchange, 39
CLTBernoulliGlobal, 332
CLTBernoulliLocal (program), 329
CLTBernoulliPlot (program), 327
CLTGeneral (program), 345
CLTIndTrialsLocal (program), 342
CLTIndTrialsPlot (program), 341
COATES, R. M., 305
CoinTosses (program), 3
Collins, People v., 153, 202
color-blindness, 424",prob.pdf
"CLTIndTrialsPlot (program), 341
COATES, R. M., 305
CoinTosses (program), 3
Collins, People v., 153, 202
color-blindness, 424
conditional density, 162
conditional distribution, 134
conditional expectation, 239
conditional probability, 133
CONDORCET, Le Marquis de, 12
condence interval, 334, 360
conjunction fallacy, 38
continuum, 41
convolution, 286, 291
of binomial distributions, 289
of Cauchy densities, 294
of exponential densities, 292, 300
of geometric distributions, 289
of normal densities, 294
of standard normal densities, 299
of uniform densities, 292, 299
CONWAY, J., 432
CRAMER, G., 227
craps, 235, 240, 468
Craps (program), 235
CROSSEN, C., 161
CROWELL, R., 468
cumulative distribution function, 61
joint, 165
customer branching process, 393
cut, 120
Dartmouth, 27
darts, 56, 57, 59, 60, 64, 71, 163, 164
Darts (program), 58
DAVID, F. N., 86, 337, 489
DAVID, F. N., 32
de MERE, CHEVALIER, 4, 31, 37
de MOIVRE, A., 37, 88, 148, 336, 489
de MONTMORT, P. R., 85
degrees of freedom, 217",prob.pdf
"DAVID, F. N., 32
de MERE, CHEVALIER, 4, 31, 37
de MOIVRE, A., 37, 88, 148, 336, 489
de MONTMORT, P. R., 85
degrees of freedom, 217
DeMere1 (program), 4
DeMere2 (program), 4
density function, 56, 59
beta, 168
Cauchy, 218, 400
chi-squared, 216, 296
conditional, 162
exponential, 53, 66, 163, 205
gamma, 207
joint, 165
log normal, 224
Maxwell, 215",prob.pdf
"INDEX 505
normal, 212
Rayleigh, 215, 295
t-, 360
uniform, 60, 205
derangement, 85
DIACONIS, P., 120, 251
Die (program), 225
DieTest (program), 297
distribution function, 1, 19
properties of, 22
Benford, 195
binomial, 99, 184
geometric, 184
hypergeometric, 193
joint, 142
marginal, 143
negative binomial, 186
Poisson, 187
uniform, 183
DNA, 348
DOEBLIN, W., 449
DOYLE, P. G., 87, 452, 470
Drunkard's Walk example, 416, 419{
421, 423, 427, 443
Dry Gulch, 280
EDWARDS, A. W. F., 108
Egypt, 30
Ehrenfest model, 410, 433, 441, 460,
461
EHRENFEST, P., 410
EHRENFEST, T, 410
EhrenfestUrn (program), 462
EISENBERG, B., 160
elevator, 89, 116
Emile's restaurant, 75
ENGLE, A., 445
envelopes, 179, 180
EPSTEIN, R., 287
equalization, 472
equalizations
expected number of, 479
ergodic Markov chain, 433
ESP, 250, 251
EUCLID, 85
Euler's formula, 202
Eulerian number, 127
event, 18
events
attraction of, 160
independent, 139, 164
repulsion of, 160
existence of God, 245
expected value, 226, 268",prob.pdf
"Eulerian number, 127
event, 18
events
attraction of, 160
independent, 139, 164
repulsion of, 160
existence of God, 245
expected value, 226, 268
exponential density, 53, 66, 163, 205
extinction, problem of, 378
factorial, 80
fair game, 241
FALK, R., 161, 176
fall, 131
fallacy, 38
FELLER, W., 11, 106, 107, 191, 201,
218, 254, 344
FERMAT, P., 4, 32{35, 112{113, 156
Fermi-Dirac statistics, 107
gurate numbers, 108
nancial records
suspicious, 196
nite additivity property, 23
FINN, J., 178
First Fundamental Mystery of Proba-
bility, 232
rst maximum of a random walk, 496
rst return to the origin, 473
Fisher's Exact Test, 193
FISHER, R. A., 252
xed column vector, 435
xed points, 82
xed row vector, 435
FixedPoints (program), 82
FixedVector (program), 437

ying bombs, 191, 201
Fourier transform, 397
FRECHET, M., 466
frequency concept of probability, 70
frustration solitaire, 86
Fundamental Limit Theorem for Reg-
ular Markov Chains, 448
fundamental matrix, 419",prob.pdf
"506 INDEX
for a regular Markov chain, 457
for an ergodic Markov chain, 458
GALAMBOS, J., 303
GALILEO, G., 13
Gallup Poll, 14, 335
Galton board, 99, 351
GALTON, F., 282, 345, 350, 376
GaltonBoard (program), 99
Gambler's Ruin, 426, 486, 487
gambling systems, 241
gamma density, 207
GARDNER, M., 181
gas diusion
Ehrenfest model of, 410, 433, 441,
460, 461
GELLER, S., 176
GeneralSimulation (program), 9
generating function
for continuous density, 393
moment, 366, 394
ordinary, 369
genes, 348, 411
genetics, 345
genotypes, 348
geometric distribution, 184
geometric series, 29
GHOSH, B. K., 160
goat, 137
GONSHOR, H., 425
GOSSET, W. S., 360
grade point average, 343
GRAHAM, R., 251
GRANBERG, D., 161
GRAUNT, J., 246
Greece, 30
GRIDGEMAN, N. T., 51, 181
GRINSTEAD, C. M., 87
GUDDER, S., 160
HACKING, I., 30, 148
HAMMING, R. W., 284
HANES data, 345
Hangtown, 280
Hanover Inn, 65
hard drive, Warp 9, 66
Hardy-Weinberg Law, 349
harmonic function, 428
Harvard, 27
hat check problem, 82, 85, 105
heights",prob.pdf
"Hanover Inn, 65
hard drive, Warp 9, 66
Hardy-Weinberg Law, 349
harmonic function, 428
Harvard, 27
hat check problem, 82, 85, 105
heights
distribution of, 345
helium, 107
HEYDE, C., 377
HILL, T., 196
Holmes, Sherlock, 91
HorseRace (program), 6
hospital, 14, 250
HOWARD, R. A., 406
HTSimulation (program), 6
HUDDE, J., 148
HUIZINGA, F., 388
HUYGENS, C., 147, 243{245
hypergeometric distribution, 193
hypotheses, 145
hypothesis testing, 101
Inclusion-Exclusion Principle, 104
independence of events, 139, 164
mutual, 141
independence of random variables
mutual, 143, 165
independence of random
variables, 143, 165
independent trials process, 144, 168
interarrival time, average, 208
interleaving, 120
irreducible Markov chain, 433
Isle Royale, 202
JAYNES, E. T., 49
JOHNSONBOUGH, R., 153
joint cumulative distribution
function, 165
joint density function, 165
joint distribution function, 142
joint random variable, 142
KAHNEMAN, D., 38
Kemeny's constant, 469, 470
KEMENY, J. G., 200, 406, 466",prob.pdf
"INDEX 507
KENDALL, D. G., 377
KEYFITZ, N., 382
KILGOUR, D. M., 179, 182
KINGSTON, J. G., 157
KONOLD, C., 161
KOZELKA, R. M., 344
Labouchere betting system, 12, 13
LABOUCHERE, H. du P., 12
LAMPERTI, J., 267, 324
LAPLACE, P. S., 51, 53, 350
last return to the origin, 482
Law (program), 310
Law of Averages, 70
Law of Large Numbers, 307, 316
for Ergodic Markov Chains, 439
Strong, 70
LawContinuous (program), 318
lead change, 482
LEONARD, B., 256
LEONTIEF, W. W., 426
LEVASSEUR, K., 485
library problem, 82
life table, 39
light bulb, 66, 72, 172
Linda problem, 38
LINDEBERG, J. W., 344
LIPSON, A., 161
Little's law for queues, 276
Lockhorn, Mr. and Mrs., 65
log normal density, 224
lottery
Powerball, 204
LUCAS, E., 119
LUSINCHI, D., 12
MAISTROV, L., 150, 310
MANN, B., 120
margin of error, 335
marginal distribution function, 143
Markov chain, 405
absorbing, 416
ergodic, 433
irreducible, 433
regular, 433
Markov Chains
Central Limit Theorem for, 464
Fundamental Limit Theorem for
Regular, 448",prob.pdf
"absorbing, 416
ergodic, 433
irreducible, 433
regular, 433
Markov Chains
Central Limit Theorem for, 464
Fundamental Limit Theorem for
Regular, 448
MARKOV, A. A., 464
martingale, 241, 242, 428
origin of word, 11
martingale betting system, 11, 14, 248
matrix
fundamental, 419
MatrixPowers (program), 407
maximum likelihood
estimate, 198, 202
Maximum Likelihood
Principle, 91, 117
Maxwell density, 215
maze, 440, 453
McCRACKEN, D., 10
mean, 226
mean rst passage matrix, 455
mean rst passage time, 453
mean recurrence matrix, 455
mean recurrence time, 454
memoryless property, 68, 164, 206
milk, 252
modular arithmetic, 10
moment generating function, 366, 394
moment problem, 368, 397
moments, 365, 393
Monopoly, 469
MonteCarlo (program), 42
Monty Hall problem, 136, 161
moose, 202
mortality table, 246
mule kicks, 201
MULLER, M. E., 213
multiple-gene hypothesis, 348
mustache, 153
mutually independent events, 141
mutually independent random
variables, 143
negative binomial distribution, 186",prob.pdf
"mustache, 153
mutually independent events, 141
mutually independent random
variables, 143
negative binomial distribution, 186
New York Times, 340
New York Yankees, 118, 253",prob.pdf
"508 INDEX
New-Age Solitaire, 130
NEWCOMB, S., 196
NFoldConvolution (program), 287
NIGRINI, M., 196
normal density, 47, 212
NormalArea (program), 322
nursery rhyme, 84
odds, 27
ordering, random, 127
ordinary generating function, 369
ORE, O., 30, 31
outcome, 18
Oz, Land of, 406, 439
POLYA, G., 15, 17, 475
Pascal's triangle, 94, 103, 108
PASCAL, B., 4, 32{35, 107, 112{113,
156, 242, 245
paternity suit, 222
PEARSON, K., 9, 351
PENNEY, W., 432
People v. Collins, 153, 202
PERLMAN, M. D., 45
permutation, 79
xed points of, 82
Philadelphia 76ers, 15
photons, 106
Pickwick, Mr., 153
Pilsdor Beer Company, 280
PITTEL, B., 256
point count, 287
Poisson approximation to the
binomial distribution, 189
Poisson distribution, 187
variance of, 263
poker, 95
polls, 333
Polya urn model, 152, 174
ponytail, 153
posterior probabilities, 145
Powerball lottery, 204
PowerCurve (program), 102
Presidential election, 335
PRICE, C., 86
prior probabilities, 145
probability
Bayes, 136
conditional, 133",prob.pdf
"PowerCurve (program), 102
Presidential election, 335
PRICE, C., 86
prior probabilities, 145
probability
Bayes, 136
conditional, 133
frequency concept of, 2
of an event, 19
transition, 406
vector, 407
problem of points, 32, 112, 147, 156
process, random, 128
PROPP, J., 256
PROSSER, R., 200
protons, 106
quadratic equation, roots of, 73
quantum mechanics, 107
QUETELET, A., 350
Queue (program), 208
queues, 186, 208, 275
quincunx, 351
RENYI, A., 167
RABELAIS, F., 12
racquetball, 157
radioactive isotope, 66, 71
RAND Corporation, 10
random integer, 39
random number generator, 2
random ordering, 127
random process, 128
random variable, 1, 18
continuous, 58
discrete, 18
functions of a, 210
joint, 142
random variables
independence of, 143
mutual independence of, 143
random walk, 471
in n dimensions, 17
RandomNumbers (program), 3
RandomPermutation (program), 82
rank event, 160
raquetball, 13
rat, 440, 453",prob.pdf
"INDEX 509
Rayleigh density, 215, 295
records, 83, 234
Records (program), 84
regression on the mean, 282
regression to the mean, 345, 352
regular Markov chain, 433
reliability of a system, 154
restricted choice, principle of, 182
return to the origin, 472
rst, 473
last, 482
probability of eventual, 475
reversibility, 463
reversion, 352
rie shue, 120
RIORDAN, J., 86
rising sequence, 120
rnd, 42
ROBERTS, F., 426
Rome, 30
ROSS, S., 270, 276
roulette, 13, 237, 432
run, 229
SAGAN, H., 237
sample, 333
sample mean, 265
sample space, 18
continuous, 58
countably innite, 28
innite, 28
sample standard deviation, 265
sample variance, 265
SAWYER, S., 412
SCHULTZ, H., 255
SENETA, E., 377, 444
service time, average, 208
SHANNON, C. E., 465
SHOLANDER, M., 39
shuing, 120
SHULTZ, H., 256
SimulateChain (program), 439
simulating a random variable, 211
snakeeyes, 27
SNELL, J. L., 87, 175, 406, 466
snowfall in Hanover, 83
spike graph, 6
Spikegraph (program), 6
spinner, 41, 55, 59, 162
spread, 266",prob.pdf
"snakeeyes, 27
SNELL, J. L., 87, 175, 406, 466
snowfall in Hanover, 83
spike graph, 6
Spikegraph (program), 6
spinner, 41, 55, 59, 162
spread, 266
St. Ives, 84
St. Petersburg Paradox, 227
standard deviation, 257
standard normal random
variable, 213
standardized random variable, 264
standardized sum, 326
state
absorbing, 416
of a Markov chain, 405
transient, 416
statistics
applications of the Central Limit
Theorem to, 333
stepping stones, 412
SteppingStone (program), 413
stick of unit length, 73
STIFEL, M., 110
STIGLER, S., 350
Stirling's formula, 81
STIRLING, J., 88
StirlingApproximations
(program), 81
stock prices, 241
StockSystem (program), 241
Strong Law of Large
Numbers, 70, 314
suit event, 160
SUTHERLAND, E., 182
t-density, 360
TARTAGLIA, N., 110
tax returns, 196
tea, 252
telephone books, 256
tennis, 157, 424
tetrahedral numbers, 108
THACKERAY, W. M., 14
THOMPSON, G. L., 406
THORP, E., 247, 253",prob.pdf
"510 INDEX
time to absorption, 419
TIPPETT, L. H. C., 10
traits, independence of, 216
transient state, 416
transition matrix, 406
transition probability, 406
tree diagram, 24, 76
innite binary, 69
Treize, 85
triangle
acute, 73
triangular numbers, 108
trout, 198
true-false exam, 267
Tunbridge, 154
TVERSKY, A., 14, 38
Two aces problem, 181
two-armed bandit, 170
TwoArm (program), 171
type 1 error, 101
type 2 error, 101
typesetter, 189
ULAM, S., 11
unbiased estimator, 266
uniform density, 205
uniform density function, 60
uniform distribution, 25, 183
uniform random variables
sum of two continuous, 63
unshue, 122
USPENSKY, J. B., 299
utility function, 227
VANDERBEI, R., 175
variance, 257, 271
calculation of, 258
variation distance, 128
VariationList (program), 128
volleyball, 158
von BORTKIEWICZ, L., 201
von MISES, R., 87
von NEUMANN, J., 10, 11
vos SAVANT, M., 40, 86, 136, 176,
181
Wall Street Journal, 161
watches, counterfeit, 91
WATSON, H. W., 377
WEAVER, W., 465",prob.pdf
"von NEUMANN, J., 10, 11
vos SAVANT, M., 40, 86, 136, 176,
181
Wall Street Journal, 161
watches, counterfeit, 91
WATSON, H. W., 377
WEAVER, W., 465
Weierstrass Approximation Theorem,
315
WELDON, W. F. R., 9
Wheaties, 118, 253
WHITAKER, C., 136
WHITEHEAD, J. H. C., 181
WICHURA, M. J., 45
WILF, H. S., 91, 474
WOLF, R., 9
WOLFORD, G., 159
Woodstock, 154
Yang, 130
Yin, 130
ZAGIER, D., 485
Zorg, planet of, 90",prob.pdf
"ELECTROMAGNETICS
VOLUME 1",Electromagnetics_Vol1.pdf
"Publication of this book was made possible in part by the Virginia Tech 
University Libraries’ Open Education Faculty Initiative Grant program: 
http://guides.lib.vt.edu/oer/grants",Electromagnetics_Vol1.pdf
"VT Publishing
Blacksburg, Virginia
ELECTROMAGNETICS
STEVEN W. ELLINGSON
VOLUME 1",Electromagnetics_Vol1.pdf
"Copyright © 2018, Steven W. Ellingson
This textbook is licensed with a Creative Commons Attribution Share-Alike 4.0 license 
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Suggested citation: Ellingson, Steven W. (2018) Electromagnetics, V ol. 1. Blacksburg, V A: VT 
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Ellingson, Steven W., author
     Electromagnetics (V olume 1) / Steven W. Ellingson
     Pages cm
     ISBN 978-0-9979201-8-5 (print)
     ISBN 978-0-9979201-9-2 (ebook)
     DOI: https://doi.org/10.21061/electromagnetics-vol-1
     1.  Electromagnetism.  2. Electromagnetic theory. 
     I.   Title
     QC760.E445 2018
     621.3",Electromagnetics_Vol1.pdf
"1.  Electromagnetism.  2. Electromagnetic theory. 
     I.   Title
     QC760.E445 2018
     621.3
Cover Design: Robert Browder
Cover Image: © Michelle Yost. Total Internal Reflection https://flic.kr/p/dWAhx5 is licensed with a Creative 
Commons Attribution-ShareAlike 2.0 license https://creativecommons.org/licenses/by-sa/2.0/ (cropped by 
Robert Browder) 
iv",Electromagnetics_Vol1.pdf
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Open textbooks are complete textbooks that have been funded, published, and licensed to be freely used, 
adapted, and distributed. As a particular type of Open Educational Resource (OER), this open textbook is 
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"Contents
Preface xii
1 Preliminary Concepts 1
1.1 What is Electromagnetics? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Fundamentals of W aves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Guided and Unguided W aves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.5 Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.6 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.7 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2 Electric and Magnetic Fields 17
2.1 What is a Field? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17",Electromagnetics_Vol1.pdf
"2 Electric and Magnetic Fields 17
2.1 What is a Field? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Electric Field Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Permittivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.4 Electric Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.5 Magnetic Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.6 Permeability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.7 Magnetic Field Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.8 Electromagnetic Properties of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
vi",Electromagnetics_Vol1.pdf
"CONTENTS vii
3 T ransmission Lines 30
3.1 Introduction to Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.2 T ypes of Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.3 Transmission Lines as T wo-Port Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.4 Lumped-Element Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.5 T elegrapher’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.6 W ave Equation for a TEM Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.7 Characteristic Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.8 W ave Propagation on a TEM Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.9 Lossless and Low-Loss Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41",Electromagnetics_Vol1.pdf
"3.9 Lossless and Low-Loss Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.10 Coaxial Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.11 Microstrip Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.12 V oltage Reflection Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.13 Standing W aves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.14 Standing W ave Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.15 Input Impedance of a T erminated Lossless Transmission Line . . . . . . . . . . . . . . . . . . . 52
3.16 Input Impedance for Open- and Short-Circuit T erminati ons . . . . . . . . . . . . . . . . . . . . 54
3.17 Applications of Open- and Short-Circuited Transmissi on Line Stubs . . . . . . . . . . . . . . . 55",Electromagnetics_Vol1.pdf
"3.17 Applications of Open- and Short-Circuited Transmissi on Line Stubs . . . . . . . . . . . . . . . 55
3.18 Measurement of Transmission Line Characteristics . . . . . . . . . . . . . . . . . . . . . . . . 56
3.19 Quarter-W avelength Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.20 Power Flow on Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.21 Impedance Matching: General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
3.22 Single-Reactance Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.23 Single-Stub Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4 V ector Analysis 70
4.1 V ector Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.2 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76",Electromagnetics_Vol1.pdf
"viii CONTENTS
4.3 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
4.4 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
4.5 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.6 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
4.8 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
4.9 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.10 The Laplacian Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
5 Electrostatics 93",Electromagnetics_Vol1.pdf
"4.10 The Laplacian Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
5 Electrostatics 93
5.1 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
5.2 Electric Field Due to Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.3 Charge Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.4 Electric Field Due to a Continuous Distribution of Charg e . . . . . . . . . . . . . . . . . . . . . 96
5.5 Gauss’ Law: Integral Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
5.6 Electric Field Due to an Infinite Line Charge using Gauss’ Law . . . . . . . . . . . . . . . . . . 101
5.7 Gauss’ Law: Differential Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.8 Force, Energy , and Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105",Electromagnetics_Vol1.pdf
"5.8 Force, Energy , and Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.9 Independence of Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.10 Kirchoff’s V oltage Law for Electrostatics: Integral F orm . . . . . . . . . . . . . . . . . . . . . 108
5.11 Kirchoff’s V oltage Law for Electrostatics: Different ial Form . . . . . . . . . . . . . . . . . . . 109
5.12 Electric Potential Field Due to Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.13 Electric Potential Field due to a Continuous Distribut ion of Charge . . . . . . . . . . . . . . . . 112
5.14 Electric Field as the Gradient of Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.15 Poisson’s and Laplace’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
5.16 Potential Field Within a Parallel Plate Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . 116",Electromagnetics_Vol1.pdf
"5.16 Potential Field Within a Parallel Plate Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . 116
5.17 Boundary Conditions on the Electric Field Intensity ( E) . . . . . . . . . . . . . . . . . . . . . . 118
5.18 Boundary Conditions on the Electric Flux Density ( D) . . . . . . . . . . . . . . . . . . . . . . 120",Electromagnetics_Vol1.pdf
"CONTENTS ix
5.19 Charge and Electric Field for a Perfectly Conducting Re gion . . . . . . . . . . . . . . . . . . . 122
5.20 Dielectric Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.21 Dielectric Breakdown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
5.22 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
5.23 The Thin Parallel Plate Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
5.24 Capacitance of a Coaxial Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
5.25 Electrostatic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
6 Steady Current and Conductivity 134
6.1 Convection and Conduction Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134",Electromagnetics_Vol1.pdf
"6 Steady Current and Conductivity 134
6.1 Convection and Conduction Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
6.2 Current Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
6.3 Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
6.4 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
6.5 Conductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
6.6 Power Dissipation in Conducting Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
7 Magnetostatics 146
7.1 Comparison of Electrostatics and Magnetostatics . . . . . . . . . . . . . . . . . . . . . . . . . 146
7.2 Gauss’ Law for Magnetic Fields: Integral Form . . . . . . . . . . . . . . . . . . . . . . . . . . 147",Electromagnetics_Vol1.pdf
"7.2 Gauss’ Law for Magnetic Fields: Integral Form . . . . . . . . . . . . . . . . . . . . . . . . . . 147
7.3 Gauss’ Law for Magnetism: Differential Form . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7.4 Ampere’s Circuital Law (Magnetostatics): Integral For m . . . . . . . . . . . . . . . . . . . . . 149
7.5 Magnetic Field of an Infinitely-Long Straight Current-B earing Wire . . . . . . . . . . . . . . . 150
7.6 Magnetic Field Inside a Straight Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
7.7 Magnetic Field of a T oroidal Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
7.8 Magnetic Field of an Infinite Current Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
7.9 Ampere’s Law (Magnetostatics): Differential Form . . . . . . . . . . . . . . . . . . . . . . . . 159
7.10 Boundary Conditions on the Magnetic Flux Density ( B) . . . . . . . . . . . . . . . . . . . . . . 160",Electromagnetics_Vol1.pdf
"7.10 Boundary Conditions on the Magnetic Flux Density ( B) . . . . . . . . . . . . . . . . . . . . . . 160
7.11 Boundary Conditions on the Magnetic Field Intensity ( H) . . . . . . . . . . . . . . . . . . . . . 161",Electromagnetics_Vol1.pdf
"x CONTENTS
7.12 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
7.13 Inductance of a Straight Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
7.14 Inductance of a Coaxial Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
7.15 Magnetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
7.16 Magnetic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
8 Time-V arying Fields 174
8.1 Comparison of Static and Time-V arying Electromagnetic s . . . . . . . . . . . . . . . . . . . . 174
8.2 Electromagnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
8.3 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178",Electromagnetics_Vol1.pdf
"8.3 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
8.4 Induction in a Motionless Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
8.5 Transformers: Principle of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
8.6 Transformers as T wo-Port Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
8.7 The Electric Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
8.8 The Maxwell-Faraday Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
8.9 Displacement Current and Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
9 Plane W aves in Lossless Media 194
9.1 Maxwell’s Equations in Differential Phasor Form . . . . . . . . . . . . . . . . . . . . . . . . . 194",Electromagnetics_Vol1.pdf
"9 Plane W aves in Lossless Media 194
9.1 Maxwell’s Equations in Differential Phasor Form . . . . . . . . . . . . . . . . . . . . . . . . . 194
9.2 W ave Equations for Source-Free and Lossless Regions . . . . . . . . . . . . . . . . . . . . . . 196
9.3 T ypes of W aves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
9.4 Uniform Plane W aves: Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
9.5 Uniform Plane W aves: Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
9.6 W ave Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
9.7 W ave Power in a Lossless Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
A Constitutive Parameters of Some Common Materials 213
A.1 Permittivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213",Electromagnetics_Vol1.pdf
"A.1 Permittivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
A.2 Permeability of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214",Electromagnetics_Vol1.pdf
"CONTENTS xi
A.3 Conductivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
B Mathematical Formulas 217
B.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
B.2 V ector Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
B.3 V ector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
C Physical Constants 220
Index 221",Electromagnetics_Vol1.pdf
"Preface
About This Book
[m0146]
Goals for this book. This book is intended to serve
as a primary textbook for a one-semester introductory
course in undergraduate engineering
electromagnetics, including the following topics:
electric and magnetic fields; electromagnetic
properties of materials; electromagnetic waves; and
devices that operate according to associated
electromagnetic principles including resistors,
capacitors, inductors, transformers, generators, and
transmission lines.
This book employs the “transmission lines first”
approach, in which transmission lines are introduced
using a lumped-element equivalent circuit model for a
differential length of transmission line, leading to
one-dimensional wave equations for voltage and
current.1 This is sufficient to address transmission
line concepts, including characteristic impedance,
input impedance of terminated transmission lines, and
impedance matching techniques. Attention then turns",Electromagnetics_Vol1.pdf
"input impedance of terminated transmission lines, and
impedance matching techniques. Attention then turns
to electrostatics, magnetostatics, time-varying fields,
and waves, in that order.
What’s new . This version of the book is the second
public release of this book. The first release, known
as “V olume 1 (Beta), ” was released in January 2018.
Improvements from the beta version include the
following:
• Correction of errors identified in the beta version
errata and many minor improvements.
• Addition of an index.
1 Are you an instructor who is not a fan of the “transmission
lines first” approach? Then see “ What are those little numbers
in square brackets? ” later in this section.
• Accessibility features: Figures now include “alt
text” suitable for screen reading software.
• Addition of a separate manual of examples and
solutions (see the web site).
• Addition of source files for the book (see the
web site).
T arget audience. This book is intended for electrical",Electromagnetics_Vol1.pdf
"solutions (see the web site).
• Addition of source files for the book (see the
web site).
T arget audience. This book is intended for electrical
engineering students in the third year of a bachelor of
science degree program. It is assumed that readers are
familiar with the fundamentals of electric circuits and
linear systems, which are normally taught in the
second year of the degree program. It is also assumed
that readers have received training in basic
engineering mathematics, including complex
numbers, trigonometry , vectors, partial differential
equations, and multivariate calculus. Review of the
relevant principles is provided at various points in the
text. In a few cases (e.g., phasors, vectors) this review
consists of a separate stand-alone section.
Notation, examples, and highlights. Section 1.7
summarizes the mathematical notation used in this
book. Examples are set apart from the main text as
follows:
Example 0.1. This is an example.",Electromagnetics_Vol1.pdf
"summarizes the mathematical notation used in this
book. Examples are set apart from the main text as
follows:
Example 0.1. This is an example.
“Highlight boxes” are used to identify key ideas as
follows:
This is a key idea.
What are those little numbers in square brackets?
This book is a product of the
Open Electromagnetics Project . This project provides
a large number of sections (“modules”) which are
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"xiii
assembled (“remixed”) to create new and different
versions of the book. The text “
[m0146]” that you see at
the beginning of this section uniquely identifies the
module within the larger set of modules provided by
the project. This identification is provided because
different remixes of this book may exist, each
consisting of a different subset and arrangement of
these modules. Prospective authors can use this
identification as an aid in creating their own remixes.
Why do some sections of this book seem to repeat
material presented in previous sections? In some
remixes of this book, authors might choose to
eliminate or reorder modules. For this reason, the
modules are written to “stand alone” as much as
possible. As a result, there may be some redundancy
between sections that would not be present in a
traditional (non-remixable) textbook. While this may
seem awkward to some at first, there are clear
benefits: In particular, it never hurts to review relevant",Electromagnetics_Vol1.pdf
"seem awkward to some at first, there are clear
benefits: In particular, it never hurts to review relevant
past material before tackling a new concept. And,
since the electronic version of this book is being
offered at no cost, there is not much gained by
eliminating this useful redundancy .
Why cite Wikipedia pages as additional reading?
Many modules cite Wikipedia entries as sources of
additional information. Wikipedia represents both the
best and worst that the Internet has to offer. Most
authors of traditional textbooks would agree that
citing Wikipedia pages as primary sources is a bad
idea, since quality is variable and content is subject to
change over time. On the other hand, many Wikipedia
pages are excellent, and serve as useful sources of
relevant information that is not strictly within the
scope of the curriculum. Furthermore, students
benefit from seeing the same material presented
differently , in a broader context, and with additional",Electromagnetics_Vol1.pdf
"scope of the curriculum. Furthermore, students
benefit from seeing the same material presented
differently , in a broader context, and with additional
references cited by Wikipedia pages. W e trust
instructors and students to realize the potential pitfalls
of this type of resource and to be alert for problems.
Acknowledgments. Here’s a list of talented and
helpful people who contributed to this book:
The staff of VT Publishing, University Libraries,
V irginia T ech:
Editor: Anita W alz
Advisors: Peter Potter, Corinne Guimont
Cover: Robert Browder, Anita W alz
Other VT contributors:
Assessment: Tiffany Shoop, Anita W alz
Accessibility: Christa Miller
V irginia T ech students:
Alt text writer: Stephanie Edwards
Figure designer: Michaela Goldammer
Figure designer: Kruthika Kikkeri
Figure designer: Y oumin Qin
Copyediting:
Melissa Ashman, Kwantlen Polytechnic University
External reviewers:
Samir El-Ghazaly , University of Arkansas
Stephen Gedney , University of Colorado Denver",Electromagnetics_Vol1.pdf
"External reviewers:
Samir El-Ghazaly , University of Arkansas
Stephen Gedney , University of Colorado Denver
Randy Haupt, Colorado School of Mines
Karl W arnick, Brigham Y oung University
Also, thanks are due to the students of the Spring
2018 and Summer 2018 sections of ECE3105 at
V irginia T ech who used the beta version of this book
and provided useful feedback. Thanks also to Justin
Y onker, instructor of the Summer 2018 section.",Electromagnetics_Vol1.pdf
"xiv PREF ACE
About the Open Electromagnetics
Project
[m0148]
The Open Electromagnetics Project was established
at V irginia T ech in 2017 with the goal of creating
no-cost openly-licensed textbooks for courses in
undergraduate engineering electromagnetics. While a
number of very fine traditional textbooks are available
on this topic, we feel that it has become unreasonable
to insist that students pay hundreds of dollars per
book when effective alternatives can be provided
using modern media at little or no cost to the student.
This project is equally motivated by the desire for the
freedom to adopt, modify , and improve educational
resources. This work is distributed under a Creative
Commons BY SA license which allows – and we
hope encourages – others to adopt, modify , improve,
and expand the scope of our work.
About the Author
[m0153]
Steven W . Ellingson (ellingson@vt.edu) is an
Associate Professor at V irginia T ech in Blacksburg,
V irginia in the United States. He received PhD and",Electromagnetics_Vol1.pdf
"Associate Professor at V irginia T ech in Blacksburg,
V irginia in the United States. He received PhD and
MS degrees in Electrical Engineering from the Ohio
State University and a BS in Electrical & Computer
Engineering from Clarkson University . He was
employed by the US Army , Booz-Allen & Hamilton,
Raytheon, and the Ohio State University
ElectroScience Laboratory before joining the faculty
of V irginia T ech, where he teaches courses in
electromagnetics, radio frequency electronics,
wireless communications, and signal processing. His
research includes topics in wireless communications,
radio science, and radio frequency instrumentation.
Professor Ellingson serves as a consultant to industry
and government and is the author of Radio Systems
Engineering (Cambridge University Press, 2016).",Electromagnetics_Vol1.pdf
"Chapter 1
Preliminary Concepts
1.1 What is Electromagnetics?
[m0037]
The topic of this book is applied engineering
electromagnetics. This topic is often described as “the
theory of electromagnetic fields and waves, ” which is
both true and misleading. The truth is that electric
fields, magnetic fields, their sources, waves, and the
behavior these waves are all topics covered by this
book. The misleading part is that our principal aim
shall be to close the gap between basic electrical
circuit theory and the more general theory that is
required to address certain topics that are of broad and
common interest in the field of electrical engineering.
(For a preview of topics where these techniques are
required, see the list at the end of this section.)
In basic electrical circuit theory , the behavior of
devices and systems is abstracted in such a way that
the underlying electromagnetic principles do not need
to be considered. Every student of electrical",Electromagnetics_Vol1.pdf
"the underlying electromagnetic principles do not need
to be considered. Every student of electrical
engineering encounters this, and is grateful since this
greatly simplifies analysis and design. For example, a
resistor is commonly defined as a device which
exhibits a particular voltage V = IR in response to a
current I, and the resistor is therefore completely
described by the value R. This is an example of a
“lumped element” abstraction of an electrical device.
Much can be accomplished knowing nothing else
about resistors; no particular knowledge of the
physical concepts of electrical potential, conduction
current, or resistance is required. However, this
simplification makes it impossible to answer some
frequently-encountered questions. Here are just a few:
• What determines R? How does one go about
designing a resistor to have a particular
resistance?
• Practical resistors are rated for power-handling
capability; e.g., discrete resistors are frequently",Electromagnetics_Vol1.pdf
"resistance?
• Practical resistors are rated for power-handling
capability; e.g., discrete resistors are frequently
identified as “1/8-W , ” “1/4-W , ” and so on. How
does one determine this, and how can this be
adjusted in the design?
• Practical resistors exhibit significant reactance as
well as resistance. Why? How is this
determined? What can be done to mitigate this?
• Most things which are not resistors also exhibit
significant resistance and reactance – for
example, electrical pins and interconnects. Why?
How is this determined? What can be done to
mitigate this?
The answers to the these questions must involve
properties of materials and the geometry in which
those materials are arranged . These are precisely the
things that disappear in lumped element device
models, so it is not surprising that such models leave
us in the dark on these issues. It should also be
apparent that what is true for the resistor is also going
to be true for other devices of practical interest,",Electromagnetics_Vol1.pdf
"apparent that what is true for the resistor is also going
to be true for other devices of practical interest,
including capacitors (and devices unintentionally
exhibiting capacitance), inductors (and devices
unintentionally exhibiting inductance), transformers
(and devices unintentionally exhibiting mutual
impedance), and so on. From this perspective,
electromagnetics may be viewed as a generalization
of electrical circuit theory that addresses these
considerations. Conversely basic electric circuit
theory may be viewed a special case of
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"2 CHAPTER 1. PRELIMINAR Y CONCEPTS
electromagnetic theory that applies when these
considerations are not important. Many instances of
this “electromagnetics as generalization” vs.
“lumped-element theory as special case” dichotomy
appear in the study of electromagnetics.
There is more to the topic, however. There are many
devices and applications in which electromagnetic
fields and waves are primary engineering
considerations that must be dealt with directly .
Examples include electrical generators and motors;
antennas; printed circuit board stackup and layout;
persistent storage of data (e.g., hard drives); fiber
optics; and systems for radio, radar, remote sensing,
and medical imaging. Considerations such as signal
integrity and electromagnetic compatibility (EMC)
similarly require explicit consideration of
electromagnetic principles.
Although electromagnetic considerations pertain to
all frequencies, these considerations become
increasingly difficult to avoid with increasing",Electromagnetics_Vol1.pdf
"Although electromagnetic considerations pertain to
all frequencies, these considerations become
increasingly difficult to avoid with increasing
frequency . This is because the wavelength of an
electromagnetic field decreases with increasing
frequency .1 When wavelength is large compared to
the size of the region of interest (e.g., a circuit), then
analysis and design is not much different from
zero-frequency (“DC”) analysis and design.
For example, the free space wavelength at 3 MHz is
about 100 m, so a planar circuit having dimensions
10 cm × 10 cm is just 0.1% of a wavelength across at
this frequency . Although an electromagnetic wave
may be present, it has about the same value over the
region of space occupied by the circuit. In contrast,
the free space wavelength at 3 GHz is about 10 cm, so
the same circuit is one full wavelength across at this
frequency . In this case, different parts of this circuit
observe the same signal with very different magnitude
and phase.",Electromagnetics_Vol1.pdf
"frequency . In this case, different parts of this circuit
observe the same signal with very different magnitude
and phase.
Some of the behaviors associated with non-negligible
dimensions are undesirable, especially if not taken
into account in the design process. However, these
behaviors can also be exploited to do some amazing
and useful things – for example, to launch an
electromagnetic wave (i.e., an antenna) or to create
1 Most readers have encountered the concepts of frequency and
wavelength previously , but can refer to Section 1.3, if need ed, for a
quick primer.
filters and impedance matching devices consisting
only of metallic shapes, free of discrete capacitors or
inductors.
Electromagnetic considerations become not only
unavoidable but central to analysis and design above a
few hundred MHz, and especially in the
millimeter-wave, infrared (IR), optical, and ultraviolet
(UV) bands. 2 The discipline of electrical engineering
encompasses applications in these frequency ranges",Electromagnetics_Vol1.pdf
"(UV) bands. 2 The discipline of electrical engineering
encompasses applications in these frequency ranges
even though – ironically – such applications may not
operate according to principles that can be considered
“electrical”! Nevertheless, electromagnetic theory
applies.
Another common way to answer the question “What
is electromagnetics?” is to identify the topics that are
commonly addressed within this discipline. Here’s a
list of topics – some of which have already been
mentioned – in which explicit consideration of
electromagnetic principles is either important or
essential:3
• Antennas
• Coaxial cable
• Design and characterization of common discrete
passive components including resistors,
capacitors, inductors, and diodes
• Distributed (e.g., microstrip) filters
• Electromagnetic compatibility (EMC)
• Fiber optics
• Generators
• Magnetic resonance imaging (MRI)
• Magnetic storage (of data)
• Microstrip transmission lines
• Modeling of non-ideal behaviors of discrete
components",Electromagnetics_Vol1.pdf
"• Magnetic storage (of data)
• Microstrip transmission lines
• Modeling of non-ideal behaviors of discrete
components
• Motors
2 See Section 1.2 for a quick primer on the electromagnetic spec-
trum and this terminology .
3 Presented in alphabetical order so as to avoid the appearanc e of
any bias on the part of the author!",Electromagnetics_Vol1.pdf
"1.2. ELECTROMAGNETIC SPECTRUM 3
• Non-contact sensors
• Photonics
• Printed circuit board stackup and layout
• Radar
• Radio wave propagation
• Radio frequency electronics
• Signal integrity
• Transformers
• W aveguides
In summary:
Applied engineering electromagnetics is the
study of those aspects of electrical engineering in
situations in which the electromagnetic proper-
ties of materials and the geometry in which those
materials are arranged is important. This re-
quires an understanding of electromagnetic fields
and waves, which are of primary interest in some
applications.
Finally , here are two broadly-defined learning
objectives that should now be apparent: (1) Learn the
techniques of engineering analysis and design that
apply when electromagnetic principles are important,
and (2) Better understand the physics underlying the
operation of electrical devices and systems, so that
when issues associated with these physical principles",Electromagnetics_Vol1.pdf
"operation of electrical devices and systems, so that
when issues associated with these physical principles
emerge one is prepared to recognize and grapple with
them.
1.2 Electromagnetic Spectrum
[m0075]
Electromagnetic fields exist at frequencies from DC
(0 Hz) to at least 1020 Hz – that’s at least 20 orders of
magnitude!
At DC, electromagnetics consists of two distinct
disciplines: electrostatics, concerned with electric
fields; and magnetostatics, concerned with magnetic
fields.
At higher frequencies, electric and magnetic fields
interact to form propagating waves. W aves having
frequencies within certain ranges are given names
based on how they manifest as physical phenomena.
These names are (in order of increasing frequency):
radio, infrared (IR), optical (also known as “light”),
ultraviolet (UV), X-rays, and gamma rays (γ-rays).
See T able 1.1 and Figure 1.1 for frequency ranges and
associated wavelengths.
The term electromagnetic spectrum refers to the",Electromagnetics_Vol1.pdf
"See T able 1.1 and Figure 1.1 for frequency ranges and
associated wavelengths.
The term electromagnetic spectrum refers to the
various forms of electromagnetic phenomena that
exist over the continuum of frequencies.
The speed (properly known as “phase velocity”) at
which electromagnetic fields propagate in free space
is given the symbol c, and has the value
∼
= 3.00 × 108 m/s. This value is often referred to as
the “speed of light. ” While it is certainly the speed of
light in free space, it is also speed of any
electromagnetic wave in free space. Given frequency
f, wavelength is given by the expression
λ= c
f in free space
T able 1.1 shows the free space wavelengths
associated with each of the regions of the
electromagnetic spectrum.
This book presents a version of electromagnetic
theory that is based on classical physics. This
approach works well for most practical problems.
However, at very high frequencies, wavelengths
become small enough that quantum mechanical",Electromagnetics_Vol1.pdf
"approach works well for most practical problems.
However, at very high frequencies, wavelengths
become small enough that quantum mechanical
effects may be important. This is usually the case in
the X-ray band and above. In some applications, these",Electromagnetics_Vol1.pdf
"4 CHAPTER 1. PRELIMINAR Y CONCEPTS
Regime Frequency Range W avelength Range
γ-Ray >3 × 1019 Hz <0.01 nm
X-Ray 3 × 1016 Hz – 3 × 1019 Hz 10–0.01 nm
Ultraviolet (UV) 2.5 × 1015 – 3 × 1016 Hz 120–10 nm
Optical 4.3 × 1014 – 2.5 × 1015 Hz 700–120 nm
Infrared (IR) 300 GHz – 4.3 × 1014 Hz 1 mm – 700 nm
Radio 3 kHz – 300 GHz 100 km – 1 mm
T able 1.1: The electromagnetic spectrum. Note that the indi cated ranges are arbitrary but consistent with
common usage.
effects become important at frequencies as low as the
optical, IR, or radio bands. (A prime example is the
photoelectric effect ; see “ Additional References”
below .) Thus, caution is required when applying the
classical version of electromagnetic theory presented
here, especially at these higher frequencies.
Theory presented in this book is applicable to
DC, radio, IR, and optical waves, and to a lesser
extent to UV waves, X-rays, and γ-rays. Cer-
tain phenomena in these frequency ranges – in",Electromagnetics_Vol1.pdf
"DC, radio, IR, and optical waves, and to a lesser
extent to UV waves, X-rays, and γ-rays. Cer-
tain phenomena in these frequency ranges – in
particular quantum mechanical effects – are not
addressed in this book.
The radio portion of the electromagnetic spectrum
alone spans 12 orders of magnitude in frequency (and
wavelength), and so, not surprisingly , exhibits a broad
range of phenomena. This is shown in Figure 1.1. For
this reason, the radio spectrum is further subdivided
into bands as shown in T able 1.2. Also shown in
T able 1.2 are commonly-used band identification
acronyms and some typical applications.
Similarly , the optical band is partitioned into the
familiar “rainbow” of red through violet, as shown in
Figure 1.1 and T able 1.3. Other portions of the
spectrum are sometimes similarly subdivided in
certain applications.
Additional Reading:
• “Electromagnetic spectrum” on Wikipedia.
• “Photoelectric effect” on Wikipedia.
400 nm
500 nm
600 nm
700 nm
1000 m
100 m
10 m
1 m
10 cm",Electromagnetics_Vol1.pdf
"• “Electromagnetic spectrum” on Wikipedia.
• “Photoelectric effect” on Wikipedia.
400 nm
500 nm
600 nm
700 nm
1000 m
100 m
10 m
1 m
10 cm
1 cm
1 mm
1000 µm
100 µm
10 µm
1 µm
1000 nm
100 nm
10 nm
0.1 Å
0.1 nm
1Å
1 nm
Wavelength
10
17
1016
1015
1014
1013
1012
1011
1010
109
108
106
Frequency (Hz)
1018
1019
UHF
VHF
7-13
FM
VHF
2-6
1000 MHz
500 MHz
100 MHz
50 MHz
Gamma-rays
X-rays
Ultraviolet
Visible
Near IR
Infra-red
Thermal IR
Far IR
Microwaves
Radar
Radio, TV
AM
Long-waves
10
7
c⃝ V . Blacus CC BY -SA 3.0
Figure 1.1: Electromagnetic Spectrum.",Electromagnetics_Vol1.pdf
"1.3. FUNDAMENT ALS OF W A VES 5
Band Frequencies W avelengths T ypical Applications
EHF 30-300 GHz 10–1 mm 60 GHz WLAN, Point-to-point data links
SHF 3–30 GHz 10–1 cm T errestrial & Satellite data links, Rada r
UHF 300–3000 MHz 1–0.1 m TV broadcasting, Cellular, WLAN
VHF 30–300 MHz 10–1 m FM & TV broadcasting, LMR
HF 3–30 MHz 100–10 m Global terrestrial comm., CB Radio
MF 300–3000 kHz 1000–100 m AM broadcasting
LF 30–300 kHz 10–1 km Navigation, RFID
VLF 3–30 kHz 100–10 km Navigation
T able 1.2: The radio portion of the electromagnetic spectru m, according to a common scheme for naming ranges
of radio frequencies. WLAN: Wireless local area network, LMR : Land mobile radio, RFID: Radio frequency
identification.
Band Frequencies W avelengths
V iolet 668–789 THz 450–380 nm
Blue 606–668 THz 495–450 nm
Green 526–606 THz 570–495 nm
Y ellow 508–526 THz 590–570 nm
Orange 484–508 THz 620–590 nm
Red 400–484 THz 750–620 nm
T able 1.3: The optical portion of the electromagnetic
spectrum.",Electromagnetics_Vol1.pdf
"Orange 484–508 THz 620–590 nm
Red 400–484 THz 750–620 nm
T able 1.3: The optical portion of the electromagnetic
spectrum.
1.3 Fundamentals of W aves
[m0074]
In this section, we formally introduce the concept of a
wave and explain some basic characteristics.
T o begin, let us consider not electromagnetic waves,
but rather sound waves. T o be clear, sound waves and
electromagnetic waves are completely distinct
phenomena. Sound waves are variations in pressure,
whereas electromagnetic waves are variations in
electric and magnetic fields. However, the
mathematics that govern sound waves and
electromagnetic waves are very similar, so the
analogy provides useful insight. Furthermore, sound
waves are intuitive for most people because they are
readily observed. So, here we go:
Imagine standing in an open field and that it is
completely quiet. In this case, the air pressure
everywhere is about 101 kPa (101,000 N/m 2) at sea
level, and we refer to this as the quiescent air",Electromagnetics_Vol1.pdf
"completely quiet. In this case, the air pressure
everywhere is about 101 kPa (101,000 N/m 2) at sea
level, and we refer to this as the quiescent air
pressure. Sound can be described as the differential
air pressure p(x,y,z,t ), which we define as the
absolute air pressure at the spatial coordinates
(x,y,z ) minus the quiescent air pressure. So, when
there is no sound, p(x,y,z,t ) = 0 . The function pas
an example of a scalar field .4
Let’s also say you are standing at x= y= z= 0 and
you have brought along a friend who is standing at
x= d; i.e., a distance dfrom you along the xaxis.
Also, for simplicity , let us consider only what is
happening along the xaxis; i.e., p(x,t).
At t= 0 , you clap your hands once. This forces the
air between your hands to press outward, creating a
region of increased pressure (i.e., p> 0) that travels
outward. As the region of increased pressure moves
outward, it leaves behind a region of low pressure
where p< 0. Air molecules immediately move",Electromagnetics_Vol1.pdf
"outward. As the region of increased pressure moves
outward, it leaves behind a region of low pressure
where p< 0. Air molecules immediately move
toward this region of lower pressure, and so the air
pressure quickly returns to the quiescent value, p= 0 .
The traveling disturbance in p(x,t) is the sound of the
clap. The disturbance continues to travel outward
until it reaches your friend, who then hears the clap.
At each point in time, you can make a plot of p(x,t)
versus xfor the current value of t. This is shown in
Figure 1.2. At times t< 0, we have simply
p(x,t) = 0 . A short time after t= 0 , the peak
pressure is located at slightly to the right of x= 0 .
The pressure is not a simple impulse because
interactions between air molecules constrain the
4 Although it’s not important in this section, you can read abou t
the concept of a “field” in Section 2.1.",Electromagnetics_Vol1.pdf
"6 CHAPTER 1. PRELIMINAR Y CONCEPTS
p(x,t)
x
p(x,t)
x
c⃝ Y . Qin CC BY 4.0
Figure 1.2: The differential pressure p(x,t) (top) a
short time after the clap and ( bottom) a slightly longer
time after the clap.
pressure to be continuous over space. So instead, we
see a rounded pulse representing the rapid build-up
and similarly rapid decline in air pressure. A short
time later p(x,t) looks very similar, except the pulse
is now further away .
Now: What precisely is p(x,t)? Completely skipping
over the derivation, the answer is that p(x,t) is the
solution to the acoustic wave equation (see
“ Additional References” at the end of this section):
∂2p
∂x2 − 1
c2
s
∂2p
∂t2 = 0 (1.1)
where cs is the speed of sound, which is about
340 m/s at sea level. Just to emphasize the quality of
the analogy between sound waves and
electromagnetic waves, know that the acoustic wave
equation is mathematically identical to equations that
that govern electromagnetic waves.",Electromagnetics_Vol1.pdf
"electromagnetic waves, know that the acoustic wave
equation is mathematically identical to equations that
that govern electromagnetic waves.
Although “transient” phenomena – analogous to a
clap – are of interest in electromagnetics, an even
more common case of interest is the wave resulting
from a sinusoidally-varying source. W e can
demonstrate this kind of wave in the context of sound
as well. Here we go:
In the previous scenario, you pick up a trumpet and
blow a perfect A note. The A note is 440 Hz, meaning
that the air pressure emerging from your trumpet is
varying sinusoidally at a frequency of 440 Hz. Let’s
say you can continue to blow this note long enough
for the entire field to be filled with the sound of your
trumpet. Now what does the pressure-versus-distance
curve look like? T wo simple observations will settle
that question:
• p(x,t) at any constant position xis a sinusoid as
a function of x. This is because the acoustic
wave equation is linear and time invariant, so a",Electromagnetics_Vol1.pdf
"• p(x,t) at any constant position xis a sinusoid as
a function of x. This is because the acoustic
wave equation is linear and time invariant, so a
sinusoidal excitation (i.e., your trumpet) results
in a sinusoidal response at the same frequency
(i.e., the sound heard by your friend).
• p(x,t) at any constant time tis also a sinusoid as
a function of x. This is because the sound is
propagating away from the trumpet and toward
your friend, and anyone in between will also
hear the A note, but with a phase shift
determined by the difference in distances.
This is enough information to know that the solution
must have the form:
p(x,t) = Amcos (ωt− βx+ ψ) (1.2)
where ω= 2 πf, f = 440 Hz, and Am, β, and ψ
remain to be determined.
Y ou can readily verify that Equation 1.2 satisfies the
acoustic wave equation when
β = ω
cs
(1.3)
In this problem, we find β ∼
= 8.13 rad/m. This means
that at any given time, the difference in phase
measured between any two points separated by a",Electromagnetics_Vol1.pdf
"In this problem, we find β ∼
= 8.13 rad/m. This means
that at any given time, the difference in phase
measured between any two points separated by a
distance of 1 m is 8.13 rad. The parameter βgoes by
at least three names: phase propagation constant ,
wavenumber, and spatial frequency . The term “spatial
frequency” is particularly apt, since βplays precisely
the same role for distance ( x) as ωplays for time ( t) –
This is apparent from Equation 1.2. However,
“wavenumber” is probably the more commonly-used
term.
The wavenumber β (rad/m) is the rate at which
the phase of a sinusoidal wave progresses with
distance.",Electromagnetics_Vol1.pdf
"1.3. FUNDAMENT ALS OF W A VES 7
Note that Am and ψare not determined by the wave
equation, but instead are properties of the source.
Specifically , Am is determined by how hard we blow ,
and ψis determined by the time at which we began to
blow and the location of the trumpet. For simplicity ,
let us assume that we begin to blow at time t≪ 0;
i.e., in the distant past so that the sound pressure field
has achieved steady state by t= 0 . Also, let us set
ψ= 0 and set Am = 1 in whatever units we choose to
express p(x,t). W e then have:
p(x,t) = cos ( ωt− βx) (1.4)
Now we have everything we need to make plots of
p(x) at various times.
Figure 1.3(a) shows p(x,t = 0) . As expected,
p(x,t = 0) is periodic in x. The associated period is
referred to as the wavelength λ. Since λis the
distance required for the phase of the wave to increase
by 2πrad, and because phase is increasing at a rate of
βrad/m, we find:
λ= 2π
β (1.5)
In the present example, we find λ∼",Electromagnetics_Vol1.pdf
"by 2πrad, and because phase is increasing at a rate of
βrad/m, we find:
λ= 2π
β (1.5)
In the present example, we find λ∼
= 77.3 cm.W avelength λ = 2 π/β is the distance required
for the phase of a sinusoidal wave to increase
by one complete cycle (i.e., 2πrad) at any given
time.
Now let us consider the situation at t= +1 /4f,
which is t= 568 µs and ωt= π/2. W e see in
Figure 1.3(b) that the waveform has shifted a distance
λ/4 to the right. It is in this sense that we say the
wave is propagating in the +xdirection.
Furthermore, we can now compute a phase velocity
vp: W e see that a point of constant phase has shifted a
distance λ/4 in time 1/4f, so
vp = λf (1.6)
In the present example, we find vp ∼
= 340 m/s; i.e., we
have found that the phase velocity is equal to the
speed of sound cs. It is in this sense that we say that
the phase velocity is the speed at which the wave
propagates.5
5 It is worth noting here is that “velocity” is technically a ve ctor;",Electromagnetics_Vol1.pdf
"the phase velocity is the speed at which the wave
propagates.5
5 It is worth noting here is that “velocity” is technically a ve ctor;
i.e., speed in a given direction. Nevertheless, this quanti ty is actually
just a speed, and this particular abuse of terminology is gene rally
accepted.
p(x,t)
x
λ
p(x,t)
x
p(x,t)
x
c⃝ Y . Qin CC BY 4.0
Figure 1.3: The differential pressure p(x,t) for (a)
t= 0 , (b) t= 1 /4f for “ −βx, ” as indicated in Equa-
tion 1.4 (wave traveling to right); and (c) t= 1 /4f for
“ +βx” (wave traveling to left).",Electromagnetics_Vol1.pdf
"8 CHAPTER 1. PRELIMINAR Y CONCEPTS
Phase velocity vp = λf is the speed at which a
point of constant phase in a sinusoidal waveform
travels.
Recall that in Equation 1.2 we declared that βxis
subtracted from the argument of the sinusoidal
function. T o understand why , let’s change the sign of
βxand see if it still satisfies the wave equation – one
finds that it does. Next, we repeat the previous
experiment and see what happens. The result is shown
in Figure 1.3(c). Note that points of constant phase
have traveled an equal distance, but now in the −x
direction. In other words, this alternative choice of
sign for βxwithin the argument of the cosine function
represents a wave that is propagating in the opposite
direction. This leads us to the following realization:
If the phase of the wave is decreasing with βx,
then the wave is propagating in the +xdirection.
If the phase of the wave is increasing with βx,
then the wave is propagating in the −xdirection.",Electromagnetics_Vol1.pdf
"then the wave is propagating in the +xdirection.
If the phase of the wave is increasing with βx,
then the wave is propagating in the −xdirection.
Since the prospect of sound traveling toward the
trumpet is clearly nonsense in the present situation,
we may neglect the latter possibility . However, what
happens if there is a wall located in the distance,
behind your friend? Then, we expect an echo from
the wall, which would be a second wave propagating
in the reverse direction and for which the argument of
the cosine function would contain the term “ +βx. ”
Finally , let us return to electromagnetics.
Electromagnetic waves satisfy precisely the same
wave equation (i.e., Equation 1.1) as do sound waves,
except that the phase velocity is much greater.
Interestingly , though, the frequencies of
electromagnetic waves are also much greater than
those of sound waves, so we can end up with
wavelengths having similar orders of magnitude. In
particular, an electromagnetic wave with λ= 77 .3 cm",Electromagnetics_Vol1.pdf
"those of sound waves, so we can end up with
wavelengths having similar orders of magnitude. In
particular, an electromagnetic wave with λ= 77 .3 cm
(the wavelength of the “ A ” note in the preceding
example) lies in the radio portion of the
electromagnetic spectrum.
An important difference between sound and
electromagnetic waves is that electromagnetic waves
are vectors; that is, they have direction as well as
magnitude. Furthermore, we often need to consider
multiple electromagnetic vector waves (in particular,
both the electric field and the magnetic field ) in order
to completely understand the situation. Nevertheless
the concepts of wavenumber, wavelength, phase
velocity , and direction of propagation apply in
precisely the same manner to electromagnetic waves
as they do to sound waves.
Additional Reading:
• “W ave Equation” on Wikipedia.
• “Acoustic W ave Equation” on Wikipedia.",Electromagnetics_Vol1.pdf
"1.4. GUIDED AND UNGUIDED W A VES 9
1.4 Guided and Unguided W aves
[m0040]
Broadly speaking, waves may be either guided or
unguided.
Unguided waves include those that are radiated by
antennas, as well as those that are unintentionally
radiated. Once initiated, these waves propagate in an
uncontrolled manner until they are redirected by
scattering or dissipated by losses associated with
materials.
Examples of guided waves are those that exist within
structures such as transmission lines, waveguides, and
optical fibers. W e refer to these as guided because
they are constrained to follow the path defined by the
structure.
Antennas and unintentional radiators emit un-
guided waves . Transmission lines, waveguides,
and optical fibers carry guided waves .
1.5 Phasors
[m0033]
In many areas of engineering, signals are
well-modeled as sinusoids. Also, devices that process
these signals are often well-modeled as linear
time-invariant (L TI) systems. The response of an L TI",Electromagnetics_Vol1.pdf
"these signals are often well-modeled as linear
time-invariant (L TI) systems. The response of an L TI
system to any linear combination of sinusoids is
another linear combination of sinusoids having the
same frequencies. 6 In other words, (1) sinusoidal
signals processed by L TI systems remain sinusoids
and are not somehow transformed into square waves
or some other waveform; and (2) we may calculate
the response of the system for one sinusoid at a time,
and then add the results to find the response of the
system when multiple sinusoids are applied
simultaneously . This property of L TI systems is
known as superposition.
The analysis of systems that process sinusoidal
waveforms is greatly simplified when the sinusoids
are represented as phasors. Here is the key idea:
A phasor is a complex-valued number that repre-
sents a real-valued sinusoidal waveform. Specif-
ically , a phasor has the magnitude and phase of
the sinusoid it represents.
Figures 1.4 and 1.5 show some examples of phasors",Electromagnetics_Vol1.pdf
"ically , a phasor has the magnitude and phase of
the sinusoid it represents.
Figures 1.4 and 1.5 show some examples of phasors
and the associated sinusoids.
It is important to note that a phasor by itself is not the
signal. A phasor is merely a simplified mathematical
representation in which the actual, real-valued
physical signal is represented as a complex-valued
constant.
Here is a completely general form for a physical
(hence, real-valued) quantity varying sinusoidally
with angular frequency ω= 2 πf:
A(t; ω) = Am(ω) cos (ωt+ ψ(ω)) (1.7)
where Am(ω) is magnitude at the specified frequency ,
ψ(ω) is phase at the specified frequency , and tis time.
Also, we require ∂Am/∂t = 0 ; that is, that the time
6 A “linear combination” of functions fi(t) where i = 1, 2, 3, ...
is ∑
i aifi(t) where the ai’s are constants.",Electromagnetics_Vol1.pdf
"10 CHAPTER 1. PRELIMINAR Y CONCEPTS
Ame
j

magnitude
Im{C}
Re{C}
Am
phase ( )
-jAm
Am-jBm
Bm
Figure 1.4: Examples of phasors, displayed here as
points in the real-imaginary plane.
variation of A(t) is completely represented by the
cosine function alone. Now we can equivalently
express A(t; ω) as a phasor C(ω):
C(ω) = Am(ω)ejψ(ω) (1.8)
T o convert this phasor back to the physical signal it
represents, we (1) restore the time dependence by
multiplying by ejωt, and then (2) take the real part of
the result. In mathematical notation:
A(t; ω) = Re
{
C(ω)ejωt}
(1.9)
T o see why this works, simply substitute the right
hand side of Equation 1.8 into Equation 1.9. Then:
A(t) = Re
{
Am(ω)ejψ(ω)ejωt
}
= Re
{
Am(ω)ej(ωt+ψ(ω))
}
= Re {Am(ω) [cos (ωt+ ψ(ω))
+jsin (ωt+ ψ(ω))]}
= Am(ω) cos (ωt+ ψ(ω))
as expected.
It is common to write Equation 1.8 as follows,
dropping the explicit indication of frequency
t
t
t
Am
Ame
jψ
Am-jBm
t
-jAm
note slightly larger amplitude
note now a sine function",Electromagnetics_Vol1.pdf
"dropping the explicit indication of frequency
t
t
t
Am
Ame
jψ
Am-jBm
t
-jAm
note slightly larger amplitude
note now a sine function
Figure 1.5: Sinusoids corresponding to the phasors
shown in Figure 1.4.",Electromagnetics_Vol1.pdf
"1.5. PHASORS 11
dependence:
C = Amejψ (1.10)
This does not normally cause any confusion since the
definition of a phasor requires that values of Cand ψ
are those that apply at whatever frequency is
represented by the suppressed sinusoidal dependence
ejωt.
T able 1.4 shows mathematical representations of the
same phasors demonstrated in Figure 1.4 (and their
associated sinusoidal waveforms in Figure 1.5). It is a
good exercise is to confirm each row in the table,
transforming from left to right and vice-versa.
It is not necessary to use a phasor to represent a
sinusoidal signal. W e choose to do so because phasor
representation leads to dramatic simplifications. For
example:
• Calculation of the peak value from data
representing A(t; ω) requires a time-domain
search over one period of the sinusoid. However,
if you know C, the peak value of A(t) is simply
|C|, and no search is required.
• Calculation of ψfrom data representing A(t; ω)
requires correlation (essentially , integration)",Electromagnetics_Vol1.pdf
"|C|, and no search is required.
• Calculation of ψfrom data representing A(t; ω)
requires correlation (essentially , integration)
over one period of the sinusoid. However, if you
know C, then ψis simply the phase of C, and no
integration is required.
Furthermore, mathematical operations applied to
A(t; ω) can be equivalently performed as operations
on C, and the latter are typically much easier than the
former. T o demonstrate this, we first make two
important claims and show that they are true.
Claim 1 : Let C1 and C2 be two complex-valued
constants (independent of t). Also,
Re
{
C1ejωt}
= Re
{
C2ejωt}
for all t. Then,
C1 = C2.
Proof: Evaluating at t= 0 we find
Re {C1} = Re {C2}. Since C1 and C2 are constant
with respect to time, this must be true for all t. At
t= π/(2ω) we find
Re
{
C1ejωt}
= Re {C1 · j} = −Im {C1}
and similarly
Re
{
C2ejωt}
= Re {C2 · j} = −Im {C2}
therefore Im {C1} = Im {C2}. Once again: Since C1
and C2 are constant with respect to time, this must be",Electromagnetics_Vol1.pdf
"Re
{
C2ejωt}
= Re {C2 · j} = −Im {C2}
therefore Im {C1} = Im {C2}. Once again: Since C1
and C2 are constant with respect to time, this must be
true for all t. Since the real and imaginary parts of C1
and C2 are equal, C1 = C2.
What does this mean? W e have just shown that if two
phasors are equal, then the sinusoidal waveforms that
they represent are also equal.
Claim 2 : For any real-valued linear operator T and
complex-valued quantity C,
T (Re {C}) = Re {T (C)}.
Proof: Let C = cr + jci where cr and ci are
real-valued quantities, and evaluate the right side of
the equation:
Re {T (C)} = Re {T (cr + jci)}
= Re {T (cr) + jT (ci)}
= T (cr)
= T (Re {C})
What does this mean? The operators that we have in
mind for T include addition, multiplication by a
constant, differentiation, integration, and so on.
Here’s an example with differentiation:
Re
{ ∂
∂ωC
}
= Re
{ ∂
∂ω(cr + jci)
}
= ∂
∂ωcr
∂
∂ωRe {C} = ∂
∂ωRe {(cr + jci)} = ∂
∂ωcr
In other words, differentiation of a sinusoidal signal",Electromagnetics_Vol1.pdf
"Re
{ ∂
∂ωC
}
= Re
{ ∂
∂ω(cr + jci)
}
= ∂
∂ωcr
∂
∂ωRe {C} = ∂
∂ωRe {(cr + jci)} = ∂
∂ωcr
In other words, differentiation of a sinusoidal signal
can be accomplished by differentiating the associated
phasor, so there is no need to transform a phasor back
into its associated real-valued signal in order to
perform this operation.",Electromagnetics_Vol1.pdf
"12 CHAPTER 1. PRELIMINAR Y CONCEPTS
A(t) C
Amcos (ωt) Am
Amcos (ωt+ ψ) Amejψ
Amsin (ωt) = Amcos
(
ωt− π
2
)
−jAm
Amcos (ωt) + Bmsin (ωt) = Amcos (ωt) + Bmcos
(
ωt− π
2
)
Am − jBm
T able 1.4: Some examples of physical (real-valued) sinusoi dal signals and the corresponding phasors. Am and
Bm are real-valued and constant with respect to t.
Summarizing:
Claims 1 and 2 together entitle us to perform op-
erations on phasors as surrogates for the physi-
cal, real-valued, sinusoidal waveforms they rep-
resent. Once we are done, we can transform the
resulting phasor back into the physical waveform
it represents using Equation 1.9, if desired.
However, a final transformation back to the time
domain is usually not desired, since the phasor tells us
everything we can know about the corresponding
sinusoid.
A skeptical student might question the value of phasor
analysis on the basis that signals of practical interest
are sometimes not sinusoidally-varying, and therefore",Electromagnetics_Vol1.pdf
"analysis on the basis that signals of practical interest
are sometimes not sinusoidally-varying, and therefore
phasor analysis seems not to apply generally . It is
certainly true that many signals of practical interest
are not sinusoidal, and many are far from it.
Nevertheless, phasor analysis is broadly applicable.
There are basically two reasons why this is so:
• Many signals, although not strictly sinusoidal,
are “narrowband” and therefore well-modeled as
sinusoidal. For example, a cellular
telecommunications signal might have a
bandwidth on the order of 10 MHz and a center
frequency of about 2 GHz. This means the
difference in frequency between the band edges
of this signal is just 0.5% of the center
frequency . The frequency response associated
with signal propagation or with hardware can
often be assumed to be constant over this range
of frequencies. With some caveats, doing phasor
analysis at the center frequency and assuming
the results apply equally well over the bandwidth",Electromagnetics_Vol1.pdf
"of frequencies. With some caveats, doing phasor
analysis at the center frequency and assuming
the results apply equally well over the bandwidth
of interest is often a pretty good approximation.
• It turns out that phasor analysis is easily
extensible to any physical signal, regardless of
bandwidth. This is so because any physical
signal can be decomposed into a linear
combination of sinusoids – this is known as
F ourier analysis. The way to find this linear
combination of sinusoids is by computing the
F ourier series, if the signal is periodic, or the
F ourier Transform, otherwise. Phasor analysis
applies to each frequency independently , and
(invoking superposition) the results can be added
together to obtain the result for the complete
signal. The process of combining results after
phasor analysis results is nothing more than
integration over frequency; i.e.:
∫ +∞
−∞
A(t; ω)dω
Using Equation 1.9, this can be rewritten:
∫ +∞
−∞
Re
{
C(ω)ejωt}
dω",Electromagnetics_Vol1.pdf
"integration over frequency; i.e.:
∫ +∞
−∞
A(t; ω)dω
Using Equation 1.9, this can be rewritten:
∫ +∞
−∞
Re
{
C(ω)ejωt}
dω
W e can go one step further using Claim 2:
Re
{∫ +∞
−∞
C(ω)ejωtdω
}
The quantity in the curly braces is simply the
Fourier transform of C(ω). Thus, we see that we
can analyze a signal of arbitrarily-large
bandwidth simply by keeping ωas an
independent variable while we are doing phasor
analysis, and if we ever need the physical signal,
we just take the real part of the Fourier transform
of the phasor. So not only is it possible to
analyze any time-domain signal using phasor
analysis, it is also often far easier than doing the
same analysis on the time-domain signal directly .",Electromagnetics_Vol1.pdf
"1.6. UNITS 13
Summarizing:
Phasor analysis does not limit us to sinusoidal
waveforms. Phasor analysis is not only applica-
ble to sinusoids and signals that are sufficiently
narrowband, but is also applicable to signals of
arbitrary bandwidth via Fourier analysis.
Additional Reading:
• “Phasor” on Wikipedia.
• “Fourier analysis” on Wikipedia.
1.6 Units
[m0072]
The term “unit” refers to the measure used to express
a physical quantity . For example, the mean radius of
the Earth is about 6,371,000 meters; in this case the
unit is the meter.
A number like “6,371,000” becomes a bit
cumbersome to write, so it is common to use a prefix
to modify the unit. For example, the radius of the
Earth is more commonly said to be 6371 kilometers,
where one kilometer is understood to mean
1000 meters. It is common practice to use prefixes,
such as “kilo-, ” that yield values in the range 0.001 to
10,000. A list of standard prefixes appears in
T able 1.5.
Writing out the names of units can also become",Electromagnetics_Vol1.pdf
"10,000. A list of standard prefixes appears in
T able 1.5.
Writing out the names of units can also become
tedious. For this reason, it is common to use standard
abbreviations; e.g., “6731 km” as opposed to
“6371 kilometers, ” where “k” is the standard
abbreviation for the prefix “kilo” and “m” is the
standard abbreviation for “meter. ” A list of
commonly-used base units and their abbreviations are
shown in T able 1.6.
T o avoid ambiguity , it is important to always indicate
the units of a quantity; e.g., writing “6371 km” as
opposed to “6371. ” Failure to do so is a common
source of error and misunderstandings. An example is
Prefix Abbreviation Multiply by:
exa E 1018
peta P 1015
tera T 1012
giga G 109
mega M 106
kilo k 103
milli m 10−3
micro µ 10−6
nano n 10−9
pico p 10−12
femto f 10−15
atto a 10−18
T able 1.5: Prefixes used to modify units.",Electromagnetics_Vol1.pdf
"14 CHAPTER 1. PRELIMINAR Y CONCEPTS
Unit Abbreviation Quantifies:
ampere A electric current
coulomb C electric charge
farad F capacitance
henry H inductance
hertz Hz frequency
joule J energy
meter m distance
newton N force
ohm Ω resistance
second s time
tesla T magnetic flux density
volt V electric potential
watt W power
weber Wb magnetic flux
T able 1.6: Some units that are commonly used in elec-
tromagnetics.
the expression:
l= 3 t
where lis length and tis time. It could be that lis in
meters and tis in seconds, in which case “ 3” really
means “3 m/s”. However, if it is intended that lis in
kilometers and tis in hours, then “ 3” really means
“3 km/h” and the equation is literally different. T o
patch this up, one might write “ l= 3 tm/s”; however,
note that this does does not resolve the ambiguity we
just identified – i.e., we still don’t know the units of
the constant “3. ” Alternatively , one might write
“ l= 3 twhere lis in meters and tis in seconds, ”",Electromagnetics_Vol1.pdf
"the constant “3. ” Alternatively , one might write
“ l= 3 twhere lis in meters and tis in seconds, ”
which is unambiguous but becomes quite awkward
for more complicated expressions. A better solution is
to write instead:
l= (3 m/s) t
or even better:
l= at where a= 3 m/s
since this separates this issue of units from the
perhaps more-important fact that lis proportional to t
and the constant of proportionality ( a) is known.
The meter is the fundamental unit of length in the
International System of Units, known by its French
acronym “SI” and sometimes informally referred to
as the “metric system. ”
In this work, we will use SI units exclusively .
Although SI is probably the most popular for
engineering use overall, other systems remain in
common use. For example, the English system, where
the radius of the Earth might alternatively be said to
be about 3959 miles, continues to be used in various
applications and to a lesser or greater extent in",Electromagnetics_Vol1.pdf
"be about 3959 miles, continues to be used in various
applications and to a lesser or greater extent in
various regions of the world. An alternative system in
common use in physics and material science
applications is the CGS (“centimeter-gram-second”)
system. The CGS system is similar to SI, but with
some significant differences. For example, the base
unit of energy is the CGS system is not the “joule”
but rather the “erg, ” and the values of some physical
constants become unitless. Therefore – once again –
it is very important to include units whenever values
are stated.
SI defines seven fundamental units from which all
other units can be derived. These fundamental units
are distance in meters (m), time in seconds (s),
current in amperes (A), mass in kilograms (kg),
temperature in kelvin (K), particle count in moles
(mol), and luminosity in candela (cd). SI units for
electromagnetic quantities such as coulombs (C) for
charge and volts (V) for electric potential are derived",Electromagnetics_Vol1.pdf
"electromagnetic quantities such as coulombs (C) for
charge and volts (V) for electric potential are derived
from these fundamental units.
A frequently-overlooked feature of units is their
ability to assist in error-checking mathematical
expressions. For example, the electric field intensity
may be specified in volts per meter (V/m), so an
expression for the electric field intensity that yields
units of V/m is said to be “dimensionally correct” (but
not necessarily correct), whereas an expression that
cannot be reduced to units of V/m cannot be correct.
Additional Reading:
• “International System of Units” on Wikipedia.
• “Centimetre-gram-second system of units” on
Wikipedia.",Electromagnetics_Vol1.pdf
"1.7. NOT A TION 15
1.7 Notation
[m0005]
The list below describes notation used in this book.
• V ectors: Boldface is used to indicate a vector;
e.g., the electric field intensity vector will
typically appear as E. Quantities not in boldface
are scalars. When writing by hand, it is common
to write “
E” or “ − →E” in lieu of “ E. ”
• Unit vectors : A circumflex is used to indicate a
unit vector; i.e., a vector having magnitude equal
to one. For example, the unit vector pointing in
the +xdirection will be indicated as ˆx. In
discussion, the quantity “ ˆx” is typically spoken
“ xhat. ”
• Time: The symbol tis used to indicate time.
• P osition: The symbols (x,y,z ), (ρ,φ,z ), and
(r,θ,φ ) indicate positions using the Cartesian,
cylindrical, and polar coordinate systems,
respectively . It is sometimes convenient to
express position in a manner which is
independent of a coordinate system; in this case,
we typically use the symbol r. For example,
r = ˆxx+ ˆyy+ ˆzzin the Cartesian coordinate",Electromagnetics_Vol1.pdf
"independent of a coordinate system; in this case,
we typically use the symbol r. For example,
r = ˆxx+ ˆyy+ ˆzzin the Cartesian coordinate
system.
• Phasors: A tilde is used to indicate a phasor
quantity; e.g., a voltage phasor might be
indicated as ˜V, and the phasor representation of
E will be indicated as ˜E.
• Curves, surfaces, and volumes: These
geometrical entities will usually be indicated in
script; e.g., an open surface might be indicated
as S and the curve bounding this surface might
be indicated as C. Similarly , the volume enclosed
by a closed surface S may be indicated as V.
• Integrations over curves, surfaces, and volumes
will usually be indicated using a single integral
sign with the appropriate subscript. For example:
∫
C
· · · dl is an integral over the curve C
∫
S
· · · ds is an integral over the surface S
∫
V
· · · dv is an integral over the volume V.
• Integrations over closed
curves and surfaces will
be indicated using a circle superimposed on the",Electromagnetics_Vol1.pdf
"∫
V
· · · dv is an integral over the volume V.
• Integrations over closed
curves and surfaces will
be indicated using a circle superimposed on the
integral sign. For example:
∮
C
· · · dl is an integral over the closed curve C
∮
S
··· ds is an integral over the closed surface S
A “closed curve” is one which forms an
unbroken loop; e.g., a circle. A “closed surface”
is one which encloses a volume with no
openings; e.g., a sphere.
• The symbol “ ∼
=” means “approximately equal
to. ” This symbol is used when equality exists,
but is not being expressed with exact numerical
precision. For example, the ratio of the
circumference of a circle to its diameter is π,
where π∼
= 3.14.
• The symbol “ ≈” also indicates “approximately
equal to, ” but in this case the two quantities are
unequal even if expressed with exact numerical
precision. For example, ex = 1 + x+ x2/2 + ...
as a infinite series, but ex ≈ 1 + xfor x≪ 1.
Using this approximation e0.1 ≈ 1.1, which is in",Electromagnetics_Vol1.pdf
"precision. For example, ex = 1 + x+ x2/2 + ...
as a infinite series, but ex ≈ 1 + xfor x≪ 1.
Using this approximation e0.1 ≈ 1.1, which is in
good agreement with the actual value
e0.1 ∼
= 1.1052.
• The symbol “ ∼” indicates “on the order of, ”
which is a relatively weak statement of equality
indicating that the indicated quantity is within a
factor of 10 or so the indicated value. For
example, µ∼ 105 for a class of iron alloys, with
exact values being being larger or smaller by a
factor of 5 or so.
• The symbol “ ≜ ” means “is defined as” or “is
equal as the result of a definition. ”
• Complex numbers : j ≜ √
−1.
• See Appendix C for notation used to identify
commonly-used physical constants.
[m0026]",Electromagnetics_Vol1.pdf
"16 CHAPTER 1. PRELIMINAR Y CONCEPTS
Image Credits
Fig. 1.1 c⃝ V . Blacus, https://commons.wikimedia.org/wiki/File:El ectromagnetic-Spectrum.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/).
Fig. 1.2: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0074 fClap.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 1.3: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0074 fTrumpet.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).",Electromagnetics_Vol1.pdf
"Chapter 2
Electric and Magnetic Fields
2.1 What is a Field?
[m0001]
A field is the continuum of values of a quantity
as a function of position and time.
The quantity that the field describes may be a scalar
or a vector, and the scalar part may be either real- or
complex-valued.
In electromagnetics, the electric field intensity E is a
real-valued vector field that may vary as a function of
position and time, and so might be indicated as
“ E(x,y,z,t ), ” “ E(r,t), ” or simply “ E. ” When
expressed as a phasor, this quantity is complex-valued
but exhibits no time dependence, so we might say
instead “ ˜E(r)” or simply “ ˜E. ”
An example of a scalar field in electromagnetics is the
electric potential, V; i.e., V(r,t).
A wave is a time-varying field that continues to exist
in the absence of the source that created it and is
therefore able to transport energy .
2.2 Electric Field Intensity
[m0002]
Electric field intensity is a vector field we assign the",Electromagnetics_Vol1.pdf
"therefore able to transport energy .
2.2 Electric Field Intensity
[m0002]
Electric field intensity is a vector field we assign the
symbol E and has units of electrical potential per
distance; in SI units, volts per meter (V/m). Before
offering a formal definition, it is useful to consider the
broader concept of the electric field .
Imagine that the universe is empty except for a single
particle of positive charge. Next, imagine that a
second positively-charged particle appears; the
situation is now as shown in Figure 2.1. Since like
charges repel, the second particle will be repelled by
the first particle and vice versa. Specifically , the first
particle is exerting force on the second particle. If the
second particle is free to move, it will do so; this is an
expression of kinetic energy . If the second particle is
somehow held in place, we say the second particle
possesses an equal amount of potential energy . This
potential energy is no less “real, ” since we can convert",Electromagnetics_Vol1.pdf
"possesses an equal amount of potential energy . This
potential energy is no less “real, ” since we can convert
it to kinetic energy simply by releasing the particle,
thereby allowing it to move.
Now let us revisit the original one particle scenario.
F
c⃝ M. Goldammer CC BY SA 4.0
Figure 2.1: A positively-charged particle experiences
a repulsive force F in the presence of another particle
which is also positively-charged.
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"18 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS
F
c⃝ M. Goldammer CC BY SA 4.0
Figure 2.2: A map of the force that would be expe-
rienced by a second particle having a positive charge.
Here, the magnitude and direction of the force is indi-
cated by the size and direction of the arrow .
In that scenario, we could make a map in which every
position in space is assigned a vector that describes
the force that a particle having a specified charge q
would experience if it were to appear there. The result
looks something like Figure 2.2. This map of force
vectors is essentially a description of the electric field
associated with the first particle.
There are many ways in which the electric field may
be quantified. Electric field intensity E is simply one
of these ways. W e define E(r) to be the force F(r)
experienced by a test particle having charge q, divided
by q; i.e.,
E(r) ≜ lim
q→0
F(r)
q (2.1)
Note that it is required for the charge to become
vanishingly small (as indicated by taking the limit) in",Electromagnetics_Vol1.pdf
"by q; i.e.,
E(r) ≜ lim
q→0
F(r)
q (2.1)
Note that it is required for the charge to become
vanishingly small (as indicated by taking the limit) in
order for this definition to work. This is because the
source of the electric field is charge, so the test
particle contributes to the total electric field. T o
accurately measure the field of interest, the test charge
must be small enough not to significantly perturb the
field. This makes Equation 2.1 awkward from an
engineering perspective, and we’ll address that later
in this section.
According the definition of Equation 2.1, the units of
E are those of force divided by charge. The SI units
for force and charge are the newton (N) and coulomb
(C) respectively , so E has units of N/C. However, we
typically express E in units of V/m, not N/C. What’s
going on? The short answer is that 1 V/m = 1 N/C:
N
C = N · m
C · m = J
C · m = V
m
where we have used the fact that 1 N ·m = 1 joule (J)
of energy and 1 J/C = 1 V .",Electromagnetics_Vol1.pdf
"N
C = N · m
C · m = J
C · m = V
m
where we have used the fact that 1 N ·m = 1 joule (J)
of energy and 1 J/C = 1 V .
Electric field intensity ( E, N/C or V/m) is a vec-
tor field that quantifies the force experienced by
a charged particle due to the influence of charge
not associated with that particle.
The analysis of units doesn’t do much to answer the
question of why we should prefer to express E in V/m
as opposed to N/C. Let us now tackle that question.
Figure 2.3 shows a simple thought experiment that
demonstrates the concept of electric field intensity in
terms of an electric circuit. This circuit consists of a
parallel-plate capacitor in series with a 9 V battery . 1
The effect of the battery , connected as shown, is to
force an accumulation of positive charge on the upper
plate, and an accumulation of negative charge on the
lower plate. If we consider the path from the position
labeled “ A, ” along the wire and through the battery to
the position labeled “B, ” the change in electric",Electromagnetics_Vol1.pdf
"labeled “ A, ” along the wire and through the battery to
the position labeled “B, ” the change in electric
potential is +9 V . It must also be true that the change
in electric potential as we travel from B to A through
the capacitor is −9 V , since the sum of voltages over
any closed loop in a circuit is zero. Said differently ,
the change in electric potential between the plates of
the capacitor, starting from node A and ending at
node B, is +9 V . Now , note that the spacing between
the plates in the capacitor is 1 mm. Thus, the rate of
change of the potential between the plates is 9 V
divided by 1 mm, which is 9000 V/m. This is literally
the electric field intensity between the plates. That is,
if one places a particle with an infinitesimally-small
charge between the plates (point “C”), and then
measures the ratio of force to charge, one finds it is
9000 N/C pointing toward A. W e come to the
following remarkable conclusion:",Electromagnetics_Vol1.pdf
"measures the ratio of force to charge, one finds it is
9000 N/C pointing toward A. W e come to the
following remarkable conclusion:
1 It is not necessary to know anything about capacitors to get t o
the point, so no worries!",Electromagnetics_Vol1.pdf
"2.2. ELECTRIC FIELD INTENSITY 19
A
B
+
-
9V
1mm
c⃝ Y . Qin CC BY 3.0
Figure 2.3: A simple circuit used to describe the con-
cept of electric field intensity . In this example, E at
point C is 9000 V/m directed from B toward A.
E points in the direction in which electric poten-
tial is most rapidly decreasing, and the magnitude
of E is the rate of change in electric potential with
distance in this direction.
The reader may have noticed that we have defined the
electric field in terms of what it does. W e have have
not directly addressed the question of what the
electric field is. This is the best we can do using
classical physics, and fortunately , this is completely
adequate for the most engineering applications.
However, a deeper understanding is possible using
quantum mechanics, where we find that the electric
field and the magnetic field are in fact manifestations
of the same fundamental force, aptly named the
electromagnetic force . (In fact, the electromagnetic",Electromagnetics_Vol1.pdf
"of the same fundamental force, aptly named the
electromagnetic force . (In fact, the electromagnetic
force is found to be one of just four fundamental
forces, the others being gravity, the strong nuclear
force, and the weak nuclear force .) Quantum
mechanics also facilitates greater insight into the
nature of electric charge and of the photon, which is
the fundamental constituent of electromagnetic
waves. For more information on this topic, an
excellent starting point is the video “Quantum
Invariance & The Origin of The Standard Model”
referenced at the end of this section.
Additional Reading:
• “Electric field” on Wikipedia.
• PBS Space Time video “Quantum Invariance &
The Origin of The Standard Model, ” on
Y ouTube.",Electromagnetics_Vol1.pdf
"20 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS
2.3 Permittivity
[m0008]
Permittivity describes the effect of material in
determining the electric field in response to electric
charge. For example, one can observe from laboratory
experiments that a particle having charge qgives rise
to the electric field
E = ˆR q 1
4πR2
1
ǫ (2.2)
where Ris distance from the charge, ˆR is a unit
vector pointing away from the charge, and ǫis a
constant that depends on the material. Note that E
increases with q, which makes sense since electric
charge is the source of E. Also note that E is
inversely proportional to 4πR2, indicating that E
decreases in proportion to the area of a sphere
surrounding the charge – a principle commonly
known as the inverse square law . The remaining
factor 1/ǫis the constant of proportionality , which
captures the effect of material. Given units of V/m for
E and C for Q, we find that ǫmust have units of
farads per meter (F/m). (T o see this, note that 1 F =
1 C/V .)",Electromagnetics_Vol1.pdf
"E and C for Q, we find that ǫmust have units of
farads per meter (F/m). (T o see this, note that 1 F =
1 C/V .)
Permittivity ( ǫ, F/m) describes the effect of ma-
terial in determining the electric field intensity in
response to charge.
In free space (that is, a perfect vacuum), we find that
ǫ= ǫ0 where:
ǫ0 ∼
= 8.854 × 10−12 F/m (2.3)
The permittivity of air is only slightly greater, and
usually can be assumed to be equal to that of free
space. In most other materials, the permittivity is
significantly greater; that is, the same charge results
in a weaker electric field intensity .
It is common practice to describe the permittivity of
materials relative to the permittivity of free space.
This relative permittivity is given by:
ǫr ≜ ǫ
ǫ0
(2.4)
V alues of ǫr for a few representative materials is
given in Appendix A.1. Note that ǫr ranges from 1
(corresponding to a perfect vacuum) to about 60 or so
in common engineering applications. Also note that",Electromagnetics_Vol1.pdf
"(corresponding to a perfect vacuum) to about 60 or so
in common engineering applications. Also note that
relative permittivity is sometimes referred to as
dielectric constant . This term is a bit misleading,
however, since permittivity is a meaningful concept
for many materials that are not dielectrics.
Additional Reading:
• “Permittivity” on Wikipedia.
• Appendix A.1 (“Permittivity of Some Common
Materials”).
• “Inverse square law” on Wikipedia.",Electromagnetics_Vol1.pdf
"2.4. ELECTRIC FLUX DENSITY 21
2.4 Electric Flux Density
[m0011]
Electric flux density , assigned the symbol D, is an
alternative to electric field intensity ( E) as a way to
quantify an electric field. This alternative description
offers some actionable insight, as we shall point out at
the end of this section.
First, what is electric flux density? Recall that a
particle having charge qgives rise to the electric field
intensity
E = ˆR q 1
4πR2
1
ǫ (2.5)
where Ris distance from the charge and ˆR points
away from the charge. Note that E is inversely
proportional to 4πR2, indicating that E decreases in
proportion to the area of a sphere surrounding the
charge. Now integrate both sides of Equation 2.5 over
a sphere S of radius R:
∮
S
E(r) · ds =
∮
S
[
ˆR q 1
4πR2
1
ǫ
]
· ds (2.6)
Factoring out constants that do not vary with the
variables of integration, the right-hand side becomes:
q 1
4πR2
1
ǫ
∮
S
ˆR · ds
Note that ds = ˆRdsin this case, and also that",Electromagnetics_Vol1.pdf
"variables of integration, the right-hand side becomes:
q 1
4πR2
1
ǫ
∮
S
ˆR · ds
Note that ds = ˆRdsin this case, and also that
ˆR · ˆR = 1 . Thus, the right-hand side simplifies to:
q 1
4πR2
1
ǫ
∮
S
ds
The remaining integral is simply the area of S, which
is 4πR2. Thus, we find:
∮
S
E(r) · ds = q
ǫ (2.7)
The integral of a vector field over a specified surface
is known as flux (see “ Additional Reading” at the end
of this section). Thus, we have found that the flux of
E through the sphere S is equal to a constant, namely
q/ǫ. Furthermore, this constant is the same regardless
of the radius Rof the sphere. Said differently , the flux
of E is constant with distance, and does not vary as E
itself does. Let us capitalize on this observation by
making the following small modification to
Equation 2.7: ∮
S
[ǫE] · ds = q (2.8)
In other words, the flux of the quantity ǫE is equal to
the enclosed charge, regardless of the radius of the
sphere over which we are doing the calculation. This",Electromagnetics_Vol1.pdf
"the enclosed charge, regardless of the radius of the
sphere over which we are doing the calculation. This
leads to the following definition:
The electric flux density D = ǫE, having units
of C/m 2, is a description of the electric field in
terms of flux, as opposed to force or change in
electric potential.
It may appear that D is redundant information given
E and ǫ, but this is true only in homogeneous media.
The concept of electric flux density becomes
important – and decidedly not redundant – when we
encounter boundaries between media having different
permittivities. As we shall see in Section 5.18,
boundary conditions on D constrain the component
of the electric field that is perpendicular to the
boundary separating two regions. If one ignores the
“flux” character of the electric field represented by D
and instead considers only E, then only the tangential
component of the electric field is constrained. In fact,
when one of the two materials comprising the",Electromagnetics_Vol1.pdf
"component of the electric field is constrained. In fact,
when one of the two materials comprising the
boundary between two material regions is a perfect
conductor, then the electric field is completely
determined by the boundary condition on D. This
greatly simplifies the problem of finding the electric
field in a region bounded or partially bounded by
materials that can be modeled as perfect conductors,
including many metals. In particular, this principle
makes it easy to analyze capacitors.
W e conclude this section with a warning. Even
though the SI units for D are C/m 2, D describes an
electric field and not a surface charge density . It is
certainly true that one may describe the amount of
charge distributed over a surface using units of C/m 2.
However, D is not necessarily a description of actual
charge, and there is no implication that the source of
the electric field is a distribution of surface charge.
On the other hand, it is true that D can be interpreted",Electromagnetics_Vol1.pdf
"the electric field is a distribution of surface charge.
On the other hand, it is true that D can be interpreted
as an equivalent surface charge density that would
give rise to the observed electric field, and in some
cases, this equivalent charge density turns out to be
the actual charge density .",Electromagnetics_Vol1.pdf
"22 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS
Additional Reading:
• “Flux” on Wikipedia.
2.5 Magnetic Flux Density
[m0003]
Magnetic flux density is a vector field which we
identify using the symbol B and which has SI units of
tesla (T). Before offering a formal definition, it is
useful to consider the broader concept of the magnetic
field .
Magnetic fields are an intrinsic property of some
materials, most notably permanent magnets. The
basic phenomenon is probably familiar, and is shown
in Figure 2.4. A bar magnet has “poles” identified as
“N” (“north”) and “S” (“south”). The N-end of one
magnet attracts the S-end of another magnet but
repels the N-end of the other magnet and so on. The
existence of a vector field is apparent since the
observed force acts at a distance and is asserted in a
specific direction. In the case of a permanent magnet,
the magnetic field arises from mechanisms occurring
at the scale of the atoms and electrons comprising the",Electromagnetics_Vol1.pdf
"the magnetic field arises from mechanisms occurring
at the scale of the atoms and electrons comprising the
material. These mechanisms require some additional
explanation which we shall defer for now .
Magnetic fields also appear in the presence of current.
For example, a coil of wire bearing a current is found
to influence permanent magnets (and vice versa) in
the same way that permanent magnets affect each
other. This is shown in Figure 2.5. From this, we infer
that the underlying mechanism is the same – i.e., the
vector field generated by a current-bearing coil is the
same phenomenon as the vector field associated with
a permanent magnet. Whatever the source, we are
S
N
c⃝ Y . Qin CC BY 3.0
Figure 2.4: Evidence of a vector field from observa-
tions of the force perceived by the bar magnets on the
right in the presence of the bar magnets on the left.",Electromagnetics_Vol1.pdf
"2.5. MAGNETIC FLUX DENSITY 23
c⃝ Y . Qin CC BY 4.0
Figure 2.5: Evidence that current can also create a
magnetic field.
now interested in quantifying its behavior.
T o begin, let us consider the effect of a magnetic field
on a electrically-charged particle. First, imagine a
region of free space with no electric or magnetic
fields. Next, imagine that a charged particle appears.
This particle will experience no force. Next, a
magnetic field appears; perhaps this is due to a
permanent magnet or a current in the vicinity . This
situation is shown in Figure 2.6 (top). Still, no force is
applied to the particle. In fact, nothing happens until
the particle in set in motion. Figure 2.6 (bottom)
shows an example. Suddenly , the particle perceives a
force. W e’ll get to the details about direction and
magnitude in a moment, but the main idea is now
evident. A magnetic field is something that applies a
force to a charged particle in motion, distinct from (in",Electromagnetics_Vol1.pdf
"evident. A magnetic field is something that applies a
force to a charged particle in motion, distinct from (in
fact, in addition to) the force associated with an
electric field.
Now , it is worth noting that a single charged particle
in motion is the simplest form of current. Remember
also that motion is required for the magnetic field to
influence the particle. Therefore, not only is current
the source of the magnetic field, the magnetic field
also exerts a force on current. Summarizing:
B
B
v > 
F
 = 
F 
c⃝ Y . Qin CC BY 4.0
Figure 2.6: The force perceived by a charged particle
that is ( top) motionless and ( bottom) moving with ve-
locity v = ˆzv, which is perpendicular to the plane of
this document and toward the reader.
The magnetic field describes the force exerted on
permanent magnets and currents in the presence
of other permanent magnets and currents.
So, how can we quantify a magnetic field? The
answer from classical physics involves another",Electromagnetics_Vol1.pdf
"of other permanent magnets and currents.
So, how can we quantify a magnetic field? The
answer from classical physics involves another
experimentally-derived equation that predicts force as
a function of charge, velocity , and a vector field B
representing the magnetic field. Here it is: The force
applied to a particle bearing charge qis
F = qv × B (2.9)
where v is the velocity of the particle and “ ×”
denotes the cross product. The cross product of two
vectors is in a direction perpendicular to each of the
two vectors, so the force exerted by the magnetic field
is perpendicular to both the direction of motion and
the direction in which the magnetic field points.
The reader would be well-justified in wondering why
the force exerted by the magnetic field should
perpendicular to B. For that matter, why should the
force depend on v? These are questions for which
classical physics provides no obvious answers.
Effective answers to these questions require concepts",Electromagnetics_Vol1.pdf
"force depend on v? These are questions for which
classical physics provides no obvious answers.
Effective answers to these questions require concepts
from quantum mechanics, where we find that the
magnetic field is a manifestation of the fundamental",Electromagnetics_Vol1.pdf
"24 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS
and aptly-named electromagnetic force . The
electromagnetic force also gives rises to the electric
field, and it is only limited intuition, grounded in
classical physics, that leads us to perceive the electric
and magnetic fields as distinct phenomena. For our
present purposes – and for most
commonly-encountered engineering applications –
we do not require these concepts. It is sufficient to
accept this apparent strangeness as fact and proceed
accordingly .
Dimensional analysis of Equation 2.9 reveals that B
has units of (N ·s)/(C·m). In SI, this combination of
units is known as the tesla (T).
W e refer to B as magnetic flux density , and therefore
tesla is a unit of magnetic flux density . A fair question
to ask at this point is: What makes this a flux density?
The short answer is that this terminology is somewhat
arbitrary , and in fact is not even uniformly accepted.
In engineering electromagnetics, the preference for",Electromagnetics_Vol1.pdf
"arbitrary , and in fact is not even uniformly accepted.
In engineering electromagnetics, the preference for
referring to B as a “flux density” is because we
frequently find ourselves integrating B over a
mathematical surface. Any quantity that is obtained
by integration over a surface is referred to as “flux, ”
and so it becomes natural to think of B as a flux
density; i.e., as flux per unit area. The SI unit for
magnetic flux is the weber (Wb). Therefore, B may
alternatively be described as having units of Wb/m 2,
and 1 Wb/m 2 = 1 T .
Magnetic flux density ( B, T or Wb/m 2) is a de-
scription of the magnetic field that can be defined
as the solution to Equation 2.9.
When describing magnetic fields, we occasionally
refer to the concept of a field line , defined as follows:
A magnetic field line is the curve in space traced
out by following the direction in which the mag-
netic field vector points.
This concept is illustrated in Figure 2.7 for a
permanent bar magnet and Figure 2.8 for a",Electromagnetics_Vol1.pdf
"netic field vector points.
This concept is illustrated in Figure 2.7 for a
permanent bar magnet and Figure 2.8 for a
current-bearing coil.
Magnetic field lines are remarkable for the following
reason:
A magnetic field line always forms a closed loop.
c⃝ Y . Qin CC BY 4.0
Figure 2.7: The magnetic field of a bar magnet, illus-
trating field lines.
c⃝ Y . Qin CC BY 4.0
Figure 2.8: The magnetic field of a current-bearing
coil, illustrating field lines.",Electromagnetics_Vol1.pdf
"2.6. PERMEABILITY 25
This is true in a sense even for field lines which seem
to form straight lines (for example, those along the
axis of the bar magnet and the coil in Figures 2.7 and
2.8), since a field line that travels to infinity in one
direction reemerges from infinity in the opposite
direction.
Additional Reading:
• “Magnetic Field” on Wikipedia.
2.6 Permeability
[m0009]
Permeability describes the effect of material in
determining the magnetic flux density . All else being
equal, magnetic flux density increases in proportion to
permeability .
T o illustrate the concept, consider that a particle
bearing charge qmoving at velocity v gives rise to a
magnetic flux density:
B(r) = µ qv
4πR2 × ˆR (2.10)
where ˆR is the unit vector pointing from the charged
particle to the field point r, Ris this distance, and
“ ×” is the cross product. Note that B increases with
charge and speed, which makes sense since moving
charge is the source of the magnetic field. Also note",Electromagnetics_Vol1.pdf
"charge and speed, which makes sense since moving
charge is the source of the magnetic field. Also note
that B is inversely proportional to 4πR2, indicating
that |B| decreases in proportion to the area of a
sphere surrounding the charge, also known as the
inverse square law . The remaining factor, µ, is the
constant of proportionality that captures the effect of
material. W e refer to µas the permeability of the
material. Since B can be expressed in units of Wb/m 2
and the units of v are m/s, we see that µmust have
units of henries per meter (H/m). (T o see this, note
that 1 H ≜ 1 Wb/A.)
Permeability ( µ, H/m) describes the effect of ma-
terial in determining the magnetic flux density .
In free space, we find that the permeability µ= µ0
where:
µ0 = 4 π× 10−7 H/m (2.11)
It is common practice to describe the permeability of
materials in terms of their relative permeability :
µr ≜ µ
µ0
(2.12)
which gives the permeability relative to the minimum",Electromagnetics_Vol1.pdf
"materials in terms of their relative permeability :
µr ≜ µ
µ0
(2.12)
which gives the permeability relative to the minimum
possible value; i.e., that of free space. Relative
permeability for a few representative materials is
given in Appendix A.2.
Note that µr is approximately 1 for all but a small
class of materials. These are known as magnetic",Electromagnetics_Vol1.pdf
"26 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS
materials, and may exhibit values of µr as large as
∼ 106. A commonly-encountered category of
magnetic materials is ferromagnetic material, of
which the best-known example is iron.
Additional Reading:
• “Permeability (electromagnetism)” on
Wikipedia.
• Section 7.16 (“Magnetic Materials”).
• Appendix A.2 (“Permeability of Some Common
Materials”).
2.7 Magnetic Field Intensity
[m0012]
Magnetic field intensity H is an alternative
description of the magnetic field in which the effect of
material is factored out. For example, the magnetic
flux density B (reminder: Section 2.5) due to a point
charge qmoving at velocity v can be written in terms
of the Biot-Savart Law :
B = µ qv
4πR2 × ˆR (2.13)
where ˆR is the unit vector pointing from the charged
particle to the field point r, Ris this distance, “ ×” is
the cross product, and µis the permeability of the
material. W e can rewrite Equation 2.13 as:
B ≜ µH (2.14)
with:
H = qv
4πR2 × ˆR (2.15)",Electromagnetics_Vol1.pdf
"the cross product, and µis the permeability of the
material. W e can rewrite Equation 2.13 as:
B ≜ µH (2.14)
with:
H = qv
4πR2 × ˆR (2.15)
so H in homogeneous media does not depend on µ.
Dimensional analysis of Equation 2.15 reveals that
the units for H are amperes per meter (A/m).
However, H does not represent surface current
density ,2 as the units might suggest. While it is
certainly true that a distribution of current (A) over
some linear cross-section (m) can be described as a
current density having units of A/m, H is associated
with the magnetic field and not a particular current
distribution. Said differently , H can be viewed as a
description of the magnetic field in terms of an
equivalent (but not actual) current.
The magnetic field intensity H (A/m), defined us-
ing Equation 2.14, is a description of the mag-
netic field independent from material properties.
It may appear that H is redundant information given
B and µ, but this is true only in homogeneous media.",Electromagnetics_Vol1.pdf
"It may appear that H is redundant information given
B and µ, but this is true only in homogeneous media.
The concept of magnetic field intensity becomes
important – and decidedly not redundant – when we
encounter boundaries between media having different
permeabilities. As we shall see in Section 7.11,
2 The concept of current density is not essential to understan d this
section; however, a primer can be found in Section 6.2.",Electromagnetics_Vol1.pdf
"2.8. ELECTROMAGNETIC PROPER TIES OF MA TERIALS 27
boundary conditions on H constrain the component
of the magnetic field which is tangent to the boundary
separating two otherwise-homogeneous regions. If
one ignores the characteristics of the magnetic field
represented by H and instead considers only B, then
only the perpendicular component of the magnetic
field is constrained.
The concept of magnetic field intensity also turns out
to be useful in a certain problems in which µis not a
constant, but rather is a function of magnetic field
strength. In this case, the magnetic behavior of the
material is said to be nonlinear. For more on this, see
Section 7.16.
Additional Reading:
• “Magnetic field” on Wikipedia.
• “Biot-Savart law” on Wikipedia.
2.8 Electromagnetic Properties of
Materials
[m0007]
In electromagnetic analysis, one is principally
concerned with three properties of matter. These
properties are quantified in terms of constitutive
parameters, which describe the effect of material in",Electromagnetics_Vol1.pdf
"properties are quantified in terms of constitutive
parameters, which describe the effect of material in
determining an electromagnetic quantity in response
to a source. Here are the three principal constitutive
parameters:
• P ermittivity (ǫ, F/m) quantifies the effect of
matter in determining the electric field in
response to electric charge. Permittivity is
addressed in Section 2.3.
• P ermeability (µ, H/m) quantifies the effect of
matter in determining the magnetic field in
response to current. Permeability is addressed in
Section 2.6.
• Conductivity (σ, S/m) quantifies the effect of
matter in determining the flow of current in
response to an electric field. Conductivity is
addressed in Section 6.3.
The electromagnetic properties of most common
materials in most common applications can be
quantified in terms of the constitutive parameters
ǫ, µ, and σ.
T o keep electromagnetic theory from becoming too
complex, we usually require the constitutive",Electromagnetics_Vol1.pdf
"ǫ, µ, and σ.
T o keep electromagnetic theory from becoming too
complex, we usually require the constitutive
parameters to exhibit a few basic properties. These
properties are as follows:
• Homogeneity. A material that is homogeneous is
uniform over the space it occupies; that is, the
values of its constitutive parameters are constant
at all locations within the material. A
counter-example would be a material that is
composed of multiple chemically-distinct
compounds that are not thoroughly mixed, such
as soil.",Electromagnetics_Vol1.pdf
"28 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS
• Isotropy. A material that is isotropic behaves in
precisely the same way regardless of how it is
oriented with respect to sources and fields
occupying the same space. A counter-example is
quartz, whose atoms are arranged in a
uniformly-spaced crystalline lattice. As a result,
the electromagnetic properties of quartz can be
changed simply by rotating the material with
respect to the applied sources and fields.
• Linearity. A material is said to be linear if its
properties are constant and independent of the
magnitude of the sources and fields applied to
the material. For example, capacitors have
capacitance, which is determined in part by the
permittivity of the material separating the
terminals (Section 5.23). This material is
approximately linear when the applied voltage V
is below the rated working voltage; i.e., ǫis
constant and so capacitance does not vary
significantly with respect to V. When V is",Electromagnetics_Vol1.pdf
"is below the rated working voltage; i.e., ǫis
constant and so capacitance does not vary
significantly with respect to V. When V is
greater than the working voltage, the dependence
of ǫon V becomes more pronounced, and then
capacitance becomes a function of V. In another
practical example, it turns out that µfor
ferromagnetic materials is nonlinear such that
the precise value of µdepends on the magnitude
of the magnetic field.
• Time-invariance. An example of a class of
materials that is not necessarily time-invariant is
piezoelectric materials, for which
electromagnetic properties vary significantly
depending on the mechanical forces applied to
them – a property which can be exploited to
make sensors and transducers.
Linearity and time-invariance (L TI) are particularly
important properties to consider because they are
requirements for superposition. For example, in a L TI
material, we may calculate the field E1 due to a point
charge q1 at r1 and calculate the field E2 due to a",Electromagnetics_Vol1.pdf
"material, we may calculate the field E1 due to a point
charge q1 at r1 and calculate the field E2 due to a
point charge q2 at r2. Then, when both charges are
simultaneously present, the field is E1 + E2. The
same is not necessarily true for materials that are not
L TI. Devices that are nonlinear, and therefore not L TI,
do not necessarily follow the rules of elementary
circuit theory , which presume that superposition
applies. This condition makes analysis and design
much more difficult.
No practical material is truly homogeneous, isotropic,
linear, and time-invariant. However, for most
materials in most applications, the deviation from this
ideal condition is not large enough to significantly
affect engineering analysis and design. In other cases,
materials may be significantly non-ideal in one of
these respects, but may still be analyzed with
appropriate modifications to the theory .
[m0054]",Electromagnetics_Vol1.pdf
"2.8. ELECTROMAGNETIC PROPER TIES OF MA TERIALS 29
Image Credits
Fig. 2.1: c⃝ M. Goldammer, https://commons.wikimedia.org/wiki/File :M0002 fT woChargedParticles.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 2.2: c⃝ M. Goldammer, https://commons.wikimedia.org/wiki/File :M0002 fForceMap.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 2.3: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0002 fBatCap.svg,
CC BY 3.0 (https://creativecommons.org/licenses/by/3.0 /).
Fig. 2.4: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0003 fBarMagnet.svg,
CC BY 3.0 (https://creativecommons.org/licenses/by/3.0 /).
Fig. 2.5: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0003 fCoilBarMagnet.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 2.6: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0003 fFqvB.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).",Electromagnetics_Vol1.pdf
"Fig. 2.6: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0003 fFqvB.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 2.7: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0003 fFieldLinesBarMagnet.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 2.8: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:0003 fFieldLinesCoil.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).",Electromagnetics_Vol1.pdf
"Chapter 3
T ransmission Lines
3.1 Introduction to T ransmission
Lines
[m0028]
A transmission line is a structure intended to
transport electromagnetic signals or power.
A rudimentary transmission line is simply a pair of
wires with one wire serving as a datum (i.e., a
reference; e.g., “ground”) and the other wire bearing
an electrical potential that is defined relative to that
datum. Transmission lines having random geometry ,
such as the test leads shown in Figure 3.1, are useful
only at very low frequencies and when loss,
reactance, and immunity to electromagnetic
interference (EMI) are not a concern.
by Dmitry G
Figure 3.1: These leads used to connect test equip-
ment to circuits in a laboratory are a very rudimentary
form of transmission line, suitable only for very low
frequencies.
However, many circuits and systems operate at
frequencies where the length or cross-sectional
dimensions of the transmission line may be a
significant fraction of a wavelength. In this case, the",Electromagnetics_Vol1.pdf
"frequencies where the length or cross-sectional
dimensions of the transmission line may be a
significant fraction of a wavelength. In this case, the
transmission line is no longer “transparent” to the
circuits at either end. Furthermore, loss, reactance,
and EMI are significant problems in many
applications. These concerns motivate the use of
particular types of transmission lines, and make it
necessary to understand how to properly connect the
transmission line to the rest of the system.
In electromagnetics, the term “transmission line”
refers to a structure which is intended to support a
guided wave . A guided wave is an electromagnetic
wave that is contained within or bound to the line, and
which does not radiate away from the line. This
condition is normally met if the length and
cross-sectional dimensions of the transmission line
are small relative to a wavelength – say λ/100 (i.e.,
1% of the wavelength). For example, two
randomly-arranged wires might serve well enough to",Electromagnetics_Vol1.pdf
"are small relative to a wavelength – say λ/100 (i.e.,
1% of the wavelength). For example, two
randomly-arranged wires might serve well enough to
carry a signal at f = 10 MHz over a length l= 3 cm,
since lis only 0.1% of the wavelength
λ= c/f = 30 m. However, if lis increased to 3 m,
or if f is increased to 1 GHz, then lis now 10% of the
wavelength. In this case, one should consider using a
transmission line that forms a proper guided wave.
Preventing unintended radiation is not the only
concern. Once we have established a guided wave on
a transmission line, it is important that power applied
to the transmission line be delivered to the circuit or
device at the other end and not reflected back into the
source. For the random wire f = 10 MHz, l= 3 cm
example above, there is little need for concern, since
we expect a phase shift of roughly
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"3.2. TYPES OF TRANSMISSION LINES 31
0.001 · 360◦ = 0 .36◦ over the length of the
transmission line, which is about 0.72◦ for a round
trip. So, to a good approximation, the entire
transmission line is at the same electrical potential
and thus transparent to the source and destination.
However, if lis increased to 3 m, or if f is increased
to 1 GHz, then the associated round-trip phase shift
becomes 72◦. In this case, a reflected signal traveling
in the opposite direction will add to create a total
electrical potential, which varies in both magnitude
and phase with position along the line. Thus, the
impedance looking toward the destination via the
transmission line will be different than the impedance
looking toward the destination directly . (Section 3.15
gives the details.) The modified impedance will
depend on the cross-sectional geometry , materials,
and length of the line.
Cross-sectional geometry and materials also
determine the loss and EMI immunity of the
transmission line.",Electromagnetics_Vol1.pdf
"and length of the line.
Cross-sectional geometry and materials also
determine the loss and EMI immunity of the
transmission line.
Summarizing:
Transmission lines are designed to support
guided waves with controlled impedance, low
loss, and a degree of immunity from EMI.
3.2 T ypes of T ransmission Lines
[m0144]
T wo common types of transmission line are coaxial
line (Figure 3.2) and microstrip line (Figure 3.3).
Both are examples of transverse electromagnetic
(TEM) transmission lines. A TEM line employs a
single electromagnetic wave “mode” having electric
and magnetic field vectors in directions perpendicular
to the axis of the line, as shown in Figures 3.4 and
3.5. TEM transmission lines appear primarily in radio
frequency applications.
TEM transmission lines such as coaxial lines and
microstrip lines are designed to support a single
electromagnetic wave that propagates along the
length of the transmission line with electric and
magnetic field vectors perpendicular to the direc-",Electromagnetics_Vol1.pdf
"electromagnetic wave that propagates along the
length of the transmission line with electric and
magnetic field vectors perpendicular to the direc-
tion of propagation.
Not all transmission lines exhibit TEM field structure.
In non-TEM transmission lines, the electric and
magnetic field vectors that are not necessarily
perpendicular to the axis of the line, and the structure
of the fields is complex relative to the field structure
of TEM lines. An example of a transmission line that
exhibits non-TEM field structure is the waveguide
(see example in Figure 3.6). W aveguides are most
prevalent at radio frequencies, and tend to appear in
applications where it is important to achieve very low
loss or where power levels are very high. Another
example is common “multimode” optical fiber
(Figure 3.7). Optical fiber exhibits complex field
c⃝ Tkgd2007 CC BY 3.0 (modified)
Figure 3.2: Structure of a coaxial transmission line.",Electromagnetics_Vol1.pdf
"32 CHAPTER 3. TRANSMISSION LINES
g  ound plane
dielectric slab metallic trace
c⃝ SpinningSpark CC BY SA 3.0 (modified)
Figure 3.3: Structure of a microstrip transmission
line.
Figure 3.4: Structure of the electric and magnetic
fields within coaxial line. In this case, the wave is
propagating away from the viewer.
Figure 3.5: Structure of the electric and magnetic
fields within microstrip line. (The fields outside the
line are possibly significant, complicated, and not
shown.) In this case, the wave is propagating away
from the viewer.
c⃝ A verse CC BY SA 2.0 Germany
Figure 3.6: A network of radio frequency waveguides
in an air traffic control radar.
structure because the wavelength of light is very small
compared to the cross-section of the fiber, making the
excitation and propagation of non-TEM waves
difficult to avoid. (This issue is overcome in a
different type of optical fiber, known as “single
mode” fiber, which is much more difficult and
expensive to manufacture.)",Electromagnetics_Vol1.pdf
"different type of optical fiber, known as “single
mode” fiber, which is much more difficult and
expensive to manufacture.)
Higher-order transmission lines, including radio-
frequency waveguides and multimode optical
fiber, are designed to guide waves that have rela-
tively complex structure.
Additional Reading:
• “Coaxial cable” on Wikipedia.
• “Microstrip” on Wikipedia.
• “W aveguide (electromagnetism)” on Wikipedia.
• “Optical fiber” on Wikipedia.",Electromagnetics_Vol1.pdf
"3.3. TRANSMISSION LINES AS TWO-POR T DEVICES 33
c⃝ BigRiz CC BY SA 3.0 Unported
Figure 3.7: Strands of optical fiber.
3.3 T ransmission Lines as
T wo-Port Devices
[m0077]
Figure 3.8 shows common ways to represent
transmission lines in circuit diagrams. In each case,
the source is represented using a Th ´evenin equivalent
circuit consisting of a voltage source VS in series with
an impedance ZS.1 In transmission line analysis, the
source may also be referred to as the generator. The
termination on the receiving end of the transmission
line is represented, without loss of generality , as an
impedance ZL. This termination is often referred to
as the load, although in practice it can be any circuit
that exhibits an input impedance of ZL.
The two-port representation of a transmission line is
completely described by its length lalong with some
combination of the following parameters:
• Phase propagation constant β, having units of
rad/m. This parameter also represents the",Electromagnetics_Vol1.pdf
"combination of the following parameters:
• Phase propagation constant β, having units of
rad/m. This parameter also represents the
wavelength in the line through the relationship
λ= 2 π/β. (See Sections 1.3 and 3.8 for details.)
• Attenuation constant α, having units of 1/m.
1 For a refresher on this concept, see “ Additional Reading” at the
end of this section.
This parameter quantifies the effect of loss in the
line. (See Section 3.8 for details.)
• Characteristic impedance Z0, having units of Ω .
This is the ratio of potential (“voltage”) to
current when the line is perfectly
impedance-matched at both ends. (See
Section 3.7 for details.)
These parameters depend on the materials and
geometry of the line.
Note that a transmission line is typically not
transparent to the source and load. In particular, the
load impedance may be ZL, but the impedance
presented to the source may or may not be equal to
ZL. (See Section 3.15 for more on this concept.)",Electromagnetics_Vol1.pdf
"load impedance may be ZL, but the impedance
presented to the source may or may not be equal to
ZL. (See Section 3.15 for more on this concept.)
Similarly , the source impedance may be ZS, but the
impedance presented to the load may or may not be
equal to ZS. The effect of the transmission line on the
source and load impedances will depend on the
parameters identified above.
ZL
ZS
VS
VS
VS
ZS
ZS
ZL
ZL
l
l
c⃝ Omegatron CC BY SA 3.0 Unported (modified)
Figure 3.8: Symbols representing transmission lines:
T op: As a generic two-conductor direct connection.
Middle: As a generic two-port “black box. ” Bottom:
As a coaxial cable.",Electromagnetics_Vol1.pdf
"34 CHAPTER 3. TRANSMISSION LINES
Additional Reading:
• “Th ´evenin’s theorem” on Wikipedia.
3.4 Lumped-Element Model
[m0029]
It is possible to ascertain the relevant behaviors of a
transmission line using elementary circuit theory
applied to a differential-length lumped-element model
of the transmission line. The concept is illustrated in
Figure 3.9, which shows a generic transmission line
aligned with its length along the zaxis. The
transmission line is divided into segments having
small but finite length ∆ z. Each segment is modeled
as an identical two-port having the equivalent circuit
representation shown in Figure 3.10. The equivalent
circuit consists of 4 components as follows:
• The resistance R′∆ zrepresents the
series-combined ohmic resistance of the two
conductors. This should account for both
conductors since the current in the actual
transmission line must flow through both
conductors. The prime notation reminds us that
R′ is resistance per unit length ; i.e., Ω /m, and it",Electromagnetics_Vol1.pdf
"transmission line must flow through both
conductors. The prime notation reminds us that
R′ is resistance per unit length ; i.e., Ω /m, and it
is only after multiplying by length that we get a
resistance in Ω .
• The conductance G′∆ zrepresents the leakage of
current directly from one conductor to the other.
When G′∆ z >0, the resistance between the
conductors is less than infinite, and therefore,
current may flow between the conductors. This
amounts to a loss of power separate from the loss
associated with R′ above. G′ has units of S/m.
Further note that G′ is not equal to 1/R′ as
defined above. G′ and R′ are describing entirely
different physical mechanisms (and in principle
either could be defined as either a resistance or a
conductance).
• The capacitance C′∆ zrepresents the
capacitance of the transmission line structure.
Capacitance is the tendency to store energy in
electric fields and depends on the cross-sectional
geometry and the media separating the
conductors. C′ has units of F/m.",Electromagnetics_Vol1.pdf
"electric fields and depends on the cross-sectional
geometry and the media separating the
conductors. C′ has units of F/m.
• The inductance L′∆ zrepresents the inductance
of the transmission line structure. Inductance is
the tendency to store energy in magnetic fields,
and (like capacitance) depends on the",Electromagnetics_Vol1.pdf
"3.5. TELEGRAPHER’S EQUA TIONS 35
z

z	 z",Electromagnetics_Vol1.pdf
"z
 z
Figure 3.9: Interpretation of a transmission line as a
cascade of discrete series-connected two-ports.
R'Δz L'Δz
G'Δz C'Δz
c⃝ Omegatron CC BY SA 3.0 Unported (modified)
Figure 3.10: Lumped-element equivalent circuit
model for each of the two-ports in Figure 3.9.
cross-sectional geometry and the media
separating the conductors. L′ has units of H/m.
In order to use the model, one must have values for
R′, G′, C′, and L′. Methods for computing these
parameters are addressed elsewhere in this book.
3.5 T elegrapher’s Equations
[m0079]
In this section, we derive the equations that govern
the potential v(z,t) and current i(z,t) along a
transmission line that is oriented along the zaxis. For
this, we will employ the lumped-element model
developed in Section 3.4.
T o begin, we define voltages and currents as shown in
Figure 3.11. W e assign the variables v(z,t) and
i(z,t) to represent the potential and current on the left
side of the segment, with reference polarity and",Electromagnetics_Vol1.pdf
"i(z,t) to represent the potential and current on the left
side of the segment, with reference polarity and
direction as shown in the figure. Similarly we assign
the variables v(z+ ∆ z,t) and i(z+ ∆ z,t) to
represent the potential and current on the right side of
the segment, again with reference polarity and
direction as shown in the figure. Applying Kirchoff’s
voltage law from the left port, through R′∆ zand
L′∆ z, and returning via the right port, we obtain:
v(z,t) − (R′∆ z) i(z,t) − (L′∆ z) ∂
∂ti(z,t)
− v(z+ ∆ z,t) = 0 (3.1)
Moving terms referring to current to the right side of
the equation and then dividing through by ∆ z, we
obtain
− v(z+ ∆ z,t) − v(z,t)
∆ z =
R′ i(z,t) + L′ ∂
∂ti(z,t) (3.2)
Then taking the limit as ∆ z→ 0:
− ∂
∂zv(z,t) = R′ i(z,t) + L′ ∂
∂ti(z,t) (3.3)
R'Δz L'Δz
G'Δz C'Δz
v(z,t) i(z+Δz,t)i(z,t)
+
_
+
_
v(z+Δz,t)
c⃝ Omegatron CC BY SA 3.0 Unported (modified)
Figure 3.11: Lumped-element equivalent circuit
transmission line model, annotated with sign conven-",Electromagnetics_Vol1.pdf
"c⃝ Omegatron CC BY SA 3.0 Unported (modified)
Figure 3.11: Lumped-element equivalent circuit
transmission line model, annotated with sign conven-
tions for potentials and currents.",Electromagnetics_Vol1.pdf
"36 CHAPTER 3. TRANSMISSION LINES
Applying Kirchoff’s current law at the right port, we
obtain:
i(z,t)−(G′∆ z) v(z+∆ z,t)−(C′∆ z) ∂
∂tv(z+∆ z,t)
− i(z+ ∆ z,t) = 0 (3.4)
Moving terms referring to potential to the right side of
the equation and then dividing through by ∆ z, we
obtain
− i(z+ ∆ z,t) − i(z,t)
∆ z =
G′ v(z+ ∆ z,t) + C′ ∂
∂tv(z+ ∆ z,t) (3.5)
T aking the limit as ∆ z→ 0:
− ∂
∂zi(z,t) = G′ v(z,t) + C′ ∂
∂tv(z,t) (3.6)
Equations 3.3 and 3.6 are the telegrapher’s equa-
tions. These coupled (simultaneous) differential
equations can be solved for v(z,t) and i(z,t)
given R′, G′, L′, C′ and suitable boundary con-
ditions.
The time-domain telegrapher’s equations are usually
more than we need or want. If we are only interested
in the response to a sinusoidal stimulus, then
considerable simplification is possible using phasor
representation.2 First we define phasors ˜V(z) and
˜I(z) through the usual relationship:
v(z,t) = Re
{
˜V(z) ejωt
}
(3.7)
i(z,t) = Re
{
˜I(z) ejωt
}
(3.8)
Now we see:",Electromagnetics_Vol1.pdf
"˜I(z) through the usual relationship:
v(z,t) = Re
{
˜V(z) ejωt
}
(3.7)
i(z,t) = Re
{
˜I(z) ejωt
}
(3.8)
Now we see:
∂
∂zv(z,t) = ∂
∂zRe
{
˜V(z) ejωt
}
= Re
{[ ∂
∂z
˜V(z)
]
ejωt
}
2 For a refresher on phasor analysis, see Section 1.5.
In other words, ∂v(z,t)/∂z expressed in phasor
representation is simply ∂˜V(z)/∂z; and
∂
∂ti(z,t) = ∂
∂tRe
{
˜I(z) ejωt
}
= Re
{ ∂
∂t
[
˜I(z)ejωt
] }
= Re
{[
jω˜I(z)
]
ejωt
}
In other words, ∂i(z,t)/∂texpressed in phasor
representation is jω˜I(z). Therefore, Equation 3.3
expressed in phasor representation is:
− ∂
∂z
˜V(z) = [ R′ + jωL′] ˜I(z) (3.9)
Following the same procedure, Equation 3.6
expressed in phasor representation is found to be:
− ∂
∂z
˜I(z) = [ G′ + jωC′] ˜V(z) (3.10)
Equations 3.9 and 3.10 are the telegrapher’s
equations in phasor representation.
The principal advantage of these equations over the
time-domain versions is that we no longer need to
contend with derivatives with respect to time – only",Electromagnetics_Vol1.pdf
"time-domain versions is that we no longer need to
contend with derivatives with respect to time – only
derivatives with respect to distance remain. This
considerably simplifies the equations.
Additional Reading:
• “T elegrapher’s equations” on Wikipedia.
• “Kirchhoff’s circuit laws” on Wikipedia.",Electromagnetics_Vol1.pdf
"3.6. W A VE EQUA TION FOR A TEM TRANSMISSION LINE 37
3.6 W ave Equation for a TEM
T ransmission Line
[m0027]
Consider a TEM transmission line aligned along the z
axis. The phasor form of the T elegrapher’s Equations
(Section 3.5) relate the potential phasor ˜V(z) and the
current phasor ˜I(z) to each other and to the
lumped-element model equivalent circuit parameters
R′, G′, C′, and L′. These equations are
− ∂
∂z
˜V(z) = [ R′ + jωL′] ˜I(z) (3.11)
− ∂
∂z
˜I(z) = [ G′ + jωC′] ˜V(z) (3.12)
An obstacle to using these equations is that we
require both equations to solve for either the potential
or the current. In this section, we reduce these
equations to a single equation – a wave equation –
that is more convenient to use and provides some
additional physical insight.
W e begin by differentiating both sides of
Equation 3.11 with respect to z, yielding:
− ∂2
∂z2
˜V(z) = [ R′ + jωL′] ∂
∂z
˜I(z) (3.13)
Then using Equation 3.12 to eliminate ˜I(z), we obtain
− ∂2
∂z2",Electromagnetics_Vol1.pdf
"− ∂2
∂z2
˜V(z) = [ R′ + jωL′] ∂
∂z
˜I(z) (3.13)
Then using Equation 3.12 to eliminate ˜I(z), we obtain
− ∂2
∂z2
˜V(z) = − [R′ + jωL′] [G′ + jωC′] ˜V(z)
(3.14)
This equation is normally written as follows:
∂2
∂z2
˜V(z) − γ2 ˜V(z) = 0 (3.15)
where we have made the substitution:
γ2 = ( R′ + jωL′) (G′ + jωC′) (3.16)
The principal square root of γ2 is known as the
propagation constant :
γ ≜
√
(R′ + jωL′) (G′ + jωC′) (3.17)
The propagation constant γ (units of m −1) cap-
tures the effect of materials, geometry , and fre-
quency in determining the variation in potential
and current with distance on a TEM transmission
line.
Following essentially the same procedure but
beginning with Equation 3.12, we obtain
∂2
∂z2
˜I(z) − γ2 ˜I(z) = 0 (3.18)
Equations 3.15 and 3.18 are the wave equations
for ˜V(z) and ˜I(z), respectively .
Note that both ˜V(z) and ˜I(z) satisfy the same linear
homogeneous differential equation. This does not
mean that ˜V(z) and ˜I(z) are equal. Rather, it means",Electromagnetics_Vol1.pdf
"homogeneous differential equation. This does not
mean that ˜V(z) and ˜I(z) are equal. Rather, it means
that ˜V(z) and ˜I(z) can differ by no more than a
multiplicative constant. Since ˜V(z) is potential and
˜I(z) is current, that constant must be an impedance.
This impedance is known as the characteristic
impedance and is determined in Section 3.7.
The general solutions to Equations 3.15 and 3.18 are
˜V(z) = V+
0 e−γz + V−
0 e+γz (3.19)
˜I(z) = I+
0 e−γz + I−
0 e+γz (3.20)
where V+
0 , V−
0 , I+
0 , and I−
0 are complex-valued
constants. It is shown in Section 3.8 that
Equations 3.19 and 3.20 represent sinusoidal waves
propagating in the +zand −zdirections along the
length of the line. The constants may represent
sources, loads, or simply discontinuities in the
materials and/or geometry of the line. The values of
the constants are determined by boundary conditions;
i.e., constraints on ˜V(z) and ˜I(z) at some position(s)
along the line.
The reader is encouraged to verify that the",Electromagnetics_Vol1.pdf
"i.e., constraints on ˜V(z) and ˜I(z) at some position(s)
along the line.
The reader is encouraged to verify that the
Equations 3.19 and 3.20 are in fact solutions to
Equations 3.15 and 3.18, respectively , for any values
of the constants V+
0 , V−
0 , I+
0 , and I−
0 .",Electromagnetics_Vol1.pdf
"38 CHAPTER 3. TRANSMISSION LINES
3.7 Characteristic Impedance
[m0052]
Characteristic impedance is the ratio of voltage to
current for a wave that is propagating in single
direction on a transmission line. This is an important
parameter in the analysis and design of circuits and
systems using transmission lines. In this section, we
formally define this parameter and derive an
expression for this parameter in terms of the
equivalent circuit model introduced in Section 3.4.
Consider a transmission line aligned along the zaxis.
Employing some results from Section 3.6, recall that
the phasor form of the wave equation in this case is
∂2
∂z2
˜V(z) − γ2 ˜V(z) = 0 (3.21)
where
γ ≜
√
(R′ + jωL′) (G′ + jωC′) (3.22)
Equation 3.21 relates the potential phasor ˜V(z) to the
equivalent circuit parameters R′, G′, C′, and L′. An
equation of the same form relates the current phasor
˜I(z) to the equivalent circuit parameters:
∂2
∂z2
˜I(z) − γ2 ˜I(z) = 0 (3.23)",Electromagnetics_Vol1.pdf
"equation of the same form relates the current phasor
˜I(z) to the equivalent circuit parameters:
∂2
∂z2
˜I(z) − γ2 ˜I(z) = 0 (3.23)
Since both ˜V(z) and ˜I(z) satisfy the same linear
homogeneous differential equation, they may differ
by no more than a multiplicative constant. Since ˜V(z)
is potential and ˜I(z) is current, that constant can be
expressed in units of impedance. Specifically , this is
the characteristic impedance , so-named because it
depends only on the materials and cross-sectional
geometry of the transmission line – i.e., things which
determine γ– and not length, excitation, termination,
or position along the line.
T o derive the characteristic impedance, first recall that
the general solutions to Equations 3.21 and 3.23 are
˜V(z) = V+
0 e−γz + V−
0 e+γz (3.24)
˜I(z) = I+
0 e−γz + I−
0 e+γz (3.25)
where V+
0 , V−
0 , I+
0 , and I−
0 are complex-valued
constants whose values are determined by boundary
conditions; i.e., constraints on ˜V(z) and ˜I(z) at some",Electromagnetics_Vol1.pdf
"0 , I+
0 , and I−
0 are complex-valued
constants whose values are determined by boundary
conditions; i.e., constraints on ˜V(z) and ˜I(z) at some
position(s) along the line. Also, we will make use of
the telegrapher’s equations (Section 3.5):
− ∂
∂z
˜V(z) = [ R′ + jωL′] ˜I(z) (3.26)
− ∂
∂z
˜I(z) = [ G′ + jωC′] ˜V(z) (3.27)
W e begin by differentiating Equation 3.24 with
respect to z, which yields
∂
∂z
˜V(z) = −γ
[
V+
0 e−γz − V−
0 e+γz]
(3.28)
Now we use this this to eliminate ∂˜V(z)/∂z in
Equation 3.26, yielding
γ
[
V+
0 e−γz − V−
0 e+γz]
= [ R′ + jωL′] ˜I(z) (3.29)
Solving the above equation for ˜I(z) yields:
˜I(z) = γ
R′ + jωL′
[
V+
0 e−γz − V−
0 e+γz]
(3.30)
Comparing this to Equation 3.25, we note
I+
0 = γ
R′ + jωL′ V+
0 (3.31)
I−
0 = −γ
R′ + jωL′ V−
0 (3.32)
W e now make the substitution
Z0 = R′ + jωL′
γ (3.33)
and observe
V+
0
I+
0
= −V−
0
I−
0
≜ Z0 (3.34)
As anticipated, we have found that coefficients in the
equations for potentials and currents are related by an",Electromagnetics_Vol1.pdf
"V+
0
I+
0
= −V−
0
I−
0
≜ Z0 (3.34)
As anticipated, we have found that coefficients in the
equations for potentials and currents are related by an
impedance, namely , Z0.
Characteristic impedance can be written entirely in
terms of the equivalent circuit parameters by
substituting Equation 3.22 into Equation 3.33,
yielding:
Z0 =
√
R′ + jωL′
G′ + jωC′ (3.35)",Electromagnetics_Vol1.pdf
"3.8. W A VE PROP AGA TION ON A TEM TRANSMISSION LINE 39
The characteristic impedance Z0 (Ω ) is the ratio
of potential to current in a wave traveling in a
single direction along the transmission line.
T ake care to note that Z0 is not the ratio of ˜V(z) to
˜I(z) in general; rather, Z0 relates only the potential
and current waves traveling in the same direction.
Finally , note that transmission lines are normally
designed to have a characteristic impedance that is
completely real-valued – that is, with no imaginary
component. This is because the imaginary component
of an impedance represents energy storage (think of
capacitors and inductors), whereas the purpose of a
transmission line is energy transfer.
Additional Reading:
• “Characteristic impedance” on Wikipedia.
3.8 W ave Propagation on a TEM
T ransmission Line
[m0080]
In Section 3.6, it is shown that expressions for the
phasor representations of the potential and current
along a transmission line are
˜V(z) = V+
0 e−γz + V−",Electromagnetics_Vol1.pdf
"phasor representations of the potential and current
along a transmission line are
˜V(z) = V+
0 e−γz + V−
0 e+γz (3.36)
˜I(z) = I+
0 e−γz + I−
0 e+γz (3.37)
where γis the propagation constant and it assumed
that the transmission line is aligned along the zaxis.
In this section, we demonstrate that these expressions
represent sinusoidal waves, and point out some
important features. Before attempting this section, the
reader should be familiar with the contents of
Sections 3.4, 3.6, and 3.7. A refresher on fundamental
wave concepts (Section 1.3) may also be helpful.
W e first define real-valued quantities αand βto be
the real and imaginary components of γ; i.e.,
α≜ Re {γ} (3.38)
β ≜ Im {γ} (3.39)
and subsequently
γ = α+ jβ (3.40)
Then we observe
e±γz = e±(α+jβ)z = e±αz e±jβz (3.41)
It may be easier to interpret this expression by
reverting to the time domain:
Re
{
e±γzejωt}
= e±αzcos (ωt± βz) (3.42)
Thus, e−γz represents a damped sinusoidal wave",Electromagnetics_Vol1.pdf
"reverting to the time domain:
Re
{
e±γzejωt}
= e±αzcos (ωt± βz) (3.42)
Thus, e−γz represents a damped sinusoidal wave
traveling in the +zdirection, and e+γz represents a
damped sinusoidal wave traveling in the −zdirection.
Let’s define ˜V+(z) and ˜I+(z) to be the potential and
current associated with a wave propagating in the +z
direction. Then:
˜V+(z) ≜ V+
0 e−γz (3.43)",Electromagnetics_Vol1.pdf
"40 CHAPTER 3. TRANSMISSION LINES


  
  
z
v+(z,t=0)
Figure 3.12: The potential v+(z,t) of the wave travel-
ing in the +zdirection at t= 0 for ψ= 0 .
or equivalently in the time domain:
v+(z,t) = Re
{
˜V+(z) ejωt
}
= Re
{
V+
0 e−γzejωt}
=
⏐
⏐V+
0
⏐
⏐e−αzcos (ωt− βz+ ψ) (3.44)
where ψis the phase of V+
0 . Figure 3.12 shows
v+(z,t). From fundamental wave theory we
recognize
β ≜ Im {γ} (rad/m) is the phase propaga-
tion constant , which is the rate at which phase
changes as a function of distance.
Subsequently the wavelength in the line is
λ= 2π
β (3.45)
Also we recognize:
α ≜ Re {γ} (1/m) is the attenuation constant ,
which is the rate at which magnitude diminishes
as a function of distance.
Sometimes the units of αare indicated as “Np/m”
(“nepers” per meter), where the term “neper” is used
to indicate the units of the otherwise unitless
real-valued exponent of the constant e.
Note that α= 0 for a wave that does not diminish in
magnitude with increasing distance, in which case the",Electromagnetics_Vol1.pdf
"real-valued exponent of the constant e.
Note that α= 0 for a wave that does not diminish in
magnitude with increasing distance, in which case the
transmission line is said to be lossless. If α> 0 then
the line is said to be lossy (or possibly “low loss” if
the loss can be neglected), and in this case the rate at
which the magnitude decreases with distance
increases with α.
Next let us consider the speed of the wave. T o answer
this question, we need to be a bit more specific about
what we mean by “speed. ” At the moment, we mean
phase velocity; that is, the speed at which a point of
constant phase seems to move through space. In other
words, what distance ∆ zdoes a point of constant
phase traverse in time ∆ t? T o answer this question,
we first note that the phase of v+(z,t) can be written
generally as
ωt− βz+ φ
where φis some constant. Similarly , the phase at
some time ∆ tlater and some point ∆ zfurther along
can be written as
ω(t+ ∆ t) − β(z+ ∆ z) + φ",Electromagnetics_Vol1.pdf
"where φis some constant. Similarly , the phase at
some time ∆ tlater and some point ∆ zfurther along
can be written as
ω(t+ ∆ t) − β(z+ ∆ z) + φ
The phase velocity vp is ∆ z/∆ twhen these two
phases are equal; i.e., when
ωt− βz+ φ= ω(t+ ∆ t) − β(z+ ∆ z) + φ
Solving for vp = ∆ z/∆ t, we obtain:
vp = ω
β (3.46)
Having previously noted that β = 2 π/λ, the above
expression also yields the expected result
vp = λf (3.47)
The phase velocity vp = ω/β = λf is the speed
at which a point of constant phase travels along
the line.
Returning now to consider the current associated with
the wave traveling in the +zdirection:
˜I+(z) = I+
0 e−γz (3.48)
W e can rewrite this expression in terms of the
characteristic impedance Z0, as follows:
˜I+(z) = V+
0
Z0
e−γz (3.49)",Electromagnetics_Vol1.pdf
"3.9. LOSSLESS AND LOW -LOSS TRANSMISSION LINES 41
Similarly , we find that the current ˜I−(z) associated
with ˜V−(z) for the wave traveling in the −zdirection
is
˜I−(z) = −V−
0
Z0
e−γz (3.50)
The negative sign appearing in the above expression
emerges as a result of the sign conventions used for
potential and current in the derivation of the
telegrapher’s equations (Section 3.5). The physical
significance of this change of sign is that wherever the
potential of the wave traveling in the −zdirection is
positive, then the current at the same point is flowing
in the −zdirection.
It is frequently necessary to consider the possibility
that waves travel in both directions simultaneously . A
very important case where this arises is when there is
reflection from a discontinuity of some kind; e.g.,
from a termination which is not perfectly
impedance-matched. In this case, the total potential
˜V(z) and total current ˜I(z) can be expressed as the",Electromagnetics_Vol1.pdf
"from a termination which is not perfectly
impedance-matched. In this case, the total potential
˜V(z) and total current ˜I(z) can be expressed as the
general solution to the wave equation; i.e., as the sum
of the “incident” ( +z-traveling) wave and the
reflected ( −z-traveling) waves:
˜V(z) = ˜V+(z) + ˜V−(z) (3.51)
˜I(z) = ˜I+(z) + ˜I−(z) (3.52)
The existence of waves propagating simultaneously in
both directions gives rise to a phenomenon known as
a standing wave . Standing waves and the calculation
of the coefficients V−
0 and I−
0 due to reflection are
addressed in Sections 3.13 and 3.12 respectively .
3.9 Lossless and Low-Loss
T ransmission Lines
[m0083]
Quite often the loss in a transmission line is small
enough that it may be neglected. In this case, several
aspects of transmission line theory may be simplified.
In this section, we present these simplifications.
First, recall that “loss” refers to the reduction of
magnitude as a wave propagates through space. In the",Electromagnetics_Vol1.pdf
"First, recall that “loss” refers to the reduction of
magnitude as a wave propagates through space. In the
lumped-element equivalent circuit model
(Section 3.4), the parameters R′ and G′ of the
represent physical mechanisms associated with loss.
Specifically , R′ represents the resistance of
conductors, whereas G′ represents the undesirable
current induced between conductors through the
spacing material. Also recall that the propagation
constant γis, in general, given by
γ ≜
√
(R′ + jωL′) (G′ + jωC′) (3.53)
With this in mind, we now define “low loss” as
meeting the conditions:
R′ ≪ ωL′ (3.54)
G′ ≪ ωC′ (3.55)
When these conditions are met, the propagation
constant simplifies as follows:
γ ≈
√
(jωL′) (jωC′)
=
√
−ω2L′C′
= jω
√
L′C′ (3.56)
and subsequently
α≜ Re {γ} ≈ 0 (low-loss approx.) (3.57)
β ≜ Im {γ} ≈ ω
√
L′C′ (low-loss approx.) (3.58)
vp = ω/β ≈ 1√
L′C′ (low-loss approx.) (3.59)
Similarly:
Z0 =
√
R′ + jωL′
G′ + jωC′ ≈
√
L′
C′ (low-loss approx.)
(3.60)",Electromagnetics_Vol1.pdf
"42 CHAPTER 3. TRANSMISSION LINES
Of course if the line is strictly lossless (i.e.,
R′ = G′ = 0 ) then these are not approximations, but
rather the exact expressions.
In practice, these approximations are quite commonly
used, since practical transmission lines typically meet
the conditions expressed in Inequalities 3.54 and 3.55
and the resulting expressions are much simpler. W e
further observe that Z0 and vp are approximately
independent of frequency when these conditions hold.
However, also note that “low loss” does not mean “no
loss, ” and it is common to apply these expressions
even when R′ and/or G′ is large enough to yield
significant loss. For example, a coaxial cable used to
connect an antenna on a tower to a radio near the
ground typically has loss that is important to consider
in the analysis and design process, but nevertheless
satisfies Equations 3.54 and 3.55. In this case, the
low-loss expression for βis used, but αmight not be
approximated as zero.

sσ s
b
a",Electromagnetics_Vol1.pdf
"satisfies Equations 3.54 and 3.55. In this case, the
low-loss expression for βis used, but αmight not be
approximated as zero.

sσ s
b
a
Figure 3.13: Cross-section of a coaxial transmission
line, indicating design parameters.
3.10 Coaxial Line
[m0143]
Coaxial transmission lines consists of metallic inner
and outer conductors separated by a spacer material
as shown in Figure 3.13. The spacer material is
typically a low-loss dielectric material having
permeability approximately equal to that of free space
(µ≈ µ0) and permittivity ǫs that may range from
very near ǫ0 (e.g., air-filled line) to 2–3 times ǫ0. The
outer conductor is alternatively referred to as the
“shield, ” since it typically provides a high degree of
isolation from nearby objects and electromagnetic
fields. Coaxial line is single-ended 3 in the sense that
the conductor geometry is asymmetric and the shield
is normally attached to ground at both ends. These
characteristics make coaxial line attractive for",Electromagnetics_Vol1.pdf
"is normally attached to ground at both ends. These
characteristics make coaxial line attractive for
connecting single-ended circuits in widely-separated
locations and for connecting antennas to receivers and
transmitters.
Coaxial lines exhibit TEM field structure as shown in
Figure 3.14.
Expressions for the equivalent circuit parameters C′
and L′ for coaxial lines can be obtained from basic
electromagnetic theory . It is shown in Section 5.24
that the capacitance per unit length is
C′ = 2πǫs
ln (b/a) (3.61)
where aand bare the radii of the inner and outer
conductors, respectively . Using analysis shown in
3 The references in “ Additional Reading” at the end of this sec tion
may be helpful if you are not familiar with this concept.",Electromagnetics_Vol1.pdf
"3.10. COAXIAL LINE 43
Figure 3.14: Structure of the electric and magnetic
fields within coaxial line. In this case, the wave is
propagating away from the viewer.
Section 7.14, the inductance per unit length is
L′ = µ0
2πln
( b
a
)
(3.62)
The loss conductance G′ depends on the conductance
σs of the spacer material, and is given by
G′ = 2πσs
ln (b/a) (3.63)
This expression is derived in Section 6.5.
The resistance per unit length, R′, is relatively
difficult to quantify . One obstacle is that the inner and
outer conductors typically consist of different
materials or compositions of materials. The inner
conductor is not necessarily a single homogeneous
material; instead, the inner conductor may consist of a
variety of materials selected by trading-off between
conductivity , strength, weight, and cost. Similarly , the
outer conductor is not necessarily homogeneous; for a
variety of reasons, the outer conductor may instead be
a metal mesh, a braid, or a composite of materials.",Electromagnetics_Vol1.pdf
"variety of reasons, the outer conductor may instead be
a metal mesh, a braid, or a composite of materials.
Another complicating factor is that the resistance of
the conductor varies significantly with frequency ,
whereas C′, L′, and G′ exhibit relatively little
variation from their electro- and magnetostatic values.
These factors make it difficult to devise a single
expression for R′ that is both as simple as those
shown above for the other parameters and generally
applicable. Fortunately , it turns out that the low-loss
conditions R′ ≪ ωL′ and G′ ≪ ωC′ are often
applicable,4 so that R′ and G′ are important only if it
is necessary to compute loss.
4 See Section 3.9 for a reminder about this concept.
Since the low-loss conditions are often met, a
convenient expression for the characteristic
impedance is obtained from Equations 3.61 and 3.62
for L′ and C′ respectively:
Z0 ≈
√
L′
C′ (low-loss)
= 1
2π
√ µ0
ǫs
ln b
a (3.64)
The spacer permittivity can be expressed as ǫs = ǫrǫ0",Electromagnetics_Vol1.pdf
"for L′ and C′ respectively:
Z0 ≈
√
L′
C′ (low-loss)
= 1
2π
√ µ0
ǫs
ln b
a (3.64)
The spacer permittivity can be expressed as ǫs = ǫrǫ0
where ǫr is the relative permittivity of the spacer
material. Since
√
µ0/ǫ0 is a constant, the above
expression is commonly written
Z0 ≈ 60 Ω√ǫr
ln b
a (low-loss) (3.65)
Thus, it is possible to express Z0 directly in terms of
parameters describing the geometry ( aand b) and
material ( ǫr) used in the line, without the need to first
compute the values of components in the
lumped-element equivalent circuit model.
Similarly , the low-loss approximation makes it
possible to express the phase velocity νp directly in
terms the spacer permittivity:
νp ≈ 1
√
L′C′ (low-loss)
= c√ǫr
(3.66)
since c≜ 1/√µ0ǫ0. In other words, the phase
velocity in a low-loss coaxial line is approximately
equal to the speed of electromagnetic propagation in
free space, divided by the square root of the relative
permittivity of the spacer material. Therefore, the",Electromagnetics_Vol1.pdf
"free space, divided by the square root of the relative
permittivity of the spacer material. Therefore, the
phase velocity in an air-filled coaxial line is
approximately equal to speed of propagation in free
space, but is reduced in a coaxial line using a
dielectric spacer.
Example 3.1. RG-59 Coaxial Cable.
RG-59 is a very common type of coaxial line.
Figure 3.15 shows a section of RG-59 cut away
so as to reveal its structure. The radii are
a∼
= 0.292 mm and b≈ 1.855 mm (mean),
yielding L′ ≈ 370 nH/m. The spacer material is
polyethylene having ǫr ∼
= 2.25, yielding",Electromagnetics_Vol1.pdf
"44 CHAPTER 3. TRANSMISSION LINES
C′ ≈ 67.7 pF/m. The conductivity of
polyethylene is σs ∼
= 5.9 × 10−5 S/m, yielding
G′ ≈ 200 µS/m. T ypical resistance per unit
length R′ is on the order of 0.1 Ω /m near DC,
increasing approximately in proportion to the
square root of frequency .
From the above values, we find that RG-59
satisfies the low-loss criteria R′ ≪ ωL′ for
f ≫ 43 kHz and G′ ≪ ωC′ for f ≫ 470 kHz.
Under these conditions, we find
Z0 ≈
√
L′/C′ ∼
= 74 Ω . Thus, the ratio of the
potential to the current in a wave traveling in a
single direction on RG-59 is about 74 Ω .
The phase velocity of RG-59 is found to be
vp ≈ 1/
√
L′C′ ∼
= 2 × 108 m/s, which is about
67% of c. In other words, a signal that takes 1 ns
to traverse a distance lin free space requires
about 1.5 ns to traverse a length- lsection of
RG-59. Since vp = λf, a wavelength in RG-59
is 67% of a wavelength in free space.
Using the expression
γ =
√
(R′ + jωL′) (G′ + jωC′) (3.67)",Electromagnetics_Vol1.pdf
"RG-59. Since vp = λf, a wavelength in RG-59
is 67% of a wavelength in free space.
Using the expression
γ =
√
(R′ + jωL′) (G′ + jωC′) (3.67)
with R′ = 0 .1 Ω /m, and then taking the real part
to obtain α, we find α∼ 0.01 m−1. So, for
example, the magnitude of the potential or
current is decreased by about 50% by traveling a
distance of about 70 m. In other words,
e−αl = 0 .5 for l∼ 70 m at relatively low
frequencies, and increases with increasing
frequency .
Additional Reading:
• “Coaxial cable” on Wikipedia. Includes
descriptions and design parameters for a variety
of commonly-encountered coaxial cables.
• “Single-ended signaling” on Wikipedia.
• Sec. 8.7 (“Differential Circuits”) in
S.W . Ellingson, Radio Systems Engineering ,
Cambridge Univ . Press, 2016.
c⃝ Arj CC BY SA 3.0
Figure 3.15: RG-59 coaxial line. A: Insulating jacket.
B: Braided outer conductor. C: Dielectric spacer. D:
Inner conductor.
hW
l

r
t

0
c⃝ 7head7metal7 CC BY SA 3.0 (modified)",Electromagnetics_Vol1.pdf
"B: Braided outer conductor. C: Dielectric spacer. D:
Inner conductor.
hW
l

r
t

0
c⃝ 7head7metal7 CC BY SA 3.0 (modified)
Figure 3.16: Microstrip transmission line structure
and design parameters.
3.11 Microstrip Line
[m0082]
A microstrip transmission line consists of a narrow
metallic trace separated from a metallic ground plane
by a slab of dielectric material, as shown in
Figure 3.16. This is a natural way to implement a
transmission line on a printed circuit board, and so
accounts for an important and expansive range of
applications. The reader should be aware that
microstrip is distinct from stripline, which is a very
different type of transmission line; see “ Additional
Reading” at the end of this section for disambiguation
of these terms.
A microstrip line is single-ended 5 in the sense that the
5 The reference in “ Additional Reading” at the end of this sect ion
may be helpful if you are not familiar with this concept.",Electromagnetics_Vol1.pdf
"3.11. MICROSTRIP LINE 45
conductor geometry is asymmetric and the one
conductor – namely , the ground plane – also normally
serves as ground for the source and load.
The spacer material is typically a low-loss dielectric
material having permeability approximately equal to
that of free space ( µ≈ µ0) and relative permittivity ǫr
in the range 2 to about 10 or so.
A microstrip line nominally exhibits TEM field
structure. This structure is shown in Figure 3.17.
Note that electric and magnetic fields exist both in the
dielectric and in the space above the dielectric, which
is typically (but not always) air. This complex field
structure makes it difficult to describe microstrip line
concisely in terms of the equivalent circuit parameters
of the lumped-element model. Instead, expressions
for Z0 directly in terms of h/W and ǫr are typically
used instead. A variety of these expressions are in
common use, representing different approximations
and simplifications. A widely-accepted and",Electromagnetics_Vol1.pdf
"used instead. A variety of these expressions are in
common use, representing different approximations
and simplifications. A widely-accepted and
broadly-applicable expression is: 6
Z0 ≈ 42.4 Ω
√ǫr + 1 ×
ln
[
1 + 4h
W′
(
Φ +
√
Φ 2 + 1 + 1 /ǫr
2 π2
)]
(3.68)
where
Φ ≜ 14 + 8 /ǫr
11
( 4h
W′
)
(3.69)
and W′ is W adjusted to account for the thickness t
of the microstrip line. T ypically t≪ W and t≪ h,
for which W′ ≈ W. Simpler approximations for Z0
are also commonly employed in the design and
analysis of microstrip lines. These expressions are
limited in the range of h/W for which they are valid,
and can usually be shown to be special cases or
approximations of Equation 3.68. Nevertheless, they
are sometimes useful for quick “back of the envelope”
calculations.
Accurate expressions for wavelength λ, phase
propagation constant β, and phase velocity vp are
similarly difficult to obtain for waves in microstrip
line. An approximate technique employs a result from",Electromagnetics_Vol1.pdf
"similarly difficult to obtain for waves in microstrip
line. An approximate technique employs a result from
6 This is from Wheeler 1977, cited in “ Additional Reading” at th e
end of this section.
h
W
Figure 3.17: Structure of the electric and magnetic
fields within microstrip line. (The fields outside the
line are possibly significant, complicated, and not
shown.) In this case, the wave is propagating away
from the viewer.
the theory of uniform plane waves in unbounded
media (Equation 9.38 from Section 9.2):
β = ω√
µǫ (3.70)
It turns out that the electromagnetic field structure in
the space between the conductors is
well-approximated as that of a uniform plane wave in
unbounded media having the same permeability µ0
but a different relative permittivity , which we shall
assign the symbol ǫr,eff (for “ effective relative
permittivity”). Then
β ≈ ω√
µ0 ǫr,eff ǫ0 (low-loss microstrip)
= β0
√ǫr,eff (3.71)
In other words, the phase propagation constant in a",Electromagnetics_Vol1.pdf
"permittivity”). Then
β ≈ ω√
µ0 ǫr,eff ǫ0 (low-loss microstrip)
= β0
√ǫr,eff (3.71)
In other words, the phase propagation constant in a
microstrip line can be approximated as the free-space
phase propagation β0 ≜ ω√
µ0ǫ0 times a correction
factor √ǫr,eff. Then ǫr,eff may be crudely
approximated as follows:
ǫr,eff ≈ ǫr + 1
2 (3.72)
i.e., ǫr,eff is roughly the average of the relative
permittivity of the dielectric slab and the relative
permittivity of free space. The assumption employed
here is that ǫr,eff is approximately the average of
these values because some fraction of the power in
the guided wave is in the dielectric, and the rest is
above the dielectric. V arious approximations are
available to improve on this approximation; however,
in practice variations in the value of ǫr for the
dielectric due to manufacturing processes typically
make a more precise estimate irrelevant.",Electromagnetics_Vol1.pdf
"46 CHAPTER 3. TRANSMISSION LINES
Using this concept, we obtain
λ= 2π
β = 2π
β0
√ǫr,eff
= λ0
√ǫr,eff
(3.73)
where λ0 is the free-space wavelength c/f. Similarly
the phase velocity vp, can be estimated using the
relationship
vp = ω
β = c√ǫr,eff
(3.74)
i.e., the phase velocity in microstrip is slower than c
by a factor of √ǫr,eff.
Example 3.2. 50 Ω Microstrip in FR4 Printed
Circuit Boards.
FR4 is a low-loss fiberglass epoxy dielectric that
is commonly used to make printed circuit boards
(see “ Additional Reading” at the end of this
section). FR4 circuit board material is
commonly sold in a slab having thickness
h∼
= 1.575 mm with ǫr ∼
= 4.5. Let us consider
how we might implement a microstrip line
having Z0 = 50 Ω using this material. Since h
and ǫr are fixed, the only parameter remaining to
set Z0 is W. A bit of experimentation with
Equation 3.68 reveals that h/W ≈ 1/2 yields
Z0 ≈ 50 Ω for ǫr = 4 .5. Thus, W should be
about 3.15 mm. The effective relative
permittivity is",Electromagnetics_Vol1.pdf
"Equation 3.68 reveals that h/W ≈ 1/2 yields
Z0 ≈ 50 Ω for ǫr = 4 .5. Thus, W should be
about 3.15 mm. The effective relative
permittivity is
ǫr,eff ≈ (4.5 + 1) /2 = 2 .75
so the phase velocity for the wave guided by this
line is about c/
√
2.75; i.e., 60% of c. Similarly ,
the wavelength of this wave is about 60% of the
free space wavelength.
Additional Reading:
• “Microstrip” on Wikipedia.
• “Printed circuit board” on Wikipedia.
• “Stripline” on Wikipedia.
• “Single-ended signaling” on Wikipedia.
• Sec. 8.7 (“Differential Circuits”) in
S.W . Ellingson, Radio Systems Engineering ,
Cambridge Univ . Press, 2016.
• H.A. Wheeler, “Transmission Line Properties of
a Strip on a Dielectric Sheet on a Plane, ” IEEE
Trans. Microwave Theory & T echniques , V ol. 25,
No. 8, Aug 1977, pp. 631–47.
• “FR-4” on Wikipedia.",Electromagnetics_Vol1.pdf
"3.12. VOL T AGE REFLECTION COEFFICIENT 47
3.12 V oltage Reflection
Coefficient
[m0084]
W e now consider the scenario shown in Figure 3.18.
Here a wave arriving from the left along a lossless
transmission line having characteristic impedance Z0
arrives at a termination located at z= 0 . The
impedance looking into the termination is ZL, which
may be real-, imaginary-, or complex-valued. The
questions are: Under what circumstances is a
reflection – i.e., a leftward traveling wave – expected,
and what precisely is that wave?
The potential and current of the incident wave are
related by the constant value of Z0. Similarly , the
potential and current of the reflected wave are related
by Z0. Therefore, it suffices to consider either
potential or current. Choosing potential, we may
express the incident wave as
˜V+(z) = V+
0 e−jβz (3.75)
where V+
0 is determined by the source of the wave,
and so is effectively a “given. ” Any reflected wave
must have the form
˜V−(z) = V−
0 e+jβz (3.76)",Electromagnetics_Vol1.pdf
"where V+
0 is determined by the source of the wave,
and so is effectively a “given. ” Any reflected wave
must have the form
˜V−(z) = V−
0 e+jβz (3.76)
Therefore, the problem is solved by determining the
value of V−
0 given V+
0 , Z0, and ZL.
Considering the situation at z= 0 , note that by
ZLZ0
I
=0
V

+

L
L

z
z
V

+
V


Figure 3.18: A wave arriving from the left incident on
a termination located at z= 0 .
definition we have
ZL ≜
˜VL
˜IL
(3.77)
where ˜VL and ˜IL are the potential across and current
through the termination, respectively . Also, the
potential and current on either side of the z= 0
interface must be equal. Thus,
˜V+(0) + ˜V−(0) = ˜VL (3.78)
˜I+(0) + ˜I−(0) = ˜IL (3.79)
where ˜I+(z) and ˜I−(z) are the currents associated
with ˜V+(z) and ˜V−(z), respectively . Since the
voltage and current are related by Z0, Equation 3.79
may be rewritten as follows:
˜V+(0)
Z0
−
˜V−(0)
Z0
= ˜IL (3.80)
Evaluating the left sides of Equations 3.78 and 3.80 at
z= 0 , we find:
V+
0 + V−",Electromagnetics_Vol1.pdf
"may be rewritten as follows:
˜V+(0)
Z0
−
˜V−(0)
Z0
= ˜IL (3.80)
Evaluating the left sides of Equations 3.78 and 3.80 at
z= 0 , we find:
V+
0 + V−
0 = ˜VL (3.81)
V+
0
Z0
− V−
0
Z0
= ˜IL (3.82)
Substituting these expressions into Equation 3.77 we
obtain:
ZL = V+
0 + V−
0
V+
0 /Z0 − V−
0 /Z0
(3.83)
Solving for V−
0 we obtain
V−
0 = ZL − Z0
ZL + Z0
V+
0 (3.84)
Thus, the answer to the question posed earlier is that
V−
0 = Γ V+
0 , where (3.85)
Γ ≜ ZL − Z0
ZL + Z0
(3.86)
The quantity Γ is known as the voltage reflection
coefficient . Note that when ZL = Z0, Γ = 0 and
therefore V−
0 = 0 . In other words,",Electromagnetics_Vol1.pdf
"48 CHAPTER 3. TRANSMISSION LINES
If the terminating impedance is equal to the char-
acteristic impedance of the transmission line,
then there is no reflection.
If, on the other hand, ZL ̸= Z0, then |Γ | >0,
V−
0 = Γ V+
0 , and a leftward-traveling reflected wave
exists.
Since ZL may be real-, imaginary-, or
complex-valued, Γ too may be real-, imaginary-, or
complex-valued. Therefore, V−
0 may be different
from V+
0 in magnitude, sign, or phase.
Note also that Γ is not the ratio of I−
0 to I+
0 . The ratio
of the current coefficients is actually −Γ . It is quite
simple to show this with a simple modification to the
above procedure and is left as an exercise for the
student.
Summarizing:
The voltage reflection coefficient Γ , given by
Equation 3.86, determines the magnitude and
phase of the reflected wave given the incident
wave, the characteristic impedance of the trans-
mission line, and the terminating impedance.
[m0085]
W e now consider values of Γ that arise for",Electromagnetics_Vol1.pdf
"wave, the characteristic impedance of the trans-
mission line, and the terminating impedance.
[m0085]
W e now consider values of Γ that arise for
commonly-encountered terminations.
Matched Load ( ZL = Z0). In this case, the
termination may be a device with impedance Z0, or
the termination may be another transmission line
having the same characteristic impedance. When
ZL = Z0, Γ = 0 and there is no reflection.
Open Circuit. An “open circuit” is the absence of a
termination. This condition implies ZL → ∞ , and
subsequently Γ → +1. Since the current reflection
coefficient is −Γ , the reflected current wave is 180◦
out of phase with the incident current wave, making
the total current at the open circuit equal to zero, as
expected.
Short Circuit. “Short circuit” means ZL = 0 , and
subsequently Γ = −1. In this case, the phase of Γ is
180◦, and therefore, the potential of the reflected
wave cancels the potential of the incident wave at the
open circuit, making the total potential equal to zero,",Electromagnetics_Vol1.pdf
"wave cancels the potential of the incident wave at the
open circuit, making the total potential equal to zero,
as it must be. Since the current reflection coefficient
is −Γ = +1 in this case, the reflected current wave is
in phase with the incident current wave, and the
magnitude of the total current at the short circuit
non-zero as expected.
Purely Reactive Load. A purely reactive load,
including that presented by a capacitor or inductor,
has ZL = jX where X is reactance. In particular, an
inductor is represented by X >0 and a capacitor is
represented by X <0. W e find
Γ = −Z0 + jX
+Z0 + jX (3.87)
The numerator and denominator have the same
magnitude, so |Γ | = 1 . Let φbe the phase of the
denominator ( +Z0 + jX). Then, the phase of the
numerator is π− φ. Subsequently , the phase of Γ is
(π− φ) − φ= π− 2φ. Thus, we see that the phase of
Γ is no longer limited to be 0◦ or 180◦, but can be any
value in between. The phase of reflected wave is
subsequently shifted by this amount.",Electromagnetics_Vol1.pdf
"Γ is no longer limited to be 0◦ or 180◦, but can be any
value in between. The phase of reflected wave is
subsequently shifted by this amount.
Other T erminations. Any other termination,
including series and parallel combinations of any
number of devices, can be expressed as a value of ZL
which is, in general, complex-valued. The associated",Electromagnetics_Vol1.pdf
"3.13. ST ANDING W A VES 49
value of |Γ | is limited to the range 0 to 1. T o see this,
note:
Γ = ZL − Z0
ZL + Z0
= ZL/Z0 − 1
ZL/Z0 + 1 (3.88)
Note that the smallest possible value of |Γ | occurs
when the numerator is zero; i.e., when ZL = Z0.
Therefore, the smallest value of |Γ | is zero. The
largest possible value of |Γ | occurs when
ZL/Z0 → ∞ (i.e., an open circuit) or when
ZL/Z0 = 0 (a short circuit); the result in either case
is |Γ | = 1 . Thus,
0 ≤ | Γ | ≤ 1 (3.89)
3.13 Standing W aves
[m0086]
A standing wave consists of waves moving in op-
posite directions. These waves add to make a
distinct magnitude variation as a function of dis-
tance that does not vary in time.
T o see how this can happen, first consider that an
incident wave V+
0 e−jβz, which is traveling in the +z
axis along a lossless transmission line. Associated
with this wave is a reflected wave
V−
0 e+jβz = Γ V+
0 e+jβz, where Γ is the voltage
reflection coefficient. These waves add to make the
total potential
˜V(z) = V+",Electromagnetics_Vol1.pdf
"V−
0 e+jβz = Γ V+
0 e+jβz, where Γ is the voltage
reflection coefficient. These waves add to make the
total potential
˜V(z) = V+
0 e−jβz + Γ V+
0 e+jβz
= V+
0
(
e−jβz + Γ e+jβz) (3.90)
The magnitude of ˜V(z) is most easily found by first
finding |˜V(z)|2, which is:
˜V(z) ˜V∗(z)
=|V+
0 |2 (
e−jβz + Γ e+jβz)(
e−jβz + Γ e+jβz) ∗
=|V+
0 |2 (
e−jβz + Γ e+jβz)(
e+jβz + Γ ∗e−jβz)
=|V+
0 |2
(
1 + |Γ |2 + Γ e+j2βz + Γ ∗e−j2βz
)
(3.91)
Let φbe the phase of Γ ; i.e.,
Γ = |Γ | ejφ (3.92)
Then, continuing from the previous expression:
|V+
0 |2
(
1 + |Γ |2 + |Γ | e+j(2βz+φ) + |Γ | e−j(2βz+φ)
)
=|V+
0 |2
(
1 + |Γ |2 + |Γ |
[
e+j(2βz+φ) + e−j(2βz+φ)
])
(3.93)
The quantity in square brackets can be reduced to a
cosine function using the identity
cos θ= 1
2
[
ejθ + e−jθ]
yielding:
|V+
0 |2
[
1 + |Γ |2 + 2 |Γ | cos (2βz+ φ)
]
(3.94)
Recall that this is | ˜V(z)|2. | ˜V(z)| is therefore the
square root of the above expression:
⏐
⏐
⏐˜V(z)
⏐
⏐
⏐= |V+
0 |
√
1 + |Γ |2 + 2 |Γ | cos (2βz+ φ)
(3.95)",Electromagnetics_Vol1.pdf
"square root of the above expression:
⏐
⏐
⏐˜V(z)
⏐
⏐
⏐= |V+
0 |
√
1 + |Γ |2 + 2 |Γ | cos (2βz+ φ)
(3.95)
Thus, we have found that the magnitude of the
resulting total potential varies sinusoidally along the
line. This is referred to as a standing wave because
the variation of the magnitude of the phasor resulting
from the interference between the incident and
reflected waves does not vary with time.
W e may perform a similar analysis of the current,
leading to:
⏐
⏐
⏐˜I(z)
⏐
⏐
⏐= |V+
0 |
Z0
√
1 + |Γ |2 − 2 |Γ | cos (2βz+ φ)
(3.96)
Again we find the result is a standing wave.
Now let us consider the outcome for a few special
cases.
Matched load. When the impedance of the
termination of the transmission line, ZL, is equal to
the characteristic impedance of the transmission line,
Z0, Γ = 0 and there is no reflection. In this case, the
above expressions reduce to | ˜V(z)| = |V+
0 | and
|˜I(z)| = |V+
0 |/Z0, as expected.
Open or Short-Circuit. In this case, Γ = ±1 and
we find:
⏐
⏐
⏐˜V(z)
⏐
⏐",Electromagnetics_Vol1.pdf
"0 | and
|˜I(z)| = |V+
0 |/Z0, as expected.
Open or Short-Circuit. In this case, Γ = ±1 and
we find:
⏐
⏐
⏐˜V(z)
⏐
⏐
⏐= |V+
0 |
√
2 + 2 cos (2 βz+ φ) (3.97)",Electromagnetics_Vol1.pdf
"50 CHAPTER 3. TRANSMISSION LINES
z
V
+
02
V
+
0
V(z)

 / 
z=0
(a) Potential.
I(z)

 ! ""
z=0
z
I
+
02
I
+
0
(b) Current.
Figure 3.19: Standing wave associated with an open-
circuit termination at z = 0 (incident wave arrives
from left).
⏐
⏐
⏐˜I(z)
⏐
⏐
⏐= |V+
0 |
Z0
√
2 − 2 cos (2βz+ φ) (3.98)
where φ= 0 for an open circuit and φ= πfor a short
circuit. The result for an open circuit termination is
shown in Figure 3.19(a) (potential) and 3.19(b)
(current). The result for a short circuit termination is
identical except the roles of potential and current are
reversed. In either case, note that voltage maxima
correspond to current minima, and vice versa.
Also note:
The period of the standing wave is λ/2; i.e., one-
half of a wavelength.
This can be confirmed as follows. First, note that the
frequency argument of the cosine function of the
standing wave is 2βz. This can be rewritten as
1.25
1.00
0.75
0.50
0.00
0 
#
2
$ 3%
& z
1.50
2.00
'
2
3(
2
5)
2
1.75
0.25 |V| 0/ V
[rad]
Γ* , - .
Γ",Electromagnetics_Vol1.pdf
"standing wave is 2βz. This can be rewritten as
1.25
1.00
0.75
0.50
0.00
0 
#
2
$ 3%
& z
1.50
2.00
'
2
3(
2
5)
2
1.75
0.25 |V| 0/ V
[rad]
Γ* , - .
Γ
0 1 2 3 4
Γ=1
Γ
5 6 7 8 9
Γ: ;
by Inductiveload (modified)
Figure 3.20: Standing waves associated with loads
exhibiting various reflection coefficients. In this figure
the incident wave arrives from the right.
2π(β/π) z, so the frequency of variation is β/π and
the period of the variation is π/β. Since β = 2 π/λ,
we see that the period of the variation is λ/2.
Furthermore, this is true regardless of the value of Γ .
Mismatched loads. A common situation is that the
termination is neither perfectly-matched ( Γ = 0 ) nor
an open/short circuit ( |Γ | = 1 ). Examples of the
resulting standing waves are shown in Figure 3.20.
Additional Reading:
• “Standing W ave” on Wikipedia.",Electromagnetics_Vol1.pdf
"3.14. ST ANDING W A VE RA TIO 51
3.14 Standing W ave Ratio
[m0081]
Precise matching of transmission lines to terminations
is often not practical or possible. Whenever a
significant mismatch exists, a standing wave
(Section 3.13) is apparent. The quality of the match is
commonly expressed in terms of the standing wave
ratio (SWR) of this standing wave.
Standing wave ratio (SWR) is defined as the ratio
of the maximum magnitude of the standing wave
to minimum magnitude of the standing wave.
In terms of the potential:
SWR ≜ maximum | ˜V|
minimum | ˜V|
(3.99)
SWR can be calculated using a simple expression,
which we shall now derive. In Section 3.13, we found
that:
⏐
⏐
⏐˜V(z)
⏐
⏐
⏐= |V+
0 |
√
1 + |Γ |2 + 2 |Γ | cos (2βz+ φ)
(3.100)
The maximum value occurs when the cosine factor is
equal to +1, yielding:
max
⏐
⏐
⏐˜V
⏐
⏐
⏐= |V+
0 |
√
1 + |Γ |2 + 2 |Γ | (3.101)
Note that the argument of the square root operator is
equal to (1 + |Γ |)2; therefore:
max
⏐
⏐
⏐˜V
⏐
⏐
⏐= |V+
0 | (1 + |Γ |) (3.102)",Electromagnetics_Vol1.pdf
"Note that the argument of the square root operator is
equal to (1 + |Γ |)2; therefore:
max
⏐
⏐
⏐˜V
⏐
⏐
⏐= |V+
0 | (1 + |Γ |) (3.102)
Similarly , the minimum value is achieved when the
cosine factor is equal to −1, yielding:
min
⏐
⏐
⏐˜V
⏐
⏐
⏐= |V+
0 |
√
1 + |Γ |2 − 2 |Γ | (3.103)
So:
min
⏐
⏐
⏐˜V
⏐
⏐
⏐= |V+
0 | (1 − |Γ |) (3.104)
Therefore:
SWR = 1 + |Γ |
1 − |Γ | (3.105)
0
2
4
6
8
10
0 0.2 0.4 0.6 0.8 1
Γ
<? @
Figure 3.21: Relationship between SWR and |Γ |.
This relationship is shown graphically in Figure 3.21.
Note that SWR ranges from 1 for perfectly-matched
terminations ( Γ = 0 ) to infinity for open- and
short-circuit terminations ( |Γ | = 1 ).
It is sometimes of interest to find the magnitude of the
reflection coefficient given SWR. Solving
Equation 3.105 for |Γ | we find:
|Γ | = SWR − 1
SWR + 1 (3.106)
SWR is often referred to as the voltage standing wave
ratio (VSWR), although repeating the analysis above
for the current reveals that the current SWR is equal",Electromagnetics_Vol1.pdf
"ratio (VSWR), although repeating the analysis above
for the current reveals that the current SWR is equal
to potential SWR, so the term “SWR” suffices.
SWR <2 or so is usually considered a “good match, ”
although some applications require SWR <1.1 or
better, and other applications are tolerant to SWR of 3
or greater.
Example 3.3. Reflection coefficient for various
values of SWR.
What is the reflection coefficient for the
above-cited values of SWR? Using
Equation 3.106, we find:
• SWR = 1.1 corresponds to |Γ | = 0 .0476.",Electromagnetics_Vol1.pdf
"52 CHAPTER 3. TRANSMISSION LINES
• SWR = 2.0 corresponds to |Γ | = 1 /3.
• SWR = 3.0 corresponds to |Γ | = 1 /2.
3.15 Input Impedance of a
T erminated Lossless
T ransmission Line
[m0087]
Consider Figure 3.22, which shows a lossless
transmission line being driven from the left and which
is terminated by an impedance ZL on the right. If ZL
is equal to the characteristic impedance Z0 of the
transmission line, then the input impedance Zin will
be equal to ZL. Otherwise Zin depends on both ZL
and the characteristics of the transmission line. In this
section, we determine a general expression for Zin in
terms of ZL, Z0, the phase propagation constant β,
and the length lof the line.
Using the coordinate system indicated in Figure 3.22,
the interface between source and transmission line is
located at z= −l. Impedance is defined at the ratio of
potential to current, so:
Zin(l) ≜
˜V(z= −l)
˜I(z= −l)
(3.107)
Now employing expressions for ˜V(z) and ˜I(z) from
Section 3.13 with z= −l, we find:",Electromagnetics_Vol1.pdf
"potential to current, so:
Zin(l) ≜
˜V(z= −l)
˜I(z= −l)
(3.107)
Now employing expressions for ˜V(z) and ˜I(z) from
Section 3.13 with z= −l, we find:
Zin(l) = V+
0
(
e+jβl + Γ e−jβl)
V+
0 (e+jβl − Γ e−jβl) /Z0
= Z0
e+jβl + Γ e−jβl
e+jβl − Γ e−jβl (3.108)
Multiplying both numerator and denominator by
ZLZ0
=0
z
z
C
=  lz
l
Zin
Figure 3.22: A transmission line driven by a source on
the left and terminated by an impedance ZL at z = 0
on the right.",Electromagnetics_Vol1.pdf
"3.15. INPUT IMPEDANCE OF A TERMINA TED LOSSLESS TRANSMISSIO N LINE 53
e−jβl:
Zin(l) = Z0
1 + Γ e−j2βl
1 − Γ e−j2βl (3.109)
Recall that Γ in the above expression is:
Γ = ZL − Z0
ZL + Z0
(3.110)
Summarizing:
Equation 3.109 is the input impedance of a
lossless transmission line having characteristic
impedance Z0 and which is terminated into a load
ZL. The result also depends on the length and
phase propagation constant of the line.
Note that Zin(l) is periodic in l. Since the argument
of the complex exponential factors is 2βl, the
frequency at which Zin(l) varies is β/π; and since
β = 2 π/λ, the associated period is λ/2. This is very
useful to keep in mind because it means that all
possible values of Zin(l) are achieved by varying l
over λ/2. In other words, changing lby more than
λ/2 results in an impedance which could have been
obtained by a smaller change in l. Summarizing to
underscore this important idea:
The input impedance of a terminated lossless",Electromagnetics_Vol1.pdf
"obtained by a smaller change in l. Summarizing to
underscore this important idea:
The input impedance of a terminated lossless
transmission line is periodic in the length of the
transmission line, with period λ/2.
Not surprisingly , λ/2 is also the period of the
standing wave (Section 3.13). This is because – once
again – the variation with length is due to the
interference of incident and reflected waves.
Also worth noting is that Equation 3.109 can be
written entirely in terms of ZL and Z0, since Γ
depends only on these two parameters. Here’s that
version of the expression:
Zin(l) = Z0
[ ZL + jZ0 tan βl
Z0 + jZLtan βl
]
(3.111)
This expression can be derived by substituting
Equation 3.110 into Equation 3.109 and is left as an
exercise for the student.
Finally , note that the argument βlappearing
Equations 3.109 and 3.111 has units of radians and is
referred to as electrical length . Electrical length can
be interpreted as physical length expressed with",Electromagnetics_Vol1.pdf
"referred to as electrical length . Electrical length can
be interpreted as physical length expressed with
respect to wavelength and has the advantage that
analysis can be made independent of frequency .",Electromagnetics_Vol1.pdf
"54 CHAPTER 3. TRANSMISSION LINES
3.16 Input Impedance for Open-
and Short-Circuit
T erminations
[m0088]
Let us now consider the input impedance of a
transmission line that is terminated in an open- or
short-circuit. Such a transmission line is sometimes
referred to as a stub. First, why consider such a
thing? From a “lumped element” circuit theory
perspective, this would not seem to have any
particular application. However, the fact that this
structure exhibits an input impedance that depends on
length (Section 3.15) enables some very useful
applications.
First, let us consider the question at hand: What is the
input impedance when the transmission line is open-
or short-circuited? For a short circuit, ZL = 0 ,
Γ = −1, so we find
Zin(l) = Z0
1 + Γ e−j2βl
1 − Γ e−j2βl
= Z0
1 − e−j2βl
1 + e−j2βl (3.112)
Multiplying numerator and denominator by e+jβl we
obtain
Zin(l) = Z0
e+jβl − e−jβl
e+jβl + e−jβl (3.113)
Now we invoke the following trigonometric identities:
cos θ= 1
2
[
e+jθ + e−jθ]",Electromagnetics_Vol1.pdf
"obtain
Zin(l) = Z0
e+jβl − e−jβl
e+jβl + e−jβl (3.113)
Now we invoke the following trigonometric identities:
cos θ= 1
2
[
e+jθ + e−jθ]
(3.114)
sin θ= 1
j2
[
e+jθ − e−jθ]
(3.115)
Employing these identities, we obtain:
Zin(l) = Z0
j2 (sin βl)
2 (cos βl) (3.116)
and finally:
Zin(l) = + jZ0 tan βl (3.117)
Figure 3.23(a) shows what’s going on. As expected,
Zin = 0 when l= 0 , since this amounts to a short
 0
 0  0.25  0.5  0.75  1
imaginary part of input impedance
length [wavelengths]
(a) Short-circuit termination ( ZL = 0 ).
 0
 0  0.25  0.5  0.75  1
imaginary part of input impedance
length [wavelengths]
(b) Open-circuit termination ( ZL → ∞ ).
Figure 3.23: Input reactance (Im {Zin}) of a stub.
Re{Zin} is always zero.
circuit with no transmission line. Also, Zin varies
periodically with increasing length, with period λ/2.
This is precisely as expected from standing wave
theory (Section 3.13). What is of particular interest
now is that as l→ λ/4, we see Zin → ∞ .",Electromagnetics_Vol1.pdf
"This is precisely as expected from standing wave
theory (Section 3.13). What is of particular interest
now is that as l→ λ/4, we see Zin → ∞ .
Remarkably , the transmission line has essentially
transformed the short circuit termination into an open
circuit!
For an open circuit termination, ZL → ∞ , Γ = +1 ,
and we find
Zin(l) = Z0
1 + Γ e−j2βl
1 − Γ e−j2βl
= Z0
1 + e−j2βl
1 − e−j2βl (3.118)",Electromagnetics_Vol1.pdf
"3.17. APPLICA TIONS OF OPEN- AND SHOR T -CIRCUITED TRANSMISS ION LINE STUBS 55
Following the same procedure detailed above for the
short-circuit case, we find
Zin(l) = −jZ0 cot βl
(3.119)
Figure 3.23(b) shows the result for open-circuit
termination. As expected, Zin → ∞ for l= 0 , and
the same λ/2 periodicity is observed. What is of
particular interest now is that at l= λ/4 we see
Zin = 0 . In this case, the transmission line has
transformed the open circuit termination into a short
circuit.
Now taking stock of what we have determined:
The input impedance of a short- or open-
circuited lossless transmission line is com-
pletely imaginary-valued and is given by Equa-
tions 3.117 and 3.119, respectively .
The input impedance of a short- or open-circuited
lossless transmission line alternates between
open- ( Zin → ∞ ) and short-circuit ( Zin = 0 )
conditions with each λ/4-increase in length.
Additional Reading:
• “Stub (electronics)” on Wikipedia.
3.17 Applications of Open- and",Electromagnetics_Vol1.pdf
"conditions with each λ/4-increase in length.
Additional Reading:
• “Stub (electronics)” on Wikipedia.
3.17 Applications of Open- and
Short-Circuited
T ransmission Line Stubs
[m0145]
The theory of open- and short-circuited transmission
lines – often referred to as stubs – was addressed in
Section 3.16. These structures have important and
wide-ranging applications.
In particular, these structures can be used to replace
discrete inductors and capacitors in certain
applications. T o see this, consider the short-circuited
line (Figure 3.23(b) of Section 3.16). Note that each
value of lthat is less than λ/4 corresponds to a
particular positive reactance; i.e., the transmission
line “looks” like an inductor. Also note that lengths
between λ/4 and λ/2 result in reactances that are
negative; i.e., the transmission line “looks” like a
capacitor. Thus, it is possible to replace an inductor or
capacitor with a short-circuited transmission line of",Electromagnetics_Vol1.pdf
"capacitor. Thus, it is possible to replace an inductor or
capacitor with a short-circuited transmission line of
the appropriate length. The input impedance of such a
transmission line is identical to that of the inductor or
capacitor at the design frequency . The variation of
reactance with respect to frequency will not be
identical, which may or may not be a concern
depending on the bandwidth and frequency response
requirements of the application. Open-circuited lines
may be used in a similar way .
This property of open- and short-circuited
transmission lines makes it possible to implement
impedance matching circuits (see Section 3.23),
filters, and other devices entirely from transmission
lines, with fewer or no discrete inductors or capacitors
required. Transmission lines do not suffer the
performance limitations of discrete devices at high
frequencies and are less expensive. A drawback of
transmission line stubs in this application is that the",Electromagnetics_Vol1.pdf
"frequencies and are less expensive. A drawback of
transmission line stubs in this application is that the
lines are typically much larger than the discrete
devices they are intended to replace.
Example 3.4. Emitter Induction Using
Short-Circuited Line.
In the design of low-noise amplifiers using",Electromagnetics_Vol1.pdf
"56 CHAPTER 3. TRANSMISSION LINES
bipolar transistors in common-emitter
configuration, it is often useful to introduce a
little inductance between the emitter and ground.
This is known as “inductive degeneration, ”
“emitter induction, ” or sometimes by other
names. It can be difficult to find suitable
inductors, especially for operation in the UHF
band and higher. However, a microstrip line can
be used to achieve the desired inductive
impedance. Determine the length of a stub that
implements a 2.2 nH inductance at 6 GHz using
microstrip line with characteristic impedance
50 Ω and phase velocity 0.6c.
Solution. At the design frequency , the
impedance looking into this section of line from
the emitter should be equal to that of a 2.2 nH
inductor, which is
+jωL = + j2πfL = + j82.9 Ω . The input
impedance of a short-circuited stub of length l
which is grounded (thus, short-circuited) at the
opposite end is +jZ0 tan βl(Section 3.16).
Setting this equal to +j82.9 Ω and noting that",Electromagnetics_Vol1.pdf
"which is grounded (thus, short-circuited) at the
opposite end is +jZ0 tan βl(Section 3.16).
Setting this equal to +j82.9 Ω and noting that
Z0 = 50 Ω , we find that βl ∼
= 1.028 rad. The
phase propagation constant is (Section 3.8):
β = ω
vp
= 2πf
0.6c
∼
= 209.4 rad/m (3.120)
Therefore, the length of the microstrip line is
l= ( βl) /β ∼
= 4.9 mm.
Additional Reading:
• “Stub (electronics)” on Wikipedia.
3.18 Measurement of
T ransmission Line
Characteristics
[m0089]
This section presents a simple technique for
measuring the characteristic impedance Z0, electrical
length βl, and phase velocity vp of a lossless
transmission line. This technique requires two
measurements: the input impedance Zin when the
transmission line is short-circuited and Zin when the
transmission line is open-circuited.
In Section 3.16, it is shown that the input impedance
Zin of a short-circuited transmission line is
Z(SC)
in = + jZ0 tan βl
and when a transmission line is terminated in an open",Electromagnetics_Vol1.pdf
"Zin of a short-circuited transmission line is
Z(SC)
in = + jZ0 tan βl
and when a transmission line is terminated in an open
circuit, the input impedance is
Z(OC)
in = −jZ0 cot βl
Observe what happens when we multiply these results
together:
Z(SC)
in · Z(OC)
in = Z2
0
that is, the product of the measurements Z(OC)
in and
Z(SC)
in is simply the square of the characteristic
impedance. Therefore
Z0 =
√
Z(SC)
in · Z(OC)
in (3.121)
If we instead divide these measurements, we find
Z(SC)
in
Z(OC)
in
= − tan2 βl
Therefore:
tan βl =
[
− Z(SC)
in
Z(OC)
in
] 1/2
(3.122)
If lis known in advance to be less than λ/2, then the
electrical length βlcan be determined by taking the
inverse tangent. If lis of unknown length and longer
than λ/2, one must take care to account for the
periodicity of tangent function; in this case, it may not",Electromagnetics_Vol1.pdf
"3.19. QUAR TER-W A VELENGTH TRANSMISSION LINE 57
be possible to unambiguously determine βl. Although
we shall not present the method here, it is possible to
resolve this ambiguity by making multiple
measurements over a range of frequencies.
Once βlis determined, it is simple to determine l
given β, βgiven l, and then vp. For example, the
phase velocity may be determined by first finding βl
for a known length using the above procedure,
calculating β = ( βl) /l, and then vp = ω/β.
3.19 Quarter-W avelength
T ransmission Line
[m0091]
Quarter-wavelength sections of transmission line play
an important role in many systems at radio and optical
frequencies. The remarkable properties of open- and
short-circuited quarter-wave line are presented in
Section 3.16 and should be reviewed before reading
further. In this section, we perform a more general
analysis, considering not just open- and short-circuit
terminations but any terminating impedance, and then
we address some applications.",Electromagnetics_Vol1.pdf
"analysis, considering not just open- and short-circuit
terminations but any terminating impedance, and then
we address some applications.
The general expression for the input impedance of a
lossless transmission line is (Section 3.15):
Zin(l) = Z0
1 + Γ e−j2βl
1 − Γ e−j2βl (3.123)
Note that when l= λ/4:
2βl = 2 · 2π
λ · λ
4 = π
Subsequently:
Zin(λ/4) = Z0
1 + Γ e−jπ
1 − Γ e−jπ
= Z0
1 − Γ
1 + Γ
(3.124)
Recall that (Section 3.15):
Γ = ZL − Z0
ZL + Z0
(3.125)
Substituting this expression and then multiplying
numerator and denominator by ZL + Z0, one obtains
Zin(λ/4) = Z0
(ZL + Z0) − (ZL − Z0)
(ZL + Z0) + ( ZL − Z0)
= Z0
2Z0
2ZL
(3.126)
Thus,
Zin(λ/4) = Z2
0
ZL
(3.127)
Note that the input impedance is inversely
proportional to the load impedance. For this reason, a",Electromagnetics_Vol1.pdf
"58 CHAPTER 3. TRANSMISSION LINES
  
l= /4
Figure 3.24: Impedance-matching using a quarter-
wavelength transmission line.
transmission line of length λ/4 is sometimes referred
to as a quarter-wave inverter or simply as a
impedance inverter .
Quarter-wave lines play a very important role in RF
engineering. As impedance inverters, they have the
useful attribute of transforming small impedances into
large impedances, and vice-versa – we’ll come back
to this idea later in this section. First, let’s consider
how quarter-wave lines are used for impedance
matching. Look what happens when we solve
Equation 3.127 for Z0:
Z0 =
√
Zin(λ/4) · ZL (3.128)
This equation indicates that we may match the load
ZL to a source impedance (represented by Zin(λ/4))
simply by making the characteristic impedance equal
to the value given by the above expression and setting
the length to λ/4. The scheme is shown in
Figure 3.24.
Example 3.5. 300-to-50 Ω match using an
quarter-wave section of line.",Electromagnetics_Vol1.pdf
"the length to λ/4. The scheme is shown in
Figure 3.24.
Example 3.5. 300-to-50 Ω match using an
quarter-wave section of line.
Design a transmission line segment that matches
300 Ω to 50 Ω at 10 GHz using a quarter-wave
match. Assume microstrip line for which
propagation occurs with wavelength 60% that of
free space.
Solution. The line is completely specified given
its characteristic impedance Z0 and length l. The
length should be one-quarter wavelength with
respect to the signal propagating in the line. The
free-space wavelength λ0 = c/f at 10 GHz is
∼
= 3 cm. Therefore, the wavelength of the signal
in the line is λ= 0 .6λ0 ∼
= 1.8 cm, and the length
of the line should be l= λ/4 ∼
= 4.5 mm.
The characteristic impedance is given by
Equation 3.128:
Z0 =
√
300 Ω · 50 Ω ∼
= 122.5 Ω (3.129)
This value would be used to determine the width
of the microstrip line, as discussed in
Section 3.11.
It should be noted that for this scheme to yield a",Electromagnetics_Vol1.pdf
"of the microstrip line, as discussed in
Section 3.11.
It should be noted that for this scheme to yield a
real-valued characteristic impedance, the product of
the source and load impedances must be a real-valued
number. In particular, this method is not suitable if
ZL has a significant imaginary-valued component and
matching to a real-valued source impedance is
desired. One possible workaround in this case is the
two-stage strategy shown in Figure 3.25. In this
scheme, the load impedance is first transformed to a
real-valued impedance using a length l1 of
transmission line. This is accomplished using
Equation 3.123 (quite simple using a numerical
search) or using the Smith chart (see “ Additional
Reading” at the end of this section). The characteristic
impedance Z01 of this transmission line is not critical
and can be selected for convenience. Normally , the
smallest value of l1 is desired. This value will always
be less than λ/4 since Zin(l1) is periodic in l1 with",Electromagnetics_Vol1.pdf
"smallest value of l1 is desired. This value will always
be less than λ/4 since Zin(l1) is periodic in l1 with
period λ/2; i.e., there are two changes in the sign of
the imaginary component of Zin(l1) as l1 is increased
from zero to λ/2. After eliminating the imaginary
component of ZL in this manner, the real component
of the resulting impedance may then be transformed
using the quarter-wave matching technique described
earlier in this section.
Example 3.6. Matching a patch antenna to
50 Ω .
A particular patch antenna exhibits a source
impedance of ZA = 35 + j35 Ω . (See
“Microstrip antenna” in “ Additional Reading” at
the end of this section for some optional reading
on patch antennas.) Interface this antenna to
50 Ω using the technique described above. For
the section of transmission line adjacent to the",Electromagnetics_Vol1.pdf
"3.19. QUAR TER-W A VELENGTH TRANSMISSION LINE 59
  
L
 
 
D /4
(completely
real-valued)
Figure 3.25: Impedance-matching a complex-valued
load impedance using quarter-wavelength transmis-
sion line.
patch antenna, use characteristic impedance
Z01 = 50 Ω . Determine the lengths l1 and l2 of
the two segments of transmission line, and the
characteristic impedance Z02 of the second
(quarter-wave) segment.
Solution. The length of the first section of the
transmission line (adjacent to the antenna) is
determined using Equation 3.123:
Z1(l1) = Z01
1 + Γ e−j2β1l1
1 − Γ e−j2β1l1
(3.130)
where β1 is the phase propagation constant for
this section of transmission line and
Γ ≜ ZA − Z01
ZA + Z01
∼
= −0.0059 + j0.4142 (3.131)
W e seek the value of smallest positive value of
β1l1 for which the imaginary part of Z1(l1) is
zero. This can determined using a Smith chart
(see “ Additional Reading” at the end of this
section) or simply by a few iterations of
trial-and-error. Either way we find",Electromagnetics_Vol1.pdf
"(see “ Additional Reading” at the end of this
section) or simply by a few iterations of
trial-and-error. Either way we find
Z1(β1l1 = 0 .793 rad) ∼
= 120.719 − j0.111 Ω ,
which we deem to be close enough to be
acceptable. Note that β1 = 2 π/λ, where λis the
wavelength of the signal in the transmission line.
Therefore
l1 = β1l1
β1
= β1l1
2π λ∼
= 0.126λ (3.132)
The length of the second section of the
transmission line, being a quarter-wavelength
transformer, should be l2 = 0 .25λ. Using
Equation 3.128, the characteristic impedance
Z02 of this section of line should be
Z02 ∼
=
√
(120.719 Ω) (50 Ω) ∼
= 77.7 Ω
(3.133)
Discussion. The total length of the matching
structure is l1 + l2 ∼
= 0.376λ. A patch antenna
would typically have sides of length about
λ/2 = 0 .5λ, so the matching structure is nearly
as big as the antenna itself. At frequencies where
patch antennas are commonly used, and
especially at frequencies in the UHF
(300–3000 MHz) band, patch antennas are often",Electromagnetics_Vol1.pdf
"patch antennas are commonly used, and
especially at frequencies in the UHF
(300–3000 MHz) band, patch antennas are often
comparable to the size of the system, so it is not
attractive to have the matching structure also
require a similar amount of space. Thus, we
would be motivated to find a smaller matching
structure.
Although quarter-wave matching techniques are
generally effective and commonly used, they have one
important contraindication, noted above – They often
result in structures that are large. That is, any
structure which employs a quarter-wave match will be
at least λ/4 long, and λ/4 is typically large compared
to the associated electronics. Other transmission line
matching techniques – and in particular, single stub
matching (Section 3.23) – typically result in
structures which are significantly smaller.
The impedance inversion property of
quarter-wavelength lines has applications beyond
impedance matching. The following example
demonstrates one such application:",Electromagnetics_Vol1.pdf
"quarter-wavelength lines has applications beyond
impedance matching. The following example
demonstrates one such application:
Example 3.7. RF/DC decoupling in transistor
amplifiers.
Transistor amplifiers for RF applications often
receive DC current at the same terminal which
delivers the amplified RF signal, as shown in
Figure 3.26. The power supply typically has a
low output impedance. If the power supply is
directly connected to the transistor, then the RF",Electromagnetics_Vol1.pdf
"60 CHAPTER 3. TRANSMISSION LINES
RF out
(moderate 
Z )
DC power in
(low 
E )
Figure 3.26: Use of an inductor to decouple the DC
input power from the RF output signal at the output of
a common-emitter RF amplifier.
will flow predominantly in the direction of the
power supply as opposed to following the
desired path, which exhibits a higher impedance.
This can be addressed using an inductor in series
with the power supply output. This works
because the inductor exhibits low impedance at
DC and high impedance at RF . Unfortunately ,
discrete inductors are often not practical at high
RF frequencies. This is because practical
inductors also exhibit parallel capacitance,
which tends to decrease impedance.
A solution is to replace the inductor with a
transmission line having length λ/4 as shown in
Figure 3.27. A wavelength at DC is infinite, so
the transmission line is essentially transparent to
the power supply . At radio frequencies, the line
transforms the low impedance of the power",Electromagnetics_Vol1.pdf
"the transmission line is essentially transparent to
the power supply . At radio frequencies, the line
transforms the low impedance of the power
supply to an impedance that is very large relative
to the impedance of the desired RF path.
Furthermore, transmission lines on printed
circuit boards are much cheaper than discrete
inductors (and are always in stock!).
Additional Reading:
• “ Quarter-wavelength impedance transformer ” on
Wikipedia.
• “ Smith chart ” on Wikipedia.
• “Microstrip antenna” on Wikipedia.
RF out
(moderate 
F )
DC power in
(low 
G
)
H /4
Figure 3.27: Decoupling of DC input power and
RF output signal at the output of a common-emitter
RF amplifier, using a quarter-wavelength transmission
line.
3.20 Power Flow on T ransmission
Lines
[m0090]
It is often important to know the power associated
with a wave on a transmission line. The power of the
waves incident upon, reflected by , and absorbed by a
load are each of interest. In this section we shall work",Electromagnetics_Vol1.pdf
"waves incident upon, reflected by , and absorbed by a
load are each of interest. In this section we shall work
out expressions for these powers and consider some
implications in terms of the voltage reflection
coefficient ( Γ ) and standing wave ratio (SWR).
Let’s begin by considering a lossless transmission line
that is oriented along the zaxis. The time-average
power associated with a sinusoidal wave having
potential v(z,t) and current i(z,t) is
Pav(z) ≜ 1
T
∫ t0+T
t0
v(z,t) i(z,t) dt (3.134)
where T ≜ 2π/f is one period of the wave and t0 is
the start time for the integration. Since the
time-average power of a sinusoidal signal does not
change with time, t0 may be set equal to zero without
loss of generality .
Let us now calculate the power of a wave incident
from z <0 on a load impedance ZL at z= 0 . W e
may express the associated potential and current as
follows:
v+(z,t) =
⏐
⏐V+
0
⏐
⏐cos (ωt− βz+ φ) (3.135)",Electromagnetics_Vol1.pdf
"3.20. POWER FLOW ON TRANSMISSION LINES 61
i+(z,t) =
⏐
⏐V+
0
⏐
⏐
Z0
cos (ωt− βz+ φ) (3.136)
And so the associated time-average power is
P+
av(z) = 1
T
∫ T
0
v+(z,t) i+(z,t) dt
=
⏐
⏐V+
0
⏐
⏐2
Z0
· 1
T
∫ T
0
cos2 (ωt− βz+ φ) dt
(3.137)
Employing a well-known trigonometric identity:
cos2 θ= 1
2 + 1
2 cos 2θ (3.138)
we may rewrite the integrand as follows
cos2 (ωt− βz+ φ) = 1
2 + 1
2 cos (2 [ωt− βz+ φ])
(3.139)
Then integrating over both sides of this quantity
∫ T
0
cos2 (ωt− βz+ φ) dt= T
2 + 0 (3.140)
The second term of the integral is zero because it is
the integral of cosine over two complete periods.
Subsequently , we see that the position dependence
(here, the dependence on z) is eliminated. In other
words, the power associated with the incident wave is
the same for all z <0, as expected. Substituting into
Equation 3.137 we obtain:
P+
av =
⏐
⏐V+
0
⏐
⏐2
2Z0
(3.141)
This is the time-average power associated with the
incident wave, measured at any point z <0 along the
line.",Electromagnetics_Vol1.pdf
"P+
av =
⏐
⏐V+
0
⏐
⏐2
2Z0
(3.141)
This is the time-average power associated with the
incident wave, measured at any point z <0 along the
line.
Equation 3.141 gives the time-average power as-
sociated with a wave traveling in a single direc-
tion along a lossless transmission line.
Using precisely the same procedure, we find that the
power associated with the reflected wave is
P−
av =
⏐
⏐Γ V+
0
⏐
⏐2
2Z0
= |Γ |2
⏐
⏐V+
0
⏐
⏐2
2Z0
(3.142)
or simply
P−
av = |Γ |2 P+
av (3.143)
Equation 3.143 gives the time-average power
associated with the wave reflected from an
impedance mismatch.
Now , what is the power PL delivered to the load
impedance ZL? The simplest way to calculate this
power is to use the principle of conservation of
power. Applied to the present problem, this principle
asserts that the power incident on the load must equal
the power reflected plus the power absorbed; i.e.,
P+
av = P−
av + PL (3.144)
Applying the previous equations we obtain:
PL =
(
1 − |Γ |2
)
P+
av
(3.145)",Electromagnetics_Vol1.pdf
"P+
av = P−
av + PL (3.144)
Applying the previous equations we obtain:
PL =
(
1 − |Γ |2
)
P+
av
(3.145)
Equations 3.145 gives the time-average power
transferred to a load impedance, and is equal to
the difference between the powers of the incident
and reflected waves.
Example 3.8. How important is it to match
50 Ω to 75 Ω ?
T wo impedances which commonly appear in
radio engineering are 50 Ω and 75 Ω . It is not
uncommon to find that it is necessary to connect
a transmission line having a 50 Ω characteristic
impedance to a device, circuit, or system having
a 75 Ω input impedance, or vice-versa. If no
attempt is made to match these impedances,
what fraction of the power will be delivered to
the termination, and what fraction of power will
be reflected? What is the SWR?
Solution. The voltage reflection coefficient
going from 50 Ω transmission line to a 75 Ω load
is
Γ = 75 − 50
75 + 50 = +0 .2
The fraction of power reflected is |Γ |2 = 0 .04,
which is 4%. The fraction of power transmitted",Electromagnetics_Vol1.pdf
"is
Γ = 75 − 50
75 + 50 = +0 .2
The fraction of power reflected is |Γ |2 = 0 .04,
which is 4%. The fraction of power transmitted
is 1 − |Γ |2, which is 96%. Going from a 50 Ω
transmission line to a 75 Ω termination changes
only the sign of Γ , and therefore, the fractions of
reflected and transmitted power remain 4% and",Electromagnetics_Vol1.pdf
"62 CHAPTER 3. TRANSMISSION LINES
96%, respectively . In either case (from
Section 3.14):
SWR = 1 + |Γ |
1 − |Γ | = 1 .5
This is often acceptable, but may not be good
enough in some particular applications. Suffice
it to say that it is not necessarily required to use
an impedance matching device to connect 50 Ω
to 75 Ω devices.
3.21 Impedance Matching:
General Considerations
[m0092]
“Impedance matching” refers to the problem of
transforming a particular impedance ZL into a
modified impedance Zin. The problem of impedance
matching arises because it is not convenient, practical,
or desirable to have all devices in a system operate at
the same input and output impedances. Here are just a
few of the issues:
• It is not convenient or practical to market coaxial
cables having characteristic impedance equal to
every terminating impedance that might be
encountered.
• Different types of antennas operate at different
impedances, and the impedance of most
antennas vary significantly with frequency .",Electromagnetics_Vol1.pdf
"encountered.
• Different types of antennas operate at different
impedances, and the impedance of most
antennas vary significantly with frequency .
• Different types of amplifiers operate most
effectively at different output impedances. For
example, amplifiers operating as current sources
operate most effectively with low output
impedance, whereas amplifiers operating as
voltage sources operate most effectively with
high output impedances.
• Independently of the above issue, techniques for
the design of transistor amplifiers rely on
intentionally mismatching impedances; i.e.,
matching to an impedance different than that
which maximizes power transfer or minimizes
reflection. In other words, various design goals
are met by applying particular impedances to the
input and output ports of the transistor. 7
For all of these reasons, electrical engineers
frequently find themselves with the task of
transforming a particular impedance ZL into a
modified impedance Zin.",Electromagnetics_Vol1.pdf
"frequently find themselves with the task of
transforming a particular impedance ZL into a
modified impedance Zin.
The reader is probably already familiar with many
approaches to the impedance matching problem that
7 For a concise introduction to this concept, see Chapter 10 of
S.W . Ellingson, Radio Systems Engineering , Cambridge Univ . Press,
2016.",Electromagnetics_Vol1.pdf
"3.22. SINGLE-REACT ANCE MA TCHING 63
employ discrete components and do not require
knowledge of electromagnetics. 8 T o list just a few of
these approaches: transformers, resistive (lossy)
matching, single-reactance matching, and
two-reactance (“L ” network) matching. However, all
of these have limitations. Perhaps the most serious
limitations pertain to the performance of discrete
components at high frequencies. Here are just a few
of the most common problems:
• Practical resistors actually behave as ideal
resistors in series with ideal inductors
• Practical capacitors actually behave as ideal
capacitors in series with ideal resistors
• Practical inductors behave as ideal inductors in
parallel with ideal capacitors, and in series with
ideal resistors.
All of this makes the use of discrete components
increasingly difficult with increasing frequency .
One possible solution to these types of problems is to
more precisely model each component, and then to",Electromagnetics_Vol1.pdf
"One possible solution to these types of problems is to
more precisely model each component, and then to
account for the non-ideal behavior by incorporating
the appropriate models in the analysis and design
process. Alternatively , one may consider ways to
replace particular troublesome components – or, in
some cases, all discrete components – with
transmission line devices. The latter approach is
particularly convenient in circuits implemented on
printed circuit boards at frequencies in the UHF band
and higher, since the necessary transmission line
structures are easy to implement as microstrip lines
and are relatively compact since the wavelength is
relatively small. However, applications employing
transmission lines as components in impedance
matching devices can be found at lower frequencies
as well.
8 For an overview , see Chapter 9 of S.W . Ellingson, Radio Sys-
tems Engineering , Cambridge Univ . Press, 2016.
3.22 Single-Reactance Matching
[m0093]",Electromagnetics_Vol1.pdf
"tems Engineering , Cambridge Univ . Press, 2016.
3.22 Single-Reactance Matching
[m0093]
An impedance matching structure can be designed
using a section of transmission line combined with a
discrete reactance, such as a capacitor or an inductor.
In the strategy presented here, the transmission line is
used to transform the real part of the load impedance
or admittance to the desired value, and then the
reactance is used to modify the imaginary part to the
desired value. (Note the difference between this
approach and the quarter-wave technique described in
Section 3.19. In that approach, the first transmission
line is used to zero the imaginary part.) There are two
versions of this strategy , which we will now consider
separately .
The first version is shown in Figure 3.28. The purpose
of the transmission line is to transform the load
impedance ZL into a new impedance Z1 for which
Re{Z1} = Re{Zin}. This can be done by solving the
equation (from Section 3.15)
Re {Z1} = Re
{
Z0
1 + Γ e−j2βl",Electromagnetics_Vol1.pdf
"Re{Z1} = Re{Zin}. This can be done by solving the
equation (from Section 3.15)
Re {Z1} = Re
{
Z0
1 + Γ e−j2βl
1 − Γ e−j2βl
}
(3.146)
for l, using a numerical search, or using the Smith
chart.9 The characteristic impedance Z0 and phase
propagation constant βof the transmission line are
independent variables and can be selected for
convenience. Normally , the smallest value of lthat
satisfies Equation 3.146 is desired. This value will be
≤ λ/4 because the real part of Z1 is periodic in lwith
period λ/4.
After matching the real component of the impedance
in this manner, the imaginary component of Z1 may
then be transformed to the desired value (Im {Zin})
by attaching a reactance Xs in series with the
transmission line input, yielding Zin = Z1 + jXS.
Therefore, we choose
Xs = Im {Zin − Z1} (3.147)
The sign of Xs determines whether this reactance is a
capacitor ( Xs <0) or inductor ( Xs >0), and the
value of this component is determined from Xs and
the design frequency .",Electromagnetics_Vol1.pdf
"capacitor ( Xs <0) or inductor ( Xs >0), and the
value of this component is determined from Xs and
the design frequency .
9 For more about the Smith chart, see “ Additional Reading” at the
end of this section.",Electromagnetics_Vol1.pdf
"64 CHAPTER 3. TRANSMISSION LINES
ZL
Z1
Z0
I
jXS
Zin
Figure 3.28: Single-reactance matching with a series
reactance.
Example 3.9. Single reactance in series.
Design a match consisting of a transmission line
in series with a single capacitor or inductor that
matches a source impedance of 50Ω to a load
impedance of 33.9 + j17.6 Ω at 1.5 GHz. The
characteristic impedance and phase velocity of
the transmission line are 50Ω and 0.6c
respectively .
Solution. From the problem statement:
Zin ≜ ZS = 50 Ω and ZL = 33 .9 + j17.6 Ω are
the source and load impedances respectively at
f = 1 .5 GHz. The characteristic impedance and
phase velocity of the transmission line are
Z0 = 50 Ω and vp = 0 .6crespectively .
The reflection coefficient Γ (i.e., ZL with respect
to the characteristic impedance of the
transmission line) is
Γ ≜ ZL − Z0
ZL + Z0
∼
= −0.142 + j0.239 (3.148)
The length lof the primary line (that is, the one
that connects the two ports of the matching",Electromagnetics_Vol1.pdf
"Γ ≜ ZL − Z0
ZL + Z0
∼
= −0.142 + j0.239 (3.148)
The length lof the primary line (that is, the one
that connects the two ports of the matching
structure) is determined using the equation:
Re {Z1} = Re
{
Z0
1 + Γ e−j2βl
1 − Γ e−j2βl
}
(3.149)
where here Re {Z1} = Re {ZS} = 50 Ω . So a
more-specific form of the equation that can be
solved for βl(as a step toward finding l) is:
1 = Re
{1 + Γ e−j2βl
1 − Γ e−j2βl
}
(3.150)
By trial and error (or using the Smith chart if you
prefer) we find βl ∼
= 0.408 rad for the primary
line, yielding Z1 ∼
= 50.0 + j29.0 Ω for the input
impedance after attaching the primary line.
W e may now solve for las follows: Since
vp = ω/β (Section 3.8), we find
β = ω
vp
= 2πf
0.6c
∼
= 52.360 rad/m (3.151)
Therefore l= ( βl) /β∼
= 7.8 mm
.
The impedance of the series reactance should be
jXs ∼
= −j29.0 Ω to cancel the imaginary part of
Z1. Since the sign of this impedance is negative,
it must be a capacitor. The reactance of a
capacitor is −1/ωC, so it must be true that",Electromagnetics_Vol1.pdf
"Z1. Since the sign of this impedance is negative,
it must be a capacitor. The reactance of a
capacitor is −1/ωC, so it must be true that
− 1
2πfC
∼
= −29.0 Ω (3.152)
Thus, we find the series reactance is a capacitor
of value C ∼
= 3.7 pF
.
The second version of the single-reactance strategy is
shown in Figure 3.29. The difference in this scheme
is that the reactance is attached in parallel. In this
case, it is easier to work the problem using admittance
(i.e., reciprocal impedance) as opposed to impedance;
this is because the admittance of parallel reactances is
simply the sum of the associated admittances; i.e.,
Yin = Y1 + jBp (3.153)
where Yin = 1 /Zin, Y1 = 1 /Z1, and Bp is the
discrete parallel susceptance; i.e., the imaginary part
of the discrete parallel admittance.
So, the procedure is as follows. The transmission line
is used to transform YL into a new admittance Y1 for
which Re {Y1} = Re{Yin}. First, we note that
Y1 ≜ 1
Z1
= Y0
1 − Γ e−j2βl
1 + Γ e−j2βl (3.154)",Electromagnetics_Vol1.pdf
"is used to transform YL into a new admittance Y1 for
which Re {Y1} = Re{Yin}. First, we note that
Y1 ≜ 1
Z1
= Y0
1 − Γ e−j2βl
1 + Γ e−j2βl (3.154)
where Y0 ≜ 1/Z0 is characteristic admittance.
Again, the characteristic impedance Z0 and phase",Electromagnetics_Vol1.pdf
"3.22. SINGLE-REACT ANCE MA TCHING 65
Y
L
J
1
K
0
M
j O
p
P
in
Figure 3.29: Single-reactance matching with a parallel
reactance.
propagation constant βof the transmission line are
independent variables and can be selected for
convenience. In the present problem, we aim to solve
the equation
Re {Y1} = Re
{
Y0
1 − Γ e−j2βl
1 + Γ e−j2βl
}
(3.155)
for the smallest value of l, using a numerical search or
using the Smith chart. After matching the real
component of the admittances in this manner, the
imaginary component of the resulting admittance may
then be transformed to the desired value by attaching
the susceptance Bp in parallel with the transmission
line input. Since we desire jBp in parallel with Y1 to
be Yin, the desired value is
Bp = Im {Yin − Y1} (3.156)
The sign of Bp determines whether this is a capacitor
(Bp >0) or inductor ( Bp <0), and the value of this
component is determined from Bp and the design
frequency .
In the following example, we address the same",Electromagnetics_Vol1.pdf
"component is determined from Bp and the design
frequency .
In the following example, we address the same
problem raised in Example 3.9, now using the parallel
reactance approach:
Example 3.10. Single reactance in parallel.
Design a match consisting of a transmission line
in parallel with a single capacitor or inductor
that matches a source impedance of 50Ω to a
load impedance of 33.9 + j17.6 Ω at 1.5 GHz.
The characteristic impedance and phase velocity
of the transmission line are 50Ω and 0.6c
respectively .
Solution. From the problem statement:
Zin ≜ ZS = 50 Ω and ZL = 33 .9 + j17.6 Ω are
the source and load impedances respectively at
f = 1 .5 GHz. The characteristic impedance and
phase velocity of the transmission line are
Z0 = 50 Ω and vp = 0 .6crespectively .
The reflection coefficient Γ (i.e., ZL with respect
to the characteristic impedance of the
transmission line) is
Γ ≜ ZL − Z0
ZL + Z0
∼
= −0.142 + j0.239 (3.157)
The length lof the primary line (that is, the one",Electromagnetics_Vol1.pdf
"transmission line) is
Γ ≜ ZL − Z0
ZL + Z0
∼
= −0.142 + j0.239 (3.157)
The length lof the primary line (that is, the one
that connects the two ports of the matching
structure) is the solution to:
Re {Y1} = Re
{
Y0
1 − Γ e−j2βl
1 + Γ e−j2βl
}
(3.158)
where here Re {Y1} = Re {1/ZS} = 0 .02 mho
and Y0 = 1 /Z0 = 0 .02 mho. So the equation to
be solved for βl(as a step toward finding l) is:
1 = Re
{1 − Γ e−j2βl
1 + Γ e−j2βl
}
(3.159)
By trial and error (or the Smith chart) we find
βl ∼
= 0.126 rad for the primary line, yielding
Y1 ∼
= 0.0200 − j0.0116 mho for the input
admittance after attaching the primary line.
W e may now solve for las follows: Since
vp = ω/β (Section 3.8), we find
β = ω
vp
= 2πf
0.6c
∼
= 52.360 rad/m (3.160)
Therefore, l= ( βl) /β∼
= 2.4 mm
.
The admittance of the parallel reactance should
be jBp ∼
= +j0.0116 mho to cancel the
imaginary part of Y1. The associated impedance
is 1/jBp ∼
= −j86.3 Ω . Since the sign of this
impedance is negative, it must be a capacitor.",Electromagnetics_Vol1.pdf
"imaginary part of Y1. The associated impedance
is 1/jBp ∼
= −j86.3 Ω . Since the sign of this
impedance is negative, it must be a capacitor.
The reactance of a capacitor is −1/ωC, so it",Electromagnetics_Vol1.pdf
"66 CHAPTER 3. TRANSMISSION LINES
must be true that
− 1
2πfC
∼
= −86.3 Ω (3.161)
Thus, we find the parallel reactance is a
capacitor
of value C ∼
= 1.2 pF
.
Comparing this result to the result from the series
reactance method (Example 3.9), we see that the
necessary length of transmission line is much shorter,
which is normally a compelling advantage. The
tradeoff is that the parallel capacitance is much
smaller and an accurate value may be more difficult to
achieve.
Additional Reading:
• “ Smith chart ” on Wikipedia.
3.23 Single-Stub Matching
[m0094]
In Section 3.22, we considered impedance matching
schemes consisting of a transmission line combined
with a reactance which is placed either in series or in
parallel with the transmission line. In many problems,
the required discrete reactance is not practical
because it is not a standard value, or because of
non-ideal behavior at the desired frequency (see
Section 3.21 for more about this), or because one",Electromagnetics_Vol1.pdf
"because it is not a standard value, or because of
non-ideal behavior at the desired frequency (see
Section 3.21 for more about this), or because one
might simply wish to avoid the cost and logistical
issues associated with an additional component.
Whatever the reason, a possible solution is to replace
the discrete reactance with a transmission line “stub”
– that is, a transmission line which has been open- or
short-circuited. Section 3.16 explains how a stub can
replace a discrete reactance. Figure 3.30 shows a
practical implementation of this idea implemented in
microstrip. This section explains the theory , and we’ll
return to this implementation at the end of the section.
Figure 3.31 shows the scheme. This scheme is usually
implemented using the parallel reactance approach, as
depicted in the figure. Although a series reactance
scheme is also possible in principle, it is usually not
as convenient. This is because most transmission",Electromagnetics_Vol1.pdf
"scheme is also possible in principle, it is usually not
as convenient. This is because most transmission
lines use one of their two conductors as a local datum;
e.g., the ground plane of a printed circuit board for
microstrip line is tied to ground, and the outer
conductor (“shield”) of a coaxial cable is usually tied
to ground. This is contrast to a discrete reactance
Q R T U V W tor
X [ \ ] ^ _ ` a
b c d e
Spinningspark CC BY SA 3.0
Figure 3.30: A practical implementation of a single-
stub impedance match using microstrip transmission
line. Here, the stub is open-circuited.",Electromagnetics_Vol1.pdf
"3.23. SINGLE-STUB MA TCHING 67
f
L
h
1
i
01
k
1
l m
n
o
in
stub characteristics:
q
02
r
2 2
stub may be open- or short-circuited
stub
Figure 3.31: Single-stub matching.
(such as a capacitor or inductor), which does not
require that either of its terminals be tied to ground.
This issue is avoided in the parallel-attached stub
because the parallel-attached stub and the
transmission line to which it is attached both have one
terminal at ground.
The single-stub matching procedure is essentially the
same as the single parallel reactance method, except
the parallel reactance is implemented using a short- or
open-circuited stub as opposed a discrete inductor or
capacitor. Since parallel reactance matching is most
easily done using admittances, it is useful to express
Equations 3.117 and 3.119 (input impedance of an
open- and short-circuited stub, respectively , from
Section 3.16) in terms of susceptance:
Bp = −Y02 cot (β2l2) short-circuited stub (3.162)",Electromagnetics_Vol1.pdf
"open- and short-circuited stub, respectively , from
Section 3.16) in terms of susceptance:
Bp = −Y02 cot (β2l2) short-circuited stub (3.162)
Bp = + Y02 tan (β2l2) open-circuited stub (3.163)
As in the main line, the characteristic impedance
Z02 = 1 /Y02 is an independent variable and is chosen
for convenience.
A final question is when should you use a
short-circuited stub, and when should you use an
open-circuited stub? Given no other basis for
selection, the termination that yields the shortest stub
is chosen. An example of an “other basis for
selection” that frequently comes up is whether DC
might be present on the line. If DC is present with the
signal of interest, then a short circuit termination
without some kind of remediation to prevent a short
circuit for DC would certainly be a bad idea.
In the following example we address the same
problem raised in Section 3.22 (Examples 3.9 and
3.10), now using the single-stub approach:
Example 3.11. Single stub matching.",Electromagnetics_Vol1.pdf
"problem raised in Section 3.22 (Examples 3.9 and
3.10), now using the single-stub approach:
Example 3.11. Single stub matching.
Design a single-stub match that matches a
source impedance of 50Ω to a load impedance of
33.9 + j17.6 Ω . Use transmission lines having
characteristic impedances of 50Ω throughout,
and leave your answer in terms of wavelengths.
Solution. From the problem statement:
Zin ≜ ZS = 50 Ω and ZL = 33 .9 + j17.6 Ω are
the source and load impedances respectively .
Z0 = 50 Ω is the characteristic impedance of the
transmission lines to be used. The reflection
coefficient Γ (i.e., ZL with respect to the
characteristic impedance of the transmission
line) is
Γ ≜ ZL − Z0
ZL + Z0
∼
= −0.142 + j0.239 (3.164)
The length l1 of the primary line (that is, the one
that connects the two ports of the matching
structure) is the solution to the equation (from
Section 3.22):
Re {Y1} = Re
{
Y01
1 − Γ e−j2β1l1
1 + Γ e−j2β1l1
}
(3.165)
where here Re {Y1} = Re {1/ZS} = 0 .02 mho",Electromagnetics_Vol1.pdf
"Section 3.22):
Re {Y1} = Re
{
Y01
1 − Γ e−j2β1l1
1 + Γ e−j2β1l1
}
(3.165)
where here Re {Y1} = Re {1/ZS} = 0 .02 mho
and Y01 = 1 /Z0 = 0 .02 mho. Also note
2β1l1 = 2
( 2π
λ
)
l1 = 4 πl1
λ (3.166)
where λis the wavelength in the transmission
line. So the equation to be solved for l1 is:
1 = Re
{1 − Γ e−j4πl1/λ
1 + Γ e−j4πl1/λ
}
(3.167)
By trial and error (or using the Smith chart; see
“ Additional Reading” at the end of this section)
we find l1 ∼
= 0.020λ
for the primary line,",Electromagnetics_Vol1.pdf
"68 CHAPTER 3. TRANSMISSION LINES
yielding Y1 ∼
= 0.0200 − j0.0116 mho for the
input admittance after attaching the primary line.
W e now seek the shortest stub having an input
admittance of ∼
= +j0.0116 mho to cancel the
imaginary part of Y1. For an open-circuited stub,
we need
Bp = + Y0 tan 2πl2/λ∼
= +j0.0116 mho
(3.168)
The smallest value of l2 for which this is true is
∼
= 0.084λ. For a short-circuited stub, we need
Bp = −Y0 cot 2πl2/λ∼
= +j0.0116 mho
(3.169)
The smallest positive value of l2 for which this is
true is ∼
= 0.334λ; i.e., much longer. Therefore,
we choose the open-circuited stub
with
l2 ∼
= 0.084λ
. Note the stub is attached in parallel
at the source end of the primary line.
Single-stub matching is a very common method for
impedance matching using microstrip lines at
frequences in the UHF band (300-3000 MHz) and
above. In Figure 3.30, the top (visible) traces
comprise one conductor, whereas the ground plane
(underneath, so not visible) comprises the other",Electromagnetics_Vol1.pdf
"above. In Figure 3.30, the top (visible) traces
comprise one conductor, whereas the ground plane
(underneath, so not visible) comprises the other
conductor. The end of the stub is not connected to the
ground plane, so the termination is an open circuit. A
short circuit termination is accomplished by
connecting the end of the stub to the ground plane
using a via; that is, a plated-through that electrically
connects the top and bottom layers.
Additional Reading:
• “Stub (electronics)” on Wikipedia.
• “ Smith chart ” on Wikipedia.
• “V ia (electronics)” on Wikipedia.
[m0151]",Electromagnetics_Vol1.pdf
"3.23. SINGLE-STUB MA TCHING 69
Image Credits
Fig. 3.1: Dmitry G, https://en.wikipedia.org/wiki/File:Mastech test leads.JPG,
public domain.
Fig. 3.2: c⃝ Tkgd2007, https://commons.wikimedia.org/wiki/File:Co axial cable cutaway .svg,
CC BY 3.0 (https://creativecommons.org/licenses/by/3.0 /).
Minor modifications from the original.
Fig. 3.3: c⃝ SpinningSpark, https://en.wikipedia.org/wiki/File:Mi crostrip structure.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/).
Minor modifications from the original.
Fig. 3.6: c⃝ A verse, https://en.wikipedia.org/wiki/File:Diplexer1.jpg,
CC BY SA 2.0 Germany (https://creativecommons.org/licens es/by-sa/2.0/de/deed.en).
Fig. 3.7: c⃝ BigRiz, https://en.wikipedia.org/wiki/File:Fibreopti c.jpg,
CC BY SA 3.0 Unported (https://creativecommons.org/licen ses/by-sa/3.0/deed.en).
Fig. 3.8: c⃝ Omegatron, https://commons.wikimedia.org/wiki/File:T ransmission line symbols.svg,",Electromagnetics_Vol1.pdf
"Fig. 3.8: c⃝ Omegatron, https://commons.wikimedia.org/wiki/File:T ransmission line symbols.svg,
CC BY SA 3.0 Unported (https://creativecommons.org/licen ses/by-sa/3.0/deed.en).
Modified.
Fig. 3.10: c⃝ Omegatron, https://commons.wikimedia.org/wiki/File:T ransmission line element.svg,
CC BY SA 3.0 Unported (https://creativecommons.org/licen ses/by-sa/3.0/deed.en).
Modified.
Fig. 3.11: c⃝ Omegatron, https://commons.wikimedia.org/wiki/File:T ransmission line element.svg,
CC BY SA 3.0 Unported (https://creativecommons.org/licen ses/by-sa/3.0/deed.en).
Modified.
Fig. 3.15: c⃝ Arj, https://en.wikipedia.org/wiki/File:RG-59.jpg,
CC BY SA 3.0 Unported (https://creativecommons.org/licen ses/by-sa/3.0/deed.en).
Fig. 3.16: c⃝ 7head7metal7, https://commons.wikimedia.org/wiki/Fil e:Microstrip scheme.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/).
Minor modifications from the original.
Fig. 3.20: by Inductiveload, https://commons.wikimedia. org/wiki/File:Standing W ave Ratio.svg,",Electromagnetics_Vol1.pdf
"Minor modifications from the original.
Fig. 3.20: by Inductiveload, https://commons.wikimedia. org/wiki/File:Standing W ave Ratio.svg,
public domain.
Modifications from the original.
Fig. 3.30: c⃝ Spinningspark, https://commons.wikimedia.org/wiki/Fi le:Stripline
stub matching.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/).
Minor modifications from the original: curly braces deleted .",Electromagnetics_Vol1.pdf
"Chapter 4
V ector Analysis
4.1 V ector Arithmetic
[m0006]
A vector is a mathematical object that has both a
scalar part (i.e., a magnitude and possibly a phase), as
well as a direction. Many physical quantities are best
described as vectors. For example, the rate of
movement through space can be described as speed;
i.e., as a scalar having SI base units of m/s. However,
this quantity is more completely described as velocity;
i.e., as a vector whose scalar part is speed and
direction indicates the direction of movement.
Similarly , force is a vector whose scalar part indicates
magnitude (SI base units of N), and direction
indicates the direction in which the force is applied.
Electric and magnetic fields are also best described as
vectors.
In mathematical notation, a real-valued vector A is
said to have a magnitude A= |A| and direction ˆa
such that
A = Aˆa (4.1)
where ˆa is a unit vector (i.e., a real-valued vector
having magnitude equal to one) having the same",Electromagnetics_Vol1.pdf
"such that
A = Aˆa (4.1)
where ˆa is a unit vector (i.e., a real-valued vector
having magnitude equal to one) having the same
direction as A. If a vector is complex-valued, then A
is similarly complex-valued.
Cartesian Coordinate System. Fundamentals of
vector arithmetic are most easily grasped using the
Cartesian coordinate system. This system is shown in
Figure 4.1. Note carefully the relative orientation of
the x, y, and zaxes. This orientation is important. For
example, there are two directions that are
perpendicular to the z= 0 plane (in which the x- and
y-axes lie), but the +zaxis is specified to be one of
y
x
z
s t u w x {
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.1: Cartesian coordinate system.
these in particular.
Position-Fixed vs. Position-Free V ectors. It is often
convenient to describe a position in space as a vector
for which the magnitude is the distance from the
origin of the coordinate system and for which the
direction is measured from the origin toward the",Electromagnetics_Vol1.pdf
"for which the magnitude is the distance from the
origin of the coordinate system and for which the
direction is measured from the origin toward the
position of interest. This is shown in Figure 4.2.
These position vectors are “position-fixed” in the
sense that they are defined with respect to a single
point in space, which in this case is the origin.
Position vectors can also be defined as vectors that are
defined with respect to some other point in space, in
which case they are considered position-fixed to that
position.
P osition-free vectors, on the other hand, are not
defined with respect to a particular point in space. An
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"4.1. VECTOR ARITHMETIC 71
|
x
z
r1
r
}
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.2: Position vectors. The vectors r1 and r1
are position-fixed and refer to particular locations.
v
~  €  ‚ƒ „… †
‡
v
ˆ ‰ Š ‹ Œ Ž 
‘
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Figure 4.3: T wo particles exhibiting the same veloc-
ity . In this case, the velocity vectors v1 and v2 are
position-free and equal.
example is shown in Figure 4.3. Particles 1 m apart
may both be traveling at 2 m/s in the same direction.
In this case, the velocity of each particle can be
described using the same vector, even though the
particles are located at different points in space.
Position-free vectors are said to be equal if they have
the same magnitudes and directions. Position-fixed
vectors, on the other hand, must also be referenced to
the same position (e.g., the origin) to be considered
equal.
Basis V ectors. Each coordinate system is defined in
terms of three basis vectors which concisely describe
’
x
z
y
x
z
c⃝ K. Kikkeri CC BY SA 4.0",Electromagnetics_Vol1.pdf
"Basis V ectors. Each coordinate system is defined in
terms of three basis vectors which concisely describe
’
x
z
y
x
z
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.4: Basis vectors in the Cartesian coordinate
system.
all possible ways to traverse three-dimensional space.
A basis vector is a position-free unit vector that is
perpendicular to all other basis vectors for that
coordinate system. The basis vectors ˆx, ˆy, and ˆz of
the Cartesian coordinate system are shown in
Figure 4.4. In this notation, ˆx indicates the direction
in which xincreases most rapidly , ˆy indicates the
direction in which yincreases most rapidly , and ˆz
indicates the direction in which zincreases most
rapidly . Alternatively , you might interpret ˆx, ˆy, and ˆz
as unit vectors that are parallel to the x-, y-, and
z-axes and point in the direction in which values
along each axis increase.
V ectors in the Cartesian Coordinate System. In
Cartesian coordinates, we may describe any vector A
as follows:",Electromagnetics_Vol1.pdf
"along each axis increase.
V ectors in the Cartesian Coordinate System. In
Cartesian coordinates, we may describe any vector A
as follows:
A = ˆxAx + ˆyAy + ˆzAz (4.2)
where Ax, Ay, and Az are scalar quantities describing
the components of A in each of the associated
directions, as shown in Figure 4.5. This description
makes it clear that the magnitude of A is:
|A| =
√
A2
x + A2
y + A2
z (4.3)
and therefore, we can calculate the associated unit",Electromagnetics_Vol1.pdf
"72 CHAPTER 4. VECTOR ANAL YSIS
“
x
z
”
x
•
z
–
—
A
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.5: Components of a vector A in the Carte-
sian coordinate system.
vector as
ˆa = A
|A| = A√
A2
x + A2
y + A2
z
= ˆxAx
(
A2
x + A2
y + A2
z
) −1/2
+ ˆyAy
(
A2
x + A2
y + A2
z
) −1/2
+ ˆzAz
(
A2
x + A2
y + A2
z
) −1/2
(4.4)
V ector Addition and Subtraction. It is common to
add and subtract vectors. For example, vectors
describing two forces A and B applied to the same
point can be described as a single force vector C that
is the sum of A and B; i.e., C = A + B. This
addition is quite simple in the Cartesian coordinate
system:
C = A + B
= ( ˆxAx + ˆyAy + ˆzAz) + ( ˆxBx + ˆyBy + ˆzBz)
= ˆx (Ax + Bx) + ˆy (Ay + By) + ˆz (Az + Bz)
(4.5)
In other words, the ˆx component of C is the sum of
the ˆx components of A and B, and similarly for ˆy
and ˆz. From the above example, it is clear that vector
addition is commutative; i.e.,
A + B = B + A (4.6)
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r
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"addition is commutative; i.e.,
A + B = B + A (4.6)
˜ ™ š › œ 
r
ž
r
Ÿ
r
  ¡
r
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r
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Figure 4.6: Relative position (distance and direction)
between locations identified by their position vectors.
In other words, vectors may be added in any order.
V ector subtraction is defined similarly:
D = A − B
= ( ˆxAx + ˆyAy + ˆzAz) − (ˆxBx + ˆyBy + ˆzBz)
= ˆx (Ax − Bx) + ˆy (Ay − By) + ˆz (Az − Bz)
(4.7)
In other words, the ˆx component of D is the
difference of the ˆx components of A and B, and
similarly for ˆy and ˆz. Like scalar subtraction, vector
subtraction is not commutative.
Relative Positions and Distances. A common task in
vector analysis is to describe the position of one point
in space relative to a different point in space. Let us
identify those two points using the position vectors r1
and r2, as indicated in Figure 4.6. W e may identify a
third vector r12 as the position of r2 relative to r1:
r12 ≜ r2 − r1 (4.8)",Electromagnetics_Vol1.pdf
"and r2, as indicated in Figure 4.6. W e may identify a
third vector r12 as the position of r2 relative to r1:
r12 ≜ r2 − r1 (4.8)
Now |r12| is the distance between these points, and
r12/|r12| is a unit vector indicating the direction to r2
from r1.
Example 4.1. Direction and distance between
positions.
Consider two positions, identified using the
position vectors r1 = 2 ˆx + 3 ˆy + 1ˆz and
r2 = 1 ˆx − 2ˆy + 3ˆz, both expressed in units of
meters. Find the direction vector that points",Electromagnetics_Vol1.pdf
"4.1. VECTOR ARITHMETIC 73
from r1 to r2, the distance between these points,
and the associated unit vector.
Solution. The vector that points from r1 to r2 is
R = r2 − r1
= (1 − 2)ˆx + (−2 − 3)ˆy + (3 − 1)ˆz
= −ˆx − 5ˆy + 2ˆz (4.9)
The distance between r1 and r2 is simply the
magnitude of this vector:
|R| =
√
(−1)2 + (−5)2 + (2)2 ∼
= 5.48 m
(4.10)
The unit vector ˆR is simply R normalized to
have unit magnitude:
ˆR = R/|R|
∼
= (−ˆx − 5ˆy + 2ˆz) /5.48
∼
= −0.182ˆx − 0.913ˆy + 0.365ˆz (4.11)
Multiplication of a V ector by a Scalar . Let’s say a
particular force is specified by a vector F. What is the
new vector if this force is doubled? The answer is
simply 2F – that is, twice the magnitude applied in
the same direction. This is an example of scalar
multiplication of a vector. Generalizing, the product
of the scalar αand the vector A is simply αA.
Scalar (“Dot”) Product of V ectors. Another
common task in vector analysis is to determine the",Electromagnetics_Vol1.pdf
"of the scalar αand the vector A is simply αA.
Scalar (“Dot”) Product of V ectors. Another
common task in vector analysis is to determine the
similarity in the direction in which two vectors point.
In particular, it is useful to have a metric which, when
applied to the vectors A = ˆaAand B = ˆbB, has the
following properties (see Figure 4.7):
• If A is perpendicular to B, the result is zero.
• If A and B point in the same direction, the result
is AB.
• If A and B point in opposite directions, the
result is −AB.
• Results intermediate to these conditions depend
on the angle ψbetween A and B, measured as if
A and B were arranged “tail-to-tail” as shown in
Figure 4.8.
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.7: Special cases of the dot product.
In vector analysis, this operator is known as the scalar
product (not to be confused with scalar
multiplication) or the dot product . The dot product is
written A · B and is given in general by the
expression:
A · B = ABcos ψ (4.12)",Electromagnetics_Vol1.pdf
"multiplication) or the dot product . The dot product is
written A · B and is given in general by the
expression:
A · B = ABcos ψ (4.12)
Note that this expression yields the special cases
previously identified, which are ψ= π/2, ψ= 0 , and
ψ= π, respectively . The dot product is commutative;
i.e.,
A · B = B · A (4.13)
The dot product is also distributive; i.e.,
A · (B + C) = A · B + A · C (4.14)
The dot product has some other useful properties. For
example, note:
A · A = ( ˆxAx + ˆyAy + ˆzAz) · (ˆxAx + ˆyAy + ˆzAz)
= ˆx · ˆxA2
x + ˆx · ˆyAxAy + ˆx · ˆzAxAz
+ ˆy · ˆxAxAy + ˆy · ˆyA2
y + ˆy · ˆzAyAz
+ ˆz · ˆxAxAz + ˆz · ˆyAyAz + ˆz · ˆzA2
z
(4.15)
which looks pretty bad until you realize that
ˆx · ˆx = ˆy · ˆy = ˆz · ˆz = 1 (4.16)",Electromagnetics_Vol1.pdf
"74 CHAPTER 4. VECTOR ANAL YSIS
A
B
¥
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.8: Calculation of the dot product.
and any other dot product of basis vectors is zero.
Thus, the whole mess simplifies to:
A · A = A2
x + A2
y + A2
z (4.17)
This is the square of the magnitude of A, so we have
discovered that
A · A = |A|2 = A2 (4.18)
Applying the same principles to the dot product of
potentially different vectors A and B, we find:
A · B = ( ˆxAx + ˆyAy + ˆzAz) · (ˆxBx + ˆyBy + ˆzBz)
= AxBx + AyBy + AzBz
(4.19)
This is a particularly easy way to calculate the dot
product, since it eliminates the problem of
determining the angle ψ. In fact, an easy way to
calculate ψgiven A and B is to first calculate the dot
product using Equation 4.19 and then use the result to
solve Equation 4.12 for ψ.
Example 4.2. Angle between two vectors.
Consider the position vectors C = 2 ˆx + 3ˆy + 1ˆz
and D = 3 ˆx − 2ˆy + 2ˆz, both expressed in units
of meters. Find the angle between these vectors.",Electromagnetics_Vol1.pdf
"Consider the position vectors C = 2 ˆx + 3ˆy + 1ˆz
and D = 3 ˆx − 2ˆy + 2ˆz, both expressed in units
of meters. Find the angle between these vectors.
Solution. From Equation 4.12
C · D = CDcos ψ (4.20)
where C = |C|, D= |D|, and ψis the angle we
seek. From Equation 4.19:
C · D = CxDx + CyDy + CzDz (4.21)
= 2 · 3 + 3 · (−2) + 1 · 2 m2 (4.22)
= 2 m2 (4.23)
also
C =
√
C2x + C2y + C2z ∼
= 3.742 m (4.24)
D=
√
D2x + D2y + D2z ∼
= 4.123 m (4.25)
so
cos ψ= C · D
CD
∼
= 0.130 (4.26)
T aking the inverse cosine, we find ψ= 82 .6◦.
Cross Product. The cross product is a form of vector
multiplication that results in a vector that is
perpendicular to both of the operands. The definition
is as follows:
A × B = ˆnABsin ψAB (4.27)
As shown in Figure 4.9, the unit vector ˆn is
determined by the “right hand rule. ” Using your right
hand, curl your fingers to traverse the angle ψAB
beginning at A and ending at B, and then ˆn points in
the direction of your fully-extended thumb.",Electromagnetics_Vol1.pdf
"hand, curl your fingers to traverse the angle ψAB
beginning at A and ending at B, and then ˆn points in
the direction of your fully-extended thumb.
It should be apparent that the cross product is not
commutative but rather is anticommutative; that is,
A × B = −B × A (4.28)
Y ou can confirm this for yourself using either
Equation 4.27 or by applying the right-hand rule.
The cross product is distributive:
A × (B + C) = A × B + A × C (4.29)
There are two useful special cases of the cross
product that are worth memorizing. The first is the
cross product of a vector with itself, which is zero:
A × A = 0 (4.30)
The second is the cross product of vectors that are
perpendicular; i.e., for which ψAB = π/2. In this",Electromagnetics_Vol1.pdf
"4.1. VECTOR ARITHMETIC 75
n
B
A
AB
¦
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.9: The cross product A × B.
x y
z
Figure 4.10: Cross products among basis vectors in
the Cartesian system. The cross product of any two
basis vectors is the third basis vector when the order
of operands is counter-clockwise, as shown in the dia-
gram, and is −1 times the third basis vector when the
order of operands is clockwise with respect to the ar-
rangement in the diagram.
case:
A × B = ˆnAB (4.31)
Using these principles, note:
ˆx × ˆx = ˆy × ˆy = ˆz × ˆz = 0 (4.32)
whereas
ˆx × ˆy = ˆz (4.33)
ˆy × ˆz = ˆx (4.34)
ˆz × ˆx = ˆy (4.35)
A useful diagram that summarizes these relationships
is shown in Figure 4.10.
It is typically awkward to “manually” determine ˆn in
Equation 4.27. However, in Cartesian coordinates the
cross product may be calculated as:
A × B = ˆx (AyBz − AzBy)
+ ˆy (AzBx − AxBz)
+ ˆz (AxBy − AyBx)
(4.36)
This may be easier to remember as a matrix
determinant:
A × B =
⏐
⏐
⏐
⏐
⏐
⏐
ˆx ˆy ˆz",Electromagnetics_Vol1.pdf
"+ ˆy (AzBx − AxBz)
+ ˆz (AxBy − AyBx)
(4.36)
This may be easier to remember as a matrix
determinant:
A × B =
⏐
⏐
⏐
⏐
⏐
⏐
ˆx ˆy ˆz
Ax Ay Az
Bx By Bz
⏐
⏐
⏐
⏐
⏐
⏐
(4.37)
Similar expressions are available for other coordinate
systems.
V ector analysis routinely requires expressions
involving both dot products and cross products in
different combinations. Often, these expressions may
be simplified, or otherwise made more convenient,
using the vector identities listed in Appendix B.3.
Additional Reading:
• “V ector field” on Wikipedia.
• “V ector algebra” on Wikipedia.",Electromagnetics_Vol1.pdf
"76 CHAPTER 4. VECTOR ANAL YSIS
4.2 Cartesian Coordinates
[m0004]
The Cartesian coordinate system is introduced in
Section 4.1. Concepts described in that section – i.e.,
the dot product and cross product – are described in
terms of the Cartesian system. In this section, we
identify some additional features of this system that
are useful in subsequent work and also set the stage
for alternative systems; namely the cylindrical and
spherical coordinate systems.
Integration Over Length. Consider a vector field
A = ˆxA(r), where r is a position vector. What is the
integral of A over some curve C through space? T o
answer this question, we first identify a
differential-length segment of the curve. Note that
this segment of the curve can be described as
dl = ˆxdx+ ˆydy+ ˆzdz (4.38)
The contribution to the integral for that segment of
the curve is simply A · dl. W e integrate to obtain the
result; i.e., ∫
C
A · dl (4.39)
For example, if A = ˆxA0 (i.e., A(r) is a constant)",Electromagnetics_Vol1.pdf
"the curve is simply A · dl. W e integrate to obtain the
result; i.e., ∫
C
A · dl (4.39)
For example, if A = ˆxA0 (i.e., A(r) is a constant)
and if C is a straight line from x= x1 and x= x2
along some constant yand z, then dl = ˆxdx,
A · dl = A0dx, and subsequently the above integral is
∫ x2
x1
A0dx= A0 (x2 − x1) (4.40)
In particular, notice that if A0 = 1 , then this integral
gives the length of C. Although the formalism seems
unnecessary in this simple example, it becomes very
useful when integrating over paths that vary in more
than one direction and with more complicated
integrands.
Note that the Cartesian system was an appropriate
choice for preceding example because this allowed
two of the three basis directions (i.e., yand z) to be
immediately eliminated from the calculation. Said
differently , the preceding example is expressed with
the minimum number of varying coordinates in the
Cartesian system. Here’s a counter-example. If C had",Electromagnetics_Vol1.pdf
"the minimum number of varying coordinates in the
Cartesian system. Here’s a counter-example. If C had
been a circle in the z= 0 plane, then the problem
would have required two basis directions to be
considered – namely , both xand y. In this case,
another system – namely , cylindrical coordinates
(Section 4.3) – minimizes the number of varying
coordinates (to just one, which is φ).
Integration Over Area. Now we ask the question,
what is the integral of some vector field A over some
surface S? The answer is
∫
S
A · ds (4.41)
W e refer to ds as the differential surface element,
which is a vector having magnitude equal to the
differential area ds, and is normal (perpendicular) to
each point on the surface. There are actually two such
directions. W e’ll return to clear up the ambiguity in a
moment. Now , as an example, if A = ˆz and S is the
surface bounded by x1 ≤ x≤ x2, y1 ≤ y≤ y2, then
ds = ˆz dxdy (4.42)
since dxdyis differential surface area in the z= 0",Electromagnetics_Vol1.pdf
"surface bounded by x1 ≤ x≤ x2, y1 ≤ y≤ y2, then
ds = ˆz dxdy (4.42)
since dxdyis differential surface area in the z= 0
plane and ˆz is normal to the z= 0 plane. So
A · ds = dxdy, and subsequently the integral in
Equation 4.41 becomes
∫ x2
x1
∫ y2
y1
dxdy = ( x2 − x1) (y2 − y1) (4.43)
Note that this has turned out to be a calculation of
area.
Once again, we see the Cartesian system was an
appropriate choice for this example because this
choice minimizes the number of varying coordinates;
in the above example, the surface of integration is
described by a constant value of zwith variable
values of xand y. If the surface had instead been a
cylinder or a sphere, not only would all three basis
directions be variable, but also the surface normal
would be variable, making the problem dramatically
more complicated.
Now let’s return to the issue of the direction of ds.
W e chose +ˆz, but why not choose −ˆz – also a normal
to the surface – as this direction? T o answer simply ,",Electromagnetics_Vol1.pdf
"W e chose +ˆz, but why not choose −ˆz – also a normal
to the surface – as this direction? T o answer simply ,
the resulting area would be negative. “Negative area”
is the expected (“positive”) area, except with respect
to the opposite-facing normal vector. In the present
problem, the sign of the area is not important, but in",Electromagnetics_Vol1.pdf
"4.3. CYLINDRICAL COORDINA TES 77
some problems this sign becomes important. One
example of a class of problems for which the sign of
area is important is when the quantity of interest is a
flux . If A were a flux density , then the integration over
area that we just performed indicates the magnitude
and direction of flux, and so the direction chosen for
ds defines the direction of positive flux. Section 2.4
describes the electric field in terms of a flux (i.e., as
electric flux density D), in which case positive flux
flows away from a positively-charged source.
Integration Over V olume. Another common task in
vector analysis is integration of some quantity over a
volume. Since the procedure is the same for scalar or
vector quantities, we shall consider integration of a
scalar quantity A(r) for simplicity . First, we note that
the contribution from a differential volume element
dv= dxdydz (4.44)
is A(r) dv, so the integral over the volume V is
∫
V
A(r) dv (4.45)",Electromagnetics_Vol1.pdf
"the contribution from a differential volume element
dv= dxdydz (4.44)
is A(r) dv, so the integral over the volume V is
∫
V
A(r) dv (4.45)
For example, if A(r) = 1 and V is a cube bounded by
x1 ≤ x≤ x2, y1 ≤ y≤ y2, and z1 ≤ z≤ z2, then
the above integral becomes
∫ x2
x1
∫ y2
y1
∫ z2
z1
dxdydz = ( x2 − x1) (y2 − y1) (z2 − z1)
(4.46)
i.e., this is a calculation of volume.
The Cartesian system was an appropriate choice for
this example because V is a cube, which is easy to
describe in Cartesian coordinates and relatively
difficult to describe in any other coordinate system.
Additional Reading:
• “Cartesian coordinate system” on Wikipedia.
4.3 Cylindrical Coordinates
[m0096]
Cartesian coordinates (Section 4.2) are not convenient
in certain cases. One of these is when the problem has
cylindrical symmetry . For example, in the Cartesian
coordinate system, the cross-section of a cylinder
concentric with the z-axis requires two coordinates to
describe: xand y. However, this cross section can be",Electromagnetics_Vol1.pdf
"concentric with the z-axis requires two coordinates to
describe: xand y. However, this cross section can be
described using a single parameter – namely the
radius – which is ρin the cylindrical coordinate
system. This results in a dramatic simplification of
the mathematics in some applications.
The cylindrical system is defined with respect to the
Cartesian system in Figure 4.11. In lieu of xand y,
the cylindrical system uses ρ, the distance measured
from the closest point on the zaxis,1 and φ, the angle
measured in a plane of constant z, beginning at the
+xaxis ( φ= 0 ) with φincreasing toward the +y
direction.
The basis vectors in the cylindrical system are ˆρ, ˆφ,
and ˆz. As in the Cartesian system, the dot product of
like basis vectors is equal to one, and the dot product
of unlike basis vectors is equal to zero. The cross
1 Note that some textbooks use “ r” in lieu of ρ for this coordinate.
§
x
¨
z
©
ª
z
«
z
c⃝ K. Kikkeri CC BY SA 4.0",Electromagnetics_Vol1.pdf
"1 Note that some textbooks use “ r” in lieu of ρ for this coordinate.
§
x
¨
z
©
ª
z
«
z
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.11: Cylindrical coordinate system and asso-
ciated basis vectors.",Electromagnetics_Vol1.pdf
"78 CHAPTER 4. VECTOR ANAL YSIS
¬
­
z
Figure 4.12: Cross products among basis vectors in the
cylindrical system. (See Figure 4.10 for instructions
on the use of this diagram.)
products of basis vectors are as follows:
ˆρ× ˆφ= ˆz (4.47)
ˆφ× ˆz = ˆρ (4.48)
ˆz × ˆρ= ˆφ (4.49)
A useful diagram that summarizes these relationships
is shown in Figure 4.12.
The cylindrical system is usually less useful than the
Cartesian system for identifying absolute and relative
positions. This is because the basis directions depend
on position. For example, ˆρis directed radially
outward from the ˆz axis, so ˆρ= ˆx for locations along
the x-axis but ˆρ= ˆy for locations along the yaxis.
Similarly , the direction ˆφvaries as a function of
position. T o overcome this awkwardness, it is
common to set up a problem in cylindrical
coordinates in order to exploit cylindrical symmetry ,
but at some point to convert to Cartesian coordinates.
Here are the conversions:
x= ρcos φ (4.50)
y= ρsin φ (4.51)",Electromagnetics_Vol1.pdf
"but at some point to convert to Cartesian coordinates.
Here are the conversions:
x= ρcos φ (4.50)
y= ρsin φ (4.51)
and zis identical in both systems. The conversion
from Cartesian to cylindrical is as follows:
ρ=
√
x2 + y2 (4.52)
φ= arctan ( y,x) (4.53)
where arctan is the four-quadrant inverse tangent
function; i.e., arctan(y/x) in the first quadrant
(x> 0, y >0), but possibly requiring an adjustment
· ˆρ ˆφ ˆz
ˆx cos φ − sin φ 0
ˆy sin φ cos φ 0
ˆz 0 0 1
T able 4.1: Dot products between basis vectors in the
cylindrical and Cartesian coordinate systems.
for the other quadrants because the signs of both x
and yare individually significant. 2
Similarly , it is often necessary to represent basis
vectors of the cylindrical system in terms of Cartesian
basis vectors and vice-versa. Conversion of basis
vectors is straightforward using dot products to
determine the components of the basis vectors in the
new system. For example, ˆx in terms of the basis
vectors of the cylindrical system is",Electromagnetics_Vol1.pdf
"determine the components of the basis vectors in the
new system. For example, ˆx in terms of the basis
vectors of the cylindrical system is
ˆx = ˆρ(ˆρ· ˆx) + ˆφ
(
ˆφ· ˆx
)
+ ˆz (ˆz · ˆx) (4.54)
The last term is of course zero since ˆz · ˆx = 0 .
Calculation of the remaining terms requires dot
products between basis vectors in the two systems,
which are summarized in T able 4.1. Using this table,
we find
ˆx = ˆρcos φ− ˆφsin φ (4.55)
ˆy = ˆρsin φ+ ˆφcos φ (4.56)
and of course ˆz requires no conversion. Going from
Cartesian to cylindrical, we find
ˆρ= ˆx cos φ+ ˆy sin φ (4.57)
ˆφ= −ˆx sin φ+ ˆy cos φ (4.58)
Integration Over Length. A differential-length
segment of a curve in the cylindrical system is
described in general as
dl = ˆρdρ+ ˆφρdφ+ ˆz dz (4.59)
Note that the contribution of the φcoordinate to
differential length is ρdφ, not simply dφ. This is
because φis an angle, not a distance. T o see why the
associated distance is ρdφ, consider the following.",Electromagnetics_Vol1.pdf
"because φis an angle, not a distance. T o see why the
associated distance is ρdφ, consider the following.
2 Note that this function is available in MA TLAB and Octave as
atan2(y,x).",Electromagnetics_Vol1.pdf
"4.3. CYLINDRICAL COORDINA TES 79
l
®
x
¯
°
±
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.13: Example in cylindrical coordinates: The
circumference of a circle.
The circumference of a circle of radius ρis 2πρ. If
only a fraction of the circumference is traversed, the
associated arclength is the circumference scaled by
φ/2π, where φis the angle formed by the traversed
circumference. Therefore, the distance is
2πρ· φ/2π= ρφ, and the differential distance is ρdφ.
As always, the integral of a vector field A(r) over a
curve C is ∫
C
A · dl (4.60)
T o demonstrate the cylindrical system, let us calculate
the integral of A(r) = ˆφwhen C is a circle of radius
ρ0 in the z= 0 plane, as shown in Figure 4.13. In this
example, dl = ˆφρ0 dφsince ρ= ρ0 and z= 0 are
both constant along C. Subsequently , A · dl = ρ0dφ
and the above integral is
∫ 2π
0
ρ0 dφ= 2 πρ0 (4.61)
i.e., this is a calculation of circumference.
Note that the cylindrical system is an appropriate
choice for the preceding example because the",Electromagnetics_Vol1.pdf
"i.e., this is a calculation of circumference.
Note that the cylindrical system is an appropriate
choice for the preceding example because the
problem can be expressed with the minimum number
of varying coordinates in the cylindrical system. If we
had attempted this problem in the Cartesian system,
we would find that both xand yvary over C, and in a
relatively complex way . 3
3 Nothing will drive this point home more firmly than trying it. It
can be done, but it’s a lot more work...
²
xd³
´ dµ
ds
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.14: Example in cylindrical coordinates: The
area of a circle.
Integration Over Area. Now we ask the question,
what is the integral of some vector field A over a
circular surface S in the z= 0 plane having radius
ρ0? This is shown in Figure 4.14. The differential
surface vector in this case is
ds = ˆz (dρ) (ρdφ) = ˆz ρdρdφ (4.62)
The quantities in parentheses are the radial and
angular dimensions, respectively . The direction of ds",Electromagnetics_Vol1.pdf
"ds = ˆz (dρ) (ρdφ) = ˆz ρdρdφ (4.62)
The quantities in parentheses are the radial and
angular dimensions, respectively . The direction of ds
indicates the direction of positive flux – see the
discussion in Section 4.2 for an explanation. In
general, the integral over a surface is
∫
S
A · ds (4.63)
T o demonstrate, let’s consider A = ˆz; in this case
A · ds = ρdρdφ and the integral becomes
∫ ρ0
0
∫ 2π
0
ρdρdφ =
( ∫ ρ0
0
ρdρ
)( ∫ 2π
0
dφ
)
=
( 1
2ρ2
0
)
(2π)
= πρ2
0
(4.64)
which we recognize as the area of the circle, as
expected. The corresponding calculation in the
Cartesian system is quite difficult in comparison.
Whereas the previous example considered a planar
surface, we might consider instead a curved surface.",Electromagnetics_Vol1.pdf
"80 CHAPTER 4. VECTOR ANAL YSIS
y
¶
·
¸
¹
2
º
1
dϕ
dz
0
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.15: Example in cylindrical coordinates: The
area of the curved surface of a cylinder.
Here we go. What is the integral of a vector field
A = ˆρover a cylindrical surface S concentric with
the zaxis having radius ρ0 and extending from
z= z1 to z= z2? This is shown in Figure 4.15. The
differential surface vector in this case is
ds = ˆρ(ρ0dφ) (dz) = ˆρρ0 dφdz (4.65)
The integral is
∫
S
A · ds =
∫ 2π
0
∫ z2
z1
ρ0 dφdz
= ρ0
( ∫ 2π
0
dφ
)( ∫ z2
z1
dz
)
= 2 πρ0 (z2 − z1)
(4.66)
which is the area of S, as expected. Once again, the
corresponding calculation in the Cartesian system is
quite difficult in comparison.
Integration Over V olume. The differential volume
element in the cylindrical system is
dv = dρ (ρdφ) dz = ρdρdφdz (4.67)
For example, if A(r) = 1 and the volume V is a
cylinder bounded by ρ≤ ρ0 and z1 ≤ z≤ z2, then
∫
V
A(r) dv=
∫ ρ0
0
∫ 2π
0
∫ z2
z1
ρdρdφdz
=
( ∫ ρ0
0
ρdρ
)( ∫ 2π
0
dφ
)( ∫ z2
z1",Electromagnetics_Vol1.pdf
"cylinder bounded by ρ≤ ρ0 and z1 ≤ z≤ z2, then
∫
V
A(r) dv=
∫ ρ0
0
∫ 2π
0
∫ z2
z1
ρdρdφdz
=
( ∫ ρ0
0
ρdρ
)( ∫ 2π
0
dφ
)( ∫ z2
z1
dz
)
= πρ2
0 (z2 − z1)
(4.68)
i.e., area times length, which is volume.
Once again, the procedure above is clearly more
complicated than is necessary if we are interested
only in computing volume. However, if the integrand
is not constant-valued then we are no longer simply
computing volume. In this case, the formalism is
appropriate and possibly necessary .
Additional Reading:
• “Cylindrical coordinate system” on Wikipedia.",Electromagnetics_Vol1.pdf
"4.4. SPHERICAL COORDINA TES 81
4.4 Spherical Coordinates
[m0097]
The spherical coordinate system is defined with
respect to the Cartesian system in Figure 4.16. The
spherical system uses r, the distance measured from
the origin; 4 θ, the angle measured from the +zaxis
toward the z= 0 plane; and φ, the angle measured in
a plane of constant z, identical to φin the cylindrical
system.
Spherical coordinates are preferred over Cartesian
and cylindrical coordinates when the geometry of the
problem exhibits spherical symmetry . For example, in
the Cartesian coordinate system, the surface of a
sphere concentric with the origin requires all three
coordinates ( x, y, and z) to describe. However, this
surface can be described using a single constant
parameter – the radius r– in the spherical coordinate
system. This leads to a dramatic simplification in the
mathematics in certain applications.
4 Note that some textbooks use “ R” in lieu of r for this coordi-
nate.
y
»
¼
z
r
θθ
ϕ
r",Electromagnetics_Vol1.pdf
"mathematics in certain applications.
4 Note that some textbooks use “ R” in lieu of r for this coordi-
nate.
y
»
¼
z
r
θθ
ϕ
r
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.16: Spherical coordinate system and associ-
ated basis vectors.
θ
ϕ
r
Figure 4.17: Cross products among basis vectors in the
spherical system. (See Figure 4.10 for instructions on
the use of this diagram.)
The basis vectors in the spherical system are ˆr, ˆθ, and
ˆφ. As always, the dot product of like basis vectors is
equal to one, and the dot product of unlike basis
vectors is equal to zero. For the cross-products, we
find:
ˆr × ˆθ= ˆφ (4.69)
ˆθ× ˆφ= ˆr (4.70)
ˆφ× ˆr = ˆθ (4.71)
A useful diagram that summarizes these relationships
is shown in Figure 4.17.
Like the cylindrical system, the spherical system is
often less useful than the Cartesian system for
identifying absolute and relative positions. The
reason is the same: Basis directions in the spherical
system depend on position. For example, ˆr is directed",Electromagnetics_Vol1.pdf
"reason is the same: Basis directions in the spherical
system depend on position. For example, ˆr is directed
radially outward from the origin, so ˆr = ˆx for
locations along the x-axis but ˆr = ˆy for locations
along the yaxis and ˆr = ˆz for locations along the z
axis. Similarly , the directions of ˆθand ˆφvary as a
function of position. T o overcome this awkwardness,
it is common to begin a problem in spherical
coordinates, and then to convert to Cartesian
coordinates at some later point in the analysis. Here
are the conversions:
x= rcos φsin θ (4.72)
y= rsin φsin θ (4.73)
z= rcos θ (4.74)
The conversion from Cartesian to spherical",Electromagnetics_Vol1.pdf
"82 CHAPTER 4. VECTOR ANAL YSIS
· ˆr ˆθ ˆφ
ˆx sin θcos φ cos θcos φ − sin φ
ˆy sin θsin φ cos θsin φ cos φ
ˆz cos θ − sin θ 0
T able 4.2: Dot products between basis vectors in the
spherical and Cartesian coordinate systems.
coordinates is as follows:
r=
√
x2 + y2 + z2 (4.75)
θ= arccos ( z/r) (4.76)
φ= arctan ( y,x) (4.77)
(4.78)
where arctan is the four-quadrant inverse tangent
function.5
Dot products between basis vectors in the spherical
and Cartesian systems are summarized in T able 4.2.
This information can be used to convert between basis
vectors in the spherical and Cartesian systems, in the
same manner described in Section 4.3; e.g.
ˆx = ˆr (ˆr · ˆx) + ˆθ
(
ˆθ· ˆx
)
+ ˆφ
(
ˆφ· ˆx
)
(4.79)
ˆr = ˆx (ˆx · ˆr) + ˆy (ˆy · ˆr) + ˆz (ˆz · ˆr) (4.80)
and so on.
Example 4.3. Cartesian to spherical conversion.
A vector field G = ˆx xz/y. Develop an
expression for G in spherical coordinates.
Solution: Simply substitute expressions in terms
of spherical coordinates for expressions in terms",Electromagnetics_Vol1.pdf
"expression for G in spherical coordinates.
Solution: Simply substitute expressions in terms
of spherical coordinates for expressions in terms
of Cartesian coordinates. Use T able 4.2 and
Equations 4.72–4.74. Making these substitutions
and applying a bit of mathematical clean-up
afterward, one obtains
G =
(
ˆr sin θcot φ+ ˆφcos θcot φ− ˆφ
)
· rcos θcos φ (4.81)
5 Note that this function is available in MA TLAB and Octave as
atan2(y,x).
dl
ρ
z
θ
a
½ ¾ ¿ À
Á Â Ã Ä
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.18: Example in spherical coordinates: Pole-
to-pole distance on a sphere.
Integration Over Length. A differential-length
segment of a curve in the spherical system is
dl = ˆr dr+ ˆθrdθ + ˆφr sin θdφ (4.82)
Note that θis an angle, as opposed to a distance. The
associated distance is rdθ in the θdirection. Note
also that in the φdirection, distance is rdφ in the
z= 0 plane and less by the factor sin θfor z <>0.
As always, the integral of a vector field A(r) over a
curve C is ∫
C
A · dl (4.83)",Electromagnetics_Vol1.pdf
"z= 0 plane and less by the factor sin θfor z <>0.
As always, the integral of a vector field A(r) over a
curve C is ∫
C
A · dl (4.83)
T o demonstrate line integration in the spherical
system, imagine a sphere of radius acentered at the
origin with “poles” at z= + aand z= −a. Let us
calculate the integral of A(r) = ˆθ, where C is the arc
drawn directly from pole to pole along the surface of
the sphere, as shown in Figure 4.18. In this example,
dl = ˆθadθ since r= aand φ(which could be any
value) are both constant along C. Subsequently ,
A · dl = adθ and the above integral is
∫ π
0
adθ = πa (4.84)
i.e., half the circumference of the sphere, as expected.
Note that the spherical system is an appropriate
choice for this example because the problem can be",Electromagnetics_Vol1.pdf
"4.4. SPHERICAL COORDINA TES 83
expressed with the minimum number of varying
coordinates in the spherical system. If we had
attempted this problem in the Cartesian system, we
would find that both zand either xor y(or all three)
vary over C and in a relatively complex way .
Integration Over Area. Now we ask the question,
what is the integral of some vector field A over the
surface S of a sphere of radius acentered on the
origin? This is shown in Figure 4.19. The differential
surface vector in this case is
ds = ˆr (rdθ) (rsin θdφ) = ˆr r2 sin θdθdφ (4.85)
As always, the direction is normal to the surface and
in the direction associated with positive flux. The
quantities in parentheses are the distances associated
with varying θand φ, respectively . In general, the
integral over a surface is
∫
S
A · ds (4.86)
In this case, let’s consider A = ˆr; in this case
A · ds = a2 sin θdθdφ and the integral becomes
∫ π
0
∫ 2π
0
a2 sin θdθdφ = a2
( ∫ π
0
sin θdθ
)( ∫ 2π
0
dφ
)
= a2 (2) (2π)
= 4 πa2",Electromagnetics_Vol1.pdf
"A · ds = a2 sin θdθdφ and the integral becomes
∫ π
0
∫ 2π
0
a2 sin θdθdφ = a2
( ∫ π
0
sin θdθ
)( ∫ 2π
0
dφ
)
= a2 (2) (2π)
= 4 πa2
(4.87)
which we recognize as the area of the sphere, as
expected. The corresponding calculation in the
Cartesian or cylindrical systems is quite difficult in
comparison.
Integration Over V olume. The differential volume
element in the spherical system is
dv = dr(rdθ) (rsin θdφ) = r2dr sin θdθdφ
(4.88)
For example, if A(r) = 1 and the volume V is a
a
r sinθ dϕ
y
x
z
r dθ
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.19: Example in spherical coordinates: The
area of a sphere.
sphere of radius acentered on the origin, then
∫
V
A(r) dv=
∫ a
0
∫ π
0
∫ 2π
0
r2dr sin θdθdφ
=
( ∫ a
0
r2dr
)( ∫ π
0
sin θdθ
)( ∫ 2π
0
dφ
)
=
( 1
3a3
)
(2) (2π)
= 4
3πa3
(4.89)
which is the volume of a sphere.
Additional Reading:
• “Spherical coordinate system” on Wikipedia.",Electromagnetics_Vol1.pdf
"84 CHAPTER 4. VECTOR ANAL YSIS
4.5 Gradient
[m0098]
The gradient operator is an important and useful tool
in electromagnetic theory . Here’s the main idea:
The gradient of a scalar field is a vector that
points in the direction in which the field is most
rapidly increasing, with the scalar part equal to
the rate of change.
A particularly important application of the gradient is
that it relates the electric field intensity E(r) to the
electric potential field V(r). This is apparent from a
review of Section 2.2 (“Electric Field Intensity”); see
in particular, the battery-charged capacitor example.
In that example, it is demonstrated that:
• The direction of E(r) is the direction in which
V(r) decreases most quickly , and
• The scalar part of E(r) is the rate of change of
V(r) in that direction. Note that this is also
implied by the units, since V(r) has units of V
whereas E(r) has units of V/m.
The gradient is the mathematical operation that",Electromagnetics_Vol1.pdf
"implied by the units, since V(r) has units of V
whereas E(r) has units of V/m.
The gradient is the mathematical operation that
relates the vector field E(r) to the scalar field V(r)
and is indicated by the symbol “ ∇” as follows:
E(r) = −∇V(r) (4.90)
or, with the understanding that we are interested in the
gradient as a function of position r, simply
E = −∇V (4.91)
At this point we should note that the gradient is a very
general concept, and that we have merely identified
one application of the gradient above. In
electromagnetics there are many situations in which
we seek the gradient ∇f of some scalar field f(r).
Furthermore, we find that other differential operators
that are important in electromagnetics can be
interpreted in terms of the gradient operator ∇. These
include divergence (Section 4.6), curl (Section 4.8),
and the Laplacian (Section 4.10).
In the Cartesian system:
∇f = ˆx∂f
∂x + ˆy ∂f
∂y + ˆz∂f
∂z (4.92)
Example 4.4. Gradient of a ramp function.",Electromagnetics_Vol1.pdf
"and the Laplacian (Section 4.10).
In the Cartesian system:
∇f = ˆx∂f
∂x + ˆy ∂f
∂y + ˆz∂f
∂z (4.92)
Example 4.4. Gradient of a ramp function.
Find the gradient of f = ax(a “ramp” having
slope aalong the xdirection).
Solution. Here, ∂f/∂x = aand
∂f/∂y = ∂f/∂z = 0 . Therefore ∇f = ˆxa.
Note that ∇f points in the direction in which f
most rapidly increases, and has magnitude equal
to the slope of f in that direction.
The gradient operator in the cylindrical and spherical
systems is given in Appendix B.2.
Additional Reading:
• “Gradient” on Wikipedia.",Electromagnetics_Vol1.pdf
"4.6. DIVERGENCE 85
4.6 Divergence
[m0044]
In this section, we present the divergence operator,
which provides a way to calculate the flux associated
with a point in space. First, let us review the concept
of flux.
The integral of a vector field over a surface is a scalar
quantity known as flux . Specifically , the flux F of a
vector field A(r) over a surface S is
∫
S
A · ds = F (4.93)
Note that A could be fairly described as a flux density ;
i.e., a quantity having units equal to the units of F, but
divided by area (i.e., m 2). Also worth noting is that
the flux of a vector field that has unit magnitude and
is normal to all points on S is simply the area of S.
It is quite useful to identify some electromagnetic
quantities as either fluxes or flux densities. Here are
two important examples:
• The electric flux density D, having units of
C/m2, is a description of the electric field as a
flux density . (See Section 2.4 for more about
electric flux density .) The integral of D over a",Electromagnetics_Vol1.pdf
"C/m2, is a description of the electric field as a
flux density . (See Section 2.4 for more about
electric flux density .) The integral of D over a
closed surface yields the enclosed charge Qencl,
having units of C. This relationship is known as
Gauss’ Law:
∮
S
D · ds = Qencl (4.94)
(See Section 5.5 for more about Gauss’ Law .)
• The magnetic flux density B, having units of
Wb/m2, is a description of the magnetic field as
a flux density . (See Section 2.5 for more about
magnetic flux density .) The integral of B over a
surface (open or closed) yields the magnetic flux
Φ , having units of Wb:
∫
S
B · ds = Φ (4.95)
This is important because, for example, the time
rate of change of Φ is proportional to electric
potential. (See Section 8.3 for more about this
principle, called F araday’s Law .)
Summarizing:
Flux is the scalar quantity obtained by integrating
a vector field, interpreted in this case as a flux
density , over a specified surface.
The concept of flux applies to a surface of finite size.",Electromagnetics_Vol1.pdf
"a vector field, interpreted in this case as a flux
density , over a specified surface.
The concept of flux applies to a surface of finite size.
However, what is frequently of interest is behavior at
a single point, as opposed to the sum or average over
a region of space. For example, returning to the idea
of electric flux density ( D), perhaps we are not
concerned about the total charge (units of C) enclosed
by a surface, but rather the charge density (C/m 3) at a
point. In this case, we could begin with Equation 4.94
and divide both sides of the equation by the volume V
enclosed by S:
∮
S D · ds
V = Qencl
V (4.96)
Now we let V shrink to zero, giving us an expression
that must be true at whatever point we decide to
converge upon. T aking the limit as V → 0:
lim
V→0
∮
S D · ds
V = lim
V→0
Qencl
V (4.97)
The quantity on the right hand side is by definition the
volume charge density ρv (units of C/m 3) at the point
at which we converge. The left hand side is the",Electromagnetics_Vol1.pdf
"volume charge density ρv (units of C/m 3) at the point
at which we converge. The left hand side is the
divergence of D, sometimes abbreviated “div D. ”
Thus, the above equation can be written
div D = ρv (4.98)
Summarizing:
Divergence is the flux per unit volume through an
infinitesimally-small closed surface surrounding
a point.
W e will typically not actually want to integrate and
take a limit in order to calculate divergence.
Fortunately , we do not have to. It turns out that this
operation can be expressed as the dot product ∇ · D;
where, for example,
∇ ≜ ˆx ∂
∂x + ˆy ∂
∂y + ˆz ∂
∂z (4.99)",Electromagnetics_Vol1.pdf
"86 CHAPTER 4. VECTOR ANAL YSIS
in the Cartesian coordinate system. This is the same
“ ∇” that appears in the definition of the gradient
operator (Section 4.5) and is same operator that often
arises when considering other differential operators.
If we expand D in terms of its Cartesian components:
D = ˆxDx + ˆyDy + ˆzDz (4.100)
Then
div D = ∇ · D = ∂Dx
∂x + ∂Dy
∂y + ∂Dz
∂z (4.101)
This seems to make sense for two reasons. First, it is
dimensionally correct. T aking the derivative of a
quantity having units of C/m 2 with respect to distance
yields a quantity having units of C/m 3. Second, it
makes sense that flux from a point should be related
to the sum of the rates of change of the flux density in
each basis direction. Summarizing:
The divergence of a vector field A is ∇ · A.
Example 4.5. Divergence of a uniform field.
A field A that is constant with respect to
position is said to be uniform. A completely
general description of such a field is
A = ˆxAx + ˆyAy + ˆzAz where Ax, Ay, and Az",Electromagnetics_Vol1.pdf
"position is said to be uniform. A completely
general description of such a field is
A = ˆxAx + ˆyAy + ˆzAz where Ax, Ay, and Az
are all constants. W e see immediately that the
divergence of such a field must be zero. That is,
∇ · A = 0 because each component of A is
constant with respect to position. This also
makes sense from the perspective of the “flux
through an infinitesimally-small closed surface”
interpretation of divergence. If the flux is
uniform, the flux into the surface equals the flux
out of the surface resulting in a net flux of zero.
Example 4.6. Divergence of a
linearly-increasing field.
Consider a field A = ˆxA0xwhere A0 is a
constant. The divergence of A is ∇ · A = A0. If
we interpret A as a flux density , then we have
found that the net flux per unit volume is simply
the rate at which the flux density is increasing
with distance.
T o compute divergence in other coordinate systems,
we merely need to know ∇ for those systems. In the
cylindrical system:
∇ = ˆρ1
ρ
∂
∂ρρ+ ˆφ1
ρ
∂",Electromagnetics_Vol1.pdf
"T o compute divergence in other coordinate systems,
we merely need to know ∇ for those systems. In the
cylindrical system:
∇ = ˆρ1
ρ
∂
∂ρρ+ ˆφ1
ρ
∂
∂φρ+ ˆz ∂
∂z (4.102)
and in the spherical system:
∇ = ˆr 1
r2
∂
∂rr2 + ˆθ 1
rsin θ
∂
∂θ sin θ+ ˆφ 1
rsin θ
∂
∂φ
(4.103)
Alternatively , one may use the explicit expressions for
divergence given in Appendix B.2.
Example 4.7. Divergence of a
radially-decreasing field.
Consider a vector field that is directed radially
outward from a point and which decreases
linearly with distance; i.e., A = ˆrA0/rwhere
A0 is a constant. In this case, the divergence is
most easily computed in the spherical coordinate
system since partial derivatives in all but one
direction ( r) equal zero. Neglecting terms that
include these zero-valued partial derivatives, we
find:
∇ · A = 1
r2
∂
∂r
(
r2
[ A0
r
])
= A0
r2 (4.104)
In other words, if we interpret A as a flux
density , then the flux per unit volume is
decreasing with as the square of distance from
the origin.",Electromagnetics_Vol1.pdf
"In other words, if we interpret A as a flux
density , then the flux per unit volume is
decreasing with as the square of distance from
the origin.
It is useful to know that divergence, like ∇ itself, is a
linear operator; that is, for any constant scalars aand
band vector fields A and B:
∇ · (aA + bB) = a∇ · A + b∇ · B (4.105)
Additional Reading:
• “Divergence” on Wikipedia.",Electromagnetics_Vol1.pdf
"4.7. DIVERGENCE THEOREM 87
4.7 Divergence Theorem
[m0046]
The Divergence Theorem relates an integral over a
volume to an integral over the surface bounding that
volume. This is useful in a number of situations that
arise in electromagnetic analysis. In this section, we
derive this theorem.
Consider a vector field A representing a flux density ,
such as the electric flux density D or magnetic flux
density B. The divergence of A is
∇ · A = f (4.106)
where is f(r) is the flux per unit volume through an
infinitesimally-small closed surface surrounding the
point at r. Since f is flux per unit volume, we can
obtain flux for any larger contiguous volume V by
integrating over V; i.e.,
flux through V =
∫
V
f dv (4.107)
In the Cartesian system, V can be interpreted as a
three-dimensional grid of infinitesimally-small cubes
having side lengths dx, dy, and dz, respectively . Note
that the flux out of any face of one of these cubes is
equal to the flux into the cube that is adjacent through",Electromagnetics_Vol1.pdf
"that the flux out of any face of one of these cubes is
equal to the flux into the cube that is adjacent through
that face. That is, the portion of the total flux that
flows between cubes cancels when added together. In
fact, the only fluxes which do not cancel in the
integration over V are those corresponding to faces
which lie on the bounding surface S, since the
integration stops there. Stating this mathematically:
∫
V
f dv=
∮
S
A · ds (4.108)
Thus, we have converted a volume integral into a
surface integral.
T o obtain the Divergence Theorem, we return to
Equation 4.106. Integrating both sides of that
equation over V, we obtain
∫
V
(∇ · A) dv=
∫
V
f dv (4.109)
Now applying Equation 4.108 to the right hand side:
∫
V
(∇ · A) dv=
∮
S
A · ds
(4.110)
The Divergence Theorem (Equation 4.110) states
that the integral of the divergence of a vector field
over a volume is equal to the flux of that field
through the surface bounding that volume.
The principal utility of the Divergence Theorem is to",Electromagnetics_Vol1.pdf
"over a volume is equal to the flux of that field
through the surface bounding that volume.
The principal utility of the Divergence Theorem is to
convert problems that are defined in terms of
quantities known throughout a volume into problems
that are defined in terms of quantities known over the
bounding surface and vice-versa.
Additional Reading:
• “Divergence theorem” on Wikipedia.",Electromagnetics_Vol1.pdf
"88 CHAPTER 4. VECTOR ANAL YSIS
4.8 Curl
[m0048]
Curl is an operation, which when applied to a vector
field, quantifies the circulation of that field. The
concept of circulation has several applications in
electromagnetics. T wo of these applications
correspond to directly to Maxwell’s Equations:
• The circulation of an electric field is proportional
to the rate of change of the magnetic field. This
is a statement of the Maxwell-F araday Equation
(Section 8.8), which includes as a special case
Kirchoff ’s V oltage Law for electrostatics
(Section 5.11).
• The circulation of a magnetic field is
proportional to the source current and the rate of
change of the electric field. This is a statement of
Ampere’s Law (Sections 7.9 and 8.9)
Thus, we are motivated to formally define circulation
and then to figure out how to most conveniently apply
the concept in mathematical analysis. The result is the
curl operator.
So, we begin with the concept of circulation:",Electromagnetics_Vol1.pdf
"the concept in mathematical analysis. The result is the
curl operator.
So, we begin with the concept of circulation:
“Circulation” is the integral of a vector field over
a closed path.
Specifically , the circulation of the vector field A(r)
over the closed path C is
∮
C
A · dl (4.111)
The circulation of a uniform vector field is zero for
any valid path. For example, the circulation of
A = ˆxA0 is zero because non-zero contributions at
each point on C cancel out when summed over the
closed path. On the other hand, the circulation of
A = ˆφA0 over a circular path of constant ρand zis a
non-zero constant, since the non-zero contributions to
the integral at each point on the curve are equal and
accumulate when summed over the path.
z
I
H
c⃝ K. Kikkeri CC BY SA 4.0
Figure 4.20: Magnetic field intensity due to a current
flowing along the zaxis.
Example 4.8. Circulation of the magnetic field
intensity surrounding a line current.
Consider a current I (units of A) flowing along",Electromagnetics_Vol1.pdf
"Example 4.8. Circulation of the magnetic field
intensity surrounding a line current.
Consider a current I (units of A) flowing along
the zaxis in the +zdirection, as shown in
Figure 4.20. It is known that this current gives
rise to a magnetic field intensity H = ˆφH0/ρ,
where H0 is a constant having units of A since
the units of H are A/m. (Feel free to consult
Section 7.5 for the details; however, no
additional information is needed to follow the
example being presented here.) The circulation
of H along any circular path of radius ain a
plane of constant zis therefore
∮
C
H · dl =
∫ 2π
φ=0
(
ˆφH0
a
)
·
(
ˆφadφ
)
= 2 πH0
Note that the circulation of H in this case has
two remarkable features: (1) It is independent of
the radius of the path of integration; and (2) it
has units of A, which suggests a current. In fact,
it turns out that the circulation of H in this case
is equal to the enclosed source current I.
Furthermore, it turns out that the circulation of",Electromagnetics_Vol1.pdf
"it turns out that the circulation of H in this case
is equal to the enclosed source current I.
Furthermore, it turns out that the circulation of
H along any path enclosing the source current is
equal to the source current! These findings are
consequences of Ampere’s Law , as noted above.
Curl is, in part, an answer to the question of what the
circulation at a point in space is. In other words, what",Electromagnetics_Vol1.pdf
"4.8. CURL 89
is the circulation as C shrinks to it’s smallest possible
size. The answer in one sense is zero, since the
arclength of C is zero in this limit – there is nothing to
integrate over. However, if we ask instead what is the
circulation per unit area in the limit, then the result
should be the non-trivial value of interest. T o express
this mathematically , we constrain C to lie in a plane,
and define S to be the open surface bounded by C in
this case. Then, we define the scalar part of the curl of
A to be:
lim
∆ s→0
∮
C A · dl
∆ s (4.112)
where ∆ sis the area of S, and (important!) we
require C and S to lie in the plane that maximizes the
above result.
Because S and it’s boundary C lie in a plane, it is
possible to assign a direction to the result. The chosen
direction is the normal ˆn to the plane in which C and
S lie. Because there are two normals at each point on
a plane, we specify the one that satisfies the right
hand rule . This rule, applied to the curl, states that the",Electromagnetics_Vol1.pdf
"a plane, we specify the one that satisfies the right
hand rule . This rule, applied to the curl, states that the
correct normal is the one which points through the
plane in the same direction as the fingers of the right
hand when the thumb of your right hand is aligned
along C in the direction of integration. Why is this the
correct orientation of ˆn? See Section 4.9 for the
answer to that question. For the purposes of this
section, it suffices to consider this to be simply an
arbitrary sign convention.
Now with the normal vector ˆn unambiguously
defined, we can now formally define the curl
operation as follows:
curl A ≜ lim
∆ s→0
ˆn
∮
C A · dl
∆ s (4.113)
where, once again, ∆ sis the area of S, and we select
S to lie in the plane that maximizes the magnitude of
the above result. Summarizing:
The curl operator quantifies the circulation of a
vector field at a point.
The magnitude of the curl of a vector field is the
circulation, per unit area, at a point and such that",Electromagnetics_Vol1.pdf
"vector field at a point.
The magnitude of the curl of a vector field is the
circulation, per unit area, at a point and such that
the closed path of integration shrinks to enclose
zero area while being constrained to lie in the
plane that maximizes the magnitude of the result.
The direction of the curl is determined by the
right-hand rule applied to the associated path of
integration.
Curl is a very important operator in electromagnetic
analysis. However, the definition (Equation 4.113) is
usually quite difficult to apply . Remarkably , however,
it turns out that the curl operation can be defined in
terms of the ∇ operator; that is, the same ∇ operator
associated with the gradient, divergence, and
Laplacian operators. Here is that definition:
curl A ≜ ∇ × A (4.114)
For example: In Cartesian coordinates,
∇ ≜ ˆx ∂
∂x + ˆy ∂
∂y + ˆz ∂
∂z (4.115)
and
A = ˆxAx + ˆyAy + ˆzAz (4.116)
so curl can be calculated as follows:
∇ × A =
⏐
⏐
⏐
⏐
⏐
⏐
ˆx ˆy ˆz
∂
∂x
∂
∂y
∂
∂z
Ax Ay Az
⏐
⏐
⏐
⏐
⏐
⏐
(4.117)",Electromagnetics_Vol1.pdf
"and
A = ˆxAx + ˆyAy + ˆzAz (4.116)
so curl can be calculated as follows:
∇ × A =
⏐
⏐
⏐
⏐
⏐
⏐
ˆx ˆy ˆz
∂
∂x
∂
∂y
∂
∂z
Ax Ay Az
⏐
⏐
⏐
⏐
⏐
⏐
(4.117)
or, evaluating the determinant:
∇ × A = ˆx
( ∂Az
∂y − ∂Ay
∂z
)
+ ˆy
( ∂Ax
∂z − ∂Az
∂x
)
+ ˆz
( ∂Ay
∂x − ∂Ax
∂y
)
(4.118)
Expressions for curl in each of the three major
coordinate systems is provided in Appendix B.2.
It is useful to know is that curl, like ∇ itself, is a
linear operator; that is, for any constant scalars aand
band vector fields A and B:
∇ × (aA + bB) = a∇ × A + b∇ × B (4.119)
Additional Reading:
• “Curl (mathematics)” on Wikipedia.",Electromagnetics_Vol1.pdf
"90 CHAPTER 4. VECTOR ANAL YSIS
4.9 Stokes’ Theorem
[m0051]
Stokes’ Theorem relates an integral over an open
surface to an integral over the curve bounding that
surface. This relationship has a number of
applications in electromagnetic theory . Here is the
theorem:
∫
S
(∇ × A) · ds =
∮
C
A · dl
(4.120)
where S is the open surface bounded by the closed
path C. The direction of the surface normal ds = ˆnds
is related to the direction of integration along C by the
right-hand rule , illustrated in Figure 4.21. In this
case, the right-hand rule states that the correct normal
is the one that points through the surface in the same
direction as the fingers of the right hand when the
thumb of your right hand is aligned along C in the
direction of integration.
Stokes’ Theorem is a purely mathematical result and
not a principle of electromagnetics per se . The
relevance of the theorem to electromagnetic theory is
primarily as a tool in the associated mathematical",Electromagnetics_Vol1.pdf
"relevance of the theorem to electromagnetic theory is
primarily as a tool in the associated mathematical
analysis. Usually the theorem is employed to
transform a problem expressed in terms of an
integration over a surface into an integration over a
closed path or vice-versa. For more information on
the theorem and its derivation, see “ Additional
Reading” at the end of this section.
c⃝ Cronholm144 (modified) CC BY SA 3.0
Figure 4.21: The relative orientations of the direc-
tion of integration C and surface normal ˆn in Stokes’
Theorem.
Additional Reading:
• “Stokes’ Theorem” on Wikipedia.",Electromagnetics_Vol1.pdf
"4.10. THE LAPLACIAN OPERA TOR 91
4.10 The Laplacian Operator
[m0099]
The Laplacian ∇2f of a field f(r) is the divergence
of the gradient of that field:
∇2f ≜ ∇ · (∇f) (4.121)
Note that the Laplacian is essentially a definition of
the second derivative with respect to the three spatial
dimensions. For example, in Cartesian coordinates,
∇2f = ∂2f
∂x2 + ∂2f
∂y2 + ∂2f
∂z2 (4.122)
as can be readily verified by applying the definitions
of gradient and divergence in Cartesian coordinates to
Equation 4.121.
The Laplacian relates the electric potential (i.e., V,
units of V) to electric charge density (i.e., ρv, units of
C/m3). This relationship is known as P oisson’s
Equation (Section 5.15):
∇2V = −ρv
ǫ (4.123)
where ǫis the permittivity of the medium. The fact
that V is related to ρv in this way should not be
surprising, since electric field intensity (E, units of
V/m) is proportional to the derivative of V with
respect to distance (via the gradient) and ρv is",Electromagnetics_Vol1.pdf
"V/m) is proportional to the derivative of V with
respect to distance (via the gradient) and ρv is
proportional to the derivative of E with respect to
distance (via the divergence).
The Laplacian operator can also be applied to vector
fields; for example, Equation 4.122 is valid even if the
scalar field “ f” is replaced with a vector field. In the
Cartesian coordinate system, the Laplacian of the
vector field A = ˆxAx + ˆyAy + ˆzAz is
∇2A = ˆx∇2Ax + ˆy∇2Ay + ˆz∇2Az (4.124)
An important application of the Laplacian operator of
vector fields is the wave equation ; e.g., the wave
equation for E in a lossless and source-free region is
(Section 9.2)
∇2E + β2E = 0 (4.125)
where βis the phase propagation constant.
It is sometimes useful to know that the Laplacian of a
vector field can be expressed in terms of the gradient,
divergence, and curl as follows:
∇2A = ∇ (∇ · A) − ∇ × (∇ × A) (4.126)
The Laplacian operator in the cylindrical and
spherical coordinate systems is given in
Appendix B.2.",Electromagnetics_Vol1.pdf
"∇2A = ∇ (∇ · A) − ∇ × (∇ × A) (4.126)
The Laplacian operator in the cylindrical and
spherical coordinate systems is given in
Appendix B.2.
Additional Reading:
• “Laplace operator” on Wikipedia.
[m0043]",Electromagnetics_Vol1.pdf
"92 CHAPTER 4. VECTOR ANAL YSIS
Image Credits
Fig. 4.1: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fCartesian.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.2: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fPositionFixed.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.3: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fPositionFree.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.4: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fCartesianBasis.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.5: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fV ecCart.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.6: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fRelativePosition.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).",Electromagnetics_Vol1.pdf
"CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.7: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fDotProductSpecialCases.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.8: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fDotProduct.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.9: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0006 fCrossProduct.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.11: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0096 fCylindricalCoordinates.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.13: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:C ylindrical Coordinates Example.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.14: c⃝ K. Kikkeri,
https://commons.wikimedia.org/wiki/File:Cylindrical Coordinates-Area of a Circle.svg,",Electromagnetics_Vol1.pdf
"Fig. 4.14: c⃝ K. Kikkeri,
https://commons.wikimedia.org/wiki/File:Cylindrical Coordinates-Area of a Circle.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.15: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0096 CylinderArea.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.16: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:S pherical Coordinate System.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.18: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0097 fArc.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.19: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0097 fSphereArea.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.20: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:B iot Savart Example.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).",Electromagnetics_Vol1.pdf
"CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 4.21: c⃝ Cronholm144, https://en.wikipedia.org/wiki/File:Stok es’ Theorem.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/). Heavily modified by author.",Electromagnetics_Vol1.pdf
"Chapter 5
Electrostatics
[m0116]
Electrostatics is the theory of the electric field in
conditions in which its behavior is independent of
magnetic fields, including
• The electric field associated with fixed
distributions of electric charge
• Capacitance (the ability of a structure to store
energy in an electric field)
• The energy associated with the electrostatic field
• Steady current induced in a conducting material
in the presence of an electrostatic field
(essentially , Ohm’s Law)
The term “static” refers to the fact that these aspects
of electromagnetic theory can be developed by
assuming sources are time-invariant; we might say
that electrostatics is the study of the electric field at
DC. However, many aspects of electrostatics are
relevant to AC, radio frequency , and higher-frequency
applications as well.
Additional Reading:
• “Electrostatics” on Wikipedia.
5.1 Coulomb’s Law
[m0102]
Consider two charge-bearing particles in free space,
identified as “particle 1” and “particle 2” in",Electromagnetics_Vol1.pdf
"5.1 Coulomb’s Law
[m0102]
Consider two charge-bearing particles in free space,
identified as “particle 1” and “particle 2” in
Figure 5.1. Let the charges borne by these particles be
q1 and q2, and let Rbe the distance between them. If
the particles bear charges of the same sign (i.e., if
q1q2 is positive), then the particles repel; otherwise,
they attract. This repulsion or attraction can be
quantified as a force experienced by each particle.
Physical observations reveal that the magnitude of the
force is proportional to q1q2, and inversely
proportional to R2. For particle 2 we find:
F = ˆRF0
q1q2
R2 (5.1)
where ˆR is the unit vector pointing from the particle 1
to the particle 2, and F0 is a constant. The value of F0
must have units of inverse permittivity; i.e., (F/m) −1.
This is most easily seen by dimensional analysis of
the above relationship, including the suspected factor:
C · C
F/m · m2 = C · C
F · m = C · C
C/V · m = C · V
m = J
m = N",Electromagnetics_Vol1.pdf
"the above relationship, including the suspected factor:
C · C
F/m · m2 = C · C
F · m = C · C
C/V · m = C · V
m = J
m = N
where we have used the facts that 1 F = 1 C/V , 1 V =
1 J/C, and 1 N = 1 J/m. This finding suggests that
F0 ∝ ǫ−1, where ǫis the permittivity of the medium
in which the particles exist. Observations confirm that
the force is in fact inversely proportional to the
permittivity , with an additional factor of 1/4π
(unitless). Putting this all together we obtain what is
commonly known as Coulomb’s Law :
F = ˆR q1q2
4πǫR2 (5.2)
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"94 CHAPTER 5. ELECTROST A TICS
Fq2
q1
particle 
Å
particle 
Æ
Ç
F
R
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.1: Coulomb’s Law describes the force per-
ceived by pairs of charged particles.
Subsequently , the force perceived by particle 2 is
equal and opposite; i.e., equal to −F.
Separately , it is known that F can be described in
terms of the electric field intensity E1 associated with
particle 1:
F = q2E1 (5.3)
This is essentially the definition of E1, as explained in
Section 2.2. Combining this result with Coulomb’s
Law , we obtain a means to directly calculate the field
associated with the first particle in the absence of the
second particle:
E1 = ˆR q1
4πǫR2 (5.4)
where now ˆRRis the vector beginning at the
particle 1 and ending at the point to be evaluated.
The electric field intensity associated with a point
charge (Equation 5.4) is (1) directed away from
positive charge, (2) proportional to the magni-
tude of the charge, (3) inversely proportional to",Electromagnetics_Vol1.pdf
"charge (Equation 5.4) is (1) directed away from
positive charge, (2) proportional to the magni-
tude of the charge, (3) inversely proportional to
the permittivity of the medium, and (3) inversely
proportional to distance squared.
W e have described this result as originating from
Coulomb’s Law , which is based on physical
observations. However, the same result may be
obtained directly from Maxwell’s Equations using
Gauss’ Law (Section 5.5).
Example 5.1. Electric field of a point charge at
the origin.
A common starting point in electrostatic analysis
is the field associated with a particle bearing
charge qat the origin of the coordinate system.
Because the electric field is directed radially
away from a positively-charged source particle
in all directions, this field is most conveniently
described in the spherical coordinate system.
Thus, ˆR becomes ˆr, Rbecomes r, and we have
E(r) = ˆr q
4πǫr2 (5.5)
Here’s a numerical example. What is the electric",Electromagnetics_Vol1.pdf
"Thus, ˆR becomes ˆr, Rbecomes r, and we have
E(r) = ˆr q
4πǫr2 (5.5)
Here’s a numerical example. What is the electric
field intensity at a distance 1 µm from a single
electron located at the origin, in free space? In
this case, q∼
= −1.60 × 10−19 C (don’t forget
that minus sign!), ǫ= ǫ0, r= 1 µm, and we
find:
E(r) = −ˆr (1.44 kV/m) (5.6)
This is large relative to electric field strengths
commonly encountered in engineering
applications. The strong electric field of the
electron is not readily apparent because electrons
in common materials tend to be accompanied by
roughly equal amounts of positive charge, such
as the protons of atoms. Sometimes, however,
the effect of individual electrons does become
significant in practical electronics through a
phenomenon known as shot noise .
Additional Reading:
• “Coulomb’s Law” on Wikipedia.
• “Shot Noise” on Wikipedia.",Electromagnetics_Vol1.pdf
"5.2. ELECTRIC FIELD DUE TO POINT CHARGES 95
5.2 Electric Field Due to Point
Charges
[m0103]
The electric field intensity associated with a single
particle bearing charge q1, located at the origin, is
(Section 5.1)
E(r) = ˆr q1
4πǫr2 (5.7)
If this particle is instead located at some position r1,
then the above expression may be written as follows:
E(r; r1) = r − r1
|r − r1|
q1
4πǫ|r − r1|2 (5.8)
or, combining like terms in the denominator:
E(r; r1) = r − r1
|r − r1|3
q1
4πǫ (5.9)
Now let us consider the field due to multiple such
particles. Under the usual assumptions about the
permittivity of the medium (reminder: Section 2.8),
the property of superposition applies. Using this
principle, we conclude:
The electric field resulting from a set of charged
particles is equal to the sum of the fields associ-
ated with the individual particles.
Stated mathematically:
E(r) =
N∑
n=1
E(r; rn) (5.10)
where N is the number of particles. Thus, we have
E(r) = 1
4πǫ
N∑
n=1
r − rn
|r − rn|3 qn (5.11)",Electromagnetics_Vol1.pdf
"Stated mathematically:
E(r) =
N∑
n=1
E(r; rn) (5.10)
where N is the number of particles. Thus, we have
E(r) = 1
4πǫ
N∑
n=1
r − rn
|r − rn|3 qn (5.11)
5.3 Charge Distributions
[m0100]
In principle, the smallest unit of electric charge that
can be isolated is the charge of a single electron,
which is ∼
= −1.60 × 10−19 C. This is very small, and
we rarely deal with electrons one at a time, so it is
usually more convenient to describe charge as a
quantity that is continuous over some region of space.
In particular, it is convenient to describe charge as
being distributed in one of three ways: along a curve,
over a surface, or within a volume.
Line Charge Distribution. Imagine that charge is
distributed along a curve C through space. Let ∆ qbe
the total charge along a short segment of the curve,
and let ∆ lbe the length of this segment. The line
charge density ρl at any point along the curve is
defined as
ρl ≜ lim
∆ l→0
∆ q
∆ l = dq
dl (5.12)",Electromagnetics_Vol1.pdf
"and let ∆ lbe the length of this segment. The line
charge density ρl at any point along the curve is
defined as
ρl ≜ lim
∆ l→0
∆ q
∆ l = dq
dl (5.12)
which has units of C/m. W e may then define ρl to be
a function of position along the curve, parameterized
by l; e.g., ρl(l). Then, the total charge Qalong the
curve is
Q=
∫
C
ρl(l) dl (5.13)
which has units of C. In other words, line charge
density integrated over length yields total charge.
Surface Charge Distribution. Imagine that charge is
distributed over a surface. Let ∆ qbe the total charge
on a small patch on this surface, and let ∆ sbe the
area of this patch. The surface charge density ρs at
any point on the surface is defined as
ρs ≜ lim
∆ s→0
∆ q
∆ s = dq
ds (5.14)
which has units of C/m 2. Let us define ρs to be a
function of position on this surface. Then the total
charge over a surface S is
Q=
∫
S
ρs ds (5.15)
In other words, surface charge density integrated over
a surface yields total charge.",Electromagnetics_Vol1.pdf
"charge over a surface S is
Q=
∫
S
ρs ds (5.15)
In other words, surface charge density integrated over
a surface yields total charge.
V olume Charge Distribution. Imagine that charge is
distributed over a volume. Let ∆ qbe the total charge",Electromagnetics_Vol1.pdf
"96 CHAPTER 5. ELECTROST A TICS
in a small cell within this volume, and let ∆ vbe the
volume of this cell. The volume charge density ρv at
any point in the volume is defined as
ρv ≜ lim
∆ v→0
∆ q
∆ v = dq
dv (5.16)
which has units of C/m 3. Since ρv is a function of
position within this volume, the total charge within a
volume V is
Q=
∫
V
ρv dv (5.17)
In other words, volume charge density integrated over
a volume yields total charge.
5.4 Electric Field Due to a
Continuous Distribution of
Charge
[m0104]
The electric field intensity associated with N charged
particles is (Section 5.2):
E(r) = 1
4πǫ
N∑
n=1
r − rn
|r − rn|3 qn (5.18)
where qn and rn are the charge and position of the
nth particle. However, it is common to have a
continuous distribution of charge as opposed to a
countable number of charged particles. In this
section, we extend Equation 5.18 using the concept of
continuous distribution of charge (Section 5.3) so that
we may address this more general class of problems.",Electromagnetics_Vol1.pdf
"continuous distribution of charge (Section 5.3) so that
we may address this more general class of problems.
Distribution of Charge Along a Curve. Consider a
continuous distribution of charge along a curve C.
The curve can be divided into short segments of
length ∆ l. Then, the charge associated with the nth
segment, located at rn, is
qn = ρl(rn) ∆ l (5.19)
where ρl is charge density (units of C/m) at rn.
Substituting this expression into Equation 5.18, we
obtain
E(r) = 1
4πǫ
N∑
n=1
r − rn
|r − rn|3 ρl(rn) ∆ l (5.20)
T aking the limit as ∆ l→ 0 yields:
E(r) = 1
4πǫ
∫
C
r − r′
|r − r′|3 ρl(r′) dl (5.21)
where r′ represents the varying position along C with
integration.
The simplest example of a curve is a straight line. It is
straightforward to use Equation 5.21 to determine the
electric field due to a distribution of charge along a
straight line. However, it is much easier to analyze
that particular distribution using Gauss’ Law , as",Electromagnetics_Vol1.pdf
"5.4. ELECTRIC FIELD DUE TO A CONTINUOUS DISTRIBUTION OF CHAR GE 97
shown in Section 5.6. The following example
addresses a charge distribution for which
Equation 5.21 is more appropriate.
Example 5.2. Electric field along the axis of a
ring of uniformly-distributed charge.
Consider a ring of radius ain the z= 0 plane,
centered on the origin, as shown in Figure 5.2.
Let the charge density along this ring be uniform
and equal to ρl (C/m). Find the electric field
along the zaxis.
Solution. The source charge position is given in
cylindrical coordinates as
r′ = ˆρa (5.22)
The position of a field point along the zaxis is
simply
r = ˆzz (5.23)
Thus,
r − r′ = −ˆρa+ ˆzz (5.24)
and
|r − r′| =
√
a2 + z2 (5.25)
Equation 5.21 becomes:
E(z) = 1
4πǫ
∫ 2π
0
−ˆρa+ ˆzz
[a2 + z2]3/2 ρl (adφ)
(5.26)
Pulling factors that do not vary with φout of the
integral and factoring into separate integrals for
the ˆφand ˆz components, we obtain:
ρl a
4πǫ[a2 + z2]3/2
[
−a
∫ 2π
0
ˆρdφ + ˆzz
∫ 2π
0
dφ
]
(5.27)",Electromagnetics_Vol1.pdf
"integral and factoring into separate integrals for
the ˆφand ˆz components, we obtain:
ρl a
4πǫ[a2 + z2]3/2
[
−a
∫ 2π
0
ˆρdφ + ˆzz
∫ 2π
0
dφ
]
(5.27)
The second integral is equal to 2π. The first
integral is equal to zero. T o see this, note that the
integral is simply summing values of ˆρfor all
possible values of φ. Since ˆρ(φ+ π) = −ˆρ(φ),
the integrand for any given value of φis equal
and opposite the integrand πradians later. (This
is one example of a symmetry argument.) Thus,
y
x
z
ρa
È É Ê Ë
r - r
Ì
point
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.2: Calculating the electric field along the
axis of a ring of charge.
we obtain
E(z) = ˆzρl a
2ǫ
z
[a2 + z2]3/2 (5.28)
It is a good exercise to confirm that this result is
dimensionally correct. It is also recommended to
confirm that when z≫ a, the result is
approximately the same as that expected from a
particle having the same total charge as the ring.
Distribution of Charge Over a Surface. Consider a",Electromagnetics_Vol1.pdf
"approximately the same as that expected from a
particle having the same total charge as the ring.
Distribution of Charge Over a Surface. Consider a
continuous distribution of charge over a surface S.
The surface can be divided into small patches having
area ∆ s. Then, the charge associated with the nth
patch, located at rn, is
qn = ρs(rn) ∆ s (5.29)
where ρs is the surface charge density (units of C/m 2)
at rn. Substituting this expression into Equation 5.18,
we obtain
E(r) = 1
4πǫ
N∑
n=1
r − rn
|r − rn|3 ρs(rn) ∆ s (5.30)
T aking the limit as ∆ s→ 0 yields:
E(r) = 1
4πǫ
∫
S
r − r′
|r − r′|3 ρs(r′) ds (5.31)",Electromagnetics_Vol1.pdf
"98 CHAPTER 5. ELECTROST A TICS
where r′ represents the varying position over S with
integration.
Example 5.3. Electric field along the axis of a
disk of uniformly-distributed charge.
Consider a circular disk of radius ain the z= 0
plane, centered on the origin, as shown in
Figure 5.3. Let the charge density over this disk
be uniform and equal to ρs (C/m2). Find the
electric field along the zaxis.
Solution. The source charge position is given in
cylindrical coordinates as
r′ = ˆρρ (5.32)
The position of a field point along the zaxis is
simply
r = ˆzz (5.33)
Thus,
r − r′ = −ˆρρ+ ˆzz (5.34)
and
|r − r′| =
√
ρ2 + z2 (5.35)
Equation 5.31 becomes:
E(z) = 1
4πǫ
∫ a
ρ=0
∫ 2π
φ=0
−ˆρρ+ ˆzz
[ρ2 + z2]3/2 ρs (ρdρdφ )
(5.36)
T o solve this integral, first rearrange the double
integral into a single integral over φfollowed by
integration over ρ:
ρs
4πǫ
∫ a
ρ=0
ρ
[ρ2 + z2]3/2
[ ∫ 2π
φ=0
(−ˆρρ+ ˆzz) dφ
]
dρ
(5.37)
Now we address the integration over φshown in
the square brackets in the above expression:",Electromagnetics_Vol1.pdf
"ρ=0
ρ
[ρ2 + z2]3/2
[ ∫ 2π
φ=0
(−ˆρρ+ ˆzz) dφ
]
dρ
(5.37)
Now we address the integration over φshown in
the square brackets in the above expression:
∫ 2π
φ=0
(−ˆρρ+ ˆzz) dφ= −ρ
∫ 2π
φ=0
ˆρdφ+ˆzz
∫ 2π
φ=0
dφ
(5.38)
The first integral on the right is zero for the
following reason. As the integral progresses in
φ, the vector ˆρrotates. Because the integration is
over a complete revolution (i.e., φfrom 0 to 2π),
the contribution from each pointing of ˆρis
canceled out by another pointing of ˆρthat is in
the opposite direction. Since there is an equal
number of these canceling pairs of pointings, the
result is zero. Thus:
∫ 2π
φ=0
(−ˆρρ+ ˆzz) dφ= 0 + ˆzz
∫ 2π
φ=0
dφ
= ˆz2πz (5.39)
Substituting this into Expression 5.37 we obtain:
ρs
4πǫ
∫ a
ρ=0
ρ
[ρ2 + z2]3/2 [ˆz2πz] dρ
=ˆzρsz
2ǫ
∫ a
ρ=0
ρdρ
[ρ2 + z2]3/2 (5.40)
This integral can be solved using integration by
parts and trigonometric substitution. Since the
solution is tedious and there is no particular",Electromagnetics_Vol1.pdf
"This integral can be solved using integration by
parts and trigonometric substitution. Since the
solution is tedious and there is no particular
principle of electromagnetics demonstrated by
this solution, we shall simply state the result:
∫ a
ρ=0
ρdρ
[ρ2 + z2]3/2 = −1√
ρ2 + z2
⏐
⏐
⏐
⏐
⏐
a
ρ=0
= −1
√
a2 + z2 + 1
|z| (5.41)
Substituting this result:
E(z) = ˆzρsz
2ǫ
( −1√
a2 + z2 + 1
|z|
)
= ˆzρs
2ǫ
( −z√
a2 + z2 + z
|z|
)
= ˆzρs
2ǫ
( −z√
a2 + z2 + sgn z
)
(5.42)
where “sgn” is the “signum” function; i.e.,
sgn z= +1 for z >0 and sgn z= −1 for
z <0. Summarizing:
E(z) = ˆzρs
2ǫ
(
sgn z− z√
a2 + z2
)
(5.43)
It is a good exercise to confirm that this result is
dimensionally correct and yields an electric field
vector that points in the expected direction and
with the expected dependence on aand z.",Electromagnetics_Vol1.pdf
"5.4. ELECTRIC FIELD DUE TO A CONTINUOUS DISTRIBUTION OF CHAR GE 99
y
x
z
ρρ
r - rÍ  
a
Î Ï Ð Ñ
point
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.3: Calculating the electric field along the
axis of a disk of charge.
A special case of the “disk of charge” scenario
considered in the preceding example is an infinite
sheet of charge. The electric field from an infinite
sheet of charge is a useful theoretical result. W e get
the field in this case simply by letting a→ ∞ in
Equation 5.43, yielding:
E(r) = ˆzρs
2ǫsgn z (5.44)
Again, it is useful to confirm that this is
dimensionally correct: C/m 2 divided by F/m yields
V/m. Also, note that Equation 5.44 is the electric field
at any point above or below the charge sheet – not just
on zaxis. This follows from symmetry . From the
perspective of any point in space, the edges of the
sheet are the same distance (i.e., infinitely far) away .
Distribution of Charge In a V olume. Consider a
continuous distribution of charge within a volume V.",Electromagnetics_Vol1.pdf
"Distribution of Charge In a V olume. Consider a
continuous distribution of charge within a volume V.
The volume can be divided into small cells (volume
elements) having volume ∆ v. Then, the charge
associated with the nth cell, located at rn, is
qn = ρv(rn) ∆ v (5.45)
where ρv is volume charge density (units of C/m 3) at
rn. Substituting this expression into Equation 5.18,
we obtain
E(r) = 1
4πǫ
N∑
n=1
r − rn
|r − rn|3 ρv(rn) ∆ v (5.46)
T aking the limit as ∆ v→ 0 yields:
E(r) = 1
4πǫ
∫
V
r − r′
|r − r′|3 ρv(r′) dv (5.47)
where r′ represents the varying position over V with
integration.",Electromagnetics_Vol1.pdf
"100 CHAPTER 5. ELECTROST A TICS
5.5 Gauss’ Law: Integral Form
[m0014]
Gauss’ Law is one of the four fundamental laws of
classical electromagnetics, collectively known as
Maxwell’s Equations . Before diving in, the reader is
strongly encouraged to review Section 2.4. In that
section, Gauss’ Law emerges from the interpretation
of the electric field as a flux density . Section 2.4 does
not actually identify Gauss’ Law , but here it is:
Gauss’ Law (Equation 5.48) states that the flux of
the electric field through a closed surface is equal
to the enclosed charge.
Gauss’ Law is expressed mathematically as follows:
∮
S
D · ds = Qencl
(5.48)
where D is the electric flux density ǫE, S is a closed
surface with differential surface normal ds, and Qencl
is the enclosed charge. W e can see the law is
dimensionally correct; D has units of C/m 2, thus
integrating D over a surface gives a quantity with
units of C/m 2 · m2 = C, which are the units of charge.
Gauss’ Law has a number of applications in",Electromagnetics_Vol1.pdf
"integrating D over a surface gives a quantity with
units of C/m 2 · m2 = C, which are the units of charge.
Gauss’ Law has a number of applications in
electromagnetic theory . One of them, as explored
below , is as a method to compute the electric field in
response to a distribution of electric charge. Note that
a method to do this, based on Coulomb’s Law , is
described in Sections 5.1, 5.2, and 5.4. Gauss’ Law
provides an alternative method that is easier or more
useful in certain applications.
Example 5.4. Electric field associated with a
charged particle, using Gauss’ Law .
In this example, we demonstrate the ability of
Gauss’ Law to predict the field associated with a
charge distribution. Let us do this for the
simplest possible charge distribution. A particle
of charge qlocated at the origin, for which we
already have the answer (Section 5.1).
Solution. Gauss’ Law applies to any surface that
encloses the charge, so for simplicity we chose a",Electromagnetics_Vol1.pdf
"already have the answer (Section 5.1).
Solution. Gauss’ Law applies to any surface that
encloses the charge, so for simplicity we chose a
sphere of radius rcentered at the origin. Note
that Qencl on the right hand side is just qfor any
surface having r> 0. Gauss’ Law in this case
becomes
∫ π
θ=0
∫ 2π
φ=0
D ·
( ˆrr2 sin θdθdφ
)
= q (5.49)
If we can solve for D, we can get E using
D = ǫE. The simplest way to solve for D is to
use a symmetry argument, which proceeds as
follows. In this problem, the magnitude of D
can depend only on r, and not θor φ. This is
because the charge has no particular orientation,
and the sphere is centered on the charge.
Similarly , it is clear that D must point either
directly toward or directly away from the charge.
In other words, D = ˆrD(r). Substituting this in
the above equation, we encounter the dot
product ˆr · ˆr, which is simply 1. Since D(r) and
r2 are constants with respect to the integration,
we obtain:
r2D(r)
∫ π
θ=0
∫ 2π
φ=0
sin θdθdφ = q (5.50)",Electromagnetics_Vol1.pdf
"r2 are constants with respect to the integration,
we obtain:
r2D(r)
∫ π
θ=0
∫ 2π
φ=0
sin θdθdφ = q (5.50)
The remaining integral is simply 4π, thus we
obtain:
D(r) = q
4πr2 (5.51)
Bringing the known vector orientation of D
back into the equation, we obtain
D = ˆr q
4πr2 (5.52)
and finally using D = ǫE we obtain the
expected result
E = ˆr q
4πǫr2 (5.53)
Here’s the point you should take away from the above
example:
Gauss’ Law combined with a symmetry argu-
ment may be sufficient to determine the electric
field due to a charge distribution. Thus, Gauss’
Law may be an easier alternative to Coulomb’s
Law in some applications.",Electromagnetics_Vol1.pdf
"5.6. ELECTRIC FIELD DUE TO AN INFINITE LINE CHARGE USING GAUS S’ LA W 101
Additional Reading:
• “Gauss’ Law” on Wikipedia.
5.6 Electric Field Due to an
Infinite Line Charge using
Gauss’ Law
[m0149]
Section 5.5 explains one application of Gauss’ Law ,
which is to find the electric field due to a charged
particle. In this section, we present another
application – the electric field due to an infinite line of
charge. The result serves as a useful “building block”
in a number of other problems, including
determination of the capacitance of coaxial cable
(Section 5.24). Although this problem can be solved
using the “direct” approach described in Section 5.4
(and it is an excellent exercise to do so), the Gauss’
Law approach demonstrated here turns out to be
relatively simple.
Example 5.5. Electric field associated with an
infinite line charge, using Gauss’ Law .
Use Gauss’ Law to determine the electric field
intensity due to an infinite line of charge along",Electromagnetics_Vol1.pdf
"infinite line charge, using Gauss’ Law .
Use Gauss’ Law to determine the electric field
intensity due to an infinite line of charge along
the zaxis, having charge density ρl (units of
C/m), as shown in Figure 5.4.
Solution. Gauss’ Law requires integration over
a surface that encloses the charge. So, our first
problem is to determine a suitable surface. A
cylinder of radius athat is concentric with the z
axis, as shown in Figure 5.4, is maximally
symmetric with the charge distribution and so is
likely to yield the simplest possible analysis. At
first glance, it seems that we may have a
problem since the charge extends to infinity in
the +zand −zdirections, so it’s not clear how
to enclose all of the charge. Let’s suppress that
concern for a moment and simply choose a
cylinder of finite length l. In principle, we can
solve the problem first for this cylinder of finite
size, which contains only a fraction of the
charge, and then later let l→ ∞ to capture the",Electromagnetics_Vol1.pdf
"solve the problem first for this cylinder of finite
size, which contains only a fraction of the
charge, and then later let l→ ∞ to capture the
rest of the charge. (In fact, we’ll find when the
time comes it will not be necessary to do that,
but we shall prepare for it anyway .)",Electromagnetics_Vol1.pdf
"102 CHAPTER 5. ELECTROST A TICS
Here’s Gauss’ Law:
∮
S
D · ds = Qencl (5.54)
where D is the electric flux density ǫE, S is a
closed surface with outward-facing differential
surface normal ds, and Qencl is the enclosed
charge.
The first order of business is to constrain the
form of D using a symmetry argument, as
follows. Consider the field of a point charge qat
the origin (Section 5.5):
D = ˆr q
4πr2 (5.55)
W e can “assemble” an infinite line of charge by
adding particles in pairs. One pair is added at a
time, with one particle on the +zaxis and the
other on the −zaxis, with each located an equal
distance from the origin. W e continue to add
particle pairs in this manner until the resulting
charge extends continuously to infinity in both
directions. The principle of superposition
indicates that the resulting field will be the sum
of the fields of the particles (Section 5.2). Thus,
we see that D cannot have any component in the
ˆφdirection because none of the fields of the",Electromagnetics_Vol1.pdf
"of the fields of the particles (Section 5.2). Thus,
we see that D cannot have any component in the
ˆφdirection because none of the fields of the
constituent particles have a component in that
direction. Similarly , we see that the magnitude
of D cannot depend on φbecause none of the
fields of the constituent particles depends on φ
and because the charge distribution is identical
(“invariant”) with rotation in φ. Also, note that
for any choice of zthe distribution of charge
above and below that plane of constant zis
identical; therefore, D cannot be a function of z
and D cannot have any component in the ˆz
direction. Therefore, the direction of D must be
radially outward; i.e., in the ˆρdirection, as
follows:
D = ˆρDρ(ρ) (5.56)
Next, we observe that Qencl on the right hand
side of Equation 5.54 is equal to ρll. Thus, we
obtain ∮
S
[ˆρDρ(ρ)] · ds = ρll (5.57)
The cylinder S consists of a flat top, curved side,
and flat bottom. Expanding the above equation
to reflect this, we obtain
ρll=
∫
top",Electromagnetics_Vol1.pdf
"The cylinder S consists of a flat top, curved side,
and flat bottom. Expanding the above equation
to reflect this, we obtain
ρll=
∫
top
[ˆρDρ(ρ)] · (+ˆzds)
+
∫
side
[ˆρDρ(ρ)] · (+ˆρds)
+
∫
bottom
[ˆρDρ(ρ)] · (−ˆzds) (5.58)
Examination of the dot products indicates that
the integrals associated with the top and bottom
surfaces must be zero. In other words, the flux
through the top and bottom is zero because D is
perpendicular to these surfaces. W e are left with
ρll=
∫
side
[Dρ(ρ)] ds (5.59)
The side surface is an open cylinder of radius
ρ= a, so Dρ(ρ) = Dρ(a), a constant over this
surface. Thus:
ρll=
∫
side
[Dρ(a)] ds= [ Dρ(a)]
∫
side
ds
(5.60)
The remaining integral is simply the area of the
side surface, which is 2πa· l. Solving for Dρ(a)
we obtain
Dρ(a) = ρll
2πal = ρl
2πa (5.61)
Remarkably , we see Dρ(a) is independent of l,
So the concern raised in the beginning of this
solution – that we wouldn’t be able to enclose all
of the charge – doesn’t matter.",Electromagnetics_Vol1.pdf
"So the concern raised in the beginning of this
solution – that we wouldn’t be able to enclose all
of the charge – doesn’t matter.
Completing the solution, we note the result must
be the same for any value of ρ(not just ρ= a),
so
D = ˆρDρ(ρ) = ˆρ ρl
2πρ (5.62)
and since D = ǫE:
E = ˆρ ρl
2πǫρ (5.63)
This completes the solution. W e have found that
the electric field is directed radially away from",Electromagnetics_Vol1.pdf
"5.7. GAUSS’ LA W: DIFFERENTIAL FORM 103
z
a
ρ
ρ
l
Ò
Ó
Ô
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.4: Finding the electric field of an infinite line
of charge using Gauss’ Law .
the line charge, and decreases in magnitude in
inverse proportion to distance from the line
charge. Suggestion: Check to ensure that this
solution is dimensionally correct.
5.7 Gauss’ Law: Differential
Form
[m0045]
The integral form of Gauss’ Law (Section 5.5) is a
calculation of enclosed charge Qencl using the
surrounding density of electric flux:
∮
S
D · ds = Qencl (5.64)
where D is electric flux density and S is the enclosing
surface. It is also sometimes necessary to do the
inverse calculation (i.e., determine electric field
associated with a charge distribution). This is
sometimes possible using Equation 5.64 if the
symmetry of the problem permits; see examples in
Section 5.5 and 5.6. If the problem does not exhibit
the necessary symmetry , then it seems that one must
fall back to the family of techniques presented in",Electromagnetics_Vol1.pdf
"the necessary symmetry , then it seems that one must
fall back to the family of techniques presented in
Section 5.4 requiring direct integration over the
charge, which is derived from Coulomb’s Law .
However, even the Coulomb’s Law / direct integration
approach has a limitation that is very important to
recognize: It does not account for the presence of
structures that may influence the electric field. For
example, the electric field due to a charge in free
space is different from the electric field due to the
same charge located near a perfectly-conducting
surface. In fact, these approaches do not account for
the possibility of any spatial variation in material
composition, which rules out their use in many
engineering applications.
T o address this broader scope of problems, we require
an alternative form of Gauss’ Law that applies at
individual points in space. That is, we require Gauss’
Law expressed in the form of a differential equation,",Electromagnetics_Vol1.pdf
"individual points in space. That is, we require Gauss’
Law expressed in the form of a differential equation,
as opposed to an integral equation. This facilitates the
use of Gauss’ Law even in problems that do not
exhibit sufficient symmetry and that involve material
boundaries and spatial variations in material
constitutive parameters. Given this differential
equation and the boundary conditions imposed by
structure and materials, we may then solve for the
electric field in these more complicated scenarios. In
this section, we derive the desired differential form of
Gauss’ Law . Elsewhere (in particular, in Section 5.15)",Electromagnetics_Vol1.pdf
"104 CHAPTER 5. ELECTROST A TICS
we use this equation as a tool to find electric fields in
problems involving material boundaries.
There are in fact two methods to develop the desired
differential equation. One method is via the definition
of divergence, whereas the other is via the divergence
theorem. Both methods are presented below because
each provides a different bit of insight. Let’s explore
the first method:
Derivation via the definition of divergence. Let the
geometrical volume enclosed by S be V, which has
volume V (units of m 3). Dividing both sides of
Equation 5.64 by V and taking the limit as V → 0:
lim
V→0
∮
S D · ds
V = lim
V→0
Qencl
V (5.65)
The quantity on the right hand side is the volume
charge density ρv (units of C/m 3) at the point at
which we converge after letting the volume go to
zero. The left hand side is, by definition, the
divergence of D, indicated in mathematical notation
as “ ∇ · D” (see Section 4.6). Thus, we have Gauss’
Law in differential form :
∇ · D = ρv",Electromagnetics_Vol1.pdf
"divergence of D, indicated in mathematical notation
as “ ∇ · D” (see Section 4.6). Thus, we have Gauss’
Law in differential form :
∇ · D = ρv
(5.66)
T o interpret this equation, recall that divergence is
simply the flux (in this case, electric flux) per unit
volume.
Gauss’ Law in differential form (Equation 5.66)
says that the electric flux per unit volume origi-
nating from a point in space is equal to the vol-
ume charge density at that point.
Derivation via the divergence theorem.
Equation 5.66 may also be obtained from
Equation 5.64 using the Divergence Theorem
(Section 4.7), which in the present case may be
written: ∫
V
(∇ · D) dv=
∮
S
D · ds (5.67)
From Equation 5.64, we see that the right hand side of
the equation may be replaced with the enclosed
charge: ∫
V
(∇ · D) dv= Qencl (5.68)
Furthermore, the enclosed charge can be expressed as
an integration of the volume charge density ρv over V:
∫
V
(∇ · D) dv=
∫
V
ρvdv (5.69)
The above relationship must hold regardless of the",Electromagnetics_Vol1.pdf
"an integration of the volume charge density ρv over V:
∫
V
(∇ · D) dv=
∫
V
ρvdv (5.69)
The above relationship must hold regardless of the
specific location or shape of V. The only way this is
possible is if the integrands are equal. Thus,
∇ · D = ρv, and we have obtained Equation 5.66.
Example 5.6. Determining the charge density at
a point, given the associated electric field.
The electric field intensity in free space is
E(r) = ˆxAx2 + ˆyBz+ ˆzCx2z
where A= 3 V/m3, B = 2 V/m2, and
C = 1 V/m4. What is the charge density at
r = ˆx2 − ˆy2 m?
Solution. First, we use D = ǫE to get D. Since
the problem is in free space, ǫ= ǫ0. Thus we
have that the volume charge density is
ρv = ∇ · D
= ∇ · (ǫ0E) = ǫ0∇ · E
= ǫ0
[ ∂
∂x
(
Ax2)
+ ∂
∂y (Bz) + ∂
∂z
(
Cx2z
) ]
= ǫ0
[
2Ax+ 0 + Cx2]
Now calculating the charge density at the
specified location r:
ǫ0
[
2(3 V/m3)(2 m) + 0 + (1 V/m4)(2 m)2]
= ǫ0 (16 V/m)
= 142 pC/m 3
T o obtain the electric field from the charge",Electromagnetics_Vol1.pdf
"specified location r:
ǫ0
[
2(3 V/m3)(2 m) + 0 + (1 V/m4)(2 m)2]
= ǫ0 (16 V/m)
= 142 pC/m 3
T o obtain the electric field from the charge
distribution in the presence of boundary conditions
imposed by materials and structure, we must enforce
the relevant boundary conditions. These boundary
conditions are presented in Sections 5.17 and 5.18.
Frequently , a simpler approach requiring only the
boundary conditions on the electric potential ( V(r))
is possible; this is presented in Section 5.15.
Furthermore, the reader should note the following.
Gauss’ Law does not always necessarily fully",Electromagnetics_Vol1.pdf
"5.8. FORCE, ENERGY , AND POTENTIAL DIFFERENCE 105
constrain possible solutions for the electric field. For
that, we might also need Kirchoff’s V oltage Law; see
Section 5.11.
Before moving on, it is worth noting that
Equation 5.66 can be solved in the special case in
which there are no boundary conditions to satisfy;
i.e., for charge only , in a uniform and unbounded
medium. In fact, no additional electromagnetic theory
is required to do this. Here’s the solution:
D(r) = 1
4π
∫
V
r − r′
|r − r′|3 ρv(r′) dv (5.70)
which we recognize as one of the results obtained in
Section 5.4 (after dividing both sides by ǫto get E). It
is reasonable to conclude that Gauss’ Law (in either
integral or differential form) is fundamental, whereas
Coulomb’s Law is merely a consequence of Gauss’
Law .
Additional Reading:
• “Gauss’ Law” on Wikipedia.
• “Partial differential equation” on Wikipedia.
• “Boundary value problem” on Wikipedia.
5.8 Force, Energy, and Potential
Difference
[m0061]",Electromagnetics_Vol1.pdf
"• “Partial differential equation” on Wikipedia.
• “Boundary value problem” on Wikipedia.
5.8 Force, Energy, and Potential
Difference
[m0061]
The force Fe experienced by a particle at location r
bearing charge qin an electric field intensity E is (see
Sections 2.2 and/or Section 5.1)
Fe = qE(r) (5.71)
If left alone in free space, this particle would
immediately begin to move. The resulting
displacement represents a loss of potential energy .
This loss can quantified using the concept of work,
W. The incremental work ∆ W done by moving the
particle a short distance ∆ l, over which we assume
the change in Fe is negligible, is
∆ W ≈ − Fe · ˆl∆ l (5.72)
where in this case ˆl is the unit vector in the direction
of the motion; i.e., the direction of Fe. The minus
sign indicates that potential energy of the system
consisting of the electric field and the particle is being
reduced. Like a spring that was previously
compressed and is now released, the system is
“relaxing. ”",Electromagnetics_Vol1.pdf
"reduced. Like a spring that was previously
compressed and is now released, the system is
“relaxing. ”
T o confirm that work defined in this way is an
expression of energy , consider the units. The product
of force (units of N) and distance (units of m) has
units of N ·m, and 1 N ·m is 1 J of energy .
Now , what if the motion of the particle is due to
factors other than the force associated with the
electric field? For example, we might consider
“resetting” the system to it’s original condition by
applying an external force to overcome Fe.
Equation 5.72 still represents the change in potential
energy of the system, but now ˆl changes sign. The
same magnitude of work is done, but now this work is
positive. In other words, positive work requires the
application of an external force that opposes and
overcomes the force associated with the electric field,
thereby increasing the potential energy of the system.
With respect to the analogy of a mechanical spring",Electromagnetics_Vol1.pdf
"thereby increasing the potential energy of the system.
With respect to the analogy of a mechanical spring
used above, positive work is achieved by compressing
the spring.",Electromagnetics_Vol1.pdf
"106 CHAPTER 5. ELECTROST A TICS
It is also worth noting that the purpose of the dot
product in Equation 5.72 is to ensure that only the
component of motion parallel to the direction of the
electric field is included in the energy tally . This is
simply because motion in any other direction cannot
be due to E, and therefore does not increase or
decrease the associated potential energy .
W e can make the relationship between work and the
electric field explicit by substituting Equation 5.71
into Equation 5.72, yielding
∆ W ≈ − qE(r) · ˆl∆ l (5.73)
Equation 5.73 gives the work only for a short distance
around r. Now let us try to generalize this result. If
we wish to know the work done over a larger
distance, then we must account for the possibility that
E varies along the path taken. T o do this, we may
sum contributions from points along the path traced
out by the particle, i.e.,
W ≈
N∑
n=1
∆ W(rn) (5.74)
where rn are positions defining the path. Substituting
Equation 5.73, we have
W ≈ − q",Electromagnetics_Vol1.pdf
"out by the particle, i.e.,
W ≈
N∑
n=1
∆ W(rn) (5.74)
where rn are positions defining the path. Substituting
Equation 5.73, we have
W ≈ − q
N∑
n=1
E(rn) · ˆl(rn)∆ l (5.75)
T aking the limit as ∆ l→ 0 we obtain
W = −q
∫
C
E(r) · ˆl(r)dl (5.76)
where C is the path (previously , the sequence of rn’s)
followed. Now omitting the explicit dependence on r
in the integrand for clarity:
W = −q
∫
C
E · dl (5.77)
where dl = ˆldlas usual. Now , we are able to
determine the change in potential energy for a
charged particle moving along any path in space,
given the electric field.
At this point, it is convenient to formally define the
electric potential difference V21 between the start
point (1) and end point (2) of C. V21 is defined as the
work done by traversing C, per unit of charge:
V21 ≜ W
q (5.78)
This has units of J/C, which is volts (V). Substituting
Equation 5.77, we obtain:
V21 = −
∫
C
E · dl
(5.79)
An advantage of analysis in terms of electrical",Electromagnetics_Vol1.pdf
"Equation 5.77, we obtain:
V21 = −
∫
C
E · dl
(5.79)
An advantage of analysis in terms of electrical
potential as opposed to energy is that we will no
longer have to explicitly state the value of the charge
involved.
The potential difference V21 between two points
in space, given by Equation 5.79, is the change
in potential energy of a charged particle divided
by the charge of the particle. Potential energy is
also commonly known as “voltage” and has units
of V .
Example 5.7. Potential difference in a uniform
electric field.
Consider an electric field E(r) = ˆzE0, which is
constant in both magnitude and direction
throughout the domain of the problem. The path
of interest is a line beginning at ˆzz1 and ending
at ˆzz2. What is V21? (It’s worth noting that the
answer to this problem is a building block for a
vast number of problems in electromagnetic
analysis.)
Solution. From Equation 5.79 we have
V21 = −
∫ z2
z1
(ˆzE0) · ˆzdz= −E0(z2 − z1)
(5.80)",Electromagnetics_Vol1.pdf
"vast number of problems in electromagnetic
analysis.)
Solution. From Equation 5.79 we have
V21 = −
∫ z2
z1
(ˆzE0) · ˆzdz= −E0(z2 − z1)
(5.80)
Note V21 is simply the electric field intensity
times the distance between the points. This may
seem familiar. For example, compare this to the
findings of the battery-charged capacitor
experiment described in Section 2.2. There too
we find that potential difference equals electric
field intensity times distance, and the signs
agree.",Electromagnetics_Vol1.pdf
"5.9. INDEPENDENCE OF P A TH 107
The solution to the preceding example is simple
because the direct path between the two points is
parallel to the electric field. If the path between the
points had been perpendicular to E, then the solution
is even easier – V21 is simply zero. In all other cases,
V21 is proportional to the component of the direct
path between the start and end points that is parallel
to E, as determined by the dot product.
5.9 Independence of Path
[m0062]
In Section 5.8, we found that the potential difference
(“voltage”) associated with a path C in an electric
field intensity E is
V21 = −
∫
C
E · dl (5.81)
where the curve begins at point 1 and ends at point 2.
Let these points be identified using the position
vectors r1 and r2, respectively .1 Then:
V21 = −
∫ r2
r1,along C
E · dl (5.82)
The associated work done by a particle bearing
charge qis
W21 = qV21 (5.83)
This work represents the change in potential energy
of the system consisting of the electric field and the",Electromagnetics_Vol1.pdf
"charge qis
W21 = qV21 (5.83)
This work represents the change in potential energy
of the system consisting of the electric field and the
charged particle. So, it must also be true that
W21 = W2 − W1 (5.84)
where W2 and W1 are the potential energies when the
particle is at r2 and r1, respectively . It is clear from
the above equation that W21 does not depend on C; it
depends only on the positions of the start and end
points and not on any of the intermediate points along
C. That is,
V21 = −
∫ r2
r1
E · dl , independent of C
(5.85)
Since the result of the integration in Equation 5.85 is
independent of the path of integration, any path that
begins at r1 and ends at r2 yields the same value of
W21 and V21. W e refer to this concept as
independence of path .
The integral of the electric field over a path be-
tween two points depends only on the locations
of the start and end points and is independent of
the path taken between those points.
1 See Section 4.1 for a refresher on this concept.",Electromagnetics_Vol1.pdf
"108 CHAPTER 5. ELECTROST A TICS
A practical application of this concept is that some
paths may be easier to use than others, so there may
be an advantage in computing the integral in
Equation 5.85 using some path other than the path
actually traversed.
5.10 Kirchoff’s V oltage Law for
Electrostatics: Integral
Form
[m0016]
As explained in Section 5.9, the electrical potential at
point r2 relative to r1 in an electric field E (V/m) is
V21 = −
∫ r2
r1
E · dl (5.86)
where the path of integration may be any path that
begins and ends at the specified points. Consider what
happens if the selected path through space begins and
ends at the same point; i.e., r2 = r1. In this case, the
path of integration is a closed loop. Since V21
depends only on the positions of the start and end
points and because the potential energy at those
points is the same, we conclude:
∮
E · dl = 0
(5.87)
This principle is known as Kirchoff ’s V oltage Law for
Electrostatics.",Electromagnetics_Vol1.pdf
"points is the same, we conclude:
∮
E · dl = 0
(5.87)
This principle is known as Kirchoff ’s V oltage Law for
Electrostatics.
Kirchoff’s V oltage Law for Electrostatics (Equa-
tion 5.87) states that the integral of the electric
field over a closed path is zero.
It is worth noting that this law is a generalization of a
principle of which the reader is likely already aware.
In electric circuit theory , the sum of voltages over any
closed loop in a circuit is zero. This is also known as
Kirchoff’s V oltage Law because it is precisely the
same principle. T o obtain Equation 5.87 for an
electric circuit, simply partition the closed path into
branches, with each branch representing one
component. Then, the integral of E over each branch
is the branch voltage; i.e., units of V/m times units of
m yields units of V . Then, the sum of these branch
voltages over any closed loop is zero, as dictated by
Equation 5.87.
Finally , be advised that Equation 5.87 is specific to",Electromagnetics_Vol1.pdf
"voltages over any closed loop is zero, as dictated by
Equation 5.87.
Finally , be advised that Equation 5.87 is specific to
electrostatics. In electrostatics, it is assumed that the
electric field is independent of the magnetic field.
This is true if the magnetic field is either zero or not",Electromagnetics_Vol1.pdf
"5.11. KIRCHOFF’S VOL T AGE LA W FOR ELECTROST A TICS: DIFFEREN TIAL FORM 109
time-varying. If the magnetic field is time-varying,
then Equation 5.87 must be modified to account for
the effect of the magnetic field, which is to make the
right hand size potentially different from zero. The
generalized version of this expression that correctly
accounts for that effect is known as the
Maxwell-F araday Equation (Section 8.8).
Additional Reading:
• “Maxwell’s Equations” on Wikipedia.
• “Kirchoff’s Circuit Laws” on Wikipedia.
5.11 Kirchoff’s V oltage Law for
Electrostatics: Differential
Form
[m0152]
The integral form of Kirchoff’s V oltage Law for
electrostatics (KVL; Section 5.10) states that an
integral of the electric field along a closed path is
equal to zero: ∮
C
E · dl = 0 (5.88)
where E is electric field intensity and C is the closed
curve. In this section, we derive the differential form
of this equation. In some applications, this differential
equation, combined with boundary conditions",Electromagnetics_Vol1.pdf
"of this equation. In some applications, this differential
equation, combined with boundary conditions
imposed by structure and materials (Sections 5.17 and
5.18), can be used to solve for the electric field in
arbitrarily complicated scenarios. A more immediate
reason for considering this differential equation is that
we gain a little more insight into the behavior of the
electric field, disclosed at the end of this section.
The equation we seek may be obtained using Stokes’
Theorem (Section 4.9), which in the present case may
be written:
∫
S
(∇ × E) · ds =
∮
C
E · dl (5.89)
where S is any surface bounded by C, and ds is the
normal to that surface with direction determined by
right-hand rule. The integral form of KVL tells us that
the right hand side of the above equation is zero, so:
∫
S
(∇ × E) · ds = 0 (5.90)
The above relationship must hold regardless of the
specific location or shape of S. The only way this is
possible for all possible surfaces is if the integrand is",Electromagnetics_Vol1.pdf
"specific location or shape of S. The only way this is
possible for all possible surfaces is if the integrand is
zero at every point in space. Thus, we obtain the
desired expression:
∇ × E = 0
(5.91)
Summarizing:
The differential form of Kirchoff’s V oltage Law
for electrostatics (Equation 5.91) states that the
curl of the electrostatic field is zero.",Electromagnetics_Vol1.pdf
"110 CHAPTER 5. ELECTROST A TICS
Equation 5.91 is a partial differential equation. As
noted above, this equation, combined with the
appropriate boundary conditions, can be solved for
the electric field in arbitrarily-complicated scenarios.
Interestingly , it is not the only such equation available
for this purpose – Gauss’ Law (Section 5.7) also does
this. Thus, we see a system of partial differential
equations emerging, and one may correctly infer that
that the electric field is not necessarily fully
constrained by either equation alone.
Additional Reading:
• “Maxwell’s Equations” on Wikipedia.
• “Boundary value problem” on Wikipedia.
5.12 Electric Potential Field Due
to Point Charges
[m0064]
The electric field intensity due to a point charge qat
the origin is (see Section 5.1 or 5.5)
E = ˆr q
4πǫr2 (5.92)
In Sections 5.8 and 5.9, it was determined that the
potential difference measured from position r1 to
position r2 is
V21 = −
∫ r2
r1
E · dl (5.93)",Electromagnetics_Vol1.pdf
"In Sections 5.8 and 5.9, it was determined that the
potential difference measured from position r1 to
position r2 is
V21 = −
∫ r2
r1
E · dl (5.93)
This method for calculating potential difference is
often a bit awkward. T o see why , consider an example
from circuit theory , shown in Figure 5.5. In this
example, consisting of a single resistor and a ground
node, we’ve identified four quantities:
• The resistance R
• The current I through the resistor
• The node voltage V1, which is the potential
difference measured from ground to the left side
of the resistor
• The node voltage V2, which is the potential
difference measured from ground to the right
side of the resistor
Let’s say we wish to calculate the potential difference
V21 across the resistor. There are two ways this can be
done:
• V21 = −IR
• V21 = V2 − V1
The advantage of the second method is that it is not
necessary to know I, R, or indeed anything about
what is happening between the nodes; it is only",Electromagnetics_Vol1.pdf
"The advantage of the second method is that it is not
necessary to know I, R, or indeed anything about
what is happening between the nodes; it is only
necessary to know the node voltages. The point is that
it is often convenient to have a common datum – in",Electromagnetics_Vol1.pdf
"5.12. ELECTRIC POTENTIAL FIELD DUE TO POINT CHARGES 111
R
IV1 V
Õ
Ö × Ø Ù
Ú Û
Ü ÝÞ ß à áâ
ã ä å æ
ç è
é êë ì í îï
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.5: A resistor in a larger circuit, used as an
example to demonstrate the concept of node voltages.
this example, ground – with respect to which the
potential differences at all other locations of interest
can be defined. When we have this, calculating
potential differences reduced to simply subtracting
predetermined node potentials.
So, can we establish a datum in general electrostatic
problems that works the same way? The answer is
yes. The datum is arbitrarily chosen to be a sphere
that encompasses the universe; i.e., a sphere with
radius → ∞ . Employing this choice of datum, we can
use Equation 5.93 to define V(r), the potential at
point r, as follows:
V(r) ≜ −
∫ r
∞
E · dl
(5.94)
The electrical potential at a point, given by Equa-
tion 5.94, is defined as the potential difference
measured beginning at a sphere of infinite radius",Electromagnetics_Vol1.pdf
"The electrical potential at a point, given by Equa-
tion 5.94, is defined as the potential difference
measured beginning at a sphere of infinite radius
and ending at the point r. The potential obtained
in this manner is with respect to the potential in-
finitely far away .
In the particular case where E is due to the point
charge at the origin:
V(r) = −
∫ r
∞
[
ˆr q
4πǫr2
]
· dl (5.95)
The principle of independence of path (Section 5.9)
asserts that the path of integration doesn’t matter as
long as the path begins at the datum at infinity and
ends at r. So, we should choose the easiest such path.
The radial symmetry of the problem indicates that the
easiest path will be a line of constant θand φ, so we
choose dl = ˆrdr. Continuing:
V(r) = −
∫ r
∞
[
ˆr q
4πǫr2
]
· [ˆrdr] (5.96)
= − q
4πǫ
∫ r
∞
1
r2 dr (5.97)
= + q
4πǫ
1
r
⏐
⏐
⏐
⏐
r
∞
(5.98)
so
V(r) = + q
4πǫr (5.99)
(Suggestion: Confirm that Equation 5.99 is
dimensionally correct.) In the context of the circuit",Electromagnetics_Vol1.pdf
"4πǫ
1
r
⏐
⏐
⏐
⏐
r
∞
(5.98)
so
V(r) = + q
4πǫr (5.99)
(Suggestion: Confirm that Equation 5.99 is
dimensionally correct.) In the context of the circuit
theory example above, this is the “node voltage” at r
when the datum is defined to be the surface of a
sphere at infinity . Subsequently , we may calculate the
potential difference from any point r1 to any other
point r2 as
V21 = V(r2) − V(r1) (5.100)
and that will typically be a lot easier than using
Equation 5.93.
It is not often that one deals with systems consisting
of a single charged particle. So, for the above
technique to be truly useful, we need a
straightforward way to determine the potential field
V(r) for arbitrary distributions of charge. The first
step in developing a more general expression is to
determine the result for a particle located at a point r′
somewhere other than the origin. Since Equation 5.99
depends only on charge and the distance between the
field point r and r′, we have
V(r; r′) ≜ + q′
4πǫ|r − r′| (5.101)",Electromagnetics_Vol1.pdf
"depends only on charge and the distance between the
field point r and r′, we have
V(r; r′) ≜ + q′
4πǫ|r − r′| (5.101)
where, for notational consistency , we use the symbol
q′ to indicate the charge. Now applying superposition,
the potential field due to N charges is
V(r) =
N∑
n=1
V(r; rn) (5.102)
Substituting Equation 5.101 we obtain:
V(r) = 1
4πǫ
N∑
n=1
qn
|r − rn| (5.103)",Electromagnetics_Vol1.pdf
"112 CHAPTER 5. ELECTROST A TICS
Equation 5.103 gives the electric potential at
a specified location due to a finite number of
charged particles.
The potential field due to continuous distributions of
charge is addressed in Section 5.13.
5.13 Electric Potential Field due
to a Continuous Distribution
of Charge
[m0065]
The electrostatic potential field at r associated with N
charged particles is
V(r) = 1
4πǫ
N∑
n=1
qn
|r − rn| (5.104)
where qn and rn are the charge and position of the
nth particle. However, it is more common to have a
continuous distribution of charge as opposed to a
countable number of charged particles. W e now
consider how to compute V(r) three types of these
commonly-encountered distributions. Before
beginning, it’s worth noting that the methods will be
essentially the same, from a mathematical viewpoint,
as those developed in Section 5.4; therefore, a review
of that section may be helpful before attempting this
section.
Continuous Distribution of Charge Along a Curve.",Electromagnetics_Vol1.pdf
"of that section may be helpful before attempting this
section.
Continuous Distribution of Charge Along a Curve.
Consider a continuous distribution of charge along a
curve C. The curve can be divided into short segments
of length ∆ l. Then, the charge associated with the
nth segment, located at rn, is
qn = ρl(rn) ∆ l (5.105)
where ρl is the line charge density (units of C/m) at
rn. Substituting this expression into Equation 5.104,
we obtain
V(r) = 1
4πǫ
N∑
n=1
ρl(rn)
|r − rn|∆ l (5.106)
T aking the limit as ∆ l→ 0 yields:
V(r) = 1
4πǫ
∫
C
ρl(l)
|r − r′|dl (5.107)
where r′ represents the varying position along C with
integration along the length l.
Continuous Distribution of Charge Over a
Surface. Consider a continuous distribution of charge",Electromagnetics_Vol1.pdf
"5.14. ELECTRIC FIELD AS THE GRADIENT OF POTENTIAL 113
over a surface S. The surface can be divided into
small patches having area ∆ s. Then, the charge
associated with the nth patch, located at rn, is
qn = ρs(rn) ∆ s (5.108)
where ρs is surface charge density (units of C/m 2) at
rn. Substituting this expression into Equation 5.104,
we obtain
V(r) = 1
4πǫ
N∑
n=1
ρs(rn)
|r − rn| ∆ s (5.109)
T aking the limit as ∆ s→ 0 yields:
V(r) = 1
4πǫ
∫
S
ρs(r′)
|r − r′| ds (5.110)
where r′ represents the varying position over S with
integration.
Continuous Distribution of Charge in a V olume.
Consider a continuous distribution of charge within a
volume V. The volume can be divided into small cells
(volume elements) having area ∆ v. Then, the charge
associated with the nth cell, located at rn, is
qn = ρv(rn) ∆ v (5.111)
where ρv is the volume charge density (units of C/m 3)
at rn. Substituting this expression into
Equation 5.104, we obtain
V(r) = 1
4πǫ
N∑
n=1
ρv(rn)
|r − rn| ∆ v (5.112)",Electromagnetics_Vol1.pdf
"at rn. Substituting this expression into
Equation 5.104, we obtain
V(r) = 1
4πǫ
N∑
n=1
ρv(rn)
|r − rn| ∆ v (5.112)
T aking the limit as ∆ v→ 0 yields:
V(r) = 1
4πǫ
∫
V
ρv(r′)
|r − r′| dv (5.113)
where r′ represents the varying position over V with
integration.
5.14 Electric Field as the
Gradient of Potential
[m0063]
In Section 5.8, it was determined that the electrical
potential difference V21 measured over a path C is
given by
V21 = −
∫
C
E(r) · dl (5.114)
where E(r) is the electric field intensity at each point
r along C. In Section 5.12, we defined the scalar
electric potential field V(r) as the electric potential
difference at r relative to a datum at infinity . In this
section, we address the “inverse problem” – namely ,
how to calculate E(r) given V(r). Specifically , we
are interested in a direct “point-wise” mathematical
transform from one to the other. Since Equation 5.114
is in the form of an integral, it should not come as a",Electromagnetics_Vol1.pdf
"transform from one to the other. Since Equation 5.114
is in the form of an integral, it should not come as a
surprise that the desired expression will be in the form
of a differential equation.
W e begin by identifying the contribution of an
infinitesimal length of the integral to the total integral
in Equation 5.114. At point r, this is
dV = −E(r) · dl (5.115)
Although we can proceed using any coordinate
system, the following derivation is particularly simple
in Cartesian coordinates. In Cartesian coordinates,
dl = ˆxdx+ ˆydy+ ˆzdz (5.116)
W e also note that for any scalar function of position,
including V(r), it is true that
dV = ∂V
∂xdx+ ∂V
∂y dy+ ∂V
∂z dz (5.117)
Note the above relationship is not specific to
electromagnetics; it is simply mathematics. Also note
that dx= dl · ˆx and so on for dyand dz. Making
these substitutions into the above equation, we obtain:
dV = ∂V
∂x (dl · ˆx) + ∂V
∂y (dl · ˆy) + ∂V
∂z (dl · ˆz)
(5.118)
This equation may be rearranged as follows:
dV =
([",Electromagnetics_Vol1.pdf
"dV = ∂V
∂x (dl · ˆx) + ∂V
∂y (dl · ˆy) + ∂V
∂z (dl · ˆz)
(5.118)
This equation may be rearranged as follows:
dV =
([
ˆx ∂
∂x + ˆy ∂
∂y + ˆz ∂
∂z
]
V
)
· dl (5.119)",Electromagnetics_Vol1.pdf
"114 CHAPTER 5. ELECTROST A TICS
Comparing the above equation to Equation 5.115, we
find:
E(r) = −
[
ˆx ∂
∂x + ˆy ∂
∂y + ˆz ∂
∂z
]
V (5.120)
Note that the quantity in square brackets is the
gradient operator “ ∇” (Section 4.5). Thus, we may
write
E = −∇V
(5.121)
which is the relationship we seek.
The electric field intensity at a point is the gra-
dient of the electric potential at that point after a
change of sign (Equation 5.121).
Using Equation 5.121, we can immediately find the
electric field at any point r if we can describe V as a
function of r. Furthermore, this relationship between
V and E has a useful physical interpretation. Recall
that the gradient of a scalar field is a vector that points
in the direction in which that field increases most
quickly . Therefore:
The electric field points in the direction in which
the electric potential most rapidly decreases.
This result should not come as a complete surprise;
for example, the reader should already be aware that",Electromagnetics_Vol1.pdf
"This result should not come as a complete surprise;
for example, the reader should already be aware that
the electric field points away from regions of net
positive charge and toward regions of net negative
charge (Sections 2.2 and/or 5.1). What is new here is
that both the magnitude and direction of the electric
field may be determined given only the potential field,
without having to consider the charge that is the
physical source of the electrostatic field.
Example 5.8. Electric field of a charged
particle, beginning with the potential field.
In this example, we determine the electric field
of a particle bearing charge qlocated at the
origin. This may be done in a “direct” fashion
using Coulomb’s Law (Section 5.1). However,
here we have the opportunity to find the electric
field using a different method. In Section 5.12
we found the scalar potential for this source was:
V(r) = q
4πǫr (5.122)
So, we may obtain the electric field using
Equation 5.121:
E = −∇V = −∇
( q
4πǫr
)
(5.123)",Electromagnetics_Vol1.pdf
"V(r) = q
4πǫr (5.122)
So, we may obtain the electric field using
Equation 5.121:
E = −∇V = −∇
( q
4πǫr
)
(5.123)
Here V(r) is expressed in spherical coordinates,
so we have (Section B.2):
E = −
[
ˆr ∂
∂r + ˆθ1
r
∂
∂θ + ˆφ 1
rsin θ
∂
∂φ
] ( q
4πǫr
)
(5.124)
In this case, V(r) does not vary with φor θ, so
the second and third terms of the gradient are
zero. This leaves
E = −ˆr ∂
∂r
( q
4πǫr
)
= −ˆr q
4πǫ
∂
∂r
1
r
= −ˆr q
4πǫ
(
− 1
r2
)
(5.125)
So we find
E = + ˆr q
4πǫr2 (5.126)
as was determined in Section 5.1.",Electromagnetics_Vol1.pdf
"5.15. POISSON’S AND LAPLACE’S EQUA TIONS 115
5.15 Poisson’s and Laplace’s
Equations
[m0067]
The electric scalar potential field V(r), defined in
Section 5.12, is useful for a number of reasons
including the ability to conveniently compute
potential differences (i.e., V21 = V(r2) − V(r1)) and
the ability to conveniently determine the electric field
by taking the gradient (i.e., E = −∇V). One way to
obtain V(r) is by integration over the source charge
distribution, as described in Section 5.13. This
method is awkward in the presence of material
interfaces, which impose boundary conditions on the
solutions that must be satisfied simultaneously . For
example, the electric potential on a perfectly
conducting surface is constant 2 – a constraint which is
not taken into account in any of the expressions in
Section 5.13.
In this section, we develop an alternative approach to
calculating V(r) that accommodates these boundary
conditions, and thereby facilitates the analysis of the",Electromagnetics_Vol1.pdf
"calculating V(r) that accommodates these boundary
conditions, and thereby facilitates the analysis of the
scalar potential field in the vicinity of structures and
spatially-varying material properties. This alternative
approach is based on P oisson’s Equation , which we
now derive.
W e begin with the differential form of Gauss’ Law
(Section 5.7):
∇ · D = ρv (5.127)
Using the relationship D = ǫE (and keeping in mind
our standard assumptions about material properties,
summarized in Section 2.8) we obtain
∇ · E = ρv
ǫ (5.128)
Next, we apply the relationship (Section 5.14):
E = −∇V (5.129)
yielding
∇ · ∇ V = −ρv
ǫ (5.130)
This is Poisson’s Equation, but it is not in the form in
which it is commonly employed. T o obtain the
2 This fact is probably already known to the reader from past
study of elementary circuit theory; however, this is establi shed in
the context of electromagnetics in Section 5.19.
alternative form, consider the operator ∇ · ∇ in
Cartesian coordinates:
∇ · ∇ =
[ ∂",Electromagnetics_Vol1.pdf
"the context of electromagnetics in Section 5.19.
alternative form, consider the operator ∇ · ∇ in
Cartesian coordinates:
∇ · ∇ =
[ ∂
∂xˆx + ∂
∂yˆy + ∂
∂zˆz
]
·
[ ∂
∂xˆx + ∂
∂yˆy + ∂
∂zˆz
]
= ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
= ∇2 (5.131)
i.e., the operator ∇ · ∇ is identically the Laplacian
operator ∇2 (Section 4.10). Furthermore, this is true
regardless of the coordinate system employed. Thus,
we obtain the following form of Poisson’s Equation:
∇2V = −ρv
ǫ (5.132)
Poisson’s Equation (Equation 5.132) states that
the Laplacian of the electric potential field is
equal to the volume charge density divided by the
permittivity , with a change of sign.
Note that Poisson’s Equation is a partial differential
equation, and therefore can be solved using
well-known techniques already established for such
equations. In fact, Poisson’s Equation is an
inhomogeneous differential equation, with the
inhomogeneous part −ρv/ǫrepresenting the source of
the field. In the presence of material structure, we",Electromagnetics_Vol1.pdf
"inhomogeneous differential equation, with the
inhomogeneous part −ρv/ǫrepresenting the source of
the field. In the presence of material structure, we
identify the relevant boundary conditions at the
interfaces between materials, and the task of finding
V(r) is reduced to the purely mathematical task of
solving the associated boundary value problem (see
“ Additional Reading” at the end of this section). This
approach is particularly effective when one of the
materials is a perfect conductor or can be modeled as
such a material. This is because – as noted at the
beginning of this section – the electric potential at all
points on the surface of a perfect conductor must be
equal, resulting in a particularly simple boundary
condition.
In many other applications, the charge responsible for
the electric field lies outside the domain of the
problem; i.e., we have non-zero electric field (hence,
potentially non-zero electric potential) in a region that",Electromagnetics_Vol1.pdf
"problem; i.e., we have non-zero electric field (hence,
potentially non-zero electric potential) in a region that
is free of charge. In this case, Poisson’s Equation
simplifies to Laplace’s Equation :
∇2V = 0 (source-free region)
(5.133)",Electromagnetics_Vol1.pdf
"116 CHAPTER 5. ELECTROST A TICS
Laplace’s Equation (Equation 5.133) states that
the Laplacian of the electric potential field is zero
in a source-free region.
Like Poisson’s Equation, Laplace’s Equation,
combined with the relevant boundary conditions, can
be used to solve for V(r), but only in regions that
contain no charge.
Additional Reading:
• “Poisson’s equation” on Wikipedia.
• “Boundary value problem” on Wikipedia.
• “Laplace’s equation” on Wikipedia.
5.16 Potential Field Within a
Parallel Plate Capacitor
[m0068]
This section presents a simple example that
demonstrates the use of Laplace’s Equation
(Section 5.15) to determine the potential field in a
source free region. The example, shown in Figure 5.6,
pertains to an important structure in electromagnetic
theory – the parallel plate capacitor. Here we are
concerned only with the potential field V(r) between
the plates of the capacitor; you do not need to be
familiar with capacitance or capacitors to follow this",Electromagnetics_Vol1.pdf
"the plates of the capacitor; you do not need to be
familiar with capacitance or capacitors to follow this
section (although you’re welcome to look ahead to
Section 5.22 for a preview , if desired). What is
recommended before beginning is a review of the
battery-charged capacitor experiment discussed in
Section 2.2. In this section you’ll see a rigorous
derivation of what we figured out in an informal way
in that section.
The parallel-plate capacitor in Figure 5.6 consists of
two perfectly-conducting circular disks separated by a
distance dby a spacer material having permittivity ǫ.
There is no charge present in the spacer material, so
Laplace’s Equation applies. That equation is
(Section 5.15):
∇2V = 0 (source-free region) (5.134)
Let VC be the potential difference between the plates,
which would also be the potential difference across
the terminals of the capacitor. The radius aof the
plates is larger than dby enough that we may neglect",Electromagnetics_Vol1.pdf
"the terminals of the capacitor. The radius aof the
plates is larger than dby enough that we may neglect
what is going on at at the edges of the plates – more
on this will be said as we work the problem. Under
this assumption, what is the electric potential field
V(r) between the plates?
This problem has cylindrical symmetry , so it makes
sense to continue to use cylindrical coordinates with
the zaxis being perpendicular to the plates.
Equation 5.134 in cylindrical coordinates is:
[ 1
ρ
∂
∂ρ
(
ρ ∂
∂ρ
)
+ 1
ρ2
∂2
∂φ2 + ∂2
∂z2
]
V = 0 (5.135)
or perhaps a little more clearly written as follows:
1
ρ
∂
∂ρ
(
ρ∂V
∂ρ
)
+ 1
ρ2
∂2V
∂φ2 + ∂2V
∂z2 = 0 (5.136)",Electromagnetics_Vol1.pdf
"5.16. POTENTIAL FIELD WITHIN A P ARALLEL PLA TE CAP ACITOR 117
d
z
ρ
a
V-+VC
V-
Figure 5.6: A parallel plate capacitor, as a demonstra-
tion of the use of Laplace’s Equation.
Since the problem has radial symmetry , ∂V/∂φ = 0 .
Since d≪ a, we expect the fields to be approximately
constant with ρuntil we get close to the edge of the
plates. Therefore, we assume ∂V/∂ρ is negligible and
can be taken to be zero. Thus, we are left with
∂2V
∂z2 ≈ 0 for ρ≪ a (5.137)
The general solution to Equation 5.137 is obtained
simply by integrating both sides twice, yielding
V(z) = c1z+ c2 (5.138)
where c1 and c2 are constants that must be consistent
with the boundary conditions. Thus, we must develop
appropriate boundary conditions. Let the node
voltage at the negative ( z= 0 ) terminal be V−. Then
the voltage at the positive ( z= + d) terminal is
V− + VC. Therefore:
V(z= 0) = V− (5.139)
V(z= + d) = V− + VC (5.140)
These are the relevant boundary conditions.
Substituting V(z= 0) = V− into Equation 5.138",Electromagnetics_Vol1.pdf
"V(z= 0) = V− (5.139)
V(z= + d) = V− + VC (5.140)
These are the relevant boundary conditions.
Substituting V(z= 0) = V− into Equation 5.138
yields c2 = V−. Substituting V(z= + d) = V− + VC
into Equation 5.138 yields c1 = VC/d. Thus, the
answer to the problem is
V(z) ≈ VC
d z+ V− for ρ≪ a (5.141)
Note that the above result is dimensionally correct
and confirms that the potential deep inside a “thin”
parallel plate capacitor changes linearly with distance
between the plates.
Further, you should find that application of the
equation E = −∇V (Section 5.14) to the solution
above yields the expected result for the electric field
intensity: E ≈ − ˆzVC/d. This is precisely the result
that we arrived at (without the aid of Laplace’s
Equation) in Section 2.2.
A reasonable question to ask at this point would be,
what about the potential field close to the edge of the
plates, or, for that matter, beyond the plates? The field
in this region is referred to as a fringing field . For the",Electromagnetics_Vol1.pdf
"plates, or, for that matter, beyond the plates? The field
in this region is referred to as a fringing field . For the
fringing field, ∂V/∂ρ is no longer negligible and must
be taken into account. In addition, it is necessary to
modify the boundary conditions to account for the
outside surfaces of the plates (that is, the sides of the
plates that face away from the dielectric) and to
account for the effect of the boundary between the
spacer material and free space. These issues make the
problem much more difficult. When an accurate
calculation of a fringing field is necessary , it is
common to resort to a numerical solution of Laplace’s
Equation. Fortunately , accurate calculation of
fringing fields is usually not required in practical
engineering applications.",Electromagnetics_Vol1.pdf
"118 CHAPTER 5. ELECTROST A TICS
5.17 Boundary Conditions on the
Electric Field Intensity ( E)
[m0020]
In homogeneous media, electromagnetic quantities
vary smoothly and continuously . At an interface
between dissimilar media, however, it is possible for
electromagnetic quantities to be discontinuous. These
discontinuities can be described mathematically as
boundary conditions and used to to constrain
solutions for the associated electromagnetic
quantities. In this section, we derive boundary
conditions on the electric field intensity E.
T o begin, consider a region consisting of only two
media that meet at an interface defined by the
mathematical surface S, as shown in Figure 5.7. If
either one of the materials is a perfect electrical
conductor (PEC), then S is an equipotential surface;
i.e., the electric potential V is constant everywhere on
S. Since E is proportional to the spatial rate of
change of potential (recall E = −∇V; Section 5.14),
we find:",Electromagnetics_Vol1.pdf
"S. Since E is proportional to the spatial rate of
change of potential (recall E = −∇V; Section 5.14),
we find:
The component of E that is tangent to a perfectly-
conducting surface is zero.
This is sometimes expressed informally as follows:
Etan = 0 on PEC surface (5.142)
where “ Etan” is understood to be the component of E
that is tangent to S. Since the tangential component
PEC
E
E E E
(equipotential)
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.7: At the surface of a perfectly-conducting
region, E may be perpendicular to the surface (two
leftmost possibilities), but may not exhibit a compo-
nent that is tangent to the surface (two rightmost pos-
sibilities).
B
A w
t
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.8: Use of KVL to determine the boundary
condition on E.
of E on the surface of a perfect conductor is zero, the
electric field at the surface must be oriented entirely
in the direction perpendicular to the surface, as shown
in Figure 5.7.
The following equation expresses precisely the same",Electromagnetics_Vol1.pdf
"in the direction perpendicular to the surface, as shown
in Figure 5.7.
The following equation expresses precisely the same
idea, but includes the calculation of the tangential
component as part of the statement:
E × ˆn = 0 (on PEC surface) (5.143)
where ˆn is either normal (i.e., unit vector
perpendicular to the surface) to each point on S. This
expression works because the cross product of any
two vectors is perpendicular to either vector
(Section 4.1), and any vector which is perpendicular
to ˆn is tangent to S.
W e now determine a more general boundary
condition that applies even when neither of the media
bordering S is a perfect conductor. The desired
boundary condition can be obtained directly from
Kirchoff’s V oltage Law (KVL; Section 5.10):
∮
C
E · dl = 0 (5.144)
Let the closed path of integration take the form of a
rectangle centered on S, as shown in Figure 5.8. Let
the sides A, B, C, and Dbe perpendicular or parallel
to the surface, respectively . Let the length of the",Electromagnetics_Vol1.pdf
"the sides A, B, C, and Dbe perpendicular or parallel
to the surface, respectively . Let the length of the
perpendicular sides be w, and let the length of the",Electromagnetics_Vol1.pdf
"5.17. BOUNDAR Y CONDITIONS ON THE ELECTRIC FIELD INTENSITY ( E) 119
parallel sides be l. From KVL we have
∮
C
E · dl =
∫
A
E · dl
+
∫
B
E · dl
+
∫
C
E · dl
+
∫
D
E · dl = 0 (5.145)
Now , let us reduce wand ltogether while (1)
maintaining a constant ratio w/l≪ 1 and (2) keeping
C centered on S. In this process, the contributions
from the Band Dsegments become equal in
magnitude but opposite in sign; i.e.,
∫
B
E · dl +
∫
D
E · dl → 0 (5.146)
This leaves
∮
C
E · dl →
∫
A
E · dl +
∫
C
E · dl → 0 (5.147)
Let us define the unit vector ˆt (“tangent”) as shown in
Figure 5.8. When the lengths of sides Aand C
become sufficiently small, we can write the above
expression as follows:
E1 · ˆt∆ l− E2 · ˆt∆ l→ 0 (5.148)
where E1 and E2 are the fields evaluated on the two
sides of the boundary and ∆ l→ 0 is the length of
sides Aand Cwhile this is happening. Note that the
only way Equation 5.148 can be true is if the
tangential components of E1 and E2 are equal. In
other words:",Electromagnetics_Vol1.pdf
"only way Equation 5.148 can be true is if the
tangential components of E1 and E2 are equal. In
other words:
The tangential component of E must be continu-
ous across an interface between dissimilar media.
Note that this is a generalization of the result we
obtained earlier for the case in which one of the
media was a PEC – in that case, the tangent
component of E on the other side of the interface
must be zero because it is zero in the PEC medium.
As before, we can express this idea in compact
mathematical notation. Using the same idea used to
obtain Equation 5.143, we have found
E1 × ˆn = E2 × ˆn on S (5.149)
or, as it is more commonly written:
ˆn × (E1 − E2) = 0 on S
(5.150)
W e conclude this section with a note about the
broader applicability of this boundary condition:
Equation 5.150 is the boundary condition that ap-
plies to E for both the electrostatic and the gen-
eral (time-varying) case.
Although a complete explanation is not possible",Electromagnetics_Vol1.pdf
"plies to E for both the electrostatic and the gen-
eral (time-varying) case.
Although a complete explanation is not possible
without the use of the Maxwell-Faraday Equation
(Section 8.8), the reason why this boundary condition
applies in the time-varying case can be disclosed here.
In the presence of time-varying magnetic fields, the
right-hand side of Equation 5.144 may become
non-zero and is proportional to the area defined by the
closed loop. However, the above derivation requires
the area of this loop to approach zero, in which case
the possible difference from Equation 5.144 also
converges to zero. Therefore, the boundary condition
expressed in Equation 5.150 applies generally .",Electromagnetics_Vol1.pdf
"120 CHAPTER 5. ELECTROST A TICS
5.18 Boundary Conditions on the
Electric Flux Density ( D)
[m0021]
In this section, we derive boundary conditions on the
electric flux density D. The considerations are quite
similar to those encountered in the development of
boundary conditions on the electric field intensity (E)
in Section 5.17, so the reader may find it useful to
review that section before attempting this section.
This section also assumes familiarity with the
concepts of electric flux, electric flux density , and
Gauss’ Law; for a refresher, Sections 2.4 and 5.5 are
suggested.
T o begin, consider a region at which two
otherwise-homogeneous media meet at an interface
defined by the mathematical surface S, as shown in
Figure 5.9. Let one of these regions be a perfect
electrical conductor (PEC). In Section 5.17, we
established that the tangential component of the
electric field must be zero, and therefore, the electric
field is directed entirely in the direction perpendicular",Electromagnetics_Vol1.pdf
"electric field must be zero, and therefore, the electric
field is directed entirely in the direction perpendicular
to the surface. W e further know that the electric field
within the conductor is identically zero. Therefore, D
at any point on S is entirely in the direction
perpendicular to the surface and pointing into the
non-conducting medium. However, it is also possible
to determine the magnitude of D. W e shall
demonstrate in this section that
At the surface of a perfect conductor, the magni-
tude of D is equal to the surface charge density
ρs (units of C/m 2) at that point.
PEC
D=nρs
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.9: The component of D that is perpendicular
to a perfectly-conducting surface is equal to the charge
density on the surface.
region 1
region 2
n
a
h
h
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.10: Use of Gauss’ Law to determine the
boundary condition on D.
The following equation expresses precisely the same
idea, but includes the calculation of the perpendicular",Electromagnetics_Vol1.pdf
"boundary condition on D.
The following equation expresses precisely the same
idea, but includes the calculation of the perpendicular
component as part of the statement:
D · ˆn = ρs (on PEC surface) (5.151)
where ˆn is the normal to S pointing into the
non-conducting region. (Note that the orientation of ˆn
is now important; we have assumed ˆn points into
region 1, and we must now stick with this choice.)
Before proceeding with the derivation, it may be
useful to note that this result is not surprising. The
very definition of electric flux (Section 2.4) indicates
that D should correspond in the same way to a
surface charge density . However, we can show this
rigorously , and in the process we can generalize this
result to the more-general case in which neither of the
two materials are PEC.
The desired more-general boundary condition may be
obtained from the integral form of Gauss’ Law
(Section 5.5), as illustrated in Figure 5.10. Let the
surface of integration S′ take the form of closed",Electromagnetics_Vol1.pdf
"obtained from the integral form of Gauss’ Law
(Section 5.5), as illustrated in Figure 5.10. Let the
surface of integration S′ take the form of closed
cylinder centered at a point on the interface and for
which the flat ends are parallel to the surface and
perpendicular to ˆn. Let the radius of this cylinder be
a, and let the length of the cylinder be 2h. From
Gauss’ Law we have
∮
S′
D · ds =
∫
top
D · ds
+
∫
side
D · ds
+
∫
bottom
D · ds = Qencl (5.152)",Electromagnetics_Vol1.pdf
"5.18. BOUNDAR Y CONDITIONS ON THE ELECTRIC FLUX DENSITY ( D) 121
where the “top” and “bottom” are in Regions 1 and 2,
respectively , and Qencl is the charge enclosed by S′.
Now let us reduce hand atogether while (1)
maintaining a constant ratio h/a≪ 1 and (2) keeping
S′ centered on S. Because h≪ a, the area of the side
can be made negligible relative to the area of the top
and bottom. Then as h→ 0 we are left with
∫
top
D · ds +
∫
bottom
D · ds → Qencl (5.153)
As the area of the top and bottom sides become
infinitesimal, the variation in D over these areas
becomes negligible. Now we have simply:
D1 · ˆn∆ A+ D2 · (−ˆn) ∆ A→ Qencl (5.154)
where D1 and D2 are the electric flux density vectors
in medium 1 and medium 2, respectively , and ∆ Ais
the area of the top and bottom sides. The above
expression can be rewritten
ˆn · (D1 − D2) → Qencl
∆ A (5.155)
Note that the left side of the equation must represent a
actual, physical surface charge; this is apparent from",Electromagnetics_Vol1.pdf
"ˆn · (D1 − D2) → Qencl
∆ A (5.155)
Note that the left side of the equation must represent a
actual, physical surface charge; this is apparent from
dimensional analysis and the fact that his now
infinitesimally small. Therefore:
ˆn · (D1 − D2) = ρs
(5.156)
where, as noted above, ˆn points into region 1.
Summarizing:
Any discontinuity in the normal component of
the electric flux density across the boundary be-
tween two material regions is equal to the surface
charge.
Now let us verify that this is consistent with our
preliminary finding, in which Region 2 was a PEC. In
that case D2 = 0 , so we see that Equation 5.151 is
satisfied, as expected. If neither Region 1 nor
Region 2 is PEC and there is no surface charge on the
interface, then we find ˆn · (D1 − D2) = 0 ; i.e.,
In the absence of surface charge, the normal com-
ponent of the electric flux density must be contin-
uous across the boundary .
Finally , we note that since D = ǫE, Equation 5.156",Electromagnetics_Vol1.pdf
"ponent of the electric flux density must be contin-
uous across the boundary .
Finally , we note that since D = ǫE, Equation 5.156
implies the following boundary condition on E:
ˆn · (ǫ1E1 − ǫ2E2) = ρs (5.157)
where ǫ1 and ǫ2 are the permittivities in Regions 1
and 2, respectively . The above equation illustrates one
reason why we sometimes prefer the “flux”
interpretation of the electric field to the “field
intensity” interpretation of the electric field.",Electromagnetics_Vol1.pdf
"122 CHAPTER 5. ELECTROST A TICS
5.19 Charge and Electric Field
for a Perfectly Conducting
Region
[m0025]
In this section, we consider the behavior of charge
and the electric field in the vicinity of a perfect
electrical conductor (PEC).
First, note that the electric field – both the electric
field intensity E and electric flux density D –
throughout a PEC region is zero. This is because the
electrical potential throughout a PEC region must be
constant. (This idea is explored further in
Section 6.3.) Recall that the electric field is
proportional to the spatial rate of change of electrical
potential (i.e., E = −∇V; Section 5.14). Thus, the
electric field must be zero throughout a PEC region.
Second, the electric field is oriented directly away
from (i.e., perpendicular to) the PEC surface, and the
magnitude of D is equal to the surface charge density
ρs (C/m2) (Section 5.18).
Now we address the question of charge distribution;
i.e., the location and density of charge. Consider the",Electromagnetics_Vol1.pdf
"ρs (C/m2) (Section 5.18).
Now we address the question of charge distribution;
i.e., the location and density of charge. Consider the
scenario shown in Figure 5.11. Here, a flat slab of
PEC material is embedded in dielectric material. 3 The
3 For the purposes of this section, it suffices to interpret “di elec-
tric” as a “nonconducting and well characterized entirely i n terms of
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.11: An infinite flat slab of PEC in the pres-
ence of an applied electric field.
thickness of the slab is finite, whereas the length and
width of the slab is infinite. The region above the slab
is defined as Region 1 and has permittivity ǫ1. The
region below the slab is defined as Region 2 and has
permittivity ǫ2. Electric fields E1 and E2 are present
in Regions 1 and 2, respectively , as shown in
Figure 5.11. T o begin, let us assume that these fields
are the result of some external stimulus that results in
the direction of these fields being generally upward,
as shown in Figure 5.11.",Electromagnetics_Vol1.pdf
"are the result of some external stimulus that results in
the direction of these fields being generally upward,
as shown in Figure 5.11.
Now , what do we know about E1 and E2? First, both
fields must satisfy the relevant boundary conditions.
That is, the component of E1 that is tangent to the
upper PEC surface is zero, so that E1 is directed
entirely in a direction perpendicular to the surface.
Similarly , the component of E2 that is tangent to the
lower PEC surface is zero, so that E2 is directed
entirely in a direction perpendicular to the surface. At
this point we have not determined the magnitudes or
signs of E1 and E2; we have established only that
there are no non-zero components tangential to (i.e.,
parallel to) the PEC surfaces.
Next, recall that the electric field must be zero within
the slab. This means that there must be zero net
charge within the slab, since any other distribution of
charge will result in a non-zero electric field, and",Electromagnetics_Vol1.pdf
"charge within the slab, since any other distribution of
charge will result in a non-zero electric field, and
subsequently a potential difference between locations
within the slab. Therefore:
There can be no static charge within a PEC.
It follows that
Charge associated with a PEC lies entirely on the
surface.
Outside the slab, the boundary conditions on D1 and
D2 in the dielectric regions require these fields to be
non-zero when the surface charge density on the PEC
is non-zero. The surface charge supports the
discontinuity in the normal component of the electric
fields. Specifically , D1 and D2 have the same
magnitude |ρs| because the surface charge densities
on both sides of the slab have equal magnitude.
However, the electric field intensity E1 = D1/ǫ1,
whereas E2 = D2/ǫ2; i.e., these are different. That
its permittivity . ” For more, see Section 5.20.",Electromagnetics_Vol1.pdf
"5.20. DIELECTRIC MEDIA 123
−
− − −−−−−−
+
+ + + + + + +
−−−
−
+
+
+
+ +
−−−− − − −
+ + + + ++
+
+
by C. Burks (modified)
Figure 5.12: Electric field lines due to a point charge
in the vicinity of PEC regions (shaded) of various
shapes.
is, the electric field intensities are unequal unless the
permittivities in each dielectric region are equal.
Finally , let us consider the structure of the electric
field in more general cases. Figure 5.12 shows field
lines in a homogeneous dielectric material in which a
point charge and PEC regions of various shapes are
embedded. Note that electric field lines now bend in
the dielectric so as to satisfy the requirement that the
tangential component of the electric field be zero on
PEC surfaces. Also note that the charge distribution
arranges itself on the PEC surfaces so as to maintain
zero electric field and constant potential within the
cube.
5.20 Dielectric Media
[m0107]
Dielectric is particular category of materials that",Electromagnetics_Vol1.pdf
"zero electric field and constant potential within the
cube.
5.20 Dielectric Media
[m0107]
Dielectric is particular category of materials that
exhibit low conductivity 4 because their constituent
molecules remain intact when exposed to an electric
field, as opposed to shedding electrons as is the case
in good conductors. Subsequently , dielectrics do not
effectively pass current, and are therefore considered
“good insulators” as well as “poor conductors. ” An
important application of dielectrics in electrical
engineering is as a spacer material in printed circuit
boards (PCBs), coaxial cables, and capacitors.
Examples of dielectrics include air, glass, teflon, and
fiberglass epoxy (the material used in common “FR4”
printed circuit boards). These and other dielectrics are
listed along with values of their constitutive
parameters in Section A.1.
The electromagnetic properties of dielectric materials
are quantified primarily by relative permittivity ǫr",Electromagnetics_Vol1.pdf
"parameters in Section A.1.
The electromagnetic properties of dielectric materials
are quantified primarily by relative permittivity ǫr
(Section 2.3), which ranges from very close to 1
upward to roughly 50, and is less than 6 or so for
most commonly-encountered materials having low
moisture content. The permeability of dielectric
materials is approximately equal to the free-space
value (i.e., µ≈ µ0); therefore, these materials are said
to be “non-magnetic. ”
Additional Reading:
• “Dielectric” on Wikipedia.
4 See Section 2.8 for a refresher on parameters describing prop -
erties of materials.",Electromagnetics_Vol1.pdf
"124 CHAPTER 5. ELECTROST A TICS
5.21 Dielectric Breakdown
[m0109]
The permittivity of an ideal dielectric is independent
of the magnitude of an applied electric field; the
material is said to be “linear. ” 5 However, all practical
dielectrics fail in this respect with sufficiently strong
electric field. T ypically , the failure is abrupt and is
observed as a sudden, dramatic increase in
conductivity , signaling that electrons are being
successfully dislodged from their host molecules. The
threshold value of the electric field intensity at which
this occurs is known as the dielectric strength , and the
sudden change in behavior observed in the presence
of an electric field greater than this threshold value is
known as dielectric breakdown .
Dielectric strength varies from about 3 MV/m for air
to about 200 MV/m in mica (a dielectric commonly
used in capacitors).
Dielectric breakdown is typically accompanied by
“arcing, ” which is a sudden flow of current associated",Electromagnetics_Vol1.pdf
"used in capacitors).
Dielectric breakdown is typically accompanied by
“arcing, ” which is a sudden flow of current associated
with the breakdown. A well known example of this
phenomenon is lightning, which occurs when charge
is exchanged between sky and ground when air (a
dielectric) exhibits breakdown. Dielectric breakdown
in solids typically damages the material.
Additional Reading:
• “Electrical Breakdown” on Wikipedia.
5 See Section 2.8 for a review of this concept.
5.22 Capacitance
[m0112]
When separate regions of positive and negative charge
exist in proximity , Coulomb forces (Section 5.1) will
attempt to decrease the separation between the
charges. As noted in Section 5.8, this can be
interpreted as a tendency of a system to reduce its
potential energy . If the charges are fixed in place, then
the potential energy remains constant. This potential
energy is proportional to the Coulomb force.
Referring back to Section 5.1, the Coulomb force is:",Electromagnetics_Vol1.pdf
"energy is proportional to the Coulomb force.
Referring back to Section 5.1, the Coulomb force is:
• Proportional to quantity of positive charge
squared
• Inversely proportional to the separation between
the charges squared
• Inversely proportional to the permittivity of the
material separating the charges
Therefore, the potential energy of the system is
likewise dependent on charge, separation, and
permittivity . Furthermore, we see that the ability of a
system to store energy in this manner depends on the
geometry of the charge distribution and the
permittivity of the intervening material.
Now recall that the electric field intensity E is
essentially defined in terms of the Coulomb force;
i.e., F = qE (Section 2.2). So, rather than thinking of
the potential energy of the system as being associated
with the Coulomb force, it is equally valid to think of
the potential energy as being stored in the electric
field associated with the charge distribution. It",Electromagnetics_Vol1.pdf
"the potential energy as being stored in the electric
field associated with the charge distribution. It
follows from the previous paragraph that the energy
stored in the electric field depends on the geometry of
the charge distribution and the permittivity of the
intervening media. This relationship is what we mean
by capacitance. W e summarize as follows:
Capacitance is the ability of a structure to store
energy in an electric field.
and",Electromagnetics_Vol1.pdf
"5.22. CAP ACIT ANCE 125
The capacitance of a structure depends on its ge-
ometry and the permittivity of the medium sepa-
rating regions of positive and negative charge.
Note that capacitance does not depend on charge,
which we view as either a stimulus or response from
this point of view . The corresponding response or
stimulus, respectively , is the potential associated with
this charge. This leads to the following definition:
C ≜ Q+
V (5.158)
where Q+ (units of C) is the total positive charge, V
(units of V) is the potential associated with this
charge (defined such that it is positive), and C(units
of F) is the associated capacitance. So:
In practice, capacitance is defined as the ratio
of charge present on one conductor of a two-
conductor system to the potential difference be-
tween the conductors (Equation 5.158).
In other words, a structure is said to have greater
capacitance if it stores more charge – and therefore
stores more energy – in response to a given potential
difference.",Electromagnetics_Vol1.pdf
"capacitance if it stores more charge – and therefore
stores more energy – in response to a given potential
difference.
Figure 5.13 shows the relevant features of this
definition. Here, a battery imposes the potential
difference V between two regions of
perfectly-conducting material. Q+ is the total charge
on the surface of the PEC region attached to the
positive terminal of the battery . An equal amount of
negative charge appears on the surface of the PEC
region attached to the negative terminal of the battery
(Section 5.19). This charge distribution gives rise to
an electric field. Assuming the two PEC regions are
fixed in place, Q+ will increase linearly with
increasing V, at a rate determined by the capacitance
Cof the structure.
A capacitor is a device that is designed to exhibit a
specified capacitance. W e can now make the
connection to the concept of the capacitor as it
appears in elementary circuit theory . In circuit theory ,
the behavior of devices is characterized in terms of",Electromagnetics_Vol1.pdf
"appears in elementary circuit theory . In circuit theory ,
the behavior of devices is characterized in terms of
terminal voltage VT in response to terminal current
IT, and vice versa. First, note that current does not
ð
-
ñ
ò
ó
ô
õ
ö
÷
-- -- -
EV
ø ù ú û ü
ý þ ÿ c g 
Q+
t 
   
    	 
 Q+-
PEC
Figure 5.13: Electrostatic interpretation of capaci-
tance.
normally flow through a capacitor, 6 so when we refer
to “terminal current” for a capacitor, what we really
mean is the flow of charge arriving or departing from
one of the conductors via the circuit, which is equal to
the flow of charge departing or arriving (respectively)
at the other conductor. This gives the appearance of
current flow through the capacitor when the current is
examined from outside the capacitor. With that
settled, we proceed as follows. Using Equation 5.158,
we express the voltage VT across the terminals of a
capacitor having capacitance C:
VT = Q+
C (5.159)
W e seek a relationship between VT and IT. Current is",Electromagnetics_Vol1.pdf
"capacitor having capacitance C:
VT = Q+
C (5.159)
W e seek a relationship between VT and IT. Current is
charge per unit time, so the charge on either
conductor is the integral of IT over time; i.e.:
Q+(t) =
∫ t
t0
IT(τ) dτ + Q+(t0) (5.160)
where t0 is an arbitrarily-selected start time. In other
words, amps integrated over time is charge. If we
define IT as being positive in the direction of the flow
of positive charge as is the usual convention, then we
have:
VT(t) = 1
C
∫ t
t0
IT(τ) dτ + 1
CQ+(t0) (5.161)
Again applying Equation 5.158, we see that the
second term is simply VT(t0). This is the expected
relationship from elementary circuit theory .
6 If it does, it’s probably experiencing dielectric breakdown ; see
Section 5.21.",Electromagnetics_Vol1.pdf
"126 CHAPTER 5. ELECTROST A TICS
Finally , solving for IT we obtain the differential form
of this relationship:
IT(t) = C d
dtVT(t) (5.162)
Additional Reading:
• “Capacitance” on Wikipedia.
• “Capacitor” on Wikipedia.
5.23 The Thin Parallel Plate
Capacitor
[m0070]
Let us now determine the capacitance of a common
type of capacitor known as the thin parallel plate
capacitor, shown in Figure 5.14. This capacitor
consists of two flat plates, each having area A,
separated by distance d. T o facilitate discussion, let us
place the origin of the coordinate system at the center
of the lower plate, with the +zaxis directed toward
the upper plate such that the upper plate lies in the
z= + dplane.
Below we shall find the capacitance by assuming a
particular charge on one plate, using the boundary
condition on the electric flux density D to relate this
charge density to the internal electric field, and then
integrating over the electric field between the plates to",Electromagnetics_Vol1.pdf
"charge density to the internal electric field, and then
integrating over the electric field between the plates to
obtain the potential difference. Then, capacitance is
the ratio of the assumed charge to the resulting
potential difference.
The principal difficulty in this approach is finding the
electric field. T o appreciate the problem, first consider
that if the area of the plates was infinite, then the
electric field would be very simple; it would begin at
the positively-charged plate and extend in a
perpendicular direction toward the negatively-charged
plate (Section 5.19). Furthermore, the field would be
constant everywhere between the plates. This much is
apparent from symmetry alone. However, when the
plate area is finite, then we expect a fringing field to
emerge. “Fringing field” is simply a term applied to
the non-uniform field that appears near the edge of
the plates. The field is non-uniform in this region
d
z
ρ
Figure 5.14: Thin parallel plate capacitor.",Electromagnetics_Vol1.pdf
"5.23. THE THIN P ARALLEL PLA TE CAP ACITOR 127
because the boundary conditions on the outside
(outward-facing) surfaces of the plates have a
significant effect in this region. In the central region
of the capacitor, however, the field is not much
different from the field that exists in the case of
infinite plate area.
In any parallel plate capacitor having finite plate area,
some fraction of the energy will be stored by the
approximately uniform field of the central region, and
the rest will be stored in the fringing field. W e can
make the latter negligible relative to the former by
making the capacitor very “thin, ” in the sense that the
smallest identifiable dimension of the plate is much
greater than d. Under this condition, we may obtain a
good approximation of the capacitance by simply
neglecting the fringing field, since an insignificant
fraction of the energy is stored there.
Imposing the “thin” condition leads to three
additional simplifications. First, the surface charge",Electromagnetics_Vol1.pdf
"fraction of the energy is stored there.
Imposing the “thin” condition leads to three
additional simplifications. First, the surface charge
distribution may be assumed to be approximately
uniform over the plate, which greatly simplifies the
analysis. Second, the shape of the plates becomes
irrelevant; they might be circular, square, triangular,
etc. When computing capacitance in the “thin” case,
only the plate area Ais important. Third, the
thickness of each of the plates becomes irrelevant.
W e are now ready to determine the capacitance of the
thin parallel plate capacitor. Here are the steps:
1. Assume a total positive charge Q+ on the upper
plate.
2. Invoking the “thin” condition, we assume the
charge density on the plates is uniform. Thus,
the surface charge density on bottom side of the
upper plate is ρs,+ = Q+/A(C/m2).
3. From the boundary condition on the bottom
surface of the upper plate, D on this surface is
−ˆzρs,+.
4. The total charge on the lower plate, Q−, must be",Electromagnetics_Vol1.pdf
"surface of the upper plate, D on this surface is
−ˆzρs,+.
4. The total charge on the lower plate, Q−, must be
equal and opposite the total charge on the upper
plate; i.e, Q− = −Q+. Similarly , the surface
charge density on the upper surface of the lower
plate, ρs,−, must be −ρs,+.
5. From the boundary condition on the top surface
of the lower plate (Section 5.18), D on this
surface is +ˆzρs,−. Since +ˆzρs,− = −ˆzρs,+, D
on the facing sides of the plates is equal.
6. Again invoking the “thin” condition, we assume
D between the plates has approximately the
same structure as we would see if the plate area
was infinite. Therefore, we are justified in
assuming D ≈ − ˆzρs,+ everywhere between the
plates. (Y ou might also see that this is
self-evident from the definition of D as the flux
density of electric charge (Section 2.4).)
7. With an expression for the electric field in hand,
we may now compute the potential difference V
between the plates as follows (Section 5.8):
V = −
∫
C
E · dl
= −
∫ d",Electromagnetics_Vol1.pdf
"we may now compute the potential difference V
between the plates as follows (Section 5.8):
V = −
∫
C
E · dl
= −
∫ d
0
( 1
ǫD
)
· (ˆzdz)
= −
∫ d
0
(
−ˆzρs,+
ǫ
)
· (ˆzdz)
= + ρs,+ d
ǫ (5.163)
8. Finally ,
C = Q+
V = ρs,+ A
ρs,+ d/ǫ = ǫA
d (5.164)
Summarizing:
C ≈ ǫA
d (5.165)
The capacitance of a parallel plate capacitor hav-
ing plate separation much less than the size of
the plate is given by Equation 5.165. This is an
approximation because the fringing field is ne-
glected.
It’s worth noting that this is dimensionally correct;
i.e., F/m times m 2 divided by m yields F . It’s also
worth noting the effect of the various parameters:
Capacitance increases in proportion to permittiv-
ity and plate area and decreases in proportion to
distance between the plates.",Electromagnetics_Vol1.pdf
"128 CHAPTER 5. ELECTROST A TICS
Example 5.9. Printed circuit board capacitance.
Printed circuit boards commonly include a
“ground plane, ” which serves as the voltage
datum for the board, and at least one “power
plane, ” which is used to distribute a DC supply
voltage (See “ Additional Reading” at the end of
this section). These planes are separated by a
dielectric material, and the resulting structure
exhibits capacitance. This capacitance may be
viewed as an equivalent discrete capacitor in
parallel with the power supply . The value of this
equivalent capacitor may be either negligible,
significant and beneficial, or significant and
harmful. So, it is useful to know the value of this
equivalent capacitor.
For a common type of circuit board, the
dielectric thickness is about 1.6 mm and the
relative permittivity of the material is about 4.5.
If the area in common between the ground and
power planes is 25 cm 2, what is the value of the
equivalent capacitor?",Electromagnetics_Vol1.pdf
"If the area in common between the ground and
power planes is 25 cm 2, what is the value of the
equivalent capacitor?
Solution. From the problem statement,
ǫ∼
= 4.5ǫ0, A∼
= 25 cm2 = 2.5 × 10−3 m2, and
d∼
= 1.6 mm. Using Equation 5.165, the value of
the equivalent capacitor is 62.3 pF .
Additional Reading:
• “Printed circuit board” on Wikipedia.
5.24 Capacitance of a Coaxial
Structure
[m0113]
Let us now determine the capacitance of
coaxially-arranged conductors, shown in Figure 5.15.
Among other applications, this information is useful
in the analysis of voltage and current waves on
coaxial transmission line, as addressed in Sections 3.4
and 3.10.
For our present purposes, we may model the structure
as consisting of two concentric perfectly-conducting
cylinders of radii aand b, separated by an ideal
dielectric having permittivity ǫs. W e place the +z
axis along the common axis of the concentric
cylinders so that the cylinders may be described as",Electromagnetics_Vol1.pdf
"dielectric having permittivity ǫs. W e place the +z
axis along the common axis of the concentric
cylinders so that the cylinders may be described as
constant-coordinate surfaces ρ= aand ρ= b.
In this section, we shall find the capacitance by
assuming a total charge Q+ on the inner conductor
and integrating over the associated electric field to
obtain the voltage between the conductors. Then,
capacitance is computed as the ratio of the assumed
charge to the resulting potential difference. This
strategy is the same as that employed in Section 5.23
for the parallel plate capacitor, so it may be useful to
review that section before attempting this derivation.
The first step is to find the electric field inside the
structure. This is relatively simple if we assume that
b
-
+ az

l
V
c⃝ K. Kikkeri CC BY SA 4.0
Figure 5.15: Determining the capacitance of a coaxial
structure.",Electromagnetics_Vol1.pdf
"5.24. CAP ACIT ANCE OF A COAXIAL STRUCTURE 129
the structure has infinite length (i.e., l→ ∞ ), since
then there are no fringing fields and the internal field
will be utterly constant with respect to z. In the
central region of a finite-length capacitor, however,
the field is not much different from the field that
exists in the case of infinite length, and if the energy
storage in fringing fields is negligible compared to the
energy storage in this central region then there is no
harm in assuming the internal field is constant with z.
Alternatively , we may think of the length las
pertaining to one short section of a much longer
structure and thereby obtain the capacitance per
length as opposed to the total capacitance. Note that
the latter is exactly what we need for the transmission
line lumped-element equivalent circuit model
(Section 3.4).
T o determine the capacitance, we invoke the
definition (Section 5.22):
C ≜ Q+
V (5.166)
where Q+ is the charge on the positively-charged",Electromagnetics_Vol1.pdf
"T o determine the capacitance, we invoke the
definition (Section 5.22):
C ≜ Q+
V (5.166)
where Q+ is the charge on the positively-charged
conductor and V is the potential measured from the
negative conductor to the positive conductor. The
charge on the inner conductor is uniformly-distributed
with density
ρl = Q+
l (5.167)
which has units of C/m. Now we will determine the
electric field intensity E, integrate E over a path
between conductors to get V, and then apply
Equation 5.166 to obtain the capacitance.
The electric field intensity for this scenario was
determined in Section 5.6, “Electric Field Due to an
Infinite Line Charge using Gauss’ Law , ” where we
found
E = ˆρ ρl
2πǫsρ (5.168)
The reader should note that in that section we were
considering merely a line of charge; not a coaxial
structure. So, on what basis do we claim the field is
the same? This is a consequence of Gauss’ Law
(Section 5.5)
∮
S
D · ds = Qencl (5.169)
which we used in Section 5.6 to find the field. If in",Electromagnetics_Vol1.pdf
"the same? This is a consequence of Gauss’ Law
(Section 5.5)
∮
S
D · ds = Qencl (5.169)
which we used in Section 5.6 to find the field. If in
this new problem we specify the same cylindrical
surface S with radius ρ<b , then the enclosed charge
is the same. Furthermore, the presence of the outer
conductor does not change the radial symmetry of the
problem, and nothing else remains that can change the
outcome. This is worth noting for future reference:
The electric field inside a coaxial structure com-
prised of concentric conductors and having uni-
form charge density on the inner conductor is
identical to the electric field of a line charge in
free space having the same charge density .
Next, we get V using (Section 5.8)
V = −
∫
C
E · dl (5.170)
where C is any path from the negatively-charged outer
conductor to the positively-charged inner conductor.
Since this can be any such path (Section 5.9), we may
as well choose the simplest one. This path is the one",Electromagnetics_Vol1.pdf
"Since this can be any such path (Section 5.9), we may
as well choose the simplest one. This path is the one
that traverses a radial of constant φand z. Thus:
V = −
∫ a
ρ=b
(
ˆρ ρl
2πǫsρ
)
· (ˆρdρ)
= − ρl
2πǫs
∫ a
ρ=b
dρ
ρ
= + ρl
2πǫs
∫ b
ρ=a
dρ
ρ
= + ρl
2πǫs
ln
( b
a
)
(5.171)
Wrapping up:
C ≜ Q+
V = ρll
(ρl/2πǫs) ln (b/a) (5.172)
Note that factors of ρl in the numerator and
denominator cancel out, leaving:
C = 2πǫsl
ln (b/a) (5.173)
Note that this expression is dimensionally correct,
having units of F . Also note that the expression
depends only on materials (through ǫs) and geometry
(through l, a, and b). The expression does not depend
on charge or voltage, which would imply non-linear
behavior.",Electromagnetics_Vol1.pdf
"130 CHAPTER 5. ELECTROST A TICS
T o make the connection back to lumped-element
transmission line model parameters (Sections 3.4 and
3.10), we simply divide by lto get the per-unit length
parameter:
C′ = 2πǫs
ln (b/a) (5.174)
Example 5.10. Capacitance of RG-59 coaxial
cable.
RG-59 coaxial cable consists of an inner
conductor having radius 0.292 mm, an outer
conductor having radius 1.855 mm, and a
polyethylene spacing material having relative
permittivity 2.25. Estimate the capacitance per
length of RG-59.
Solution. From the problem statement,
a= 0 .292 mm, b= 1 .855 mm, and ǫs = 2 .25ǫ0.
Using Equation 5.174 we find C′ = 67 .7 pF/m.
5.25 Electrostatic Energy
[m0114]
Consider a structure consisting of two perfect
conductors, both fixed in position and separated by an
ideal dielectric. This could be a capacitor, or it could
be one of a variety of capacitive structures that are not
explicitly intended to be a capacitor – for example, a
printed circuit board. When a potential difference is",Electromagnetics_Vol1.pdf
"explicitly intended to be a capacitor – for example, a
printed circuit board. When a potential difference is
applied between the two conducting regions, a
positive charge Q+ will appear on the surface of the
conductor at the higher potential, and a negative
charge Q− = −Q+ will appear on the surface of the
conductor at the lower potential (Section 5.19).
Assuming the conductors are not free to move,
potential energy is stored in the electric field
associated with the surface charges (Section 5.22).
W e now ask the question, what is the energy stored in
this field? The answer to this question has relevance
in several engineering applications. For example,
when capacitors are used as batteries, it is useful to
know to amount of energy that can be stored. Also,
any system that includes capacitors or has unintended
capacitance is using some fraction of the energy
delivered by the power supply to charge the
associated structures. In many electronic systems –",Electromagnetics_Vol1.pdf
"capacitance is using some fraction of the energy
delivered by the power supply to charge the
associated structures. In many electronic systems –
and in digital systems in particular – capacitances are
periodically charged and subsequently discharged at a
regular rate. Since power is energy per unit time, this
cyclic charging and discharging of capacitors
consumes power. Therefore, energy storage in
capacitors contributes to the power consumption of
modern electronic systems. W e’ll delve into that topic
in more detail in Example 5.11.
Since capacitance Crelates the charge Q+ to the
potential difference V between the conductors, this is
the natural place to start. From the definition of
capacitance (Section 5.22):
V = Q+
C (5.175)
From Section 5.8, electric potential is defined as the
work done (i.e., energy injected) by moving a charged
particle, per unit of charge; i.e.,
V = We
q (5.176)",Electromagnetics_Vol1.pdf
"5.25. ELECTROST A TIC ENERGY 131
where qis the charge borne by the particle and We
(units of J) is the work done by moving this particle
across the potential difference V. Since we are
dealing with charge distributions as opposed to
charged particles, it is useful to express this in terms
of the contribution ∆ We made to We by a small
charge ∆ q. Letting ∆ qapproach zero we have
dWe = Vdq (5.177)
Now consider what must happen to transition the
system from having zero charge ( q= 0 ) to the
fully-charged but static condition ( q= Q+). This
requires moving the differential amount of charge dq
across the potential difference between conductors,
beginning with q= 0 and continuing until q= Q+.
Therefore, the total amount of work done in this
process is:
We =
∫ Q+
q=0
dWe
=
∫ Q+
0
Vdq
=
∫ Q+
0
q
C dq
= 1
2
Q2
+
C (5.178)
Equation 5.178 can be expressed entirely in terms of
electrical potential by noting again that C = Q+/V,
so
We = 1
2CV2 (5.179)",Electromagnetics_Vol1.pdf
"= 1
2
Q2
+
C (5.178)
Equation 5.178 can be expressed entirely in terms of
electrical potential by noting again that C = Q+/V,
so
We = 1
2CV2 (5.179)
Since there are no other processes to account for the
injected energy , the energy stored in the electric field
is equal to We. Summarizing:
The energy stored in the electric field of a capac-
itor (or a capacitive structure) is given by Equa-
tion 5.179.
Example 5.11. Why multicore computing is
power-neutral.
Readers are likely aware that computers
increasingly use multicore processors as
opposed to single-core processors. For our
present purposes, a “core” is defined as the
smallest combination of circuitry that performs
independent computation. A multicore processor
consists of multiple identical cores that run in
parallel. Since a multicore processor consists of
N identical processors, you might expect power
consumption to increase by N relative to a
single-core processor. However, this is not the",Electromagnetics_Vol1.pdf
"N identical processors, you might expect power
consumption to increase by N relative to a
single-core processor. However, this is not the
case. T o see why , first realize that the power
consumption of a modern computing core is
dominated by the energy required to
continuously charge and discharge the multitude
of capacitances within the core. From
Equation 5.179, the required energy is 1
2 C0V2
0
per clock cycle, where C0 is the sum capacitance
(remember, capacitors in parallel add) and V0 is
the supply voltage. Power is energy per unit
time, so the power consumption for a single core
is
P0 = 1
2C0V2
0 f0 (5.180)
where f0 is the clock frequency . In a N-core
processor, the sum capacitance is increased by
N. However, the frequency is decreased by N
since the same amount of computation is
(nominally) distributed among the N cores.
Therefore, the power consumed by an N-core
processor is
PN = 1
2 (NC0) V2
0
( f0
N
)
= P0 (5.181)
In other words, the increase in power associated",Electromagnetics_Vol1.pdf
"Therefore, the power consumed by an N-core
processor is
PN = 1
2 (NC0) V2
0
( f0
N
)
= P0 (5.181)
In other words, the increase in power associated
with replication of hardware is nominally offset
by the decrease in power enabled by reducing
the clock rate. In yet other words, the total
energy of the N-core processor is N times the
energy of the single core processor at any given
time; however, the multicore processor needs to
recharge capacitances 1/Ntimes as often.
Before moving on, it should be noted that the
usual reason for pursuing a multicore design is
to increase the amount of computation that can
be done; i.e., to increase the product f0N.
Nevertheless, it is extremely helpful that power
consumption is proportional to f0 only , and is
independent of N.
The thin parallel plate capacitor (Section 5.23) is
representative of a large number of practical",Electromagnetics_Vol1.pdf
"132 CHAPTER 5. ELECTROST A TICS
applications, so it is instructive to consider the
implications of Equation 5.179 for this structure in
particular. For the thin parallel plate capacitor,
C ≈ ǫA
d (5.182)
where Ais the plate area, dis the separation between
the plates, and ǫis the permittivity of the material
between the plates. This is an approximation because
the fringing field is neglected; we shall proceed as if
this is an exact expression. Applying Equation 5.179:
We = 1
2
( ǫA
d
)
(Ed)2 (5.183)
where Eis the magnitude of the electric field intensity
between the plates. Rearranging factors, we obtain:
We = 1
2ǫE2 (Ad) (5.184)
Recall that the electric field intensity in the thin
parallel plate capacitor is approximately uniform.
Therefore, the density of energy stored in the
capacitor is also approximately uniform. Noting that
the product Adis the volume of the capacitor, we find
that the energy density is
we = We
Ad = 1
2ǫE2 (5.185)
which has units of energy per unit volume (J/m 3).",Electromagnetics_Vol1.pdf
"that the energy density is
we = We
Ad = 1
2ǫE2 (5.185)
which has units of energy per unit volume (J/m 3).
The above expression provides an alternative method
to compute the total electrostatic energy . Within a
mathematical volume V, the total electrostatic energy
is simply the integral of the energy density over V;
i.e.,
We =
∫
V
we dv (5.186)
This works even if Eand ǫvary with position. So,
even though we arrived at this result using the
example of the thin parallel-plate capacitor, our
findings at this point apply generally . Substituting
Equation 5.185 we obtain:
We = 1
2
∫
V
ǫE2dv (5.187)
Summarizing:
The energy stored by the electric field present
within a volume is given by Equation 5.187.
It’s worth noting that this energy increases with the
permittivity of the medium, which makes sense since
capacitance is proportional to permittivity .
[m0034]",Electromagnetics_Vol1.pdf
"5.25. ELECTROST A TIC ENERGY 133
Image Credits
Fig. 5.1: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0102 fCoulombsLaw .svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.2: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0104 fRing.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.3: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0104 fDisk.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.4: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:4 .5 m0149 commons.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.5: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0064 fResistor.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.7: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0020 fPEC.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).",Electromagnetics_Vol1.pdf
"Fig. 5.7: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0020 fPEC.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.8: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0020 fKVL.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.9: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0021 fPEC.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.10: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0021 fGL.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.11: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0025 fFlatSlab.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 5.12: by C. Burks, https://commons.wikimedia.org/wi ki/File:Electrostatic induction.svg,
in public domain. Minor modifications.
Fig. 5.15: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0113 fCapCoax.svg,",Electromagnetics_Vol1.pdf
"in public domain. Minor modifications.
Fig. 5.15: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0113 fCapCoax.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).",Electromagnetics_Vol1.pdf
"Chapter 6
Steady Current and Conductivity
6.1 Convection and Conduction
Currents
[m0110]
In practice, we deal with two physical mechanisms
for current: convection and conduction. The
distinction between these types of current is important
in electromagnetic analysis.
Convection current consists of charged particles
moving in response to mechanical forces, as opposed
to being guided by the electric field (Sections 2.2
and/or 5.1). An example of a convection current is a
cloud bearing free electrons that moves through the
atmosphere driven by wind.
Conduction current consists of charged particles
moving in response to the electric field and not
merely being carried by motion of the surrounding
material. In some materials, the electric field is also
able to dislodge weakly-bound electrons from atoms,
which then subsequently travel some distance before
reassociating with other atoms. For this reason, the
individual electrons in a conduction current do not",Electromagnetics_Vol1.pdf
"reassociating with other atoms. For this reason, the
individual electrons in a conduction current do not
necessarily travel the full distance over which the
current is perceived to exist.
The distinction between convection and conduction is
important because Ohm’s Law (Section 6.3) – which
specifies the relationship between electric field
intensity and current – applies only to conduction
current.
Additional Reading:
• “Electric current” on Wikipedia.
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"6.2. CURRENT DISTRIBUTIONS 135
6.2 Current Distributions
[m0101]
In elementary electric circuit theory , current is the
rate at which electric charge passes a particular point
in a circuit. For example, 1 A is 1 C per second. In
this view current is a scalar quantity , and there are
only two possible directions because charge is
constrained to flow along defined channels. Direction
is identified by the sign of the current with respect to
a reference direction, which is in turn defined with
respect to a reference voltage polarity . The
convention in electrical engineering defines positive
current as the flow of positive charge into the positive
voltage terminal of a passive device such as a resistor,
capacitor, or inductor. For an active device such as a
battery , positive current corresponds to the flow of
positive charge out of the positive voltage terminal.
However, this kind of thinking only holds up when
the systems being considered are well-described as",Electromagnetics_Vol1.pdf
"However, this kind of thinking only holds up when
the systems being considered are well-described as
“lumped” devices connected by infinitesimally thin,
filament-like connections representing wires and
circuit board traces. Many important problems in
electrical engineering concern situations in which the
flow of current is not limited in this way . Examples
include wire- and pin-type interconnects at radio
frequencies, circuit board and enclosure grounding,
and physical phenomena such as lightning. T o
accommodate the more general class of problems, we
must define current as a vector quantity . Furthermore,
current in these problems can spread out over surfaces
and within volumes, so we must also consider spatial
distributions of current.
Line Current. As noted above, if a current I is
constrained to follow a particular path, then the only
other consideration is direction. Thus, a line current is
specified mathematically as ˆlI, where the direction ˆl",Electromagnetics_Vol1.pdf
"other consideration is direction. Thus, a line current is
specified mathematically as ˆlI, where the direction ˆl
may vary with position along the path. For example,
in a straight wire ˆl is constant, whereas in a coil ˆl
varies with position along the coil.
Surface Current Distribution. In some cases,
current may be distributed over a surface. For
example, the radio-frequency current on a wire of
radius amade from a metal with sufficiently high
conductivity can be modeled as a uniform surface
current existing on the wire surface. In this case, the
current is best described as a surface current density
Js, which is the total current I on the wire divided by
the circumference 2πaof the wire:
Js = ˆu I
2πa (units of A/m) (6.1)
where ˆu is the direction of current flow .
V olume Current Distribution. Imagine that current
is distributed within a volume. Let ˆu∆ ibe the current
passing through a small open planar surface defined
within this volume, and let ∆ sbe the area of this",Electromagnetics_Vol1.pdf
"passing through a small open planar surface defined
within this volume, and let ∆ sbe the area of this
surface. The volume current density J at any point in
the volume is defined as
J ≜ lim
∆ s→0
ˆu∆ i
∆ s = ˆu di
ds (units of A/m 2) (6.2)
In general, J is a function of position within this
volume. Subsequently the total current passing
through a surface S is
I =
∫
S
J · ds (units of A) (6.3)
In other words, volume current density integrated over
a surface yields total current through that surface. Y ou
might recognize this as a calculation of flux, and it is. 1
Example 6.1. Current and current density in a
wire of circular cross-section.
Figure 6.1 shows a straight wire having
cross-sectional radius a= 3 mm. A battery is
connected across the two ends of the wire
resulting in a volume current density
J = ˆz8 A/m2, which is uniform throughout the
wire. Find the net current I through the wire.
Solution. The net current is
I =
∫
S
J · ds
W e choose S to be the cross-section",Electromagnetics_Vol1.pdf
"wire. Find the net current I through the wire.
Solution. The net current is
I =
∫
S
J · ds
W e choose S to be the cross-section
perpendicular to the axis of the wire. Also, we
choose ds to point such that I is positive with
respect to the sign convention shown in
Figure 6.1 , which is the usual choice in electric
1 However, it is not common to refer to net current as a flux; go
figure!",Electromagnetics_Vol1.pdf
"136 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY
J
I
a
z
-+
ds
c⃝ K. Kikkeri CC BY SA 4.0
Figure 6.1: Net current I and current density J in a
wire of circular cross-section.
circuit analysis. With these choices, we have
I =
∫ a
ρ=0
∫ 2π
φ=0
( ˆz8 A/m2)
· (ˆzρdρdφ )
=
(
8 A/m2) ∫ a
ρ=0
∫ 2π
φ=0
ρdρdφ
=
(
8 A/m2)
·
(
πa2)
= 226 µA
This answer is independent of the
cross-sectional surface S used to do the
calculation. For example, you could use an
alternative surface that is tilted 45◦ away from
the axis of the wire. In that case, the
cross-sectional area would increase, but the dot
product of J and ds would be proportionally less
and the outcome would be the same. Since the
choice of S is arbitrary – any surface with edges
at the perimeter of the wire will do – you should
make the choice that makes the problem as
simple as possible, as we have done above.
Additional Reading:
• “Electric current” on Wikipedia.
• “Current density” on Wikipedia.
6.3 Conductivity
[m0010]",Electromagnetics_Vol1.pdf
"Additional Reading:
• “Electric current” on Wikipedia.
• “Current density” on Wikipedia.
6.3 Conductivity
[m0010]
Conductivity is one of the three primary “constitutive
parameters” that is commonly used to characterize the
electromagnetic properties of materials (Section 2.8).
The key idea is this:
Conductivity is a property of materials that deter-
mines conduction current density in response to
an applied electric field.
Recall that conduction current is the flow of charge in
response to an electric field (Section 6.1). Although
the associated force is straightforward to calculate
(e.g., Section 5.1), the result is merely the force
applied, not the speed at which charge moves in
response. The latter is determined by the mobility of
charge, which is in turn determined by the atomic and
molecular structure of the material. Conductivity
relates current density to the applied field directly ,
without requiring one to grapple separately with the
issues of applied force and charge mobility .",Electromagnetics_Vol1.pdf
"without requiring one to grapple separately with the
issues of applied force and charge mobility .
In the absence of material – that is, in a true, perfect
vacuum – conductivity is zero because there is no
charge available to form current, and therefore the
current is zero no matter what electric field is applied.
At the other extreme, a good conductor is a material
that contains a supply of charge that is able to move
freely within the material. When an electric field is
applied to a good conductor, charge-bearing material
constituents move in the direction determined by the
electric field, creating current flow in that direction.
This relationship is summarized by Ohm’s Law for
Electromagnetics:
J = σE
(6.4)
where E is electric field intensity (V/m); J is the
volume current density , a vector describing the
current flow , having units of A/m 2 (see Section 6.2);
and σis conductivity . Since E has units of V/m, we
see σhas units of Ω −1m−1, which is more commonly",Electromagnetics_Vol1.pdf
"and σis conductivity . Since E has units of V/m, we
see σhas units of Ω −1m−1, which is more commonly
expressed as S/m, where 1 S (“siemens”) is defined as
1 Ω −1. Section A.3 provides values of conductivity
for a representative set of materials.",Electromagnetics_Vol1.pdf
"6.3. CONDUCTIVITY 137
Conductivity σ is expressed in units of S/m,
where 1 S = 1 Ω −1.
It is important to note that the current being addressed
here is conduction current , and not convection
current, displacement current, or some other form of
current – see Section 6.2 for elaboration.
Summarizing:
Ohm’s Law for Electromagnetics (Equation 6.4)
states that volume density of conduction current
(A/m2) equals conductivity (S/m) times electric
field intensity (V/m).
The reader is likely aware that there is also an “Ohm’s
Law” in electric circuit theory , which states that
current I (units of A) is voltage V (units of V)
divided by resistance R(units of Ω ); i.e., I = V/R.
This is in fact a special case of Equation 6.4; see
Section 6.4 for more about this.
As mentioned above, σdepends on both the
availability and mobility of charge within the
material. At the two extremes, we have perfect
insulators, for which σ= 0 , and perfect conductors ,
for which σ→ ∞ . Some materials approach these",Electromagnetics_Vol1.pdf
"material. At the two extremes, we have perfect
insulators, for which σ= 0 , and perfect conductors ,
for which σ→ ∞ . Some materials approach these
extremes, whereas others fall midway between these
conditions. Here are a few classes of materials that
are frequently encountered:
• A perfect vacuum – “free space” – contains no
charge and therefore is a perfect insulator with
σ= 0 .
• Good insulators typically have conductivities
≪ 10−10 S/m, which is sufficiently low that the
resulting currents can usually be ignored. The
most important example is air, which has a
conductivity only slightly greater than that of
free space. An important class of good insulators
is lossless dielectrics ,2 which are
well-characterized in terms of permittivity ( ǫ)
alone, and for which µ= µ0 and σ= 0 may
usually be assumed.
• P oor insulators have conductivities that are low ,
but nevertheless sufficiently high that the
2 See Section 5.20 for a discussion of dielectric materials.",Electromagnetics_Vol1.pdf
"but nevertheless sufficiently high that the
2 See Section 5.20 for a discussion of dielectric materials.
resulting currents cannot be ignored. For
example, the dielectric material that is used to
separate the conductors in a transmission line
must be considered a poor insulator as opposed
to a good (effectively lossless) insulator in order
to characterize loss per length along the
transmission line, which can be significant. 3
These lossy dielectrics are well-characterized in
terms of ǫand σ, and typically µ= µ0 can be
assumed.
• Semiconductors such as those materials used in
integrated circuits have intermediate
conductivities, typically in the range 10−4 to
10+1 S/m.
• Good conductors are materials with very high
conductivities, typically greater than 105 S/m.
An important category of good conductors
includes metals, with certain metals including
alloys of aluminum, copper, and gold reaching
conductivities on the order of 108 S/m. In such",Electromagnetics_Vol1.pdf
"includes metals, with certain metals including
alloys of aluminum, copper, and gold reaching
conductivities on the order of 108 S/m. In such
materials, minuscule electric fields give rise to
large currents. There is no significant storage of
energy in such materials, and so the concept of
permittivity is not relevant for good conductors.
The reader should take care to note that terms such as
good conductor and poor insulator are qualitative and
subject to context. What may be considered a “good
insulator” in one application may be considered to be
a “poor insulator” in another.
One relevant category of material was not included in
the above list – namely , perfect conductors . A perfect
conductor is a material in which σ→ ∞ . It is
tempting to interpret this as meaning that any electric
field gives rise to infinite current density; however,
this is not plausible even in the ideal limit. Instead,
this condition is interpreted as meaning that",Electromagnetics_Vol1.pdf
"this is not plausible even in the ideal limit. Instead,
this condition is interpreted as meaning that
E = J/σ→ 0 throughout the material; i.e., E is zero
independently of any current flow in the material. An
important consequence is that the potential field V is
equal to a constant value throughout the material.
(Recall E = −∇V (Section 5.14), so constant V
means E = 0 .) W e refer to the volume occupied by
such a material as an equipotential volume . This
concept is useful as an approximation of the behavior
3 Review Sections 3.4, 3.9, and associated sections for a refr esher
on this issue.",Electromagnetics_Vol1.pdf
"138 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY
of good conductors; for example, metals are often
modeled as perfectly-conducting equipotential
volumes in order to simplify analysis.
A perfect conductor is a material for which σ →
∞, E → 0, and subsequently V (the electric po-
tential) is constant.
One final note: It is important to remain aware of the
assumptions we have made about materials in this
book, which are summarized in Section 2.8. In
particular, we continue to assume that materials are
linear unless otherwise indicated. For example,
whereas air is normally considered to be a good
insulator and therefore a poor conductor, anyone who
has ever witnessed lightning has seen a demonstration
that under the right conditions – i.e., sufficiently large
potential difference between earth and sky – current
will readily flow through air. This particular situation
is known as dielectric breakdown (see Section 5.21).
The non-linearity of materials can become evident",Electromagnetics_Vol1.pdf
"is known as dielectric breakdown (see Section 5.21).
The non-linearity of materials can become evident
before reaching the point of dielectric breakdown, so
one must be careful to consider this possibility when
dealing with strong electric fields.
Additional Reading:
• “Electrical Resistivity and Conductivity” on
Wikipedia.
• “Ohm’s Law” on Wikipedia.
6.4 Resistance
[m0071]
The concept of resistance is most likely familiar to
readers via Ohm’s Law for Devices ; i.e., V = IR
where V is the potential difference associated with a
current I. This is correct, but it is not the whole story .
Let’s begin with a statement of intent:
Resistance R(Ω ) is a characterization of the con-
ductivity of a device (as opposed to a material) in
terms of Ohms Law for Devices; i.e., V = IR.
Resistance is a property of devices such as resistors,
which are intended to provide resistance, as well as
being a property of most practical electronic devices,
whether it is desired or not.",Electromagnetics_Vol1.pdf
"which are intended to provide resistance, as well as
being a property of most practical electronic devices,
whether it is desired or not.
Resistance is a manifestation of the conductivity of
the materials comprising the device, which
subsequently leads to the “ V = IR” relationship.
This brings us to a very important point and a
common source of confusion. Resistance is not
necessarily the real part of impedance. Let’s take a
moment to elaborate. Impedance ( Z) is defined as the
ratio of voltage to current; i.e., V/I; or equivalently in
the phasor domain as ˜V/˜I. Most devices – not just
devices exhibiting resistance – can be characterized in
terms of this ratio. Consider for example the input
impedance of a terminated transmission line
(Section 3.15). This impedance may have a non-zero
real-valued component even when the transmission
line and the terminating load are comprised of perfect
conductors. Summarizing:
Resistance results in a real-valued impedance.",Electromagnetics_Vol1.pdf
"line and the terminating load are comprised of perfect
conductors. Summarizing:
Resistance results in a real-valued impedance.
However, not all devices exhibiting a real-valued
impedance exhibit resistance. Furthermore, the
real component of a complex-valued impedance
does not necessarily represent resistance.
Restating the main point in yet other words:
Resistance pertains to limited conductivity, not simply
to voltage-current ratio.
Also important to realize is that whereas conductivity
σ(units of S/m) is a property of materials, resistance
depends on both conductivity and the geometry of the",Electromagnetics_Vol1.pdf
"6.4. RESIST ANCE 139
E
I
a
z
- +
V
z=0 z=
σ
c⃝ K. Kikkeri CC BY SA 4.0
Figure 6.2: Analysis of the resistance of straight wire
of circular cross-section.
device. In this section, we address the question of
how the resistance of a device can be is determined.
The following example serves this purpose.
Figure 6.2 shows a straight wire of length lcentered
on the zaxis, forming a cylinder of material having
conductivity σand cross-sectional radius a. The ends
of the cylinder are covered by perfectly-conducting
plates to which the terminals are attached. A battery
is connected to the terminals, resulting in a uniform
internal electric field E between the plates.
T o calculate Rfor this device, let us first calculate V,
then I, and finally R= V/I. First, we compute the
potential difference using the following result from
Section 5.8:
V = −
∫
C
E · dl (6.5)
Here we can view V as a “given” (being the voltage
of the battery), but wish to evaluate the right hand",Electromagnetics_Vol1.pdf
"Section 5.8:
V = −
∫
C
E · dl (6.5)
Here we can view V as a “given” (being the voltage
of the battery), but wish to evaluate the right hand
side so as to learn something about the effect of the
conductivity and geometry of the wire. The
appropriate choice of C begins at z= 0 and ends at
z= l, following the axis of the cylinder. (Remember:
C defines the reference direction for increasing
potential, so the resulting potential difference will be
the node voltage at the end point minus the node
voltage at the start point.) Thus, dl = ˆzdzand we
have
V = −
∫ l
z=0
E · (ˆzdz)
W e do not yet know E; however, we know it is
constant throughout the device and points in the −ˆz
direction since this is the direction of current flow and
Ohm’s Law for Electromagnetics (Section 6.3)
requires the electric field to point in the same
direction. Thus, we may write E = −ˆzEz where Ez
is a constant. W e now find:
V = −
∫ l
z=0
(−ˆzEz) · (ˆzdz)
= + Ez
∫ l
z=0
dz
= + Ezl
The current I is given by (Section 6.2)",Electromagnetics_Vol1.pdf
"is a constant. W e now find:
V = −
∫ l
z=0
(−ˆzEz) · (ˆzdz)
= + Ez
∫ l
z=0
dz
= + Ezl
The current I is given by (Section 6.2)
I =
∫
S
J · ds
W e choose the surface S to be the cross-section
perpendicular to the axis of the wire. Also, we choose
ds to point such that I is positive with respect to the
sign convention shown in Figure 6.2. With these
choices, we have
I =
∫ a
ρ=0
∫ 2π
φ=0
J · (−ˆz ρdρdφ )
From Ohm’s Law for Electromagnetics, we have
J = σE = −ˆzσEz (6.6)
So now
I =
∫ a
ρ=0
∫ 2π
φ=0
(−ˆzσEz) · (−ˆzρdρdφ )
= σEz
∫ a
ρ=0
∫ 2π
φ=0
ρdρdφ
= σEz
(
πa2)
Finally:
R= V
I = Ezl
σEz(πa2) = l
σ(πa2)
This is a good-enough answer for the problem posed,
but it is easily generalized a bit further. Noting that
πa2 in the denominator is the cross-sectional area A
of the wire, so we find:
R= l
σA (6.7)",Electromagnetics_Vol1.pdf
"140 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY
i.e., the resistance of a wire having cross-sectional
area A– regardless of the shape of the cross-section,
is given by the above equation.
The resistance of a right cylinder of material,
given by Equation 6.7, is proportional to length
and inversely proportional to cross-sectional area
and conductivity .
It is important to remember that Equation 6.7
presumes that the volume current density J is uniform
over the cross-section of the wire. This is an excellent
approximation for thin wires at “low” frequencies
including, of course, DC. At higher frequencies it
may not be a good assumption that J is
uniformly-distributed over the cross-section of the
wire, and at sufficiently high frequencies one finds
instead that the current is effectively limited to the
exterior surface of the wire. In the “high frequency”
case, Ain Equation 6.7 is reduced from the physical
area to a smaller value corresponding to the reduced",Electromagnetics_Vol1.pdf
"case, Ain Equation 6.7 is reduced from the physical
area to a smaller value corresponding to the reduced
area through which most of the current flows.
Therefore, Rincreases with increasing frequency . T o
quantify the high frequency behavior of R(including
determination of what constitutes “high frequency” in
this context) one requires concepts beyond the theory
of electrostatics, so this is addressed elsewhere.
Example 6.2. Resistance of 22 A WG hookup
wire.
A common type of wire found in DC
applications is 22A WG (“ American Wire
Gauge”; see “ Additional Resources” at the end
of this section) copper solid-conductor hookup
wire. This type of wire has circular cross-section
with diameter 0.644 mm. What is the resistance
of 3 m of this wire? Assume copper conductivity
of 58 MS/m.
Solution. From the problem statement, the
diameter 2a= 0 .644 mm, σ= 58 × 106 S/m,
and l= 3 m. The cross-sectional area is
A= πa2 ∼
= 3.26 × 10−7 m2. Using
Equation 6.7 we obtain R= 159 mΩ .",Electromagnetics_Vol1.pdf
"diameter 2a= 0 .644 mm, σ= 58 × 106 S/m,
and l= 3 m. The cross-sectional area is
A= πa2 ∼
= 3.26 × 10−7 m2. Using
Equation 6.7 we obtain R= 159 mΩ .
Example 6.3. Resistance of steel pipe.
A pipe is 3 m long and has inner and outer radii
of 5 mm and 7 mm respectively . It is made from
steel having conductivity 4 MS/m. What is the
DC resistance of this pipe?
Solution. W e can use Equation 6.7 if we can
determine the cross-sectional area Athrough
which the current flows. This area is simply the
area defined by the outer radius, πb2, minus the
area defined by the inner radius πa2. Thus,
A= πb2 − πa2 ∼
= 7.54 × 10−5 m2. From the
problem statement, we also determine that
σ= 4 × 106 S/m and l= 3 m. Using
Equation 6.7 we obtain R∼
= 9.95 mΩ .
Additional Reading:
• “Resistor” on Wikipedia.
• “Ohm’s Law” on Wikipedia.
• “American wire gauge” on Wikipedia.",Electromagnetics_Vol1.pdf
"6.5. CONDUCT ANCE 141
6.5 Conductance
[m0105]
Conductance, like resistance (Section 6.4), is a
property of devices. Specifically:
Conductance G (Ω −1 or S) is the reciprocal of
resistance R.
Therefore, conductance depends on both the
conductivity of the materials used in the device, as
well as the geometry of the device.
A natural question to ask is, why do we require the
concept of conductance, if it simply the reciprocal of
resistance? The short answer is that the concept of
conductance is not required; or, rather, we need only
resistance or conductance and not both. Nevertheless,
the concept appears in engineering analysis for two
reasons:
• Conductance is sometimes considered to be a
more intuitive description of the underlying
physics in cases where the applied voltage is
considered to be the independent “stimulus” and
current is considered to be the response. This is
why conductance appears in the lumped element
model for transmission lines (Section 3.4), for
example.",Electromagnetics_Vol1.pdf
"why conductance appears in the lumped element
model for transmission lines (Section 3.4), for
example.
• Characterization in terms of conductance may be
preferred when considering the behavior of
devices in parallel, since the conductance of a
parallel combination is simply the sum of the
conductances of the devices.
Example 6.4. Conductance of a coaxial
structure.
Let us now determine the conductance of a
structure consisting of coaxially-arranged
conductors separated by a lossy dielectric, as
shown in Figure 6.3. The conductance per unit
length G′ (i.e., S/m) of this structure is of
interest in determining the characteristic
impedance of coaxial transmission line, as
addressed in Sections 3.4 and 3.10.
For our present purposes, we may model the
structure as two concentric perfectly-conducting
cylinders of radii aand b, separated by a lossy
dielectric having conductivity σs. W e place the
+zaxis along the common axis of the concentric
cylinders so that the cylinders may be described",Electromagnetics_Vol1.pdf
"dielectric having conductivity σs. W e place the
+zaxis along the common axis of the concentric
cylinders so that the cylinders may be described
as constant-coordinate surfaces ρ= aand ρ= b.
There are at least 2 ways to solve this problem.
One method is to follow the procedure that was
used to find the capacitance of this structure in
Section 5.24. Adapting that approach to the
present problem, one would assume a potential
difference V between the conductors, from that
determine the resulting electric field intensity E,
and then using Ohm’s Law for Electromagnetics
(Section 6.3) determine the density J = σsE of
the current that leaks directly between
conductors. From this, one is able to determine
the total leakage current I, and subsequently the
conductance G≜ I/V. Although highly
recommended as an exercise for the student, in
this section we take an alternative approach so as
to demonstrate that there are a variety of
approaches available for such problems.",Electromagnetics_Vol1.pdf
"this section we take an alternative approach so as
to demonstrate that there are a variety of
approaches available for such problems.
The method we shall use below is as follows: (1)
Assume a leakage current I between the
conductors; (2) Determine the associated current
density J, which is possible using only
geometrical considerations; (3) Determine the
associated electric field intensity E using J/σs;
(4) Integrate E over a path between the
conductors to get V. Then, as before,
conductance G≜ I/V.
The current I is defined as shown in Figure 6.3,
with reference direction according to the
engineering convention that positive current
flows out of the positive terminal of a source.
The associated current density must flow in the
same direction, and the circular symmetry of the
problem therefore constrains J to have the form
J = ˆρI
A (6.8)
where Ais the area through which I flows. In
other words, current flows radially outward from
the inner conductor to the outer conductor, with",Electromagnetics_Vol1.pdf
"142 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY
density that diminishes inversely with the area
through which the total current flows. (It may be
helpful to view J as a flux density and I as a
flux, as noted in Section 6.2.) This area is simply
circumference 2πρtimes length l, so
J = ˆρ I
2πρl (6.9)
which exhibits the correct units of A/m 2.
Now from Ohm’s Law for Electromagnetics we
find the electric field within the structure is
E = J
σs
= ˆρ I
2πρlσs
(6.10)
Next we get V using (Section 5.8)
V = −
∫
C
E · dl (6.11)
where C is any path from the negatively-charged
outer conductor to the positively-charged inner
conductor. Since this can be any such path
(Section 5.9), we should choose the simplest
one. The simplest path is the one that traverses a
radial of constant φand z. Thus:
V = −
∫ a
ρ=b
(
ˆρ I
2πρlσs
)
· (ˆρdρ)
= − I
2πlσs
∫ a
ρ=b
dρ
ρ
= + I
2πlσs
∫ b
ρ=a
dρ
ρ
= + I
2πlσs
ln
( b
a
)
(6.12)
Wrapping up:
G≜ I
V = I
(I/2πlσs) ln (b/a) (6.13)
Note that factors of I in the numerator and",Electromagnetics_Vol1.pdf
"= + I
2πlσs
∫ b
ρ=a
dρ
ρ
= + I
2πlσs
ln
( b
a
)
(6.12)
Wrapping up:
G≜ I
V = I
(I/2πlσs) ln (b/a) (6.13)
Note that factors of I in the numerator and
denominator cancel out, leaving:
G= 2πlσs
ln (b/a) (6.14)
b
a
z
ρ
l
I

s
J
Figure 6.3: Determining the conductance of a struc-
ture consisting of coaxially-arranged conductors sepa-
rated by a lossy dielectric.
Note that Equation 6.14 is dimensionally
correct, having units of S = Ω −1. Also note that
this is expression depends only on materials
(through σs) and geometry (through l, a, and b).
Notably it does not depend on current or voltage,
which would imply non-linear behavior.
T o make the connection back to lumped-element
transmission line model parameters
(Sections 3.4 and 3.10), we simply divide by lto
get the per-length parameter:
G′ = 2πσs
ln (b/a) (6.15)
Example 6.5. Conductance of RG-59 coaxial
cable.
RG-59 coaxial cable consists of an inner
conductor having radius 0.292 mm, an outer
conductor having radius 1.855 mm, and a",Electromagnetics_Vol1.pdf
"cable.
RG-59 coaxial cable consists of an inner
conductor having radius 0.292 mm, an outer
conductor having radius 1.855 mm, and a
polyethylene spacing material exhibiting
conductivity of about 5.9 × 10−5 S/m. Estimate
the conductance per length of RG-59.
Solution. From the problem statement,
a= 0 .292 mm, b= 1 .855 mm, and
σs ∼
= 5.9 × 10−5 S/m. Using Equation 6.15, we
find G′ ∼
= 200 µS/m.",Electromagnetics_Vol1.pdf
"6.6. POWER DISSIP A TION IN CONDUCTING MEDIA 143
6.6 Power Dissipation in
Conducting Media
[m0106]
The displacement of charge in response to the force
exerted by an electric field constitutes a reduction in
the potential energy of the system (Section 5.8). If the
charge is part of a steady current, there must be an
associated loss of energy that occurs at a steady rate.
Since power is energy per unit time, the loss of energy
associated with current is expressible as power
dissipation. In this section, we address two questions:
(1) How much power is dissipated in this manner, and
(2) What happens to the lost energy?
First, recall that work is force times distance traversed
in response to that force (Section 5.8). Stated
mathematically:
∆ W = + F · ∆ l (6.16)
where the vector F is the force (units of N) exerted by
the electric field, the vector ∆ l is the direction and
distance (units of m) traversed, and ∆ W is the work
done (units of J) as a result. Note that a “ +” has been",Electromagnetics_Vol1.pdf
"distance (units of m) traversed, and ∆ W is the work
done (units of J) as a result. Note that a “ +” has been
explicitly indicated; this is to emphasize the
distinction from the work being considered in
Section 5.8. In that section, the work “ ∆ W”
represented energy from an external source that was
being used to increase the potential energy of the
system by moving charge “upstream” relative to the
electric field. Now , ∆ W represents this internal
energy as it is escaping from the system in the form of
kinetic energy; therefore, positive ∆ W now means a
reduction in potential energy , hence the sign change. 4
The associated power ∆ P (units of W) is ∆ W
divided by the time ∆ t(units of s) required for the
distance ∆ l to be traversed:
∆ P = ∆ W
∆ t = F · ∆ l
∆ t (6.17)
Now we’d like to express force in terms of the electric
field exerting this force. Recall that the force exerted
by an electric field intensity E (units of V/m) on a",Electromagnetics_Vol1.pdf
"field exerting this force. Recall that the force exerted
by an electric field intensity E (units of V/m) on a
4 It could be argued that it is bad form to use the same variable
to represent both tallies; nevertheless, it is common practic e and so
we simply remind the reader that it is important to be aware of the
definitions of variables each time they are (re)introduced.
particle bearing charge q(units of C) is qE
(Section 2.2). However, we’d like to express this
force in terms of a current, as opposed to a charge. An
expression in terms of current can be constructed as
follows. First, note that the total charge in a small
volume “cell” is the volume charge density ρv (units
of C/m 3) times the volume ∆ vof the cell; i.e.,
q= ρv∆ v(Section 5.3). Therefore:
F = qE = ρv ∆ vE (6.18)
and subsequently
∆ P = ρv ∆ vE · ∆ l
∆ t = E ·
(
ρv
∆ l
∆ t
)
∆ v (6.19)
The quantity in parentheses has units of C/m 3 · m ·
s−1, which is A/m 2. Apparently this quantity is the",Electromagnetics_Vol1.pdf
"∆ t = E ·
(
ρv
∆ l
∆ t
)
∆ v (6.19)
The quantity in parentheses has units of C/m 3 · m ·
s−1, which is A/m 2. Apparently this quantity is the
volume current density J, so we have
∆ P = E · J ∆ v (6.20)
In the limit as ∆ v→ 0 we have
dP = E · J dv (6.21)
and integrating over the volume V of interest we
obtain
P =
∫
V
dP =
∫
V
E · J dv (6.22)
The above expression is commonly known as Joule’s
Law. In our situation, it is convenient to use Ohm’s
Law for Electromagnetics ( J = σE; Section 6.3) to
get everything in terms of materials properties ( σ),
geometry ( V), and the electric field:
P =
∫
V
E · (σE) dv (6.23)
which is simply
P =
∫
V
σ|E|2 dv (6.24)
Thus:
The power dissipation associated with current
is given by Equation 6.24. This power is pro-
portional to conductivity and proportional to the
electric field magnitude squared.",Electromagnetics_Vol1.pdf
"144 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY
This result facilitates the analysis of power dissipation
in materials exhibiting loss; i.e., having finite
conductivity . But what is the power dissipation in a
perfectly conducting material? For such a material,
σ→ ∞ and E → 0 no matter how much current is
applied (Section 6.3). In this case, Equation 6.24 is
not very helpful. However, as we just noted, being a
perfect conductor means E → 0 no matter how much
current is applied, so from Equation 6.22 we have
found that:
No power is dissipated in a perfect conductor.
When conductivity is finite, Equation 6.24 serves as a
more-general version of a concept from elementary
circuit theory , as we shall now demonstrate. Let
E = ˆzEz, so |E|2 = E2
z. Then Equation 6.24
becomes:
P =
∫
V
σE2
z dv= σE2
z
∫
V
dv (6.25)
The second integral in Equation 6.25 is a calculation
of volume. Let’s assume V is a cylinder aligned along
the zaxis. The volume of this cylinder is the",Electromagnetics_Vol1.pdf
"of volume. Let’s assume V is a cylinder aligned along
the zaxis. The volume of this cylinder is the
cross-sectional area Atimes the length l. Then the
above equation becomes:
P = σE2
z Al (6.26)
For reasons that will become apparent very shortly ,
let’s reorganize the above expression as follows:
P = ( σEzA) (Ezl) (6.27)
Note that σEz is the current density in A/m 2, which
when multiplied by Agives the total current.
Therefore, the quantity in the first set of parentheses
is simply the current I. Also note that Ezlis the
potential difference over the length l, which is simply
the node-to-node voltage V (Section 5.8). Therefore,
we have found:
P = IV (6.28)
as expected from elementary circuit theory .
Now , what happens to the energy associated with this
dissipation of power? The displacement of charge
carriers in the material is limited by the conductivity ,
which itself is finite because, simply put, other
constituents of the material get in the way . If charge is",Electromagnetics_Vol1.pdf
"which itself is finite because, simply put, other
constituents of the material get in the way . If charge is
being displaced as described in this section, then
energy is being used to displace the charge-bearing
particles. The motion of constituent particles is
observed as heat – in fact, this is essentially the
definition of heat. Therefore:
The power dissipation associated with the flow of
current in any material that is not a perfect con-
ductor manifests as heat.
This phenomenon is known as joule heating , ohmic
heating, and by other names. This conversion of
electrical energy to heat is the method of operation for
toasters, electric space heaters, and many other
devices that generate heat. It is of course also the
reason that all practical electronic devices generate
heat.
Additional Reading:
• “Joule Heating” on Wikipedia.
[m0057]",Electromagnetics_Vol1.pdf
"6.6. POWER DISSIP A TION IN CONDUCTING MEDIA 145
Image Credits
Fig. 6.1: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0101 fMatEx1.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 6.2: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0071 fMatEx1.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).",Electromagnetics_Vol1.pdf
"Chapter 7
Magnetostatics
[m0117]
Magnetostatics is the theory of the magnetic field in
conditions in which its behavior is independent of
electric fields, including
• The magnetic field associated with various
spatial distributions of steady current
• The energy associated with the magnetic field
• Inductance, which is the ability of a structure to
store energy in a magnetic field
The word ending “-statics” refers to the fact that these
aspects of electromagnetic theory can be developed
by assuming that the sources of the magnetic field are
time-invariant; we might say that magnetostatics is
the study of the magnetic field at DC. However, many
aspects of magnetostatics are applicable at “ AC” as
well.
7.1 Comparison of Electrostatics
and Magnetostatics
[m0115]
Students encountering magnetostatics for the first
time have usually been exposed to electrostatics
already . Electrostatics and magnetostatics exhibit
many similarities. These are summarized in T able 7.1.",Electromagnetics_Vol1.pdf
"already . Electrostatics and magnetostatics exhibit
many similarities. These are summarized in T able 7.1.
The elements of magnetostatics presented in this table
are all formally introduced in other sections; the sole
purpose of this table is to point out the similarities.
The technical term for these similarities is duality.
Duality also exists between voltage and current in
electrical circuit theory . For more about the concept
of duality , see “ Additional Reading” at the end of this
section.
Additional Reading:
• “Duality (electricity and magnetism)” on
Wikipedia.
• “Duality (electrical circuits)” on Wikipedia.
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"7.2. GAUSS’ LA W FOR MAGNETIC FIELDS: INTEGRAL FORM 147
electrostatics magnetostatics
Sources static charge steady current,
magnetizable material
Field intensity E (V/m) H (A/m)
Flux density D (C/m2) B (Wb/m2=T)
Material relations D = ǫE B = µH
J = σE
Force on charge q F = qE F = qv × B
Maxwell’s Eqs.
∮
S D · ds = Qencl
∮
S B · ds = 0
(integral)
∮
C E · dl = 0
∮
C H · dl = Iencl
Maxwell’s Eqs. ∇ · D = ρv ∇ · B = 0
(differential) ∇ × E = 0 ∇ × H = J
Boundary Conditions ˆn × [E1 − E2] = 0 ˆn × [H1 − H2] = Js
ˆn · [D1 − D2] = ρs ˆn · [B1 − B2] = 0
Energy storage Capacitance Inductance
Energy density we = 1
2 ǫE2 wm = 1
2 µH2
Energy dissipation Resistance
T able 7.1: A summary of the duality between electrostatics a nd magnetostatics.
7.2 Gauss’ Law for Magnetic
Fields: Integral Form
[m0018]
Gauss’ Law for Magnetic Fields (GLM) is one of the
four fundamental laws of classical electromagnetics,
collectively known as Maxwell’s Equations . Before",Electromagnetics_Vol1.pdf
"four fundamental laws of classical electromagnetics,
collectively known as Maxwell’s Equations . Before
diving in, the reader is strongly encouraged to review
Section 2.5. In that section, GLM emerges from the
“flux density” interpretation of the magnetic field.
GLM is not identified in that section, but now we are
ready for an explicit statement:
Gauss’ Law for Magnetic Fields (Equation 7.1)
states that the flux of the magnetic field through
a closed surface is zero.
This is expressed mathematically as follows:
∮
S
B · ds = 0
(7.1)
where B is magnetic flux density and S is a closed
surface with outward-pointing differential surface
normal ds. It may be useful to consider the units. B
has units of Wb/m 2; therefore, integrating B over a
surface gives a quantity with units of Wb, which is
magnetic flux, as indicated above.
GLM can also be interpreted in terms of magnetic
field lines. For the magnetic flux through a closed
surface to be zero, every field line entering the",Electromagnetics_Vol1.pdf
"GLM can also be interpreted in terms of magnetic
field lines. For the magnetic flux through a closed
surface to be zero, every field line entering the
volume enclosed by S must also exit this volume –
field lines may not begin or end within the volume.
The only way this can be true for every possible
surface S is if magnetic field lines always form closed
loops. This is in fact what we find in practice, as
shown in Figure 7.1.
Following this argument one step further, GLM
implies there can be no particular particle or structure
that can be the source of the magnetic field (because
then that would be a start point for field lines). This is
one way in which the magnetic field is very different
from the electrostatic field, for which every field line
begins at a charged particle. So, when we say that
current (for example) is the source of the magnetic
field, we mean only that the field coexists with
current, and not that the magnetic field is somehow
attached to the current. Summarizing, there is no",Electromagnetics_Vol1.pdf
"field, we mean only that the field coexists with
current, and not that the magnetic field is somehow
attached to the current. Summarizing, there is no
“localizable” quantity , analogous to charge for
electric fields, associated with magnetic fields. This is
just another way in which magnetic fields are weird!",Electromagnetics_Vol1.pdf
"148 CHAPTER 7. MAGNETOST A TICS
c⃝ Y ouming / K. Kikkeri CC BY SA 4.0
Figure 7.1: Gauss’ Law for Magnetostatics applied to
a two-dimensional bar magnet. For the surface S =
SA, every field line entering S also leaves S, so the
flux through S is zero. For the surface S = SB, every
field line within S remains in S, so the flux through S
is again zero.
Summarizing:
Gauss’ Law for Magnetic Fields requires that
magnetic field lines form closed loops. Further-
more, there is no particle that can be identified as
the source of the magnetic field.
Additional Reading:
• “Gauss’ Law for Magnetism” on Wikipedia.
• “Maxwell’s Equations” on Wikipedia.
7.3 Gauss’ Law for Magnetism:
Differential Form
[m0047]
The integral form of Gauss’ Law (Section 7.2) states
that the magnetic flux through a closed surface is
zero. In mathematical form:
∮
S
B · ds = 0 (7.2)
where B is magnetic flux density and S is the
enclosing surface. Just as Gauss’s Law for
electrostatics has both integral (Sections 5.5) and",Electromagnetics_Vol1.pdf
"where B is magnetic flux density and S is the
enclosing surface. Just as Gauss’s Law for
electrostatics has both integral (Sections 5.5) and
differential (Section 5.7) forms, so too does Gauss’
Law for Magnetic Fields. Here we are interested in
the differential form for the same reason. Given a
differential equation and the boundary conditions
imposed by structure and materials, we may then
solve for the magnetic field in very complicated
scenarios.
The equation we seek may be obtained from
Equation 7.2 using the Divergence Theorem
(Section 4.7), which in the present case may be
written: ∫
V
(∇ · B) dv=
∮
S
B · ds (7.3)
Where V is the mathematical volume bounded by the
closed surface S. From Equation 7.2 we see that the
right hand side of the equation is zero, leaving:
∫
V
(∇ · B) dv= 0 (7.4)
The above relationship must hold regardless of the
specific location or shape of V. The only way this is
possible is if the integrand is everywhere equal to
zero. W e conclude:
∇ · B = 0
(7.5)",Electromagnetics_Vol1.pdf
"specific location or shape of V. The only way this is
possible is if the integrand is everywhere equal to
zero. W e conclude:
∇ · B = 0
(7.5)
The differential (“point”) form of Gauss’ Law
for Magnetic Fields (Equation 7.5) states that the
flux per unit volume of the magnetic field is al-
ways zero.",Electromagnetics_Vol1.pdf
"7.4. AMPERE’S CIRCUIT AL LA W (MAGNETOST A TICS): INTEGRAL FO RM 149
This is another way of saying that there is no point in
space that can be considered to be the source of the
magnetic field, for if it were, then the total flux
through a bounding surface would be greater than
zero. Said yet another way , the source of the magnetic
field is not localizable.
Additional Reading:
• “Gauss’ Law for Magnetism” on Wikipedia.
7.4 Ampere’s Circuital Law
(Magnetostatics): Integral
Form
[m0019]
Ampere’s circuital law (ACL) relates current to the
magnetic field associated with the current. In the
magnetostatic regime, the law is (see also Figure 7.2):
∮
C
H · dl = Iencl
(7.6)
That is, the integral of the magnetic field intensity H
over a closed path C is equal to the current enclosed
by that path, Iencl. Before proceeding to interpret this
law , it is useful to see that it is dimensionally correct.
That is, H (units of A/m) integrated over a distance",Electromagnetics_Vol1.pdf
"law , it is useful to see that it is dimensionally correct.
That is, H (units of A/m) integrated over a distance
(units of m) yields a quantity with units of current
(i.e., A).
In general, Iencl may be either positive or negative.
The direction corresponding to positive current flow
must be correctly associated with C. The relationship
follows the right-hand rule of Stokes’ Theorem
(Section 4.9) summarized as follows. The direction of
positive Iencl is the direction in which the fingers of
the right hand intersect any surface S bordered by C
when the thumb of the right hand points in the
direction of integration. The connection to Stokes’
Theorem is not a coincidence. See Section 7.9 for
more about this.
Iencl
c⃝ K. Kikkeri CC BY SA 4.0
Figure 7.2: Reference directions for Ampere’s Cir-
cuital Law (Equation 7.6).",Electromagnetics_Vol1.pdf
"150 CHAPTER 7. MAGNETOST A TICS
Note that S can be any surface that is bounded by C –
not just the taut surface implied in Figure 7.2.
The integral form of Ampere’s Circuital Law for
magnetostatics (Equation 7.6) relates the mag-
netic field along a closed path to the total cur-
rent flowing through any surface bounded by that
path.
ACL plays a role in magnetostatics that is very
similar to the role played by the integral form of
Gauss’ Law in electrostatics (Section 5.5). That is,
ACL provides a means calculate the magnetic field
given the source current. ACL also has a similar
limitation. Generally , symmetry is required to
simplify the problem sufficiently so that the integral
equation may be solved. Fortunately , a number of
important problems fall in this category . Examples
include the problems addressed in Sections 7.5
(magnetic field of a line current), 7.6 (magnetic field
inside a straight coil), 7.7 (magnetic field of a toroidal",Electromagnetics_Vol1.pdf
"(magnetic field of a line current), 7.6 (magnetic field
inside a straight coil), 7.7 (magnetic field of a toroidal
coil), 7.8 (magnetic field of a current sheet), and 7.11
(boundary conditions on the magnetic field intensity).
For problems in which the necessary symmetry is not
available, the differential form of ACL may be
required (Section 7.9).
Finally , be aware that the form of ACL addressed here
applies to magnetostatics only . In the presence of a
time-varying electric field, the right side of ACL
includes an additional term known as the
displacement current (Section 8.9).
Additional Reading:
• “Maxwell’s Equations” on Wikipedia.
• “Ampere’s Circuital Law” on Wikipedia.
7.5 Magnetic Field of an
Infinitely-Long Straight
Current-Bearing Wire
[m0119]
In this section, we use the magnetostatic form of
Ampere’s Circuital Law (ACL) (Section 7.4) to
determine the magnetic field due to a steady current I
(units of A) in an infinitely-long straight wire. The",Electromagnetics_Vol1.pdf
"determine the magnetic field due to a steady current I
(units of A) in an infinitely-long straight wire. The
problem is illustrated in Figure 7.3. The wire is an
electrically-conducting circular cylinder of radius a.
Since the wire is a cylinder, the problem is easiest to
work in cylindrical coordinates with the wire aligned
along the zaxis.
Here’s the relevant form of ACL:
∮
C
H · dl = Iencl (7.7)
where Iencl is the current enclosed by the closed path
C. ACL works for any closed path, so to exploit the
z
y
x
a
I
ρ
c⃝ K. Kikkeri CC BY SA 4.0
Figure 7.3: Determination of the magnetic field due
to steady current in an infinitely-long straight wire.",Electromagnetics_Vol1.pdf
"7.5. MAGNETIC FIELD OF AN INFINITEL Y -LONG STRAIGHT CURRENT -BEARING WIRE 151
symmetry of the cylindrical coordinate system we
choose a circular path of radius ρin the z= 0 plane,
centered at the origin. With this choice we have
Iencl = I for ρ≥ a (7.8)
For ρ<a , we see that Iencl <I . a steady (DC)
current will be distributed uniformly throughout the
wire (Section 6.4). Since the current is uniformly
distributed over the cross section, Iencl is less than the
total current I by the same factor that the area
enclosed by C is less than πa2, the cross-sectional
area of the wire. The area enclosed by C is simply
πρ2, so we have
Iencl = Iπρ2
πa2 = Iρ2
a2 for ρ<a (7.9)
For the choice of C made above, Equation 7.7
becomes
∫ 2π
φ=0
H ·
(
ˆφρdφ
)
= Iencl (7.10)
Note that we have chosen to integrate in the +φ
direction. Therefore, the right-hand rule specifies that
positive Iencl corresponds to current flowing in the
+zdirection, which is consistent with the direction",Electromagnetics_Vol1.pdf
"positive Iencl corresponds to current flowing in the
+zdirection, which is consistent with the direction
indicated in Figure 7.3. (Here’s an excellent exercise
to test your understanding. Change the direction of
the path of integration and confirm that you get the
same result obtained at the end of this section.
Changing the direction of integration should not
change the magnetic field associated with the
current!)
The simplest way to solve for H from Equation 7.10
is to use a symmetry argument, which proceeds as
follows:
• Since the distribution of current is uniform and
infinite in the z-dimension, H can’t depend on z,
and so H · ˆz must be zero everywhere.
• The problem is identical after any amount of
rotation in φ; therefore, the magnitude of H
cannot depend on φ. This is a form of radial
symmetry . Since we determined above that H
can’t depend on zeither, it must be that the
magnitude of H can depend only on ρ.
• The radial symmetry of the problem also",Electromagnetics_Vol1.pdf
"can’t depend on zeither, it must be that the
magnitude of H can depend only on ρ.
• The radial symmetry of the problem also
requires that H · ˆρbe equal to zero. If this were
not the case, then the field would not be radially
symmetric. Since we determined above that H · ˆz
is also zero, H must be entirely ± ˆφ-directed.
From the above considerations, the most general form
of the magnetic field intensity can be written
H = ˆφH(ρ). Substituting this into Equation 7.10, we
obtain
Iencl =
∫ 2π
φ=0
[
ˆφH(ρ)
]
·
(
ˆφρdφ
)
= ρH(ρ)
∫ 2π
φ=0
dφ
= 2 πρH(ρ) (7.11)
Therefore, H(ρ) = Iencl/2πρ. Reassociating the
known direction, we obtain:
H = ˆφIencl
2πρ (7.12)
Therefore, the field outside of the wire is:
H = ˆφ I
2πρ for ρ≥ a (7.13)
whereas the field inside the wire is:
H = ˆφ Iρ
2πa2 for ρ<a (7.14)
(By the way , this is a good time for a units check.)
Note that as ρincreases from zero to a(i.e., inside the
wire), the magnetic field is proportional to ρand",Electromagnetics_Vol1.pdf
"Note that as ρincreases from zero to a(i.e., inside the
wire), the magnetic field is proportional to ρand
therefore increases. However, as ρcontinues to
increase beyond a(i.e., outside the wire), the
magnetic field is proportional to ρ−1 and therefore
decreases.
If desired, the associated magnetic flux density can be
obtained using B = µH.
Summarizing:
The magnetic field due to current in an infinite
straight wire is given by Equations 7.13 (outside
the wire) and 7.14 (inside the wire). The mag-
netic field is + ˆφ-directed for current flowing in
the +zdirection, so the magnetic field lines form
concentric circles perpendicular to and centered
on the wire.",Electromagnetics_Vol1.pdf
"152 CHAPTER 7. MAGNETOST A TICS
I
H
c⃝ Jfmelero CC BY SA 4.0 (modified)
Figure 7.4: Right-hand rule for the relationship be-
tween the direction of current and the direction of the
magnetic field.
Finally , we point out another “right-hand rule” that
emerges from this solution, shown in Figure 7.4 and
summarized below:
The magnetic field due to current in an infinite
straight wire points in the direction of the curled
fingers of the right hand when the thumb of the
right hand is aligned in the direction of current
flow .
This simple rule turns out to be handy in quickly
determining the relationship between the directions of
the magnetic field and current flow in many other
problems, and so is well worth committing to
memory .
7.6 Magnetic Field Inside a
Straight Coil
[m0120]
In this section, we use the magnetostatic integral form
of Ampere’s Circuital Law (ACL) (Section 7.4) to
determine the magnetic field inside a straight coil of
the type shown in Figure 7.5 in response to a steady",Electromagnetics_Vol1.pdf
"determine the magnetic field inside a straight coil of
the type shown in Figure 7.5 in response to a steady
(i.e., DC) current. The result has a number of
applications, including the analysis and design of
inductors, solenoids (coils that are used as magnets,
typically as part of an actuator), and as a building
block and source of insight for more complex
problems.
The present problem is illustrated in Figure 7.6. The
coil is circular with radius aand length l, and consists
of N turns (“windings”) of wire wound with uniform
winding density . Since the coil forms a cylinder, the
problem is easiest to work in cylindrical coordinates
with the axis of the coil aligned along the zaxis.
T o begin, let’s take stock of what we already know
about the answer, which is actually quite a bit. The
magnetic field deep inside the coil is generally aligned
with axis of the coil as shown in Figure 7.7. This can
be explained using the result for the magnetic field",Electromagnetics_Vol1.pdf
"with axis of the coil as shown in Figure 7.7. This can
be explained using the result for the magnetic field
due to a straight line current (Section 7.5), in which
we found that the magnetic field follows a “right-hand
rule. ” The magnetic field points in the direction of the
fingers of the right hand when the thumb of the right
hand is aligned in the direction of current flow . The
wire comprising the coil is obviously not straight, but
we can consider just one short segment of one turn
and then sum the results for all such segments. When
we consider this for a single turn of the coil, the
situation is as shown in Figure 7.8. Summing the
results for many loops, we see that the direction of the
magnetic field inside the coil must be generally in the
by Zureks (public domain)
Figure 7.5: A straight coil.",Electromagnetics_Vol1.pdf
"7.6. MAGNETIC FIELD INSIDE A STRAIGHT COIL 153
Figure 7.6: Determination of the magnetic field due
to DC current in a coil.c⃝ Geek3 CC BY SA 3.0
Figure 7.7: Magnetic field lines inside a straight coil
with closely-spaced windings. (Dotted circles repre-
sent current flowing up/out from the page; crossed cir-
cles represent current flowing down/into the page.)
+ˆz direction when the current I is applied as shown
in Figures 7.6 and 7.7. However, there is one caveat.
The windings must be sufficiently closely-spaced that
the magnetic field lines can only pass through the
openings at the end of the coil and do not take any
“shortcuts” between individual windings.
Figure 7.7 also indicates that the magnetic field lines
near the ends of the coil diverge from the axis of the
coil. This is understandable since magnetic field lines
form closed loops. The relatively complex structure
of the magnetic fields near the ends of the coil (the
“fringing field”) and outside of the coil make them",Electromagnetics_Vol1.pdf
"of the magnetic fields near the ends of the coil (the
“fringing field”) and outside of the coil make them
relatively difficult to analyze. Therefore, here we
shall restrict our attention to the magnetic field deep
inside the coil. This restriction turns out to be of little
consequence as long as l≫ a.
Also, it is apparent from the radial symmetry of the
coil that the magnitude of the magnetic field cannot
depend on φ. Putting these findings together, we find
I
c⃝ Chetvorno CC0 1.0 (modified)
Figure 7.8: Magnetic field due to a single loop.
that the most general form for the magnetic field
intensity deep inside the coil is H ≈ ˆzH(ρ). That is,
the direction of H is ±ˆz and the magnitude of H
depends, at most, on ρ. In fact, we will soon find with
the assistance of ACL that the magnitude of H
doesn’t depend on ρeither.
Here’s the relevant form of ACL:
∮
C
H · dl = Iencl (7.15)
where Iencl is the current enclosed by the closed path
C. ACL works for any closed path that encloses the",Electromagnetics_Vol1.pdf
"∮
C
H · dl = Iencl (7.15)
where Iencl is the current enclosed by the closed path
C. ACL works for any closed path that encloses the
current of interest. Also, for simplicity , we prefer a
path that lies on a constant-coordinate surface. The
selected path is shown in Figure 7.9. The benefits of
this particular path will soon become apparent.
However, note for now that this particular choice is
consistent with the right-hand rule relating the
direction of C to the direction of positive I. That is,
when I is positive, the current in the turns of the coil
pass through the surface bounded by C in the same
direction as the fingers of the right hand when the
thumb is aligned in the indicated direction of C.
Let’s define N to be the number of windings in the
coil. Then, the winding density of the coil is N/l
(turns/m). Let the path length in the zdirection be l′,
as indicated in Figure 7.9. Then the enclosed current
is
Iencl = N
l l′I (7.16)
That is, the number of turns per unit length times",Electromagnetics_Vol1.pdf
"154 CHAPTER 7. MAGNETOST A TICS
  
z
l'
ρ2
ρ1
A
D
C
B
z1
 
z2
cylindrical 
form of coil
c⃝ K. Kikkeri CC BY SA 4.0
Figure 7.9: Selected path of integration.
length gives number of turns, and this quantity times
the current through the wire is the total amount of
current crossing the surface bounded by C.
For the choice of C made above, and taking our
approximation for the form of H as exact,
Equation 7.15 becomes
∮
C
[ˆzH(ρ)] · dl = N
l l′I (7.17)
The integral consists of segments A, B, C, and D, as
shown in Figure 7.9. Let us consider the result for
each of these segments individually:
• The integral over segments Band Dis zero
because dl = ˆρdρfor these segments, and so
H · dl = 0 for these segments.
• It is also possible to make the contribution from
Segment Cgo to zero simply by letting
ρ2 → ∞ . The argument is as follows. The
magnitude of H outside the coil must decrease
with distance from the coil, so for ρsufficiently
large, H(ρ) becomes negligible. If that’s the",Electromagnetics_Vol1.pdf
"magnitude of H outside the coil must decrease
with distance from the coil, so for ρsufficiently
large, H(ρ) becomes negligible. If that’s the
case, then the integral over Segment Calso
becomes negligible.
With ρ2 → ∞ , only Segment Acontributes
significantly to the integral over C and Equation 7.17
becomes:
N
l l′I =
∫ z2
z1
[ˆzH(ρ1)] · (ˆzdz)
= H(ρ1)
∫ z2
z1
dz
= H(ρ1) [z2 − z1] (7.18)
Note z2 − z1 is simply l′. Also, we have found that
the result is independent of ρ1, as anticipated earlier.
Summarizing:
H ≈ ˆzNI
l inside coil (7.19)
Let’s take a moment to consider the implications of
this remarkably simple result.
• Note that it is dimensionally correct; that is,
current divided by length gives units of A/m,
which are the units of H.
• W e have found that the magnetic field is simply
winding density ( N/l) times current. T o increase
the magnetic field, you can either use more turns
per unit length or increase the current.
• W e have found that the magnetic field is uniform",Electromagnetics_Vol1.pdf
"the magnetic field, you can either use more turns
per unit length or increase the current.
• W e have found that the magnetic field is uniform
inside the coil; that is, the magnetic field along
the axis is equal to the magnetic field close to the
cylinder wall formed by the coil. However, this
does not apply close to ends of the coil, since we
have neglected the fringing field.
These findings have useful applications in more
complicated and practical problems, so it is
worthwhile taking note of these now . Summarizing:
The magnetic field deep inside an ideal straight
coil (Equation 7.19) is uniform and proportional
to winding density and current.
Additional Reading:
• “Electromagnetic Coil” on Wikipedia.
• “Solenoid” on Wikipedia.",Electromagnetics_Vol1.pdf
"7.7. MAGNETIC FIELD OF A TOROIDAL COIL 155
7.7 Magnetic Field of a T oroidal
Coil
[m0049]
A toroid is a cylinder in which the ends are joined to
form a closed loop. An example of a toroidal coil is
shown in Figure 7.10. T oroidal coils are commonly
used to form inductors and transformers. The
principal advantage of toroidal coils over straight
coils in these applications is magnetic field
containment – as we shall see in this section, the
magnetic field outside of a toroidal coil can be made
negligibly small. This reduces concern about
interactions between this field and other fields and
structures in the vicinity .
In this section, we use the magnetostatic form of
Ampere’s Circuital Law (ACL) (Section 7.4) to
determine the magnetic field due to a steady (DC)
current flowing through a toroidal coil. The problem
is illustrated in Figure 7.11. The toroid is circular
with inner and outer radii aand b, respectively . The
coil consists of N windings (turns) of wire wound",Electromagnetics_Vol1.pdf
"with inner and outer radii aand b, respectively . The
coil consists of N windings (turns) of wire wound
with uniform winding density . This problem is easiest
to work in cylindrical coordinates with the toroid
centered on the origin in z= 0 plane.
T o begin, let’s take stock of what we already know
about the answer, which is actually quite a bit. First, a
review of Section 7.6 (“Magnetic Field Inside a
Straight Coil”) is recommended. There it is shown
that the magnetic field deep inside a straight coil is
aligned with axis of the coil. This can be explained
c⃝ Slick CC0 1.0
Figure 7.10: A toroidal coil used as a large-value in-
ductor in the power supply of a wireless router.
x
y
a
b
I
H
Figure 7.11: Geometry of a toroidal coil.
using the result for the magnetic field due to a straight
line current (Section 7.5), in which we found that the
magnetic field follows a “right-hand rule” – The
magnetic field points in the direction of the curled",Electromagnetics_Vol1.pdf
"magnetic field follows a “right-hand rule” – The
magnetic field points in the direction of the curled
fingers of the right hand when the thumb of the right
hand is aligned in the direction of current flow . The
wire comprising the coil is obviously not straight, but
we can consider just one short segment of one turn
and then sum the results for all such segments. When
we do this, we see that the direction of the magnetic
field inside the coil must be in the + ˆφdirection when
the current Iis applied as shown in Figure 7.11. Also,
because the problem is identical after any amount of
rotation around the zaxis, the magnitude of the
magnetic field cannot depend on φ. Putting these
findings together, we find that the most general form
for the magnetic field intensity inside or outside the
coil is H = ˆφH(ρ,z).
Here’s the relevant form of ACL:
∮
C
H · dl = Iencl (7.20)
where Iencl is the current enclosed by the closed path
C. ACL works for any closed path, but we need one",Electromagnetics_Vol1.pdf
"∮
C
H · dl = Iencl (7.20)
where Iencl is the current enclosed by the closed path
C. ACL works for any closed path, but we need one
that encloses some current so as to obtain a
relationship between I and H. Also, for simplicity ,
we prefer a path that lies on a constant-coordinate
surface. The selected path is a circle of radius ρ",Electromagnetics_Vol1.pdf
"156 CHAPTER 7. MAGNETOST A TICS
x
y
a
b
I
ρ
C
Figure 7.12: Selected path of integration.
centered on the origin in the z= z0 plane, as shown
in Figure 7.12. W e further require C to lie entirely
inside the coil, which ensures that the enclosed
current includes the current of all the windings as they
pass through the hole at the center of the toroid. W e
choose the direction of integration to be in the +φ
direction, which is consistent with the indicated
direction of positive I. That is, when I is positive, the
current in the windings of the coil pass through the
surface bounded by C in the same direction as the
curled fingers of the right hand when the thumb is
aligned in the indicated direction of C.
In terms of the variables we have defined, the
enclosed current is simply
Iencl = NI (7.21)
Equation 7.20 becomes
∮
C
[
ˆφH(ρ,z0)
]
· dl = NI (7.22)
Now evaluating the integral:
NI =
∫ 2π
0
[
ˆφH(ρ,z0)
]
·
(
ˆφρdφ
)
= ρH(ρ,z0)
∫ 2π
0
dφ
= 2 πρH(ρ,z0) (7.23)",Electromagnetics_Vol1.pdf
"∮
C
[
ˆφH(ρ,z0)
]
· dl = NI (7.22)
Now evaluating the integral:
NI =
∫ 2π
0
[
ˆφH(ρ,z0)
]
·
(
ˆφρdφ
)
= ρH(ρ,z0)
∫ 2π
0
dφ
= 2 πρH(ρ,z0) (7.23)
It is now clear that the result is independent of z0.
Summarizing:
H = ˆφNI
2πρ inside coil (7.24)
Let’s take a moment to consider the implications of
this result.
• Note that it is dimensionally correct; that is,
current divided by the circumference of C (2πρ)
gives units of A/m, which are the units of H.
• W e have found that the magnetic field is
proportional to winding density (i.e., number of
windings divided by circumference) times
current. T o increase the magnetic field you can
either use more windings or increase the current.
• Remarkably , we have found that the magnitude
of the magnetic field inside the coil depends only
on ρ; i.e., the distance from the central (here, z)
axis. It is independent of z.
Summarizing:
The magnetic field inside a toroidal coil (Equa-
tion 7.24) depends only on distance from the cen-",Electromagnetics_Vol1.pdf
"axis. It is independent of z.
Summarizing:
The magnetic field inside a toroidal coil (Equa-
tion 7.24) depends only on distance from the cen-
tral axis and is proportional to winding density
and current.
Now let us consider what happens outside the coil.
For this, we consider any path of integration ( C) that
lies completely outside the coil. Note that any such
path encloses no current and therefore Iencl = 0 for
any such path. In this case we have:
∮
C
H · dl = 0 (7.25)
There are two ways this could be true. Either H could
be zero everywhere along the path, or H could be
non-zero along the path in such a way that the integral
windings out to be zero. The radial symmetry of the
problem rules out the second possibility – if H is
radially symmetric and C is radially symmetric, then
the sign of H · dl should not change over C.
Therefore:",Electromagnetics_Vol1.pdf
"7.8. MAGNETIC FIELD OF AN INFINITE CURRENT SHEET 157
The magnetic field everywhere outside an ideal
toroidal coil is zero.
Note the caveat signaled by the use of the adjective
“ideal. ” In a practical toroidal coil, we expect there
will be some leakage of magnetic flux between the
windings. In practice, this leakage can be made
negligibly small by using a sufficiently high winding
density and winding the wire on material on a toroidal
form (a “core”) having sufficiently large permeability .
The use of a high-permeability core, as shown in
Figure 7.10, will dramatically improve the already
pretty-good containment. In fact, the use of such a
core allows the spacing between windings to become
quite large before leakage becomes significant.
One final observation about toroidal coils is that at no
point in the derivation of the magnetic field did we
need to consider the cross-sectional shape of the coil;
we merely needed to know whether C was inside or
outside the coil. Therefore:",Electromagnetics_Vol1.pdf
"need to consider the cross-sectional shape of the coil;
we merely needed to know whether C was inside or
outside the coil. Therefore:
The magnetic field inside an ideal toroidal coil
does not depend on the cross-sectional shape of
the coil.
Additional Reading:
• “T oroidal Inductors and Transformers” on
Wikipedia.
7.8 Magnetic Field of an Infinite
Current Sheet
[m0121]
W e now consider the magnetic field due to an infinite
sheet of current, shown in Figure 7.13. The solution
to this problem is useful as a building block and
source of insight in more complex problems, as well
as being a useful approximation to some practical
problems involving current sheets of finite extent
including, for example, microstrip transmission line
and ground plane currents in printed circuit boards.
The current sheet in Figure 7.13 lies in the z= 0
plane and the current density is Js = ˆxJs (units of
A/m); i.e., the current is uniformly distributed such
that the total current crossing any segment of width",Electromagnetics_Vol1.pdf
"A/m); i.e., the current is uniformly distributed such
that the total current crossing any segment of width
∆ yalong the ydirection is Js∆ y.
T o begin, let’s take stock of what we already know
about the answer, which is actually quite a bit. For
example, imagine the current sheet as a continuum of
thin strips parallel to the xaxis and very thin in the y
dimension. Each of these strips individually behaves
like a straight line current I = Js∆ y(units of A). The
magnetic field due to each of these strips is
determined by a “right-hand rule” – the magnetic field
points in the direction of the curled fingers of the right
hand when the thumb of the right hand is aligned in
the direction of current flow . (Section 7.5). It is
apparent from this much that H can have no ˆy
component, since the field of each individual strip has
z
C
JsLz
Ly
y
Figure 7.13: Analysis of the magnetic field due to an
infinite thin sheet of current.",Electromagnetics_Vol1.pdf
"158 CHAPTER 7. MAGNETOST A TICS
no ˆy component. When the magnetic field due to each
strip is added to that of all the other strips, the ˆz
component of the sum field must be zero due to
symmetry . It is also clear from symmetry
considerations that the magnitude of H cannot
depend on xor y. Summarizing, we have determined
that the most general form for H is ˆyH(z), and
furthermore, the sign of H(z) must be positive for
z <0 and negative for z >0.
It’s possible to solve this problem by actually
summing over the continuum of thin current strips as
imagined above. 1 However, it’s far easier to use
Ampere’s Circuital Law (ACL; Section 7.4). Here’s
the relevant form of ACL:
∮
C
H · dl = Iencl (7.26)
where Iencl is the current enclosed by a closed path C.
ACL works for any closed path, but we need one that
encloses some current so as to obtain a relationship
between Js and H. Also, for simplicity , we prefer a
path that lies on a constant-coordinate surface. A",Electromagnetics_Vol1.pdf
"between Js and H. Also, for simplicity , we prefer a
path that lies on a constant-coordinate surface. A
convenient path in this problem is a rectangle lying in
the x= 0 plane and centered on the origin, as shown
in Figure 7.13. W e choose the direction of integration
to be counter-clockwise from the perspective shown
in Figure 7.13, which is consistent with the indicated
direction of positive Js according to the applicable
right-hand rule from Stokes’ Theorem. That is, when
Js is positive (current flowing in the +ˆx direction),
the current passes through the surface bounded by C
in the same direction as the curled fingers of the right
hand when the thumb is aligned in the indicated
direction of C.
Let us define Ly to be the width of the rectangular
path of integration in the ydimension and Lz to be
the width in the zdimension. In terms of the variables
we have defined, the enclosed current is simply
Iencl = JsLy (7.27)
Equation 7.26 becomes
∮
C
[ˆyH(z)] · dl = JsLy (7.28)",Electromagnetics_Vol1.pdf
"we have defined, the enclosed current is simply
Iencl = JsLy (7.27)
Equation 7.26 becomes
∮
C
[ˆyH(z)] · dl = JsLy (7.28)
Note that H · dl = 0 for the vertical sides of the path,
since H is ˆy-directed and dl = ˆzdzon those sides.
1 In fact, this is pretty good thing to try , if for no other reaso n than
to see how much simpler it is to use ACL instead.
Therefore, only the horizontal sides contribute to the
integral and we have:
∫ +Ly /2
−Ly /2
[
ˆyH
(
−Lz
2
)]
· (ˆydy)
+
∫ −Ly /2
+Ly /2
[
ˆyH
(
+Lz
2
)]
· (ˆydy) = JsLy (7.29)
Now evaluating the integrals:
H
(
−Lz
2
)
Ly − H
(
+Lz
2
)
Ly = JsLy (7.30)
Note that all factors of Ly cancel in the above
equation. Furthermore, H(−Lz/2) = −H(+Lz/2)
due to (1) symmetry between the upper and lower
half-spaces and (2) the change in sign between these
half-spaces, noted earlier. W e use this to eliminate
H(+Lz/2) and solve for H(−Lz/2) as follows:
2H(−Lz/2) = Js (7.31)
yielding
H(−Lz/2) = + Js
2 (7.32)
and therefore
H(+Lz/2) = −Js
2 (7.33)",Electromagnetics_Vol1.pdf
"H(+Lz/2) and solve for H(−Lz/2) as follows:
2H(−Lz/2) = Js (7.31)
yielding
H(−Lz/2) = + Js
2 (7.32)
and therefore
H(+Lz/2) = −Js
2 (7.33)
Furthermore, note that H is independent of Lz; for
example, the result we just found indicates the same
value of H(+Lz/2) regardless of the value of Lz.
Therefore, H is uniform throughout all space, except
for the change of sign corresponding for the field
above vs. below the sheet. Summarizing:
H = ±ˆy Js
2 for z≶ 0 (7.34)
The magnetic field intensity due to an infinite
sheet of current (Equation 7.34) is spatially uni-
form except for a change of sign corresponding
for the field above vs. below the sheet.",Electromagnetics_Vol1.pdf
"7.9. AMPERE’S LA W (MAGNETOST A TICS): DIFFERENTIAL FORM 159
7.9 Ampere’s Law
(Magnetostatics): Differential
Form
[m0118]
The integral form of Amperes’ Circuital Law (ACL;
Section 7.4) for magnetostatics relates the magnetic
field along a closed path to the total current flowing
through any surface bounded by that path. In
mathematical form:∮
C
H · dl = Iencl (7.35)
where H is magnetic field intensity , C is the closed
curve, and Iencl is the total current flowing through
any surface bounded by C. In this section, we derive
the differential form of this equation. In some
applications, this differential equation, combined with
boundary conditions associated with discontinuities in
structure and materials, can be used to solve for the
magnetic field in arbitrarily complicated scenarios. A
more direct reason for seeking out this differential
equation is that we gain a little more insight into the
relationship between current and the magnetic field,
disclosed at the end of this section.",Electromagnetics_Vol1.pdf
"equation is that we gain a little more insight into the
relationship between current and the magnetic field,
disclosed at the end of this section.
The equation we seek may be obtained using Stokes’
Theorem (Section 4.9), which in the present case may
be written: ∫
S
(∇ × H) · ds =
∮
C
H · dl (7.36)
where S is any surface bounded by C, and ds is the
differential surface area combined with the unit vector
in the direction determined by the right-hand rule
from Stokes’ Theorem. ACL tells us that the right
side of the above equation is simply Iencl. W e may
express Iencl as the integral of the volume current
density J (units of A/m 2; Section 6.2) as follows:
Iencl =
∫
S
J · ds (7.37)
so we may rewrite Equation 7.36 as follows:
∫
S
(∇ × H) · ds =
∫
S
J · ds (7.38)
The above relationship must hold regardless of the
specific location or shape of S. The only way this is
possible for all possible surfaces in all applicable
scenarios is if the integrands are equal. Thus, we",Electromagnetics_Vol1.pdf
"possible for all possible surfaces in all applicable
scenarios is if the integrands are equal. Thus, we
obtain the desired expression:
∇ × H = J
(7.39)
That is, the curl of the magnetic field intensity at a
point is equal to the volume current density at that
point. Recalling the properties of the curl operator
(Section 4.8) – in particular, that curl involves
derivatives with respect to direction – we conclude:
The differential form of Ampere’s Circuital Law
for magnetostatics (Equation 7.39) indicates that
the volume current density at any point in space
is proportional to the spatial rate of change of the
magnetic field and is perpendicular to the mag-
netic field at that point.
Additional Reading:
• “Ampere’s circuital law” on Wikipedia.
• “Boundary value problem” on Wikipedia.",Electromagnetics_Vol1.pdf
"160 CHAPTER 7. MAGNETOST A TICS
7.10 Boundary Conditions on the
Magnetic Flux Density (B)
[m0022]
In homogeneous media, electromagnetic quantities
vary smoothly and continuously . At an interface
between dissimilar media, however, it is possible for
electromagnetic quantities to be discontinuous.
Continuities and discontinuities in fields can be
described mathematically by boundary conditions and
used to constrain solutions for fields away from these
interfaces.
In this section, we derive the boundary condition on
the magnetic flux density B at a smooth interface
between two material regions, as shown in
Figure 7.14. 2 The desired boundary condition may be
obtained from Gauss’ Law for Magnetic Fields
(GLM; Section 7.2):
∮
S
B · ds = 0 (7.40)
where S is any closed surface. Let S take the form of
cylinder centered at a point on the interface, and for
which the flat ends are parallel to the surface and
perpendicular to ˆn, as shown in Figure 7.14. Let the",Electromagnetics_Vol1.pdf
"which the flat ends are parallel to the surface and
perpendicular to ˆn, as shown in Figure 7.14. Let the
radius of this cylinder be a, and let the length of the
2 It may be helpful to note the similarity (duality , in fact) betw een
this derivation and the derivation of the associated bounda ry condi-
tion on D presented in Section 5.18.
region 1
region 2
n
a
h
h
c⃝ K. Kikkeri CC BY SA 4.0
Figure 7.14: Determination of the boundary condition
on B at the interface between material regions.
cylinder be 2h. From GLM, we have
∮
S
B · ds =
∫
top
B · ds
+
∫
side
B · ds
+
∫
bottom
B · ds = 0 (7.41)
Now let us reduce hand atogether while (1)
maintaining a constant ratio h/a≪ 1 and (2) keeping
S centered on the interface. Because h≪ a, the area
of the side can be made negligible relative to the area
of the top and bottom. Then, as h→ 0, we are left
with
∫
top
B · ds +
∫
bottom
B · ds → 0 (7.42)
As the area of the top and bottom sides become
infinitesimal, the variation in B over these areas",Electromagnetics_Vol1.pdf
"with
∫
top
B · ds +
∫
bottom
B · ds → 0 (7.42)
As the area of the top and bottom sides become
infinitesimal, the variation in B over these areas
becomes negligible. Now we have simply:
B1 · ˆn∆ A+ B2 · (−ˆn) ∆ A→ 0 (7.43)
where B1 and B2 are the magnetic flux densities at
the interface but in regions 1 and 2, respectively , and
∆ Ais the area of the top and bottom sides. Note that
the orientation of ˆn is important – we have assumed ˆn
points into region 1, and we must now stick with this
choice. Thus, we obtain
ˆn · (B1 − B2) = 0
(7.44)
where, as noted above, ˆn points into region 1.
Summarizing:
The normal (perpendicular) component of B
across the boundary between two material re-
gions is continuous.
It is worth noting what this means for the magnetic
field intensity H. Since B = µH, it must be that
The normal (perpendicular) component of H
across the boundary between two material re-
gions is discontinuous if the permeabilities are
unequal.",Electromagnetics_Vol1.pdf
"7.11. BOUNDAR Y CONDITIONS ON THE MAGNETIC FIELD INTENSITY ( H) 161
7.11 Boundary Conditions on the
Magnetic Field Intensity (H)
[m0023]
In homogeneous media, electromagnetic quantities
vary smoothly and continuously . At a boundary
between dissimilar media, however, it is possible for
electromagnetic quantities to be discontinuous.
Continuities and discontinuities in fields can be
described mathematically by boundary conditions and
used to constrain solutions for fields away from these
boundaries. In this section, we derive boundary
conditions on the magnetic field intensity H.
T o begin, consider a region consisting of only two
media that meet at a smooth boundary as shown in
Figure 7.15. The desired boundary condition can be
obtained directly from Ampere’s Circuital Law (ACL;
Section 7.4): ∮
C
H · dl = Iencl (7.45)
where C is any closed path and Iencl is the current
that flows through the surface bounded by that path in
the direction specified by the “right-hand rule” of",Electromagnetics_Vol1.pdf
"that flows through the surface bounded by that path in
the direction specified by the “right-hand rule” of
Stokes’ theorem. Let C take the form of a rectangle
centered on a point on the boundary as shown in
Figure 7.15, perpendicular to the direction of current
flow at that location. Let the sides A, B, C, and Dbe
perpendicular and parallel to the boundary . Let the
length of the parallel sides be l, and let the length of
  
region 1
A
D w
tn
Figure 7.15: Determining the boundary condition on
H at the smooth boundary between two material re-
gions.
the perpendicular sides be w. Now we apply ACL.
W e must integrate in a counter-clockwise direction in
order to be consistent with the indicated reference
direction for Js. Thus:
∮
H · dl =
∫
A
H · dl
+
∫
B
H · dl
+
∫
C
H · dl
+
∫
D
H · dl = Iencl (7.46)
Now we let wand lbecome vanishingly small while
(1) maintaining the ratio l/wand (2) keeping C
centered on the boundary . In this process, the
contributions from the Band Dsegments become",Electromagnetics_Vol1.pdf
"(1) maintaining the ratio l/wand (2) keeping C
centered on the boundary . In this process, the
contributions from the Band Dsegments become
equal in magnitude but opposite in sign; i.e.,
∫
B
H · dl +
∫
D
H · dl → 0 (7.47)
This leaves
∫
A
H · dl +
∫
C
H · dl → Iencl (7.48)
Let us define the unit vector ˆt (“tangent”) as shown in
Figure 7.15. Now we have simply:
H1 ·
(
−ˆt∆ l
)
+ H2 ·
(
+ˆt∆ l
)
= Iencl (7.49)
where H1 and H2 are the fields evaluated on the two
sides of the boundary , and ∆ l→ 0 is the length of
sides Aand C. As always, Iencl (units of A) may be
interpreted as the flux of the current density Js (units
of A/m) flowing past a line on the surface having
length ∆ l(units of m) perpendicular to ˆt × ˆn, where
ˆn is the normal to the surface, pointing into Region 1.
Stated mathematically:
Iencl → Js ·
(
∆ lˆt × ˆn
)
(7.50)
Before proceeding, note this is true regardless of the
particular direction we selected for ˆt; it is only
necessary that ˆt be tangent to the boundary . Thus,",Electromagnetics_Vol1.pdf
"particular direction we selected for ˆt; it is only
necessary that ˆt be tangent to the boundary . Thus,
ˆt × ˆn need not necessarily be in the same direction as
Js. Now Equation 7.49 can be written:
H2 · ˆt∆ l− H1 · ˆt∆ l= Js ·
( ˆt × ˆn
)
∆ l (7.51)",Electromagnetics_Vol1.pdf
"162 CHAPTER 7. MAGNETOST A TICS
Eliminating the common factor of ∆ land arranging
terms on the left:
(H2 − H1) · ˆt = Js ·
( ˆt × ˆn
)
(7.52)
The right side may be transformed using a vector
identity (Equation B.18 of Section B.3) to obtain:
(H2 − H1) · ˆt = ˆt · (ˆn × Js) (7.53)
Equation 7.53 is the boundary condition we seek. W e
have found that the component of H2 − H1 (the
difference between the magnetic field intensities at
the boundary) in any direction tangent to the
boundary is equal to the component of the current
density flowing in the perpendicular direction ˆn × Js.
Said differently:
A discontinuity in the tangential component of
the magnetic field intensity at the boundary must
be supported by surface current flowing in a di-
rection perpendicular to this component of the
field.
An important consequence is that:
If there is no surface current, then the tangential
component of the magnetic field intensity is con-
tinuous across the boundary .",Electromagnetics_Vol1.pdf
"If there is no surface current, then the tangential
component of the magnetic field intensity is con-
tinuous across the boundary .
It is possible to obtain a mathematical form of the
boundary condition that is more concise and often
more useful than Equation 7.53. This form may be
obtained as follows. First, we note that the dot
product with respect to ˆt on both sides of
Equation 7.53 means simply “any component that is
tangent to the boundary . ” W e need merely to make
sure we are comparing the same tangential
component on each side of the equation. For example
ˆn × (H2 − H1) is tangential to the boundary , since ˆn
is perpendicular to the boundary and therefore any
cross product involving ˆn will be perpendicular to ˆn.
The corresponding component of the current density
is ˆn × (ˆn × Js), so Equation 7.53 may be
equivalently written as follows:
ˆn × (H2 − H1) = ˆn × ˆn × Js (7.54)
Applying a vector identity (Equation B.19 of
Section B.3) to the right side of Equation 7.54 we",Electromagnetics_Vol1.pdf
"ˆn × (H2 − H1) = ˆn × ˆn × Js (7.54)
Applying a vector identity (Equation B.19 of
Section B.3) to the right side of Equation 7.54 we
obtain:
ˆn × ˆn × Js = ˆn (ˆn · Js) − Js(ˆn · ˆn)
= ˆn (0) − Js(1)
= −Js (7.55)
Therefore:
ˆn × (H2 − H1) = −Js (7.56)
The minus sign on the right can be eliminated by
swapping H2 and H1 on the left, yielding
ˆn × (H1 − H2) = Js
(7.57)
This is the form in which the boundary condition is
most commonly expressed.
It is worth noting what this means for the magnetic
field intensity B. Since B = µH:
In the absence of surface current, the tangential
component of B across the boundary between
two material regions is discontinuous if the per-
meabilities are unequal.",Electromagnetics_Vol1.pdf
"7.12. INDUCT ANCE 163
7.12 Inductance
[m0123]
Current creates a magnetic field, which subsequently
exerts force on other current-bearing structures. For
example, the current in each winding of a coil exerts a
force on every other winding of the coil. If the
windings are fixed in place, then this force is unable
to do work (i.e., move the windings), so instead the
coil stores potential energy . This potential energy can
be released by turning off the external source. When
this happens, charge continues to flow , but is now
propelled by the magnetic force. The potential energy
that was stored in the coil is converted to kinetic
energy and subsequently used to redistribute the
charge until no current flows. At this point, the
inductor has expended its stored energy . T o restore
energy , the external source must be turned back on,
restoring the flow of charge and thereby restoring the
magnetic field.
Now recall that the magnetic field is essentially",Electromagnetics_Vol1.pdf
"restoring the flow of charge and thereby restoring the
magnetic field.
Now recall that the magnetic field is essentially
defined in terms of the force associated with this
potential energy; i.e., F = qv × B where qis the
charge of a particle comprising the current, v is the
velocity of the particle, and B is magnetic flux density
(Section 2.5). So, rather than thinking of the potential
energy of the system as being associated with the
magnetic force applied to current, it is equally valid to
think of the potential energy as being stored in the
magnetic field associated with the current distribution.
The energy stored in the magnetic field depends on
the geometry of the current-bearing structure and the
permeability of the intervening material because the
magnetic field depends on these parameters.
The relationship between current applied to a
structure and the energy stored in the associated
magnetic field is what we mean by inductance. W e
may fairly summarize this insight as follows:",Electromagnetics_Vol1.pdf
"structure and the energy stored in the associated
magnetic field is what we mean by inductance. W e
may fairly summarize this insight as follows:
Inductance is the ability of a structure to store en-
ergy in a magnetic field.
The inductance of a structure depends on the ge-
ometry of its current-bearing structures and the
permeability of the intervening medium.
Note that inductance does not depend on current,
which we view as either a stimulus or response from
this point of view . The corresponding response or
stimulus, respectively , is the magnetic flux associated
with this current. This leads to the following
definition:
L= Φ
I (single linkage) (7.58)
where Φ (units of Wb) is magnetic flux, I (units of
A) is the current responsible for this flux, and L(units
of H) is the associated inductance. (The “single
linkage” caveat will be explained below .) In other
words, a device with high inductance generates a
large magnetic flux in response to a given current, and",Electromagnetics_Vol1.pdf
"words, a device with high inductance generates a
large magnetic flux in response to a given current, and
therefore stores more energy for a given current than a
device with lower inductance.
T o use Equation 7.58 we must carefully define what
we mean by “magnetic flux” in this case. Generally ,
magnetic flux is magnetic flux density (again, B, units
of Wb/m 2) integrated over a specified surface S, so
Φ =
∫
S
B · ds (7.59)
where ds is the differential surface area vector, with
direction normal to S. However, this leaves
unanswered the following questions: Which S, and
which of the two possible normal directions of ds?
For a meaningful answer, S must uniquely associate
the magnetic flux to the associated current. Such an
association exists if we require the current to form a
closed loop. This is shown in Figure 7.16. Here C is
the closed loop along which the current flows, S is a
surface bounded by C, and the direction of ds is
defined according to the right-hand rule of Stokes’",Electromagnetics_Vol1.pdf
"surface bounded by C, and the direction of ds is
defined according to the right-hand rule of Stokes’
Theorem (Section 4.9). Note that C can be a closed
loop of any shape; i.e., not just circular, and not
restricted to lying in a plane. Further note that S used
in the calculation of Φ can be any surface bounded by
C. This is because magnetic field lines form closed
loops such that any one magnetic field line intersects
any open surface bounded by C exactly once. Such an
intersection is sometimes called a “linkage. ” So there
we have it – we require the current I to form a closed
loop, we measure the magnetic flux through this loop
using the sign convention of the right-hand rule, and
the ratio is the inductance.
Many structures consist of multiple such loops – the
coil is of course one of these. In a coil, each winding",Electromagnetics_Vol1.pdf
"164 CHAPTER 7. MAGNETOST A TICS
I
B
ds
c⃝ K. Kikkeri CC BY SA 4.0
Figure 7.16: Association between a closed loop of
current and the associated magnetic flux.
carries the same current, and the magnetic fields of the
windings add to create a magnetic field, which grows
in proportion to the winding density (Section 7.6).
The magnetic flux density inside a coil is proportional
to the number of windings, N, so the flux Φ in
Equation 7.58 should properly be indicated as NΦ .
Another way to look at this is that we are counting the
number of times the same current is able to generate a
unique set of magnetic field lines that intersect S.
Summarizing, our complete definition for inductance
is
L= NΦ
I (identical linkages) (7.60)
An engineering definition of inductance is Equa-
tion 7.60, with the magnetic flux defined to be
that associated with a single closed loop of cur-
rent with sign convention as indicated in Fig-
ure 7.16, and Ndefined to be the number of times",Electromagnetics_Vol1.pdf
"that associated with a single closed loop of cur-
rent with sign convention as indicated in Fig-
ure 7.16, and Ndefined to be the number of times
the same current I is able to create that flux.
What happens if the loops have different shapes? For
example, what if the coil is not a cylinder, but rather
cone-shaped? (Y es, there is such a thing – see
“ Additional Reading” at the end of this section.) In
this case, one needs a better way to determine the
factor NΦ since the flux associated with each loop of
current will be different. However, this is beyond the
scope of this section.
An inductor is a device that is designed to exhibit a
specified inductance. W e can now make the
connection to the concept of the inductor as it appears
in elementary circuit theory . First, we rewrite
Equation 7.60 as follows:
I = NΦ
L (7.61)
T aking the derivative of both sides of this equation
with respect to time, we obtain:
d
dtI = N
L
d
dtΦ (7.62)
Now we need to reach beyond the realm of",Electromagnetics_Vol1.pdf
"with respect to time, we obtain:
d
dtI = N
L
d
dtΦ (7.62)
Now we need to reach beyond the realm of
magnetostatics for just a moment. Section 8.3
(“Faraday’s Law”) shows that the change in Φ
associated with a change in current results in the
creation of an electrical potential equal to −NdΦ /dt
realized over the loop C. In other words, the terminal
voltage V is +NdΦ /dt, with the change of sign
intended to keep the result consistent with the sign
convention relating current and voltage in passive
devices. Therefore, dΦ /dtin Equation 7.62 is equal
to V/N. Making the substitution we find:
V = Ld
dtI (7.63)
This is the expected relationship from elementary
circuit theory .
Another circuit theory concept related to inductance
is mutual inductance . Whereas inductance relates
changes in current to instantaneous voltage in the
same device (Equation 7.63), mutual inductance
relates changes in current in one device to
instantaneous voltage in a different device. This can",Electromagnetics_Vol1.pdf
"same device (Equation 7.63), mutual inductance
relates changes in current in one device to
instantaneous voltage in a different device. This can
occur when the two devices are coupled (“linked”) by
the same magnetic field. For example, transformers
(Section 8.5) typically consist of separate coils that
are linked by the same magnetic field lines. The
voltage across one coil may be computed as the
time-derivative of current on the other coil times the
mutual inductance.
Let us conclude this section by taking a moment to
dispel a common misconception about inductance.
The misconception pertains to the following question.
If the current does not form a closed loop, what is the
inductance? For example, engineers sometimes refer",Electromagnetics_Vol1.pdf
"7.13. INDUCT ANCE OF A STRAIGHT COIL 165
to the inductance of a pin or lead of an electronic
component. A pin or lead is not a closed loop, so the
formal definition of inductance given above – ratio of
magnetic flux to current – does not apply . The
broader definition of inductance – the ability to store
energy in a magnetic field – does apply , but this is not
what is meant by “pin inductance” or “lead
inductance. ” What is actually meant is the imaginary
part of the impedance of the pin or lead – i.e., the
reactance – expressed as an equivalent inductance. In
other words, the reactance of an inductive device is
positive, so any device that also exhibits a positive
reactance can be viewed from a circuit theory
perspective as an equivalent inductance. This is not
referring to the storage of energy in a magnetic field;
it merely means that the device can be modeled as an
inductor in a circuit diagram. In the case of “pin
inductance, ” the culprit is not actually inductance, but",Electromagnetics_Vol1.pdf
"inductor in a circuit diagram. In the case of “pin
inductance, ” the culprit is not actually inductance, but
rather skin effect (see “ Additional References” at the
end of this section). Summarizing:
Inductance implies positive reactance, but posi-
tive reactance does not imply the physical mech-
anism of inductance.
Additional Reading:
• “Inductance” on Wikipedia.
• “Inductor” on Wikipedia.
• T .A. Winslow ,
“Conical Inductors for Broadband Applications, ”
IEEE Microwave Mag. , V ol. 6, No. 1, Mar 2005,
pp. 68–72.
• “Skin Effect” on Wikipedia.
7.13 Inductance of a Straight
Coil
[m0124]
In this section, we determine the inductance of a
straight coil, as shown in Figure 7.17. The coil is
circular with radius aand length land consists of N
windings of wire wound with uniform winding
density . Also, we assume the winding density N/lis
large enough that magnetic field lines cannot enter or
exit between windings but rather must traverse the
entire length of the coil. Since the coil forms a",Electromagnetics_Vol1.pdf
"exit between windings but rather must traverse the
entire length of the coil. Since the coil forms a
cylinder, the problem is easiest to work in cylindrical
coordinates with the axis of the coil aligned along the
zaxis.
Inductance Lin this case is given by (Section 7.12)
L= NΦ
I (7.64)
where I is current and Φ is the magnetic flux
associated with one winding of the coil. Magnetic
flux in this case is given by
Φ =
∫
S
B · ds (7.65)
where B is the magnetic flux density (units of T =
Wb/m2), S is the surface bounded by a single current
loop, and ds points in the direction determined by the
right hand rule with respect to the direction of
positive current flow .
First, let’s determine the magnetic field. The magnetic
Figure 7.17: Determination of the inductance of a
straight coil.",Electromagnetics_Vol1.pdf
"166 CHAPTER 7. MAGNETOST A TICS
flux density deep inside the coil is (Section 7.6):
B ≈ ˆzµNI
l (7.66)
Is it reasonable to use this approximation here? Since
inductance pertains to energy storage, the question is
really what fraction of the energy is stored in a field
that is well-described by this approximation, as
opposed to energy stored in the “fringing field” close
to the ends of the coil. If we make lsufficiently large
relative to a, then presumably energy storage in the
fringing field will be negligible in comparison. Since
the alternative leads to a much more complicated
problem, we shall assume that Equation 7.66 is valid
for the interior of the coil.
Next, we determine Φ . In this case, a natural choice
for S is the interior cross-section of the coil in a plane
perpendicular to the axis. The direction of ds must be
+ˆz since this is the direction in which the fingers of
the right hand point when the current flows in the
direction indicated in Figure 7.17. Thus, we have
Φ ≈
∫
S
(",Electromagnetics_Vol1.pdf
"the right hand point when the current flows in the
direction indicated in Figure 7.17. Thus, we have
Φ ≈
∫
S
(
ˆzµNI
l
)
· (ˆzds)
= µNI
l
∫
S
ds
= µNI
l A (7.67)
where Ais the cross-sectional area of the coil.
Finally from Equation 7.64 we obtain
L≈ µN2A
l (l≫ a) (7.68)
Note that this is dimensionally correct; that is,
permeability (units of H/m) times area (units of m 2)
divided by length (units of m) gives units of H, as
expected. Also, it is worth noting that inductance is
proportional to permeability and cross-sectional area,
and inversely proportional to length. Interestingly the
inductance is proportional to N2 as opposed to N;
this is because field strength increases with N, and
independently there are N flux linkages. Finally , we
note that the inductance does not depend on the shape
of the coil cross-section, but only on the area of the
cross-section. Summarizing:
The inductance of a long straight coil is given ap-
proximately by Equation 7.68.",Electromagnetics_Vol1.pdf
"cross-section. Summarizing:
The inductance of a long straight coil is given ap-
proximately by Equation 7.68.
Again, this result is approximate because it neglects
the non-uniform fringing field near the ends of the
coil and the possibility that magnetic field lines
escape between windings due to inadequate winding
density . Nevertheless, this result facilitates useful
engineering analysis and design.
Additional Reading:
• “Inductance” on Wikipedia.",Electromagnetics_Vol1.pdf
"7.14. INDUCT ANCE OF A COAXIAL STRUCTURE 167
7.14 Inductance of a Coaxial
Structure
[m0125]
Let us now determine the inductance of coaxial
structure, shown in Figure 7.18. The inductance of
this structure is of interest for a number of reasons –
in particular, for determining the characteristic
impedance of coaxial transmission line, as addressed
in Section 3.10.
For our present purpose, we may model the structure
as shown in Figure 7.18. This model consists of two
concentric perfectly-conducting cylinders of radii a
and b, separated by a homogeneous material having
permeability µ. T o facilitate analysis, let us place the
+zaxis along the common axis of the concentric
cylinders, so that the cylinders may be described as
the surfaces ρ= aand ρ= b.
Below we shall find the inductance by assuming a
current I on the inner conductor and integrating over
the resulting magnetic field to obtain the magnetic
flux Φ between the conductors. Then, inductance can",Electromagnetics_Vol1.pdf
"the resulting magnetic field to obtain the magnetic
flux Φ between the conductors. Then, inductance can
be determined as the ratio of the response flux to the
source current.
Before we get started, note the derivation we are
about to do is similar to the derivation of the
capacitance of a coaxial structure, addressed in
Section 5.24. The reader may benefit from a review of
that section before attempting this derivation.
b
a
z
ρ
z=l
I
μ
z=0
H
Figure 7.18: Determining the inductance of coaxial
line.
The first step is to find the magnetic field inside the
structure. This is relatively simple if we may neglect
fringing fields, since then the internal field may be
assumed to be constant with respect to z. This
analysis will also apply to the case where the length l
pertains to one short section of a much longer
structure; in this case we will obtain the inductance
per length as opposed to the total inductance for the
structure. Note that the latter is exactly what we need",Electromagnetics_Vol1.pdf
"per length as opposed to the total inductance for the
structure. Note that the latter is exactly what we need
for the transmission line lumped-element equivalent
circuit model (Section 3.4).
T o determine the inductance, we invoke the definition
(Section 7.12):
L≜ Φ
I (7.69)
A current I flowing in the +zdirection on the inner
conductor gives rise to a magnetic field inside the
coaxial structure. The magnetic field intensity for this
scenario was determined in Section 7.5 where we
found
H = ˆφ I
2πρ , a≤ ρ≤ b (7.70)
The reader should note that in that section we were
considering merely a line of current; not a coaxial
structure. So, on what basis do we claim the field for
inside the coaxial structure is the same? This is a
consequence of Ampere’s Law (Section 7.4):
∮
C
H · dl = Iencl (7.71)
If in this new problem we specify the same circular
path C with radius greater than aand less than b, then
the enclosed current is simply I. The presence of the",Electromagnetics_Vol1.pdf
"path C with radius greater than aand less than b, then
the enclosed current is simply I. The presence of the
outer conductor does not change the radial symmetry
of the problem, and nothing else remains that can
change the outcome. This is worth noting for future
reference:
The magnetic field inside a coaxial structure
comprised of concentric conductors bearing cur-
rent Iis identical to the magnetic field of the line
current I in free space.
W e’re going to need magnetic flux density ( B) as
opposed to H in order to get the magnetic flux. This
is simple since they are related by the permeability of
the medium; i.e., B = µH. Thus:
B = ˆφ µI
2πρ , a≤ ρ≤ b (7.72)",Electromagnetics_Vol1.pdf
"168 CHAPTER 7. MAGNETOST A TICS
Next, we get Φ by integrating over the magnetic flux
density
Φ =
∫
S
B · ds (7.73)
where S is any open surface through which all
magnetic field lines within the structure must pass.
Since this can be any such surface, we may as well
choose the simplest one. The simplest such surface is
a plane of constant φ, since such a plane is a
constant-coordinate surface and perpendicular to the
magnetic field lines. This surface is shown as the
shaded area in Figure 7.18. Using this surface we
find:
Φ =
∫ b
ρ=a
∫ l
z=0
(
ˆφ µI
2πρ
)
·
(
ˆφdρdz
)
= µI
2π
( ∫ l
z=0
dz
)( ∫ b
ρ=a
dρ
ρ
)
= µIl
2π ln
( b
a
)
(7.74)
Wrapping up:
L≜ Φ
I = (µIl/2π) ln (b/a)
I (7.75)
Note that factors of I in the numerator and
denominator cancel out, leaving:
L= µl
2πln
( b
a
)
(7.76)
Note that this is dimensionally correct, having units of
H. Also note that this is expression depends only on
materials (through µ) and geometry (through l, a, and",Electromagnetics_Vol1.pdf
"H. Also note that this is expression depends only on
materials (through µ) and geometry (through l, a, and
b). Notably , it does not depend on current, which
would imply non-linear behavior.
T o make the connection back to lumped-element
transmission line model parameters (Sections 3.4 and
3.10), we simply divide by lto get the per-unit length
parameter:
L′ = µ
2πln
( b
a
)
(7.77)
which has the expected units of H/m.
Example 7.1. Inductance of RG-59 coaxial
cable.
RG-59 coaxial cable consists of an inner
conductor having radius 0.292 mm, an outer
conductor having radius 1.855 mm, and
polyethylene (a non-magnetic dielectric) spacing
material. Estimate the inductance per length of
RG-59.
Solution. From the problem statement,
a= 0 .292 mm, b= 1 .855 mm, and µ∼
= µ0
since the spacing material is non-magnetic.
Using Equation 7.77, we find L′ ∼
= 370 nH/m.",Electromagnetics_Vol1.pdf
"7.15. MAGNETIC ENERGY 169
7.15 Magnetic Energy
[m0127]
Consider a structure exhibiting inductance; i.e., one
that is able to store energy in a magnetic field in
response to an applied current. This structure could
be a coil, or it could be one of a variety of inductive
structures that are not explicitly intended to be an
inductor; for example, a coaxial transmission line.
When current is applied, the current-bearing elements
of the structure exert forces on each other. Since these
elements are not normally free to move, we may
interpret this force as potential energy stored in the
magnetic field associated with the current
(Section 7.12).
W e now want to know how much energy is stored in
this field. The answer to this question has relevance in
several engineering applications. One issue is that any
system that includes inductance is using some
fraction of the energy delivered by the power supply
to energize this inductance. In many electronic",Electromagnetics_Vol1.pdf
"system that includes inductance is using some
fraction of the energy delivered by the power supply
to energize this inductance. In many electronic
systems – in power systems in particular – inductors
are periodically energized and de-energized at a
regular rate. Since power is energy per unit time, this
consumes power. Therefore, energy storage in
inductors contributes to the power consumption of
electrical systems.
The stored energy is most easily determined using
circuit theory concepts. First, we note that the
electrical potential difference v(t) (units of V) across
an inductor is related to the current i(t) (units of A)
through the inductor as follows (Section 7.12):
v(t) = Ld
dti(t) (7.78)
where L(units of H) is the inductance. The
instantaneous power associated with the device is
p(t) = v(t)i(t) (7.79)
Energy (units of J) is power (units of J/s) integrated
over time. Let Wm be the energy stored in the
inductor. At some time t0 in the past, i(t0) = 0 and",Electromagnetics_Vol1.pdf
"over time. Let Wm be the energy stored in the
inductor. At some time t0 in the past, i(t0) = 0 and
Wm = 0 . As current is applied, Wm increases
monotonically . At the present time t, i(t) = I. Thus,
the present value of the magnetic energy is:
Wm =
∫ t0+t
t0
p(τ)dτ (7.80)
Now evaluating this integral using the relationships
established above:
Wm =
∫ t+t0
t0
v(τ)i(τ)dτ
=
∫ t+t0
t0
[
L d
dτi(τ)
]
i(τ)dτ
= L
∫ t+t0
t0
[ d
dτi(τ)
]
i(τ)dτ (7.81)
Changing the variable of integration from τ (and dτ)
to i(and di) we have
Wm = L
∫ t+t0
t0
di
dτ idτ
= L
∫ I
0
idi (7.82)
Evaluating the integral we obtain the desired
expression
Wm = 1
2LI2 (7.83)
The energy stored in an inductor in response to
a steady current I is Equation 7.83. This energy
increases in proportion to inductance and in pro-
portion to the square of current.
The long straight coil (Section 7.13) is representative
of a large number of practical applications, so it is
useful to interpret the above findings in terms of this",Electromagnetics_Vol1.pdf
"of a large number of practical applications, so it is
useful to interpret the above findings in terms of this
structure in particular. For this structure we found
L= µN2A
l (7.84)
where µis the permeability , N is the number of
windings, Ais cross-sectional area, and lis length.
The magnetic field intensity inside this structure is
related to I by (Section 7.6):
H = NI
l (7.85)",Electromagnetics_Vol1.pdf
"170 CHAPTER 7. MAGNETOST A TICS
Substituting these expressions into Equation 7.83, we
obtain
Wm = 1
2
[ µN2A
l
][ Hl
N
] 2
= 1
2µH2Al (7.86)
Recall that the magnetic field inside a long coil is
approximately uniform. Therefore, the density of
energy stored inside the coil is approximately
uniform. Noting that the product Alis the volume
inside the coil, we find that this energy density is
Wm/Al; thus:
wm = 1
2µH2 (7.87)
which has the expected units of energy per unit
volume (J/m 3).
The above expression provides an alternative method
to compute the total magnetostatic energy in any
structure. Within a mathematical volume V, the total
magnetostatic energy is simply the integral of the
energy density over V; i.e.,
Wm =
∫
V
wm dv (7.88)
This works even if the magnetic field and the
permeability vary with position. Substituting
Equation 7.87 we obtain:
Wm = 1
2
∫
V
µH2dv (7.89)
Summarizing:
The energy stored by the magnetic field present
within any defined volume is given by Equa-
tion 7.89.",Electromagnetics_Vol1.pdf
"Wm = 1
2
∫
V
µH2dv (7.89)
Summarizing:
The energy stored by the magnetic field present
within any defined volume is given by Equa-
tion 7.89.
It’s worth noting that this energy increases with the
permeability of the medium, which makes sense since
inductance is proportional to permeability .
Finally , we reiterate that although we arrived at this
result using the example of the long straight coil,
Equations 7.87 and 7.89 are completely general.
7.16 Magnetic Materials
[m0058]
As noted in Section 2.5, magnetic fields arise in the
presence of moving charge (i.e., current) and in the
presence of certain materials. In this section, we
address these “magnetic materials. ”
A magnetic material may be defined as a substance
that exhibits permeability µ(Section 2.6) that is
significantly different from the permeability of free
space µ0. Since the magnetic flux density B is related
to the magnetic field intensity H via B = µH,
magnetic materials may exhibit magnetic flux density",Electromagnetics_Vol1.pdf
"to the magnetic field intensity H via B = µH,
magnetic materials may exhibit magnetic flux density
in response to a given magnetic field intensity that is
significantly greater than that of other materials.
Magnetic materials are also said to be
“magnetizable, ” meaning that the application of a
magnetic field causes the material itself to become a
source of the magnetic field.
Magnetic media are typically metals, semiconductors,
or heterogeneous media containing such materials.
An example is ferrite, which consists of iron particles
suspended in a ceramic. Magnetic media are
commonly classified according to the physical
mechanism responsible for their magnetizability .
These mechanisms include paramagnetism,
diamagnetism, and ferromagnetism. All three of these
mechanisms involve quantum mechanical processes
operating at the atomic and subatomic level, and are
not well-explained by classical physics. These
processes are beyond the scope of this book (but",Electromagnetics_Vol1.pdf
"operating at the atomic and subatomic level, and are
not well-explained by classical physics. These
processes are beyond the scope of this book (but
information is available via “ Additional References”
at the end of this section). However, it is possible to
identify some readily-observable differences between
these categories of magnetic media.
Paramagnetic and diamagnetic materials exhibit
permeability that is only very slightly different than
µ0 and typically by much less than 0.01%. These
materials exhibit very weak and temporary
magnetization. The principal distinction between
paramagnetic and diamagnetic media is in the
persistence and orientation of induced magnetic
fields. Paramagnetic materials – including aluminum,
magnesium, and platinum – exhibit a very weak
persistent magnetic field, and the magnetic field",Electromagnetics_Vol1.pdf
"7.16. MAGNETIC MA TERIALS 171
induced in the material is aligned in the same
direction as the impressed (external) magnetic field.
Diamagnetic materials – including copper, gold, and
silicon – do not exhibit a persistent magnetic field,
and the magnetic field induced in the material is
(counter to intuition!) aligned in the opposite
direction as the impressed magnetic field. The
magnetization of paramagnetic and diamagnetic
media is typically so weak that it is not often a
consideration in engineering analysis and design.
Paramagnetic and diamagnetic media exhibit per-
meability only very slightly different than that of
free space, with little or no magnetization.
Ferromagnetic materials , on the other hand, exhibit
permeability that can be many orders of magnitude
greater than µ0. (See Appendix A.2 for some example
values.) These materials can be readily and
indefinitely magnetized, thus, permanent magnets are
typically comprised of ferromagnetic materials.",Electromagnetics_Vol1.pdf
"values.) These materials can be readily and
indefinitely magnetized, thus, permanent magnets are
typically comprised of ferromagnetic materials.
Commonly-encountered ferromagnetic materials
include iron, nickel, and cobalt.
Ferromagnetic materials are significantly non-linear
(see definition in Section 2.8), exhibiting saturation
and hysteresis. This is illustrated in Figure 7.19. In
this plot, the origin represents a ferromagnetic
material that is unmagnetized and in a region free of
an external magnetic field. The external magnetic
field is quantified in terms of H, plotted along the
horizontal axis. As the external field is increased, so
to is B in the material, according to the relationship
B = µH. Right away we see the material is
non-linear, since the slope of the curve – and hence µ
– is not constant.
Once the external magnetizing field H exceeds a
certain value, the response field B no longer
significantly increases. This is saturation. Once",Electromagnetics_Vol1.pdf
"Once the external magnetizing field H exceeds a
certain value, the response field B no longer
significantly increases. This is saturation. Once
saturated, further increases in the external field result
do not significantly increase the magnetization of the
material, so there is no significant increase in B.
From this state of saturation, let us now reduce the
external field. W e find that the rate of decrease in B
with respect to H is significantly less than the rate
that B originally increased with respect to H. In fact,
B is still greater than zero even when H has been
reduced to zero. At this point, the magnetization of
B
saturation
H
c⃝ Ndthe (modified) CC BY SA 3.0
Figure 7.19: Non-linearity in a ferromagnetic material
manifesting as saturation and hysteresis.
the material is obvious, and a device comprised of this
material could be used as a magnet.
If we now apply an external field in the reverse
direction, we find that we are eventually able to zero",Electromagnetics_Vol1.pdf
"material could be used as a magnet.
If we now apply an external field in the reverse
direction, we find that we are eventually able to zero
and then redirect the response field. As we continue
to decrease H (that is, increase the magnitude in the
reverse direction), we once again reach saturation.
The same behavior is observed when we once again
increase H. The material is eventually demagnetized,
remagnetized in the opposite direction and then
saturated in that direction. At this point, it is apparent
that a return to the start condition ( H = B = 0 ; i.e.,
demagnetized when there is no external field) is not
possible.
Hysteresis is the name that we apply to this particular
form of non-linear behavior. Hysteresis has important
implications in engineering applications. First, as
identified above, it is an important consideration in
the analysis and design of magnets. In applications
where a ferromagnetic material is being used because
high permeability is desired – e.g., in inductors",Electromagnetics_Vol1.pdf
"where a ferromagnetic material is being used because
high permeability is desired – e.g., in inductors
(Section 7.12) and transformers (Section 8.5) –
hysteresis complicates the design and imposes limits
on the performance of the device.
Hysteresis may also be exploited as a form of
memory . This is apparent from Figure 7.19. If B >0,",Electromagnetics_Vol1.pdf
"172 CHAPTER 7. MAGNETOST A TICS
then recent values of H must have been relatively
large and positive. Similarly , If B <0, then recent
values of H must have been relatively large and
negative. Furthermore, the most recent sign of H can
be inferred even if the present value of H is zero. In
this sense, the material “remembers” the past history
of its magnetization and thereby exhibits memory .
This is the enabling principle for a number of digital
data storage devices, including hard drives (see
“ Additional Reading” at the end of this section).
Summarizing:
Ferromagnetic media exhibit permeability µthat
is orders of magnitude greater than that of free
space and are readily magnetizable. These mate-
rials are also nonlinear in µ, which manifests as
saturation and hysteresis.
Additional Reading:
• Section A.2 (“Permeability of Some Common
Materials”)
• “Magnetism” on Wikipedia.
• “Ferrite (magnet)” on Wikipedia.
• “Paramagnetism” on Wikipedia.
• “Diamagnetism” on Wikipedia.",Electromagnetics_Vol1.pdf
"Materials”)
• “Magnetism” on Wikipedia.
• “Ferrite (magnet)” on Wikipedia.
• “Paramagnetism” on Wikipedia.
• “Diamagnetism” on Wikipedia.
• “Ferromagnetism” on Wikipedia.
• “Magnetic hysteresis” on Wikipedia.
• “Magnetic storage” on Wikipedia.
• “Hard disk drive” on Wikipedia.
[m0150]",Electromagnetics_Vol1.pdf
"7.16. MAGNETIC MA TERIALS 173
Image Credits
Fig. 7.1: c⃝ Y ouming / K. Kikkeri,
https://commons.wikimedia.org/wiki/File:M0018 fGLMBarMagnet (2).svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 7.2: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0019 fACL.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 7.3: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0119 fLSW .svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 7.4: c⃝ Jfmelero, https://et.wikipedia.org/wiki/Fail:Manoder echa.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/), modified.
Fig. 7.5: Zureks, https://en.wikipedia.org/wiki/File:S olenoid-1.png,
public domain.
Fig. 7.7: c⃝ Geek3, https://commons.wikimedia.org/wiki/File:VFPt Solenoid correct.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/).
Fig. 7.8: Chetvorno, https://en.wikipedia.org/wiki/Fil e:Magnetic field of loop 3.svg,",Electromagnetics_Vol1.pdf
"Fig. 7.8: Chetvorno, https://en.wikipedia.org/wiki/Fil e:Magnetic field of loop 3.svg,
public domain via CC0 1.0, modified.
Fig. 7.9: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0120 fPath.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 7.10: Slick, https://en.wikipedia.org/wiki/File:
3Com OfficeConnect ADSL Wireless 11g Firewall Router 2012-10-28-0869.jpg,
public domain via CC0 1.0 (https://creativecommons.org/p ublicdomain/zero/1.0/deed.en).
Fig. 7.14: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0022 fBoundary .svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 7.16: c⃝ K. Kikkeri, https://commons.wikimedia.org/wiki/File:M 0123 fFluxCurrent.svg,
CC BY SA 4.0 (https://creativecommons.org/licenses/by-s a/4.0/).
Fig. 7.19: c⃝ Ndthe, https://commons.wikimedia.org/wiki/File:Hyste resisSVG.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/deed.en), modified by author.",Electromagnetics_Vol1.pdf
"Chapter 8
Time-V arying Fields
8.1 Comparison of Static and
Time-V arying
Electromagnetics
[m0013]
Students encountering time-varying electromagnetic
fields for the first time have usually been exposed to
electrostatics and magnetostatics already . These
disciplines exhibit many similarities as summarized in
T able 8.1. The principles of time-varying
electromagnetics presented in this table are all
formally introduced in other sections; the sole
purpose of this table is to point out the differences.
W e can summarize the differences as follows:
Maxwell’s Equations in the general (time-
varying) case include extra terms that do not ap-
pear in the equations describing electrostatics and
magnetostatics. These terms involve time deriva-
tives of fields and describe coupling between
electric and magnetic fields.
The coupling between electric and magnetic fields in
the time-varying case has one profound consequence
in particular. It becomes possible for fields to",Electromagnetics_Vol1.pdf
"the time-varying case has one profound consequence
in particular. It becomes possible for fields to
continue to exist even after their sources – i.e.,
charges and currents – are turned off. What kind of
field can continue to exist in the absence of a source?
Such a field is commonly called a wave. Examples of
waves include signals in transmission lines and
signals propagating away from an antenna.
Additional Reading:
• “Maxwell’s Equations” on Wikipedia.
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"8.2. ELECTROMAGNETIC INDUCTION 175
Electrostatics / Time-V arying
Magnetostatics (Dynamic)
Electric & magnetic independent possibly coupled
fields are...
Maxwell’s Eqns.
∮
S D · ds = Qencl
∮
S D · ds = Qencl
(integral)
∮
C E · dl = 0
∮
C E · dl = − ∂
∂t
∫
S B · ds∮
S B · ds = 0
∮
S B · ds = 0∮
C H · ds = Iencl
∮
C H · dl = Iencl+
∫
S
∂
∂tD · ds
Maxwell’s Eqns. ∇ · D = ρv ∇ · D = ρv
(differential) ∇ × E = 0 ∇ × E = − ∂
∂tB
∇ · B = 0 ∇ · B = 0
∇ × H = J ∇ × H = J+ ∂
∂tD
T able 8.1: Comparison of principles governing static and ti me-varying electromagnetic fields. Differences in
the time-varying case relative to the static case are highli ghted in blue.
8.2 Electromagnetic Induction
[m0129]
When an electrically-conducting structure is exposed
to a time-varying magnetic field, an electrical
potential difference is induced across the structure.
This phenomenon is known as electromagnetic
induction. A convenient introduction to
electromagnetic induction is provided by Lenz’s Law .",Electromagnetics_Vol1.pdf
"This phenomenon is known as electromagnetic
induction. A convenient introduction to
electromagnetic induction is provided by Lenz’s Law .
This section explains electromagnetic induction in the
context of Lenz’s Law and provides two examples.
Let us begin with the example depicted in Figure 8.1,
involving a cylindrical coil. Attached to the terminals
of the coil is a resistor for which we may identify an
electric potential difference V and current I. In this
particular case, the sign conventions indicated for V
and I are arbitrary , but it is important to be consistent
once they are established.
Now let us introduce a bar magnet as shown in
Figure 8.1. The magnet is centered along the axis of
coil, to the right of the coil, and with its north pole
facing toward the coil. The magnet is responsible for
the magnetic flux density Bimp. W e refer to Bimp as
an impressed magnetic field because this field exists
independently of any response that may be induced",Electromagnetics_Vol1.pdf
"an impressed magnetic field because this field exists
independently of any response that may be induced
by interaction with the coil. Note that Bimp points to
the left inside the coil.
The experiment consists of three tests. W e will find in
Y . Qin (modified) CC BY 4.0
Figure 8.1: An experiment demonstrating electro-
magnetic induction and Lenz’s Law .
two of these tests that current flows (i.e., |I| >0), and
subsequently , there is an induced magnetic field Bind
due to this current. It is the direction of the current
and subsequently the direction of Bind inside the coil
that we wish to observe. The findings are summarized
below and in T able 8.2.
• When the magnet is motionless, we have the
unsurprising result that there is no current in the
coil. Therefore, no magnetic field is induced, and
the total magnetic field is simply equal to Bimp.
• When the magnet moves toward the coil, we
observe current that is positive with respect to
the reference direction indicated in Figure 8.1.",Electromagnetics_Vol1.pdf
"• When the magnet moves toward the coil, we
observe current that is positive with respect to
the reference direction indicated in Figure 8.1.
This current creates an induced magnetic field
Bind that points to the right, as predicted by
magnetostatic considerations from the right-hand",Electromagnetics_Vol1.pdf
"176 CHAPTER 8. TIME-V AR YING FIELDS
rule. Since Bimp points to the left, it appears
that the induced current is opposing the increase
in the magnitude of the total magnetic field.
• When the magnet moves away from the coil, we
observe current that is negative with respect to
the reference direction indicated in Figure 8.1.
This current yields Bind that points to the left.
Since Bimp points to the left, it appears that the
induced current is opposing the decrease in the
magnitude of the total magnetic field.
The first conclusion one may draw from this
experiment is that changes in the magnetic field can
induce current. This was the claim made in the first
paragraph of this section and is a consequence of
Faraday’s Law , which is tackled in detail in
Section 8.3. The second conclusion – also associated
with Faraday’s Law – is the point of this section: The
induced magnetic field – that is, the one due to the
current induced in the coil – always opposes the",Electromagnetics_Vol1.pdf
"induced magnetic field – that is, the one due to the
current induced in the coil – always opposes the
change in the impressed magnetic field. Generalizing:
Lenz’s Law states that the current that is induced
by a change in an impressed magnetic field cre-
ates an induced magnetic field that opposes (acts
to reduce the effect of) the change in the total
magnetic field.
When the magnet moves, three things happen: (1) A
current is induced, (2) A magnetic field is induced
(which adds to the impressed magnetic field), and (3)
the value of V becomes non-zero. Lenz’s Law does
not address which of these are responding directly to
the change in the impressed magnetic field, and which
of these are simply responding to changes in the other
quantities. Lenz’s Law may leave you with the
incorrect impression that it is I that is induced, and
that Bind and V are simply responding to this current.
In truth, the quantity that is induced is actually V.
This can be verified in the above experiment by",Electromagnetics_Vol1.pdf
"In truth, the quantity that is induced is actually V.
This can be verified in the above experiment by
replacing the resistor with a high-impedance
voltmeter, which will indicate that V is changing even
though there is negligible current flow . Current flow
is simply a response to the induced potential.
Nevertheless, it is common to say informally that “ I
is induced, ” even if it is only indirectly through V.
So, if Lenz’s Law is simply an observation and not an
explanation of the underlying physics, then what is it
good for? Lenz’s Law is often useful for quickly
determining the direction of current flow in practical
electromagnetic induction problems, without
resorting to the mathematics associated with
Faraday’s Law . Here’s an example:
Example 8.1. Electromagnetic induction
through a transformer.
Figure 8.2 shows a rudimentary circuit
consisting of a battery and a switch on the left, a
voltmeter on the right, and a transformer linking
the two. It is not necessary to be familiar with",Electromagnetics_Vol1.pdf
"voltmeter on the right, and a transformer linking
the two. It is not necessary to be familiar with
transformers to follow this example; suffice it to
say , the transformer considered here consists of
two coils wound around a common toroidal
core, which serves to contain magnetic flux. In
this way , the flux generated by either coil is
delivered to the other coil with negligible loss.
The experiment begins with the switch on the
left in the open state. Thus, there is no current
and no magnetic field apparent in the coil on the
left. The voltmeter reads 0 V . When the switch is
closed, what happens?
Solution. Closing the switch creates a current in
the coil on the left. Given the indicated polarity
of the battery , this current flows
counter-clockwise through the circuit on the left,
with current arriving in the left coil through the
bottom terminal. Given the indicated direction
of the winding in the left coil, the impressed
magnetic field Bimp is oriented",Electromagnetics_Vol1.pdf
"bottom terminal. Given the indicated direction
of the winding in the left coil, the impressed
magnetic field Bimp is oriented
counter-clockwise through the toroidal core. The
coil on the right “sees” Bimp increase from zero
to some larger value. Since the voltmeter
presumably has input high impedance,
negligible current flows. However, if current
were able to flow , Lenz’s Law dictates that it
would be induced to flow in a counter-clockwise
direction around the circuit on the right, since
the induced magnetic field Bind would then be
clockwise-directed so as to oppose the increase
in Bimp. Therefore, the potential measured at
the bottom of the right coil would be higher than
the potential at the top of the right coil. The",Electromagnetics_Vol1.pdf
"8.2. ELECTROMAGNETIC INDUCTION 177
Magnet is ... |Bimp| in coil is ... Circuit Response Bind inside coil
Motionless constant V = 0 , I = 0 none
Moving toward coil increasing V >0, I >0 Pointing right
Moving away from coil decreasing V <0, I <0 Pointing left
T able 8.2: Results of the experiment associated with Figure 8.1.
+ -+ - +-
- +V
E. Bach CC0 1.0 (modified)
Figure 8.2: Electromagnetic induction through a
transformer.
figure indicates that the voltmeter measures the
potential at its right terminal relative to its left
terminal, so the needle will deflect to the right.
This deflection will be temporary , since the
current provided by the battery becomes
constant at a new non-zero value and Bind
responds only to the change in Bimp. The
voltmeter reading will remain at zero for as long
as the switch remains closed and the current
remains steady .
Here are some follow up exercises to test your
understanding of what is going on: (1) Now
open the switch. What happens? (2) Repeat the",Electromagnetics_Vol1.pdf
"Here are some follow up exercises to test your
understanding of what is going on: (1) Now
open the switch. What happens? (2) Repeat the
original experiment, but before starting, swap
the terminals on the battery .
Finally , it is worth noting that Lenz’s law can also be
deduced from the principle of conservation of energy .
The argument is that if the induced magnetic field
reinforced the change in the impressed magnetic field,
then the sum magnetic field would increase. This
would result in a further increase in the induced
magnetic field, leading to a positive feedback
situation. However, positive feedback cannot be
supported without an external source of energy ,
leading to a logical contradiction. In other words, the
principle of conservation of energy requires the
negative feedback described by Lenz’s law .
Additional Reading:
• “Electromagnetic Induction” on Wikipedia.
• “Lenz’s Law” on Wikipedia.",Electromagnetics_Vol1.pdf
"178 CHAPTER 8. TIME-V AR YING FIELDS
8.3 Faraday’s Law
[m0055]
Faraday’s Law describes the generation of electric
potential by a time-varying magnetic flux. This is a
form of electromagnetic induction , as discussed in
Section 8.2.
T o begin, consider the scenario shown in Figure 8.3.
A single loop of wire in the presence of an impressed
magnetic field B. For reasons explained later, we
introduce a small gap and define VT to be the
potential difference measured across the gap
according to the sign convention indicated. The
resistance Rmay be any value greater than zero,
including infinity; i.e., a literal gap.
As long as Ris not infinite, we know from Lenz’s
Law (Section 8.2) to expect that a time-varying
magnetic field will cause a current to flow in the wire.
Lenz’s Law also tells us the direction in which the
current will flow . However, Lenz’s Law does not tell
us the magnitude of the current, and it sidesteps some
important physics that has profound implications for",Electromagnetics_Vol1.pdf
"us the magnitude of the current, and it sidesteps some
important physics that has profound implications for
the analysis and design of electrical devices,
including generators and transformers.
The more complete picture is given by Faraday’s Law .
In terms of the scenario of Figure 8.3, Faraday’s Law
relates the potential VT induced by the time variation
of B. VT then gives rise to the current identified in
Lenz’s Law . The magnitude of this current is simply
VT/R. Without further ado, here’s Faraday’s Law for
V
T
+ -
R

B
Figure 8.3: A single loop of wire in the presence of
an impressed magnetic field.
this single loop scenario:
VT = − ∂
∂tΦ (single loop) (8.1)
Here Φ is the magnetic flux (units of Wb) associated
with any open surface S bounded by the loop:
Φ =
∫
S
B · ds (8.2)
where B is magnetic flux density (units of T or
Wb/m2) and ds is the differential surface area vector.
T o make headway with Faraday’s Law , one must be
clear about the meanings of S and ds. If the wire loop",Electromagnetics_Vol1.pdf
"T o make headway with Faraday’s Law , one must be
clear about the meanings of S and ds. If the wire loop
in the present scenario lies in a plane, then a good
choice for S is the simply the planar area bounded by
the loop. However, any surface that is bounded by the
loop will work, including non-planar surfaces that
extend above and/or below the plane of the loop. All
that is required is that every magnetic field line that
passes through the loop also passes through S. This
happens automatically if the curve C defining the
edge of the open surface S corresponds to the loop.
Subsequently , the magnitude of ds is the differential
surface element dsand the direction of ds is the unit
vector ˆn perpendicular to each point on S, so
ds = ˆnds.
This leaves just one issue remaining – the orientation
of ˆn. This is sorted out in Figure 8.4. There are two
possible ways for a vector to be perpendicular to a
surface, and the direction chosen for ˆn will affect the",Electromagnetics_Vol1.pdf
"possible ways for a vector to be perpendicular to a
surface, and the direction chosen for ˆn will affect the
sign of VT. Therefore, ˆn must be somehow related to
the polarity chosen for VT. Let’s consider this
relationship. Let C begin at the “ −” terminal of VT
and follow the entire perimeter of the loop, ending at
the “ +” terminal. Then, ˆn is determined by the
following “right-hand rule:” ˆn points in the direction
of the curled fingers of the right hand when the thumb
of the right hand is aligned in the direction of C. It’s
worth noting that this convention is precisely the
convention used to relate ˆn and C in Stokes’ Theorem
(Section 4.9).
Now let us recap how Faraday’s Law is applied to the
single loop scenario of Figure 8.3:
1. Assign “ +” and “ −” terminals to the gap voltage
VT.",Electromagnetics_Vol1.pdf
"8.3. F ARADA Y’S LA W 179
V

+ -

S
n
C
Figure 8.4: Relationship between the polarity of VT
and orientations of C and ˆn in the planar single-loop
scenario.
2. The orientation of ˆn is determined by the right
hand rule, taking the direction of C to be the
perimeter of the loop beginning at “ −” and
ending at “ +”
3. B yields a magnetic flux Φ associated with the
loop according to Equation 8.2. S is any open
surface that intersects all the magnetic field lines
that pass through the loop (so you might as well
choose S in a way that results in the simplest
possible integration).
4. By Faraday’s Law (Equation 8.1), VT is the time
derivative of Φ with a change of sign.
5. The current I flowing in the loop is VT/R, with
the reference direction (i.e., direction of positive
current) being from “ +” to “ −” through the
resistor. Think of the loop as a voltage source,
and you’ll get the correct reference direction for
I.
At this point, let us reiterate that the electromagnetic",Electromagnetics_Vol1.pdf
"and you’ll get the correct reference direction for
I.
At this point, let us reiterate that the electromagnetic
induction described by Faraday’s Law induces
potential (in this case, VT) and not current. The
current in the loop is simply the induced voltage
divided by the resistance of the loop. This point is
easily lost, especially in light of Lenz’s Law , which
seems to imply that current, as opposed to potential,
is the thing being induced.
W ondering about the significance of the minus sign in
Equation 8.1? That is specifically Lenz’s Law: The
current Ithat ends up circulating in the loop generates
its own magnetic field (“ Bind” in Section 8.2), which
is distinct from the impressed magnetic field B and
which tends to oppose change in B. Thus, we see that
Faraday’s Law subsumes Lenz’s Law .
Frequently one is interested in a structure that consists
of multiple identical loops. W e have in mind here
something like a coil with N ≥ 1 windings tightly",Electromagnetics_Vol1.pdf
"of multiple identical loops. W e have in mind here
something like a coil with N ≥ 1 windings tightly
packed together. In this case, Faraday’s Law is
VT = −N ∂
∂tΦ (8.3)
Note that the difference is simply that the gap
potential VT is greater by N. In other words, each
winding of the coil contributes a potential given by
Equation 8.1, and these potentials add in series.
Faraday’s Law , given in general by Equation 8.3,
states that the potential induced in a coil is pro-
portional to the time derivative of the magnetic
flux through the coil.
The induced potential VT is often referred to as
“emf, ” which is a contraction of the term
electromotive force – a misnomer to be sure, since no
actual force is implied, only potential. The term
“emf” is nevertheless frequently used in the context
of Faraday’s Law applications for historical reasons.
Previously in this section, we considered the
generation of emf by time variation of B. However,
Equation 8.1 indicates that what actually happens is",Electromagnetics_Vol1.pdf
"Previously in this section, we considered the
generation of emf by time variation of B. However,
Equation 8.1 indicates that what actually happens is
that the emf is the result of time variation of the
magnetic flux, Φ . Magnetic flux is magnetic flux
density integrated over area, so it appears that emf can
also be generated simply by varying S, independently
of any time variation of B. In other words, emf may
be generated even when B is constant, by instead
varying the shape or orientation of the coil. So, we
have a variety of schemes by which we can generate
emf. Here they are:
1. Time-varying B (as we considered previously in
this section). For example, B might be due to a
permanent magnet that is moved (e.g., translated
or rotated) in the vicinity of the coil, as described
in Section 8.2. Or, more commonly , B might be
due to a different coil that bears a time-varying
current. These mechanisms are collectively",Electromagnetics_Vol1.pdf
"180 CHAPTER 8. TIME-V AR YING FIELDS
referred to as transformer emf . Transformer emf
is the underlying principle of operation of
transformers; for more on this see Section 8.5.
2. The perimeter C – and thus the surface S over
which Φ is determined – can be time-varying.
For example, a wire loop might be rotated or
changed in shape in the presence of a constant
magnetic field. This mechanism is referred to as
motional emf and is the underlying principle of
operation of generators (Section 8.7).
3. Transformer and motional emf can exist in
various combinations. From the perspective of
Faraday’s Law , transformer and motional emf
are the same in the sense in either case Φ is
time-varying, which is all that is required to
generate emf.
Finally , a comment on the generality of Faraday’s
Law . Above we have introduced Faraday’s Law as if
it were specific to loops and coils of wire. However,
the truth of the matter is that Faraday’s Law is
fundamental physics. If you can define a closed path",Electromagnetics_Vol1.pdf
"the truth of the matter is that Faraday’s Law is
fundamental physics. If you can define a closed path
– current-bearing or not – then you can compute the
potential difference achieved by traversing that path
using Faraday’s Law . The value you compute is the
potential associated with electromagnetic induction,
and exists independently and in addition to the
potential difference associated with the static electric
field (e.g., Section 5.12). In other words:
Faraday’s Law indicates the contribution of elec-
tromagnetic induction (the generation of emf by
a time-varying magnetic flux) to the potential dif-
ference achieved by traversing a closed path.
In Section 8.8, this insight is used to transform the
static form of Kirchoff’s V oltage Law (Section 5.10)
– which gives the potential difference associated with
electric field only – into the Maxwell-F araday
Equation (Section 8.8), which is a general statement
about the relationship between the instantaneous",Electromagnetics_Vol1.pdf
"electric field only – into the Maxwell-F araday
Equation (Section 8.8), which is a general statement
about the relationship between the instantaneous
value of the electric field and the time derivative of
the magnetic field.
Additional Reading:
• “Faraday’s law of induction” on Wikipedia.
8.4 Induction in a Motionless
Loop
[m0056]
In this section, we consider the problem depicted in
Figure 8.5, which is a single motionless loop of wire
in the presence of a spatially-uniform but
time-varying magnetic field. A small gap is
introduced in the loop, allowing us to measure the
induced potential VT. Additionally , a resistance Ris
connected across VT in order to allow a current to
flow . This problem was considered in Section 8.3 as
an introduction to Faraday’s Law; in this section, we
shall actually work the problem and calculate some
values. This is intended to serve as an example of the
application of Faraday’s Law , a demonstration of
transformer emf, and will serve as a first step toward",Electromagnetics_Vol1.pdf
"application of Faraday’s Law , a demonstration of
transformer emf, and will serve as a first step toward
an understanding of transformers as devices.
In the present problem, the loop is centered in the
z= 0 plane. The magnetic flux density is
B = ˆbB(t); i.e., time-varying magnitude B(t) and a
constant direction ˆb. Because this magnetic field is
spatially uniform (i.e., the same everywhere), we will
find that only the area of the loop is important, and
not it’s specific shape. For this reason, it will not be
necessary to specify the radius of the loop or even
require that it be a circular loop. Our task is to find
expressions for VT and I.
V

+ -
B
Figure 8.5: A single loop of wire in the presence of
an impressed spatially-uniform but time-varying mag-
netic field.",Electromagnetics_Vol1.pdf
"8.4. INDUCTION IN A MOTIONLESS LOOP 181
T o begin, remember that Faraday’s Law is a
calculation of electric potential and not current. So,
the approach is to first find VT, and then find the
current I that flows through the gap resistance in
response.
The sign convention for VT is arbitrary; here, we have
selected “ +” and “ −” terminals as indicated in
Figure 8.5. 1 Following the standard convention for
the reference direction of current through a passive
device, I should be directed as shown in Figure 8.5. It
is worth repeating that these conventions for the signs
of VT and I are merely references; for example, we
may well find that I is negative, which means that
current flows in a clockwise direction in the loop.
W e now invoke Faraday’s Law:
VT = −N ∂
∂tΦ (8.4)
The number of windings N in the loop is 1, and Φ is
the magnetic flux through the loop. Thus:
VT = − ∂
∂t
∫
S
B · ds (8.5)
where S is any open surface that intersects all of the
magnetic field lines that pass through the loop. The",Electromagnetics_Vol1.pdf
"VT = − ∂
∂t
∫
S
B · ds (8.5)
where S is any open surface that intersects all of the
magnetic field lines that pass through the loop. The
simplest such surface is simply the planar surface
defined by the perimeter of the loop. Then ds = ˆnds,
where dsis the differential surface element and ˆn is
the normal to the plane of the loop. Which of the two
possible normals to the loop? This is determined by
the right-hand rule of Stokes’ Theorem. From the
“ −” terminal, we point the thumb of the right hand in
the direction that leads to the “ +” terminal by
traversing the perimeter of the loop. When we do this,
the curled fingers of the right hand intersect S in the
same direction as ˆn. T o maintain the generality of
results derived below , we shall not make the
substitution ˆn = + ˆz; nevertheless we see this is the
case for a loop parallel to the z= 0 plane with the
polarity of VT indicated in Figure 8.5.
T aking this all into account, we have
VT = − ∂
∂t
∫
S
(
ˆbB(t)
)
· (ˆnds)
= −
(
ˆb · ˆn",Electromagnetics_Vol1.pdf
"polarity of VT indicated in Figure 8.5.
T aking this all into account, we have
VT = − ∂
∂t
∫
S
(
ˆbB(t)
)
· (ˆnds)
= −
(
ˆb · ˆn
) ∂
∂t
∫
S
B(t) ds (8.6)
1 A good exercise for the student is to repeat this problem with
the terminal polarity reversed; one should obtain the same ans wer.
Since the magnetic field is uniform, B(t) may be
extracted from the integral. Furthermore, the shape
and the orientation of the loop are time-invariant, so
the remaining integral may be extracted from the time
derivative operation. This leaves:
VT = −
(
ˆb · ˆn
) ( ∂
∂tB(t)
) ∫
S
ds (8.7)
The integral in this expression is simply the area of
the loop, which is a constant; let the symbol A
represent this area. W e obtain
VT = −
(
ˆb · ˆnA
) ∂
∂tB(t) (8.8)
which is the expression we seek. Note that the
quantity ˆb · ˆnAis the projected area of the loop. The
projected area is equal to Awhen the the magnetic
field lines are perpendicular to the loop (i.e., ˆb = ˆn),",Electromagnetics_Vol1.pdf
"projected area is equal to Awhen the the magnetic
field lines are perpendicular to the loop (i.e., ˆb = ˆn),
and decreases to zero as ˆb · ˆn → 0. Summarizing:
The magnitude of the transformer emf induced
by a spatially-uniform magnetic field is equal to
the projected area times the time rate of change of
the magnetic flux density , with a change of sign.
(Equation 8.8).
A few observations about this result:
• As promised earlier, we have found that the
shape of the loop is irrelevant; i.e., a square loop
having the same area and planar orientation
would result in the same VT. This is because the
magnetic field is spatially uniform, and because
it is the magnetic flux ( Φ ) and not the magnetic
field or shape of the loop alone that determines
the induced potential.
• The induced potential is proportional to A; i.e.,
VT can be increased by increasing the area of the
loop.
• The peak magnitude of the induced potential is
maximized when the plane of the loop is",Electromagnetics_Vol1.pdf
"VT can be increased by increasing the area of the
loop.
• The peak magnitude of the induced potential is
maximized when the plane of the loop is
perpendicular to the magnetic field lines.
• The induced potential goes to zero when the
plane of the loop is parallel to the magnetic field
lines. Said another way , there is no induction
unless magnetic field lines pass through the loop.",Electromagnetics_Vol1.pdf
"182 CHAPTER 8. TIME-V AR YING FIELDS
• The induced potential is proportional to the rate
of change of B. If B is constant in time, then
there is no induction.
Finally , the current in the loop is simply
I = VT
R (8.9)
Again, electromagnetic induction induces potential,
and the current flows only in response to the induced
potential as determined by Ohm’s Law . In particular,
if the resistor is removed, then R→ ∞ and I → 0,
but VT is unchanged.
One final comment is that even though the current Iis
not a direct result of electromagnetic induction, we
can use I as a check of the result using Lenz’s Law
(Section 8.2). W e’ll demonstrate this in the example
below .
Example 8.2. Induction in a motionless circular
loop by a linearly-increasing magnetic field
Let the loop be planar in the z= 0 plane and
circular with radius a= 10 cm. Let the
magnetic field be ˆzB(t) where
B(t) = 0 , t< 0
= B0t/t0 , 0 ≤ t≤ t0
= B0 , t>t 0 (8.10)
i.e., B(t) begins at zero and increases linearly to",Electromagnetics_Vol1.pdf
"magnetic field be ˆzB(t) where
B(t) = 0 , t< 0
= B0t/t0 , 0 ≤ t≤ t0
= B0 , t>t 0 (8.10)
i.e., B(t) begins at zero and increases linearly to
B0 at time t0, after which it remains constant at
B0. Let B0 = 0 .2 T , t0 = 3 s, and let the loop be
closed by a resistor R= 1 kΩ . What current I
flows in the loop?
Solution. Adopting the sign conventions of
Figure 8.5 we first note that ˆn = + ˆz; this is
determined by the right-hand rule with respect to
the indicated polarity of VT. Thus, Equation 8.8
becomes
VT = −
(
ˆb · ˆzA
) ∂
∂tB(t)
Note ˆb · ˆzA= Asince ˆb = ˆz; i.e., because the
plane of the loop is perpendicular to the
magnetic field lines. Since the loop is circular,
A= πa2. Also
∂
∂tB(t) = 0 , t< 0
= B0/t0 , 0 ≤ t≤ t0
= 0 , t>t 0
Putting this all together:
VT = −πa2 B0
t0
= −2.09 mV , 0 ≤ t≤ t0
and VT = 0 before and after this time interval,
since B is constant during those times.
Subsequently the induced current is
I = VT
R = −2.09 µA , 0 ≤ t≤ t0",Electromagnetics_Vol1.pdf
"since B is constant during those times.
Subsequently the induced current is
I = VT
R = −2.09 µA , 0 ≤ t≤ t0
and I = 0 before and after this time interval. W e
have found that the induced current is a constant
clockwise flow that exists only while B is
increasing.
Finally , let’s see if the result is consistent with
Lenz’s Law . The current induced while B is
changing gives rise to an induced magnetic field
Bind. From the right-hand rule that relates the
direction of I to the direction of Bind
(Section 7.5), the direction of Bind is generally
−ˆz inside the loop. In other words, the magnetic
field associated with the induced current opposes
the increasing impressed magnetic field that
induced the current, in accordance with Lenz’s
Law .
Example 8.3. Induction in a motionless circular
loop by a sinusoidally-varying magnetic field.
Let us repeat the previous example, but now with
B(t) = B0 sin 2πf0t
with f0 = 1 kHz.
Solution. Now
∂
∂tB(t) = 2 πf0B0 cos 2πf0t",Electromagnetics_Vol1.pdf
"8.5. TRANSFORMERS: PRINCIPLE OF OPERA TION 183
So
VT = −2π2f0a2B0 cos 2πf0t
Subsequently
I = VT
R = −2π2f0a2B0
R cos 2πf0t
Substituting values, we have:
I = −(39.5 mA) cos [(6.28 krad/s)t]
It should be no surprise that VT and I vary
sinusoidally , since the source ( B) varies
sinusoidally . A bit of useful trivia here is that VT
and I are 90 ◦ out of phase with the source. It is
also worth noting what happens when B(t) = 0 .
This occurs twice per period, at t= n/2f0
where nis any integer, including t= 0 . At these
times B(t) is zero, but VT and hence IR are
decidedly non-zero; in fact, they are at their
maximum magnitude. Again, it is the change in
B that induces voltage and subsequently current,
not B itself.
8.5 T ransformers: Principle of
Operation
[m0031]
A transformer is a device that connects two electrical
circuits through a shared magnetic field.
Transformers are used in impedance transformation,
voltage level conversion, circuit isolation, conversion",Electromagnetics_Vol1.pdf
"circuits through a shared magnetic field.
Transformers are used in impedance transformation,
voltage level conversion, circuit isolation, conversion
between single-ended and differential signal modes,
and other applications. 2 The underlying
electromagnetic principle is Faraday’s Law
(Section 8.3) – in particular, transformer emf.
The essential features of a transformer can be derived
from the simple experiment shown in Figures 8.6 and
8.7. In this experiment, two coils are arranged along a
common axis. The winding pitch is small, so that all
magnetic field lines pass through the length of the
coil, and no lines pass between the windings. T o
further contain the magnetic field, we assume both
coils are wound on the same core, consisting of some
material exhibiting high permeability . The upper coil
has N1 turns and the lower coil has N2 turns.
In Part I of this experiment (Figure 8.6), the upper
coil is connected to a sinusoidally-varying voltage
source V(1)",Electromagnetics_Vol1.pdf
"In Part I of this experiment (Figure 8.6), the upper
coil is connected to a sinusoidally-varying voltage
source V(1)
1 in which the subscript refers to the coil
and the superscript refers to “Part I” of this
experiment. The voltage source creates a current in
the coil, which in turn creates a time-varying
magnetic field B in the core.
The lower coil has N2 turns wound in the opposite
direction and is open-circuited. Given the
closely-spaced windings and use of a
high-permeability core, we assume that the magnetic
field within the lower coil is equal to B generated in
the upper coil. The potential induced in the lower coil
is V(1)
2 with reference polarity indicated in the figure.
From Faraday’s Law we have
V(1)
2 = −N2
∂
∂tΦ 2 (8.11)
where Φ 2 is the flux through a single turn of the lower
2 See “ Additional Reading” at the end of this section for more on
these applications.",Electromagnetics_Vol1.pdf
"184 CHAPTER 8. TIME-V AR YING FIELDS
turnsV
1
(1)
+
−
V2
(1)
+
−
N1
turnsN2
Figure 8.6: Part I of an experiment demonstrating the
linking of electric circuits using a transformer.
turnsV

(2)
+
−
V2
(2)
+
−
N1
turnsN2
Figure 8.7: Part II of an experiment demonstrating the
linking of electric circuits using a transformer.
coil. Thus:
V(1)
2 = −N2
∂
∂t
∫
S
B · (−ˆzds) (8.12)
Note that the direction of ds = −ˆzdsis determined
by the polarity we have chosen for V(1)
2 .
In Part II of the experiment (Figure 8.7), we make
changes as follows. W e apply a voltage V(2)
2 to the
lower coil and open-circuit the upper coil. Further, we
adjust V(2)
2 such that the induced magnetic flux
density is again B – that is, equal to the field in Part I
of the experiment. Now we have
V(2)
1 = −N1
∂
∂tΦ 1 (8.13)
which is
V(2)
1 = −N1
∂
∂t
∫
S
B · (+ˆzds) (8.14)
For reasons that will become apparent in a moment,
let’s shift the leading minus sign into the integral. W e
then have
V(2)
1 = + N1
∂
∂t
∫
S",Electromagnetics_Vol1.pdf
"For reasons that will become apparent in a moment,
let’s shift the leading minus sign into the integral. W e
then have
V(2)
1 = + N1
∂
∂t
∫
S
B · (−ˆzds) (8.15)
Comparing this to Equations 8.11 and 8.12, we see
that we may rewrite this in terms of the flux in the
lower coil in Part I of the experiment:
V(2)
1 = + N1
∂
∂tΦ 2 (8.16)
In fact, we can express this in terms of the potential in
Part I of the experiment:
V(2)
1 =
(
−N1
N2
)(
−N2
∂
∂tΦ 2
)
=
(
−N1
N2
)
V(1)
2 (8.17)
W e have found that the potential in the upper coil in
Part II is related in a simple way to the potential in the
lower coil in Part I of the experiment. If we had done
Part II first, we would have obtained the same result
but with the superscripts swapped. Therefore, it must
be generally true – regardless of the arrangement of
terminations – that
V1 = −N1
N2
V2 (8.18)",Electromagnetics_Vol1.pdf
"8.6. TRANSFORMERS AS TWO-POR T DEVICES 185
This expression should be familiar from elementary
circuit theory – except possibly for the minus sign.
The minus sign is a consequence of the fact that the
coils are wound in opposite directions. W e can make
the above expression a little more general as follows:
V1
V2
= pN1
N2
(8.19)
where pis defined to +1 when the coils are wound in
the same direction and −1 when coils are wound in
opposite directions. (It is an excellent exercise to
confirm that this is true by repeating the above
analysis with winding direction changed for either the
upper or lower coil, for which pwill then turn out to
be +1.) This is the “transformer law” of basic electric
circuit theory , from which all other characteristics of
transformers as two-port circuit devices can be
obtained (See Section 8.6 for follow-up on that).
Summarizing:
The ratio of coil voltages in an ideal transformer
is equal to the ratio of turns with sign determined",Electromagnetics_Vol1.pdf
"Summarizing:
The ratio of coil voltages in an ideal transformer
is equal to the ratio of turns with sign determined
by the relative directions of the windings, per
Equation 8.19.
A more familiar transformer design is shown in
Figure 8.8 – coils wound on a toroidal core as
opposed to a cylindrical core. Why do this? This
arrangement confines the magnetic field linking the
two coils to the core, as opposed to allowing field
lines to extend beyond the device. This confinement
is important in order to prevent fields originating
outside the transformer from interfering with the
magnetic field linking the coils, which would lead to
electromagnetic interference (EMI) and
electromagnetic compatibility (EMC) problems. The
principle of operation is in all other respects the same.
Additional Reading:
• “Transformer” on Wikipedia.
• “Balun” on Wikipedia.
V1
V2
N1 turns
N2 turns
Φ
Magnetic
Flux,
Transformer
Core
+
+
−
−
c⃝ BillC (modified) CC BY SA 3.0
Figure 8.8: Transformer implemented as coils sharing",Electromagnetics_Vol1.pdf
"V2
N1 turns
N2 turns
Φ
Magnetic
Flux,
Transformer
Core
+
+
−
−
c⃝ BillC (modified) CC BY SA 3.0
Figure 8.8: Transformer implemented as coils sharing
a toroidal core. Here p= +1 .
8.6 T ransformers as T wo-Port
Devices
[m0032]
Section 8.5 explains the principle of operation of the
ideal transformer. The relationship governing the
terminal voltages V1 and V2 was found to be
V1
V2
= pN1
N2
(8.20)
where N1 and N2 are the number of turns in the
associated coils and pis either +1 or −1 depending
on the relative orientation of the windings; i.e.,
whether the reference direction of the associated
fluxes is the same or opposite, respectively .
W e shall now consider ratios of current and
impedance in ideal transformers, using the two-port
model shown in Figure 8.9. By virtue of the reference
current directions and polarities chosen in this figure,
the power delivered by the source V1 is V1I1, and the
power absorbed by the load Z2 is −V2I2. Assuming
the transformer windings have no resistance, and",Electromagnetics_Vol1.pdf
"power absorbed by the load Z2 is −V2I2. Assuming
the transformer windings have no resistance, and
assuming the magnetic flux is perfectly contained
within the core, the power absorbed by the load must
equal the power provided by the source; i.e.,",Electromagnetics_Vol1.pdf
"186 CHAPTER 8. TIME-V AR YING FIELDS
V1I1 = −V2I2. Thus, we have 3
I1
I2
= −V2
V1
= −pN2
N1
(8.21)
W e can develop an impedance relationship for ideal
transformers as follows. Let Z1 be the input
impedance of the transformer; that is, the impedance
looking into port 1 from the source. Thus, we have
Z1 ≜ V1
I1
= +p(N1/N2) V2
−p(N2/N1) I2
= −
( N1
N2
) 2 ( V2
I2
)
(8.22)
In Figure 8.9, Z2 is the the output impedance of port
2; that is, the impedance looking out port 2 into the
load. Therefore, Z2 = −V2/I2 (once again the minus
sign is a result of our choice of sign conventions in
Figure 8.9). Substitution of this result into the above
equation yields
Z1 =
( N1
N2
) 2
Z2 (8.23)
and therefore
Z1
Z2
=
( N1
N2
) 2
(8.24)
3 The minus signs in this equation are a result of the reference
polarity and directions indicated in Figure 8.9. These are mo re-or-
less standard in electrical two-port theory (see “ Addition al Reading”
at the end of this section), but are certainly not the only rea son-",Electromagnetics_Vol1.pdf
"less standard in electrical two-port theory (see “ Addition al Reading”
at the end of this section), but are certainly not the only rea son-
able choice. If you see these expressions without the minus si gns,
it probably means that a different combination of reference di rec-
tions/polarities is in effect.
I
2
V

+
−
I1
V1
+
−
+_ Z2
Z1
Figure 8.9: The transformer as a two-port circuit de-
vice.
Thus, we have demonstrated that a transformer scales
impedance in proportion to the square of the turns
ratio N1/N2. Remarkably , the impedance
transformation depends only on the turns ratio, and is
independent of the relative direction of the windings
(p).
The relationships developed above should be viewed
as AC expressions, and are not normally valid at DC.
This is because transformers exhibit a fundamental
limitation in their low-frequency performance. T o see
this, first recall Faraday’s Law:
V = −N ∂
∂tΦ (8.25)
If the magnetic flux Φ isn’t time-varying, then there is",Electromagnetics_Vol1.pdf
"this, first recall Faraday’s Law:
V = −N ∂
∂tΦ (8.25)
If the magnetic flux Φ isn’t time-varying, then there is
no induced electric potential, and subsequently no
linking of the signals associated with the coils. At
very low but non-zero frequencies, we encounter
another problem that gets in the way – magnetic
saturation. T o see this, note we can obtain an
expression for Φ from Faraday’s Law by integrating
with respect to time, yielding
Φ( t) = − 1
N
∫ t
t0
V(τ)dτ + Φ( t0) (8.26)
where t0 is some earlier time at which we know the
value of Φ( t0). Let us assume that V(t) is
sinusoidally-varying. Then the peak value of Φ after
t= t0 depends on the frequency of V(t). If the
frequency of V(t) is very low , then Φ can become
very large. Since the associated cross-sectional areas
of the coils are presumably constant, this means that
the magnetic field becomes very large. The problem
with that is that most practical high-permeability
materials suitable for use as transformer cores exhibit",Electromagnetics_Vol1.pdf
"with that is that most practical high-permeability
materials suitable for use as transformer cores exhibit
magnetic saturation; that is, the rate at which the
magnetic field is able to increase decreases with
increasing magnetic field magnitude (see
Section 7.16). The result of all this is that a
transformer may work fine at (say) 1 MHz, but at
(say) 1 Hz the transformer may exhibit an apparent
loss associated with this saturation. Thus, practical
transformers exhibit highpass frequency response.
It should be noted that the highpass behavior of
practical transistors can be useful. For example, a
transformer can be used to isolate an undesired DC
offset and/or low-frequency noise in the circuit",Electromagnetics_Vol1.pdf
"8.7. THE ELECTRIC GENERA TOR 187
DifferentialSingle-
Ended
Single-
Ended
c⃝ SpinningSpark CC BY SA 3.0 (modified)
Figure 8.10: Transformers used to convert a single-
ended (“unbalanced”) signal to a differential (bal-
anced) signal, and back.
attached to one coil from the circuit attached to the
other coil.
The DC-isolating behavior of a transformer also
allows the transformer to be used as a balun, as
illustrated in Figure 8.10. A balun is a two-port
device that transforms a single-ended (“unbalanced”)
signal – that is, one having an explicit connection to a
datum (e.g., ground) – into a differential (“balanced”)
signal, for which there is no explicit connection to a
datum. Differential signals have many benefits in
circuit design, whereas inputs and outputs to devices
must often be in single-ended form. Thus, a common
use of transformers is to effect the conversion
between single-ended and differential circuits.
Although a transformer is certainly not the only",Electromagnetics_Vol1.pdf
"use of transformers is to effect the conversion
between single-ended and differential circuits.
Although a transformer is certainly not the only
device that can be used as a balun, it has one very big
advantage, namely bandwidth.
Additional Reading:
• “Transformer” on Wikipedia.
• “T wo-port network” on Wikipedia.
• “Saturation (magnetic)” on Wikipedia.
• “Balun” on Wikipedia.
• S.W . Ellingson, “Differential Circuits” (Sec. 8.7)
in Radio Systems Engineering , Cambridge Univ .
Press, 2016.
8.7 The Electric Generator
[m0030]
A generator is a device that transforms mechanical
energy into electrical energy , typically by
electromagnetic induction via Faraday’s Law
(Section 8.3). For example, a generator might consist
of a gasoline engine that turns a crankshaft to which
is attached a system of coils and/or magnets. This
rotation changes the relative orientations of the coils
with respect to the magnetic field in a time-varying
manner, resulting in a time-varying magnetic flux and",Electromagnetics_Vol1.pdf
"with respect to the magnetic field in a time-varying
manner, resulting in a time-varying magnetic flux and
subsequently induced electric potential. In this case,
the induced potential is said to be a form of motional
emf, as it is due entirely to changes in geometry as
opposed to changes in the magnitude of the magnetic
field. Coal- and steam-fired generators, hydroelectric
generators, wind turbines, and other energy
generation devices operate using essentially this
principle.
Figure 8.11 shows a rudimentary generator, which
serves as to illustrate the relevant points. This
generator consists of a planar loop that rotates around
the zaxis; therefore, the rotation can be parameterized
in φ. In this case, the direction of rotation is specified
to be in the +φdirection. The frequency of rotation is
f0; that is, the time required for the loop to make one
complete revolution is 1/f0. W e assume a
time-invariant and spatially-uniform magnetic flux",Electromagnetics_Vol1.pdf
"f0; that is, the time required for the loop to make one
complete revolution is 1/f0. W e assume a
time-invariant and spatially-uniform magnetic flux
density B = ˆbB0 where ˆb and B0 are both constants.
For illustration purposes, the loop is indicated to be
circular. However, because the magnetic field is
time-invariant and spatially-uniform, the specific
shape of the loop is not important, as we shall see in a
moment. Only the area of the loop will matter.
The induced potential is indicated as VT with
reference polarity as indicated in the figure. This
potential is given by Faraday’s Law:
VT = − ∂
∂tΦ (8.27)
Here Φ is the magnetic flux associated with an open
surface S bounded by the loop:
Φ =
∫
S
B · ds (8.28)
As usual, S can be any surface that intersects all",Electromagnetics_Vol1.pdf
"188 CHAPTER 8. TIME-V AR YING FIELDS
z
y
x
V

+
-
direction of
rotation
Figure 8.11: A rudimentary single-loop generator,
shown at time t= 0 .
magnetic field lines passing through the loop, and
also as usual, the simplest choice is simply the planar
area bounded by the loop. The differential surface
element ds is ˆnds, where ˆn is determined by the
reference polarity of VT according to the “right hand
rule” convention from Stokes’ Theorem. Making
substitutions, we have
Φ =
∫
S
(
ˆbB0
)
· (ˆnds)
=
[
ˆb · ˆn
]
B0
∫
S
ds
=
[
ˆb · ˆnA
]
B0 (8.29)
where Ais the area of the loop.
T o make headway , an expression for ˆn is needed. The
principal difficulty here is that ˆn is rotating with the
loop, so it is time-varying. T o sort this out, first
consider the situation at time t= 0 , which is the
moment illustrated in Figure 8.11. Beginning at the
“ −” terminal, we point the thumb of our right hand in
the direction that leads to the “ +” terminal by
traversing the loop; ˆn is then the direction",Electromagnetics_Vol1.pdf
"“ −” terminal, we point the thumb of our right hand in
the direction that leads to the “ +” terminal by
traversing the loop; ˆn is then the direction
perpendicular to the plane of the loop in which the
fingers of our right hand pass through the loop. W e
see that at t= 0 , ˆn = + ˆy. A bit later, the loop will
have rotated by one-fourth of a complete rotation, and
at that time ˆn = −ˆx. This occurs at t= 1 /(4f0).
Later still, the loop will have rotated by one-half of a
compete rotation, and then ˆn = −ˆy. This occurs at
t= 1 /(2f0). It is apparent that
ˆn(t) = −ˆx sin 2πf0t+ ˆy cos 2πf0t (8.30)
as can be confirmed by checking the three special
cases identified above. Now applying Faraday’s Law ,
we find
VT = − ∂
∂tΦ
= − ∂
∂t
[
ˆb · ˆn(t)
]
AB0
= −
[
ˆb · ∂
∂tˆn(t)
]
AB0 (8.31)
For notational convenience we make the following
definition:
ˆn′(t) ≜ − 1
2πf0
∂
∂tˆn(t) (8.32)
which is simply the time derivative of ˆn divided by
2πf0 so as to retain a unit vector. The reason for",Electromagnetics_Vol1.pdf
"definition:
ˆn′(t) ≜ − 1
2πf0
∂
∂tˆn(t) (8.32)
which is simply the time derivative of ˆn divided by
2πf0 so as to retain a unit vector. The reason for
including a change of sign will become apparent in a
moment. Applying this definition, we find
ˆn′(t) = + ˆx cos 2πf0t+ ˆy sin 2πf0t (8.33)
Note that this is essentially the definition of the radial
basis vector ˆρfrom the cylindrical coordinate system
(which is why we applied the minus sign in
Equation 8.32). Recognizing this, we write
ˆρ(t) = + ˆx cos 2πf0t+ ˆy sin 2πf0t (8.34)
and finally
VT = +2 πf0AB0 ˆb · ˆρ(t)
(8.35)
If the purpose of this device is to generate power, then
presumably we would choose the magnetic field to be
in a direction that maximizes the maximum value of
ˆb · ˆρ(t). Therefore, power is optimized for B
polarized entirely in some combination of ˆx and ˆy,
and with B · ˆz = 0 . Under that constraint, we see that
VT varies sinusoidally with frequency f0 and exhibits
peak magnitude
max |VT(t)| = 2 πf0AB0
(8.36)",Electromagnetics_Vol1.pdf
"8.8. THE MAXWELL-F ARADA Y EQUA TION 189
It’s worth noting that the maximum voltage
magnitude is achieved when the plane of the loop is
parallel to B; i.e., when ˆb · ˆn(t) = 0 so that
Φ( t) = 0 . Why is that? Because this is when Φ( t) is
most rapidly increasing or decreasing. Conversely ,
when the plane of the loop is perpendicular to B (i.e.,
ˆb · ˆn(t) = 1 ), |Φ( t)| is maximum but its
time-derivative is zero, so VT(t) = 0 at this instant.
Example 8.4. Rudimentary electric generator.
The generator in Figure 8.11 consists of a
circular loop of radius a= 1 cm rotating at 1000
revolutions per second in a static and
spatially-uniform magnetic flux density of 1 mT
in the +ˆx direction. What is the induced
potential?
Solution. From the problem statement,
f0 = 1 kHz, B0 = 1 mT , and ˆb = + ˆx.
Therefore ˆb · ˆρ(t) = ˆx · ˆρ(t) = cos 2 πf0t. The
area of the loop is A= πa2. From
Equation 8.35 we obtain
VT(t) ∼
= (1.97 mV) cos [(6.28 krad/s) t]",Electromagnetics_Vol1.pdf
"area of the loop is A= πa2. From
Equation 8.35 we obtain
VT(t) ∼
= (1.97 mV) cos [(6.28 krad/s) t]
Finally , we note that it is not really necessary for the
loop to rotate in the presence of a magnetic field with
constant ˆb; it works equally well for the loop to be
stationary and for ˆb to rotate – in fact, this is
essentially the same problem. In some practical
generators, both the potential-generating coils and
fields (generated by some combination of magnets
and coils) rotate.
Additional Reading:
• “Electric Generator” on Wikipedia.
8.8 The Maxwell-Faraday
Equation
[m0050]
In this section, we generalize Kirchoff’s V oltage Law
(KVL), previously encountered as a principle of
electrostatics in Sections 5.10 and 5.11. KVL states
that in the absence of a time-varying magnetic flux,
the electric potential accumulated by traversing a
closed path C is zero. Here is that idea in
mathematical form:
V =
∮
C
E · dl = 0 (8.37)
Now recall Faraday’s Law (Section 8.3):
V = − ∂
∂tΦ = − ∂
∂t
∫
S",Electromagnetics_Vol1.pdf
"mathematical form:
V =
∮
C
E · dl = 0 (8.37)
Now recall Faraday’s Law (Section 8.3):
V = − ∂
∂tΦ = − ∂
∂t
∫
S
B · ds (8.38)
Here, S is any open surface that intersects all
magnetic field lines passing through C, with the
relative orientations of C and ds determined in the
usual way by the Stokes’ Theorem convention
(Section 4.9). Note that Faraday’s Law agrees with
KVL in the magnetostatic case. If magnetic flux is
constant, then Faraday’s Law says V = 0 . However,
Faraday’s Law is very clearly not consistent with
KVL if magnetic flux is time-varying. The correction
is simple enough; we can simply set these expressions
to be equal. Here we go:
∮
C
E · dl = − ∂
∂t
∫
S
B · ds (8.39)
This general form is known by a variety of names;
here we refer to it as the Maxwell-F araday Equation
(MFE).
The integral form of the Maxwell-Faraday Equa-
tion (Equation 8.39) states that the electric poten-
tial associated with a closed path C is due entirely
to electromagnetic induction, via Faraday’s Law .",Electromagnetics_Vol1.pdf
"tial associated with a closed path C is due entirely
to electromagnetic induction, via Faraday’s Law .
Despite the great significance of this expression as
one of Maxwell’s Equations, one might argue that all
we have done is simply to write Faraday’s Law in a
slightly more verbose way . This is true. The real
power of the MFE is unleashed when it is expressed",Electromagnetics_Vol1.pdf
"190 CHAPTER 8. TIME-V AR YING FIELDS
in differential, as opposed to integral form. Let us
now do this.
W e can transform the left-hand side of Equation 8.39
into a integral over S using Stokes’ Theorem.
Applying Stokes’ theorem on the left, we obtain
∫
S
(∇ × E) · ds = − ∂
∂t
∫
S
B · ds (8.40)
Now exchanging the order of integration and
differentiation on the right hand side:
∫
S
(∇ × E) · ds =
∫
S
(
− ∂
∂tB
)
· ds (8.41)
The surface S on both sides is the same, and we have
not constrained S in any way . S can be any
mathematically-valid open surface anywhere in space,
having any size and any orientation. The only way the
above expression can be universally true under these
conditions is if the integrands on each side are equal
at every point in space. Therefore,
∇ × E = − ∂
∂tB (8.42)
which is the MFE in differential form.
What does this mean? Recall that the curl of E is a
way to take a directive of E with respect to position
(Section 4.8). Therefore the MFE constrains spatial",Electromagnetics_Vol1.pdf
"way to take a directive of E with respect to position
(Section 4.8). Therefore the MFE constrains spatial
derivatives of E to be simply related to the rate of
change of B. Said plainly:
The differential form of the Maxwell-Faraday
Equation (Equation 8.42) relates the change in
the electric field with position to the change in
the magnetic field with time.
Now that is arguably new and useful information. W e
now see that electric and magnetic fields are coupled
not only for line integrals and fluxes, but also at each
point in space.
Additional Reading:
• “Faraday’s Law of Induction” on Wikipedia.
• “Maxwell’s Equations” on Wikipedia.
8.9 Displacement Current and
Ampere’s Law
[m0053]
In this section, we generalize Ampere’s Law ,
previously encountered as a principle of
magnetostatics in Sections 7.4 and 7.9. Ampere’s
Law states that the current Iencl flowing through
closed path C is equal to the line integral of the
magnetic field intensity H along C. That is:
∮
C
H · dl = Iencl (8.43)",Electromagnetics_Vol1.pdf
"closed path C is equal to the line integral of the
magnetic field intensity H along C. That is:
∮
C
H · dl = Iencl (8.43)
W e shall now demonstrate that this equation is
unreliable if the current is not steady; i.e., not DC.
First, consider the situation shown in Figure 8.12.
Here, a current I flows in the wire, subsequently
generating a magnetic field H that circulates around
the wire (Section 7.5). When we perform the
integration in Ampere’s Law along any path C
enclosing the wire, the result is I, as expected. In this
case, Ampere’s Law is working even when I is
time-varying.
Now consider the situation shown in Figure 8.13, in
which we have introduced a parallel-plate capacitor.
In the DC case, this situation is simple. No current
flows, so there is no magnetic field and Ampere’s Law
is trivially true. In the AC case, the current I can be
non-zero, but we must be clear about the physical
origin of this current. What is happening is that for",Electromagnetics_Vol1.pdf
"non-zero, but we must be clear about the physical
origin of this current. What is happening is that for
one half of a period, a source elsewhere in the circuit
is moving positive charge to one side of the capacitor
and negative charge to the other side. For the other
half-period, the source is exchanging the charge, so
that negative charge appears on the previously
positively-charged side and vice-versa. Note that at
no point is current flowing directly from one side of
the capacitor to the other; instead, all current must
flow through the circuit in order to arrive at the other
plate. Even though there is no current between the
plates, there is current in the wire, and therefore there
is also a magnetic field associated with that current.
Now we are ready to shine a light on the problem.
Recall that from Stokes’ Theorem (Section 4.9), the
line integral over C is mathematically equivalent to an",Electromagnetics_Vol1.pdf
"8.9. DISPLACEMENT CURRENT AND AMPERE’S LA W 191
I
I
S1
H
C
C. Burks (modified)
Figure 8.12: Ampere’s Law applied to a continuous
line of current.
E
I
I
S1
H
C
C. Burks (modified)
Figure 8.13: Ampere’s Law applied to a parallel plate
capacitor.
integral over any open surface S that is bounded by C.
T wo such surfaces are shown in Figure 8.12 and
Figure 8.13, indicated as S1 and S2. In the wire-only
scenario of Figure 8.12, the choice of S clearly
doesn’t matter; any valid surface intersects current
equal to I. Similarly in the scenario of Figure 8.13,
everything seems fine if we choose S = S1. If, on the
other hand, we select S2 in the parallel-plate capacitor
case, then we have a problem. There is no current
flowing through S2, so the right side of Equation 8.43
is zero even though the left side is potentially
non-zero. So, it appears that something necessary for
the time-varying case is missing from Equation 8.43.
T o resolve the problem, we postulate an additional",Electromagnetics_Vol1.pdf
"the time-varying case is missing from Equation 8.43.
T o resolve the problem, we postulate an additional
term in Ampere’s Law that is non-zero in the above
scenario. Specifically , we propose:
∮
C
H · dl = Ic + Id (8.44)
where Ic is the enclosed current (formerly identified
as Iencl) and Id is the proposed new term. If we are to
accept this postulate, then here is a list of things we
know about Id:
• Id has units of current (A).
• Id = 0 in the DC case and is potentially
non-zero in the AC case. This implies that Id is
the time derivative of some other quantity .
• Id must be somehow related to the electric field.
How do we know Id must be related to the electric
field? This is because the Maxwell-Faraday Equation
(Section 8.8) tells us that spatial derivatives of E are
related to time derivatives of H; i.e., E and H are
coupled in the time-varying (here, AC) case. This
coupling between E and H must also be at work here,
but we have not yet seen E play a role. This is pretty",Electromagnetics_Vol1.pdf
"coupling between E and H must also be at work here,
but we have not yet seen E play a role. This is pretty
strong evidence that Id depends on the electric field.
Without further ado, here’s Id:
Id =
∫
S
∂D
∂t · ds (8.45)
where D is the electric flux density (units of C/m 2)
and is equal to ǫE as usual, and S is the same open
surface associated with C in Ampere’s Law . Note that",Electromagnetics_Vol1.pdf
"192 CHAPTER 8. TIME-V AR YING FIELDS
this expression meets our expectations: It is
determined by the electric field, it is zero when the
electric field is constant (i.e., not time varying), and
has units of current.
The quantity Id is commonly known as displacement
current. It should be noted that this name is a bit
misleading, since Id is not a current in the
conventional sense. Certainly , it is not a conduction
current – conduction current is represented by Ic, and
there is no current conducted through an ideal
capacitor. It is not unreasonable to think of Id as
current in a more general sense, for the following
reason. At one instant, charge is distributed one way
and at another, it is distributed in another way . If you
define current as a time variation in the charge
distribution relative to S – regardless of the path taken
by the charge – then Id is a current. However, this
distinction is a bit philosophical, so it may be less
confusing to interpret “displacement current” instead",Electromagnetics_Vol1.pdf
"distinction is a bit philosophical, so it may be less
confusing to interpret “displacement current” instead
as a separate electromagnetic quantity that just
happens to have units of current.
Now we are able to write the general form of
Ampere’s Law that applies even when sources are
time-varying. Here it is:
∮
C
H · dl = Ic +
∫
S
∂D
∂t · ds (8.46)
As is the case in the Maxwell-Faraday Equation, most
of the utility of Ampere’s Law is unleashed when
expressed in differential form. T o obtain this form the
first step is to write Ic as an integral of over S; this is
simply (see Section 6.2):
Ic =
∫
S
J · ds (8.47)
where J is the volume current density (units of
A/m2). So now we have
∮
C
H · dl =
∫
S
J · ds +
∫
S
∂D
∂t · ds
=
∫
S
(
J + ∂D
∂t
)
· ds (8.48)
W e can transform the left side of the above equation
into a integral over S using Stokes’ Theorem
(Section 4.9). W e obtain
∫
S
(∇ × H) · ds =
∫
S
(
J + ∂D
∂t
)
· ds (8.49)
The surface S on both sides is the same, and we have",Electromagnetics_Vol1.pdf
"(Section 4.9). W e obtain
∫
S
(∇ × H) · ds =
∫
S
(
J + ∂D
∂t
)
· ds (8.49)
The surface S on both sides is the same, and we have
not constrained S in any way . S can be any
mathematically-valid open surface anywhere in space,
having any size and any orientation. The only way the
above expression can be universally true under these
conditions is if the integrands on each side are equal
at every point in space. Therefore:
∇ × H = J + ∂
∂tD (8.50)
which is Ampere’s Law in differential form.
What does Equation 8.50 mean? Recall that the curl
of H is a way to describe the direction and rate of
change of H with position. Therefore, this equation
constrains spatial derivatives of H to be simply
related to J and the time derivative of D
(displacement current). Said plainly:
The differential form of the general (time-
varying) form of Ampere’s Law (Equation 8.50)
relates the change in the magnetic field with po-
sition to the change in the electric field with time,
plus current.",Electromagnetics_Vol1.pdf
"relates the change in the magnetic field with po-
sition to the change in the electric field with time,
plus current.
As is the case in the Maxwell-Faraday Equation
(Section 8.8), we see that electric and magnetic fields
become coupled at each point in space when sources
are time-varying.
Additional Reading:
• “Displacement Current” on Wikipedia.
• “Ampere’s Circuital Law” on Wikipedia.
• “Maxwell’s Equations” on Wikipedia.
[m0126]",Electromagnetics_Vol1.pdf
"8.9. DISPLACEMENT CURRENT AND AMPERE’S LA W 193
Image Credits
Fig. 8.1: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0003 fCoilBarMagnet.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /), modified from the original.
Fig. 8.2: E. Bach, https://commons.wikimedia.org/wiki/File:Faraday emf experiment.svg,
public domain via CC0 1.0 (https://creativecommons.org/p ublicdomain/zero/1.0/deed.en), modified from
the original.
Fig. 8.8: c⃝ BillC, https://commons.wikimedia.org/wiki/File:Transformer3d col3.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/). Modified by author.
Fig. 8.10: c⃝ SpinningSpark, https://en.wikipedia.org/wiki/File:Transformer balance.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/). Modified by author.
Fig. 8.12: C. Burks, https://en.wikipedia.org/wiki/File:Displacement current in capacitor.svg,
public domain. Heavily modified by author.",Electromagnetics_Vol1.pdf
"Fig. 8.12: C. Burks, https://en.wikipedia.org/wiki/File:Displacement current in capacitor.svg,
public domain. Heavily modified by author.
Fig. 8.13: C. Burks, https://en.wikipedia.org/wiki/File:Displacement current in capacitor.svg,
public domain. Heavily modified by author.",Electromagnetics_Vol1.pdf
"Chapter 9
Plane W aves in Lossless Media
9.1 Maxwell’s Equations in
Differential Phasor Form
[m0042]
In this section, we derive the phasor form of
Maxwell’s Equations from the general time-varying
form of these equations. Here we are interested
exclusively in the differential (“point”) form of these
equations. It is assumed that the reader is comfortable
with phasor representation and its benefits; if not, a
review of Section 1.5 is recommended before
attempting this section.
Maxwell’s Equations in differential time-domain
form are Gauss’ Law (Section 5.7):
∇ · D = ρv (9.1)
the Maxwell-Faraday Equation (MFE; Section 8.8):
∇ × E = − ∂
∂tB (9.2)
Gauss’ Law for Magnetism (GSM; Section 7.3):
∇ · B = 0 (9.3)
and Ampere’s Law (Section 8.9):
∇ × H = J + ∂
∂tD (9.4)
W e begin with Gauss’s Law (Equation 9.1). W e
define ˜D and ˜ρv as phasor quantities through the
usual relationship:
D = Re
{
˜Dejωt
}
(9.5)
and
ρv = Re
{
˜ρvejωt}
(9.6)
Substituting these expressions into Equation 9.1:
∇ ·
[
Re
{",Electromagnetics_Vol1.pdf
"usual relationship:
D = Re
{
˜Dejωt
}
(9.5)
and
ρv = Re
{
˜ρvejωt}
(9.6)
Substituting these expressions into Equation 9.1:
∇ ·
[
Re
{
˜Dejωt
}]
= Re
{
˜ρvejωt}
(9.7)
Divergence is a real-valued linear operator. Therefore,
we may exchange the order of the “Re” and “ ∇·”
operations (this is “Claim 2” from Section 1.5):
Re
{
∇ ·
[
˜Dejωt
]}
= Re
{
˜ρvejωt}
(9.8)
Next, we note that the differentiation associated with
the divergence operator is with respect to position and
not with respect to time, so the order of operations
may be further rearranged as follows:
Re
{[
∇ · ˜D
]
ejωt
}
= Re
{
˜ρvejωt}
(9.9)
Finally , we note that the equality of the left and right
sides of the above equation implies the equality of the
associated phasors (this is “Claim 1” from
Section 1.5); thus,
∇ · ˜D = ˜ρv
(9.10)
In other words, the differential form of Gauss’ Law
for phasors is identical to the differential form of
Gauss’ Law for physical time-domain quantities.",Electromagnetics_Vol1.pdf
"for phasors is identical to the differential form of
Gauss’ Law for physical time-domain quantities.
The same procedure applied to the MFE is only a
little more complicated. First, we establish the phasor
representations of the electric and magnetic fields:
E = Re
{
˜Eejωt
}
(9.11)
B = Re
{
˜Bejωt
}
(9.12)
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"9.1. MAXWELL ’S EQUA TIONS IN DIFFERENTIAL PHASOR FORM 195
After substitution into Equation 9.2:
∇ ×
[
Re
{
˜Eejωt
}]
= − ∂
∂t
[
Re
{
˜Bejωt
}]
(9.13)
Both curl and time-differentiation are real-valued
linear operations, so we are entitled to change the
order of operations as follows:
Re
{
∇ ×
[
˜Eejωt
]}
= −Re
{ ∂
∂t
[
˜Bejωt
] }
(9.14)
On the left, we note that the time dependence ejωt can
be pulled out of the argument of the curl operator,
since it does not depend on position:
Re
{[
∇ × ˜E
]
ejωt
}
= −Re
{ ∂
∂t
[
˜Bejωt
] }
(9.15)
On the right, we note that ˜B is constant with respect
to time (because it is a phasor), so:
Re
{[
∇ × ˜E
]
ejωt
}
= −Re
{
˜B ∂
∂tejωt
}
= −Re
{
˜Bjωejωt
}
= Re
{[
−jω˜B
]
ejωt
}
(9.16)
And so we have found:
∇ × ˜E = −jω˜B (9.17)
Let’s pause for a moment to consider the above result.
In the general time-domain version of the MFE, we
must take spatial derivatives of the electric field and
time derivatives of the magnetic field. In the phasor",Electromagnetics_Vol1.pdf
"must take spatial derivatives of the electric field and
time derivatives of the magnetic field. In the phasor
version of the MFE, the time derivative operator has
been replaced with multiplication by jω. This is a
tremendous simplification since the equations now
involve differentiation over position only .
Furthermore, no information is lost in this
simplification – for a reminder of why that is, see the
discussion of Fourier Analysis at the end of
Section 1.5. Without this kind of simplification, much
of what is now considered “basic” engineering
electromagnetics would be intractable.
The procedure for conversion of the remaining two
equations is very similar, yielding:
∇ · ˜B = 0
(9.18)
∇ × ˜H = ˜J + jω˜D (9.19)
The details are left as an exercise for the reader.
The differential form of Maxwell’s Equations
(Equations 9.10, 9.17, 9.18, and 9.19) involve
operations on the phasor representations of the
physical quantities. These equations have the ad-",Electromagnetics_Vol1.pdf
"(Equations 9.10, 9.17, 9.18, and 9.19) involve
operations on the phasor representations of the
physical quantities. These equations have the ad-
vantage that differentiation with respect to time
is replaced by multiplication by jω.",Electromagnetics_Vol1.pdf
"196 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
9.2 W ave Equations for
Source-Free and Lossless
Regions
[m0036]
Electromagnetic waves are solutions to a set of
coupled differential simultaneous equations – namely ,
Maxwell’s Equations. The general solution to these
equations includes constants whose values are
determined by the applicable electromagnetic
boundary conditions. However, this direct approach
can be difficult and is often not necessary . In
unbounded homogeneous regions that are “source
free” (containing no charges or currents), a simpler
approach is possible. In this section, we reduce
Maxwell’s Equations to wave equations that apply to
the electric and magnetic fields in this simpler
category of scenarios. Before reading further, the
reader should consider a review of Section 1.3 (noting
in particular Equation 1.1) and Section 3.6 (wave
equations for voltage and current on a transmission
line). This section seeks to develop the analogous",Electromagnetics_Vol1.pdf
"equations for voltage and current on a transmission
line). This section seeks to develop the analogous
equations for electric and magnetic waves.
W e can get the job done using the differential “point”
phasor form of Maxwell’s Equations, developed in
Section 9.1. Here they are:
∇ · ˜D = ˜ρv (9.20)
∇ × ˜E = −jω˜B (9.21)
∇ · ˜B = 0 (9.22)
∇ × ˜H = ˜J + jω˜D (9.23)
In a source-free region, there is no net charge and no
current, hence ˜ρv = 0 and ˜J = 0 in the present
analysis. The above equations become
∇ · ˜D = 0 (9.24)
∇ × ˜E = −jω˜B (9.25)
∇ · ˜B = 0 (9.26)
∇ × ˜H = + jω˜D (9.27)
Next, we recall that ˜D = ǫ˜E and that ǫis a
real-valued constant for a medium that is
homogeneous, isotropic, and linear (Section 2.8).
Similarly , ˜B = µ˜H and µis a real-valued constant.
Thus, under these conditions, it is sufficient to
consider either ˜D or ˜E and either ˜B or ˜H. The
choice is arbitrary , but in engineering applications it is
customary to use ˜E and ˜H. Eliminating the",Electromagnetics_Vol1.pdf
"choice is arbitrary , but in engineering applications it is
customary to use ˜E and ˜H. Eliminating the
now-redundant quantities ˜D and ˜B, the above
equations become
∇ · ˜E = 0 (9.28)
∇ × ˜E = −jωµ˜H (9.29)
∇ · ˜H = 0 (9.30)
∇ × ˜H = + jωǫ˜E (9.31)
It is important to note that requiring the region of
interest to be source-free precludes the possibility of
loss in the medium. T o see this, let’s first be clear
about what we mean by “loss. ” For an
electromagnetic wave, loss is observed as a reduction
in the magnitude of the electric and magnetic field
with increasing distance. This reduction is due to the
dissipation of power in the medium. This occurs
when the conductivity σis greater than zero because
Ohm’s Law for Electromagnetics ( ˜J = σ˜E;
Section 6.3) requires that power in the electric field be
transferred into conduction current, and is thereby lost
to the wave (Section 6.6). When we required J to be
zero above, we precluded this possibility; that is, we",Electromagnetics_Vol1.pdf
"to the wave (Section 6.6). When we required J to be
zero above, we precluded this possibility; that is, we
implicitly specified σ= 0 . The fact that the
constitutive parameters µand ǫappear in
Equations 9.28–9.31, but σdoes not, is further
evidence of this.
Equations 9.28–9.31 are Maxwell’s Equations for
a region comprised of isotropic, homogeneous,
and source-free material. Because there can be no
conduction current in a source-free region, these
equations apply only to material that is lossless
(i.e., having negligible σ).
Before moving on, one additional disclosure is
appropriate. It turns out that there actually is a way to
use Equations 9.28–9.31 for regions in which loss is
significant. This requires a redefinition of ǫas a
complex-valued quantity . W e shall not consider this
technique in this section. W e mention this simply
because one should be aware that if permittivity",Electromagnetics_Vol1.pdf
"9.2. W A VE EQUA TIONS FOR SOURCE-FREE AND LOSSLESS REGIONS 197
appears as a complex-valued quantity , then the
imaginary part represents loss.
T o derive the wave equations we begin with the MFE,
Equation 9.29. T aking the curl of both sides of the
equation we obtain
∇ ×
(
∇ × ˜E
)
= ∇ ×
(
−jωµ˜H
)
= −jωµ
(
∇ × ˜H
)
(9.32)
On the right we can eliminate ∇ × ˜H using
Equation 9.31:
−jωµ
(
∇ × ˜H
)
= −jωµ
(
+jωǫ˜E
)
= + ω2µǫ˜E (9.33)
On the left side of Equation 9.32, we apply the vector
identity
∇ × ∇ × A = ∇ (∇ · A) − ∇2A (9.34)
which in this case is
∇ × ∇ × ˜E = ∇
(
∇ · ˜E
)
− ∇2 ˜E (9.35)
W e may eliminate the first term on the right using
Equation 9.28, yielding
∇ × ∇ × ˜E = −∇2 ˜E (9.36)
Substituting these results back into Equation 9.32 and
rearranging terms we have
∇2 ˜E + ω2µǫ˜E = 0 (9.37)
This is the wave equation for ˜E. Note that it is a
homogeneous (in the mathematical sense of the word)
differential equation, which is expected since we have",Electromagnetics_Vol1.pdf
"homogeneous (in the mathematical sense of the word)
differential equation, which is expected since we have
derived it for a source-free region.
It is common to make the following definition
β ≜ ω√
µǫ (9.38)
so that Equation 9.37 may be written
∇2 ˜E + β2 ˜E = 0 (9.39)
Why go the the trouble of defining β? One reason is
that βconveniently captures the contribution of the
frequency , permittivity , and permeability all in one
constant. Another reason is to emphasize the
connection to the parameter βappearing in
transmission line theory (see Section 3.8 for a
reminder). It should be clear that βis a phase
propagation constant , having units of 1/m (or rad/m,
if you prefer), and indicates the rate at which the
phase of the propagating wave progresses with
distance.
The wave equation for ˜H is obtained using essentially
the same procedure, which is left as an exercise for
the reader. It should be clear from the duality
apparent in Equations 9.28-9.31 that the result will be",Electromagnetics_Vol1.pdf
"the reader. It should be clear from the duality
apparent in Equations 9.28-9.31 that the result will be
very similar. One finds:
∇2 ˜H + β2 ˜H = 0
(9.40)
Equations 9.39 and 9.40 are the wave equations
for ˜E and ˜H, respectively , for a region comprised
of isotropic, homogeneous, lossless, and source-
free material.
Looking ahead, note that ˜E and ˜H are solutions to the
same homogeneous differential equation.
Consequently , ˜E and ˜H cannot be different by more
than a constant factor and a direction. In fact, we can
also determine something about the factor simply by
examining the units involved: Since ˜E has units of
V/m and ˜H has units of A/m, this factor will be
expressible in units of the ratio of V/m to A/m, which
is Ω . This indicates that the factor will be an
impedance. This factor is known as the wave
impedance and will be addressed in Section 9.5. This
impedance is analogous the characteristic impedance
of a transmission line (Section 3.7).
Additional Reading:",Electromagnetics_Vol1.pdf
"impedance is analogous the characteristic impedance
of a transmission line (Section 3.7).
Additional Reading:
• “W ave Equation” on Wikipedia.
• “Electromagnetic W ave Equation” on Wikipedia.",Electromagnetics_Vol1.pdf
"198 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
9.3 T ypes of W aves
[m0142]
Solutions to the electromagnetic wave equations
(Section 9.2) exist in a variety of forms, representing
different types of waves. It is useful to identify three
particular geometries for unguided waves. Each of
these geometries is defined by the shape formed by
surfaces of constant phase, which we refer to as
phasefronts. (Keep in mind the analogy between
electromagnetic waves and sound waves (described in
Section 1.3), and note that sound waves also exhibit
these geometries.)
A spherical wave has phasefronts that form
concentric spheres, as shown in Figure 9.1. W aves are
well-modeled as spherical when the dimensions of the
source of the wave are small relative to the scale at
which the wave is observed. For example, the wave
radiated by an antenna having dimensions of 10 cm,
when observed in free space over a scale of 10 km,
appears to have phasefronts that are very nearly",Electromagnetics_Vol1.pdf
"when observed in free space over a scale of 10 km,
appears to have phasefronts that are very nearly
spherical. Note that the magnitude of the field on a
phasefront of a spherical wave may vary significantly ,
but it is the shape of phasefronts that make it a
spherical wave.
A cylindrical wave exhibits phasefronts that form
concentric cylinders, as shown in Figure 9.2. Said
differently , the phasefronts of a cylindrical wave are
circular in one dimension, and planar in the
perpendicular direction. A cylindrical wave is often a
good description of the wave that emerges from a
line-shaped source.
A plane wave exhibits phasefronts that are planar,
with planes that are parallel to each other as shown in
Figure 9.3. There are two conditions in which waves
are well-modeled as plane waves. First, some
structures give rise to waves that appear to have
planar phasefronts over a limited area; a good
example is the wave radiated by a parabolic reflector,
as shown in Figure 9.4. Second, all waves are",Electromagnetics_Vol1.pdf
"planar phasefronts over a limited area; a good
example is the wave radiated by a parabolic reflector,
as shown in Figure 9.4. Second, all waves are
well-modeled as plane waves when observed over a
small region located sufficiently far from the source.
In particular, spherical waves are “locally planar” in
the sense that they are well-modeled as planar when
observed over a small portion of the spherical
c⃝ Y . Qin CC BY 4.0
Figure 9.1: The phasefronts of a spherical wave form
concentric spheres.
c⃝ Y . Qin CC BY 4.0
Figure 9.2: The phasefronts of a cylindrical wave
form concentric cylinders.",Electromagnetics_Vol1.pdf
"9.4. UNIFORM PLANE W A VES: DERIV A TION 199
c⃝ Y . Qin CC BY 4.0
Figure 9.3: The phasefronts of a plane wave form
parallel planes.
feed
reflector
planar p      onts
c⃝ Y . Qin CC BY 4.0
Figure 9.4: Plane waves formed in the region in front
of a parabolic reflector antenna.
phasefront, as shown in Figure 9.5. An analogy is that
the Earth seems “locally flat” to an observer on the
ground, even though it is clearly spherical to an
observer in orbit. The “locally planar” approximation
is often employed because it is broadly applicable and
simplifies analysis.
Most waves are well-modeled as spherical, cylin-
drical, or plane waves.
Plane waves (having planar phasefronts) are of
particular importance due to wide applicability of
the “locally planar” approximation.
""locally-planar"" phasefronts
spherical phasefront
c⃝ Y . Qin CC BY 4.0
Figure 9.5: “Locally planar” approximation of a
spherical wave over a limited area.
9.4 Uniform Plane W aves:
Derivation
[m0038]",Electromagnetics_Vol1.pdf
"c⃝ Y . Qin CC BY 4.0
Figure 9.5: “Locally planar” approximation of a
spherical wave over a limited area.
9.4 Uniform Plane W aves:
Derivation
[m0038]
Section 9.2 showed how Maxwell’s Equations could
be reduced to a pair of phasor-domain “wave
equations, ” namely:
∇2 ˜E + β2 ˜E = 0 (9.41)
∇2 ˜H + β2 ˜H = 0 (9.42)
where β = ω√
µǫ, assuming unbounded
homogeneous, isotropic, lossless, and source-free
media. In this section, we solve these equations for
the special case of a uniform plane wave. A uniform
plane wave is one for which both ˜E and ˜H have
constant magnitude and phase in a specified plane.
Despite being a special case, the solution turns out to
be broadly applicable, appearing as a common
building block in many practical and theoretical
problems in unguided propagation (as explained in
Section 9.3), as well as in more than a few
transmission line and waveguide problems.
T o begin, let us assume that the plane over which ˜E
and ˜H have constant magnitude and phase is a plane",Electromagnetics_Vol1.pdf
"transmission line and waveguide problems.
T o begin, let us assume that the plane over which ˜E
and ˜H have constant magnitude and phase is a plane
of constant z. First, note that we may make this",Electromagnetics_Vol1.pdf
"200 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
assumption with no loss of generality . For example,
we could alternatively select a plane of constant y,
solve the problem, and then simply exchange
variables to get a solution for planes of constant z(or
x).1 Furthermore, the solution for any planar
orientation not corresponding to a plane of constant x,
y, or zmay be similarly obtained by a rotation of
coordinates, since the physics of the problem does not
depend on the orientation of this plane – if it does,
then the medium is not isotropic!
W e may express the constraint that the magnitude and
phase of ˜E and ˜H are constant over a plane that is
perpendicular to the zaxis as follows:
∂
∂x
˜E = ∂
∂y
˜E = ∂
∂x
˜H = ∂
∂y
˜H = 0 (9.43)
Let us identify the Cartesian components of each of
these fields as follows:
˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez , and (9.44)
˜H = ˆx ˜Hx + ˆy ˜Hy + ˆz ˜Hz (9.45)
Now Equation 9.43 may be interpreted in detail for ˜E
as follows:
∂
∂x
˜Ex = ∂
∂x
˜Ey = ∂
∂x",Electromagnetics_Vol1.pdf
"˜H = ˆx ˜Hx + ˆy ˜Hy + ˆz ˜Hz (9.45)
Now Equation 9.43 may be interpreted in detail for ˜E
as follows:
∂
∂x
˜Ex = ∂
∂x
˜Ey = ∂
∂x
˜Ez = 0 (9.46)
∂
∂y
˜Ex = ∂
∂y
˜Ey = ∂
∂y
˜Ez = 0 (9.47)
and for ˜H as follows:
∂
∂x
˜Hx = ∂
∂x
˜Hy = ∂
∂x
˜Hz = 0 (9.48)
∂
∂y
˜Hx = ∂
∂y
˜Hy = ∂
∂y
˜Hz = 0 (9.49)
The wave equation for ˜E (Equation 9.41) written
explicitly in Cartesian coordinates is
[ ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
] (
ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez
)
+β2
(
ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez
)
= 0
(9.50)
1 By the way , this is a highly-recommended exercise for the stu-
dent.
Decomposing this equation into separate equations
for each of the three coordinates, we obtain the
following:
[ ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
]
˜Ex + β2 ˜Ex = 0 (9.51)
[ ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
]
˜Ey + β2 ˜Ey = 0 (9.52)
[ ∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
]
˜Ez + β2 ˜Ez = 0 (9.53)
Applying the constraints of Equations 9.46 and 9.47,
we note that many of the terms in
Equations 9.51–9.53 are zero. W e are left with:
∂2
∂z2
˜Ex + β2 ˜Ex = 0 (9.54)
∂2
∂z2",Electromagnetics_Vol1.pdf
"we note that many of the terms in
Equations 9.51–9.53 are zero. W e are left with:
∂2
∂z2
˜Ex + β2 ˜Ex = 0 (9.54)
∂2
∂z2
˜Ey + β2 ˜Ey = 0 (9.55)
∂2
∂z2
˜Ez + β2 ˜Ez = 0 (9.56)
Now we will show that Equation 9.43 also implies
that ˜Ez must be zero. T o show this, we use Ampere’s
Law for a source-free region (Section 9.2):
∇ × ˜H = + jωǫ˜E (9.57)
and take the dot product with ˆz on both sides:
ˆz ·
(
∇ × ˜H
)
= + jωǫ˜Ez
∂
∂y
˜Hx − ∂
∂x
˜Hy = + jωǫ˜Ez (9.58)
Again applying the constraints of Equation 9.43, the
left side of Equation 9.58 must be zero; therefore,
˜Ez = 0 . The exact same procedure applied to ˜H
(using the Maxwell-Faraday Equation; also given in
Section 9.2) reveals that ˜Hz is also zero. 2 Here is
what we have found:
If a wave is uniform over a plane, then the electric
and magnetic field vectors must lie in this plane.
This conclusion is a direct consequence of the fact
that Maxwell’s Equations require the electric field to",Electromagnetics_Vol1.pdf
"This conclusion is a direct consequence of the fact
that Maxwell’s Equations require the electric field to
2 Showing this is a highly-recommended exercise for the reader.",Electromagnetics_Vol1.pdf
"9.4. UNIFORM PLANE W A VES: DERIV A TION 201
be proportional to the curl of the magnetic field and
vice-versa.
The general solution to Equation 9.54 is:
˜Ex = E+
x0e−jβz + E−
x0e+jβz (9.59)
where E+
x0 and E−
x0 are complex-valued constants.
The values of these constants are determined by
boundary conditions – possibly sources – outside the
region of interest. Since in this section we are limiting
our scope to source-free and homogeneous regions,
we may for the moment consider the values of E+
x0
and E−
x0 to be arbitrary , since any values will satisfy
the associated wave equation. 3
Similarly we have for ˜Ey:
˜Ey = E+
y0e−jβz + E−
y0e+jβz (9.60)
where E+
y0 and E−
y0 are again arbitrary constants.
Summarizing, we have found
˜E = ˆx ˜Ex + ˆy ˜Ey (9.61)
where ˜Ex and ˜Ey are given by Equations 9.59 and
9.60, respectively .
Note that Equations 9.59 and 9.60 are essentially the
same equations encountered in the study of waves in
lossless transmission lines; for a reminder, see",Electromagnetics_Vol1.pdf
"same equations encountered in the study of waves in
lossless transmission lines; for a reminder, see
Section 3.6. Specifically , factors containing e−jβz
describe propagation in the +zdirection, whereas
factors containing e+jβz describe propagation in the
−zdirection. W e conclude:
If a wave is uniform over a plane, then possible
directions of propagation are the two directions
perpendicular to the plane.
Since we previously established that the electric and
magnetic field vectors must lie in the plane, we also
conclude:
The direction of propagation is perpendicular to
the electric and magnetic field vectors.
This conclusion turns out to be generally true; i.e., it
is not limited to uniform plane waves. Although we
3 The reader is encouraged to confirm that these are solutions b y
substitution into the associated wave equation.
will not provide a rigorous proof of this, one way to
see that this is true is to imagine that any type of wave
can be interpreted as the sum (formally , a linear",Electromagnetics_Vol1.pdf
"see that this is true is to imagine that any type of wave
can be interpreted as the sum (formally , a linear
combination) of uniform plane waves, so
perpendicular orientation of the field vectors with
respect to the direction of propagation is inescapable.
The same procedure yields the uniform plane wave
solution to the wave equation for ˜H, which is
˜H = ˆx ˜Hx + ˆy ˜Hy (9.62)
where
˜Hx = H+
x0e−jβz + H−
x0e+jβz (9.63)
˜Hy = H+
y0e−jβz + H−
y0e+jβz (9.64)
and where H+
x0, H−
x0, H+
y0 and H−
y0 are arbitrary
constants. Note that the solution is essentially the
same as that for ˜E, with the sole difference being that
the arbitrary constants may apparently have different
values.
T o this point, we have seen no particular relationship
between the electric and magnetic fields, and it may
appear that the electric and magnetic fields are
independent of each other. However, Maxwell’s
Equations – specifically , the two curl equations –
make it clear that there must be a relationship",Electromagnetics_Vol1.pdf
"independent of each other. However, Maxwell’s
Equations – specifically , the two curl equations –
make it clear that there must be a relationship
between these fields. Subsequently the arbitrary
constants in the solutions for ˜E and ˜H must also be
related. In fact, there are two considerations here:
• The magnitude and phase of ˜E must be related to
the magnitude and phase of ˜H. Since both fields
are solutions to the same differential (wave)
equation, they may differ by no more than a
multiplicative constant. Since the units of ˜E and
˜H are V/m and A/m respectively , this constant
must be expressible in units of V/m divided by
A/m; i.e., in units of Ω , an impedance.
• The direction of ˜E must be related to direction
of ˜H.
Let us now address these considerations. Consider an
electric field that points in one of the cardinal
directions – let’s say +ˆx – and make the definition
E0 ≜ E+
x0 for notational convenience. Then the
electric field intensity may be written as follows:",Electromagnetics_Vol1.pdf
"E0 ≜ E+
x0 for notational convenience. Then the
electric field intensity may be written as follows:
˜E = ˆxE0e−jβz (9.65)",Electromagnetics_Vol1.pdf
"202 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
Again, there is no loss of generality here since the
coordinate system could be rotated in such a way that
any uniform plane could be described in this way .
W e may now determine ˜H from ˜E using the
Maxwell-Faraday Equation (Section 9.2):
∇ × ˜E = −jωµ˜H (9.66)
Solving this equation for ˜H, we find:
˜H = ∇ × ˜E
−jωµ = ∇ ×
[ ˆxE0e−jβz]
−jωµ (9.67)
Now let us apply the curl operator. The complete
expression for the curl operator in Cartesian
coordinates is given in Section B.2. Here let us
consider one component at a time, starting with the ˆx
component:
ˆx ·
(
∇ × ˜E
)
= ∂˜Ez
∂y − ∂˜Ey
∂z (9.68)
Since ˜Ey = ˜Ez = 0 , the above expression is zero and
subsequently ˜Hx = 0 . Next, the ˆy component:
ˆy ·
(
∇ × ˜E
)
= ∂˜Ex
∂z − ∂˜Ez
∂x (9.69)
Here ˜Ez = 0 , so we have simply
ˆy ·
(
∇ × ˜E
)
= ∂˜Ex
∂z (9.70)
It is not necessary to repeat this procedure for ˜Hz,
since we know in advance that ˜H must be",Electromagnetics_Vol1.pdf
"ˆy ·
(
∇ × ˜E
)
= ∂˜Ex
∂z (9.70)
It is not necessary to repeat this procedure for ˜Hz,
since we know in advance that ˜H must be
perpendicular to the direction of propagation and
subsequently ˜Hz = 0 . Returning to Equation 9.67,
we obtain:
˜H = ˆy 1
−jωµ
∂
∂z
˜Ex
= ˆy 1
−jωµ
∂
∂z
(
E0e−jβz)
= ˆy 1
−jωµ
(
−jβE0e−jβz)
= ˆy β
ωµE0e−jβz (9.71)
Note that ˜H points in the +ˆy direction. So, as
expected, both ˜E and ˜H are perpendicular to the
H
E
E x  
direction of pr
o






!
""
Figure 9.6: Relationship between the electric field di-
rection, magnetic field direction, and direction of prop-
agation.
direction of propagation. However, we have now
found a more specific relationship: ˜E and ˜H are
perpendicular to each other . Just as ˆx × ˆy = ˆz, we
see that ˜E × ˜H points in the direction of propagation.
This is illustrated in Figure 9.6. Summarizing:
˜E, ˜H, and the direction of propagation ˜E× ˜H are
mutually perpendicular.
Now let us resolve the question of the factor relating",Electromagnetics_Vol1.pdf
"˜E, ˜H, and the direction of propagation ˜E× ˜H are
mutually perpendicular.
Now let us resolve the question of the factor relating
˜E and ˜H. The factor is now seen to be β/ωµ in
Equation 9.71, which can be simplified as follows:
β
ωµ = ω√µǫ
ωµ = 1√
µ/ǫ
(9.72)
The factor
√
µ/ǫappearing above has units of Ω is
known variously as the wave impedance or the
intrinsic impedance of the medium . Assigning this
quantity the symbol “ η, ” we have:
η≜
˜Ex
˜Hy
=
√ µ
ǫ (9.73)
The ratio of the electric field intensity to the
magnetic field intensity is the wave impedance η
(Equation 9.73; units of Ω ). In lossless media, η
is determined by the ratio of permeability of the
medium to the permittivity of the medium.",Electromagnetics_Vol1.pdf
"9.5. UNIFORM PLANE W A VES: CHARACTERISTICS 203
The wave impedance in free space, assigned the
symbol η0, is
η0 ≜
√
µ0
ǫ0
∼
= 377 Ω . (9.74)
Wrapping up our solution, we find that if ˜E is as
given by Equation 9.65, then
˜H = ˆy E0
η e−jβz (9.75)
9.5 Uniform Plane W aves:
Characteristics
[m0039]
In Section 9.4, expressions for the electric and
magnetic fields are determined for a uniform plane
wave in lossless media. If the planar phasefront is
perpendicular to the zaxis, then waves may propagate
in either the +ˆz direction of the −ˆz direction. If we
consider only the former, and select ˜E to point in the
+ˆx direction, then we find
˜E = + ˆxE0e−jβz (9.76)
˜H = + ˆy E0
η e−jβz (9.77)
where β = ω√µǫis the phase propagation constant,
η=
√
µ/ǫis the wave impedance, and E0 is a
complex-valued constant associated with the
magnitude and phase of the source. This result is in
fact completely general for uniform plane waves,
since any other possibility may be obtained by simply",Electromagnetics_Vol1.pdf
"fact completely general for uniform plane waves,
since any other possibility may be obtained by simply
rotating coordinates. In fact, this is pretty easy
because (as determined in Section 9.4) ˜E, ˜H, and the
direction of propagation are mutually perpendicular,
with the direction of propagation pointing in the same
direction as ˜E × ˜H.
In this section, we identify some important
characteristics of uniform plane waves, including
wavelength and phase velocity . Chances are that
much of what appears here will be familiar to the
reader; if not, a quick review of Sections 1.3
(“Fundamentals of W aves”) and 3.8 (“W ave
Propagation on a Transmission Line”) are
recommended.
First, recall that ˜E and ˜H are phasors representing
physical (real-valued) fields and are not the field
values themselves. The actual, physical electric field
intensity is
E = Re
{
˜Eejωt
}
(9.78)
= Re
{ˆxE0e−jβzejωt}
= ˆx |E0| cos (ωt− βz+ ψ)
where ψis the phase of E0. Similarly:
H = ˆy |E0|
η cos (ωt− βz+ ψ) (9.79)",Electromagnetics_Vol1.pdf
"204 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
x
z
y
E
H
c⃝ E. Boutet (Modified) CC BY SA 3.0
Figure 9.7: Relationship between the electric field di-
rection, magnetic field direction, and direction of prop-
agation (here, +ˆz).
This result is illustrated in Figure 9.7. Note that both
E and H (as well as their phasor representations)
have the same phase and have the same frequency and
position dependence cos (ωt− βz+ ψ). Since βis a
real-valued constant for lossless media, only the
phase, and not the magnitude, varies with z. If ω→ 0,
then β → 0 and the fields no longer depend on z; in
this case, the field is not propagating. For ω >0,
cos (ωt− βz+ ψ) is periodic in z; specifically , it has
the same value each time zincreases or decreases by
2π/β. This is, by definition, the wavelength λ:
λ≜ 2π
β (9.80)
If we observe this wave at some fixed point (i.e., hold
zconstant), we find that the electric and magnetic
fields are also periodic in time; specifically , they have",Electromagnetics_Vol1.pdf
"zconstant), we find that the electric and magnetic
fields are also periodic in time; specifically , they have
the same value each time tincreases by 2π/ω. W e
may characterize the speed at which the wave travels
by comparing the distance required to experience 2π
of phase rotation at a fixed time, which is 1/β; to the
time it takes to experience 2πof phase rotation at a
fixed location, which is 1/ω. This is known as the
phase velocity 4 vp:
vp ≜ 1/β
1/ω = ω
β (9.81)
Note that vp has the units expected from its definition,
namely (rad/s)/(rad/m) = m/s. If we make the
4 W e acknowledge that this is a misnomer, since velocity is prop-
erly defined as speed in a specified direction, and vp by itself does
not specify direction. In fact, the phase velocity in this ca se is prop-
erly said to be +ˆzvp. Nevertheless, we adopt the prevailing termi-
nology .
substitution β = ω√µǫ, we find
vp = ω
ω√µǫ = 1√µǫ (9.82)
Note that vp, like the wave impedance η, depends",Electromagnetics_Vol1.pdf
"nology .
substitution β = ω√µǫ, we find
vp = ω
ω√µǫ = 1√µǫ (9.82)
Note that vp, like the wave impedance η, depends
only on material properties. For example, the phase
velocity of an electromagnetic wave in free space,
given the special symbol c, is
c≜ vp|µ=µ0,ǫ=ǫ0
= 1
√µ0ǫ0
∼
= 3.00 × 108 m/s
(9.83)
This constant is commonly referred to as the speed of
light, but in fact it is the phase velocity of an
electromagnetic field at any frequency (not just
optical frequencies) in free space. Since the
permittivity ǫand permeability µof any material is
greater than that of a vacuum, vp in any material is
less than the phase velocity in free space.
Summarizing:
Phase velocity is the speed at which any point of
constant phase appears to travel along the direc-
tion of propagation.
Phase velocity is maximum ( = c) in free space,
and slower by a factor of 1/√µrǫr in any other
lossless medium.
Finally , we note the following relationship between
frequency f, wavelength, and phase velocity:
λ= 2π",Electromagnetics_Vol1.pdf
"lossless medium.
Finally , we note the following relationship between
frequency f, wavelength, and phase velocity:
λ= 2π
β = 2π
ω/vp
= 2π
2πf/vp
= vp
f (9.84)
Thus, given any two of the parameters f, λ, and vp,
we may solve for the remaining quantity . Also note
that as a consequence of the inverse-proportional
relationship between λand vp, we find:
At a given frequency , the wavelength in any ma-
terial is shorter than the wavelength in free space.
Example 9.1. W ave propagation in a lossless
dielectric.
Polyethylene is a low-loss dielectric having
ǫr ∼
= 2.3. What is the phase velocity in
polyethylene? What is wavelength in",Electromagnetics_Vol1.pdf
"9.5. UNIFORM PLANE W A VES: CHARACTERISTICS 205
polyethylene? The frequency of interest is
1 GHz.
Solution. Low-loss dielectrics exhibit µr ∼
= 1
and σ≈ 0. Therefore the phase velocity is
vp = 1
√µ0ǫrǫ0
= c√ǫr
∼
= 1.98 × 108 m/s
(9.85)
i.e., very nearly two-thirds of the speed of light
in free space. The wavelength at 1 GHz is
λ= vp
f
∼
= 19.8 cm (9.86)
Again, about two-thirds of a wavelength at the
same frequency in free space.
Returning to polarization and magnitude, it is useful
to note that Equation 9.79 could be written in terms of
E as follows:
H = 1
ηˆz × E (9.87)
i.e., H is perpendicular to both the direction of
propagation (here, +ˆz) and E, with magnitude that is
different from E by the wave impedance η. This
simple expression is very useful for quickly
determining H given E when the direction of
propagation is known. In fact, it is so useful that it
commonly explicitly generalized as follows:
H = 1
η
ˆk × E (9.88)
where ˆk (here, +ˆz) is the direction of propagation.",Electromagnetics_Vol1.pdf
"commonly explicitly generalized as follows:
H = 1
η
ˆk × E (9.88)
where ˆk (here, +ˆz) is the direction of propagation.
Similarly , given H, ˆk, and η, we may find E using
E = −ηˆk × H (9.89)
These spatial relationships can be readily verified
using Figure 9.7.
Equations 9.88 and 9.89 are known as the plane
wave relationships .
The plane wave relationships apply equally well to
the phasor representations of E and H; i.e.,
˜E = −ηˆk × ˜H (9.90)
˜H = 1
η
ˆk × ˜E (9.91)
These equations can be readily verified by applying
the definition of the associated phasors (e.g.,
Equation 9.78). It also turns out that these
relationships apply at each point in space, even if the
waves are not planar, uniform, or in lossless media.
Example 9.2. Analysis of a radially-directed
plane wave.
Consider the scenario illustrated in Figure 9.8.
Here a uniform plane wave having frequency
f = 3 GHz is propagating along a path of
constant φ, where is φis known but not
specified. The phase of the electric field is",Electromagnetics_Vol1.pdf
"f = 3 GHz is propagating along a path of
constant φ, where is φis known but not
specified. The phase of the electric field is
π/2 radians at ρ= 0 (the origin) and t= 0 . The
material is an effectively unbounded region of
free space. The electric field is oriented in the
+ˆz direction and has peak magnitude of
1 mV/m. Find (a) the electric field intensity in
phasor representation, (b) the magnetic field
intensity in phasor representation, and (c) the
actual, physical electric field along the radial
path.
Solution: First, realize that this
“radially-directed” plane wave is in fact a plane
wave, and not a cylindrical wave. It may well we
be that if we “zoom out” far enough, we are able
to perceive a cylindrical wave (for more on this
idea, see Section 9.3), or it might simply be that
wave is exactly planar, and cylindrical
coordinates just happen to be a convenient
coordinate system for this application. In either
case, the direction of propagation ˆk = +ˆρand",Electromagnetics_Vol1.pdf
"coordinates just happen to be a convenient
coordinate system for this application. In either
case, the direction of propagation ˆk = +ˆρand
the solution to this example will be the same.
Here’s the phasor representation for the electric
field intensity of a uniform plane wave in a
lossless medium, analogous to Equation 9.76:
˜E = + ˆzE0e−jβρ (9.92)
From the problem statement, |E0| = 1 mV/m.
Also from the problem statement, the phase of
E0 is π/2 radians; in fact, we could just write
E0 = + j|E0|. Thus, the answer to (a) is
˜E = + ˆzj|E0| e−jβρ (9.93)",Electromagnetics_Vol1.pdf
"206 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
where the propagation constant
β = ω√
µǫ= 2 πf√µ0ǫ0 ∼
= 62.9 rad/m.
The answer to (b) is easiest to obtain from the
plane wave relationship:
˜H = 1
η
ˆk × E
= 1
ηˆρ×
(
+ˆzj|E0| e−jβρ)
= − ˆφj|E0|
η e−jβρ (9.94)
where ηhere is
√
µ0/ǫ0 ∼
= 377 Ω . Thus, the
answer to part (b) is
˜H = − ˆφjH0e−jβρ (9.95)
where H0 ∼
= 2.65 µA/m. At this point you
should check vector directions: ˜E × ˜H should
point in the direction of propagation +ˆρ. Here
we find ˆz ×
(
− ˆφ
)
= +ˆρ, as expected.
The answer to (c) is obtained by applying the
defining relationship for phasors to the answer to
part (a):
E = Re
{
˜Eejωt
}
= Re
{(
+ˆzj|E0| e−jβρ)
ejωt}
= + ˆz |E0| Re
{
ejπ/2e−jβρejωt
}
= + ˆz |E0| cos
(
ωt− βρ+ π
2
)
(9.96)
Additional Reading:
• “Electromagnetic radiation” on Wikipedia.
y
x
E
ϕ
Figure 9.8: A radially-directed plane wave.
9.6 W ave Polarization
[m0131]
P olarization refers to the orientation of the electric",Electromagnetics_Vol1.pdf
"y
x
E
ϕ
Figure 9.8: A radially-directed plane wave.
9.6 W ave Polarization
[m0131]
P olarization refers to the orientation of the electric
field vector. For waves, the term “polarization” refers
specifically to the orientation of this vector with
increasing distance along the direction of propagation,
or, equivalently , the orientation of this vector with
increasing time at a fixed point in space. The relevant
concepts are readily demonstrated for uniform plane
waves, as shown in this section. A review of
Section 9.5 (“Uniform Plane W aves: Characteristics”)
is recommended before reading further.
T o begin, consider the following uniform plane wave,
described in terms of the phasor representation of its
electric field intensity:
˜Ex = ˆxExe−jβz (9.97)
Here Ex is a complex-valued constant representing
the magnitude and phase of the wave, and βis the
positive real-valued propagation constant. Therefore,
this wave is propagating in the +ˆz direction in",Electromagnetics_Vol1.pdf
"positive real-valued propagation constant. Therefore,
this wave is propagating in the +ˆz direction in
lossless media. This wave is said to exhibit linear
polarization (and “linearly polarized”) because the
electric field always points in the same direction,
namely +ˆx. Now consider the wave
˜Ey = ˆyEye−jβz (9.98)
Note that ˜Ey is identical to ˜Ex except that the electric
field vector now points in the +ˆy direction and has
magnitude and phase that is different by the factor",Electromagnetics_Vol1.pdf
"9.6. W A VE POLARIZA TION 207
x
z
by RJB1 (Modified)
Figure 9.9: Linear polarization. Here Ex is shown in
blue, Ey is shown in green, and E is shown in red with
black vector symbols. In this example the phases of
Ex and Ey are zero and φ= −π/4.
Ey/Ex. This wave too is said to exhibit linear
polarization, because, again, the direction of the
electric field is constant with both time and position.
In fact, all linearly-polarized uniform plane waves
propagating in the +ˆz direction in lossless media can
be described as follows:
˜E = ˆρEρe−jβz (9.99)
This is so because ˆρcould be ˆx, ˆy, or any other
direction that is perpendicular to ˆz. If one is
determined to use Cartesian coordinates, the above
expression may be rewritten using (Section 4.3)
ˆρ= ˆx cos φ+ ˆy cos φ (9.100)
yielding
˜E = ( ˆx cos φ+ ˆy cos φ) Eρe−jβz (9.101)
When written in this form, φ= 0 corresponds to
˜E = ˜Ex, φ= π/2 corresponds to ˜E = ˜Ey, and any
other value of φcorresponds to some other constant",Electromagnetics_Vol1.pdf
"When written in this form, φ= 0 corresponds to
˜E = ˜Ex, φ= π/2 corresponds to ˜E = ˜Ey, and any
other value of φcorresponds to some other constant
orientation of the electric field vector; see Figure 9.9
for an example.
A wave is said to exhibit linear polarization if the
direction of the electric field vector does not vary
with either time or position.
Linear polarization arises when the source of the wave
is linearly polarized. A common example is the wave
radiated by a straight wire antenna, such as a dipole
or a monopole. Linear polarization may also be
created by passing a plane wave through a polarizer;
this is particularly common at optical frequencies (see
“ Additional Reading” at the end of this section).
A commonly-encountered alternative to linear
polarization is circular polarization . For an
explanation, let us return to the linearly-polarized
plane waves ˜Ex and ˜Ey defined earlier. If both of
these waves exist simultaneously , then the total",Electromagnetics_Vol1.pdf
"plane waves ˜Ex and ˜Ey defined earlier. If both of
these waves exist simultaneously , then the total
electric field intensity is simply the sum:
˜E = ˜Ex + ˜Ey
= ( ˆxEx + ˆyEy) e−jβz (9.102)
If the phase of Ex and Ey is the same, then
Ex = Eρcos φ, Ey = Eρsin φ, and the above
expression is essentially the same as Equation 9.101.
In this case, ˜E is linearly polarized. But what if the
phases of Ex and Ey are different? In particular, let’s
consider the following case. Let Ex = E0, some
complex-valued constant, and let Ey = + jE0, which
is E0 phase-shifted by +π/2 radians. With no further
math, it is apparent that ˜Ex and ˜Ey are different only
in that one is phase-shifted by π/2 radians relative to
the other. For the physical (real-valued) fields, this
means that Ex has maximum magnitude when Ey is
zero and vice versa. As a result, the direction of
E = Ex+ Ey will rotate in the x− yplane, as shown
in Figure 9.10
The rotation of the electric field vector can also be",Electromagnetics_Vol1.pdf
"E = Ex+ Ey will rotate in the x− yplane, as shown
in Figure 9.10
The rotation of the electric field vector can also be
identified mathematically . When Ex = E0 and
Ey = + jE0, Equation 9.102 can be written:
˜E = ( ˆx + jˆy) E0e−jβz (9.103)
Now reverting from phasor notation to the physical
field:
E = Re
{
˜Eejωt
}
= Re
{
(ˆx + jˆy) E0e−jβzejωt}
= ˆx |E0| cos (ωt− βz) + ˆy |E0| cos
(
ωt− βz+ π
2
)
(9.104)
As anticipated, we see that both Ex and Ey vary
sinusoidally , but are π/2 radians out of phase",Electromagnetics_Vol1.pdf
"208 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
by Dave3457
Figure 9.10: A circularly-polarized wave (in red, with
black vector symbols) resulting from the addition of
orthogonal linearly-polarized waves (shown in green
and blue) that are phase-shifted by π/2 radians.
resulting in rotation in the plane perpendicular to the
direction of propagation.
In the example above, the electric field vector rotates
either clockwise or counter-clockwise relative to the
direction of propagation. The direction of this
rotation can be identified by pointing the thumb of the
left hand in the direction of propagation; in this case,
the fingers of the left hand curl in the direction of
rotation. For this reason, this particular form of
circular polarization is known as left circular (or
”left-hand” circular) polarization (LCP). If we instead
had chosen Ey = −jE0 = −jEx, then the direction
of E rotates in the opposite direction, giving rise to
right circular (or “right-hand” circular) polarization",Electromagnetics_Vol1.pdf
"of E rotates in the opposite direction, giving rise to
right circular (or “right-hand” circular) polarization
(RCP). These two conditions are illustrated in
Figure 9.11.
A wave is said to exhibit circular polarization
if the electric field vector rotates with constant
magnitude. Left- and right-circular polarizations
may be identified by the direction of rotation with
respect to the direction of propagation.
In engineering applications, circular polarization is
useful when the relative orientations of transmit and
receive equipment is variable and/or when the
medium is able rotate the electric field vector. For
example, radio communications involving satellites in
by Dave3457
Figure 9.11: Left-circular polarization (LCP; top) and
right-circular polarization (RCP; bottom).",Electromagnetics_Vol1.pdf
"9.7. W A VE POWER IN A LOSSLESS MEDIUM 209
non-geosynchronous orbits typically employ circular
polarization. In particular, satellites of the U.S.
Global Positioning System (GPS) transmit circular
polarization because of the variable geometry of the
space-to-earth radio link and the tendency of the
Earth’s ionosphere to rotate the electric field vector
through a mechanism known F araday rotation
(sometimes called the “Faraday effect”). If GPS were
instead to transmit using a linear polarization, then a
receiver would need to continuously adjust the
orientation of its antenna in order to optimally receive
the signal. Circularly-polarized radio waves can be
generated (or received) using pairs of
perpendicularly-oriented dipoles that are fed the same
signal but with a 90◦ phase shift, or alternatively by
using an antenna that is intrinsically
circularly-polarized, such as a helical antenna (see
“ Additional Reading” at the end of this section).",Electromagnetics_Vol1.pdf
"using an antenna that is intrinsically
circularly-polarized, such as a helical antenna (see
“ Additional Reading” at the end of this section).
Linear and circular polarization are certainly not the
only possibilities. Elliptical polarization results when
Ex and Ey do not have equal magnitude. Elliptical
polarization is typically not an intended condition, but
rather is most commonly observed as a degradation in
a system that is nominally linearly- or
circularly-polarized. For example, most antennas that
are said to be “circularly polarized” instead produce
circular polarization only in one direction and various
degrees of elliptical polarization in all other
directions.
Additional Reading:
• “Polarization (waves)” on Wikipedia.
• “Dipole antenna” on Wikipedia.
• “Polarizer” on Wikipedia.
• “Faraday effect” on Wikipedia.
• “Helical antenna” on Wikipedia.
9.7 W ave Power in a Lossless
Medium
[m0041]
In many applications involving electromagnetic",Electromagnetics_Vol1.pdf
"• “Helical antenna” on Wikipedia.
9.7 W ave Power in a Lossless
Medium
[m0041]
In many applications involving electromagnetic
waves, one is less concerned with the instantaneous
values of the electric and magnetic fields than the
power associated with the wave. In this section, we
address the issue of how much power is conveyed by
an electromagnetic wave in a lossless medium. The
relevant concepts are readily demonstrated in the
context of uniform plane waves, as shown in this
section. A review of Section 9.5 (“Uniform Plane
W aves: Characteristics”) is recommended before
reading further.
Consider the following uniform plane wave,
described in terms of the phasor representation of its
electric field intensity:
˜E = ˆxE0e−jβz (9.105)
Here E0 is a complex-valued constant associated with
the source of the wave, and βis the positive
real-valued propagation constant. Therefore, the wave
is propagating in the +ˆz direction in lossless media.
The first thing that should be apparent is that the",Electromagnetics_Vol1.pdf
"is propagating in the +ˆz direction in lossless media.
The first thing that should be apparent is that the
amount of power conveyed by this wave is infinite.
The reason is as follows. If the power passing through
any finite area is greater than zero, then the total
power must be infinite because, for a uniform plane
wave, the electric and magnetic field intensities are
constant over a plane of infinite area. In practice, we
never encounter this situation because all practical
plane waves are only “locally planar” (see Section 9.3
for a refresher on this idea). Nevertheless, we seek
some way to express the power associated with such
waves.
The solution is not to seek total power, but rather
power per unit area. This quantity is known as the
spatial power density , or simply “ power density , ” and
has units of W/m 2.5 Then, if we are interested in total
power passing through some finite area, then we may
5 Be careful: The quantities power spectral density (W/Hz) and",Electromagnetics_Vol1.pdf
"power passing through some finite area, then we may
5 Be careful: The quantities power spectral density (W/Hz) and
power flux density (W/(m2·Hz)) are also sometimes referred to as
“power density . ” In this section, we will limit the scope to spatial
power density (W/m 2).",Electromagnetics_Vol1.pdf
"210 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
simply integrate the power density over this area.
Let’s skip to the answer, and then consider where this
answer comes from. It turns out that the
instantaneous power density of a uniform plane wave
is the magnitude of the P oynting vector
S ≜ E × H (9.106)
Note that this equation is dimensionally correct; i.e.
the units of E (V/m) times the units of H (A/m) yield
the units of spatial power density (V ·A/m2, which is
W/m2). Also, the direction of E × H is in the
direction of propagation (reminder: Section 9.5),
which is the direction in which the power is expected
to flow . Thus, we have some compelling evidence that
|S| is the power density we seek. However, this is not
proof – for that, we require the P oynting Theorem,
which is a bit outside the scope of the present section,
but is addressed in the “ Additional Reading” at the
end of this section.
A bit later we’re going to need to know S for a",Electromagnetics_Vol1.pdf
"but is addressed in the “ Additional Reading” at the
end of this section.
A bit later we’re going to need to know S for a
uniform plane wave, so let’s work that out now . From
the plane wave relationships (Section 9.5) we find that
the magnetic field intensity associated with the
electric field in Equation 9.105 is
˜H = ˆy E0
η e−jβz (9.107)
where η=
√
µ/ǫis the real-valued impedance of the
medium. Let ψbe the phase of E0; i.e.,
E0 = |E0|ejψ. Then
E = Re
{
˜Eejωt
}
= ˆx |E0| cos (ωt− βz+ ψ) (9.108)
and
H = Re
{
˜Hejωt
}
= ˆy |E0|
η cos (ωt− βz+ ψ) (9.109)
Now applying Equation 9.106,
S = ˆz|E0|2
η cos2 (ωt− βz+ ψ) (9.110)
As noted earlier, |S| is only the instantaneous power
density , which is still not quite what we are looking
for. What we are actually looking for is the
time-average power density Save – that is, the average
value of |S| over one period T of the wave. This may
be calculated as follows:
Save = 1
T
∫ t0+T
t=t0
|S|dt
= |E0|2
η
1
T
∫ t0+T
t=t0
cos2 (ωt− ks+ ψ) dt
(9.111)",Electromagnetics_Vol1.pdf
"be calculated as follows:
Save = 1
T
∫ t0+T
t=t0
|S|dt
= |E0|2
η
1
T
∫ t0+T
t=t0
cos2 (ωt− ks+ ψ) dt
(9.111)
Since ω= 2 πf = 2 π/T, the definite integral equals
T/2. W e obtain
Save = |E0|2
2η (9.112)
It is useful to check units again at this point. Note
(V/m)2 divided by Ω is W/m 2, as expected.
Equation 9.112 is the time-average power density
(units of W/m 2) associated with a sinusoidally-
varying uniform plane wave in lossless media.
Note that Equation 9.112 is analogous to a
well-known result from electric circuit theory . Recall
the time-average power Pave (units of W) associated
with a voltage phasor ˜V across a resistance Ris
Pave = | ˜V|2
2R (9.113)
which closely resembles Equation 9.112. The result is
also analogous to the result for a voltage wave on a
transmission line (Section 3.20), for which:
Pave = |V+
0 |2
2Z0
(9.114)
where V+
0 is a complex-valued constant representing
the magnitude and phase of the voltage wave, and Z0",Electromagnetics_Vol1.pdf
"Pave = |V+
0 |2
2Z0
(9.114)
where V+
0 is a complex-valued constant representing
the magnitude and phase of the voltage wave, and Z0
is the characteristic impedance of the transmission
line.
Here is a good point at which to identify a common
pitfall. |E0| and | ˜V| are the peak magnitudes of the
associated real-valued physical quantities. However,
these quantities are also routinely given as root mean
square (“rms”) quantities. Peak magnitudes are
greater by a factor of
√
2, so Equation 9.112",Electromagnetics_Vol1.pdf
"9.7. W A VE POWER IN A LOSSLESS MEDIUM 211
expressed in terms of the rms quantity lacks the factor
of 1/2.
Example 9.3. Power density of a typical radio
wave.
A radio wave transmitted from a distant location
may be perceived locally as a uniform plane
wave if there is no nearby structure to scatter the
wave; a good example of this is the wave
arriving at the user of a cellular telephone in a
rural area with no significant terrain scattering.
The range of possible signal strengths varies
widely , but a typical value of the electric field
intensity arriving at the user’s location is
10 µV/m rms. What is the corresponding power
density?
Solution: From the problem statement,
|E0| = 10 µV/m rms. W e assume propagation
occurs in air, which is indistinguishable from
free space at cellular frequencies. If we use
Equation 9.112, then we must first convert |E0|
from rms to peak magnitude, which is done by
multiplying by
√
2. Thus:
Save = |E0|2
2η
∼
=
( √
2 · 10 × 10−6 V/m
) 2
2 · 377 Ω
∼",Electromagnetics_Vol1.pdf
"from rms to peak magnitude, which is done by
multiplying by
√
2. Thus:
Save = |E0|2
2η
∼
=
( √
2 · 10 × 10−6 V/m
) 2
2 · 377 Ω
∼
= 2.65 × 10−13 W/m2
Alternatively , we can just use a version of
Equation 9.112, which is appropriate for rms
units:
Save = |E0,rms|2
η
∼
=
(
10 × 10−6 V/m
) 2
377 Ω
∼
= 2.65 × 10−13 W/m2
Either way , we obtain the correct answer,
0.265 pW/m 2 (that’s picowatts per square
meter).
Considering the prevalence of phasor representation,
it is useful to have an alternative form of the Poynting
vector which yields time-average power by operating
directly on field quantities in phasor form. This is
Save, defined as:
Save ≜ 1
2Re
{
˜E × ˜H∗
}
(9.115)
(Note that the magnetic field intensity phasor is
conjugated.) The above expression gives the expected
result for a uniform plane wave. Using
Equations 9.105 and 9.107, we find
Save = 1
2Re
{( ˆxE0e−jβz)
×
(
ˆy E0
η e−jβz
) ∗}
(9.116)
which yields
Save = ˆz|E0|2
2η (9.117)
as expected.
Additional Reading:",Electromagnetics_Vol1.pdf
"Save = 1
2Re
{( ˆxE0e−jβz)
×
(
ˆy E0
η e−jβz
) ∗}
(9.116)
which yields
Save = ˆz|E0|2
2η (9.117)
as expected.
Additional Reading:
• “Poynting vector” on Wikipedia.
• “Poynting’s Theorem” on Wikipedia.
[m0035]",Electromagnetics_Vol1.pdf
"212 CHAPTER 9. PLANE W A VES IN LOSSLESS MEDIA
Image Credits
Fig. 9.1: c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0142 fSphericalPhasefront.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 9.2 c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0142 fCylindricalPhasefront.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 9.3 c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0142 fPlanarPhasefront.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 9.4 c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0142 fPlaneW avesReflector.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 9.5 c⃝ Y . Qin, https://commons.wikimedia.org/wiki/File:M0142 fLocallyPlanar.svg,
CC BY 4.0 (https://creativecommons.org/licenses/by/4.0 /).
Fig. 9.7: c⃝ E. Boutet, https://en.wikipedia.org/wiki/File:Onde electromagnetique.svg,
CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/). Heavily modified by author.",Electromagnetics_Vol1.pdf
"CC BY SA 3.0 (https://creativecommons.org/licenses/by-s a/3.0/). Heavily modified by author.
Fig. 9.9: RJB1, https://commons.wikimedia.org/wiki/Fil e:Linearly Polarized W ave.svg,
public domain. Modified by author.
Fig. 9.10: Dave3457, https://en.wikipedia.org/wiki/Fil e:Circular.Polarization.Circularly .Polarized.Light
With.Components Left.Handed.svg,
public domain.
Fig. 9.11: Dave3457, https://en.wikipedia.org/wiki/Fil e:Circular.Polarization.Circularly .Polarized.Light
Without.Components Left.Handed.svg
and https://en.wikipedia.org/wiki/File:Circular.Pola rization.Circularly .Polarized.Light
Without.Components Right.Handed.svg,
public domain.",Electromagnetics_Vol1.pdf
"Appendix A
Constitutive Parameters of Some
Common Materials
A.1 Permittivity of Some
Common Materials
[m0135]
The values below are relative permittivity ǫr ≜ ǫ/ǫ0
for a few materials that are commonly encountered in
electrical engineering applications, and for which
permittivity emerges as a consideration. Note that
“relative permittivity” is sometimes referred to as
dielectric constant .
Here we consider only the physical (real-valued)
permittivity , which is the real part of the complex
permittivity (typically indicated as ǫ′ or ǫ′
r) for
materials exhibiting significant loss.
Permittivity varies significantly as a function of
frequency . The values below are representative of
frequencies from a few kHz to about 1 GHz. The
values given are also representative of optical
frequencies for materials such as silica that are used
in optical applications. Permittivity also varies as a
function of temperature. In applications where
precision better than about 10% is required, primary",Electromagnetics_Vol1.pdf
"function of temperature. In applications where
precision better than about 10% is required, primary
references accounting for frequency and temperature
should be consulted. The values presented here are
gathered from a variety of references, including those
indicated in “ Additional References. ”
Free Space (vacuum): ǫr ≜ 1
Solid Dielectrics :
Material ǫr Common uses
Styrofoam1 1.1
T eflon 2 2.1
Polyethylene 2.3 coaxial cable
Polypropylene 2.3
Silica 2.4 optical fiber 3
Polystyrene 2.6
Polycarbonate 2.8
Rogers RO3003 3.0 PCB substrate
FR4 (glass epoxy laminate) 4.5 PCB substrate
1 Properly known as extruded polystyrene foam
(XPS).
2 Properly known as polytetrafluoroethylene (PTFE).
3 T ypically doped with small amounts of other
materials to slightly raise or lower the index of
refraction ( = √
ǫr).
Non-conducting spacing materials used in discrete
capacitors exhibit ǫr ranging from about 5 to 50.
Semiconductors commonly appearing in electronics",Electromagnetics_Vol1.pdf
"Non-conducting spacing materials used in discrete
capacitors exhibit ǫr ranging from about 5 to 50.
Semiconductors commonly appearing in electronics
– including carbon, silicon, geranium, indium
phosphide, and so on – typically exhibit ǫr in the
range 5–15.
Glass exhibits ǫr in the range 4–10, depending on
composition.
Gasses, including air, typically exhibit ǫr ∼
= 1 to
within a tiny fraction of a percent.
Liquid water typically exhibits ǫr in the range
72–81. Distilled water exhibits ǫr ≈ 81 at room
temperature, whereas sea water tends to be at the
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"214 APPENDIX A. CONSTITUTIVE P ARAMETERS OF SOME COMMON MA TERIA LS
lower end of the range.
Other liquids typically exhibit ǫr in the range 10–90,
with considerable variation as a function of
temperature and frequency . Animal flesh and blood
consists primarily of liquid matter and so also exhibits
permittivity in this range.
Soil typically exhibits ǫr in the range 2.5–3.5 when
dry and higher when wet. The permittivity of soil
varies considerably depending on composition.
Additional Reading:
• CRC Handbook of Chemistry and Physics .
• “Relative permittivity” on Wikipedia.
• “Miscellaneous Dielectric Constants” on
microwaves101.com.
A.2 Permeability of Some
Common Materials
[m0136]
The values below are relative permeability
µr ≜ µ/µ0 for a few materials that are commonly
encountered in electrical engineering applications,
and for which µr is significantly different from 1.
These materials are predominantly ferromagnetic
metals and (in the case of ferrites) materials",Electromagnetics_Vol1.pdf
"and for which µr is significantly different from 1.
These materials are predominantly ferromagnetic
metals and (in the case of ferrites) materials
containing significant ferromagnetic metal content.
Nearly all other materials exhibit µr that is not
significantly different from that of free space.
The values presented here are gathered from a variety
of references, including those indicated in “ Additional
References” at the end of this section. Be aware that
permeability may vary significantly with frequency;
values given here are applicable to the frequency
ranges for applications in which these materials are
typically used. Also be aware that materials
exhibiting high permeability are also typically
non-linear; that is, permeability depends on the
magnitude of the magnetic field. Again, values
reported here are those applicable to applications in
which these materials are typically used.
Free Space (vacuum): µr ≜ 1.
Iron (also referred to by the chemical notation “Fe”)",Electromagnetics_Vol1.pdf
"which these materials are typically used.
Free Space (vacuum): µr ≜ 1.
Iron (also referred to by the chemical notation “Fe”)
appears as a principal ingredient in many materials
and alloys employed in electrical structures and
devices. Iron exhibits µr that is very high, but which
decreases with decreasing purity . 99.95% pure iron
exhibits µr ∼ 200,000. This decreases to ∼ 5000 at
99.8% purity and is typically below 100 for purity
less than 99%.
Steel is an iron alloy that comes in many forms, with
a correspondingly broad range of permeabilites.
Electrical steel , commonly used in electrical
machinery and transformers when high permeability
is desired, exhibits µr ∼ 4000. Stainless steel ,
encompassing a broad range of alloys used in
mechanical applications, exhibits µr in the range
750–1800. Carbon steel , including a broad class of
alloys commonly used in structural applications,
exhibits µr on the order of 100.",Electromagnetics_Vol1.pdf
"A.3. CONDUCTIVITY OF SOME COMMON MA TERIALS 215
Ferrites include a broad range of ceramic materials
that are combined with iron and various combinations
of other metals and are used as magnets and magnetic
devices in various electrical systems. Common
ferrites exhibit µr in the range 16–640.
Additional Reading:
• Section 7.16 (“Magnetic Materials”)
• CRC Handbook of Chemistry and Physics .
• “Magnetic Materials” on microwaves101.com.
• “Permeability (electromagnetism)” on
Wikipedia.
• “Iron” on Wikipedia.
• “Electrical steel” on Wikipedia.
• “Ferrite (magnet)” on Wikipedia.
A.3 Conductivity of Some
Common Materials
[m0137]
The values below are conductivity σfor a few
materials that are commonly encountered in electrical
engineering applications, and for which conductivity
emerges as a consideration.
Note that materials in some applications are described
instead in terms of resistivity, which is simply the
reciprocal of conductivity .
Conductivity may vary significantly as a function of",Electromagnetics_Vol1.pdf
"instead in terms of resistivity, which is simply the
reciprocal of conductivity .
Conductivity may vary significantly as a function of
frequency . The values below are representative of
frequencies from a few kHz to a few GHz.
Conductivity also varies as a function of temperature.
In applications where precise values are required,
primary references accounting for frequency and
temperature should be consulted. The values
presented here are gathered from a variety of
references, including those indicated in “ Additional
References” at the end of this section.
Free Space (vacuum): σ≜ 0.
Commonly encountered elements :
Material σ(S/m)
Copper 5.8 × 107
Gold 4.4 × 107
Aluminum 3.7 × 107
Iron 1.0 × 107
Platinum 0.9 × 107
Carbon 1.3 × 105
Silicon 4.4 × 10−4
W ater exhibits σranging from about 6 µS/m for
highly distilled water (thus, a very poor conductor) to
about 5 S/m for seawater (thus, a relatively good
conductor), varying also with temperature and",Electromagnetics_Vol1.pdf
"about 5 S/m for seawater (thus, a relatively good
conductor), varying also with temperature and
pressure. T ap water is typically in the range
5–50 mS/m, depending on the level of impurities
present.
Soil typically exhibits σin the range 10−4 S/m for
dry soil to about 10−1 S/m for wet soil, varying also
due to chemical composition.",Electromagnetics_Vol1.pdf
"216 APPENDIX A. CONSTITUTIVE P ARAMETERS OF SOME COMMON MA TERIA LS
Non-conductors. Most other materials that are not
well-described as conductors or semiconductors and
are dry exhibit σ <10−12 S/m. Most materials that
are considered to be insulators, including air and
common dielectrics, exhibit σ <10−15 S/m, often by
several orders of magnitude.
Additional Reading:
• CRC Handbook of Chemistry and Physics .
• “Conductivity (electrolytic)” on Wikipedia.
• “Electrical resistivity and conductivity” on
Wikipedia.
• “Soil resistivity” on Wikipedia.",Electromagnetics_Vol1.pdf
"Appendix B
Mathematical Formulas
B.1 T rigonometry
[m0138]
ejθ = cos θ+ jsin θ (B.1)
cos θ= 1
2
(
ejθ + e−jθ)
(B.2)
sin θ= 1
j2
(
ejθ − e−jθ)
(B.3)
cos2 θ= 1
2 + 1
2 cos 2θ (B.4)
sin2 θ= 1
2 − 1
2 cos 2θ (B.5)
B.2 V ector Operators
[m0139]
This section contains a summary of vector operators
expressed in each of the three major coordinate
systems:
• Cartesian ( x,y,z)
• cylindrical ( ρ,φ,z)
• spherical ( r,θ,φ)
Associated basis vectors are identified using a caret ( ˆ)
over the symbol. The vector operand A is expressed
in terms of components in the basis directions as
follows:
• Cartesian: A = ˆxAx + ˆyAy + ˆzAz
• cylindrical: A = ˆρAρ + ˆφAφ + ˆzAz
• spherical: A = ˆrAr + ˆθAθ + ˆφAφ
Gradient
Gradient in Cartesian coordinates:
∇f = ˆx∂f
∂x + ˆy ∂f
∂y + ˆz∂f
∂z (B.6)
Gradient in cylindrical coordinates:
∇f = ˆρ∂f
∂ρ + ˆφ1
ρ
∂f
∂φ + ˆz∂f
∂z (B.7)
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"218 APPENDIX B. MA THEMA TICAL FORMULAS
Gradient in spherical coordinates:
∇f = ˆr∂f
∂r + ˆθ1
r
∂f
∂θ + ˆφ 1
rsin θ
∂f
∂φ (B.8)
Divergence
Divergence in Cartesian coordinates:
∇ · A = ∂Ax
∂x + ∂Ay
∂y + ∂Az
∂z (B.9)
Divergence in cylindrical coordinates:
∇ · A = 1
ρ
∂
∂ρ (ρAρ) + 1
ρ
∂Aφ
∂φ + ∂Az
∂z (B.10)
Divergence in spherical coordinates:
∇ · A = 1
r2
∂
∂r
(
r2Ar
)
+ 1
rsin θ
∂
∂θ (Aθsin θ)
+ 1
rsin θ
∂Aφ
∂φ (B.11)
Curl
Curl in Cartesian coordinates:
∇ × A = ˆx
( ∂Az
∂y − ∂Ay
∂z
)
+ ˆy
( ∂Ax
∂z − ∂Az
∂x
)
+ ˆz
( ∂Ay
∂x − ∂Ax
∂y
)
(B.12)
Curl in cylindrical coordinates:
∇ × A = ˆ ρ
( 1
ρ
∂Az
∂φ − ∂Aφ
∂z
)
+ ˆφ
( ∂Aρ
∂z − ∂Az
∂ρ
)
+ ˆz 1
ρ
[ ∂
∂ρ (ρAφ) − ∂Aρ
∂φ
]
(B.13)
Curl in spherical coordinates:
∇ × A = ˆr 1
rsin θ
[ ∂
∂θ (Aφsin θ) − ∂Aθ
∂φ
]
+ ˆθ1
r
[ 1
sin θ
∂Ar
∂φ − ∂
∂r (rAφ)
]
+ ˆφ1
r
[ ∂
∂r (rAθ) − ∂Ar
∂θ
]
(B.14)
Laplacian
Laplacian in Cartesian coordinates:
∇2f = ∂2f
∂x2 + ∂2f
∂y2 + ∂2f
∂z2 (B.15)
Laplacian in cylindrical coordinates:
∇2f = 1
ρ
∂
∂ρ
(
ρ∂f
∂ρ
)
+ 1
ρ2
∂2f",Electromagnetics_Vol1.pdf
"∇2f = ∂2f
∂x2 + ∂2f
∂y2 + ∂2f
∂z2 (B.15)
Laplacian in cylindrical coordinates:
∇2f = 1
ρ
∂
∂ρ
(
ρ∂f
∂ρ
)
+ 1
ρ2
∂2f
∂φ2 + ∂2f
∂z2 (B.16)
Laplacian in spherical coordinates:
∇2f = 1
r2
∂
∂r
(
r2 ∂f
∂r
)
+ 1
r2 sin θ
∂
∂θ
( ∂f
∂θ sin θ
)
+ 1
r2 sin2 θ
∂2f
∂φ2 (B.17)",Electromagnetics_Vol1.pdf
"B.3. VECTOR IDENTITIES 219
B.3 V ector Identities
[m0140]
Algebraic Identities
A · (B × C) = B · (C × A) = C · (A × B)
(B.18)
A × (B × C) = B (A · C) − C (A · B) (B.19)
Identities Involving Differential Operators
∇ · (∇ × A) = 0 (B.20)
∇ × (∇f) = 0 (B.21)
∇ × (fA) = f(∇ × A) + ( ∇f) × A (B.22)
∇ · (A × B) = B · (∇ × A) − A · (∇ × B)
(B.23)
∇ · (∇f) = ∇2f (B.24)
∇ × ∇ × A = ∇ (∇ · A) − ∇2A (B.25)
∇2A = ∇ (∇ · A) − ∇ × (∇ × A) (B.26)
Divergence Theorem: Given a closed surface S
enclosing a contiguous volume V,
∫
V
(∇ · A) dv=
∮
S
A · ds (B.27)
where the surface normal ds is pointing out of the
volume.
Stokes’ Theorem: Given a closed curve C bounding
a contiguous surface S,
∫
S
(∇ × A) · ds =
∮
C
A · dl (B.28)
where the direction of the surface normal ds is related
to the direction of integration along C by the “right
hand rule. ”",Electromagnetics_Vol1.pdf
"Appendix C
Physical Constants
[m0141]
The speed of light in free space (c), which is the
phase velocity of any electromagnetic radiation in
free space, is ∼
= 2.9979 × 108 m/s. This is commonly
rounded up to 3 × 108 m/s. This rounding incurs error
of ∼
= 0.07%, which is usually much less than other
errors present in electrical engineering calculations.
The charge of an electron is ∼
= −1.602 × 10−19 C.
The constant e≜ +1.602176634 × 10−19 C is known
as the “elementary charge, ” so the charge of the
electron is said to be −e.
The permittivity of free space (ǫ0) is
∼
= 8.854 × 10−12 F/m.
The permeability of free space (µ0) is
4π× 10−7 H/m.
The wave impedance of free space (η0) is the ratio
of the magnitude of the electric field intensity to that
of the magnetic field intensity in free space and is
√
µ0/ǫ0 ∼
= 376.7 Ω . This is also sometimes referred
to as the intrinsic impedance of free space .",Electromagnetics_Vol1.pdf
"√
µ0/ǫ0 ∼
= 376.7 Ω . This is also sometimes referred
to as the intrinsic impedance of free space .
Electromagnetics V ol 1 . c⃝ 2018 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061 /electromagnetics- vol- 1",Electromagnetics_Vol1.pdf
"Index
acoustic wave equation, see wave equation, acoustic
admittance, 64–66
characteristic, see characteristic admittance
air, 123, 124
aluminum, 170, 215
Ampere’s law
general form, 190–192
magnetostatics, 88, 149–150, 152, 155, 158, 159,
167
amplifier, 59, 62
antenna, 207
dipole, 207
helical, 209
impedance matching, 58, 62
monopole, 207
patch, 58
arcing, 124
attenuation constant, see propagation constant
balun, 187
Biot-Savart Law , 26
boundary conditions, 118, 120, 160, 161
capacitance, 34, 124–126
definition, 124
capacitor, 125–126
energy storage, 130
parallel plate, 18, 116–117, 126–128
carbon, 215
Cartesian coordinate system, 70, 76–77
ceramic, 170
CGS (system of units), 14
characteristic admittance, 64
characteristic impedance, 33, 37–40, 48, 63, 210
coaxial cable, 43
definition, 39
measurement, 56
charge density
line, 95, 96, 101, 112
surface, 95, 97, 112
volume, 85, 95, 99, 113
charge mobility , 136
circulation, 88
coax, see transmission line
cobalt, 171
coil, 152–154, 163, 165",Electromagnetics_Vol1.pdf
"surface, 95, 97, 112
volume, 85, 95, 99, 113
charge mobility , 136
circulation, 88
coax, see transmission line
cobalt, 171
coil, 152–154, 163, 165
toroidal, see toroidal coil
conductance, 34, 141–142
definition, 141
conductivity , 27, 136–138
definition, 136
of common materials, 215–216
conductor
good, 136, 137
perfect, 21, 118, 120, 122, 137, 144
perfect (definition), 138
poor, 123
constitutive parameters, 27
copper, 171, 215
Coulomb’s law , 93–94, 100, 103, 105
cross product, 74–75
curl, 88–89, 218
definition, 89
current, 22
conduction, 134
convection, 134
displacement, see displacement current
current density , 135–136
line, 135
surface, 135, 157
volume, 135
cylindrical coordinate system, 77–80
datum, 110
diamagnetic material, 170
dielectric (material), 42, 45, 123, 204
examples, 213
lossless, 137
lossy , 137
dielectric breakdown, 124, 138
dielectric constant, 20, 213
dielectric strength, 124
221",Electromagnetics_Vol1.pdf
"222 INDEX
dipole, see antenna
displacement current, 150, 192
divergence, 85–86, 218
definition, 85
divergence theorem, 87, 104, 148, 219
dot product, 73–74
duality , 146, 197
electric field, see field, electric
electric field intensity , 17–19, 94, 124
boundary conditions, 118
definition, 18
related to current, 136
electric flux density , 21–22, 85
boundary conditions, 120
definition, 21
electrical length, 53
measurement, 56
electromagnetic compatibility (EMC), 185
electromagnetic force, 19, 24
electromagnetic interference (EMI), 30, 185
electromagnetic spectrum, see spectrum
electromotive force, see emf
electron, 220
electrostatics (description), 93
emf, 179
motional, 180, 187
transformer, 180–183
emitter induction, 55
energy
electrostatic, 130–132
kinetic, 17
magnetostatic, 169–170
potential, 17, 105–107, 130
English system of units, 14
Faraday rotation, 209
Faraday’s law , 85, 176, 178–181, 189
ferrite, 170, 215
ferromagnetic material, 171
field
definition, 17
electric, 17–19, 21–22",Electromagnetics_Vol1.pdf
"Faraday rotation, 209
Faraday’s law , 85, 176, 178–181, 189
ferrite, 170, 215
ferromagnetic material, 171
field
definition, 17
electric, 17–19, 21–22
magnetic, 22–27
scalar, 5, 17
vector, 17
field intensity , magnetic, see magnetic field intensity
field line (magnetic), 24
filter, 55
flux, 21, 24, 77, 79, 83, 85, 135, 161
definition, 85
magnetic, 163, 178
flux density , electric, see electric flux density
flux density , magnetic, see magnetic flux density
flux linkage
seelinkage, 163
force, 18, 23, 70, 105, 124
electromotive, see emf
Fourier series, 12
Fourier transform, 12
FR4, 46, 123, 213
fringing (field), 117, 126, 153, 166, 167
gamma ray , 3
Gauss’ law
electric field, 85, 100–101, 103
magnetic field, 147–148
generator, 33, 180, 187–189
geranium, 213
glass, 123, 213
Global Positioning System (GPS), 209
gold, 171, 215
gradient, 84, 114, 217
ground, 111
ground plane, 128
heat, 144
homogeneity , 27
homogeneous (media), 21, 26, 27, 196
hysteresis, 171
impedance",Electromagnetics_Vol1.pdf
"gradient, 84, 114, 217
ground, 111
ground plane, 128
heat, 144
homogeneity , 27
homogeneous (media), 21, 26, 27, 196
hysteresis, 171
impedance
characteristic, see characteristic impedance
matching, 55, 62–68
of resistance, 138
wave, see wave impedance
impedance inverter, 58
independence of path, 107–108
index of refraction, 213
indium phosphide, 213
inductance, 34, 163–165
definition, 163, 164
induction, 175–178, 180
inductor, 152, 164
coaxial, 167–168
hysteresis, 171
straight coil, 165–166
infrared, 3
insulator
good, 123",Electromagnetics_Vol1.pdf
"INDEX 223
perfect, 137
poor, 137
inverse square law , 20, 25
iron, 171, 214, 215
isotropic (media), 28, 196
isotropy , 28
joule heating, 144
Joule’s law , 143
kinetic energy , see energy , kinetic
Kirchoff’s voltage law
electrostatics, 88, 105, 108–110, 118, 180, 189
Laplace’s equation, 115, 116
Laplacian (operator), 91, 218
Lenz’s law , 175, 178, 182
lightning, 124, 138
linear (media), 28, 124, 196
linear and time-invariant (L TI), 28
linearity , 28
linkage, 163
load
matched, 48
reactive, 48
lumped-element model, see transmission line
magnesium, 170
magnetic field intensity , 26–27
definition, 26
magnetic flux, see flux, magnetic
magnetic flux density , 22–25, 85
definition, 24
magnetostatics (description), 146
materials (properties), 27–28, 170–172
materials, magnetic, 170–172
Maxwell’s equations, 175, 196
differential phasor form, 194–196
source-free lossless region, 196
Maxwell-Faraday equation, 88, 180, 189–191, 202
metric system, 14
mica, 124
microstrip, 31, 44–46, 56, 63
mode, 31",Electromagnetics_Vol1.pdf
"source-free lossless region, 196
Maxwell-Faraday equation, 88, 180, 189–191, 202
metric system, 14
mica, 124
microstrip, 31, 44–46, 56, 63
mode, 31
monopole, see antenna
motional emf, see emf
mutual inductance, 164
neper, 40
nickel, 171
non-magnetic (material), 123
nonlinear (media), 27
notation, 15
Ohm’s law , 134, 136, 143, 196
ohmic heating, 144
open circuit, 48, 49, 54
optical (frequency), 3
optical fiber
multimode, 31
single mode, 32
paramagnetic material, 170
patch antenna, see antenna
perfect conductor, see conductor, perfect
permanent magnet, 22
permeability , 25–27, 123, 170
of common materials, 214–215
relative, 25, 214
permittivity , 20, 27, 124, 137
complex-valued, 197
effective, 45
of common materials, 213–214
relative, 20, 123, 213
phase velocity , 3, 8, 40, 204
coaxial cable, 43
in microstrip, 46
measurement, 57
phasefront, 198, 203
phasor, 9–13, 194, 203, 207
phasor (definition), 9
photoelectric effect, 4
photon, 19
pipe, 140
plane wave relationships, 205",Electromagnetics_Vol1.pdf
"phasefront, 198, 203
phasor, 9–13, 194, 203, 207
phasor (definition), 9
photoelectric effect, 4
photon, 19
pipe, 140
plane wave relationships, 205
platinum, 170, 215
Poisson’s equation, 91, 115–116
polarization, 206
circular, 207
elliptical, 209
linear, 206
polarizer, 207
polycarbonate, 213
polyethylene, 213
polypropylene, 213
polystyrene, 213
polytetrafluoroethylene, 213
potential difference, 106
potential energy , see energy , potential
power density , see waves",Electromagnetics_Vol1.pdf
"224 INDEX
power dissipation, 143–144
Poynting theorem, 210
Poynting vector, 210
printed circuit board (PCB), 44, 46, 123
capacitance, 127–128
material, 213
propagation constant, 37
attenuation, 33, 40
phase, 6, 33, 40, 197
quantum mechanics, 19, 23
quarter-wave inverter, 58
radio (frequency), 3
reactance, 165
reflection coefficient, 47–48
definition, 48
relative permeability , see permeability , relative
relative permittivity , see permittivity , relative
resistance, 34, 138–140
definition, 138
of wire (DC), 139–140
resistivity , 215
resistor, 1
RG-59, 43, 130, 142
right-hand rule
cross product, 74
magnetic field of a wire, 152, 155, 157, 182
Stokes’ theorem, 90, 151, 158, 159, 161, 163,
178, 181, 219
saturation (magnetic), 171, 186
scalar field, see field, scalar
scalar product, see dot product
semiconductor (material), 137, 213
shield, 42
short circuit, 48, 49, 54
shot noise, 94
SI (system of units), 14
signum (sgn) function, 98
silica, 213
silicon, 171, 213, 215
single-stub matching, 66–68",Electromagnetics_Vol1.pdf
"shot noise, 94
SI (system of units), 14
signum (sgn) function, 98
silica, 213
silicon, 171, 213, 215
single-stub matching, 66–68
skin effect, 165
soil, 214, 215
solenoid, 152
sound (waves), 5
spatial frequency , 6
spectrum, 3
speed of light, 3, 204, 220
spherical coordinate system, 81–83
standing wave, 41, 49–50
definition, 49
standing wave ratio (SWR), 51–52
definition, 51
steel, 140, 214
Stokes’ theorem, 90, 109, 163, 178, 181, 189, 190,
219
stripline, 44
stub
applications, 55, 66–68
definition, 54
open circuit, 54
short circuit, 54
styrofoam, 213
superposition, 9, 28
susceptance, 64, 67
symmetry , 100, 102, 151
teflon, 123, 213
telegrapher’s equations, 35–37
TEM line, see transmission line
Th´evenin equivalent circuit, 33
time-invariance (media), 28
toroid, 155
toroidal coil, 155–157
transformer, 164, 183–187
emf, see emf
hysteresis, 171
induction, 176
transmission line
coaxial, 31, 42–44, 62, 130, 141–142
definition, 30
input impedance, 52–55
lossless, 40
low loss, 40, 41, 43",Electromagnetics_Vol1.pdf
"induction, 176
transmission line
coaxial, 31, 42–44, 62, 130, 141–142
definition, 30
input impedance, 52–55
lossless, 40
low loss, 40, 41, 43
lumped-element model, 34–35, 41
microstrip, see microstrip
quarter-wavelength, 57–60
stripline, 44
stub, see stub
time-average power, 61
transverse electromagnetic (TEM), 31, 37, 42, 45
two-port representation, 33–34
wave equations, 37
two-port, 33
ultraviolet, 3
unit vector, 70
units, 13–14",Electromagnetics_Vol1.pdf
"INDEX 225
vector
arithmetic, 70–75, 219
definition, 70
identity , 219
position-fixed, 70
position-free, 70
unit, 70
velocity , 70
voltage reflection coefficient, see reflection coefficient
voltage standing wave ratio (VSWR), 51
water, 213, 215
wave equation
acoustic, 6
electromagnetic, 91
source-free lossless region, 196–197
TEM transmission line, 37
wave impedance, 197, 202, 220
waveguide, 31
wavelength, 2, 3, 7, 204
in microstrip, 46
wavenumber, 6
waves
cylindrical, 198
definition, 17
fundamentals, 5–8
guided, 9, 30
plane, 198–206, 209
polarization, 206–209
power density , 209–211
spherical, 198
standing, see standing wave
transmission line analogy , 210
unguided, 9
wire
DC resistance of, 139–140
work, 105
X-ray , 3",Electromagnetics_Vol1.pdf
"Computer Science I
Dr. Chris Bourke
cbourke@cse.unl.edu
Department of Computer Science & Engineering
University of Nebraska–Lincoln
Lincoln, NE 68588, USA
2018/10/12 08:09:12
Version 1.3.7
φ
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"Copyleft (Copyright)
The entirety of this book is free and is released under aCreative Commons Attribution-
ShareAlike 4.0 International License(see http://creativecommons.org/licenses/
by-sa/4.0/ for details).
i",ComputerScienceOne.pdf
"Draft Notice
This book is a draft that has been released for evaluation and comment. Some of the later
chapters are included as placeholders and indicators for the intended scope of the final
draft, but are intentionally left blank. The author encourages people to send feedback
including suggestions, corrections, and reviews to inform and influence the final draft.
Thank you in advance to anyone helping out or sending constructive criticisms.
iii",ComputerScienceOne.pdf
"Preface
“If you really want to understand something, the best way is to try and
explain it to someone else. That forces you to sort it out in your own mind...
that’s really the essence of programming. By the time you’ve sorted out a
complicated idea into little steps that even a stupid machine can deal with,
you’ve certainly learned something about it yourself.” —Douglas Adams,
Dirk Gently’s Holistic Detective Agency [8]
“The world of A.D. 2014 will have few routine jobs that cannot be done better
by some machine than by any human being. Mankind will therefore have
become largely a race of machine tenders. Schools will have to be oriented in
this direction. All the high-school students will be taught the fundamentals
of computer technology, will become proficient in binary arithmetic and will
be trained to perfection in the use of the computer languages that will have
developed out of those like the contemporary Fortran” —Isaac Asimov 1964",ComputerScienceOne.pdf
"be trained to perfection in the use of the computer languages that will have
developed out of those like the contemporary Fortran” —Isaac Asimov 1964
I’ve been teaching Computer Science since 2008 and was a Teaching Assistant long
before that. Before that I was a student. During that entire time I’ve been continually
disappointed in the value (note, not quality) of textbooks, particularly Computer Science
textbooks and especially introductory textbooks. Of primary concern are the costs,
which have far outstripped inflation over the last decade [ 30] while not providing any
real additional value. New editions with trivial changes are released on a regular basis in
an attempt to nullify the used book market. Publishers engage in questionable business
practices and unfortunately many institutions are complicit in this process.
In established fields such as mathematics and physics, new textbooks are especially
questionable as the material and topics don’t undergo many changes. However, in",ComputerScienceOne.pdf
"questionable as the material and topics don’t undergo many changes. However, in
Computer Science, new languages and technologies are created and change at breakneck
speeds. Faculty and students are regularly trying to give away stacks of textbooks
(“Learn Java 4!,” “Introduction to Cold Fusion,” etc.) that are only a few years old and
yet are completely obsolete and worthless. The problem is that such books have built-in
obsolescence by focusing too much on technological specifics and not enough on concepts.
There are dozens of introductory textbooks for Computer Science; add in the fact that
there are multiple languages and many gimmicks (“Learn Multimedia Java,” “Gaming
with JavaScript,” “Build a Robot with C!”), it is a publisher’s paradise: hundreds of
variations, a growing market, and customers with few alternatives.
v",ComputerScienceOne.pdf
"Preface
That’s why I like organizations like OpenStax ( http://openstaxcollege.org/) that
attempt to provide free and “open” learning materials. Though they have textbooks for
a variety of disciplines, Computer Science is not one of them (currently, that is). This
might be due to the fact that there are already a huge amount of resources available
online such as tutorials, videos, online open courses, and even interactive code learning
tools. With such a huge amount of resources, why write this textbook then? Firstly,
layoff. Secondly, I don’t really expect this book to have much impact beyond my own
courses or department. I wanted a resource that presented an introduction to Computer
Science how I teach it in my courses and it wasn’t available. However, if it does find its
way into another instructor’s classes or into the hands of an aspiring student that wants
to learn, then great!
Several years ago our department revamped our introductory courses in a “Renaissance",ComputerScienceOne.pdf
"to learn, then great!
Several years ago our department revamped our introductory courses in a “Renaissance
in Computing” initiative in which we redeveloped several different “flavors” of Computer
Science I (one intended for Computer Science majors, one for Computer Engineering
majors, one for non-CE engineering majors, one for humanities majors, etc.). The courses
are intended to be equivalent in content but have a broader appeal to those in different
disciplines. The intent was to provide multiple entry points into Computer Science. Once
a student had a solid foundation, they could continue into Computer Science II and pick
up a second programming language with little difficulty.
This basic idea informed how I structured this book. There is a separation of concepts
and programming language syntax. The first part of this book uses pseudocode with
a minimum of language-specific elements. Subsequent parts of the book recapitulate",ComputerScienceOne.pdf
"a minimum of language-specific elements. Subsequent parts of the book recapitulate
these concepts but in the context of a specific programming language. This allows for a
“plug-in” style approach to Computer Science: the same book could theoretically be used
for multiple courses or the book could be extended by adding another part for a new
language with minimal effort.
Another inspiration for the structure of this book is the Computer Science I Honors course
that I developed. Usually Computer Science majors take CS1 using Java as the primary
language while CE students take CS1 using C. Since the honors course consists of both
majors (as well as some of the top students), I developed the Honors version to cover
both languages at the same time in parallel. This has led to many interesting teaching
moments: by covering two languages, it provides opportunities to highlight fundamental
differences and concepts in programming languages. It also keeps concepts as the focus of",ComputerScienceOne.pdf
"differences and concepts in programming languages. It also keeps concepts as the focus of
the course emphasizing that syntax and idiosyncrasies of individual languages are only of
secondary concern. Finally, actively using multiple languages in the first class provides a
better opportunity to extend knowledge to other programming languages–once a student
has a solid foundation in one language learning a new one should be relatively easy.
The exercises in this book are a variety of exercises I’ve used in my courses over the
years. They have been made as generic as possible so that they could be assigned using
any language. While some have emphasized the use of “real-world” exercises (whatever
that means), my exercises have focused more on solving problems of a mathematical
vi",ComputerScienceOne.pdf
"nature (most of my students have been Engineering students). Some of them are more
easily understood if students have had Calculus but it is not absolutely necessary.
It may be clich´ e, but the two quotes above exemplify what I believe a Computer Science
I course is about. The second is from Isaac Asimov who was asked at the 1964 World’s
Fair what he though the world of 2014 would look like. His prediction didn’t become
entirely true, but I do believe we are on the verge of a fundamental social change that
will be caused by more and more automation. Like the industrial revolution, but on
a much smaller time scale and to a far greater extent, automation will fundamentally
change how we live and not work (I say “not work” because automation will very easily
destroy the vast majority of today’s jobs–this represents a huge economic and political
challenge that will need to be addressed). The time is quickly approaching where being",ComputerScienceOne.pdf
"challenge that will need to be addressed). The time is quickly approaching where being
able to program and develop software will be considered a fundamental skill as essential
as arithmetic. I hope this book plays some small role in helping students adjust to that
coming world.
The first quote describes programming, or more fundamentally Computer Science and
“problem solving.” Computers do not solve problems, humans do. Computers only make
it possible to automate solutions on a large scale. At the end of the day, the human race
is still responsible for tending the machines and will be for some time despite what Star
Trek and the most optimistic of AI advocates think.
I hope that people find this book useful. If value is a ratio of quality vs cost then this
book has already succeeded in having infinite value. 1 If you have suggestions on how to
improve it, please feel free to contact me. If you end up using it and finding it useful,
please let me know that too!",ComputerScienceOne.pdf
"improve it, please feel free to contact me. If you end up using it and finding it useful,
please let me know that too!
1or it might be undefined, or NaN, or this book is Exceptional depending on which language sections
you read
vii",ComputerScienceOne.pdf
"Acknowledgements
I’d like to thank the Department of Computer Science & Engineering at the University
of Nebraska–Lincoln for their support during my writing and maintaining this book.
This book is dedicated to my family.
ix",ComputerScienceOne.pdf
"Contents
Copyleft (Copyright) i
Draft Notice iii
Preface v
Acknowledgements ix
1. Introduction 1
1.1. Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2. Computing Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3. Basic Program Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4. Syntax Rules & Pseudocode . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.5. Documentation, Comments, and Coding Style . . . . . . . . . . . . . . . 14
2. Basics 17
2.1. Control Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.1.1. Flowcharts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2. Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2.1. Naming Rules & Conventions . . . . . . . . . . . . . . . . . . . . 19
2.2.2. Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22",ComputerScienceOne.pdf
"2.2.2. Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2.3. Declaring Variables: Dynamic vs. Static Typing . . . . . . . . . . 31
2.2.4. Scoping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.3. Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.3.1. Assignment Operators . . . . . . . . . . . . . . . . . . . . . . . . 33
2.3.2. Numerical Operators . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.3.3. String Concatenation . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.3.4. Order of Precedence . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3.5. Common Numerical Errors . . . . . . . . . . . . . . . . . . . . . . 38
2.3.6. Other Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.4. Basic Input/Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.4.1. Standard Input & Output . . . . . . . . . . . . . . . . . . . . . . 42",ComputerScienceOne.pdf
"2.4.1. Standard Input & Output . . . . . . . . . . . . . . . . . . . . . . 42
2.4.2. Graphical User Interfaces . . . . . . . . . . . . . . . . . . . . . . . 42
2.4.3. Output Using printf()-style Formatting . . . . . . . . . . . . . 43
2.4.4. Command Line Input . . . . . . . . . . . . . . . . . . . . . . . . . 44
xi",ComputerScienceOne.pdf
"Contents
2.5. Debugging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.5.1. Types of Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.5.2. Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.6. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.6.1. Temperature Conversion . . . . . . . . . . . . . . . . . . . . . . . 50
2.6.2. Quadratic Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
2.7. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3. Conditionals 65
3.1. Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
3.1.1. Comparison Operators . . . . . . . . . . . . . . . . . . . . . . . . 66
3.1.2. Negation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
3.1.3. Logical And . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69",ComputerScienceOne.pdf
"3.1.3. Logical And . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.1.4. Logical Or . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
3.1.5. Compound Statements . . . . . . . . . . . . . . . . . . . . . . . . 71
3.1.6. Short Circuiting . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3.2. The If Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
3.3. The If-Else Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.4. The If-Else-If Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
3.5. Ternary If-Else Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.6. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.6.1. Meal Discount . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.6.2. Look Before You Leap . . . . . . . . . . . . . . . . . . . . . . . . 83
3.6.3. Comparing Elements . . . . . . . . . . . . . . . . . . . . . . . . . 84",ComputerScienceOne.pdf
"3.6.3. Comparing Elements . . . . . . . . . . . . . . . . . . . . . . . . . 84
3.6.4. Life & Taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
3.7. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
4. Loops 95
4.1. While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
4.1.1. Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.2. For Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
4.2.1. Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.3. Do-While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.4. Foreach Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.5. Other Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.5.1. Nested Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103",ComputerScienceOne.pdf
"4.5.1. Nested Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.5.2. Infinite Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.5.3. Common Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.5.4. Equivalency of Loops . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.6. Problem Solving With Loops . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.7. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4.7.1. For vs While Loop . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4.7.2. Primality Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
4.7.3. Paying the Piper . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
xii",ComputerScienceOne.pdf
"Contents
4.8. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
5. Functions 133
5.1. Defining & Using Functions . . . . . . . . . . . . . . . . . . . . . . . . . 134
5.1.1. Function Signatures . . . . . . . . . . . . . . . . . . . . . . . . . . 134
5.1.2. Calling Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
5.1.3. Organizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
5.2. How Functions Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
5.2.1. Call By Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5.2.2. Call By Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5.3. Other Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
5.3.1. Functions as Entities . . . . . . . . . . . . . . . . . . . . . . . . . 142
5.3.2. Function Overloading . . . . . . . . . . . . . . . . . . . . . . . . . 144",ComputerScienceOne.pdf
"5.3.2. Function Overloading . . . . . . . . . . . . . . . . . . . . . . . . . 144
5.3.3. Variable Argument Functions . . . . . . . . . . . . . . . . . . . . 145
5.3.4. Optional Parameters & Default Values . . . . . . . . . . . . . . . 145
5.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
6. Error Handling 151
6.1. Error Handling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
6.2. Error Handling Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . 153
6.2.1. Defensive Programming . . . . . . . . . . . . . . . . . . . . . . . 153
6.2.2. Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
6.3. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
7. Arrays, Collections & Dynamic Memory 159
7.1. Basic Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
7.2. Static & Dynamic Memory . . . . . . . . . . . . . . . . . . . . . . . . . . 162",ComputerScienceOne.pdf
"7.2. Static & Dynamic Memory . . . . . . . . . . . . . . . . . . . . . . . . . . 162
7.2.1. Dynamic Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
7.2.2. Shallow vs. Deep Copies . . . . . . . . . . . . . . . . . . . . . . . 166
7.3. Multidimensional Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
7.4. Other Collections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
7.5. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
8. Strings 177
8.1. Basic Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
8.2. Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
8.3. Tokenizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
8.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
9. File Input/Output 183",ComputerScienceOne.pdf
"8.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
9. File Input/Output 183
9.1. Processing Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
9.1.1. Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
9.1.2. Error Handling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
9.1.3. Buffered and Unbuffered . . . . . . . . . . . . . . . . . . . . . . . 187
xiii",ComputerScienceOne.pdf
"Contents
9.1.4. Binary vs Text Files . . . . . . . . . . . . . . . . . . . . . . . . . 187
9.2. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
10.Encapsulation & Objects 197
10.1. Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
10.1.1. Defining . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
10.1.2. Creating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
10.1.3. Using Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
10.2. Design Principles & Best Practices . . . . . . . . . . . . . . . . . . . . . 200
10.3. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
11.Recursion 203
11.1. Writing Recursive Functions . . . . . . . . . . . . . . . . . . . . . . . . . 204
11.1.1. Tail Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205",ComputerScienceOne.pdf
"11.1.1. Tail Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
11.2. Avoiding Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
11.2.1. Memoization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
11.3. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
12.Searching & Sorting 211
12.1. Searching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
12.1.1. Linear Search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
12.1.2. Binary Search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
12.1.3. Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
12.2. Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
12.2.1. Selection Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
12.2.2. Insertion Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224",ComputerScienceOne.pdf
"12.2.2. Insertion Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
12.2.3. Quick Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
12.2.4. Merge Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
12.2.5. Other Sorts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
12.2.6. Comparison & Summary . . . . . . . . . . . . . . . . . . . . . . . 237
12.3. Searching & Sorting In Practice . . . . . . . . . . . . . . . . . . . . . . . 238
12.3.1. Using Libraries and Comparators . . . . . . . . . . . . . . . . . . 238
12.3.2. Preventing Arithmetic Errors . . . . . . . . . . . . . . . . . . . . 239
12.3.3. Avoiding the Difference Trick . . . . . . . . . . . . . . . . . . . . 241
12.3.4. Importance of a Total Order . . . . . . . . . . . . . . . . . . . . . 242
12.3.5. Artificial Ordering . . . . . . . . . . . . . . . . . . . . . . . . . . 242
12.3.6. Sorting Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 243",ComputerScienceOne.pdf
"12.3.6. Sorting Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
12.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
13.Graphical User Interfaces & Event Driven Programming 247
14.Introduction to Databases & Database Connectivity 249
xiv",ComputerScienceOne.pdf
"Contents
I. The C Programming Language 251
15.Basics 253
15.1. Getting Started: Hello World . . . . . . . . . . . . . . . . . . . . . . . . 253
15.2. Basic Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
15.2.1. Basic Syntax Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 255
15.2.2. Preprocessor Directives . . . . . . . . . . . . . . . . . . . . . . . . 255
15.2.3. Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
15.2.4. The main() Function . . . . . . . . . . . . . . . . . . . . . . . . 259
15.3. Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260
15.3.1. Declaration & Assignment . . . . . . . . . . . . . . . . . . . . . . 260
15.4. Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
15.5. Basic I/O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263",ComputerScienceOne.pdf
"15.5. Basic I/O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
15.6. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
15.6.1. Converting Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
15.6.2. Computing Quadratic Roots . . . . . . . . . . . . . . . . . . . . . 267
16.Conditionals 271
16.1. Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
16.1.1. Order of Precedence . . . . . . . . . . . . . . . . . . . . . . . . . 273
16.1.2. Comparing Strings and Characters . . . . . . . . . . . . . . . . . 273
16.2. If, If-Else, If-Else-If Statements . . . . . . . . . . . . . . . . . . . . . . . 274
16.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
16.3.1. Computing a Logarithm . . . . . . . . . . . . . . . . . . . . . . . 276
16.3.2. Life & Taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277",ComputerScienceOne.pdf
"16.3.2. Life & Taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
16.3.3. Quadratic Roots Revisited . . . . . . . . . . . . . . . . . . . . . . 279
17.Loops 283
17.1. While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
17.2. For Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
17.3. Do-While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
17.4. Other Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
17.5. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
17.5.1. Normalizing a Number . . . . . . . . . . . . . . . . . . . . . . . . 287
17.5.2. Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
17.5.3. Nested Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
17.5.4. Paying the Piper . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
18.Functions 291",ComputerScienceOne.pdf
"17.5.4. Paying the Piper . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
18.Functions 291
18.1. Defining & Using Functions . . . . . . . . . . . . . . . . . . . . . . . . . 291
18.1.1. Declaration: Prototypes . . . . . . . . . . . . . . . . . . . . . . . 291
18.1.2. Void Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
18.1.3. Organizing Functions . . . . . . . . . . . . . . . . . . . . . . . . . 294
18.1.4. Calling Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
xv",ComputerScienceOne.pdf
"Contents
18.2. Pointers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
18.2.1. Passing By Reference . . . . . . . . . . . . . . . . . . . . . . . . . 297
18.2.2. Function Pointers . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
18.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
18.3.1. Generalized Rounding . . . . . . . . . . . . . . . . . . . . . . . . 301
18.3.2. Quadratic Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
19.Error Handling 305
19.1. Language Supported Error Codes . . . . . . . . . . . . . . . . . . . . . . 305
19.1.1. POSIX Error Codes . . . . . . . . . . . . . . . . . . . . . . . . . 306
19.2. Error Handling By Design . . . . . . . . . . . . . . . . . . . . . . . . . . 308
19.3. Enumerated Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
19.4. Using Enumerated Types for Error Codes . . . . . . . . . . . . . . . . . 310
20.Arrays 313",ComputerScienceOne.pdf
"19.4. Using Enumerated Types for Error Codes . . . . . . . . . . . . . . . . . 310
20.Arrays 313
20.1. Basic Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
20.2. Dynamic Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
20.3. Using Arrays with Functions . . . . . . . . . . . . . . . . . . . . . . . . . 318
20.4. Multidimensional Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
20.4.1. Contiguous 2-D Arrays . . . . . . . . . . . . . . . . . . . . . . . . 322
20.5. Dynamic Data Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 324
21.Strings 325
21.1. Character Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
21.2. String Library . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
21.3. Arrays of Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
21.4. Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331",ComputerScienceOne.pdf
"21.4. Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
21.5. Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
21.6. Tokenizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
22.File I/O 335
22.1. Opening Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
22.2. Reading & Writing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
22.2.1. Plaintext Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
22.2.2. Binary Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
22.3. Closing Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
23.Structures 341
23.1. Defining Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
23.1.1. Alternative Declarations . . . . . . . . . . . . . . . . . . . . . . . 342",ComputerScienceOne.pdf
"23.1.1. Alternative Declarations . . . . . . . . . . . . . . . . . . . . . . . 342
23.1.2. Nested Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
23.2. Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
23.2.1. Declaration & Initialization . . . . . . . . . . . . . . . . . . . . . 344
23.2.2. Selection Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 346
xvi",ComputerScienceOne.pdf
"Contents
23.3. Arrays of Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
23.4. Using Structures With Functions . . . . . . . . . . . . . . . . . . . . . . 351
23.4.1. Factory Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
23.4.2. To String Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 354
23.4.3. Passing Arrays of Structures . . . . . . . . . . . . . . . . . . . . . 355
24.Recursion 357
25.Searching & Sorting 361
25.1. Comparator Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
25.2. Function Pointers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
25.3. Searching & Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
25.3.1. Searching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
25.3.2. Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
25.3.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374",ComputerScienceOne.pdf
"25.3.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
25.4. Other Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
25.4.1. Sorting Pointers to Elements . . . . . . . . . . . . . . . . . . . . . 377
II. The Java Programming Language 381
26.Basics 383
26.1. Getting Started: Hello World . . . . . . . . . . . . . . . . . . . . . . . . 384
26.2. Basic Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
26.2.1. Basic Syntax Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 385
26.2.2. Program Structure . . . . . . . . . . . . . . . . . . . . . . . . . . 386
26.2.3. The main() Method . . . . . . . . . . . . . . . . . . . . . . . . . 389
26.2.4. Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
26.3. Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390
26.3.1. Declaration & Assignment . . . . . . . . . . . . . . . . . . . . . . 391",ComputerScienceOne.pdf
"26.3.1. Declaration & Assignment . . . . . . . . . . . . . . . . . . . . . . 391
26.4. Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
26.5. Basic I/O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
26.6. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396
26.6.1. Converting Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 396
26.6.2. Computing Quadratic Roots . . . . . . . . . . . . . . . . . . . . . 400
27.Conditionals 403
27.1. Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403
27.1.1. Order of Precedence . . . . . . . . . . . . . . . . . . . . . . . . . 405
27.1.2. Comparing Strings and Characters . . . . . . . . . . . . . . . . . 406
27.2. If, If-Else, If-Else-If Statements . . . . . . . . . . . . . . . . . . . . . . . 407
27.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408",ComputerScienceOne.pdf
"27.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408
27.3.1. Computing a Logarithm . . . . . . . . . . . . . . . . . . . . . . . 408
27.3.2. Life & Taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410
xvii",ComputerScienceOne.pdf
"Contents
27.3.3. Quadratic Roots Revisited . . . . . . . . . . . . . . . . . . . . . . 411
28.Loops 415
28.1. While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
28.2. For Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
28.3. Do-While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
28.4. Enhanced For Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418
28.5. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
28.5.1. Normalizing a Number . . . . . . . . . . . . . . . . . . . . . . . . 419
28.5.2. Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
28.5.3. Nested Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420
28.5.4. Paying the Piper . . . . . . . . . . . . . . . . . . . . . . . . . . . 421
29.Methods 423
29.1. Defining Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424",ComputerScienceOne.pdf
"29.Methods 423
29.1. Defining Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424
29.1.1. Void Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426
29.1.2. Using Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426
29.1.3. Passing By Reference . . . . . . . . . . . . . . . . . . . . . . . . . 427
29.2. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428
29.2.1. Generalized Rounding . . . . . . . . . . . . . . . . . . . . . . . . 428
30.Error Handling & Exceptions 431
30.1. Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431
30.1.1. Catching Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . 431
30.1.2. Throwing Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . 433
30.1.3. Creating Custom Exceptions . . . . . . . . . . . . . . . . . . . . . 433
30.1.4. Checked Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . 434",ComputerScienceOne.pdf
"30.1.4. Checked Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . 434
30.2. Enumerated Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
30.2.1. More Tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
31.Arrays 439
31.1. Basic Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439
31.2. Dynamic Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
31.3. Using Arrays with Methods . . . . . . . . . . . . . . . . . . . . . . . . . 442
31.4. Multidimensional Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . 443
31.5. Dynamic Data Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 444
32.Strings 449
32.1. Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449
32.2. String Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
32.3. Arrays of Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452",ComputerScienceOne.pdf
"32.3. Arrays of Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452
32.4. Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453
32.5. Tokenizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455
xviii",ComputerScienceOne.pdf
"Contents
33.File I/O 457
33.1. File Input . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
33.2. File Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
34.Objects 461
34.1. Data Visibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462
34.2. Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
34.2.1. Accessor & Mutator Methods . . . . . . . . . . . . . . . . . . . . 464
34.3. Constructors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
34.4. Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
34.5. Common Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
34.6. Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
34.7. Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474
35.Recursion 479
36.Searching & Sorting 483",ComputerScienceOne.pdf
"34.7. Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474
35.Recursion 479
36.Searching & Sorting 483
36.1. Comparators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
36.2. Searching & Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
36.2.1. Searching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
36.2.2. Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
36.3. Other Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
36.3.1. Sorted Collections . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
36.3.2. Handling null values . . . . . . . . . . . . . . . . . . . . . . . . 490
36.3.3. Importance of equals() and hashCode() Methods . . . . . . . 491
36.3.4. Java 8: Lambda Expressions . . . . . . . . . . . . . . . . . . . . . 493
III. The PHP Programming Language 495
37.Basics 497",ComputerScienceOne.pdf
"36.3.4. Java 8: Lambda Expressions . . . . . . . . . . . . . . . . . . . . . 493
III. The PHP Programming Language 495
37.Basics 497
37.1. Getting Started: Hello World . . . . . . . . . . . . . . . . . . . . . . . . 498
37.2. Basic Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
37.2.1. Basic Syntax Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 499
37.2.2. PHP Tags . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
37.2.3. Libraries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
37.2.4. Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
37.2.5. Entry Point & Command Line Arguments . . . . . . . . . . . . . 502
37.3. Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
37.3.1. Using Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
37.4. Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504",ComputerScienceOne.pdf
"37.4. Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504
37.4.1. Type Juggling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505
37.4.2. String Concatenation . . . . . . . . . . . . . . . . . . . . . . . . . 508
37.5. Basic I/O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508
xix",ComputerScienceOne.pdf
"Contents
37.6. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509
37.6.1. Converting Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 509
37.6.2. Computing Quadratic Roots . . . . . . . . . . . . . . . . . . . . . 512
38.Conditionals 515
38.1. Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515
38.1.1. Order of Precedence . . . . . . . . . . . . . . . . . . . . . . . . . 517
38.2. If, If-Else, If-Else-If Statements . . . . . . . . . . . . . . . . . . . . . . . 517
38.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519
38.3.1. Computing a Logarithm . . . . . . . . . . . . . . . . . . . . . . . 519
38.3.2. Life & Taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520
38.3.3. Quadratic Roots Revisited . . . . . . . . . . . . . . . . . . . . . . 522
39.Loops 527
39.1. While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527",ComputerScienceOne.pdf
"39.Loops 527
39.1. While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
39.2. For Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528
39.3. Do-While Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
39.4. Foreach Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
39.5. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530
39.5.1. Normalizing a Number . . . . . . . . . . . . . . . . . . . . . . . . 530
39.5.2. Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530
39.5.3. Nested Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531
39.5.4. Paying the Piper . . . . . . . . . . . . . . . . . . . . . . . . . . . 531
40.Functions 535
40.1. Defining & Using Functions . . . . . . . . . . . . . . . . . . . . . . . . . 535
40.1.1. Declaring Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 535",ComputerScienceOne.pdf
"40.1.1. Declaring Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 535
40.1.2. Organizing Functions . . . . . . . . . . . . . . . . . . . . . . . . . 537
40.1.3. Calling Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 537
40.1.4. Passing By Reference . . . . . . . . . . . . . . . . . . . . . . . . . 538
40.1.5. Optional & Default Parameters . . . . . . . . . . . . . . . . . . . 539
40.1.6. Function Pointers . . . . . . . . . . . . . . . . . . . . . . . . . . . 540
40.2. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540
40.2.1. Generalized Rounding . . . . . . . . . . . . . . . . . . . . . . . . 540
40.2.2. Quadratic Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . 541
41.Error Handling & Exceptions 543
41.1. Throwing Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543
41.2. Catching Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543",ComputerScienceOne.pdf
"41.2. Catching Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543
41.3. Creating Custom Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . 544
42.Arrays 547
42.1. Creating Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547
xx",ComputerScienceOne.pdf
"Contents
42.2. Indexing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547
42.2.1. Strings as Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
42.2.2. Non-Contiguous Indices . . . . . . . . . . . . . . . . . . . . . . . 549
42.2.3. Key-Value Initialization . . . . . . . . . . . . . . . . . . . . . . . 549
42.3. Useful Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550
42.4. Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551
42.5. Adding Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552
42.6. Removing Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553
42.7. Using Arrays in Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 554
42.8. Multidimensional Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . 555
43.Strings 557
43.1. Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557",ComputerScienceOne.pdf
"43.Strings 557
43.1. Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557
43.2. String Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
43.3. Arrays of Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560
43.4. Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560
43.5. Tokenizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561
44.File I/O 563
44.1. Opening Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563
44.2. Reading & Writing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564
44.2.1. Using URLs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565
44.2.2. Closing Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565
45.Objects 567
45.1. Data Visibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568",ComputerScienceOne.pdf
"45.Objects 567
45.1. Data Visibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568
45.2. Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568
45.2.1. Accessor & Mutator Methods . . . . . . . . . . . . . . . . . . . . 570
45.3. Constructors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571
45.4. Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572
45.5. Common Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
45.6. Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
45.7. Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574
46.Recursion 577
47.Searching & Sorting 581
47.1. Comparator Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581
47.1.1. Searching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584",ComputerScienceOne.pdf
"47.1.1. Searching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584
47.1.2. Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584
Glossary 587
Acronyms 599
xxi",ComputerScienceOne.pdf
"Contents
Index 610
References 613
xxii",ComputerScienceOne.pdf
"List of Algorithms
1.1. An example of pseudocode: finding a minimum value . . . . . . . . . . . . 13
2.1. Assignment Operator Demonstration . . . . . . . . . . . . . . . . . . . . . 34
2.2. Addition and Subtraction Demonstration . . . . . . . . . . . . . . . . . . 35
2.3. Multiplication and Division Demonstration . . . . . . . . . . . . . . . . . 36
2.4. Temperature Conversion Program . . . . . . . . . . . . . . . . . . . . . . 50
2.5. Quadratic Roots Program . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.1. An if-statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.2. An if-else Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
3.3. Example If-Else-If Statement . . . . . . . . . . . . . . . . . . . . . . . . . 79
3.4. General If-Else-If Statement . . . . . . . . . . . . . . . . . . . . . . . . . . 81
3.5. If-Else-If Statement With a Bug . . . . . . . . . . . . . . . . . . . . . . . 81",ComputerScienceOne.pdf
"3.5. If-Else-If Statement With a Bug . . . . . . . . . . . . . . . . . . . . . . . 81
3.6. A simple receipt program . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
3.7. Preventing Division By Zero Using an If Statement . . . . . . . . . . . . . 83
3.8. Comparing Students by Name . . . . . . . . . . . . . . . . . . . . . . . . . 84
3.9. Computing Tax Liability with If-Else-If . . . . . . . . . . . . . . . . . . . 86
3.10. Computing Tax Credit with If-Else-If . . . . . . . . . . . . . . . . . . . . . 86
4.1. Counter-Controlled While Loop . . . . . . . . . . . . . . . . . . . . . . . . 97
4.2. Normalizing a Number With a While Loop . . . . . . . . . . . . . . . . . 99
4.3. A General For Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
xxiii",ComputerScienceOne.pdf
"LIST OF ALGORITHMS
4.4. Counter-Controlled For Loop . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.5. Summation of Numbers in a For Loop . . . . . . . . . . . . . . . . . . . . 100
4.6. Counter-Controlled Do-While Loop . . . . . . . . . . . . . . . . . . . . . . 101
4.7. Flag-Controlled Do-While Loop . . . . . . . . . . . . . . . . . . . . . . . . 102
4.8. Example Foreach Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.9. Foreach Loop Computing Grades . . . . . . . . . . . . . . . . . . . . . . . 103
4.10. Nested For Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.11. Infinite Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.12. Computing the Geometric Series Using a For Loop . . . . . . . . . . . . . 107
4.13. Computing the Geometric Series Using a While Loop . . . . . . . . . . . . 108
4.14. Determining if a Number is Prime or Composite . . . . . . . . . . . . . . 109",ComputerScienceOne.pdf
"4.14. Determining if a Number is Prime or Composite . . . . . . . . . . . . . . 109
4.15. Counting the number of primes. . . . . . . . . . . . . . . . . . . . . . . . . 109
4.16. Computing a loan amortization schedule . . . . . . . . . . . . . . . . . . . 111
4.17. Scaling a Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.1. A function in pseudocode . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
5.2. Using a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
11.1. Recursive CountDown(n) Function . . . . . . . . . . . . . . . . . . . . . 203
11.2. Recursive Fibonacci(n) Function . . . . . . . . . . . . . . . . . . . . . . 204
11.3. Recursive Fibonacci(n) Function With Memoization . . . . . . . . . . . 208
12.1. Linear Search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
12.2. Recursive Binary Search Algorithm, BinarySearch(A,l,r,e k) . . . . . . 214",ComputerScienceOne.pdf
"12.2. Recursive Binary Search Algorithm, BinarySearch(A,l,r,e k) . . . . . . 214
12.3. Iterative Binary Search Algorithm, BinarySearch(A,ek) . . . . . . . . . 215
12.4. Selection Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
12.5. Insertion Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
12.6. QuickSort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
xxiv",ComputerScienceOne.pdf
"LIST OF ALGORITHMS
12.7. In-Place Partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
12.8. MergeSort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
12.9. Merge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
xxv",ComputerScienceOne.pdf
"List of Code Samples
1.1. A simple program in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2. A simple program in C, compiled to assembly . . . . . . . . . . . . . . . 10
1.3. A simple program in C, resulting machine code formatted in hexadecimal
(partial) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1. Example of variable scoping in C . . . . . . . . . . . . . . . . . . . . . . 32
2.2. Compound Assignment Operators in C . . . . . . . . . . . . . . . . . . . 41
2.3. printf() examples in C . . . . . . . . . . . . . . . . . . . . . . . . . . 45
2.4. Output Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.1. Zune Bug . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
15.1. Hello World Program in C . . . . . . . . . . . . . . . . . . . . . . . . . . 254
15.2. Fahrenheit-to-Celsius Conversion Program in C . . . . . . . . . . . . . . 268",ComputerScienceOne.pdf
"15.2. Fahrenheit-to-Celsius Conversion Program in C . . . . . . . . . . . . . . 268
15.3. Quadratic Roots Program in C . . . . . . . . . . . . . . . . . . . . . . . 269
16.1. Examples of Conditional Statements in C . . . . . . . . . . . . . . . . . . 275
16.2. Logarithm Calculator Program in C . . . . . . . . . . . . . . . . . . . . . 280
16.3. Tax Program in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
16.4. Quadratic Roots Program in C With Error Checking . . . . . . . . . . . 282
17.1. While Loop in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
17.2. Flag-controlled While Loop in C . . . . . . . . . . . . . . . . . . . . . . . 284
17.3. For Loop in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
17.4. Do-While Loop in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
17.5. Normalizing a Number with a While Loop in C . . . . . . . . . . . . . . 287",ComputerScienceOne.pdf
"17.5. Normalizing a Number with a While Loop in C . . . . . . . . . . . . . . 287
17.6. Summation of Numbers using a For Loop in C . . . . . . . . . . . . . . . 287
17.7. Nested For Loops in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
17.8. Loan Amortization Program in C . . . . . . . . . . . . . . . . . . . . . . 290
19.1. Using the errno.h library . . . . . . . . . . . . . . . . . . . . . . . . . 307
23.1. A Student structure declaration . . . . . . . . . . . . . . . . . . . . . . 344
25.1. C Function Pointer Syntax Examples . . . . . . . . . . . . . . . . . . . . 371
25.2. C Search Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
25.3. C Sort Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376
25.4. C Comparator Function for Strings . . . . . . . . . . . . . . . . . . . . . 377
xxvii",ComputerScienceOne.pdf
"List of Code Samples
25.5. Sorting Structures via Pointers . . . . . . . . . . . . . . . . . . . . . . . 378
25.6. Handling Null Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
26.1. Hello World Program in Java . . . . . . . . . . . . . . . . . . . . . . . . 384
26.2. Basic Input/Output in Java . . . . . . . . . . . . . . . . . . . . . . . . . 396
26.3. Fahrenheit-to-Celsius Conversion Program in Java . . . . . . . . . . . . . 399
26.4. Quadratic Roots Program in Java . . . . . . . . . . . . . . . . . . . . . . 402
27.1. Examples of Conditional Statements in Java . . . . . . . . . . . . . . . . 407
27.2. Logarithm Calculator Program in Java . . . . . . . . . . . . . . . . . . . 412
27.3. Tax Program in Java . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413
27.4. Quadratic Roots Program in Java With Error Checking . . . . . . . . . . 414
28.1. While Loop in Java . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415",ComputerScienceOne.pdf
"28.1. While Loop in Java . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
28.2. Flag-controlled While Loop in Java . . . . . . . . . . . . . . . . . . . . . 416
28.3. For Loop in Java . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
28.4. Do-While Loop in Java . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418
28.5. Enhanced For Loops in Java Example 1 . . . . . . . . . . . . . . . . . . . 418
28.6. Enhanced For Loops in Java Example 2 . . . . . . . . . . . . . . . . . . . 419
28.7. Normalizing a Number with a While Loop in Java . . . . . . . . . . . . . 419
28.8. Summation of Numbers using a For Loop in Java . . . . . . . . . . . . . 420
28.9. Nested For Loops in Java . . . . . . . . . . . . . . . . . . . . . . . . . . . 420
28.10. Loan Amortization Program in Java . . . . . . . . . . . . . . . . . . . . . 422
34.1. The completed Java Student class. . . . . . . . . . . . . . . . . . . . . 477",ComputerScienceOne.pdf
"34.1. The completed Java Student class. . . . . . . . . . . . . . . . . . . . . 477
36.1. Java Search Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
36.2. Using Java Collection’s Sort Method . . . . . . . . . . . . . . . . . . . . 490
36.3. Handling Null Values in Java Comparators . . . . . . . . . . . . . . . . . 491
37.1. Hello World Program in PHP . . . . . . . . . . . . . . . . . . . . . . . . 498
37.2. Hello World Program in PHP with HTML . . . . . . . . . . . . . . . . . 498
37.3. Type Juggling in PHP . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506
37.4. Fahrenheit-to-Celsius Conversion Program in PHP . . . . . . . . . . . . . 511
37.5. Quadratic Roots Program in PHP . . . . . . . . . . . . . . . . . . . . . . 513
38.1. Examples of Conditional Statements in PHP . . . . . . . . . . . . . . . . 518
38.2. Logarithm Calculator Program in C . . . . . . . . . . . . . . . . . . . . . 523",ComputerScienceOne.pdf
"38.2. Logarithm Calculator Program in C . . . . . . . . . . . . . . . . . . . . . 523
38.3. Tax Program in PHP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
38.4. Quadratic Roots Program in PHP With Error Checking . . . . . . . . . 525
39.1. While Loop in PHP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
39.2. Flag-controlled While Loop in PHP . . . . . . . . . . . . . . . . . . . . . 528
39.3. For Loop in PHP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
39.4. Do-While Loop in PHP . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
39.5. Normalizing a Number with a While Loop in PHP . . . . . . . . . . . . . 530
xxviii",ComputerScienceOne.pdf
"List of Code Samples
39.6. Summation of Numbers using a For Loop in PHP . . . . . . . . . . . . . 531
39.7. Nested For Loops in PHP . . . . . . . . . . . . . . . . . . . . . . . . . . 531
39.8. Loan Amortization Program in PHP . . . . . . . . . . . . . . . . . . . . 533
44.1. Processing a file line-by-line in PHP . . . . . . . . . . . . . . . . . . . . . 564
45.1. The completed PHP Student class. . . . . . . . . . . . . . . . . . . . . 576
47.1. Using PHP’s usort() Function . . . . . . . . . . . . . . . . . . . . . . . 585
xxix",ComputerScienceOne.pdf
"List of Figures
1.1. Depiction of Computer Memory . . . . . . . . . . . . . . . . . . . . . . . 6
1.2. A Compiling Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.1. Types of Flowchart Nodes . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2. Example of a flowchart for a simple ATM process . . . . . . . . . . . . . 19
2.3. Elements of a printf() statement in C . . . . . . . . . . . . . . . . . . 44
2.4. Intersection of two circles. . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.1. Control flow diagrams for sequential control flow and an if-statement. . . 77
3.2. An if-else Flow Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
3.3. Control Flow for an If-Else-If Statement . . . . . . . . . . . . . . . . . . 80
3.4. Quadrants of the Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . 87
3.5. Three types of triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 90",ComputerScienceOne.pdf
"3.5. Three types of triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
3.6. Intersection of Two Rectangles . . . . . . . . . . . . . . . . . . . . . . . . 91
3.7. Examples of Floor Tiling . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.1. A Typical Loop Flow Chart . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.2. A Do-While Loop Flow Chart. The continuation condition is checked after
the loop body. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4.3. Plot of f(x) = sin x
x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
4.4. A rectangle for the interval [ −5,5]. . . . . . . . . . . . . . . . . . . . . . 117
4.5. Follow the bouncing ball . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.6. Sampling points in a circle . . . . . . . . . . . . . . . . . . . . . . . . . . 119
4.7. Regular polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120",ComputerScienceOne.pdf
"4.7. Regular polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
4.8. A polygon and its centroid. Whoo! . . . . . . . . . . . . . . . . . . . . . 130
5.1. A function declaration (prototype) in the C programming language with
the return type, identifier, and parameter list labeled. . . . . . . . . . . . 135
5.2. Program Stack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
5.3. Demonstration of Pass By Value . . . . . . . . . . . . . . . . . . . . . . . 141
5.4. Demonstration of Pass By Reference . . . . . . . . . . . . . . . . . . . . 143
7.1. Example of an Array . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
7.2. Example returning a static array . . . . . . . . . . . . . . . . . . . . . . 163
7.3. Pitfalls of Returning Static Arrays . . . . . . . . . . . . . . . . . . . . . . 174
7.4. Depiction of Application Memory. . . . . . . . . . . . . . . . . . . . . . . 175",ComputerScienceOne.pdf
"7.4. Depiction of Application Memory. . . . . . . . . . . . . . . . . . . . . . . 175
7.5. Shallow vs. Deep Copies . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
xxxi",ComputerScienceOne.pdf
"List of Figures
9.1. Linux Tree Directory Structure . . . . . . . . . . . . . . . . . . . . . . . 186
9.2. An example polygon for n= 5 . . . . . . . . . . . . . . . . . . . . . . . . 188
9.3. A Word Search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
9.4. A solved Sudoku puzzle . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
9.5. A DNA Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
9.6. Codon Table for RNA to Protein Translation . . . . . . . . . . . . . . . . 195
11.1. Recursive Fibonacci Computation Tree . . . . . . . . . . . . . . . . . . . 207
12.1. Array of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
12.2. A Sorted Array . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
12.3. Binary Search Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
12.4. Example of the benefit of ordered (indexed) elements in Windows 7 . . . 220",ComputerScienceOne.pdf
"12.4. Example of the benefit of ordered (indexed) elements in Windows 7 . . . 220
12.5. Selection Sort Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
12.6. Insertion Sort Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
12.7. Partitioning Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
12.8. Partitioning Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
12.9. Partitioning Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
12.10. Merge Sort Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
12.11. Merge Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
12.12. Generalized Sorting with a Comparator . . . . . . . . . . . . . . . . . . . 240
18.1. Pointer Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
20.1. Dynamically Allocating Multidimensional Arrays . . . . . . . . . . . . . 321",ComputerScienceOne.pdf
"20.1. Dynamically Allocating Multidimensional Arrays . . . . . . . . . . . . . 321
20.2. Contiguous Two Dimensional Array . . . . . . . . . . . . . . . . . . . . . 324
21.1. Example of a character array (string) in C. . . . . . . . . . . . . . . . . . 325
23.1. Contiguous Structure Array . . . . . . . . . . . . . . . . . . . . . . . . . 350
23.2. Array of Structure Pointers . . . . . . . . . . . . . . . . . . . . . . . . . 350
23.3. Hybrid Array of Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 352
xxxii",ComputerScienceOne.pdf
"1. Introduction
Computers are awesome. The human race has seen more advancements in the last 50 years
than in the entire 10,000 years of human history. Technology has transformed the way we
live our daily lives, how we interact with each other, and has changed the course of our
history. Today, everyone carries smart phones which have more computational power than
supercomputers from even 20 years ago. Computing has become ubiquitous, the “internet
of things” will soon become a reality in which every device will become interconnected
and data will be collected and available even about the smallest of minutiae.
However, computers are also dumb. Despite the most fantastical of depictions in science
fiction and and hopes of Artificial Intelligence, computers can only do what they are told
to do. The fundamental art of Computer Science is problem solving. Computers are
not good at problem solving; you are the problem solver. It is still up to you, the user,",ComputerScienceOne.pdf
"not good at problem solving; you are the problem solver. It is still up to you, the user,
to approach a complex problem, study it, understand it, and develop a solution to it.
Computers are only good at automating solutions once you have solved the problem.
Computational sciences have become a fundamental tool of almost every discipline.
Scholars have used textual analysis and data mining techniques to analyze classical
literature and historic texts, providing new insights and opening new areas of study.
Astrophysicists have used computational analysis to detect dozens of new exoplanets.
Complex visualizations and models can predict astronomical collisions on a galactic scale.
Physicists have used big data analytics to push the boundaries of our understanding of
matter in the search for the Higgs boson and study of elementary particles. Chemists
simulate the interaction of millions of combinations of compounds without the need for",ComputerScienceOne.pdf
"simulate the interaction of millions of combinations of compounds without the need for
expensive and time consuming physical experiments. Biologists use massively distributed
computing models to simulate protein folding and other complex processes. Meteorologists
can predict weather and climactic changes with ever greater accuracy.
Technology and data analytics have changed how political campaigns are run, how
products are marketed and even delivered. Social networks can be data mined to track
and predict the spread of flu epidemics. Computing and automation will only continue
to grow. The time is soon coming where basic computational thinking and the ability
to develop software will be considered a basic skill necessary to every discipline, a
requirement for many jobs and an essential skill akin to arithmetic.
Computer Science is not programming. Programming is a necessary skill, but it is only
the beginning. This book is intended to get you started on your journey.
1",ComputerScienceOne.pdf
"1. Introduction
1.1. Problem Solving
At its heart, Computer Science is about problem solving. That is not to say that only
Computer Science is about problem solving. It would be hubris to think that Computer
Science holds a monopoly on “problem solving.” Indeed, it would be hard to find any
discipline in which solving problems was not a substantial aspect or motivation if not
integral. Instead, Computer Science is the study of computers and computation. It
involves studying and understanding computational processes and the development of
algorithms and techniques and how they apply to problems.
Problem solving skills are not something that can be distilled down into a single step-
by-step process. Each area and each problem comes with its own unique challenges and
considerations. General problem solving techniques can be identified, studied and taught,
but problem solving skills are something that come with experience, hard work, and most",ComputerScienceOne.pdf
"but problem solving skills are something that come with experience, hard work, and most
importantly, failure. Problem solving is part and parcel of the human experience.
That doesn’t mean we can’t identify techniques and strategies for approaching problems,
in particular problems that lend themselves to computational solutions. A prerequisite
to solving a problem is understanding it. What is the problem? Who or what entities
are involved in the problem? How do those entities interact with each other? What are
the problems or deficiencies that need to be addressed? Answering these questions, we
get an idea of where we are.
Ultimately, what is desired in a solution? What are the objectives that need to be
achieved? What would an ideal solution look like or what would it do? Who would use
the solution and how would they use it? By answering these questions, we get an idea of
where we want to be . Once we know where we are and where we want to be, the problem",ComputerScienceOne.pdf
"where we want to be . Once we know where we are and where we want to be, the problem
solving process can begin: how do we get from point A to point B?
One of the first things a good engineer asks is: does a solution already exist? If a solution
already exists, then the problem is already solved! Ideally the solution is an “off-the-shelf”
solution: something that already exists and which may have been designed for a different
purpose but that can be repurposed for our problem. However, there may be exceptions
to this. The existing solution may be infeasible: it may be too resource intensive or
expensive. It may be too difficult or too expensive to adapt to our problem. It may solve
most of our problem, but may not work in some corner cases. It may need to be heavily
modified in order to work. Still, this basic question may save a lot of time and effort in
many cases.
In a very broad sense, the problem solving process is one that involves
1. Design
2. Implementation
2",ComputerScienceOne.pdf
"1.1. Problem Solving
3. Testing
4. Refinement
After one has a good understanding of a problem, they can start designing a solution. A
design is simply a plan on the construction of a solution. A design “on paper” allows
you to see what the potential solution would look like before investing the resources in
building it. It also allows you to identify possible impediments or problems that were
not readily apparent. A design allows you to an opportunity to think through possible
alternative solutions and weigh the advantages and disadvantages of each. Designing a
solution also allows you to understand the problem better. Design can involve gathering
requirements and developing use cases. How would an individual use the proposed
solution? What features would they need or want?
Implementations can involve building prototype solutions to test the feasibility of the
design. It can involve building individual components and integrating them together.",ComputerScienceOne.pdf
"design. It can involve building individual components and integrating them together.
Testing involves finding, designing, and developing test cases: actual instances of the
problem that can be used to test your solution. Ideally, the a test case instance involves
not only the “input” of the problem, but also the “output” of the problem: a feasible or
optimal solution that is known to be correct via other means. Test cases allow us to test
our solution to see if it gives correct and perhaps optimal solutions.
Refinement is a process by which we can redesign, reimplement and retest our solution.
We may want to make the solution more efficient, cheaper, simpler or more elegant. We
may find there are components that are redundant or unnecessary and try to eliminate
them. We may find errors or bugs in our solution that fail to solve the problem for some
or many instances. We may have misinterpreted requirements or there may have been",ComputerScienceOne.pdf
"or many instances. We may have misinterpreted requirements or there may have been
miscommunication, misunderstanding or differing expectations in the solution between
the designers and stakeholders. Situations may change or requirements may have been
modified or new requirements created and the solution needs to be adapted. Each of
these steps may need to be repeated many times until an ideal solution, or at least
acceptable, solution is achieved.
Yet another phase of problem solving is maintenance. The solution we create may need
to be maintained in order to remain functional and stay relevant. Design flaws or bugs
may become apparent that were missed in previous phases. The solution may need to be
updated to adapt to new technology or requirements.
In software design there are two general techniques for problem solving; top-down and
bottom-up design. A top-down design strategy approaches a problem by breaking it",ComputerScienceOne.pdf
"bottom-up design. A top-down design strategy approaches a problem by breaking it
down into smaller and smaller problems until either a solution is obvious or trivial or a
preexisting solution (the aforementioned “off-the-shelf” solution) exists. The solutions to
the subproblems are combined and interact to solve the overall problem.
A bottom-up strategy attempts to first completely define the smallest components or
3",ComputerScienceOne.pdf
"1. Introduction
entities that make up a system first. Once these have been defined and implemented,
they are combined and interactions between them are defined to produce a more complex
system.
1.2. Computing Basics
Everyone has some level of familiarity with computers and computing devices just as
everyone has familiarity with automotive basics. However, just because you drive a car
everyday doesn’t mean you can tell the difference between a crankshaft and a piston. To
get started, let’s familiarize ourselves with some basic concepts.
A computer is a device, usually electronic, that stores, receives, processes, and outputs
information. Modern computing devices include everything from simple sensors to mobile
devices, tablets, desktops, mainframes/servers, supercomputers and huge grid clusters
consisting of multiple computers networked together.
Computer hardware usually refers to the physical components in a computing system",ComputerScienceOne.pdf
"consisting of multiple computers networked together.
Computer hardware usually refers to the physical components in a computing system
which includes input devices such as a mouse/touchpad, keyboard, or touchscreen, output
devices such as monitors, storage devices such as hard disks and solid state drives, as
well as the electronic components such as graphics cards, main memory, motherboards
and chips that make up the Central Processing Unit (CPU).
Computer processors are complex electronic circuits (referred to as Very Large Scale Inte-
gration (VLSI)) which contain thousands of microscopic electronic transistors–electronic
“gates” that can perform logical operations and complex instructions. In addition to
the CPU a processor may contain an Arithmetic and Logic Unit (ALU) that performs
arithmetic operations such as addition, multiplication, division, etc.
Computer Software usually refers to the actual machine instructions that are run on a",ComputerScienceOne.pdf
"Computer Software usually refers to the actual machine instructions that are run on a
processor. Software is usually written in a high-level programming language such as C or
Java and then converted to machine code that the processor can execute.
Computers “speak” in binary code. Binary is nothing more than a structured collection
of 0s and 1s. A single 0 or 1 is referred to as a bit. Bits can be collected to form larger
chunks of information: 8 bits form a byte, 1024 bytes is referred to as a kilobyte, etc.
Table 1.1 contains a several more binary units. Each unit is in terms of a power of 2
instead of a power of 10. As humans, we are more familiar with decimal–base-10 numbers
and so units are usually expressed as powers of 10, kilo- refers to 10 3, mega- is 10 6, etc.
However, since binary is base-2 (0 or 1), units are associated with the closest power
of 2. Computers are binary machines because it is the most practical to implement in",ComputerScienceOne.pdf
"of 2. Computers are binary machines because it is the most practical to implement in
electronic devices. 0s and 1s can be easily represented by low/high voltage; low/high
frequency; on-off; etc. It is much easier to design and implement systems that switch
between only two states.
4",ComputerScienceOne.pdf
"1.3. Basic Program Structure
Unit 2n Number of bytes
Kilobyte (KB) 210 1,024
Megabyte (MB) 220 1,048,576
Gigabyte (GB) 230 1,073,741,824
Terabyte (TB) 240 1,099,511,627,776
Petabyte (PB) 250 1,125,899,906,842,624
Exabyte (EB) 260 1,152,921,504,606,846,976
Zettabyte (ZB) 270 1,180,591,620,717,411,303,424
Yottabyte (YB) 280 1,208,925,819,614,629,174,706,176
Table 1.1.: Various units of digital information with respect to bytes. Memory is usually
measured using powers of two.
Computer memory can refer to secondary memory which are typically longterm storage
devices such as hard disks, flash drives, SD cards, optical disks (CDs, DVDs), etc. These
generally have a large capacity but are slower (the time it takes to access a chunk of data
is longer). Or, it can refer to main memory (or primary memory): data stored on chips
that is much faster but also more expensive and thus generally smaller.
The first hard disk (IBM 350) was developed in 1956 by IBM and had a capacity of",ComputerScienceOne.pdf
"The first hard disk (IBM 350) was developed in 1956 by IBM and had a capacity of
3.75MB and cost $3,200 ($27,500 in 2015 dollars) per month to lease. For perspective,
the first commercially available TB hard drive was released in 2007. As of 2015, terabyte
hard disks can be commonly purchased for $50–$100.
Main memory, sometimes referred to as Random Access Memory (RAM) consists of
a collection of addresses along with contents. An address usually refers to a single
byte of memory (called byte-addressing). The content, that is the byte of data that is
stored at an address, can be anything. It can represent a number, a letter, etc. To the
computer it is all just a bunch of 0s and 1s. For convenience, memory addresses are
represented using hexadecimal, which is a base-16 counting system using the symbols
0,1,..., 9,a,b,c,d,e,f . Numbers are prefixed with a 0x to indicate they represent
hexadecimal numbers. Figure 1.1 depicts memory and its address/contents.",ComputerScienceOne.pdf
"hexadecimal numbers. Figure 1.1 depicts memory and its address/contents.
Separate computing devices can be connected to each other through a network. Networks
can be wired with electrical signals or light as in fiber optics which provide large bandwidth
(the amount of data that can be sent at any one time), but can be expensive to build and
maintain. They can also be wireless, but provide shorter range and lower bandwidth.
1.3. Basic Program Structure
Programs start out as source code, a collection of instructions usually written in a high-
level programming language. A source file containing source code is nothing more than
5",ComputerScienceOne.pdf
"1. Introduction
Address Contents
... ...
0x7fff58310b8f
0x7fff58310b8b 0x32
0x7fff58310b8a 0x3e
0x7fff58310b89 0xcf
0x7fff58310b88 0x23
0x7fff58310b87 0x01
0x7fff58310b86 0x32
0x7fff58310b85 0x7c
0x7fff58310b84 0xff
0x7fff58310b83
0x7fff58310b82
0x7fff58310b81
0x7fff58310b80
0x7fff58310b7f
0x7fff58310b7e
0x7fff58310b7d
0x7fff58310b7c 3.14159265359
0x7fff58310b7b
0x7fff58310b7a
0x7fff58310b79
0x7fff58310b78 32,321,231
0x7fff58310b77
0x7fff58310b76
0x7fff58310b75
0x7fff58310b74 1,458,321
0x7fff58310b73 \0
0x7fff58310b72 o
0x7fff58310b71 l
0x7fff58310b70 l
0x7fff58310b6f e
0x7fff58310b6e H
0x7fff58310b88 0xfa
0x7fff58310b87 0xa8
0x7fff58310b86 0xba
... ...
Figure 1.1.: Depiction of Computer Memory. Each address refers to a byte, but different
types of data (integers, floating-point numbers, characters) may require
different amounts of memory. Memory addresses and some data is represented
in hexadecimal.
6",ComputerScienceOne.pdf
"1.3. Basic Program Structure
a plain text file that can be edited by any text editor. However, many developers and
programmers utilize modern Integrated Development Environment (IDE) that provide a
text editor with code highlighting: various elements are displayed in different colors to
make the code more readable and elements can be easily identified. Mistakes such as
unclosed comments or curly brackets can be readily apparent with such editors. IDEs can
also provide automated compile/build features and other tools that make the development
process easier and faster.
Some languages are compiled languages meaning that a source file must be translated
into machine code that a processor can understand and execute. This is actually a
multistep process. A compiler may first preprocess the source file(s) and perform some
pre-compiler operations. It may then transform the source code into another language
such as an assembly language, a lower-level more machine-like language. Ultimately, the",ComputerScienceOne.pdf
"such as an assembly language, a lower-level more machine-like language. Ultimately, the
compiler transforms the source code into object code, a binary format that the machine
can understand.
To produce an executable file that can actually be run, a linker may then take the object
code and link in any other necessary objects or precompiled library code necessary to
produce a final program. Finally, an executable file (still just a bunch of binary code) is
produced.
Once an executable file has been produced we can run the program. When a program is
executed, a request is sent to the operating system to load and run the program. The
operating system loads the executable file into memory and may setup additional memory
for its variables as well as its call stack (memory to enable the program to make function
calls). Once loaded and setup, the operating system begins executing the instructions at
the program’s entry point.",ComputerScienceOne.pdf
"calls). Once loaded and setup, the operating system begins executing the instructions at
the program’s entry point.
In many languages, a program’s entry point is defined by a main function or method.
A program may contain many functions and pieces of code, but this special function is
defined as the one that gets invoked when a program starts. Without a main function,
the code may still be useful: libraries contain many useful functions and procedures so
that you don’t have to write a program from scratch. However, these functions are not
intended to be run by themselves. Instead, they are written so that other programs can
use them. A program becomes executable only when a main entry point is provided.
This compile-link-execute process is roughly depicted in Code Sample 1.2. An example of
a simple C program can be found in Code Sample 1.1 along with the resulting assembly
code produced by a compiler in Figure 1.2 and the final machine code represented in
hexadecimal in Code Sample 1.3.",ComputerScienceOne.pdf
"code produced by a compiler in Figure 1.2 and the final machine code represented in
hexadecimal in Code Sample 1.3.
In contrast, some languages are interpreted, not compiled. The source code is contained
in a file usually referred to as a script. Rather than being run directly by an operating
system, the operating system loads and execute another program called an interpreter.
The interpreter then loads the script, parses, and execute its instructions. Interpreted
7",ComputerScienceOne.pdf
"1. Introduction
Text Editor
or IDE Source File
CompilerSyntax
Error(s)
Object File Linker
Other Object
Files &
Libraries
Executable
File
Results &
Output
success
run
Input
Figure 1.2.: A Compiling Process
8",ComputerScienceOne.pdf
"1.3. Basic Program Structure
1 #include<stdlib.h>
2 #include<stdio.h>
3 #include<math.h>
4
5 int main(int argc, char **argv) {
6
7 if(argc != 2) {
8 fprintf(stderr, ""Usage: %s x\n"", argv[0]);
9 exit(1);
10 }
11
12 double x = atof(argv[1]);
13 double result = sqrt(x);
14
15 if(x < 0) {
16 fprintf(stderr, ""Cannot handle complex roots\n"");
17 exit(2);
18 }
19
20 printf(""square root of %f = %f\n"", x, result);
21
22 return 0;
23 }
Code Sample 1.1.: A simple program in C
9",ComputerScienceOne.pdf
"1. Introduction
.section __TEXT,__text,regular,pure_instructions
.globl _main
.align 4, 0x90
_main: ## @main
.cfi_startproc
## BB#0:
pushq %rbp
Ltmp2:
.cfi_def_cfa_offset 16
Ltmp3:
.cfi_offset %rbp, -16
movq %rsp, %rbp
Ltmp4:
.cfi_def_cfa_register %rbp
subq $48, %rsp
movl $0, -4(%rbp)
movl %edi, -8(%rbp)
movq %rsi, -16(%rbp)
cmpl $2, -8(%rbp)
je LBB0_2
## BB#1:
leaq L_.str(%rip), %rsi
movq ___stderrp@GOTPCREL(%rip), %rax
movq (%rax), %rdi
movq -16(%rbp), %rax
movq (%rax), %rdx
movb $0, %al
callq _fprintf
movl $1, %edi
movl %eax, -36(%rbp) ## 4-byte Spill
callq _exit
LBB0_2:
movq -16(%rbp), %rax
movq 8(%rax), %rdi
callq _atof
xorps %xmm1, %xmm1
movsd %xmm0, -24(%rbp)
movsd -24(%rbp), %xmm0
sqrtsd %xmm0, %xmm0
movsd %xmm0, -32(%rbp)
ucomisd -24(%rbp), %xmm1
jbe LBB0_4
## BB#3:
leaq L_.str1(%rip), %rsi
movq ___stderrp@GOTPCREL(%rip), %rax
movq (%rax), %rdi
movb $0, %al
callq _fprintf
movl $2, %edi
movl %eax, -40(%rbp) ## 4-byte Spill
callq _exit
LBB0_4:
leaq L_.str2(%rip), %rdi",ComputerScienceOne.pdf
"movq (%rax), %rdi
movb $0, %al
callq _fprintf
movl $2, %edi
movl %eax, -40(%rbp) ## 4-byte Spill
callq _exit
LBB0_4:
leaq L_.str2(%rip), %rdi
movsd -24(%rbp), %xmm0
movsd -32(%rbp), %xmm1
movb $2, %al
callq _printf
movl $0, %ecx
movl %eax, -44(%rbp) ## 4-byte Spill
movl %ecx, %eax
addq $48, %rsp
popq %rbp
retq
.cfi_endproc
.section __TEXT,__cstring,cstring_literals
L_.str: ## @.str
.asciz ""Usage: %s x\n""
L_.str1: ## @.str1
.asciz ""Cannot handle complex roots\n""
L_.str2: ## @.str2
.asciz ""square root of %f = %f\n""
.subsections_via_symbols
Code Sample 1.2.: A simple program in C, compiled to assembly
10",ComputerScienceOne.pdf
"1.3. Basic Program Structure
00000e40 55 48 89 e5 48 83 ec 30 c7 45 fc 00 00 00 00 89 |UH..H..0.E......|
00000e50 7d f8 48 89 75 f0 81 7d f8 02 00 00 00 0f 84 2c |}.H.u..}.......,|
00000e60 00 00 00 48 8d 35 f2 00 00 00 48 8b 05 9f 01 00 |...H.5....H.....|
00000e70 00 48 8b 38 48 8b 45 f0 48 8b 10 b0 00 e8 94 00 |.H.8H.E.H.......|
00000e80 00 00 bf 01 00 00 00 89 45 dc e8 81 00 00 00 48 |........E......H|
00000e90 8b 45 f0 48 8b 78 08 e8 6e 00 00 00 0f 57 c9 f2 |.E.H.x..n....W..|
00000ea0 0f 11 45 e8 f2 0f 10 45 e8 f2 0f 51 c0 f2 0f 11 |..E....E...Q....|
00000eb0 45 e0 66 0f 2e 4d e8 0f 86 25 00 00 00 48 8d 35 |E.f..M...%...H.5|
00000ec0 a5 00 00 00 48 8b 05 45 01 00 00 48 8b 38 b0 00 |....H..E...H.8..|
00000ed0 e8 41 00 00 00 bf 02 00 00 00 89 45 d8 e8 2e 00 |.A.........E....|
00000ee0 00 00 48 8d 3d 9d 00 00 00 f2 0f 10 45 e8 f2 0f |..H.=.......E...|
00000ef0 10 4d e0 b0 02 e8 22 00 00 00 b9 00 00 00 00 89 |.M...."".........|",ComputerScienceOne.pdf
"00000ef0 10 4d e0 b0 02 e8 22 00 00 00 b9 00 00 00 00 89 |.M...."".........|
00000f00 45 d4 89 c8 48 83 c4 30 5d c3 ff 25 08 01 00 00 |E...H..0]..%....|
00000f10 ff 25 0a 01 00 00 ff 25 0c 01 00 00 ff 25 0e 01 |.%.....%.....%..|
00000f20 00 00 00 00 4c 8d 1d dd 00 00 00 41 53 ff 25 cd |....L......AS.%.|
00000f30 00 00 00 90 68 00 00 00 00 e9 e6 ff ff ff 68 0c |....h.........h.|
00000f40 00 00 00 e9 dc ff ff ff 68 18 00 00 00 e9 d2 ff |........h.......|
00000f50 ff ff 68 27 00 00 00 e9 c8 ff ff ff 55 73 61 67 |..h'........Usag|
00000f60 65 3a 20 25 73 20 78 0a 00 43 61 6e 6e 6f 74 20 |e: %s x..Cannot |
00000f70 68 61 6e 64 6c 65 20 63 6f 6d 70 6c 65 78 20 72 |handle complex r|
00000f80 6f 6f 74 73 0a 00 73 71 75 61 72 65 20 72 6f 6f |oots..square roo|
00000f90 74 20 6f 66 20 25 66 20 3d 20 25 66 0a 00 00 00 |t of %f = %f....|
00000fa0 01 00 00 00 1c 00 00 00 00 00 00 00 1c 00 00 00 |................|
00000fb0 00 00 00 00 1c 00 00 00 02 00 00 00 40 0e 00 00 |............@...|",ComputerScienceOne.pdf
"00000fb0 00 00 00 00 1c 00 00 00 02 00 00 00 40 0e 00 00 |............@...|
00000fc0 34 00 00 00 34 00 00 00 0b 0f 00 00 00 00 00 00 |4...4...........|
00000fd0 34 00 00 00 03 00 00 00 0c 00 01 00 10 00 01 00 |4...............|
00000fe0 00 00 00 00 00 00 00 01 14 00 00 00 00 00 00 00 |................|
00000ff0 01 7a 52 00 01 78 10 01 10 0c 07 08 90 01 00 00 |.zR..x..........|
00001000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
00001010 00 00 00 00 00 00 00 00 34 0f 00 00 01 00 00 00 |........4.......|
00001020 3e 0f 00 00 01 00 00 00 48 0f 00 00 01 00 00 00 |>.......H.......|
00001030 52 0f 00 00 01 00 00 00 00 00 00 00 00 00 00 00 |R...............|
00001040 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
...(cut for room)...
00002000 11 22 18 54 00 00 00 00 11 40 5f 5f 5f 73 74 64 |."".T.....@___std|
00002010 65 72 72 70 00 51 72 10 90 40 64 79 6c 64 5f 73 |errp.Qr..@dyld_s|",ComputerScienceOne.pdf
"00002010 65 72 72 70 00 51 72 10 90 40 64 79 6c 64 5f 73 |errp.Qr..@dyld_s|
00002020 74 75 62 5f 62 69 6e 64 65 72 00 80 e8 ff ff ff |tub_binder......|
00002030 ff ff ff ff ff 01 90 00 72 18 11 40 5f 61 74 6f |........r..@_ato|
00002040 66 00 90 00 72 20 11 40 5f 65 78 69 74 00 90 00 |f...r .@_exit...|
00002050 72 28 11 40 5f 66 70 72 69 6e 74 66 00 90 00 72 |r(.@_fprintf...r|
00002060 30 11 40 5f 70 72 69 6e 74 66 00 90 00 00 00 00 |0.@_printf......|
00002070 00 01 5f 00 05 00 02 5f 6d 68 5f 65 78 65 63 75 |.._...._mh_execu|
00002080 74 65 5f 68 65 61 64 65 72 00 21 6d 61 69 6e 00 |te_header.!main.|
00002090 25 02 00 00 00 03 00 c0 1c 00 00 00 00 00 00 00 |%...............|
000020a0 c0 1c 00 00 00 00 00 00 fa de 0c 05 00 00 00 14 |................|
000020b0 00 00 00 01 00 00 00 00 00 00 00 00 00 00 00 00 |................|
000020c0 02 00 00 00 0f 01 10 00 00 00 00 00 01 00 00 00 |................|
000020d0 16 00 00 00 0f 01 00 00 40 0e 00 00 01 00 00 00 |........@.......|",ComputerScienceOne.pdf
"000020d0 16 00 00 00 0f 01 00 00 40 0e 00 00 01 00 00 00 |........@.......|
000020e0 1c 00 00 00 01 00 00 01 00 00 00 00 00 00 00 00 |................|
000020f0 27 00 00 00 01 00 00 01 00 00 00 00 00 00 00 00 |'...............|
00002100 2d 00 00 00 01 00 00 01 00 00 00 00 00 00 00 00 |-...............|
00002110 33 00 00 00 01 00 00 01 00 00 00 00 00 00 00 00 |3...............|
00002120 3c 00 00 00 01 00 00 01 00 00 00 00 00 00 00 00 |<...............|
00002130 44 00 00 00 01 00 00 01 00 00 00 00 00 00 00 00 |D...............|
00002140 03 00 00 00 04 00 00 00 05 00 00 00 06 00 00 00 |................|
00002150 07 00 00 00 00 00 00 40 02 00 00 00 03 00 00 00 |.......@........|
00002160 04 00 00 00 05 00 00 00 06 00 00 00 20 00 5f 5f |............ .__|
00002170 6d 68 5f 65 78 65 63 75 74 65 5f 68 65 61 64 65 |mh_execute_heade|
00002180 72 00 5f 6d 61 69 6e 00 5f 5f 5f 73 74 64 65 72 |r._main.___stder|
00002190 72 70 00 5f 61 74 6f 66 00 5f 65 78 69 74 00 5f |rp._atof._exit._|",ComputerScienceOne.pdf
"00002190 72 70 00 5f 61 74 6f 66 00 5f 65 78 69 74 00 5f |rp._atof._exit._|
000021a0 66 70 72 69 6e 74 66 00 5f 70 72 69 6e 74 66 00 |fprintf._printf.|
000021b0 64 79 6c 64 5f 73 74 75 62 5f 62 69 6e 64 65 72 |dyld_stub_binder|
000021c0 00 00 00 00 |....|
000021c4
Code Sample 1.3.: A simple program in C, resulting machine code formatted in hexadec-
imal (partial)
11",ComputerScienceOne.pdf
"1. Introduction
languages may still have a predefined main function, but in general, a script starts
executing starting with the first instruction in the script file. Adhering to the syntax
rules is still important, but since interpreted languages are not compiled, syntax errors
become runtime errors. A program may run fine until its first syntax error at which
point it fails.
There are other ways of compiling and running programs. Java for example represents a
compromise between compiled and interpreted languages. Java source code is compiled
into Java bytecode which is not actually machine code that the operating system and
hardware can run directly. Instead, it is compiled code for a Java Virtual Machine (JVM).
This allows a developer to write highly portable code, compile it once and it is runnable
on any JVM on any system (write-once, compile-once, run-anywhere).
In general, interpreted languages are slower than compiled languages because they are",ComputerScienceOne.pdf
"In general, interpreted languages are slower than compiled languages because they are
being run through another program (the interpreter) instead of being executed directly
by the processor. Modern tools have been introduced to solve this problem. Just In Time
(JIT) compilers have been developed that take scripts that are not usually compiled,
and compile them to a native machine code format which has the potential to run much
faster than when interpreted. Modern web browsers typically do this for JavaScript code
(Google Chrome’s V8 JavaScript engine for example).
Another related technology are transpilers. Transpilers are source-to-source compilers.
They don’t produce assembly or machine code, instead they translate code in one
high-level programming language to another high-level programming language. This
is sometimes done to ensure that scripting languages like JavaScript are backwards
compatible with previous versions of the language. Transpilers can also be used to",ComputerScienceOne.pdf
"compatible with previous versions of the language. Transpilers can also be used to
translate one language into the same language but with different aspects (such as parallel
or synchronized code) automatically added. They can also be used to translate older
languages such as Pascal to more modern languages as a first step in updating a legacy
system.
1.4. Syntax Rules & Pseudocode
Programming languages are a lot like human languages in that they have syntax rules.
These rules dictate the appropriate arrangements of words, punctuation, and other
symbols that form valid statements in the language. For example, in many programming
languages, commands or statements are terminated by semicolons (just as most sentences
are ended with a period). This is an example of “punctuation” in a programming language.
In English paragraphs are separated by lines, in programming languages blocks of code
are separated by curly brackets. Variables are comparable to nouns and operations and",ComputerScienceOne.pdf
"are separated by curly brackets. Variables are comparable to nouns and operations and
functions are comparable to verbs. Complex documents often have footnotes that provide
additional explanations; code has comments that provide documentation and explanation
for important elements. English is read top-to-bottom, left-to-right. Programming
12",ComputerScienceOne.pdf
"1.4. Syntax Rules & Pseudocode
languages are similar: individual executable commands are written one per line. When a
program executes, each command executes one after the other, top-to-bottom. This is
known as sequential control flow.
A block of code is a section of code that has been logically grouped together. Many
languages allow you to define a block by enclosing the grouped code around opening and
closing curly brackets. Blocks can be nested within each other to form sub-blocks.
Most languages also have reserved words and symbols that have special meaning. For
example, many languages assign special meaning to keywords such as for, if, while,
etc. that are used to define various control structures such as conditionals and loops.
Special symbols include operators such as + and * for performing basic arithmetic.
Failure to adhere to the syntax rules of a particular language will lead to bugs and
programs that fail to compile and/or run. Natural languages such as English are very",ComputerScienceOne.pdf
"programs that fail to compile and/or run. Natural languages such as English are very
forgiving: we can generally discern what someone is trying to say even if they speak
in broken English (to a point). However, a compiler or interpreter isn’t as smart as a
human. Even a small syntax error (an error in the source code that does not conform
to the language’s rules) will cause a compiler to completely fail to understand the code
you have written. Learning a programming language is a lot like learning a new spoken
language (but, fortunately a lot easier).
In subsequent parts of this book we focus on particular languages. However, in order
to focus on concepts, we’ll avoid specific syntax rules by using pseudocode, informal,
high-level descriptions of algorithms and processes. Good pseudocode makes use of plain
English and mathematical notation, making it more readable and abstract. A small
example can be found in Algorithm 1.1.
Input : A collection of numbers, A= {a1,a2,...,a n}",ComputerScienceOne.pdf
"example can be found in Algorithm 1.1.
Input : A collection of numbers, A= {a1,a2,...,a n}
Output :The minimal element in A
1 Let min be equal to a1
2 foreach element ai in A do
3 if ai <min then
4 ai is less than the smallest element we’ve found so far
5 Update min to be equal to ai
6 end
7 end
8 output min
Algorithm 1.1:An example of pseudocode: finding a minimum value
13",ComputerScienceOne.pdf
"1. Introduction
1.5. Documentation, Comments, and Coding Style
Good code is not just functional, it is also beautiful. Good code is organized, easy to
read, and well documented. Organization can be achieved by separating code into useful
functions and collecting functions into modules or libraries. Good organization means
that at any one time, we only need to focus on a small part of a program.
It would be difficult to read an essay that contained random line breaks, paragraphs
were not indented, it contained different spacing or different fonts, etc. Likewise, code
should be legible. Well written code is consistent and makes good use of whitespace
and indentation. Code within the same code block should be indented at the same level.
Nested blocks should be further indented just like the outline of an essay or table of
contents.
Code should also be well documented. Each line or segment of code should be clear",ComputerScienceOne.pdf
"contents.
Code should also be well documented. Each line or segment of code should be clear
enough that it tells the user what the code does and how it does it. This is referred to as
“self-documenting” code. A person familiar with the particular language should be able
to read your code and immediately understand what it does. In addition, well-written
code should contain sufficient and clear comments. A comment in a program is intended
for a human user to read. A comment is ultimately ignored by the compiler/interpreter
and has no effect on the actual program. Good comments tell the user why the code was
written or why it was written the way it was. Comments provide a high-level description
of what a block of code, function, or program does. If the particular method or algorithm
is of interest, it should also be documented.
There are typically two ways to write comments. Single line comments usually begin",ComputerScienceOne.pdf
"is of interest, it should also be documented.
There are typically two ways to write comments. Single line comments usually begin
with two forward slashes, //.1 Everything after the slashes until the next line is ignored.
Multiline comments begin with a /* and end with a */; everything between them is
ignored even if it spans multiple lines. This syntax is shared among many languages
including C, Java, PHP and others. Some examples:
1 double x = sqrt(y); //this is a single line comment
2
3 /*
4 This is a multiline comment
5 each line is ignored, but allows
6 for better formatting
7 */
8
9 /**
10 * This is a doc-style comment, usually placed in
1You can remember the difference between a forward slash / and a backslash \ by thinking of a
person facing right and either leaning backwards (backslash) or forwards (forward slash).
14",ComputerScienceOne.pdf
"1.5. Documentation, Comments, and Coding Style
11 * front of major portions of code such as a function
12 * to provide documentation
13 * It begins with a forward-slash-star-star
14 */
The last example above is a doc-style comment. It originated with Java, but has since
been adopted by many other programming languages. Syntactically it is a normal
multiline comment, but begins with a /**. Asterisks are aligned together on each
line. Certain commenting systems allow you to place other marked up data inside these
comments such as labeling parameters ( @param x) or use HTML code to provide links
and style. These doc-style comments are used to provide documentation for major parts
of the code especially functions and data structures. Though not part of the language,
other documentation tools can be used to gather the information in doc-style comments
to produce formatted documentation such as web pages or Portable Document Format
(PDF) documents.",ComputerScienceOne.pdf
"to produce formatted documentation such as web pages or Portable Document Format
(PDF) documents.
Comments should not be trivial: they should not explain something that should be
readily apparent to an experienced user or programmer. For example, if a piece of code
adds two numbers together and stores the result, there should not be a comment that
explains the process. It is a simple and common enough operation that is self-evident.
However, if a function uses a particular process or algorithm such as a Fourier Transform
to perform an operation, it would be appropriate to document it in a series of comments.
Comments can also detail how a function or piece of code should be used. This is
typically done when developing an Application Programmer Interface (API) for use
by other programmers. The API’s available functions should be well documented so
that users will know how and when to use a particular function. It can document the",ComputerScienceOne.pdf
"that users will know how and when to use a particular function. It can document the
function’s expectations and behavior such as how it handles bad input or error situations.
15",ComputerScienceOne.pdf
"2. Basics
2.1. Control Flow
The flow of control (or simply control flow) is how a program processes its instructions.
Typically, programs operate in a linear orsequential flow of control. Executable statements
or instructions in a program are performed one after another. In source code, the order
that instructions are written defines their order. Just like English, a program is “read”
top to bottom. Each statement may modify the state of a program. The state of a
program is the value of all its variables and other information/data stored in memory
at a given moment during its execution. Further, an executable statement may instead
invoke (or call or execute) another procedure (also called subroutine, function, method,
etc.) which is another unit of code that has been encapsulated into one unit so that it
can be reused.
This type of control flow is usually associated with a procedural programming paradigm
(which is closely related to imperative or structured programming paradigms). Though",ComputerScienceOne.pdf
"(which is closely related to imperative or structured programming paradigms). Though
this text will mostly focus on languages that are procedural (or that have strong procedural
aspects), it is important to understand that there are other programming language
paradigms. Functional programming languages such as Scheme and Haskell achieve
computation through the evaluation of mathematical functions with as little or no (“pure”
functional) state at all. Declarative languages such as those used in database languages
like SQL or in spreadsheets like Excel specify computation by expressing the logic of
computation rather than explicitly specifying control flow. For a more formal introduction
to programming language paradigms, a good resource is Seven Languages in Seven Weeks:
A Pragmatic Guide to Learning Programming Languages by Tate [36].
2.1.1. Flowcharts
Sometimes processes are described using diagrams called flowcharts. A flowchart is a",ComputerScienceOne.pdf
"2.1.1. Flowcharts
Sometimes processes are described using diagrams called flowcharts. A flowchart is a
visual representation of an algorithm or process consisting of boxes or “nodes” connected
by directed edges. Boxes can represent an individual step or a decision to be made. The
edges establish an order of operations in the diagram.
Some boxes represent decisions to be made which may have one or more alternate routes
(more than one directed edge going out of the box) depending on the the result of the
17",ComputerScienceOne.pdf
"2. Basics
Decision
Node
(a) Decision Node
Control to
Perform
(b) Control Node
Action to
Perform
(c) Action Node
Figure 2.1.: Types of Flowchart Nodes. Control and action nodes are distinguished
by color. Control nodes are automated steps while action nodes are steps
performed as part of the algorithm being depicted.
decision. Decision boxes are usually depicted with a diamond shaped box.
Other boxes represent a process, operation, or action to be performed. Boxes representing
a process are usually rectangles. We will further distinguish two types of processes using
two different colorings: we’ll use green to represent boxes that are steps directly related
to the algorithm being depicted. We’ll use blue for actions that are necessary to the
control flow of the algorithm such as assigning a value to a variable or incrementing a
value as part of a loop. Figure 2.1 depicts the three types of boxes we’ll use. Figure 2.2
depicts a simple ATM (Automated Teller Machine) process as an example.",ComputerScienceOne.pdf
"depicts a simple ATM (Automated Teller Machine) process as an example.
2.2. Variables
In mathematics, variables are used as placeholders for values that aren’t necessarily
known. For example, in the equation,
x= 3y+ 5
the variables x and y represent numbers that can take on a number of different values.
Similarly, in a computer program, we also use variables to store values. A variable is
essentially a memory location in which a value can be stored. Typically, a variable is
referred to by a name or identifier (like x,y,z in mathematics). In mathematics variables
are usually used to hold numerical values. However, in programming, variables can
usually hold different types of values such as numbers, strings (a collection of characters),
Booleans (true or false values), or more complex types such as arrays or objects.
18",ComputerScienceOne.pdf
"2.2. Variables
Input PIN Is PINcorrect?
Eject Card
Getamount ofwithdraw
User Input
SufficientFunds? Dispenseamount
no
yes
amount
no
yes
Figure 2.2.: Example of a flowchart for a simple ATM process
2.2.1. Naming Rules & Conventions
Most programming languages have very specific rules as to what you can use as variable
identifiers (names). For example, most programming languages do not allow you to use
whitespace characters (space, tab, etc.) in a variable’s identifier. Allowing spaces would
make variable names ambiguous: where does the variable’s name end and the rest of the
program continue? How could you tell the difference between “average score” and two
separate variables named “average” and “score”? Many programming languages also
have reserved words–words or terms that are used by the programming language itself
and have special meaning. Variable names cannot be the same as any reserved word as
the language wouldn’t be able to distinguish between them.",ComputerScienceOne.pdf
"and have special meaning. Variable names cannot be the same as any reserved word as
the language wouldn’t be able to distinguish between them.
For similar reasons, many programming languages do not allow you to start a variable
name with a number as it would make it more difficult for a compiler or interpreter to
parse a program’s source code. Yet other languages require that variables begin with a
specific character (PHP for example requires that all variables begin with a dollar sign,
$).
In general, most programming languages allow you to use a combination of uppercase
A-Z and lowercase a-z letters as well as numbers, [0-9] and certain special characters
such as underscores _ or dollar signs, $. Moreover, most programming languages (like
English) are case sensitive meaning that a variable name using lowercase letters is not
the same variable as one that uses uppercase letters. For example, the variables x and",ComputerScienceOne.pdf
"the same variable as one that uses uppercase letters. For example, the variables x and
X are different; the variables average, Average and AVERAGE are all different as well.
A few languages are case-insensitive meaning that they do not recognize differences in
lower and uppercase letters when used in variable identifiers. Even in these languages,
19",ComputerScienceOne.pdf
"2. Basics
however, using a mixture of lowercase and uppercase letters to refer to the same variable
is discouraged: it is difficult to read, inconsistent, and just plain ugly.
Beyond the naming rules that languages may enforce, most languages have established
naming conventions; a set of guidelines and best-practices for choosing identifier names
for variables (as well as functions, methods, and class names). Conventions may be
widely adopted on a per-language basis or may be established within a certain library,
framework or by an organization that may have official style guides. Naming conventions
are intended to give source code consistency which ultimately improves readability and
makes it easier to understand. Following a consistent convention can also greatly reduce
the chance for errors and mistakes. Good naming conventions also has an aesthetic
appeal; code should be beautiful.
There are several general conventions when it comes to variables. An early convention,",ComputerScienceOne.pdf
"appeal; code should be beautiful.
There are several general conventions when it comes to variables. An early convention,
but still in common use is underscore casing in which variable names consisting of more
than one word have words separated by underscore characters with all other characters
being lowercase. For example:
average_score, number_of_students, miles_per_hour
A variation on this convention is to use all uppercase letters such as MILES_PER_HOUR.
A more modern convention is to use lower camel casing (or just camel casing) in which
variable names with multiple words are written as one long word with the first letter in
each new word capitalized but with the first word’s first letter lowercase. For example:
averageScore, numberOfStudents, milesPerHour
The convention refers to the capitalized letters resembling the humps of a camel. One
advantage that camel casing has over underscore casing is that you’re not always straining",ComputerScienceOne.pdf
"advantage that camel casing has over underscore casing is that you’re not always straining
to type the underscore character. Yet another similar convention is upper camel casing,
also known as PascalCase1 which is like camel casing, but the first letter in the first word
is also capitalized:
AverageScore, NumberOfStudents, MilesPerHour
Each of these conventions is used in various languages in different contexts which we’ll
explore more fully in subsequent sections (usually underscore lowercasing and camel
casing are used to denote variables and functions, PascalCase is used to denote user
defined types such as classes or structures, and underscore uppercasing is used to denote
static and constant variables). However, for our purposes, we’ll use lower camel casing
for variables in our pseudocode.
1Rarely, this is referred to as DromedaryCase; a Dromedary is an Arabian camel.
20",ComputerScienceOne.pdf
"2.2. Variables
There are exceptions and special cases to each of these conventions such as when a variable
name involves an acronym or a hyphenated word, etc. In such cases sensible extensions or
compromises are employed. For example, xmlString or priorityXMLParser (involving
the acronym Extensible Markup Language (XML)) may be used which keep all letters in
the acronym consistent (all lowercase or all uppercase).
In addition to these conventions, there are several best-practice principles when deciding
on identifiers.
•Be descriptive, but not verbose – Use variable names that describe what the
variable represents. The examples above, averageScore, numberOfStudents,
milesPerHour clearly indicate what the variable is intended to represent. Using
good, descriptive names makes it your code self-documenting (a reader can make
sense of it without having to read extensive supplemental documentation).
Avoid meaningless variable names such as value, aVariable, or some cryptic",ComputerScienceOne.pdf
"sense of it without having to read extensive supplemental documentation).
Avoid meaningless variable names such as value, aVariable, or some cryptic
combination of v10 (its the 10th variable I’ve used!). Ambiguous variables such
as name should also be avoided unless the context makes its clear what you are
referring to (as when used inside of a Person object).
Single character variables are commonly used, but used in a context in which their
meaning is clearly understood. For example, variable names such as x, y are okay
if they are used to refer to points in the Euclidean plane. Single character variables
such as i, j are often used as index variables when iterating over arrays. In this
case, terseness is valued over descriptiveness as the context is very well understood.
As a general rule, the more a variable is used, the shorter it should be. For
example, the variable numStudents may be preferred over the full variable
numberOfStudents.",ComputerScienceOne.pdf
"example, the variable numStudents may be preferred over the full variable
numberOfStudents.
•Avoid abbreviations (or at least use them sparingly) – You’re not being charged by
the character in your code; you can afford to write out full words. Abbreviations can
help to write shorter variable names, but not all abbreviations are the same. The
word “abbreviation” itself could be abbreviated as “abbr.”, “abbrv.” or “abbrev.”
for example. Abbreviations are not always universally understood by all users,
may be ambiguous and non-standard. Moreover, modern IDEs provide automatic
code completion, relieving you of the need to type longer variable names. If the
abbreviation is well-known or understood from context, then it may make sense to
use it.
•Avoid acronyms (or at least use them sparingly) – Using acronyms in variable
names come with many of the same problems as abbreviations. However, if it makes
sense in the context of your code and has little chance of being misunderstood or",ComputerScienceOne.pdf
"sense in the context of your code and has little chance of being misunderstood or
mistaken, then go for it. For example, in the context of a financial application,
APR (Annual Percentage Rate) would be a well-understood acronym in which case
21",ComputerScienceOne.pdf
"2. Basics
the variable apr may be preferred over the longer annualPercentageRate.
•Avoid pluralizations, use singular forms – English is not a very consistent language
when it comes to rules like pluralizations. For most cases you simply add “s”; for
others you add “es” or change the “y” to “i” and add “es”. Some words are the
same form for singular and plural such as “glasses.” 2 Other words have completely
different forms (“focus” becomes “foci”). Still yet there are instances in which
multiple words are acceptable: the plural of “person” can be “persons” or “people”.
Avoiding plural forms keeps things simple and consistent: you don’t need to be
a grammarian in order easily read code. One potential exception to this is when
using a collection such as an array to hold more than one element or the variable
represents a quantity that is pluralized (as with numberOfStudents above).
Though the guidelines above provide a good framework from which to write good variable",ComputerScienceOne.pdf
"Though the guidelines above provide a good framework from which to write good variable
names, reasonable people can and do disagree on best practice because at some point as
you go from generalities to specifics, conventions become more of a matter of personal
preference and subjective aesthetics. Sometimes an organization may establish its own
coding standards or style guide that must be followed which of course trumps any of the
guidelines above.
In the end, a good balance must be struck between readability and consistency. Rules
and conventions should be followed, until they get in the way of good code that is.
2.2.2. Types
A variable’s type (or data type ) is the characterization of the data that it represents. As
mentioned before, a computer only “speaks” in 0s and 1s (binary). A variable is merely
a memory location in which a series of 0s and 1s is stored. That binary string could
represent a number (either an integer or a floating point number), a single alphanumeric",ComputerScienceOne.pdf
"represent a number (either an integer or a floating point number), a single alphanumeric
character or series of characters (string), a Boolean type or some other, more complex
user-defined type.
The type of a variable is important because it affects how the raw binary data stored
at a memory location is interpreted. Moreover, some types take a different amount of
memory to store. For example, an integer type could take 32 bits while a floating point
type could take 64 bits. Programming languages may support different types and may
do so in different ways. In the next few sections we’ll describe some common types that
are supported by many languages.
2These are called plurale tantum (nouns with no singular form) and singular tantum (nouns with no
plural form) for you grammarians. Words like “sheep” are unchanging irregular plurals; words whose
singular and plural forms are the same.
22",ComputerScienceOne.pdf
"2.2. Variables
Numeric Types
At their most basic, computers are number crunching machines. Thus, the most basic
type of variable that can be used in a computer program is a numeric type. There are
several numeric types that are supported by various programming languages. The most
simple is an integer type which can represent whole numbers 0, 1, 2, etc. and their
negations, −1,−2,... . Floating point numeric types represent decimal numbers such
as 0.5,3.14,4.0, etc. However, neither integer nor floating point numbers can represent
every possible number since they use a finite number of bits to represent the number. We
will examine this in detail below. For now, let’s understand how a computer represents
both integers and floating point numbers in memory.
As humans, we “think” in base-10 (decimal) because we have 10 fingers and 10 toes.
When we write a number with multiple digits in base-10 we do so using “places” (ones",ComputerScienceOne.pdf
"When we write a number with multiple digits in base-10 we do so using “places” (ones
place, tens place, hundreds place, etc.). Mathematically, a number in base-10 can be
broken down into powers of ten; for example:
3,201 = 3 ×103 + 2 ×102 + 0 ×101 + 1 ×100
In general, any number in base-10 can be written as the summation of powers of 10
multiplied by numbers 0–9,
ck ×10k + ck−1 ×10k−1 + ···c1 ·101 + c0
In binary, numbers are represented in the same way, but in base-2 in which we only have
0 and 1 as symbols. To illustrate, let’s consider counting from 0: in base-10, we would
count 0,1,2,..., 9 at which point we “carry-over” a 1 to the tens spot and start over at
0 in the ones spot, giving us 10 ,11,12,..., 19 and repeat the carry-over to 20.
With only two symbols, the carry-over occurs much more frequently, we count 0 ,1 and
then carry over and have 10. It is important to understand, this is not “ten”: we are",ComputerScienceOne.pdf
"then carry over and have 10. It is important to understand, this is not “ten”: we are
counting in base-2, so 10 is actually equivalent to 2 in base-10. Continuing, we have 11
and again carry over, but we carry it over twice giving us 100 (just like we’d carry over
twice when going from 99 to 100 in base-10). A full count from 0 to 16 in binary can be
found in Table 2.1. In many programming languages, a prefix of 0b is used to denote a
number represented in binary. We use this convention in the table.
As a fuller example, consider again the number 3,201. This can be represented in binary
23",ComputerScienceOne.pdf
"2. Basics
as follows.
0b110010000001 = 1×211 + 1 ×210 + 0 ×29 + 0 ×28+
1 ×27 + 0 ×26 + 0 ×25 + 0 ×24+
0 ×23 + 0 ×22 + 0 ×21 + 1 ×20
= 211 + 210 + 27 + 20
= 2,048 + 1,024 + 128 + 1
= 3,201
Base-10 Binary
0 0b0
1 0b1
2 0b10
3 0b11
4 0b100
5 0b101
6 0b110
7 0b111
8 0b1000
9 0b1001
10 0b1010
11 0b1011
12 0b1100
13 0b1101
14 0b1110
15 0b1111
16 0b10000
Table 2.1.: Counting in Binary
Representing negative numbers is a bit more com-
plicated and is usually done using a scheme called
two’s complement. We omit the details, but essen-
tially the first bit in the representation serves as a
sign bit : zero indicates positive, while 1 indicates
negative. Negative values are represented as a com-
plement with respect to 2 n (a complement is where
0s and 1s are “flipped” to 1s and 0s).
When represented using two’s complement, binary
numbers with nbits can represent numbers xin the
range
−2n−1 ≤x≤2n−1 −1
Note that the upper bound follows from the fact
that
0b11 ... 011
  
n bits
=
n−2∑
i=0",ComputerScienceOne.pdf
"range
−2n−1 ≤x≤2n−1 −1
Note that the upper bound follows from the fact
that
0b11 ... 011
  
n bits
=
n−2∑
i=0
2i = 2n−1 −1
The −1 captures the idea that we start at zero. The
exponent in the upper bound is n−1 since we need
one bit to represent the sign. The lower bound
represents the idea that we have 2 n−1 possible values (n−1 since we need one bit for
the sign bit) and we don’t need to start at zero, we can start at −1. Table 2.2 contains
ranges for common integer types using various number of bits.
n (number of bits) minimum maximum
8 -128 127
16 -32,768 32,767
32 -2,147,483,648 2,147,483,647
64 -9,223,372,036,854,775,808 9,223,372,036,854,775,807
128 ≈−3.4028 ×1038 ≈3.4028 ×1038
Table 2.2.: Ranges for various signed integer types
24",ComputerScienceOne.pdf
"2.2. Variables
Some programming languages allow you to define variables that are unsigned in which
the sign bit is not used to indicate positive/negative. With the extra bit we can represent
numbers twice as big; using n bits we can represent numbers x in the range
0 ≤x≤2n −1
Floating point numbers in binary are represented in a manner similar to scientific notation.
Recall that in scientific notation, a number is normalized by multiplying it by some power
of 10 so that it its most significant digit is between 1 and 9. The resulting normalized
number is called the significand while the power of ten that the number was scaled by is
called the exponent (and since we are base-10, 10 is the base). In general, a number in
scientific notation is represented as:
significand ×baseexponent
For example,
14326.123 = 1.4326123
  
significand
× 10

base
exponent

4
Sometimes the notations 1.4326123e+4, 1.4326123e4 or 1.4326123E4 are used. As before,",ComputerScienceOne.pdf
"  
significand
× 10

base
exponent

4
Sometimes the notations 1.4326123e+4, 1.4326123e4 or 1.4326123E4 are used. As before,
we can see that a fractional number in base-10 can be seen as a summation of powers of
10:
1.4326123 = 1×101 + 4 ×10−1 + 3 ×10−2 + 2 ×10−3+
6 ×10−4 + 1 ×10−5 + 2 ×10−6 + 3 ×10−7
In binary, floating point numbers are represented in a similar way, but the base is 2,
consequently a fractional number in binary is a summation of powers of 2. For example,
110.011 =1 ×22 + 1 ×21 + 0 ×20 + 0 ×2−1 + 1 ×2−2 + 1 ×2−3
=1 ×4 + 1×2 + 0×1 + 0×1
2 + 1 ×1
4 + 1 ×1
8
=4 + 2 + 0 + 0 +1
4 + 1
8
=6.375
In binary, the significand is often referred to as a mantissa. We also normalize a binary
floating point number so that the mantissa is between 1
2 and 1. This is where the term
floating point comes from: the decimal point (more generally called a radix point) “floats”
left and right to ensure that the number is always normalized. The example above would
normalized to",ComputerScienceOne.pdf
"left and right to ensure that the number is always normalized. The example above would
normalized to
0.110011 ×23
Here, 0.110011 is the mantissa and 3 is the exponent (which in binary would be 0b11).
25",ComputerScienceOne.pdf
"2. Basics
Name Bits Exponent Bits Mantissa Bits Significant Digits
of Precision
Approximate
Range
Half 16 5 10 ≈3.3 103 ∼104.5
Single 32 8 23 ≈7.2 10−38 ∼1038
Double 64 11 52 ≈15.9 10−308 ∼10308
Quadruple 128 15 112 ≈34.0 10−4931 ∼104931
Table 2.3.: Summary of Floating-point Precisions in the IEEE 754 Standard. Half and
quadruple are not widely adopted.
Most modern programming languages implement floating point numbers according to the
Institute of Electrical and Electronics Engineers (IEEE) 754 Standard [ 20] (also called
the International Electrotechnical Commission (IEC) 60559 [ 19]). When represented
in binary, a fixed number of bits must be used to represent the sign, mantissa and
exponent. The standard defines several precisions that each use a fixed number of bits
with a resulting number of significant digits (base-10) of precision. Table 2.3 contains a
summary of a few of the most commonly implemented precisions.",ComputerScienceOne.pdf
"summary of a few of the most commonly implemented precisions.
Just as with integers, the finite precision of floating point numbers results in several limi-
tations. First, irrational numbers such as π= 3.14159 ... can only be approximated out
to a certain number of digits. For example, with single precision π≈3.1415927 which is
accurate only to the 6th decimal place and with double precision,π≈3.1415926535897931
approximate to only 15 decimal places. 3 In fact, regardless of how many bits we allow in
our representation, an irrational number like π (that never repeats and never terminates)
will only ever be an approximation. Real numbers like π require an infinite precision,
but computers are only finite machines.
Even numbers that have a finite representation (rational numbers) such as 1
3 = 0.333 are
not represented exactly when using floating point numbers. In double precision binary,
1
3 = 0b1.0101010101010101010101010101010101010101010101010101×2−2",ComputerScienceOne.pdf
"1
3 = 0b1.0101010101010101010101010101010101010101010101010101×2−2
which when represented in scientific notation in decimal is
3.3333333333333330 ×10−1
That is, there are only 16 digits of precision, after which the remaining (infinite) sequence
of 3s get cut off.
Programming languages usually only support the common single and double precisions
defined by the IEEE 754 standard as those are commonly supported by hardware.
3The first 80 digits of π are
3.14159265358979323846264338327950288419716939937510582097494459230781640628620899
though only 39 digits of π are required to accurately calculate the volume of the known universe to
within one atom.
26",ComputerScienceOne.pdf
"2.2. Variables
However, there are languages that support arbitrary precision (also called multiprecision)
numbers and yet other languages that have many libraries to support “big number”
arithmetic. Arbitrary precision is still not infinite: instead, as more digits are needed,
more memory is allocated. If you want to compute 10 more digits of π, you can but at a
cost. To support the additional digits, more memory is allocated. Also, operations are
performed in software using many operations which can be much slower than performing
fixed-precision arithmetic directly in hardware. Still, there are many applications where
such accuracy or large numbers are absolutely essential.
Characters & Strings
Another type of data is textual data which can either be single characters or a sequence
of characters which are called strings. Strings are sometimes used for human readable
data such as messages or output, but may also model general data. For example, DNA",ComputerScienceOne.pdf
"data such as messages or output, but may also model general data. For example, DNA
is usually encoded using strings consisting of the characters C, G, A, T (corresponding
to the nucleases cytosine, guanine, adenine, and thymine). Numerical characters and
punctuation can also be used in strings in which case they do not represent numbers,
but instead may represent textual versions of numerical data.
Different programming languages implement characters and strings in different ways (or
may even treat them the same). Some languages implement strings by defining arrays
of characters. Other languages may treat strings as dynamic data types. However, all
languages use some form of character encoding to represent strings. Recall that computers
only speak in binary: 0s and 1s. To represent a character like the capital letter “A”, the
binary sequence 0b1000001 is used. In fact, the most common alphanumeric characters
are encoded according to the American Standard Code for Information Interchange",ComputerScienceOne.pdf
"are encoded according to the American Standard Code for Information Interchange
(ASCII) text standard. The basic ASCII text standard assigns characters to the decimal
values 0–127 using 7 bits to encode each character as a number. Table 2.4 contains a
complete listing of standard ASCII character set.
The ASCII table was designed to enforce a lexicographic ordering: letters are in alphabetic
order, uppercase precede lowercase versions, and numbers precede both. This design
allows for an easy and natural comparison among strings, “alpha” would come before
“beta” because they differ in the first letter. The characters have numerical values 97
and 98 respectively; since 97 < 98, the order follows. Likewise, “Alpha” would come
before “alpha” (since 65 < 97), and “alpha” would come before “alphanumeric”: the
sixth character is empty in the first string (usually treated as the null character with
value 0) while it is “n” in the second (value of 110). This is the ordering that we would",ComputerScienceOne.pdf
"value 0) while it is “n” in the second (value of 110). This is the ordering that we would
expect in a dictionary.
There are several other nice design features built into the ASCII table. For example, to
convert between uppercase and lowercase versions, you only need to “flip” the second
bit (0 for uppercase, 1 for lowercase). There are also several special characters that
27",ComputerScienceOne.pdf
"2. Basics
Binary Dec Character
0b000 0000 0 \0 Null character
0b000 0001 1 Start of Header
0b000 0010 2 Start of Text
0b000 0011 3 End of Text
0b000 0100 4 End of Transmission
0b000 0101 5 Enquiry
0b000 0110 6 Acknowledgment
0b000 0111 7 \a Bell
0b000 1000 8 \b Backspace
0b000 1001 9 \t Horizontal Tab
0b000 1010 10 \n Line feed
0b000 1011 11 \v Vertical Tab
0b000 1100 12 \f Form feed
0b000 1101 13 \r Carriage return
0b000 1110 14 Shift Out
0b000 1111 15 Shift In
0b001 0000 16 Data Link Escape
0b001 0001 17 Device Control 1
0b001 0010 18 Device Control 2
0b001 0011 19 Device Control 3
0b001 0100 20 Device Control 4
0b001 0101 21 Negative Ack
0b001 0110 22 Synchronous idle
0b001 0111 23 End of Trans. Block
0b001 1000 24 Cancel
0b001 1001 25 End of Medium
0b001 1010 26 Substitute
0b001 1011 27 Escape
0b001 1100 28 File Separator
0b001 1101 29 Group Separator
0b001 1110 30 Record Separator
0b001 1111 31 Unit Separator
0b010 0000 32 (space)
0b010 0001 33 !
0b010 0010 34 ""
0b010 0011 35 #",ComputerScienceOne.pdf
"0b001 1110 30 Record Separator
0b001 1111 31 Unit Separator
0b010 0000 32 (space)
0b010 0001 33 !
0b010 0010 34 ""
0b010 0011 35 #
0b010 0100 36 $
0b010 0101 37 %
0b010 0110 38 &
0b010 0111 39 ’
0b010 1000 40 (
0b010 1001 41 )
0b010 1010 42 *
Binary Dec Character
0b010 1011 43 +
0b010 1100 44 ,
0b010 1101 45 -
0b010 1110 46 .
0b010 1111 47 /
0b011 0000 48 0
0b011 0001 49 1
0b011 0010 50 2
0b011 0011 51 3
0b011 0100 52 4
0b011 0101 53 5
0b011 0110 54 6
0b011 0111 55 7
0b011 1000 56 8
0b011 1001 57 9
0b011 1010 58 :
0b011 1011 59 ;
0b011 1100 60 <
0b011 1101 61 =
0b011 1110 62 >
0b011 1111 63 ?
0b100 0000 64 @
0b100 0001 65 A
0b100 0010 66 B
0b100 0011 67 C
0b100 0100 68 D
0b100 0101 69 E
0b100 0110 70 F
0b100 0111 71 G
0b100 1000 72 H
0b100 1001 73 I
0b100 1010 74 J
0b100 1011 75 K
0b100 1100 76 L
0b100 1101 77 M
0b100 1110 78 N
0b100 1111 79 O
0b101 0000 80 P
0b101 0001 81 Q
0b101 0010 82 R
0b101 0011 83 S
0b101 0100 84 T
0b101 0101 85 U
Binary Dec Character
0b101 0110 86 V",ComputerScienceOne.pdf
"0b100 1111 79 O
0b101 0000 80 P
0b101 0001 81 Q
0b101 0010 82 R
0b101 0011 83 S
0b101 0100 84 T
0b101 0101 85 U
Binary Dec Character
0b101 0110 86 V
0b101 0111 87 W
0b101 1000 88 X
0b101 1001 89 Y
0b101 1010 90 Z
0b101 1011 91 [
0b101 1100 92 \
0b101 1101 93 ]
0b101 1110 94 ^
0b101 1111 95
0b110 0000 96 ‘
0b110 0001 97 a
0b110 0010 98 b
0b110 0011 99 c
0b110 0100 100 d
0b110 0101 101 e
0b110 0110 102 f
0b110 0111 103 g
0b110 1000 104 h
0b110 1001 105 i
0b110 1010 106 j
0b110 1011 107 k
0b110 1100 108 l
0b110 1101 109 m
0b110 1110 110 n
0b110 1111 111 o
0b111 0000 112 p
0b111 0001 113 q
0b111 0010 114 r
0b111 0011 115 s
0b111 0100 116 t
0b111 0101 117 u
0b111 0110 118 v
0b111 0111 119 w
0b111 1000 120 x
0b111 1001 121 y
0b111 1010 122 z
0b111 1011 123 {
0b111 1100 124 |
0b111 1101 125 }
0b111 1110 126 ~
0b111 1111 127 Delete
Table 2.4.: ASCII Character Table. The first and second column indicate the binary and",ComputerScienceOne.pdf
"0b111 1101 125 }
0b111 1110 126 ~
0b111 1111 127 Delete
Table 2.4.: ASCII Character Table. The first and second column indicate the binary and
decimal representation respectively. The third column visualizes the resulting
character when possible. Characters 0–31 and 127 are control characters that
are not printable or print whitespace. The encoding is designed to impose a
lexicographic ordering: A–Z are in order, uppercase letters precede lowercase
letters, numbers precede letters and are also in order.
28",ComputerScienceOne.pdf
"2.2. Variables
need to be escaped to be defined. For example, though your keyboard has a tab and an
enter key, if you wanted to code those characters, you would need to specify them in
some way other than using those keys (since typing those keys will affect what you are
typing rather than specifying a character). The standard way to escape characters is to
use a backslash along with another, single character. The three most common are the
(horizontal) tab, \t, the endline character, \n, and the null terminating character, \0.
The tab and endline character are used to specify their whitespace characters respectively.
The null character is used in some languages to denote the end of a string and is not
printable.
ASCII is quite old, originally developed in the early sixties. President Johnson first
mandated that all computers purchased by the federal government support ASCII in 1968.
However, it is quite limited with only 128 possible characters. Since then, additional",ComputerScienceOne.pdf
"However, it is quite limited with only 128 possible characters. Since then, additional
extensions have been developed. The Extended ASCII character set adds support for
128 additional characters (numbered 128 through 255) by adding 1 more bit (8 total).
Included in the extension are support for common international characters with diacritics
such as ¨ u, ~ nand £ (which are characters 129, 164, and 156 respectively).
Even 256 possible characters are not enough to represent the wide array of international
characters when you consider languages like Chinese Japanese Korean (CJK). Unicode
was developed to solve this problem by establishing a standard encoding that supports
1,112,064 possible characters, though only a fraction of these are actually currently
assigned.4 Unicode is backward compatible, so it works with plain ASCII characters. In
fact, the most common encoding for Unicode, UTF-8 uses a variable number of bytes to",ComputerScienceOne.pdf
"fact, the most common encoding for Unicode, UTF-8 uses a variable number of bytes to
encode characters. 1-byte encodings correspond to plain ASCII, there are also 2, 3, and
4-byte encodings.
In most programming languages, strings literals are defined by using either single or
double quotes to indicate where the string begins and ends. For example, one may be
able to define the string ""Hello World"". The double quotes are not part of the string,
but instead specify where the string begins and ends. Some languages allow you to use
either single or double quotes. PHP for example would allow you to also define the same
string as 'Hello World'. Yet other languages, such as C distinguish the usage of single
and double quotes: single quotes are for single characters such as 'A' or '\n' while
double quotes are used for full strings such as ""Hello World"".
In any case, if you want a single or double quote to appear in your string you need to",ComputerScienceOne.pdf
"double quotes are used for full strings such as ""Hello World"".
In any case, if you want a single or double quote to appear in your string you need to
escape it similar to how the tab and endline characters are escaped. For example, in C
'\'' would refer to the single quote character and ""Dwayne \""The Rock\"" Johnson""
would allow you to use double quotes within a string. In our pseudocode we’ll use the
stylized double quotes, “Hello World” in any strings that we define. We will examine
4As of 2012, 110,182 are assigned to characters, 137,468 are reserved for private use (they are valid
characters, but not defined so that organizations can use them for their own purposes), with 2,048
surrogates and 66 non-character control codes. 864,348 are left unassigned meaning that we are
well-prepared for encoding alien languages when they finally get here.
29",ComputerScienceOne.pdf
"2. Basics
string types more fully in Chapter 8.
Boolean Types
A Boolean is another type of variable that is used to hold a truth value, either true or
false, of a logical statement. Some programming languages explicitly support a built-in
Boolean type while others implicitly support them. For languages that have explicit
Boolean types, typically the keywords true and false are used, but logical expressions
such as x≤10 can also be evaluated and assigned to Boolean variables.
Some languages do not have an explicit Boolean type and instead support Booleans
implicitly, sometimes by using numeric types. For example, in C, false is associated with
zero while any non-zero value is associated with true. In either case, Boolean values are
used to make decisions and control the flow of operations in a program (see Chapter 3).
Object & Reference Types
Not everything is a number or string. Often, we wish to model real-world entities such",ComputerScienceOne.pdf
"Object & Reference Types
Not everything is a number or string. Often, we wish to model real-world entities such
as people, locations, accounts, or even interactions such as exchanges or transactions.
Most programming languages allow you to create user-defined types by using objects or
structures. Objects and structures allow you to group multiple pieces of data together
into one logical entity; this is known as encapsulation. For example, a Student object may
consist of a first-name, last-name, GPA, year, major, etc. Grouping these separate pieces
of data together allows us to define a more complex type. We explore these concepts in
more depth in Chapter 10.
In contrast to the built-in numeric, character/string, and Boolean types (also called
primitive data types) user-defined types do not necessarily take a fixed amount of memory
to represent. Since they are user-defined, it is up to the programmer to specify how",ComputerScienceOne.pdf
"to represent. Since they are user-defined, it is up to the programmer to specify how
they get created and how they are represented in memory. A variable that refers to an
object or structure is usually a reference or pointer: a reference to where the object is
stored in memory on a computer. Many programming languages use the keyword null
(or sometimes NULL or a some variation) to indicate an invalid reference. The null
keyword is often used to refer to uninitialized or “missing” data.
Another common user-defined type is an enumerated type which allows a user to define a
list of keywords associated with integers. For example, the cardinal directions, “north”,
“south”, “east”, and “west” could be associated with the integers 0, 1, 2, 3 respectively.
Defining an enumerated type then allows you to use these keywords in your program
directly without having to rely on mysterious numerical values, making a program more
readable and less prone to error.
30",ComputerScienceOne.pdf
"2.2. Variables
2.2.3. Declaring Variables: Dynamic vs. Static Typing
In some languages, variables must be declared before they can be referred to or used.
When you declare a variable, you not only give it an identifier, but also define its type.
For example, you can declare a variable named numberOfStudents and define it to be
an integer. For the life of that variable, it will always be an integer type. You can only
give that variable integer values. Attempts to assign, say, a string type to an integer
variable may either result in a syntax error or a runtime error when the program is
executed or lead to unexpected or undefined behavior. A language that requires you to
declare a variable and its type is a statically typed language.
The declaration of a variable is typically achieved by writing a statement that includes
the variable’s type (using a built-in keyword of the language) along with the variable
name. For example, in C-style languages, a line like
int x;",ComputerScienceOne.pdf
"the variable’s type (using a built-in keyword of the language) along with the variable
name. For example, in C-style languages, a line like
int x;
would create an integer variable associated with the identifier x.
In other languages, typically interpreted languages, you do not have to declare a variable
before using it. Such languages are generally referred to as dynamically typed languages.
Instead of declaring a variable to have a particular type, the type of a variable is
determined by the type of value that is assigned to it. If you assign an integer to a
variable it becomes an integer. If you assign a string to it, it becomes a string type.
Moreover, a variable’s type can change during the execution of a program. If you reassign
a value to a variable, it dynamically changes its type to match the type of the value
assigned.
In PHP for example, a line like
$x = 10;
would create an integer variable associated with the identifier $x. In this example, we",ComputerScienceOne.pdf
"assigned.
In PHP for example, a line like
$x = 10;
would create an integer variable associated with the identifier $x. In this example, we
did not declare that $x was an integer. Instead, it was inferred by the value that we
assigned to it (10).
At first glance it may seem that dynamically typed languages are better. Certainly
they are more flexible (and allow you to write less so-called “boilerplate” code), but
that flexibility comes at a cost. Dynamically typed variables are generally less efficient.
Moreover, dynamic typing opens the door to a lot of potential type mismatching errors.
For example, you may have a variable that is assumed to always be an integer. In a
dynamically typed language, no such assumption is valid as a reassignment can change
the variable’s type. It is impossible to enforce this assumption by the language itself and
may require a lot of extra code to check a variable’s type and deal with “type safety”",ComputerScienceOne.pdf
"may require a lot of extra code to check a variable’s type and deal with “type safety”
issues. The advantages and disadvantages of each continue to be debated.
31",ComputerScienceOne.pdf
"2. Basics
1 {
2 int a;
3 {
4 //this is a new code block inside the outer block
5 int b;
6 //at this point in the code, both a and b are in-scope
7 }
8 //at this point, only a is in-scope, b is out-of-scope
9 }
Code Sample 2.1.: Example of variable scoping in C
2.2.4. Scoping
The scope of a variable is the section of code in which a variable is valid or “known.”
In a statically typed language, a variable must be declared before it can be used. The
code block in which the variable is declared is therefore its scope. Outside of this code
block, the variable is invalid. Attempts to reference or use a variable that is out-of-scope
typically result in a syntax error. An example using the C programming language is
depicted in Code Sample 2.1.
Scoping in a dynamically typed language is similar, but since you don’t declare a variable,
the scope is usually defined by the block of code where you first use or reference the",ComputerScienceOne.pdf
"the scope is usually defined by the block of code where you first use or reference the
variable. In some languages using a variable may cause that variable to become globally
scoped.
A globally scoped variable is valid throughout the entirety of a program. A global variable
can be accessed and referenced on every line of code. Sometimes this is a good thing:
for example, we could define a variable to represent π and then use it anywhere in our
program. We would then be assured that every computation involving π would be using
the same definition of π (rather than one line of coding using the approximation 3.14
while another uses 3.14159).
On the same token, however, global variables make the state and execution of a program
less predictable: if any piece of code can access a global variable, then potentially any
piece of code could change that variable. Imagine some questionable code changing the
value of our global π variable to 3. For this reason, using global variables is generally",ComputerScienceOne.pdf
"value of our global π variable to 3. For this reason, using global variables is generally
considered bad practice. 5 Even if no code performs such an egregious operation, the
fact that anything can change the value means that when testing, you must test for
5Coders often say “globals are evil” and indeed have often demonstrated that they have low moral
standards. Global variables that is. Coders are always above reproach.
32",ComputerScienceOne.pdf
"2.3. Operators
the potential that anything will change the value, greatly increasing the complexity
of software testing. To capture the advantages of a global variable while avoiding the
disadvantages, it is common to only allow global constants; variables whose values cannot
be changed once set.
Another argument against globally scoped variables is that once the identifier has been
used, it cannot be reused or redefined for other purposes (a floating point variable with
the identifier pi means we cannot use the identifier pi for any other purpose) as
it would lead to conflicts. Defining many globally scoped variables (or functions, or
other elements) starts to pollute the namespace by reserving more and more identifiers.
Problems arise when one attempts to use multiple libraries that have both used the same
identifiers for different variables or functions. Resolving the conflict can be difficult or
impossible if you have no control over the offending libraries.
2.3. Operators",ComputerScienceOne.pdf
"impossible if you have no control over the offending libraries.
2.3. Operators
Now that we have variables, we need a way to work with variables. That is, given two
variables we may wish to add them together. Or we may wish to take two strings and
combine them to form a new string. In programming languages this is accomplished
through operators which operate on one or more operands. An operator takes the values
of its operands and combines them in some way to produce a new value. If an operator
is applied to variable(s), then the values used in the operation are the values stored in
the variable at the time that the operator is evaluated.
Many common operators are binary in that they operate on two operands such as common
arithmetic operations like addition and multiplication. Some operators are unary in that
they only operate on one variable. The first operator that we look at is a unary operator
and allows us to assign values to variables.
2.3.1. Assignment Operators",ComputerScienceOne.pdf
"and allows us to assign values to variables.
2.3.1. Assignment Operators
The assignment operator is a unary operator that allows you to take a value and assign
it to a variable. The assignment operator usually takes the following form: the value is
placed on the right-hand-side of the operator while the variable to which we are assigning
the value is placed on the left-hand-side of the operator. For our pseudocode, we’ll use a
generic “left-arrow” notation:
a←10
which should be read as “place the value 10 into the variable a.” Many C-style pro-
gramming languages commonly use a single equal sign for the assignment operator. The
example above might be written as
33",ComputerScienceOne.pdf
"2. Basics
a = 10;
It is important to realize that when this notation is used, it is not an algebraic declaration
like a = b which is an algebraic assertion that the variables a and b are equal. An
assignment operator is different: it means place the value on the right-hand-side into the
variable on the left-hand-side. For that reason, writing something like
10 = a;
is invalid syntax. The left-hand-side must be a variable.
The right-hand-side, however, may be a literal, another variable, or even a more complex
expression. In the example before,
a←10
the value 10 was acting as a numerical literal: a way of expressing a (human-readable)
value that the computer can then interpret as a binary value. In code, we can conveniently
write numbers in base-10; when compiled or interpreted, the numerical literals are
converted into binary data that the computer understands and placed in a memory
location corresponding to the variable. This entire process is automatic and transparent",ComputerScienceOne.pdf
"location corresponding to the variable. This entire process is automatic and transparent
to the user. Literals can also be strings or other values. For example:
message←“hello world”
We can also “copy” values from one variable to another. Assuming that we’ve assigned
the value 10 to the variable a, we can then copy it to another variable b:
b←a
This does not mean that a and b are the same variable. The value that is stored in the
variable aat the time that this statement is executed is copied into the variable b. There
are now two different variables with the same value. If we reassign the value in a, the
value in b is unaffected. This is illustrated in Algorithm 2.1
1 a←10
2 b←a
//a and b both store the value 10 at this point
3 a←20
//now a has the value 20, but b still has the value 10
4 b←25
//a still stores a value of 20, b now has a value of 25
Algorithm 2.1:Assignment Operator Demonstration
The right-hand-side can also be a more complex expression, for example the result of",ComputerScienceOne.pdf
"Algorithm 2.1:Assignment Operator Demonstration
The right-hand-side can also be a more complex expression, for example the result of
summing two numbers together.
34",ComputerScienceOne.pdf
"2.3. Operators
2.3.2. Numerical Operators
Numerical operators allow you to create complex expressions involving either numerical
literals and/or numerical variables. For most numerical operators, it doesn’t matter if
the operands are integers or floating point numbers. Integers can be added to floating
point numbers without much additional code for example.
The most basic numerical operator is the unary negation operator. It allows you to
negate a numerical literal or variable. For example,
a←−10
or
a←−b
The usage of a negation is so common that it is often not perceived to be an operator
but it is.
Addition & Subtraction
You can also add (sum) two numbers using the + (plus) operator and subtract using the
−(minus) operator in a straightforward way. Note that most languages can distinguish
the minus operator and the negation operator by how you use it just like a mathematical
expression. If applied to one operand, it is interpreted as a negation operator. If applied",ComputerScienceOne.pdf
"expression. If applied to one operand, it is interpreted as a negation operator. If applied
to two operands, it represents subtraction. Some examples can be found in Algorithm
2.2.
1 a←10
2 b←20
3 c←a+ b
4 d←a−b
//c has the value 30 while d has the value −10
5 c←a+ 10
6 d←−d
//c now has the value 20 and d now has the value 10
Algorithm 2.2:Addition and Subtraction Demonstration
Multiplication & Division
You can also multiply and divide literals and variables. In mathematical expressions
multiplication is represented as a×bor a·bor simply just aband division is represented
35",ComputerScienceOne.pdf
"2. Basics
as a÷bor a/bor a
b. In our pseudocode, we’ll generally use a·band a
b, but in programming
languages it is difficult to type these symbols. Usually programming languages use *
for multiplication and / for division. Similar examples are provided in Algorithm 2.3.
1 a←10
2 b←20
3 c←a·b
4 d←a
b
//c has the value 200 while d has the value 0.5
Algorithm 2.3:Multiplication and Division Demonstration
Careful! Some languages specify that the result of an arithmetic operation on variables
of a certain type must match. That is, an integer plus an integer results in an integer. A
floating point number divided by a floating point number results a floating point number.
When we mix types, say an integer and a floating point number, the result is generally a
floating point number. For the most part this is straightforward. The one tricky case is
when we have an integer divided by another integer, 3 /2 for example.",ComputerScienceOne.pdf
"when we have an integer divided by another integer, 3 /2 for example.
Since both operands are integers, the result must be an integer. Normally, 3 /2 = 1.5,
but since the result must be an integer, the fractional part gets truncated (cut-off) and
only the integral part is kept for the final result. This can lead to weird results such as
1/3 = 0 and 99/100 = 0. The result is not rounded down or up; instead the fractional
part is completely thrown out. Care must be taken when dividing integer variables in a
statically typed language. Type casting can be used to force variables to change their
type for the purposes of certain operations so that the full answer is preserved. For
example, in C we can write
1 int a = 10;
2 int b = 20;
3 double c;
4 int d;
5 c = (double) a / (double) b;
6 d = a / b;
7 //the value in c is correctly 0.5 but the value in d is 0
36",ComputerScienceOne.pdf
"2.3. Operators
Integer Division
Recall that in arithmetic, when you divide integers a/b, b might not go into a evenly in
which case you get a remainder. For example, 13 /5 = 2 with a remainder r= 3. More
generally we have that
a= qb+ r
Where a is the dividend, b is the divisor, q is the quotient (the result) and r is the
remainder. We can also perform integer division in most programming languages. In
particular, the integer division operator is the operator that gives us the remainder of
the integer division operation in a/b. In mathematics this is the modulo operator and is
denoted
amod b
For example,
13 mod 5 = 3
It is possible that the remainder is zero, for example,
10 mod 5 = 0
Many programming languages support this operation using the percent sign. For example,
c = a % b;
2.3.3. String Concatenation
Strings can also be combined to form new strings. In fact, strings can often be combined
with non-string variables to form new strings. You would typically do this in order to",ComputerScienceOne.pdf
"with non-string variables to form new strings. You would typically do this in order to
convert a numerical value to a string representation so that it can be output to the user
or to a file for longterm storage. The operation of combining strings is referred to as
string concatenation. Some languages support this through the same plus operator that
is used with addition. For example,
message←“hello ” + “world!”
which combines the two strings to form one string containing the characters “hello world!”,
storing the value into the message variable. For our pseudocode we’ll adopt the plus
operator for string concatenation.
The string concatenation operator can also sometimes be combined with non-string
types; numerical types for example. This allows you to easily convert numbers to a
human-readable, base-10 format so that they can be printed to the output. For example
suppose that the variable b contains the value 20, then
message←“the answer is ” + b
37",ComputerScienceOne.pdf
"2. Basics
might result in the string “the answer is 20” being stored in the variable message.
Other languages use different symbols to distinguish concatenation and addition. Still
yet other languages do not directly support an operator for string concatenation which
must instead be done using a function.
2.3.4. Order of Precedence
In mathematics, when you write an expression such as:
a+ b·c
you interpret it as “multiply band cand then add a.” This is because multiplication has
a higher order of precedence than addition. The order of precedence (sometimes referred
to as order of operations ) is a set of rules which define the order in which operations
should be evaluated. In this case, multiplication is performed before addition. If, instead,
we had written
(a+ b) ·c
we would have a different interpretation: “add a and b and then multiply the result by
c.” That is, the inclusion of parentheses changes the order in which we evaluate the",ComputerScienceOne.pdf
"c.” That is, the inclusion of parentheses changes the order in which we evaluate the
operations. Adding parentheses can have no effect (if we wrote a+ (bc) for example), or
it can cause operations with a lower order of precedence to be evaluated first as in the
example above.
Numerical operators are similar when used in most programming languages. The same
order of precedence is used and parentheses can be used to change the order of evaluation.
2.3.5. Common Numerical Errors
When dealing with numeric types it is important to know and understand their limitations.
In mathematics, the following operations might be considered invalid.
•Division by zero: a
b where b= 0. This is an undefined operation in mathematics
and also in programming languages. Depending on the language, any number of
things may happen. It may be a fatal error or exception; the program may continue
executing but give “garbage” results from then on; the result may be a special value",ComputerScienceOne.pdf
"executing but give “garbage” results from then on; the result may be a special value
such as null, “NaN” (not-a-number) or “INF” (a special representation of infinity).
It is best to avoid such an operation entirely using conditionals statements and
defensive programming (see Chapter 3).
•Other potentially invalid operations involve common mathematical functions. For
example, √−1 would be a complex result, i which some languages do support.
38",ComputerScienceOne.pdf
"2.3. Operators
However, many do not. Similarly, the natural logarithm of zero, ln (0) and negative
values, ln (−1) is undefined. In either case you could expect a result like “NaN” or
“INF.”
•Still other operations seem like they should be valid, but because of how numbers
are represented in binary, the results are invalid. Recall that for a 32-bit signed,
two’s complement number, the maximum representable value is 2,147,483,647.
Suppose this maximum value is stored in a variable, b. Now suppose we attempt
to add one more,
c←b+ 1
Mathematically we’d expect the result to be 2,147,483,648, but that is more than
the maximum representable integer. What happens is something called arithmetic
overflow. The actual number stored in binary in memory for 2,147,483,647 is
0b011 ... 11  
31 1s
When we add 1 to this, it is carried over all the way to the 32nd bit, giving the
result
0b100 ... 00  
31 0s
in binary. However, the 32nd bit is the sign bit, so this is a negative number.",ComputerScienceOne.pdf
"result
0b100 ... 00  
31 0s
in binary. However, the 32nd bit is the sign bit, so this is a negative number.
In particular, if this is a two’s complement integer, it has the decimal value
−2,147,483,648 which is obviously wrong. Another example would be if we have a
“large number, say 2 billion and attempt to double it (multiply by 2). We would
expect 4 billion as a result, but again overflow occurs and the result (using 32-bit
signed two’s complement integers) is −294,967,296.
•A similar phenomenon can happen with floating point numbers. If an operation
(say multiplying two “small” numbers together) results in a number that is smaller
than the smallest floating point number that can be represented, the result is said
to have resulted in underflow. The result can essentially be zero, or an error can
be raised to indicate that underflow has occurred. The consequences of underflow
can be very complex.
•Floating-point operations can also result in a loss of precision even if no overflow",ComputerScienceOne.pdf
"can be very complex.
•Floating-point operations can also result in a loss of precision even if no overflow
or underflow occurs. For example, when adding a very large number a and a very
small number b, the result might be no different from the value of a. This is because
(for example) double precision floating point numbers only have about 16 significant
digits of precision with the least significant digits being cutoff in order to preserve
the magnitude.
As another example, suppose we compute
√
2 = 1.41421356 ... . If we squared the
result, mathematically we would expect to get 2. However, since we only have a
certain number of digits of precision, squaring the result in a computer may result
in a value slightly different from 2 (either 1.9999998 or 2.0000001).
39",ComputerScienceOne.pdf
"2. Basics
2.3.6. Other Operators
Many programming languages support other “convenience” operators that allow you to
perform common operations using less code. These operators are generally syntactic
sugar: the don’t add any functionality. The same operation could be achieved using
other operators. However, they do add simpler or more terse syntax for doing so.
Increment Operators
Adding or subtracting one to a variable is a very common operation. So common, that
most programming languages define increment operators such as i++ and i-- which
add one and subtract one from the variables applied. The same effect could be achieved
by writing
i←(i+ 1) and i←(i−1)
but the increment operators provide a shorthand way of expressing the operation.
The operators i++ and i-- are postfix operators: the operator is written after (post)
the operand. Some languages define similar prefix increment operators, ++i and --i.
The effect is similar: each adds or subtracts one from the variable i. However, the",ComputerScienceOne.pdf
"The effect is similar: each adds or subtracts one from the variable i. However, the
difference is when the operator is used in a larger expression. A postfix operator retains
the original value for the expression, a prefix operator takes on the new, incremented
value in the expression.
To illustrate, suppose the variable i has the value 10. In the following line of code, i is
incremented and used in an expression that adds 5 and stores the result in a variable x:
x = 5 + (i++);
The value of x after this code is 15 while the value of i is now 11. This is because
the postfix operator increments i, but i++ retains the value 10 in the expression. In
contrast, with the line
x = 5 + (++i);
the variable i again now has the value 11, but the value of x is 16 since ++i takes
on the new, incremented value of 11. Appropriately using each can lead to some very
concise code, but it is important to remember the difference.
Compound Assignment Operators",ComputerScienceOne.pdf
"concise code, but it is important to remember the difference.
Compound Assignment Operators
If we want to increment or decrement a variable by an amount other than 1 we can do
so using compound assignment operators that combine an arithmetic operator and an
assignment operator into one. For example, a += 10 would add 10 to the variable a.
40",ComputerScienceOne.pdf
"2.4. Basic Input/Output
1 int a = 10;
2 a += 5; //adds 5 to a
3 a -= 3; //subtracts 3 from a
4 a *= 2; //multiplies a by 2
5 a /= 4; //divides a by 4
6
7 //you can also use compound assignment operators with variables:
8 int b = 5;
9 a += b; //adds the value stored in b to a
10 a -= b; //subtracts the value stored in b from a
11 a *= b; //multiplies a by b
12 a /= b; //divides a by b
Code Sample 2.2.: Compound Assignment Operators in C
The same could be achieved by coding a = a + 10, but the former is a bit shorter as
we don’t have to repeat the variable.
You can do the same with subtraction, multiplication, and division. More examples
using the C programming language can be found in Code Snippet 2.2. It is important to
note that these operators are not, strictly speaking, equivalent. That is, a += 10 is not
equivalent to a = a + 10. They have the same effect, but the first involves only one
operator while the second involves two operators.
2.4. Basic Input/Output",ComputerScienceOne.pdf
"operator while the second involves two operators.
2.4. Basic Input/Output
Not all variables can be coded using literals. Sometimes a program needs to read in
values as input from a user who can give different values on different runs of a program.
Likewise, a computer program often needs to produce output to the user to be of any
use.
The most basic types of programs are interactive programs that interact with a human
user. Generally, the program may interactive the user to enter some input value(s) or
make some choices. It may then compute some values and respond to the user with some
output. In the following sections we’ll overview the various types of input and output
(I/O for short) that are available.
41",ComputerScienceOne.pdf
"2. Basics
2.4.1. Standard Input & Output
The standard input (stdin for short), standard output (stdout) and standard error (stderr)
are three standard communication streams that are defined by most computer systems.
Though perhaps an over simplification, the keyboard usually serves as a standard input
device while the monitor (or the system console) serves as a standard output device. The
standard error is usually displayed in the same display but may be displayed differently
on some systems (it is typeset in red in some consoles that support color to indicate that
the output is communicating an error).
As a program is executing, it may prompt a user to enter input. A program may wait
(called blocking) until a user has typed whatever input they want to provide. The user
typically hits the enter key to indicate their input is done and the program resumes,
reading the input provided via the standard input. The program may also produce
output which is displayed to the user.",ComputerScienceOne.pdf
"reading the input provided via the standard input. The program may also produce
output which is displayed to the user.
The standard input and output are generally universal: almost any language, and
operating system will support them and they are the most basic types of input/output.
However, the type of input and output is somewhat limited (usually limited to text-based
I/O) and doesn’t provide much in the way of input validation. As an example, suppose
that a program prompts a user to enter a number. Since the input device (keyboard) is
does not really restrict the user, a more obstinate user may enter a non-numeric value,
say “hello”. The program may crash or provide garbage output with such input.
2.4.2. Graphical User Interfaces
A much more user-oriented way of reading input and displaying output is to use a
Graphical User Interface (GUI). GUIs can be implemented as traditional “thick-client”
applications (programs that are installed locally on your machine) or as “thin-client”",ComputerScienceOne.pdf
"applications (programs that are installed locally on your machine) or as “thin-client”
applications such as a web application. They typically support general “widgets” such as
input boxes, buttons, sliders, etc. that allow a user to interact with the program in a
more visual way. They also allow the programmer to do better input validation. Widgets
could be design so that only good input is allowed by creating modal restrictions: the
user is only allowed to select one of several “radio” buttons for example. GUIs also
support visual feedback cues to the user: popups, color coding, and other elements can
be used to give feedback on errors and indicate invalid selections.
Graphical user interfaces can also make use of more modern input devices: mice, touch
screens with gestures, even gaming devices such as the Kinect allow users to use a full
body motion as an input mechanism. We discuss GUIs in more detail in Chapter 13. To
begin, we’ll focus more on plain textual input and output.
42",ComputerScienceOne.pdf
"2.4. Basic Input/Output
Language Standard Output String Output
C printf() sprintf()
Java System.out.printf() String.format()
PHP printf() sprintf()
Table 2.5.: printf()-style Methods in Several Languages. Languages support format-
ting directly to the Standard Output as well as to strings that can be further
used or manipulated. Most languages also support printf()-style formatting
to other output mechanisms (streams, files, etc.).
2.4.3. Output Using printf()-style Formatting
Recall that many languages allow you to concatenate a string and a non-string type in
order to produce a string that can then be output to the standard output. However,
concatenation doesn’t provide much in the way of customizability when it comes to
formatting output. We may want to format a floating point number so that it only prints
two decimal places (as with US currency). We may want to align a column of data so
that number places match up. Or we may want to justify text either left or right.",ComputerScienceOne.pdf
"that number places match up. Or we may want to justify text either left or right.
Such data formatting can be achieved through the use of a printf()-style formatting
function. The ideas date back to the mid-60s, but the modern printf() comes from
the C programming language. Numerous programming languages support this style
of formatted output ( printf() stands for print formatted). Most support either
printing the resulting formatted output to the standard output as well as to strings and
other output mechanisms (files, streams, etc.). Table 2.5 contains a small sampling of
printf()-style functions supported in several languages. We’ll illustrate this usage
using the C programming language for our examples, but the concepts are generally
universal across most languages.
The function works by providing it a number of arguments. The first argument is always
a string that specifies the formatting of the result using several placeholders (flags that",ComputerScienceOne.pdf
"a string that specifies the formatting of the result using several placeholders (flags that
begin with a percent sign) which will be replaced with values stored in variables but
in a formatted manner. Subsequent arguments to the function are the list of variables
to be printed; each argument is delimited by a comma. Figure 2.3 gives an example
of of a printf() statement with two placeholders. The placeholders are ultimately
replaced with the values stored in the provided variables a,b. If a,b held the values 10
and 2.718281, the code would end up printing
The value of a = 10, the value of b is 2.718281
Though there are dozens of placeholders that are supported, we will focus only on a few:
• %d formats an integer variable or literal
43",ComputerScienceOne.pdf
"2. Basics
printf(""The value of a = %d, the value of b is %f\n"", a, b);
Placeholders
Format String
Print List
Figure 2.3.: Elements of a printf() statement in C
• %f formats a floating point variable or literal
• %c formats a single character variable or literal
• %s formats a string variable or literal
Misuse of placeholders may result in garbage output. For example, using an integer
placeholder, %d, but providing a string argument; since strings cannot be (directly)
converted to integers, the output will not be correct.
In addition to these placeholders, you can also add modifiers. A number n between
the percent sign and character ( %nd, %nf, %ns)) specifies that the result should be
formatted with a minimum of n columns. If the output takes less than n columns,
printf() will pad out the result with spaces so that there are ncolumns. If the output
takes n or more columns, then the modifier will have no effect (it specifies a minimum
not a maximum).",ComputerScienceOne.pdf
"takes n or more columns, then the modifier will have no effect (it specifies a minimum
not a maximum).
Floating-point numbers have a second modifier that allows you to specify the number of
digits of precision to be formatted. In particular, you can use the placeholder %n.mf in
which n has the same meaning, but m specifies the number of decimals to be displayed.
By default, 6 decimals of precision are displayed. If mis greater than the precision of the
number, zeros are usually used for subsequent digits; if mis smaller than the precision of
the number, rounding may occur. Note that the n modifier includes the decimal point
as a column. Both modifiers are optional.
Finally, each of these modifiers can be made negative (example: %-20d) to left-justify
the result. By default, justification is to the right. Several examples are illustrated in
Code Sample 2.3 with the results in Code Sample 2.4.
2.4.4. Command Line Input",ComputerScienceOne.pdf
"Code Sample 2.3 with the results in Code Sample 2.4.
2.4.4. Command Line Input
Not all programs are interactive. In fact, the vast majority of software is developed to
interact with other software and does not expect that a user is sitting at the console
44",ComputerScienceOne.pdf
"2.4. Basic Input/Output
1 int a = 4567;
2 double b = 3.14159265359;
3
4 printf(""a=%d\n"", a);
5 printf(""a=%2d\n"", a);
6 printf(""a=%4d\n"", a);
7 printf(""a=%8d\n"", a);
8
9 //by default, prints 6 decimals of precision
10 printf(""b=%f\n"", b);
11 //the .m modifier is optional:
12 printf(""b=%10f\n"", b);
13 //the n modifier is also optional:
14 printf(""b=%.2f\n"", b);
15 //note that this rounds!
16 printf(""b=%10.3f\n"", b);
17 //zeros are added so that 15 decimals are displayed
18 printf(""b=%20.15f\n"", b);
Code Sample 2.3.: printf() examples in C
a=4567
a=4567
a=4567
a= 4567
b=3.141593
b= 3.141593
b=3.14
b= 3.142
b= 3.141592653590000
Code Sample 2.4.: Result of computation in Code Sample 2.3. Spaces are highlighted
with a for clarity.
45",ComputerScienceOne.pdf
"2. Basics
constantly providing it with input. Most languages and operating systems support
non-interactive input from the Command Line Interface (CLI). This is input that is
provided at the command line when the program is executed. Input provided from the
command line are usually referred to as command line arguments. For example, if we
invoke a program named myProgram from the command line prompt using something
like the following:
~>./myProgram a 10 3.14
Then we would have provided 4 command line arguments. The first argument is usually
the program’s name, all subsequent arguments are separated by whitespace. Command
line arguments are provided to the program as strings and it is the program’s responsibility
to convert them if needed and to validate them to ensure that the correct expected
number and type of arguments were provided.
Within a program, command line arguments are usually referred to as an argument vector",ComputerScienceOne.pdf
"number and type of arguments were provided.
Within a program, command line arguments are usually referred to as an argument vector
(sometimes in a variable named argv) and argument count (sometimes in a variable
named argc). We explore how each language supports this in subsequent chapters.
2.5. Debugging
Making mistakes in programming is inevitable. Even the most expert of software
developers make mistakes. 6 Errors in computer programs are usually referred to as
bugs. The term was popularized by Grace Hopper in 1947 while working on a Mark
II Computer at a US Navy research lab. Literally, a moth stuck in the computer was
impeding its operation. Removing the moth or “debugging” the computer fixed it. In
this section we will identify general types of errors and outline ways to address them.
2.5.1. Types of Errors
When programming, there are several types of errors that can occur. Some can be easily
detected (or even easily fixed) by compilers and other modern code analysis tools such",ComputerScienceOne.pdf
"detected (or even easily fixed) by compilers and other modern code analysis tools such
as IDEs.
6A severe security bug in the popular unix bash shell utility went undiscovered for 25 years before it
was finally fixed in September 2014, missed by thousands of experts and some of the best coders in
the world.
46",ComputerScienceOne.pdf
"2.5. Debugging
Syntax Errors
Syntax errors are errors in the usage of a programming language itself. A syntax error
can be a failure to adhere to the rules of the language such as misspelling a keyword or
forgetting proper “punctuation” (such as missing an ending semicolon). When you have a
syntax error, you’re essentially not “speaking the same language.” You wouldn’t be very
comprehensible if you started injecting non-sense words or words from different language
when speaking to someone in English. Similarly, a computer can’t understand what
you’re trying to say (or what directions you’re trying to give it) if you’re not speaking
the same language.
Typically syntax errors prevent you from even compiling a program, though syntax
errors can be a problem at runtime with interpreted languages. When a syntax error is
encountered, a compiler will fail to complete the compilation process and will generally",ComputerScienceOne.pdf
"encountered, a compiler will fail to complete the compilation process and will generally
quit. Ideally, the compiler will give reasons for why it was unable to compile and will
hopefully identify the line number where the syntax error was encountered with a hint on
what was wrong. Unfortunately, many times a compiler’s error message isn’t too helpful
or may indicate a problem on one line where the root cause of the problem is earlier
in the program. One cannot expect too much from a compiler after all. If a compiler
were able to correctly interpret and fix our errors for us, we’d have “natural language”
programming where we could order the computer to execute our commands in plain
English. If we had this science fiction-level of computer interaction we wouldn’t need
programming languages at all.
Fixing syntax errors involves reading and interpreting the compiler error messages,
reexamining the program and fixing any and all issues to conform to the syntax of",ComputerScienceOne.pdf
"reexamining the program and fixing any and all issues to conform to the syntax of
the programming language. Fixing one syntax error may enable the compiler to find
additional syntax errors that it had not found before. Only once all syntax errors have
been resolved can a program actually compile. For interpreted languages, the program
may be able to run up to where it encounters a syntax error and then exits with a fatal
error. It may take several test runs to resolve such errors.
Runtime Errors
Once a program is free of syntax errors it can be compiled and be run. However, that
doesn’t mean that the program is completely free of bugs, just that it is free of the types
of bugs (syntax errors) that the compiler is able to detect. A compiler is not able to
predict every action or possible event that could occur when a program is actually run.
A runtime error is an error that occurs while a program is being executed. For example,",ComputerScienceOne.pdf
"A runtime error is an error that occurs while a program is being executed. For example,
a program could attempt to access a file that does not exist, or attempt to connect to a
remote database, but the computer has lost its network connection, or a user could enter
bad data that results in an invalid arithmetic operation, etc.
47",ComputerScienceOne.pdf
"2. Basics
A compiler cannot be expected to detect such errors because, by definition, the conditions
under which runtime errors occur occur at runtime, not at compile time. One run of
a program could execute successfully, while another subsequent run could fail because
the system conditions have changed. That doesn’t mean that we should not attempt to
mitigate the consequences of runtime errors.
As a programmer it is important to think about the potential problems and runtime
errors that could occur and make contingency plans accordingly. We can make reasonable
assumptions that certain kinds of errors may occur in the execution of our program and
add code to handle those errors if they occur. This is known as error handling (which
we discuss in detail in Chapter 6). For example, we could add code that checks if a user
enters bad input and then re-prompt them to enter good input. If a file is missing, we",ComputerScienceOne.pdf
"enters bad input and then re-prompt them to enter good input. If a file is missing, we
could add code to create it as needed. By checking for these errors and preventing illegal,
potentially fatal operations, we practice defensive programming.
Logic Errors
Other errors may be a result of bad code or bad design. Computers do exactly as they
are told to do. Logic errors can occur if we tell the computer to do something that we
didn’t intend for them to do. For example, if we tell the computer to execute command
A under condition X, but we meant to have the computer execute command B under
condition Y, we have caused a logical error. The computer will perform the first set of
instructions, not the second as we intended. The program may be free of syntax errors
and may execute without any problems, but we certainly don’t get the results that we
expected.
Logic errors are generally only detected and addressed by rigorous software testing. When",ComputerScienceOne.pdf
"expected.
Logic errors are generally only detected and addressed by rigorous software testing. When
developing software, we can also design a collection of test cases : a set of inputs along
with correct outputs that we would expect the program of code to produce. We can then
test the program with these inputs to see if they produce the same output as in the test
cases. If they don’t, then we’ve uncovered a logical error that needs to be addressed.
Rigorous testing can be just as complex (or even more complex) than writing the program
itself. Testing alone cannot guarantee that a program is free of bugs (in general, the
number of possible inputs is infinite; it is impossible to test all possibilities). However,
the more test cases that we design and pass the higher the confidence we have that the
program is correct.
Testing can also be very tedious. Modern software engineering techniques can help
streamline the process. Many testing frameworks have been developed and built that",ComputerScienceOne.pdf
"streamline the process. Many testing frameworks have been developed and built that
attempt to automate the testing process. Test cases can be randomly generated and
test suites can be repeatedly run and verified throughout the development process.
Frameworks can perform regression testing to see if fixing one bug caused or uncovered
another, etc.
48",ComputerScienceOne.pdf
"2.5. Debugging
2.5.2. Strategies
A common beginner’s way of debugging a program is to insert temporary print statements
throughout their program to see what values variables have at certain points in an attempt
to isolate where an error is occurring. This is an okay strategy for extremely simple
programs, but its the “poor man’s” way of debugging. As soon as you start writing
more complex programs you quickly realize that this strategy is slow, inefficient, and
can actually hide the real problems. The standard output is not guaranteed to work as
expected if an error has occurred, so print statements may actually mislead you into
thinking the problem occurs at one point in the program when it actually occurs in a
different part.
Instead, it is much better to use a proper debugging tool in order to isolate the problem.
A debugger is a program, that allows you to “simulate” an execution of your program.
You can set break points in your program on certain lines and the debugger will execute",ComputerScienceOne.pdf
"You can set break points in your program on certain lines and the debugger will execute
your program up to those points. It then pauses and allows you to look at the program’s
state: you can examine the contents of memory, look at the values stored in certain
variables, etc. Debuggers will also allow you to resume the execution of your program
to the next break point or allow you to “step” through the execution of your program
line by line. This allows you to examine the execution of a program at human speed in
order to diagnose the exact point in execution where the problem occurs. IDEs allow
you to do this visually with a graphical user interface and easy visualization of variables.
However, there are command line debuggers such as GDB (GNU’s Not Unix! (GNU)
Debugger) that you interact with using text commands.
In general, debugging strategies attempt to isolate a problem to the smallest possible code",ComputerScienceOne.pdf
"Debugger) that you interact with using text commands.
In general, debugging strategies attempt to isolate a problem to the smallest possible code
segment. Thus, it is best practice to design your code using good procedural abstraction
and place your code into functions and methods (see Chapter 5). It is also good practice
to create test cases and test suites as you develop these small pieces of code.
It can also help to diagnose a problem by looking at the nature of the failure. If some
test cases pass and others fail you can get a hint as to what’s wrong by examining the
key differences between the test cases. If one value passes and another fails, you can
trace that value as it propagates through the execution of your program to see how it
affects other values.
In the end, good debugging skills, just like good coding skills, come from experience. A
seasoned expert may be able to look at an error message and immediately diagnose the",ComputerScienceOne.pdf
"seasoned expert may be able to look at an error message and immediately diagnose the
problem. Or, a bug can escape the detection of hundreds of the best developers and
software tools and end up costing millions of dollars and thousands of man-hours.
49",ComputerScienceOne.pdf
"2. Basics
2.6. Examples
Let’s apply these concepts by developing several prompt-and-compute style programs.
That is, the programs will prompt the user for input, perform some calculations, and
then output a result.
To write these programs, we’ll use pseudocode, an informal, abstract description of a
program/algorithm. Pseudocode does not use any language-specific syntax. Instead,
it describes processes at a high-level, making use of plain English and mathematical
notation. This allows us to focus on the actual process/program rather than worrying
about the particular syntax of a specific language. Good pseudocode is easily translated
into any programming language.
2.6.1. Temperature Conversion
Temperature can be measured in several different scales. The most common for everyday
use is Celsius and Fahrenheit. Let’s write a program to convert from Fahrenheit to
Celsius using the following formula:
C = 5
9 ·(F −32)
The basic outline of the program will be three simple steps:",ComputerScienceOne.pdf
"Celsius using the following formula:
C = 5
9 ·(F −32)
The basic outline of the program will be three simple steps:
1. Read in a Fahrenheit value from the user
2. Compute a Celsius value using the formula above
3. Output the result to the user
This is actually pretty good pseudocode already, but let’s be a little more specific using
some of the operators and notation we’ve established above. The full program can be
found in Algorithm 2.4.
1 prompt the user to enter a temperature in Fahrenheit
2 F ←read input from user
3 C ←5
9 ·(F −32)
4 Output C to the user
Algorithm 2.4:Temperature Conversion Program
50",ComputerScienceOne.pdf
"2.7. Exercises
2.6.2. Quadratic Roots
A common math exercise is to find the roots of a quadratic equation with coefficients,
a,b,c ,
ax2 + bx+ c= 0
using the quadratic formula,
x= −b±
√
b2 −4ac
2a
Following the same basic outline, we’ll read in the coefficients from the user, compute
each of the roots, and output the results to the user. We need two computations, one for
each of the roots which we label r1,r2. The full procedure is presented in Algorithm 2.5.
1 prompt the user to enter a
2 a←read input from user
3 prompt the user to enter b
4 b←read input from user
5 prompt the user to enter c
6 c←read input from user
7 r1 ←−b+
√
b2−4ac
2a
8 r2 ←−b−
√
b2−4ac
2a
9 Output “the roots of ax2 + bx+ c are r1,r2”
Algorithm 2.5:Quadratic Roots Program
2.7. Exercises
Exercise 2.1. Write a program that calculates mileage deduction for income tax using
the standard rate of $0.575 per mile. Your program will read in a beginning and ending",ComputerScienceOne.pdf
"the standard rate of $0.575 per mile. Your program will read in a beginning and ending
odometer reading and calculate the difference and total deduction. Take care that your
output is in whole cents. An example run of the program may look like the following.
INCOME TAX MILEAGE CALCULATOR
Enter beginning odometer reading--> 13505.2
Enter ending odometer reading--> 13810.6
You traveled 305.4 miles. At $.575 per mile,
your reimbursement is $175.61
51",ComputerScienceOne.pdf
"2. Basics
Exercise 2.2. Write a program to compute the total “cost” C of a loan. That is, the
total amount of interest paid over the life of a loan. To compute this value, use the
following formula.
C = p·i·(1 + i)12n
(1 + i)12n −1 ∗12n−p
where
•p is the starting principle amount
•i= r
12 where r is the APR on the interval [0 ,1]
•n is the number of years the loan is to be paid back
Exercise 2.3. Write a program to compute the annualized appreciation of an asset (say
a house). The program should read in a purchase price p, a sale price s and compute
their difference d= s−p (it should support a loss or gain). Then, it should compute an
appreciation rate: r= d
p along with an (average) annualized appreciation rate (that is,
what was the appreciation rate in each year that the asset was held that compounded ):
(1 + r)
1
y −1
Where y is the number of years (possibly fractional) the asset was held (and r is on the
scale [0,1]).",ComputerScienceOne.pdf
"(1 + r)
1
y −1
Where y is the number of years (possibly fractional) the asset was held (and r is on the
scale [0,1]).
Exercise 2.4. The annual percentage yield (APY) is a much more accurate measure of
the true cost of a loan or savings account that compounds interest on a monthly or daily
basis. For a large enough number of compounding periods, it can be calculated as:
APY = ei −1
where i is the nominal interest rate (6% = 0 .06). Write a program that prompts the user
for the nominal interest rate and outputs the APY.
Exercise 2.5. Write a program that calculates the speed of sound ( v, feet-per-second)
in the air of a given temperature T (in Fahrenheit). Use the formula,
v= 1086
√
5T + 297
247
Be sure your program does not lose the fractional part of the quotient in the formula
shown and format the output to three decimal places.
Exercise 2.6. Write a program to convert from radians to degrees using the formula
deg= 180 ·rad
π",ComputerScienceOne.pdf
"Exercise 2.6. Write a program to convert from radians to degrees using the formula
deg= 180 ·rad
π
However, radians are on the scale [0 ,2π). After reading input from the user be sure to
do some error checking and give an error message if their input is invalid.
52",ComputerScienceOne.pdf
"2.7. Exercises
Exercise 2.7. Write a program to compute the Euclidean Distance between two points,
(x1,y2) and (x2,y2) using the formulate:
√
(x1 −x2)2 + (y1 −y2)2
Exercise 2.8. Write a program that will compute the value of sin(x) using the first 4
terms of the Taylor series:
sin(x) ≈x−x3
3! + x5
5! −x7
7!
In addition, your program will compute the absolute difference between this calculation
and a standard implementation of the sine function supported in your language. Your
program should prompt the user for an input value xand display the appropriate output.
Your output should looks something like the following.
Sine Approximation
===================
Enter x: 1.15
Sine approximation: 0.912754
Sine value: 0.912764
Difference: 0.000010
Exercise 2.9. Write a program to compute the roots of a quadratic equation:
ax2 + bx+ c= 0
using the well-known quadratic formula:
−b±
√
b2 −4ac
2a
Your program will prompt the user for the values, a,b,c and output each real root.",ComputerScienceOne.pdf
"using the well-known quadratic formula:
−b±
√
b2 −4ac
2a
Your program will prompt the user for the values, a,b,c and output each real root.
However, for “invalid” input (a= 0 or values that would result in complex roots), the
program will instead output a message that informs the user why that the inputs are
invalid (with a specific reason).
Exercise 2.10. One of Ohm’s laws can be used to calculate the amount of power in
Watts (the rate of energy conversion; 1 joule per second) in terms of Amps (a measure of
current, 1 amp = 6 .241 ×1018 electrons per second) and Ohms (a measure of electrical
resistance). Specifically:
W = A2 ·O
Develop a simple program to read in two of the terms from the user and output the third.
53",ComputerScienceOne.pdf
"2. Basics
Exercise 2.11. Ohm’s Law models the current through a conductor as follows:
I = V
R
where V is the voltage (in volts), R is the resistance (in Ohms) and I is the current (in
amps). Write a program that, given two of these values computes the third using Ohm’s
Law.
The program should work as follows: it prompts the user for units of the first value:
the user should be prompted to enter V, R, or I and should then be prompted for the
value. It should then prompt for the second unit (same options) and then the value. The
program will then output the third value depending on the input. An example run of
the program:
Current Calculator
==============
Enter the first unit type (V, R, I): V
Enter the voltage: 25.75
Enter the second unit type (V, R, I): I
Enter the current: 72
The corresponding resistance is 0.358 Ohms
Exercise 2.12. Consider the following linear system of equations in two unknowns:
ax+ by = c
dx+ ey = f",ComputerScienceOne.pdf
"The corresponding resistance is 0.358 Ohms
Exercise 2.12. Consider the following linear system of equations in two unknowns:
ax+ by = c
dx+ ey = f
Write a program that prompts the user for the coefficients in such a system (prompt for
a,b,c,d,e,f ). Then output a solution to the system (the values for x,y). Take care to
handle situations in which the system is inconsistent.
Exercise 2.13. The surface area of a sphere of radius r is
4πr2
and the volume of a sphere with radius r is
4
3πr3
Write a program that prompts the user for a radius r and outputs the surface area
and volume of the corresponding sphere. If the radius entered is invalid, print an error
message and exit. Your output should look something like the following.
54",ComputerScienceOne.pdf
"2.7. Exercises
Sphere Statistics
=================
Enter radius r: 2.5
area: 78.539816
volume: 65.449847
Exercise 2.14. Write a program that prompts for the latitude and longitude of two
locations (an origin and a destination) on the globe. These numbers are in the range
[−180,180] (negative values correspond to the western and southern hemispheres). Your
program should then compute the air distance between the two points using the Spherical
Law of Cosines. In particular, the distance d is
d= arccos (sin(ϕ1) sin(ϕ2) + cos(ϕ1) cos(ϕ2) cos(∆))·R
•ϕ1 is the latitude of location A, ϕ2 is the latitude of location B
•∆ is the difference between location B’s longitude and location A’s longitude
•R is the (average) radius of the earth, 6,371 kilometers
Note: the formula above assumes that latitude and longitude are measured in radians r,
−π≤r≤π. See Exercise 2.6 for how to convert between them. Your program output
should look something like the following.
City Distance
========================",ComputerScienceOne.pdf
"should look something like the following.
City Distance
========================
Enter latitude of origin: 41.9483
Enter longitude of origin: -87.6556
Enter latitude of destination: 40.8206
Enter longitude of destination: -96.7056
Air distance is 764.990931
Exercise 2.15. Write a program that prompts the user to enter in a number of days.
Your program should then compute the number of years, weeks, and days that number
represents. For this exercise, ignore leap years (thus all years are 365 days). Your output
should look something like the following.
Day Converter
=============
55",ComputerScienceOne.pdf
"2. Basics
Enter number of days: 1000
That is
2 years
38 weeks
4 days
Exercise 2.16. The derivative of a function f(x) can be estimated using the difference
function:
f′(x) ≈f(x+ ∆x) −f(x)
∆x
That is, this gives us an estimate of the slope of the tangent line at the point x. Write a
program that prompts the user for an x value and a ∆x value and outputs the value of
the difference function for all three of the following functions:
f(x) = x2
f(x) = sin( x)
f(x) = ln( x)
Your output should looks something like the following.
Derivative Approximation
===================
Enter x: 2
Enter delta-x: 0.1
(x^2)' ~= 4.100000
sin'(x) ~= -0.460881
ln'x(x) ~= 0.487902
In addition, your program should check for invalid inputs: ∆ x cannot be zero, and ln(x)
is undefined for x≤0. If given invalid inputs, appropriate error message(s) should be
output instead.
Exercise 2.17.Write a program that prompts the user to enter two points in the plane,",ComputerScienceOne.pdf
"output instead.
Exercise 2.17.Write a program that prompts the user to enter two points in the plane,
(x1,y1) and (x2,y2) which define a line segment ℓ. Your program should then compute and
output an equation for the perpendicular line intersecting the midpoint of ℓ. You should
take care that invalid inputs (horizontal or vertical lines) are handled appropriately. An
example run of your program would look something like the following.
56",ComputerScienceOne.pdf
"2.7. Exercises
Perpendicular Line
====================
Enter x1: 2.5
Enter y1: 10
Enter x2: 3.5
Enter y2: 11
Original Line:
y = 1.0000 x + 7.5000
Perpendicular Line:
y = -1.0000 x + 13.5000
Exercise 2.18.Write a program that computes the total for a bill. The program should
prompt the user for a sub-total. It should then prompt whether or not the customer is
entitled to an employee discount (of 15%) by having them enter 1 for yes, 2 for no. It
should then compute the new sub-total and apply a 7.35% sales tax, and print the receipt
details along with the grand total. Take care that you properly round each operation.
An example run of the program should look something like the following.
Please enter a sub-total: 100
Apply employee discount (1=yes, 2=no)? 1
Receipt
========================
Sub-Total $ 100.00
Discount $ 15.00
Taxes $ 6.25
Total $ 91.25
Exercise 2.19.The ROI (Return On Investment) is computed by the following formula:
ROI = Gain from Investment −Cost of Investment",ComputerScienceOne.pdf
"Total $ 91.25
Exercise 2.19.The ROI (Return On Investment) is computed by the following formula:
ROI = Gain from Investment −Cost of Investment
Cost of Investment
Write a program that prompts the user to enter the cost and gain (how much it was sold
for) from an investment and computes and outputs the ROI. For example, if the user
enters $100,000 and $120,000 respectively, the output look similar to the following.
57",ComputerScienceOne.pdf
"2. Basics
Cost of Investment: $100000.00
Gain of Investment: $120000.00
Return on Investment: 20.00%
Exercise 2.20. Write a program to compute the real cost of driving. Gas mileage (in
the US) is usually measured in miles per gallon but the real cost should be measured in
how much it costs to drive a mile, that is, dollars per mile. Write a program to assist a
user in figuring out the real cost of driving. Prompt the user for the following inputs.
•Beginning odometer reading
•Ending odometer reading
•Number of gallons it took to fill the tank
•Cost of gas in dollars per gallon
For example, if the user enters 50,125, 50,430, 10 (gallons), and $3.25 (per gallon), then
your output should be something like the following.
Miles driven: 305
Miles per gallon: 30.50
Cost per mile: $0.11
Exercise 2.21. A bearing can be measured in degrees on the scale of [0 ,360) with 0◦
being due north, 90 ◦ due east, etc. The (initial) directional bearing from location A to",ComputerScienceOne.pdf
"being due north, 90 ◦ due east, etc. The (initial) directional bearing from location A to
location B can be computed using the following formula.
θ= atan2
(
sin(∆) ·cos(ϕ2), cos(ϕ1) ·sin(ϕ2) −sin(ϕ1) ·cos(ϕ2) cos(∆)
)
Where
•ϕ1 is the latitude of location A
•ϕ2 is the latitude of location B
•∆ is the difference between location B’s longitude and location A’s longitude
•atan2 is the two-argument arctangent function
Note: the formula above assumes that latitude and longitude are measured in radians r,
−π <r <π. To convert from degrees d (−180 <d< 180) to radians r, you can use the
58",ComputerScienceOne.pdf
"2.7. Exercises
simple formula:
r= d
180π
Write a program to prompt a user for a latitude/longitude of two locations (an origin and
a destination) and computes the directional bearing (in degrees) from the origin to the
destination. For example, if the user enters: 40 .8206,−96.7056 (40.8206◦ N, 96.7056◦ W)
and 41.9483,−87.6556 (41.9483◦N, 87.6556◦W), your program should output something
like the following.
From (40.8206, -96.7056) to (41.9483, -87.6556):
bearing 77.594671 degrees
Exercise 2.22. General relativity tells us that time is relative to your velocity. As you
approach the speed of light ( c = 299,792 km/s), time slows down relative to objects
traveling at a slower velocity. This time dilation is quantified by the Lorentz equation
t′= t√
1 −v2
c2
Where t is the time duration on the traveling space ship and t′ is the time duration on
the (say) Earth.
For example, if we were traveling at 50% the speed of light relative to Earth, one hour in",ComputerScienceOne.pdf
"the (say) Earth.
For example, if we were traveling at 50% the speed of light relative to Earth, one hour in
our space ship ( t= 1) would correspond to
t′= 1√
1 −(.5)2
= 1.1547
hours on Earth (about 1 hour, 9.28 minutes).
Write a program that prompts the user for a velocity which represents the percentage
p of the speed of light (that is, p= v
c) and a time duration t in hours and outputs the
relative time duration on Earth.
For example, if the user enters 0 .5 and 1 respectively as in our example, it should output
something like the following:
Traveling at 1 hour(s) in your space ship at
50.00% the speed of light, your friends on
Earth would experience:
1 hour(s)
9.28 minute(s)
59",ComputerScienceOne.pdf
"2. Basics
Your output should be able to handle years, weeks, days, hours, and minutes. So if the
user inputs something like 0 .9999 and 168, your output should look something like:
Traveling at 168.00 hour(s) in your space ship at
99.99% the speed of light, your friends on
Earth would experience:
1 year(s)
18 week(s)
3 day(s)
17 hour(s)
41.46 minute(s)
Exercise 2.23.Radioactive isotopes decay into other isotopes at a rate that is measured
by a half-life, H. For example, Strontium-90 has a half-life of 28.79 years. If you started
with 10 kilograms of Strontium-90, 28.79 years later you would have only 5 kilograms
(with the remaining 5 kilograms being Yttrium-90 and Zirconium-90, Strontium-90’s
decay products).
Given a mass mof an isotope with half-life H we can determine how much of the isotope
remains after y years using the formula,
r= m·
(1
2
)(y/H)
For example, if we have m= 10 kilograms of Strontium-90 with H = 28.79, after y= 2
years we would have
r= 10 ·
(1
2
)(2/28.79)
= 9.5298",ComputerScienceOne.pdf
"(1
2
)(y/H)
For example, if we have m= 10 kilograms of Strontium-90 with H = 28.79, after y= 2
years we would have
r= 10 ·
(1
2
)(2/28.79)
= 9.5298
kilograms of Strontium-90 left.
Write a program that prompts the user for an amount m (mass, in kilograms) of an
isotope and its half-life H as well as a number of years y and outputs the amount of
the isotope remaining after y years. For the example above your output should look
something like the following.
Starting with 10.00kg of an isotope with half-life
28.79 years, after 2.00 years you would have
9.5298 kilograms left.
60",ComputerScienceOne.pdf
"2.7. Exercises
Exercise 2.24.In sports, the magic number is a number that indicates the combination
of the number of games that a leader in a division must win and/or the 2nd place team
must lose for the leader to clinch the division. The magic number can be computed using
the following formula:
G+ 1 −WA −LB
where G is the number of games played in the season, WA is the number of wins the
leader currently has and LB is the number of losses the 2nd place team currently has.
Write a program that prompts the user to enter:
•G, the total number of games in the season
•WA the number of wins of the leading team
•LA the number of losses of the leading team
•WB the number of wins of the second place team
•LB the number of losses of the second place
Your program will then output the current win percentages of both teams, the magic
number of the leading team as well as the percentage of the remaining games that must",ComputerScienceOne.pdf
"number of the leading team as well as the percentage of the remaining games that must
go in team A’s favor to satisfy the magic number (for this, we will assume A and B do
not play each other).
For example, if a user enters the values 162, 96, 58, 93, 62, the output should look
something like the following.
Team Wins Loss Perc Magic Number
A 96 58 62.34% 5
B 93 62 60.00%
With 15 total games remaining, 33.33% must go in Team A's favor
Exercise 2.25. The Doppler Effect is a change in the observed spectral frequency of
light when objects in space are moving toward or away from us. If the spectral frequency
of the object is known, then the relative velocity can be estimated based on either the
blue-shift (for objects traveling toward the observer) or the red-shift (for objects traveling
away from the observer) of the observed frequency.
The blue-shift equation to determine velocity is given by
vb = c
(λ
λb
−1
)
61",ComputerScienceOne.pdf
"2. Basics
The red-shift equation to determine velocity is given by
va = c
(
1 − λ
λr
)
where
•c is the speed of light (299,792.458 km/s)
•λ is the actual spectral line of the object (ex: hydrogen is 434nm)
•λr is the observed (red-shifted) spectral line and λb is the observed (blue-shifted)
spectral line
Write a program that prompts the user to enter which spectral shift they want to compute
(1 for blue-shift, 2 for red-shift). The program should then prompt for the spectral line
of an object and the observed (shifted) frequency and output the velocity of the distant
object. For example, if a user enters the values 1 (blue-shifted), 434 (nm) and 487 (nm),
the output should look something like the following.
Spectral Line: 434nm
Observed Line: 487nm
Relative Velocity: 32626.28 km/s
Exercise 2.26. Radiometric dating is a technique by which the age of rocks, minerals,
and other material can be estimated by measuring the proportion of radioactive isotopes",ComputerScienceOne.pdf
"and other material can be estimated by measuring the proportion of radioactive isotopes
it still has to its decay products. It can be computed with the following formula:
D= D0 + N(eλt −1)
where
•t is age of the sample,
•D is number of atoms of the daughter isotope in the sample,
•D0 is number of atoms of the daughter isotope in the original composition,
•N is number of atoms of the parent isotope in the sample
•λ is the decay constant of the parent isotope,
λ= ln 2
t1/2
where t1/2 is the half-life of the parent isotope (in years).
62",ComputerScienceOne.pdf
"2.7. Exercises
Write a program that prompts the user to enter D,D0,N, and t1/2 and computes the
approximate age of the material, t.
For example, if the user were to enter 150, 50, 300, 28.8 (Strontium-90’s half-life) then
the program should output something like the following.
The sample appears to be 11.953080 years old.
Exercise 2.27.Suppose you have two circles each centered at (0,0) and (d,0) with radii
of R,r respectively. These circles may intersect at two points, forming an asymmetric
“lens” as in Figure 2.4.
The area of this lens can be computed using the following formula:
A= r2 cos−1
(d2 + r2 −R2
2dr
)
+ R2 cos−1
(d2 + R2 −r2
2dR
)
−
1
2
√
(−d+ r+ R)(d+ r−R)(d−r+ R)(d+ r+ R)
Write a program that prompts the user for the two radii and the x-offset dand computes
the area of this lens. Your program should handle the special cases where the two circles
do not intersect and when they intersect at a single point (the area is zero).
Figure 2.4.: Intersection of two circles.
63",ComputerScienceOne.pdf
"3. Conditionals
When writing code, its important to be able to distinguish between one or more situations.
Based on some condition being true or false, you may want to perform some action if
its true, while performing another, different action if it is false. Alternatively, you may
simply want to perform one action if and only if the condition is true, and do nothing
(move forward in your program) if it is false.
Normally, the control flow of a program is sequential: each statement is executed top-to-
bottom one after the other. A conditional statement (sometimes called selection control
structures) interrupts this normal control flow and executes statements only if some
specified condition holds. The usual way of achieving this in a programming language is
through the use conditional statements such as the if statement, if-else statement, and
if-else-if statement.
By using conditional statements, we can design more expressive programs whose behavior",ComputerScienceOne.pdf
"if-else-if statement.
By using conditional statements, we can design more expressive programs whose behavior
depends on their state: if the value of some variable is greater than some threshold, we
can perform action A, otherwise, we can perform action B. You do this on a daily basis
as you make decisions for yourself. At a cafe you may want to purchase the grande coffee
which costs $2. If you have $2 or more, then you’ll buy it. Otherwise, if you have less
than $2, you can settle for the normal coffee which costs $1. Yet still, if you have less
than $1 you’ll not be able to make a purchase. The value of your pocket book determines
your decision and subsequent actions that you take.
Similarly, our programs need to be able to “make decisions” based on various conditions
(they don’t actually make decisions for themselves as computer are not really “intelligent”,
we are simply specifying what should occur based on the conditions). Conditions in a",ComputerScienceOne.pdf
"we are simply specifying what should occur based on the conditions). Conditions in a
program are specified by coding logical statements using logical operators.
3.1. Logical Operators
In logic, everything is black and white: a logical statement is an assertion that is either
true or it is false. As previously discussed, some programming languages allow you to
define and use Boolean variables that can be assigned the value true or false. We can
also formulate statements that involve other types of variables whose truth values are
determined by the values of the variables at run time.
65",ComputerScienceOne.pdf
"3. Conditionals
3.1.1. Comparison Operators
Suppose we have a variable age representing the age of an individual. Suppose we wish
to execute some code if the person is an adult, age≥18 and a different piece of code if
they are not an adult, age< 18. To achieve this, we need to be able to make comparisons
between variables, constants, and even more complex expressions. Such logical statements
may not have a fixed truth value. That is, they could be true or false depending on the
value of the variables involved when the program is run.
Such comparisons are common in mathematics and likewise in programming languages.
Comparison operators are usually binary operators in that they are applied to two
operands: a left operand and a right operand. For example, if a,b are variables (or
constants or expressions), then the comparison,
a≤b
is true if the value stored in a is less than or equal to the value stored in b. Otherwise, if",ComputerScienceOne.pdf
"constants or expressions), then the comparison,
a≤b
is true if the value stored in a is less than or equal to the value stored in b. Otherwise, if
the value stored in b is strictly less than the value stored in a, the expression is false.
Further, a,b are the operands and ≤is the binary operator.
In general, operators do not commute. That is,
a≤b and b≤a
are not equivalent, just as they are not in mathematics. However,
a≤b and b≥a
are equivalent. Thus, the order of operands is important and can change the meaning
and truth value of an expression.
A full listing of binary operators can be found in Table 3.1. In this table, we present both
the mathematical notation used in our pseudocode examples as well as the most common
ways of representing these comparison operators in most programming languages. The
need for alternative representations is because the mathematical symbols are not part of
the ASCII character set common to most keyboards.",ComputerScienceOne.pdf
"need for alternative representations is because the mathematical symbols are not part of
the ASCII character set common to most keyboards.
When using comparison operators, either operand can be variables, constants, or even
more complex expressions. For example, you can make comparisons between two variables,
a<b, a>b, a ≤b, a ≥b, a = b, a ̸= b
or they can be between a variable and a constant
a< 10, a> 10, a ≤10, a ≥10, a = 10, a ̸= 10
or
10 <b, 10 >b, 10 ≤b, 10 ≥b, 10 = b, 10 ̸= b
66",ComputerScienceOne.pdf
"3.1. Logical Operators
Psuedocode Code Meaning Type
< < less than relational
> > greater than relational
≤ <= less than or equal to relational
≥ >= greater than or equal to relational
= == equal to equality
̸= != not equal to equality
Table 3.1.: Comparison Operators
Comparisons can also be used with more complex expressions such as
√
b2 −4ac <0
which could commonly be expressed in code as
sqrt(b*b - 4*a*c) < 0
Observe that both operands could be constants, such as 5 ≤10 but there would be little
point. Since both are constants, the truth value of the expression is already determined
before the program runs. Such an expression could easily be replaced with a simple true
or false variable. These are referred to as tautologies and contradictions respectively.
We’ll examine them in more detail below.
Pitfalls
Sometimes you may want to check that a variable falls within a certain range. For
example, we may want to test that xlies in the interval [0,10] (between 0 and 10 inclusive",ComputerScienceOne.pdf
"example, we may want to test that xlies in the interval [0,10] (between 0 and 10 inclusive
on both ends). Mathematically we could express this as
0 ≤x≤10
and in code, we may try to do something like
0 <= x <= 10
However, when used in code, the operators <= are binary and must be applied to two
operands. In a language the first inequality, 0 <= x would be evaluated and would
result in either true or false. The result is then used in the second comparison which
results in a question such as true ≤10 or false ≤10.
Some languages would treat this as a syntax error and not allow such an expression to
be compiled since you cannot compare a Boolean value to a numerical value. However,
other languages may allow this, typically representing true with some nonzero value such
as 1 and false with 0. In either case, the expression would evaluate to true since both
67",ComputerScienceOne.pdf
"3. Conditionals
0 ≤10 and 1 ≤10. However, this is clearly wrong: if xhad a value of 20 for example, the
first expression would evaluate to false, making the entire expression true, but 20 ̸≤10.
The solution is to use logical operators to express the same logic using two comparison
operators (see Section 3.1.3).
Another common pitfall when programming is to mistake the assignment operator
(typically only one equals sign, =) and the equality operator (typically two equal signs,
==). As before, some languages will not allow it. The expression a = 10 would not have
a truth value associated with it. Attempts to use the expression in a logical statement
would be a syntax error. Other languages may permit the expression and would give it a
truth value equal to the value of the variable. For example, a = 10 would take on the
value 10 and be treated as true (nonzero value) while a = 0 would take on the value 0",ComputerScienceOne.pdf
"value 10 and be treated as true (nonzero value) while a = 0 would take on the value 0
and be treated as false (zero). In either case, we probably do not get the result that we
want. Take care that you use proper equality comparison operators.
Other Considerations
The comparison operators that we’ve examined are generally used for comparing numerical
types. However, sometimes we wish to compare non-numeric types such as single
characters or strings. Some languages allow you to use numeric operators with these
types as well.
Some dynamically typed languages (PHP, JavaScript, etc.) have additional rules when
comparison operators are used with mixed types (that is, we compare a string with a
numeric type). They may even have additional “strict” comparison operators such as
(a === b) and (a !== b) which are true only if the values and types match. So, for
example, (10 == ""10"") may be true because the values match, but (10 === ""10"")",ComputerScienceOne.pdf
"example, (10 == ""10"") may be true because the values match, but (10 === ""10"")
would be false since the types do not match (one is an integer, the other a string). We
discuss specifics in subsequent chapters are they pertain to specific languages.
3.1.2. Negation
The negation operator is an operator that “flips” the truth value of the expression that
it is applied to. It is very much like the numerical negation operator which when applied
to positive numbers results in their negation and vice versa. When the logical negation
operator is applied to a variable or statement, it negates its truth value. If the variable
or statement was true, its negation is false and vice versa.
Also like the numerical negation operator, the logical negation operator is a unary
operator as it applies to only one operand. In modern logic, the symbol ¬is used to
68",ComputerScienceOne.pdf
"3.1. Logical Operators
a ¬a
false true
true false
Table 3.2.: Logical Negation, ¬Operator
denote the negation operator 1, examples:
¬p, ¬(a> 10), ¬(a≤b)
We will adopt this notation in our pseudocode, however most programming languages use
the exclamation mark, ! for the negation operator, similar to its usage in the inequality
comparison operator, !=. The negation operator applies to the variable or statement
immediately following it, thus
¬(a≤b) and ¬a≤b
are not the same thing (indeed, the second expression may not even be valid depending
on the language). Further, when used with comparison operators, it is better to use the
“opposite” comparison. For example,
¬(a≤b) and ( a>b )
are equivalent, but the second expression is preferred as it is simpler. Likewise,
¬(a= b) and ( a̸= b)
are equivalent, but the second expression is preferred.
3.1.3. Logical And
The logical and operator (also called a conjunction) is a binary operator that is true if",ComputerScienceOne.pdf
"3.1.3. Logical And
The logical and operator (also called a conjunction) is a binary operator that is true if
and only if both of its operands is true. If one of its operands is false, or if both of them
are false, then the result of the logical and is false.
Many programming languages use two ampersands, a && b to denote the logical And
operator.2 However, for our pseudocode we will adopt the notation And and we will
use expressions such as aAnd b. Table 3.3 contains a truth table representation of the
logical And operator.
1 This notation was first used by Heyting, 1930 [ 16]; prior to that the tilde symbol was used ( ∼p for
example) by Peano [33] and Whitehead & Russell [ 37]. However, the tilde operator has been adopted
to mean bit-wise negation in programming languages.
2In logic, the “wedge” symbol, p∧q is used to denote the logical And. It was first used again by
Heyting, 1930 [16] but should not be confused for the keyboard caret, ˆ, symbol. Many programming",ComputerScienceOne.pdf
"Heyting, 1930 [16] but should not be confused for the keyboard caret, ˆ, symbol. Many programming
languages do use the caret as an operator, but it is usually the exclusive-or operator which is true if
and only if exactly one of its operands is true.
69",ComputerScienceOne.pdf
"3. Conditionals
a b aAnd b
false false false
false true false
true false false
true true true
Table 3.3.: Logical And Operator
The logical And is used to combine logical statements to form more complex logical
statements. Recall that we couldn’t directly use two comparison operators to check that
a variable falls within a range, 0 ≤x≤10. However, we can now use a logical And to
express this:
(0 ≤x) And (x≤10)
This expression is true only if both comparisons are true.
Though the And operator is a binary operator, we can write statements that involve
more than one variable or expression by using multiple instances of the operator. For
example,
b2 −4ac≥0 And a̸= 0 And c> 0
The above statement would be evaluated left-to-right; the first two operands would be
evaluated and the result would be either true or false. Then the result would be used as
the first operand of the second logical And. In this case, if any of the operands evaluated",ComputerScienceOne.pdf
"the first operand of the second logical And. In this case, if any of the operands evaluated
to false, the entire expression would be false. Only if all three were true would the
statement be true.
3.1.4. Logical Or
The logical or operator is the binary operator that is true if at least one of its operands
is true. If both of its operands are false, then the logical or is false. This is in contrast
to what is usually meant by “or” colloquially. If someone says “you can have cake
or ice-cream,” usually they implicitly also mean, “but not both.” With the logical or
operator, if both operands are true, the result is still true.
Many programming languages use two vertical bars (also referred to as Sheffer strokes ),
|| to denote the logical Or operator.3. However, for our pseudocode we will adopt the
notation Or, thus the logical or can be expressed as aOr b. Table 3.4 contains a truth
table representation of the logical Or operator.",ComputerScienceOne.pdf
"notation Or, thus the logical or can be expressed as aOr b. Table 3.4 contains a truth
table representation of the logical Or operator.
As with the logical And, the logical Or is used to combine logical statements to make
3In logic, the “vee” symbol, p∨q is used to denote the logical Or. It was first used by Russell, 1906
[35].
70",ComputerScienceOne.pdf
"3.1. Logical Operators
a b aOr b
false false false
false true true
true false true
true true true
Table 3.4.: Logical Or Operator
more complex statements. For example,
(age≥18) Or (year = “senior”)
which is true if the individual is aged 18 or older, is a senior, or is both 18 or older and
a senior. If the individual is aged less than 18 and is not a senior, then the statement
would be false.
We can also write statements with multiple Or operators,
a>b Or b>c Or a>c
which will be evaluated left-to-right. If any of the three operands is true, the statement
will be true. The statement is only false when all three of the operands is false.
3.1.5. Compound Statements
The logical And and Or operators can be combined to express even more complex logical
statements. For example, you can express the following statements involving both of the
operators:
aAnd (bOr c) aOr (bAnd c)
As an example, consider the problem of deciding whether or not a given year is a leap",ComputerScienceOne.pdf
"operators:
aAnd (bOr c) aOr (bAnd c)
As an example, consider the problem of deciding whether or not a given year is a leap
year. The Gregorian calendar defines a year as a leap year if it is divisible by 4. However,
every year that is divisible by 100 is not a leap year unless it is also divisible by 400.
Thus, 2012 is a leap year (4 goes into 2012 503 times), however, 1900 was not a leap year:
though it is divisible by 4 (1900 /4 = 475 with no remainder), it is also divisible by 100.
The year 2000 was a leap year: it was divisible by 4 and 100 thus it was divisible by 400.
When generalizing these rules into logical statements we can follow a similar process: A
year is a leap year if it is divisible by 400 or it is divisible by 4 and not by 100. This
logic can be modeled with the following expression.
year mod 400 = 0 Or (year mod 4 = 0 And year mod 100 ̸= 0)
When writing logical statements in programs it is generally best practice to keep things",ComputerScienceOne.pdf
"year mod 400 = 0 Or (year mod 4 = 0 And year mod 100 ̸= 0)
When writing logical statements in programs it is generally best practice to keep things
simple. Logical statements should be written in the most simple and succinct (but
correct) way possible.
71",ComputerScienceOne.pdf
"3. Conditionals
Tautologies and Contradictions
Some logical statements have the same meaning regardless of the variables involved. For
example,
aOr ¬a
is always true regardless of the value of a. To see this, suppose that a is true, then the
statement becomes
aOr ¬a= true Or false
which is true. Now suppose that a is false, then the statement is
aOr ¬a= false Or true
which again is true. A statement that is always true regardless of the truth values of its
variables is a tautology.
Similarly, the statement
aAnd ¬a
is always false (at least one of the operands will always be false). A statement that is
always false regardless of the truth values of its variables is a contradiction.
In most cases, it is pointless to program a conditional statement with tautologies or
contradictions: if an if-statement is predicated on a tautology it will always be executed.
Likewise, an if-statement involved with a contradiction will never be executed. In either",ComputerScienceOne.pdf
"Likewise, an if-statement involved with a contradiction will never be executed. In either
case, many compilers or code analysis tools may indicate and warn about these situations
and encourage you to modify the code or to remove “dead code.” Some languages may
not even allow you write such statements.
There are always exceptions to the rule. Sometimes you may wish to intentionally write
an infinite loop (see Section 4.5.2) for example in which case a statement similar to the
following may be written.
1 while true do
//some computation
2 end
De Morgan’s Laws
Another tool to simplify your logic is De Morgan’s Laws. When a logical And statement
is negated, it is equivalent to an unnegated logical Or statement and vice versa. That is,
¬(aAnd b) and ¬aOr ¬b
72",ComputerScienceOne.pdf
"3.1. Logical Operators
Order Operator
1 ¬
2 And
3 Or
Table 3.5.: Logical Operator Order of Precedence
are equivalent to each other;
¬(aOr b) and ¬aAnd ¬b
are also equivalent to each other. Though equivalent, it is generally preferable to write
the simpler statement. From one of our previous examples, we could write
¬((0 ≤x) And (x≤10))
or we could apply De Morgan’s Law and simplify this to
(0 >x) Or (x> 10)
which is more concise and arguably more readable.
Order of Precedence
Recall that numerical operators have a well defined order of precedence that is taken from
mathematics (multiplication is performed before addition for example, see Section 2.3.4).
When working with logical operators, we also have an order of precedence that somewhat
mirrors those of numerical operators. In particular, negations are always applied first,
followed by And operators, and then lastly Or operators.
For example, the statement
aOr bAnd c",ComputerScienceOne.pdf
"followed by And operators, and then lastly Or operators.
For example, the statement
aOr bAnd c
is somewhat ambiguous. We don’t just evaluate it left-to-right since the And operator
has a higher order of precedence (this is similar to the mathematical expression a+ b·c
where the multiplication would be evaluated first). Instead, this statement would be
evaluated by evaluating the And operator first and then the result would be applied to
the Or operator. Equivalently,
aOr (bAnd c)
If we had meant that the Or operator should be evaluated first, then we should have
explicitly written parentheses around the operator and its operands like
(aOr b) And c
73",ComputerScienceOne.pdf
"3. Conditionals
In fact, its best practice to write parentheses even if it is not necessary. Writing
parentheses is often clearer and easier to read and more importantly communicates intent.
By writing
aOr (bAnd c)
the intent is clear: we want the And operator to be evaluated first. By not writing the
parentheses we leave our meaning somewhat ambiguous and force whoever is reading the
code to recall the rules for order of precedence. By explicitly writing parentheses, we
reduce the chance for error both in writing and in reading. Besides, its not like we’re
paying by the character.
For similar operators of the same precedence, they are evaluated left-to-right, thus
aOr bOr c is equivalent to (( aOr b) Or c)
and
aAnd bAnd c is equivalent to (( aAnd b) And c)
3.1.6. Short Circuiting
Consider the following statement:
aAnd b
As we evaluate this statement, suppose that we find that a is false. Do we need to",ComputerScienceOne.pdf
"3.1.6. Short Circuiting
Consider the following statement:
aAnd b
As we evaluate this statement, suppose that we find that a is false. Do we need to
examine the truth value of b? The answer is no: since a is false, regardless of the truth
value of b, the statement is false because it is a logical And. Both operands must be
true for an And to be true. Since the first is false, the second is irrelevant.
Now imagine evaluating this statement in a computer. If the first operand of an And
statement is false, we don’t need to examine/evaluate the second. This has some potential
for improved efficiency: if the second operand does not need to be evaluated, a program
could ignore it and save a few CPU cycles. In general, the speed up for most operations
would be negligible, but in some cases the second operand could be very “expensive” to
compute (it could be a complex function call, require a database query to determine,
etc.) in which case it could make a substantial difference.",ComputerScienceOne.pdf
"compute (it could be a complex function call, require a database query to determine,
etc.) in which case it could make a substantial difference.
Historically, avoiding even a few operations in old computers meant a difference on
the order of milliseconds or even seconds. Thus, it made sense to avoid unnecessary
operations. This is now known as short circuiting and to this day is still supported in
most programming languages.4 Though the differences are less stark in terms of CPU
resources, most developers and programmers have come to expect this behavior and write
statements under the assumption that short-circuiting will occur.
4 Historically, the short-circuited version of the And operator was known as McCarthy’s sequential
conjunction operationwhich was formally defined by John McCarthy (1962) as “if p then q, else
false”, eliminating the evaluation of q if p is false [23].
74",ComputerScienceOne.pdf
"3.2. The If Statement
Short circuiting is commonly used to “check” for invalid operations. This is commonly
used to prevent invalid operations. For example, consider the following statement:
(d̸= 0 And 1/d> 1)
The first operand is checking to see if d is not zero and the second checks to see if its
reciprocal is greater than 1. With short-circuiting, if d = 0, then the second operand
will not be evaluated and the division by zero will be prevented. If d ̸= 0 then the
first operand is true and so the second operand will be evaluated as normal. Without
short-circuiting, both operands would be evaluated leading to a division by zero error.
There are many other common patterns that rely on short-circuiting to avoid invalid or
undefined operations. For example, short-circuiting is used to check that a variable is
valid (defined or not Null) before using it, or to check that an index variable is within
the range of an array’s size before accessing a value.",ComputerScienceOne.pdf
"valid (defined or not Null) before using it, or to check that an index variable is within
the range of an array’s size before accessing a value.
Because of short-circuiting, the logical And is effectively not commutative. An operator
is commutative if the order of its operands is irrelevant. For example, addition and
multiplication are both commutative,
x+ y= y+ x x ·y= y·x
but subtraction and division are not,
x−y̸= y−x x/y ̸= y/x
In logic, the And and Or operators are commutative, but when used in most programming
languages they are not,
(aAnd b) ̸= (bAnd a) and ( aOr b) ̸= (bOr a)
It is important to emphasize that they are still logically equivalent, but they are not
effectively equivalent: because of short-circuiting, each of these statements have a
potentially different effect.
The Or operator is also short-circuited: if the first operand is true, then the truth value
of the expression is already determined to be true and so the second operand will not be
evaluated. In the expression,",ComputerScienceOne.pdf
"of the expression is already determined to be true and so the second operand will not be
evaluated. In the expression,
aOr b
if a evaluates to true, then b is not evaluated (since if either operand is true, the entire
expression is true).
3.2. The If Statement
Normally, the flow of control (or control flow) in a program is sequential. Each instruction
is executed, one after the other, top-to-bottom and in individual statements left-to-right
75",ComputerScienceOne.pdf
"3. Conditionals
just as one reads in English. Moreover, in most programming languages, each statement
executes completely before the next statement begins. A visualization of this sequential
control flow can be found in the control flow diagram in Figure 3.1(a).
However, it is often necessary for a program to “make decisions.” Some segments of code
may need to be executed only if some condition is satisfied. The if statement is a control
structure that allows us to write a snippet of code predicated on a logical statement.
The code executes if the logical statement is true, and does not execute if the logical
statement is false. This control flow is featured in Figure 3.1(b)
An example using pseudocode can be found in Algorithm 3.1. The use of the keyword
“if” is common to most programming languages. The logical statement associated with
the if-statement immediately follows the “if” keyword and is usually surrounded by",ComputerScienceOne.pdf
"the if-statement immediately follows the “if” keyword and is usually surrounded by
parentheses. The code block immediately following the if-statement is bound to the
if-statement.
1 if (⟨condition⟩) then
2 Code Block
3 end
Algorithm 3.1:An if-statement
As in the flow chart, if the ⟨condition⟩evaluates to true, then the code block bound to
the statement executes in its entirety. Otherwise, if the condition evaluates to false, the
code block bound to the statement is skipped in its entirety.
A simple if-statement can be viewed as a “do this if and only if the condition holds.”
Alternatively, “if this condition holds do this, otherwise don’t.” In either case, once the
if-statement finishes execution, the program returns to the normal sequential control
flow.
3.3. The If-Else Statement
An if-statement allows you to specify a code segment that is executed or is not executed.
An if-else statement allows you to specify an alternative. An if-else statement allows you",ComputerScienceOne.pdf
"An if-else statement allows you to specify an alternative. An if-else statement allows you
to define a condition such that if the condition is true, one code block executes and if
the condition is false, an entirely different code block executes.
The control flow of an if-else statement is presented in Figure 3.2. Note that Code
Block A and Code Block B are mutually exclusive . That is, one and only one of them is
executed depending on the truth value of the ⟨condition⟩. A presentation of a generic
if-else statement in our pseudocode can be found in Algorithm 3.2
76",ComputerScienceOne.pdf
"3.3. The If-Else Statement
Statement 1
Statement 2
Statement 3
(a) Sequential Flow Chart
⟨condition⟩ Code
Block
Remaining
Program
true
false
(b) If-Statement Flow Chart
Figure 3.1.: Control flow diagrams for sequential control flow and an if-statement. In
sequential control, statements are executed one after the other as they are
written. In an if-statement, the normal flow of control is interrupted and a
Code Block is only executed if the given condition is true, otherwise it is
not. After the if-statement, normal sequential control flow resumes.
77",ComputerScienceOne.pdf
"3. Conditionals
⟨condition⟩ Code
BlockA
Code
BlockB
Remaining
Program
true
false
Figure 3.2.: An if-else Flow Chart
Just as with an if-statement, the keyword “if” is used. In fact, the if-statement is simply
just an if-else statement with the else block omitted (equivalently, we could have defined
an empty else block, but since it would have no effect, a simple if-statement with no
else block is preferred). It is common to most programming languages to use the “else”
keyword to denote the else block of code. Since there is only one ⟨condition⟩to evaluate
and it can only be true or false, it is not necessary to specify the conditions under which
the else block executes. It is assumed that if the ⟨condition⟩evaluates to false, the else
block executes.
As with an if-statement, the block of code associated with the if-statement as well as the
block of code associated with the else-statement are executed in their entirety or not at",ComputerScienceOne.pdf
"block of code associated with the else-statement are executed in their entirety or not at
all. Whichever block of code executes, normal flow of control returns and the remaining
program continues executing sequentially.
3.4. The If-Else-If Statement
An if-statement allows you to define a “do this or do not” and an if-else statement allows
you to define a “do this or do that” statement. Yet another generalization is an if-else-if
statement. Using such a statement you can define any number of mutually exclusive
code blocks.
78",ComputerScienceOne.pdf
"3.4. The If-Else-If Statement
1 if (⟨condition⟩) then
2 Code Block A
3 else
4 Code Block B
5 end
Algorithm 3.2:An if-else Statement
To illustrate, consider the case in which we have exactly three mutually exclusive
possibilities. At a particular university, there are three possible semesters depending
on the month. January through May is the Spring semester, June/July is the Summer
semester, and August through December is the Fall semester. These possibilities are
mutually exclusive because it cannot be both Spring and Summer at the same time
for example. Suppose we have the current month stored in a variable named month.
Algorithm 3.3 expresses the logic for determining which semester it is using an if-else-if
statement.
1 if (month≥January) And (month≤May) then
2 semester←“Spring”
3 else if(month> May) And (month≤July) then
4 semester←“Summer”
5 else
6 semester←“Fall”
7 end
Algorithm 3.3:Example If-Else-If Statement",ComputerScienceOne.pdf
"3 else if(month> May) And (month≤July) then
4 semester←“Summer”
5 else
6 semester←“Fall”
7 end
Algorithm 3.3:Example If-Else-If Statement
Let’s understand how this code works. First, the “if” and “else” keywords are used
just as the two previous control structures, but we are now also using the “else if”
keyword combination to specify an additional condition. Each condition, starting with
the condition associated with the if-statement is checked in order. If and when one of the
conditions is satisfied (evaluates to true), the code block associated with that condition
is executed and all other code blocks are ignored .
Each of the code blocks in an if-else-if control structure are mutually exclusive. One
and only one of the code blocks will ever execute. Similar to the sequential control flow,
the first condition that is satisfied is the one that is executed. If none of the conditions
is satisfied, then the code block associated with the else-statement is the one that is
executed.",ComputerScienceOne.pdf
"is satisfied, then the code block associated with the else-statement is the one that is
executed.
In our example, we only identified three possibilities. You can generalize an if-else-if
statement to specify as many conditions as you like. This generalization is depicted in
79",ComputerScienceOne.pdf
"3. Conditionals
⟨condition1⟩
⟨condition2⟩
⟨condition3⟩
...
⟨conditionn⟩
CodeBlockA
CodeBlockB
CodeBlockC
...
CodeBlockN
CodeBlockM
RemainingProgram
if(⟨condition1⟩)
else if(⟨condition2⟩)
else if(⟨condition3⟩)
else if(⟨conditionn⟩)
else
true
true
true
true
false
false
false
false
false
Figure 3.3.: Control Flow for an If-Else-If Statement. Each condition is evaluated in
sequence. The first condition that evaluates to true results in the corre-
sponding code block being executed. After executing, the program continues.
Thus, each code block is mutually exclusive : at most one of them is executed.
80",ComputerScienceOne.pdf
"3.4. The If-Else-If Statement
Algorithm 3.4 and visualized in Figure 3.3. Similar to the if-statement, the else-statement
and subsequent code block is optional. If omitted, then it may be possible that none of
the code blocks is executed.
1 if (⟨condition1⟩) then
2 Code Block A
3 else if(⟨condition2⟩) then
4 Code Block B
5 else if(⟨condition3⟩) then
6 Code Block C
7 ...
8 else
9 Code Block
10 end
Algorithm 3.4:General If-Else-If Statement
The design of if-else-if statements must be done with care to ensure that your statements
are each mutually exclusive and capture the logic you intend. Since the first condition
that evaluates to true is the one that is executed, the order of the conditions is important.
A poorly designed if-else-if statement can lead to bugs and logical errors.
As an example, consider describing the loudness of a sound by itsdecibellevel in Algorithm
3.5.
1 if decibel≤70 then
2 comfort ←“intrusive”
3 else ifdecibel≤50 then
4 comfort ←“quiet”
5 else ifdecibel≤90 then",ComputerScienceOne.pdf
"3.5.
1 if decibel≤70 then
2 comfort ←“intrusive”
3 else ifdecibel≤50 then
4 comfort ←“quiet”
5 else ifdecibel≤90 then
6 comfort ←“annoying”
7 else
8 comfort ←“dangerous”
9 end
Algorithm 3.5:If-Else-If Statement With a Bug
Suppose that decibel= 20 which should be described as a “quite” sound. However, in the
algorithm, the first condition, decibel≤70 evaluates to true and the sound is categorized
as “intrusive”. The bug is that the second condition, decibel≤50 should have come first
in order to capture all decibel levels less than or equal to 50.
Alternatively, we could have followed the example in Algorithm 3.3 and completely
81",ComputerScienceOne.pdf
"3. Conditionals
specified both lower bounds and upper bounds in our condition. For example, the
condition for “intrusive” could have been
(decibel> 50) And (decibel≤70)
However, doing this is unnecessary if we order our conditions appropriately and we can
potentially write simpler conditions if we remember the fact that the if-else-if statement
is mutually exclusive.
3.5. Ternary If-Else Operator
Another conditional operator is the ternary if-then-else operator. It is often used to
write an expression that can take on one of two values depending on the truth value of a
logical expression. Most programming languages support this operator which has the
following syntax:
E ? X : Y
Here, E is a Boolean expression. If E evaluates to true, the statement takes on the
value X which does not need to be a Boolean value: it can be anything (an integer,
string, etc.). If E evaluates to false, the statement takes on the value Y.
A simple usage of this expression is to find the minimum of two values:",ComputerScienceOne.pdf
"string, etc.). If E evaluates to false, the statement takes on the value Y.
A simple usage of this expression is to find the minimum of two values:
min = ( (a < b) ? a : b );
If a<b is true, then min will take on the value a. Otherwise it will take on the value b
(in which case a≥b and so b is minimal). Most programming languages support this
special syntax as it provides a nice convenience (yet another example of syntactic sugar).
3.6. Examples
3.6.1. Meal Discount
Consider the problem of computing a receipt for a meal. Suppose we have the subtotal
cost of all items in the meal. Further, suppose that we want to compute a discount
(senior citizen discount, student discount, or employee discount, etc.). We can then apply
the discount, compute the sales tax, and sum a total, reporting each detail to the user.
To do this, we first prompt the user to enter a subtotal. We can then ask the user if
there is a discount to be computed. If the user answers yes, then we again prompt them",ComputerScienceOne.pdf
"there is a discount to be computed. If the user answers yes, then we again prompt them
for an amount (to allow different types of discounts). Otherwise, the discount will be
82",ComputerScienceOne.pdf
"3.6. Examples
zero. We can then proceed to calculate each of the amounts above. To do this we’ll need
an if-statement. We could also use a conditional statement to check to see if the input
makes sense: we wouldn’t want a discount amount that is greater than 100%. The full
algorithm is presented in Algorithm 3.6.
1 Prompt the user for a subtotal
2 subTotal ←read input from user
3 discountPercent ←0
4 Ask the user if they want to apply a discount
5 hasDiscount←get user input if hasDiscount= “yes” then
6 Prompt the user for a discount amount
7 discountPercent ←read user input
8 end
9 if discountPercent> 100 then
10 Error! Discount cannot be more than 100%
11 end
12 discount←subTotal ×discountPercent
13 discountTotal ←subTotal −discount
14 tax←taxRate×discountTotal
15 grandTotal ←discountTotal + tax
16 output subTotal,discountTotal,tax,grandTotal to user
Algorithm 3.6:A simple receipt program
3.6.2. Look Before You Leap",ComputerScienceOne.pdf
"16 output subTotal,discountTotal,tax,grandTotal to user
Algorithm 3.6:A simple receipt program
3.6.2. Look Before You Leap
Recall that dividing by zero is an invalid operation in most programming languages (see
Section 2.3.5). Now that we have a means by which numerical values can be checked, we
can prevent such errors entirely.
Suppose that we were going to compute a quotient of two variables x/y. If y= 0, this
would be an invalid operation and lead to undefined, unexpected or erroneous behavior.
However, if we checked whether or not the denominator is zero before we compute the
quotient then we could prevent such errors. We present this idea in Algorithm 3.7.
1 if y̸= 0 then
2 q←x/y
3 end
Algorithm 3.7:Preventing Division By Zero Using an If Statement
83",ComputerScienceOne.pdf
"3. Conditionals
This approach to programming is known as defensive programming. We are essentially
checking the conditions for an invalid operation before performing that operation. In
the example above, we simply chose not to perform the operation. Alternatively, we
could use an if-else statement to perform alternate operations or handle the situation
differently. Defensive programming is akin to “looking before leaping”: before taking a
potentially dangerous step, you look to see if you are at the edge of a cliff, and if so you
don’t take that dangerous step.
3.6.3. Comparing Elements
Suppose we have two students, student A and student B and we want to compare them:
we want to determine which one should be placed first in a list and which should be
placed second. For this exercise let’s suppose that we want to order them first by their
last names (so that Anderson comes before Zadora). What if they have the same last",ComputerScienceOne.pdf
"last names (so that Anderson comes before Zadora). What if they have the same last
name, like Jane Smith and John Smith? If the last names are equal, then we’ll want to
order them by their first names (Jane before John). If both their first names and last
names are the same, we’ll say either order is okay.
Names will likely be represented using strings, so let’s say that <, = and > apply to
strings, ordering them lexicographically (which is consistent with alphabetic ordering).
We’ll first need to compare their last names. If equal, then we’ll need another conditional
construct. This is achieved by nesting conditional statements as in Algorithm 3.8.
1 if A’s last name < B’s last name then
2 output A comes first
3 else ifA’s last name > B’s last name then
4 output B comes first
5 else
//last names are equal, so compare their first names
6 if A’s first name < B’s first name then
7 output A comes first
8 else ifA’s first name > B’s first name then
9 output B comes first
10 else",ComputerScienceOne.pdf
"6 if A’s first name < B’s first name then
7 output A comes first
8 else ifA’s first name > B’s first name then
9 output B comes first
10 else
11 Either ordering is fine
12 end
13 end
Algorithm 3.8:Comparing Students by Name
84",ComputerScienceOne.pdf
"3.6. Examples
3.6.4. Life & Taxes
Another example in which there are several cases that have to be considered is computing
an income tax liability using marginal tax brackets. Table 3.6 contains the 2014 US
Federal tax margins and marginal rates for a married couple filing jointly based on the
Adjusted Gross Income (income after deductions).
AGI is over But not over Tax
0 $18,150 10% of the AGI
$18,150 $73,800 $1,815 plus 15% of the AGI in excess of
$18,150
$73,800 $148,850 $10,162.50 plus 25% of the AGI in excess of
$73,800
$148,850 $225,850 $28,925 plus 28% of the AGI in excess of
$148,850
$225,850 $405,100 $50,765 plus 33% of the AGI in excess of
$225,850
$405,100 $457,600 $109,587.50 plus 35% of the AGI in excess
of $405,100
$457,600 — $127,962.50 plus 39.6% of the AGI in excess
of $457,600
Table 3.6.: 2014 Tax Brackets for Married Couples Filing Jointly
In addition, one of the tax credits (which offsets tax liability) tax payers can take is the",ComputerScienceOne.pdf
"In addition, one of the tax credits (which offsets tax liability) tax payers can take is the
child tax credit. The rules are as follows:
•If the AGI is $110,000 or more, they cannot claim a credit (the credit is $0)
•Each child is worth a $1,000 credit, however at most $3,000 can be claimed
•The credit is not refundable: if the credit results in a negative tax liability, the tax
liability is simply $0
As an example: suppose that a couple has $35,000 AGI (placing them in the second tax
bracket) and has two children. Their tax liability is
$1,815 + 0.15 ×($35,000 −$18,150) = $4,342.50
However, the two children represent a $2,000 refund, so their total tax liability would be
$2,342.50.
Let’s first design some code that computes the tax liability based on the margins and
rates in Table 3.6. We’ll assume that the AGI is stored in a variable named income.
Using a series of if-else-if statements as presented in Algorithm 3.9, the variable tax will
85",ComputerScienceOne.pdf
"3. Conditionals
contain our initial tax liability.
1 if income≤18,150 then
2 tax←.10 ·income
3 else ifincome> 18,150 And income≤73,800 then
4 tax←1,815 + .15 ·(income−18,150)
5 else ifincome> 73,800 And income≤148,850 then
6 tax←10,162.50 + .25 ·(income−73,800)
7 else ifincome> 148850 And income≤225,850 then
8 tax←28,925 + .28 ·(income−148,850)
9 else ifincome> 225,850 And income≤405,100 then
10 tax←50,765 + .33 ·(income−225,850)
11 else ifincome> 405,100 And income≤457,600 then
12 tax←109,587.50 + .35 ·(income−405,100)
13 else
14 tax←127,962.50 + .396 ·(income−457,600)
15 end
Algorithm 3.9:Computing Tax Liability with If-Else-If
We can then compute the amount of a tax credit and adjust the tax accordingly by using
similar if-else-if and if-else statements as in Algorithm 3.10.
1 if income≥110,000 then
2 credit←0
3 else ifnumberOfChildren ≤3 then
4 credit←numberOfChildren ∗1,000
5 else
6 credit←3000
7 end
//Now adjust the tax, taking care that its a nonrefundable credit",ComputerScienceOne.pdf
"4 credit←numberOfChildren ∗1,000
5 else
6 credit←3000
7 end
//Now adjust the tax, taking care that its a nonrefundable credit
8 if credit>tax then
9 tax←0
10 else
11 tax←(tax−credit)
12 end
Algorithm 3.10:Computing Tax Credit with If-Else-If
86",ComputerScienceOne.pdf
"3.7. Exercises
3.7. Exercises
Exercise 3.1. Write a program that prompts the user for an x and a y coordinate in
the Cartesian plane and prints out a message indicating if the point ( x,y) lies on an axis
(x or y axis, or both) or what quadrant it lies in (see Figure 3.4).
x
y
Quadrant IQuadrant II
Quadrant III Quadrant IV
Figure 3.4.: Quadrants of the Cartesian Plane
Exercise 3.2. A BOGO (Buy-One, Get-One) sale is a promotion in which a person
buys two items and receives a 50% discount on the less expensive one. Write a program
that prompts the user for the cost of two items, computes a 50% discount on the less
expensive one, and then computes a grand total.
Exercise 3.3. Price Per Mile. Write a program to determine which type of gas is the
better deal with respect to price-per-mile driven. For example, suppose unleaded costs
$2.50 per gallon and your vehicle is able to get an average of 30 miles per gallon. The",ComputerScienceOne.pdf
"$2.50 per gallon and your vehicle is able to get an average of 30 miles per gallon. The
true cost of unleaded is thus 8.33 cents per mile. Now suppose that the ethanol fuel costs
only $2.25 per gallon but only yields 25 miles per gallon, thus 9 cents per mile, a worse
deal.
Write a program that prompts the user to enter the price (per gallon) of one type of gas
as well as the miles-per-gallon. Then prompt for the same two values for a second type
of gas and compute the true cost of each type. Print a message indicating which type of
gas is the better deal. For example, if the user enters the values above the output would
look something like:
87",ComputerScienceOne.pdf
"3. Conditionals
Gas A: $0.0833 per mile
Gas B: $0.0900 per mile
Gas A is the better deal.
Exercise 3.4.Various substances have different boiling points. A selection of substances
and their boiling points can be found in Table 3.7. Write a program that prompts the
user for the observed boiling point of a substance in degrees Celsius and identifies the
substance if the observed boiling point is within 5% of the expected boiling point. If the
data input is more than 5% higher or lower than any of the boiling points in the table, it
should output Unknown substance.
Substance Boiling Point (C)
Methane -161.7
Butane -0.5
Water 100
Nonane 150.8
Mercury 357
Copper 1187
Silver 2193
Gold 2660
Table 3.7.: Expected Boiling Points
Exercise 3.5. Electrical resistance in various metals can be measured using nano-ohm
metres (nΩ ·m). Table 3.8 gives the resistivity of several metals.
Material Resistivity (nΩ ·m)
Copper 16.78
Aluminum 26.50
Beryllium 35.6
Potassium 72.0
Iron 96.10",ComputerScienceOne.pdf
"Material Resistivity (nΩ ·m)
Copper 16.78
Aluminum 26.50
Beryllium 35.6
Potassium 72.0
Iron 96.10
Table 3.8.: Resistivity of several metals
Write a program that prompts the user for an observed resistivity of an unknown material
(as nano-ohm metres) and identifies the substance if the observed resistivity is within
±3% of the known resistivity of any of the materials in Table 3.8. If the input value lies
outside the ±3% range, output Unknown substance.
Exercise 3.6. The visible light spectrum is measured in nanometer (nm) frequencies.
Ranges roughly correspond to visible colors as depicted in Table 3.9.
88",ComputerScienceOne.pdf
"3.7. Exercises
Color Wave length range (nm)
Violet 380 – 450
Blue 450 – 475
Indigo 476 – 495
Green 495 – 570
Yellow 570 – 590
Orange 590 – 620
Red 620 - 750
Table 3.9.: Visible Light Spectrum Ranges
Write a program that takes an integer corresponding to a wavelength and outputs the
corresponding color. If the value lies outside the ranges it should output Not a visible
wavelength. If a value lies within multiple color ranges it should print all that apply
(for example, a wavelength of 495 is “Indigo-green”).
Exercise 3.7. A certain production of steel is graded according to the following condi-
tions:
(i) Hardness must be greater than 50
(ii) Carbon content must be less than 0.7
(iii) Tensile strength must be greater than 5600
A grade of 5 thru 10 is is assigned to the steel according to the conditions in Table 3.10.
Write a program that will read in the hardness, carbon content, and tensile strength as
Grade Conditions
10 All three conditions are met
9 Conditions (i) and (ii) are met",ComputerScienceOne.pdf
"Grade Conditions
10 All three conditions are met
9 Conditions (i) and (ii) are met
8 Conditions (ii) and (iii) are met
7 Conditions (i) and (iii) are met
6 If only 1 of the three conditions is met
5 If none of the conditions are met
Table 3.10.: Grades of Steel
inputs and output the corresponding grade of the steel.
Exercise 3.8. A triangle can be characterized in terms of the length of its three sides.
In particular, an equilateral triangle is a triangle with all three sides being equal. A
triangle such that two sides have the same length is isosceles and a triangle with all three
sides having a different length is scalene. Examples of each can be found in Figure 3.5.
In addition, the three sides of a triangle are valid only if the sum of any two sides is
strictly greater than the third length.
89",ComputerScienceOne.pdf
"3. Conditionals
Write a program to read in three numbers as the three sides of a triangle. If the three
sides do not form a valid triangle, you should indicate so. Otherwise, if valid, your
program should output whether or not the triangle is equilateral, isosceles or scalene.
(a) Equilateral Triangle
 (b) Isosceles Triangle
 (c) Scalene Triangle
Figure 3.5.: Three types of triangles
Exercise 3.9. Body Mass Index (BMI) is a healthy statistic based on a person’s mass
and height. For a healthy adult male BMI is calculated as
BMI = m
h2 ·703.069579
where m is the person’s mass (in lbs) and h is the person’s height (in whole inches).
Write a program that reads in a person’s mass and height as input and outputs a
characterization of the person’s health with respect to the categories in Table 3.11.
Range Category
BMI <15 Very severely underweight
15 ≤BMI <16 Severely underweight
16 ≤BMI <18.5 Underweight
18.5 ≤BMI <25 Normal
25 ≤BMI <30 Overweight
30 ≤BMI <35 Obese Class I",ComputerScienceOne.pdf
"15 ≤BMI <16 Severely underweight
16 ≤BMI <18.5 Underweight
18.5 ≤BMI <25 Normal
25 ≤BMI <30 Overweight
30 ≤BMI <35 Obese Class I
35 ≤BMI <40 Obese Class II
BMI ≥40 Obese Class III
Table 3.11.: BMI Categories
Exercise 3.10.Let R1 and R2 be rectangles in the plane defined as follows. Let ( x1,y1)
be point corresponding to the lower-left corner of R1 and let (x2,y2) be the point of its
upper-right corner. Let ( x3,y3) be point corresponding to the lower-left corner of R2 and
let (x4,y4) be the point of its upper-right corner.
Write a program to determine the intersection of these two rectangles. In general, the
intersection of two rectangles is another rectangle. However, if the two rectangles abut
each other, the intersection could be a horizontal or vertical line segment (or even a
point). It is also possible that the intersection is empty. Your program will need to
distinguish between these cases.
90",ComputerScienceOne.pdf
"3.7. Exercises
x
y
(2,1)
(6,7.5)
(4,5.5)
(8.5,8.25)
Figure 3.6.: Intersection of Two Rectangles
If the intersection of R1,R2 is a rectangle, R3, your program should output two points
(the lower-left and upper-right corners of R3) as well as the area of R3. If the intersection
is a line segment, your program should output the two end-points and whether it is a
vertical or horizontal line segment. Finally, if the intersection is empty your program
should output “empty intersection”. Your program should also be robust enough to
check that the input is valid (it should not accept empty or “reversed” rectangles).
Your program should read in x1,y1,x2,y2,x3,y3,x4,y4 from the user and perform the
computation above. As an example, the values 2,1,6,7.5,4,5.5,8.5,8.25 would correspond
to the two rectangles in Figure 3.6.
The output for this instance should look something like the following.
Intersecting rectangle: (4, 5.5), (6, 7.5)
Area: 4.00",ComputerScienceOne.pdf
"The output for this instance should look something like the following.
Intersecting rectangle: (4, 5.5), (6, 7.5)
Area: 4.00
Exercise 3.11. Write an app to help people track their cell phone usage. Cell phone
plans for this particular company give you a certain number of minutes every 30 days
which must be used or they are lost (no rollover). We want to track the average number
of minutes used per day and inform the user if they are using too many minutes or can
afford to use more.
Write a program that prompts the user to enter the following pieces of data:
91",ComputerScienceOne.pdf
"3. Conditionals
•Number of minutes in the plan per 30 day period, m
•The current day in the 30 day period, d
•The total number of minutes used so far u
The program should then compute whether the user is over, under, or right on the average
daily usage under the plan. It should also inform them of how many minutes are left
and how many, on average, they can use per day for the rest of the month. Of course, if
they’ve run out of minutes, it should inform them of that too.
For example, if the user enters m= 250, d= 10, and u= 150, your program should print
out something similar to the following.
10 days used, 20 days remaining
Average daily use: 15 min/day
You are EXCEEDING your average daily use (8.33 min/day),
continuing this high usage, you'll exceed your minute plan by
200 minutes.
To stay below your minute plan, use no more than 5 min/day.
Of course, if the user is under their average daily use, a different message should be",ComputerScienceOne.pdf
"To stay below your minute plan, use no more than 5 min/day.
Of course, if the user is under their average daily use, a different message should be
presented. You are allowed/encouraged to compute any other stats for the user that you
feel would be useful.
Exercise 3.12. Write a program to help a floor tile company determine how many tiles
they need to send to a work site to tile a floor in a room. For simplicity, assume that all
rooms are perfectly rectangular with no obstructions; we will also omit any additional
measurements related to grouting.
Further, we will assume that all tile is laid in a grid pattern centered at the center of the
room. That is, four tiles will meet at their corners at the center of the room with tiles
laid out to the edge of the room. Thus, it may be the case that the final row and/or
column at the edge may need to be cut. Also note that if the cut is short enough, the
remaining tile can be used on the other end of the room (same goes for the corners).",ComputerScienceOne.pdf
"remaining tile can be used on the other end of the room (same goes for the corners).
The program will take the following input:
•w - the width of the room
•l - the length of the room
92",ComputerScienceOne.pdf
"3.7. Exercises
0.9 0.9
9.8
10.0
Center of the room
(a) Example 1
0.4 0.4
8.8
10.0
Center of the room
(b) Example 2
Figure 3.7.: Examples of Floor Tiling
•t - width/length of the tile (all tiles are perfectly square)
If we can use whole tiles to perfectly fit the room, then we do so. For example, on
the input (10,10,1), we could perfectly tile a 10 ×10 room with 100 1 ×1 tiles. If the
tiles don’t perfectly fit, then we have to consider the possibility of waste and/or reuse.
Consider the examples in Figure 3.7.
The first example is from the input (9 .8,100,1). In this case, we lay the tiles from the
center of the room (8 full tile lengths) but are left with 0 .9 on either side. If we cut a
tile to fit the left side, we are left with only .1 tile which is too short for the right side.
Therefore, we are forced to waste the 0 .1 length and cut a full tile for the right side. In
all, 100 tiles are required.
The second example is from the input (8 .8,100,1). In this case, we again lay tiles from",ComputerScienceOne.pdf
"all, 100 tiles are required.
The second example is from the input (8 .8,100,1). In this case, we again lay tiles from
the center of the room (8 full tile lengths) and are left with 0 .4 lengths on either side.
Here, we can reuse the cut tile: cut a tile on one side 0 .4 with 0.6 remaining, and cut 0 .4
on the other side of the tile (with the center 0 .2 length of the tile being waste). Thus,
both sides can be tiled with a single tile, meaning only 90 full tiles are needed to tile this
room.
You may further assume that tiles used on the length-side end of the room cannot be
used to tile the width-side of the room (and vice versa). Your program will compute and
output the number of tiles required.
93",ComputerScienceOne.pdf
"4. Loops
Computers are really good at automation. A key aspect of automation is the ability to
repeat a process over and over on different pieces of data until some condition is met.
For example, if we have a collection of numbers and we want to find their sum we would
iterate over each number, adding it to a total, until we have examined every number.
Another example may include sending an email message to each student in a course. To
automate the process, we could iterate over each student record and for each student we
would generate and send the email.
Automated repetition is where loops come in handy. Computers are perfectly suited for
performing such repetitive tasks. We can write a single block of code that performs some
action or processes a single piece of data, then we can write a loop around that block of
code to execute it a number of times.
Loops provide a much better alternative than repeating (cut-paste-cut-paste) the same",ComputerScienceOne.pdf
"code to execute it a number of times.
Loops provide a much better alternative than repeating (cut-paste-cut-paste) the same
code over and over with different variables. Indeed, we wouldn’t even do this in real
life. Suppose that you took a 100 mile trip. How would you describe it? Likely, you
wouldn’t say, “I drove a mile, then I drove a mile, then I drove a mile, ... ” repeated 100
times. Instead, you would simply state “I drove 100 miles” or maybe even, “I drove until
I reached my destination.”
Loops allow us to write concise, repeatable code that can be applied to each element in
a collection or perform a task over and over again until some condition is met. When
writing a loop, there are three essential components:
•An initialization statement that specifies how the loop begins
•A continuation (or termination) condition that specifies whether the loop should
continue to execute or terminate
•An iteration statement that makes progress toward the termination condition",ComputerScienceOne.pdf
"continue to execute or terminate
•An iteration statement that makes progress toward the termination condition
The initialization statement is executed before the loop begins and serves as a way to set
the loop up. Typically, the initialization statement involves setting the initial value of
some variable.
The continuation statement is a logical statement (that evaluates to true or false) that
specifies if the loop should continue (if the value is true) or should terminate (if the value
is false). Upon termination, code returns to a sequential control flow and the program
95",ComputerScienceOne.pdf
"4. Loops
Initialization:
i ←1
Continuation:
i≤10?
loop body
Iteration:
i←(i+ 1)
remaining
program
true
repeat
false
Figure 4.1.: A Typical Loop Flow Chart
96",ComputerScienceOne.pdf
"4.1. While Loops
continues.
The iteration statement is intended to update the state of a program to make progress
toward the termination condition. If we didn’t make such progress, the loop would
continue on forever as the termination condition would never be satisfied. This is known
as an infinite loop , and results in a program that never terminates.
As a simple example, consider the following outline.
•Initialize the value of a variable i to 1
•While the value of i is less than or equal to 10 ... (continuation condition)
•Perform some action (this is sometimes referred to as the loop body
•Iterate the variable i by adding one to its value
The code outline above specifies that some action is to be performed once for each value:
i = 1,i = 2,...,i = 10, after which the loop terminates. Overall, the loop executes a
total of 10 times. Prior to each of the 10 executions, the value of iis checked; as it is less",ComputerScienceOne.pdf
"total of 10 times. Prior to each of the 10 executions, the value of iis checked; as it is less
than or equal to 10, the action is performed. At the end of each of the 10 iterations, the
variable i is incremented by 1 and the termination condition is checked again, repeating
the process. There are several different types of loops that vary in syntax and style but
they all have the same three basic components.
4.1. While Loops
A while loop is a type of loop that places the three components in their logical order. The
initialization statement is written before the loop code. Typically the keyword while is
used to specify the continuation/termination condition. Finally, the iteration statement
is usually performed at the end of the loop inside the code block associated with the loop.
A small, counter-controlled while loop is presented in Algorithm 4.1 which illustrates the
previous example of iterating a variable i from 1 to 10.
1 i←1 //Initialization statement
2 while (i≤10) do",ComputerScienceOne.pdf
"previous example of iterating a variable i from 1 to 10.
1 i←1 //Initialization statement
2 while (i≤10) do
3 Perform some action
4 i←(i+ 1) //Iteration statement
5 end
Algorithm 4.1:Counter-Controlled While Loop
Prior to the while statement, the variable i is initialized to 1. This action is only
97",ComputerScienceOne.pdf
"4. Loops
performed once and it is done so before the loop code. Then, before the loop code is
executed, the continuation condition is checked. Since i= 1 ≤10, the condition evaluates
to true and the loop code block is executed. The last line of the code block is the iteration
statement, where i is incremented by 1 and now has a value of 2. The code returns to
the top of the loop and again evaluates the continuation condition (which is still true as
i= 2 ≤10).
On the 10th iteration of the loop when i= 10, the loop will execute for the last time. At
the end of the loop, i is incremented to 11. The loop still returns to the top and the
continuation condition is still checked one last time. However, since i= 11 ̸≤10, the
condition is now false and the loop terminates. Regular sequential control flow returns
and the program continues executing whatever code is specified after the loop.
4.1.1. Example
In the previous example we knew that we wanted the loop to execute ten times. Though",ComputerScienceOne.pdf
"4.1.1. Example
In the previous example we knew that we wanted the loop to execute ten times. Though
you can use a while loop in counter-controlled situations, while loops are typically used
in scenarios when you may not know how many iterations you want the loop to execute
for. Instead of a straightforward iteration, the loop itself may update a variable in a
less-than-predictable manner.
As an example, consider the problem of normalizing a number as is typically done in
scientific notation. Given a number x (for simplicity, we’ll consider x≥1), we divide
it by 10 until its value is in the interval [1 ,10), keeping track of how many times we’ve
divided by 10. For example, if we have the number x= 32,145.234, we would divide by
10 four times, resulting in 3 .2145234 so that we could express it as
3.2145234 ×104
A simple realization of this process is presented in Algorithm 4.2. Rather than some
fixed number of iterations, the number of times the loop executes depends on how large",ComputerScienceOne.pdf
"fixed number of iterations, the number of times the loop executes depends on how large
x is. For the example mentioned, it executes 4 times; for an input of x= 10,000,000 it
would execute 7 times. A while loop allows us to specify the repetition process without
98",ComputerScienceOne.pdf
"4.2. For Loops
having to know up front how many times it will execute.
Input : A number x, x≥0
Output :x normalized, k its exponent
1 k←0
2 while x> 10 do
3 x←(x/10)
4 k←(k+ 1)
5 end
6 output x,k
Algorithm 4.2:Normalizing a Number With a While Loop
4.2. For Loops
A for loop is similar to a while loop but allows you to specify the three components on
the same line. In many cases, this results in a loop that is more readable; if the code
block in a while loop is long it may be difficult to see the initialization, continuation,
and iteration statements clearly. For loops are typically used to iterate over elements
stored in a collection such as an array (see Chapter 7). Usually the keyword for is used
to identify all three components. A general example is given in Algorithm 4.3.
1 for ( ⟨initialization⟩; ⟨continuation⟩; ⟨iteration⟩) do
2 Perform some action
3 end
Algorithm 4.3:A General For Loop
Note the additional syntax: in many programming languages, semicolons are used at",ComputerScienceOne.pdf
"2 Perform some action
3 end
Algorithm 4.3:A General For Loop
Note the additional syntax: in many programming languages, semicolons are used at
the end of executable statements. Semicolons are also used to delimit each of the three
loop components in a for-loop (otherwise there may be some ambiguity as to where
each of the components begins and ends). However, the semicolons are typically only
placed after the initialization statement and continuation condition and are omitted after
the iteration statement. A more concrete example is given in Algorithm 4.4 which is
equivalent to the counter-controlled while loop we examined earlier.
Though all three components are written on the same line, the initialization statement
is only ever executed once; at the beginning of the loop. The continuation condition is
checked prior to each and every execution of the loop. Only if it evaluates to true does
the loop body execute. The iteration condition is performed at the end of each loop
iteration.",ComputerScienceOne.pdf
"the loop body execute. The iteration condition is performed at the end of each loop
iteration.
99",ComputerScienceOne.pdf
"4. Loops
1 for ( i←1; i≤10; i←(i+ 1) ) do
2 Perform some action
3 end
Algorithm 4.4:Counter-Controlled For Loop
4.2.1. Example
As a more concrete example, consider Algorithm 4.5 in which we do the same iteration
(i will take on the values 1 ,2,3,..., 10), but in each iteration we add the value of i for
that iteration to a running total, sum.
1 sum←0
2 for ( i←1; i≤10; i←(i+ 1) ) do
3 sum←(sum+ i)
4 end
Algorithm 4.5:Summation of Numbers in a For Loop
Again, the initialization of i= 1 is only performed once. On the first iteration of the loop,
i= 1 and so sumwill be given the value sum+i= 0+1 = 1 At the end of the loop, iwill
be incremented and will have a value of 2. The continuation condition is still satisfied, so
once again the loop body executes and sum will be given the value sum+ i= 1 + 2 = 3.
On the 10th (last) iteration, sum will have a value 1 + 2 + 3 +··· + 9 = 45 and i= 10.
Thus sum+ i= 45 + 10 = 55 after which i will be incremented to 11. The continuation",ComputerScienceOne.pdf
"Thus sum+ i= 45 + 10 = 55 after which i will be incremented to 11. The continuation
condition is still checked, but since 11 ̸≤10, the loop body will not be executed and the
loop will terminate, resulting in a final sum value of 55.
4.3. Do-While Loops
Yet another type of loop is the do-while loop. One major difference between this type
of loop and the others is that it is always executed at least once. The way that this is
achieved is that the continuation condition is checked at the end of the loop rather than
prior to is execution. The same counter-controlled example can be found in Algorithm
4.6.
In contrast to the previous examples, the loop body is executed on the first iteration
without checking the continuation condition. Only after the loop body, including the
incrementing of the iteration variable iis the continuation condition checked. If true, the
loop repeats at the beginning of the loop body.
100",ComputerScienceOne.pdf
"4.3. Do-While Loops
1 i←1
2 do
3 Perform some action
4 i←(i+ 1)
5 while i≤10
Algorithm 4.6:Counter-Controlled Do-While Loop
Initialization:i←1
loop body
Iteration:i←(i+ 1)
Continuation:i≤10?
remainingprogram
false
true
Figure 4.2.: A Do-While Loop Flow Chart. The continuation condition is checked after
the loop body.
Do-while loops are typically used in scenarios in which the first iteration is used to “setup”
the continuation condition (thus, it needs to be executed at least once). A common
example is if the loop body performs an operation that may result in an error code (or
flag) that is either true (an error occurred) or false (no error occurred).
From this perspective, a do-while loop can also be seen as a do-until loop: perform a
task until some condition is no longer satisfied. The subtle wording difference implies
101",ComputerScienceOne.pdf
"4. Loops
1 do
2 Read some data
3 isError ←result of reading
4 while isError
Algorithm 4.7:Flag-Controlled Do-While Loop
that we’ll perform the action before checking to see if it should be performed again.
4.4. Foreach Loops
Many languages support a special type of loop for iterating over individual elements
in a collection (such as a set, list, or an array). In general, such loops are referred to
as foreach loops. These types of loops are essentially syntactic sugar: iterating over a
collection could be achieved with a for loop or a while loop, but foreach loops provide a
more convenient way to iterate over a collections. We will revisit these loops when we
examine arrays in Chapter 7. For now, we look at a simple example in Algorithm 4.8.
1 foreach element a in the collection A do
2 process the element a
3 end
Algorithm 4.8:Example Foreach Loop
How the elements are stored in the collection and how they are iterated over is not our",ComputerScienceOne.pdf
"3 end
Algorithm 4.8:Example Foreach Loop
How the elements are stored in the collection and how they are iterated over is not our
(primary) concern. We simply want to apply the same block of code to each element,
the foreach loop handles the details on how each element is iterated over. The syntax
also provides a way to refer to each element (the a variable in the algorithm). On each
iteration of the loop, the foreach loop updates the reference a to the next element in
the array. The loop terminates after it has iterated through each and every one of the
elements. In this way, a foreach loop simplifies the syntax: we don’t have to specify any
of the three components ourselves. As a more concrete example, consider iterating over
each student in a course roster. For each student, we wish to compute their grade and
then email them the results. The foreach loop allows us to do this without worrying
about the iteration details (see Algorithm 4.9).
102",ComputerScienceOne.pdf
"4.5. Other Issues
1 foreach (student s in the class C) do
2 g←compute a’s grade
3 send a an email informing them of their grade g
4 end
Algorithm 4.9:Foreach Loop Computing Grades
4.5. Other Issues
4.5.1. Nested Loops
Just as with conditional statements, we can nest loops within loops to perform more
complex processes. Though you can do this with any type of loop, we present a simple
example using for loops in Algorithm 4.10.
1 n←10
2 m←20
3 for (i←1; i≤m; i←(i+ 1)) do
4 for (j ←1; j ≤n; j ←(j+ 1)) do
5 output (i,j)
6 end
7 end
Algorithm 4.10:Nested For Loops
The outer for loop executes a total of 20 times while the inner for loop executes 10
times. Since the inner for loop is nested inside the outer loop, the entire inner loop
executes all 10 iterations for each of the 20 iterations of the outer loop. Thus, in
total the inner most output operation executes 10 ×20 = 200 times. Specifically, it
outputs (1,1),(1,2),..., (1,10),(2,1),(2,2),..., (2,10),(3,1),..., (20,10). Nested loops",ComputerScienceOne.pdf
"outputs (1,1),(1,2),..., (1,10),(2,1),(2,2),..., (2,10),(3,1),..., (20,10). Nested loops
are commonly used when iterating over elements in two-dimensional arrays such as
tabular data or matrices. Nested loops can also be used to process all pairs in a collection
of elements.
4.5.2. Infinite Loops
Sometimes a simple mistake in the design of a loop can make it execute forever. For
example, if we accidentally iterate a variable in the wrong direction or write the opposite
103",ComputerScienceOne.pdf
"4. Loops
termination/continuation condition. Such a loop is referred to as an infinite loop. As an
example, suppose we forgot the increment operation from a previous example.
1 sum←0
2 i←1
3 while i≤10 do
4 sum←(sum+ i)
5 end
Algorithm 4.11:Infinite Loop
In Algorithm 4.11 we never make progress toward the terminating condition! Thus, the
loop will execute forever, i will continue to have the value 0 and since 0 ≤10, the loop
body will continue to execute. Care is needed in the design of your loops to ensure that
they make progress toward the termination condition.
Most of the time an infinite loop is not something you want and usually you must terminate
your buggy program externally (sometimes referred to as “killing” it). However, infinite
loops do have their uses. A poll loop is a loop that is intended to not terminate. At a
system level, for example, a computer may poll devices (such as input/output devices)",ComputerScienceOne.pdf
"system level, for example, a computer may poll devices (such as input/output devices)
one-by-one to see if there is any active input/output request. Instead of terminating,
the poll loop simply repeats itself, returning back to the first device. As long as the
computer is in operation, we don’t want this process to stop. This can be viewed as an
infinite loop as it doesn’t have any termination condition.
The Zune Bug
Though proper testing and debugging should reduce the likelihood of such bugs, there
are several notable instances in which an infinite loop impacted real software. One
such instance was the Microsoft Zune bug. The Zune was a portable music player, a
competitor to the iPod. At about midnight on the night of December 31st, 2008, Zunes
everywhere failed to startup properly. A firmware clock driver designed by a 3rd party
company contained the following code.
2008 was a leap year, so the check on line 2 evaluated to true. However, though December",ComputerScienceOne.pdf
"company contained the following code.
2008 was a leap year, so the check on line 2 evaluated to true. However, though December
31st, 2008 was the 366th day of the year ( days = 366) the third line evaluated to false
and the loop was repeated without any of the program state being updated. The problem
was “fixed” 24 hours later when it was the 367th day and line 3 worked. The problem
was that line 3 should have been days >= 366).
The failure was that this code was never tested on the “corner cases” that it was designed
for. No one thought to test the driver to see if it worked on the last day of a leap year.
104",ComputerScienceOne.pdf
"4.5. Other Issues
1 while(days > 365) {
2 if(IsLeapYear(year)) {
3 if(days > 366) {
4 days -= 366;
5 year += 1;
6 }
7 } else {
8 days -= 365;
9 year += 1;
10 }
11 }
Code Sample 4.1.: Zune Bug
The code worked the vast majority of the time, but this illustrates the need for rigorous
testing.
4.5.3. Common Errors
When writing loops its important to use the proper syntax in the language in which you
are coding. Many languages use semicolons to terminate executable statements. However,
the while statements are not executable: they are part of the control structure of the
language and do not have semicolons at the end. A misplaced semicolon could be a
syntax error, or it could be syntactically correct but lead to incorrect results. A common
error is to place a semicolon at the end of a while statement as in
while(count <= 10); //WRONG!!!
In this example, the while loop binds to an empty executable statement and results in
an infinite loop!",ComputerScienceOne.pdf
"while(count <= 10); //WRONG!!!
In this example, the while loop binds to an empty executable statement and results in
an infinite loop!
Other common errors are the result of misidentifying either the initialization statement
or the continuation condition. Starting a counter at 1 instead of zero, or using a ≤
comparison instead of a < , etc. These can lead to a loop being off-by-one resulting in a
logic error.
Other errors are the result of using improper variable types. Recall that operations
involving floating-point numbers can have round off and precision errors, 1
3 + 1
3 + 1
3 may
not be equal to one for example. It is best to avoid using floating-point numbers or
comparisons in the control of your loops. Boolean and integer types are much less error
105",ComputerScienceOne.pdf
"4. Loops
prone.
Finally, you must always ensure that your loops are making progress toward the termina-
tion condition. A failure to properly increment a counter can lead to incorrect results or
even an infinite loop.
4.5.4. Equivalency of Loops
It might not seem obvious at first, but in fact, any type of loop can be re-written as
another type of loop and perform equivalent operations. That is, any while loop can be
rewritten as an equivalent for loop. Any do-while loop can be rewritten as an equivalent
while loop!
So why do we have different types of loops? The short answer is that we want our
programming languages to be flexible. We could design a language in which every loop
had to be a while loop for example, but there are some situations in which it would be
more “natural” to write code with a for loop. By providing several options, programmers
have the choice of which type of loop to write.
In general, there are no “rules” as to which loop to apply to which situation. There",ComputerScienceOne.pdf
"have the choice of which type of loop to write.
In general, there are no “rules” as to which loop to apply to which situation. There
are general trends, best practices, and situations where it is more common to use one
loop rather than another, but in the end it does come down to personal choice and style.
Some software projects or organizations may have established guidelines or style guide
that establishes such guidelines in the interest of consistency and uniformity.
4.6. Problem Solving With Loops
Loops can be applied to any problem that requires repetition of some sort or to simplify
repeated code. When designing loops, it is important to identify the three components
by asking the questions:
•Where does the loop start? What variables or other state may need to be initialized
or setup prior to the beginning of the loop?
•What code needs to be repeated? How can it be generalized to depend on loop
control variables? This helps you to identify and write the loop body.",ComputerScienceOne.pdf
"•What code needs to be repeated? How can it be generalized to depend on loop
control variables? This helps you to identify and write the loop body.
•When should the loop end? How many times do we want it to execute? This helps
you to identify the continuation and/or termination condition.
•How do we make progress toward the termination condition? What variable(s)
need to be incremented and how?
106",ComputerScienceOne.pdf
"4.7. Examples
4.7. Examples
4.7.1. For vs While Loop
Let’s consider how to write a loop to compute the classic geometric series,
1
1 −x =
∞∑
k=0
xk = 1 + x+ x2 + x3 + ···
Obviously a computer cannot compute an infinite series as it is required to terminate in
a finite number of steps. Thus, we can approach this problem in a number of different
ways.
One way we could approximate the series is to compute it out to a fixed number of terms.
To do so, we could initialize a sum variable to zero, then iteratively compute and add
terms to the sum until we have computed n terms. To keep track of the terms, we can
define a counter variable, k as in the summation.
Following our strategy, we can identify the initialization:kshould start at 0. The iteration
is also easy: k should be incremented by 1 each time. The continuation condition should
continue the loop until we have computed n terms. However, since k starts at 0, we",ComputerScienceOne.pdf
"continue the loop until we have computed n terms. However, since k starts at 0, we
would want to continue while k <n. We would not want to continue the iteration when
k = n as that would make n+ 1 iterations (again since k starts at 0). Further, since
we know the number of iterations we want to execute, a for loop is arguably the most
appropriate loop for this problem. Our solution is presented in Algorithm 4.12.
Input : x, n≥0
Output :An approximated value of 1
1−x using a geometric series
1 sum←0
2 for (k= 0; k <n; k←(k+ 1)) do
3 sum←(sum+ xk)
4 end
5 output sum
Algorithm 4.12:Computing the Geometric Series Using a For Loop
As an alternative, consider the following approach: instead of computing a predefined
number of terms, what if we computed terms until the difference between the value in the
previous iteration and the value in the current iteration is negligible, say less than some
small ϵ amount. We could stop our computation because any further iterations would",ComputerScienceOne.pdf
"small ϵ amount. We could stop our computation because any further iterations would
only affect the summation less and less. That is, the current value represents a “good
enough” approximation. That way, if someone wanted an even better approximation,
they could specify a smaller ϵ.
107",ComputerScienceOne.pdf
"4. Loops
This approach will be more straightforward with a while loop since the continuation
condition will be more along the lines of “while the estimation is not yet good enough,
continue the summation.” This approach will also be easier if we keep track of both
a current and a previous value of the summation, then computing and checking the
difference will be easier.
Input : x, ϵ> 0
1 sumprev ←0
2 sumcurr ←1
3 k←1
4 while |sumprev −sumcurr|≥ ϵ do
5 sumprev ←sumcurr
6 sumcurr ←(sumcurr + xk)
7 k←(k+ 1)
8 end
9 output sum
Algorithm 4.13:Computing the Geometric Series Using a While Loop
On lines 1–2 we initialize our values to ensure that the while loop will execute at least
once. In the continuation condition, we use the absolute value of the difference as the
series can oscillate between negative and positive values.
4.7.2. Primality Testing
An integer n> 1 is called prime if the only integers that divide it evently are 1 and itself.",ComputerScienceOne.pdf
"4.7.2. Primality Testing
An integer n> 1 is called prime if the only integers that divide it evently are 1 and itself.
Otherwise it is called composite. For example, 30 is composite as it is divisible by 2, 3,
and 5 among others. However, 31 is prime as it is only divisible by 1 and 31.
Consider the problem of determining whether or not a given integer n is prime or
composite, referred to as primality testing. A straightforward way of determining this is
to simply try dividing by every integer 2 up to √n: if any of these integers divides n,
then n is composite. Otherwise, if none of them do, n is prime. Observe that we only
need to go up to √n since any prime divisor greater than that will correspond to some
prime divisor less than √n.
A simple for loop can be constructed to capture this idea. Our initialization clearly starts
at i= 2, incrementing by 1 each time until ihas exceeded √n. This solution is presented",ComputerScienceOne.pdf
"at i= 2, incrementing by 1 each time until ihas exceeded √n. This solution is presented
in Algorithm 4.14. Of course this is certainly not the most efficient way to solve this
problem, but we will not go into more advanced algorithms here.
Now consider this more general problem: given an integer m> 1, determine how many
108",ComputerScienceOne.pdf
"4.7. Examples
Input : n> 1
1 for (i←2; i≤√n; i←(i+ 1)) do
2 if i divides n then
3 output composite
4 end
5 end
6 output prime
Algorithm 4.14:Determining if a Number is Prime or Composite
prime numbers ≤m there are. A key observation is that we’ve already solved part of
the problem: determining if a given number is prime in the previous exercise. To solve
this more general problem, we could reuse or adapt our previous solution. In particular,
we could surround the previous solution in an outer loop and iterate over integers from
2 up to m. The inner loop would then determine if the integer is prime and instead of
outputting a result, could increment a counter of the number of primes it has found so
far. This solution is presented in Algorithm 4.15.
Input : m> 1
1 numberOfPrimes ←0
2 for (j = 2; j ≤m; j ←(j+ 1)) do
3 isPrime ←true
4 for (i←2; i≤√j; i←(i+ 1)) do
5 if (i divides j) then
6 isPrime ←false
7 end
8 end
9 if (isPrime) then
10 numberOfPrimes ←(numberOfPrimes + 1)
11 end
12 end",ComputerScienceOne.pdf
"5 if (i divides j) then
6 isPrime ←false
7 end
8 end
9 if (isPrime) then
10 numberOfPrimes ←(numberOfPrimes + 1)
11 end
12 end
13 output numberOfPrimes
Algorithm 4.15:Counting the number of primes.
4.7.3. Paying the Piper
Banks issue loans to customers as one lump sum called a principle P that the borrower
must pay back over a number of terms. Usually payments are made on a monthly basis.
109",ComputerScienceOne.pdf
"4. Loops
Further, banks charge an amount of interest on a loan measured as an Annual Percentage
Rate (APR). Given these conditions, the borrower makes monthly payments determined
by the following formula.
monthlyPayment = iP
1 −(1 + i)−n
Where i= apr
12 is the monthly interest rate, and n is the number of terms (in months).
For simplicity, suppose we borrow P = $1,000 at 5% interest ( apr = 0.05) to be paid
back over a term of 2 years ( n= 24). Our monthly payment would (rounded) be
monthlyPayment =
.05
12 ·1000
1 −(1 + .05
12 )−24 = $43.87
When the borrower makes the first month’s payment, some of it goes to interest, some of
it goes to paying down the balance. Specifically, one month’s interest on $1,000 is
$1,000 ·0.05
12 = $4.17
and so $43.87 −$4.17 = $39.70 goes to the balance, making the new balance $960.30.
The next month, this new balance is used to compute the new interest payment,
$960.30 ·0.05
12 = $4.00",ComputerScienceOne.pdf
"The next month, this new balance is used to compute the new interest payment,
$960.30 ·0.05
12 = $4.00
And so on until the balance is fully paid. This process is known as loan amortization.
Let’s write a program that will calculate a loan amortization schedule given the inputs as
described above. To start, we’ll need to compute the monthly payment using the formula
above and for that we’ll need a monthly interest rate. The balance will be updated
month-to-month, so we’ll use another variable to represent the balance. Finally, we’ll
want to track the current month in the loan schedule process.
Once we have these variables setup, we can start a loop that will repeat once for each
month in the loan schedule. We could do this using either type of loop, but for this
exercise, let’s use a while loop. Using our month variable, we’ll start by initializing it to
1 and run the loop through the last month, n.
On each iteration we compute that month’s interest and principle payments as above,",ComputerScienceOne.pdf
"1 and run the loop through the last month, n.
On each iteration we compute that month’s interest and principle payments as above,
update the balance, and also be sure to update our month counter variable to ensure
we’re making progress toward the termination condition. On each iteration we’ll also
output each of these variables to the user. The full program can be found in Algorithm
4.16.
If we were to actually implement this we’d need to be more careful. This outlines the basic
process, but keep in mind that US dollars are only accurate to cents. A monthly payment
110",ComputerScienceOne.pdf
"4.8. Exercises
can’t be $43.871 cents. We’ll need to take care to round properly. This introduces another
issue: by rounding the final month’s payment may not match the expected monthly
payment (we may over or under pay in the final month). An actual implementation may
need to handle the final month’s payment separately with different logic and operations
than are outside the loop.
Input : A principle, P, a number of terms, n, an APR, apr
Output :A loan amortization schedule
1 balance←P //The initial balance is the principle
2 i←apr
12 //monthly interest rate
3 monthlyPayment ← iP
1−(1+i)−n
4 month←1 //A month counter
5 while (month≤n) do
6 monthInterest ←i·balance
7 monthPrinciple ←monthlyPayment −monthInterest
8 balance←balance−monthPrinciple
9 month= (month+ 1)
10 output month,monthInterest,monthPrinciple,balance
11 end
Algorithm 4.16:Computing a loan amortization schedule
4.8. Exercises
Exercise 4.1.Write a for-loop and a while-loop that accomplishes each of the following.",ComputerScienceOne.pdf
"4.8. Exercises
Exercise 4.1.Write a for-loop and a while-loop that accomplishes each of the following.
(a) Prints all integers 1 thru 100 on the same line delimited by a single space
(b) Prints all even integers 0 up to n in reverse order
(c) A list of integers divisible by 3 between aand bwhere a,b are parameters or inputs
(d) Prints all positive powers of two up to 2 30 : 1 ,2,4,..., 1073741824 one value per
line (try computing up to 2 31 and 232 and discern reasons for why it may fail)
(e) Prints all even integers 2 thru 200 on 10 different lines (10 numbers on each line)
in reverse order
(f) Prints the following pattern of numbers (hint: use two nested loops; the result can
be computed using some value of i+ 10j)
111",ComputerScienceOne.pdf
"4. Loops
11 21 31 41 51 61 71 81 91 101
12 22 32 42 52 62 72 82 92 102
13 23 33 43 53 63 73 83 93 103
14 24 34 44 54 64 74 84 94 104
15 25 35 45 55 65 75 85 95 105
16 26 36 46 56 66 76 86 96 106
17 27 37 47 57 67 77 87 97 107
18 28 38 48 58 68 78 88 98 108
19 29 39 49 59 69 79 89 99 109
20 30 40 50 60 70 80 90 100 110
Exercise 4.2. Civil engineers have come up with two different models on how a city’s
population will grow over the next several years. The first projection assumes a 10%
annual growth rate while the second projection assumes a linear growth rate of 50,000
additional citizens per year. Write a program to project the population growth under
both models. Take, as input, the initial population of the city along with a number of
years to project the population.
In addition, compute how many years it would take to double the population under each
model.
Exercise 4.3. Write a loan program similar to the amortization schedule program we",ComputerScienceOne.pdf
"model.
Exercise 4.3. Write a loan program similar to the amortization schedule program we
developed in Section 4.7.3. However, give the user an option to specify an extra monthly
payment amount in order to pay off the loan early. Calculate how much quicker the loan
gets paid off and how much they save in interest.
Exercise 4.4. The rate of decay of a radioactive isotope is given in terms of its half-life
H, the time lapse required for the isotope to decay to one-half of its original mass.
For example, the isotope Strontium-90 ( 90Sr) has a half-life of 28.9 years. If we start
with 10kg of Strontium-90 then 28.9 years later you would expect to have only 5kg of
Strontium-90 (and 5kg of Yttrium-90 and Zirconium-90, isotopes which Strontium-90
decays into).
Write a program that takes the following input:
•Atomic Number (integer)
•Element Name
•Element Symbol
•H (half-life of the element)
•m, an initial mass in grams
112",ComputerScienceOne.pdf
"4.8. Exercises
Your program will then produce a table detailing the amount of the element that remains
after each year until less than 50% of the original amount remains. This amount can be
computed using the following formula:
r= m×
(1
2
)(y/H)
y is the number of years elapsed, and H is the half-life of the isotope in years.
For example, using your program on Strontium-90 (symbol: Sr, atomic number: 38)
with a half-life of 28.9 years and an initial amount of 10 grams would produce a table
something like the following.
Strontium-90 (38-Sr)
Elapsed Years Amount
-------------------------
- 10g
1 9.76g
2 9.53g
3 9.30g
...
28 5.11g
29 4.99g
Exercise 4.5. Write a program that computes various statistics on a collection of
numbers that can be read in from the command line, as command line arguments, or via
other means. In particular, given n numbers,
x1,x2,...,x n
your program should compute the following statistics:
•The minimum number
•The maximum number
•The mean,
µ= 1
n
n∑
i=1
xi",ComputerScienceOne.pdf
"x1,x2,...,x n
your program should compute the following statistics:
•The minimum number
•The maximum number
•The mean,
µ= 1
n
n∑
i=1
xi
•The variance,
σ2 = 1
n
n∑
i=1
(xi −µ)2
113",ComputerScienceOne.pdf
"4. Loops
•And the standard deviation,
σ=
√1
n
n∑
i=1
(xi −µ)2
where n is the number of numbers that was provided. For example, with the numbers,
3.14,2.71,42,3,13
your output should look something like:
Minimum: 2.71
Maximum: 42.00
Mean: 12.77
Variance: 228.77
Standard
Deviation: 15.13
Exercise 4.6. The ancient Greek mathematician Euclid developed a method for finding
the greatest common divisor of two positive integers, a and b. His method is as follows:
1. If the remainder of a/b is 0 then b is the greatest common divisor.
2. If it is not 0, then find the remainder r of a/b and assign b to a and the remainder
r to b.
3. Return to step (1) and repeat the process.
Write a program that uses a function to perform this procedure. Display the two integers
and the greatest common divisor.
Exercise 4.7.Write a program to estimate the value of e≈2.718281 ... using the series:
e=
∞∑
k=0
1
k!
Obviously, you will need to restrict the summation to a finite number of n terms.",ComputerScienceOne.pdf
"e=
∞∑
k=0
1
k!
Obviously, you will need to restrict the summation to a finite number of n terms.
Exercise 4.8. The value of π can be expressed by the following infinite series:
π= 4 ·
(
1 −1
3 + 1
5 −1
7 + 1
9 − 1
11 + 1
13 −···
)
114",ComputerScienceOne.pdf
"4.8. Exercises
An approximation can be made by taking the first n terms of the series. For n= 4, the
approximation is
π≈4 ·
(
1 −1
3 + 1
5 −1
7
)
= 2.8952
Write a program that takes n as input and outputs an approximation of π according to
the series above.
Exercise 4.9. The sine function can be approximated using the following Taylor series.
sin (x) =
∞∑
i=0
(−1)i
(2i+ 1)!x2i+1 = x−x3
3! + x5
5! −···
Write a function that takes x and n as inputs and approximates sin x by computing the
first n terms in the series above.
Exercise 4.10. One way to compute π is to use Machin’s formula:
π
4 = 4 arctan 1
5 −arctan 1
239
To compute the arctan function, you could use the following series:
arctan x= x
1 −x3
3 + x5
5 −x7
7 + ··· =
∞∑
i=0
(−1)i
2k+ 1x2i+1
Write a program to estimate π using these formulas but allowing the user to specify how
many terms to use in the series to compute it. Compare the estimate with the built-in
definition of π in your language.",ComputerScienceOne.pdf
"many terms to use in the series to compute it. Compare the estimate with the built-in
definition of π in your language.
Exercise 4.11.The arithmetic-geometric mean of two numbers x,y, denoted M(x,y) (or
agm(x,y)) can be computed iteratively as follows. Initially, a1 = 1
2 (x+ y) and g1 = √xy
(i.e. the normal arithmetic and geometric means). Then, compute
an+1 = 1
2 (an + gn)
gn+1 = √angn
The two sequences will converge to the same number which is the arithmetic-geometric
mean of x,y. Obviously we cannot compute an infinite sequence, so we compute until
|an −gn|<ϵ for some small number ϵ.
Exercise 4.12. The integral of a function is a measure of the area under its curve. One
numerical method for computing the integral of a function f(x) on an interval [a,b] is
the rectangle rule. Specifically, an interval [ a,b] is split up into n equal subintervals of
size h= b−a
n . Then the integral is approximated by computing:
∫ b
a
f(x)dx≈
n−1∑
i=0
f(a+ ih) ·h
115",ComputerScienceOne.pdf
"4. Loops
Write a program to approximate an integral using the rectangle method. For this
particular exercise you will integrate the function
f(x) = sin x
x
For reference, the function is depicted in Figure 4.3. Write a program that will read the
end points a,b and the number of subintervals n and computes the integral of f using
the rectangle method. It should then output the approximation.
Figure 4.3.: Plot of f(x) = sin x
x
Exercise 4.13. Another way to compute an integral is to a technique called Monte
Carlo Integration, a randomized numerical integration method.
Given the interval [a,b], we enclose the function in a region of interest with a rectangle
of a known area Ar. We then randomly select n points within the rectangle and count
the number of random points that are within the function’s curve. If m of the n points
are within the curve, we can estimate the integral to be
∫ b
a
f(x) dx≈m
nAr
Consider again the function f(x) = sin(x)
x . Note that the global maximum and minimum",ComputerScienceOne.pdf
"∫ b
a
f(x) dx≈m
nAr
Consider again the function f(x) = sin(x)
x . Note that the global maximum and minimum
of this function are 1 and ≈−0.2172 respectively. Therefore, we can also restrict the
rectangle along the y-axis from −.25 to 1. That is, the lower left of the rectangle will be
(a,−.25) and the upper right will be ( b,1) for a known area of
Ar = |a−b|×1.25
Figure 4.4 illustrates the rectangle for the interval [ −5,5].
Write a program that will takes as input interval valuesa,b, and an integer nand perform
a Monte Carlo estimate of the integral of the function above. Realize that this is just an
approximation and it is randomized so your answers may not match exactly and may be
different on various executions of your program. Take care that you handle points within
the curve but under the x-axis correctly.
116",ComputerScienceOne.pdf
"4.8. Exercises
Figure 4.4.: A rectangle for the interval [ −5,5].
Exercise 4.14. Consider a ball trapped in a 2-D box. Suppose that it has an initial
position (x,y) within the box (the box’s dimensions are specified by its lower left ( xℓ,yℓ)
and an upper right ( xr,yr) points) along with an initial angle of travel θ in the range
[0,2π). As the ball travels in this direction it will eventually collide with one of the sides
of the box and bounce off. For this model, we will assume no loss of velocity (it keeps
going) and its angle of reflection is perfect.
Write a program that takes as input, x,y,θ,x ℓ,yℓ,xr,yr, and an integer n and computes
the first n−1 Euclidean points on the box’s perimeter that the ball bounces off of in its
travel (include the initial point in your printout for a total of npoints). You may assume
that the input will always be “good” (the ball will always begin somewhere inside the
box and the lower left and upper right points will not be reversed).",ComputerScienceOne.pdf
"box and the lower left and upper right points will not be reversed).
As an example, consider the inputs:
x= 1,y = 1,θ = .392699,xℓ = 0,yℓ = 0,xr = 4,yr = 3,n = 20
Starting at (1,1), the ball travels up and to the right bouncing off the right wall. Figure
4.5 illustrates this and the subsequent bounces back and forth.
Your output should simply be the points and should look something like the following.
(1.000000, 1.000000)
(4.000000, 2.242640)
(2.171572, 3.000000)
(0.000000, 2.100506)
(4.000000, 0.443652)
(2.928929, 0.000000)
(0.000000, 1.213202)
(4.000000, 2.870056)
(3.686287, 3.000000)
(0.000000, 1.473090)
117",ComputerScienceOne.pdf
"4. Loops
y
x
(1,1)
(1.284,3)
Figure 4.5.: Follow the bouncing ball
(3.556355, 0.000000)
(4.000000, 0.183764)
(0.000000, 1.840617)
(2.798998, 3.000000)
(4.000000, 2.502529)
(0.000000, 0.845675)
(2.041640, 0.000000)
(4.000000, 0.811179)
(0.000000, 2.468033)
(1.284282, 3.000000)
Exercise 4.15. An integer n ≥2 is prime if its only divisors are 1 and itself, n. For
example, 2,3,5,7,11,... are primes. Write a program that outputs all prime numbers 2
up to m where m is read as input.
Exercise 4.16.An integer n≥2 is prime if the only integers that evenly divide it are 1
and n itself, otherwise it is composite. The prime factorization of an integer is a list of
its prime divisors along with their multiplicities. For example, the prime decomposition
of 188,760 is:
188,760 = 2 ·2 ·2 ·3 ·5 ·11 ·11 ·13
Write a program that takes an integer n as input and outputs the prime factorization of
n. If nis invalid, an appropriate error message should be displayed instead. Your output",ComputerScienceOne.pdf
"n. If nis invalid, an appropriate error message should be displayed instead. Your output
should look something like the following.
118",ComputerScienceOne.pdf
"4.8. Exercises
1001 = 7 * 11 * 13
Exercise 4.17. One way of estimating π is to randomly sample points within a 2 ×2
square centered at the origin. If the distance between the randomly chosen point ( x,y)
and the origin is less than or equal to 1, then the point lies inside the unit circle centered
at the origin and we count it. If the point lies outside the circle then we can ignore it. If
y
x
Figure 4.6.: Sampling points in a circle
we sample n points and m of them lie within the circle, then π can be estimated as
π≈4m
n
Given a point (x,y), its distance from the origin is simply
√
x2 + y2
This idea is illustrated in Figure 4.6. Example code is given to randomly generate
numbers within a bound. Write a program that takes an integer nas input and randomly
samples n points within the 2 ×2 square and outputs an approximation of π.
Of course, you’ll need a way to generate random numbers within the range [ −1,1]. Since",ComputerScienceOne.pdf
"Of course, you’ll need a way to generate random numbers within the range [ −1,1]. Since
you are using some randomization, the result is just an approximation and may not
match exactly or even be the same between two different runs of your program.
119",ComputerScienceOne.pdf
"4. Loops
Figure 4.7.: Regular polygons
Exercise 4.18. A regular polygon is a polygon that is equiangular. That is, it has n
sides and n points whose angle from the center are all equal in measure. Examples for
n= 3 through n= 8 can be found in Figure 4.7.
Write a program that takes n and a radius r as inputs and computes the points of a
regular n-sided polygon centered at the origin (0 ,0). Each point should be distance r
from the origin and the first point should lie on the positive x-axis. Each subsequent
point should be at an angle θ equal to 2π
n from the previous point. Recall that given the
polar coordinates θ,r we can convert to cartesian coordinates ( x,y) using the following.
x = r·cos θ
y = r·sin θ
Your program should be robust enough to check for invalid inputs. If invalid, an error
message should be printed and the program should exit.
For example, running your program with n= 5,r = 6 should produce the points of a",ComputerScienceOne.pdf
"message should be printed and the program should exit.
For example, running your program with n= 5,r = 6 should produce the points of a
pentagon with “radius” 6. The output should look something like:
Regular 5-sided polygon with radius 6.0:
(6.0000, 0.0000)
(1.8541, 5.7063)
(-4.8541, 3.5267)
(-4.8541, -3.5267)
(1.8541, -5.7063)
Exercise 4.19. Let p1 = (x1,y1) and p2 = (x2,y2) be two points in the cartesian plane
which define a line segment. Suppose we travel along this line starting at p1 taking n
steps that are an equal distance apart until we reach p2. We wish to know which points
correspond to each of these steps and which step along this path is closest to anther
point p3 = (x3,y3). Recall that the distance between two points can be computed using
the Euclidean distance formula:
δ=
√
(x1 −x2)2 + (y1 −y2)2
Write a program that takes three points and an integernas inputs and outputs a sequence
of points along the line defined by p1,p2 that are distance δ
n apart from each other. It",ComputerScienceOne.pdf
"of points along the line defined by p1,p2 that are distance δ
n apart from each other. It
120",ComputerScienceOne.pdf
"4.8. Exercises
should also indicate which of these computed points is the closest to the third point.
For example, the execution of your program with inputs 0 ,2,−5.5,7.75,−2,3,10 should
produce output that looks something like:
(0.00, 2.00) to (-5.50, 7.75) distance: 7.9569
(0.00, 2.00)
(-0.55, 2.58)
(-1.10, 3.15)
(-1.65, 3.72) <-- Closest point to (-2, 3)
(-2.20, 4.30)
(-2.75, 4.88)
(-3.30, 5.45)
(-3.85, 6.02)
(-4.40, 6.60)
(-4.95, 7.17)
(-5.50, 7.75)
Exercise 4.20.The natural log of a number xis usually computed using some numerical
approximation method. One such method is to use the following Taylor series.
ln x= (x−1) −(x−1)2
2 + (x−1)3
3 −(x−1)4
4 + ···
However, this only works for |x−1|≤ 1 (except for x= 0) and diverges otherwise. For
x such that |x|>1, we can use the series
ln y
y−1 = 1
y + 1
2y2 + 1
3y3 + ···
where y = x
x−1 . Of course such an infinite computation cannot be performed by a
computer. Instead, we approximate ln x by computing the series out to a finite number",ComputerScienceOne.pdf
"computer. Instead, we approximate ln x by computing the series out to a finite number
of terms, n. Your program should print an error message and exit for x≤0; otherwise it
should use the first series for 0 <x ≤1 and the second for x> 1.
Another series that has better convergence properties and works for any range of x is as
follows
ln x= ln 1 + y
1 −y = 2y
(
1 + 1
3y2 + 1
5y4 + 1
7y6 ···
)
where y= (x−1)
(x+1) .
You will write a program that approximates ln xusing these two methods computed to n
terms. You will also compute the error of each method by comparing the approximated
value to the standard math library’s log function.
121",ComputerScienceOne.pdf
"4. Loops
Your program should accept x and n as inputs. It should be robust enough to reject
any invalid inputs ( ln x is not defined for x = 0 you may also print an error for any
negative value; n must be at least one). It will then compute an approximation using
both methods and print the relative error of each method.
For example, the execution of your program with inputs 3 .1415,6 should produce output
that looks something like:
Taylor Series: ln(3.1415) ~= 1.11976
Error: 0.02494
Other Series: ln(3.1415) ~= 1.14466
Error: 0.00004
Exercise 4.21.There are many different numerical methods to compute the square root
of a number. In this exercise, you will implement several of these methods.
(a) The Exponential Identity Method involves the following identity:
√x= e
1
2 ln (x)
Which assumes the use of built-in (or math-library) functions for eand the natural
log, ln.
(b) The Babylonian Method involves iteratively computing the following recurrence:
ai = 1
2
(
ai−1 + x
ai−1
)",ComputerScienceOne.pdf
"log, ln.
(b) The Babylonian Method involves iteratively computing the following recurrence:
ai = 1
2
(
ai−1 + x
ai−1
)
where a1 = 1.0. Computation is repeated until |ai −ai−1|≤ δ where δ is some
small constant value.
(c) A method developed for one of the first electronic computers (EDSAC [ 29]) involves
the following iteration. Let a0 = x, c0 = x−1. Then compute
ai+1 = ai −aici
2
ci+1 = c2
i (ci−3)
4
The iteration is performed for as many iterations as specified ( n), or until the
change in a is negligible. The resulting value for a is used as an approximation for√x≈a. However, this method only works for values of xsuch that 0 <x< 3. We
can easily overcome this by scaling x by some power of 4 so that the scaled value
of x satisfies 1
2 ≤x< 2. After applying the method we can then scale back up by
the appropriate value of 2 (since
√
4 = 2). Algorithm 4.17 describes how to scale x.
122",ComputerScienceOne.pdf
"4.8. Exercises
Write a program to compute the square root of an input number using these methods
and compare your results.
1 power ←0
2 while x< 1
2 do
//Scale up
3 x←(x·4)
4 power ←(power−1)
5 end
6 while x≥2 do
//Scale down
7 x←x
4
8 power ←(power+ 1)
9 end
Algorithm 4.17:Scaling a value xso that it satisfies 1
2 ≤x< 2. After execution,
power indicates what power of 2 the value x was scaled by.
Exercise 4.22. There are many different numerical methods to compute the natural
logarithm of a number, ln x. In this exercise, you will implement several of these methods.
(a) A formula version approximates the natural logarithm as:
ln(x) ≈ π
2M(1,4/s) −mln(2)
Where M(a,b) is the arithmetic-geometric mean and s= x2m. In this formula, m
is a parameter (a larger m provides more precision).
(b) The standard Taylor Series for the natural logarithm is:
ln(x) =
∞∑
n=1
(−1)n+1
n (x−1)n
As we cannot compute an infinite series, we will simply compute the series to the",ComputerScienceOne.pdf
"ln(x) =
∞∑
n=1
(−1)n+1
n (x−1)n
As we cannot compute an infinite series, we will simply compute the series to the
first m terms. Also note that this series is not convergent for values x> 1
(c) Borchardt’s algorithm is an iterative method that works as follows. Let
a0 = 1 + x
2 b0 = √x
Then repeat:
ak+1 = ak+bk
2
bk+1 =
√
ak+1bk
123",ComputerScienceOne.pdf
"4. Loops
until the absolute difference between ak,bk is small; that is |ak −bk|<ϵ. Then the
logarithm is approximated as
ln(x) ≈2 x−1
ak + bk
(d) Newton’s method works if x is sufficiently close to 1. It works by setting y0 = 1
and then computing
yn+1 = yn + 2x−eyn
x+ eyn
The iteration is performed m times.
To ensure that some of the methods above work, you may need to scale the number x to
be as close as possible to 1. One way to do this is to divide or multiply by e until x is
close to 1. Suppose we divided by e ktimes; that is x= z·ek where z is close to 1. Then
ln(x) = ln(z·ek) = ln(z) + ln(ek) = ln(z) + k
Thus, we can apply the methods above to Newton’s method to z and add k to the result
to get ln(x). A similar trick can be used to ensure that the Taylor Series method is
convergent.
Exercise 4.23. Write a program that takes a positive integer n as a command line
argument and outputs a list of all pairs of integers x,y such that 1 ≤x,y ≤n whose",ComputerScienceOne.pdf
"argument and outputs a list of all pairs of integers x,y such that 1 ≤x,y ≤n whose
sum, x+ y is prime. For example, if n= 5, the output should look something like the
following.
1 + 2 = 3 is prime
1 + 4 = 5 is prime
2 + 3 = 5 is prime
2 + 5 = 7 is prime
3 + 4 = 7 is prime
Exercise 4.24. Consider the following variation on the classical “FizzBuzz” challenge.
Write a program that will print out numbers 1 through n where n is provided as a
command line argument. However, if the number is a perfect square (that is, the square
of some integer; for example 1 = 1 2,4 = 22,9 = 32, etc.) print “Go Huskers” instead. If
the number is a prime (2 ,3,5,7, etc.) print “Go Cubs” instead.
Exercise 4.25. Implement a program to solve the classic “Rainfall Problem” (which
has been used in CS education studies to assess programming knowledge and skills).
Write an interactive program that repeatedly prompts the user to enter an integer (which
124",ComputerScienceOne.pdf
"4.8. Exercises
represents the amount of rainfall on a particular day) until it reads the integer 99999.
After 99999 is entered, it should print out the correct average. That is, it should not count
the final 99999. Negative values should also be ignored. For example, if the user entered
the sequence 4 0 -1 10 99999 the output should look something like the following.
Average rainfall: 4.66
Exercise 4.26.Write a program that takes an integer nand a subsequent list of integers
as command line arguments and determines which number(s) between 1 and n are
missing from the list. For example, if the following numbers are given to the program:
10 5 2 3 9 2 8 8 your output should look something like:
Missing numbers 1 thru 10:
1, 4, 6, 7, 10
Exercise 4.27. Write a program that takes a list of pairs of numbers representing
latitudes/longitudes (on the scale [−180,180] (negative values correspond to the southern",ComputerScienceOne.pdf
"latitudes/longitudes (on the scale [−180,180] (negative values correspond to the southern
and western hemispheres). Then, starting with the first pair, calculate the intermediate
air distances between each location as well as a final total distance.
To compute air distance from location A to a location B, use the Spherical Law of
Cosines:
d= arccos (sin(ϕ1) sin(ϕ2) + cos(ϕ1) cos(ϕ2) cos(∆))·R
where
•ϕ1 is the latitude of location A, ϕ2 is the latitude of location B
•∆ is the difference between location B’s longitude and location A’s longitude
•R is the (average) radius of the earth, 6,371 kilometers
Note: the formula above assumes that latitude and longitude are measured in radians r,
−π≤r≤π. To convert from degrees deg (−180 ≤deg≤180) to radians r, you can use
the simple formula:
r= deg
180π
For example, if the command line arguments were
40.8206 -96.756 41.8806 -87.6742 41.9483 -87.6556 28.0222 -81.7329
125",ComputerScienceOne.pdf
"4. Loops
your output should look something like:
(40.8206, -96.7560) to (41.8806, -87.6742): 766.8053km
(41.8806, -87.6742) to (41.9483, -87.6556): 7.6836km
(41.9483, -87.6556) to (28.0222, -81.7329): 1638.7151km
Total Distance: 2413.2040
Exercise 4.28. A DNA sequence is made up of a sequence of four nucleotide bases, A,
C, G, T (adenine, cytosine, guanine, thymine). One particularly interesting statistic of a
DNA sequence is finding a CG island : a subsequence that contains the highest frequency
of guanine and cytosine.
For simplicity, we will be interested in subsequences of a particular length, nthat will be
provided as part of the input.
Write a program that takes, as command line arguments, an integer n and a DNA
sequence. The program should then find all subsequences of the given DNA string of
length nwith the maximal frequency of C and G in it. For example, if the DNA sequence
is
ACAAGATGCCATTGTCCCCCGGCCTCCTGCTGCTGCTGCTCTCCGGGGCCACGGC",ComputerScienceOne.pdf
"length nwith the maximal frequency of C and G in it. For example, if the DNA sequence
is
ACAAGATGCCATTGTCCCCCGGCCTCCTGCTGCTGCTGCTCTCCGGGGCCACGGC
and the “window” size that we’re interested in is n= 5 then you would scan the sequence
and find every subsequence with the maximum number of C or G bases. Your output
should include all CG Islands (by indices) in the sequence similar to the following.
n = 5
highest frequency: 5 / 5 = 100.00%
CG Islands:
15 thru 20: CCCCC
16 thru 21: CCCCG
17 thru 22: CCCGG
18 thru 23: CCGGC
19 thru 24: CGGCC
42 thru 47: CCGGG
43 thru 48: CGGGG
44 thru 49: GGGGC
45 thru 50: GGGCC
126",ComputerScienceOne.pdf
"4.8. Exercises
Exercise 4.29. Write a program that will assist people in saving for retirement using a
tax-deferred 401k program.
Your program will read the following inputs as command line arguments.
•An initial starting balance
•A monthly contribution amount (we’ll assume its the same over the life of the
savings plan)
•An (average) annual rate of return (on the scale [0 ,1])
•An (average) annual rate of inflation (on the scale [0 ,1])
•A number of years until retirement
Your program will then compute a monthly savings table detailing the (inflation-adjusted)
interest earned each month, contribution, and new balance. The inflation-adjusted rate
of return can be computed with the following formula.
1 + rate of return
1 + inflation rate −1
To get the monthly rate, simply divide by 12. Each month, interest is applied to the
balance at this rate (prior to the monthly deposit) and the monthly contribution is added.
Thus, the earnings compound month to month.",ComputerScienceOne.pdf
"balance at this rate (prior to the monthly deposit) and the monthly contribution is added.
Thus, the earnings compound month to month.
Be sure that your program handles bad inputs as well as it can. It should also round to
the nearest cent for every figure. Finally, as of 2014, annual 401k contributions cannot
exceed $17,500. If the user’s proposed savings schedule violates this limit, display an
error message instead of the savings table.
For inputs 10000 500 0.09 0.012 10 your output should look something like the
following:
Month Interest Balance
1 $ 64.23 $ 10564.23
2 $ 67.85 $ 11132.08
3 $ 71.50 $ 11703.58
4 $ 75.17 $ 12278.75
5 $ 78.87 $ 12857.62
6 $ 82.58 $ 13440.20
7 $ 86.33 $ 14026.53
8 $ 90.09 $ 14616.62
9 $ 93.88 $ 15210.50
127",ComputerScienceOne.pdf
"4. Loops
...
116 $ 678.19 $ 106767.24
117 $ 685.76 $ 107953.00
118 $ 693.37 $ 109146.37
119 $ 701.04 $ 110347.41
120 $ 708.75 $ 111556.16
Total Interest Earned: $ 41556.16
Total Nest Egg: $ 111556.16
Exercise 4.30. An affine cipher is an encryption scheme that encrypts messages using
the following function:
ek(x) = (ax+ b) mod n
Where n is some integer and 0 ≤a,b,x ≤n−1. That is, we fix n, which will be used to
encode an alphabet as in Table 4.1.
x character
0 (space)
1 A
2 B
3 C
... ...
25 Y
26 Z
27 .
28 !
Table 4.1.: Character Mapping for n= 29
Then we choose integers a,b to define the encryption function. Suppose a= 10,b = 13,
then
ek(x) = (10x+ 13) mod 29
So to encrypt “HELLO!” we would encode it as 8 ,5,12,12,15,27, then encrypt them,
ek(8) = (10 ·8 + 13) mod 29 = 6
ek(5) = (10 ·5 + 13) mod 29 = 5
ek(12) = (10 ·12 + 13) mod 29 = 17
ek(12) = (10 ·12 + 13) mod 29 = 17
ek(15) = (10 ·15 + 13) mod 29 = 18
128",ComputerScienceOne.pdf
"4.8. Exercises
ek(28) = (10 ·28 + 13) mod 29 = 3
Which, when mapped back to characters using our encoding is “FEQQRC.”
To decrypt a message we need to invert the decryption function, that is,
dk(y) =
(
a−1 ·(y−b)
)
mod n
where a−1 is the inverse of amodulo n. The inverse of an integer ais the value such that
(a·a−1) mod n= 1
so for a= 10,n = 29, the inverse, 10−1 mod 29 = 3 since 3 ·10 mod 29 = 1. Given a and
n, how can we find an inverse, a−1? Obviously it cannot be zero, nor can it be 1 (1 is its
own inverse). There is a simple algorithm (the Extended Euclidean Algorithm) that can
solve this problem, but n= 29 is small enough that a brute-force strategy of testing all
possibilities will suffice.
Write a program that takes a,b and an encrypted message as command line arguments
and decrypts the message. Your program should print the decrypted message and other
cipher information to the standard output. For example:
a = 10
b = 13
a^-1 = 3
Encrypted Message: FEQQRC
Decrypted Message: HELLO!",ComputerScienceOne.pdf
"cipher information to the standard output. For example:
a = 10
b = 13
a^-1 = 3
Encrypted Message: FEQQRC
Decrypted Message: HELLO!
Exercise 4.31.A centroid (or barycenter) of a plane figure is the arithmetic mean of all
the points in the shape. The centroid of a non-self-intersecting closed polygon defined by
n vertices (x0,y0),(x1,y1),..., (xn1 ,yn−1) is the point ( Cx,Cy) which can be computed
using:
Cx = 1
6A
n−1∑
i=0
(xi + xi+1)(xi yi+1 −xi+1 yi)
Cy = 1
6A
n−1∑
i=0
(yi + yi+1)(xi yi+1 −xi+1 yi)
with A being the polygon’s signed area:
129",ComputerScienceOne.pdf
"4. Loops
(1,1.5)
(2,3.6)
(4.25,2.1)
(3.7,1.5)
(2.25,0.25)
(2.468, 1.902)
Figure 4.8.: A polygon and its centroid. Whoo!
A= 1
2
n−1∑
i=0
(xi yi+1 −xi+1 yi)
In these formulas, the vertices are assumed to be numbered in order of their occurrence
along the polygon’s perimeter. Furthermore, the vertex ( xn,yn) = (x0,y0).
Write a program that takes n pairs of x-y coordinates as command line arguments that
represent a polygon and computes its centroid. For example, the an input such as
1 1.5 2 3.6 4.25 2.1 3.7 1.5 2.25 0.25
would correspond to the polygon in Figure 4.8 with a centroid of (2 .46,1.90). The output
should look something like the following.
Centroid: (2.468405, 1.902874)
Exercise 4.32. Suppose we are given a set of n points (x0,y0),(x1,y1),..., (xn1 ,yn−1)
in the plane. Write a program that outputs instructions for someone to “walk” to each
point in the order that they are given starting at the origin by only going in the cardinal",ComputerScienceOne.pdf
"point in the order that they are given starting at the origin by only going in the cardinal
directions (not “as the bird flies”). For example, if the following input is given,
3 5 2 7 9 5
Then the output should look like the following.
Go north 5.00 units
Go east 3.00 units
Go north 2.00 units
130",ComputerScienceOne.pdf
"4.8. Exercises
Go west 1.00 units
Go south 2.00 units
Go east 7.00 units
Exercise 4.33.A histogram is a graphical representation of the distribution of numerical
data. Typically, a histogram is represented as a (vertical) bar graph. However, we’ll limit
our attention to graphing a horizontal ASCII histogram.
In particular, write a program that takes a sequence of numerical values representing
grades (on a 0–100 scale; reject any invalid values). The program will then display
a horizontal graph of the distribution of these values over 6 fixed ranges that looks
something like the following.
0 - 49 (2, 11.11%) ******
50 - 59 (1, 5.56%) ***
60 - 69 (1, 5.56%) ***
70 - 79 (5, 27.78%) **************
80 - 89 (4, 22.22%) ************
90 - 100 (5, 27.78%) **************
In the diagram, each star, * represents (at least) 2%
Exercise 4.34. Shannon entropy is a measure of information in a piece of data; specifi-
cally it is the expected value (average) of the information the data contains. A higher",ComputerScienceOne.pdf
"cally it is the expected value (average) of the information the data contains. A higher
entropy value corresponds to more random data while 0 indicates highly structured data.
Entropy is calculated using the following formula.
H(X) = −
n∑
i=1
p(xi) ·log2 p(xi)
where x1,x2,...,x n are the distinct symbols in the data sequence X and p(xi) is the
probability of the xi-th symbol. If the length of X is n, then this is simply
p(xi) = c(xi)
n
where c(xi) is a count of the number of times xi appears in X. For example, the string
“Hello World” has 8 distinct symbols with the probability of “l” being .2727, “o” being
.1818 and the remaining being .0909. Thus the entropy would be
H(’Hello World’) = 2.845351
131",ComputerScienceOne.pdf
"4. Loops
Write a program that takes a string as command line input (which may contain spaces)
and computes its entropy.
132",ComputerScienceOne.pdf
"5. Functions
In mathematics, a function is a mapping from a set of inputs to a set of outputs such
that each input is mapped to exactly one output. For example, the function
f(x) = x2
maps numeric values to their squares. The input is a variable x. When we assign an
actual value to xand evaluate the function, then the function has a value, its output. For
example, setting x= 2 as input, the output would be 2 2 = 4. Mathematical functions
can have multiple inputs,
f(x,y) = x2 + y2 f(x,y,z ) = 2x+ 3y−4z
However, a function will only ever have a single output value.
In programming languages, a function (sometimes called subroutine or procedure) can
take multiple inputs and produce one output value. We’ve already seen some examples
of these functions. For example, most languages provide a math library that you can
use to evaluate the square root or sine of a value x. We’ve also seen some functions
with multiple input values such as the “power” function that allows you to compute",ComputerScienceOne.pdf
"with multiple input values such as the “power” function that allows you to compute
f(x,y) = xy. Finally, the main entry point to many programs is defined by a main
function.
More formally, a function is a sequence of instructions (code) that is packaged into a
unit that can be reused. A function performs a specific task: given a number of inputs,
it executes some sequence of operations (executes some code) and “returns” (outputs) a
result. The output can be captured into a variable or used in an expression by whatever
code invoked or “called” the function.
Defining and using functions in programming has numerous advantages. The most
obvious advantage is that it allows you a way to organize code. By separating a program
it into distinct units of code it is more organized and it is clearer what each piece or
segment of code does. This also facilitates top-down design: one way to approach a
problem is to split it up into a series of subproblems until each subproblem is either",ComputerScienceOne.pdf
"problem is to split it up into a series of subproblems until each subproblem is either
trivial to deal with, or an existing, “off-the-shelf” solution already exists for it. Functions
may serve as the logical unit for each subproblem.
Another advantage is that by organizing code into functions, those functions can bereused
in multiple places either in your program/project or even in other programs/projects.
133",ComputerScienceOne.pdf
"5. Functions
A prime example of this are the standard libraries available in most programming
languages that provide functions to perform standard input/output or mathematical
functions. These standard libraries provide functions that are used by thousands of
different programs across multiple different platforms.
Functions also form an isolated unit of code. This allows for better and easier testing.
By isolating pieces of code, we can rigorously test those pieces of code by themselves
without worrying about the larger program or contexts.
Functions facilitates procedural abstraction. Placing code into functions allows you to
abstract the details of how the function computes its answer. As an example: consider
a standard math library’s square root function: it may use some interpolation method,
a Taylor series, or some other method entirely to compute the square root of a given
number. However, by putting this functionality into a function, we, as programmers, do",ComputerScienceOne.pdf
"number. However, by putting this functionality into a function, we, as programmers, do
not need to concern ourselves about these details. Instead, we simply use the function,
allowing us to focus on the larger issues at hand in our program.
5.1. Defining & Using Functions
Like variables, many programming languages may require that you declare a function
before you can use it. A function declaration may simply include a description of the
function’s input/output and name. A function declaration may require you to define the
function’s body at the same time or separately. Functions can also have scope: some
areas of the code may be able to “see” the function or know about it and be able to
invoke the function while other areas of the code may not be able to see the function
and therefore may not be able to invoke it.
Some interpreted programming languages use function hoisting which allows you to
use/invoke functions before you declare them. This works because the interpreter does",ComputerScienceOne.pdf
"use/invoke functions before you declare them. This works because the interpreter does
an initial scan of the code and identifies all function declarations. Only after it has
“hoisted” all functions into scope does it start to execute the program. Thus, a function
declaration can appear after it has been used and it will still work.
5.1.1. Function Signatures
A function is usually defined by its signature: every function can be identified by its
name (also called an identifier), its list of input parameters, and its output. A function
signature allows the programming language to uniquely identify each function so that
134",ComputerScienceOne.pdf
"5.1. Defining & Using Functions
when you invoke a function there is no ambiguity in which function you are calling.
1 Function sum(a, b)
2 x←a+ b
3 return x
4 end
Algorithm 5.1:A function in pseudocode. In this case, the name (identifier) of
the function is sum and it has two parameters, a and b. Its body is contained in
lines 2–3. Its return value is indicated by the return statement on line 3.
A function declaration in pseudocode is presented in Algorithm 5.1. In the pseudocode,
explicit variable types are omitted, and thus the return type is inferred from the return
statement. In Figure 5.1 we have provided an example of a function declaration in the C
programming language with each element labeled.
double getDistance(double x1, double y1, double x2, double y2);
Parameters
Return
Type
Identifier
(name)
Figure 5.1.: A function declaration (prototype) in the C programming language with the
return type, identifier, and parameter list labeled.",ComputerScienceOne.pdf
"(name)
Figure 5.1.: A function declaration (prototype) in the C programming language with the
return type, identifier, and parameter list labeled.
Some languages only allow you to use one identifier for one function (like variables) while
other languages allow you to define multiple functions with the same identifier as long
as the parameter list is different (see Section 5.3.2 below). In general, like variables,
function names are case sensitive. Also similar to variables, modern lower camel casing
is used with function names.
When defining the parameters to a function (its input), you usually provide a comma
delimited list of variable names. In the case of statically typed languages, the types of
the variable parameters are also specified. The order is important as when you invoke
the function, the number of inputs must match the number of parameters in the function
declaration. The variable types may also need to match. In some dynamically typed",ComputerScienceOne.pdf
"declaration. The variable types may also need to match. In some dynamically typed
languages, you may be able to call functions with different types or you may be able to
omit some of the parameters (see Section 5.3.4 below).
Similarly, the return type of the function may need to be specified in statically typed
languages while with dynamic languages, functions may conditionally return different
types. We generally refer to the “return value” or “return type” because when a function
135",ComputerScienceOne.pdf
"5. Functions
is done executing, it “returns” the control flow back to the line of code that invoked it,
returning its computed value.
You can also define functions that may not have any inputs or may not have any output.
Some languages use the keyword void to indicate no return value and such functions are
known as “void functions.” When a function doesn’t have any input values, its parameter
list is usually empty.
The function signature may be accompanied by the functionbody which contains the actual
code that specifies what the function does. Typically the function body is demarcated
with a code block using opening and closing curly brackets, { ... }. Within the
function you can generally write any valid code including declaring variables. When you
declare a variable inside a function, however, it is local to that function. That is, the
variable’s scope is only defined within the function. A local variable cannot be accessed",ComputerScienceOne.pdf
"variable’s scope is only defined within the function. A local variable cannot be accessed
outside the function, indeed the local variable does not usually survive when the function
ends its execution and returns control back to line of code that called it. Function
parameters are essentially locally scoped variables as well and can usually be treated as
such.
5.1.2. Calling Functions
When a function has been defined and is in scope, you can invoke or “call” the function
by coding the function name and providing the input parameters which can be either
variables or literals. When provided as inputs, parameters are referred to as arguments
to the function. The arguments are typically provided as a comma delimited list and
placed inside parentheses.
Invoking a function changes the usual flow of control. When invoked, control flow is
transferred over to the function. When the function finishes executing the code in its",ComputerScienceOne.pdf
"transferred over to the function. When the function finishes executing the code in its
body, control flow returns to the point in the code that invoked it. It is common for a
program to be written so that a function calls another function and that function calls
another. This can form a deep chain of function calls in which the flow of control is
transferred multiple times. Upon the completion of each function, control is returned
back to the function that called it, referred to as the calling function.
If a function returns a value it can either be captured in a variable using an assignment
136",ComputerScienceOne.pdf
"5.2. How Functions Work
operator or by using it in an expression.
1 a←10
2 b←20
3 c←sum(a,b)
Algorithm 5.2:Using a function. We invoke a function by indicating its name
(identifier) and passing it arguments.
5.1.3. Organizing
Functions provide code organization, but functions themselves should also be organized.
We’ve seen this with standard libraries. Functions that provide basic input/output are
all grouped together into one library. Functions that involve math functions are grouped
together into a separate math library.
Some languages allow you to define and “import” individual libraries which organize
similar functions together. Some languages do this by collecting functions into “utility”
classes or modules. Only when you import these modules do the functions come into
scope and can be used in your code. If you do not import these modules, then the
functions are out of scope and cannot be used.
In some languages once a function is imported it is part of the global scope and can be",ComputerScienceOne.pdf
"functions are out of scope and cannot be used.
In some languages once a function is imported it is part of the global scope and can be
“seen” by any part of the code. This can cause conflicts: if you import modules from two
different libraries each with different functions that have the same name or signature,
then the two function definitions may be in conflict or it may make your code ambiguous
as to which function you intend to invoke. This is sometimes referred to as “polluting the
namespace.” There are several techniques that can avoid this situation. Some languages
allow you to place functions into a namespace to keep functions with the same name in
different “spaces.” Other languages allow you to place functions into different classes and
then invoke them by explicitly specifying which class’s function you want to call. Yet
other languages don’t have great support for function organization and it is the library",ComputerScienceOne.pdf
"other languages don’t have great support for function organization and it is the library
designer’s responsibility to avoid naming conflicts, typically by adding a library-specific
prefix to every function.
5.2. How Functions Work
To understand how functions work in practice, it is necessary to understand how a
program operates at a lower level. In particular, each program has a program stack (also
called a call stack). A stack is a data structure that holds elements in a Last-In First-Out
137",ComputerScienceOne.pdf
"5. Functions
(LIFO) manner. Elements are added to the “top” of the stack in an operation called
push and elements can be removed from the top of the stack in an operation called pop.
In general, elements cannot be inserted or removed from the middle or “bottom” of the
stack.
In the context of a program, a call stack is used to keep track of the flow of control.
Depending on the operating system, compiler and architecture, the details of how
elements are stored in the program stack may vary. In general when a program begins,
the operating system loads it into memory at the bottom of the call stack. Global
variables (static data) are stored on top of the main program. Each time a function is
called, a new stack frame is created and pushed on top of the stack. This frame contains
enough space to hold values for the arguments passed to the function, local variables
declared and used by the function, as well as a space for a return value and a return",ComputerScienceOne.pdf
"declared and used by the function, as well as a space for a return value and a return
address. The return address is a memory location that the program should return to after
the execution of the function. That way, when the function finishes its execution, the
stack frame can be removed (popped) and the lower stack frame of the calling function
is preserved. This is a very efficient way to keep track of the flow of control in a program.
As each function calls another function, each stack frame is preserved by pushing a new
one on top of the program stack. Each time a function terminates execution and returns,
the removal of the stack frame means that all local variables go out of scope . Thus,
variables that are local to a function are not accessible outside the function.
To illustrate, consider the following snippet of C code. The main() function invokes
the average() function which in turn invokes the sum() function. Each invocation",ComputerScienceOne.pdf
"the average() function which in turn invokes the sum() function. Each invocation
creates a new stack frame on top of the last in the program stack which is depicted in
Figure 5.2.
1 double sum(double a, double b) {
2 double x = a + b;
3 return x;
4 }
5
6 double average(double a, double b) {
7 double y = sum(a, b) / 2.0;
8 return y;
9 }
10
11 int main(int argc, char **argv) {
12 double n = 10.0;
13 double m = 16.0;
14 double ave = average(n, m);
15 printf(""average = %f\n"", ave);
16 return 0;
17 }
138",ComputerScienceOne.pdf
"5.2. How Functions Work
Bottom of the stack
(high memory)
Top of the stack
(low memory)
Unused stack space
Program Code
Global Variables
Static Content
argumentsargc, argv
return address, value (0)
local variablesn, m, ave
main()stack frame
argumentsa, b
return address, value (13.0)
local variablesy
average()stack frame
argumentsa, b
return address, value (26.0)
local variablesx
sum()stack frame
Figure 5.2.: Program Stack. At the bottom we have the program’s code, followed by
static content such as global variables. Each function call has its own stack
frame along with its own arguments and local variables. In particular, the
variable arguments a and b in two different stack frames are completely
different variables. Upon returning from the sum() function call, the top-
most stack frame would be popped and removed, returning to the code
for the average() function via the return address. The stack is depicted
bottom-up with high memory at the bottom and low memory at the top,",ComputerScienceOne.pdf
"for the average() function via the return address. The stack is depicted
bottom-up with high memory at the bottom and low memory at the top,
but this may differ depending on the architecture.
139",ComputerScienceOne.pdf
"5. Functions
5.2.1. Call By Value
When a function is invoked, arguments are passed to it. When you invoke a function you
can pass it variables as arguments. However, the variables themselves are not passed to
the function, but instead the values stored in the variables at the time that you call the
function are passed to the function. This mechanism is known as call by value and the
variable values are passed by value to the function.
Recall that the arguments passed to a function are placed in a new stack frame for that
function. Thus, in reality copies of the values of the variables are passed to the function.
Any changes to the parameters inside the function have no effect on the original variables
that were “passed” to the function when it was invoked.
To illustrate, consider the following C code. We have a function sum that takes two
integer parameters a and b which are passed by value. Inside sum, we create another",ComputerScienceOne.pdf
"integer parameters a and b which are passed by value. Inside sum, we create another
variable x which is the sum of the two passed variables. We then change the value
of the first variable, a to 10. Elsewhere in the code we call sum on two variables,
n, m with values 5 and 15 respectively. The invocation of the function sum means
that the two values, 5 and 15, stored in the variables are copied into a new stack frame.
Thus, changing the value to the first parameter changes the copy and has no effect on
the variable n. At the end of this code snippet n retains its original value of 5. The
program stack frames are depicted in Figure 5.3.
1 int sum(int a, int b) {
2 int x = a + b;
3 a = 10;
4 return x;
5 }
6
7 ...
8
9 int n = 5;
10 int m = 15;
11 int k = sum(n, m);
5.2.2. Call By Reference
Some languages allow you to pass a parameter to a function by providing its memory
address instead of the value stored in it. Since the memory address is being provided",ComputerScienceOne.pdf
"address instead of the value stored in it. Since the memory address is being provided
to the function, the function is able to access the original variable and manipulate the
140",ComputerScienceOne.pdf
"5.2. How Functions Work
0x0010 n = 5
0x0014 m = 10
0x0018 k
calling function
stack frame
0x0080 a = 5
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(a) Upon invocation of the sum() function, a new
stack frame is created which holds the parameters
and local variable. The parameter variables a
and b are distinct from the original argument
variables n and m .
0x0010 n = 5
0x0014 m = 10
0x0018 k
calling function
stack frame
0x0080 a = 10
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(b) The change to the variable a in the sum()
function changes the parameter variable, but the
original variable n is unaffected.
0x0010 n = 5
0x0014 m = 10
0x0018 k
calling function
stack frame
0x0080 a = 10
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(c) When the sum() function finishes execution,
its stack frame is removed and the variables a ,
b , and x are no longer valid. The return value
20 is stored in another return value location.
0x0010 n = 5
0x0014 m = 10
0x0018 k = 20
calling function",ComputerScienceOne.pdf
"20 is stored in another return value location.
0x0010 n = 5
0x0014 m = 10
0x0018 k = 20
calling function
stack frame
0x0080 a = 10
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(d) The returned value is stored in the variable k
and the variable n retains its original value.
Figure 5.3.: Demonstration of Pass By Value. Passing variables by value means that
copies of the values stored in the variables are provided to the function.
Changes to parameter variables do not affect the original variables.
141",ComputerScienceOne.pdf
"5. Functions
contents stored at that memory address. In particular, the function is now able to make
changes to the original variable. This mechanism is known as call by reference and the
variables are passed by reference.
To illustrate, consider the following C code. Here, the variable a is passed by reference
( b is still passed by value, the *a and &n in the following code are dereferencing and
referencing operators respectively. For details, see Section 18.2). Below when we invoke
the sum() function, we pass not the value stored in n, but the memory address of the
variable n. Thus, when we change the value of the variable a in the function, we are
actually changing the value of n (since we have access to its memory location). At the
conclusion of this snippet of code, the value stored in n has been changed to 10. The
program stack frames are depicted in Figure 5.4.
1 int sum(int *a, int b) {
2 int x = *a + b;
3 *a = 10;
4 return x;
5 }
6
7 ...
8
9 int n = 5;
10 int m = 15;",ComputerScienceOne.pdf
"1 int sum(int *a, int b) {
2 int x = *a + b;
3 *a = 10;
4 return x;
5 }
6
7 ...
8
9 int n = 5;
10 int m = 15;
11 int k = sum(&n, m);
Whether or not a variable is passed by value or by reference depends on the language,
type of variable, and the syntax used.
5.3. Other Issues
5.3.1. Functions as Entities
In programming languages, any entity that can be stored in a variable or passed as
an argument to a function or returned as a value from a function is referred to as a
“first-class citizen.” Numerical values for example are usually first-class citizens as they
can be stored in variables and passed to functions.
Functional Programming is a programming language paradigm in which functions them-
selves are first-class citizens. That is, functions can be assigned to variables, functions
142",ComputerScienceOne.pdf
"5.3. Other Issues
0x0010 n = 5
0x0014 m = 10
0x0018 k
calling function
stack frame
0x0080 a = 0x0010
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(a) Upon invocation of the sum() function, a new
stack frame is created which holds the parameters
and local variable. The parameter variable a
holds the memory location of the original variable
n .
0x0010 n = 10
0x0014 m = 10
0x0018 k
calling function
stack frame
0x0080 a = 0x0010
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(b) The change to the variable a in the sum()
function actually changes what the variable a
references. That is, the original variable n .
0x0010 n = 10
0x0014 m = 10
0x0018 k
calling function
stack frame
0x0080 a = 0x0010
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(c) When the sum() function finishes execution,
its stack frame is removed and the variables a ,
b , and x are no longer valid. The return value
20 is stored in another return value location.
0x0010 n = 10
0x0014 m = 10
0x0018 k = 20",ComputerScienceOne.pdf
"b , and x are no longer valid. The return value
20 is stored in another return value location.
0x0010 n = 10
0x0014 m = 10
0x0018 k = 20
calling function
stack frame
0x0080 a = 0x0010
0x0084 b = 15
0x0088 x = 20sum()
stack frame
...
(d) The returned value is stored in the variable k
and the value in the variable n has now changed.
Figure 5.4.: Demonstration of Pass By Reference. Passing variables by reference means
that the memory address of the variables are provided to the function. The
function is able to make changes to the original variable because it knows
where it is stored.
143",ComputerScienceOne.pdf
"5. Functions
can be passed to other functions as arguments, and functions can even return functions
as a result. This is done as a matter of course in functional programming languages
such as Haskell and Clojure, but many programming languages contain some functional
aspects.
For example, some languages support the same concept by using function pointers which
are essentially references to where the function is stored in memory. As a memory
location is essentially a number, it can be passed around in functions and be stored in a
variable. Purists would argue that this is not sufficient to call a function a “first-class
citizen” in such a language. They may argue that a language must be able to create
new functions at runtime for it to be considered a language in which functions are “true”
first-class citizens.
In any case, there are several advantages to being able to pass functions around as
arguments or store them in variables. Passing a function to another function as an",ComputerScienceOne.pdf
"arguments or store them in variables. Passing a function to another function as an
argument gives you the ability to provide a callback. A callback is simply a function
that gets passed to another function as an argument. The idea is that the function that
receives the callback will execute or “call back” the passed function at some point.
Using callbacks enables us to program a “generic” function that provides some generalized
functionality. Then more specific behavior can be be implemented in the callback function.
For example, we could create a generic sort function that sorts elements in a collection.
We could make the sort function generic so that it could sort any type of data: numbers,
strings, objects, etc. A callback would provide more specific behavior on how to order
individual elements in the sorted array.
As another example, consider GUI Programming in which we want to design a user
interface. In particular, we may be able to create a button element in our interface. We",ComputerScienceOne.pdf
"interface. In particular, we may be able to create a button element in our interface. We
need to be able to specify what happens when the user clicks the button. This could be
achieved by passing in a function as a callback to “register” it with the click “event.”
A related issue is anonymous functions (also known as lambda expressions). Typically, we
simply want to create a function so that we can pass it as a callback to another function.
We may have no intention of actually calling this function directly as it may not be of
much use other than passing it as a callback. Some languages allow you to define a
function “inline” without an identifier so that it can be passed to another function. Since
the function has no name and cannot be invoked by other sections of the code (other
than the function we passed it to), it is known as an anonymous function.
5.3.2. Function Overloading
Some languages do not allow you to define more than one function with the same name",ComputerScienceOne.pdf
"5.3.2. Function Overloading
Some languages do not allow you to define more than one function with the same name
in the same scope. This is to prevent ambiguity in the code. When you write code to
invoke a function and there are several functions with that name, which one are you
144",ComputerScienceOne.pdf
"5.3. Other Issues
actually calling?
Some languages do allow you to define multiple functions with the same name as long as
they differ in either the number (also called arity) or type of parameters. For example,
you could define two absolute value, |x|functions with the same name, but one of them
takes a floating point number while the other takes an integer as its parameter. This
is known as function overloading because you are “overloading” the code by defining
multiple functions with the same name.
The ambiguity problem is solved by requiring that each function with the same name
differs in their parameters. If you invoke the absolute value function and pass it a floating
point number, clearly you meant to call the first version. If you passed it an integer, it is
clear that you intended to invoke the second version. Depending on the type and number
of arguments you pass to a function, the compiler or interpreter is able to determine",ComputerScienceOne.pdf
"of arguments you pass to a function, the compiler or interpreter is able to determine
which version you intend to call and is able to make the right function call. This process
is known as static dispatch.
In a language without function overloading, we would be forced to use different names
for functions that perform the same operation but on different types.
5.3.3. Variable Argument Functions
Many languages allow you to define special functions that take a variable number of
parameters. Often they are referred to as “vararg” (short for variable argument) functions.
The syntax for doing so varies as does how you can write a function to operate on a
variable number of arguments (usually through some array or collection data structure).
The standard printf (print formatted) function found in many languages is a good
example of a vararg function. The printf function allows you to use one function to
print any number of variables or values. Without a vararg function, you would have to",ComputerScienceOne.pdf
"print any number of variables or values. Without a vararg function, you would have to
implement a printf version for no arguments, 1 argument, 2 arguments, etc. Even
then, you would only be able to support up to n arguments for as many functions as
you defined. By using a vararg function, we can write a single function that operates
on all of these possibilities. It is import to note, a vararg function is not an example
of function overloading. There is still only one function defined, it just takes a variable
number of arguments.
5.3.4. Optional Parameters & Default Values
Suppose that you define a function which has, say, three parameters. Now suppose you
invoke the function but only provide it 2 of the 3 arguments that it expects. Some
languages would not allow this and it would be considered a syntax or runtime error. Yet
145",ComputerScienceOne.pdf
"5. Functions
other languages may have very complex rules about what happens when an argument is
omitted. Some languages allow you to omit some arguments when calling functions as a
feature of the language. That is, the parameters to a function are optional.
When a language allows parameters to be optional, it usually also allows you to define
default values to the parameters if the calling function does not provide them. If a
user calls the function without specifying a parameter, it takes on the default value.
Alternatively, the default could be a non-value like “null” or “undefined.” Inside the
function you could implement logic that determined whether or not a parameter was
passed to the function and alter the behavior of the function accordingly.
5.4. Exercises
Exercise 5.1. Recall that the greatest common divisor (gcd) of two positive integers, a
and b is the largest positive integer that divides both a and b. Adapt the solution from",ComputerScienceOne.pdf
"and b is the largest positive integer that divides both a and b. Adapt the solution from
Exercise 4.6 into a function. If the language you use supports it, return the gcd via a
pass by reference variable.
Exercise 5.2. Write a function that scales an input x to to its scientific notation scale
so that 1 ≤x <10. If you language supports pass by reference, the amount that x is
shifted should be stored in a pass-by-reference parameter. For example, a call to this
function with x= 314.15 should return 3 .1415 and the amount it is scaled by is n= −2.
Exercise 5.3.Write a function that returns the most significant digit of a floating point
number. The function should only return an integer in the range 1–9 (it should return
zero only if x= 0).
Exercise 5.4. Write a function that, given an integer x, sums the values of its digits.
That is, for x= 29423 the sum 2 + 9 + 4 + 2 + 3 = 20.
Exercise 5.5. Write a function to convert radians to degrees using the formula,
deg= 180 ·rad
π",ComputerScienceOne.pdf
"That is, for x= 29423 the sum 2 + 9 + 4 + 2 + 3 = 20.
Exercise 5.5. Write a function to convert radians to degrees using the formula,
deg= 180 ·rad
π
Write another function to covert degrees to radians.
Exercise 5.6. Write functions to compute the diameter, circumference and area of a
circle given its radius. If your language supports pass by reference, compute all three of
these with one function.
Exercise 5.7.The arithmetic-geometric mean of two numbers x,y, denoted M(x,y) (or
agm(x,y)) can be computed iteratively as follows. Initially, a1 = 1
2 (x+ y) and g1 = √xy
(i.e. the normal arithmetic and geometric means). Then, compute
an+1 = 1
2 (an + gn)
gn+1 = √angn
146",ComputerScienceOne.pdf
"5.4. Exercises
The two sequences will converge to the same number which is the arithmetic-geometric
mean of x,y. Obviously we cannot compute an infinite sequence, so we compute until
|an −gn|<ϵ for some small number ϵ.
Exercise 5.8.Write a function to compute the annual percentage yield (APY) given an
annual percentage rate (APR) using the formula
APY = eAPR −1
Exercise 5.9.Write a function that will compute the air distance between two locations
given their latitudes and longitudes. Use the formula as in Exercise 2.14.
Exercise 5.10. Write a function to convert a color represented in the RGB (red-green-
blue) color model (used in digital monitors) to a CMYK (cyan-magenta-yellow-key) used
in printing. RGB values are integers in the range [0 ,255] while CMYK are fractional
numbers in the range [0 ,1]. To convert to CMYK, you first need to scale each integer
value to the range [0 ,1] by simply computing
r′= r
255, g ′= g
255, b ′= b
255
and then using the following formulas:",ComputerScienceOne.pdf
"value to the range [0 ,1] by simply computing
r′= r
255, g ′= g
255, b ′= b
255
and then using the following formulas:
K = 1 −max{r′,g′,b′}
C = (1 −r′−k)/(1 −k)
M = (1 −g′−k)/(1 −k)
Y = (1 −b′−k)/(1 −k)
Exercise 5.11. Write a function to convert from CMYK to RGB using the following
formulas.
r= 255 ·(1 −C) ·(1 −K)
g= 255 ·(1 −M) ·(1 −K)
b= 255 ·(1 −Y) ·(1 −K)
Exercise 5.12.Write some functions to convert an RGB color to a gray scale, “removing”
the color values. An RGB color value is grayscale if all three components have the same
value. To transform a color value to grayscale, there there are several possible techniques.
The average method simply sets all three values to the average:
r+ g+ b
3
The lightness method averages the most prominent and least prominent colors:
max{r,g,b }+ min{r,g,b }
2
147",ComputerScienceOne.pdf
"5. Functions
The luminosity technique uses a weighted average to account for a human perceptual
preference toward green:
0.21r+ 0.72g+ 0.07b
Exercise 5.13. Adapt the methods to compute a square root in Exercise 4.21 into
functions.
Exercise 5.14. Adapt the methods to compute the natural logarithm in Exercise 4.22
into functions.
Exercise 5.15. Weight (mass in the presence of gravity) can be measured in several
scales: kilogram force (kgf), pounds (lbs), ounces (oz), or Newtons (N). To convert
between these scales, you can use the following facts:
•1 kgf is equal to 2.20462 pounds
•There are 16 ounces in a pound
•1 kgf is equal to 9.80665 Newtons
Write a collection of functions to convert between these scales.
Exercise 5.16. Length can be measured by several different units. We will concern
ourselves with the following scales: kilometer, mile, nautical mile, and furlong. A measure
in each one of these scales can be converted to another using the following facts.",ComputerScienceOne.pdf
"in each one of these scales can be converted to another using the following facts.
•One mile is equivalent to 1.609347219 kilometers
•One nautical mile is equivalent to 1.15078 miles
•A furlong is 1
8 -th of a mile
Write a collection of functions to convert between these scales.
Exercise 5.17.Temperature can be measured in several scales: Celsius, Kelvin, Fahren-
heit, and Newton. To convert between these scales, you can use the following conversion
table.
From/To Celsius Kelvin Fahrenheit Newton
Celsius – c+ 273.15 c9
5 + 32 c33
100
Kelvin k−273.15 – 9
5 k−459.67 .33k−90.1395
Fahrenheit (f −32)5
9
5
9 f + 255.372 – 11
60 f −88
15
Newton n100
33
100
33 n+ 273.15 60
11 n+ 32 –
Table 5.1.: Conversion Chart
Write a collection of functions to convert between these scales.
148",ComputerScienceOne.pdf
"5.4. Exercises
Exercise 5.18. Energy can be measured in several different scales: calories ( c), joules
(J), ergs ( erg) and foot-pound force (ft-lbf) among others. To convert between these
scales, you can use the following facts:
•1 erg equals 1 .0 ×10−7J
•1 ft-lbs equals 1 .3558 joules
•1 calorie is equal to 4 .184 joules
Write a collection of functions to convert between these scales.
Exercise 5.19. Pressure is a measure of force applied to the surface of an object per
unit area. There are several units that can be used to measure pressure:
•Pascal (Pa) which is one Newton per square meter
•Pound-force Per Square Inch (psi)
•Atmosphere (atm) or standard atmospheric pressure
•The torr, an absolute scale for pressure
To convert between these units, you can use the following formulas.
•1 psi is equal to 6,894.75729 Pascals, 1 psi is equal to 0 .06804596 atmospheres
•1 atmosphere is equal to 101,325 Pascals
•1 torr is equal to 1
760 atmosphere and 101,325
760 Pascals",ComputerScienceOne.pdf
"•1 atmosphere is equal to 101,325 Pascals
•1 torr is equal to 1
760 atmosphere and 101,325
760 Pascals
Write a collection of functions to convert between these scales.
149",ComputerScienceOne.pdf
"6. Error Handling
Writing perfect code is difficult. The more complex a system or code base, the more
likely it is to have bugs. That is, flaws or mistakes in a program that result in incorrect
behavior or unintended consequences. The term “bug” has been used in engineering
for quite a while. The term was popularized in the context of computer systems by
Grace Hopper who, when working on the Naval Mark II computer in 1946, tracked a
malfunction to a literal bug, a moth, trapped in a relay [2].
Some of the biggest modern engineering failures can be tracked to simple software bugs.
For example, on September 26th, 1983 a newly installed Soviet early warning system
gave indication that nuclear missiles had been launched on the Soviet Union by the
United States. Stanislav Petrov, a lieutenant colonel in the Soviet Air Defense Forces and
duty officer at the time, did not trust the new system and did not report the incident to",ComputerScienceOne.pdf
"duty officer at the time, did not trust the new system and did not report the incident to
superiors who may have ordered a counter strike. Petrov was correct as the false alarm
was caused by sunlight reflections off of high altitude clouds as well as other bugs in the
newly deployed system [28].
In September 1999 the Mars Climate Orbiter, a project intended to study the Martian
climate and atmosphere was lost after it entered into the upper atmosphere of Mars and
disintegrated. The error was due to a subsystem that measured the craft’s momentum
in a non-standard pound force per second unit when all other systems expected the
standard newton second unit [ 1]. The loss of the project was calculated at over $125
million.
There are numerous other examples, some that have caused inconvenience to users (such
as the Zune bug mentioned in Section 4.5.2) to bugs in medical devices that have cost
dozens of lives to those resulting in the loss of millions of dollars [6].",ComputerScienceOne.pdf
"dozens of lives to those resulting in the loss of millions of dollars [6].
In some sense, Software Engineering and programming is unique. If you build a bridge
and forget one bolt its likely not going to cause the bridge to collapse. If you draw
up plans for a development and the land survey is a few inches off overall, its not a
catastrophic problem. However, if you forget one character or are off by one number in a
program, it can cause a complete system failure.
There are a variety of reasons for why bugs make it into systems. Bugs could be the result
of a fundamental misunderstanding of the problem or requirements. Poor management
and the pressure of time constraints to deliver a project may make developers more
careless. A lack of proper testing may mean many more bugs survive the development
151",ComputerScienceOne.pdf
"6. Error Handling
process than otherwise should have. Even expert programmers can overlook a simple
mistake when writing thousands of lines of code.
Given the potential for error, it is important to have good software development method-
ologies that emphasize testing a system at all levels. Working in teams where regular
code reviews are held so that colleagues can examine, critique, and catch potential bugs
are essential for writing robust code.
Modern coding tools and techniques can also help to improve the robustness of code. For
example, debuggers are tools that help a developer debug (that is, find and fix the cause
of an error) a program. Debuggers allow you to simulate the execution of a program
statement-by-statement and view the current state of the program such as variable values.
You can “step through” the execution line by line to find where an error occurs in order
to localize and identify a bug.",ComputerScienceOne.pdf
"You can “step through” the execution line by line to find where an error occurs in order
to localize and identify a bug.
Other tools allow you to perform static analysis on source code to search for potential
problems. That is, problems that are not syntax errors and are not necessarily bugs
that are causing problems, but instead are anti-patterns or code smells. Anti-patterns
are essentially common bad-habits that can be found in code. They are an attempted
solution to a commonly encountered problem but which don’t actually solve the problem
or introduces new problems. Code smells are “symptoms” in a source code that indicate
a possible deeper design or implementation flaw. Failure to adhere to good programming
principles such as properly initializing variables or failure to check for null values are
examples of smells. Static analysis tools automatically examine the code base for potential
issues like these. For example, a lint (or linter) is a tool that can examine source code",ComputerScienceOne.pdf
"issues like these. For example, a lint (or linter) is a tool that can examine source code
suspicious or non-portable code or code that does not comply with generally accepted
standards or ways of doing things.
Even if code contains no bugs, it is still susceptible to errors. For example, a program
could connect to a remote database to query and process data. However, if the network
connection is temporarily unavailable, the program will not be able to execute properly.
Because of the potential of such errors, it is important to write code that is not only bug
free but is also robust and resilient. We must anticipate possible error conditions and
write code to detect, prevent, or recover from such errors. Generally, this is referred to
as error handling.
Much of what we now consider Software Engineering was pioneered by people like
Margaret Hamilton who was the lead Apollo flight software designer at NASA. During",ComputerScienceOne.pdf
"Margaret Hamilton who was the lead Apollo flight software designer at NASA. During
the Apollo 11 Moon landing (1969), an error in one system caused the lander’s computer
to become overworked with data. However, because the system was designed with a
robust architecture, it could detect and handle such situations by prioritizing more
important tasks (those related to landing) over lower priority tasks. The resilience that
was built into the system is credited with its success [11].
152",ComputerScienceOne.pdf
"6.1. Error Handling
6.1. Error Handling
In general, errors are potential conditions or situations that can reasonably be anticipated
by a developer. For example, if we write code to open and process a file, there are several
things that could go wrong. The file may not exist, or we may not have permission on
the system to read it, or the formatting in the file may be corrupted or not as expected.
Still yet, everything could be fine with the file, but it may contain erroneous or invalid
values.
If an error can be anticipated, we may be able to write code that detects the particular
error condition and handles it by executing code that may be able to recover from the
error condition. In the case of a missing file for example, we may be able to prompt the
user for an alternate file.
We may be able to detect but not necessarily recover from certain errors. For example, if
the file has been corrupted in example above, there may not be a way to properly “fix”",ComputerScienceOne.pdf
"the file has been corrupted in example above, there may not be a way to properly “fix”
it. If it contains invalid data, we may not even want the program to fix it as it may
indicate a bug or other issue that needs to be addressed. Still yet, there may be some
error conditions that we cannot recover from at all as they are completely unexpected.
In such instances, we may want the error to result in the termination of the program in
which case the error is considered fatal.
6.2. Error Handling Strategies
There are several general strategies for performing error handling. We’ll look at two
general methods here: defensive programming and exceptions.
6.2.1. Defensive Programming
Defensive programming is a “look before you leap” strategy. Suppose we have a potentially
“dangerous” section of code; that is, a line or block of code whose execution could encounter
or result in an error condition. Before we execute the code, we perform a check to see",ComputerScienceOne.pdf
"or result in an error condition. Before we execute the code, we perform a check to see
if the error condition is present (usually using a conditional statement). If the error
condition does not hold, then we proceed with the code as normal. However, if the error
condition does hold, instead of executing the code, we execute alternative code that
handles the error.
Suppose we are about to divide by a number. To prevent a division by zero error, we
can check if our denominator is zero or not. If it is, then we raise or handle the error
instead of performing the division. What should be done in such a case? We could, as an
alternative, use a predefined value as a result instead. Or we could notify the user and
153",ComputerScienceOne.pdf
"6. Error Handling
ask for an alternative. Or we could log the error and proceed as normal. Or we could
decide that the error is so egregious that it should be fatal and terminate the execution
of the program.
Which is the right way to handle this error? It really depends on your design requirements
really. This raises the question, though: “who” is responsible for making these decisions?
Suppose we’re designing a function for a library that is not just for our project but others
as well (as is the case with the standard library functions). Further, the function we’re
designing could have multiple different error conditions that it checks for. In this scenario
there are two entities that could handle the errors: the function itself and the code that
invokes the function.
Suppose that we decide to handle the errors inside the function. As designers of the
function, we’ve made the decision to handle the errors for the user (the code that invokes",ComputerScienceOne.pdf
"function, we’ve made the decision to handle the errors for the user (the code that invokes
our function). Regardless of how we decide to handle the errors, this design decision has
essentially taken any decision making ability away from users. This is not very flexible
for someone using our code. If they have different design considerations or requirements,
they may need or want to handle the errors in a different way than we did.
Now suppose that we decide not to handle the errors inside our function. Defensive
programming may still be used to prevent the execution of code that results in an error.
However, we now need a way to communicate the error condition to the calling function
so that it can know what type of error happened and handle it appropriately.
Error Codes
One common pattern to communicate errors to a calling function is to use the return
type as an error code. Usually this is an integer type. By convention 0 is used to indicate",ComputerScienceOne.pdf
"type as an error code. Usually this is an integer type. By convention 0 is used to indicate
“no error” and various other non-zero values are used to indicate various types of errors.
Depending on the system and standard used, error codes may have a predefined value or
may be specific to an application or library.
One problem with using the return type to indicate errors is that functions are no longer
able to use the return type to return an actual computed value. If a language supports
pass by reference, then this is not generally a problem. However, even with such languages
there are situations where the return type must be used to return a value. In such cases,
the function can still communicate a general error message by returning some flag value
such as null.
Alternatively, a language may support error codes by using a shared global variable that
can be set by a function to indicate an error. The calling function can then examine the",ComputerScienceOne.pdf
"can be set by a function to indicate an error. The calling function can then examine the
variable to see if an error occurred during the invocation of the function.
154",ComputerScienceOne.pdf
"6.2. Error Handling Strategies
Limitations
Defensive programming has its limitations. Let’s return to the example of processing a
file. To check for all four of the error conditions we identified, we would need a series of
checks similar to the following.
1 if file does not exists then
2 return an error code
3 end
4 if we do not have permissions then
5 return an error code
6 end
7 if the file is corrupted then
8 return an error code
9 end
10 if the file contains invalid values then
11 return an error code
12 end
13 process file data
A problem arises when an error condition is checked and does not hold. Then, later in
the execution, circumstances change and the error condition does hold. However, since
it was already checked for, the program remains under the assumption that the error
condition does not hold. For example, suppose that another process or program deletes
the file that we wish to process after its existence has been checked but before we start",ComputerScienceOne.pdf
"the file that we wish to process after its existence has been checked but before we start
processing it. Because of the sequential nature of our program, this type of error checking
is susceptible to these issues.
6.2.2. Exceptions
An exception is an event or occurrence of an anomalous, erroneous or “exceptional”
condition that requires special handling. Exceptions interrupt the normal flow of control
in a program by handing the flow of control over to exception handlers.
Languages usually support exception handling using a try-catch control structure such
as the following.
try {
//potentially dangerous code here
155",ComputerScienceOne.pdf
"6. Error Handling
} catch(Exception e) {
//exception handling code here
}
The try is used to encapsulate potentially dangerous code, or simply code that would
fail if an error condition occurs. If an error occurs at some point within the try block,
control flow is immediately transferred to the catch block. The catch block is where
you specify how to handle the exception. If the code in the try block does not result in
an exception, then control flow will skip over the catch statement and resume normally
after.
It is important to understand how exceptions interrupt the normal control flow. For
example, consider the following pseudocode.
try {
statement1;
statement2;
statement3;
} catch(Exception e) {
//exception handling code here
}
Suppose statement1 executes with no error but that when statement2 executes, it
results an exception. Control flow is then transferred to the catch block, skipping
statement3 entirely. In general, there may not be a mechanism for your catch block",ComputerScienceOne.pdf
"statement3 entirely. In general, there may not be a mechanism for your catch block
to recover and execute statement3. Therefore, it may be necessary to make your
try-catch blocks fine-grained, perhaps having only a single statement within the try
statement.
Some languages only support a generic Exception and the type of error may need to be
communicated through other means such as a string error message. Still other languages
may support many different types of exceptions and you may be able to provide multiple
catch statements to handle each one differently. In such languages, the order in which
you place your catch statements may be important. Similar to an if-else-if statement,
the first one that matches the exception will be the one that executes. Thus, it is best
practice to order your catch blocks from the most specific to the most general.
Some languages also support a third finally control statement as in the following
example.
156",ComputerScienceOne.pdf
"6.3. Exercises
try {
//potentially dangerous code here
} catch(Exception e) {
//exception handling code here
} finally {
//unconditionally executed code here
}
The try-catch block operates as previously described. However, the finally block
will execute regardless of whether or not an exception was raised. If no exception
was raised, then the try block will fully execute and the finally block will execute
immediately after. If an exception was raised, control flow will be transferred to the
catch block. After the catch block has executed, the finally block will execute.
finally blocks are generally used to handle resources that need to be “cleaned up”
whether or not an exception occurs. For example, opening a connection to a database
to retrieve and process data. Whether or not an exception occurs during this process
the connection will need to be properly closed as it represents a substantial amount of
resources (a network connection, memory and processing time on both the server and",ComputerScienceOne.pdf
"resources (a network connection, memory and processing time on both the server and
client machines, etc.). Failure to properly close the connection may result in wasted
resources. By placing the clean up code inside a finally statement, we can be assured
that it will execute regardless of an error or exception.
In addition to handling exceptions, a language may allow you to “throw” your own
exceptions by using the keyword throw. In this way you can also practice defensive
programming. You could write a conditional statement to check for an error condition
and then throw an appropriate exception.
6.3. Exercises
Exercise 6.1. Rewrite the function to compute the GCD in Exercise 5.1 to handle
invalid inputs.
Exercise 6.2. Rewrite the function to compute statistics of a circle in Exercise 5.6 to
handle invalid input (negative radius).
Exercise 6.3. Rewrite the function to compute the annual percentage yield in Exercise
5.8 to handle invalid input.",ComputerScienceOne.pdf
"Exercise 6.3. Rewrite the function to compute the annual percentage yield in Exercise
5.8 to handle invalid input.
Exercise 6.4. Rewrite the function to compute air distance in Exercise 5.9 to handle
invalid input (latitude/longitude values outside the range [ −180,180]).
157",ComputerScienceOne.pdf
"6. Error Handling
Exercise 6.5. Rewrite the function to convert from RGB to CMYK in Exercise 5.10 to
handle invalid inputs (values outside the range [0 ,255]).
Exercise 6.6. Rewrite the function to convert from CMYK to RGB in Exercise 5.11 to
handle invalid inputs.
Exercise 6.7. Rewrite the square root functions from Exercise 5.13 to handle invalid
inputs.
Exercise 6.8. Rewrite the natural logarithm functions from Exercise 5.14 to handle
invalid inputs.
Exercise 6.9. Rewrite the weight conversion functions from Exercise 5.15 to handle
invalid inputs.
Exercise 6.10. Rewrite the length conversion functions from Exercise 5.16 to handle
invalid inputs.
Exercise 6.11. Rewrite the temperature conversion functions from Exercise 5.17 to
handle invalid inputs.
Exercise 6.12. Rewrite the energy conversion functions from Exercise 5.18 to handle
invalid inputs.
Exercise 6.13. Rewrite the pressure conversion functions from Exercise 5.19 to handle
invalid inputs.
158",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic
Memory
Rarely do we ever deal with a single piece of data in a program. Instead, most data is
made up of a collection of similar elements. A program to compute grades would be
designed to operate on an entire roster of students. Scientific data represents a collection
of many different samples or measurements. A bank maintains and processes many
different accounts, etc.
Arrays are a way to collect similar pieces of data together in an ordered collection. The
pieces of data stored in an array are referred to as elements and are stored in an ordered
manner. That is, there is a “first” element, “second” element, etc. and a “last” element.
Though the elements are ordered, they are not necessarily sorted in any particular manner.
Instead, the order is usually determined by the order in which you place elements into
the array.
Some languages only allow you to store the same types of elements in an array. For",ComputerScienceOne.pdf
"the array.
Some languages only allow you to store the same types of elements in an array. For
example, an integer array would only be able to store integers, an array of floating-point
numbers would only be able to store floating-point numbers. Other languages allow you
to create arrays that can hold mixed elements. A mixed array would be able to hold
elements of any type, so it could hold integers, floating-point numbers, strings, objects,
or even other arrays.
Some languages treat arrays as full-fledged object types that not only hold elements, but
have methods that can be called to manipulate or transform the contents of the array.
Other languages treat arrays as a primitive type, simply using arrays as storage data
structures.
Languages can vary greatly in how they implement and represent arrays of elements, but
many have the same basic usage patterns allowing you to create arrays and manipulate
their contents.
159",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
index
contents 2
0
3
1
5
2
7
3
11
4
13
5
17
6
19
7
23
8
29
9
Figure 7.1.: An integer array of size 10. Using zero-indexing, the first element is at index
0, the last at index 9.
7.1. Basic Usage
Creating Arrays
Though there can be great variation in how a language uses arrays, there are some
commonalities. Languages may allow you to create static arrays or dynamically allocated
arrays (see Section 7.2 below for a detailed discussion). Static arrays are generally created
using the program stack space while dynamically allocated arrays are stored in the heap.
In either case you generally declare an array by specifying its size. In statically typed
languages, you must also declare the array’s type (integer, floating-point, etc.).
Indexing Arrays
Once an array has been created you can use it by assigning values to it or by retrieving
values from it. Because there is more than one element, you must specify which element",ComputerScienceOne.pdf
"values from it. Because there is more than one element, you must specify which element
you are assigning or retrieving. This is generally done through indexing. An index is
an integer that specifies an element in the array. The index is used in conjunction with
(usually) square brackets and the array’s identifier. For example, if the array’s identifier
is arr and the index is an integer value stored in the variable i, we would refer to the
i-th element using the syntax arr[i]. An example is presented in Figure 7.1.
For most programming languages, indices start at zero. This is known as zero-indexing. 1
Thus, the first element is at arr[0], the second at arr[1] etc. When an array is
stored in memory, each element is usually stored one after the other in one contiguous
space in memory. Further, each element is of a specific type which is represented using a
fixed number of bytes in memory. Thus the index actually acts as an offset in memory",ComputerScienceOne.pdf
"fixed number of bytes in memory. Thus the index actually acts as an offset in memory
from the beginning of the array. For example, if we have an array of integers which each
take 4 bytes each, then the 5th element would be stored at index 4, which is an an offset
equal to 4 ×4 = 16 bytes away from the beginning of the array. The first element, being
at index 0 is 4 ×0 = 0 bytes from the beginning of the array (that is, the first element is
at the beginning of the array).
Once an element has been indexed, it can essentially be treated as a regular variable,
1Some languages do use 1-indexing but there are very strong arguments in favor of zero-indexing [ 13].
160",ComputerScienceOne.pdf
"7.1. Basic Usage
assigning and retrieving values as you would regular variables. Care must be taken so
that you do not make a reference to an element that does not exist. For example, using a
negative index or an index i≥n in an array of n elements. Depending on the language,
indexing an array element that is out-of-bounds may result in undefined behavior, an
exception being thrown, or a corruption of memory.
Iteration
Since we can simply index elements in an array, it is natural to iterate over elements
in an array using a for loop. We can create a for loop using an index variable i which
starts at 0, and increments by one on each iteration, accessing the i-th element using the
syntax described above, arr[i]. One issue is that such a loop would need to know how
large the array is in order to define a terminating condition.
1 n←size of array arr
2 for (i←0; i< (n−1); i←(i+ 1)) do
3 process element arr[i]
4 end",ComputerScienceOne.pdf
"large the array is in order to define a terminating condition.
1 n←size of array arr
2 for (i←0; i< (n−1); i←(i+ 1)) do
3 process element arr[i]
4 end
Some languages build the size of the array into a property that can be accessed. Java,
for example, has a arr.length property. Other languages provide functions that you
can use to determine their size. Still other languages (such as C), place the burden of
“bookkeeping” the size of an array on you, the programmer. Whenever you pass an array
to a function you need to also pass a size parameter that informs the function of how
many elements are in the array. Yet other functions may also require you to pass the
size of each element in the array.
Some languages also support a basic foreach loop (cf. Section 4.4). A foreach loop is
syntactic sugar that allows you to iterate over the elements in an array (usually in order)
without the need for boilerplate code that creates and increments an index variable.
1 foreach element a in arr do",ComputerScienceOne.pdf
"without the need for boilerplate code that creates and increments an index variable.
1 foreach element a in arr do
2 process element a
3 end
The actual syntax may vary if a language supports such a loop.
161",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
Using Arrays in Functions
Most programming languages allow you to use arrays as both function parameters and as
return types. You can pass arrays to functions and functions can be defined that return
arrays. Typically, when arrays are passed to functions, they are passed by reference so
as to avoid making an entirely new copy of the array. As a consequence, if the function
makes changes to the array elements, those changes may be realized in the calling function.
Sometimes you may want this behavior. However, sometimes you may not want the
function to make changes to the passed array. Some languages allow you to use various
keywords to prevent any changes to the passed array.
If a function is designed to return an array, care must be taken to ensure that the correct
type of array is returned. Recall that static arrays are allocated on the call stack. It
would be inappropriate to return static arrays from a function as the array is part of",ComputerScienceOne.pdf
"would be inappropriate to return static arrays from a function as the array is part of
the stack frame that is destroyed when the function returns control back to the calling
function (we discuss this in detail below). Instead, if a function returns an array, it
should be an array that has been dynamically allocated (on the heap).
7.2. Static & Dynamic Memory
Recall that arrays can be declared as static arrays, meaning that when you declare them,
they are allocated and stored on the program’s call stack within the stack frame in which
they are declared. For example, if a function foo() creates a static array of 5 integers,
each requiring 4 bytes, then 20 contiguous bytes are allocated on the stack frame for
foo() to store the static array.
This can cause several potential problems. First, the typical stack space allocated for a
program is relatively small. It can be as large as a couple of Megabytes (MBs) or on",ComputerScienceOne.pdf
"program is relatively small. It can be as large as a couple of Megabytes (MBs) or on
some older systems or those with limited resources, even on the order of a few hundred
Kilobytes (KBs). When dealing with data of any substantial size, you could quickly run
out of stack space, resulting in a stack overflow.
Another problem arises if we want to return a static array from a function. Recall that
when a function returns control to the calling function, its stack frame is popped off the
top and goes out-of-scope (see Section 5.2). Since the array is part of the stack frame of
the function, it too goes out of scope. Depending on how the system works, the contents
of the frame may be completely changed in the “bookkeeping” process of returning from
the function. Even if the contents remain unchanged when the function returns, they
will almost certainly be overwritten when another function call is made and a new stack",ComputerScienceOne.pdf
"will almost certainly be overwritten when another function call is made and a new stack
frame overwrites the old one. For these reasons, static arrays are of very limited use.
They must be kept small and cannot be returned from a function.
162",ComputerScienceOne.pdf
"7.2. Static & Dynamic Memory
Demonstration in C
To make this concept a bit more clear, we’ll use a concrete example in the C programming
language. Consider the program code in Figure 7.2. Here, we have a function foo()
that creates a static integer array of size 5, int b[5];. This memory is allocated on
the stack frame and then assigned values. Printing them in the function will give the
expected results,
b[0] = 5
b[1] = 10
b[2] = 15
b[3] = 20
b[4] = 25
1 #include<stdlib.h>
2 #include<stdio.h>
3
4 int * foo(int n) {
5 int i;
6 int b[5];
7 for(i=0; i<5; i++) {
8 b[i] = n*i;
9 printf(""b[%d] = %d\n"", i, b[i]);
10 }
11 return b;
12 }
13
14 int main(int argc, char **argv) {
15 int i, m = 5;
16 int *a = foo(m);
17 for(i=0; i<5; i++) {
18 printf(""a[%d] = %d\n"", i, a[i]);
19 }
20 return 0;
21 }
Figure 7.2.: Example returning a static array
163",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
However, when the function foo() ends execution and returns control back to the
main() function, (sometimes called unwinding), the contents of foo()’s stack frame
are altered as part of the process. Some of the contents are the same, but elements have
been completely altered. Printing the “returned” contents of the array gives us garbage:
a[0] = 1564158624
a[1] = 32767
a[2] = 15
a[3] = 20
a[4] = -626679356
This is not an issue when returning primitive types as the return value is placed in a
special memory location available to the calling function. Even in our example, the
return value is properly communicated to the calling function: its just that the returned
value is a pointer to the array’s location (which happens to be a memory address in the
“stale” stack frame). The stack frames are depicted in Figure 7.3.
7.2.1. Dynamic Memory
The solution to the problems presented by static arrays is to use dynamically allocated",ComputerScienceOne.pdf
"7.2.1. Dynamic Memory
The solution to the problems presented by static arrays is to use dynamically allocated
arrays. Such arrays are not allocated and stored in the program call stack, but instead
are stored in another area called the heap. In fact, because of the shortcomings of static
arrays, some languages only allow you to use dynamically allocated arrays even if it is
done so implicitly.
Recall that when a program is loaded into memory, its code is placed on the program call
stack in memory. On top of that is any static data (global variables or data for example).
Subsequent function calls place new stack frames on top of that. In addition, at the
“other end” of the program’s memory space, the heap is allocated. The heap is a “pile” of
memory like the stack is, but it is less structured. The stack is one contiguous piece of
memory, but the heap may have “holes” in it as various chunks of it are allocated and",ComputerScienceOne.pdf
"memory, but the heap may have “holes” in it as various chunks of it are allocated and
deallocated for the program. In general, the heap is larger but allocation and deallocation
is a more involved and more costly process while the stack is smaller and faster to
allocate/deallocate stack frames from.
Initially, a program is allocated a certain amount of memory for its heap. When a
program attempts to dynamically allocate memory, say for a new array of integers, it
makes a request to the operating system for a certain amount of memory (a certain
number of bytes). The system responds by allocating a chunk of the heap memory which
the program can then use to store elements in the array. Usually, the memory space is
164",ComputerScienceOne.pdf
"7.2. Static & Dynamic Memory
stored as a pointer or reference. The reference is stored in a variable in a stack frame,
but the actual contents of the array are stored in the heap space.
Depending on the language and system, if a program uses all of its heap space and runs
out, the operating system may terminate the program or it may attempt to reserve even
more memory for the program.
Memory Management
If a program no longer needs a dynamically allocated memory space, it should “clean up”
after itself by deallocating or “freeing” the memory, releasing it back to the heap space
so that it can be reused either by the program or some other program or process on the
system. The process of allocating and deallocating memory is generally referred to as
memory management. If a program does not free up memory, it may eventually run out
and be forced to terminate. Even if it does not necessarily run out of available memory,
its performance may degrade or it may impact system performance.",ComputerScienceOne.pdf
"its performance may degrade or it may impact system performance.
If a program has poor memory management and fails to deallocate memory when it is
no longer needed, the memory “leaks”: the available memory gradually runs out because
it is not released back to the heap for reallocation. Programs which such poor memory
management are said to have a memory leak. Sometimes this is a consequence of a
dangling pointer: when a program dynamically allocates a chunk of memory but then
due to carelessness, loses the reference to the memory chunk, making it impossible to
free up.
Some languages have automatic garbage collection that handle memory management for
us. The language itself is able to monitor the dynamically allocated pieces of memory
and determine if any variable in the program still references it. If the memory is no
longer referenced, it is “garbage” and becomes eligible to be “collected.” The system",ComputerScienceOne.pdf
"longer referenced, it is “garbage” and becomes eligible to be “collected.” The system
itself then frees the memory and makes it available to the program or operating system.
In such “memory managed” languages, we are responsible for allocating memory, but are
not (necessarily) responsible for deallocating it.
Even if a language offers automated memory management, it is still possible to have
memory leaks and other memory allocation issues. Automated memory management
does not solve all of our memory management problems. Moreover, it comes at a cost.
The additional resources and overhead required to monitor memory can have a significant
performance cost. However, with modern garbage collection systems and algorithms,
the performance gap between garbage collected languages and user-managed memory
languages has been shrinking. In any case, all program memory is reclaimed by the
operating system when the program terminates.
165",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
7.2.2. Shallow vs. Deep Copies
In most languages, an array variable is actually a reference to the array in memory.
We could create an array referred to by a variable A and then create another reference
variable B and set it “equal” to A. However, this is simply a shallow copy. Both the
reference variables refer to the same data in memory. Consequently, if we change the
value of an element in one, the change is realized in both.
Often, we want a completely different copy, referred to as a deep copy. With a deep copy,
A and B would refer to different memory blocks. Changes to one would not affect the
other. This distinction is illustrated in Figure 7.5.
7.3. Multidimensional Arrays
A normal array is usually one dimensional. One can think an array as a single “row” in a
table that contains a certain number of entries. Most programming languages allow you
to define multidimensional arrays. For example, two dimensional arrays would model",ComputerScienceOne.pdf
"to define multidimensional arrays. For example, two dimensional arrays would model
having multiple rows in a full table. You can also view two dimensional arrays as matrices
in mathematics. A matrix is a rectangular array of numbers that have a certain number
of rows and a certain number of columns.
As an example, consider the following 2 ×3 matrix (it has 2 rows and 3 columns):
[1 9 −8
2.5 3 5
]
In mathematics, entries in a matrix are indexed via their row and column. For example,
ai,j would refer to the element in the i-th row and j-th column. Referring to the row
first and column second is referred to as row major ordering. If the number of rows and
the number of columns are the same, the matrix is referred to as a square matrix. For
example, the following is a square, 10 ×10 matrix.


2 68 9 44 80 79 77 59 27 2
3 86 22 42 58 24 45 39 7 47
5 7 17 12 29 56 68 14 65 3
7 35 64 69 79 56 52 77 82 85
11 55 36 5 25 6 22 25 72 37
13 20 37 74 3 53 87 70 3 78",ComputerScienceOne.pdf
"3 86 22 42 58 24 45 39 7 47
5 7 17 12 29 56 68 14 65 3
7 35 64 69 79 56 52 77 82 85
11 55 36 5 25 6 22 25 72 37
13 20 37 74 3 53 87 70 3 78
17 72 68 26 11 6 63 70 29 16
19 59 6 26 87 18 82 27 75 19
23 73 30 80 51 14 34 67 59 58
29 48 2 39 18 21 33 28 40 34


166",ComputerScienceOne.pdf
"7.4. Other Collections
We can do something similar in most programming languages. First, languages may
vary in how you can create multidimensional arrays, but you usually have to provide a
size for each dimension when you create them. Once created, you can index them by
providing multiple indices. For example, with a two dimensional array, we could provide
two indices each in their own square brackets arr[i][j] referring to the i-th row and
j-th column. Multidimensional arrays usually use the same 0-indexing scheme as single
dimensional arrays.
You can further generalize this and create 3-dimensional arrays, 4-dimensional arrays, etc.
However, the use cases for arrays of dimension 3 and certainly for arrays of dimension
4 or greater are rare. If you need to store such data, it may be more appropriate to
define a custom, user-defined structure or object (see Chapter 10) instead of a higher
dimensional array.
We usually think of 2-dimensional arrays as having the same number of elements in",ComputerScienceOne.pdf
"dimensional array.
We usually think of 2-dimensional arrays as having the same number of elements in
each “row”. In the example above, both of the matrix’s rows had 3 elements in it. Some
languages allow you to create 2-dimensional arrays with a different number of elements
in each row. Special care must be taken to ensure that you do not index an element that
does not exist.
7.4. Other Collections
Aside from basic arrays, many languages have rich libraries of other dynamic collections.
Dynamic collections are not the same thing as dynamically allocated arrays. Once an
array is created, its size is fixed and cannot, in general, be changed. However, dynamic
collections can grow (and shrink) as needed when you add or remove elements from them.
Lists are ordered collections that are essentially dynamic arrays. Lists are ordered and
are usually zero-indexed just like arrays. Lists are generally objects and provide methods",ComputerScienceOne.pdf
"are usually zero-indexed just like arrays. Lists are generally objects and provide methods
that can be used to add, remove, and retrieve elements from the list. If you add an
element to a list, the list will automatically grow to accommodate it, so its size is not
fixed when created. Two common implementations of lists are array-based lists and
linked lists. Array-based lists indexarray-based list use an array to store elements. When
the array fills up, the list allocates a new, larger array to hold more elements, copying the
original contents over to the new array with a larger capacity. Linked lists hold elements
in nodes that are linked together. Adding a new element simply involves creating a new
node and linking it to the last element in the list.
Some languages also define what are called sets. Sets allow you to store elements
dynamically just like lists, but sets are generally unordered. There is no concept of a",ComputerScienceOne.pdf
"dynamically just like lists, but sets are generally unordered. There is no concept of a
first, second, or last element in a set. Iterating over the elements in a set could result
in a different enumeration of the elements each time. Elements in sets are also usually
unique. For example, a set containing integers would only ever contain one instance of
167",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
each integer. The value 10, for example, would only ever appear once. If you added 10
to a set that already contained it, the operation would have no effect on the set.
Another type of dynamic array are associative arrays (sometimes called dictionaries).
An associative array holds elements, but may not be restricted in how they are indexed.
In particular, a language that supports associative arrays may allow you to use integers
or strings as indices, or even any arbitrary object. Further, when using integers to index
elements, indices need not be fully defined nor contiguous. In an associative array you
could define an element at index 5 and then place the next element at index 10, skipping
index 6 through 9 which would remain undefined.
One way to look at associative arrays is as a map. A map is a data structure that
stores elements as key-value pairs. Both they keys and values could be any arbitrary",ComputerScienceOne.pdf
"stores elements as key-value pairs. Both they keys and values could be any arbitrary
type (integers or strings) or object depending on the language. You could map account
numbers (stored as strings) to account objects, or vice versa. Using a smart data structure
like a map can make data manipulation a lot easier and more straightforward.
7.5. Exercises
Exercise 7.1.Write a function to return the index of the maximum element in an array
of numbers.
Exercise 7.2. Write a function to return the index of the minimum element in an array
of numbers.
Exercise 7.3. Write a function to compute the mean (average) of an array of numbers.
Exercise 7.4.Write a function to compute the standard deviation of an array of numbers,
σ=
√1
N
N∑
i=1
(xi −µ)2
where µ is the mean of the array of numbers.
Exercise 7.5. Write a function that takes two arrays of numbers that are sorted and
merges them into one array (retuning a new array as a result).",ComputerScienceOne.pdf
"Exercise 7.5. Write a function that takes two arrays of numbers that are sorted and
merges them into one array (retuning a new array as a result).
Exercise 7.6.Write a function that takes an integer nand produces a new array of size
n filled with 1s.
Exercise 7.7. Write a function that takes an array of numbers are computes returns
the median element. The median is defined as follows:
•If n is odd, the median is the n+1
2 -th largest element
168",ComputerScienceOne.pdf
"7.5. Exercises
•If n is even, the median is the average of the n
2 and the ( n
2 + 1)-th largest elements
Exercise 7.8. The dot product of two arrays (or vectors) of the same dimension is
defined as the sum of the product of each of their entries. That is,
n∑
i=1
ai ×bi
Write a function to compute the dot product of two arrays (you may assume that they
are of the same dimension)
Exercise 7.9. The norm of an n-dimensional vector, ⃗ x= (x1,x2,...,x n) captures the
notion of “distance” in a higher dimensional space and is defined as
∥⃗ x∥=
√
x2
1 + ··· + x2
n
Write a function that takes an array of numbers that represents an n-dimensional vector
and computes its norm.
Exercise 7.10. Write a function that takes two arrays A,B and creates and returns a
new array that is the concatenation of the two. That is, the new array will contain all
elements a followed by all elements in b.
Exercise 7.11. Write a function that takes an array of numbers A and an element x",ComputerScienceOne.pdf
"elements a followed by all elements in b.
Exercise 7.11. Write a function that takes an array of numbers A and an element x
and returns true/false if A contains x
Exercise 7.12. Write a function that takes an array of numbers A, an element x and
two indices i,j and returns true/false if A contains x somewhere between index i and j.
Exercise 7.13. Write a function that takes an array of numbers A and an element x
and returns the multiplicity of x; that is the number of times x appears in A.
Exercise 7.14.Write a function to compute asliding window mean. That is, it computes
the average of the first m numbers in the array. The next value is the average of the
values index from 1 to m, then 2 to m+ 1, etc. The last window is the average of the
last m elements. Obviously, m≤n (for m= n, this is the usual mean). Since there is
more than one value, your function will return a (new) array of means of size n−m+ 1.
Exercise 7.15. Write a function to compute the mode of an array of numbers. The",ComputerScienceOne.pdf
"Exercise 7.15. Write a function to compute the mode of an array of numbers. The
mode is the most common value that appears in the array. For example, if the array
contained the elements 2,9,3,4,2,1,8,9,2, the mode would be 2 as it appears more than
any other element. The mode may not be unique; multiple elements could appear the
same, maximal number of times. Your function should simply return a mode.
Exercise 7.16. Write a function to find all modes of an array. That is, it should find
all modes in an array and return a new array containing all the mode values.
169",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
Exercise 7.17.Write a function to filter out certain elements from an array. Specifically,
the function will create a new array containing only elements that are greater than or
equal to a certain threshold δ.
Exercise 7.18. Write a function that takes an array of numbers and creates a new
“deep” copy of the array. In addition, the function should take a new “size” parameter
which will be the size of the copy. If the new size is less than the original, then the new
array will be a truncated copy. If the new size is greater then the copy will be padded
with zero values at the end.
Exercise 7.19. Write a function that takes an array A and two indices i,j and returns
a new array that is a subarray of A consisting of elements i through j.
Exercise 7.20. Write a function that takes two arrays A,B and creates and returns a
new array that represents the unique intersection of A and B. That is, an array that",ComputerScienceOne.pdf
"new array that represents the unique intersection of A and B. That is, an array that
contains elements that are in both A and B. However, elements should not be included
more than once.
Exercise 7.21. Write a function that takes two arrays A,B and creates and returns a
new array that represents the unique union of A and B. That is, an array that contains
elements that are either in A or B (or both). However, elements should not be included
more than once.
Exercise 7.22. Write a function that takes an array of numbers and returns the sum of
its elements.
Exercise 7.23. Write a function that takes an array of numbers and two indices i,j
and computes the sum of its elements between i and j inclusive.
Exercise 7.24. Write a function that takes two arrays of numbers and determines if
they are equal: that is, each index contains the same element.
Exercise 7.25. Write a function that takes two arrays of numbers and determines if",ComputerScienceOne.pdf
"they are equal: that is, each index contains the same element.
Exercise 7.25. Write a function that takes two arrays of numbers and determines if
they both contain the same elements (though are not necessarily equal) regardless of
their multiplicity. That is, the function should return true even if the arrays’ elements
appear a different number of times or in a different order. For example [2 ,2,3] would be
equal to an array containing [3 ,2,3,3,3,2,2,2].
Exercise 7.26.A suffix array is a lexicographically sorted array of all suffixes of a string.
Suffix arrays have many applications in text indexing, data compression algorithms and
in bioinformatics. Write a program that takes a string and produces its suffix array.
For example, the suffixes of science are science, cience, ience, ence, nce, ce,
and e. The suffix array (sorted) would look something like the following.
170",ComputerScienceOne.pdf
"7.5. Exercises
ce
cience
e
ence
ience
nce
science
Exercise 7.27. An array of size n represents a permutation if it contains all integers
0,1,2,..., (n−1) exactly once. Write a function to determine if an array is a permutation
or not.
Exercise 7.28. The k-th order statistic of an array is the k-th largest element. For our
purposes, k starts at 0, thus the minimum element is the 0-th order statistic and the
largest element is the n−1-th order statistic.
Another way to view it is: suppose we were to sort the array, then the k-th order statistic
would be the element at index k in the sorted array. Write a function to find the k-th
order statistic.
Exercise 7.29.Write a function that takes two n×nsquare matrices A,B and returns
a new n×n matrix C which is the product, A×B. The product of two matrices of
dimension n×n is defined as follows:
cij =
n∑
k=1
aikbkj
Where 1 ≤i,j ≤n and cij is the (i,j)-th entry of the matrix C.",ComputerScienceOne.pdf
"dimension n×n is defined as follows:
cij =
n∑
k=1
aikbkj
Where 1 ≤i,j ≤n and cij is the (i,j)-th entry of the matrix C.
Exercise 7.30.We can multiply a matrix by a single scalar valuexby simply multiplying
each entry in the matrix by x. Write a function that takes a matrix of numbers and an
element x and performs scalar multiplication.
Exercise 7.31. Write a function that takes two matrices and determines if they are
equal (all of their elements are the same).
Exercise 7.32. Write a function that takes a matrix and an index i and returns a new
array that contains the elements in the i-th row of the matrix.
Exercise 7.33. Write a function that takes a matrix and an index j and returns a new
array that contains the elements in the j-th column of the matrix.
Exercise 7.34. Iterated Matrix Multiplication is where you take a square matrix, A
and multiply it by itself k times,
Ak = A×A×···× A  
ktimes
Write a function to compute the k-th power of a matrix A.
171",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
Exercise 7.35.The transpose of a square matrix is an operation that “flips” the matrix
along the diagonal from the upper left to the lower right. In particular the values mi,j
and mj,i are swapped. Write a function to transpose a given matrix
Exercise 7.36. Write a function that takes a matrix, a row index i, and a number x
and adds x to each value in the i-th row.
Exercise 7.37. Write a function that takes a matrix, a row index i, and a number x
and multiples each value in the i-th row with x.
Exercise 7.38.Write a function that takes a matrix, and two row indices i,j and swaps
each value in row i with the value in row j
Exercise 7.39. A special matrix that is often used is the identity matrix. The identity
matrix is an n×n matrix with 1s on its diagonal and zeros everywhere else. Write a
function that, given n, creates a new n×n identity matrix.
Exercise 7.40.Write a function to convert a matrix of integers to floating point numbers.",ComputerScienceOne.pdf
"function that, given n, creates a new n×n identity matrix.
Exercise 7.40.Write a function to convert a matrix of integers to floating point numbers.
Exercise 7.41. Write a function to determine if a given matrix is a permutation matrix.
A permutation matrix is a matrix that represents a permutation of the rows of an identity
matrix. That is, A is a permutation matrix if every row and every column has exactly
one 1 and the rest are zeros.
Exercise 7.42. Write a function to determine if a given square matrix is symmetric. A
matrix is symmetric if mi,j = mj,i for all i,j.
Exercise 7.43.Write a function to compute the trace of a matrix. The trace of a matrix
is the sum of its elements along its diagonal.
Exercise 7.44.Write a function that returns a “resized” copy of a matrix. The function
takes a matrix of size n×m(not necessarily square) and creates a copy that is p×q (p,q
are part of the input to the function). If p<n and/or q <m, the values are “cut off”. If",ComputerScienceOne.pdf
"are part of the input to the function). If p<n and/or q <m, the values are “cut off”. If
p>n and/or q >m, the matrix is padded (to the right and to the bottom) with zeros.
Exercise 7.45. A submatrix is a matrix formed by selecting certain rows and columns
from a larger matrix. Write a function that constructs a submatrix from a larger matrix.
To do so, the function will take a matrix as well as two row indices i,j and two column
indices k,ℓ and it will return a new matrix which consists of entries from the i-th row
through the j-th row and k-th column through the ℓ-th column.
For example, if A is
A =


8 2 4 1
10 4 2 3
12 42 1 0

.
then a call to this function with i= 1,j = 2,k = 2,ℓ = 3) should result in
A =
[2 3
1 0
]
.
172",ComputerScienceOne.pdf
"7.5. Exercises
Exercise 7.46.The Kronecker product (http://en.wikipedia.org/wiki/Kronecker_
product) is a matrix operation on two matrices that produces a larger block matrix.
Specifically, if Ais an m×nmatrix and B is a p×q matrix, then the Kronecker product
A⊗B is the mp×nq block matrix:
A ⊗B =


a11B ··· a1nB
... ... ...
am1B ··· amnB


more explicitly:
A ⊗B =


a11b11 a11b12 ··· a11b1q ··· ··· a1nb11 a1nb12 ··· a1nb1q
a11b21 a11b22 ··· a11b2q ··· ··· a1nb21 a1nb22 ··· a1nb2q
... ... ... ... ... ... ... ...
a11bp1 a11bp2 ··· a11bpq ··· ··· a1nbp1 a1nbp2 ··· a1nbpq
... ... ... ... ... ... ...
... ... ... ... ... ... ...
am1b11 am1b12 ··· am1b1q ··· ··· amnb11 amnb12 ··· amnb1q
am1b21 am1b22 ··· am1b2q ··· ··· amnb21 amnb22 ··· amnb2q
... ... ... ... ... ... ... ...
am1bp1 am1bp2 ··· am1bpq ··· ··· amnbp1 amnbp2 ··· amnbpq


.
Write a function that computes the Kronecker product.",ComputerScienceOne.pdf
"am1bp1 am1bp2 ··· am1bpq ··· ··· amnbp1 amnbp2 ··· amnbpq


.
Write a function that computes the Kronecker product.
Exercise 7.47. The Hadamard product is an entry-wise product of two matrices of
equal size. Let A,B be two n×m matrices, then the Hadamard product is defined as
follows.
A◦B =


a11 a12 ··· a1m
a21 a22 ··· a2m
... ... ... ...
an1 an2 ··· anm

◦


b11 b12 ··· b1m
b21 b22 ··· b2m
... ... ... ...
bn1 bn2 ··· bnm

=


a11b11 a12b12 ··· a1mb1m
a21b21 a22b22 ··· a2mb2m
... ... ... ...
an1bn1 an2bn2 ··· anmbnm


Write a function to compute the Hadamard product of two n×m matrices.
173",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
Stack Frame Variable Address Content
... ...
b[4] 0x5c44cb76 25
b[3] 0x5c44cb72 20
b[2] 0x5c44cb68 15
b[1] 0x5c44cb64 10
b[0] 0x5c44cb60 5
i 0x5c44cb56 5
foo n 0x5c44cb52 5
... ...
a 0x5c44cb34 NULL
m 0x5c44cb30 5
main i 0x5c44cb26 0
(a) Program stack at the end of the execution of foo() prior to
returning control back to main() .
Stack Frame Variable Address Content
... ...
b[4] 0x5c44cb76 -626679356
b[3] 0x5c44cb72 20
b[2] 0x5c44cb68 15
b[1] 0x5c44cb64 32767
b[0] 0x5c44cb60 1564158624
i 0x5c44cb56 5
foo n 0x5c44cb52 5
... ...
a 0x5c44cb34 0x5c44cb60
m 0x5c44cb30 5
main i 0x5c44cb26 0
(b) Upon returning, the stack frame is no longer valid; the pointer
variable a points to a stack memory address but the frame and
its local variables are no longer valid. Some have been overwritten
with other values. Subsequent usage or access of the values in a are
undefined behavior.",ComputerScienceOne.pdf
"with other values. Subsequent usage or access of the values in a are
undefined behavior.
Figure 7.3.: Illustration of the pitfalls of returning a static array in C. Static arrays
are locally scoped and exist only within the function/block in which they
are declared. The program stack frame in which the variables are stored
is invalid when the function returns control back to the calling function.
Depending on how the system/compiler/language handles this unwinding
process, values may be changed, unavailable, etc.
174",ComputerScienceOne.pdf
"7.5. Exercises
Program Code
Static Content
Allocated Stack
Available Stack
Available Heap
Allocated Heap
Stack Growth
Heap Growth
Figure 7.4.: Depiction of Application Memory. The details of how application memory
is allocated and how the stack/heap “grow” may vary depending on the
architecture. The figure shows stack memory growing “upward” while heap
allocation grows “downward.” Allocation and deallocation may fragment
the heap space though.
175",ComputerScienceOne.pdf
"7. Arrays, Collections & Dynamic Memory
A
B
2
0
3
1
5
2
7
3
11
4
13
5
17
6
19
7
23
8
29
9
(a) A shallow copy. B refers to A which refers to the array. Thus, B implicitly
refers to the same array.
A
B
2
0
3
1
5
2
7
3
11
4
13
5
-1
6
19
7
23
8
29
9
(b) When an element in a shallow copy is changed, A[6] = -1; , it is changed
from the perspective of both A and B.
A
B
2
0
3
1
5
2
7
3
11
4
13
5
17
6
19
7
23
8
29
9
2
0
3
1
5
2
7
3
11
4
13
5
17
6
19
7
23
8
29
9
(c) A deep copy. B refers to its own copy of the array distinct from A. Both are
stored in separate memory locations.
A
B
2
0
3
1
5
2
7
3
11
4
13
5
-1
6
19
7
23
8
29
9
2
0
3
1
5
2
7
3
11
4
13
5
17
6
19
7
23
8
29
9
(d) When an element in a deep copy is changed, A[6] = -1; , it is changed
only in the array A. The element in B is unaffected.
Figure 7.5.: Shallow copies are when two references refer to the same data in memory
(a) and (b). Changes to one affect the other. Deep copies (c) and (d) are",ComputerScienceOne.pdf
"(a) and (b). Changes to one affect the other. Deep copies (c) and (d) are
distinct data in memory, changes to one do not affect the other.
176",ComputerScienceOne.pdf
"8. Strings
A string is an ordered sequence of characters. We’ve previously seen string data types
as literals. Most languages allow you to define and use static string literals using the
double quote syntax. We used strings to specify output formatting using printf()-style
functions for example. When reading input from a user, we read it as a string and
converted the input to numbers. We also described some basic operations on strings
including concatenation. We now examine strings in more depth.
Programming languages vary greatly in how they represent string data types. Some
languages have string types built-in to the language and others require that you use
arrays and yet others treat strings as a special type of object. One issue with string
representations is determining where and how the string ends. Some languages use a
length prefix string representation. The length (the number characters in the string)",ComputerScienceOne.pdf
"length prefix string representation. The length (the number characters in the string)
is stored in a special location at the beginning of a string. Then the string characters
are stored as an array. Other Object-Oriented Programming (OOP) languages use a
special character, the null-terminating character to denote the end of a string. Still
other languages store strings as arrays or dynamic arrays and the “bookkeeping” is done
internally as part of an object representation.
Other details vary as well. Most languages support the basic ASCII characters, others
have full Unicode support or support Unicode through a library. Most languages also
provide large libraries of functions and operations that make working with strings easier.
8.1. Basic Operations
Depending on how a language supports strings, it may support various basic operations
to create strings and assign them to variables. Usually languages allow you to create",ComputerScienceOne.pdf
"to create strings and assign them to variables. Usually languages allow you to create
and use string literals using the double quote syntax. Modifying a string or copying
one string into another may be supported by the built-in assignment operator or it may
require the use of a copy function. When copying strings, similar issues come into play
as with arrays. The “copy” could be a shallow copy or a deep copy (see Section 7.2.2).
Other basic operations may include accessing and/or modifying individual characters
in a string. In order to do so, we may also need to know a string’s length (so that we
do not access invalid characters). A language could provide this as part of the string
177",ComputerScienceOne.pdf
"8. Strings
itself (usually called a property of the string) or through a function call. We can further
use such functionality to iterate over the individual characters in a string using an
index-controlled for-loop.
More advanced operations on strings include concatenation which is the operation of
combining one or more strings to create a new string. Concatenation simply appends one
string to the end of another string. Another common operation is to extract a substring
from a string, that is create a new string from a portion of another string. Commonly,
this is done via some standard function that may operate by specifying indices and/or
the length of the desired substring. Finally, it is also common to deal with collections
of strings. Some languages allow you to create arrays of strings or dynamic collections
(lists or sets) of strings. for languages in which strings are arrays of characters, an array
of strings might be implemented with a 2-dimensional array of characters.",ComputerScienceOne.pdf
"of strings might be implemented with a 2-dimensional array of characters.
8.2. Comparisons
When processing strings there are several other standard operations. In particular, we
often have need to make comparisons between two string variables or between a string
variable and a literal. Some languages allow you to use the same operators such as ==
or even < to make comparisons between strings. The implied behavior would compare
strings for equality (case sensitive) or for lexicographic order. For example “Apple” <
“Banana” might evaluate to true because “Apple” precedes “Banana” in alphabetic order.
Many languages, however, require that you make string comparisons using a function.
Using the equality operator == may be correct syntactically, but is usually making a
pointer or reference comparison which evaluates to true if and only if the two variables
represent the same memory address. Even if two string variables have the same content,",ComputerScienceOne.pdf
"represent the same memory address. Even if two string variables have the same content,
the equality operator may evaluate to false if they are distinct (deep) copies of the same
string. Likewise, the inequality operators <,≤, etc. may only be comparing memory
addresses which is meaningless for comparing strings.
The solution that many languages provide is the use of a comparator, which is either a
function or an object that facilitates the comparison of strings (and more generally, any
object). Generally, a comparator function takes two arguments, a,b and compares them,
not just for equality, but for their relative order: does a“come before” bor does b“come
before” a, or are they equal. To distinguish between these three cases, a comparator
returns an integer value with the following general contract: it returns
•Something negative if a<b
•Zero if a= b
•Something positive if a>b
178",ComputerScienceOne.pdf
"8.3. Tokenizing
Using this contract we can determine the relative ordering of any two strings. In general
we cannot make any assumptions about the actual value that a comparator returns, only
that it returns something negative or positive. The actual magnitude of the returned
value need not be −1 or +1, and it may not even have any predefined meaning.
8.3. Tokenizing
It is common to store different pieces of data as a string such that each individual piece
of data is demarcated by some delimiter. For example, Comma Separated Values (CSV)
or Tab Separated Values (TSV) data use commas and tabs to delimit data. For example,
the string
Smith,Joe,12345678,1985-09-08
is a CSV string holding data about a particular person (last name, first name, ID, date
of birth). Often we need to process such strings to extract each individual piece of data.
Processing such strings is usually referred to as parsing. In particular, a string is “split”",ComputerScienceOne.pdf
"Processing such strings is usually referred to as parsing. In particular, a string is “split”
into a collection of individual strings called tokens (thus the process is also sometimes
referred to as tokenizing). In the example above, the string would be processed into
4 individual strings, Smith, Joe, 12345678, and 1985-09-08. Each string could
further be tokenized if needed, such as parsing the date of birth to extract the year,
month, and date.
Most languages provide a function to facilitate tokenization. Some do so by directly
returning an array or collection of the resulting tokens (usually with the delimiter
removed). Others have a more manual process that requires a loop structure to iterate
over each token.
8.4. Exercises
Exercise 8.1. Write functions to reverse a string. If appropriate, write versions to do
so by manipulating a given string and returning a new string that is a reversed copy.
Exercise 8.2. Write a function to replace all spaces in a string with two spaces.",ComputerScienceOne.pdf
"Exercise 8.2. Write a function to replace all spaces in a string with two spaces.
Exercise 8.3. Write a program to take a phrase (International Business Machines) and
“acronymize” it by producing a string that is an upper-cased string of the first letter of
each word in the phrase (IBM).
Exercise 8.4. Write a function that takes a string containing a word and returns a
pluralized version according to the following rules.
179",ComputerScienceOne.pdf
"8. Strings
1. If the noun ends in “y,” remove the “y” and add “ies”
2. If the noun ends in “s,” “ch,” or “sh,” add “es”
3. In all other cases, just add “s”
Exercise 8.5. Write a function that takes a string and determines if it is a palindrome
or not. A palindrome is a word that is spelled exactly the same when the letters are
reversed.
Exercise 8.6. Write a function to compute the longest common prefix of two strings.
For example, the longest common prefix of “global” and “glossary” is “glo”. If two
strings have no common prefix, then the longest common prefix is the empty string.
Exercise 8.7.Write a function to remove any whitespace from a given string. For exam-
ple, if the string passed to the function contains ""Hello World How are you? ""
then it should result in the string ""HelloWorldHowareyou?""
Exercise 8.8. Write a function that takes a string and flips the case of each alphabetic
character in it. For example, if the input string is ""GNU Image Processing Tool-Kit""",ComputerScienceOne.pdf
"character in it. For example, if the input string is ""GNU Image Processing Tool-Kit""
then it should output ""gnu iMAGE pROCESSING tOOL-kIT""
Exercise 8.9.Write a function to validate a variable name according to the rules that it
must begin with an alphabetic character, a–z or A–Z but may contain any alphanumeric
character a–z, A–Z, 0–9, or underscores _. Your function should take a string with a
possible variable name and return true or false depending on whether or not it is valid.
Exercise 8.10.Write a function to convert a string that represents a variable name using
under_score_casing to lowerCamelCasing. That is, it should remove all underscores,
and replace the first letter of each word with an uppercase (except the first word).
Exercise 8.11. Write a function that takes a string and another character c and counts
the number of times that c appears in the string.
Exercise 8.12.Write a function that takes a string and another character cand removes",ComputerScienceOne.pdf
"the number of times that c appears in the string.
Exercise 8.12.Write a function that takes a string and another character cand removes
all instances of c from the string. For example, a call to this function on the string
""Hello World"" with c being equal to 'o' would result in the string ""Hell Wrld"".
Exercise 8.13. Write a function that takes a string and two characters, c and d and
replaces all instances of c with d.
Exercise 8.14. Write a function to determine if a given string s contains a substring t.
The function should return true if t appears anywhere inside s and false otherwise.
Exercise 8.15. Write a function that takes a string s and returns a new string that
contains the first character of each word in s capitalized. You may assume that words
are separated by a single space. For example, if we call this function with the string
""International Business Machines"" it should return ""IBM"". If we call it with the",ComputerScienceOne.pdf
"""International Business Machines"" it should return ""IBM"". If we call it with the
string ""Flint Lockwood Diatonic Super Mutating Dynamic Food Replicator"" it
should return ""FLDSMDFR""
180",ComputerScienceOne.pdf
"8.4. Exercises
Exercise 8.16. Write a function that trims leading and trailing white space from a
string. Inner whitespace should not be modified.
Exercise 8.17. Write a function that splits a string containing a unix path/file into its
three components: the directory path, the file base name and the file extension. For
example, if the input string is /usr/home/message.txt then the three components
would be /usr/home/, message and txt respectively. For the purposes of this function,
you may assume that the path ends with the last forward slash (or is empty if none) and
that the extension is always after the last period. That is, you should be able to handle
inputs such as ../foo/bar/baz.old.txt.
Exercise 8.18. Write a function that (re)formats a string representing a telephone
number. Phone numbers can be written using a variety of formats, for example
1-402-555-1234 or +4025551234 or 402 555-1234, etc. Assume that you will only",ComputerScienceOne.pdf
"1-402-555-1234 or +4025551234 or 402 555-1234, etc. Assume that you will only
deal with 10 digit US phone numbers. Create a new string that uses the “standard”
format of (402) 555-1234.
Exercise 8.19. Write a function that takes a string and splits it up to an array of
strings. The split will be length-based: the function will also take an integer n and will
split the given string up into strings of length n. It is possible that the last string will
not be of length n.
For example, if we pass ""Hello World, how are you?"" with n = 3 then it should
return an array of size 9 containing the strings ""Hel"", ""lo "", ""Wor"", ""ld,"", "" ho"",
""w a"", ""re "", ""you"", ""?""
Exercise 8.20. HyperText Markup Language (HTML) (Hypertext Markup Language)
is the primary document description language for the World Wide Web (WWW). Certain
characters are not rendered in browsers as they are special characters used in HTML; in
particular tags which begin and end with the < and >.",ComputerScienceOne.pdf
"characters are not rendered in browsers as they are special characters used in HTML; in
particular tags which begin and end with the < and >.
To display such characters correctly they need to be escaped (similar to how you need
to escape tabs \t and endline \n characters). Properly escaping these characters
is not only important for proper rendering, but there are also security issues involved
(Cross-Site Scripting Attacks).
Write a function that takes a string and escapes the HTML characters in Table 8.1.
Replace the following with this
& &amp;
< &lt;
> &gt;
"" &quot;
Table 8.1.: Replacement HTML Characters
181",ComputerScienceOne.pdf
"9. File Input/Output
A file is a block of data used for storing information. Normally, we think of a file as
something that is stored on a hard drive (or memory stick or other physical media),
but the concept of a file is much more general. For example, when a file is loaded
(“read”) by a program it then exists in main memory. An executable program itself is a
file (containing instructions to be executed), both stored on the hard drive and run in
memory.
In a typical unix-based system, everything is a file. Directories are files, executables are
files, devices are files, etc. Even the familiar standard input and standard output are
buffers that are treated as files that can be read from or written to.
Data files may be stored as binary data or as plaintext files. Plaintext files are still stored
as binary data, but are stored in an encoding using the ASCII text values. Binary files
will also have structure, but it depends on the application that produced the file to give",ComputerScienceOne.pdf
"will also have structure, but it depends on the application that produced the file to give
meaning to the data. For example, an image file in a Joint Photographic Experts Group
(JPEG) format is essentially just binary data but it has a very specific format that an
image viewer would be able to process, but, say, a text editor would not. Further, if the
binary format is corrupted, the image viewer might not be able to display the image
correctly
Typical programs are short lived, anywhere from a few milliseconds to maybe a user
“session.” We often want data to be saved across multiple runs of a program. We need
to save it or persist it in some durable storage medium (disk). Files provide a way to
achieve data persistence.
9.1. Processing Files
In general, processing data in a file involves three basic steps:
1. Open (or create) the file
2. Read from (or write to) the file
3. Close the file
183",ComputerScienceOne.pdf
"9. File Input/Output
Depending on the language, the act of opening a file may determine if it will be read
from or written to. When read from, the file is referred to as an input file while a file
that is written to is an output file. Languages may also have different a different API or
functions to read/write or append to a file.
A file may be read line by line until the end of the end of the file has been reached.
Languages usually support this by using a special End Of File (EOF) flag or value to
indicate the end of a file has been reached.
A file is a resource just like memory and we need to properly manage it. We need to
make sure that we close a file once we are finished processing it. Depending on the
language and other factors such as the operating system, failure to close a file may result
in corrupted data. Though a file may be closed automatically for us when the program
terminates, its still best practice to properly close it.
9.1.1. Paths",ComputerScienceOne.pdf
"terminates, its still best practice to properly close it.
9.1.1. Paths
When opening a file on a file system, it is necessary to specify which file you want to open.
This is typically done by specifying at least the name of the file. Often files will have
“extensions” which indicate the type of file it is such as .txt for text files or .html for
HTML files. However, file extensions are only for organizational purposes and have no
real bearing on what data is stored in the file.
More important is the path of the file. Usually, if no path is specified, then implicitly
we are opening the file in the Current Working Directory (CWD). For example, if we
open the file data.txt then we are opening the file in the same directory in which our
program is executing. When specifying a path we can either specify an absolute path
or a relative path. An absolute path specifies each and every subdirectory in the file
system from the root to the directory that the file is located in. The root directory is the",ComputerScienceOne.pdf
"system from the root to the directory that the file is located in. The root directory is the
top-most directory in the file system. Each subdirectory is separated by some delimiter.
Windows systems use a backslash as a directory delimiter while the root directory is
specified using a “volume” name such as C:\. For example, an absolute path on a
Windows system may look something like:
C:\applications\users\data\data.txt
On a Unix-based system, a forward slash is used as a directory delimiter and the root
directory is simply a single forward slash. The same directory structure in a Unix-based
system would look like the following.
/applications/users/data/data.txt
A path may also be relative to the current working directory. In most systems (Windows
and Unix-based) the current directory is denoted using a single period, .. You can use
184",ComputerScienceOne.pdf
"9.1. Processing Files
this to specify directories deeper in the directory tree from the current directory. For
example (in Unix),
./app/data/data.txt
would refer to the directory app in the current working directory, the directory data
within that, and finally the file data.txt within that directory.
We can also refer to files further up the directory tree using the “parent” directory symbol
which is two periods, ... For example,
../../system/data.txt
would refer to a file two levels up in the subdirectory system.
9.1.2. Error Handling
When dealing with files there are many potential error conditions that may be anticipated
and may need to be dealt with. Some types of errors that can occur include the following.
•Permission errors – Your program may not have the proper permissions to read or
write to a particular directory or file. For example, user-level processes typically
do not have permissions to read system-level files (such as password files) nor can",ComputerScienceOne.pdf
"do not have permissions to read system-level files (such as password files) nor can
they write to them lest they corrupt critical system data.
•File Not Found errors – Your program may attempt to open a file that does not
exist. Sometimes, particularly when writing, the file will be created if it does not
exist. However, for reading, this may be a serious error.
•I/O Errors – We may be able to successfully find and open a file, but while
processing it something might go wrong with the file system that results in a
general input/output error.
•Formatting errors – As previously mentioned, the format of a file is highly dependent
on the application that created it (though there are universal data formats such as
XML or JavaScript Object Notation (JSON)). If the format is not as expected the
file may be corrupted and the program may not be able to successfully read data
from it.
Depending on the language, these errors may be fatal or may result in error codes or",ComputerScienceOne.pdf
"from it.
Depending on the language, these errors may be fatal or may result in error codes or
error values that can be dealt with. Or they may result in exceptions that can be caught
and handled.
185",ComputerScienceOne.pdf
"9. File Input/Output
/proc/self/
|-- attr
|-- cwd -> /proc
|-- fd
| `-- 3 -> /proc/15589/fd
|-- fdinfo
|-- net
| |-- dev_snmp6
| |-- netfilter
| |-- rpc
| | |-- auth.rpcsec.context
| | |-- auth.rpcsec.init
| | |-- auth.unix.gid
| | |-- auth.unix.ip
| | |-- nfs4.idtoname
| | |-- nfs4.nametoid
| | |-- nfsd.export
| | `-- nfsd.fh
| `-- stat
|-- root -> /
`-- task
`-- 15589
|-- attr
|-- cwd -> /proc
|-- fd
| `-- 3 -> /proc/15589/task/15589/fd
|-- fdinfo
`-- root -> /
Figure 9.1.: The Linux file system (like most file systems), defines a tree directory
structure. Each file and directory is contained in subdirectories all contained
within the root directory, /. This diagram was generated by the Tree
command, http://mama.indstate.edu/users/ice/tree/.
186",ComputerScienceOne.pdf
"9.1. Processing Files
9.1.3. Buffered and Unbuffered
When processing files the input/output may be either buffered or unbuffered. A buffered
input or output “stream” is one in which data that is read/written is actually stored in
memory in a “buffer” until such a time as the buffer is “flushed” and the accumulated
data is passed to/from the actual file.
For example, in a buffered output file, our program could write several kilobytes of data
to the output file, but it might not actually be written to the file right away. Instead,
those kilobytes of data are stored in memory until the buffer fills up or some other event
takes place to cause the buffer to be flushed. At that point, the data stored in the buffer
is emptied and written to the file.
Buffered input/output is used because I/O operations are expensive in terms of system
resources and can slow the system down. Because of this, it is better to keep I/O
operations as infrequent as possible. Buffers help to reduce the number of I/O operations",ComputerScienceOne.pdf
"operations as infrequent as possible. Buffers help to reduce the number of I/O operations
performed by a program by making them less frequent.
There are some instances in which we want unbuffered I/O. When error messages are
written to the standard error output for example, we would prefer to know about errors
as soon as possible rather than waiting for error messages to accumulate in a buffer.
Using an unbuffered output means that data is written to the standard error (which is a
file) immediately. However, because errors are (hopefully) infrequent and (likely) fatal,
this is not a performance issue.
9.1.4. Binary vs Text Files
As previously mentioned, files can be stored as pure binary data or as plaintext (ASCII).
Depending on our application and the nature of the data being written to files, the choice
of which to use may be clear. If we want the data in our files to be human-readable, then
we need to store them as plaintext. However, in general, we should prefer storing data in",ComputerScienceOne.pdf
"we need to store them as plaintext. However, in general, we should prefer storing data in
a binary format. The reason for this is that binary generally requires less space and is
more efficient to process.
Consider as an example, storing a collection of integers in a file. Each integer requires 4
bytes when represented in binary. However, when represented as a string, it requires as
many bytes as there are digits in the number. For example, the maximum representable
value of such an integer would be 231 −1 = 2,147,483,647, requiring 10 ASCII characters
and thus 10 bytes to represent, two and a half times as much memory as with the binary
representation. Further, if a lot of numbers are stored, each number (as a string) would
need to be delimited by yet another character. With a binary representation, no delimiter
would be necessary as we would know that each 4 byte block represents a single number.
This disparity can be even more stark with other types such as floating-point numbers.
187",ComputerScienceOne.pdf
"9. File Input/Output
There are additional performance issues when reading/writing the data and converting
binary numbers to their string representations. With binary data no such parsing is
necessary. As long as the data does not need to be human-readable, binary formats
should be preferred.
9.2. Exercises
Exercise 9.1. Write a function that takes a string representing a file name and opens
and processes the file, returning all of its contents as a single string.
Exercise 9.2. Consider an irregular, 2-D simple polygon with n points,
(x1,y1),(x2,y2),..., (xn,yn)
The area A of the polygon can be computed as
A= 1
2
n−1∑
i=0
(xiyi+1 −xi+1yi)
Note, that the initial and end point will be the same, ( x0,y0) = (xn,yn). An example
polygon for n= 5 can be found in Figure 9.2.
Figure 9.2.: An example polygon for n= 5
Write a program to open and process a text file containing n coordinates. In particular,
the first line is a single integer nthat indicates how many points should be read in. Each",ComputerScienceOne.pdf
"the first line is a single integer nthat indicates how many points should be read in. Each
line after that has the x,y coordinates of each point separated by a single space.
4
1.0 0.0
13.2 1.25
20.5 18.4
16.37 24.54
188",ComputerScienceOne.pdf
"9.2. Exercises
After reading the file in, it will compute the area of the polygon according to the formula
above and output it to the user. For example, the output for the above file may be
something like Area of the polygon: 197.9135
Exercise 9.3. Write a program that processes an input text file and scrubs it of any
HTML characters that need to be escaped (see Exercise 8.20 for details). It should
produce a new output file with all special characters escaped.
Exercise 9.4. Write a program that spell checks a plain text file. The program will
open a text file and process each word separately, checking for proper spelling against a
standard dictionary.
You may assume that each word is separated by some whitespace (you may assume that
there are no multi-line hyphenated words). However, you should ignore all punctuation
(periods, question marks, etc.).
Use a standard American dictionary provided on your unix system which stores words",ComputerScienceOne.pdf
"(periods, question marks, etc.).
Use a standard American dictionary provided on your unix system which stores words
one per line. Your output should include all misspelled or unrecognized words (words
not contained in the dictionary file).
Exercise 9.5. A standard word search consists of an n×n grid in which there are a
number of words hidden, some intersecting, with dummy letters filling in the blanks. An
example is provided in Figure 9.3.
Figure 9.3.: A Word Search
Write a program to solve a word search. Your program will read in an input file with
the following format: the first line will contain a single integer n which is followed by n
lines with n characters (not including the end line character) corresponding to the word
search.
Once you read in the word search, you will iterate through all possible words running
down, right, or diagonally down-right. You will attempt to match each possibility against",ComputerScienceOne.pdf
"down, right, or diagonally down-right. You will attempt to match each possibility against
a standard English dictionary. If the word matches a word in the dictionary, output it to
the standard output, otherwise ignore it. To simplify, you may restrict your attention to
words that have a length between 3 and 8 (inclusive).
189",ComputerScienceOne.pdf
"9. File Input/Output
Exercise 9.6. Write a crossword puzzle cheater. The program will take, as input, a
“partial” word in a crossword puzzle. That is, some of the letters are known (from other
solved clues) while some of the letters are not known. For the purposes of this exercise,
we’ll use a hyphen as a placeholder for missing letters.
Your program will match the partial word against words in a standard English dictionary
and list all possible matches. For example, if the user provided foo- as input it might
match food, fool, and foot.
Exercise 9.7. Bridge is a four player (2 team) game played with a standard 52-card
deck. Prior to play, a round of bidding is performed to determine which team is playing
for or against the contract, the trump suit, and at what level. Understanding the rules of
the game or the bidding conventions involved are not necessary for this exercise. Instead,
write a program to assist players in how they should bid based on the following point
system.",ComputerScienceOne.pdf
"write a program to assist players in how they should bid based on the following point
system.
A standard 52-card deck is dealt evenly to 4 different hands (Players 1 thru 4, 13 cards
each). Each player’s hand is worth a number of points based on the following rules:
•Each Ace in the hand is worth 4 points
•Each King is worth 3
•Each Queen is worth 2
•Each Jack is worth 1
•For each suit (Diamond, Spade, Club, Heart) such that the hand has only 2 cards
(a “doubleton”) an additional point is added
•For each suit that the hand has only 1 card in (a “singleton”) two additional points
are added
•For each suit that the hand has no cards (a “void”) 3 additional points are added.
Write a program that reads in a text file containing a deal. The formatting is as
follows: the input file will have 4 lines, one for each player. Each line contains the cards
dealt to that player delimited by a single space. The cards are indicated by the rank
(A,K,Q,J, 10,9,..., 2) and the suit (D, S, C, H). An example:",ComputerScienceOne.pdf
"dealt to that player delimited by a single space. The cards are indicated by the rank
(A,K,Q,J, 10,9,..., 2) and the suit (D, S, C, H). An example:
3C 3D 7S QD KC AS 6S AC JS 4S JD 7H 6D
5D 8C 7D AH 3H QC 8D JH 5H 9D 7C 9C 4D
2H 10D 8H KS QH 4C 10S 9S 6H 8S KD AD QS
2D 10C 6C 2C 10H 4H 2S 3S 5C 9H KH JC 5S
190",ComputerScienceOne.pdf
"9.2. Exercises
Your program should process the file and output the total number of points each hand
represents. You should not make any assumptions about the ordering of the input.
Hand 1 Points: 17
Hand 2 Points: 10
Hand 3 Points: 16
Hand 4 Points: 6
Exercise 9.8. The game of Sudoku is played on a 9 ×9 grid in which entries consist of
the numbers 1 thru 9. Initially, the board is presented with some values filled in and
others blank. The player has to fill in the remaining values until all grid boxes are filled
and the following constraints are satisfied.
•In each of the 9 rows, each number, 1–9 must appear exactly once
•In each of the 9 columns, each number 1–9 must appear exactly once
•In each of the 3 ×3 sub-grids, each number 1–9 must appear exactly once
A full example is presented in Figure 9.4.
Figure 9.4.: A solved Sudoku puzzle
Write a program that processes a text file containing a possible sudoku solution and",ComputerScienceOne.pdf
"Figure 9.4.: A solved Sudoku puzzle
Write a program that processes a text file containing a possible sudoku solution and
determine if it is a valid or invalid solution. The file will have the following format: it
will contain 9 lines with 9 numbers on each line delimited by a single space. If the input
represents a valid solution, output ”Valid Solution”, otherwise output at least one reason
why the input is not a valid solution.
Exercise 9.9. Write a program that parses and processes a data file containing Comma
Separated Values (CSV) and produce an equivalent JSON (JavaScript Object Notation)
output file containing the same data.
191",ComputerScienceOne.pdf
"9. File Input/Output
The input file will have the following format. The first line is a CSV list of column names.
Each subsequent line is an individual record with values for each column. The number
of columns and rows may vary from file to file. The following is an example containing
data about students, which has four columns and 3 records.
lastName,firstName,NUID,GPA
Castro,Starlin,11223344,3.48
Rizzo,Anthony,55667788,3.95
Bryant,Chris,01234567,2.7
The output file will be formatted in JSON where each “object” (record) is denoted with
opening and closing curly brackets, each record is separated by a comma, and all records
are enclosed in square brackets (putting them in an array). For each record, each value is
denoted with a key (the column name) and a value. For this exercise, treat all values as
strings even if they are numbers. For example, the input file above would be formatted
as follows.
1 [
2 {
3 ""lastName"": ""Castro"",
4 ""firstName"": ""Starlin"",
5 ""NUID"": ""11223344"",
6 ""GPA"": ""3.48""",ComputerScienceOne.pdf
"as follows.
1 [
2 {
3 ""lastName"": ""Castro"",
4 ""firstName"": ""Starlin"",
5 ""NUID"": ""11223344"",
6 ""GPA"": ""3.48""
7 },
8 {
9 ""lastName"": ""Rizzo"",
10 ""firstName"": ""Anthony"",
11 ""NUID"": ""55667788"",
12 ""GPA"": ""3.95""
13 },
14 {
15 ""lastName"": ""Bryant"",
16 ""firstName"": ""Chris"",
17 ""NUID"": ""01234567"",
18 ""GPA"": ""2.7""
19 }
20 ]
Exercise 9.10. Ranked voting elections are elections where each voter ranks each
candidate rather than just voting for a single candidate. If there are n candidates, then
192",ComputerScienceOne.pdf
"9.2. Exercises
each voter will rank them 1 (best) through n (worst). Usually, the winner of such an
election is determined by a Condorcet method (the candidate that would win in by a
majority in all head-to-head contests). However, we’ll use an alternative method, a Borda
count.
In a Borda count, points are awarded to each candidate for each ballot. For every number
1 ranking, a candidate receives n points, for every 2 ranking, a candidate gets n−1
points, and so on. For a rank of n, the candidate only receives 1 point. The candidates
are then ordered by their total points and the one with the highest point count wins
the election. Such a system usually leads to a “consensus” candidate rather than one
preferred by a majority.
Implement a Borda-count based ranked voting program. Your program will read in a file
in the following format. The first line will contain an ordered list of candidates delimited",ComputerScienceOne.pdf
"in the following format. The first line will contain an ordered list of candidates delimited
by commas. Each line after that will represent a single ballot’s ranking of the candidates
and will contain comma delimited integers 1 through n. The order of the rankings will
correspond to the order of the candidates on the first line.
Your program will take an input file name as a command line argument, open the file
and process it. It will then report the results including the point total for each candidate
(in order) as well as the overall winner. It will also report the total number of ballots.
You may assume each ballot is valid and all rankings are provided.
An example input:
Alice,Bob,Charlie,Deb
2,1,4,3
3,4,2,1
4,2,3,1
3,2,1,4
3,1,4,2
An example output:
Election Results
Number of ballots: 5
Candidate Points
Bob 15
Deb 14
Charlie 11
Alice 10
193",ComputerScienceOne.pdf
"9. File Input/Output
Winner is Bob
Exercise 9.11. A DNA sequence is a sequence of some combination of the characters
A (adenine), C (cytosine), G (guanine), and T (thymine) which correspond to the
four nucleobases that make up DNA. Given a long DNA sequence, its often useful to
compute the frequency of n-grams. An n-gram is a DNA subsequence of length n. Since
there are four bases, there are 4 n possible n-grams.
Write a program that processes a DNA sequence from a plaintext file and, given n,
computes the relative frequency of each n-gram as it appears in the sequence. As an
example, consider the sequence in Figure 9.5.
GGAAGTAGCAGGCCGCATGCTTGGAGGTAAAGTTCATGGTTCCCTGGCCC
Figure 9.5.: A DNA Sequence
To compute the frequency of all n= 2 n-grams, we would consider all 16 combinations
of length-two DNA sequences. We would then go through the sequence and count up the
number of times each 2-gram appears. We then compute the relative frequency (note:",ComputerScienceOne.pdf
"number of times each 2-gram appears. We then compute the relative frequency (note:
if a sequence is length L, then the total number of n-grams in it is L−(n−1)). The
relative frequency of each such 2-gram is calculated below.
AA 6.1224%
AC 0.0000%
AG 10.2041%
AT 4.0816%
CA 6.1224%
CC 10.2041%
CG 2.0408%
CT 4.0816%
GA 4.0816%
GC 10.2041%
GG 12.2449%
GT 8.1633%
TA 4.0816%
TC 4.0816%
TG 8.1633%
TT 6.1224%
Exercise 9.12. Given a long DNA sequence, it is often useful to compute the number
of instances of a certain subsequence. As an example, if we were to search for the
194",ComputerScienceOne.pdf
"9.2. Exercises
subsequence GTA in the DNA sequence in Figure 9.5, it appears twice. As another
example, in the sequence CCCC, the subsequence CC appears three times.
Write a program that processes a text file containing a DNA sequence and, given a
subequence s, searches the DNA sequence and counts the number of times s appears.
Exercise 9.13. Protein sequencing in an organism consists of a two step process. First
the DNA is translated into RNA by replacing each thymine (T) nucleotide with uracil
(U). Then, the RNA sequence is translated into a protein according to the following rules.
The RNA sequence is processed 3 bases at a time. Each trigram is translated into a single
amino acid according to known encoding rules. There are 20 such amino acids, each
represented by a single letter in (A,C,D,E,F,G,H,I,K,L,M,N,P,Q,R,S,T,V,W,Y ).
The rules for translating trigrams are presented in Figure 9.6. Each triple defines a",ComputerScienceOne.pdf
"The rules for translating trigrams are presented in Figure 9.6. Each triple defines a
protein, but we’re only interested in the first letter of each protein. Moreover, the
trigrams UAA, UAG, and UGA are special markers that indicate a (premature) end to
the protein sequencing (there may be additional nucleotides left in the RNA sequence,
but they are ignored and the translation ends).
Figure 9.6.: Codon Table for RNA to Protein Translation
As an example, suppose we start with the DNA sequence AAATTCCGCGTACCC ;
it would be encoded into RNA as AAAUUCCGCGUACCC ; and into an amino acid
sequence KFRVP .
Write a program that processes a file containing a DNA sequence and outputs the
translated proteins (only the first letter of each protein) to an output file.
195",ComputerScienceOne.pdf
"9. File Input/Output
Exercise 9.14. Recently, researchers have successfully inserted two new artificial nucle-
ases into simple bacteria that successfully reproduced the artificial bases through several
generations. The artificial bases d5SICS and dNaM, ( X and Y for short) mimic the
natural G, and C nucleobases respectively.
Write a program that takes a normal DNA sequence and replace some of its G,C pairs
with X,Y respectively. DNA is translated into RNA which is then translated into 20
different amino acids. Each amino acid produced depends on a 3 nucleobase codon. For
this exercise, we will change G,C pairs with X,Y pairs but only in codons that represent
the amino acids Threonine (an essential amino acid) and Alanine (a non-essential amino
acid). Table 9.1 contains the codons corresponding to these amino acids and the codons
you should translate each one to. All other codons should not be modified.
Your program should open and process a DNA sequence contained in a file and modify",ComputerScienceOne.pdf
"Your program should open and process a DNA sequence contained in a file and modify
the DNA sequence as described above and output the artificial DNA sequence to a new
output file.
Amino Acid Codon Artificial Codon
Threonine
ACT AYT
ACC AYY
ACA AYA
ACG AYX
Alanine
GCT XYT
GCC XYY
GCA XYA
GCG XYX
Table 9.1.: Amino Acid Codons
196",ComputerScienceOne.pdf
"10. Encapsulation & Objects
One reason we prefer to write programs in high-level programming languages is that we
can use syntax that is closer to plain English. Though programming language syntax
is a far cry from “natural” language, it is far closer than lower level languages such as
assembly or binary machine code. However, from what we’ve seen so far, when writing
programs we are still forced to utilize the primitive variable types built-in to the language
we’re using, which is still quite limiting.
As a motivating example, suppose we were to write a program that involved organizing
the enrollment of students into courses. To model a particular student, we would need
a collection of variables, say a first name, last name, GPA, and a unique identification
number (likely a lot more, but let’s keep it simple). Each of these pieces of data could
be modeled by strings, a floating-point number and perhaps an integer. 1 Each of these",ComputerScienceOne.pdf
"be modeled by strings, a floating-point number and perhaps an integer. 1 Each of these
pieces of data are stored in separate, unrelated variables even though they represent a
single entity.
Even worse, suppose that we needed to keep track of a collection of students. Each piece
of data would need to be stored in a separate array. If we wanted to rearrange the data
(say, sort it), we would need to do a lot of manual bookkeeping to make sure that the
separate pieces of data that represented a single entity were all aligned at the same index
in each of the arrays. If we wanted to pass the data around to functions, we’d be forced
to pass multiple arrays to each function. This becomes all the more complex when we
attempt to model entities with more pieces of data.
The solution is to encapsulate the pieces of data into one logical entity, sometimes referred
to as an object. More formally, encapsulation is a mechanism by which pieces of data can",ComputerScienceOne.pdf
"to as an object. More formally, encapsulation is a mechanism by which pieces of data can
be grouped together along with the functions that operate on that data. Encapsulation
may also provides a means to protect that data by controlling the visibility of that data
from code outside the object.
Contrast this with an array which is also a collection of data. However, an array usually
contains pieces of similar data (all elements are integers or all elements are floating
point numbers) while an object may collect pieces of dissimilar data that make up a
larger entity. It is like the difference between rows and columns in a table. Consider the
student data in Table 10.1. Each row represents a record while each column represents a
1Depending on the identification number, it may be more appropriately modeled with a string. Social
Security Numbers for example are not purely numeric; they include dashes and may begin with zeros.
197",ComputerScienceOne.pdf
"10. Encapsulation & Objects
First Name Last Name ID GPA
Tom Baker 74 3.75
Christopher Eccleston 5 3.5
David Tennant 10 4.0
Matt Smith 29 3.2
Peter Capaldi 13 2.9
Table 10.1.: Student Data
collection of data from each record. A single column is comparable to an array while
each row is comparable to an object. In this example, each object has four pieces of data
encapsulated in it, a first name, last name, an ID, and a GPA.
To represent this data in code without objects we would need at least 4 separate arrays,
more if we wanted to model more data for a student. Moreover, data in separate
arrays or collections have no real logical relationship to each other. The solution that
most programming languages provide is allowing you to define an object or structure
that collects pieces of data into one logical unit, allowing you to name the object (say
“Student”) so that you can deal with the data in a more abstract way. With objects, we",ComputerScienceOne.pdf
"“Student”) so that you can deal with the data in a more abstract way. With objects, we
can treat each row in the table as a single, distinct entity allowing us to collect Student
objects into a single array or collection rather than many separate ones.
10.1. Objects
Though languages differ in how they support objects, they all have some commonalities.
A language needs to provide ways to define objects, create instances of objects, and to
use them in code.
10.1.1. Defining
Most object oriented programming languages such as C++ and Java are class-based
languages. Meaning that they allow you to define objects by declaring and defining a
“class.” A class is essentially a blueprint for what the object is and how it is defined.
Generally, a class declaration allows you to specify member variables and member methods
which are part of the class. Further, full encapsulation is achieved by using visibility
keywords such as public or private to either allow or restrict access to variables and",ComputerScienceOne.pdf
"keywords such as public or private to either allow or restrict access to variables and
methods from code outside the object.
Non-object-oriented languages may not support full encapsulation. Instead they may
allow you to define structures which support the grouping of data, but make it difficult
198",ComputerScienceOne.pdf
"10.1. Objects
or impossible to achieve the other two aspects of encapsulation (the grouping of methods
that act on that data and the protection of data).
In either case, a language allows you to define the member variables and to name the
class or structure so that instances can be referred to by that type. Built-in types such
as numbers or strings already have a type name defined by the language. However, an
object is a user-defined type that is not built-in to the language. Once defined, however,
the class or structure can be referred to just like any built-in variable type.
It is not unusual to create objects that are made of other objects. For example, a student
object may be defined by using two strings for its first and last name. In the language,
a string may also be an object. As a more complex example, suppose that we wanted
an additional member variable to model a student’s date of birth. A date may itself",ComputerScienceOne.pdf
"an additional member variable to model a student’s date of birth. A date may itself
be an object as it consists of several pieces of information (a year, month, and date
at least). When an object “owns” an instance of another object it is referred to as
composition as the object is composed of other objects. Further, an object may consist of
a collection of other objects (suppose that a student object owned an array of course
objects representing their schedule). This is a form of composition known as aggregation
(multiple objects have been aggregated by the object).
10.1.2. Creating
Once a blueprint for an object (or structure) has been declared and defined, we need a
way to create instances of the object. The concept of an “object” is general and abstract.
It is more like the idea of a student. Only once we have created an entity that exists in
memory do we have an actual instance of the class. Creating instances of an object is
usually referred to as instantiation.",ComputerScienceOne.pdf
"memory do we have an actual instance of the class. Creating instances of an object is
usually referred to as instantiation.
Languages may be able to automatically create instances of your object with default
values. After all, your object is likely composed of built-in types. The student example
above for example could be modeled with two strings, an integer, and a floating point
number. The language/compiler/interpreter “knows” how to deal with these built-in
types, so it can extend that knowledge to create instances of your object which are
essentially just collections of types that it already knows how to deal with.
Object-oriented languages usually provide a special method for you to be able to specify
the details of how to create an instance. These are called constructor methods. Sometimes
you can define multiple constructors methods that take different number(s) of arguments
and/or have different behavior. Constructor methods typically have special syntax or",ComputerScienceOne.pdf
"and/or have different behavior. Constructor methods typically have special syntax or
have the same name as the class.
In other languages that do not fully support object-oriented programming, you must
define utility functions that can be used to create instances of your object. Sometimes
these are referred to as factory functions as they are responsible for “manufacturing”
199",ComputerScienceOne.pdf
"10. Encapsulation & Objects
instances of your object.
10.1.3. Using Objects
After defining and creating an object, you can usually use it like any regular variable. In
a strongly typed language you would declare a variable whose type matches the declared
class or structure. The variable type can usually be passed and returned from functions,
assigned to other variables, etc.
A language also provides ways to access the member variables or methods that are visible
to the outside world. Languages usually allow you to do this through the “dot operator”
or the “arrow operator.” Suppose we have an instance of a student object stored in
a variable s. To access the first name of this instance, we may be able to use either
s.firstName or s->firstName. We can access and invoke visible methods likewise.
The dot/arrow operators are how code outside the object interacts with the object.
Outside code is able to do this because it holds a reference, s to the object. However,",ComputerScienceOne.pdf
"Outside code is able to do this because it holds a reference, s to the object. However,
inside the object, we may not have a reference (the variable s was ostensibly declared
and used outside the object and so is not in scope inside the object). However, we
still have need to reference member variables or methods from inside the object. Many
languages use open recursion , a mechanism by which we can write code so that an
instance is able to refer to itself. Languages use keywords such as this or self or
something similar. The keyword is essentially a self-reference to the object itself so that
you can refer to “this” object from within the object.
10.2. Design Principles & Best Practices
Using objects in your code follows more of a bottom-up design rather than a top-down
design approach. In a top-down design, a program is designed by breaking a program
or problem down into smaller and smaller components. Bottom-up design approaches a",ComputerScienceOne.pdf
"or problem down into smaller and smaller components. Bottom-up design approaches a
problem differently. First, real-world entities involved with the problem are modeled by
defining objects. Then objects are used as building blocks that can be combined and
made to interact to solve a problem.
Object design is usually a straightforward process. Typically an object is modeling a
real-world entity, so it is simple enough to decompose the entity into its constituent
components. We do this until the component can either be modeled by a built-in type
such as a string or number or by an existing object. In general, you want to keep things
as simple as possible. Any time you need to associate pieces of data together into one
logical unit, it is appropriate to encapsulate them into an object.
A good design principle is to utilize composition as much as possible. If you have multiple
200",ComputerScienceOne.pdf
"10.3. Exercises
pieces of data that define a logical entity or unit, it is good design to create another
object. For example, suppose a student object needs to model a mailing address; think
about what an address is: it is a street address, city, state, zip, etc. Rather than having
these as member fields to your object, it is probably more appropriate to define an
“address” object, especially if such an object would be useful elsewhere in a program.
10.3. Exercises
Exercise 10.1. A complex number consists of two real numbers: a real component a
and an imaginary component biwhere bis a real number and i= √−1. Define an object
or structure to model a complex number. Write functions to:
•Create a complex number
•Print a complex number
•Perform basic arithmetic operations on two complex numbers including addition,
subtraction and multiplication as defined by:
c1 + c2 = (a1 + b1i) + (a2 + b1i) = (a1 + a2) + (b1 + b2)i
c1 −c2 = (a1 + b1i) −(a2 + b1i) = (a1 −a2) + (b1 −b2)i",ComputerScienceOne.pdf
"c1 + c2 = (a1 + b1i) + (a2 + b1i) = (a1 + a2) + (b1 + b2)i
c1 −c2 = (a1 + b1i) −(a2 + b1i) = (a1 −a2) + (b1 −b2)i
c1 ·c2 = (a1 + b1i) ·(a2 + b1i) = (a1a1 −b1b2) + (a1b2 + b1a1)i
Exercise 10.2. Design an object (or structure) that models an album. Include at least
the album title, artist (or band) and release year. Include any other data that you think
is relevant and write functions to support your object.
Exercise 10.3. Design an object (or structure) that models a bank savings account.
Include at least the balance, APR, an account “number” and customer information
(which may be another object or structure). Include any other data that you think is
relevant and write functions to support your object.
Exercise 10.4.Design an object (or structure) that models a sports stadium. Include at
least the stadium name, the team that plays there, its city, state, and year built. Include
both latitude and longitude data. Include any other data that you think is relevant and",ComputerScienceOne.pdf
"both latitude and longitude data. Include any other data that you think is relevant and
write functions to support your object. Write a parser to process a flat file data of all
stadiums (in your chosen sport) and build instances of all of them.
Exercise 10.5. Design an object (or structure) that models a network-connected elec-
tronic device. Include at least a unique ID, a human-readable name for the device, a
Media Access Control (MAC) address and Internet Protocol (IP) address as well as the
device’s bandwidth in megabits per second. Include any other data that you think is
relevant and write functions to support your object.
201",ComputerScienceOne.pdf
"10. Encapsulation & Objects
Exercise 10.6. Design an object (or structure) that models an airport. Include at least
the name, FAA designation, its city, state, and latitude/longitude data. Include any
other data that you think is relevant and write functions to support your object.
202",ComputerScienceOne.pdf
"11. Recursion
Suppose we wanted to write a simple program that performed a countdown, printing 10,
9, 8, . . . , 2, 1 and when it reached zero it printed a “Happy New Year” message. Likely
our first instinct would be to write a very simple for loop using an increment variable.
But suppose we lived in a world without the usual loop control structures that we are
now familiar with. How might we write such a program?
After thinking about it for a while, we might think: well, we don’t have loops, but we
still have functions. In particular what if we had a function that took the “current” value
of our counter variable and decremented it, passing it to another function, which did the
same thing. For example, we could pass 10 to such a function, which would then subtract
1, passing 9 to another function and so on. A check could be made to see if the value was
zero, in which case we print our special message and no longer call any more functions.",ComputerScienceOne.pdf
"zero, in which case we print our special message and no longer call any more functions.
In fact, we would not need to define 10 different functions to do so. Instead, we could
define one function that called itself. It might look something like Algorithm 11.1.
Input : An integer n≥0
Output :A countdown of integers n,... 0
1 if n= 0 then
2 output “Happy New Year!!!”
3 else
4 output n
5 CountDown(n−1)
6 end
Algorithm 11.1:Recursive CountDown(n) Function
The function in this case is called CountDown(). In Line 5 the function calls itself on
a decremented value. When a function calls itself, it is a recursive function. When a
language allows functions to call themselves they support recursion.
This is not that odd of a concept. We’ve seen many examples where functions invoke (call)
other functions. Each function call simply creates a new stack frame on the program
stack. There is nothing particularly special about which functions call which other",ComputerScienceOne.pdf
"stack. There is nothing particularly special about which functions call which other
functions, so there is little difference when a function calls itself.
203",ComputerScienceOne.pdf
"11. Recursion
This was not just a toy example. There are many programming languages in which
recursion is used as a matter of course. Functional programming languages tend to avoid
control structures like loops and even (mutable) variables. Instead, control flow is defined
by evaluating a series of functions, making recursion a fundamental technique.
Recursion is extensively used in mathematics. Recurrence relations or recursive functions
are common. The Fibonacci sequence is a common, if not overused 1 example. It has a
simple definition: the next value in the sequence is simply the sum of the two previous
values. The sequence starts with the initial values of 1. The first few terms in the
sequence:
1,1,2,3,5,8,13,21,34,55,89,...
The more formal mathematical definition can be stated as follows.
Fn =



1 if n= 0
1 if n= 1
Fn−1 + Fn−2 otherwise
The Fibonacci sequence is the cliche example for recursion. We can define an algorithmic
function to compute the n-th Fibonacci number as follows.",ComputerScienceOne.pdf
"The Fibonacci sequence is the cliche example for recursion. We can define an algorithmic
function to compute the n-th Fibonacci number as follows.
Input : An integer n≥0
Output :The n-th Fibonacci number, Fn
1 if n≤1 then
2 output 1
3 else
4 output Fibonacci(n−1) + Fibonacci(n−2)
5 end
Algorithm 11.2:Recursive Fibonacci(n) Function
Though hackneyed, it does provide a good example for how recursive functions work.
We’ll also utilize it as an example of why you should avoid recursion in practice . We use
it to illustrate how the problems with recursion can be mitigated or avoided altogether.
11.1. Writing Recursive Functions
When writing a recursive function, there are several key elements that we need to take
care of to ensure that it executes correctly. In particular, every recursive function requires
1The Fibonacci sequence is nothing special; its simply a second order linear homogenous recurrence",ComputerScienceOne.pdf
"1The Fibonacci sequence is nothing special; its simply a second order linear homogenous recurrence
relation with coefficients of 1. The near reverence that so many people attribute to it borders on
mysticism.
204",ComputerScienceOne.pdf
"11.1. Writing Recursive Functions
at least one base case or base condition which serves as a terminating condition for the
recursion. A base case is a condition which, instead of making a recursive call, processes
and returns a value. Without a base case, the recursion would continue unbounded: the
function would call itself over and over again, creating new stack frame after stack frame
until we run out of stack space. If a program makes too many function calls and runs
out of stack memory, it may lead to a stack overflow and the termination of the program.
Even if we don’t have unbounded recursion, it is still possible to run out of stack space
even with simple recursion.
The other key element that we need is to ensure that every recursive call makes progress
toward one of the terminating conditions. If no progress is made, then again we may have
an unbounded recursion. In the Fibonacci example in Algorithm 11.2, the base case can",ComputerScienceOne.pdf
"an unbounded recursion. In the Fibonacci example in Algorithm 11.2, the base case can
be found in the first if-statement: when n reaches 1 or less, no recursive calls are made.
In the else-statement, we make two recursive calls, but both of them make progress
toward this base case. The first decrements n by 1 and the second by 2, eventually
reaching n= 1.
11.1.1. Tail Recursion
Making many function calls can be costly in terms of stack space. One optimization
that can be made is to use tail recursion. The last operation that a function executes
is referred to as the tail operation. If a function invokes another function as its tail
operation, its a tail call. For example, consider the following snippet of code:
1 int foo(int x) {
2 ...
3 return bar(x-1) + 1;
4 }
Here, foo() calls bar() but it is not the last operation before it returns. Instead, it
invokes bar(), takes the result and adds one then returns to the calling function. Note",ComputerScienceOne.pdf
"invokes bar(), takes the result and adds one then returns to the calling function. Note
that decrementing x is performed before the invocation of bar(). In contrast, consider
the following modified code:
1 int foo(int x) {
2 ...
3 return bar(x-1);
4 }
205",ComputerScienceOne.pdf
"11. Recursion
Here, the invocation of bar() is the last operation performed by foo(). Thus, this is
a tail call.
Tail calls have the advantage that a language or compiler can generally optimize the
function call with respect to the stack frame. Since the function foo() is essentially
done with its computation, its stack frame is no longer needed. The system, therefore, can
reuse the stack frame. Tail recursion is such an important optimization, some languages
require it or “guarantee” it in other ways.
11.2. Avoiding Recursion
Recursion is not essential; some languages do not even support recursion. In fact,
any recursive function can be rewritten to not use recursion. Usually, you can write
an equivalent loop structure or use an in-memory stack data structure to replace the
recursion. So why use it? Proponents would argue that recursion allows you to write
simple code that more closely matches mathematical functions and expressions. Recursion",ComputerScienceOne.pdf
"simple code that more closely matches mathematical functions and expressions. Recursion
is also a natural way to think about certain problem solving techniques such as divide-
and-conquer (see Chapter 12). It is also a natural way to code in functional programming
languages.
These arguments, however, are subjective. One person’s “cleaner” or “more understand-
able” code is another person’s spaghetti code hack. What is “natural” for one person may
be “weird” and “odd” for another. However, there are many other arguments against
recursion, many of which are objective reasons: that recursion is more expensive and can
easily lead to inefficient, exponential algorithms.
In general, recursion requires lots of function calls which requires creating and removing
lots of stack frames. This usually results in a lot of overhead and resources being used
to perform the computation. Unless you are using a language in which recursion is",ComputerScienceOne.pdf
"to perform the computation. Unless you are using a language in which recursion is
optimized and made to be more efficient (such as functional programming languages),
this is a lot more expensive than using simple loops and iteration.
Another reason to avoid recursion is that it can lead to a lot of extraneous re-computations.
The cliched example of the Fibonacci recursion is a prime example of this. 2 Consider
the computation of Fibonacci(5). This results in two recursive calls, each of those calls
results in 2 recursive calls and so on as depicted in Figure 11.1.
As depicted in the figure, several function calls are repeated: Fibonacci(3) is called
twice, Fibonacci(2) is called 3 times, etc. 15 total function calls are made to compute
2Its overuse as an example of recursion is even less explicable as it solves a problem that no one cares
about.
206",ComputerScienceOne.pdf
"11.2. Avoiding Recursion
Fibonacci(5)
Fibonacci(4)
Fibonacci(3)
Fibonacci(2)
Fibonacci(1)Fibonacci(0)
Fibonacci(1)
Fibonacci(2)
Fibonacci(1)Fibonacci(0)
Fibonacci(3)
Fibonacci(2)
Fibonacci(1)Fibonacci(0)
Fibonacci(1)
Figure 11.1.: Recursive Fibonacci Computation Tree
Fibonacci(5). In general, the computation of Fibonacci(n) will result in anexponential
number of function calls. The number of function calls to compute Fn with this recursive
solution will be equal to
Fn−2 +
n−1∑
i=0
Fi
That is more than the first n−1 Fibonacci numbers combined! It should come as no
surprise that the Fibonacci sequence grows exponentially and thus so would the number
of operations with this recursive solution.
To put this in perspective, consider computing F45 = 1 ,836,311,903 ( n = 45), the
maximum representable value for a 32-bit signed two’s complement integer. Executing
a C implementation of this recursive algorithm took about 8 seconds 3 and required
3,672,623,805 function calls!",ComputerScienceOne.pdf
"a C implementation of this recursive algorithm took about 8 seconds 3 and required
3,672,623,805 function calls!
What if we wanted to compute F100 = 573,147,844,013,817,084,101 (573 quintillion)
it would result in 1,146,295,688,027,634,168,201 (1.146 sextillion) function calls. Using
the same hardware, at 4 .59 ×108 (459 million) function calls per second, it would take
2.497 ×1012 seconds to compute. That would be more than 79,191 years! Even if we
performed these (useless) calculations on hardware that was 1 million times faster than
my laptop, it would still take over 4 weeks!
3On a 2.7GHz Intel Core i7.
207",ComputerScienceOne.pdf
"11. Recursion
11.2.1. Memoization
The inefficiency in the example above comes from the fact that we make the same function
calls on the same values over and over. One way to avoid recomputing the same values is
to store them into a table (or tableau if you prefer being fancy). Then, when you need
to compute a value, you look at the table to see if it has already been computed. If
it has, we reuse the value stored in the table, otherwise we compute it by making the
appropriate recursive calls. Once computed, we place the value into the table so that it
can be looked up on subsequent function calls. This approach is usually referred to as
memoization.
The “table” in this scenario is very general: it can be achieved using a number of different
data structures including simple arrays, or even maps (mapping input value(s) to output
values). The table is essentially serving as a cache for the previously computed values. An",ComputerScienceOne.pdf
"values). The table is essentially serving as a cache for the previously computed values. An
illustration of how this might work can be found in Algorithm 11.3. Here, the recursion
only occurs if the value Fn is not yet defined.
Input : An integer n≥0, a global map M that maps n values to Fn
Output :The n-th Fibonacci number, Fn
1 if Fn is defined in M then
2 output M(n)
3 else
4 a←Fibonacci(n−1)
5 b←Fibonacci(n−2)
6 Define M(n) = a+ b
7 output (a+ b)
8 end
Algorithm 11.3:Recursive Fibonacci(n) Function With Memoization
In many functional programming languages, memoization is implicit and provided by
the language itself to ensure that we do not have the same problems with recomputing
values as we observed above. In languages such as C and Java, memoization becomes
our responsibility and is not an optimization provided by the language itself.
However, if we are filling in a table of values anyway, we really don’t need to make",ComputerScienceOne.pdf
"However, if we are filling in a table of values anyway, we really don’t need to make
recursive calls at all. We simply need to figure out in what order to fill out the values
in the table. This is actually the basis of a powerful programming technique known
as dynamic programming which is a “bottom-up” approach to solving problems by
combining solutions to smaller ones.
208",ComputerScienceOne.pdf
"11.3. Exercises
11.3. Exercises
Exercise 11.1. The binomial coefficients, C(n,k) or
(n
k
)
(“n choose k”), are defined as
the number of ways you can select k distinct items from a collection of n items. A direct
combinatorial definition is (n
k
)
= n!
k!(n−k)!
An alternative is Pascal’s identity, which gives a recurrence to compute this value:
(n
k
)
=
(n−1
k
)
+
(n−1
k−1
)
Where
(n
0
)
= 1 for any n and for all k >n,
(n
k
)
= 0. Finally,
(n
1
)
= n.
1. Write a recursive function using Pascal’s identity to compute
(n
k
)
. Benchmark its
performance.
2. Write a recursive version that uses memoization to avoid recomputing values
3. Modify your functions to utilize an arbitrary precision numeric type so that you
can compute arbitrarily large values.
Exercise 11.2. The Jacobsthal sequence is very similar to the Fibonacci sequence in
that it is defined by its two previous terms. The difference is that the second term is
multiplied by two.
Jn =



0 if n= 0
1 if n= 1
Jn−1 + 2Jn−2 otherwise",ComputerScienceOne.pdf
"multiplied by two.
Jn =



0 if n= 0
1 if n= 1
Jn−1 + 2Jn−2 otherwise
1. Write a recursive function that computes the n-th Jacobsthal number. Benchmark
its performance.
2. Write a recursive version that uses memoization to avoid recomputing values
3. Modify your functions to utilize an arbitrary precision numeric type so that you
can compute arbitrarily large values.
209",ComputerScienceOne.pdf
"12. Searching & Sorting
Searching and sorting are two fundamental operations when dealing with collections of
data. Both operations are not only important in and of themselves, but they also form
the basis of many algorithms and other more complex operations. These operations are
so essential that a wide variety of algorithms and techniques have been developed to
solve them, each with their own advantages and disadvantages. This variety provides a
good framework from which to study the relative efficiency and complexity of algorithms
through algorithm analysis.
12.1. Searching
Searching is a very basic operation. Given a collection of data, we wish to find a particular
element or elements that match a certain criteria. More formally, we have the following.
Problem 1(Searching).
Given: a collection of elements, A= {a1,a2,...,a n}and a key element ek
Output: The element ai in A such that ai = ek
The “equality” in this problem statement is not explicitly specified. In fact, this is a",ComputerScienceOne.pdf
"Output: The element ai in A such that ai = ek
The “equality” in this problem statement is not explicitly specified. In fact, this is a
very general, abstract statement of the basic search problem. We didn’t specify that the
“collection” was an array, a list, a set, or any other particular data structure. Nor did we
specify what type of elements were in the collection. They could be numbers, they could
be strings, they could be objects.
There are many variations of this general search problem that we could consider. For
example, we could generalize it to find the “first” or “last” such element if our collection is
ordered. We could find all elements that match our criteria. Some basic operations that
we’ve already considered such as finding the minimum or maximum ( extremal elements),
or median element are also variations on this search problem.
When designing a solution to any of these variations additional considerations must be",ComputerScienceOne.pdf
"or median element are also variations on this search problem.
When designing a solution to any of these variations additional considerations must be
made. We may wish our search to be index-based (that is, output the index i rather
than the element ai). We may need to think about how to handle unsuccessful searches
(return null? A special flag value? Throw an exception?, etc.).
211",ComputerScienceOne.pdf
"12. Searching & Sorting
index
contents 42
0
4
1
9
2
4
3
102
4
34
5
12
6
2
7
0
8
Figure 12.1.: Array of Integers
When implementing a solution in a programming language, we of course will need to be
more specific about the type of collection being searched and the type of elements in
the collection. However, we will still want to keep our solution as general as possible.
As we’ll see, most programming languages facilitate some sort of generic programming
so that we do not need to reimplement the solution for each type of collection or for
each type of variable. Instead, we can write one solution, then configure it to allow for
comparisons of any type of variable (numeric, string, object, etc.).
12.1.1. Linear Search
The first solution that we’ll look at is thelinear search algorithm (also known as sequential
search). This is a basic, straightforward solution to the search problem that works by
simply iterating through each element ai, testing for equality, and outputting the first",ComputerScienceOne.pdf
"simply iterating through each element ai, testing for equality, and outputting the first
element that matches the criteria. The pseudocode is presented as Algorithm 12.1.
Input : A collection of elements A= {a1,...,a n}and a key ek
Output :An element a in A such that a= ek according to some criteria; φ if no
such element exists
1 foreach ai in the collection A do
2 if ai = ek then
3 output ai
4 end
5 end
6 output φ
Algorithm 12.1:Linear Search
To illustrate, consider the following example searches. Suppose we wish to search the
0-indexed array of integers in Figure 12.1.
A search for the key ek = 102 would start at the first element. 42 ̸= 102 so the search
would continue; it would compare it against 4, then 9, then 5, and finally find 102 at
index i = 4, making a total of 5 comparisons (including the final comparison to the
matched element).
212",ComputerScienceOne.pdf
"12.1. Searching
a1 ··· an
2 −1 an
2
an
2 +1 ··· an
m<m >m
Figure 12.2.: When an array is sorted, all elements in the left half are less than the
middle element m, all elements in the right half are greater than m.
A search for the key ek = 42 would get lucky. It would find it after only one comparison as
the first element is a match. A search for the element 20 would result in an unsuccessful
search with a total of 10 comparisons being made. Finally a search for ek = 4 would
only require two comparisons as we find 4 at the second index. There is a duplicate
element at index 3, but the way we’ve defined linear search is to find the “first” such
element. Again, we could design any number of variations on this solution. We give a
more detailed analysis of this algorithm below.
12.1.2. Binary Search
An alternative search algorithm is binary search. This is a clever algorithm that requires
that the array being searched is sorted in ascending order. Though it works on any type",ComputerScienceOne.pdf
"that the array being searched is sorted in ascending order. Though it works on any type
of data, let’s again use an integer array as an example. Suppose we’re searching for the
key element ek. We start by looking at the element in the middle of the array, call it m.
Since the array is sorted, everything in the left-half of the array is <m and everything
in the right-half of the array is >m.1 We will now make one comparison between ek and
m. There are three cases to consider.
1. If ek = m, then we’ve found an element that matches our key and search criteria
and we are done. We can output m and stop the algorithm.
2. If ek <m then we know that if a matching element exists, it must lie in the left-half
of the list. This is because all elements in the right-half are >m.
3. If ek > mthen we know that if a matching element exists, it must lie in the
right-half of the list. This is because all elements in the left-half are <m.",ComputerScienceOne.pdf
"right-half of the list. This is because all elements in the left-half are <m.
In either of the second two cases, we have essentially cut the array in half, halving the
number of elements we need to consider. Suppose that the second case applies. Then
we can consider elements indexed from 1 to n
2 −1 (we need not consider an
2
as the first
case would have applied if we found a match). We can then do the same trick: check the
middle element among the remaining elements and determine which half to cutout and
1If duplicate elements are in the array, then elements in the left/right half could be less than or equal
to and greater than or equal to m, but this will not affect how our algorithm works.
213",ComputerScienceOne.pdf
"12. Searching & Sorting
which half to consider. We repeat this process until we’ve either found the element we
are looking for or the range in which we are searching becomes “empty” indicating an
unsuccessful search.
This description suggests a recursive solution. Given two indices l,r, we can compute
the index of the middle element, m= l+r
2 and make one of two recursive calls depending
on the cases identified above. Of course, we will need to make sure that our base case is
taken care of: if the two indices are invalid, that is if the left is greater than the right,
l>r , then we know that the search was unsuccessful.
Input : A sorted collection of elements A= {a1,...,a n}, bounds 1 ≤l,r ≤n,
and a key ek
Output :An element a in A such that a= ek according to some criteria; φ if no
such element exists
1 if l>r then
2 output φ
3 end
4 m←⌊l+r
2 ⌋
5 if am = ek then
6 output am
7 else ifam <ek then
8 BinarySearch(A,m + 1,r,ek)
9 else
10 BinarySearch(A,l,m −1,ek)
11 end",ComputerScienceOne.pdf
"3 end
4 m←⌊l+r
2 ⌋
5 if am = ek then
6 output am
7 else ifam <ek then
8 BinarySearch(A,m + 1,r,ek)
9 else
10 BinarySearch(A,l,m −1,ek)
11 end
Algorithm 12.2:Recursive Binary Search Algorithm, BinarySearch(A,l,r,e k)
As discussed in Chapter 11, non-recursive solutions are generally better than recursive
ones. We can design a straightforward iterative version of binary search using a while loop.
We initialize two index variables, l,r and update them on each iteration depending on
the three cases above. The loop stops when we’ve found our element or l>r resulting in
an unsuccessful search. The iterative version is presented in Algorithm 12.3, an example
214",ComputerScienceOne.pdf
"12.1. Searching
run of the algorithm is shown in Figure 12.3.
Input : A sorted collection of elements A= {a1,...,a n}and a key ek
Output :An element a∈A such that a= ek according to some criteria; φ if no
such element exists
1 l←1
2 r←n
3 while l≤r do
4 m←⌊l+r
2 ⌋
5 if am = ek then
6 output am
7 else ifam <ek then
8 l←(m+ 1)
9 else
10 r←(m−1)
11 end
12 end
13 output φ
Algorithm 12.3:Iterative Binary Search Algorithm, BinarySearch(A,ek)
12.1.3. Analysis
When algorithms are implemented and run on a computer, they require a certain amount
of resources. In general, we could consider a lot of different resources such as computation
time and memory. Algorithm analysis involves quantifying how many resource(s) an
algorithm requires to execute with respect to the size of the input it is run on.
When analyzing algorithms, we want to keep the analysis as abstract and general as
possible, independent of any particular language, framework or hardware. We could",ComputerScienceOne.pdf
"possible, independent of any particular language, framework or hardware. We could
always update the hardware on which we run our implementation, but that does not
necessarily make the algorithm faster. The algorithm would still execute the same number
of operations. Faster machines just mean that more steps can be performed in less time.
In fact, the concept of an algorithm itself is a mathematical concept that predates modern
computers by thousands of years. One of the oldest algorithms, for example, Euler’s
GCD (greatest common divisor) algorithm dates to 300 BCE. Whether or not you’re
“running” it on a piece of papyrus 2,300 years ago or on a modern supercomputer, the
same number of divisions and subtractions are performed.
To keep things abstract, we analyze an algorithm by identifying an elementary operation.
215",ComputerScienceOne.pdf
"12. Searching & Sorting
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(a) Initially, l= 0,r = 10 and so we examine the middle element at index m= 5
which is 12.
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(b) Since 64 >12, we update our left index variable l to m+ 1, thus l= 6 and
we’ve eliminated the left half of the list from consideration.
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(c) Our new middle index is m= l+r
2 = 6+10
2 = 8, corresponding to the element
102.
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(d) Since 64 <102, we update the right index variable r to m−1 = 7, eliminating
the right half of the subarray.
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(e) Here, l = 6,r = 7, and so our new middle index is m = ⌊6+7
2 ⌋= 6. Since
64 >34, we update our left index variable l to m+ 1 = 7
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9",ComputerScienceOne.pdf
"2 ⌋= 6. Since
64 >34, we update our left index variable l to m+ 1 = 7
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(f) Since 64 >42 we again update our left index variable l to m+ 1 = 8.
index
contents -3
0
2
1
4
2
4
3
9
4
12
5
34
6
42
7
102
8
157
9
180
10
(g) Since l = 8 and r = 7, l > rand the loop terminates, resulting in an
unsuccessful search.
Figure 12.3.: The worst case scenario for binary search, resulting in an unsuccessful
search for ek = 64. This example is run on a 0-indexed array with an array
of integers of size 11.
216",ComputerScienceOne.pdf
"12.1. Searching
This is generally the most common or most “expensive” operation that the algorithm
performs. Sometimes there may be more than one reasonable choice for an elementary
operation which may give different results in our analysis. However, we generally do not
consider basic operations that are necessary to the control flow of an algorithm. For
example, variable assignments or the iteration of index variables.
Once we have identified an elementary operation, we can quantify the complexity of
an algorithm by analyzing the number of times the elementary operation is executed
with respect to the input size. For a collection, the input size is generally the number
of elements in the collection, n. We can then characterize the number of elementary
operations and thus the complexity of the algorithm itself as a function of the input size.
We illustrate this process by analyzing and comparing the two search algorithms.
Linear Search Analysis",ComputerScienceOne.pdf
"We illustrate this process by analyzing and comparing the two search algorithms.
Linear Search Analysis
When considering the linear search algorithm, the input size is clearly the number of
elements in the collection, n. The best candidate for the elementary operation is the
comparison (Line 2, Algorithm 12.1). To analyze this algorithm, we need to determine
how many comparisons are made with respect to the size of the collection, n.
As we saw in the examples, the number of comparisons made by linear search can vary
depending on the element we’re searching and the configuration of the collection being
searched. Because of this variability, we can analyze the algorithm in one of three ways:
by looking at the best case scenario, worst case scenario, and average case scenario.
The best case scenario is when the number of operations is minimized. For linear search,
the best case scenario happens when we get lucky and the first element that we examine",ComputerScienceOne.pdf
"the best case scenario happens when we get lucky and the first element that we examine
matches our criteria, requiring only a single comparison operation. In general, it is not
reasonable to assume that the best case scenario will be commonly encountered.
The worst case scenario is when the number of operations is maximized. This happens
when we get “unlucky” and have to search the entire collection finding a match at the
last element or not finding a match at all. In either case, we make n comparisons to
search the collection.
A formal average case analysis is not difficult, but is a bit beyond the scope of the
present analysis. However, informally, we could expect to make about n
2 comparisons for
successful searches if we assume that all elements have a uniform probability of being
searched for.
Both the worst-case and average-case are reasonable scenarios from which to analyze
the linear search algorithm. In the end, however, the only difference between the two",ComputerScienceOne.pdf
"the linear search algorithm. In the end, however, the only difference between the two
analyses is a constant factor. Both analyses result in two linear functions,
f1(n) = n f 2(n) = 1
2n
217",ComputerScienceOne.pdf
"12. Searching & Sorting
The only difference being the constant factor 1
2 . In fact, this is why the algorithm is called
linear search. The number of comparison operations performed by the algorithm grows
linearly with respect to the input size. For example, if we were to double the input size
from n to 2n, then we would expect the number of comparisons to search the collection
would also double (this applies in either the worst-case and average-case scenarios).
This is what is most important in algorithm analysis: quantifying the complexity of
an algorithm by the rate of growth of the operations (and thus resources) required to
execute the algorithm as the input size n grows.2
Binary Search Analysis
Like linear search, the input size is the size of the collection n and the elementary
operation is the comparison. As presented in the pseudocode, binary search would seem
to perform two comparisons (Lines 5 and 7 in Algorithm 12.3). However, to make the",ComputerScienceOne.pdf
"to perform two comparisons (Lines 5 and 7 in Algorithm 12.3). However, to make the
analysis simpler, we will instead count one comparison operation per iteration of the while
loop. This is a reasonable simplification; in practice the comparison operation would
likely involve a single function call, after which distinguishing between the three cases is
a simple matter of control flow. Further, even if we were to consider both comparisons,
it would only contribute a constant factor of 2 to the final analysis.
Since we perform one comparison for each iteration of the while loop, we need to determine
how many times the while loop executes. In the worst case, the number of iterations is
maximized when we fail to find an element that does not match our criteria. However
each iteration essentially cuts the array in half each time. That is, if we start with an
array of size n, then after the first iteration, it is of size n
2 . After the second iteration we",ComputerScienceOne.pdf
"array of size n, then after the first iteration, it is of size n
2 . After the second iteration we
have cut it in half again, so it is of size n
4 , after the third iteration it is of size n
8 and so
on. More generally, after k iterations, the size of the array is
n
2k
The loop terminates one iteration after the the index variables are equal, l= r. Equiv-
alently, when l = r, the size of the subarray under consideration is 1. That is, the
algorithm stops when the array size has been cut down to 1:
n
2k = 1
Solving for k gives us the number of iterations:
k= log2 (n)
Adding one additional comparison for the final iteration gives us a total of
k= log2 (n)
2This is the basis of Big-O analysis, something that we will not discuss in detail here, but is of prime
importance when analyzing algorithms.
218",ComputerScienceOne.pdf
"12.1. Searching
comparisons (see Table 12.1). Thus, binary search performs a logarithmic number of
comparisons in the worst case. As we will see, this is exponentially better than linear
search.
Iteration Array Size Comparisons
1 n
2 1
2 n
4 1
3 n
8 1
4 n
16 1
... ... ...
k n
2k 1
... ... ...
log2 (n) 1 1
log2 (n) + 1 0 1
Total log2 (n) + 1
Table 12.1.: Number of comparisons and array size after the iteration during the execution
of binary search.
Comparative Analysis
Binary search presents a clear advantage over linear search. There is an exponential
difference between a linear function, n
2 and a logarithmic function, log2 (n). To put this
in perspective, consider searching a moderately large database of 1 trillion (1012) records.3
Using linear search, even the average-case scenario would require about
1012
2 = 5 ×1011
or about 500 billion comparisons. However, using binary search would only require at
most
log2 (1012) = 12 ·log2 (10) <40",ComputerScienceOne.pdf
"1012
2 = 5 ×1011
or about 500 billion comparisons. However, using binary search would only require at
most
log2 (1012) = 12 ·log2 (10) <40
comparisons to search. This is a huge difference in performance.
As another comparison, let’s consider how each algorithm’s complexity grows as we
increase the size of the collection being searched. As observed earlier, if we double the
3In the era of “big data,” 1 trillion records only qualifies as moderately large.
219",ComputerScienceOne.pdf
"12. Searching & Sorting
input size, n →2n, we would expect the number of comparisons performed by linear
search to also double. However, if we double the input size for binary search, we get the
following.
log2 (2n) = log2 (2) + log (n) = log (n) + 1
That is, only a single additional comparison is necessary to search an array of twice the
size.
The difference between these two algorithms shows up in many different instances. Figure
12.4 contains a screen shot of the search feature in Windows 7. When searching for
particular files or content in particular files, the search can be greatly increased if the
files have been indexed, that is, sorted. As the dialog indicates, non-indexed (unsorted)
records will take far longer to search.
Figure 12.4.: Example of the benefit of ordered (indexed) elements in Windows 7
Though binary search presents a clear advantage over linear search, it only works if the
collection has been sorted. 4 Thus, we now turn our attention to the problem of sorting a",ComputerScienceOne.pdf
"collection has been sorted. 4 Thus, we now turn our attention to the problem of sorting a
collection.
12.2. Sorting
Sorting a collection of data is another fundamental data operation. It is conceptually
simple, but is ubiquitous. There are a large variety of algorithms, data structures and
applications built around the problem of sorting. As we’ve already seen, being able
to sort a collection provides a huge speed up when searching for a particular element.
Sorting provides a natural way to store and organize data.
Problem 2(Sorting).
Given: a collection of orderable elements, A= {a1,a2,...,a n}
Output: A, sorted in ascending order
The requirement that the collection be made of “orderable” elements can be a bit
technical5, but essentially we need to be guaranteed that given two elements, a,b in the
4Binary search also only works when searching an array with random access to its elements. The",ComputerScienceOne.pdf
"4Binary search also only works when searching an array with random access to its elements. The
performance of binary search cannot generally be realized with data structures such as linked lists or
unordered sets.
5We require thatAbe a total ordera partially ordered binary relation such that all pairs are comparable.
220",ComputerScienceOne.pdf
"12.2. Sorting
collection, we can determine whether a < b, a = b or a > b. If such a determination
cannot be made, then sorting is impossible.
Again, we can consider variations on this problem. We may want our collection to be
sorted in descending order instead of ascending. 6 We may also want the collection itself
to be permuted (that is, reordered) or we may instead want a copy of the collection to be
created and sorted so that the original is unchanged.
We will examine several standard sorting algorithms. Though there are dozens of sorting
algorithms, we will only focus on a few of the more common ones. As with searching,
we can analyze a sorting algorithm based on the number of comparisons it makes in the
worst, best, or average cases. We may also look at alternative resources or operations:
how many swaps does the algorithm make? How much extra memory is required? Etc.
Though we will examine, analyze and compare several sorting algorithms, most pro-",ComputerScienceOne.pdf
"Though we will examine, analyze and compare several sorting algorithms, most pro-
gramming languages provide standard functionality to sort a collection of elements.
It is generally preferable to use the functionality built into whatever language you’re
using rather that reimplementing your own. Typically, these functions are well-designed,
well-tested, optimized and more efficient than any custom alternatives.
12.2.1. Selection Sort
The first sorting algorithm we’ll examine is Selection Sort which is similar to the way
a human would sort a collection of objects. The basic idea is that you search through
the collection and find the minimal element (we say minimal and not minimum because
there could be elements with the same value). Once we’ve found a minimal element, we
can swap it with the “first” element, a1, placing the minimal element where it belongs in
the collection.
We repeat this process for the remaining elements, a2,...,a n, finding the minimal element",ComputerScienceOne.pdf
"the collection.
We repeat this process for the remaining elements, a2,...,a n, finding the minimal element
among these and swapping with a2. In general, if we have already sorted the first i−1
elements, a1,...,a i−1 then we only need to examine the elements ai,...,a n for the
minimal element, swapping it with ai. We end this process after we have sorted the first
n−1 elements, a1,...,a n−1 since the last element an will already be where it needs to
be. We present Selection Sort as Algorithm 12.4. We illustrate the execution of Selection
6Technically, these are referred to asnon-decreasing and non-increasing respectively. This is because the
collection could contain duplicate elements and not lead to a strictly increasing or strictly decreasing
ordering.
221",ComputerScienceOne.pdf
"12. Searching & Sorting
Sort in Figure 12.5.
Input : A collection A= {a1,...,a n}
Output :An array A′ containing all elements of A in nondecreasing order
1 for i= 1,..., (n−1) do
2 amin ←ai
3 for j = (i+ 1),...,n do
4 if amin >aj then
5 min←aj
6 end
7 end
8 swap amin and ai
9 end
Algorithm 12.4:Selection Sort
Analysis
We now analyze Selection Sort to determine how complex it is. First, the elementary
operation is the comparison on line 4. We need to determine how many times this line is
executed with respect to the size of the input, n. Observe that on the i-th iteration, the
first i−1 elements are sorted and we need not make any comparisons among them. We
start by assuming that the ai is the minimum element and compare it to the remaining
n−i elements, requiring n−i comparisons, to find the minimal element. For example,
in the first iteration we make n−1 comparisons, the second we make n−2, etc. The
last iteration we make only 1 comparison. Totaling these all up gives us",ComputerScienceOne.pdf
"last iteration we make only 1 comparison. Totaling these all up gives us
(n−1) + (n−2) + (n−3) + ··· + 3 + 2 + 1
Which can be rewritten as
n−1∑
i=1
i= 1 + 2 + 3 +··· + (n−2) + (n−1)
We can use Gauss’s Formula to solve this summation.
Theorem 1(Gauss’s Formula). The sum of integers 1 up to ncan be written as follows.
n∑
i=1
i= 1 + 2 + 3 +··· + (n−1) + n= n(n+ 1)
2
In Selection Sort, the number of comparisons doesn’t sum up ton, only n−1. Substituting
this in gives us
222",ComputerScienceOne.pdf
"12.2. Sorting
42 4 9 4 10234 12 2 0
swap
0 4 9 4 10234 12 2 42
(a) First iteration. We find the minimal element, 0, at the last index,
swapping it with the first element. At this point, the first element is sorted.
0 4 9 4 10234 12 2 42
swap
0 2 9 4 10234 12 4 42
(b) Second Iteration. Now starting with the second element, the minimal
element among the remaining is found at the second to last element. 4 and
2 are swapped. At this point, the first two elements are sorted.
0 2 9 4 10234 12 4 42
swap
0 2 4 9 10234 12 4 42
(c) Third iteration. Since we are using the strictly-less than comparison, the
first 4 is the minimal element and swapped with 9.
0 2 4 9 10234 12 4 42
swap
0 2 4 4 10234 12 9 42
(d) Fourth iteration. At this point, the first 3 elements are sorted. We find
the minimal element (the other 4) and swap it with 9. At the end of this
iteration, the first 4 elements are sorted.
0 2 4 4 10234 12 9 42
swap
0 2 4 4 9 34 12 10242",ComputerScienceOne.pdf
"iteration, the first 4 elements are sorted.
0 2 4 4 10234 12 9 42
swap
0 2 4 4 9 34 12 10242
(e) Fifth iteration. The 9 is swapped with 102, sorting the first 5 elements.
0 2 4 4 9 34 12 10242
swap
0 2 4 4 9 12 34 10242
(f) Sixth iteration. 12 is swapped with 34.
0 2 4 4 9 12 34 10242
swap
0 2 4 4 9 12 34 10242
(g) Seventh iteration. 34 ends up being the minimal element and we essen-
tially swap it with itself. Even though the “current” element was also the
minimal element, we still had to compare the current element with all other
elements; in this case we made 2 comparisons.
0 2 4 4 9 12 34 10242
swap
0 2 4 4 9 12 34 42 102
(h) Eighth iteration. This is the final iteration, we swap 42 and 102. After
this iteration, the final element, 102 is already where it belongs.
Figure 12.5.: Example execution of Selection Sort.
223",ComputerScienceOne.pdf
"12. Searching & Sorting
n−1∑
i=1
i= n(n−1)
2
Another way to analyze the code is to count the number of comparisons with respect
to the for loop index variables. In particular, there is one comparison made on line 4.
Line 4 itself is executed once for each time the inner for loop on line 3 executes which
executes for j running from i+ 1 up to n. Line 3 and the entire inner for loop executes
once for each time the outer for loop executes, that is for i running from 1 up to n−1.
This gives us the following summation.
n−1∑
i=1
n∑
j=i+1
1

line 4  
line 3  
line 1
Solving this summation gives us:
n−1∑
i=1
n∑
j=i+1
1 =
n−1∑
i=1
n−i
=
n−1∑
i=1
n−
n−1∑
i=1
i
= n(n−1) −n(n−1)
2
= n(n−1)
2
This illustrates that Selection Sort is a quadratic sorting algorithm, requiring roughly n2
comparisons to sort an array of n elements. We analyze this further below.
12.2.2. Insertion Sort
Another basic sorting algorithm is Insertion Sort which works slightly differently than",ComputerScienceOne.pdf
"12.2.2. Insertion Sort
Another basic sorting algorithm is Insertion Sort which works slightly differently than
Selection Sort. Rather than finding a minimal element and swapping with the “current”
element, the current element is inserted in the list where it belongs.
If we consider just the first element, the collection is sorted by definition. Now consider
the second element: either its greater than the first element and so is already sorted,
or it is less than the first element and needs to be swapped. In either case, the first
224",ComputerScienceOne.pdf
"12.2. Sorting
two elements are now sorted. If we continue this, then on the i-th iteration, the first i
elements, a1,...,a i are sorted (just as with Selection Sort). Now consider the ( i+ 1)-th
element: we will insert it amongst the elements a1,...,a i where it needs to be.
We insert the “current” element, ai+1 by comparing it to ai: if ai >ai+1, then we swap
them. We keep comparing the current element to the one immediately to its left, shifting
the current element down the collection until we find an element that is less than or
equal to the current element at which point we stop. Now the elements a1,...,a i+1 are
now sorted. In contrast to Selection Sort, we do need to process the last element as it
might not be where it needs to be. The full pseudocode is presented in Algorithm 12.5.
An example of an execution of Insertion Sort can be found in Figure12.6.
Input : A collection A= {a1,...,a n}
Output :An array A′ containing all elements of A in nondecreasing order
1 for i= 2,...,n do",ComputerScienceOne.pdf
"Input : A collection A= {a1,...,a n}
Output :An array A′ containing all elements of A in nondecreasing order
1 for i= 2,...,n do
2 x←ai
3 j ←i
4 while j >1 and aj−1 >x do
5 aj ←aj−1
6 decrement j
7 end
8 aj ←x
9 end
Algorithm 12.5:Insertion Sort
Analysis
As we can see in the example run of Insertion Sort, not every iteration needs to make
comparisons with all elements in the sorted part of the collection. For example, in
iteration 4 we only had to make one comparison and we were done. In other iterations
such as the last two, we had to make comparisons to every element in the sorted part of
the collection. This illustrates that Insertion Sort is adaptive and may have a different
complexity depending on the structure of the collection it sorts.
Let’s consider the best case in which the number of comparisons is minimized. Suppose,
for example, we ran Insertion Sort on a collection that was already sorted. Each iteration",ComputerScienceOne.pdf
"for example, we ran Insertion Sort on a collection that was already sorted. Each iteration
would only need one comparison to determine that the element was already where it
needed to be. As there are only n−1 iterations, in the best case, Insertion Sort makes
n−1 comparisons.
In contrast, the worst case would occur if the list was already sorted, but in reverse order.
225",ComputerScienceOne.pdf
"12. Searching & Sorting
42 4 9 4 10234 12 2 0 4 42 9 4 10234 12 2 0
(a) First iteration. We insert 4 in front of 42, requiring 1 comparison.
4 42 9 4 10234 12 2 0
2
1
4 9 42 4 10234 12 2 0
(b) Second iteration. The first two elements are sorted, we insert 9 by making
two comparisons: to find that it is less than 42, but greater than 4. At the
end of the iteration, the first three elements are sorted.
4 9 42 4 10234 12 2 0
3 2 1
4 4 9 42 10234 12 2 0
(c) Third iteration. The first three elements are sorted, we insert the second
four by making 3 comparisons.
4 4 9 42 10234 12 2 0
1
4 4 9 42 10234 12 2 0
(d) Fourth iteration. Here, only one comparison is necessary to find that 102
is already sorted.
4 4 9 42 10234 12 2 0
123
4 4 9 34 42 10212 2 0
(e) Fifth iteration. Here, 3 comparisons are necessary to insert 34 between 9
and 42.
4 4 9 34 42 10212 2 0
1234
4 4 9 12 34 42 102 2 0
(f) Sixth iteration. Here, 4 comparisons are necessary to insert 12 between 9
and 34.
4 4 9 12 34 42 102 2 0",ComputerScienceOne.pdf
"1234
4 4 9 12 34 42 102 2 0
(f) Sixth iteration. Here, 4 comparisons are necessary to insert 12 between 9
and 34.
4 4 9 12 34 42 102 2 0
1234567
2 4 4 9 12 34 42 102 0
(g) Seventh iteration. Here, the number of comparisons is maximized as we
shift 2 all the way to the front of the array.
Figure 12.6.: Example execution of Insertion Sort. Each iteration depicts the comparisons
to previous elements; the last comparison is dashed indicating a comparison
was made, but not a swap. The final iteration is omitted for space, but
would require 8 comparisons to insert 0 at the front of the collection.
226",ComputerScienceOne.pdf
"12.2. Sorting
The i-th iteration would require i comparisons to move the current element all the way
to the front of the collection. Again, this gives us a summation:
n−1∑
i=1
i= n(n−1)
2
matching the complexity of Selection Sort.
We could also analyze Insertion Sort with respect to average case. On average, we would
expect to make about 1
2 of the total possible comparisons on each iteration. That is,
n−1∑
i=1
i
2
Moving the constant outside the summation and applying Gauss’s Formula would give
us an expected number of comparisons to be
n(n−1)
4
This is still a quadratic function, but the constant involved and the fact that Insertion
Sort is more adaptive to the input make Insertion Sort a much better algorithm in
practice than Selection Sort. In fact, Insertion Sort is very efficient on “small” arrays in
practice and is used in many hybrid algorithm implementations (see Section 12.2.5).
12.2.3. Quick Sort",ComputerScienceOne.pdf
"practice and is used in many hybrid algorithm implementations (see Section 12.2.5).
12.2.3. Quick Sort
Selection Sort and Insertion Sort are simple algorithms, but inefficient, making a quadratic
number of comparisons in even the average case. A more sophisticated and efficient
algorithm is Quick Sort, first developed by Tony Hoare in the early 1960s [ 17, 18]. Quick
Sort is a divide-and-conquer style algorithm that attempts to solve the sorting problem
by splitting a collection up into two parts, then recursively sorting each part.
The basic idea is as follows. First, choose a pivot element p in the collection. We will
then partition all the elements in the collection around this pivot element by placing all
elements less than p in the “left” partition and all elements greater than p in the “right”
partition. This concept is similar to binary search. After partitioning all the elements,
we place p between them so that p is where it needs to be. We then repeat this process",ComputerScienceOne.pdf
"we place p between them so that p is where it needs to be. We then repeat this process
on the two partitions recursively. The recursion stops when the size of the sub-collection
is trivially sorted (it is empty or consists of a single element).
Quick Sort is presented as Algorithm 12.6 with a partitioning subroutine presented in
Algorithm 12.7. The partitioning chooses the first element in the sub-collection as the
pivot element. Further, the partitioning is done in-place: we maintain two index variables,
227",ComputerScienceOne.pdf
"12. Searching & Sorting
i,j and increment/decrement them respectively to find a pair that are both in the wrong
partitions, swapping them. The partitioning continues until the two index variables meet
each other. As a final step, the pivot element is placed between the two partitions and
the index at which it is placed is returned to the main QuickSort routine so that it
can make two recursive calls on the two partitions.
We depict a few examples of the partitioning algorithm. Figure 12.7 depicts the first
partitioning on the same example array while Figures 12.8 and 12.9 depict subsequent
228",ComputerScienceOne.pdf
"12.2. Sorting
partitioning operations on subarrays as part of the recursion.
Input : A collection A= {a1,...,a n}, indices l,r
Output :A, sorted in ascending order
1 if l<r then
2 p←Partition(A,l,r )
3 QuickSort(A,l,p −1)
4 QuickSort(A,p + 1,r)
5 end
Algorithm 12.6:QuickSort
Input : A collection A= {a1,...,a n}, indices l,r
Output :An index s such that the sub-collection A from l to r has been
partitioned around s so that all elements al,...a s−1 are less than as
and all elements as+1,...,a r are greater than as
1 pivot←al
2 i←(l+ 1)
3 j ←r
4 while i<j do
5 while i<j And ai ≤pivot do
6 i←(i+ 1)
7 end
8 while i<j And aj ≥pivot do
9 j ←(j−1)
10 end
11 swap ai, aj
12 end
//Swap the pivot
13 if ai ≤pivot then
14 swap pivot, ai
15 output i
16 else
17 swap pivot, ai−1
18 output (i−1)
19 end
Algorithm 12.7:In-Place Partition
229",ComputerScienceOne.pdf
"12. Searching & Sorting
42 4 9 4 10234 12 2 0
ji
swap
42 4 9 4 0 34 12 2 102
i j
(a) First iteration. 42 is chosen as the pivot element. The index variable
i moves over to 102, the first element that is greater than 42 and on the
“wrong” side of the partition. The j index variable does not move as 0 is less
than the pivot element and on the wrong side; these are swapped.
42 4 9 4 0 34 12 2 102
j
i
2 4 9 4 0 34 12 42 102
s
(b) Second iteration. The index variable i moves all the way to the right
as all remaining elements are less than the pivot, 42. The index variable j
remains at 102 as i is now equal to j. The pivot, 42 is placed between the
two partitions and its resulting index, s is returned.
Figure 12.7.: Example execution of the Partition subroutine in Quick Sort. A total
of 8 comparisons are made, 9 if you count the last swap outside the while
loop. The partition returns s as the pivot position; the right-partition is
sorted as it only consists of 1 element.
2 4 9 4 0 34 12
ji",ComputerScienceOne.pdf
"loop. The partition returns s as the pivot position; the right-partition is
sorted as it only consists of 1 element.
2 4 9 4 0 34 12
ji
swap
2 0 9 4 4 34 12
i j
(a) First iteration. In this partitioning, 2 is the pivot element.
The index variable i is not incremented as 4 is larger than
the pivot. The index variable j decrements to 0 and they are
swapped.
2 0 9 4 4 34 12
i,j
0 2 9 4 4 34 12
s
(b) Second iteration. The index variable i is incremented
once, while j is decremented to meet it. After the while loop
the second case applies as ai = 9 > 2 = pivot, and so we
swap ai−1 = 2 with the pivot.
Figure 12.8.: Example execution of the Partition subroutine in Quick Sort on the first
recursive call on the left partition. A total of 6 comparisons are made, 7 if
you count the last swap outside the while loop. The partition returns s as
the pivot position; in this case the left-partition is sorted as it only consists
of 1 element.
230",ComputerScienceOne.pdf
"12.2. Sorting
9 4 4 34 12
i,j
4 4 9 34 12
s
(a) First (and only) iteration. In this parti-
tioning, 9 is the pivot. The index variable i
is incremented to 34 while j decrements to
match. 34 is swapped with itself. After this
iteration, the second condition applies and
the pivot is swapped with ai−1 = 4
Figure 12.9.: Example execution of the Partition subroutine in Quick Sort on the
second recursive call on the right partition. A total of 5 comparisons are
made, 6 if you count the last swap outside the while loop. The partition
returns s as the pivot position.
Analysis
We can easily analyze how many comparisons are made by the Partition subroutine.
Suppose that we are given a (sub)array of n elements. Since the pivot element must be
compared to every other element in the (sub)array, it must make n−1 comparisons to
partition the elements around the pivot. If we count the last comparison to determine
where to place the pivot, it would be n comparisons (but the difference is trivial).",ComputerScienceOne.pdf
"where to place the pivot, it would be n comparisons (but the difference is trivial).
The analysis of Quick Sort itself is a bit more involved and is highly dependent on how
well the Partition subroutine splits the array. In the worst case, our pivot choice
will always partition the array into one “empty” subarray and one subarray with n−1
elements (we do not count the pivot element). One such example of this would be if our
collection is already sorted. In this case, one of the recursive calls would result in no
comparisons and the other would result in n−1 comparisons. This lopsided recursion
would result in the following number of comparisons:
n+ (n−1) + (n−2) + ··· + 3 + 2 + 1
which is exactly Gauss’s Formula, meaning that Quick Sort would perform
n(n+ 1)
2
in the worst case.
However, in the sorting scenario, we cannot always assume the worst case. If we are given
random data, then the likelihood that we will always have such an extremely lopsided",ComputerScienceOne.pdf
"random data, then the likelihood that we will always have such an extremely lopsided
partitioning is extremely small. A more reasonable analysis would involve the average
case: where each partition is roughly an equal size, about n
2 . This happens when our
pivot choice is the median (or close to the median) element.
231",ComputerScienceOne.pdf
"12. Searching & Sorting
To analyze the number of comparisons in this case, we can setup a recurrence relation:
C(n) = 2C
(n
2
)
+ n
Here, C(n) represents the number of comparisons made by Quick Sort on an array of size
n. The first term on the right hand side represents the fact that we make 2 recursive calls
on subarrays of size (roughly) n
2 . The second term captures the number of comparisons
we make in the Partition subroutine. Solving this recurrence is a bit beyond the scope
of the current discussion. However, it can be shown that this is roughly equivalent to
nlog (n)
comparisons in the best case.
The difference between the best and worst case scenarios is in our pivot choice. Though
the ideal pivot choice is the median element, to find the median the usual strategy is to
sort the collection first. This puts the cart before the horse. If the data we are sorting
is in more-or-less random order, then the choice of the first element as a pivot is good",ComputerScienceOne.pdf
"is in more-or-less random order, then the choice of the first element as a pivot is good
enough as it is unlikely that we will always have a bad paritioning. However, there are
other strategies.
One strategy is to randomly choose a pivot element. A random choice would make the
worst-case scenario very unlikely over many runs of the algorithm. Another strategy
is to choose the median of three elements (either fixed or randomized), ensuring that
neither partition will ever be empty. This strategy can be taken further to find the
median-of-three amongst each third of the array and then take the median of these (or the
“ninther”). The more effort you put into finding the median the more the performance
degrades in practice.7 A more sophisticated analysis of these strategies yields a complexity
similar to the best case [9].
12.2.4. Merge Sort
Another “fast” sorting algorithm is Merge Sort, due to John von Neumann, 1945 (as",ComputerScienceOne.pdf
"similar to the best case [9].
12.2.4. Merge Sort
Another “fast” sorting algorithm is Merge Sort, due to John von Neumann, 1945 (as
reported by Knuth [27]). The performance is similar to Quick Sort’s best/average case,
making
nlog (n)
comparisons. However, Merge Sort’s performance does not depend on the structure of the
input array or a pivot choice, guaranteeing this performance in the best/average/worst
case.8
7For example, there does exist a median-finding algorithm that runs in linear time which would
guarantee the theoretically best running time, but the algorithm is recursive and has a large overhead,
making its use slow in practice.
8Still, Quick Sort is still more common in practice because it has less memory requirements and the
pivot choice strategies can mitigate the risk of the worst-case scenario.
232",ComputerScienceOne.pdf
"12.2. Sorting
Merge Sort works by first dividing the list into two (roughly) equal partitions. It then
recursively sorts each partition. The recursion stops when the subarray is of size ≤1 just
as with Quick Sort. The difference, however, is what Merge Sort does after the recursion.
After having sorted the left partition, L and the right partition R, Merge Sort merges
the two sorted partitions into one.
The Merge subroutine works by maintaining two index variables, i,j, one for each
partition. Suppose that the index variables correspond to the elements Li and Rj in
the left/right partition. If Li ≤Rj, we add Li to the end of a temporary array and
increment i. Otherwise we place Rj into the temporary array and increment j. We
continue until we have examined every element in one or both partitions (if one partition
still has elements in it, we can simply copy the rest over in order). This merge operation
works because each subarray is sorted.",ComputerScienceOne.pdf
"still has elements in it, we can simply copy the rest over in order). This merge operation
works because each subarray is sorted.
Merge Sort is presented as Algorithm 12.8 with the Merge subroutine as Algorithm
12.9.
Input : A (sub)collection A= {al,...,a r}, indices l,r
Output :A collection A′ which is A sorted
1 if l<r then
2 m←⌊l+r
2 ⌋
3 L←MergeSort(A,l,m )
4 R←MergeSort(A,m + 1,r)
5 output Merge(L,R)
6 else
7 output A
8 end
Algorithm 12.8:MergeSort
233",ComputerScienceOne.pdf
"12. Searching & Sorting
Input : Two sorted collections, L, R of size n,m respectively.
Output :A sorted collection A consisting of all elements of L and R
1 A←a new, empty collection
2 i←1
3 j ←1
4 k←1 //index variable for A
5
6 while i≤nAnd j ≤m do
7 if Li ≤Ri then
8 Ak ←Li
9 i←(i+ 1)
10 else
11 Ak ←Lj
12 j ←(j+ 1)
13 end
14 k←(k+ 1)
15 end
//At least one collection is empty, we can blindly copy the other
16 while i≤n do
17 Ak ←Li
18 i←(i+ 1)
19 k←(k+ 1)
20 end
21 while j ≤m do
22 Ak ←Rj
23 j ←(j+ 1)
24 k←(k+ 1)
25 end
26 output A
Algorithm 12.9:Merge
We present an example run of Merge Sort in Figure 12.10 (here, we have made the
collection of size 8 to emphasize the even split). An example run of the Merge
subroutine on the last merge operation of this algorithm is presented in Figure 12.11.
234",ComputerScienceOne.pdf
"12.2. Sorting
42 4 9 4 102 34 12 2
42 4 9 4 102 34 12 2
42 4 9 4 102 34 12 2
42 4 9 4 102 34 12 2
4 42 4 9 34 102 2 12
4 4 9 42 2 12 34 102
2 4 4 9 12 34 42 102
merge merge merge merge
merge merge
merge
Figure 12.10.: Illustration of Merge Sort’s recursion and merge operations.
235",ComputerScienceOne.pdf
"12. Searching & Sorting
4 4 9 42 2 12 34 102
i j
2
k
(a) Iteration One. L1 = 4 > R1 = 2, so
the element in the right partition is copied
into A1. Both j,k are incremented.
4 4 9 42 2 12 34 102
i j
2 4
k
(b) Iteration Two. Now the element in
the left partition is lesser and is copied,
incrementing i,k
4 4 9 42 2 12 34 102
i j
2 4 4
k
(c) Iteration Three. Again the element in
the left partition is less.
4 4 9 42 2 12 34 102
i j
2 4 4 9
k
(d) Iteration Four. And so on.
4 4 9 42 2 12 34 102
i j
2 4 4 9 12
k
(e) Iteration Five
4 4 9 42 2 12 34 102
i j
2 4 4 9 12 34
k
(f) Iteration Six
4 4 9 42 2 12 34 102
i j
2 4 4 9 12 34 42
k
(g) Iteration Seven
4 4 9 42 2 12 34 102
j
2 4 4 9 12 34 42 102
k
(h) Final Copy. After one sub-collection
has been exhausted the rest (in this case
only one) of the elements in the other are
blindly copied over.
Figure 12.11.: Demonstration of the merge operation in Merge Sort. Here we depict the
final Merge subroutine invocation from the previous example.
236",ComputerScienceOne.pdf
"12.2. Sorting
Analysis
Because Merge Sort divides the list first, an even split is guaranteed. After the recursion,
the Merge subroutine requires at most n−1 comparisons to merge the two collections.
This leads to a recurrence relation similar to Quick Sort,
C(n) = 2C
(n
2
)
+ (n−1)
A similar analysis yields a complexity of nlog (n).
12.2.5. Other Sorts
There are dozens of other sorting algorithms and many more variations on sorting
algorithms, each with some unique properties and advantages. For example, Heap Sort,
due to J. W. J. Williams, 1964 [ 38], uses a data structure called a heap. A heap stores
elements in a tree structure and offers efficient insertion of elements and retrieval of the
minimal (or maximal for a “max-heap”) element. The sorting algorithm then simply
inserts each element into the heap, and retrieves each element out of the heap. Because
of the the heap data structure, elements come out of the heap in order.",ComputerScienceOne.pdf
"of the the heap data structure, elements come out of the heap in order.
Other sorting algorithms are variations on those we’ve already seen or hybrid sorting
algorithms. For example, we’ve already noted that Insertion Sort is efficient on “small”
arrays. One common technique is to use a fast sorting algorithm such as Quick Sort or
Merge Sort, but then switch over to Insertion Sort at some point in the recursion rather
than recursing to the point that the collection is of size 1 (or 0). This switch point is
usually tuned using empirical experiments.
A more recent sorting algorithm is Tim Sort, due to Tim Peters in 2002 [ 34] who
originally developed it as the default sorting algorithm for the Python language. It has
since been adopted in Java (version 7 for arrays of non-primitive types). Tim Sort is a
sort of “bottom-up” Merge Sort/Insertion Sort hybrid. Instead of recursing, it looks for
subsequences of elements that are already in order or nearly in order, then merges them.",ComputerScienceOne.pdf
"subsequences of elements that are already in order or nearly in order, then merges them.
It was designed to have many of the desirable properties of a sorting algorithm including
stability (see Section 12.3.6).
12.2.6. Comparison & Summary
To illustrate just how much more efficient an nlog (n) sorting algorithm is over a naive
n2 algorithm consider the same scenarios as with searching. Suppose we want to sort a
huge collection of 1 trillion, 10 12, elements. Doing so with Selection Sort or Insertion
Sort would require about
n2 = (1012)2 = 1024
237",ComputerScienceOne.pdf
"12. Searching & Sorting
or 1 septillion comparisons. The same sorting operation using either Quick Sort or Merge
Sort would require only
nlog (n) = 1012 log 1012 ≈4 ×1013
or just under 40 trillion comparisons. This is 25 billion times fewer operations.
As another example, suppose that we sort a collection with n elements and then want to
double the size of our collection to 2 n. How do each of these types of algorithms perform?
For the slow, n2 algorithms we have
(2n)2 = 4n2
Doubling the input size quadruples the number of operations Selection Sort and Insertion
Sort perform. However, for Quick Sort and Merge Sort, we only have
2nlog (2n) = 2n(log (2) + log (n)) = 2nlog (n) + 2n
That is, only about twice as many operations (with an additive term of 2 n). Table 12.2
gives a summary of the complexity and other properties of the sorting algorithms we’ve
seen.
Algorithm Complexity Stable? NotesBest Average Worst
Selection Sort ∼n2 ∼n2 ∼n2 No",ComputerScienceOne.pdf
"seen.
Algorithm Complexity Stable? NotesBest Average Worst
Selection Sort ∼n2 ∼n2 ∼n2 No
Insertion Sort ∼n ∼n2 ∼n2 Yes Best in practice for
small collections
Quick Sort ∼nlog n ∼nlog n ∼n2 No Performance depends
on pivot choices
Merge Sort ∼nlog n ∼nlog n ∼nlog n Yes Requires extra space
Table 12.2.: Summary of Sorting Algorithms. See Section 12.3.6 for a discussion on
stability.
12.3. Searching & Sorting In Practice
12.3.1. Using Libraries and Comparators
In practice you do not write your own searching and sorting algorithm implementations.
Instead, you use the functionality provided by the language, a framework, or a library.
First, there is rarely a good reason to “reinvent the wheel” and write your own if the
functionality already exists. Second, the algorithms and code provided by a language
or library are typically optimized and have been well-tested. Established libraries have
238",ComputerScienceOne.pdf
"12.3. Searching & Sorting In Practice
been developed with thousands of man-hours and have proven themselves over millions
of computing hours.
Typically, searching and sorting functions in a language are made to be generic: they
don’t just search a collection of numbers or strings. Instead, they accept collections of
any type. The basic algorithms for searching and sorting are generic themselves (after all
we presented them as pseudocode). The only real difference between an implementation
for, say, numbers versus a sorting algorithm for strings is how pairs of elements are
compared and ordered.
This genericness extends to objects. It would be extremely inefficient, development-wise,
to have to rewrite Quick Sort for a collection of student objects to sort them by name,
then another to sort them by GPA, then another for ID, and repeat them all again for a
reversed ordering. Instead, a single, generic sorting algorithm is implemented that can
be configured by passing in a comparator.",ComputerScienceOne.pdf
"reversed ordering. Instead, a single, generic sorting algorithm is implemented that can
be configured by passing in a comparator.
A comparator is a function or object that provides functionality to determine how to
order two elements. In general, a comparator takes two arguments, a,b which are the
elements to be compared and returns an integer value with the following “contract.” It
returns:
•something negative, <0 if a comes before b, a<b
•zero if a and b are equivalent, a= b
•something positive, >0 if a comes after b, a>b
We’ve previously seen this type of functionality when comparing strings in Chapter 8.
Using comparators allows us to reuse the generic search and sorting algorithms. Now for
each sorting that we want, we only need to create a comparator to define the ordering
rather than reimplementing the whole algorithm every single time. This relationship is
depicted in Figure 12.12.
12.3.2. Preventing Arithmetic Errors",ComputerScienceOne.pdf
"rather than reimplementing the whole algorithm every single time. This relationship is
depicted in Figure 12.12.
12.3.2. Preventing Arithmetic Errors
Recall that in binary search, we need to compute the index of the middle element.9 Given
a left l and right r index variable, we computed the middle index as
m←
⌊l+ r
2
⌋
Though mathematically correct, when implementing an expression like this, care must
be taken. We may initially translate this pseudocode into a line that looks something
like the following.
9Some implementations of Quick Sort will do something similar when choosing the “middle” element
as a pivot.
239",ComputerScienceOne.pdf
"12. Searching & Sorting
Sorting
Algorithmarray
Input sorted arrayOutput
Comparator
a,b order =



<0 if a<b
0 if a= b
>0 if a>b
Figure 12.12.: Generalized Sorting with a Comparator. A sorting algorithm doesn’t
need to know what it is sorting or how they are ordered as long as it has
access to a comparator that does know how to order elements. By using a
comparator, the sorting function can be kept general and generic so that
one implementation can be used for any type of data.
int middle_index = (left + right) / 2;
However, this is prone to arithmetic errors in certain situations, particularly when dealing
with large arrays. If the variables left and right have a sum that exceeds the
maximum value that a signed 32-bit integer can hold (2 31 −1 = 2,147,483,647), then
overflow will occur before the division by 2, leading to a (potentially) negative index
value and thus an invalid out-of-bounds error or exception.",ComputerScienceOne.pdf
"overflow will occur before the division by 2, leading to a (potentially) negative index
value and thus an invalid out-of-bounds error or exception.
One solution would be to use a variable type that supports larger values. For example,
a 64-bit signed integer would be able to handle arrays of 9.223 quintillion elements.
However, depending on the language, you may not be able to use such variable types as
index values. Another solution is to use operations that do not introduce this potential
for overflow. For example,
l+ r
2 = l+ (r−l)
2
but the expression on the right hand side will not be prone to overflow. Thus the code,
int middle_index = left + (right - left) / 2;
would be safer to use. In fact, this bug is quite common [ 32] and was in the Java
implementation of binary search, unreported for nearly a decade [10].
240",ComputerScienceOne.pdf
"12.3. Searching & Sorting In Practice
12.3.3. Avoiding the Difference Trick
Another issue related to arithmetic overflow is a common “trick” used in comparators
when ordering integer values. Consider the following example: we want to order integer
values in ascending order, the basic logic would look something like the following.
1 if a<b then
2 output −1
3 else ifa= b then
4 output 0
5 else
6 output +1
7 end
For example, if a= 5 and b= 10 then our comparator would output −1. If the values
were a= 10,b = 5 it would output +1, and if they were equal, a= b= 10 then it would
output zero. A common “trick” is to instead compute the difference between these values,
a−b. Observe:
a b a −b
5 10 −5
10 5 5
5 5 0
The sign of the difference in each of these examples matches the logic of our if-else-if
statement. The lazy programmer may be tempted to write a one liner, “output ( a−b)”
instead of the if-else-if statement. However, this would fail for certain values of a and b.",ComputerScienceOne.pdf
"instead of the if-else-if statement. However, this would fail for certain values of a and b.
Suppose both of them are 32-bit 2s complement integers. And suppose that a= 231 −1,
the maximum representable value, and b= −10. In the logic above, b would come before
a and so we would expect a positive result. However, our arithmetic trick would give the
following result:
(231 −1) −(−10) = 231 + 9
which, mathematically, is positive, but exceeds the maximum representable value, leading
to overflow. In most systems, the result of this arithmetic is −2,147,483,639, a negative
value. There are many other input values that could cause arithmetic overflow. Only if
you are absolutely sure that no arithmetic overflow is possible should you even consider
using this “trick.”
Another issue with this trick is when comparing floating point values. GPAs for example:
suppose that a= 4.0 and b= 3.9. Their difference would be a−b= 0.1. However, a",ComputerScienceOne.pdf
"suppose that a= 4.0 and b= 3.9. Their difference would be a−b= 0.1. However, a
comparator returns an integer value. In some languages this result would be casted to
241",ComputerScienceOne.pdf
"12. Searching & Sorting
an integer, truncating the fractional value, so that 0 .1 →0, meaning that a GPA of 3.9
is “equivalent” to the 4.0. Given the potential for errors, it is best to avoid this trick
altogether.
12.3.4. Importance of a Total Order
In many applications it is important to design your comparator to provide a total order.
For example, suppose we have students “Gary Busey” with an ID of 1234 and another
student with the same name, “Gary Busey” but with an ID of 4321. It is entirely possible
(and common) for different people to have the same names. However, a comparator that
ordered student objects based only on the name fields would interpret these as the same
person.
The distinction is especially important when using a comparator in a sorted collection
(such as a Binary Search Tree, or Ordered Hash Map data structure) that maintains
an ordering on the elements. Many of these data structures do not allow “duplicate”",ComputerScienceOne.pdf
"an ordering on the elements. Many of these data structures do not allow “duplicate”
elements where duplication is determined by the comparator. Attempting to insert both
of these students will likely result in only one of them being in the collection as the
second will be interpreted as a a duplicate. In such scenarios it is important to design
your comparator to break ties by using some unique data field of an object such as an ID.
12.3.5. Artificial Ordering
Many languages recognize what is called a “natural ordering” on certain elements. For
example, numbers have a natural ordering: from least to greatest. Strings also have a
natural ordering: lexicographic ordering. However, we often wish to order elements in
an artificial ordering. For example, suppose we wish to include the year of a student,
Freshman, Sophomore, Junior, or Senior. If we modeled these as strings, the natural
ordering would be alphabetical:
Freshman,Junior,Senior,Sophomore",ComputerScienceOne.pdf
"Freshman, Sophomore, Junior, or Senior. If we modeled these as strings, the natural
ordering would be alphabetical:
Freshman,Junior,Senior,Sophomore
However, if we order students by year, we would want the artificial order of Freshman,
Sophomore, Junior, or Senior.
To do this, we might have a whole series of if-else-if statements to order these elements
242",ComputerScienceOne.pdf
"12.3. Searching & Sorting In Practice
Input : Two student objects, a,b
1 if a.year = b.year then
2 output 0
3 else ifa.year = “Freshman′′ then
4 output −1
5 else ifb.year = “Freshman′′ then
6 output 1
7 else ifa.year = “Sophomore′′ then
8 output −1
9 ...
However, this logic is complex and does not provide a good solution if we want to then
add support for “Pre-freshman” or “Graduate”, etc.
Another solution would be to model the year using a data type that has a natural
ordering. For example, we could define an enumerated type and associate each of the
years with 0, 1, 2, 3; giving them a natural ordering. Alternatively, we could use a data
structure, such as a map, to model the artificial ordering. For example, we could map
“Freshman” to 0, “Sophomore” to 1, etc. Then, when we wanted to order two elements,
we could look up the “natural value” via the map and use their natural ordering to order
our elements.
12.3.6. Sorting Stability",ComputerScienceOne.pdf
"we could look up the “natural value” via the map and use their natural ordering to order
our elements.
12.3.6. Sorting Stability
One desirable property of sorting algorithms is stability. A sorting algorithm is stable if
the relative order of “equal” elements is preserved. For example, suppose we have the
following integer values:
10,2a,5,2b
The subscripts on the two 2 values are for demonstration purposes only. A stable sorting
algorithm would sort the elements as:
2a,2b,5,10
preserving the relative ordering of the two 2 elements. The element 2 a came before 2b
prior to sorting and remains so after sorting. However, an unstable sorting algorithm
may instead produce
2b,2a,5,10
The collection is still sorted, but the two equal elements have been reversed from their
original ordering.
243",ComputerScienceOne.pdf
"12. Searching & Sorting
Sorting stability is often desirable for data presentation. A user can typically sort table
data by clicking on a column header. Suppose we sorted a table of students first by GPA
then by year. We would expect that all Freshman would be grouped together and within
that group, would be ordered by GPA (because the first ordering would be preserved).
An unstable sorting algorithm would result in all Freshman being grouped together, but
within that grouping, the students would not necessarily be ordered by GPA.
As indicated in Table 12.2, some algorithms (as presented) are stable and others are not.
For example, Selection Sort is not stable. Consider the following input:
2a,2b,1,5
In the first iteration, 1 and 2 a would be swapped, resulting in a sorted collection and no
further swaps, but with 2 b preceding 2a.
Insertion Sort, however, is stable: it only moves an element down the collection if it is",ComputerScienceOne.pdf
"further swaps, but with 2 b preceding 2a.
Insertion Sort, however, is stable: it only moves an element down the collection if it is
strictly less than its adjacent element. Likewise, Merge Sort is stable since we “prefer”
elements from the left partition when equal.
Quick Sort is not stable as the partitioning can easily place things out of their original
relative ordering. Usually, sorting algorithms can be made to be stable by doing some
extra work, but it may impact the performance.
12.4. Exercises
Exercise 12.1. Give an input example input demonstrating that the Quick Sort al-
gorithm, as presented, is unstable. Run through the algorithm to demonstrate how it
results in an unstable sort.
Exercise 12.2. Implement the sorting algorithms described in this chapter in the
programming language of your choice. Then setup experiments: run each on randomly
ordered collections many times to get an average run time. Graph these empirical results",ComputerScienceOne.pdf
"ordered collections many times to get an average run time. Graph these empirical results
and see if they match the theoretical results of our analysis. Be sure to also include any
built-in sorting functionality your language provides as a benchmark.
Exercise 12.3. Ternary search is similar to binary search in that it searches a sorted
array. However, instead of splitting the array in two, it splits it into three partitions
and makes two comparisons to determine which third the element lies in (if it is in
the collection). Implement ternary search and benchmark it against a binary search
implementation.
Exercise 12.4. Data has a “natural” ordering: numbers are ordered in nondecreasing
order, strings are ordered lexicographically (according to the ASCII text table). However,
in many situations, the natural ordering isn’t the expected ordering. For example, class
244",ComputerScienceOne.pdf
"12.4. Exercises
years are usually ordered Freshman, Sophomore, Junior, Senior whereas the natural
ordering would order them Freshman, Junior, Senior, Sophomore.
Write a program to sort a collection of strings according to an arbitrary artificial ordering.
That is, instead of the A–Z alphabetic ordering, we will order them according to an
alternative ordering of the English alphabet. As an example, one possible ordering would
be:
n e d c r h a l g k m z f w j o b v x q y i p u s t
Your job will be to sort a list of English language words using this artificial ordering of
the English alphabet. Specifically, you will read in an input file in the following format.
The first line is the new ordering of English letters, each separated by a space. Each line
after that contains a single string. You will read in this file, process it and produce a
reordered list of the words sorted according to the new ordering.
An example input:
n e d c r h a l g k m z f w j o b v x q y i p u s t
vcawufotrb",ComputerScienceOne.pdf
"reordered list of the words sorted according to the new ordering.
An example input:
n e d c r h a l g k m z f w j o b v x q y i p u s t
vcawufotrb
laencfuesw
gvtkwekfom
vrsfqictqc
wmcvmjmtet
qetegyqelu
newaxdtjlt
nfrfrwkknj
fzqrvgblov
gkkmgwwwpa
The result ordering:
Artificial Ordering:
newaxdtjlt
nfrfrwkknj
laencfuesw
gkkmgwwwpa
gvtkwekfom
fzqrvgblov
wmcvmjmtet
vcawufotrb
vrsfqictqc
245",ComputerScienceOne.pdf
"12. Searching & Sorting
qetegyqelu
For simplicity, you can assume that all words will be lower case and no non-alphabetic
characters are used. However, you may not assume that all words will be the same length.
Words of a shorter length that are a prefix of another word should be ordered first. For
example, “newax” should come before “newaxn” in the ordering above.
Exercise 12.5. Write comparators for all the member fields of the album object in
Exercise 10.2.
Exercise 12.6. Write comparators for all the member fields of the savings account
object in Exercise 10.3.
Exercise 12.7. Write comparators for all the member fields of the stadium object in
Exercise 10.4.
Exercise 12.8.Write comparators for all the member fields of the network device object
in Exercise 10.5.
Exercise 12.9. Write comparators for all the member fields of the airport object in
Exercise 10.6.
246",ComputerScienceOne.pdf
"13. Graphical User Interfaces & Event
Driven Programming
This chapter may appear in a future version.
247",ComputerScienceOne.pdf
"14. Introduction to Databases &
Database Connectivity
This chapter may appear in a future version.
249",ComputerScienceOne.pdf
"Part I.
The C Programming Language
251",ComputerScienceOne.pdf
"15. Basics
The C programming language is a relatively old language, but still widely used. It is
universal in that nearly every system, platform, and operating system has a C compiler
that produces machine code for that system. C is used extensively in systems programming
for operating system kernels, embedded systems, microcontrollers, and supercomputers. It
is generally an “imperative” language which is a paradigm that characterizes computation
in terms of executable statements and functions that change a program’s state (variable
values). C has been highly influential in the design and syntax of other languages
including C++, Objective-C, C#, Java, and PHP, among many others. Many languages
have adopted the basic syntactic elements and structured programming approach of the
C language.
C was originally developed by Dennis Ritchie while at AT&T Bell Labs 1969–1972. C
was born out of the need for a new language for the PDP-11 minicomputer that used",ComputerScienceOne.pdf
"was born out of the need for a new language for the PDP-11 minicomputer that used
the Unix operating system (written by Ken Thompson). From its inception, C has had
a close relation to Unix; in fact the operating system was subsequently rewritten in C,
making it the first OS to be written in a language other than assembly. The language
was dubbed “C” as its predecessor was named “B,” a simplified version of BCPL (Basic
Combined Programming Language). The first formal specification was published as The
C Programming Language by Kernighan and Ritchie (1978) [ 26] often referred to as
“The K&R Book” which would later become the American National Standards Institute
(ANSI) C standard.
C gained in popularity and directly influenced object-oriented variations of it. Bjarne
Stroustrup developed C++ while at Bell Labs from 1979–1983. Brad Cox and Tom
Love developed Objective-C from 1981–1983 at their company Stepstone. Subsequent",ComputerScienceOne.pdf
"Love developed Objective-C from 1981–1983 at their company Stepstone. Subsequent
standards of the C language have added and extended features. In 1990, the International
Organization for Standardization (ISO)/IEC 9899:1990 standard, referred to as C89
or C90, was adopted. About every 10 years since, a new standard has been adopted;
ISO/IEC 9899:1999 (referred to as C99) in 1999 and ISO/IEC 9899:2011 (C11) in 2011.
15.1. Getting Started: Hello World
The hallmark of an introduction to a new programming language is the Hello World!
program. It consists of a simple program whose only purpose is to print out the message
253",ComputerScienceOne.pdf
"15. Basics
1 #include<stdlib.h>
2 #include<stdio.h>
3
4 /**
5 * Basic Hello World program in C
6 * Prints ""Hello World"" to the standard output and exits
7 */
8 int main(int argc, char **argv) {
9
10 printf(""Hello World\n"");
11
12 return 0;
13 }
Code Sample 15.1.: Hello World Program in C
“Hello World!” to the user in some manner. The simplicity of the program allows the focus
to be on the basic syntax of the language. It is also typically used to ensure that your
development environment, compiler, runtime environment, etc. are functioning properly
with a Minimum Working Example (MWE). The Hello World! program is generally
attributed to Brian Kernighan who used it as an example of programming in C in 1974
[25]. A basic Hello World! program in C can be found in Code Sample 15.1.
We will not focus on any particular development environment, code editor, or any
particular operating system, compiler, or ancillary standards in our presentation. However,",ComputerScienceOne.pdf
"particular operating system, compiler, or ancillary standards in our presentation. However,
as a first step, you should be able to write, compile, and run the above program on
the environment you intend to use for the rest of this book. This may require that you
download and install a basic C compiler/development environment (such as GCC, the
GNU Compiler Collection on OSX/Unix/Linux, cygwin or MinGW for Windows) or a
full IDE (such as Xcode for OSX, or Code::Blocks, http://www.codeblocks.org/ for
Windows).
15.2. Basic Elements
Using the Hello World! program as a starting point, we will now examine the basic
elements of the C language.
254",ComputerScienceOne.pdf
"15.2. Basic Elements
15.2.1. Basic Syntax Rules
C is a highly influential programming language. Many modern programming languages
have adopted syntactic elements that originated in C. Usually such languages are referred
to as “C-style syntax” languages. These elements include the following.
•C is a statically typed language so variables must be declared along with their
types before using them.
•Strings are delimited with double quotes. Single characters, including special
escaped characters are delimited by single quotes; ""this is a string"", and
these are characters: 'A', '4', '$' and '\n'
•Executable statements are terminated by a semicolon, ;
•Code blocks are defined by using opening and closing curly brackets, { ... }.
Moreover, code blocks can be nested: code blocks can be defined within other code
blocks.
•Variables are scoped to the code block in which they are declared and are only
valid within that code block.
•In general, whitespace between coding elements is ignored.",ComputerScienceOne.pdf
"valid within that code block.
•In general, whitespace between coding elements is ignored.
Though not a syntactic requirement, the proper use of whitespace is important for good,
readable code. Code inside code blocks is indented at the same indentation level. Nested
code blocks are indented further. Think of a typical table of contents or the outline of a
formal paper or essay. Sections and subsections or points and subpoints all follow proper
indentation with elements at the same level at the same indentation. This convention is
used to organize code and make it more readable.
15.2.2. Preprocessor Directives
The lines,
1 #include<stdlib.h>
2 #include<stdio.h>
are preprocessor directives. Preprocessor directives are instructions to the compiler to
modify the source code before it starts to compile it. These particular lines are “including”
standard libraries of functions so that the program can use functionality that has already
255",ComputerScienceOne.pdf
"15. Basics
been implemented for us.
The first, stdlib.h represents the C standard ( std) library ( lib). This library is so
essential that many compilers will automatically include it even if you do not explicitly
do so in your program. Still, it is best practice to include it in your code.
The second, stdio.h is the standard ( std) input/output ( io) library which contains
basic input/output (I/O) functions that we can use. In particular, the standard output
function printf() is part of this library.
Failure to include a library means that you will not be able to use the functions it
provides in your program. Using the functions without including the library may result in
a compiler error. The .h in the library names stands for “header”; function declarations
are typically contained in a header file while their definitions are placed in a source file
of the same name. We’ll explore this convention in detail when we look at functions in C
(Chapter 18).",ComputerScienceOne.pdf
"of the same name. We’ll explore this convention in detail when we look at functions in C
(Chapter 18).
There are many other important standard libraries that we’ll touch on as needed; of
immediate interest is the standard mathematics library, math.h. It includes many
useful functions to compute common mathematical functions such as the square root and
natural logarithm. Table 15.1 highlights several of these functions. To use them you’d
include the math library in your source file #include<math.h> and then “call” them
by providing input and assigning the output value to a variable. For example:
1 double x = 1.5;
2 double y, z;
3 y = sqrt(x); //y now has the value √x=
√
1.5
4 z = sin(x); //z now has the value sin (x) = sin (1.5)
In both of the function calls above, the value of the variable x is “passed” to the math
function which computes and “returns” the result which then gets assigned to another
variable.
Macros",ComputerScienceOne.pdf
"function which computes and “returns” the result which then gets assigned to another
variable.
Macros
Another preprocessor directive establishes macros using the #define keyword. A macro
is a single instruction that specifies a more complex set of instructions. The macro can
be used to define constants to be used throughout your program. To illustrate, consider
the following example.
256",ComputerScienceOne.pdf
"15.2. Basic Elements
Function Description
abs(x) Absolute value for int variables, |x|a
fabs(x) Absolute value for double variables
ceil(x) Ceiling function, ⌈46.3⌉= 47.0
floor(x) Floor function, ⌊46.3⌋= 46.0
cos(x) Cosine functionb
sin(x) Sine functionb
tan(x) Tangent functionb
exp(x) Exponential function, ex, e= 2.71828 ...
log(x) Natural logarithm, ln (x)c
log10(x) Logarithm base 10, log 10 (x)c
pow(x,y) The power function, computes xy
sqrt(x) Square root functionc
Table 15.1.: Several functions defined in the C standard math library. aThe absolute value
function is actually in the standard library, stdlib.h. ball trigonometric
functions assume input is in radians, not degrees. cInput is assumed to be
positive, x> 0.
257",ComputerScienceOne.pdf
"15. Basics
1 #define MILES_PER_KM 1.609
The macro defines an “alias” for the MILES_PER_KM identifier as the value 1 .609. Essen-
tially, the C preprocessor will go through the code and any instance of MILES_PER_KM
will be replaced with 1.609. The advantage of using a macro like this is that we can use
the identifier MILES_PER_KM throughout our program instead of magic numbers whose
meaning and intent may not be immediately clear. Moreover, if we want to change the
definition (say make it more precise, using 1 .60934 instead) then we only need to change
the macro instead of making the same change throughout our program.
As a stylistic note: macro constants in C are usually associated with uppercase underscore
casing as in our example. Also, the math standard library defines several macros for
common mathematical constants such as π, e, and
√
2 ( M_PI, M_E, and M_SQRT2
respectively) among others.
15.2.3. Comments
Comments can be written in a C program either as a single line using two forward",ComputerScienceOne.pdf
"respectively) among others.
15.2.3. Comments
Comments can be written in a C program either as a single line using two forward
slashes, //comment or as a multiline comment using a combination of forward slash
and asterisk: /* comment */ . With a single line comment, everything on the line
after the forward slashes is ignored. With a multiline comment, everything between the
forward slash/asterisk is ignored. Comments are ultimately ignored by the compiler so
the amount of comments do not have an effect on the final executable code. Consider
the following example.
1 //this is a single line comment
2 int x; //this is also a single line comment, but after some code
3
4 /*
5 This is a comment that can
6 span multiple lines to format the comment
7 message more clearly
8 */
9 double y;
Most code editors and IDEs will present comments in a special color or font to distinguish
them from the rest of the code (just as our example above does). Failure to close a
258",ComputerScienceOne.pdf
"15.2. Basic Elements
multiline comment will likely result in a compiler error but with color-coded comments
its easy to see the mistake visually.
15.2.4. The main() Function
Every executable program starts its execution somewhere. In C, the starting point is the
main() function. When a program is compiled to an executable and the program is
invoked, the code in the main() function starts executing. In our example, the code
between the two curly brackets defines the main() function. At the end of the function
we have a return statement. We’ll examine return statements in detail when we
examine functions. The program is “returning” a zero value to the operating system.
The convention in C is that zero indicates “no error occurred” while a non-zero value
is used to indicate that “some” error occurred (the value used is determined by system
standards such as the Portable Operating System Interface (POSIX) standard).",ComputerScienceOne.pdf
"standards such as the Portable Operating System Interface (POSIX) standard).
In addition, our main() function has two arguments: argc and argv which serve to
communicate any command line arguments provided to the program (review Section
2.4.4 for details). The first, argc is an integer that indicates the number of arguments
provided including the executable file name itself. The second, argv actually stores
the arguments as strings. We will understand the syntax later on, but for now we can
at least understand how we might convert these arguments to different types such as
integers and floating point numbers.
Recall that argv is the argument vector: it is an array (see Chapter 20) of the command
line arguments. To access them, you can index them starting at zero, the first being
argv[0], the second argv[1], etc. (the last one of course would be at argv[argc-1]).
The first one is always the name of the executable file being run. The remaining are the",ComputerScienceOne.pdf
"The first one is always the name of the executable file being run. The remaining are the
command line arguments provided by the user.
To convert them you can use two different functions, atoi() and atof() which are
short for alphanumeric to integer and floafing-point number respectively. An example:
1 //prints the first command line argument:
2 printf(""%s\n"", argv[0]);
3 //converts the ""second"" command line argument to an integer
4 int x = atoi(argv[1]);
5 //converts the ""third"" command line argument to a double:
6 double y = atof(argv[2]);
259",ComputerScienceOne.pdf
"15. Basics
15.3. Variables
As previewed, the three primary primitive types supported in C are int, double, and
char which support integers, floating point numbers, and single ASCII characters.
Integer ( int) types are only guaranteed to “be at least” 16 bytes by the C standard but
are usually 32-bit signed integers on most modern systems.1 With a 32-bit signed int we
can represent integers in the range −2,147,483,648 to 2,147,483,647. Doubles ( double)
types are double-precision floating point numbers as per the IEEE 754 standard and
provide about 16 digits of precision.
C provides a float (single precision floating point number) type and there are various
modifiers such as short, long, unsigned and signed that can be used, but these
are either system-dependent or rely on later versions of the C standard (such as C99).
We will restrict our focus to more portable, interoperable code and stick with the basic
int and double types in our code.",ComputerScienceOne.pdf
"We will restrict our focus to more portable, interoperable code and stick with the basic
int and double types in our code.
Finally, the char type is typically a single byte that represents a single ASCII character.
For all intents and purposes a char can be treated as an integer in the range 0 to 127
(or 255) as defined by the ASCII text table (see Table 2.4).
15.3.1. Declaration & Assignment
C is a statically typed language meaning that all variables must be declared before you
can use them or refer to them. In addition, when declaring a variable, you must specify
both its type and its identifier. For example:
1 int numUnits;
2 double costPerUnit;
3 char firstInitial;
Each declaration specifies the variable’s type followed by the identifier and ending with
a semicolon. The identifier rules are fairly standard: a name can consist of lower and
uppercase alphabetic characters, numbers, and underscores but may not begin with a",ComputerScienceOne.pdf
"uppercase alphabetic characters, numbers, and underscores but may not begin with a
numeric character. We adopt the modern camelCasing naming convention for variables
in our code.
The assignment operator is a single equal sign, = and is a right-to-left assignment. The
1You may have to deal with 16-bit int types in legacy systems/compilers or in modern embedded
systems.
260",ComputerScienceOne.pdf
"15.3. Variables
variable that we wish to assign the value to appears on the left-hand-side while the value
(literal, variable or expression) is on the right-hand-size. Using our variables from before,
we can assign them values:
1 numUnits = 42;
2 costPerUnit = 32.79;
3 firstInitial = 'C';
An important thing to understand and to keep in mind is: if you declare a variable but
do not assign it a value, its value is undefined. If we code something like int a;, the
value of the variable a is not necessarily zero; depending on the system, it could contain
a special value that indicates “uninitialized memory” or it could contain garbage, or it
could have the value zero. The C standard does not specify default values for variables.
The default value of variables is highly system dependent–on the compiler, the libraries,
and even the operating system. You should not make any assumptions on the initial or
default values of variables. If you need such assumptions, then values must be assigned.",ComputerScienceOne.pdf
"default values of variables. If you need such assumptions, then values must be assigned.
For brevity, C allows you to declare a variable and immediately assign it a value on the
same line. Our example could be written more compactly written as follows.
1 int numUnits = 42;
2 double costPerUnit = 32.79;
3 char firstInitial = 'C';
As another shorthand, we can declare multiple variables on the same line by delimiting
them with a comma. However, they must be of the same type. We can also use an
assignment with them.
1 int numOrders, numUnits = 42, numCustomers = 10, numItems;
2 double costPerUnit = 32.79, salesTaxRate;
Another convenient keyword is const, short for “constant.” We can apply it to any
variable to indicate that it is a read-only variable. Of course, we must assign it a value
at declaration. For example:
261",ComputerScienceOne.pdf
"15. Basics
1 const int secret = 42;
2 const double salesTaxRate = 0.075;
Any attempt to reassign the values of const variables will result in a compiler error.
15.4. Operators
C supports the standard arithmetic operators for addition, subtraction, multiplication,
and division using +, -, *, and / respectively. Each of these operators is a binary
operator that acts on two operands which can be literals, variables or expressions and
follow the usual rules of arithmetic when it comes to order of precedence (multiplication
and division before addition and subtraction).
1 int a = 10, b = 20, c = 30, d;
2 d = a + 5;
3 d = a + b;
4 d = a - b;
5 d = a + b * c;
6 d = a * b;
7 d = a / b; //integer division and truncation! See below
8
9 double x = 1.5, y = 3.4, z = 10.5, w;
10 w = x + 5.0;
11 w = x + y;
12 w = x - y;
13 w = x + y * z;
14 w = x * y;
15 w = x / y;
16
17 //you can do arithmetic with both types:
18 w = a + x;
19 d = b + y; //truncation also occurs here!",ComputerScienceOne.pdf
"14 w = x * y;
15 w = x / y;
16
17 //you can do arithmetic with both types:
18 w = a + x;
19 d = b + y; //truncation also occurs here!
Special care must be taken when dealing with int types. For all four operators, if
both operands are integers, the result will be an integer. For addition, subtraction, and
multiplication this isn’t a big deal, but for division it means that when we divide, say
262",ComputerScienceOne.pdf
"15.5. Basic I/O
10 / 20, the result is not 0 .5 as expected. The number 0.5 is a floating point number.
As such, the fractional part gets truncated (cut off and thrown out) leaving only zero. In
the code above, d = a / b; the variable d ends up getting the value zero because of
this.
Similarly, attempting to assign a floating point number to an integer also results in
truncation because an int type cannot handle the fractional part. In the line d = b + y
above, b + y correctly evaluates as 20 + 3 .4 = 23.4, but when assigned to the int
variable d the .4 gets truncated and d is assigned the value 23. Assigning an int
value to a double variable is not a problem as the integer 2 implicitly becomes the
floating point number 2.0.
A solution to this problem is to use explicit type casting to force at least one of the
operands in an integer division to become a double type. For example:
1 int a = 10, b = 20;
2 double x;
3
4 x = (double) a / b;",ComputerScienceOne.pdf
"operands in an integer division to become a double type. For example:
1 int a = 10, b = 20;
2 double x;
3
4 x = (double) a / b;
results in x getting the “correct” value of 0 .5. This works because the (double) code
forces the int variable a to temporarily be treated as a double variable (in this case
10.0) for the purposes of division (so that truncation does not occur).
C also supports the integer remainder operator using the % symbol. This operator gives
the remainder of the result of dividing two integers. Examples:
1 int x;
2
3 x = 10 % 5; //x is 0
4 x = 10 % 3; //x is 1
5 x = 29 % 5; //x is 4
15.5. Basic I/O
The standard I/O library ( stdio.h) contains many functions that facilitate input and
output including printf() for standard output and scanf() (scan formatted input)
263",ComputerScienceOne.pdf
"15. Basics
for standard input.
The printf() function works exactly as discussed in Section 2.4.3. The scanf()
function works using similar placeholders as printf(). To illustrate how it works,
consider the following lines of code:
1 int a;
2 printf(""Please enter a number: "");
3 scanf(""%d"", &a);
The printf() statement prompts the user for an input. The scanf() then executes
and the program waits for the user to enter input. The user is free to start typing. When
the user is done, they hit the enter key at which point the program resumes and reads
the input from the standard input buffer, converts the value entered by the user into
an integer and places the result in the variable a where it can now be used by the
remainder of the program.
A few points of interest. First, the same placeholder as printf() was used, %d for
int values. However, when we “passed” the variable a to scanf() we placed an
ampersand, & in front of it. This is passing the variable by reference and we’ll explore",ComputerScienceOne.pdf
"ampersand, & in front of it. This is passing the variable by reference and we’ll explore
that concept further in Chapter 18, but for now just know that when using variables
with scanf(), an ampersand is required. Failure to place an ampersand in front of
a variable with scanf() will likely result in a segmentation fault (an illegal memory
access).
You can use the same placeholder, %c with scanf() to read in a single character as
well. However, for floating point numbers, in particular double types, the placeholder
%lf must be used (which stands for long float, a double precision number). Failure to
use the correct placeholder may result in garbage results as the input will be interpreted
incorrectly. Another example:
1 double x;
2 printf(""Please enter a fractional number: "");
3 scanf(""%lf"", &x);
Another potential problem is that scanf() expects a certain format (thus its name).
If we prompt the user for a number but they just start mashing the keyboard giving",ComputerScienceOne.pdf
"If we prompt the user for a number but they just start mashing the keyboard giving
non-numerical input, we may get incorrect results. scanf() will interpret the input as
zero. It may be very difficult to distinguish between the case of a user actually entering
264",ComputerScienceOne.pdf
"15.6. Examples
in zero as a legitimate input versus bad input. In general, scanf() is not a good
mechanism for reading input (and in fact can be very dangerous), but it does provide a
good starting point.
15.6. Examples
15.6.1. Converting Units
Let’s write a program that will prompt the user to enter a temperature in degrees
Fahrenheit and convert it to degrees Celsius using the formula
C = (F −32) ·5
9
We begin with the basic program outline which will include preprocessor directives to
bring in the standard library and the standard input/output library (we’ll need to prompt
for input and print the result as output to the user). Further, we want our program
to be executable, so we need to put our code into the main() function. Finally, we’ll
document our program to indicate its purpose.
1 #include<stdlib.h>
2 #include<stdio.h>
3
4 /**
5 * This program converts Fahrenheit temperatures to
6 * Celsius
7 */
8 int main(int argc, char **argv) {
9
10 //TODO: implement this
11
12 return 0;",ComputerScienceOne.pdf
"5 * This program converts Fahrenheit temperatures to
6 * Celsius
7 */
8 int main(int argc, char **argv) {
9
10 //TODO: implement this
11
12 return 0;
13 }
It is common for programmers to use a comment along with a TODO note to themselves
as a reminder of things that they still need to implement in a program.
Let’s first outline the basic steps that our program will go through:
1. We’ll first prompt the user for input, asking them for a temperature in Fahrenheit
265",ComputerScienceOne.pdf
"15. Basics
2. Next we’ll read the user’s input, likely into a floating point number as degrees can
be fractional
3. Once we have the input, we can calculate the degrees Celsius by using the formula
above
4. Lastly, we will want to print the result to the user to inform them of the value
Sometimes its helpful to write an outline of such a program directly in the code using
comments to provide a step-by-step process. For example:
1 #include<stdlib.h>
2 #include<stdio.h>
3
4 /**
5 * This program converts Fahrenheit temperatures to
6 * Celsius
7 */
8 int main(int argc, char **argv) {
9
10 //TODO: implement this
11 //1. Prompt the user for input in Fahrenheit
12 //2. Read the Fahrenheit value from the standard input
13 //3. Compute the degrees Celsius
14 //4. Print the result to the user
15
16 return 0;
17 }
As we read each step it becomes apparent that we’ll need a couple of variables: one to
hold the Fahrenheit (input) value and one for the Celsius (output) value. It also makes",ComputerScienceOne.pdf
"hold the Fahrenheit (input) value and one for the Celsius (output) value. It also makes
sense that each of these should be double variables as we want to support fractional
values. So at the top of our main() function, we’ll add the variable declarations:
double fahrenheit, celsius;
Each of the steps is now straightforward; we’ll use a printf() statement in the first
step to prompt the user for input:
printf(""Please enter degrees in Fahrenheit: "");
In the second step, we’ll use the standard input to read the fahrenheit variable value
from the user. Recall that we use the placeholder %lf for reading double values and
266",ComputerScienceOne.pdf
"15.6. Examples
use an ampersand when using scanf():
scanf(""%lf"", &fahrenheit);
We can now compute celsius using the formula provided:
celsius = (fahrenheit - 32) * (5 / 9);
Finally, we use printf() again to output the result to the user:
printf(""%f Fahrenheit is %f Celsius\n"", fahrenheit, celsius);
Try typing and running the program as defined above and you’ll find that you don’t
get correct answers. In fact, you’ll find that no matter what values you enter, you get
zero. This is because of the calculation using 5 / 9: recall what happens with integer
division: truncation! This will always end up being zero.
One way we could fix it would be to pull out our calculators and find that 5
9 = 0.55555 ...
and replace 5/9 with 0.555555. But, how many fives? It may be difficult to tell how
accurate we can make this floating point number by hardcoding it ourselves. A much
better approach would be to let the compiler take care of the optimal computation for us",ComputerScienceOne.pdf
"better approach would be to let the compiler take care of the optimal computation for us
by making at least one of the numbers a double to prevent integer truncation. That is,
we should instead use 5.0 / 9. The full program can be found in Code Sample 15.2.
15.6.2. Computing Quadratic Roots
Some programs require the user to enter multiple inputs. The prompt-input process can
be repeated. In this example, consider asking the user for the coefficients, a,b,c to a
quadratic polynomial,
ax2 + bx+ c
and computing its roots using the quadratic formula,
x= −b±
√
b2 −4ac
2a
As before, we can create a basic program with a main() function and start filling in
the details. In particular, we’ll need to prompt for the input a, then read it in; then
prompt for b, read it in and repeat for c. We’ll also need several variables: three for the
coefficients a,b,c and two more; one for each root. Thus, we have
1 double a, b, c, root1, root2;
2
3 printf(""Please enter a: "");
4 scanf(""%lf"", &a);
267",ComputerScienceOne.pdf
"15. Basics
1 #include<stdlib.h>
2 #include<stdio.h>
3
4 /**
5 * This program converts Fahrenheit temperatures to
6 * Celsius
7 */
8 int main(int argc, char **argv) {
9
10 double fahrenheit, celsius;
11
12 //1. Prompt the user for input in Fahrenheit
13 printf(""Please enter degrees in Fahrenheit: "");
14
15 //2. Read the Fahrenheit value from the standard input
16 scanf(""%lf"", &fahrenheit);
17
18 //3. Compute the degrees Celsius
19 celsius = (fahrenheit - 32) * 5.0/9;
20
21 //4. Print the result to the user
22 printf(""%f Fahrenheit is %f Celsius\n"", fahrenheit, celsius);
23
24 return 0;
25 }
Code Sample 15.2.: Fahrenheit-to-Celsius Conversion Program in C
5 printf(""Please enter b: "");
6 scanf(""%lf"", &b);
7 printf(""Please enter c: "");
8 scanf(""%lf"", &c);
Now to compute the roots: we need to take care that we correctly adapt the formula so
it accurately reflects the order of operations. We also need to use the standard math",ComputerScienceOne.pdf
"it accurately reflects the order of operations. We also need to use the standard math
library’s square root function (unless you want to write your own! 2 Carefully adapting
2We will write several square root methods as exercises later!
268",ComputerScienceOne.pdf
"15.6. Examples
the formula leads to
1 root1 = (-b + sqrt(b*b - 4*a*c) ) / (2*a);
2 root2 = (-b - sqrt(b*b - 4*a*c) ) / (2*a);
Finally, we print the output using printf(). The full program can be found in Code
Sample 15.3.
1 #include<stdlib.h>
2 #include<stdio.h>
3 #include<math.h>
4
5 /**
6 * This program computes the roots to a quadratic equation
7 * using the quadratic formula.
8 */
9 int main(int argc, char **argv) {
10
11 double a, b, c, root1, root2;
12
13 printf(""Please enter a: "");
14 scanf(""%lf"", &a);
15 printf(""Please enter b: "");
16 scanf(""%lf"", &b);
17 printf(""Please enter c: "");
18 scanf(""%lf"", &c);
19
20 root1 = (-b + sqrt(b*b - 4*a*c) ) / (2*a);
21 root2 = (-b - sqrt(b*b - 4*a*c) ) / (2*a);
22
23 printf(""The roots of %fx^2 + %fx + %f are: \n"", a, b, c);
24 printf("" root1 = %f\n"", root1);
25 printf("" root2 = %f\n"", root2);
26 return 0;
27 }
Code Sample 15.3.: Quadratic Roots Program in C
This program was interactive. As an alternative, we could have read all three of the",ComputerScienceOne.pdf
"27 }
Code Sample 15.3.: Quadratic Roots Program in C
This program was interactive. As an alternative, we could have read all three of the
269",ComputerScienceOne.pdf
"15. Basics
inputs as command line arguments, converting them to floating point numbers. Lines
12–17 in the program could have been changed to
1 a = atof(argv[1]);
2 b = atof(argv[2]);
3 c = atof(argv[3]);
Finally, think about the possible inputs a user could provide that may cause problems
for this program. For example:
•What if the user entered zero for a?
•What if the user entered some combination such that b2 <4ac?
•What if the user entered non-numeric values?
•For the command line argument version, what if the user provided less than three
arguments? Or more?
How might we prevent the consequences of such bad inputs? How might we handle the
event that a user enters bad input and how do we communicate these errors to the user?
To begin to resolve these issues, we’ll need conditionals.
270",ComputerScienceOne.pdf
"16. Conditionals
C supports the basic if, if-else, and if-else-if conditional structures as well as switch
statements. Logical statements are built using the standard logical operators for numeric
comparisons as well as logical operators such as negation, And, and Or. However, there
are a few idiosyncrasies that need to be understood.
16.1. Logical Operators
C has no built-in Boolean type, nor does it have any keywords associated with true and
false. Instead, C uses numeric types as implicit Boolean types with the convention that
zero is associated with false and any non-zero value is associated with true. So the values
1, 2.5, and even negatives, −1, −123.2421, etc. are all interpreted as true when used in
logical statements.
The standard numeric comparison operators are also supported. Consider the following
code snippet:
1 int a = 10;
2 int b = 20;
3 int c = 10;
4 int d = 0;
The six standard comparison operators are presented in Table 16.1 using these variables",ComputerScienceOne.pdf
"1 int a = 10;
2 int b = 20;
3 int c = 10;
4 int d = 0;
The six standard comparison operators are presented in Table 16.1 using these variables
as examples. The comparison operators are the same when used with double types as
well and int types can be compared with each other without type casting. The three
basic logical operators are also supported as described in Table 16.2 using the same code
snippet variable values as examples.
271",ComputerScienceOne.pdf
"16. Conditionals
Name Operator Syntax Examples Value
Equals == a == 10 true
b == 10 false
a == b false
a == c true
Not Equals != a != 10 false
b != 10 true
a != b true
a != c false
Strictly Less Than < a < 15 true
a < 5 false
a < b true
a < c false
Less Than Or Equal To <= a <= 15 true
a <= 5 false
a <= b true
a <= c true
Strictly Greater Than > a > 15 false
a > 5 true
a > b false
a > c false
Greater Than Or Equal To >= a >= 15 false
a >= 5 true
a >= b false
a >= c true
Table 16.1.: Comparison Operators in C
Operator Operator Syntax Examples Values
Negation ! !a false
!d true
And && a && b true
a && d false
Or || a || b true
!a || d false
Table 16.2.: Logical Operators in C
272",ComputerScienceOne.pdf
"16.1. Logical Operators
Operator(s) Associativity Notes
Highest ++, -- left-to-right increment operators
-, ! right-to-left unary negation operator, logical
not
*, /, % left-to-right
+, - left-to-right addition, subtraction
<, <=, >, >= left-to-right comparison
==, != left-to-right equality, inequality
&& left-to-right logical And
|| left-to-right logical Or
Lowest =, +=, -=, *=, /= right-to-left assignment and compound assign-
ment operators
Table 16.3.: Operator Order of Precedence in C. Operators on the same level have
equivalent order and are performed in the associative order specified.
16.1.1. Order of Precedence
At this point it is worth summarizing the order of precedence of all the operators that
we’ve seen so far including assignment, arithmetic, comparison, and logical. Since all of
these operators could be used in one statement, for example,
(b*b < 4*a*c || a == 0 || argc != 4)
it is important to understand the order in which each one gets evaluated. Table 16.3",ComputerScienceOne.pdf
"(b*b < 4*a*c || a == 0 || argc != 4)
it is important to understand the order in which each one gets evaluated. Table 16.3
summarizes the order of precedence for the operators seen so far. This is not an exhaustive
list of C operators.
16.1.2. Comparing Strings and Characters
The comparison operators in Table 16.1 can also be used for single characters because
of the nature of the ASCII text table (see Table 2.4). Each alphanumeric character,
including the various symbols and whitespace characters, is associated with an integer 0–
127. We can therefore write statements like ('A' < 'a'), which is true since uppercase
letters are ordered before lowercase letters in the ASCII table ( 'A' is 65 and 'a' is 97
and so 65 <97 is true). Several more examples can be found in Table 16.4.
Numeric comparison operators cannot be used to compare strings in C. For example,
we could write (""aardvark"" < ""zebra"") which would be valid C, and it would even",ComputerScienceOne.pdf
"we could write (""aardvark"" < ""zebra"") which would be valid C, and it would even
have a result. However, that result wouldn’t necessarily be true or false. The reason for
this is that strings in C are actually represented as arrays, which in turn are represented
as memory locations. We’ll explore these issues in greater depth later on, but for now
273",ComputerScienceOne.pdf
"16. Conditionals
Comparison Example Result
('A' < 'a') true
('A' == 'a') false
('A' < 'Z') true
('0' < '9') true
('\n' < 'A') true
(' ' < '\n') false
Table 16.4.: Character comparisons in C
understand that you can write this code, it will compile, and it will even run. However,
the results will not be as expected.
16.2. If, If-Else, If-Else-If Statements
Conditional statements in C utilize the keywords if, else, and else if. Condi-
tions are placed inside parentheses immediately after the if and else if keywords.
Examples of all three can be found in Code Sample 16.1.
Some observations about the syntax: the statement, if(x < 10) does not have a
semicolon at the end. This is because it is a conditional statement that determines
the flow of control and not an executable statement. Therefore, no semicolon is used.
Suppose we made a mistake and did include a semicolon:
1 int x = 15;
2 if(x < 10); {
3 printf(""x is less than 10\n"");
4 }",ComputerScienceOne.pdf
"Suppose we made a mistake and did include a semicolon:
1 int x = 15;
2 if(x < 10); {
3 printf(""x is less than 10\n"");
4 }
Some compilers may give a warning, but this is valid C; it will compile and it will run.
However, it will end up printing x is less than 10, even though x= 15! Recall that
a conditional statement binds to the executable statement or code block immediately
following it. In this case, we’ve provided an empty executable statement ended by the
semicolon. The code is essentially equivalent to
1 int x = 15;
2 if(x < 10) {
3 }
274",ComputerScienceOne.pdf
"16.2. If, If-Else, If-Else-If Statements
1 //example of an if statement:
2 if(x < 10) {
3 printf(""x is less than 10\n"");
4 }
5
6 //example of an if-else statement:
7 if(x < 10) {
8 printf(""x is less than 10\n"");
9 } else {
10 printf(""x is 10 or more \n"");
11 }
12
13 //example of an if-else-if statement:
14 if(x < 10) {
15 printf(""x is less than 10\n"");
16 } else if(x == 10) {
17 printf(""x is equal to ten\n"");
18 } else {
19 printf(""x is greater than 10\n"");
20 }
Code Sample 16.1.: Examples of Conditional Statements in C
275",ComputerScienceOne.pdf
"16. Conditionals
4 printf(""x is less than 10\n"");
This is obviously not what we wanted. The semicolon ended up binding to the empty
executable statement. The code block containing the print statement immediately
followed, but it was not bound to the conditional statement which is why the print
statement executed regardless of the value of x.
Another convention that we’ve used in our code is where we have placed the curly brackets.
First, if a conditional statement is bound to only one statement, the curly brackets are
not necessary. However, it is best practice to include them even if they are not necessary
and we’ll follow this convention. Second, the opening curly bracket is on the same line as
the conditional statement while the closing curly bracket is indented to the same level
as the start of the conditional statement. Moreover, the code inside the code block is
indented. If there were more statements in the block, they would have all been at the
same indentation level.",ComputerScienceOne.pdf
"indented. If there were more statements in the block, they would have all been at the
same indentation level.
16.3. Examples
16.3.1. Computing a Logarithm
The logarithm of x is the exponent that some base must be raised to get x. The most
common logarithm is the natural logarithm, ln (x) which is base e= 2.71828 ... . But
logarithms can be in any base b >1.1 What if we wanted to compute log2 (x)? Or
logπ (x)? Let’s write a program that will prompt the user for a number x and a base b
and computes logb(x).
Arbitrary bases can be computed using the change of base formula:
logb(x) = loga(x)
loga(b)
If we can compute some base a, then we can compute any base b. Fortunately we have
such a solution. Recall that the standard library provides a function to compute the
natural logarithm, log()). This is one of the fundamentals of problems solving: if a
solution already exists, use it. In this case, a solution exists for a different, but similar",ComputerScienceOne.pdf
"solution already exists, use it. In this case, a solution exists for a different, but similar
problem (computing the natural logarithm), but we can adapt the solution using the
change of base formula. In particular, if we have variables b (base) and x, we can
compute logb(x) using
1Bases can also be 0 <b< 1, but we’ll restrict our attention to increasing functions only.
276",ComputerScienceOne.pdf
"16.3. Examples
log(x) / log(b)
But wait: we have a problem similar to the examples in the previous section. The user
could enter invalid values such as b = −10 or x = −2.54 (logarithms are undefined
for non-positive values in any base). We want to ensure that b >1 and x >0. With
conditionals, we can now do this. Once we have read in the input from the user we can
make a check for good input using an if statement.
1 if(x <= 0 || b <= 1) {
2 printf(""Error: bad input!\n"");
3 exit(1);
4 }
This code has something new: exit(1). The exit() function immediately terminates
the program regardless of the rest of the code that it may remain. The argument passed
to exit is an integer that represents an error code. The convention is that zero indicates
“no error” while non-zero values indicate some error. This is a simple way of performing
error handling : if the user provides bad input, we inform them and quit the program,",ComputerScienceOne.pdf
"error handling : if the user provides bad input, we inform them and quit the program,
forcing them to run it again and provide good input. By prematurely terminating the
program we avoid any illegal operation that would give a bad result.
Alternatively, we could have split the conditions into two statements and given more
descriptive error messages. We use this design in the full program which can be found in
Code Sample 16.2. The program also takes the input as command line arguments. Now
that we have conditionals, we can actually check that the correct number of arguments
was provided by the user and quit in the event that they don’t provide the correct
number.
16.3.2. Life & Taxes
Let’s adapt the conditional statements we developed in Section 3.6.4 into a full C program.
The first thing we need to do is establish the variables we’ll need and read them in from
the user. At the same time we can check for bad input (negative values) for both the
inputs.",ComputerScienceOne.pdf
"the user. At the same time we can check for bad input (negative values) for both the
inputs.
1 double income, baseTax, numChildren, credit, totalTax;
2
3 printf(""Please enter your Adjusted Gross Income: "");
4 scanf(""%lf"", &income);
5
277",ComputerScienceOne.pdf
"16. Conditionals
6 printf(""How many children do you have?"");
7 scanf(""%d"", &numChildren);
8
9 if(income < 0 || numChildren < 0) {
10 printf(""Invalid inputs"");
11 exit(1);
12 }
Next, we can code a series of if-else-if statements for the income range. By placing the
ranges in increasing order, we only need to check the upper bounds just as in the original
example.
1 if(income <= 18150) {
2 baseTax = income * .10;
3 } else if(income <= 73800) {
4 baseTax = 1815 + (income - 18150) * .15;
5 } else if(income <= 148850) {
6 ...
7 } else {
8 baseTax = 127962.50 + (income - 457600) * .396;
9 }
Next we compute the child tax credit, taking care that it does not exceed $3,000. A
conditional based on the number of children suffices as at this point in the program we
already know it is zero or greater.
1 if(numChildren <= 3) {
2 credit = numChildren * 1000;
3 } else {
4 credit = 3000;
5 }
Finally, we need to ensure that the credit does not exceed the total tax liability (the",ComputerScienceOne.pdf
"3 } else {
4 credit = 3000;
5 }
Finally, we need to ensure that the credit does not exceed the total tax liability (the
credit is non-refundable, so if the credit is greater, the tax should only be zero, not
negative).
278",ComputerScienceOne.pdf
"16.3. Examples
1 if(baseTax - credit >= 0) {
2 totalTax = baseTax - credit;
3 } else {
4 totalTax = 0;
5 }
The full program is presented in Code Sample 16.3.
16.3.3. Quadratic Roots Revisited
Let’s return to the quadratic roots program we previously designed that uses the quadratic
equation to compute the roots of a quadratic polynomial by reading coefficients a,b,c
in from the user. One of the problems we had previously identified is if the user enters
“bad” input: if a= 0, we would end up dividing by zero; if b2 −4ac< 0 then we would
have complex roots. With conditionals, we can now check for these issues and exit with
an error message.
Another potential case we might want to handle differently is when there is only one
distinct root ( b2 −4ac= 0). In that case, the quadratic formula simplifies to −b
2a and we
can print a different, more specific message to the user. The full program can be found
in Code Sample 16.4.
279",ComputerScienceOne.pdf
"16. Conditionals
1 #include<stdlib.h>
2 #include<stdio.h>
3 #include<math.h>
4
5 /**
6 * This program computes the logarithm base b (b > 1)
7 * of a given number x > 0
8 */
9 int main(int argc, char **argv) {
10
11 double b, x, result;
12 if(argc != 3) {
13 printf(""Usage: %s b x \n"", argv[0]);
14 exit(1);
15 }
16
17 b = atof(argv[1]);
18 x = atof(argv[2]);
19
20 if(x <= 0) {
21 printf(""Error: x must be greater than zero\n"");
22 exit(1);
23 }
24 if(b <= 1) {
25 printf(""Error: base must be greater than one\n"");
26 exit(1);
27 }
28
29 result = log(x) / log(b);
30 printf(""log_(%f)(%f) = %f\n"", b, x, result);
31 return 0;
32 }
Code Sample 16.2.: Logarithm Calculator Program in C
280",ComputerScienceOne.pdf
"16.3. Examples
1 #include<stdlib.h>
2 #include<stdio.h>
3
4 int main(int argc, char **argv) {
5
6 double income, baseTax, credit, totalTax;
7 int numChildren;
8
9 //prompt for income from the user
10 printf(""Please enter your Adjusted Gross Income: "");
11 scanf(""%lf"", &income);
12
13 //prompt for children
14 printf(""How many children do you have?"");
15 scanf(""%d"", &numChildren);
16
17 if(income < 0 || numChildren < 0) {
18 printf(""Invalid inputs"");
19 exit(1);
20 }
21
22 if(income <= 18150) {
23 baseTax = income * .10;
24 } else if(income <= 73800) {
25 baseTax = 1815 + (income -18150) * .15;
26 } else if(income <= 148850) {
27 baseTax = 10162.50 + (income - 73800) * .25;
28 } else if(income <= 225850) {
29 baseTax = 28925.00 + (income - 148850) * .28;
30 } else if(income <= 405100) {
31 baseTax = 50765.00 + (income - 225850) * .33;
32 } else if(income <= 457600) {
33 baseTax = 109587.50 + (income - 405100) * .35;
34 } else {
35 baseTax = 127962.50 + (income - 457600) * .396;
36 }
37",ComputerScienceOne.pdf
"33 baseTax = 109587.50 + (income - 405100) * .35;
34 } else {
35 baseTax = 127962.50 + (income - 457600) * .396;
36 }
37
38 if(numChildren <= 3) {
39 credit = numChildren * 1000;
40 } else {
41 credit = 3000;
42 }
43
44 if(baseTax - credit >= 0) {
45 totalTax = baseTax - credit;
46 } else {
47 totalTax = 0;
48 }
49
50 printf(""AGI: $%10.2f\n"", income);
51 printf(""Tax: $%10.2f\n"", baseTax);
52 printf(""Credit: $%10.2f\n"", credit);
53 printf(""Tax Liability: $%10.2f\n"", totalTax);
54
55 return 0;
56 }
Code Sample 16.3.: Tax Program in C
281",ComputerScienceOne.pdf
"16. Conditionals
1 #include<stdlib.h>
2 #include<stdio.h>
3 #include<math.h>
4
5 /**
6 * This program computes the roots to a quadratic equation
7 * using the quadratic formula.
8 */
9 int main(int argc, char **argv) {
10
11 double a, b, c, root1, root2;
12
13 if(argc !=4) {
14 printf(""Usage: %s a b c\n"", argv[0]);
15 exit(1);
16 }
17
18 a = atof(argv[1]);
19 b = atof(argv[2]);
20 c = atof(argv[3]);
21
22 if(a == 0) {
23 printf(""Error: a cannot be zero\n"");
24 exit(1);
25 } else if(b*b < 4*a*c) {
26 printf(""Error: cannot handle complex roots\n"");
27 exit(1);
28 } else if(b*b == 4*a*c) {
29 root1 = -b / (2*a);
30 printf(""Only one distinct root: %f\n"", root1);
31 } else {
32 root1 = (-b + sqrt(b*b - 4*a*c) ) / (2*a);
33 root2 = (-b - sqrt(b*b - 4*a*c) ) / (2*a);
34
35 printf(""The roots of %fx^2 + %fx + %f are: \n"", a, b, c);
36 printf("" root1 = %f\n"", root1);
37 printf("" root2 = %f\n"", root2);
38 }
39 return 0;
40 }
Code Sample 16.4.: Quadratic Roots Program in C With Error Checking
282",ComputerScienceOne.pdf
"17. Loops
C supports while loops, for loops, and do-while loops using the keywords while, for,
and do (along with another while). Continuation conditions for loops are enclosed
in parentheses, (...) and the blocks of code associated with the loop are enclosed in
curly brackets.
17.1. While Loops
Code Sample 17.1 contains an example of a basic while loop in C. Just as with conditional
statements, our code styling places the opening curly bracket on the same line as the
while keyword and continuation condition. The inner block of code is also indented
and all lines in the block are indented to the same level.
In addition, the continuation condition does not contain a semicolon since it is not an
executable statement. Just as with an if-statement, if we had placed a semicolon it would
have led to unintended results. Consider the following:
1 while(i <= 10); {
2 //perform some action
3 i++; //iteration
4 }
1 int i = 1; //Initialization
2 while(i <= 10) { //continuation condition",ComputerScienceOne.pdf
"1 while(i <= 10); {
2 //perform some action
3 i++; //iteration
4 }
1 int i = 1; //Initialization
2 while(i <= 10) { //continuation condition
3 //perform some action
4 i++; //iteration
5 }
Code Sample 17.1.: While Loop in C
283",ComputerScienceOne.pdf
"17. Loops
1 int i = 1;
2 int flag = 1;
3 while(flag) {
4 //perform some action
5 i++; //iteration
6 if(i>10) {
7 flag = 0;
8 }
9 }
Code Sample 17.2.: Flag-controlled While Loop in C
A similar problem occurs: the while keyword and continuation condition bind to
the next executable statement or code block. As a consequence of the semicolon, the
executable statement that gets bound to the while loop is empty. What happens is
even worse: the program will enter an infinite loop. To see this, the code is essentially
equivalent to the following:
1 while(i <= 10) {
2 }
3 {
4 //perform some action
5 i++; //iteration
6 }
In the while loop, we never increment the counter variable i, the loop does nothing,
and so the computation will continue on forever! Some compilers may warn you about
this, others will not. It is valid C and it will compile and run, but obviously won’t work
as intended. Avoid this problem by using proper syntax and good style.",ComputerScienceOne.pdf
"as intended. Avoid this problem by using proper syntax and good style.
Another common use case for a while loop is a flag-controlled loop in which we use a
Boolean flag rather than an expression to determine if a loop should continue or not.
Recall that in C, zero is treated as false and any non-zero numeric value is treated as
true. We can thus create an implicit Boolean flag by using an integer variable and setting
it to 1 for true and 0 for false (when we want the loop to terminate). An example can
be found in Code Sample 17.2.
284",ComputerScienceOne.pdf
"17.2. For Loops
17.2. For Loops
For loops in C use the familiar syntax of placing the initialization, continuation condition,
and iteration on the same line as the keyword for. An example can be found in Code
Sample 17.3.
1 int i;
2 for(i=1; i<=10; i++) {
3 //perform some action
4 }
Code Sample 17.3.: For Loop in C
Semicolons are placed at the end of the initialization and continuation condition, but not
the iteration statement. Just as with while loops, the opening curly bracket is placed on
the same line as the for keyword. Code within the loop body is indented, all at the
same indentation level.
Another observation is that we declared the index variable i prior to the for loop.
Some languages allow you to declare the index variable in the initialization statement,
for example for(int i=1; i<=10; i++). Doing so scopes the index variable to the
loop and so i would be out-of-scope before and after the loop body. This is a nice",ComputerScienceOne.pdf
"loop and so i would be out-of-scope before and after the loop body. This is a nice
convenience and is generally good practice. However, C89 and prior standards do not
allow you to do this; the variable must be declared prior to the loop structure. C99 and
newer standards do allow you to do this and some compilers will be somewhat forgiving
when you use the newer syntax (by supporting their own non-standard extensions to C).
For maximum portability, we’ll follow the older convention.
17.3. Do-While Loops
C supports do-while loops. Recall that the difference between a while loop and a do-while
loop is when the continuation condition is checked. For a while loop it is prior to the
beginning of the loop body and in a do-while loop it is at the end of the loop. This
means that a do-while always executes at least once. An example can be found in Code
Sample 17.4.
Note the syntax and style: the opening curly bracket is again on the same line as the",ComputerScienceOne.pdf
"Sample 17.4.
Note the syntax and style: the opening curly bracket is again on the same line as the
keyword do. The while keyword and continuation condition are on the same line
as the closing curly bracket. In a slight departure from prior syntax, a semicolon does
285",ComputerScienceOne.pdf
"17. Loops
1 int i;
2 do {
3 //perform some action
4 i++;
5 } while(i<=10);
Code Sample 17.4.: Do-While Loop in C
appear at the end of the continuation condition even though it is not an executable
statement.
17.4. Other Issues
C does not support a traditional foreach loop. When iterating over a collection like an
array, you iterate an index variable, typically with a for loop. However, there are some
syntactic tricks that you can use to get the same effect. Some of this will be a preview of
Chapter 20 where we discuss arrays in C, but in short, you can assign a variable to an
element in an array in iteration statement:
1 double arr[] = {1.41, 2.71, 3.14};
2 int n = 3;
3 int i = 0;
4 double x;
5 for(x=arr[0]; i<n; x=arr[++i]) {
6 //x now holds the i-th element in arr
7 }
The initialization sets the variable x to the first element in the array, arr[0]. The
loop continues for as many elements as there are in the array, n. The iteration does two",ComputerScienceOne.pdf
"loop continues for as many elements as there are in the array, n. The iteration does two
things: it assigns x to the next element in the array while at the same time incrementing
the index variable using the prefix increment operator (see Section 2.3.6).
286",ComputerScienceOne.pdf
"17.5. Examples
17.5. Examples
17.5.1. Normalizing a Number
Let’s revisit the example from Section 4.1.1 in which wenormalize a number by continually
dividing it by 10 until it is in the range [1 ,10). The code in Code Sample 17.5 specifically
refers to the value 32145 .234 but would work equally well with any non-negative value of
x.
1 double x = 32145.234;
2 int k = 0;
3 while(x > 10) {
4 x = x / 10; //or: x /= 10;
5 k++;
6 }
Code Sample 17.5.: Normalizing a Number with a While Loop in C
17.5.2. Summation
Let’s revisit the example from Section 4.2.1 in which we computed the sum of integers
1 + 2 +··· + 10. The code is presented in Code Sample 17.6
1 int i;
2 int sum = 0;
3 for(i=1; i<=10; i++) {
4 sum += i;
5 }
Code Sample 17.6.: Summation of Numbers using a For Loop in C
Of course we could easily have generalized the code somewhat. Instead of computing a
sum up to a particular number, we could have written it to sum up to another variable",ComputerScienceOne.pdf
"sum up to a particular number, we could have written it to sum up to another variable
n, in which case the for loop would instead look like the following.
287",ComputerScienceOne.pdf
"17. Loops
1 for(i=1; i<=n; i++) {
2 sum += i;
3 }
17.5.3. Nested Loops
Recall that you can write loops within loops. The inner loop will execute fully for each
iteration of the outer loop. An example of two nested of loops in C can be found in Code
Sample 17.7.
1 int i, j;
2 int n = 10;
3 int m = 20;
4 for(i=0; i<n; i++) {
5 for(j=0; j<m; j++) {
6 printf(""(i, j) = (%d, %d)\n"", i, j);
7 }
8 }
Code Sample 17.7.: Nested For Loops in C
The inner loop executes for j = 0 ,1,2,..., 19 < m = 20 for a total of 20 itera-
tions. However, it executes 20 times for each iteration of the outer loop. Since
the outer loop executes for i = 0 ,1,2,..., 9 < n = 10, the total number of times
the printf() statement execute is 10 ×20 = 200. In this example, the sequence
(0,0),(0,1),(0,2),..., (0,19),(1,0),..., (9,19) will be printed.
17.5.4. Paying the Piper
Let’s adapt the solution for the loan amortization schedule we developed in Section 4.7.3.",ComputerScienceOne.pdf
"17.5.4. Paying the Piper
Let’s adapt the solution for the loan amortization schedule we developed in Section 4.7.3.
First, we’ll read the principle, terms, and interest as command line inputs. Adapting the
formula for the monthly payment and using the standard math library’s pow() function,
we get
288",ComputerScienceOne.pdf
"17.5. Examples
1 double monthlyPayment = (monthlyInterestRate * principle) /
2 (1 - pow( (1 + monthlyInterestRate), -n));
However, recall that we may have problems due to accuracy. The monthly payment
could come out to be a fraction of a cent, say $43.871. For accuracy, we need to ensure
that all of the figures for currency are rounded to the nearest cent. The standard math
library does have a round() function, but it only rounds to the nearest whole number,
not the nearest 100th.
However, we can adapt the “off-the-shelf” solution to fit our needs. If we take the number,
multiply it by 100, we get (say) 4387.1 which we can now round to the nearest whole
number, giving us 4387. We can then divide by 100 to get a number that has been
rounded to the nearest 100th! In C, we could simply do the following.
monthlyPayment = round(monthlyPayment * 100.0) / 100.0;
We can use the same trick to round the monthly interest payment and any other number",ComputerScienceOne.pdf
"monthlyPayment = round(monthlyPayment * 100.0) / 100.0;
We can use the same trick to round the monthly interest payment and any other number
expected to be whole cents. To output our numbers, we use printf() and take care
to align our columns to make it look nice. To finish our adaptation, we handle the
final month separately to account for an over/under payment due to rounding. The full
solution can be found in Code Sample 17.8.
289",ComputerScienceOne.pdf
"17. Loops
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<math.h>
4
5 int main(int argc, char **argv) {
6
7 if(argc != 4) {
8 printf(""Usage: %s principle apr terms\n"", argv[0]);
9 exit(1);
10 }
11
12 double principle = atof(argv[1]);
13 double apr = atof(argv[2]);
14 int n = atoi(argv[3]);
15
16 double balance = principle;
17 double monthlyInterestRate = apr / 12.0;
18 int i;
19
20 //monthly payment
21 double monthlyPayment = (monthlyInterestRate * principle) /
22 (1 - pow( (1 + monthlyInterestRate), -n));
23 //round to the nearest cent
24 monthlyPayment = round(monthlyPayment * 100.0) / 100.0;
25
26 printf(""Principle: $%.2f\n"", principle);
27 printf(""APR: %.4f%%\n"", apr*100.0);
28 printf(""Months: %d\n"", n);
29 printf(""Monthly Payment: $%.2f\n"", monthlyPayment);
30
31 //for the first n-1 payments in a loop:
32 for(i=1; i<n; i++) {
33 // compute the monthly interest, rounded:
34 double monthlyInterest =
35 round( (balance * monthlyInterestRate) * 100.0) / 100.0;",ComputerScienceOne.pdf
"33 // compute the monthly interest, rounded:
34 double monthlyInterest =
35 round( (balance * monthlyInterestRate) * 100.0) / 100.0;
36 // compute the monthly principle payment
37 double monthlyPrinciplePayment = monthlyPayment - monthlyInterest;
38 // update the balance
39 balance = balance - monthlyPrinciplePayment;
40 // print i, monthly interest, monthly principle, new balance
41 printf(""%d\t$%10.2f $%10.2f $%10.2f\n"", i, monthlyInterest,
42 monthlyPrinciplePayment, balance);
43 }
44
45 //handle the last month and last payment separately
46 double lastInterest = round( (balance * monthlyInterestRate) * 100.0) / 100.0;
47 double lastPayment = balance + lastInterest;
48
49 printf(""Last payment = $%.2f\n"", lastPayment);
50
51 return 0;
52 }
Code Sample 17.8.: Loan Amortization Program in C290",ComputerScienceOne.pdf
"18. Functions
As a procedural-style language, functions are essential in C programming. As we’ve already
seen, C provides a large library of standard functions to perform basic input/output,
math, and many other functions. C also provides the ability to define and use your own
functions.
When you define functions in C, careful thought must be made as to the naming of your
functions. This is because C does not support function overloading. When you name a
function, that is the only function that can have that name. Consequently, you cannot,
in general, use the same function names as defined in the standard libraries or any other
3rd party library that you would like to use in your programs.
C supports both call by value and call by reference using pointers (see Section 18.2). C
also supports vararg functions ( printf() being a prime example) and allows you to
define vararg functions, but we will not cover them in depth here. Finally, parameters",ComputerScienceOne.pdf
"define vararg functions, but we will not cover them in depth here. Finally, parameters
are not, in general, optional. For modern versions of C the omission of parameters is a
syntax error. For older versions, complex rules dictate what happens when arguments
are omitted, but doing so usually results in garbage.
18.1. Defining & Using Functions
For modern C, defining functions is a two step process. First, you declare a function and
its signature using a prototype. Then you define the function by providing a function
body that defines what the function does.
18.1.1. Declaration: Prototypes
Just as with variables, functions in C must be declared before they can be used. The
modern way to declare a function in C is to use a prototype declaration which specifies
the function’s signature including its return type, identifier, and parameters. However, a
prototype does not include the actual body of function. Instead, the function’s definition,",ComputerScienceOne.pdf
"prototype does not include the actual body of function. Instead, the function’s definition,
which includes the function body is included later in the program. Consequently, a
prototype always ends with a semicolon.
291",ComputerScienceOne.pdf
"18. Functions
Typically, the documentation for functions is included with the prototype but is not
repeated with the function definition. This is a principle known as Don’t Repeat Yourself
(DRY). Consider the following examples. In these examples we use a commenting style
known as “doc comments.” This style was originally developed for Java but has since
been adopted by many other languages.
1 /**
2 * Computes the sum of the two arguments.
3 */
4 int sum(int a, int b);
5
6 /**
7 * Computes the Euclidean distance between the 2-D points,
8 * (x1,y1) and (x2,y2).
9 */
10 double getDistance(double x1, double y1, double x2, double y2);
11
12 /**
13 * Computes a monthly payment for a loan with the given
14 * principle at the given APR (annual percentage rate) which
15 * is to be repaid over the given number of terms (usually
16 * months).
17 */
18 double getMonthlyPayment(double principle, double apr, int terms);",ComputerScienceOne.pdf
"16 * months).
17 */
18 double getMonthlyPayment(double principle, double apr, int terms);
In each of these, the return type is the first thing specified. The function identifier (name)
is then specified. Function names must follow the same naming rules as variables: they
must begin with an alphabetic character and may contain alphanumeric characters as
well as underscores. Using modern coding conventions we usually name functions using
lower camel casing.
Each prototype ends with a semicolon. Further, prototypes do not specify what the
function does, they only specify its signature. Later in the program, we can provide
the actual definition of each function by using the following syntax. We repeat the
signature, but instead of using a semicolon, we provide a code block, enclosed using
opening/closing curly brackets, that specifies the function body. Here are the definitions
from the prototype examples above:
1 int sum(int a, int b) {
2 return (a + b);
3 }
292",ComputerScienceOne.pdf
"18.1. Defining & Using Functions
4
5 double getDistance(double x1, double y1, double x2, double y2) {
6 double xDiff = (x1-x2);
7 double yDiff = (y1-y2);
8 return sqrt( xDiff * xDiff + yDiff * yDiff);
9 }
10
11 double getMonthlyPayment(double principle, double apr, int terms) {
12 double rate = (apr / 12.0);
13 double payment = (principle * rate) / (1-pow(1+rate, -terms));
14 return payment;
15 }
The keyword return is used to specify the value that is returned to the calling function.
18.1.2. Void Functions
The keyword void can be used in C to indicate a function does not return a value,
in which case it is called a “void function.” Though it is not necessary, it is still good
practice to include a return statement.
1 //prototype:
2 void printCopyright();
3
4 //definition:
5 void printCopyright() {
6 printf(""(c) Bourke 2015\n"");
7 return;
8 }
In the example above, we’ve also illustrated how to define a function that has no inputs.",ComputerScienceOne.pdf
"6 printf(""(c) Bourke 2015\n"");
7 return;
8 }
In the example above, we’ve also illustrated how to define a function that has no inputs.
Some sources may include an explicit void keyword as a parameter to indicate the
function takes no parameters as in void printCopyright(void);.
293",ComputerScienceOne.pdf
"18. Functions
18.1.3. Organizing Functions
The separation of a function declaration (prototype) and a function definition provides a
natural way to organize functions in C. We place prototypes into a header file which has
a file extension .h and then place the corresponding function definitions into a source
file with the file extension, .c.
We’ve seen this before with the standard libraries: we use #include<math.h> to
“include” the math library’s header file in our code. This essentially brings in the math
library function prototypes so that we can write calls to functions like sqrt() or sin().
Only when we compile do we actually need to link our code to the function definitions.
When we separate prototypes into header files and definitions into source files we also
need to “include” the prototypes in our source file just as we would need to include them
in any other file in which we use one of the functions. Suppose our functions above have",ComputerScienceOne.pdf
"in any other file in which we use one of the functions. Suppose our functions above have
their prototypes in a file named utils.h and their definitions in a file named utils.c.
In the utils.c source file we would typically use the following syntax to include the
header file:
#include ""utils.h""
We use the double quote syntax with user-defined libraries while the usual less-than/greater-
than syntax is used with standard libraries. With the less-than/greater-than syntax, the
compiler will attempt to look for the header file(s) in a specified system directory which
it will fail to find if it is a user-defined library. Furthermore, other elements are usually
included in header files such as preprocessor directives and other declarations (such as
enumerated types and structures which we introduce later).
18.1.4. Calling Functions
Once a function has been defined, or at least a prototype has been brought into scope
via an #include statement, you can write code to call your function(s). The syntax",ComputerScienceOne.pdf
"via an #include statement, you can write code to call your function(s). The syntax
for doing so is to simply provide the function name followed by parentheses containing
values or variables to pass to the function. Some examples:
1 int a = 10, b = 20;
2 int c = sum(a, b); //c contains the value 30
3
4 //invoke a function with literal values:
5 double dist = getDistance(0.0, 0.0, 10.0, 20.0);
6
7 //invoke a function with a combination:
8 double p = 1500.0;
294",ComputerScienceOne.pdf
"18.2. Pointers
9 double r = 0.05;
10 double monthlyPayment = getMonthlyPayment(p, r, 60);
By default, all primitive types including int, double, and char are passed by value.
To be able to pass arguments by reference, we need to use pointers.
18.2. Pointers
Consider the following line of C code.
int a = 10;
This line creates an integer variable and sets it equal to 10. In more detail, this line
creates a spot in memory (typically 32 bits) and stores a binary representation of the
value 10 at that location. In many instances, we don’t care where the variable is stored
in memory. However, we may have need to communicate that memory location to other
functions. To do so, we can use pointers.
A pointer in C is a reference to a memory location. Because different types ( int,
double, char) take a different amount of memory, it is necessary to have a pointer for
each type. That is, a pointer that points to a memory location that stores an int or a",ComputerScienceOne.pdf
"each type. That is, a pointer that points to a memory location that stores an int or a
pointer that points to a memory location that stores a double, etc.
The syntax for declaring a pointer is to use an asterisk.
1 //regular variable declarations
2 int a;
3 double b;
4
5 //pointer variable declarations
6 int *ptrA;
7 double *ptrB;
If ptrA represents a memory location, what values can it take on? A memory location
is just a number, so you could do something like the following.
ptrA = 10;
295",ComputerScienceOne.pdf
"18. Functions
Though syntactically this makes sense (and generally the compiler will let you do this with
at most a warning), it is not really what you want. This assigns to the pointer variable
ptrA the value 10, which will be interpreted as the memory address 10. This memory
address may not belong to your program, or it may not even exist as a valid memory
address. Attempts to access the value stored at an arbitrary memory location may be
illegal and may result in the operating system killing the program with a segmentation
fault or similar error.
There are many reasons why a program should not be allowed access to arbitrary memory
locations, but one of the prime reasons is security. Imagine if the operating system
allowed a program access to any part of memory; in particular memory that contained
sensitive information such as passwords or secret Secure Sockets Layer (SSL) keys. To
prevent this, operating systems generally only allow a program to access its own memory.",ComputerScienceOne.pdf
"prevent this, operating systems generally only allow a program to access its own memory.
Referencing Operator
So how do we assign a valid value to a pointer? We do so using a referencing operator.
Given a normal variable as above, we place an ampersand in front of it to get the memory
address of the variable. For example:
1 ptrA = &a;
2 ptrB = &b;
The operation &a results in the memory address of the variable a and we can assign it
to a pointer value. The pointer type and variable type should match; an integer pointer
should point to an int, a double pointer should point to a double. Making a double
pointer point to an int variable type such as
ptrB = &a;
is valid syntax, but since the two types use different amounts of memory you may get
garbage results.
There is a special value used in C called NULL which is a (case-sensitive) keyword used
for an uninitialized, undefined, empty or otherwise invalid or meaningless value. In the",ComputerScienceOne.pdf
"for an uninitialized, undefined, empty or otherwise invalid or meaningless value. In the
context of memory locations, NULL “points” to nothing. As with regular variables, its
best practice to initialize pointer values to NULL. For example,
int *ptrA = NULL;
Without an initialization, the pointer may point to a random memory address which may
be dangerous to attempt to access. You can also test whether or not a pointer points to
296",ComputerScienceOne.pdf
"18.2. Pointers
NULL using the usual equality operator.
1 int *ptrA = NULL;
2 ...
3 if(ptrA == NULL) {
4 printf(""Error: invalid memory location\n"");
5 }
Dereferencing Operator
Once we have a valid pointer to a memory location, we may want to manipulate the
contents of the memory it references. To do this we use the inverse operation, the
dereferencing operator which again uses an asterisk. Given a pointer variable ptrA, we
apply an asterisk in front of it to turn it into a regular variable. Consider the following
example.
1 //declare a normal integer variable
2 int a = 10;
3 //declare a pointer and initialize it to NULL
4 int *ptrA = NULL;
5
6 //point ptrA to a's memory location
7 ptrA = &a;
8
9 //change the value of the variable a using its pointer
10 *ptrA = 20;
11
12 //now the variable a has a value of 20:
13 printf(""a = %d\n"", a); //prints ""a = 20""
Figure 18.1 depicts how these lines of code operate in memory.
18.2.1. Passing By Reference",ComputerScienceOne.pdf
"13 printf(""a = %d\n"", a); //prints ""a = 20""
Figure 18.1 depicts how these lines of code operate in memory.
18.2.1. Passing By Reference
Now that we have the ability to reference a memory location using pointers, we can write
functions that pass variables by reference. To do so, we use the same asterisk syntax
297",ComputerScienceOne.pdf
"18. Functions
0xc260ec80
0xc260ec84 10a
0xc260ec88
0xf289fb14
0xf289fb18ptrA NULL
0xc260ec88
...
Address Contents
(a) After the first two lines memory
has been dedicated for the variable a
and the pointer variable ptrA and
their values have been initialized.
0xc260ec80
0xc260ec84 10a
0xc260ec88
0xf289fb14
0xf289fb18ptrA 0xc260ec84
0xc260ec88
...
Address Contents
ptrA = &a;
(b) Making ptrA point to the variable a ’s
memory location. The value stored in the vari-
able ptrA is a memory address.
0xc260ec80
0xc260ec84 20a
0xc260ec88
0xf289fb14
0xf289fb18ptrA 0xc260ec84
0xc260ec88
...
Address Contents
*ptrA = 20;
(c) Dereferencing ptrA and assigning a value
changes the value stored at what it points to.
Figure 18.1.: Pointer Operations. Pointers can be made to point to other variable’s
memory locations. You can manipulate/access values of variables via their
pointers using dereferencing.
298",ComputerScienceOne.pdf
"18.2. Pointers
used with pointer variables.
1 //prototypes
2 /**
3 * This function sums the first two variables (passed by
4 * value) and places the result into the third variable
5 * (passed by reference).
6 */
7 void sum(int a, int b, int *c);
8
9 /**
10 * This function swaps the values stored in the
11 * two variables passed by reference.
12 */
13 void swap(int *a, int *b);
In the function definitions, we can use the dereferencing operator to access or modify the
value stored in the variable pointed to by the pointer.
1 void sum(int a, int b, int *c) {
2 int x = a + b;
3 *c = x;
4 return;
5 }
6
7 void swap(int *a, int *b) {
8 int temp = *a;
9 *a = *b;
10 *b = temp;
11 return;
12 }
To invoke these functions, we need to pass pointers to the functions appropriately. We
could do this by creating pointer variables or using the referencing operator directly.
1 int x = 10;
2 int y = 20;
3 int c;
299",ComputerScienceOne.pdf
"18. Functions
4 int *ptrC = &c;
5 sum(x, y, ptrC);
6 //at this point c contains the value 30
7
8 swap(&x, &y);
9 //at this point, the values in x and y have been swapped
10 // x contains 20 and y contains 10
This should look familiar. We saw this same syntax when we used scanf() to read input
from the standard input. We needed to place an ampersand in front of each variable in
order to pass the variable by reference so that scanf() could place the results into the
respective memory locations. If the variables had been passed by value, then scanf()
would not have been able to manipulate their values.
You can also specify functions to return pointers which we discuss in detail in Chapter
20.
18.2.2. Function Pointers
Functions are just pieces of code that reside somewhere in memory just as variables do.
Since we can create pointers that point to variables, it makes sense to be able to create
variables that point to functions too! These are referred to as function pointers.",ComputerScienceOne.pdf
"variables that point to functions too! These are referred to as function pointers.
The syntax for declaring function pointers is similar to variable pointers. However, since
a function’s signature involves a return type and parameter list, these need to be specified.
For example, suppose we wanted to create a function pointer that could point to the
math library’s sqrt() function which takes a single double parameter and returns a
double value.
double (*ptrToSqrt)(double) = NULL;
The above line creates a function pointer that can point to any function that takes
a single double parameter and returns a double value. As is good practice, we’ve
initialized it to point to NULL. The function pointer itself is named ptrToSqrt. To
make it point to the sqrt() function we can use the following syntax.
ptrToSqrt = sqrt;
This is because a function’s identifier acts as a pointer as well! Once we have a pointer
to a function, we can invoke the function via its pointer as we would any other function",ComputerScienceOne.pdf
"to a function, we can invoke the function via its pointer as we would any other function
call.
300",ComputerScienceOne.pdf
"18.3. Examples
double x = ptrToSqrt(2.0);
Some more examples:
1 //this pointer can point any function that takes
2 //three arguments: an int, double, and a char
3 //and returns an int value
4 int (*ptrToFunc)(int, double, char)= NULL;
5
6 double x;
7 double (*ptr)(double) = NULL;
8 //we can make it point to sqrt:
9 ptr = sqrt;
10 x = ptr(2.0); //x contains 1.4142...
11 //or we can make it point to fabs
12 ptr = fabs;
13 x = ptr(-10.5); //x contains 10.5
You generally want to create and use function pointers when passing and returning
functions as arguments to other functions as callbacks. We discuss this in further detail
in Chapter 25.
18.3. Examples
18.3.1. Generalized Rounding
Recall that the standard math library provides a round() function that rounds a number
to the nearest whole number. We’ve had need to round to cents as well. We now have
the ability to write a function to do this for us. Before we do, however, let’s think more",ComputerScienceOne.pdf
"the ability to write a function to do this for us. Before we do, however, let’s think more
generally. What if we wanted to round to the nearest tenth? Or what if we wanted to
round to the nearest 10s or 100s place? Let’s write a general purpose rounding function
that allows us to specify which decimal place to round to.
The most natural input values would be to specify the place using an integer exponent.
That is, if we wanted to round to the nearest tenth, then we would pass it −1 as
0.1 = 10−1, −2 if we wanted to round to the nearest 100th, etc. In the other direction,
passing in 0 would correspond to the usual round function, 1 to the nearest 10s spot,
and so on. Moreover, we could demonstrate good code reuse (as well as procedural
abstraction) by scaling the input value and reusing the functionality already provided
301",ComputerScienceOne.pdf
"18. Functions
in the math library’s round() function. We could further define a roundToCents()
function that used our generalized round function.
Let’s also think about organization. We could place the prototypes into a round.h
header file and the corresponding definitions in a round.c source file. The contents of
these two files are presented here:
1 /**
2 * Rounds to the nearest digit specified by the place
3 * argument. In particular to the (10^place)-th digit
4 */
5 double roundToPlace(double x, int place);
6
7 /**
8 * Rounds to the nearest cent
9 */
10 double roundToCents(double x);
1 #include<math.h>
2 #include ""round.h""
3
4 double roundToPlace(double x, int place) {
5 double scale = pow(10, -place);
6 double rounded = round(x * scale) / scale;
7 return rounded;
8 }
9
10 double roundToCents(double x) {
11 return roundToPlace(x, -2);
12 }
Observe that neither of these files contains a main() function. By themselves they",ComputerScienceOne.pdf
"11 return roundToPlace(x, -2);
12 }
Observe that neither of these files contains a main() function. By themselves they
would not be able to be compiled into an executable program. We’ve essentially built a
small library of rounding functions. We could compile them though into a binary object
file using gcc (something like gcc -c round.c). We could then link into the object file
when compiling an executable program that uses these functions.
302",ComputerScienceOne.pdf
"18.3. Examples
18.3.2. Quadratic Roots
Another advantage of passing variables by reference is that we can “return” multiple
values with one function call. Functions are limited in that they can only return at most
one value. But if we pass multiple parameters by reference, the function can manipulate
the contents of them, thereby communicating (though not strictly returning) multiple
values.
Consider again the problem of computing the roots of a quadratic equation,
ax2 + bx+ c= 0
using the quadratic formula,
−b±
√
b2 −4ac
2a
Since there are two roots, we may have to write two functions, one for the “plus” root
and one for the “minus” root both of which take the coefficients, a,b,c as arguments.
However, if we wrote a single function that took the coefficients as parameters by value
as well as two other parameters by reference, we could compute both root values and
place each one in two pass by reference variables.
1 void quadraticRoots(double a, double b, double c,",ComputerScienceOne.pdf
"place each one in two pass by reference variables.
1 void quadraticRoots(double a, double b, double c,
2 double *root1, double *root2) {
3 double discriminant = sqrt(b*b - 4*a*c);
4 *root1 = (-b + discriminant) / (2*a);
5 *root2 = (-b - discriminant) / (2*a);
6 return;
7 }
By using pass by reference variables, we avoid multiple functions. We also note that the
return value in this case is unused since we are “returning” the root values in the two
pass by reference variables. This frees up the return value to be used to communicate
errors to the calling function. Recall that there could be several “bad” inputs to this
function. The roots could be complex values, the coefficient a could be zero, etc. And
now that we are dealing with pointers, the pointers could be invalid (point to NULL). In
the next chapter, we examine how we can use the return value to communicate different
errors to the calling function, letting it handle those errors.
303",ComputerScienceOne.pdf
"19. Error Handling
The C language does not support exceptions or exception handling. Instead, the usual
method of error handling is done through defensive programming. As a user, it is your
responsibility to write code that checks for invalid or unsafe operations before executing
them and handle the error appropriately.
One way that we’ve already seen to handle errors in C is to use the exit() function in
the standard library. This function immediately terminates the execution of our program
and “returns” an integer valued “exit code.” This exit code can be used to communicate
errors between different processes. The process that executed the program can use it to
determine if the program exited with or without an error. By convention, 0 indicates an
error while a non-zero value indicates an error. Using the exit() function should only
be done when an error should be considered fatal or severe enough that the program
should immediately terminate.
19.1. Language Supported Error Codes",ComputerScienceOne.pdf
"be done when an error should be considered fatal or severe enough that the program
should immediately terminate.
19.1. Language Supported Error Codes
In the standard libraries, some functions are designed to indicate an error by returning
a special “flag” value such as −1 or NULL. In general, though, the standard library
functions communicate an error code represented as an integer value. As previously
mentioned, C uses the convention that 0 indicates “no error” while a non-zero value
indicates an error.
Error codes are communicated through a special global variable called errno which is
defined in the errno.h standard header file. This header file also defines several macros
that are identified with various error codes. The C standard actually only requires the
following three error codes to be supported.
• EDOM indicates an error in the domain of a function; that is, an error with respect
to the function’s input value(s). For example, calling the math library’s sqrt(-1)",ComputerScienceOne.pdf
"to the function’s input value(s). For example, calling the math library’s sqrt(-1)
on a negative value results in an EDOM error as C does not support complex
numbers.
• ERANGE indicates an error in the range of a function; that is, an error with respect
to the function’s output value(s). For example, calling the math library’s log(0)
305",ComputerScienceOne.pdf
"19. Error Handling
with a zero value results in an ERANGE error as C does not support −∞as an
actual number.
• EILSEQ indicates an illegal byte sequences in characters on systems that use
UTF-8.
All three of these are defined in the errno.h header file. Depending on the system,
additional error codes may also be defined and supported (see POSIX Error Codes
below).
When an error occurs, a function will set the global variable errno to one of these error
code values. Upon returning from a function, you can check for these error codes. Since
these error codes are represented as integers, you simply use the numerical comparison
operator, ==. You can check for no error by making a comparison to zero.
In addition, the standard string library, defined in the header file, string.h provides a
function, char * strerror(errno) that can be used to map the value in errno to a
human-readable error message. We discuss strings in detail later on, but we can see how",ComputerScienceOne.pdf
"human-readable error message. We discuss strings in detail later on, but we can see how
to use this function in Code Sample 19.1. The output of this program is as follows.
result: 1.4142, error: 0
result: -nan, error: 33
it was an EDOM error
Error Message: Numerical argument out of domain
result: -inf, error: 34
it was an ERANGE error
Error Message: Numerical result out of range
For this particular system, the EDOM and ERANGE error codes were associated with the
integer values 33 and 34 respectively. These numbers are not necessarily the same on all
systems so comparisons must be made against the macro names for portability.
19.1.1. POSIX Error Codes
POSIX is an IEEE set of standards for maintaining compatibility between various
operating systems. It defines several rules and expectations that operating systems must
adhere to in order to be POSIX compliant. Such standards allow developers to develop
code that should be interoperable among a collection of different operating systems,",ComputerScienceOne.pdf
"code that should be interoperable among a collection of different operating systems,
reducing the need for rewrites and reimplementations for different systems.
In particular, POSIX compliant systems define many other error codes (see [ 3]) that
306",ComputerScienceOne.pdf
"19.1. Language Supported Error Codes
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<math.h>
4 #include<string.h>
5 #include<errno.h>
6
7 int main(int argc, char **argv) {
8
9 double a = -1, b = 2, c = 0.0;
10 double x;
11
12 //okay
13 x = sqrt(b);
14 printf(""result: %.4f, error: %d\n"", x, errno);
15
16 //NaN and EDOM error...
17 x = sqrt(a);
18 printf(""result: %.4f, error: %d\n"", x, errno);
19
20 //make a comparison
21 if(errno == EDOM) {
22 printf(""it was an EDOM error\n"");
23 }
24
25 //error messages can be accessed via:
26 char *errMsg = strerror(errno);
27 printf(""Error Message: %s\n"", errMsg);
28
29 //ERANGE error
30 x = log(c);
31 printf(""result: %.4f, error: %d\n"", x, errno);
32
33 if (errno == ERANGE) {
34 printf(""it was an ERANGE error\n"");
35 }
36
37 //error messages can be accessed via:
38 errMsg = strerror(errno);
39 printf(""Error Message: %s\n"", errMsg);
40
41 return 0;
42
43 }
Code Sample 19.1.: Using the errno.h library
307",ComputerScienceOne.pdf
"19. Error Handling
can be used beyond the three mentioned above. For example, the ENOENT error code
corresponds to “No such file or directory” and EACCES corresponds to a “Permission
denied” error.
19.2. Error Handling By Design
In our own code we could communicate errors to calling functions by setting the errno
variable, however, we may run into compatibility issues with the standard error codes
or POSIX error codes. Instead, it may be more appropriate to do error handling by
utilizing the return value of a function to communicate an error code. We could design
our functions to always return an int value to indicate an error: 0 for no error and some
non-zero value to indicate various different types of errors. Of course, as a consequence
any value that needs to be “returned” to the calling function would need to be done so
via a pass by reference variable.
As an example, let’s revisit the quadratic roots example in Section 18.3.2. By returning",ComputerScienceOne.pdf
"via a pass by reference variable.
As an example, let’s revisit the quadratic roots example in Section 18.3.2. By returning
the two output values via pass by reference variables, we freed up the return value. We can
now modify our function to return an integer indicating an error code instead. Previously
we had identified several different types of errors: division by zero (if a= 0), complex
roots (if b2 −4ac <0) and a NULL pointer error if the variables passed by reference
were NULL. We can now modify our function to check for these errors and return an
appropriate error code. We return zero in the event that no error was encountered.
1 int quadraticRoots(double a, double b, double c,
2 double *root1, double *root2) {
3 if(a == 0) {
4 return 1;
5 } else if( b*b - 4*a*c < 0 ) {
6 return 2;
7 } else if(root1 == NULL || root2 == NULL) {
8 return 3;
9 }
10 double discriminant = sqrt(b*b - 4*a*c);
11 *root1 = (-b + discriminant) / (2*a);
12 *root2 = (-b - discriminant) / (2*a);
13 return 0;",ComputerScienceOne.pdf
"9 }
10 double discriminant = sqrt(b*b - 4*a*c);
11 *root1 = (-b + discriminant) / (2*a);
12 *root2 = (-b - discriminant) / (2*a);
13 return 0;
14 }
Now when a function invokes our quadraticRoots() it can check to see what kind of
308",ComputerScienceOne.pdf
"19.3. Enumerated Types
error code it returned and handle the error in whatever way it wants.
There is still an issue, however. The usage of the integers 1, 2, 3 to indicate the various
errors was arbitrary. These are essentially magic numbers that the calling function
would have to deal with by making comparisons with various integers. The numbers
themselves are meaningless and someone using them would need to constantly refer to
documentation to understand which integer corresponded to which error condition.
It would be much better if we could follow the strategy of the errno and define human-
readable identifiers for each error code. We could accomplish this by defining macros,
but another solution is to use an enumerated type.
19.3. Enumerated Types
C allows you to define enumerated types which allow you to define a fixed list of possible
values. For example, the days of the week or months of the year are possible values used",ComputerScienceOne.pdf
"values. For example, the days of the week or months of the year are possible values used
in a date. However, they are more-or-less fixed (no one will be adding a new day of the
week or month any time soon). An enumerated type allows us to define these values
using human-readable keywords (much like the #define macro).
To declare an enumerated type we use the keywords typedef enum (short for type
definition and enumeration). We then provide a comma delimited list of keywords inside
a code block. At the end of the code block we provide the type’s identifier. That is, the
name of the type itself. For example, the names of integer and floating-point types in C
are int and double respectively. This identifier gives our type a name that we can
later use to declare variables of that type. Consider the following example.
1 typedef enum {
2 SUNDAY,
3 MONDAY,
4 TUESDAY,
5 WEDNESDAY,
6 THURSDAY,
7 FRIDAY,
8 SATURDAY
9 } DayOfWeek;",ComputerScienceOne.pdf
"1 typedef enum {
2 SUNDAY,
3 MONDAY,
4 TUESDAY,
5 WEDNESDAY,
6 THURSDAY,
7 FRIDAY,
8 SATURDAY
9 } DayOfWeek;
In this example we’ve defined an enumeration of the days of the week. The name of the
type itself is DayOfWeek and we can now declare variables of this type. The possible
309",ComputerScienceOne.pdf
"19. Error Handling
values it can take are SUNDAY, MONDAY, etc. and we can use these keywords in our
program. For example,
1 DayOfWeek today = MONDAY;
2
3 if(today == SUNDAY || today == SATURDAY) {
4 printf(""It is the weekend!\n"");
5 }
Note the modern naming conventions: the type identifier uses upper camel casing while
the enumerated values follow an uppercase underscore convention. Though our example
does not contain a value with multiple words, if it had, we would have used an underscore
to separate them. Furthermore, enumerated type declarations are usually placed in
separate header files along with function prototype declarations.
Care must be taken when using enumerated types, however. Internally, C simply associates
integers with the values. Thus, in our example, SUNDAY is actually 0, MONDAY is 1,
and SATURDAY is 6. When we do assignments or equality comparisons, we’re actually
just comparing integers. Consequently, a DayOfWeek variable may be assigned values",ComputerScienceOne.pdf
"just comparing integers. Consequently, a DayOfWeek variable may be assigned values
that do not correspond to our enumeration. For example,
DayOfWeek today = 1000;
is valid code and will not (in general) result in any compiler errors or warnings, even
though it is assigning an invalid value to the variable. You should only assign valid values
to an enumerated type variable. Proper error checking should also be done.
Despite this limitation, using enumerated types in C provides an obvious advantage.
Without an enumerated type we’d be forced to use a collection of magic numbers to
indicate values. Even for something as simple as the days of the week we’d be constantly
trying to remember: which day is Wednesday again? I forget, does our week start with
Monday or Sunday? Etc. By using an enumerated type these questions are mostly moot
as we can use the more human-readable keywords and eliminate the guess work.
19.4. Using Enumerated Types for Error Codes",ComputerScienceOne.pdf
"as we can use the more human-readable keywords and eliminate the guess work.
19.4. Using Enumerated Types for Error Codes
Let’s apply enumerated types to our quadraticRoots() function example from before.
First, we define our enumerated type which includes all types of errors including a
NO_ERROR type. Note that we start with NO_ERROR as its value will be zero following
our convention.
310",ComputerScienceOne.pdf
"19.4. Using Enumerated Types for Error Codes
1 typedef enum {
2 NO_ERROR,
3 DIV_BY_ZERO_ERROR,
4 COMPLEX_ROOT_ERROR,
5 NULL_POINTER_ERROR
6 } ErrorCode;
Now in the quadraticRoots() function, we can return the appropriate error code as
an enumerated type value.
1 ErrorCode quadraticRoots(double a, double b, double c,
2 double *root1, double *root2) {
3 if(a == 0) {
4 return DIV_BY_ZERO_ERROR;
5 } else if( b*b - 4*a*c < 0 ) {
6 return COMPLEX_ROOT_ERROR;
7 } else if(root1 == NULL || root2 == NULL) {
8 return NULL_POINTER_ERROR;
9 }
10 double discriminant = sqrt(b*b - 4*a*c);
11 *root1 = (-b + discriminant) / (2*a);
12 *root2 = (-b - discriminant) / (2*a);
13 return NO_ERROR;
14 }
311",ComputerScienceOne.pdf
"20. Arrays
C allows you to declare and use arrays. Since C is statically typed, arrays must also
be typed when they are declared and may only hold that particular type of element. C
supports the use of both static arrays and dynamic arrays through standard library calls.
20.1. Basic Usage
To declare a static array, you use syntax similar to declaring a regular variable, but you
use square brackets to designate the variable as an array. Within the square brackets
you indicate the size (number of elements) in the array. For example,
1 int arr[5];
2 double values[10];
The two declarations above create arrays of size 5 and 10 respectively. The array types
are also defined using the usual keywords. In this case, arr can only hold int values
and values can only hold double values. Arrays follow the same naming rules and
conventions as regular variables. Many times, identifiers are made plural as an array
naturally holds more than one value.",ComputerScienceOne.pdf
"conventions as regular variables. Many times, identifiers are made plural as an array
naturally holds more than one value.
Another way to declare an array is to use the compound declaration and assignment
syntax whereby you can initialize an array to hold a certain list of values.
1 int a[] = { 2, 3, 5, 7, 11 };
Using this syntax we do not need to specify the size of the array as the compiler is smart
enough to count the number of elements we’ve provided. The elements themselves are
denoted inside curly brackets and delimited with commas.
313",ComputerScienceOne.pdf
"20. Arrays
Variable Length Arrays
C99 introduced Variable Length Arrays (VLAs, which are also supported in GNU C89)
which allow you to declare a static array whose size is determined by a variable. For
example,
1 int n = 5;
2 int arr[n];
or within a function,
1 void foo(int n) {
2 int arr[n];
3 ...
4 }
In either case, care must be taken as static arrays are allocated on the stack. The stack
is generally small and allocating even a moderately large array on the stack may lead to
a stack overflow. In addition, VLAs are not supported in any C++ standard and should
be avoided if portable code is desired.
Indexing
Once an array has been created, its elements can be accessed using indexing. C uses the
standard 0-indexing scheme so the first element is at index 0, the second at index 1, etc.
Indexing an element involves using the square bracket notation and providing an index.
Once indexed, an array element can be treated as a normal variable and can be used",ComputerScienceOne.pdf
"Once indexed, an array element can be treated as a normal variable and can be used
with other operators such as the assignment operator or comparison operators.
1 arr[0] = 42;
2 if(arr[4] < 0) {
3 printf(""negative!\n"");
4 }
5 printf(""arr[1] = %d\n"", arr[1]);
314",ComputerScienceOne.pdf
"20.1. Basic Usage
Recall that an index is actually an offset. The compiler and system know exactly how
many bytes each int element takes and so an index i calculates exactly how many
bytes from the first element the i-th element is located at. Consequently it is possible
to index elements that are beyond the range of the array. For example, arr[-1] or
arr[5] would attempt to access an element immediately before the first element and
immediately after the last element. Obviously, these elements are not part of the array.
If these out-of-bound elements represent a memory space that does not belong to our
program, then it is likely that a segmentation fault will occur and terminate our program.
However, it is also likely that the memory space surrounding our array does belongs to
our program. Still, accessing those blocks of memory may not give us meaningful values
and modifying them could corrupt other variable values or generally lead to undefined",ComputerScienceOne.pdf
"and modifying them could corrupt other variable values or generally lead to undefined
behavior. It is our responsibility as programmers to write code that does not go beyond
the bounds of an array.
How can we keep track of the size of the array so that we do not access elements after the
last element? With C the responsibility again falls to us. We must always have secondary
variables that store the size of an array. If we pass an array to a function (see below), we
must also pass a “size” parameter so that the function knows how big the array is.
Iteration
C provides no foreach loop to iterate over an array. Instead, the most natural way to
iterate over the elements is a normal for loop that increments an index variable.
1 int i, n = 10;
2 int arr[n];
3 for(i=0; i<n; i++) {
4 arr[i] = 5 * i;
5 }
The for loop above initializes the variable i to zero, corresponding to the first element
in the array. The continuation condition is specifies that the loop continues while i is",ComputerScienceOne.pdf
"in the array. The continuation condition is specifies that the loop continues while i is
strictly less than the size of the array. This iteration for loop is idiomatic when dealing
with arrays. An alternative was presented in Section 17.4.
315",ComputerScienceOne.pdf
"20. Arrays
20.2. Dynamic Memory
Recall that static arrays have many shortcomings (see Section 7.2). In general they
should be avoided since stack space is limited and they cannot be returned from functions.
Fortunately, C provides several standard library functions that facilitate the creation
and management of dynamically allocated arrays.
The primary function to allocate memory is the memory allocation function, malloc():
void * malloc(size_t size);
which takes a single parameter, the number of bytes that you wish to allocate. The type,
size_t is an alias for an unsigned integer type, so you can think of it as an integer.
Thus, malloc(200) would allocate 200 bytes and return a pointer to the memory space.
In some instances, the allocation may fail. For example, if the program or system has
run out of available memory or you simply request too much. In the event of a failure,
malloc() returns a NULL pointer. The returned value can thus be checked to see if",ComputerScienceOne.pdf
"malloc() returns a NULL pointer. The returned value can thus be checked to see if
the allocation was successful or not.
We don’t have to manually calculate how many bytes we need for an array of a particular
size. The macro sizeof() can be used to determine the number of bytes any type of
variable requires on a system. For example, sizeof(int) gives the number of bytes an
int takes while sizeof(double) gives the number of bytes for a double, etc. Thus,
if we want to allocate an array of 100 integers, we could call malloc() as
malloc(100 * sizeof(int));
Using sizeof() is actually preferable as some systems may use a different number of
bytes for various types.
Finally, note the return type of malloc(): it is a void pointer. The malloc() function
simply allocates chunks of memory. It doesn’t care that you intend to use the memory to
store integers or floating-point numbers. Thus, malloc() returns a “generic” pointer,",ComputerScienceOne.pdf
"store integers or floating-point numbers. Thus, malloc() returns a “generic” pointer,
simply an address in memory. Once we have that pointer we can treat it as an integer
pointer, int * or a floating-point pointer, double * depending on what we want to
store.
One way of doing this is to explicitly cast the void pointer as the pointer that we want.
Some examples:
1 int *arr = NULL;
2 double *values = NULL;
3
4 arr = (int *) malloc(sizeof(int) * 10);
316",ComputerScienceOne.pdf
"20.2. Dynamic Memory
5 values = (double *) malloc(sizeof(double) * 100);
The pointer cast is just like when we casted int types as double types so that we
could perform division without truncation. In this case, we convert the returned generic
void pointer into a int pointer and double pointer respectively.1
Once created, a dynamic array can be used just like a static array. You use the array’s
identifier as well as an index to access or modify each element. The same rules and
pitfalls apply, so care must be taken to not access elements outside the bounds of the
array. Finally, when using malloc() it is important to understand that the memory
that is allocated is uninitialized. Just as with variables you cannot make any assumptions
as to the contents of the memory space that is allocated. It may contain garbage values,
it may contain the contents that occupied the memory last time it was used, etc. A full
example:
1 int n = 100;
2 int *arr = NULL;",ComputerScienceOne.pdf
"it may contain the contents that occupied the memory last time it was used, etc. A full
example:
1 int n = 100;
2 int *arr = NULL;
3 arr = (int *) malloc(sizeof(int) * n);
4 if(arr == NULL) {
5 fprintf(stderr, ""unable to allocate memory!\n"");
6 exit(1);
7 }
8 for(i=0; i<n; i++) {
9 arr[i] = (i+1) * 10;
10 printf(""a[%d] = %d\n"", i, arr[i]);
11 }
There are other functions that can be used to allocate memory in C. For example,
calloc() is similar to malloc() but initializes the contents to zero (null bytes).
realloc() can be used to resize an existing memory space (though it may fail if it
cannot be expanded).
1Performing an explicit pointer cast is actually not necessary in C as the type system will do an implicit
cast for us. Whether explicit or implicit types casts are “better” can evoke debates akin to nerd
holy wars. Though there are advantages and disadvantages to both [ 15], we perform explicit pointer",ComputerScienceOne.pdf
"holy wars. Though there are advantages and disadvantages to both [ 15], we perform explicit pointer
casts in this book as clarity and intent is more important than brevity. Even more important is that
explicit pointer casts are necessary in C++ so doing so in our C code makes our code more portable.
317",ComputerScienceOne.pdf
"20. Arrays
Deallocation
Once dynamically allocated memory is no longer needed, we should release it so that
it can be reused by the program or the operating system. The free() function in the
standard library does this for us. All we need to do is provide the pointer to free()
and it deallocates the memory block.
1 free(arr);
Once freed, accessing old memory pointed to by the arr pointer is undefined behavior
and may lead to unexpected or fatal results.
20.3. Using Arrays with Functions
Since arrays are represented using pointers, we can pass and return arrays to and from
functions simply by passing in a pointer. When passing an array to a function, we also
need to make sure that we communicate to the function the size of the array as there is
no reliable way for the function to determine this. By passing both a pointer and an
integer representing the size, we are implicitly indicating that we are passing an array
and not just a single value.",ComputerScienceOne.pdf
"integer representing the size, we are implicitly indicating that we are passing an array
and not just a single value.
As an example, the following function takes an array of integers and computes their sum.
1 /**
2 * This function computes the sum of elements in the
3 * given array which contains n elements
4 */
5 int computeSum(int *arr, int n) {
6 int i;
7 int sum = 0;
8 for(i=0; i<size; i++) {
9 sum += arr[i];
10 }
11 return sum;
12 }
318",ComputerScienceOne.pdf
"20.4. Multidimensional Arrays
In this example we had no need to make changes to any of the elements in the array.
However, the array was still passed by reference, meaning we could have. When passing
arrays, we can use the keyword const (short for constant) to explicitly indicate that no
changes will be made to the array elements. For example,
int computeSum(const int *arr, int n)
This is enforced by the compiler: if we do attempt to make changes to a const array, it
will be a compiler error.
We can also create an array in a function and return it as a value. As previously discussed,
we will need to do so by creating a dynamically allocated array. For example, the following
function creates a deep copy of the given integer array. That is, a completely new array
that is a distinct copy of the old array. In contrast, a shallow copy would be if we simply
made one reference point to another reference.
1 /**
2 * This function creates a new copy of the given integer",ComputerScienceOne.pdf
"made one reference point to another reference.
1 /**
2 * This function creates a new copy of the given integer
3 * array which contains n elements and returns a pointer
4 * to the new copy.
5 */
6 int * makeCopy(const int *a, int n) {
7 int *copy = (int *)malloc(sizeof(int) * n);
8 int i;
9 for(i=0; i<n; i++) {
10 copy[i] = a[i];
11 }
12 return copy;
13 }
The function returns an integer pointer. Here we have a similar problem with respect to
the size of the array. We only have one return value, which must be the pointer. In this
particular example the calling function knows how big the array is because it specified
the size by passing in n. In general we could have communicated the size of the returned
array through another pass by reference integer variable.
20.4. Multidimensional Arrays
C supports multidimensional arrays both static and dynamically allocated; though we’ll
only focus on dynamically allocated 2-dimensional arrays. Conceptually, a 2-dimensional
319",ComputerScienceOne.pdf
"20. Arrays
array of, say, integers can be modeled as an array of pointers that point to an array of
integers. That is, a pointer to pointers, for example, int **. The initial pointer points
to an array of integer pointers, int * and each integer pointer points to an array of
int variables.
To understand this better, let’s look at the details of allocation. Consider the following
code.
1 int i;
2 int **myMatrix = NULL;
3 myMatrix = (int **)malloc(n * sizeof(int*));
4 for(i=0; i<n; i++) {
5 myMatrix[i] = (int *)malloc(n * sizeof(int));
6 }
Line 3 invokes malloc() to create an array of n integer pointers, that is int * types.
We then go into a loop to iterate over each of these pointers and invoke malloc()
again to setup each array. This process is visualized in Figure 20.1. Note the syntax:
when we invoke malloc() to create an array of pointers, we use sizeof(int*) to
determine how many bytes each integer pointer takes. We also do an explicit type cast",ComputerScienceOne.pdf
"determine how many bytes each integer pointer takes. We also do an explicit type cast
of (int **) to match our pointer-to-pointer(s) variable myMatrix.
Once the allocation has completed, we can treat myMatrix as a normal 2-dimensional
array and index each element with two index variables.
1 int i, j;
2 for(i=0; i<n; i++) {
3 for(j=0; j<n; j++) {
4 myMatrix[i][j] = 0;
5 }
6 }
To deallocate and free multidimensional arrays, we need to work backwards. If we imme-
diately freed the pointer-to-pointers, free(myMatrix);, we would lose all references to
the individual array “rows” and would not be able to free them, resulting in a memory
leak. We need to free up each row first, then we can free the pointer-to-pointers.
320",ComputerScienceOne.pdf
"20.4. Multidimensional Arrays
**myMatrix
*myMatrix[0] ?
*myMatrix[1] ?
*myMatrix[2] ?
...
*myMatrix[n-1]?
(a) Initialization of the pointer-to-pointers. The first invocation of malloc() sets up an
array of integer pointers, int * that are uninitialized (where they point to is undefined).
**myMatrix
*myMatrix[0] myMatrix[0][0]myMatrix[0][1]myMatrix[0][2]··· myMatrix[0][n-1]
*myMatrix[1] ?
*myMatrix[2] ?
...
*myMatrix[n-1]?
(b) Initialization of the first pointer. On the first iteration of the for loop when i = 0, the
first “row” is initialized when malloc() is invoked.
**myMatrix
*myMatrix[0] myMatrix[0][0]myMatrix[0][1]myMatrix[0][2]··· myMatrix[0][n-1]
*myMatrix[1] myMatrix[1][0]myMatrix[1][1]myMatrix[1][2]··· myMatrix[1][n-1]
*myMatrix[2] ?
...
*myMatrix[n-1]?
(c) Initialization of the second pointer. On the second iteration of the for loop when i = 1,
the second “row” is initialized.
**myMatrix
*myMatrix[0] myMatrix[0][0]myMatrix[0][1]myMatrix[0][2]··· myMatrix[0][n-1]",ComputerScienceOne.pdf
"the second “row” is initialized.
**myMatrix
*myMatrix[0] myMatrix[0][0]myMatrix[0][1]myMatrix[0][2]··· myMatrix[0][n-1]
*myMatrix[1] myMatrix[1][0]myMatrix[1][1]myMatrix[1][2]··· myMatrix[1][n-1]
*myMatrix[2] myMatrix[2][0]myMatrix[2][1]myMatrix[2][2]··· myMatrix[2][n-1]
...
*myMatrix[n-1]myMatrix[n-1][0]myMatrix[n-1][1]myMatrix[n-1][2]··· myMatrix[n-1][n-1]
(d) After the termination of the for loop, each row has been initialized and the pointer-to-
pointers can be treated like a 2-dimensional array.
Figure 20.1.: The process of dynamically allocating a 2-dimensional array of integers
using malloc() in a for-loop.
321",ComputerScienceOne.pdf
"20. Arrays
1 for(i=0; i<n; i++) {
2 free(myMatrix[i]);
3 }
4 free(myMatrix);
20.4.1. Contiguous 2-D Arrays
The example depicted in Figure 20.1 constructs a two dimensional array. However, each
“row” of the array was created using an independent call to malloc() which may result
in non-contiguous memory blocks (each row may not be located in the memory address
immediately following the previous row). In general, this is not a problem for most
situations (we can expect most implementations of malloc() to be efficient). However,
it might be better in some scenarios if the two dimensional array were all one big block
of contiguous memory.
We can achieve this by a single call to malloc() and some clever pointer manipulation.
We do this very similar to the previous example. Our first call to malloc() to set up
the pointer-to-pointers is the same. However, instead of calling malloc() over and over
in a for loop, we make one single call asking for a number of bytes to accommodate the",ComputerScienceOne.pdf
"in a for loop, we make one single call asking for a number of bytes to accommodate the
entire n×m array. We then store the result at the “beginning” of the array (at index
zero). In the following example, we will create a 4 ×3 sized matrix.
1 int **m = (int **) malloc(sizeof(int *) * 4);
2 m[0] = (int *) malloc(sizeof(int) * (4 * 3));
We’re not done yet, however. We still need to initialize all of the other pointers. To do
so, we dereference the array (once) to get the address of the start of the memory block
and then compute an offset from this beginning on where the next “row” should be.
Since each row has 3 elements, this arithmetic is simple.
1 for(i=1; i<4; i++) {
2 arr[i] = (*arr + (3 * i));
3 }
322",ComputerScienceOne.pdf
"20.4. Multidimensional Arrays
Now we can treat the array like we would any other two dimensional array by specifying
two indices.
1 for(i=0; i<4; i++) {
2 for(j=0; j<3; j++) {
3 arr[i][j] = 10 * i + j;
4 }
5 }
Which would result in an array that, conceptually, looks something like the following.
[ 0 1 2 ]
[ 10 11 12 ]
[ 20 21 22 ]
[ 30 31 32 ]
Which are all stored in one large chunk of memory. To see this, we can print out the
memory address of each “row” during a particular run of the program:
arr[0] = 0x7fe51b403270
arr[1] = 0x7fe51b40327c
arr[2] = 0x7fe51b403288
arr[3] = 0x7fe51b403294
Note that each hexadecimal value differs by 0x0c (that is, 12 bytes):
0x...70 + 0x0c = 0x...7c
0x...7c + 0x0c = 0x...88
0x...88 + 0x0c = 0x...94
This is exactly how many bytes the 3 integers (4 bytes each) take in memory in each
“row.” The pointer setup is depicted in Figure 20.2.
323",ComputerScienceOne.pdf
"20. Arrays
**m
m[0] m[1] m[2] m[3]
m[0][0]m[0][1]m[0][2]m[1][0]m[1][1]m[1][2]m[2][0]m[2][1]m[2][2]m[3][0]m[3][1]m[3][2]
Figure 20.2.: Contiguous Two Dimensional Array. The initial pointer-to-pointers m is
initialized to an array of pointers, m[0]..m[3]. A 4 ×3 contiguous array
is created and each of the pointers is initialized to the proper location so it
can be treated as a two-dimensional array.
20.5. Dynamic Data Structures
The C standard library does not provide a variety of advanced, dynamic data structures
such as linked lists or sets. As of C89 and POSIX.1 the C standard library does provide
hash tables and binary search trees, but does not require implementations of other dynamic
data structures. Instead, you must either implement your own or utilize a third-party
library such as glibc, the GNU C library ( http://www.gnu.org/software/libc/).
324",ComputerScienceOne.pdf
"21. Strings
C has no built-in string type. Instead, strings are represented as arrays of char elements.
They differ from, say arrays of int or double types, however, in that they are null
terminated arrays. The end of the string must always be denoted with a null-terminating
character, '\0' (the 0 valued character in the ASCII table). Failure to properly
null-terminate a string may lead to undefined behavior or fatal errors.
C provides a good variety of functions in its standard string library (included in the
string.h header). All of these functions operate under the assumption that the strings
passed to it are null-terminated. It is your responsibility to ensure that they are. Moreover,
some of the functions that operate on strings will inset the null-terminating character for
us, but others will not. It is important to understand the expectations and guarantees of
each function.
21.1. Character Arrays",ComputerScienceOne.pdf
"us, but others will not. It is important to understand the expectations and guarantees of
each function.
21.1. Character Arrays
We’ve previously used string literals when using the standard input and output functions,
printf() and scanf(). You can also create static strings (as static arrays) using the
standard assignment operator. Some examples:
1 char firstName[] = ""Tom"";
2 char lastName[] = ""Waits"";
index
char *s H
0
e
1
l
2
l
3
o
4
\0
5
?
6
?
7
?
8
?
9
Figure 21.1.: A string in C is achieved by using a char array. However, the string is
terminated by a null-terminating character, \0. Though an array may have
space for additional characters, they are irrelevant if the null terminator
precedes them.
325",ComputerScienceOne.pdf
"21. Strings
This syntax can only be used when creating static strings (they are allocated on the
stack and locally scoped). The compiler is able to scan the string literal and determine
how many characters are needed and even inserts the null-terminating character for us.
Thus, the length of the two strings in the example are 3 and 5 respectively, but the array
size created for them will be 4 and 6 respectively to accommodate the null-terminating
character.
Static strings have the same limitations as static arrays. Since they are allocated on the
stack, they cannot be returned from a function. Just as with int and double types,
however, we can dynamically allocate memory to hold char types using malloc(). The
important difference being that we need to always allocate at least one more character
to accommodate the null-terminating character.
1 char *fullName = NULL;
2 fullName = (char *) malloc(sizeof(char) * 10);",ComputerScienceOne.pdf
"to accommodate the null-terminating character.
1 char *fullName = NULL;
2 fullName = (char *) malloc(sizeof(char) * 10);
This example creates a dynamically allocated char array that is able to hold 9 characters
(since one will be needed for the null-terminating character). However, once we have
created the array, we cannot simply assign an entire string to it using the usual assignment
operator.
1 //THIS IS *WRONG*:
2 fullName = ""Tom Waits"";
This will compile, but doesn’t give us what we want. Recall that fullName is a character
pointer. Using the assignment operator simply makes it point to the static, literal string
""Tom Waits"". In fact, we lose our reference to the dynamically allocated array that
was created with malloc(), resulting in a memory leak.
Instead, we need to use a function in the standard library to copy a string. The function
in the standard library that allows you to copy strings is as follows.
char *strcpy(char *dest, const char *src);",ComputerScienceOne.pdf
"in the standard library that allows you to copy strings is as follows.
char *strcpy(char *dest, const char *src);
The name, strcpy is short for “string copy.”1 Both arguments are character pointers.
In keeping with the use of the assignment operator, the second argument (“source”)
is copied into the first, “destination” (just as with an assignment operator, the value
1The abbreviations are mostly historic: in the 70s and 80s when memory was measured in kilobytes,
saving a few characters made a significant difference.
326",ComputerScienceOne.pdf
"21.2. String Library
on the right-hand-side is copied into the variable on the left-hand-side). Moreover, the
second argument has been marked as const indicating that it will not be changed.
The contents of the first argument will be changed since we are copying a string into
it, erasing whatever contents it had prior. Finally, the function returns a pointer to the
dest argument, mostly so that it can be used in nested function calls (though we’ll
avoid such confusingly terse “tricks”). From our example:
1 //""assign"" (copy) ""Tom Waits"" into the string fullName:
2 strcpy(fullName, ""Tom Waits"");
In addition, we can access and modify individual characters in a string using the usual
indexing and assignment operator with char literals.
1 //access individual characters:
2 char firstInitial = fullName[0]; //'T'
3 char lastInitial = fullName[4]; //'W'
4
5 //printing:
6 printf(""First Initial: %c\n"", fullName[0]);
7
8 //modifying:
9 firstInitial[0] = 't';
10 firstInitial[4] = 'w';",ComputerScienceOne.pdf
"4
5 //printing:
6 printf(""First Initial: %c\n"", fullName[0]);
7
8 //modifying:
9 firstInitial[0] = 't';
10 firstInitial[4] = 'w';
11 //fullName is now ""tom waits""
21.2. String Library
The standard string library provides many other convenient functions that allow you to
process and modify strings. We highlight a few of the more common ones here. A full
list of supported functions can be found in standard documentation.
Length
As with regular arrays, we are responsible for ensuring that we do not access characters
outside the character array of a string. Since a string is null-terminated, there is a nice
327",ComputerScienceOne.pdf
"21. Strings
function provided by the string library to determine its length,
size_t strlen(const char *s);
Recall that size_t can essentially be treated as an integer, indicating the number of
bytes in the passed string. Since a character is a single byte, this function tells us how
many character are in the given string not including the null-terminating string. This
function is an abbreviation for “string length.” Using this function we can easily iterate
over each character in a string.
1 int i;
2 for(i=0; i<strlen(fullName); i++) {
3 printf(""fullName[%d] = %c\n"", i, fullName[i]);
4 }
Concatenation
A concatenation function is provided that allows you to append one string to the end of
another.
char *strcat(char *dest, const char *src);
Similar to the strcpy() function, strcat() appends the “source” ( src) string to
the end of the “destination” ( dest) string. It is your responsibility to ensure that the",ComputerScienceOne.pdf
"the end of the “destination” ( dest) string. It is your responsibility to ensure that the
destination string is large enough to accommodate the source string. If it is not, then it
could lead to undefined behavior as strcat() overwrites memory after the end of the
destination string. Further, strcat() will copy the null-terminating character for you
so that the resulting string (now stored in dest) is a valid null-terminated string.
1 char *formattedName = (char *)malloc(11 * sizeof(char));
2 //copy ""Waits"" into formattedName
3 strcpy(formattedName, lastName);
4 //append a "", "" to formattedName
5 strcat(formattedName, "", "");
6 //append the firstName to the formattedName
7 strcat(formattedName, firstName);
8 //formattedName now contains ""Waits, Tom""
328",ComputerScienceOne.pdf
"21.2. String Library
Byte-Limited Versions
C also provides several byte-limited versions of the copy and concatenation functions:
char *strncpy(char *dest, const char *src, size_t n);
char *strncat(char *dest, const char *src, size_t n);
They work similarly in that they copy/concatenate the source, src string into the
destination, dest string. However, there is a third parameter, n which specifies at
most how many bytes to copy/concatenate. The parameter, n allows you to limit the
number of characters that the operation uses. If either of these functions encounters the
null-terminating character before n bytes have been copied/concatenated, they stop and
copy the null-terminating character for us.
However, these functions do not always handle the null-terminating character for us. If
the null-terminating character appears in the first n bytes, it will be copied for us, but
if it is not, we need to be sure to handle it ourselves. For example,
1 char email[] = ""twaits@cse.unl.edu"";",ComputerScienceOne.pdf
"if it is not, we need to be sure to handle it ourselves. For example,
1 char email[] = ""twaits@cse.unl.edu"";
2 char *login = (char *) malloc(7 * sizeof(char));
3
4 //only copy the first 6 characters:
5 strncpy(login, email, 6);
6 //at this point, login contains ""twaits"" but is
7 //not null-terminated, so we manually terminate it
8 login[6] = '\0';
Computing a Substring
The byte-limited versions of the copy function can be used to compute a substring of
another string. The parameter n can be used to limit the length of the string, but how
might we specify where the substring should start?
For example, in the string ""Thomas Alan Waits"" we may want to get the substring
representing his middle name, ""Alan"". The length is 4 and we can specify that the
copying should start by using an index. In this case, we want the copying to start at
index 7, the 8th character. If the string is stored in an array named name, this would",ComputerScienceOne.pdf
"index 7, the 8th character. If the string is stored in an array named name, this would
be name[7]. However, indexing an array like this results in a single character and
strncpy() expects a string (a character pointer). Fortunately, we know how to do this:
329",ComputerScienceOne.pdf
"21. Strings
using the referencing operator, we can turn the 8th character into a character pointer,
&name[7]. A full example:
1 char name[] = ""Thomas Alan Waits"";
2 char *middleName = (char *) malloc(sizeof(char) * 5);
3 strncpy(middleName, &name[7], 5);
4 middleName[4] = '\0';
5 //middleName now holds ""Alan""
In the call to malloc(), we included space for a null-terminating character. Furthermore,
strncpy() does not insert the null-terminating character for us in this case. In line 4
we manually insert it.
21.3. Arrays of Strings
We often need to deal with collections of strings. An array of strings can be viewed as a
2-dimensional character array. Indeed, we’ve seen this before. In the typical C program’s
main() function, command line arguments are passed as the second parameter:
int main(int argc, char **argv)
which is represented as a double character pointer, char **. Conceptually, each row
is a string, char *. The first parameter, argc indicates how many rows there are.",ComputerScienceOne.pdf
"is a string, char *. The first parameter, argc indicates how many rows there are.
The main() function doesn’t need to be told how big each “row” is since strings are
null-terminated.
We can create our own arrays of strings similar to how we created 2-dimensional arrays
of int and double types.
1 //create an array that can hold 5 strings
2 char **names = (char **) malloc(5 * sizeof(char*));
3 int i;
4 for(i=0; i<5; i++) {
5 //each string can hold at most 19 characters
6 names[i] = (char *) malloc(sizeof(char) * 20);
7 }
8 strcpy(names[0], ""Margaret Hamilton"");
9 strcpy(names[1], ""Ada Lovelace"");
10 strcpy(names[2], ""Grace Hopper"");
330",ComputerScienceOne.pdf
"21.4. Comparisons
11 strcpy(names[3], ""Marie Curie"");
12 strcpy(names[4], ""Hedy Lamarr"");
21.4. Comparisons
When comparing strings in C, we cannot use the numerical comparison operators such
as ==, or <. Because strings are represented as arrays, using these operators actually
compares the variable’s memory addresses.
1 char *a = (char *) malloc(sizeof(char) * 13);
2 char *b = (char *) malloc(sizeof(char) * 13);
3 strcpy(a, ""Hello World!"");
4 strcpy(b, ""Hello World!"");
5
6 if(a == b) {
7 printf(""strings match!\n"");
8 }
The code above will not print anything even though the strings a and b have the
same content. This is because a == b is comparing the memory address of the two
variables. Since they point to different memory addresses (created by two separate calls
to malloc()) they are not equal.
The C string library provides a standard comparator function to compare strings based
on their content:
int strcmp(const char *a, const char *b);",ComputerScienceOne.pdf
"The C string library provides a standard comparator function to compare strings based
on their content:
int strcmp(const char *a, const char *b);
The function takes two strings and returns an integer based on the lexicographic ordering
of a and b. If a precedes b, strcmp() returns something negative. It returns zero if
a and b have the same content. Otherwise it returns something positive if b precedes
a. Some examples:
1 int x;
2 x = strcmp(""apple"", ""banana""); //x is negative
3 x = strcmp(""zelda"", ""mario""); //x is positive
331",ComputerScienceOne.pdf
"21. Strings
4 x = strcmp(""Hello"", ""Hello""); //x is zero
5
6 //shorter strings precede longer strings:
7 x = strcmp(""apple"", ""apples""); //x is negative
8 //uppercase precede lowercase:
9 x = strcmp(""Apple"", ""apple""); //x is negative
In the last example, ""Apple"" precedes ""apple"" since uppercase letters are ordered
before lowercase letters according to the ASCII table. We can also make comparisons
ignoring case if we need to using the alternative:
int strcasecmp(const char *s1, const char *s2);
which is a case-insensitive version. Here, strcasecmp(""Apple"", ""apple"") will return
zero as the two strings are the same ignoring the cases.
The comparison functions also have byte-limited versions,
int strncmp(const char *s1, const char *s2, size_t n);
and
int strncasecmp(const char *s1, const char *s2, size_t n);
Both will only make comparisons in the first n bytes of the strings. Thus, the comparison,
strncmp(""apple"", ""apples"", 5) will result in zero as the two strings are equal in",ComputerScienceOne.pdf
"strncmp(""apple"", ""apples"", 5) will result in zero as the two strings are equal in
the first 5 bytes.
21.5. Conversions
We’ve previously examined the functions atoi() and atof() that allow you to convert
strings that hold numeric values to int and double values respectively. Another way to
convert strings to numbers is to use a function similar to the familiar scanf() function
which reads its string from the standard input. The sscanf() function reads its “input”
from a string instead of the standard input.
1 char s[] = ""103212.3214"";
2 double x;
3 sscanf(s, ""%lf"", &x);
4 //x now has the value 103212.3214
332",ComputerScienceOne.pdf
"21.6. Tokenizing
The sscanf() function differs in its first argument: the string that contains the value
you want to parse. Otherwise, the second two arguments are as in scanf(): the format
(as a string) and the variable(s) that the results should be stored in (passed by reference).
Likewise, there is a companion sprintf() which is similar to the printf() function,
but instead of printing to the standard output, it “prints” to the string. That is, the
result is placed in a string.
1 int x = 10;
2 double y = 3.14;
3 char *s = (char *) malloc(sizeof(char) * 50);
4 sprintf(s, ""The value of x is %d, y = %f."", x, y);
5 //s now contains ""The value of x is 10, y = 3.140000.""
21.6. Tokenizing
Recall that tokenizing is the process of splitting up a string along some delimiter. For
example, the comma delimited string, ""Smith,Joe,12345678,1985-09-08"" contains
four pieces of data delimited by a comma. Our aim is to split this string up into four",ComputerScienceOne.pdf
"four pieces of data delimited by a comma. Our aim is to split this string up into four
separate strings so that we can process each one. The C string library provides a
tokenizing function:
char *strtok(char *str, const char *delim);
which works as follows. The first argument is the string that you want to tokenize and
the second contains the delimiter that you want to split along. The second argument is
actually a string and allows you to specify more than one delimiter, but we’ll restrict our
attention to single character delimiters. The function returns a pointer to the first token
in the string. To get the second and all subsequent tokens, we call strtok() again, but
we pass it NULL as the first argument to continue parsing the same string.
If we pass a new string as the first argument to strtok() the tokenization process
will start over on the new string. Note that the first argument does not have the
const keyword. This is because strtok() will make changes to the string during",ComputerScienceOne.pdf
"const keyword. This is because strtok() will make changes to the string during
the tokenization process. If the string needs to be preserved, tokenization should be
performed on a deep copy of the string.
When there are no more tokens in the string, strtok() returns NULL to indicate no
333",ComputerScienceOne.pdf
"21. Strings
more tokens are in the string. This logic can be used to write a while loop to iterate over
each token. Consider the following example.
1 char data[] = ""Smith,Joe,12345678,1985-09-08"";
2 char *token = NULL;
3 //make the initial call to strtok:
4 token = strtok(data, "","");
5 while(token != NULL) {
6 printf(""token: %s\n"", token);
7 //get the next token:
8 token = strtok(NULL, "","");
9 }
This outputs the following:
token: Smith
token: Joe
token: 12345678
token: 1985-09-08
It should be noted that strtok() is not reentrant. It can only work on one string at a
time. In a multithreaded application, two threads cannot both use strtok() or they
would end up processing each other’s strings. Even in a non-threaded application, we
need to be careful. For example, we cannot process a string using strtok() in one
function and then call another function that does the same. C does provide a reentrant
version, strtok_r() that can be used.
334",ComputerScienceOne.pdf
"22. File I/O
C provides several functions to manipulate and process files. Like other I/O functions,
these are all defined in the standard input/output library, stdio.h. Writing binary
or plaintext data is determined by which functions you use. Whether or not a file
input/output stream is buffered or unbuffered is determined by the system configuration.
There are some ways in which this can be changed, but we will not cover them in detail.
22.1. Opening Files
Files are represented in C by a FILE * pointer type defined in the standard input/output
library. As a pointer, it essentially points to the file stored in memory. To open a file,
you use the fopen() function (short for file open) which takes two arguments and
returns a FILE * pointer:
FILE *fopen(const char *path, const char *mode);
The first argument is the file path/name that you want to open for processing. The
second argument is a string representing the “mode” that you want to open the file in.",ComputerScienceOne.pdf
"second argument is a string representing the “mode” that you want to open the file in.
There are several supported modes, but the two we will be interested in are reading, in
which case you pass it ""r"" and writing in which case you pass it ""w"". The path can
be an absolute path, relative path, or may be omitted if the file is in the current working
directory.
1 //open a file for reading (input):
2 FILE *input = fopen(""/user/apps/data.txt"", ""r"");
3 //open a file for writing (output):
4 FILE *output = fopen(""./results.txt"", ""w"");
5
6 if(input == NULL) {
7 fprintf(stderr, ""Unable to open input file"");
8 exit(1);
9 }
10
11 if(output == NULL) {
335",ComputerScienceOne.pdf
"22. File I/O
12 fprintf(stderr, ""Unable to open output file"");
13 exit(1);
14 }
The two checks above check that the file opened successfully. If the file opening failed,
fopen() returns NULL. Opening a file can fail for a number of reasons. On POSIX
systems for example, additional information can be obtained by accessing the standard
error number, errno (see Section 19.1). Some errors that can result:
• ENOENT – No such file or directory
• EACCES – Permission denied
• ENOMEM – Insufficient storage space is available
among many other possibilities. These error codes can be used to implement more specific
error handling code if desired.
22.2. Reading & Writing
When a file is opened, the file pointer returned by fopen() initially points to the
beginning of the file. As you read from it or write to it, the pointer advances through
the file content.
22.2.1. Plaintext Files
To process a plaintext file we use two functions, fprintf() for output and fscanf()",ComputerScienceOne.pdf
"the file content.
22.2.1. Plaintext Files
To process a plaintext file we use two functions, fprintf() for output and fscanf()
for input. These two functions should look familiar. They work almost exactly the same
as printf() and scanf() that work with the standard output and standard input
respectively. In fact, the standard in/out are both files so it makes sense that they can
be read from/written to. The f prefixed versions are simply generalizations that can be
used for any file. The only difference is that the first argument for these two functions is
the file you want to output to or read from.
1 int x = 10;
2 double y = 3.14;
3
4 //write to a plaintext file
336",ComputerScienceOne.pdf
"22.2. Reading & Writing
5 fprintf(output, ""Hello World!\n"");
6 fprintf(output, ""x = %d, y = %f\n"", x, y);
7
8 //read from a plaintext file
9 fscanf(input, ""%d"", &x);
10 fscanf(input, ""%lf"", &y);
11
12 //these are equivalent to printf, scanf:
13 fprintf(stdout, ""Please enter an integer:"");
14 fscanf(stdin, ""%d"", &x);
Using fscanf() for arbitrary string input is potentially dangerous as there is limited
bounds checking. We must store the input value from the file into a string (character
array), but if the file or line contains more characters than the array can accommodate
we may have a buffer overflow.
A better way to read input is to use fgets() which allows us to limit the number of
bytes that are read.
char *fgets(char *s, int size, FILE *stream);
The first argument is the string that the input data will be read into. The second
parameter is how we limit the number of characters that will be read. It actually reads",ComputerScienceOne.pdf
"parameter is how we limit the number of characters that will be read. It actually reads
in one fewer character, size-1 to account for the null-terminating character which
fgets() automatically inserts for us. The last argument is the file pointer that we wish
to read from.
The behavior of fgets() is that it reads up to size-1 characters from the input
file and places the results into s. If fgets() encounters either an EOF symbol or
an endline character, '\n' it stops reading. In the case of an endline character, it is
included in the result and may need to be chomped out (that is, removed).
The fgets() function can be used to process a file line by line until the end of the
file is reached. Each line can be processed (perhaps tokenized) individually to extract
particular pieces of data. This is typically how a CSV or similar file may be processed.
To determine if the end of the file has been reached, you can use the return value of the",ComputerScienceOne.pdf
"To determine if the end of the file has been reached, you can use the return value of the
function: it returns NULL when no more characters have been read.
1 FILE *input = fopen(""data.txt"", ""r"");
2 //this assumes that no line is more than 999 characters
3 char line[1000];
337",ComputerScienceOne.pdf
"22. File I/O
4 //read the first line
5 char *s = fgets(line, 1000, input);
6 while(s != NULL) {
7
8 //chomp the endline character from line:
9 line[strlen(line)-1] = '\0';
10
11 //process the current buffer
12 //for demonstration, we simply print it:
13 printf(""line = %s\n"", line);
14
15 //read the next line
16 s = fgets(line, 1000, input);
17 }
A similar function, int fgetc(FILE *stream) allows us to get a single character from
the input file (returned as an int) if we prefer to read character by character.
22.2.2. Binary Files
To read and write binary data to files we use two different functions:
size_t fread(void *ptr, size_t size, size_t nmemb, FILE *stream);
for reading from a file and
size_t fwrite(const void *ptr, size_t size, size_t nmemb, FILE *stream);
to write to a file. Both functions have the exact same arguments:
1. ptr is a pointer to the item that is being read/written (possibly an array of
elements)",ComputerScienceOne.pdf
"1. ptr is a pointer to the item that is being read/written (possibly an array of
elements)
2. size is the number of bytes that each element stored at ptr take, usually
determined by using sizeof()
3. nmemb is the number of elements to be written/read
4. stream is the file stream to be read from/written to
These functions allow you to read/write an arbitrary amount of binary data. When
reading, it is your responsibility, as usual, to ensure that the ptr is large enough to
338",ComputerScienceOne.pdf
"22.3. Closing Files
accommodate the data you are reading into it.
1 int x = 10;
2
3 FILE *binaryOutputFile = fopen(""demo.bin"", ""w"");
4
5 //write a single int to a file:
6 fwrite(&x, sizeof(int), 1, binaryOutputFile);
7
8 FILE *binaryInputFile = fopen(""input.bin"", ""r"");
9
10 //read a single int from the file:
11 fread(&x, sizeof(int), 1, binaryInputFile);
12
13 //read an entire array from the file:
14 int *a = (int *) malloc(10 * sizeof(int));
15 fread(a, sizeof(int), 10, binaryInputFile);
22.3. Closing Files
Once you are done processing a file, you should close it using the fclose() function:
int fclose(FILE *fp);
which takes a single argument, the FILE pointer of the file you wish to close. Closing
an invalid file may result in an error or segmentation fault. In any case, after a file has
been closed, it cannot be read from or written to, doing so is undefined behavior. Failure
to close a file may result in a corrupted file.
339",ComputerScienceOne.pdf
"23. Structures
Strictly speaking, C is not an object-oriented programming language, it is an imperative
(or, relatedly a structured or procedural) programming language. This means that C can
be characterized as a language that changes a program’s state through statements and
the use of function calls. Though C does not have objects, it does support a “weak” form
of encapsulation through the use of structures. Structures are a user-defined type that
collects multiple pieces of data together into one logical unit. Once defined, structures
can be used in a program just like any other variable type; structure variables can be
declared, passed and returned from functions, pointers to structures can be used, etc.
Structures form a “weak” form of encapsulation in that they only provide the grouping of
data. The protection of data through visibility keywords is not supported. The grouping
of functions that act on that data is also not readily supported. 1 However, even this",ComputerScienceOne.pdf
"of functions that act on that data is also not readily supported. 1 However, even this
weak form provides a very useful and convenient way to collect related pieces of ?ata.
23.1. Defining Structures
To define a structure, we use the keyword struct along with the previously introduced
keyword typedef. To motivate an example, let’s reconsider a student entity, which has
an ID (integer), first and last name (strings), and a GPA (a floating point number). Each
of these pieces of data can be encapsulated into a structure as follows.
1 typedef struct {
2 int id;
3 char *firstName;
4 char *lastName;
5 double gpa;
6 } Student;
Note the syntax and naming conventions:
1You can define a structure with function pointers as elements, so structures could technically include
“methods” but this is not really what most people think of when considering object-oriented paradigms.
341",ComputerScienceOne.pdf
"23. Structures
•The elements of the structure (also referred to as components or members) are
included inside curly brackets, delimited using semicolons (in contrast to an enu-
merated type which is a list, these elements do not constitute a list).
•A structure may contain any number of elements of any type.
•The name (identifier) of the structure is provided at the end, ended with a semicolon.
•We use a modern naming convention: each element is named using lower camel
casing while the name of the structure itself uses upper camel casing.
In addition, structure declarations are generally placed in a header file along with any
any function prototypes that use the structure as a parameter or a return type. Once
you have defined a structure you can use it as you would a built-in variable type. For
example,
Student s;
declares a Student structure with the variable name s.
23.1.1. Alternative Declarations
In some examples and code bases you may find an alternative way of declaring a structure",ComputerScienceOne.pdf
"23.1.1. Alternative Declarations
In some examples and code bases you may find an alternative way of declaring a structure
that looks like the following.
1 struct Student {
2 ...
3 };
This syntax omits the keyword typedef and places the structure name at the beginning.
The difference is a bit technical (the original syntax creates an anonymous structure with
the Student identifier placed in the global namespace, while this declaration places the
Student identifier in the structure namespace), but one of the consequences of using
this type of declaration is that the Student identifier is not in the global scope, so to
declare a variable of the type Student you need to further specify that it is a structure
using the following syntax.
struct Student s;
Further, you may also see another style,
342",ComputerScienceOne.pdf
"23.1. Defining Structures
1 typedef struct Student {
2 ...
3 } Student;
Which places the Student identifier in both the global space and in the “structure”
space. Which style of declaration you use depends on several factors, but for simplicity
we’ll stick with the first style.
In addition, you may see some older naming conventions use lowercase underscore
naming for structure names. In particular, POSIX systems use names that end with
_t (indicating a structure type). You should avoid such naming conventions to avoid
conflicts.
Finally, though C does not provide a mechanism for the protection of data, many
libraries and code bases will begin certain member variables with underscore(s) to
indicate that they are “internal” variables used by the library and should not be accessed
or modified. For example, _someVariable or __anotherVariable would indicate
“private” variables that should not be tampered with. The C language itself would still",ComputerScienceOne.pdf
"“private” variables that should not be tampered with. The C language itself would still
allow you access and modify the variables, but doing so may violate assumptions and
expectations of the library code, leading to undefined behavior.
23.1.2. Nested Structures
Consider adding another member variable to the Student structure to model a student’s
date of birth. How might we model a date? Unix systems model time as a single
integer value that represents the number of milliseconds that have passed since the
Unix epoch, January 1st, 1970 (also known as Coordinated Universal Time (UTC)).
For example, the New Horizons space probe was launched on January 19th, 2006 at
19:00:00 UTC. This corresponds to 1,137,697,200, roughly 1.137 billion milliseconds since
the epoch. Another way to model time is as a string. There are several standards for
time representations, but the most common is ISO standard 8601 [ 22] which specifies",ComputerScienceOne.pdf
"time representations, but the most common is ISO standard 8601 [ 22] which specifies
time using a format that includes a date, time and an offset from Greenwich Mean
Time (GMT). For example, the New Horizons launch date/time would be represented as
""2006-01-19T19:00:00+00:00"".
To keep things simple, we’ll model our date using three numbers: a year, a month and
a date. But how should we implement this? Conceptually, a date is a single entity,
separating these three numbers doesn’t make much sense. This is a perfect opportunity
to define another structure.
343",ComputerScienceOne.pdf
"23. Structures
1 typedef struct {
2 int id;
3 char *firstName;
4 char *lastName;
5 double gpa;
6 Date dateOfBirth;
7 } Student;
Code Sample 23.1.: A Student structure declaration
1 typedef struct {
2 int year;
3 int month;
4 int date;
5 } Date;
Once we have defined a structure we can use it as we would a normal variable, so it
makes sense that we could include it in another structure.
This is a good illustration of a form of composition where one structure may be composed
of other structures. Since the Student structure “owns” an instance of the Date
structure, it is necessary to ensure that the Date structure is declared before the
Student structure (just as we need to declare variables before we use them.
23.2. Usage
23.2.1. Declaration & Initialization
Once we have defined a structure (and included the header file it has been defined in),
we can create instances using the usual syntax.
1 Student s;
2 Student t;
344",ComputerScienceOne.pdf
"23.2. Usage
These static declarations will allocate enough space on the stack to hold all of the data
associated with the structures (the two char * pointers, int, and double and the
three int variables in the Date structure). However, the values stored in each of the
structure’s member variables are undefined. With this style of declaration, C does not
define default values.
Another way to declare instances of our structures is to use the following syntax.
1 Student s = {};
2 Student t = {
3 12345678,
4 ""Grace"",
5 ""Hopper"",
6 4.0,
7 {
8 1906,
9 1,
10 1
11 }
12 };
The first declaration creates a Student structure with default values (zero for any
numeric types, null for any pointers). The second creates a Student structure
initialized with the values provided in the curly brackets. The order matters here and
will match the ordering of the original structure declaration. Since the dateOfBirth
is a structure itself, a nested set of values within curly brackets is necessary. One draw",ComputerScienceOne.pdf
"is a structure itself, a nested set of values within curly brackets is necessary. One draw
back to this type of declaration is that the character pointers are statically declared. The
strings ""Grace"" and ""Hopper"" are initialized in a read-only segment of memory, any
attempts to change the contents of these strings are undefined behavior. The pointers
themselves, however, can be reassigned.
This static declaration allocates the structure on the stack. Since structures consist of
multiple pieces of data, their memory footprint is larger. Allocating larger and larger
structures on the stack runs the risk of running out of stack memory resulting in a stack
overflow. To solve this, we can instead use dynamically allocated structures.
To dynamically allocate a structure, we use a pointer to a structure and a call to
malloc() to allocate enough space for the structure. Even though our structure is
user-defined, we can still use the sizeof() macro to determine the number of bytes",ComputerScienceOne.pdf
"user-defined, we can still use the sizeof() macro to determine the number of bytes
required by the structure. The compiler and macro are “smart” enough to look at the
structure declaration and determine the size of each of its variables and add up the total
number of bytes required.
345",ComputerScienceOne.pdf
"23. Structures
Student *s = (Student *) malloc(sizeof(Student) * 1);
The multiplication by 1 in this example is not strictly necessary, but emphasizes the fact
that we are allocating space for one structure and not an array of structures. Initializing
a dynamically allocated structure like this does not initialize any of its variables as
there are no default values defined by C. Moreover, it does not initialize memory for any
pointer member variables. For example the firstName and lastName pointers need
to be manually initialized with additional malloc() calls.
23.2.2. Selection Operators
Once we have a declared structure, we need to access its member variables and perhaps
update them. If we have a statically declared structure, such as
Student s;
we can use the direct component selector operator, which is simply just a period followed
by the member variable we wish to access. Commonly, this is referred to simply as the
dot operator.
1 Student s;
2
3 //set values:
4 s.id = 87654321;",ComputerScienceOne.pdf
"by the member variable we wish to access. Commonly, this is referred to simply as the
dot operator.
1 Student s;
2
3 //set values:
4 s.id = 87654321;
5 s.gpa = 3.9;
6
7 //access values:
8 printf(""Name: %s, %s\n"", s.lastName, s.firstName);
9 printf(""GPA: %.2f\n"", s.gpa);
When structures are nested, we can use the dot operator multiple times to access members
variables of member variables.
1 s.dateOfBirth.year = 2525;
2 s.dateOfBirth.month = 12;
3 s.dateOfBirth.date = 25;
When we have a pointer to a structure, we cannot directly use the dot operator. Instead,
346",ComputerScienceOne.pdf
"23.2. Usage
we have to change the pointer into a “normal” structure by dereferencing it, then we can
use the dot operator. However, the dot operator has a higher order of precedence than
the dereferencing operator, thus parentheses are required:
1 Student *s = ...;
2 (*s).id = 87654321;
This can be a bit unwieldy, so C provides a convenience operator, the indirect component
selector operator, or more commonly, the arrow operator that allows us to select a member
variable with a single operator that resembles a right-pointing arrow.
1 Student *s = (Student *) malloc(sizeof(Student));
2
3 //set values:
4 s->id = 87654321;
5 s->gpa = 3.9;
6
7 //initialize the string variables:
8 s->firstName = (char *) malloc(sizeof(char) * 6);
9 strcpy(s->firstName, ""Grace"");
10 s->lastName = (char *) malloc(sizeof(char) * 7);
11 strcpy(s->lastName, ""Hopper"");
12
13 //access values:
14 printf(""Name: %s, %s\n"", s->lastName, s->firstName);
15 printf(""GPA: %.2f\n"", s->gpa);
16
17 s->dateOfBirth.year = 2525;",ComputerScienceOne.pdf
"12
13 //access values:
14 printf(""Name: %s, %s\n"", s->lastName, s->firstName);
15 printf(""GPA: %.2f\n"", s->gpa);
16
17 s->dateOfBirth.year = 2525;
Note the last line: the dateOfBirth member variable was another structure, but not a
pointer to a structure, so we use the dot operator to access the year member variable. Al-
ternatively, we could have defined dateOfBirth to be a pointer, Date *dateOfBirth;
in the Student structure. If we had, then to access the year member variable using
two arrow operators, s->dateOfBirth->year.
347",ComputerScienceOne.pdf
"23. Structures
23.3. Arrays of Structures
Just as we can create arrays of built-in types such as integers, we can also create arrays of
our user-defined structures. As an example, the following creates an array of 10 Student
structures. Once created, we can treat them like any other array.
1 Student *roster = (Student *) malloc(sizeof(Student) * 10);
2
3 ...
4
5 double sum = 0.0;
6 for(i=0; i<10; i++) {
7 sum += roster[i].gpa;
8 }
9 double averageGpa = sum / 10;
As in the example, we can index each element in the array, roster. Once indexed, each
element is a regular structure and so we use the dot operator to access each of its member
variables. As with any other array, each element takes up a number of bytes, equal to
sizeof(Student). We can swap and reassign each element just like any other variable.
For example, the following code swaps the first two elements using a temporary variable.
1 Student temp = roster[0];
2 roster[0] = roster[1];
3 roster[1] = temp;",ComputerScienceOne.pdf
"1 Student temp = roster[0];
2 roster[0] = roster[1];
3 roster[1] = temp;
Each of these operations copies over every byte that makes up the structure. For small
structures, this isn’t that big of a deal. However, for larger structures, this may become
an issue, especially if we do this often or pass structures around to functions.
As an alternative, we could instead deal indirectly with structures by creating an array
of pointers to structures. Swapping elements then involves only copying pointer values
rather than every byte that makes up the structure. To do this, we use the familiar
pointer-to-pointer syntax.
1 Student **roster = (Student **) malloc(sizeof(Student *) * 10);
2 for(i=0; i<10; i++) {
348",ComputerScienceOne.pdf
"23.3. Arrays of Structures
3 roster[i] = (Student *) malloc(sizeof(Student) * 1);
4 }
5
6 //access each as pointers and use the arrow operator
7 roster[0]->id = 87654321;
8 roster[0]->gpa = 4.0;
9
10 //swap the first two *pointers*:
11 Student *temp = roster[0];
12 roster[0] = roster[1];
13 roster[1] = temp;
As in the example above, if each element in the array is a pointer to a structure, then we
use the arrow operator to access each member variable.
The differences between these two approaches is illustrated in Figures 23.1 and 23.2. As
presented, each Student structure takes 40 bytes.2 With the first approach (as in Figure
23.1, each structure instance is stored contiguously in memory. Swapping two records
involves copying entire blocks of 40 bytes. In contrast, using pointers to structures as in
Figure 23.2 means that structures may be stored non-contiguously in different memory
locations. Swapping two structures is done indirectly by swapping pointers (only 8 bytes).",ComputerScienceOne.pdf
"locations. Swapping two structures is done indirectly by swapping pointers (only 8 bytes).
The difference in this particular example is not that great (8 vs. 40 bytes), but with
larger structures it can become an issue. Each approach has its own advantages and
disadvantages and one may be more appropriate than the other in different situations.
Hybrid Approach
It is also possible to take a “hybrid” approach to storing structures in arrays that stores
them both contiguously, but also allows you to reference them indirectly through a pointer
array. This is essentially what we did in Section 20.4.1 when we created a contiguous
2-dimensional array of integers. We made a single call to malloc and then had to setup
the pointers to each “row.” To do this with structures, we can do something similar.
First, we declare and allocate a dynamic array as before.
1 int n = 10;
2 Student *rosterData = (Student *) malloc(n * sizeof(Student));",ComputerScienceOne.pdf
"First, we declare and allocate a dynamic array as before.
1 int n = 10;
2 Student *rosterData = (Student *) malloc(n * sizeof(Student));
2This is just an estimate and may vary on different systems and compilers. Usually, compilers use
alignment and may pad structures with extra bytes in order to make it more efficient to store in
memory.
349",ComputerScienceOne.pdf
"23. Structures
Student *roster roster[0]
(Student)
roster[1]
(Student)
roster[2]
(Student)
...
roster[n-1]
(Student)
40 bytes
40 bytes
40 bytes
40 bytes
Figure 23.1.: An array of structures. Each record is stored in a contiguous manner one
after the other.
Student **roster roster[0]
(Student*)
roster[1]
(Student*)
roster[2]
(Student*)
...
roster[n-1]
(Student*)
Student
(40 bytes)
Student
(40 bytes)
Student
(40 bytes)
Student
(40 bytes)
Figure 23.2.: An array of structure pointers. Each record is a pointer that refers to a
structure which may be stored non-contiguously in completely different
memory locations.
350",ComputerScienceOne.pdf
"23.4. Using Structures With Functions
Then, we can declare an array of Student pointers as well.
1 Student **roster = (Student **) malloc(n * sizeof(Student *));
But now we need to make eachroster[i] pointer point to thei-th record in rosterData.
Each record, rosterData[i] is a regular structure, but we need a pointer to it. In
order to get a pointer, we use the referencing operator, &rosterData[i] and make each
pointer point to it.
1 for(i=0; i<n; i++) {
2 roster[i] = &rosterData[i];
3 }
This approach is illustrated in Figure 23.3. Now we can indirectly reference each record
in rosterData via a pointer in the roster array.
1 roster[0]->nuid = 12345678;
If we wanted to “swap” two records, we simply swap their pointers instead of their data.
Indeed, care must be taken not to swap the data in rosterData otherwise the pointers
would become invalid and out-of-synch with the data array.
23.4. Using Structures With Functions",ComputerScienceOne.pdf
"would become invalid and out-of-synch with the data array.
23.4. Using Structures With Functions
As with built-in types, we can use structures as parameters to functions as well as the
return type of functions. Consider the following examples.
1 void foo(Student s);
2 Student bar();
351",ComputerScienceOne.pdf
"23. Structures
Student **roster roster[0]
(Student*)
roster[1]
(Student*)
roster[2]
(Student*)
...
roster[n-1]
(Student*)
Student *rosterData
roster[0]
(Student)
roster[1]
(Student)
roster[2]
(Student)
...
roster[n-1]
(Student)
40 bytes
40 bytes
40 bytes
40 bytes
8 bytes
Figure 23.3.: Hybrid Array of Structures. The rosterData is an array of contiguous
structures. The roster array is an array of pointers that refer to each
record. Accessing elements in rosterData is done indirectly through a
pointer in roster
In the first prototype, we would pass a Student structure by value to the function. As
we’ve already seen, this would mean that every byte of the structure would be copied
onto the stack. For larger structures, we risk a stack overflow while for even smaller
structures this can be very inefficient. Likewise, the second prototype would return a
structure. When assigned to a variable, every byte is copied. A better solution is to use",ComputerScienceOne.pdf
"structure. When assigned to a variable, every byte is copied. A better solution is to use
dynamically allocated structures, passing them by reference to functions, and returning
pointers to dynamically allocated instances as return types. For example, the functions
above would be better implemented as follows.
1 void foo(const Student *s);
2 Student * bar();
In the first example, we’ve used the const keyword which prevents any changes to
the structure’s values (we may omit this if we need to design a function that changes
its values). We will now consider several idiomatic examples of using structure with
functions.
352",ComputerScienceOne.pdf
"23.4. Using Structures With Functions
23.4.1. Factory Functions
Properly creating and initializing structure instances can be a complex and tedious
task. However, it is likely that we will need to repeat this operation over and over. We
can simplify our task if we write a utility function that creates a structure instance
for us. We provide the function the values we want initialized to. Such functions are
sometimes referred to as factory functions as they can be used to manufacture as many
instances as we want (in object-oriented programming languages, such methods are called
constructors).
We will need to take care that we make deep copies of any dynamically allocated elements
such as strings. Shallow copies where references are shared may lead to unexpected
behavior as changes to one string may affect multiple structures.
1 /**
2 * This function creates a new student structure with the
3 * given values.
4 */
5 Student * createStudent(const char *firstName,
6 const char *lastName,",ComputerScienceOne.pdf
"3 * given values.
4 */
5 Student * createStudent(const char *firstName,
6 const char *lastName,
7 int id,
8 double gpa) {
9
10 Student *s = NULL;
11 s = (Student *) malloc(sizeof(Student) * 1);
12
13 //make a deep copy of the strings
14 s->firstName = (char *)malloc(sizeof(char)*(strlen(firstName)+1));
15 strcpy(s->firstName, firstName);
16 s->lastName = (char *)malloc(sizeof(char)*(strlen(lastName)+1));
17 strcpy(s->lastName, lastName);
18
19 s->id = id;
20 s->gpa = gpa;
21
22 return s;
23 }
Another common operation is to create a copy of a given structure. We will want to
again ensure that everything is a deep copy so that the two structures are not sharing
references. We can easily do this by reusing the functionality we just wrote in the factory
353",ComputerScienceOne.pdf
"23. Structures
function.
1 /**
2 * This function creates a new, deep copy of a Student
3 * structure .
4 */
5 Student * copyStudent(const Student *s) {
6 return createStudent(s->firstName, s->lastName, s->id, s->gpa);
7 }
23.4.2. To String Functions
Another common operation with structures is to output their data as a human-readable
string representation. We could write a function that output a structure’s member
variables to the standard output using printf(). However, it would more useful if we
created a general function that returned a string representation of the structure. That
way, we could decide what to do with the string: output it to the standard output, to a
file, etc. Not all structure instances will require the same length string. Some students
for example may have long names, while others have short. We can handle this by first
calculating the length of the string necessary for whatever formatting we’ve chosen.
1 /**
2 * Returns a string representation of the given",ComputerScienceOne.pdf
"calculating the length of the string necessary for whatever formatting we’ve chosen.
1 /**
2 * Returns a string representation of the given
3 * Student structure.
4 */
5 char * studentToString(const Student *s) {
6
7 int n = strlen(s->firstName) +
8 strlen(s->lastName) +
9 8 + //id, assumed to always be at most 8 digits
10 4 + //gpa, assumed to be 0.0 - 4.0
11 19; //other formatting characters
12
13 char *str = (char *) malloc(sizeof(char) * n);
14
15 sprintf(str, ""%s, %s, ID = %d (GPA = %.2f)"",
16 s->lastName, s->firstName, s->id, s->gpa);
17
18 return str;
354",ComputerScienceOne.pdf
"23.4. Using Structures With Functions
19 }
Here, we’ve utilized a variation on the familiar printf() function, sprintf() which
“prints” the result not to the standard output or a file, but to a string, specified as the
first argument. This function would end up returning a string similar to the following for
our previous example: ""Hopper, Grace, ID = 87654321 (GPA = 3.90)""
23.4.3. Passing Arrays of Structures
We often also have need to pass arrays of structures to functions. As an example, consider
passing an entire roster of students to a function in order to compute the average GPA.
If we have an array of structures, we would have a function that looked something like
the following.
1 /**
2 * Computes the average GPA of the Student structures in
3 * the given roster (which is of size n).
4 */
5 double computeAverageGpa(const Student *roster, int n) {
6 double sum = 0.0;
7 int i;
8 for(i=0; i<n; i++) {
9 sum += roster[i].gpa;
10 }
11 return sum / n;
12 }",ComputerScienceOne.pdf
"6 double sum = 0.0;
7 int i;
8 for(i=0; i<n; i++) {
9 sum += roster[i].gpa;
10 }
11 return sum / n;
12 }
When we pass in the array of structures, it is not passed by value. That is, the total
number of bytes for each student is not copied onto the call stack. Nevertheless, as we’ve
previously seen it is sometimes preferable to maintain an array of pointers to structures.
If we had an array of pointers, we would have a function that looks something like the
following.
1 /**
2 * Computes the average GPA of the Student structures in
3 * the given roster (which is of size n).
355",ComputerScienceOne.pdf
"23. Structures
4 */
5 double computeAverageGpa(const Student **roster, int n) {
6 double sum = 0.0;
7 int i;
8 for(i=0; i<n; i++) {
9 sum += roster[i]->gpa;
10 }
11 return sum / n;
12 }
The only difference here is in how we access the gpa member variable using the arrow
operator instead of the dot operator.
356",ComputerScienceOne.pdf
"24. Recursion
C supports recursion with no special syntax necessary. However, as a structured,
procedural language, recursion is generally expensive and iterative or other non-recursive
solutions are generally preferred. We present a few examples to demonstrate how to
write recursive functions in C.
The first example of a recursive function we gave was the toy count down example. In C
it could be implemented as follows.
1 void countDown(int n) {
2 if(n==0) {
3 printf(""Happy New Year!\n"");
4 } else {
5 printf(""%d\n"", n);
6 countDown(n-1);
7 }
8 }
As another example that actually does something useful, consider the following recursive
summation function that takes an array, its size and an index variable. The recursion
works as follows: if the index variable has reached the size of the array, it stops and returns
zero (the base case). Otherwise, it makes a recursive call to recSum(), incrementing
the index variable by 1. When the function returns, it adds its result to the i-th element",ComputerScienceOne.pdf
"the index variable by 1. When the function returns, it adds its result to the i-th element
in the array. To invoke this function we would call it with an initial value of 0 for the
index variable: recSum(arr, n, 0).
1 int recSum(const int *arr, int size, int i) {
2 if(i == size) {
3 return 0;
4 } else {
5 return recSum(arr, size, i+1) + arr[i];
6 }
7 }
357",ComputerScienceOne.pdf
"24. Recursion
This example was not tail-recursive as the recursive call was not the final operation (the
sum was the final operation). To make this function tail recursive, we can carry the
summation through to each function call ensuring that the summation is done prior to
the recursive function call.
1 int recSumTail(const int *arr, int size, int i, int sum) {
2 if(i == size) {
3 return sum;
4 } else {
5 return recSumTail(arr, size, i+1, sum + arr[i]);
6 }
7 }
As a final example, consider the following C implementation of the naive recursive
Fibonacci sequence. An additional condition has been included to check for “invalid”
negative values of n for which zero is returned.
1 int fibonacci(int n) {
2 if(n < 0) {
3 return 0;
4 } else if(n <= 1) {
5 return 1;
6 } else {
7 return fibonacci(n-1) + fibonacci(n-2);
8 }
9 }
C is not a language that provides implicit memoization. Instead, we need to explicitly",ComputerScienceOne.pdf
"6 } else {
7 return fibonacci(n-1) + fibonacci(n-2);
8 }
9 }
C is not a language that provides implicit memoization. Instead, we need to explicitly
keep track of values using a table. In the following example, the table is passed to the
function as an argument.
1 int fibonacciMemoization(int n, int *table) {
2 if(n < 0) {
3 return 0;
4 } else if(n <= 1) {
5 return 1;
358",ComputerScienceOne.pdf
"6 } else if(table[n] > 0) {
7 return table[n];
8 } else {
9 int a = fibonacciMemoization(n-1, table);
10 int b = fibonacciMemoization(n-2, table);
11 int result = (a + b);
12 table[n] = result;
13 return result;
14 }
15 }
It is the responsibility of the calling function to ensure that the table array is large
enough to accommodate all values. In this case should be at least of size ( n+ 1) to
compute the n-th Fibonacci number.
359",ComputerScienceOne.pdf
"25. Searching & Sorting
The standard C library provides several functions to search and sort arrays of any type of
element including int, double, or even user-defined structures such as our Student
example from Chapter 23. These functions are able to operate on any type of array
because they take generic void * pointers. However, these functions still need a way to
order the generic elements. To understand how this all works we need to understand
both comparator functions and function pointers.
25.1. Comparator Functions
Let’s consider a “generic” Quick Sort algorithm as was presented in Algorithm 12.6. The
algorithm itself specifies how to sort elements, but it doesn’t specify how they are ordered.
The difference is subtle but important. Essentially, Quick Sort needs to know when two
elements, a,b are in order, out of order, or equivalent in order to decide which partition
each element goes in. However, it doesn’t “know” anything about the elements a and b",ComputerScienceOne.pdf
"each element goes in. However, it doesn’t “know” anything about the elements a and b
themselves. They could be numbers, they could be strings, they could be user-defined
objects.
A sorting algorithm still needs to be able to determine the proper ordering in order to sort.
In C this is achieved through a comparator function, which is a function that is responsible
for comparing two elements and determining their proper order. A comparator function
has the following signature and behavior:
int cmp(const void *a, const void *b);
•The function takes two generic void * pointers which refer to the elements being
compared.
•Moreover, the const keyword is used to indicate that no changes will be made to
the elements.
•The function returns an integer indicating the relative ordering of the two elements:
– It returns something negative, < 0 if a comes before b (that is, a<b )
– It returns zero if a and b are equal (a= b)
361",ComputerScienceOne.pdf
"25. Searching & Sorting
– It returns something positive, > 0 if a comes after b (that is, a>b )
Note that there is no guarantee on the value’s magnitude, it doesnot necessarily return −1
or +1; it just returns something negative or positive. We’ve previously seen this pattern
when comparing strings. The standard string library provides a function, strcmp()
that has the same basic contract: it takes two strings and returns something negative,
zero or something positive depending on the lexicographic ordering of the two strings.
Strictly speaking, however, strcmp() is not a comparator function. It is defined to take
two const char * parameters, not const void * pointers.
The C language (and thus the compiler) “knows” how to compare built-in primitive types
like int and double using the built-in comparison operators. However, to generalize
the comparison operation, we use void * pointers so that we can write and use general",ComputerScienceOne.pdf
"the comparison operation, we use void * pointers so that we can write and use general
comparator functions that can be used in generic searching and sorting functions. As an
example, let’s write a comparator function that orders integers in ascending order. We’ll
call our function cmpInt() and use the signature above.
1 int cmpInt(const void *a, const void *b) {
2
3 }
But how might we implement this function? Obviously we want to compare the values
stored in the pointer variables, so we need to dereference them. However, the comparison,
(*a < *b) for example is not comparing integer values. The variables a and b are
void pointers, not int pointers. In general, you cannot dereference a void pointer.
Dereferencing an int * or double * is possible because the compiler knows how many
bytes each type takes. However, the number of bytes a void type takes is undefined.
Thus, the first step is to make them int pointers by doing an explicit type cast:
1 const int *x = (const int *) a;",ComputerScienceOne.pdf
"Thus, the first step is to make them int pointers by doing an explicit type cast:
1 const int *x = (const int *) a;
2 const int *y = (const int *) b;
Now the variables x and y are int pointers which can be dereferenced and compared
(we preserve the keyword const to ensure we do not make changes to the variables).
1 int cmpInt(const void *a, const void *b) {
2 const int *x = (const int *) a;
3 const int *y = (const int *) b;
362",ComputerScienceOne.pdf
"25.1. Comparator Functions
4 if(*x < *y) {
5 return -1;
6 } else if(*x == *y) {
7 return 0;
8 } else {
9 return 1;
10 }
11 }
What if we wanted to order integers in the opposite order? We could write another
comparator in which the comparisons or values are reversed. Even simpler, we could
reuse the comparator above and “flip” the sign by multiplying by −1 (the purpose of
writing functions is code reuse, after all). Even simpler, we could flip the arguments we
pass to cmpInt() to reverse the order.
1 int cmpIntDesc(const void *a, const void *b) {
2 return cmpInt(b, a);
3 }
The examples above illustrate the standard implementation pattern of comparator
functions:
1. Make the general const void * pointers into a const pointer to the specific
type you want to compare by making an explicit type cast.
2. Use the state of the type (one or more components if you are comparing structures)
to determine their order.
3. Return an integer that expresses the desired order of elements.",ComputerScienceOne.pdf
"to determine their order.
3. Return an integer that expresses the desired order of elements.
The cautious observer may be asking: what happens if I call a comparator and pass it
pointers to mismatched elements? For example, what if I pass two double pointers to
cmpInt()? Obviously, the code will not work as intended (it will end up comparing the
upper 32 bits of the double values, which leads to unintended behavior). A comparator
function is designed with certain expectations. Namely, that a user will always pass it
valid pointers to int values. If a user violates these expectations, it is their fault, not
ours. This is no different from many other functions in the standard library. Passing
non-null terminated strings to most string functions for example is undefined behavior.
To illustrate some more examples, consider the Student structure we defined in Code
363",ComputerScienceOne.pdf
"25. Searching & Sorting
Sample 23.1. The following code samples demonstrate various ways of ordering Student
structures based on one or more of their components.
1 /**
2 * A comparator function to order Student structures by
3 * last name/first name in alphabetic order
4 */
5 int studentByNameCmp(const void *s1, const void *s2) {
6 const Student *a = (const Student *)s1;
7 const Student *b = (const Student *)s2;
8 int result = strcmp(a->lastName, b->lastName);
9 if(result == 0) {
10 return strcmp(a->firstName, b->firstName);
11 } else {
12 return result;
13 }
14 }
1 /**
2 * A comparator function to order Student structures by
3 * last name/first name in reverse alphabetic order
4 */
5 int studentByNameCmpDesc(const void *s1, const void *s2) {
6 return studentByNameCmp(s2, s1);
7 }
1 /**
2 * A comparator function to order Student structures by
3 * id in ascending numerical order
4 */
5 int studentIdCmp(const void *s1, const void *s2) {
6 const Student *a = (const Student *)s1;",ComputerScienceOne.pdf
"3 * id in ascending numerical order
4 */
5 int studentIdCmp(const void *s1, const void *s2) {
6 const Student *a = (const Student *)s1;
7 const Student *b = (const Student *)s2;
8 if(a->nuid < b->nuid) {
9 return -1;
10 } else if(a->nuid == b->nuid) {
11 return 0;
364",ComputerScienceOne.pdf
"25.2. Function Pointers
12 } else {
13 return 1;
14 }
15 }
1 /**
2 * A comparator function to order Student structures by
3 * GPA in descending order
4 */
5 int studentGpaCmp(const void *s1, const void *s2) {
6 const Student *a = (const Student *)s1;
7 const Student *b = (const Student *)s2;
8 if(a->gpa > b->gpa) {
9 return -1;
10 } else if(a->gpa == b->gpa) {
11 return 0;
12 } else {
13 return 1;
14 }
15 }
25.2. Function Pointers
Now that we have comparator functions to order elements, we need a way of passing a
comparator to a generic search or sort function so that it can be used by that function.
To achieve this in C, we use function pointers.
Recall that a pointer is simply a reference to some memory location. As we’ve already
seen, pointers can point to variables of simple data types such as int or double,
arrays of these types or even user-defined structures. We’ve also seen that pointers can
be generic using the void keyword. The malloc() function for example, returned a",ComputerScienceOne.pdf
"be generic using the void keyword. The malloc() function for example, returned a
generic void pointer, void *, to allow it to be used to allocate memory for any type.
Generic void pointers simply point to the start of a memory block, not necessarily to
a memory location of any particular type of variable. As we’ve also already seen, the
program code itself lives in memory (the stack). It thus makes sense that we could
365",ComputerScienceOne.pdf
"25. Searching & Sorting
reference a memory location that, instead of containing variables, contains executable
code, in particular a function. This is what a function pointer does: it points to a memory
location where the code for the function is stored.
To declare a function pointer, we need to specify more information than a typical int *
or double * pointer. Since it points to a function, we need to specify the function’s
signature: its return type and parameter list. Consider the following example:
int (*ptrToFunc)(int, double, char) = NULL;
In this declaration, we’ve created a function pointer named ptrToFunc. This pointer
is capable of pointing to any function whose return type is int (indicated by the first
keyword) and which takes three parameters: an int, a double and a char. The
assignment operation has initialized this pointer to NULL.
Consider another example: let’s declare a function pointer that is capable of pointing to",ComputerScienceOne.pdf
"assignment operation has initialized this pointer to NULL.
Consider another example: let’s declare a function pointer that is capable of pointing to
the math library’s sqrt() function. The sqrt() function returns a double and takes
a single double parameter. Adapting the syntax above, we would create this pointer as
follows.
double (*ptrToSqrt)(double) = NULL;
Again, we have initialized it to NULL; it does not yet reference the sqrt() function.
To make it reference this function, we need to assign it a value. Recall that to get the
memory location of a regular variable, say int x;, we use the referencing operator, &x.
Similarly, with functions we can use the referencing function, in this case &sqrt to get
the memory address of the sqrt() function, however this is not necessary. Similar to
arrays, the identifier (name) of a function serves as its memory address! The identifier
sqrt itself is the memory location of the function. Thus to assign ptrToSqrt to point",ComputerScienceOne.pdf
"sqrt itself is the memory location of the function. Thus to assign ptrToSqrt to point
to sqrt(), we simply need to do the following:
ptrToSqrt = sqrt;
Note the difference: usually we invoke sqrt() by writing parentheses and providing an
argument. When referencing the function itself, we omit the parentheses. Additional
examples of this syntax can be found in Code Sample 25.1.
Callbacks
The main use for function pointers is so that references to functions can be passed
as parameters to other functions. The passed function is known as a callback. This
gives us the ability to write a more abstract and generic function. For example, in GUI
programming, we frequently need to associate a particular function with a particular
366",ComputerScienceOne.pdf
"25.2. Function Pointers
event. Suppose we create a button; we need to be able to specify what happens when
that button gets clicked. We do so by providing a function as a callback to a registration
function that associates the “click” event with the provided function. Thus, whenever a
user clicks the button, the callback is invoked (“called back”).
Let’s illustrate some syntax usage with another example. Suppose we want to create a
getMax() function. We could write one function for arrays of integers, another for arrays
of doubles, another for Student arrays that gets the student with the maximum GPA,
then another for the ID, then another for the name and so on. Or, we could program
a generic getMax() function that could be used for any type by taking a comparator
function as a callback. To illustrate, consider the following non-generic function for
integers.
1 int getMax(const int *arr, int n) {
2 int i, maxIndex = 0;
3 for(i=1; i<n; i++) {
4 if(arr[maxIndex} < arr[i] {",ComputerScienceOne.pdf
"integers.
1 int getMax(const int *arr, int n) {
2 int i, maxIndex = 0;
3 for(i=1; i<n; i++) {
4 if(arr[maxIndex} < arr[i] {
5 //we've found something larger, update the maxIndex:
6 maxIndex = i;
7 }
8 }
9 return maxIndex;
10 }
This simple function iterates through the array, keeping track of the maximum value
found “so far” and updating it when it finds something larger. Because we’ve specified
that arr is an array if int values, we can use the less-than comparison operator (line
4). Now let’s make it more generic: rather than taking an array of int values, it will
now take a generic void array. Further, we will pass a function pointer to this function
that references a generic comparator function that can be used to replace the less-than
comparison operator.
1 int getMax(const void *arr, int n,
2 int(*cmp)(const void *, const void *)) {
3 int i, maxIndex = 0;
4 for(i=1; i<n; i++) {
5 ...
6 }
7 return maxIndex;
8 }
367",ComputerScienceOne.pdf
"25. Searching & Sorting
There are a couple of issues here that we have to deal with. When working with generic
void * pointers in C and using arrays, you cannot simply index using the usual 0, 1,
2, etc. indices. Recall that when elements are stored in an array, the index represents
an offset of a memory address. If the array is an array of integers or double or some
other built-in type, the compiler knows how large each one is and is able to compute the
appropriate offset given the usual 0, 1, 2, etc. indices.
However, when dealing with void * elements, a function must be told how many bytes
each element takes. C uses an unsigned integer type, size_t to indicate a size in bytes,
so we’ll use it as well. We modify the function signature above to pass in a size_t size
parameter.
We now need a way to access each element in the array. The array itself is generic. We
cannot simply use indices such as arr[i] to access elements. Since it is a void *",ComputerScienceOne.pdf
"cannot simply use indices such as arr[i] to access elements. Since it is a void *
pointer, we cannot simply dereference it using an index. The compiler doesn’t know how
many bytes each element takes, so it cannot compute an offset using the index variable
i. Instead, we need to do the pointer arithmetic ourselves.
We could use the size parameter, say arr[i*size] to compute the offset of the i-th
element, but this still dereferences a void pointer, which we generally do not want to
do. Instead, we can use the pointer arr and do some simple arithmetic; arr is the
starting memory location, so if we simply add i*size to it, that gives us the memory
location of the i-th element as a void pointer !
1 void *first = arr + 0 * size; //first element
2 void *second = arr + 1 * size; //second element
3 void *third = arr + 2 * size; //third element
4 ...
5 void *x = arr + i * size; //i-th element
We can use this in our getMax() function to iterate over each element in the array",ComputerScienceOne.pdf
"4 ...
5 void *x = arr + i * size; //i-th element
We can use this in our getMax() function to iterate over each element in the array
and pass it to our comparator. The comparator expects generic void pointers, and that
is what our pointer arithmetic is computing. Passing two memory addresses to the
comparator determines which is the larger of two elements in the array. To do this, we
call the comparator on the maximum element we’ve found so far and the i-th element
in the loop. If it returns something negative, then we know that the “max” element is
less than the i-th element and so update our maxIndex variable. Making these changes
results in the this final version.
368",ComputerScienceOne.pdf
"25.2. Function Pointers
1 int getMax(const void *arr, int n, size_t size,
2 int(*cmp)(const void *, const void *)) {
3 int i, maxIndex = 0;
4 for(i=1; i<n; i++) {
5 if(cmp(arr + maxIndex * size, arr + i * size) < 0) {
6 //we've found something larger, update the max_index:
7 maxIndex = i;
8 }
9 }
10 return maxIndex;
11 }
Suppose we have an array of integers and the cmpInt() comparator function above.
We can use our getMax() function as follows.
1 int arr[] = {8, 2, 9, 10, 4, 2, 2, 5, 6, 7};
2 int maxIndex = getMax(arr, 10, sizeof(int), cmpInt);
3 printf(""maximum value: %d\n"", arr[maxIndex]);
In this example, the getMax() function would return the index 3 and print the maximum
value stored there, 10. The getMax() function returns the index corresponding to
the “maximum” element according to the comparator used. If instead we had used
cmpIntDesc() the “maximum” element would have been the least element, (2 in the",ComputerScienceOne.pdf
"cmpIntDesc() the “maximum” element would have been the least element, (2 in the
example above) because it would have been the element ordered “last” by the descending
comparator.
Consider another example with our Student structure.
1 int n = 10;
2 Student *roster = (Student *) malloc(sizeof(Student) * n);
3 ...
4 int maxIndex = getMax(roster, n, sizeof(Student), studentByNameCmp);
Since studentByNameCmp() orders by lexicographic ordering, a student with the last
name “Zadora” would have been “larger” than someone with a last name “Anderson.”
369",ComputerScienceOne.pdf
"25. Searching & Sorting
Thus the code above will return an index corresponding to thelast student in lexicographic
ordering of their name. Similarly, if we had used the studentGpaCmp() comparator
instead, getMax() would have returned an index for the student with the lowest GPA
as this comparator ordered highest to lowest.
Another variation on this function would be to return a pointer to the maximum element
rather than an index. The same logic would have applied, but the return type would be
a void * pointer and the return statement would return the memory address of the
maximum element instead. To avoid warnings, in the final return statement we cast arr
as a void pointer (instead of a const void pointer) to make the return value compatible
with the return type.
1 void * getMax(const void *arr, int n, size_t size,
2 int(*cmp)(const void *, const void *)) {
3 int i, maxIndex = 0;
4 for(i=1; i<n; i++) {
5 if(cmp(arr + maxIndex * size, arr + i * size) < 0) {",ComputerScienceOne.pdf
"2 int(*cmp)(const void *, const void *)) {
3 int i, maxIndex = 0;
4 for(i=1; i<n; i++) {
5 if(cmp(arr + maxIndex * size, arr + i * size) < 0) {
6 //we've found something larger, update the max_index:
7 maxIndex = i;
8 }
9 }
10 return (void *)arr + maxIndex * size;
11 }
370",ComputerScienceOne.pdf
"25.2. Function Pointers
1 #include<stdio.h>
2 #include<stdlib.h>
3
4 int function01(int a, double b);
5 void function02(double x, char y);
6
7 void runAFunction(int (*theFunc)(int, double));
8
9 int main(int argc, char **arg) {
10
11 int i = 5;
12 double d = 3.14;
13 char c = 'Q';
14
15 //calling a function normally...
16 int j = function01(i, d);
17 function02(d, c);
18
19 //function pointer declaration
20 int (*pt2Func01)(int, double) = NULL;
21 void (*pt2Func02)(double, char) = NULL;
22
23 //assignment
24 pt2Func01 = function01;
25 //or:
26 pt2Func01 = &function01;
27 pt2Func02 = &function02;
28
29 //you can invoke a function using a pointer to it:
30 j = pt2Func01(i, d);
31 pt2Func02(d, c);
32
33 //alternatively, you can invoke a function by dereferencing it:
34 j = (*pt2Func01)(i, d);
35 (*pt2Func02)(d, c);
36
37 //With function pointers, you can now pass entire functions as arguments to another function!
38 printf(""Calling runAFunction...\n"");
39 runAFunction(pt2Func01);",ComputerScienceOne.pdf
"38 printf(""Calling runAFunction...\n"");
39 runAFunction(pt2Func01);
40 //we should not pass in the second pointer as it would not match the signature:
41 //syntactically okay, compiler warning, undefined behavior
42 //runAFunction(pt2Func02);
43 }
44
45 void runAFunction(int (*theFunc)(int, double)) {
46
47 printf(""calling within runAfunction...\n"");
48 int result = theFunc(20, .5);
49 printf(""the result was %d\n"", result);
50
51 return;
52 }
53
54 int function01(int a, double b) {
55 printf(""You called function01 on a = %d, b = %f\n"", a, b);
56 return a + 10;
57 }
58
59 void function02(double x, char y) {
60
61 printf(""You called function02 on x = %f, y = %c\n"", x, y);
62
63 }
Code Sample 25.1.: C Function Pointer Syntax Examples
371",ComputerScienceOne.pdf
"25. Searching & Sorting
25.3. Searching & Sorting
We now turn our attention to the search and sorting functions provided by the standard
library. Each function is a generic implementation that takes advantage of function
pointers and comparator functions.
25.3.1. Searching
Linear Search
The C search library, search.h provides a linear search function to search arrays, named
lfind() (linear find). This function does not require that the array be sorted and
performs a linear search algorithm, returning a pointer to the first element such that
the comparator returns 0 (indicating equality). The full signature of the function is as
follows.
1 void *lfind(const void *key,
2 const void *base,
3 size_t *nmemb,
4 size_t size,
5 int(*compar)(const void *, const void *));
Each argument represents:
• key – a “dummy” structure instance matching the element you are searching for.
For example, if you are searching for a Student with the last name ""Smith""",ComputerScienceOne.pdf
"For example, if you are searching for a Student with the last name ""Smith""
then you can construct an instance with the same last name, ignoring the value of
all other components.
• base – a pointer to the array to be searched
• nmemb – the size of the array (number of “members”)
• size – the size, in bytes, of each element in the array (generally you use use
sizeof() to determine this)
• compar – a pointer to a comparator function to use in the search.
As with our getMax() example, lfind() returns a pointer to the element it finds. If
no such element is found, this function will return NULL.
372",ComputerScienceOne.pdf
"25.3. Searching & Sorting
In the same library, there is another linear search function:
1 void *lsearch(const void *key,
2 void *base,
3 size_t *nmemb,
4 size_t size,
5 int(*compar)(const void *, const void *));
It differs in that if it does not find a matching element, it still returns NULL but also
attempts to insert the element being searched for at the end of the array. The function
will assume there is enough room at the end of the array to accommodate the inserted
element (if not, the behavior is undefined). This behavior is hinted at by the fact that
base is not const. Moreover, nmemb is passed by reference. If the function inserts
the element, it will also increment nmemb variable to reflect this new element. It is your
responsibility to ensure that there is enough valid memory to accommodate any inserts
when using lsearch().
Binary Search
In the standard library ( stdlib.h) there is an additional binary search function that",ComputerScienceOne.pdf
"when using lsearch().
Binary Search
In the standard library ( stdlib.h) there is an additional binary search function that
can be used to more efficiently search a sorted array. The prototype:
1 void *bsearch(const void *key,
2 const void *base,
3 size_t nmemb,
4 size_t size,
5 int (*compar)(const void *, const void *));
All parameters are exactly as with lfind() as is the behavior: it returns a pointer to
the first element that it finds (though first does not necessarily mean the first in the order
of the array) and NULL if no matching element is found. It is an essential requirement
that the array be sorted with the same comparator as was used to sort the array or NULL
may be returned erroneously.
373",ComputerScienceOne.pdf
"25. Searching & Sorting
25.3.2. Sorting
The standard library also provides a generic sorting function, qsort(). Though the
name suggests a Quick Sort implementation, it does not necessarily have to be (it was
when the function was originally designed). Modern implementations of qsort() may
implement alternatives such as Merge Sort or non-recursive hybrid Quick Sort algorithms.
The prototype and parameters are similar to the search functions but do not include
a key. The array is also not const indicating that it will be changed (which is the
whole point of calling the function). The function will sort elements in ascending order
according to the provided comparator function.
1 void qsort(void *base,
2 size_t nmemb,
3 size_t size,
4 int(*compar)(const void *, const void *))
• base – pointer an array of elements
• nmemb – the size of the array (number of members)
• size – the size (in bytes) of each element (use sizeof())
• compar – a comparator function used to order elements",ComputerScienceOne.pdf
"• size – the size (in bytes) of each element (use sizeof())
• compar – a comparator function used to order elements
The advantages to using qsort() (as well as lfind() and bsearch()) should be
clear. There is no need to write a new function that reimplements the same algorithm for
every possible ordering of every possible user-defined structure. We need only to create a
comparator function and pass it to qsort(). There is less code, and less chance of bugs.
The qsort() function is well-designed, optimized, and most importantly well-tested
and proven.
These functions represent a sort of “weak” form of polymorphic behavior found in
more modern object-oriented programming languages and other languages that support
“generic programming.” Polymorphism is the characteristic that the same code can be
executed on different types, greatly reducing the need for duplicate code.
25.3.3. Examples
We illustrate the usage of these functions in Code Samples 25.2 and 25.3.
374",ComputerScienceOne.pdf
"25.3. Searching & Sorting
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<search.h>
4
5 #include ""student.h""
6
7 int main(int argc, char **argv) {
8
9 int n = 0;
10 Student *roster = loadStudents(""student.data"", &n);
11 int i;
12 size_t numElems = n;
13
14
15 printf(""Roster: \n"");
16 printStudents(roster, n);
17
18 /* Searching */
19 Student *castro = NULL;
20 Student *castroKey = NULL;
21 Student *sandberg = NULL;
22 char *str = NULL;
23
24 castro = linearSearchStudentByNuid(roster, 10, 131313);
25 str = studentToString(castro);
26 printf(""castro: %s\n"", str);
27 free(str);
28
29 //create a key that will match according to the NUID
30 int nuid = 23232323;
31 Student * key = createEmptyStudent();
32 key->nuid = nuid;
33
34 //use lfind to find the first such instance:
35 //sandberg =
36 sandberg = lfind(key, roster, &numElems, sizeof(Student), studentIdCmp);
37 str = studentToString(sandberg);
38 printf(""sandberg: %s\n"", str);
39 free(str);
40",ComputerScienceOne.pdf
"37 str = studentToString(sandberg);
38 printf(""sandberg: %s\n"", str);
39 free(str);
40
41 //create a key with only the necessary fields
42 castroKey = createStudent(""Starlin"", ""Castro"", 0, 0.0);
43 //sort according to a comparator function
44 qsort(roster, n, sizeof(Student), studentLastNameCmp);
45
46 castro = bsearch(castroKey, roster, n, sizeof(Student), studentLastNameCmp);
47 str = studentToString(castro);
48 printf(""castro (via binary search): %s\n"", str);
49 free(str);
50
51 //create a key with only the necessary fields
52 castroKey = createStudent(NULL, NULL, 131313, 0.0);
53 //sort according to a comparator function
54 qsort(roster, n, sizeof(Student), studentIdCmp);
55
56 castro = bsearch(castroKey, roster, n, sizeof(Student), studentIdCmp);
57 str = studentToString(castro);
58 printf(""castro (via binary search): %s\n"", str);
59 free(str);
60
61 return 0;
62 }
.
Code Sample 25.2.: C Search Examples
375",ComputerScienceOne.pdf
"25. Searching & Sorting
1 #include<stdio.h>
2 #include<stdlib.h>
3
4 #include ""student.h""
5
6
7 int main(int argc, char **argv) {
8
9 int n = 0;
10 Student *roster = loadStudents(""student.data"", &n);
11 int i;
12 size_t numElems = n;
13
14 printf(""Roster: \n"");
15 printStudents(roster, n);
16
17 printf(""\n\n\nSorted by last name/first name: \n"");
18 qsort(roster, numElems, sizeof(Student), studentLastNameCmp);
19 printStudents(roster, n);
20
21 printf(""\n\n\nSorted by ID: \n"");
22 qsort(roster, numElems, sizeof(Student), studentIdCmp);
23 printStudents(roster, n);
24
25 printf(""\n\n\nSorted by ID, descending: \n"");
26 qsort(roster, numElems, sizeof(Student), studentIdCmpDesc);
27 printStudents(roster, n);
28
29 printf(""\n\n\nSorted by GPA: \n"");
30 qsort(roster, numElems, sizeof(Student), studentGPACmp);
31 printStudents(roster, n);
32
33 return 0;
34 }
.
Code Sample 25.3.: C Sort Examples
376",ComputerScienceOne.pdf
"25.4. Other Considerations
25.4. Other Considerations
25.4.1. Sorting Pointers to Elements
Recall that it is sometimes preferable to maintain an array of pointers to structures
rather than an array of structures. Sorting is a scenario where this is particularly true.
When sorting an array of structure elements, the entire structure is copied back and forth
as elements are swapped. Depending on the number of bytes of a structure, this can be
quite expensive. It is generally more efficient to sort an array of pointers to structures
instead. A prime example of this is sorting an array of strings.
An array of strings can be thought of as a 2-dimensional array of chars. Specifically,
an array of strings is a char ** type. That is, an array of pointers to pointers of
chars. We may be tempted to use strcmp() in the standard string library, passing it
to qsort(). Unfortunately this will not work. qsort() requires two const void *",ComputerScienceOne.pdf
"to qsort(). Unfortunately this will not work. qsort() requires two const void *
types, while strcmp() takes two const char * types. This difference is subtle but
important.1 The recommended way of doing this is to define a different comparator
function as in Code Sample 25.4.
1 /* compare strings via pointers */
2 int pstrcmp(const void *p1, const void *p2)
3 {
4 return strcmp(*(char * const *)p1, *(char * const *)p2);
5 }
Code Sample 25.4.: C Comparator Function for Strings
Observe the behavior of this function: it uses the standard strcmp() function, but
makes the proper explicit type casting before doing so. The *(char * const *) casts
the generic void pointers as pointers to strings (or pointers to pointers to characters),
then dereferences it to be compatible with strcmp().
Another case is when we wish to sort user-defined structures. The Student structure
presented earlier is “small” in that it only has a few fields. When structures are stored",ComputerScienceOne.pdf
"presented earlier is “small” in that it only has a few fields. When structures are stored
in an array and sorted, there may be many swaps of individual elements which involves
a lot of memory copying. If the structures are small this is not too bad, but for “larger”
structures this could be potentially expensive. Instead, it may be preferred to have
an array of pointers to structures. Swapping elements involves only swapping pointers
1A full discussion can be found on the c-faq, http://c-faq.com/lib/qsort1.html.
377",ComputerScienceOne.pdf
"25. Searching & Sorting
instead of the entire structure. This is far cheaper as a memory address is likely to be
far smaller than the actual structure it points to. This is essentially equivalent to the
string scenario: we have an array of pointers to be sorted, our comparator function then
needs to deal with pointers to pointers. 2 An example appears in Code Sample 25.5.
1 /**
2 * Orders two Student pointers according to the last name/first name
3 */
4 int studentPtrLastNameCmp(const void *s1, const void *s2) {
5 //we receive a pointer to an individual element in the array
6 //but individual elements are POINTERS to students thus we cast
7 //them as (const Student **) then dereference to get a pointer
8 //to a Student!
9 const Student *a = *(const Student **)s1;
10 const Student *b = *(const Student **)s2;
11 int result = strcmp(a->lastName, b->lastName);
12 if(result == 0) {
13 return strcmp(a->firstName, b->firstName);
14 } else {
15 return result;
16 }
17 }
18
19
20",ComputerScienceOne.pdf
"12 if(result == 0) {
13 return strcmp(a->firstName, b->firstName);
14 } else {
15 return result;
16 }
17 }
18
19
20
21 Student **roster = (Student **) malloc(sizeof(Student *) * n);
22 ...
23 qsort(roster, n, sizeof(Student *), studentPtrLastNameCmp);
Code Sample 25.5.: Sorting Structures via Pointers
Another issue when sorting arrays of pointers is that we may now have to deal with NULL
elements. When sorting arrays of elements this is not an issue as a properly initialized
array will contain non-null elements (though elements could still be uninitialized, the
memory space will be valid).
How we handle NULL pointers is more of a design decision. We could ignore it and any
attempt to access a NULL structure will result in undefined behavior (or segmentation
faults, etc.). Or we could give NULL values an explicit ordering with respect to other
2Again, a full discussion can be found on c-faq, http://c-faq.com/lib/qsort2.html.
378",ComputerScienceOne.pdf
"25.4. Other Considerations
elements. That is, we could order all NULL pointers before non- NULL elements (and
consider all NULL pointers to be equal). An example with respect to our Student
structure is given in Code Snippet 25.6.
1 int studentPtrLastNameCmpWithNulls(const void *s1, const void *s2) {
2 const Student *a = *(const Student **)s1;
3 const Student *b = *(const Student **)s2;
4 if(a == NULL && b == NULL) {
5 return 0;
6 } else if(a == NULL && b != NULL) {
7 return -1;
8 } else if (a != NULL && b == NULL) {
9 return 1;
10 }
11 int result = strcmp(a->lastName, b->lastName);
12 if(result == 0) {
13 return strcmp(a->firstName, b->firstName);
14 } else {
15 return result;
16 }
17 }
Code Sample 25.6.: Handling Null Values
379",ComputerScienceOne.pdf
"Part II.
The Java Programming Language
381",ComputerScienceOne.pdf
"26. Basics
The Java programming language was developed in the early 1990s at Sun Microsystems
by James Gosling, Mike Sheridan, and Patrick Naughton. Its original intention was
to enable cable box sets to be more interactive. By the mid-90s, Java was retargeted
toward the WWW. The first public release came on May 23, 1995 with the first Java
Development Kit (JDK), Java 1.0 on January 23rd, 1996. A new, updated release has
come about every other year. As of 2014, Java 8 is the current stable version.
Today, Java is one of the most popular programming languages, consistently ranked
as one of the top 2 languages (see http://www.tiobe.com). It is now owned and
maintained by Oracle, but there are many open source tools, compilers and runtime
environments available. Java is used in everything from mobile devices (Android) and
desktop applications to enterprise application servers.
From its inception, Java was designed with 5 basic principles:
1. Simple, Object-oriented, familiar",ComputerScienceOne.pdf
"From its inception, Java was designed with 5 basic principles:
1. Simple, Object-oriented, familiar
2. Robust and secure
3. Architecture-neutral and portable
4. High performance
5. Interpreted, threaded and dynamic
Java offers many key features that have made it popular. It is unique in that it is
not entirely compiled nor interpreted. Instead, Java source code is compiled into an
intermediate form, called Java bytecode. This bytecode is not directly runnable on
a processor. Instead, a JVM, an application that was written and compiled for a
particular system, interprets the bytecode and runs the application. This added layer
of abstraction means that Java source code can be written once (and compiled once)
and then run anywhere on any device that has a JVM. The added layer of abstraction
makes development easier, but comes at a cost in performance. However, the most recent
JVMs have offered performance that is comparable to native machine code in many
applications.",ComputerScienceOne.pdf
"JVMs have offered performance that is comparable to native machine code in many
applications.
Another key feature is that Java has its own automated garbage collection. Some
languages require manual memory management, meaning that requesting, managing,
383",ComputerScienceOne.pdf
"26. Basics
1 package unl.cse; //package declaration
2
3 //imports would go here
4
5 /**
6 * A basic hello world program in Java
7 */
8 public class HelloWorld {
9
10 //static main method
11 public static void main(String args[]) {
12 System.out.println(""Hello World!"");
13 }
14
15 }
Code Sample 26.1.: Hello World Program in Java
and freeing memory is part of the code that you write as a developer. Failure to handle
memory management properly can lead to wasted resources (memory leaks), poor or
unstable performance, and even more serious security issues (buffer overflows). In Java,
there is no manual memory management. The JVM handles the allocation and clean up
of memory automatically.
In following with the five design principles, Java is similar in syntax to C (called “C-style
syntax”). Executable statements are terminated by semicolons, code blocks are defined
by opening/closing curly brackets, etc. Java is also fundamentally a class-based OOP",ComputerScienceOne.pdf
"by opening/closing curly brackets, etc. Java is also fundamentally a class-based OOP
language. With the exception of a few primitive types, in Java everything is a class or
belongs to an class.
26.1. Getting Started: Hello World
The hallmark of an introduction to a new programming language is the Hello World!
program. It consists of a simple program whose only purpose is to print out the message
“Hello World!” to the user. The simplicity of the program allows the focus to be on the
basic syntax of the language. It is also typically used to ensure that your development
environment, compiler, runtime environment, etc. are functioning properly with a minimal
example. A basic Hello World! program in Java can be found in Code Sample 26.1.
384",ComputerScienceOne.pdf
"26.2. Basic Elements
We will not focus on any particular development environment, code editor, or any
particular operating system, compiler, or ancillary standards in our presentation. However,
as a first step, you should be able to write, compile, and run the above program
on the environment you intend to use for the rest of this book. This may require
that you download and install a JDK and IDE. Eclipse ( http://eclipse.org/) is the
industry standard, though IntelliJ (https://www.jetbrains.com/idea/) and NetBeans
(https://netbeans.org/) are also popular.
26.2. Basic Elements
Using the Hello World! program as a starting point, we will now examine the basic
elements of the Java language.
26.2.1. Basic Syntax Rules
Java’s syntax is adopted from C, referred to as “C-style syntax.” These elements include
the following.
•Java is a statically typed language so variables must be declared along with their
types before using them.",ComputerScienceOne.pdf
"the following.
•Java is a statically typed language so variables must be declared along with their
types before using them.
•Strings are delimited with double quotes. Single characters, including special
escaped characters are delimited by single quotes; ""this is a string"", and
these are characters: 'A', '4', '$' and '\n'
•In addition, Java uses Unicode (UTF-16 encoding) to represent characters. This is
fully back-compatible with ASCII, but also allows you to specify Unicode characters
using special escape sequences and hexadecimal encodings. For example, '\u4FFA'
represents a Japanese character:
The string ""\u4FFA\u306F\u6700\u9AD8\u3060\u305C\uFF01"" represents the
phrase
•Executable statements are terminated by a semicolon, ;
•Code blocks are defined using opening and closing curly brackets, { ... }. More-
over, code blocks can be nested: code blocks can be defined within other code
385",ComputerScienceOne.pdf
"26. Basics
blocks.
•Variables are scoped to the code block in which they are declared and are only
valid within that code block.
•In general, whitespace between coding elements is ignored.
Though not a syntactic requirement, the proper use of whitespace is important for good,
readable code. Code inside code blocks is indented at the same indentation. Nested
code blocks are indented further. Think of a typical table of contents or the outline of a
formal paper or essay. Sections and subsections or points and subpoints all follow proper
indentation with elements at the same level at the same indentation. This convention is
used to organize code and make it more readable.
26.2.2. Program Structure
Classes
In Java, everything is a class or belongs to a class. A class is an extensible program or
blueprint for creating objects. Objects are an integral part of OOP that we’ll explore later.
However, to start out our programs will be simple enough that they can be contained in",ComputerScienceOne.pdf
"However, to start out our programs will be simple enough that they can be contained in
a single class. To declare a class, you use the following syntax.
public class HelloWorld { ... }
where the contents of the class are placed between the opening/closing curly brackets.
In addition, a class must be placed in a Java source file with the same name. In our
previous example, the class must be in a file named HelloWorld.java.
When naming classes, the most commonly accepted naming convention is to use uppercase
camel casing (also called PascalCase) in which each word in the name is capitalized
including the first, Employee, SavingsAccount, ImageFile, etc. Class names (as
well as their source file names) are case sensitive.
Packages
Java code is organized into modules called packages. Packages are essentially directories
(or folders) which follow a directory tree structure which allows subdirectories and
separate directories at the same level. It all starts at the root directory called the",ComputerScienceOne.pdf
"separate directories at the same level. It all starts at the root directory called the
“default” package.
Within a source file, we declare which package the file belongs to using the keyword
package followed by a fully qualified package declaration which is essentially just the
386",ComputerScienceOne.pdf
"26.2. Basic Elements
names of the directories that the file is located in, separated by a period. The declaration
is terminated by a semicolon. For example, the package declaration,
package unl.cse;
would indicate that the file belongs in the directory cse which is a subdirectory of the
directory unl. The absence of a package declaration will mean that the file is associated
with the default directory.
Packages allow you to organize source files and code functionality. Mathematics related
classes can be placed in one package while image related classes can be placed in another,
etc. It also provides separate name spaces for classes. There could be 3 or 4 different
classes named List for example. They could not be located in the same directory; if
you wanted to use one which one would you be referring to? By separating them into
packages, they can all exist without conflict. When we want to use a particular one, we
import that class with its fully qualified package name.
Imports",ComputerScienceOne.pdf
"import that class with its fully qualified package name.
Imports
An import statement essentially “brings in” another class so that its methods and
functionality can be used. For example, there is a class named Scanner (located in
the package java.util) that makes it easy to read input from the standard input. To
include it in our program so that we can use its functionality, we would need 1 to import
it:
import java.util.Scanner;
Classes in the package java.lang (such as String and Math) are considered standard
and are imported by default without an explicit import statement.
You may see some code that uses a wildcard like import java.util.*; which ends
up importing every class in that package. This is generally considered bad practice. In
general, code should be intentional and specific, importing every class even if they are
not used goes against this principle.
When naming packages, you must follow the general naming rules for identifiers (see below).",ComputerScienceOne.pdf
"not used goes against this principle.
When naming packages, you must follow the general naming rules for identifiers (see below).
Package names cannot begin with a number, no whitespace, etc. Moreover, the general con-
vention for package names is to use lowercase underscore casing, here_is_an_example.
Moreover, packages and subpackages follow the same convention as directories: the top
most directory is the most general and subdirectories are more and more specific.
In many of our examples we’ll use unl.cse (UNL, University of Nebraska–Lincoln; CSE,
1You can still use it without importing it, but you’d need to use a fully qualified path name at
declaration/instantiation.
387",ComputerScienceOne.pdf
"26. Basics
Function Description
Math.abs(x) Absolute value function, |x|a
Math.ceil(x) Ceiling function, ⌈46.3⌉= 47.0
Math.floor(x) Floor function, ⌊46.3⌋= 46.0
Math.cos(x) Cosine functionb
Math.sin(x) Sine functionb
Math.tan(x) Tangent functionb
Math.exp(x) Exponential function, ex, e= 2.71828 ...
Math.log(x) Natural logarithm, ln (x)c
Math.log10(x) Logarithm base 10, log 10 (x)c
Math.pow(x,y) The power function, computes xy
Math.sqrt(x) Square root functionc
Table 26.1.: Several methods defined in the Java Math library. aThere are several versions
of the absolute value method, one for each numeric type but all with the
same name. ball trigonometric functions assume input is in radians, not
degrees. cInput is assumed to be positive, x> 0.
Department of Computer Science & Engineering) which illustrates this general-to-specific
organization.
There are many other important classes and packages provided by the standard JDK",ComputerScienceOne.pdf
"organization.
There are many other important classes and packages provided by the standard JDK
that we’ll examine as needed. Of immediate interest is the Math class and its library
of common mathematical functions such as the square root and the natural logarithm.
Table 26.1 highlights several of these functions. To use them you’d need to invoke them
by using the class name and dot operator. For example:
1 double x = 1.5;
2 double y, z;
3 y = Math.sqrt(x); //y now has the value √x=
√
1.5
4 z = Math.sin(x); //z now has the value sin (x) = sin (1.5)
In both of the method calls above, the value of the variable x is “passed” to the math
function which computes and “returns” the result which then gets assigned to another
variable.
388",ComputerScienceOne.pdf
"26.2. Basic Elements
26.2.3. The main() Method
Every executable program has to have a beginning: a point at which the program starts
to execute. In Java, a class may contain many variables and methods, but a class is only
executable if it contains a main() method. When a Java class is compiled and the JVM
is started, the JVM loads the class into memory and starts executing code contained in
the main() method.
In addition, our main() method takes an array of String types which serve to
communicate any command line arguments provided to the program (review Section 2.4.4
for details). The array, args stores the arguments as strings. The number of arguments
provided can be determined using the length property of the array. Specifically,
args.length is an integer indicating how many arguments were provided. This does
not include the name of the program (class) itself as that is already be known to the
programmer.",ComputerScienceOne.pdf
"not include the name of the program (class) itself as that is already be known to the
programmer.
To access any one argument, it will be necessary to index the array. The index for the
first argument is zero, thus the first argument is args[0], the second is args[1], etc.
The last one would be at args[args.length-1].
If a user is expected to provide numbers as input, they’ll need to be converted as the
args array are only String types. To convert the arguments you can use parsing
methods provided by the Integer and Double classes. An example:
1 //converts the ""first"" command line argument to an integer
2 int x = Integer.parseInt(args[0]);
3 //converts the ""third"" command line argument to a double:
4 double y = Double.parseDouble(args[1]);
26.2.4. Comments
Comments can be written in a Java program either as a single line using two forward
slashes, //comment or as a multiline comment using a combination of forward slash and",ComputerScienceOne.pdf
"slashes, //comment or as a multiline comment using a combination of forward slash and
asterisk: /* comment */ . With a single line comment, everything on the line after the
forward slashes is ignored. With a multiline comment, everything in between the forward
slash/asterisk is ignored. Comments are ultimately ignored so the amount of comments
do not have an effect on the final executable code. Consider the following example.
389",ComputerScienceOne.pdf
"26. Basics
1 //this is a single line comment
2 int x; //this is also a single line comment, but after some code
3
4 /*
5 This is a comment that can
6 span multiple lines to format the comment
7 message more clearly
8 */
9 double y;
Most code editors and IDEs will present comments in a special color or font to distinguish
them from the rest of the code (just as our example above does). Failure to close a
multiline comment will likely result in a compiler error but with color-coded comments
its easy to see the mistake visually.
Another common comment style convention is the Javadoc (Java Documentation) style
of comments. Javadoc style comments are multiline comments that begin with /**.
The Javadoc framework allows you to markup your comments with tags and links so that
documentation can be automatically generated and published. We will sometimes use
this style, but we will not cover the details.
26.3. Variables",ComputerScienceOne.pdf
"documentation can be automatically generated and published. We will sometimes use
this style, but we will not cover the details.
26.3. Variables
Java has 8 built-in primitive types supporting numbers (integers and floating point
numbers), Booleans, and characters. Table 26.2 contains a complete description of these
types. Each of these primitive types also has a corresponding wrapper class defined in
the java.lang package. Wrapper classes provide object versions of each of these classes.
The object versions have many utility methods that can be used in relation to their type.
For example, the aforementioned Integer.parseInt() method is part of the Integer
wrapper class.
The wrapper classes, however, are different. These are objects, so when a reference is
declared for them, by default, that reference refers to null. The keyword null is used
to indicate a special memory address that represents “nothing.” In fact, the default value",ComputerScienceOne.pdf
"to indicate a special memory address that represents “nothing.” In fact, the default value
for any object type is null. Care must be taken when mixing primitive types and their
wrapper classes (see below) as null references may result in a NullPointerException.
Finally, instances of the wrapper classes are immutable. Once they are created, they
cannot be changed. References can be made to refer to a different object, but the object’s
value cannot be changed.
390",ComputerScienceOne.pdf
"26.3. Variables
Type Description Wrapper Class
byte 8-bit signed 2s complement integer Byte
short 16-bit signed 2s complement integer Short
int 32-bit signed 2s complement integer Integer
long 64-bit signed 2s complement integer Long
float 32-bit IEEE 754 floating point number Float
double 64-bit floating point number Double
boolean may be set to true or false Boolean
char 16-bit Unicode (UTF-16) character Character
Table 26.2.: Primitive types in Java
26.3.1. Declaration & Assignment
Java is a statically typed language meaning that all variables must be declared before
you can use them or refer to them. In addition, when declaring a variable, you must
specify both its type and its identifier. For example:
1 int numUnits;
2 double costPerUnit;
3 char firstInitial;
4 boolean isStudent;
Each declaration specifies the variable’s type followed by the identifier and ending with a
semicolon. The identifier rules are fairly standard: a name can consist of lowercase and",ComputerScienceOne.pdf
"semicolon. The identifier rules are fairly standard: a name can consist of lowercase and
uppercase alphabetic characters, numbers, and underscores but may not begin with a
numeric character. We adopt the modern camelCasing naming convention for variables
in our code. In general, variables must be assigned a value before you can use them in
an expression. You do not have to immediately assign a value when you declare them
(though it is good practice), but some value must be assigned before they can be used or
the compiler will issue an error. 2
The assignment operator is a single equal sign, = and is a right-to-left assignment. That
is, the variable that we wish to assign the value to appears on the left-hand-side while
the value (literal, variable or expression) is on the right-hand-size. Using our variables
from before, we can assign them values:
2Instance variables, that is variables declared as part of an object do have default values. For objects,",ComputerScienceOne.pdf
"from before, we can assign them values:
2Instance variables, that is variables declared as part of an object do have default values. For objects,
the default is null , for all numeric types, zero is the default value. For the boolean type, false
is the default, and the default char value is \0 , the null-terminating character (zero in the ASCII
table).
391",ComputerScienceOne.pdf
"26. Basics
1 numUnits = 42;
2 costPerUnit = 32.79;
3 firstInitial = 'C';
4 isStudent = true;
For brevity, Java allows you to declare a variable and immediately assign it a value on
the same line. So these two code blocks could have been more compactly written as:
1 int numUnits = 42;
2 double costPerUnit = 32.79;
3 char firstInitial = 'C';
4 boolean isStudent = true;
As another shorthand, we can declare multiple variables on the same line by delimiting
them with a comma. However, they must be of the same type. We can also use an
assignment with them.
1 int numOrders, numUnits = 42, numCustomers = 10, numItems;
2 double costPerUnit = 32.79, salesTaxRate;
Another useful keyword is final. Though it has several uses, when applied to a variable
declaration, it makes it a read-only variable. After a value has been assigned to a final
variable, its value cannot be changed.
1 final int secret = 42;
2 final double salesTaxRate = 0.075;",ComputerScienceOne.pdf
"variable, its value cannot be changed.
1 final int secret = 42;
2 final double salesTaxRate = 0.075;
Any attempt to reassign the values of final variables will result in a compiler error.
392",ComputerScienceOne.pdf
"26.4. Operators
26.4. Operators
Java supports the standard arithmetic operators for addition, subtraction, multiplication,
and division using +, -, *, and / respectively. Each of these operators is a binary
operator that acts on two operands which can either be literals or other variables and
follow the usual rules of arithmetic when it comes to order of precedence (multiplication
and division before addition and subtraction).
1 int a = 10, b = 20, c = 30, d;
2 d = a + 5;
3 d = a + b;
4 d = a - b;
5 d = a + b * c;
6 d = a * b;
7 d = a / b; //integer division and truncation! See below
8
9 double x = 1.5, y = 3.4, z = 10.5, w;
10 w = x + 5.0;
11 w = x + y;
12 w = x - y;
13 w = x + y * z;
14 w = x * y;
15 w = x / y;
16
17 //you can do arithmetic with both types:
18 w = a + x;
19
20 //however you CANNOT assign a double to an integer:
21 d = b + y; //compilation error
22 //but you can do so with an explicit type cast:
23 d = (int) (b + y); //though truncation occurs, d is 23",ComputerScienceOne.pdf
"21 d = b + y; //compilation error
22 //but you can do so with an explicit type cast:
23 d = (int) (b + y); //though truncation occurs, d is 23
In addition, you can mix the wrapper classes with their primitive types. You must be
careful though. The wrapper classes are object references which can be null. If a null
reference is used in an arithmetic expression, it will result in a NullPointerException
which can be caught and handled (see Chapter 30). If not caught, it will end up being a
fatal error. Some examples:
1 int a = 10, c;
2 Integer b = 20;
393",ComputerScienceOne.pdf
"26. Basics
3
4 //int and Integer can be mixed:
5 c = a + b;
6
7 double x = 3.14, z;
8 Double y = 2.71;
9 //double and Double can be mixed:
10 z = x + y;
11
12 //all types can be mixed:
13 double u = a + x + b + y;
14
15 //Be careful:
16 Integer d = null;
17 c = a + d; //NullPointerException
This works because of a mechanism called autoboxing (or autounboxing in this case). The
wrapper class is acting like a “box”: it is an object that stores the value of a primitive
type. When it gets used in an arithmetic expression, it gets “unboxed” and converted to
a primitive type so that the arithmetic operation is performed on compatible primitive
types. This is all done by the compiler and is completely transparent to us. However,
that is the reason that we may get a NullPointerExcpetion. Our code actually gets
converted from c = a + d; to c = a +d.doubleValue();. The doubleValue()
method returns a double primitive value. However, if d is null, you can’t call a",ComputerScienceOne.pdf
"method returns a double primitive value. However, if d is null, you can’t call a
method on it; thus the NullPointerException is thrown as a consequence.
Special care must be taken when dealing with int types. For all four operators, if
both operands are integers, the result will be an integer. For addition, subtraction, and
multiplication this isn’t an issue, but for division it means that when we divide, say
(10 / 20), the result is not 0.5 as expected. The number 0.5 is a floating point number.
As such, the fractional part gets truncated (cut off and thrown out) leaving only zero. In
the code above, d = a / b; the variable d ends up getting the value zero because of
this.
A solution to this problem is to use explicit type casting to force at least one of the
operands in an integer division to become a double type. For example:
1 int a = 10, b = 20;
2 double x;
3
4 x = (double) a / b;
394",ComputerScienceOne.pdf
"26.5. Basic I/O
Assigning a floating point number to an integer is not allowed in Java and attempting to
do so will be treated as a compiler error. This is because Java does not support implicit
type casts. However, you can do so if you provide an explicit type cast as in the code
above,
d = (int) (b + y);
In this code, b + y is correctly computed as 20 + 3 .4 = 23.4, but the explicit type cast
(down to an integer) results in truncation. The .4 gets cutoff and d gets the value 23.
Assigning an int value to a double variable is not a problem as the integer 2 becomes
the floating point number 2 .0.
Java also supports the integer remainder operator using the % symbol. This operator
gives the remainder of the result of dividing two integers. Examples:
1 int x;
2
3 x = 10 % 5; //x is 0
4 x = 10 % 3; //x is 1
5 x = 29 % 5; //x is 4
26.5. Basic I/O
Java provides several ways to perform input and output operations with the standard",ComputerScienceOne.pdf
"4 x = 10 % 3; //x is 1
5 x = 29 % 5; //x is 4
26.5. Basic I/O
Java provides several ways to perform input and output operations with the standard
input/output as part of the System class. The System class contains a standard output
stream which can be accessed using System.out. It also has a standard input stream
which can be accessed using System.in.
For output, there are several methods that can be used but we’ll focus onSystem.out.println()
and System.out.printf(). The first is for printing strings on a single line. The ln
at the end indicates that the method will insert an endline character, \n for you. The
second is a printf-style output method that takes placeholders like %d and %f (review
Section 2.4.3 for details).
The easiest way to read input is through the Scanner class. You can create a new
instance of the Scanner class and associate it with the standard input using the following
395",ComputerScienceOne.pdf
"26. Basics
1 Scanner s = new Scanner(System.in);
2 int a;
3 System.out.println(""Please enter a number: "");
4 a = s.nextInt();
5 System.out.printf(""Great, you entered %d\n"", a);
Code Sample 26.2.: Basic Input/Output in Java
code.
Scanner s = new Scanner(System.in);
The variable s is now active and can be read from. You can get specific values
from the Scanner by calling various methods such as s.nextInt() to get an int,
s.nextDouble() to get a double, etc. When these methods are called, the program
blocks until the user enters her input and presses the enter/return key. The conversion to
the type you requested is automatic. A full example is depicted in Code Sample 26.2.
One potential problem with using Scanner is that the methods cannot force a user to
enter good input. In the example above, if the user, instead of entering a number, entered
""Hello"", the conversion to a number would fail and result in aInputMismatchException.
26.6. Examples
26.6.1. Converting Units",ComputerScienceOne.pdf
"""Hello"", the conversion to a number would fail and result in aInputMismatchException.
26.6. Examples
26.6.1. Converting Units
Let’s write a program that will prompt the user to enter a temperature in degrees
Fahrenheit and convert it to degrees Celsius using the formula
C = (F −32) ·5
9
We begin with the basic program outline which will include a package and class declaration.
We’ll also need to read from the standard input, so we’ll import the Scanner class.
We’ll want want our class to be executable, so we need to put a main() method in our
class. Finally, we’ll document our program to indicate its purpose.
396",ComputerScienceOne.pdf
"26.6. Examples
1 package unl.cse;
2
3 import java.util.Scanner;
4
5 /**
6 * This program converts Fahrenheit temperatures to
7 * Celsius
8 */
9 public class TemperatureConverter {
10
11 public static void main(String args[]) {
12
13 //TODO: implement this
14
15 }
16 }
It is common for programmers to use a comment along with a TODO note to themselves
as a reminder of things that they still need to do with the program.
Let’s first outline the basic steps that our program will go through:
1. We’ll first prompt the user for input, asking them for a temperature in Fahrenheit
2. Next we’ll read the user’s input, likely into a floating point number as degrees can
be fractional
3. Once we have the input, we can calculate the degrees Celsius by using the formula
above
4. Lastly, we will want to print the result to the user to inform them of the value
Sometimes it is helpful to write an outline of such a program directly in the code using
comments to provide a step-by-step process. For example:",ComputerScienceOne.pdf
"Sometimes it is helpful to write an outline of such a program directly in the code using
comments to provide a step-by-step process. For example:
1 package unl.cse;
2
3 import java.util.Scanner;
4
5 /**
6 * This program converts Fahrenheit temperatures to
7 * Celsius
397",ComputerScienceOne.pdf
"26. Basics
8 */
9 public class TemperatureConverter {
10
11 public static void main(String args[]) {
12
13 //TODO: implement this
14 //1. Prompt the user for input in Fahrenheit
15 //2. Read the Fahrenheit value from the standard input
16 //3. Compute the degrees Celsius
17 //4. Print the result to the user
18
19 }
20 }
As we read each step it becomes apparent that we’ll need a couple of variables: one to
hold the Fahrenheit (input) value and one for the Celsius (output) value. It also makes
sense that each of these should be double variables as we want to support fractional
values. At the top of our main() method, we’ll add the variable declarations:
double fahrenheit, celsius;
We’ll also need a Scanner, initialized to read from the standard input:
Scanner s = new Scanner(System.in);
Each of the steps is now straightforward; we’ll use a System.out.println() statement
in the first step to prompt the user for input:
System.out.println(""Please enter degrees in Fahrenheit: "");",ComputerScienceOne.pdf
"in the first step to prompt the user for input:
System.out.println(""Please enter degrees in Fahrenheit: "");
In the second step, we’ll use our Scanner to read in a value from the user for the
fahrenheit variable. Recall that we use the method s.nextDouble() to read a
double value from the user.
fahrenheit = s.nextDouble();
We can now compute celsius using the formula provided:
celsius = (fahrenheit - 32) * (5 / 9);
Finally, we use System.out.printf() to output the result to the user:
System.out.printf(""%f Fahrenheit is %f Celsius\n"", fahrenheit, celsius);
398",ComputerScienceOne.pdf
"26.6. Examples
Try typing and running the program as defined above and you’ll find that you don’t get
correct answers. In fact, you’ll find that no matter what values you enter, you get zero.
This is because of the calculation using (5 / 9): recall what happens with integer
division: truncation! This will always end up being zero.
One way we could fix it would be to pull out our calculators and find that 5
9 = 0.55555 ...
and replace (5 / 9) with 0.555555. But, how many fives? It may be difficult to tell
how accurate we can make this floating point number by hardcoding it ourselves. A much
better approach would be to let the compiler take care of the optimal computation for us
by making at least one of the numbers a double to prevent integer truncation. That is,
we should instead use 5.0 / 9. The full program can be found in Code Sample 26.3.
1 package unl.cse;
2
3 import java.util.Scanner;
4
5 public class TemperatureConverter {
6
7 public static void main(String args[]) {
8",ComputerScienceOne.pdf
"1 package unl.cse;
2
3 import java.util.Scanner;
4
5 public class TemperatureConverter {
6
7 public static void main(String args[]) {
8
9 double fahrenheit, celsius;
10 Scanner s = new Scanner(System.in);
11
12 //1. Prompt the user for input in Fahrenheit
13 System.out.println(""Please enter degrees in Fahrenheit: "");
14
15 //2. Read the Fahrenheit value from the standard input
16 fahrenheit = s.nextDouble();
17
18 //3. Compute the degrees Celsius
19 celsius = (fahrenheit - 32) * 5.0 / 9;
20
21 //4. Print the result to the user
22 System.out.printf(""%f Fahrenheit is %f Celsius\n"",
23 fahrenheit, celsius);
24
25 }
26 }
Code Sample 26.3.: Fahrenheit-to-Celsius Conversion Program in Java
399",ComputerScienceOne.pdf
"26. Basics
26.6.2. Computing Quadratic Roots
Some programs require the user to enter multiple inputs. The prompt-input process can
be repeated. In this example, consider asking the user for the coefficients, a,b,c to a
quadratic polynomial,
ax2 + bx+ c
and computing its roots using the quadratic formula,
x= −b±
√
b2 −4ac
2a
As before, we can create a basic program with a main() method and start filling in
the details. In particular, we’ll need to prompt for the input a, then read it in; then
prompt for b, read it in and repeat for c. We’ll also need several variables: three for the
coefficients a,b,c and two more; one for each root.
1 double a, b, c, root1, root2;
2 Scanner s = new Scanner(System.in);
3
4 System.out.println(""Please enter a: "");
5 a = s.nextDouble();
6 System.out.println(""Please enter b: "");
7 b = s.nextDouble();
8 System.out.println(""Please enter c: "");
9 c = s.nextDouble();
Now to compute the roots: we need to adapt the formula so it accurately reflects the",ComputerScienceOne.pdf
"8 System.out.println(""Please enter c: "");
9 c = s.nextDouble();
Now to compute the roots: we need to adapt the formula so it accurately reflects the
order of operations. We also need to use the standard math library’s square root method.
1 root1 = (-b + Math.sqrt(b*b - 4*a*c) ) / (2*a);
2 root2 = (-b - Math.sqrt(b*b - 4*a*c) ) / (2*a);
Finally, we print the output using System.out.printf(). The full program can be
found in Code Sample 26.4.
This program was interactive. As an alternative, we could have read all three of the
inputs as command line arguments, taking care that we need to convert them to floating
point numbers. Lines 16–21 in the program could have been changed to
400",ComputerScienceOne.pdf
"26.6. Examples
1 a = Double.parseDouble(args[0]);
2 b = Double.parseDouble(args[1]);
3 c = Double.parseDouble(args[2]);
Finally, think about the possible inputs a user could provide that may cause problems
for this program. For example:
•What if the user entered zero for a?
•What if the user entered some combination such that b2 <4ac?
•What if the user entered non-numeric values?
•For the command line argument version, what if the user provided less than three
arguments? Or more?
How might we prevent the consequences of such bad input? How might we handle the
event that a user enters bad input and how do we communicate these errors to the user?
To start to resolve these issues, we’ll need conditionals.
401",ComputerScienceOne.pdf
"26. Basics
1 package unl.cse;
2
3 import java.util.Scanner;
4
5 /**
6 * This program computes the roots to a quadratic equation
7 * using the quadratic formula.
8 */
9 public class QuadraticRoots {
10
11 public static void main(String args[]) {
12
13 double a, b, c, root1, root2;
14 Scanner s = new Scanner(System.in);
15
16 System.out.println(""Please enter a: "");
17 a = s.nextDouble();
18 System.out.println(""Please enter b: "");
19 b = s.nextDouble();
20 System.out.println(""Please enter c: "");
21 c = s.nextDouble();
22
23 root1 = (-b + Math.sqrt(b*b - 4*a*c) ) / (2*a);
24 root2 = (-b - Math.sqrt(b*b - 4*a*c) ) / (2*a);
25
26 System.out.printf(""The roots of %fx^2 + %fx + %f are: \n"",
27 a, b, c);
28 System.out.printf("" root1 = %f\n"", root1);
29 System.out.printf("" root2 = %f\n"", root2);
30
31 }
32
33 }
Code Sample 26.4.: Quadratic Roots Program in Java
402",ComputerScienceOne.pdf
"27. Conditionals
Java supports the basic if, if-else, and if-else-if conditional structures as well as switch
statements. Java has Boolean types and logical statements are built using the standard
logical operators for numeric comparisons as well as logical operators such as negations,
And, and Or that can be used with Boolean types.
27.1. Logical Operators
Recall that Java has Boolean types built-in to the language using either the primitive type,
boolean or its wrapper class Boolean. Moreover, the keywords true and false
can be used to assign and check values. Because Java has Boolean types, it does not
allow you to mix logical operators with numeric types. That is, code like the following is
invalid.
1 int a = 10;
2 boolean b = true;
3 boolean result = (a || b); //compilation error
The standard numeric comparison operators are also supported. Consider the following
code snippet:
1 int a = 10;
2 int b = 20;
3 int c = 10;
4 boolean x = true;
5 boolean y = false;",ComputerScienceOne.pdf
"code snippet:
1 int a = 10;
2 int b = 20;
3 int c = 10;
4 boolean x = true;
5 boolean y = false;
The six standard comparison operators are presented in Table 27.1 using these variables
as examples. The comparison operators are the same when used with double types as
403",ComputerScienceOne.pdf
"27. Conditionals
Name Operator Syntax Examples Value
Equals == a == 10 true
b == 10 false
a == b false
a == c true
Not Equals != a != 10 false
b != 10 true
a != b true
a != c false
Strictly Less Than < a < 15 true
a < 5 false
a < b true
a < c false
Less Than Or Equal To <= a <= 15 true
a <= 5 false
a <= b true
a <= c true
Strictly Greater Than > a > 15 false
a > 5 true
a > b false
a > c false
Greater Than Or Equal To >= a >= 15 false
a >= 5 true
a >= b false
a >= c true
Table 27.1.: Comparison Operators in Java
well and int types and can be compared with each other without type casting.
Furthermore, because of autoboxing and unboxing, the wrapper classes for numeric types
can be compared using the same operators. For example:
1 int a = 10;
2 Integer b = 20;
3 Double x = 3.14;
4 boolean r;
5 r = (a < b);
6 r = (a >= b);
7 r = (x == 2.71);
404",ComputerScienceOne.pdf
"27.1. Logical Operators
Operator Operator Syntax Examples Values
Negation ! !x false
!y true
And && x && true true
x && y false
Or || x || false true
x || y true
!x || y false
Table 27.2.: Logical Operators in Java with x = true and y = false both being
Boolean variables.
Operator(s) Associativity Notes
Highest ++, -- left-to-right postfix increment operators
-, ! right-to-left unary negation operator, logical
not
*, /, % left-to-right
+, - left-to-right addition, subtraction
<, <=, >, >= left-to-right comparison
==, != left-to-right equality, inequality
&& left-to-right logical And
|| left-to-right logical Or
Lowest =, +=, -=, *=, /= right-to-left assignment and compound assign-
ment operators
Table 27.3.: Operator Order of Precedence in Java. Operators on the same level have
equivalent order and are performed in the associative order specified.
The three basic logical operators are also supported as described in Table 27.2.
27.1.1. Order of Precedence",ComputerScienceOne.pdf
"The three basic logical operators are also supported as described in Table 27.2.
27.1.1. Order of Precedence
At this point it is worth summarizing the order of precedence of all the operators that
we’ve seen so far including assignment, arithmetic, comparison, and logical. Since all of
these operators could be used in one statement, for example,
(b*b < 4*a*c || a == 0 || args.length != 4)
It is important to understand the order in which each one gets evaluated. Table 27.3
summarizes the order of precedence for the operators seen so far. This is not an exhaustive
list of Java operators.
405",ComputerScienceOne.pdf
"27. Conditionals
27.1.2. Comparing Strings and Characters
The comparison operators in Table 27.1 can also be used for single characters because
of the nature of the ASCII text table (see Table 2.4). Each alphanumeric character,
including the various symbols and whitespace characters, is associated with an integer 0–
127. We can therefore write statements like ('A' < 'a'), which is true since uppercase
letters are ordered before lowercase letters in the ASCII table ( 'A' is 65 and 'a' is 97
and so 65 <97 is true). Several more examples can be found in Table 27.4.
Comparison Example Result
('A' < 'a') true
('A' == 'a') false
('A' < 'Z') true
('0' < '9') true
('\n' < 'A') true
(' ' < '\n') false
Table 27.4.: Character comparisons in Java
Numeric comparison operators cannot be used to compare strings in Java. For example,
we could not code something like (""aardvark"" < ""zebra""). The Java compiler would
not allow you to do this because the comparison operator is for numeric types only.",ComputerScienceOne.pdf
"not allow you to do this because the comparison operator is for numeric types only.
However, the following code would compile and run:
1 String s = ""aardvark"";
2 String t = ""zebra"";
3 boolean b = (s == t);
but it wouldn’t necessarily give you what you want. To understand why this is okay,
recall that a String is an object; the s and t variables are references to that object
in memory. When we use the equality comparison, == we’re asking if s and t are
the same memory address . In this case, likely they are not and so the result is false.
However, similar code,
1 String s = new String(""liger"");
2 String t = new String(""liger"");
3 boolean b = (s == t);
406",ComputerScienceOne.pdf
"27.2. If, If-Else, If-Else-If Statements
1 //example of an if statement:
2 if(x < 10) {
3 System.out.println(""x is less than 10"");
4 }
5
6 //example of an if-else statement:
7 if(x < 10) {
8 System.out.println(""x is less than 10"");
9 } else {
10 System.out.println(""x is 10 or more"");
11 }
12
13 //example of an if-else-if statement:
14 if(x < 10) {
15 System.out.println(""x is less than 10"");
16 } else if(x == 10) {
17 System.out.println(""x is equal to ten"");
18 } else {
19 System.out.println(""x is greater than 10"");
20 }
Code Sample 27.1.: Examples of Conditional Statements in Java
would also result in false because s and t represent different strings in memory,
even though they have the same sequence of characters. We’ll explore how to properly
compare strings later. For now, avoid using the equality operators with strings.
27.2. If, If-Else, If-Else-If Statements
Conditional statements in Java utilize the keywords if, else, and else if. Condi-",ComputerScienceOne.pdf
"27.2. If, If-Else, If-Else-If Statements
Conditional statements in Java utilize the keywords if, else, and else if. Condi-
tions are placed inside parentheses immediately after the if and else if keywords.
Examples of all three can be found in Code Sample 27.1.
The statement, if(x < 10) does not have a semicolon at the end. This is because it
is a conditional statement that determines the flow of control and not an executable
statement. Therefore, no semicolon is used. Suppose we made a mistake and did include
a semicolon:
407",ComputerScienceOne.pdf
"27. Conditionals
1 int x = 15;
2 if(x < 10); {
3 System.out.println(""x is less than 10"");
4 }
Some compilers may give a warning, but this is valid Java; it will compile and it will run.
However, it will end up printing x is less than 10, even though x= 15! Recall that
a conditional statement binds to the executable statement or code block immediately
following it. In this case, we’ve provided an empty executable statement ended by the
semicolon. The code is essentially equivalent to
1 int x = 15;
2 if(x < 10) {
3 }
4 System.out.println(""x is less than 10"");
Which is obviously not what we wanted. The semicolon ended up binding to the empty
executable statement, and the code block containing the print statement immediately
followed, but wasnot bound to the conditional statement which is why the print statement
executed regardless of the value of x.
Another convention that we’ve used in our code is where we have placed the curly brackets.",ComputerScienceOne.pdf
"executed regardless of the value of x.
Another convention that we’ve used in our code is where we have placed the curly brackets.
First, if a conditional statement is bound to only one statement, the curly brackets are
not necessary. However, it is best practice to include them even if they are not necessary
and we’ll follow this convention. Second, the opening curly bracket is on the same line as
the conditional statement while the closing curly bracket is indented to the same level
as the start of the conditional statement. Moreover, the code inside the code block is
indented. If there were more statements in the block, they would have all been at the
same indentation level.
27.3. Examples
27.3.1. Computing a Logarithm
The logarithm of x is the exponent that some base must be raised to get x. The most
common logarithm is the natural logarithm, ln (x) which is base e= 2.71828 ... . But
408",ComputerScienceOne.pdf
"27.3. Examples
logarithms can be in any base b >1.1 What if we wanted to compute log2 (x)? Or
logπ (x)? Let’s write a program that will prompt the user for a number x and a base b
and computes logb(x).
Arbitrary bases can be computed using the change of base formula:
logb(x) = loga(x)
loga(b)
If we can compute some base a, then we can compute any base b. Fortunately we have
such a solution. Recall that the standard library provides a function to compute the
natural logarithm, Math.log(). This is one of the fundamentals of problems solving: if
a solution already exists, use it. In this case, a solution exists for a different, but similar
problem (computing the natural logarithm), but we can adapt the solution using the
change of base formula. In particular, if we have variables b (base) and x, we can
compute logb(x) using
Math.log(x) / Math.log(b)
However, we have a problem similar to the examples in the previous section. The user",ComputerScienceOne.pdf
"compute logb(x) using
Math.log(x) / Math.log(b)
However, we have a problem similar to the examples in the previous section. The user
could enter invalid values such as b = −10 or x = −2.54 (logarithms are undefined
for non-positive values in any base). We want to ensure that b >1 and x >0. With
conditionals, we can now do this. Once we have read in the input from the user we can
make a check for good input using an if statement.
1 if(x <= 0 || b <= 1) {
2 System.out.println(""Error: bad input!"");
3 System.exit(1);
4 }
This code has something new: System.exit(1). The exit() function immediately
terminates the program regardless of the rest of the code that it may remain. The
argument passed to exit() is an integer that represents an error code. The convention
is that zero indicates “no error” while non-zero values indicate some error. This is a
simple way of performing error handling : if the user provides bad input, we inform",ComputerScienceOne.pdf
"simple way of performing error handling : if the user provides bad input, we inform
them and quit the program, forcing them to run it again and provide good input. By
prematurely terminating the program we avoid any illegal operation that would give a
bad result.
Alternatively, we could have split the conditions into two statements and given a more
descriptive error message. We use this design in the full program which can be found in
1Bases can also be 0 <b< 1, but we’ll restrict our attention to increasing functions only.
409",ComputerScienceOne.pdf
"27. Conditionals
Code Sample 27.2. The program also takes the input as command line arguments. Now
that we have conditionals, we can actually check that the correct number of arguments
was provided by the user and quit in the event that they don’t provide the correct
number.
27.3.2. Life & Taxes
Let’s adapt the conditional statements we developed in Section 3.6.4 into a full Java
program. The first thing we need to do is establish the variables we’ll need and read
them in from the user. At the same time we can check for bad input (negative values)
for both the inputs.
1 Scanner s = new Scanner(System.in);
2 double income, baseTax, numChildren, credit, totalTax;
3
4 System.out.println(""Please enter your Adjusted Gross Income: "");
5 income = s.nextDouble();
6
7 System.out.println(""How many children do you have?"");
8 numChildren = s.nextDouble();
9
10 if(income < 0 || numChildren < 0) {
11 System.out.println(""Invalid inputs"");
12 System.exit(1);
13 }",ComputerScienceOne.pdf
"8 numChildren = s.nextDouble();
9
10 if(income < 0 || numChildren < 0) {
11 System.out.println(""Invalid inputs"");
12 System.exit(1);
13 }
Next, we can code a series of if-else-if statements for the income range. By placing the
ranges in increasing order, we only need to check the upper bounds just as in the original
example.
1 if(income <= 18150) {
2 baseTax = income * .10;
3 } else if(income <= 73800) {
4 baseTax = 1815 + (income - 18150) * .15;
5 } else if(income <= 148850) {
6 ...
7 } else {
8 baseTax = 127962.50 + (income - 457600) * .396;
9 }
410",ComputerScienceOne.pdf
"27.3. Examples
Next we compute the child tax credit, taking care that it does not exceed $3,000. A
conditional based on the number of children should suffice as at this point in the program
we already know it is zero or greater.
1 if(numChildren <= 3) {
2 credit = numChildren * 1000;
3 } else {
4 credit = 3000;
5 }
Finally, we need to ensure that the credit does not exceed the total tax liability (the
credit is non-refundable, so if the credit is greater, the tax should only be zero, not
negative).
1 if(baseTax - credit >= 0) {
2 totalTax = baseTax - credit;
3 } else {
4 totalTax = 0;
5 }
The full program is presented in Code Sample 27.3.
27.3.3. Quadratic Roots Revisited
Let’s return to the quadratic roots program we previously designed that uses the quadratic
equation to compute the roots of a quadratic polynomial by reading coefficients a,b,c
in from the user. One of the problems we had previously identified is if the user enters",ComputerScienceOne.pdf
"in from the user. One of the problems we had previously identified is if the user enters
“bad” input: if a= 0, we would end up dividing by zero; if b2 −4ac< 0 then we would
have complex roots. With conditionals, we can now check for these issues and exit with
an error message.
Another potential case we might want to handle differently is when there is only one
distinct root ( b2 −4ac= 0). In that case, the quadratic formula simplifies to −b
2a and we
can print a different, more specific message to the user. The full program can be found
in Code Sample 27.4.
411",ComputerScienceOne.pdf
"27. Conditionals
1 /**
2 * This program computes the logarithm base b (b > 1)
3 * of a given number x > 0
4 */
5 public class Logarithm {
6
7 public static void main(String args[]) {
8
9 double b, x, result;
10 if(args.length != 2) {
11 System.out.println(""Usage: b x"");
12 System.exit(1);
13 }
14
15 b = Double.parseDouble(args[0]);
16 x = Integer.parseInt(args[1]);
17
18 if(x <= 0) {
19 System.out.println(""Error: x must be greater than zero"");
20 System.exit(1);
21 }
22 if(b <= 1) {
23 System.out.println(""Error: base must be greater than one"");
24 System.exit(1);
25 }
26
27 result = Math.log(x) / Math.log(b);
28 System.out.printf(""log_(%f)(%f) = %f\n"", b, x, result);
29
30 }
31
32 }
Code Sample 27.2.: Logarithm Calculator Program in Java
412",ComputerScienceOne.pdf
"27.3. Examples
1 import java.util.Scanner;
2
3 public class Taxes {
4
5 public static void main(String args[]) {
6
7 Scanner s = new Scanner(System.in);
8 double income, baseTax, totalTax, numChildren, credit;
9
10 System.out.println(""Please enter your Adjusted Gross Income: "");
11 income = s.nextDouble();
12
13 System.out.println(""How many children do you have?"");
14 numChildren = s.nextDouble();
15
16 if(income < 0 || numChildren < 0) {
17 System.out.println(""Invalid inputs"");
18 System.exit(1);
19 }
20
21 if(income <= 18150) {
22 baseTax = income * .10;
23 } else if(income <= 73800) {
24 baseTax = 1815 + (income -18150) * .15;
25 } else if(income <= 148850) {
26 baseTax = 10162.50 + (income - 73800) * .25;
27 } else if(income <= 225850) {
28 baseTax = 28925.00 + (income - 148850) * .28;
29 } else if(income <= 405100) {
30 baseTax = 50765.00 + (income - 225850) * .33;
31 } else if(income <= 457600) {
32 baseTax = 109587.50 + (income - 405100) * .35;
33 } else {",ComputerScienceOne.pdf
"30 baseTax = 50765.00 + (income - 225850) * .33;
31 } else if(income <= 457600) {
32 baseTax = 109587.50 + (income - 405100) * .35;
33 } else {
34 baseTax = 127962.50 + (income - 457600) * .396;
35 }
36
37 if(numChildren <= 3) {
38 credit = numChildren * 1000;
39 } else {
40 credit = 3000;
41 }
42
43 if(baseTax - credit >= 0) {
44 totalTax = baseTax - credit;
45 } else {
46 totalTax = 0;
47 }
48
49 System.out.printf(""AGI: $%10.2f\n"", income);
50 System.out.printf(""Tax: $%10.2f\n"", baseTax);
51 System.out.printf(""Credit: $%10.2f\n"", credit);
52 System.out.printf(""Tax Liability: $%10.2f\n"", totalTax);
53
54 }
55
56 }
Code Sample 27.3.: Tax Program in Java
413",ComputerScienceOne.pdf
"27. Conditionals
1 /**
2 * This program computes the roots to a quadratic equation
3 * using the quadratic formula.
4 */
5 public class Roots {
6
7 public static void main(String args[]) {
8 double a, b, c, root1, root2;
9
10 if(args.length != 3) {
11 System.err.println(""Usage: a b c\n"");
12 System.exit(1);
13 }
14
15 a = Double.parseDouble(args[0]);
16 b = Double.parseDouble(args[1]);
17 c = Double.parseDouble(args[2]);
18
19 if(a == 0) {
20 System.err.println(""Error: a cannot be zero"");
21 System.exit(1);
22 } else if(b*b < 4*a*c) {
23 System.err.println(""Error: cannot handle complex roots\n"");
24 System.exit(1);
25 } else if(b*b == 4*a*c) {
26 root1 = -b / (2*a);
27 System.out.printf(""Only one distinct root: %f\n"", root1);
28 } else {
29 root1 = (-b + Math.sqrt(b*b - 4*a*c) ) / (2*a);
30 root2 = (-b - Math.sqrt(b*b - 4*a*c) ) / (2*a);
31
32 System.out.printf(""The roots of %fx^2 + %fx + %f are: \n"",
33 a, b, c);
34 System.out.printf("" root1 = %f\n"", root1);",ComputerScienceOne.pdf
"31
32 System.out.printf(""The roots of %fx^2 + %fx + %f are: \n"",
33 a, b, c);
34 System.out.printf("" root1 = %f\n"", root1);
35 System.out.printf("" root2 = %f\n"", root2);
36 }
37 }
38
39 }
Code Sample 27.4.: Quadratic Roots Program in Java With Error Checking
414",ComputerScienceOne.pdf
"28. Loops
Java supports while loops, for loops, and do-while loops using the keywords while, for,
and do (along with another while). Continuation conditions for loops are enclosed
in parentheses, (...) and the blocks of code associated with the loop are enclosed in
curly brackets.
28.1. While Loops
Code Sample 28.1 contains an example of a basic while loop in C. Just as with conditional
statements, our code styling places the opening curly bracket on the same line as the
while keyword and continuation condition. The inner block of code is also indented
and all lines in the block are indented to the same level.
In addition, the continuation condition does not have a semicolon since it is not an
executable statement. Just as with an if-statement, if we had included a semicolon it
would have led to unintended results. Consider the following:
1 while(i <= 10); {
2 //perform some action
3 i++; //iteration
4 }
1 int i = 1; //Initialization
2 while(i <= 10) { //continuation condition",ComputerScienceOne.pdf
"1 while(i <= 10); {
2 //perform some action
3 i++; //iteration
4 }
1 int i = 1; //Initialization
2 while(i <= 10) { //continuation condition
3 //perform some action
4 i++; //iteration
5 }
Code Sample 28.1.: While Loop in Java
415",ComputerScienceOne.pdf
"28. Loops
1 int i = 1;
2 boolean flag = true;
3 while(flag) {
4 //perform some action
5 i++; //iteration
6 if(i>10) {
7 flag = false;
8 }
9 }
Code Sample 28.2.: Flag-controlled While Loop in Java
A similar problem occurs: the while keyword and continuation condition bind to
the next executable statement or code block. As a consequence of the semicolon, the
executable statement that gets bound to the while loop is empty. What happens is
even worse: the program will enter an infinite loop. To see this, the code is essentially
equivalent to the following:
1 while(i <= 10) {
2 }
3 {
4 //perform some action
5 i++; //iteration
6 }
In the while loop, we never increment the counter variable i, the loop does nothing,
and so the computation will continue on forever! Some compilers will warn you about
this, others will not. It is valid Java and it will compile and run, but obviously won’t
work as intended. Avoid this problem by using proper syntax.",ComputerScienceOne.pdf
"this, others will not. It is valid Java and it will compile and run, but obviously won’t
work as intended. Avoid this problem by using proper syntax.
Another common use case for a while loop is a flag-controlled loop in which we use a
Boolean flag rather than an expression to determine if a loop should continue or not. An
example can be found in Code Sample 28.2.
416",ComputerScienceOne.pdf
"28.2. For Loops
28.2. For Loops
For loops in Java use the familiar syntax of placing the initialization, continuation
condition, and iteration on the same line as the keyword for. An example can be found
in Code Sample 28.3.
1 for(int i=1; i<=10; i++) {
2 //perform some action
3 }
Code Sample 28.3.: For Loop in Java
Semicolons are placed at the end of the initialization and continuation condition, but not
the iteration statement. Just as with while loops, the opening curly bracket is placed on
the same line as the for keyword. Code within the loop body is indented, all at the
same indentation level.
Another observation: the declaration of the counter variable i was done in the initializa-
tion statement. This scopes the variable to the loop itself. The variable i is valid inside
the loop body, but will be out-of-scope after the loop body. It is possible to declare the
variable prior to the loop, but the variable i would have a much larger scope. It is best",ComputerScienceOne.pdf
"variable prior to the loop, but the variable i would have a much larger scope. It is best
practice to limit the scope of variables only to where they are needed. Thus, we will
write our loops as above.
28.3. Do-While Loops
Finally, Java supports do-while loops. Recall that the difference between a while loop
and a do-while loop is when the continuation condition is checked. For a while loop it is
prior to the beginning of the loop body and in a do-while loop it is at the end of the
loop. This means that a do-while always executes at least once . An example can be
found in Code Sample 28.4.
The opening curly bracket is again on the same line as the keyword do. The while
keyword and continuation condition are on the same line as the closing curly bracket.
In a slight departure from previous syntax, a semicolon does appear at the end of the
continuation condition even though it is not an executable statement.
417",ComputerScienceOne.pdf
"28. Loops
1 int i;
2 do {
3 //perform some action
4 i++;
5 } while(i <= 10);
Code Sample 28.4.: Do-While Loop in Java
28.4. Enhanced For Loops
Java also supports foreach loops (which were introduced in JDK 1.5.0) which Java refers
to as “Enhanced For Loops.” Foreach loops allow you to iterate over each element in a
collection without having to define an index variable or otherwise “get” each element.
We’ll revisit these concepts in detail in Chapter 31, but let’s take a look at a couple of
examples.
An enhanced for loop in Java still uses the keyword for but uses different syntax for
its control. The example in Code Sample 28.5 illustrates this syntax: (int a : arr).
The last element of this syntax is a reference to the collection that we want to iterate
over. The first part is the type and local reference variable that the loop will use.
1 int arr[] = {10, 20, 8, 42};
2 int sum = 0;
3 for(int a : arr) {
4 sum += a;
5 }
Code Sample 28.5.: Enhanced For Loops in Java Example 1",ComputerScienceOne.pdf
"1 int arr[] = {10, 20, 8, 42};
2 int sum = 0;
3 for(int a : arr) {
4 sum += a;
5 }
Code Sample 28.5.: Enhanced For Loops in Java Example 1
The code (int a : arr) should be read as “for each integer element a in the collection
arr...” Within the enhanced for loop, the variable a will be automatically updated for
you on each iteration. Outside the loop body, the variable a is out-of-scope.
Java allows you to use an enhanced for loop with any array or collection (technically,
anything that implements the Iterable interface). One example is a List, an ordered
collection of elements. Code Sample 28.6 contains an example.
418",ComputerScienceOne.pdf
"28.5. Examples
1 List<Integer> list = Arrays.asList(10, 20, 8, 42);
2 int sum = 0;
3 for(Integer a : list) {
4 sum += a;
5 }
Code Sample 28.6.: Enhanced For Loops in Java Example 2
28.5. Examples
28.5.1. Normalizing a Number
Let’s revisit the example from Section 4.1.1 in which wenormalize a number by continually
dividing it by 10 until it is less than 10. The code in Code Sample 28.7 specifically refers
to the value 32145.234 but would work equally well with any value of x.
1 double x = 32145.234;
2 int k = 0;
3 while(x > 10) {
4 x = x / 10; //or: x /= 10;
5 k++;
6 }
Code Sample 28.7.: Normalizing a Number with a While Loop in Java
28.5.2. Summation
Let’s revisit the example from Section 4.2.1 in which we computed the sum of integers
1 + 2 +··· + 10. The code is presented in Code Sample 28.8
We could have easily generalized the code somewhat. Instead of computing a sum up to
a particular number, we could have written it to sum up to another variable n, in which",ComputerScienceOne.pdf
"a particular number, we could have written it to sum up to another variable n, in which
case the for loop would instead look like the following.
419",ComputerScienceOne.pdf
"28. Loops
1 int sum = 0;
2 for(int i=1; i<=10; i++) {
3 sum += i;
4 }
Code Sample 28.8.: Summation of Numbers using a For Loop in Java
1 for(int i=1; i<=n; i++) {
2 sum += i;
3 }
28.5.3. Nested Loops
Recall that you can write loops within loops by nesting them. The inner loop will execute
fully for each iteration of the outer loop. An example of two nested of loops in Java can
be found in Code Sample 28.9.
1 int i, j;
2 int n = 10;
3 int m = 20;
4 for(i=0; i<n; i++) {
5 for(j=0; j<m; j++) {
6 System.out.printf(""(i, j) = (%d, %d)\n"", i, j);
7 }
8 }
Code Sample 28.9.: Nested For Loops in Java
The inner loop execute for j = 0,1,2,..., 19 <m = 20 for a total of 20 iterations. How-
ever, it executes 20 times for each iteration of the outer loop. Since the outer loop execute
for i = 0,1,2,..., 9 < n= 10, the total number of times the System.out.printf()
statement execute is 10 ×20 = 200. In this example, the sequence
(0,0),(0,1),(0,2),..., (0,19),(1,0),..., (9,19)
will be printed.
420",ComputerScienceOne.pdf
"28.5. Examples
28.5.4. Paying the Piper
Let’s adapt the solution for the loan amortization schedule we developed in Section 4.7.3.
First, we’ll read the principle, terms, and interest as command line inputs. Adapting the
formula for the monthly payment and using the math library’s Math.pow() function,
we have the following.
1 double monthlyPayment = (monthlyInterestRate * principle) /
2 (1 - pow( (1 + monthlyInterestRate), -n));
However, recall that we may have problems due to accuracy. The monthly payment
could come out to be a fraction of a cent, say $43.871. For accuracy, we need to ensure
that all of the figures for currency are rounded to the nearest cent. The standard math
library does have a Math.round() function, but it only rounds to the nearest whole
number, not the nearest 100th.
However, we can adapt the “off-the-shelf” solution to fit our needs. If we take the number,
multiply it by 100, we get 4387.1 which we can now round to the nearest whole number,",ComputerScienceOne.pdf
"multiply it by 100, we get 4387.1 which we can now round to the nearest whole number,
giving us 4387. We can then divide by 100 to get a number that has been rounded to
the nearest 100th! In Java, we could simply do the following.
monthlyPayment = Math.round(monthlyPayment * 100.0) / 100.0;
We can use the same trick to round the monthly interest payment and any other number
expected to be whole cents. To output our numbers, we use System.out.printf() and
take care to align our columns to make make it look nice. To finish our adaptation, we
handle the final month separately to account for an over/under payment due to rounding.
The full solution can be found in Code Sample 28.10.
421",ComputerScienceOne.pdf
"28. Loops
1 public class LoanAmortization {
2
3 public static void main(String args[]) {
4
5 if(args.length != 4) {
6 System.err.println(""Usage: principle apr terms"");
7 System.exit(1);
8 }
9
10 double principle = Double.parseDouble(args[0]);
11 double apr = Double.parseDouble(args[1]);
12 int n = Integer.parseInt(args[2]);
13
14 double balance = principle;
15 double monthlyInterestRate = apr / 12.0;
16
17 //monthly payment
18 double monthlyPayment = (monthlyInterestRate * principle) /
19 (1 - Math.pow( (1 + monthlyInterestRate), -n));
20 //round to the nearest cent
21 monthlyPayment = Math.round(monthlyPayment * 100.0) / 100.0;
22
23 System.out.printf(""Principle: $%.2f\n"", principle);
24 System.out.printf(""APR: %.4f%%\n"", apr*100.0);
25 System.out.printf(""Months: %d\n"", n);
26 System.out.printf(""Monthly Payment: $%.2f\n"", monthlyPayment);
27
28 //for the first n-1 payments in a loop:
29 for(int i=1; i<n; i++) {
30 // compute the monthly interest, rounded:
31 double monthlyInterest =",ComputerScienceOne.pdf
"27
28 //for the first n-1 payments in a loop:
29 for(int i=1; i<n; i++) {
30 // compute the monthly interest, rounded:
31 double monthlyInterest =
32 Math.round( (balance * monthlyInterestRate) * 100.0) / 100.0;
33 // compute the monthly principle payment
34 double monthlyPrinciplePayment = monthlyPayment - monthlyInterest;
35 // update the balance
36 balance = balance - monthlyPrinciplePayment;
37 // print i, monthly interest, monthly principle, new balance
38 System.out.printf(""%d\t$%10.2f $%10.2f $%10.2f\n"", i, monthlyInterest,
39 monthlyPrinciplePayment, balance);
40 }
41
42 //handle the last month and last payment separately
43 double lastInterest = Math.round(
44 (balance * monthlyInterestRate) * 100.0) / 100.0;
45 double lastPayment = balance + lastInterest;
46
47 System.out.printf(""Last payment = $%.2f\n"", lastPayment);
48
49 }
50
51 }
Code Sample 28.10.: Loan Amortization Program in Java
422",ComputerScienceOne.pdf
"29. Methods
As an object-oriented programming language, functions in Java are usually referred to
as methods and are essential to writing programs. The distinction is that a function is
usually a standalone element while methods are functions that are members of a class.
In Java, since everything is a class or belongs to a class, standalone functions cannot be
defined.
In Java you can define your own methods, but they need to be placed within a class.
Usually methods that act on data in the class (or instances of the class, see Chapter 34)
or have common functionality are placed into one class. For example, all the basic math
methods are part of the java.lang.Math class. It is not uncommon to place similar
methods together into one “utility” class.
Java supports method overloading, so within the same class you can define multiple
methods with the same name as long as they differ in either the number (also called",ComputerScienceOne.pdf
"methods with the same name as long as they differ in either the number (also called
arity) or type of parameters. For example, in the java.lang.Math class, there are
3 versions of the absolute value method, abs(), one that takes/returns an int, one
that takes/returns a double and one for float types. Naming conflicts can easily be
solved by ensuring that you place your methods in a class/package that is unique to your
application.
In Java, the 8 primitive types ( int, double, char, boolean, etc.) are always passed
by value. All object types, however, such as the wrapper classes Integer, Double as
well as String, etc. are passed by reference. That is, the memory address in the JVM
is passed to the method. This is done for efficiency. For objects that are “large” it would
be inefficient to copy the entire object into the call stack in order to pass it to a method.
Though object types are passed by reference, the method cannot necessarily change",ComputerScienceOne.pdf
"Though object types are passed by reference, the method cannot necessarily change
them. Recall that the wrapper classes Integer, Double and the String class are all
immutable, meaning that once created they cannot be modified. Though they are passed
by reference, the method that receives them cannot change them.
There are many mutable objects in Java. The StringBuilder class for example is a
mutable object. If you pass a StringBuilder instance to a method, that method is free
to invoke mutator methods on the object such as .append() that change the object’s
state. Since it is the same object as in the calling method, the calling method can “see”
those changes.
423",ComputerScienceOne.pdf
"29. Methods
As of Java 5, you can write and use vararg methods. The System.out.printf()
method is a prime example of this. However, we will not discuss in detail how to do this.
Instead, refer to standard Java documentation. Finally, parameters are not optional in
Java. This is because Java supports method overloading. You can write multiple versions
of the same method that each take a different number of arguments. You can even design
them so that the more specific versions (with fewer arguments) invoke the more general
versions (with more arguments), passing in sensible “defaults.”
29.1. Defining Methods
Defining methods is fairly straightforward. First you create a class to place them in.
Then you provide the method signature along with the body of the method. In addition,
there are several modifiers that you can place in the method signature to specify its
visibility and whether or not the method “belongs” to the class or to instances of the",ComputerScienceOne.pdf
"visibility and whether or not the method “belongs” to the class or to instances of the
class. This is a concept we’ll explore in Chapter 34. For now, we’ll only focus on what is
needed to get started.
Typically, the documentation for methods is included with the method definition using
“Javadoc” style comments. Consider the following examples.
1 /**
2 * Computes the sum of the two arguments.
3 * @param a
4 * @param b
5 * @return the sum, <code>a + b</code>
6 */
7 public static int sum(int a, int b) {
8 return (a + b);
9 }
10
11 /**
12 * Computes the Euclidean distance between the 2-D points,
13 * (x1,y1) and (x2,y2).
14 * @param x1
15 * @param y1
16 * @param x2
17 * @param y2
18 * @return
19 */
20 public static double getDistance(double x1, double y1,
21 double x2, double y2) {
22 double xDiff = (x1-x2);
424",ComputerScienceOne.pdf
"29.1. Defining Methods
23 double yDiff = (y1-y2);
24 return Math.sqrt( xDiff * xDiff + yDiff * yDiff);
25 }
26
27 /**
28 * Computes a monthly payment for a loan with the given
29 * principle at the given APR (annual percentage rate) which
30 * is to be repaid over the given number of terms.
31 * @param principle - the amount borrowed
32 * @param apr - the annual percentage rate
33 * @param terms - number of terms (usually months)
34 * @return
35 */
36 public static double getMonthlyPayment(double principle,
37 double apr, int terms) {
38 double rate = (apr / 12.0);
39 double payment = (principle * rate) / (1-Math.pow(1+rate, -terms));
40 return payment;
41 }
In each of the examples above, the first modifier keyword we used was public. This
makes the method visible to all other parts of the code base. Any other piece of code can
invoke the method and take advantage of the functionality it provides. Alternatively, we",ComputerScienceOne.pdf
"invoke the method and take advantage of the functionality it provides. Alternatively, we
could have used the keywords private (which makes it visible only to other methods in
the same class) protected or “package protected” by omitting the modifier altogether.
We will want our methods to be available to other classes, so we’ll make most of them
public. The second modifier is static which makes it so that the method belongs
to the class itself rather than instances of the class. We discuss visibility keywords and
instances in detail in Chapter 34. For now, we will make all of our methods static.
After the modifiers, we provide the method signature including the return type, its
identifier (name), and its parameter list. Method names must follow the same naming rules
as variables: they must begin with an alphabetic character and may contain alphanumeric
characters as well as underscores. However, using modern coding conventions we name",ComputerScienceOne.pdf
"characters as well as underscores. However, using modern coding conventions we name
methods using lower camel casing. Immediately after the signature we provide a method
body which contains the code that will be run upon invocation of the method. The
method body is enclosed using opening/closing curly brackets.
425",ComputerScienceOne.pdf
"29. Methods
29.1.1. Void Methods
The keyword void can be used in Java to indicate a method does not return a value,
in which case it is called a “void method.” Though it is not necessary, it is still good
practice to include a return statement.
1 public static void printCopyright()
2 System.out.println(""(c) Bourke 2015"");
3 return;
4 }
In the example above, we’ve also illustrated how to define a method that has no input
parameters.
29.1.2. Using Methods
Once a method has been defined in a class, you can make use of the method as follows.
First, you may need to import the class itself depending on where it is. For example,
suppose that the examples we’ve presented so far are contained in a class named Utils
(short for “utilities”) which is in a package named unl.cse. Then in the class in which
we want to call some of these functions we would import it using
import unl.cse.Utils;
prior to the class declaration. Once the class has been imported, we can invoke a method",ComputerScienceOne.pdf
"import unl.cse.Utils;
prior to the class declaration. Once the class has been imported, we can invoke a method
in the class by first referencing the class and using the dot operator to access one of its
methods. For example,
1 int a = 10, b = 20;
2 int c = Utils.sum(a, b); //c contains the value 30
3
4 //invoke a method with literal values:
5 double dist = Utils.getDistance(0.0, 0.0, 10.0, 20.0);
6
7 //invoke a method with a combination:
8 double p = 1500.0;
9 double r = 0.05;
10 double monthlyPayment = Utils.getMonthlyPayment(p, r, 60);
426",ComputerScienceOne.pdf
"29.1. Defining Methods
The Utils.methodName() syntax is used because the methods are static–they belong
to the class and so must be invoked through the class using the class’s name. We’ve
previously seen this syntax when using System. or Math. with the standard JDK
library functions.
29.1.3. Passing By Reference
Java does not allow you the ability to specify if a variable is passed by reference or by
value. Instead, all primitive types are passed by value while all object types are passed
by reference. Moreover, most of the built-in types such as Integer and String are
immutable, even though they are passed by reference, any method that receives them
cannot change them. Only if the passed object is mutable can the method make changes
to it (by invoking its methods).
As an example, consider the following piece of code. The StringBuilder class is a
mutable string object. You can change the string contents stored in a StringBuilder",ComputerScienceOne.pdf
"mutable string object. You can change the string contents stored in a StringBuilder
by calling one of its many methods such as append(), which will add whatever string
you give it to the end.
In the main method, we create two objects, a String and a StringBuilder and
pass it to a method that makes changes to both by appending "" world!"" to them.
Understand what happens here though. The first line in change() actually creates a
new string and then changes what the parameter variable s references. The reference
to the original string, ""Hello"" is lost and replaced with the new string. In contrast,
the StringBuilder instance is actually changed via its append() method but is still
the same object.
1 public class Mutability {
2
3 public static void change(String s, StringBuilder sb) {
4 s = s + "" world!"";
5 sb.append("" world!"");
6
7 System.out.println(""change: s = "" + s);
8 System.out.println(""change: sb = "" + sb);
9 }
10
11 public static void main(String args[]) {
12 String a = ""Hello"";",ComputerScienceOne.pdf
"8 System.out.println(""change: sb = "" + sb);
9 }
10
11 public static void main(String args[]) {
12 String a = ""Hello"";
13 StringBuilder b = new StringBuilder(""Hello"");
14
15 System.out.println(""main: s = "" + a);
427",ComputerScienceOne.pdf
"29. Methods
16 System.out.println(""main: b = "" + b);
17
18 change(a, b);
19
20 System.out.println(""main after: s = "" + a);
21 System.out.println(""main after: b = "" + b);
22
23 }
24 }
To see this, observe the following output. When we return to the main method, the
original string s is unchanged (since it was immutable). However, the StringBuilder
has been changed by the method.
main: s = Hello
main: b = Hello
change: s = Hello world!
change: sb = Hello world!
main after: s = Hello
main after: b = Hello world!
29.2. Examples
29.2.1. Generalized Rounding
Recall that the math library provides a Math.round() method that rounds a number
to the nearest whole number. We’ve had need to round to cents as well. We now have
the ability to write a method to do this for us. Before we do, however, let’s think more
generally. What if we wanted to round to the nearest tenth? Or what if we wanted to
round to the nearest 10s or 100s place? Let’s write a general purpose rounding method",ComputerScienceOne.pdf
"round to the nearest 10s or 100s place? Let’s write a general purpose rounding method
that allows us to specify which decimal place to round to.
The most natural input would be to specify the place using an integer exponent. That
is, if we wanted to round to the nearest tenth, then we would pass it −1 as 0.1 = 10−1,
−2 if we wanted to round to the nearest 100th, etc. In the other direction, passing
in 0 would correspond to the usual round function, 1 to the nearest 10s spot, and so
428",ComputerScienceOne.pdf
"29.2. Examples
on. We could demonstrate good code reuse (as well as good procedural abstraction)
by scaling the input value and reusing the functionality already provided in the math
library’s Math.round() method. We could further define a roundToCents() method
that used our generalized round method. Finally, we could place all of these methods
into RoundUtils Java class for good organization.
1 package unl.cse;
2
3 /**
4 * A collection of rounding utilities
5 *
6 */
7 public class RoundUtils {
8
9 /**
10 * Rounds to the nearest digit specified by the place
11 * argument. In particular to the (10^place)-th digit
12 *
13 * @param x the number to be rounded
14 * @param place the place to be rounded to
15 * @return
16 */
17 public static double roundToPlace(double x, int place) {
18 double scale = Math.pow(10, -place);
19 double rounded = Math.round(x * scale) / scale;
20 return rounded;
21 }
22
23 /**
24 * Rounds to the nearest cent (100th place)
25 *
26 * @param x
27 * @return
28 */",ComputerScienceOne.pdf
"20 return rounded;
21 }
22
23 /**
24 * Rounds to the nearest cent (100th place)
25 *
26 * @param x
27 * @return
28 */
29 public static double roundToCents(double x) {
30 return RoundUtils.roundToPlace(x, -2);
31 }
32
33 }
Observe that this class does not contain a main() method. That means that this class
429",ComputerScienceOne.pdf
"29. Methods
is not executable itself. It only provides functionality to other classes in the code base.
430",ComputerScienceOne.pdf
"30. Error Handling & Exceptions
Java supports error handling through the use of exceptions. Java has many different
predefined types of exceptions that you can use in your own code. It also allows
you to define your own exception types by creating new classes that inherit from the
predefined classes. Java uses the standard try-catch-finally control structure to
handle exceptions and allows you to throw your own exceptions.
30.1. Exceptions
Java defines a base class named Throwable. There are two majors subtypes of
Throwable: Error and Exception. The Error class is used primarily for fatal
errors such as the JVM running out of memory or some other extreme case that your
code cannot reasonably be expected to recover from. There are dozens of types of
exceptions that are subclasses of the standard Java Exception class defined by the
JDK including IOException (and its subclasses such as FileNotFoundException) or
SQLException (when working with Structured Query Language (SQL) databases). An",ComputerScienceOne.pdf
"SQLException (when working with Structured Query Language (SQL) databases). An
important subclass of Exception is RuntimeException which represent unchecked
exceptions that do not need to be explicitly caught (see Section 30.1.4). We’ll mostly
focus on this type of exception.
30.1.1. Catching Exceptions
To catch an exception in Java you can use the standard try-catch control block (and
optionally use the finally block to clean up any resources). Let’s take, for example,
the simple task of reading input from a user using Scanner and manually parsing its
value into an integer. If the user enters a non-numeric value, parsing will fail and result
in a NumberFormatException that we can then catch and handle. For example,
1 Scanner s = new Scanner(System.in);
2 int n = 0;
3
431",ComputerScienceOne.pdf
"30. Error Handling & Exceptions
4 try {
5 String input = s.next();
6 n = Integer.parseInt(input);
7 } catch (NumberFormatException nfe) {
8 System.err.println(""You entered invalid data!"");
9 System.exit(1);
10 }
In this example, we’ve simply displayed an error message to the standard error output
and exited the program. That is, we’ve made the design decision that this error should
be fatal. We could have chosen to handle this error differently in the catch block.
The code above could have resulted in other exceptions. For example if the Scanner
had failed to read the next token from the standard input, it would have thrown a
NoSuchElementException. We can add as many catch blocks as we want to handle
each exception differently.
1 Scanner s = new Scanner(System.in);
2 int n = 0;
3
4 try {
5 String input = s.next();
6 n = Integer.parseInt(input);
7 } catch (NumberFormatException nfe) {
8 System.err.println(""You entered invalid data!"");
9 System.exit(1);",ComputerScienceOne.pdf
"6 n = Integer.parseInt(input);
7 } catch (NumberFormatException nfe) {
8 System.err.println(""You entered invalid data!"");
9 System.exit(1);
10 } catch (NoSuchElementException nsee) {
11 System.err.println(""Input reading failed, using default..."");
12 n = 20; //a default value
13 } catch(Exception e) {
14 System.err.println(""A general exception occurred"");
15 e.printStackTrace();
16 System.exit(1);
17 }
Each catch block catches a different type of exception. Thus, the name of the variable
that holds each exception must be different in the chain of catch blocks; nfe, nsee,
e.
432",ComputerScienceOne.pdf
"30.1. Exceptions
Note that the last catch block was written to catch a generic Exception. This last
block will essentially catch any other type of exception. Much like an if-else-if
statement, the first type of exception that is caught is the block that will be executed
and they are all mutually exclusive. Thus, a “catch all” block like this should always be
the last catch block. The most specific types of exceptions should be caught first and
the most general types should be caught last.
30.1.2. Throwing Exceptions
We can also manually throw an exception. For example, we can throw a generic
RuntimeException using the following.
1 throw new RuntimeException(""Something went wrong"");
By using a generic RuntimeException, we can attach a message to the exception (which
can be printed by code that catches the exception). If we want more fine-grained control
over the type of exceptions, we need to define our own exceptions.
30.1.3. Creating Custom Exceptions",ComputerScienceOne.pdf
"over the type of exceptions, we need to define our own exceptions.
30.1.3. Creating Custom Exceptions
To create your own exceptions, you need to create a new class to represent the exception
and make it extend RuntimeException by using the keyword extends. This makes
your exception a subclass or subtype of the RuntimeException. This is a concept known
as inheritance in OOP. Consider the example in the previous chapter of computing
the roots of a quadratic polynomial. One possible error situation is when the roots are
complex numbers. We could define a new Java class as follows.
1 public class ComplexRootException extends RuntimeException {
2
3 /**
4 * Constructor that takes an error message
5 */
6 public ComplexRootException(String errorMessage) {
7 super(errorMessage);
8 }
9 }
433",ComputerScienceOne.pdf
"30. Error Handling & Exceptions
Now in our code we can throw and catch this new type of exception.
1 //throw this exception:
2 if( b*b - 4*a*c < 0) {
3 throw new ComplexRootException(""Cannot Handle complex roots"");
4 }
1 try {
2 r1 = getComplexRoot01(a, b, c);
3 } catch(ComplexRootException cre) {
4 //handle the exception here...
5 }
30.1.4. Checked Exceptions
A checked exception in Java is an exception that must be explicitly caught and han-
dled. For example, the generic Exception class is a checked exception (others include
IOException, SQLException, etc.). If a checked exception is thrown within a block of
code then it must either be caught and handled within that block of code or the method
must be specified to explicitly throw the exception. For example, the method
1 public static void processFile() {
2 Scanner s = new Scanner(new File(""data.csv""));
3 }
would not actually compile because Scanner throws a checked FileNotFoundException.",ComputerScienceOne.pdf
"2 Scanner s = new Scanner(new File(""data.csv""));
3 }
would not actually compile because Scanner throws a checked FileNotFoundException.
Either we would have to explicitly catch the exception:
1 public static void processFile() {
2 Scanner s = null;
3 try {
4 s = new Scanner(new File(""data.csv""));
5 } catch (FileNotFoundException e) {
434",ComputerScienceOne.pdf
"30.1. Exceptions
6 //handle the exception here
7 }
8 }
or we would need to specify that the method processFile() explicitly throws the
exception:
1 public static void processFile() throws FileNotFoundException {
2 Scanner s = new Scanner(new File(""data.csv""));
3 }
Doing this, however, would force any code that called the processFile() method to
surround it in a try-catch block and explicitly handle it (or once again, throw it back
to the calling method).
The point of a checked exception is to force code to deal with potential issues that can be
reasonably anticipated (such as the unavailability of a file). However, from another point
of view checked exceptions represent the exact opposite goal of error handling. Namely,
that a function or code block can and should inform the calling function that an error
has occurred, but not explicitly make a decision on how to handle the error. A checked
exception doesn’t make the full decision for the calling function, but it does eliminate",ComputerScienceOne.pdf
"exception doesn’t make the full decision for the calling function, but it does eliminate
ignoring the error as an option from the calling function.
Java also supports unchecked exceptions which do not need to be explicitly caught. For
example, NumberFormatException or NullPointerException are unchecked excep-
tions. If an unchecked exception is thrown and not caught, it bubbles up through the
call stack until some piece of code does catch it. If no code catches it, it results in a fatal
error and terminates the execution of the JVM.
The RuntimeException class and any of its subclasses are unchecked exceptions. In
our ComplexRootException example above, because we extended RuntimeException
we made it an unchecked exception, allowing the calling function to decide not only how
to handle it, but whether or not to handle it at all . If we had instead decided to extend
Exception we would have made our exception a checked exception.",ComputerScienceOne.pdf
"Exception we would have made our exception a checked exception.
There is considerable debate as to whether or not checked exceptions are a good thing
(and as to whether or not unchecked exceptions are a good thing). Many1 feel that checked
1The author included.
435",ComputerScienceOne.pdf
"30. Error Handling & Exceptions
exceptions were a mistake and their usage should be avoided. The rationale behind
checked exceptions is summed up in the following quote from the Java documentation [ 7].
Here’s the bottom line guideline: If a client can reasonably be expected to
recover from an exception, make it a checked exception. If a client cannot do
anything to recover from the exception, make it an unchecked exception
The problem is that the JDK’s own design violates this principle. For example,
FileNotFoundException is a checked exception; the reasoning being that a program
could reprompt the user for a different file. The problem is the assumption that the
program we are writing is always interactive. In fact most software is not interactive and
is instead designed to interact with other software. Reprompting is not appropriate or
even an option in the vast majority of cases.
As another example, consider Java’s SQL library which allows you to programmatically",ComputerScienceOne.pdf
"even an option in the vast majority of cases.
As another example, consider Java’s SQL library which allows you to programmatically
connect to an SQL database. Nearly every method in the API explicitly throws a checked
SQLException. It stretches the imagination to understand how our code or even a user
would be able to recover (programmatically at least) from a lost internet connection or a
bad password, etc. In general, you should use unchecked exceptions.
30.2. Enumerated Types
Another way to prevent errors is by restricting the values of a variable that a programmer
is allowed to use. Java allows you to define a special class called an enumerated type
which allow you to define a fixed list of possible values. For example, the days of the week
or months of the year are possible values used in a date. However, they are more-or-less
fixed (no one will be adding a new day of the week or month any time soon). An
enumerated class allows us to define these values using human-readable keywords.",ComputerScienceOne.pdf
"enumerated class allows us to define these values using human-readable keywords.
To create an enumerated type class we use the keyword enum (short for enumeration)
instead of class. Since an enumerated type is a class, it must follow the same rules.
It must be in a .java source file of the same name. Inside the enum we provide a
comma-delimited list of keywords to define our enumeration. Consider the following
example.
1 public enum Day {
2 SUNDAY,
3 MONDAY,
4 TUESDAY,
5 WEDNESDAY,
6 THURSDAY,
7 FRIDAY,
436",ComputerScienceOne.pdf
"30.2. Enumerated Types
8 SATURDAY;
9 }
In the example, since the name of the enumeration is Day this declaration must be in a
source file named Day.java. We can now declare variables of this type. The possible
values it can take are restricted to SUNDAY, MONDAY, etc. and we can use these keywords
in our program. However these values belong to the class Day and must be accessed
statically. For example,
1 Day today = Day.MONDAY;
2
3 if(today == Day.SUNDAY || today == Day.SATURDAY) {
4 System.out.println(""Its the weekend!"");
5 }
Note the naming conventions: the name of the enumerated type follows the same
upper camel casing that all classes use while the enumerated values follow an uppercase
underscore convention. Though our example does not contain a value with multiple
words, if it had, we would have used an underscore to separate them.
Using an enumerated type has two advantages. First, it enforces that only a particular",ComputerScienceOne.pdf
"Using an enumerated type has two advantages. First, it enforces that only a particular
set of predefined values can be used. You would not be able to assign a value to a
Day variable that was not already defined by the Day enumeration. Second, without
an enumerated type we’d be forced to use a collection of magic numbers to indicate
values. Even for something as simple as the days of the week we’d be constantly trying to
remember: which day is Wednesday again? Do our weeks start with Monday or Sunday?
By using an enumerated type these questions are mostly moot as we can use the more
human-readable keywords and eliminate the guess work.
30.2.1. More Tricks
Every enum type has some additional built-in methods that you can use. For example,
there is a values() method that can be called on the enum that returns an array of
the enum’s values. You can use an enhanced for loop to iterate over them, for example,
437",ComputerScienceOne.pdf
"30. Error Handling & Exceptions
1 for(Day d : Day.values() {
2 System.out.println(d.name());
3 }
In the example above, we used another feature: each enum value has a name() method
that returns the value as a String. This example would end up printing the following.
SUNDAY
MONDAY
TUESDAY
WEDNESDAY
THURSDAY
FRIDAY
SATURDAY
Of course, we may want more human-oriented representations. To do this we could
override the class’s toString() method to return a better string representation. For
example:
1 public String toString() {
2 if(this == SUNDAY) {
3 return ""Sunday"";
4 } else if(this == MONDAY) {
5 return ""Monday"";
6 }
7 ...
8 } else {
9 return ""Saturday"";
10 }
11 }
Because enum types are full classes in Java, many more tricks can be used that leverage
the power of classes including using additional state and constructors. We will cover
these topics later.
438",ComputerScienceOne.pdf
"31. Arrays
Java allows you to declare and use arrays. Since Java is statically typed, arrays must
also be typed when they are declared and may only hold that particular type of element.
Since Java has automated garbage collection, memory management is greatly simplified.
Finally, in Java, only locally scoped primitives and references are allocated on the program
call stack. As such, there are no static arrays; all arrays are allocated in the heap space.
31.1. Basic Usage
To declare an array reference, you use syntax similar to declaring a regular variable, but
you use square brackets to designate the variable as an array. Arrays for any type of
variable can be created including primitives and objects.
1 int arr[] = null;
2 double values[] = null;
3 String names[] = null;
This example1 only declares 3 references that can refer to an array, it doesn’t actually
create them as the references are all set to null. To create arrays of a particular size,",ComputerScienceOne.pdf
"create them as the references are all set to null. To create arrays of a particular size,
we need to initialize them using the keyword new.
1 arr = new int[10];
2 values = new double[20];
3 names = new String[5];
1You may see code examples that use the alternative notation int[] arr = null; . Both are
acceptable. Some would argue that this way is more correct because it keeps the brackets with
the type, avoiding the type-variable name-type separation in our example. Those preferring the
int arr[] = null; notation would argue that it is “more natural” because that is how we
ultimately index the array. Six of one, half-dozen of the other.
439",ComputerScienceOne.pdf
"31. Arrays
Each of these initializations creates a new array (allocated on the heap) of the specified
size (10, 20, and 5 respectively). These arrays can only hold values of the specified type,
int, double, and String respectively. The default value for each element in these
new arrays will be zero for the numeric types and null for object types. Alternatively,
you could use a compound declaration/assignment syntax to initialize specific values:
1 int arr[] = { 2, 3, 5, 7, 11 };
Using this syntax we do not need to specify the size of the array as the compiler is smart
enough to count the number of elements we’ve provided. The elements themselves are
denoted inside curly brackets and delimited with commas. For both types of syntax, the
actual array is always allocated on the heap while the reference variables, arr, values,
and names are stored in the program call stack.
Indexing
Once an array has been created, its elements can be accessed by indexing them. Java",ComputerScienceOne.pdf
"and names are stored in the program call stack.
Indexing
Once an array has been created, its elements can be accessed by indexing them. Java
uses the standard 0-indexing scheme so the first element is at index 0, the second at index
1, etc. Indexing an element involves using the square bracket notation and providing
an integer index. Once indexed, an array element can be treated as a normal variable
and can be used with other operators such as the assignment operator or comparison
operators.
1 arr[0] = 42;
2 if(arr[4] < 0) {
3 System.out.println(""negative!"");
4 }
5 System.out.println(""arr[1] = "" + arr[1]);
Recall that an index is actually an offset. The compiler and system know exactly how
many bytes each element takes and so an index i calculates exactly how many bytes
from the first element the i-th element is located at. Consequently it is possible to index
elements that are beyond the range of the array. For example, arr[-1] or arr[5]",ComputerScienceOne.pdf
"elements that are beyond the range of the array. For example, arr[-1] or arr[5]
would attempt to access an element immediately before the first element and immediately
after the last element. Obviously, these elements are not part of the array. If you
attempt to access an element outside the bounds of an array, the JVM will throw an
IndexOutOfBoundsException which is an unchecked RuntimeException that you can
440",ComputerScienceOne.pdf
"31.1. Basic Usage
catch and handle if you choose. To prevent such an exception you can write code that
does not exceed the bounds of the array. Java arrays have a special length property that
gives you the size of the array. You can access the property using the dot operator, so
arr.length would give the value 5 for the example above.
Iteration
Using the .length property you can design a for loop that increments an index variable
to iterate over the elements in an array.
1 int arr[] = new int[5];
2 for(int i=0; i<arr.length; i++) {
3 arr[i] = 5 * i;
4 }
The for loop above initializes the variable i to zero, corresponding to the first element
in the array. The continuation condition specifies that the loop continues while i is
strictly less than the size of the array denoted using the arr.length property. This for
loop iteration is idiomatic when processing arrays.
In addition, Java (as of version 5) supports foreach loops that allow you to iterate over",ComputerScienceOne.pdf
"loop iteration is idiomatic when processing arrays.
In addition, Java (as of version 5) supports foreach loops that allow you to iterate over
the elements of an array without using an index variable. Java refers to these loops as
“enhanced for loops,” but they are essentially foreach loops.
1 for(int a : arr) {
2 System.out.println(a);
3 }
The syntax still uses the keyword for, but instead of an index variable, it specifies the
type and a the loop-scoped variable identifier followed by a colon and the array that you
want to iterate over. Each iteration of the loop updates the variable a to the next value
in the array.
441",ComputerScienceOne.pdf
"31. Arrays
31.2. Dynamic Memory
The use of the keyword new dynamically allocates memory space (on the heap) for the
array. Because Java has automated garbage collection, if the reference to the array goes
out of scope, it is automatically cleaned up by the JVM. This process is automated and
essentially transparent to us. There are sophisticated algorithms and heuristics at work
that determine the “optimal” time to perform garbage collection.
1 int arr[];
2 //create a new array of size 10:
3 arr = new arr[10];
4 //lose the reference by explicitly setting it to null:
5 arr = null;
6 //all references to the old memory are now lost and it is
7 //eligible for garbage collection
8
9 //we can also allocate new memory:
10 arr = new arr[20];
31.3. Using Arrays with Methods
We can pass and return arrays to and from methods in Java. We use the same syntax as
when we define them by specifying a type and using square brackets. As an example, the",ComputerScienceOne.pdf
"when we define them by specifying a type and using square brackets. As an example, the
following method takes an array of integers and computes its sum.
1 /**
2 * This method computes the sum of elements in the
3 * given array which contains n elements
4 */
5 public static int computeSum(int arr[]) {
6 int sum = 0;
7 for(int i=0; i<size; i++) {
8 sum += arr[i];
9 }
10 return sum;
11 }
442",ComputerScienceOne.pdf
"31.4. Multidimensional Arrays
In Java, arrays are always passed by reference. Though we did not make any changes
to the contents of the passed array in the particular example, in general we could have.
Any such changes would be realized in the calling method. Unfortunately, there is no
mechanism by which we can prevent changes to arrays when passed to methods. 2
We can also create an array in a method and return it as a value. For example, the
following method creates a deep copy of the given integer array. That is, a completely
new array that is a distinct copy of the old array. In contrast, a shallow copy would be if
we simply made one reference point to another reference.
1 /**
2 * This method creates a new copy of the given array
3 * and returns it.
4 */
5 public static int [] makeCopy(int a[]) {
6 int copy[] = new int[a.length];
7 for(int i=0; i<n; i++) {
8 copy[i] = a[i];
9 }
10 return copy;
11 }",ComputerScienceOne.pdf
"5 public static int [] makeCopy(int a[]) {
6 int copy[] = new int[a.length];
7 for(int i=0; i<n; i++) {
8 copy[i] = a[i];
9 }
10 return copy;
11 }
The method returns an integer array. In fact, this method is so useful, that it is already
provided as part of the Java Software Development Kit (SDK). The class Arrays
contains dozens of utility methods that process arrays. In particular there is a copyOf()
method that allows you to create deep copies of arrays and even allows you to expand or
shrink their size.
31.4. Multidimensional Arrays
Java supports multidimensional arrays, though we’ll only focus on 2-dimensional arrays.
To declare and allocate 2-dimensional arrays, we use two square brackets with two
specified dimensions.
1 int matrix[][] = new int[10][20];
2The use of the keyword final only prevents us from changing the array reference, not modifying its
contents.
443",ComputerScienceOne.pdf
"31. Arrays
This creates a 2-dimensional array of integers with 10 rows and 20 columns. Once created,
we can index individual elements by specifying row and column indices.
1 for(int i=0; i<matrix.length; i++) {
2 for(int j=0; j<matrix[i].length; j++) {
3 matrix[i][j] = 10;
4 }
5 }
31.5. Dynamic Data Structures
The Java SDK provides a rich assortment of dynamic data structures as alternatives to
arrays including lists, sets and maps. The classes that support and implement these data
structures are defined in the Collections library under the java.util package. We will
cover how to use some of these data structures, but we will not go into the details of how
they are implemented nor the OOP concepts that underly them.
The Java List is an interface that defines a dynamic list data structure. This data
structure provides a dynamic collection that can grow and shrink automatically as you
add and remove elements from it. It is an interface, so it doesn’t actually provide an",ComputerScienceOne.pdf
"add and remove elements from it. It is an interface, so it doesn’t actually provide an
implementation, just a specification for the publicly available methods that you can use.
Two common implementations are ArrayList, which uses an array to hold elements,
and LinkedList which stores elements in linked nodes.
To create an instance of either of these lists, you use the new keyword and the following
syntax.
1 List<Integer> a = new ArrayList<Integer>();
2 List<String> b = new LinkedList<String>();
The first line creates a new instance of an ArrayList that is parameterized to hold
Integer types. The second creates a new instance of a LinkedList that has been
parameterized to only hold String types. The parameterization is specified by placing
the parameterized type inside the angle brackets. Because of this parameterization, it
444",ComputerScienceOne.pdf
"31.5. Dynamic Data Structures
would be a compiler error to attempt to add anything other than Integers to the first
list or anything other than Strings to the second. Once these lists have been created,
you can add and remove elements using the add() method.
1 a.add(42);
2 a.add(81);
3 a.add(17);
4
5 b.add(""Hello"");
6 b.add(""World"");
7 b.add(""Computers!"");
The order that you add elements is preserved, so in the first list, the first element would
be 42, the second 81, and the last 17. You can remove elements by specifying an index
of the element to remove. Like arrays, lists are 0-indexed.
1 a.remove(0);
2
3 b.remove(2);
4 b.remove(0);
As you remove elements, the indices are “shifted” down, so that after removing the first
element in the list a, 81 becomes the new first element. Removing the last then the
first element in the list b leaves it with only one element, ""World"" as the first element
(at index 0).
You can also retrieve elements from a list using 0-indexing and the get() method.",ComputerScienceOne.pdf
"(at index 0).
You can also retrieve elements from a list using 0-indexing and the get() method.
1 List<Double> values = new ArrayList<Double>();
2 values.add(3.14);
3 values.add(2.71);
4 values.add(42.0);
5
6 double x = values.get(1); //get the second value, 2.71
7 double y = values.get(2); //get the third value, 42.0
445",ComputerScienceOne.pdf
"31. Arrays
Any attempt to access an element that lies outside the bounds of the List, will result
in an IndexOutOfBoundsException just as with arrays. To stay within bounds you
can use the size() method to determine how many elements are in the collection. In
this example, values.size() would return an integer value of 3.
Finally, most collections implement the Iterable interface which allows you iterate
over the elements using an enhanced for loop just as with arrays.
1 for(Double x : values) {
2 System.out.println(x);
3 }
There are dozens of other methods that allow you to insert, remove, and retrieve elements
from a Java List; refer to the documentation for details.
Another type of collection supported by the Collections library is a Set. A set differs
from a list in that it is unordered. There is no concept of the first element or last element.
Moreover, a Set does not allow duplicate elements. 3 A commonly used implementation
of the Set interface is the HashSet.",ComputerScienceOne.pdf
"Moreover, a Set does not allow duplicate elements. 3 A commonly used implementation
of the Set interface is the HashSet.
1 Set<String> names = new HashSet<String>();
2
3 names.add(""Robin"");
4 names.add(""Little John"");
5 names.add(""Marian"");
6
7 //this has no effect since Robin is already part of the set:
8 names.add(""Robin"");
Since the elements in a Set are unordered, we cannot use an index-based get() method
as we did with a List. Fortunately, we can still use an enhanced for loop to iterate over
the elements.
1 for(String name : names) {
2 System.out.println(name);
3Duplicates are determined by how the equals() and possibly the hashCode() methods are imple-
mented in your particular objects.
446",ComputerScienceOne.pdf
"31.5. Dynamic Data Structures
3 }
When this code executes we cannot expect any particular order of the three names. Any
permutation of the three may be printed. If we executed the loop more than once we
may even observe a different enumeration of the names!
Finally, Java also supports a Map data structure which allows you to store key-value
pairs. The keys and values can be any object type and are specified by two parameters
when you create the map. A common implementation is the HashMap.
1 //define a map that maps integers (keys) to strings (values):
2 Map<Integer, String> numbers = new HashMap<Integer, String>();
3
4 //add key-value pairs:
5 numbers.add(10, ""ten"");
6 numbers.add(20, ""twenty"");
7
8 //retrieve values given a key:
9 String name = numbers.get(20); //name equals ""twenty""
10
11 //invalid mappings result in null:
12 name = numbers.get(30); //name is null
The Collections library is much more extensive than what we’ve presented here. It",ComputerScienceOne.pdf
"12 name = numbers.get(30); //name is null
The Collections library is much more extensive than what we’ve presented here. It
includes implementations of stacks, queues, hash tables, balanced binary search trees,
and many other dynamic data structure implementations.
447",ComputerScienceOne.pdf
"32. Strings
As we’ve previously seen, Java has a String class in the standard JDK. Internally, Java
strings are stored as arrays of characters. However, because of the String class, we
never directly interact with this representation. Making using strings much easier than
in other languages. Java strings have also supported Unicode since version 1.
Java provides many methods as part of the String class that can be used to process
and manipulate strings. These methods do not change the strings since strings in Java
are immutable. Instead, these methods operate by returning a new modified string that
can then be stored in a variable.
32.1. Basics
We can declare String variables and assign them values using the regular assignment
operator.
1 String firstName = ""Thomas"";
2 String lastName = ""Waits"";
3
4 //we can also reassign values
5 firstName = ""Tom"";
Note that the reassignment in the last line in the example does not change the original",ComputerScienceOne.pdf
"3
4 //we can also reassign values
5 firstName = ""Tom"";
Note that the reassignment in the last line in the example does not change the original
string. It just makes the variable firstName point to a new string. If there are no other
references to the old string, it becomes eligible for garbage collection and the JVM takes
care of it. Strings can be copied using the String class’s copy constructor:
String copy = new String(firstName);
However, since strings are immutable, there is rarely reason to create such a deep copy.
Though you can’t change individual characters in a string, you can access them using the
449",ComputerScienceOne.pdf
"32. Strings
charAt() method and providing an index. Characters in a string are 0-indexed just as
with elements in arrays.
1 String fullName = ""Tom Waits"";
2 //access individual characters:
3 char firstInitial = fullName.charAt(0); //'T'
4 char lastInitial = fullName.charAt(4); //'W'
5
6 if(fullName.charAt(8) == 's') {
7 ...
8 }
32.2. String Methods
The String class provides dozens of convenient methods that allow you to process
and modify strings. We highlight a few of the more common ones here. A full list of
supported functions can be found in standard documentation.
Length
When accessing individual characters in a string, it is necessary to know the length of
the string so that we do not access invalid characters. The length() method returns
an int that represents the number of characters in the string.
1 String s = ""Hello World!"";
2 char last = s.charAt(s.length()-1);
3 //last is '!'
Using this method we can easily iterate over each character in a string.",ComputerScienceOne.pdf
"2 char last = s.charAt(s.length()-1);
3 //last is '!'
Using this method we can easily iterate over each character in a string.
1 for(int i=0; i<fullName.length(); i++) {
2 System.out.printf(""fullName[%d] = %c\n"", i, fullName.charAt(i));
3 }
4
450",ComputerScienceOne.pdf
"32.2. String Methods
5 //or we can use toCharArray and an enhanced for loop:
6 for(char c : fullName.toCharArray()) {
7 System.out.println(c);
8 }
Concatenation
Java has a concatenation operator built into the language. The familiar plus sign, + can
be used to combine one or more strings by appending them to each other. Concatenation
results in a new string.
1 String firstName = ""Tom"";
2 String lastName = ""Waits"";
3
4 String formattedName = lastName + "", "" + firstName;
5 //formattedName now contains ""Waits, Tom""
Concatenation also works with numeric types and even objects.
1 int x = 10;
2 double y = 3.14;
3
4 String s = ""Hello, x is "" + x + "" and y = "" + y;
5 //s contains ""Hello, x is 10 and y = 3.14""
When used with objects, the concatenation operator internally invokes the object’s
toString() method. The plus operator as a concatenation operator is actually syntactic
sugar. When the code is compiled, it is actually replaced with equivalent code that uses a",ComputerScienceOne.pdf
"sugar. When the code is compiled, it is actually replaced with equivalent code that uses a
series of the StringBuilder class’s (a mutable version of the String class) append()
method. The compiler will actually replace the code using the concatenation operator
with code that does not use the concatenation operator. The example above would be
replaced with something like the following.
451",ComputerScienceOne.pdf
"32. Strings
1 String firstName = ""Tom"";
2 String lastName = ""Waits"";
3
4 String formattedName = new StringBuilder(lastName)
5 .append("", "")
6 .append(firstName)
7 .toString();
You can also use the StringBuilder class yourself directly.
Computing a Substring
There are two methods that allow you to compute a substring of a string. The first allows
you to specify a beginning index with the entire remainder of the string being included
in the returned string. The second allows you to specify a beginning index as well as an
ending index. In both cases, the beginning index is inclusive (that is the resulting string
includes the character at that index), but in the second, the ending index is exclusive
(the character at the end index is not included).
1 String name = ""Thomas Alan Waits"";
2
3 String firstName = name.substring(0, 6); //""Thomas""
4 String middleName = name.substring(7, 11); //""Alan""
5 String lastName = name.substring(12); //""Waits""",ComputerScienceOne.pdf
"4 String middleName = name.substring(7, 11); //""Alan""
5 String lastName = name.substring(12); //""Waits""
The result of the two argument substring() method will always have length equal to
endIndex - beginIndex.
32.3. Arrays of Strings
We often need to deal with collections of strings. In Java, we can define arrays of strings.
Indeed, we’ve seen arrays of strings before. In a Java class’s main() method, command
line arguments are passed as an array of strings:
public static void main(String args[])
452",ComputerScienceOne.pdf
"32.4. Comparisons
We can create our own arrays of strings similar to how we created arrays of int and
double types.
1 //create an array that can hold 5 strings
2 String names[] = new String[5];
3
4 names[0] = ""Margaret Hamilton"";
5 names[1] = ""Ada Lovelace"";
6 names[2] = ""Grace Hopper"";
7 names[3] = ""Marie Curie"";
8 names[4] = ""Hedy Lamarr"";
Better yet, we can use dynamic collections such as a List or a Set of strings.
1 List<String> names = new ArrayList<String>();
2
3 names.add(""Margaret Hamilton"");
4 names.add(""Ada Lovelace"");
5 names.add(""Grace Hopper"");
6 names.add(""Marie Curie"");
7 names.add(""Hedy Lamarr"");
8
9 System.out.println(names.get(2)); //prints ""Grace Hopper""
32.4. Comparisons
When comparing strings in Java, we cannot use the numerical comparison operators such
as ==, or <. Because strings are objects, using these operators actually compares the
variables’ memory addresses.
1 String a = new String(""Hello World!"");
2 String b = new String(""Hello World!"");
3
4 if(a == b) {",ComputerScienceOne.pdf
"variables’ memory addresses.
1 String a = new String(""Hello World!"");
2 String b = new String(""Hello World!"");
3
4 if(a == b) {
5 System.out.println(""strings match!"");
453",ComputerScienceOne.pdf
"32. Strings
6 }
The code above will not print anything even though the strings a and b have the same
content. This is because a == b is comparing the memory address of the two variables.
Since they point to different memory addresses (created by two separate calls to the
constructors) they will not be equal.
Instead, the String class provides a compareTo() method1 that takes another string
and returns something negative, zero, or something positive depending on the relative
lexicographic ordering of the strings.
1 int x;
2 String a = ""apple"";
3 String b = ""zelda"";
4 String c = ""Hello"";
5 x = a.compareTo(""banana""); //x is negative
6 x = b.compareTo(""mario""); //x is positive
7 x = c.compareTo(""Hello""); //x is zero
8
9 //shorter strings precede longer strings:
10 x = a.compareTo(""apples""); //x is negative
11
12 String d = ""Apple"";
13 x = d.compareTo(""apple""); //x is negative
In the last example, ""Apple"" precedes ""apple"" since uppercase letters are ordered",ComputerScienceOne.pdf
"12 String d = ""Apple"";
13 x = d.compareTo(""apple""); //x is negative
In the last example, ""Apple"" precedes ""apple"" since uppercase letters are ordered
before lowercase letters according to the ASCII table. We can also make case-insensitive
comparisons using the compareToIgnoreCase() method which works the same way.
Here, d.compareToIgnoreCase(""apple"") will return zero as the two strings are the
same ignoring the cases.
Note that the commonly used equals() method only returns true or false depending
on whether or not two strings are the same or different. They cannot be used to provide
a relative ordering of two strings.
1This method is part of the String class due to the fact that strings implement the Comparable
interface which defines a lexicographic ordering.
454",ComputerScienceOne.pdf
"32.5. Tokenizing
1 String a = ""apple"";
2 String b = ""apple"";
3 String c = ""Hello"";
4
5 boolean result;
6 result = a.equals(b); //true
7 result = a.equals(c); //false
32.5. Tokenizing
Recall that tokenizing is the process of splitting up a string using some delimiter. For
example, the comma delimited string, ""Smith,Joe,12345678,1985-09-08"" contains
four pieces of data delimited by a comma. Our aim is to split this string up into four
separate strings so that we can process each one.
Java provides a very simple method to do this called split() that takes a string
delimiter and returns an array of strings containing the tokens. For example,
1 String data = ""Smith,Joe,12345678,1985-09-08"";
2
3 String tokens[] = data.split("","");
4 //tokens is { ""Smith"", ""Joe"", ""12345678"", ""1985-09-08"" }
5
6 String dateTokens[] = tokens[3].split(""-"");
7 //dateTokens is now { ""1985"", ""09"", ""08"" }
The delimiter this method uses is actually a regular expression; a sequence of characters",ComputerScienceOne.pdf
"7 //dateTokens is now { ""1985"", ""09"", ""08"" }
The delimiter this method uses is actually a regular expression; a sequence of characters
that define a search pattern in which special characters can be used to define complex
patterns. For example, the complex expression,
^[+-]?(\d+(\.\d+)?|\.\d+)([eE][+-]?\d+)?$
will match any valid numerical value including scientific notation. We will not cover
regular expressions in depth, but to demonstrate their usefulness, here’s an example by
which you can split a string along any and all whitespace:
455",ComputerScienceOne.pdf
"32. Strings
1 String s = ""Alpha Beta \t Gamma \n Delta \t\nEpsilon"";
2 String tokens[] = s.split(""[\\s]+"");
3 //tokens is now { ""Alpha"", ""Beta"", ""Gamma"", ""Delta"", ""Epsilon"" }
456",ComputerScienceOne.pdf
"33. File I/O
Java provides several different classes and utilities that support manipulating and pro-
cessing files. In general, most file operations may result in an IOException, a checked
exception that must be caught and handled.
33.1. File Input
Though there are several ways that you can do file input, the easiest is to use the
familiar Scanner class. We’ve previously used this class to read from the standard
input, System.in, but it can also be used to read from a file using the File class
(in the java.io package). The File class is just a representation of a file resource
and does not represent the actual file (it cannot be opened and closed). To initialize a
Scanner to read from a file you can use the following.
1 Scanner s = null;
2 try {
3 s = new Scanner(new File(""/user/apps/data.txt""));
4 } catch (FileNotFoundException e) {
5 //handle the exception here
6 }
When initializing a Scanner to read from a file, the exception, FileNotFoundException",ComputerScienceOne.pdf
"5 //handle the exception here
6 }
When initializing a Scanner to read from a file, the exception, FileNotFoundException
must be caught and handled as it is a checked exception. Otherwise, once created, the
Scanner class does not throw any checked exceptions.
Once created, you can use the usual methods and functionality provided by the Scanner
class including reading the nextInt(), nextDouble(), next string using next() or
even the entire nextLine(). This last one can be used in conjunction with hasNext()
to read an entire file, line-by-line.
457",ComputerScienceOne.pdf
"33. File I/O
1 String line;
2 while(s.hasNext()) {
3 line = s.nextLine();
4 //process the line
5 }
Once we are done reading the file, we can close the Scanner to free up resources:
s.close();. We could have placed all this code within one large try-catch block
with perhaps a finally block to close the Scanner once were were done to ensure
that it would be closed regardless of any exceptions. However, Java 7 introduced a new
construct, the try-with-resources statement.
The try-with-resources statement allows us to place the initialization of a “closeable”
resource (defined by the AutoCloseable interface) into the try statement. The JVM
will then automatically close this resource upon the conclusion of the try-catch block.
We can still provide a finally block if we wish, but this relieves us of the need to
explicitly close the resource in a finally block. A full example:
1 File f = new File(""/user/apps/data.txt"");
2 //initialize a Scanner in the try statement",ComputerScienceOne.pdf
"1 File f = new File(""/user/apps/data.txt"");
2 //initialize a Scanner in the try statement
3 try (Scanner s = new Scanner(f)) {
4 String line;
5 while(s.hasNext()) {
6 line = s.nextLine();
7 //process the line
8 }
9 } catch (Exception e) {
10 //handle the exception here
11 }
Using the Scanner class to do file input offers a more abstract interaction with a file.
It also uses a buffered input stream for performance. Binary files can still be read using
nextByte(). However, when dealing with binary files, the better solution is to use a
class that models and abstracts the underlying file. For example, if you are reading or
writing an image file, you should use the java.awt.Image class to read and write to
files. The JDK and other libraries offer a wide variety of classes to model all kinds of
data.
458",ComputerScienceOne.pdf
"33.2. File Output
33.2. File Output
There are several ways to achieve file output, but we’ll look at the simplest and most
useful way. Java provides a PrintWriter class which offers many convenient methods for
writing primitive and String types in a formatted manner. It has the usual print()
and println() methods that we are familiar with as well as a printf() style method
for formatting.
PrintWriter is also buffered, depending on how you instantiate it, it might “flush” its
buffer after every endline character or you may do it manually by invoking the flush()
method.
1 int x = 10;
2 double pi = 3.14;
3 PrintWriter pw = null;
4 try {
5 pw = new PrintWriter(new File(""data.txt""));
6 } catch(FileNotFoundException fnfe) {
7 throw new RuntimeException(fnfe);
8 }
9
10 pw.println(""Hello World!"");
11 pw.println(x);
12 pw.printf(""x = %d, pi = %f\n"", x, pi);
13
14 pw.close();
The close() method will conveniently close all the underlying resources and flush the",ComputerScienceOne.pdf
"12 pw.printf(""x = %d, pi = %f\n"", x, pi);
13
14 pw.close();
The close() method will conveniently close all the underlying resources and flush the
buffer for us, writing any buffered data to the file. In addition, PrintWriter implements
AutoCloseable and so it can be used in a try-catch-with resources statement.
Another “convenience” of PrintWriter is that is “swallows” exceptions (just as the
Scanner class did). That means we don’t have to deal explicitly with the checked
IOExceptions that the underlying class(es) throw as the PrintWriter silently catches
them (though doesn’t handle them). However, this can also be viewed as a disadvantage
in that if we want to do error handling, we need to manually check if there was an
error (using checkError()). The PrintWriter class is intended mostly for formatted
output. It does not provide a way to write binary data to an output file. Just as with
binary input, it is best to use a class that abstracts the file type and data so that we",ComputerScienceOne.pdf
"binary input, it is best to use a class that abstracts the file type and data so that we
don’t have to deal with the low-level details of the binary data.
459",ComputerScienceOne.pdf
"33. File I/O
However, if necessary, binary output can be done using a FileOutputStream. Typically,
you can load all your data into a byte array and dump it all at once.
1 byte data[] = ...;
2 try (FileOutputStream fos =
3 new FileOutputStream(new File(""outfile.bin"")) ){
4 fos.write(data);
5 } catch(IOException ioe) {
6 throw new RuntimeException(ioe);
7 }
This example uses a try-with-resources statement so the FileOutputStream will auto-
matically be closed for us. You could wrap a FileOutputStream in a BufferedWriter,
but it will likely not gain you anything in terms of performance. A buffered output
stream is better if your writes are frequent and “small.” Here small means smaller than
the size of the buffer ( BufferedWriter is typically 8KB). Writing several individual
integer values for example would be better done by buffering them and writing them
all at once. In our example, we simply wanted to dump a bunch of data, likely more",ComputerScienceOne.pdf
"all at once. In our example, we simply wanted to dump a bunch of data, likely more
than 8KB in practice, all at once. Moreover, using a BufferedWriter would lose us
the ability to write raw byte data.
460",ComputerScienceOne.pdf
"34. Objects
Java is a class-based object-oriented programming language, meaning that it facilitates
the creation of objects through the use of classes. Classes are essentially “blueprints”
for creating instances of objects. We’ve been implicitly using classes all along since
everything in Java is a class or belongs to a class. Now, however, we will start using
classes in more depth rather than simply using static methods.
An object is an entity that is characterized by identity, state and behavior. The identity
of an object is an aspect that distinguishes it from other objects. The variables and
values that a variable takes on within an object is its state. Typically the variables that
belong to an object are referred to as member variables. Finally, an object may also
have functions that operate on the data of an object. In the context of object-oriented
programming, a function that belongs to an object is referred to as a member method.",ComputerScienceOne.pdf
"programming, a function that belongs to an object is referred to as a member method.
As a class-based object-oriented language, Java implements objects using classes. A
class is essentially a blueprint for creating instances of the class. A class simply specifies
the member variables and member methods that belong to instances of the class. We
can create actual instances through the use of special constructor methods. To begin,
let’s define a class that models a student by defining member variables to support a first
name, last name, a unique identifier, and GPA.
To declare a class, we use the class keyword. Inside the class (denoted by curly
brackets), we place any code that belongs to the class. To declare member variables
within a class, we use the normal variable declaration syntax, but we do so outside any
methods.
1 package unl.cse;
2
3 public class Student {
4
5 //member variables:
6 String firstName;
7 String lastName;
8 int id;
9 double gpa;
10
11 }
461",ComputerScienceOne.pdf
"34. Objects
Recall that a package declaration allows you to organize classes and code within a package
(directory) hierarchy. Moreover, source code for a class must be in a source file with the
same name (and is case sensitive) with the .java extension. Our Student class would
need to be in a file named Student.java and would be compiled to a class named
Student.class.
34.1. Data Visibility
Recall that encapsulation involves not only the grouping of data, but the protection
of data. To provide for the protection of data, Java defines several visibility keywords
that specify what segments of code can “see” the variables. Visibility in this context
determines whether or not a segment of code can access and/or modify the variable’s
value. Java defines four levels of visibility using they keywords public, protected
and private (a fourth level is defined by the absence of any of these keywords). Each
of these keywords can be applied to both member variables and member methods.",ComputerScienceOne.pdf
"of these keywords can be applied to both member variables and member methods.
• public – This is the least restrictive visibility level and makes the member variable
visible to every class.
• protected – This is a bit more restrictive and makes it so that the member
variable is only visible to the code in the same class, same package, or any subclass
of the class.1
•No modifier – the absence of any modifier means that the member variable is visible
to any class in the same package. This is also referred to as the default or package
protected visibility level.
• private – this is the most restrictive visibility level, private member variables
are only visible to instances of the class itself. As we’ll see later this also means
that the member variables are visible between instances. That is, one instance can
see another instance’s variables.
Table 34.1 summarizes these four keywords with respect to their access levels. It is",ComputerScienceOne.pdf
"see another instance’s variables.
Table 34.1 summarizes these four keywords with respect to their access levels. It is
important to understand that protection in the context of encapsulation and does not
involve protection in the sense of “security.” The protection is this context is a design
principle. Limiting the access of variables only affects how the rest of the code base
interacts with our class and its data. Encapsulation can easily be “broken” by other
1Subclasses are involved with inheritance, another object-oriented programming concept that we will
not discuss here).
462",ComputerScienceOne.pdf
"34.2. Methods
code (through reflection 2 or other means) and the values of variables can be accessed or
modified.
Modifier Class Package Subclass World
public Y Y Y Y
protected Y Y Y N
none (default) Y Y N N
private Y N N N
Table 34.1.: Java Visibility Keywords & Access Levels
We now update our class declaration to incorporate these visibility keywords. In general,
it is best practice to make member variables private and control access to them via
accessor and mutator methods (see below) unless there is a compelling design reason to
increase their visibility.
1 package unl.cse;
2
3 public class Student {
4
5 //member variables:
6 private String firstName;
7 private String lastName;
8 private int id;
9 private double gpa;
10
11 }
34.2. Methods
The third aspect of encapsulation involves the grouping of methods that act on an object’s
data (its state). Within a class, we can declare member methods using the syntax we’re",ComputerScienceOne.pdf
"data (its state). Within a class, we can declare member methods using the syntax we’re
already familiar with. We declare a member method by providing a signature and body.
We can use the same visibility keywords as with member variables in order to allow or
restrict access to the methods. With methods, visibility and access determine whether or
not the method may be invoked.
2Reflection is a mechanism by which you can write code that modifies other code at runtime. It is
more of an aspect-oriented programming paradigm.
463",ComputerScienceOne.pdf
"34. Objects
In contrast to the methods we defined in Chapter 29, when defining a member method,
we do not use the static keyword. Making a variable or a method static means that
the method belongs to the class and not to instances of the class. We add to our example
by providing two public methods that compute and return a result on the member
variables. We also use Javadoc style comments to document each member method.
1 package unl.cse;
2
3 public class Student {
4
5 //member variables:
6 private String firstName;
7 private String lastName;
8 private int id;
9 private double gpa;
10
11 /**
12 * Returns a formatted String of the Student's
13 * name as Last, First.
14 */
15 public String getFormattedName() {
16 return lastName + "", "" + firstName;
17 }
18
19 /**
20 * Scales the GPA, which is assumed to be on a
21 * 4.0 scale to a percentage.
22 */
23 public double getGpaAsPercentage() {
24 return gpa / 4.0;
25 }
26
27 }
34.2.1. Accessor & Mutator Methods",ComputerScienceOne.pdf
"21 * 4.0 scale to a percentage.
22 */
23 public double getGpaAsPercentage() {
24 return gpa / 4.0;
25 }
26
27 }
34.2.1. Accessor & Mutator Methods
Since we have made all the member variables private, no code outside the class may
access or modify their values. It is generally good practice to make member variables
private to restrict access. However, if we still want code outside the object to access
464",ComputerScienceOne.pdf
"34.2. Methods
or mutate (that is, change) an instance’s variables, we can define accessor and mutator
methods (or just simply getter and setter methods) to facilitate this.
Each getter method returns the value of the instance’s variable while each setter method
takes a value and sets the instance’s variable to the new value. It is common to name
each getter/setter by prefixing a get and set to the variable’s name using lower camel
casing. For example:
1 public String getFirstName() {
2 return firstName;
3 }
4
5 public void setFirstName(String firstName) {
6 firstName = firstName;
7 }
In the setter example, there is a problem: the code has no effect. There are two variables
named firstName: the instance’s member variable and the variable in the method
parameter. The scoping rules of Java mean that the parameter variable name(s) take
precedent. This code has no effect because it is setting the parameter variable to itself.
It is essentially doing the following.
1 int a = 10;
2 a = a;",ComputerScienceOne.pdf
"It is essentially doing the following.
1 int a = 10;
2 a = a;
To solve this, we use something called open recursion. When an instance of a class is
created, for example,
Student s = ...;
the reference variable s is how we can refer to it. This variable, however, exists outside
the class. Inside the class, we need a way to refer to the instance itself. In Java we use
the keyword this to refer to the instance inside the class. For example, the member
variables of an instance can be accessed using the this keyword along with the dot
operator (more below). In our example, this.firstName would refer to the instance’s
firstName and not to the parameter variable. Even when it is not necessary to use the
this keyword (as in the getter example above) it is still best practice to do so. Our
updated getters and setter methods would thus look like the following.
465",ComputerScienceOne.pdf
"34. Objects
1 public String getFirstName() {
2 return this.firstName;
3 }
4
5 public void setFirstName(String firstName) {
6 this.firstName = firstName;
7 }
One advantage to using getters and setters (as opposed to naively making everything
public) is that you can have greater control over the values that your variables can
take. For example, we may want to do some data validation by rejecting null values or
invalid values. For example:
1 public void setFirstName(String firstName) {
2 if(firstName == null) {
3 throw new IllegalArgumentException(""names cannot be null"");
4 } else {
5 this.firstName = firstName;
6 }
7 }
8
9 public void setGpa(double gpa) {
10 if(gpa < 0.0 || gpa > 4.0) {
11 throw new IllegalArgumentException(""GPAs must be in [0, 4.0]"");
12 } else {
13 this.gpa = gpa;
14 }
15 }
Controlling access of member variables through getters and setters is good encapsulation.
Doing so makes your code more predictable and more testable. Making your member",ComputerScienceOne.pdf
"Doing so makes your code more predictable and more testable. Making your member
variables public means that any piece of code can change their values. There is no way
to do validation or prevent bad values.
In fact, it is good practice to not even have setter methods. If the value of member
variables cannot be changed, it makes the object immutable. We’ve seen this before with
the built-in wrapper classes ( Integer, String, etc.). Immutability is a nice property
because it makes instances of the class thread-safe. That is, we can use instances of
466",ComputerScienceOne.pdf
"34.3. Constructors
the class in a multithreaded program without having to worry about threads changing
the state of instances on one another. Immutable classes are also safer to use in certain
collections such as a Set. Elements in a Set are unique; attempting to add a duplicate
element will have no effect on the Set. However, if the elements we add are mutable,
we could end up with duplicates (see Section 34.5).
34.3. Constructors
If we make the (good) design decision to make our class immutable, we still need a
way to initialize the object’s member variables. This is where constructors come in. A
constructor is a special method that specifies how an object is constructed. With built-in
primitive variables such as an int, the Java language (compiler and JVM) “know” how
to interpret and assign a value to such a variable. However, with user-defined objects
such as our Student class, we need to specify how the object is created.",ComputerScienceOne.pdf
"such as our Student class, we need to specify how the object is created.
A constructor method has special syntax. Though it may still one of the visibility
keywords, it has no return type and its name is the same as the class. A constructor may
take any number of parameters. For example, the following constructor allows someone
to construct a Student instance and specify all four member variables.
1 public Student(String firstName, String lastName,
2 int id, double gpa) {
3 this.firstName = firstName;
4 this.lastName = lastName;
5 this.id = id;
6 this.gpa = gpa;
7 }
Java allows us to define multiple constructors, or even no constructor at all. If we
do not specify a constructor, Java provides a default, no argument constructor. This
constructor uses default values for all member variables (zero for numerical types, false
for boolean types, and null for objects). If we do specify a constructor, this default",ComputerScienceOne.pdf
"for boolean types, and null for objects). If we do specify a constructor, this default
constructor becomes unavailable. However, we can always “restore” it by explicitly
defining it.
1 public Student() {
2 this.firstName = null;
3 this.lastName = null;
4 this.id = 0;
467",ComputerScienceOne.pdf
"34. Objects
5 this.gpa = 0.0;
6 }
Alternatively, we can define constructors that accept a subset of variable values.
1 public Student(String firstName, String lastName) {
2 this.firstName = firstName;
3 this.lastName = lastName;
4 this.id = 0;
5 this.gpa = 0.0;
6 }
In both of these examples, we repeated a lot of code. One shortcut is to make all your
constructors call the most general constructor. To invoke another constructor, we use
the this keyword as a method call. For example:
1 public Student() {
2 this(null, null, 0, 0.0);
3 }
4
5 public Student(String firstName, String lastName) {
6 this(firstName, lastName, 0, 0.0);
7 }
Another, very useful type of constructor is the copy constructor that allows you to make
a copy of an instance by passing it to a constructor. The following example copies each
of the member variables of s into this.
1 public Student(Student s) {
2 this(s.firstName, s.lastName, s.id, s.gpa);
3 }
468",ComputerScienceOne.pdf
"34.4. Usage
34.4. Usage
Once we have defined our class and its constructors, we can create and use instances of
it. Just as with regular variables, we need to declare instances of a class by providing
the type and a variable name. For example:
1 Student s = null;
2 Student t = null;
Both of these declarations are simply just reference variables. They may refer to an
instance of the class Student, but we have initialized them to null. To make these
variables refer to valid instances, we invoke a constructor by using the new keyword and
providing arguments to the constructor.
1 Student s = new Student(""Alan"", ""Turing"", 1234, 3.4);
2 Student t = new Student(""Margaret"", ""Hamilton"", 4321, 3.9);
The process of creating a new instance by invoking a constructor is referred to as
instantiation. Once instances have been instantiated, they can be used by invoking their
methods via the dot operator. The dot operator is used to select a member variable (if
public) or member method.",ComputerScienceOne.pdf
"methods via the dot operator. The dot operator is used to select a member variable (if
public) or member method.
1 System.out.println(s.getFormattedName());
2
3 if(s.getGpa() < t.getGpa()) {
4 System.out.println(t.getFirstName() + "" has a better GPA"");
5 }
34.5. Common Methods
Every class in Java has several methods that they inherit from the java.lang.Object
class. Here, we will highlight three of these important methods.
469",ComputerScienceOne.pdf
"34. Objects
The toString() method returns a String representation of the object. However, the
default behavior that all classes inherit from the Object class is that it returns a string
containing the fully qualified class name (package and class name) along with a hexadeci-
mal representation of the JVM memory address at which the instance is stored. An exam-
ple with our Student class may produce something like: unl.cse.Student@272d7a10.
We can override this functionality and change the behavior of the toString() method
to return whatever we want. To do so, we simply define the function (with a signature
that matches the toString() method in the Object class) in our class.
1 public String toString() {
2 return String.format(""%s, %s (ID = %d); %.2f"",
3 this.lastName,
4 this.firstName,
5 this.id,
6 this.gpa);
7 }
This example would return a String containing something similar to
""Hamilton, Margaret (ID = 4321); 3.90""",ComputerScienceOne.pdf
"5 this.id,
6 this.gpa);
7 }
This example would return a String containing something similar to
""Hamilton, Margaret (ID = 4321); 3.90""
The toString() method is a very convenient way to print instances of your class.
Two other methods, the equals() and the hashCode() method are used extensively
by other classes such as the Collections library. The equals() method:
public boolean equals(Object obj)
takes another object obj and returns true or false depending on whether or not
the instance is “equal” to obj. Recall that identity is one of the defining characteristics
of objects. This method is how Java achieves identity; this method allows you to define
what “equality” means. By default, the behavior inherited from the Object class simply
checks if the object, this and the passed object, obj are located at the same memory
address, essentially (this == obj).
However, conceptually we may want different behavior. For example, two Student",ComputerScienceOne.pdf
"address, essentially (this == obj).
However, conceptually we may want different behavior. For example, two Student
objects may be the same if they have the same id value (it is, after all supposed to be
a unique identifier). Alternatively, we may consider two objects to be the same if every
member variable holds the same value. Even with only a four member variables, the
logic can get quite complicated, especially if we have to account for null values. For
470",ComputerScienceOne.pdf
"34.5. Common Methods
this reason, many IDEs provide functionality to automatically generate such methods.
The following example was generated by an IDE.
1 public boolean equals(Object obj) {
2 if (this == obj) {
3 return true;
4 }
5 if (obj == null) {
6 return false;
7 }
8 if (!(obj instanceof Student)) {
9 return false;
10 }
11 Student other = (Student) obj;
12 if (firstName == null) {
13 if (other.firstName != null) {
14 return false;
15 }
16 } else if (!firstName.equals(other.firstName)) {
17 return false;
18 }
19 if (Double.doubleToLongBits(gpa) !=
20 Double.doubleToLongBits(other.gpa)) {
21 return false;
22 }
23 if (lastName == null) {
24 if (other.lastName != null) {
25 return false;
26 }
27 } else if (!lastName.equals(other.lastName)) {
28 return false;
29 }
30 if (nuid != other.nuid) {
31 return false;
32 }
33 return true;
34 }
Related, the hashCode() method,
public int hashCode()
returns an integer representation of the object. As with the equals() method, the",ComputerScienceOne.pdf
"34 }
Related, the hashCode() method,
public int hashCode()
returns an integer representation of the object. As with the equals() method, the
default behavior is to return a value associated with the memory address of the instance.
In general, however, the behavior should be overridden to be based on the entire state
of the object. A hash is simply a function that maps data to a “small” set of values.
In this context, we are mapping object instances to integers so that they can be used
in hash table-based data structures such as HashSet or HashMap. The hashCode()
method is used to map an instance to an integer so that it can be used as an index in
these data structures. Again, most IDEs will provide functionality to generate a good
implementation for the hashCode() method (as the example below is).
How ever you design the equals() and hashCode() method, there is a requirement:
471",ComputerScienceOne.pdf
"34. Objects
if two instances are equal (that is, equals() returns true) then they must have the
same hashCode() value. This requirement is necessary to ensure that hash table-based
data structures operate properly. It is okay if unequal objects have equal or unequal
hash values. This rule only applies when the objects are equal.
1 public int hashCode() {
2 final int prime = 31;
3 int result = 1;
4 result = prime * result + ((firstName == null) ? 0 :
5 firstName.hashCode());
6 long temp;
7 temp = Double.doubleToLongBits(gpa);
8 result = prime * result + (int) (temp ^ (temp >>> 32));
9 result = prime * result + ((lastName == null) ? 0 :
10 lastName.hashCode());
11 result = prime * result + nuid;
12 return result;
13 }
34.6. Composition
Another important concept when designing classes is composition. Composition is a
mechanism by which an object is made up of other objects. One object is said to “own”
an instance of another object. We’ve already seen this with our Student example: the",ComputerScienceOne.pdf
"an instance of another object. We’ve already seen this with our Student example: the
Student class owns two instances of the String class.
To illustrate the importance of composition, we could extend the design of our Student
class to include a date of birth. However, a date of birth is also made up of multiple
pieces of data (a year, a month, a date, and maybe even a time and/or locale). We could
design our own date/time class to model this, but its generally best to use what the
language already provides. Java 8 introduced the java.time package in which there
are many updated and improved classes for dealing with dates and times. The class
LocalDate for example, could be used to model a date of birth:
1 private LocalDate dateOfBirth;
472",ComputerScienceOne.pdf
"34.6. Composition
We can take this concept further and design our classes to own collections of other
classes. For example, we could define a Course class and then update our Student
class to own a collection of Course objects representing a student’s class schedule (this
type of collection ownership is sometimes referred to as aggregation rather than strict
composition).
1 private Set<Course> schedule;
Both of these updates beg the question: who is responsible for instantiating the instances
of dateOfBirth and the schedule? Should we force the “outside” user of our Student
class to build the LocalDate instance and pass it to a constructor? Should we allow
the outside code to simply provide us a date of birth as a string? Both of these design
choices have advantages and disadvantages that have to be considered.
What about the course schedule? We could require that a user provide the constructor
with a pre-computed Set of courses, but care must be taken. Consider the following",ComputerScienceOne.pdf
"with a pre-computed Set of courses, but care must be taken. Consider the following
(partial) example.
1 public Student(..., Set<Course> schedule) {
2 ...
3 this.schedule = schedule;
4 }
This is an example of a shallow copy: the instance’s schedule variable is referencing
the same collection as the parameter variable. If a change is made, say a course is added,
via one of these references, then the change is effectively made for both. If we are going
to design it this way, it would be better to make a deep copy of the set of courses:
1 public Student(..., Set<Course> schedule) {
2 ...
3 this.schedule = new HashSet<Course>(schedule);
4 }
This is the same pattern we described above: almost every data structure in the Java
Collections library has a (deep) copy constructor.
473",ComputerScienceOne.pdf
"34. Objects
Alternatively, we could make our design a bit more flexible by allowing the construction of
a Student instance without having to provide a course schedule. Instead, we could add
a method that allowed the outside code to add a course to the schedule. Something
like the following.
1 public void addCourse(Course c) {
2 this.schedule.add(c);
3 }
This adds some flexibility to our object, but removes the immutability property. Design
is always a balance and compromise between competing considerations.
34.7. Example
We present the full and completed Student class in Code Sample 34.1.
474",ComputerScienceOne.pdf
"34.7. Example
1 package unl.cse;
2
3 public class Student {
4
5 private String firstName;
6 private String lastName;
7 private int id;
8 private double gpa;
9
10 public Student(String firstName, String lastName, int id, double gpa) {
11 this.firstName = firstName;
12 this.lastName = lastName;
13 this.id = id;
14 this.gpa = gpa;
15 }
16
17 public Student() {
18 this(null, null, 0, 0.0);
19 }
20
21 public Student(String firstName, String lastName) {
22 this(firstName, lastName, 0, 0.0);
23 }
24
25 public String getFirstName() {
26 return firstName;
27 }
28
29 public String getLastName() {
30 return lastName;
31 }
32
33 public int getId() {
34 return id;
35 }
36
37 public double getGpa() {
38 return gpa;
39 }
40
41 /**
42 * Returns a formatted String of the Student's
43 * name as Last, First.
44 */
45 public String getFormattedName() {
46 return lastName + "", "" + firstName;
475",ComputerScienceOne.pdf
"34. Objects
47 }
48
49 /**
50 * Scales the GPA, which is assumed to be on a
51 * 4.0 scale to a percentage.
52 */
53 public double getGpaAsPercentage() {
54 return gpa / 4.0;
55 }
56
57 @Override
58 public String toString() {
59 return String.format(""%s, %s (ID = %d); %.2f"",
60 this.lastName,
61 this.firstName,
62 this.id,
63 this.gpa);
64 }
65
66 @Override
67 public int hashCode() {
68 final int prime = 31;
69 int result = 1;
70 result = prime * result + ((firstName == null) ? 0 : firstName.hashCode());
71 long temp;
72 temp = Double.doubleToLongBits(gpa);
73 result = prime * result + (int) (temp ^ (temp >>> 32));
74 result = prime * result + id;
75 result = prime * result + ((lastName == null) ? 0 : lastName.hashCode());
76 return result;
77 }
78
79
80 @Override
81 public boolean equals(Object obj) {
82 if (this == obj) {
83 return true;
84 }
85 if (obj == null) {
86 return false;
87 }
88 if (!(obj instanceof Student)) {
89 return false;
90 }
91 Student other = (Student) obj;",ComputerScienceOne.pdf
"84 }
85 if (obj == null) {
86 return false;
87 }
88 if (!(obj instanceof Student)) {
89 return false;
90 }
91 Student other = (Student) obj;
92 if (firstName == null) {
476",ComputerScienceOne.pdf
"34.7. Example
93 if (other.firstName != null) {
94 return false;
95 }
96 } else if (!firstName.equals(other.firstName)) {
97 return false;
98 }
99 if (Double.doubleToLongBits(gpa) != Double.doubleToLongBits(other.gpa)) {
100 return false;
101 }
102 if (id != other.id) {
103 return false;
104 }
105 if (lastName == null) {
106 if (other.lastName != null) {
107 return false;
108 }
109 } else if (!lastName.equals(other.lastName)) {
110 return false;
111 }
112 return true;
113 }
114
115
116
117 }
Code Sample 34.1.: The completed Java Student class.
477",ComputerScienceOne.pdf
"35. Recursion
Java supports recursion with no special syntax necessary. However, as an object-oriented
language, recursion is generally expensive and iterative or other non-recursive solutions
are generally preferred. We present a few examples to demonstrate how to write recursive
methods in Java. The first example of a recursive method we gave was the toy count
down example. In Java it could be implemented as follows.
1 public static void countDown(int n) {
2 if(n == 0) {
3 System.out.println(""Happy New Year!"");
4 } else {
5 System.out.println(n);
6 countDown(n-1);
7 }
8 }
As another example that actually does something useful, consider the following recursive
summation method that takes an array and an index variable. The recursion works as
follows: if the index variable has reached the size of the array, it stops and returns zero
(the base case). Otherwise, it makes a recursive call to recSum(), incrementing the",ComputerScienceOne.pdf
"(the base case). Otherwise, it makes a recursive call to recSum(), incrementing the
index variable by 1. When the method returns, it adds its result to the i-th element in
the array. To invoke this method we would call it with an initial value of 0 for the index
variable: recSum(arr, 0).
1 public static int recSum(int arr[], int i) {
2 if(i == arr.length) {
3 return 0;
4 } else {
5 return recSum(arr, i+1) + arr[i];
6 }
7 }
This example was not tail-recursive as the recursive call was not the final operation (the
sum was the final operation). To make this method tail recursive, we can carry the
summation through to each method call ensuring that the summation is done prior to
479",ComputerScienceOne.pdf
"35. Recursion
the recursive method call.
1 public static int recSumTail(int arr[], int i, int sum) {
2 if(i == arr.length) {
3 return sum;
4 } else {
5 return recSumTail(arr, i+1, sum + arr[i]);
6 }
7 }
As another example, consider the following Java implementation of the naive recursive
Fibonacci sequence. An additional condition has been included to check for “invalid”
negative values of n for which an exception is thrown.
1 public static int fibonacci(int n) {
2 if(n < 0) {
3 throw new IllegalArgumentException(""Undefined for n < 0"");
4 } else if(n <= 1) {
5 return 1;
6 } else {
7 return fibonacci(n-1) + fibonacci(n-2);
8 }
9 }
Java is not a language that provides implicit memoization. Instead, we need to explicitly
keep track of values using a table. In the following example, we use a Map data structure
to store previously computed values.
1 public static int fibMemoization(int n, Map<Integer, Integer> m) {
2 if(n < 0) {
3 throw new IllegalArgumentException(""Undefined for n < 0"");",ComputerScienceOne.pdf
"1 public static int fibMemoization(int n, Map<Integer, Integer> m) {
2 if(n < 0) {
3 throw new IllegalArgumentException(""Undefined for n < 0"");
4 } else if(n <= 1) {
5 return 1;
6 } else {
7 Integer result = m.get(n);
8 if(result == null) {
9 Integer a = fibMemoization(n-1, m);
10 Integer b = fibMemoization(n-2, m);
11 result = a + b;
12 m.put(n, result);
13 }
480",ComputerScienceOne.pdf
"14 return result;
15 }
16 }
Java provides an arbitrary precision data type, BigInteger that can be used to compute
arbitrarily large integer values. Since Fibonacci numbers grow exponentially, using an
int we could only represent up to to F45. Using BigInteger we can support much
larger values. An example:
1 public static BigInteger fibMem(int n, Map<Integer, BigInteger> m) {
2 if(n < 0) {
3 throw new IllegalArgumentException(""Undefined for n < 0"");
4 } else if(n <= 1) {
5 return BigInteger.ONE;
6 } else {
7 BigInteger result = m.get(n);
8 if(result == null) {
9 BigInteger a = fibMem(n-1, m);
10 BigInteger b = fibMem(n-2, m);
11 result = a.add(b);
12 m.put(n, result);
13 }
14 return result;
15 }
16 }
481",ComputerScienceOne.pdf
"36. Searching & Sorting
Java provides several methods to search and sort arrays as well as Lists of elements of
any type. These methods are able to operate on collections of any type because there are
several overloaded versions of these functions as well as versions that take a Comparator
object that specifies how the elements should be ordered.
36.1. Comparators
Let’s consider a “generic” Quick Sort algorithm as was presented in Algorithm 12.6. The
algorithm itself specifies how to sort elements, but it doesn’t specify how they are ordered.
The difference is subtle but important. Quick Sort needs to know when two elements, a,b
are in order, out of order, or equivalent in order to decide which partition each element
goes in. However, it doesn’t “know” anything about the elements a and b themselves.
They could be numbers, they could be strings, they could be user-defined objects.
A sorting algorithm still needs to be able to determine the proper ordering in order to",ComputerScienceOne.pdf
"A sorting algorithm still needs to be able to determine the proper ordering in order to
sort. In Java this is achieved by using a Comparator object, which is responsible for
comparing two elements and determining their proper order. A Comparator<T> is an
interface in Java that specifies a single method:
public int compare(T a, T b)
•A Comparator<T> is parameterized to operate on any type T. The parameteriza-
tion of T is basically a variable for a type. When you create a Comparator, you
specify an actual type just as we did with with ArrayList<Integer>. However,
the implementation itself is generic and can operate on any type.
•The method takes two instances, a and b whose type matches the parameterized
type T
•The method returns an integer indicating the relative ordering of the two elements:
– It returns something negative, < 0 if a comes before b (that is, a<b )
– It returns zero if a and b are equal (a= b)
– It returns something positive, > 0 if a comes after b (that is, a>b )",ComputerScienceOne.pdf
"– It returns zero if a and b are equal (a= b)
– It returns something positive, > 0 if a comes after b (that is, a>b )
There is no guarantee on the value’s magnitude, it does not necessarily return −1 or +1;
it just returns something negative or positive. We’ve previously seen this pattern when
comparing strings and other wrapper classes. Each of the standard types implements
483",ComputerScienceOne.pdf
"36. Searching & Sorting
something similar, the Comparable interface, that specifies a compareTo() method
with the same basic contract. Strings for example, are ordered in lexicographic ordering.
Numeric types such as Integer and Double also have compareTo() methods that
order elements in ascending order. Java refers to these built-in orderings as “natural”
orderings.
For user-defined classes, however, we need to specify how to order them. We could use the
Comparable<T> interface as the built-in wrapper classes have, but using a Comparator
is more flexible. Making a class Comparable locks you into one “natural” ordering. If
you wanted a different ordering, you would still have to use a Comparator. Even for the
wrapper classes, if we wanted a descending order, we would need to use a Comparator.
As an example, let’s write a Comparator that orders Integer types in ascending order,
matching the “natural” ordering already defined.
1 Comparator<Integer> cmpInt = new Comparator<Integer>(){",ComputerScienceOne.pdf
"matching the “natural” ordering already defined.
1 Comparator<Integer> cmpInt = new Comparator<Integer>(){
2 public int compare(Integer a, Integer b) {
3 if(a < b) {
4 return -1;
5 } else if(a == b) {
6 return 0;
7 } else {
8 return 1;
9 }
10 }
11 };
This is new syntax. When we create a Comparator in Java we are creating a new instance
of a class. However, we didn’t define a class. We didn’t use public class... nor did
we place it into a .java source file. Instead, this is an anonymous class declaration.
The class we’ve created has no name; the cmpInt is the variable’s name, but the class
itself has no name, its “anonymous.” This syntax allows us to define and instantiate a
class inline without having to create a separate class source file. This is an advantage
because we generally do not need multiple instances of a Comparator; they would all
have the same functionality anyway. An anonymous class allows us to create ad-hoc
classes with a one-time (or a single purpose) use.",ComputerScienceOne.pdf
"have the same functionality anyway. An anonymous class allows us to create ad-hoc
classes with a one-time (or a single purpose) use.
Another issue with this method is that it may not be able to handle null values. When
a and b get unboxed for the comparisons, if they are null, a NullPointerException
will be thrown. We discuss how to deal with this below in Section 36.3.2.
Another issue with this Comparator is the second case where we use the equality operator
== to compare values. With the less-than operator, the two integers get unboxed and
their values are compared. However, when we use the == operator on object instances,
it is comparing memory addresses in the JVM, not the actual values. We could solve
484",ComputerScienceOne.pdf
"36.1. Comparators
this issue by rearranging our cases so that the equality is our final case, avoiding the
use of the equality operator. Even better, however, we can exploit the built-in natural
ordering of the integers by using the compareTo() method.
1 Comparator<Integer> cmpInt = new Comparator<Integer>(){
2 public int compare(Integer a, Integer b) {
3 return a.compareTo(b);
4 }
5 };
What if we wanted to order integers in the opposite, descending order? We could simply
write another Comparator that reversed the ordering:
1 Comparator<Integer> cmpIntDesc = new Comparator<Integer>(){
2 public int compare(Integer a, Integer b) {
3 return b.compareTo(a);
4 }
5 };
We might be tempted to instead reuse the original comparator and write line 3 above
simply as
return cmpInt(b, a);
However, Java will not allow you to reference another “outside” variable like this inside a
Comparator object. An anonymous class in Java is a sort-of closure; a function that has",ComputerScienceOne.pdf
"Comparator object. An anonymous class in Java is a sort-of closure; a function that has
a scope in which variables are bound. In this case, the anonymous class has a reference to
the cmpInt variable, but it is not necessarily “bound” as the variable could be reassigned
outside the Comparator. If we want to do something like this, Java forces us to make
the cmpInt variable final so that it cannot be changed.
To illustrate some more examples, consider the Student object we defined in Code
Sample 34.1. The following Code Samples demonstrate various ways of ordering Student
instances based on one or more of their member variable values.
1 /**
2 * This Comparator orders Student objects by
3 * last name/first name
4 */
5 Comparator<Student> byName = new Comparator<Student>() {
6 @Override
7 public int compare(Student a, Student b) {
8 if(a.getLastName().equals(b.getLastName())) {
485",ComputerScienceOne.pdf
"36. Searching & Sorting
9 return a.getFirstName().compareTo(b.getFirstName());
10 } else {
11 return a.getLastName().compareTo(b.getLastName());
12 }
13 }
14 };
1 /**
2 * This Comparator orders Student objects by
3 * last name/first name in descending (Z-to-A) order
4 */
5 Comparator<Student> byNameDesc = new Comparator<Student>() {
6 @Override
7 public int compare(Student a, Student b) {
8 if(b.getLastName().equals(a.getLastName())) {
9 return b.getFirstName().compareTo(a.getFirstName());
10 } else {
11 return b.getLastName().compareTo(a.getLastName());
12 }
13 }
14 };
1 /**
2 * This Comparator orders Student objects by
3 * id in ascending order
4 */
5 Comparator<Student> byId = new Comparator<Student>() {
6 @Override
7 public int compare(Student a, Student b) {
8 return a.getId().compareTo(b.getId());
9 }
10 };
1 /**
2 * This Comparator orders Student objects by
3 * GPA in descending order
486",ComputerScienceOne.pdf
"36.2. Searching & Sorting
4 */
5 Comparator<Student> byGpa = new Comparator<Student>() {
6 @Override
7 public int compare(Student a, Student b) {
8 return b.getGpa().compareTo(a.getGpa());
9 }
10 };
36.2. Searching & Sorting
We now turn our attention to the search and sorting methods provided by the JDK.
Most of these methods are generic implementations that either sort an array or collection
using the natural ordering or take a take a parameterized Comparator<T> to determine
the ordering.
36.2.1. Searching
Linear Search
Java Set and List collections provide several linear search methods. Both have a
public boolean contains(Object o) method that returns true or false depend-
ing on whether or not an object matching o is in the collection. The List collection has
an additional public int indexOf(Object o) method that returns the index of the
first matched element and −1 if no such element is found ( Sets are unordered and have",ComputerScienceOne.pdf
"first matched element and −1 if no such element is found ( Sets are unordered and have
no indices, so this method would not apply). All of these functions are equality-based
and they do not take a Comparator object. Instead, the elements are compared using
the object’s equals() (and possibly hashCode()) method. The first element x such
that x.equals(o) returns true is the element that is determined to match. For this
reason, it is important to override both of these methods when designing objects (see
Section 36.3.3).
Binary Search
The Arrays and Collections classes provide many variations on methods that imple-
ment binary search. All of these methods assume that the array or List being searched
are sorted appropriately (you cannot use binary search on a Set as it is an unordered
collection).
The Arrays class provides several binarySearch() methods, one for each primitive
type as well as variations that allow you to search within a specified range of elements.",ComputerScienceOne.pdf
"type as well as variations that allow you to search within a specified range of elements.
The most generally useful version is as follows:
487",ComputerScienceOne.pdf
"36. Searching & Sorting
public static <T> int binarySearch(T[] a, T key, Comparator<T> c)
That is, it takes an array of elements as well as key and a Comparator all of the same
type T. It returns an integer representing the index at which it finds the first matching
element (there is no guarantee that the first element in the sorted list is returned). If
no match is found, then the method returns something negative. 1 Another version has
the same behavior but can be called without a Comparator, relying instead on the
natural ordering of elements. For this variation, the type of elements must implement
the Comparable interface.
The Collections class provides a similar method,
public static <T> int binarySearch(List<T> list, T key, Comparator<T> c)
The only difference is that the method takes a List instead of an array. Otherwise, the
behavior is the same. We present several examples in Code Sample 36.1.
36.2.2. Sorting",ComputerScienceOne.pdf
"behavior is the same. We present several examples in Code Sample 36.1.
36.2.2. Sorting
As with searching, the Arrays and Collections classes provide parameterized sorting
methods to sort arrays and Lists. In Java 6 and prior, the implementation was a hybrid
merge/insertion sort. Java 7 switched the implementation to Tim Sort. Here too, there
are several variations that rely on the natural ordering or allow you to sort a subpart of
the collection.
The Arrays provides the following method, which sorts arrays of a particular type with
a Comparator for that type.
public static <T> void sort(T[] a, Comparator<T> c)
Likewise, Collections provides the following method.
public static <T> void sort(List<T> list, Comparator<T> c)
Several examples of the usage of these methods are presented in Code Sample 36.2.
36.3. Other Considerations
36.3.1. Sorted Collections
The Java Collections framework also provides several sorted data structures. Though",ComputerScienceOne.pdf
"36.3. Other Considerations
36.3.1. Sorted Collections
The Java Collections framework also provides several sorted data structures. Though
an ordered collection such as a List can be sorted calling the Collections.sort()
method, a sorted data structure maintains an ordering. As you add elements to the data
structure, they are inserted in the appropriate spot. Such data structures also prevent
you from arbitrarily inserting elements in any order.
Java defines the SortedSet<T> interface for sorted collections with a TreeSet<T>
implementation.2 You can construct a TreeSet<T> by providing it a Comparator<T>
1It actually returns a negative value corresponding to the insertion point at which the element would
be if it had existed.
2As the name implies, the implementation is a balanced binary search tree, in particular a red-black
tree.
488",ComputerScienceOne.pdf
"36.3. Other Considerations
1 ArrayList<Student> roster = ...
2
3 Student castroKey = null;
4 int castroIndex;
5
6 //create a ""key"" that will match according to the
7 // Student.equals() method
8 castroKey = new Student(""Starlin"", ""Castro"", 131313, 3.95);
9 castroIndex = roster.indexOf(castroKey);
10 System.out.println(""at index "" + castroIndex + "": "" +
11 roster.get(castroIndex));
12
13 //create a key with only the necessary fields to match
14 / the comparator
15 castroKey = new Student(""Starlin"", ""Castro"", 0, 0.0);
16 //sort the list according to the comparator
17 Collections.sort(roster, byName);
18 castroIndex = Collections.binarySearch(roster, castroKey, byName);
19 System.out.println(""at index "" + castroIndex + "": "" +
20 roster.get(castroIndex));
21
22 //create a key with only the necessary fields to match
23 // the comparator
24 castroKey = new Student(null, null, 131313, 0.0);
25 //sort the list according to the comparator
26 Collections.sort(roster, byId);",ComputerScienceOne.pdf
"24 castroKey = new Student(null, null, 131313, 0.0);
25 //sort the list according to the comparator
26 Collections.sort(roster, byId);
27 castroIndex = Collections.binarySearch(roster, castroKey, byId);
28 System.out.println(""at index "" + castroIndex + "": "" +
29 roster.get(castroIndex));
Code Sample 36.1.: Java Search Examples
to use to maintain the ordering (alternatively, you may omit the Comparator and the
TreeSet will use the natural ordering of elements).
1 Comparator<Integer> cmpIntDesc = new Comparator<Integer>() { ... };
2 SortedSet<Integer> sortedInts = new TreeSet<Integer>(cmpIntDesc);
3 sortedInts.add(10);
4 sortedInts.add(30);
5 sortedInts.add(20);
489",ComputerScienceOne.pdf
"36. Searching & Sorting
1 List<Student> roster = ...
2 Student rosterArr[] = ...
3 Comparator byName = ...
4 Comparator byGPA = ...
5
6 //sort by name:
7 Collections.sort(roster, byName);
8 Arrays.sort(rosterArr, byName);
9
10 //sort by GPA:
11 Collections.sort(roster, byGPA);
12 Arrays.sort(rosterArr, byGPA);
Code Sample 36.2.: Using Java Collection’s Sort Method
6 //sortedInts has elements 30, 20, 10 in descending order
7
8 SortedSet<Student> sortedStudents = new TreeSet<Student>(byName);
9 //add students, they will be sorted by name
10 for(Student s : sortedStudents) {
11 System.out.println(s);
12 }
When using a SortedSet it is important that the Comparator properly orders all
elements. A SortedSet is still a Set: it does not allow duplicate elements. If the
Comparator you use returns zero for any element that you attempt to insert, it will
not be inserted. To demonstrate how this might fail, consider our byName Comparator",ComputerScienceOne.pdf
"not be inserted. To demonstrate how this might fail, consider our byName Comparator
for Student objects. Suppose two students have the same first name and last name,
“John Doe” and “John Doe,” but have different IDs (as they are different people). The
Comparator would return 0 for these two objects because they have the same first/last
name, even though they are distinct people. Only one of these objects could exist in the
SortedSet. To solve this problem, it is important to define Comparators that “break
ties” appropriately. In this case, we would want to modify the Comparator to return a
comparison of IDs if the first and last name are the same.
36.3.2. Handling null values
When sorting collections or arrays of objects, we may need to consider the possibility
of uninitialized null objects. Some collections allow null element(s) while others do
490",ComputerScienceOne.pdf
"36.3. Other Considerations
not. How we handle these is a design decision. We could ignore it, in which case such
elements would likely result in a NullPointerException and expect the user to prevent
or handle such instances. This may be the preferable choice in most instances, in fact.
Alternatively, we could handle null objects in the design of our Comparator. Code
Sample 36.3 presents a Comparator for our Student class that orders null instances
first.
1 Comparator<Student> byNameWithNulls = new Comparator<Student>() {
2 @Override
3 public int compare(Student a, Student b) {
4 if(a == null && b == null) {
5 return 0;
6 } else if(a == null && b != null) {
7 return -1;
8 } else if(a != null && b == null) {
9 return 1;
10 } else {
11 if(a.getLastName().equals(b.getLastName())) {
12 return a.getFirstName().compareTo(b.getFirstName());
13 } else {
14 return a.getLastName().compareTo(b.getLastName());
15 }
16 }
17 }
18 };
Code Sample 36.3.: Handling Null Values in Java Comparators",ComputerScienceOne.pdf
"13 } else {
14 return a.getLastName().compareTo(b.getLastName());
15 }
16 }
17 }
18 };
Code Sample 36.3.: Handling Null Values in Java Comparators
This solution only handles null Student values not null values within the Student
object. If the getters used in the Comparator return null values, then we could still
have NullPointerExceptions.
36.3.3. Importance of equals() and hashCode() Methods
Recall that in Chapter 34 we briefly discussed the importance of overloading the
equals() and hashCode() methods of our objects. We reiterate this in the con-
text of searching. Recall that the Set and List collections provide linear search
methods that do not require the use of a Comparator. This is because the collections
use the object’s equals() to determine when a particular instance has been found. In
the case of hash-based data structures such as HashSet, the hashCode() method is
491",ComputerScienceOne.pdf
"36. Searching & Sorting
also important. To illustrate the importance of these methods, consider the following
code.
1 Student a = new Student(""Joe"", ""Smith"", 1234, 3.5);
2 Student b = new Student(""Joe"", ""Smith"", 1234, 3.5);
3
4 Set<Student> s = new HashSet<Student>();
5 s.add(a);
6 s.add(b);
If we do not override the equals() and hashCode() methods, at the end of this
code snippet, the set will contain both of the Student instances even though they are
conceptually the same object (all of their member variables will have the same values).
However, if we do override the equals() and hashCode() methods as described in
Section 34.5, then the code snippet above would result in the Set only having one object
(the first one added). This is because during the attempt to add the second instance, the
equals() method would have returned true when compared to the first instance and
the duplicate element would not have been added to the Set (as Sets do not allow
duplicates).",ComputerScienceOne.pdf
"the duplicate element would not have been added to the Set (as Sets do not allow
duplicates).
Also recall that we saw some of the advantages of immutable objects. Again, we point
out another advantage. Suppose that we properly override the equals() and the
hashCode() methods in our Student object. Further, suppose that we make our
Student class mutable: allowing a user to change the ID via a setter. Consider the
following code snippet.
1 Student a = new Student(""Joe"", ""Smith"", 1234, 3.5);
2 Student b = new Student(""Joe"", ""Smith"", 5679, 3.5);
3
4 Set<Student> s = new HashSet<Student>();
5 s.add(a);
6 s.add(b);
7 //s now contains both students
8 b.setId(1234);
When we instantiate the two Student instances, they are distinct as their IDs are
different. So when we add them to the set, their equals() method returns false and
they are both added to the Set. However, when we change the ID using the setter",ComputerScienceOne.pdf
"they are both added to the Set. However, when we change the ID using the setter
on the last line, both objects are now identical, violating the no-duplicates property of
the Set. This is not a failing of the Set class, just one of the many consequences of
designing mutable objects.
492",ComputerScienceOne.pdf
"36.3. Other Considerations
36.3.4. Java 8: Lambda Expressions
Java 8 introduced a lot of functional-style syntax, including lambda expressions. Lambda
expressions are essentially anonymous functions that can be passed around to other
methods or objects. One use for lambda expressions is if we want to sort a List with
respect to one data field, we need not build a full Comparator. Instead we can use the
following syntax.
1 List<Student> roster = ...
2
3 roster.sort((a, b) -> a.getLastName().compareTo(b.getLastName()));
The new syntax is the use of the arrow operator as a lambda expres-
sion. In this case, it maps a pair, (a, b) to the value of the expression
a.getLastName().compareTo(b.getLastName()) (which takes advantage of the fact
that strings implement the Comparable interface). With respect to Comparators, we
can use the following syntax to build a more complex ordering.
1 Comparator<Student> c =
2 (a, b) -> a.getLastName().compareTo(b.getLastName());",ComputerScienceOne.pdf
"can use the following syntax to build a more complex ordering.
1 Comparator<Student> c =
2 (a, b) -> a.getLastName().compareTo(b.getLastName());
3 c = c.thenComparing( (a, b) ->
4 a.getFirstName().compareTo(b.getFirstName()));
5 c = c.thenComparing( (a, b) -> a.getGpa().compareTo(b.getGpa()));
6
7 //pass the comparator to the sort method:
8 roster.sort(c);
We can make this even more terse using method references, another new feature in Java
8.
1 //using getters as key extractors
2 myList.sort(
3 Comparator.comparing(Student::getLastName)
4 .thenComparing(Student::getFirstName)
5 .thenComparing(Student::getGpa));
There are several other convenient methods provided by the updated Comparator
interface. For example, the reversed() member method returns a new Comparator
that defines the reversed order. The static methods, nullsFirst() and nullsLast()
493",ComputerScienceOne.pdf
"36. Searching & Sorting
can be used to modify a Comparator to order null values.
494",ComputerScienceOne.pdf
"Part III.
The PHP Programming Language
495",ComputerScienceOne.pdf
"37. Basics
In the mid-1990s the World Wide Web was in its infancy but becoming more and more
popular. For the most part, web pages contained static content: articles and text that
was “just-there.” Web pages were far from the fully interactive and dynamic applications
that they’ve become. Rasmus Lerdorf had a home page containing his resume and he
wanted to track how many visitors were coming to his page. With purely static pages,
this was not possible. So, in 1994 he developed PHP/FI which stood for Personal Home
Page tools and Forms Interpreter. This was a series of binary tools written in C that
operated through a web server’s Common Gateway Interface. When a user visited a
webpage, instead of just retrieving static content, a script was invoked: it could do a
number of operations and send back HTML formatted as a response. This made web
pages much more dynamic. By serving it through a script, a counter could be maintained",ComputerScienceOne.pdf
"pages much more dynamic. By serving it through a script, a counter could be maintained
that tracked how many people had visited the site. Lerdorf eventually released his source
code in 1995 and it became widely used.
Today, PHP is used in a substantial number of pages on the web and is used in many
Content Management System (CMS) applications such as Drupal and WordPress. Because
of its history, much of the syntax and aspects of PHP (now referred to as “PHP: Hypertext
Preprocessor”) are influenced or inspired by C. In fact, many of the standard functions
available in the language are directly from the C language.
PHP is a scripting language and is not compiled. Instead, PHP code is interpreted by
a PHP interpreter. Though there are several interpreters available, the de facto PHP
interpreter is the Zend Engine, a free and open source project that is widely available on
many platforms, operating systems, and web servers. Because it was originally intended",ComputerScienceOne.pdf
"many platforms, operating systems, and web servers. Because it was originally intended
to serve web pages, PHP code can be interleaved with static HTML tags. This allows
PHP code to be embedded in web pages and dynamically interpreted/rendered when
a user requests a webpage through a web browser. Though rendering web pages is its
primary purpose, PHP can be used as a general scripting language from the command
line.
PHP is a dirty, ugly language held together by duct tape, cracked asbestos-filled spackle
and sadness (http://phpsadness.com/). It is the cockroach of languages: it is invasive
and it is a survivor. It is a hydra; rewrite one PHP app and two more pop up. It is
much maligned and a source of ridicule. It is the source of hundreds of bad practices,
thousands of bad programmers, millions of bugs and an infinite abyss of security holes
and vulnerabilities. It is a language that was born in the wild and raised in the darkness",ComputerScienceOne.pdf
"and vulnerabilities. It is a language that was born in the wild and raised in the darkness
by two schizophrenic monkeys. But its got character; and in this world, that’s enough.
497",ComputerScienceOne.pdf
"37. Basics
1 <?php
2
3 printf(""Hello World\n"");
4
5 ?>
Code Sample 37.1.: Hello World Program in PHP
1 <html>
2 <head>
3 <title>Hello World PHP Page</title>
4 </head>
5 <body>
6 <h1>A Simple PHP Script</h1>
7
8 <?php printf(""<p>Hello World</p>""); ?>
9
10 </body>
11 </html>
Code Sample 37.2.: Hello World Program in PHP with HTML
37.1. Getting Started: Hello World
The hallmark of an introduction to a new programming language is the Hello World!
program. It consists of a simple program whose only purpose is to print the message
“Hello World!” to the user. The simplicity of the program allows the focus to be on the
basic syntax of the language. It is also typically used to ensure that your development
environment, interpretrer, runtime environment, etc. are functioning properly with a
minimal example. A basic Hello World! program in PHP can be found in Code Sample
37.1. A version in which the PHP code is interleaved in HTML is presented in Code
Sample 37.2.",ComputerScienceOne.pdf
"37.1. A version in which the PHP code is interleaved in HTML is presented in Code
Sample 37.2.
We will not focus on any particular development environment, code editor, or any
particular operating system, interpreter, or ancillary standards in our presentation.
However, as a first step, you should be able to write and run the above program on
the environment you intend to use for the rest of this book. This may require that
you download and install a basic PHP interpreter/development environment (such as
a standard LAMP, WAMP or MAMP technology stack) or a full IDE (such as Eclipse,
PhpStorm, etc.).
498",ComputerScienceOne.pdf
"37.2. Basic Elements
37.2. Basic Elements
Using the Hello World! program as a starting point, we will examine the basic elements
of the PHP language.
37.2.1. Basic Syntax Rules
PHP has adopted many aspects of the C programming language (the interpreter itself is
written in C). However, there are some major aspects in which it differs from C. The
major aspects and syntactic elements of PHP include the following.
•PHP is an interpreted language: it is not compiled. Instead it is interpreted through
an interpreter that parses commands in a script file line-by-line. Any syntax errors,
therefore, become runtime errors.
•PHP is dynamically typed : you do not declare variables or specify a variable’s
type. Instead, when you assign a variable a value, the variable’s type dynamically
changes to accommodate the assigned value. Variable names always begin with a
dollar sign, $.
•Strings and characters are essentially the same thing (characters are strings of",ComputerScienceOne.pdf
"dollar sign, $.
•Strings and characters are essentially the same thing (characters are strings of
length 1). Strings and characters can be delimited by either single quotes or double
quotes. ""this is a string"", 'this is also a string'; 'A' and ""a"" are
both single character strings.
•Executable statements are terminated by a semicolon, ;
•Code blocks are defined by using opening and closing curly brackets, { ... }.
Moreover, code blocks can be nested: code blocks can be defined within other code
blocks.
•A complete list of reserved and keywords can be found in the PHP Manual: http://
php.net/manual/en/reserved.php and http://php.net/manual/en/reserved.
keywords.php
PHP also supports aspects of OOP including classes and inheritance. There is also no
memory management in PHP: it has automated garbage collection.
Though not a syntactic requirement, the proper use of whitespace is important for good,
readable code. Code inside code blocks is indented at the same indentation. Nested",ComputerScienceOne.pdf
"readable code. Code inside code blocks is indented at the same indentation. Nested
code blocks are indented further. Think of a typical table of contents or the outline of a
formal paper or essay. Sections and subsections or points and subpoints all follow proper
indentation with elements at the same level at the same indentation. This convention is
used to organize code and make it more readable.
499",ComputerScienceOne.pdf
"37. Basics
37.2.2. PHP Tags
PHP code can be interleaved with static HTML or text. Because of this, we need a way
to indicate what should be interpreted as PHP and what should be treated as static text.
We can do this using PHP tags: the opening tag is <?php and the closing tag is ?>.
Anything placed between these tags will be interpreted as PHP code which must adhere
to the syntax rules of the language.
A file can contain multiple opening/closing PHP tags to allow you to interleave multiple
sections of HTML or text. When an interpreter runs a PHP script, it will start processing
the script file. Whenever it sees the opening PHP tag, it begins to execute the commands
and stops executing commands when it sees the closing tag. The text outside the PHP
tags is treated as raw data. The interpreter includes it as part of its output without
modifying or processing it.
37.2.3. Libraries
PHP has many built-in functions that you can use. These standard libraries are loaded",ComputerScienceOne.pdf
"modifying or processing it.
37.2.3. Libraries
PHP has many built-in functions that you can use. These standard libraries are loaded
and available to use without any special command to import or include them. Full docu-
mentation on each of these functions is maintained in the PHP manual, available online
at http://php.net/. The manual’s web pages also contain many curated comments
from PHP developers which contain further explanations, tips & tricks, suggestions, and
sample code to provide further assistance.
There are many useful functions that we’ll mention as we progress through each topic.
One especially useful collection of functions is the math library. This library is directly
adapted from C’s standard math library so all the function names are the same. It
provides many common mathematical functions such as the square root and natural
logarithm. Table 37.1 highlights several of these functions; full documentation can be",ComputerScienceOne.pdf
"logarithm. Table 37.1 highlights several of these functions; full documentation can be
found in the PHP manual ( http://php.net/manual/en/ref.math.php). To use these,
you simply “call” them by providing input and assigning the output to a variable.
1 $x = 1.5;
2 $y = sqrt($x); //y now has the value √x=
√
1.5
3 $z = sin($x); //z now has the value sin (x) = sin (1.5)
In both of the function calls above, the value of the variable $x is “passed” to the math
function which computes and “returns” the result which then gets assigned to another
variable.
37.2.4. Comments
Comments can be written in PHP code either as a single line using two forward slashes,
//comment or as a multiline comment using a combination of forward slash and asterisk:
/* comment */ . With a single line comment, everything on the line after the forward
500",ComputerScienceOne.pdf
"37.2. Basic Elements
Function Description
abs($x) Absolute value, |x|
ceil($x) Ceiling function, ⌈46.3⌉= 47.0
floor($x) Floor function, ⌊46.3⌋= 46.0
cos($x) Cosine functiona
sin($x) Sine functiona
tan($x) Tangent functiona
exp($x) Exponential function, ex, e= 2.71828 ...
log($x) Natural logarithm, ln (x)b
log10($x) Logarithm base 10, log 10 (x)b
pow($x,$y) The power function, computes xyc
sqrt($x) Square root functionb
Table 37.1.: Several functions defined in the PHP math library. aAll trigonometric
functions assume input is in radians, not degrees. bInput is assumed to
be positive, x> 0. cAlternatively, PHP supports exponentiation by using
x ** y.
501",ComputerScienceOne.pdf
"37. Basics
slashes is ignored. With a multiline comment, everything in between the forward
slash/asterisk is ignored. Comments are ultimately ignored by the interpreter. Consider
the following example.
1 //this is a single line comment
2 $x = 10; //this is also a single line comment, but after some code
3
4 /*
5 This is a comment that can
6 span multiple lines to format the comment
7 message more clearly
8 */
9 $y = 3.14;
Most code editors and IDEs will present comments in a special color or font to distinguish
them from the rest of the code (just as our example above does). Failure to close a
multiline comment will likely result in a fatal interpreter error but with color-coded
comments its easy to see the mistake visually.
37.2.5. Entry Point & Command Line Arguments
Every PHP script begins executing at the top of the script file and proceed in a linear,
sequential manner top-to-bottom. In addition, you can provide a PHP script with inputs",ComputerScienceOne.pdf
"sequential manner top-to-bottom. In addition, you can provide a PHP script with inputs
when you run it from the command line. These command line arguments are stored in
two variables, $argc and $argv. The first variable, $argc is an integer that indicates
the number of arguments provided including the name of the script file being executed.
The second, $argv is an array (see Section 42) of strings consisting of the command
line arguments. To access them, you can index them starting at zero, the first being
$argv[0], the second $argv[1], the last one would be $argv[$argc-1]. The first
one is always the name of the script file being run. The remaining are the command line
arguments provided by the user when the script is executed. We’ll see several examples
later.
37.3. Variables
PHP is a dynamically typed language. As a consequence, you do not declare variables
before you start using them. If you need to store a value into a variable, you simply",ComputerScienceOne.pdf
"before you start using them. If you need to store a value into a variable, you simply
name the variable and use an assignment operator to assign the value. Since you do not
declare variables, you also do not specify a variable’s type. If you assign a string value
to a variable, its type becomes a string. If you assign an integer to a variable, its type
becomes an integer. If you reassign the value of a variable to a value with a different
type, the variable’s type also changes.
502",ComputerScienceOne.pdf
"37.3. Variables
Internally PHP supports several different types: Booleans, integers, floating point
numbers, strings, arrays, and objects. The way that integers are represented may be
platform dependent, but are usually 32-bit signed two’s complement integers, able to
represent integers between −2,147,483,648 and 2,147,483,647. Floating point numbers
are also platform dependent, but are usually 64-bit double precision numbers defined
by the IEEE 754 standard, providing about 16 digits of precision. Strings and single
characters are the same thing in PHP. Strings are represented as sequences of characters
from the extended ASCII text table (see Table 2.4) which includes all characters in the
range 0–255. PHP does not have native Unicode support for international characters.
37.3.1. Using Variables
To use a variable in PHP, you simply need to assign a value to a named variable identifier
and the variable is in scope. Variable names always begin with a single dollar sign, $.",ComputerScienceOne.pdf
"and the variable is in scope. Variable names always begin with a single dollar sign, $.
The assignment operator is a single equal sign, = and is a right-to-left assignment. The
variable that we wish to assign the value to appears on the left-hand-side while the value
(literal, variable or expression) is on the right-hand-size. For example:
1 $numUnits = 42;
2 $costPerUnit = 32.79;
3 $firstInitial = ""C"";
Each assignment also implicitly changes the variable’s type. Each of the variables above
becomes an integer, floating point number, and string respectively. Assignment statements
are terminated by a semicolon like most executable statements in PHP. The identifier
rules are fairly standard: a variable’s name can consist of lowercase and uppercase
alphabetic characters, numbers, and underscores. You can also use the extended ASCII
character set in variable names but it is not recommended (umlauts and other diacritics",ComputerScienceOne.pdf
"character set in variable names but it is not recommended (umlauts and other diacritics
can easily be confused). Variable names are case sensitive. As previously mentioned,
variable names must always begin with a dollar sign, $. Stylistically, we adopt the
modern lower camel casing naming convention for variables in our code.
If you do not assign a value to a variable, that variable remains undefined or “unset.”
Undefined variables are treated as null in PHP. The concept of “null” refers to
uninitialized, undefined, empty, missing, or meaningless values. In PHP the keyword
null is used which is case insensitive ( null, Null and NULL are all the same), but
for consistency, we’ll use null. When null values are used in arithmetic expressions,
null is treated as zero. So, (10 + null) is equal to 10. When null is used in the
context of strings, it is treated as an empty string and ignored. When used in a Boolean
expression or conditional, null is treated as false.",ComputerScienceOne.pdf
"context of strings, it is treated as an empty string and ignored. When used in a Boolean
expression or conditional, null is treated as false.
PHP also allows you to define constants: values that cannot be changed once set. To
define a constant, you invoke a function named define and providing a name and value.
503",ComputerScienceOne.pdf
"37. Basics
1 define(""PI"", 3.14159);
2 define(""INSTITUTION"", ""University of Nebraska-Lincoln"");
3 define(""COST_PER_UNIT"", 2.50);
Constant names are case sensitive. By convention, we use uppercase underscore casing.
An attempt to redefine a constant value will raise a script warning, but will ultimately
have no effect. When referring to constants later in the script, you use the constant’s
name. You do not treat it as a string, nor do you use a dollar sign. For example:
$area = $r * $r * PI;
37.4. Operators
PHP supports the standard arithmetic operators for addition, subtraction, multiplication,
and division using +, -, *, and / respectively. Each of these operators is a binary op-
erator that acts on two operands which can either be literals, variables or expressions and
follow the usual rules of arithmetic when it comes to order of precedence (multiplication
and division before addition and subtraction).
1 $a = 10, $b = 20, $c = 30;
2 $d = $a + 5;
3 $d = $a + $b;
4 $d = $a - $b;",ComputerScienceOne.pdf
"and division before addition and subtraction).
1 $a = 10, $b = 20, $c = 30;
2 $d = $a + 5;
3 $d = $a + $b;
4 $d = $a - $b;
5 $d = $a + $b * $c;
6 $d = $a * $b; //d becomes a floating point number with a value .5
7
8 $x = 1.5, $y = 3.4, $z = 10.5;
9 $w = $x + 5.0;
10 $w = $x + $y;
11 $w = $x - $y;
12 $w = $x + $y * $z;
13 $w = $x * $y;
14 $w = $x / $y;
15
16 //mixing integers and floating point numbers is no problem
17 $w = $a + $x;
PHP also supports the integer remainder operator using the % symbol. This operator
gives the remainder of the result of dividing two integers. Examples:
504",ComputerScienceOne.pdf
"37.4. Operators
1 $x = 10 % 5; //x is 0
2 $x = 10 % 3; //x is 1
3 $x = 29 % 5; //x is 4
37.4.1. Type Juggling
The expectations of an arithmetic expression involving two variables that are either
integers or floating point numbers are straightforward. We expect the sum/product/etc.
as a result. However, since PHP is dynamically typed, a variable involved in an arithmetic
expression could be anything, including a Boolean, object, array, or string. When a
Boolean is involved in an arithmetic expression, true is treated as 1 and false is
treated as zero. If an array is involved in an expression, it is usually a fatal error. 1
Non-null objects are treated as 1 and null is treated as 0.
However, when string values are involved in an arithmetic expression, PHP does something
called type juggling. When juggled, an attempt is made to convert a string variable into
a numeric value by parsing it. The parsing goes over each numeric character, converting",ComputerScienceOne.pdf
"a numeric value by parsing it. The parsing goes over each numeric character, converting
the value to a numeric type (either an integer or floating point number). The first type a
non-numeric character is encountered, the parsing stops and the value parsed so far is
the value used in the expression.
Consider the examples in Code Sample 37.3. In the first segment, $a is type juggled to
the value 10, then added to 5, resulting in 15. In the second example, $a represents a
floating point number, and is converted to 3 .14, the result of adding to 5 is thus 8 .14. In
the third example, the string does not contain any numerical values. In this case, the
parsing stops at the first character and what has been parsed so far is zero! Finally, in
the last example, the first two characters in $a are numeric, so the parsing ends at the
third character, and what has been parsed so far is 10.
Relying on type juggling to convert values can be ugly and error prone. You can write",ComputerScienceOne.pdf
"third character, and what has been parsed so far is 10.
Relying on type juggling to convert values can be ugly and error prone. You can write
much more intentional code by using the several conversion functions provided by PHP.
For example:
1 $a = intval(""10"");
2 $b = floatval(""3.14"");
3 $c = intval(""ten""); //c has the value zero
In all three of the examples above, the strings are converted just as they are when type
juggled. However, the variables are guaranteed to have the type indicated (integer or
floating point number).
1PHP does allow you to “add” two arrays together which results in their union.
505",ComputerScienceOne.pdf
"37. Basics
1 $a = ""10"";
2 $b = 5 + $a; //b = 15
3
4 $a = ""3.14"";
5 $b = 5 + $a; //b = 8.14
6
7 $a = ""ten"";
8 $b = 5 + $a; //b = 5
9
10 //partial conversions also occur:
11 $a = ""10ten"";
12 $b = 5 + $a; //b = 15
Code Sample 37.3.: Type Juggling in PHP
There are several utility functions that can be used to help determine the type of variable.
The function is_numeric($x) returns true if $x is a numeric (integer or floating
point number) or represents a pure numeric string. The functions is_int($x) and
is_float($x) each return true or false depending on whether or not $x is of that
type.
1 $a = 10;
2 $b = ""10"";
3 $c = 3.14;
4 $d = ""3.14"";
5 $e = ""hello"";
6 $f = ""10foo"";
7
8 is_numeric($a); //true
9 is_numeric($b); //true
10 is_numeric($c); //true
11 is_numeric($d); //true
12 is_numeric($e); //false
13 is_numeric($f); //false
14
15 is_int($a); //true
16 is_int($b); //false
17 is_int($c); //false
18 is_int($d); //false
19 is_int($e); //false
506",ComputerScienceOne.pdf
"37.4. Operators
Value of $var isset($var) empty($var) is_null($var)
42 bool(true) bool(false) bool(false)
"""" (an empty string) bool(true) bool(true) bool(false)
"" "" (space) bool(true) bool(false) bool(false)
false bool(true) bool(true) bool(false)
true bool(true) bool(false) bool(false)
array() (an empty array) bool(true) bool(true) bool(false)
null bool(false) bool(true) bool(true)
""0"" (0 as a string) bool(true) bool(true) bool(false)
0 (0 as an integer) bool(true) bool(true) bool(false)
0.0 (0 as a float) bool(true) bool(true) bool(false)
var $var; (declared with no value) bool(false) bool(true) bool(true)
NULL byte ( ""\0"") bool(true) bool(false) bool(false)
Table 37.2.: Results for various variable values
20 is_int($f); //false
21
22 is_float($a); //false
23 is_float($b); //false
24 is_float($c); //true
25 is_float($d); //false
26 is_float($e); //false
27 is_float($f); //false
A more general way to determine the type of a variable is to use the functiongettype($x)",ComputerScienceOne.pdf
"26 is_float($e); //false
27 is_float($f); //false
A more general way to determine the type of a variable is to use the functiongettype($x)
which returns a string representation of the type of the variable $x. The string returned
by this function is one of the following depending on the type of $x: ""boolean"",
""integer"", ""double"", ""string"", ""array"", ""object"", ""resource"", ""NULL"", or
""unknown type"".
Other checker functions allow you to determine if a variable has been set, if its null,
“empty” etc. For example, is_null($x) returns true if $x is not set or is set, but
has been set to null. The function isset($x) returns true only if $x is set and it
is not null. The function empty($x) returns true if $x represents an empty entity:
an empty string, false, an empty array, null, ""0"", 0, or an unset variable. Several
examples are presented in Table 37.2.
507",ComputerScienceOne.pdf
"37. Basics
37.4.2. String Concatenation
Strings in PHP can be concatenated (combined) in several different ways. One way
you can combine strings is by using the concatenation operator. In PHP the string
concatenation operator is a single period.
1 $s = ""Hello"";
2 $t = ""World!"";
3 $msg = $s . "" "" . $t; //msg contains ""Hello World!""
Another way you can combine strings is by placing variables directly inside a string. The
variables inside the string are replaced with the variable’s values. Example:
1 $x = 13;
2 $name = ""Starlin"";
3 $msg = ""Hello, $name, your number is $x"";
4 //msg contains the string ""Hello, Starlin, your number is 13""
37.5. Basic I/O
Recall the main purpose of PHP is as a scripting language to serve dynamic webpages.
However, it does support input and output from the standard input/output. There are
several ways to print output to the standard output. The keywords print and echo
(which are aliases of each other) allow you to print any variable or string. PHP also has",ComputerScienceOne.pdf
"(which are aliases of each other) allow you to print any variable or string. PHP also has
a printf() function to allow formatted output. Some examples:
1 $a = 10;
2 $b = 3.14;
3
4 print $a;
5 print ""The value of a is $a\n"";
6
7 echo $a;
8 echo ""The value of a is $a\n"";
9
10 printf(""The value of a is %d, and b is %f\n"", $a, $b);
There are also several ways to perform standard input, but the easiest is to use fgets
508",ComputerScienceOne.pdf
"37.6. Examples
(short for file get string) using the keyword STDIN (Standard Input). This function
will return, as a string, everything the user enters up to and including the enter key
(interpreted as the endline character, \n). To remove the endline character, you can use
another function, trim which removes leading and trailing whitespace from a string. A
full example:
1 //prompt the user to enter input
2 printf(""Please enter a number: "");
3 $a = fgets(STDIN);
4 $a = trim($a);
Alternatively, lines 3–4 could be combined into one: $a = trim(fgets(STDIN)); The
call to fgets waits (referred to as “blocking”) for the user to enter input. The user is
free to start typing. When the user is done, they hit the enter key at which point the
program resumes and reads the input from the standard input buffer, and returns it as a
string value which we assign to the variable $a.
The standard input is unstructured. The user is free to type whatever they want. If",ComputerScienceOne.pdf
"string value which we assign to the variable $a.
The standard input is unstructured. The user is free to type whatever they want. If
we prompt the user for a number but they just start mashing the keyboard giving
non-numerical input, we may get incorrect results. We can use the conversion functions
mentioned above to attempt to properly convert the values. However, this only guarantees
that the resulting variable is of the type we want (integer or floating point value for
example). The standard input is not a good mechanism for reading input, but it provides
a good starting point to develop a few simple programs.
37.6. Examples
37.6.1. Converting Units
Let’s write a program that prompts the user to enter a temperature in degrees Fahrenheit
and convert it to degrees Celsius using the formula
C = (F −32) ·5
9
We begin with the basic script shell with the opening and closing PHP tags and some
comments documenting the purpose of our script.
1 <?php
2
3 /**",ComputerScienceOne.pdf
"9
We begin with the basic script shell with the opening and closing PHP tags and some
comments documenting the purpose of our script.
1 <?php
2
3 /**
4 * This program converts Fahrenheit temperatures to
5 * Celsius
509",ComputerScienceOne.pdf
"37. Basics
6 */
7
8 //TODO: implement this
9
10 ?>
It is common for programmers to use a comment along with a TODO note to themselves
as a reminder of things that they still need to do with the program. Let’s first outline
the basic steps that our program will go through:
1. We’ll first prompt the user for input, asking them for a temperature in Fahrenheit
2. Next we’ll read the user’s input and use a conversion function to ensure the input
is a floating point number
3. Once we have the input, we can calculate the degrees Celsius by using the formula
above
4. Lastly, we will want to print the result to the user to inform them of the value
Sometimes it is helpful to write an outline of such a program directly in the code using
comments to provide a step-by-step process. For example:
1 <?php
2
3 /**
4 * This program converts Fahrenheit temperatures to
5 * Celsius
6 */
7
8 //TODO: implement this
9
10 //1. Prompt the user for input in Fahrenheit",ComputerScienceOne.pdf
"4 * This program converts Fahrenheit temperatures to
5 * Celsius
6 */
7
8 //TODO: implement this
9
10 //1. Prompt the user for input in Fahrenheit
11 //2. Read the Fahrenheit value from the standard input
12 //3. Compute the degrees Celsius
13 //4. Print the result to the user
14
15 ?>
As we read each step it becomes apparent that we’ll need a couple of variables: one to
hold the Fahrenheit (input) value and one for the Celsius (output) value. We’ll want to
ensure that these are floating point numbers which we can do by making some explicit
conversion. We use a printf() statement in the first step to prompt the user for input.
510",ComputerScienceOne.pdf
"37.6. Examples
printf(""Please enter degrees in Fahrenheit: "");
In the second step, we’ll use the standard input to read the $fahrenheit variable value
from the user. Recall that we can use fgets to read from the standard input, but may
have to trim the trailing whitespace.
$fahrenheit = trim(fgets(STDIN));
If we want to ensure that the variable $fahrenheit is a floating point value, we can
use floatval():
$fahrenheit = floatval($fahrenheit);
We can now compute $celsius using the formula provided:
$celsius = ($fahrenheit - 32) * (5 / 9);
Finally, we use printf() again to output the result to the user:
printf(""%f Fahrenheit is %f Celsius\n"", $fahrenheit, $celsius);
The full program can be found in Code Sample 37.4.
1 <?php
2
3 /**
4 * This program converts Fahrenheit temperatures to
5 * Celsius
6 */
7
8 //1. Prompt the user for input in Fahrenheit
9 printf(""Please enter degrees in Fahrenheit: "");
10
11 //2. Read the Fahrenheit value from the standard input",ComputerScienceOne.pdf
"9 printf(""Please enter degrees in Fahrenheit: "");
10
11 //2. Read the Fahrenheit value from the standard input
12 $fahrenheit = trim(fgets(STDIN));
13 $fahrenheit = floatval($fahrenheit);
14
15 //3. Compute the degrees Celsius
16 $celsius = ($fahrenheit - 32) * (5/9);
17
18 //4. Print the result to the user
19 printf(""%f Fahrenheit is %f Celsius\n"", $fahrenheit, $celsius);
20
21 ?>
Code Sample 37.4.: Fahrenheit-to-Celsius Conversion Program in PHP
511",ComputerScienceOne.pdf
"37. Basics
37.6.2. Computing Quadratic Roots
Some programs require the user to enter multiple inputs. The prompt-input process can
be repeated. In this example, consider asking the user for the coefficients, a,b,c to a
quadratic polynomial,
ax2 + bx+ c
and computing its roots using the quadratic formula,
x= −b±
√
b2 −4ac
2a
As before, we can create a basic program with PHP tags and start filling in the details.
In particular, we’ll need to prompt for the input a, then read it in; then prompt for b,
read it in and repeat for c. Thus, we have
1 printf(""Please enter a: "");
2 $a = floatval(trim(fgets(STDIN)));
3 printf(""Please enter b: "");
4 $b = floatval(trim(fgets(STDIN)));
5 printf(""Please enter c: "");
6 $c = floatval(trim(fgets(STDIN)));
We need to take care that we correctly adapt the formula so it accurately reflects the
order of operations. We also need to use the math library’s square root function.
1 $root1 = (-$b + sqrt($b*$b - 4*$a*$c) ) / (2*$a);",ComputerScienceOne.pdf
"order of operations. We also need to use the math library’s square root function.
1 $root1 = (-$b + sqrt($b*$b - 4*$a*$c) ) / (2*$a);
2 $root2 = (-$b - sqrt($b*$b - 4*$a*$c) ) / (2*$a);
Finally, we print the output using printf(). The full program can be found in Code
Sample 37.5.
This program was interactive. As an alternative, we could have read all three of the
inputs as command line arguments, taking care to convert them to floating point numbers.
Lines 8–13 in the program could have been changed to
1 $a = floatval($argv[1]);
2 $b = floatval($argv[2]);
3 $c = floatval($argv[3]);
Finally, think about the possible input a user could provide that may cause problems for
this program. For example:
512",ComputerScienceOne.pdf
"37.6. Examples
1 <?php
2
3 /**
4 * This program computes the roots to a quadratic equation
5 * using the quadratic formula.
6 */
7
8 printf(""Please enter a: "");
9 $a = floatval(trim(fgets(STDIN)));
10 printf(""Please enter b: "");
11 $b = floatval(trim(fgets(STDIN)));
12 printf(""Please enter c: "");
13 $c = floatval(trim(fgets(STDIN)));
14
15 $root1 = (-$b + sqrt($b*$b - 4*$a*$c) ) / (2*$a);
16 $root2 = (-$b - sqrt($b*$b - 4*$a*$c) ) / (2*$a);
17
18 printf(""The roots of %fx^2 + %fx + %f are: \n"", $a, $b, $c);
19 printf("" root1 = %f\n"", $root1);
20 printf("" root2 = %f\n"", $root2);
21
22 ?>
Code Sample 37.5.: Quadratic Roots Program in PHP
•What if the user entered zero for a?
•What if the user entered some combination such that b2 <4ac?
•What if the user entered non-numeric values?
•For the command line argument version, what if the user provided less than three
arguments? Or more?
How might we prevent the consequences of such bad input? How might we handle the",ComputerScienceOne.pdf
"arguments? Or more?
How might we prevent the consequences of such bad input? How might we handle the
event that a user enters bad input and how do we communicate these errors to the user?
To begin to resolve these issues, we’ll need conditionals.
513",ComputerScienceOne.pdf
"38. Conditionals
PHP supports the basic if, if-else, and if-else-if conditional structures as well as switch
statements. Logical statements are built using the standard logical operators for numeric
comparisons as well as logical operators such as negations, And, and Or.
38.1. Logical Operators
PHP has a built-in Boolean type and supports the keywords true and false. However,
any variable can be treated as a Boolean if used in a logical expression. Depending on
the variable, it could evaluate to true or false! For example, an empty string, """", null,
or a numeric value of zero, 0 are all considered false. A non-empty string, a non-zero
numeric value, or a non-empty array all evaluate to true. It is best to avoid these issues
by writing clean code that uses clear, explicit statements.
Because PHP is dynamically typed, comparison operators work differently depending on
how they are used. First, let’s consider the four basic inequality operators, <, <=, >,",ComputerScienceOne.pdf
"how they are used. First, let’s consider the four basic inequality operators, <, <=, >,
and >=. When used to compare numeric types to numeric types, these operators work
as expected and the value of the numbers are compared.
1 $a = 10;
2 $b = 20;
3 $c = 20;
4
5 $r = ($a < $b); //true
6 $r = ($a <= $b); //false
7 $r = ($b <= $c); //true
8 $r = ($a > $b); //false
9 $r = ($a >= $b); //false
10 $r = ($b >= $c); //true
When these operators are used to compare strings to strings, the strings are compared
lexicographically according to the standard ASCII text table. Some examples follow,
but it is better to use a function (in particular strcmp see Chapter 43) to do string
comparisons.
515",ComputerScienceOne.pdf
"38. Conditionals
1 $s = ""aardvark"";
2 $t = ""zebra"";
3
4 $r = ($s < $t); //true
5 $r = ($s <= $t); //true
6 $r = ($s >= $t); //false
7 $r = ($s > $t); //false
However, when these operators are used to compare strings to numeric types, the strings
are converted to numbers using the same type juggling that happens when strings are
mixed with arithmetic operators. In the following example, $b gets converted to a
numeric type when compared to $a which give the results indicated in the comments.
1 $a = 10;
2 $b = ""10"";
3
4 $r = ($a <= $b); //true
5 $r = ($a < $b); //false
6 $r = ($a >= $b); //true
7 $r = ($a > $b); //false
With the equality operators, == and !=, something similar happens. When the types of
the two operands match, the expected comparison is made: when numbers are compared
to numbers their values are compared; when strings are compared to strings, their content
is compared (case sensitively). However, when the types are different they are type",ComputerScienceOne.pdf
"is compared (case sensitively). However, when the types are different they are type
juggled and strings are converted to numbers for the purpose of comparison. Thus, a
comparison like (10 == ""10"") ends up being true! The operators are == and != are
referred to as loose comparison operators because of this.
What if we want to ensure that we’re comparing apples to apples? To rectify this, PHP
offers another set of comparison operators, strict comparison operators, === and !==
(each has an extra equals sign, =). These operators will make a comparison without
type juggling either operand first. Now a similar comparison, (10 === ""10"") ends up
evaluating to false. The operator === will only evaluate to true if the both the operands’
type and value are the same.
1 $a = 10;
2 $b = ""10"";
3
4 $r = ($a == $b); //true
5 $r = ($a != $b); //false
516",ComputerScienceOne.pdf
"38.2. If, If-Else, If-Else-If Statements
Operator(s) Associativity Notes
Highest ++, -- left-to-right increment operators
-, ! right-to-left unary negation operator, logical
not
*, /, % left-to-right
+, - left-to-right addition, subtraction
<, <=, >, >= left-to-right comparison
==, !=, ===, !== left-to-right equality, inequality
&& left-to-right logical And
|| left-to-right logical Or
Lowest =, +=, -=, *=, /= right-to-left assignment and compound assign-
ment operators
Table 38.1.: Operator Order of Precedence in PHP. Operators on the same level have
equivalent order and are performed in the associative order specified.
6 $r = ($a === $b); //false
7 $r = ($a !== $b); //true
The three basic logical operators, not !, and &&, and or || are also supported in PHP.
38.1.1. Order of Precedence
At this point it is worth summarizing the order of precedence of all the operators that
we’ve seen so far including assignment, arithmetic, comparison, and logical. Since all of",ComputerScienceOne.pdf
"we’ve seen so far including assignment, arithmetic, comparison, and logical. Since all of
these operators could be used in one statement, for example,
($b*$b < 4*$a*$c || $a === 0 || $argc != 4)
it is important to understand the order in which each one gets evaluated. Table 38.1
summarizes the order of precedence for the operators seen so far. This is not an exhaustive
list of PHP operators.
38.2. If, If-Else, If-Else-If Statements
Conditional statements in PHP utilize the keywords if, else, and else if. Condi-
tions are placed inside parentheses immediately after the if and else if keywords.
Examples of all three can be found in Code Sample 38.1.
Observe that the statement, if($x < 10) does not have a semicolon at the end. This
is because it is a conditional statement that determines the flow of control and not an
executable statement. Therefore, no semicolon is used. Suppose we made a mistake and
did include a semicolon:
517",ComputerScienceOne.pdf
"38. Conditionals
1 //example of an if statement:
2 if($x < 10) {
3 printf(""x is less than 10\n"");
4 }
5
6 //example of an if-else statement:
7 if($x < 10) {
8 printf(""x is less than 10\n"");
9 } else {
10 printf(""x is 10 or more \n"");
11 }
12
13 //example of an if-else-if statement:
14 if($x < 10) {
15 printf(""x is less than 10\n"");
16 } else if($x === 10) {
17 printf(""x is equal to ten\n"");
18 } else {
19 printf(""x is greater than 10\n"");
20 }
Code Sample 38.1.: Examples of Conditional Statements in PHP
518",ComputerScienceOne.pdf
"38.3. Examples
1 $x = 15;
2 if($x < 10); {
3 printf(""x is less than 10\n"");
4 }
This PHP code will run without error or warning. However, it will end up printing
x is less than 10, even though x= 15! Recall that a conditional statement binds
to the executable statement or code block immediately following it. In this case, we’ve
provided an empty executable statement ended by the semicolon. The code is essentially
equivalent to
1 $x = 15;
2 if($x < 10) {
3 }
4 printf(""x is less than 10\n"");
This is obviously not what we wanted. The semicolon was bound to the empty executable
statement and the code block containing the print statement immediately followed, but
was not bound to the conditional statement which is why the print statement executed
regardless of the value of x.
Another convention that we’ve used in our code is where we have placed the curly brackets.
If a conditional statement is bound to only one statement, the curly brackets are not",ComputerScienceOne.pdf
"If a conditional statement is bound to only one statement, the curly brackets are not
necessary. However, it is best practice to include them even if they are not necessary
and we’ll follow this convention. Second, the opening curly bracket is on the same line as
the conditional statement while the closing curly bracket is indented to the same level
as the start of the conditional statement. Moreover, the code inside the code block is
indented. If there were more statements in the block, they would have all been at the
same indentation level.
38.3. Examples
38.3.1. Computing a Logarithm
The logarithm of x is the exponent that some base must be raised to get x. The most
common logarithm is the natural logarithm, ln (x) which is base e= 2.71828 ... . But
logarithms can be in any base b >1.1 What if we wanted to compute log2 (x)? Or
logπ (x)? Let’s write a program that will prompt the user for a number x and a base",ComputerScienceOne.pdf
"logπ (x)? Let’s write a program that will prompt the user for a number x and a base
b and computes logb(x). Arbitrary bases can be computed using the change of base
1Bases can also be 0 <b< 1, but we’ll restrict our attention to increasing functions only.
519",ComputerScienceOne.pdf
"38. Conditionals
formula:
logb(x) = loga(x)
loga(b)
If we can compute some base a, then we can compute any base b. Fortunately we have
such a solution. Recall that the standard library provides a function to compute the
natural logarithm, log()). This is one of the fundamentals of problems solving: if a
solution already exists, use it. In this case, a solution exists for a different, but similar
problem (computing the natural logarithm), but we can adapt the solution using the
change of base formula. In particular, if we have variables b (base) and x, we can
compute logb(x) using
log(x) / log(b)
We have a problem similar to the examples in the previous section. The user could enter
invalid values such as b= −10 or x= −2.54 (logarithms are undefined for non-positive
values in any base). We want to ensure that b >1 and x >0. With conditionals, we
can now do this. Once we have read in the input from the user we can make a check for
good input using an if statement.",ComputerScienceOne.pdf
"can now do this. Once we have read in the input from the user we can make a check for
good input using an if statement.
1 if($x <= 0 || $b <= 1) {
2 printf(""Error: bad input!\n"");
3 exit(1);
4 }
This code has something new: exit(1). The exit() function immediately terminates
the script regardless of the rest of the code that may remain. The argument passed to
exit is an integer that represents an error code. The convention is that zero indicates
“no error” while non-zero values indicate some error. This is a simple way of performing
error handling : if the user provides bad input, we inform them and quit the program,
forcing them to run it again and provide good input. By prematurely terminating the
program we avoid any illegal operation that would give a bad result.
Alternatively, we could have split the conditions into two statements and given a more
descriptive error message. We use this design in the full program which can be found",ComputerScienceOne.pdf
"descriptive error message. We use this design in the full program which can be found
in Code Sample 38.2. The program also takes the input as command line arguments.
Now that we have conditionals, we can check that the correct number of arguments was
provided by the user and quit in the event that they don’t provide the correct number.
38.3.2. Life & Taxes
Let’s adapt the conditional statements we developed in Section 3.6.4 into a full PHP
script. The first thing we need to do is establish the variables we’ll need and read them
in from the user. At the same time we can check for bad input (negative values) for both
the inputs.
520",ComputerScienceOne.pdf
"38.3. Examples
1 //prompt for income from the user
2 printf(""Please enter your Adjusted Gross Income: "");
3
4 $income = floatval(trim(fgets(STDIN)));
5
6 //prompt for children
7 printf(""How many children do you have? "");
8 $numChildren = intval(trim(fgets(STDIN)));
9
10 if($income < 0 || $numChildren < 0) {
11 printf(""Invalid inputs"");
12 exit(1);
13 }
Next, we can code a series of if-else-if statements for the income range. By placing the
ranges in increasing order, we only need to check the upper bounds just as in the original
example.
1 if($income <= 18150) {
2 $baseTax = $income * .10;
3 } else if($income <= 73800) {
4 $baseTax = 1815 + ($income -18150) * .15;
5 } else if($income <= 148850) {
6 ...
7 } else {
8 $baseTax = 127962.50 + ($income - 457600) * .396;
9 }
Next we compute the child tax credit, taking care that it does not exceed $3,000. A
conditional based on the number of children suffices as at this point in the program we
already know it is zero or greater.",ComputerScienceOne.pdf
"conditional based on the number of children suffices as at this point in the program we
already know it is zero or greater.
1 if($numChildren <= 3) {
2 $credit = $numChildren * 1000;
3 } else {
4 $credit = 3000;
5 }
Finally, we need to ensure that the credit does not exceed the total tax liability (the
credit is non-refundable, so if the credit is greater, the tax should only be zero, not
521",ComputerScienceOne.pdf
"38. Conditionals
negative).
1 if($baseTax - $credit >= 0) {
2 $totalTax = $baseTax - $credit;
3 } else {
4 $totalTax = 0;
5 }
The full program is presented in Code Sample 38.3.
38.3.3. Quadratic Roots Revisited
Let’s return to the quadratic roots program we previously designed that uses the quadratic
equation to compute the roots of a quadratic polynomial by reading coefficients a,b,c
in from the user. One of the problems we had previously identified is if the user enters
“bad” input: if a= 0, we would end up dividing by zero; if b2 −4ac< 0 then we would
have complex roots. With conditionals, we can now check for these issues and exit with
an error message.
Another potential case we might want to handle differently is when there is only one
distinct root ( b2 −4ac= 0). In that case, the quadratic formula simplifies to −b
2a and we
can print a different, more specific message to the user. The full program can be found
in Code Sample 38.4.
522",ComputerScienceOne.pdf
"38.3. Examples
1 <?php
2
3 /**
4 * This program computes the logarithm base b (b > 1)
5 * of a given number x > 0
6 */
7
8 if($argc != 3) {
9 printf(""Usage: %s b x \n"", $argv[0]);
10 exit(1);
11 }
12
13 $b = floatval($argv[1]);
14 $x = floatval($argv[2]);
15
16 if($x <= 0) {
17 printf(""Error: x must be greater than zero\n"");
18 exit(1);
19 }
20 if($b <= 1) {
21 printf(""Error: base must be greater than one\n"");
22 exit(1);
23 }
24
25 $result = log($x) / log($b);
26 printf(""log_(%f)(%f) = %f\n"", $b, $x, $result);
27
28 ?>
Code Sample 38.2.: Logarithm Calculator Program in C
523",ComputerScienceOne.pdf
"38. Conditionals
1 <?php
2 //prompt for income from the user
3 printf(""Please enter your Adjusted Gross Income: "");
4
5 $income = floatval(trim(fgets(STDIN)));
6
7 //prompt for children
8 printf(""How many children do you have? "");
9 $numChildren = intval(trim(fgets(STDIN)));
10
11 if($income < 0 || $numChildren < 0) {
12 printf(""Invalid inputs"");
13 exit(1);
14 }
15
16 if($income <= 18150) {
17 $baseTax = $income * .10;
18 } else if($income <= 73800) {
19 $baseTax = 1815 + ($income -18150) * .15;
20 } else if($income <= 148850) {
21 $baseTax = 10162.50 + ($income - 73800) * .25;
22 } else if($income <= 225850) {
23 $baseTax = 28925.00 + ($income - 148850) * .28;
24 } else if($income <= 405100) {
25 $baseTax = 50765.00 + ($income - 225850) * .33;
26 } else if($income <= 457600) {
27 $baseTax = 109587.50 + ($income - 405100) * .35;
28 } else {
29 $baseTax = 127962.50 + ($income - 457600) * .396;
30 }
31
32 if($numChildren <= 3) {
33 $credit = $numChildren * 1000;
34 } else {",ComputerScienceOne.pdf
"28 } else {
29 $baseTax = 127962.50 + ($income - 457600) * .396;
30 }
31
32 if($numChildren <= 3) {
33 $credit = $numChildren * 1000;
34 } else {
35 $credit = 3000;
36 }
37
38 if($baseTax - $credit >= 0) {
39 $totalTax = $baseTax - $credit;
40 } else {
41 $totalTax = 0;
42 }
43
44 printf(""AGI: $%10.2f\n"", $income);
45 printf(""Tax: $%10.2f\n"", $baseTax);
46 printf(""Credit: $%10.2f\n"", $credit);
47 printf(""Tax Liability: $%10.2f\n"", $totalTax);
48
49 ?>
Code Sample 38.3.: Tax Program in PHP
524",ComputerScienceOne.pdf
"38.3. Examples
1 <?php
2
3 /**
4 * This program computes the roots to a quadratic equation
5 * using the quadratic formula.
6 */
7
8 if($argc != 4) {
9 printf(""Usage: %s a b c\n"", $argv[0]);
10 exit(1);
11 }
12
13 $a = floatval($argv[1]);
14 $b = floatval($argv[2]);
15 $c = floatval($argv[3]);
16
17 if($a === 0) {
18 printf(""Error: a cannot be zero\n"");
19 exit(1);
20 } else if($b*$b < 4*$a*$c) {
21 printf(""Error: cannot handle complex roots\n"");
22 exit(1);
23 } else if($b*$b === 4*$a*$c) {
24 $root1 = -$b / (2*$a);
25 printf(""Only one distinct root: %f\n"", $root1);
26 } else {
27 $root1 = (-$b + sqrt($b*$b - 4*$a*$c) ) / (2*$a);
28 $root2 = (-$b - sqrt($b*$b - 4*$a*$c) ) / (2*$a);
29
30 printf(""The roots of %fx^2 + %fx + %f are: \n"", $a, $b, $c);
31 printf("" root1 = %f\n"", $root1);
32 printf("" root2 = %f\n"", $root2);
33 }
34
35 ?>
Code Sample 38.4.: Quadratic Roots Program in PHP With Error Checking
525",ComputerScienceOne.pdf
"39. Loops
PHP supports while loops, for loops, and do-while loops using the keywords while, for,
and do (along with another while). Continuation conditions for loops are enclosed in
parentheses, (...) and blocks of code associated with the loop are enclosed in curly
brackets.
39.1. While Loops
Code Sample 39.1 contains an example of a basic while loop in PHP. As with conditional
statements, our code style places the opening curly bracket on the same line as the
while keyword and continuation condition. The inner block of code is also indented
and all lines in the block are indented to the same level.
In addition, the continuation condition does not contain a semicolon since it is not an
executable statement. Just as with an if-statement, if we had placed a semicolon it would
have led to unintended results. Consider the following:
1 while($i <= 10); {
2 //perform some action
3 $i++; //iteration
4 }
A similar problem occurs. The while keyword and continuation condition bind to",ComputerScienceOne.pdf
"1 while($i <= 10); {
2 //perform some action
3 $i++; //iteration
4 }
A similar problem occurs. The while keyword and continuation condition bind to
the next executable statement or code block. As a consequence of the semicolon, the
executable statement that gets bound to the while loop is empty. What happens is
1 $i = 1; //Initialization
2 while($i <= 10) { //continuation condition
3 //perform some action
4 $i++; //iteration
5 }
Code Sample 39.1.: While Loop in PHP
527",ComputerScienceOne.pdf
"39. Loops
1 $i = 1;
2 $flag = true;
3 while($flag) {
4 //perform some action
5 $i++; //iteration
6 if($i>10) {
7 $flag = false;
8 }
9 }
Code Sample 39.2.: Flag-controlled While Loop in PHP
even worse: the program will enter an infinite loop. To see this, the code is essentially
equivalent to the following:
1 while($i <= 10) {
2 }
3 {
4 //perform some action
5 $i++; //iteration
6 }
In the while loop, we never increment the counter variable $i, the loop does nothing,
and so the computation will continue on forever! It is valid PHP and will run, but
obviously won’t work as intended. Avoid this problem by using proper syntax.
Another common use for a while loop is a flag-controlled loop in which we use a Boolean
flag rather than an expression to determine if a loop should continue or not. Since PHP
has built-in Boolean types, we can use a variable along with the keywords true and
false. An example can be found in Code Sample 39.2.
39.2. For Loops",ComputerScienceOne.pdf
"false. An example can be found in Code Sample 39.2.
39.2. For Loops
For loops in PHP use the familiar syntax of placing the initialization, continuation
condition, and iteration on the same line as the for keyword. An example can be found
in Code Sample 39.3.
Semicolons are placed at the end of the initialization and continuation condition, but not
the iteration statement. With while loops, the opening curly bracket is placed on the
same line as the for keyword. Code within the loop body is indented, all at the same
indentation level.
528",ComputerScienceOne.pdf
"39.3. Do-While Loops
1 $i;
2 for($i=1; $i<=10; $i++) {
3 //perform some action
4 }
Code Sample 39.3.: For Loop in PHP
1 $i;
2 do {
3 //perform some action
4 $i++;
5 } while($i <= 10);
Code Sample 39.4.: Do-While Loop in PHP
39.3. Do-While Loops
PHP also supports do-while loops. Recall that the difference between a while loop and a
do-while loop is when the continuation condition is checked. For a while loop it is prior
to the beginning of the loop body and in a do-while loop it is at the end of the loop.
This means that a do-while always executes at least once. An example can be found in
Code Sample 39.4.
The opening curly bracket is again on the same line as the keyword do. The while
keyword and continuation condition are on the same line as the closing curly bracket.
In a slight departure from previous syntax, a semicolon does appear at the end of the
continuation condition even though it is not an executable statement.
39.4. Foreach Loops",ComputerScienceOne.pdf
"continuation condition even though it is not an executable statement.
39.4. Foreach Loops
Finally, PHP supports foreach loops using the keyword foreach. Some of this will be a
preview of Section 42 where we discuss arrays in PHP, 1 In short you can iterate over the
elements of an array as follows.
1 $arr = array(1.41, 2.71, 3.14);
2 foreach($arr as $x) {
3 //x now holds the ""current"" element in arr
1Actually, PHP supports associative arrays, which are not the same thing as traditional arrays.
529",ComputerScienceOne.pdf
"39. Loops
4 }
In the foreach syntax we specify the array we want to iterate over, $arr and use the
keyword as. The last element in the statement is the variable name that we want to use
within the loop. This should be read as “ foreach element $x in the array $arr...”.
Inside the loop, the variable $x will be automatically updated on each iteration to the
next element in $arr.
39.5. Examples
39.5.1. Normalizing a Number
Let’s revisit the example from Section 4.1.1 in which wenormalize a number by continually
dividing it by 10 until it is less than 10. The code in Code Sample 39.5 specifically refers
to the value 32145.234 but would work equally well with any value of $x.
1 $x = 32145.234;
2 $k = 0;
3 while($x > 10) {
4 $x = $x / 10;
5 $k++;
6 }
Code Sample 39.5.: Normalizing a Number with a While Loop in PHP
39.5.2. Summation
Let’s revisit the example from Section 4.2.1 in which we computed the sum of integers
1 + 2 +··· + 10. The code is presented in Code Sample 39.6",ComputerScienceOne.pdf
"Let’s revisit the example from Section 4.2.1 in which we computed the sum of integers
1 + 2 +··· + 10. The code is presented in Code Sample 39.6
We could easily generalize this code. Instead of computing a sum up to a particular
number, we could have written it to sum up to another variable $n, in which case the
for loop would instead look like the following.
1 for($i=1; $i<=$n; $i++) {
2 $sum += $i;
3 }
530",ComputerScienceOne.pdf
"39.5. Examples
1 $sum = 0;
2 for($i=1; $i<=10; $i++) {
3 $sum += $i;
4 }
Code Sample 39.6.: Summation of Numbers using a For Loop in PHP
39.5.3. Nested Loops
Recall that you can write loops within loops. The inner loop will execute fully for each
iteration of the outer loop. An example of two nested of loops in PHP can be found in
Code Sample 39.7.
1 $n = 10;
2 $m = 20;
3 for($i=0; $i<$n; $i++) {
4 for($j=0; $j<$m; $j++) {
5 printf(""(i, j) = (%d, %d)\n"", $i, $j);
6 }
7 }
Code Sample 39.7.: Nested For Loops in PHP
The inner loop executes for j = 0 ,1,2,..., 19 < m = 20 for a total of 20 itera-
tions. However, it executes 20 times for each iteration of the outer loop. Since
the outer loop executes for i = 0 ,1,2,..., 9 < n = 10, the total number of times
the printf() statement executes is 10 ×20 = 200. In this example, the sequence
(0,0),(0,1),(0,2),..., (0,19),(1,0),..., (9,19) will be printed.
39.5.4. Paying the Piper",ComputerScienceOne.pdf
"(0,0),(0,1),(0,2),..., (0,19),(1,0),..., (9,19) will be printed.
39.5.4. Paying the Piper
Let’s adapt the solution for the loan amortization schedule we developed in Section 4.7.3.
First, we’ll read the principle, terms, and interest as command line inputs. Adapting the
formula for the monthly payment and using the standard math library’s pow() function,
we get
1 $monthlyPayment = ($monthlyInterestRate * $principle) /
2 (1 - pow( (1 + $monthlyInterestRate), -$n));
531",ComputerScienceOne.pdf
"39. Loops
The monthly payment may come out to be a fraction of a cent, say $43.871. For accuracy,
we need to ensure that all of the figures for currency are rounded to the nearest cent.
The standard math library does have a round() function, but it only rounds to the
nearest whole number, not the nearest 100th.
However, we can adapt the “off-the-shelf” solution to fit our needs. If we take the number,
multiply it by 100, we get 4387.1 which we can now round to the nearest whole number,
giving us 4387. We can then divide by 100 to get a number that has been rounded to
the nearest 100th!
$monthlyPayment = round($monthlyPayment * 100) / 100;
We can use the same trick to round the monthly interest payment and any other number
expected to be whole cents. To output our numbers, we use printf() and take care to
align our columns to make make it look nice. To finish our adaptation, we handle the
final month separately to account for an over/under payment due to rounding. The full",ComputerScienceOne.pdf
"final month separately to account for an over/under payment due to rounding. The full
solution can be found in Code Sample 39.8.
532",ComputerScienceOne.pdf
"39.5. Examples
1 <?php
2 if($argc != 4) {
3 printf(""Usage: %s principle apr terms\n"", $argv[0]);
4 exit(1);
5 }
6
7 $principle = floatval($argv[1]);
8 $apr = floatval($argv[2]);
9 $n = intval($argv[3]);
10
11 $balance = $principle;
12 $monthlyInterestRate = $apr / 12;
13
14 //monthly payment
15 $monthlyPayment = ($monthlyInterestRate * $principle) /
16 (1 - pow( (1 + $monthlyInterestRate), -$n));
17 //round to the nearest cent
18 $monthlyPayment = round($monthlyPayment * 100) / 100;
19
20 printf(""Principle: $%.2f\n"", $principle);
21 printf(""APR: %.4f%%\n"", $apr * 100.0);
22 printf(""Months: %d\n"", $n);
23 printf(""Monthly Payment: $%.2f\n"", $monthlyPayment);
24
25 //for the first n-1 payments in a loop:
26 for($i=1; $i<$n; $i++) {
27 // compute the monthly interest, rounded:
28 $monthlyInterest =
29 round( ($balance * $monthlyInterestRate) * 100) / 100;
30 // compute the monthly principle payment
31 $monthlyPrinciplePayment = $monthlyPayment - $monthlyInterest;
32 // update the balance",ComputerScienceOne.pdf
"30 // compute the monthly principle payment
31 $monthlyPrinciplePayment = $monthlyPayment - $monthlyInterest;
32 // update the balance
33 $balance = $balance - $monthlyPrinciplePayment;
34 // print i, monthly interest, monthly principle, new balance
35 printf(""%d\t$%10.2f $%10.2f $%10.2f\n"", $i, $monthlyInterest,
36 $monthlyPrinciplePayment, $balance);
37 }
38
39 //handle the last month and last payment separately
40 $lastInterest = round( ($balance * $monthlyInterestRate) * 100) / 100;
41 $lastPayment = $balance + $lastInterest;
42
43 printf(""Last payment = $%.2f\n"", $lastPayment);
44 ?>
Code Sample 39.8.: Loan Amortization Program in PHP
533",ComputerScienceOne.pdf
"40. Functions
Functions are essential in PHP programming. PHP provides a large library of standard
functions to perform basic input/output, math, and many other functions. PHP also
provides the ability to define and use your own functions.
PHP does not support function overloading, so when you define a function and give it a
name, that name cannot be in conflict with any other function name in the standard
library or any other code that you might use. Careful thought should go into the design
and naming of your functions.
PHP supports both call by value and call by reference. As of PHP 5.6, vararg functions
are also supported (though earlier versions supported some vararg-like functions such as
printf()). However, we will not go into detail here. Finally, another feature of PHP
is that function parameters are all optional. You may invoke a function with a subset
of the parameters. Depending on your PHP setup, the interpreter may issue a warning",ComputerScienceOne.pdf
"of the parameters. Depending on your PHP setup, the interpreter may issue a warning
that a parameter was omitted. However, PHP allows you to define default values for
optional parameters.
40.1. Defining & Using Functions
In general, you can define functions anywhere in your PHP script or codebase. They can
even appear after code that invokes them because PHP hoists the function definitions
by doing two passes of the script. However, it is good style to include function definitions
at the top of your script or in a separate PHP file for organization.
40.1.1. Declaring Functions
In PHP, to declare a function you use the keyword function. Because PHP is dynami-
cally typed, a function can return any type. Therefore, you do not declare the return
type (just as you do not declare a variable’s type). After the function keyword you
provide an identifier and parameters as the function signature. Immediately following,
you provide the function body enclosed with opening/closing curly brackets.",ComputerScienceOne.pdf
"you provide the function body enclosed with opening/closing curly brackets.
Typically, the documentation for functions is included with its declaration. Consider
the following examples. In these examples we use a commenting style known as “doc
comments.” This style was originally developed for Java but has since been adopted by
many other languages.
535",ComputerScienceOne.pdf
"40. Functions
1 /**
2 * Computes the sum of the two arguments.
3 */
4 function sum($a, $b) {
5 return ($a + $b);
6 }
7
8 /**
9 * Computes the Euclidean distance between the 2-D points,
10 * (x1,y1) and (x2,y2).
11 */
12 function getDistance($x1, $y1, $x2, $y2) {
13 $xDiff = ($x1-$x2);
14 $yDiff = ($y1-$y2);
15 return sqrt( $xDiff * $xDiff + $yDiff * $yDiff);
16 }
17
18 /**
19 * Computes a monthly payment for a loan with the given
20 * principle at the given APR (annual percentage rate) which
21 * is to be repaid over the given number of terms (usually
22 * months).
23 */
24 function getMonthlyPayment($principle, $apr, $terms) {
25 $rate = ($apr / 12.0);
26 $payment = ($principle * $rate) / (1-pow(1+$rate, -$terms));
27 return $payment;
28 }
Function identifiers (names) follow similar naming rules as variables, however they do
not begin with a dollar sign. Function names must begin with an alphabetic character",ComputerScienceOne.pdf
"not begin with a dollar sign. Function names must begin with an alphabetic character
and may contain alphanumeric characters as well as underscores. However, using modern
coding conventions we usually name functions using lower camel casing. Another quirk
of PHP is that function names are case insensitive. Though we declared a function,
getDistance() above, it could be invoked with either getdistance(), GETDISTANCE
or any other combination of uppercase/lowercase letters. However, good code will use
consistent naming and your function calls should match their declaration.
The keyword return is used to specify the value that is returned to the calling function.
Whatever value you end up returning is the return type of the function. Since you do not
specify variable or return types, functions are usually referred to as returning a “mixed”
type. You could design a function that, given one set of inputs, returns a number while",ComputerScienceOne.pdf
"type. You could design a function that, given one set of inputs, returns a number while
another set of inputs returns a string. You can use the syntax return; to return no
536",ComputerScienceOne.pdf
"40.1. Defining & Using Functions
value (you do not use the keyword void). In practice, however, the function ends up
returning null when doing this.
40.1.2. Organizing Functions
There are many coding standards that guide how PHP code should be organized. We’ll
only discuss a simple mechanism here. One way to organize functions is to collect
functions with similar functionality into separate PHP source files.
Suppose the functions above are in a PHP source file named utils.php. We could
include them in another source file (our “main” source file) using an include_once
function invocation. An example:
1 <?php
2
3 include_once(""utils.php"");
4
5 //we can now use the functions in utils.php:
6 $p = getMonthlyPayment(1000, 0.05, 12);
7
8 ?>
The include_once function loads and evaluates the given PHP source file at the point
in the code in which it is invoked. The “once” in the function refers to the fact that if
the source file was already included in the script/code before, it will not be included",ComputerScienceOne.pdf
"the source file was already included in the script/code before, it will not be included
a second time. This allows you to include the same source file in multiple source files
without a conflict.
40.1.3. Calling Functions
The syntax for calling a function is to simply provide the function name followed by
parentheses containing values or variables to pass to the function. Some examples:
1 $a = 10, $b = 20;
2 $c = sum($a, $b); //c contains the value 30
3
4 //invoke a function with literal values:
5 $dist = getDistance(0.0, 0.0, 10.0, 20.0);
6
7 //invoke a function with a combination:
8 $p = 1500.0;
9 $r = 0.05;
10 $monthlyPayment = getMonthlyPayment($p, $r, 60);
537",ComputerScienceOne.pdf
"40. Functions
40.1.4. Passing By Reference
By default, all types (including numbers, strings, etc.) are passed by value. To be able
to pass arguments by reference, we need to use slightly different syntax when defining
our functions.
To specify that a parameter is to be passed by reference, we place an ampersand, & in
front of it in the function signature. 1 No other syntax is necessary. When you call the
function, PHP automatically takes care of the referencing/dereferencing for you.
1 <?php
2
3 function swap($a, $b) {
4 $t = $a;
5 $a = $b;
6 $b = $t;
7 }
8
9 function swapByRef(&$a, &$b) {
10 $t = $a;
11 $a = $b;
12 $b = $t;
13 }
14
15 $x = 10;
16 $y = 20;
17
18 printf(""x = %d, y = %d\n"", $x, $y);
19 swap($x, $y);
20 printf(""x = %d, y = %d\n"", $x, $y);
21 swapByRef($x, $y);
22 printf(""x = %d, y = %d\n"", $x, $y);
23
24 ?>
The first function, swap() passes both variables by value. Swapping the values only",ComputerScienceOne.pdf
"22 printf(""x = %d, y = %d\n"", $x, $y);
23
24 ?>
The first function, swap() passes both variables by value. Swapping the values only
affects the copies of the parameters. The original variables $x and $y will be unaffected.
In the second function, swapByRef(), both variables are passed by reference as there
are ampersands in front of them. Swapping them inside the function swaps the original
1Those familiar with pointers in C will note that this is the exact opposite of the C operator.
538",ComputerScienceOne.pdf
"40.1. Defining & Using Functions
variables. The output to this code is as follows.
x = 10, y = 20
x = 10, y = 20
x = 20, y = 10
Observe that when we invoked the function, swapByRef($x, $y); we used the same
syntax as the pass by value version. The only syntax needed to pass by reference is in
the function signature itself.
40.1.5. Optional & Default Parameters
Parameters in PHP functions are optional. You can invoke a function without providing
a subset of parameters. However, if a parameter is not provided, PHP will treat
the parameter as null. If we invoked the getDistance() function with only two
parameters:
getDistance(10.0, 20.0);
then inside the function, $x1 and $y1 would take on the values 10 and 20, but $x2
and $y2 would be null. When used in the distance formula calculations, both would
be treated as zero. If we had invoked the function with no arguments,
getDistance();
then all four parameters would be treated as null (and thus zero in the calculations).",ComputerScienceOne.pdf
"getDistance();
then all four parameters would be treated as null (and thus zero in the calculations).
PHP also allows you to define alternative default values for function parameters. Consider
the following example.
1 function getMonthlyPayment($principle, $apr = 0.05, $terms = 60) {
2 $rate = ($apr / 12);
3 $mp = (($principle * $rate) / (1-pow(1+$rate, -$terms)));
4 return round($mp * 100) / 100;
5 }
In this example, the second and third parameter have been given default values of 0.05
and 60 respectively. If a call to this function omits these parameters, they are not treated
as null, but take on these defaults instead.
1 $x = getMonthlyPayment(10000, 0.07, 72);
2 print $x.""\n""; //170.49
3
4 //terms will be 60
5 $x = getMonthlyPayment(10000, 0.07);
539",ComputerScienceOne.pdf
"40. Functions
6 print $x.""\n""; //198.01
7
8 //apr will be 0.05, terms will be 60
9 $x = getMonthlyPayment(10000);
10 print $x.""\n""; //188.71
11
12 //balance will be null (0), apr will be 0.05, terms will be 60
13 $x = getMonthlyPayment();
14 print $x.""\n""; //0 but also a warning
It would not be possible to invoke getMonthlyPayment() by omitting only the second
argument. Providing n arguments will match the first n parameters. Thus, in your
function design you should place any optional/default parameters last.
40.1.6. Function Pointers
Functions are just pieces of code that reside somewhere in memory just as variables do.
Since we can pass variables by reference, it also makes sense that we would do the same
with functions.
In PHP, functions are first-class citizens 2 meaning that you can assign a function to a
variable just as you would a numeric value. For example, you can do the following.
1 $func = swapByRef;
2
3 $func($x, $y);",ComputerScienceOne.pdf
"variable just as you would a numeric value. For example, you can do the following.
1 $func = swapByRef;
2
3 $func($x, $y);
In the example above, we assigned the function swapByRef() to the variable $func
by using its identifier. The variable essentially holds a reference to the swapByRef()
function. Since it refers to a function, we can also invoke the function using the variable
as in the last line. This allows you to treat functions as callbacks to other functions. We
will revisit this concept in Chapter 47.
40.2. Examples
40.2.1. Generalized Rounding
Recall that the standard math library provides a round() function that rounds a number
to the nearest whole number. We’ve had need to round to cents as well. We now have
2Some would use a much more restrictive definition of first-class and would not consider them first-class
citizens in this sense
540",ComputerScienceOne.pdf
"40.2. Examples
the ability to write a function to do this for us. Before we do, however, let’s think more
generally. What if we wanted to round to the nearest tenth? Or what if we wanted to
round to the nearest 10s or 100s place? Let’s write a general purpose rounding function
that allows us to specify which decimal place to round to.
The most natural input would be to specify the place using an integer exponent. That
is, if we wanted to round to the nearest tenth, then we would pass it −1 as 0.1 = 10−1,
−2 if we wanted to round to the nearest 100th, etc. In the other direction, passing in 0
would correspond to the usual round function, 1 to the nearest 10s spot, and so on.
Moreover, we could demonstrate good code reuse (as well as procedural abstraction)
by scaling the input value and reusing the functionality already provided in the math
library’s round() function. We could further define a roundToCents() function that
used our generalized round function. Consider the following.",ComputerScienceOne.pdf
"library’s round() function. We could further define a roundToCents() function that
used our generalized round function. Consider the following.
1 <?php
2
3 /**
4 * Rounds to the nearest digit specified by the place
5 * argument. In particular to the (10^place)-th digit
6 */
7 function roundToPlace($x, $place) {
8 $scale = pow(10, -$place);
9 $rounded = round(x * $scale) / $scale;
10 return $rounded;
11 }
12
13 /**
14 * Rounds to the nearest cent
15 */
16 function roundToCents($x) {
17 return roundToPlace($x, -2);
18 }
19
20 ?>
We could place these functions into a file named round.php and include them in another
PHP source file.
40.2.2. Quadratic Roots
Another advantage of passing variables by reference is that we can “return” multiple
values with one function call. Functions are limited in that they can only return at most
one value. But if we pass multiple parameters by reference, the function can manipulate
541",ComputerScienceOne.pdf
"40. Functions
the contents of them, thereby communicating (though not strictly returning) multiple
values.
Consider again the problem of computing the roots of a quadratic equation,
ax2 + bx+ c= 0
using the quadratic formula,
x= −b±
√
b2 −4ac
2a
Since there are two roots, we may have to write two functions, one for the “plus” root
and one for the “minus” root both of which take the coefficients, a,b,c as arguments.
However, if we wrote a single function that took the coefficients as parameters by value
as well as two other parameters by reference, we could compute both root values, one in
each of the by-reference variables.
1 function quadraticRoots($a, $b, $c, &$root1, &$root2) {
2 $discriminant = sqrt($b*$b - 4*$a*$c);
3 $root1 = (-$b + $discriminant) / (2*$a);
4 $root2 = (-$b - $discriminant) / (2*$a);
5 return;
6 }
By using pass by reference variables, we avoid multiple functions. Recall that there
could be several “bad” inputs to this function. The roots could be complex values, the",ComputerScienceOne.pdf
"could be several “bad” inputs to this function. The roots could be complex values, the
coefficient acould be zero, etc. In the next chapter, we examine how we can handle these
errors.
542",ComputerScienceOne.pdf
"41. Error Handling & Exceptions
Modern versions of PHP support error handling through the use of exceptions. PHP has
several different predefined types of exceptions and also allows you to define your own
exception types by creating new classes that inherit from the generic Exception class.
PHP uses the standard try-catch-finally control structure to handle exceptions and
allows you to throw your own exceptions.
41.1. Throwing Exceptions
Though PHP defines several different types of exceptions, we’ll only cover the generic
Exception class. We can throw an exception in PHP by using the keyword throw
and creating a new Exception with an error message.
throw new Exception(""Something went wrong"");
By using a generic Exception, we can only attach a message to the exception (which
can be printed by code that catches the exception). If we want more fine-grained control
over the type of exceptions, we need to define our own exceptions.
41.2. Catching Exceptions",ComputerScienceOne.pdf
"over the type of exceptions, we need to define our own exceptions.
41.2. Catching Exceptions
To catch an exception in PHP you can use the standard try-catch control block.
Optionally (and as of PHP version 5.5.0) you can use the finally block to clean up any
resources or execute code regardless of whether or not an exception was raised. Consider
the simple task of reading input from a user and manually parsing its value into an
integer. If the user enters a non-numeric value, parsing will fail and we should instead
throw an exception. Consider the following function.
1 function readNumber() {
2 $input = readline(""Please enter a number: "");
3 if( is_numeric($input) ) {
4 $value = floatval($input);
5 } else {
6 throw new Exception(""Invalid input!"");
7 }
8 }
543",ComputerScienceOne.pdf
"41. Error Handling & Exceptions
Elsewhere in the code, we can surround a call to readNumber() in a try-catch
statement.
1 try {
2 readNumber();
3 } catch(Exception $e) {
4 printf(""Error: exception encountered: "" . $e->getMessage());
5 exit(1);
6 }
In this example, we’ve simply displayed an error message to the standard output and exited
the program. We’ve made the design decision that this error should be fatal. We could
have chosen to handle this error differently in the catch block. The $e->getMessage()
prints the message that the exception was created with. In this case, ""Invalid input!"".
41.3. Creating Custom Exceptions
As of PHP 5.3.0 it is possible to define custom exceptions by extending the Exception
class. To do this, you need to declare a new class. We cover classes and objects in
Chapter 45. For now, we present a simple example. Consider the example in the previous
chapter of computing the roots of a quadratic polynomial. One possible error situation",ComputerScienceOne.pdf
"chapter of computing the roots of a quadratic polynomial. One possible error situation
is when the roots are complex numbers. We could define a new PHP exception class as
follows.
1 /**
2 * Defines a ComplexRoot exception class
3 */
4 class ComplexRootException extends Exception
5 {
6 public function __construct($message = null,
7 $code = 0,
8 Exception $previous = null) {
9 // call the parent constructor
10 parent::__construct($message, $code, $previous);
11 }
12
13 // custom string representation of object
14 public function __toString() {
15 return __CLASS__ . "": [{$this->code}]: {$this->message}\n"";
16 }
17 }
544",ComputerScienceOne.pdf
"41.3. Creating Custom Exceptions
Now in our code we can catch and even throw this new type of exception.
1 if( $b*$b - 4*$a*$c < 0) {
2 throw new ComplexRootException(""Cannot Handle complex roots"");
3 }
1 try {
2 $r1 = getRoot($a, $b, $c);
3 } catch(ComplexRootException $e) {
4 //handle here
5 } catch(Exception $e) {
6 //handle all other types of exceptions here
7 }
In the code above we had two catch blocks. Since we can have multiple types of
exceptions, we can catch each different type and handle them differently if we choose.
Each catch block catches a different type of exception. The last catch block was
written to catch a generic Exception. Much like an if-else-if statement, the first
type of exception that is caught is the block that will be executed and they are all
mutually exclusive. Thus, a generic “catch all” block like this should always be the last
catch block. The most specific types of exceptions should be caught first and the most
general types should be caught last.
545",ComputerScienceOne.pdf
"42. Arrays
PHP allows you to use arrays, but PHP arrays are actually associative arrays. Though
you can treat them as regular arrays and use contiguous integer indices, they are more
flexible than that. Integer indices need to be contiguous or start at zero and you can
use strings as indices. In addition, since PHP is dynamically typed, PHP arrays allow
mixed types. An array has no fixed type and you can place different mixed types
into the same array. PHP arrays are dynamic, so there is no memory management or
allocation/deallocation of memory space. Arrays will grow and shrink automatically as
you add and remove elements.
42.1. Creating Arrays
Since PHP is dynamically typed, you do not need to declare an array or specify a
particular type of variable that it holds. However, there are several ways that you
can initialize an array. To create an empty array, you can call the array() function.
Optionally, you can provide an initial list of elements to insert into the array by providing",ComputerScienceOne.pdf
"Optionally, you can provide an initial list of elements to insert into the array by providing
the list of elements as arguments to the function.
1 //create an empty array:
2 $arr = array();
3
4 //create an array with elements 10, 20, 30:
5 $arr = array(10, 20, 30);
6
7 //create an array with mixed types:
8 $arr = array(10, 3.14, ""Hello"");
42.2. Indexing
By default, when inserting elements, 0-indexing is used. In the three examples above,
each element would be located at indices 0, 1, and 2 respectively. The usual square
bracket syntax can be used to access and assign elements.
547",ComputerScienceOne.pdf
"42. Arrays
1 //create an array with elements 10, 20, 30:
2 $arr = array(10, 20, 30);
3
4 //get the first element:
5 $x = $arr[0]; //x has value 10
6
7 //change the 3rd element to 5:
8 $arr[2] = 5;
9
10 //print the 2nd element:
11 printf(""$arr[1] = %d\n"", $arr[1]);
Attempting to access an element at an invalid index does not result in an error or an
exception (though a warning may be issued depending on how PHP is setup). Instead, if
you attempt to access an invalid element, it will be treated as a null value.
1 $arr = array(10, 20, 30);
2 $x = $arr[10]; //x is null
However, an array could have null values in it as elements. How do we distinguish
whether or not an accessed value was actually null or if it is not part of the array?
PHP provides a function, array_key_exists() to distinguish between these two cases.
It returns true or false depending on whether or not a particular index has been set
or not.
1 $arr = array(10, null, 30);
2 if(array_key_exists(1, $arr)) {",ComputerScienceOne.pdf
"or not.
1 $arr = array(10, null, 30);
2 if(array_key_exists(1, $arr)) {
3 print ""index 1 contains an element\n"";
4 }
5 if(!array_key_exists(10, $arr)) {
6 print ""index 10 does not contain an element\n"";
7 }
8
9 if($arr[1] === $arr[10]) {
10 print ""but they are both null\n"";
11 }
548",ComputerScienceOne.pdf
"42.2. Indexing
42.2.1. Strings as Indices
Since arrays in PHP are associative arrays, keys are not limited to integers. You can also
use strings as keys to index elements.
1 $arr = array();
2 $arr[0] = 5;
3 $arr[""foo""] = 10;
4 $arr[""hello""] = ""world"";
5
6 print ""value = "" . $arr[""hello""];
Note that strings that contain integer values will be type-juggled into their numeric
values. For example, $arr[""10""] = 3; will be equivalent to $arr[10] = 3;. However,
strings containing floating point values will not be coerced but will remain as strings,
$arr[""3.15""] = 7; for example.
42.2.2. Non-Contiguous Indices
Another aspect of PHP associative arrays is that integer indices need not be contiguous.
For example,
1 $arr = array();
2 $arr[0] = 10;
3 $arr[5] = 20;
The values at indices 1 through 4 are undefined and the array contains some “holes” in
its indices.
42.2.3. Key-Value Initialization
Since associative arrays in PHP can be indexed by either integers or strings and need not",ComputerScienceOne.pdf
"its indices.
42.2.3. Key-Value Initialization
Since associative arrays in PHP can be indexed by either integers or strings and need not
be ordered or contiguous, we can use a special key-value initialization syntax to define
not only the values, but the keys that map to those values when we initialize an array.
The “double arrow,” => is used to denote the mapping.
1 $arr = array(
2 ""foo"" => 5,
3 4 => ""bar"",
4 0 => 3.14,
549",ComputerScienceOne.pdf
"42. Arrays
5 ""baz"" => ""ten""
6 );
42.3. Useful Functions
There are dozens of useful functions PHP defines that can be used with arrays. We’ll
only highlight a few of the more useful ones. First, the count() function can be used
to compute how many elements are stored in the array.
1 $arr = array(10, 20, 30);
2 $n = count($arr); //n is 3
For convenience and debugging, a special function, print_r() allows you to print the
contents of an array in a human-readable format that resembles the key-value initialization
syntax above. For example,
1 $arr = array(
2 ""foo"" => 5,
3 4 => ""bar"",
4 0 => 3.14,
5 ""baz"" => ""ten""
6 );
7 print_r($arr);
would end up printing
Array
(
[foo] => 5
[4] => bar
[0] => 3.14
[baz] => ten
)
Two other functions, array_keys() and array_values() return new zero-indexed
arrays containing the keys and the values of an array respectively. Reusing the example
above,
550",ComputerScienceOne.pdf
"42.4. Iteration
1 $keys = array_keys($arr);
2 $vals = array_values($arr);
3 print_r($keys);
4 print_r($vals);
would print
Array
(
[0] => foo
[1] => 4
[2] => 0
[3] => baz
)
Array
(
[0] => 5
[1] => bar
[2] => 3.14
[3] => ten
)
Finally, you can use the equality operators, == and === to compare arrays. The first is
the loose equality operator and evaluates to true if the two compared arrays have the
same key-value pairs while the second is the strict equality operator and is true only if
the arrays have the same key/value pairs in the same order and are of the same type.
42.4. Iteration
If we have an array in PHP that we know is 0-indexed and all elements are contiguous, we
can use a normal for loop to iterate over its elements by incrementing an index variable.
1 for($i=0; $i<count($arr); $i++) {
2 print $arr[$i] . ""\n"";
3 }
This fails, however, when we have an associative array that has a mix of integer and",ComputerScienceOne.pdf
"1 for($i=0; $i<count($arr); $i++) {
2 print $arr[$i] . ""\n"";
3 }
This fails, however, when we have an associative array that has a mix of integer and
string keys or “holes” in the indexing of integer keys. For this reason, it is more reliable
to use foreach loops. There are several ways that we can use a foreach loop. The
most general usage is to use the double arrow notation to iterate over each key-value
pair.
551",ComputerScienceOne.pdf
"42. Arrays
1 //for each key value pair:
2 foreach($arr as $key => $val) {
3 print ""$key maps to $val \n"";
4 }
This syntax gives you access to both the key and the value for each element in the array
$arr. The keyword as is used to denote the variable names $key and $val that
will be changed on each iteration of the loop. You need not use the identifiers $key and
$val; you can use any legal variable names for the key/value variables. If you do not
need the keys when iterating, you can use the following shorthand syntax.
1 //for each value:
2 foreach($arr as $val) {
3 print ""$val \n"";
4 }
42.5. Adding Elements
You can easily add elements to an array by providing an index and using the assignment
operator as in the previous examples. There are also several functions PHP defines that
can insert elements to the beginning ( array_unshift()), end ( array_push()) or at
an arbitrary position ( array_splice()).
1 $arr = array(10, 20, 30);
2
3 array_unshift($arr, 15);",ComputerScienceOne.pdf
"an arbitrary position ( array_splice()).
1 $arr = array(10, 20, 30);
2
3 array_unshift($arr, 15);
4 //arr is now [15, 10, 20, 30]
5
6 array_push($arr, 25);
7 //arr is now [15, 10, 20, 30, 25]
8
9 //insert 35 at index 3:
10 array_splice($arr, 3, 0, 35);
11 //arr is now [15, 10, 20, 35, 30, 25]
Another, simpler way of adding elements is to use the following syntax:
552",ComputerScienceOne.pdf
"42.6. Removing Elements
1 $arr = array(10, 20, 30);
2 $arr[] = 5;
3 $arr[] = 15;
4 $arr[] = 25;
5 print_r($arr);
By using the assignment operator but not specifying the index, the element will be added
to the next available integer index. Since there were already 3 elements in the array, each
subsequent element is inserted at index 3, 4, and 5 respectively. In general, the element
will be inserted at the maximum index value already used plus one. The example above
results in the following.
Array
(
[0] => 10
[1] => 20
[2] => 30
[3] => 5
[4] => 15
[5] => 25
)
42.6. Removing Elements
You can remove elements from an array using the unset() function. This function only
removes the element from the array, it does not shift other elements down to fill in the
unused index.
1 $arr = array(10, 20, 30);
2 unset($arr[1]);
3 print_r($arr);
This example would result in the following.
Array
(
[0] => 10
[2] => 30
)
Further, you can use unset() to destroy all elements in the array:
553",ComputerScienceOne.pdf
"42. Arrays
unset($arr);
destroys the entire array. It does not merely empty the array, but it unsets the variable
$arr itself.
42.7. Using Arrays in Functions
By default, all arguments to a function in PHP are passed by value; this includes arrays.
Thus, if you make any changes to an array passed to a function, the changes are not
realized in the calling function. You can explicitly specify that the array parameter is
passed by reference so that any changes to the array are realized in the calling function.
To illustrate, consider the following example.
1 function setFirst($a) {
2 $a[0] = 5;
3 }
4
5 $arr = array(10, 20, 30);
6 print_r($arr);
7 setFirst($arr);
8 print_r($arr);
This example results in the following.
Array
(
[0] => 10
[1] => 20
[2] => 30
)
Array
(
[0] => 10
[1] => 20
[2] => 30
)
That is, the change to the first element does not affect the original array. However, if we
specify that the array is passed by reference, then the change is realized. For example,",ComputerScienceOne.pdf
"specify that the array is passed by reference, then the change is realized. For example,
1 function setFirst(&$a) {
2 $a[0] = 5;
3 }
4
5 $arr = array(10, 20, 30);
554",ComputerScienceOne.pdf
"42.8. Multidimensional Arrays
6 print_r($arr);
7 setFirst($arr);
8 print_r($arr);
This now results in the original array being changed:
Array
(
[0] => 10
[1] => 20
[2] => 30
)
Array
(
[0] => 5
[1] => 20
[2] => 30
)
42.8. Multidimensional Arrays
PHP supports multidimensional arrays in the sense that elements in an array can be of
any type, including other arrays. We can use the same syntax and operations for single
dimensional arrays. For example, we can use the double arrow syntax and assign arrays
as values to create a 2-dimensional array.
1 $mat = array(
2 0 => array(10, 20, 30),
3 1 => array(40, 50, 60),
4 2 => array(70, 80, 90)
5 );
6 print_r($mat);
Which results in the following:
Array
(
[0] => Array
(
[0] => 10
[1] => 20
[2] => 30
)
555",ComputerScienceOne.pdf
"42. Arrays
[1] => Array
(
[0] => 40
[1] => 50
[2] => 60
)
[2] => Array
(
[0] => 70
[1] => 80
[2] => 90
)
)
Alternatively, you can use two indices to get and set values from a 2-dimensional array.
1 for($i=0; $i<3; $i++) {
2 for($j=0; $j<4; $j++) {
3 $mat[$i][$j] = ($i+$j)*3;
4 }
5 }
which results in:
Array
(
[0] => Array
(
[0] => 0
[1] => 3
[2] => 6
[3] => 9
)
[1] => Array
(
[0] => 3
[1] => 6
[2] => 9
[3] => 12
)
[2] => Array
(
[0] => 6
[1] => 9
[2] => 12
[3] => 15
)
)
556",ComputerScienceOne.pdf
"43. Strings
As we’ve previously seen, PHP has a built-in string type. Internally, PHP strings are
simply a sequence of bytes, but for our purposes we can treat it as a 0-indexed character
array. PHP strings are mutable and can be changed, but it is considered best practice to
treat them as immutable and rely on the many functions PHP provides to manipulate
strings.
43.1. Basics
We can create strings by assigning a string literal value to a variable. Strings can be
specified by either single quotes or double quotes (there are no individual characters in
PHP, only single character strings), but we will use the double quote syntax.
1 $firstName = ""Thomas"";
2 $lastName = ""Waits"";
3
4 //we can also reassign values
5 $firstName = ""Tom"";
The reassignment in the last line in the example effectively destroys the old string. The
assignment operator can also be used to make copies of strings.
1 $firstName = ""Thomas"";
2 $alias = $firstName;",ComputerScienceOne.pdf
"assignment operator can also be used to make copies of strings.
1 $firstName = ""Thomas"";
2 $alias = $firstName;
It is important to understand that this assignment makes a deep copy of the string.
Changes to the first do not affect the second one. You can make changes to individual
characters in a string by treating it like a zero-indexed array.
1 $a = ""hello"";
2 $a[0] = ""H"";
3 $a[5] = ""!"";
4 //a is now ""Hello!""
557",ComputerScienceOne.pdf
"43. Strings
The last line extends the string by adding an additional character. You can even remove
characters by setting them to the empty string.
1 $a = ""Apples!"";
2 $a[5] = """";
3 //a is now ""Apple!""
43.2. String Functions
PHP provides dozens of convenient functions that allow you to process and modify strings.
We highlight a few of the more common ones here. A full list of supported functions can
be found in standard documentation. Because of the history of PHP, many of the same
functions defined in the C string library can also be used in PHP.
Length
When accessing individual characters in a string, it is necessary that we know the length
of the string so that we do not access invalid characters (though doing so is not an error,
it just results in null). The strlen() function returns an integer that represents the
number of characters in the string.
1 $s = ""Hello World!"";
2 $x = strlen($s); //x is 12
3 $s = """";
4 $x = strlen($s); //x is 0
5
6 //careful:
7 $s = NULL",ComputerScienceOne.pdf
"1 $s = ""Hello World!"";
2 $x = strlen($s); //x is 12
3 $s = """";
4 $x = strlen($s); //x is 0
5
6 //careful:
7 $s = NULL
8 $x = strlen($s); //x is 0
As demonstrated in the last example, strlen() will return 0 for null strings. Recall
that we can distinguish between these two situations by using is_null(). Using this
function we can iterate over each individual character in a string.
1 $fullName = ""Tom Waits"";
2 for($i=0; $i<strlen($fullName); $i++) {
3 printf(""fullName[%d] = %s\n"", $i, $fullName[$i]);
4 }
558",ComputerScienceOne.pdf
"43.2. String Functions
This would print the following
fullName[0] = T
fullName[1] = o
fullName[2] = m
fullName[3] =
fullName[4] = W
fullName[5] = a
fullName[6] = i
fullName[7] = t
fullName[8] = s
Concatenation
PHP has a concatenation operator built into the language. To concatenate one or more
strings together, you can use a simple period between them as the concatenation operator.
Concatenation results in a new string.
1 $firstName = ""Tom"";
2 $lastName = ""Waits"";
3
4 $formattedName = $lastName . "", "" . $firstName;
5 //formattedName now contains ""Waits, Tom""
Concatenation also works with other variable types.
1 $x = 10;
2 $y = 3.14;
3
4 $s = ""Hello, x is "" . $x . "" and y = "" . $y;
5 //s contains ""Hello, x is 10 and y = 3.14""
Computing a Substring
PHP provides a simple function, substr() to compute a substring of a string. It takes
at at least 2 arguments: the string to operate on and the starting index. There is a third,",ComputerScienceOne.pdf
"at at least 2 arguments: the string to operate on and the starting index. There is a third,
optional parameter that allows you to specify the length of the resulting substring.
559",ComputerScienceOne.pdf
"43. Strings
1 $name = ""Thomas Alan Waits"";
2
3 $firstName = substr($name, 0, 6); //""Thomas""
4 $middleName = substr($name, 7, 4); //""Alan""
5 $lastName = substr($name, 12); //""Waits""
In the final example, omitting the optional length parameter results in the entire remainder
of the string being returned as the substring.
43.3. Arrays of Strings
We often need to deal with collections of strings. In PHP we can define arrays of strings.
Indeed, we’ve seen arrays of strings before. When processing command line arguments,
PHP defines an array of strings, $argv. Each string can be accessed using an index,
$argv[0] for example is always the name of the script.
We can create our own arrays of strings using the same syntax as with other arrays.
1 $names = array(
2 ""Margaret Hamilton"",
3 ""Ada Lovelace"",
4 ""Grace Hopper"",
5 ""Marie Curie"",
6 ""Hedy Lamarr"");
43.4. Comparisons
When comparing strings in PHP, we can use the usual comparison operators such",ComputerScienceOne.pdf
"4 ""Grace Hopper"",
5 ""Marie Curie"",
6 ""Hedy Lamarr"");
43.4. Comparisons
When comparing strings in PHP, we can use the usual comparison operators such
as ===, <, or <= which will compare the strings lexicographically. However, this is
generally discouraged because of type juggling issues and strict vs loose equality/inequality
comparisons. Instead, there are several comparator methods that PHP provides to
compare strings based on their content. strcmp($a, $b) takes two strings and returns
an integer based on the lexicographic ordering of $a and $b. If $a precedes $b,
strcmp() returns something negative. It returns zero if $a and $b have the same
content. Otherwise it returns something positive if $b precedes $a.
1 $x = strcmp(""apple"", ""banana""); //x is negative
2 $x = strcmp(""zelda"", ""mario""); //x is positive
560",ComputerScienceOne.pdf
"43.5. Tokenizing
3 $x = strcmp(""Hello"", ""Hello""); //x is zero
4
5 //shorter strings precede longer strings:
6 $x = strcmp(""apple"", ""apples""); //x is negative
7
8 $x = strcmp(""Apple"", ""apple""); //x is negative
In the last example, ""Apple"" precedes ""apple"" since uppercase letters are ordered
before lowercase letters according to the ASCII table. We can also make case insen-
sitive comparisons if we need to using the alternative, strcasecmp($a, $b). Here,
strcasecmp(""Apple"", ""apple"") will return zero as the two strings are the same ig-
noring the cases.
The comparison functions also have length-limited versions, strncmp($a, $b, $n)
and strncasecmp($a, $b, $n). Both will only make comparisons in the first $n
characters of the strings. Thus, strncmp(""apple"", ""apples"", 5) will result in zero
as the two strings are equal in the first 5 characters.
43.5. Tokenizing
Recall that tokenizing is the process of splitting up a string along some delimiter. For",ComputerScienceOne.pdf
"43.5. Tokenizing
Recall that tokenizing is the process of splitting up a string along some delimiter. For
example, the comma delimited string, ""Smith,Joe,12345678,1985-09-08"" contains
four pieces of data delimited by a comma. Our aim is to split this string up into four
separate strings so that we can process each one. PHP provides several functions to to
this, explode() and preg_split().
The simpler one, explode() takes two arguments: the first one is a string delimiter and
the second is the string to be processed. It then returns an array of strings.
1 $data = ""Smith,Joe,12345678,1985-09-08"";
2
3 $tokens = explode("","", $data);
4 //tokens is [ ""Smith"", ""Joe"", ""12345678"", ""1985-09-08"" ]
5
6 $dateTokens = explode(""-"", $tokens[3]);
7 //dateTokens is now [ ""1985"", ""09"", ""08"" ]
The more sophisticated preg_split() also takes two arguments, 1 but instead of a
simple delimiter, it uses a regular expression; a sequence of characters that define a",ComputerScienceOne.pdf
"simple delimiter, it uses a regular expression; a sequence of characters that define a
search pattern in which special characters can be used to define complex patterns. For
1The “preg” stands for Perl Compatible Regular Expression.
561",ComputerScienceOne.pdf
"43. Strings
example, the complex expression ^[+-]?(\d+(\.\d+)?|\.\d+)([eE][+-]?\d+)?$ will
match any valid numerical value including scientific notation. We will not cover regular
expressions in depth, but to demonstrate their usefulness, here’s an example by which
you can split a string along any and all whitespace:
1 $s = ""Alpha Beta \t Gamma \n Delta \t\nEpsilon"";
2 $tokens = preg_split(""/[\s]+/"", $s);
3 //tokens is now [ ""Alpha"", ""Beta"", ""Gamma"", ""Delta"", ""Epsilon"" ]
562",ComputerScienceOne.pdf
"44. File I/O
Because of the history of PHP, file functions, just like string functions, were mostly
influenced by the C standard library functions and have very similar naming and usage.
Writing binary or plaintext data is determined by which functions you use. In general
whether or not a file input/output stream is buffered or unbuffered is determined by the
system configuration. There are some ways in which this can be changed, but we will
not cover them in detail.
44.1. Opening Files
Files are represented in PHP as a “resource” (also referred to as a handle) that can
be passed around to other functions to read and write to the file. For our purposes, a
resource is simply a variable that can be stored and passed to other functions.
To open a file, you use the fopen() function (short for file open) which requires two
arguments and returns a file handle. The first argument is the file path/name that you
want to open for processing. The second argument is a string representing the “mode”",ComputerScienceOne.pdf
"want to open for processing. The second argument is a string representing the “mode”
that you want to open the file in. There are several supported modes, but the two we
will be interested in are reading, in which case you pass it ""r"" and writing in which case
you pass it ""w"". The path can be an absolute path, relative path, or may be omitted if
the file is in the current working directory.
1 //open a file for reading (input):
2 $input = fopen(""/user/apps/data.txt"", ""r"");
3 //open a file for writing (output):
4 $output = fopen(""./results.txt"", ""w"");
5
6 if(!$input) {
7 printf(""Unable to open input file\n"");
8 exit(1);
9 }
10
11 if(!$output) {
12 printf(""Unable to open output file\n"");
13 exit(1);
14 }
563",ComputerScienceOne.pdf
"44. File I/O
1 $h = fopen(""input.data"", ""r"");
2 while(!feof($h)) {
3 //read the next line:
4 $line = fgets($h);
5 //trim it:
6 $line = trim($line);
7 //process it, we'll just print it
8 print $line;
9 }
Code Sample 44.1.: Processing a file line-by-line in PHP
The two conditionals above check that the file opened successfully. If opening the file
failed, fopen() returns false (and the interpreter issues warning).
44.2. Reading & Writing
When a file is opened, the file handle returned by fopen() initially points to the
beginning of the file. As you read or write from it, the resource advances through the file
content.
Reading
There are a couple of ways to read input from a file. To read a file line by line, we
can use fgets() to get each line. It takes a single argument: the file handle that
was returned by fopen() and returns a string containing the entire line including the
endline character. If necessary, leading and trailing whitespace can be removed using",ComputerScienceOne.pdf
"endline character. If necessary, leading and trailing whitespace can be removed using
trim(). To determine the end of a file you can use feof() which returns true when
the handle has reached the end of the file. Code Sample 44.1 contains a full example of
processing a file line-by-line.
Alternatively there is a convenient function, file_get_contents() that will retrieve
the entire file as a string.
$fileContents = file_get_contents(""./inputFile.txt"");
Writing
There are several ways that we can output to files, but the easiest is to simply use
fwrite() (short for file write). This is a “binary-safe” file output function that takes
two arguments: the file handle to write to and a string of data to be written. It also
returns an integer representing the number of bytes written to the file (or false on
564",ComputerScienceOne.pdf
"44.2. Reading & Writing
error).
1 $x = 10;
2 $y = 3.14;
3
4 //write to a plaintext file
5 fwrite($output, ""Hello World!\n"");
6 fwrite($output, ""x = $x, y = $y\n"");
44.2.1. Using URLs
A nice feature of PHP is that you can use URLs as file names to read and write to a URL.
“Reading” from a URL mean connecting to a remote resource, such as a webservice and
downloading its contents (which may be HTML, XML or JSON data). Writing to a URL
may be used to post data to a web page in order to receive a response.
As an example, we can download a webpage in PHP as follows.
1 $h = fopen(""http://cse.unl.edu"", ""r"");
2 $contents = """";
3 while(!feof($h)) {
4 $contents .= fgets($h);
5 }
6
7 //or just:
8 $contents = file_get_contents(""http://cse.unl.edu"");
44.2.2. Closing Files
Once you are done processing a file, you should close it using the fclose() function.
fclose($h);
which takes a single argument, the file handle that you wish to close.
565",ComputerScienceOne.pdf
"45. Objects
Object-oriented features have been continually added to PHP with each new version.
Starting with version 5, PHP has had a full, class-based object-oriented programming
support, meaning that it facilitates the creation of objects through the use of classes and
class declarations. Classes are essentially “blueprints” for creating instances of objects.
An object is an entity that is characterized by identity, state and behavior. The identity
of an object is an aspect that distinguishes it from other objects. The variables and
values that a variable takes on within an object is its state. Typically the variables that
belong to an object are referred to as member variables. Finally, an object may also
have functions that operate on the data of an object. In the context of object-oriented
programming, a function that belongs to an object is referred to as a member method.
A class declaration specifies the member variables and member methods that belong",ComputerScienceOne.pdf
"A class declaration specifies the member variables and member methods that belong
to instances of the class. We discuss how to create and use instances of a class below.
However, to begin, let’s define a class that models a student by defining member variables
to support a first name, last name, a unique identifier, and GPA.
To declare a class, we use the class keyword. Inside the class (denoted by curly
brackets), we place any code that belongs to the class. To declare member variables in a
class, we specify the variable names and their visibility inside the class, but outside any
methods in the class.
1 class Student {
2
3 //member variables:
4 private $firstName;
5 private $lastName;
6 private $id;
7 private $gpa;
8
9 }
To organize code, it is common practice to place class declarations in separate files with
the same name as the class. For example, this Student class declaration would be
placed in a file named Student.php and included in any other script files that utilized
the class.
567",ComputerScienceOne.pdf
"45. Objects
45.1. Data Visibility
Recall that encapsulation involves not only the grouping of data, but the protection of
data. The class declaration above achieves the grouping of data. To provide for the
protection of data, PHP defines several visibility keywords that specify what segments
of code can “see” the variables. Visibility in this context determines whether or not a
segment of code can access and/or modify the variable’s value. PHP defines three levels
of visibility using they keywords public, protected and private. Each of these
keywords can be applied to both member variables and member methods.
• public – This is the least restrictive visibility level and makes the member variable
visible to any code segment.
• protected – This is a bit more restrictive and makes it so that the member
variable is only visible to the code in the same class, or any subclass of the class.1
• private – this is the most restrictive visibility level, private member variables",ComputerScienceOne.pdf
"• private – this is the most restrictive visibility level, private member variables
are only visible to instances of the class itself.
Table 45.1 summarizes these four keywords with respect to their access levels. It is
important to understand that protection is in the context of encapsulation and does
not involve protection in the sense of “security.” Protection is this context is a design
principle. Limiting the access of variables only affects how the rest of the code base
interacts with our class and its data.
Modifier Class Subclass World
public Y Y Y
protected Y Y N
private Y N N
Table 45.1.: PHP Visibility Keywords & Access Levels
In general, it is best practice to make member variables private and control access to
them via accessor and mutator methods (see below) unless there is a compelling design
reason to increase their visibility.
45.2. Methods
The third aspect of encapsulation involves the grouping of methods that act on an object’s",ComputerScienceOne.pdf
"reason to increase their visibility.
45.2. Methods
The third aspect of encapsulation involves the grouping of methods that act on an object’s
data. Within a class, we can declare member methods using the syntax we’re already
familiar with. We declare a member method by using the keyword function and
providing a signature and body. We can use the same visibility keywords as with member
1Subclasses are involved with inheritance, another object-oriented programming concept that we will
not discuss here.
568",ComputerScienceOne.pdf
"45.2. Methods
variables in order allow or restrict access to the methods. With methods, visibility and
access determine whether or not the method may be invoked.
We add to our example by providing two public methods that compute and return a
result on the member variables. We also use javadoc-style comments to document each
member method.
1 class Student {
2
3 //member variables:
4 private $firstName;
5 private $lastName;
6 private $id;
7 private $gpa;
8
9 /**
10 * Returns a formatted String of the Student's
11 * name as Last, First.
12 */
13 public function getFormattedName() {
14 return $this->lastName . "", "" . $this->firstName;
15 }
16
17 /**
18 * Scales the GPA, which is assumed to be on a
19 * 4.0 scale to a percentage.
20 */
21 public function getGpaAsPercentage() {
22 return $this->gpa / 4.0;
23 }
24
25 }
There is some new syntax in the example above. In the member methods, we need a way
to refer to the instance’s member variables. The keyword $this is used to refer to the",ComputerScienceOne.pdf
"to refer to the instance’s member variables. The keyword $this is used to refer to the
instance, this is known as open recursion.
When an instance of a class is created, for example,
$s = new Student();
the reference variable $s is how we can refer to it. This variable, however, exists outside
the class. Inside the class, we need a way to refer to the instance itself. Since we don’t
have a variable inside the class to reference the instance itself, PHP provides the keyword
$this in order to do so. Then, to access the member variables we use the arrow operator
569",ComputerScienceOne.pdf
"45. Objects
(more below) and reference the member variable via its identifier but with no dollar sign .2
45.2.1. Accessor & Mutator Methods
Since we have made all the member variables private, no code outside the class may
access or modify their values. It is generally good practice to make member variables
private to restrict access. However, if we still want code outside the object to access or
mutate (that is, change) the variable values, we can define accessor and mutator methods
(or simply getter and setter methods) to facilitate this.
Each getter method returns the value of the instance’s variable while each setter method
takes a value and sets the instance’s variable to the new value. It is common to name
each getter/setter by prefixing a get and set to the variable’s name using lower camel
casing. For example:
1 public function getFirstName() {
2 return $this->firstName;
3 }
4
5 public function setFirstName($firstName) {
6 $this->firstName = $firstName;
7 }",ComputerScienceOne.pdf
"1 public function getFirstName() {
2 return $this->firstName;
3 }
4
5 public function setFirstName($firstName) {
6 $this->firstName = $firstName;
7 }
One advantage to using getters and setters (as opposed to naively making everything
public) is that you can have greater control over the values that your variables can
take. For example, we may want to do some data validation by rejecting null values or
invalid values. For example:
1 public function setFirstName($firstName) {
2 if($firstName === null) {
3 throw new Exception(""names cannot be null"");
4 } else {
5 $this->firstName = $firstName;
6 }
7 }
8
9 public function setGpa($gpa) {
10 if($gpa < 0.0 || $gpa > 4.0) {
2You can use the syntax $this->$foo but it will assume that $foo is a string that contains the
name of another variable, for example, if $foo = ""firstName""; then $this->$foo would resolve
to the instance’s $firstName variable. This is useful if your object has been dynamically created",ComputerScienceOne.pdf
"to the instance’s $firstName variable. This is useful if your object has been dynamically created
by adding variables at runtime that were not part of the original class declaration.
570",ComputerScienceOne.pdf
"45.3. Constructors
11 throw new Exception(""GPAs must be in [0, 4.0]"");
12 } else {
13 $this->gpa = $gpa;
14 }
15 }
Controlling access of member variables through getters and setters is good encapsulation.
Doing so makes your code more predictable and more testable. Making your member
variables public means that any piece of code can change their values. There is no way
to do validation or prevent bad values.
In fact, it is good practice to not even have setter methods. If the value of member
variables cannot be changed, it makes the object immutable. Immutability is a nice
property because it makes instances of the class thread-safe. That is, we can use instances
of the class in a multithreaded program without having to worry about threads changing
the values of the instance on one another.
45.3. Constructors
If we make the (good) design decision to make our class immutable, we still need a way
to initialize the values. This is where a constructor comes in. A constructor is a special",ComputerScienceOne.pdf
"to initialize the values. This is where a constructor comes in. A constructor is a special
method that specifies how an object is constructed. With built-in variable types such as
an numbers or strings, PHP “knows” how to interpret and assign a value. However, with
user-defined objects such as our Student class, we need to specify how the object is
created.
Just as with functions outside of classes, PHP does not support function overloading
inside classes. That is, you can have one and only one function with a given identifier
(name). Thus, there is only one possible constructor. Moreover, PHP reserves the
name __construct for the constructor method. The two underscores are a naming
convention used by PHP to denote “magic methods” that are reserved and have a special
purpose in the language. Further, magic methods must be made public. Some magic
methods provide default behavior while others do not. For example, if you do not define",ComputerScienceOne.pdf
"methods provide default behavior while others do not. For example, if you do not define
a constructor method, the default behavior will be to create an object whose member
variables all have null values.
The following constructor allows a user to construct an instance of our Student instance
and specify all four member variables.
1 public function __construct($firstName, $lastName, $id, $gpa) {
2 $this->firstName = $firstName;
3 $this->lastName = $lastName;
4 $this->id = $id;
5 $this->gpa = $gpa;
571",ComputerScienceOne.pdf
"45. Objects
6 }
Though we cannot define multiple constructors, we can use the default value feature
of PHP functions to allow a user to call our constructor with a different number of
parameters. For example,
1 public function __construct($firstName, $lastName,
2 $id = 0, $gpa = 0.0) {
3 $this->firstName = $firstName;
4 $this->lastName = $lastName;
5 $this->id = $id;
6 $this->gpa = $gpa;
7 }
45.4. Usage
Once we have defined our class and its constructors, we can create and use instances of
it. With regular variables, we simply need to assign them to an instance of an object
and their type will dynamically change to match. To create new instances, we invoke a
constructor by using the new keyword and providing arguments to the constructor.
1 $s = new Student(""Alan"", ""Turing"", 1234, 3.4);
2 $t = new Student(""Margaret"", ""Hamilton"", 4321, 3.9);
3 $u = new Student(""John"", ""Smith"");
The process of creating a new instance by invoking a constructor is referred to as",ComputerScienceOne.pdf
"3 $u = new Student(""John"", ""Smith"");
The process of creating a new instance by invoking a constructor is referred to as
instantiation. Once instances have been instantiated, they can be used by invoking their
methods via the same arrow operator we used to access member variables. Outside the
class, however this will only work if the member method is public.
1 print $t->getFormattedName() . ""\n"";
2
3 if($s->getGpa() < $t->getGpa()) {
4 print $t->getFirstName() . "" has a better GPA\n"";
5 }
572",ComputerScienceOne.pdf
"45.5. Common Methods
45.5. Common Methods
Another useful magic method is the __toString() method which returns a string
representation of the object. Unlike the constructor method, there is no default behavior
with the __toString() method. If you do not define this function, it cannot be used
(and any attempts to do so will result in a fatal error). We can define the method to
return the values of all or some of the class’s variables in whatever format we want.
1 public function __toString() {
2 return sprintf(""%s, %s (ID = %d); %.2f"",
3 $this->lastName,
4 $this->firstName,
5 $this->id,
6 $this->gpa);
7 }
This would return a string containing something similar to
""Hamilton, Margaret (ID = 4321); 3.90""
45.6. Composition
Another important concept when designing classes is composition. Composition is a
mechanism by which an object is made up of other objects. One object is said to “own”
an instance of another object.",ComputerScienceOne.pdf
"mechanism by which an object is made up of other objects. One object is said to “own”
an instance of another object.
To illustrate the importance of composition, we could extend the design of our Student
class to include a date of birth. However, a date of birth is also made up of multiple
pieces of data (a year, a month, a date, and maybe even a time and/or locale). We could
design our own date/time class to model this, but its generally best to use what the
language already provides. PHP 5.2 introduced the DateTime object in which there is
a lot of functionality supporting the representation and comparison of dates and time.
We can take this concept further and have our own user-defined classes own instances of
each other. For example, we could define a Course class and then update our Student
class to own a collection of Course objects representing a student’s class schedule
(this type of collection ownership is sometimes referred to as aggregation rather than
composition).",ComputerScienceOne.pdf
"(this type of collection ownership is sometimes referred to as aggregation rather than
composition).
Both of these design updates beg the question: who is responsible for instantiating the
instances of $dateOfBirth and the $schedule? Should we force the “outside” user of
our Student class to build a DateTime instance and pass it to a constructor? Should
we allow the outside code to simply provide us a date of birth as a string and make the
constructor responsible for creating the proper DateTime instance? Do we require that
a user create a complete array of Course instances and provide it to the constructor at
instantiation?
573",ComputerScienceOne.pdf
"45. Objects
A more flexible approach might be to allow the construction of a Student instance
without having to provide a course schedule. Instead, we could add a method that
allowed the outside code to add a course to the student. For example,
1 public function addCourse($c) {
2 $this->schedule[] = $c;
3 }
This adds some flexibility to our object, but removes the immutability property. Design
is always a balance and compromise between competing considerations.
45.7. Example
We present the full and completed Student class in Code Sample 45.1.
574",ComputerScienceOne.pdf
"45.7. Example
1 <?php
2 class Student {
3
4 private $firstName;
5 private $lastName;
6 private $id;
7 private $gpa;
8 private $dateOfBirth;
9 private $schedule;
10
11 public function __construct($firstName, $lastName, $id = 0, $gpa = 0.0,
12 $dateOfBirth = null, $schedule = array()) {
13 $this->firstName = $firstName;
14 $this->lastName = $lastName;
15 $this->id = $id;
16 $this->gpa = $gpa;
17 $this->dateOfBirth = new DateTime($dateOfBirth);
18 $this->schedule = $schedule;
19 }
20
21 public function __toString() {
22 return $this->getFormattedName() . "" born "" .
23 $this->dateOfBirth->format(""Y-m-d"");
24 }
25
26 /**
27 * Returns a formatted String of the Student's
28 * name as Last, First.
29 */
30 public function getFormattedName() {
31 return $this->lastName . "", "" . $this->firstName;
32 }
33
34 /**
35 * Scales the GPA, which is assumed to be on a
36 * 4.0 scale to a percentage.
37 */
38 public function getGpaAsPercentage() {
39 return $this->gpa / 4.0;
40 }
41",ComputerScienceOne.pdf
"36 * 4.0 scale to a percentage.
37 */
38 public function getGpaAsPercentage() {
39 return $this->gpa / 4.0;
40 }
41
42 public function getFirstName() {
43 return $this->firstName;
44 }
45
46 public function getLastName() {
575",ComputerScienceOne.pdf
"45. Objects
47 return $this->lastName;
48 }
49
50 public function getId() {
51 return $this->id;
52 }
53
54 public function getGpa() {
55 return $this->gpa;
56 }
57
58 public function addCourse($c) {
59 $this->schedule[] = $c;
60 }
61
62 }
63
64 ?>
Code Sample 45.1.: The completed PHP Student class.
576",ComputerScienceOne.pdf
"46. Recursion
PHP supports recursion with no special syntax necessary. However, recursion is generally
expensive and iterative or other non-recursive solutions are generally preferred. We
present a few examples to demonstrate how to write recursive functions in PHP.
The first example of a recursive function we gave was the toy count down example. In
PHP it could be implemented as follows.
1 function countDown($n) {
2 if($n===0) {
3 printf(""Happy New Year!\n"");
4 } else {
5 printf(""%d\n"", $n);
6 countDown($n-1);
7 }
8 }
As another example that actually does something useful, consider the following recursive
summation function that takes an array, its size and an index variable. The recursion
works as follows: if the index variable has reached the size of the array, it stops and returns
zero (the base case). Otherwise, it makes a recursive call to recSum(), incrementing
the index variable by 1. When the function returns, it adds the result to the i-th element",ComputerScienceOne.pdf
"the index variable by 1. When the function returns, it adds the result to the i-th element
in the array. To invoke this function we would call it with an initial value of 0 for the
index variable: recSum($arr, 0).
1 function recSum($arr, $i) {
2 if($i === count($arr)) {
3 return 0;
4 } else {
5 return recSum($arr, $i+1) + $arr[$i];
6 }
7 }
This example was not tail-recursive as the recursive call was not the final operation (the
sum was the final operation). To make this function tail recursive, we can carry the
summation through to each function call ensuring that the summation is done prior to
577",ComputerScienceOne.pdf
"46. Recursion
the recursive function call.
1 function recSumTail($arr, $i, $sum) {
2 if($i === count($arr)) {
3 return $sum;
4 } else {
5 return recSumTail($arr, $i+1, $sum + $arr[$i]);
6 }
7 }
As a final example, consider the following PHP implementation of the naive recursive
Fibonacci sequence. An additional condition has been included to check for “invalid”
negative values of n for which we throw an exception.
1 function fibonacci($n) {
2 if($n < 0) {
3 throw new Exception(""Undefined for n<0."");
4 } else if($n <= 1) {
5 return 1;
6 } else {
7 return fibonacci($n-1) + fibonacci($n-2);
8 }
9 }
PHP is not a language that provides implicit memoization. Instead, we need to explicitly
keep track of values using a table. In the following example, the table is passed through
as an argument.
1 function fibonacciMemoization($n, $table) {
2 if($n < 0) {
3 return 0;
4 } else if($n <= 1) {
5 return 1;
6 } else if(isset($table[$n])) {
7 return $table[$n];
8 } else {",ComputerScienceOne.pdf
"2 if($n < 0) {
3 return 0;
4 } else if($n <= 1) {
5 return 1;
6 } else if(isset($table[$n])) {
7 return $table[$n];
8 } else {
9 $a = fibonacciMemoization($n-1, $table);
10 $b = fibonacciMemoization($n-2, $table);
11 $result = ($a + $b);
12 $table[$n] = $result;
13 return $result;
578",ComputerScienceOne.pdf
"14 }
15 }
579",ComputerScienceOne.pdf
"47. Searching & Sorting
PHP provides over a dozen different sorting functions each with different properties and
behavior. Recall that PHP has associative arrays which store elements as key-value
pairs. Some functions sort by keys, others sort by the value (with some in ascending
order, others in descending order). Some functions preserve the key-value mapping
and others do not. For simplicity, we will focus on two of the most common sorting
functions, sort(), which sorts the elements in ascending order and usort() which
sorts according to a comparator function.
47.1. Comparator Functions
Consider a “generic” Quick Sort algorithm as was presented in Algorithm 12.6. The
algorithm itself specifies how to sort elements, but it doesn’t specify how they are ordered.
The difference is subtle but important. Quick Sort needs to know when two elements, a,b
are in order, out of order, or equivalent in order to decide which partition each element",ComputerScienceOne.pdf
"are in order, out of order, or equivalent in order to decide which partition each element
goes in. However, it doesn’t “know” anything about the elements a and b themselves.
They could be numbers, they could be strings, they could be user-defined objects.
A sorting algorithm still needs to be able to determine the proper ordering in order
to sort. In PHP this is achieved through a comparator function, which is a function
that is responsible for comparing two elements and determining their proper order. A
comparator function has the following signature and behavior.
•The function takes two elements $a and $b to be compared.
•The function returns an integer indicating the relative ordering of the two elements:
– It returns something negative if $a comes before $b (that is, a<b )
– It returns zero if $a and $b are equal (a= b)
– It returns something positive if $a comes after $b (that is, a>b )
There is no guarantee on the value’s magnitude, it does not necessarily return −1 or +1;",ComputerScienceOne.pdf
"There is no guarantee on the value’s magnitude, it does not necessarily return −1 or +1;
it just returns something negative or positive. We’ve previously seen this pattern when
comparing strings. The standard PHP library provides a function, strcmp($a, $b)
that has the same contract: it takes two strings and returns something negative, zero or
something positive depending on the lexicographic ordering of the two strings.
The PHP language “knows” how to compare built-in types like numbers and strings.
However, to generalize the comparison operation, we can define comparator functions
581",ComputerScienceOne.pdf
"47. Searching & Sorting
that encapsulate more complex logic. As a simple first example, let’s write a comparator
function that orders numbers in ascending order.
1 function cmpInt($a, $b) {
2 if($a < $b) {
3 return -1;
4 } else if($a === $b) {
5 return 0;
6 } else {
7 return -1;
8 }
9 }
What if we wanted to order integers in the opposite order? We could write another
comparator in which the comparisons or values are reversed. Even simpler, we could
reuse the comparator above and “flip” the sign by multiplying by −1 (this is one of the
purposes of writing functions: code reuse). Even simpler still, we could just flip the
arguments we pass to cmpInt() to reverse the order.
1 function cmpIntDesc($a, $b) {
2 return cmpInt($b, $a);
3 }
To illustrate some more examples, consider the Student class we defined in Code
Sample 45.1. The following code samples demonstrate various ways of ordering Student
instances based on one or more of their components.
1 /**",ComputerScienceOne.pdf
"Sample 45.1. The following code samples demonstrate various ways of ordering Student
instances based on one or more of their components.
1 /**
2 * A comparator function to order Student instances by
3 * last name/first name in alphabetic order
4 */
5 function studentByNameCmp($a, $b) {
6 int result = strcmp($a->getLastName(), $b->getLastName());
7 if(result == 0) {
8 return strcmp($a->getFirstName(), $b->getFirstName());
9 } else {
10 return result;
11 }
12 }
582",ComputerScienceOne.pdf
"47.1. Comparator Functions
1 /**
2 * A comparator function to order Student instances by
3 * last name/first name in reverse alphabetic order
4 */
5 function studentByNameCmpDesc($a, $b) {
6 return studentByNameCmp($b, $a);
7 }
1 /**
2 * A comparator function to order Student instances by
3 * id in ascending numerical order
4 */
5 function studentIdCmp($a, $b) {
6 if($a->getId() < $b->getId()) {
7 return -1;
8 } else if($a->getId() === $b->getId()) {
9 return 0;
10 } else {
11 return 1;
12 }
13 }
1 /**
2 * A comparator function to order Student instances by
3 * GPA in descending order
4 */
5 function studentGpaCmp($a, $b) {
6 if($a->getGpa() > $b->getGpa()) {
7 return -1;
8 } else if($a->getGpa() == $b->getGpa()) {
9 return 0;
10 } else {
11 return 1;
12 }
13 }
583",ComputerScienceOne.pdf
"47. Searching & Sorting
47.1.1. Searching
PHP provides a linear search function, array_search() that can be used to search for
an element in an array. The array can be specified to use loose comparisons (default) or
strict comparisons. It returns the key (i.e. index) of the first matching element it finds
and false if the search was unsuccessful. For example:
1 $arr = array(10, 8, 3, 12, 4, 42, 7, 108);
2 $index = array_search(12, $arr); //index is now 3
3
4 $arr = array(""hello"", 10, ""mixed"", 12, ""20"");
5 //a loose search:
6 $index = array_search(20, $arr); //index is now 4
7 //a strict search:
8 $index = array_search(20, $arr, false); //index is now false.
PHP does not provide a standard binary search function. Though you can write your
own binary search implementation, likely the reason that that PHP does not provide
one is because one is not needed. The purpose of binary search is to search a sorted
array efficiently. However, PHP arrays are not usual arrays: they are associative arrays,",ComputerScienceOne.pdf
"array efficiently. However, PHP arrays are not usual arrays: they are associative arrays,
essentially key-value maps. Retrieving an element via its key is essentially a constant-time
operation, even more efficient that binary search. A better solution may be to simply
store the elements using a proper key which can be used to retrieve the element later on.
47.1.2. Sorting
PHP’s sort() function can be used to sort elements in ascending order. This is useful
if you have arrays of numbers or strings, but doesn’t work very well if you have an array
of mixed types or objects.
1 $arr = array(""banana"", ""Apple"", ""zebra"", ""apple"", ""bananas"");
2 sort($arr);
3 //arr is now (""Apple"", ""apple"", ""banana"", ""bananas"", ""zebra"")
4
5 $arr = array(10, 8, 3, 12, 4, 42, 7, 108);
6 sort($arr);
7 //arr is now (3, 4, 7, 8, 10, 12, 42, 108)
PHP provides a more versatile sorting function, usort() (user defined sort) that
accepts a comparator function that it uses to define the ordering of elements. To pass",ComputerScienceOne.pdf
"accepts a comparator function that it uses to define the ordering of elements. To pass
a comparator function to the usort() function, we pass a string value containing the
584",ComputerScienceOne.pdf
"47.1. Comparator Functions
name of the comparator function we wish to use. Recall that function names in PHP are
case insensitive, though it is still best practice to match the naming. Several examples of
the usage of this function are presented in Code Sample 47.1.
1 $arr = array(10, 8, 3, 12, 4, 42, 7, 108);
2 usort($arr, ""cmpIntDesc"");
3 //arr is now 108, 42, 12, 10, 8, 7, 4, 3
4
5 //roster is an array of Student instances
6 $roster = ...
7
8 //sort by name:
9 usort($roster, ""studentByNameCmp"");
10
11 //sort by ID:
12 usort($roster, ""studentIdCmp"");
13
14 //sort by GPA:
15 usort($roster, ""studentGpaCmp"");
Code Sample 47.1.: Using PHP’s usort() Function
585",ComputerScienceOne.pdf
"Glossary
abstraction a technique for managing complexity whereby levels of complexity are
established so that higher levels do not see or have to worry about details at lower
levels.
acceptance testing a phase of software testing that is tested, usually by humans, for
acceptability and whether or not it fulfills the business requirements.
ad-hoc testing informal software testing method that is done without much planning
usually as a “sanity check” to see if a piece of code works and intended for immediate
feedback.
algorithm a process or method that consists of a specified step-by-step set of operations.
17
anonymous class a class that is defined “inline” without declaring a named class; typi-
cally created because the instance has a single use and there is no reason to create
multiple instances. 484
anonymous function a function that has no identifier or name, typically created so that
it can be passed as an argument to another function as a callback. 144",ComputerScienceOne.pdf
"it can be passed as an argument to another function as a callback. 144
anti-pattern a common software pattern that is used as a solution to recurring problems
that is usually ineffective in solving the problem or introduces risks and other
problems; a technical term for common “bad-habits” that can be found in software.
152
array an ordered collection of pieces of data, usually of the same type.
assignment operator an operator that allows a user to assign a value to a variable. 33
backward compatible a program, code, library, or standard that is compatible with
previous versions so that current and older versions of it can coexist and successfully
operate without breaking anything. 29
bar a placeholder name. , see foo
baz a placeholder name. , see foo
587",ComputerScienceOne.pdf
"Glossary
bike shedding a phenomenon in projects where a disproportionate amount of time is
spent on trivial matters at the expense of spending adequate resources on more
important matters. The term was applied to software development by Kamp [ 24]
but derives from Parkinson’s law of triviality [31].
bit the basic unit of information in a digital computer. A bit can be either 1 or 0
(alternatively, true/false, on/off, high voltage/low voltage, etc.). Originally a
portmanteau (mash up) of binary digit. 4, 22, 23
Boolean a data type that represents the truth value of a logical statement. Booleans
typically have only two values: true or false. 30
bug A flaw or mistake in a computer program that results in incorrect behavior that may
have unintended such as errors or failure. The term predates modern computer
systems but was popularized by Grace Hopper who, when working with the Mark
II computer in 1946 traced a system failure to a moth stuck in a relay. 46, 81, 151",ComputerScienceOne.pdf
"II computer in 1946 traced a system failure to a moth stuck in a relay. 46, 81, 151
byte a unit of information in a digital computer consisting of 8 bits. 4
cache a component or data structure that stores data in an efficiently retrievable manner
so that future requests for the data are fast. 208
call by reference when a variable’s memory address is passed as a parameter to a
function, enabling the function to manipulate the contents of the memory address
and change the original variable’s value. 142
call by value when a copy of a variable’s value is passed as a parameter to a function;
the function has no reference to the original variable and thus changes to the copy
inside the function have no effect on the original variable. 140
callback a function or executable unit of code that is passed as an argument to another
function with the intention that the function that it is passed to will execute or
“call back” the passed function at some point. 144, 366",ComputerScienceOne.pdf
"function with the intention that the function that it is passed to will execute or
“call back” the passed function at some point. 144, 366
cargo cult programming refers to the act of including code, patterns, or processes with
no real purpose due to a lack of understanding of a bug or of the problem they
were attempting to solve or a lack of understanding of the copied code/pattern
itself or an inability to adapt it properly..
case sensitive a language is case sensitive if it recognizes differences between lower and
upper case characters in identifier names. A language is case insensitive if it does
not.. 19
chomp the operation of removing any endline characters from a string (especially when
read from a file); may also refer more generally to removing leading and trailing
whitespace from a string or “trimming” it. 337,
588",ComputerScienceOne.pdf
"Glossary
closure a function with its own environment in which variables exist. 485
code smell a symptom or common pattern in source code that is usually indicative of
a deeper problem or design flaw; smells are usually not bugs and may not cause
problems in and of themselves, but instead indicate a pattern of carelessness or low
quality of software design or implementation. 152
comparator a function or object that allows you to pass in two elements a,b for com-
parison and returns an integer indicating their relative order: something negative,
zero, or something positive if a<b , a= b or a>b respectively. 178, 331, 560
compile the process of translating code in a high-level programming language to a low
level language such as assembly or machine code.
computer engineering a discipline integrating electrical engineering and computer sci-
ence that tends to focus on the development of hardware and its interaction with
software.",ComputerScienceOne.pdf
"ence that tends to focus on the development of hardware and its interaction with
software.
computer science the mathematical modeling and scientific study of computation.
concatenation the process of combining two (or more) strings to create a new string by
appending one of them to the end of the other. 178
constant a variable whose value cannot be changed once set.
continuous integration a software development/engineering process by which developer
code is merged and shared on a frequent basis.
contradiction a logical statement that is always false regardless of the truth values of
the statement’s variables. 72
control flow the order in which individual statements in a program are executed or
evaluated. 75
Conway’s Law an observation that organizations that build software will inevitably
design it as modeling the structure of the organization itself. Attributed to Melvin
Conway (1967) who originally stated it as “organizations which design systems",ComputerScienceOne.pdf
"Conway (1967) who originally stated it as “organizations which design systems
. . . are constrained to produce designs which are copies of the communication
structures of these organizations”.
copy pasta bad coding practice of copy-pasting code and making slight changes when a
function or other control structure should be written instead. Copy-pasta practices
often lead to poorly designed code, bugs or spaghetti code..
cruft anything that is left over, redundant or getting in the way; in the context of code
cruft is code that is no longer needed, legacy or simply poorly written source code.
589",ComputerScienceOne.pdf
"Glossary
dangling pointer when a reference to dynamically allocated memory is lost and the
memory can no longer be deallocated, resulting in a memory leak. Alternatively,
when a reference points to memory that gets deallocated or reallocated but the
pointer remains unmodified, still referencing the deallocated memory. 165
dead code a code segment that has no effect on a program either because it is unused
or unreachable (the conditions involving the code will never be satisfied). 72
debug the process of analyzing a program to find a fault or error with the code that
leads to bad or unexpected results. 152
debugger a software tool that facilitates debugging; usually a debugger simulates the
execution of a program allowing a developer to view the contents of a program as
it executes and to “walk” through the execution step by step. 152
deep copy in contrast to a shallow copy, a deep copy is a copy of an array or other piece",ComputerScienceOne.pdf
"deep copy in contrast to a shallow copy, a deep copy is a copy of an array or other piece
of data that is distinct from the original. Changes to one copy do not affect the
other. 166, 319, 443, 449, 557
defensive programming an approach to programming in which error conditions are
checked and handled, preventing undefined or erroneous operations from happening
in a program. 38, 48, 84
dynamic programming a technique for solving problems that involves iteratively com-
puting values to subproblems, storing them in a table so that they can be used to
solve larger versions of the problem. 208
dynamic typing a variable whose type can change during runtime based on the value it
is assigned. 31,
edge case a situation or input that represents an extreme value or parameter.
encapsulation the grouping and protection of data together into one logical entity along
with the functionality (functions or methods) that act on that data. 30",ComputerScienceOne.pdf
"with the functionality (functions or methods) that act on that data. 30
enumerated type a data type (usually user defined) that consists of a list of named
values. 309, 436
exception an event or occurrence of an erroneous or “exceptional” condition that inter-
rupts the normal flow of control in a program, handing control over to exception
handler(s). 155
expression a combination of values, constants, literals, variables, operators and possibly
function calls such that when evaluated, produce a resulting value. 34
file a resource on a computer stored in memory that holds data. 183
590",ComputerScienceOne.pdf
"Glossary
flowchart a diagram that represents an algorithm or process, showing steps as boxes
connected by arrows which establish an order or flow. 17
foo along with “bar,” “baz,” and the full “foobar”, foo is a commonly used placeholder
name in programming used to denote generic variables, functions, etc. Usually
these terms are used as examples when demonstrating a concept. The terms have
a long history [21].
foobar a placeholder name. , see foo
function a sequence of program instructions that perform a specific task, packaged as a
unit, also known as a subroutine. 133
function overloading the ability to define multiple functions with the same name but
with with a different number of or different types of parameters. 145
fuzzing An automated software testing technique that involves providing invalid or
random data as input to a program or piece of code to test that it fails in an
expected manner..
garbage collection automated memory management in which a garbage collector at-",ComputerScienceOne.pdf
"expected manner..
garbage collection automated memory management in which a garbage collector at-
tempts to reclaim memory (garbage) that is no longer being used by a program so
that it can be reallocated for other purposes. 165, 383
global scope a variable, function, or other element in a program has global scope if it is
visible or has effect throughout the entire program. 32,
grok slang, meaning to understand something; originally from Stranger in a Strange
Land by Robert A. Heinlein..
hardware (computer hardware) the physical components that make up a computer
system such as the processor, motherboard, storage devices, input and output
devices, etc..
heisenbug a bug in a program that fails to manifest itself under testing or debugging
conditions. Derived from the Heisenberg principle which states that you cannot
observe something without affecting it, thus tainting your observations..
hexadecimal base-16 number system using the symbols 0, 1, . . . , 9, A, B, C, D, E, F;",ComputerScienceOne.pdf
"hexadecimal base-16 number system using the symbols 0, 1, . . . , 9, A, B, C, D, E, F;
usually denoted with a prefix 0x such as 0xff1321ab01. 470
hoisting usually used in interpreted languages, hoisting involves processing code to find
variable or function declarations and processing them before actually executing the
code or script. 134
identifier a symbol, token, or label that is used to refer to a variable. Essentially, a
variable’s name. 18
591",ComputerScienceOne.pdf
"Glossary
idiom in the context of programming, an idiom is a commonly used pattern, expression
or way of structuring code that is well-understood for users of the language. For
example, a for-loop structure that iterates over elements in an array. May also refer
to a programming design pattern.. 315,
immutable an object whose internal state cannot be changed once created, alternatively,
one whose internal state cannot be observably changed once created. 390, 423, 449,
466, 571
inheritance an object oriented programming principle that allows you to derive an object
from another object, usually to allow for more specificity.
input data or information that is provided to a computer program for processing. 41
integration testing a type of software testing used to determine if the integration of
different components or modules works or not.
interactive a program that is designed to interface with humans by prompting them for
input and displaying output directly to them. 41",ComputerScienceOne.pdf
"interactive a program that is designed to interface with humans by prompting them for
input and displaying output directly to them. 41
interactive an informal, abstract, high-level description of a process or algorithm. 41
keyword a word in a programming language with a special meaning in a particular
context. In contrast to a reserved word, a keyword may be used for an identifier
(variable or function name) but it is strongly discouraged to do so as the keyword
already has an intended meaning.
kilobyte a unit of information in a digital computer consisting of 1024 bytes (equivalently,
210 bytes), KB for short.
kludge a poorly designed or “thrown-together” solution; a design that is a collection of
ill-fitting parts that may be functional, but is fragile and not easily maintained.
lambda expression another term for anonymous functions. 144, 493
lexicographic a generalization of the usual dictionary order as codified with the ASCII
character table. 178",ComputerScienceOne.pdf
"lexicographic a generalization of the usual dictionary order as codified with the ASCII
character table. 178
linking the process of generating an executable file from (multiple) object files.
lint (or linter) a static code analysis tool that analyzes code for suspicious or error-prone
code that is likely to cause problems. 152
literal in a programming language, a literal is notation for specifying a value such as a
number of string that can be directly assigned to a variable. 29, 34
load testing a form of performance testing used to determine to what extent a piece of
software or a system can handle a high demand or “load”.
592",ComputerScienceOne.pdf
"Glossary
magic number a value used in a program with unexplained, undocumented, or am-
biguous meaning, usually making the code less understandable. 258, 309, 310,
437
mantissa the part of a floating-point number consisting of its significant digits (called a
significand in scientific notation). 25
map a data structure that allows you to store elements as key-value pairs with the key
mapping to a value.
memoization a technique which uses a table to store previously computed values of a
function so that they do not need to be recomputed, essentially the table serves as
a cache. 208
memory leak the gradual loss of memory when a program fails to deallocate or free up
unused memory, degrading performance and possibly resulting in the termination
of the program when memory runs out. 165, 320, 326
naming convention a set of guidelines for choosing identifier names for variables, func-
tions, etc. in a programming language. Conventions may be generally accepted",ComputerScienceOne.pdf
"tions, etc. in a programming language. Conventions may be generally accepted
by all developers of a particular language or they may be established for use in a
particular library, framework, or organization. 20
network a collection of two or more computer systems linked together through a physical
connection over which data can be transmitted using some protocol.
octal base-8 number system using the symbols 0, 1, 2, . . . , 6, 7; usually denoted with a
single leading zero such as 0123742.
open recursion a mechanism by which an object is able to refer to itself usually using a
keyword such as this or self.. 200, 465, 569
operand the arguments that an operator applies to. 33
operator a symbol used to denote some transformation that combines or changes the
operands it is applied to to produce a new value. 33
order of precedence the order in which operators are evaluated, multiplication is per-
formed before addition for example. 38",ComputerScienceOne.pdf
"order of precedence the order in which operators are evaluated, multiplication is per-
formed before addition for example. 38
output data or information that is produced as the result of the execution of a program.
41
overflow when an arithmetic operation results in a number that is larger than the
specified type can represent overflow occurs resulting in an invalid result. 39
parse to process data to identify its individual components or elements. 179,
593",ComputerScienceOne.pdf
"Glossary
persistence the characteristic of data that outlives the process or program that created
it; the saving of data across multiple runs of a program. 183
pointer a reference to a particular memory location in a computer. 30, 295
polymorphism an object oriented programming concept that allows you to treat a
variable, method, or object as different types.
primitive a basic data type that is defined and provided by a programming language.
Typically numeric and character types are primitive types in a language for example.
Generally, the user doesn’t need to define the operations involving primitive types
as they are defined by the language. Primitive data types are used as the basic
building blocks in a program and used to compose more complex user-defined types.
30, 390
procedural abstraction the concept that a procedure or sequence of operations can be
encapsulated into one logical unit (function, subroutine, etc.) so that a user need",ComputerScienceOne.pdf
"encapsulated into one logical unit (function, subroutine, etc.) so that a user need
not concern themselves with the low-level details of how it operates. 134
program stack also referred to as a call stack , it is an area of memory where stack
frames are stored for each function call containing memory for arguments, local
variables and return values/addresses. 137
protocol a set of rules or procedures that define how communication takes place.
pseudocode the act of a program asking a user to enter input and subsequently waiting
for the user to enter data. 13
query an operation that retrieves data, typically from a database. 152
queue a data structure that store elements in a FIFO (First-In First-Out) manner;
elements can be added to the end of a queue by an enqueue operation and removed
from the start of a queue by a dequeue operation.
radix the base of a number system. Binary, octal, decimal, hexadecimal would be base
2, 8, 10, and 16 respectively.",ComputerScienceOne.pdf
"radix the base of a number system. Binary, octal, decimal, hexadecimal would be base
2, 8, 10, and 16 respectively.
reentrant a function that can be interrupted during its execution while another thread
can successfully invoke the function without the two functions interfering with the
data used in either function call. 334
refactor the process of modifying, updating or restructuring code without changing
its external behavior; refactoring may be done to make code more efficient, more
readable, more reliable, or simply to bring it into compliance with style or coding
conventions.
reference a reference in a computer program is a variable that refers to an object or
function in memory. 30
594",ComputerScienceOne.pdf
"Glossary
regression testing a type of software testing that tests software that previously passed
testing but that was changed or refactored in some way. Regression testing tests
to see if the software still passes the tests or not in which case it is said to have
“regressed”.
regular expression a sequence of characters in which special characters and directives
can be used to define a complex pattern that can be searched and matched in
another string or data. 455, 561
reserved word a word or identifier in a language that has a special meaning to the
syntax of the language and therefore cannot be used as an identifier in variables,
functions, etc.. 19
scope the scope of a variable, method, or other entity in a program is the part of the
program in which the name or reference of the entity is bound. That is, the part of
the program that “knows” about the variable in which the variable can be accessed,
changed, or used. 32, 255, 386",ComputerScienceOne.pdf
"the program that “knows” about the variable in which the variable can be accessed,
changed, or used. 32, 255, 386
segmentation fault a fault or error that arises when a program attempts to access a
segment of memory that it is not allowed access to, usually resulting in the program
being terminated by the operating system. 296, 315
shallow copy in contrast to a deep copy, a shallow copy is merely a reference that refers
to the original array or piece of data. The two references point to the same data,
so if the data is modified, both references will realize it.. 166, 319, 443
short circuiting the process by which the second operand in a logical statement is not
evaluated if the value of the expression is determined by the first operand. 74
signature a function signature is how a function is uniquely identified. A signature
includes the name (identifier) of the function, its parameter list (and maybe types)
and the return type. 134",ComputerScienceOne.pdf
"includes the name (identifier) of the function, its parameter list (and maybe types)
and the return type. 134
software any set of machine-readable instructions that can be executed in a computer
processor.
software engineering the study and application of engineering principles to the design,
development, and maintenance of complex software systems.
spaghetti code a negative term used for code that is overly complex, disorganized or
unstructured code.
stack a data structure that stores elements in a LIFO (last-in first-out) manner; elements
can be added to a stack via a push operation which places the element on the “top”
of the stack; elements can be removed from the top of the stack via a pop operation.
137, 206
595",ComputerScienceOne.pdf
"Glossary
stack overflow when a program runs out of stack space, it may result in a stack overflow
and the termination of the program. 205
static analysis the analysis of software that is performed on source (or object) code
without actually running or compiling a program usually by using an automated
tool that can detect actual or potential problems with the source code (other than
syntactic problems that could easily be found by a compiler). 152
static dispatch when function overloading is supported in a language, this is the mecha-
nism by which the compiler determines which function should be called based on
the number and type of arguments passed to the function when it is called. 145
static typing a variable whose type is specified when it is created (declared) and does
not change while the variable remains in scope. 31,
string a data type that consists of a sequence of characters which are encoded under
some encoding standard such as ASCII or Unicode. 27, 177",ComputerScienceOne.pdf
"string a data type that consists of a sequence of characters which are encoded under
some encoding standard such as ASCII or Unicode. 27, 177
string concatenation an operation by which a string and another data type are combined
to form a new string. 37
syntactic sugar syntax in a language or program that is not absolutely necessary (that
is, the same thing can be achieved using other syntax), but may be shorter, more
convenient, or easier to read/write. In general, such syntax makes the language
“sweeter” for the humans reading and writing it. 40, 82, 102, 161, 451
tautology a logical statement that is always true regardless of the truth values of the
statement’s variables. 72
test case an input/output pair that is known to be correct that is to be used to test a
unit of software.
test case a collection of tests that are used to test a piece or collection of software.
token when something (usually a string) is parsed, the individual components or elements",ComputerScienceOne.pdf
"token when something (usually a string) is parsed, the individual components or elements
are referred to as tokens. 179
top-down design an approach to problem solving where a problem is broken down into
smaller parts. 3, 133
transpile to (automatically) translate code in one programming language into code
in another programming language, usually between two high-level programming
languages.
truncation removing the fractional part of a floating-point number to make it an integer.
Truncation is not a rounding operation. 36, 263, 394,
596",ComputerScienceOne.pdf
"Glossary
two’s complement A way of representing signed (positive and negative) integers using
the first bit as a sign bit (0 for positive, 1 for negative) and where negative numbers
are represented as the complement with respect to 2 n (the result of subtracting the
number from 2n) . 24
type a variable’s type is the classification of the data it represents which could be
numeric, string, boolean, or a user defined type. 22
type casting converting or variable’s type into another type, for example, converting an
integer into a more general floating-point number, or converting a floating-point
number into an integer, truncating and losing the fractional part. 36, 263, 394
underflow when an arithmetic operation involving floating-point numbers results in a
number that is smaller than the smallest representable number underflow occurs
resulting in an invalid result. 39
Unicode an international character encoding standard used in programming languages
and data formats. 449",ComputerScienceOne.pdf
"resulting in an invalid result. 39
Unicode an international character encoding standard used in programming languages
and data formats. 449
unit test software testing method that tests an individual “unit” of code; depending on
the context, a unit may refer to an individual function, class, or module.
unwinding the process of removing a stack frame when returning from a function. 164
validation the process of verifying that data is correct or conforms to certain expectations
including formatting, type, range of values, represents a valid value, etc.. 42
variable a memory location which stores a value that may be set using an assignment
operator. Typically a variable is referred to using a name or identifier. 18
widget a generic term for a graphical user interface component such as a button or text
box.
yak shaving an ostensibly pointless activity that despite its uselessness allows you to
overcome a a difficulty and solve a larger problem. Basically, screwing around",ComputerScienceOne.pdf
"overcome a a difficulty and solve a larger problem. Basically, screwing around
until you become familiar enough with the problem that the solution presents itself.
Attributed to Carlin Vieri who took it from an episode of Ren and Stimpy ..
597",ComputerScienceOne.pdf
"Acronyms
ACID Atomicity Consistency Isolation Durability.
ACM Association for Computing Machinery.
ALU Arithmetic and Logic Unit. 4
ANSI American National Standards Institute. 253
API Application Programmer Interface. 15, 184, 436
ASCII American Standard Code for Information Interchange. 27, 66, 183, 187, 260, 325,
332, 385, 391, 454, 561
BYOD Bring Your Own Device.
CE Computer Engineering.
CI Continuous Integration.
CJK Chinese Japanese Korean. 29
CLA Command Line Arguments.
CLI Command Line Interface. 46
CMS Content Management System. 497
CMYK Cyan-Magenta-Yellow-Key.
CPU Central Processing Unit. 4, 74
CRUD Create Retrieve Update Destroy.
CS Computer Science.
CSS Cascading Style Sheets.
CSV Comma Separated Values. 179, 337
CWD Current Working Directory. 184
CYA Cover Your Ass.
599",ComputerScienceOne.pdf
"Acronyms
DRY Don’t Repeat Yourself. 292
EB Exabyte.
ECMA European Computer Manufacturers Association.
EDI Electronic Data Interchange.
EOF End Of File. 184
FIFO First-In First-Out.
FOSS Free and Open Source Software.
FUD Fear Uncertainty Doubt.
GB Gigabyte.
GCC GNU Compiler Collection.
GDB GNU Debugger.
GIF Graphics Interchange Format.
GIGO Garbage-In Garbage-Out.
GIMP GNU Image Manipulation Program.
GIS Geographic Information System.
GNU GNU’s Not Unix!. 49, 324
GUI Graphical User Interface. 42, 144, 366
HTML HyperText Markup Language. 181, 184, 497, 565
IDE Integrated Development Environment. 7, 21, 46, 49, 258, 385, 390, 471, 502
IEC International Electrotechnical Commission. 26, 253
IEEE Institute of Electrical and Electronics Engineers. 26, 260, 306, 503
IP Internet Protocol. 201
ISO International Organization for Standardization. 253, 343
JDBC Java Database Connectivity.
JDK Java Development Kit. 383, 385, 388, 427, 431, 436, 449, 458, 487
600",ComputerScienceOne.pdf
"Acronyms
JEE Java Enterprise Edition.
JIT Just In Time. 12
JPEG Joint Photographic Experts Group. 183
JRE Java Runtime Environment.
JSON JavaScript Object Notation. 185, 565
JVM Java Virtual Machine. 12, 383, 384, 389, 423, 431, 435, 440, 442, 449, 467, 470,
484
KB Kilobyte. 162, 460
KISS Keep It Simple, Stupid.
LIFO Last-In First-Out. 137
LOC Lines Of Code.
MAC Media Access Control. 201
MB Megabyte. 162
MP3 MPEG-2 Audio Layer III.
MPEG Moving Picture Experts Group.
MVP Minimum Viable Product.
MWE Minimum Working Example. 254
NIST National Institute of Standards and Technology.
ODBC Open Database Connectivity.
OEM Original Equipment Manufacturer.
OOP Object-Oriented Programming. 177, 384, 386, 433, 444, 499
PB Petabyte.
PDF Portable Document Format. 15
PHP PHP: Hypertext Preprocessor (a recursive backronym; used to stand for Personal
Home Page).
PNG Portable Network Graphics.
601",ComputerScienceOne.pdf
"Acronyms
POJO Plain Old Java Object.
POSIX Portable Operating System Interface. 259, 306, 336, 343
RAM Random Access Memory. 5
REPL Read-Eval-Print Loop.
RGB Red-Green-Blue.
ROM Read-Only Memory.
RTFM Read The “Freaking” Manual.
RTM Read The Manual.
SD Software Development.
SDK Software Development Kit. 443, 444
SE Software Engineering.
SEO Search Engine Optimization.
SQL Structured Query Language. 431, 436
SSL Secure Sockets Layer. 296
STEAM Science, Technology, Engineering, Art, and Math.
STEM Science, Technology, Engineering, and Math.
TB Terabyte.
TCP Transmission Control Protocol.
TDD Test-Driven Development.
TSV Tab Separated Values. 179
TUI Text User Interface.
UI User Interface.
UML Unified Modeling Language.
URL Uniform Resource Locator.
UTC Coordinated Universal Time. 343
UTF-8 Universal (Character Set) Transformation Format–8-bit.
602",ComputerScienceOne.pdf
"Acronyms
UX User Experience.
VLSI Very Large Scale Integration. 4
W3C World Wide Web Consortium.
WWW World Wide Web. 181, 383, 497
XML Extensible Markup Language. 21, 185, 565
YAGNI You Ain’t Gonna Need It.
603",ComputerScienceOne.pdf
"Index
aggregation, 199
algorithm, 215
anonymous class, 484
arrays, 159
in C, 313
in Java, 439
in PHP, 547
indexing, 160
iteration, 161
multidimensional, 166
static, see static array160
arrow operator, 347
assignment operator, 33
associative arrays, 168
bandwidth, 5
basic input, 41
basic output, 41
binary, 4, 23
counting, 23
binary search, 213
analysis, 218
C, 373
bit, 4
block, see code block
Boolean, 30
bottom-up design, 3, 200
buffered, 187
bug, 151
byte, 4
C
arrays, 313
comparators, 361
conditionals, 271
error handling, 305
file I/O, 335
functions, 291
loops, 283
recursion, 357
strings, 325
cache, 208
call by reference, 140
call by value, 140
call stack, 7
callback, 144, 366
calling functions, see invoking functions
case sensitive, 19
class, 198, 386
class-based, 198
code block, 13
command line arguments, 44
comment, 14
comparator, 178, 238, 239, 361
in Java, 483
comparison operators, 66
compiled language, 7
composition, 199, 344, 472
compound logic, 71
computer, 4
conditionals, 65",ComputerScienceOne.pdf
"in Java, 483
comparison operators, 66
compiled language, 7
composition, 199, 344, 472
compound logic, 71
computer, 4
conditionals, 65
if statement, 75
if-else statement, 76
if-else-if statement, 78
in C, 271
in Java, 403
in PHP, 515
conjunction, see logical operators–and
constants, 33
605",ComputerScienceOne.pdf
"Index
constructor, 199, 467
contradiction, 72
control flow, 17
copy constructor, 468
De Morgan’s laws, 72
debugger, 152
debugging, 46, 49
deep copy, 166, 353
defensive programming, 48, 84, 153
disjunction, see logical operators–or
do-while loop, 100
dot operator, 346
dynamic memory, 164
dynamic programming, 208
dynamic typing, 31
elementary operation, 215
encapsulation, 30, 197
C, 341
enumerated type, 30
enumerated types
in Java, 436
error code, 154
error handling, 151
exception, 155
in C, 305
in Java, 431
in PHP, 543
exception, 155
fatal error, 153
file, 183
file I/O, 183
in C, 335
in Java, 457
in PHP, 563
first-class citizen, 142
floating point number, 25
flowchart, 17
for loop, 99
foreach loop, 102, 161
function hoisting, 535
function pointers, 365
function prototype, 291
functional programming, 142
functions, 133
in C, 291
in Java, 423
in PHP, 535
garbage collection, 165
Gauss’s Formula, 222
generic programming, 212
global scope, 32
graphical user interface, 42
hardware, 4
heap, 160",ComputerScienceOne.pdf
"in PHP, 535
garbage collection, 165
Gauss’s Formula, 222
generic programming, 212
global scope, 32
graphical user interface, 42
hardware, 4
heap, 160
Heap Sort, 237
Hello World
C, 253
hello world
Java, 384
hexadecimal, 5
hoisting, 134, 535
identifier, 18
if statement, 75
if-else statement, 76
if-else-if statement, 78
immutable, 423
index, 160
infinite loop, 97, 103
Insertion Sort, 224
Integrated Development Environment, 7
interpreted language, 7
invoking functions, 136
Java
arrays, 439
Comparators, 483
conditionals, 403
enumerated types, 436
error handling, 431
file I/O, 457
loops, 415
methods, 423
objects, 461
recursion, 479
606",ComputerScienceOne.pdf
"Index
strings, 449
kilobyte, 4
lambda expression, 493
linear search, 212
C, 372
linked list, 167
linter, 152
list
in Java, 444
lists, 167
literal, 34
logic errors, 48
logical operators, 65
and, 69
negation, 68
or, 70
loops, 95
do-while loop, 100
for loop, 99
foreach loop, 102
in C, 283
in Java, 415
in PHP, 527
infinite loop, 103
while loop, 97
lower camel casing, 20
main memory, 5
mantissa, 25
maps, 168
member methods, 198
member variables, 198
memoization, 208, 358
memory, 5
memory leak, 165, 326
memory management, 165
Merge Sort, 232
method overloading, 423
multidimensional arrays, 166, 443
natural order, 242
nested loops, 103
nesting, 84
null, 30
object, 30, 197
objects, 197
defining, 198
in Java, 461
using, 200
open recursion, 200, 465
in PHP, 569
operator, 33
logical, 65
operators
addition, 35
arithmetic, 35
assignment, see asignment opera-
tor33
C, 262
compound assignment, 40
division, 35
increment, 40
integer division, 37
multiplication, 35
negation, 35
string concatenation, 37",ComputerScienceOne.pdf
"tor33
C, 262
compound assignment, 40
division, 35
increment, 40
integer division, 37
multiplication, 35
negation, 35
string concatenation, 37
subtraction, 35
optional parameters, 145
order of precedence, 38
logic, 73
overflow, 39
overloading, 483
parse, 179
pass by reference, 297, 423
pass by value, 423
paths, 184
absolute, 184
relative, 184
persistence, 183
PHP
arrays, 547
conditionals, 515
error handling, 543
file I/O, 563
functions, 535
607",ComputerScienceOne.pdf
"Index
loops, 527
magic method, 571
recursion, 577
strings, 557
pointers, 295
pollute the namespace, 33
polymorphism, 374
preprocessor directive, 255
primitive types, 30
printf, 43
problem solving, 2
procedural abstraction, 134, 301
program stack, 137
pseudocode, 13, 50
Quick Sort, 227
recursion, 203
C, 357
in Java, 479
in PHP, 577
reentrant, 334
regression testing, 48
runtime errors, 47
scope, 32, 134
global, 137
local, 136
search
binary search, 213
linear search, 212
search algorithm
analysis, 215
searching, 211
segmentation fault, 264, 296
Selection Sort, 221
analysis, 222
sequential control flow, 13
set, 167
in Java, 446
sets, 167
shallow copy, 166, 353
short circuiting, 74
software, 4
software testing, 48
Sorting
C, 374
sorting, 211, 220
Heap Sort, 237
Insertion Sort, 224
Merge Sort, 232
Quick Sort, 227
Selection Sort, 221
stability, 243
Tim Sort, 237
stack overflow, 162, 345
standard input, 42
standard output, 42
static array, 160
static typing, 31
static vs dynamic memory, 162",ComputerScienceOne.pdf
"Tim Sort, 237
stack overflow, 162, 345
standard input, 42
standard output, 42
static array, 160
static typing, 31
static vs dynamic memory, 162
strings, 27, 177
comparison, 178
in C, 325
in Java, 449
in PHP, 557
tokenizing, 179
structures, 341
arrays, 348
defining, 341
style guide, 20, 22
syntactic sugar, 40
syntax error, 13, 47
tail recursion, 205
tautology, 72
ternary if-else operator, 82
test case, 49
test cases, 48
testing, 3
Tim Sort, 237
top-down design, 3, 133
transpiler, 12
truncation, 36
try-with-resources, 458
two’s complement, 24
type casting, 36, 263
type juggling, 505
unbuffered, 187
608",ComputerScienceOne.pdf
"Index
underflow, 39
underscore casing, 20
unicode, 29
upper camel casing, 20
use case, 3
vararg function, 145, 424
variable, 18
identifier, 18
naming conventions, 19
naming rules, 19
scope, 32
types, 22
variable argument function, see vararg
function
variables
C, 260
visibility, 197, 424, 462
while loop, 97
609",ComputerScienceOne.pdf
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613",ComputerScienceOne.pdf
"DC Electrical Circuit AnalysisDC Electrical Circuit Analysis
A Practical ApproachA Practical Approach
James M. FioreJames M. Fiore",DCElectricalCircuitAnalysis.pdf
2,DCElectricalCircuitAnalysis.pdf
"DC Electrical Circuit AnalysisDC Electrical Circuit Analysis
A Practical ApproachA Practical Approach
 
by
 
James M. Fiore
 
 
   
Version 1.0.11, 11 January 2024
3",DCElectricalCircuitAnalysis.pdf
"This DC Electrical Circuit Analysis, by James M. Fiore is copyrighted under the terms of a 
Creative Commons license: 
 
  
This work is freely redistributable for non-commercial use, share-alike with attribution
  Device data sheets and other product information are copyright by their respective owners
and have been obtained through publicly accessible manufacturer's web sites.
Published by James M. Fiore via dissidents 
 
ISBN13: 978-1654515478 
   
For more information or feedback, contact:
James Fiore, Professor
Electrical Engineering Technology
Mohawk Valley Community College
1101 Sherman Drive
Utica, NY 13501
jfiore@mvcc.edu
oer@jimfiore.org
For the latest revisions, related titles, and links to low cost print versions, go to: 
www.mvcc.edu/jfiore or my mirror sites www.dissidents.com and www.jimfiore.org
YouTube Channel: Electronics with Professor Fiore
Cover art, Chapman's Contribution, by the author 
4",DCElectricalCircuitAnalysis.pdf
"PrefacePreface
Welcome to DC Electrical Circuit Analysis, an open educational resource (OER). The goal of this text is to 
introduce the theory and practical application of analysis of DC electrical circuits. It is offered free of charge 
under a Creative Commons non-commercial, share-alike with attribution license. For your convenience, along 
with the free pdf and odt files, print copies are available at a very modest charge. Check my web sites for links.
This text is based on the earlier Workbook for DC Electrical Circuits, which it replaces. The original expository 
text has been greatly expanded and includes many examples along with computer simulations. For the 
convenience of those who used the Workbook, many of the problem sets are the same, with some re-ordering 
depending on the chapter. 
The text begins with coverage of scientific and engineering notation along with the metric system. Also included",DCElectricalCircuitAnalysis.pdf
"depending on the chapter. 
The text begins with coverage of scientific and engineering notation along with the metric system. Also included
is a discussion of the scientific method, the underpinning of our entire system of investigation and technology. 
From there, basic concepts and quantities are introduced such as charge, current, energy, power and voltage. 
Subsequent chapters introduce resistance, series circuits, parallel circuits and series-parallel circuits. The text 
continues with chapters covering network theorems, more advanced techniques such as nodal and mesh analysis,
and finally finishes with introductions to capacitors, inductors and magnetic circuits. The companion AC 
Electrical Circuit Analysis text picks up after this point. Each chapter begins with a set of learning objectives 
and concludes with practice exercises that are generally divided into four major types: analysis, design,",DCElectricalCircuitAnalysis.pdf
"and concludes with practice exercises that are generally divided into four major types: analysis, design, 
challenge and simulation. Many SPICE-based circuit simulators are available, both free and commercial, that can
be used with this text. The answers to most odd-numbered exercises can be found in the Appendix. A table of 
standard resistor sizes is also in the Appendix, which is useful for real-world design problems. Finally, the 
Appendix includes a section reviewing simultaneous equation solutions. If you have any questions regarding 
this text, or are interested in contributing to the project, do not hesitate to contact me. 
This text is part of a series of OER titles in the areas of electricity, electronics, audio and computer 
programming. It includes three other textbooks covering semiconductor devices, operational amplifiers, and 
embedded programming using the C language with the Arduino platform. There is a text covering AC electrical",DCElectricalCircuitAnalysis.pdf
"embedded programming using the C language with the Arduino platform. There is a text covering AC electrical 
circuits similar to this one, and also seven laboratory manuals; one for each of the five texts plus individual titles
covering computer programming using the Python language, and the science of sound. The most recent versions 
of all of my OER texts and manuals may be found at my MVCC   web site as well as my mirror site: 
www.dissidents.com 
This text was created using several free and open software applications including Open Office, Dia, SciDA Vis, 
and XnView.
Special thanks to the following individuals for their efforts in reviewing and proofreading the DC and AC 
Electrical Circuit Analysis texts: Glenn Ballard, John Markham, João Nuno Carvalho, Mark Steffka and Jim 
Noon.
5",DCElectricalCircuitAnalysis.pdf
"For Those Without and Those Within

“All we can do...is to run over several instances, and examine carefully the principle, 
which binds the different thoughts to each other, 
never stopping till we render the principle as general as possible.”
— David Hume, An Enquiry Concerning Human Understanding (1748)
6",DCElectricalCircuitAnalysis.pdf
"Table of ContentsTable of Contents
Chapter 1: Fundamentals . . . . . . . 10
1.0 Chapter Objectives . . . . . . . .  10
1.1 Introduction . . . . . . . . .  10
1.2 Significant Digits and Resolution . . . . . .  11
1.3 Scientific and Engineering Notation . . . . . .  14
1.4 The Metric System . . . . . . . .  16
1.5 The Scientific Method. . . . . . . .  18
1.6 Critical Thinking . . . . . . . .  21
1.7 RoHS . . . . . . . . .  24
Summary . . . . . . . . .  25
Exercises . . . . . . . . .  26
Chapter 2: Basic Quantities . . . . . . 30
2.0 Chapter Objectives . . . . . . . .  30
2.1 Introduction . . . . . . . . .  30
2.2 An Atomic Model . . . . . . . .  30
2.3 Charge and Current . . . . . . . .  34
2.4 Energy and Voltage . . . . . . . .  36
2.5 Power and Efficiency . . . . . . . .  39
2.6 Energy Cost and Battery Life . . . . . . .  42
2.7 Resistance and Conductance . . . . . . .  47
2.8 Instrumentation and Laboratory . . . . . .  63
Summary . . . . . . . . .  67
Exercises . . . . . . . . .  68",DCElectricalCircuitAnalysis.pdf
"2.8 Instrumentation and Laboratory . . . . . .  63
Summary . . . . . . . . .  67
Exercises . . . . . . . . .  68
Chapter 3: Series Resistive Circuits . . . . .  74
3.0 Chapter Objectives . . . . . . . .   74
3.1 Introduction . . . . . . . . .   74
3.2 Conventional Current Flow and Electron Flow . . . . .   74
3.3 The Series Connection . . . . . . .   75
3.4 Combining Series Components . . . . . .   77
3.5 Ohm's Law . . . . . . . . .   78
3.6 Kirchhoff's Voltage Law . . . . . . .   82
3.7 Series Analysis . . . . . . . .   83
3.8 Potentiometers and Rheostats . . . . . .   92
Summary . . . . . . . . .   96
Exercises . . . . . . . . .   97
7",DCElectricalCircuitAnalysis.pdf
"Chapter 4: Parallel Resistive Circuits . . . . . 110
4.0 Chapter Objectives . . . . . . . .  110
4.1 Introduction . . . . . . . . .  110
4.2 The Parallel Connection . . . . . . .  111 
4.3 Combining Parallel Components . . . . . .  111
4.4 Kirchhoff's Current Law . . . . . . .  115
4.5 Parallel Analysis . . . . . . . .  117
4.6 Current Limiting: Fuses and Circuit Breakers . . . . .  125
Summary . . . . . . . . .  127
Exercises . . . . . . . . .  128
Chapter 5: Series-Parallel Resistive Circuits . . . . 138
5.0 Chapter Objectives . . . . . . . .  138
5.1 Introduction . . . . . . . . .  138
5.2 Series-Parallel Connections . . . . . . .  139
5.3 Simplifying Series-Parallel Components . . . . .  139
5.4 Series-Parallel Analysis . . . . . . .  143
Summary . . . . . . . . .  155
Exercises . . . . . . . . .  157
Chapter 6: Analysis Theorems and Techniques . . . . 170
6.0 Chapter Objectives . . . . . . . .  170
6.1 Introduction . . . . . . . . .  170
6.2 Source Conversions . . . . . . . .  171",DCElectricalCircuitAnalysis.pdf
"6.0 Chapter Objectives . . . . . . . .  170
6.1 Introduction . . . . . . . . .  170
6.2 Source Conversions . . . . . . . .  171
6.3 Superposition Theorem . . . . . . .  177
6.4 Thévenin's Theorem . . . . . . . .  181
6.5 Norton's Theorem . . . . . . . .  187
6.6 Maximum Power Transfer Theorem . . . . . .  188
6.7 Delta-Y Conversions . . . . . . . .  191
Summary . . . . . . . . .  197
Exercises . . . . . . . . .  199
Chapter 7: Nodal & Mesh Analysis, Dependent Sources . . 214
7.0 Chapter Objectives . . . . . . . .  214
7.1 Introduction . . . . . . . . .  214
7.2 Nodal Analysis . . . . . . . .  215
7.3 Mesh Analysis . . . . . . . .  229
7.4 Dependent Sources . . . . . . . .  238
Summary . . . . . . . . .  243
Exercises . . . . . . . . .  244
8",DCElectricalCircuitAnalysis.pdf
"Chapter 8: Capacitors . . .. .. .. .. .. 260
8.0 Chapter Objectives . . . . . . . .  260
8.1 Introduction . . . . . . . . .  260
8.2 Capacitance and Capacitors . . . . . . .  261
8.3 Initial and Steady-State Analysis of RC Circuits . . . . .  272
8.4 Transient Response of RC Circuits . . . . . .  274
Summary . . . . . . . . .  285
Exercises . . . . . . . . .  287
Chapter 9: Inductors . . .. .. .. .. .. 294
9.0 Chapter Objectives . . . . . . . .  294
9.1 Introduction . . . . . . . . .  294
9.2 Inductance and Inductors . . . . . . .  295
9.3 Initial and Steady-State Analysis of RL Circuits . . . . .  303
9.4 Initial and Steady-State Analysis of RLC Circuits . . . . .  306
9.5 Transient Response of RL Circuits . . . . . .  307
Summary . . . . . . . . .  317
Exercises . . . . . . . . .  318
Chapter 10: Magnetic Circuits and Transformers .. .. .. 324
10.0 Chapter Objectives . . . . . . . .  324
10.1 Introduction . . . . . . . .  324
10.2 Electromagnetic Induction. . . . . . . .  325",DCElectricalCircuitAnalysis.pdf
"10.0 Chapter Objectives . . . . . . . .  324
10.1 Introduction . . . . . . . .  324
10.2 Electromagnetic Induction. . . . . . . .  325
10.3 Magnetic Circuits . . . . . . . .  329
10.4 Transformers . . . . . . . .  342
Summary . . . . . . . . .  348
Exercises . . . . . . . . .  350
Appendices
A: Standard Component Sizes . . . . . . . 355
B: Methods of Solution of Linear Simultaneous Equations . . . 356
C: Equation Proofs . . . . . . . . 361
D: Answers to Selected Odd-Numbered Problems . . . . 363
E: Base Units . . . . . . . . 373
F: Appropriate Commentary . . . . . . . 374
9",DCElectricalCircuitAnalysis.pdf
"1 1 FundamentalsFundamentals
1.0 Chapter Learning Objectives1.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe significant digits and resolution.
• Express and compute numeric values using scientific and engineering notation.
• Describe the metric system and detail its advantages.
• Define the scientific method.
• Give examples of cognitive biases and logical fallacies that affect critical thinking.
• Describe the RoHS directive.
1.1 Introduction 1.1 Introduction 
A Little Background and PerspectiveA Little Background and Perspective
This text focuses on the analysis of DC (direct current) electrical circuits. It assumes no prior knowledge of 
electrical quantities, systems or circuit theory. As with any new endeavor, it is important to define the 
terminology and tools to be used at the outset. We shall be examining the basic electrical quantities, their",DCElectricalCircuitAnalysis.pdf
"terminology and tools to be used at the outset. We shall be examining the basic electrical quantities, their 
relationships, proper terminology, and a variety of analysis techniques and theorems that have broad application 
in the field. In this regard, our analytical “tools” are the appropriate mathematics and standards, and the 
scientific method, which are detailed in this chapter. The definition of specific electrical quantities and their 
relationships begins in Chapter Two. Various analysis techniques and theorems are detailed in subsequent 
chapters.
The initial research into electricity occurred in the late eighteenth and early nineteenth centuries by individuals 
such as Alessandro V olta, André-Marie Ampère, Michael Faraday and Georg Ohm. This work was expanded 
later in the nineteenth century by Gustav Kirchhoff, James Clerk Maxwell, Léon Charles Thévenin and others.",DCElectricalCircuitAnalysis.pdf
"later in the nineteenth century by Gustav Kirchhoff, James Clerk Maxwell, Léon Charles Thévenin and others. 
The late 1800s and early 1900s saw the practical application of electrical theory to solve practical problems (i.e.,
the field of electrical engineering). Perhaps the two names most associated with this period are Thomas Edison 
and Nikola Tesla. Numerous other individuals made contributions as well, eventually leading to the age of 
electronics, fully coming into its own by the mid-twentieth century with the introduction of solid state devices 
such as the bipolar junction transistor. The pace of these developments has been quite rapid. For example, a little
over a century ago the average person did not have ready access to something as simple as a modern flashlight. 
To put this into perspective, if we were to scale all of human history from the emergence of modern homo",DCElectricalCircuitAnalysis.pdf
"To put this into perspective, if we were to scale all of human history from the emergence of modern homo 
sapiens to today into a single year; radio, television, digital computers and all the rest would show up only 
within the last couple of hours before midnight on the last day of the year.
10",DCElectricalCircuitAnalysis.pdf
"At this point we need to distinguish between electricity and electronics. The term 
electricity tends to refer to the general relationships between electrical quantities 
such as voltage and current. In practical use, an electrical system tends to refer to a 
system where electrical energy is used directly to perform some manner of physical 
work. Examples include commercial and residential wiring systems for lighting, 
heating and the like. Electrical power generation and transmission would normally 
fall into this category, such as the high voltage transmission line seen in Figure 1.1. 
In contrast, electronic systems, or simply electronics, tends to refer to systems where
electrical signals are used to represent, store and/or manipulate some kind of 
information. This runs the gamut from radio and television to computing devices, 
cell phones, non-acoustic musical instruments, etc. Some of these applications may",DCElectricalCircuitAnalysis.pdf
"cell phones, non-acoustic musical instruments, etc. Some of these applications may 
be obvious, where the individual interacts directly with the device such as a cell 
phone or tablet. On the other hand, the human interaction may be minimal such as 
with the engine management system of a modern car. A good example of a modern 
device packed with electronics is the DSLR, or Digital Single Lens Reflex camera. 
These devices include numerous electronic sensors and actuators to adjust to 
ambient light, for automatic focus, and the like. An example is shown in Figure 1.2. 
The usage and scale of power transmission lines versus digital cameras may seem to 
be wildly separated but, ultimately, they are both made possible by the application
of the basic laws of electrical circuit theory.
Another good example of the shear scale of difference is to look at a single
electrical device such as a DC motor. Figure 1.3 shows a simple DC hobby motor,",DCElectricalCircuitAnalysis.pdf
"electrical device such as a DC motor. Figure 1.3 shows a simple DC hobby motor,
the kind of device that might be used for a motorized toy. This device runs on 12
volts, as available from a battery, and draws only a few hundredths of an amp of
current (to put this in perspective, the typical old-fashioned incandescent light bulb
draws between one half and one amp of current from a 120 volt socket). 
Compare this to the industrial DC motor seen in Figure 1.4. This is a 4000 
horsepower motor used to reduce eight inch thick copper slabs down to ½ inch 
thickness. It draws 4680 amps maximum at 700 volts. It clearly dwarfs the 
technician standing nearby.
Now that we have a little background and perspective, it's time to look at 
mathematics, specifically, how we represent and manipulate values that may be
very, very large or very, very small, all while keeping appropriate accuracy.
1.2 Significant Digits and Resolution1.2 Significant Digits and Resolution",DCElectricalCircuitAnalysis.pdf
"1.2 Significant Digits and Resolution1.2 Significant Digits and Resolution
A key element of any measurement or derived value is the resulting resolution.
Resolution refers to the finest change or variation that can be discerned by a
measurement system. For digital measurement systems, this is typically the last or 
11
Figure 1.1
345,000 volt transmission 
lines.
Figure 1.2
A digital SLR camera.
Figure 1.3
A DC hobby motor.",DCElectricalCircuitAnalysis.pdf
"lowest level digit displayed. For example, a bathroom scale may show weights in 
whole pounds. Thus, one pound would be the resolution of the measurement. Even if
the scale was otherwise perfectly accurate, we could not be assured of a person's 
weight to within better than one pound using this scale as there is no way of 
indicating fractions of a pound. 
Related to resolution is a value's number of significant digits. Significant digits can 
be thought of as representing potential percentage accuracy in measurement or 
computation. Continuing with the bathroom scale example, consider what happens 
when weighing a 156 pound adult versus a small child of 23 pounds. As the scale
only resolves to one pound, that presents us with an uncertainty of one pound out of
156 for the adult, but a much larger uncertainty of one pound out of 23 for the child.
The 156 pound measurement has three significant digits (i.e., units, tens and",DCElectricalCircuitAnalysis.pdf
"The 156 pound measurement has three significant digits (i.e., units, tens and 
hundreds) while the 23 pound measurement has but two significant digits (units and
tens). 
In general, leading and trailing zeroes are not considered significant. For example,
the value 173.58 has five significant digits while the value 0.00143 has only three
significant digits (the “143” portion) as does 0.000000143. Similarly, if we compute
the value 63/3.0, we arrive at 21, with two significant digits. If your calculator
shows 21.0 or 21.00, those extra trailing zeroes do not increase accuracy and are not 
considered significant. An exception to this rule is when measuring values in the 
laboratory. If a high resolution voltmeter indicates a value of, say, 120.0 volts, those 
last two zeroes are considered significant in that they reflect the resolution of the 
measurement (i.e, the meter is capable of reading down to tenths of volts).",DCElectricalCircuitAnalysis.pdf
"measurement (i.e, the meter is capable of reading down to tenths of volts). 
When performing calculations, the results will generally be no more accurate than 
the accuracy of the initial measurements. Consequently, it is senseless to divide two 
measured values obtained with three significant digits and report the result with ten 
significant digits, even if that's what shows up on the calculator. The same would be 
true for multiplication. For these sorts of calculations, we can't expect the result to 
be any better than the “weakest link” in terms of resolution and resulting significant 
digits. For example, consider the value 3.5 divided by 2.3. Both of these values have
two significant digits. Using a standard calculator, we find an answer of 1.52173913.
The result has nine significant digits implying much greater accuracy and resolution 
than we started with, and thus is misleading. To two significant digits, the answer",DCElectricalCircuitAnalysis.pdf
"than we started with, and thus is misleading. To two significant digits, the answer 
would be rounded to 1.5. On a long chain of calculations it may be advisable to 
round the intermediate results to a further digit and then round the final answer as 
indicated previously. This will help to mitigate accumulated errors. 
When it comes to addition and subtraction, the larger value will tend to indicate the 
number of significant digits available, particularly in the laboratory. This is due to 
the resolution of the measurement (that is, its finest digit). For example, suppose we 
have to add two distances. We drive a car from a parking lot to its exit, a distance we
record with a tape measure as being 51.17 feet. We then drive from the exit to the 
next city, which the car's odometer records as being 60.0 miles. Is it fair to say the 
12
Figure 1.4
An industrial DC motor.",DCElectricalCircuitAnalysis.pdf
"total distance is 60 miles plus 51.17 feet? No, it is not. Why? Because we cannot 
expect the car's odometer to be accurate to within 0.01 feet, like the tape measure. In
fact, the odometer is reading out a value with one-tenth mile resolution. One-tenth of
a mile is 528 feet, or more than ten times the entire measurement given by the tape. 
We would simply ignore the tape measurement because it is smaller than the 
resolution of the odometer. The proper result is 60.0 miles. Now in contrast, if the 
tape measure had indicated 552.7 feet, or slightly more than 0.1 miles, we could 
safely say the total distance was 60.1 miles, rounding the result to finest resolution 
digit of the more coarsely measured value. Of course, this example is a bit contrived 
but it is designed to show the limitations of measurement devices. It is also 
complicated by the fact that we are using using USA customary unit (miles, feet),",DCElectricalCircuitAnalysis.pdf
"complicated by the fact that we are using using USA customary unit (miles, feet), 
complete with their odd conversion factors. Fortunately, virtually all current work in 
science and technology uses the much simpler metric system, which we will be 
examining shortly.
In summary, 
• When multiplying and dividing, the value with the least number of 
significant digits indicates the number of significant digits in the result.
• When adding and subtracting, the value with the least resolution (i.e., the 
most coarse) will set the limit of resolution in the result.
Example 1.1
Determine the number of significant digits in the following values.
A. 12.6
B. 0.0034
C. 43.001
D. 5400
Answers:
A. All three digits are significant.
B. Two. Only the 3 and 4 count because leading zeroes are not significant.
C. Five. Embedded zeroes are significant.
D. Two. Trailing zeroes are not generally considered significant.
13",DCElectricalCircuitAnalysis.pdf
"Example 1.2
Perform the following computations, leaving the answer with the 
appropriate number of significant digits.
A. 55 ∙ 10.1
B. 2312.5 / 16.2
C. 1756.2 + 345.1
D. 750.2 − 0.004
Answers:
A. 555.5 which rounds to 560 (55 has only two significant digits).
B. 142.7469136... which rounds to 143 (16.2 has three significant digits).
C. 2101.3 (both values have resolution to tenths place).
D. 750.2 (do not extend the answer beyond the more coarse resolution of 
tenths place).
1.3 Scientific and Engineering Notation1.3 Scientific and Engineering Notation
Scientific and engineering notations are ways to express numbers without a lot of 
trailing or leading zeroes. They also simplify calculations. Before going any further, 
it would be a good idea to obtain a proper scientific calculator, such as the ones 
shown in Figure 1.5. The simplest can be had for $10 to $20, while more
powerful graphing types will run ten times as much (although good buys",DCElectricalCircuitAnalysis.pdf
"shown in Figure 1.5. The simplest can be had for $10 to $20, while more
powerful graphing types will run ten times as much (although good buys
can often be found on the used market). Some of the brands to consider are
Casio, Sharp and Texas Instruments. If you plan to continue study into AC
electrical circuits, it would be wise to make sure the calculator can solve 
simultaneous equations with complex coefficients.
The idea behind scientific notation is to represent the value in two parts: a precision 
portion, or mantissa; and the magnitude, a power of ten called the exponent. Thus, 
360 can be written as 3.6 times 100, or 3.6 ∙ 102, where 3.6 is the mantissa and 2 is 
the exponent. Similarly, 0.00275 can be written as the value 2.75 times 0.001, or 
2.75 ∙ 10−3. As the base of the exponent is always 10, a more compact form replaces 
“∙ 10” with “E”. Thus, these two values can be written as 3.6E2 and 2.75E−3. When",DCElectricalCircuitAnalysis.pdf
"“∙ 10” with “E”. Thus, these two values can be written as 3.6E2 and 2.75E−3. When 
adding or subtracting values in this form, the first step is to make sure that all of the 
values have the same exponent. Then, the precision portions are simply added 
together. Thus, 3.6E2 + 1.1E2 is equal to 4.7E2. Similarly, 3.6E2 + 5E1 is converted 
as 3.6E2 + 0.5E2 yielding 4.1E2. Alternately, it can be converted as 36E1 + 5E1 
yielding 41E1 (the same answer, or 410 in ordinary form). Where this notation is 
14
Figure 1.5
Scientific calculators.",DCElectricalCircuitAnalysis.pdf
"particularly useful is when multiplying or dividing. For multiplying, multiply the 
mantissas and add the exponents. For dividing, divide the mantissas and subtract the 
exponents. For example, multiply 20000 by 360000. This is equivalent to 2E4 times 
3.6E5. The result is 7.2E9 (i.e., 7200000000). Similarly, dividing the value 0.006 by 
50000 yields 6E−3 divided by 5E4, or 1.2E−7 (or in ordinary form 0.00000012). 
Notice the cumbersome and error-prone quantities of trailing and leading zeroes in 
these examples when using ordinary form. With scientific notation, we no longer 
have to deal with them.
Engineering notation is the same as scientific notation with the caveat that the 
exponent must by a multiple of 3. Thus, 390000 would be written as either 390E3 or
possibly as 0.39E6. Each multiple of 3 has a name and abbreviating letter to make 
the written representation even more compact. You are probably already familiar",DCElectricalCircuitAnalysis.pdf
"the written representation even more compact. You are probably already familiar 
with many of them. For example, 103 or 1000, is kilo which is abbreviated as k. 
Also, 106, or one million, is Mega, and abbreviated M (note capital letter).
Exponent Name Abbreviation
12 Tera T
9 Giga G
6 Mega M
3 kilo k
−3 milli m (note lower case)
−6 micro μ (Greek letter mu) 
−9 nano n
−12 pico p
A partial list of exponents, their names, and abbreviations is shown in Figure 1.6. 
There are standards for both larger and smaller exponents, but this table is 
something that should be committed to memory as they are in wide use.
Example 1.3
Convert the following into engineering notation using appropriate prefixes.
A. 2100 grams 
B. 0.005 meters
C. 32,000,000 bits per second (bps)
D. 0.0000741 seconds
15
Figure 1.6
Common engineering notation 
prefixes and abbreviations.",DCElectricalCircuitAnalysis.pdf
"Answers:
A. 2.1 kilograms (2.1 kg)
B. 5 millimeters (5 mm)
C. 32 Megabits per second (32 Mbps)
D. 74.1 microseconds (74.1 μs)
Notice that the final answers are much more compact and less error prone.
After some initial effort, certain common shortcuts will become second nature. For 
example (and keeping it generic), micro times kilo will yield milli (i.e., millionths 
times thousands yields thousandths). Similarly, the reciprocal of kilo is milli and the 
reciprocal of micro is Mega. Further, milli divided by kilo yields micro, Mega 
divided by kilo yields kilo, and so on. Gaining facility with these kinds of shortcuts 
will allow you to make estimates in your head very quickly, a much desired skill.
1.4 The Metric System1.4 The Metric System
In order to make consistent measurements of physical phenomena, some system of 
measurement needs to be standardized. While almost any scheme can be made to 
work, some systems are more logical and easier to deal with than others. The current",DCElectricalCircuitAnalysis.pdf
"work, some systems are more logical and easier to deal with than others. The current
standard in the world of science and technology is the metric system, or more 
properly, the International System of Units (abbreviated SI, from the French Système
International). This standard has found wide adoption across the globe and is the 
system used by roughly 95% of the human population. This is in contrast to the 
system of USA customary units which is based on a prior system of units originating 
in England. It should be noted that the USA, comprising some 5% of the global 
population, is the only country of reasonable economic power left that has not 
adopted the metric system for general use. This is only true in the consumer realm 
though as virtually all scientists, engineers and technicians in the USA routinely use 
the metric system1. Even many products built and sold in the USA use metric 
components internally; it's just not noticed by the consumer.",DCElectricalCircuitAnalysis.pdf
"the metric system1. Even many products built and sold in the USA use metric 
components internally; it's just not noticed by the consumer. 
Why has the metric system seen such wide adoption compared to USA customary 
units? The simple answer is because it is an easier system to learn and manipulate. 
First, a main strength of the metric system is that everything is based on powers of 
ten. Second, customary units often consist of many different variations on a single 
parameter while the metric system uses just a single parameter. For example, 
consider the measurement of length. For human-size lengths we might measure 
something in inches or feet. If it's a bit larger we might use yards. If it's even farther, 
we would use miles. In the metric system there is just one unit for length and that's 
1 It is also true of consumers who buy soda and similar beverages that are routinely sold in 
metric two or three liter bottles. Go figure.
16",DCElectricalCircuitAnalysis.pdf
"the meter (roughly 39.37 inches). If we want to talk about particularly small or large 
values we would simply add our engineering notation prefixes to come up with 
millimeters or kilometers for everyday conversion (although if a value happened to 
be particularly large or small we would say something like 2.3E9 meters). 
But customary units complicate the system even more because the various versions 
of a parameter are usually not based on a power of ten. For example, there are 12 
inches in a foot and 5280 feet in a mile. This is further complicated by the fact that 
there is more than one kind of mile (the common statute mile is 5280 feet but the 
nautical mile is roughly 6076 feet). Similarly, if we measure the weight of something
we could use ounces, pounds and tons (more than one kind), and again, the 
conversion factors are not convenient (16 ounces to the pound, 2000 pounds to the 
common ton). 
These odd conversion factors unnecessarily complicate matters. For example,",DCElectricalCircuitAnalysis.pdf
"common ton). 
These odd conversion factors unnecessarily complicate matters. For example, 
suppose we have a length measurement of 2 feet, 8 ½ inches and we need to 
multiply this by a factor of 3. Multiplying each part by three yields 6 feet, 25 ½ 
inches, but that needs to be simplified to 8 feet, 1 ½ inches. This extra step can be 
avoided if we base everything on powers of ten. 
For individuals only familiar with USA customary units, the ease of the metric 
system can be illustrated with currency. USA currency is, in effect, metric: 100 cents 
make up one dollar. Simply imagine how difficult it would be if, instead of cents, a 
dollar consisted of 13 flarkneks and 85 dollars made up a skroon. Further, imagine 
you decide to buy a new computer that costs 21 skroon, 10 dollars and 12 flarkneks. 
Now compute the total with a 6.5% local tax rate. Will 22 skroon be enough? What's
the change? Ouch! It's not impossible but it's extra error prone work. That's precisely",DCElectricalCircuitAnalysis.pdf
"the change? Ouch! It's not impossible but it's extra error prone work. That's precisely
the issue with feet, miles, ounces, pounds, pints, gallons and so on. 
In the metric system typically we use the MKS variation: meters for length or 
distance, kilograms for mass and seconds for time. Further, we typically use liters for
volume (liquid measure) and Celsius (AKA Centigrade) for temperature, although it 
is sometimes more convenient to use the absolute temperature scale of kelvin. 
Fortunately, the thermal energy indicated by a one degree Celsius change is identical
to that for kelvin. The two schemes only differ in their zero reference point (absolute
zero for kelvin and the point at which water freezes for Celsius).
As virtually all of the calculations presented in this text are metric, there is little 
need to concern ourselves with conversions back and forth between the two systems.
Indeed, if the USA were to finally make the switch to metric, no one would ever",DCElectricalCircuitAnalysis.pdf
"Indeed, if the USA were to finally make the switch to metric, no one would ever 
need to concern themselves with conversions, with the possible exception of 
historians, and perhaps people trying to recreate old family recipes and the like. 
Sometimes, careless people tend to think of units of measure as an afterthought. 
Don't get caught in this trap. Failure to pay attention to proper units can have 
catastrophic results. One example is the Mars Climate Orbiter. In 1999 this mission 
17",DCElectricalCircuitAnalysis.pdf
"to study the planet Mars suffered a spectacular failure. A subcontractor had used 
USA customary units for its software which then fed values to other systems that 
were expecting metric/SI units (as specified in the system contract). The result was 
the destruction of the orbiter as it attempted orbital insertion. The cost of the mission
was nearly $330 million, completely wasted. 
A table of commonly used quantities, their SI units and typical equivalents is shown 
in Figure 1.7.  Further details on units can be found in Appendix E.
Quantity
(abbreviation)
Unit
(abbreviation) Approximate Equivalent
Length (l) meter (m) 1 m ≈ 39.37 inches
Mass (m) kilogram (kg) 1 kg ≈ 2.2 pounds (on Earth)
Time (t) second (s) NA
Force (F) newton (N) 1 N ≈ 0.225 pound-force
Energy (w) joule (J) 3.6E6 J ≈ 1 kWh
Power (P) watt (W) 746 W ≈ 1 horsepower
Temperature (T) kelvin (K) or
Celsius (°C)
Fahrenheit = 32 + 9/5 ∙ °C
kelvin = °C + 273.15
Electric Charge (Q) coulomb (C) NA
Current (I) amp (A) NA",DCElectricalCircuitAnalysis.pdf
"Temperature (T) kelvin (K) or
Celsius (°C)
Fahrenheit = 32 + 9/5 ∙ °C
kelvin = °C + 273.15
Electric Charge (Q) coulomb (C) NA
Current (I) amp (A) NA
V oltage (V or E) volt (V) NA
1.5 The Scientific Method1.5 The Scientific Method
The scientific method is a means of uncovering and explaining physical phenomena.
It relies on observation and logical reasoning to achieve insight into the actions and 
relations of the phenomena, and a means of relating it to similar items. The method 
starts with observations and measurements. Based on these and background 
knowledge such as mathematics, physics, the behavior of similar phenomena, and so
on, a tentative explanation, or hypothesis, is derived. The hypothesis is then tested 
using experimental or field data. The precise nature of the tests depend on the area of
study, but the idea will always be the same: namely to see if the predictions of the 
hypothesis are borne out under alternate applicable conditions or data sets. A proper",DCElectricalCircuitAnalysis.pdf
"hypothesis are borne out under alternate applicable conditions or data sets. A proper 
hypothesis must be testable and falsifiable. That is, the hypothesis must be able to be
proven wrong by subsequent observation. If it is not, then the hypothesis is better 
classified as philosophy. For example, Newtonian gravitation could be proven false 
by letting go of a suspended hammer and watching it remain motionless or fall 
upwards, rather than fall down toward the Earth. Similarly, Evolution could be 
proven false by the discovery of “fossil rabbits in the Precambrian”, to quote famous
18
Figure 1.7
Common quantities, SI units 
and equivalents. 
Note: As a general rule, when 
units are abbreviated they are 
capitalized. When they are 
spelled out they are treated as 
common nouns and therefore 
not capitalized, even when 
named in honor of someone (an
exception being some 
temperature scales).",DCElectricalCircuitAnalysis.pdf
"biologist J. B. S. Haldane (Haldane was being somewhat snippy, but in general, he 
meant anything that would be clearly out of the expected time-line, in this case a 
small mammal predating even the most simple creatures with backbones). 
A hypothesis is tested by repeated cycles of prediction, observation and 
measurement, and also subject to peer review, that is, the critique of others in the 
field of study who may point out inconsistencies in the explanations or logic, errors 
in experimental design, and so on. This cycle continues until the weight of data and 
scientific opinion elevates the hypothesis to a theory. One might say that a theory is 
a hypothesis that has withstood the rigors of the scientific method. This cycle was 
well expressed by the Marquise du Châtelet2. She explained hypotheses as “probable
propositions”. While it would only take one observation to falsify a hypothesis, 
several would be required to vindicate it: “each non-contradictory result would add",DCElectricalCircuitAnalysis.pdf
"several would be required to vindicate it: “each non-contradictory result would add 
to the probability of the hypothesis and ultimately…we would arrive at a point 
where its ‘certitude’ and even its ‘truth’, was so probable that we could not refuse 
our assent”. 
It is important to note that the scientific usage of the word theory is entirely different
from its popular usage, which is perhaps closer to “hunch” or “seat-of-the-pants 
guess”. Also, a scientific theory is not true in the same sense as a fact. Facts come in 
three main varieties: direct observations, indirect observations and those that may be
logically deduced. A direct observation is something that you have measured 
yourself, such as noting the time it takes for a ball to reach the ground when released
from a given height. An indirect observation is something that may be inferred from 
other known quantities or proper historical data, such as “George Washington was",DCElectricalCircuitAnalysis.pdf
"other known quantities or proper historical data, such as “George Washington was 
the first president of the United States of America”. An example of the third variety 
would be “If x is an even integer larger than four and smaller than eight, then x must 
be six”. At first glance, it may seem that facts are highest on the pecking order, but 
scientific theories are much more useful than facts because isolated facts have very 
little predictive capacity. It is the predictive and explanatory ability of theories that 
make them so powerful.
Consider the following. Suppose you hold out a stone at arm’s length and let go. It 
drops to the ground. That’s a fact. You saw it happen. Unfortunately, by itself, it 
doesn’t tell you very much. Suppose you repeat this several times and each time the 
stone drops in precisely the same way as it did initially. This is beginning to get 
useful because you’re noticing a pattern, and patterns can be predictive. Now,",DCElectricalCircuitAnalysis.pdf
"useful because you’re noticing a pattern, and patterns can be predictive. Now, 
suppose you pick up stones of differing sizes, say 100 grams, 200 grams, half a 
kilogram and a kilogram, and drop each of them in turn. You observe that they each 
hit the ground in the same amount of time. Further, you drop them from different 
heights and you notice that the higher up they are, the longer it takes for them to hit 
the ground, but they all take the same amount of time to reach the ground. 
2  Du Châtelet was that most rare of 18th century women: a mathematician and physicist. 
She translated Newton’s Principia Mathematica into French and was a companion of 
V oltaire. Unfortunately, after an affair with the poet Jean François de Saint-Lambert in 
her early 40’s, she became pregnant and died six days after giving birth.
19",DCElectricalCircuitAnalysis.pdf
"You might now formulate a hypothesis: namely that the mass of a stone doesn’t have
an effect on how fast it falls from a given height and that height and fall time are 
directly related. Your hypothesis is predictive. Although you used only four sizes of 
stones and a few heights, your broadened hypothesis should apply to any stone 
dropped from any height. So now you (and a bunch friends) starting picking up 
random pairs of stones and drop them from random heights, and sure enough, you 
see the same effect again and again. If you do this enough and it is continually 
verified without exception, you might even make a “law of falling stones”, 
particularly if you were able to quantify the times and heights through careful 
measurement and reduce the relation to a nice formula. It is useful because you can 
now predict what will happen with any stone dropped from any height. But this law 
is rather limited. It only applies to stones because you may have noticed that stones",DCElectricalCircuitAnalysis.pdf
"is rather limited. It only applies to stones because you may have noticed that stones 
drop much faster than pieces of cork. While you might then proceed to make a “law 
of falling cork”, that would unnecessarily complicate things. Instead, you could take 
a step back and try to figure out why stones and cork both fall, but not at the same 
rate. Eventually, you might discover that the difference has to do with air friction 
and you can now create a law governing falling bodies in a frictionless environment.
That’s even more useful than the original “law of falling stones”. 
But even this new and improved “law of falling bodies” doesn’t offer a lot of insight 
into what is really going on in the larger scheme of things. Through repeated 
observations and experiments this could be extended to cover not just falling bodies 
on the Earth, but the interactions between any bodies, including falling stones and",DCElectricalCircuitAnalysis.pdf
"on the Earth, but the interactions between any bodies, including falling stones and 
cork on the moon, or the interaction between the Earth and the Sun, the Sun and the 
other planets, the Sun and other stars, and so on. What you’ll have arrived at is a 
full-blown theory of gravitation (Newtonian gravitation). Now that is an extremely 
useful tool. It helps us design airplanes, get satellites into orbit, even get people to 
the moon and back safely. 
Thus a theory is a “best estimate so far”, a model to explain some observable aspect 
of the universe. It is understood that as our view of what is observable widens and 
our knowledge extends, so too a given theory may be refined. The Newtonian 
gravitation model was sufficient to describe the movements of the planets around the
sun and is still used to plan the flight of spacecraft. In the early 1900’s, however, 
Einstein’s Theory of Relativity refined Newtonian Gravitation to include more",DCElectricalCircuitAnalysis.pdf
"Einstein’s Theory of Relativity refined Newtonian Gravitation to include more 
extreme (relativistic) effects. It is important to note that we did not “throw out” 
Newton’s equations; rather, we now have a better understanding of the set of 
conditions under which they apply. While this trek towards more and more 
refinement is not truth in and of itself, to paraphrase the late Harvard paleontologist, 
Stephen Jay Gould, to ignore the weight of scientific data behind an established 
theory would be perverse.
20",DCElectricalCircuitAnalysis.pdf
"1.6 Critical Thinking: Avoiding Being Fooled1.6 Critical Thinking: Avoiding Being Fooled
As humans, we need to recognize that we are fallible. No matter how good our 
intentions, we make mistakes and can be fooled. The first step toward reducing and 
ultimately eliminating these sources of error is to understand them. We will lump 
these into two broad categories: cognitive bias and logical fallacies. Understanding 
these will enhance our ability to think critically and avoid being fooled (or fooling 
ourselves).
Cognitive BiasCognitive Bias
A cognitive bias is an inclination toward looking at a situation in an unbalanced or 
unfair manner. Generally, people are not aware of these biases. One example is 
confirmation bias (AKA, confirmation of expected outcomes). That is, we expect (or 
hope) to see a certain result and thus we tend to overvalue evidence that confirms it 
while discounting evidence that contradicts it. One way to avoid this is through the",DCElectricalCircuitAnalysis.pdf
"while discounting evidence that contradicts it. One way to avoid this is through the 
use of a double-blind test. Suppose we wish to test a new drug to see if it is effective 
and safe. As we may have invested a lot of time and money developing the drug, it is
only natural that we want it to work, and this may skew our analysis 
(unintentionally, of course). What we do is have a third party create two sets of pills;
one is the drug under test and the other is a placebo (it looks like the other pill but 
does nothing). These sets are identified using codes known only to the third party. 
The sets are then given to the researchers who, in turn, give them to the patients. The
important thing is that neither the patients nor the researchers know which pills are 
which. When the trial has run its course, the researchers (us) analyze the data to 
determine if any set of pills was successful. Only after the analysis is completed",DCElectricalCircuitAnalysis.pdf
"determine if any set of pills was successful. Only after the analysis is completed 
does the third party tell the researchers which set was real and which set was the 
placebo.
Another cognitive bias is the Dunning-Kruger effect, named after the two social 
psychologists who studied it. This states that the knowledge needed to determine if 
someone is competent in a certain field is competence in that same field. Thus, 
individuals who have low competence are not in a position to accurately evaluate 
their own level of competence. Consequently, these individuals often over estimate 
their competence. This is known as illusory superiority. To put it crudely, these 
individuals are too ignorant of the subject to understand just how ignorant they are. 
Among the highly competent, two other effects may be seen. First, the advanced 
individual may be keenly aware of any shortcomings or gaps in their competence",DCElectricalCircuitAnalysis.pdf
"individual may be keenly aware of any shortcomings or gaps in their competence 
and may undervalue their level as a result. Second, they may assume that their level 
of competence is typical, and that most people are therefore “at their level”. 
It is useful to remember that in our complex and interdependent society, no one can 
be an expert at everything, or even at most things. Instead, it is likely that we are all 
largely ignorant of a majority of subjects and/or incompetent at a variety of skills.
21",DCElectricalCircuitAnalysis.pdf
"Logical FallaciesLogical Fallacies
Logical fallacies represent faulty reasoning. They are “thinking traps” that people 
sometimes fall into. Familiarity with them will help reduce their occurrence. There 
are dozens of logical fallacies but we shall only investigate a representative few. 
To help explain the process, we'll begin with a syllogism. This is, in essence, a 
simple argument. It starts with a major premise (a generalization) which is followed 
by a minor premise (a more specific statement). From these, we derive a conclusion. 
For example:
All humans breathe air. (major premise)
Alice is a human. (minor premise)
Therefore, Alice breathes air. (conclusion)
Errors can occur when either premise is false or when the conclusion does not 
follow (the latter being referred to as a non-sequitur). For example:
All fish live in water.
Lobsters live in water.
Therefore, lobsters are fish. 
The problem with this is the linkage between the major and minor premises. Saying",DCElectricalCircuitAnalysis.pdf
"Lobsters live in water.
Therefore, lobsters are fish. 
The problem with this is the linkage between the major and minor premises. Saying 
“all fish live in water” does not preclude something else (like a lobster or a sea snake
or kelp) from living in water. Compare the prior example to this version:
All fish live in water.
Trout are fish.
Therefore, trout live in water. 
While these examples may seem obvious, there are trickier versions. For example:
I am made of nothing but atoms.
Atoms are not alive.
Therefore, I am not alive.
Nope. Doesn't work. This error is called the fallacy of composition. Basically, it says
that what is true of the parts must be true of the whole, and vice versa. It ignores the 
concept of emergent properties (consider the behavior of a single bird to a flock, or a
single fish to a school). 
The fallacy of composition can be illustrated without using a syllogism. Suppose 
you are in a crowded movie theater. If you stand up, you will have a better view of",DCElectricalCircuitAnalysis.pdf
"you are in a crowded movie theater. If you stand up, you will have a better view of 
the screen. In contrast, it is not true that if everyone stands up, everyone will have a
better view. In fact, everyone will most likely have a worse view. If one person 
22",DCElectricalCircuitAnalysis.pdf
"stands, they are unique. That unique character is lost when everyone stands. 
Turning to a different fallacy, the Latin phrase post hoc ergo propter hoc can be 
translated as “Before this, therefore because of this”. This fallacy is sometimes 
referred to as the post hoc fallacy or the causation fallacy. It is an error regarding 
causality; the assumption being that because event A occurred before event B, then 
event A must have caused event B. On the surface, it seems logical enough. For 
example, you might see a lightning strike and then hear a clap of thunder. It seems 
reasonable to assume that the lightning caused the thunder (generally speaking, that 
is the case). On the other hand, you might wake up some morning when it's dark 
outside. Shortly thereafter, the sun rises. Obviously, your waking did not cause the 
sun to rise. 
Another error involves proportional contribution to an outcome. Relative size is",DCElectricalCircuitAnalysis.pdf
"sun to rise. 
Another error involves proportional contribution to an outcome. Relative size is 
mistakenly seen as a determiner. That is, the error assumes that only large 
contributors have any sway in the outcome. Basically, this fallacy proposes that if 
something makes up only a small percentage of the total, then its effect must be 
minimal. This is easily proven wrong. As an example, the atmosphere of the Earth is
comprised largely of nitrogen (78%) and oxygen (21%) along with a number of trace
gases such as argon, carbon dioxide and so forth. If the atmosphere was suddenly 
altered so that it included just 0.1% hydrogen sulfide, every human likely would be 
dead after their first breath of this new combination. 
Along with size, there is a similar issue regarding linearity of effect. A linear 
function is one that can be plotted as a straight line. More to the point, if we have a 
linear system, then doubling an input to that system doubles its effect. To wit, if you",DCElectricalCircuitAnalysis.pdf
"linear system, then doubling an input to that system doubles its effect. To wit, if you 
order two pieces of pie for dessert, it will cost twice as much as one piece. The 
reality is that many systems do not behave in a linear fashion. Systems or 
relationships can be logarithmic, square law, cubic, or follow some other 
characteristic. For instance, the braking distance of a car does not vary linearly with 
its speed, it varies in accordance with the square of its speed. Therefore, if you're 
traveling twice as fast, it doesn't take twice as far to come to a stop, it takes four 
(two squared) times farther to stop. Remember this the next time you're speeding 
down a highway.
The next two items sometimes appear in arguments for or against a proposition. 
They are the excluded middle and the ad hominem. The excluded middle presents a 
false set of choices. Essentially, it falsely reduces the set of possible outcomes and",DCElectricalCircuitAnalysis.pdf
"false set of choices. Essentially, it falsely reduces the set of possible outcomes and 
then proceeds to disprove all but one of them. By process of elimination, the 
remaining outcome should be true. For example, someone might complain that a 
particular politician would only support a particular bill if said politician was either 
stupid or a communist (or a fascist, take your pick). They then show that the 
politician is not stupid, so by their logic the politician must be a communist. Of 
course, there are any number of other possible scenarios that have been excluded; 
for instance, the politician might have taken a hefty bribe to vote for the bill or the 
analysis of the bill by the complainer might be faulty.
23",DCElectricalCircuitAnalysis.pdf
"Finally, ad hominem is a Latin term meaning “to the person”. The ad hominem 
attempts to disprove a point by arguing against the person making a claim, not the 
claim itself. For example, suppose Doug makes a claim in favor of a new theory of 
gravity. Fran's counterargument is that Doug is an evil person because he likes to 
spray paint foul words on other people's pet cats, and therefore he can't be trusted. 
The reality is that, in spite of his proclivity for penning feline profanity, Doug's ideas
regarding gravity might be spot on. Those ideas need to be addressed directly.
1.7 RoHS1.7 RoHS
Electrical components are not without a downside. Some devices potentially may 
contain hazardous substances such as toxic metals. Even if the devices are 
constructed and used with care, these substances can still create life cycle issues; that
is, when their useful life is over, simply “throwing them away” can create 
environmental contamination.",DCElectricalCircuitAnalysis.pdf
"is, when their useful life is over, simply “throwing them away” can create 
environmental contamination. 
In the early 2000s the European Union ratified the RoHS Directive. This stands for 
the Restriction of Hazardous Substances, and controls the use of several materials 
used in the construction of electrical and electronic systems. These materials include
lead, mercury, cadmium, hexavalent chromium (used in chrome plating), and other 
specific chemicals such as certain flame retardants and plasticizers. The RoHS 
Directive specifies the amount of these materials allowed in any homogeneous 
material that makes up the device or product. Homogeneous materials, are, in 
essence, the smallest parts that could be separated mechanically. Examples would be
items such as a machine screw, the toggle lever of a switch, or the magnet of a 
loudspeaker. Each of these items must individually meet the limits set by the",DCElectricalCircuitAnalysis.pdf
"loudspeaker. Each of these items must individually meet the limits set by the 
directive. Most of these substances have limits of 1000 ppm (parts per million) 
although cadmium (used in rechargeable Ni-Cd batteries and certain light detectors) 
has a limit of 100 ppm.
The RoHS Directive has been updated periodically. The third variation (RoHS 3) 
became effective in the summer of 2019. Related to RoHS is WEEE, which stands 
for Waste from Electrical and Electronic Equipment. This controls the recovery and 
recycling of these devices. Generally, products sold in the EU market must meet the 
WEEE requirements. Several countries outside of the EU have adopted similar 
requirements including Japan, China, South Korea and Turkey. The USA has not 
adopted RoHS per se, however, several US states have adopted “RoHS-like” 
requirements, including California, Colorado and New York. With increasing 
environmental consciousness, it is likely that more countries will adopt similar 
measures.",DCElectricalCircuitAnalysis.pdf
"environmental consciousness, it is likely that more countries will adopt similar 
measures.
 
To help electronic and electrical system designers make their products meet the 
RoHS requirements, electrical component manufacturers often use some manner of 
“RoHS compliant” logo or statement on their data sheets. There is no standard for 
24
Figure 1.8
RoHS compliance label.",DCElectricalCircuitAnalysis.pdf
"these logos, however, a finished product will bear the standard RoHS compliance 
label, as shown in Figure 1.8.
1.8 Summary1.8 Summary
An electrical system is one that refers to the direct use of electrical energy in terms 
of power generation, transmission or application. In contrast, an electronic system is 
one that tends to use electrical signals as a means to represent, store and manipulate 
information. 
When measuring physical quantities and in computations, care must be taken to 
maintain accuracy. Of particular importance are the concepts of resolution and 
significant digits. Resolution refers to the finest digit place that may be discerned 
(e.g., tenths or hundredths place). Generally, accuracy increases with the number of 
significant digits used to represent a value. Leading and trailing zeroes are not 
considered significant, the exception being trailing zeroes as reported by a 
measurement device such as a digital multimeter. In computations, results can be no",DCElectricalCircuitAnalysis.pdf
"measurement device such as a digital multimeter. In computations, results can be no 
more accurate or have greater resolution than the original values being manipulated. 
Thus, if the division of two values, each with three significant digits, yields a 
calculated result with much more than three significant digits, those extra digits only
lead to a false sense of accuracy and resolution. As a general rule, the results of 
multiplication and division will have as many usable significant digits as the least 
precise of the original values. For addition and subtraction, the resolution of the 
result can be no greater than that of the original value with the least resolution.
Scientific and engineering notations are ways of representing values in a compact 
and less error prone form. They consist of two parts: a mantissa consisting of the 
significant digits and an exponent, or power of ten, for scaling. Engineering notation",DCElectricalCircuitAnalysis.pdf
"significant digits and an exponent, or power of ten, for scaling. Engineering notation 
further stipulates that the exponent must be a multiple of three, and these multiples 
use prefix names such as kilo and milli for further simplification. When multiplying 
or dividing values, the mantissas are multiplied (or divided) while the exponents are 
added (or subtracted). For addition and subtraction the values are adjusted so that 
they have the same exponent and the mantissas are then added or subtracted as 
needed. 
The metric system (or SI) is a measurement system based on powers of ten. 
Conversions between units of the same type are no longer necessary which 
simplifies computations. It is a global standard that is ubiquitous in the field of 
science, engineering and technology.
The scientific method is a technique used to uncover the reality behind the physical 
world. It begins with observations of phenomena which then lead to a tentative",DCElectricalCircuitAnalysis.pdf
"world. It begins with observations of phenomena which then lead to a tentative 
hypothesis. The hypothesis is tested experimentally with the results either 
confirming or denying the hypothesis. The process is repeated creating a feedback 
25",DCElectricalCircuitAnalysis.pdf
"loop. If the hypothesis is repeatedly verified and not rejected, it may be elevated to a
scientific theory. It is important during this process to be aware of logical fallacies 
and cognitive biases that can lead to false interpretations of the experimental results. 
Finally, RoHS, or the Restriction of Hazardous Substances, is a European Union 
directive aimed at reducing environmental contamination by restricting the use of 
certain substances such as lead and mercury. RoHS has a direct impact on the 
construction and disposal of electrical and electronic products. Other countries and 
states have similar directives and restrictions in place to help protect the 
environment.
Review QuestionsReview Questions
1. Describe the differences between scientific notation and engineering 
notation.
2. List the terms and abbreviations (i.e, words and symbols) for engineering 
notation from 10−12 through 1012. 
3. Give at least one advantage of the metric system over the",DCElectricalCircuitAnalysis.pdf
"notation from 10−12 through 1012. 
3. Give at least one advantage of the metric system over the 
customary/Imperial system of measurement.
4. Outline the process of the scientific method.
5. Explain the differences between a scientific theory, a hypothesis and a fact.
6. Describe at least three examples of cognitive biases and logical fallacies. 
7. What is RoHS?
1.9 Exercises 1.9 Exercises 
AnalysisAnalysis
1. Round the following to four significant digits: a) 14.5423  b) 30056            
c)  76.90032  d) 0.00084754
2. Round the following to three significant digits: a) 354.005  b) 9100.46         
c) 1.0054  d) 0.000052753
3. Round the following to five significant digits: a) 5.100237  b) 1020.765        
c) 1.00540  d) 0.00004578053
4. Round the following to four significant digits: a) 5.100237  b) 1020.765       
c) 1.00540  d) 0.00004578053
5. Convert the following to scientific notation: a) 23.61  b) 12000  c) 7632      
d) 0.00509",DCElectricalCircuitAnalysis.pdf
"c) 1.00540  d) 0.00004578053
5. Convert the following to scientific notation: a) 23.61  b) 12000  c) 7632      
d) 0.00509
6. Convert the following to scientific notation: a) 4253  b) 640000000  c) 2.03  
d) 0.00000658
26",DCElectricalCircuitAnalysis.pdf
"7. Convert the following to scientific notation: a) 41.56  b) 954000  c) 84.035  
d) 0.0001632
8. Convert the following to scientific notation: a) 11200  b) 30000000  c) 325.2
d) 0.00002504
9. Convert the following to engineering notation: a) 12000  b) 470  c) 6.5       
d) 0.00198
10. Convert the following to engineering notation: a) 3500  b) 17.9  c) 5601000 
d) 0.0000355
11. Convert the following to engineering notation: a) 33.2  b) 313.6  c) 43000  
d) 0.000076
12. Convert the following to engineering notation: a) 0.23  b) 76.95  c) 45500  
d) 0.00890 
For problems 13 through 18 express the results using engineering notation 
without rounding or truncating. 
13. Compute the following: a) 1.2E2 times 3.0E4  b) 5.4 times 3.1E3                 
c) 6.01E3 times 2.0E−1  d) 5.3E9 times 4.1E−5 
14. Compute the following: a) 7.1E1 times 4.0E2  b) 2.32 times 5.6E3               
c) 6.01E3 times 3.0E−2  d) 5.2E5 times 8.2E−7",DCElectricalCircuitAnalysis.pdf
"14. Compute the following: a) 7.1E1 times 4.0E2  b) 2.32 times 5.6E3               
c) 6.01E3 times 3.0E−2  d) 5.2E5 times 8.2E−7
15. Compute the following: a) 8.2E1 / 4E2  b) 2.42 / 2.0E3  c) 6.09E3 / 3.0E−2  
d) 9.6E5 / 3.1E−7
16. Compute the following: a) 4.6E2 / 2.0E2  b) 2.35 / 4.0E6                             
c) 5.15E4 / 3.0E−3  d) 6.8E−2 / 5.0E−4
17. Compute the following: a) 8.2E1 + 4E2  b) 2.42 + 2.0E1                              
c) 6.09E3 − 3.0E−2  d) 9.6E5 − 5.1E6
18. Compute the following: a) 4.2E3 + 9E2  b) 3.53 + 4.2E−1                              
c) 1.05E−2 − 5.0E−2  d) 9.4E5 − 3.2E4 
For problems 18 through 22 express the results using engineering notation with 
proper rounding to reflect the resulting resolution. 
19. Compute the following: a) 16.2 − 0.4  b) 4356 + 378  c) 0.012 − 0.005 
20. Compute the following: a) 4.5 ∙ 43.1  b) 1201 / 23.6  c) 890.1 ∙ 0.172
21. Compute the following: a) 51E3 ∙ 62.7E2  b) 6.733E2 / 1.01E−3",DCElectricalCircuitAnalysis.pdf
"21. Compute the following: a) 51E3 ∙ 62.7E2  b) 6.733E2 / 1.01E−3                   
c) 20.12E6 / 65.6E9  
22. Compute the following: a) 17.9E3 − 10.4E6  b) 81E3 + 12E3                       
c) 76.0E3 + 1465  
27",DCElectricalCircuitAnalysis.pdf
"For problems 23 through 26 determine the result of the computation in SI base 
units (meters, kilograms, seconds) with appropriate engineering notation units 
(kilo, milli, etc.).
23. Determine the result of 20 meters times 0.10 kilograms in kg·m.
24. Determine the result of 34 kilometers divided by 10 millimeters.
25. Determine the resulting velocity in m/s if a distance of 3.6 meters is covered
in the space of 1.2 milliseconds.
26. Determine the result of 24 μs divided by 3.0 km. 
28",DCElectricalCircuitAnalysis.pdf
"NotesNotes
    ♫♫    ♫♫
29",DCElectricalCircuitAnalysis.pdf
"2 2 Basic QuantitiesBasic Quantities
2.0 Chapter Learning Objectives2.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe a basic, functional atomic model.
• Describe fundamental quantities and relations including charge, current, energy, voltage, power,  
resistance and conductance; and perform basic computations using them.
• Compute efficiency, energy cost, battery life and DMM accuracy.
• Utilize the resistor color code.
2.1 Introduction 2.1 Introduction 
In this chapter we shall examine the major quantities of interest in electrical systems and describe their 
relationships. These include voltage, current, power, energy and resistance. We begin with basic definitions and 
proceed to example computations. We shall go beyond the theoretical and investigate practical aspects as found 
in electrical laboratories. This includes measurement devices and power sources, as well as various resistive",DCElectricalCircuitAnalysis.pdf
"in electrical laboratories. This includes measurement devices and power sources, as well as various resistive 
devices. We shall examine batteries from a functional perspective, and perform computations regarding 
efficiency and cost of energy. We will also introduce some of the device symbols used to create electrical circuit 
schematics. This material presents the foundation upon which we will build circuit analysis techniques in 
subsequent chapters. Treat it accordingly.
2.2 An Atomic Model2.2 An Atomic Model
In order to understand electrical circuit behavior we must first define what it is that we are chasing. Ultimately, 
electrical circuits are all about the movement of electric charge. What do we mean by charge? To answer that 
question, we need to come up with a usable model of an atom, in other words, a functional description of its 
internal structure. To be sure, atoms are far smaller than most people imagine. There are far more atoms in a",DCElectricalCircuitAnalysis.pdf
"internal structure. To be sure, atoms are far smaller than most people imagine. There are far more atoms in a 
gallon of water than there are gallons of water in all of the oceans, lakes and rivers of planet Earth combined. It 
would be pointless to ask what an atom might “look like” because its components are all smaller than the 
shortest wavelengths of light that humans can see. That would be a little like asking a human what a dog whistle 
sounds like, when in fact it's at a pitch that's higher than humans can hear. Ultimately though, all we really need 
is a model to explain its observed behavior. 
Perhaps the most prolific model in the popular imagination is the planetary model shown in Figure 2.1, so 
named because it is reminiscent of simple models of our solar system. In this model, the core, or nucleus, is 
drawn at the center and contains positively charged protons and non-charged neutrons. Revolving around this 
30",DCElectricalCircuitAnalysis.pdf
"core are negatively charged electrons, each following a nice, regular, planar path 
much like a planet around the sun. Unfortunately, this model is not accurate to say 
the least, although its popularity can sometimes lead to wildly erroneous and darkly 
humorous conclusions. In spite of its inaccuracy, the components (nucleus, protons, 
neutrons and electrons) are perfectly valid. Before we come up with a more accurate 
and useful model, let's take a closer look at these items.
Unlike people, all particles of a type are indistinguishable: every proton is identical 
to every other proton and every electron is identical to every other electron. The 
number of protons determine what the element is. The simplest element, hydrogen, 
has a single proton in its nucleus. In contrast, helium has two protons in its nucleus 
and copper has 29. The nucleus may also contain a number of neutrons as well (not
necessarily equal to the number of protons). In any stable atom, the number of",DCElectricalCircuitAnalysis.pdf
"necessarily equal to the number of protons). In any stable atom, the number of
electrons will equal the number of protons. If they aren't the same, the atom is
called an ion (technically, a cation has fewer electrons than protons and an anion is
its opposite). With an external energy source, it is possible to separate an electron
from an atom. This leaves us with an electron and an atom that now has a net
positive charge. It is this ability to separate these particles, and thus separating
charge, that gives rise to our concept of electricity including such items as current
and voltage, as we shall see.
Most of the mass of any given atom is from the protons and neutrons. Protons and
neutrons have similar masses, about 1.67E−24 grams each. The mass of an
electron is roughly 2000 times smaller. The radius of a proton is approximately 
0.87E−15 meters and the mean distance to the nearest electron is about 5.3E−11",DCElectricalCircuitAnalysis.pdf
"0.87E−15 meters and the mean distance to the nearest electron is about 5.3E−11 
meters. Therefore an electron is about 60,000 times farther away from the proton 
than the size of said proton. To put this into perspective, that's roughly the same as 
the ratio between a golf ball and a sphere with a radius of 3/4ths of a mile or 1200 
meters. This would be the case for a hydrogen atom as it consists of a single proton 
and electron. The magnitude of this ratio is not much different for other substances, 
including things like diamond and quartz that are very hard and solid. If you think 
about that for a moment, you realize that the idea of “solidity” is in some ways an 
illusion because the vast majority of what we call “something” is really just empty 
space (like a golf ball rattling around inside a sphere that's a mile and a half wide). 
In reality, the feeling of solidity when you grab something with your hand is just the",DCElectricalCircuitAnalysis.pdf
"In reality, the feeling of solidity when you grab something with your hand is just the 
result of the interaction of atomic forces between your hand and the object. Indeed, 
if we could somehow get rid of that “empty space”, we could fit a mountain inside 
of a car (although it would probably require a severely upgraded suspension).
For our immediate work, perhaps the most important issue is that of charge, an 
essential part of the theory of electromagnetism. It was mentioned that protons are 
positively charged and electrons are negatively charged. What then is charge? 
Charge is not an obvious physical attribute like someone's height or eye color. It's 
closer to a behavioral characteristic. Saying a particle is charged is like saying 
someone has an attractive personality. The magnitude of the charge on an electron is 
the same as the magnitude of the charge on a proton. The only difference is the 
31
Figure 2.1
Planetary atomic model: 
Popular but sadly incorrect.",DCElectricalCircuitAnalysis.pdf
"the same as the magnitude of the charge on a proton. The only difference is the 
31
Figure 2.1
Planetary atomic model: 
Popular but sadly incorrect.
Image source (modified)",DCElectricalCircuitAnalysis.pdf
"polarity: protons are positively charged while electrons are negatively charged. An 
important thing to remember is that particles with the same polarity repel each other 
while opposites attract3. 
Charge is measured in coulombs, named after Charles-Augustin de Coulomb a 
French physicist of the eighteenth century. The charge on a single electron is tiny, a 
mere 1.602E−19 coulombs. Alternately, it takes 6.242E18 electrons to create a 
charge of one coulomb. More on this in a moment.
One of the major issues with the planetary model is the idea that electrons whirl 
around the nucleus in stable, planet-like orbits. That's simply not true. First, the 
electron inhabits a region of 3D space and does not simply move through a plane. 
Second, due to the Heisenberg Uncertainty Principle, we can't precisely plot
the position and trajectory of a given electron. The best we can do is make a
plot of where the electron is likely to be. This is called a probability contour.",DCElectricalCircuitAnalysis.pdf
"plot of where the electron is likely to be. This is called a probability contour.
To understand this concept, imagine that you could record the position of a
specific electron relative to the nucleus at a specific time. A moment later you
record its new position, a moment after that you record the next position, and
so on for thousands of measurements. If you attempted to plot them all, you
would wind up with a cloud of dots around the nucleus. This cloud is referred
to as an orbital. You wouldn't know how the electron got from one position to
the next but you would get a general idea of where it was likely to be.
Ultimately, an orbital looks nothing like a planetary orbit. 
A good analogy to an electron probability contour is the mapping of the flight of 
migratory birds. For example, a certain group of Canada geese might migrate from 
their summer range in the far north of Canada to their summer abode in the southern",DCElectricalCircuitAnalysis.pdf
"their summer range in the far north of Canada to their summer abode in the southern 
USA via s section of the Atlantic Flyway, perhaps through central New York State, as
shown in purple, in Figure 2.2. The flyway tells you where these birds are likely to 
be found while migrating. If they are taking the route through central New York, at 
some point it is likely that any particular goose in that flock will be found near the 
cities of Utica, Syracuse and so forth. It is much less likely that any particular goose 
from that flock will be found in Ohio or New Hampshire. Texas, of course, is right 
out. The flyway gives a good bit of data regarding where these birds are likely to be 
but in no way can it predict with any accuracy precisely where any individual bird 
will be on a specific date, nor precisely where it will end its migration.
Returning from our avian diversion, we note that there are several potential orbitals.",DCElectricalCircuitAnalysis.pdf
"Returning from our avian diversion, we note that there are several potential orbitals. 
Due to quantum physics, only certain orbitals are allowed. The permissible electron 
energy levels are first grouped into shells, then subshells and finally orbitals. It is 
important to remember that orbitals indicate the electron energy level. That is, a 
higher orbital implies a higher energy level. Further, orbitals fill in first from lowest 
energy level to highest energy level. 
3 The obvious question might be “Why doesn't the nucleus break apart?” The answer is that
the sub-atomic strong nuclear force has a greater effect at atomic distances than 
electromagnetism, and thus binds the protons and neutrons together in the nucleus.
32
Figure 2.2
Electron behavior versus 
migratory birds.
Figure 2.3
Electron probability contour 
for innermost orbital, 1s.
Image source",DCElectricalCircuitAnalysis.pdf
"Shells are denoted by their principal quantum number, n; 1, 2, 3, etc. The higher the 
number, the more subshells it can contain. Subshells are designated by letters, the 
first four being s, p, d, and f. Shell 1 contains only subshell s while shell 2 contains 
subshell types s and p. Shell 3 contains subshell types s, p and d, and so on. These 
subshells may also have variations within them. There is one variation on s, three 
variations on p, five variations on d, etc. These variations are the orbitals and each 
orbital can hold a maximum of two electrons. 
Putting this all together, we find that the first shell can contain a maximum of two 
electrons: two in the single s subshell orbital (1s). The second shell can contain a 
maximum of eight electrons: two in the s subshell (2s) plus two in each of the three 
p subshell orbitals (2p). In like manner the third shell can contain a maximum of 18 
electrons: two in 3s, six in 3p and two in each of the the five d subshell orbitals (3d).",DCElectricalCircuitAnalysis.pdf
"electrons: two in 3s, six in 3p and two in each of the the five d subshell orbitals (3d).
You can condense this into a simple formula:
Number of electrons per shell = 2n2, where n is the shell number
Figure 2.3 shows the electron probability contour of the innermost orbital, namely 
1s (i.e., principle quantum number 1, subshell s). As you can see, it is spherical in 
shape. The nucleus is located at the center, obscured here. All s orbitals are similarly 
spherically shaped although the internals change. 1s is the lowest energy orbital.
Orbitals are not limited to simple spherical shapes. Higher order orbitals can take on 
a variety of forms. Figure 2.4 shows a much more complex electron probability 
contour. The nucleus is situated in the center of the concentric rings. Obviously, this 
is nothing like the well-behaved elliptical orbits of planets around the sun (nor the 
paths of migratory birds, unless they are very confused).",DCElectricalCircuitAnalysis.pdf
"is nothing like the well-behaved elliptical orbits of planets around the sun (nor the 
paths of migratory birds, unless they are very confused). 
As visually interesting as these graphics are, they are cumbersome to work with. 
Consequently, a more functional graphic is called for. Such a device is the Bohr 
model, named after Danish physicist Niels Bohr. An example is shown in Figure 2.5.
It is important to understand that the Bohr model is an energy description of the
atom, not an attempt to mimic its physical appearance or structure. The nucleus is
placed at the center. It is surrounded by concentric rings that represent the electron
shells. The higher the number, the larger the ring and the greater the energy level. If
an electron were to move from a higher level to a lower level, the energy difference
is radiated out. This could be in the form of heat or light. This is a point worth
remembering. For example, this transition is what makes light emitting diodes",DCElectricalCircuitAnalysis.pdf
"remembering. For example, this transition is what makes light emitting diodes
(LEDs) function. The inverse is also possible, namely that by absorbing energy, an
electron can move into a higher orbital. This is an equally powerful concept.
Using the Bohr model we can create diagrams to represent individual elements. For 
example, copper has an atomic number of 29 meaning that it has 29 protons and 29 
electrons. The electron shell configuration is 2-8-18-1. That is, the first three shells 
are completely filled and there is a single electron in the fourth shell. The Bohr 
33
Figure 2.4
A higher order electron 
probability contour.
Image source
Figure 2.5
Generic Bohr model.
Image source",DCElectricalCircuitAnalysis.pdf
"model for copper would simply show four rings, the first three being filled and with 
a single electron in the fourth ring. This is shown in Figure 2.6. In this version, the 
individual electrons are drawn in each shell and the atomic number is indicated at 
the nucleus. Again, please do not imagine this representing individual electrons 
orbiting the nucleus in lanes. This is an energy level depiction.
It is worth noting that a lone outer electron is only loosely bound to the nucleus thus 
making the movement of charge through said material relatively easy, given the 
application of some external energy source. Consequently, copper is a very good 
conductor. This is a characteristic shared by common metals: they are good 
conductors because they only have one or two loosely held electrons in their outer 
shell. Examples include silver, gold and aluminum (one outer electron for each, all 
of them being very good conductors); and iron, tin and nickel (two outer electrons",DCElectricalCircuitAnalysis.pdf
"of them being very good conductors); and iron, tin and nickel (two outer electrons 
for each, and this group being not quite as conductive as those in the first group).
2.3 Charge and Current2.3 Charge and Current
As already noted, charge is an attractive force. It is denoted by the letter Q and has 
units of coulombs. Electrons are negatively charged and protons are positively 
charged. All electrons and protons exhibit the same magnitude of charge, roughly 
1.602E−19 coulombs. Thus, one coulomb is equivalent to the charge exhibited by 
approximately 1/1.602E−19, or 6.242E18 electrons. Further, opposite charges attract
while like charges repel, similar to the poles of a magnet. 
It is possible to move charge from one point to another. The rate of charge 
movement over time is called current. It is denoted by the letter I and has units of 
amps (or amperes)4. One amp of current is defined as one coulomb per second. 
1 amp ≡ 1 coulomb / 1 second (2.1)",DCElectricalCircuitAnalysis.pdf
"amps (or amperes)4. One amp of current is defined as one coulomb per second. 
1 amp ≡ 1 coulomb / 1 second (2.1)
That is, one amp can be visualized as approximately 6.242E18 electrons passing 
4 I stands for Intensity (of current), and was so named by André-Marie Ampère.
34
Figure 2.6
Bohr model of copper.
+29
Figure 2.7
Defining current as charge 
flow through a wire.",DCElectricalCircuitAnalysis.pdf
"through a wire in a period of one second. Consider Figure 2.7. Here we have a wire 
with electrons flowing through it in the direction of the arrow. We cut this wire with 
an imaginary plane, leaving us with the highlighted disk. Now imagine that we could
count the number of electrons passing through this disk over the course of one 
second. Because we know the charge possessed by one electron, we simply multiply 
the number of electrons by the charge for each to yield the total charge, and thus we 
arrive at the current. As a formula:
I = Q / t (2.2)
Where
I is the current in amps,
Q is the the charge in coulombs,
t is the time in seconds.
A common analogy for electric current is the flow of water through a pipe or river. 
Just as we can imagine water flow as “gallons or liters per minute”, we imagine 
electric current as “coulombs per second”. 
Example 2.1
In the course of one-half second, a certain battery delivers a charge of three 
coulombs. Determine the resulting current.
I = Q",DCElectricalCircuitAnalysis.pdf
"Example 2.1
In the course of one-half second, a certain battery delivers a charge of three 
coulombs. Determine the resulting current.
I = Q
t
I = 3C
0.5s
I = 6 A
Example 2.2
A device delivers a current of 25 mA. Determine the charge transferred in 
two seconds along with the equivalent total number of electrons.
I = Q
t
Q = I ×t
Q = 25 mA×2s    from Definition 2.1, amp-seconds is coulombs
Q = 50 mC
As one coulomb is equivalent to 6.242E18 electrons, just multiply to find 
the total number of electrons transferred.
35",DCElectricalCircuitAnalysis.pdf
"Total electrons = Q×number of electrons per coulomb
Total electrons = 50 mC×6.242E18
Total electrons = 3.121E17electrons
In sum, the larger the charge transferred within a given time period, the greater the 
current. Modern electrical and electronic systems might deal with currents of under a
picoamp or, at the other extreme, thousands of amps. That is an astonishing range. It 
is roughly equivalent to a single drop of water dripping from a leaky faucet each 
second compared to 1000 times the flow of water over Niagara Falls.
2.4 Energy and Voltage 2.4 Energy and Voltage 
Energy is defined as the ability to do work. It is denoted by the letter W. The basic 
unit is the joule although other units are sometimes used (for example, the calorie or 
the kilowatt-hour, kWh). 
If we were to move a charge from one point to another (for example, separating an 
electron from an atom), we would have to expend energy to do so. This is illustrated",DCElectricalCircuitAnalysis.pdf
"electron from an atom), we would have to expend energy to do so. This is illustrated 
in Figure 2.8. In this Figure, we would say that B has a higher electric potential than 
A. In other words, there is a potential difference between B and A. We refer to this 
change as voltage. It is denoted by the letter V (or sometimes E5) and has units of 
volts, in honor of Alessandro V olta. One volt is defined as one joule per coulomb. 
1 volt ≡ 1 joule / 1 coulomb (2.3)
 As you might guess, the bigger the charge to be moved, the greater the energy 
required. Expressed as a formula,
V = W / Q (2.4)
Where
V is the voltage in volts,
W is the energy in joules,
Q is the charge in coulombs.
Unlike current, voltage always implies two points for measurement because it 
involves a difference. Often, one of the points is a common reference, such as earth 
ground or a circuit ground (i.e., chassis ground). Sometimes people will refer to a",DCElectricalCircuitAnalysis.pdf
"ground or a circuit ground (i.e., chassis ground). Sometimes people will refer to a 
point in a circuit as having a certain voltage, as in “point X is 12 volts”. Although 
common in use, this is somewhat sloppy and not strictly correct. It is important to 
always remember that this value is relative to some reference point. As a general 
rule, a voltage will be denoted using the two points as subscripts, for example, VAB, 
that is, the voltage at point A relative to point B. If only a single subscript is used, as 
5 E is used for voltage sources such as batteries. It is short for EMF, or electromotive force.
36
Figure 2.8
Defining voltage as work to 
move charge.
A
B
Energy",DCElectricalCircuitAnalysis.pdf
"in VA, then the second, or reference, point is assumed to be the system common or 
ground. In this case, we're referring to the voltage at point A relative to the system 
common point. Finally, by definition, VAB = VA − VB, as they have the same reference.
Example 2.3
100 joules are expended to move a 20 coulomb charge from point A to point 
B. Determine the resulting voltage.
V BA = W
Q
V BA = 100 J
20C
V BA = 5 V
Note that it is possible for a voltage to be negative. This simply means that the 
potential at the point of interest is less than that of the reference point. In Example 
2.3 we discovered that point B is five volts above point A. We could just as easily 
say that point A is five volts below point B, or VAB = −5 V . Further, we can state the 
difference in terms of the individual ground-referenced voltages, or VBA = VB − VA. 
Static Electricity and ESDStatic Electricity and ESD
While it is obviously true that higher voltages imply higher associated energies",DCElectricalCircuitAnalysis.pdf
"Static Electricity and ESDStatic Electricity and ESD
While it is obviously true that higher voltages imply higher associated energies 
(charge being held constant), it is not true that a particularly high voltage is 
necessarily lethal. This is because a very high voltage can be achieved by moving a 
small charge with only modest energy input. A good example of this is static 
electricity, so called because it is not associated with a moving current. 
Static electricity is commonly generated through the triboelectric effect which 
involves the transfer of electrons from one material to another via physical contact 
such as rubbing or scraping. If said materials are good electrical insulators, charges 
will remain on the materials and can build to very high levels, creating a large 
voltage. Many plastics, such as polystyrene and polyester, are good candidates. The 
effect can be noticed with certain fabrics, especially under low humidity. For",DCElectricalCircuitAnalysis.pdf
"effect can be noticed with certain fabrics, especially under low humidity. For 
example, removing a polyester fleece pullover or jacket can elicit a certain crackling
sound. The sliding of the fleece builds up the charge and eventually the voltage will 
be become so large that it will arc through the air to surrounding objects which have 
a lower voltage. This happens quickly over many parts of the garment, each crackle 
being an individual arc. In fact, if tried in darkness, it is possible to see a cascade of 
small sparks. This is the same phenomenon that causes a spark when you touch a car
after sliding off the seat on a cold and dry winter's day, or a small shock when you 
touch an object (or another person) after walking across a carpet in a dry library. 
37",DCElectricalCircuitAnalysis.pdf
"Anyone who has opened a box filled with polystyrene packing peanuts can
attest to the troublesome nature of the triboelectric effect, as the very light
peanuts can easily adhere to other objects due to the electric charge generated
through their displacement. No amount of manic brushing or throwing of the
pieces will reduce the effect and may, in fact, make it worse. A simple
solution in some instances is to spray a fine mist of water on the packing
peanuts as the water will provide a conduction path, draining off the
charges. Of course, this will not be appropriate in all situations, particularly
in the one shown in Figure 2.9. In the prior examples, the static voltage
may be on the order of a few thousand volts but the associated energy may
be just a few microjoules. In spite of the high voltage, this is not enough to
kill someone. On the other hand, the same voltage achieved with a much
higher charge and energy could be lethal.",DCElectricalCircuitAnalysis.pdf
"kill someone. On the other hand, the same voltage achieved with a much
higher charge and energy could be lethal.
Beyond its simple inconvenience and inadvertent feline entertainment capabilities, 
high static potential can damage sensitive electronic devices. Care must be taken    
to prevent the accidental buildup of damaging charges. In the electronics industry 
this is commonly referred to as ESD, or electrostatic discharge. Steps to reduce   
ESD include humidity control and the use of conductive devices such as resistive 
wrist straps for technicians to continually bleed off the charges, thus preventing     
the creation of a high static potential.
The Height AnalogyThe Height Analogy
Just as the flow of water can be seen as an analogy for electrical current, a 
serviceable analogy for voltage involves pressure or height. Indeed, sometimes 
voltage is referred to as “electrical pressure”. The height analogy ties together the",DCElectricalCircuitAnalysis.pdf
"voltage is referred to as “electrical pressure”. The height analogy ties together the 
concepts of voltage and energy. In this analogy, height corresponds to voltage and 
mass corresponds to charge.
To begin, we note that there are two kinds of energy: kinetic energy, or energy of 
motion; and potential energy, or energy by virtue of position. Potential energy is the 
product of mass, gravity and height, or w = mgh. Keeping gravity equal, we see that 
the more mass something has or the higher up it is, the greater its potential energy. 
We might think of potential energy as the object's potential to inflict damage when 
released.
To illustrate, we shall call upon 18th century Scottish philosopher and noted soccer 
fan6, David Hume. If a soccer ball is held stationary over Mr. Hume's head, as shown
in Figure 2.10, the ball has energy by virtue of its position. Releasing the ball from 
the position shown will scarcely bother Mr. Hume as there is little energy associated",DCElectricalCircuitAnalysis.pdf
"the position shown will scarcely bother Mr. Hume as there is little energy associated 
with this position relative to the top of his noggin. In fact, he would be hard pressed 
to head the ball to another player. If, on the other hand, the ball were held 
6 Hume, author of An Enquiry Concerning Human Understanding, died in 1776. The 
modern game of soccer was established in 1863. Let's not let that deter our own inquiry.
38
Figure 2.9
Triboelectric effect and static 
electricity: Cat fur meets 
polystyrene. Image source.
Figure 2.10
Understanding voltage: 
David Hume does a header.",DCElectricalCircuitAnalysis.pdf
"considerably higher, its potential energy would be much greater. Therefore, the 
impact on Mr. Hume's head would be increased dramatically and he would have 
little difficulty heading the ball down field (i.e., the transformation of potential 
energy into kinetic energy), although the chances of him gaining a concussion are 
greatly increased. 
Thus, we note that the height of the ball gives us some insight into its potential 
energy, although these terms are not synonymous. That is to say, if, instead of a 
soccer ball, we had used a ping-pong ball, even when dropped from an extreme 
height, the chances of a concussion are non-existent7. On the other hand, if the ball 
had been replaced with one made of solid iron, a drop from even a modest height 
could render our most excellent philosopher seriously dead. So it is with voltage. If 
the charge associated with the voltage is small, even a relatively high voltage will",DCElectricalCircuitAnalysis.pdf
"the charge associated with the voltage is small, even a relatively high voltage will 
not be lethal (as in the case of simple static electricity on clothing), however, if 
associated with a sufficiently large charge, a much lower voltage can be deadly.
2.5 Power and Efficiency2.5 Power and Efficiency
The terms power and energy are often used incorrectly as synonyms. Although 
related, they are not the same thing. As already mentioned, energy is the ability to do
work. In contrast, power is the rate of energy usage. Power is denoted by the letter P
and has units of watts, although other units are sometimes used such as the 
horsepower (1 horsepower ≈ 746 watts). One watt is defined as one joule of energy 
consumed per second.
1 watt ≡ 1 joule / 1 second (2.5)
As a formula,
P = W / t (2.6)
Where
P is the power in watts,
W is the energy in joules,
t is the time in seconds.
To better understand the concept, consider for a moment a delicious peanut butter",DCElectricalCircuitAnalysis.pdf
"W is the energy in joules,
t is the time in seconds.
To better understand the concept, consider for a moment a delicious peanut butter 
and banana sandwich. This sandwich contains a certain number of food calories, let's
say 300 in total. A food calorie refers to a certain amount of energy that humans can 
extract from an item of food. That energy enables us to do some manner of work 
such as walking, swimming or just breathing. The sandwich can be seen as an 
energy storage medium, a battery for biological units called humans. The question 
7 The construction of the referred sentence owes a certain debt to the writing style of Mr. 
Hume.
39",DCElectricalCircuitAnalysis.pdf
"is, what do we do with the energy, and more to the point, how fast do we use it? 
For example, that sandwich might be sufficient to allow someone to run a 5k (3.1 
mile) road race in 17 minutes. In contrast, it also might be sufficient to allow that 
same person to watch television for three hours. It's the same amount of energy 
being used, it's just being used at a much faster rate in the former case. That rate is 
power. The 5k runner has a much higher power output than the TV watcher. 
Example 2.4
100 joules are consumed by a device in 0.1 seconds. Determine the power in
watts and in horsepower.
P = W
t
P = 100 J
0.1s
P = 1000W
As one horsepower is approximately 746 watts, this is equivalent to
Php = PW
746 W/hp
Php = 1000W
746 W/hp
Php = 1.34 hp
Power can also be found by multiplying a current by the associated voltage. To 
begin, we note the definitions of current and voltage, Equations 2.2 and 2.4 
respectively, and then combine them.
I = Q
t
V = W
Q
I ×V = Q
t × W
Q = W
t",DCElectricalCircuitAnalysis.pdf
"respectively, and then combine them.
I = Q
t
V = W
Q
I ×V = Q
t × W
Q = W
t
From equation 2.6, we know that P = W/t, thus P = IV. This is known as power law. 
P = I ×V (2.7)
Where
P is the power in watts,
I is the current in amps,
V is the voltage in volts.
40",DCElectricalCircuitAnalysis.pdf
"Example 2.5
If a 9 volt battery delivers a current of 0.1 amps, determine the power 
delivered in watts. 
P = I ×V
P = 0.1amps × 9volts
P = 0.9W
EfficiencyEfficiency
Efficiency is the ratio of useful output power to applied power expressed as a 
percentage. It is denoted by the Greek letter η (eta) and is always less than 100%. 
Expressed as a formula,
η = Pout
Pin
×100 % (2.8)
Where
η is the efficiency in percent,
Pout is the output power,
Pin is the input power.
Generally speaking, the higher the efficiency, the better. This implies less
waste. In other words, if a system is 30% efficient, then 70% of the input
power is wasted, whereas if a system is 99% efficient, then only 1% of the
input power is wasted. The concept is illustrated graphically in Figure 2.11.
In most systems, waste power turns into heat which is not a desired
commodity, and in fact often reduces the lifespan of electrical components.
Example 2.6",DCElectricalCircuitAnalysis.pdf
"commodity, and in fact often reduces the lifespan of electrical components.
Example 2.6
If a device draws 200 watts of power and has a useful output of 120 watts, 
determine the efficiency. 
η = Pout
Pin
× 100 %
η = 120 W
200 W × 100 %
η = 60%
In this case, the device is wasting 40% of the input power, or 80 watts. 
41
Figure 2.11
Basic concept of efficiency.
Input
Useful 
Output
Waste",DCElectricalCircuitAnalysis.pdf
"Example 2.7
An audio amplifier has a maximum rated output of 100 watts to a 
loudspeaker. If it exhibits an efficiency of 70%, determine the input power 
required and the amount of power wasted. 
η = Pout
Pin
× 100 %
Pin = Pout
η × 100%
Pin = 100 W
70 % × 100 %
Pin = 142.9 watts
As 142.9 watts are drawn by the amplifier and only 100 watts are delivered 
to the loudspeaker, then the difference, or 42.9 watts, is wasted power (most 
likely just making the amplifier hot). 
2.6 Energy Cost and Battery Life2.6 Energy Cost and Battery Life
As we have seen, knowing the voltage and current demands of a given device allows
us to determine its power rating and energy consumption. The next steps are to 
determine the cost of operating a device and, if it's battery powered, how long the 
device will last before needing new batteries.
Computing Energy CostComputing Energy Cost
Once we know the power drawn by the device and for how long it will be used, the",DCElectricalCircuitAnalysis.pdf
"Computing Energy CostComputing Energy Cost
Once we know the power drawn by the device and for how long it will be used, the 
cost to operate a device can be determined given the per unit cost of energy. In spite 
of the fact that many people refer to their local electricity supplier as “the power 
company”, we do not buy “power”, per se. Rather, we are billed for energy. 
Although it would be possible to determine the cost per joule (or more practically, 
per megajoule), suppliers normally bill based on kilowatt-hours, or kWh, the product
of the power and time8. This unit is used because most residential and commercial 
devices and appliances are rated in terms of power consumption in watts. 
Multiplying the power consumption in watts by the length of time the device is used 
in hours yields a watt-hour value. This is then scaled by a factor of one thousand to 
arrive at kWh. For example, a 1500 watt toaster oven used for 30 minutes (i.e., 0.5",DCElectricalCircuitAnalysis.pdf
"arrive at kWh. For example, a 1500 watt toaster oven used for 30 minutes (i.e., 0.5 
hours) yields 750 watt-hours, or 0.75 kWh. Finally, knowing the per kWh cost, a 
simple multiplication yields the cost of electricity. Thus, if the utility charges 10 
8 Remember, power is the rate of energy usage per unit time, and thus multiplying power 
by time yields energy. One kWh is approximately equal to 3.6 megajoules.
42",DCElectricalCircuitAnalysis.pdf
"cents per kWh, then the cost to run that toaster oven is 7.5 cents. If it is used for a 
full hour then it costs 15 cents, and so on.
To put usage in perspective, a typical home in the USA or Canada uses about 900 
kWh per month while households in many countries in Europe might use one half to 
one quarter of that amount. Global electricity generation is around 25 million 
gigawatt hours per year. Typical electricity rates in the USA are between ten and 
twenty cents per kWh, depending on geographic region and sector (e.g., residential 
or commercial).
Example 2.8
A 100 watt incandescent light bulb is left on for 24 hours. If the cost of 
electricity is 15 cents per kWh, determine the cost to run the light. 
Cost = P× t × rate
Cost = 100 W × 24 hours × 0.15  $/kWh
Cost = $0.36
At this point, it should be obvious that the more efficient a device is, the less 
expensive it will be to run. In fact, it is quite possible that in the long run, a device",DCElectricalCircuitAnalysis.pdf
"expensive it will be to run. In fact, it is quite possible that in the long run, a device 
that is more expensive than a similar, though less efficient, device may be 
considerably less expensive to use over the course of its lifespan. A good example is 
a comparison between an ordinary incandescent light bub and an LED light. LED 
lighting can be an order of magnitude more efficient than incandescent lighting. 
Indeed, incandescent bulbs may convert less than five percent of their input into 
usable light output, the remaining 95 percent just turning into heat. While they are 
less expensive to purchase initially, their operating cost is much higher. A proper 
way to compare lights is to examine their light output in lumens, not their power 
consumption. For example, a 60 watt incandescent light produces about 800 lumens 
of illumination. That same level of illumination can be obtained with an LED 
drawing just 9 watts. Further, typically LED lights last ten to twenty times longer",DCElectricalCircuitAnalysis.pdf
"drawing just 9 watts. Further, typically LED lights last ten to twenty times longer 
than incandescent bulbs. 
This will be illustrated in the next example.
 
Example 2.9
A certain 14 watt LED light produces the same illumination as a 75 watt 
incandescent light bulb. Assume the LED costs $11 and has an expected life 
of 15,000 hours. The incandescent costs 50 cents each and has an expected 
life of 1000 hours. If the cost of electricity is 12 cents per kWh, determine 
the cost to run each version for 15,000 hours. 
43",DCElectricalCircuitAnalysis.pdf
"First, it should be noted that 15 incandescent bulbs will be needed. At 50 
cents each, that's $7.50 for the bulbs. The cost to run them is,
Cost = P× t × rate
Cost = 75W × 15000 hours × 0.12  $/kWh
Cost = $135.00
The total is $ 142.50. Now for the single LED required:
Cost = P× t × rate
Cost = 14W × 15000 hours × 0.12  $/kWh
Cost = $25.20
The total is $36.20, a considerable savings, not to mention other positive 
factors including only having to change the light once instead of fifteen 
times; a considerable reduction in wasted energy, thus lowering demand and 
environmental impact; and finally, a reduction in burned out light bulbs for a
further lowering of environmental impact (waste stream reduction).
BatteriesBatteries
A battery is a device used to store electrical energy, generally in the form of a 
chemical cell. Ideally, it presents a constant voltage, its current varying according to 
what it drives. In reality, as the battery is used, its voltage will begin to decrease.",DCElectricalCircuitAnalysis.pdf
"what it drives. In reality, as the battery is used, its voltage will begin to decrease. 
Eventually, the energy stored in the battery will be exhausted and its voltage will 
drop to zero. The storage capacity of a battery is measured in amp-hours, Ah (or 
milliamp-hours, mAh, for smaller batteries). All other factors being equal, the 
battery with the higher amp-hour rating will last longer before being depleted. This 
is true for the same size of battery using different compositions (e.g., zinc-carbon 
versus alkaline), as well as batteries of different physical sizes but having the same 
voltage. For example, all AAA, AA, C and D cell batteries have a nominal voltage of 
1.5 volts. If they are all of the same type, such as alkaline, then the immediate 
practical difference is that the larger the physical size, the greater its energy storage, 
and thus the longer it will last. 
Over a modest range of currents, the expected lifespan of a battery can be computed",DCElectricalCircuitAnalysis.pdf
"and thus the longer it will last. 
Over a modest range of currents, the expected lifespan of a battery can be computed 
based on its amp-hour rating and the current drawn from it. 
Battery life ≈ amp-hour rating / current draw (2.9)
This equation is best used as a rough guide. If the current draw is considerably 
higher than the current at which the battery was tested, the predicted lifespan will be 
overly optimistic. On the other hand, if the current draw is considerably lower, it is 
likely that the battery will last longer than predicted. 
44",DCElectricalCircuitAnalysis.pdf
"The maximum current output of a battery is likewise limited. If such were not the 
case, we might expect even very small batteries to produce phenomenally large 
currents for very shorts periods of time. This is not the case. 
A graph of discharge curves at various load currents is shown in Figure 2.12. Notice 
how the battery voltage begins at the rated 1.5 volts and then begins to fall. After a 
certain point, the rate of decrease accelerates and “drops off of a cliff”. The actual 
service life will also be application dependent in that some devices can tolerate a 
lower voltage than others. For example, an old-fashioned flashlight will still work 
with largely depleted batteries, it simply won't be that bright. If we were to consider 
75% of rated voltage as the useful lower limit (a little over 1.1 volts), we see that at 
a 50 mA draw, the battery will last around 18 hours, achieving about 900 mAh. At",DCElectricalCircuitAnalysis.pdf
"a 50 mA draw, the battery will last around 18 hours, achieving about 900 mAh. At 
100 mA, it will last around 7 hours, yielding 700 mAh. If we used 1.0 volts as our 
lower usable limit, we would arrive at 1.05 Ah and 910 mAh, respectively.   
Example 2.10
A certain battery is rated at 10 Ah. Approximately how long will it last with 
a 0.5 amp draw?
Lifespan ≈ Ah
I
Lifespan ≈ 10 Ah
500 mA
Lifespan ≈ 20 hours
45
Figure 2.12
Battery discharge curves for 
various load currents at room 
temperature.
Courtesy of Duracell",DCElectricalCircuitAnalysis.pdf
"Remember, 20 hours is only an approximation. For a more accurate rendering, 
consider the following example.
Example 2.11
Using the graph of Figure 2.12, determine the expected lifespan with a 100 
milliamp draw for a lower voltage limit of 1.2 volts. Also determine the 
effective amp-hour rating at this point.
The 50 mA curve (purple) passes through 1.2 volts at approximately 5 hours.
This is the expected lifespan. The corresponding amp-hour rating is
Ah ≈I ×t
Ah ≈50 mA×5hours
Ah ≈250 mAh
Batteries also tend to show decreased performance with reductions in temperature. 
This is shown in Figure 2.13. For this particular battery, the capacity at freezing 
(0°C) is roughly half of what it is at room temperature (21°C). Peak current capacity 
may also be reduced with temperature. This is true of many kinds of batteries, 
including large 12 volt automotive batteries. This is a major reason why cars are 
often much more difficult to start on very cold days; their current capacity is",DCElectricalCircuitAnalysis.pdf
"often much more difficult to start on very cold days; their current capacity is 
reduced, thereby reducing the output of the starter motor.
46
Figure 2.13
Battery discharge curves for 
different temperatures.
Courtesy of Duracell",DCElectricalCircuitAnalysis.pdf
"A table of typical amp hour ratings for common battery sizes is shown in Figure 
2.14. These values are appropriate for good quality alkaline batteries. Rechargeable 
NiMH (nickel-metal hydride) would be around the same. Remember, these are just 
approximations.
2.7 Resistance and Conductance2.7 Resistance and Conductance
We have seen that both current and voltage deal with the movement of charge. 
Consequently, in any electrical system, voltage and current are interrelated. Let us 
consider the most simple case. This would involve a single voltage source, such as a 
battery, driving a single, homogeneous item such as a length of wire or a block of a 
given material. The physical characteristics of this item will dictate how much 
current will flow. In general terms,
 Effect = Cause
Oppostion
In this case the cause is the voltage source and the effect is the resulting current. The 
opposition is the characteristic of the item in question, in other words, its ability to",DCElectricalCircuitAnalysis.pdf
"opposition is the characteristic of the item in question, in other words, its ability to 
restrict current flow. We call this characteristic resistance. In other words, resistance 
is a measure of how difficult it is to establish a flow of current (i.e., “resistance to 
current flow”) under a given set of circumstances. It is denoted by the letter R and 
has units of ohms, in honor of Georg Ohm, a researcher from the early 1800s. The 
unit is denoted by the capital Greek letter omega, Ω. 
Sometimes it is more convenient to use the inverse view of this phenomenon, and 
instead of referring to how difficult it is to establish a current, we would be 
interested in how easy it is to establish a current. This is called conductance and it is 
the reciprocal of resistance. Conductance is denoted by the letter G and has units of 
siemens, named after Werner von Siemens. The unit is abbreviated as S. 
R = 1
G
G = 1
R
(2.10)
47
Battery Size Capacity (mAh)
AAA 1000
AA 2500
C 5000
D 10000",DCElectricalCircuitAnalysis.pdf
"R = 1
G
G = 1
R
(2.10)
47
Battery Size Capacity (mAh)
AAA 1000
AA 2500
C 5000
D 10000
9 V olt 500
Figure 2.14
Typical battery capacities.",DCElectricalCircuitAnalysis.pdf
"In this simple scenario, resistance is a function of the material the current is passing
through along with its shape. This is illustrated in Figure 2.15 where the arrow
shows the direction of current flow. An obvious question is “What is this block
made of?” It should come as no surprise that the material chosen has a great
impact on the current. We have already seen that metals such as copper are good
conductors of electricity. Other materials, such as certain plastics and ceramics,
are not. These materials are referred to as insulators. 
The measure of a material's inherent and general ability to restrict current flow is
called resistivity. Resistivity is denoted by the Greek letter ρ (rho). All other
factors being equal, the higher the resistivity, the greater will be the resistance.
Further, the greater the length of the material that the current must pass through, the 
greater the resistance. Finally, as the surface area of the cross section (i.e, the face)",DCElectricalCircuitAnalysis.pdf
"greater the resistance. Finally, as the surface area of the cross section (i.e, the face) 
grows, the resistance decreases. Expressed as a formula
R =ρl
A (2.11)
Where
R is the resistance in ohms,
ρ is the resistivity,
l is the length of the material,
A is area of the face (h times w).
Resistivity is often specified in ohm-centimeters with the length and area similarly 
specified in centimeters and square centimeters, respectively. A table of resistivity 
values for a variety of materials is shown in Figure 2.16. Note that resistivity is not 
necessarily constant across temperature. Indeed, this change can be exploited as a 
means of measuring temperature. In other applications, we might need it to be as 
stable as possible across temperature. This need led to the creation of the alloys 
Constantan and Manganin in the late 1800s which exhibit very high stability across 
temperature.
From this table we can see that silver has lower resistivity than copper which in turn",DCElectricalCircuitAnalysis.pdf
"temperature.
From this table we can see that silver has lower resistivity than copper which in turn 
is lower than gold. This means that if we made identically sized wires of these three 
materials, the silver version would have the least resistance and gold the highest. 
Why then, do audio and video cables often feature gold plating? Certainly, it isn't 
due to lower resistivity and enhanced conductivity. The reason is that gold is a noble 
metal, meaning that it does not tarnish. In contrast, the surface of both silver and 
copper will oxidize, creating a patina (the dark “stain” noticeable on old silver and 
copper implements). The oxide will create a high resistance layer and reduce the 
integrity of the connection.
It is important to note that Formula 2.11 does not include a term for the mass of the 
material. It is only concerned with the shape of the item. This is an important 
distinction. If we were to alter the mass but keep the ratio of the length versus the 
48
Figure 2.15",DCElectricalCircuitAnalysis.pdf
"distinction. If we were to alter the mass but keep the ratio of the length versus the 
48
Figure 2.15
Defining resistance.
w
l
h",DCElectricalCircuitAnalysis.pdf
"area the same, the resulting resistance would be unchanged. In general, the increase 
in mass by itself does not necessarily alter the resistance but it may have an impact 
on the power handling capability of the device. 
In contrast, if we take that that original mass and reshape it, or just apply the current 
to a difference face, such that the effective surface area and length change, then the 
resulting resistance will also change. This is illustrated in Figure 2.17. Here we have 
taken the item shown in Figure 2.15 and directed the current flow from top to 
bottom rather than through the side. In this orientation, the surface area is much 
increased and the length through which the current must travel is greatly reduced. 
49
Figure 2.16
Resistivity values for various 
materials.
Material Resistivity
ρ (Ω·cm) at 20 °C 
Temperature
coefficient
(K−1) 
Silver 1.59×10−6 0.0038
Copper 1.68×10−6 0.00404
Gold 2.44×10−6 0.0034
Aluminium 2.65×10−6 0.0039
Zinc 5.90×10−6 0.0037",DCElectricalCircuitAnalysis.pdf
"coefficient
(K−1) 
Silver 1.59×10−6 0.0038
Copper 1.68×10−6 0.00404
Gold 2.44×10−6 0.0034
Aluminium 2.65×10−6 0.0039
Zinc 5.90×10−6 0.0037
Nickel 6.99×10−6 0.006
Iron 9.7×10−6 0.005
Platinum 1.06×10−5 0.00392
Tin 1.09×10−5 0.0045
Titanium 4.20×10−5 0.0038
Manganin 4.82×10−5 0.000002 
Constantan 4.90×10−5 0.000008
Stainless steel 6.90×10−5 0.00094
Nichrome 1.10×10−4 0.0004
Carbon (amorphous) 5×10−4 to 8×10−3 −0.0005
Silicon 6.4×104 −0.075
Glass 1×1013 to 1×1017
Carbon (diamond) 1×1014
Hard rubber 1×1015
Air 1011 to 1017
PET 1×1023
Teflon 1×1025 to 1×1027
Figure 2.17
Resistance will change due to 
to altering area and length 
with mass unchanged.",DCElectricalCircuitAnalysis.pdf
"Consequently, the effective resistance in this orientation will be considerably less 
than that seen in the original.
Example 2.12
A certain material has a resistivity of 0.2 ohm-centimeters. Determine the 
resistance of a piece that is 0.3 cm wide, 0.5 cm high and 4 cm long.
A =h×w
A =0.5cm×0.3cm
A =0.15 cm2
R = ρl
A
R =0.2Ωcm×4cm
0.15cm2
R =5.333Ω
For our next example, let's consider a spool of wire. In many cases we treat wire  
ideally, as if it has no resistance. While this can be a good approximation in many 
instances, especially with relatively short runs of wire, such is not always the case.
Example 2.13
A certain gauge of copper wire has a diameter of 0.6 mm. Determine the 
resistance if the spool is 200 meters long. 
The table of Figure 2.16 indicates that the resistivity of copper is 1.68E−6 
ohm-centimeters. The trick here is that we must keep the units consistent. As
there are 100 centimeters to the meter, the length is 200 meters times 100, or",DCElectricalCircuitAnalysis.pdf
"there are 100 centimeters to the meter, the length is 200 meters times 100, or
20,000 centimeters (i.e., 20E3). Given that there are 10 millimeters to the 
centimeter, the diameter must be decreased by a factor of ten, yielding a 
diameter of 0.06 cm and thus a radius of 0.03 cm. 
A = π r2
A = π(0.03cm)2
A =2.83E-3cm2
R = ρl
A
R =1.68E-6Ωcm×20E3 cm
2.83E-3cm2
R =11.9Ω
50",DCElectricalCircuitAnalysis.pdf
"The amount of resistance seen in Example 2.13 would be considered excessive if the
item to be connected is something as simple as a loudspeaker, which typically would
be around 8 Ω. And while no one would likely need 200 meters of cable to connect a
loudspeaker in their home, that sort of distance would be unremarkable in a large 
stadium or airport terminal. Don't forget, we'd need wire to and from the 
loudspeaker, achieving a total separation of 100 meters at most.
Various thicknesses of cables and wires are used for a wide variety of purposes. To 
make this easier, wire thicknesses have been standardized. The most common 
standard in North America is the American Wire Gauge, or AWG. This is a non-
metric specification with origins in the mid nineteenth century. The larger the gauge 
number, the smaller the wire diameter, and the less current it safely can carry. To put 
the gauge numbers in perspective, typical small home appliances use 16 or 18 gauge",DCElectricalCircuitAnalysis.pdf
"the gauge numbers in perspective, typical small home appliances use 16 or 18 gauge 
wire, basic home wiring uses 12 gauge (with a 20 amp circuit breaker), and hook-up 
wire used in an electrical circuits or electronics laboratory solderless breadboard is 
commonly 22 or 24 gauge. 
For general purpose wiring, copper is by far the most common metal used because it
is highly conductive and relatively inexpensive. Some other metals are used in 
special cases, for instance, aluminum is used for long distance power transmission 
lines. 
51
AWG 
Diameter Resistance/length
(in) (mm) (Ω/km) (Ω/1000ft) 
0000 (4/0) 0.4600 11.684 0.1608 0.04901
00 (2/0) 0.3648 9.266 0.2557 0.07793
0 (1/0) 0.3249 8.251 0.3224 0.09827
2 0.2576 6.544 0.5127 0.1563
4 0.2043 5.189 0.8152 0.2485
6 0.1620 4.115 1.296 0.3951
8 0.1285 3.264 2.061 0.6282
10 0.1019 2.588 3.277 0.9989
12 0.0808 2.053 5.211 1.588
14 0.0641 1.628 8.286 2.525
16 0.0508 1.291 13.17 4.016
18 0.0403 1.024 20.95 6.385
20 0.0320 0.812 33.31 10.15",DCElectricalCircuitAnalysis.pdf
"12 0.0808 2.053 5.211 1.588
14 0.0641 1.628 8.286 2.525
16 0.0508 1.291 13.17 4.016
18 0.0403 1.024 20.95 6.385
20 0.0320 0.812 33.31 10.15
22 0.0253 0.644 52.96 16.14
24 0.0201 0.511 84.22 25.67
26 0.0159 0.405 133.9 40.81
32 0.00795 0.202 538.3 164.1
40 0.00314 0.0799 3441 1049
Figure 2.18
AWG wire data for copper 
conductors.",DCElectricalCircuitAnalysis.pdf
"A table of gauge sizes and associated parameters is shown in Figure 2.18. This table 
assumes copper is being used for the wire. Note that as the diameter of the wire 
decreases, the amount of resistance for a particular length increases, as expected 
from Equation 2.11.
Using this table we can perform a quick crosscheck of Example 2.13. The wire 
diameter used in that example was 0.6 mm which is just a little smaller than AWG 22
as listed in the table. Further, #22 wire is listed as having a resistance of 
approximately 53 Ω per km. For 200 meters, as used in the example problem, this 
works out to 10.6 Ω. As #22 wire is slightly larger in diameter, we expect it to show 
slightly less resistance than the calculated value of 11.9 Ω, which it does.
While it may not be immediately apparent, gauge numbers proceed in a logarithmic 
fashion based on diameter. Stepping up one gauge number (e.g., from #10 to #11)",DCElectricalCircuitAnalysis.pdf
"fashion based on diameter. Stepping up one gauge number (e.g., from #10 to #11) 
decreases the diameter by a factor of approximately 0.89. As resistance is inversely 
proportional to the square of the diameter (i.e., the area), the resistance increases by 
over 25%. Even numbered sizes are particularly common in use and a jump to the 
next higher even gauge number (e.g., #18 to #20) produces a resistance increase of 
nearly 60%.  
Resistors and the Resistor Color CodeResistors and the Resistor Color Code
Resistors are devices used to control the currents and voltages in a circuit. They are 
available in many shapes and sizes, and are normally designed to maintain stable 
ohmic values in spite of environmental changes such as temperature and humidity. A
sample of different resistor styles is shown in Figure 2.19. As a general rule, the 
larger the resistor, the more power it can handle. At the back of the figure is a large",DCElectricalCircuitAnalysis.pdf
"larger the resistor, the more power it can handle. At the back of the figure is a large 
ceramic power resistor using a rectangular aluminum housing. This device is rated 
for 200 watts of dissipation. Immediately in front and to the right of it are several 
smaller ceramic power resistors with ratings in the 5 to 20 watt range. Along the left 
side is a set of carbon composition and carbon film resistors ranging from 1 watt 
down to one-tenth watt in dissipation. Toward the center is a multi-lead chip resistor 
that contains several resistors in one package. With few exceptions, all of these items
are classified as “through-hole” components, that is, their leads are designed to go 
through pre-drilled holes in a printed circuit board. These are also the most 
commonly used components in an electrical lab as the leads fit into solderless 
breadboards and the components are a convenient size. 
52",DCElectricalCircuitAnalysis.pdf
"With the increasing desire to shrink components, modern production designs         
use surface mounting techniques in place of through-hole construction. At the 
bottom-center is a small dot which is, in fact, a surface mount resistor capable
of ½ watt of power dissipation. A close-up is shown in Figure 2.20. Obviously,
these devices are too small to be practical to use without special equipment.
Resistors are linear bilateral devices. Being linear, their current-voltage
relation   can be drawn as a straight line. Bilateral means that the polarity of
voltage or direction of current will not matter. In other words, unlike a battery,
these devices cannot be inserted into a circuit backwards because either
orientation works the same way. If the horizontal axis is voltage and the vertical
is current, then the slope of the line yields the conductance. Thus, the steeper
the line, the lower its resistance. This is illustrated in Figure 2.21.",DCElectricalCircuitAnalysis.pdf
"the line, the lower its resistance. This is illustrated in Figure 2.21. 
Not all electrical components are linear and bilateral. A good example is the
semiconductor diode, a commonly used device in electronic systems. The
current-voltage plot of a diode is shown in Figure 2.22 for comparison.
Note that the plot is not a simple straight line, thus it is not linear. Further,
the first and third quadrant responses are wildly different, indicating that
polarity is of utmost importance. Clearly, it matters which way these kinds
of devices are inserted into a circuit.
53
Figure 2.19
A variety of different resistors.
Figure 2.20
Close up of a surface mount 
resistor.
Figure 2.21
Current-voltage plot of two 
resistors.
V
I
-I
-V",DCElectricalCircuitAnalysis.pdf
"Resistors are available in standardized ohmic values and at standardized power 
ratings (see Appendix A). Along with their resistance value, resistors also have a 
specified tolerance. This specifies an allowable range of variation of the stated value.
For example, a 220 ohm resistor may have a tolerance of 10%. This means that the 
actual value of any given specimen from a box of these resistors may be off of the
nameplate or nominal value by 10% or 22 ohms. Thus, any particular resistor
might be as high as 242 ohms or as low as 198 ohms. 
General purpose resistors use a color code to denote their value and tolerance.
Typically, this will involve four color stripes: two for the precision/mantissa, one
for the power of ten, and the fourth to indicate the tolerance. For higher precision,
a five stripe version may be used with the first three denoting the precision/
mantissa. Alternately, high precision resistors may have their nominal value",DCElectricalCircuitAnalysis.pdf
"mantissa. Alternately, high precision resistors may have their nominal value
printed directly on them. Refer to Figure 2.23 for an example of the basic variety. 
The first two bands, here yellow and violet, indicate the precision or leading digits. 
The third band, here orange, indicates the power of ten or “number of zeroes” to 
add. The fourth band, silver in this example, indicates the tolerance. Note that the 
fourth band is spaced away from the others to avoid accidentally reversing the order.
The color code is shown in Figure 2.24. To assist in remembering the sequence, a 
number of mnemonic aids have been used, the first letter of each word starting with 
the same letter as the corresponding color. One example is the “Picnic Basket 
Mnemonic” which is: Black Bears Robbed Our Yummy Goodies Beating Various 
Gray Wolves. Another possibility is to note that the middle section follows a 
rainbow with black and white at the extreme ends.
Color Numeral
Black 0
Brown 1
Red 2
Orange 3",DCElectricalCircuitAnalysis.pdf
"rainbow with black and white at the extreme ends.
Color Numeral
Black 0
Brown 1
Red 2
Orange 3
Yellow 4
Green 5
Blue 6
Violet 7
Gray 8
White 9
For small resistor values
If the multiplier band is gold then 
divide by10; if silver, divide by 100.
54
Figure 2.22
Current-voltage plot of a 
diode: neither linear nor 
bilateral.
V
I
-I
-V
Figure 2.23
Example of basic four stripe 
resistor color code.
Figure 2.24
Resistor color code.",DCElectricalCircuitAnalysis.pdf
"The tolerance band colors are as follows: For basic parts silver is ±10% and gold is 
±5%. If the fourth band is omitted, this indicates a tolerance of ±20%, although it is 
seldom used in modern designs. For precision parts some colors are reused but 
follow the color code numerals: Brown is ±1% tolerance and red is ±2% tolerance. 
Tighter tolerances are also available. Sometimes an extra band is added that 
indicates a reliability rating or temperature coefficient. In such cases, it is best to 
refer to the manufacturer's data sheets for details. We will not pursue these further.
Example 2.14
Determine the nominal, maximum and minimum acceptable resistance 
values for the resistor pictured in Figure 2.17.
The colors are yellow-violet-orange. This translates to 4, 7 and 3. The value 
is “47 with 3 zeroes”, or 47000 ohms. The silver fourth band indicates 10% 
tolerance. Thus, the resistor pictured is nominally 47 kΩ with ±10%",DCElectricalCircuitAnalysis.pdf
"is “47 with 3 zeroes”, or 47000 ohms. The silver fourth band indicates 10% 
tolerance. Thus, the resistor pictured is nominally 47 kΩ with ±10% 
variation around the nominal value being acceptable. The tolerance yields 
±4.7 kΩ, so the acceptable range is from 42.3 kΩ to 51.7 kΩ.
Example 2.15
A precision resistor has a color code of orange-blue-green-brown-brown. 
Determine its value and acceptable range.
The final band indicates that this is a ±1% tolerance component. The first 
four colors translate to 3, 6, 5 and 1. The value is “365 with 1 zero”, and 
thus 3650 ohms, or 3.65 kΩ. The allowable range is 3.65 kΩ ± 36.5 Ω.
Example 2.16
Determine the nominal value and tolerance of a resistor with the color code 
green-blue-gold-silver.
The first two colors translate to 5 and 6. Gold in the third band means 
“multiply by 0.1”. The final band indicates that this is a ±10% tolerance 
component. Therefore, nominal value is 5.6 Ω. The allowable range is 5.6 Ω 
± 0.56 Ω.
55",DCElectricalCircuitAnalysis.pdf
"An example of a resistor data sheet is shown in Figure 2.25. This data sheet is for a 
series of surface mount chip resistors. The available tolerance grades range from 
0.5% to 20%. Also, for each variant there are two temperature coefficients available 
with the most stable being 100 ppm/°C (ppm is short for “parts per million”, thus 
100 ppm is equivalent to 0.01%). Note that these devices are too small to use the 
resistor color code. Instead, a numeric code is printed directly on them that follows
the rules for color bands, meaning that the final digit is the multiplier. Thus “150”
is 15 Ω (1-5 with no zeroes). The exception is very small values where the letter
“R” is used in place of a decimal point. Consequently, “2R4” is 2.4 Ω.
A particularly important characteristic of most electronic devices is their power 
handling capacity. For this series there are several variations with power handling 
from 0.1 watts to 1.5 watts, and maximum working voltages up to 500 volts. Power",DCElectricalCircuitAnalysis.pdf
"from 0.1 watts to 1.5 watts, and maximum working voltages up to 500 volts. Power 
handling is also a function of ambient temperature. In general, devices have a 
maximum internal temperature that they can reach before they are damaged. Control
of heat tends to be a major issue in many system designs. If the ambient temperature 
increases, there is less “headroom” for the device's temperature rise, and thus the 
device will not be able to dissipate as much power. This is illustrated in the graph of 
Figure 2.26.
56
Figure 2.25
Example resistor data sheet.
Courtesy of Stackpole Electronics",DCElectricalCircuitAnalysis.pdf
"Note how the power dissipation is constant at temperatures at and below 70 °C. This
temperature is considered the maximum normal operating temperature for this 
device and the power dissipation at this temperature is the one quoted in the data 
sheet. If the device is operated in a warmer environment, the power dissipation is 
derated by the percentage given in the graph. For example, at 100 °C the dissipation 
is only about 65% of the nominal rating. Of particular importance, at 155 °C the 
device can no longer dissipate any power, and therefore this temperature serves as 
the absolute ceiling. After all, a resistor that cannot dissipate any power is a resistor 
that cannot have any current through it or voltage across it without burning up. It is 
essentially non-functional.
Of course, just like a skillet on a stove, these devices do not heat up instantly, that is,
they exhibit a thermal time constant. They take a certain amount of time to heat (and",DCElectricalCircuitAnalysis.pdf
"they exhibit a thermal time constant. They take a certain amount of time to heat (and
also to cool). For short periods, devices can handle considerably more power than 
their long term rating. The graph of Figure 2.27, shows the power dissipation for 
single pulses. In some cases, these resistors can handle powers of an order of 
magnitude greater. For example, consider the type RPC0402 (bottom-most line). 
From the data sheet of Figure 2.25 this device is specified as having a long term 
rating of 0.2 watts. In contrast, the graph of Figure 2.27 shows it is capable of 
withstanding a single pulse of greater than 20 watts for 100 microseconds, just under
10 watts for a millisecond pulse and roughly 2 watts for a one-tenth second pulse.
 
57
Figure 2.26
Resistor power derating graph.
Courtesy of Stackpole Electronics",DCElectricalCircuitAnalysis.pdf
"Other Resistive DevicesOther Resistive Devices
Along with standard fixed resistors, there are several other kinds of resistive devices 
that have been designed to be sensitive to changes in their environment. Thus, they 
can serve as sensors because as their resistance changes, it impacts the flow of 
current and the resulting voltage. We will look at examples of this in upcoming 
work. Some of the environmental inputs include temperature, force and light levels.
Force Sensing Resistor (FSR)
The force sensing resistor consists of two layers of material with a nominal 
separation distance. It is presented as a flat membrane that is usually round or 
rectangular, and perhaps with an adhesive backing. An example of an FSR is shown 
in Figure 2.28. With no force applied, the device shows an extremely high 
resistance, well into the megohms. As force is applied to the surface, the two layers 
come into better contact which decreases the net resistance. This is borne out in the",DCElectricalCircuitAnalysis.pdf
"come into better contact which decreases the net resistance. This is borne out in the 
graph of Figure 2.29. The graph shows roughly a straight line response between 
resistance and force when plotted on a log-log scale. At the highest force levels, the 
resistance may drop to just a few hundred ohms.
58
Figure 2.27
Pulse power dissipation.
Courtesy of Stackpole Electronics
Figure 2.28
A force sensing resistor (FSR).",DCElectricalCircuitAnalysis.pdf
"Photoresistor
As their name implies, photoresistors are sensitive to changes in light level. They
are also called LDRs, short for Light Dependent Resistor. Different materials may
be used in their construction, but the most common is cadmium sulfide, CdS. As a
consequence, photoresistors are sometimes generically referred to as “CdS cells”.  
A photoresistor is shown in Figure 2.30 along with its corresponding resistance 
graph in Figure 2.31. In total darkness the device exhibits a very high resistance. As 
light levels increase, the resistance decreases. As with the FSR, we see an reverse 
relation between the resistance and the environmental factor: as the environmental 
input increases (force, light level), the resistance does the opposite and decreases. 
And once again, we see a straight line when plotted with log-log scales. Technically, 
we refer to this as a negative relation because the slope of the plot line is negative. 
59
Figure 2.29
FSR response curve.",DCElectricalCircuitAnalysis.pdf
"we refer to this as a negative relation because the slope of the plot line is negative. 
59
Figure 2.29
FSR response curve.
Courtesy of Interlink Electronics
Figure 2.30
A photoresitor, or LDR.
Figure 2.31
Photoresitor response curve.
Courtesy of Advanced Photonix",DCElectricalCircuitAnalysis.pdf
"To put the brightness of the light into common terms, 0.01 foot-candles (roughly 0.1 
lux) is equivalent to a clear moon-lit evening. At this level, the photoresistor is 
showing over 1 megohm of resistance. In contrast, 100 foot-candles (roughly 1000 
lux) is equivalent to an overcast day (for reference, direct sunlight is perhaps 100 
times stronger). At this level, the photoresistor's value has dropped to about 1000 
ohms. 
A point worth noting is that the light that the photoresistor “sees” is not necessarily 
the same as what a human sees. The sensitivity of the device at various wavelengths 
(i.e., colors) may differ starkly from human vision. In fact, differing materials 
exhibit different sensitivities at various wavelengths. Some of these may be 
wavelengths that the unaided human eye cannot see at all (infrared or ultraviolet, for 
example). 
An example of sensitivity curves is shown in Figure 2.32. Note the variations",DCElectricalCircuitAnalysis.pdf
"example). 
An example of sensitivity curves is shown in Figure 2.32. Note the variations 
between the different materials, CdS being one of the three (left side). The peak 
sensitivities vary as do the precise shapes. In practice, this means that some of these 
units will be more or less sensitive to certain colors than other units. The CdS curve 
indicates peak sensitivity of about 540 nm, which corresponds to green. In 
comparison, the CdSe (cadmium selenide) cell exhibits a peak of just over 700 nm, 
which corresponds to red. At some wavelengths, the relative response of one 
material may be no more than 10% of the response of a different material.
It is important to note that the use of cadmium, such as in CdS cells and the like, is 
severely restricted by the RoHS directive (see Chapter One). 
Thermistor
A thermistor is a device whose resistance is a function of temperature. These
devices are available in two basic types. Either PTC, for Positive Temperature",DCElectricalCircuitAnalysis.pdf
"devices are available in two basic types. Either PTC, for Positive Temperature
Coefficient; or NTC, for Negative Temperature Coefficient. PTC devices show an 
increase in resistance as temperature increases and NTC devices show a decrease in 
resistance as temperatures rise. Ideally, these are linear relationships with the plots 
showing straight lines. The reality is that linearity can only be assumed across fairly 
60
Figure 2.32
Photoresitor sensitivity curves.
Courtesy of Token Electronics
Figure 2.33
NTC thermistor.",DCElectricalCircuitAnalysis.pdf
"narrow ranges of temperature. For wider ranges, there will be noticeable deviation 
from a straight line as the curve is logarithmic in nature. A basic NTC thermistor is 
shown in Figure 2.33. 
Generic thermistor performance graphs are shown in Figure 2.34 with the idealized
straight line response at top and the more realistic non-linear response below.
Thermistors will be specified in terms of their room temperature resistance (usually
taken to be 25 °C) along with their sensitivity which is denoted as beta (β). The
larger the value of beta, the steeper the curve and the greater the sensitivity.
The curves labeled Beta 2 show increased sensitivity, meaning that there will
be a greater change in resistance for a certain change in temperature. 
The following equation can be used to determine the resistance of a thermistor
at some other temperature of interest with greater accuracy than using a simple
linear approximation. All that is needed is a reference temperature and",DCElectricalCircuitAnalysis.pdf
"linear approximation. All that is needed is a reference temperature and
corresponding resistance, the beta value and the new temperature of interest.
Using a reference temperature of 25 °C:
RT =R25 e
β( 1
T − 1
298.15) (2.12)
Where
RT is the resistance at the new kelvin temperature, T,
R25 is the resistance at 25 °C,
T is the temperature of interest in kelvin,
β is the device beta.
Note that the constant “298.15” in Equation 2.12 is equivalent to the reference
temperature of 25 °C (0 K is −273.15 °C). Consequently, if a different
reference point is used, simply insert the new reference temperature in its place
and use the corresponding resistance in place of R25. 
Varistor
Varistors are used as limiting devices, primarily to suppress unwanted voltage spikes
in electronic equipment. The varistor is a unique device in that it has a highly non-
linear current-voltage characteristic. This is shown in Figure 2.35. Remember, when",DCElectricalCircuitAnalysis.pdf
"linear current-voltage characteristic. This is shown in Figure 2.35. Remember, when 
plotted with the voltage on the horizontal axis, the slope of the line represents the 
conductance. Consequently, the varistor shows a region of near zero conductance or 
extremely high resistance (the horizontal section), and two sections that are nearly 
vertical, which indicate extremely high conductance or near zero resistance. This 
characteristic allows the varistor to act as a limiting device. 
Imagine that a lightning strike effects a local power line. This will create a sudden 
but short-lived spike in the voltage. A normal 120 volt wall outlet normally produces
regular peaks of approximately 170 volts. A lightning strike might add several 
61
Figure 2.34
NTC thermistor response 
curves: Ideal (top) and real.
R (log)
T (C)
25
25
R
-25 0 50 75
Beta 1
Beta 2
R (log)
T (C)
25
25
R
-25 0 50 75
Beta 1
Beta 2",DCElectricalCircuitAnalysis.pdf
"hundred volts to this. The resulting voltage could be so high that it would damage 
electronic equipment attached to the outlet. To alleviate this problem, a varistor can 
be placed across the incoming voltage lines. The vertical break point voltages would 
be set for a value just over 170 volts, the normal maximum. Under typical 
circumstances the varistor would see a voltage in its horizontal region and thus 
behave as a very high resistance. It would drain virtually no current from the
outlet, and consequently it would have no impact on the rest of the circuitry. On
the other hand, if a large spike hits the line, the varistor will swing into the
vertical region, show a much reduced resistance, and act as a shunting path for
the spike's current. It will effectively clamp the voltage to some maximum rated
value. Of course, the varistor has to absorb the energy presented by that spike,
and important parameters of a varistor include the amount of energy that it can",DCElectricalCircuitAnalysis.pdf
"and important parameters of a varistor include the amount of energy that it can
absorb (in joules) and its maximum current capacity, along with the maximum
clamping voltage.  
Strain gauge
The strain gauge is device used to measure mechanical strain on some item. Strain 
occurs when an item is under compression (its length gets shortened or squished) or 
when it's under tension (it gets elongated or stretched). Both of these things can 
happen simultaneously, for example when a bar is experiencing bending or torque 
(one side is under compression while the other is under tension). Engineering strain 
is defined as the change in length over the initial length. If an item experiences too 
much strain, it can become permanently deformed or fail (for example, the landing 
gear in an aircraft or the suspension components in a car). Simply put, strain gauges 
are used to measure this effect.
A strain gauge is made typically of very thin metal foil in a specific pattern. See",DCElectricalCircuitAnalysis.pdf
"are used to measure this effect.
A strain gauge is made typically of very thin metal foil in a specific pattern. See
Figure 2.36 for example shapes. Typically, the shape is that of a series of fine lines
connected in a back-and-forth pattern, the two ends terminating into larger pads for
soldering on connecting leads. In operation, the strain gauge is glued to the
material being investigated, for example, a metal bar that is part of a
suspension system. As they are glued together, the strain gauge experiences
the same deformation as the metal bar. Any deformation will create changes in
the length of the strain gauge's foil wires as well as their frontal surface area.
For example, under tension, the length increases while the surface area
decreases (the surface area must decrease because the foil wire has finite
mass). Recalling the basic resistance relation, Equation 2.11, both of these
effects will cause the resistance to rise. The greater the strain under tension,",DCElectricalCircuitAnalysis.pdf
"effects will cause the resistance to rise. The greater the strain under tension,
the greater the rise in resistance. The opposite will occur under compression and the 
resistance will decrease. These changes in resistance are not large but they are 
sufficient to alter the associated voltage or current which can then be calibrated to 
determine the applied strain.
62
Figure 2.35
Current-voltage characteristic
of a varistor.
I
V +
+
-
-
Figure 2.36
Strain guages.
Courtesy of Zemic",DCElectricalCircuitAnalysis.pdf
"2.8 Instrumentation and Laboratory2.8 Instrumentation and Laboratory
At this point we need to shift gears and focus on a few practical aspects, the sort of 
things we might deal with in an electrical laboratory. It is one thing to discuss 
abstract concepts such as voltage and current, and quite another to deal with them in 
the real world. 
SourcesSources
For starters, let's consider the idea of power sources, that is, devices that can produce
reliable, stable and constant voltages and/or currents. As mentioned previously, a 
battery approximates an ideal voltage source. The problem, of course, is that 
batteries only last a finite period of time and their voltage begins to sag after being 
used for a while. Also, their voltage is fixed and not adjustable. An adjustable 
voltage source would be much more flexible in a laboratory. 
To address this problem, most electrical labs use a variable DC voltage supply in",DCElectricalCircuitAnalysis.pdf
"voltage source would be much more flexible in a laboratory. 
To address this problem, most electrical labs use a variable DC voltage supply in
place of simple batteries. These sources typically are adjustable from 0 to 25 volts
or higher and can produce one amp of current or more. An example is shown in
Figure 2.37. This particular unit has two main outputs plus a third auxiliary
output. All outputs are independent, meaning that the voltage of each can be set
independent of the others. Further, the two main outputs offer programmable
current limiting. This feature can be thought of as a programmable fuse or
circuit breaker: it limits the amount of current a circuit can draw to some pre-
selected level. It is worth noting that even if a source is rated for, say, two amps,
that only refers to the maximum value that can be drawn; that is not the current
that it would produce by default. 
Multiple output power supplies are usually wired in a “floating” form, meaning",DCElectricalCircuitAnalysis.pdf
"that it would produce by default. 
Multiple output power supplies are usually wired in a “floating” form, meaning
that the negative terminal is not tied back to earth ground. A separate connector
is presented (usually green) that is tied back to earth ground if it is needed. Having 
floating supplies gives the user much more flexibility. For example, two floating 
supplies can be wired in sequence to create a single higher voltage. They can also be
wired to present a positive voltage and a negative voltage. The unit shown in Figure 
2.37 also features twin LED displays for voltage and current for each of the main 
outputs. It is worthwhile to point out that the color coding used for electronic 
equipment is not the same as that used for residential wiring in North America. The 
electronics standard is that the common or ground terminal is black while the “hot” 
or positive terminal is red.
While there are a large number of makes and models of affordable laboratory DC",DCElectricalCircuitAnalysis.pdf
"or positive terminal is red.
While there are a large number of makes and models of affordable laboratory DC 
voltage sources available, the same is not true of DC current sources. This is not 
usually a problem. As we shall see in upcoming work, it is possible to emulate a 
constant current source using a voltage source and some attached components. 
63
Figure 2.37
A multi-output adjustable DC 
voltage supply.",DCElectricalCircuitAnalysis.pdf
"Schematic SymbolsSchematic Symbols
Clearly, it would not be practical to create circuit drawings (schematics) using 
pictures of the actual devices. Instead, we use simple schematic symbols to represent
them. There are two widely used schemes: ANSI (American National Standards 
Institue) and IEC (International Electrotechnical Commission). For many electronic 
components the ANSI and IEC versions are the same, but there are notable 
differences. ANSI tends to dominate in North America (surprise!) while IEC tends to 
dominate elsewhere. 
Figure 2.38 shows the symbols for DC current and voltage sources. The long bar at 
the top of the voltage source is the positive terminal while the shorter bar at the 
bottom denotes the negative terminal. For the current source, the arrow points in the 
direction of conventional current flow. If the source is adjustable, it will often be 
shown with a diagonal arrow drawn through it. In some schematics, a DC voltage",DCElectricalCircuitAnalysis.pdf
"shown with a diagonal arrow drawn through it. In some schematics, a DC voltage 
source will simply be drawn as a “connection dot” or node with the voltage labeled. 
It is assumed that the other end of the source is tied back to the system common or 
ground.
Speaking of ground, there are three different symbols used for a circuit's common 
connection point. These are: earth ground, chassis ground and signal or digital
ground. These are shown in Figure 2.39. Earth ground is used when the circuit
common is tied back to true earth (e.g., to the third pin on an AC receptacle).
Chassis ground is a more generic term which would represent a common reference
point that did not go back to true earth (e.g., a portable device). Finally, digital or
signal ground is used to distinguish between common points where they might be 
separated for noise or interference concerns (e.g., a sensitive analog signal common 
separate from a digital logic common).",DCElectricalCircuitAnalysis.pdf
"separated for noise or interference concerns (e.g., a sensitive analog signal common 
separate from a digital logic common). 
The symbols for resistors and other resistive devices vary considerably between 
ANSI and IEC versions. ANSI uses a zig-zag line while IEC uses a simple rectangle. 
These are illustrated in Figure 2.40. This text will focus on using the ANSI standard 
symbols for resistors. One final note about component values: As decimal points are 
easy to lose on photocopies or small display screens, engineering notation 
multipliers are sometimes placed in the position of a decimal point. For example, a 
4.7 k Ω resistor may be listed as simply 4k7. 
64
Figure 2.38
Voltage (left) and current 
source schematic symbols.
Figure 2.39
Schematic symbols for ground.
(L-R) earth, chassis, signal.
Figure 2.40
Schematic symbols for resistive 
devices. (L-R) resistor (ANSI), 
resistor (IEC), photoresistor, 
thermistor.",DCElectricalCircuitAnalysis.pdf
"Measurement – The Digital MultimeterMeasurement – The Digital Multimeter
Perhaps the most handy measurement device in the electrical laboratory is the
digital multimeter, or DMM for short. These handheld devices are used to measure
voltage, current and resistance; and depending on model may have other
measurement capabilities as well. A pair of typical units are shown in Figure 2.41. 
Beyond the measurement functions of the DMM, perhaps the single most important
characteristic is its accuracy. It is important to remember that no measurement
device is perfect and there is always some uncertainly between what the device
indicates and the true value of the parameter being measured. As a DMM uses a
digital display, it has both finite range and resolution. As you might guess, the
more digits the meter can display, the greater its potential accuracy. DMMs are
commonly described as having 3 ½ digits, 3 ¾ digits, 4 ½ digits, and so on for their",DCElectricalCircuitAnalysis.pdf
"commonly described as having 3 ½ digits, 3 ¾ digits, 4 ½ digits, and so on for their
displays. A fractional digit is simply a leading digit that cannot go up to nine. By
common use, the term “½ digit” means that the leading digit can be no more than 
one. A ¾ digit specification typically means that the first digit can't be larger than 
three (it might also be four or even five as this terminology is not standardized)9. To 
clarify, a 3 ½ digit display is also referred to as a “2000 count” display. This is 
because there are 2000 possible values, from 0000 to 1999. Similarly, a 4 ¾ digit 
display is called a 40,000 count display because it has 40,000 possible values, from 
00,000 to 39,000. 
Using a 2000 count display as an example, the next question to ask is where to place
the decimal point. This display could be set up to read from 0 to 1999 volts. For 
convenience, we would call that a “2000 volt scale”, meaning that the maximum",DCElectricalCircuitAnalysis.pdf
"convenience, we would call that a “2000 volt scale”, meaning that the maximum 
voltage that can be displayed is approximately 2000 volts. With an otherwise perfect
meter, the error can be as large as 1 volt because we have no way of indicating 
fractions of volts. To solve this limitation we could add other scales by shifting the 
decimal point. For instance, we could make a nominal “20 volt scale” which would 
range up to 19.99 volts, giving us resolution down to 10 millivolts. We could go 
further and make a 2 volt scale with a maximum of 1.999 volts and 1 millivolt 
resolution as well as a 200 millivolt scale with a maximum of 199.9 millivolts 
yielding a resolution of one-tenth millivolt. We might need to measure very large as 
well as very small voltages, so the scale setting is user adjustable. 
It is important to use the lowest scale that can display the desired value otherwise a 
loss of resolution and accuracy will occur. For example, if we need to measure a",DCElectricalCircuitAnalysis.pdf
"loss of resolution and accuracy will occur. For example, if we need to measure a 
voltage somewhere around five or six volts with this 2000 count meter, it should be 
set for the 20 volt scale. A lower setting, such as a 2 volt or 200 millivolt scale, will 
not be able to display the measurement and instead will indicate an overload 
condition (usually by flashing an abbreviated error message such as “Err” or “OL”). 
On the other hand, if the higher 2000 volt scale is used, the meter will only be able 
9 You might rightly ask how it is that having just one numeral out of nine counts as “half” 
and three out of nine counts as “three fourths”. Well, that's marketing for you.
65
Figure 2.41
Digital Multimeters.",DCElectricalCircuitAnalysis.pdf
"to resolve the measurement to the nearest volt. Indeed, this is so important that some
meters have auto-ranging, meaning that they will automatically choose the scale 
setting to give the best result.
The accuracy specification for a DMM is in two parts. The first part is the percent 
deviation around the measured value. The second part is the added number of 
counts. The accuracy specification for a typical 3 ½ digit meter might be ± 2% of 
reading plus 3 counts. To determine the range of possible values for a reading, first 
determine the percentage of the reading and then add in the number of counts (i.e., 
one count represents the resolution of the meter on that particular scale). The 
resulting value represents the uncertainty of the reading. It can also be visualized as 
an “error envelope” surrounding the displayed value. Somewhere in that envelope is 
the true value measured. This is shown in the following examples.
Example 2.17",DCElectricalCircuitAnalysis.pdf
"the true value measured. This is shown in the following examples.
Example 2.17
A 3 ½ digit (2000 count) DMM has an accuracy specification of ± 2% of 
reading plus 3 counts. On its 20 volt scale it measures 5.01 volts. Determine 
the uncertainty in this measurement. 
On its 20 volt scale this meter can display up to 19.99 volts. Therefore its 
resolution, or single count value, is 0.01 volts. Thus, 3 counts is 0.03 volts. 
To this we add 2% of the reading of 5.01 volts, or 0.1002 volts. The total is 
0.1302 volts or 130.2 millivolts. This represents the error envelope on either 
side of the reading. That is, the true value is within the range of 5.01 volts ± 
130.2 millivolts, or somewhere between 4.8798 volts and 5.1402 volts. In 
other words, the ambiguity is about 130 millivolts out of about 5 volts.
Example 2.18
A 4 ¾ digit (40,000 count) DMM has an accuracy specification of ± 0.1% of 
reading plus 8 counts. On its 4 volt scale it measures 3.0035 volts.",DCElectricalCircuitAnalysis.pdf
"A 4 ¾ digit (40,000 count) DMM has an accuracy specification of ± 0.1% of 
reading plus 8 counts. On its 4 volt scale it measures 3.0035 volts. 
Determine the uncertainty in this measurement. 
On its 4 volt scale this meter can display up to 3.9999 volts. Therefore its 
resolution is 0.0001 volts or 0.1 millivolts. 8 counts is 0.8 millivolts. To this 
we add 0.1% of the reading of 3.0035 volts, or 3.0035 millivolts. The total is
3.8035 millivolts. This represents the error envelope on either side of the 
reading and the true value is somewhere between 2.9996965 volts and 
3.0073035 volts. The ambiguity here is just a few millivolts out of about 3 
volts. Clearly, this meter's level of ambiguity is much reduced compared to 
the meter used in Example 2.17.   
66",DCElectricalCircuitAnalysis.pdf
"2.9 Summary2.9 Summary
In this chapter we have examined the basic quantities that make up an electrical 
circuit. Charge, measured in coulombs, is a characteristic of subatomic particles; 
protons being positively charged and electrons being negatively charged. Like 
charges repel and opposite charges attract. The Bohr model of an atom shows that 
electrons are contained in energy shells and a single electron in the outermost shell is
only loosely bound to an atom. With an applied external energy source these 
electrons (and thus their charge) can be moved from place to place. The rate of 
charge movement over time is called current and is measured in amps. Energy is 
defined as the ability to do work and is measured in joules. V oltage is defined as the 
energy required to move a charge divided by that charge, with units of volts. Power 
is the rate of energy usage over time and is measured in watts.",DCElectricalCircuitAnalysis.pdf
"energy required to move a charge divided by that charge, with units of volts. Power 
is the rate of energy usage over time and is measured in watts. 
Some materials allow the easy flow of current and are called conductors. In contrast,
materials that inhibit the flow of electrical current are called insulators. The measure 
of this inhibition is called resistance and has units of ohms. Its reciprocal is called 
conductance and is measured in siemens. 
Resistors are devices designed to restrict the flow of current. They are available in a 
wide range of resistive values, power ratings, and physical sizes and configurations. 
Common resistors use a series of colored stripes, or color code, to signify their 
resistance value and tolerance. 
Efficiency is the ratio of useful power output to applied power input, normally 
expressed as a percentage. The higher a system's efficiency, the less energy it will",DCElectricalCircuitAnalysis.pdf
"expressed as a percentage. The higher a system's efficiency, the less energy it will 
use, the less it will cost to run, and generally the less heat it will generate. If the 
system is battery powered, higher efficiency leads to longer battery life. Battery 
capacity is measured in amp-hours. All other factors being equal, the higher the 
amp-hour rating, the longer the battery will last. 
The digital multimeter, or DMM, is a versatile tool for measuring voltage, current, 
resistance, and possibly other circuit or electrical device parameters. The accuracy of
a DMM is dependent on its resolution which is in turn set by the number of digits 
used, or its total count. The count specification is added to a percent tolerance to 
arrive at a worst case error value for some specific reading.
67",DCElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. Describe the Bohr atomic model.
2. How does charge relate to current and voltage?
3. What is the relationship between energy and power?
4. What is the relationship between resistance and conductance?
5. Define efficiency. How does it relate to operating cost?
6. Define count in terms of a digital multimeter accuracy specification.
7. Describe the resistor color code and its use.
2.10 Exercises 2.10 Exercises 
AnalysisAnalysis
1. A 2000 count digital multimeter has an accuracy specification of ±1%. If it 
reads a voltage of 34.2 volts, what is the possible range of the true voltage? 
2. A 4000 count digital multimeter has an accuracy specification of ±0.5%. If it
reads a current of 142.0 mA, what is the possible range of the true current? 
3. Determine the expected voltage scales for a standard “three and a half” digit 
meter between 1 and 1000 volts.
4. Determine the expected current scales for a standard “three and a half” digit",DCElectricalCircuitAnalysis.pdf
"meter between 1 and 1000 volts.
4. Determine the expected current scales for a standard “three and a half” digit 
meter between 1 mA and 1 amp.
5. A 2000 count digital multimeter has an accuracy specification of ±2% and 
±5 counts. If it reads a voltage of 7.00 volts, what is the possible range of 
the true voltage? 
6. A 50000 count digital multimeter has an accuracy specification of ±0.1% 
and ±4 counts. If it reads a voltage of 7.00 volts, what is the possible range 
of the true voltage? 
7. What is the charge in coulombs of a million million (1012) electrons?
8. What is the charge in coulombs of 1015 electrons?
9. How many electrons would be needed for a charge of 20 coulombs?
10. How many electrons would be needed for a charge of 1 microcoulomb?
11. If a charge of 2 coulombs passes through a wire in 5 seconds, what is the 
current?
12. If a charge of 300 millicoulombs passes through a wire in 0.1 seconds, what 
is the current?
68",DCElectricalCircuitAnalysis.pdf
"13. How much charge must be transferred in 0.1 seconds in order to achieve a 
current of 5 amps?
14. How much charge must be transferred in 20 seconds in order to achieve a 
current of 10 microamps?
15. Determine the resulting voltage if it takes 2 joules to move 10 coulombs of 
charge. 
16. Determine the voltage if 15 joules is used to move 0.5 coulombs of charge. 
17. How much energy is required to create a 10 volt potential difference with a 
2 coulomb charge?
18. How much energy is required to create a 50 millivolt potential difference 
with a 0.1 coulomb charge?
19. What is the wattage equivalent of two horsepower (2 hp)?
20. What is the horsepower equivalent of 1000 watts?
21. If a device draws 2 amps of current from a 12 volt battery, determine the 
power delivered.
22. If a device draws 10 milliamps of current from a 1.5 volt battery, determine 
the power delivered.
23. A 2 hp motor draws 1800 watts from its source. Determine its efficiency.",DCElectricalCircuitAnalysis.pdf
"the power delivered.
23. A 2 hp motor draws 1800 watts from its source. Determine its efficiency.
24. A 5 hp motor draws 4.5 kw from its source. Determine its efficiency.
25. An audio power amplifier is rated for 500 watts of maximum output at an 
efficiency of 80%. Determine the amount of wasted power.
26. A compressor draws 10 amps of current from a 120 volt source. Its rated 
output is 1 hp. Determine the efficiency.
27. An application requires a battery to deliver 15 mA for at least 200 hours. 
Determine the required amp-hour rating.
28. Determine the required rating for a battery to deliver 0.8 A for at least 30 
hours. 
29. A certain 12 volt battery has a rating of 6 Ah. Determine the expected 
battery life using a 5 mA draw.
30. A certain AA battery has a rating of 800 mAh. Determine the expected 
battery life using a 20 mA draw.
31. Assume that a certain piece of material has a resistance of 80 ohms. 
Determine the new resistance if the length of the piece is doubled and no",DCElectricalCircuitAnalysis.pdf
"31. Assume that a certain piece of material has a resistance of 80 ohms. 
Determine the new resistance if the length of the piece is doubled and no 
other parameters are changed.
69",DCElectricalCircuitAnalysis.pdf
"32. Assume that a certain piece of material has a resistance of 2 k ohms. 
Determine the new resistance if the width and height of the piece are 
doubled and no other parameters are changed.
33. Assume that a certain piece of material has a resistance of 4 ohms. 
Determine the new resistance if the resistivity is doubled and no other 
parameters are changed.
34. Assume that a certain piece of material has a resistance of 10 k ohms. 
Determine the new resistance if the length, width and height of the piece are 
all halved.
35. A certain material has a resistivity of 100 ohm-centimeters. Determine the 
resistance of a piece that is 1 cm wide, 0.5 cm high and 6 cm long.
36. A certain material has a resistivity of 2000 ohm-centimeters. Determine the 
resistance of a piece that is 2 mm wide, 4 mm high and 10 mm long.
37. A 40 ohm resistor has dimension of 0.4 cm wide by 0.2 cm high by 1 cm 
long. Determine the resistivity in ohm-centimeters.",DCElectricalCircuitAnalysis.pdf
"37. A 40 ohm resistor has dimension of 0.4 cm wide by 0.2 cm high by 1 cm 
long. Determine the resistivity in ohm-centimeters.
38. A 5000 ohm resistor has dimension of 5 mm wide by 3 mm high by 6 mm 
long. Determine the resistivity in ohm-centimeters.
39. A resistor with the color code yellow-violet-red-silver has a
measured value of 4806 ohms. Is this resistor within tolerance? As a
percentage, how far is it from the nominal value? 
40. A resistor with the color code orange-orange-yellow-gold has a
measured value of 33.9 k ohms. Is this resistor within tolerance? As
a percentage, how far is it from the nominal value? 
41. A resistor with the color code brown-black-orange-silver has a
measured value of 9980 ohms. Is this resistor within tolerance? As a
percentage, how far is it from the nominal value? 
42. A resistor with the color code green-blue-black-gold has a measured value 
of 50 ohms. Is this resistor within tolerance? As a percentage, how far is it 
from the nominal value?",DCElectricalCircuitAnalysis.pdf
"of 50 ohms. Is this resistor within tolerance? As a percentage, how far is it 
from the nominal value? 
43. Determine the value of the resistors pictured in Figure 2.42 (left-to-
right: red-black-yellow, blue-gray-orange, red-violet-red-silver).
44. Determine the value of the resistors pictured in Figure 2.43 (left-to-
right: red-red-orange-gold, brown-black-red-gold, yellow-violet-
orange-silver).
45. Determine the value of the resistors pictured in Figure 2.44 (left-to-
right: orange-orange-green-silver, white-brown-black-silver, orange-
white-brown-gold).
70
Figure 2.42
a
b
c
Figure 2.43
a
b
c",DCElectricalCircuitAnalysis.pdf
"46. Determine the value of the resistors pictured in Figure 2.45 (left-to-right: 
silver-orange-violet-yellow, green-blue-yellow-gold, gold-black-orange-
yellow).
47. Determine the maximum and minimum allowed values of the
resistors pictured in Figure 2-42.
48. Determine the maximum and minimum allowed values of the
resistors pictured in Figure 2-43.
49. Determine the maximum and minimum allowed values of the
resistors pictured in Figure 2-44.
50. Determine the maximum and minimum allowed values of the
resistors pictured in Figure 2-45.
DesignDesign
51. Determine the resistor color code for the following ohmic values
using 10% tolerance: a) 56 Ω  b) 33 kΩ  c) 470 kΩ  d) 1.2 kΩ
e) 750 Ω
52. Determine the resistor color code for the following ohmic values
using 5% tolerance: a) 47 Ω  b) 22 kΩ  c) 390 kΩ  d) 2.2 kΩ
e) 560 Ω
ChallengeChallenge
53. A radio transmitter is rated for 100 watts of maximum output at an",DCElectricalCircuitAnalysis.pdf
"e) 560 Ω
ChallengeChallenge
53. A radio transmitter is rated for 100 watts of maximum output at an 
efficiency of 90%. If it is fed from a 120 volt source, determine the current 
draw.
54. A certain 75 watt incandescent bulb produces 71 watts worth of heat and the
remainder in the form of light. Determine its efficiency as a lighting device 
and its efficiency as a heating device. 
55. Assume it takes an 1800 watt toaster 3 minutes to toast a bagel “just right”. 
If you toast a bagel every morning for a year and electricity costs 15 
cents/kWh, how much will you have spent in that year to toast bagels?
71
Figure 2.44
a
b
c
Figure 2.45
a
b
c",DCElectricalCircuitAnalysis.pdf
"56. Assume that you can buy standard 60 watt incandescent light bulbs for 50 
cents each and that each has an expected life span of 1000 hours. In 
comparison, you can buy an LED light bulb that produces the same amount 
of light but only consumes 7 watts. The LED bulbs cost $5.50 each and have
an expected life of 20,000 hours. Assuming electricity costs 14 cents/kWh, 
determine the total cost of running incandescent lights versus LEDs for 
40,000 hours. 
57. Given a 1.5 volt battery with a 500 mAh rating, how much current can it 
produce continuously for 25 hours?. 
58. Given a 9 volt battery with a 100 mAh rating, determine the total energy 
storage in joules. 
59. A 2000 count DMM with ±1% accuracy is used to measure the resistance of 
a spool of AWG 22 wire. If the measurement is 4.50 Ω, determine the length 
of the wire and the possible inaccuracy (in feet or meters).
60. A length of AWG 26 wire is attached to one end of a length of AWG 32 wire.",DCElectricalCircuitAnalysis.pdf
"of the wire and the possible inaccuracy (in feet or meters).
60. A length of AWG 26 wire is attached to one end of a length of AWG 32 wire. 
The total length of the two wires is 200 meters. Determine the length and 
resistance of each piece if the total resistance is 75 Ω. 
72",DCElectricalCircuitAnalysis.pdf
"NotesNotes
    ♫♫    ♫♫
73",DCElectricalCircuitAnalysis.pdf
"3 3 Series Resistive CircuitsSeries Resistive Circuits
3.0 Chapter Learning Objectives3.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe the differences between conventional current flow and electron flow.
• Identify series resistive circuits that include one or more voltage sources or a single current source.
• Compute equivalent resistance, and component and node voltages for series resistive circuits. 
• Compute circulating current and component powers for series resistive circuits.
• Utilize Ohm's law, Kirchhoff's voltage law (KVL) and the voltage divider rule (VDR) to aid in the 
analysis of series resistive circuits.
• Identify and describe the usage of potentiometers and rheostats.
• Utilize computer simulation tools to investigate and verify basic electric circuit quantities such as 
component voltages.
3.1 Introduction3.1 Introduction",DCElectricalCircuitAnalysis.pdf
"component voltages.
3.1 Introduction3.1 Introduction
With the requisite background information of the first two chapters now behind us, in this chapter we shall begin
our analysis of electrical circuits. Here we will introduce the most simple configuration, the series loop. It can 
contain any number of resistors and voltage sources, or in place of the voltage sources, a single current source. 
We shall examine how to determine the current flow through each component, the voltage across each 
component and the power either dissipated or generated by each component. Other practical issues will also be 
examined, such as the effect of component tolerance on the circuit parameters. 
3.2 Conventional Current Flow and Electron Flow3.2 Conventional Current Flow and Electron Flow
Before we dive into series circuits we need to consider an interesting question involving the direction of current",DCElectricalCircuitAnalysis.pdf
"Before we dive into series circuits we need to consider an interesting question involving the direction of current 
flow. Does it flow from positive to negative or from negative to positive? For that matter, does it even make a 
difference as far as our analyses will be concerned? 
Benjamin Franklin (pictured in Figure 3.1) began experimenting with the phenomenon of electricity in 1746. In 
1752 he performed his famous kite experiment proving that lightning is a form of electricity by capturing charge
from storm clouds in a leyden jar (an early form of an electrical charge storage device)10. At this time the 
10 It is worth noting that Franklin's kite was not struck by lightning. If it had been, he likely would have been killed. The 
hemp string that was used for the kite was sufficiently wet from rain that it was possible to transfer charge from the 
atmosphere to the leyden jar, and subsequently to a metal key which would emit a spark. 
74",DCElectricalCircuitAnalysis.pdf
"modern concept of an atomic model with electrons and protons did not exist and
electricity was conceived of as a sort of fluid. Franklin surmised that the “electrical
flow” moved from positive to negative. This idea was accepted and became the
conventional view. Today we call this idea conventional current flow. In this model,
current flows from a more positive voltage to a less positive voltage. We know now
that the electron is the charge carrier in metals and the electrons travel in the reverse 
direction. Essentially, Franklin guessed wrong. Electrons move from a lower 
potential to a higher potential. We call this model electron flow. For most work, 
engineers and technicians use conventional flow, although in some cases, such as the
explanation of semiconductors, electron flow is easier to visualize for some people. 
In short, conventional flow exists for historical reasons, and it is the model used for 
most analyses, including this text.",DCElectricalCircuitAnalysis.pdf
"In short, conventional flow exists for historical reasons, and it is the model used for 
most analyses, including this text. 
You might think the current direction would make a big difference in an analysis; 
after all, it certainly makes a big difference if you drive a car in the wrong direction. 
It turns out that both forms will achieve the desired results, we just have to be 
consistent with the usage. 
To better understand this, consider that the movement of a net negative charge in
one direction can be thought of as a movement of a net positive charge in the other
direction. That is, the movement of an electron creates a “hole” where it used to be
and that hole is net positive. This is illustrated in Figure 3.2. Here we start at the
top with a tube of identical marbles, all pushed to the right. In each step below it we 
move a marble to the left, mimicking the flow of electrons in a circuit. When we",DCElectricalCircuitAnalysis.pdf
"move a marble to the left, mimicking the flow of electrons in a circuit. When we 
reach the bottom, each marble has been pushed left by one place. We can also arrive 
at the bottom drawing by simply taking the right-most marble in the top drawing and
inserting it to the extreme left by jumping over the other three marbles. Here's the 
important bit: instead of imagining the marbles moving left, we can also think in 
terms of “negative space” and imagine the empty slot moving to the right. That's 
hole flow. The two views are functionally identical as they lead to the same 
outcome. 
3.3 The Series Connection3.3 The Series Connection
The word circuit comes from the Latin root circ, meaning “ring” or “around”. An
electrical circuit consists of at least one ring or loop through which current flows.
For example, if we have a battery attached to a lamp as in Figure 3.3, the current
exits the battery, flows through the lamp, and then returns to the other side of the",DCElectricalCircuitAnalysis.pdf
"exits the battery, flows through the lamp, and then returns to the other side of the
battery creating a loop or completed circuit. Without a path back to the battery, 
current will not flow. Thus, if we cut one of the wires connecting the battery and 
lamp, there is no path for current and no current flows. This is referred to as an open
circuit and is a common fault that occurs when electronic systems are dropped or 
struck forcefully. Obviously, this will tend to render the circuit unusable. In this 
example, the lamp will not illuminate. The opposite of the open circuit is the short 
75
Figure 3.1
Benjamin Franklin: 
Technically incorrect but it 
doesn't really matter.
Figure 3.2
Electron versus hole flow.
Figure 3.3
Battery and lamp circuit.",DCElectricalCircuitAnalysis.pdf
"circuit. In a short circuit, an unintended alternate path for current flow exists and this
also can create a malfunction. In the case of our battery and lamp, a short circuit can 
occur if a piece of wire or metal accidentally fell across the terminals of the lamp. 
The current would then have a high conductance (i.e., low resistance) path around 
the lamp. The vast majority of the current would take this low resistance path instead
of the higher resistance path presented by the lamp. The result would be that the 
lamp would not light. In either the open or short case, the light does not function but 
there is an important difference: for the short circuit, excessive current will flow out 
of the battery because there is little to resist the flow of current. Thus, the battery 
will be drained very quickly. For the open, no current flows and thus the battery is 
not drained. 
Series connections are not limited to just two components. In general, a series",DCElectricalCircuitAnalysis.pdf
"not drained. 
Series connections are not limited to just two components. In general, a series
connection is any connection of components configured such that the current
through each component must be the same. This is illustrated in Figure 3.4. Inside
each of the lettered boxes would be a component such as a resistor or a
voltage source. Note that for each component, there is one entrance point and
one exit point. No matter which box you pick, the current flowing through it
must be the same as the current flowing into the next box or out of the
preceding box, and it doesn't matter if you follow this path in a clockwise or
counterclockwise fashion. Thus, this entire configuration is a series
connection. The idea that current is consistent throughout should be self-
evident. After all, the only way the current through, say, item B could be
different from that flowing through item C or D is if some of it somehow",DCElectricalCircuitAnalysis.pdf
"evident. After all, the only way the current through, say, item B could be
different from that flowing through item C or D is if some of it somehow 
“disappeared” along the way by magic. It is important to remember that consistent 
current is the hallmark feature defining a series connection:
The current is the same everywhere in a series connection. (2.1)
It is possible that only a portion of a circuit exhibits a series connection. Consider
the more complex diagram presented in Figure 3.5. Some of these items are in
series and some are not. For example, items A and B are in series with each other
but not in series with the remaining items. Why? Because if we imagine
the current flowing through A and then through B they must be the
same, however, once beyond B, the current could split and flow down
other branches: a portion entering C, a portion entering D and the
remainder flowing into E. On the other hand, items C and F are in series",DCElectricalCircuitAnalysis.pdf
"other branches: a portion entering C, a portion entering D and the
remainder flowing into E. On the other hand, items C and F are in series
with each other because whatever current is flowing through one of
them must be flowing through the other. Thus, items A and B are in
series with each other and items C and F are in series with each other,
although all four are not in series as a group. 
It is possible that no two items in a circuit are strictly in series. We will see examples
of this in upcoming chapters. Further, just because the currents through two items 
happen to be the same, that does not necessarily mean they are in series. Identical 
currents could be just a by-product of the component values chosen. For example, 
76
Figure 3.4
A generic series configuration.
A B
C
D
E
Figure 3.5
A more complex configuration.
A B
C
D
F
E",DCElectricalCircuitAnalysis.pdf
"even if items D and E in Figure 3.5 have the same numeric value for current, we 
would not say that they are in series, anymore than we would say that any two 
people with the same last name would have to be siblings.
3.4 Combining Series Components3.4 Combining Series Components
Typically, a series connection will include multiple resistors. Referring back to the 
resistance equation presented in Chapter 2, Equation 2.11, we can see that resistors 
in series add. 
R =ρl
A
If we consider two identical resistors placed in series, one after the other as in
Figure 3.6, the effective length would double while keeping the resistivity and area
unchanged. The combined result would be a doubling of the resistance of just one
of them. If we then generalize this to two arbitrary resistors of identical resistivity
and area, the lengths would dictate the resistance of each, and the combined lengths 
would then reflect the resistance of the pair. We can generalize this further for N",DCElectricalCircuitAnalysis.pdf
"would then reflect the resistance of the pair. We can generalize this further for N 
resistors. Thus we find that the equivalent resistance of a group of series resistors is 
their sum:
RTotal = R1+R2+R3+...+RN (3.2)
Consequently, as resistors in series add, total resistance may be found by summing 
the individual resistors. 
Example 3.1
A string of resistors is placed in series as shown in Figure 3.7. Their values
are: 120 Ω, 390 Ω, 560 Ω and 470 Ω. Determine the equivalent series
value.
RT = R1+R2+R3+R4
RT = 120 Ω+390 Ω+560Ω+470Ω
RT =1540Ω
Multiple voltage sources in series may also be added, however, polarities must be 
considered as opposing voltages partially cancel each other (i.e., adding a negative). 
This concept is presented in the next example.
77
Figure 3.6
Resistors in series.
A
B
Figure 3.7
Resistor string for Example 
3.1.",DCElectricalCircuitAnalysis.pdf
"Example 3.2
Determine the equivalent series value of the voltage sources presented in
Figure 3.8.
If we use point b as our reference, by inspection the top of the 12 volt source
is 12 volts above point b (reminder, the long bar denotes the positive 
terminal). Also, by inspection, the right side of the 3 volt source (point a) is 
negative with respect to its left side. As the left side of this source is 
connected to the positive terminal of the 12 volt source, then it too must be 
12 volts above point b. As its right side is 3 volts less than this side, point a 
must be 3 volts less than 12 volts, or 9 volts above point b. 
Thus Vab = 9 volts. 
Chasing this further, if the 3 volt source had been flipped so that it had the 
same polarity as the 12 volt source (positive toward the right) then Vab = 15 
volts. If the 12 volt source had been flipped (positive toward bottom) with 
the 3 volt source as drawn originally, then Vab = −15 volts. If both sources",DCElectricalCircuitAnalysis.pdf
"the 3 volt source as drawn originally, then Vab = −15 volts. If both sources 
had been flipped then Vab = −9 volts. Finally, if point a had been taken as the 
reference then these four potentials would have the opposite polarity 
because, by definition, Vba = −Vab.
In contrast to voltage sources, differing current sources are not placed in series as
they would each attempt to establish a different series current, a practical
impossibility. Refer to Figure 3.9 as a reminder!
3.5 Ohm's Law3.5 Ohm's Law
In Chapter 2, the concept of resistance was introduced using the following 
generality:
 Effect = Cause
Oppostion
To review, the cause is a voltage source, the opposition is the resistance, and the 
effect is the resulting current. If the item the current passes through is linear, such
as the simple resistors depicted in Figure 3.10, then this relationship can be
rewritten as:
 I = V
R
78
Figure 3.8
Voltage sources in series.
Figure 3.9
Placing current sources in",DCElectricalCircuitAnalysis.pdf
"rewritten as:
 I = V
R
78
Figure 3.8
Voltage sources in series.
Figure 3.9
Placing current sources in 
series generally is evil.
Figure 3.10
Current-voltage plots for 
simple resistors.
V
I
-I
-V",DCElectricalCircuitAnalysis.pdf
"or more commonly, solved for the voltage and expressed as:
V = I ×R (3.3)
This is called Ohm's law, and is named in honor of Georg Ohm, a researcher from 
the early nineteenth century. It is, along with Kirchhoff's laws that we shall see 
shortly, one of the most important and useful equations available for the analysis of 
DC electrical circuits. For the sake of completeness, Ohm's law can also be 
visualized in terms of resistance:
R = V
I
All three forms of this equation are useful. For example, we can use the first version 
if we have a known voltage applied across a resistor and we want to determine the 
resulting current. Similarly, Equation 3.3 can be used to find the voltage drop across 
a resistor if we know the current through it. Finally, we can use the last version if we
need to set the current to a certain value and have to find the resistance required to 
do so.
Example 3.3
A 9 volt battery is used in series to power a 40 Ω resistor as shown in",DCElectricalCircuitAnalysis.pdf
"do so.
Example 3.3
A 9 volt battery is used in series to power a 40 Ω resistor as shown in
Figure 3.11. Determine the circulating current. Also determine the
minimum power rating for the resistor.
I = V
R
I = 9 V
40 Ω
I =0.225amps
From power law, Equation 2.7,
P = I ×V
P = 0.225 A×9 V
P =2.025 watts
Power could also be determined using V2/R or I2R. Try it. You should get the 
same result. 
79
Figure 3.11
Circuit for Example 3.3.",DCElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
To verify the results of the preceding example, the circuit is entered into a simulator, 
as shown in Figure 3.12. Multisim™ is used here, although any quality circuit 
simulator will do. Virtual DMMs are used to measure the current and voltage. 
Remember, current is the rate of charge flow and is measured at a single point; the 
middle of a connecting wire if you will. Consequently, the ammeter is inserted 
between the battery and resistor. In contrast, voltage is a potential difference which 
implies two points for the measurement, and thus, the voltmeter is placed across the 
resistor.
Note the voltage is exactly 9 volts as expected. The current is just shy of the 
expected 225 milliamps. This is due to the fact that all real world ammeters exhibit 
some finite internal resistance. This extra resistance, though very much smaller than 
the resistor, adds to the total resistance and thus slightly decreases the current. The",DCElectricalCircuitAnalysis.pdf
"the resistor, adds to the total resistance and thus slightly decreases the current. The 
simulator has been programmed to mimic this behavior and thus we see a very slight
decrease in the current in the simulation, which is precisely the effect we would see 
in a proper physical lab.
This simulation uses virtual instruments because they echo the layout of a physical 
lab and offer some familiarity. Unfortunately, they become cumbersome in larger 
circuits. We shall look at other, more direct, simulation techniques later in this 
chapter.
80
Figure 3.12
The circuit of Example 3.3 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"Example 3.4
Refer to the battery and lamp circuit shown back in Figure 3.3. Determine 
the resistance of the lamp if the current flowing through it is 300 mA and the
battery voltage is 6 volts. Also determine the power dissipated by the lamp.
R = V
I
R = 6 V
0.3A
R =20Ω
From power law,
P = I 2
×R
P = 0.3A2
×20Ω
P =1.8watts
Let's continue with a couple of examples utilizing current sources.
Example 3.5
Determine the voltage developed across the resistor in the circuit of
Figure 3.13. Also determine the power dissipated.
V = I ×R
V = 10  mA×2.2k Ω
V =22 volts
From power law,
P = I 2
×R
P = (10  mA)2×2.2 kΩ
P =0.22  watts
Computational hint: don't forget the “milli” part of the current (milli squared
is micro).
81
Figure 3.13
Circuit for Example 3.5.",DCElectricalCircuitAnalysis.pdf
"Example 3.6
Determine the resistor value required in the circuit of Figure 3.14 in order
for the source to generate 24 watts of power.
First, note that power generated must equal power dissipated. Thus, the 
power generated by the source must equal the power dissipated by the 
resistor. From power law,
P = I 2
×R
R = P
I 2
R = 24  W
(2  A)2
R = 6Ω
As a crosscheck, note that Ohm's law indicates that a 6 Ω resistor would 
produce a 12 volt drop given a 2 amp current, and that 12 volts times 2 amps
yields the expected 24 watts.
It is now time to move onto series circuits using multiple resistors and/or current 
sources. 
3.6 Kirchhoff's Voltage Law (KVL)3.6 Kirchhoff's Voltage Law (KVL)
Along with Ohm's law, the key law governing series circuits is Kirchhoff's voltage 
law, or KVL. Named after nineteenth century German physicist Gustav Kirchhoff, 
this law states that the sum of voltage rises and voltage drops around a series loop",DCElectricalCircuitAnalysis.pdf
"this law states that the sum of voltage rises and voltage drops around a series loop 
must equal zero (the rises and drops having opposite polarities). Alternately, it may 
be reworded as the sum of voltage rises around a series loop must equal the sum of 
voltage drops. As a pseudo formula: 
∑ V↑ =  ∑ V↓ (3.4)
The Voltage Divider Rule (VDR)The Voltage Divider Rule (VDR)
An outgrowth of KVL is the voltage divider rule (VDR). In a series connection, the 
current is the same through each component. Thus, the voltage drops in a series 
connection must be directly proportional to the size of the resistances: the larger the 
resistor, the larger its voltage, and the larger its share of the total voltage applied to 
the series connection. Thus, the voltage across any resistor must equal the net 
82
Figure 3.14
Circuit for Example 3.6.",DCElectricalCircuitAnalysis.pdf
"supplied voltage times the ratio of the resistor of interest to the total resistance:
VRx = E ∙ RX /RTOTAL (3.5)
In fact, Equation 3.5 is just a combination of two Ohm's law calculations into a 
single formula. The circulating current is equal to E/RTOTAL. This current is then 
multiplied by the resistor of interest (RX) to arrive at the voltage across that resistor 
(VRx). It is worth pointing out that “the resistor of interest” can, in fact, be the sum of 
multiple resistors in series. While VDR is not required for any particular analysis, it 
serves two purposes: first, it saves some time because it skips over the computation 
of current, and second, it reinforces the ideal of a proportional division of voltage in 
a series connection. For example, if there are two resistors in series and one of them 
is twice the size of the other, then it must be the case that the larger resistor sees 
twice the voltage of the smaller resistor.
3.7 Series Analysis3.7 Series Analysis",DCElectricalCircuitAnalysis.pdf
"twice the voltage of the smaller resistor.
3.7 Series Analysis3.7 Series Analysis
Ohm's law, Kirchhoff's voltage law, the voltage divider rule and series component 
combinations are the tools we will use to solve general series circuit problems. There
are multiple techniques for analyzing these circuits:
• If all voltage source and resistor values are given, the circulating current can
be found by dividing the equivalent voltage by the sum of the resistances. 
Once the current is found, Ohm's law can be used to find the voltage drops 
across individual resistors. Alternately, the individual voltage drops can be 
found using the voltage divider rule. At that point, power law can be used to 
find the power dissipation in each resistor or the power developed by the 
source(s). 
• If the circuit uses a current source instead of a voltage source, then the 
circulating current is known and the voltage drop across any resistor may be 
determined directly using Ohm's law.",DCElectricalCircuitAnalysis.pdf
"circulating current is known and the voltage drop across any resistor may be 
determined directly using Ohm's law. 
• If the problem concerns determining resistance values, the basic idea will be
to use these rules in reverse. For example, if a resistor value is needed to set 
a specific current, the total required resistance can be determined from this 
current and the given voltage supply. The values of the other series resistors 
can then be subtracted from the total, yielding the required resistor value. 
Similarly, if the voltages across two resistors are known, as long as one of 
the resistance values is known, the other resistance can be determined using 
either the voltage divider rule or Ohm's law. 
83",DCElectricalCircuitAnalysis.pdf
"Often, there will be more than one way to solve a given problem. We shall refer to 
these as solution paths. No particular path is “more correct” than any other, although
some paths may be shorter or easier in a given case, as will be illustrated next.
Example 3.7
Determine the voltage Vb (i.e., the voltage across the 100 Ω resistor) in
the circuit of Figure 3.15. Also determine the circulating current.
First, note that there is a single voltage source that will create a clockwise 
current flowing through the two resistors. KVL says that the combined drops 
across those resistors must equal the rise of the source, or 12 volts. One 
solution path involves first finding the total resistance. Dividing this into the 
source voltage will yield the current. Ohm's law can then be used to 
determine the resistor voltage.
RT = R1+R2
RT = 200 Ω+100 Ω
RT = 300 Ω
I = E
RT
I = 12 V
300Ω
I = 40  mA
V b = I ×R2
V b = 40  mA×100Ω
V b = 4  volts",DCElectricalCircuitAnalysis.pdf
"determine the resistor voltage.
RT = R1+R2
RT = 200 Ω+100 Ω
RT = 300 Ω
I = E
RT
I = 12 V
300Ω
I = 40  mA
V b = I ×R2
V b = 40  mA×100Ω
V b = 4  volts
A second solution path involves first finding the resistor voltage via the 
voltage divider rule. The current can then be found via Ohm's law.
V b = E R2
RT
V b = 12V 100 Ω
300Ω
V b = 4  volts
I = V b
R2
I = 4V
100 Ω
I = 40  mA
84
Figure 3.15
Circuit for Example 3.7.",DCElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
Let's verify the preceding example using a computer simulation. Instead of using 
virtual instruments, we shall use a more direct route, namely a listing of node 
voltages. A node voltage is the voltage at a given point with respect to some other 
point, normally ground, as in Vb in the prior example. Note that as we treat 
connecting wires ideally (having no resistance), a node is the entire connection wire,
not a specific location on it. Normally, simulators will automatically number each 
node in a circuit with ground being node 0. This method skips the entire business of 
inserting virtual meters in the circuit and reduces schematic clutter.
Note that ground is node 0, the power supply connection to the first resistor is node 
1 and the connection between the two resistors is node 2. Thus, node 2 is Vb. A DC 
operating point analysis is performed. The results are shown in a separate output",DCElectricalCircuitAnalysis.pdf
"operating point analysis is performed. The results are shown in a separate output 
window, as depicted in Figure 3.17. The values are as expected and with no loading 
effects as might be seen with virtual instruments.
85
Figure 3.16
The circuit of Example 3.7 in 
the simulator.
Figure 3.17
Simulation results for the 
circuit of Figure 3.16.",DCElectricalCircuitAnalysis.pdf
"Example 3.8
Determine the voltage Va (i.e., the voltage across the current source) in the
circuit of Figure 3.18.
Given the direction of the current source, the current will flow 
counterclockwise. This produces voltage drops (plus to minus) left to right 
across the 10 Ω, bottom to top across the 3 Ω, and right to left across the 1 Ω
resistor. This means that node a is negative with respect to ground (i.e., the 
voltage keeps dropping across the resistors as we travel counterclockwise 
from ground, indicating that the resulting potential keeps getting more and 
more negative with respect to ground). To find Va, we may simply sum the 
resistors and then use Ohm's law to determine their combined voltage drop, 
which of course, must equal the voltage rise from the current source. 
RT = R1+R2+R3
RT = 1Ω+3Ω+10Ω
RT = 14Ω
V a =−I ×RT
V a =−5 A×14 Ω
V a =−70  volts
One practical point when designing circuits is to appreciate the variations in current",DCElectricalCircuitAnalysis.pdf
"RT = 14Ω
V a =−I ×RT
V a =−5 A×14 Ω
V a =−70  volts
One practical point when designing circuits is to appreciate the variations in current 
and voltage produced by component tolerances. For example, in the preceding 
problem, we determined that there will be 70 volts across the current source, but this
assumes that the resistor values are exactly as stated. In the previous chapter we 
discovered that all resistor values will vary somewhat, and we grade this variation as
a percent tolerance. Thus, if the true resistor values are off a bit, we would expect 
this to impact the associated voltages. Often, we would like to determine worst case 
values such as a maximum voltage or a minimum current. Sometimes these 
conditions occur when all resistors are at their maximum or minimum value. For 
example, the maximum value of source voltage in the previous problem occurs when
all of the resistor values are at their maximum. This is not always the case, though.",DCElectricalCircuitAnalysis.pdf
"all of the resistor values are at their maximum. This is not always the case, though. 
Sometimes the minima and maxima are achieved with odd combinations, such as 
one resistor being at maximum and another being at minimum. For instance, in the 
circuit used for Example 3.7, Vb will be at minimum when R2 is at minimum and 
when R1 is at maximum. This can be verified easily using VDR.
Computer SimulationComputer Simulation
Along with worst case, we might also like to know the sort of typical spread we get 
on a particular parameter using normal component variations. This can be performed
86
Figure 3.18
Circuit for Example 3.8.",DCElectricalCircuitAnalysis.pdf
"in a simulator and it is usually called Monte Carlo Analysis or something similar. 
Basically, the simulator computes a series of simulation runs, each with component 
values chosen randomly within the specified tolerances of the chosen components, 
mimicking the effect of pulling the components out of a parts bins. To illustrate the 
Monte Carlo technique, the circuit of Example 3.8 is entered in a simulator as shown
in Figure 3.19.
The resistors are given 10% tolerances. Five runs are specified for node voltage 1 
which is equivalent to Va in the original circuit. The results are shown in Figure 3.20.
Note that the nominal value run indicates −70 volts as expected. The other runs 
show variations on this voltage, some above and some below the nominal result.
87
Figure 3.19
The circuit of Example 3.8 in 
the simulator.
Figure 3.20
Simulation results for the 
circuit of Figure 3.19.",DCElectricalCircuitAnalysis.pdf
"The Importance of PolarityThe Importance of Polarity
A key item of importance when analyzing series circuits, or indeed any electrical 
circuit, is noting the polarity of the voltages. This was apparent in Example 3.8.  
Determining polarity for voltage sources is easy as their polarity is fixed (positive is 
always at the “long bar” and negative at the “short bar”). The polarity of any resistor
will depend on the direction of current flow. To avoid confusion we will use the 
following standard: 
• Where conventional current enters a resistor, we label this point with a plus 
sign, and where the current leaves, a minus sign. 
• During analysis, moving from plus to minus is a voltage drop. That is, we 
are moving from a more positive or higher potential to a less positive or 
lower potential, thus “dropping” voltage. 
• Similarly, traversing from minus to plus is a voltage rise.
Remember, Kirchhoff's voltage law states that the sum of voltage rises around a",DCElectricalCircuitAnalysis.pdf
"• Similarly, traversing from minus to plus is a voltage rise.
Remember, Kirchhoff's voltage law states that the sum of voltage rises around a 
series loop must equal the sum of voltage drops.
Along with determining the voltage across single resistors, we are often interested
in determining the voltage between arbitrary points in a circuit. These may span
several components. It is imperative that we know the polarities of the individual
voltages in order to determine the voltage between any two circuit points. The
foregoing is illustrated using the series circuit schematic of Figure 3.21.
Here, the two voltage sources, E1 and E2, aid each other because they are both trying 
to establish a clockwise current. This produces a net voltage of E1+E2. Their 
polarities are fixed. The current direction will be as drawn because conventional 
current flows out of the positive terminal of a voltage source. The value of this",DCElectricalCircuitAnalysis.pdf
"current flows out of the positive terminal of a voltage source. The value of this 
current will be (E1+E2) / (R1+R2) via Ohm's law. Knowing the direction of current, 
the polarities of the voltage drops across the two resistors may be found. The point 
where the current enters the resistor is positive, and negative where it exits. Once 
these are labeled, either Ohm's law or the voltage divider rule can be used to 
determine the voltages developed across the two resistors. Note that KVL states that 
the sum of these two resistor voltages must equal the net supplied voltage, or E1+E2 
in this case. 
To determine the voltage from point a to point c, or Vac, we start at point a and sum 
the voltage drops and rises until we get to point c. Here, the voltage across R1 shows 
up as a drop (+ to − and taken as positive) and the voltage across E2 shows up as a 
rise (− to +, and thus negative). The result is the voltage across R1 minus the voltage",DCElectricalCircuitAnalysis.pdf
"rise (− to +, and thus negative). The result is the voltage across R1 minus the voltage 
across E2. Depending on the specific values, the sum may wind up being either a 
positive or negative value. If it's positive, this signifies that point a is at a higher 
potential than is point c. Conversely, if it's negative, this signifies that point a is at a 
lower potential than is point c. If we reverse the order, starting at point c and moving
88
Figure 3.21
A multi-source, multi-resistor 
series circuit.",DCElectricalCircuitAnalysis.pdf
"to point a, or Vca, we will wind up with the same voltage magnitude but the sign
will flip. In the laboratory, the first point letter (the a in Vac) is where the red lead
of the voltmeter is connected and the second letter is where the black lead is
connected. As a reminder, if a single connection point is used, as in Va, the second
letter is assumed to be ground. Thus, Va is the voltage from point a to ground. 
To illustrate the importance of voltage source polarity, consider the series circuit 
shown in Figure 3.22. This circuit is identical to the previous circuit with the 
exception that the polarity of source E2 has been flipped. This can have a drastic 
change on the resulting current and voltage drops. For example, if E2 is larger than 
E1, the net supplied voltage will be E2−E1 and the direction of conventional current 
will be as drawn. Effectively, E2 will be charging E1. Note that this is opposite to the",DCElectricalCircuitAnalysis.pdf
"will be as drawn. Effectively, E2 will be charging E1. Note that this is opposite to the 
prior circuit. Because both the direction and the magnitude of the current have 
changed, the voltage drops and polarities of R1 and R2 will change. Consequently, 
any point-to-point voltage, such as Vac, will also change. 
Example 3.9
Determine the voltages Vb and Vca in the circuit of Figure 3.23.
The first step here would be to note that the two sources are in opposition. 
The result is an effective source of 10.5 volts with the same polarity as the 
12 volt source, meaning it creates a clockwise current (rather the opposite of
Figure 3.22). The total resistance will be the sum of the resistors, or 500 Ω. 
From this we can compute the current and then the voltage drops across each
resistor. The polarity across the 100 Ω will be + to − left to right, and across 
the 400 Ω it will be + to − right to left. The polarities for both sources are 
top to bottom + to −. This is shown in Figure 3.24.",DCElectricalCircuitAnalysis.pdf
"the 400 Ω it will be + to − right to left. The polarities for both sources are 
top to bottom + to −. This is shown in Figure 3.24.
I = ETotal
RTotal
I = 10.5 V
500 Ω
I = 21  mA
V 100 = I×R100
V 100 = 21  mA×100Ω
V 100 = 2.1 volts
V 400 = I ×R400
V 400 = 21  mA×400Ω
V 400 = 8.4 volts
To find Vb we start at node b and work our way to ground, summing the 
potentials as we go. First we see the 1.5 volt source. Even though this is a 
89
Figure 3.22
Series circuit with altered 
polarity.
Figure 3.23
Circuit for Example 3.9.
Figure 3.24
Circuit for Example 3.9 with 
current direction and voltage 
polarities shown.",DCElectricalCircuitAnalysis.pdf
"source, it is still considered a voltage drop because the polarity is + to −. The
polarity on the 400 Ω resistor is the same, so we add in its 8.4 volts for 9.9 
volts total. That is, node b is 9.9 volts above ground. 
For Vca we proceed similarly. Starting at node c we move through the 1.5 
volt source, but this time the polarity is − to +, or a rise of 1.5 volts (i.e., 
taken as negative). Now at node b, we continue through the 100 Ω resistor, 
also seeing a − to + polarity, or a rise of 2.1 volts. That adds up to 3.6 volts 
in magnitude, however, the starting node is lower than the ending node, thus 
Vca = −3.6 volts. We could also say Vac = 3.6 volts.
Back in Chapter 2 the height analogy for voltage was presented. This can be applied 
nicely in the prior example. You can think of ground as literally ground: the earth 
under your feet. V oltage rises would be like climbing up a ladder and voltage drops",DCElectricalCircuitAnalysis.pdf
"under your feet. V oltage rises would be like climbing up a ladder and voltage drops 
would be like climbing down a ladder. Think of each volt as being analogous to one 
foot or one meter. Thus, for Vb, you climb up the ladder 8.4 feet for the 400 Ω and 
another 1.5 feet for the 1.5 volt source, leaving you 9.9 feet (volts) above the 
ground. Therefore, Vb must be positive with respect to ground by 9.9 volts. This 
analogy works for voltages below ground as well: just think in terms of digging a 
hole. For voltages that are not referenced to ground, think of climbing up and down 
ladders, and then comparing the height of where you stopped to the height of where 
you started.
Precisely labeling every voltage polarity may seem like unnecessary work, but as 
circuit complexity increases so does the opportunity for polarity errors. This 
important step should not be omitted. Consider the following example which is only 
slightly more complex than the one preceding. 
Example 3.10",DCElectricalCircuitAnalysis.pdf
"important step should not be omitted. Consider the following example which is only 
slightly more complex than the one preceding. 
Example 3.10
Determine the voltage Vbe in the circuit of Figure 3.25.
The 24 volt and 6 volt sources are aiding each other while the 4 volt
source is in opposition. The result is an effective supplied potential of
26 volts with a current circulating counterclockwise. The current
direction and voltage polarities are shown in Figure 3.26. 
First, find the circulating current and then find the voltage drops across
the resistors of interest. The total resistance is the sum of the three, or
13 kΩ.
90
Figure 3.25
Circuit for Example 3.10.",DCElectricalCircuitAnalysis.pdf
"I = ETotal
RTotal
I = 26 V
13 k Ω
I = 2  mA
V 10k = I ×R1
V 10k = 2  mA×10 k Ω
V 10k = 20  volts
Similarly, V2k = 4 volts and V1k = 2 volts with the polarities shown. To
determine Vbe, we can start at node b and move to node e by either
going to the left or right side. We shall do both and compare the results. 
Going to the left, we see + to − 20 volts across the 10 kΩ, − to + 24 across 
the left side source, and then + to − 4 volts across the 2 kΩ. This adds up to 
0 volts, meaning that nodes b and e are at the same potential! Double 
checking by going down the right side we see + to − 6 volts for the top 
source, − to + 4 volts for the right side source, and finally − to + 2 volts for 
the 1 kΩ. This is also 0 volts, as expected. 
It is important to understand that saying Vbe = 0 volts does not imply that Vb 
or Ve are 0 volts. In fact, we can see that Ve is below ground by the voltage 
dropped across the 2 kΩ, or −4 volts. Similarly, Vb is above node a by 20",DCElectricalCircuitAnalysis.pdf
"dropped across the 2 kΩ, or −4 volts. Similarly, Vb is above node a by 20 
volts, and node a itself is 24 volts below ground given the polarity of the left
side source. The result, once again, is −4 volts.
One more example to better illustrate the voltage divider rule in a larger circuit.
Example 3.11
Determine the voltages, Vb, Vd, and Vbd in the circuit of Figure 3.27.
While we could find the circulating current, then use Ohm's law to find the 
individual voltage drops, and finally add these pieces to find the potentials 
we're looking for, we shall instead focus on using VDR which may be faster. 
The voltage divider rule states that the voltage across any resistor or group 
of resistors in a series loop is proportional to their resistance compared to the
total resistance in that loop. The total resistance in this circuit is 12 kΩ. To 
find the voltage across any resistor or group of resistors, we simply make a",DCElectricalCircuitAnalysis.pdf
"find the voltage across any resistor or group of resistors, we simply make a 
ratio of the resistance of interest to the total resistance and multiply by the 
applied source voltage. Numbering the resistors R1 at top to R4 at bottom:
91
Figure 3.26
Circuit for Example 3.10 with 
current direction and voltage 
polarities shown.
Figure 3.27
Circuit for Example 3.11.",DCElectricalCircuitAnalysis.pdf
"V d = E R4
RT
V d = 36 V 4.5 kΩ
12 kΩ
V d = 13.5 volts
V b = E R2+R3+R4
RT
V b = 36V 8k Ω
12 k Ω
V b = 24  volts
V bd = E R2 +R3
RT
V bd = 36 V 3.5k Ω
12 kΩ
V bd = 10.5  volts
There is another way to find Vbd here, and that's to notice that by definition 
Vbd = Vb − Vd. Therefore, Vbd = 24 volts − 13.5 volts, or 10.5 volts.
3.8 Potentiometers and Rheostats3.8 Potentiometers and Rheostats
Some applications require the use of variable resistances. One example is a basic 
volume control, but other examples include equipment calibration and user 
adjustable parameters such as the bass, treble and balance controls on a home audio 
system. Adjustable resistances fall under two categories, potentiometers and 
rheostats. Generally, these would be the same physical device, with the application 
and usage determining its name. 
A potentiometer is a resistor with three leads. Along with the typical end leads, there",DCElectricalCircuitAnalysis.pdf
"and usage determining its name. 
A potentiometer is a resistor with three leads. Along with the typical end leads, there 
is a third lead referred to as the wiper or tap. This third lead is connected to a small 
metallic arm that makes contact with the resistive material and which can be moved 
along the length of the material. Potentiometers, or pots for short, often are formed 
in a circular shape (rotary travel) but they can also be in the form of a straight line 
(linear travel). Several different types are shown in Figure 3.28. The larger circular 
units with shafts and the straight unit at the top are designed to be adjusted by the 
end user (appropriately stylish knobs being attached to said shafts, of course), while 
the smaller units are intended to be adjusted only by a technician (note the small 
adjustment screw slots intended to receive fine screwdrivers). The smaller units are 
designed to be mounted directly on a printed circuit board while the user adjustable",DCElectricalCircuitAnalysis.pdf
"designed to be mounted directly on a printed circuit board while the user adjustable 
units are mounted on a front panel, and have wiring lugs instead of straight leads. 
92",DCElectricalCircuitAnalysis.pdf
"A cutaway view of a rotary panel mount pot with its back casing removed is shown 
in Figure 3.29.
Most circular potentiometers utilize a 270° rotation from end to end, also known
as “3/4 turn”. Some precision pots used for calibration may use an internal worm
gear to move the wiper and achieve 20 or more turns to shuttle the wiper from
one end to the other. This results in the ability to make very fine
adjustments. The rectangular units in the right-center of Figure 3.28 are
examples of this type.
The schematic symbol for a basic potentiometer is shown in Figure 3.30.
For convenience, its terminals are labeled A and B for the two ends, and W
for the wiper arm. 
A pot can be visualized as two resistors in sequence with a tap between
them. The total resistance never changes. What does change is the ratio of
the two pieces. 
RAB = RAW + RWB (3.6)
At the extreme left of travel, RAW will be 0 and RWB will be equal to RAB. At the",DCElectricalCircuitAnalysis.pdf
"the two pieces. 
RAB = RAW + RWB (3.6)
At the extreme left of travel, RAW will be 0 and RWB will be equal to RAB. At the  
extreme right, RAW will equal RAB and RWB will be 0. If the wiper is precisely in the 
middle, then RAW = RWB = RAB/2, and so on11. 
11 This 50/50 division is true if the pot has a linear taper, meaning that the resistance change
is proportional to the wiper position. While this is typical, it is possible to have a taper 
that is not linear for special applications, such as audio taper. In these devices the 
resistance ratio at mid-position might be something like 10/90 instead of 50/50. This 
would be desired for a volume control because the human ear responds to changes in 
sound pressure in a logarithmic fashion. Using a linear taper pot would result in a volume
control that is too sensitive at one end and yet produces hardly no change at the other end.
93
Figure 3.28
A variety of potentiometers.
Figure 3.29
Cutaway view of a rotary 
potentiometer.",DCElectricalCircuitAnalysis.pdf
"As a result, potentiometers are very useful as controllable voltage dividers (hence
the name). Sometimes it is desirable to control two signals simultaneously from a
single knob (e.g., one volume knob for a hi-fi stereo controls both the left and
right channels). For this application, multiple resistive elements can be stacked
and use a common shaft to control all of the wiper arms in tandem. These are known
as ganged potentiometers. The tall, round unit in the left-center of Figure 3.28 is a 
dual gang pot. Note that it has a total of six solder lugs, three for the top element and
three for the back element. Finally, unlike fixed resistors, pots are available in 
limited sizes. Typical values use multiples of 1 or 5 for the total resistance (for 
example, 1 kΩ, 5 kΩ, 10 kΩ, etc.) although a limited selection of other sizes are also
available.
If only the wiper and only one of the two end terminals is used, then the device is",DCElectricalCircuitAnalysis.pdf
"available.
If only the wiper and only one of the two end terminals is used, then the device is
referred to as a rheostat. Rheostats are often used to limit current. The value of the
rheostat ranges from 0 up to the nominal value of the potentiometer. The schematic
symbols for a rheostat are shown in Figure 3.31. The ANSI symbol is shown at the
left and the IEC version at the right.
Time for a couple of examples.
Example 3.12
For the circuit depicted in Figure 3.32, determine the maximum and
minimum voltages for Vc, as well as its value when the pot is adjusted to
midway (assume linear taper).
The voltage divider rule states that the voltage across any resistor or group 
of resistors in a series loop is proportional to their resistance compared to the
total resistance in that loop. The total resistance in this circuit is 8 kΩ. The 
voltage at node c will depend on the setting of the pot. The maximum 
voltage occurs when the wiper is all the way up, contacting node b.
V c (max) = E Ri",DCElectricalCircuitAnalysis.pdf
"voltage occurs when the wiper is all the way up, contacting node b.
V c (max) = E Ri
RT
V c (max) = 16 V 6 kΩ
8 kΩ
V c (max) = 12  volts
For the minimum case, the wiper is set all the way down, contacting node d.
94
Figure 3.30
Schematic symbol for a 
potentiometer, with labels.
Figure 3.31
Schematic symbols for a 
rheostat: ANSI (right) and IEC 
(left).
Figure 3.32
Circuit for Example 3.12.",DCElectricalCircuitAnalysis.pdf
"V c (min) = E Ri
RT
V c (min) = 16V 1 kΩ
8 kΩ
V c (min) = 2  volts
For the midway case, the pot can be treated as two 2.5 kΩ resistors.
V c (mid )= E Ri
RT
V c (mid )= 16 V 3.5k Ω
8 kΩ
V c (mid )= 7  volts
Note that the mid-position achieves a voltage midway between the extremes.
Thus the pot allows a range of voltage adjustment between a minimum of 2 
volts and a maximum of 12 volts. The upper and lower fixed resistors serve 
to set these limits. The larger the values of these resistors compared to the 
value of the pot, the more narrow the range of adjustment. If these two 
resistors were excluded, the pot would allow the entire range of adjustment, 
from ground (0 volts) up to the power supply value of 16 volts.
Example 3.13
The potentiometer in the circuit of Figure 3.33 is connected as a rheostat.
Determine the maximum and minimum current in the circuit.
The rheostat will present a resistance value between 0 Ω and 10 kΩ. The",DCElectricalCircuitAnalysis.pdf
"Determine the maximum and minimum current in the circuit.
The rheostat will present a resistance value between 0 Ω and 10 kΩ. The 
maximum current case occurs when the wiper is set to the extreme left, 
producing 0 Ω. 
I (max) = E
RT
I (max) = 50 V
1 kΩ
I (max) = 50  mA
For the minimum case, the wiper is set all the way to the right, producing the
full 10 kΩ.
95
Figure 3.33
Circuit for Example 3.13.",DCElectricalCircuitAnalysis.pdf
"I (min)= E
RT
I (min) = 50 V
11 kΩ
I (min) = 4.545  mA
3.9 Summary3.9 Summary
Electrons are the charge carrier in metals. Their direction of travel, called electron 
flow, runs from the negative terminal of a voltage source to the positive terminal. 
The more commonly used conventional current flow runs from the positive terminal 
to the negative terminal. As conventional current enters a resistor, we mark this point
as positive, and where it exits, negative. Thus, any voltage with a + to − polarity is 
deemed a voltage drop (which dissipates energy in a resistor). A polarity of − to + is 
deemed a voltage rise. 
A series connection is any connection in which the current through one component 
must be identical to the current flowing through any other component in that 
connection. In other words, the current exiting any component must be the only 
current flowing into the next component in line; there can no intervening",DCElectricalCircuitAnalysis.pdf
"current flowing into the next component in line; there can no intervening 
connections for current flow to or from other parts of the circuit. The equivalent 
resistance of a string of resistors placed in series is simply the sum of their resistance
values. Consequently, the larger resistors dominate the equivalent value.
Ohm's law states that the voltage across a resistor must equal the product of its 
resistance and the current flowing through it. 
Kirchhoff's voltage law (KVL) states that the sum of voltage rises around any series 
loop must equal the sum of voltage drops around that loop. Alternately, it may be 
stated as the sum of voltage rises and drops around a series loop must be zero, as the 
rises and drops have opposite polarity. 
The voltage divider rule (VDR) is a convenient computational shortcut based on 
Ohm's law and KVL. It states that the voltage across any given resistor or group of",DCElectricalCircuitAnalysis.pdf
"Ohm's law and KVL. It states that the voltage across any given resistor or group of 
resistors in a series connection is proportional to their resistance as a percentage of 
the total resistance. For example, if a resistor makes up one quarter of the total 
resistance in a series string of resistors, then that resistor will get one quarter of the 
total voltage applied to that string.
A potentiometer, or pot, is a three terminal variable resistance. The resistance of the 
entire device is fixed, however, a third lead is attached internally to an adjustable 
wiper, effectively creating two resistors. The sum of the two parts always equals the 
nominal value of the pot with the values of each piece being determined by the 
position of the wiper. Pots may be either rotary or linear (straight line) in motion and
96",DCElectricalCircuitAnalysis.pdf
"are used typically to adjust a voltage level. If the wiper and only one of the end 
terminals is used, the device is called a rheostat. A common application for a 
rheostat is the control of current.
Finally, computer simulation tools may be used to build virtual circuits and 
determine various parameters such as voltages and currents. They may use virtual 
instruments that echo real world instruments or they may use more direct means, 
offering results in the form of tables.
Review QuestionsReview Questions
1. Describe the differences between electron flow and conventional current 
flow.
2. Define the term series connection.
3. How is the equivalent resistance for a string of series connected resistors 
computed? Is the process identical for series connected voltage sources?
4. Why are current sources generally not placed in series?
5. Define Ohm's law.
6. Define Kirchhoff's voltage law.
7. Define the voltage divider rule.
8. What is meant by a Monte Carlo analysis?",DCElectricalCircuitAnalysis.pdf
"5. Define Ohm's law.
6. Define Kirchhoff's voltage law.
7. Define the voltage divider rule.
8. What is meant by a Monte Carlo analysis?
9. What is the difference between a potentiometer and a rheostat? 
3.10 Exercises3.10 Exercises
AnalysisAnalysis
1. For the circuit of Figure 3.34, determine the circulating current.
2. For the circuit of Figure 3.35, determine the circulating current.
97
Figure 3.34
Figure 3.35",DCElectricalCircuitAnalysis.pdf
"3. For the circuit of Figure 3.34, determine the power dissipated in the resistor.
4. For the circuit of Figure 3.35, determine the power dissipated in the resistor.
5. Determine the voltage at the open terminals of Figure 3.36.
6. Determine the voltage at the open terminals of Figure 3.37.
7. Determine the voltage at the open terminals of Figure 3.38.
8. Determine the equivalent resistance of circuit shown in Figure 3.39.
98
Figure 3.36
Figure 3.37
Figure 3.38
Figure 3.39",DCElectricalCircuitAnalysis.pdf
"9. Determine the equivalent resistance of circuit shown in Figure 3.40.
10. Determine the equivalent resistance of circuit shown in Figure 3.41.
11. For the circuit of Figure 3.42, determine the circulating current.
12. For the circuit of Figure 3.42, determine the voltages across each resistor 
and find Vab.
13. Given the circuit of Figure 3.42, determine the power dissipated by each 
resistor and the power delivered by the source.
99
Figure 3.40
Figure 3.41
Figure 3.42",DCElectricalCircuitAnalysis.pdf
"14. For the circuit of Figure 3.43, determine the circulating current.
15. Given the circuit of Figure 3.43, determine the voltages across each resistor 
and find Vba.
16. For the circuit of Figure 3.43, determine the power dissipated by each 
resistor and the power delivered by the source.
17. For the circuit of Figure 3.44, determine the circulating current.
18. For the circuit of Figure 3.44, determine the voltages across each resistor 
and find Va.
19. For the circuit of Figure 3.45, determine the circulating current and indicate 
all voltage polarities.
20. Given the circuit of Figure 3.45, determine the voltages across each resistor 
and find Vb, Vbc, and Vca.
21. For the circuit of Figure 3.45, determine the power delivered by the source.
100
Figure 3.43
Figure 3.44
Figure 3.45",DCElectricalCircuitAnalysis.pdf
"22. For the circuit of Figure 3.46, determine the circulating current and indicate 
all voltage polarities.
23. Given the circuit of Figure 3.46, determine the voltages across each resistor 
and find Vc, Vac, and Va.
24. For the circuit of Figure 3.46, determine the power dissipated by the 10 Ω 
resistor.
25. For the circuit of Figure 3.47, determine the circulating current and indicate 
all voltage polarities.
26. For the circuit of Figure 3.47, determine the voltages across each resistor 
and find Vb, Vc, and Vca.
27. Referring to the circuit of 3.48, determine the voltages across each resistor 
and find Vb, Vc, and Vac.
28. Referring to the circuit of Figure 3.48, determine the circulating current and 
indicate all voltage polarities.
101
Figure 3.46
Figure 3.47
Figure 3.48",DCElectricalCircuitAnalysis.pdf
"29. Given the circuit of 3.49, determine the voltages across each resistor and 
find Vb, Vc, and Vac.
30. Referring to the circuit of Figure 3.49, determine the circulating current and 
indicate all voltage polarities.
31. Given the circuit of 3.50, determine the circulating current and indicate all 
voltage polarities.
32. Referring to the circuit of Figure 3.50, determine the voltages across each 
resistor and find Vb, Vc, and Vac.
33. Given the circuit of 3.51, determine the circulating current and indicate all 
voltage polarities.
34. Referring to the circuit of Figure 3.51, determine the voltages across each 
resistor and find Vb, Vc, and Vac.
102
Figure 3.49
Figure 3.50
Figure 3.51",DCElectricalCircuitAnalysis.pdf
"35. Using the voltage divider rule, determine the voltages Vb, Vc and Vac for the 
circuit shown in Figure 3.52.
36. Using the voltage divider rule, determine the voltages Vb, Vc and Vbd for the 
circuit shown in Figure 3.53.
37. For the circuit of Figure 3.53, determine Vb if the 4 kΩ resistor is 
accidentally shorted. How does this compare to the original circuit?
38. For the circuit of Figure 3.53, determine Vb if the 4 kΩ resistor is 
accidentally opened. How does this compare to the original circuit?
103
Figure 3.52
Figure 3.53",DCElectricalCircuitAnalysis.pdf
"39. Given the circuit shown in Figure 3.54, find the voltage drop across the 
resistor.
40. Given the circuit shown in Figure 3.55, find the voltage drops across the 
resistor.
41. Find the voltage drops across the resistors in the circuit of Figure 3.56.
42. Find the voltage drops across the resistors in the circuit of Figure 3.57.
104
Figure 3.54
Figure 3.55
Figure 3.56
Figure 3.57",DCElectricalCircuitAnalysis.pdf
"43. Find the voltage drops across the resistors in the circuit of Figure 3.58.
44. Find the voltage drops across the resistors in the circuit of Figure 3.59.
45. The circuit of Figure 3.60 uses a linear taper potentiometer. Determine Vb 
when the wiper arm is at position a, position b, and at the halfway point.
46. What is the maximum current flowing through the potentiometer of Figure 
3.60? At what position(s) does this occur?
DesignDesign
47. Redesign the circuit of Figure 3.34 using a new resistor such that the current
from the 12 volt battery is 0.1 A.
48. Redesign the circuit of Figure 3.35 using a new resistor such that the current
from the 9 volt battery is 2 mA.
49. For the circuit of Figure 3.39, find the value of a series voltage source that 
would generate 1 mA of current if it was connected across the terminals.
105
Figure 3.58
Figure 3.59
Figure 3.60",DCElectricalCircuitAnalysis.pdf
"50. For the circuit of Figure 3.41, find the value of a series voltage source that 
would generate 1 mA of current if it was connected across the terminals.
51. Determine values for the resistors in Figure 3.61 such that R1 is four times 
the size of R2 and R2 is three times the size of R3, with the total resistance 
equaling 8 kΩ.
52. For the circuit shown in Figure 3.62, determine values for R1 and R2 such 
that Vab is 6 volts if E is a 9 volt battery and the total current draw is 20 mA.
53. Consider the circuit shown in Figure 3.63. If all resistors have the same 
value, determine that value if E, a 24 volt source, generates a total power of 
10 watts.
106
Figure 3.61
Figure 3.62
Figure 3.63",DCElectricalCircuitAnalysis.pdf
"54. For the circuit shown in Figure 3.64, determine values for R1 and R2 such 
that Vab is 6 volts if I is a 2 mA source and the total voltage drop is 24 volts.
55. Consider the circuit of Figure 3.53. Is it possible to add a fifth resistor such 
that the circulating current is 0.1 mA? If so, what is that resistor value?
56. Consider the circuit of Figure 3.53. Is it possible to add a fifth resistor such 
that the circulating current is 2 mA? If so, what is that resistor value?
ChallengeChallenge
57. Assume that two AA cells, E1 and E2, rated at 900 mAh each are used to 
drive a 2 watt lamp as shown in Figure 3.65. Determine the expected life of 
the batteries.
58. Given the circuit of Figure 3.66, determine the required values of E, R1, R2 
and R3 if there is one volt across R3, the total current draw is 10 mA, the 
voltage across R1 is twice the size of voltage across R2 and the power 
dissipation in R2 is 100 mW.
107
Figure 3.64
Figure 3.65
Figure 3.66",DCElectricalCircuitAnalysis.pdf
"59. Given the circuit of Figure 3.66, determine the required source voltage if R1 
is 1 kΩ, R2 is 1 kΩ, the power dissipation in R1 is 4 mW and the power 
dissipation in R3 is 2 mW.
60. Given the circuit of Figure 3.67, determine Vc, Vdb and Vce.
61. Given the circuit of Figure 3.67, determine Vac, Veb and Vd.
62. Refer to the circuit of Figure 3.53. Assuming each resistor has a 10% 
tolerance, determine the maximum and minimum values for Vc.
SimulationSimulation
63. Simulate the solution of design problem 43 and determine if the values 
produce the required results.
64. Perform a DC simulation on the circuit of Figure 3.46 and find all of the 
node voltages along with the circulating current.
65. Perform a DC simulation on the circuit of Figure 3.47 and find all of the 
node voltages along with the circulating current.
66. Perform a DC simulation on the circuit of Figure 3.59 and find all of the 
node voltages.",DCElectricalCircuitAnalysis.pdf
"node voltages along with the circulating current.
66. Perform a DC simulation on the circuit of Figure 3.59 and find all of the 
node voltages.
67. Perform a DC simulation on the circuit of Problem 34 and find all of the 
node voltages.
68. Perform a DC simulation on the circuit of Problem 36 and find all of the 
node voltages.
69. Simulate the solution of Challenge problem 58 and determine if the values 
produce the required results.
70. Simulate the circuit of Figure 3.67 (Challenge problems 60 and 61) and 
determine if the node voltages produced match the expected results.
71. Perform a Monte Carlo simulation on the circuit of Figure 3.52. Set each 
resistor to 5% tolerance and run at least ten variations for Vb to determine a 
typical spread of values.
108
Figure 3.67",DCElectricalCircuitAnalysis.pdf
"Source: www.xkcd.com
This page intentionally left not blank.
109",DCElectricalCircuitAnalysis.pdf
"4 4 Parallel Resistive CircuitsParallel Resistive Circuits
4.0 Chapter Learning Objectives4.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Identify parallel resistive circuits that include a single voltage source or one or more current sources.
• Compute equivalent resistance of parallel resistive networks. 
• Determine the equivalent of multiple parallel current sources.
• Compute component and total current for parallel resistive networks. 
• Compute system voltage and component powers for parallel resistive networks.
• Utilize Ohm's law, Kirchhoff's current law (KCL) and the current divider rule (CDR) to aid in the 
analysis of parallel resistive circuits.
• Describe the function of fuses and circuit breakers.
4.1 Introduction 4.1 Introduction 
In the preceding chapter we examined series circuits, starting with series configurations of resistors and voltage",DCElectricalCircuitAnalysis.pdf
"In the preceding chapter we examined series circuits, starting with series configurations of resistors and voltage 
sources. In this chapter we shall examine the “mirror twin” of series circuits, namely parallel circuits. Again, we 
shall begin by defining the parallel configuration, and explore how to combine sources and resistors in parallel. 
From there we shall introduce new laws and rules unique to parallel configurations. For everything said about 
series circuits, there are corresponding statements regarding parallel circuits. Just as there were specific laws and
techniques useful in the series case, such as Kirchhoff's voltage law and the voltage divider rule, there are 
corresponding laws and techniques useful in the series case (e.g., Kirchhoff's current law and the current divider 
rule). 
A parallel circuit may contain any number of resistors and current sources, or in place of the current sources, a",DCElectricalCircuitAnalysis.pdf
"rule). 
A parallel circuit may contain any number of resistors and current sources, or in place of the current sources, a 
single voltage source. We shall examine how to determine the current flow through each component, the voltage
across each component and the power either dissipated or generated by each component. As usual, other 
practical issues will also be examined along with appropriate computer simulations. 
Parallel circuits are in many ways the complement of series circuits. The hallmark of a series connection is that 
all components in that connection see the same current. Similarly, the most notable characteristic of a parallel 
connection is that all components see the same voltage. This implies that parallel connections have only two 
nodes, or two points of connection upon which everything is connected. Just as a series connection can be 
envisioned as a sort of chain with every element a link in that chain, a parallel connection can be envisioned as a",DCElectricalCircuitAnalysis.pdf
"envisioned as a sort of chain with every element a link in that chain, a parallel connection can be envisioned as a
sort of ladder with every element a rung on that ladder.
110",DCElectricalCircuitAnalysis.pdf
"4.2 The Parallel Connection4.2 The Parallel Connection
Consider the generic layout of Figure 4.1, each component represented by a box.
Although it may be drawn oddly, there are only two common connection nodes in
this configuration: one along the top and the other along the bottom (if you're
having difficulty seeing the bottom node, simply slide box E to the right and then
rotate it counterclockwise 180 degrees so that its bottom winds up on top). Each of 
these blocks is like a rung on a ladder. If we were to take a voltmeter and place it 
across any of these blocks, the meter would read the same voltage across each 
because the probes are always contacting the same two nodes (remember, ideally the
connecting wires have negligible resistance). Thus, we come to the first observation 
of parallel circuits:
      In a parallel connection the voltage is the same across each component.     (4.1)
A more complex configuration is shown in Figure 4.2. In this layout some",DCElectricalCircuitAnalysis.pdf
"A more complex configuration is shown in Figure 4.2. In this layout some
elements are in series and some are in parallel. Only elements D and E are strictly
in parallel here as they are the only two that connect to the same two nodes, and
thus must see the same voltage. Items C and F might at first appear to be in
parallel with them but upon closer inspection it should be noted that C and F are in 
series with each other. That is, C and F will split the voltage that D or E sees, in 
accordance with the voltage divider rule. In other words, it is not true that the 
voltage across C or across F must be the same as the voltage across D or E. The 
same can be said for items A and B.12 
4.3 Combining Parallel Components4.3 Combining Parallel Components
Our first step is to determine how to combine parallel components in order to create 
a single equivalent component. Unlike series connections, this can be a little more 
time consuming.
Sources in ParallelSources in Parallel",DCElectricalCircuitAnalysis.pdf
"a single equivalent component. Unlike series connections, this can be a little more 
time consuming.
Sources in ParallelSources in Parallel
First, voltage sources are not placed in parallel as a general rule, see Figure 4.3.
The reason is because a parallel connection requires the same voltage across each
component. This would be impossible to achieve with each source trying to
maintain a different voltage across the same two nodes. This may result in
damage to the sources (e.g., exploding batteries). The main exception to this rule is 
if the sources have the same potential and the goal is to extend operational life (e.g., 
multiple D cell batteries in parallel). 
12 What can be said is that the series combination of C and F together is in parallel with D 
which is, in turn, in parallel with E. This is the basic idea behind series-parallel networks,
the theme of the next chapter. 
111
Figure 4.1
A parallel configuration.
A B
C
D
E
Figure 4.2
A more complex configuration.
A B
C
D
F
E",DCElectricalCircuitAnalysis.pdf
"the theme of the next chapter. 
111
Figure 4.1
A parallel configuration.
A B
C
D
E
Figure 4.2
A more complex configuration.
A B
C
D
F
E
Figure 4.3
Do not place voltage sources 
in parallel.",DCElectricalCircuitAnalysis.pdf
"When it comes to current sources, they may simply be added together, however,
like series voltage sources, polarity is important. Referring to the three parallel
current sources in Figure 4.4, the middle and right sources are pumping current
into the top node while the left source is feeding the bottom node. You can also
think of the left source as draining the top node. Thus, these three sources together 
behave like a single source of five amps feeding the top node (eight amps in, three 
amps out). 
Resistors in ParallelResistors in Parallel
When placed in parallel, resistor values do not add they way they do in series 
connections. The reason for this is obvious if we look at the basic resistance
equation from Chapter 2, Equation 2.11: 
R =ρl
A
If we consider two identical resistors placed in parallel, side by side as in Figure 4.5,
the effective area would double while keeping the resistivity and length unchanged.",DCElectricalCircuitAnalysis.pdf
"the effective area would double while keeping the resistivity and length unchanged. 
The combined result would be a halving of the resistance of just one of them. Thus 
we see that placing resistors in parallel results in a decrease in net resistance, the 
opposite of the series case.
This situation is simplified if we consider conductance instead of resistance. 
Recalling that conductance, G, is the reciprocal of resistance, we can rewrite the 
resistance equation:
G = A
ρl
 
Now we can see that the increase in area creates a proportional increase in 
conductivity. If we generalize this for N resistors we find that the equivalent 
conductance of a group of parallel resistors is their sum:
GTotal = G1+G2+G3 +...+GN (4.2)
As we deal normally with component resistance values instead of conductance 
values, Equation 4.2 may be rewritten in terms of equivalent resistance:
1
RParallel
= 1
R1
+ 1
R2
+ 1
R3
+...+ 1
RN
RParallel = 1
1
R1
+ 1
R2
+ 1
R3
+...+ 1
RN
(4.3)
112
Figure 4.4",DCElectricalCircuitAnalysis.pdf
"1
RParallel
= 1
R1
+ 1
R2
+ 1
R3
+...+ 1
RN
RParallel = 1
1
R1
+ 1
R2
+ 1
R3
+...+ 1
RN
(4.3)
112
Figure 4.4
Current sources in parallel.
Figure 4.5
Resistors in parallel.
A
B",DCElectricalCircuitAnalysis.pdf
"Among other things, Equation 4.3 tells us that the equivalent resistance of a group of
parallel resistances will always be smaller than the smallest resistance in that group. 
It also indicates that if we have N identical resistors of value R, the equivalent will 
be R/N (e.g., 3 parallel 3.6 kΩ resistors are equivalent to a single 1.2 kΩ resistor). As
a convenience, it is common to use two parallel lines, ||, as a shortcut way of saying 
“in parallel with”, as in R1 || R2. Also, || has higher operational precedence than +. 
Product-Sum RuleProduct-Sum Rule
A handy shortcut for two parallel resistors may be created from Equation 4.3. We 
start by writing the version for just two parallel resistors,
RParallel = 1
1
R1
+ 1
R2
Multiply the fraction through by R1 R2 and simplify:
RParallel = R1 R2
R1+R2
(4.4)
For obvious reasons, Equation 4.4 is referred to as the product-sum rule. It is only 
applicable to two resistors, however, it can be used repeatedly on pairs of resistors",DCElectricalCircuitAnalysis.pdf
"applicable to two resistors, however, it can be used repeatedly on pairs of resistors 
within larger groups instead of using Equation 4.3. For example, a group of four 
resistors can be reduced to two pairs, and then those two equivalents can be reduced 
as a third pair. This can be a handy technique for obtaining quick estimates.
We can take the product-sum rule a step further by writing R2 as a multiple of R1:
   R2 = N R1
Substituting this into Equation 4.4 yields:
RParallel = R1 N R1
R1+N R1
RParallel = N R1
2
(N +1)R1
RParallel = N
N +1 R1 (4.5)
Equation 4.5 is useful for quick estimates of parallel resistor pairs. For example, if 
we have a 24 kΩ resistor in parallel with an 8 kΩ resistor, that's a ratio (N) of 3:1.  
113",DCElectricalCircuitAnalysis.pdf
"Thus the equivalent will be N/(N+1), or 3/4ths, of the smaller resistor, yielding 6 kΩ. 
Similarly, a 180 Ω in parallel with a 90 Ω would yield 2/3rds of 90 Ω, or 60 Ω.  
Example 4.1
A group of resistors is placed in parallel as shown in Figure 4.6. Determine
the equivalent parallel value.
There are several solution paths. We'll try them all and compare. First, let's 
try the conductance form:
RParallel = 1
1
R1
+ 1
R2
+ 1
R3
+...+ 1
RN
RParallel = 1
1
2k Ω+ 1
6k Ω+ 1
12 kΩ
RParallel ≈ 1.333 kΩ
We could also use the product-sum rule twice. We'll use the middle and 
right-most resistor first (although any pairing would do, but this pairing is 
convenient): 
R23 = R2 R3
R2+R3
R23 = 6k Ω12 k Ω
6 kΩ+12k Ω
R23 = 4 k Ω
RParallel = R1 R23
R1+R23
RParallel = 2 kΩ4 k Ω
2 kΩ+4k Ω
RParallel ≈ 1.333 kΩ
Finally, we could use the ratio technique. First, consider the 6 kΩ and 12 kΩ 
resistor pair. That's a 2:1 ratio, so the result will be 2/3rds of the smaller",DCElectricalCircuitAnalysis.pdf
"resistor pair. That's a 2:1 ratio, so the result will be 2/3rds of the smaller 
resistor, and 2/3rds of 6 kΩ is 4 kΩ. The ratio between this and the 2 kΩ 
resistor is also 2:1, and 2/3rds of 2 kΩ is approximately 1.333 kΩ. 
The ratio technique happened to work out well here because of the 
convenient resistor values with perfect integer ratios. That will not always be
the case.
114
Figure 4.6
Resistor arrangement for 
Example 4.1.",DCElectricalCircuitAnalysis.pdf
"4.4 Kirchhoff's Current Law (KCL)4.4 Kirchhoff's Current Law (KCL)
Just as Kirchhoff's voltage law is a key element in understanding series circuits, 
Kirchhoff's current law (KCL) is the operative rule for parallel circuits. It states that 
the sum of all currents entering and exiting a node must sum to zero. Alternately, it 
can be stated as the sum of currents entering a node must equal the sum of currents 
exiting that node. As a pseudo formula: 
∑ I→ =  ∑ I← (4.6)
Recalling that a node is a connection area wherein the voltage is the same
(ignoring the resistance of connecting wires), we see that there are two nodes in a
parallel configuration. For each of these nodes, whatever currents flow in must be
balanced by the same total current flowing out. For example, referring to the
circuit shown in Figure 4.7, there is a current source feeding the top node. At the 
three-way connection with R1 and R2, KCL along with Ohm's law dictate that the",DCElectricalCircuitAnalysis.pdf
"three-way connection with R1 and R2, KCL along with Ohm's law dictate that the 
incoming current from the source must split with one portion exiting by flowing 
down through R1 and the remainder exiting by flowing down through R2. At the 
bottom node (ground) these two resistor currents must combine and equal the 
current that is being drawn back into the current source, which must be the same as 
the current originally produced by the source. This is perfectly sensible because it 
would be impossible for the current source to establish two different currents; one at 
its input and the other at its output. To assert otherwise would be a little like saying 
you're two different heights depending on whether you measure from your toes up to
your head versus from your head down to your toes. Thus, a simplistic way of 
stating KCL is “What goes in must come out”.13
The Current Divider Rule (CDR)The Current Divider Rule (CDR)",DCElectricalCircuitAnalysis.pdf
"stating KCL is “What goes in must come out”.13
The Current Divider Rule (CDR)The Current Divider Rule (CDR)
Just as series circuits follow the voltage divider rule (voltage dividing in proportion 
to resistance), parallel circuits follow the current divider rule which states that 
current divides in reverse proportion to resistance (i.e., in direct proportion to 
conductance). This can be reduced to a simple formula when only two resistors are 
involved. Consider two parallel resistors, R1 and R2, fed by a current, ITotal, as in 
Figure 4.7. First, we note that the voltage across the pair must equal the entering 
current times the effective resistance of the pair.
V = I Total
R1 R2
R1+R2
13 For details on this corollary to Fudd's First Law of Opposition, see Firesign Theatre, I 
Think We're All Bozos On This Bus. Remember, the future can't wait.
115
Figure 4.7
A simple parallel circuit.",DCElectricalCircuitAnalysis.pdf
"The current through each of the resistors must be this voltage divided by the 
corresponding resistance. For the current through R1:
I R1 = V
R1
I R1 = I Total
R1 R2
R1(R1+R2)
I R1 = I Total
R2
R1+R2
Similarly, for the current through R2:
I R2 = V
R2
I R2 = I Total
R1 R2
R2(R1+R2)
I R2 = I Total
R1
R1+R2
Thus, the current through one of the resistors will equal the total current times the 
ratio of the opposite resistor over the sum of the two resistors. In general:
I X = ITotal
RY
R X +RY
(4.7)
This rule is convenient in that you don't have to compute the parallel equivalent 
resistance, but remember, it is valid only when there are just two resistors involved. 
A more general version that can be used for any number of resistors is Ii = ITotal ∙ 
RP/Ri, where RP is the total equivalent parallel resistance and Ri is the resistor of 
interest. This is, in essence, merely a rewriting of the fact that all components must",DCElectricalCircuitAnalysis.pdf
"interest. This is, in essence, merely a rewriting of the fact that all components must 
see the same voltage: Ii ∙ Ri  = ITotal ∙ RP. This is not as much of a shortcut as the two 
resistor version because the parallel resistance still must be calculated.
Example 4.2
A simple parallel network is shown in Figure 4.8. Determine the current
through each resistor.
The current divider rule can be used to find the currents through the two 
resistors. At the top node the total entering current is 10 amps.
116
Figure 4.8
Circuit for Example 4.2.",DCElectricalCircuitAnalysis.pdf
"I R1 = I Total
R2
R1+R2
I 15 = 10 A 5Ω
15Ω+5Ω
I 15 = 2.5 A
I R2 = ITotal
R1
R1+R2
I 5 = 10 A 15Ω
15Ω+5Ω
I 5 = 7.5  A
At this point it is worth taking a moment to verify the voltage polarities and 
current directions. Given the direction of the source, the currents through the
two resistors must be flowing from top to bottom. Consequently, the voltage 
polarities must be + to − from top to bottom. This is illustrated in Figure 4.9.
At the top node, entering current is in red with exiting current in blue. Note 
that KCL is satisfied as 10 amps enters while a total of 10 amps (2.5 amps 
plus 7.5 amps) exits.
4.5 Parallel Analysis4.5 Parallel Analysis
Keeping in mind that the voltage across each element in a parallel configuration is 
constant, Ohm's law dictates that currents divide among parallel resistors in 
proportion to their conductance (i.e., in inverse proportion to their resistance). As a 
consequence, Ohm's law, Kirchhoff's current law, the current divider rule and",DCElectricalCircuitAnalysis.pdf
"consequence, Ohm's law, Kirchhoff's current law, the current divider rule and 
parallel component combinations are the tools we will use to solve general parallel 
circuit problems. There are multiple techniques for analyzing these circuits:
• If the circuit uses a voltage source and its value along with the resistor 
values are given, the resistor currents can be found by dividing the source 
voltage by each resistance. Once these currents are found, KCL can be used 
to determine the source current. At that point, power law can be used to find 
the power dissipation in each resistor or the power developed by the source, 
as needed. 
• If the circuit uses current sources, then the total circulating current can be 
found by combining their values (taking into account the current directions, 
of course). The individual resistor branch currents then can be found using 
the current divider rule (repeatedly, if needed). Alternately, the effective",DCElectricalCircuitAnalysis.pdf
"the current divider rule (repeatedly, if needed). Alternately, the effective 
parallel resistance may be found first. Multiplying this by the total source 
current via Ohm's law will determine the system voltage, and from there the 
individual resistor branch currents may be found using Ohm's law. 
117
Figure 4.9
Polarities and directions for 
Example 4.2.",DCElectricalCircuitAnalysis.pdf
"• If the problem concerns determining resistance values, the basic idea will be
to use these rules in reverse. For example, if a resistor value is needed to set 
a specific voltage, the equivalent parallel resistance can be determined from 
this voltage and the given current source. The conductance values of the 
other parallel resistors can then be subtracted from the total conductance 
(i.e., the reciprocal of the equivalent parallel resistance), yielding the 
remaining required conductance value, the reciprocal being the required 
resistance. Similarly, if the currents through two parallel resistors are 
known, as long as one of the resistance values is known, the other resistance
can be determined using either the current divider rule in reverse (i.e., the 
currents divide in inverse proportion to the resistances) or via Ohm's law. 
Example 4.3
A simple parallel network is shown in Figure 4.10. Determine the current",DCElectricalCircuitAnalysis.pdf
"Example 4.3
A simple parallel network is shown in Figure 4.10. Determine the current
through each resistor as well as the total current exiting the voltage source.
The most direct solution is to use Ohm's law to determine each of the 
resistor branch currents. KCL can then be used to determine the current 
flowing from the source. Recalling the voltage is the same across parallel 
branches:
I 600 = E
R1
I 600 = 12 V
600 Ω
I 600 = 20  mA
I 400 = E
R2
I 400 = 12 V
400Ω
I 400 = 30  mA
KCL dictates that the entering current must equal the sum of the exiting 
currents, or 50 mA. 
An alternate technique would be to determine the parallel resistance and 
divide this into the source voltage to determine the exiting source current.
 
RParallel = R1 R2
R1+R2
RParallel = 400 Ω600Ω
400 Ω+600 Ω
RParallel = 240Ω
118
Figure 4.10
Circuit for Example 4.3.",DCElectricalCircuitAnalysis.pdf
"I Total = E
RParallel
I Total = 12 V
240 Ω
I Total = 50  mA
Now CDR can be used to find the currents through the two resistors.
I R1 = I Total
R2
R1+R2
I 400 = 50 mA 600Ω
400 Ω+600 Ω
I 400 = 30  mA
I R2 = ITotal
R1
R1+R2
I 600 = 50 mA 400Ω
400 Ω+600 Ω
I 600 = 20  mA
The voltage polarities and current directions are illustrated in Figure 4.11. 
The voltage polarity is + to − from top to bottom, as set by the voltage 
source. With these polarities, the currents through the two resistors must be 
flowing from top to bottom and the current from the source is flowing right, 
from the positive terminal. At the top node, entering current is in red with 
exiting current in blue.
Computer SimulationComputer Simulation
The circuit of Example 4.3 is entered into a simulator as shown in Figure 4.12. In 
this example, virtual instruments are used. Three ammeters are inserted in line to 
measure the current exiting the voltage source as well as the currents flowing",DCElectricalCircuitAnalysis.pdf
"measure the current exiting the voltage source as well as the currents flowing 
through the two resistors. The polarities of the ammeters are set to match those 
illustrated in Figure 4.11. As a consequence, we expect to see all positive currents.
The result is pretty much as expected. The current through the 600 Ω resistor is 
precisely 20 mA and the source current is precisely the sum of the two resistor 
currents. The only wrinkle is that the current through the 400 Ω resistor is very 
slightly less than expected, 29.999 mA versus 30 mA calculated. This is due to meter
loading effects. In Chapter 3 we saw that ammeters present a very low internal 
resistance, but this cannot always be ignored, especially when measuring in series 
with resistors having very small values. This slight increase in resistance forces a 
slight decrease in current due to Ohm's law. 
119
Figure 4.11
Polarities and directions for 
Example 4.3.",DCElectricalCircuitAnalysis.pdf
"It turns out that a similar situation exists for voltmeters. Ideally, voltmeters present a 
very high internal resistance which, when placed across a resistor, has minimal 
impact. This effect cannot always be ignored, especially when measuring across 
resistors having large values, as the current divider rule will come into play. 
Fortunately, the internal resistance of many virtual instruments is adjustable and can 
be set to extreme values to minimize any impact on measurements. In the world of 
physical instruments this is not possible so the internal resistance of any real world 
meter is something that must always be kept in mind. 
Example 4.4
A parallel network is shown in Figure 4.13. Determine the current through
each resistor.
The current divider rule can be used to find the currents through the two 
resistors. At the bottom node (ground) the total entering current is 2 mA. 
I R1 = I Total
R2
R1+R2
I 12k = 2mA 100 kΩ
12 kΩ+100 kΩ
I 12k ≈ 1.7857  mA",DCElectricalCircuitAnalysis.pdf
"I R1 = I Total
R2
R1+R2
I 12k = 2mA 100 kΩ
12 kΩ+100 kΩ
I 12k ≈ 1.7857  mA
The current through the 100 kΩ can be found via KCL as follows:
120
Figure 4.12
The circuit of Example 4.3 in a 
simulator.
Figure 4.13
Circuit for Example 4.4.",DCElectricalCircuitAnalysis.pdf
"I R2 = I Total −I R1
I 100k = 2 mA −1.7857 mA
I 100k ≈ 0.2143 mA
The polarities and directions are shown in Figure 4.14. Note that the currents
are flowing up through the resistors producing voltage drops of + to − from 
ground up to the top. This means that the voltage of the top node is negative 
with respect to ground.
Computer SimulationComputer Simulation
In order to get a better handle on the idea of meter loading, the circuit shown in 
Example 4.4 is entered into a simulator. This is shown in Figure 4.15 with the 
simulation using virtual instruments. Using Ohm's law, the voltage drop across the 
system should be 100 kΩ times its current, or approximately 21.4286 volts if we 
carry the digits out a little further.
The virtual voltmeter has been set for an internal resistance of 1 GΩ. The results are 
nearly the same as those calculated previously. To test the impact of the voltmeter's 
internal resistance the simulation is run a second time with the internal resistance set",DCElectricalCircuitAnalysis.pdf
"internal resistance the simulation is run a second time with the internal resistance set
to 1 MΩ, a value seen commonly in general purpose digital multimeters. The result 
is shown in Figure 4.16. The voltage magnitude has dropped more than 200 mV , 
entirely due to meter loading. 
121
Figure 4.14
Polarities and directions for 
Example 4.4.
Figure 4.15
The circuit of Example 4.4 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"Depending on component values, the results could be much worse or hardly 
noticeable at all. As a general rule, make sure the voltmeter's internal resistance is at 
least 10 times (and preferably, 100 times) larger than any resistance it is placed 
across in order to avoid loading errors caused by unwanted current division. In the 
case of an ammeter, its internal resistance should be at least 10 times (and 
preferably, 100 times) smaller than any resistance with which it is placed in series.
Example 4.5
A parallel network is shown in Figure 4.17. Determine the current through
the 200 Ω resistor.
The first step is to simplify the circuit. The two current sources
aid each other as they both feed current into the top node. Thus
they are equivalent to a single 1.4 amp current source with the
same direction. Further, the 60 Ω and 120 Ω resistors are in
parallel and can be treated as a single unit. Note this is a 2:1
ratio, and thus the result will be 2/3rds of 60 Ω, or 40 Ω. The",DCElectricalCircuitAnalysis.pdf
"parallel and can be treated as a single unit. Note this is a 2:1
ratio, and thus the result will be 2/3rds of 60 Ω, or 40 Ω. The
basic product-sum rule or conductance formula could also be used.
We have simplified the circuit down to a single 1.4 amp source feeding a 
200 Ω resistor in parallel with a 40 Ω resistor. The current divider rule can 
be used to find the current through the 200 Ω resistor as follows:
122
Figure 4.16
Simulation repeated using a 
voltmeter internal resistance of 
1 MΩ.
Figure 4.17
Circuit for Example 4.5.",DCElectricalCircuitAnalysis.pdf
"I R1 = I Total
R2
R1+R2
I 200 = 1.4 A 40Ω
200Ω+40Ω
I 200 ≈ 233.33  mA
As a crosscheck, Ohm's law indicates the system voltage must be 200 Ω 
times 233.33 mA, or approximately 46.667 volts. Via Ohm's law, the current
through the 60 Ω must be 46.667 volts divided by 60 Ω, or approximately 
777.78 mA. Similarly, the current through the 120 Ω can be determined to 
be approximately 388.89 mA. Via KCL, these three currents should add up 
to the total supplied current of 1.4 amps, which they do.
An alternate solution path would be to find the equivalent of the three 
parallel resistors, or 200 || 60 || 120, which is 33.333 Ω. This is fed by the 
combined sources yielding 1.4 amps. Ohm's law is then used to find the 
system voltage of approximately 46.667 volts (1.4 amps times 33.333 Ω). 
From there Ohm's law is used again to find the current flowing through the 
200 Ω (46.667 volts divided by 200 Ω, or approximately 233.33 mA).",DCElectricalCircuitAnalysis.pdf
"From there Ohm's law is used again to find the current flowing through the 
200 Ω (46.667 volts divided by 200 Ω, or approximately 233.33 mA).
Let's continue with the prior example, this time with a design twist.
Example 4.6
Using the circuit of Figure 4.17, determine the value of an additional fourth 
parallel resistor such that the system voltage will drop to 42 volts.
The total source current in this circuit is fixed at 1.4 amps. If the resistance it
feeds is reduced, then by Ohm's law, the system voltage must also be 
reduced. One way to solve this problem is to first determine the required 
parallel equivalent resistance. Then the conductances of the three known 
resistors can be subtracted out, leaving the conductance of the new resistor, 
from which we determine its resistance.  
RParallel = V
I
RParallel = 42 V
1.4A
RParallel = 30 Ω
G4 = GParallel −G1 −G2 −G3
G4 = 1
30 Ω − 1
200 Ω − 1
60 Ω − 1
120Ω
G4 ≈ 3.333 mS
123",DCElectricalCircuitAnalysis.pdf
"And finally, R4 = 1/G4, thus R4 = 300 Ω.
The smaller the value of this fourth resistor, the more current it will siphon 
off from the other three resistors, thus reducing the system voltage further. 
Time for an example involving power.
Example 4.7
A parallel network is shown in Figure 4.18. Determine the total power
dissipated by the four resistors.
While the powers for each resistor can be calculated
individually and then added, it is worthwhile to note that the
total power dissipated must be equal to the power generated.
The power generated by the battery is dictated by power law,
in other words, the battery voltage times its output current. To
find this current, we can determine the effective resistance and
then apply Ohm's law. 
The total resistance can be found directly by using the conductance formula, 
however, the values can be divided into convenient pairs first. The pair of 
200 Ω resistors is equivalent to a single 100 Ω. Similarly, the two 600 Ω",DCElectricalCircuitAnalysis.pdf
"200 Ω resistors is equivalent to a single 100 Ω. Similarly, the two 600 Ω 
resistors are equivalent to a single 300 Ω. Product-sum rule may be used 
next or 100 || 300 can be determined using the ratio rule. This is a 3:1 ratio, 
so the result will 3/4ths of the smaller resistor, or 75 Ω.
I Total = E
RTotal
I Total = 9 V
75Ω
I Total = 120  mA
P = E×I
P = 9V×120 mA
P = 1.08 W
Slightly quicker, we could also use the following approach:
P = E2
R
P = (9 V)2
75Ω
P = 1.08 W
124
Figure 4.18
Circuit for Example 4.7.",DCElectricalCircuitAnalysis.pdf
"4.6 Current Limiting: Fuses and Circuit Breakers4.6 Current Limiting: Fuses and Circuit Breakers
Most electrical and electronic devices are designed to run off of a constant voltage 
source. Examples include battery powered devices and the numerous items that 
connect to standard residential and commercial electrical systems (e.g., the 120 V AC 
system in North America). As such, parallel connections are ideal when a variable 
number of devices are to be powered because, unlike a series connection, each 
device will see the same voltage rather than having that voltage split among them. 
There is another advantage to a parallel connection. Consider the two lighting 
schemes presented in Figure 4.19; one utilizing a series configuration and the other a
parallel configuration.
Typically, when a lamp fails, it fails creating an open. If any of the lamps in the 
series connection fails, current in the loop is interrupted and all of the lamps go out.",DCElectricalCircuitAnalysis.pdf
"series connection fails, current in the loop is interrupted and all of the lamps go out. 
In contrast, if one of the lamps in the parallel connection fails, the remaining lights 
will stay on. Further, in the series connection, there is no immediate way to 
determine which lamp has failed as none of them will be lit. Each lamp has to be 
tested in turn to determine the faulty unit. In the parallel system, it is obvious which 
lamp has failed because that's the one that's out. 
In a home, school or commercial setting, any number of devices may be plugged
into wall sockets with the expectation that each unit will receive the same voltage
and operate correctly due to the parallel nature of the system. As convenient as this
is, there is a practical problem. Every device that is added to the system
presents another path for current flow. Due to KCL, this means that the
total current drawn from the source must increase. This creates a",DCElectricalCircuitAnalysis.pdf
"presents another path for current flow. Due to KCL, this means that the
total current drawn from the source must increase. This creates a
potential problem because if too much current is drawn, there can be an
appreciable voltage drop across the wires feeding the outlets as they
can no longer be assumed to be zero ohms. This lowered voltage may
present a problem for the attached devices, but more importantly, the
large current results in unexpected heating of the wires because of the
power dissipation in them (I2R) and creates a possible fire hazard.
Consequently, we must include some method to limit the current to a
safe maximum value. 
125
Figure 4.19
Lamp connections:
Series (left) and parallel 
(right).
Figure 4.20
Various styles of fuses.",DCElectricalCircuitAnalysis.pdf
"While some electronic systems use active current limiting schemes14, the more 
common approach is a fuse or circuit breaker. These devices are placed between the 
voltage source and the various loads, and see their combined current. If this current 
becomes too great, the device interrupts the current flow by opening the circuit. 
Nothing will operate but nothing catches fire, either.
Fuses are used typically in electronic systems, motors, automotive subsystems and
the like. An array of various types of fuses is shown in Figure 4.20. Whether they're
the cylindrical type found in electronic instrumentation or the blade type common
in automobiles, fuses are one-shot devices. They consist of a fine wire or metal link
that will heat up and melt if it experiences too high of a current. Upon melting,
circuit continuity is lost and current flow stops. This is called a “blown” fuse. The 
fuse will then need to be replaced once the problem has been fixed (i.e., whatever it",DCElectricalCircuitAnalysis.pdf
"fuse will then need to be replaced once the problem has been fixed (i.e., whatever it 
was that caused excessive current to flow in the first place). 
Fuses have a nominal current rating, for example, two amps. This does not mean that
as soon as two amps is reached, the fuse blows. It will take some time for the fuse to 
heat up. The higher the current is above the nominal rating, the faster the fuse will 
blow. For example, it might take this nominal two amp fuse several seconds before it
blows if the current is, say, 2.5 amps, but less than a second if the current is five 
amps. Fuses are also available in fast-blow and slow-blow forms. Fast-blow fuses 
will activate many times faster than standard fuses while slow-blow types will 
require much long to activate (this prevents the fuse from activating accidentally 
with certain kinds of loads that exhibit a high initial startup current but that falls 
back to a more modest level once running smoothly, such as a motor). The",DCElectricalCircuitAnalysis.pdf
"back to a more modest level once running smoothly, such as a motor). The
schematic symbols for the fuse are shown in Figure 4.21. The top two are IEC
standard and the bottom version is the ANSI (North American) standard.
Fuses are generally inexpensive and effective. The obvious downside to a fuse is
that they need to be replaced once they do their job. A circuit breaker solves this
problem. A circuit breaker, or just breaker for short, can be thought of as a
resettable fuse. Breakers do not have a melting link that interrupts current flow.
Instead, a breaker operates like an intelligent switch: if the current is too large,
the switch is thrown open. The operator can then fix the problem and reset the
breaker (i.e., return the switch to the operating position). A selection of breakers
is shown in Figure 4.22. Circuit breakers are available in a variety of sizes. For
residential wiring, 15 and 20 amp sizes are common for general use although",DCElectricalCircuitAnalysis.pdf
"residential wiring, 15 and 20 amp sizes are common for general use although
higher sizes such as 30 and 50 amps are available for circuits feeding high power
appliances such as electric stoves, water heaters and clothes dryers. 
14 For details, see Fiore, J., Semiconductor Devices as well as Operational Amplifiers and 
Linear Integrated Circuits, both OER titles.
126
Figure 4.21
Schematic symbols for fuses: 
IEC (top and middle), ANSI 
(bottom).
Figure 4.22
Residential circuit breakers 
including cutaway view.",DCElectricalCircuitAnalysis.pdf
"4.7 Summary4.7 Summary
In this chapter we have introduced and examined parallel circuits using either one or
more current sources or a single voltage source, along with two or more resistors. 
The hallmark of a parallel configuration is that all components are connected to just 
two nodes. This means that all elements in a parallel configuration see the same 
voltage. If multiple current sources are present, they may be combined into a single 
equivalent current source by adding their values, taking care to watch for the 
directions of source currents; like directions add while opposing directions subtract, 
in much the same manner as series-connected voltage sources. V oltage sources 
generally are not placed in parallel as it would be a practical impossibility to 
establish different voltages across the same two nodes. The major exception to this 
rule is if the sources use the same voltage and the goal is to extend battery life.",DCElectricalCircuitAnalysis.pdf
"rule is if the sources use the same voltage and the goal is to extend battery life. 
Dissimilar parallel voltage sources will most likely cause excessive heat and an 
unstable voltage.
The equivalent resistance of a group of parallel resistors will always be smaller than 
the smallest resistor in the group because each resistor adds another path for current 
flow, thus enhancing conductivity and reducing resistance. In general, the effective 
resistance is found by summing the individual conductances to find the total 
conductance of the group and then taking the reciprocal of this value. Two shortcut 
techniques when just two resistors are involved are the product-sum rule and the 
ratio rule (which is particularly handy when one resistor is the integer multiple of the
other). 
Kirchhoff's current law (KCL) states that the sum of currents entering a node must 
equal the sum of currents leaving that node. Another way of stating this is that the",DCElectricalCircuitAnalysis.pdf
"equal the sum of currents leaving that node. Another way of stating this is that the 
sum of all currents entering and exiting a node must be zero (obeying polarities). 
The current divider rule (CDR) is useful for determining the division of current 
between two parallel resistors. It states that the current through one resistor must 
equal the current entering the pair times the ratio of the other resistance value to the 
summed resistance of the pair. 
To determine individual branch currents in a circuit driven by a voltage source, 
Ohm's law may be used by dividing the source voltage by the individual resistor 
values. These branch currents must sum to the total current delivered by the voltage 
source due to KCL. If the parallel network is driven by current sources, the 
individual branch currents can be found by determining the effective parallel 
resistance and then using Ohm's law to find the system voltage. Once the voltage is",DCElectricalCircuitAnalysis.pdf
"resistance and then using Ohm's law to find the system voltage. Once the voltage is 
known, Ohm's law is used again on each resistor to find the associated branch 
current. Alternately, the resistors can be simplified into successive pairings and then 
CDR can be applied repeatedly. 
Real world ammeters and voltmeters exhibit internal resistances that can load down 
circuit elements and cause errors in measurement. It is best if voltmeters have an 
127",DCElectricalCircuitAnalysis.pdf
"internal resistance that is lat east an order of magnitude (preferably two) larger than 
the resistors they are placed across in order to limit undesired current divider effects.
Similarly, ammeters should have an internal resistance at least an order of magnitude
(preferably two) smaller than the resistors with which they are placed in-line to limit
undesired Ohm's law effects. 
Fuses and circuit breakers are used to limit potentially damaging current in a system.
Both devices will activate by opening the circuit if the current passing through them 
exceeds their rated value for a specified period of time. This results in zero current 
flow. Fuses are one-shot devices that must be replaced after being activated. In 
contrast, circuit breakers can be activated numerous times and are simply reset via a 
switch or button after being tripped.
Review QuestionsReview Questions
1. Define the term parallel connection.",DCElectricalCircuitAnalysis.pdf
"switch or button after being tripped.
Review QuestionsReview Questions
1. Define the term parallel connection.
2. How is the equivalent resistance for a group of parallel connected resistors 
computed? 
3. How is the equivalent value for parallel connected current sources 
computed?
4. Why are voltage sources generally not placed in parallel?
5. Define the product-sum rule and its use.
6. Define Kirchhoff's current law.
7. Define the current divider rule.
8. Explain the basic operation and usage of fuses and circuit breakers. 
4.8 Exercises4.8 Exercises
AnalysisAnalysis
1. Determine the effective resistance of the network shown in Figure 4.23.
128
Figure 4.23",DCElectricalCircuitAnalysis.pdf
"2. Determine the effective resistance of the network shown in Figure 4.24.
3. Determine the effective resistance of the network shown in Figure 4.25.
4. Find the effective source current of the network shown in Figure 4.26.
5. Determine the effective source current of the network shown in Figure 4.27. 
129
Figure 4.24
Figure 4.25
Figure 4.26
Figure 4.27",DCElectricalCircuitAnalysis.pdf
"6. Find the source and resistor currents for the circuit of Figure 4.28.
7. Determine the source and resistor currents for the circuit of Figure 4.29.
8. Find the source and resistor currents for the circuit of Figure 4.30.
9. For the circuit of Figure 4.31, determine the source and resistor currents.
130
Figure 4.28
Figure 4.29
Figure 4.30
Figure 4.31",DCElectricalCircuitAnalysis.pdf
"10. Determine the current through each resistor in the circuit of Figure 4.32. 
Also determine the total power generated by the source.
11. Consider the circuit shown in Figure 4.32. Assume that the 100 kΩ is 
replaced with another resistor ten times as large. Will this have a major 
impact on the current exiting source? Why/why not?
12. Consider the circuit shown in Figure 4.32. Assume that the 1 kΩ is replaced 
with another resistor ten times smaller. Will this have a major impact on the 
current exiting source? Why/why not?
13. Find the current through each resistor in the circuit of Figure 4.33. 
14. Determine the current through each resistor in the circuit of Figure 4.34. 
Also determine the total current exiting by the source.
131
Figure 4.32
Figure 4.33
Figure 4.34",DCElectricalCircuitAnalysis.pdf
"15. Determine the current through each resistor in the circuit of Figure 4.35. 
16. For the circuit shown in Figure 4.36, determine the current through each 
resistor and the source voltage.
17. For the circuit shown in Figure 4.37, determine the current through each 
resistor and the source voltage.
18. For the circuit shown in Figure 4.38, determine the current through each 
resistor and the source voltage.
132
Figure 4.35
Figure 4.36
Figure 4.37
Figure 4.38",DCElectricalCircuitAnalysis.pdf
"19. For the circuit shown in Figure 4.39, determine the current through each 
resistor and the source voltage.
20. For the circuit shown in Figure 4.40, determine the current through each 
resistor and the source voltage.
21. For the circuit shown in Figure 4.41, determine the current through each 
resistor and the source voltage.
22. For the circuit shown in Figure 4.42, determine the current through each 
resistor and the source voltage.
133
Figure 4.39
Figure 4.40
Figure 4.41
Figure 4.42",DCElectricalCircuitAnalysis.pdf
"23. Referring to the circuit of Figure 4.42, determine the resistor currents if the 
right-most 75 kΩ resistor is accidentally opened (i.e., unconnected). How do
these results compare to those of problem 22?
24. Referring to the circuit of Figure 4.42, determine the resistor currents if the 
right-most 75 kΩ resistor is accidentally shorted. How do these results 
compare to those of problems 22 and 23?
25. For the circuit shown in Figure 4.43, determine the current through each 
resistor and the source voltage.
26. Given the circuit of Figure 4.44, find the currents through the two resistors.
 
27. If the 5 mA current source shown in Figure 4.44 is accidentally wired in 
upside down, does the voltage across the 12 k Ω resistor become more 
positive or more negative with respect to ground?
28. Given the circuit of Figure 4.45, find the voltages across the three resistors.
 
29. Find the currents through the three resistors in Figure 4.45.
134
Figure 4.43
Figure 4.44
Figure 4.45",DCElectricalCircuitAnalysis.pdf
"DesignDesign
30. For the network shown in Figure 4.46, determine a for values for R1 given 
that R2 is 12 kΩ and the equivalent combination is 8 kΩ.
31. Add a third parallel resistor to the circuit of Figure 4.30 such that the source 
current is 10 mA.
32. Add a fourth parallel resistor to the circuit of Figure 4.32 such that the 
source current is 20 mA.
33. Consider the circuit shown in Figure 4.36. Determine a new value for the 
current source such that the source voltage equals 10 volts.
34. Consider the circuit shown in Figure 4.38. Determine a new value for the 
current source such that the source voltage equals 20 volts.
35. Given the circuit of Figure 4.47, if the source is 6 volts and R1 is 2 kΩ, what 
must be the value of R2 if the total current exiting the source is 10 mA?
36. For the circuit shown in Figure 4.48, determine values for resistors R2 and 
R3 such that the current through R2 is twice the current through R1 and the",DCElectricalCircuitAnalysis.pdf
"R3 such that the current through R2 is twice the current through R1 and the 
current through R3 is half the current through R1. The source is 6 volts and 
R1 is 2 kΩ. 
135
Figure 4.46
Figure 4.47
Figure 4.48",DCElectricalCircuitAnalysis.pdf
"37. For the circuit shown in Figure 4.49, determine values for resistors R1 and 
R2 such that the current through R2 is twice the current through R1. The 
source is 10 mA and R1 is 6 kΩ. 
ChallengeChallenge
38. For the circuit shown in Figure 4.34, determine a new value for the 11 kΩ 
resistor such that the supply current is 50 mA.
39. For the circuit shown in Figure 4.36, determine a new value for the 2 kΩ 
resistor such that the voltage drop across the 6 kΩ is 15 volts.
40. Consider the circuit shown in Figure 4.43. If the current source was replaced
with a voltage source, what value is needed so that the same currents flow 
through the resistors as in the original circuit?
41. For the network shown in Figure 4.46, determine values for R1 and R2 such 
that R2 is twice the size of R1 and the equivalent combination is 6 kΩ.
42. Given the network of Figure 4.25, is it possible to replace the 60 Ω resistor 
with another value such that the equivalent combination of the three",DCElectricalCircuitAnalysis.pdf
"with another value such that the equivalent combination of the three 
resistors is 25 Ω? If so, what is that value?
43. Given the network of Figure 4.33, is it possible to add a fifth parallel resistor
such that the source current is 1 mA? If so, what is that value?
44. For the circuit shown in Figure 4.48, determine values for the three resistors 
such that the current through R1 is twice the current through R2 and four 
times the current through R3. The source is 12 volts and should produce a 
total of 9 mA of current. 
45. For the circuit shown in Figure 4.49 determine values for the two resistors 
such that the current through R1 is half the current through R2. The source is 
24 mA and should produce a drop of 16 volts across R1. 
136
Figure 4.49",DCElectricalCircuitAnalysis.pdf
"46. Given three current sources with values of 1 mA, 2 mA and 7 mA; how 
would they need to be connected in order to deliver 4 volts across a 1 kΩ 
load resistor?
47. Consider the circuit of Figure 4.50. Assume I is a 4 mA source. Using only 
5% standard resistor values (see Appendix A), pick values for the three 
resistors such that the voltage across R1 is within 10% of 10 volts as long as 
the resistors are within tolerance.
SimulationSimulation
48. Verify the currents found in problem 11 via a DC simulation.
49. Verify the currents found in problem 15 via a DC simulation.
50. Verify the currents and voltages found in problem 17 via a DC simulation.
51. Verify the results found in problem 25 via a DC simulation.
52. Verify the results found in problem 27 via a DC simulation.
53. Perform a DC simulation on the design of problem 44 to verify its 
performance.
54. Perform a DC simulation on the design of problem 45 to verify its 
performance.",DCElectricalCircuitAnalysis.pdf
"performance.
54. Perform a DC simulation on the design of problem 45 to verify its 
performance.
55. Perform a DC simulation on the design of problem 46 to verify its 
performance.
56. Perform a Monte Carlo or worst-case simulation on the design of problem 
47 to verify its performance.
137
Figure 4.50",DCElectricalCircuitAnalysis.pdf
"5 5 Series-Parallel Resistive CircuitsSeries-Parallel Resistive Circuits
5.0 Chapter Learning Objectives5.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Identify series-parallel resistive circuits that include a single effective voltage or current source.
• Compute component and node voltages for single-source series-parallel resistive networks. 
• Compute branch currents for single-source series-parallel resistive networks.
• Break down series-parallel networks into smaller series-only and parallel-only sub-units, and utilize 
KVL, KCL and other previously established techniques to aid in the analysis of these circuits.
5.1 Introduction 5.1 Introduction 
In the preceding two chapters, series circuits and parallel circuits were examined. Each offered unique laws and 
rules for their configuration, such as Kirchhoff's voltage law and the current divider rule. Distinct techniques of",DCElectricalCircuitAnalysis.pdf
"rules for their configuration, such as Kirchhoff's voltage law and the current divider rule. Distinct techniques of 
analysis were offered to determine system current and voltage, and the currents and voltages associated with 
individual components. This almost mechanical approach was successful mainly because there are so few 
variations on the theme. In the case of a series circuit, more resistors might be added in the “daisy chain”, and 
likewise multiple voltage sources, but once you have experience with a few of these types of circuits, there is 
not much concern when simply adding more of the same because a pattern becomes apparent. The same holds 
true with parallel circuits: more resistors might be added as further “rungs on the ladder”, as well as multiple 
current sources similarly arranged, but again, once familiarity has been gained, obvious patterns emerge. This is",DCElectricalCircuitAnalysis.pdf
"current sources similarly arranged, but again, once familiarity has been gained, obvious patterns emerge. This is 
not the case with series-parallel circuits, and thus, there are no simple recipes for our solution paths. Fortunately,
laws and rules such as Ohm's law, KCL and KVL still hold true for these circuits, thus the focus shifts to their 
appropriate application.
To begin with, there are infinite variations of series-parallel circuits. This chapter deals with a subset, namely 
those that are driven by a single current or voltage source (after reducing trivial combinations), and which may 
be simplified using series and parallel resistor combinations. More complex configurations using multiple 
sources await in subsequent chapters. The key to analyzing basic series-parallel circuits is in recognizing 
portions of the circuit, or sub-circuits, that exhibit a series or parallel configuration by themselves, and then",DCElectricalCircuitAnalysis.pdf
"portions of the circuit, or sub-circuits, that exhibit a series or parallel configuration by themselves, and then 
applying the series and parallel analysis rules to those sections. Ohm's law, KVL and KCL may be used in turn to 
“chip away” at the problem until all currents and voltages are found. As individual voltages and currents are 
determined, this makes it easier to apply these rules to determine other values. Given this observation, the 
number of potential solution paths tends to grow exponentially as the number of components increases. As a 
consequence, when faced with the same circuit, six people may solve it six different ways, no particular way 
being more or less correct than any other. The only thing we can say is that some solution paths might be more 
computationally efficient than others, meaning they take less work. Do not let this bother you. The fact that these
138",DCElectricalCircuitAnalysis.pdf
"circuits can be solved in a diverse number of ways is a strength, not a weakness. 
After all, you only need to recognize one of those ways, not all of them, in order to 
be successful. In this chapter's examples, various competing methods will be 
explored, but not every solution path will be spelled out for each one. Flexibility of 
thought and view will prove to be an asset.
5.2 Series-Parallel Connections5.2 Series-Parallel Connections
A basic series-parallel system is shown in Figure 5.1. It should be obvious that
this system is neither a simple series connection nor a simple parallel connection,
but rather some combination of the two. In order for it to be series, the current
through each component would have to be the same. This does not have to be true
here. Consider the connection node directly above block C. This also
connects to items B, D and E. All we really know about that node is that
there are four paths leading to it and that the four associated currents must",DCElectricalCircuitAnalysis.pdf
"there are four paths leading to it and that the four associated currents must
sum to zero, thanks to KCL. There is nothing that requires the current
through, say, block D to be the same as that through block E. Similarly, there
is nothing that stipulates that the voltage across block C must be the same as
that across block F. After all, that is the hallmark of a parallel configuration,
and that would require those two blocks to be connected to the same two nodes, 
which they aren't. 
What we can say about the configuration of Figure 5.1 is that some of the 
components are in series among themselves, and some are in parallel among 
themselves. We might recognize these groupings as sub-circuits. For example, it is 
true that blocks A and B are in series with each other because the current through 
one of them must be the same as the current through the other. The same is true of 
blocks C and F; they are in series with each other. Similarly, we can say that blocks",DCElectricalCircuitAnalysis.pdf
"blocks C and F; they are in series with each other. Similarly, we can say that blocks 
D and E are in parallel with each other because they must see the same voltage as 
they are each connected to the same two nodes. 
5.3 Simplifying Series-Parallel Components5.3 Simplifying Series-Parallel Components
We can go beyond the simple observations presented above. Precisely how we 
approach this will depend on the components represented by the blocks (most often 
resistors, but possibly a voltage source or a current source) and how they are wired 
together. For starters, if all of the items in a sub-circuit are of the same type, they can
just be combined using the techniques discussed in the previous chapters. Resistors 
in series are added, as are voltage sources in series-being wary of polarity. Resistors 
in parallel are combined using conductance or the product-sum rule while parallel 
current sources are added, again being wary of polarity. Things get a little trickier",DCElectricalCircuitAnalysis.pdf
"current sources are added, again being wary of polarity. Things get a little trickier 
when there is a mix, such as two resistors in series with a voltage source, or a 
resistor in parallel with a couple of current sources. Precisely how this sort of 
139
A B
C
D
F
E
Figure 5.1
A series-parallel 
configuration.",DCElectricalCircuitAnalysis.pdf
"situation is handled will become apparent shortly.
To illustrate, let us consider the resistor configuration presented in Figure 5.2.
Imagine that an ohmmeter is connected to the two open terminals. What would it
read? The ohmmeter would apply a small current to the circuit to determine the
effective resistance of the group. The three resistors pictured here are not a simple
series configuration because we can identify a current split at the 40 Ω and 80 Ω 
resistors. Similarly, it is not a parallel network because the 40 Ω and 160 Ω resistors 
are not both tied to the same two nodes. What we can say is that the 40 Ω and 160 Ω 
resistors are in series with each other as they must each see the same current. Thus, 
we can combine the two into a single equivalent resistance of 200 Ω. This equivalent
is in parallel with the 80 Ω resistor and 80 || 200 is approximately 57.14 Ω. That's 
what the ohmmeter should read. Indeed, wherever we see this configuration with",DCElectricalCircuitAnalysis.pdf
"what the ohmmeter should read. Indeed, wherever we see this configuration with 
these precise values, we can replace it with a single resistor of 57.14 Ω. 
The simplification technique can be outlined as follows:
• Identify subgroups of resistors that exhibit series or parallel configurations 
within themselves.
• Replace the subgroups with a single equivalent resistance.
• Repeat the above steps until the circuit is reduced to a single resistance, or 
alternately, a simple series-only or parallel-only configuration.
The looping nature of the technique outlined above may need to be invoked many 
times, depending on the complexity of the network. In simpler networks, a single 
pass may be sufficient. Initially, it may be wise to redraw the circuit for each 
iteration of the loop. Eventually, with practice, this will not be needed except with 
the most complex networks. Let's look at a somewhat more involved example.
Example 5.1",DCElectricalCircuitAnalysis.pdf
"the most complex networks. Let's look at a somewhat more involved example.
Example 5.1
A series-parallel resistor network is shown in Figure 5.3. Determine its
equivalent resistance.
The first step is to recognize those subgroups that are in series or in
parallel with themselves. One obvious candidate is that the two 100 Ω
resistors are in parallel with each other. Two resistors of equal value in
parallel are equivalent to half the resistance, or 50 Ω in this case.
The other candidate is the 40 Ω, 200 Ω pair. These are in series. The
equivalent resistance of the pair is the series combination, or 240 Ω. 
There are no other subsets of resistors that can be reduced (yet). The 
network is redrawn in Figure 5.4 with the equivalent resistances. The 
process is now repeated.
140
Figure 5.2
A simple series-parallel 
resistor network.
Figure 5.3
Network for Example 5.1.",DCElectricalCircuitAnalysis.pdf
"In the newly reduced network, the 50 Ω and 30 Ω resistances are in series,
yielding 80 Ω. 
Without redrawing, the process may be repeated again, this time the 240 Ω
being in parallel with the new 80 Ω equivalent. 240 || 80 is equal to 60 Ω. 
This sequence has reduced the three right-most resistances down to one: the 
new 60 Ω equivalent.
Finally, the 60 Ω equivalent is in series with the 10 Ω resistor, yielding a 
final equivalent resistance of 70 Ω.
Ladders Ladders 
A ladder is a unique series-parallel configuration. It is arranged in a cascade of series
and parallel connections. Ladders are used in a variety of applications, one of which 
will be examined in the next section. 
The name is obvious based on the appearance of the network. An example resistive 
ladder network is shown in Figure 5.5. The main thing to note is that the ladder 
network consists of “sections loading sections” repeatedly. By loading, we mean that",DCElectricalCircuitAnalysis.pdf
"network consists of “sections loading sections” repeatedly. By loading, we mean that
this element (the load) draws off current and power from the prior section. As a 
result, simple two-resistor voltage dividers cannot be used. For example, it is a 
common error to assume that R1 and R2 create a voltage divider which can be used to
determine Vbf if Vaf is known. Similarly, the uninitiated may believe incorrectly that 
R3 and R4, or R5 and R6 create voltage dividers. 
In fact, the only proper voltage divider in this configuration is between R7 and R8. 
The reason for this becomes apparent if we imagine determining the equivalent 
resistance of the network by placing an ohmmeter at terminals a and f. The 
ohmmeter will apply a current to node a which passes through R1. At node b this 
current splits, part going down R2 and part continuing through R3. Clearly then, R1 
and R2 are not in series as they do not see the same current. Further, the current that",DCElectricalCircuitAnalysis.pdf
"and R2 are not in series as they do not see the same current. Further, the current that 
flows through R3 enters node c where it splits again, part down R4 with the remainder
flowing through R5. Thus, R3 and R4 are not in series, either. The same sort of thing 
happens again at node d. This would continue for as many rungs as are on the ladder,
the sole exception being the final two resistors which are in series (R7 and R8 here). 
141
Figure 5.4
Partially simplified network.
Figure 5.5
A resistive ladder network.",DCElectricalCircuitAnalysis.pdf
"How then do we find the equivalent resistance of this network? We begin at the end 
farthest from the open terminals. We have already noted that R7 and R8 are in series. 
This pair, if treated as a single resistance, is in parallel with R6, or R6||(R7+R8). This 
group of three is in series with R5, yielding R5+R6||(R7+R8)15. This new group of four 
is in parallel with R4, yielding R4||(R5+R6||(R7+R8)). This group is in series with R3, 
and that new group is in turn in parallel with R2. Finally, that penultimate grouping is
in series with R1. Therefore, the equivalent resistance of the network is 
R1+R2||(R3+ R4||(R5+R6||(R7+R8))). Surely, it is a tad tedious to compute the value, but 
particularly difficult. A word to the wise: if the goal is determine the voltages at the 
various nodes or the myriad branch currents, it will be helpful to keep track of the 
equivalent resistances of the various groupings. For example, if Vcf is known and the",DCElectricalCircuitAnalysis.pdf
"equivalent resistances of the various groupings. For example, if Vcf is known and the 
goal is to find Vdf, the voltage divider would be R6||(R7+R8) / (R5+R6||(R7+R8)). 
Example 5.2
A resistive ladder network is shown in Figure 5.6. Determine its equivalent
resistance.
Starting at the far end from the open terminals, we note that the 7 kΩ
and 5 kΩ resistors are in series, yielding 12 kΩ. This is in parallel with
the 8 kΩ. 12 k || 8 k is 4.8 kΩ. This is in series with the 1.2 kΩ yielding
6 kΩ. The 6 kΩ equivalent is in parallel with the 1.5 kΩ which yields
1.2 kΩ.
Finally, the 1.2 kΩ equivalent is in series with the 3 kΩ, yielding a final 
equivalent resistance of 4.2 kΩ.
Remember, always start at the far end and work toward the open terminals, i.e., the 
terminals into which you are trying to find the equivalent resistance.  
Example 5.3
A resistive ladder network is shown in Figure 5.7. Determine its equivalent
resistance.
The far end from the open terminals is the extreme left.",DCElectricalCircuitAnalysis.pdf
"A resistive ladder network is shown in Figure 5.7. Determine its equivalent
resistance.
The far end from the open terminals is the extreme left.
First, the 6 kΩ and 4 kΩ resistors are in series, yielding
10 kΩ. This is in parallel with the 8 kΩ, achieving
approximately 4.444 kΩ. In turn, this is in series with the
2 kΩ yielding 6.444 kΩ. That result is in parallel with
the 12 kΩ which yields 4.193 kΩ. Placing this in series 
15 Recall that || has higher precedence than + or –, therefore || is performed first, rather like 
multiplication when mixed with addition and subtraction.
142
Figure 5.6
Network for Example 5.2.
Figure 5.7
Network for Example 5.3.",DCElectricalCircuitAnalysis.pdf
"with the 10 kΩ resistor yields 14.193 kΩ. This is in parallel with the 20 kΩ 
yielding 8.302 kΩ.
Finally, the 8.302 kΩ equivalent is in series with the 5 kΩ, yielding a final 
equivalent resistance of approximately 13.3 kΩ.
5.4 Series-Parallel Analysis 5.4 Series-Parallel Analysis 
Analysis of series-parallel networks involves recognizing those sub-circuits that are 
in series or that are in parallel among themselves, performing simplifications as 
needed, and winding up with a simple series-only or parallel-only equivalent. Then 
the various laws such as Ohm's law, KVL, KCL, VDR and CDR are applied to the 
various simplified networks to determine the parameters of interest. There is no 
single solution technique, each circuit being unique.
Let's begin by considering the circuit of Figure 5.8. To review, this is neither just
a series circuit nor just a parallel circuit. If it was a series circuit then the current",DCElectricalCircuitAnalysis.pdf
"a series circuit nor just a parallel circuit. If it was a series circuit then the current
through all components would have to be same, that is, there would no nodes
where the current could divide. This is clearly not the case as the current
flowing through R1 can divide at node b, with one portion flowing down
through R2 and the remainder through R3. On the other hand, if it was strictly
parallel, then all of the components would have to exhibit the same voltage and
therefore there would be only two connection points in the circuit. This is also
not the case as there are three such points: a, b and ground.
What is true is that resistors R2 and R3 are in parallel. We know this because both 
components are attached to the same two nodes, b and ground, and must exhibit the 
same voltage, Vb. As such, we can find the equivalent resistance of this pair and treat
the result as a single resistance, let's call it Rx. In this newly simplified circuit, Rx is",DCElectricalCircuitAnalysis.pdf
"the result as a single resistance, let's call it Rx. In this newly simplified circuit, Rx is 
in series with R1 and the source, E. We have reduced the original circuit down to a 
simple series circuit and thus the series analysis rules may be applied. 
There are many solution paths at this point. For example, we could find the total 
resistance, Rt, by adding R1 to Rx. Dividing this by E yields the total current flowing 
out of the source, Itotal. This current must flow through R1 so Ohm's law can be used 
to find the voltage drop across R1. This same current must be flowing through Rx, so 
Ohm's law can be used to find the associated voltage (Vb). The currents through R2 
and R3 may then be found using Ohm's law for each resistor (e.g., the current 
through R2 must be Vb/R2). Alternately, these currents may be found by using the 
current divider rule between R2 and R3 (e.g., the current through R2 must be",DCElectricalCircuitAnalysis.pdf
"current divider rule between R2 and R3 (e.g., the current through R2 must be 
Itotal ∙ R1/(R1 + R2); remember, the current divider rule uses the ratio of the opposite 
resistor over the sum). 
143
Figure 5.8
A simple series-parallel 
circuit.",DCElectricalCircuitAnalysis.pdf
"Another solution path would be to apply the voltage divider rule to R1 and Rx in 
order to derive the two voltage drops (or the rule can be applied to find just one of 
the drops and the other voltage may be found by subtracting that from the source, an 
application of KVL). Once the voltages are determined, Ohm's law can be used to 
find the currents. If powers are needed, they are easily determined once the voltages 
and currents are found.
Example 5.4
A series-parallel circuit is shown in Figure 5.9. Determine Vb.
As we are only interested in finding one voltage, the voltage divider
rule is a good candidate. The 1 kΩ is in parallel with the 3 kΩ, yielding
an equivalent resistance of 750 Ω. From here we apply VDR.
V b = E Req
R1+Req
V b = 6V 750 Ω
250Ω+750Ω
V b = 4.5V
To clarify the current directions and voltage polarities, the circuit is
redrawn and appropriately labeled in Figure 5.10. We can crosscheck the
answer in a variety of ways. For example, Figure 5.10 indicates, via KVL,",DCElectricalCircuitAnalysis.pdf
"redrawn and appropriately labeled in Figure 5.10. We can crosscheck the
answer in a variety of ways. For example, Figure 5.10 indicates, via KVL,
that the voltage drop across the 250 Ω resistor plus Vb must equal the
source voltage of 6 volts. Consequently, V250 must be 6 volts minus 4.5
volts, or 1.5 volts. Ohm's law then indicates that the current through the
250 Ω must be 1.5 V/250 Ω, or 6 milliamps. Based on the earlier
resistive equivalence, the total resistance as seen by the source must be
250 Ω in series with 750 Ω, or 1 kΩ. This indicates the source current
must be 6 V/1 kΩ, or 6 milliamps. As the source and the 250 Ω are in
series, they must see the same current. This crosscheck shows that they
do: happy happy joy joy.
We can go further and verify the currents via KCL. For the first parallel 
resistor we find:
I 1k = V b
1 kΩ
I 1k = 4.5V
1 kΩ
I 1k = 4.5 mA
144
Figure 5.9
Circuit for Example 5.4.
Figure 5.10
Circuit for Example 5.4 with 
polarities and directions 
included.",DCElectricalCircuitAnalysis.pdf
"And for the second resistor we see:
I 3k = V b
1 kΩ
I 3k = 4.5V
3 kΩ
I 3k = 1.5mA
Obviously, these sum to the entering current of 6 mA.
Example 5.5
Determine Vb and the source current in the circuit of Figure 5.11.  
In this circuit the 3 kΩ resistor is in parallel with the series combination
of the 2 kΩ and 1 kΩ. This leads to an equivalent resistance of 3 kΩ in
parallel with 3 kΩ, or 1.5 kΩ. From here we can find the source current.
I s = E
RTotal
I s = 30 V
1.5k Ω
I s = 20 mA
This current should split evenly down the two vertical paths as they each 
present 3 kΩ of resistance (10 mA each achieves 30 volts for Va, which is the
source).
The voltage divider rule is a good choice for Vb as we know the applied
voltage.
V b = E Req
R1+Req
V b = 30 V 1k Ω
1kΩ+2k Ω
V b = 10 V
For clarity, the circuit is redrawn and relabeled with current directions and
voltage polarities in Figure 5.12. As a further exercise, try to verify the",DCElectricalCircuitAnalysis.pdf
"voltage polarities in Figure 5.12. As a further exercise, try to verify the
answers above using alternate means, as illustrated in Exercise 5.4.
145
Figure 5.11
Circuit for Example 5.5.
Figure 5.12
Circuit for Example 5.5 with 
polarities and directions 
included.",DCElectricalCircuitAnalysis.pdf
"As the circuit grows, more and more solution paths exist. Consider the circuit of
Figure 5.13. In this case, R3 and R4 are in parallel. This parallel combination is in
series with R2. Finally, this set of three resistors is in parallel with R1 and E,
reducing to a parallel circuit. Consequently, we know that the voltage across 
R1 must be equal to E. Also, the currents through R1 and R2 must add up to the
current exiting the source (KCL). Further, the currents through R3 and R4 must
add up to the current flowing through R2 (KCL). Also, the voltages across R3
and R4 must be the same (they are in parallel) and that this voltage plus the
voltage across R2 must equal E (KVL). One solution path would be to find the
total resistance as seen by the source, R1 || (R2 + (R3 || R4)), and use this to find
the total current flowing out of the source. The current divider rule can then be used 
between R1 and the equivalent resistance (R2 + (R3 || R4)). Ohm's law can be used",DCElectricalCircuitAnalysis.pdf
"between R1 and the equivalent resistance (R2 + (R3 || R4)). Ohm's law can be used 
subsequently to find various voltage drops. Alternately, the voltage divider rule can 
be used between R2 and the parallel equivalent (R3 || R4) as this combination is driven
by E. Knowing the voltages, the currents may be determined. With so many possible
solution paths for large circuits, it is often worthwhile to take a moment to map out a
strategy rather than just “diving in” and hoping it will all work out. 
Example 5.6
The circuit of Figure 5.14 is the same as the resistive network presented in
Example 5.1 but with the addition of a voltage source.
Determine Vb, Vc, Vd, the source current Is, and the current
flowing through the 40 Ω resistor, I40. 
Thanks to Example 5.1, the equivalent resistances of various
portions have already been determined, saving some time. To
reiterate, the pair of 100 Ω resistors are in parallel for 50 Ω.
That's in series with the 30 Ω for 80 Ω. In the middle path, the",DCElectricalCircuitAnalysis.pdf
"reiterate, the pair of 100 Ω resistors are in parallel for 50 Ω.
That's in series with the 30 Ω for 80 Ω. In the middle path, the
200 Ω and 40 Ω are in series, yielding 240 Ω. 240 || 80 = 60 Ω.
This is in series with the 10 Ω, yielding a total resistance loading
the source of 70 Ω.
I s = E
RTotal
I s = 16V
70Ω
I s ≈ 228.6 mA
The three voltages can be quickly found via VDR as we know all of the 
associated sub-circuit resistances.
146
Figure 5.13
A slightly more complex 
series-parallel circuit.
Figure 5.14
Circuit for Example 5.6.",DCElectricalCircuitAnalysis.pdf
"V b = E Rx
Rx +Ry
V b = 16V 60 Ω
60Ω+10Ω
V b ≈ 13.71V
V c = V b
Rx
Rx+Ry
V c = 13.71 V 50Ω
50 Ω+30Ω
V c ≈ 8.57 V
V d = V b
Rx
Rx+Ry
V d = 13.71 V 200 Ω
200 Ω+40 Ω
V d ≈ 11.43 V
Lastly, the current through the 40 Ω resistor can be found by dividing its
voltage by its resistance. Its voltage is Vb − Vd. A quicker method is to note
that because it is in series with the 240 Ω, their currents are the same.
I 40 = V b
RSeries
I 40 = 13.71 V
240 Ω
I 40 ≈ 57.1mA
Once again the circuit is redrawn in Figure 5.15 to illustrate the
current directions and voltage polarities. The usefulness of this will
become even more apparent in the next example.
Example 5.6
Determine Vac and the voltages across each resistor in the circuit of Figure
5.16.  
This circuit is perhaps the most complex offered so far. It may
not be immediately apparent what the voltage polarities are for
the various resistors or the proper current directions. Without
these data, it will be impossible to determine Vac. To complicate",DCElectricalCircuitAnalysis.pdf
"the various resistors or the proper current directions. Without
these data, it will be impossible to determine Vac. To complicate
matters, the voltage of interest is not ground referenced, but
instead refers to two nodes located in arbitrary points of the
circuit. It is worth recognizing that, by definition, Vac = Va − Vc, 
147
Figure 5.15
Circuit for Example 5.6 with 
directions and polarities.
Figure 5.16
Circuit for Example 5.6.",DCElectricalCircuitAnalysis.pdf
"but how are those voltages determined? 
To assist with this minor quandary, the circuit has been redrawn in Figure 
5.17. Note that the current source and associated 1 kΩ resistor have been 
flipped to the side, essentially trading places with the 2 kΩ and 7 kΩ series 
combination. This is entirely acceptable as both subgroups share node b and 
ground, meaning they are in parallel. Of course, the order of parallel 
elements in a situation like this does not matter. This orientation, having the 
source at one end of the circuit, tends to make visualization of current flows 
a little easier.
In any case, the current exits the source and then splits at node b, some
heading through the 10 kΩ with the rest flowing down the 7 kΩ with 2 kΩ
series combo. At node c the current that entered the 10 kΩ splits again,
flowing down into the parallel subgroup of the 12 kΩ and 24 kΩ. Knowing
the current directions, the voltage polarities are determined directly.",DCElectricalCircuitAnalysis.pdf
"flowing down into the parallel subgroup of the 12 kΩ and 24 kΩ. Knowing
the current directions, the voltage polarities are determined directly.
Now that the circuit has been redrawn, several solution
paths might come to mind. For starters, the voltage drop
across the 1 kΩ is trivial as both its current and resistance
are known. From there, a current divider can be used to find
the current flowing down into the series 7 kΩ plus 2 kΩ
combo. Ohm's law could then be used to find their voltages.
Following that, KCL can be used to find the current through
the 10 kΩ (and thus its voltage), and then Ohm's law can be
applied to find Vc by treating the two right-most parallel
resistors as a single unit. 
Another solution path would be to determine the equivalent resistance that 
loads the current source. This allows the source voltage to be determined, 
after which a voltage divider may be applied between the 1 kΩ and the",DCElectricalCircuitAnalysis.pdf
"loads the current source. This allows the source voltage to be determined, 
after which a voltage divider may be applied between the 1 kΩ and the 
remaining five resistors to find Vb. From there, two more voltage dividers 
can be used to find Va and Vc which will allow determination of the 
remaining resistor voltages via KVL. There are other possible paths as well 
but for lack of a better reason, we shall chase the first path that was outlined.
Before getting too deep, it would be wise to determine the equivalent 
resistances of a few subgroups. First, the 12 kΩ || 24 kΩ combo reduces 
down to 8 kΩ. This is in series with the 10 kΩ, leaving 18 kΩ for the three 
right-most resistors. In the middle branch, the 2 kΩ + 7 kΩ series combo is 
equivalent to 9 kΩ. The parallel combination of 9 kΩ || 18 kΩ yields 6 kΩ. 
Finally, this last bit is in series with the 1 kΩ producing 7 kΩ for the 
equivalent resistance of all resistors.  
148
Figure 5.17
Circuit for Example 5.6",DCElectricalCircuitAnalysis.pdf
"equivalent resistance of all resistors.  
148
Figure 5.17
Circuit for Example 5.6 
redrawn and with polarities 
and current directions added.",DCElectricalCircuitAnalysis.pdf
"The source voltage and the voltage across the 1 kΩ resistor are determined 
via Ohm's law: 
V source = I source×Requivalent
V source = 10 mA×7k Ω
V source = 70 V
V 1k = I source×R
V 1k = 10 mA×1k Ω
V 1k = 10 V
As the source produces 70 volts and the drop on the 1 kΩ is 10 volts, then by
KVL, Vb must be 70 volts − 10 volts, or 60 volts (which will be useful for a 
crosscheck in a moment). Using a current divider, the 10 mA source splits 
between the middle and right branches. The equivalent resistances of those 
branches are 9 kΩ and 18 kΩ, as found earlier.
I middle = I source
Rright
Rright+Rmiddle
I middle = 10 mA 18k Ω
18 k Ω+9k Ω
I middle ≈ 6.667 mA
Via KCL, the right branch current must be 10 mA − 6.667 mA, or 
approximately 3.333 mA. As a crosscheck, if 6.667 mA passes through a 
total of 9 kΩ, that produces a voltage drop of 60 volts, exactly as expected 
from the earlier calculation for node b. At this point, the current through",DCElectricalCircuitAnalysis.pdf
"from the earlier calculation for node b. At this point, the current through 
each resistor is known (treating the 12 kΩ || 24 kΩ combo as a unit) and 
therefore Ohm's law may be used to determine their voltage drops. They are 
summarized below.
V7k ≈ 46.67 V 
V2k ≈ 13.33 V 
V10k ≈ 33.33 V 
V12k = V24k ≈ 26.67 V 
Finally, to determine Vac we know that Vac = Va − Vc, and by observation Va is 
the drop across the 2 kΩ while Vc is the drop across the 12 kΩ || 24 kΩ 
combo.
V ac = V a−V c
V ac ≈ 13.33 V−26.67 V
V ac ≈−13.33 V
149",DCElectricalCircuitAnalysis.pdf
"Do not let the negative sign be bothersome. All it indicates is that node a is 
at a lower potential than node c, i.e., that node a is 13.33 volts below node c.
It is also purely coincidental that Vac and Va have the same magnitude. In the 
lab, if a voltmeter is connected such that the red (positive) lead is attached to
node a and the black (negative or reference) lead is attached to node c, the 
meter will show a negative value. If the leads are connected in reverse, thus 
measuring Vca, the voltage will show up as positive (meaning that node c is 
higher in potential than node a, the reverse way of looking at the situation)16.
Any Route WorksAny Route Works
In the example just completed it was noted that, by definition, Vac = Va − Vc, and this 
dictated the method employed to find said voltage. It turns out that this is just one of 
many ways of determining a voltage from one point to another. In general:
To find some voltage VXY, start at point X and proceed along any convenient",DCElectricalCircuitAnalysis.pdf
"To find some voltage VXY, start at point X and proceed along any convenient 
route to Y, subtracting voltage rises (polarities of − to +) and adding voltage 
drops (polarities of + to −) along the way.
In the prior example, an obvious route for Vac would be to go down the 2 kΩ and 
then up the 12 kΩ. That gives a drop of approximately13.33 volts (positive). Going 
up the 12 kΩ produces a rise of approximately 26.67 volts (negative). The result is 
approximately −13.33 volts, as previously calculated. The technique states that any 
route may be used, no matter how inefficient or wacky it might be. To verify this, 
one odd route might be to go up the 7 kΩ, down through the 1 kΩ and source, and 
then up through the 24 kΩ. The summation would be − 46.67 − 10 + 70 − 26.67, or 
approximately −13.33 volts once again. An easy way to remember whether the 
potential is added or subtracted is to just look at the polarity sign where the 
component is entered along your chosen route.",DCElectricalCircuitAnalysis.pdf
"potential is added or subtracted is to just look at the polarity sign where the 
component is entered along your chosen route. 
R-2R LadderR-2R Ladder
As promised earlier in the chapter, it's time to look at a ladder network. While 
ladders can use any resistance values desired, certain ratios turn out to be eminently 
practical. Of particular note is the R-2R ladder, an example of which is shown in 
Figure 5.18, and which is being driven by a current source. In this configuration, 
only two different resistor values are used; some initial value and a second value of 
precisely twice that size. As drawn in Figure 5.18, all horizontal resistors use value 
R (6 kΩ here) while all of the vertical resistors use 2R (12 kΩ here) with the 
exception of the very last, or terminating, resistor which uses R.
16 Or, if you prefer this situation posed as a question: When standing normally, is your head 
above your feet or are your feet below your head? Answer: Yes!
150",DCElectricalCircuitAnalysis.pdf
"It turns out that this arrangement offers something very unique: a halving of 
successive node voltages and branch currents. This is extremely useful in digital to 
analog conversion circuits; the sort of hardware that would turn the digital bits of an 
MP3 or WA V file into listenable analog signals to feed loudspeakers or headphones. 
Here is how it works. Starting at the far right end, it should be obvious that the last 
two 6 kΩ resistors comprise a 50% voltage divider, and thus Ve will half of Vd. The 
combination of the final three resistors works out to 12 kΩ || (6 kΩ + 6 kΩ), or 6 kΩ 
again. Thus, the divider from c to d is also 50%. This process repeats as we move to 
the left, each equivalent working out to R and resulting in a 50% voltage divider. It 
will not matter what value is chosen for R. As long as 2R is used for the other set of 
resistors, the result will always be successive 50% voltage dividers.
Example 5.7",DCElectricalCircuitAnalysis.pdf
"resistors, the result will always be successive 50% voltage dividers.
Example 5.7
Determine all lettered node voltages in the circuit of Figure 5.18.  
Continuing the prior analysis, the entire resistor network presents an 
equivalent resistance of 12 kΩ. The node a voltage is determined by Ohm's 
law:
V a = I ×R
V a = 24 mA×12 kΩ
V a = 288 V
The prior analysis indicated that the subsequent node voltages should be, 
from left to right, 144 volts, 72 volts, 36 volts and 18 volts. A quick check 
follows. The source of 24 mA flows into the left-most 6 kΩ, producing a 
drop of 144 volts by Ohm's law. This can be subtracted from Va (288 V) to 
arrive at Vb, or 144 volts, or half of the prior node voltage, as expected. The 
current through the left-most vertical resistor must be Vb/12 kΩ, or 12 mA. 
That's half of the entering current, as expected. KCL indicates that the 
remaining current, 24 mA − 12 mA, or 12 mA, must be flowing into the",DCElectricalCircuitAnalysis.pdf
"That's half of the entering current, as expected. KCL indicates that the 
remaining current, 24 mA − 12 mA, or 12 mA, must be flowing into the 
second 6 kΩ between nodes b and c. This will produce a voltage drop of 72 
volts across that resistor. In other words, Vc must be 72 volts less than Vb 
(144 volts), or 72 volts. This is half of Vb, as expected. Likewise, the current 
151
Figure 5.18
An R-2R ladder network.",DCElectricalCircuitAnalysis.pdf
"flowing down through the second 12 kΩ must be Vc/12 kΩ, or 6 mA, once 
again half of the preceding current, as expected. The process continues as we
move to the right, each subsequent node seeing half the voltage of the prior 
node and each current getting cut in half as well.  
Computer SimulationComputer Simulation
In order to verify the results of Example 5.7, the R-2R ladder circuit is entered into a
simulator, as shown in Figure 5.19. The numbered nodes 1 through 5 in the 
simulation schematic correspond to the lettered nodes of the original circuit, a 
through e, respectively. This is one example where the use of virtual instruments 
would result in a very cluttered display with five multimeters. Consequently, the 
simulation uses a DC operating point analysis with a single output window, as shown
in Figure 5.20. 
Although the node numbers are not in ascending order, the voltages are a perfect 
match to the results computed previously. Each successive node sees half the voltage",DCElectricalCircuitAnalysis.pdf
"match to the results computed previously. Each successive node sees half the voltage
of the prior node. Although not explicitly listed here, it should be obvious that the 
currents will also be cut in half when progressing from left to right, via a quick 
application of Ohm's law.
152
Figure 5.19
The circuit of Example 5.6 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"Bridges Bridges 
A bridge is a particular series-parallel configuration consisting of two pairs of series 
connected elements placed in parallel. An example of a resistive bridge being driven 
by a voltage source is shown in Figure 5.21. In this circuit, R1 and R2 create one 
series connection while R2 and R3 create the other. The two pairs are then placed in 
parallel. While this circuit is shown with a voltage source, it could also be driven by 
a current source. Other elements might also be added. 
A typical use for a bridge is the measurement of some environmental quantity. As
stated in Chapter 2, there are certain resistive devices that are sensitive to
environmental change. Examples include photoresistors, whose resistance is a
function of light level; and thermistors, whose resistance is a function of
temperature.
To understand the operation, the voltages Vb and Vc are established by the voltage
dividers comprised of R1 and R2; and R3 and R4, respectively. With ordinary",DCElectricalCircuitAnalysis.pdf
"dividers comprised of R1 and R2; and R3 and R4, respectively. With ordinary
resistors, these voltages would be unchanging and therefore Vbc would also be a
fixed value. We shall set it up such that Vb and Vc are the same voltage, and thus Vbc
will be zero. Now assume that we replace R1 with a photoresistor. As the light level
increases, its resistance decreases. This will cause Vb to rise which will cause Vbc to
go positive. If the light level decreases, the resistance of R1 will increase, forcing Vb 
to drop which will cause Vbc to go negative. The greater the change in light, the 
larger Vbc will be, and its sign indicates whether the light level has increased or 
decreased compared to the original set point. By simply placing some manner of 
voltmeter between points b and c, we can create a display that indicates light levels. 
153
Figure 5.20
Simulation results of the circuit 
of Example 5.7. 
Figure 5.21
A resistive bridge network.",DCElectricalCircuitAnalysis.pdf
"If we want to flip the sign, we would place the photoresistor in the position of R2 
instead of R1. Further, it is possible to increase the sensitivity by using sensors at 
opposite corners (e.g., R1 and R4), and comparative or differential measurements are 
possible by using both sides (e.g., R1 with R3).
Example 5.8
Determine Vbc in the circuit of Figure 5.22.  
Perhaps the most straightforward way to do this is to treat each half as a
voltage divider and then subtract Vc from Vb.
V b = E R2
R2+R1
V b = 12V 2k Ω
2k Ω+1k Ω
V b = 8V
V c = E R4
R4+R3
V c = 12 V 4k Ω
4k Ω+3k Ω
V c ≈ 6.857 V
Thus, Vbc is 8 volts minus 6.857 volts, or 1.143 volts.
Computer SimulationComputer Simulation
In order to verify the results of Example 5.8, the bridge network of Figure 5.22 is 
entered into a simulator, as depicted in Figure 5.23. 
154
Figure 5.22
Circuit for Example 5.8.
Figure 5.23
The circuit of Example 5.8 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"A DC operating point analysis is run and the results are shown in Figure 5.24. The 
voltages are precisely as expected.
Final CommentsFinal Comments
Series-parallel simplification techniques will not work for all circuits. Complex 
multi-source circuits and some resistive networks such as delta configurations or 
five-element bridge configurations require other techniques that will be addressed in 
later chapters.
5.5 Summary5.5 Summary
In this chapter we have determined how to identify basic series-parallel networks 
driven by a single effective voltage or current source. A key element here is to 
identify sub-circuits or subgroups of resistors that are comprised of either series-only
or parallel-only configurations within themselves. These groupings can then be 
reduced to equivalent resistances using series and parallel combination techniques 
examined in earlier chapters. This process may be repeated until the entire circuit is",DCElectricalCircuitAnalysis.pdf
"examined in earlier chapters. This process may be repeated until the entire circuit is 
simplified down to either a single series loop or parallel arrangement of resistors 
driven by a voltage or current source. 
Once a circuit has been simplified, series and parallel analysis techniques, and laws 
such as Ohm's law, KVL, KCL, VDR and CDR, may be employed to determine 
various voltages and currents in the simplified equivalent. Given these results, the 
circuit may be expanded back into its original form in stages, reapplying these rules 
and techniques to determine voltages and currents within the sub-circuits. The 
process may be iterated until every current and voltage in the original circuit is 
155
Figure 5.24
Simulation results for the 
bridge circuit of Example 5.8.",DCElectricalCircuitAnalysis.pdf
"discovered, if desired. Once these values are determined, power calculations are 
trivial.
There is an infinite variety of series-parallel configurations and consequently no 
single solution technique will work for all of them. Indeed, the more complex the 
circuit, the more solution paths there are for that circuit. It is therefore wise to plan 
some strategy for a solution instead of randomly “diving in” in order to ease the 
ultimate effort. 
Two series-parallel configurations of note are the ladder network and the bridge 
network. A resistive ladder may be used to derive several different currents or 
voltages from a single source; the R-2R variation being particularly useful in that it 
divides the source down by powers of two. A basic resistive bridge is comprised of 
four elements arranged as a pair of series resistors in parallel with another pair of 
series resistors. Bridges can be used as part of a measurement scheme. For example,",DCElectricalCircuitAnalysis.pdf
"series resistors. Bridges can be used as part of a measurement scheme. For example, 
one or more of the resistors could be environmentally sensitive, such as a 
photoresistor, force sensing resistor or thermistor. As the associated variable 
changes, the resistance of the sensor changes which unbalances the bridge, and the 
resulting voltage indicates both the magnitude and sign of said environmental 
change.
Review QuestionsReview Questions
1. In general, describe the process of reducing a series-parallel resistive 
network down to a single equivalent resistance.
2. Do Ohm's law, KVL and KCL still apply in series-parallel networks? Why?
3. Is there a finite number of variations of series-parallel networks? Why/why 
not?
4. Describe the primary functional characteristic of an R-2R ladder network.
5. What is meant by the terms “load” and “loading”, as in “R1 loads R2”?
6. How might a bridge network be used to sense changes in temperature?",DCElectricalCircuitAnalysis.pdf
"5. What is meant by the terms “load” and “loading”, as in “R1 loads R2”?
6. How might a bridge network be used to sense changes in temperature?
7. Describe a general procedure to find the voltage between two arbitrary 
points in a series-parallel circuit.
156",DCElectricalCircuitAnalysis.pdf
"5.6 Exercises 5.6 Exercises 
AnalysisAnalysis
1. In the circuit of Figure 5.1, which individual resistors are strictly in series 
and which are in parallel?
2. In the circuit of Figure 5.2, which individual resistors are strictly in series 
and which are in parallel?
3. In the circuit of Figure 5.3, which individual resistors are strictly in series 
and which are in parallel?
157
Figure 5.1
Figure 5.2
Figure 5.3",DCElectricalCircuitAnalysis.pdf
"4. In the circuit of Figure 5.4, which individual resistors are strictly in series 
and which are in parallel?
5. In the circuit of Figure 5.5, which individual resistors are strictly in series 
and which are in parallel?
6. In the circuit of Figure 5.6, which individual resistors are strictly in series 
and which are in parallel?
7. In the circuit of Figure 5.7, which individual resistors are strictly in series 
and which are in parallel?
158
Figure 5.4
Figure 5.5
Figure 5.6
Figure 5.7",DCElectricalCircuitAnalysis.pdf
"8. In the circuit of Figure 5.8, which individual resistors are strictly in series 
and which are in parallel?
9. Determine the equivalent resistance of the network shown in Figure 5.1 (i.e.,
as if an ohmmeter is connected to the two terminals). 
10. Determine the equivalent resistance of the network shown in Figure 5.2. 
11. Determine the equivalent resistance of the network shown in Figure 5.3. 
12. Determine the equivalent resistance of the network shown in Figure 5.4. 
13. Determine the equivalent resistance of the network shown in Figure 5.5. 
14. Determine the equivalent resistance of the network shown in Figure 5.6. 
15. Determine the equivalent resistance of the network shown in Figure 5.7. 
16. Determine the equivalent resistance of the network shown in Figure 5.8. 
17. For the circuit of Figure 5.9, find voltages Va, Vb and Vab.
18. For the circuit of Figure 5.9, find the current through each resistor.",DCElectricalCircuitAnalysis.pdf
"17. For the circuit of Figure 5.9, find voltages Va, Vb and Vab.
18. For the circuit of Figure 5.9, find the current through each resistor.
19. For the circuit of Figure 5.10, find the current through each resistor.
20. For the circuit of Figure 5.10, find voltages Va, Vb and Vab.
159
Figure 5.8
Figure 5.9
Figure 5.10",DCElectricalCircuitAnalysis.pdf
"21. For the circuit of Figure 5.11, find voltages Va, Vb and Vab.
22. For the circuit of Figure 5.11, find the current through each resistor.
23. For the circuit of Figure 5.12, find the current through each resistor.
24. For the circuit of Figure 5.12, find voltages Va, Vb and Vab.
25. In the circuit of Figure 5.13, find voltages Va, Vb and Vab.
26. In the circuit of Figure 5.13, find the current through each resistor.
27. In the circuit of Figure 5.14, find the current through each resistor.
160
Figure 5.11
Figure 5.12
Figure 5.13",DCElectricalCircuitAnalysis.pdf
"28. In the circuit of Figure 5.14, find voltages Va, Vb and Vab.
29. For the circuit of Figure 5.15, find voltages Vb, Vc and Vcb.
30. For the circuit of Figure 5.15, find the current through the 470 Ω and 3.6 kΩ
resistors along with the source current.
31. For the circuit of Figure 5.16, find the current through the 2 kΩ and 5.1 kΩ 
resistors along with the source current.
32. For the circuit of Figure 5.16, find voltages Vc, Vb, Vbc and Vab.
161
Figure 5.14
Figure 5.15
Figure 5.16",DCElectricalCircuitAnalysis.pdf
"33. For the circuit of Figure 5.17, find voltages Va and Vb.
34. For the circuit of Figure 5.17, find the current through the 10 kΩ and 4 kΩ 
resistors along with the source current.
35. For the circuit of Figure 5.18, find the current through the 30 kΩ and the 
left-most 36 kΩ resistors along with the source current. 
36. For the circuit of Figure 5.18, find voltages Va, Vb and Vab.
37. In the circuit of Figure 5.18, must the currents through the two 36 kΩ 
resistors be the same? Why/why not?
38. In the circuit of Figure 5.18, must the currents through the two 48 kΩ 
resistors be the same? Further, must these currents be the same as the 
currents through the two 36 kΩ resistors Why/why not?
39. For the circuit of Figure 5.19, find voltages Va, Vb and Vab.
40. For the circuit of Figure 5.19, find the current through the 3 kΩ, 11 kΩ, and 
both 9 kΩ resistors.
162
Figure 5.17
Figure 5.18
Figure 5.19",DCElectricalCircuitAnalysis.pdf
"41. Given the circuit of Figure 5.20, find the current through the 47 kΩ, 5.1 kΩ, 
and 3.9 kΩ resistors.
42. Given the circuit of Figure 5.20, find voltages Va, Vb and Vab.
43. For the circuit of Figure 5.21, find voltages Vb, Vc and Vd.
44. For the circuit of Figure 5.21, find the current through the 6.8 kΩ, 8.5 kΩ, 
and 20 kΩ resistors.
45. The circuit of Figure 5.22 is called an R-2R ladder. Find the current through 
each of the 60 kΩ resistors.
46. Given the circuit of Figure 5.22, find voltages Vb, Vc, Vd and Ve. What is 
unique about this configuration?
163
Figure 5.20
Figure 5.21
Figure 5.22",DCElectricalCircuitAnalysis.pdf
"47. For the circuit of Figure 5.23, find voltages Va, Vb and Vab.
48. For the circuit of Figure 5.23, find the current through each of the resistors.
49. For the circuit of Figure 5.24, find the current through each of the resistors.
50. For the circuit of Figure 5.24, find voltages Va, Vb and Vab.
51. Given the circuit of Figure 5.25, find voltages Va, Vb and Vab.
52. Given the circuit of Figure 5.25, find the current through each of the 
resistors.
53. For the circuit of Figure 5.26, find the current through the 200 Ω, 400 Ω, 
and 100 Ω resistors.
164
Figure 5.23
Figure 5.24
Figure 5.25
Figure 5.26",DCElectricalCircuitAnalysis.pdf
"54. For the circuit of Figure 5.26, find voltages Va, Vb and Vab.
55. Given the circuit of Figure 5.27, find voltages Va, Vb and Vab.
56. Given the circuit of Figure 5.27, find the current through the 5 kΩ, and 6 kΩ
resistors.
57. For the circuit of Figure 5.28, find the current through the 9 kΩ and right-
most 82 kΩ resistors.
58. For the circuit of Figure 5.28, find voltages Va, Vb and Vab.
59. Must the current through the 82 kΩ resistors be identical in the circuit of 
4.28? Why/why not?
60. Must the current through the 25 kΩ resistors be identical in the circuit of 
4.28? Must they be the same as the currents through the 82 kΩ resistors? 
Why/why not?
61. For the circuit of Figure 5.29, find voltages Va, Vb and Vab.
62. For the circuit of Figure 5.29, find the current through the 30 kΩ, 5 kΩ, and 
1 kΩ resistors.
165
Figure 5.27
Figure 5.28
Figure 5.29",DCElectricalCircuitAnalysis.pdf
"63. For the circuit of Figure 5.30, find the current through the 1 kΩ, 2.2 kΩ, and
18 kΩ resistors.
64. For the circuit of Figure 5.30, find voltages Va, Vb and Vc.
65. Given the circuit of Figure 5.31, find voltages Va, Vb and Vab.
66. Given the circuit of Figure 5.31, find the current through the 20 kΩ, 15 kΩ, 
and 3.3 kΩ resistors.
67. The circuit of Figure 5.32 is referred to as an R-2R ladder network. Find the 
current through each of the 4 kΩ resistors.
68. For the circuit of Figure 5.32, find voltages Va, Vb and Vab. What is the unique
characteristic of this configuration?
166
Figure 5.30
Figure 5.31
Figure 5.32",DCElectricalCircuitAnalysis.pdf
"DesignDesign
69. Determine a new value for the 33 Ω resistor in Figure 5.9 such that the 
source current is 10 mA.
70. Determine a new value for the 20 kΩ resistor in Figure 5.12 such that Vb is 4
volts.
71. Determine a new value for the 470 Ω resistor in Figure 5.15 such that Vb is 6
volts.
72. Determine a new value for the 1 kΩ resistor in Figure 5.14 such that the 
source current is 20 mA.
73. Determine a new value for the 12 kΩ resistor in Figure 5.16 such that Vbc is 
0 volts.
74. Given the circuit of Figure 5.24, determine a new value for the 100 k Ω 
resistor such that Va is 75 volts.
75. Given the circuit of Figure 5.23, determine a new value for the current 
source such that Vb is 8 volts.
76. Determine a new value for the 2.2 kΩ resistor in Figure 5.30 such that Vbc is 
0 volts.
77. Given the circuit of Figure 5.32, determine a new value for the current 
source such that Ve is 1 volt.
ChallengeChallenge",DCElectricalCircuitAnalysis.pdf
"0 volts.
77. Given the circuit of Figure 5.32, determine a new value for the current 
source such that Ve is 1 volt.
ChallengeChallenge
78. Utilizing only 1 kΩ resistors, create a series-parallel combination that 
achieves 1.25 kΩ of total resistance.
79. Utilizing only 12 kΩ resistors, create a series-parallel combination that 
achieves 9 kΩ of total resistance.
80. Consider the circuit of Figure 5.22. Alter the values of the two right-most 30
kΩ resistors such that Ve is 1.2 volts. The remaining node voltages should be 
unchanged from the original circuit.
81. Alter the circuit shown in Figure 5.22 such that another “wrung” is added to 
the ladder creating a new right-most node f. If the former terminating 30 kΩ 
resistor (i.e., the vertical one) is reset to 60 kΩ, determine the values of the 
resistors on the new wrung such that Vf is 0.5 volts.
167",DCElectricalCircuitAnalysis.pdf
"82. Given the circuit of Figure 5.25, determine a new value for the 2 k Ω 
resistor such that Vb is 12 volts.
83. Given the circuit of Figure 5.32, determine new values for the resistors such 
that all of the node voltages are twice the value of the original circuit's node 
voltages.
SimulationSimulation
84. Perform a DC simulation on the result of problem 24 to verify the node 
voltages.
85. Perform a DC simulation on the result of problem 25 to verify the node 
voltages.
86. Perform a DC simulation on the result of problem 43 to verify the node 
voltages.
87. Perform a DC simulation on the result of problem 46 to verify the node 
voltages.
88. Perform a DC simulation on the result of problem 51 to verify the node 
voltages.
89. Perform a DC simulation on the result of problem 68 to verify the node 
voltages.
90. Perform a DC simulation on the proposed solution to problem 69 to verify 
the new design.
91. Perform a DC simulation of the alteration presented in Challenge problem",DCElectricalCircuitAnalysis.pdf
"the new design.
91. Perform a DC simulation of the alteration presented in Challenge problem 
80. Does the new design meet all of the requirements?
92. Perform a DC simulation of the alteration requested in Challenge problem 
83. Does the new design meet all of the requirements?
168",DCElectricalCircuitAnalysis.pdf
"NotesNotes
    ♫♫    ♫♫
169",DCElectricalCircuitAnalysis.pdf
"6 6 Analysis Theorems and TechniquesAnalysis Theorems and Techniques
6.0 Chapter Learning Objectives6.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Define whether or not a circuit is a linear bilateral network.
• Find the voltage source equivalent of a current source and vice versa.
• Compute voltages and currents in multi-source networks using superposition.
• Simplify networks using Thévenin's and Norton's theorems.
• Determine conditions for maximum power transfer and compute the maximum power.
• Utilize delta-Y and Y-delta conversions for circuit simplification and analysis.
6.1 Introduction6.1 Introduction
As presented in the previous chapter, series-parallel analysis broadened our ability to analyze circuits beyond 
simple series and parallel networks, and leveraged the laws and techniques learned in earlier chapters. In this 
chapter we shall examine a number of theorems and techniques to help us analyze yet more complex circuits and",DCElectricalCircuitAnalysis.pdf
"chapter we shall examine a number of theorems and techniques to help us analyze yet more complex circuits and
address specialized applications. Specifically, we will address a method of analyzing circuits that contain 
multiple current and/or voltage sources that are connected in a non-trivial fashion (i.e., not just series voltage 
sources or parallel current sources). It is called the superposition theorem and can be applied to any circuit or 
parameter that meets certain requirements, including circuits that have both current sources and voltage sources 
together. 
We shall also examine several ways of simplifying circuits or making functional equivalents for them. These are
more tools to aid in the process of circuit simplification and analysis. The first item in this category is to make 
more accurate models of our idealized constant voltage and current sources. Once this is completed, it will be",DCElectricalCircuitAnalysis.pdf
"more accurate models of our idealized constant voltage and current sources. Once this is completed, it will be 
possible to convert from one type of source to another, such as creating a current source that is the functional 
equivalent of a voltage source. By this we mean that if we swap one for the other in any circuit, the remainder of
the circuit will behave identically, producing the same component voltage drops and branch currents. This 
technique is useful in a number of ways, not the least of which is that it can help reduce more complex circuits 
to ease analysis. 
The concept of equivalence can be extended beyond just a single source to an entire network. For this we will 
explore Thévenin's and Norton's theorems. Using these theorems, entire circuits utilizing dozens of components 
can be modeled as a single source with an associated resistance. When coupled with the maximum power",DCElectricalCircuitAnalysis.pdf
"can be modeled as a single source with an associated resistance. When coupled with the maximum power 
transfer theorem, these theorems will allow us to determine component values that produce the maximum 
amount of load power. 
170",DCElectricalCircuitAnalysis.pdf
"Finally, we will examine how to find equivalent circuits for certain resistor 
arrangements that use three connecting points, in other words, resistor arrangements 
shaped like the letter Y or like a triangle. These are known as delta and Y 
configurations. These configurations are difficult to address with basic series-
parallel simplification techniques but the conversion equivalences will help solve 
that issue.
6.2 Source Conversions6.2 Source Conversions
We begin by considering a more realistic model for constant voltage and constant 
current sources. The ideal voltage source produces a stated potential forever, and 
without regard to what it is connected to. The ideal current source behaves similarly;
it will always produce the same current regardless of its load. These expectations are
not realistic. For example, if we were to place a solid bar of copper across the 
terminals of a voltage source, the bar may exhibit a resistance of mere milliohms,",DCElectricalCircuitAnalysis.pdf
"terminals of a voltage source, the bar may exhibit a resistance of mere milliohms, 
implying an output current of thousands of amps. Similarly, if we were to disconnect
a current source from any load, its effective load would be the resistance of the air 
between its terminals and Ohm's law would dictate an output voltage of perhaps 
thousands or even millions of volts. Real world sources do not behave like this. 
Realistic Models for SourcesRealistic Models for Sources
A highly precise model for any voltage source or current source could be fairly 
complex but for general purpose work, we can greatly improve our ideal sources by 
simply adding a resistor to them. This resistance is referred to as the internal 
resistance of the source. It is important to understand that this is not a resistor, as in 
an internal component that can be altered, but rather a mathematical addition to the 
source that better predicts how it will behave. Further, the value of this effective",DCElectricalCircuitAnalysis.pdf
"source that better predicts how it will behave. Further, the value of this effective 
resistance can be deduced in a laboratory via proper measurements.
The model for a voltage source adds a resistance in series, as shown in Figure 6.1.
This resistance sets an upper limit on the source's current output. Even if the output
terminals are shorted, the maximum current will be dictated via Ohm's law to be
the source voltage divided by the internal resistance, or E/R. Obviously, this
internal resistance will create some voltage divider effect with the attached load. To 
minimize this effect, the internal resistance should be as small as practicably 
possible. Thus,
The ideal internal resistance of a voltage source is zero ohms (a short). 
Given a zero ohm internal resistance, this improved model reverts back to the 
original ideal source. In the case of a lab power supply, the internal resistance 
typically will be a small fraction of an ohm. 
171
Figure 6.1
Practical voltage source 
model.",DCElectricalCircuitAnalysis.pdf
"For a current source, the improved model adds a resistance in parallel, as shown in
Figure 6.2. This resistance sets an upper limit on the source's voltage output. If the
output terminals are opened, the maximum voltage will no longer produce a huge
voltage. Instead, it is dictated by Ohm's law to be the source current times the
internal resistance, or I∙R. This internal resistance will create some current divider 
effect with the attached load. To minimize this effect, the internal resistance should 
be as large as practicably possible. Thus,
The ideal internal resistance of a current source is infinite ohms. 
Given an infinite ohm internal resistance (i.e., an open), this improved model reverts
back to the original ideal source. From here on, whenever we deal with practical 
voltage and current sources, we understand that these sources have some associated 
internal resistance, even if they're not shown explicitly in a schematic diagram.",DCElectricalCircuitAnalysis.pdf
"internal resistance, even if they're not shown explicitly in a schematic diagram. 
Further, whenever we talk about ideal sources, we simply use a short for the internal 
resistance of a voltage source and an open for the internal resistance of a current 
source. 
Source EquivalencesSource Equivalences
For any simple voltage source consisting of an ideal voltage source with a series 
internal resistance, an equivalent current source may be created. Similarly, for any 
simple current source consisting of an ideal current source with a parallel internal 
resistance, an equivalent voltage source may be created. By “equivalent”, we mean 
the load currents must be identical for both circuits given any value of load 
resistance. (Note that if the currents are the same then the voltages must also be the 
same due to Ohm’s Law.) Consider the simple voltage source of Figure 6.1. Its 
equivalent current source would be that shown in Figure 6.2. The opposite is also 
true.",DCElectricalCircuitAnalysis.pdf
"equivalent current source would be that shown in Figure 6.2. The opposite is also 
true.
For reasons that will become apparent under the section on Thévenin's theorem 
following, the internal resistances of these two circuits must be identical if they are 
to behave identically. Knowing that, it is a straightforward process to find the 
required value of the other source. As the current/voltage characteristic is linear for 
these circuits, a plot line can be defined by just two points. The two obvious points 
to use are the opened and shorted load cases. In other words, if it's equivalent for 
these two situations, it must work for any load. The shorted load case produces the 
maximum load current with zero load voltage, while in contrast, the opened load 
case produces the maximum load voltage with zero load current.
Given a voltage source, the maximum current is developed when the load is shorted 
producing a current of E/R. Under that same load condition, all of the current from",DCElectricalCircuitAnalysis.pdf
"producing a current of E/R. Under that same load condition, all of the current from 
the current source version must be flowing through the load. Therefore, the value of 
the equivalent current source must be the maximum current of E/R. It would be 
nonsensical to use a current source that was larger or smaller than this value. 
172
Figure 6.2
Practical current source 
model.",DCElectricalCircuitAnalysis.pdf
"To continue, if we look at the open load case, for the voltage source the load current 
would be zero and the load voltage would be the entire source voltage of E. For the 
current source, the load would also see no current and its voltage would be the 
voltage appearing across its internal resistance which is R times the current E/R, or 
just E. Thus, the two behave identically at the load limits. 
Similarly, if we start with a current source, an open load produces the maximum 
load voltage of I∙R. Therefore, the equivalent voltage source must have a value of 
I∙R. For the current source, a shorted load would produce a load current equal to the 
source value, or I. The voltage source version would produce a current of E/R, where
the value of E was just found to be equal to I∙R, and thus the load current would be 
I∙R/R, or just I. Once again, the two versions behave identically at the load limits. 
To summarize the process of source conversion:",DCElectricalCircuitAnalysis.pdf
"I∙R/R, or just I. Once again, the two versions behave identically at the load limits. 
To summarize the process of source conversion:
• The internal resistance will be the same for both versions.
• If converting from a voltage source to a current source, the value of the 
current source will be the maximum current available from the voltage 
source (shorted load case), and is equal to E/R.
• If converting from a current source to a voltage source, the value of the 
voltage source will be the maximum voltage available from the current 
source (opened load case), and is equal to I∙R.
If a multi-source is being converted (i.e., voltage sources in series or current sources 
in parallel), first combine the sources to arrive at the simplest source and then do the 
conversion. Do not convert the sources first and then combine them as you will wind
up with series-parallel configurations rather than simple sources.",DCElectricalCircuitAnalysis.pdf
"up with series-parallel configurations rather than simple sources. 
Judicious use of source conversions can sometimes simplify multi-source circuits by
allowing converted sources to be combined, resulting in a single source.
Example 6.1
Determine the current source equivalent for the source of Figure 6.3.  
First off, the resistance value is simply copied over, therefore the internal 
resistance of the current source is 50 Ω. The value of the current source is 
computed using Ohm's law based on the maximum current produced under 
the shorted load case. Thus, all 15 volts drops across the 50 Ω resistance.
I s = E
Rs
I s = 15V
50Ω
I s = 0.3A
173
Figure 6.3
Source for Example 6.1.",DCElectricalCircuitAnalysis.pdf
"The equivalent current source is shown in Figure 6.4. We know that this
will work for the shorted an opened cases, but if any doubt is left as to its
universal nature, simply substitute any other resistance value and compare
the outcomes of the two circuits. For no particular reason, let's try using a
load of 200 Ω and see if the load currents are identical.
For the original voltage source we can use Ohm's law:
I L = E
Rs+RL
I L = 15 V
50Ω+200Ω
I L = 60 mA
For the equivalent current source we can use CDR:
I L = I s
Rs
Rs+RL
I L = 0.3 A 50 Ω
50Ω+200Ω
I L = 60 mA
Try this with any other load resistor. The results should always be identical.
Now let's try one going the other way.
Example 6.2
Determine the voltage source equivalent for the source of Figure 6.5.  
Once again, the resistance value is simply copied over, therefore the internal 
resistance of the voltage source is 22 kΩ. The value of the voltage source is",DCElectricalCircuitAnalysis.pdf
"resistance of the voltage source is 22 kΩ. The value of the voltage source is 
based on the maximum voltage produced under the opened load case and is 
computed using Ohm's law. For the open case, all 15 milliamps flows 
through the 22 kΩ resistance.
Es = I ×Rs
Es = 15 mA×22k Ω
Es = 330 V
The equivalent voltage source is shown in Figure 6.6. Again, let's try using 
another value of load resistor to see if the load current and voltage results 
are identical between the two sources. This time, we'll match the load to the 
internal resistance of 22 kΩ. 
174
Figure 6.4
Equivalent current source for 
the source of Figure 6.3.
Figure 6.5
Source for Example 6.2.",DCElectricalCircuitAnalysis.pdf
"For a voltage source with matched resistances we wind up with a simple
50% voltage divider, thus the load voltage will be half the source voltage,
or 165 volts. The original current source sees the current split in half due to
the current divider rule. Thus, the load current should be 7.5 mA. Given
this current, the load voltage will be,
V L = I ×Rs
V L = 7.5mA×22 k Ω
V L = 165 V
Once again the results turn out to be identical.
Now that we have it is possible to substitute one kind of source for another, it is time
to investigate how we might put this to use beyond just giving us a different way of 
driving a circuit. If applied intelligently, source conversions can be used to simplify 
and reduce complex circuits, and thus ease computational difficulty. For example, 
consider the circuit of Figure 6.7. 
This circuit is unlike any circuit we have seen so far. Although we have analyzed
circuits using multiple voltage sources, they have always been in a simple series",DCElectricalCircuitAnalysis.pdf
"circuits using multiple voltage sources, they have always been in a simple series
loop. As such, their voltages may be added together to find a single equivalent
voltage source. Such is not the case here. In this circuit the voltage sources are
part of a series-parallel network and thus their potentials cannot be merely
added. In fact, there are no further simplifications to be performed on this
circuit using basic series-parallel techniques. We appear to be stuck. 
But we're not. This circuit can be simplified into a straight all-parallel network
through the use of source conversions. The E1, R1 combo can be converted into
one current source while the E2, R2 combo can be converted into a second source. 
Once these are converted, the resulting circuit will consist of two current sources 
and three resistors, all in parallel. This is the sort of circuit we solved back in 
Chapter 4.
Example 6.3
Determine Vb for the circuit of Figure 6.8.",DCElectricalCircuitAnalysis.pdf
"Chapter 4.
Example 6.3
Determine Vb for the circuit of Figure 6.8.  
The first step will be to convert the voltage sources into current
sources. We will treat the resistors attached to their positive
terminals as their internal resistances. In other words, we have a 15
volt source with 1 kΩ resistance and a 6 volt source with a 4 kΩ
resistance. 
175
Figure 6.6
Equivalent current source for 
the source of Figure 6.5.
Figure 6.7
A dual source circuit.
Figure 6.8
Circuit for Example 6.3.",DCElectricalCircuitAnalysis.pdf
"For the first source, the current will be:
I s = E
Rs
I s = 15 V
1k Ω
I s = 15mA
And for the second source: 
I s = E
Rs
I s = 6V
4 kΩ
I s = 1.5mA
The equivalent converted circuit is shown in Figure 6.9. Before continuing, 
it is worth noting that connection nodes a and c no longer exist in this 
circuit. More on this in a moment. This new circuit consists of a pair of 
current sources that sum to 16.5 mA and which drive three parallel resistors, 
1 kΩ || 4 kΩ || 5 kΩ, or approximately 689.7 Ω. Ohm's law tells us Vb is:
V b = I Total×REquivalent
V b ≈ 16.5mA×689.7Ω
V b = 11.38 V
We can now take this voltage and apply it back into the original 
(unconverted) circuit. With this potential known, it is relatively easy to 
determine the other currents and voltages using KVL and Ohm's law. 
Looking at the 1 kΩ resistor, the voltage across it must be 15 V − 11.38 V , or
3.62 V . Therefore, the current through it must be 3.62 mA. Similarly, the",DCElectricalCircuitAnalysis.pdf
"3.62 V . Therefore, the current through it must be 3.62 mA. Similarly, the 
voltage across the 4 kΩ resistor must be 11.38 V − 6 V , or 5.38 V , which 
yields a current of 1.345 mA. Both of these currents are flowing left to right.
Then third current, flowing down through the 5 kΩ, is 11.38 V/5 kΩ, or 
2.276 mA. KCL states that the current entering node b must equal the 
currents leaving. The entering current is 3.62 mA. The exiting currents are 
1.345 mA and 2.276 mA, or 3.62 mA when rounded to three places (like the 
entering current). 
As mentioned, nodes a and c disappeared in the converted circuit in Example 6.3. 
This brings up an important point. Equivalent circuits are equivalent in terms that 
the items connected to the equivalent behave the same way as they do with the 
original circuit. That does not mean that the items within the equivalent see the same
current or voltage. We do not expect the voltage across the 1 kΩ or 4 kΩ resistors in",DCElectricalCircuitAnalysis.pdf
"current or voltage. We do not expect the voltage across the 1 kΩ or 4 kΩ resistors in 
the converted version to be the same as those in the original version.
176
Figure 6.9
Equivalent current source 
version of the circuit depicted 
in Figure 6.8.",DCElectricalCircuitAnalysis.pdf
"6.3 Superposition Theorem6.3 Superposition Theorem
As useful as the source conversion technique proved to be in Example 6.3, it will not
work for all circuits. Thus, more general approaches are needed. One of these 
methods is superposition. 
Superposition allows the analysis of multi-source series-parallel circuits. 
Superposition can only be applied to networks that are linear and bilateral. Further, it
cannot be used to find values for non-linear functions, such as power, directly. 
Fortunately, if the circuit contains nothing but resistors, and ordinary voltage sources
and current sources, the circuit will be a linear bilateral network. Also, although 
power is a square law function (i.e., it is proportional to the square of voltage or 
current), it can be computed from the resulting voltage or current values so this 
presents no limits to analysis.
The basic idea is to determine the contribution of each source by itself, and then",DCElectricalCircuitAnalysis.pdf
"presents no limits to analysis.
The basic idea is to determine the contribution of each source by itself, and then
adding the results to get the final answer(s). Let's start with an almost trivial
example to illustrate the idea. Consider the circuit of Figure 6.10. Here we have
two voltage sources driving a pair of resistors. It is a basic series circuit. The way
this was approached in Chapter 3 was to combine the sources and the resistors into a
single source and resistor. The circulating current could then be found using Ohm's 
law. As the two sources are in opposition, the net voltage is 10.5 volts while the total
resistance is 500 Ω. This yields a circulating current of 21 mA. 
Instead of this approach, let's consider the contribution of each source by itself. To 
do that, we will replace all other sources in the circuit with their ideal internal 
resistance. From earlier work we have discovered that the ideal internal resistance of",DCElectricalCircuitAnalysis.pdf
"resistance. From earlier work we have discovered that the ideal internal resistance of
a voltage source is a short, or zero ohms. Thus, we wind up with two circuits: one 
with a 12 volt source that drives 100 Ω, 400 Ω and a short; and a second circuit with 
a 1.5 volt source that drives pretty much the same thing, but with an opposing 
current direction. The first circuit produces a clockwise current of 12 V/ 500 Ω, or 
24 milliamps. Meanwhile, the second circuit produces a counterclockwise current of 
1.5 V / 500 Ω, or 3 mA. As these two currents oppose each other, the resulting 
current is 24 mA − 3 mA, or 21 mA, the same value originally computed. Granted, 
this second method is less efficient than the original method but it illustrates the 
process. More importantly, this process can be applied to a variety of multi-source 
series-parallel circuits. A key element is to remember the directions of the currents",DCElectricalCircuitAnalysis.pdf
"series-parallel circuits. A key element is to remember the directions of the currents 
and polarities of the voltages created in the sub-circuits. Without these data, it will 
be impossible to determine whether the various pieces of the puzzle are added to or 
subtracted from the total.
177
Figure 6.10
A dual source circuit.",DCElectricalCircuitAnalysis.pdf
"To summarize the superposition technique:
• For every voltage or current source in the original circuit, create a new sub-
circuit. The sub-circuits will be identical to the original except that all 
sources other than the one under consideration will be replaced by their 
ideal internal resistance. This means that all remaining voltage sources will 
be shorted and all remaining current sources will be opened.
• Label the current directions and voltage polarities on each of the new sub-
circuits, as generated by the source under consideration.
• Solve each of the sub-circuits for the desired voltages and/or currents using 
standard series-parallel analysis techniques. Make sure to note the voltage 
polarities and current directions for these items.
• Add all of the contributions from each of the sub-circuits to arrive at the 
final values, being sure to account for current directions and voltage 
polarities in the process.",DCElectricalCircuitAnalysis.pdf
"final values, being sure to account for current directions and voltage 
polarities in the process.
In order to get a better handle on the superposition technique, let's reexamine the the 
dual source circuit shown in Figure 6.8 (repeated in Figure 6.11 for ease of 
reference). We will solve this using superposition. As the circuit has two sources, it 
will require two sub-circuits. 
Example 6.4
Determine Vb for the circuit of Figure 6.11 using superposition.  
As this circuit has two voltage sources, two sub-circuits will be
needed. The first sub-circuit will use the 15 volt source.
Consequently, the 6 volt source will be replaced with its ideal
internal resistance, a short. This new circuit is shown in Figure 6.12. 
The current directions are as follows: current exits the source and
travels through the 1 kΩ producing a voltage drop + to − from left to
right. At node b the current splits. The 4 kΩ and 5 kΩ are in parallel so",DCElectricalCircuitAnalysis.pdf
"travels through the 1 kΩ producing a voltage drop + to − from left to
right. At node b the current splits. The 4 kΩ and 5 kΩ are in parallel so
they both see the same voltage. Some of the current flows down through
the 5 kΩ resistor producing a voltage drop + to − from top to bottom.
The remainder of the current flows to the right, through the 4 kΩ,
producing a voltage drop + to − from left to right. Vb can be
determined via voltage divider between the 1 kΩ and the parallel
combo of the 5 kΩ and 4 kΩ. 5 kΩ in parallel with 4 kΩ is
approximately 2.222 kΩ.  
178
Figure 6.11
Circuit for Example 6.4.
Figure 6.12
First sub-circuit for the circuit
of Figure 6.11.",DCElectricalCircuitAnalysis.pdf
"V b = E Rx
Rx +Ry
V b = 15V 2.222 kΩ
2.222 kΩ+1 kΩ
V b ≈ 10.34 V
The second sub-circuit will use the 6 volt source. Therefore, the 15 volt
source will be replaced with its ideal internal resistance, a short. This new
circuit is shown in Figure 6.13. 
The current directions are as follows: current exits the source and
travels through the 4 kΩ producing a voltage drop + to − from right to
left. At node b the current splits. Now the 1 kΩ and 5 kΩ are in parallel
so they both see the same voltage. Some of the current flows down
through the 5 kΩ resistor producing a voltage drop + to − from top to
bottom, just as it did in the first sub-circuit. The remainder of the
current flows to the left, through the 1 kΩ, producing a voltage drop +
to − from right to left. Once again, Vb can be determined via voltage divider, 
this time between the 4 kΩ and the parallel combo of the 5 kΩ and 1 kΩ is 
approximately 833.3 Ω.  
V b = E Rx
Rx +Ry
V b = 6V 833.3Ω
833.3Ω+4 kΩ
V b ≈ 1.034 V",DCElectricalCircuitAnalysis.pdf
"approximately 833.3 Ω.  
V b = E Rx
Rx +Ry
V b = 6V 833.3Ω
833.3Ω+4 kΩ
V b ≈ 1.034 V
Both sub-circuits show polarities of + to − from top to bottom for Vb, 
therefore these two voltages simply add, for a total of approximately 11.37 
volts. This agrees nicely with the result obtained using source conversions. 
The slight deviation in the final digit is no doubt due to the rounding of 
intermediate values, such as resistor combinations. 
It is instructive to pursue this example further and investigate a current. In 
the source conversion version it was discovered that the current through the 
1 kΩ resistor is 3.62 mA and flowing left to right. Using superposition, the 
corresponding current in the first sub-circuit can be found using KVL and 
Ohm's law, (15 V − 10.34 V)/ 1 kΩ, or 4.66 mA flowing left to right. In the 
second sub-circuit, this current can be obtained using Ohm's law alone,  
1.034 V / 1 kΩ, or 1.034 mA flowing right to left. As this current opposes",DCElectricalCircuitAnalysis.pdf
"second sub-circuit, this current can be obtained using Ohm's law alone,  
1.034 V / 1 kΩ, or 1.034 mA flowing right to left. As this current opposes 
the current from the first sub-circuit, the two values are subtracting leaving 
approximately 3.63 mA flowing left to right, agreeing with the original 
answer within rounding error.
179
Figure 6.13
Second sub-circuit for the 
circuit of Figure 6.11.",DCElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
For further verification, the circuit of Example 6.4 is entered into a simulator as 
shown in Figure 6.14.
A DC operating point analysis is performed, as shown in Figure 6.15. 
The voltage at node 2, which is Vb in the original circuit, agrees nicely with the 
values computed previously.
This exploration shows that the superposition technique offers a fairly 
straightforward method of solving a variety of multi-source series-parallel circuits. It
is not without its limitations, though. First, if a circuit has a large number of sources,
superposition can become a bit tedious as it requires as many sub-circuits as there 
are sources, and every one of these sub-circuits requires separate solutions. Second, 
there are some series-parallel configurations that superposition along with series-
parallel analysis techniques will not solve. We will investigate methods of dealing 
with these issues later in this chapter and in the following chapter. 
180",DCElectricalCircuitAnalysis.pdf
"with these issues later in this chapter and in the following chapter. 
180
Figure 6.14
The circuit of Example 6.4 in a 
simulator.
Figure 6.15
Simulation results for the  
circuit of Example 6.4.",DCElectricalCircuitAnalysis.pdf
"6.4 Thévenin's Theorem6.4 Thévenin's Theorem
Thévenin's theorem, named after Léon Charles Thévenin, is a powerful analysis
tool. For DC, it states:
Any single port linear network can be reduced to a simple voltage 
source, Eth, in series with an internal resistance, Rth.
An example is shown in Figure 6.16. The phrase “single port network” means that 
the circuit is cut in such a way that only two connections exist to the remainder of 
the circuit (two connection points makes up one port). The remainder of the circuit 
may be a single component or a large multi-component sub-circuit. As there are 
many ways to cut a typical circuit, there are many possible Thévenin equivalents. 
The important thing is that there are only two connection points between the two 
portions of the circuit and neither point has to be ground.
Consider the circuit shown in Figure 6.17. Suppose we cut the circuit
immediately to the left of R4. That is, we will find the Thévenin equivalent that",DCElectricalCircuitAnalysis.pdf
"immediately to the left of R4. That is, we will find the Thévenin equivalent that
drives R4. The first step is to make the cut, removing the remainder of the circuit
(in this case, just R4). We then determine the open circuit output voltage. This is
the maximum voltage that could appear between the cut points and is called the
Thévenin voltage, Eth. This is shown in Figure 6.18. In a circuit such as this,
basic series-parallel analysis may be used to find Eth. This process turns out to
be quite straightforward in this particular circuit. Due to the open, no current
flows through R3, thus no voltage is developed across R3, and therefore Eth must
equal the voltage developed across R2 which may be obtained via a voltage
divider with resistor R1 and source E.
The second part is finding the Thévenin resistance, Rth. Beginning with the “cut”
circuit, replace all sources with their ideal internal resistance (thus shorting voltage",DCElectricalCircuitAnalysis.pdf
"circuit, replace all sources with their ideal internal resistance (thus shorting voltage
sources and opening current sources). From the perspective of the cut point, look
back into the circuit and simplify by making appropriate series and parallel
combinations to determine its equivalent resistance. This is shown in Figure
6.19. Looking in from where the cut was made (right-to-left here), we find that 
R1 and R2 are in parallel, and this combination is then in series with R3. Thus, 
Rth = R3 + (R1 || R2). A common error is to determine the equivalent resistance
that the source drives. Remember, everything is determined from the vantage
point of the cut or port. 
Thévenin equivalents are not limited to single source circuits. It is possible to find 
the equivalent of a network with several sources. Finding the open circuit output 
voltage will undoubtedly require some extra work, for example the use of 
superposition of source conversions. Finding the Thévenin resistance is unchanged,",DCElectricalCircuitAnalysis.pdf
"superposition of source conversions. Finding the Thévenin resistance is unchanged, 
just remember to replace every source with its ideal internal resistance before 
simplifying the network.
181
Figure 6.16
Thévenin equivalent circuit.
Figure 6.17
Example circuit.
Figure 6.18
Finding Eth..",DCElectricalCircuitAnalysis.pdf
"As noted earlier, the original circuit could be cut in a number of different ways. We 
might, for example, want to determine the Thévenin equivalent that drives R2 in the 
circuit of Figure 6.17. This cut appears in Figure 6.20. Clearly, this will result in 
different values for both Eth and Rth. For example, Rth is now R1 || (R3 + R4).
Finding Thévenin Equivalents in LabFinding Thévenin Equivalents in Lab
The procedure for experimentally determining an equivalent in the laboratory 
mimics the analytical approach. The first step is to figuratively cut the circuit and 
isolate the section that is to be converted. At this point we have the open circuit 
version of the circuit and all we need do is connect a multimeter to the open port to 
determine Eth. 
There are two methods for determining Rth. The first method works well for simple
circuits that use only resistors and current and/or voltage sources. The second",DCElectricalCircuitAnalysis.pdf
"circuits that use only resistors and current and/or voltage sources. The second
method is more generally applicable and will work for circuits with active devices
such as transistors. To use the first method, the sources are physically removed
from the circuit and replaced with their ideal internal resistance. Thus voltage
sources are replaced with a shorting wire and current sources are left as opens.
Do not simply place shorting wires across the terminals of voltage sources as
doing so will cause an overload and potentially damage the sources. Once this is
completed, a multimeter is placed at the open port and set to read resistance. The
indicated value is Rth.
The second method uses a variable resistance, namely either a rheostat or a decade 
box, and exploits the voltage divider rule. In this version, the sources are left active 
(powered up) and are not replaced with opens or shorts. Once Eth is measured, the",DCElectricalCircuitAnalysis.pdf
"(powered up) and are not replaced with opens or shorts. Once Eth is measured, the 
variable resistor is placed across the port connections. This resistance is adjusted 
until the port voltage drops to exactly half of Eth. In the equivalent circuit, there are 
only two resistances of concern, Rth and this variable load resistance, and they are in 
series. Consequently, if the load voltage is now half of the open circuit voltage (Eth), 
then the other half of the voltage must be dropping across the equivalent internal 
resistance (Rth). For this to be true in a series circuit, the two resistances must have 
the same value. Thus, we simply remove the rheostat from the circuit and use a 
multimeter to determine its precise value. If a decade box is used instead, the value 
is determined directly by reading the knob settings.
182
Figure 6.19
Finding Rth..
Figure 6.20
An alternate cut or port 
location.",DCElectricalCircuitAnalysis.pdf
"Whether determined analytically or empirically, the Thévenin equivalent circuit can 
replace the original single port network regardless of what the original was 
connected to. The same voltages and currents will be seen in this other portion, and 
it won't matter if the other portion is comprised of a single resistor, multiple 
resistors, multiple resistors and multiple sources, or even multiple resistors and 
sources wired to a selection of piquant cheeses. The Thévenin equivalent is a true 
functional equivalent and can be used on any linear bilateral network.
Example 6.5
Determine the Thévenin equivalent of the circuit driving the 1 kΩ in
Figure 6.21. Verify that the equivalent produces the same voltage
across this resistor as the original circuit.
First, we'll redraw the circuit showing the portion to be 
Théveninized, as shown in Figure 6.22. The open circuit output
voltage will be the voltage across the 800 Ω resistor as there will be",DCElectricalCircuitAnalysis.pdf
"Théveninized, as shown in Figure 6.22. The open circuit output
voltage will be the voltage across the 800 Ω resistor as there will be
no voltage drop across the 200 Ω resistor (as no current flows
through it into the open). This is found via a simple voltage divider.
Eth = E Rx
Rx +R y
Eth = 10 V 800 Ω
800Ω+100 Ω
Eth ≈ 8.889 V
The equivalent resistance is found by shorting the voltage source and then
simplifying the circuit. The result is 200 Ω in series with the parallel
combination of the 100 Ω and 800 Ω. 200 + 100 || 800 ≈ 288.89 Ω.
To determine Vc, we can use a voltage divider between the 1 kΩ and
the 288.89 Ω along with the equivalent source voltage of 8.889 volts.
V c = Eth
RL
RL +Rth
V c = 8.889 V 1k Ω
1 kΩ+288.89Ω
V c ≈ 6.897 V
Now let's determine Vc in the original circuit using series-parallel analysis 
techniques. Perhaps the quickest approach is a pair of voltage dividers. Vb is 
found via a divider between the parallel combination of the 1 kΩ + 200 Ω",DCElectricalCircuitAnalysis.pdf
"found via a divider between the parallel combination of the 1 kΩ + 200 Ω 
and the 800 Ω, against the 100 Ω. Vc is is then found using that voltage with 
a divider between the 1 kΩ and 200 Ω. Using this approach, Vb is 
approximately 8.276 volts and Vc is approximately 6.897 volts, providing an 
excellent match. 
183
Figure 6.21
Circuit for Example 6.5.
Figure 6.22
Finding Eth.",DCElectricalCircuitAnalysis.pdf
"It might seem that the Thévenin method is the “long way home” in this 
example, and it is, but it has the advantage of being more efficient if several 
different loads are being considered. For example, suppose we decided to 
determine the output voltage not just for the 1 kΩ, but for a group of a half 
dozen different resistors. The double voltage divider would have to be 
determined for each load resistor using the straight series-parallel method 
but only a single divider needs to be computed for each load when using the 
Thévenin method. The Thévenin method is also of great use when 
determining maximum power transfer, as we shall see a later in this chapter.
Computer SimulationComputer Simulation
To verify the results of Example 6.5, the original circuit is recreated in a simulator 
along its Thévenin equivalent, as shown in Figure 6.23.
A DC operating point analysis is performed and the results are shown in Figure 6.24.",DCElectricalCircuitAnalysis.pdf
"along its Thévenin equivalent, as shown in Figure 6.23.
A DC operating point analysis is performed and the results are shown in Figure 6.24.
Note that the voltages across the identical 1 kΩ loads (nodes 3 and 5) are virtually 
the same, indicating functional equivalence between the two. The slight deviation is 
due to the rounding of the Thévenin voltage and resistance.
184
Figure 6.23
The original circuit of Example 
6.5 in a simulator along with 
the equivalent.",DCElectricalCircuitAnalysis.pdf
"As a further check, the simulation is run again, but this time using an alternate load 
resistor value. The value chosen is the Thévenin resistance value. By matching the 
resistances, this should produce a 50/50 voltage divider and the load voltage should 
equal half of the Thévenin source voltage, or approximately 4.444 volts. That is 
precisely what we find, as seen in Figure 6.25. 
As mentioned previously, Thévenin's theorem can be applied to much more complex
multi-source circuits, and the item being driven need not be just a single resistor. 
This is illustrated in the next example.
185
Figure 6.24
Results of the simulation 
showing equivalence.
Figure 6.25
Results of the simulation for 
matched resistance.",DCElectricalCircuitAnalysis.pdf
"Example 6.6
Determine the Thévenin equivalent of the circuit driving the
resistor/voltage source combo in Figure 6.26. Verify that the
equivalent produces the same voltage across this resistor as
the original circuit.
Determining Rth is not particularly difficult here. After
shorting the 10 volt source and open the current source, it
can be seen that the 1 kΩ and 3 kΩ are in parallel, yielding
750 Ω. This is in series with the 1250 Ω for a total of 2 kΩ,
which is in turn in parallel with the 6 kΩ. The final result is
1.5 kΩ.
Finding Eth is a little more involved. One possibility is to use a source 
conversion on the 10 volt source as that will leave us with two parallel 
current sources that may be combined directly. The converted source will be 
10 V/ 1kΩ, or 10 mA, in parallel with 1 kΩ. This results in a total of 25 mA 
feeding upwards with 1 kΩ || 3 kΩ, or 750 Ω. We can use the current divider 
rule to determine the current flowing through the 1250 Ω plus 6 kΩ branch,",DCElectricalCircuitAnalysis.pdf
"rule to determine the current flowing through the 1250 Ω plus 6 kΩ branch, 
and then use Ohm's law to find the open circuit voltage (i.e., the voltage 
across the 6 kΩ).
I 6k = I S
Rx
Rx +Ry
I 6k = 25 mA 750Ω
750Ω+7250Ω
I 6k = 2.34375 mA
Eth = I 6k×R6k
Eth = 2.34375 mA×6 k Ω
Eth = 14.0625V
Another approach to determine this value would be to use superposition. 
This is left as an additional exercise. 
Turning our attention to the voltage produced across the 6 volt with 500 Ω 
sub-circuit, we have a simple series loop with Eth opposing this 6 volt 
source, leaving 8.0625 volts to produce a clockwise current through Rth in 
series with the 500 Ω (a total of 2 kΩ). That current will be 8.0625 V / 2 kΩ,
or 4.03125 mA. This will produce a drop across the 500 Ω of 2.015625 volts
with a polarity of + to − from top to bottom. This will add to the 6 volt 
source resulting in a final potential of 8.015625 volts. The results are",DCElectricalCircuitAnalysis.pdf
"with a polarity of + to − from top to bottom. This will add to the 6 volt 
source resulting in a final potential of 8.015625 volts. The results are 
verified via the simulation technique used previously. The node voltages for 
both the original and equivalent circuits are shown in Figure 6.27.
186
Figure 6.26
Circuit for Example 6.6.",DCElectricalCircuitAnalysis.pdf
"6.5 Norton's Theorem6.5 Norton's Theorem
Norton's theorem is credited to Edward Lawry Norton. In a nutshell, Norton's 
theorem is the current source version of Thévenin's theorem. That is, a single port 
DC network can be reduced to a single current source, IN, with parallel internal 
resistance, RN. Indeed, once you understand one of them, the other is but a minor 
extension, so we will not need to spend a great deal of time here once the basics are 
outlined sufficiently. 
The process of determining the Norton equivalent current and resistance is very 
similar to that employed for the Thévenin equivalent. First, the Norton resistance is 
found the same way as is the Thévenin resistance: replace all sources with their ideal
internal resistance and then perform appropriate series and parallel combinations to 
reduce this to a single resistance value. Consequently, the Norton and Thévenin 
resistance values are identical, RN = Rth. Second, instead of finding the open circuit",DCElectricalCircuitAnalysis.pdf
"resistance values are identical, RN = Rth. Second, instead of finding the open circuit 
output voltage, we find the short circuit output current. This is the Norton current. 
Instead of thinking in terms of a voltmeter at the opened load, we think in terms of 
connecting a shorting ammeter across the load. Either way, we're looking for the 
maximum value that can be obtained.
Perhaps the most useful thing to remember here is that if we can create a Thévenin 
equivalent for a network then it must be possible to create a Norton equivalent. 
Indeed, once a Thévenin equivalent is found, a source conversion can be performed 
on it to yield the Norton equivalent! The opposite, of course, is also true.
187
Figure 6.27
Results of the simulation using 
the original and equivalent 
circuits.",DCElectricalCircuitAnalysis.pdf
"6.6 Maximum Power Transfer Theorem6.6 Maximum Power Transfer Theorem
Given a simple voltage source with internal resistance, a useful question to ask is
“What value of load resistance will yield the maximum amount of power in the
load?” While it is not true that maximizing load power is a goal of all circuit
designs, it is a goal of a portion of them and thus worth a closer look. Consider the
basic circuit depicted in Figure 6.28 with source E, source internal resistance Ri and 
load resistance R.
We would like to describe the load power in terms of the load resistance. To make 
the job easier, we may normalize the voltage source E to 1 volt and the source 
resistance Ri to 1 Ohm. By doing this, R also becomes a normalized value, that is, it 
no longer represents a simple resistance value but rather represents a ratio in 
comparison to Ri. In this way the analysis will work for any set of source values.",DCElectricalCircuitAnalysis.pdf
"comparison to Ri. In this way the analysis will work for any set of source values. 
Note that the value of E will equally scale the power in both Ri and R, so a precise 
value is not needed, and thus, we may as well chose 1 volt for convenience. 
The power in the load can be determined by using I2R where I = E / (Ri+R). Using 
our normalized values I = 1 / (1+R) and thus the load power is:
P =(
1
1+R )
2
R    or after expanding,  P = R
R2
+2 R+1
We now have an equation that describes the load power in terms of the load 
resistance. Before we go any further, take a look at what this equation tells you, in 
general. It is obvious that maximum power will not occur at the extremes. If R = 0 or
R = ∞ (i.e., shorted or opened load) the load power is zero. While shorting the load 
will yield the maximum load current, it will also yield zero load voltage and thus no 
power. Similarly, opening the load will yield maximum load voltage but it will also",DCElectricalCircuitAnalysis.pdf
"power. Similarly, opening the load will yield maximum load voltage but it will also 
yield zero load current, and again, no load power. To find the precise value that 
produces the maximum load power, the proof can be divided into two major 
portions. The first involves graphing the function and the second requires differential
calculus to solve for a precise value. We shall proceed with the graphing portion 
which will lead us to the answer. The more rigorous proof of the second method is 
detailed in Appendix C. 
Graphing the Power Function Graphing the Power Function P = R / P = R / ((RR22++22R+1R+1))
 This may be done in parts, looking at the contribution of each term, and then 
combining to form the final result. First, consider the denominator R2 + 2R + 1. It 
consists of three segments. The most simple is the horizontal line at +1. The 2R term
creates a straight line with slope 2 starting at the origin. The R2 term creates a simple",DCElectricalCircuitAnalysis.pdf
"creates a straight line with slope 2 starting at the origin. The R2 term creates a simple
curve with increasing slope that starts at the origin, crosses the horizontal +1 line at 
R = 1 and also crosses the 2R line at R = 2. These items are drawn individually and 
188
Figure 6.28
Defining maximum power 
transfer.",DCElectricalCircuitAnalysis.pdf
"then added together as illustrated in Figure 6.29.
Of course, we must remember that we really want 1 / ( R2 + 2R + 1) and so we plot 
the reciprocal of the combo as shown in Figure 6.30 (red curve). We also include the
numerator term, R. This is shown as a straight line (blue) with a slope of 1. Finally, 
these two curves are multiplied together to produce the load power equation (black). 
189
Normalized Components and Total
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
 
Normalized Load Resistance
0 0.5 1 1.5 2
 
2R
R2+2R+1
R2
1
Figure 6.29
The three terms of the power 
equation plotted individually 
and then summed.
Figure 6.30
The power equation and 
components plotted.
Normalized Components
0.0
0.5
1.0
1.5
2.0
2.5
Normalized Load Power
0
0.05
0.1
0.15
0.2
0.25
Normalized Load Resistance
0 0.5 1 1.5 2 2.5
 
R
1/(R2+2R+1)
R/(R2+2R+1)
(right axis)",DCElectricalCircuitAnalysis.pdf
"A close examination of the power curve will show that the peak occurs at R = 1. 
This is easier to see if we plot the completed power curve using a logarithmic 
horizontal axis and also scale the vertical axis to 100%, as shown in Figure 6.31. The
peak is more apparent and the curve is symmetrical in shape rather than lopsided. 
This reinforces the idea that the ratio of the resistances is what matters.
Finally, we can state:
Maximum load power will be achieved when the load resistance is equal to 
the internal resistance of the driving source. 
No other value of load resistance will produce a higher load power. For the circuit of
Figure 6.28, this means that R must equal Ri. As an exercise, we can try substituting 
a few values around the peak to verify this. For example, given E = 1 V and Ri = 1 Ω,
we calculate P for R = 0.5 Ω, 1 Ω, and 2 Ω:
P0.5 = 0.5/(0.52+2∙0.5+1) = 2/9
P1 = 1/(12+2∙1+1) = 2/8 = ¼ 
P2 = 2/(22+2∙2+1) = 2/9",DCElectricalCircuitAnalysis.pdf
"we calculate P for R = 0.5 Ω, 1 Ω, and 2 Ω:
P0.5 = 0.5/(0.52+2∙0.5+1) = 2/9
P1 = 1/(12+2∙1+1) = 2/8 = ¼ 
P2 = 2/(22+2∙2+1) = 2/9
You can also try this with extremely small variations such as R = 0.9999 Ω along 
with R = 1.0001 Ω and you will not achieve a value equal to or greater than ¼ Watt. 
While matching the resistance produces the maximum load power, it does not 
produce maximum load current or maximum load voltage. In fact, this condition 
produces a load voltage and a load current that are half of their maximums. Their 
190
Load Power as a Percentage of Maximum
0
20
40
60
80
100
  
Normalized Load Resistance
0.01 0.1 1 10 100
 
Figure 6.31
The load power curve with 
logarithmic axis showing 
symmetry.",DCElectricalCircuitAnalysis.pdf
"product, however, is at the maximum. Further, efficiency at maximum load power is 
only 50% (i.e., only half of all generated power goes to the load with the other half 
being wasted internally). Values of R greater than Ri will achieve higher efficiency 
but at reduced load power. Sometimes we favor efficiency over maximal load power.
As any linear two point network can be reduced to something like Figure 6.28 by 
using Thévenin's theorem, combining the two theorems allows us to determine 
maximum power conditions for any resistor in a complex circuit.
6.7 Delta-Y Conversions6.7 Delta-Y Conversions
Certain component configurations, such as bridged networks, cannot be reduced to a
single resistance using basic series-parallel conversion techniques. One method for 
simplification involves converting sections into more convenient forms. The 
configurations in question are three-point networks containing three resistors. Due to",DCElectricalCircuitAnalysis.pdf
"configurations in question are three-point networks containing three resistors. Due to
the manner in which they drawn, they are referred to as delta networks and Y 
networks17. These configurations are shown in Figures 6.32. In this particular case 
the delta version is drawn upside down so that its terminal designations match those 
of the Y configuration.
Alternately, if they are slightly redrawn they are known as pi (also called “π”) 
networks and T (also called “tee”) networks. These configurations are shown in 
Figure 6.33.
17 In some sources the capital Greek letter delta (Δ) is used instead of spelling out “delta” 
and the letter Y is spelled out as “wye”. Thus, you may come across discussion of “Δ-Y ”,
“Δ-wye” or “delta-wye” networks. It's all the same stuff. 
191
Figure 6.32
Delta and Y (Δ-Y) networks.
Figure 6.33
Alternate form: Pi and T(π-T) 
networks.",DCElectricalCircuitAnalysis.pdf
"It is possible to convert back and forth between delta and Y networks. That is, for 
every delta network, there exists a Y network such that the resistances seen between 
the X, Y and Z terminals are identical, and vice versa. Consequently, one 
configuration can replace another in order to simplify a larger circuit. 
Δ-Y ConversionΔ-Y Conversion
A true equivalent circuit would present the same resistance between any two 
terminals as the original circuit. Consider the unloaded case for the circuits of Figure
6.32 (i.e., just these networks with nothing else connected to them). The equivalent 
resistances seen between each pair of terminals for the delta and Y respectively are:
RXY = Ra || (Rb+Rc) = Rd + Re (6.1)
RXZ = Rb || (Ra+Rc) = Rd + Rf (6.2)
RZY = Rc || (Rb+Ra) = Re + Rf (6.3)
Assuming we have the delta and are looking for the Y equivalent, note that we have 
three equations with three unknowns (Rd, Re and Rf). Thus, they can be solved using",DCElectricalCircuitAnalysis.pdf
"three equations with three unknowns (Rd, Re and Rf). Thus, they can be solved using 
a term elimination process. If we subtract Equation 6.3 from Equation 6.1 we 
eliminate the second resistance (Re) and arrive at a difference between the first and 
third unknown resistances (Rd − Rf). This quantity can then be added to Equation 6.2 
to eliminate the third resistance (Rf), leaving just the first unknown resistance (Rd). 
(Rd + Re) − (Re + Rf) = (Rd − Rf) = Ra || (Rb+Rc) − Rb || (Ra+Rc)
(Rd + Rf) + (Rd − Rf) = 2Rd = 2( Rb || (Ra+Rc) + Ra || (Rb+Rc) − Rc || (Ra+Rb) )
Therefore,
Rd = Rb || (Ra+Rc) + Ra || (Rb+Rc) − Rc || (Ra+Rb) 
which, after simplifying, is:
Rd = Ra Rb
Ra+Rb+Rc
(6.4)
Similarly, we can show that
Re = Ra Rc
Ra +Rb +Rc
(6.5)
R f = Rb Rc
Ra +Rb +Rc
(6.6)
Note that if three identical resistors are used, the values of the Y equivalent will all 
be one-third of that value.
192",DCElectricalCircuitAnalysis.pdf
"Y-Δ ConversionY-Δ Conversion
For the reverse process of converting Y to delta, start by noting the similarities of the
expressions for Rd, Re and Rf. If two of these expressions are divided, a single 
equation for Ra, Rb or Rc will result. For example, using Equations 6.4 and 6.5:
Rd
Re
=
Ra Rb
Ra+Rb+Rc
Ra Rc
Ra+Rb+Rc
 
Rd
Re
= Ra Rb
Ra Rc
Rd
Re
= Rb
Rc
Therefore,
Rb
Rc
= Rd
Re
Rb = Rc Rd
Re
This process can be repeated for Equations 6.4 and 6.6 to obtain an expression for 
Ra. The two expressions for Ra and Rb can then be substituted into Equation 6.4 to 
obtain an expression for Rc that utilizes only Rd, Re and Rf. A similar process is 
followed for Ra and Rb resulting in:
Ra = Rd Re +Re R f +Rd R f
R f
(6.7)
Rb = Rd Re +Re R f +Rd R f
Re
(6.8)
Rc = Rd Re +Re R f +Rd R f
Rd
(6.9)
If three identical resistors are used, the values of the delta equivalent will all be three
times that value, the inverse of the situation when converting from delta to Y .",DCElectricalCircuitAnalysis.pdf
"times that value, the inverse of the situation when converting from delta to Y .
Thus, equations 6.4, 6.5 and 6.6 can be used to convert a delta network into a Y 
network, and equations 6.7, 6.8 and 6.9 can be used to convert a Y network into a 
delta network. An example of how this will tame an otherwise incorrigible series-
parallel network is next.
193",DCElectricalCircuitAnalysis.pdf
"Example 6.7
Determine the equivalent of the bridge circuit shown in Figure 6.34. 
This circuit uses a five resistor bridge which cannot be further
simplified using basic series-parallel combinations. As a result, the
techniques presented in Chapter 5 will not be sufficient to obtain a
solution. A delta-Y conversion is proposed to simplify the circuit
instead. We begin by defining a delta configuration using the top three
resistors (shown in blue). When replaced with a Y configuration, we
will have resistors in series with the 2 kΩ and 4 kΩ, and a third resistor
in series with the source. This new network can solved using basic
series-parallel techniques.
Note that a Y-delta is not as good of a choice. For example, using the 1 kΩ, 
2 kΩ and the 5kΩ as a sideways Y , the delta version leaves resistors in 
parallel with the 3 kΩ and the 4 kΩ, and another in parallel with the source. 
This is straightforward to solve, but unfortunately, node b disappears in the 
process.",DCElectricalCircuitAnalysis.pdf
"This is straightforward to solve, but unfortunately, node b disappears in the 
process.
We shall use Equations 6.4, 6.5 and 6.6 to perform the delta-Y conversion. 
The orientation in Figure 6.34 is skewed compared to the reference of Figure
6.32. To make them match, imagine rotating the networks of Figure 6.32 
slightly clockwise so that point X is at the very top (matching node a), 
leaving Y and Z at the bottom (matching c and b, respectively). Given this 
new orientation, Ra is the 3 kΩ, Rb is the 1 kΩ and Rc is the 5 kΩ.
Rd = Ra Rb
Ra+Rb+Rc
Rd = 3k Ω1 kΩ
3 kΩ+1 kΩ+5kΩ
Rd ≈ 333.333Ω
Re = Ra Rc
Ra +Rb +Rc
Re = 3 kΩ5k Ω
3k Ω+1k Ω+5k Ω
Re ≈ 1.666667 k Ω
R f = Rb Rc
Ra +Rb +Rc
R f = 1 kΩ5k Ω
3 kΩ+1 kΩ+5 kΩ
R f ≈ 555.556Ω
194
Figure 6.35
Equivalent for the circuit of 
Figure 6.34.
Figure 6.34
Circuit for Example 6.7.",DCElectricalCircuitAnalysis.pdf
"The equivalent network is placed back into the original circuit as shown in 
Figure 6.35 (red replaces blue). We now have a 333.333 Ω in series with 
2.555556 kΩ || 5.666667 kΩ. To determine Vb and Vc the total equivalent 
resistance can be used to find the source current. From there, a current 
divider may be employed along with Ohm's law to find the node voltages.
RTotal = 333.333Ω+2.555556 kΩ||5.666667 k Ω
RTotal ≈ 2.0946 kΩ
I s = E
RTotal
I s ≈ 12 V
2.0946 kΩ
I s ≈ 5.729 mA
I b = I s
Rright
Rleft +Rright
I b ≈ 5.729 mA 5.666667 kΩ
2.555556 k Ω+5.666667 kΩ
I b ≈ 3.9484 mA
V b = I b×R
V b ≈ 3.948 mA×2k Ω
V b ≈ 7.8968V
By similar process, Vc is approximately 7.1226 volts.
To verify the results of Example 6.7, we can take the two node voltages, apply them 
back to the original circuit, and determine whether or not they satisfy KVL or KCL. 
For example, examining node b, the current flowing through the 1 kΩ resistor has to",DCElectricalCircuitAnalysis.pdf
"For example, examining node b, the current flowing through the 1 kΩ resistor has to 
equal the sum of the currents flowing through the 2 kΩ and 5 kΩ resistors, via KCL.
I 1k = V a−V b
R
I 1k ≈ 12 V−7.8968V
1kΩ
I 1k ≈ 4.1032 mA
I 5k = V b−V c
R
I 5k ≈ 7.8968 V−7.1226 V
5 kΩ
I 5k ≈ 0.1548mA
195",DCElectricalCircuitAnalysis.pdf
"I 2k = I 1k −I5k
I 2k ≈ 4.1032 mA −0.1548mA
I 2k ≈ 3.9484 mA
This result matches the current Ib calculated previously. For completeness sake, the 
process can be replicated for node c.
Computer SimulationComputer Simulation
For further verification, both the original circuit of Figure 6.34 and the converted 
circuit of Figure 6.35 are entered into a simulator, as shown in Figure 6.36.
Both resistive networks are connected to a common power supply. Nodes 2 and 4 
correspond to Vb, while nodes 3 and 5 correspond to Vc. The results of a DC 
operating point simulation are shown in Figure 6.37.
196
Figure 6.36
Original and converted bridge 
circuits in simulator.",DCElectricalCircuitAnalysis.pdf
"The voltages match each other perfectly, along with the manual calculation. In 
closing, we see that it is possible to swap delta networks with Y networks, and 
achieve identical results. 
6.8 Summary6.8 Summary
In this chapter we have examined several new techniques and theorems to assist with
the analysis of DC electrical circuits. Beginning with more practical models for 
voltage and current sources, we added an internal resistance which sets limits on the 
source's maximum output. For a voltage source, this resistance is in series, its ideal 
value being a short. For current sources, the resistance is in parallel, its ideal value 
being an open. 
Source conversions allow us to create an equivalent current source for any practical 
voltage source and vice versa. An equivalent source is one that will create the same 
voltage across (and current into) whatever the new source is connected to as did the 
original source. In some cases, this swap allows differing sources to be combined",DCElectricalCircuitAnalysis.pdf
"original source. In some cases, this swap allows differing sources to be combined 
into a single source, simplifying analysis.
The superposition theorem states that, for any multi-source linear bilateral network, 
the contributions of each source may be determined independent of all other sources,
the final result being the summation of the contributions, cognizant of current 
directions and voltage polarities. Thus, the original circuit of N sources generates N 
new circuits, one for each source under consideration and with all other sources 
replaced by their ideal internal resistance.
197
Figure 6.37
Simulation results for original 
and converted bridges.",DCElectricalCircuitAnalysis.pdf
"Thévenin's and Norton's theorems allow the simplification of complex linear single 
port (i.e., two connecting points) networks. The Thévenin equivalent consists of a 
voltage source with series a resistance while the Norton equivalent consists of a 
current source with a parallel resistance. These equivalents, when replacing the 
original sub-circuit, will create the same voltage across the remainder of the circuit 
with the same current draw. That is, the remainder of the circuit will see no 
difference between being driven by the original sub-circuit or by either the Thévenin
or Norton equivalents.
The maximum power transfer theorem states that for a simple voltage source with an
internal resistance driving a single resistor load, the maximum load power will be 
achieved when the load resistance equals the internal resistance. At this point, 
efficiency will be 50%. If the load resistance is higher than the internal resistance,",DCElectricalCircuitAnalysis.pdf
"efficiency will be 50%. If the load resistance is higher than the internal resistance, 
the load power will not be as great, however, the system efficiency will increase.
Delta-Y conversions allow the generation of equivalent “three connection point” 
resistor networks. Resistor networks with three elements in the shape of a triangle or
delta (with one connection point at each corner) can be converted into a three 
element network in the shape of a Y or T, or vice versa. The two versions will 
behave identically to the remainder of the circuit. This allows the simplification of 
some circuits and eases analysis.
Review QuestionsReview Questions
1. Define the term linear bilateral network.
2. What are the ideal internal resistances of voltage sources and current 
sources?
3. Outline the process of converting a voltage source into a current source, and 
vice versa.
4. In general, describe the process of using superposition to analyze a multi-
source circuit.",DCElectricalCircuitAnalysis.pdf
"vice versa.
4. In general, describe the process of using superposition to analyze a multi-
source circuit.
5. What do Thévenin's and Norton's theorems state? How are they related?
6. What does the maximum power transfer theorem state? How might it be 
used with Thévenin's or Norton's theorems?
7. What are delta and Y configurations? How are they related?
198",DCElectricalCircuitAnalysis.pdf
"6.9 Exercises6.9 Exercises
AnalysisAnalysis
1. For the circuit shown in Figure 6.38, determine the equivalent current 
source.
2. Given the circuit shown in Figure 6.39, determine the equivalent current 
source.
3. Determine the equivalent current source for the circuit shown in Figure 6.40.
4. For the circuit shown in Figure 6.41, determine the equivalent current 
source.
199
Figure 6.38
Figure 6.39
Figure 6.41
Figure 6.40",DCElectricalCircuitAnalysis.pdf
"5. For the circuit shown in Figure 6.42, determine the equivalent voltage 
source.
6. Given the circuit shown in Figure 6.43, determine the equivalent voltage 
source.
7. Determine the equivalent voltage source for the circuit shown in Figure 
6.44.
8. For the circuit shown in Figure 6.45, determine the equivalent voltage 
source.
200
Figure 6.42
Figure 6.43
Figure 6.44
Figure 6.45",DCElectricalCircuitAnalysis.pdf
"9. Given the circuit shown in Figure 6.46, determine the equivalent voltage 
source.
10. Using source conversion, find Vb for the circuit shown in Figure 6.47.
11. Using source conversion, find the current through the 3 kΩ resistor in the 
circuit of Figure 6.48.
12. Using source conversion, find Vb for the circuit shown in Figure 6.49.
201
Figure 6.46
Figure 6.47
Figure 6.48
Figure 6.49",DCElectricalCircuitAnalysis.pdf
"13. Using source conversion, find Va for the circuit shown in Figure 6.50.
14. Using source conversion, find Vb for the circuit shown in Figure 6.51.
15. Using superposition, determine Vb for the circuit shown in Figure 6.47.
16. Using superposition, find the current through the 3 kΩ resistor for the circuit
of Figure 6.47.
17. Using superposition, find the current through the 3 kΩ resistor for the circuit
of Figure 6.48.
18. Using superposition, determine Vab for the circuit shown in Figure 6.48.
19. Using superposition, determine Vb for the circuit shown in Figure 6.49.
20. Using superposition, find the current through the 4 kΩ resistor for the circuit
of Figure 6.49.
21. Using superposition, find the current through the 30 kΩ resistor for the 
circuit of Figure 6.50.
22. Using superposition, determine Va for the circuit shown in Figure 6.50.
23. Using superposition, determine Vba for the circuit shown in Figure 6.51.",DCElectricalCircuitAnalysis.pdf
"23. Using superposition, determine Vba for the circuit shown in Figure 6.51.
24. Using superposition, find the current through the 10 kΩ resistor for the 
circuit of Figure 6.51.
202
Figure 6.50
Figure 6.51",DCElectricalCircuitAnalysis.pdf
"25. Using superposition, find the current through the 1.5 kΩ resistor for the 
circuit of Figure 6.52.
26. Using superposition, determine Vab for the circuit shown in Figure 6.52.
27. Using superposition, determine Vb for the circuit shown in Figure 6.53.
28. Using superposition, find the current through the 200 Ω resistor for the 
circuit of Figure 6.53.
29. Using superposition, find the current through the 4 kΩ resistor for the circuit
of Figure 6.54.
30. Using superposition, determine Vb for the circuit shown in Figure 6.54.
31. Is it possible to determine Vb in Figure 6.54 by using source conversions 
instead of superposition? Why/why not?
203
Figure 6.52
Figure 6.53
Figure 6.54",DCElectricalCircuitAnalysis.pdf
"32. Using superposition, determine Vbc for the circuit shown in Figure 6.55.
33. Using superposition, find the current through the 10 kΩ resistor for the 
circuit of Figure 6.55.
34. Using superposition, find the currents through the 100 Ω and 700 Ω resistors
for the circuit shown in Figure 6.56.
35. Using superposition, determine Vbd for the circuit shown in Figure 6.56.
36. Is it possible to determine Vbd in Figure 6.56 by using source conversions 
instead of superposition? Why/why not?
37. Using superposition, determine Vad for the circuit shown in Figure 6.57.
204
Figure 6.55
Figure 6.56
Figure 6.57",DCElectricalCircuitAnalysis.pdf
"38. Using superposition, find the current through the 20 kΩ resistor for the 
circuit shown in Figure 6.57.
39. Using superposition, find the current through the 12 kΩ resistor for the 
circuit shown in Figure 6.58.
40. Using superposition, determine Vb for the circuit shown in Figure 6.58.
41. Using superposition, determine Vb for the circuit shown in Figure 6.59.
42. Using superposition, find the current through the 100 Ω resistor for the 
circuit of Figure 6.59.
43. Using superposition, find the current through the 5 kΩ resistor for the circuit
of Figure 6.60.
44. Using superposition, determine Vc for the circuit shown in Figure 6.60.
205
Figure 6.58
Figure 6.59
Figure 6.60",DCElectricalCircuitAnalysis.pdf
"45. Given the circuit shown in Figure 6.61, determine the Thévenin equivalent 
circuit that is driving the 4 kΩ resistor.
46. Given the circuit shown in Figure 6.61, determine the Norton equivalent 
circuit driving the 4 kΩ resistor.
47. Given the circuit shown in Figure 6.61, determine the Norton equivalent 
circuit driving the 12 kΩ resistor.
48. Determine the Thévenin equivalent circuit driving the 12 kΩ resistor for the 
circuit shown in Figure 6.62.
49. Given the circuit shown in Figure 6.63, determine the Thévenin equivalent 
circuit that is driving the 20 kΩ resistor.
50. For the circuit shown in Figure 6.63, determine the Norton equivalent circuit
driving the 4 kΩ resistor.
206
Figure 6.61
Figure 6.62
Figure 6.63",DCElectricalCircuitAnalysis.pdf
"51. Given the circuit shown in Figure 6.64, determine the Norton equivalent 
circuit driving the 40 Ω resistor.
52. Determine the Thévenin equivalent circuit driving the 10 Ω resistor for the 
circuit shown in Figure 6.64.
53. Given the circuit shown in Figure 6.65, determine the Norton equivalent 
circuit driving the 12 kΩ resistor.
54. Given the circuit of Figure 6.61, determine the power in the 4 kΩ resistor. If 
this resistor can be replaced with any other value, is it possible to achieve a 
higher power? Why/why not? 
55. Given the circuit of Figure 6.62, determine the power in the 12 kΩ resistor. 
If this resistor can be replaced with any other value, is it possible to achieve 
a higher power? Why/why not? 
56. Given the circuit of Figure 6.64, determine the power in the 40 Ω resistor. If 
this resistor can be replaced with any other value, is it possible to achieve a 
higher power? Why/why not? 
57. Given the circuit of Figure 6.65, determine the power in the 6 kΩ resistor. If",DCElectricalCircuitAnalysis.pdf
"higher power? Why/why not? 
57. Given the circuit of Figure 6.65, determine the power in the 6 kΩ resistor. If 
this resistor can be replaced with any other value, is it possible to achieve a 
higher power? Why/why not? 
207
Figure 6.64
Figure 6.65",DCElectricalCircuitAnalysis.pdf
"DesignDesign
58. Consider the 4 kΩ resistor to be the load in Figure 6.61. Determine a new 
value for the load in order to achieve maximum load power. Also determine 
the maximum load power.
59. Consider the 12 kΩ resistor to be the load in Figure 6.62. Determine a new 
value for the load in order to achieve maximum load power. Also determine 
the maximum load power.
60. Consider the 40 Ω resistor to be the load in Figure 6.64. Determine a new 
value for the load in order to achieve maximum load power. Also determine 
the maximum load power.
61. Consider the 6 kΩ resistor to be the load in Figure 6.65. Determine a new 
value for the load in order to achieve maximum load power. Also determine 
the maximum load power.
62. Redesign the circuit of Figure 6.52 so that it uses only current sources and 
produces the same component currents and voltages as the original circuit.
63. Redesign the circuit of Figure 6.54 so that it uses only current sources and",DCElectricalCircuitAnalysis.pdf
"63. Redesign the circuit of Figure 6.54 so that it uses only current sources and 
produces the same component currents and voltages as the original circuit.
64. Redesign the circuit of Figure 6.58 so that it uses only voltage sources and 
produces the same component currents and voltages as the original circuit.
65. Redesign the circuit of Figure 6.60 so that it uses only voltage sources and 
produces the same component currents and voltages as the original circuit.
66. Convert the delta network of Figure 6.66 into a Y network.
208
Figure 6.66",DCElectricalCircuitAnalysis.pdf
"67. Convert the pi network of Figure 6.67 into a T network.
68. Convert the Y network of Figure 6.68 into a delta network.
69. Convert the T network of Figure 6.69 into a pi network.
209
Figure 6.67
Figure 6.68
Figure 6.69",DCElectricalCircuitAnalysis.pdf
"ChallengeChallenge
70. Redesign the circuit of Figure 6.70 so that it uses only current sources and 
produces the same node voltages as the original circuit.
71. Using any combination of techniques, find the current through the 9 kΩ 
resistor for the circuit of Figure 6.70.
72. Using any combination of techniques, determine Vbc for the circuit shown in 
Figure 6.70.
73. Is it possible to determine Vbc in Figure 6.70 by using just source 
conversions? Why/why not?
74. Using superposition, find the current through the 25 Ω resistor for the circuit
of Figure 6.71.
210
Figure 6.70
Figure 6.71",DCElectricalCircuitAnalysis.pdf
"75. Using superposition, find Vab in the circuit of Figure 6.71.
76. Redesign the circuit of Figure 6.71 using only voltage sources so that it 
achieves the same node voltages as the original.
77. Is it possible to determine Vbc in Figure 6.72 by using just source 
conversions or just superposition? Why/why not?
78. Using any combination of techniques, find the current through the 100 Ω 
resistor for the circuit shown in Figure 6.72.
79. Redesign the circuit of Figure 6.72 so that it uses only voltage sources and 
produces the same node voltages as the original circuit.
80. Determine the Thévenin and Norton equivalents driving the 40 Ω resistor for
the circuit shown in Figure 6.53.
81. Determine the Thévenin and Norton equivalents driving the 12 kΩ resistor 
for the circuit shown in Figure 6.54.
82. Given the circuit of Figure 6.63, determine if the 4 kΩ resistor is the optimal
value to achieve maximum power dissipation in that resistor. If it is not,",DCElectricalCircuitAnalysis.pdf
"value to achieve maximum power dissipation in that resistor. If it is not, 
determine the value that will produce maximum power in the resistor along 
with the resulting power. 
83. For the circuit of Figure 6.73, determine an equivalent circuit using just a 
single voltage source.
211
Figure 6.72
Figure 6.73",DCElectricalCircuitAnalysis.pdf
"SimulationSimulation
84. Using DC bias simulations, compare the original circuit of problem 1 to its 
converted equivalent. Do this by connecting a resistor to the output 
terminals, trying several different resistance values and checking to see if 
the two circuits always produce the same voltage across this resistor.
85. Using DC bias simulations, compare the original circuit of problem 5 to its 
converted equivalent. Do this by connecting a resistor to the output 
terminals, trying several different resistance values and checking to see if 
the two circuits always produce the same voltage across this resistor.
86. Perform a DC bias simulation on the circuit of problem 11 to verify the node
voltages.
87. Perform a DC bias simulation on the circuit of problem 13 to verify the node
voltages.
88. Perform a DC bias simulation on the circuit of problem 19 to verify the node
voltages.
89. Perform a DC bias simulation on the circuit of problem 21 to verify the 
resistor current.",DCElectricalCircuitAnalysis.pdf
"voltages.
89. Perform a DC bias simulation on the circuit of problem 21 to verify the 
resistor current.
90. Create DC bias simulations of the original and equivalent circuits generated 
in problem 45 to determine if the equivalent circuit is truly equivalent. Do 
this by substituting several different values for the 4 kΩ resistor in both 
circuits to see if the same load voltage is obtained for both circuits.
91.  Create DC bias simulations of the original and equivalent circuits generated 
in problem 49 to determine if the equivalent circuit is truly equivalent. Do 
this by substituting several different values for the 20 kΩ resistor in both 
circuits to see if the same load voltage is obtained for both circuits.
92. Create DC bias simulations of the original and equivalent circuits generated 
in problem 53 to determine if the equivalent circuit is truly equivalent. Do 
this by substituting several different values for the 12 kΩ resistor in both",DCElectricalCircuitAnalysis.pdf
"this by substituting several different values for the 12 kΩ resistor in both 
circuits to see if the same load voltage is obtained for both circuits.
93. Perform a DC bias simulation on the circuit of problem 63 to verify that the 
node voltages of the new design match those of the original.
94. Perform a DC bias simulation on the circuit of problem 65 to verify that the 
node voltages of the new design match those of the original.
95. Using either Monte Carlo simulation or multiple DC bias simulations, verify
that the resistance calculated in problem 55 achieves maximum power. Do 
this by trying several resistor values near the calculated value, and 
determining the power for each based on the squared load voltages.
212",DCElectricalCircuitAnalysis.pdf
"96. Using either Monte Carlo simulation or multiple DC bias simulations, verify
that the resistance calculated in problem 56 achieves maximum power. Do 
this by trying several resistor values near the calculated value, and 
determining the power for each based on the squared load voltages.
This XKCD comic is too large to fit on this page. Just go to this link instead:
https://xkcd.com/1732/
213",DCElectricalCircuitAnalysis.pdf
"7 7 Nodal & Mesh AnalysisNodal & Mesh Analysis, Dependent Sources, Dependent Sources
7.0 Chapter Learning Objectives7.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Utilize nodal analysis techniques to solve for voltages in multi-source series-parallel networks.
• Utilize mesh analysis techniques to solve for currents in multi-source series-parallel networks.
• Analyze networks using dependent voltage and current sources.
7.1 Introduction 7.1 Introduction 
This chapter presents the culmination of the analysis of DC circuits that employ resistors along with any number 
of voltage and/or current sources. Here we shall focus on nodal analysis and mesh analysis. Both nodal analysis 
and mesh analysis generate a system of simultaneous linear equations that are used to solve the circuit for 
various voltages or currents. There are several methods that can be used to solve the simultaneous equations",DCElectricalCircuitAnalysis.pdf
"various voltages or currents. There are several methods that can be used to solve the simultaneous equations 
including substitution, Gauss-Jordan elimination and expansion by minors. These methods are reviewed in 
Appendix B and are not covered in this chapter. Instead, to keep clutter to a minimum and focus on the circuit 
analysis aspects, the explanations and examples will simply detail the process of examining the circuit and 
applying basic circuit laws to create the system of equations. From there, the specific technique used to solve 
these simultaneous equations is up to you based on your personal preferences. 
It is particularly efficient to obtain a more advanced scientific calculator that can solve these equations directly 
rather than plowing through the solution manually. By doing so, you can spend your precious time mastering the
process of circuit analysis and creating the equations. Manual solution techniques, though not necessarily",DCElectricalCircuitAnalysis.pdf
"process of circuit analysis and creating the equations. Manual solution techniques, though not necessarily 
difficult, can be tedious, time consuming and error prone. If you plan on continuing your study into AC electrical
circuits, you should consider obtaining a calculator that can solve simultaneous equations with complex 
coefficients. Such calculators can be expensive when purchased new, such as the Texas Instruments TI-89 and 
Nspire models. On the used market, perfectly satisfactory older models such as the TI-85 and TI-86 can be 
found at considerable discount. Another model to consider is the Casio FX-9750GII, although it is not quite as 
powerful as some of the other units mentioned.  
Along with nodal and mesh, we shall also introduce the concept of dependent sources. These are current and 
voltage sources whose value is not fixed to some particular value, but rather is dependent on some other current",DCElectricalCircuitAnalysis.pdf
"voltage sources whose value is not fixed to some particular value, but rather is dependent on some other current 
or voltage in the circuit. What makes this interesting is that this controlling current or voltage may itself be 
affected by the value produced by the dependent source. Dependent sources are often used to model the 
behavior of active electronic devices such as bipolar and field effect transistors. 
214",DCElectricalCircuitAnalysis.pdf
"7.2 Nodal Analysis7.2 Nodal Analysis
Nodal analysis is a technique that can be applied to virtually any circuit. In general 
use, it might be considered a universal solution technique as there are no practical 
circuit configurations that it cannot handle. Nodal analysis relies on the application 
of Kirchhoff's current law to create a series of node equations that can be solved for 
node voltages. These equations are based on Ohm's law and will be of the form        
I = V/R, or more generally, I = (1/RX)∙VA + (1/RY)∙VB... Once the node voltages are 
obtained, finding any branch currents or component powers becomes an almost 
trivial exercise. We will examine two variations; a general version that can be used 
with both voltage and current sources, and a second somewhat quicker version that 
can be used with circuits only driven by current sources. 
General MethodGeneral Method
Consider the circuit shown in Figure 7.1. We begin by labeling connection nodes",DCElectricalCircuitAnalysis.pdf
"General MethodGeneral Method
Consider the circuit shown in Figure 7.1. We begin by labeling connection nodes 
and assigning current directions. We are particularly interested in current junctions, 
that is, places where currents can combine or split. These are also known as 
summing nodes and are circled in green on the figure. The current directions are 
chosen arbitrarily and may be the opposite of reality. This is not a problem. If we 
assign directions that are wrong, the resulting directions will ultimately show up 
reversed but the computed node voltages will be just fine.
One node is chosen as the reference. This is the point to which all other node 
voltages will be measured against. Typically, the reference node is ground, although 
it does not have to be.
We now write a current summation equation for each summing node, except for the 
reference node. In this circuit there is only one node where currents combine (other",DCElectricalCircuitAnalysis.pdf
"reference node. In this circuit there is only one node where currents combine (other 
than ground) and that's node b. Points a and c are places where components connect,
but they are not summing nodes, so we can ignore them for now. Using KCL on 
node b we can say:
   I1 + I2 = I3
215
Figure 7.1
Basic dual voltage source 
circuit with currents and nodes 
defined.",DCElectricalCircuitAnalysis.pdf
"Next, we describe these currents in terms of the node voltages and associated 
components via Ohm's law. For example, I3 is the node b voltage divided by R3 
while I1 is the voltage across R1 divided by R1. This voltage is Va − Vb. Therefore,
V a −V b
R1
+V c −V b
R2
= V b
R3
Noting that Va = E1 and Vc = E2, with a little algebra this can be reduced to a series of
products of conductances and voltages:
(
1
R1 )E1 +(
1
R2 )E2 =(
1
R1
+ 1
R2
+ 1
R3 )V b
All quantities are known except for Vb and thus it is easily found with a little more 
algebra. As we shall see, this conductance-voltage product format turns out to be a 
convenient way of writing these equations. Also, note that the first two terms on the 
left reduce to fixed current values. If there had been more nodes, there would have 
been more equations, one for each node. 
For current sources, a more direct approach is possible. Consider the circuit of",DCElectricalCircuitAnalysis.pdf
"been more equations, one for each node. 
For current sources, a more direct approach is possible. Consider the circuit of 
Figure 7.2. We start as before, identifying nodes and labeling currents. This is shown
in Figure 7.3. We then write current summation equations at each node (except for 
ground). We consider currents entering a node as positive, and exiting as negative.
Node a:   I1 = I3 + I4
Node b:   I3 = I2 + I5, and rearranging in terms of the fixed source,
Node b: −I2 = −I3 + I5
216
Figure 7.2
Basic dual current source 
circuit.
Figure 7.3
Basic dual current source 
circuit with currents and nodes 
defined.",DCElectricalCircuitAnalysis.pdf
"The currents are then described by their Ohm's law equivalents:
Node a:   I 1 = V a −V b
R3
+V a
R1
Node b: −I 2 =−V a −V b
R3
+V b
R2
Expanding and collecting terms yields:
Node a:   I 1 = (
1
R1
+ 1
R3 )V a −(
1
R3 )V b
Node b: −I 2 =−(
1
R3 )V a +(
1
R3
+ 1
R2 )V b
As the resistor values and currents are known, simultaneous equation solution 
techniques may be used to solve for the node voltages. Once again, there will be as 
many equations as node voltages. It is very important that the terms “line up” when 
the final system of equations is written out. That is, there should be a column for the 
Va terms, a column for the Vb terms, and so on. They should not be written out in 
random order. This format will make it much it easier to enter the coefficients into a 
calculator or solve manually.
Example 7.1
Determine Vb in the circuit of Figure 7.4.
This circuit has two current summing nodes, ground and node b. Let
us assume that the currents from the two sources flow into node b",DCElectricalCircuitAnalysis.pdf
"This circuit has two current summing nodes, ground and node b. Let
us assume that the currents from the two sources flow into node b
and the exiting current flows down through the 5 kΩ resistor. It will
be convenient if we find the conductance equivalents of the resistors
before continuing. Taking the reciprocals we find: 1 kΩ = 1 mS,
4 kΩ = 0.250 mS, and 5 kΩ = 0.2 mS. Through KCL we can say:
   I1k + I4k = I5k
Replacing these currents with their Ohm's law equivalents yields, 
15 V −V b
1k Ω + 6V −V b
4 kΩ = V b
5 kΩ
collecting terms,
15mA +1.5mA = (
1
1kΩ + 1
4k Ω + 1
5k Ω)V b
217
Figure 7.4
Circuit for Example 7.1.",DCElectricalCircuitAnalysis.pdf
"and solving for Vb,
V b = 16.5 mA
1mS +0.25mS +0.2mS
V b = 11.379volts
As Vb is higher than the 6 volt source, our assumed current direction for the 
4 kΩ resistor was incorrect; we assumed right to left but it is in fact left to 
right, flowing from 11.379 volts to 6 volts. As you can see, this did not 
present a problem.
For comparison sake, this circuit was solved in Chapter 6 using source 
conversions and also using superposition. 
Inspection MethodInspection Method
The system of equations can be obtained directly through inspection if the circuit
contains no voltage sources. For the node under inspection, sum all of the current
sources connected to it to obtain the current constant. The conductance term for
that node will be the sum of all of the conductances connected to that node. For
the other node conductances, determine the conductances between the node
under inspection and these other nodes. These terms will all be negative. As a",DCElectricalCircuitAnalysis.pdf
"under inspection and these other nodes. These terms will all be negative. As a
crosscheck, the set of equations produced must exhibit diagonal symmetry, that
is, if a diagonal is drawn from the upper left to the lower right through the
conductance-voltage pairs, then the coefficients found above the diagonal will
have to match those found below the diagonal as we move perpendicular from
the diagonal. This is illustrated in Figure 7.5. Diagonal symmetry can be seen in
Example 7.1 by noting the matching “−1/R3” coefficients in the final pair of
equations. As an example of the inspection method, focusing on node a in Figure
7.2, we find the fixed current source I1 feeding it (entering, therefore positive).
The conductances directly connected to node a are 1/R1 and 1/R3, yielding the
coefficient for Va. The only conductance common between nodes a and b is 1/R3, 
yielding the Vb coefficient.
The inspection method is summarized as follows:",DCElectricalCircuitAnalysis.pdf
"yielding the Vb coefficient.
The inspection method is summarized as follows:
1. Verify that the circuit uses only current sources with resistors and no voltage
sources. If voltage sources exist, they must be converted to current sources 
before proceeding.
2. Find all of the current summing nodes and number them. Also decide on the 
reference node (usually ground).
3. To generate an equation, locate the first node. This is the node of interest and
the next few steps will be associated with it.
218
1 V a
-2 Va
-6 Va
-9 Va
-2 Vb
5 V b
-4 Vb
-7 Vb
-6 Vc
-4 Vc
8 V c
0 V c
-9 Vd
-7 Vd
0 V d
3 V d
Figure 7.5
Diagonal symmetry.",DCElectricalCircuitAnalysis.pdf
"4.  Sum the current sources feeding the node of interest. Entering is deemed 
positive while exiting is deemed negative. The sum is placed on one side of 
the equals sign.
5. Next, find all of the resistors connected to the node of interest and write 
them as a sum of conductances on the other side of the equals sign, the 
group being multiplied by this node's voltage (e.g., V1).
6. Now find all of the resistors that are connected to the node of interest and to 
other nodes (except the ground reference). For each of these other nodes,  
multiply the sum of the conductances between the node of interest and this 
other node by this other node's voltage, and then subtract that product from 
the equation built so far. Once all other nodes are considered, this equation 
is finished.
7. Find the next node and treat this as the new node of interest. 
8. Repeat steps 4 through 7 until all nodes have been treated as the node of",DCElectricalCircuitAnalysis.pdf
"7. Find the next node and treat this as the new node of interest. 
8. Repeat steps 4 through 7 until all nodes have been treated as the node of 
interest. Each iteration creates a new equation. There will be as many 
equations as there are nodes, less the reference node. Check for diagonal 
symmetry and solve.
The inspection method is best observed in action; as in the following example. 
Example 7.2
Determine Va and Vb in the circuit of Figure 7.6. Also determine the current
flowing through the 100 Ω resistor.
This circuit has three current summing nodes; ground, node a and
node b. Also, it uses only current sources and no voltage sources.
Therefore, we are able to use the inspection method to obtain the
system of equations (in this case, two equations).
We begin by focusing on node a, our first node of interest. We
will build the first expression piece by piece. First we find the
current sources feeding this node:
800 mA − 2A = ...",DCElectricalCircuitAnalysis.pdf
"will build the first expression piece by piece. First we find the
current sources feeding this node:
800 mA − 2A = ...
Next, we find all of the resistors connected to node a and write
them as conductances, the group being multiplied by node voltage a:
800 mA − 2A =(
1
10 Ω + 1
50 Ω + 1
100Ω)V a  ...
Now find all of the resistors that are connected to this node and to the other 
nodes. Multiply those resistors (expressed as conductances) by the other 
associated node voltages and subtract the products from the expression built 
219
Figure 7.6
Circuit for Example 7.2.",DCElectricalCircuitAnalysis.pdf
"so far. Repeat for all remaining nodes except the ground reference. In this 
example there is only one other node, node b, and thus only one iteration.
800 mA −2 A = (
1
10Ω + 1
50Ω + 1
100Ω)V a −(
1
50Ω + 1
100 Ω)V b
Finally, simplify the constants and coefficients, and the first expression is 
complete:
−1.2 A = 130 mSV a − 30 mSV b
Now we repeat the entire process for the next equation. Node b is our new 
node of interest. The fixed current sources are:
−800 mA −300 mA = ...
Next, we find all of the resistors connected to node b and write them as 
conductances, the group being multiplied by node voltage b:
−800 mA −300 mA = (
1
25Ω + 1
50Ω + 1
100 Ω)V b  ...
Now we find all of the resistors that are connected to this node and to the 
other nodes. Multiply those resistors (expressed as conductances) by these 
other node voltages and subtract those products from the expression built so 
far. Make sure the terms align vertically based on the node voltages.
−800 mA −300 mA = −(
1",DCElectricalCircuitAnalysis.pdf
"far. Make sure the terms align vertically based on the node voltages.
−800 mA −300 mA = −(
1
50Ω + 1
100 Ω)V a+(
1
25Ω + 1
50Ω + 1
100Ω)V b
Finally, simplify the constants and coefficients, and this expression is 
complete:
−1.1 A =−30 mSV a +70 mSV b
We now have two equations with two unknowns. Let's check for diagonal 
symmetry:
−1.2 A = 130 mSV a − 30 mSV b
−1.1 A =−30 mSV a +70 mSV b
The perpendicular coefficient is −30 mS on each side. We can now solve the 
system. The results are Va = −14.27 volts and Vb = −21.83 volts. The current 
flowing through the 100 resistor is (−14.27 V − (−21.83 V))/100 Ω, or 
approximately 75.6 mA flowing left to right. 
220",DCElectricalCircuitAnalysis.pdf
"Let's verify that these values are correct. We can perform a KCL summation 
at node a and see if it balances. We already know that 2 amps and 75.6 
milliamps exit while 800 milliamps enters. We only need to find the currents
through the 10 Ω and 50 Ω resistors. First, notice that the 50 Ω sees the same
voltage as the 100 Ω resistor. As it is half the resistance, it must produce 
twice the current, or 151.2 milliamps exiting. The current through the 10 Ω 
is found via Ohm's law, or −14.27 V/10 Ω, which is 1.427 amps entering. 
Entering: 0.8 A + 1.427 A = 2.227 A
Exiting: 2 A + 0.0756 A + 0.1512 A ≈ 2.227 A 
KCL is satisfied. To complete the verification we perform the same 
summation on node b. This is left as an exercise.
Example 7.3
Write the node equations for the circuit of Figure 7.7.
This circuit has four current summing nodes; ground and nodes a,  
b, and c. Also, it uses only current sources and no voltage sources.",DCElectricalCircuitAnalysis.pdf
"This circuit has four current summing nodes; ground and nodes a,  
b, and c. Also, it uses only current sources and no voltage sources.
Therefore, we are able to use the inspection method to obtain the
system of equations (three equations).
As we have already seen, we will wind up with conductance values
in the final equations. We can save some work later by finding the
conductance value of each resistor right now. The result is shown in
Figure 7.8. 
We begin by focusing on node a, our first node of interest. We will build
the first expression piece by piece. First we find the current sources
feeding this node. That's just the 1.5 A source, entering is positive.
1.5A = ...
Next, we find all of the conductances connected to node a and
multiply by node voltage a:
1.5A = (0.2S +0.25S)V a  ...
Find all of the conductances that are connected to this node and to
the other nodes. Multiply those conductances by the other associated node",DCElectricalCircuitAnalysis.pdf
"Find all of the conductances that are connected to this node and to
the other nodes. Multiply those conductances by the other associated node 
voltages and subtract those products from the expression built so far. Repeat 
for all remaining nodes except the ground reference.
221
Figure 7.7
Circuit for Example 7.3.
Figure 7.8
Circuit of Figure 7.7 shown 
with conductances instead of 
resistances.",DCElectricalCircuitAnalysis.pdf
"1.5A = (0.2S +0.25S)V a − (0.25S)V b − (0)V c
Finally, simplify the constants and coefficients, and the first expression is 
complete:
1.5A = 0.45S V a − 0.25SV b − 0V c
We leave the zero term in just for alignment. Now we repeat the entire 
process for the next equation. The node of interest is now node b. Here is the
result:
0 =−0.25 SV a +0.8S V b − 0.5V c
And finally for node c we have:
1A = 0V a − 0.5SV b +0.6V c
Our three equations are (with some padding to see the columns better):
1.5A = 0.45S V a −0.25SV b −0V c
      0 =−0.25 SV a +0.8SV b −0.5V c
   1A = −0V a −0.5S V b +0.6V c
Check for diagonal symmetry: we have pairs of −0.25, 0, and −0.5 
straddling the diagonal.  
Now the system of equations can be solved for the three node voltages and 
verified using KCL summations at each node. This is left as an exercise.
Converting Sources and Other SimplificationsConverting Sources and Other Simplifications",DCElectricalCircuitAnalysis.pdf
"Converting Sources and Other SimplificationsConverting Sources and Other Simplifications
Given circuits with voltage sources, it may be easier to convert them to current
sources and then apply the inspection technique rather than using the general
approach outlined initially. There is one trap to watch out for when using source
conversions: the voltage across or current through a converted component
will most likely not be the same as the voltage or current in the original
circuit. This is because the location of the converted component will have
changed. For example, the circuit in Figure 7.9 could be solved using nodal
analysis by converting the voltage source and the associated resistance into a
current source. That is, E/R1 would be converted into a source I3 with a
parallel resistor R1. This is shown in Figure 7.10. 
222
Figure 7.9
Circuit with both current and 
voltage sources.",DCElectricalCircuitAnalysis.pdf
"Here is the trap: in the converted circuit, although R1 still connects to node a, the
other end no longer connects to the voltage source. Rather, the right side now
connects to node c. Therefore, the voltage drop across R1 in the converted circuit is
not likely to equal the voltage drop seen across R1 in the original circuit (the
only way they would be equal is if E was 0 volts). In the converted circuit,
nodes a and c have not changed from the original, so the original voltage
across R1 can be determined via Va, Vc and E in the original circuit.
If a circuit uses voltage sources exclusively, or even a very large proportion of
voltage sources, an alternate technique called mesh analysis may be a better
choice than using numerous conversions. Mesh analysis will be detailed later
in this chapter.
A final item to consider is to simplify resistor networks in order to reduce the 
number of nodes. Fewer nodes means fewer equations and a quicker solution. For",DCElectricalCircuitAnalysis.pdf
"number of nodes. Fewer nodes means fewer equations and a quicker solution. For 
example, a portion of a network might consist of two parallel resistors which are in 
series with a third resistor. Ordinarily, the common junction of these three resistors 
would constitute a current summing node and an equation would need to be derived 
for it. It may be easier to simply replace the trio with a single resistor that equals the 
series-parallel combination value. This would remove the node and simplify creating
the system of equations, but not alter the remainder of the circuit.
SupernodeSupernode
On occasion you may come across a circuit like the one shown in Figure 7.11 that
has a voltage source without a series resistance associated with it. Without that
resistance, it becomes impossible to create an expression for the current passing
through the source using the general method, and impossible to convert the
voltage source into a current source in order to use the inspection method.",DCElectricalCircuitAnalysis.pdf
"voltage source into a current source in order to use the inspection method.
One possible way out of this quandary is to simply add a very small resistor
in series with it so that a source conversion is possible. The resistor in
question would have to be much smaller than any surrounding resistors in
order to have minimal impact on the results. A reduction by two orders of
magnitude would generally yield a variation smaller than that produced by
resistor tolerances in all but high precision circuits. The other way out is to
use a supernode. 
A supernode is, in effect, the combination of two nodes. It relies on a simple 
observation. If we examine the circuit of Figure 7.11, the path of the voltage source 
produces identical currents flowing into and out of nodes a and b. As a consequence,
if we treat the two nodes as one big node, then when we write a KCL summation, 
these two terms will cancel. To see just how this works, refer to Figure 7.12.
223
Figure 7.11
Circuit for supernode",DCElectricalCircuitAnalysis.pdf
"these two terms will cancel. To see just how this works, refer to Figure 7.12.
223
Figure 7.11
Circuit for supernode 
analysis.
Figure 7.10
Circuit converted to current 
sources only.",DCElectricalCircuitAnalysis.pdf
"In this version we have replaced the voltage source with its ideal internal resistance, 
a short. We have also labeled the two nodes of interest, a and b, and labeled the 
currents, drawn with convenient directions. 
Due to the shorted voltage source, nodes a and b are now the same node. Consider 
the currents entering and exiting this combined or “super” node. On the left side 
(formerly node a) we see a constant current Ix entering while I1, I2 and I3 are exiting. 
On the right side (formerly node b) we see Iy exiting along with I4, and entering we 
see I1 and I2. Now let's set up the entering currents on the left side of the equals sign 
with the exiting currents on the right:
Σ Iin = Σ I out
I x+I 1+I 2 = I y +I 1+I 2+I3 +I 4
This can be simplified to:
I x − I y = I 3 +I 4
Writing this in terms of Ohm's law we have:
I x − I y = 1
R1
V a + 1
R3
V b
We also know that Va − Vb = E from the original circuit. Assuming all sources and",DCElectricalCircuitAnalysis.pdf
"I x − I y = 1
R1
V a + 1
R3
V b
We also know that Va − Vb = E from the original circuit. Assuming all sources and 
resistors are known, that makes two equations with two unknowns, solvable using 
simultaneous equation techniques. This is illustrated in the following example.
224
Figure 7.12
Circuit modified for supernode 
analysis.",DCElectricalCircuitAnalysis.pdf
"Example 7.4
Find Va and Vb for the circuit of Figure 7.13.
As shown in Figure 7.14, we short the 60 volt source and write a
current summation at the a b supernode:
Σ Iin = Σ I out
1A+I 1+I 2 = 2.5 A+I 1+I 2+I 3+I 4
This can be simplified to:
−1.5 A = I 3 +I 4
Writing this in terms of Ohm's law we have:
−1.5A = 1
4ΩV a + 1
10ΩV b
−1.5 A = 0.25SV a +0.1S V b
We also know that Va − Vb = 60 volts. Therefore Vb = Va − 60 volts.
Substituting this into the prior equation yields:
 
−1.5 A = 0.25SV a +0.1S(V a −60 V)
−1.5 A = 0.25SV a +0.1S V a −6A
4.5 A = 0.35SV a
 
V a = 12.857 V
We know that Vb is 60 volts below Va, so Vb = −47.143 volts. 
To verify, we will perform a KCL summation at each node. For
node a, assuming I1 exits as drawn:
I 1 = 1A − V a
4Ω −V a−V b
20Ω
I 1 = 1 A−12.857V
4Ω −12.857V−(−47.143 V)
20Ω
I 1 =−5.2143A (negative exit means it's entering)
225
Figure 7.14
Circuit modified for supernode
analysis.
Figure 7.13
Circuit for Example 7.4.",DCElectricalCircuitAnalysis.pdf
"Doing likewise for node b, and assuming I1 enters as drawn:
I 1 = 2.5 A+ V b
10Ω −V a−V b
20Ω
I 1 = 2.5 A+−47.143 V
10Ω − 12.857V−(−47.143V)
20Ω
I 1 =−5.2143A (negative enter means it's exiting)
These currents match meaning that the current through the voltage source is 
verified to be the same at both terminals, as it must be.
An alternative to the basic supernode technique is to simply describe one node 
voltage in terms of another at the outset. This is illustrated in the example following.
Example 7.5
Find Va, Vb and Vc for the circuit of Figure 7.15.
Once again we have a situation of a voltage source without a series
resistance. Without having to short it and thus treating nodes a and c as an
explicit supernode, we can instead note by observation that the currents
entering and exiting the voltage source must be identical. 
The circuit is redrawn in Figure 7.16 with currents labeled and using
equivalent conductances in place of the resistances. This time the voltage",DCElectricalCircuitAnalysis.pdf
"The circuit is redrawn in Figure 7.16 with currents labeled and using
equivalent conductances in place of the resistances. This time the voltage
source is left in. 
We start with the observation that Vc = Va − 8 V . In other words, Vc is locked 
to Va and if we find one of them, we can determine the other. Therefore, 
instead of writing equations using three nodes, we shall instead refer to node
c in reference to node a, that is we shall write Va − 8 V wherever we need Vc.
Thus, this three node circuit will only need two equations.
We begin at node a and apply KCL as usual.
Σ Iin = Σ I out
I 1+I 3 = I 2
This is expanded using Ohm's law and we solve for I1:
I 1 = I 2 −I 3
I 1 = 0.2S V a −0.5S(V b −V a )
I 1 = 0.7SV a −0.5S V b
226
Figure 7.15
Circuit for Example 7.5.",DCElectricalCircuitAnalysis.pdf
"On to node b:
1A = I 3+I 4
1A = 0.5S(V b −V a )+0.25S(V b −V c)
1A = 0.5S(V b −V a )+0.25S(V b −(V a −8V))
1A = 0.5S(V b −V a )+0.25S(V b −V a+8 V)
 
−1 A =−0.75V a +0.75V b
And finally node c:
I 4 = I1+I5
I 1 = I 4 −I 5
I 1 = 0.25S(V b −V c)−0.1SV c
I 1 = 0.25S(V b −(V a−8V))−0.1S(V a −8 V)
I 1 = 0.25S(V b −V a +8 V)−0.1S(V a −8V)
 
I 1 =−0.35V a+0.25V b+2.8A
The final equations for nodes a and c both equal I1, meaning they equal each
other. Thus,
0.7SV a −0.5SV b = −0.35S V a +0.25S V b +2.8 A
2.8 A = 1.05SV a −0.75SV b
The final equations are:
2.8 A = 1.05S V a −0.75S V b
−1 A =−0.75SV a +0.75SV b
The solution is Va = 6 volts and Vb ≈ 4.6667 volts. As Vc is 8 volts less than 
Va, then Vc = −2 volts.
227
Figure 7.16
Circuit of Example 7.5 with 
currents labeled and using 
conductances.",DCElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
In order to verify the result of Example 7.5, the circuit is entered into a simulator as 
shown in Figure 7.17.
A DC operating point simulation is run. The results are shown in Figure 7.18 and 
match the calculated values perfectly. Node 1 corresponds to Vb, node 2 corresponds 
to Va and node 3 corresponds to Vc.
228
Figure 7.17
Circuit of Example 7.5 in the 
simulator.
Figure 7.18
Simulation results for the 
circuit of Example 7.5.",DCElectricalCircuitAnalysis.pdf
"7.3 Mesh Analysis7.3 Mesh Analysis
In some respects mesh analysis is a mirror of nodal analysis. While nodal analysis 
leverages KCL to create a series of node equations that are used to solve for node 
voltages, mesh analysis uses KVL to create a series of loop equations that can be 
solved for mesh currents. A mesh current should not be confused with a branch 
current. While branch currents represent the current flowing through a particular 
component, mesh currents combine to create branch currents. That is, the current 
through any particular component may be an individual mesh current or a 
combination of two mesh currents. Circuits using complex series-parallel 
arrangement with multiple voltage and/or current sources may be solved using this 
technique. It is, however, limited to planar circuits. Planar circuits are those that can 
be drawn on a flat plane without having any of their wires cross. In essence, these",DCElectricalCircuitAnalysis.pdf
"be drawn on a flat plane without having any of their wires cross. In essence, these 
circuits can be drawn so that they appear like a series of window panes. Non-planar 
circuits may be 3D in appearance and in such a circuit it becomes impossible to 
define the loop equations as any given component might have more than two 
meshing currents. 
General MethodGeneral Method
Consider the circuit of Figure 7.19. We begin by designating a series of loops. These
loops should be minimal in size and cover all components at least once. By 
convention, the loops are drawn clockwise. There is nothing magic about them being
clockwise, it is just a matter of consistency. The current path along a loop is referred 
to as a mesh current. In the circuit of Figure 7.19 we have two loops, and thus, two 
mesh currents, I1 and I2. Note that all components exist in at least one loop (and 
sometimes in more than one loop, like R3). Depending on circuit values, one or more",DCElectricalCircuitAnalysis.pdf
"sometimes in more than one loop, like R3). Depending on circuit values, one or more
of these loop's directions may in fact be opposite of reality. This is not a problem. If 
this is the case, the currents will show up as negative values, and thus we know that 
they're really flowing counter-clockwise.
We begin by writing KVL equations for each loop.
Loop 1: E1 = voltage across R1 + voltage across R3
Loop 2: −E2 = voltage across R2 + voltage across R3  
(E2 is negative as I2 is drawn flowing out of its negative terminal.)
229
Figure 7.19
Circuit for mesh analysis.",DCElectricalCircuitAnalysis.pdf
"Expand the voltage terms using Ohm's law.
Loop 1: E1 = I1 ∙ R1 + (I1 − I2) R3
Loop 2: −E2 = I2 ∙ R2 + (I2 − I1) R3
Expanding and collecting terms yields:
Loop 1: E1 = (R1 + R3) I1 − R3 ∙ I2
Loop 2: −E2 = R3 ∙ I1 + (R2 + R3) I2
Assuming that the resistor values and source voltages are known, we have two 
equations with two unknowns. These can be solved for I1 and I2 using simultaneous 
equation solution techniques such as determinants or Gauss-Jordan elimination. 
Example 7.6
Find Vb and Vbc for the circuit of Figure 7.20.
The first step is to define a set of clockwise loops that are of minimum
size and which cover all components. These loops are drawn in Figure
7.21. Also, the polarities of the voltage drops produced by these currents
are drawn next to the components in the same color for easy
identification. Note that some components see only one current, such as
the 3 kΩ resistor that sees only I2; and some see two currents in",DCElectricalCircuitAnalysis.pdf
"identification. Note that some components see only one current, such as
the 3 kΩ resistor that sees only I2; and some see two currents in
opposing directions, such the 5 kΩ resistor that sees both I2 and I3. With
multiple current directions flowing through a single component,
obeying the resulting voltage polarities becomes very important.
We now write KVL summations around each loop. If the current passing 
through a component sees a polarity of + to − then this is counted as a 
voltage drop while the reverse counts as a voltage rise. This is true even for 
voltage sources (i.e., a voltage source may appear negative in one loop and 
positive in another). 
ΣV rises = ΣV drops
 
Loop 1: 12 V = V 1k +V 2k
Loop 2:   0V = V 3k +V 5k+V 1k
Loop 3:   0V = V 5k +V 4k+V 2k
These are expanded using Ohm's law. The current for the loop under 
consideration is deemed positive while any opposing (i.e., meshing) current 
is deemed negative:
230
Figure 7.20
Circuit for Example 7.6.",DCElectricalCircuitAnalysis.pdf
"Loop 1: 12 V = 1k(I 1 −I2)+2 k(I1 −I 2)
Loop 2:   0V = 3 k(I2)+5k(I 2 −I 3)+1k(I 2 −I 1)
Loop 3:   0V = 5 k(I 3 −I 2)+4k(I 3)+2k(I 3 −I1)
The terms are expanded:
Loop 1: 12 V = 1k I 1 −1 k I 2+2 k I 1 −2 k I 3
Loop 2:   0V = 3 k I 2 +5k I 2 −5k I3+1k I 2 −1 k I 1
Loop 3:   0V = 5 k I 3 −5k I 2+4k I 3+2k I 3 −2k I1
Like terms are grouped, once again creating aligned columns for
the current coefficients:
Loop 1: 12 V = (1k+2k)I 1 −1 k I 2 −2 k I3
Loop 2:   0V = −1k I 1+(3 k+5k+1 k) I2 −5k I 3
Loop 3:   0V = −2k I1 −5k I 2+(5k+4k+2k)I 3
Which simplifies to:
Loop 1: 12 V = 3k I1 −1k I2 −2 k I3
Loop 2:   0V = −1k I 1 +9 k I 2 −5k I 3
Loop 3:   0V = −2k I1 −5k I2 +11 k I 3
Note that we have diagonal symmetry. The solution is I1 = 5.729 mA,  
I2 = 1.6258 mA and I3 = 1.7806 mA. To find Vb we need to find the net 
current through the 2 kΩ resistor. That's I1 − I3, or 5.729 mA − 1.7806 mA, 
which is 3.9484 mA. When multiplied by 2 kΩ we find Vb = 7.8968 volts.",DCElectricalCircuitAnalysis.pdf
"current through the 2 kΩ resistor. That's I1 − I3, or 5.729 mA − 1.7806 mA, 
which is 3.9484 mA. When multiplied by 2 kΩ we find Vb = 7.8968 volts. 
Similarly, Vbc = (I3 − I2) 5 kΩ = (1.7806 mA − 1.6258 mA ) 5 kΩ, or 774 mV .
Computer Simulation Computer Simulation 
In order to verify the results of the analysis, the circuit is entered into a simulator 
and a virtual voltmeter is placed across the 5 kΩ resistor. This is shown in Figure 
7.22. The results agree nicely with the original analysis.
As nice as this is, in a practical circuit we need to be concerned about the effects of 
component tolerance. It would be extraordinarily odd if during production each 
resistor was equal to precisely the nominal value specified. In reality, each resistor 
will have a stated tolerance. Thus, with all five resistors changing slightly from unit 
to unit, we should expect the currents and voltages to vary as well. The question is,",DCElectricalCircuitAnalysis.pdf
"to unit, we should expect the currents and voltages to vary as well. The question is, 
what would be a typical spread? This can be addressed via a Monte Carlo analysis. 
This analysis is, in fact, a series of simulations. Each simulation uses randomized 
component values that fall within the stated tolerance spread. For example, if we 
specify that the 1 kΩ resistor has a 5% tolerance, then the actual resistor used for a 
231
Figure 7.21
Circuit of Example 7.6 with 
current loops and polarities 
added.",DCElectricalCircuitAnalysis.pdf
"simulation will be a randomly generated value between 950 Ω and 1050 Ω. We then 
state how many trials we'd like, each one using their own unique randomized values 
for each component.  
The results of a Monte Carlo analysis using the DC operating point simulation is 
shown in Figure 7.23. Ten randomized trials were generated for the value of Vb. 
Each of the five resistors was assigned a tolerance of 5%. As can be seen, voltage 
values both above and below the nominal simulation can result.
232
Figure 7.22
Circuit of Figure 7.20 in a 
simulator with a virtual 
voltmeter.",DCElectricalCircuitAnalysis.pdf
"Inspection MethodInspection Method
Like nodal analysis, there is a method to generate the equations by inspection. 
Simply focus on one loop, which we'll call the loop under inspection, and ask the 
following questions: what is the total source voltage in this loop? This yields the 
voltage constant for the equation. Then sum all of the resistance values in the loop 
under inspection. This yields the coefficient for that current term. For the remaining 
current coefficients in this equation, sum the resistances that are in common between
the loop under inspection and the other loops, respectively. These values will always
be negative. Repeat this process for all loops, with each loop in turn becoming the 
loop under inspection. As was the case with nodal analysis, the set of equations 
produced must exhibit diagonal symmetry, that is, if a diagonal is drawn from the 
upper left to the lower right through the IR pairs, then the coefficients found above",DCElectricalCircuitAnalysis.pdf
"upper left to the lower right through the IR pairs, then the coefficients found above 
the diagonal will have to match those found below the diagonal.
While it is possible to extend this technique to include current sources, it is often 
easier and less error-prone to convert the current sources into voltage sources and 
continue with the direct inspection method outlined above. Either way, it is 
important to remember that the number of loops determines the number of equations
to be solved. 
233
Figure 7.23
Monte Carlo analysis results 
for the circuit of Figure 7.20.",DCElectricalCircuitAnalysis.pdf
"Example 7.7
Find Vc for the circuit of Figure 7.24.  
First, we'll label the loops, as shown in Figure 7.25.
Now, beginning with loop 1, sum all of the voltage sources in this
loop. That's +50 volts. Secondly, sum all of the resistors in this
loop. That's 100 Ω + 200 Ω, or 300 Ω. This is the coefficient for
the I1 term. The coefficient for the I2 term is the sum of all
resistances that are in common between loops 1 and 2. That's just
the 200 Ω resistor. The coefficient for the I3 term would be the sum of all
resistances that are in common between loop 1 and loop 3. In this circuit,
there aren't any so the coefficient is zero. The first equation is: 
Loop 1: 50 V = 300I1 −200I2 −0I3
Repeat the process for loops 2 and 3:
Loop 2:   0V = −200I1 +900I2 −400I3
Loop 3: 20V = −0I1 −400I2 +900I3
Note that the 20 volt source shows up positive because of the direction of I3 
(into the negative terminal and out of the positive terminal).",DCElectricalCircuitAnalysis.pdf
"Note that the 20 volt source shows up positive because of the direction of I3 
(into the negative terminal and out of the positive terminal). 
As required, we have diagonal symmetry. The solution is I1 = 214.47 mA,  
I2 = 71.698 mA and I3 = 54.088 mA. Vc is the net current through the 400 Ω 
resistor times 400 Ω. The net current is I2 − I3, or 71.698 mA − 54.088 mA, 
which is 17.61 mA. When multiplied by 400 Ω we find Vb = 7.044 volts. 
Note that we use I2 − I3 and not I3 − I2. The reason is because we are trying 
to find the voltage from node c to ground, and I2 is the current flowing in 
that direction. In contrast, using I3 − I2 would yield the voltage from ground 
to node c; the same magnitude but opposite sign.
As a crosscheck, KVL states that Vc + V500 has to equal the −20 volt source. 
The drop across the 500 Ω is 54.088 mA times 500 Ω, or 27.044 volts from 
left to right. Starting at node c of 7.044 volts and then dropping 27.044 volts 
yields − 20 volts as expected. 
234",DCElectricalCircuitAnalysis.pdf
"left to right. Starting at node c of 7.044 volts and then dropping 27.044 volts 
yields − 20 volts as expected. 
234
Figure 7.24
Circuit for Example 7.7.
Figure 7.25
Circuit of Example 7.7 with 
current loops identified.",DCElectricalCircuitAnalysis.pdf
"SupermeshSupermesh
Sometimes you may run across a current source which has no associated internal 
resistance, such as found in the circuit of Figure 7.26. This is similar to the situation 
seen previously under nodal analysis where a voltage source does not have a 
specified internal resistance. There are two ways of solving this predicament. The 
first is add a very large resistance in parallel with the current source and then 
perform a source conversion on the pair so that the inspection method of mesh can 
be followed. The larger the value of this resistor, the more accuracy that is obtained. 
As a general rule it should be, at minimum, at least a couple of orders of magnitude 
larger than any surrounding resistor, and preferably larger. The second method is to 
use supermesh. A supermesh is a larger mesh loop than contains other mesh loops 
inside of it.
Consider the circuit shown in Figure 7.26. In the center we have a current source, Is,",DCElectricalCircuitAnalysis.pdf
"inside of it.
Consider the circuit shown in Figure 7.26. In the center we have a current source, Is, 
which lacks an associated internal resistance. Two traditional mesh loops, I1 and I2, 
are labeled as usual. The problem here is that we cannot use an Ohm's law based IR 
voltage drop for Vb. We have no way to express this as the voltage across Is is an 
unknown. On the other hand, what we do know is that Is must equal the combination
of the traditional mesh currents I1 and I2. That is, from the perspective of the first 
loop, Is = I1 − I2. Remember, one or both of the mesh currents could be negative, and 
thus rotating counterclockwise.
At this point we invoke the idea of a supermesh loop. First, we replace the offending
current source with its ideal internal resistance (an open). The supermesh loop is 
drawn which encompasses the original two loops. This is shown in Figure 7.27. The 
supermesh loop is drawn in red and labeled.
235
Figure 7.26
Circuit for supermesh.
Figure 7.27",DCElectricalCircuitAnalysis.pdf
"supermesh loop is drawn in red and labeled.
235
Figure 7.26
Circuit for supermesh.
Figure 7.27
Supermesh labeled.",DCElectricalCircuitAnalysis.pdf
"We now perform a KVL summation around the supermesh loop, similar to what we 
have done previously. The difference this time is that we need to recognize that 
components each see one of the original mesh currents; namely I1 or I2 in this case. 
We do not solve for a supermesh current, we simply use the supermesh to define the 
loop for the KVL summation. The summation follows:
ΣV rises = ΣV drops
 
E1 = V R1 +V R2 +E2
The voltage drops across the resistors can be expanded using Ohm's law, using the 
original mesh current associated with each resistor.
E1 −E 2 = I 1 R1+I2 R2
 
By inspection, 
I s = I 1 −I 2   or
I 2 = I 1 −I s
We now have two equations with two unknowns and can solve for I1 and I2. This 
procedure is illustrated in the following example.
Example 7.8
Find Vb for the circuit of Figure 7.28.  
First, we'll label the loops, as shown in Figure 7.29.
Next, we'll perform a KVL summation around the supermesh
loop.
ΣV rises = ΣV drops
 
20 V = V R1 +V R2+12V",DCElectricalCircuitAnalysis.pdf
"Next, we'll perform a KVL summation around the supermesh
loop.
ΣV rises = ΣV drops
 
20 V = V R1 +V R2+12V
Expand using Ohm's law and rearrange:
20 V −12 V = 5ΩI1+8ΩI 2
8V = 5ΩI 1+8ΩI 2
236
Figure 7.28
Circuit for Example 7.8.",DCElectricalCircuitAnalysis.pdf
"By inspection we know
3A = I2 −I 1   or
I 2 = I 1+3A
We can substitute this expression into the prior supermesh
expression and solve for I1: 
8 V = 5ΩI 1+8ΩI 2
8 V = 5ΩI 1+8Ω(I1 +3A)
8 V = 5ΩI 1+8ΩI1 +24 V
−16 V = 13Ω I 1
 
I 1 ≈−1.231A
Thus, I2 = −1.231 A + 3 A, or 1.769 A. To determine Vb we simply subtract 
the drop across the 5 Ω resistor from the 20 volt source. 
V b = 20 V − I 1 5Ω
V b = 20 V − (−1.231 A)5Ω
V b ≈ 26.15 V
As a crosscheck, we could also add the voltage across the 8 Ω resistor to the 
12 volt source:
V b = 12V +I2 8Ω
V b = 12V +1.769 A 8Ω
V b ≈26.15 V
Comparison of Nodal and MeshComparison of Nodal and Mesh
Compared to nodal analysis, mesh analysis has the advantage of dealing with 
resistances rather than conductances when writing the system of equations. Further, 
the mesh inspection method works with voltage sources which tends to be 
convenient for many circuits, while the nodal inspection method requires current",DCElectricalCircuitAnalysis.pdf
"convenient for many circuits, while the nodal inspection method requires current 
sources. On the down side, the resulting set of mesh currents requires further 
processing in order to find either branch currents or node voltages, while nodal 
analysis produces node voltages directly with no further processing. Mesh also has 
the disadvantage of being limited to planar circuits while there are no such limits to 
nodal. 
237
Figure 7.29
Circuit of Figure 7.28 with 
supermesh labeled.",DCElectricalCircuitAnalysis.pdf
"7.4 Dependent Sources7.4 Dependent Sources
A dependent source is a current or voltage source whose value is not fixed (i.e., 
independent) but rather which depends on some other circuit current or voltage. The 
general form for the value of a dependent source is Y = kX where X and Y are 
currents and/or voltages and k is the proportionality factor. For example, the value of
a dependent voltage source may be a function of a current, so instead of the source 
being equal to, say, 10 volts, it could be equal to twenty times the current passing 
through a particular resistor, or V=20I. 
There are four possible dependent sources: the voltage-controlled voltage source 
(VCVS), the current-controlled voltage source (CCVS), the voltage-controlled current
source (VCCS), and the current-controlled current source (CCCS). The source and 
control parameters are the same for both the VCVS and the CCCS so k is unitless",DCElectricalCircuitAnalysis.pdf
"control parameters are the same for both the VCVS and the CCCS so k is unitless 
(although it may be given as volts/volt and amps/amp, respectively). For the VCCS 
and CCVS, k has units of amps/volt and volts/amp, respectively. These are referred to
as the transresistance and transconductance of the sources with units of ohms and 
siemens.
The schematic symbols for dependent or controlled sources are usually drawn using 
a diamond. Also, there may be a secondary connection for the controlling current or 
voltage. Examples of a voltage-controlled voltage source, current-controlled voltage 
source, voltage-controlled current source and a current-controlled current source are 
shown in Figure 7.30, left-to-right. On each of these symbols, the control element is 
shown to the left of the source. This portion is not always drawn on a schematic. 
Instead, the source simply may be labeled as a function, for example, V = 0.02 IX 
where IX is the controlling current.",DCElectricalCircuitAnalysis.pdf
"Instead, the source simply may be labeled as a function, for example, V = 0.02 IX 
where IX is the controlling current.
Dependent sources are not “off-the-shelf” items in the same way that a battery is. 
Rather, dependent sources usually are used to model the behavior of more complex 
devices. For example, a bipolar junction transistor commonly is modeled as a CCCS 
while a field effect transistor may be modeled as a VCCS18. Similarly, many op amp 
circuits are modeled as VCVS systems. Solutions for circuits using dependent 
sources follow along the lines of those established for independent sources (i.e., the 
application of Ohm's law, KVL, KCL, etc.), however, the sources are now dependent 
on the remainder of the circuit which tends to complicate the analysis. 
18 For details, see Fiore, J.M., Semiconductor Devices: Theory and Application, and 
Operational Amplifiers and Linear Integrated Circuits: Theory and Application, both free 
OER titles.
238
Figure 7.30",DCElectricalCircuitAnalysis.pdf
"Operational Amplifiers and Linear Integrated Circuits: Theory and Application, both free 
OER titles.
238
Figure 7.30
Dependent sources (L to R): 
VCVS, CCVS, VCCS, CCCS.",DCElectricalCircuitAnalysis.pdf
"In general, there are two possible configurations: isolated and coupled. A example of
the isolated form is shown in Figure 7.31.
In this configuration, the dependent source (center) does not interact with the sub-
circuit on the left driven by the independent source, thus it can be analyzed as two 
separate circuits. Solutions for this form are relatively straightforward in that the 
control value for the dependent source can be computed directly. This value is then 
substituted into the dependent source and the analysis continues as is usual. 
Sometimes it is convenient if the solution for a particular voltage or current is 
defined in terms of the control parameter rather than as a specific value (e.g., the 
voltage across a particular resistor might be expressed as 8 VA instead of 12 volts, 
where VA is 1.5 volts). 
The second type of circuit (coupled) is somewhat more complex in that the 
dependent source can affect the parameter that controls the dependent source. In",DCElectricalCircuitAnalysis.pdf
"dependent source can affect the parameter that controls the dependent source. In 
other words, the dependent source(s) will contribute terms that include the 
controlling parameter(s), thus partly controlling itself. Some additional effort will be
in order to solve these circuits. To illustrate, consider the circuit of Figure 7.32. 
In this example it should be obvious that the current from the dependent source can 
affect the voltage at node a, and it is this very voltage that in turn sets up the value 
of the current source. Circuits of this type can be analyzed using mesh or nodal 
analysis. Nodal analysis works well here and is illustrated following.
239
Figure 7.31
Dependent source: isolated 
configuration.
Figure 7.32
Dependent source: coupled 
configuration.",DCElectricalCircuitAnalysis.pdf
"We begin by defining current directions. Assume that the currents through R1 and R3 
are flowing into node a, the current through R2 is flowing out of node a, and the 
current through R4 is flowing out of node b. We shall number the branch currents to 
reflect the associated resistor. The resulting KCL equations are:
Σ I in = Σ I out
 
Node a: I 1+I3 = I 2
Nodeb: k V a = I 3+I 4
The currents are then described by their Ohm's law equivalents:
Node a: E−V a
R1
+ V b−V a
R3
= V a
R2
Nodeb: k V a = V b−V a
R3
+ V b
R4
Expanding terms yields:
Node a: E
R1
−V a
R1
+V b
R3
−V a
R3
= V a
R2
Nodeb: k V a = V b
R3
−V a
R3
+V b
R4
Collecting terms and simplifying yields:
Node a : E
R1
= (
1
R1
+ 1
R2
+ 1
R3 )V a − 1
R3
V b
Nodeb:   0 =−(k+ 1
R3)V a + (
1
R3
+ 1
R4 )V b
Values for the resistors, k, and E normally are known, so the analysis proceeds 
directly.
Also, it is worth remembering that it is possible to perform source conversions on",DCElectricalCircuitAnalysis.pdf
"directly.
Also, it is worth remembering that it is possible to perform source conversions on 
dependent sources, within limits. The same procedure is followed as for independent
sources. The new source will remain a dependent source (e.g., converting VCVS to 
VCCS). This process is not applicable if the control parameter directly involves the 
internal impedance (i.e., is its voltage or current).
Time for an example, this one utilizing a simplified model of a transistor amplifier.
240",DCElectricalCircuitAnalysis.pdf
"Example 7.9
Find Vb and Vc for the circuit of Figure 7.33. 
This CCCS would be typical of a simple model of a bipolar
junction transistor (nodes a, b, and c). Ideally, the output voltage, 
Vc, would be equal to the input voltage (1 V) times the resistor
ratio of 15 kΩ / 2 kΩ and inverted, or approximately −7.5 volts.
In reality, it usually comes up a little short of this. Let's see how
well this works out.
This circuit is a good candidate for nodal analysis using the
general method. Note that the points labeled a and b are the same
node so we can write just two KCL summations. Using the
current directions as draw on the schematic, for node b we have:
Σ I in = Σ I out
 
I x +100 I x = V b
2 kΩ
1011V−V b
10 k Ω = V b
2 kΩ
10.1 mA = (
1
2 kΩ + 101
10 kΩ )V b
10.1 mA = 10.6 mSV b
 
V b = 0.95283 V
For node c we have:
Σ I in = Σ Iout
 
− V c
15 kΩ = 100 I x
− V c
15 kΩ = 100 1V−V b
10 k Ω
 
V c = −7.0755 V
Alternately, instead of writing the second KCL summation we could have",DCElectricalCircuitAnalysis.pdf
"− V c
15 kΩ = 100 I x
− V c
15 kΩ = 100 1V−V b
10 k Ω
 
V c = −7.0755 V
Alternately, instead of writing the second KCL summation we could have 
used Vb to determine Ix, i.e., (1 − Vb)/10 kΩ. As the current through the        
15 k Ω resistor is 100Ix, we could then use Ohm's law to find Vc. In any case,
we do see that Vc is inverted and just shy of the 7.5 volt estimate.
241
Figure 7.33
Circuit for Example 7.9.",DCElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
For verification, the dependent source circuit of Example 7.9 is entered into a 
simulator as shown in Figure 7.34. A DC operating point analysis is run, the results 
of which are shown in Figure 7.35. The output voltage shows roughly −7.08 volts 
and a Vb of just under 1 volt, thus achieving excellent agreement with the manual 
calculation.
While this simulation is effective in this case, just using a dependent source to stand 
in for a transistor is rather limited. There are many other, perhaps more subtle, 
elements to a proper transistor model that will ensure both greater accuracy and 
proper results under a wide range of operating conditions. Any quality simulator will
include a parts library using simulation models that are fine-tuned to particular 
manufacturer part numbers.
242
Figure 7.34
Circuit of Figure 7.33 in a 
simulator.
Figure 7.35
Simulation results for the 
circuit of Figure 7.33.",DCElectricalCircuitAnalysis.pdf
"7.5 Summary7.5 Summary
Nodal analysis can be used to solve virtually any complex multi-source DC electrical
circuit. It is based on KCL, writing expressions involving each node in the circuit. A 
system of equations results, there being as many equations as there are nodes in the 
circuit, minus the reference node (which usually is taken as ground). The set of 
equations will exhibit diagonal symmetry, which can be used as a crosscheck before 
setting out to solve them. The solution will be a complete set of node voltages. From
these, any branch current may be determined as needed. 
There are two different methods of creating the system of equations. The first 
method is deemed the general method and will work for a mix of current sources and
voltage sources. Individual currents are defined based on the node voltages and any 
known current sources. KCL is then applied at each node, followed by simplification",DCElectricalCircuitAnalysis.pdf
"known current sources. KCL is then applied at each node, followed by simplification 
and combination of terms to arrive at the end equations. The second approach uses 
the inspection method. If the circuit contains only current sources (or if the voltage 
sources are converted), this method allows direct generation of the system of 
equations without the need for simplification and thus is less prone to error. 
Mesh analysis can be used to solve any planar complex multi-source DC electrical 
circuit. It is based on KVL, writing expressions involving each closed loop in the 
circuit. The loops are minimally sized and the set of loops must cover every 
component in the circuit. A system of equations results, there being as many 
equations as there are loops. As with nodal analysis, the set of equations will exhibit 
diagonal symmetry. The solution will be a complete set of mesh currents. From 
these, any node voltage may be determined.",DCElectricalCircuitAnalysis.pdf
"diagonal symmetry. The solution will be a complete set of mesh currents. From 
these, any node voltage may be determined. 
Like nodal, mesh offers two different methods of creating the system of equations. 
The general method will work for a mix of current sources and voltage sources. 
Individual loops are defined based on the meshing currents passing through each 
component. KVL is then applied around each loop, followed by simplification and 
combination of terms to arrive at the end equations. In contrast, if the circuit 
contains only voltage sources (or if the current sources are converted), then the 
inspection method may be used. This method allows direct generation of the system 
of equations and is faster and less error prone. 
Dependent sources are current or voltage sources whose value depends on the 
current or voltage developed in some other part of the circuit. There are four types: 
current controlled current source, current controlled voltage source, voltage",DCElectricalCircuitAnalysis.pdf
"current controlled current source, current controlled voltage source, voltage 
controlled current source and voltage controlled voltage source. These sources are 
used commonly to model the characteristics of active devices such bipolar and field 
effect transistors. Techniques for solution tend to be a bit more involved than when 
using constant sources, however, nodal analysis in particular tends to work well.
243",DCElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. Describe the practical differences between nodal analysis and mesh analysis.
2. What is diagonal symmetry? Of what use is it?
3. What are the differences between the general method and the inspection 
method of nodal analysis?
4. What are the differences between the general method and the inspection 
method of mesh analysis?
5. What is a supernode?
6. What is a supermesh?
7. Describe the concept of dependent sources and how they differ from 
independent or constant sources. 
7.6 Exercises7.6 Exercises
AnalysisAnalysis
1. Given the circuit in Figure 7.36, write the mesh loop equations. 
2. Using mesh analysis, determine the value of Vb for the circuit shown in 
Figure 7.36.
3. For the circuit shown in Figure 7.36, use mesh analysis to determine the 
current through the 1 kΩ resistor.
4. Given the circuit in Figure 7.37, write the mesh loop equations and the 
associated determinants. 
244
Figure 7.36
Figure 7.37",DCElectricalCircuitAnalysis.pdf
"5. Using mesh analysis, determine the value of Vb for the circuit shown in 
Figure 7.37.
6. For the circuit shown in Figure 7.37, use mesh analysis to determine the 
current through the 500 Ω resistor.
7. Given the circuit in Figure 7.38, write the mesh loop equations. 
8. Using mesh analysis, determine the value of Vb for the circuit shown in 
Figure 7.38.
9. For the circuit shown in Figure 7.38, use mesh analysis to determine the 
current through the 75 Ω resistor.
10. Given the circuit in Figure 7.39, write the mesh loop equations and the 
associated determinants. 
11. Using mesh analysis, determine the value of Vbc for the circuit shown in 
Figure 7.39.
12. For the circuit shown in Figure 7.39, use mesh analysis to determine the 
current through the 10 kΩ resistor.
245
Figure 7.38
Figure 7.39",DCElectricalCircuitAnalysis.pdf
"13. Given the circuit in Figure 7.40, write the mesh loop equations. 
14. Using mesh analysis, determine the value of Vac for the circuit shown in 
Figure 7.40.
15. For the circuit shown in Figure 7.40, use mesh analysis to determine the 
current through the 4 kΩ resistor.
16. Given the circuit in Figure 7.41, write the mesh loop equations and the 
associated determinants. 
17. Using mesh analysis, determine the value of Vc for the circuit shown in 
Figure 7.41.
18. For the circuit shown in Figure 7.41, use mesh analysis to determine the 
current through the 10 kΩ resistor.
19. Given the circuit in Figure 7.42, write the mesh loop equations. 
246
Figure 7.40
Figure 7.41
Figure 7.42",DCElectricalCircuitAnalysis.pdf
"20. Using mesh analysis, determine the value of Vbd for the circuit shown in 
Figure 7.42.
21. For the circuit shown in Figure 7.42, use mesh analysis to determine the 
current through the 500 Ω resistor.
22. Given the circuit in Figure 7.43, write the mesh loop equations. 
23. Using mesh analysis, determine the value of Vad for the circuit shown in 
Figure 7.43.
24. For the circuit shown in Figure 7.43, use mesh analysis to determine the 
current through the 3 kΩ resistor.
25. Given the circuit in Figure 7.44, write the mesh loop equations. 
247
Figure 7.44
Figure 7.43",DCElectricalCircuitAnalysis.pdf
"26. Using mesh analysis, determine the value of Ve for the circuit shown in 
Figure 7.44.
27. For the circuit shown in Figure 7.44, use mesh analysis to determine the 
current through the 9 kΩ resistor.
28. Given the circuit in Figure 7.45, write the mesh loop equations and the 
associated determinants. 
29. Using mesh analysis, determine the value of Vbc for the circuit shown in 
Figure 7.45.
30. Given the circuit shown in Figure 7.45, use mesh analysis to determine the 
current through the 600 Ω resistor.
31. Given the circuit in Figure 7.46, write the mesh loop equations. 
32. Using mesh analysis, determine the value of Vbc for the circuit shown in 
Figure 7.46.
33. Given the circuit shown in Figure 7.46, use mesh analysis to determine the 
current through the 800 Ω resistor.
248
Figure 7.45
Figure 7.46",DCElectricalCircuitAnalysis.pdf
"34. For the circuit in Figure 7.47, write the mesh loop equations. 
35. Using mesh analysis, determine the value of Va for the circuit shown in 
Figure 7.47.
36. For the circuit shown in Figure 7.47, use mesh analysis to determine the 
current through the 3 kΩ resistor.
37. Given the circuit in Figure 7.48, write the mesh loop equations. 
38. Using mesh analysis, determine the value of Vc for the circuit shown in 
Figure 7.48.
39. For the circuit shown in Figure 7.48, use mesh analysis to determine the 
current passing through the 8.5 kΩ resistor.
40. Given the circuit in Figure 7.49, write the mesh loop equations (consider 
using source conversion).
249
Figure 7.47
Figure 7.48
Figure 7.49",DCElectricalCircuitAnalysis.pdf
"41. Using mesh analysis, determine the value of Va for the circuit shown in 
Figure 7.49.
42. For the circuit shown in Figure 7.49, use mesh analysis to determine the 
current passing through the 30 kΩ resistor.
43. Given the circuit in Figure 7.50, write the mesh loop equations (consider 
using source conversion). 
44. Using mesh analysis, determine the value of Vb for the circuit shown in 
Figure 7.50.
45. For the circuit shown in Figure 7.50, use mesh analysis to determine the 
current through the 5 kΩ resistor.
46. Given the circuit in Figure 7.51, write the node equations. 
47. Using nodal analysis, determine the value of Vb for the circuit shown in 
Figure 7.51.
48. For the circuit shown in Figure 7.51, use nodal analysis to determine the 
current through the 3 kΩ resistor.
250
Figure 7.50
Figure 7.51",DCElectricalCircuitAnalysis.pdf
"49. Given the circuit in Figure 7.52, write the node equations. 
50. Using nodal analysis, determine the value of Vb for the circuit shown in 
Figure 7.52.
51. For the circuit shown in Figure 7.52, use nodal analysis to determine the 
current passing through the 12 kΩ resistor.
52. Given the circuit in Figure 7.53, write the node equations. 
53. Using nodal analysis, determine the value of Vba for the circuit shown in 
Figure 7.53.
54. For the circuit shown in Figure 7.53, use nodal analysis to determine the 
current passing through the 100 Ω resistor.
55. Given the circuit in Figure 7.54, write the node equations. 
251
Figure 7.52
Figure 7.53
Figure 7.54",DCElectricalCircuitAnalysis.pdf
"56. Using nodal analysis, determine the value of Vac for the circuit shown in 
Figure 7.54.
57. For the circuit shown in Figure 7.54, use nodal analysis to determine the 
current passing through the 20 kΩ resistor.
58. Given the circuit in Figure 7.55, write the node equations. 
59. Using nodal analysis, determine the value of Vb for the circuit shown in 
Figure 7.55.
60. For the circuit shown in Figure 7.55, use nodal analysis to determine the 
current through the 100 Ω resistor.
61. Given the circuit in Figure 7.56, write the node equations. 
62. Using nodal analysis, determine the value of Vb for the circuit shown in 
Figure 7.56.
63. For the circuit shown in Figure 7.56, use nodal analysis to determine the 
current through the 40 Ω resistor.
252
Figure 7.55
Figure 7.56",DCElectricalCircuitAnalysis.pdf
"64. Given the circuit in Figure 7.48, write the node equations. 
65. Using nodal analysis, determine the value of Vd for the circuit shown in 
Figure 7.48.
66. For the circuit shown in Figure 7.48, use nodal analysis to determine the 
current passing through the 20 kΩ resistor.
67. Given the circuit in Figure 7.49, write the node equations using the general 
approach. Do not use source conversions. 
68. Using nodal analysis, determine the value of Vab for the circuit shown in 
Figure 7.49.
69. For the circuit shown in Figure 7.49, use nodal analysis to determine the 
current passing through the 30 kΩ resistor.
70. Given the circuit in Figure 7.50, write the node equations. 
71. Using nodal analysis, determine the value of Vb for the circuit shown in 
Figure 7.50.
72. For the circuit shown in Figure 7.50, use nodal analysis to determine the 
current through the 4 kΩ resistor.
73. For the circuit of Figure 7.57, determine Vb.
74. For the circuit of Figure 7.58, determine Vc.
253",DCElectricalCircuitAnalysis.pdf
"current through the 4 kΩ resistor.
73. For the circuit of Figure 7.57, determine Vb.
74. For the circuit of Figure 7.58, determine Vc.
253
Figure 7.57
Figure 7.58",DCElectricalCircuitAnalysis.pdf
"75. Given the circuit of Figure 7.59, determine Vb.
76. Find the current through the 10 kΩ resistor given the circuit of Figure 7.60.
77. Given the circuit of Figure 7.61, determine Vc.
78. In the circuit of Figure 7.62, determine Va.
254
Figure 7.59
Figure 7.60
Figure 7.61
Figure 7.62",DCElectricalCircuitAnalysis.pdf
"79. For the circuit of Figure 7.63, determine Va.
80. For the circuit of Figure 7.64, determine Va.
ChallengeChallenge
81. Given the circuit in Figure 7.43, write the node equations. 
82. Using nodal analysis, determine the value of Vb for the circuit shown in 
Figure 7.43.
83. For the circuit shown in Figure 7.43, use nodal analysis to determine the 
current through the 3 kΩ resistor.
84. Given the circuit in Figure 7.41, write the node equations. 
85. Using nodal analysis, determine the value of Vc for the circuit shown in 
Figure 7.41.
86. For the circuit shown in Figure 7.41, use nodal analysis to determine the 
current through the 8 kΩ resistor.
87. Given the circuit in Figure 7.45, write the node equations and the associated 
determinants. 
88. Using nodal analysis, determine the value of Vbc for the circuit shown in 
Figure 7.45.
255
Figure 7.63
Figure 7.64",DCElectricalCircuitAnalysis.pdf
"89. For the circuit shown in Figure 7.45, use nodal analysis to determine the 
current through the 2 kΩ resistor.
90. Given the circuit of Figure 7.65, determine Vc.
91. Find the current through the 10 kΩ resistor in the circuit of Figure 7.66.
92. Given the circuit of Figure 7.67, determine Vc.
256
Figure 7.65
Figure 7.66
Figure 7.67",DCElectricalCircuitAnalysis.pdf
"93. Given the circuit of Figure 7.68, determine the current through the 5 kΩ 
resistor.
94. For the circuit of Figure 7.69, determine Vb.
95. For the circuit of Figure 7.70, determine Vc.
257
Figure 7.68
Figure 7.69
Figure 7.70",DCElectricalCircuitAnalysis.pdf
"96. For the circuit of Figure 7.71, determine Va.
97. For the circuit of Figure 7.72, determine Vb.
SimulationSimulation
98. Perform a DC bias simulation on the circuit depicted in Figure 7.42 to verify
the component currents.
99. Perform a DC bias simulation on the circuit depicted in Figure 7.44 to verify
the component currents.
100. Perform a DC bias simulation on the circuit depicted in Figure 7.42 to 
verify the loop currents and node voltages.
101. Perform a DC bias simulation on the circuit depicted in Figure 7.51 to 
verify the node voltages.
102. Perform a DC bias simulation on the circuit depicted in Figure 7.54 to 
verify the node voltages.
103. Perform a DC bias simulation on the circuit depicted in Figure 7.55 to 
verify the node voltages.
258
Figure 7.71
Figure 7.72",DCElectricalCircuitAnalysis.pdf
"NotesNotes
     ♫♫     ♫♫
259",DCElectricalCircuitAnalysis.pdf
"8 8 CapacitorsCapacitors
8.0 Chapter Learning Objectives8.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe the theoretical and practical aspects of capacitor construction.
• Describe the current-voltage characteristic behavior of capacitors.
• Utilize component data sheets to determine operating characteristics of capacitors.
• Determine the initial and steady-state equivalents of resistor-capacitor networks.
• Determine the transient response of basic RC networks.
8.1 Introduction8.1 Introduction
This chapter introduces another passive device, the capacitor. Capacitors are fundamentally different from 
resistors in terms of both their construction and their operation. For starters, when placed in DC circuits, 
capacitors are not ohmic, unlike resistors. Their current-voltage characteristic does not respond to Ohm's law. 
Instead, their current-voltage characteristic is dynamic in nature. Further, in the ideal case, capacitors do not",DCElectricalCircuitAnalysis.pdf
"Instead, their current-voltage characteristic is dynamic in nature. Further, in the ideal case, capacitors do not 
dissipate power. Indeed, capacitors are energy storage devices. In a way, you could imagine them to be a little 
like rechargeable batteries, but unlike batteries these devices could be charged or discharged in tiny fractions of 
a second and last for billions upon billions of cycles. 
Capacitors are an integral part of modern electronic systems. They are used in AC-to-DC power supplies to help 
smooth and stabilize the output voltage. In audio and communications systems they are used in filters, for 
example to control the high and low frequency response of amplifiers and similar equipment, or for tuning 
purposes. Essentially, in any application that needs to smooth out a varying voltage, store electric charge or filter
a signal; a capacitor is likely to be used.",DCElectricalCircuitAnalysis.pdf
"a signal; a capacitor is likely to be used.
Much of the original work on capacitors was done in the 18th century. One of the earliest examples of a capacitor
was the Leyden jar, which consists of a glass bottle or jar lined with metal foil on the inside and the outside. 
Benjamin Franklin's famous kite experiment made use of a Leyden jar. In the early days, capacitors were known
as condensers (not to be confused with the condensers used in air conditioning systems).
260",DCElectricalCircuitAnalysis.pdf
"8.2 Capacitance and Capacitors8.2 Capacitance and Capacitors
A capacitor is a device that stores energy. Capacitors store energy in the form of an 
electric field. At its most simple, a capacitor can be little more than a pair of metal 
plates separated by air. As this constitutes an open circuit, DC current will not flow 
through a capacitor. If this simple device is connected to a DC voltage source, as 
shown in Figure 8.1, negative charge will build up on the bottom plate while 
positive charge builds up on the top plate. This process will continue until the 
voltage across the capacitor is equal to that of the voltage source. In the process, a 
certain amount of electric charge will have accumulated on the plates.
The ability of this device to store charge with regard to the voltage appearing across 
it is called capacitance. Its symbol is C and it has units of farads (F), in honor of 
Michael Faraday, a 19th century English scientist who did early work in",DCElectricalCircuitAnalysis.pdf
"Michael Faraday, a 19th century English scientist who did early work in 
electromagnetism. By definition, if a total charge of 1 coulomb is associated with a 
potential of 1 volt across the plates, then the capacitance is 1 farad.
1 farad ≡ 1 coulomb / 1 volt (8.1)
or more generally,
C= Q
V (8.2)
Where
C is the capacitance in farads,
Q is the charge in coulombs,
V is the voltage in volts.
From Equation 8.2 we can see that, for any given voltage, the greater the 
capacitance, the greater the amount of charge that can be stored. We can also see 
that, given a certain size capacitor, the greater the voltage, the greater the charge that
is stored. These observations relate directly to the amount of energy that can be 
stored in a capacitor.
261
Figure 8.1
Basic capacitor with voltage 
source.+
-
+
-
+
-
+ -
+
-
+
-+
-
electron flow
-
+",DCElectricalCircuitAnalysis.pdf
"Unsurprisingly, the energy stored in capacitor is proportional to the capacitance. It is
also proportional to the square of the voltage across the capacitor.
W =1
2 CV 2
(8.3)
Where
W is the energy in joules,
C is the capacitance in farads,
V is the voltage in volts.
The basic capacitor consists of two conducting plates separated by an insulator, or 
dielectric. This material can be air or made from a variety of different materials such
as plastics and ceramics. This is depicted in Figure 8.2.
For practical capacitors, the plates may be stacked alternately or even made of foil 
and formed into a rolled tube. However it is constructed, the characteristics of the 
dielectric will play a major role in the performance of the device, as we shall see.
In general, capacitance increases directly with plate area, A, and inversely with plate 
separation distance, d. Further, it is also proportional to a physical characteristic of",DCElectricalCircuitAnalysis.pdf
"separation distance, d. Further, it is also proportional to a physical characteristic of 
the dielectric; the permittivity, ε. Thus, capacitance is equal to:
C=ε A
d (8.4)
Where
C is the capacitance in farads,
A is the plate area in square meters,
d is the plate separation distance in meters,
ε is permittivity of the dielectric between the plates.
It should be noted that the effective plate area is somewhat larger than the precise 
physical area of the plates. This is due to a phenomenon called fringing. Essentially, 
the electric field lines bulge outward at the plate edges rather than maintain uniform 
parallel orientation. This is illustrated in Figure 8.3
262
l
w
d
Plate 1
Plate 2
Dielectric
Figure 8.2
Components of a generic 
capacitor.",DCElectricalCircuitAnalysis.pdf
"From Equation 8.4 it is obvious that the permittivity of the dielectric plays a major
role in determining the volumetric efficiency of the capacitor, in other words, the
amount of capacitance that can be packed into a given sized component. Some
dielectrics are notably more efficient than others. To make comparisons easier, 
relative permittivity is often used, that is, the ratio of the dielectric's permittivity to 
that of a vacuum, ε0. 
A table of relative permittivity for a variety of dielectrics is shown in Figure 8.4. A 
number of common dielectrics, such as various poly plastic films and mica, exhibit 
permittivities two to six times that of air, but there are also ceramic dielectrics whose
dielectrics are hundreds to thousands of times that of air. 
Material Relative permittivity, εr = ε/ε0
Vacuum 1 (ε0=8.85E−12 farads/meter)
Air 1.00058986 (at STP)
PTFE/Teflon 2.1 
Polyethylene/XLPE 2.25 
Polyimide 3.4 
Polypropylene 2.2–2.36 
Polystyrene 2.4–2.7 
Polyester (Mylar) 3.1",DCElectricalCircuitAnalysis.pdf
"Air 1.00058986 (at STP)
PTFE/Teflon 2.1 
Polyethylene/XLPE 2.25 
Polyimide 3.4 
Polypropylene 2.2–2.36 
Polystyrene 2.4–2.7 
Polyester (Mylar) 3.1
Paper 1.4
Mica 3–6
Silicon dioxide 3.9
Rubber 7 
Diamond 5.5–10 
Silicon 11.68 
Titanium dioxide 86–173 
Strontium titanate 310 
Calcium copper titanate >250,000
At quick glance, it might seem that choosing the dielectric with the highest 
permittivity would be the best choice but this is not necessarily the case. There are 
several other factors that go into this decision including temperature stability, 
leakage resistance (effective parallel resistance), ESR (equivalent series resistance) 
and breakdown strength. For an ideal capacitor, leakage resistance would be infinite 
and ESR would be zero. 
Unlike resistors, capacitors do not have maximum power dissipation ratings. Instead,
they have maximum voltage ratings. The breakdown strength of the dielectric will 
263
+ + + + + + + +
- - - - - - - -
Figure 8.3
Capacitor electric field with",DCElectricalCircuitAnalysis.pdf
"263
+ + + + + + + +
- - - - - - - -
Figure 8.3
Capacitor electric field with 
fringing.
Figure 8.4
Relative permittivity of various 
dielectrics.
Data derived from Wikipedia and other 
sources.",DCElectricalCircuitAnalysis.pdf
"set an upper limit on how large of a voltage may be placed across a capacitor before 
it is damaged. Breakdown strength is measured in volts per unit distance, thus, the 
closer the plates, the less voltage the capacitor can withstand. For example, halving 
the plate distance doubles the capacitance but also halves its voltage rating. Figure 
8.5 lists the breakdown strengths of a variety of different dielectrics. Comparing the 
tables of Figures 8.4 and 8.5 hints at the complexity of the situation. For instance, 
consider polystyrene versus polypropylene. Polystyrene offers modestly increased 
permittivity yet polypropylene has a considerable advantage in terms of breakdown 
strength. As a consequence, the plates can be placed much closer together when 
using polypropylene while achieving the same voltage rating as a capacitor using 
polystyrene. Therefore, the polypropylene capacitor will require less volume for the",DCElectricalCircuitAnalysis.pdf
"polystyrene. Therefore, the polypropylene capacitor will require less volume for the 
same capacitance. As an added benefit, polypropylene exhibits high temperature 
stability and low moisture absorption, among other characteristics. Comparing 
polypropylene to polyester, we find that polyester's improved permittivity along with
a similar breakdown strength yields improved volumetric efficiency over 
polypropylene. Unfortunately, polyester suffers from greater temperature 
dependence.
Substance Breakdown Strength
(kV/mm)
Air 3.0 
Borosilicate glass 20–40 
PTFE (Teflon, insulating film) 60–173 
Polyethylene 19–160 
Polypropylene 650
Polystyrene 19.7 
PEEK (Polyether ether ketone) 23 
Polyester (Mylar) 580
Neoprene rubber 15.7–26.7 
Distilled water 65–70 
Waxed paper 40–60 
Mica 118 
Diamond 2,000 
PZT (ceramic) 10–25
Capacitor Styles and PackagingCapacitor Styles and Packaging
Capacitors are available in a wide range of capacitance values, from just a few",DCElectricalCircuitAnalysis.pdf
"Capacitor Styles and PackagingCapacitor Styles and Packaging
Capacitors are available in a wide range of capacitance values, from just a few 
picofarads to well in excess of a farad, a range of over 1012. Unlike resistors, whose 
physical size relates to their power rating and not their resistance value, the physical 
size of a capacitor is related to both its capacitance and its voltage rating (a 
264
Figure 8.5
Dielectric strength of various 
dielectrics.
Data derived from Wikipedia and other 
sources.c",DCElectricalCircuitAnalysis.pdf
"consequence of Equation 8.4. Modest surface mount capacitors can be quite small 
while the power supply filter capacitors commonly used in consumer electronics 
devices such as an audio amplifier can be considerably larger than a D cell battery. A
sampling of capacitors is shown in Figure 8.6.
Toward the front and left side of the photo are a variety of plastic film capacitors.
The disk-shaped capacitor uses a ceramic dielectric. The small square device
toward the front is a surface mount capacitor, and to its right is a teardrop-shaped
tantalum capacitor, commonly used for power supply bypass applications in
electronic circuits. The medium sized capacitor to the right with folded leads is a 
paper capacitor, at one time very popular in audio circuitry. A number of capacitors 
have a crimp ring at one side, including the large device with screw terminals. These
are aluminum electrolytic capacitors. These devices tend to exhibit high volumetric",DCElectricalCircuitAnalysis.pdf
"are aluminum electrolytic capacitors. These devices tend to exhibit high volumetric 
efficiency but generally do not offer top performance in other areas such as absolute 
accuracy and leakage current. They usually are polarized, meaning that the leads 
must match the polarity of the applied voltage. Inserting them into a circuit 
backwards can result in catastrophic failure. The polarity is usually identified by a 
series of minus signs and/or a stripe that indicates the negative lead. Tantalum
capacitors are also polarized but are typically denoted with a plus sign next to the
positive lead. A variable capacitor used for tuning radios is shown in Figure 8.7.
One set of plates is fixed to the frame while an intersecting set of plates is affixed
to a shaft. Rotating the shaft changes the amount of plate area that overlaps, and
thus changes the capacitance.
For large capacitors, the capacitance value and voltage rating are usually printed",DCElectricalCircuitAnalysis.pdf
"thus changes the capacitance.
For large capacitors, the capacitance value and voltage rating are usually printed 
directly on the case. Some capacitors use “MFD” which stands for “microfarads”. 
While a capacitor color code exists, rather like the resistor color code, it has 
generally fallen out of favor. For smaller capacitors a numeric code is used that 
echoes the color code. Typically it consists of a three digit number such as “152”. 
265
Figure 8.6
A variety of capacitor styles 
and packages.
Figure 8.7
A variable capacitor.
Figure 8.8
Capacitor schematic symbols 
(top-bottom): non-polarized, 
polarized, variable.",DCElectricalCircuitAnalysis.pdf
"The first two digits are the precision portion and the third digit is the power of ten 
multiplier. The result is in picofarads. Thus, 152 is 1500 pf. 
The schematic symbols for capacitors are shown in Figure 8.8. There are three
symbols in wide use. The first symbol, using two parallel lines to echo the two
plates, is for standard non-polarized capacitors. The second symbol represents
polarized capacitors. In this variant, the positive lead is drawn with a straight line
for that plate and often denoted with a plus sign. The negative terminal is drawn
with a curved line. The third symbol is used for variable capacitors and is drawn 
with an arrow through it, rather like a rheostat. 
In order to obtain accurate measurements of capacitors, an LCR meter, such as the 
one shown in Figure 8.9, may be used. These devices are designed to measure the 
three common passive electrical components: resistors, capacitors and inductors19.",DCElectricalCircuitAnalysis.pdf
"three common passive electrical components: resistors, capacitors and inductors19. 
Unlike a simple digital multimeter, an LCR meter can also measure the values at 
various AC frequencies instead of just DC, and also determine secondary 
characteristics such as equivalent series resistance and effective parallel leakage 
resistance.
Capacitor Data SheetCapacitor Data Sheet
A portion of a typical capacitor data sheet is shown in Figure 8.10. This is for a 
series of through-hole style metallized film capacitors using polypropylene for the 
dielectric. First we see a listing of general features. For starters, we find that the 
capacitors use a flame retardant epoxy coating and are also RoHS compliant. We 
then move to a set of electrical performance specifications. For example, we see that
this series is available in two variants, one rated at 800 volts DC and the other rated 
at 1600 volts DC. Further, tolerance is available as either ±3% or ±5%. Dissipation",DCElectricalCircuitAnalysis.pdf
"at 1600 volts DC. Further, tolerance is available as either ±3% or ±5%. Dissipation 
factor (tan δ) is a measure of particular importance for AC operation and is 
proportional to the ESR (equivalent series resistance, ideally 0), smaller being better. 
The insulation resistance indicates the value of an effective parallel leakage 
resistance (higher is better), here, some 30,000 MΩ. Finally, we see physical size 
data, essential for printed circuit board layouts.
Capacitors in Series and in ParallelCapacitors in Series and in Parallel
Multiple capacitors placed in series and/or parallel do not behave in the same 
manner as resistors. Placing capacitors in parallel increases overall plate area, and 
thus increases capacitance, as indicated by Equation 8.4. Therefore capacitors in 
parallel add in value, behaving like resistors in series. In contrast, when capacitors 
are placed in series, it is as if the plate distance has increased, thus decreasing",DCElectricalCircuitAnalysis.pdf
"are placed in series, it is as if the plate distance has increased, thus decreasing 
capacitance. Therefore capacitors in series behave like resistors in parallel. Their  
value is found via the reciprocal of summed reciprocals or the product-sum rule.
19 Inductors are the subject of the next chapter.
266
Figure 8.9
An LCR meter, designed to 
read capacitance, resistance 
and inductance.",DCElectricalCircuitAnalysis.pdf
"Example 8.1
Find the equivalent capacitance of the network shown in Figure 8.11.  
These capacitors are all in parallel, and thus, the equivalent value is the
sum of the three capacitances:
CTotal = C1+C2+C 3
CTotal = 1μ F+100 nF+560 nF
CTotal = 1.66μ F
267
Figure 8.11
Circuit for Example 8.1.
Figure 8.10
Capacitor data sheet.
Courtesy of Panasonic",DCElectricalCircuitAnalysis.pdf
"Example 8.2
Find the equivalent capacitance of the network shown in Figure 8.12.  
In this circuit, we find that the left and middle capacitors are in parallel.
This combination is in series with the capacitor to the right:
Cleft = C1+C 2
Cleft = 3.3μF+4.7μF
Cleft = 8μ F
 
CTotal = Cleft C3
Cleft +C3
CTotal = 8μ F 16μ F
8μ F+16μ F
CTotal ≈ 5.33μ F
If a circuit contains nothing but a voltage source in parallel with a group of
capacitors, the voltage will be the same across all of the capacitors, just as it is in a
resistive parallel circuit. If the circuit instead consists of multiple capacitors that
are in series with a voltage source, as shown in Figure 8.13, the voltage will divide 
between them in inverse proportion. In other words, the larger the capacitance, the 
smaller its share of the applied voltage. The voltages can also be found by first 
determining the series equivalent capacitance. The total charge may then be",DCElectricalCircuitAnalysis.pdf
"determining the series equivalent capacitance. The total charge may then be 
determined using the applied voltage. Finally, the individual voltages are computed 
from Equation 8.2, V = Q/C, where Q is the total charge and C is the capacitance of 
interest. This is illustrated in the following example.
Example 8.3
Find the voltages across the capacitors in Figure 8.14.  
The first step is to determine the total capacitance. As these are in series, we
can use the reciprocal rule:
CTotal = 1
1
C1
+ 1
C2
+ 1
C3
CTotal = 1
1
2μ F + 1
4μ F + 1
8μ F
CTotal ≈ 1.143μ F
268
Figure 8.12
Circuit for Example 8.2.
Figure 8.13
A simple capacitors-only 
series circuit.
Figure 8.14
Circuit for Example 8.3.",DCElectricalCircuitAnalysis.pdf
"From here we determine the total charge:
Q = V C
Q = 12 V1.143μ F
Q = 13.71μC
Charge is constant across all of the series capacitors, therefore:
V 2uF = Q
C
V 2uF = 13.71μC
2μ F
V 2uF = 6.855 V
V 4uF = Q
C
V 4uF = 13.71μC
4μ F
V 4uF = 3.427 V
V 8uF = Q
C
V 8uF = 13.71μC
8μ F
V 8uF = 1.714 V
The sum of the three voltages is 12 volts (within rounding error) and verifies
KVL as expected.
A Practical TipA Practical Tip
While it may be tempting to try, do not attempt to verify the operation of Example 
8.3 in the laboratory using a standard DMM. The reason is because the internal 
resistance of a typical digital voltmeter is many orders of magnitude lower than the 
leakage resistance of the capacitors. As a result, charge will be transferred to the 
meter, ruining the measurement. It would be akin to trying to measure the voltages 
across a string of resistors, each in excess of 100 MΩ, with a meter whose internal",DCElectricalCircuitAnalysis.pdf
"across a string of resistors, each in excess of 100 MΩ, with a meter whose internal 
resistance is 1 MΩ. The meter's resistance dominates the parallel combination and 
causes excessive loading which ruins the measurement. A special sort of voltmeter, 
an electrostatic voltmeter or electrometer, is needed for these types of measurements.
These are sometimes referred to as non-charge transfer meters.
269",DCElectricalCircuitAnalysis.pdf
"Current-Voltage RelationshipCurrent-Voltage Relationship
The fundamental current-voltage relationship of a capacitor is not the same as that of
resistors. Capacitors do not so much resist current; it is more productive to think in 
terms of them reacting to it. The current through a capacitor is equal to the 
capacitance times the rate of change of the capacitor voltage with respect to time 
(i.e., its slope). That is, the value of the voltage is not important, but rather how 
quickly the voltage is changing. Given a fixed voltage, the capacitor current is zero 
and thus the capacitor behaves like an open. If the voltage is changing rapidly, the 
current will be high and the capacitor behaves more like a short. Expressed as a 
formula:
i=C d v
d t (8.5)
Where
i is the current flowing through the capacitor,
C is the capacitance,
dv/dt is the rate of change of capacitor voltage with respect to time.
A particularly useful form of Equation 8.5 is:
d v
d t = i
C (8.6)",DCElectricalCircuitAnalysis.pdf
"dv/dt is the rate of change of capacitor voltage with respect to time.
A particularly useful form of Equation 8.5 is:
d v
d t = i
C (8.6)
An alternate way of looking at Equation 8.5 indicates that if a capacitor is fed by a
constant current source, the voltage will rise at a constant rate (dv/dt). It is
continuously depositing charge on the plates of the capacitor at a rate of I, which is
equivalent to Q/t. As long as the current is present, feeding the capacitor, the
voltage across the capacitor will continue to rise. A good analogy is if we had a pipe 
pouring water into a tank, with the tank's level continuing to rise. This process of 
depositing charge on the plates is referred to as charging the capacitor. For example,
considering the circuit in Figure 8.15, we see a current source feeding a single 
capacitor. If we were to plot the capacitor's voltage over time, we would see 
something like the graph of Figure 8.16.
270
Figure 8.15
Capacitor with current source.
V
t
Figure 8.16",DCElectricalCircuitAnalysis.pdf
"something like the graph of Figure 8.16.
270
Figure 8.15
Capacitor with current source.
V
t
Figure 8.16
Capacitor voltage versus time.",DCElectricalCircuitAnalysis.pdf
"As time progresses, the voltage across the capacitor increases with a positive 
polarity from top to bottom. With a theoretically perfect capacitor and source, this 
would continue forever, or until the current source was turned off. In reality, this line
would either begin to deflect horizontally as the source reached its limits, or the 
capacitor would fail once its breakdown voltage was reached. The slope of this line 
is dictated by the size of the current source and the capacitance. 
Example 8.4
Determine the rate of change of voltage across the capacitor in the circuit
of Figure 8.17. Also determine the capacitor's voltage 10 milliseconds after
power is switched on.
First, note the direction of the current source. This will produce a negative 
voltage across the capacitor from top to bottom. The rate of change is:
dv
dt = i
C
dv
dt =−5μ A
30 nF
dv
dt ≈−166.7 volts per second
Thus, for every second, the voltage rises another −166.7 volts. Assuming it",DCElectricalCircuitAnalysis.pdf
"dv
dt = i
C
dv
dt =−5μ A
30 nF
dv
dt ≈−166.7 volts per second
Thus, for every second, the voltage rises another −166.7 volts. Assuming it 
is completely uncharged when power is applied, after 10 milliseconds it will 
have risen to −166.7 V/s times 10 ms, or −1.667 volts. 
Equation 8.6 provides considerable insight into the behavior of capacitors. As just 
noted, if a capacitor is driven by a fixed current source, the voltage across it rises at 
the constant rate of i/C. There is a limit to how quickly the voltage across the 
capacitor can change. An instantaneous change means that dv/dt is infinite, and thus,
the current driving the capacitor would also have to be infinite (an impossibility). 
This is not an issue with resistors, which obey Ohm's law, but it is a limitation of 
capacitors. Therefore we can state a particularly important characteristic of 
capacitors:
The voltage across a capacitor cannot change instantaneously. (8.7)",DCElectricalCircuitAnalysis.pdf
"capacitors:
The voltage across a capacitor cannot change instantaneously. (8.7)
This observation will be key to understanding the operation of capacitors in DC 
circuits.
271
Figure 8.17
Circuit for Example 8.4.",DCElectricalCircuitAnalysis.pdf
"8.3 Initial and Steady-State Analysis of RC Circuits8.3 Initial and Steady-State Analysis of RC Circuits
When analyzing resistor-capacitor circuits, always remember that capacitor voltage 
cannot change instantaneously. If we assume that a capacitor in a circuit is not 
initially charged, then its voltage must be zero. The instant the circuit is energized, 
the capacitor voltage must still be zero. If there is no voltage across the device, then 
it is behaving like a short circuit. We call this the initial state. Thus, we have our 
first rule regarding RC circuits:
For DC analysis, initially capacitors appear as shorts. (8.8)
Consider the circuit of Figure 8.18. Assume that C1 and C2 are initially uncharged 
and there is no voltage across them. 
The instant power is applied, the two capacitors appear as short circuits. If we 
redraw the circuit for this instant in time, we arrive at the equivalent circuit shown in
Figure 8.19.",DCElectricalCircuitAnalysis.pdf
"redraw the circuit for this instant in time, we arrive at the equivalent circuit shown in
Figure 8.19. 
Given this equivalent, we can see that shorting C2 places R2 and R3 in parallel, 
however, they are both shorted out by C1. This leaves only R1 left in the circuit along
with the source, E. At this point, currents will begin to flow, and thus begin charging 
up the capacitors. As the capacitor voltages rise, the current will begin to decrease, 
and eventually the capacitors will stop charging. At that point no further current will 
be flowing, and thus the capacitor will behave like an open. We call this the steady-
state condition and we can state our second rule:
At steady-state, capacitors appear as opens. (8.9)
272
Figure 8.18
A basic resistor-capacitor (RC) 
circuit.
Figure 8.19
A basic RC circuit, initial state.",DCElectricalCircuitAnalysis.pdf
"Continuing with the example, at steady-state both capacitors behave as opens. This 
is shown in Figure 8.20. This leaves E to drop across R1 and R2. This will create a 
simple voltage divider. The steady-state voltage across C1 will equal that of R2. As 
C2 is also open, the voltage across R3 will be zero while the voltage across C2 will be
the same as that across R2.
In reality, practical capacitors can be thought of as an ideal capacitance in parallel 
with a very large (leakage) resistance, so there will be a limit to this performance.
Example 8.5
Given the circuit of Figure 8.21, find the voltage across the 6 kΩ resistor
for both the initial and steady-state conditions assuming the capacitor is
initially uncharged. 
For the initial state the capacitor is treated as a short. The initial state
equivalent circuit is drawn below in Figure 8.22. Immediately apparent
is the parallel connection between the 6 kΩ and 3 kΩ resistors. This",DCElectricalCircuitAnalysis.pdf
"equivalent circuit is drawn below in Figure 8.22. Immediately apparent
is the parallel connection between the 6 kΩ and 3 kΩ resistors. This
combination is equivalent to 2 kΩ. Therefore, we can perform a voltage
divider to find the potential across the 6 kΩ (i.e., the 2 kΩ combo).
V 6k = E Rx
Rx+Ry
V 6k = 24 V 2 kΩ
2 kΩ+1 kΩ
V 6k = 16 V
For the steady-state condition the capacitor will be fully charged, its current 
will be zero, and we treat it as an open. The steady-state equivalent circuit is 
273
Figure 8.22
Circuit of Figure 8.20, initial 
state.
Figure 8.20
A basic RC circuit, steady-state.
Figure 8.21
Circuit for Example 8.4.",DCElectricalCircuitAnalysis.pdf
"drawn below in Figure 8.23.
The 3 kΩ resistor is now out of the picture, leaving us with the 6 kΩ in 
series with the 1 kΩ resistor. Once again, a voltage divider may be used to 
determine the voltage across the 6 kΩ.
V 6k = E Rx
Rx+Ry
V 6k = 24 V 6k Ω
6k Ω+1 kΩ
V 6k = 20.57 V
8.4 Transient Response of RC Circuits8.4 Transient Response of RC Circuits
The question remains, “What happens between the time the circuit is powered up 
and when it reaches steady-state?” This is known as the transient response. Consider
the circuit shown in Figure 8.24. Note the use of a voltage source rather than a fixed 
current source, as examined earlier.
The key to the analysis is to remember that capacitor voltage cannot change
instantaneously. Assuming the capacitor is uncharged, the instant power is applied,
the capacitor voltage must be zero. Therefore all of the source voltage drops across
the resistor. This creates the initial current, and this current starts to charge the",DCElectricalCircuitAnalysis.pdf
"the resistor. This creates the initial current, and this current starts to charge the
capacitor (the initial rate being equal to i/C as dictated by Equation 8.6). According 
to Kirchhoff's voltage law, as the capacitor voltage begins to increase, the resistor 
voltage must decrease because the sum of the two must equal the fixed source 
voltage. This means that the circulating current must also decrease. This, in turn, 
means that the rate of capacitor voltage increase begins to slow. As the capacitor 
voltage continues to increase, less voltage is available for the resistor, causing 
further reductions in current, and a further slowing of the rate of capacitor voltage 
change. Eventually, the capacitor voltage will be nearly equal to the source voltage. 
This will result in a very small potential across the resistor and an equally small 
current, slowing subsequent capacitor voltage increases to a near standstill.",DCElectricalCircuitAnalysis.pdf
"current, slowing subsequent capacitor voltage increases to a near standstill. 
Theoretically, the capacitor voltage approaches the source voltage but never quite 
equals it. Similarly, the current drops to near zero, but never completely turns off. 
This is illustrated in Figure 8.25.
274
Figure 8.23
Circuit of Figure 8.20, steady-
state.
Figure 8.24
A simple RC circuit.",DCElectricalCircuitAnalysis.pdf
"The dashed red line represents the initial rate of change of capacitor voltage. This 
trajectory is what would be expected if an ideal current source drove the capacitor, 
as in Example 8.4. As noted previously, the rate of voltage change versus times is 
equal to i/C, and therefore in this case, E/RC. If the initial rate of change were to 
continue unabated, the source voltage, E, would be reached in RC seconds. 
Consequently, RC is referred to as the charge time constant and is denoted by τ 
(Greek letter tau). Thus,
Time constant, τ = RC (8.10)
As noted, once the capacitor begins to charge, the current begins to decrease and the 
capacitor voltage curve begins to fall away from the initial trajectory. The solid red 
curve represents the capacitor voltage. Notice that after five time constants the 
capacitor is nearly fully charged and the circuit is considered to be in steady-state 
(i.e., the capacitor behaves as an open).",DCElectricalCircuitAnalysis.pdf
"capacitor is nearly fully charged and the circuit is considered to be in steady-state 
(i.e., the capacitor behaves as an open). 
Steady-state is reached in approximately five time constants. (8.11)
275
Figure 8.25
Normalized charge and 
discharge curves.
Normalized Charge/Discharge Curves
Percent of Maximum
0
20
40
60
80
100
 
Time Constants
0 1 2 3 4 5",DCElectricalCircuitAnalysis.pdf
"This sort of recursively dependent operation is characteristic of exponential 
functions. The equation for the capacitor's voltage charging curve is:
V C (t ) = E (1 − ϵ−t
τ ) (8.12)
Where
VC(t) is the capacitor voltage at time t,
E is the source voltage,
t is the time of interest,
τ is the time constant,
ε (also written e) is the base of natural logarithms, approximately 2.718.
The dashed blue line shows the initial slope of current change. The solid blue curve 
shows the circulating current (and by extension of Ohm's law, the resistor voltage). 
The equation for this curve, which follows the general shape ε−t, is:
I (t )= E
R ϵ− t
τ (8.13)
and
V R(t) = E ϵ−t
τ  (8.14)
Once power is removed or bypassed, the stored charge on the capacitor will dissipate
through any associated resistor(s) creating a discharge current which will end with 
the capacitor voltage drained back to zero. During the discharge phase, both the",DCElectricalCircuitAnalysis.pdf
"the capacitor voltage drained back to zero. During the discharge phase, both the 
capacitor's voltage and current will follow the solid blue curve; Equations 8.13 and 
8.14 being appropriate. The discharge time constant may be different from the 
charge times constant, depending on the associated resistances.
A precise derivation of the exponential charge/discharge equations is given in 
Appendix C. 
Example 8.6
Given the circuit of Figure 8.26, assume the switch is closed at time t = 0.
Determine the charging time constant, the amount of time after the switch
is closed before the circuit reaches steady-state, and the capacitor voltage
at t = 0, t = 50 milliseconds and t = 1 second. Assume the capacitor is
initially uncharged. 
First, the time constant:
τ = RC
τ = 50 kΩ2μF
τ = 100 ms
276
Figure 8.26
Circuit for Example 8.6.",DCElectricalCircuitAnalysis.pdf
"Steady-state will be reached in five time constants, or 500 milliseconds. 
Therefore we know that VC(0) = 0 volts and VC(1) = 100 volts. To find 
VC(50 ms) we simply solve Equation 8.12.
V C (t ) = E (1 − ϵ−t
τ )
V C (50 ms) = 100 V(1 − ϵ
− 50 ms
100 ms)
V C (50 ms) ≈39.35 V
This value can also be determined graphically from Figure 8.25. The time of
50 milliseconds represents one-half time constant. Find this value on the 
horizontal axis and then track straight up to the solid red curve that 
represents the charging capacitor voltage. The point of intersection is at  
approximately 40% of the maximum value on the vertical axis. The 
maximum value here is the source voltage of 100 volts. Therefore the 
capacitor will have reached approximately 40% of 100 volts, or just about 
40 volts. 
Computer SimulationComputer Simulation
The circuit of Figure 8.26 is entered into a simulator, as shown in Figure 8.27. In",DCElectricalCircuitAnalysis.pdf
"40 volts. 
Computer SimulationComputer Simulation
The circuit of Figure 8.26 is entered into a simulator, as shown in Figure 8.27. In 
order to reflect the notion of a time-varying circuit with a switch, the 100 volt DC 
voltage source has been replaced with a rectangular pulse voltage source. This 
source starts at 0 volts and then immediately steps up to 100 volts. It stays at this 
level for 500 milliseconds before dropping back to 0 volts.
A transient (i.e., time domain) simulation is run, plotting the capacitor voltage over 
time for the first 500 milliseconds. This is seen in Figure 8.28. There are several 
items to note here. First, we see that the overall shape perfectly echoes the generic 
curve presented in Figure 8.25. Second, we see that after the predicted five time 
constants, or 500 milliseconds, the capacitor voltage has plateaued, indicating 
277
Figure 8.27
Circuit of Figure 8.26 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"steady-state. Third, at steady-state the capacitor voltage has virtually reached the 
maximum value set be the source, or 100 volts. Finally, at 50 milliseconds, we see 
that the capacitor voltage has reached roughly 40 volts, just as predicted.
As mentioned previously, it is possible for a circuit to have different charge and 
discharge time constants. This could occur if a switch introduces or removes 
component(s) between the charge versus discharge phases. Further, because the 
capacitor is discharging, its current direction will be opposite to that of the charging 
current. KVL still must be satisfied, but because the capacitor is now behaving as a 
source, the magnitude of the discharge resistance's voltage must equal the capacitor 
voltage magnitude. Therefore, its curve will take the same shape (the solid blue 
curve of Figure 8.25). These issues will be illustrated in the following example.
Example 8.7",DCElectricalCircuitAnalysis.pdf
"curve of Figure 8.25). These issues will be illustrated in the following example.
Example 8.7
For the circuit of Figure 8.29, assume the capacitor is initially uncharged.
At time t = 0 the switch contacts position 1. The switch is thrown to
position 2 at time t = 50 milliseconds.
Determine the charging time constant, the amount of time after the
switch is closed before the circuit reaches steady-state, the maximum
charging and discharging currents, and the capacitor voltage at t = 0, 
t = 50 milliseconds, t = 90 milliseconds, and t = 1 second. 
278
Figure 8.28
Simulation results for the 
circuit of Figure 8.25.
Figure 8.29
Circuit for Example 8.7.",DCElectricalCircuitAnalysis.pdf
"We begin with the charge time constant:
τcharge = RC
τcharge = 20 kΩ220 nF
τcharge = 4.4 ms
Steady-state will be reached in 5 times 4.4 milliseconds, or 22 milliseconds. 
The capacitor is initially uncharged, so VC(0) = 0 volts. As the capacitor will 
have reached steady-state in 22 milliseconds, VC(50 ms) = 12 volts.
The maximum charging current will occur at t = 0 when all of the 12 volt 
source drops across the 20 kΩ resistor, or 600 μamps, flowing left to right.
At 50 milliseconds the switch is thrown to position 2. The 12 volt source and
20 kΩ resistor are no longer engaged. At this point the capacitor has 12 volts
across it, positive to negative, top to bottom. As the capacitor voltage cannot
change instantaneously, the capacitor now acts as a voltage source and 
discharges through the 120 kΩ resistor. Note that the discharge current is 
flowing counterclockwise, the opposite of the charging current. The 
discharge time constant is:
τdischarge = RC
τdischarge = 120 kΩ220 nF",DCElectricalCircuitAnalysis.pdf
"flowing counterclockwise, the opposite of the charging current. The 
discharge time constant is:
τdischarge = RC
τdischarge = 120 kΩ220 nF
τdischarge = 26.4 ms
The capacitor will fully discharge down to 0 volts in 5 time constants, or 
some 132 milliseconds after the switch is thrown to position 2. Thus steady-
state occurs at t = 182 milliseconds. The maximum discharge current occurs 
the instant the switch is thrown to position 2 when all of the capacitor's 12 
volts drops across the 120 kΩ resistor, yielding 100 μamps, flowing top to 
bottom.
Clearly, at t = 90 milliseconds the capacitor is in the discharge phase. The 
capacitor's voltage and current during the discharge phase follow the solid 
blue curve of Figure 8.25. The elapsed time for discharge is 90 milliseconds 
minus 50 milliseconds, or 40 milliseconds net. We can use a slight variation 
on Equation 8.14 to find the capacitor voltage at this time.
V C (t) = E ϵ−t
τ
V C (40 ms) = 12 V− ϵ
− 40 ms
26.4 ms",DCElectricalCircuitAnalysis.pdf
"on Equation 8.14 to find the capacitor voltage at this time.
V C (t) = E ϵ−t
τ
V C (40 ms) = 12 V− ϵ
− 40 ms
26.4 ms
V C (40 ms) ≈2.637 V
The shape of the capacitor's voltage will appear somewhat like a rounded 
pulse, rising with a curve and then falling back to zero with a 
complementary curve (the red and then blue curves of Figure 8.25).
279",DCElectricalCircuitAnalysis.pdf
"Basic single resistor-capacitor circuits prove to be fairly easy to solve given a little 
practice, but what if a more complex circuit is used? In this situation the section 
feeding the capacitor may be simplified using Thévenin's theorem to determine the 
effective source voltage and charging resistance. The circuit then reverts back to a 
simple RC network which may be solved directly.
If power is interrupted before the capacitor is fully charged, the equations presented 
previously may be used to determine the precise voltage(s) and current(s) reached. 
The capacitor will then behave as a voltage source and begin to discharge, its 
voltage curve following the blue plot line of Figure 8.25, with its maximum voltage 
being what the capacitor charged to, not the associated driving voltage. The 
following example and simulations address these issues.
Example 8.8
For this example we shall revisit the circuit of Example 8.5. The circuit is",DCElectricalCircuitAnalysis.pdf
"following example and simulations address these issues.
Example 8.8
For this example we shall revisit the circuit of Example 8.5. The circuit is
redrawn in Figure 8.30 for convenience. Assume the capacitor is
initially uncharged. 
Determine the charging time constant, the amount of time after the
switch is closed before the circuit reaches steady-state, and the
capacitor voltage at t = 0, 100 milliseconds, and 200 milliseconds. At
200 milliseconds, the switch is opened. Determine how long it takes
for the capacitor to fully discharge and the voltage across the 6 kΩ
resistor at t = 275 milliseconds (i.e., 75 milliseconds after the switch is 
opened). 
The first step is to determine the Thévenin equivalent driving the capacitor. 
If we remove the capacitor and determine the open circuit voltage at those 
points, we see that it is just a voltage divider between the 24 volt source, the 
6 kΩ resistor and the 1 kΩ resistor (the 3 kΩ resistor has no current through",DCElectricalCircuitAnalysis.pdf
"6 kΩ resistor and the 1 kΩ resistor (the 3 kΩ resistor has no current through 
it and thus produces no voltage drop). This works out to 20.57 volts. The 
Thévenin resistance will be 3 kΩ in series with 1 kΩ || 6 kΩ, or roughly 
3.857 kΩ. The equivalent charging circuit is drawn in Figure 8.31.
We can now determine the charging time constant:
τcharge = RC
τcharge = 3.857 kΩ10μ F
τcharge = 38.57 ms
Steady-state will be reached in 5 time constants, or 192.8 ms. Thus, we
know that VC(0) = 0 volts and VC(200 ms) = 20.57 volts. For the capacitor
voltage at 100 milliseconds, we simply use the charge equation.
280
Figure 8.30
Circuit for Example 8.8.
Figure 8.31
Thévenin equivalent for the 
circuit of Figure 8.30 driving 
the capacitor.",DCElectricalCircuitAnalysis.pdf
"V C (t )= E (1−ϵ−t
τ )
V C (100 ms) = 20.57 V(1− ϵ
− 100 ms
38.57 ms )
V C (100 ms) ≈19.03 V
For the discharge phase, we need to determine the time constant. With the 
voltage source removed, the capacitor will discharge through the now series 
combination of the 3 kΩ resistor and 6 kΩ resistor.
τdischarge = RC
τdischarge = 9k Ω10μ F
τdischarge = 90 ms
Steady-state will be reached 450 milliseconds later at t = 650 milliseconds. 
To find V6k at t = 275 milliseconds, we can find the voltage across the 
capacitor and then perform a voltage divider between the 6 kΩ and 3 kΩ 
resistors. Remembering that t = 275 milliseconds is 75 milliseconds into the 
discharge phase, we have:
V C (t ) = E ϵ− t
τ
V C (75 ms) = 20.57 V(1− ϵ
−75 ms
90 ms)
V C (75 ms) ≈8.94 V
Finally, this voltage splits between the 6 kΩ and 3 kΩ resistors. Using the 
voltage divider rule, we find:
V 6k = V C
Rx
Rx+Ry
V 6k = 8.94 V 6 kΩ
6 kΩ+3kΩ
V 6k = 5.96 V
Computer SimulationComputer Simulation",DCElectricalCircuitAnalysis.pdf
"voltage divider rule, we find:
V 6k = V C
Rx
Rx+Ry
V 6k = 8.94 V 6 kΩ
6 kΩ+3kΩ
V 6k = 5.96 V
Computer SimulationComputer Simulation
In order to verify the analysis of Example 8.8, the circuit of Figure 8.30 is entered 
into a simulator, as shown in Figure 8.32. In place of a DC source, a pulse generator 
is used to mimic the on-off nature of the switch. This starts starts at zero volts and 
then immediately jumps up to 24 volts. It stays at this level for 200 milliseconds 
before returning back to zero. This is sufficient time to check whether or not the 
capacitor voltage has reached steady-state (predicted to take 192.8 milliseconds).
281",DCElectricalCircuitAnalysis.pdf
"A transient analysis is run on this circuit, plotting the capacitor voltage (i.e., the 
difference between the node 2 and node 3 voltages). The result is shown in Figure 
8.33. This plot confirms nicely the charge phase of the capacitor. After 
approximately 200 milliseconds, the voltage has leveled out at just over 20 volts, 
precisely as predicted. 
What about the discharge phase? In order to investigate that portion, the simulation 
circuit is modified. The pulse voltage source is disconnected from the remainder of 
the circuit, just as it would be if the switch in Figure 8.30 had been opened again. 
Also, the capacitor is modified to have an initial voltage of 20.57 volts, the precise 
value at had reached after it attained steady-state. A second transient analysis is run, 
282
Figure 8.32
Circuit of Figure 8.30 in a 
simulator.
Figure 8.33
Simulation results for the 
charge phase of the circuit of 
Figure 8.30.",DCElectricalCircuitAnalysis.pdf
"again plotting the capacitor voltage. The results of this simulation are shown in 
Figure 8.34. It is worth noting that the time axis is relative to the switch being 
opened, not the original timing. That is, the horizontal origin of 0 milliseconds 
corresponds to t = 200 milliseconds.
Note that the time required to reach the new steady-state value of zero volts has 
stretched out to some 450 milliseconds after the switch is opened, precisely as 
predicted. 
At this point, a fair question to ask is, “Couldn't we leave the pulse source in place in
order to investigate the discharge phase?” Although the pulse source does go back 
down to zero volts at t = 200 milliseconds, that's not the same as opening the switch 
back in Figure 8.30. If the source is still connected but producing zero volts, it 
becomes part of the Thévenin equivalent. As such, the 1 kΩ resistor is back in the 
circuit, producing a discharge time constant identical to the charge time constant.",DCElectricalCircuitAnalysis.pdf
"circuit, producing a discharge time constant identical to the charge time constant.
To prove the point, the simulation is run again. The pulse remains at 24 volts for 200
milliseconds and then jumps down to zero, as before. The simulation time is 
extended out to 400 milliseconds total, enough to see both the charge and discharge 
phases. Further, instead of just plotting the capacitor voltage; the voltages of the 
source (blue trace, node 1), the 6 kΩ resistor (green trace, node 2) and the 3 kΩ 
resistor (red trace, node 5) are plotted. The results are shown in Figure 8.35.
283
Figure 8.34
Simulation results for the 
discharge phase of the circuit of
Figure 8.30.",DCElectricalCircuitAnalysis.pdf
"The first item to note is that steady-state appears to be reached in just under 200 
milliseconds for both the charge and discharge phases, as expected. Second, note 
that the node 2 voltage (green) minus the node 3 voltage (red) starts at zero and 
winds up at a little over 20 volts at 200 milliseconds. This is, of course, the capacitor
voltage, but what is interesting here is that this plot shows how the voltage across 
the 3 kΩ resistor shrinks as the capacitor voltage grows, the sum equaling the node 2
voltage. This makes perfect sense because, as the capacitor voltage increases, the 
current through it must be decreasing, and as this same current is flowing through 
the 3 kΩ resistor, the resistor's voltage must also be decreasing due to Ohm's law. 
The other interesting part of this plot is what happens at 200 milliseconds when the 
source goes back to zero. Note that the node 3 voltage immediately jumps to a",DCElectricalCircuitAnalysis.pdf
"source goes back to zero. Note that the node 3 voltage immediately jumps to a 
negative value. This is because the voltage across the capacitor cannot change 
instantaneously. It must still have 20.57 volts across it the instant the source goes 
back to zero. In this situation, because the source is essentially a short, the capacitor 
winds up in series with the 3 kΩ resistor and the parallel combination of the 1 kΩ 
and 6 kΩ resistors, or about 857 Ω. Calculating the voltage divider between the 3 kΩ
and 857 Ω resistors with 20.57 volt source shows the 3 kΩ resistor receiving 
approximately 16 volts. Further, the discharge current will be flowing out of the 
capacitor in a counterclockwise direction, meaning it flows from ground up through 
the 3 kΩ resistor. Thus, we expect node 3 to be at approximately −16 volts, which is 
precisely what the plot indicates. How cool is that?
What if we didn't wait for steady-state? How would these plots change? Essentially,",DCElectricalCircuitAnalysis.pdf
"precisely what the plot indicates. How cool is that?
What if we didn't wait for steady-state? How would these plots change? Essentially, 
the trajectories of the curves would not change. After all, how would the circuit 
“know” that the switch would open early or the pulse would flip prematurely? 
284
Figure 8.35
Simulation results using pulse 
voltage source for discharge 
phase.",DCElectricalCircuitAnalysis.pdf
"What happens is that the curves are followed to the point in time where the circuit is 
interrupted. From there, the next phase occurs with the present voltages as the 
starting points. To test this, the simulation is run yet again, but this time the source 
pulse width is shortened to just 50 milliseconds, well short of steady-state. The 
results of the simulation are shown in Figure 8.36.
Comparing the plot of Figure 8.36 to that of Figure 8.35 shows that the two are 
identical up to 50 milliseconds. At that point, the input pulse returns to zero and the 
capacitor begins to discharge. The shapes and timings of the node 2 and node 3 
voltages are the same as they were in Figure 8.35, however, the amplitudes are 
reduced. This is because the capacitor did not have time to reach the steady-state 
voltage of 20.57 volts. In fact, it only reaches about 14.94 volts using Equation 8.12. 
Applying the voltage divider on this potential as before shows that the 3 kΩ resistor",DCElectricalCircuitAnalysis.pdf
"Applying the voltage divider on this potential as before shows that the 3 kΩ resistor 
should jump to approximately −11.6 volts, which is confirmed by the simulation. 
8.5 Summary8.5 Summary
The capacitor is a device that is used to store electric charge. Capacitance, C, is 
measured in farads, F. The idealized device consists of two conductive plates 
separated by some distance, that space being filled by an insulating dielectric. 
Capacitance is directly proportional to the plate area and the dielectric's permittivity, 
and inversely proportional to the plate distance. Another important characteristic of 
the dielectric is its breakdown strength. Along with the plate spacing, this will 
285
Figure 8.36
Simulation results for 
interrupted discharge phase 
using pulse voltage source.",DCElectricalCircuitAnalysis.pdf
"establish the capacitor's voltage rating. The permittivity will also help to determine 
the capacitor's volumetric efficiency, a measure of how much capacitance can be 
achieved within a given volume. Non-ideal parameters include the ESR, or 
equivalent series resistance, which is ideally zero; and the effective parallel leakage 
resistance, ideally infinity. Absolute accuracy, temperature stability and similar 
parameters round out the distinguishing features of one kind of capacitor against 
another. When placed in parallel, capacitors add in the same manner as resistors in 
series. When placed in series, capacitors behave like resistors in parallel.
Perhaps the most important operational characteristic regarding capacitors is that 
voltage across a capacitor cannot change instantaneously. It will take some finite 
amount of time before the charge on the capacitor builds, leading to a predictable",DCElectricalCircuitAnalysis.pdf
"amount of time before the charge on the capacitor builds, leading to a predictable 
rise in voltage across it. Because of this, for DC circuits capacitors initially behave 
like shorts, but after sufficient time has passed, they behave like opens. The amount 
of time required to reach steady-state is five time constants, where one time constant
is defined as the product of the circuit's effective resistance and its capacitance. The 
current charge curve is of the shape ε−t. The current starts at a maximum and 
eventually approaches zero as time passes. The corresponding voltage shape is of the
form 1−ε−t. Here, the capacitor's voltage starts at zero and rises to some maximum 
value. The capacitor's voltage discharge curve effectively is swapped compared to 
the charge curve (e.g. voltage follows ε−t).
Review QuestionsReview Questions
1. What are the physical characteristics of capacitors and how do they affect 
capacitance?
2. Define the voltage-current characteristic for capacitors.",DCElectricalCircuitAnalysis.pdf
"capacitance?
2. Define the voltage-current characteristic for capacitors.
3. What is meant by a capacitor's volumetric efficiency?
4. How do the permittivity and breakdown strength of the dielectric affect the 
overall capacitance and voltage rating?
5. How do capacitors combine when placed in series and how do they combine
when placed in parallel?
6. Define the initial and steady-state behavior of capacitors.
7. Define time constant for an RC circuit.
8. Describe the charge and discharge characteristics of RC circuits.
286",DCElectricalCircuitAnalysis.pdf
"8.6 Exercises8.6 Exercises
AnalysisAnalysis
1. For the circuit shown in Figure 8.37, determine the effective capacitance.
2. Determine the effective capacitance of the configuration shown in Figure 
8.38.
3. Given the capacitor network shown in Figure 8.39, determine the effective 
value.
4. Determine the effective capacitance of network shown in Figure 8.40.
287
Figure 8.37
Figure 8.40
Figure 8.38
Figure 8.39",DCElectricalCircuitAnalysis.pdf
"5. Determine the voltage across each capacitor for the circuit shown in Figure 
8.41.
6. Determine the voltage across each capacitor for the circuit shown in Figure 
8.42.
7. Determine the initial voltage across each component for the circuit shown in
Figure 8.43.
8. Given the network shown in Figure 8.43, determine the steady-state voltage 
across each component.
288
Figure 8.41
Figure 8.42
Figure 8.43",DCElectricalCircuitAnalysis.pdf
"9. For the circuit shown in Figure 8.44, determine the capacitor voltage 3 
microseconds after the power is switched on.
10. For the circuit shown in Figure 8.45, determine the capacitor voltage 5 
seconds after the power is switched on.
11. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 8.46. 
12. For the circuit shown in Figure 8.46, determine the capacitor voltage and 
circulating current 20 microseconds and 100 milliseconds after the switch is 
thrown.
13. Given the circuit shown in Figure 8.47, determine the capacitor voltage and 
circulating current 200 milliseconds and 10 seconds after the switch is 
thrown.
289
Figure 8.44
Figure 8.45
Figure 8.46
Figure 8.47",DCElectricalCircuitAnalysis.pdf
"14. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 8.47.
15. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 8.48, switch position 1.
16. Determine the charge and discharge time constants for the circuit shown in 
Figure 8.49.
17. Given the circuit shown in Figure 8.48, determine the capacitor voltage 10 
milliseconds after the power is turned on. At this point, the switch is thrown 
to position 2. Determine how long it will take the capacitor to discharge to 
nearly zero volts.
18. For the circuit shown in Figure 8.49, determine the capacitor voltage 400 
microseconds after the power is turned on. At this point, the switch is 
thrown to position 2. Determine how long it will take the capacitor to 
discharge to nearly zero volts.
19. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 8.50, switch position 1.
290",DCElectricalCircuitAnalysis.pdf
"19. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 8.50, switch position 1.
290
Figure 8.48
Figure 8.49
Figure 8.50",DCElectricalCircuitAnalysis.pdf
"20. Given the circuit shown in Figure 8.50, determine the capacitor voltage 12 
milliseconds after the power is turned on. At this point, the switch is thrown 
to position 2. Determine how long it will take the capacitor to discharge to 
nearly zero volts.
DesignDesign
21. Given the circuit of Figure 8.46, determine a new resistor value such that 
steady-state is reached in 2 milliseconds.
22. Given the circuit of Figure 8.47, determine a new resistor value such that 
steady-state is reached in 5 seconds.
23. When an audio amplifier is turned on, the power surge can cause an audible 
pop from the loudspeaker. To prevent this, amplifiers often connect to the 
loudspeaker via a relay. The relay is energized to connect the loudspeaker 
once the output has settled, typically a few seconds after power is applied. 
This delay may be created via an RC network. Suppose the driving circuit is 
5 volts and the relay trips at 4 volts. Further, the associated charging",DCElectricalCircuitAnalysis.pdf
"5 volts and the relay trips at 4 volts. Further, the associated charging 
resistance is 10 kΩ. Determine the capacitance required to achieve a 2 
second delay time.
ChallengeChallenge
24. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 8.51.
25. For the circuit shown in Figure 8.51, determine the capacitor voltage 1 
second after the power is turned on. At this point, the switch is thrown to 
position 2. Determine how long it will take the capacitor to discharge to 
nearly zero volts.
291
Figure 8.51",DCElectricalCircuitAnalysis.pdf
"SimulationSimulation
26. Perform a transient analysis to verify the time to steady-state of Figure 8.46 
(problem 11).
27. Perform a transient analysis to verify the time to steady-state of Figure 8.47 
(problem 14).
28. Use a transient analysis to verify the design of problem 21.
29. Use a transient analysis to verify the design of problem 22.
30. Use a transient analysis to verify the operation of the circuit shown in Figure
8.51 as specified in problem 25. You may wish to do this as two separate 
simulations, one for each switch position, with the second position using a 
pre-charged capacitor.
292",DCElectricalCircuitAnalysis.pdf
"NotesNotes
    ♫♫    ♫♫
293",DCElectricalCircuitAnalysis.pdf
"99  InductorsInductors
9.0 Chapter Learning Objectives9.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe the theoretical and practical aspects of inductor construction.
• Describe the current-voltage characteristic behavior of inductors.
• Utilize component data sheets to determine operating characteristics of inductors.
• Determine the initial and steady-state equivalents of resistor-inductor networks.
• Determine the initial and steady-state equivalents of resistor-capacitor-inductor networks.
• Determine the transient response of basic RL networks.
9.1 Introduction 9.1 Introduction 
This chapter introduces our third passive device, the inductor. Inductors are fundamentally different from both 
resistors and capacitors in terms of their construction and their operation. Inductors do, however, share certain 
broad traits with capacitors. First, they are energy storage devices. In the case of the inductor, energy is stored in",DCElectricalCircuitAnalysis.pdf
"broad traits with capacitors. First, they are energy storage devices. In the case of the inductor, energy is stored in
a magnetic field, similar to the case of the capacitor which utilizes an electric field. Further, in the ideal case, 
inductors do not dissipate power. Also like capacitors, when inductors are placed in DC circuits they are not 
ohmic, meaning that their current-voltage characteristic does not respond to Ohm's law. Instead, their current-
voltage characteristic is dynamic in nature. In some respects, though, their current-voltage behavior is opposite 
to the way in which capacitors behave, and thus they offer their own unique performance characteristics.
Inductors have a long history of use in electronic systems. In fact, one of the most common uses in the home is 
as an integral part of a typical loudspeaker system. They are used in modern switch-mode DC power supplies in",DCElectricalCircuitAnalysis.pdf
"as an integral part of a typical loudspeaker system. They are used in modern switch-mode DC power supplies in 
products such as personal computers and televisions. In audio and communications systems they are used in 
filters and tuning circuits alongside capacitors. Like a capacitor, for any application that needs to smooth out a 
varying voltage, store energy or filter a signal; an inductor is a likely candidate. 
Unfortunately, real-world inductors generally do not behave as close to their desired ideal operation as do real-
world capacitors. The secondary effects of inductor construction limit their performance; perhaps the most 
notable factor being their potentially large equivalent series resistance. They are also susceptible to external 
magnetic fields which can introduce noise and interference, degrading signal quality. For these reasons, there are
areas where, given a choice, capacitors would be preferred over inductors. But this is by no means a broad",DCElectricalCircuitAnalysis.pdf
"areas where, given a choice, capacitors would be preferred over inductors. But this is by no means a broad 
condemnation and there are areas where the use of inductors is essential. Beyond this, the very concept of 
inductance is important in that informs designers of the practical limits of performance of their circuits.
294",DCElectricalCircuitAnalysis.pdf
"The concept of electromagnetic induction was first discovered by English scientist 
Michael Faraday in the early 19th century. He noticed that if he wrapped two wires 
around an iron ring and introduced a current in one of them, then a transient (short-
lived) current would appear in the second coil of wire. He noticed a similar effect 
when he slid a bar magnet through a wire coil. Around the same time, American 
Joseph Henry discovered much the same independently of Faraday and performed 
considerable research in this area. In his honor, the unit of inductance, the henry 
(abbreviated H), is named after him. The symbol for inductance is L, named after 
physicist Heinrich Emil Lenz. Inductors are commonly referred to as coils or 
chokes, for reasons that will be apparent shortly.
9.2 Inductance and Inductors9.2 Inductance and Inductors
To begin, we need to examine the interrelation between electric current and",DCElectricalCircuitAnalysis.pdf
"9.2 Inductance and Inductors9.2 Inductance and Inductors
To begin, we need to examine the interrelation between electric current and 
magnetic fields in a conductor. When a current passes through a conductor, such as a
wire, a magnetic field is created around the conductor that is proportional to the 
strength of the current. This is illustrated in Figure 9.1.
The magnetic field can be thought of as sets of concentric rings around the 
conductor, although for clarity only single loops are drawn in the Figure. The 
number of magnetic lines in a given area is known as the magnetic flux and is given 
the symbol Φ (the Greek letter phi). The unit of magnetic flux is the weber, Wb, 
named after Wilhelm Weber, a 19th century German physicist. 
Magnetic flux ≡ the number of magnetic lines enclosed
 in a given area. (9.1)
Note that the magnetic field runs the length of the conductor. The direction of the
field lines follows the right hand rule: if you grasp the wire with your right hand",DCElectricalCircuitAnalysis.pdf
"field lines follows the right hand rule: if you grasp the wire with your right hand
such that your thumb is pointing in the direction of conventional current flow, then
your fingers wrap in the direction of the magnetic field. This rule is illustrated in
Figure 9.2. 
If we form the conductor into a loop, the field lines are corralled into the center of 
the loop. This is illustrated in Figure 9.3. In this diagram it can be seen that the lines 
effectively collect in the center, going into the page. 
295
Figure 9.1
Magnetic field around a 
conductor.I
Figure 9.2
Right hand rule.
Image source (modified)",DCElectricalCircuitAnalysis.pdf
"The enhancement effect can be magnified by adding more loops in tandem. This is
known as a solenoid and is shown in Figure 9.4. It is the most basic form of an
inductor.
The concentrating effect of the magnetic field is shown in Figure 9.5. In this figure, 
the coil is shown from the side, as a cross-section of the individual loops. The dots 
inside the conductors indicate that the current is flowing toward you, out of the page;
while the the crosses indicate that the current is flowing into the page. The lines of 
flux exit out the right, loop around, and reenter from the left. Due to limited space, 
the entire loop for each line is not drawn, and it is important to remember that 
magnetic flux lines do not end, but always create a loop. Further, although it is 
shown as a plane here, this field is three dimensional, with lines looping back into 
the page as well as in front of it.
This is how electromagnets can be created20. The north pole is the exiting end (right",DCElectricalCircuitAnalysis.pdf
"the page as well as in front of it.
This is how electromagnets can be created20. The north pole is the exiting end (right 
side) of Figure 9.5 while the south pole is the entering end (left side). 
20 Truly one of the coolest inventions of all time: a magnet with an on/off switch.
296
Figure 9.3
Magnetic field around a loop.
Figure 9.4
Solenoid.
Figure 9.5
Magnetic field in a solenoid.
Image source.",DCElectricalCircuitAnalysis.pdf
"If the current changes, there will be a commensurate change in the magnetic field. 
Further, this change in the field will induce a current in the conductor that creates a 
magnetic field that opposes the original change in the field. This is known as Lenz's 
law. Alternately, it can be stated that the induced current caused by a changing 
magnetic field will oppose the change in the original current that created that change
in the original magnetic field.  
At this point we can offer a proper definition of the weber:
1 weber ≡ the magnetic flux that, acting on a single loop
    of a conductor, produces a potential of 1 volt 
    if the flux is reduced to zero at an even rate 
    over 1 second. (9.2)
In magnetic circuits we are also interested in the magnetic flux density which is the 
magnetic flux per unit area. The symbol for flux density is B and has units of teslas 
(T), named after Nicola Tesla, the Serbo-Croatian-American engineer and inventor.",DCElectricalCircuitAnalysis.pdf
"(T), named after Nicola Tesla, the Serbo-Croatian-American engineer and inventor. 
One tesla is defined as one weber per square meter.
1 tesla ≡ 1 weber / meter2 (9.3)
To provide a reference, the magnetic flux density of the Earth near the equator is 
approximately 31 μT, while the value of the voice coil gap in a loudspeaker is 
around 1 to 2 T, with medical MRI scanners being a little higher still. 
Finally, we come to the definition of inductance and its unit, the henry:
Inductance is a measure of the tendency of a conductor to 
oppose a change in the current flowing through it. (9.4)
1 henry ≡ 1 weber / 1 amp (9.5)
Unsurprisingly, the energy stored in the magnetic field of an inductor is proportional 
to the inductance. It is also proportional to the square of the current through the 
inductor.
W =1
2 L I 2
(9.6)
Where
W is the energy in joules,
L is the inductance in henries,
I is the current in amps.
297",DCElectricalCircuitAnalysis.pdf
"An inductor in its simplest form consists of a series of wire loops. These might be
wound around an iron core, although a non-ferrous core might also be used. For a
simple single layer inductor, such as the one drawn in Figure 9.6, the inductance is 
described by the following formula:
L=μ A N 2
l (9.7)
Where
L is the inductance in henries,
μ is the permeability of the core material,
A is the cross sectional area of the coil,
N is the number of coils or turns,
l is the length of the coil.
Inductors may also be wound using multiple layers or around a toroidal core, and 
these designs utilize alternate formulas. 
Inductor Styles and PackagingInductor Styles and Packaging
Equation 9.7 indicates that, in order to achieve high inductance, we would like a 
core with high permeability, permeability being a measure of how easy it is to 
establish magnetic flux in said material. Substances such as iron or ferrite have a",DCElectricalCircuitAnalysis.pdf
"establish magnetic flux in said material. Substances such as iron or ferrite have a 
much greater permeability than air and are used commonly for cores. They do have a
disadvantage in that they will saturate sooner than an air core, and this can lead to 
distortion. 
Another approach is to pack as many turns as is possible within a given length. One 
way to do this is to minimize the thickness of the insulation around the wire21. This 
can be achieved by using a thin enamel coating instead of the typical plastic 
insulation. A second method is to use a very fine wire. This leads to two problems, 
namely an undesirable increase in equivalent series resistance (known colloquially 
as coil resistance or Rcoil), and limited current carrying capacity. All of these effects 
have to be balanced in order to achieve the best performance for a given application.
Commercial inductors range in value from a fraction of a nanohenry for small",DCElectricalCircuitAnalysis.pdf
"Commercial inductors range in value from a fraction of a nanohenry for small
surface mount “chip” inductors up to several henries. Some devices exhibit large
internal inductances even though they are not specifically used as inductors. One
common example is a transformer. Another example is an electric guitar or bass
pickup, such as the one shown in Figure 9.7 with its cover removed. Units such as
this may be constructed of several thousand turns of very fine wire (typically AWG
41 to 44) and achieve inductances in excess of one henry. 
21 The wound wire must be insulated otherwise each loop will short to the loops next to it 
and we'd be left with a tube instead of a series of loops.
298
Figure 9.6
Simple air-core inductor 
dimensions.
Figure 9.7
Electric bass guitar pickup.",DCElectricalCircuitAnalysis.pdf
"A variety of inductors is shown in Figure 9.9, all of which are of the through-hole 
type (surface mount inductors do not appear considerably different from their 
surface mount resistor and capacitor cousins).  
The two units toward the left are molded inductors and use a standard color code, 
similar to the ones used for resistors and capacitors. The unit at the top (yellow) is a 
high current inductor that features low Rcoil. The three inductors in the center use 
obvious ferrite cores, two wound on straight cores and the third wound on a toroidal 
core. The unit to the right uses a high permeability material at the very top and is 
wrapped in a plastic sheath for protection. Variable inductors are also a possibility 
and can be made by using a ferrite core that can be slid within the coils, effectively 
changing the permeability of the core (part ferrite, part air).
The schematic symbols for inductors are shown in Figure 9.10. The standard",DCElectricalCircuitAnalysis.pdf
"changing the permeability of the core (part ferrite, part air).
The schematic symbols for inductors are shown in Figure 9.10. The standard
symbol is at the top. The variable inductor symbol is in the middle and is a two-
lead device, somewhat reminiscent of the symbol for a rheostat. At the bottom is
the symbol for an inductor with an iron, ferrite, or similar high permeability core.
In general, like resistors, single inductors are not polarized and cannot be inserted
into a circuit backwards. There are, however, special applications where multiple 
coils can be wound on a common core, and for these, the polarity of their 
interconnection can make a difference.
Inductor Data SheetInductor Data Sheet
A portion of an inductor data sheet is shown in Figure 9.12. This page lists the 
available sizes of this particular model, each with corresponding quantities. We can 
see this model is available in inductance values ranging from 1 μH up to 100 mH.",DCElectricalCircuitAnalysis.pdf
"see this model is available in inductance values ranging from 1 μH up to 100 mH. 
Tolerance of the smaller values is ±10% while values at and above 33 μH are at 
±5%. Q is the quality factor and is particularly important in AC circuits (higher being
better), along with its associated frequency, fQ. Continuing across we find IR. This is 
the maximum rated current. For the smaller values, we find they can withstand in 
excess of 2 amps while the larger units can withstand only tens of milliamps. 
299
Figure 9.9
A collection of inductors.
Figure 9.10
Inductor schematic symbols
(top-bottom): standard, 
variable, iron/ferrite core.",DCElectricalCircuitAnalysis.pdf
"Finally, we come to Rmax. This is also known as Rcoil. It represents the equivalent 
series resistance of the inductor. In general, smaller is better. For this model, it 
ranges from a fraction of an ohm to a few hundred ohms. This trend is typical of 
inductors; all else being equal, the larger the inductance, the larger the value of the 
associated series resistance. In many circuits, the value of Rcoil cannot be ignored.
300
Figure 9.12
Inductor data sheet.
Courtesy of TDK",DCElectricalCircuitAnalysis.pdf
"Inductors in Series and in ParallelInductors in Series and in Parallel
Suppose we take two identical inductors and place them in series. This effectively 
doubles both the length and the number of loops. From Equation 9.7 we can see that 
doubling both the number of loops and the length would double the inductance. This
is because N2 goes up by a factor of four which is then halved be the increased 
length. Consequently, inductors in series add values just like resistors in series. By 
extension, inductors in parallel behave like resistors in parallel. The equivalent of 
parallel inductors can be found by using either the product-sum rule or by taking the 
reciprocal of the sum of their reciprocals. 
Example 9.1
Find the equivalent inductance of the network shown in Figure 9.11.  
The 6 mH and 12 mH inductors are in parallel. The equivalent value of the 
pair is:
Lparallel = L2 L3
L2+L3
Lparallel = 6 mH12 mH
6 mH+12 mH
Lparallel = 4mH",DCElectricalCircuitAnalysis.pdf
"pair is:
Lparallel = L2 L3
L2+L3
Lparallel = 6 mH12 mH
6 mH+12 mH
Lparallel = 4mH
This combination is in series with the 5 mH inductor. Therefore the total 
equivalent inductance is 4 mH + 5 mH, or 9 mH.
Current-Voltage RelationshipCurrent-Voltage Relationship
The fundamental current-voltage relationship of the inductor is the mirror image of 
that of the capacitor:
v=L di
dt (9.8)
This states that the voltage across the inductor is a function of how quickly the 
current is changing. If the current is not changing (i.e., in steady-state), then the 
voltage across the inductor is zero. In this case, the inductor behaves like a short, or 
more accurately, like its Rcoil value. In contrast, during a rapid initial current change, 
the inductor voltage can be large, and thus the inductor behaves like an open.
301
Figure 9.11
Circuit for Example 9.1.",DCElectricalCircuitAnalysis.pdf
"If we rearrange Equation 9.8 and solve for the rate of change of current, we find
that:
di
dt = v
L (9.9)
Thus if an inductor is fed by a constant voltage source, the current will rise at a 
constant rate equal to v/L. For example, considering the circuit in Figure 9.11, we 
see a voltage source feeding a single inductor. If we were to plot the inductor's 
current over time, we would see something like the graph of Figure 9.12.
As time progresses, the current through the inductor increases, flowing from top to 
bottom. With a theoretically perfect inductor and source, this would continue as long
as the circuit was energized. In reality, this line would either begin to deflect 
horizontally as the source reached its limits, or the inductor would fail once its 
maximum current or power handling was reached. The slope of this line is dictated 
by the size of the applied voltage source and the inductance. 
Example 9.2",DCElectricalCircuitAnalysis.pdf
"by the size of the applied voltage source and the inductance. 
Example 9.2
Determine the rate of change of current through the inductor in the circuit
of Figure 9.13. Also determine the inductor's current 10 microseconds after
power is switched on.
From Equation 9.9, the rate of change of current is:
di
dt = v
L
di
dt = 10V
50 mH
di
dt = 200 A per second
In other words, for every second, the current rises another 200 amps. Thus, 
after just 10 microseconds it will have risen to 200 A/s times 10 μs, or 2 mA.
302
I
t
Figure 9.11
Inductor with voltage source.
Figure 9.12
Inductor current versus time.
Figure 9.13
Circuit for Example 9.2.",DCElectricalCircuitAnalysis.pdf
"Equation 9.8 is key to understanding the behavior of inductors. As noted previously, 
if an inductor is driven by a fixed voltage source and ignoring Rcoil, the current 
through it rises at the constant rate of v/L. This change in current through the 
inductor is not limitless. An instantaneous change requires that di/dt is infinite, and 
thus, the voltage driving the inductor would also have to be infinite, which is a clear 
impossibility. Therefore we can state a particularly important characteristic of 
capacitors:
The current through an inductor cannot change instantaneously. (9.10)
This observation will be central to analyzing the operation of inductors in DC 
circuits. 
9.3 Initial and Steady-State Analysis of RL Circuits9.3 Initial and Steady-State Analysis of RL Circuits
When analyzing resistor-inductor circuits, remember that current through an 
inductor cannot change instantaneously as this would require an infinite voltage",DCElectricalCircuitAnalysis.pdf
"inductor cannot change instantaneously as this would require an infinite voltage 
source. When a circuit is first energized, the current through the inductor will still be
zero, which is characteristic of opens. Once at steady-state, the current has leveled 
out and therefore the voltage across the inductor will approach zero, which is 
characteristic of shorts. Thus, we can state the general behavior of inductors at the 
beginning and ending of the charge cycle:
For DC analysis, initially inductors appear as opens. (9.11)
At steady-state, inductors appear as shorts. (9.12)
This is the opposite of what was seen with capacitors. For example, in the circuit of 
Figure 9.14, initially L is open, leaving us with R1 and R2 in series with the source, E.
At steady-state, L shorts out, leaving R1 in series with the parallel combination of R2 
and R3. All practical inductors will exhibit some internal resistance, so it is often best",DCElectricalCircuitAnalysis.pdf
"and R3. All practical inductors will exhibit some internal resistance, so it is often best
to think of an inductor as an ideal inductance with a small resistance (Rcoil) in series 
with it. 
303
Figure 9.14
Basic RL circuit.",DCElectricalCircuitAnalysis.pdf
"Example 9.3
Assuming the initial current through the inductor is zero in the circuit of
Figure 9.15, determine the voltage across the 2 kΩ resistor when power
is applied and after the circuit has reached steady-state. Draw each of the
equivalent circuits.
First, we'll redraw the circuit for the initial-state equivalent. To do so,
open the inductor. The new equivalent is shown in Figure 9.16. By
opening the inductor, the 6 kΩ resistor has been removed from the circuit
and sees no voltage. What we are left with is a voltage divider between
the source and the 1 kΩ and 2 kΩ resistors. 
Using the voltage divider rule,
V 2k = E Rx
Rx+R y
V 2k = 25 V 2k Ω
2k Ω+1 kΩ
V 2k ≈ 16.67 V
For steady-state, we redraw using a short in place of the inductor, as shown 
in Figure 9.17. Here we have another voltage divider, this time between the 
1 kΩ resistor and the parallel combination of 2 kΩ and 6 kΩ, or 1.5 kΩ.
V 2k = E Rx
Rx +R y
V 2k = 25 V 1.5kΩ
1.5kΩ+1kΩ
V 2k = 15V
304
Figure 9.15",DCElectricalCircuitAnalysis.pdf
"1 kΩ resistor and the parallel combination of 2 kΩ and 6 kΩ, or 1.5 kΩ.
V 2k = E Rx
Rx +R y
V 2k = 25 V 1.5kΩ
1.5kΩ+1kΩ
V 2k = 15V
304
Figure 9.15
Circuit for Example 9.3.
Figure 9.16
Initial-state equivalent of the 
circuit of Figure 9.15. 
Figure 9.17
Steady-state equivalent of the 
circuit of Figure 9.15.",DCElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
To verify the results of Example 9.3, the circuit is entered into a simulator as shown 
in Figure 9.18. A DC operating point analysis is run and the results are shown in 
Figure 9.19. 
The steady-state potential at node 2 corresponds to the voltage across the 2 kΩ 
resistor and agrees with the theoretical calculation of 15 volts. Note that node 3 is 
also 15 volts, indicating that the steady-state voltage across the inductor is zero, 
meaning it is behaving as a short, exactly as expected.
305
Figure 9.18
Circuit of Figure 9.15 in a 
simulator.
Figure 9.19
Simulation results for the 
circuit of Figure 9.15.",DCElectricalCircuitAnalysis.pdf
"9.4 Initial and Steady-State Analysis of RLC Circuits9.4 Initial and Steady-State Analysis of RLC Circuits
When analyzing resistor-inductor-capacitor circuits, remember that capacitor voltage
cannot change instantaneously, thus, initially, capacitors behave as a short circuit. 
Once the capacitor has been charged and is in a steady-state condition, it behaves 
like an open. This is opposite of the inductor. As we have seen, initially an inductor 
behaves like an open, but once steady-state is reached, it behaves like a short. For 
example, in the circuit of Figure 9.20, initially L is open and C is a short, leaving us 
with R1 and R2 in series with the source, E. At steady-state, L shorts out both C and 
R2, leaving all of E to drop across R1. For improved accuracy, replace the inductor 
with an ideal inductance in series with the corresponding Rcoil value. Similarly, 
practical capacitors can be thought of as an ideal capacitance in parallel with a very 
large (leakage) resistance.",DCElectricalCircuitAnalysis.pdf
"practical capacitors can be thought of as an ideal capacitance in parallel with a very 
large (leakage) resistance.
Example 9.4
Assuming the initial current through the inductor is zero and the capacitor
is uncharged in the circuit of Figure 9.21, determine the current
through the 2 kΩ resistor when power is applied and after the circuit
has reached steady-state. Draw each of the equivalent circuits.
For the initial-state equivalent we open the inductor and short the
capacitor. The new equivalent is shown in Figure 9.22. The shorted
capacitor removes everything to its right from the circuit. All that's left
is the source and the 2 kΩ resistor. 
We can find the current through the 2 kΩ resistor using Ohm's law.
306
Figure 9.20
Basic RLC circuit.
Figure 9.21
Circuit for Example 9.4.
Figure 9.22
Initial-state equivalent of the 
circuit of Figure 9.21.",DCElectricalCircuitAnalysis.pdf
"I 2k = E
R
I 2k = 14 V
2k Ω
I 2k = 7 mA
Steady-state is redrawn in Figure 9.23, using a short in place of the inductor,
and an open for the capacitor. We are left with a resistance of 2 kΩ in series 
with the parallel combination of 1 kΩ and 4 kΩ, or 2.8 kΩ in total.
I2k = E
R
I 2k = 14 V
2.8k Ω
I 2k = 5mA
9.5 Transient Response of RL Circuits9.5 Transient Response of RL Circuits
The transient response of RL circuits is nearly the mirror image of that for RC 
circuits. To appreciate this, consider the circuit of Figure 9.24. 
Again, the key to this analysis is to remember that inductor current cannot change 
instantaneously. When power is first applied, the circulating current must remain at 
zero. Therefore no voltage drop is produced across the resistor, and by KVL, the 
voltage across the inductor must equal the source, E. This establishes the initial rate 
of change of current via Equation 9.9 (di/dt = E/L) and is represented by the dashed",DCElectricalCircuitAnalysis.pdf
"of change of current via Equation 9.9 (di/dt = E/L) and is represented by the dashed 
red line in the graph of Figure 9.25. As the current starts to increase, the voltage drop
across the resistor begins to increase. This reduces the voltage available for the 
inductor, thus slowing the rate of change of current. This is depicted by the solid red 
307
Figure 9.23
Steady-state equivalent of the 
circuit of Figure 9.21. 
Figure 9.24
RL circuit for transient 
response analysis.",DCElectricalCircuitAnalysis.pdf
"curve on the graph. Meanwhile, the solid blue curve represents the decreasing 
inductor voltage. Thus, in the RL circuit, the inductor's voltage curve echoes the RC 
circuit's current curve (or resistor voltage curve), and the RL current curve echoes 
the RC circuit's capacitor voltage curve. 
The curves presented in Figure 9.25 are identical to those presented in Chapter 8 
when we discussed capacitors. They are reproduced here for your convenience.
As noted before, the rate of current change versus time is equal to v/L, and therefore 
in this case, E/L. If the initial rate of change were to continue unabated, the 
maximum (steady-state) current, E/R, would be reached in L/R seconds22. Therefore 
the time constant for an RL circuit is:
τ = L
R (9.13)
Once again, five constants will achieve steady-state. 
22 Which is to say, E/L amps per second times L/R seconds yields E/R amps. 
308
Normalized Charge/Discharge Curves
Percent of Maximum
0
20
40
60
80
100
 
Time Constants
0 1 2 3 4 5",DCElectricalCircuitAnalysis.pdf
"308
Normalized Charge/Discharge Curves
Percent of Maximum
0
20
40
60
80
100
 
Time Constants
0 1 2 3 4 5
Figure 9.25
Normalized charge and 
discharge curves.",DCElectricalCircuitAnalysis.pdf
"Following the prior work on capacitors, the relevant equations for the RL circuit can 
be shown to be23:
V L (t ) = E ϵ− t
τ (9.14)
V R(t) = E (1 − ϵ− t
τ)  (9.15)
 I (t )= E
R
(1 − ϵ−t
τ)  (9.16)
Where
VL(t) is the inductor voltage at time t,
VR(t) is the resistor voltage at time t,
I(t) is the current at time t,
E is the source voltage,
R is the series resistance,
t is the time of interest,
τ is the time constant,
ε (also written e) is the base of natural logarithms, approximately 2.718.
Time for an example.
Example 9.5
Given the circuit of Figure 9.26, assume the switch is closed at time t = 0.
Determine the charging time constant, the amount of time after the switch
is closed before the circuit reaches steady-state, and the inductor voltage and
current at t = 0, t = 2 microseconds and t = 1 millisecond. Assume the 
inductor is initially uncharged. 
First, the time constant:
τ = L
R
τ = 400μ H
150Ω
τ ≈ 2.667μs",DCElectricalCircuitAnalysis.pdf
"inductor is initially uncharged. 
First, the time constant:
τ = L
R
τ = 400μ H
150Ω
τ ≈ 2.667μs
Steady-state will be reached in five time constants, or approximately 13.33 
microseconds. Thus we know that VL(0) = 9 volts and VL(1 ms) = 0 volts. 
Because the inductor is an open initially, IL(0) = 0 amps. At IL(1 ms), the 
circuit is in steady-state and the inductor acts like a short. Therefore, all of 
the 9 volt source drops across the 150 Ω resistor, for 60 mA. 
23 See also Appendix C.
309
Figure 9.26
Circuit for Example 9.5.",DCElectricalCircuitAnalysis.pdf
"To find VL(2 μs) we simply solve Equation 9.14.
V L (t )= E ϵ−t
τ
V L (2μ s) = 9V ϵ
− 2μs
2.667μ s
V L (2μ s) ≈4.251 V
This value can also be determined graphically from Figure 9.25. The time of
2 microseconds represents 75% of a time constant. Find this value on the 
horizontal axis and then track straight up to the solid blue curve that 
represents the charging inductor voltage. The point of intersection is right 
around 47% of the maximum value on the vertical axis. The maximum value
here is the source voltage of 9 volts. Therefore the inductor will have 
reached approximately 47% of 9 volts, or just over 4.2 volts. 
The current can be found in a similar manner using Equation 9.16.
I L (t)= E
R
(1 − ϵ− t
τ)
I L (2μs)= 9V
150 Ω
(1 − ϵ
− 2μs
2.667μ s)
I L (2μs)= 60 mA (1 − ϵ−0.75
)
I L (2μs)= 31.66 mA
Computer SimulationComputer Simulation
To verify our analysis, the circuit of Figure 9.26 is entered into a simulator, as shown",DCElectricalCircuitAnalysis.pdf
")
I L (2μs)= 31.66 mA
Computer SimulationComputer Simulation
To verify our analysis, the circuit of Figure 9.26 is entered into a simulator, as shown
in Figure 9.27. In order to reflect the notion of a time-varying circuit with a switch, 
the 9 volt DC voltage source has been replaced with a rectangular pulse voltage 
source. This source starts at 0 volts and then immediately steps up to 9 volts. It stays 
at this level for 20 microseconds before dropping back to 0 volts.
310
Figure 9.27
The circuit of Figure 9.26 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"The results of a transient analysis are shown in Figure 9.28. The waveform shown 
tracks the inductor's voltage at node 2 with respect to ground. 
We can see that the voltage starts at 9 volts as expected. It then falls back to zero and
is at steady-state in less than 15 microseconds, just as predicted. At 20 
microseconds, the pulse source returns to zero volts. At this instant, the current 
through the inductor must still be the steady-state current of 60 milliamps. This 
current will still be flowing in a clockwise direction, thus it will produce a 9 volt 
drop across the 150 Ω resistor with a + to − polarity from left to right. This 
effectively places node 2 negative with respect to ground. The result is that the 
polarity of the inductor's voltage flips, with the inductor now acting as a short-lived 
source. We see this on the transient analysis as a negative 9 volt spike. The discharge
time constant is identical to the charge constant, and thus we see the inductor's",DCElectricalCircuitAnalysis.pdf
"time constant is identical to the charge constant, and thus we see the inductor's 
voltage fall back to zero in the same amount of time.
Example 9.6
Given the circuit of Figure 9.29, find VL at t = 1 microsecond after the
circuit is energized. Assume the inductor is initially uncharged. 
First, the time constant:
τ = L
R
τ = 6 mH
15 kΩ
τ = 400 ns
311
Figure 9.28
Simulation results for the 
circuit of Figure 9.26.
Figure 9.29
Circuit for Example 9.6.",DCElectricalCircuitAnalysis.pdf
"Steady-state will be reached in five time constants, or 2 microseconds, at 
which point the inductor voltage will be zero as it will be behaving as a 
short. In contrast, as the inductor is initially an open (IL(0) = 0), all of the 
current from the source will flow into the 15 kΩ resistor, producing 30 volts 
across this parallel network. Therefore, we can state that VL(0) = 30 volts 
and VL(2 μs) = 0 volts, defining the extremes. In order to find VL(1 μs), we 
can use Equation 9.14.
V L (t)= E ϵ− t
τ
V L (1μs) = 30 V ϵ
− 1μs
0.4μ s
V L (1μs) ≈2.463 V
There are a few different ways to crosscheck this value. For starters, we can 
determine the inductor current using a slight modification of Equation 9.16 
(the current source value is used in place of E/R as the equation effectively 
requires the maximum or steady-state current).
I L (t) = I (1 − ϵ− t
τ )
I L (1μs) = 2mA(1 − ϵ
− 1μs
0.4μ s)
I L (1μs) = 1.836 mA
The inductor voltage of 2.463 volts would also have to appear across the",DCElectricalCircuitAnalysis.pdf
"τ )
I L (1μs) = 2mA(1 − ϵ
− 1μs
0.4μ s)
I L (1μs) = 1.836 mA
The inductor voltage of 2.463 volts would also have to appear across the 
parallel 15 kΩ resistor. This produces 2.463 V / 15 kΩ, or 0.164 mA of 
resistor current. By KCL, the remainder of the 2 mA source current must be 
flowing down through the inductor. This yields a net inductor current of 
2 mA − 0.164 mA, or 1.836 mA, verifying our prior result.
Example 9.6 also reinforces the concept of the time constant being inversely 
proportional to the resistance, rather than directly proportional as in the RC case. In 
the circuit of Figure 9.29, it should be obvious that the larger the resistance value, 
the larger the resulting initial-state voltage. From Equation 9.9 it can be seen that if 
the voltage across the inductor is increased, then the initial rate of change of current 
with respect to time will increase, and that implies a shorter time constant.",DCElectricalCircuitAnalysis.pdf
"with respect to time will increase, and that implies a shorter time constant.
For more complex circuits, Thévenin's theorem may be used in order to determine 
the effective source voltage and charging resistance. As we saw with RC circuits, it's 
also possible that the discharge resistance may be considerably different from the 
charging resistance. In such a case, the charge and discharge curves can be highly 
asymmetric in both time and amplitude. Generally, the larger the discharge 
resistance is when compared to the charge resistance, the larger in voltage and the 
shorter in time the discharge spike will be (think in terms of the area under the curve
staying constant).
312",DCElectricalCircuitAnalysis.pdf
"Example 9.7
Assume the initial current through the inductor is zero in Figure 9.30.
Determine the time constant. Also, determine the inductor voltage
and the voltage across the 6 kΩ resistor 200 nanoseconds after the
switch is closed.
This circuit is based on the circuit presented in Figure 9.15 as used
in Example 9.3. In that analysis it was discovered that the steady-
state voltage for the 6 kΩ and 2 kΩ resistors was 15 volts, the pair
being in parallel. Further, the initial voltage across the 2 kΩ resistor
and the inductor was 16.67 volts and for the 6 kΩ resistor, 0 volts.
For the Thévenin circuit, the open circuit voltage at the inductor would be 
the potential across the 2 kΩ resistor, which is obtained from a voltage 
divider between it and the 1 kΩ resistor, or 16.67 volts. The equivalent 
resistance is obtained by shorting the voltage source which leaves the 1 kΩ 
and 2 kΩ resistors in parallel, which is then in series with the 6 kΩ resistor,",DCElectricalCircuitAnalysis.pdf
"and 2 kΩ resistors in parallel, which is then in series with the 6 kΩ resistor, 
yielding approximately 6.667 kΩ. The equivalent is shown in Figure 9.31. 
Immediately, we can see that the initial-state voltage across the inductor is 
confirmed in the equivalent circuit. We can now determine the time constant.
τ = L
R
τ = 1mH
6.667 k Ω
τ = 150 ns
Steady-state will be reached in 750 nanoseconds. To find the inductor 
voltage we can use Equation 9.14. 
V L (t )= E ϵ− t
τ
V L (200 ns)= 16.67 V ϵ
−200 ns
150 ns
V L (200 ns) ≈4.39 V
313
Figure 9.30
Circuit for Example 9.7.
Figure 9.31
Thévenin equivalent of the 
circuit of Figure 9.30 driving 
the inductor.",DCElectricalCircuitAnalysis.pdf
"Referring back to the original circuit, in order to determine the voltage 
across the 6 kΩ resistor we can find the current through it and use Ohm's 
law. The current would be the same as the inductor's current as the two are 
in series. Thus, Equation 9.16 would do the trick.
I L (t) = E
R
(1 − ϵ− t
τ )
I L (200 ns) = 16.67 V
6.667 k Ω
(1 − ϵ
−200 ns
150 ns )
I L (200 ns)= 2.5mA (1 − ϵ−1.333)
I L (200 ns) ≈ 1.841mA
And finally,
V 6k (200 ns) = I R
V 6k (200 ns) = 1.841 mA 6k Ω
V 6k (200 ns) ≈ 11.05 V
Further, the voltage across the 2 kΩ resistor must be the sum, or 
approximately 15.44 volts.
Computer SimulationComputer Simulation
The results of Example 9.7 are crosschecked in a simulator. Once again the circuit is 
built using a pulse generator, as shown in Figure 9.32. 
A transient analysis is run out to 1 microsecond which is modestly into steady-state. 
Node voltages 2 and 3 are plotted, as shown in Figure 9.33. The initial voltage",DCElectricalCircuitAnalysis.pdf
"Node voltages 2 and 3 are plotted, as shown in Figure 9.33. The initial voltage 
across the 2 kΩ resistor (node 2) is as predicted, approximately 16.7 volts, and falls 
to 15 volts at steady-state, approximately 750 nanoseconds later. The voltage across 
the 6 kΩ resistor (node 2) starts at zero volts and also winds up at 15 volts in steady-
state, just as predicted. Further, note that the predicted voltages across the 2 kΩ and 
6 kΩ resistors at 200 nanoseconds are verified. 
314
Figure 9.32
Circuit of Figure 9.30 in a 
simulator.",DCElectricalCircuitAnalysis.pdf
"The voltage across the inductor is node 2 minus node 3. This differential is plotted 
separately in Figure 9.34. The expected initial, steady-state and 200 nanosecond 
voltages are as predicted.
315
Figure 9.33
Transient analysis simulation of
the circuit of Figure 9.30.
Figure 9.34
Simulation of the inductor 
voltage versus time for the 
circuit of Figure 9.30.",DCElectricalCircuitAnalysis.pdf
"One very important observation is that if an RL circuit is abruptly altered or opened, 
very large voltage spikes may occur. This is due to the fact that inductor current 
cannot change instantaneously. If the circuit is opened, the open represents a very 
large resistance. Ohm's law indicates that the inductor current times this very large 
resistance may produce a very large voltage across the new open. In fact, the 
potential may be sufficient to cause a spark or arc. Note that because the current 
cannot change instantaneously (both magnitude and direction), the inductor now 
behaves as a voltage source of very high magnitude and with reverse polarity. This 
phenomenon is used to create the ignition spark in internal combustion engines. In 
short, the ignition coil is charged, creating some current flow. The circuit is then 
interrupted leaving just the coil in series with the spark plug, the spark plug being",DCElectricalCircuitAnalysis.pdf
"interrupted leaving just the coil in series with the spark plug, the spark plug being 
little more than a precisely sized gap between two electrodes. This results in a large 
voltage being developed across the spark plug gap, typically in the neighborhood of 
20,000 volts, which is sufficient to create a small arc (i.e., the spark) that then ignites
the air-fuel mixture in the piston. 
An example of producing a discharge voltage spike that is considerably larger than 
the source voltage can be illustrated with the circuit of Figure 9.35. We shall assume 
that the inductor is initially uncharged when power is applied, and that the switch is 
in position 1. In this case, the circuit consists of just the 12 volt source, the 2.2 kΩ 
resistor, and the inductor. The circuit reaches steady-state in roughly 227 
nanoseconds. At that point the inductor behaves as a short, leaving the full 12 volt 
source to drop across the 2.2 kΩ resistor. This produces a clockwise current of",DCElectricalCircuitAnalysis.pdf
"source to drop across the 2.2 kΩ resistor. This produces a clockwise current of 
approximately 5.455 mA. 
If we now move the switch to position 2, this current must be maintained because 
the current through the inductor cannot change instantaneously. The new discharge 
resistance is now the series combination of the two resistors, or 49.2 kΩ. Ohm's law 
and KVL dictate that the resulting inductor voltage must be 49.2 kΩ times 5.455 mA,
or just beyond −268 volts. This potential is negative because the clockwise current is
flowing up through the 47 kΩ resistor, producing a drop + to − from ground up. The 
increased resistance also shortens the time constant and now steady-state will be 
reached in only 10.1 nanoseconds. Thus, we see a much larger magnitude spike 
(over twenty times that of the source voltage) with a much shorter time duration 
(less than one-twentieth of the charge time).
316
Figure 9.35
Circuit illustrating large 
discharge voltage spike.",DCElectricalCircuitAnalysis.pdf
"Depending on the kind of switch that is used, things can be even more extreme than 
what has been just described. Switches come in two basic varieties: make-before-
break and break-before-make. The former makes contact with the second position 
before it breaks contact with the first position, while the latter does the opposite. The
behavior just described assumes that a make-before-break switch is being used. In 
contrast, if a break-before-make switch is used, we'd see a drastically different 
result.
Let us pick up back at steady-state with 5.455 mA flowing through the inductor. We 
now throw the switch to position 2. This new switch breaks contact with the wire 
leading back to voltage source prior to it making contact with the 47 kΩ resistor. For
a short instant of time the switch is not making contact with anything and the 
resulting resistance in the loop is determined by the air gap between the switch",DCElectricalCircuitAnalysis.pdf
"resulting resistance in the loop is determined by the air gap between the switch 
contacts. Even if this was a mere 10 MΩ, the resulting potential would be over 
50,000 volts. This will almost certainly create a spark and we will have inadvertently
recreated the spark plug scenario. Hopefully, there are no combustible gases nearby.
9.6 Summary9.6 Summary
The inductor is a device that stores energy in the form of a magnetic field. 
Inductance, L, is measured in henries, H. The idealized device consists of several 
loops or coils of wire. These may or may not be wrapped around a magnetic core 
material. Inductance is directly proportional to the permeability of the core material 
and the cross sectional area of the loops, and inversely proportional to the length of 
the coil. It is also proportional to the square of the number of loops. The most 
important non-ideal parameter is Rcoil, or equivalent coil resistance. Ideally, this value",DCElectricalCircuitAnalysis.pdf
"important non-ideal parameter is Rcoil, or equivalent coil resistance. Ideally, this value
is zero. Absolute accuracy, temperature stability and similar parameters round out 
the distinguishing features of one kind of inductor against another. When placed in 
series, inductors add in the same manner as resistors in series. When placed in 
parallel, inductors combine like resistors in parallel.
Perhaps the most important operational characteristic regarding inductors is that 
current through an inductor cannot change instantaneously. It will take some finite 
amount of time before the magnetic field reacts, leading to a predictable rise in 
current through it. Because of this, for DC circuits inductors initially behave like 
opens, but after sufficient time has passed, they behave like shorts. The amount of 
time required to reach steady-state is five time constants, where one time constant is 
defined as the inductance divided by the circuit's effective resistance. The current",DCElectricalCircuitAnalysis.pdf
"defined as the inductance divided by the circuit's effective resistance. The current 
charge curve is of the shape 1−ε−t. The inductor's current starts at zero and rises to 
some maximum value. The corresponding voltage shape is of the form ε−t. Here, the 
voltage starts at a maximum and eventually approaches zero as time passes.
317",DCElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. What are the physical characteristics of inductors and how do they affect 
inductance?
2. Define the voltage-current characteristic for inductors.
3. What is Rcoil?
4. How do inductors combine when placed in series? How do they combine 
when placed in parallel?
5. Define the initial and steady-state behavior of inductors.
6. Define time constant for an RL circuit.
7. How do the charge and discharge characteristics of RL circuits compare to 
those of RC circuits?
9.7 Exercises9.7 Exercises
AnalysisAnalysis
1. For the circuit shown in Figure 9.36, determine the effective inductance.
2. Determine the effective inductance of the network shown in Figure 9.37.
3. Given the inductor network shown in Figure 9.38, determine the effective 
value.
318
Figure 9.36
Figure 9.37
Figure 9.38",DCElectricalCircuitAnalysis.pdf
"4. Determine the effective inductance of network shown in Figure 9.39.
5. Determine the initial voltage across each component for the circuit shown in
Figure 9.40.
6. Given the network shown in Figure 9.40, determine the steady-state voltage 
across each component.
7. Given the network shown in Figure 9.41, determine the steady-state voltage 
across each component.
8. Determine the initial voltage across each component for the circuit shown in
Figure 9.41.
9. Given the network shown in Figure 9.42, determine the steady-state voltage 
across each component.
319
Figure 9.39
Figure 9.40
Figure 9.41
Figure 9.42",DCElectricalCircuitAnalysis.pdf
"10. Determine the initial voltage across each component for the circuit shown in
Figure 9.42.
11. Determine the initial voltage across each component for the circuit shown in
Figure 9.43.
12. Given the network shown in Figure 9.43, determine the steady-state voltage 
across each component.
13. Given the network shown in Figure 9.44, determine the steady-state voltage 
across each component.
14. Determine the initial voltage across each component for the circuit shown in
Figure 9.44.
15. For the circuit shown in Figure 9.45, determine the inductor current 20 
microseconds after the power is switched on. Assume this is an ideal 
inductor with no internal resistance.
320
Figure 9.43
Figure 9.44
Figure 9.45",DCElectricalCircuitAnalysis.pdf
"16. For the circuit shown in Figure 9.46, determine the inductor current 100 
milliseconds after the power is switched on. Assume this is an ideal inductor
with no internal resistance.
17. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 9.47.
18. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 9.48. 
19. Given the circuit shown in Figure 9.37, determine the inductor voltage and 
circulating current 200 milliseconds and 2 seconds after the switch is 
thrown. How does this change if we include 5 Ω of internal resistance to the 
inductor?
20. For the circuit shown in Figure 9.48, determine the inductor voltage and 
circulating current 50 nanoseconds and 100 milliseconds after the switch is 
thrown.
321
Figure 9.46
Figure 9.47
Figure 9.48",DCElectricalCircuitAnalysis.pdf
"21. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 9.49, switch position 1.
22. Determine the charge and discharge time constants for the circuit shown in 
Figure 9.50.
23. Given the circuit shown in Figure 9.49, determine the inductor current 1 
millisecond after the power is turned on. At this point, the switch is thrown 
to position 2. Determine the inductor voltage the instant of switch contact 
(assume ideal switch).
24. For the circuit shown in Figure 9.50, determine the inductor current 400 
microseconds after the power is turned on. At this point, the switch is 
thrown to position 2. Determine the inductor voltage the instant of switch 
contact (assume ideal switch).
25. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 9.51, switch position 1.
322
Figure 9.49
Figure 9.50
Figure 9.51",DCElectricalCircuitAnalysis.pdf
"26. Given the circuit shown in Figure 9.51, determine both the inductor current 
and voltage 10 milliseconds after the power is turned on. At this point, the 
switch is thrown to position 2. Determine how long it will take the inductor 
to discharge to nearly zero amps. Assume ideal switch.
DesignDesign
27. For Figure 9.47, determine a new resistor value such that steady-state is 
reached in 2 milliseconds.
28. For Figure 9.48, determine a new inductor value such that steady-state is 
reached in 5 microseconds. 
ChallengeChallenge
29. Determine the time constant and the time required to reach steady-state for 
the circuit shown in Figure 9.52.
30. For the circuit shown in Figure 9.52, determine the inductor voltage 400 
microseconds after the power is turned on. At this point, the switch is 
thrown to position 2. Determine how long it will take the inductor to 
discharge to nearly zero amps. Assume ideal switch.
SimulationSimulation",DCElectricalCircuitAnalysis.pdf
"thrown to position 2. Determine how long it will take the inductor to 
discharge to nearly zero amps. Assume ideal switch.
SimulationSimulation
31. Use a transient analysis to verify the time constant and time to steady-state 
of Figure 9.48 (problem 18).
32. Use a transient analysis to verify the inductor voltage waveform of       
Figure 9.47 as specified in problem 19.  
323
Figure 9.52",DCElectricalCircuitAnalysis.pdf
"10 10 Magnetic Circuits and TransformersMagnetic Circuits and Transformers
10.0 Chapter Learning Objectives10.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe the basic quantities of a magnetic circuit including magnetic flux, flux density and 
magnetizing force.
• Outline a BH curve and define aspects including hysteresis, coercivity and retentivity.
• Analyze basic magnetic circuits using BH curves.
• Describe the uses and operation of transformers. 
• Analyze basic transformer circuits.
10.1 Introduction10.1 Introduction
This chapter extends the material presented in the chapter on inductors and introduces the concept of magnetic 
circuits. Basic applications are also illustrated. Here we shall take a closer look at the parameters of magnetic 
circuits such as permeability and flux density, and introduce new parameters including reluctance and",DCElectricalCircuitAnalysis.pdf
"circuits such as permeability and flux density, and introduce new parameters including reluctance and 
magnetizing force. We shall also examine a key graphic that helps define the magnetic properties of materials, 
namely the BH curve. In the process we shall define new terms including coercivity and retentivity, and 
introduce the concept of hysteresis. 
As we shall see, there are certain parallels between magnetic circuits and electrical circuits. For example, there 
is an “Ohm's law for magnetic circuits”, sometimes referred to as Hopkinson's law. There is also a version of 
Kirchhoff's voltage law dealing with the magnetic equivalents of voltage rises and drops. 
Perhaps the two most common applications of magnetic circuits for the new student involve the use of 
transformers and relays, both of which will be examined in this chapter. Other common applications include 
dynamic loudspeakers and microphones, magnetic resonance imaging (i.e., medical MRI), pick-up cartridges for",DCElectricalCircuitAnalysis.pdf
"dynamic loudspeakers and microphones, magnetic resonance imaging (i.e., medical MRI), pick-up cartridges for 
vinyl phonographs and the various applications made available by magnetic Hall effect sensors such as 
advanced anti-lock braking systems, timing control systems for internal combustion engines, and other instances
requiring position and motion sensing. These magnetic devices tend to be much more reliable and capable than 
simple devices such as mechanical switches. This does not mean that magnetic systems are in some way “taking
over”, indeed, there are some applications where they have become obsolete. One example is the use of output 
transformers to couple a home audio power amplifier to a loudspeaker. At one time, these were an integral part 
of every design, however, today they generally are not used. With this in mind, it is perhaps best to think of 
magnetic circuits as another item in the engineer's or technician's toolkit to help solve practical problems.
324",DCElectricalCircuitAnalysis.pdf
"10.2 Electromagnetic Induction10.2 Electromagnetic Induction
Perhaps the most important observation regarding magnetic systems is Faraday's law
of electromagnetic induction. Briefly, it states:
If a conductor is cut by changing magnetic lines of force, 
a voltage will be induced in the conductor. (10.1)
More specifically, 
e =− d Φ
dt (10.2)
Where
e is the induced voltage,
dΦ/dt is the rate of change of magnetic flux with respect to time.
Therefore, the larger the flux and the more quickly it fluctuates, the greater the 
induced voltage. What is important here is a relative change with respect to the 
conductor and field. This can be accomplished in two basic ways: by a magnetic 
field that is itself fluctuating around a fixed conductor, or by a conductor moving 
through a magnetic field. 
The concept of electromagnetic induction is a key element of a variety of 
transducers. Simply put,
A transducer is a device that transforms energy 
from one type to another. (10.3)",DCElectricalCircuitAnalysis.pdf
"transducers. Simply put,
A transducer is a device that transforms energy 
from one type to another. (10.3)
Although many people don't think of them as such, electric motors and generators 
are perfect examples of transducers. An electric motor transforms electrical energy 
into mechanical energy and a generator does the precise opposite, transforming 
mechanical energy into electrical energy. More commonly, the word transducer is 
associated with devices such as loudspeakers and microphones. A loudspeaker 
transforms its electrical input into an acoustic pressure wave (sound) whereas the 
microphone performs the complementary function of transforming an acoustic 
pressure wave into an electrical signal. While there are different ways of 
constructing these devices, the most common designs are the dynamic loudspeaker 
and dynamic microphone, both of which rely on the principle of electromagnetic 
induction. The designs of the two devices are strikingly similar, mostly varying in",DCElectricalCircuitAnalysis.pdf
"induction. The designs of the two devices are strikingly similar, mostly varying in 
size and parts optimization. Let's take a closer look at how they operate.
325",DCElectricalCircuitAnalysis.pdf
"Dynamic Loudspeakers and MicrophonesDynamic Loudspeakers and Microphones
Regardless of their output capabilities and frequency range, all dynamic 
loudspeakers share a common set of elements. Corresponding elements can be found
in dynamic microphones. Indeed, the devices are so similar that in some 
applications, small loudspeakers might pull double duty and switch over to a 
microphone mode of operation. A cutaway view of a loudspeaker designed to 
reproduce low frequencies is shown in Figure 10.1.  
The heart of the system is a voice coil (H) which is a coil of tightly wound magnet 
wire. This sits inside of a magnetic field that is created by a powerful permanent 
magnet (E). The voice coil is connected to a diaphragm (F) that is is usually made of
paper or plastic. This assembly is connected to the frame by the spider (D) at the top 
of the voice coil, and a suspension at the end of the diaphragm (B). The voice coil",DCElectricalCircuitAnalysis.pdf
"of the voice coil, and a suspension at the end of the diaphragm (B). The voice coil 
and diaphragm can move back and forth relative to the frame, like a piston. 
To understand the operation, recall that passing a current through a coil of wire
creates a magnetic field around the coil, as illustrated in Figure 10.2. This magnetic
field is the same as is created by a normal magnet. In this case the north pole is on
the left side (i.e., lines of force exiting). Consequently, this coil will interact with a
permanent magnet just like any other magnet would. In the case of the
loudspeaker, the voice coil is fed a current from the amplifier that echoes the
music or voice signal. This current creates a magnetic field around the voice coil
which interacts with the field of the permanent magnet. As the direction of the
current changes, the poles of the voice coil's field reverse. Thus, sometimes the
voice coil is pushed outward and sometimes it is pulled inward. Further, the larger",DCElectricalCircuitAnalysis.pdf
"voice coil is pushed outward and sometimes it is pulled inward. Further, the larger
the current, the greater the field it creates, and the greater the push or pull. 
326
Figure 10.1
Cutaway view of a dynamic 
loudspeaker.
A. Frame
B. Suspension
C. Lead wire
D. Spider
E. Magnet
F . Diaphragm
G. Voice coil former
H. Voice coil
I. Dust cap
Image courtesy of Audio Technology.
Figure 10.2
Magnetic field around a coil.
Image ©, courtesy of HyperPhysics",DCElectricalCircuitAnalysis.pdf
"This results in a back-and-forth motion of the diaphragm that echoes the shape of the
waveform being fed from the amplifier. As the diaphragm pushes against the air, it 
sets up a pressure wave, and we have sound. 
A dynamic microphone runs the sequence in reverse. First, the diaphragm will move 
back and forth in accordance to an applied sound wave. This causes the voice coil to 
move back and forth within the field of the permanent magnet. We now have a coil 
of wire being cut by magnetic lines of force, and by Faraday's law of induction, this 
means that a voltage will be induced in the coil. As long as the diaphragm can 
respond to the subtle changes in the acoustic pressure wave, then the resulting 
induced voltage should be of high fidelity. Clearly, to do this with great accuracy, the
diaphragm/voice coil/suspension assembly will need to very light and nimble, 
meaning that the components of the microphone will tend to be much smaller than 
those of a loudspeaker.",DCElectricalCircuitAnalysis.pdf
"meaning that the components of the microphone will tend to be much smaller than 
those of a loudspeaker.
Electric Guitar PickupElectric Guitar Pickup
A set of pickups for an electric guitar is shown in Figure 10.3. Although it might be
tempting to think that the pickups are just microphones, they are not. If you have
any doubt of this statement, just try screaming into a guitar pickup and listen to
what comes out24. The operation of the guitar pickup does, however, rely on
Faraday's law. 
As discussed in the prior chapter covering inductance, an electric guitar pickup
consists of thousands of turns of very fine wire around a permanent magnet. The
pickup is placed immediately below the guitar strings. In this position, the metal
guitar strings (generally various combinations of steel, nickel and/or cobalt), are
within the field generated by the magnet. When a string is plucked, it vibrates in a 
complex pattern that depends on the note it is tuned to and the overtones that are",DCElectricalCircuitAnalysis.pdf
"complex pattern that depends on the note it is tuned to and the overtones that are 
present. This pattern distorts the magnetic field because the strings have a much 
higher permeability than the air around them. We now have a changing magnetic 
field, in the middle of which is large coil of wire. Faraday's law states that a voltage 
must be induced in this coil, and that it will follow the motions of the strings. This 
induced voltage is then fed to an amplifier, hopefully, set to 11. 
The astute observer might ask, “Why are there multiple pickups?” Perhaps 
surprisingly, it is not to get a larger, and thus louder, signal. It has to do with the 
timbre, or tone quality, of the sound. When a guitar string vibrates, its motion can be 
thought of as containing the myriad motions of the fundamental pitch and all of the 
overtones that go with it. Towards the bridge, where the strings are attached, the 
lowest pitched elements produce little motion, and thus their strength in the overall",DCElectricalCircuitAnalysis.pdf
"lowest pitched elements produce little motion, and thus their strength in the overall 
signal is reduced. The result is that a pickup placed closer to the bridge sounds 
“thin” or “cutting” while one placed further up sounds “full” or “thick”.
24 A little string resonance if you're lucky, and some irritated band-mates if you're not.
327
Figure 10.3
Pickups on an electric guitar.",DCElectricalCircuitAnalysis.pdf
"Bicycle ComputerBicycle Computer
Our third and final illustrative example is that of a bicycle computer. These handy
little devices consist of a small sensor system on the front wheel which is
connected to a display unit mounted on the handlebars. Typically, these units will
display current speed, average speed, elapsed time, distance covered, and other
attributes of interest. 
The sensing apparatus of a typical bicycle computer is shown in Figure 10.4. This
apparatus consists of two parts: a permanent magnet mounted to one of the wheel
spokes, and a weatherproof sensing unit mounted to the front fork. The principle of
operation is fairly straightforward: each time the while rotates, the magnet attached
to the spoke swings by the sensor. The moving magnetic field creates a short-lived
voltage spike in the sensor, an example shown in Figure 10.5. The computer records
these spikes over time. Knowing the circumference of the wheel, a simple",DCElectricalCircuitAnalysis.pdf
"these spikes over time. Knowing the circumference of the wheel, a simple 
multiplication yields the accumulated distance traversed. Given internal clocking 
circuitry, the time recorded between the pulses can be turned into a velocity. Given 
these data, other attributes such as average or maximum speed are easily obtained 
with further computation.
Using the oscilloscope trace shown in Figure 10.5 as an example, we can see that the
time between the pulses is about 3.5 divisions, with each division being 200 
milliseconds in length. This gives one rotation every 0.7 seconds, or roughly 5140 
revolutions per hour. The sensor was mounted on a bike with 700 by 25 mm tires 
which yields a circumference of about 2.1 meters. Thus, the velocity would be 5140 
revolutions per hour at 2.1 meters per revolution, or about 10.8 km/h (≈ 6.73 MPH).25
25 In reality, the bike was mounted on a maintenance stand so its velocity was, in fact, zero.",DCElectricalCircuitAnalysis.pdf
"25 In reality, the bike was mounted on a maintenance stand so its velocity was, in fact, zero. 
Running alongside a bike with an oscilloscope and probes is no easy feat, especially 
when the scope needs AC power.
328
Figure 10.4
Bicycle computer sensor and 
magnet.
Figure 10.5
Sensor signal feeding bicycle 
computer.",DCElectricalCircuitAnalysis.pdf
"10.3 Magnetic Circuits10.3 Magnetic Circuits
Magnetic circuits include applications such as transformers and relays. A very
simple magnetic circuit is shown in Figure 10.6. First, it consists of a magnetic
core. The core may be comprised of a single material such as sheet steel but can
also use multiple sections and air gap(s). Around the core is at least one set of turns
of wire, i.e., a coil formed around the core. Multiple sets of turns are used for 
transformers (in the simplest case, one for the primary and another for the 
secondary). As we have seen already, passing current through the windings generates
a magnetic flux, Φ, in the core. As this flux is constrained within the cross sectional 
area of the core, A, we can derive a flux density, B.
B = Φ
A (10.4)
Where
B is the magnetic flux density in teslas,
Φ is the magnetic flux in webers,
A is the area in square meters.
Recall from Chapter 9 that one tesla is defined as one weber per square meter. An",DCElectricalCircuitAnalysis.pdf
"Φ is the magnetic flux in webers,
A is the area in square meters.
Recall from Chapter 9 that one tesla is defined as one weber per square meter. An 
alternate unit that is sometimes used is gauss (cgs system of units), named after Carl 
Friedrich Gauss, the German mathematician and scientist.
1 tesla = 10,000 gauss (10.5)
Example 10.1
A magnetic flux of 6E−5 webers exists in a core whose cross section has  
dimensions of 1 centimeter by 2 centimeters. Determine the flux density in 
teslas.
First, convert the dimensions to meters to find the area. There are 100 
centimeters to the meter.
A = width×height
A = 0.01 m×0.02 m
A = 2E-4 m2
B = Φ
A
B = 6E-5Wb
2E-4 m2
B = 0.3T
329
Figure 10.6
Simple magnetic circuit.",DCElectricalCircuitAnalysis.pdf
"Ohm's Law for Magnetic Circuits (Hopkinson's or Rowland's Law)Ohm's Law for Magnetic Circuits (Hopkinson's or Rowland's Law)
There is a common parallel drawn between magnetic circuits and electrical circuits, 
namely Hopkinson's law (Rowland's law). For electrical circuits, Ohm's law states:
V = I R
In like manner, for magnetic circuits, we have:
F = Φ R (10.6)
Where
F is the magnetomotive force (or MMF) in amp-turns,
Φ is the magnetic flux in webers,
R is the reluctance of the material in amp-turns/weber.
The magnetomotive force compares to a source voltage or electromotive force 
(EMF), the magnetic flux is likened to the flow of current, and the reluctance stands 
in for resistance (i.e., on the one hand we have a material that resists the flow of 
current, and on the other, a material in which there is “reluctance” to establish 
magnetic flux). Further, magnetomotive force is the product of the current flowing 
through a coil and the number of loops or turns in the coil:",DCElectricalCircuitAnalysis.pdf
"magnetic flux). Further, magnetomotive force is the product of the current flowing 
through a coil and the number of loops or turns in the coil:
F = N I (10.7)
Where
F is the magnetomotive force in amp-turns,
N is the number of loops or turns in the coil,
I is the current in the coil in amps.
The equation for reluctance has a nice parallel with the equation for resistance 
(Equation 2.11 from Chapter 2): 
R = ρl
A
R = l
μ A (10.8)
Where
R is the reluctance in amp-turns/weber,
l is the length of the material in meters,
A is the cross sectional area of the material in square meters,
μ is the permeability of the material in henries/meter.
330",DCElectricalCircuitAnalysis.pdf
"Given the characteristics of the coil and the path length of the magnetic circuit, the 
magnetic flux gives rise to a magnetizing force, H. 
H = N I
l (10.9)
Where
H is the magnetizing force in amp-turns/meter,
N is the number of turns or loops in the coil,
I is the coil current in amps,
l is the length of the magnetic path in meters.
Equation 10.8 reveals that ferromagnetic materials (i.e., materials that have high
permeability, such as steel) produce a low reluctance. The practical problem here is
that μ, unlike the resistivity ρ of resistors, is not a constant for these sorts of 
materials. It can vary considerably, as seen in the general diagram presented in 
Figure 10.7. As a result, it is impractical to find reluctance in same manner that we 
find resistance. All is not lost, though.
The flux density and corresponding magnetizing force for any given material are 
related by the following equation:
B = μ H (10.10)
Where
B is the flux density in teslas,",DCElectricalCircuitAnalysis.pdf
"related by the following equation:
B = μ H (10.10)
Where
B is the flux density in teslas,
μ is the permeability of the material in henries/meter,
H is the magnetizing force in amp-turns/meter.
Once again, the tricky bit here is the permeability of the core material. For air, we 
can use the permeability of free space, μ0.
μ0 = 4π × 10−7 H/m ≈ 1.257×10−6 H/m (10.11)
For other materials, such as sheet steel or cast steel, we shall take another path;
namely an empirically derived curve that plots flux density B against magnetizing
force H. Such graphs generically are referred to as “BH curves”. An example is
shown in Figure 10.8. Clearly, this curve is not a nice straight line, or even an
obvious, predictable function. The immediately apparent attributes are the initial
steep rise which is followed by a flattening of the curve. This flattening
corresponds to saturation of the magnetic material. In contrast, a plot for air would",DCElectricalCircuitAnalysis.pdf
"corresponds to saturation of the magnetic material. In contrast, a plot for air would
reveal a straight line with a very shallow slope. As we shall see, being able to
achieve a high flux density for a given magnetizing force will result in an effective
and efficient magnetic circuit. So while air has the positive attribute of not
saturating, the resulting flux density is low, usually leading to lower performance.
331
B
H
Figure 10.7
Typical permeability curve for 
a high permeability material.
H
μ
Figure 10.8
Generic BH curve.",DCElectricalCircuitAnalysis.pdf
"The The BHBH Curve Curve
The process of generating a BH curve is as follows. First, we create a core of the
material to be investigated. A coil of wire is then wrapped around this core. An
example is shown in Figure 10.9. Here we have a basic toroid with a coil of N
turns. 
We begin with the system at rest and not energized. A small current, I, is applied to 
the coil. This produces a magnetizing force via Equation 10.9. There will be a 
corresponding flux density, as per Equation 10.10. 
The current is then increased. This produces an increase in magnetizing force and a 
corresponding change in flux density. The current is increased further, until the 
curve flattens, indicating that saturation has been reached. This trajectory is shown 
as the dashed line in Figure 10.10, starting at point a with point b indicating 
saturation. The current is then reduced. This causes a reduction in magnetizing force,
but while the flux density decreases, it does not trace back perfectly along the",DCElectricalCircuitAnalysis.pdf
"but while the flux density decreases, it does not trace back perfectly along the 
original trajectory. Instead, the curve is displaced above the original. 
Eventually, the current will be reduced to zero. This corresponds to point c on the 
graph of Figure 10.10. At this point, even though there is no current in the coil, there
is still flux in the core. The resulting flux density is referred to as retentivity and is a 
measure of residual magnetism. This phenomenon is what makes it possible to 
magnetize materials. 
If we now reverse the direction of the coil current and begin to increase its 
magnitude, the flux density will continue to drop. At point d, it will have reached 
332
Figure 10.9
Toroidal core with coil.
I
N turns
B
H
-B
-H
b
a
c
d
e
f
g
Retentivity 
(residual magnetism)
Coercivity 
(coercive force)
Saturation
Figure 10.10
Generation of BH curve.",DCElectricalCircuitAnalysis.pdf
"zero. As we have effectively coerced the flux back to zero, we call the magnetizing 
force required to do this the coercivity or coercive force. 
As the current magnitude increases, the flux density also increases but with opposite 
sign. Eventually, saturation is reached again at point e. Once again, if the current's 
magnitude is decreased, the magnitude of the flux density will also decrease but will 
not perfectly trace back along the original trajectory. This time it will follow a lower 
path. When the current is reduced to zero at point f, a mirror retentivity is evident. 
Further positive increases in current show a mirror coercivity at point g. Finally, as 
current is increased to maximum, we again reach saturation at point b. 
If the current is cycled in this manner again, the process is repeated, and the outer 
path indicated by the arrows is taken again. Thus, a specific value of magnetizing",DCElectricalCircuitAnalysis.pdf
"path indicated by the arrows is taken again. Thus, a specific value of magnetizing 
force can give rise to different values of flux density: it depends on the recent history
of the material. This effect is known as hysteresis and is found in other areas as well.
Effectively, published BH curves follow the middle of the hysteresis loop. An 
example of BH curves for three different core materials is shown in Figure 10.1126.  
Curve A is for sheet steel (which is common in transformers), curve B is for cast
steel, and curve C is for cast iron. We shall make use of these in upcoming
examples. Curves for other materials are also available.
26 Curves based on https://en.wikipedia.org/wiki/Saturation_(magnetic); Steinmetz (1917), 
Theory and Calculation of Electric Circuits; and Boylestad (2010), Introductory Circuit 
Analysis.
333
Figure 10.11
BH curves for:
A. Sheet steel
B. Cast steel
C. Cast ironB (T)
H (At/m)
1.0
1.5
0.5
100 200 300 400 500
A
B
C
600 700",DCElectricalCircuitAnalysis.pdf
"Example 10.2
Assume the toroid of Figure 10.9 is made of cast steel, has a 500 turn coil, a 
cross section of 2 cm by 2 cm, and an average path length of 50 cm. 
Determine the flux in webers if a current of 0.3 amps feeds the coil.
We shall use Equation 10.9 to find the magnetizing force and from the BH 
curve, find the flux density. 
H = N I
l
H = 500 turns×0.3 A
0.5 m
H = 300 At/m (amp-turns per meter)
Cast steel corresponds to curve B (green) in Figure 10.10. A decent estimate 
for the flux density is 0.52 teslas. This is the corresponding flux density. In 
order to find the flux, we need to find the area of the core.
A = width×height
A = 0.02 m×0.02 m
A = 4E-4m2
Φ = B×A
Φ = 0.52 T×4E-4 m2
Φ = 2.08E-4 Wb
KVL for Magnetic CircuitsKVL for Magnetic Circuits
A cursory examination of Equations 10.7 and 10.9 shows that:
F = H l = N I (10.12)
Continuing the Ohm's law analogy, the amp-turns product of the coil, NI, is",DCElectricalCircuitAnalysis.pdf
"F = H l = N I (10.12)
Continuing the Ohm's law analogy, the amp-turns product of the coil, NI, is
analogous to a voltage rise. Further, the Hl product is analogous to a voltage drop.
If we then extend the analogy to include the concept of Kirchhoff's voltage law, it
should come as no surprise that the sum of NI “rises” must equal the sum of the Hl
“drops”. In the circuit of Figure 10.9, there is a single “rise” and a single “drop”.
Thus, the magnetic circuit is analogous to the minimal electric circuit shown in 
Figure 10.12. F is the magnetomotive force, NI, while R is the reluctance of the 
toroidal core. This reluctance will experience a “drop” of Hl.
The core could consist of two or more different materials, creating the equivalent of 
a series circuit. In this case, a table such as the one found in Figure 10.13 can be 
334
Figure 10.12
Electrical circuit analogy for 
the magnetic system of Figure 
10.9.",DCElectricalCircuitAnalysis.pdf
"used to aid in computation. In this table, each section of the core gets its own row.
The table is divided into two sides (note the thick separating line in the center). In 
general, we will be working through problems where we know the data on the left 
side and need to find something on the right side, or vice versa. The bridge between 
the two sides (i.e., traversing the thick line) is a BH curve for the material of that
particular section. The exception to this rule is if the section is an air gap. In that
case we can use Definition 10.11 which shows that for air, H ≈ 7.958E5 B or
alternately, B ≈ 1.257E−6 H.  
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
1
2
Time for a few illustrative examples. We shall consider a simple system like that of 
Figure 10.9, a two section core, a core with an air gap, and a core with two coils.
Example 10.3
Assume the core of Figure 10.14 is made of sheet steel, has a 200 turn coil,",DCElectricalCircuitAnalysis.pdf
"Example 10.3
Assume the core of Figure 10.14 is made of sheet steel, has a 200 turn coil,
a cross section of 1 cm by 1 cm, and an average path length of 12 cm.
Determine the coil current required to achieve a flux of 1E−4 webers.
The analogous circuit consists of a single source and reluctance, like
that of Figure 10.12. Thus,
N I = H sheet l sheet
We shall begin by filling in the portions of the table that can be
addressed directly, such as the path length, area and flux. Don't forget to 
convert the centimeter values into meters.
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1E−4 1E−4 0.12
Now we determine the flux density to complete the left side of the table. 
B = Φ
A
B = 1E-4 Wb
1E-4 m2
B = 1T
335
Figure 10.13
Table format used for 
magnetic circuit analysis.
Figure 10.14
Magnetic system for Example 
10.3.
N turns
I",DCElectricalCircuitAnalysis.pdf
"Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1E−4 1E−4 1 0.12
We use the BH curve of Figure 10.11 to find H from this flux density. From 
the blue curve (A) the value is about 190 amp-turns per meter.
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1E−4 1E−4 1 190 0.12
 
At this point we find the Hl “drop” with a simple multiplication. 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1E−4 1E−4 1 190 0.12 22.8
This “drop” is 22.8 amp-turns and the coil has 200 turns. Therefore the 
required current is:
I = H l
N
I = 22.8At
200 t
I = 114 mA
Next, let's consider a core with two sections.
Example 10.4
Given the magnetic system shown in Figure 10.15, assume section A is
made of sheet steel and section B is made of cast steel. Each part has a",DCElectricalCircuitAnalysis.pdf
"Given the magnetic system shown in Figure 10.15, assume section A is
made of sheet steel and section B is made of cast steel. Each part has a
cross section of 2 cm by 2 cm. The path length of A is 12 cm and the
path length of B is 4 cm. If the coil has 50 turns, determine the coil
current required to achieve a flux of 2E−4 webers.
The analogous circuit consists of a single source and two reluctances.
This is shown in Figure 10.16. 
336
Figure 10.15
Magnetic system for Example 
10.4.
N turns
I
A
B",DCElectricalCircuitAnalysis.pdf
"Therefore:
N I = H sheet l sheet+H cast lcast
For our table there will be two rows, one for sheet steel and the second for 
cast steel. The first row will require using curve A (sheet steel) from Figure 
10.11 while the second row will require using curve B (cast steel).
The flux, like the current in a series loop, will be the same in both sections. 
From here we can fill out a series of other values in the table to arrive at: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 2E−4 4E−4 0.5 0.12
Cast Steel 2E−4 4E−4 0.5 0.04
The BH curves are used to transition to the right side, and we arrive at:
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 2E−4 4E−4 0.5 70 0.12
Cast Steel 2E−4 4E−4 0.5 290 0.04
 And now we fill in the Hl “drops”: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)",DCElectricalCircuitAnalysis.pdf
"And now we fill in the Hl “drops”: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 2E−4 4E−4 0.5 70 0.12 8.4
Cast Steel 2E−4 4E−4 0.5 290 0.04 11.6
337
Figure 10.16
Analogous electrical circuit for 
the system of Figure 10.15.",DCElectricalCircuitAnalysis.pdf
"Using our KVL analogy, the total “drop” is 8.4 At + 11.6 At, or 20 amp-
turns. The coil was specified as having 50 turns. Therefore:
I = H l
N
I = 20 At
50 t
I = 400 mA
Notice that even though the cast steel section is shorter than the sheet steel 
section, it produces a larger “drop”; just like a larger resistor in an electrical 
circuit. This requires a larger current in the coil than if the entire core had 
been made of sheet steel.
Finally, it should be apparent that the current demand can be reduced if the 
coil has more turns. There is a practical limit here because all of the turns 
have to fit within the opening of the core. Once that space is full, the only 
way to increase the number of turns is to use finer gauge wire but doing so 
increases wire resistance and power loss, and also lowers maximum current 
capacity.
Dealing With Air GapsDealing With Air Gaps
Some systems include an air gap in the core. If there is only a single material for",DCElectricalCircuitAnalysis.pdf
"capacity.
Dealing With Air GapsDealing With Air Gaps
Some systems include an air gap in the core. If there is only a single material for
the remainder, this will create a system like that depicted in Figure 10.17. The main
observation here is that the permeability of air is far below that of ferromagnetic
materials and thus the gap will exhibit a relatively large reluctance compared to its
path length. The obvious question then is, why would we use a gap? One possibility 
is an electro-magnetic relay, the internals of which are illustrated in Figure 10.18. 
338
Figure 10.17
Analogous electrical circuit 
for a system with an air gap.
Figure 10.18
Internals of an electrical relay.",DCElectricalCircuitAnalysis.pdf
"To create a relay, we place one portion of the core on a hinge which can be held 
open with a small spring. This creates the air gap (to the immediate left of the core in
the Figure). If we apply a sufficiently large current, the resulting magnetic flux will 
be enough to overcome the spring tension and close the second piece onto the first 
(i.e., magnetic attraction). Once this happens, the gap disappears, reducing the 
reluctance around the loop. The system will stay closed until the coil current is 
turned off and the restoring spring pulls apart the two sections again. We add 
insulated metal contacts to the moving section and what we end up with is a heavy 
duty switch that is “thrown” by a controlling current rather than a mechanical lever 
moved by a human. 
It is worth repeating that for air gaps, instead of a BH curve we can use Definition 
10.11 which shows that for air, H ≈ 7.958E5 B or alternately, 
B ≈ 1.257E−6 H.  
Example 10.5",DCElectricalCircuitAnalysis.pdf
"10.11 which shows that for air, H ≈ 7.958E5 B or alternately, 
B ≈ 1.257E−6 H.  
Example 10.5
Given the magnetic system shown in Figure 10.19, assume the core is
made of sheet steel. The cross section throughout is 3E−4 m2. The path
length of the main core is 8 cm and the path length of the gap is 1 mm.
How many turns will the coil need in order for a coil current of 400 mA
to achieve a flux of 1.2E−4 webers?
The analogous circuit consists of a single source and two reluctances,
one for the sheet steel core and a second for the air gap. This is shown
in Figure 10.17. Further, the analogous KVL relation is:
N I = H sheet l sheet+H gap l gap
For this table there will be two rows, one for the sheet steel core and one for 
the air gap. As we have seen, the flux will be the same in both sections. 
From there we can fill out some of the other values in the table resulting in: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)",DCElectricalCircuitAnalysis.pdf
"Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1.2E−4 3E−4 0.4 8E−2
Gap 1.2E−4 3E−4 0.4 1E−3
339
Figure 10.19
Magnetic system for Example 
10.5.
N turns
I
gap",DCElectricalCircuitAnalysis.pdf
"The BH curve for sheet steel and Definition 10.11 for the air gap are used to 
transition to the right side, yielding:
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1.2E−4 3E−4 0.4 63 8E−2
Gap 1.2E−4 3E−4 0.4 3.183E5 1E−3
The Hl “drops” are filled in: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 1.2E−4 3E−4 0.4 62 8E−2 5.0
Gap 1.2E−4 3E−4 0.4 3.183E5 1E−3 318.3
Using our KVL analogy, the total “drop” is 5.0 At + 318.3 At, or 323.3 amp-
turns. The coil current was specified as being 400 mA. Therefore:
N = H l
I
N = 323.3 At
400 mA
N ≈ 808 turns
And finally, an example using two coils.
Example 10.6
Given the magnetic system shown in Figure 10.20, assume the core is
made of sheet steel. The cross section throughout is 3E−4 m2 and the
length of the core is 20 cm. Coil one consists of 2000 turns and coil two",DCElectricalCircuitAnalysis.pdf
"made of sheet steel. The cross section throughout is 3E−4 m2 and the
length of the core is 20 cm. Coil one consists of 2000 turns and coil two
consists of 500 turns. If a current of 120 mA into coil one achieves a
flux of 2.4E−4 webers, determine the current of coil 2.
The analogous circuit consists of two voltage sources and a single
reluctance for the sheet steel core. This is illustrated in Figure 10.21. 
340
Figure 10.20
Magnetic system for Example 
10.6.
N1
I1
N2
I2",DCElectricalCircuitAnalysis.pdf
"The analogous KVL relation is:
N 1 I1 −N 2 I 2 = H sheet l sheet
As we are seeking I2, we can rearrange this equation into a more useful
form:
N 2 I 2 = N1 I 1 −H sheet l sheet
Only one row will be required for this table. We fill out the obvious values 
in the table resulting in: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 2.4E−4 3E−4 0.8 0.2
The BH curve for sheet steel is used to transition to the right side:
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 2.4E−4 3E−4 0.8 100 0.2
The Hl “drop” is computed: 
Section Flux
Φ (Wb)
Area
A (m2)
Flux Density
B (T)
Magnetizing Force
H (At/m)
Length
l (m)
“Drop”
Hl (At)
Sheet Steel 2.4E−4 3E−4 0.8 100 0.2 20
Using our KVL analogy, we find:
N 2 I 2 = N1 I 1 −H sheet l sheet
N 2 I 2 = 2000turns×120 mA −20 At
N 2 I 2 = 240 At −20 At
N 2 I 2 = 220 At",DCElectricalCircuitAnalysis.pdf
"Using our KVL analogy, we find:
N 2 I 2 = N1 I 1 −H sheet l sheet
N 2 I 2 = 2000turns×120 mA −20 At
N 2 I 2 = 240 At −20 At
N 2 I 2 = 220 At
Coil two was specified as having having 500 turns, therefore its current is:
 I 2 = 220 At
500 turns
I 2 = 440 mA
341
Figure 10.21
Analogous electrical circuit 
for the system of Figure 10.20.",DCElectricalCircuitAnalysis.pdf
"An item of key importance is that the preceding example must make use of an AC 
current in order to function as described. A DC current will not produce the predicted
results. The reason for this goes back to Definition 10.1 and Equation 10.2: Unless 
the flux is changing relative to the conducting coil, no voltage will be induced in a 
coil. So, while a DC input current will create a flux in the core, that flux will be 
static and unchanging. Consequently, the second coil will not produce an output. In 
contrast, an AC current smoothly varies in amplitude and polarity. This creates a 
smoothly varying magnetic flux which in turn allows a voltage to be induced in the 
second coil.
Assuming for a moment that an AC current was used in the preceding example, 
notice that the current is nearly four times higher in coil two than in coil one. If we 
had been able to reduce the core “drop” to zero, perhaps by reducing the core's",DCElectricalCircuitAnalysis.pdf
"had been able to reduce the core “drop” to zero, perhaps by reducing the core's 
reluctance to a negligible value, then the current would have been increased 
precisely by a factor four (to 480 mA versus 440 mA). This is the same as the ratio 
of the number of turns of coil one to the number of turns of coil two, and is known 
as the turns ratio. It is a key parameter describing transformers, which just happens 
to be the subject of the next section.
10.4 Transformers10.4 Transformers
Transformers have been in wide use in the electrical industry for over a century. 
Beginning with Michael Faraday and working through a sequence of scientists and 
inventors, the fundamentals of modern transformer design had been established by 
the late 1800s. 
Transformer have three basic uses: 
• V oltage scaling. 
• Isolation. 
• Impedance matching. 
Transformers range from very small units used in communications systems up to",DCElectricalCircuitAnalysis.pdf
"• V oltage scaling. 
• Isolation. 
• Impedance matching. 
Transformers range from very small units used in communications systems up to 
large power distribution units many times the size of an adult human. A couple of  
transformers typical of the type used in consumer electronics devices is shown in 
Figure 10.22. These devices are designed to handle standard residential AC input 
voltages (120 volts in North America, 220 to 240 volts in much of the remainder of 
the planet) and feeding circuits that demand perhaps 10 to 30 watts of power. The 
unit on the left is designed to be mounted directly onto a printed circuit board while 
the unit on the right is designed for chassis mount. 
342",DCElectricalCircuitAnalysis.pdf
"In contrast, Figure 10.23 shows pole mounted transformers used to reduce typical 
North American residential distribution lines (in the vicinity of 15 kV) down to 
home voltage (120 volts). These transformers are capable of delivering 1000 or more
times as much power as the transformers of Figure 10.22. The transformers in these 
figures by no means present the extremes. Transformers are available that are 
considerably smaller or larger than the ones pictured here.
Transformers exploit magnetic circuit theory for their operation, as illustrated
previously in Example 10.6. In its most basic form, the device has two coils or 
windings; one for the input side (called the primary winding) and one for the output
side (the secondary winding). It is possible to have multiple input and/or output
windings. Such is the case for the transformer on the left side of Figure 10.22; it
has two input primary coils and two secondary coils. This allows for a variety of",DCElectricalCircuitAnalysis.pdf
"has two input primary coils and two secondary coils. This allows for a variety of
interconnection options, as we shall see. A typical schematic symbol is shown in
Figure 10.24. This symbol is appropriate for the magnetic system of Figure 10.20
which featured two coils. The vertical lines in the symbol represent the magnetic
core.
Voltage Scaling and IsolationVoltage Scaling and Isolation
A particularly important transformer parameter is the ratio of the number of turns
on the primary coil (NP) to the number of turns on the secondary coil (NS). This is
denoted N, and referred to as the turns ratio. 
N = N P
N S
(10.13)
343
Figure 10.22
Transformers suitable for 
consumer electronics devices. 
The unit on the left features 
four separate coils for 
enhanced flexibility.
Figure 10.23
A cluster of three residential 
power transformers.",DCElectricalCircuitAnalysis.pdf
"As the two windings are on the same core and ideally see the same flux, the voltage 
ratio of the primary to secondary will be equal to N for an ideal lossless transformer. 
Thus, if N=10 and the primary voltage (VP) is 60 volts, the secondary voltage (VS) 
will be 6 volts. Thus:
V S = V P
N (10.14)
The primary and secondary appear as “NI” potentials in the magnetic circuit,
therefore, the current will change inversely with N. For example, if N=10 and the
primary current is 1 amp, the secondary current will be 10 amps. Thus:
I S = N I P (10.15)
It is important to note that the product of primary current and voltage must equal the
product of secondary current and voltage for an ideal transformer. Such a device
does not dissipate power but rather transforms it (hence the name). 
V S I S = V P I P (10.16)
Real world devices will exhibit some loss due to finite winding wire resistance, non-
ideal behavior of the magnetic circuit and the like. Instead of a power dissipation",DCElectricalCircuitAnalysis.pdf
"ideal behavior of the magnetic circuit and the like. Instead of a power dissipation 
rating, power transformers are given a V A (volt-amps) rating, typically thought of as 
secondary voltage times maximum secondary current. Transformers like the ones 
pictured in Figure 10.22 have ratings in the range of 10 to 30 V A. Generally, the 
higher the V A rating, the larger and heavier the transformer. On the other hand, 
applications using signal transformers in communications systems tend to focus on 
the turns ratio and not on the V A rating as the associated voltages and currents tend 
to be small. Their goal generally is to increase the amplitude of a signal voltage.
Also, because there is no direct electrical connection between the primary and 
secondary, the transformer may be used solely for isolation. This is a safety feature. 
If N=1, then there is no change in the voltage being presented to the device, ideally. 
 
Example 10.7",DCElectricalCircuitAnalysis.pdf
"If N=1, then there is no change in the voltage being presented to the device, ideally. 
 
Example 10.7
A microphone feeds a small transformer which is used as the first stage of a 
preamplifier. The transformer has a primary winding of 50 turns and a 
secondary winding of 250 turns. Assuming an ideal transformer, determine 
the output voltage of the transformer if the signal from the microphone is 35 
microvolts.
As this is an ideal transformer, the turns ratio indicates how much the input 
voltage is scaled. First we use Equation 10.13 to determine the turns ratio. 
344
Figure 10.24
Schematic symbol for a simple
transformer.",DCElectricalCircuitAnalysis.pdf
"N = N P
N S
N = 50
250
N = 0.2
Now we use Equation 10.14 to determine the secondary voltage.
V S = V P
N
V S = 35μ V
0.2
V S = 175μV
Power applications will make use of N to scale the voltage in order to create a more 
appropriate or efficient level. If the voltage is brought down (N>1), it is referred to 
as a step-down transformer. If N<1, it is referred to as a step-up transformer. For 
example, a consumer electronics product may require a modest DC voltage for its 
operation, say 15 or 20 volts. To achieve this, the standard North American 
residential “wall” voltage of 120 volts may be reduced using a step-down
transformer with N=5. This would result in a secondary voltage of 24 volts that
could then be rectified, filtered, and regulated to the desired DC level. Conversely,
a step-up transformer could be used to create a much higher secondary potential
(for example, in long distance power transmission to reduce I2R losses).",DCElectricalCircuitAnalysis.pdf
"(for example, in long distance power transmission to reduce I2R losses). 
For some applications, split secondaries or multiple secondaries may be used. A
center tap is relatively common and is used to divide the secondary into two equal 
voltage sections (e.g., a 24 volt center-tapped secondary would behave as two 12 
volt secondaries in series). The schematic symbol for a transformer with a center-
tapped secondary is shown in Figure 10.25. 
Multiple secondaries are also common and are more flexible than a center-tapped 
secondary. Multiple secondaries can be combined in series to increase the secondary 
voltage or combined in parallel to increase secondary current capacity (i.e, load 
current).
Example 10.8
The power transformer shown in Figure 10.26 has turns ratio of 10. If the
primary is connected to a 120 volt source and the secondary is connected in
series with a 32 Ω load resistor, determine the current through the load
resistor. Assume this is an ideal transformer.",DCElectricalCircuitAnalysis.pdf
"series with a 32 Ω load resistor, determine the current through the load
resistor. Assume this is an ideal transformer.
The turns ratio indicates secondary voltage via Equation 10.4.
345
Figure 10.25
Schematic symbol for a 
transformer with a center- 
tapped secondary.
Figure 10.26
Circuit for Example 10.8.",DCElectricalCircuitAnalysis.pdf
"V S = V P
N
V S = 120 V
10
V S = 12V
Ohm's law can be used to find the secondary current.
I S = V S
RL
I S = 12V
32Ω
I S = 375 mA
As mentioned earlier, transformers work with AC voltages and currents. Unlike the 
consistent polarity of DC sources, AC (literally, Alternating Current) flips back and 
forth in voltage polarity and current direction. In order to determine the phase 
relationship between the primary and secondary, transformers use “dot notation”. 
The dot indicates the positive polarity of voltage. This is illustrated in the circuit of 
Figure 10.27.
On the primary side, the current flows into the dot and establishes the positive 
reference. In this case, the primary is seen as the load for the source, E. On the 
secondary side, current flows out of the dot and also notates the positive reference 
because the secondary is seen as the source for the load, R. 
Impedance Matching
The third use of the transformer is impedance matching. Impedance is an AC",DCElectricalCircuitAnalysis.pdf
"Impedance Matching
The third use of the transformer is impedance matching. Impedance is an AC 
phenomenon that consists of both resistance and reactance (an ohmic value 
associated with capacitors and inductors). The symbol for impedance is Z. Although 
impedance is measured in ohms, it differs from ordinary resistance in that it is a 
vector quantity and has a phase angle. For our present purposes, we can ignore this 
distinction.27 
27 The concept of impedance is covered in great detail in the companion text, AC Electrical 
Circuit Analysis, another free OER title.
346
Figure 10.27
Transformer dot convention.",DCElectricalCircuitAnalysis.pdf
"Because both the voltage and current are being scaled, the source's “view” of the 
load changes. The impedance seen by the source can be defined via Ohm's law as 
the voltage it produces divided by its output current.
ZP = V P
I P
(10.17)
From Equation 10.14, we know that VP = VS ∙ N. Substituting this into Equation 
10.17 yields:
Z P = V S N
I P
(10.18)
From Equation 10.15, we know that IP = IS / N. Substituting this into Equation 10.18 
yields:
Z P = V S N 2
I S
(10.18)
By definition,
Z S = V S
I S
(10.19)
Combining Equations 10.18 and 10.19 leaves us with:
Z P = N 2 Z S (10.20)
Thus, the impedance seen by the source is the secondary impedance times the square
of the turns ratio. This is called the reflected impedance.
Example 10.9
Assume that the transformer shown in Figure 10.27 has turns ratio of 5. If 
the primary is connected to an 20 volt source and the secondary is connected
to an 8 Ω load resistor, determine the primary current and the reflected",DCElectricalCircuitAnalysis.pdf
"to an 8 Ω load resistor, determine the primary current and the reflected 
impedance. Assume this is an ideal transformer.
First we'll use Equation 10.20 to find the reflected impedance seen in the 
primary. 
347",DCElectricalCircuitAnalysis.pdf
"Z P = N 2
Z S
Z P = 52
×8Ω
Z P = 200 Ω
Ohm's law can be used to find the primary current.
I P = V P
RP
I P = 20V
200Ω
I P = 100 mA
Here is where impedance matching comes in. In the preceding example, if the 
source, E, had a large internal resistance compared to the load resistance, R, and 
there was no transformer, there would be considerable loss in the system. For 
example, if the source impedance was 10 Ω and it was directly connected to the 8 Ω 
load without the transformer, the resulting voltage divider would cause a great deal 
of internal power dissipation in the source leading to inefficiency. On the other hand,
in the given circuit with transformer, the source sees a reflected impedance of 200 Ω.
The resulting voltage divider between 200 Ω and the 10 Ω internal resistance of the 
source will cause very little loss, yielding a much more efficient transfer.
10.5 Summary10.5 Summary
This chapter has extended the material presented in Chapter 9 on inductance into the",DCElectricalCircuitAnalysis.pdf
"10.5 Summary10.5 Summary
This chapter has extended the material presented in Chapter 9 on inductance into the
realm of magnetic circuits. We began with the concept of electromagnetic induction, 
which in simple terms states that if a conductor experiences changing magnetic lines
of force, then a voltage will be induced in that conductor. This concept is exploited 
by an array of different devices including electric motors, generators, relays, 
loudspeakers, microphones, and other transducers and sensors. For example, 
dynamic loudspeakers and microphones both make use of a coil of wire suspended 
in a fixed magnetic field. In the case of the loudspeaker, the signal fed into the coil 
creates a changing magnetic field that interacts with the permanent magnet's field 
and causes an attached diaphragm to vibrate against the air, which in turn creates 
sound. A dynamic microphone works in an opposite manner. Sound waves cause a",DCElectricalCircuitAnalysis.pdf
"sound. A dynamic microphone works in an opposite manner. Sound waves cause a 
diaphragm to vibrate back and forth. This diaphragm is attached to a coil, and the 
motion of the coil within the permanent magnet's field causes a voltage to be 
induced in the coil. This voltage can then be amplified by electronic circuitry. 
Magnetic circuits are constructed by creating one or more coils of wire wrapped 
around a core that is made of a ferromagnetic material such as steel. A key part of 
analyzing magnetic circuits is Hopkinson's law (also called Rowland's law), which is
the magnetic circuit version of Ohm's law. In this analogy, magnetic flux is likened 
348",DCElectricalCircuitAnalysis.pdf
"to current flow, magnetic reluctance stands in for resistance, and magnetomotive 
force is analogous to electromotive force (i.e., voltage). Further, the magnetomotive 
force is the product of the number of turns in a coil and the current through said coil,
or NI. When current is passed through the coil, a magnetizing force, H, and an 
associated magnetic flux are created. Given this, the core material will have a 
corresponding flux density, B. The linkage between B and H is found through a BH 
curve for that particular material. BH curves are non-linear for ferromagnetic 
materials. They also exhibit hysteresis, meaning that the recent operational history of
the material plays a role in its current state. Every element of the core has a certain 
reluctance and thus exhibits an analogous “voltage drop”. These “drops” must add 
up to the NI “rises” of the coil(s), in a KVL analogy.
Transformers use one or more primary coils to transform an input voltage into one",DCElectricalCircuitAnalysis.pdf
"up to the NI “rises” of the coil(s), in a KVL analogy.
Transformers use one or more primary coils to transform an input voltage into one 
more output voltages appearing on secondary coils. The ratio between the number of
turns of the primary coil to the number of turns of the secondary coil is called the 
turns ratio, N. The secondary voltage is equal to the primary voltage divided by N 
while the secondary current is equal to the primary current times N. If N is greater 
than one, we have a step-down transformer (reducing the voltage), and if N is less 
than one, we have a step-up transformer (increasing the voltage). Ideally, 
transformers do not dissipate power. Thus, they do not have a power rating, but 
instead have a V A (volt-amps) rating to indicate their capacity. Transformers can also
be used for impedance matching to increase the efficiency of a system. The 
impedance seen by the source on the primary is equal to the secondary impedance",DCElectricalCircuitAnalysis.pdf
"impedance seen by the source on the primary is equal to the secondary impedance 
times the square of the turns ratio. This is called the reflected impedance.
Review QuestionsReview Questions
1. Describe the concept of electromagnetic induction.
2. Describe Ohm's law for magnetic circuits (Hopkinson's or Rowland's law).
3. Describe hysteresis.
4. Define residual magnetism and coercivity.
5. Outline the operation of a relay.
6. What are some of the uses of transformers?
7. What is turns ratio? How does it affect voltage and current in the primary 
and secondary?
8. What is the difference between a step-up and a step-down transformer?
9. What is reflected impedance?
349",DCElectricalCircuitAnalysis.pdf
"10.6 Exercises 10.6 Exercises 
AnalysisAnalysis
1. In the magnetic circuit shown in Figure 10.1, assume the cross section is 1 
cm by 1 cm with a path length of 8 cm. The entire core is made of sheet 
steel and there are 100 turns on the winding. Determine the current to 
establish a flux of 8E−5 webers.
2. Repeat problem 1 using cast steel for the core.
3. Given the core shown in Figure 10.1, assume the cross section is 2 cm by 2 
cm with a path length of 10 cm. The entire core is made of cast steel and 
there are 200 turns on the winding. Determine the current to establish a flux 
of 4E−4 webers.
4. Repeat problem 3 using sheet steel for the core.
350
N turns
I Figure 10.1",DCElectricalCircuitAnalysis.pdf
"5. In the magnetic circuit shown in Figure 10.1, assume the cross section is 1 
cm by 1 cm with a path length of 8 cm. The entire core is made of sheet 
steel. Determine the number of turns required to establish a flux of 8E−5 
webers given a current of 50 mA.
6. Given the core shown in Figure 10.1, assume the cross section is 2 cm by     
2 cm with a path length of 10 cm. The entire core is made of sheet steel 
Determine the number of turns required to establish a flux of 4E−4 webers 
given a current of 200 mA.
7. In the magnetic circuit shown in Figure 10.2, assume the cross section is      
1 cm by 1 cm. Section A is sheet steel with a path length of 6 cm. Section B 
is cast steel with a length of 2 cm. There are 500 turns on the winding. 
Determine the current to establish a flux of 5E−5 webers.
8. In the magnetic circuit shown in Figure 10.2, assume the cross section is      
1 cm by 1 cm. Section A is sheet steel with a path length of 6 cm. Section B",DCElectricalCircuitAnalysis.pdf
"1 cm by 1 cm. Section A is sheet steel with a path length of 6 cm. Section B 
is cast steel with a length of 2 cm. Determine the number of turns on the coil
to establish a flux of 6E−5 webers with a current of 50 mA.
9. A transformer is shown in Figure 10.3, assume the cross section is 5 cm by   
5 cm. The core is sheet steel with a path length of 20 cm. N1 is 500 turns 
and N2 is 200 turns. Determine the secondary current (I2) if a primary 
current (I1) of 1 amp establishes a flux of 1.5E−3 webers.
10. Given the same conditions of problem 9, alter the secondary turns (N2) so 
that the secondary current (I2) is 3 amps given the original primary current 
of 1 amp.
351
Figure 10.2
N turns
I
A
B
Figure 10.3
N1
I1
N2
I2",DCElectricalCircuitAnalysis.pdf
"11. In general, how would the performance noted in problem 9 change if cast 
steel was substituted for sheet steel?
12. Given the results of problems 9 through 11, what does the ratio of N1 to N2 
represent in terms of idealized performance, and what steps should be taken 
to make the transformer operate as close to ideal as possible?
13. Given the magnetic circuit shown in Figure 10.4, assume the cross section is
1 cm by 2 cm with a path length of 6 cm. The entire core is made of sheet 
steel with the exception of a 1 mm air gap. Determine the current required to
establish a flux of 8E−5 webers if N = 1000 turns.
14. Using the data given in problem 13, determine the number of turns required 
to establish the same flux when the input current is 200 mA.
15. An ideal transformer has a 6:1 voltage step-down ratio. If the primary is 
driven by 24 V AC and the load is 100 Ω, determine the load voltage and 
current, and the primary side current.",DCElectricalCircuitAnalysis.pdf
"driven by 24 V AC and the load is 100 Ω, determine the load voltage and 
current, and the primary side current.
16. A 120 V AC transformer is specified as having a 36 volt center-tapped 
secondary. If each side of the secondary is connected to its own 50 Ω load, 
determine the load currents and the primary side current.
17. An ideal transformer has a 12:1 voltage step-down ratio. If the secondary is 
connected to a 10 Ω load, what impedance is seen from the primary side?
18. An ideal transformer has a 1:5 voltage step-up ratio. If the secondary is 
connected to a 2 Ω load, what impedance is seen from the primary side?
DesignDesign
19. A step-down transformer with N=8 has a 15 volt RMS secondary which is 
connected to a load with an effective value of 5 Ω. Determine the minimum 
acceptable V A rating of the transformer.
352
N turns
I
gap
Figure 10.4",DCElectricalCircuitAnalysis.pdf
"20. A step-up transformer with N=.5 is driven from a 120 V AC source. The 
secondary is connected to a load with an effective value of 150 Ω. 
Determine the minimum acceptable V A rating of the transformer.
ChallengeChallenge
21. A transformer specified as having a 120 V AC primary with an 18 volt 
secondary is accidentally connected backwards, with its secondary 
connected to the source and its primary connected to a 16 Ω load. Determine
the load current in both the normal and reversed connections. Also 
determine the required transformer V A rating for both connections.
22. Consider the distributed public address system for an airport as shown in 
Figure 10.5. It consists of an audio power amplifier with a nominal 70 volt 
RMS output that is connected to four remote loudspeakers, each separate 
from the others and some 150 meters away from the amplifier. Each 
loudspeaker assembly includes a 10:1 voltage step-down transformer that",DCElectricalCircuitAnalysis.pdf
"from the others and some 150 meters away from the amplifier. Each 
loudspeaker assembly includes a 10:1 voltage step-down transformer that 
feeds the loudspeaker impedance of 8 Ω (resistive) off its secondary. These 
four lines are fed in parallel by the amplifier. Determine the power delivered
to each loudspeaker and the total current delivered by the power amplifier. 
Assume the transformers are ideal and ignore any cable resistance.
23. Continuing with the preceding problem, assume that the wiring connecting 
each transformer back to the amplifier is AWG 22. Determine the power lost 
in each of the 150 meter long sections of dual cable. Further, suppose the 
system is reconfigured without the transformers and the output of the 
amplifier is lowered to 7 volts RMS to compensate. Determine the power lost
in each of the cable feeds under the new configuration.
353
Figure 10.5",DCElectricalCircuitAnalysis.pdf
"SimulationSimulation
24. Use a transient analysis to verify the results of problem 15.
25. Use a transient analysis to verify the results of problem 16.
354",DCElectricalCircuitAnalysis.pdf
"Appendix AAppendix A
Standard Component SizesStandard Component Sizes
Passive components (resistors, capacitors and inductors) are available in standard sizes. The tables below are for
resistors. The same digits are used in subsequent decades up to at least 1 Meg ohm (higher decades are not 
shown). Capacitors and inductors are generally not available in as many standard values as are resistors. 
Capacitors below 10 nF (.01 μF) are usually available at the 5% standard digits while larger capacitances tend to
be available at the 20% standards.
5% and 10% standard values, EIA E24 and EIA E125% and 10% standard values, EIA E24 and EIA E12
10% values (EIA E12) are bold
20% values (seldom used) are every fourth value starting from 10 
(i.e., every other 10% value)
10 11 12 13 15 16 18 20 22 24 27 30
33 36 39 43 47 51 56 62 68 75 82 91
1% and 2% standard values, EIA E96 and EIA E481% and 2% standard values, EIA E96 and EIA E48
2% values (EIA E48) are bold",DCElectricalCircuitAnalysis.pdf
"1% and 2% standard values, EIA E96 and EIA E481% and 2% standard values, EIA E96 and EIA E48
2% values (EIA E48) are bold
10.0 10.2 10.5 10.7 11.0 11.3 11.5 11.8 12.1 12.4 12.7 13.0
13.3 13.7 14.0 14.3 14.7 15.0 15.4 15.8 16.2 16.5 16.9 17.4
17.8 18.2 18.7 19.1 19.6 20.0 20.5 21.0 21.5 22.1 22.6 23.2
23.7 24.3 24.9 25.5 26.1 26.7 27.4 28.0 28.7 29.4 30.1 30.9
31.6 32.4 33.2 34.0 34.8 35.7 36.5 37.4 38.3 39.2 40.2 41.2
42.2 43.2 44.2 45.3 46.4 47.5 48.7 49.9 51.1 52.3 53.6 54.9
56.2 57.6 59.0 60.4 61.9 63.4 64.9 66.5 68.1 69.8 71.5 73.2
75.0 76.8 78.7 80.6 82.5 84.5 86.6 88.7 90.9 93.1 95.3 97.6
355",DCElectricalCircuitAnalysis.pdf
"Appendix BAppendix B  
Methods of Solution of Linear Simultaneous EquationsMethods of Solution of Linear Simultaneous Equations
Some circuit analysis methods, such as nodal analysis and mesh analysis, yield a set of linear simultaneous 
equations. There will be as many equations as there are unknowns. For example, a particular circuit might yield 
three equations with three unknown currents (often referred to as a “3 by 3” for the matrix it creates). There are 
several techniques that may be used to solve this system of equations. The methods include graphical, 
substitution, Guass-Jordan elimination and determinants (determinants may be solved via Cramer's Rule/Sarrus’ 
Rule or via expansion by minors). 
GraphicalGraphical
Graphical solutions involve plotting the individual equations on graph paper. The location of where the lines 
cross is the solution to the system (i.e., values that satisfy all of the equations). This technique will not be",DCElectricalCircuitAnalysis.pdf
"cross is the solution to the system (i.e., values that satisfy all of the equations). This technique will not be 
discussed further because it is only practical for two unknowns. It would be very difficult to draw something 
like a four dimensional graph for four equations with four unknowns! 
SubstitutionSubstitution
The idea here is to write one of the equations in terms of one of the unknowns and then substitute this back into 
one of the other equations resulting in a simplified version. This process is iterated for as many unknowns as the
system includes. Take for example the following 2x2:
10 = 20I1 + 8I2
 2 =  8I1 + 4I2
Solve the second equation for I2.
 2 =  8I1 + 4I2
4I2 =   2 − 8I1
 I2 = 0.5 − 2I1
Substitute this back into the first equation and expand/simplify/solve.
10 = 20I1 + 8I2
10 = 20I1 + 8(.5 − 2I1)
10 = 20I1 + 4 − 16I1
10 =  4I1 + 4 
 I1 =  1.5
Finally, substitute this value back into one of the two original equations to determine I2. 
356",DCElectricalCircuitAnalysis.pdf
"2 = 8I1 + 4I2
  2 = 8(1.5) + 4I2
  2 = 12 + 4I2
4I2 = −10 
 I2 = −2.5 
For a 3x3, this process is iterated as follows: Equation 2 would be solved for I3 and this would be substituted 
back into equation 1 yielding a new equation (let’s call it A) with only I1 and I2 terms. Similarly, Equation 3 
would be solved for I3 and this would be substituted back into equation 2 yielding a new equation (let’s call it B)
with only I1 and I2 terms. Equations A and B now make a 2x2 with I1 and I2 as the unknowns and can be solved 
as outlined above. This would yield values for I1 and I2 which could then be substituted into one of the three 
original equations to obtain I3. While the substitution method is perfectly valid for an arbitrarily sized system, it 
proves cumbersome as the system gets larger. 
Gauss-Jordan EliminationGauss-Jordan Elimination
In some respects, Gauss-Jordan is similar to substitution but it tends to involve less overhead for larger systems",DCElectricalCircuitAnalysis.pdf
"In some respects, Gauss-Jordan is similar to substitution but it tends to involve less overhead for larger systems 
and thus is generally preferred. This method involves multiplying one equation by a constant such that when it is
subtracted from another equation, one of the unknown terms disappears. The process is then iterated for as many
unknowns exist in the system. Using the same example from before:
10 = 20I1 + 8I2
 2 =  8I1 + 4I2
Multiply the second equation by the ratio of the coefficients for I2 (8/4 = 2).
 2 =  8I1 + 4I2
 4 = 16I1 + 8I2
Subtract this new equation from the first equation. The I2 terms will cancel leaving just I1.
10 = 20I1 + 8I2
 4 = 16I1 + 8I2
 6 = 4I1
 I1 = 1.5 
Substitute this result back into one of the original equations to obtain I2. For a 3x3, iterate as follows: Using 
equations 1 and 2, multiply equation 2 by the ratio of the coefficients for I3. Subtract this equation from equation",DCElectricalCircuitAnalysis.pdf
"equations 1 and 2, multiply equation 2 by the ratio of the coefficients for I3. Subtract this equation from equation
1 to generate a new equation (let’s call it equation A) that only has I1 and I2 as unknowns. Using equations 2 and 
3, multiply equation 3 by the ratio of the coefficients for I3. Subtract this equation from equation 2 to generate a 
new equation (let’s call it equation B) that only has I1 and I2 as unknowns. Equations A and B now make a 2x2 
with I1 and I2 as the unknowns and can be solved as outlined previously. This would yield values for I1 and I2 
which could then be substituted into one of the three original equations to obtain I3. Like the substitution 
method, Gauss-Jordan grows rapidly as the system size increases. The process tends to be formulaic though, and
thus easier to handle.  
357",DCElectricalCircuitAnalysis.pdf
"DeterminantsDeterminants
Determinants revolve around the concept of a matrix which itself is little more than an ordered collection of 
coefficients and/or constants. It is imperative that the unknowns be in the same order in each equation (i.e., I1 
ascending to IX left to right) A simple coefficient matrix for the original 2x2 example is:
20  8
 8  4
The resultant value (properly referred to as the determinant) for a 2x2 matrix such as this may be solved using 
Sarrus’ Rule: Simply multiply the two values along the upper right-lower left diagonal and then subtract that 
product from the product of the two values found along on the upper left-lower right diagonal. In this example 
that’s: 
20*4 − 8*8 = 16
A solution involves dividing one determinant by another determinant (Cramer's Rule). That is, each matrix is 
solved for its resultant value and then these two values are divided to determine the final answer. One of these",DCElectricalCircuitAnalysis.pdf
"solved for its resultant value and then these two values are divided to determine the final answer. One of these 
matrices will be the coefficient matrix just discussed. This will be placed in the denominator. The numerator 
matrix is a modified version of the basic coefficient matrix. It is created by replacing one column of coefficients 
with the constant values from the original system of equations. For example, the numerator matrix used to find 
I1 would replace the first column (the I1 coefficients 20 and 8) with the constants 10 and 2:
10  8
 2  4
The resultant value is 40 − 16 or 24.
Similarly, the numerator matrix for I2 would replace the I2 coefficients in the second column (8 and 4) with the 
constants 10 and 2:
20 10
 8  2
The resultant value is 40 − 80 or −40. To find any particular unknown, simply divide the modified matrix by the 
basic coefficient matrix.
      10  8
       2  4
I1 = --------
      20  8
       8  4
      24
I1 = -----
      16
358",DCElectricalCircuitAnalysis.pdf
"I1 =  1.5
In like fashion I2 is found:
      20 10
       8  2
I2 = --------
      20  8
       8  4
     −40
I2 = -----
      16
I2 = −2.5
Sarrus’ Rule may also be used with a 3x3 matrix. This is achieved by extending the matrix. Fourth and fifth 
columns are added to the right of the 3x3 matrix by simply making copies of the first two columns. This creates 
three right to left diagonals with three values each and three left to right diagonals with three values each. The 
three values along each diagonal are multiplied together. The three right to left products are then subtracted from
the sum of the three left to right products yielding a single resultant value (the determinant). To create the 
modified numerator matrix, replace the coefficient column of interest with the constant terms and then replicate 
columns one and two. For example, given these three equations:
10 = 20I1 + 8I2 + 3I3
 2 =  8I1 + 4I2 + 5I3
 7 =  3I1 + 5I2 + 6I3",DCElectricalCircuitAnalysis.pdf
"columns one and two. For example, given these three equations:
10 = 20I1 + 8I2 + 3I3
 2 =  8I1 + 4I2 + 5I3
 7 =  3I1 + 5I2 + 6I3
The basic coefficient matrix (i.e., denominator) is:
20  8  3
 8  4  5
 3  5  6
The extended matrix is:
20  8  3  20  8
 8  4  5   8  4
 3  5  6   3  5
The result is: 
20*4*6 + 8*5*3 + 3*8*5 − 3*4*3 − 20*5*5 − 8*8*6
Sarrus’ Rule does not work beyond 3x3. 
359",DCElectricalCircuitAnalysis.pdf
"Expansion by Minors Expansion by Minors 
Expansion by Minors is another method that may be used to generate a determinant solution. This involves 
breaking the matrix into a series of smaller matrices (minors) that are combined using row-column coefficients. 
The position of these coefficients will also indicate whether the determinant of any particular sub-matrix is 
added or subtracted to the total. 
The first step is to establish a single row or column from which to derive the coefficients. This can be any 
horizontal row or vertical column (no diagonals). Each element of the chosen row or column determines the 
associated minor (essentially, that which is left over). Consider the 3x3 used previously:
20  8  3
 8  4  5
 3  5  6
Choosing the top row yields coefficients of 20, 8 and 3. For each of these, blot out its row and column and see 
what is left. This leaves three 2x2 matrices, one for each coefficient. Multiply each coefficient by the",DCElectricalCircuitAnalysis.pdf
"what is left. This leaves three 2x2 matrices, one for each coefficient. Multiply each coefficient by the 
determinant of its 2x2 matrix. To determine whether this result is added or subtracted to the others, the sign may 
be found using the following map for the coefficients:
+  −  +  − etc.
−  +  −  + etc.
+  −  +  − etc.
The origin in the upper left is positive and the signs continually alternate across from it and down from it. The 
result using the top row for the coefficients is found thus (the 2x2 matrices are bold red for clarity):
     4 5       8 5       8 4
20 * 5 6 − 8 * 3 6 + 3 * 3 5
If the second column was used instead (8, 4, 5), the result is found like so:
     8 5      20 3      20 3
−8 * 3 6 + 4 * 3 6 − 5 * 8 5
In closing, whichever method is used, always look for null coefficient terms (that is, places in the equations and 
matrices where the coefficients are zero). Smart use of these can considerably simplify the computations as there",DCElectricalCircuitAnalysis.pdf
"matrices where the coefficients are zero). Smart use of these can considerably simplify the computations as there
are few mathematical operations easier than multiplying by zero. For example, if a particular row of a matrix 
contains a few zeros, that would be a good candidate for the coefficient row when using expansion by minors 
because some 2x2 minors need not be computed (they will just be multiplied by the zero coefficient).
360",DCElectricalCircuitAnalysis.pdf
"Appendix CAppendix C
Equation ProofsEquation Proofs
Maximum Power Transfer: Refining the Maximizing Value of P = R/(R2+2R+1)
While the algebraic and graphing technique explored in Chapter 6 leads to a proper answer, it is incomplete. For 
the ultimate value we can use a little differential calculus. Examining the curve of the function indicates a single 
peak, and thus we may find the corresponding value by taking the first derivative of the function, setting it to 
zero (i.e., find the point where the slope goes to zero), and solving for R. There are three basic techniques for 
determining the derivative of the equation above: using quotient rule, using product rule, and using chain rule. 
The quotient rule is perhaps the most obvious route although not necessarily the most efficient.
The quotient rule states that the derivative of a quotient is 
2)(
)(')()(')(
)(
)'(
xg
xgxfxfxg
xg
xf 
f(x) is R and therefore f '(x) is 1.
g(x) is R2+2R+1 and therefore g′(x) is 2R+2.",DCElectricalCircuitAnalysis.pdf
"2)(
)(')()(')(
)(
)'(
xg
xgxfxfxg
xg
xf 
f(x) is R and therefore f '(x) is 1.
g(x) is R2+2R+1 and therefore g′(x) is 2R+2.
Note that for simplicity, g(x) may also be written as (R+1)2.
Plugging these values into the quotient formula yields
4
2
)1(
)22(1)1(

 R
RRR
dR
dP
This may be simplified to
4
2
)1(
1

 R
R
dR
dP
For dP/dR to be zero, R must equal 1, meaning that maximum power is developed in the load when the load 
resistance matches the resistance of the driving source.
QED
361",DCElectricalCircuitAnalysis.pdf
"Exponential Charge and Discharge Equations
Utilizing a simple series RC circuit, such as depicted in the accompanying figure, we may write an expression 
for the natural discharge of the capacitor using KCL. We shall assume that the capacitor has some initial starting 
voltage, vC(0) = v0. Looking at node A, the total current entering and leaving this node must be 0. Expressing this
using the fundamental current/voltage relationships for capacitors and resistors gives us:
C dvC
dt + v0
R = 0
Applying a little algebra:
dvC
dt + v0
R C = 0
dvC
v0
= − 1
RC dt
Remember, the voltage at t = 0 is v0 and at time t is vC(t). Integrating will allow us to obtain an expression for 
vC(t). Using x and y as variables of integration yields:
∫
v0
vC (t)
dx
x =− 1
R C ∫
0
t
dy
 
Using the rule regarding natural logs, this can be written as:
ln(
vC (t )
v0 )=− 1
RC t
And solving for vC(t) we discover:
vC (t) = v0 ϵ
− t
RC",DCElectricalCircuitAnalysis.pdf
"dy
 
Using the rule regarding natural logs, this can be written as:
ln(
vC (t )
v0 )=− 1
RC t
And solving for vC(t) we discover:
vC (t) = v0 ϵ
− t
RC
Finally, given that RC = τ and that the starting voltage for the capacitor normally would be its fully charged 
value; that of the associated voltage source E, we find:
vC (t) = E ϵ−t
τ  (discharge phase)
The equation above also describes the shape of the current (and hence the resistor voltage) during the charge 
phase, and therefore to satisfy KVL during this phase we find: 
vC (t) = E (1 −ϵ− t
τ) (charge phase)
362",DCElectricalCircuitAnalysis.pdf
"Appendix D Appendix D 
Answers to Selected Odd-Numbered ProblemsAnswers to Selected Odd-Numbered Problems
1 Fundamentals1 Fundamentals
  1. 14.54, 30060, 76.90, 0.0008475  3. 5.1002, 1020.8, 1.0054, 0.000045781
  5. 2.361E1, 1.2E4, 7.632E3, 5.09E−3  7. 4.156E1, 9.54000E5, 8.4035E1, 1.632E−4
  9. 12E3 (or 12 k), 470, 6.5, 1.98E−3 (or 1.98 m) 11. 33.2, 313.6, 43E3 (or 43 k), 76E−6 (or 76 μ)
13. 3.6E6, 16.74E3, 1.202E3, 217.3E3 15.  0.205, 1.21E−3, 301E3, 3.2E12
17. 482, 22.42, approx. 6.09E3 (6.08997E3), −4.14E6 19. 15.8, 4.734E3, 7E−3
21. 320E6, 667E3, 307E−6 23. 2 kg·m
25. 3 km/s
363",DCElectricalCircuitAnalysis.pdf
"2 Basic Quantities2 Basic Quantities
  1. 33.858 volts to 34.542 volts  3. 2 volt, 20 volt and 200 volt scales
  5. 6.81 volts to 7.19 volts  7. 1.602E−7 coulombs
  9. 1.248E20 electrons 11. 0.4 amps
13. 0.5 coulombs 15. 0.2 volts
17. 20 joules 19. 1492 watts
21. 24 watts 23. 82.9%
25. 125 watts 27. 3 Ah
29. 1200 hours 31. 160 ohms
33. 8 ohms 35. 1200 ohms
37. 3.2 ohm-cm 39. yes, +2.26%
41. yes, −0.2% 43. 200 k, 68 k, 2.7 k
45. 3.3 M, 91, 390 47. 160 k to 240 k, 54.4 k to 81.6 k, 2.43 k to 2.97 k 
49. 2.97 M to 3.63 M, 81.9 to 100.1, 370.5 to 409.5 51. yellow-violet-black-gold, red-red-orange-gold,
      orange-white-yellow-gold, brown-red-orange-gold,
violet-green-brown-gold.
364",DCElectricalCircuitAnalysis.pdf
"3 Series Resistive Circuits3 Series Resistive Circuits
  1. 120 mA  3. 1.44 W
  5. 15 V  7. 33 V
  9. 20.6 k 11. 40 mA
13. 320 mW (200), 160 mW (100),
      480 mW (source)
15. 6 V (2 k), 12 V (4 k), −6 V
17. 0.25 mA 19. 15 mA, + − left to right on 200, + − top to bottom
       on 100, + − right to left on 500, + − top to bottom
       on source.
21. 180 mW 23. 2.5 V (50), 0.5 V (10), 3 V (60), Vc = 3 V ,
      Vac = 3 V , Va = 6 V
25. 200 mA, + − left to right on 25, + − top to bottom
      on 10, + − right to left on 5, + − top to bottom on
      source.
27. 2 V (400), 1 V (200), Vb = 10 V , Vc = 1 V ,
      Vac = 11 V
29. 3.6 V (2 k), 14.4 V (8 k), Vb = 9.6 V ,
      Vc = −14.4 V , Vac = 20.4 V
31. 35 mA, + − left to right on 400,  + − right to left
      on 200, + − top to bottom on 12 V source, 
      + − bottom to top on 9 V source.
33. 1.8 mA, + − left to right on 2 k,  + − right to left
      on 8 k, + − top to bottom on sources. 
35. Vb = 14.4 V , Vc = 9 V , Vac = 6 V",DCElectricalCircuitAnalysis.pdf
"33. 1.8 mA, + − left to right on 2 k,  + − right to left
      on 8 k, + − top to bottom on sources. 
35. Vb = 14.4 V , Vc = 9 V , Vac = 6 V
37. 1 V , it's lower 39. 20 V
41. 40 V (20), 100 V (50) 43. 12 V (4), 30 V (10), 15 V (5)
45. 20 V at a, 0 V at b, 10 V halfway 47. Change the 100 to 120
49. 5.5 V 51. R1 = 6 k, R2 = 1.5 k, R3 = 500
53. 19.2 55. Yes, 10 k
365",DCElectricalCircuitAnalysis.pdf
"4 Parallel Resistive Circuits4 Parallel Resistive Circuits
 1. 80 Ω  3. 14.3 Ω
 5. 4 mA down  7. 120 mA (200), 480 mA (50), 600 mA (source)
 9. 219.5 mA (82), 264.7 mA (68),
     484.2 mA (source)
11. No. It is already the smallest current by an
      order of magnitude and this makes it smaller.
13. 333.3 μA (each 36 k), 250 μA (each 48 k) 15. 0.1915 mA (47 k), 1.765 mA (5.1 k),
      5 mA (1.8 k)
17. 1.6 A (10), 0.4 A (40), 16 V 19. 50 mA (200), 25 mA (each 400), 10 V
21. 145.5 mA (3 k), 218.2 μA (2 k),
      36.37 μA (12 k), 0.4364 V
23. 2.571 mA (each 25 k), 0.8571 mA (75 k), higher 
because there is one less path for current flow.
25. 5.714 mA (5 k), 2.857 mA (each 10 k),
      28.57 mA (1 k), 28.57 V
27. The voltage is larger but now negative, so it
       becomes more negative.
29. 11.25 mA (2 k), 3.75 mA (6 k), 5 mA (4.5 k) 31. 1.714 k
33. 6.667 mA 35. 857 Ω
37. 3 k
366",DCElectricalCircuitAnalysis.pdf
"5 Series-Parallel Resistive Circuits5 Series-Parallel Resistive Circuits
  1. 10 k with 30 k  3. None
  5. 200 with 300  7. None
  9. 12.5 k 11. 23.3 k
13. 106 Ω 15. 3 k
17. Va = 2 V , Vb = 1.34 V , Vab = 0.662 V 19. 0.4 mA (12 k), 0.4 mA (3 k), 0.8 mA (7.5 k)
21. Va = −18 V , Vb = −9.35 V , Vab = −8.65 V 23. 903 μA (3.3 k), 602 μA (10 k), 301 μA (20 k)
25. Va = 100 V , Vb = 59.5 V , Vab = 40.5 V 27. 15 mA (1 k), 3.75 mA (2.2 k), 3.75 mA (1.8 k)
29. Vb = 2.87 V , Vc = 5.1 V , Vcb = 2.23 V 31. 4 mA (2 k), 0.702 mA (5.1 k), 4.702 mA (source)
33. Va = 10 V , Vb = 5.263 V 35. 0.2222 mA (30 k), 0.3333 mA (36 k),
      0.8889 mA (source)
37. Yes, because they also have the same voltage. 39. Va = −22.4 V , Vb = −36 V , Vab = 13.6 V
41. 191.5 μA (47 k), 448.9 μA (5.1 k),
      1.72 mA (3.9 k)
43. Vb = 10.53 V , Vc = 7.643 V , Vd = 5.945 V
45. 133.3 μA (left), 66.67 μA (middle),
      33.33 μA (right)
47. Va = 22 V , Vb = 16 V , Vab = 6 V
49. 375 mA (100 k), 125 mA (60 k),
      125 mA (240 k)",DCElectricalCircuitAnalysis.pdf
"33.33 μA (right)
47. Va = 22 V , Vb = 16 V , Vab = 6 V
49. 375 mA (100 k), 125 mA (60 k),
      125 mA (240 k)
51. Va = 4.91 V , Vb = 3.27 V , Vab = 1.64 V
53. 50 mA (200), 25 mA (400), 25 mA (100) 55. Va = −1.2 V , Vb = −3.2 V , Vab = 2 V
57. 1.2 mA (9 k), 0.6 mA (82 k) 59. Yes, because they also have the same voltage.
61. Va = 3.54 V , Vb = 35.4 V , Vab = −31.86 V 63.  878 μA (1 k), 878 μA (2.2 k), 122 μA (18 k)
65. Va = −5.29 V , Vb = −3.31 V , Vab = −1.98 V 67. 4 mA (left), 2 mA (middle), 1 mA (right)
69. 133.3 Ω 71. 110 Ω
73. 2.55 k 75. 1 A
77. 4 mA
367",DCElectricalCircuitAnalysis.pdf
"6 Analysis Theorems and Techniques6 Analysis Theorems and Techniques
 1. 4.26 mA in parallel with 4.7 k  3. 4.56 mA in parallel with 5.7 k
 5. 26.4 V in series with 2.2 k  7. −40 V in series with 800 Ω
 9. −18 V in series with 9 k 11. 2.55 mA left to right
13. 49.5 V 15. 8.18 V
17. 2.55 mA 19. 4.2 V
21. 1.25 mA 23. 0.7 V
25. 8.09 mA 27. 42.7 V
29. 0.696 mA 31. Yes, it's only current sources and resistors in 
parallel.
33. 1.17 mA 35. −15.65 V
37. −11.6 V 39. 4.03 mA (up)
41. 38.7 mV 43. 1.19 mA left to right
45. 7.2 V with 6.2 k 47. 2 mA with 3 k
49. 5.71 V with 1.71 k 51. 1.538 A with 13 Ω
53. 66.7 μA with 3 k 55. 1.92 mW. Yes, match Rthev.
57. 4.27 μW. Yes, match Rthev. 59. 3 k, 3 mW
61. 4 k, 4.44 μW 63. Replace 10 V/2 k with 5 mA||2 k, and 24 V/10 k 
with 2.4 mA||10 k
65. Replace 2 mA||20 k with 40 V + 20 k, and 0.4 mA||
3 k with 1.2 V + 3 k
368",DCElectricalCircuitAnalysis.pdf
"7 Nodal & Mesh Analysis, Dependent Sources7 Nodal & Mesh Analysis, Dependent Sources
 1. Loop ordering is left to right
      9 = (1 k + 3 k)I1 − (3 k)I2
     −12 = − (3 k)I1 + (3 k + 2 k)I2
 3. 0.818 mA left to right
 5. 3.14 V
 7. 50 = (200 +  40)I1 − (40)I2
    −120 = − (40)I1 + (40 + 75)I2
 9. 1.03 A right to left
11. −2.3 V
13.  First, combine the 4 k||12 k = 3 k
    −34 = (2 k + 10 k)I1 − (10 k)I2
     24 = − (10 k)I1 + (10 k + 3 k)I2
15. 0.696 mA (up)
17. 12.27 V
19. 60 = (100 + 200)I1 − (200)I2 − (0) I3 − (0) I4 
      0 = − (200)I1 + (200 + 300 + 400)I2 − (400) I3 − (0) I4
      0 = − (0)I1 − (400)I2 + (400 + 500 + 600) I3 − (600) I4
     −150 = − (0)I1 − (0)I2 − (600) I3 + (600 + 700) I4
21. 47.7 mA right to left
23. −9.34 V
25. Loop 1 at top, then left to right. 
     30 = (15 k + 10 k + 9 k + 3 k)I1 − (3 k)I2 − (9 k) I3 − (10 k) I4 
     25 = − (3 k)I1 + (2 k + 3 k + 4 k)I2 − (4 k) I3 − (0) I4
     10 = − (9 k)I1 − (4 k)I2 + (4 k + 9 k + 5 k) I3 − (5 k) I4",DCElectricalCircuitAnalysis.pdf
"25 = − (3 k)I1 + (2 k + 3 k + 4 k)I2 − (4 k) I3 − (0) I4
     10 = − (9 k)I1 − (4 k)I2 + (4 k + 9 k + 5 k) I3 − (5 k) I4
     −10 = − (10 k)I1 − (0)I2 − (5 k) I3 + (5 k + 10 k + 7 k) I4
27. 0.86 mA left to right
29. −3.26 V
369",DCElectricalCircuitAnalysis.pdf
"31. Loop ordering: left, top, bottom.
      24 = (2 k + 8 k)I1 − (2 k)I2 − (8 k) I3
       0 = − (2 k)I1 + (2 k + 22 k + 800)I2 − (22 k) I3
       0 = − (8 k)I1 − (22 k)I2 + (8 k + 22 k + 400) I3 
33. 19.8 mA (down)
35. −22.4 V
37. 20  = (6.8 k + 10 k)I1 − (10 k)I2 − (0 k) I3
       0 = − (10 k)I1 + (10 k + 8.5 k + 30 k)I2 − (30 k) I3
       0 = − (0)I1 − (30 k)I2 + (30 k + 20 k + 70 k) I3 
39. 0.34 mA left to right
41. 49.5 V
43. Current source converts to 10 V + 20 k.
     8 = (5 k + 4 k)I1 − (4 k)I2
    −10 = − (4 k)I1 + (4 k + 20 k)I2
45. 0.76 mA left to right
47. −2.73 V
49. −6 mA = (1/24 k + 1/12 k + 1/9 k)Va − (1/9 k)Vb 
      −2 mA = − (1/9 k)Va + (1/9 k + 1/25 k)Vb 
51. 4.03 mA (up)
53. −774 mV
55. Combine 15 k + 25 k  = 40 k
       2 mA = (1/20 k + 1/40 k + 1/5 k)Va − (1/5 k)Vb 
      0.4 mA = − (1/5 k)Va + (1/5 k + 1/3 k)Vb 
57. 0.538 mA
59. 324 mV
61. (300 − 20)mA = (1/40 + 1/10 + 1/100)Va − (1/10)Vb − (1/40)Vc 
      10 mA = − (1/10)Va + (1/10 + 1/50 + 1/250)Vb − (1/250)Vc",DCElectricalCircuitAnalysis.pdf
"59. 324 mV
61. (300 − 20)mA = (1/40 + 1/10 + 1/100)Va − (1/10)Vb − (1/40)Vc 
      10 mA = − (1/10)Va + (1/10 + 1/50 + 1/250)Vb − (1/250)Vc
      (−300 − 50)mA = − (1/40)Va − (1/250)Vb + (1/250 + 1/25 + 1/40)Vc
63. 177 mV
65. 5.94 V
370",DCElectricalCircuitAnalysis.pdf
"67. This is deceptively simple. First combine the parallel 36 k resistors to a single 18 k. There is only one node 
of concern, node a. The 4 mA source enters this node. Define an exiting current, I1, down through the 18 k. 
I1 = Va/18 k. Define an entering current from the voltage source, I2. 
I2 = (12 − Va)/30 k. From KCL, 4 mA + I2 = I1. Substitute the current expressions into the KCL equation, 
expand and simplify. This results in a single node equation: 4.4 mA = (1/30 k + 1/18 k) Va.
69. 1.25 mA left to right
71. 4.2 V
73. 240 V
75. 720 V
77. 441 mV
79. 6 V
8 Capacitors8 Capacitors
 1. 5 μF  3. 2 μF
 5. 8 V across 10 μF, 4 V across 20 μF  7. 2 k gets 5.38 V , 5 μF gets 0 V , 
      4 k and 3 k get 4.62 V
 9. 23. 1.5 V 11. 20 μs,  100 μs
13. 13.2 V , 24 V 15. 1 ms, 5 ms
17. 10 V , 15 ms 19. 3 ms, 15 ms
21. Replace 100 Ω with 2 k 23. 125 μF
371",DCElectricalCircuitAnalysis.pdf
"9 Inductors9 Inductors
 1.  6 mH  3.  0.75 mH
 5. 2 k gets 3.33 V , 5 μF gets 0 V , 
      4 k and 3 mH get 6.67 V
 7. 2 k gets 3.33 V , 3 mH gets 0 V , 
      4 k and 5 μF get 6.67 V
 9.  2 k gets 5.38 V , 5 mH gets 0 V , 
      4 k and 3 k get 4.62 V
11. 2 k, 10 k and 20 μF get 0 V , 
      8 k and 12 mH get 20 V
13. 33 k gets 15 V , all other components get 0 V 15.  0.2 A
17. 20 ms, 100 ms 19. 0 V and 480 mA for both times. 
      With 5 Ω, 2 V and 400 mA for both times. 
21.  40 ns, 200 ns 23.  3 mA, −60 V
25. 833 ns, 4.17 μs 27.  Replace 25 Ω with 250 Ω.
10 Magnetic Circuits and Transformers10 Magnetic Circuits and Transformers
 1. 80 mA 3. 95 mA
 5. 160 turns 7. 20 mA
 9. 2.42 A 11. Cast steel causes greater core losses so the
      secondary current would be reduced.
13. 324 mA 15. vload = 4 volts, iload = 40 mA, iprimary = 6.67 mA
17. 1440 Ω 19. 45 V A
372",DCElectricalCircuitAnalysis.pdf
"Appendix EAppendix E
Base UnitsBase Units
In Chapters 1 and 2, a variety of measurement units were introduced and defined, along with the metric system. 
It is useful to understand that all of the units used are built from three foundational units. These units describe 
the conceptual building blocks of the system. They are mass (kilograms, kg), length (meters, m) and time 
(seconds, s). They are the foundation of the MKS system. Every other unit used, such as watts, volts or amps, 
can be built from these three fundamental elements. For example, it is not arbitrary that ohms times amps yields 
volts.
In Chapter 2 it was noted that power equals current times voltage, and that this can be derived from the 
associated units. Current was defined as the rate of charge movement per unit time, and one amp equals one 
coulomb per second. Similarly, voltage was defined as the energy required to move a certain charge, and one",DCElectricalCircuitAnalysis.pdf
"coulomb per second. Similarly, voltage was defined as the energy required to move a certain charge, and one 
volt equals one joule per coulomb. If we multiply the units together we get:
P = I ×V
P = C
s × J
C
P = J
s
Joules per second is energy per unit time and that is the very definition of power, with one watt being equal to 
one joule per second. But this brings up the question, “Then what is a joule in terms of the three fundamental 
units?” Energy can be defined as a force working over a specified distance. One joule is defined as the force of 
one newton working over a distance of one meter. This creates a new question, “Then what is a newton?” 
Newton's second law of motion states that force is equal to mass times acceleration. Acceleration, in turn, is the 
rate of change of velocity, or distance per unit time, per unit time. For these units, we find:
force = mass×acceleration
force = kg× m
s2
force = kg m
s2
energy = force×distance
energy = kg m
s2 ×m
energy = kg m2
s2",DCElectricalCircuitAnalysis.pdf
"force = mass×acceleration
force = kg× m
s2
force = kg m
s2
energy = force×distance
energy = kg m
s2 ×m
energy = kg m2
s2
Thus, we verify that one joule is equal to one kilogram-meter2 per second2. If we apply this back to the 
definition for power, we find that one watt is equal to one kilogram-meter2 per second3.
373",DCElectricalCircuitAnalysis.pdf
"Appendix FAppendix F
“Could I do one more immediately?”
– Bill Bruford
374",DCElectricalCircuitAnalysis.pdf
"MULTIVARIABLE
CALCULUS
Don Shimamoto",Multivariable_Calculus_Shimamoto.pdf
"Multivariable Calculus
Don Shimamoto
Swarthmore College",Multivariable_Calculus_Shimamoto.pdf
"ii
©2019 Don Shimamoto
ISBN: 978-1-7082-4699-0
This work is licensed under the Creative Commons Attribution 4.0 International
License. To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/.
Unless noted otherwise, the graphics in this work were created using the software packages: Cin-
derella, Mathematica, and, in a couple of instances, Adobe Illustrator and Canvas.",Multivariable_Calculus_Shimamoto.pdf
"Contents
Preface vii
I Preliminaries 1
1 Rn 3
1.1 Vector arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 The matrix of a linear transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4 Matrix multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.5 The geometry of the dot product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.6 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.7 Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
II Vector-valued functions of one variable 23
2 Paths and curves 25
2.1 Parametrizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25",Multivariable_Calculus_Shimamoto.pdf
"2 Paths and curves 25
2.1 Parametrizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.2 Velocity, acceleration, speed, arclength . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.3 Integrals with respect to arclength . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.4 The geometry of curves: tangent and normal vectors . . . . . . . . . . . . . . . . . . 31
2.5 The cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.6 The geometry of space curves: Frenet vectors . . . . . . . . . . . . . . . . . . . . . . 38
2.7 Curvature and torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.8 The Frenet-Serret formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.9 The classification of space curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42",Multivariable_Calculus_Shimamoto.pdf
"2.9 The classification of space curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.10 Exercises for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
III Real-valued functions 53
3 Real-valued functions: preliminaries 55
3.1 Graphs and level sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.2 More surfaces in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.3 The equation of a plane in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.4 Open sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
3.5 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
iii",Multivariable_Calculus_Shimamoto.pdf
"iv CONTENTS
3.6 Some properties of continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.7 The Cauchy-Schwarz and triangle inequalities . . . . . . . . . . . . . . . . . . . . . . 71
3.8 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
3.9 Exercises for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4 Real-valued functions: differentiation 83
4.1 The first-order approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.2 Conditions for differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
4.3 The mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.4 The C1 test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.5 The Little Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92",Multivariable_Calculus_Shimamoto.pdf
"4.5 The Little Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4.6 Directional derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4.7 ∇f as normal vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
4.8 Higher-order partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.9 Smooth functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.10 Max/min: critical points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.11 Classifying nondegenerate critical points . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.11.1 The second-order approximation . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.11.2 Sums and differences of squares . . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.12 Max/min: Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105",Multivariable_Calculus_Shimamoto.pdf
"4.12 Max/min: Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.13 Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
5 Real-valued functions: integration 121
5.1 Volume and iterated integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
5.2 The double integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
5.3 Interpretations of the double integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
5.4 Parametrizations of surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
5.4.1 Polar coordinates ( r, θ) in R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
5.4.2 Cylindrical coordinates ( r, θ, z) in R3 . . . . . . . . . . . . . . . . . . . . . . . 137
5.4.3 Spherical coordinates ( ρ, ϕ, θ) in R3 . . . . . . . . . . . . . . . . . . . . . . . 138",Multivariable_Calculus_Shimamoto.pdf
"5.4.3 Spherical coordinates ( ρ, ϕ, θ) in R3 . . . . . . . . . . . . . . . . . . . . . . . 138
5.5 Integrals with respect to surface area . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5.6 Triple integrals and beyond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
5.7 Exercises for Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
IV Vector-valued functions 155
6 Differentiability and the chain rule 157
6.1 Continuity revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
6.2 Differentiability revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
6.3 The chain rule: a conceptual approach . . . . . . . . . . . . . . . . . . . . . . . . . . 161
6.4 The chain rule: a computational approach . . . . . . . . . . . . . . . . . . . . . . . . 162
6.5 Exercises for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
7 Change of variables 173",Multivariable_Calculus_Shimamoto.pdf
"6.5 Exercises for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
7 Change of variables 173
7.1 Change of variables for double integrals . . . . . . . . . . . . . . . . . . . . . . . . . 174
7.2 A word about substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
7.3 Examples: linear changes of variables, symmetry . . . . . . . . . . . . . . . . . . . . 178",Multivariable_Calculus_Shimamoto.pdf
"CONTENTS v
7.4 Change of variables for n-fold integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 182
7.5 Exercises for Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
V Integrals of vector fields 195
8 Vector fields 197
8.1 Examples of vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
8.2 Exercises for Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
9 Line integrals 203
9.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
9.2 Another word about substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
9.3 Conservative fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
9.4 Green’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
9.5 The vector field W . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219",Multivariable_Calculus_Shimamoto.pdf
"9.5 The vector field W . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
9.6 The converse of the mixed partials theorem . . . . . . . . . . . . . . . . . . . . . . . 222
9.7 Exercises for Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
10 Surface integrals 235
10.1 What the surface integral measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
10.2 The definition of the surface integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
10.3 Stokes’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
10.4 Curl fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
10.5 Gauss’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
10.6 The inverse square field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257",Multivariable_Calculus_Shimamoto.pdf
"10.6 The inverse square field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
10.7 A more substitution-friendly notation for surface integrals . . . . . . . . . . . . . . . 259
10.8 Independence of parametrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
10.9 Exercises for Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
11 Working with differential forms 273
11.1 Integrals of differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
11.2 Derivatives of differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
11.3 A look back at the theorems of multivariable calculus . . . . . . . . . . . . . . . . . 279
11.4 Exercises for Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
Answers to selected exercises 285
Index 307",Multivariable_Calculus_Shimamoto.pdf
vi CONTENTS,Multivariable_Calculus_Shimamoto.pdf
"Preface
This book is based on a course that I taught several times at Swarthmore College. The material is
standard, though this particular course is geared towards students who enjoy learning mathematics
for its own sake. As a result, there is a priority placed on understanding why things are true and
a recognition that, when details are sketched or omitted, that should be acknowledged. Otherwise,
the level of rigor is fairly normal. The course has a prerequisite of a semester of linear algebra.
That is not necessary for this book, though it helps. Many of the students go on to more advanced
courses in mathematics, and the course is calibrated to be part of the path on the way to the more
theoretical, proof-oriented nature of upper-level material. Indeed, the hope is that the course will
help inspire the students to continue on.
Roughly speaking, the book is organized into three main parts depending on the type of function",Multivariable_Calculus_Shimamoto.pdf
"help inspire the students to continue on.
Roughly speaking, the book is organized into three main parts depending on the type of function
being studied: vector-valued functions of one variable, real-valued functions of many variables, and,
finally, the general case of vector-valued functions of many variables. The table of contents gives a
pretty good idea of the topics that are covered, but here are a few notes:
• Chapter 1 contains the basics of working with vectors in Rn. For students who have studied
linear algebra, this is review.
• Chapter 2 is concerned with curves in Rn. This belongs to the study of functions of one
variable, and many students will have studied parametric equations in their first-year calculus
courses, at least for curves in the plane. One of the appealing aspects of the topic is that it is
possible to give a reasonably complete proof of a substantial result—the classification of space",Multivariable_Calculus_Shimamoto.pdf
"possible to give a reasonably complete proof of a substantial result—the classification of space
curves—in a way that reinforces the basic vector concepts that have just been introduced.
Many of the proofs along the way and in the related exercises feel like calculations. This
allows the students to ease into the mindset of proving things before encountering the more
argument-based proofs to come.
• Chapters 3, 4, and 5 study real-valued functions of many variables. This includes the topics
most closely associated with multivariable calculus: partial derivatives and multiple integrals.
The discussion of differentiation emphasizes first-order approximations and the notion of
differentiability. For integration, the focus is almost entirely on functions of two variables
and, after the change of variables theorem is introduced later on, functions of three variables.
• Chapters 6 and 7 begin the study of vector-valued functions of many variables. The main",Multivariable_Calculus_Shimamoto.pdf
"• Chapters 6 and 7 begin the study of vector-valued functions of many variables. The main
results here are the chain rule and the change of variables theorem. Their importance to the
subsequent theory is reiterated throughout the rest of the book.
• Chapters 8, 9, and 10 introduce vector fields and their integrals over curves inRn and surfaces
in R3, in other words, line and surface integrals. This leads to the theorems of Green, Stokes,
and Gauss, which are often taken as the target destinations of a multivariable calculus course.
• The book closes in Chapter 11 with a brief discussion of differential forms and their relation
to what has come before. The treatment is superficial but hopefully still illuminating. The
vii",Multivariable_Calculus_Shimamoto.pdf
"viii PREFACE
chapter is a success if the students come away believing that something interesting is going
on and curious enough to want to learn more.
As is always the case, the most useful way for students to learn the material is by doing problems,
and this book is written to get to the exercises as quickly as possible. In some cases, proofs are
sketched in the text and the details are left for the exercises. I have tried to make sure that there is
enough information in the text for the students to be able to do all the problems, but some students
and instructors may find it helpful to see more worked examples. I believe that there are enough
exercises in the book that instructors can choose to include some of them—ranging from routine
calculations to proofs—as part of their own presentations of the material. Or students can try some
of the exercises on their own and check their answers against those in the back of the book. The",Multivariable_Calculus_Shimamoto.pdf
"of the exercises on their own and check their answers against those in the back of the book. The
exercises are written with these possibilities in mind. I hope that this also gives instructors the
flexibility to integrate the approach taken in the book more easily with their personal perspectives
on the subject.
In writing the book, it has become clear that my own view of the material is heavily influenced
by the multivariable calculus course I took as a student. It was taught by Greg Brumfiel, and I
thank him for getting me excited about the subject in such a lasting way. I also thank my colleague
Ralph Gomez for reading an entire draft of the manuscript and making numerous suggestions
that improved the book considerably. I wish I had thought of them myself. Lastly, I thank the
students who took my courses over the years for their responsiveness and feedback, which shaped
my approach to the material as it evolved to its current state, such as it is. The students did not",Multivariable_Calculus_Shimamoto.pdf
"my approach to the material as it evolved to its current state, such as it is. The students did not
know that the course they were taking might someday become the basis for a textbook, but then
I wasn’t aware of it either.",Multivariable_Calculus_Shimamoto.pdf
"Part I
Preliminaries
1",Multivariable_Calculus_Shimamoto.pdf
"Chapter 1
Rn
Let R denote the set of real numbers. Its elements are also called scalars. If n is a positive integer,
then Rn is defined to be the set of all sequences x of n real numbers:
x = (x1, x2, . . . , xn). (1.1)
The elements of Rn are called points, vectors, or n-tuples. We follow the convention of indicating
vectors in boldface and scalars in plainface. For a vector x, the individual scalar entries xi for
i = 1, 2, . . . , nare called coordinates or components.
Multivariable calculus studies functions between these sets, that is, functions of the form
f : Rn → Rm, or, more accurately, of the form f : A → Rm, where A is a subset of Rn. In
this context, if x represents a typical point of Rn, the coordinates x1, x2, . . . , xn are referred to as
variables. For example, first-year calculus studies real-valued functions of one variable, functions
of the form f : R → R.
This chapter collects some of the background information about Rn that we use throughout",Multivariable_Calculus_Shimamoto.pdf
"of the form f : R → R.
This chapter collects some of the background information about Rn that we use throughout
the book. The presentation is meant to be self-contained, though readers who have studied linear
algebra are likely to have a greater perspective on how the pieces fit together as part of a bigger
picture.
1.1 Vector arithmetic
There are two basic algebraic operations in Rn. Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) be
elements of Rn, and let c be a real number. The aforementioned operations are defined as follows.
• Addition: x + y = (x1 + y1, x2 + y2, . . . , xn + yn).
• Scalar multiplication: cx = (cx1, cx2, . . . , cxn).
Because of our familiarity with R2, we illustrate these concepts there in some detail. An element
of R2 is an ordered pair x = (x1, x2). Geometrically, x is a point in the plane plotted in the usual
way. In particular, the origin (0 , 0) is called the zero vector and is denoted by 0. Alternatively,",Multivariable_Calculus_Shimamoto.pdf
"way. In particular, the origin (0 , 0) is called the zero vector and is denoted by 0. Alternatively,
we may visualize x by drawing the arrow starting at (0 , 0) and ending at ( x1, x2). We’ll go back
and forth freely between the point/arrow viewpoints.
Given two vectors x = (x1, x2) and y = (y1, y2) in R2, the sum x + y as defined above is the
point that results from adding the displacements in each of the horizontal and vertical directions,
respectively. For instance, if x = (1, 2) and y = (3, 4), then x + y = (4, 6). Thinking of x and y as
arrows emanating from 0, this places x + y at the vertex opposite the origin in the parallelogram
3",Multivariable_Calculus_Shimamoto.pdf
"4 CHAPTER 1. Rn
Figure 1.1: Vector addition
determined by x and y, as on the left of Figure 1.1. If we think of x + y as an arrow as well, it is
one of the diagonals of the parallelogram, as shown on the right.
Another way to reach x + y is to move the arrow representing y so that it retains the same
length and direction but begins at the endpoint of x. This is called a “translation” of the original
vector y. Then x + y is the destination if you go along x followed by the translated version of
y, as illustrated in Figure 1.2, like following two displacements x and y in succession. It is often
Figure 1.2: Using the translation of a vector
convenient to translate vectors, especially when they represent quantities for which length and
direction are the most relevant characteristics. Nevertheless, it’s important to remember that the
translations are only copies: the real vector starts at the origin.",Multivariable_Calculus_Shimamoto.pdf
"translations are only copies: the real vector starts at the origin.
Similarly, if c is a real number, then cx results from multiplying each of the coordinate displace-
ments by a factor of c. For instance, if x = (1, 2), then 3 x = (3, 6). In general, cx is an arrow |c|
times as long as x and in the same direction as x if c >0, the opposite direction if c <0. See Figure
1.3. In particular, ( −1)x = (−x1, −x2) is a copy of x rotated by 180◦ to reverse the direction. It
is usually denoted by −x since it satisfies x + (−1)x = 0.
Going back to the parallelogram used to visualizex+y, we could also look at the other diagonal,
say drawn as an arrow from y to x, as indicated in Figure 1.4. We sometimes denote this arrow by− →yx. It is what you would add to y to get to x. In other words, it is the difference x − y:
− →yx = x − y.
Again, the real x − y should be returned so that it starts at 0.",Multivariable_Calculus_Shimamoto.pdf
"1.1. VECTOR ARITHMETIC 5
Figure 1.3: Scalar multiplication
Figure 1.4: The difference of two vectors
In the case of R2, the coordinates are usually denoted by x, y, rather than x1, x2. Then R2 is
the usual xy-plane. The notation is potentially confusing since x is also often used to denote the
generic vector in Rn, as in equation (1.1) above. Hopefully, the context and the use of boldface
will clarify whether a coordinate or vector is intended. Similarly, R3 represents three-dimensional
space. Its coordinates are often denoted by x, y, z.
Returning to the general case, every vector x = (x1, x2, . . . , xn) in Rn can be decomposed as a
sum along the coordinate directions:
x = (x1, 0, . . . ,0) + (0, x2, . . . ,0) + ··· + (0, 0, . . . , xn)
= x1(1, 0, . . . ,0) + x2(0, 1, . . . ,0) + ··· + xn(0, 0, . . . ,1).
The vectors e1 = (1, 0, . . . ,0), e2 = (0, 1, . . . ,0), . . . ,en = (0, 0, . . . ,1), with a 1 in a single compo-",Multivariable_Calculus_Shimamoto.pdf
"The vectors e1 = (1, 0, . . . ,0), e2 = (0, 1, . . . ,0), . . . ,en = (0, 0, . . . ,1), with a 1 in a single compo-
nent and zeros everywhere else, are called the standard basis vectors. Thus:
x = x1e1 + x2e2 + ··· + xnen. (1.2)
The scalar coefficients xi are the coordinates of x.
In general, a set of n vectors {v1, v2, . . . ,vn} in Rn is called a basis if every x in Rn can be
written in a unique way as a combinationx = c1v1 +c2v2 +··· +cnvn for some scalars c1, c2, . . . , cn.
Any basis can be used to define its own coordinate system in Rn, and Rn has many different bases.
Apart from a few occasions, however, we’ll stick with the standard basis.
In R2, the standard basis vectors are usually denoted by i = (1, 0) and j = (0, 1). In R3, the
corresponding names are i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1).",Multivariable_Calculus_Shimamoto.pdf
"6 CHAPTER 1. Rn
1.2 Linear transformations
Linear transformations are functions that respect vector addition and scalar multiplication. More
precisely:
Definition. A function T : Rn → Rm is called a linear transformation if:
• T(x + y) = T(x) + T(y) and
• T(cx) = c T(x).
The conditions must be satisfied for all x, y in Rn and all c in R.
This definition may seem austere, but many familiar functions are linear transformations. We
just may not be used to thinking of them that way.
Example 1.1. Let T : R2 → R2 be counterclockwise rotation by π
3 about the origin. That is, if
x ∈ R2, then T(x) is the point that is reached after x is rotated counterclockwise by π
3 about 0. We
give a geometric argument that T satisfies the two requirements for being a linear transformation.
First, consider the parallelogram determined by vectors x and y in R2. Then T rotates this
parallelogram to another parallelogram, the one determined by T(x) and T(y). In particular, the",Multivariable_Calculus_Shimamoto.pdf
"parallelogram to another parallelogram, the one determined by T(x) and T(y). In particular, the
vertex x + y is rotated to the vertex T(x) +T(y). This says exactly that T(x + y) = T(x) +T(y).
This is illustrated at the left of Figure 1.5.
Figure 1.5: Rotations are linear transformations.
This geometric argument is so simple that it may not be clear that there is any actual reasoning
behind it. The reader is encouraged to go through it carefully to pin down how it proves what
we want. Also, the case where x and y are collinear, so that they don’t determine an honest
parallelogram, requires a separate argument. We won’t give it, though see the reasoning in the
next paragraph.
Similarly, T rotates the line through the origin and x to the line through the origin and T(x).
The point on the first line c times as far from the origin as x is rotated to the point c times as far
from the origin as T(x), as indicated on the right of the figure. In other words, T(cx) = c T(x).",Multivariable_Calculus_Shimamoto.pdf
"from the origin as T(x), as indicated on the right of the figure. In other words, T(cx) = c T(x).
Example 1.2. Likewise, the function T : R2 → R2 that reflects a point x = (x1, x2) in the x1-axis
is a linear transformation. The appropriate supporting diagrams are shown in Figure 1.6.
Example 1.3. Let T : R3 → R2 be the function that projects a point x = (x1, x2, x3) in R3 onto
the x1x2-plane, that is, T(x1, x2, x3) = (x1, x2). Rather than use pictures, this time we show that
T is linear by calculating.
For instance:
T(x + y) = T(x1 + y1, x2 + y2, x3 + y3) = (x1 + y1, x2 + y2).",Multivariable_Calculus_Shimamoto.pdf
"1.3. THE MATRIX OF A LINEAR TRANSFORMATION 7
Figure 1.6: So are reflections.
On the other hand:
T(x) + T(y) = (x1, x2) + (y1, y2) = (x1 + y1, x2 + y2).
Both expressions equal the same thing, so T(x + y) = T(x) + T(y).
Similarly, T(cx) = T(cx1, cx2, cx3) = ( cx1, cx2) while c T(x) = c(x1, x2) = ( cx1, cx2). Thus
T(cx) = c T(x) as well.
1.3 The matrix of a linear transformation
Let T : Rn → Rm be a linear transformation. One of the things that make such a function easy to
work with is that, althoughT(x) is defined for allx in Rn, the transformation is actually determined
by a finite amount of input. To make sense of this, recall from equation (1.2) that every x in Rn
can be expressed in terms of the standard basis vectors: x = x1e1 + x2e2 + ··· + xnen. Thus, by
the defining properties of a linear transformation:
T(x) = T(x1e1 + x2e2 + ··· + xnen)
= T(x1e1) + T(x2e2) + ··· + T(xnen)
= x1T(e1) + x2T(e2) + ··· + xnT(en)
= x1a1 + x2a2 + ··· + xnan, (1.3)",Multivariable_Calculus_Shimamoto.pdf
"T(x) = T(x1e1 + x2e2 + ··· + xnen)
= T(x1e1) + T(x2e2) + ··· + T(xnen)
= x1T(e1) + x2T(e2) + ··· + xnT(en)
= x1a1 + x2a2 + ··· + xnan, (1.3)
where aj = T(ej) for j = 1 , 2, . . . , n. Conversely, given any n vectors a1, a2, . . . ,an in Rm, it’s
straightforward to check by calculation that the formula T(x) = x1a1 + x2a2 + ··· + xnan satisfies
the two requirements for being a linear transformation. This is Exercise 2.3 at the end of the
chapter. Moreover, T(ej) = T(0, 0, . . . ,1, . . . ,0) = 0 · a1 + 0· a2 + ··· + 1· aj + ··· + 0· an = aj for
each j. In other words, a linear transformation T : Rn → Rm is completely determined, as in (1.3),
once you know the values T(e1) = a1, T(e2) = a2, . . . , T(en) = an, and there are no restrictions on
what these values can be.
To understand how this might be used, we consider first the case of a linear transformation
T : Rn → R whose values are real numbers. Then every T(ej) is a scalar, say T(ej) = aj, and
equation (1.3) becomes:",Multivariable_Calculus_Shimamoto.pdf
"T : Rn → R whose values are real numbers. Then every T(ej) is a scalar, say T(ej) = aj, and
equation (1.3) becomes:
T(x) = x1a1 + x2a2 + ··· + xnan. (1.4)
Sums of this type are an indispensable part of vector algebra.
Definition. Given x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) in Rn, the dot product, denoted
by x · y, is defined by:
x · y = x1y1 + x2y2 + ··· + xnyn.",Multivariable_Calculus_Shimamoto.pdf
"8 CHAPTER 1. Rn
For example, inR3, if x = (1, 2, 3) and y = (4, 5, 6), then x·y = 1·4+2·5+3·6 = 4+10+18 = 32.
The dot product satisfies a variety of elementary properties, such as x · y = y · x. The ones
we shall use are pretty obvious, so we won’t bother listing them out, though please see Exercises
5.5–5.10 if you’d like to see some of them collected together.
Returning to the main point, we have shown in equation (1.4) that every real-valued linear
transformation T : Rn → R has the form:
T(x) = a · x
for some vector a = (a1, a2, . . . , an) in Rn.
The analysis for the general case of a linear transformation T : Rn → Rm follows the same
pattern except that now the values aj = T(ej) are vectors in Rm. We record these values by
putting them in the columns of a rectangular table. That is, for each j = 1, 2, . . . , n, say aj is the
vector (a1j, a2j, . . . , amj) in Rm, and let A be the table:
A =


a11 a12 ··· a1j ··· a1n
a21 a22 ··· a2j ··· a2n
... ... ... ...",Multivariable_Calculus_Shimamoto.pdf
"vector (a1j, a2j, . . . , amj) in Rm, and let A be the table:
A =


a11 a12 ··· a1j ··· a1n
a21 a22 ··· a2j ··· a2n
... ... ... ...
am1 am2 ··· amj ··· amn

,
where aj is highlighted in red in the jth column. Such a table is called a matrix. In fact, A is
called an m by n matrix, which means that it has m rows and n columns. The subscripting has
been chosen so that aij is the entry in row i, column j, where the rows are numbered starting from
the top and the columns starting from the left.
The matrix obtained in this way is called the matrix of T with respect to the standard bases.
Since, by equation (1.3), the transformation T is completely determined by the columns aj, the
matrix contains all the data we need to find T(x) for all x in Rn.
We illustrate this with the three examples of linear transformations considered earlier.
Example 1.4. Let T : R2 → R2 be the counterclockwise rotation by π
3 about the origin. Then T",Multivariable_Calculus_Shimamoto.pdf
"Example 1.4. Let T : R2 → R2 be the counterclockwise rotation by π
3 about the origin. Then T
rotates the vector e1 = (1, 0) to the vector on the unit circle that makes an angle of π
3 with the
positive x1-axis. That is, T(e1) = (cos π
3 , sin π
3 ) = ( 1
2 ,
√
3
2 ). Similarly, T(e2) = (cos 5π
6 , sin 5π
6 ) =
(−
√
3
2 , 1
2 ). Hence the matrix of T with respect to the standard bases is:
A =
""
1
2 −
√
3
2√
3
2
1
2
#
. (1.5)
Example 1.5. If T : R2 → R2 is the reflection in the x1-axis, then T(e1) = T(1, 0) = (1 , 0) and
T(e2) = T(0, 1) = (0, −1). Hence:
A =
1 0
0 −1

. (1.6)
Example 1.6. Lastly, if T : R3 → R2 is the projection of x1x2x3-space onto the x1x2-plane, then
T(e1) = T(1, 0, 0) = (1, 0), T(e2) = T(0, 1, 0) = (0, 1), and T(e3) = T(0, 0, 1) = (0, 0), so:
A =
1 0 0
0 1 0

. (1.7)",Multivariable_Calculus_Shimamoto.pdf
"1.3. THE MATRIX OF A LINEAR TRANSFORMATION 9
To use the matrix A to compute T(x) in a systematic way, we observe the convention that
vectors are identified with matrices having a single column. Thus:
x =


x1
x2
...
xn


and aj =


a1j
a2j
...
amj


represent vectors in Rn and Rm, respectively. Sometimes they’re referred to as column vectors.
With this notation, equation (1.3) becomes:
T(x) = x1


a11
a21
...
am1

 + x2


a12
a22
...
am2

 + ··· + xn


a1n
a2n
...
amn


=


a11x1 + a12x2 + ··· + a1nxn
a21x1 + a22x2 + ··· + a2nxn
...
am1x1 + am2x2 + ··· + amnxn

. (1.8)
This is similar to the real-valued case (1.4) in that each component is a dot product. Specifically,
the ith component is the dot product of the ith row of A and x. This expression is given a name.
Definition (Preliminary version of matrix multiplication) . Let A be an m by n matrix, and let",Multivariable_Calculus_Shimamoto.pdf
"Definition (Preliminary version of matrix multiplication) . Let A be an m by n matrix, and let
x be a column vector in Rn. Then the product Ax is defined to be the column vector y in Rm
whose ith component is the dot product of the ith row of A and x:
yi = ai1x1 + ai2x2 + ··· + ainxn.
In other words, Ax is the column vector given by equation (1.8) above.
We’ve labeled this as a “preliminary” version, because we define a more general matrix multi-
plication shortly that includes this as a special case.
Using this definition, equation (1.8) can be written in the simple formT(x) = Ax. The preceding
discussion is summarized in the following result.
Proposition 1.7.
1. Given any linear transformation T : Rn → Rm, there is an m by n matrix A such that:
• the jth column of A is T(ej) for j = 1, 2, . . . , n, and
• T(x) = Ax for all x in Rn.
2. Conversely, given any m by n matrix A, the formula T(x) = Ax defines a linear transforma-
tion T : Rn → Rm.",Multivariable_Calculus_Shimamoto.pdf
"• T(x) = Ax for all x in Rn.
2. Conversely, given any m by n matrix A, the formula T(x) = Ax defines a linear transforma-
tion T : Rn → Rm.
We apply this to the examples previously considered. For instance, if T is the counterclockwise
rotation of R2 by π
3 about the origin, then by (1.5):
T(x) =
""
1
2 −
√
3
2√
3
2
1
2
#x1
x2

=
""
1
2 x1 −
√
3
2 x2√
3
2 x1 + 1
2 x2
#
=
1
2x1 −
√
3
2 x2,
√
3
2 x1 + 1
2x2


.",Multivariable_Calculus_Shimamoto.pdf
"10 CHAPTER 1. Rn
Once you get the hang of it, this may be the simplest way to find a formula for a rotation.
For the reflection in the x1-axis, (1.6) gives:
T(x) =
1 0
0 −1
x1
x2

=
1 · x1 + 0 · x2
0 · x1 − 1 · x2

=
 x1
−x2

= (x1, −x2).
We didn’t need matrix methods to come up with this formula, but at least it’s correct.
Lastly, for the projection of R3 onto the x1x2-plane, we find from (1.7) that:
T(x) =
1 0 0
0 1 0


x1
x2
x3

 =
x1 + 0 + 0
0 + x2 + 0

=
x1
x2

= (x1, x2),
as expected.
1.4 Matrix multiplication
Let T : Rn → Rm and S : Rm → Rp be functions, where, as indicated, the codomain of T equals
the domain of S. The composition S ◦ T : Rn → Rp is defined to be the function given by
(S ◦ T)(x) = S(T(x)) for all x in Rn.
If S and T are linear transformations, it’s easy to check that S ◦ T is a linear transformation,
too:
• S(T(x + y)) = S(T(x) + T(y)) = S(T(x)) + S(T(y)),
• S(T(cx)) = S(c T(x)) = c S(T(x)).",Multivariable_Calculus_Shimamoto.pdf
"too:
• S(T(x + y)) = S(T(x) + T(y)) = S(T(x)) + S(T(y)),
• S(T(cx)) = S(c T(x)) = c S(T(x)).
In each case, the first equation follows from the definition of linear transformation applied to T
and the second applied to S. As a result, S ◦ T is represented by a matrix C with respect to the
standard bases. Let A and B be the matrices of S and T, respectively, with respect to the standard
bases. We shall describe C in terms of A and B.
By Proposition 1.7, the jth column cj of C is S(T(ej)). For the same reason, T(ej) is the jth
column of B. Let’s call it bj, so cj = S(bj). On the other hand, S is represented by the matrix A,
so cj is the matrix product cj = Abj. Its ith component is the ith row of A dotted with bj, that
is, the dot product of the ith row of A and the jth column of B. Doing this for all j = 1, 2, . . . , n,
fills out the matrix C. The matrix obtained in this way is called the product of A and B.",Multivariable_Calculus_Shimamoto.pdf
"fills out the matrix C. The matrix obtained in this way is called the product of A and B.
Definition (Final version of matrix multiplication) . Let A be a p by m matrix and B an m by
n matrix. Their product, denoted by AB, is the p by n matrix C whose (i, j)th entry is the dot
product of the ith row of A and the jth column of B:
cij = ai1b1j + ai2b2j + ··· + aimbmj =
mX
k=1
aikbkj.
Visually, the action is something like this:


...
··· cij ···
...

 =


... ... ...
ai1 ai2 ··· aim
... ... ...




··· b1j ···
··· b2j ···
...
··· bmj ···

.
This only makes sense when the number of columns of A equals the number of rows of B.
The definitions have been rigged so that the following statement is an immediate consequence.",Multivariable_Calculus_Shimamoto.pdf
"1.5. THE GEOMETRY OF THE DOT PRODUCT 11
Proposition 1.8. Composition of linear transformations corresponds to matrix multiplication.
That is, let S : Rm → Rp and T : Rn → Rm be linear transformations with matrices A and B,
respectively, with respect to the standard bases. Then S ◦ T is a linear transformation, and its
matrix with respect to the standard bases is the product AB.
Example 1.9. Let T : R2 → R2 be the counterclockwise rotation by π
3 about the origin, S : R2 →
R2 the reflection in the x1-axis, and consider the composition S◦T (first rotate, then reflect). Using
the matrices from (1.5) and (1.6), the matrix of S ◦ T with respect to the standard bases is the
product:
1 0
0 −1
""
1
2 −
√
3
2√
3
2
1
2
#
=
""
1 · 1
2 + 0 ·
√
3
2 1 · (−
√
3
2 ) + 0· 1
2
0 · 1
2 + (−1) ·
√
3
2 0 · (−
√
3
2 ) + (−1) · 1
2
#
=
""
1
2 −
√
3
2
−
√
3
2 −1
2
#
.
Working backwards and looking at the columns of this matrix, this tells us that ( S ◦ T)(e1) =
(1
2 , −
√
3
2 ) = (cos(−π
3 ), sin(−π",Multivariable_Calculus_Shimamoto.pdf
"√
3
2 −1
2
#
.
Working backwards and looking at the columns of this matrix, this tells us that ( S ◦ T)(e1) =
(1
2 , −
√
3
2 ) = (cos(−π
3 ), sin(−π
3 )) and (S ◦T)(e2) = (−
√
3
2 , −1
2 ) = (cos(7π
6 ), sin(7π
6 )). These points are
plotted in Figure 1.7. A linear transformation that has the same effect on e1 and e2 is the reflection
Figure 1.7: The composition of a rotation and a reflection
in the line ℓ that passes through the origin and makes an angle of −π
6 with the positive x1-axis.
But linear transformations are completely determined by what they do to the standard basis: two
transformations that do the same thing must be the same transformation. Thus we conclude that
S ◦ T is the reflection in ℓ. This can be verified with geometric reasoning as well.
By comparison, the composition T ◦ S (first reflect, then rotate) is represented by the same
matrices multiplied in the opposite order:
""
1
2 −
√
3
2√
3
2
1
2
#1 0
0 −1

=
""
1
2
√
3
2√
3
2 −1
2
#
. (1.9)",Multivariable_Calculus_Shimamoto.pdf
"matrices multiplied in the opposite order:
""
1
2 −
√
3
2√
3
2
1
2
#1 0
0 −1

=
""
1
2
√
3
2√
3
2 −1
2
#
. (1.9)
Note that this is different from the matrix of S ◦ T. Matrix multiplication need not obey the
commutative law AB = BA. (Can you describe geometrically the linear transformation that (1.9)
represents?)
1.5 The geometry of the dot product
We now discuss some geometric properties of the dot product, or, perhaps more accurately, how
the dot product can be used to develop geometric intuition about Rn when n is large enough to be
outside direct experience.",Multivariable_Calculus_Shimamoto.pdf
"12 CHAPTER 1. Rn
Definition. The norm, or magnitude, of a vector x = (x1, x2, . . . , xn) in Rn, denoted by ∥x∥, is
defined to be:
∥x∥ =
q
x2
1 + x2
2 + ··· + x2n.
For instance, in R3, if x = (1, 2, 3), then ∥x∥ = √1 + 4 + 9 =
√
14.
The following simple property gets used a lot.
Proposition 1.10. If x ∈ Rn, then x · x = ∥x∥2.
Proof. Both sides equal x2
1 + x2
2 + ··· + x2
n.
We return to the familiar setting of the plane and examine these notions there. For instance,
if x = ( x1, x2), then ∥x∥ =
p
x2
1 + x2
2. By the Pythagorean theorem, this is the length of the
hypotenuse of a right triangle with legs |x1| and |x2|. If we think of x as an arrow emanating from
the origin, then ∥x∥ is the length of the arrow. If we think of x as a point, then ∥x∥ is the distance
from x to the origin.
Given two points x and y in R2, the distance between them is the length of the arrow that
connects them, − →yx = x − y. Hence:
Distance between x and y = ∥x − y∥.",Multivariable_Calculus_Shimamoto.pdf
"connects them, − →yx = x − y. Hence:
Distance between x and y = ∥x − y∥.
Next, let x = (x1, x2) and y = (y1, y2) be nonzero elements of R2, regarded as arrows emanating
from the origin. Suppose that the arrows are perpendicular. If x and y do not form a horizon-
tal/vertical pair, then the slopes of the lines through the origin that contain them are defined and
are negative reciprocals. The slope is the ratio of vertical displacement to horizontal displacement,
so this gives x2
x1
= −y1
y2
. See Figure 1.8. This is easily rearranged to become x1y1 + x2y2 = 0, or
Figure 1.8: Perpendicular vectors in the plane: for instance, note that x has slope x2/x1.
x · y = 0. If x and y do form a horizontal/vertical pair, say x = (x1, 0) and y = (0, y2), then
x · y = 0 once again. Conversely, if x · y = 0, the preceding reasoning can be reversed to conclude
that x and y are perpendicular. Thus:
In R2, x and y are perpendicular if and only if x · y = 0.",Multivariable_Calculus_Shimamoto.pdf
"that x and y are perpendicular. Thus:
In R2, x and y are perpendicular if and only if x · y = 0.
For vectors in the plane in general, let θ denote the angle between two given vectors x and y,
where 0 ≤ θ ≤ π. To study the relationship between the dot product and θ, assume for the moment
that neither x nor y is a scalar multiple of the other, so θ ̸= 0, π, and consider the triangle whose",Multivariable_Calculus_Shimamoto.pdf
"1.5. THE GEOMETRY OF THE DOT PRODUCT 13
Figure 1.9: Vectors x and y in R2 and the angle θ between them
vertices are 0, x, and y. Two of the sides of this triangle have lengths ∥x∥ and ∥y∥, and the length
of the third side is the length of the arrow − →yx = x−y. See Figure 1.9. Thus, by the law of cosines,
∥x − y∥2 = ∥x∥2 + ∥y∥2 − 2∥x∥∥y∥cos θ. By Proposition 1.10, this is the same as:
(x − y) · (x − y) = x · x + y · y − 2∥x∥∥y∥cos θ. (1.10)
Multiplying out the left-hand side gives (x−y)·(x−y) = x·x−x·y−y·x+y·y = x·x−2x·y+y·y.
This expansion uses some of the elementary, but obvious, algebraic properties of the dot product
that we declined to list. After substitution into (1.10), we obtain:
x · x − 2x · y + y · y = x · x + y · y − 2∥x∥∥y∥cos θ.
Hence, after some cancellation:
x · y = ∥x∥∥y∥cos θ. (1.11)
In the case that x or y is a scalar multiple of the other, then θ = 0 or π. Say, for instance, that",Multivariable_Calculus_Shimamoto.pdf
"x · y = ∥x∥∥y∥cos θ. (1.11)
In the case that x or y is a scalar multiple of the other, then θ = 0 or π. Say, for instance, that
y = cx. Then x·y = x·(cx) = c∥x∥2. Meanwhile, ∥y∥ = |c|∥x∥, so ∥x∥∥y∥cos θ = |c|∥x∥2 cos θ =
±c∥x∥2 cos θ. The plus sign occurs when |c| = c, i.e., the scalar multiple c is nonnegative, in which
case θ = 0 and cos θ = 1. The minus sign occurs when c <0, whence cos θ = cos π = −1. Either
way, ∥x∥∥y∥cos θ = c∥x∥2, the same as x · y. Thus (1.11) is valid for all x, y in R2.
We use our knowledge of the plane to try to create some intuition about Rn when n ≥ 3. This
is especially important when n > 3, since visualization in those spaces is almost completely an
act of imagination. For instance, if v is a nonzero vector in Rn, we think of the set of all scalar
multiples cv as a “line.” If w is a second vector, not a scalar multiple of v, and c and d are scalars,
the parallelogram law of addition suggests that the combination cv + dw should lie in a “plane”",Multivariable_Calculus_Shimamoto.pdf
"the parallelogram law of addition suggests that the combination cv + dw should lie in a “plane”
determined by v and w and that, as c and d range over all possible scalars, cv + dw should sweep
out this plane. As a result, the set of all vectors cv + dw, where c, d∈ R, is called the plane
spanned by v and w. We denote it by P. The only reason that it’s a plane is because that’s what
we have chosen to call it.
We would like to make P into a replica of the familiar plane R2. We sketch one approach for
doing this, without filling in many technical details. We begin by choosing two vectors u1 and u2
in P such that u1 · u1 = u2 · u2 = 1 and u1 · u2 = 0. These vectors play the role of the standard
basis vectors e1 and e2, which of course satisfy the same relations. (It’s not hard to show that such
vectors u1 and u2 exist. In fact, in terms of the spanning vectors v and w, one can check that
u1 = 1
∥v∥v and u2 = 1
∥z∥z, where z = w − v·w",Multivariable_Calculus_Shimamoto.pdf
"vectors u1 and u2 exist. In fact, in terms of the spanning vectors v and w, one can check that
u1 = 1
∥v∥v and u2 = 1
∥z∥z, where z = w − v·w
v·v v, is one possibility, though there are many others.)
We useu1 and u2 to establish a two-dimensional coordinate system internal toP. Every element
of P can be written in the form x = a1u1 +a2u2 for some scalars a1 and a2. In terms of our budding
intuition, we think of
p
a2
1 + a2
2 as representing the distance from the origin, or the length, of x.",Multivariable_Calculus_Shimamoto.pdf
"14 CHAPTER 1. Rn
If y = b1u2 + b2u2 is another element of P, then:
x · y = (a1u1 + a2u2) · (b1u1 + b2u2)
= a1b1u1 · u1 + (a1b2 + a2b1)u1 · u2 + a2b2u2 · u2
= a1b1 + a2b2.
Thus the dot product in Rn agrees with the result we would expect in terms of the newly created
internal coordinates in P. In particular, ∥x∥2 = x · x = a2
1 + a2
2, so the norm in Rn represents our
intuitive notion of length in P. This is true even though the coordinates of x = (x1, x2, . . . , xn) in
Rn don’t really have anything to do with P.
Continuing in this way, we can build upP as a copy of R2 and transfer over the familiar concepts
of plane geometry, such as distance and angle. The following terminology reflects this intuition.
Definition. A vector x in Rn is called a unit vector if ∥x∥ = 1.
For example, the standard basis vectors ei = (0, 0, . . . ,0, 1, 0, . . . ,0) are unit vectors.
Corollary 1.11. If x is a nonzero vector in Rn, then u = 1
∥x∥x is a unit vector. It is called the",Multivariable_Calculus_Shimamoto.pdf
"Corollary 1.11. If x is a nonzero vector in Rn, then u = 1
∥x∥x is a unit vector. It is called the
unit vector in the direction of x .
Proof. We calculate thatu·u =
 1
∥x∥x


·
 1
∥x∥x


= 1
∥x∥2

x·x


= 1
∥x∥2 ∥x∥2 = 1. Thus, by Proposition
1.10, ∥u∥ = 1.
Our main result in this section is that the relation between the dot product and angles that we
derived for the plane in equation (1.11) is true in Rn for all n.
Theorem 1.12. If x and y are vectors in Rn, then:
x · y = ∥x∥∥y∥cos θ,
where θ is the angle between x and y in the plane that they span.
Proof. The case that x or y is a scalar multiple of the other is proved in the same way as in R2.
Otherwise, x and y span a plane P, and the points 0, x, and y are the vertices of a “triangle” in
P. Since we can make P into a geometric replica of R2, the law of cosines remains true, and, since
the norm in Rn represents length in P, this takes the form:
∥x − y∥2 = ∥x∥2 + ∥y∥2 − 2∥x∥∥y∥cos θ.",Multivariable_Calculus_Shimamoto.pdf
"the norm in Rn represents length in P, this takes the form:
∥x − y∥2 = ∥x∥2 + ∥y∥2 − 2∥x∥∥y∥cos θ.
From here, the argument is identical to the one for R2.
Lastly, we introduce the standard terminology for the case of perpendicular vectors, that is,
when θ = π
2 , so cos θ = 0.
Definition. Two vectors x and y in Rn are called orthogonal if x · y = 0.
For instance, the standard basis vectors in Rn are orthogonal: ei · ej = 0 whenever i ̸= j.",Multivariable_Calculus_Shimamoto.pdf
"1.6. DETERMINANTS 15
1.6 Determinants
The determinant is a function that assigns a real number to an n by n matrix. There’s a separate
function for each n. We shall focus almost exclusively on the cases n = 2 and n = 3, since those are
the cases we really need later. The determinant is not defined for matrices in which the numbers
of rows and columns are unequal.
The determinant of a 2 by 2 matrix is defined to be:
det
a11 a12
a21 a22

= a11a22 − a12a21.
It’s the product of the diagonal entries minus the product of the off-diagonal entries. For instance,
det [1 2
3 4] = 1 · 4 − 2 · 3 = 4 − 6 = −2.
One often thinks of the determinant as a function of the rows of the matrix. If x = (x1, x2) and
y = (y1, y2), let

x y

denote the matrix whose rows are x and y:
det

x y

= det
x1 x2
y1 y2

= x1y2 − x2y1.
The main geometric property of 2 by 2 determinants is the following.
Proposition 1.13. Let x and y be vectors in R2, neither a scalar multiple of the other. Then:
det

x y",Multivariable_Calculus_Shimamoto.pdf
"Proposition 1.13. Let x and y be vectors in R2, neither a scalar multiple of the other. Then:
det

x y
 = Area of the parallelogram determined by x and y.
Proof. We show the equivalent result that

det

x y

2 = (Area)2. The area of the parallelogram
is (base)(height). As base, we use the arrow x, which has length ∥x∥. For the height h, we drop a
Figure 1.10: Area of a parallelogram
perpendicular from the point y to the base. See Figure 1.10. Let θ be the angle between x and y,
where as usual 0 ≤ θ ≤ π. Then the height is given by h = ∥y∥sin θ, so Area = ∥x∥∥y∥sin θ and:
(Area)2 = ∥x∥2∥y∥2 sin2 θ. (1.12)",Multivariable_Calculus_Shimamoto.pdf
"16 CHAPTER 1. Rn
Meanwhile:

det

x y

2 = (x1y2 − x2y1)2
= x2
1y2
2 − 2x1x2y1y2 + x2
2y2
1
= (x2
1 + x2
2)(y2
1 + y2
2) − x2
1y2
1 − x2
2y2
2 − 2x1x2y1y2
= (x2
1 + x2
2)(y2
1 + y2
2) − (x1y1 + x2y2)2
= ∥x∥2∥y∥2 − (x · y)2
= ∥x∥2∥y∥2 − ∥x∥2∥y∥2 cos2 θ
= ∥x∥2∥y∥2(1 − cos2 θ)
= ∥x∥2∥y∥2 sin2 θ. (1.13)
Comparing equations (1.12) and (1.13) gives the result.
If x or y is a scalar multiple of the other, for instance, if y = cx for some scalar c, then the
“parallelogram” they determine degenerates into a line segment, which has area 0. At the same
time, det

x y

=
 x1 x2
cx1 cx2

= cx1x2 − cx1x2 = 0, so the proposition remains valid when things
degenerate, too.
Determinants satisfy a great many algebraic properties. We list only the ones that we shall use,
which barely begins to scratch the surface. In part (c) of the following proposition, the transpose
of a matrix A, denoted by At, refers to the matrix obtained by turning the rows into columns and",Multivariable_Calculus_Shimamoto.pdf
"of a matrix A, denoted by At, refers to the matrix obtained by turning the rows into columns and
the columns into rows. So the ( i, j)th entry of At is the (j, i)th entry of A. For instance:
1 2
3 4
t
=
1 3
2 4

,
1 2 3
4 5 6
t
=


1 4
2 5
3 6

,


1
2
3


t
=

1 2 3

, and so on.
If A is an m by n matrix, then At is n by m.
Proposition 1.14.
(a) If two of the rows are equal, the determinant is 0.
(b) Interchanging two rows flips the sign of the determinant.
(c) det( At) = det A.
(d) det( AB) = (det A)(det B).
Proof. For 2 by 2 determinants, the proofs are easy.
(a) det

x x

= det [x1 x2
x1 x2 ] = x1x2 − x2x1 = 0.
(b) det

y x

= det [y1 y2
x1 x2 ] = y1x2 − y2x1 = −(x1y2 − x2y1) = −det

x y

.
(c) det(At) = det [a11 a21
a12 a22 ] = a11a22 − a21a12 = det [a11 a12
a21 a22 ] = det A.
(d) det(AB) = det

[ a11 a12
a21 a22 ]
h
b11 b12
b21 b22
i
= det
h
a11b11+a12b21 a11b12+a12b22
a21b11+a22b21 a21b12+a22b22
i
= (a11b11 + a12b21)(a21b12",Multivariable_Calculus_Shimamoto.pdf
"
[ a11 a12
a21 a22 ]
h
b11 b12
b21 b22
i
= det
h
a11b11+a12b21 a11b12+a12b22
a21b11+a22b21 a21b12+a22b22
i
= (a11b11 + a12b21)(a21b12
+a22b22)−(a11b12 +a12b22)(a21b11 +a22b21). After multiplying out, the terms involving a11a21b11b12
and a12a22b21b22 cancel in pairs, leaving:
det(AB) = a11b11a22b22 + a12b21a21b12 − a11b12a22b21 − a12b22a21b11.
One can check that (det A)(det B) =

a11a22 − a12a21


b11b22 − b12b21


expands to the same thing.",Multivariable_Calculus_Shimamoto.pdf
"1.6. DETERMINANTS 17
For 3 by 3 matrices, the determinant is defined in terms of the 2 by 2 case:
det


a11 a12 a13
a21 a22 a23
a31 a32 a33

 = a11 · det
a22 a23
a32 a33

− a12 · det
a21 a23
a31 a33

+ a13 · det
a21 a22
a31 a32

. (1.14)
The signs in the sum alternate, and the pattern is that the terms run along the entries of the first
row, a1j, each multiplied by the 2 by 2 determinant obtained by deleting the first row and jth
column of the original matrix. The process is known as expansion along the first row . For
instance:
det


1 2 3
4 5 6
7 8 9

 = 1 · det
5 6
8 9

− 2 · det
4 6
7 9

+ 3 · det
4 5
7 8

= 1 · (45 − 48) − 2 · (36 − 42) + 3· (32 − 35) = −3 + 12− 9 = 0.
One can show that 3 by 3 determinants satisfy all four of the algebraic properties of Proposition
1.14. The calculations are longer than in the 2 by 2 case but still straightforward, except for the
product formula det(AB) = (det A)(det B) which calls for a new approach.",Multivariable_Calculus_Shimamoto.pdf
"product formula det(AB) = (det A)(det B) which calls for a new approach.
One consequence of the properties is that there’s a formula for expanding the determinant along
any row. To expand along the ith row, interchange row i with the row above it repeatedly until it
reaches the top, then expand using equation (1.14). The result has the same form as (1.14), except
that the leading scalar factors come from the ith row and sometimes the signs might alternate
beginning with a minus sign depending on how many sign flips were introduced in getting the ith
row to the top.
In addition, since det(At) = det A, any general statement about rows applies to columns as well,
so there are formulas for expanding along any column, too. We won’t write down those formulas
precisely.
There is also a geometric interpretation of 3 by 3 determinants in terms of three-dimensional
volume. This is discussed in the next chapter.",Multivariable_Calculus_Shimamoto.pdf
"There is also a geometric interpretation of 3 by 3 determinants in terms of three-dimensional
volume. This is discussed in the next chapter.
For larger matrices, the same pattern continues, that is, n by n determinants can be defined in
terms of (n − 1) by (n − 1) determinants using expansion along the first row. For the 4 by 4 case,
the formula is:
det


a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44

 = a11 · det


a22 a23 a24
a32 a33 a34
a42 a43 a44

 − a12 · det


a21 a23 a24
a31 a33 a34
a41 a43 a44


+ a13 · det


a21 a22 a24
a31 a32 a34
a41 a42 a44

 − a14 · det


a21 a22 a23
a31 a32 a33
a41 a42 a43

.
The algebraic properties of Proposition 1.14 remain true for n by n determinants whatever the
value of n, though, to prove this, one should really develop the algebraic structure of determinants
in a systematic way rather than hope for success with brute force calculation. We leave this level",Multivariable_Calculus_Shimamoto.pdf
"in a systematic way rather than hope for success with brute force calculation. We leave this level
of generality for a course in linear algebra. In what follows, we work for the most part with the
cases n = 2 and n = 3.",Multivariable_Calculus_Shimamoto.pdf
"18 CHAPTER 1. Rn
1.7 Exercises for Chapter 1
Section 1 Vector arithmetic
In Exercises 1.1–1.4, let x and y be the vectors x = (1, 2, 3) and y = (4, −5, 6) in R3. Also, 0
denotes the zero vector, 0 = (0, 0, 0).
1.1. Find x + y, 2x, and 2x − 3y.
1.2. Find − →yx and y + − →yx.
1.3. If x + y + z = 0, find z.
1.4. If x − 2y + 3z = 0, find z.
1.5. Let x and y be points in Rn.
(a) Show that the midpoint of line segment xy is given by m = 1
2 x + 1
2 y. ( Hint: What do
you add to x to get to the midpoint?)
(b) Find an analogous expression in terms of x and y for the point p that is 2/3 of the way
from x to y.
(c) Let x = (1, 1, 0) and y = (0, 1, 1). Find the point z in R3 such that y is the midpoint of
line segment xz.
Section 2 Linear transformations
2.1. Let T : R3 → R2 be the projection of x1x2x3-space onto the x1x2-plane, T(x1, x2, x3) =
(x1, x2). Draw pictures that show that T satisfies the requirements for being a linear trans-
formation.",Multivariable_Calculus_Shimamoto.pdf
"(x1, x2). Draw pictures that show that T satisfies the requirements for being a linear trans-
formation.
2.2. Let T : R2 → R3 be the function defined by T(x1, x2) = (x1, x2, 0). Show that T is a linear
transformation.
2.3. Let a1, a2, . . . ,an be vectors in Rm. Verify that the function T : Rn → Rm given by
T(x1, x2, . . . , xn) = x1a1 + x2a2 + ··· + xnan
is a linear transformation.
Section 3 The matrix of a linear transformation
3.1. Let T : R2 → R2 be the linear transformation such that T(e1) = (1, 2) and T(e2) = (3, 4).
(a) Find the matrix of T with respect to the standard bases.
(b) Find T(5, 6).
(c) Find a general formula for T(x1, x2).
3.2. Let T : R3 → R3 be the linear transformation such that T(e1) = (1, 0, −1), T(e2) = (−1, 1, 0),
and T(e3) = (0, −1, 1).
(a) Find the matrix of T with respect to the standard bases.
(b) Find T(1, −1, 1).",Multivariable_Calculus_Shimamoto.pdf
"1.7. EXERCISES FOR CHAPTER 1 19
(c) Find the set of all points x = (x1, x2, x3) in R3 such that T(x) = 0.
In Exercises 3.3–3.6, find the matrix of the given linear transformationT : R2 → R2 with respect
to the standard bases.
3.3. T is the counterclockwise rotation by π
2 about the origin.
3.4. T is the reflection in the x2-axis.
3.5. T is the identity function, T(x) = x.
3.6. T is the dilation about the origin by a factor of 2, T(x) = 2x.
3.7. Let ρθ : R2 → R2 be the rotation about the origin counterclockwise by an angle θ. Show that
the matrix of ρθ with respect to the standard bases is:
Rθ =
cos θ −sin θ
sin θ cos θ

. (1.15)
The matrix Rθ is called a rotation matrix.
3.8. Let T : R2 → R2 be the linear transformation whose matrix with respect to the standard
bases is A =
0 1
1 0

. Describe T geometrically.
3.9. Let T : R2 → R2 be the linear transformation whose matrix with respect to the standard
bases is A =
1 2
2 4

.
(a) Find T(e1) and T(e2).",Multivariable_Calculus_Shimamoto.pdf
"3.9. Let T : R2 → R2 be the linear transformation whose matrix with respect to the standard
bases is A =
1 2
2 4

.
(a) Find T(e1) and T(e2).
(b) Describe the image of T, that is, the set of all y in R2 such that y = T(x) for some x in
R2.
3.10. Let v1 = (1, 1) and v2 = (−1, 1).
(a) Find scalars c1, c2 such that c1v1 + c2v2 = e1.
(b) Find scalars c1, c2 such that c1v1 + c2v2 = e2.
(c) Let T : R2 → R2 be the linear transformation such that T(v1) = (−1, −1) and T(v2) =
(−2, 2). Find the matrix of T with respect to the standard bases.
(d) Let S : R2 → R2 be the linear transformation such that S(v1) = e1 and S(v2) = e2.
Find the matrix of S with respect to the standard bases.
3.11. Let T : R2 → R3 be the linear transformation given by T(x1, x2) = ( x1, x2, 0). Find the
matrix of T with respect to the standard bases.
3.12. Let T : R3 → R3 be the rotation about the x3-axis by π
2 counterclockwise as viewed looking",Multivariable_Calculus_Shimamoto.pdf
"matrix of T with respect to the standard bases.
3.12. Let T : R3 → R3 be the rotation about the x3-axis by π
2 counterclockwise as viewed looking
down from the positive x3-axis. Find the matrix of T with respect to the standard bases.
3.13. Let T : R3 → R3 be the rotation by π about the line x1 = x2, x3 = 0, in the x1x2-plane. Find
the matrix of T with respect to the standard bases.
3.14. Let A and B be m by n matrices. If Ax = Bx for all column vectors x in Rn, show that
A = B.",Multivariable_Calculus_Shimamoto.pdf
"20 CHAPTER 1. Rn
Section 4 Matrix multiplication
In Exercises 4.1–4.6, find the indicated matrix products.
4.1. AB and BA, where A =
1 −1
1 1

and B =
 2 4
−1 3

4.2. AB and BA, where A =
1 2
3 4

and B =
1 0
0 1

4.3. AB and BA, where A =


0 0 1
1 0 0
0 1 0

 and B =


0 1 0
0 0 1
1 0 0


4.4. AB and BA, where A =


1 2 3
0 4 5
0 0 6

 and B =


1 1 1
0 1 1
0 0 1


4.5. AB and BA, where A =
2 −1 −4
3 2 1

and B =


2 3
−1 2
−4 1


4.6. AB and BA, where A =

1 2 3

and B =


4
5
6


4.7. We have seen that the commutative law AB = BA does not hold in general for matrix
multiplication. In fact, the situation is worse than that: it’s rare for AB and BA even to
both be defined. In that sense, the preceding half dozen exercises are misleading.
(a) Find an example of matrices A and B such that neither AB nor BA is defined.
(b) Find an example where AB is defined but BA is not.
4.8. (a) Let T : Rn → Rm, S : Rm → Rp and R: Rp → Rq be functions. Show that:
",Multivariable_Calculus_Shimamoto.pdf
"(b) Find an example where AB is defined but BA is not.
4.8. (a) Let T : Rn → Rm, S : Rm → Rp and R: Rp → Rq be functions. Show that:

(R ◦ S) ◦ T


(x) =

R ◦ (S ◦ T)


(x) for all x in Rn.
(b) Show that matrix multiplication is associative, that is, if A is a q by p matrix, B a p by
m matrix, and C an m by n matrix, show that:
(AB)C = A(BC).
Since it does not matter how the terms are grouped, we shall write the product henceforth
simply as ABC, without parentheses. ( Hint: See Exercise 3.14.)
In Exercises 4.9–4.10:
• let Rθ =
cos θ −sin θ
sin θ cos θ

be the matrix (1.15) that represents the rotation ρθ : R2 → R2 about
the origin counterclockwise by angle θ, and
• let S =
1 0
0 −1

be the matrix (1.6) that represents the reflection r: R2 → R2 in the x1-axis.",Multivariable_Calculus_Shimamoto.pdf
"1.7. EXERCISES FOR CHAPTER 1 21
4.9. Use matrices to prove that r ◦ ρθ ◦ r = ρ−θ.
4.10. (a) Compute the product A = RθSR−θ.
(b) By thinking about the corresponding composition of linear transformations, give a ge-
ometric description of the linear transformation T : R2 → R2 that is represented with
respect to the standard bases by A.
Section 5 The geometry of the dot product
5.1. Let x = (−1, 1, −2) and y = (4, −1, −1).
(a) Find x · y.
(b) Find ∥x∥ and ∥y∥.
(c) Find the angle between x and y.
5.2. Find the unit vector in the direction of x = (2, −1, −2).
5.3. Find all unit vectors in R2 that are orthogonal to x = (1, 2).
5.4. What does the sign of x · y tell you about the angle between x and y?
In Exercises 5.5–5.10, show that the dot product satisfies the given property. The properties
are true for vectors in Rn, though you may assume in your arguments that the vectors are in R2,
i.e., x = (x1, x2), y = (y1, y2), and so on. The proofs for Rn in general are similar.
5.5. x · y = y · x",Multivariable_Calculus_Shimamoto.pdf
"i.e., x = (x1, x2), y = (y1, y2), and so on. The proofs for Rn in general are similar.
5.5. x · y = y · x
5.6. (cx) · y = c(x · y) for any scalar c
5.7. ∥cx∥ = |c|∥x∥ for any scalar c
5.8. w · (x + y) = w · x + w · y
5.9. (v + w) · (x + y) = v · x + v · y + w · x + w · y (Hint: Make use of the preceding exercises.)
5.10. (x + y) · (x − y) = ∥x∥2 − ∥y∥2
5.11. Recall that a rhombus is a planar quadrilateral whose sides all have the same length. Use the
dot product to show that the diagonals of a rhombus are perpendicular to each other.
5.12. The Pythagorean theorem states that, for a right triangle in the plane, a2 + b2 = c2, where
a and b are the lengths of the legs and c is the length of the hypotenuse. Use vector algebra
and the dot product to verify that the theorem remains true for right triangles in Rn. ( Hint:
The hypotenuse is a diagonal of a rectangle.)
Section 6 Determinants
In Exercises 6.1–6.4, find the given determinant.
6.1. det
 2 5
−3 4

6.2. det
 0 5
−3 4
",Multivariable_Calculus_Shimamoto.pdf
"22 CHAPTER 1. Rn
6.3. det


1 −2 3
−4 5 −6
7 −8 9


6.4. det


1 0 0
2 3 0
4 5 6


6.5. Find the area of the parallelogram in R2 determined by x = (4, 0) and y = (1, 3).
6.6. Find the area of the parallelogram in R2 determined by x = (−2, −3) and y = (−3, 2).
6.7. Let A =
1 −2
2 −4

. Use the product rule for determinants to show that there is no 2 by 2
matrix B such that AB =
1 0
0 1

.
6.8. Let A be an n by n matrix, and let At be its transpose. Show that det( AtA) = (det A)2.
6.9. If (x1, x2, x3) ∈ R3, let:
T(x1, x2, x3) = det


x1 x2 x3
1 2 3
4 5 6

. (1.16)
(a) Expand the determinant to find a formula for T(x1, x2, x3).
(b) Show that equation (1.16) defines a linear transformation T : R3 → R.
(c) Find the matrix of T with respect to the standard bases.
(d) The matrix you found in part (c) should be a 1 by 3 matrix. Thinking of it as a vector
in R3, show that it is orthogonal to both (1 , 2, 3) and (4, 5, 6), the second and third rows",Multivariable_Calculus_Shimamoto.pdf
"in R3, show that it is orthogonal to both (1 , 2, 3) and (4, 5, 6), the second and third rows
of the matrix used to define T(x1, x2, x3).",Multivariable_Calculus_Shimamoto.pdf
"Part II
Vector-valued functions of one
variable
23",Multivariable_Calculus_Shimamoto.pdf
"Chapter 2
Paths and curves
This chapter is concerned with curves in Rn. While we may have an intuitive sense of what a
curve is, at least in R2 or R3, the formal description here is somewhat indirect in that, rather than
requiring a curve to have a defining equation, we describe it by how it is swept out, like the trace
of a skywriter. Thus, in addition to studying geometric features of the curve, such as its length,
we also look at quantities related to the skywriter’s motion, such as its velocity and acceleration.
The goal of the chapter is a remarkable result about curves in R3 that describes measurements
that characterize the geometry of a space curve completely. Along the way, we shall gain valuable
experience applying vector methods.
The functions in this chapter take their values in Rn, but they are functions of one variable. As
a result, we treat the material as a continuation of first-year calculus without going back to redefine",Multivariable_Calculus_Shimamoto.pdf
"a result, we treat the material as a continuation of first-year calculus without going back to redefine
or redevelop concepts that are introduced there. At times, this assumed familiarity may lead to a
rather loose treatment of certain basic topics, such as continuity, derivatives, and integrals. When
we study functions of more than one variable, we shall go back and define these concepts carefully,
and what we say then applies retroactively to what we cover here. We hope that any reader who
becomes anxious about a possible lack of rigor will be willing to wait.
2.1 Parametrizations
Let I be an interval of real numbers, typically, I = [a, b], (a, b), or R.
Definition. A continuous function α: I → Rn is called a path. As t varies over I, α(t) traces out
a curve C. More precisely:
C = {x ∈ Rn : x = α(t) for some t in I}.
This is also known as the image of α. We say that α is a parametrization of the curve. We often",Multivariable_Calculus_Shimamoto.pdf
"C = {x ∈ Rn : x = α(t) for some t in I}.
This is also known as the image of α. We say that α is a parametrization of the curve. We often
refer to the input variable t as time and think of α(t) as the position of a moving object at time
t. See Figure 2.1.
A path is a vector-valued function of one variable. To follow our notational convention, we
should write the value in boldface as α(t), but for the most part we continue to use plainface,
usually reserving boldface for a particular kind of vector-valued function that we begin studying in
Chapter 8. For each t in I, α(t) is a point of Rn, so we may write α(t) = (x1(t), x2(t), . . . , xn(t)),
where each of the n coordinates is a real number that depends on t, i.e., a real-valued function of
one variable.
Here are some of the standard examples of parametrized curves that we shall refer to frequently.
25",Multivariable_Calculus_Shimamoto.pdf
"26 CHAPTER 2. PATHS AND CURVES
Figure 2.1: A parametrization α: [a, b] → Rn of a curve C
Example 2.1. Circles in R2: x2 + y2 = a2, where a is the radius of the circle.
The equation of the circle can be rewritten as
x
a

2 +
y
a

2 = 1. To parametrize the circle, we
use the identity cos 2 t + sin2 t = 1 and let x
a = cos t and y
a = sin t, or x = a cos t and y = a sin t.
These expressions show that t has a geometric interpretation as the angle that ( x, y) makes with
the positive x-axis, as shown in Figure 2.2. The substitutions x = a cos t and y = a sin t satisfy the
defining equation x2 + y2 = a2 of the circle, so α: R → R2,
α(t) = (a cos t, asin t), a constant, a >0,
parametrizes the circle, traversing it in the counterclockwise direction.
Figure 2.2: A circle of radius a
Example 2.2. Graphs y = f(x) in R2.
As a concrete example, consider the curve described by y = sin x in R2, where 0 ≤ x ≤ π. See",Multivariable_Calculus_Shimamoto.pdf
"Example 2.2. Graphs y = f(x) in R2.
As a concrete example, consider the curve described by y = sin x in R2, where 0 ≤ x ≤ π. See
Figure 2.3. It consists of points of the form ( x, sin x), and it is traced out as x goes from x = 0 to
x = π. Hence x serves as a natural parameter, and α: [0, π] → R2,
α(x) = (x, sin x),
is a parametrization of the curve.
Figure 2.3: The graph y = sin x, 0 ≤ x ≤ π",Multivariable_Calculus_Shimamoto.pdf
"2.1. PARAMETRIZATIONS 27
Example 2.3. Helices in R3.
A helix winds around a circular cylinder, say of radius a. If the axis of the cylinder is the z-axis,
then the x and y-coordinates along the helix satisfy the equation of the circle as in Example 2.1,
while the z-coordinate changes at a constant rate. Thus we set x = a cos t, y = a sin t, and z = bt
for some constant b. That is, the helix is parametrized by α: R → R3, where:
α(t) = (a cos t, asin t, bt), a, b constant, a >0.
If b >0, the helix is said to be “right-handed” and if b <0 “left-handed.” Each type is shown in
Figure 2.4.
Figure 2.4: Helices: right-handed (at left) and left-handed (at right)
Example 2.4. Lines in Rn.
Given a point a and a nonzero vector v in Rn, the line through a and parallel to v is
parametrized by α: R → Rn, where:
α(t) = a + tv.
See Figure 2.5. If a = ( a1, a2, . . . , an) and v = ( v1, v2, . . . , vn), then α(t) = ( a1, a2, . . . , an) +",Multivariable_Calculus_Shimamoto.pdf
"α(t) = a + tv.
See Figure 2.5. If a = ( a1, a2, . . . , an) and v = ( v1, v2, . . . , vn), then α(t) = ( a1, a2, . . . , an) +
t(v1, v2, . . . , vn) = (a1 + tv1, a2 + tv2, . . . , an + tvn), so in terms of coordinates:



x1 = a1 + tv1
x2 = a2 + tv2
...
xn = an + tvn.
If the domain of α is restricted to an interval of finite length, then α parametrizes a finite segment
of the line.
Figure 2.5: As t varies, α(t) = a + tv traces out a line.",Multivariable_Calculus_Shimamoto.pdf
"28 CHAPTER 2. PATHS AND CURVES
Example 2.5. Find a parametrization of the line in R3 that passes through the points a = (1, 2, 3)
and b = (4, 5, 6).
The line passes through the point a = (1, 2, 3) and is parallel to v = − →ab = b − a = (4, 5, 6) −
(1, 2, 3) = (3, 3, 3). The setup is indicated in Figure 2.6. Therefore one parametrization is:
α(t) = (1, 2, 3) + t(3, 3, 3) = (1 + 3t, 2 + 3t, 3 + 3t).
Figure 2.6: Parametrizing the line through two given points a and b
Any nonzero scalar multiple of v = (3, 3, 3) is also parallel to the line and could be used as
part of a parametrization as well. For instance, w = (1 , 1, 1) is such a multiple, and β(t) =
(1, 2, 3) + t(1, 1, 1) = (1 + t, 2 + t, 3 + t) is another parametrization of the same line.
2.2 Velocity, acceleration, speed, arclength
We now introduce some basic aspects of calculus for paths and curves. We discuss derivatives of
paths in this section and integrals of real-valued functions over paths in the next.",Multivariable_Calculus_Shimamoto.pdf
"paths in this section and integrals of real-valued functions over paths in the next.
Definition. Given a path α: I → Rn, the derivative of α is defined by:
α′(t) = lim
h→0
1
h

α(t + h) − α(t)


= lim
h→0
1
h

x1(t + h), x2(t + h), . . . , xn(t + h)


−

x1(t), x2(t), . . . , xn(t)


= lim
h→0
x1(t + h) − x1(t)
h , x2(t + h) − x2(t)
h , . . . ,xn(t + h) − xn(t)
h

= (x′
1(t), x′
2(t), . . . , x′
n(t)),
provided the limit exists. The derivative is also called the velocity v(t) of α.
Likewise, α′′(t) = v′(t) is called the acceleration, denoted by a(t).
Example 2.6. For the helix parametrized by α(t) = (cos t, sin t, t):
• velocity: v(t) = α′(t) = (−sin t, cos t, 1),
• acceleration: a(t) = v′(t) = (−cos t, −sin t, 0).
Now, choose some time t0, and, for any time t, let s(t) be the distance traced out along the
path from α(t0) to α(t). Then s(t) is called the arclength function, and we think of the rate of",Multivariable_Calculus_Shimamoto.pdf
"2.2. VELOCITY, ACCELERATION, SPEED, ARCLENGTH 29
change of distance ds
dt as the speed. Unfortunately, these terms should be defined more carefully,
and, to get everything in the right logical order, it seems best to define the speed first.
To define what we expect ds
dt to be, we make the intuitive approximation that the distance △s
along the curve between two nearby points α(t) and α(t + △t) is approximately the straight line
distance between them, as indicated in Figure 2.7. We assume that △t is positive. If △t is small:
“distance along curve” ≈ straight line distance, or
△s ≈ ∥α(t + △t) − α(t)∥.
Thus △s
△t ≈ ∥α(t+△t)−α(t)
△t ∥. In the limit as △t goes to 0, this approaches ∥α′(t)∥, the magnitude of
Figure 2.7: △s ≈ ∥α(t + △t) − α(t)∥
the velocity. We take this as the definition of speed.
Definition. If α: I → Rn is a differentiable path, then its speed, denoted by v(t), is defined to
be:
v(t) = ∥v(t)∥.",Multivariable_Calculus_Shimamoto.pdf
"Definition. If α: I → Rn is a differentiable path, then its speed, denoted by v(t), is defined to
be:
v(t) = ∥v(t)∥.
Note that the speed v(t) is a scalar quantity, whereas the velocity v(t) is a vector.
Now, to define the length of the path from t = a to t = b, we integrate the speed.
Definition. The arclength from t = a to t = b is defined to be
Rb
a v(t) dt =
Rb
a ∥v(t)∥dt.
The arclength function s(t) we considered above is then given by integrating the speed v(t),
so, by the fundamental theorem of calculus, ds
dt = v(t). Thus the definitions realize the intuitive
relations with which we began.
Example 2.7. For the helix parametrized by α(t) = (cos t, sin t, t), find:
(a) the speed and
(b) the arclength from t = 0 to t = 4π, i.e., two full twists around the helix.
First, as just shown in Example 2.6, the velocity is v(t) = (−sin t, cos t, 1). Thus:
(a) speed: v(t) = ∥v(t)∥ =
p
sin2 t + cos2 t + 1 =
√
2.
(b) arclength:
R4π
0 v(t) dt =
R4π
0
√
2 dt =
√
2 t
4π
0 = 4π
√
2.",Multivariable_Calculus_Shimamoto.pdf
"(a) speed: v(t) = ∥v(t)∥ =
p
sin2 t + cos2 t + 1 =
√
2.
(b) arclength:
R4π
0 v(t) dt =
R4π
0
√
2 dt =
√
2 t
4π
0 = 4π
√
2.
Summary of definitions
velocity v(t) = α′(t)
acceleration a(t) = α′′(t)
speed v(t) = ∥v(t)∥ = ds
dt
arclength =
Rb
a v(t) dt",Multivariable_Calculus_Shimamoto.pdf
"30 CHAPTER 2. PATHS AND CURVES
2.3 Integrals with respect to arclength
Let C be a curve in Rn, and let f : C → R be a real-valued function defined on C. To formulate
the definition of the integral of f over C, we adapt the usual approach of first-year calculus for
integrating a function over an interval, namely:
• Chop up the domain C into a sequence of short pieces of lengths △s1, △s2, . . . ,△sk.
• Choose a point x1, x2, . . . ,xk in each piece.
• Consider sums of the form P
j f(xj) △sj.
• Take the limit of the sums as △sj goes to 0.
See Figure 2.8.
Figure 2.8: Defining the integral with respect to arclength
We can rewrite the sums in a way that leads to a normal first-year calculus integral over an
interval by using a parametrization α: [a, b] → Rn of C. We assume that [ a, b] can be subdivided
into a sequence of consecutive subintervals of lengths △t1, △t2, . . . ,△tk such that the subinterval",Multivariable_Calculus_Shimamoto.pdf
"into a sequence of consecutive subintervals of lengths △t1, △t2, . . . ,△tk such that the subinterval
of length △tj gets mapped to the curve segment of length △sj for each j = 1, 2, . . . , k. Then the
sum that models the integral can be written as:
X
j
f(xj) △sj =
X
j
f(α(cj)) △sj
△tj
△tj,
where cj is a value of the parameter for which α(cj) = xj. In the limit as △tj goes to 0, these sums
approach the integral
Rb
a f(α(t))ds
dt dt. In this last expression, ds
dt is the speed, also denoted by v(t).
This intuition is formalized in the following definition.
Definition. Let α: [a, b] → Rn be a differentiable parametrization of a curve C, and let v(t) denote
its speed. If f : C → R is a continuous real-valued function, the integral of f with respect to
arclength, denoted by
R
α f ds, is defined to be:
Z
α
f ds=
Z b
a
f(α(t)) v(t) dt.
We also denote this integral by
R
C f ds, though this raises some issues that are addressed later when",Multivariable_Calculus_Shimamoto.pdf
"Z
α
f ds=
Z b
a
f(α(t)) v(t) dt.
We also denote this integral by
R
C f ds, though this raises some issues that are addressed later when
we have more practice with parametrizations and their effect on calculations. (See page 208.)
Example 2.8. If f = 1 (constant function), then the definition of the integral says
R
C 1 ds =Rb
a v(t) dt. On the other hand, this is also precisely the definition of arclength. Hence:
Z
C
1 ds = arclength of C.",Multivariable_Calculus_Shimamoto.pdf
"2.4. THE GEOMETRY OF CURVES: TANGENT AND NORMAL VECTORS 31
Example 2.9. Consider again the portion of the helix parametrized by α(t) = (cos t, sin t, t),
0 ≤ t ≤ 4π. Find
R
C (x + y + z) ds.
Here, f(x, y, z) = x + y + z, so f(α(t)) = f(cos t, sin t, t) = cos t + sint + t. In other words, we
read off the components of the parametrization to substitute x = cos t, y = sin t, and z = t into
the formula for f. From Example 2.7, v(t) = ∥v(t)∥ =
√
2. Thus:
Z
C
(x + y + z) ds =
Z 4π
0
(cos t + sint + t)
√
2 dt
=
√
2 (sint − cos t + 1
2t2)
4π
0
=
√
2

(0 − 1 + 8π2) − (0 − 1 + 0)


= 8
√
2 π2.
2.4 The geometry of curves: tangent and normal vectors
With the basic notions involving paths and curves in Rn now in place, we direct most of the
remainder of the chapter towards the special case of curves in R3. It turns out that we are already
in a position to derive a fairly substantial result about the geometry of such curves: we shall",Multivariable_Calculus_Shimamoto.pdf
"in a position to derive a fairly substantial result about the geometry of such curves: we shall
describe scalar measurements that determine whether two curves in R3 are congruent in the same
spirit that one learns in high school to test whether triangles in the plane are congruent. In a sense,
this gives a classification of curves in three-dimensional space.
The main idea is to attach a three-dimensional coordinate system that travels along the curve.
By measuring how fast the coordinate directions are changing, one recovers the geometry of the
curve. In order for this to work, the coordinate directions need to have some intrinsic connection
with the geometry of the curve.
We begin with the most natural direction related to the motion along the curve, namely, the
tangent direction. Let α: I → R3 be a differentiable parametrization of a curve, and consider the
difference α(t + h) − α(t) between two nearby points on the curve. As h goes to 0, this vector",Multivariable_Calculus_Shimamoto.pdf
"difference α(t + h) − α(t) between two nearby points on the curve. As h goes to 0, this vector
approaches the tangent direction at α(t), and hence so does the scalar multiple 1
h(α(t + h) − α(t)).
See Figure 2.9. In other words, the derivative α′(t) = lim h→0 1
h(α(t + h) − α(t)) points in the
Figure 2.9: As h → 0, α(t + h) − α(t) approaches the tangent direction.
direction tangent to the curve.
Definition. If α′(t) ̸= 0, define:
T(t) = 1
∥α′(t)∥α′(t).
This is a unit vector in the tangent direction at α(t), called the unit tangent vector (Figure
2.10).",Multivariable_Calculus_Shimamoto.pdf
"32 CHAPTER 2. PATHS AND CURVES
Figure 2.10: The unit tangent vector T(t), translated to start at α(t)1
The remaining two curve-related coordinate directions are orthogonal to this first one. In R3,
there is a whole plane of orthogonal possibilities, but one of the possibilities turns out to be naturally
distinguished. Identifying it requires some preliminary work.
Proposition 2.10 (Product rule for the dot product) . If α, β: I → Rn are differentiable paths,
then:
(α · β)′ = α′ · β + α · β′.
Proof. Note that α · β is real-valued, so by definition of the derivative in first-year calculus:
(α · β)′(t) = lim
h→0
1
h

α(t + h) · β(t + h) − α(t) · β(t)


.
The expression within the limit can be put into more manageable form by using the trick of “adding
zero,” suitably disguised, in the middle:
1
h

α(t + h) · β(t + h) − α(t) · β(t)


= 1
h

α(t + h) · β(t + h)−α(t) · β(t + h)
+ α(t) · β(t + h) − α(t) · β(t)


= α(t + h) − α(t)
h · β(t + h) + α(t) · β(t + h) − β(t)
h .",Multivariable_Calculus_Shimamoto.pdf
"

= 1
h

α(t + h) · β(t + h)−α(t) · β(t + h)
+ α(t) · β(t + h) − α(t) · β(t)


= α(t + h) − α(t)
h · β(t + h) + α(t) · β(t + h) − β(t)
h .
Now, taking the limit as h goes to 0 gives the desired result: (α·β)′(t) = α′(t)·β(t)+ α(t)·β′(t).
Here are two immediate consequences.
Corollary 2.11. (∥α∥2)′ = 2α · α′.
Proof. By Proposition 1.10 of Chapter 1, ∥α∥2 = α · α, so by the product rule, ( ∥α∥2)′ = (α · α)′ =
α′ · α + α · α′ = 2α · α′.
Corollary 2.12. ∥α∥ is constant if and only if α · α′ = 0.
In words, a vector-valued function of one variable has constant magnitude if and only if it is
orthogonal to its derivative at all times.
Proof. ∥α∥ is constant if and only if ∥α∥2 is constant. This in turn is true if and only if (∥α∥2)′ = 0.
By the preceding corollary, this is equivalent to saying that α · α′ = 0.
Returning to the geometry of curves, the unit tangent vector T(t) satisfies ∥T(t)∥ = 1, which",Multivariable_Calculus_Shimamoto.pdf
"Returning to the geometry of curves, the unit tangent vector T(t) satisfies ∥T(t)∥ = 1, which
is constant. Therefore T and T′ are orthogonal. The unit vector in the direction of T′ will be
orthogonal to T, too. This gives us our second coordinate direction.
1From now on, we won’t always state that a vector has been translated. Any vector that starts at a point other
than the origin should be understood to be a translated copy whether there is an explicit note to that effect or not.",Multivariable_Calculus_Shimamoto.pdf
"2.5. THE CROSS PRODUCT 33
Definition. If T′(t) ̸= 0, then:
N(t) = 1
∥T′(t)∥T′(t)
is called the principal normal vector. See Figure 2.11.
Figure 2.11: The unit tangent and principal normal vectors, T(t) and N(t)
Example 2.13. For the helix parametrized by α(t) = (cos t, sin t, t), find the unit tangent T(t)
and the principal normal N(t).
As calculated in Examples 2.6 and 2.7, α′(t) = (−sin t, cos t, 1) and ∥α′(t)∥ = v(t) =
√
2. Thus:
T(t) = 1
∥α′(t)∥α′(t) = 1√
2(−sin t, cos t, 1).
Continuing with this, T′(t) = 1√
2 (−cos t, −sin t, 0), and ∥T′(t)∥ = 1√
2
p
cos2 t + sin2 t + 02 = 1√
2 .
Hence:
N(t) = 1
∥T′(t)∥T′(t)
= 1
1√
2
· 1√
2(−cos t, −sin t, 0)
= (−cos t, −sin t, 0).
As a check, we verify that the unit tangent and principal normal are orthogonal, as predicted
by the theory:
T(t) · N(t) = 1√
2

(−sin t)(−cos t) + (cost)(−sin t) + 1· 0


= 1√
2

sin t cos t − cos t sin t + 0


= 0.
The remaining step is to find a third and final coordinate direction, that is, a unit vector in",Multivariable_Calculus_Shimamoto.pdf
"

= 1√
2

sin t cos t − cos t sin t + 0


= 0.
The remaining step is to find a third and final coordinate direction, that is, a unit vector in
R3 orthogonal to both T(t) and N(t). There are two choices for such a vector, and, in the next
section, we describe a systematic way to single out one of them, denoted by B(t). Once this is
done, we will have a basis of orthogonal unit vectors {T(t), N(t), B(t)} at each point of the curve
that is naturally adapted to the motion along the curve.
2.5 The cross product
We remain focused on R3. Recall that the standard basis vectors are called i = (1, 0, 0), j = (0, 1, 0),
and k = (0, 0, 1) and that every vector x = (x1, x2, x3) in R3 can be written in terms of them:
x = (x1, x2, x3) = (x1, 0, 0) + (0, x2, 0) + (0, 0, x3) = x1i + x2j + x3k.",Multivariable_Calculus_Shimamoto.pdf
"34 CHAPTER 2. PATHS AND CURVES
Our goal for the moment is this: given vectors v and w in R3, find a vector, written v × w,
that is orthogonal to both v and w. We concentrate initially not so much on what v × w actually
is but rather on requiring it to have a certain critical property. Namely, whatever v × w is, we
insist that it satisfy the following.
Key Property. For all x in R3, x · (v × w) = det


x1 x2 x3
v1 v2 v3
w1 w2 w3

, that is:
x · (v × w) = det

x v w

(P)
where

x v w

denotes the 3 by 3 matrix whose rows are x, v, w.
For instance, i × j must satisfy:
x · (i × j) = det

x i j

= det


x1 x2 x3
1 0 0
0 1 0


= x1 · (0 − 0) − x2 · (0 − 0) + x3 · (1 − 0)
= x3 for all x = (x1, x2, x3) in R3.
But x · k = (x1, x2, x3) · (0, 0, 1) = x3 for all x as well, so k satisfies the key property (P) required
of i × j. We would expect that i × j = k.
Assuming for the moment that v × w can be defined in general to satisfy (P), our main goal
follows immediately.",Multivariable_Calculus_Shimamoto.pdf
"Assuming for the moment that v × w can be defined in general to satisfy (P), our main goal
follows immediately.
Proposition 2.14. v × w is orthogonal to v and w (Figure 2.12).
Proof. We take dot products and use the key property:
v · (v × w) = det

v v w

= 0 (two equal rows ⇒ det = 0).
Thus v × w is orthogonal to v. For the same reason, w · (v × w) = det

w v w

= 0.
Figure 2.12: The cross product v × w is orthogonal to v and w.
It remains to define v ×w in a way that satisfies (P). It turns out that there is no choice about
how to do this. Set v × w = ( c1, c2, c3). The components c1, c2, c3 are determined by (P). For
instance:
c1 = (1, 0, 0) · (c1, c2, c3) = i · (v × w) = det

i v w

= det


1 0 0
v1 v2 v3
w1 w2 w3

 = det
v2 v3
w2 w3

.",Multivariable_Calculus_Shimamoto.pdf
"2.5. THE CROSS PRODUCT 35
Similarly, c2 and c3 are determined by looking at j · (v × w) and k · (v × w), respectively.
Definition. Given vectors v and w in R3, their cross product is defined by:
v × w =

det
v2 v3
w2 w3

, −det
v1 v3
w1 w3

, det
v1 v2
w1 w2

.
One can check that this is the same as:
v × w = det


i j k
v1 v2 v3
w1 w2 w3

. (×)
Working backwards, it is not hard to verify that the formula given by ( ×) does indeed satisfy
(P) (Exercise 5.8) so that we have actually accomplished something.
The odd-looking determinant in ( ×) is somewhat illegitimate in that the entries in the top row
are vectors, not scalars. It is meant to be calculated in the usual way by expansion along the top
row. To understand how it works, perhaps it is best to do some examples.
Example 2.15. First, we reproduce our result for i ×j, putting i = (1, 0, 0) and j = (0, 1, 0) in the
second and third rows of ( ×) and expanding along the first row:
i × j = det


i j k
1 0 0
0 1 0
",Multivariable_Calculus_Shimamoto.pdf
"second and third rows of ( ×) and expanding along the first row:
i × j = det


i j k
1 0 0
0 1 0

 = i · (0 − 0) − j · (0 − 0) + k · (1 − 0) = k,
as expected.
Similarly, j × k = i and k × i = j.
Example 2.16. If v = (1, 2, 3) and w = (4, 5, 6), then:
v × w = det


i j k
1 2 3
4 5 6


= i · (12 − 15) − j · (6 − 12) + k · (5 − 8)
= −3i + 6j − 3k
= (−3, 6, −3).
As a check against possible computational error, one can go back and confirm that the result is
orthogonal to both v and w.
Example 2.17. With the same v and w as in the previous example, let u = (1, −2, 3). Then:
u · (v × w) = (1, −2, 3) · (−3, 6, −3) = −3 − 12 − 9 = −24. (2.1)
Suppose that we use the same three factors but in a permuted order, say w · (v × u). Taking
advantage of the result of (2.1), what can you say about the value of this product without calculating
the individual dot and cross products involved?
We use (P) and properties of the determinant:
w · (v × u) = det

w v u

(property (P))
= −det

u v w
",Multivariable_Calculus_Shimamoto.pdf
"We use (P) and properties of the determinant:
w · (v × u) = det

w v u

(property (P))
= −det

u v w

(row switch)
= −u · (v × w) (property (P))
= −(−24) (by (2.1))
= 24.",Multivariable_Calculus_Shimamoto.pdf
"36 CHAPTER 2. PATHS AND CURVES
The basic properties of the cross product are summarized below.
1. w × v = −v × w
(Justification. Interchanging rows in (×) changes the sign of the determinant.)
2. v × v = 0
(Justification. Equal rows in (×) implies the determinant is zero.)
3. ( The length of the cross product. )
∥v × w∥ = ∥v∥∥w∥sin θ, where θ is the angle between v and w (2.2)
(Justification. This appears as Exercise 5.9. It uses v · w = ∥v∥∥w∥cos θ.)
The length formula (2.2) for v × w has some useful consequences:
(a) v × w = 0 if and only if v = 0, w = 0, or θ = 0 or π. This follows immediately from (2.2).
In other words, v × w = 0 if and only if v or w is a scalar multiple of the other.
(b) Consider the parallelogram determined by v and w. We can find its area by thinking of v as
base and dropping a perpendicular from w to get the height, as in the case of Figure 1.10 in
Chapter 1. See Figure 2.13.
Area of parallelogram = (base)(height)
= ∥v∥(∥w∥sin θ)
= ∥v × w∥.",Multivariable_Calculus_Shimamoto.pdf
"Chapter 1. See Figure 2.13.
Area of parallelogram = (base)(height)
= ∥v∥(∥w∥sin θ)
= ∥v × w∥.
Figure 2.13: The parallelogram determined by v and w
In other words:
∥v × w∥ is the area of the parallelogram determined by v and w.
(c) We can now describe the geometric interpretation of 3 by 3 determinants as volume that was
mentioned in Chapter 1. Let u, v, and w be points in R3 such that 0, u, v, and w are
not coplanar. This determines a paralellepiped, which is the three-dimensional analogue of
a parallelogram in the plane, as shown in Figure 2.14. It consists of all points of the form
au + bv + cw, where a, b, care scalars such that 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, and 0 ≤ c ≤ 1. Then:
det

u v w
 is the volume of the parallelepiped determined by u, v, and w.",Multivariable_Calculus_Shimamoto.pdf
"2.5. THE CROSS PRODUCT 37
Figure 2.14: The parallelepiped determined by u, v, w in R3
This follows from property (P), which we can use to reverse course and take what we’ve
learned about dot and cross products to tell us about determinants. We leave the details for
the exercises (namely, Exercise 5.13), though the relevant ingredients are labeled in Figure
2.14. In the figure, φ denotes the angle between u and v × w.
4. ( The direction of the cross product. )
We have seen that the cross productv×w is orthogonal to v and w, but this does not pin down
its direction completely. There are two possible orthogonal directions, each opposite the other. We
present a criterion that distinguishes the possibilities in terms of v and w.
Definition. Let (v1, v2, . . . ,vn) be a basis of Rn, where the vectors have been listed in a particular
order. We say that (v1, v2, . . . ,vn) is right-handed if det

v1 v2 . . .vn

> 0 and left-handed
if det

v1 v2 . . .vn

< 0. (As usual,

v1 v2 . . .vn
",Multivariable_Calculus_Shimamoto.pdf
"
v1 v2 . . .vn

> 0 and left-handed
if det

v1 v2 . . .vn

< 0. (As usual,

v1 v2 . . .vn

is the n by n matrix whose rows are
v1, v2, . . . ,vn.)
Example 2.18. Determine the orientation of the standard basis ( e1, e2, . . . ,en).
det

e1 e2 . . .en

= det


1 0 . . .0
0 1 . . .0
... ... ... ...
0 0 . . .1

 = 1.
This is positive, so ( e1, e2, . . . ,en) is right-handed. By comparison, ( e2, e1, . . . ,en) is left-handed
(switching two rows flips the sign of the determinant). The orientation of a basis depends on the
order in which the basis vectors are presented.
In R3, given v and w, what is the orientation of ( v, w, v × w)? Note that:
det

v w v × w

= + det

v × w v w

(two row switches)
= (v × w) · (v × w) (property (P) with x = v × w)
= ∥v × w∥2. (2.3)
As shown earlier, v × w ̸= 0 provided that v and w are not scalar multiples of one another. If
this is the case, then equation (2.3) implies that det

v w v × w

is positive. Thus:",Multivariable_Calculus_Shimamoto.pdf
"this is the case, then equation (2.3) implies that det

v w v × w

is positive. Thus:
Proposition 2.19. If v and w are vectors in R3, neither a scalar multiple of the other, then the
triple (v, w, v × w) is right-handed.
In R3, the right-handed orientation gets its name from a rule of thumb known as the “right-hand
rule.” This is based on a convention regarding the standard basis ( i, j, k), namely, if you rotate the",Multivariable_Calculus_Shimamoto.pdf
"38 CHAPTER 2. PATHS AND CURVES
fingers of your right hand from i to j, your thumb points in the direction of k. This is not so
much a fact as it is an agreement: whenever we draw R3, we agree to orient the positive x, y, and
z-axes so that the right-hand rule for ( i, j, k) is satisfied, as in Figure 2.15. As we continue with
further material, figures in R3 become increasingly prominent, so it is probably good to state this
convention explicitly.
More generally, given a basis ( v, w, z) of R3, the plane P spanned by v and w separates R3
into two half-spaces, or “sides.” Once we adopt the right-hand rule for ( i, j, k), it turns out that
(v, w, z) is right-handed in the sense of the definition if and only if, when you rotate the fingers
of your right hand from v to w, your thumb lies on the same side of P as z. In particular, by
Proposition 2.19, your thumb lies on the same side as v × w. In this way, your right thumb gives
the direction of v × w.",Multivariable_Calculus_Shimamoto.pdf
"Proposition 2.19, your thumb lies on the same side as v × w. In this way, your right thumb gives
the direction of v × w.
Figure 2.15: The circle of life: i × j = k, j × k = i, k × i = j
Proving the connection between the sign of a determinant and the right-hand rule would be
too great a digression. One approach uses the continuity of the determinant and the geometry of
rotations to show that, if the rule works for ( i, j, k), then it works in general. In any case, the
right-hand rule is sometimes a convenient way to find cross products geometrically. For instance,
in addition to i × j = k, it allows us to see that j × k = i, not −i, and that k × i = j, as pictured
in Figure 2.15.
2.6 The geometry of space curves: Frenet vectors
We take up the following question: How can one tell when two curves in R3 are congruent, that is,
either curve can be translated, rotated, and/or reflected so that it fits exactly on top of the other?",Multivariable_Calculus_Shimamoto.pdf
"either curve can be translated, rotated, and/or reflected so that it fits exactly on top of the other?
As mentioned earlier, the main idea is to construct a basis of orthogonal unit vectors that moves
along the curve and then to study how the basis vectors change during the motion.
So let C be a curve in R3, parametrized by a differentiable path α: I → R3, where I is an
interval in R. In Section 2.4, we defined two of the vectors in our moving basis:
• the unit tangent vector T(t) = 1
∥α′(t)∥α′(t) = 1
v(t) v(t) and
• the principal normal vector N(t) = 1
∥T′(t)∥T′(t).
These definitions make sense only if α′(t) ̸= 0 and T′(t) ̸= 0, so we assume that this is the case
from now on. Both T(t) and N(t) are constructed to be unit vectors, and we showed before that
T(t) · N(t) = 0. At this point, the third basis vector is easily defined.
Definition. The cross product B(t) = T(t) × N(t) is called the binormal vector.",Multivariable_Calculus_Shimamoto.pdf
"2.6. THE GEOMETRY OF SPACE CURVES: FRENET VECTORS 39
As a cross product, B(t) is orthogonal to T(t) and N(t). Moreover, its length is ∥B(t)∥ =
∥T(t)∥∥N(t)∥sin θ = 1 · 1 · sin π
2 = 1. Thus ( T(t), N(t), B(t)) is a collection of orthogonal unit
vectors. The vectors are called the Frenet vectors of α and are illustrated in Figure 2.16.
Figure 2.16: The Frenet vectors T(t), N(t), B(t)
Example 2.20 (The Frenet vectors of a helix) . We shall use the helix with parametrization
α(t) = (a cos t, asin t, bt)
as a running example to illustrate new concepts as they arise. Here, a and b are constant, and
a >0. Recall that the helix is called right-handed if b >0 and left-handed if b <0. If b = 0, the
curve collapses to a circle of radius a in the xy-plane.
To find the Frenet vectors T(t), N(t), and B(t) of the helix, we simply and carefully follow the
definitions.
Unit tangent: α′(t) = (−a sin t, acos t, b), so ∥α′(t)∥ =
p
a2 sin2 t + a2 cos2 t + b2 =
√
a2 + b2. Thus:
T(t) = 1
∥α′(t)∥α′(t) = 1√",Multivariable_Calculus_Shimamoto.pdf
"definitions.
Unit tangent: α′(t) = (−a sin t, acos t, b), so ∥α′(t)∥ =
p
a2 sin2 t + a2 cos2 t + b2 =
√
a2 + b2. Thus:
T(t) = 1
∥α′(t)∥α′(t) = 1√
a2 + b2 (−a sin t, acos t, b).
Principal normal: From the last calculation, T′(t) = 1√
a2+b2 (−a cos t, −a sin t, 0), so ∥T′(t)∥ =
1√
a2+b2
p
a2 cos2 t + a2 sin2 t + 02 = a√
a2+b2 . Hence:
N(t) = 1
∥T′(t)∥T′(t) = 1
a√
a2+b2
· 1√
a2 + b2 (−a cos t, −a sin t, 0)
= 1
a(−a cos t, −a sin t, 0)
= (−cos t, −sin t, 0).
Binormal: Lastly:
B(t) = T(t) × N(t) = det


i j k
− a sin t√
a2+b2
a cos t√
a2+b2
b√
a2+b2
−cos t −sin t 0


= i ·
 b sin t√
a2 + b2


− j ·
 b cos t√
a2 + b2


+ k ·
 a sin2 t√
a2 + b2 + a cos2 t√
a2 + b2


= 1√
a2 + b2

b sin t, −b cos t, a).",Multivariable_Calculus_Shimamoto.pdf
"40 CHAPTER 2. PATHS AND CURVES
2.7 Curvature and torsion
We use the Frenet vectors (T, N, B) to make two geometric measurements:
• curvature, the rate of turning, and
• torsion, the rate of wobbling.
First, curvature measures how fast the tangent direction is changing, which is represented by
∥T′(t)∥, the magnitude of the rate of change of the unit tangent vector. A given curve can be traced
out in many different ways, however, so, if we want curvature to reflect purely on the geometry of
the curve, we must take into account the effect of the parametrization. This is an important point,
and it is an issue that we address more carefully later when we discuss the effect of parametrization
on vector integrals over curves. (See Exercise 1.13 in Chapter 9.) For the time being, let’s agree
informally that we can normalize for the effect of the parametrization by dividing by the speed.
For instance, if you traverse the same curve a second time, twice as fast before, the unit tangent",Multivariable_Calculus_Shimamoto.pdf
"For instance, if you traverse the same curve a second time, twice as fast before, the unit tangent
will change twice as fast, too. Dividing ∥T′(t)∥ by the speed maintains a constant ratio.
Definition. The curvature is defined by:
κ(t) = ∥T′(t)∥
v(t) ,
where v(t) is the speed. 2 Note that, by definition, curvature is a nonnegative quantity.
Example 2.21 (The curvature of a helix) . For the helix α(t) = (a cos t, asin t, bt), the ingredients
that go into the definition of curvature were obtained as part of the work of Example 2.20, namely,
∥T′(t)∥ = a√
a2+b2 and v(t) = ∥α′(t)∥ =
√
a2 + b2. Thus:
Curvature of a helix: κ(t) =
a√
a2+b2
√
a2 + b2 = a
a2 + b2 .
In particular, for a circle of radius a, in which case b = 0, the curvature is κ = a
a2+02 = 1
a.
Next, to measure wobbling, we look at the binormalB, which is orthogonal to the plane spanned
by T and N. This plane is the “plane of motion” in the sense that it contains the velocity and",Multivariable_Calculus_Shimamoto.pdf
"by T and N. This plane is the “plane of motion” in the sense that it contains the velocity and
acceleration vectors (see Exercise 9.4), so B′ can be thought of as representing the rate of wobble
of that plane.
We prove in Lemma 2.24 below that B′(t) is always a scalar multiple of N(t), that is:
B′(t) = c(t)N(t),
where c(t) is a scalar that depends on t. For the moment, let’s accept this as true. Then c(t)
represents the rate of wobble. We again normalize by dividing by the speed. Moreover, the
convention is to throw in a minus sign.
Definition. Given that B′(t) = c(t)N(t) as above, then the torsion is defined by:
τ(t) = −c(t)
v(t),
where v(t) is the speed.
2What Exercise 1.13 in Chapter 9 shows is that the curvature is essentially a function of the points on the curve
C in the sense that, if x ∈ C and if there is only one value of the parameter t such that α(t) = x, then the quantity
∥T′(t)∥",Multivariable_Calculus_Shimamoto.pdf
"C in the sense that, if x ∈ C and if there is only one value of the parameter t such that α(t) = x, then the quantity
∥T′(t)∥
v(t) is the same for all such parametrizations of C. Thus we could denote the curvature by κ(x) and calculate it
without worrying about which parametrization we use.",Multivariable_Calculus_Shimamoto.pdf
"2.7. CURVATURE AND TORSION 41
Example 2.22 (The torsion of a helix) . To find the torsion of the helix α(t) = (a cos t, asin t, bt),
we again piggyback on the calculations of Example 2.20. For instance, using the result that B(t) =
1√
a2+b2 (b sin t, −b cos t, a), we obtain B′(t) = 1√
a2+b2 (b cos t, bsin t, 0). To find c(t), we want to
manipulate this to be a scalar multiple of N(t) = (−cos t, −sin t, 0):
B′(t) = 1√
a2 + b2 (b cos t, bsin t, 0) = − b√
a2 + b2 (−cos t, −sin t, 0) = − b√
a2 + b2 N(t).
Therefore c(t) = − b√
a2+b2 , so:
Torsion of a helix: τ(t) = −c(t)
v(t) = −
− b√
a2+b2
√
a2 + b2 = b
a2 + b2 .
Note that the torsion is positive if the helix is right-handed ( b > 0) and negative if left-handed
(b <0). In the case of a circle ( b = 0), the torsion is zero. This reflects the fact that planar curves
don’t wobble in space at all.
To complete the discussion of torsion, it remains to justify the claim that B′ is a scalar multiple",Multivariable_Calculus_Shimamoto.pdf
"don’t wobble in space at all.
To complete the discussion of torsion, it remains to justify the claim that B′ is a scalar multiple
of N. The argument uses a product rule for the cross product.
Proposition 2.23 (Product rules). Let I be an interval of real numbers, and let α, β: I → Rn be
differentiable paths and f : I → R a differentiable real-valued function. Then:
1. (Dot product) ( α · β)′ = α′ · β + α · β′.
2. (Cross product) For paths α, βin R3, (α × β)′ = α′ × β + α × β′.
3. (Scalar multiplication) (fα)′ = f′α + fα′.
Proof. The first of these was proven earlier (Proposition 2.10), and the others are the same, swap-
ping in the appropriate type of product. We leave them as exercises (Exercise 7.3).
Lemma 2.24. The derivative B′ of the binormal is always a scalar multiple of the principal normal
N.
Proof. We show that B′ is orthogonal (a) to B and (b) to T. The only vectors orthogonal to both
are precisely the scalar multiples of N, so the lemma follows.",Multivariable_Calculus_Shimamoto.pdf
"are precisely the scalar multiples of N, so the lemma follows.
(a) B is a unit vector, so ∥B∥ is constant. Hence B and its derivative B′ are always orthogonal
(Corollary 2.12).
(b) By definition, B = T × N, so using the product rule:
B′ = T′ × N + T × N′
= (∥T′∥N) × N + T × N′ (definition of N)
= ∥T′∥0 + T × N′ (v × v = 0 always)
= T × N′.
As a cross product, B′ is orthogonal to T (and to N′ for that matter).",Multivariable_Calculus_Shimamoto.pdf
"42 CHAPTER 2. PATHS AND CURVES
2.8 The Frenet-Serret formulas
We saw in Examples 2.21 and 2.22 that, for the helix α(t) = (a cos t, asin t, bt), a >0, the curvature
and torsion are given by:
κ = a
a2 + b2 and τ = b
a2 + b2 ,
respectively. For example, if α(t) = (2 cost, 2 sint, t), i.e., a = 2 and b = 1, then κ = 2
5 and τ = 1
5 .
In particular, helices have constant curvature and torsion.
We are about to see that the geometry of a curve in R3 is determined by its curvature and
torsion. For instance, it turns out that any curve with constant curvature and torsion is necessarily
congruent to a helix. The proof is based on the following formulas, which describe precisely how
the Frenet vectors change within the coordinate systems that they determine.
Proposition 2.25 (Frenet-Serret formulas). Let α: I → R3 be a differentiable path in R3 with
speed v, Frenet vectors (T, N, B), curvature κ, and torsion τ. Assume that v ̸= 0 and T′ ̸= 0 so",Multivariable_Calculus_Shimamoto.pdf
"speed v, Frenet vectors (T, N, B), curvature κ, and torsion τ. Assume that v ̸= 0 and T′ ̸= 0 so
that the Frenet vectors are defined for all t. Then:
(1) T′ = κvN,
(2) N′ = −κvT + τv B,
(3) B′ = −τv N.
Proof. The first and third of these are basically the definitions of curvature and torsion, so we prove
them first. To prove (1), we know that, by definition, N = 1
∥T′∥T′ and κ = ∥T′∥
v . Thus:
T′ = ∥T′∥N = κvN.
For (3), we know that B′ = cN and τ = −c
v for some scalar-valued function c. Hence c = −τv ,
and B′ = −τv N.
Lastly, for (2), the right-hand rule gives N = B × T (see Figure 2.16). Therefore, according to
the product rule and the two Frenet-Serret formulas that were just proven:
N′ = B′ × T + B × T′
= −τv N × T + B × (κvN).
But, again by the right-hand rule, N × T = −B and B × N = −T. Substituting into the last
expression gives N′ = −κvT + τv B, completing the proof.
2.9 The classification of space curves",Multivariable_Calculus_Shimamoto.pdf
"expression gives N′ = −κvT + τv B, completing the proof.
2.9 The classification of space curves
Our discussion of curves in R3 culminates with a result on the connection between curvature and
torsion—that is, turning and wobbling—and the intrinsic geometry of a curve. On the one hand,
moving a curve rigidly in space does not change geometric features of the curve such as its length.
It is plausible that this extends to not changing the curvature and torsion as well, and we begin by
elaborating on this point.
For example, suppose that a path α(t) is translated by a constant amount c to obtain a new
path β(t) = α(t) + c. Then β′(t) = α′(t), so α and β have the same velocity and therefore the
same speed. It follows from the definition that they also have the same unit tangent vector T(t).
Following along with further definitions, they then have the same principal normal N(t), hence the
same binormal B(t), and finally the same curvature κ(t) and torsion τ(t). In other words, speed,",Multivariable_Calculus_Shimamoto.pdf
"same binormal B(t), and finally the same curvature κ(t) and torsion τ(t). In other words, speed,
curvature, and torsion are preserved under translations.",Multivariable_Calculus_Shimamoto.pdf
"2.9. THE CLASSIFICATION OF SPACE CURVES 43
The same conclusion applies to rotations, though we don’t really have the tools to give a rigorous
proof at this point. So we try an intuitive explanation. Suppose that α is rotated in R3 to obtain
a new path β. The velocities α′(t) and β′(t) are related by the same rotation, and then so are
the respective Frenet vectors (T(t), N(t), B(t)). So the Frenet vectors of the two paths are not the
same, but, since they are rotated versions of one another, the corresponding vectors change at the
same rates. In particular, the speed, curvature, and torsion are the same.
Our main theorem is a converse to these considerations. That is, we show that, if two paths α
and β have the same speed, curvature, and torsion, then they differ by a translation and/or rotation
in the sense that there is a composition of translations and rotations F : R3 → R3 that transforms",Multivariable_Calculus_Shimamoto.pdf
"in the sense that there is a composition of translations and rotations F : R3 → R3 that transforms
α into β, i.e., F(α(t)) = β(t) for all t. Hence measuring the three scalar quantities v, κ, τ suffices to
determine the geometry of a space curve. It is striking and satisfying that such a complete answer
is possible. In the spirit of the theorems in plane geometry about congruent triangles ( SAS, ASA,
SSS , etc.), we call the theorem the “ vκτ theorem.”
Theorem 2.26 (vκτ theorem). Let I be an interval of real numbers, and let α, β: I → R3 be
differentiable paths in R3 such that v and T′ are nonzero so that the Frenet vectors are defined for
all t. Then either path can be translated and rotated so that it fits exactly on top of the other if and
only if they have:
• the same speed v(t),
• the same curvature κ(t), and
• the same torsion τ(t).
In particular, two paths with the same speed, curvature, and torsion are congruent to each other.",Multivariable_Calculus_Shimamoto.pdf
"• the same torsion τ(t).
In particular, two paths with the same speed, curvature, and torsion are congruent to each other.
Proof. That translations and rotations preserve speed, curvature, and torsion was discussed above.
For the converse, assume that α and β have the same speed, curvature, and torsion. We proceed
in three stages to move α on top of β.
Step 1. A translation.
Choose now and for the rest of the proof a point a in the interval I. We first translate α so
that α(a) moves to β(a). In other words, we shift α by the constant vector d = β(a) − α(a) to get
a new path γ(t) = α(t) + d = α(t) + β(a) − α(a). Then:
γ(a) = α(a) + β(a) − α(a) = β(a).
Moreover, as a translate, γ has the same speed, curvature, and torsion as α and hence as β as well.
Step 2. Two rotations.
Let x0 denote the common point γ(a) = β(a). The unit tangents to γ and β at x0 may not
be equal, but we can rotate γ about x0 until they are. Then, rotate γ again using this tangent",Multivariable_Calculus_Shimamoto.pdf
"be equal, but we can rotate γ about x0 until they are. Then, rotate γ again using this tangent
direction as axis of rotation until the principal normals at x0 coincide, too. The binormals are then
automatically the same, since they are the cross products of T and N. Moreover, neither rotation
changes the speed, curvature, or torsion.
Call this final path eα. In this way, we have translated and rotated α into a path eα satisfying
the following three conditions:
• eα(a) = β(a) = x0,
• at the point x0, eα and β have the same Frenet vectors, and
• eα and β have the same speed, curvature, and torsion for all t.",Multivariable_Calculus_Shimamoto.pdf
"44 CHAPTER 2. PATHS AND CURVES
Step 3. A calculation.
We show that eα = β. Since eα has been obtained from α by translation and rotation, the
theorem follows.
The argument uses the Frenet-Serret formulas, which we repeat for convenience:
T′ = κvN, N′ = −κvT + τv B, B′ = −τv N.
We denote the Frenet vectors of eα by ( eT, eN, eB) and those of β by (T, N, B). Using similar
notation for the speed, curvature, and torsion, we have ev = v, eκ = κ, and eτ = τ by construction.
Let θ be the angle between eT(t) and T(t). Then eT(t)·T(t) = ∥eT(t)∥∥T(t)∥cos θ = 1·1·cos θ =
cos θ. Therefore eT(t) · T(t) is less than or equal to 1 always and is equal to 1 if and only if θ = 0,
that is, if and only if eT(t) = T(t). The same comments apply to eN(t) · N(t) and eB(t) · B(t).
Thus, if we define f : I → R by
f(t) = eT(t) · T(t) + eN(t) · N(t) + eB(t) · B(t),
then the previous remarks imply:
f(t)
(
≤ 3 for all t,
= 3 if and only if eT(t) = T(t), eN(t) = N(t), and eB(t) = B(t).",Multivariable_Calculus_Shimamoto.pdf
"then the previous remarks imply:
f(t)
(
≤ 3 for all t,
= 3 if and only if eT(t) = T(t), eN(t) = N(t), and eB(t) = B(t).
For instance, f(a) = 3 by Step 2.
Now, by the product rule, f′ = (eT′ · T + eT · T′) + (eN′ · N + eN · N′) + (eB′ · B + eB · B′). We use
the Frenet-Serret formulas to substitute for every derivative in sight, keeping in mind that ev = v,
eκ = κ, and eτ = τ:
f′ =

κv eN · T + eT · κvN

+

(−κv eT + τv eB) · N + eN · (−κvT + τv B)

+

(−τv eN) · B + eB · (−τv N)

= (κv eN · T + κv eT · N) + (−κv eT · N + τv eB · N − κv eN · T + τv eN · B)
−(τv eN · B + τv eB · N).
Miraculously, the terms cancel in pairs so that f′ = 0 for all t.
Hence f is a constant function. Since f(a) = 3, it follows that f(t) = 3 for all t. As noted
above, this in turn implies that eT(t) = T(t) for all t (and eN(t) = N(t), eB(t) = B(t) as well, but
we don’t need them).
By definition of the unit tangent, this gives 1
v(t) eα′(t) = 1
v(t) β′(t), so eα′(t) = β′(t) for all t.",Multivariable_Calculus_Shimamoto.pdf
"we don’t need them).
By definition of the unit tangent, this gives 1
v(t) eα′(t) = 1
v(t) β′(t), so eα′(t) = β′(t) for all t.
Consequently eα and β differ by a constant vector: eα(t) = β(t) + c. Since eα(a) = x0 = β(a), the
constant is c = 0. Thus eα(t) = β(t) for all t. We have succeeded in translating and rotating α to
reach a path eα that lies on top of β.
Note that the theorem does not say what happens when a path is reflected in a plane, that is,
how the speed, curvature, and torsion of a path are related to those of its mirror image. For a clue
as to what happens in this case, see Exercise 9.14.",Multivariable_Calculus_Shimamoto.pdf
"2.10. EXERCISES FOR CHAPTER 2 45
2.10 Exercises for Chapter 2
Section 1 Parametrizations
In Exercises 1.1–1.8, sketch the curve parametrized by the given path. Indicate on the curve
the direction in which it is being traversed.
1.1. α: [0, 2] → R2, α(t) = (t, t2)
1.2. α: [0, 6π] → R2, α(t) = (t cos t, tsin t)
1.3. α: R → R2, α(t) = (et, e−t) (Hint: How are the x and y-coordinates related?)
1.4. α: R → R3, α(t) = (cos t, sin t, t)
1.5. α: R → R3, α(t) = (cos t, sin t, −t)
1.6. α: R → R3, α(t) = (sin t, cos t, t)
1.7. α: R → R3, α(t) = (sin t, cos t, −t)
1.8. α: [0, 2π] → R3, α(t) = (cos t, sin t, cos 2t)
In Exercises 1.9–1.12, find a parametrization of the given curve.
1.9. The portion of the parabola y = x2 in R2 with −1 ≤ x ≤ 1
1.10. The portion of the curve xy = 1 in R2 with 1 ≤ x ≤ 2
1.11. The circle x2 + y2 = a2 in R2 traversed once in the clockwise direction
1.12. A right-handed helix in R3 that:
• lies on the circular cylinder of radius 2 whose axis is the z-axis,",Multivariable_Calculus_Shimamoto.pdf
"1.12. A right-handed helix in R3 that:
• lies on the circular cylinder of radius 2 whose axis is the z-axis,
• completes one complete twist as it rises from z = 0 to z = 1.
1.13. Find a parametrization of the line in R3 that passes through the point a = (1, 2, 3) and is
parallel to v = (4, 5, 6).
1.14. Let ℓ be the line in R3 parametrized by α(t) = (1 + 2t, −3t, 4 + 5t). Find a point that lies on
ℓ and a vector that is parallel to ℓ.
1.15. Find a parametrization of the line in R3 that passes through the points a = (1 , 0, 0) and
b = (0, 0, 1).
1.16. Find a parametrization of the line in R3 that passes through the points a = (1, −2, 3) and
b = (−4, 5, −6).
1.17. Let ℓ be the line in Rn parametrized by α(t) = a + tv, and let p be a point in Rn. Suppose
that q is the foot of the perpendicular dropped from p to ℓ (Figure 2.17).
(a) Since q lies on ℓ, there is a value of t such that q = a+ tv. Find a formula for t in terms",Multivariable_Calculus_Shimamoto.pdf
"(a) Since q lies on ℓ, there is a value of t such that q = a+ tv. Find a formula for t in terms
of a, v, and p. ( Hint: ℓ lies in a direction perpendicular to p − q.)
(b) Find a formula for the point q.",Multivariable_Calculus_Shimamoto.pdf
"46 CHAPTER 2. PATHS AND CURVES
Figure 2.17: The foot of the perpendicular from p to ℓ
1.18. Let ℓ be the line in R3 parametrized by α(t) = (1 + t, 1 + 2t, 1 + 3t), and let p = (0, 0, 4).
Find the foot of the perpendicular dropped from p to ℓ.
1.19. Let ℓ and m be lines in R3 parametrized by:
α(t) = a + tv and β(t) = b + tw,
respectively, where a = (a1, a2, a3), v = (v1, v2, v3), b = (b1, b2, b3), and w = (w1, w2, w3).
(a) Explain why the problem of determining whether ℓ and m intersect in R3 amounts to
solving a system of three equations in two unknowns t1 and t2.
(b) Determine whether the lines parametrized by α(t) = (−1, 0, 1) + t(1, −3, 2) and β(t) =
(1, 5, 4) + t(−1, 1, 6) intersect by writing down the corresponding system of equations,
as described in part (a), and solving the system.
(c) Repeat for the lines parametrized by α(t) = (2 , 1, 0) + t(0, 0, 1) and β(t) = (0 , 1, 3) +
t(1, 0, 0).
Section 2 Velocity, acceleration, speed, arclength
2.1. Let α(t) = (t3, 3t2, 6t).",Multivariable_Calculus_Shimamoto.pdf
"t(1, 0, 0).
Section 2 Velocity, acceleration, speed, arclength
2.1. Let α(t) = (t3, 3t2, 6t).
(a) Find the velocity, speed, and acceleration.
(b) Find the arclength from t = 0 to t = 4.
2.2. Let α(t) = (1
2 t2, 2t, 4
3 t3/2).
(a) Find the velocity, speed, and acceleration.
(b) Find the arclength from t = 2 to t = 4.
2.3. Let α(t) = (cos 2t, sin 2t, 4t3/2).
(a) Find the velocity, speed, and acceleration.
(b) Find the arclength from t = 1 to t = 2.
2.4. Let a and b be real numbers, a >0, and let α(t) = (a cos t, asin t, bt).
(a) Find the velocity, speed, and acceleration.
(b) Find the arclength from t = 0 to t = 2π.
2.5. Find a parametrization of the line through the points a = (1 , 1, 1) and b = (2 , 3, 4) that
traverses the line with constant speed 1.",Multivariable_Calculus_Shimamoto.pdf
"2.10. EXERCISES FOR CHAPTER 2 47
Section 3 Integrals with respect to arclength
In Exercises 3.1–3.2, evaluate the integral with respect to arclength for the curve C with the
given parametrization α.
3.1.
R
C xyz ds, where α(t) = (t3, 3t2, 6t), 0 ≤ t ≤ 1
3.2.
R
C z ds, where α(t) = (cos t + t sin t, sin t − t cos t, t), 0 ≤ t ≤ 4
3.3. Let C be the line segment in R2 from (1, 0) to (0 , 1). With as little calculation as possible,
find the values of the following integrals.
(a)
R
C 5 ds
(b)
R
C (x + y) ds
3.4. Let C be the line segment in R2 parametrized by α(t) = (t, t), 0 ≤ t ≤ 1.
(a) Evaluate
R
C (x + y) ds.
(b) Find a triangle in R3 whose base is the segment C (in the xy-plane) and whose area is
represented by the integral in part (a). ( Hint: Area under a curve.)
Section 4 The geometry of curves: tangent and normal vectors
Note: Further exercises on this material appear in Section 9.
In Exercises 4.1–4.2, find the unit tangent T(t) and the principal normal N(t) for the given",Multivariable_Calculus_Shimamoto.pdf
"In Exercises 4.1–4.2, find the unit tangent T(t) and the principal normal N(t) for the given
path α. Then, verify that T(t) and N(t) are orthogonal.
4.1. α(t) = (sin 4t, cos 4t, 3t)
4.2. α(t) = (t, t2, 0)
4.3. Let α be a differentiable path in Rn.
(a) If the speed is constant, show that the velocity and acceleration vectors are always
orthogonal. ( Hint: Consider ∥v(t)∥2.)
(b) Conversely, if the velocity and acceleration are always orthogonal, show that the speed
is constant.
4.4. Let C be a curve in Rn traced out by a differentiable path α: (a, b) → Rn defined on an open
interval (a, b). Assume that there is a point α(t0) on C that is closer to the origin than any
other point of C. Prove that the tangent vector α′(t0) is orthogonal to α(t0). Draw a sketch
that illustrates the conclusion.
Section 5 The cross product
In Exercises 5.1–5.4, (a) find the cross product v × w and (b) verify that it is orthogonal to v
and w.
5.1. v = (1, 1, 1), w = (1, 3, 5)
5.2. v = (1, 2, 3), w = (2, 4, 6)",Multivariable_Calculus_Shimamoto.pdf
"and w.
5.1. v = (1, 1, 1), w = (1, 3, 5)
5.2. v = (1, 2, 3), w = (2, 4, 6)
5.3. v = (1, −2, 4) , w = (−2, 1, 3)",Multivariable_Calculus_Shimamoto.pdf
"48 CHAPTER 2. PATHS AND CURVES
5.4. v = (2, 0, −1), w = (0, −2, 1)
5.5. Find a nonzero vector inR3 that points in a direction perpendicular to the plane that contains
the origin and the points p = (1, 1, 1) and q = (2, 1, −3).
5.6. Find a nonzero vector inR3 that points in a direction perpendicular to the plane that contains
the points p = (1, 0, 0), q = (0, 1, 0), and r = (0, 0, 1).
5.7. Let v = (−1, 1, −2), w = (4, −1, −1), and u = (−3, 2, 1).
(a) Find v × w.
(b) What is the area of the parallelogram determined by v and w?
(c) Find u · (v × w). Then, use your answer to find v · (u × w) and (u × v) · w.
(d) What is the volume of the parallelepiped determined by u, v, and w?
5.8. Verify that the cross product v ×w as defined in equation (×) satisfies the key property that
x · (v × w) = det

x v w

for all x in R3.
5.9. Let v = (v1, v2, v3) and w = (w1, w2, w3) be vectors in R3.
(a) Use the definitions of the dot and cross products in terms of coordinates to prove that:",Multivariable_Calculus_Shimamoto.pdf
"(a) Use the definitions of the dot and cross products in terms of coordinates to prove that:
∥v × w∥2 = ∥v∥2∥w∥2 − (v · w)2.
(b) Use part (a) to give a proof that ∥v × w∥ = ∥v∥∥w∥sin θ, where θ is the angle between
v and w.
5.10. Do there exist nonzero vectors v and w in R3 such that v · w = 0 and v × w = 0? Explain.
5.11. In R3, let ℓ be the line parametrized by α(t) = a + tv, and let p be a point.
Figure 2.18: The distance from a point to a line
(a) Show that the distance from p to ℓ is given by the formula:
d = ∥(p − a) × v∥
∥v∥ .
See Figure 2.18.
(b) Find the distance from the point p = (1 , 1, −1) to the line parametrized by α(t) =
(2, 1, 2) + t(1, −2, 2).",Multivariable_Calculus_Shimamoto.pdf
"2.10. EXERCISES FOR CHAPTER 2 49
5.12. True or false: u × (v × w) = (u × v) × w for all u, v, w in R3. Either give a proof or find a
counterexample.
5.13. Let u, v, and w be points in R3 such that 0, u, v, and w are not coplanar. Prove that the
volume of the parallelepiped determined by u, v, and w is equal to
det

u v w
. ( Hint:
See Figure 2.14, where φ is the angle between u and v × w. Keep in mind that u and v × w
may lie on opposite sides of the plane spanned by v and w.)
5.14. Let (v, w, z) be a basis of R3, and let θ be the angle between v × w and z, where 0 ≤ θ ≤ π,
as usual. Use the definition of handedness to show that θ <π
2 if (v, w, z) is right-handed and
θ >π
2 if left-handed. Draw pictures to illustrate both situations. ( Hint: Consider z·(v×w).)
5.15. Strictly speaking, the cross product is defined only in R3, but there is an analogous operation
in R4 if three factors are allowed. That is, given vectors u, v, w in R4, the product u ×v ×w",Multivariable_Calculus_Shimamoto.pdf
"in R4 if three factors are allowed. That is, given vectors u, v, w in R4, the product u ×v ×w
is a vector in R4 that satisfies the property:
x · (u × v × w) = det

x u v w

for all x in R4, where the determinant is of the 4 by 4 matrix whose rows are x, u, v, w.
(a) Assuming for the moment that such a productu×v×w exists, show that it is orthogonal
to each of u, v, and w.
(b) If u × v × w ̸= 0, is the quadruple ( u, v, w, u × v × w) right-handed or left-handed?
(c) Describe how you might find a formula for u×v×w, and use it to compute the product
if u = (1, −1, 1, −1), v = (1, 1, −1, −1), and w = (−1, 1, 1, −1).
5.16. In general, the “cross product” in Rn is a function of n − 1 factors. In R2, this means that
there is only one factor.
(a) Given a vector v = (a, b) in R2, find a reasonable formula for “ v×”, and explain how
you came up with it.
(b) Draw a sketch that illustrates v and your candidate for v×.
Section 6 The geometry of space curves: Frenet vectors",Multivariable_Calculus_Shimamoto.pdf
"you came up with it.
(b) Draw a sketch that illustrates v and your candidate for v×.
Section 6 The geometry of space curves: Frenet vectors
Note: Further exercises on this material appear in Section 9.
In Exercises 6.1–6.2, find the binormal B(t) for the given path α. These problems are continu-
ations of Exercises 4.1–4.2.
6.1. α(t) = (sin 4t, cos 4t, 3t)
6.2. α(t) = (t, t2, 0)
6.3. Consider the line in R3 parametrized by α(t) = (1 + t, 1 + 2t, 1 + 3t). What happens when
you try to find its Frenet vectors ( T(t), N(t), B(t))?
Section 7 Curvature and torsion
Note: Further exercises on this material appear in Section 9.
7.1. Find a parametrization of a curve that has constant curvature 1
2 and constant torsion 0.
7.2. Find a parametrization of a curve that has constant curvature 1
2 and constant torsion 1
4 .
7.3. Prove the product rule for cross products: If α and β are differentiable paths in R3, then:
(α × β)′ = α′ × β + α × β′.",Multivariable_Calculus_Shimamoto.pdf
"50 CHAPTER 2. PATHS AND CURVES
Section 9 The classification of space curves
9.1. Consider the curve parametrized by:
α(t) = (cos t, −sin t, t).
(a) Find the speed v(t).
(b) Find the unit tangent T(t).
(c) Find the principal normal N(t).
(d) Find the binormal B(t).
(e) Find the curvature κ(t).
(f) Find the torsion τ(t).
(g) The curve traced out by α is congruent to a curve with a parametrization of the form
β(t) = ( a cos t, asin t, bt), where a and b are constant and a >0. Find the values of a
and b. Then, sketch the curves traced out by α and β, and describe a motion of R3 that
takes the path α and puts it right on top of β.
9.2. Let α be the path defined by:
α(t) = (6 cost, 8 cost, 10 sint).
(a) Find the speed v(t).
(b) Find the unit tangent T(t).
(c) Find the principal normal N(t).
(d) Find the binormal B(t).
(e) Find the curvature κ(t).
(f) Find the torsion τ(t).
(g) Based on your answers to parts (e) and (f), what can you say about the type of curve
traced out by α?",Multivariable_Calculus_Shimamoto.pdf
"(f) Find the torsion τ(t).
(g) Based on your answers to parts (e) and (f), what can you say about the type of curve
traced out by α?
9.3. Let α be the path defined by:
α(t) = (t +
√
3 sint,
√
3 t − sin t, 2 cost).
The following calculations are kind of messy, but the conclusion may be unexpected.
(a) Find the speed v(t).
(b) Find the unit tangent T(t).
(c) Find the principal normal N(t).
(d) Find the binormal B(t).
(e) Find the curvature κ(t).
(f) Find the torsion τ(t).
(g) Based on your answers to parts (e) and (f), what can you say about the type of curve
traced out by α?",Multivariable_Calculus_Shimamoto.pdf
"2.10. EXERCISES FOR CHAPTER 2 51
In Exercises 9.4–9.8, α is a path in R3 with velocity v(t), speed v(t), acceleration a(t), and
Frenet vectors T(t), N(t), and B(t). You may assume that v(t) ̸= 0 and T′(t) ̸= 0 for all t so that
the Frenet vectors are defined.
9.4. Prove the following statements.
(a) v = vT
(b) a = v′T + κv2N (Hence v′ is the tangential component of acceleration and κv2 the
normal component.)
9.5. Prove that v · a = vv′.
9.6. Show that the curvature is given by κ = ∥v×a∥
v3 .
9.7. Let C be the curve in R3 parametrized by α(t) = (t, t2, t3). Use the formula from the previous
exercise to find the curvature at the point (1 , 1, 1).
9.8. According to Exercise 9.4, the velocity and acceleration vectors lie in the plane spanned by
T and N. Thus, like the binormal vector B, the cross product v × a is orthogonal to that
plane. It follows that the unit vector in the direction of v ×a must be ±B. Show that in fact
it is B. In other words, show that B = v×a
∥v×a∥.",Multivariable_Calculus_Shimamoto.pdf
"plane. It follows that the unit vector in the direction of v ×a must be ±B. Show that in fact
it is B. In other words, show that B = v×a
∥v×a∥.
9.9. Let α(t) = (2 cost, 2 sint, 2 cost).
(a) Find the curvature κ(t). (You may use the result of Exercise 9.6 if you wish.)
(b) For this curve, what does v × a tell you about the torsion τ(t)?
9.10. Let α be a path in R3 such that v ̸= 0 and T′ ̸= 0, and let:
w = τv T + κvB.
Prove the Darboux formulas:
w × T = T′,
w × N = N′,
w × B = B′.
9.11. Suppose that the Darboux vector w = τv T + κvB from the previous exercise is a constant
vector.
(a) Prove that τv and κv are both constant. ( Hint: Calculate w′, and use the Frenet-Serret
formulas. Note that T and B need not be constant even if w is.)
(b) If α has constant speed, prove that α is congruent to a helix.
9.12. Let α be a path in R3 that lies on the sphere of radius a centered at the origin, that is:
∥α(t)∥ = a for all t.",Multivariable_Calculus_Shimamoto.pdf
"9.12. Let α be a path in R3 that lies on the sphere of radius a centered at the origin, that is:
∥α(t)∥ = a for all t.
Let T(t) be the unit tangent,N(t) the principal normal,v(t) the speed, andκ(t) the curvature.
Assume that v(t) ̸= 0 and T′(t) ̸= 0 for all t so that the Frenet vectors are defined.
(a) Show that α(t) · T(t) = 0 for all t.
(b) Show that κ(t)α(t) · N(t) = −1 for all t. ( Hint: Differentiate part (a).)",Multivariable_Calculus_Shimamoto.pdf
"52 CHAPTER 2. PATHS AND CURVES
(c) Show that κ(t) ≥ 1
a for all t.
9.13. Let α be a path in R3 that has:
• constant speed v = 1,
• positive torsion τ(t), and
• binormal vector B(t) =

−
√
2
2 ,
√
2
2 sin 2t,
√
2
2 cos 2t


.
(a) Find, in whatever order you wish, the unit tangent T(t), the principal normal N(t), the
curvature κ(t), and the torsion τ(t).
(b) In addition, if α(0) = (0, 0, 0), find a formula for α(t) as a function of t. ( Hint: Use your
formula for T(t).)
9.14. In this problem, we determine the effect of a reflection, specifically, the reflection in the xy-
plane, on the Frenet vectors, the speed, the curvature, and the torsion of a path. To fix
some notation, if p = (x, y, z) is a point in R3, let p∗ = (x, y,−z) denote its reflection in the
xy-plane.
Now, let α(t) = ( x(t), y(t), z(t)) be a path in R3 whose Frenet vectors are defined for all t,
and let β(t) be its reflection:
β(t) = α(t)∗ = (x(t), y(t), −z(t)).
(a) Show that α and β have the same speed v.",Multivariable_Calculus_Shimamoto.pdf
"and let β(t) be its reflection:
β(t) = α(t)∗ = (x(t), y(t), −z(t)).
(a) Show that α and β have the same speed v.
(b) Show the unit tangent vectors of α and β are related by Tβ = T∗
α, i.e., they are also
reflections of one another.
(c) Similarly, show that the principal normals are reflections: Nβ = N∗
α. ( Hint: To organize
the calculation, start by writing Tα = (t1, t2, t3). Then, what does Tβ look like?)
(d) Show that α and β have the same curvature: κα = κβ.
(e) Show that the binormals are related in the following way: if Bα = ( b1, b2, b3), then
Bβ = (−b1, −b2, b3) = −B∗
α.
(f) Describe how the torsions τα and τβ of the paths are related. ( Hint: Use parts (c) and
(e).)",Multivariable_Calculus_Shimamoto.pdf
"Part III
Real-valued functions
53",Multivariable_Calculus_Shimamoto.pdf
"Chapter 3
Real-valued functions: preliminaries
Thus far, we have studied functions α for which the input is a real number t and the output is a
vector α(t) = ( x1(t), x2(t), . . . , xn(t)). The techniques from calculus that we used were basically
familiar from first-year calculus. Now, we reverse the roles and consider functions where the input is
a vectorx = (x1, x2, . . . , xn) and the output is a real numbery = f(x) = f(x1, x2, . . . , xn). Thinking
of the coordinates x1, x2, . . . , xn as variables, these are real-valued functions of n variables. More
formally, they are functions f : A → R, where the domain A is a subset of Rn. When more than
one variable is involved, we shall need methods that go beyond first-year calculus.
Example 3.1. We have seen that a helix can be parametrized by α(t) = (a cos t, asin t, bt). The
values of a and b give the radius of the cylinder around which the helix winds and the rate at which",Multivariable_Calculus_Shimamoto.pdf
"values of a and b give the radius of the cylinder around which the helix winds and the rate at which
it rises, respectively. If a and b vary, the size and shape of the helix change.
The curvature of the helix is given by κ = a
a2+b2 . We can think of this as a real-valued function
that describes how the curvature depends on the geometric parameters a and b. In a way, it’s like
a function on the set of helices, but, strictly speaking, it is a function κ: A → R whose domain
as far as helices go is A = {(a, b) ∈ R2 : a >0}, a subset of R2. Likewise, the torsion of a helix,
τ = b
a2+b2 , is a real-valued function of the same two variables with the same domain.
3.1 Graphs and level sets
In order to analyze a function, it seems like a good idea to try to visualize its behavior somehow.
It is easy enough to come up with ways of doing this in principle, but it takes a lot of practice to
do it effectively when there are two or more variables.",Multivariable_Calculus_Shimamoto.pdf
"do it effectively when there are two or more variables.
For instance, to graph a real-valued function of n variables, we replace y = f(x) in the one-
variable case with xn+1 = f(x1, x2, . . . , xn).
Definition. Let A be a subset of Rn, and let f : A → R be a function. The graph of f is the set:
{(x1, x2, . . . , xn+1) ∈ Rn+1 : (x1, x2, . . . , xn) ∈ A and xn+1 = f(x1, x2, . . . , xn)}.
Note that the graph is a subset of Rn+1, so we can hope to draw it only if n equals 1 or 2, i.e.,
when there are one or two variables. In the case of two variables, which is our focus, the domain
A is a subset of R2, and the graph is the surface in R3 that satisfies the equation z = f(x, y) and
whose projection on the xy-plane is A.
Example 3.2. Let f : R2 → R be given by f(x, y) =
p
x2 + y2. To sketch the graph z =
p
x2 + y2,
we slice with various well-chosen planes and then try to reconstruct an image of the surface from
the cross-sections.
55",Multivariable_Calculus_Shimamoto.pdf
"56 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
For instance, consider the cross-sections with the three coordinate planes. Theyz-plane is where
x = 0, so the equation of the intersection of the graph with this plane is z =
p
02 + y2 = |y| and
x = 0, a V -shaped curve. Similarly, the cross-section with the xz-plane is z = |x|, y= 0, another
V -shaped curve at right angles to the first. These cross-sections are shown in Figure 3.1. Lastly,
Figure 3.1: Two cross-sections with coordinate planes: z = |y|, x= 0 (left) and z = |x|, y= 0
(right)
the cross-section with the xy-plane, where z = 0, is 0 =
p
x2 + y2, z= 0. This consists of the single
point (0, 0, 0).
In general, cross-sections with horizontal planes z = c are also useful. Here, they are described
by c =
p
x2 + y2, or x2 + y2 = c2, where c ≥ 0. This is a circle of radius c in the plane z = c. As c
increases, the circles get bigger. Some of them are shown in Figure 3.2.
Figure 3.2: Some horizontal cross-sections",Multivariable_Calculus_Shimamoto.pdf
"increases, the circles get bigger. Some of them are shown in Figure 3.2.
Figure 3.2: Some horizontal cross-sections
By putting all this information together, we can assemble a pretty good picture of the whole
graph z =
p
x2 + y2. It is the circular cone with vertex at the origin pictured in Figure 3.3.
Figure 3.3: A circular cone: the graph z =
p
x2 + y2",Multivariable_Calculus_Shimamoto.pdf
"3.1. GRAPHS AND LEVEL SETS 57
The information about horizontal cross-sections in Figure 3.2 can also be presented by projecting
the cross-sections onto the xy-plane and presenting them like a topographical map. As we saw
above, the cross-section with z = c is described by the points ( x, y) such that x2 + y2 = c2, a
circle of radius c. When projected onto the xy-plane, this is called a level curve. Sketching a
representative sample of level curves in one picture and labeling them with the corresponding value
of c is called a contour map. In this case, it consists of concentric circles about the origin. See
Figure 3.4. The function f is constant on each circle and increases uniformly as you move out.
0
0.5
1
1.5
2
2.52.5
2.52.5
-2 -1 0 1 2
-2
-1
0
1
2
Figure 3.4: Level curves of f(x, y) =
p
x2 + y2 corresponding to c = 0, 0.5, 1, 1.5, 2, 2.5
Definition. Let A be a subset of Rn, and let f : A → R be a real-valued function. Given a real",Multivariable_Calculus_Shimamoto.pdf
"p
x2 + y2 corresponding to c = 0, 0.5, 1, 1.5, 2, 2.5
Definition. Let A be a subset of Rn, and let f : A → R be a real-valued function. Given a real
number c, the level set corresponding to c is {x ∈ A : f(x) = c}. It is the set of all points in the
domain at which f has a value of c. For functions of two variables, level sets are also called level
curves and, for functions of three variables, level surfaces.
A level set is a subset of the domain A, which is contained in Rn. So we can hope to draw level
sets only for functions of one, two, or three variables.
Example 3.3. Let f : R2 → R be given by f(x, y) = x2 −y2. Sketch some level sets and the graph
of f.
Level sets: We set f(x, y) = x2 − y2 = c and choose a few values of c. For example:
Level set
c = 1 x2 − y2 = 1, a hyperbola
c = 0 x2 − y2 = 0, or y = ±x, two lines
c = −1 x2 − y2 = −1, or y2 − x2 = 1, a hyperbola
These curves and a couple of others are sketched in Figure 3.5.
-2
-2
-1
-1
0 0
0
1 1 22
0
-2 -1 0 1 2
-2
-1
0
1
2",Multivariable_Calculus_Shimamoto.pdf
"These curves and a couple of others are sketched in Figure 3.5.
-2
-2
-1
-1
0 0
0
1 1 22
0
-2 -1 0 1 2
-2
-1
0
1
2
Figure 3.5: Level curves of f(x, y) = x2 − y2 corresponding to c = −2, −1, 0, 1, 2",Multivariable_Calculus_Shimamoto.pdf
"58 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
The graph: For the graph, the level sets are taken out of the xy-plane and raised to height z = c.
Furthermore, identifying the cross-sections with the coordinate planes provides some additional
framework.
Coordinate plane Equation of cross-section in that plane
yz-plane: x = 0 z = −y2, a downward parabola
xz-plane: y = 0 z = x2, an upward parabola
These are shown in Figure 3.6.
Figure 3.6: Two cross-sections with coordinate planes: z = −y2, x= 0 (left) and z = x2, y= 0
(right)
Reconstructing the graph from these pieces gives a saddle-like surface, also known as a hy-
perbolic paraboloid. See Figure 3.7. It has the distinctive feature that, in the cross-section with
Figure 3.7: A saddle surface: the graph z = x2 − y2
the yz-plane, the origin looks like a local maximum, while, in the xz-plane, it looks like a local
minimum. (This last information can be deduced from the contour map, too. See Figure 3.5.)
3.2 More surfaces in R3",Multivariable_Calculus_Shimamoto.pdf
"minimum. (This last information can be deduced from the contour map, too. See Figure 3.5.)
3.2 More surfaces in R3
Not all surfaces in R3 are graphs z = f(x, y). We look at a few examples of some other surfaces
and the equations that define them.
Example 3.4. A sphere of radius a centered at the origin.
A point ( x, y, z) lies on the sphere if and only if ∥(x, y, z)∥ = a, in other words, if and only ifp
x2 + y2 + z2 = a. Hence:
Equation of a sphere: x2 + y2 + z2 = a2
The sphere of radius 1 is shown in Figure 3.8, left.",Multivariable_Calculus_Shimamoto.pdf
"3.2. MORE SURFACES IN R3 59
Figure 3.8: The unit sphere x2 + y2 + z2 = 1 (left) and the cylinder x2 + y2 = 1 (right)
Example 3.5. A circular cylinder of radius a whose axis is the z-axis.
Here, as long as (x, y) satisfies the condition for being on the circle of radiusa, z can be anything.
Thus:
Equation of a circular cylinder: x2 + y2 = a2
An example is shown on the right of Figure 3.8.
Example 3.6. Sketch the surface x2 + y2 − z2 = 1.
This is not the graphz = f(x, y) of a function: given an input (x, y), there can be two choices ofz
that satisfy the equation. In fact, the surface is the union of two graphs, namely, z =
p
x2 + y2 − 1
and z = −
p
x2 + y2 − 1. Nevertheless, as was the case with graphs, we can try to construct a
sketch by looking at cross-sections with various well-chosen planes. For example:
Cross-sectional plane Equation of cross-section in that plane
yz-plane: x = 0 y2 − z2 = 1, a hyperbola
xz-plane: y = 0 x2 − z2 = 1, a hyperbola",Multivariable_Calculus_Shimamoto.pdf
"Cross-sectional plane Equation of cross-section in that plane
yz-plane: x = 0 y2 − z2 = 1, a hyperbola
xz-plane: y = 0 x2 − z2 = 1, a hyperbola
horizontal plane: z = c x 2 + y2 = 1 + c2, a circle of radius
√
1 + c2
See Figure 3.9.
Figure 3.9: Cross-sections with the yz-plane (left), xz-plane (middle), and horizontal planes (right)
These cross-sections fit together to form a surface that looks like a nuclear cooling tower, as
shown in Figure 3.10.
In fact, each of these last three surfaces is a level set of a function of three variables. The
sphere is the level set of f(x, y, z) = x2 + y2 + z2 corresponding to c = a2; the cylinder the",Multivariable_Calculus_Shimamoto.pdf
"60 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
Figure 3.10: The cooling tower x2 + y2 − z2 = 1
level set of f(x, y, z) = x2 + y2 corresponding to c = a2; and the cooling tower the level set of
f(x, y, z) = x2 + y2 − z2 corresponding to c = 1.
This suggests a way to visualize the behavior of functions of three variables, which would require
four dimensions to graph. Namely, determine the level sets given by f(x, y, z) = c. Then the way
these surfaces morph into one another as c varies reflects which points get mapped to which values.
For example, the function f(x, y, z) = x2 + y2 + z2 is constant on its level sets, which are spheres,
and the larger the sphere, the greater the value of f.
Example 3.7. A graph of a function f(x, y) of two variables can be regarded as a level set, too, but
of a function of three variables. For let F(x, y, z) = f(x, y) − z. Then the graph z = f(x, y) is the",Multivariable_Calculus_Shimamoto.pdf
"of a function of three variables. For let F(x, y, z) = f(x, y) − z. Then the graph z = f(x, y) is the
level set of F corresponding to c = 0. Both are defined by the condition F(x, y, z) = f(x, y) − z =
0.
3.3 The equation of a plane in R3
Suppose that a real-valued function of three variables T : R3 → R happens to be a linear transfor-
mation. As with any linear transformation, T is represented by a matrix, in this case, a 1 by 3
matrix:
T(x, y, z) =

A B C



x
y
z

 = Ax + By + Cz.
The level set of T corresponding to a value D is the set of all ( x, y, z) in R3 such that:
Ax + By + Cz = D. (3.1)
To understand what this level set is, we rewrite (3.1) as a dot product, ( A, B, C) ·(x, y, z) = D, or,
writing the typical point ( x, y, z) of R3 as x:
n · x = D,
where n = (A, B, C). Suppose that we happen to know one particular point p on the level set.
Then n · p = D, so all points on the level set satisfy:
n · x = D = n · p. (3.2)",Multivariable_Calculus_Shimamoto.pdf
"3.3. THE EQUATION OF A PLANE IN R3 61
In other words, n·(x−p) = 0. This says that x lies on the level set if and only if x−p is orthogonal
to n. Geometrically, this is true if and only if x lies in the plane that both passes through p and
is perpendicular to n. This is illustrated in Figure 3.11. We say that n is a normal vector to the
Figure 3.11: The plane through p with normal vector n
plane. Thus the level sets of the linear transformation T are planes, all having as normal vector
the vector n whose components are the entries of the matrix that represents T.
For future reference, we record the results of (3.1) and (3.2).
Proposition 3.8.
1. In R3, the equation of the plane through p with normal vector n is:
n · x = n · p, where x = (x, y, z).
2. Alternatively, the equation Ax + By + Cz = D is the equation of a plane with normal vector
n = (A, B, C).
Example 3.9. Find an equation of the plane containing the point p = (1 , 2, 3) and the line",Multivariable_Calculus_Shimamoto.pdf
"n = (A, B, C).
Example 3.9. Find an equation of the plane containing the point p = (1 , 2, 3) and the line
parametrized by α(t) = (4, 5, 6) + t(7, 8, 9). See Figure 3.12.
Figure 3.12: The plane determined by a given point and line
We use the form n · x = n · p. As a point in the desired plane, we can take p = (1, 2, 3). From
its parametrization, the line α passes through the point a = (4, 5, 6) and is parallel to v = (7, 8, 9).
Hence, as normal vector to the plane, we can use the cross product:
− →ap × v = (p − a) × v =

(1, 2, 3) − (4, 5, 6)


× (7, 8, 9)
= (−3, −3, −3) × (7, 8, 9)
= det


i j k
−3 −3 −3
7 8 9


= (−3, 6, −3)
= −3(1, −2, 1).",Multivariable_Calculus_Shimamoto.pdf
"62 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
The scalar multiple n = (1, −2, 1) is also a normal vector, and, since it’s a little simpler, it’s the
one we use. Substituting into n · x = n · p gives:
(1, −2, 1) · (x, y, z) = (1, −2, 1) · (1, 2, 3),
x − 2y + z = 1 − 4 + 3,
or x − 2y + z = 0.
Example 3.10. Are the planes x + y − z = 5 and x − 2y + 3z = 6 perpendicular?
The angle between two planes is the same as the angle between their normal vectors (Figure
3.13), so it suffices to check if the normals are orthogonal. Reading off the coefficients from the
Figure 3.13: The angle between two planes
equations that define the planes, the normals are n1 = (1 , 1, −1) and n2 = (1 , −2, 3). Then
n1 · n2 = 1 − 2 − 3 = −4, which is nonzero. The planes are not perpendicular.
3.4 Open sets
We prepare for the important concept of continuity. In first-year calculus, a real-valued function f
of one variable is said to be continuous if its graph has no holes or breaks. Intuitively, whenever x",Multivariable_Calculus_Shimamoto.pdf
"of one variable is said to be continuous if its graph has no holes or breaks. Intuitively, whenever x
approaches a point a, the value f(x) approaches the value f(a). This can be written succinctly as
limx→a f(x) = f(a).
Unfortunately, this idea is difficult to formulate in a rigorous way. The notions of continuity and
limit are closely related. In many respects, limits are more fundamental, but we shall work more
directly with continuity. Hence we base our discussion on the definition of a continuous function
and modify it later to incorporate limits.
We begin by introducing the most natural type of subset of Rn for taking limits. These are sets
in which there is a cushion around every point so that one can approach the point from all possible
directions.
Definition. Let a be a point of Rn, and assume that r >0. The open ball with center a and
radius r, denoted by B(a, r), is defined to be:
B(a, r) = {x ∈ Rn : ∥x − a∥ < r}.",Multivariable_Calculus_Shimamoto.pdf
"radius r, denoted by B(a, r), is defined to be:
B(a, r) = {x ∈ Rn : ∥x − a∥ < r}.
In other words, it is the set of all points within r units of a.
Example 3.11. In R, an open ball is an open interval ( a − r, a+ r). In R2, it is a disk, the region
in the interior of a circle. These are shown in Figure 3.14. In R3, an open ball is a solid ball, the
region in the interior of a sphere.",Multivariable_Calculus_Shimamoto.pdf
"3.4. OPEN SETS 63
Figure 3.14: Open balls in R and R2
Definition. A subset U of Rn is called open if, given any point a of U, there exists a positive real
number r such that B(a, r) ⊂ U.
So to show that a set U is open, one starts with an arbitrary point a of U and finds a value of r
so that the open ball about a of radius r stays entirely within U. The value of r depends typically
on the conditions that define the set U as well as on the point a. For the time being, we shall be
content to find a concrete expression for r and accept geometric intuition as justification that it
works. If, for some point a of U, there is no value of r that works, then U is not open.
Example 3.12. In R2, let U be the open ball B((0, 0), 1) of radius 1 centered at the origin. This
is the set of all x in R2 such that ∥x∥ < 1. We verify that this open ball is in fact an open set.
Let a be a point of U, and let r = 1 − ∥a∥. Note that r >0 since ∥a∥ < 1. Then B(a, r) ⊂ U,",Multivariable_Calculus_Shimamoto.pdf
"Let a be a point of U, and let r = 1 − ∥a∥. Note that r >0 since ∥a∥ < 1. Then B(a, r) ⊂ U,
so, given any a, we’ve found an r that works. See Figure 3.15. Hence U is an open set.
Figure 3.15: Open balls are open sets.
To those who would like a more detailed justification that B(a, r) ⊂ U, please see Exercise
7.2. Note that the expression r = 1 − ∥a∥ gets smaller as a approaches the boundary of the disk.
That this is necessary makes sense from the geometry of the situation. By modifying the argument
slightly, one can show that every open ball in Rn is an open set in Rn (Exercise 7.3).
Example 3.13. Prove that the set U = {(x, y) ∈ R2 : x >0, y <0} is open in R2.
This is the fourth quadrant of the plane, not including the coordinate axes. Let a be a point
of U. Write a = (c, d), so c >0 and d <0. The closest point on the boundary of U lies on one of
the axes, either c or |d| units away (Figure 3.16). We choose r to be whichever of these is smaller,",Multivariable_Calculus_Shimamoto.pdf
"the axes, either c or |d| units away (Figure 3.16). We choose r to be whichever of these is smaller,
that is, r = min{c, |d|}. Then B(a, r) ⊂ U. Thus U is open.
Example 3.14. Is U = {(x, y) ∈ R2 : x >0, y≤ 0} open in R2?
This is the same set as in the previous example with the positive x-axis added in. The change is
enough to keep the set from being open. For instance, the point a = (1, 0) is in U, but every open",Multivariable_Calculus_Shimamoto.pdf
"64 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
Figure 3.16: An open quadrant
ball about a includes points outside of U, namely, points in the upper half-plane. This is shown in
Figure 3.17. Thus no value of r works for this choice of a.
Figure 3.17: A nonopen quadrant
Example 3.15. If U and V are open sets in Rn, prove that their intersection U ∩ V is open, too.
Let a be a point of U ∩ V . Then a ∈ U, and, since U is open, there exists a positive number
r1 such that B(a, r1) ⊂ U. Likewise, a ∈ V , and there exists an r2 such that B(a, r2) ⊂ V . Let
r = min {r1, r2} (Figure 3.18). Then B(a, r) ⊂ B(a, r1) ⊂ U and B(a, r) ⊂ B(a, r2) ⊂ V , so
Figure 3.18: The intersection of two open sets
B(a, r) ⊂ U ∩ V . In other words, given any a in U ∩ V , there’s an r that works.
By similar reasoning, the intersection U1 ∩ U2 ∩ ··· ∩Uk of a finite number of open sets in Rn
is open in Rn. On the other hand, an infinite intersection U1 ∩ U2 ∩ ···need not be open. For",Multivariable_Calculus_Shimamoto.pdf
"is open in Rn. On the other hand, an infinite intersection U1 ∩ U2 ∩ ···need not be open. For
instance, in R2, consider the sequence of concentric open balls U1 = B((0, 0), 1), U2 = B((0, 0), 1
2 ),",Multivariable_Calculus_Shimamoto.pdf
"3.4. OPEN SETS 65
U3 = B((0, 0), 1
3 ), etc. Each Un is open, but the only point common to all of them is the origin.
Thus U1 ∩ U2 ∩ ···= {(0, 0)}. This is no longer an open set.
There is also a notion of closed set, though the definition may not be what one would guess.
Definition. A subset K of Rn is called closed if Rn − K = {x ∈ Rn : x /∈ K} is open. The set
Rn − K is called the complement of K in Rn.
Example 3.16. Let K = {x ∈ R2 : ∥x∥ ≤1}. This is the open ball centered at the origin of radius
1 together with its boundary, the unit circle, as shown in Figure 3.19. Its complement R2 − K is
the set of all points x such that ∥x∥ > 1, which is an open set. (Briefly, given a in R2 − K, then
r = ∥a∥ −1 works in the definition of open set.) Hence K is closed.
Figure 3.19: Closed balls are closed sets.
More generally, if a ∈ Rn and r ≥ 0, then K(a, r) = {x ∈ Rn : ∥x − a∥ ≤r} is called a closed
ball. These are closed subsets of Rn as well by a similar argument.",Multivariable_Calculus_Shimamoto.pdf
"ball. These are closed subsets of Rn as well by a similar argument.
Example 3.17. Let K = {x = (x, y) ∈ R2 : either ∥x∥ < 1 or x2 + y2 = 1 where y ≥ 0}. This
is the same as the closed ball of the preceding example except that the lower semicircle has been
deleted. See Figure 3.20.
Figure 3.20: A set that is neither open nor closed
K is no longer closed: for instance, the point a = (0, −1) belongs to R2 − K, but no open ball
about a is contained entirely within R2 − K. At the same time, K is not open either: the point
a = (0, 1) belongs to K, but no open ball about a stays within K.",Multivariable_Calculus_Shimamoto.pdf
"66 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
3.5 Continuity
We are ready for continuity. Intuitively, the idea is that a function f is continuous at a point a if
limx→a f(x) = f(a). That is, as x gets close to a, f(x) gets close to f(a). We make this precise by
expressing the requirement in terms of open balls.
Definition. Let U be an open set in Rn, and let f : U → R be a real-valued function. We say that
f is continuous at a point a of U if, given any open ball B(f(a), ϵ) about f(a), there exists an
open ball B(a, δ) about a such that:
f(B(a, δ)) ⊂ B(f(a), ϵ).
In other words, if x ∈ B(a, δ), then f(x) ∈ B(f(a), ϵ) (Figure 3.21). Equivalently, writing out the
definition of open ball, f is continuous at a if, given any ϵ >0, there exists a δ >0 such that:
if ∥x − a∥ < δ,then |f(x) − f(a)| < ϵ.
Figure 3.21: f maps the open ball B(a, δ) about a into the open ball ( f(a) − ϵ, f(a) + ϵ) about
f(a).",Multivariable_Calculus_Shimamoto.pdf
"if ∥x − a∥ < δ,then |f(x) − f(a)| < ϵ.
Figure 3.21: f maps the open ball B(a, δ) about a into the open ball ( f(a) − ϵ, f(a) + ϵ) about
f(a).
The definition says that f is continuous at a if you can guarantee that f(x) will be as close
to f(a) as you want (“within ϵ”) by making sure that x is close enough to a (“within δ”). The
strategy for proving that a function is continuous is similar formally to proving that a set is open.
There, one starts with a point a and tries to find a radius r. Here, one starts with an ϵ and tries
to find a δ.
It may seem that using the open ball B(f(a), ϵ) in the definition is an unnecessarily confusing
way to write the interval (f(a) − ϵ, f(a) +ϵ), but we have in mind the generalization of continuity
to vector-valued functions in Chapter 6. The definition phrased in terms of open balls extends
naturally to the more general case, as we shall see.
Definition. A function f : U → R is simply called continuous if it is continuous at every point",Multivariable_Calculus_Shimamoto.pdf
"naturally to the more general case, as we shall see.
Definition. A function f : U → R is simply called continuous if it is continuous at every point
of its domain U.
Example 3.18. Let f : R2 → R be defined by f(x, y) = x. In other words, f is the projection of
the xy-plane onto the x-axis. Is f a continuous function?
Let a be a point of R2. To get a sense of what answer to expect, we consider lim x→a f(x).
(Of course, we haven’t defined rigorously what a limit is yet, so this is meant to be completely
informal.) Write x = (x, y) and a = (c, d). Then:
lim
x→a
f(x) = lim
(x,y)→(c,d)
f(x, y) = lim
(x,y)→(c,d)
x = c.",Multivariable_Calculus_Shimamoto.pdf
"3.5. CONTINUITY 67
At the same time, f(a) = f(c, d) = c, which agrees. Thus we expect intuitively that f is continuous
at a.
To prove this rigorously, let ϵ >0 be given. We want to find a radius δ so that the open ball
B

(c, d), δ


is projected inside the open interval B(f(c, d), ϵ) = B(c, ϵ) = ( c − ϵ, c+ ϵ). From the
geometry of the projection, taking δ = ϵ works. See Figure 3.22. That is, if ( x, y) ∈ B((c, d), ϵ),
then its x-coordinate satisfies c − ϵ < x < c+ ϵ. In other words, c − ϵ < f(x, y) < c+ ϵ, or
f(x, y) ∈ (c − ϵ, c+ ϵ). Therefore δ = ϵ works in the definition of continuity.
Actually, δ could be anything less than ϵ, too, but the definition requires only that we find one
δ that works, not all of them.
Figure 3.22: For the projection, choosing δ = ϵ works.
Since f is continuous at every point a of R2, it is a continuous function.
Example 3.19. Let f(x, y) = xy
x2+y2 for all (x, y) ̸= (0, 0). Is it possible to assign a value to f(0, 0)",Multivariable_Calculus_Shimamoto.pdf
"Example 3.19. Let f(x, y) = xy
x2+y2 for all (x, y) ̸= (0, 0). Is it possible to assign a value to f(0, 0)
so that f is continuous at (0, 0)? See Figure 3.23.
Figure 3.23: The graph z = xy
x2+y2 : can it be made continuous at (0 , 0)?
Say we set f(0, 0) = c. First, the intuitive approach: we want to know if it’s possible to choose
c so that:
lim
(x,y)→(0,0)
f(x, y) = c.
To get a feel for this, we try approaching the origin in a couple of different ways. For instance,",Multivariable_Calculus_Shimamoto.pdf
"68 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
suppose that we approach along the x-axis. Then f(x, 0) = x·0
x2+02 = 0, so:
lim
(x,0)→(0,0)
f(x, 0) = lim
(x,0)→(0,0)
0 = 0.
On the other hand, if we approach along the line y = x, then f(x, x) = x·x
x2+x2 = 1
2 , so:
lim
(x,x)→(0,0)
f(x, x) = lim
(x,x)→(0,0)
1
2 = 1
2.
Thus f(x, y) can approach different values depending on how you approach the origin. As a result,
intuitively, the limit does not exist, so f cannot be continuous regardless of what c is.
To prove this rigorously, we need to articulate what it means for the criterion for continuity to
fail. The definition says that a function is continuous if, for every ϵ, there exists a δ. The negation
of this is that there is some ϵ for which there is no δ. We try to find such a bad ϵ.
Our intuitive calculations above showed that there are points arbitrarily close to the origin
where f = 0 and other points arbitrarily close where f = 1
2 . We choose ϵ so that these values of f",Multivariable_Calculus_Shimamoto.pdf
"where f = 0 and other points arbitrarily close where f = 1
2 . We choose ϵ so that these values of f
cannot both be within ϵ of f(0, 0) = c.
For this, let ϵ = 1
8 . Then B(f(0, 0), ϵ) = B(c, 1
8 ) = (c− 1
8 , c+ 1
8 ), an interval of length 1
4 . As just
noted, in any open ball B((0, 0), δ) about the origin, there will be points ( x, 0) where f(x, 0) = 0
and points ( x, x) where f(x, x) = 1
2 . It’s impossible to fit both these values inside an interval of
length 1
4 . So for ϵ = 1
8 , there is no δ >0 such that f

B((0, 0), δ)


⊂ (c − 1
8 , c+ 1
8 ). Thus f is not
continuous at (0, 0) no matter what the value of f(0, 0) is.
The ϵδ-definition of continuity is awkward and difficult to work with. One of its virtues, how-
ever, is that, not only can it be used to verify that a function is continuous, but it also gives an
unambiguous condition for proving that a function is not. That is the moral of the preceding
example.
Example 3.20. In the same spirit, let f(x, y) = x3+y3",Multivariable_Calculus_Shimamoto.pdf
"example.
Example 3.20. In the same spirit, let f(x, y) = x3+y3
x2+y2 for all ( x, y) ̸= (0, 0). The graph of f is
shown in Figure 3.24. Is it possible to assign a value to f(0, 0) so that f is continuous at (0, 0)?
Figure 3.24: The graph z = x3+y3
x2+y2
As before, intuitively, in order to be continuous, f(0, 0) should equal lim (x,y)→(0,0)
x3+y3
x2+y2 , if the
limit exists. We try approaching the origin along various lines:
Along the x-axis f(x, 0) = x3+03
x2+02 = x → 0 as x → 0
Along the y-axis f(0, y) = 03+y3
02+y2 = y → 0 as y → 0
Along the line y = mx f (x, mx) = x3+m3x3
x2+m2x2 = 1+m3
1+m2 x → 0 as x → 0",Multivariable_Calculus_Shimamoto.pdf
"3.6. SOME PROPERTIES OF CONTINUOUS FUNCTIONS 69
We get a consistent answer of 0, but this does not prove that the limit is 0. We have not
exhausted all possible ways of approaching the origin, and being close to the origin is different
from being close to it along any individual curve. Nevertheless, the evidence suggests that choosing
f(0, 0) = 0 might make f continuous, and this gives us something concrete to shoot for.
So set f(0, 0) = 0, and let ϵ >0 be given. We want to find a δ >0 such that, if ∥(x, y)−(0, 0)∥ <
δ, then |f(x, y) −0| < ϵ. This is satisfied automatically when ( x, y) = (0, 0) for any value of δ since
f(0, 0) = 0, so we assume (x, y) ̸= (0, 0) and look for a δ such that, if ∥(x, y)∥ < δ, then
x3+y3
x2+y2
 < ϵ.
To hunt for a connection between the quantities ∥(x, y)∥ and
x3+y3
x2+y2
, we introduce polar coor-
dinates: x = r cos θ and y = r sin θ, where r =
p
x2 + y2 = ∥(x, y)∥. See Figure 3.25.
Figure 3.25: Polar coordinates r and θ
Then, if (x, y) ̸= (0, 0):",Multivariable_Calculus_Shimamoto.pdf
"p
x2 + y2 = ∥(x, y)∥. See Figure 3.25.
Figure 3.25: Polar coordinates r and θ
Then, if (x, y) ̸= (0, 0):
x3 + y3
x2 + y2 = r3(cos3 θ + sin3 θ)
r2 = r(cos3 θ + sin3 θ).
Since |cos3 θ + sin3 θ| ≤2, this implies that:

x3 + y3
x2 + y2
 = |r(cos3 θ + sin3 θ)| ≤2r = 2
p
x2 + y2 = 2∥(x, y)∥.
Thus, if ∥(x, y)∥ < a, then
x3+y3
x2+y2
 ≤ 2∥(x, y)∥ < 2a.
So let δ = ϵ
2 . With this choice, if ∥(x, y)∥ < δ, then |f(x, y)| < 2δ = ϵ, i.e., f maps B((0, 0), δ)
inside (−ϵ, ϵ). In other words, given any ϵ >0, we’ve found a δ >0 that satisfies the definition of
continuity, namely,δ = ϵ
2 . Therefore setting f(0, 0) = 0 makes f continuous at (0, 0).
3.6 Some properties of continuous functions
The definition of continuity makes for unimpeachable arguments, but, as we learn more about
continuous functions, we might prefer to build on what we learn and not have to start from scratch
with the definition every time. We present a couple of general properties that are especially useful.",Multivariable_Calculus_Shimamoto.pdf
"with the definition every time. We present a couple of general properties that are especially useful.
These results are behind the working principle that most functions that look continuous really are
continuous. One need not bring out the ϵ’s and δ’s.
Proposition 3.21. Let f, g: U → R be real-valued functions defined on an open set U in Rn. If f
and g are continuous at a point a of U, then so are:



f + g,
cf for any scalar c,
fg, the product of f and g,
f
g , the quotient, assuming that g(a) ̸= 0.",Multivariable_Calculus_Shimamoto.pdf
"70 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
Before proving this result, let’s apply it to some examples.
Example 3.22. We showed earlier in Example 3.18 that the projection onto thex-axis f : R2 → R,
f(x, y) = x, is continuous. Likewise, the projection onto the y-axis g : R2 → R, g(x, y) = y, is
continuous. It then follows immediately from the proposition that functions like x + y, 3 x, xy,
x2 = x · x, x3 − 4x2y2 + 5y, and, at points other than (0 , 0), xy
x2+y2 are all continuous.
Proof. We prove the statement about the sum f + g and leave the rest for the exercises. (See
Exercises 7.4–7.7.) For this, we want to argue that, as x gets close to a, f(x) + g(x) gets close
to f(a) + g(a). Since f(x) and g(x) get close to f(a) and g(a) individually, this conclusion seems
clear, and it’s just a matter of feeding the intuition into the formal definition. The key ingredient
in the argument below is the “triangle inequality.”",Multivariable_Calculus_Shimamoto.pdf
"in the argument below is the “triangle inequality.”
Let ϵ >0 be given. We need to come up with a δ >0 such that f(x) +g(x) ∈ B(f(a) +g(a), ϵ)
whenever x ∈ B(a, δ), in other words, a δ such that:
|(f(x) + g(x)) − (f(a) + g(a))| < ϵwhenever x ∈ B(a, δ).
Note that:
|(f(x) + g(x)) − (f(a) + g(a))| = |(f(x) − f(a)) + (g(x) − g(a))|. (3.3)
The triangle inequality says that, for any real numbers r and s, |r + s| ≤ |r| + |s|. This seems
reasonable, but it has a generalization to Rn that is important enough that we discuss it separately
in the next section. Accepting that it is true for the time being, we find from equation (3.3) that:
|(f(x) + g(x)) − (f(a) + g(a))| ≤ |f(x) − f(a)| + |g(x) − g(a)|. (3.4)
This separates the f contribution and the g contribution into terms that can be manipulated
independently.
Since f is continuous at a, taking the positive quantity ϵ
2 as the given input in the definition
of continuity, there is a δ1 > 0 such that |f(x) − f(a)| < ϵ",Multivariable_Calculus_Shimamoto.pdf
"2 as the given input in the definition
of continuity, there is a δ1 > 0 such that |f(x) − f(a)| < ϵ
2 whenever x ∈ B(a, δ1). Similarly, there
is a δ2 > 0 such that |g(x) − g(a)| < ϵ
2 whenever x ∈ B(a, δ2). Let δ = min{δ1, δ2}. Then the last
two inequalities remain true for all x in B(a, δ), and substituting them into (3.4) gives:
(f(x) + g(x)) − (f(a) + g(a))
 < ϵ
2 + ϵ
2 = ϵ whenever x ∈ B(a, δ).
This shows that f + g is continuous at a.
Proposition 3.23. Compositions of continuous functions are continuous. That is, let U be an
open set in Rn and V an open set in R, and let f : U → R and g : V → R be functions such
that f(x) ∈ V for all x in U. (This assumption guarantees that the composition g ◦ f : U → R is
defined.) Then, if f and g are continuous, so is g ◦ f.
Again, we look at some examples first.
Example 3.24. We accept without comment that the familiar functions of one variable from first-",Multivariable_Calculus_Shimamoto.pdf
"Again, we look at some examples first.
Example 3.24. We accept without comment that the familiar functions of one variable from first-
year calculus that were said to be continuous there really are continuous. This includes |x|, n√x,
sin x, cos x, ex, and ln x. The proposition then implies that functions like
p
x2 + y2, sin(x + y),
and exy are continuous. For example, the first is the composition ( x, y) 7→ x2 + y2 7→
p
x2 + y2,
which is a composition of continuous steps.",Multivariable_Calculus_Shimamoto.pdf
"3.7. THE CAUCHY-SCHWARZ AND TRIANGLE INEQUALITIES 71
Proof. Let a be a point of U, and let ϵ >0 be given. We want to find an open ball about a that
is mapped by g ◦ f inside the interval B(g(f(a)), ϵ) =

g(f(a)) − ϵ, g(f(a)) + ϵ


. We work our
way backwards from g(f(a)) to a using the definition of continuity twice to find such a ball. The
ingredients are depicted in Figure 3.26.
First, since g is continuous at the point f(a), there exists a δ′ > 0 such that:
g(B(f(a), δ′)) ⊂ B(g(f(a)), ϵ).
But then since f is continuous at a, treating δ′ as the given input, there exists a δ >0 such that:
f(B(a, δ)) ⊂ B(f(a), δ′).
Linking these two steps:
(g ◦ f)(B(a, δ)) = g(f(B(a, δ))) ⊂ g(B(f(a), δ′)) ⊂ B(g(f(a)), ϵ).
Thus, given an ϵ, we’ve found a δ that works.
Figure 3.26: The composition of continuous functions: given an ϵ, a δ′ exists (g is continuous), then
a δ exists (f is continuous).
3.7 The Cauchy-Schwarz and triangle inequalities",Multivariable_Calculus_Shimamoto.pdf
"a δ exists (f is continuous).
3.7 The Cauchy-Schwarz and triangle inequalities
We now discuss two fundamental inequalities about vectors in Rn, one involving the dot product
and the other involving vector addition.
Theorem 3.25 (Cauchy-Schwarz inequality).
|v · w| ≤ ∥v∥∥w∥
for all v, w in Rn.
Proof. Recall that v·w = ∥v∥∥w∥cos θ, where θ is the angle between v and w (see Theorem 1.12).
Thus, since |cos θ| ≤1, we have |v · w| = ∥v∥∥w∥| cos θ| ≤ ∥v∥∥w∥.
From this follows a generalization of the property of R that was alluded to in the proof of
Proposition 3.21.
Theorem 3.26 (Triangle inequality).
∥v + w∥ ≤ ∥v∥ + ∥w∥
for all v, w in Rn.",Multivariable_Calculus_Shimamoto.pdf
"72 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
Proof. It’s equivalent to square both sides and prove that ∥v + w∥2 ≤ (∥v∥ + ∥w∥)2. But:
∥v + w∥2 = (v + w) · (v + w)
= v · v + v · w + w · v + w · w
= ∥v∥2 + 2v · w + ∥w∥2. (3.5)
Also, v · w ≤ |v · w| ≤ ∥v∥∥w∥, where the second inequality is Cauchy-Schwarz. Substituting this
into (3.5) gives:
∥v + w∥2 ≤ ∥v∥2 + 2∥v∥∥w∥ + ∥w∥2 = (∥v∥ + ∥w∥)2.
The connection with triangles is illustrated in Figure 3.27.
Figure 3.27: The geometry of the triangle inequality: the length ∥v + w∥ of one side of a triangle
is less than or equal to the sum ∥v∥ + ∥w∥ of the lengths of the other two sides.
3.8 Limits
We used our intuition about limits as a way to think about continuity. Now that continuity has
been defined rigorously, we turn the tables and use continuity to give a formal definition of limits.
One technical, though important, point is that, in determining how f(x) behaves as x ap-",Multivariable_Calculus_Shimamoto.pdf
"One technical, though important, point is that, in determining how f(x) behaves as x ap-
proaches a, what is happening right at a does not matter. The value of f(a), or even whether f is
defined there, is irrelevant.
Definition. Let U be an open set in Rn, and let a be a point of U. If f is a real-valued function
that is defined on U, except possibly at the point a, then we say that limx→a f(x) exists if there is
a number L such that the function ef : U → R defined by
ef(x) =
(
f(x) if x ̸= a,
L if x = a
is continuous at a. When this happens, we write lim x→a f(x) = L.
By carefully applying the definition of continuity to the function ef, the definition of limit can
be restated using ϵ’s and δ’s, which is its more customary form. Namely, lim x→a f(x) = L means:
Given any ϵ >0, there exists a δ >0 such that |f(x) −L| < ϵwhenever ∥x −a∥ < δ,
except possibly when x = a.
We shall never use this formulation, however.",Multivariable_Calculus_Shimamoto.pdf
"except possibly when x = a.
We shall never use this formulation, however.
If f is defined at a, then it’s basically a tautology from the definition that limx→a f(x) = f(a) if
and only if f is continuous at a, thus completing the intuitive connection between the two concepts
with which we began.",Multivariable_Calculus_Shimamoto.pdf
"3.9. EXERCISES FOR CHAPTER 3 73
Example 3.27. In Example 3.20, we showed that the function given by f(x, y) = x3+y3
x2+y2 , which
is defined for ( x, y) ̸= (0, 0), could be made continuous at (0 , 0) by setting f(0, 0) = 0. (Strictly
speaking, this extended function is the one called ef in the definition.) Hence, by definition:
lim
(x,y)→(0,0)
x3 + y3
x2 + y2 = 0.
Similarly, in Example 3.19, for the function f(x, y) = xy
x2+y2 , there is no value that can be
assigned to f(0, 0) that makes f continuous at (0, 0). In this case:
lim
(x,y)→(0,0)
xy
x2 + y2 does not exist.
Because of the close connection between the two concepts, properties of continuous functions
have analogues for limits. Here is the version of Proposition 3.21 for limits.
Proposition 3.28. Assume that limx→a f(x) and limx→a g(x) both exist. Then:
1. limx→a

f(x) + g(x)


=

limx→a f(x)


+

limx→a g(x)


.
2. limx→a

cf(x)


= c

limx→a f(x)


for any scalar c.
3. limx→a

f(x)g(x)


=

limx→a f(x)


limx→a g(x)

",Multivariable_Calculus_Shimamoto.pdf
"
limx→a f(x)


+

limx→a g(x)


.
2. limx→a

cf(x)


= c

limx→a f(x)


for any scalar c.
3. limx→a

f(x)g(x)


=

limx→a f(x)


limx→a g(x)


.
4. limx→a
f(x)
g(x) = limx→a f(x)
limx→a g(x) , provided that limx→a g(x) ̸= 0.
We leave the proofs for the exercises (Exercise 8.1).
3.9 Exercises for Chapter 3
Section 1 Graphs and level sets
For the functions f(x, y) in Exercises 1.1–1.7: (a) sketch the cross-sections of the graph z =
f(x, y) with the coordinate planes, (b) sketch several level curves of f, labeling each with the
corresponding value of c, and (c) sketch the graph of f.
1.1. f(x, y) = x2 + y2
1.2. f(x, y) = x2 + y2 + 1
1.3. f(x, y) = y − x2
1.4. f(x, y) = x + y
1.5. f(x, y) = |x| + |y|
1.6. f(x, y) = xy
1.7. f(x, y) = y
x2+1
1.8. The graph of the function f(x, y) = x3 − 3xy2 is called a monkey saddle. You will need to
bring one along whenever you invite a monkey to go riding with you. The level sets of f are",Multivariable_Calculus_Shimamoto.pdf
"bring one along whenever you invite a monkey to go riding with you. The level sets of f are
not easy to sketch directly, but it is still possible to get a reasonable idea of what the graph
looks like.",Multivariable_Calculus_Shimamoto.pdf
"74 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
(a) Sketch the level set corresponding to c = 0. ( Hint: x3 − 3xy2 = x(x +
√
3y)(x −
√
3y).)
(b) Draw the region of the xy-plane in which f(x, y) > 0 and the region in which f(x, y) < 0.
(Hint: Part (a) might help.)
(c) Use the information from parts (a) and (b) to make a rough sketch of the monkey saddle.
1.9. It is possible to construct an origami approximation of the saddle surface z = x2 − y2 of
Example 3.3 by folding a piece of paper. The resulting model is called a hypar. (The
saddle surface itself is known as a hyperbolic paraboloid.) See Figure 3.28. Construct your
own hypar by starting with a square piece of paper and following the instructions appended
to the end of the chapter (page 80). 3
Figure 3.28: A hypar
Section 2 More surfaces in R3
In Exercises 2.1–2.4, sketch the surface in R3 described by the given equation.
2.1. x2 + y2 + z2 = 4
2.2. x2 + y2 = 4
2.3. x2 + z2 = 4
2.4. x2 + y2
9 + z2
4 = 1",Multivariable_Calculus_Shimamoto.pdf
"2.1. x2 + y2 + z2 = 4
2.2. x2 + y2 = 4
2.3. x2 + z2 = 4
2.4. x2 + y2
9 + z2
4 = 1
In Exercises 2.5–2.9, sketch the level sets corresponding to the indicated values ofc for the given
function f(x, y, z) of three variables. Make a separate sketch for each individual level set.
2.5. f(x, y, z) = x2 + y2 + z2, c = 0, 1, 2
2.6. f(x, y, z) = x2 + y2, c = 0, 1, 2
2.7. f(x, y, z) = x2 + y2 − z, c = −1, 0, 1
2.8. f(x, y, z) = x2 + y2 − z2, c = −1, 0, 1
2.9. f(x, y, z) = x2 − y2 − z2, c = −1, 0, 1
2.10. Can the saddle surface z = x2 −y2 in R3 be described as a level set? If so, how? What about
the parabola y = x2 in R2?
3For more information about hypars, see http://erikdemaine.org/hypar/.",Multivariable_Calculus_Shimamoto.pdf
"3.9. EXERCISES FOR CHAPTER 3 75
Section 3 The equation of a plane
3.1. Find an equation of the plane through the point p = (1, 2, 3) with normal vector n = (4, 5, 6).
3.2. Find an equation of the plane through the point p = (2 , −1, 3) with normal vector n =
(1, −4, 5).
3.3. Find a point that lies on the plane x + 3y + 5z = 9 and a normal vector to the plane.
3.4. Find a point that lies on the plane x + z = 1 and a normal vector to the plane.
3.5. Are the planes x − y + z = 8 and 2x + y − z = −1 perpendicular? Are they parallel?
3.6. Are the planes x − y + z = 8 and 2x − 2y + 2z = −1 perpendicular? Are they parallel?
3.7. Find an equation of the plane that contains the points p = (1 , 0, 0), q = (0 , 2, 0), and
r = (0, 0, 3).
3.8. Find an equation of the plane that contains the points p = (1 , 1, 1), q = (2 , −1, 2), and
r = (−1, 2, 0).
3.9. Find an equation of the plane that contains the point p = (1, 1, 1) and the line parametrized
by α(t) = (1, 0, 1) + t(2, 3, −4).",Multivariable_Calculus_Shimamoto.pdf
"r = (−1, 2, 0).
3.9. Find an equation of the plane that contains the point p = (1, 1, 1) and the line parametrized
by α(t) = (1, 0, 1) + t(2, 3, −4).
3.10. Find a parametrization of the line of intersection of the planes x+y +z = 3 and x−y +z = 1.
3.11. What point on the plane x − y + 2z = 3 is closest to the point q = (2, 1, −1)? ( Hint: Find a
parametrization of an appropriate line through q.)
3.12. Two planes in R3 are perpendicular to each other. Their line of intersection is described by
the parametric equations:
x = 2 − t, y = −1 + t, z = −4 + 2t.
If one of the planes has equation 2 x + 4y − z = 4, find an equation for the other plane.
3.13. Consider the plane x + y + z = 1.
(a) Sketch the cross-sections with the three coordinate planes.
(b) Sketch and describe in words the portion of the plane that lies in the “first octant” of
R3, that is, the part where x ≥ 0, y ≥ 0, and z ≥ 0.
3.14. Sketch and describe some of the level sets of the function f(x, y, z) = x + y + z.",Multivariable_Calculus_Shimamoto.pdf
"R3, that is, the part where x ≥ 0, y ≥ 0, and z ≥ 0.
3.14. Sketch and describe some of the level sets of the function f(x, y, z) = x + y + z.
3.15. Let π1 and π2 be parallel planes in R3 given by the equations:
π1 : Ax + By + Cz = D1 and π2 : Ax + By + Cz = D2.
(a) If p1 and p2 are points on π1 and π2, respectively, show that:
n · (p2 − p1) = D2 − D1,
where n = (A, B, C) is a normal vector to the two planes.",Multivariable_Calculus_Shimamoto.pdf
"76 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
(b) Show that the perpendicular distance d between π1 and π2 is given by:
d = |D2 − D1|√
A2 + B2 + C2 .
(c) Find the perpendicular distance between the planes x + y + z = 1 and x + y + z = 5.
3.16. Let α be a path in R3 that has constant torsion τ = 0. You may assume that α′(t) ̸= 0 and
T′(t) ̸= 0 for all t so that the Frenet vectors of α are always defined.
(a) Prove that the binormal vector B is constant.
(b) Fix a time t = t0, and let x0 = α(t0). Show that the function
f(t) = B · (α(t) − x0)
is a constant function. What is the value of the constant?
(c) Prove that α(t) lies in a single plane for all t. In other words, curves that have constant
torsion 0 must be planar. ( Hint: Use part (b) to identify a point and a normal vector
for the plane in which the path lies.)
Section 4 Open sets
In Exercises 4.1–4.7, let U be the set of points ( x, y) in R2 that satisfy the given conditions.",Multivariable_Calculus_Shimamoto.pdf
"Section 4 Open sets
In Exercises 4.1–4.7, let U be the set of points ( x, y) in R2 that satisfy the given conditions.
Sketch U, and determine whether it is an open set. Your arguments should be at a level of rigor
comparable to those given in the text.
4.1. 0 < x <1 and 0 < y <1
4.2. |x| < 1 and |y| < 1
4.3. |x| ≤1 and |y| < 1
4.4. x2 + y2 = 1
4.5. 1 < x2 + y2 < 4
4.6. y > x
4.7. y ≥ x
4.8. Is R2 − {(0, 0)}, the plane with the origin removed, an open set in R2? Justify your answer.
4.9. Prove that the union of any collection of open sets in Rn is open. That is, if {Uα} is a
collection of open sets in Rn, then S
α Uα is open.
Section 5 Continuity
5.1. Let f : R2 → R be the function:
f(x, y) =
( x3y
x2+y2 if (x, y) ̸= (0, 0),
0 if ( x, y) = (0, 0).
(a) Let a be a positive real number. Show that, if ∥(x, y)∥ < a, then |f(x, y)| < a2.",Multivariable_Calculus_Shimamoto.pdf
"3.9. EXERCISES FOR CHAPTER 3 77
(b) Use the ϵδ-definition of continuity to determine whether f is continuous at (0, 0).
5.2. Consider the function f(x, y) defined by:
f(x, y) = x2y4
(x2 + y4)2 if (x, y) ̸= (0, 0).
Determine what happens to the value of f(x, y) as (x, y) approaches the origin along:
(a) the x-axis,
(b) the y-axis,
(c) the line y = mx,
(d) the parabola x = y2.
(e) Is it possible to assign a value to f(0, 0) so that f is continuous at (0 , 0)? Justify your
answer using the ϵδ-definition of continuity.
5.3. Let f(x, y) = x2−y2
x2+y2 if (x, y) ̸= (0, 0). Is it possible to assign a value to f(0, 0) so that f is
continuous at (0, 0)? Justify your answer using the ϵδ-definition.
5.4. Let f(x, y) = x3+2y3
x2+y2 if (x, y) ̸= (0, 0). Is it possible to assign a value to f(0, 0) so that f is
continuous at (0, 0)? Justify your answer using the ϵδ-definition.
Section 6 Some properties of continuous functions",Multivariable_Calculus_Shimamoto.pdf
"continuous at (0, 0)? Justify your answer using the ϵδ-definition.
Section 6 Some properties of continuous functions
6.1. Let a = (1, 2). Show that the function f : R2 → R given by f(x) = a · x is continuous.
In Exercises 5.1–5.2, you determined the continuity of the following two functions at (0 , 0).
Now, consider points (x, y) other than (0 , 0). Determine all such points at which f is continuous.
6.2. f(x, y) = x3y
x2+y2 if (x, y) ̸= (0, 0), f(0, 0) = 0
6.3. f(x, y) = x2y4
(x2+y4)2 if (x, y) ̸= (0, 0)
6.4. Let U be an open set in Rn, and let f : U → R be a function that is continuous at a point a
of U.
(a) If f(a) > 0, show that there exists an open ball B = B(a, r) centered at a such that
f(x) > f(a)
2 for all x in B. ( Hint: Let ϵ = f(a)
2 .)
(b) Similarly, if f(a) < 0, show that there exists an open ball B = B(a, r) such that
f(x) < f(a)
2 for all x in B.
In particular, if f(a) ̸= 0, there is an open ball B centered at a throughout which f(x) has
the same sign as f(a).",Multivariable_Calculus_Shimamoto.pdf
"f(x) < f(a)
2 for all x in B.
In particular, if f(a) ̸= 0, there is an open ball B centered at a throughout which f(x) has
the same sign as f(a).
6.5. Let U be the set of all points ( x, y) in R2 such that y >sin x, that is:
U = {(x, y) ∈ R2 : y >sin x}.
Prove that U is an open set in R2. ( Hint: Apply the previous exercise to an appropriate
continuous function.)",Multivariable_Calculus_Shimamoto.pdf
"78 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
Section 7 The Cauchy-Schwarz and triangle inequalities
7.1. Let v and w be vectors in Rn.
(a) Show that ∥v∥ − ∥w∥ ≤ ∥v − w∥. ( Hint: v = (v − w) + w.)
(b) Show that
∥v∥ − ∥w∥
 ≤ ∥v − w∥.
7.2. (a) Let a be a point of Rn. If r >0, show that B(a, r) ⊂ B(0, ∥a∥+r). Draw a picture that
illustrates the result in R2. ( Hint: To prove a set inclusion of this type, you must show
that, if x ∈ B(a, r), then x ∈ B(0, ∥a∥+r), that is, if ∥x−a∥ < r, then ∥x−0∥ < ∥a∥+r.)
(b) In R2, if a ∈ B((0, 0), 1) and r = 1 − ∥a∥, show that B(a, r) ⊂ B((0, 0), 1). (Recall that
this inclusion was used in Example 3.12 to prove that B((0, 0), 1) is an open set.)
7.3. Let a be a point of Rn, and let r be a positive real number. Prove that the open ball B(a, r)
is an open set in Rn.
Exercises 7.4–7.7 provide the missing proofs of parts of Proposition 3.21 about properties of",Multivariable_Calculus_Shimamoto.pdf
"is an open set in Rn.
Exercises 7.4–7.7 provide the missing proofs of parts of Proposition 3.21 about properties of
continuous functions. Here, U is an open set in Rn, f, g: U → R are real-valued functions defined
on U, and a is a point of U.
7.4. If f is continuous at a and c is a scalar, prove that cf is continuous at a as well. ( Hint: One
approach is to consider the two cases c = 0 and c ̸= 0 separately. Another is to start by
showing that |cf(x) − cf(a)| ≤(|c| + 1)|f(x) − f(a)|.)
7.5. If f is continuous at a, prove that there exists a δ > 0 such that, if x ∈ B(a, δ), then
|f(x)| < |f(a)| + 1. (Hint: Use Exercise 7.1.)
7.6. If f and g are continuous at a, prove that their product fg is continuous at a. ( Hint: Use
an “add zero” trick and the previous exercise to show that there is a δ1 > 0 such that, if
x ∈ B(a, δ1), then
f(x)g(x)−f(a)g(a)
 ≤

|f(a)|+1

g(x)−g(a)
+

|g(a)|+1

f(x)−f(a)
.)
7.7. Assume that f and g are continuous at a and that g(a) ̸= 0.",Multivariable_Calculus_Shimamoto.pdf
"f(x)g(x)−f(a)g(a)
 ≤

|f(a)|+1

g(x)−g(a)
+

|g(a)|+1

f(x)−f(a)
.)
7.7. Assume that f and g are continuous at a and that g(a) ̸= 0.
(a) Prove that 1
g is continuous at a. ( Hint: Use Exercise 6.4 to show that there exists a
δ1 > 0 such that, if x ∈ B(a, δ1), then
 1
g(x) − 1
g(a)
 ≤ 2
|g(a)|2 |g(x) − g(a)|.)
(b) Prove that the quotient f
g is continuous at a. ( Hint: f
g = f · 1
g .)
7.8. Let f : U → R be a real-valued function defined on an open set U in Rn, and let g : U → R
be given by g(x) = |f(x)|.
(a) If f is continuous at a point a of U, is g necessarily continuous at a as well? Either give
a proof or find a counterexample.
(b) Conversely, if g is continuous at a, is f as well? Again, give a proof or find a counterex-
ample.",Multivariable_Calculus_Shimamoto.pdf
"3.9. EXERCISES FOR CHAPTER 3 79
Section 8 Limits
8.1. Prove parts 1 and 3 of Proposition 3.28 about limits of sums and products. ( Hint: By using
the corresponding properties of continuous functions in Proposition 3.21, you should be able
to avoid ϵ’s and δ’s entirely.)
8.2. Let U be an open set in Rn, x0 a point of U, and f a real-valued function defined on U,
except possibly at x0, such that lim x→x0 f(x) = L. Let I be an open interval in R, and let
α: I → U be a continuous path in U that passes through x0, i.e., α(t0) = x0 for some t0 in
I. Consider the function g : I → R given by:
g(t) =
(
f(α(t)) if α(t) ̸= x0,
L if α(t) = x0.
Show that limt→t0 g(t) = L.
8.3. (Sandwich principles.) Let U be an open set in Rn, and let a be a point of U.
(a) Let f, g, h: U → R be real-valued functions such that f(x) ≤ g(x) ≤ h(x) for all x in
U. If f(a) = h(a)—let’s call the common value c—and if f and h are continuous at a,
prove that g(a) = c and that g is continuous at a as well.",Multivariable_Calculus_Shimamoto.pdf
"U. If f(a) = h(a)—let’s call the common value c—and if f and h are continuous at a,
prove that g(a) = c and that g is continuous at a as well.
(b) Let f, g, hbe real-valued functions defined on U, except possibly at the point a, such
that f(x) ≤ g(x) ≤ h(x) for all x in U, except possibly when x = a. If lim x→a f(x) =
limx→a h(x) = L, prove that limx→a g(x) = L, too.",Multivariable_Calculus_Shimamoto.pdf
"80 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES
Addendum: Folding a hypar (Exercise 1.9)
1. Take your square piece of paper, and fold and unfold along each of the diagonals.
Step 1
2. Turn the paper over.
3. Fold the bottom edge of the paper to the center point, but only crease the part between the
diagonals. Unfold back to the square.
Step 3 Step 4
4. Repeat for the other three sides.
5. Now, fold the bottom edge up to the upper crease line that you just made, 3/4 of the way
up, again only creasing the part between the diagonals. This is a fairly short crease. Unfold
back to the square.
Step 5 Step 6
6. Fold the bottom edge to the lower crease line (the one you made in step 3), once again only
creasing between the diagonals. Unfold back to the square.
7. Repeat for all four sides. At this point, the square should be divided into four concentric
square rings. See the figure on the next page.
8. Turn the paper over.",Multivariable_Calculus_Shimamoto.pdf
"square rings. See the figure on the next page.
8. Turn the paper over.
9. The goal now is to create additional creases halfway between the ones constructed so far, for
a total of eight concentric square rings. You can do this by folding the bottom edge up to
previously constructed folds, but use only every other fold. That is, fold the bottom edge
up to the existing creases 7/8, 5/8, 3/8, and 1/8 of the way to the top, again only creasing
between the diagonals. Do this for all four sides.",Multivariable_Calculus_Shimamoto.pdf
"3.9. EXERCISES FOR CHAPTER 3 81
Step 7
Step 9
10. Turn the paper over. The creases should alternate as you move from one concentric square
to the next: mountain, valley, mountain, valley, . . . .
11. Now, try to fold all the creases. It’s easiest to start from the outer ring and work your way
in, pinching in at the corners and trying to compress the faces of adjacent rings together as
you move towards the center. Half of the short diagonal creases between the rings need to
be reversed so that the rings nest inside each other, though the paper can be coaxed to do
this without too much trouble. In the end, you should get something shaped like an X, or a
four-armed starfish.
12. Open up the arms of the starfish. The paper will naturally assume the hypar shape.",Multivariable_Calculus_Shimamoto.pdf
82 CHAPTER 3. REAL-VALUED FUNCTIONS: PRELIMINARIES,Multivariable_Calculus_Shimamoto.pdf
"Chapter 4
Real-valued functions: differentiation
We now extend the notion of derivative from real-valued functions of one variable, as in first-year
calculus, to real-valued functions of n variables f : U → R, where U is an open set in Rn. The
most natural choice would be to copy the one-variable definition verbatim and let the derivative
at a point a be limx→a
f(x)−f(a)
x−a . Unfortunately, this involves dividing by a vector, which does not
make sense.
One potential remedy would be to consider lim x→a
f(x)−f(a)
∥x−a∥ instead. Here, it turns out that
the limit may fail to exist even for very simple functions. For example, for the function f(x) = x of
one variable, this would be limx→a x−a
|x−a| which approaches +1 from the right and −1 from the left.
So we must look harder to find a formulation of the one-variable derivative that can be generalized.
4.1 The first-order approximation
In first-year calculus, we think of a function f as being differentiable at a point a if it has a good",Multivariable_Calculus_Shimamoto.pdf
"4.1 The first-order approximation
In first-year calculus, we think of a function f as being differentiable at a point a if it has a good
tangent line approximation:
f(x) ≈ ℓ(x) when x is near a,
where y = ℓ(x) = mx + b is the tangent line to the graph of f at the point ( a, f(a)). See Figure
4.1.
Figure 4.1: Approximating a differentiable function f(x) by one of its tangent lines ℓ(x)
In fact, the tangent line is given by
ℓ(x) = f(a) + f′(a) · (x − a), (4.1)
so m = f′(a) and b = f(a) − f′(a) · a.
83",Multivariable_Calculus_Shimamoto.pdf
"84 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
By definition, f′(a) = lim x→a
f(x)−f(a)
x−a , assuming the limit exists. This can be rewritten as
0 = limx→a
f(x)−f(a)
x−a − f′(a)


= limx→a
f(x)−f(a)−f′(a)·(x−a)
x−a , or, using equation (4.1), as:
lim
x→a
f(x) − ℓ(x)
x − a = 0. (4.2)
In particular, when x is near a, f(x) − ℓ(x) must be much much smaller than x − a in order for
f(x)−ℓ(x)
x−a to be near 0. Note that f(x) − ℓ(x) is the error in using the tangent line to approximate
f, so, in order for f to be differentiable at a, not only must this error go to 0 as x approaches a, it
must go to 0 much faster than x − a. This is the principle we generalize in moving to functions of
more than one variable.
Definition. A function ℓ: Rn → Rm is called an affine function if it has the formℓ(x) = T(x)+b,
where:
• T : Rn → Rm is a linear transformation and
• b is a vector in Rm.
Let U be an open set in Rn and f : U → R a real-valued function. Provisionally, we’ll say",Multivariable_Calculus_Shimamoto.pdf
"• b is a vector in Rm.
Let U be an open set in Rn and f : U → R a real-valued function. Provisionally, we’ll say
that f is differentiable at a point a of U if there exists a real-valued affine function ℓ: Rn → R,
ℓ(x) = T(x) + b, such that:
lim
x→a
f(x) − ℓ(x)
∥x − a∥ = 0. (4.3)
This is the generalization of equation (4.2).
We can pin down the best candidate for ℓ much more precisely. For instance, as ℓ is meant
to approximate f near a, it should be true that f(a) = ℓ(a), that is, f(a) = T(a) + b. Thus
b = f(a) − T(a), and ℓ(x) = T(x) + f(a) − T(a). Using the fact that T is linear, this becomes:
ℓ(x) = f(a) + T(x − a).
This happens to have the same form as the tangent line (4.1), which seems like a good sign.
Substituting into equation (4.3), we say, still provisionally, that f is differentiable at a if there
exists a linear transformation T : Rn → R such that:
lim
x→a
f(x) − f(a) − T(x − a)
∥x − a∥ = 0. (4.4)",Multivariable_Calculus_Shimamoto.pdf
"exists a linear transformation T : Rn → R such that:
lim
x→a
f(x) − f(a) − T(x − a)
∥x − a∥ = 0. (4.4)
There is really only one choice for what T can be as well. For, as a linear transformation, T is
represented by a matrix, in this case, a 1 by n matrix A =

m1 m2 . . . mn

, where mj = T(ej)
for j = 1, 2, . . . , n. (See Proposition 1.7.) We shall determine the values of the entries mj.
Suppose that x approaches a in the x1-direction, i.e., let x = a + he1, and let h go to 0. Then
x − a = he1, and ∥x − a∥ = |h|. Assume for a moment that h >0. In order for equation (4.4) to
hold, it must be true that:
lim
h→0
f(a + he1) − f(a) − T(he1)
h = 0;
or, using the linearity of T:
lim
h→0
f(a + he1) − f(a) − hT(e1)
h = 0. (4.5)",Multivariable_Calculus_Shimamoto.pdf
"4.1. THE FIRST-ORDER APPROXIMATION 85
If h <0, then the denominator in equation (4.4) is |h| = −h. The minus sign can be factored out
and canceled, so equation (4.5) still holds. This can be solved for T(e1):
m1 = T(e1) = lim
h→0
f(a + he1) − f(a)
h
= lim
h→0
f(a1 + h, a2, . . . , an) − f(a1, a2, . . . , an)
h .
This limit of a difference quotient is the definition of the ordinary one-variable derivative where
x2, . . . , xn are held fixed atx2 = a2, . . . , xn = an and only x1 varies. It is called the partial derivative
∂f
∂x1
(a).
Similarly, by approaching a in the x2, . . . , xn directions, we find that m2 = ∂f
∂x2
(a), . . . , mn =
∂f
∂xn (a) and hence that A =
h
∂f
∂x1
(a) ∂f
∂x2
(a) . . . ∂f
∂xn (a)
i
.
Definition. For a real-valued function f : U → R defined on an open set U in Rn and a point a of
U:
1. If j = 1, 2, . . . , n, the partial derivative of f at a with respect to xj is defined by:
∂f
∂xj
(a) = lim
h→0
f(a + hej) − f(a)
h .",Multivariable_Calculus_Shimamoto.pdf
"U:
1. If j = 1, 2, . . . , n, the partial derivative of f at a with respect to xj is defined by:
∂f
∂xj
(a) = lim
h→0
f(a + hej) − f(a)
h .
Note that a+hej = (a1, . . . , aj +h, . . . , an), so a+hej and a differ only in the jth coordinate.
Thus the partial derivative is defined by the one-variable difference quotient for the derivative
with variable xj.
Other common notations for the partial derivative are fxj (a) and (Djf)(a).
2. Df(a) is defined to be the 1 by n matrix Df(a) =
h
∂f
∂x1
(a) ∂f
∂x2
(a) . . . ∂f
∂xn (a)
i
. The
corresponding vector ∇f(a) =
 ∂f
∂x1
(a), ∂f
∂x2
(a), . . . ,∂f
∂xn (a)


in Rn is called the gradient of
f. (The intent of the notation might be clearer if these objects were denoted by ( Df)(a) and
(∇f)(a), respectively, but the extra parentheses are usually omitted.)
Example 4.1. Let f(x, y) = x3 + 2x2y − 3y2, and let a = (2, 1). To find ∂f
∂x (a), we fix y = 1 and
differentiate with respect to x. Since f(x, 1) = x3 + 2x2 − 3, this gives ∂f",Multivariable_Calculus_Shimamoto.pdf
"∂x (a), we fix y = 1 and
differentiate with respect to x. Since f(x, 1) = x3 + 2x2 − 3, this gives ∂f
∂x (x, 1) = 3 x2 + 4x and
then ∂f
∂x (2, 1) = 12 + 8 = 20.
Similarly, for the partial derivative at a with respect to y, f(2, y) = 8 + 8y − 3y2, so ∂f
∂y (2, y) =
8 − 6y and ∂f
∂y (2, 1) = 8 − 6 = 2.
In practice, this is not how partial derivatives are calculated usually. Instead, one works directly
with the general formula for f. For instance, to find the partial derivative with respect to x,
differentiate with respect to x, thinking of all other variables—in this case, only y—as constant:
∂f
∂x = 3x2 + 4xy − 0 = 3x2 + 4xy, so ∂f
∂x (2, 1) = 12 + 8 = 20 as before. Likewise, ∂f
∂y = 2x2 − 6y, and
∂f
∂y (2, 1) = 8 − 6 = 2.
In any case, Df(a) =

20 2

, and ∇f(a) = (20, 2).
We now have all the ingredients needed to state the formal definition of differentiability, as
motivated by equation (4.4).",Multivariable_Calculus_Shimamoto.pdf
"86 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Definition. Let U be an open set in Rn, f : U → R a real-valued function, and a a point of U.
Then f is said to be differentiable at a if:
lim
x→a
f(x) − f(a) − Df(a) · (x − a)
∥x − a∥ = 0.
When this happens, the matrix Df(a) is called the derivative of f at a. It is also known as the
Jacobian matrix. The affine function ℓ(x) = f(a)+ Df(a)·(x−a) is called the first-order affine
approximation of f at a. (Note the resemblance to the one-variable tangent line approximation
(4.1).)
The product in the numerator of the definition is the matrix product:
Df(a) · (x − a) =
h
∂f
∂x1
(a) ∂f
∂x2
(a) . . . ∂f
∂xn (a)
i


x1 − a1
x2 − a2
...
xn − an

.
It could also be written as a dot product: ∇f(a) · (x − a).
The definition generalizes equation (4.2) for one variable in that it says thatf is differentiable at
a if the error in using the first-order approximation ℓ goes to 0 faster than ∥x−a∥ as x approaches",Multivariable_Calculus_Shimamoto.pdf
"a if the error in using the first-order approximation ℓ goes to 0 faster than ∥x−a∥ as x approaches
a. In particular, the first-order approximation is a good approximation. This is worth emphasizing:
the existence of a good first-order approximation is often the most important way to think of what
it means for a function to be differentiable. Note that the derivative Df(a) itself is no longer just
a number the way it is in first-year calculus, but rather a matrix or, even better, the linear part of
a good affine approximation of f near a.
Example 4.2. Let f : R2 → R be f(x, y) = x2 + y2, and let a = (1, 2). Is f differentiable at a?
We calculate the various elements that go into the definition. First, ∂f
∂x = 2x and ∂f
∂y = 2y, so
Df(x, y) =

2x 2y

and Df(a) = Df(1, 2) =

2 4

. Also, f(a) = 5, and x − a =
x − 1
y − 2

. Thus
we need to check if:
lim
(x,y)→(1,2)
x2 + y2 − 5 −

2 4
x − 1
y − 2

p
(x − 1)2 + (y − 2)2 = 0.",Multivariable_Calculus_Shimamoto.pdf
"x − 1
y − 2

. Thus
we need to check if:
lim
(x,y)→(1,2)
x2 + y2 − 5 −

2 4
x − 1
y − 2

p
(x − 1)2 + (y − 2)2 = 0.
Let’s simplify the numerator, completing the square at an appropriate stage:
x2 + y2 − 5 −

2 4
x − 1
y − 2

= x2 + y2 − 5 −

2(x − 1) + 4(y − 2)


= x2 + y2 − 5 − (2x − 2 + 4y − 8)
= x2 − 2x + y2 − 4y + 5
= (x2 − 2x + 1) − 1 + (y2 − 4y + 4) − 4 + 5
= (x − 1)2 + (y − 2)2.
Therefore:
lim
(x,y)→(1,2)
x2 + y2 − 5 −

2 4
x − 1
y − 2

p
(x − 1)2 + (y − 2)2 = lim
(x,y)→(1,2)
(x − 1)2 + (y − 2)2
p
(x − 1)2 + (y − 2)2
= lim
(x,y)→(1,2)
p
(x − 1)2 + (y − 2)2.",Multivariable_Calculus_Shimamoto.pdf
"4.1. THE FIRST-ORDER APPROXIMATION 87
The function in this last expression is continuous, as it is a composition of sums and products of
continuous pieces. Thus the value of the limit is simply the value of the function at the point (1, 2).
In other words:
lim
(x,y)→(1,2)
x2 + y2 − 5 −

2 4
x − 1
y − 2

p
(x − 1)2 + (y − 2)2 =
p
(1 − 1)2 + (2 − 2)2 = 0.
This shows that the answer is: yes, f is differentiable at a = (1, 2).
In addition, the preceding calculations show that the first-order approximation of f(x, y) =
x2 + y2 at a = (1, 2) is:
ℓ(x, y) = f(a) + Df(a) · (x − a)
= 5 +

2 4
x − 1
y − 2

= 2x + 4y − 5. (4.6)
For instance, ( x, y) = (1 .05, 1.95) is near a, and f(1.05, 1.95) = 1 .052 + 1.952 = 4 .905 while
ℓ(1.05, 1.95) = 2(1.05) + 4(1.95) − 5 = 4.9. The values are pretty close.
The same sort of reasoning can be used to show that, more generally, the function f(x, y) =
x2 + y2 is differentiable at every point of R2. This is Exercise 1.16.
Example 4.3. Let f(x, y) =
( xy",Multivariable_Calculus_Shimamoto.pdf
"x2 + y2 is differentiable at every point of R2. This is Exercise 1.16.
Example 4.3. Let f(x, y) =
( xy
x2+y2 if (x, y) ̸= (0, 0),
0 if ( x, y) = (0, 0)
. Its graph is shown in Figure 4.2. Is f
differentiable at a = (0, 0)?
Figure 4.2: The graph z = xy
x2+y2
Using the definition requires knowing the partial derivatives at (0 , 0). To calculate them, we
could apply the quotient rule to the formula that defines f, but the results would be valid only
when (x, y) ̸= (0, 0), which is precisely what we don’t need. So instead we go back to the basic idea
behind partial derivatives, namely, that they are one-variable derivatives where all variables but
one are held constant. For example, to find ∂f
∂x (0, 0), fix y = 0 in f(x, y) and look at the resulting
function of x. By the formula for f, f(x, 0) = x·0
x2+02 = 0 if x ̸= 0. This expression also holds when
x = 0: by definition, f(0, 0) = 0. Thus f(x, 0) = 0 for all x. Differentiating this with respect to x
gives ∂f
∂x (x, 0) = d
dx

0

",Multivariable_Calculus_Shimamoto.pdf
"x = 0: by definition, f(0, 0) = 0. Thus f(x, 0) = 0 for all x. Differentiating this with respect to x
gives ∂f
∂x (x, 0) = d
dx

0


= 0, whence ∂f
∂x (0, 0) = 0. A symmetric calculation results in ∂f
∂y (0, 0) = 0.
Hence Df(0, 0) =

0 0

.",Multivariable_Calculus_Shimamoto.pdf
"88 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Moreover, x − a =
x − 0
y − 0

=
x
y

, and ∥x − a∥ =
p
x2 + y2. Thus, according to the definition
of differentiability, we need to check if:
lim
(x,y)→(0,0)
xy
x2+y2 − 0 −

0 0
x
y

p
x2 + y2 = 0,
i.e., if lim(x,y)→(0,0)
xy
(x2+y2)3/2 = 0.
For this, suppose we approach the origin along the line y = x. Then we are looking at the
behavior of x·x
(x2+x2)3/2 = x2
23/2|x|3 = 1
23/2|x|. This blows up as x goes to 0. It certainly doesn’t
approach 0. Hence there’s no way that lim (x,y)→(0,0)
xy
(x2+y2)3/2 can equal 0 (in fact, the limit does
not exist), so f is not differentiable at (0 , 0).
This example shows that a function of more than one variable can fail to be differentiable at a
point even though all of its partial derivatives exist there. This differs from the one-variable case.
4.2 Conditions for differentiability
We now look for ways to tell if a function is differentiable without having to do all the work of",Multivariable_Calculus_Shimamoto.pdf
"4.2 Conditions for differentiability
We now look for ways to tell if a function is differentiable without having to do all the work of
going through the definition. As we saw in the last example, calculating the partial derivatives is
not enough. That’s bad news, because calculating partial derivatives is pretty easy. The good news
is that we are about to obtain results that show that the examples in the last section could have
been solved much more quickly using means other than the definition.
For the remainder of this section:
U denotes an open set in Rn, f : U → R is a real-valued function defined on U, and
a is a point of U.
We won’t keep repeating this.
For instance, in Example 4.3 above, we asked if the function
f(x, y) =
( xy
x2+y2 if (x, y) ̸= (0, 0),
0 if ( x, y) = (0, 0)
(4.7)
were differentiable at (0 , 0). In fact, we had shown before that f is not continuous at (0 , 0) (see",Multivariable_Calculus_Shimamoto.pdf
"0 if ( x, y) = (0, 0)
(4.7)
were differentiable at (0 , 0). In fact, we had shown before that f is not continuous at (0 , 0) (see
Example 3.19 in Chapter 3). In first-year calculus, discontinuous functions cannot be differentiable.
If the same were true for functions of more than one variable, we could have saved ourselves a lot
of work. We state the result in contrapositive form.
Proposition 4.4. If f is differentiable at a, then f is continuous at a.
Proof. We just outline the idea and leave the details for the exercises (Exercise 2.3). If f is
differentiable at a, then limx→a(f(x) − ℓ(x)) = 0, where ℓ is the first-order approximation at a. In
other words:
lim
x→a
(f(x) − f(a) − Df(a) · (x − a)) = 0.
The matrix product term Df(a) ·(x −a) goes to 0 as x approaches a (it’s a continuous function of
x), so the last line becomes lim x→a(f(x) − f(a) − 0) = 0. Hence lim x→a f(x) = f(a). As a result,
f is continuous at a.",Multivariable_Calculus_Shimamoto.pdf
"4.3. THE MEAN VALUE THEOREM 89
Example 4.5. Returning to function (4.7) above, we could have simply said from the start that f
is not continuous at (0, 0) and therefore is not differentiable there either, avoiding the definition of
differentiability entirely.
Conversely, we know from first-year calculus that a function can be continuous without being
differentiable. The absolute value function f(x) = |x| is the standard example. It is continuous,
but, because its graph has a corner, it is not differentiable at the origin. Sometimes, we can use
the failure of a one-variable derivative to tell us about the differentiability of a function of more
than one variable.
Proposition 4.6. If some partial derivative ∂f
∂xj
does not exist at a, then f is not differentiable at
a.
Proof. This follows simply because if ∂f
∂xj
(a) doesn’t exist, then neither does Df(a), so the condition
for differentiability cannot be satisfied.",Multivariable_Calculus_Shimamoto.pdf
"Proof. This follows simply because if ∂f
∂xj
(a) doesn’t exist, then neither does Df(a), so the condition
for differentiability cannot be satisfied.
Example 4.7. One analogue of the absolute value function is the function f : R2 → R, f(x, y) =
∥(x, y)∥ =
p
x2 + y2. We have seen that its graph is a cone with vertex at the origin (Example
3.2).
To find ∂f
∂x at the origin, we fix y = 0 to get f(x, 0) =
√
x2 = |x|. The derivative with respect to
x does not exist at x = 0, so, by definition, ∂f
∂x (0, 0) does not exist either. Thus f is not differentiable
at (0, 0) (though it is continuous there).
We have saved the most useful criterion for differentiability for last. To understand why it is
true, however, we take a slight detour.
4.3 The mean value theorem
We devote this section to reviewing an important, and possibly underappreciated, result from
first-year calculus.
Theorem 4.8 (Mean Value Theorem). Let f : [a, b] → R be a continuous function that is differen-",Multivariable_Calculus_Shimamoto.pdf
"first-year calculus.
Theorem 4.8 (Mean Value Theorem). Let f : [a, b] → R be a continuous function that is differen-
tiable on the open interval (a, b). Then there exists a point c in (a, b) such that:
f(b) − f(a) = f′(c) · (b − a).
The theorem relates the difference in function values f(b) − f(a) to the difference in domain
values b −a. The relation can be useful even if one cannot completely pin down the location of the
point c (in fact, one rarely tries). For instance, if f(x) = sin x, then |f′(c)| = |cos c| ≤1 regardless
of what c is. Hence the mean value theorem gives |sin b − sin a| ≤ |b − a| for all a, b. By the same
reasoning, |cos b − cos a| ≤ |b − a| as well.
There is an n-variable version of the mean value theorem that follows from results we prove
later (see Exercise 5.5), but for now we show how the one-variable theorem can be used to study a
difference in function values for a function of two variables by allowing only one variable to change",Multivariable_Calculus_Shimamoto.pdf
"difference in function values for a function of two variables by allowing only one variable to change
at a time and using partial derivatives.
Example 4.9. Let f(x, y) be a continuous real-valued function of two variables that is defined
on an open ball B in R2 and whose partial derivatives ∂f
∂x and ∂f
∂y exist at every point of B. Let
a = (a1, a2) and b = (b1, b2) be points of B. We show that there exist points p and q in B such
that ∥p − a∥ ≤ ∥b − a∥, ∥q − a∥ ≤ ∥b − a∥, and:
f(b) − f(a) = ∂f
∂x (p) · (b1 − a1) + ∂f
∂y (q) · (b2 − a2). (4.8)",Multivariable_Calculus_Shimamoto.pdf
"90 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Figure 4.3: Going from a to b by way of c
See Figure 4.3.
To prove this, we introduce the intermediate point c = (b1, a2) and think of going from a to b
by first going from a to c and then from c to b, as in Figure 4.3. Only one variable varies along
each of the segments, so by writing
f(b) − f(a) = (f(b) − f(c)) + (f(c) − f(a)), (4.9)
the mean value theorem applies to each of the differences on the right.
That is, in the second difference, using x as variable and y = a2 fixed, the mean value theorem
implies that there exists a point p = (p1, a2) on the line segment between a and c such that:
f(c) − f(a) = ∂f
∂x (p) · (b1 − a1). (4.10)
Similarly, for the first difference, usingy as variable andx = b1 fixed, there exists a pointq = (b1, q2)
on the line segment between c and b such that:
f(b) − f(c) = ∂f
∂y (q) · (b2 − a2). (4.11)
Substituting (4.10) and (4.11) into equation (4.9) results in f(b)−f(a) = ∂f
∂y (q)·(b2 −a2)+ ∂f
∂x (p)·",Multivariable_Calculus_Shimamoto.pdf
"f(b) − f(c) = ∂f
∂y (q) · (b2 − a2). (4.11)
Substituting (4.10) and (4.11) into equation (4.9) results in f(b)−f(a) = ∂f
∂y (q)·(b2 −a2)+ ∂f
∂x (p)·
(b1 − a1), which is the desired conclusion (4.8).
4.4 The C1 test
We present the result that often gives the simplest way to show that a function is differentiable. It
uses partial derivatives.
Definition. Let U be an open set in Rn. A real-valued function f : U → R all of whose partial
derivatives ∂f
∂x1
, ∂f
∂x2
, . . . ,∂f
∂xn are continuous on U is said to be of class C1, or continuously
differentiable, on U.
Theorem 4.10 (The C1 test). If f is of class C1 on U, then f is differentiable at every point of
U.
Proof. We give the proof in the case of two variables, that is, U ⊂ R2. This contains the main new
idea, and we leave for the reader the generalization to n variables, which is a relatively straightfor-
ward extension.",Multivariable_Calculus_Shimamoto.pdf
"4.4. THE C1 TEST 91
We use the definition of differentiability. Let a be a point of U. We first analyze the error
f(x) −f(a) −Df(a) ·(x−a) in the first-order approximation of f at a. Since U is an open set and
we are interested only in what happens when x is near a, we may assume that x lies in an open
ball about a that is contained in U. Say a = (c, d) and x = (x, y). By the result we just obtained
in Example 4.9, there are points p and q such that ∥p − a∥ ≤ ∥x − a∥, ∥q − a∥ ≤ ∥x − a∥, and:
f(x) − f(a) = ∂f
∂x (p) · (x − c) + ∂f
∂y (q) · (y − d).
Also:
Df(a) · (x − a) =
h
∂f
∂x (a) ∂f
∂y (a)
ix − c
y − d

= ∂f
∂x (a) · (x − c) + ∂f
∂y (a) · (y − d).
Hence:
f(x) − f(a) − Df(a) · (x − a) =
∂f
∂x (p) − ∂f
∂x (a)


· (x − c) +
∂f
∂y (q) − ∂f
∂y (a)


· (y − d),
so:
|f(x) − f(a) − Df(a) · (x − a)|
∥x − a∥
=
|
∂f
∂x (p) − ∂f
∂x (a)


· (x − c) +
∂f
∂y (q) − ∂f
∂y (a)


· (y − d)|
∥x − a∥
≤

∂f
∂x (p) − ∂f
∂x (a)
 · |x − c|
∥x − a∥ +

∂f
∂y (q) − ∂f
∂y (a)
 · |y − d|",Multivariable_Calculus_Shimamoto.pdf
"∂f
∂y (q) − ∂f
∂y (a)


· (y − d)|
∥x − a∥
≤

∂f
∂x (p) − ∂f
∂x (a)
 · |x − c|
∥x − a∥ +

∂f
∂y (q) − ∂f
∂y (a)
 · |y − d|
∥x − a∥,
where the last step uses the triangle inequality. Note that |x−c|
∥x−a∥ =
√
(x−c)2
√
(x−c)2+(y−d)2 ≤ 1. Similarly,
|y−d|
∥x−a∥ ≤ 1. Therefore:
|f(x) − f(a) − Df(a) · (x − a)|
∥x − a∥ ≤

∂f
∂x (p) − ∂f
∂x (a)
 +

∂f
∂y (q) − ∂f
∂y (a)
. (4.12)
At last, we bring in the assumption that f has continuous partial derivatives. Since ∥p − a∥ ≤
∥x−a∥ and ∥q−a∥ ≤ ∥x−a∥, it follows that, as x approaches a, so do p and q. By the continuity
of the partial derivatives, this means that ∂f
∂x (p) approaches ∂f
∂x (a) and ∂f
∂y (q) approaches ∂f
∂y (a).
Hence, in the limit as x goes to a, the right side of (4.12) approaches 0, and therefore so must the
left. By definition, f is differentiable at a.
Example 4.11. In Example 4.2, we showed that the function f : R2 → R, f(x, y) = x2 + y2, is",Multivariable_Calculus_Shimamoto.pdf
"left. By definition, f is differentiable at a.
Example 4.11. In Example 4.2, we showed that the function f : R2 → R, f(x, y) = x2 + y2, is
differentiable at the point a = (1, 2) by using the definition of differentiability. It is much simpler to
note that ∂f
∂x = 2x and ∂f
∂y = 2y are both continuous on R2. Hence, by the C1 test, f is differentiable
at every point of R2.
Example 4.12. The other example we looked at carefully is the function:
f(x, y) =
( xy
x2+y2 if (x, y) ̸= (0, 0),
0 if ( x, y) = (0, 0)",Multivariable_Calculus_Shimamoto.pdf
"92 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
of (4.7). We saw that f is not differentiable at the origin. On the other hand, at points other than
the origin, the partial derivatives can be found using the quotient rule. This gives:
∂f
∂x = (x2 + y2) · y − xy · 2x
(x2 + y2)2 = y3 − x2y
(x2 + y2)2 = y(y2 − x2)
(x2 + y2)2
and ∂f
∂y = (x2 + y2) · x − xy · 2y
(x2 + y2)2 = x(x2 − y2)
(x2 + y2)2 .
As algebraic combinations of x and y, both of these partial derivatives are continuous on the open
set R2 − {(0, 0)}. As a result, the C1 test says that f is differentiable at all points other than the
origin.
4.5 The Little Chain Rule
In one-variable calculus, the derivative measures the rate of change of a function. When there is
more than one variable, the derivative is no longer just a number, but it still has something to say
about rates of change.
Let U be an open set in Rn, and let f : U → R be a real-valued function defined on U. Suppose",Multivariable_Calculus_Shimamoto.pdf
"about rates of change.
Let U be an open set in Rn, and let f : U → R be a real-valued function defined on U. Suppose
that α: I → U is a path in U defined on an interval I in R. The composition f ◦ α: I → R is a
real-valued function of one variable that describes how f behaves along the path. We examine the
rate of change of this composition.
Let t0 be a point of I, and let a = α(t0). If f is differentiable at a, then its first-order
approximation applies: f(x) ≈ f(a) + ∇f(a) · (x − a), or:
f(x) − f(a) ≈ ∇f(a) · (x − a) when x is near a.
Here, we use the gradient form of the derivative.
Assume in addition that α is differentiable at t0. Then α is continuous at t0, so, when t is near
t0, α(t) is near α(t0). Substituting x = α(t) and a = α(t0) into the previous approximation gives:
f(α(t)) − f(α(t0)) ≈ ∇f(α(t0)) · (α(t) − α(t0)) when t is near t0.
Thus f(α(t))−f(α(t0))
t−t0
≈ ∇f(α(t0))· α(t)−α(t0)
t−t0
. In the limit as t approaches t0, the left side approaches",Multivariable_Calculus_Shimamoto.pdf
"Thus f(α(t))−f(α(t0))
t−t0
≈ ∇f(α(t0))· α(t)−α(t0)
t−t0
. In the limit as t approaches t0, the left side approaches
the one-variable derivative ( f ◦ α)′(t0) and the right approaches ∇f(α(t0)) · α′(t0). Writing t in
place of t0, this gives the following result.
Theorem 4.13 (Little Chain Rule) . Let U be an open set in Rn, α: I → U a path in U defined
on an open interval I in R, and f : U → R a real-valued function. If α is differentiable at t and f
is differentiable at α(t), then f ◦ α is differentiable at t and:
(f ◦ α)′(t) = ∇f(α(t)) · α′(t).
In words, the derivative of the composition is “gradient dot velocity.” While we outlined the
idea why this is true, a complete proof of the result as stated entails a more careful treatment of
the approximations involved. The details appear in the exercises (see Exercise 5.6).
4.6 Directional derivatives
In this section and the next, we present some interpretations of the gradient that follow from the",Multivariable_Calculus_Shimamoto.pdf
"4.6 Directional derivatives
In this section and the next, we present some interpretations of the gradient that follow from the
Little Chain Rule. The first concerns the rate of change of a function of n variables in a given",Multivariable_Calculus_Shimamoto.pdf
"4.6. DIRECTIONAL DERIVATIVES 93
direction. Let a be a point of an open set U in Rn. To represent a direction going away from a, we
choose a unit vector u pointing in that direction. We travel in that direction on the line through
a and parallel to u, which is parametrized by α(t) = a + tu. The parametrization has velocity
α′(t) = u and hence constant speed ∥u∥ = 1.
If f : U → R is a real-valued function, then the derivative ( f ◦ α)′(t) is the rate of change of f
along this line. In particular, since α(0) = a, we think of ( f ◦ α)′(0) as the rate of change at a in
the direction of u.
Definition. Given the input described above, ( f ◦ α)′(0) is called the directional derivative of
f at a in the direction of the unit vector u , denoted by (Duf)(a). If v is any nonzero vector
in Rn, we define the directional derivative in the direction of v to be (Duf)(a), where u = 1
∥v∥v is
the unit vector in the direction of v.
Directional derivatives are easy to compute, for by the Little Chain Rule:",Multivariable_Calculus_Shimamoto.pdf
"∥v∥v is
the unit vector in the direction of v.
Directional derivatives are easy to compute, for by the Little Chain Rule:
(f ◦ α)′(0) = ∇f(α(0)) · α′(0) = ∇f(a) · u.
In other words:
Proposition 4.14. If f is differentiable at a and u is a unit vector, then the directional derivative
is given by:
(Duf)(a) = ∇f(a) · u.
As a result, ( Duf)(a) = ∥∇f(a)∥∥u∥cos θ = ∥∇f(a)∥cos θ, where θ is the angle between
∇f(a) and the direction vector u. Keeping a fixed but varying the direction, we see that (Duf)(a)
is maximized when θ = 0, i.e., when u points in the same direction as ∇f(a), and that, in this
direction, (Duf)(a) = ∥∇f(a)∥.
Corollary 4.15. The maximum directional derivative of f at a:
• occurs in the direction of ∇f(a) and
• has a value of ∥∇f(a)∥.
Sometimes, weather reports mention the “temperature gradient.” Usually, this does not refer to
the gradient vector that we have been studying but rather to one of the aspects of the gradient in",Multivariable_Calculus_Shimamoto.pdf
"the gradient vector that we have been studying but rather to one of the aspects of the gradient in
the corollary, that is, the magnitude and/or direction of the maximum rate of temperature increase.
Example 4.16. Let f : R3 → R be f(x, y, z) = x2 + y2 + z2, and let a = (1, 2, 3).
(a) Find the directional derivative of f at a in the direction of v = (2, −1, 1).
(b) Find the direction and rate of maximum increase at a.
(a) First, ∇f = (2x, 2y, 2z), so ∇f(a) = (2 , 4, 6). Next, the unit vector in the direction of v is
u = 1
∥v∥v = 1√
6 (2, −1, 1). Thus:
(Duf)(a) = ∇f(a) · u
= (2, 4, 6) · 1√
6(2, −1, 1) = 1√
6(4 − 4 + 6) = 6√
6 =
√
6.
(b) The direction of maximum increase is the direction of ∇f(a) = (2, 4, 6), that is, the direction of
the unit vector u = 1√4+16+36 (2, 4, 6) = 1√
56 (2, 4, 6) = 1√
14 (1, 2, 3). The rate of maximum increase
is ∥∇f(a)∥ =
√
56 = 2
√
14.",Multivariable_Calculus_Shimamoto.pdf
"94 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
4.7 ∇f as normal vector
Let f : U → R be a differentiable real-valued function of n variables defined on an open set U in
Rn, and let S be the level set of f corresponding to f = c. Let a be a point of S, and let v be a
vector tangent to S at a. We take this to mean that v can be realized as the velocity vector of a
path in S, that is, there is a differentiable path α: I → Rn that:
• lies in S, i.e., α(t) ∈ S for all t,
• passes through a, say a = α(t0) for some t0 in I, and
• has velocity v at a, i.e., α′(t0) = v.
This is illustrated in Figure 4.4. Then f(α(t)) = c for all t, so ( f ◦ α)′(t) = c′ = 0. By the
Figure 4.4: A level set S of f and a vector v tangent to it
Little Chain Rule, this can be written ∇f(α(t)) · α′(t) = 0. In particular, when t = t0, we obtain
∇f(a)·v = 0. This is true for all vectors v tangent to S at a, which has the following consequence.",Multivariable_Calculus_Shimamoto.pdf
"∇f(a)·v = 0. This is true for all vectors v tangent to S at a, which has the following consequence.
Proposition 4.17. If a is a point on a level set S of f, then ∇f(a) is a normal vector to S at a.
Example 4.18. Find an equation for the plane tangent to the graph z = x2 + y2 at the point
(1, 2, 5).
As with the equation of any plane, it suffices to find a point p on the plane and a normal vector
n. As a point, we can use the point of tangency, p = (1, 2, 5). To find a normal vector, we rewrite
the graph as a level set. That is, rewrite z = x2 + y2 as x2 + y2 −z = 0. Then the graph is the level
set of the function of three variables F(x, y, z) = x2 + y2 − z corresponding to c = 0. In particular,
∇F(p) is a normal vector to the level set, that is, to the graph, at p, so it is normal to the tangent
plane as well. See Figure 4.5.
The rest is calculation: ∇F = (2x, 2y, −1), so n = ∇F(p) = (2 , 4, −1) is our normal vector.
Substituting into the equation n · x = n · p for a plane gives:",Multivariable_Calculus_Shimamoto.pdf
"Substituting into the equation n · x = n · p for a plane gives:
(2, 4, −1) · (x, y, z) = (2, 4, −1) · (1, 2, 5) = 2 + 8− 5 = 5,
or 2 x + 4y − z = 5.
Note that we just studied the graph of the function f(x, y) = x2 + y2 of two variables, and
we considered the point p = (1, 2, 5) = ( a, f(a)) on the graph, where a = (1, 2). In the remarks
following Example 4.2, we saw that the first-order approximation of f at a is given by ℓ(x, y) =
2x+4y−5 (see equation (4.6)). Thus the graph of the approximation ℓ has equation z = 2x+4y−5,
which is exactly the tangent plane we just found.",Multivariable_Calculus_Shimamoto.pdf
"4.7. ∇f AS NORMAL VECTOR 95
Figure 4.5: The tangent plane to z = x2 + y2 at p = (1, 2, 5)
In other words, geometrically, the approximation we get by using the tangent plane at (a, f(a))
to approximate the graph coincides with the first-order approximation ata. This extends a familiar
idea from first-year calculus, where tangent lines are used to obtain the first-order approximation
of a differentiable function. In fact, it is the idea we used to motivate our multivariable definition
of differentiability in the first place.
Example 4.19. Suppose that T(x, y) represents the temperature at the point ( x, y) in an open
set U of the plane. There are two natural families of curves associated with this function.
The first is the curves along which the temperature is constant. They are called isotherms
and are the level sets of T: all points on a single isotherm have the same temperature. See Figure
4.6.4",Multivariable_Calculus_Shimamoto.pdf
"and are the level sets of T: all points on a single isotherm have the same temperature. See Figure
4.6.4
Figure 4.6: Isotherms are level curves of temperature. Heat travels along curves that are always
orthogonal to the isotherms.
The second is the curves along which heat flows. The principle here is that, at every point, heat
travels in the direction in which the temperature is decreasing as rapidly as possible, from hot to
cold.
From our work on directional derivatives, we know that the maximum rate of decrease occurs
in the direction of −∇T. And we’ve just seen that ∇T is always normal to the level sets, the
isotherms. In other words, the isotherms and the curves of heat flow are mutually orthogonal
families of curves. If you have a map of the isotherms, you can visualize the flow of heat: always
move at right angles to the isotherms.
4Figure taken from https://www.noaa.gov/jetstream/surface-temperature-map by National Weather Service,
public domain.",Multivariable_Calculus_Shimamoto.pdf
"96 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
4.8 Higher-order partial derivatives
Once they start partial differentiating, most people can’t stop.
Example 4.20. Let f(x, y) = x3 − 3x2y4 + exy. Then:
∂f
∂x = 3x2 − 6xy4 + yexy
and ∂f
∂y = −12x2y3 + xexy.
Both of these are again real-valued functions of x and y, so they have partial derivatives of their
own. In the case of ∂f
∂x :
∂2f
∂x2 = ∂
∂x
∂f
∂x

= 6x − 6y4 + y2exy
and ∂2f
∂y ∂x= ∂
∂y
∂f
∂x

= −24xy3 + exy + xyexy.
Similarly, for ∂f
∂y :
∂2f
∂x ∂y= ∂
∂x
∂f
∂y

= −24xy3 + exy + xyexy
and ∂2f
∂y2 = ∂
∂y
∂f
∂y

= −36x2y2 + x2exy.
These are the second-order partial derivatives . They too have partial derivatives, and the
process could continue indefinitely to obtain partial derivatives of higher and higher order.
Perhaps the most striking part of the preceding calculation is that the two partial derivatives
∂2f
∂y ∂x and ∂2f
∂x ∂y turned out to be equal. They are called the mixed partials . To see whether",Multivariable_Calculus_Shimamoto.pdf
"∂2f
∂y ∂x and ∂2f
∂x ∂y turned out to be equal. They are called the mixed partials . To see whether
they might be equal in general, we return to the definition of a partial derivative as a limit of
difference quotients and try to describe the second-order partials in terms of the original function
f. In fact, what we are about to discuss is a method one might use to numerically approximate
the second-order partials.
Let a = (c, d). First, consider ∂2f
∂x ∂y = ∂
∂x
∂f
∂y


. The function being differentiated is ∂f
∂y . It is
being differentiated with respect to x, so we let x vary and hold y constant:
∂2f
∂x ∂y(c, d) =
 ∂
∂x
∂f
∂y


(c, d) ≈
∂f
∂y (c + h, d) − ∂f
∂y (c, d)
h .
The approximation is meant to hold when h is small. The terms in the numerator are both partial
derivatives of f with respect to y, so in each case we differentiate f, letting y vary and holding x
constant:
∂2f
∂x ∂y(c, d) ≈
f(c+h,d+k)−f(c+h,d)
k − f(c,d+k)−f(c,d)
k
h",Multivariable_Calculus_Shimamoto.pdf
"constant:
∂2f
∂x ∂y(c, d) ≈
f(c+h,d+k)−f(c+h,d)
k − f(c,d+k)−f(c,d)
k
h
≈ f(c + h, d+ k) − f(c + h, d) − f(c, d+ k) + f(c, d)
hk (4.13)
when h and k are small.",Multivariable_Calculus_Shimamoto.pdf
"4.8. HIGHER-ORDER PARTIAL DERIVATIVES 97
This expression is symmetric enough in h and k that perhaps you are willing to believe that
you would get the same thing when the order of differentiation is reversed. But if you are nervous
about it, here is the parallel calculation for ∂2f
∂y ∂x:
∂2f
∂y ∂x(c, d) =
 ∂
∂y
∂f
∂x


(c, d)
≈
∂f
∂x (c, d+ k) − ∂f
∂x (c, d)
k
≈
f(c+h,d+k)−f(c,d+k)
h − f(c+h,d)−f(c,d)
h
k
≈ f(c + h, d+ k) − f(c, d+ k) − f(c + h, d) + f(c, d)
hk . (4.14)
Indeed, comparing the approximations shows that they are the same, so:
∂2f
∂x ∂y(c, d) ≈ ∂2f
∂y ∂x(c, d).
A proper proof that the two sides are actually equal requires tightening up the approximations that
are involved. We leave the details for the exercises (Exercise 8.7), but the main idea is to use the
mean value theorem to show that the quotient Q = f(c+h,d+k)−f(c+h,d)−f(c,d+k)+f(c,d)
hk common to
approximations (4.13) and (4.14) is equal on the nose to the values of the mixed partials at nearby
points, that is:",Multivariable_Calculus_Shimamoto.pdf
"hk common to
approximations (4.13) and (4.14) is equal on the nose to the values of the mixed partials at nearby
points, that is:
∂2f
∂x ∂y(p) = Q = ∂2f
∂y ∂x(q), (4.15)
where p and q are points that are close to a but depend on h and k. As h and k go to 0, p and q
approach a. If the mixed partials were known to be continuous at a, then, in the limit, equation
(4.15) becomes:
∂2f
∂x ∂y(a) = ∂2f
∂y ∂x(a).
Thus the mixed partials are equal provided one invokes a requirement of continuity. In fact, there
are functions for which the mixed partials are not equal, so some sort of restriction of this type is
unavoidable.
For functions of more than two variables, the second-order mixed partials treat all but two vari-
ables as fixed, so we remain essentially in the two-variable case. In other words, if the mixed partials
are equal for functions of two variables, they are also equal for more than two. We summarize the
discussion with the following result.",Multivariable_Calculus_Shimamoto.pdf
"are equal for functions of two variables, they are also equal for more than two. We summarize the
discussion with the following result.
Theorem 4.21 (Equality of mixed partials) . Let f : U → R be a real-valued function defined on
an open set U in Rn, all of whose first and second-order partial derivatives are continuous on U.
(Such a function is said to be of class C2.) Then:
∂2f
∂xi ∂xj
= ∂2f
∂xj ∂xi
at all points of U and for all i, j.",Multivariable_Calculus_Shimamoto.pdf
"98 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Example 4.22. Find ∂4f
∂x ∂y2 ∂z if f(x, y, z) =

x2 + y3 + sin(xz)


(cos x − xz5).
The order of differentiation doesn’t matter, so we may as well choose one that simplifies the
bookkeeping. Differentiating with respect to x or z would involve the product rule, and who wants
that, so let’s differentiate first with respect to y: ∂f
∂y = 3y2(cos x − xz5). The next choices don’t
make much difference, but let’s try differentiating with respect to z: ∂2f
∂z ∂y = −15xy2z4. Then, with
respect to x: ∂3f
∂x ∂z ∂y= −15y2z4. And lastly with respect to y again:
∂4f
∂y ∂x ∂z ∂y= −30yz4.
By the equality of mixed partials, this is the same as the originally requested ∂4f
∂x ∂y2 ∂z .
4.9 Smooth functions
The C1 test states that functions with continuous partial derivatives are differentiable. The converse
is false, however. There are plenty of functions that are differentiable but don’t have continuous",Multivariable_Calculus_Shimamoto.pdf
"is false, however. There are plenty of functions that are differentiable but don’t have continuous
partials, though actually writing one down takes some determination. In the one-variable case,
f(x) = x2 sin(1/x) is an example. If f(0) is defined to be 0, then f is differentiable at a = 0 but f′
is not continuous there. Similarly, there are functions whose first-order partials are continuous but
whose second-order partials are not. There is an example in the exercises (Exercise 8.6), though
again it is somewhat contrived to make the point.
Most of the differentiable functions that we work with have continuous partial derivatives of
all possible orders. Such functions are said to be of class C∞. We refer to them simply as
smooth. We shall usually take them as the default and state our results involving derivatives for
smooth functions. We may not actually need all those higher-order partial derivatives to draw the",Multivariable_Calculus_Shimamoto.pdf
"smooth functions. We may not actually need all those higher-order partial derivatives to draw the
conclusion we are trying to reach, but keeping track of the degree of differentiability that is really
necessary in every case is not something we want to dwell on.
In addition, so far we have defined continuity and differentiability only for functions defined on
open sets. This is so that every point in the domain can be approached within the domain from
all possible directions, making the corresponding limiting processes easier to handle. Sometimes,
however, we are interested in domains that are not open. If D is a subset of Rn, not necessarily
open, we say that a function f : D → R is differentiable at a point a of D if it agrees with
a differentiable function near a, that is, there is an open ball B containing a and a function
g : B → R such that g(x) = f(x) for all x in B ∩ D and g is differentiable at a in the sense",Multivariable_Calculus_Shimamoto.pdf
"g : B → R such that g(x) = f(x) for all x in B ∩ D and g is differentiable at a in the sense
previously defined. We make the analogous modification for smoothness. For continuity, we keep
the definition as before but restrict our attention to points where f is defined: f is continuous at a
if, given any open ball B(f(a), ϵ) about f(a), there exists an open ball B(a, δ) about a such that
f(B(a, δ) ∩ D) ⊂ B(f(a), ϵ). We won’t dwell on these points either. In most concrete examples, f
is defined by a well-behaved formula that applies to an open set larger than the domain in which
we happen to be interested, so there isn’t a serious problem.
4.10 Max/min: critical points
In first-year calculus, one of the featured applications of the derivative is finding the largest and/or
smallest values that a function attains and where it attains them. Whenever one wants to do
something in an optimal way—for instance, the most productive or the least costly—and the",Multivariable_Calculus_Shimamoto.pdf
"something in an optimal way—for instance, the most productive or the least costly—and the
objective can be modeled by a smooth function, calculus is likely to be of great use. We discuss",Multivariable_Calculus_Shimamoto.pdf
"4.10. MAX/MIN: CRITICAL POINTS 99
this for functions of more than one variable. This is a big subject that overflows with applications.
Our treatment is not comprehensive. Instead, we have selected a couple of topics that illustrate
some of the similarities and differences that pertain to moving to the multivariable case.
Definition. Let f : U → R be a function defined on an open set U in Rn. f is said to have a local
maximum at a point a of U if there is an open ball B(a, r) centered at a such that f(a) ≥ f(x)
for all x in B(a, r). It is said to have a local minimum at a if instead f(a) ≤ f(x) for all x in
B(a, r). See Figure 4.7. It has a global maximum or global minimum at a if the associated
inequalities are true for all x in U, not just in B(a, r).
Figure 4.7: Local maximum (left) and local minimum (right) at a
If f is smooth and has a local maximum or local minimum at a, then it has a local maximum or",Multivariable_Calculus_Shimamoto.pdf
"If f is smooth and has a local maximum or local minimum at a, then it has a local maximum or
local minimum in every direction, in particular, in each of the coordinate directions. The derivatives
in these directions are the partial derivatives. They are essentially one-variable derivatives, so from
first-year calculus, at a local maximum or minimum, all of them must be 0, that is, ∂f
∂xj
(a) = 0 for
j = 1, 2, . . . , n.
Definition. A point a is called a critical point of f if ∂f
∂xj
(a) = 0 for j = 1, 2, . . . , n. Equivalently,
Df(a) =

0 0 . . .0

, or ∇f(a) = 0.
The preceding discussion proves the following.
Proposition 4.23. Let U be an open set in Rn, and let f : U → R be a smooth function. If f has
a local maximum or a local minimum at a, then a is a critical point of f.
Hence, if one is looking for local maxima or minima of a smooth function on an open set, the
critical points are the only possible candidates. This is just like the one-variable case. Conversely,",Multivariable_Calculus_Shimamoto.pdf
"critical points are the only possible candidates. This is just like the one-variable case. Conversely,
as in the one-variable case, not every critical point is necessarily a local max or min, though there
are new ways in which this failure can occur. A good example to keep in mind are functions whose
graphs are saddles, such as f(x, y) = x2 − y2 from Example 3.3 in Chapter 3. At a saddle point,
the function has a local maximum in one direction and a local minimum in another, so there is no
open ball about the point in which it is exclusively one or the other.
Even so, in first-year calculus, a given critical point can often be classified as a local maximum
or minimum using the second derivative. There is an analogue of this for more than one variable
that is similar in spirit while reflecting the greater number of possibilities. We discuss the situation
for two variables.
Let U be an open set in R2, and let f : U → R be a smooth function. If a ∈ U, consider the",Multivariable_Calculus_Shimamoto.pdf
"for two variables.
Let U be an open set in R2, and let f : U → R be a smooth function. If a ∈ U, consider the
matrix of second-order partial derivatives:
H(a) =
"" ∂2f
∂x2 (a) ∂2f
∂y ∂x(a)
∂2f
∂x ∂y(a) ∂2f
∂y2 (a)
#
.",Multivariable_Calculus_Shimamoto.pdf
"100 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
It is called the Hessian matrix of f at a. By the equality of mixed partials, the two off-diagonal
terms, ∂2f
∂x ∂y(a) and ∂2f
∂y ∂x(a), are actually equal.
Theorem 4.24 (Second derivative test for functions of two variables) . Let a be a critical point of
f such that det H(a) ̸= 0. Under this assumption, a is called a nondegenerate critical point.
1. If det H(a) > 0 and:
(a) if ∂2f
∂x2 (a) > 0, then f has a local minimum at a,
(b) if ∂2f
∂x2 (a) < 0, then f has a local maximum at a.
2. If det H(a) < 0, then f has a saddle point at a.
If det H(a) = 0, the test is inconclusive.
We discuss the ideas behind the test in the next section, but, before that, we present some
examples. The first involves three simple functions for which we already know the answers, but it
is meant to illustrate how the test works and confirm that it gives the right results. In a way, these
three functions are prototypes of the general situation.",Multivariable_Calculus_Shimamoto.pdf
"three functions are prototypes of the general situation.
Example 4.25. Find all critical points of the following functions and, if possible, classify each as
a local maximum, local minimum, or saddle point.
(a) f(x, y) = x2 + y2
(b) f(x, y) = −x2 − y2
(c) f(x, y) = x2 − y2
Intuitively, it’s clear that the first two of these attain their minimum and maximum values,
respectively, at the origin, while, as noted earlier, the third has a saddle point there. The graphs
of the functions are shown in Figure 4.8.
Figure 4.8: Prototypical nondegenerate critical points at (0 , 0): f(x, y) = x2 + y2 (left), f(x, y) =
−x2 − y2 (middle), f(x, y) = x2 − y2 (right)
The second derivative test reaches these conclusions as follows.
(a) To find the critical points, we set ∂f
∂x = 2x = 0 and ∂f
∂y = 2y = 0. The only solution is x = 0
and y = 0. Hence the only critical point is (0 , 0). To classify its type, we look at the second-order
partials: ∂2f
∂x2 = 2, ∂2f
∂x ∂y = ∂2f
∂y ∂x = 0, and ∂2f",Multivariable_Calculus_Shimamoto.pdf
"partials: ∂2f
∂x2 = 2, ∂2f
∂x ∂y = ∂2f
∂y ∂x = 0, and ∂2f
∂y2 = 2. Therefore:
H(0, 0) =
2 0
0 2

.
Since det H(0, 0) = 4 > 0 and ∂2f
∂x2 (0, 0) = 2 > 0, the second derivative test implies that f has a
local minimum at (0 , 0), as expected.",Multivariable_Calculus_Shimamoto.pdf
"4.10. MAX/MIN: CRITICAL POINTS 101
(b) By the same calculation, (0 , 0) is the only critical point, only this time:
H(0, 0) =
−2 0
0 −2

.
Hence det H(0, 0) = 4 > 0 and ∂2f
∂x2 (0, 0) = −2 < 0, so f has a local maximum at (0 , 0).
(c) Again, (0, 0) is the only critical point, but:
H(0, 0) =
2 0
0 −2

.
Since det H(0, 0) = −4 < 0, f has a saddle point at (0 , 0).
Example 4.26. We repeat for the function f(x, y) = x3 + 8y3 − 3xy. A portion of the graph of f
is shown in Figure 4.9.
Figure 4.9: The graph of f(x, y) = x3 + 8y3 − 3xy for −1 ≤ x ≤ 1, −1 ≤ y ≤ 1
First, to find the critical points, we set:
(∂f
∂x = 3 x2 − 3y = 0
∂f
∂y = 24 y2 − 3x = 0.
The first equation implies that y = x2, and substituting this into the second gives 24 x4 − 3x = 0,
or 3x(8x3 − 1) = 0. Hence x = 0, in which case y = 02 = 0, or x3 = 1
8 , i.e., x = 1
2 , in which case
y =
1
2

2 = 1
4 . As a result, f has two critical points, (0 , 0) and ( 1
2 , 1
4 ).
To classify their types, we consider the Hessian:",Multivariable_Calculus_Shimamoto.pdf
"y =
1
2

2 = 1
4 . As a result, f has two critical points, (0 , 0) and ( 1
2 , 1
4 ).
To classify their types, we consider the Hessian:
H(x, y) =
 6x −3
−3 48 y

.
For instance, at the critical point (0 , 0), det H(0, 0) = det
 0 −3
−3 0

= −9. Since this is negative,
(0, 0) is a saddle point.
Similarly, det H(1
2 , 1
4 ) = det
 3 −3
−3 12

= 36 − 9 = 27. This is positive and ∂2f
∂x2 (0, 0) = 3 is
positive (it’s the upper left entry of the Hessian), so, by the second derivative test, ( 1
2 , 1
4 ) is a local
minimum.
As a check on the plausibility of these conclusions, see Figure 4.10 for what the graph of f looks
like near each of the critical points.",Multivariable_Calculus_Shimamoto.pdf
"102 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Figure 4.10: The graph of f(x, y) = x3 + 8y3 − 3xy near (0, 0) (left) and near ( 1
2 , 1
4 ) (right)
4.11 Classifying nondegenerate critical points
We now say a few words about why the second derivative test for functions of two variables works
(Theorem 4.24 of the preceding section). We won’t give a complete proof, but, since the test may
appear to come out of nowhere, a word or two of explanation is warranted.
We begin by reviewing the corresponding situation for functions of one variable. Let f : I → R
be a smooth function defined on an open interval I in R. The critical points of f are the points a
where f′(a) = 0. At such points, the usual first-order approximation f(x) ≈ f(a) + f′(a) · (x − a)
degenerates to f(x) ≈ f(a), which gives us little information about the behavior of f(x) when x is
near a.
As a refinement, we add an additional term by considering the second-order Taylor approxima-
tion at a:",Multivariable_Calculus_Shimamoto.pdf
"near a.
As a refinement, we add an additional term by considering the second-order Taylor approxima-
tion at a:
f(x) ≈ f(a) + f′(a) · (x − a) + f′′(a)
2 · (x − a)2. (4.16)
If a is a critical point, this reduces to f(x) ≈ f(a) +f′′(a)
2 ·(x−a)2, a parabola whose general shape
is determined by the sign of the coefficient f′′(a)
2 . If f′′(a) > 0, then the approximating parabola
is concave up and f has a local minimum at a. If f′′(a) < 0, the parabola is concave down and f
has a local maximum. If f′′(a) = 0, the approximation degenerates again. Further information is
required to determine the type of critical point in this case.
4.11.1 The second-order approximation
Let f : U → R be a smooth function of two variables defined on an open set U in R2, and let a be
a point of U. The situation here is that the second-order approximation of f(x) for x near a is:
f(x) ≈ f(a) + Df(a) · (x − a) + 1
2(x − a)t · H(a) · (x − a). (4.17)",Multivariable_Calculus_Shimamoto.pdf
"f(x) ≈ f(a) + Df(a) · (x − a) + 1
2(x − a)t · H(a) · (x − a). (4.17)
The various dots are matrix products, where, writing x = (x, y) and a = (c, d):
• H(a) =
"" ∂2f
∂x2 (a) ∂2f
∂x ∂y(a)
∂2f
∂x ∂y(a) ∂2f
∂y2 (a)
#
is the Hessian matrix at a,
• x − a =
x − c
y − d

, and
• (x − a)t =

x − c y − d

, the transpose of x − a.
To justify the approximation, let x be a point near a, and let v = x − a. Consider the real-
valued function g defined by g(t) = f(a + tv). Then g(0) = f(a) and g(1) = f(a + v) = f(x), so",Multivariable_Calculus_Shimamoto.pdf
"4.11. CLASSIFYING NONDEGENERATE CRITICAL POINTS 103
finding an approximation for f(x) is the same as approximating g(1). As a real-valued function of
one variable, g(t) has a second-order Taylor approximation (4.16) at 0, which for g(1) gives:
f(x) = g(1) ≈ g(0) + g′(0) · (1 − 0) + g′′(0)
2 · (1 − 0)2
≈ f(a) + g′(0) + g′′(0)
2 . (4.18)
To find the derivatives of g, note that, by construction, g is the composition g(t) = f(α(t)),
where α(t) = a + tv. Hence, by the Little Chain Rule:
g′(t) = ∇f(α(t)) · α′(t) = ∇f(a + tv) · v. (4.19)
In particular, g′(0) = ∇f(a) · v, or in matrix form, g′(0) = Df(a) · v. Substituting this into (4.18)
and recalling that v = x − a gives:
f(x) ≈ f(a) + Df(a) · (x − a) + g′′(0)
2 . (4.20)
So far, we have succeeded only in reconstructing the first-order approximation of f at a. The
really new information comes from determining g′′(0). From equation (4.19), note that g′ is also",Multivariable_Calculus_Shimamoto.pdf
"really new information comes from determining g′′(0). From equation (4.19), note that g′ is also
a composition, namely, g′(t) = u(α(t)), where u(x) = ∇f(x) · v and α(t) = a + tv as before.
Therefore, by the same Little Chain Rule calculation:
g′′(0) = (u ◦ α)′(0) = ∇u(a) · v. (4.21)
To find ∇u, we let v = (h, k) and write out u(x) = ∇f(x) ·v =
∂f
∂x , ∂f
∂y


·(h, k) = h∂f
∂x + k∂f
∂y . Then
∇u = (h∂2f
∂x2 + k ∂2f
∂x ∂y, h∂2f
∂y ∂x+ k∂2f
∂y2 ). From equation (4.21) and the equality of mixed partials, we
obtain:
g′′(0) =

h∂2f
∂x2 (a) + k ∂2f
∂x ∂y(a), h ∂2f
∂y ∂x(a) + k∂2f
∂y2 (a)

· (h, k)
= ∂2f
∂x2 (a) h2 + 2 ∂2f
∂x ∂y(a) hk + ∂2f
∂y2 (a) k2. (4.22)
To simplify the notation somewhat, let A = ∂2f
∂x2 (a), B = ∂2f
∂x ∂y(a), and C = ∂2f
∂y2 (a). We do some
algebraic rearrangement in order to get the quadratic term of the approximation into the desired
form (4.17):
g′′(0) = Ah2 + 2Bhk + Ck2 (4.23)
= (Ah2 + Bhk) + (Bhk + Ck2)
=

h k
Ah + Bk
Bh + Ck

=

h k
A B
B C
h
k
",Multivariable_Calculus_Shimamoto.pdf
"form (4.17):
g′′(0) = Ah2 + 2Bhk + Ck2 (4.23)
= (Ah2 + Bhk) + (Bhk + Ck2)
=

h k
Ah + Bk
Bh + Ck

=

h k
A B
B C
h
k

= (x − a)t · H(a) · (x − a).
Substituting this into (4.20) gives the second-order approximation f(x) ≈ f(a) +Df(a) ·(x −a) +
1
2 (x − a)t · H(a) · (x − a) as stated in (4.17). This accomplishes what we set out to do.",Multivariable_Calculus_Shimamoto.pdf
"104 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
We remark that one can continue in this way to obtain higher-order approximations of f by
using higher-order Taylor approximations of g. The coefficients involve the higher-order derivatives
g(n)(0) of g. The idea is that, by the Little Chain Rule, each additional derivative of g corresponds
to applying the “operator” v·∇ = (h, k)·( ∂
∂x , ∂
∂y ) = h ∂
∂x +k ∂
∂y . Thus g(n)(0) =

h ∂
∂x +k ∂
∂y

nf


(a),
where

h ∂
∂x + k ∂
∂y

n denotes the n-fold composition

h ∂
∂x + k ∂
∂y


◦

h ∂
∂x + k ∂
∂y


◦ ··· ◦

h ∂
∂x + k ∂
∂y


.
We leave for the curious and ambitious reader the work of checking that this formula is consistent
with our results for the first two derivatives and of figuring out a general formula for the nth-order
approximation of f. For example, see Exercise 11.3 for the third-order approximation.
4.11.2 Sums and differences of squares",Multivariable_Calculus_Shimamoto.pdf
"approximation of f. For example, see Exercise 11.3 for the third-order approximation.
4.11.2 Sums and differences of squares
Still assuming that f is a smooth function of two variables, if a is a critical point, then Df(a) =
0 0

, so the second-order approximation (4.17) becomes:
f(x) = f(a + v) ≈ f(a) + 1
2(x − a)t · H(a) · (x − a)
≈ f(a) + 1
2vt · H(a) · v,
where v = x − a. The approximation holds when v is near 0.
We assume that det H(a) ̸= 0 and determine the nature of the critical point a. To do this,
it is simpler to write the quadratic term of the approximation using the notation of (4.23) with
v = (h, k) so that:
f(a + v) ≈ f(a) + 1
2(Ah2 + 2Bhk + Ck2) (4.24)
and H(a) =
A B
B C

.
We consider the case A ̸= 0. Similar arguments apply if A = 0 (see Exercise 11.5). Completing
the square on the expression in parentheses in (4.24) gives:
Ah2 + 2Bhk + Ck2 = A(h2 + 2B
A hk) + Ck2
= A(h2 + 2B
A hk + B2
A2 k2) − B2
A k2 + Ck2
= A(h + B
A k)2 + (C − B2
A )k2
= A(h + B
A k)2 + AC−B2",Multivariable_Calculus_Shimamoto.pdf
"Ah2 + 2Bhk + Ck2 = A(h2 + 2B
A hk) + Ck2
= A(h2 + 2B
A hk + B2
A2 k2) − B2
A k2 + Ck2
= A(h + B
A k)2 + (C − B2
A )k2
= A(h + B
A k)2 + AC−B2
A k2
= A˜h2 + AC−B2
A k2,
where ˜h = h + B
A k. Then (4.24) becomes:
f(a + v) ≈ f(a) + 1
2

A˜h2 + AC−B2
A k2

.
The important point is that the expression in parentheses is now a combination of squares,
and we know from Example 4.25 what to expect of expressions of this type. For instance, if both
coefficients A and AC−B2
A are positive, then f has a local minimum at v = 0, like x2 + y2. If
both coefficients are negative, then f has a local maximum, like −x2 − y2. If the coefficients have
opposite signs, then f has a saddle point, like x2 −y2. Note that this last case arises precisely when
the product of the coefficients is negative, that is, when A
AC−B2
A


= AC − B2 < 0.
Recall that A = ∂2f
∂x2 (a), and observe that the expression AC − B2 is the determinant of the
Hessian: det H(a) = det
A B
B C
",Multivariable_Calculus_Shimamoto.pdf
"A


= AC − B2 < 0.
Recall that A = ∂2f
∂x2 (a), and observe that the expression AC − B2 is the determinant of the
Hessian: det H(a) = det
A B
B C

= AC − B2. Collecting the information in the previous paragraph",Multivariable_Calculus_Shimamoto.pdf
"4.12. MAX/MIN: LAGRANGE MULTIPLIERS 105
about the signs of the coefficients A and AC−B2
A then gives the second-derivative test as stated in
Theorem 4.24.
The key idea of the argument was to complete the square to turn f into one of the prototypes
of being a sum and/or difference of squares, at least up to second order. This is not as ad hoc as it
might seem. A theorem in linear algebra states that, if M is a symmetric matrix in the sense that
Mt = M, then it is always possible to complete the square and convert the expression vt ·M ·v into
a sum and/or difference of squares. This is one interpretation of a powerful result known as the
spectral theorem. Thanks to the equality of mixed partials, the Hessian matrix H(a) is symmetric,
so the spectral theorem applies. In fact, the spectral theorem is true for n by n symmetric matrices
for all n. Using similar ideas, this leads to a corresponding classification of nondegenerate critical",Multivariable_Calculus_Shimamoto.pdf
"for all n. Using similar ideas, this leads to a corresponding classification of nondegenerate critical
points for real-valued functions of n variables when n >2.
4.12 Max/min: Lagrange multipliers
We now consider the problem of maximizing or minimizing a function subject to a “constraint.”
The constraint is often what gives an optimization problem substance. For instance, trying to find
the maximum value of a function like f(x, y, z) = x + 2y + 3z is silly: the function can be made
arbitrarily large by going far away from the origin where x, y, and z are all large. But if ( x, y, z)
is confined to a limited region of space, then there’s something to think about. We illustrate with
a specific example.
Example 4.27. Let S be the surface described by x2
4 + y2
9 + z2 = 1 in R3. It is an ellipsoid, as
sketched in Figure 4.11. If f(x, y, z) = x + 2y + 3z, at what points of S does f attain its maximum
and minimum values, and what are those values?
Figure 4.11: The ellipsoid x2
4 + y2",Multivariable_Calculus_Shimamoto.pdf
"and minimum values, and what are those values?
Figure 4.11: The ellipsoid x2
4 + y2
9 + z2 = 1
Here, we want to maximize and minimize f = x + 2y + 3z subject to the constraint x2
4 +
y2
9 + z2 = 1. One approach is to convert this to a two-variable problem, for instance, writing
z = ±
q
1 − x2
4 − y2
9 on S, so f = x + 2y + 3
q
1 − x2
4 − y2
9 or f = x + 2y − 3
q
1 − x2
4 − y2
9 . These
are functions of two variables, so we could find the critical points, i.e., set the partial derivatives
with respect to x and y equal to 0, and proceed as in the previous two sections.
Instead, we try a new approach based on organizing the values of f on S according to the level
sets of f. Each level set f = x + 2y + 3z = c is a plane that intersects S typically in a curve, if at
all. As c varies, we get a family of level curves on S with f constant on each. See Figure 4.12.
Let’s say we want to maximize f on S. Then, starting at some point, we move on S in the",Multivariable_Calculus_Shimamoto.pdf
"Let’s say we want to maximize f on S. Then, starting at some point, we move on S in the
direction of increasing c. We continue doing this until we reach a point a where we can’t go any
farther. Let cmax denote the value of f at this point a. Observe that at a:",Multivariable_Calculus_Shimamoto.pdf
"106 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Figure 4.12: Intersecting S with a level set f = c (left) and a family of intersections (right)
• The level set f = cmax must be tangent to S. Otherwise we would be able to cross the level
curve at a and make the value of f larger while remaining on S.
• ∇f(a) is a normal vector to the level set f = cmax. This is because gradients are always
normal to level sets (Proposition 4.17).
• Similarly, the surface S is itself a level set, namely, the set where:
g(x, y, z) = x2
4 + y2
9 + z2 = 1.
Hence ∇g(a) is a normal vector to S.
The situation is shown in Figure 4.13.
Figure 4.13: The level set f = cmax and the ellipsoid S are tangent at a.
Since the level set f = cmax and the surface S are tangent at a, their normal vectors are scalar
multiples of each other. In other words, there is a scalar λ such that ∇f(a) = λ∇g(a). This is the
idea behind the following general principle.",Multivariable_Calculus_Shimamoto.pdf
"multiples of each other. In other words, there is a scalar λ such that ∇f(a) = λ∇g(a). This is the
idea behind the following general principle.
Proposition 4.28 (Method of Lagrange multipliers) . Let U be an open set in Rn, and suppose
that you want to maximize or minimize a smooth function f : U → R subject to the constraint
g(x) = c for some smooth function g : U → R and some constant c. If f has a local maximum or
local minimum at a point a and if ∇g(a) ̸= 0, then there is a scalar λ such that:
∇f(a) = λ∇g(a).
The scalar λ is called a Lagrange multiplier.
We return to Example 4.27 and find the maximum and minimum of f = x + 2y + 3z subject to
the constraint g = x2
4 + y2
9 +z2 = 1. According to the method of Lagrange multipliers, the solutions",Multivariable_Calculus_Shimamoto.pdf
"4.12. MAX/MIN: LAGRANGE MULTIPLIERS 107
occur at points where ∇f = λ∇g, that is, where (1 , 2, 3) = λ(1
2 x, 2
9 y, 2z). This gives a system of
equations: 


1 = 1
2λx
2 = 2
9λy
3 = 2λz.
There are four unknowns and only three equations, which may seem like not enough information,
but we must remember that the constraint provides a fourth equation that can be included as part
of the system. In this case, the equations in the system imply that λ ̸= 0, so we can solve for each
of x, y, and z in terms of λ, then substitute into the constraint: x = 2
λ, y = 9
λ, z = 3
2λ, and:
2
λ

2
4 +
9
λ

2
9 +
 3
2λ

2 = 1.
This simplifies to 1
λ2 + 9
λ2 + 9
4λ2 = 1, or 4+36+9
4λ2 = 1. Hence 4 λ2 = 49, so λ = ±7
2 .
If λ = 7
2 , then:
x = 2
λ = 2
7
2
= 4
7, y = 9
7
2
= 18
7 , z = 3
2 · 7
2
= 3
7,
and f = x + 2y + 3z = 4
7 + 36
7 + 9
7 = 7. If λ = −7
2 , it’s the same thing with minus signs. Thus, on
S, f attains a maximum value of 7 at ( 4
7 , 18
7 , 3
7 ) and a minimum value of −7 at (−4",Multivariable_Calculus_Shimamoto.pdf
"2 , it’s the same thing with minus signs. Thus, on
S, f attains a maximum value of 7 at ( 4
7 , 18
7 , 3
7 ) and a minimum value of −7 at (−4
7 , −18
7 , −3
7 ).
Example 4.29. Suppose that, when a manufacturer spends x dollars on labor, y dollars on equip-
ment, and z dollars on research, the total number of units that it produces is:
f(x, y, z) = 60x1/6y1/3z1/2.
This is an example of what is called a Cobb-Douglas production function. If the manufacturer has
a total budget of B dollars, how should it distribute its spending among the three categories in
order to maximize production?
Here, we want to maximize the functionf above subject to the budget constraintg = x+y+z =
B. Following the method of Lagrange multipliers, we set ∇f = λ∇g:
(10x−5/6y1/3z1/2, 20x1/6y−2/3z1/2, 30x1/6y1/3z−1/2) = λ(1, 1, 1).
In other words: 


10x−5/6y1/3z1/2 = λ (4.25)
20x1/6y−2/3z1/2 = λ (4.26)
30x1/6y1/3z−1/2 = λ. (4.27)
Both (4.25) and (4.26) equal λ, and setting them equal gives:",Multivariable_Calculus_Shimamoto.pdf
"

10x−5/6y1/3z1/2 = λ (4.25)
20x1/6y−2/3z1/2 = λ (4.26)
30x1/6y1/3z−1/2 = λ. (4.27)
Both (4.25) and (4.26) equal λ, and setting them equal gives:
10x−5/6y1/3z1/2 = 20x1/6y−2/3z1/2. (4.28)
If z = 0, then f = 0, which is clearly not the maximum value. Thus we may assume z ̸= 0 so that
the factors of z1/2 in equation (4.28) can be canceled, leaving 10x−5/6y1/3 = 20x1/6y−2/3, or y = 2x.
Similarly, setting (4.25) and (4.27) equal and assuming y ̸= 0 gives 10 x−5/6z1/2 = 30x1/6z−1/2, or
z = 3x.
Thus the budget constraint becomes x + y + z = x + 2x + 3x = 6x = B, so x = B
6 . We conclude
that the optimal level of production is achieved by spending x = B
6 dollars on labor, y = 2x = B
3
dollars on equipment, and z = 3x = B
2 dollars on research.",Multivariable_Calculus_Shimamoto.pdf
"108 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
It’s worth stepping back to review this calculation, as there is more here than meets the eye.
A partial derivative can be viewed as a one-variable rate of change. For instance, in the previous
example, ∂f
∂x is the rate at which the output changes per dollar spent on labor. This is called the
marginal productivity with respect to labor. There are similar interpretations of ∂f
∂y and ∂f
∂z
with respect to equipment and research, respectively.
The Lagrange multiplier condition ∇f = λ∇g with g = x + y + z says that ( ∂f
∂x , ∂f
∂y , ∂f
∂z ) =
λ(1, 1, 1) = (λ, λ, λ), or:
∂f
∂x = ∂f
∂y = ∂f
∂z = λ. (4.29)
In other words, the marginal productivities with respect to labor, equipment, and research are
equal. This part of the analysis applies to any production function f, not just the Cobb-Douglas
model of the example. Thus, at the optimal level of production, an extra dollar spent on labor,",Multivariable_Calculus_Shimamoto.pdf
"model of the example. Thus, at the optimal level of production, an extra dollar spent on labor,
equipment, or research all increase production by the same amount. This makes sense since, if
there were an advantage to increasing the amount spent on one of them, then the manufacturer
could increase production by spending more on that one and less on the others. At the optimal
level, no such advantage can exist.
Moreover, according to (4.29), the common value of these marginal productivities is the La-
grange multiplier λ. In the example, by (4.25):
λ = 10y1/3z1/2
x5/6
= 10
B
3

1/3B
2

1/2
B
6

5/6
= 10
1
3

1/31
2

1/2B5/6 · 65/6
B5/6
= 60
1
3

1/31
2

1/21
6

1/6
≈ 21.82.
So, at the optimal level, an additional dollar spent on any combination of labor, equipment, or
research increases production by approximately 21.82 units. The seemingly extraneous Lagrange
mutliplier has told us something interesting.
4.13 Exercises for Chapter 4
Section 1 The first-order approximation",Multivariable_Calculus_Shimamoto.pdf
"mutliplier has told us something interesting.
4.13 Exercises for Chapter 4
Section 1 The first-order approximation
1.1. Let f(x, y) = x2 − y2, and let a = (2, 1).
(a) Find ∂f
∂x (a) and ∂f
∂y (a).
(b) Find Df(a) and ∇f(a).
(c) Find the first-order approximation ℓ(x, y) of f(x, y) at a. (You may assume that f is
differentiable at a.)
(d) Compare the values of f(2.01, 1.01) and ℓ(2.01, 1.01).
1.2. Let f(x, y, z) = x2y + y2z + z2x, and let a = (1, −1, 1).",Multivariable_Calculus_Shimamoto.pdf
"4.13. EXERCISES FOR CHAPTER 4 109
(a) Find ∂f
∂x (a), ∂f
∂y (a), and ∂f
∂z (a).
(b) Find Df(a) and ∇f(a).
(c) Find the first-order approximation ℓ(x, y, z) of f(x, y, z) at a. (You may assume that f
is differentiable at a.)
(d) Compare the values of f(1.05, −1.1, 0.95) and ℓ(1.05, −1.1, 0.95).
In Exercises 1.3–1.8, find (a) the partial derivatives ∂f
∂x and ∂f
∂y and (b) the matrix Df(x, y).
1.3. f(x, y) = x3 − 2x2y + 3xy2 − 4y3
1.4. f(x, y) = sin x sin 2y
1.5. f(x, y) = xexy
1.6. f(x, y) = x2−y2
x2+y2
1.7. f(x, y) = 1
x2+y2
1.8. f(x, y) = xy
In Exercises 1.9–1.13, find (a) the partial derivatives ∂f
∂x , ∂f
∂y , and ∂f
∂z and (b) the gradient
∇f(x, y, z).
1.9. f(x, y, z) = x + 2y + 3z + 4
1.10. f(x, y, z) = xy + xz + yz − xyz
1.11. f(x, y, z) = e−x2−y2−z2
sin(x + 2y)
1.12. f(x, y, z) = x+y
y+z
1.13. f(x, y, z) = ln(x2 + y2)
1.14. Let:
f(x, y) = xyxyxy
sin(xy) + x3y + x4
y2 + xy + 1 arctan
 ln x
x5 + y6

.
Evaluate ∂f
∂y (1, π). ( Hint: Do not try to find a general formula for ∂f",Multivariable_Calculus_Shimamoto.pdf
"f(x, y) = xyxyxy
sin(xy) + x3y + x4
y2 + xy + 1 arctan
 ln x
x5 + y6

.
Evaluate ∂f
∂y (1, π). ( Hint: Do not try to find a general formula for ∂f
∂y . Ever.)
1.15. Let f(x, y) = x + 2y. Use the definition of differentiability to prove that f is differentiable at
every point a = (c, d) of R2.
1.16. Use the definition of differentiability to show that the function f(x, y) = x2 + y2 is differen-
tiable at every point a = (c, d) of R2.
1.17. Let f : R2 → R be the function:
f(x, y) =
p
x4 + y4.
(a) Find formulas for the partial derivatives ∂f
∂x and ∂f
∂y at all points ( x, y) other than the
origin.
(b) Find the values of the partial derivatives ∂f
∂x (0, 0) and ∂f
∂y (0, 0). (Note that the formulas
you found in part (a) probably do not apply at (0 , 0).)",Multivariable_Calculus_Shimamoto.pdf
"110 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
(c) Use the definition of differentiability to determine whether f is differentiable at (0 , 0).
(Hint: Polar coordinates might be useful.)
1.18. Consider the function f : R2 → R defined by:
f(x, y) =
( x3+2y3
x2+y2 if (x, y) ̸= (0, 0),
0 if ( x, y) = (0, 0).
(a) Find the values of the partial derivatives ∂f
∂x (0, 0) and ∂f
∂y (0, 0). (This shouldn’t require
much calculation.)
(b) Use the definition of differentiability to determine whether f is differentiable at (0, 0).
1.19. Let f and g be differentiable real-valued functions defined on an open set U in Rn. Prove
that:
∇(fg) = f ∇g + g ∇f.
(Hint: Partial derivatives are one-variable derivatives.)
1.20. Let f, g: U → R be real-valued functions defined on an open set U in Rn, and let a be a point
of U. Use the definition of differentiability to prove the following statements.
(a) If f and g are differentiable at a, so is f + g. (It’s reasonably clear that D(f + g)(a) =",Multivariable_Calculus_Shimamoto.pdf
"(a) If f and g are differentiable at a, so is f + g. (It’s reasonably clear that D(f + g)(a) =
Df(a) +Dg(a), so the problem is to show that the limit in the definition of differentia-
bility for f + g is 0.)
(b) If f is differentiable at a, so is cf for any scalar c.
Section 2 Conditions for differentiability
2.1. Let f : R2 → R be the function defined by:
f(x, y) =
(
x2 − y2 if (x, y) ̸= (0, 0),
π if (x, y) = (0, 0).
Is f differentiable at (0, 0)? If so, find Df(0, 0).
2.2. Is the function f(x, y) = x2/3 + y2/3 differentiable at (0, 0)? If so, find Df(0, 0).
2.3. This exercise gives the proof of Proposition 4.4, that differentiability implies continuity. Let
U be an open set in Rn, f : U → R a real-valued function, and a a point of U. Define a
function Q: U → R by:
Q(x) =
( f(x)−f(a)−∇f(a)·(x−a)
∥x−a∥ if x ̸= a,
0 if x = a.
Note that the quotient is the one that appears in the definition of differentiability using the
gradient form of the derivative.",Multivariable_Calculus_Shimamoto.pdf
"0 if x = a.
Note that the quotient is the one that appears in the definition of differentiability using the
gradient form of the derivative.
(a) If f is differentiable at a, show that Q is continuous at a.
(b) If f is differentiable at a, show that there exists a δ >0 such that, if x ∈ B(a, δ), then
|Q(x)| < 1.",Multivariable_Calculus_Shimamoto.pdf
"4.13. EXERCISES FOR CHAPTER 4 111
(c) If f is differentiable at a, prove that f is continuous at a. ( Hint: Use the triangle
inequality and the Cauchy-Schwarz inequality to show that:
|f(x) − f(a)| ≤ |Q(x)|∥x − a∥ + ∥∇f(a)∥∥x − a∥.)
Section 3 The mean value theorem
3.1. Let B = B(a, r) be an open ball in R2 centered at the point a = (c, d), and let f : B → R be
a differentiable function such that:
Df(x) =

0 0

for all x in B.
Prove that f(x) = f(a) for all x in B. In other words, if the derivative of f is zero everywhere,
then f is a constant function.
Section 4 The C1 test
4.1. We showed in Example 4.7 that the functionf(x, y) =
p
x2 + y2 is not differentiable at (0, 0).
Is f differentiable at points other than the origin? If so, which points?
In Exercises 4.2–4.6, use the various criteria for differentiability discussed so far to determine
the points at which the function is differentiable and the points at which it is not. Your reasons",Multivariable_Calculus_Shimamoto.pdf
"the points at which the function is differentiable and the points at which it is not. Your reasons
may be brief as long as they are clear and precise. It should not be necessary to use the definition
of differentiability.
4.2. f : R2 → R, f(x, y) = sin(x + y)
4.3. f : R2 → R, f(x, y) =
( x2−y2
x2+y2 if (x, y) ̸= (0, 0),
1 if ( x, y) = (0, 0)
4.4. f : R2 → R, f(x, y) = |x| + |y|
4.5. f : R3 → R, f(x, y, z) = x4 + 2x2yz − y3z3
4.6. f : R4 → R, f(w, x, y, z) = det
w x
y z

4.7. Let f(x, y) = 3
p
x3 + 8y3.
(a) Find formulas for the partial derivatives∂f
∂x and ∂f
∂y at all points (x, y) where x3+8y3 ̸= 0.
(b) Do ∂f
∂x (0, 0) and ∂f
∂y (0, 0) exist? If so, what are their values? If not, why not?
(c) Is f differentiable at (0, 0)?
(d) More generally, at which points of R2 is f differentiable? ( Hint: If a = ( c, d) is a
point other than the origin where c3 + 8d3 = 0, note that
3√
x3 + 8d3 =
3√
x3 − c3 =
3
p
(x − c)(x2 + cx + c2) = 3√x − c ·
3√",Multivariable_Calculus_Shimamoto.pdf
"point other than the origin where c3 + 8d3 = 0, note that
3√
x3 + 8d3 =
3√
x3 − c3 =
3
p
(x − c)(x2 + cx + c2) = 3√x − c ·
3√
x2 + cx + c2 where x2 + cx + c2 ̸= 0 when x = c.)
Section 5 The Little Chain Rule
5.1. Let f(x, y) = x2 + y2, and let α(t) = (t2, t3). Calculate ( f ◦ α)′(1) in two different ways:",Multivariable_Calculus_Shimamoto.pdf
"112 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
(a) by using the Little Chain Rule,
(b) by substituting for x and y in terms of t in the formula for f to obtain (f ◦α)(t) directly
and differentiating the result.
5.2. Let f(x, y, z) = xyz, and let α(t) = (cos t, sin t, t). Calculate ( f ◦α)′(π
6 ) in two different ways:
(a) by using the Little Chain Rule,
(b) by substituting for x, y, and z in terms of t in the formula for f to obtain ( f ◦ α)(t)
directly and differentiating the result.
5.3. Let f(x, y, z) be a differentiable real-valued function of three variables, and let α(t) =
(x(t), y(t), z(t)) be a differentiable path in R3. If w = f(α(t)), use the Little Chain Rule
to find a formula for dw
dt in terms of the partial derivatives of f and the derivatives with
respect to t of x, y, and z.
5.4. The Little Chain Rule often lurks in the background of a type of first-year calculus problem
known as “related rates.” For instance, as an (admittedly artificial) example, suppose that",Multivariable_Calculus_Shimamoto.pdf
"known as “related rates.” For instance, as an (admittedly artificial) example, suppose that
the length ℓ and width w of a rectangular region in the plane are changing as functions of
time t, so that the area A = ℓw also changes.
(a) Find a formula for dA
dt in terms of ℓ, w, and their derivatives by differentiating the formula
A = ℓw directly with respect to t.
(b) Obtain the same result using the Little Chain Rule. ( Hint: Let α(t) = (ℓ(t), w(t)).)
(c) Suppose that, at the instant that the length of the region is 100 inches and its width is 40
inches, the length is increasing at a rate of 2 inches/second and the width is increasing
at a rate of 3 inches/second. How fast is the area changing?
5.5. Let a be a point of Rn, and let B = B(a, r) be an open ball centered at a. Let f : B → R be
a differentiable function. If b is any point of B, show that there exists a point c on the line
segment connecting a and b such that:
f(b) − f(a) = ∇f(c) · (b − a).",Multivariable_Calculus_Shimamoto.pdf
"segment connecting a and b such that:
f(b) − f(a) = ∇f(c) · (b − a).
Note that this is a generalization of the mean value theorem to real-valued functions of more
than one variable. ( Hint: Explain why the line segment can be parametrized by α(t) =
a+ t(b−a), 0 ≤ t ≤ 1. Then, note that the composition f ◦α is a real-valued function of one
variable, so the one-variable mean value theorem applies.)
5.6. In this exercise, we prove the Little Chain Rule (Theorem 4.13). Let U be an open set in Rn,
α: I → U a path in U defined on an open interval I, and f : U → R a real-valued function.
Let t0 be a point of I. Assume that α is differentiable at t0 and f is differentiable at α(t0).
Let a = α(t0), and let Q: U → R be the function defined in Exercise 2.3:
Q(x) =
( f(x)−f(a)−∇f(a)·(x−a)
∥x−a∥ if x ̸= a,
0 if x = a.
(a) Prove that limt→t0
f(α(t))−f(α(t0))−∇f(α(t0))·(α(t)−α(t0))
t−t0
 = 0. ( Hint: f(x)−f(a)−∇f(a)·
(x − a) = ∥x − a∥Q(x). Or see Exercise 8.2 in Chapter 3.)",Multivariable_Calculus_Shimamoto.pdf
"f(α(t))−f(α(t0))−∇f(α(t0))·(α(t)−α(t0))
t−t0
 = 0. ( Hint: f(x)−f(a)−∇f(a)·
(x − a) = ∥x − a∥Q(x). Or see Exercise 8.2 in Chapter 3.)
(b) Prove the Little Chain Rule: ( f ◦ α)′(t0) = ∇f(α(t0)) · α′(t0).",Multivariable_Calculus_Shimamoto.pdf
"4.13. EXERCISES FOR CHAPTER 4 113
5.7. Let f : R3 → R be a differentiable function with the property that ∇f(x) points in the same
direction as x for all nonzero x in R3. If a > 0, prove that f is constant on the sphere
x2 +y2 +z2 = a2. ( Hint: If p and q are any two points on the sphere, there is a differentiable
path α on the sphere from p to q.)
Section 6 Directional derivatives
In Exercises 6.1–6.4, find the directional derivative of f at the point a in the direction of the
vector v.
6.1. f(x, y) = e−2xy3, a = (0, 1), v = (4, 1)
6.2. f(x, y) = x2 + y2, a = (1, 2), v = (−2, 1)
6.3. f(x, y, z) = x3 − 2x2yz + xz − 3, a = (1, 0, −1), v = (1, −1, 2)
6.4. f(x, y, z) = sin x sin y cos z, a = (0, π
2 , π), v = (π, 2π, 2π)
6.5. For a certain differentiable real-valued function f : R2 → R, the maximum directional deriva-
tive at the point a = (0, 0) has a value of 6 and occurs in the direction from a to the point
b = (4, 1). Find ∇f(a).
6.6. Let f(x, y) = x2 − y2.
(a) At the point a = (
√",Multivariable_Calculus_Shimamoto.pdf
"b = (4, 1). Find ∇f(a).
6.6. Let f(x, y) = x2 − y2.
(a) At the point a = (
√
2, 1), find a unit vector u that points in the direction in which f is
increasing most rapidly. What is the rate of increase in this direction?
(b) Describe the set of all points a = (x, y) in R2 such that f increases most rapidly at a in
the direction that points directly towards the origin.
6.7. A certain function has the form:
f(x, y) = rx2 + sy2,
where r and s are constant. Find values of r and s so that the maximum directional derivative
of f at the point a = (1, 2) has a value of 10 and occurs in the direction from a to the point
b = (2, 3).
6.8. Let f(x, y) = xy, and let a = (5, −5).
(a) Find all directions in which the directional derivative of f at a is equal to 1.
(b) Is there a direction in which the directional derivative at a is equal to 10?
6.9. The temperature in a certain region of space, in degrees Celsius, is modeled by the function
T(x, y, z) = 20e−x2−2y2−4z2",Multivariable_Calculus_Shimamoto.pdf
"6.9. The temperature in a certain region of space, in degrees Celsius, is modeled by the function
T(x, y, z) = 20e−x2−2y2−4z2
, where x, y, z are measured in meters. At the point a = (2, −1, 3):
(a) In which direction is the temperature increasing most rapidly?
(b) In which direction is it decreasing most rapidly?
(c) If you travel in the direction described in part (a) at a speed of 10 meters/second, how
fast is the observed temperature changing at a in degrees Celsius per second?
6.10. Let c = (c1, c2, . . . , cn) be a vector in Rn, and define f : Rn → R by f(x) = c · x.
(a) Find the derivative Df(x).",Multivariable_Calculus_Shimamoto.pdf
"114 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
(b) If u is a unit vector in Rn, find a formula for the directional derivative ( Duf)(x).
6.11. You discover that your happiness is a function of your location in the plane. At the point
(x, y), your happiness is given by the formula:
H(x, y) = x3e2y happs,
where “happ” is a unit of happiness. For instance, at the point (1 , 0), you are H(1, 0) = 1
happ happy.
(a) At the point (1, 0), in which direction is your happiness increasing most rapidly? What
is the value of the directional derivative in this direction?
(b) Still at the point (1 , 0), in which direction is your happiness decreasing most rapidly?
What is the value of the directional derivative in this direction?
(c) Suppose that, starting at (1 , 0), you follow a curve so that you are always traveling in
the direction in which your happiness increases most rapidly. Find an equation for the
curve, and sketch the curve. An equation describing the relationship between x and y",Multivariable_Calculus_Shimamoto.pdf
"curve, and sketch the curve. An equation describing the relationship between x and y
along the curve is fine. No further parametrization is necessary. ( Hint: What can you
say about the slope at each point of the curve?)
(d) Suppose instead that, starting at (1, 0), you travel along a curve such that the directional
derivative in the tangent direction is always equal to zero. Find an equation describing
this curve, and sketch the curve.
6.12. Does there exist a differentiable function f : U → R defined on an open subset U of Rn with
the property that, for some point a in U, the directional derivatives at a satisfy (Duf)(a) > 0
for all unit vectors u in Rn, i.e., the directional derivative at a is positive in every direction?
Either find such a function and point a, or explain why none exists.
Section 7 ∇f as normal vector
7.1. Find an equation of the tangent plane to the surface z = x2 − y2 at the point a = (1, 2, −3).",Multivariable_Calculus_Shimamoto.pdf
"Section 7 ∇f as normal vector
7.1. Find an equation of the tangent plane to the surface z = x2 − y2 at the point a = (1, 2, −3).
7.2. Find an equation of the tangent plane to the surfacex2+y2−z2 = 1 at the pointa = (1, −1, 1).
7.3. The surface in R3 given by
(9x2 + y2 + z2 − 1)3 − y2z3 − 2
5x2z3 = 0
was featured in an article in the February 14, 2019 issue of The New York Times.5 See Figure
4.14. Nowhere in the article was the tangent plane at the point (0 , 1, 1) mentioned. Scoop
the Times by finding an equation of the tangent plane.
7.4. Let S be the ellipsoid x2 + y2
2 + z2
4 = 4. Let a = (1, 2, 2), and let n be the unit normal vector
to S at a that points outward from S. If f(x, y, z) = xyz, find the directional derivative
(Dnf)(a).
7.5. The saddle surface y2 − x2 − z = 0 and the sphere x2 + y2 + z2 = 14 intersect in a curve C
that passes through the point a = (1, 2, 3). Find a parametrization of the line tangent to C
at a.",Multivariable_Calculus_Shimamoto.pdf
"that passes through the point a = (1, 2, 3). Find a parametrization of the line tangent to C
at a.
5https://www.nytimes.com/2019/02/14/science/math-algorithm-valentine.html. Actually, the article con-
cerns a family of such surfaces, parametrized by two real numbers. The surface in the exercise is a specific example.",Multivariable_Calculus_Shimamoto.pdf
"4.13. EXERCISES FOR CHAPTER 4 115
Figure 4.14: The surface (9 x2 + y2 + z2 − 1)3 − y2z3 − 2
5 x2z3 = 0
7.6. Find all values of c such that, at every point of intersection of the spheres
(x − c)2 + y2 + z2 = 3 and x2 + (y − 1)2 + z2 = 1,
the respective tangent planes are perpendicular to one another.
Section 8 Higher-order partial derivatives
In Exercises 8.1–8.4, find all four second-order partial derivatives of the given function f.
8.1. f(x, y) = x4 − 2x3y + 3x2y2 − 4xy3 + 5y4
8.2. f(x, y) = sin x cos y
8.3. f(x, y) = e−x2−y2
8.4. f(x, y) = y
x+y
8.5. Let f(x, y) = x4y3. Find i and j so that the ( i + j)th-order partial derivative ∂i+jf
∂xi ∂yj (0, 0) is
nonzero. What is the value of this partial derivative?
8.6. Consider the function f : R2 → R defined by:
f(x, y) =
( x3y
x2+y2 if (x, y) ̸= (0, 0),
0 if ( x, y) = (0, 0).
(a) Find formulas for the partial derivatives ∂f
∂x (x, y) and ∂f
∂y (x, y) if (x, y) ̸= (0, 0).
(b) Find the values of ∂f
∂x (0, 0) and ∂f
∂y (0, 0).",Multivariable_Calculus_Shimamoto.pdf
"∂x (x, y) and ∂f
∂y (x, y) if (x, y) ̸= (0, 0).
(b) Find the values of ∂f
∂x (0, 0) and ∂f
∂y (0, 0).
(c) Use your answers to parts (a) and (b) to evaluate the second-order partial derivatives:
∂2f
∂y ∂x(0, 0) =
 ∂
∂y
∂f
∂x


(0, 0) and ∂2f
∂x ∂y(0, 0) =
 ∂
∂x
∂f
∂y


(0, 0)
at the origin (and only at the origin). ( Warning: Do not assume that the second-order
partial derivatives are continuous.)
(d) Is f of class C2 on R2? Is it of class C1?",Multivariable_Calculus_Shimamoto.pdf
"116 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Figure 4.15: The rectangle with vertices a, u, v, w
8.7. This exercise gives the details of a full proof of the equality of mixed partials (Theorem 4.21).
Let a = (c, d) be a point of R2, and let f : B → R be a real-valued function defined on an open
ball B = B(a, r) centered at a. Assume that the first and second-order partial derivatives of
f are continuous on B. Let R be the rectangle whose vertices are a = (c, d), u = (c + h, d),
v = (c + h, d+ k), and w = (c, d+ k), where h and k are nonzero real numbers small enough
that R ⊂ B. See Figure 4.15.
Let:
∆ = f(v) − f(u) − f(w) + f(a)
= f(c + h, d+ k) − f(c + h, d) − f(c, d+ k) + f(c, d).
(a) Show that ∆ =
∂f
∂y (c + h, y1) − ∂f
∂y (c, y1)


k for some number y1 between d and d + k.
(Hint: Note that ∆ = g(d + k) − g(d), where g is the function defined by g(y) =
f(c + h, y) − f(c, y). Apply the mean value theorem.)
(b) Show that there is a point p = (x1, y1) in R such that ∆ = ∂2f",Multivariable_Calculus_Shimamoto.pdf
"f(c + h, y) − f(c, y). Apply the mean value theorem.)
(b) Show that there is a point p = (x1, y1) in R such that ∆ = ∂2f
∂x ∂y(p) hk. ( Hint: Apply
the mean value theorem again, this time to the difference in the result of part (a).)
(c) Similarly, show that there is a point q = (x2, y2) in R such that ∆ = ∂2f
∂y ∂x(q) hk. ( Hint:
Consider j(x) = f(x, d+ k) − f(x, d).)
(d) Deduce that the mixed partials at a are equal: ∂2f
∂x ∂y(a) = ∂2f
∂y ∂x(a). ( Hint: What hap-
pens to R as h, k→ 0?)
Section 10 Max/min: critical points
In Exercises 10.1–10.6, find all critical points of f, and, if possible, classify their type.
10.1. f(x, y) = 2 − 10x − 3x2 − 20y + 8xy + 3y2
10.2. f(x, y) = 5 − 2x − 2x2 − 2xy + 2y − y2
10.3. f(x, y) = x4 + 2y4 + 3x2y2 + 4x2 + 5y2
10.4. f(x, y) = x3 + x2 + 4xy − 2y2 + 9
10.5. f(x, y) = 2x3 + y2 − 6xy
10.6. f(x, y) = 4x3 + y3 + (1 − x − y)3
10.7. Let k be a real number, and let f(x, y) = x3 + y3 − kxy. Find all critical points of f, and",Multivariable_Calculus_Shimamoto.pdf
"10.6. f(x, y) = 4x3 + y3 + (1 − x − y)3
10.7. Let k be a real number, and let f(x, y) = x3 + y3 − kxy. Find all critical points of f, and
classify each one as a local maximum, local minimum, or neither. (The answers may depend
on the value of k.)",Multivariable_Calculus_Shimamoto.pdf
"4.13. EXERCISES FOR CHAPTER 4 117
10.8. Let f(x, y) = 3x4 − 4x2y + y2.
(a) Make a sketch in R2 indicating the set of points ( x, y) where f(x, y) > 0 and the set
where f(x, y) < 0. ( Hint: Factor f.)
(b) Explain why f does not have a local maximum or minimum at (0 , 0).
(c) On the other hand, show that f does have a local minimum at (0 , 0) when restricted to
any line y = mx.
10.9. (a) Given a point a in Rn, define f : Rn → R by f(x) = ∥x − a∥2. Show that ∇f(x) =
2(x − a).
(b) Let p1, p2, . . . ,pm be m distinct points in Rn, and let F(x) = Pm
k=1 ∥x−pk∥2. Assuming
that F has a global minimum, show that it must occur when:
x = 1
m
mX
k=1
pk (the “centroid”).
10.10. Let (x1, y1), (x2, y2), . . . ,(xn, yn) be a collection of n distinct points in R2, where n ≥ 2. We
think of the points as “data.” The line y = mx + b that best fits the data in the sense of
least squares is defined to be the one that minimizes the quantity:
E =
nX
i=1
(mxi + b − yi)2.",Multivariable_Calculus_Shimamoto.pdf
"least squares is defined to be the one that minimizes the quantity:
E =
nX
i=1
(mxi + b − yi)2.
Note that using a line y = mx + b to predict the data value y = yi when x = xi gives an error
of magnitude |mxi + b − yi|. Consequently E is called the total squared error. It depends on
the line, which we identify here by its slope m and y-intercept b. Thus E is a function of the
two variables m and b, and we can apply the methods we have been studying to find where
it assumes its minimum.
Calculate the partial derivatives ∂E
∂m and ∂E
∂b , and show that the critical points (m, b) of E are
the solutions of the system of equations:



 nP
i=1
x2
i


m +
 nP
i=1
xi


b =
nP
i=1
xiyi
 nP
i=1
xi


m + nb =
nP
i=1
yi.
(4.30)
In Exercises 10.11–10.12, find the best fitting line in the sense of least squares for the given
data points by solving the system (4.30). Then, plot the points and sketch the line.
10.11. (1, 1), (2, 3), (3, 3)
10.12. (1, 2), (2, 1), (3, 4), (4, 3)",Multivariable_Calculus_Shimamoto.pdf
"10.11. (1, 1), (2, 3), (3, 3)
10.12. (1, 2), (2, 1), (3, 4), (4, 3)
10.13. In this exercise, we use the second derivative test to verify that, for the best fitting line in the
sense of least squares, the critical point ( m, b) given by the system (4.30) is a local minimum
of the total squared error E.
(a) If x1, x2, . . . , xn are real numbers, show that (
nP
i=1
xi)2 ≤ n(
nP
i=1
x2
i ) and that equality
holds if and only if x1 = x2 = ··· = xn. ( Hint: Consider the dot product x · v, where
x = (x1, x2, . . . , xn) and v = (1, 1, . . . ,1).)",Multivariable_Calculus_Shimamoto.pdf
"118 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
(b) Let ( x1, y1), (x2, y2), . . . ,(xn, yn) be given data points. Show that the Hessian of the
total squared error E is given by:
H(m, b) =


2
 nP
i=1
x2
i


2
 nP
i=1
xi


2
 nP
i=1
xi


2n

.
(c) Assuming that the data points don’t all have the same x-coordinate, show that the
critical point of E given by (4.30) is a local minimum. (In fact, it is a global minimum,
as follows once one factors in that E is a quadratic polynomial in m and b, though we
won’t go through the details to justify this.)
Section 11 Classifying nondegenerate critical points
11.1. Find the second-order approximation of f(x, y) = cos(x + y) at a = (0, 0).
11.2. Find the second-order approximation of f(x, y) = x2 + y2 at a = (1, 2).
11.3. Let U be an open set in R2, and let f : U → R be a smooth real-valued function defined on
U. Let a be a point of U, and let x = a + v be a nearby point, where v = (h, k).",Multivariable_Calculus_Shimamoto.pdf
"U. Let a be a point of U, and let x = a + v be a nearby point, where v = (h, k).
Due to the equality of mixed partials, the powers of the operator v · ∇= h ∂
∂x + k ∂
∂y under
composition expand in the same way as ordinary binomial expressions. For instance:

h ∂
∂x + k ∂
∂y

2f =

(h ∂
∂x + k ∂
∂y ) ◦ (h ∂
∂x + k ∂
∂y )

f
=

h2 ∂
∂x ◦ ∂
∂x + hk ∂
∂x ◦ ∂
∂y + hk ∂
∂y ◦ ∂
∂x + k2 ∂
∂y ◦ ∂
∂y


f
=

h2 ∂2
∂x2 + 2hk ∂2
∂x ∂y+ k2 ∂2
∂y2


f
= h2 ∂2f
∂x2 + 2hk ∂2f
∂x ∂y+ k2 ∂2f
∂y2 .
Note that this agrees with the expression derived in (4.22) as part of the quadratic term in
the second-order approximation of f.
(a) Find the corresponding expansion of

h ∂
∂x + k ∂
∂y

3f.
(b) Find a formula for the third-order approximation of f at a. (Recall from first-year
calculus that the third-order Taylor approximation of a function of one variable isf(x) =
f(a + h) ≈ f(a) + f′(a) h + f′′(a)
2 h2 + f′′′(a)
3! h3.)",Multivariable_Calculus_Shimamoto.pdf
"calculus that the third-order Taylor approximation of a function of one variable isf(x) =
f(a + h) ≈ f(a) + f′(a) h + f′′(a)
2 h2 + f′′′(a)
3! h3.)
11.4. Let f(x, y) = ax2 + by2, where a and b are nonzero constants. For which values of a and b
is the origin a local maximum? For which is it a local minimum? For which is it a saddle
point?
11.5. Complete the description of the behavior of the second-order approximation near a critical
point a by considering the case that det H(a) ̸= 0 and A = 0, where H(a) =
A B
B C

.
(a) Show that det H(a) < 0.",Multivariable_Calculus_Shimamoto.pdf
"4.13. EXERCISES FOR CHAPTER 4 119
(b) Show that the quadratic term Ah2 + 2Bhk + Ck2 = 2Bhk + Ck2 in the second-order
approximation can be written as a difference of two perfect squares. (Hence, based on
our prototypical models, we predict a to be a saddle point. Hint: In the case that C = 0,
too, consider the expansions of ( a ± b)2.)
Section 12 Max/min: Lagrange multipliers
12.1. Consider the problem of finding the maximum and minimum values of the function f(x, y) =
xy subject to the constraint x + 2y = 1.
(a) Explain why a minimum value does not exist.
(b) On the other hand, you may assume that a maximum does exist. Use Lagrange multi-
pliers to find the maximum value and the point at which it is attained.
12.2. (a) Find the points on the curve x2 − xy + y2 = 4 at which the function f(x, y) = x2 + y2
attains its maximum and minimum values. You may assume that these maximum and
minimum values exist.",Multivariable_Calculus_Shimamoto.pdf
"attains its maximum and minimum values. You may assume that these maximum and
minimum values exist.
(b) Use your answer to part (a) to help sketch the curve x2 −xy + y2 = 4. ( Hint: Note that
f(x, y) = x2 + y2 represents the square of the distance from the origin.)
12.3. At what points on the unit sphere x2 + y2 + z2 = 1 does the function f(x, y, z) = 2x−3y + 4z
attain its maximum and minimum values?
12.4. Find the minimum value of the function f(x, y, z) = x2 + y2 + z2 subject to the constraint
2x − 3y + 4z = 1. At what point is the minimum attained?
12.5. Find the positive values of x, y, and z for which the function f(x, y, z) = xy2z3 attains its
maximum value subject to the constraint 3x+4y+5z = 12. You may assume that a maximum
exists.
12.6. (a) Use the method of Lagrange multipliers to find the positive values of x, y, and z such
that xyz = 1 and x + y + z is as small as possible. What is the minimum value? You
may assume that a minimum exists.",Multivariable_Calculus_Shimamoto.pdf
"that xyz = 1 and x + y + z is as small as possible. What is the minimum value? You
may assume that a minimum exists.
(b) Use part (a) to show that, for all positive real numbers a, b, c:
a + b + c
3 ≥
3√
abc.
This is called the inequality of the arithmetic and geometric means . ( Hint: Let
k =
3√
abc, and consider a/k, b/k, and c/k.)
12.7. Let A be a 3 by 3 symmetric matrix, A =


a b c
b d e
c e f

, and consider the quadratic polynomial:
Q(x) = xtAx =

x y z



a b c
b d e
c e f




x
y
z

,
where x =


x
y
z

. Use Lagrange multipliers to show that, when Q is restricted to the unit
sphere x2+y2+z2 = 1, any point x at which it attains its maximum or minimum value satisfies",Multivariable_Calculus_Shimamoto.pdf
"120 CHAPTER 4. REAL-VALUED FUNCTIONS: DIFFERENTIATION
Ax = λx for some scalar λ and that the maximum or minimum value is the corresponding
value of λ. (In the language of linear algebra, x is called an eigenvector of A and λ is
called the corresponding eigenvalue.) In fact, maximum and minimum values do exist, so
the conclusion is not an empty one.
12.8. You discover that your happiness is a function of your daily routine. After a great deal of
soul-searching, you decide to focus on three activities to which you will devote your entire
day: if x, y, and z are the number of hours a day that you spend eating, sleeping, and
studying multivariable calculus, respectively, then, based on data provided by the federal
government, your happiness is given by the function:
h(x, y, z) = 1000 − 23x2 − 23y2 − 46y − z2 happs.6
There is a mix of eating, sleeping, and studying that leads to the greatest possible daily
happiness (you may assume this). What is it? Answer in two ways, as follows.",Multivariable_Calculus_Shimamoto.pdf
"happiness (you may assume this). What is it? Answer in two ways, as follows.
(a) Substitute for z in terms of x and y in the formula for h, and find the critical points of
the resulting function of x and y.
(b) Use Lagrange multipliers.
12.9. After conducting extensive market research, a manufacturer of monkey saddles discovers that,
if it produces a saddle consisting of x ounces of leather, y ounces of copper, and z ounces of
premium bananas, the monkey rider experiences a total of:
S = 2xy + 3xz + 4yz units of satisfaction.
Leather costs $1 per ounce, copper $2 per ounce, and premium bananas $3 per ounce, and the
manufacturer is willing to spend at most $1000 per saddle. What combination of ingredients
yields the most satisfied monkey? You may assume that a maximum exists.
6See Exercise 6.11, page 114.",Multivariable_Calculus_Shimamoto.pdf
"Chapter 5
Real-valued functions: integration
5.1 Volume and iterated integrals
We now study the integral of real-valued functions of more than one variable. Most of our time
is devoted to functions of two variables. In this section, we introduce the integral informally by
thinking about how to visualize it. This uses an idea likely to be familiar from first-year calculus.
It won’t be until the next section that we define what the integral actually is.
Let D be a subset of R2. At this point, we won’t be too careful about putting any conditions on
D, but it’s best to think of it as some sort of two-dimensional blob, as opposed to a set of discrete
points or a curve. Let f : A → R be a real-valued function whose domain A is a subset of R2 that
contains D. For the moment, assume that f(x) ≥ 0 for all x in D.
Let W be the region in R3 that lies below the graph z = f(x, y) and above D, as shown in
Figure 5.1. That is:
W = {(x, y, z) ∈ R3 : (x, y) ∈ D and 0 ≤ z ≤ f(x, y)}.",Multivariable_Calculus_Shimamoto.pdf
"Figure 5.1. That is:
W = {(x, y, z) ∈ R3 : (x, y) ∈ D and 0 ≤ z ≤ f(x, y)}.
Figure 5.1: The region W under the graph z = f(x, y) and above D
Consider the volume of W, which we denote variously by
RR
D f(x, y) dA,
RR
D f(x, y) dx dy, or
just
RR
D f dA. To calculate it, we apply the following principle from first-year calculus:
Volume = Integral of cross-sectional area.
For instance, in first-year calculus, the cross-sections may turn out to be disks, washers, or, in a
slight variant, cylinders.
Example 5.1. Find the volume of the region under the plane z = 4 −x + 2y and above the square
1 ≤ x ≤ 3, 2 ≤ y ≤ 4, in the xy-plane. See Figure 5.2.
121",Multivariable_Calculus_Shimamoto.pdf
"122 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.2: The region under z = 4 − x + 2y and above the square 1 ≤ x ≤ 3, 2 ≤ y ≤ 4
Here, f(x, y) = 4 − x + 2y, and, if D denotes the specified square, we are looking for
RR
D(4 −
x + 2y) dA. As above, let W denote the region in question, below the plane and above the square.
We consider first cross-sections perpendicular to the x-axis, that is, intersections of W with planes
of the form x = constant. There is such a cross-section for each value of x from x = 1 to x = 3, so:
Volume =
Z 3
1
A(x) dx, (5.1)
where A(x) is the area of the cross-section at x. A typical cross-section is shown in Figure 5.3. The
Figure 5.3: A cross-section perpendicular to the x-axis
base is a line segment from y = 2 to y = 4, and the top is a curve lying in the graph z = 4 −x + 2y.
(In fact, the cross-sections in this example are trapezoids, but let’s not worry about that.) In other
words, A(x) is an area under a curve, so it too is an integral: A(x) =
R4",Multivariable_Calculus_Shimamoto.pdf
"words, A(x) is an area under a curve, so it too is an integral: A(x) =
R4
2 (4 − x + 2y) dy, where,
for the given cross-section, x is fixed. Integrating the cross-sectional area as in equation (5.1) and
evaluating gives:
Volume =
Z 3
1
Z 4
2

4 − x + 2y


dy

dx
=
Z 3
1

4y − xy + y2

y=4
y=2
dx
=
Z 3
1

(16 − 4x + 16) − (8 − 2x + 4)


dx
=
Z 3
1
(20 − 2x) dx
= 20x − x23
1
= (60 − 9) − (20 − 1)
= 32.",Multivariable_Calculus_Shimamoto.pdf
"5.1. VOLUME AND ITERATED INTEGRALS 123
Alternatively, we could have used cross-sections perpendicular to the y-axis, in which case the
cross-sections occur from y = 2 to y = 4 (Figure 5.4):
Volume =
Z 4
2
A(y) dy. (5.2)
Figure 5.4: A cross-section perpendicular to the y-axis
Each cross-section lies below the graph and above a line segment that goes from x = 1 to x = 3,
so A(y) =
R3
1 (4 − x + 2y) dx. This time, the integral of the cross-sectional area (5.2) gives:
Volume =
Z 4
2
Z 3
1

4 − x + 2y


dx

dy
=
Z 4
2

4x − 1
2x2 + 2yx


x=3
x=1
dy
=
Z 4
2

(12 − 9
2 + 6y) − (4 − 1
2 + 2y)


dy
=
Z 4
2
(4 + 4y) dy
= 4y + 2y24
2
= 48 − 16
= 32.
According to this approach, the volume is the integral of an integral, or an iterated integral.
Evaluating the integral consists of a sequence of “partial antidifferentiations” with respect to one
variable at a time, all other variables being held constant. Once you integrate with respect to a",Multivariable_Calculus_Shimamoto.pdf
"variable at a time, all other variables being held constant. Once you integrate with respect to a
particular variable, that variable disappears from the remainder of the calculation.
Example 5.2. Find the volume of the solid that lies inside the cylindersx2 +z2 = 1 and y2 +z2 = 1
and above the xy-plane.
The cylinders have axes along the y and x-axis, respectively, and intersect one another at right
angles, as in Figure 5.5. Perhaps it’s simplest to focus on the quarter of the solid that lies in the
“first octant,” i.e., the region of R3 in which x, y, and z are all nonnegative. The base of this portion
is the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (Figure 5.6, left). Half of the solid lies under x2 + z2 = 1
(Figure 5.6, right), running in the y-direction above the triangle in the xy-plane bounded by the
x-axis, the line x = 1, and the line y = x. Call this triangle D1. The other half runs in the
x-direction, under y2 + z2 = 1 and above a complementary triangle D2 within the unit square.",Multivariable_Calculus_Shimamoto.pdf
"124 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.5: The cylinders x2 + z2 = 1 (left), y2 + z2 = 1 (middle), and the region contained in both
(right)
Figure 5.6: The portion in the first octant (left) and the portion under only x2 + z2 = 1 (right)
The volume of the original solid is 8 times the volume of the portion over D1. This lies below
x2 + z2 = 1, i.e., below the graph z =
√
1 − x2. Hence:
Volume = 8
ZZ
D1
p
1 − x2 dA.
To calculate the volume over D1, let’s try cross-sections perpendicular to the x-axis. These
exist from x = 0 to x = 1:
Volume = 8
Z 1
0
A(x) dx.
The base of each cross-section is a line segment perpendicular to the x-axis within D1. The lower
endpoint is always at y = 0, but the upper endpoint depends on x. Indeed, the upper endpoint lies
on the line y = x. See Figure 5.7.
Figure 5.7: The base of the cross-section at x",Multivariable_Calculus_Shimamoto.pdf
"5.1. VOLUME AND ITERATED INTEGRALS 125
Hence A(x) =
Rx
0
√
1 − x2 dy, so:
Volume = 8
Z 1
0
Z x
0
p
1 − x2 dy

dx
= 8
Z 1
0
p
1 − x2 · y

y=x
y=0

dx
= 8
Z 1
0
x
p
1 − x2 dx (let u = 1 − x2, du= −2x dx)
= 8

−1
2 · 2
3(1 − x2)3/21
0

= −8
3(03/2 − 13/2)
= 8
3.
Alternatively, suppose we had used cross-sections perpendicular to they-axis. The cross-sections
exist from y = 0 to y = 1, and, for each such y, the base of the cross-section is a line segment in
the x-direction that goes from x = y to x = 1, as illustrated in Figure 5.8.
Figure 5.8: The base of the cross-section at y
Consequently:
Volume = 8
Z 1
0
A(y) dy = 8
Z 1
0
Z 1
y
p
1 − x2 dx

dy.
While not impossible, it is a little less obvious how to do the first partial antidifferentiation with
respect to x by hand. In other words, our original order of antidifferentiation might be preferable.
Example 5.3. Consider the iterated integral
Z 2
0
Z 4
x2
x3ey3
dy

dx.
(a) Sketch the domain of integration D in the xy-plane.",Multivariable_Calculus_Shimamoto.pdf
"Example 5.3. Consider the iterated integral
Z 2
0
Z 4
x2
x3ey3
dy

dx.
(a) Sketch the domain of integration D in the xy-plane.
(b) Evaluate the integral.
(a) The domain of integration can be reconstructed from the endpoints of the integrals. The
outermost integral says that x goes from x = 0 to x = 2. Geometrically, we are integrating areas
of cross-sections perpendicular to the x-axis. Then, the inner integral says that, for each x, y goes
from y = x2 to y = 4. Figure 5.9 shows the relevant input. Hence D is described by the conditions:
D = {(x, y) ∈ R2 : 0 ≤ x ≤ 2, x2 ≤ y ≤ 4}.
This is the region in the first quadrant bounded by the parabola y = x2, the line y = 4, and the
y-axis.",Multivariable_Calculus_Shimamoto.pdf
"126 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.9: The domain of integration: cross-sections perpendicular to the x-axis
(b) To evaluate the integral as presented, we would antidifferentiate first with respect toy, treating
x as constant: Z 2
0
Z 4
x2
x3ey3
dy

dx =
Z 2
0

x3
Z 4
x2
ey3
dy

dx.
The innermost antiderivative looks hard. So, having nothing better to do, we try switching the
order of antidifferentiation. Changing to cross-sections perpendicular to the y-axis, we see from the
description of D that y goes from y = 0 to y = 4, and, for each y, x goes from x = 0 to x = √y.
The thinking behind the switched order is illustrated in Figure 5.10.
Figure 5.10: Reversing the order of antidifferentiation
Therefore:
Z 2
0
Z 4
x2
x3ey3
dy

dx =
Z 4
0
Z √y
0
x3ey3
dx

dy
=
Z 4
0
1
4x4ey3

x=√y
x=0

dy
=
Z 4
0
1
4y2ey3
− 0


dy (let u = y3, du= 3y2 dy)
= 1
4 · 1
3ey3 4
0
= 1
12(e64 − 1).
5.2 The double integral
We now define the integral
RR",Multivariable_Calculus_Shimamoto.pdf
"
dy
=
Z 4
0
1
4y2ey3
− 0


dy (let u = y3, du= 3y2 dy)
= 1
4 · 1
3ey3 4
0
= 1
12(e64 − 1).
5.2 The double integral
We now define the integral
RR
D f(x, y) dA of a real-valued function f of two variables whose domain
is a subset D of R2. We assume the following:",Multivariable_Calculus_Shimamoto.pdf
"5.2. THE DOUBLE INTEGRAL 127
• D is a bounded subset of R2: this means that there is a rectangle R in R2 such that D ⊂ R.
• f is a bounded function: this means that there is a scalar M such that |f(x)| ≤M for all
x in D.
To define the integral, we take as a guiding principle that it is a limit of weighted sums. This is
often what gives the integral its power.
We begin by reviewing the one-variable case. Let f : [a, b] → R be defined on a closed interval
[a, b]. We subdivide the interval inton subintervals of widths△x1, △x2, . . . ,△xn and choose a point
pi in the ith subinterval for i = 1, 2, . . . , n. See Figure 5.11. We then form the sum P
i f(pi) △xi.
This is called a Riemann sum , and we refer to the pi as sample points . We think of the
Figure 5.11: Subdividing [ a, b] into n subintervals of widths △x1, △x2, . . . ,△xn with sample points
p1, p2, . . . , pn
Riemann sum as the sum of the values f(pi) at the sample points weighted by the width of the",Multivariable_Calculus_Shimamoto.pdf
"p1, p2, . . . , pn
Riemann sum as the sum of the values f(pi) at the sample points weighted by the width of the
subintervals (though it may be equally appropriate on occasion to think of it as the widths of the
subintervals weighted by the function values). The integral is defined as a limit of the Riemann
sums: Z b
a
f(x) dx = lim
△xi→0
X
i
f(pi) △xi.
The limit is a different type from the ones we encountered before. Here, it means that, given any
ϵ >0, there exists a δ >0 such that, for all Riemann sums P
i f(pi) △xi based on subdivisions of
[a, b] with △xi < δfor all i, we have |P
i f(pi) △xi −
Rb
a f(x) dx| < ϵ. This is extremely cumbersome
to work with.
Moving on to functions of two variables f : D → R, where D ⊂ R2, we do our best to mimic
the one-variable case and proceed in two steps.
Step 1. Integrals over rectangles.
We first introduce some notation that makes it easier to specify the kind of rectangles we have
in mind.",Multivariable_Calculus_Shimamoto.pdf
"Step 1. Integrals over rectangles.
We first introduce some notation that makes it easier to specify the kind of rectangles we have
in mind.
Definition. If X and Y are sets, then the Cartesian product X × Y is defined to be the set of
all ordered pairs:
X × Y = {(x, y) : x ∈ X, y∈ Y }.
For instance, the product of two closed intervals R = [a, b] × [c, d] = {(x, y) : x ∈ [a, b], y∈
[c, d]} = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d} is a rectangle in R2 whose sides are parallel to the
coordinate axes (Figure 5.12). We define how to integrate a function f(x, y) over such a rectangle
R.
We chop up R into a grid of subrectangles with sides parallel to the axes and of dimensions
△xi by △yj. A simple example is shown in Figure 5.13. We choose a sample point pij in each
subrectangle and consider the Riemann sum P
i,j f(pij) △xi △yj. This is a weighted sum in which
the function values at the sample points are weighted by the areas of the subrectangles.
Now, let △xi and △yj go to 0.",Multivariable_Calculus_Shimamoto.pdf
"128 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.12: The rectangle R = [a, b] × [c, d] in R2
Figure 5.13: A subdivision of a rectangle into 4 subrectangles
Definition. Let R = [ a, b] × [c, d] be a rectangle in R2. The integral of a bounded function
f : R → R is defined by:
ZZ
R
f(x, y) dA = lim
△xi→0
△yj→0
X
i,j
f(pij) △xi △yj,
assuming that the limit exists. When it does, we say that f is integrable on R.
There are other versions of integrals, too, and the one defined here is known technically as the
Riemann integral. Integrals of functions of two variables are called double integrals.
We begin with a standard somewhat artificial example that is constructed to make a point.
Example 5.4. Let R = [0, 1] × [0, 1], and define f : R → R by:
f(x, y) =
(
0 if x and y are both rational,
1 otherwise .
For instance, f(1
2 , 3
5 ) = 0, while f(1
2 ,
√
2
2 ) = 1 and f(π
6 ,
√
2
2 ) = 1. The graph z = f(x, y) consists of",Multivariable_Calculus_Shimamoto.pdf
"1 otherwise .
For instance, f(1
2 , 3
5 ) = 0, while f(1
2 ,
√
2
2 ) = 1 and f(π
6 ,
√
2
2 ) = 1. The graph z = f(x, y) consists of
two 1 by 1 squares, one at height z = 0 and one at height z = 1, both with lots of missing points.
To evaluate the integral
RR
R f(x, y) dA using the definition, consider a subdivision of R into
subrectangles. Regardless of the subdivision, each subrectangle contains points where f = 0 as well
as points where f = 1. Thus, depending on the choice of sample points, some Riemann sums have
the form P
i.j 0·△xi △yj = 0, and some have the formP
i,j 1·△xi △yj = Area of R = 1. Still others
have values in between. This is true no matter what △xi and △yj are, so P
i,j f(pij) △xi △yj does
not approach a limit as △xi and △yj go to 0. Consequently f is not integrable on R.
This raises the question of whether it’s possible to tell when a function is integrable. For the
preceding function, in some sense the problem is that it is too discontinuous.",Multivariable_Calculus_Shimamoto.pdf
"5.2. THE DOUBLE INTEGRAL 129
Theorem 5.5. Let f : R → R be a bounded function defined on a rectangle R = [a, b] × [c, d]. If
f is continuous on R, except possibly at a finite number of points or on a finite number of smooth
curves, then f is integrable on R.
This is actually a special case of a more general theorem that characterizes precisely when a
function is Riemann integrable. Unfortunately, the proof uses concepts that are best left for a
course in real analysis, so we do not go into it further. One consequence, however, is the useful
fact that continuous functions are integrable. This is worth remembering. It covers most of the
functions that we work with.
A second question is that, once a function is known to be integrable, is there a practical way to
evaluate the integral? Here again, we only indicate why the answer is plausible without attempting
anything like a correct proof. The intuition is that, after you subdivide a rectangle and choose",Multivariable_Calculus_Shimamoto.pdf
"anything like a correct proof. The intuition is that, after you subdivide a rectangle and choose
sample points, you can organize the Riemann sum P
i,j f(pij) △xi △yj either:
• column-by-column: P
i
P
j f(pij) △yj


△xi, or
• row-by-row: P
j
P
i f(pij) △xi


△yj.
These are Riemann sums of Riemann sums. The first sum is an approximation of the iterated
integral
Rb
a
Rd
c f(x, y) dy


dx, while the second sum approximates
Rd
c
Rb
a f(x, y) dx


dy. Refer to
Figure 5.14. As △xi and △yj go to 0, these approximations become exact, and, in the limit, the
iterated integrals coincide with the double integral.
Figure 5.14: Organizing Riemann sums row-by-row or column-by-column. For instance, a typical
column sum is P
j f(pij) △yj, where i is fixed.
Theorem 5.6 (Fubini’s theorem). If f is integrable on R = [a, b] × [c, d], then
RR
R f(x, y) dA is
equal to both of the iterated integrals:
ZZ
R
f(x, y) dA =
Z b
a
Z d
c
f(x, y) dy

dx =
Z d
c
Z b
a
f(x, y) dx

dy.",Multivariable_Calculus_Shimamoto.pdf
"RR
R f(x, y) dA is
equal to both of the iterated integrals:
ZZ
R
f(x, y) dA =
Z b
a
Z d
c
f(x, y) dy

dx =
Z d
c
Z b
a
f(x, y) dx

dy.
In particular, the iterated integrals that we computed in Section 5.1 were actually examples of
double integrals.
Step 2. Integrals over bounded sets in general.
Let f : D → R be a bounded function defined on a bounded subset D of R2. By definition, D is
contained in a rectangle R. By taking an even larger rectangle, if necessary, we may assume that
R has the form [ a, b] × [c, d]. Define a function ef : R → R by:
ef(x, y) =
(
f(x, y) if ( x, y) ∈ D,
0 otherwise .",Multivariable_Calculus_Shimamoto.pdf
"130 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
An example of a graph of ef is shown in Figure 5.15.
Figure 5.15: The general case: the graph z = ef(x, y)
The integral of ef, as a function defined on a rectangle, has been covered in Step 1.
Definition. We say that f : D → R is integrable on D if ef is integrable on R. If so, we defineRR
D f(x, y) dA =
RR
R
ef(x, y) dA.
Thus the integral of f over D is defined as a limit of Riemann sums for ef, P
i,j ef(pij) △xi △yj,
based on subdivisions of the larger rectangle R into subrectangles. Since ef = 0 away from D,
however, only those subrectangles that intersect D contribute to the Riemann sum. Hence we
might write: ZZ
D
f(x, y) dA = lim
△xi→0
△yj→0
X
subrectangles
that intersectD
f(pij) △xi △yj.
The subdivision is indicated in Figure 5.16.
Figure 5.16: The general case: subdividing a rectangle containing D into subrectangles. Only those
subrectangles that intersect D may contribute to a Riemann sum for ef.",Multivariable_Calculus_Shimamoto.pdf
"subrectangles that intersect D may contribute to a Riemann sum for ef.
As Figure 5.15 indicates, even iff is continuous onD, the function ef might well be discontinuous
on the boundary of D. Nevertheless, by Theorem 5.5, ef will still be integrable on R as long as the
boundary of D consists of finitely many smooth curves. In other words, the integral can tolerate
a certain amount of bad behavior on the boundary. Having this latitude is something we take
advantage of later on.
Here are some basic properties of the integral. They are quite reasonable and can be proven
using the definition of the integral as a limit of sums.
Proposition 5.7. Let f and g be integrable functions on a bounded set D in R2. Then:
1.
RR
D(f + g) dA =
RR
D f dA+
RR
D g dA.",Multivariable_Calculus_Shimamoto.pdf
"5.3. INTERPRETATIONS OF THE DOUBLE INTEGRAL 131
2.
RR
D cf dA= c
RR
D f dAfor any scalar c.
3. If f(x, y) ≤ g(x, y) for all (x, y) in D, then
RR
D f dA≤
RR
D g dA.
5.3 Interpretations of the double integral
We have defined the integral of a bounded real-valued function f(x, y) of two variables over a
bounded subset D of R2 as:
ZZ
D
f(x, y) dA = lim
△xi→0
△yj→0
X
subrectangles
that intersectD
f(pij) △xi △yj,
where the subrectangles come from subdividing a rectangle R that contains D. Now, we go about
trying to understand why this might be useful.
First, the integral is a limit of sums. Therefore, if a typical summand f(pij) △xi △yj represents
“something,” then
RR
D f(x, y) dA represents the total “something.” Also, we have noted before that
a Riemann sum is the sum of the sample values f(pij) weighted by the area of the subrectangles.
We consider a few situations where such a weighted sum might arise.",Multivariable_Calculus_Shimamoto.pdf
"We consider a few situations where such a weighted sum might arise.
Example 5.8. The first is when f(x, y) represents some sort of density, or quantity per unit
area, at the point ( x, y), for instance, mass density or population density. Then f(pij) △xi △yj is
approximately the quantity on a typical subrectangle, and:
ZZ
D
(density)dA = total quantity in D,
for instance, the total mass or total population.
Example 5.9. Suppose that f is the constant function f(x, y) = 1 for all ( x, y) in D. Then
f(pij) △xi △yj = 1 · △xi △yj is the area of a subrectangle, so:
ZZ
D
1 dA = Area (D).
Example 5.10. Let’s return to the graph of a function z = f(x, y), where f(x, y) ≥ 0 for all (x, y)
in D. Then f(pij) is the height of the graph above the sample point pij, and f(pij) △xi △yj is
approximately the volume below the graph and above a typical subrectangle (Figure 5.17). Thus:
ZZ
D
f(x, y) dA = volume below the graph of f and above D.",Multivariable_Calculus_Shimamoto.pdf
"ZZ
D
f(x, y) dA = volume below the graph of f and above D.
As a result, volume can be regarded either as an iterated integral, as in Section 5.1, or as a
double integral, i.e., a limit of Riemann sums. That these two approaches give the same thing is
another way of understanding Fubini’s theorem.
Example 5.11. Average value of a function.
To average a list of numbers, say 2 , 4, 7, 4, 4, we find the sum and divide by the number of
entries:
average = 2 + 4 + 7 + 4 + 4
5 = 21
5 = 4.2.",Multivariable_Calculus_Shimamoto.pdf
"132 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.17: Visualizing a Riemann sum as approximating the volume under a graph
This could also be written as:
2 · 1
5 + 4 · 1
5 + 7 · 1
5 + 4 · 1
5 + 4 · 1
5 or 2 · 1
5 + 4 · 3
5 + 7 · 1
5.
In other words, it’s a sum in which each value is weighted by its relative frequency.
Similarly, to find the average value of a function f(x, y) over a region D, we take a sampling
of points pij in D and consider the sum in which each function value f(pij) is weighted by the
fraction of D that it represents. If the sampling comes from a subdivision into subrectangles and
we think of each sample point as representing its subrectangle, this gives:
Average value ≈
X
f(pij) △xi △yj
Area (D) = 1
Area (D)
X
f(pij) △xi △yj.
Taking the limit as △xi and △yj go to 0 leads to the following definition.
Definition.
1. The average value of a bounded function f(x, y) on a bounded subset D of R2, denoted by
f, is defined to be:
f = 1
Area (D)
ZZ
D
f dA.",Multivariable_Calculus_Shimamoto.pdf
"Definition.
1. The average value of a bounded function f(x, y) on a bounded subset D of R2, denoted by
f, is defined to be:
f = 1
Area (D)
ZZ
D
f dA.
2. For example, the average x and y-coordinates on D are:
x = 1
Area (D)
ZZ
D
x dA and y = 1
Area (D)
ZZ
D
y dA.
The point (x, y) is called the centroid of D.
Example 5.12. Let D be the triangular region in R2 with vertices (0, 0), (2, 2), and (−2, 2).
(a) Find the average value of f(x, y) = y2 − x2 on D. The graph of f is in Figure 5.18.
(b) Find the centroid of D.
(a) Here, f = 1
Area (D)
RR
D(y2 − x2) dA, where the region D is shown in Figure 5.19. We could find
the area of D using Area (D) =
RR
D 1 dA as in Example 5.9, but, since D is a triangle, we can use
geometry instead:
Area (D) = 1
2(base)(height) = 1
2 · 4 · 2 = 4.",Multivariable_Calculus_Shimamoto.pdf
"5.3. INTERPRETATIONS OF THE DOUBLE INTEGRAL 133
Figure 5.18: The graph of f(x, y) = y2 − x2 over the triangle D
Figure 5.19: The triangle D
To set up the double integral
RR
D(y2 −x2) dA as an iterated integral, note that D is bounded on
the left by y = −x, on the right by y = x, and on top by y = 2. We use cross-sections perpendicular
to the y-axis. These cross-sections exist from y = 0 to y = 2, and, for each y, x goes from x = −y
to x = y. Hence:
ZZ
D
(y2 − x2) dA =
Z 2
0
Z y
−y
(y2 − x2) dx

dy
=
Z 2
0

y2x − 1
3x3

x=y
x=−y
dy
=
Z 2
0

(y3 − 1
3y3) − (−y3 + 1
3y3)


dy
=
Z 2
0
4
3y3 dy
= 1
3y42
0
= 16
3 .
Therefore f = 1
4 · 16
3 = 4
3 .
As a side note, suppose we had tried to calculate the double integral using cross-sections per-
pendicular to the x-axis instead. These cross-sections exist from x = −2 to x = 2, but the y
endpoints of a given cross-section are different for the left and right halves of the triangle. When x",Multivariable_Calculus_Shimamoto.pdf
"endpoints of a given cross-section are different for the left and right halves of the triangle. When x
is between −2 and 0, the cross-section at x goes from y = −x to y = 2, whereas when x is between
0 and 2, it goes from y = x to y = 2 (Figure 5.20). Thus, to describe D this way, the integral must
be split into two pieces:
ZZ
D
(y2 − x2) dA =
Z 0
−2
Z 2
−x
(y2 − x2) dy

dx +
Z 2
0
Z 2
x
(y2 − x2) dy

dx.",Multivariable_Calculus_Shimamoto.pdf
"134 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.20: Reversing the order of integration to integrate over D
Actually, we could try to take advantage of symmetry properties of D and f to avoid having
to evaluate both halves. We take up this idea more fully when we discuss the change of variables
theorem in Chapter 7.
(b) Because of the symmetry of D in the y-axis, it’s intuitively clear that the average x-coordinate
is 0: x = 0. Again, we justify this more carefully later (see Example 7.7 in Chapter 7).
For the average y-coordinate, by definition, y = 1
Area (D)
RR
D y dA= 1
4
RR
D y dA. We describe D
using the same limits of integration as in part (a) with cross-sections perpendicular to the y-axis:
ZZ
D
y dA=
Z 2
0
Z y
−y
y dx

dy
=
Z 2
0

yx

x=y
x=−y

dy
=
Z 2
0
(y2 − (−y2)) dy
=
Z 2
0
2y2 dy
= 2
3y32
0
= 16
3 .
Thus y = 1
4 · 16
3 = 4
3 , and the centroid is the point ( x, y) = (0, 4
3 ).
5.4 Parametrizations of surfaces",Multivariable_Calculus_Shimamoto.pdf
"=
Z 2
0
2y2 dy
= 2
3y32
0
= 16
3 .
Thus y = 1
4 · 16
3 = 4
3 , and the centroid is the point ( x, y) = (0, 4
3 ).
5.4 Parametrizations of surfaces
Thus far, we have studied surfaces that arise as:
• graphs z = f(x, y) (or, analogously, y = g(x, z) and x = h(y, z)) or
• level sets f(x, y, z) = c.
We next describe surfaces by how they are traced out. This is like the way we described curves using
parametrizations, except that now, since surfaces are two-dimensional, we need two parameters. In
other words, surfaces will be described by functions σ : D → Rn, where D ⊂ R2, as illustrated in
Figure 5.21. We want the domain of the parametrization to be two-dimensional, so we assume that
D is an open set U in R2 together with all of its boundary points. The technical term is that D
is the closure of U. Actually, the theory carries over if some of the boundary points are missing,",Multivariable_Calculus_Shimamoto.pdf
"5.4. PARAMETRIZATIONS OF SURFACES 135
but, in our examples, D will often be a rectangle or a closed disk in R2, in which case all boundary
points are included.
The coordinates in D are the parameters. There are often natural choices for what to call them
depending on the surface, though generically we call them s and t. For every ( s, t) in D, σ(s, t)
is a point in Rn, so we may write σ(s, t) = ( x1(s, t), x2(s, t), . . . , xn(s, t)). We shall focus almost
exclusively on surfaces in R3, in which case σ(s, t) = (x(s, t), y(s, t), z(s, t)).
Figure 5.21: A parametrization σ of a surface S in R3
The reason for introducing the topic now is that this is a good time to talk about how to integrate
real-valued functions over surfaces. Because surfaces are two-dimensional objects, in principle, these
integrals should be extensions of the double integrals we are in the midst of studying. For example,
the interpretations of limits of weighted sums that apply to integrals over regions in the plane",Multivariable_Calculus_Shimamoto.pdf
"the interpretations of limits of weighted sums that apply to integrals over regions in the plane
should then carry over to integrals over surfaces as well.
In the remainder of this section, we look at several examples of parametrizations of surfaces
and, in particular, some common geometric parameters. Integrals we leave for the next section.
Example 5.13. Consider the portion of the cone z =
p
x2 + y2 in R3 where 0 ≤ z ≤ 2. See Figure
5.22. The surface is already described in terms of x and y, so we use ( x, y) as parameters with
z =
p
x2 + y2. That is, let:
σ(x, y) = (x, y,
p
x2 + y2).
The domain D of σ is the set of all ( x, y) such that 0 ≤
p
x2 + y2 ≤ 2, or x2 + y2 ≤ 4. This is a
disk of radius 2 in the xy-plane. As ( x, y) ranges over D, σ(x, y) sweeps out the given portion of
the cone.
Figure 5.22: A parametrization of a cone using x and y as parameters",Multivariable_Calculus_Shimamoto.pdf
"136 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
In this example, we used the x and y coordinates of the xyz-coordinate system as parameters
for the surface. This is always an option for surfaces that are graphs z = f(x, y) of functions of
two variables. We next introduce some other coordinate systems that can be used to locate points
in the plane or in three-dimensional space. These coordinate systems provide the most convenient
way to approach a variety of situations, though, for the time being, we just show how two of the
coordinates are natural parameters for certain common surfaces.
5.4.1 Polar coordinates ( r, θ) in R2
We referred to polar coordinates in Chapter 3, but perhaps it is best to introduce them formally.
To plot a point in R2 with polar coordinates ( r, θ), go out r units along the positive x-axis, then
rotate counterclockwise about the origin by the angle θ (Figure 5.23). This brings you to the
Figure 5.23: Polar coordinates ( r, θ)",Multivariable_Calculus_Shimamoto.pdf
"rotate counterclockwise about the origin by the angle θ (Figure 5.23). This brings you to the
Figure 5.23: Polar coordinates ( r, θ)
point (r cos θ, rsin θ) in rectangular coordinates. Thus the conversions from polar to rectangular
coordinates are: (
x = r cos θ
y = r sin θ.
Also, by the Pythagorean theorem, r =
p
x2 + y2. All of R2 can be described by choosing r and θ
in the ranges r ≥ 0, 0 ≤ θ ≤ 2π. Actually, θ = 0 and θ = 2π represent the same points, along the
positive x-axis, but it is convenient to allow this degree of redundancy.
For example, the disk D of radius 2 about the origin in Example 5.13 can be described in polar
coordinates by the conditions 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, that is, ( r, θ) ∈ [0, 2] × [0, 2π]. We think of
imposing polar coordinates on D as a transformation T : [0, 2] × [0, 2π] → R2 from the rθ-plane to
the xy-plane, where T(r, θ) = (r cos θ, rsin θ). Horizontal segments 0 ≤ r ≤ 2, θ = constant, in the",Multivariable_Calculus_Shimamoto.pdf
"the xy-plane, where T(r, θ) = (r cos θ, rsin θ). Horizontal segments 0 ≤ r ≤ 2, θ = constant, in the
rθ-plane get mapped to radial segments emanating from the origin in the xy-plane, and vertical
segments r = constant, 0 ≤ θ ≤ 2π, get mapped to circles centered at the origin. This is illustrated
in Figure 5.24.
Example 5.14. We return to the cone z =
p
x2 + y2, 0 ≤ z ≤ 2, from Example 5.13. In polar
coordinates, the equation of the cone is z = r, and combining this with the polar description of the
disk D gives a parametrization eσ of the cone with r and θ as parameters:
eσ : [0, 2] × [0, 2π] → R3, eσ(r, θ) = (r cos θ, rsin θ, r).
See Figure 5.25. The actual cone is the same one as before, but one advantage of this parametriza-
tion is that, for the purposes of integration, rectangles are nicer domains than disks.",Multivariable_Calculus_Shimamoto.pdf
"5.4. PARAMETRIZATIONS OF SURFACES 137
Figure 5.24: The polar coordinate transformation from the rθ-plane to the xy-plane
Figure 5.25: A parametrization of a cone using polar coordinates r and θ as parameters
5.4.2 Cylindrical coordinates ( r, θ, z) in R3
Cylindrical coordinates are polar coordinates in the xy-plane together with the usual z-coordinate.
To plot a point with cylindrical coordinates (r, θ, z), go out r units along the positive x-axis, rotate
counterclockwise about the positive z-axis by the angle θ, and then go vertically z units. These
coordinates are shown in Figure 5.26. This brings you to the point ( r cos θ, rsin θ, z) in xyz-space.
Figure 5.26: Cylindrical coordinates ( r, θ, z)
Thus the conversions from cylindrical to rectangular coordinates are:



x = r cos θ
y = r sin θ
z = z.",Multivariable_Calculus_Shimamoto.pdf
"138 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Again r =
p
x2 + y2. All of R3 is covered by choosing r ≥ 0, 0 ≤ θ ≤ 2π, −∞ < z <∞.
Example 5.15 (Circular cylinders). We parametrize the circular cylinder of radius a and height
h given by the conditions:
x2 + y2 = a2, 0 ≤ z ≤ h.
The axis of the cylinder is the z-axis.
In cylindrical coordinates, along the cylinder, r is fixed at the radius a, but θ and z can vary
independently. The surface is traced out by setting r = a and letting θ and z range over 0 ≤ θ ≤ 2π,
0 ≤ z ≤ h. In other words, we can parametrize the cylinder using θ and z as parameters. With
r = a fixed, the conversions from cylindrical to rectangular coordinates give the parametrization:
σ : [0, 2π] × [0, h] → R3, σ (θ, z) = (a cos θ, asin θ, z).
See Figure 5.27.
Figure 5.27: A parametrization of a cylinder using cylindrical coordinates θ and z as parameters
It is instructive to think about how σ transforms the rectangle D = [0, 2π] × [0, h] to become",Multivariable_Calculus_Shimamoto.pdf
"It is instructive to think about how σ transforms the rectangle D = [0, 2π] × [0, h] to become
the cylinder S. In polar/cylindrical coordinates, θ = 0 and θ = 2π represent the same thing in
xyz-space. Thus σ maps the points (0 , z) and (2π, z) in the θz-parameter plane to the same point
in R3. Apart from this, distinct points in D get mapped to distinct points in xyz-space. One way
to construct a cylinder is to take a rectangle like D and glue together, or identify, points (0 , z) on
the left edge with points (2 π, z) on the right. It is clear geometrically that the result is a cylinder,
and σ gives a formula that carries it out.
5.4.3 Spherical coordinates ( ρ, ϕ, θ) in R3
The three spherical coordinates of a point in R3 are described geometrically as follows.
• ρ is the distance from the origin to the point.
• ϕ is the angle from the positive z-axis to the point. It is like an angle of latitude, except",Multivariable_Calculus_Shimamoto.pdf
"• ρ is the distance from the origin to the point.
• ϕ is the angle from the positive z-axis to the point. It is like an angle of latitude, except
measured down from the positive z-axis rather than up or down from the equatorial plane.
• θ is the usual polar/cylindrical angle.
To plot a point in R3 with spherical coordinates ( ρ, ϕ, θ), first go out ρ units along the positive z-
axis. This brings you to the point (0, 0, ρ) in rectangular coordinates. Then, rotate counterclockwise
about the positive y-axis by the angle ϕ, bringing you to the point ( ρ sin ϕ, 0, ρcos ϕ) in the xz-
plane, a distance ρ from the origin and r = ρ sin ϕ from the z-axis. Finally, rotate counterclockwise",Multivariable_Calculus_Shimamoto.pdf
"5.4. PARAMETRIZATIONS OF SURFACES 139
Figure 5.28: Spherical coordinates ( ρ, ϕ, θ)
about the positive z-axis by the angle θ. This lands at the point ( ρ sin ϕ cos θ, ρsin ϕ sin θ, ρcos ϕ).
The coordinates are labeled in Figure 5.28.
Thus the conversions from spherical to rectangular coordinates are:



x = ρ sin ϕ cos θ
y = ρ sin ϕ sin θ
z = ρ cos ϕ.
From the Pythagorean theorem, the distance from the origin is ρ =
p
x2 + y2 + z2. All of R3 is
covered by spherical coordinates in the intervals ρ ≥ 0, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π. (Note that the
angular interval 0 ≤ ϕ ≤ π ranges from the direction of the positive z-axis to the direction of the
negative z-axis and the interval 0≤ θ ≤ 2π sweeps all the way around thez-axis. This is why values
of ϕ in the interval π < ϕ≤ 2π are not needed—they would duplicate points already covered.)
Example 5.16 (Spheres). We parametrize the sphere of radius a centered at the origin:
x2 + y2 + z2 = a2.",Multivariable_Calculus_Shimamoto.pdf
"Example 5.16 (Spheres). We parametrize the sphere of radius a centered at the origin:
x2 + y2 + z2 = a2.
In spherical coordinates, the sphere is traced out by keeping the value of ρ fixed at the radius a,
while ϕ and θ vary independently. More precisely, set ρ = a, and let ϕ and θ range over 0 ≤ ϕ ≤ π,
0 ≤ θ ≤ 2π. Hence we use ϕ and θ as parameters, and, from the spherical to rectangular conversions
with ρ = a fixed, obtain the parametrization:
σ : [0, π] × [0, 2π] → R3, σ (ϕ, θ) = (a sin ϕ cos θ, asin ϕ sin θ, acos ϕ).
This is illustrated in Figure 5.29.
Again the parametrization gives precise instructions for how to transform the rectangle D =
[0, π] × [0, 2π] into a sphere. Here, all points where ϕ = 0 get mapped to the north pole. These are
the points along the left edge of D. Similarly, the right edge of D, where ϕ = π, gets mapped to
the south pole. Finally, points where θ = 0 and θ = 2π, that is, along the top and bottom edges,",Multivariable_Calculus_Shimamoto.pdf
"the south pole. Finally, points where θ = 0 and θ = 2π, that is, along the top and bottom edges,
are identified in pairs along a meridian on the sphere in the xz-plane. Thus, to construct a sphere
from a rectangle, collapse each of the left and right sides to points and zip together the remaining
two sides to close up the surface.
Example 5.17. We return one last time to the cone z =
p
x2 + y2, 0 ≤ z ≤ 2, which we
have parametrized twice already, once using rectangular coordinates ( x, y) and once using po-
lar/cylindrical coordinates (r, θ) as parameters. It can be parametrized using spherical coordinates
as well.",Multivariable_Calculus_Shimamoto.pdf
"140 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.29: A parametrization of a sphere using spherical coordinates ϕ and θ as parameters
Here, along the cone, the angle ϕ down from the positive z-axis is fixed at π
4 . To see this, note,
for instance, that the cross-section with the yz-plane is z = |y|. On the other hand, ρ and θ can
vary independently. For example, ρ ranges from 0 to 2
√
2. (To determine the upper endpoint, one
can either use trigonometry or solve for ρ using z = ρ cos ϕ with z = 2 and ϕ = π
4 .) Also, θ rotates
all the way around from 0 to 2 π. From the spherical conversions to rectangular coordinates, we
obtain a parametrization:
bσ : [0, 2
√
2] × [0, 2π] → R3,
bσ(ρ, θ) =

ρ sin π
4 cos θ, ρsin π
4 sin θ, ρcos π
4


=
√
2
2 ρ cos θ,
√
2
2 ρ sin θ,
√
2
2 ρ


.
We leave as an exercise the instructions that bσ gives for turning a rectangle into a cone. See Figure
5.30.
Figure 5.30: A parametrization of a cone using spherical coordinates ρ and θ as parameters",Multivariable_Calculus_Shimamoto.pdf
"5.30.
Figure 5.30: A parametrization of a cone using spherical coordinates ρ and θ as parameters
As the cone illustrates, it may be possible to parametrize a given surface in several different
ways.
5.5 Integrals with respect to surface area
We now discuss how to integrate a real-valued function f : S → R defined on a surface S in R3. To
formulate a reasonable definition, we follow the usual procedure for integrating: chop up S into a",Multivariable_Calculus_Shimamoto.pdf
"5.5. INTEGRALS WITH RESPECT TO SURFACE AREA 141
two-dimensional partition of small pieces of surface area △Sij, choose a sample point pij in each
piece, form a sum P
i,j f(pij) △Sij, and take the limit as the size of the pieces goes to zero.
We use a parametrization of S to convert the calculation into a double integral over a region in
the plane. The approach is similar to what we did when we defined integrals over curves with respect
to arclength, though that was a long time ago (Section 2.3 to be precise). Thus let σ : D → R3 be
a parametrization of S, where D ⊂ R2. We write σ(s, t) = ( x(s, t), y(s, t), z(s, t)). We subdivide
D (or, really, a rectangle that contains D) into small subrectangles of dimensions △si by △tj, and
choose a sample point aij in each subrectangle. Intuitively, the expectation is that σ transforms
each subrectangle into a small “curvy quadrilateral” in S. This is one of the small pieces of S",Multivariable_Calculus_Shimamoto.pdf
"each subrectangle into a small “curvy quadrilateral” in S. This is one of the small pieces of S
alluded to earlier. We denote its area by △Sij and take pij = σ(aij) as the corresponding sample
point. This is depicted in Figure 5.31.
Figure 5.31: The integral with respect to surface area
Let ∂σ
∂s and ∂σ
∂t be the vectors given by:
∂σ
∂s =
∂x
∂s , ∂y
∂s , ∂z
∂s

and ∂σ
∂t =
∂x
∂t , ∂y
∂t , ∂z
∂t

.
We think of these vectors as velocities with respect to s and t, respectively. Then σ transforms
the △si side of a typical subrectangle of D approximately to the vector ∂σ
∂s △si and the △tj side
approximately to ∂σ
∂t △tj. This gives an approximation of the area of the curvy quadrilaterals:
△Sij ≈ area of the parallelogram determined by ∂σ
∂s △si and ∂σ
∂t △tj
≈




∂σ
∂s △si × ∂σ
∂t △tj




 (by an old property about the cross product and areas, page 36)
≈




∂σ
∂s × ∂σ
∂t




△si △tj.
See Figure 5.32.
Hence P
i,j f(pij) △Sij ≈ P
i,j f(σ(aij)) ∥∂σ
∂s × ∂σ",Multivariable_Calculus_Shimamoto.pdf
"≈




∂σ
∂s × ∂σ
∂t




△si △tj.
See Figure 5.32.
Hence P
i,j f(pij) △Sij ≈ P
i,j f(σ(aij)) ∥∂σ
∂s × ∂σ
∂t ∥△si △tj. In the limit as △si and △tj go to
zero, this becomes the double integral
RR
D f(σ(s, t)) ∥∂σ
∂s × ∂σ
∂t ∥ds dt. With this as motivation, we
adopt the following definition.
Definition. Let S be a surface in R3, and let f : S → R be a continuous real-valued function.
Then the integral of f with respect to surface area , denoted by
RR
S f dS, is defined to be:
ZZ
S
f dS=
ZZ
D
f(σ(s, t))




∂σ
∂s × ∂σ
∂t




ds dt,
where σ : D → R3 is a parametrization of S.",Multivariable_Calculus_Shimamoto.pdf
"142 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
Figure 5.32: A blowup of the previous figure: σ sends a small △si by △tj subrectangle to a curvy
quadrilateral whose area △Sij is approximately


∂σ
∂s × ∂σ
∂t


△si △tj.
We should acknowledge a point that was also glossed over in connection with integrals with re-
spect to arclength. Namely, we motivated the surface integral using Riemann sumsP
i,j f(pij) △Sij
based on the surface, but the actual definition relies on a parametrization σ. Perhaps a better nota-
tion for the integral as defined would be
RR
σ f dS. The issue is whether the choice of parametrization
affects the value of the integral. Once we have a few more tools available, we shall see that there is
no need for concern: using the definition of the integral with two different parametrizations of the
same surface gives the same value. The effect of parametrizations on surface integrals is discussed
at the end of Chapter 10.",Multivariable_Calculus_Shimamoto.pdf
"same surface gives the same value. The effect of parametrizations on surface integrals is discussed
at the end of Chapter 10.
Example 5.18. If f(x, y, z) = 1 for all ( x, y, z) in S, then P
i,j f(pij) △Sij = P
i,j 1 · △Sij, the
sum of the small pieces of surface area. This is the total area of S, that is:
Surface area of S =
ZZ
S
1 dS. (5.3)
Example 5.19. Let S be the sphere x2 + y2 + z2 = a2 of radius a.
(a) Find the surface area of S.
(b) Find
RR
S z2 dS.
(c) Find
RR
S (x2 + y2 + z2) dS.
We use the parametrization of a sphere derived in Example 5.16 with spherical coordinates
(ϕ, θ) as parameters and ρ = a fixed:
σ : [0, π] × [0, 2π] → R3, σ (ϕ, θ) = (a sin ϕ cos θ, asin ϕ sin θ, acos ϕ).
Let D denote the domain of the parametrization, D = [0, π] × [0, 2π].
(a) Applying the previous example, we integrate the function f = 1. Then, using the definition of
the integral:
Area (S) =
ZZ
S
1 dS =
ZZ
D
1 ·




∂σ
∂ϕ × ∂σ
∂θ




dϕ dθ.
We calculate ∂σ
∂ϕ and ∂σ",Multivariable_Calculus_Shimamoto.pdf
"the integral:
Area (S) =
ZZ
S
1 dS =
ZZ
D
1 ·




∂σ
∂ϕ × ∂σ
∂θ




dϕ dθ.
We calculate ∂σ
∂ϕ and ∂σ
∂θ and place them in the rows of the determinant for a cross product:
∂σ
∂ϕ × ∂σ
∂θ = det


i j k
a cos ϕ cos θ a cos ϕ sin θ −a sin ϕ
−a sin ϕ sin θ a sin ϕ cos θ 0


=

a2 sin2 ϕ cos θ, a2 sin2 ϕ sin θ, a2 cos ϕ sin ϕ(cos2 θ + sin2 θ)


= a2 sin ϕ

sin ϕ cos θ, sin ϕ sin θ, cos ϕ


,",Multivariable_Calculus_Shimamoto.pdf
"5.5. INTEGRALS WITH RESPECT TO SURFACE AREA 143
so: 



∂σ
∂ϕ × ∂σ
∂θ




 = a2 sin ϕ
q
sin2 ϕ cos2 θ + sin2 ϕ sin2 θ + cos2 ϕ
= a2 sin ϕ
q
sin2 ϕ + cos2 ϕ
= a2 sin ϕ. (5.4)
Hence:
Area (S) =
ZZ
D
a2 sin ϕ dϕ dθ
=
Z 2π
0
Z π
0
a2 sin ϕ dϕ

dθ
=
Z 2π
0

−a2 cos ϕ

ϕ=π
ϕ=0

dθ
=
Z 2π
0

a2 − (−a2)


dθ
=
Z 2π
0
2a2 dθ
= 2a2θ
2π
0
= 4πa2.
This formula gets used enough that we put it in a box:
The surface area of a sphere of radius a is 4πa2.
(b) Here, f(x, y, z) = z2, and substituting for z in terms of the parameters gives f(σ(ϕ, θ)) =
a2 cos2 ϕ. Using the result of (5.4) for ∥∂σ
∂ϕ × ∂σ
∂θ ∥, we obtain:
ZZ
S
z2 dS =
ZZ
D
f(σ(ϕ, θ))




∂σ
∂ϕ × ∂σ
∂θ




dϕ dθ
=
ZZ
D
(a2 cos2 ϕ) · (a2 sin ϕ) dϕ dθ
= a4
ZZ
D
cos2 ϕ sin ϕ dϕ dθ
= a4
Z 2π
0
Z π
0
cos2 ϕ sin ϕ dϕ

dθ (let u = cos ϕ, du= −sin ϕ dϕ)
= a4
Z 2π
0

−1
3 cos3 ϕ

ϕ=π
ϕ=0

dθ
= a4
Z 2π
0
1
3 − (−1
3)


dθ
= a4
Z 2π
0
2
3 dθ
= 2
3a4θ
2π
0
= 4
3πa4.",Multivariable_Calculus_Shimamoto.pdf
"144 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
(c) This time, f(x, y, z) = x2 + y2 + z2. We use a trick! On the sphere S, x2 + y2 + z2 = a2, soRR
S (x2 + y2 + z2) dS =
RR
S a2 dS = a2 RR
S 1 dS = a2Area (S). Using the formula for the area from
part (a) then gives: ZZ
S
(x2 + y2 + z2) dS = a2 · 4πa2 = 4πa4.
The trick enabled us to evaluate the integral with almost no calculation. This is a highly
desirable situation. In retrospect, we can get even more mileage out of it by going back and using
the result of part (c) to redo
RR
S z2 dS in part (b). For the sphere S is symmetric in x, y, and z,
whence
RR
S x2 dS =
RR
S y2 dS =
RR
S z2 dS. Therefore
RR
S (x2 + y2 + z2) dS = 3
RR
S z2 dS, or:
ZZ
S
z2 dS = 1
3
ZZ
S
(x2 + y2 + z2) dS = 1
3 · 4πa4 = 4
3πa4,
where the next-to-last step used part (c). Happily, this is the same as the answer we obtained in
part (b) originally by grinding it out with a parametrization.
5.6 Triple integrals and beyond",Multivariable_Calculus_Shimamoto.pdf
"part (b) originally by grinding it out with a parametrization.
5.6 Triple integrals and beyond
Integrals of real-valued functions of three or more variables are conceptually the same as double
integrals, though the notation is more encumbered. We merely outline the definition. Given:
• a bounded subset W of Rn (that is, there is an n-dimensional box R = [a1, b1] × [a2, b2] ×
··· ×[an, bn] such that W ⊂ R) and
• a bounded real-valued function f : W → R (that is, there is a scalar M such that |f(x)| ≤M
for all x in W),
one subdivides W into small n-dimensional subboxes of dimensions △x1 by △x2 by ··· by △xn,
chooses a sample point p in each subbox, and considers sums of the form Pf(p) △x1 △x2 ···△ xn,
where the summation is over all the subboxes. The integral is the limit of these sums as the
dimensions of the subboxes go to zero. It is denoted by:
ZZ
···
Z
W
f dV or
ZZ
···
Z
W
f dx1 dx2 ··· dxn.
For instance, if W ⊂ R3 and f = f(x, y, z) is a function of three variables, then
RRR",Multivariable_Calculus_Shimamoto.pdf
"ZZ
···
Z
W
f dV or
ZZ
···
Z
W
f dx1 dx2 ··· dxn.
For instance, if W ⊂ R3 and f = f(x, y, z) is a function of three variables, then
RRR
W f dVis called
a triple integral.
The higher-dimensional integral has the same sort of weighted sum intepretations as the double
integral, such as:
• n-dimensional volume: Vol (W) =
RR
···
R
W 1 dV ,
• average value: f = 1
Vol(W)
RR
···
R
W f dV.
For example, if W ⊂ R3, the average x-coordinate on W is x = 1
Vol(W)
RRR
W x dV, similarly for y
and z, and the point ( x, y, z) is called the centroid of W.
To calculate the integral, one usually evaluates a sequence ofn one-dimensional partial integrals.
This can be complicated, because it may be difficult to determine the nested limits of integration
that describe the domain W. This is true even in three dimensions, where, in principle, it is still
possible to visualize what W looks like. We illustrate this by examining in detail a single example
of a triple integral.",Multivariable_Calculus_Shimamoto.pdf
"5.6. TRIPLE INTEGRALS AND BEYOND 145
Figure 5.33: Three of the planes that bound W: z = y (left), y = x (middle), and x = 1 (right).
The fourth is z = 0, the xy-plane.
Example 5.20. Find
RRR
W (y + z) dV if W is the solid region in R3 bounded by the planes z = y,
y = x, x = 1, and z = 0. See Figure 5.33.
The planes in question slice through one another in such a way that the solid W that they
bound is a tetrahedron. Each of its four triangular faces is contained in one of the planes. Looking
at W towards the origin from the first octant, the z = y face is the top, z = 0 is the bottom, x = 1
is the front, and y = x is the back. This is shown on the left of Figure 5.34.
Figure 5.34: The region of integration W (left) and its projection D on the xy-plane (right)
To find the endpoints that describe W in an iterated integral, our approach is to focus on
choosing which variable to integrate with respect to first. Once that variable is integrated out of",Multivariable_Calculus_Shimamoto.pdf
"choosing which variable to integrate with respect to first. Once that variable is integrated out of
the picture, what remains is a double integral, which by now is a much more familiar situation.
For instance, say we decide to integrate first with respect to z, treating x and y as constant.
We need to know the values of x and y for which this process makes sense, that is, the pairs ( x, y)
for which there is at least one z such that (x, y, z) ∈ W. These pairs are precisely the projection of
W on the xy-plane. In this example, this is the triangle D at the base of the tetrahedron, which is
shown on the right of Figure 5.34. In the xy-plane, it is bounded by the x-axis and the lines x = 1
and y = x.
For each (x, y) in D, the values of z for which (x, y, z) ∈ W go from z = 0 to z = y. Thus:
ZZZ
W
(y + z) dV =
ZZ
D
Z y
0
(y + z) dz

dx dy.
The remaining limits of integration come from describing D as a double integral in the usual way,",Multivariable_Calculus_Shimamoto.pdf
"ZZZ
W
(y + z) dV =
ZZ
D
Z y
0
(y + z) dz

dx dy.
The remaining limits of integration come from describing D as a double integral in the usual way,
for instance, using the order of integration dy dx. After that, the integral can be evaluated one",Multivariable_Calculus_Shimamoto.pdf
"146 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
antidifferentiation at a time. Thus:ZZZ
W
(y + z) dV =
Z 1
0
Z x
0
Z y
0
(y + z) dz

dy

dx
=
Z 1
0
Z x
0
(yz + 1
2z2)

z=y
z=0
dy

dx
=
Z 1
0
Z x
0
3
2y2 dy

dx
=
Z 1
0
1
2y3

y=x
y=0

dx
=
Z 1
0
1
2x3 dx
= 1
8x41
0
= 1
8.
Example 5.21. Express the same integral
RRR
W (y+z) dV from the previous example as an iterated
integral in which:
(a) the first integration is with respect to x,
(b) the first integration is with respect to y.
(a) To integrate first with respect to x, we treat y and z as constant. Proceeding as in the previous
example, the relevant values of ( y, z) are the points in the projection D of W on the yz-plane.
As before, this is a triangle, though no longer one of the faces of the tetrahedron. The triangle is
bounded in the yz-plane by the y-axis and the lines y = 1 and z = y. See Figure 5.35 on the left.
For each (y, z) in D, the point ( x, y, z) is in W from x = y to x = 1. Hence:
ZZZ
W
(y + z) dV =",Multivariable_Calculus_Shimamoto.pdf
"For each (y, z) in D, the point ( x, y, z) is in W from x = y to x = 1. Hence:
ZZZ
W
(y + z) dV =
ZZ
D
Z 1
y
(y + z) dx

dy dz.
Describing the remaining double integral over D using the order of integration, say dz dy, gives:
ZZZ
W
(y + z) dV =
Z 1
0
Z y
0
Z 1
y
(y + z) dx

dz

dy.
We won’t carry out the iterated antidifferentiation, though we are optimistic that the final answer
is again 1
8 .
(b) This time, the integral will be a double integral of an integral with respect to y over the
projection of W on the xz-plane . Again, the projection D is a triangle. Two sides of the triangle
are contained in the x-axis and the line x = 1 in the xz-plane. The third side is the projection of
the line where the planes z = y and y = x intersect. This line of intersection in R3 is described by
z = y = x, and its projection on the xz-plane is z = x. This is the remaining side of D. See Figure
5.35 on the right.
For each (x, z) in D, y goes from y = z to y = x. Therefore:
ZZZ
W
(y + z) dV =",Multivariable_Calculus_Shimamoto.pdf
"5.35 on the right.
For each (x, z) in D, y goes from y = z to y = x. Therefore:
ZZZ
W
(y + z) dV =
ZZ
D
Z x
z
(y + z) dy

dx dz.
Finally, integrating over D using the order dz dxresults in:
ZZZ
W
(y + z) dV =
Z 1
0
Z x
0
Z x
z
(y + z) dy

dz

dx.",Multivariable_Calculus_Shimamoto.pdf
"5.7. EXERCISES FOR CHAPTER 5 147
Figure 5.35: Integrating first with respect to x (left) or y (right)
5.7 Exercises for Chapter 5
Section 1 Volume and iterated integrals
In Exercises 1.1–1.4, evaluate the given iterated integral.
1.1.
Z 2
1
Z 4
3
x2y dy

dx
1.2.
Z 1
0
Z x
x2
(x + y) dy

dx
1.3.
Z π/2
0
Z sin y
0
(x cos y + 1)dx

dy
1.4.
Z 2
1
Z y
−y
sin(3x + 2y) dx

dy
1.5. Evaluate
RR
D
x2
y dA if D is the rectangle described by 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.
1.6. Evaluate
RR
D(x + y) dA if D is the quarter-disk x2 + y2 ≤ 1, x ≥ 0, y ≥ 0.
1.7. Evaluate
RR
D ex+y dA if D is the triangular region with vertices (0 , 0), (0, 2), (1, 0).
1.8. Evaluate
RR
D xy dAif D is the triangular region with vertices (0 , 0), (1, 1), (2, 0).
In Exercises 1.9–1.14, (a) sketch the domain of integration D in the xy-plane and (b) write an
equivalent expression with the order of integration reversed.
1.9.
Z 2
1
Z 4
3
f(x, y) dy

dx
1.10.
Z 1
0
Z e
ex
f(x, y) dy

dx
1.11.
Z 1
0
Z y
y2
f(x, y) dx

dy",Multivariable_Calculus_Shimamoto.pdf
"1.9.
Z 2
1
Z 4
3
f(x, y) dy

dx
1.10.
Z 1
0
Z e
ex
f(x, y) dy

dx
1.11.
Z 1
0
Z y
y2
f(x, y) dx

dy
1.12.
Z π/2
0
Z sin y
0
f(x, y) dx

dy",Multivariable_Calculus_Shimamoto.pdf
"148 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
1.13.
Z 1
−1
Z √
1−x2
0
f(x, y) dy

dx
1.14.
Z 6
−3
Z −1+√3+y
1
3 y
f(x, y) dx

dy
1.15. Consider the iterated integral
Z 1
1
4
Z 4
1
x
yexy dy

dx.
(a) Sketch the domain of integration D in the xy-plane.
(b) Write an equivalent expression with the order of integration reversed.
(c) Evaluate the integral using whatever method seems best.
1.16. Evaluate the integral
Z 6
0
Z 3
y
2
sin(x2) dx

dy.
1.17. Evaluate the integral
ZZ
D
p
1 − x2 dA, where D is the quarter-disk in the first quadrant
described by x2 + y2 ≤ 1, x ≥ 0, and y ≥ 0.
1.18. Find the volume of the wedge-shaped solid that lies above the xy-plane, below the plane
z = y, and inside the cylinder x2 + y2 = 1.
1.19. Find the volume of the pyramid-shaped solid in the first octant bounded by the three coor-
dinate planes and the planes x + z = 1 and y + 2z = 2.
1.20. (a) If c is a positive constant, find the volume of the tetrahedron in the first octant bounded",Multivariable_Calculus_Shimamoto.pdf
"1.20. (a) If c is a positive constant, find the volume of the tetrahedron in the first octant bounded
by the plane x + y + z = c and the three coordinate planes.
(b) Consider the tetrahedron W in the first octant bounded by the plane x + y + z = 1 and
the three coordinate planes. Suppose that you want to divide W into three pieces of
equal volume by slicing it with two planes parallel to x + y + z = 1, i.e., with planes of
the form x + y + z = c. How should the slices be made?
1.21. Let D be the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and let f : D → R be the function given by
f(x, y) = min{x, y}. Find
RR
D f(x, y) dx dy.
Section 2 The double integral
2.1. Let R = [0, 1] × [0, 1], and let f : R → R be the function given by:
f(x, y) =
(
1 if ( x, y) = (1
2 , 1
2 ),
0 otherwise .
(a) Let R be subdivided into a 3 by 3 grid of subrectangles of equal size (Figure 5.36). What
are the possible values of Riemann sums based on this subdivision? (Different choices",Multivariable_Calculus_Shimamoto.pdf
"are the possible values of Riemann sums based on this subdivision? (Different choices
of sample points may give different values.)
(b) Repeat for a 4 by 4 grid of subrectangles of equal size.
(c) How about for an n by n grid of subrectangles of equal size?",Multivariable_Calculus_Shimamoto.pdf
"5.7. EXERCISES FOR CHAPTER 5 149
Figure 5.36: A 3 by 3 subdivision of R = [0, 1] × [0, 1]
(d) Let δ be a positive real number. If R is subdivided into a grid of subrectangles of
dimensions △xi by △yj, where △xi < δand △yj < δfor all i, j, show that any Riemann
sum based on the subdivision lies in the range 0 ≤ P
i,j f(pij) △xi △yj < 4δ2.
(e) Show that
RR
R f(x, y) dA = 0, i.e., that lim
△xi→0
△yj→0
P
i,j f(pij) △xi △yj = 0. Technically,
this means you need to show that, given any ϵ >0, there exists a δ >0 such that, for
any subdivision of R into subrectangles with △xi < δ and △yj < δ for all i, j, every
Riemann sum based on the subdivision satisfies
P
i,j f(pij) △xi △yj − 0
 < ϵ.
2.2. Let R = [0, 1] × [0, 1], and let f : R → R be the function given by:
f(x, y) =
(
1 if x = 0 or y = 0,
0 otherwise .
(a) Let R be subdivided into a 3 by 3 grid of subrectangles of equal size (Figure 5.36). What
are the possible values of Riemann sums based on this subdivision?",Multivariable_Calculus_Shimamoto.pdf
"are the possible values of Riemann sums based on this subdivision?
(b) Repeat for a 4 by 4 grid of subrectangles of equal size.
(c) How about for an n by n grid of subrectangles of equal size?
(d) Is f integrable on R? If so, what is the value of
RR
R f(x, y) dA?
2.3. Here is a result that will come in handy increasingly as we calculate more integrals. Let
R = [a, b] × [c, d] be a rectangle, and let F : R → R be a real-valued function such that the
variables x and y separate into two continuous factors, that is, F(x, y) = f(x)g(y), where
f : [a, b] → R and g : [c, d] → R are continuous. Show that:
ZZ
R
f(x)g(y) dx dy=
Z b
a
f(x) dx
Z d
c
g(y) dy

.
2.4. Let f : U → R be a continuous function defined on an open set U in R2, and let a be a point
of U.
(a) If f(a) > 0, prove that there exists a rectangle R containing a such that:
ZZ
R
f dA >0.
(Hint: Use Exercise 6.4 of Chapter 3.)",Multivariable_Calculus_Shimamoto.pdf
"150 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
(b) Let g : U → R be another continuous function on U. If f(a) > g(a), prove that there
exists a rectangle R containing a such that:
ZZ
R
f dA >
ZZ
R
g dA.
(Hint: Consider f − g.)
2.5. In this exercise, we use the double integral to give another proof of the equality of mixed
partial derivatives. Let f(x, y) be a real-valued function defined on an open set U in R2
whose first and second-order partial derivatives are continuous on U.
(a) Let R = [a, b] × [c, d] be a rectangle that is contained in U. Show that:
ZZ
R
∂2f
∂x ∂y(x, y) dA =
ZZ
R
∂2f
∂y ∂x(x, y) dA.
(Hint: Use Fubini’s theorem, with appropriate orders of integration, and the fundamen-
tal theorem of calculus to show that both sides are equal to f(b, d) − f(b, c) − f(a, d) +
f(a, c).)
(b) Show that ∂2f
∂x ∂y(a) = ∂2f
∂y ∂x(a) for all a in U. ( Hint: Use Exercise 2.4 to show that
something goes wrong if they are not equal.)
Section 3 Interpretations of the double integral",Multivariable_Calculus_Shimamoto.pdf
"something goes wrong if they are not equal.)
Section 3 Interpretations of the double integral
3.1. The population density of birds in a wildlife refuge decreases at a uniform rate with the
distance from a river. If the river is modeled as the x-axis in the plane, then the density at
the point (x, y) is given by f(x, y) = 5 −|y| hundred birds per square mile, where x and y are
measured in miles. Find the total number of birds in the rectangle R = [−2, 2] × [0, 1].
3.2. A small cookie has the shape of the region in the first quadrant bounded by the curves y = x2
and x = y2, where x and y are measured in inches. Chocolate is poured unevenly on top
of the cookie in such a way that the density of chocolate at the point ( x, y) is given by
f(x, y) = 100(x + y) grams per square inch. Find the total mass of chocolate on the cookie.
3.3. The eye of a tornado is positioned directly over the origin in the plane. Suppose that the",Multivariable_Calculus_Shimamoto.pdf
"3.3. The eye of a tornado is positioned directly over the origin in the plane. Suppose that the
wind speed on the ground at the point ( x, y) is given by v(x, y) = 30(x2 + y2) miles per hour.
(a) Find the average wind speed on the square R = [0, 2] × [0, 2].
(b) Find all points ( x, y) of R at which the wind speed equals the average.
3.4. Find the centroid of the half-disk of radius a in R2 described by x2 + y2 ≤ a2, y ≥ 0.
3.5. Find the centroid of the triangular region in R2 with vertices (0, 0), (1, 2), and (1, 3).
3.6. Find the centroid of the triangular region in R2 with vertices (0, 0), (2, 2), and (1, 3).
3.7. Let D be the fudgsicle-shaped region of R2 that consists of the rectangle [ −1, 1] × [0, h] of
height h topped by a half-disk of radius 1, as shown in Figure 5.37. Find the value of h such
that the point (0 , h) is the centroid of D.",Multivariable_Calculus_Shimamoto.pdf
"5.7. EXERCISES FOR CHAPTER 5 151
Figure 5.37: The fudgsicle
3.8. Let R be a rectangle in R2, and let f : R → R be an integrable function. If f is continuous,
then it is true that there is a point of R at which f assumes its average value, that is, there
exists a point c of R such that f(c) = f. The proof is not hard, but it depends on subtle
properties of the real numbers that are beyond what we cover here. The same conclusion
need not hold, however, for discontinuous functions. Find an example of a rectangle R and
an integrable function f : R → R such that there is no point c of R at which f(c) = f.
Section 4 Parametrization of surfaces
4.1. Find a parametrization of the upper hemisphere of radius a described by x2 + y2 + z2 = a2,
z ≥ 0, using the following coordinates as parameters. Be sure to state what the domain of
your parametrization is.
(a) the rectangular coordinates x, y
(b) the polar/cylindrical coordinates r, θ
(c) the spherical coordinates ϕ, θ",Multivariable_Calculus_Shimamoto.pdf
"your parametrization is.
(a) the rectangular coordinates x, y
(b) the polar/cylindrical coordinates r, θ
(c) the spherical coordinates ϕ, θ
4.2. Let S be the surface described in cylindrical coordinates by z = θ, 0 ≤ r ≤ 1, and 0 ≤ θ ≤ 4π.
It is called a helicoid. Converting to rectangular coordinates gives a parametrization:
σ : [0, 1] × [0, 4π] → R3, σ (r, θ) = (r cos θ, rsin θ, θ).
(a) Describe the curve that is traced out by σ if r = 1
2 is held fixed and θ varies, that is, the
curve parametrized by the path α(θ) = σ(1
2 , θ), 0 ≤ θ ≤ 4π.
(b) Similarly, describe the curve that is traced out by σ if θ = π
4 is held fixed and r varies,
that is, the curve parametrized by β(r) = σ(r, π
4 ), 0 ≤ r ≤ 1.
(c) Describe S in a few words, and draw a sketch.
Section 5 Integrals with respect to surface area
5.1. Let S be the triangular surface in R3 whose vertices are (1, 0, 0), (0, 1, 0), and (0, 0, 1).
(a) Find a parametrization of S using x and y as parameters. What is the domain of your",Multivariable_Calculus_Shimamoto.pdf
"(a) Find a parametrization of S using x and y as parameters. What is the domain of your
parametrization? ( Hint: The triangle is contained in a plane.)
(b) Use your parametrization and formula (5.3) to find the area of S. Check that your
answer agrees with what you would get using the formula: Area = 1
2 (base)(height).
(c) Find
RR
S 7 dS. ( Hint: Use your answer to part (b).)",Multivariable_Calculus_Shimamoto.pdf
"152 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION
5.2. Let S be the cylindrical surface x2 + y2 = 4, 0 ≤ z ≤ 3. Find the integral
RR
S z2 dS.
5.3. Consider the graph z = f(x, y) of a smooth real-valued function f defined on a bounded
subset D of R2. Show that the surface area of the graph is given by the formula:
Surface area =
ZZ
D
s
1 +
∂f
∂x
2
+
∂f
∂y
2
dx dy.
5.4. Let f : [a, b] → R be a smooth real-valued function of one variable, where a ≥ 0, and let S be
the surface of revolution that is swept out when the curve z = f(x) in the xz-plane is rotated
all the way around the z-axis. Show that the surface area of S is given by the formula:
Surface area = 2 π
Z b
a
x
p
1 + f′(x)2 dx.
(Hint: Parametrize S using cylindrical coordinates.)
5.5. Let W be the solid bounded on top by the plane z = y + 5, on the sides by the cylinder
x2 + y2 = 4, and on the bottom by the plane z = 0.
(a) Sketch W.
(b) Let S be the surface that bounds W (top, sides, and bottom). Find the surface area of
S.",Multivariable_Calculus_Shimamoto.pdf
"(a) Sketch W.
(b) Let S be the surface that bounds W (top, sides, and bottom). Find the surface area of
S.
(c) Find
RR
S z dS, where S is the surface in part (b).
5.6. Let S be the helicoid from Exercise 4.2, parametrized by:
σ : [0, 1] × [0, 4π] → R3, σ (r, θ) = (r cos θ, rsin θ, θ).
Find
RR
S z
p
x2 + y2 dS.
5.7. Let W be the solid region of Example 5.2 that lies inside the cylinders x2 + z2 = 1 and
y2 + z2 = 1 and above the xy-plane, and let S be the exposed part of the surface that bounds
W, that is, the cylindrical surfaces but not the base.
(a) Find the surface area of S.
(b) Find
RR
S z dS.
5.8. The sphere x2 + y2 + z2 = a2 of radius a sits inscribed in the circular cylinder x2 + y2 = a2.
Given real numbers c and d, where −a ≤ c < d≤ a, show that the portions of the sphere
and the cylinder lying between the planes z = c and z = d have equal surface area. ( Hint:
Parametrize both surfaces using the cylindrical coordinates θ and z as parameters.)",Multivariable_Calculus_Shimamoto.pdf
"Parametrize both surfaces using the cylindrical coordinates θ and z as parameters.)
Section 6 Triple integrals and beyond
In Exercises 6.1–6.2, evaluate the given iterated integral.
6.1.
Z 1
0
Z x
0
Z x
y
(x + y − z) dz

dy

dx
6.2.
Z 1
0
Z 2x
x
Z xy
0
x3y2z dz

dy

dx",Multivariable_Calculus_Shimamoto.pdf
"5.7. EXERCISES FOR CHAPTER 5 153
6.3. Find
RRR
W xyz dVif W is the solid region in R3 below the surface z = x2 + y2 and above the
square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
6.4. Find
RRR
W (xz + y) dV if W is the solid region in R3 above the xy-plane bounded by the
surface y = x2 and the planes z = y, y = 1, and z = 0.
6.5. Find
RRR
W x dVif W is the region in the first octant bounded by the saddle surfacez = x2−y2
and the planes x = 2, y = 0, and z = 0.
6.6. Find
RRR
W (1 + x + y + z)2 dV if W is the solid tetrahedron in the first octant bounded by
the plane x + y + z = 1 and the three coordinate planes.
6.7. Let a be a positive real number, and let W = [0, a] × [0, a] × [0, a] in R3. Find the average
value of f(x, y, z) = x2 + y2 + z2 on W.
For each of the triple integrals in Exercises 6.8–6.11, (a) sketch the domain of integration in
R3, (b) write an equivalent expression in which the first integration is with respect to x, and (c)",Multivariable_Calculus_Shimamoto.pdf
"R3, (b) write an equivalent expression in which the first integration is with respect to x, and (c)
write an equivalent expression in which the first integration is with respect to y.
6.8.
Z 2
−2
Z
q
2−x2
2
0
Z
q
1−x2
4 −y2
2
0
f(x, y, z) dz

dy

dx
6.9.
Z 1
−1
Z √
1−x2
0
Z 1
0
f(x, y, z) dz

dy

dx
6.10.
Z 1
0
Z 1
x
Z x
0
f(x, y, z) dz

dy

dx
6.11.
Z 1
0
Z 1
x
Z y
0
f(x, y, z) dz

dy

dx",Multivariable_Calculus_Shimamoto.pdf
154 CHAPTER 5. REAL-VALUED FUNCTIONS: INTEGRATION,Multivariable_Calculus_Shimamoto.pdf
"Part IV
Vector-valued functions
155",Multivariable_Calculus_Shimamoto.pdf
"Chapter 6
Differentiability and the chain rule
So far, we have studied functions of which at least one of the domain or codomain is a subset of R,
in other words:
• vector-valued functions of one variable, also known as paths, α: I → Rn, where I ⊂ R, or
• real-valued functions of n variables f : U → R, where U ⊂ Rn.
Now, we consider the general case of vector-valued functions of n variables, that is, functions of the
form f : U → Rm, where U ⊂ Rn and both n and m are allowed to be greater than 1.
Given such a vector-valued function, we can leverage what we know about real-valued functions
because, if x ∈ U, then, as an element of Rm, f(x) has m coordinates:
f(x) = (f1(x), f2(x), . . . , fm(x)),
where each fi(x) is a real number. In other words, each component is a real-valued function
fi : U → R. Thus a vector-valued function is also a sequence of real-valued functions. We use
the notation above for the components consistently from now on. For example, if f : R2 → R2 is",Multivariable_Calculus_Shimamoto.pdf
"the notation above for the components consistently from now on. For example, if f : R2 → R2 is
the polar coordinate transformation f(r, θ) = (r cos θ, rsin θ), then f1(r, θ) = r cos θ and f2(r, θ) =
r sin θ. Also, as with paths, we denote f in plainface type, reserving boldface for a special kind of
vector-valued function that we study beginning in Chapter 8.
We treated the case of real-valued functions fairly rigorously, and we shall see that much of the
theory carries over to vector-valued functions without incident. On the other hand, we discussed
paths back in Chapter 2 more informally, so what we are about to do here in the vector-valued case
can be taken as establishing the theory behind what we did then.
6.1 Continuity revisited
The motivation and definition for continuity of vector-valued functions carry over almost verbatim
from the real-valued case.
Definition. Let U be an open set in Rn, and let f : U → Rm be a function. We say that f is",Multivariable_Calculus_Shimamoto.pdf
"from the real-valued case.
Definition. Let U be an open set in Rn, and let f : U → Rm be a function. We say that f is
continuous at a point a of U if, given any open ball B(f(a), ϵ) about f(a), there exists an open
ball B(a, δ) about a such that:
f(B(a, δ)) ⊂ B(f(a), ϵ).
In other words, if x ∈ B(a, δ), then f(x) ∈ B(f(a), ϵ). See Figure 6.1. Equivalently, given any
ϵ >0, there exists a δ >0 such that:
if ∥x − a∥ < δ,then ∥f(x) − f(a)∥ < ϵ.
157",Multivariable_Calculus_Shimamoto.pdf
"158 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
Figure 6.1: Continuity of a vector-valued function: given any ball B(f(a), ϵ) about f(a), there is a
ball B(a, δ) about a such that f(B(a, δ)) ⊂ B(f(a), ϵ).
We say that f is a continuous function if it is continuous at every point of its domain.
Proposition 6.1. If f(x) = ( f1(x), f2(x), . . . , fm(x)), then f is continuous at a if and only if
f1, f2, . . . , fm are all continuous at a.
Proof. Intuitively, this is because f(x)−f(a) can be made small if and only if all of its components
fi(x)−fi(a) can be made small, but this needs to be made more precise. There are two implications
that need to be proven here, that each condition implies the other. We give the details for one of
the implications and leave the other for the exercises.
Namely, we show that, if f1, f2, . . . , fm are continuous at a, then so is f. Let ϵ >0 be given.
We want to ensure that ∥f(x) − f(a)∥ < ϵ. Note that:
∥f(x) − f(a)∥ =
q
f1(x) − f1(a)

2 +

f2(x) − f2(a)",Multivariable_Calculus_Shimamoto.pdf
"We want to ensure that ∥f(x) − f(a)∥ < ϵ. Note that:
∥f(x) − f(a)∥ =
q
f1(x) − f1(a)

2 +

f2(x) − f2(a)

2 + ··· +

fm(x) − fm(a)

2.
If we could arrange that |fi(x) − fi(a)| < ϵ√m for i = 1 , 2, . . . , m, then ∥f(x) − f(a)∥ <q
ϵ2
m + ϵ2
m + ··· + ϵ2
m =
√
ϵ2 = ϵ would follow.
This we can do. For taking ϵ√m as the given input in the definition of continuity, the fact thatfi is
continuous at a for i = 1, 2, . . . , mmeans that there exists a δi > 0 such that |fi(x) − fi(a)| < ϵ√m
whenever ∥x − a∥ < δi. Let δ = min {δ1, δ2, . . . , δm}. If ∥x − a∥ < δ, it remains true that
|fi(x) − fi(a)| < ϵ√m for all i, and hence, as noted above, ∥f(x) − f(a)∥ < ϵ. We have found a δ
that satisfies the definition of continuity for f.
The proof of the converse is Exercise 1.1.
The proposition allows us to determine iff is continuous by looking at its real-valued component
functions, which is something we have studied before. For instance, for the function f(r, θ) =",Multivariable_Calculus_Shimamoto.pdf
"functions, which is something we have studied before. For instance, for the function f(r, θ) =
(r cos θ, rsin θ), both component functions f1(r, θ) = r cos θ and f2(r, θ) = r sin θ are continuous as
real-valued algebraic combinations and compositions of continuous pieces. Thereforef is continuous
as well.
Similarly, the definition of limit for vector-valued functions carries over from the real-valued
case essentially unchanged.
Definition. Let U be an open set in Rn, and let a be a point of U. If f : U − {a} →Rm is a
vector-valued function defined on U, except possibly at the point a, then we say that lim x→a f(x)
exists if there is a vector L in Rm such that the function ef : U → Rm defined by
ef(x) =
(
f(x) if x ̸= a,
L if x = a",Multivariable_Calculus_Shimamoto.pdf
"6.2. DIFFERENTIABILITY REVISITED 159
is continuous at a. When this happens, we write lim x→a f(x) = L.
As before, it is immediate that f : U → Rm is continuous at a if and only if limx→a f(x) = f(a).
One consequence of this is that Proposition 6.1 then implies that a vector-valued limit can also be
viewed as a sequence of real-valued limits.
Proposition 6.2. If f(x) = (f1(x), f2(x), . . . , fm(x)) and L = (L1, L2, . . . , Lm), then limx→a f(x)
= L if and only if limx→a fi(x) = Li for all i = 1, 2, . . . , m.
6.2 Differentiability revisited
Recall that, in the real-valued case, f : U → R is differentiable at a if it has a first-order approxi-
mation f(x) ≈ ℓ(x) = f(a) + ∇f(a) · (x − a) such that lim x→a
f(x)−ℓ(x)
∥x−a∥ = 0. For a vector-valued
function f : U → Rm, we find an approximation, again denoted by ℓ, by compiling the approxima-
tions of the component functions f1, f2, . . . , fm into one vector:
ℓ(x) =


ℓ1(x)
ℓ2(x)
...
ℓm(a)

 =


f1(a) + ∇f1(a) · (x − a)",Multivariable_Calculus_Shimamoto.pdf
"tions of the component functions f1, f2, . . . , fm into one vector:
ℓ(x) =


ℓ1(x)
ℓ2(x)
...
ℓm(a)

 =


f1(a) + ∇f1(a) · (x − a)
f2(a) + ∇f2(a) · (x − a)
...
fm(a) + ∇fm(a) · (x − a)


=


f1(a)
f2(a)
...
fm(a)

 +


∇f1(a) · (x − a)
∇f2(a) · (x − a)
...
∇fm(a) · (x − a)


= f(a) +


∇f1(a)
∇f2(a)
...
∇fm(a)

 · (x − a)
= f(a) + Df(a) · (x − a),
where, in the last two lines, Df(a) =


∇f1(a)
∇f2(a)
...
∇fm(a)

 is the matrix whose rows are the gradients of the
component functions and x − a is a column vector.
We consider the error f(x) − ℓ(x) = f(x) − f(a) − Df(a) · (x − a) in using the approximation
and proceed as in the real-valued case.
Definition. Let U be an open set in Rn, and let f : U → Rm be a function. If a ∈ U, consider the
matrix:
Df(a) =


∂f1
∂x1
(a) ··· ∂f1
∂xn (a)
... ... ...
∂fm
∂x1
(a) ··· ∂fm
∂xn (a)

.
We say that f is differentiable at a if:
lim
x→a
f(x) − f(a) − Df(a) · (x − a)
∥x − a∥ = 0. (6.1)",Multivariable_Calculus_Shimamoto.pdf
"... ... ...
∂fm
∂x1
(a) ··· ∂fm
∂xn (a)

.
We say that f is differentiable at a if:
lim
x→a
f(x) − f(a) − Df(a) · (x − a)
∥x − a∥ = 0. (6.1)
When this happens, Df(a) is called the derivative, or Jacobian matrix, of f at a.",Multivariable_Calculus_Shimamoto.pdf
"160 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
To repeat a point made in the real-valued case, the derivative is not a single number, but rather
a matrix or, even better, the linear part of a good affine approximation of f near a.
We begin by calculating a few simple examples of Df(a). In general, it is an m by n matrix
whose entries are various partial derivatives. The ith row is the gradient ∇fi, and the jth column
is the “velocity” ∂
∂xj
of f with respect to xj.
Example 6.3. Let f : R2 → R2 be f(r, θ) = (r cos θ, rsin θ). Then:
Df(r, θ) =
cos θ −r sin θ
sin θ r cos θ

.
The formula above gives what the matrix looks like at a general point a = (r, θ). At a specific
point, such as a = (2, π
4 ), we have Df(a) = Df(2, π
4 ) =
""√
2
2 −2 ·
√
2
2√
2
2 2 ·
√
2
2
#
=
""√
2
2 −
√
2√
2
2
√
2
#
.
Example 6.4. Let g : R3 → R2 be the projection of xyz-space onto the xy-plane, g(x, y, z) = (x, y).
Then:
Dg(x, y, z) =
1 0 0
0 1 0

.
In other words, Dg(a) =
1 0 0
0 1 0

for all a in R3.",Multivariable_Calculus_Shimamoto.pdf
"Then:
Dg(x, y, z) =
1 0 0
0 1 0

.
In other words, Dg(a) =
1 0 0
0 1 0

for all a in R3.
Example 6.5. Let h: R2 → R3 be h(x, y) = (x2 + y3, x4y5, e6x+7y). Then, if a = (x, y):
Dh(a) = Dh(x, y) =


2x 3y2
4x3y5 5x4y4
6e6x+7y 7e6x+7y

.
To determine whether a function is differentiable without going through the definition, we may
use criteria similar to those in the real-valued case. For instance, the same reasoning as before
shows that a differentiable function must be continuous and that it is necessary for all the partial
derivatives ∂fi
∂xj
to exist in order for f to be differentiable.
In addition, the condition for differentiability in the definition (6.1) requires that a certain limit
be the zero vector. This can happen if and only if the limit of each of the components is zero, too.
This follows from Proposition 6.2. In other words, we have the following componentwise criterion
for differentiability.",Multivariable_Calculus_Shimamoto.pdf
"This follows from Proposition 6.2. In other words, we have the following componentwise criterion
for differentiability.
Proposition 6.6. If f(x) = ( f1(x), f2(x), . . . , fm(x)), then f is differentiable at a if and only if
f1, f2, . . . , fm are all differentiable at a.
Example 6.7. Let I be an open interval in R, and let α: I → Rn be a path in Rn, where
α(t) = ( x1(t), x2(t), . . . , xn(t)). By the last proposition, α is differentiable if and only if each of
its components x1, x2, . . . , xn is differentiable. The latter condition is essentially how we defined
differentiability of paths in Chapter 2, so what we did there is supported by the theory we now
have in place. Moreover, by definition of the derivative matrix:
Dα(t) =


x′
1(t)
x′
2(t)
...
x′
n(t)

.
This agrees with α′(t), the velocity, under the usual identification of a vector in Rn with a column
matrix.",Multivariable_Calculus_Shimamoto.pdf
"6.3. THE CHAIN RULE: A CONCEPTUAL APPROACH 161
According to Proposition 6.6, we may apply the C1 test for real-valued functions to each of the
components of a vector-valued function to obtain the following criterion.
Theorem 6.8 (The C1 test). If all the partial derivatives ∂fi
∂xj
(i.e., the entries of Df(x)) exist
and are continuous on U, then f is differentiable at every point of U.
To illustrate this, in each of Examples 6.3–6.5 above, all entries of theD matrices are continuous,
so those three functions f, g, hare differentiable at every point of their domains.
A function f is called smooth if each of its component functions f1, f2, . . . , fm is smooth in the
sense we have discussed previously (see Section 4.9), that is, they have continuous partial derivatives
of all orders. As before, we generally state our results about differentiability for smooth functions
without worrying about whether this is a stronger assumption than necessary.
6.3 The chain rule: a conceptual approach",Multivariable_Calculus_Shimamoto.pdf
"without worrying about whether this is a stronger assumption than necessary.
6.3 The chain rule: a conceptual approach
We now discuss the all-important rule for the derivative of a composition. We first take a fairly
high-level approach to understand where the rule comes from.
Let U and V be open sets in Rn and Rm, respectively, and let f : U → V and g : V → Rp
be functions. The composition g ◦ f : U → Rp is then defined, and by definition, if a ∈ U, the
composition is differentiable at a if there is a good first-order approximation of the form:
g(f(x)) ≈ g(f(a)) + D(g ◦ f)(a) · (x − a) (6.2)
when x is near a. The challenge is to figure out the expression that goes into the box forD(g◦f)(a).
The idea is that the first-order approximation of the compositiong◦f should be the composition
of the first-order approximations of g and f individually. Let b = f(a), and assume that f is
differentiable at a and g is differentable at b. Then there are good first-order approximations:",Multivariable_Calculus_Shimamoto.pdf
"differentiable at a and g is differentable at b. Then there are good first-order approximations:
• for f : f(x) ≈ ℓf (x)
≈ f(a) + Df(a) · (x − a) when x is near a, (6.3)
• for g : g(y) ≈ ℓg(y)
≈ g(b) + Dg(b) · (y − b) when y is near b
≈ g(f(a)) + Dg(f(a)) · (y − f(a)). (6.4)
Note that, by continuity, when x is near a, f(x) is near f(a) = b, so substituting y = f(x) in (6.4)
gives:
g(f(x)) ≈ g(f(a)) + Dg(f(a)) · (f(x) − f(a)).
Then, using (6.3) to substitute f(x) − f(a) ≈ Df(a) · (x − a) yields:
g(f(x)) ≈ g(f(a)) + Dg(f(a)) · Df(a) · (x − a).
If you like, you can check that the right side of this last approximation is also the composition
ℓg(ℓf (x)) of the first-order approximations of g and f. In any case, the upshot is that we have
identified something that fits naturally in the box in equation (6.2).
Theorem 6.9 (Chain rule). Let U and V be open sets in Rn and Rm, respectively, and let f : U → V",Multivariable_Calculus_Shimamoto.pdf
"Theorem 6.9 (Chain rule). Let U and V be open sets in Rn and Rm, respectively, and let f : U → V
and g : V → Rp be functions. Let a be a point of U. If f is differentiable at a and g is differentiable
at f(a), then g ◦ f is differentiable at a and:
D(g ◦ f)(a) = Dg(f(a)) · Df(a),
where the product on the right is matrix multiplication.",Multivariable_Calculus_Shimamoto.pdf
"162 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
In terms of the sizes of the various parties involved, the left side of the chain rule is a p by n
matrix, while the right side is a product ( p by m) × (m by n). This need not be memorized. In
practice, it usually takes care of itself.
The chain rule says that the derivative of a composition is the product of the derivatives, or,
to be even more sophisticated, the composition of the linear parts of the associated first-order
approximations. Though the objects involved are more complicated, this is actually the same as
what happens back in first-year calculus. There, if y is a function of u, say y = g(u), and u is a
function of x, say u = f(x), then y becomes a function of x via composition: y = g(f(x)). The
one-variable chain rule says:
dy
dx = dy
du
du
dx.
Again, the derivative of a composition is the product of the derivatives of the composed steps.",Multivariable_Calculus_Shimamoto.pdf
"one-variable chain rule says:
dy
dx = dy
du
du
dx.
Again, the derivative of a composition is the product of the derivatives of the composed steps.
The line of reasoning we used to obtain the chain rule is natural and reasonably straightforward,
but a proper proof requires greater attention to the various approximations involved. The letter ϵ
is likely to appear. The arguments are a little like those needed for a correct proof of the Little
Chain Rule back in Exercise 5.6 of Chapter 4, though the details here are more complicated. We
shall say more about the proof of the chain rule in the next section.
The following example is meant to illustrate how the pieces of the chain rule fit together.
Example 6.10. Let f : R2 → R2 be given by f(r, θ) = (r cos θ, rsin θ) and g : R2 → R3 by g(x, y) =
(x2 + y2, 2x + 3y, xy2). Find D(g ◦ f)(2, π
2 ).
By the chain rule,
D(g ◦ f)(2, π
2 ) = Dg

f(2, π
2 )


· Df(2, π
2 ) = Dg(0, 2) · Df(2, π
2 ),
since f(2, π
2 ) = (2 cos π
2 , 2 sin π",Multivariable_Calculus_Shimamoto.pdf
"2 ).
By the chain rule,
D(g ◦ f)(2, π
2 ) = Dg

f(2, π
2 )


· Df(2, π
2 ) = Dg(0, 2) · Df(2, π
2 ),
since f(2, π
2 ) = (2 cos π
2 , 2 sin π
2 ) = (0, 2). Calculating the necessary derivatives is straightforward:
Dg(x, y) =


2x 2y
2 3
y2 2xy

 ⇒ Dg(0, 2) =


0 4
2 3
4 0

,
Df(r, θ) =
cos θ −r sin θ
sin θ r cos θ

⇒ Df(2, π
2 ) =
0 −2
1 0

.
Therefore:
D(g ◦ f)(2, π
2 ) =


0 4
2 3
4 0


0 −2
1 0

=


4 0
3 −4
0 −8

.
6.4 The chain rule: a computational approach
The chain rule states that the derivative of a composition is the product of the derivatives:
D(g ◦ f)(a) = Dg(f(a)) · Df(a).
This is remarkably concise and elegant. Nevertheless, under the hood, there is a lot going on. Each
entry of a matrix product involves considerable information: the ( i, j)th entry of the product is the",Multivariable_Calculus_Shimamoto.pdf
"6.4. THE CHAIN RULE: A COMPUTATIONAL APPROACH 163
dot product of the ith row and jth column:


...
··· (i, j) ···
...

 =

∗ ∗ ··· ∗ ∗




∗
∗
...
∗
∗


.
There is one such dot product for each entry of D(g ◦ f)(a). We shall write out these entries in
a few cases in a moment and discover that we have seen this type of product before. You can
think of the discovery either as an alternative way to derive the chain rule or as confirmation of
the consistency of the theory.
We take a simple situation that we have studied previously and then tweak it a couple of times.
Example 6.11. Let α: R → R3 be a smooth path in R3, α(t) = (x(t), y(t), z(t)), and let g : R3 → R
be a smooth real-valued function of three variables, w = g(x, y, z). Then w becomes a function of
t by composition: w = g(α(t)) = g(x(t), y(t), z(t)). As a matter of notation, from here on, we use
w to denote the value of the composition, reserving g to stand for the original function of x, y, z.",Multivariable_Calculus_Shimamoto.pdf
"w to denote the value of the composition, reserving g to stand for the original function of x, y, z.
The composition is a real-valued function of one variable, so the Little Chain Rule applies. It
says that the derivative of the compositon is “gradient dot velocity”:
dw
dt = ∇g(α(t)) · α′(t) =
∂g
∂x , ∂g
∂y , ∂g
∂z

·
dx
dt , dy
dt , dz
dt

= ∂g
∂x
dx
dt + ∂g
∂y
dy
dt + ∂g
∂z
dz
dt , (6.5)
where the partial derivatives ∂g
∂x , ∂g
∂y , ∂g
∂z are evaluated at the point α(t) = ( x(t), y(t), z(t)). This
calculation can be visualized using a “dependence diagram,” as shown in Figure 6.2. It illustrates
w = g
x y z
t
Figure 6.2: A dependence diagram for the Little Chain Rule
the dependence of the various variables involved: g is a function of x, y, z, and in turn x, y, z are
functions of t. The Little Chain Rule (6.5) says that, to find the derivative of the top with respect
to the bottom, multiply the derivatives along each possible path from top to bottom and add these
products.",Multivariable_Calculus_Shimamoto.pdf
"to the bottom, multiply the derivatives along each possible path from top to bottom and add these
products.
Tweak 1. Suppose that x, y, z are functions of two variables s and t. In other words, replace
α: R → R3 with a smooth function f : R2 → R3, where f(s, t) = (x(s, t), y(s, t), z(s, t)). Then the
composition g◦f : R2 → R3 → R is a function of s and t: w = (g◦f)(s, t) = g(x(s, t), y(s, t), z(s, t)).
The corresponding dependence diagram is shown in Figure 6.3.
To compute ∂w
∂s (or ∂w
∂t ), only one variable actually varies. Because the composition is again
real-valued, the Little Chain Rule applies with respect to that variable, just as in the previous case.
Notationally, because w, x, y, and z are no longer functions of just one variable, we need to replace",Multivariable_Calculus_Shimamoto.pdf
"164 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
w = g
x y z
s t
Figure 6.3: Dependence diagram for tweak 1
the ordinary derivative terms d
ds (or d
dt) in equation (6.5) with partial derivatives ∂
∂s (or ∂
∂t ). For
instance, for ∂w
∂s , the Little Chain Rule says:
∂w
∂s = ∂g
∂x
∂x
∂s + ∂g
∂y
∂y
∂s + ∂g
∂z
∂z
∂s . (6.6)
The relevant paths in the dependence diagram are shown in bold in the figure. Similarly:
∂w
∂t = ∂g
∂x
∂x
∂t + ∂g
∂y
∂y
∂t + ∂g
∂z
∂z
∂t . (6.7)
In fact, equations (6.6) and (6.7) can be rewritten as a single matrix equation:
∂w
∂s
∂w
∂t

=
h
∂g
∂x
∂g
∂y
∂g
∂z
i


∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t
∂z
∂s
∂z
∂t

. (6.8)
The entries corresponding to equation (6.6) are shown in red. The point is that the matrix on the
left of equation (6.8) is the derivativeD(g◦f), while those on the right areDg and Df, respectively.
Tweak 2. We alter g so that it is vector-valued, say g : R3 → R2, where g(x, y, z) = ( g1(x, y, z),",Multivariable_Calculus_Shimamoto.pdf
"Tweak 2. We alter g so that it is vector-valued, say g : R3 → R2, where g(x, y, z) = ( g1(x, y, z),
g2(x, y, z)), and leave f : R2 → R3 as in tweak 1. The composition g ◦f : R2 → R2 is vector-valued:
w = g

f(s, t)


=

g1(f(s, t)), g2(f(s, t))


. Each of the components is a real-valued function of s
and t: w1 = g1(f(s, t)) = g1(x(s, t), y(s, t), z(s, t)) and w2 = g2(f(s, t)) = g2(x(s, t), y(s, t), z(s, t)).
The dependence diagram is shown in Figure 6.4.
w1 = g1 w2 = g2
x y z
s t
Figure 6.4: Dependence diagram for tweak 2
Both w1 and w2 have partial derivatives with respect to s and t, and each of them fits exactly
into the context of tweak 1. For example:
∂w2
∂t = ∂g2
∂x
∂x
∂t + ∂g2
∂y
∂y
∂t + ∂g2
∂z
∂z
∂t ,",Multivariable_Calculus_Shimamoto.pdf
"6.4. THE CHAIN RULE: A COMPUTATIONAL APPROACH 165
as highlighted in bold in the figure. Again, the collection of the four individual partial derivatives
of the composition computed in this way can be organized into a single matrix equation:
""∂w1
∂s
∂w1
∂t
∂w2
∂s
∂w2
∂t
#
=
""∂g1
∂x
∂g1
∂y
∂g1
∂z
∂g2
∂x
∂g2
∂y
∂g2
∂z
#

∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t
∂z
∂s
∂z
∂t

.
Once again, D(g ◦ f) appears on the left, while the product of Dg and Df is on the right.
The same pattern persists in general for any composition. The individual partial derivatives of
the composition can be calculated using the Little Chain Rule and then combined as the matrix
equation D(g ◦ f)(a) = Dg(f(a)) · Df(a). In a way, the matrix version of the chain rule is a
bookkeeping device that keeps track of many, many Little Chain Rule calculations!
Indeed, since we presented a rigorous proof of the Little Chain Rule in Exercise 5.6 of Chapter",Multivariable_Calculus_Shimamoto.pdf
"Indeed, since we presented a rigorous proof of the Little Chain Rule in Exercise 5.6 of Chapter
4, we can think of this approach as supplying a proof of the general chain rule as well.
Example 6.12. The substitutions x = r cos θ and y = r sin θ convert a smooth real-valued function
f(x, y) of x and y into a function of r and θ: w = f(r cos θ, rsin θ). For instance, if f(x, y) = x2y3,
then w = (r cos θ)2(r sin θ)3 = r5 cos2 θ sin3 θ.
We find general formulas for ∂w
∂r and ∂w
∂θ in terms of ∂f
∂x and ∂f
∂y . As just discussed, we can
compute these individual partials using the Little Chain Rule and a dependence diagram (Figure
6.5).
w = f
x y
r θ
Figure 6.5: Finding partial derivatives with respect to polar coordinates
Hence:
∂w
∂r = ∂f
∂x
∂x
∂r + ∂f
∂y
∂y
∂r = ∂f
∂x cos θ + ∂f
∂y sin θ
and ∂w
∂θ = ∂f
∂x
∂x
∂θ + ∂f
∂y
∂y
∂θ = −∂f
∂x r sin θ + ∂f
∂y r cos θ.
Example 6.13 (Implicit differentiation) . Let f : R2 → R be a smooth real-valued function of",Multivariable_Calculus_Shimamoto.pdf
"∂x
∂x
∂θ + ∂f
∂y
∂y
∂θ = −∂f
∂x r sin θ + ∂f
∂y r cos θ.
Example 6.13 (Implicit differentiation) . Let f : R2 → R be a smooth real-valued function of
two variables, and let S be the level set corresponding to f = c, where c is a constant. That is,
S = {(x, y) ∈ R2 : f(x, y) = c}. Assume that the condition f(x, y) = c defines y implicitly as a
smooth function of x, y = y(x), for some interval of x values. See Figure 6.6. Find the derivative
dy
dx in terms of f and its partial derivatives.
We start with the condition f(x, y(x)) = c that defines y(x). The expression on the left is the
composition x 7→ (x, y(x)) 7→ f(x, y(x)). Call this last value w. The dependence diagram is in
Figure 6.7.
The derivative dw
dx can be computed using the chain rule:
dw
dx = ∂f
∂x
dx
dx + ∂f
∂y
dy
dx = ∂f
∂x + ∂f
∂y
dy
dx.",Multivariable_Calculus_Shimamoto.pdf
"166 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
Figure 6.6: The level set f(x, y) = c defines y implicitly as a function of x on an interval of x values.
w = f
x y
x
Figure 6.7: A dependence diagram for implicit differentiation
On the other hand, w has a constant value of c, so dw
dx = 0. Hence:
∂f
∂x + ∂f
∂y
dy
dx = 0. (6.9)
Solving for dy
dx gives dy
dx = −∂f/∂x
∂f/∂y .
For example, consider the ellipse x2
9 + y2
4 = 1. It’s the level set of f(x, y) = x2
9 + y2
4 corresponding
to c = 1. Hence, by the previous example, along the ellipse, we have:
dy
dx = −∂f/∂x
∂f/∂y = −
2
9 x
1
2 y = −4x
9y .
Of course, implicit differentiation problems of this type appear in first-year calculus, and the
multivariable chain rule is never mentioned. There, to find the derivative along the ellipse, one
simply differentiates both sides of the equation of the ellipse and uses the one-variable chain rule.
In other words, one computes d
dx
x2
9 + y2
4


= d
dx(1), or:
2
9x + 1
2y dy
dx = 0.",Multivariable_Calculus_Shimamoto.pdf
"In other words, one computes d
dx
x2
9 + y2
4


= d
dx(1), or:
2
9x + 1
2y dy
dx = 0.
This is exactly the same as the result given by equation (6.9), from which the general formula
for dy
dx followed immediately. The multivariable chain rule gives us a way of articulating what the
one-variable calculation is doing.
6.5 Exercises for Chapter 6
Section 1 Continuity revisited
1.1. (a) If x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) are elements of Rn, show that |xi − yi| ≤
∥x − y∥ for i = 1, 2, . . . , n.",Multivariable_Calculus_Shimamoto.pdf
"6.5. EXERCISES FOR CHAPTER 6 167
(b) Let U be an open set in Rn, and let f : U → Rm be a function, where f(x) =
(f1(x), f2(x), . . . , fm(x)). Show that, if f is continuous at a point a of U, then so
are f1, f2, . . . , fm.
1.2. Let U be an open set in Rn, a a point of U, and f : U − {a} →Rm a function defined
on U, except possibly at a. Write out in terms of ϵ’s and δ’s what it means to say that
limx→a f(x) = L.
Section 2 Differentiability revisited
2.1. Let f : R2 → R2 be given by f(x, y) = (x2 − y2, 2xy).
(a) Find Df(x, y).
(b) Find Df(1, 2).
2.2. Let f : R3 → R2 be given by f(x, y, z) =

xy2z3 + 2, xcos(yz)


.
(a) Find Df(x, y, z).
(b) Find Df(−π
2 , 1, π
2 ).
2.3. Find Df(x, y, z) if f(x, y, z) = x + y2 + 3z3.
2.4. Find Df(x, y, z) if f(x, y, z) = (x, y2, 3z3).
2.5. Find Df(x, y) if f(x, y) = (x, y,0).
2.6. Find Df(x) if f(x) = (x2, xsin x, 2x + 3).
2.7. Let f : R2 → R3 be given by f(x, y) = (ex+y, e−x cos y, e−x sin y).
(a) Find Df(x, y).",Multivariable_Calculus_Shimamoto.pdf
"2.6. Find Df(x) if f(x) = (x2, xsin x, 2x + 3).
2.7. Let f : R2 → R3 be given by f(x, y) = (ex+y, e−x cos y, e−x sin y).
(a) Find Df(x, y).
(b) Show that f is differentiable at every point of R2.
2.8. Let f : R3 → R3 be the spherical coordinate transformation from ρϕθ-space to xyz-space:
f(ρ, ϕ, θ) = (ρ sin ϕ cos θ, ρsin ϕ sin θ, ρcos ϕ).
(a) Find Df(ρ, ϕ, θ).
(b) Show that f is differentiable at every point of R3.
(c) Show that det Df(ρ, ϕ, θ) = ρ2 sin ϕ. (This result will be used in the next chapter.)
2.9. Let T : R3 → R3 be a linear transformation, and let A be its matrix with respect to the
standard bases. Hence T(x) = Ax, where A is a 3 by 3 matrix, A =


a b c
d e f
g h i

.
(a) Find DT (x, y, z).
(b) Use the definition of differentiability to show that T is differentiable at every point a of
R3.
2.10. Given points x = (x1, x2, x3) and y = (y1, y2, y3) in R3, let us think of ( x, y) as the point",Multivariable_Calculus_Shimamoto.pdf
"R3.
2.10. Given points x = (x1, x2, x3) and y = (y1, y2, y3) in R3, let us think of ( x, y) as the point
(x1, x2, x3, y1, y2, y3) in R6. Let f : R6 → R and g : R6 → R3 be the dot and cross products in
R3, respectively, that is:
f(x, y) = x · y and g(x, y) = x × y.",Multivariable_Calculus_Shimamoto.pdf
"168 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
(a) Find Df(x, y) and Dg(x, y).
(b) Show that f and g are differentiable at every point ( x, y) of R6.
Section 3 The chain rule: a conceptual approach
3.1. Let f : R2 → R3 and g : R3 → R2 be given by:
f(s, t) =

s − t + 2, e2s+3t, scos t


and g(x, y, z) = (xyz, x2 + z3).
(a) Find Df(s, t) and Dg(x, y, z).
(b) Find D(g ◦ f)(0, 0).
(c) Find D(f ◦ g)(0, 0, 0).
3.2. Let f : R2 → R3 and g : R3 → R be the functions:
f(s, t) =

2s − t2, st− 1, 2s2 + st − t2

and g(x, y, z) = (x + 1)2eyz.
If h(s, t) = g(f(s, t)), find Dh(1, 2).
3.3. Let U and V be open sets in R2, and let f : U → V and g : V → U be inverse functions. That
is:
• g(f(x, y)) = (x, y) for all ( x, y) in U and
• f(g(x, y)) = (x, y) for all ( x, y) in V .
Let a be a point of U, and let b = f(a). If f is differentiable at a and g is differentiable at
b, what can you say about the following matrix products?
(a) Dg(b) · Df(a)
(b) Df(a) · Dg(b)",Multivariable_Calculus_Shimamoto.pdf
"b, what can you say about the following matrix products?
(a) Dg(b) · Df(a)
(b) Df(a) · Dg(b)
3.4. Exercise 5.5 of Chapter 4 described a generalization of the mean value theorem for real-valued
functions of n variables. Does the same generalization apply to vector-valued functions? More
precisely, let a be a point of Rn, and let B = B(a, r) be an open ball centered at a. Is it
necessarily true that, if f : B → Rm is a differentiable function on B and b is any point of B,
then there exists a point c on the line segment connecting a and b such that:
f(b) − f(a) = Df(c) · (b − a)?
If true, give a proof. Otherwise, find a counterexample.
Section 4 The chain rule: a computational approach
4.1. Let f(x, y, z) be a smooth real-valued function of x, y, and z. The substitutions x = s + 2t,
y = 3s+4t, and z = 5s+6t convert f into a function of s and t: w = f(s+2t, 3s+4t, 5s+6t).
Find expressions for ∂w
∂s and ∂w
∂t in terms of ∂f
∂x , ∂f
∂y , and ∂f
∂z .",Multivariable_Calculus_Shimamoto.pdf
"Find expressions for ∂w
∂s and ∂w
∂t in terms of ∂f
∂x , ∂f
∂y , and ∂f
∂z .
4.2. Let w = f(x, y) be a smooth real-valued function, where:
x = s2 − t2 + u2 + 2s − 2t and y = stu.
Find expressions for ∂w
∂s , ∂w
∂t , and ∂w
∂u in terms of ∂f
∂x and ∂f
∂y .",Multivariable_Calculus_Shimamoto.pdf
"6.5. EXERCISES FOR CHAPTER 6 169
4.3. The spherical substitutions x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, and z = ρ cos ϕ convert a smooth
real-valued function f(x, y, z) into a function of ρ, ϕ, and θ:
w = f(ρ sin ϕ cos θ, ρsin ϕ sin θ, ρcos ϕ).
(a) Find formulas for ∂w
∂ρ , ∂w
∂ϕ , and ∂w
∂θ in terms of ∂f
∂x , ∂f
∂y , and ∂f
∂z .
(b) If f(x, y, z) = x2 + y2 + z2, use your answer to part (a) to find ∂w
∂ϕ , and verify that it is
the same as the result obtained if you first write w in terms of ρ, ϕ, and θ directly, say
by substituting for x, y, and z, and then differentiate that expression with respect to ϕ.
4.4. The substitutions x = s2 − t2 and y = 2st convert a smooth real-valued function f(x, y) into
a function of s and t: w = f(s2 − t2, 2st). Find a formula for


∇w


 =


∂w
∂s , ∂w
∂t



 in terms
of ∂f
∂x and ∂f
∂y .
4.5. Let f(x, y) be a smooth real-valued function of x and y. The substitutions
x = e2s+t − 1 and y = s2 − 3st + 2t
convert f into a function w of s and t. If ∂w",Multivariable_Calculus_Shimamoto.pdf
"x = e2s+t − 1 and y = s2 − 3st + 2t
convert f into a function w of s and t. If ∂w
∂s = 5 and ∂w
∂t = 4 when s and t are both 0, find
the values of ∂f
∂x and ∂f
∂y when x and y are both 0. (Note that when s and t are both 0, so are
x and y.)
4.6. The substitutions x = r cos θ and y = r sin θ convert a smooth real-valued function f(x, y)
into a function of r and θ: w = f(r cos θ, rsin θ).
(a) Show that r∂w
∂r = x∂f
∂x + y∂f
∂y .
(b) Find a similar expression for ∂w
∂θ in terms of x, y, ∂f
∂x , and ∂f
∂y .
4.7. (a) Suppose that w = f(u, v) is a smooth real-valued function, where u = x/z and v = y/z.
Show that:
x∂w
∂x + y∂w
∂y + z ∂w
∂z = 0. (6.10)
(b) Without calculating any partial derivatives, deduce that w = x2+xy+y2
z2 satisfies equation
(6.10).
4.8. Let f : U → R be a smooth real-valued function of three variables defined on an open set U
in R3, and let c be a constant. Assume that the condition f(x, y, z) = c defines z implicitly",Multivariable_Calculus_Shimamoto.pdf
"in R3, and let c be a constant. Assume that the condition f(x, y, z) = c defines z implicitly
as a smooth function of x and y, z = z(x, y), on some open set of points ( x, y) in R2. Show
that, on this open set:
∂z
∂x = −∂f/∂x
∂f/∂z and ∂z
∂y = −∂f/∂y
∂f/∂z .
4.9. The condition xy + xz + yz + ex+2y+3z = 1 defines z implicitly as a smooth function of x and
y on some open set of points ( x, y) containing (0, 0) in R2. Find ∂z
∂x and ∂z
∂y when x = 0 and
y = 0.
4.10. (a) Let f : R3 → R2 be a smooth function. By definition, the level set S of f corresponding
to the value c = (c1, c2) is the set of all ( x, y, z) such that f(x, y, z) = c or, equivalently,
where f1(x, y, z) = c1 and f2(x, y, z) = c2. Typically, the last two equations define a
curve in R3, the curve of intersection of the surfaces defined by f1 = c1 and f2 = c2.",Multivariable_Calculus_Shimamoto.pdf
"170 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE
Assume that the condition f(x, y, z) = c defines x and y implicitly as smooth functions
of z on some interval of z values in R, that is, x = x(z) and y = y(z) are functions
that satisfy f(x(z), y(z), z) = c. Then the corresponding portion of the level set is
parametrized by the path α(z) = (x(z), y(z), z).
Show that x′(z) and y′(z) satisfy the matrix equation:
""∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y
#x′(z)
y′(z)

= −
""∂f1
∂z
∂f2
∂z
#
,
where the partial derivatives are evaluated at the point ( x(z), y(z), z). ( Hint: Let w =
(w1, w2) = f(x(z), y(z), z).)
(b) Let f(x, y, z) = ( x − cos z, y− sin z), and let C be the level set of f corresponding to
c = (0 , 0). Find the functions x = x(z) and y = y(z) and the parametrization α of
C described in part (a), and describe C geometrically. Then, use part (a) to help find
Dα(z), and verify that your answer makes sense.",Multivariable_Calculus_Shimamoto.pdf
"C described in part (a), and describe C geometrically. Then, use part (a) to help find
Dα(z), and verify that your answer makes sense.
4.11. Let f(x, y) be a smooth real-valued function of x and y defined on an open set U in R2. The
partial derivative ∂f
∂x is also a real-valued function of x and y (as is ∂f
∂y ). If α(t) = (x(t), y(t))
is a smooth path in U, then substituting for x and y in terms of t converts ∂f
∂x into a function
of t, say w = ∂f
∂x (x(t), y(t)).
(a) Show that:
dw
dt = ∂2f
∂x2
dx
dt + ∂2f
∂y ∂x
dy
dt ,
where the partial derivatives are evaluated at α(t).
(b) Now, consider the composition g(t) = (f ◦α)(t). Find a formula for the second derivative
d2g
dt2 in terms of the partial derivatives of f as a function of x and y and the derivatives
of x and y as functions of t. ( Hint: Differentiate the Little Chain Rule. Note that this
involves differentiating some products.)
(c) Let a be a point of U. If α: I → U, α(t) = a+tv = (a1 +tv1, a2 +tv2), is a parametriza-",Multivariable_Calculus_Shimamoto.pdf
"involves differentiating some products.)
(c) Let a be a point of U. If α: I → U, α(t) = a+tv = (a1 +tv1, a2 +tv2), is a parametriza-
tion of a line segment containing a and if g(t) = (f ◦ α)(t), use your formula from part
(b) to show that:
d2g
dt2 = v2
1
∂2f
∂x2 + 2v1v2
∂2f
∂x ∂y+ v2
2
∂2f
∂y2 ,
where the partial derivatives are evaluated at α(t). (Compare this with equation (4.22)
in Chapter 4 regarding the second-order approximation of f at a.)
4.12. Let f(x, y) be a smooth real-valued function defined on an open set in R2. The polar co-
ordinate substitutions x = r cos θ and y = r sin θ convert f into a function of r and θ:
w = f(r cos θ, rsin θ).
(a) Show that:
∂2w
∂θ2 = −r cos θ ∂f
∂x − r sin θ ∂f
∂y + r2 sin2 θ ∂2f
∂x2 − 2r2 sin θ cos θ ∂2f
∂x ∂y
+ r2 cos2 θ ∂2f
∂y2 .
(Hint: Start with the results of Example 6.12.)",Multivariable_Calculus_Shimamoto.pdf
"6.5. EXERCISES FOR CHAPTER 6 171
(b) Find an analogous expression for the mixed partial derivative ∂2w
∂r ∂θ.
(c) Find an analogous expression for ∂2w
∂r2 .
(d) If f(x, y) = x2 +y2 +xy, use your answer to part (b) to find ∂2w
∂r ∂θ. Then, verify that it is
the same as the result obtained by writing w in terms of r and θ directly by substituting
for x and y and then differentiating that expression with respect to r and θ.",Multivariable_Calculus_Shimamoto.pdf
172 CHAPTER 6. DIFFERENTIABILITY AND THE CHAIN RULE,Multivariable_Calculus_Shimamoto.pdf
"Chapter 7
Change of variables
We now turn to the counterpart for integrals of the chain rule. It is called the change of variables
theorem. For functions of one variable, the corresponding notion is the method of substitution.
The one-variable and multivariable versions are expressed in equations that are similar formally,
but the approaches one might take to understand where they come from are quite different. It’s
exciting when a change in perspective still leads to a result of recognizable form. We make a few
remarks comparing the two versions later, after the theorem has been stated.
In many respects, the change of variables theorem and the chain rule hold together the foun-
dation of the more advanced theory of multivariable calculus. We begin, however, with a concrete
example where changing variables just seems like the right thing to do.
Example 7.1. Let D be the disk x2 + y2 ≤ 4 in the xy-plane. Find
RR
D
p
x2 + y2 dx dy. For",Multivariable_Calculus_Shimamoto.pdf
"Example 7.1. Let D be the disk x2 + y2 ≤ 4 in the xy-plane. Find
RR
D
p
x2 + y2 dx dy. For
instance, if one wanted to know the average distance to the origin on D, one would evaluate this
integral and divide by Area (D) = 4π.
Figure 7.1: The disk x2 + y2 ≤ 4
Setting up the integral with the order of integration dy dxin the usual way (Figure 7.1) gives:
ZZ
D
p
x2 + y2 dx dy=
Z 2
−2
Z √
4−x2
−
√
4−x2
p
x2 + y2 dy

dx.
While not impossible, this looks potentially messy. On the other hand, expressing the problem in
terms of polar coordinates seems promising for two reasons.
• The function to be integrated is simpler:
p
x2 + y2 = r.
• The region of integration is simpler: in polar coordinates, D is described by 0 ≤ r ≤ 2,
0 ≤ θ ≤ 2π, in other words, by a rectangle in the rθ-plane.
173",Multivariable_Calculus_Shimamoto.pdf
"174 CHAPTER 7. CHANGE OF VARIABLES
The obstacle is what to do about the dA = dx dypart of the integral. For this, we need to see how
small pieces of area in the xy-plane are related to those in the rθ-plane. We return to this example
later.
7.1 Change of variables for double integrals
Let
RR
D f(x, y) dx dybe a given double integral. We want to describe how to convert the integral to
an equivalent one with a different domain of integration and a correspondingly different integrand.
The setup is to assume that there is a smooth function T : D∗ → D, where D∗ is a subset of
R2. We think of T as transforming D∗ onto D, so we assume that T(D∗) = D and that T is
one-to-one, except possibly along the boundary of D∗. This last part means that whenever a and
b are distinct points of D∗, not on the boundary, then the values T(a) and T(b) are distinct, too:
a ̸= b ⇒ T(a) ̸= T(b). (For the purposes of integration, anomalous behavior confined to the",Multivariable_Calculus_Shimamoto.pdf
"a ̸= b ⇒ T(a) ̸= T(b). (For the purposes of integration, anomalous behavior confined to the
boundary can be allowed typically without messing things up. See the remarks on page 130.)
The new integral will be an integral over D∗. It helps keep things straight if the coordinates
in the domain and codomain have different names, so we think of T as a transformation from the
uv-plane to the xy-plane and write T(u, v) = (x(u, v), y(u, v)).
Thus we want to convert the given integral over D with respect to x and y to an integral over
D∗ with respect to u and v. We outline the main idea. It is similar to what we did in Section 5.5 to
define surface integrals with respect to surface area using a parametrization, though of course there
is a big difference between motivating a definition and justifying a theorem. Another difference is
that, in the interim, we have learned about the derivative of a vector-valued function.",Multivariable_Calculus_Shimamoto.pdf
"that, in the interim, we have learned about the derivative of a vector-valued function.
Figure 7.2: Change of variables: x and y as functions of u and v via transformation T
To begin, subdivide D∗ into small subrectangles of dimensions △ui by △vj, as in Figure 7.2.
The transformation T sends a typical subrectangle of area △ui △vj in the uv-plane to a small curvy
quadrilateral in the xy-plane. Let △Aij denote the area of the curvy quadrilateral. If we choose
a sample point pij in each curvy quadrilateral, we can form a Riemann sum-like approximation of
the integral over D: ZZ
D
f(x, y) dx dy≈
X
i,j
f(pij) △Aij. (7.1)
To turn this into a Riemann sum over D∗, we need to relate the areas of the subrectangles and the
curvy quadrilaterals.
To do so, we use the first-order approximation of T. Choose a point aij in the (i, j)th subrect-
angle of D∗ such that T(aij) = pij. Then the first-order approximation says that, for points u in",Multivariable_Calculus_Shimamoto.pdf
"7.1. CHANGE OF VARIABLES FOR DOUBLE INTEGRALS 175
D∗ near aij:
T(u) ≈ T(aij) + DT (aij) · (u − aij)
≈ pij + DT (aij) · (u − aij)
≈

pij − DT (aij) · aij


+ DT (aij) · u. (7.2)
It may be difficult to get a foothold on the behavior ofT itself, but describing how the approximation
transforms the (i, j)th subrectangle is quite tractable.
For this, we bring in some linear algebra. Note that the derivative DT (aij) is a 2 by 2 ma-
trix. As shown in Chapter 1, any 2 by 2 matrix A =
a b
c d

determines a linear transformation
L: R2 → R2 given by matrix multiplication: L(x) = Ax. This transformation sends the unit
square determined by e1 and e2 to the parallelogram determined by L(e1) = [ ac ] and L(e2) =
 b
d

,
which are the columns of A (Figure 7.3). We saw in Proposition 1.13 of Chapter 1 that the area
Figure 7.3: A linear transformation L sends the square determined bye1 and e2 to the parallelogram
determined by L(e1) and L(e2).
of this parallelogram is |det

L(e1) L(e2)

|, where",Multivariable_Calculus_Shimamoto.pdf
"determined by L(e1) and L(e2).
of this parallelogram is |det

L(e1) L(e2)

|, where

L(e1) L(e2)

is the matrix whose rows are
L(e1) and L(e2). This is precisely the matrix transpose At. Thus the area of the parallelogram is
|det(At)| = |det A|, where we have used the general fact that an n by n matrix and its transpose
have the same determinant (Proposition 1.14, Chapter 1). In particular, L has changed the area
by a factor of |det A|.
This principle extends easily to a couple of slightly more general cases.
• Given nonzero scalars ℓ and w, let R be the |ℓ| by |w| rectangle determined by ℓe1 and we2.
Then L sends R to the parallelogram P determined by L(ℓe1) = ℓL(e1) and L(we2) = wL(e2).
Compared to the original case, the areas of both rectangle and parallelogram have changed
by a common factor of |ℓw|, hence they still differ by a factor of |det A|.
• Suppose that we translate the rectangle R in the previous case by a constant vector a. This",Multivariable_Calculus_Shimamoto.pdf
"• Suppose that we translate the rectangle R in the previous case by a constant vector a. This
results in another |ℓ| by |w| rectangle R′ consisting of all points of the forma+x, where x ∈ R.
By linearity,L(a+x) = L(a)+L(x), so L transforms R′ to the translation of the parallelogram
L(R) = P by L(a). Let’s call the translated parallelogram P′. Since translations don’t affect
area, the areas of R′ and its image P′ again differ by a factor of |det A|.
This shows the following.
Lemma 7.2. A linear transformation L: R2 → R2 represented with respect to the standard bases by
a 2 by 2 matrix A sends rectangles whose sides are parallel to the coordinate axes to parallelograms
and alters the area of the rectangles by a factor of |det A|.",Multivariable_Calculus_Shimamoto.pdf
"176 CHAPTER 7. CHANGE OF VARIABLES
Returning to the first-order approximation (7.2) of T, we conclude that multiplication by
DT (aij) transforms the subrectangle of area △ui △vj that contains aij to a parallelogram of area
|det DT (aij)|△ui △vj. Hence, after additional translation, the first-order approximation (7.2)
sends the subrectangle containing aij to a parallelogram that:
• approximates the curvy quadrilateral of area △Aij that contains pij and
• has area |det DT (aij)|△ui △vj.
Thus the Riemann sum approximation (7.1) becomes:
ZZ
D
f(x, y) dx dy≈
X
i,j
f(pij) △Aij ≈
X
i,j
f(T(aij)) |det DT (aij)|△ui △vj.
Letting △ui and △vj go to zero, this becomes an integral over D∗, giving the following major
result.
Theorem 7.3 (Change of variables theorem for double integrals). Let D and D∗ be bounded subsets
of R2, and let T : D∗ → D be a smooth function such that T(D∗) = D and that is one-to-one, except
possibly on the boundary of D∗. If f is integrable on D, then:
ZZ
D
f(x, y) dx dy=
ZZ",Multivariable_Calculus_Shimamoto.pdf
"possibly on the boundary of D∗. If f is integrable on D, then:
ZZ
D
f(x, y) dx dy=
ZZ
D∗
f(T(u, v)) |det DT (u, v)|du dv.
In other words, x and y are replaced in terms of u and v in the function f using (x, y) = T(u, v),
and dx dyis replaced by |det DT (u, v)|du dv=
det
∂x
∂u
∂x
∂v∂y
∂u
∂y
∂v
du dv. The determinant of the
derivative, detDT (u, v), is sometimes referred to as the Jacobian determinant.
By comparison, in first-year calculus, if f(x) is a real-valued function of one variable and x is
expressed in terms of another variable u, say as x = T(u), then the method of substitution says
that: Z T(b)
T(a)
f(x) dx =
Z b
a
f(T(u)) T′(u) du.
In this case, it seems simplest to understand the result in terms of antiderivatives using the chain
rule and the fundamental theorem of calculus rather than through Riemann sums, though we
won’t present the details of the argument. In particular, substituting for x in terms of u includes
the substitution dx = T′(u) du = dx",Multivariable_Calculus_Shimamoto.pdf
"won’t present the details of the argument. In particular, substituting for x in terms of u includes
the substitution dx = T′(u) du = dx
du du. The substitution dx dy= |det DT (u, v)|du dvin the
previous paragraph is the analogue in the double integral case. We say more about the principle
of substitution in the next section.
Returning to double integrals, in the case of polar coordinates, the substitutions for x and y
are given by (x, y) = T(r, θ) = (r cos θ, rsin θ). Then DT (r, θ) =
cos θ −r sin θ
sin θ r cos θ

, and:
|det DT (r, θ)| = |r cos2 θ + r sin2 θ| = |r| = r.
By the change of variables theorem, this means that “ dx dy= r dr dθ” and that polar coordinates
transform integrals as follows.
Corollary 7.4. In polar coordinates:
ZZ
D
f(x, y) dx dy=
ZZ
D∗
f(r cos θ, rsin θ) r dr dθ,
where D∗ is the region in the rθ-plane that describes D in polar coordinates.",Multivariable_Calculus_Shimamoto.pdf
"7.2. A WORD ABOUT SUBSTITUTION 177
We now complete Example 7.1, which asked to evaluate
RR
D
p
x2 + y2 dx dy, where D is the
disk x2 + y2 ≤ 4 in the xy-plane. As noted earlier, D is described in polar coordinates by the
rectangle D∗ given by 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π in the rθ-plane. The polar coordinate transformation
T is shown in Figure 7.4.
Figure 7.4: A change of variables in polar coordinates
We verify that T satisfies the hypotheses of the theorem, though we shall grow less meticulous
about writing out the details on this point after a couple more examples. All the points on the
left side of D∗, where r = 0, are mapped to the origin, that is, T(0, θ) = (0 , 0) for 0 ≤ θ ≤ 2π.
Also, along the top and bottom of D∗, θ = 0 and θ = 2 π correspond to the same thing in the
xy-plane: T(r,0) = T(r,2π) for 0 ≤ r ≤ 2. Apart from this, distinct points in D∗ are mapped to
distinct values in D. In other words, T is one-to-one away from the boundary of D∗, so the change",Multivariable_Calculus_Shimamoto.pdf
"distinct values in D. In other words, T is one-to-one away from the boundary of D∗, so the change
of variables theorem applies.
Since
p
x2 + y2 = r, this gives:
ZZ
D
p
x2 + y2 dx dy=
ZZ
D∗
r · r dr dθ
=
Z 2π
0
Z 2
0
r2 dr

dθ
=
Z 2π
0
1
3r3

r=2
r=0

dθ
=
Z 2π
0
8
3 dθ
= 8
3θ
2π
0
= 16
3 π.
7.2 A word about substitution
The change of variables theorem concerns how to convert an integral with respect to x and y into
an integral with respect to two other variables, say u and v. Another name for this is substitution:
we require expressions to substitute for x, y, and dA = dx dyin terms of u and v. Substituting for
x and y means writing them as functions of u and v: x = x(u, v) and y = y(u, v). This is equivalent
to having a function T(u, v) = ( x(u, v), y(u, v)) from the uv-plane to the xy-plane. The logistics
can be confusing and perhaps counterintuitive: the transformation of the integral goes from being",Multivariable_Calculus_Shimamoto.pdf
"can be confusing and perhaps counterintuitive: the transformation of the integral goes from being
with respect to x and y to being with respect to u and v, but, in order to accomplish this, the",Multivariable_Calculus_Shimamoto.pdf
"178 CHAPTER 7. CHANGE OF VARIABLES
geometric transformation goes the other way, from ( u, v) to ( x, y). This is an inherent aspect of
substitution.
Sometimes, the original integrand is denoted by an expression like η = f(x, y) dx dy, and then
the transformed integrandf(T(u, v)) |det DT (u, v)|du dvis denoted byT∗(η). Another appropriate
notation for it might be η∗. With this notation, the change of variables theorem becomes:
ZZ
T(D∗)
η =
ZZ
D∗
T∗(η). (7.3)
Actually, what we just said is not entirely accurate and the situation is a little more complicated, 7
but it serves to illustrate the general principle: in a substitution, the geometry is “pushed forward,”
and the integral is “pulled back.”
7.3 Examples: linear changes of variables, symmetry
We illustrate the change of variables theorem with three further examples.
Example 7.5. Let D be the region in R2 given by x2
9 + y2
4 ≤ 1, y ≥ 0. Evaluate
RR
D(x2 +y2) dx dy.",Multivariable_Calculus_Shimamoto.pdf
"Example 7.5. Let D be the region in R2 given by x2
9 + y2
4 ≤ 1, y ≥ 0. Evaluate
RR
D(x2 +y2) dx dy.
Here, the integrand f(x, y) = x2 + y2 is fairly simple, but the region of integration poses some
problems. D is the region inside the upper half of an ellipse. The endpoints of the iterated integral
with respect to x and y are somewhat hard to work with, and polar coordinates don’t help either—
the relationship between r and θ needed to describe D is a little complicated. On the other hand,
we can think of D as a transformed semicircular disk, which is easy to describe in polar coordinates.
Thus we attempt to map the half-disk D∗ of radius 1 given in the uv-plane by u2 + v2 ≤ 1, v ≥ 0,
over to D and hope that this does not complicate the function to be integrated too much.
Figure 7.5: Changing an integral over a half-ellipse to an integral over a half-disk
The x and y-intercepts of the ellipse are ±3 and ±2, respectively, so to achieve the transforma-",Multivariable_Calculus_Shimamoto.pdf
"The x and y-intercepts of the ellipse are ±3 and ±2, respectively, so to achieve the transforma-
tion, as illustrated in Figure 7.5, we stretch horizontally by a factor of 3 and vertically by a factor
of 2. In other words, define:
T(u, v) = (3u, 2v).
In effect, we are making the substitutions x = 3u and y = 2v. Then T transforms D∗ to D, that
is, if ( u, v) ∈ D∗, then T(u, v) ∈ D. This is because if ( u, v) satisfies u2 + v2 ≤ 1 and v ≥ 0, then
(x, y) = T(u, v) satisfies x2
9 + y2
4 = (3u)2
9 + (2v)2
4 = u2 + v2 ≤ 1 and y = 2v ≥ 0.
Moreover, T is one-to-one on the whole uv-plane. This seems reasonable geometrically, or, in
terms of equations, if ( u1, v1) ̸= (u2, v2), then (3 u1, 2v1) ̸= (3u2, 2v2). Perhaps it is more readable
7The correct expressions are η = f(x, y) dx ∧dy and T∗(η) = f(T(u, v)) detDT (u, v) du ∧dv. We shall learn more
about integrands like this and, more importantly, how to integrate them in the last two chapters. In particular, for",Multivariable_Calculus_Shimamoto.pdf
"about integrands like this and, more importantly, how to integrate them in the last two chapters. In particular, for
the correct version of equation (7.3), see Exercise 4.20 in Chapter 11.",Multivariable_Calculus_Shimamoto.pdf
"7.3. EXAMPLES: LINEAR CHANGES OF VARIABLES, SYMMETRY 179
to phrase this in the contrapositive: if (3u1, 2v1) = (3u2, 2v2), then (u1, v1) = (u2, v2). This is clear.
For instance, 3u1 = 3u2 ⇒ u1 = u2.
Finally, DT (u, v) = [ 3 0
0 2], so |det DT (u, v)| = 6, and f(T(u, v)) = (3u)2 + (2v)2 = 9u2 + 4v2.
Thus, by the change of variables theorem:
ZZ
D
(x2 + y2) dx dy=
ZZ
D∗
(9u2 + 4v2) · 6 du dv.
We convert this to polar coordinates in the uv-plane using u = r cos θ, v = r sin θ, and du dv=
r dr dθ. The half-disk D∗ is described in polar coordinates by 0 ≤ r ≤ 1, 0 ≤ θ ≤ π, in other words,
by the rectangle [0, 1] × [0, π] in the rθ-plane. Thus:
ZZ
D
(x2 + y2) dx dy=
ZZ
D∗in
polar
(9r2 cos2 θ + 4r2 sin2 θ) · 6 · r dr dθ
= 6
Z π
0
Z 1
0
r3(9 cos2 θ + 4 sin2 θ) dr

dθ. (7.4)
At this point, we pause to put on the record a couple of facts from the exercises that we shall
use freely from now on (see Exercise 2.3 in Chapter 5 and Exercise 1.1 in this chapter).",Multivariable_Calculus_Shimamoto.pdf
"use freely from now on (see Exercise 2.3 in Chapter 5 and Exercise 1.1 in this chapter).
• For a function F whose variables separate into two independent continuous factors,F(x, y) =
f(x)g(y), the integral over a rectangle [ a, b] × [c, d] is given by:
Z d
c
Z b
a
f(x)g(y) dx

dy =
Z b
a
f(x) dx
Z d
c
g(y) dy

.
Note that integrals over rectangles are characterized by the property that all the limits of
integration are constant.
•
Rπ
0 cos2 θ dθ=
Rπ
0 sin2 θ dθ= π
2 . (An easy way to remember this is to note that
Rπ
0 cos2 θ dθ=Rπ
0 sin2 θ dθfrom the graphs of the sine and cosine functions and that
Rπ
0 (cos2 θ + sin2 θ) dθ =Rπ
0 1 dθ = θ
π
0 = π.)
Applying these facts to the present integral (7.4) gives:
ZZ
D
(x2 + y2) dx dy= 6
Z 1
0
r3 dr
Z π
0
(9 cos2 θ + 4 sin2 θ) dθ

= 6 ·
1
4r41
0


·

9 · π
2 + 4 · π
2


= 6 · 1
4 · 13
2 π
= 39
4 π.
Example 7.6. If D is the parallelogram with vertices (0, 0), (3, 2), (2, 3), (−1, 1), find
RR
D xy dx dy.",Multivariable_Calculus_Shimamoto.pdf
"2 + 4 · π
2


= 6 · 1
4 · 13
2 π
= 39
4 π.
Example 7.6. If D is the parallelogram with vertices (0, 0), (3, 2), (2, 3), (−1, 1), find
RR
D xy dx dy.
The difficulty here is that setting up an iterated integral in terms of x and y involves splitting
into three integrals in order to describe D. To simplify the domain, we use the fact previously
noted that a linear transformation T : R2 → R2 transforms the unit square into the parallelogram
determined by T(e1) and T(e2). Here, parallelogram D is determined by the vectors (3 , 2) and
(−1, 1), so we want T(e1) = (3, 2) and T(e2) = (−1, 1). Using Proposition 1.7 from Chapter 1, we
put these vectors into the columns of a matrix A =
3 −1
2 1

and define:
T(u, v) =
3 −1
2 1
u
v

=
3u − v
2u + v

= (3u − v, 2u + v).",Multivariable_Calculus_Shimamoto.pdf
"180 CHAPTER 7. CHANGE OF VARIABLES
Figure 7.6: A linear change of variables
Then T maps the unit square D∗ = [0, 1] × [0, 1] in the uv-plane onto D. See Figure 7.6. We leave
for the exercises the verification that T is one-to-one on R2 (Exercise 3.3), so we can use the change
of variables theorem to pull back the original integral to an integral over D∗.
The substitutions given by T are x = 3u − v and y = 2u + v. Also, DT (u, v) =
3 −1
2 1

, and
|det DT (u, v)| = |3 + 2| = 5. Hence:
ZZ
D
xy dx dy=
ZZ
D∗
(3u − v)(2u + v) · 5 du dv
= 5
Z 1
0
Z 1
0
(6u2 + uv − v2) dv

du
= 5
Z 1
0
(6u2v + 1
2uv2 − 1
3v3)

v=1
v=0
du
= 5
Z 1
0
(6u2 + 1
2u − 1
3) du
= 5

2u3 + 1
4u2 − 1
3u
1
0)
= 5 (2 + 1
4 − 1
3)
= 5 · 24 + 3− 4
12
= 115
12 .
Example 7.7. A subset D of the xy-plane is said to be symmetric in the y-axis if, whenever
(x, y) is in D, so is ( −x, y) (Figure 7.7). Let D be such a subset.
(a) Show that
RR
D x dx dy= 0.",Multivariable_Calculus_Shimamoto.pdf
"(x, y) is in D, so is ( −x, y) (Figure 7.7). Let D be such a subset.
(a) Show that
RR
D x dx dy= 0.
(b) Let L = {(x, y) ∈ D : x ≤ 0} and R = {(x, y) ∈ D : x ≥ 0} be the left and right halves of D,
respectively. Show that
RR
L y dx dy=
RR
R y dx dy.
(a) The rough intuition here is that, in the Riemann sumsPx △x △y for
RR
D x dx dy, the symmetry
of D means that each contribution for x >0 is balanced by an opposite contribution for x <0,
and vice versa. For a more polished argument, let T : R2 → R2 be the reflection in the y-axis,
T(x, y) = (−x, y). Since D is symmetric in the y-axis, T transforms D into itself, i.e., T(D) = D.
Hence, in the notation of the change of variables theorem, D∗ = D. Since distinct points have
distinct reflections, T is one-to-one. Also DT (x, y) =
−1 0
0 1

, which gives|det DT (x, y)| = |−1| = 1.",Multivariable_Calculus_Shimamoto.pdf
"7.3. EXAMPLES: LINEAR CHANGES OF VARIABLES, SYMMETRY 181
Figure 7.7: Symmetry in the y-axis
We apply the change of variables theorem with f(x, y) = x so that f(T(x, y)) = f(−x, y) = −x.
Then:
ZZ
D
x dx dy=
ZZ
D∗=D
f(T(x, y)) |det DT (x, y)|dx dy
=
ZZ
D
−x · 1 dx dy
= −
ZZ
D
x dx dy.
As a result, 2
RR
D x dx dy= 0, so
RR
D x dx dy= 0.
(b) Here, the intuition is that the contributions to the Riemann sums Py △x △y for
RR
D y dx dy
are the same for ( x, y) and ( −x, y), hence the sums converge to the same value on L and on R.
More formally, again let T(x, y) = (−x, y). Then T(L) = R, i.e., R∗ = L, and |det DT (x, y)| = 1
as before. Now, f(x, y) = y, so f(T(x, y)) = f(−x, y) = y. Consequently:
ZZ
R
y dx dy=
ZZ
R∗=L
f(T(x, y)) |det DT (x, y)|dx dy
=
ZZ
L
y · 1 dx dy
=
ZZ
L
y dx dy.
This is exactly what we wanted to show.
This example can be generalized. Namely, let D be symmetric in the y-axis, and let f : D → R
be an integrable real-valued function.",Multivariable_Calculus_Shimamoto.pdf
"This example can be generalized. Namely, let D be symmetric in the y-axis, and let f : D → R
be an integrable real-valued function.
• If f(−x, y) = −f(x, y) for all ( x, y) in D, then:
ZZ
D
f(x, y) dx dy= 0.
• If f(−x, y) = f(x, y) for all ( x, y) in D, then:
ZZ
L
f(x, y) dx dy=
ZZ
R
f(x, y) dx dy.
The justifications use essentially the same reasoning as in the example.",Multivariable_Calculus_Shimamoto.pdf
"182 CHAPTER 7. CHANGE OF VARIABLES
7.4 Change of variables for n-fold integrals
The change of variables theorem for double integrals has a natural analogue for functions of n
variables. Let T be a smooth transformation that sends a subset W∗ of Rn onto a subset W of Rn
and that is one-to-one, except possibly on the boundary of W∗. See Figure 7.8. We think of T as
a function from ( u1, u2, . . . , un)-space to (x1, x2, . . . , xn)-space, so:
T(u1, u2, . . . , un) =

x1(u1, u2, . . . , un), x2(u1, u2, . . . , un), . . . , xn(u1, u2, . . . , un)


.
Then the change of variables formula reads:
ZZ
···
Z
W
f(x) dx1 dx2 ··· dxn =
ZZ
···
Z
W∗
f(T(u))
det DT (u)
du1 du2 ··· dun.
Figure 7.8: A three-dimensional change of variables
More informally, one uses T to substitute for x1, x2, . . . , xn in terms of u1, u2, . . . , un and sets
dx1 dx2 ··· dxn =
det DT (u)
du1 du2 ··· dun. This pulls back the integral over W to an integral
over W∗.",Multivariable_Calculus_Shimamoto.pdf
"dx1 dx2 ··· dxn =
det DT (u)
du1 du2 ··· dun. This pulls back the integral over W to an integral
over W∗.
The intuition behind this generalization is the same as for double integrals: A linear transfor-
mation L: Rn → Rn represented by a matrix A alters n-dimensional volume by a factor of |det A|.
So, to obtain the integral, one chops up W∗ into small pieces and, on each one, approximates T by
its first-order approximation, which alters volume by a factor of |det DT (u)|.
For triple integrals, there are two standard sets of alternate variables.
• Cylindrical coordinates. For this, the change of variables is T(r, θ, z) = ( x, y, z), where,
as we saw in Section 5.4.2, the substitutions are given by:



x = r cos θ
y = r sin θ
z = z.
Therefore DT (r, θ, z) =


cos θ −r sin θ 0
sin θ r cos θ 0
0 0 1

. After expanding along the third column, this
gives |det DT (r, θ, z)| =
det
cos θ −r sin θ
sin θ r cos θ
 = r, i.e.:
dx dy dz= r dr dθ dz.",Multivariable_Calculus_Shimamoto.pdf
"7.4. CHANGE OF VARIABLES FOR n-FOLD INTEGRALS 183
• Spherical coordinates. This time, T(ρ, ϕ, θ) = (x, y, z) with conversions:



x = ρ sin ϕ cos θ
y = ρ sin ϕ sin θ
z = ρ cos ϕ.
See Section 5.4.3. Then DT (ρ, ϕ, θ) =


sin ϕ cos θ ρ cos ϕ cos θ −ρ sin ϕ sin θ
sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ
cos ϕ −ρ sin ϕ 0

. We leave as
an exercise the calculation that |det DT (ρ, ϕ, θ)| = ρ2 sin ϕ (Exercise 2.8 in Chapter 6), i.e.:
dx dy dz= ρ2 sin ϕ dρ dϕ dθ.
Example 7.8. Find the volume of a closed ball of radius a in R3, that is, the volume of:
W = {(x, y, z) ∈ R3 : x2 + y2 + z2 ≤ a2}.
See Figure 7.9.
Figure 7.9: A ball of radius a in R3
We begin with Vol (W) =
RRR
W 1 dx dy dz. (See page 144.) As we are about to discover, W
is easy to describe in spherical coordinates, so we convert to an integral in spherical coordinates.
The endpoints depend on the order of integration, and here we choose dρ dϕ dθ. Thus the first",Multivariable_Calculus_Shimamoto.pdf
"The endpoints depend on the order of integration, and here we choose dρ dϕ dθ. Thus the first
integration is with respect to ρ. For fixed ϕ and θ, ρ varies along a radial line segment from ρ = 0
to ρ = a. Then, for fixed θ, ϕ rotates all the way down from the positive z-axis, ϕ = 0, to the
negative z-axis, ϕ = π. Finally, θ varies from θ = 0 to θ = 2π. In other words, W is described in
ρϕθ-space by the three-dimensional box W∗ = [0, a] × [0, π] × [0, 2π]. Hence:
Vol (W) =
ZZZ
W
1 dx dy dz
=
ZZZ
W∗
1 · ρ2 sin ϕ dρ dϕ dθ
=
Z 2π
0
Z π
0
Z a
0
ρ2 sin ϕ dρ

dϕ

dθ.
The variables of the integrand separate into independent factors and all endpoints are constant, so,",Multivariable_Calculus_Shimamoto.pdf
"184 CHAPTER 7. CHANGE OF VARIABLES
as in the two-dimensional case, we can separate the integrals as well:
Vol (W) =
Z a
0
ρ2 dρ
Z π
0
sin ϕ dϕ
Z 2π
0
1 dθ

=
1
3ρ3a
0

−cos ϕ
π
0

θ
2π
0

= 1
3a3 ·

1 − (−1)


· 2π
= 4
3πa3.
That is:
The volume of a three-dimensional ball of radius a is 4
3 πa3.
This is often referred to as the volume of a sphere, though that is a misnomer. A sphere is a
surface, not a solid.
For those who would like more practice setting up these types of integrals, this example could
have been solved using cylindrical coordinates as well, where dx dy dz= r dr dθ dz. We use the
order of integration dz dr dθ. For fixed r and θ, z varies from z = −
p
a2 − x2 − y2 = −
√
a2 − r2
to z = +
√
a2 − r2. Moreover, the possible values of r and θ come from the two-dimensional disk of
radius a in the xy-plane, which in polar coordinates is described by 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π. As a
result:
Vol (W) =
Z 2π
0
Z a
0
Z √
a2−r2
−
√
a2−r2
1 · r dz

dr

dθ
=
Z 2π
0",Multivariable_Calculus_Shimamoto.pdf
"result:
Vol (W) =
Z 2π
0
Z a
0
Z √
a2−r2
−
√
a2−r2
1 · r dz

dr

dθ
=
Z 2π
0
Z a
0
rz

z=
√
a2−r2
z=−
√
a2−r2
dr

dθ
=
Z 2π
0
Z a
0
2r
p
a2 − r2 dr

dθ.
Since we already calculated the answer, we won’t complete the evaluation.
Example 7.9. Let W be the region in R3 inside both the circular cylinder x2 + y2 = 4 and the
sphere x2 + y2 + z2 = 9 and above the xy-plane. Find
RRR
W (x2 + y2)z dx dy dz.
The region of integration W is a silo-shaped solid bounded laterally by a circular cylinder and
capped with a sphere, as drawn in Figure 7.10. Because the base is a disk in the xy-plane, perhaps
the simplest way to describe W is with polar coordinates together with the z-coordinate, that is,
with cylindrical coordinates. Substituting x = r cos θ, y = r sin θ, and dx dy dz= r dr dθ dzgives:
ZZZ
W
(x2 + y2)z dx dy dz=
ZZZ
Win
cylindrical
r2z · r dr dθ dz=
ZZZ
Win
cylindrical
r3z dr dθ dz.",Multivariable_Calculus_Shimamoto.pdf
"ZZZ
W
(x2 + y2)z dx dy dz=
ZZZ
Win
cylindrical
r2z · r dr dθ dz=
ZZZ
Win
cylindrical
r3z dr dθ dz.
We describe W in cylindrical coordinates using the order of integration dz dr dθ. For fixed r and
θ, z goes from z = 0 to z =
p
9 − x2 − y2 =
√
9 − r2. The domain of values of ( r, θ) comes from",Multivariable_Calculus_Shimamoto.pdf
"7.4. CHANGE OF VARIABLES FOR n-FOLD INTEGRALS 185
Figure 7.10: The region bounded by a circular cylinder of radius 2, a sphere of radius 3, and the
xy-plane
the base of W, which is a disk of radius 2. Thus 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. Therefore:
ZZZ
W
(x2 + y2)z dx dy dz=
Z 2π
0
Z 2
0
Z √
9−r2
0
r3z dz

dr

dθ
=
Z 2π
0
Z 2
0
1
2r3z2

z=
√
9−r2
z=0
dr

dθ
=
Z 2π
0
Z 2
0
1
2r3(9 − r2) dr

dθ
= 1
2
Z 2
0
r3(9 − r2) dr
Z 2π
0
1 dθ

= 1
2
9
4r4 − 1
6r62
0

θ
2π
0

= 1
2 ·

36 − 32
3


· 2π
= 1
2 · 76
3 · 2π
= 76
3 π.
Example 7.10. Find the centroid (x, y, z) of the solidW bounded above by the spherex2+y2+z2 =
8 and below by the cone z =
p
x2 + y2, shown in Figure 7.11.
First, by symmetry, x = 0 and y = 0. More precisely, consider:
x = 1
Vol (W)
ZZZ
W
x dx dy dz.
Note that W is symmetric in the yz-plane in the sense that ( −x, y, z) is in W whenever (x, y, z)
is in W. In addition, the function being integrated, namely, f(x, y, z) = x, satisfies f(−x, y, z) =",Multivariable_Calculus_Shimamoto.pdf
"is in W. In addition, the function being integrated, namely, f(x, y, z) = x, satisfies f(−x, y, z) =
−x = −f(x, y, z). The same sort of change of variables argument that we applied in Example 7.7
to study symmetry for double integrals shows that
RRR
W x dx dy dz= 0, and hence x = 0. The
argument that y = 0 is similar using symmetry in the xz-plane.
To find z = 1
Vol(W)
RRR
W z dx dy dz, we actually must do some calculation. We begin with
Vol (W) =
RRR
W 1 dx dy dz. It is simplest to describe W using spherical coordinates. Using the",Multivariable_Calculus_Shimamoto.pdf
"186 CHAPTER 7. CHANGE OF VARIABLES
Figure 7.11: An ice cream cone-shaped solid, capped by a sphere of radius 2
√
2
order of integration dρ dϕ dθ, for fixed ϕ and θ, ρ goes along a radial segment fromρ = 0 to ρ = 2
√
2,
the radius of the spherical cap. Then, for fixed θ, the angle ϕ varies from ϕ = 0 to ϕ = π
4 . Lastly,
θ goes all the way around from θ = 0 to θ = 2π. Hence:
Vol (W) =
ZZZ
W
1 dx dy dz=
ZZZ
Win
spherical
1 · ρ2 sin ϕ dρ dϕ dθ
=
Z 2π
0
Z π
4
0
Z 2
√
2
0
ρ2 sin ϕ dρ

dϕ

dθ
=
Z 2
√
2
0
ρ2 dρ
Z π
4
0
sin ϕ dϕ
Z 2π
0
1 dθ

=
1
3ρ32
√
2
0

−cos ϕ

π
4
0

θ
2π
0

= 16
√
2
3 ·

−
√
2
2 − (−1)

· 2π
= 32
3 (
√
2 − 1)π.
Similarly, substituting the conversion z = ρ cos ϕ:
ZZZ
W
z dx dy dz=
ZZZ
Win
spherical
ρ cos ϕ · ρ2 sin ϕ dρ dϕ dθ
=
Z 2π
0
Z π
4
0
Z 2
√
2
0
ρ3 cos ϕ sin ϕ dρ

dϕ

dθ
=
Z 2
√
2
0
ρ3 dρ
Z π
4
0
cos ϕ sin ϕ dϕ
Z 2π
0
1 dθ

=
1
4ρ42
√
2
0
1
2 sin2 ϕ

π
4
0

θ
2π
0

= 16 · 1
4 · 2π
= 8π.",Multivariable_Calculus_Shimamoto.pdf
"7.4. CHANGE OF VARIABLES FOR n-FOLD INTEGRALS 187
As a result, z = 1
32
3 (
√
2−1)π · 8π = 3
4(
√
2−1) , and the centroid of W is the point:
(x, y, z) =

0, 0, 3
4(
√
2−1)


.
As a reality check, note that this is approximately the point (0 , 0, 1.8), whereas the radius of the
sphere is 2
√
2 ≈ 2.8. Hence the centroid is roughly two-thirds of the way to the spherical top of
the region. Given the top-heavy shape of W, this is plausible.
We close with a quadruple integral.
Example 7.11. Find the volume of the four-dimensional closed unit ball:
W = {(x1, x2, x3, x4) ∈ R4 : x2
1 + x2
2 + x2
3 + x2
4 ≤ 1}.
As usual, we begin with Vol ( W) =
RRRR
W 1 dx1 dx2 dx3 dx4. This can be analyzed with tech-
niques similar to those we used for triple integrals. For instance, say we integrate first with respect
to x4, treating x1, x2, and x3 as fixed. The projection of W on the x1x2x3-subspace is the three-
dimensional ball B given by x2
1 + x2
2 + x3",Multivariable_Calculus_Shimamoto.pdf
"to x4, treating x1, x2, and x3 as fixed. The projection of W on the x1x2x3-subspace is the three-
dimensional ball B given by x2
1 + x2
2 + x3
3 ≤ 1. For each ( x1, x2, x3) in B, the values of x4 for
which (x1, x2, x3, x4) ∈ W range from x4 = −
p
1 − x2
1 − x2
2 − x2
3 to x4 = +
p
1 − x2
1 − x2
2 − x2
3.
This is indicated in Figure 7.12, which is supposed to be a drawing in four-dimensional space R4.
Of course, this should be taken with a grain of salt, but we’re doing the best we can.
Figure 7.12: The volume of a four-dimensional ball as a triple integral of a single integral,RRR
(
R
. . . dx4) dx1 dx2 dx3
Hence:
Vol (W) =
ZZZ
B
Z √
1−x2
1−x2
2−x2
3
−
√
1−x2
1−x2
2−x2
3
1 dx4

dx1 dx2 dx3 (7.5)
=
ZZZ
B
2
q
1 − x2
1 − x2
2 − x2
3 dx1 dx2 dx3.
This triple integral can be evaluated, say using spherical coordinates, though it gets a little messy.
Here is an alternative. One can think of equation (7.5) above as expressing the volume of W as",Multivariable_Calculus_Shimamoto.pdf
"Here is an alternative. One can think of equation (7.5) above as expressing the volume of W as
the triple integral of a single integral. We look at it instead as a double integral of a double integral,
along the lines of Vol ( W) =
RR
x3,x4
RR
x1,x2
1 dx1 dx2

dx3 dx4. More precisely, the projection of
W on the x3x4-plane is the set of all pairs ( x3, x4) for which there is at least one ( x1, x2) such that",Multivariable_Calculus_Shimamoto.pdf
"188 CHAPTER 7. CHANGE OF VARIABLES
(x1, x2, x3, x4) is in W. This is satisfied by all points in the unit disk x2
3 + x2
4 ≤ 1. In fact, given
such a ( x3, x4), the corresponding points ( x1, x2) are those such that x2
1 + x2
2 ≤ 1 − x2
3 − x2
4. We
write this way of describing W as:
Vol (W) =
ZZ
x2
3+x2
4≤1
ZZ
x2
1+x2
2≤1−x2
3−x2
4
1 dx1 dx2

dx3 dx4.
See Figure 7.13 (more nonsense).
Figure 7.13: The volume of a four-dimensional ball as a double integral of a double integral,RR
(
RR
. . . dx1 dx2) dx3 dx4
The inner integral represents the area of a two-dimensional disk of radius a =
p
1 − x2
3 − x2
4 in
the x1x2-plane. Hence
RR
x2
1+x2
2≤1−x2
3−x2
4
1 dx1 dx2 = πa2 = π(1 − x2
3 − x2
4), whence:
Vol (W) =
ZZ
x2
3+x2
4≤1
π(1 − x2
3 − x2
4) dx3 dx4.
This is an integral over the unit disk in thex3x4-plane, so it can be evaluated using polar coordinates
with dx3 dx4 = r dr dθ:
Vol (W) =
Z 2π
0
Z 1
0
π(1 − r2) · r dr

dθ
= π
Z 1
0
r(1 − r2) dr
Z 2π
0
1 dθ

= π
1
2r2 − 1
4r41
0
",Multivariable_Calculus_Shimamoto.pdf
"with dx3 dx4 = r dr dθ:
Vol (W) =
Z 2π
0
Z 1
0
π(1 − r2) · r dr

dθ
= π
Z 1
0
r(1 − r2) dr
Z 2π
0
1 dθ

= π
1
2r2 − 1
4r41
0

θ
2π
0

= π · 1
4 · 2π
= π2
2 .
We leave for the exercises the problems of finding the volume of a five-dimensional ball and,
more broadly, a strategy for approaching the volume of an n-dimensional ball in general. See
Exercises 4.9 and 4.10.",Multivariable_Calculus_Shimamoto.pdf
"7.5. EXERCISES FOR CHAPTER 7 189
7.5 Exercises for Chapter 7
Section 1 Change of variables for double integrals
1.1. This exercise involves two standard one-variable integrals that appear regularly enough that it
seems like a good idea to get out into the open how they can be evaluated quickly. Namely, we
compute the integrals
Rnπ
0 cos2 x dxand
Rnπ
0 sin2 x dx, where n is a positive integer. They can
be found using trigonometric identities, but there is another way that is easier to reproduce
on the spot.
(a) First, sketch the graphs y = cos2 x and y = sin2 x for x in the interval [0 , nπ], and use
them to illustrate that
Rnπ
0 cos2 x dx=
Rnπ
0 sin2 x dx.
(b) This is an optional exercise for those who are uneasy about drawing the conclusion in
part (a) based only on a picture.
(i) Show that
Rnπ
0 sin2 x dx=
Rnπ+ π
2π
2
cos2 x dx. ( Hint: Use the substitution u = x + π
2
and the identity sin(θ − π
2 ) = −cos θ.)
(ii) Show that
R π
2
0 cos2 x dx=
Rnπ+ π
2",Multivariable_Calculus_Shimamoto.pdf
"Rnπ+ π
2π
2
cos2 x dx. ( Hint: Use the substitution u = x + π
2
and the identity sin(θ − π
2 ) = −cos θ.)
(ii) Show that
R π
2
0 cos2 x dx=
Rnπ+ π
2
nπ cos2 x dx. ( Hint: cos(θ − π) = −cos θ.)
Deduce that
Rnπ
0 cos2 x dx=
Rnπ
0 sin2 x dx.
(c) Integrate the identity cos2 x + sin2 x = 1, and use part (a) (or (b)) to show that:
Z nπ
0
cos2 x dx= nπ
2 and
Z nπ
0
sin2 x dx= nπ
2 .
1.2. Find
RR
D y2 dx dy, where D is the upper half-disk described by x2 + y2 ≤ 1, y ≥ 0.
1.3. Find
RR
D xy dx dy, where D is the region in the first quadrant lying inside the circlex2+y2 = 4
and below the line y = x.
1.4. Find
RR
D cos(x2 + y2) dx dy, where D is the region described by 4 ≤ x2 + y2 ≤ 16.
1.5. Let D be the region in the xy-plane satisfying x2 + y2 ≤ 2, y ≥ x, and x ≥ 0. See Figure
7.14.
Figure 7.14: The region x2 + y2 ≤ 2, y ≥ x, and x ≥ 0
Consider the integral
ZZ
D
p
2 − x2 dx dy.
(a) Write an expression for the integral using the order of integration dy dx.",Multivariable_Calculus_Shimamoto.pdf
"Consider the integral
ZZ
D
p
2 − x2 dx dy.
(a) Write an expression for the integral using the order of integration dy dx.
(b) Write an expression for the integral using the order of integration dx dy.",Multivariable_Calculus_Shimamoto.pdf
"190 CHAPTER 7. CHANGE OF VARIABLES
(c) Write an expression for the integral in polar coordinates in whatever order of integration
you prefer.
(d) Evaluate the integral using whichever approach seems best.
1.6. Let W be the region in R3 lying above the xy-plane, inside the cylinder x2 + y2 = 1, and
below the plane x + y + z = 2. Find the volume of W.
1.7. Let W be the region in R3 lying above the xy-plane, inside the cylinder x2 + y2 = 1, and
below the plane x + y + z = 1. Find the volume of W.
1.8. In this exercise, we evaluate the improper one-variable integral
R∞
−∞ e−x2
dx by following the
unlikely strategy of relating it to an improper double integral that turns out to be more
tractable. Let a be a positive real number.
(a) Let Ra be the rectangle Ra = [−a, a] × [−a, a]. Show that:
ZZ
Ra
e−x2−y2
dx dy=
Z a
−a
e−x2
dx
2
.
(b) Let Da be the disk x2 + y2 ≤ a2. Use polar coordinates to evaluate
RR
Da
e−x2−y2
dx dy.",Multivariable_Calculus_Shimamoto.pdf
"ZZ
Ra
e−x2−y2
dx dy=
Z a
−a
e−x2
dx
2
.
(b) Let Da be the disk x2 + y2 ≤ a2. Use polar coordinates to evaluate
RR
Da
e−x2−y2
dx dy.
(c) Note that, as a goes to ∞, both Ra and Da fill out all of R2. It is true that both
lim
a→∞
RR
Ra
e−x2−y2
dx dyand lim
a→∞
RR
Da
e−x2−y2
dx dyexist and that they are equal. Their
common value is the improper integral
RR
R2 e−x2−y2
dx dy. Use this information along
with your answers to parts (a) and (b) to show that:
Z ∞
−∞
e−x2
dx = √π.
Section 3 Examples: linear changes of variables, symmetry
3.1. Let D be the set of points ( x, y) in R2 such that (x − 3)2 + (y − 2)2 ≤ 4.
(a) Sketch and describe D.
(b) Let T : R2 → R2 be the translation T(u, v) = (u, v) + (3, 2) = (u + 3, v+ 2). Describe
the region D∗ in the uv-plane such that T(D∗) = D.
(c) Use the change of variables in part (b) to convert the integral
ZZ
D
(x + y) dx dyover D
to an integral over D∗. Then, evaluate the integral using whatever techniques seem best.",Multivariable_Calculus_Shimamoto.pdf
"ZZ
D
(x + y) dx dyover D
to an integral over D∗. Then, evaluate the integral using whatever techniques seem best.
3.2. Let a and b be positive real numbers. Find the area of the region x2
a2 + y2
b2 ≤ 1 inside an
ellipse by using the change of variables T(u, v) = ( au, bv). (Recall from Chapter 5 that
Area (D) =
RR
D 1 dA.)
3.3. Let T : R2 → R2 be the linear change of variables T(u, v) = (3u − v, 2u + v) of Example 7.6.
(a) Show that the only point ( u, v) that satisfies T(u, v) = 0 is (u, v) = 0.
(b) Show that T is one-to-one on R2. ( Hint: Start by assuming that T(a) = T(b), and use
the linearity of T and part (a).)
3.4. Find
RR
D(2x + 4y) dx dyif:",Multivariable_Calculus_Shimamoto.pdf
"7.5. EXERCISES FOR CHAPTER 7 191
(a) D is the parallelogram with vertices (0 , 0), (3, 1), (5, 5), and (2, 4),
(b) D is the triangular region with vertices (3 , 1), (5, 5), and (2, 4). ( Hint: Take advantage
of the work you’ve already done in part (a). You should be able to use quite a bit of it.)
3.5. Find
ZZ
D
(x + y) ex2−y2
dx dy, where D is the rectangle with vertices (0 , 0), (1 , −1), (3 , 1),
and (2, 2).
3.6. Find
ZZ
D
p
12 + x2 + 3y2 dx dy, where D is the region in R2 described by x2 + 3y2 ≤ 12 and
0 ≤ y ≤ 1√
3 x.
3.7. This problem concerns the integral:
ZZ
D
y
x exy dx dy, (7.6)
where D is the region in the first quadrant of the xy-plane that is bounded by the curves
y = x, y = 4x, xy = 1, and xy = 2. See Figure 7.15, left. The main idea is to simplify the
integrand by making the substitutions u = xy and v = y
x.
(a) With these substitutions, solve for x and y as functions of u and v: ( x, y) = T(u, v).
(b) Describe the region D∗ in the uv-plane such that T(D∗) = D.",Multivariable_Calculus_Shimamoto.pdf
"(b) Describe the region D∗ in the uv-plane such that T(D∗) = D.
(c) Evaluate the integral (7.6).
D
Figure 7.15: The region bounded by y = x, y = 4 x, xy = 1, and xy = 2 (left) and the region
x2 − 4xy + 8y2 ≤ 4 (right)
3.8. Let D be the region in R2 described by x2 − 4xy + 8y2 ≤ 4. See Figure 7.15, right. FindZZ
D
y2 dx dy. ( Hint: x2 − 4xy + 8y2 = (x − 2y)2 + 4y2.)
3.9. Find
ZZ
D
(x − y)2
(x + y)2 dx dy, where D is the square with vertices (2 , 0), (4, 2), (2, 4), and (0, 2).
3.10. Let T : R2 → R2 be a one-to-one linear transformation from the uv-plane to the xy-plane
represented by a matrix A =
a b
c d

, i.e., (x, y) = T(u, v) = A ·[ uv ]. Let D∗ and D be bounded
regions in R2 of positive area such that T maps D∗ onto D, as shown in Figure 7.16.
(a) Let k = |det A|. Use the change of variables theorem to show that:
Area (D) = k ·

Area (D∗)


.",Multivariable_Calculus_Shimamoto.pdf
"192 CHAPTER 7. CHANGE OF VARIABLES
Figure 7.16: A linear change of variables
(b) Let (u, v) and (x, y) denote the centroids of D∗ and D, respectively. Show that:
(x, y) = T(u, v).
In other words, linear transformations map centroids to centroids.
3.11. A region D in the xy-plane is called symmetric in the line y = x if, whenever a point
(x, y) is in D, so is its reflection ( y, x). If D is such a region, prove that its centroid lies on
the line y = x.
3.12. Let D be a subset of R2 that is symmetric about the origin, that is, whenever ( x, y) is in D,
so is (−x, −y).
(a) Let f : D → R be an integrable function on D. Find a condition on f that ensures thatRR
D f(x, y) dx dy= 0. Justify your answer.
(b) Find an example of such a function other than the constant function f = 0.
Section 4 Change of variables for n-fold integrals
4.1. Find
ZZZ
W
(x2 + y2 + z2) dx dy dzif W is the region in R3 that lies above the cone z =
p
x2 + y2 and below the plane z = 2.
4.2. Find
ZZZ
W",Multivariable_Calculus_Shimamoto.pdf
"4.1. Find
ZZZ
W
(x2 + y2 + z2) dx dy dzif W is the region in R3 that lies above the cone z =
p
x2 + y2 and below the plane z = 2.
4.2. Find
ZZZ
W
x2z dx dy dzif W is the region in R3 described by x2 + y2 + z2 ≤ 4, y ≥ 0, and
z ≥ 0.
4.3. Find
ZZZ
W
z
(3 + x2 + y2)2 dx dy dzif W is the region in R3 given by x2 + y2 + z2 ≤ 1, z ≥ 0.
(Hint: The proper coordinate system and order of integration can make a difference.)
4.4. A class of enthusiastic multivariable calculus students celebrates the change of variables theo-
rem by drilling a cylindrical hole of radius 1 straight through the center of the earth. Assuming
that the earth is a three-dimensional closed ball of radius 2, find the volume of the portion
of the earth that remains.
4.5. Find the centroid of the half-ball of radius a in R3 that lies inside the sphere x2 +y2 +z2 = a2
and above the xy-plane.
4.6. Let W be the wedge-shaped solid in R3 that lies under the plane z = y, inside the cylinder",Multivariable_Calculus_Shimamoto.pdf
"and above the xy-plane.
4.6. Let W be the wedge-shaped solid in R3 that lies under the plane z = y, inside the cylinder
x2 + y2 = 1, and above the xy-plane. Find the centroid of W.",Multivariable_Calculus_Shimamoto.pdf
"7.5. EXERCISES FOR CHAPTER 7 193
4.7. Let W be the region in R3 that lies inside the three cylinders x2 + z2 = 1, y2 + z2 = 1,
x2 +y2 = 1, and above the xy-plane. Find the volume of W. (Hints: The integral
R 1
cos2 θ dθ =R
sec2θ dθ= tan θ + C might be helpful. Also, sin 3 θ = sin2 θ · sin θ = (1 − cos2 θ) sinθ.)
4.8. Within the solid ball x2 + y2 + z2 ≤ a2 of radius a in R3, find the average distance to the
origin.
4.9. Find the volume of the five-dimensional closed unit ball:
W = {(x1, x2, x3, x4, x5) ∈ R5 : x2
1 + x2
2 + x2
3 + x2
4 + x2
5 ≤ 1}.
(Hint: Organize the calculation as a double integral of a triple integral:
RR
(
RRR
. . .).)
4.10. If a >0, let Wn(a) denote the closed ball x2
1 + x2
2 + ··· + x2
n ≤ a2 in Rn of radius a centered
at the origin, and let Vol (Wn(a)) denote its n-dimensional volume. For instance, we showed
in Example 7.11 that Vol (W4(1)) = π2
2 .
(a) Use the change of variables theorem to show that Vol ( Wn(a)) = an Vol (Wn(1)).",Multivariable_Calculus_Shimamoto.pdf
"in Example 7.11 that Vol (W4(1)) = π2
2 .
(a) Use the change of variables theorem to show that Vol ( Wn(a)) = an Vol (Wn(1)).
(b) By thinking of Vol (Wn(1)) as a double integral of an ( n − 2)-fold integral, find a rela-
tionship between Vol (Wn(1)) and Vol (Wn−2(1)).
(c) Verify that your answer to part (b) predicts a correct relationship between Vol ( W4(1))
and Vol (W2(1)).
(d) Find Vol (W6(1)), the volume of the six-dimensional closed unit ball.",Multivariable_Calculus_Shimamoto.pdf
194 CHAPTER 7. CHANGE OF VARIABLES,Multivariable_Calculus_Shimamoto.pdf
"Part V
Integrals of vector fields
195",Multivariable_Calculus_Shimamoto.pdf
"Chapter 8
Vector fields
Now that we know what it means to differentiate vector-valued functions of more than one variable,
we turn to integrating them. The functions that we integrate, however, are of a special type. The
integrals we consider are rather specialized, too. The functions are called vector fields, and this
short chapter is devoted to introducing them. The integration begins in the next chapter.
8.1 Examples of vector fields
The domain of a vector field F is a subset U of Rn. The values of F are in Rn as well, but we don’t
think so much of F as transforming U into its image. Rather, for a point x of U, we visualize the
value F(x) as an arrow translated to start at x.
For example, F might represent a force exerted on an object, and the force might depend on
where the object is. Then F(x) could represent the force exerted if the object is at the point x. Or
some medium might be flowing through U, and F(x) could represent the velocity of the medium
at x.",Multivariable_Calculus_Shimamoto.pdf
"some medium might be flowing through U, and F(x) could represent the velocity of the medium
at x.
By taking a sample of points throughout U and drawing the corresponding arrows translated to
start at those points, one can see how some vector quantity acts on U. For instance, with the aid
of the arrows, one might be able to visualize the path an object would follow under the influence
of a force or a flowing medium.
With this prelude, the actual definition is quite simple.
Definition. Let U be a subset of Rn. A function F: U → Rn is called a vector field on U.
A distinctive feature of vector fields is that both the elements x of the domain and their values
F(x) are in Rn, as opposed to a function in general from Rn to Rm.
Example 8.1. Let F: R2 → R2 be given by F(x, y) = (1, 0) for all ( x, y) in R2.
This vector field assigns the vector (1, 0) = i to every point of the plane. To visualize the vector",Multivariable_Calculus_Shimamoto.pdf
"This vector field assigns the vector (1, 0) = i to every point of the plane. To visualize the vector
field, we draw the vector (1 , 0), or rather a translated copy of it, at a representative collection of
points. That is, at each such point, we draw an arrow of length 1 pointing to the right. This is a
constant vector field. It is sketched on the left of Figure 8.1.
As the figure shows, the sample vectors often end up running into each other, making the picture
a little muddled. As a result, we usually depict a vector field by drawing a constant positive scalar
multiple of the vectors F(x), where the scalar factor is chosen to improve the readability of the
picture. For the vector field in the last example, this is done on the right of the figure. The lengths
of the vectors are scaled down, but the arrows still point in the correct direction.
197",Multivariable_Calculus_Shimamoto.pdf
"198 CHAPTER 8. VECTOR FIELDS
Figure 8.1: The constant vector field F(x, y) = (1, 0) = i literally (left) and scaled down (right)
Example 8.2. Let F: R2 → R2 be given by F(x, y) = (−y, x).
For example, F(1, 0) = (−0, 1) = (0, 1) and F(0, 1) = (−1, 0), as in Figure 8.2.
Figure 8.2: The vector field F(x, y) = (−y, x) at the points (1 , 0) and (0, 1)
We can get some qualitative information of how F acts by noting, for example, that in the
first quadrant, where both x and y are positive, F(x, y) = (−y, x) has a negative first component
and positive second component, i.e., F(+, +) = ( −, +), so the arrow points to the left and up.
Similarly, in the second quadrant F(−, +) = (−, −) (to the left and down), in the third quadrant
F(−, −) = (+, −) (to the right and down), and in the fourth F(+, −) = (+, +) (to the right and
up).
Figure 8.3: The vector field F(x, y) = (−y, x)
In fact, we can identify both the length and direction of F(x, y) more precisely:",Multivariable_Calculus_Shimamoto.pdf
"8.1. EXAMPLES OF VECTOR FIELDS 199
• ∥F(x, y)∥ =
p
(−y)2 + x2 = ∥(x, y)∥,
• F(x, y) · (x, y) = (−y, x) · (x, y) = −yx + xy = 0.
In other words, the length of F(x, y) equals the distance of ( x, y) to the origin, and its direction
is orthogonal to ( x, y) in the counterclockwise direction. After drawing in a sample of arrows, one
can imagine F as describing a circular counterclockwise flow, or vortex, around the origin, growing
in magnitude as one moves further out, as in Figure 8.3.
Example 8.3. Let W: R2 −{(0, 0)} →R2 be given by W(x, y) =

− y
x2+y2 , x
x2+y2


, where (x, y) ̸=
(0, 0).
This is a nonconstant positive scalar multiple of the previous example. The scalar factor is
1
x2+y2 . Thus the vectors point in the same direction as before, but now:
∥W(x, y)∥ = 1
x2 + y2 ∥(x, y)∥ = ∥(x, y)∥
∥(x, y)∥2 = 1
∥(x, y)∥.
Hence the vector field also circulates about the origin, but its length is inversely proportional to",Multivariable_Calculus_Shimamoto.pdf
"∥(x, y)∥2 = 1
∥(x, y)∥.
Hence the vector field also circulates about the origin, but its length is inversely proportional to
the distance to the origin. See Figure 8.4. We shall see later that this vector field has some notable
features.
Figure 8.4: The vector field W(x, y) =

− y
x2+y2 , x
x2+y2


Example 8.4 (The inverse square field) . An important vector field G from classical physics is
characterized by the following properties.
• G is defined at every point of R3 except the origin.
• The length of G is inversely proportional to the square of the distance to the origin. In other
words, ∥G(x, y, z)∥ = c
∥(x,y,z)∥2 for some positive constant c.
• The direction of G(x, y, z) is from ( x, y, z) to the origin. That is, G(x, y, z) is a negative
scalar multiple of (x, y, z).
See Figure 8.5. For instance, in physics, the gravitational force due to a mass at the origin and the
electrostatic force due to a charged particle at the origin are both modeled by such a field.",Multivariable_Calculus_Shimamoto.pdf
"200 CHAPTER 8. VECTOR FIELDS
Figure 8.5: An inverse square vector field
To find a formula for G, we write G(x, y, z) = c
∥(x,y,z)∥2 u, where u is the unit vector in the
direction from (x, y, z) to (0, 0, 0), that is, u = − (x,y,z)
∥(x,y,z)∥. Hence G(x, y, z) = − c
∥(x,y,z)∥2 · (x,y,z)
∥(x,y,z)∥ =
− c
∥(x,y,z)∥3 (x, y, z).
This is called aninverse square field. To summarize, it is the vector fieldG: R3−{(0, 0, 0)} →
R3 given variously by:
G(x, y, z) = − c
∥(x, y, z)∥3 (x, y, z)
= − c
(x2 + y2 + z2)3/2 (x, y, z)
=

− cx
(x2 + y2 + z2)3/2 , − cy
(x2 + y2 + z2)3/2 , − cz
(x2 + y2 + z2)3/2

,
for some positive constant c.
In the future, we take the constant of proportionality to be c = 1 and refer to the resulting
vector field G as “the” inverse square field.
These examples shall often serve as test cases for the concepts we are about to develop.
8.2 Exercises for Chapter 8
In Exercises 1.1–1.6, sketch the given vector field F. To get started, it may be helpful to get an",Multivariable_Calculus_Shimamoto.pdf
"8.2 Exercises for Chapter 8
In Exercises 1.1–1.6, sketch the given vector field F. To get started, it may be helpful to get an
idea of the general direction of the arrows F(x, y) at points on the x and y-axes and in each of the
four quadrants.
1.1. F (x, y) = (x, y)
1.2. F (x, y) = (2x, y)
1.3. F (x, y) = (y, −x)
1.4. F (x, y) = (−2y, x)
1.5. F (x, y) = (y, x)
1.6. F (x, y) = (−x, y)",Multivariable_Calculus_Shimamoto.pdf
"8.2. EXERCISES FOR CHAPTER 8 201
1.7. (a) Find a smooth vector field F on R2 such that, at each point ( x, y), F(x, y) is a unit
vector normal to the parabola of the form y = x2 + c that passes through that point.
(b) Find a smooth vector field F on R2 such that, at each point ( x, y), F(x, y) is a unit
vector tangent to the parabola of the form y = x2 + c that passes through that point.
1.8. (a) Find a smooth vector field F on R2 such that ∥F(x, y)∥ = ∥(x, y)∥ and, at each point
(x, y) other than the origin, F(x, y) is a vector normal to the curve of the form xy = c
that passes through that point.
(b) Find a smooth vector field F on R2 such that ∥F(x, y)∥ = ∥(x, y)∥ and, at each point
(x, y) other than the origin, F(x, y) is a vector tangent to the curve of the form xy = c
that passes through that point.
Let F be a vector field on an open set U in Rn. If we think of F as the velocity of a medium",Multivariable_Calculus_Shimamoto.pdf
"that passes through that point.
Let F be a vector field on an open set U in Rn. If we think of F as the velocity of a medium
flowing through U, then an object dropped into U will follow a path along which the velocity at
each point is the value of the vector field at the point. More precisely, a smooth path α : I → U
defined on an interval I is called an integral path of F if α′(t) = F(α(t)) for all t in I. The curve
traced out by α is called an integral curve.
For example, let F(x, y) = (−y, x) be the circulating vector field from Example 8.2. If a >0,
let α: R → R2 be the parametrization of a circle of radius a given by:
α(t) = (a cos t, asin t). (8.1)
Then α′(t) = ( −a sin t, acos t), and F(α(t)) = F(a cos t, asin t) = ( −a sin t, acos t). Since the two
are equal, α is an integral path of F. The integral curves are circles centered at the origin. This
reinforces our sense that F describes counterclockwise circular flow around the origin.",Multivariable_Calculus_Shimamoto.pdf
"reinforces our sense that F describes counterclockwise circular flow around the origin.
1.9. Let F(x, y) = (−2y, x). (This vector field also appears in Exercise 1.4.)
(a) If a is a real number, show that α(t) =
√
2 a cos(
√
2 t), asin(
√
2 t)


is an integral path
of F.
(b) Show that the integral curves satisfy the equation x2
2 + y2 = a2, and sketch a represen-
tative sample of them. Indicate the direction in which the curves are traversed by the
integral paths.
1.10. Let F(x, y) = (y, x). (This vector field also appears in Exercise 1.5.)
(a) If a is a real number, show that α(t) =

a(et + e−t), a(et − e−t)


and α(t) =

a(et −
e−t), a(et + e−t)


are integral paths of F.
(b) Describe the integral curves, and sketch a representative sample of them. Indicate the
direction in which the curves are traversed by the integral paths. ( Hint: To find a
relationship between x and y on the integral paths, compute x(t)2 and y(t)2.)",Multivariable_Calculus_Shimamoto.pdf
"relationship between x and y on the integral paths, compute x(t)2 and y(t)2.)
1.11. Find integral paths for the vector field W(x, y) = (− y
x2+y2 , x
x2+y2 ), where (x, y) ̸= (0, 0), from
Example 8.3 by modifying the integral paths for F(x, y) = (−y, x) given in equation (8.1).
Finding explicit formulas for the integral paths of a vector field may be difficult in prac-
tice. To simplify the discussion, we restrict ourselves to two-dimensional vector fields F(x, y) =
(F1(x, y), F2(x, y)) defined on open sets in R2. If α(t) = (x(t), y(t)) is a path, then the condition
α′(t) = F(α(t)) that it must satisfy to be an integral path can be written in terms of coordinates",Multivariable_Calculus_Shimamoto.pdf
"202 CHAPTER 8. VECTOR FIELDS
as
dx
dt , dy
dt


=

F1(x(t), y(t)), F2(x(t), y(t))


. In other words, the components x(t) and y(t) of α(t)
must satisfy the differential equations:



dx
dt = F1(x, y)
dy
dt = F2(x, y).
(8.2)
Let (x0, y0) be a point of R2. For the vector fields in Exercises 1.12–1.13, (a) solve the differ-
ential equations (8.2) to find the integral path α(t) = (x(t), y(t)) of F that satisfies the condition
α(0) = (x0, y0) and (b) describe the corresponding integral curves of F geometrically and sketch a
representative sample of them, including the direction in which they are traversed by the integral
paths. (Note that these vector fields also appear in Exercises 1.1 and 1.6, respectively.)
1.12. F (x, y) = (x, y)
1.13. F (x, y) = (−x, y)",Multivariable_Calculus_Shimamoto.pdf
"Chapter 9
Line integrals
We begin integrating vector fields in the case that the domain of integration is a curve. The integrals
are called line integrals. By comparison, we learned in Section 2.3 about the integral over a curve
C of a real-valued function f. By definition,
R
C f ds=
Rb
a f(α(t)) ∥α′(t)∥dt, where α : [a, b] → Rn
is a smooth parametrization of C. This is called the integral with respect to arclength.
The integral of a vector field is a new concept—for example, we don’t simply integrate each of
the real-valued component functions of the vector field independently using the integral with respect
to arclength. In the grand scheme of things, line integrals are a richer subject than integrals with
respect to arclength. In part, this is because of the ways in which the line integral can be interpreted
and applied, many of which originate in physics (for instance, see Exercise 3.19 at the end of the",Multivariable_Calculus_Shimamoto.pdf
"and applied, many of which originate in physics (for instance, see Exercise 3.19 at the end of the
chapter). But in addition, on a more abstract level, line integrals provide us with a first peek into
a line of unifying results that relate various types of integrals to one another. By the end of the
book, we shall be in a position to see that these connections run quite deep.
We open the chapter by defining the line integral and discussing what it represents. There can
be more than one option for how to compute a given line integral, and we identify the circumstances
under which the different approaches apply. These computational techniques, however, follow from
theoretical principles that are interesting in their own right. For instance, for vector fields in the
plane, some line integrals can be expressed as double integrals. This result is known as Green’s
theorem. Throughout the chapter, the discussion reveals an active interplay between geometric",Multivariable_Calculus_Shimamoto.pdf
"theorem. Throughout the chapter, the discussion reveals an active interplay between geometric
features of the curve, the vector field, and, on occasion, the geometry of the domain on which the
vector field is defined as well.
9.1 Definitions and examples
Let F: U → Rn be a vector field on an open set U in Rn, and let C be a curve in U. Roughly
speaking, the integral of F over C is meant to represent the “effectiveness” of F in contributing
to motion along C. By this, we mean the degree of alignment of F with the direction of motion.
For instance, suppose that F is a force field and that α(t) is the position of a moving object at
time t. Then the line integral of F over C is called the work done by F. If F points in the same
direction as the direction of motion, then F helps move the object along. If it points in the opposite
direction, then it impedes the motion. If it is orthogonal, it neither helps nor hinders.",Multivariable_Calculus_Shimamoto.pdf
"direction, then it impedes the motion. If it is orthogonal, it neither helps nor hinders.
In general, we measure the contribution of F to the motion at each point of C by finding the
scalar component Ftan of F in the direction tangent to the curve. See Figure 9.1. Then, we integrate
with respect to arclength, obtaining
R
C Ftan ds.
Note that Ftan = ∥F∥cos θ, where θ is the angle between F and the tangent direction. Hence,
203",Multivariable_Calculus_Shimamoto.pdf
"204 CHAPTER 9. LINE INTEGRALS
Figure 9.1: The component Ftan of a vector field in the tangent direction
if T = α′
∥α′∥ is the unit tangent vector, then:
Ftan = ∥F∥cos θ = ∥F∥∥T∥cos θ = F · T.
By the definition of the integral with respect to arclength with f = Ftan , we obtain:
Z
C
Ftan ds =
Z
C
F · T ds =
Z b
a

F(α(t)) · T(t)


∥α′(t)∥dt
=
Z b
a

F(α(t)) · α′(t)
∥α′(t)∥


∥α′(t)∥dt
=
Z b
a
F(α(t)) · α′(t) dt.
This motivates the following definition.
Definition. Let U be an open set in Rn, and let F: U → Rn be a continuous vector field on U. If
α: [a, b] → U is a smooth path in U, then the line integral, denoted by
R
α F · ds, is defined to be:
Z
α
F · ds =
Z b
a
F(α(t)) · α′(t) dt.
Written out in coordinates, say as F = (F1, F2, . . . , Fn) and α = (x1, x2, . . . , xn), the definition
becomes:
Z
α
F · ds =
Z b
a
(F1 ◦ α, F2 ◦ α, . . . , Fn ◦ α) ·
dx1
dt , dx2
dt , . . . ,dxn
dt


dt
=
Z b
a

(F1 ◦ α) dx1
dt + (F2 ◦ α) dx2
dt + ··· + (Fn ◦ α) dxn
dt


dt. (9.1)",Multivariable_Calculus_Shimamoto.pdf
"dx1
dt , dx2
dt , . . . ,dxn
dt


dt
=
Z b
a

(F1 ◦ α) dx1
dt + (F2 ◦ α) dx2
dt + ··· + (Fn ◦ α) dxn
dt


dt. (9.1)
This suggests an alternative notation for the line integral:
Z
α
F · ds =
Z
α
F1 dx1 + F2 dx2 + ··· + Fn dxn. (9.2)
The integrand F1 dx1 + F2 dx2 + ··· + Fn dxn is called a differential form, or, more precisely,
a differential 1-form. This notation is very convenient, because it is easy to remember how to
get from (9.2) to (9.1) in order to evaluate the integral. One simply uses the parametrization to
substitute for x1, x2, . . . , xn in terms of the parameter t wherever they appear in the integrand.
This includes the substitution dxi = dxi
dt dt for i = 1, 2, . . . , n. We use differential form notation for
line integrals whenever possible.",Multivariable_Calculus_Shimamoto.pdf
"9.1. DEFINITIONS AND EXAMPLES 205
One final word about notation: we have chosen to write the line integral as an integral over
α rather than over the underlying curve C to acknowledge that the definition makes use of the
specific parametrization α. This is an important point, and we elaborate on it momentarily.
Just as in first-year calculus a function can be integrated over many intervals, so can a vector
field be integrated over many paths. The relation between the vector field and the path is reflected
in the value of the integral.
Example 9.1. Let F be the vector field on R2 given by F(x, y) = (−y, x). This is the vector field
from Chapter 8 that circulates around the origin, shown again in Figure 9.2.
Figure 9.2: The circulating vector field F(x, y) = (−y, x)
Integrate F over the following paths.
(a) α1(t) = (cos t, sin t), 0 ≤ t ≤ 2π
(b) α2(t) = (t, t), 0 ≤ t ≤ 3
(c) α3(t) = (cos t, sin t), 0 ≤ t ≤ π/2
(d) α4(t) = (1 − t, t), 0 ≤ t ≤ 1
(e) α5(t) = (cos2 t, sin2 t), 0 ≤ t ≤ π/2",Multivariable_Calculus_Shimamoto.pdf
"(b) α2(t) = (t, t), 0 ≤ t ≤ 3
(c) α3(t) = (cos t, sin t), 0 ≤ t ≤ π/2
(d) α4(t) = (1 − t, t), 0 ≤ t ≤ 1
(e) α5(t) = (cos2 t, sin2 t), 0 ≤ t ≤ π/2
In differential form notation, the integrals
R
α F · ds we seek are written
R
α −y dx+ x dy.
Figure 9.3: The vector field F(x, y) = (−y, x) along the unit circle (left), the line y = x (middle),
and the line x + y = 1 (right)
(a) The path α1 traces out the unit circle counterclockwise. The direction of the vector field and
the direction of motion are aligned at every point. See Figure 9.3 at left. As a result, F · T is
always positive, and we expect the integral to be positive as well. In fact, for this parametrization,",Multivariable_Calculus_Shimamoto.pdf
"206 CHAPTER 9. LINE INTEGRALS
x = cos t and y = sin t, so substitution gives:
Z
α1
−y dx+ x dy=
Z 2π
0

−y(t) dx
dt + x(t) dy
dt


dt
=
Z 2π
0

−sin t · (−sin t) + cost · cos t


dt
=
Z 2π
0
1 dt
= t
2π
0
= 2π.
(b) The path α2 traverses a line segment radiating out from (0 , 0) to (3 , 3) along the line y = x.
At every point of this path, the vector field is orthogonal to the direction of motion (Figure 9.3,
middle). In other words, F · T = 0 at every point, so we would expect the integral to be zero, too.
To confirm this: Z
α2
−y dx+ x dy=
Z 3
0

−t · 1 + t · 1) dt =
Z 3
0
0 dt = 0.
(c) This again moves along the unit circle, but only over the arc from (1 , 0) to (0 , 1). Using the
work from part (a): Z
α3
−y dx+ x dy=
Z π
2
0
1 dt = t

π
2
0 = π
2 .
(d) The path α4 also goes from (1 , 0) to (0 , 1), but here the parametrization satisfies x + y =
(1 − t) +t = 1 so the curve is a segment contained in the line x + y = 1. The vector field is neither",Multivariable_Calculus_Shimamoto.pdf
"(1 − t) +t = 1 so the curve is a segment contained in the line x + y = 1. The vector field is neither
perfectly aligned with the direction of motion nor perpendicular to it, but it does have a positive
component in the direction of motion at every point (Figure 9.3, right). Thus we expect the line
integral to be positive. Indeed:
Z
α4
−y dx+ x dy=
Z 1
0

−t · (−1) + (1− t) · 1)


dt =
Z 1
0
1 dt = 1.
(e) α5 is another path from (1 , 0) to (0 , 1), and, again, x + y = cos2 t + sin2 t = 1. Thus the path
traces out the same line segment as in part (d), though in a different way. The value of the line
integral is:
Z
α5
−y dx+ x dy=
Z π
2
0

−sin2 t ·

2 cost · (−sin t)


+ cos2 t · 2 sint cos t

dt
=
Z π
2
0

2 sin3 t cos t + 2 cos3 t sin t


dt
=
Z π
2
0
2 sint cos t (sin2 t + cos2 t) dt
=
Z π
2
0
2 sint cos t dt
= sin2 t

π
2
0
= 1.",Multivariable_Calculus_Shimamoto.pdf
"9.1. DEFINITIONS AND EXAMPLES 207
Note that each of the paths α3, α4, and α5 above goes from (1 , 0) to (0 , 1). The values of the
line integral are not all equal, however. Thus changing the path between the endpoints can change
the integral. On the other hand, α4 and α5 not only go between the same two points but also trace
out the same curve, just parametrized in different ways. The integrals over these two paths are
equal. Is this a coincidence?
In general, suppose that α: [a, b] → U and β : [c, d] → U are smooth parametrizations of the
same curve C. To help distinguish between them, we use t to denote the parameter of α and u for
the parameter of β. It simplifies the argument somewhat to assume that α is one-to-one, except
possibly at the endpoints, where α(a) = α(b) is allowed, and likewise for β. This is not strictly
necessary, though it has been true of all the examples of parametrizations we have encountered
thus far.
We want to compare the integrals
R
α F · ds =
Rb",Multivariable_Calculus_Shimamoto.pdf
"thus far.
We want to compare the integrals
R
α F · ds =
Rb
a F(α(t)) · α′(t) dt and
R
β F · ds =
Rd
c F(β(u)) ·
β′(u) du obtained using the respective parametrizations. Consider first the case that α and β trace
out C in the same direction, that is, both start at a point p = α(a) = β(c) and end at a point
q = α(b) = β(d), as in Figure 9.4.
Let x be a point of C other than the endpoints. Then x = α(t) for some value of t and x = β(u)
for some value of u. We write this as t = g(u), that is, for a given value of u, g(u) is the value of t
such that α(t) = β(u). We also set g(c) = a and g(d) = b. Thus:
β(u) = α(g(u)) for all u in [c, d]. (9.3)
For instance, in the case of the line segments α4 and α5 above, α5(u) = (cos 2 u, sin2 u) =

1 −
sin2 u, sin2 u) = α4(sin2 u), so g(u) = sin2 u.
Figure 9.4: Two parametrizations α and β that trace out the same curve in the same direction:
g(c) = a and g(d) = b
By equation (9.3) and the chain rule: 8
β′(u) = α′(g(u)) g′(u). (9.4)
Hence:
Z
β
F · ds =",Multivariable_Calculus_Shimamoto.pdf
"g(c) = a and g(d) = b
By equation (9.3) and the chain rule: 8
β′(u) = α′(g(u)) g′(u). (9.4)
Hence:
Z
β
F · ds =
Z d
c
F(β(u)) · β′(u) du (by definition)
=
Z d
c
F(α(g(u))) · α′(g(u)) g′(u) du (let t = g(u), dt= g′(u) du)
=
Z g(d)
g(c)
F(α(t)) · α′(t) dt =
Z b
a
F(α(t)) · α′(t) dt =
Z
α
F · ds.
8We assume here that g is smooth. Indeed, equation (9.3) may be the best way to define what it means in the
first place for α and β to parametrize the same curve C, namely, that there exists a smooth function g : [c, d] → [a, b]
such that β(u) = α(g(u)) and g′(u) ̸= 0 for all u in [c, d]. Then, we say that α and β traverse C in the same direction
if g′ > 0 and in opposite directions if g′ < 0.",Multivariable_Calculus_Shimamoto.pdf
"208 CHAPTER 9. LINE INTEGRALS
In other words, both parametrizations give the same value of the line integral.
If the parametrizations traverse C in opposite directions so that, say p = α(a) = β(d) and
q = α(b) = β(c), then the only difference is that the endpoints are reversed. That is, g(d) = a and
g(c) = b, so:
Z
β
F · ds =
Z g(d)
g(c)
F(α(t)) · α′(t) dt (just as before, but . . . )
=
Z a
b
F(α(t)) · α′(t) dt
= −
Z b
a
F(α(t)) · α′(t) dt (9.5)
= −
Z
α
F · ds.
The integrals differ by a sign.
We summarize this as follows.
Proposition 9.2. If α and β are smooth parametrizations of the same curve C, then:
Z
β
F · ds = ±
Z
α
F · ds,
where the choice of sign depends on whether they traverse C in the same or opposite direction.
This allows us to define a line integral over a curve, as opposed to over a parametrization.
Definition. An oriented curve is a curve C with a specified direction of traversal. Given such a
C, define: Z
C
F · ds =
Z
α
F · ds,",Multivariable_Calculus_Shimamoto.pdf
"Definition. An oriented curve is a curve C with a specified direction of traversal. Given such a
C, define: Z
C
F · ds =
Z
α
F · ds,
where α is any smooth parametrization that traverses C in the given direction. By the preceding
proposition, any two such parametrizations give the same answer.
The point is that the line integral depends only on the underlying curve and the direction
in which it is traversed. This is critical in that it allows us to think of the line integral as an
integral over an oriented geometric set, independent of the parametrization that is used to trace
it out. This is an issue that we overlooked previously when we discussed integrals over curves
with respect to arclength and integrals over surfaces with respect to surface area. In both of those
cases, the integrals were defined using parametrizations, but we did not check that the answers
were independent of the particular parametrization used to trace out the curve or surface.",Multivariable_Calculus_Shimamoto.pdf
"were independent of the particular parametrization used to trace out the curve or surface.
In the case of the integral with respect to arclength, the story is that
R
α f ds=
R
β f dsfor any
smooth parametrizations α and β of C. The new twist is that the integral is not only independent
of the parametrization but also of the direction in which it traverses C. The proof proceeds as
before, but a difference arises because the definition
R
β f ds=
Rd
c f(β(u)) ∥β′(u)∥du involves the
norm ∥β′(u))∥, not β′(u) itself. This introduces a factor of |g′(u)| after applying the chain rule as
in equation (9.4) and taking norms. If α and β traverse C in the same direction, then g′(u) > 0,
so |g′(u)| = g′(u) and the argument proceeds essentially unchanged. If they traverse C in opposite
directions, then g′(u) < 0, so |g′(u)| = −g′(u). This minus sign is counteracted by the minus
sign that appears in this case when reversing the limits of integration in (9.5), so the two minus",Multivariable_Calculus_Shimamoto.pdf
"sign that appears in this case when reversing the limits of integration in (9.5), so the two minus
signs cancel. We leave the details for the exercises (Exercise 1.12). The role of parametrizations in
surface integrals is discussed in Chapter 10.
Having cleared the air on that lingering matter, we now return to studying line integrals.",Multivariable_Calculus_Shimamoto.pdf
"9.1. DEFINITIONS AND EXAMPLES 209
Example 9.3. Evaluate
R
C x dx+2y dy+3z dzif C is the oriented curve consisting of three quarter-
circular arcs on the unit sphere in R3 from (1, 0, 0) to (0, 1, 0) to (0, 0, 1) and then back to (1 , 0, 0).
See Figure 9.5.
Figure 9.5: A curve consisting of three arcs C1, C2, and C3
The vector field being integrated here is given by F(x, y, z) = ( x, 2y, 3z). Since C consists of
three arcs C1, C2, C3, we can imagine traversing each of them in succession and breaking up the
integral as
R
C F · ds =
R
C1
F · ds +
R
C2
F · ds +
R
C3
F · ds. Moreover, we may parametrize each arc
however we like as long as it is traversed in the proper direction. In this way, we can avoid having
to parametrize all of C over a single parameter interval.
For C1, we use the usual parametrization of the unit circle in the xy-plane from (1 , 0, 0) to
(0, 1, 0): α1(t) = (cos t, sin t, 0), 0 ≤ t ≤ π
2 .
Z
C1
x dx+ 2y dy+ 3z dz=
Z π
2
0
",Multivariable_Calculus_Shimamoto.pdf
"(0, 1, 0): α1(t) = (cos t, sin t, 0), 0 ≤ t ≤ π
2 .
Z
C1
x dx+ 2y dy+ 3z dz=
Z π
2
0

cos t · (−sin t) + 2 sint · cos t + 3 · 0 · 0


dt
=
Z π
2
0
sin t cos t dt= 1
2 sin2 t

π
2
0
= 1
2.
For C2, we use a similar parametrization, only now in the yz-plane from (0 , 1, 0) to (0 , 0, 1):
α2(t) = (0, cos t, sin t), 0 ≤ t ≤ π
2 .
Z
C2
x dx+ 2y dy+ 3z dz=
Z π
2
0

0 · 0 + 2 cost · (−sin t) + 3 sint · cos t


dt
=
Z π
2
0
sin t cos t dt= 1
2 sin2 t

π
2
0
= 1
2.
Lastly, for C3, we travel in the xz-plane from (0 , 0, 1) to (1 , 0, 0): α3(t) = (sin t, 0, cos t), 0 ≤
t ≤ π
2 .
Z
C3
x dx+ 2y dy+ 3z dz=
Z π
2
0

sin t · cos t + 2 · 0 · 0 + 3 cost · (−sin t)


dt
= −2
Z π
2
0
sin t cos t dt= −sin2 t

π
2
0
= −1.
Adding these three calculations gives
R
C x dx+ 2y dy+ 3z dz= 1
2 + 1
2 − 1 = 0.
This example illustrates a general circumstance that we should make explicit. In addition to
integrating over smooth oriented curves, we can integrate over curves C that are concatenations of",Multivariable_Calculus_Shimamoto.pdf
"210 CHAPTER 9. LINE INTEGRALS
smooth oriented curves, that is, curves of the form C = C1 ∪C2 ∪···∪ Ck, where C1, C2, . . . , Ck are
traversed in succession and each of them has a smooth parametrization. Such a curve C is called
piecewise smooth. Under these conditions, we define:
Z
C
F · ds =
Z
C1
F · ds +
Z
C2
F · ds + ··· +
Z
Ck
F · ds.
This allows us to deal with curves that have corners, as in the example, without worrying about
whether the entire curve can be parametrized in a smooth way.
9.2 Another word about substitution
Before moving on, we comment on the approach we took to defining the line integral. A curve is
one-dimensional in the sense that it can be traced out using one parameter. Thus we might think
of an integral over a curve as essentially a one-variable integral. The one-variable integrals that we
are used to are those from first-year calculus, integrals over intervals in the real line.",Multivariable_Calculus_Shimamoto.pdf
"are used to are those from first-year calculus, integrals over intervals in the real line.
The integral of a vector field F in Rn over a curve C is a new type of integral, however. F itself
is a function of n variables x1, x2, . . . , xn, but, along C, each of these variables can be expressed
in terms of the parameter t. This is exactly the information provided by the parametrization
α(t) = ( x1(t), x2(t), . . . , xn(t)), a ≤ t ≤ b. So the idea is that substituting for x1, x2, . . . , xn in
terms of t converts an integral over C to an integral over the interval I = [a, b], where things are
familiar.
This fits into a typical pattern. Whereas the parametrization α transforms the interval I into
the curve C, substitution has the effect of pulling back an integral over C to an integral over I.
For instance, after substitution, the component function Fi(x1, x2, . . . , xn) of F(x1, x2, . . . , xn) is",Multivariable_Calculus_Shimamoto.pdf
"For instance, after substitution, the component function Fi(x1, x2, . . . , xn) of F(x1, x2, . . . , xn) is
converted to the function Fi(x1(t), x2(t), . . . , xn(t)) = Fi(α(t)). We say that α pulls back Fi to
Fi ◦ α and write:
α∗(Fi) = Fi ◦ α.
Similarly, the expression dxi is pulled back to dxi
dt dt, and we write:
α∗(dxi) = dxi
dt dt.
Actually, the xi’s on each side of this equation mean slightly different things (coordinate label
vs. function of t) and it would be better to write α∗(dxi) = dαi
dt dt, but hopefully our cavalier
attitude will not lead us astray.
For a line integral, by pulling back the individual terms, the entire integrand F1 dx1 + F2 dx2 +
··· + Fn dxn in terms of x1, x2, . . . , xn is pulled back to the expression:
α∗(F1 dx1 + F2 dx2 + ··· + Fn dxn) =

(F1 ◦ α) dx1
dt + (F2 ◦ α) dx2
dt + ··· + (Fn ◦ α) dxn
dt

dt
in terms of t. If we denote F1 dx1 + F2 dx2 + ··· + Fn dxn by ω, then the definition of the line
integral reads: Z
α(I)
ω =
Z
I
α∗(ω). (9.6)",Multivariable_Calculus_Shimamoto.pdf
"dt

dt
in terms of t. If we denote F1 dx1 + F2 dx2 + ··· + Fn dxn by ω, then the definition of the line
integral reads: Z
α(I)
ω =
Z
I
α∗(ω). (9.6)
This looks a lot like the change of variables theorem. For instance, compare with equation (7.3) of
Chapter 7. We verified that the value of the integral is independent of the parametrization α so",Multivariable_Calculus_Shimamoto.pdf
"9.3. CONSERVATIVE FIELDS 211
that it is well-defined as an integral over the oriented curve C = α(I). The chain rule played an
essential, though easily missed, role in the verification. See equation (9.4).
This approach recurs more generally. In order to define a new type of integral, we use substitu-
tion to pull the integral back to a familiar situation. This is done in a way that allows the change
of variables theorem and the chain rule to conspire to make the process legitimate.
9.3 Conservative fields
If C is a smooth oriented curve in an open set U in Rn from a point p to a point q as in Figure
9.6, the value of the line integral
R
C F · ds does not depend on how C is parametrized as long as
it is traversed in the correct direction. We have seen, however, that the value does depend on the
actual curve C: choosing a different curve from p to q can change the integral. Are there situations
where even changing the curve doesn’t matter?
Figure 9.6: An oriented curve C from p to q",Multivariable_Calculus_Shimamoto.pdf
"where even changing the curve doesn’t matter?
Figure 9.6: An oriented curve C from p to q
By definition,
R
C F · ds =
Rb
a F(α(t)) · α′(t) dt, where α: [a, b] → U is a smooth parametrization
that traces out C with the proper orientation. Now, suppose—and this is a big assumption—that
F is the gradient of a smooth real-valued function f, that is, F = ∇f, or, in terms of components,
(F1, F2, . . . , Fn) = ( ∂f
∂x1
, ∂f
∂x2
, . . . ,∂f
∂xn ). Then F(α(t)) · α′(t) = ∇f(α(t)) · α′(t) = (f ◦ α)′(t), where
the last equality is the Little Chain Rule. In other words, the integrand in the definition of the line
integral is a derivative. So:
Z
C
F · ds =
Z b
a
(f ◦ α)′(t) dt = (f ◦ α)(t)
b
a = f(α(b)) − f(α(a)) = f(q) − f(p).
This depends only on the endpoints p and q and not at all on the curve that connects them.
In order to apply the Little Chain Rule, this argument requires thatC have a smooth parametri-",Multivariable_Calculus_Shimamoto.pdf
"In order to apply the Little Chain Rule, this argument requires thatC have a smooth parametri-
zation. If C is piecewise smooth instead, say C = C1 ∪ C2 ∪ ··· ∪Ck, where each Ci is smooth and
C1 goes from p to p1, C2 from p1 to p2, and so on until Ck goes from pk−1 to q, then applying the
previous result to the pieces gives a telescoping sum in which the intermediate endpoints appear
twice with opposite signs:
Z
C
F · ds =
Z
C1
F · ds +
Z
C2
F · ds + ··· +
Z
Ck
F · ds
=

f(p1) − f(p)


+

f(p2) − f(p1)


+ ··· +

f(q) − f(pk−1)


= −f(p) + f(q)
= f(q) − f(p).
In other words, in the end, the same result holds, and again only p and q matter.
Definition. Let U be an open set in Rn. A vector field F: U → Rn is called a conservative, or
a gradient, field on U if there exists a smooth real-valued function f : U → R such that F = ∇f.
Such an f is called a potential function for F.",Multivariable_Calculus_Shimamoto.pdf
"212 CHAPTER 9. LINE INTEGRALS
We have proven the following.
Theorem 9.4 (Conservative vector field theorem) . If F is a conservative vector field on U with
potential function f, then for any piecewise smooth oriented curve C in U:
Z
C
F · ds = f(q) − f(p),
where p and q are the starting and ending points, respectively, of C.
A couple of other ways to write this are:
Z
C
∇f · ds = f(q) − f(p)
or
Z
C
∂f
∂x1
dx1 + ∂f
∂x2
dx2 + ··· + ∂f
∂xn
dxn = f(q) − f(p).
We shall give some examples to illustrate the theorem shortly, but first we present an immediate
general consequence that also helps explain in part why the word conservative is used. (See Exercise
3.19 for another reason.)
Definition. A curve C is called closed if it starts and ends at the same point. See Figure 9.7 for
some examples.
Figure 9.7: Examples of closed curves: an ellipse (left) and a figure eight (middle) in R2 and a knot
in R3 (right)",Multivariable_Calculus_Shimamoto.pdf
"some examples.
Figure 9.7: Examples of closed curves: an ellipse (left) and a figure eight (middle) in R2 and a knot
in R3 (right)
Corollary 9.5. If F is conservative on U, then for any piecewise smooth oriented closed curve C
in U: Z
C
F · ds = 0.
Example 9.6. Let F be the vector field on R3 given by F(x, y, z) = (x, 2y, 3z). Is F a conservative
field?
We look to see if there is a real-valued function f such that ∇f = F, that is, ∂f
∂x = x, ∂f
∂y = 2y,
and ∂f
∂z = 3z. By inspection, it’s easy to see that f(x, y, z) = 1
2 x2 + y2 + 3
2 z2 works. Thus F is
conservative with f as potential function.
It follows that
R
C x dx+ 2y dy+ 3z dz= 0 for any piecewise smooth oriented closed curve C in
R3. This explains the answer of 0 that we obtained in Example 9.3 when we integrated F around
the closed curve consisting of three circular arcs (Figure 9.5).
Similarly, if C is any curve from (1 , 0, 0) to (0 , 1, 0), such as the arc C1 in the xy-plane of",Multivariable_Calculus_Shimamoto.pdf
"Similarly, if C is any curve from (1 , 0, 0) to (0 , 1, 0), such as the arc C1 in the xy-plane of
Example 9.3, then by the conservative vector field theorem:
Z
C
x dx+ 2y dy+ 3z dz= f(0, 1, 0) − f(1, 0, 0) = (0 + 1 + 0)− (1
2 + 0 + 0) = 1
2.
Again, this agrees with what we got before. More importantly, it is a much simpler way to evaluate
the integral.",Multivariable_Calculus_Shimamoto.pdf
"9.3. CONSERVATIVE FIELDS 213
Example 9.7. Let G be the inverse square field:
G(x, y, z) =

− x
(x2 + y2 + z2)3/2 , − y
(x2 + y2 + z2)3/2 , − z
(x2 + y2 + z2)3/2

,
where (x, y, z) ̸= (0, 0, 0). Find
R
C G · ds if C is the portion of the helix parametrized by α(t) =
(cos t, sin t, t), 0 ≤ t ≤ 3π.
We could calculate the integral explicitly using the definition and the parametrization, but,
inspired by the previous example, we check first to see if G is conservative. After a little trial-
and-error, we find that g(x, y, z) = ( x2 + y2 + z2)−1/2 = 1
∥(x,y,z)∥ works as a potential funcion.
For instance, ∂g
∂x = −1
2 (x2 + y2 + z2)−3/2 · 2x = − x
(x2+y2+z2)3/2 , and so on. Since C goes from
p = α(0) = (1, 0, 0) to q = α(3π) = (−1, 0, 3π), the conservative vector field theorem implies:
Z
C
G · ds = g(−1, 0, 3π) − g(1, 0, 0) = 1√
1 + 9π2 − 1.
The truth is that evaluating this integral directly using the parametrization is actually fairly",Multivariable_Calculus_Shimamoto.pdf
"G · ds = g(−1, 0, 3π) − g(1, 0, 0) = 1√
1 + 9π2 − 1.
The truth is that evaluating this integral directly using the parametrization is actually fairly
straightforward. Try it—it turns out that it’s not as bad as it might look. This is often the case
with conservative fields. Instead, perhaps the most meaningful takeaway from the approach we
took is that we can say that
R
C G · ds = 1√
1+9π2 − 1 for any piecewise smooth curve in the domain
of G, R3 − {(0, 0, 0)}, from p = (1 , 0, 0) to q = ( −1, 0, 3π). That the given curve is a helix is
irrelevant. Indeed, more generally, if C is any piecewise smooth curve in R3 − {(0, 0, 0)} from a
point p to a point q, then using the potential function g(x) = 1
∥x∥ above gives:
Z
C
G · ds = 1
∥q∥ − 1
∥p∥.
In particular, the integral of the inverse square field around any piecewise smooth oriented closed
curve that avoids the origin is 0.
Example 9.8. Let F be the counterclockwise circulation field F(x, y) = ( −y, x) on R2 (Figure",Multivariable_Calculus_Shimamoto.pdf
"curve that avoids the origin is 0.
Example 9.8. Let F be the counterclockwise circulation field F(x, y) = ( −y, x) on R2 (Figure
9.2). Is F a conservative field?
This time, the answer is no, and we give a couple of ways of looking at it. First, if F were
conservative, then its integral around any closed curve would be zero. But just by looking at the
way the field circulates around the origin, one can see that the integral around a circle centered
at the origin and oriented counterclockwise is positive. In fact, we calculated earlier in Example
9.1 that, if C is the unit circle traversed counterclockwise, then
R
C −y dx+ x dy= 2π. Since C is
closed and the integral is not zero, F is not conservative.
A more direct approach is to try to guess a potential function f. Here, we want ∇f = F =
(−y, x), that is: ( ∂f
∂x = −y
∂f
∂y = x. (9.7)
To satisfy the first condition, we might try f(x, y) = −xy, but then ∂f
∂y = −x, not x, and there
doesn’t seem to be any way to fix it.",Multivariable_Calculus_Shimamoto.pdf
"∂y = x. (9.7)
To satisfy the first condition, we might try f(x, y) = −xy, but then ∂f
∂y = −x, not x, and there
doesn’t seem to be any way to fix it.
This shows that our first guess didn’t work, but it’s possible that we are simply bad guessers.
To show that (9.7) is truly inconsistent, we bring in the second-order partials. For if there were a
function f that satisfied (9.7), then:
∂2f
∂y ∂x= ∂
∂y
∂f
∂x


= ∂
∂y

−y) = −1 and ∂2f
∂x ∂y= ∂
∂x
∂f
∂y


= ∂
∂x

x) = 1.
This contradicts the equality of mixed partials. Hence no such f exists.",Multivariable_Calculus_Shimamoto.pdf
"214 CHAPTER 9. LINE INTEGRALS
This last argument is easily generalized.
Theorem 9.9 (Mixed partials theorem) . Let F be a smooth vector field on an open set U in Rn,
written in terms of components as F = (F1, F2, . . . , Fn). If F is conservative, then:
∂Fi
∂xj
= ∂Fj
∂xi
for all i, j= 1, 2, . . . , n.
Equivalently, if ∂Fi
∂xj
̸= ∂Fj
∂xi
for some i, j, then F is not conservative.
Proof. Assume that F is conservative, and let f be a potential function. Then F = ∇f, that is,
(F1, F2, . . . , Fn) = ( ∂f
∂x1
, ∂f
∂x2
, . . . ,∂f
∂xn ). Therefore Fi = ∂f
∂xi
and Fj = ∂f
∂xj
, and so ∂Fi
∂xj
= ∂2f
∂xj ∂xi
and
∂Fj
∂xi
= ∂2f
∂xi ∂xj
. By the equality of mixed partials, these last two partial derivatives are equal.
We look at what the mixed partials theorem says for R2 and R3. For a vector field F = (F1, F2)
on a subset of R2, there is just one mixed partials pair: ∂F1
∂y and ∂F2
∂x . Thus:
In R2, if F = (F1, F2) is conservative, then ∂F1
∂y = ∂F2
∂x .",Multivariable_Calculus_Shimamoto.pdf
"on a subset of R2, there is just one mixed partials pair: ∂F1
∂y and ∂F2
∂x . Thus:
In R2, if F = (F1, F2) is conservative, then ∂F1
∂y = ∂F2
∂x .
Next, for F = ( F1, F2, F3) on a subset of R3, there are three mixed partials pairs: (i) ∂F1
∂y
and ∂F2
∂x , (ii) ∂F1
∂z and ∂F3
∂x , and (iii) ∂F2
∂z and ∂F3
∂y . We describe a slick way to compare the partial
derivatives in each pair.
Let ∇ = ( ∂
∂x , ∂
∂y , ∂
∂z ). This is not a legitimate vector, but we think of it as an “operator.” For
instance, it acts on a real-valued function f(x, y, z) by scalar multiplication: ∇f = ( ∂
∂x , ∂
∂y , ∂
∂z ) f =
(∂f
∂x , ∂f
∂y , ∂f
∂z ). This agrees with the usual notation for the gradient.
For a vector field F = (F1, F2, F3), ∇ acts by the cross product:
∇ ×F = det


i j k
∂
∂x
∂
∂y
∂
∂z
F1 F2 F3


=
∂F3
∂y − ∂F2
∂z , −
∂F3
∂x − ∂F1
∂z


, ∂F2
∂x − ∂F1
∂y

=
∂F3
∂y − ∂F2
∂z , ∂F1
∂z − ∂F3
∂x , ∂F2
∂x − ∂F1
∂y

.",Multivariable_Calculus_Shimamoto.pdf
"∂
∂x
∂
∂y
∂
∂z
F1 F2 F3


=
∂F3
∂y − ∂F2
∂z , −
∂F3
∂x − ∂F1
∂z


, ∂F2
∂x − ∂F1
∂y

=
∂F3
∂y − ∂F2
∂z , ∂F1
∂z − ∂F3
∂x , ∂F2
∂x − ∂F1
∂y

.
This is called the curl of F. Note that its components are precisely the differences within the three
mixed partials pairs. Hence:
In R3, if F = (F1, F2, F3) is conservative, then ∇ ×F = 0.
Example 9.10. Let F(x, y, z) = (2xy3z4 + x, 3x2y2z4 + 3, 4x2y3z3 + y2). Is F conservative?
We begin by trying to guess a potential function. In order to get ∂f
∂x = 2xy3z4 + x, we guess
f = x2y3z4 + 1
2 x2. Then, to get ∂f
∂y = 3x2y2z4 + 3, we modify this to f = x2y3z4 + 1
2 x2 + 3y. Then,
to get ∂f
∂z = 4x2y3z3 + y2, we modify again to f = x2y3z4 + 1
2 x2 + 3y + y2z. Unfortunately, this
spoils ∂f
∂y . It seems unlikely that a potential function exists.",Multivariable_Calculus_Shimamoto.pdf
"9.4. GREEN’S THEOREM 215
Having raised the suspicion, we compute the curl:
∇ ×F = det


i j k
∂
∂x
∂
∂y
∂
∂z
2xy3z4 + x 3x2y2z4 + 3 4 x2y3z3 + y2


=

12x2y2z3 + 2y − 12x2y2z3, 8xy3z3 − 8xy3z3, 6xy2z4 − 6xy2z4

= (2y, 0, 0).
This is not identically 0, so F is not conservative.
We remark that it is not true in general that, if ∂Fi
∂xj
= ∂Fj
∂xi
for all i, j, then F is conservative.
In other words, the converse of the mixed partials theorem does not hold. This is a point to which
we return in Sections 9.5 and 9.6.
9.4 Green’s theorem
We now discuss the first of three major theorems of multivariable calculus, each of which is named
after somebody. The one at hand applies to smooth vector fields F = (F1, F2) on subsets of R2.
Suppose that such a vector field F is defined on a rectangle R = [ a, b] × [c, d] and that we
integrate F around the boundary of R, oriented counterclockwise. Let’s call this oriented curve C.",Multivariable_Calculus_Shimamoto.pdf
"integrate F around the boundary of R, oriented counterclockwise. Let’s call this oriented curve C.
We break up C into the four sides C1, C2, C3, C4 of the rectangle but orient the horizontal sides
C1 and C3 to the right and the vertical sides C2 and C4 upward. See Figure 9.8.
Figure 9.8: The boundary of a rectangle R = [a, b] × [c, d]
Taking into account the orientations, we write C = C1 ∪ C2 ∪ (−C3) ∪ (−C4), where the minus
signs indicate that the segment has been given the orientation opposite the direction in which it is
actually traversed in C. Then the line integral over C can be written as:
Z
C
F1 dx + F2 dy =
Z
C1
+
Z
C2
−
Z
C3
−
Z
C4

F1 dx + F2 dy. (9.8)
Hopefully, the notation, where the integral signs are meant to distribute over the integrand, is
self-explanatory.
For the horizontal segment C1, we use x as parameter, keeping y = c fixed: α1(x) = ( x, c),
a ≤ x ≤ b. Then, along C1, dx
dx = 1 and dy
dx = d
dx

c


= 0, so:
Z
C1
F1 dx + F2 dy =
Z b
a
",Multivariable_Calculus_Shimamoto.pdf
"a ≤ x ≤ b. Then, along C1, dx
dx = 1 and dy
dx = d
dx

c


= 0, so:
Z
C1
F1 dx + F2 dy =
Z b
a

F1(x, c) · 1 + F2(x, c) · 0


dx =
Z b
a
F1(x, c) dx.",Multivariable_Calculus_Shimamoto.pdf
"216 CHAPTER 9. LINE INTEGRALS
Similarly, for C3: Z
C3
F1 dx + F2 dy =
Z b
a
F1(x, d) dx.
For the vertical segments C2 and C4, we use y as parameter, c ≤ y ≤ d, keeping x fixed. Hence
dx
dy = 0 and dy
dy = 1, which gives:
Z
C2
F1 dx + F2 dy =
Z d
c
F2(b, y) dy and
Z
C4
F1 dx + F2 dy =
Z d
c
F2(a, y) dy.
Substituting these four calculations into equation (9.8) results in:
Z
C
F1 dx + F2 dy =
Z b
a
F1(x, c) dx +
Z d
c
F2(b, y) dy −
Z b
a
F1(x, d) dx −
Z d
c
F2(a, y) dy
=
Z d
c

F2(b, y) − F2(a, y)


dy −
Z b
a

F1(x, d) − F1(x, c)


dx.
Somewhat remarkably, the fundamental theorem of calculus applies to each of these terms:
Z
C
F1 dx + F2 dy =
Z d
c

F2(x, y)

x=b
x=a

dy −
Z b
a

F1(x, y)

y=d
y=c

dx
=
Z d
c
Z b
a
∂F2
∂x (x, y) dx

dy −
Z b
a
Z d
c
∂F1
∂y (x, y) dy

dx
=
ZZ
R
∂F2
∂x − ∂F1
∂y


dx dy. (9.9)
This expresses the line integral around the boundary of the rectangle as a certain double integral
over the rectangle itself.",Multivariable_Calculus_Shimamoto.pdf
"∂y


dx dy. (9.9)
This expresses the line integral around the boundary of the rectangle as a certain double integral
over the rectangle itself.
Next, suppose that D is a rectangle with a rectangular hole, as on the left of Figure 9.9. We
examine the relationship (9.9), working from right to left, for this new domain. We write D as the
union of four rectangles R1, R2, R3, R4 that overlap at most along portions of their boundaries, as
on the right of Figure 9.9, so that the double integral over D may be decomposed:
ZZ
D
∂F2
∂x − ∂F1
∂y


dx dy=
ZZ
R1
+
ZZ
R2
+
ZZ
R3
+
ZZ
R4
∂F2
∂x − ∂F1
∂y


dx dy.
Figure 9.9: A rectangle with a rectangular hole
Let Cj be the boundary of Rj, j = 1, 2, 3, 4, oriented counterclockwise. Equation (9.9) applies
to each of the preceding integrals over the rectangles R1, R2, R3, R4, which yields:
ZZ
D
∂F2
∂x − ∂F1
∂y


dx dy=
Z
C1
+
Z
C2
+
Z
C3
+
Z
C4

F1 dx + F2 dy.",Multivariable_Calculus_Shimamoto.pdf
"9.4. GREEN’S THEOREM 217
Note that, for the boundaries of R1, R2, R3, R4, the portions in the interior of D appear as part of
two different Cj traversed in opposite directions. Thus these portions of the line integrals cancel
out, leaving only the parts left exposed around the boundary of the original region D. Hence, if C
denotes the boundary of D, we have:
ZZ
D
∂F2
∂x − ∂F1
∂y


dx dy=
Z
C
F1 dx + F2 dy.
This is the same relation as we obtained for the filled-in rectangle in equation (9.9). As the
argument shows, it is valid provided that C is oriented so that the outer perimeter is traversed
counterclockwise and the inner perimeter is traversed clockwise, as shown in Figure 9.10.
Figure 9.10: The oriented boundary of a rectangle with a rectangular hole
Before generalizing, we need a further definition.
Definition. We have said that a curve C is called closed if it starts and ends at the same point.",Multivariable_Calculus_Shimamoto.pdf
"Before generalizing, we need a further definition.
Definition. We have said that a curve C is called closed if it starts and ends at the same point.
In addition, C is said to be simple if, apart from this, it does not intersect itself.
For example, a circle or ellipse is simple, but a figure eight is not.
Now, let D be a subset of R2 whose boundary consists of a finite number of simple closed curves.
We adopt the following notation:
∂D denotes the boundary of D oriented so that, as
you traverse the boundary, D stays on the left.
For instance, this describes the orientation of the boundary of the rectangle-with-rectangular-hole
in Figure 9.10. Note that oriented boundaries are completely different from partial derivatives,
even though the same symbol “ ∂ ” happens to be used for both.
The relation we found for line integrals around the oriented boundaries of rectangles or rectan-
gles with holes holds in greater generality.",Multivariable_Calculus_Shimamoto.pdf
"The relation we found for line integrals around the oriented boundaries of rectangles or rectan-
gles with holes holds in greater generality.
Theorem 9.11 (Green’s theorem). Let U be an open set in R2, and let D be a bounded subset of
U such that the boundary of D consists of a finite number of piecewise smooth simple closed curves.
If F = (F1, F2) is a smooth vector field on U, then:
Z
∂D
F1 dx + F2 dy =
ZZ
D
∂F2
∂x − ∂F1
∂y


dx dy.
For example, consider the special case that F happens to be a conservative field. ThenR
∂D F1 dx + F2 dy = 0, since ∂D consists of closed curves and the integral of a conservative field
around any closed curve is 0. On the other hand, by the mixed partials theorem, the mixed partials
∂F2
∂x and ∂F1
∂y are equal, so
RR
D
∂F2
∂x − ∂F1
∂y


dx dy= 0 as well. Thus, for conservative fields, Green’s
theorem confirms the generally accepted belief that 0 = 0.",Multivariable_Calculus_Shimamoto.pdf
"218 CHAPTER 9. LINE INTEGRALS
We have proven Green’s theorem only in very special cases. The argument given for a rectangle
with a hole, however, can be extended easily to any domain D that is a union of finitely many
rectangles that intersect at most along common portions of their boundaries. For a general regionD,
one strategy for giving a proof is to show that D can be approximated by such a union of rectangles
so that Green’s theorem is approximately true for D. By taking a limit of such approximations, the
theorem can be proven. This requires considerable technical expertise which we leave for another
course.
Example 9.12. Let F(x, y) = (x − y2, xy). Find
R
C F · ds if C is the oriented simple closed curve
consisting of the semicircular arc in the upper half-plane from (2 , 0) to (−2, 0) followed by the line
segment from (−2, 0) to (2, 0). See Figure 9.11.
Figure 9.11: An integral around the boundary of a half-disk",Multivariable_Calculus_Shimamoto.pdf
"segment from (−2, 0) to (2, 0). See Figure 9.11.
Figure 9.11: An integral around the boundary of a half-disk
Let D be the filled-in half-disk bounded by C so that C = ∂D. By Green’s theorem:
Z
C=∂D
(x − y2) dx + xy dy=
ZZ
D
 ∂
∂x (xy) − ∂
∂y (x − y2)


dx dy
=
ZZ
D

y − (−2y)


dx dy
=
ZZ
D
3y dx dy.
This last double integral can be evaluated using polar coordinates:
Z
C
(x − y2) dx + xy dy=
Z π
0
Z 2
0
3r sin θ · r dr

dθ
= 3
Z 2
0
r2 dr
Z π
0
sin θ dθ

= 3
1
3r32
0

−cos θ
π
0

= 3 · 8
3 · 2
= 16.
Example 9.13. Once again, consider the vector field F(x, y) = ( −y, x) that circulates counter-
clockwise about the origin. Let C be a piecewise smooth simple closed curve in R2, oriented
counterclockwise. For example, it could be the one shown in Figure 9.12. Can we say anything
about the integral of F over C?
There are portions of C that go counterclockwise about the origin, for instance, the parts",Multivariable_Calculus_Shimamoto.pdf
"about the integral of F over C?
There are portions of C that go counterclockwise about the origin, for instance, the parts
furthest from the origin. The tangential component of F along these portions is positive. But there",Multivariable_Calculus_Shimamoto.pdf
"9.5. THE VECTOR FIELD W 219
Figure 9.12: Integrating the circulating vector field F(x, y) = (−y, x) around a simple closed curve
C
may also be places where C moves in the clockwise direction around the origin. For instance, if C
does not encircle the origin, this happens at points nearest the origin. The tangental component
is negative along such places. This means that there may be two types of contributions to the
integral that counteract one another. The magnitude of F increases as you move away from the
origin, however, so we might expect the positive contribution to be greater. In fact, using Green’s
theorem, we can measure precisely how much greater it is.
Let D be the filled-in region bounded by C. Then, by Green’s theorem:
Z
C
−y dx+ x dy=
ZZ
D
 ∂
∂x (x) − ∂
∂y (−y)


dx dy=
ZZ
D
2 dx dy= 2 Area (D).
In general, the same reasoning shows the following fact.
Corollary 9.14. Let D be a subset of R2 that satisfies the assumptions of Green’s theorem. Then:
Area (D) = 1
2
Z
∂D",Multivariable_Calculus_Shimamoto.pdf
"Corollary 9.14. Let D be a subset of R2 that satisfies the assumptions of Green’s theorem. Then:
Area (D) = 1
2
Z
∂D
−y dx+ x dy.
This illustrates the striking principle that information about all of D can be recovered from
measurements taken only along its boundary.
9.5 The vector field W
One of the early examples we looked at when we started studying vector fields was the vector field
W: R2 − {(0, 0)} →R2 given by:
W(x, y) =

− y
x2 + y2 , x
x2 + y2

.
This was Example 8.3 in Chapter 8. Like our usual circulating vector field, W circulates counter-
clockwise around the origin, but, due to the factor of 1
x2+y2 , the arrows get shorter as you move
away (Figure 9.13). We are about to see that W is an instructive example to keep in mind.
First, we integrate W around the circle Ca of radius a centered at the origin and oriented
counterclockwise. We evaluate the integral in two different ways.",Multivariable_Calculus_Shimamoto.pdf
"counterclockwise. We evaluate the integral in two different ways.
Approach A. We use Green’s theorem. Let Da be the filled-in disk of radius a so that Ca = ∂Da.
See Figure 9.14.",Multivariable_Calculus_Shimamoto.pdf
"220 CHAPTER 9. LINE INTEGRALS
Figure 9.13: The vector field W
Figure 9.14: The circle and disk of radius a: Ca = ∂Da
By the quotient rule:
∂W2
∂x = ∂
∂x
 x
x2 + y2

= (x2 + y2) · 1 − x · 2x
(x2 + y2)2 = y2 − x2
(x2 + y2)2
and ∂W1
∂y = ∂
∂y

− y
x2 + y2

= −(x2 + y2) · 1 − y · 2y
(x2 + y2)2 = y2 − x2
(x2 + y2)2 .
Hence ∂W2
∂x = ∂W1
∂y , so Green’s theorem gives:
Z
Ca
− y
x2 + y2 dx + x
x2 + y2 dy =
ZZ
Da
∂W2
∂x − ∂W1
∂y

dx dy=
ZZ
Da
0 dx dy= 0.
Approach B. We calculate the line integral directly using a parametrization of Ca, say α(t) =
(a cos t, asin t), 0 ≤ t ≤ 2π. This gives:
Z
Ca
− y
x2 + y2 dx + x
x2 + y2 dy
=
Z 2π
0

− a sin t
a2 cos2 t + a2 sin2 t · (−a sin t) + a cos t
a2 cos2 t + a2 sin2 t · (a cos t)

dt
=
Z 2π
0
a2 sin2 t + a2 cos2 t
a2 dt
=
Z 2π
0
a2
a2 dt
=
Z 2π
0
1 dt
= 2π.",Multivariable_Calculus_Shimamoto.pdf
"9.5. THE VECTOR FIELD W 221
In other words, 0 = 2π.
Evidently, there’s been a mistake. The truth is that Approach A is flawed. The diskDa contains
the origin, and the origin is not in the domain of W. Thus Green’s theorem does not apply to W
and Da. Instead, the correct answer is B:
Z
Ca
− y
x2 + y2 dx + x
x2 + y2 dy = 2π. (9.10)
This calculation has some noteworthy consequences.
• The value of the integral is 2 π regardless of the radius of the circle.
• W is not conservative, as integrating it around a closed curve does not always give 0.
• Nevertheless, all possible mixed partials pairs are equal: ∂W2
∂x = ∂W1
∂y . Hence the converse of
the mixed partials theorem can fail.
We broaden the analysis by considering any piecewise smooth simple closed curve C in the
plane, oriented counterclockwise, that does not pass through the origin, meaning that (0 , 0) /∈ C.
This doesn’t seem like very much to go on, but it turns out that we can say a lot about the integral
of W over C.",Multivariable_Calculus_Shimamoto.pdf
"This doesn’t seem like very much to go on, but it turns out that we can say a lot about the integral
of W over C.
There are two cases, corresponding to the fact that C separates the plane into two regions,
which we refer to as the interior and the exterior.
Case 1. (0, 0) lies in the exterior of C. See Figure 9.15.
Figure 9.15: The origin lies in the exterior of C: ∂D = C.
Then the region D bounded by C does not contain the origin, so Approach A above is valid and
Green’s theorem applies with C = ∂D. Since the mixed partials of W are equal, it follows that:
Z
C
− y
x2 + y2 dx + x
x2 + y2 dy =
ZZ
D
∂W2
∂x − ∂W1
∂y

dx dy= 0.
Case 2. (0, 0) lies in the interior of C. See Figure 9.16.
As we saw with the flaw in Approach A, we can no longer fill in C and use Green’s theorem.
Instead, let Cϵ be a circle of radius ϵ centered at the origin and oriented counterclockwise, where ϵ
is chosen small enough that Cϵ is contained entirely in the interior of C.",Multivariable_Calculus_Shimamoto.pdf
"is chosen small enough that Cϵ is contained entirely in the interior of C.
Let D be the region inside of C but outside of Cϵ. The origin does not belong to D, so Green’s
theorem applies. Since the mixed partials of W are equal, this gives:
Z
∂D
− y
x2 + y2 dx + x
x2 + y2 dy =
ZZ
D
∂W2
∂x − ∂W1
∂y


dx dy= 0.",Multivariable_Calculus_Shimamoto.pdf
"222 CHAPTER 9. LINE INTEGRALS
Figure 9.16: The origin lies in the interior of C: ∂D = C ∪ (−Cϵ).
On the other hand, taking orientation into account, ∂D = C ∪ (−Cϵ), so:
Z
∂D
− y
x2 + y2 dx + x
x2 + y2 dy =
Z
C
− y
x2 + y2 dx + x
x2 + y2 dy −
Z
Cϵ
− y
x2 + y2 dx + x
x2 + y2 dy.
Combining these results gives:
Z
C
− y
x2 + y2 dx + x
x2 + y2 dy =
Z
Cϵ
− y
x2 + y2 dx + x
x2 + y2 dy.
But we showed in equation (9.10) that the integral of W around any counterclockwise circle about
the origin, such as Cϵ, is 2π. Hence the value around C must be the same.
To summarize, we have proven:
Z
C
− y
x2 + y2 dx + x
x2 + y2 dy =
(
0 if (0 , 0) lies in the exterior of C,
2π if (0, 0) lies in the interior of C. (9.11)
If C is oriented clockwise instead, then the change in orientation reverses the sign of the integral,
so the corresponding values in the two cases are 0 and −2π, respectively. That there are exactly",Multivariable_Calculus_Shimamoto.pdf
"so the corresponding values in the two cases are 0 and −2π, respectively. That there are exactly
three possible values of the integral of W over all piecewise smooth oriented simple closed curves
in R2 − {(0, 0)} may be somewhat surprising.
9.6 The converse of the mixed partials theorem
We now discuss briefly and informally conditions regarding when a vector fieldF = (F1, F2, . . . , Fn)
is conservative. On the one hand, we have seen that, if F is conservative, then:
C1. the integral of F over any piecewise smooth oriented curve C depends only on the endpoints
of C,
C2. the integral of F over any piecewise smooth oriented closed curve is 0, and
C3. all mixed partial pairs are equal: ∂Fi
∂xj
= ∂Fj
∂xi
for all i, j.
So these conditions are necessary for F to be conservative. Are any of them sufficient? It turns
out that the converses corresponding to C1 and C2 are true, but we know from the example of W",Multivariable_Calculus_Shimamoto.pdf
"out that the converses corresponding to C1 and C2 are true, but we know from the example of W
in the previous section that the converse of C3 is not. Computing the mixed partials and seeing
if they are equal is the simplest of the three conditions to check, however, so we could ask if there
are special circumstances under which the converse of C3 does hold. We investigate this next and
toss in a few remarks about the other two conditions along the way.",Multivariable_Calculus_Shimamoto.pdf
"9.6. THE CONVERSE OF THE MIXED PARTIALS THEOREM 223
Definition. A subset U of Rn is called simply connected if:
• every pair of points in U can be joined by a piecewise smooth curve in U and
• every closed curve in U can be continuously contracted within U to a point.
For example, Rn itself is simply connected. So are rectangles and balls. A sphere in R3 is simply
connected as well. A closed curve in any of them can be collapsed to a point within the set rather
like a rubber band shrinking in on itself. See Figure 9.17.
Figure 9.17: Examples of simply connected sets: a disk, a rectangle, and a sphere
The punctured plane R2 − {(0, 0)} is not simply connected, however. A closed curve that goes
around the origin cannot be shrunk to a point without passing through the origin, i.e., without
leaving the set. Similarly, an annulus in R2, that is, the ring between two concentric circles, is not
simply connected. An example of a non-simply connected surface is the surface of a donut, which is",Multivariable_Calculus_Shimamoto.pdf
"simply connected. An example of a non-simply connected surface is the surface of a donut, which is
called a torus. A loop that goes through the hole cannot be contracted to a point while remaining
on the surface. These examples are shown in Figure 9.18.
Figure 9.18: Examples of non-simply connected sets and closed curves that cannot be contracted
within the set: the punctured plane, an annulus, and a torus
Theorem 9.15 (Converse of the mixed partials theorem) . Let F = (F1, F2, . . . , Fn) be a smooth
vector field on an open set U in Rn. If
• ∂Fi
∂xj
= ∂Fj
∂xi
for all i, j= 1, 2, . . . , nand
• U is simply connected,
then F is a conservative vector field on U.
A rigorous proof of this theorem is quite hard, so we postpone even any discussion of it until
after we look at an example.
Example 9.16. Consider again the vector field W(x, y) =

− y
x2+y2 , x
x2+y2


, ( x, y) ̸= (0 , 0), of
Section 9.5. We established that W is not conservative on R2 − {(0, 0)}. On the other hand,",Multivariable_Calculus_Shimamoto.pdf
"
− y
x2+y2 , x
x2+y2


, ( x, y) ̸= (0 , 0), of
Section 9.5. We established that W is not conservative on R2 − {(0, 0)}. On the other hand,
the mixed partials of W are equal; therefore Theorem 9.15 implies that W is conservative when
restricted to any simply connected subset of R2 − {(0, 0)}.",Multivariable_Calculus_Shimamoto.pdf
"224 CHAPTER 9. LINE INTEGRALS
For example, letU be the right half-plane, U = {(x, y) ∈ R2 : x >0}. This is a simply connected
set, so a potential function for W must exist on U. Indeed, one can check that w(x, y) = arctan(y
x)
works. For instance, ∂w
∂x = 1
1+

y
x

2 ·

− y
x2


= − y
x2+y2 .
To illustrate how this might be used, suppose that C is a piecewise smooth curve in the right
half-plane that goes from a point p, say on the line y = −x in the fourth quadrant, to a point q
on the line y = x in the first quadrant, as in Figure 9.19.
Figure 9.19: A curve in the right half-plane from the line y = −x to the line y = x
Using w as potential function:
Z
C
− y
x2 + y2 dx + x
x2 + y2 dy = w(q) − w(p) = arctan(1) − arctan(−1) = π
4 − (−π
4 ) = π
2 .
Note that this difference in the arctangent measures the angle about the origin swept out by C.
The same potential function works in the left half-plane x <0. Similarly, on the upper half-",Multivariable_Calculus_Shimamoto.pdf
"The same potential function works in the left half-plane x <0. Similarly, on the upper half-
plane y >0 and the lower half-plane y <0, v(x, y) = −arctan(x
y ) works as a potential function for
W. Integrating W over a curve in any of these half-planes amounts to finding the counterclockwise
change in angle traced out by the curve.
We can use these observations to go beyond the result of (9.11) and give a more complete
description of the integral of W over curves in R2 − {(0, 0)}. Any such curve C can be divided
into pieces, each of which lies in one of the four half-planes above. By summing the integrals over
each of these pieces, we find that the integral of W over all of C is the total counterclockwise angle
about the origin swept out by C from start to finish.
If C is a closed curve so that it comes back to where it started, the total angle is an integer
multiple of 2 π:
R
C − y
x2+y2 dx + x
x2+y2 dy = 2 πn. This conclusion does not require that C be",Multivariable_Calculus_Shimamoto.pdf
"multiple of 2 π:
R
C − y
x2+y2 dx + x
x2+y2 dy = 2 πn. This conclusion does not require that C be
simple. The integer n represents the net number of times that C goes around the origin in the
counterclockwise sense and is given a special name.
Definition. Let C be a piecewise smooth oriented closed curve in R2 −{(0, 0)}. Then the winding
number of C with respect to (0 , 0) is the integer defined by the equation:
winding # = 1
2π
Z
C
− y
x2 + y2 dx + x
x2 + y2 dy.
See Figure 9.20 for some examples.
For instance, ifCa is the circle of radiusa centered at the origin, traversed once counterclockwise,
then we showed in Section 9.5 that
R
Ca
− y
x2+y2 dx + x
x2+y2 dy = 2π (see equation (9.10)). Hence,
according to the definition, Ca has winding number 1.",Multivariable_Calculus_Shimamoto.pdf
"9.6. THE CONVERSE OF THE MIXED PARTIALS THEOREM 225
Figure 9.20: Oriented closed curves with winding number 1 (left), 2 (middle), and −1 (right)
One reason that this discussion is interesting is that, in first-year calculus, we tend to focus on
the integral of a function as telling us mainly about the function. The winding number, on the
other hand, is an instance where the integral is telling us about the geometry of the domain, i.e.,
the curve, over which the integral is taken.
We close by going over some of the ideas behind the proof of Theorem 9.15, the converse of
the mixed partials theorem. We do this only in the case of a vector field F = (F1, F2) on an open
set U in R2. Thus we assume that ∂F2
∂x = ∂F1
∂y and that U is simply connected, and we want to
prove that F is conservative on U, that is, that it has a potential function f. We think of f as an
“anti-gradient” of F. Then the method of finding f is a lot like what is done in first-year calculus",Multivariable_Calculus_Shimamoto.pdf
"“anti-gradient” of F. Then the method of finding f is a lot like what is done in first-year calculus
for the fundamental theorem, where an antiderivative is constructed by integrating.
Choose a point a of U. If x is any point of U, then, as part of the definition of simply connected,
there is a piecewise smooth curve C in U from a to x. The idea is to show that the line integralR
C F · ds =
R
C F1 dx + F2 dy is the same regardless of which curve C is chosen. For suppose that
C1 is another piecewise smooth curve in U from a to x. See Figure 9.21. We want to prove that:
Z
C
F1 dx + F2 dy =
Z
C1
F1 dx + F2 dy.
Figure 9.21: Two curves C and C1 from a to x and the region D that fills in the closed curve
C2 = C ∪ (−C1)
By reversing the orientation on C1 and appending it to C, we obtain a closed curve C2 =
C ∪ (−C1) in U. Here’s the hard part. Since U is simply connected, a closed curve such as C2 can",Multivariable_Calculus_Shimamoto.pdf
"C ∪ (−C1) in U. Here’s the hard part. Since U is simply connected, a closed curve such as C2 can
be shrunk within U to a point. The idea is that, in the process of the contraction, C2 sweeps out
a filled-in region D such that C2 = ±∂D. It is difficult to turn this intuition into a rigorous proof.
Assuming it has been done, we apply Green’s theorem together with the assumption of equal mixed
partials: Z
C2
F1 dx + F2 dy = ±
ZZ
D
∂F2
∂x − ∂F1
∂y

dx dy= ±
ZZ
D
0 dx dy= 0. (9.12)",Multivariable_Calculus_Shimamoto.pdf
"226 CHAPTER 9. LINE INTEGRALS
At the same time, since C2 = C ∪ (−C1), we have:
Z
C2
F1 dx + F2 dy =
Z
C
F1 dx + F2 dy −
Z
C1
F1 dx + F2 dy. (9.13)
By combining equations (9.12) and (9.13), we find that
R
C F1 dx + F2 dy =
R
C1
F1 dx + F2 dy, as
desired.
As a result, if x ∈ U, we may define f(x) =
R
C F1 dx + F2 dy, where C is any piecewise smooth
curve in U from a to x. We write this more suggestively as:
f(x) =
Z x
a
F1 dx + F2 dy. (9.14)
Note that, whether U is simply connected or not, as long as condition C1 above holds—namely,
the line integral depends only on the endpoints of a curve—then the formula in equation (9.14)
describes a well-defined function f. The reasoning above also contains the argument that, if con-
dition C2 holds—that the integral around any closed curve is 0—that too is enough to show that
the function f in (9.14) is well-defined. This basically follows from equation (9.13).",Multivariable_Calculus_Shimamoto.pdf
"the function f in (9.14) is well-defined. This basically follows from equation (9.13).
Thus, to show that any of C1, C2, or, on simply connected domains, C3 suffice to imply that F
is conservative, it remains only to confirm that f in (9.14) is a potential function for F: ∂f
∂x = F1
and ∂f
∂y = F2. This part is actually not so bad, and we leave it for the exercises (Exercise 6.4).
9.7 Exercises for Chapter 9
Section 1 Definitions and examples
1.1. Let F be the vector field on R2 given by F(x, y) = (x, y).
(a) Sketch the vector field.
(b) Based on your drawing, describe some paths over which you expect the integral of F to
be positive. Find an example of such a path, and calculate the integral over it.
(c) Describe some paths over which you expect the integral to be 0. Find an example of
such a path, and calculate the integral over it.
1.2. Find
R
C F · ds if F(x, y) = ( −y, x) and C is the circle x2 + y2 = a2 of radius a, traversed
counterclockwise.
1.3. Find
R",Multivariable_Calculus_Shimamoto.pdf
"1.2. Find
R
C F · ds if F(x, y) = ( −y, x) and C is the circle x2 + y2 = a2 of radius a, traversed
counterclockwise.
1.3. Find
R
C F·ds if F(x, y) = (cos2 x, sin y cos y) and C is the curve parametrized byα(t) = (t, t2),
0 ≤ t ≤ π.
1.4. Find
R
C F · ds if F(x, y, z) = (x + y, y+ z, x+ z) and C is the curve parametrized by α(t) =
(t, t2, t3), 0 ≤ t ≤ 1.
1.5. Find
R
C F·ds if F(x, y, z) = (−y, x, z) and C is the curve parametrized byα(t) = (cos t, sin t, t),
0 ≤ t ≤ 2π.
1.6. Consider the line integral
R
C (x + y) dx + (y − x) dy.
(a) What is the vector field that is being integrated?
(b) Evaluate the integral if C is the line segment parametrized by α(t) = (t, t), 0 ≤ t ≤ 1,
from (0, 0) to (1, 1).",Multivariable_Calculus_Shimamoto.pdf
"9.7. EXERCISES FOR CHAPTER 9 227
(c) Evaluate the integral if C consists of the line segment from (0 , 0) to (1 , 0) followed by
the line segment from (1 , 0) to (1, 1).
(d) Evaluate the integral if C consists of the line segment from (0 , 0) to (0 , 1) followed by
the line segment from (0 , 1) to (1, 1).
1.7. Consider the line integral
R
C xy dx+ yz dy+ xz dz.
(a) What is the vector field that is being integrated?
(b) Evaluate the integral if C is the line segment from (1 , 2, 3) to (4, 5, 6).
(c) Evaluate the integral if C is the curve from (1, 2, 3) to (4, 5, 6) consisting of three consec-
utive line segments, each parallel to one of the coordinate axes: from (1 , 2, 3) to (4, 2, 3)
to (4, 5, 3) to (4, 5, 6).
1.8. Find
R
C (x − y2) dx + 3xy dyif C is the portion of the parabola y = x2 from (−1, 1) to (2, 4).
1.9. Find
R
C y dx+ z dy+ x dzif C is the curve of intersection in R3 of the cylinder x2 + y2 = 1",Multivariable_Calculus_Shimamoto.pdf
"1.9. Find
R
C y dx+ z dy+ x dzif C is the curve of intersection in R3 of the cylinder x2 + y2 = 1
and the plane z = x + y, oriented counterclockwise as viewed from high above the xy-plane,
looking down.
For the line integrals in Exercises 1.10–1.11, (a) sketch the curve C over which the integral
is taken and (b) evaluate the integral. ( Hint: Look for ways to reduce, or eliminate, messy
calculation.)
1.10.
R
C 8z e(x+y)2
dx + (2x2 + y2)2 dy + 6xy2z dz, where C is parametrized by:
α(t) =

(sin t)t+12, (sin t)t+12, (sin t)t+12

, 0 ≤ t ≤ π/2
1.11.
R
C e8xy dx − ln

cos2(x + y) + πxy1,000,000

dy, where C is parametrized by:
α(t) =
(
(cos t, sin t) if 0 ≤ t ≤ π,
(cos t, −sin t) if π ≤ t ≤ 2π
1.12. If α: [a, b] → Rn is a smooth parametrization of a curve C and f is a continuous real-valued
function defined on C, let us write
R
α f dsto denote the quantity
Rb
a f(α(t)) ∥α′(t)∥dt that
appears in the definition of the integral with respect to arclength
R
C f ds. Write out the details",Multivariable_Calculus_Shimamoto.pdf
"Rb
a f(α(t)) ∥α′(t)∥dt that
appears in the definition of the integral with respect to arclength
R
C f ds. Write out the details
that the integral is independent of the parametrization in the sense that, if β : [c, d] → Rn is
another smooth parametrization of the same curve C, then:
Z
α
f ds=
Z
β
f ds.
Consider both cases that α and β traverse C in the same or opposite direction. You may
assume α and β are one-to-one, except possibly at the endpoints of their respective domains.
1.13. In this exercise, we return to the geometry of curves in R3 and study the role of the
parametrization in defining curvature. Recall that, if C is a curve in R3 with a smooth
parametrization α: I → R3, then the curvature is defined by the equation:
κα(t) = ∥T′
α(t)∥
vα(t) ,",Multivariable_Calculus_Shimamoto.pdf
"228 CHAPTER 9. LINE INTEGRALS
where Tα(t) is the unit tangent vector andvα(t) is the speed. We have embellished our original
notation with the subscript α to emphasize our interest in the effect of the parametrization.
We assume that vα(t) ̸= 0 for all t so that κα(t) is defined.
Let β : J → R3 be another such parametrization of C. As in the text, we assume that α and
β are one-to-one, except possibly at the endpoints of their respective domains. As in equation
(9.3), there is a function g : J → I, assumed to be smooth, such that β(u) = α(g(u)) for all
u in J. In other words, g(u) is the value of t such that β(u) = α(t).
(a) Show that the velocities of α and β are related by vβ(u) = vα

g(u)


g′(u).
(b) Show that the unit tangents are related by Tβ(u) = ±Tα

g(u)


.
(c) Show that κβ(u) = κα

g(u)


.
In other words, if x = β(u) = α(t) is any point of C other than an endpoint, then
κβ(u) = κα(t). We denote this common value by κ(x), i.e., κ(x) is defined to be κα(t)",Multivariable_Calculus_Shimamoto.pdf
"κβ(u) = κα(t). We denote this common value by κ(x), i.e., κ(x) is defined to be κα(t)
for any smooth parametrization α of C, where t is the value of the parameter such that
α(t) = x. In other words, the curvature at x depends only on the point x and not on
how C is parametrized.
Section 3 Conservative fields
3.1. Let F: R2 → R2 be the vector field F(x, y) = (y, x).
(a) Is F a conservative vector field? If so, find a potential function for it. If not, explain
why not.
(b) Find
R
C F · ds if C is the line segment from the point (4 , 1) to the point (2 , 3).
3.2. Let F: R2 → R2 be the vector field F(x, y) = (y, xy− x).
(a) Show that F is not conservative.
(b) Find an oriented closed curve C in R2 such that
R
C F · ds ̸= 0.
In Exercises 3.3–3.4, (a) find the curl of F and (b) determine if F is conservative. If it is
conservative, find a potential function for it, and, if not, explain why not.
3.3. F (x, y, z) = (3y4z2, 4x3z2, −3x2y2)
3.4. F (x, y, z) = (y − z, z− x, y− x)",Multivariable_Calculus_Shimamoto.pdf
"3.3. F (x, y, z) = (3y4z2, 4x3z2, −3x2y2)
3.4. F (x, y, z) = (y − z, z− x, y− x)
In Exercises 3.5–3.8, determine whether the vector fieldF is conservative. If it is, find a potential
function for it. If not, explain why not.
3.5. F (x, y) = (x2 − y2, 2xy)
3.6. F (x, y) = (x2 + y2, 2xy)
3.7. F (x, y, z) = (x + yz, y+ xz, z+ xy)
3.8. F (x, y, z) = (x − yz, −y + xz, z− xy)",Multivariable_Calculus_Shimamoto.pdf
"9.7. EXERCISES FOR CHAPTER 9 229
3.9. Let p be a positive real number, and let F be the vector field on R3 − {(0, 0, 0)} given by:
F(x, y, z) =

− x
(x2 + y2 + z2)p , − y
(x2 + y2 + z2)p , − z
(x2 + y2 + z2)p

.
For instance, the inverse square field is the case p = 3/2.
(a) The norm of F is given by ∥F(x, y, z)∥ = ∥(x, y, z)∥q for some exponent q. Find q in
terms of p.
(b) For which values ofp is F a conservative vector field? The case p = 1 may require special
attention.
3.10. Find
R
C ey dx + xey dy if C is the curve parametrized by α(t) = (et2
, t3), 0 ≤ t ≤ 1.
3.11. Let a and b be real numbers. If C is a piecewise smooth oriented closed curve in the punctured
plane R2 − {(0, 0)}, show that:
Z
C
ax
x2 + y2 dx + by
x2 + y2 dy =
Z
C
(b − a)y
x2 + y2 dy.
3.12. Let G(x, y, z) =

− x
(x2+y2+z2)3/2 , − y
(x2+y2+z2)3/2 , − z
(x2+y2+z2)3/2


be the inverse square field.
Find the integral of G over the curve parametrized by α(t) = (1 +t + t2, t+ t2 + t3, t2 + t3 +
t4), 0 ≤ t ≤ 1.",Multivariable_Calculus_Shimamoto.pdf
"

be the inverse square field.
Find the integral of G over the curve parametrized by α(t) = (1 +t + t2, t+ t2 + t3, t2 + t3 +
t4), 0 ≤ t ≤ 1.
3.13. Find
R
C (y + z) dx + (x + z) dy + (x + y) dz if C is the curve in R3 from the origin to (1 , 1, 1)
that consists of the sequence of line segments, each parallel to one of the coordinate axes,
from (0, 0, 0) to (1, 0, 0) to (1, 1, 0) and finally to (1 , 1, 1).
3.14. Let C be a piecewise smooth oriented curve in R3 from a point p = (x1, y1, z1) to a point
q = (x2, y2, z2). Show that
R
C 1 dx = x2 −x1. (Similar formulas apply to
R
C 1 dy and
R
C 1 dz.)
3.15. (a) Find a vector field F = (F1, F2) on R2 with the property that, whenever C is a piecewise
smooth oriented curve in R2, then:
Z
C
F1 dx + F2 dy = ∥q∥2 − ∥p∥2,
where p and q are the starting and ending points, respectively, of C.
(b) Sketch the vector field F that you found in part (a).
3.16. Does there exist a vector field F on Rn with the property that, whenever C is a piecewise",Multivariable_Calculus_Shimamoto.pdf
"3.16. Does there exist a vector field F on Rn with the property that, whenever C is a piecewise
smooth oriented curve in Rn, then:
Z
C
F · ds = p · q,
where p and q are the starting and ending points, respectively, of C? Either describe such a
vector field as precisely as you can, or explain why none exists.
3.17. Let f, g: Rn → R be smooth real-valued functions. Prove that, for all piecewise smooth
oriented closed curves C in Rn:Z
C
(f ∇g) · ds = −
Z
C
(g ∇f) · ds.
(Hint: See Exercise 1.19 in Chapter 4.)",Multivariable_Calculus_Shimamoto.pdf
"230 CHAPTER 9. LINE INTEGRALS
3.18. For a vector field F = (F1, F2, F3) on an open set in R3, is the cross product ∇ ×F, i.e., the
curl, necessarily orthogonal to F? Either prove that it is, or find an example where it isn’t.
3.19. Newton’s second law of motion, F = ma, relates the force F acting on an object to the
object’s mass m and acceleration a. Let F be a smooth force field on an open set of R3,
and assume that, under the influence of F, an object of mass m travels so that its motion is
described by a smooth path α: [a, b] → R3 with velocity v(t) and acceleration a(t).
(a) The quantity K(t) = 1
2 m∥v(t)∥2 is called the kinetic energy of the object. Use New-
ton’s second law to show that:
Z
C
F · ds = K(b) − K(a),
where C is the oriented curve parametrized by α. In other words, the work done by the
force equals the change in the object’s kinetic energy.
(b) Assume, in addition, that F is a conservative field with potential function f. Show that
the function
E(t) = 1",Multivariable_Calculus_Shimamoto.pdf
"(b) Assume, in addition, that F is a conservative field with potential function f. Show that
the function
E(t) = 1
2m∥v(t)∥2 − f(α(t))
is constant. (The function −f is called a potential energy for F. The fact that, for a
conservative force field F, the sum of the kinetic and potential energies along an object’s
path is constant is known as the principle of conservation of energy.)
3.20. Let F be a smooth vector field on an open set U in Rn. A smooth path α: I → U defined
on an interval I is called an integral path of F if α′(t) = F(α(t)) for all t. In other words, at
time t, when the position is the point α(t), the velocity equals the value of the vector field at
that point. (See page 201.)
(a) If α: [a, b] → U is a nonconstant integral path of F and C is the oriented curve
parametrized by α, show that
R
C F · ds > 0. ( Hint: The assumption that α is not
constant means that there is some point t in [a, b] where α′(t) ̸= 0.)",Multivariable_Calculus_Shimamoto.pdf
"R
C F · ds > 0. ( Hint: The assumption that α is not
constant means that there is some point t in [a, b] where α′(t) ̸= 0.)
(b) Let f : U → R be a smooth real-valued function on U, and let F = ∇f. If there is a
nonconstant integral path ofF going from a point p to a point q, show that f(q) > f(p).
(c) Show that a conservative vector field F cannot have a nonconstant closed integral path.
3.21. Let U be an open set in Rn with the property that every pair of points in U can be joined by
a piecewise smooth curve in U. Let f : U → R be a smooth real-valued function defined on
U. If ∇f(x) = 0 for all x in U, use line integrals to give a simple proof that f is a constant
function.
Section 4 Green’s theorem
4.1. Find
R
C (x2 − y2) dx + 2xy dyif C is the boundary of the square with vertices (0 , 0), (1, 0),
(1, 1), and (0, 1), oriented counterclockwise.
4.2. Find
R
C (2xy2 − y3) dx + (x2y + x3) dy if C is the unit circle x2 + y2 = 1, oriented counter-
clockwise.
4.3. Find
R",Multivariable_Calculus_Shimamoto.pdf
"4.2. Find
R
C (2xy2 − y3) dx + (x2y + x3) dy if C is the unit circle x2 + y2 = 1, oriented counter-
clockwise.
4.3. Find
R
C −y cos x dx+ xy dyif C is the boundary of the parallelogram with vertices (0 , 0),
(2π, π), (3π, 3π), and (π, 2π), oriented counterclockwise.",Multivariable_Calculus_Shimamoto.pdf
"9.7. EXERCISES FOR CHAPTER 9 231
Figure 9.22: An oval track
4.4. Find
R
C (ex+2y − 3y) dx + (4x + 2ex+2y) dy if C is the track shown in Figure 9.22 consisting
of two straightaways joined by semicircles at each end, oriented counterclockwise.
4.5. (a) Use the parametrization α(t) = (a cos t, asin t), 0 ≤ t ≤ 2π, and Corollary 9.14 to verify
that the area of the disk x2 + y2 ≤ a2 of radius a is πa2.
(b) Let a and b be positive real numbers. By modifying the parametrization in part (a), use
Corollary 9.14 to find the area of the region x2
a2 + y2
b2 ≤ 1 inside an ellipse.
In Exercises 4.6–4.11, evaluate the line integral using whatever methods seem best.
4.6.
R
C (x+y) dx+(x−y) dy, where C is the curve in R2 consisting of the line segment from (0, 1)
to (1, 0) followed by the line segment from (1 , 0) to (2, 1)
4.7.
R
C
ex
x2+y2 dx− x
x2+y2 dy, where C is the unit circle x2 +y2 = 1 in R2, oriented counterclockwise
4.8.
R",Multivariable_Calculus_Shimamoto.pdf
"4.7.
R
C
ex
x2+y2 dx− x
x2+y2 dy, where C is the unit circle x2 +y2 = 1 in R2, oriented counterclockwise
4.8.
R
C (3x2y − xy2 − 3x2y2) dx + (x3 − 2x3y + x2y) dy, where C is the closed triangular curve in
R2 with vertices (0, 0), (1, 1), and (0, 1), oriented counterclockwise
4.9.
R
C yz2 dx+x4z dy+x2y2 dz, where C is the curve parametrized by α(t) = (t, t2, t3), 0 ≤ t ≤ 1
4.10.
R
C (6x2−4y+2xy) dx+(2x−2 siny+3x2) dy, where C is the diamond-shaped curve|x|+|y| = 1
in R2, oriented counterclockwise
4.11.
R
C yz dx+xz dy+xy dz, where C is the curve of intersection in R3 of the cylinder x2 +y2 = 1
and the saddle surface z = x2 − y2, oriented counterclockwise as viewed from high above the
xy-plane, looking down
Let U be an open set in R2. A smooth real-valued function h: U → R is called harmonic on
U if:
∂2h
∂x2 + ∂2h
∂y2 = 0 at all points of U.
Exercises 4.12–4.15 concern harmonic functions.
4.12. Show that h(x, y) = e−x sin y is a harmonic function on R2.",Multivariable_Calculus_Shimamoto.pdf
"∂y2 = 0 at all points of U.
Exercises 4.12–4.15 concern harmonic functions.
4.12. Show that h(x, y) = e−x sin y is a harmonic function on R2.
4.13. Let D be a bounded subset of U whose boundary consists of a finite number of piecewise
smooth simple closed curves. If h is harmonic on U, prove that:
Z
∂D
−h∂h
∂y dx + h∂h
∂x dy =
ZZ
D
∥∇h(x, y)∥2 dx dy.",Multivariable_Calculus_Shimamoto.pdf
"232 CHAPTER 9. LINE INTEGRALS
4.14. Let D be a region as in the preceding exercise, and suppose that every pair of points in D
can be joined by a piecewise smooth curve in D. If h is harmonic on U and if h = 0 at all
points of ∂D, prove that h = 0 on all of D. ( Hint: In addition to the preceding exercise, see
Exercises 2.4 of Chapter 5 and 3.21 of this chapter.)
4.15. Continuing with the previous exercise, prove that a harmonic function is completely deter-
mined on D by its values on the boundary in the sense that, if h1 and h2 are harmonic on U
and if h1 = h2 at all points of ∂D, then h1 = h2 on all of D.
Section 5 The vector field W
5.1. Let F = (F1, F2) be a smooth vector field that:
• is defined everywhere in R2 except at three points p, q, r and
• satisfies the mixed partials condition ∂F2
∂x = ∂F1
∂y everywhere in its domain.
Let C1, C2, C3 be small circles centered at p, q, r, respectively, oriented counterclockwise as
shown in Figure 9.23. Assume that:
Z
C1",Multivariable_Calculus_Shimamoto.pdf
"Let C1, C2, C3 be small circles centered at p, q, r, respectively, oriented counterclockwise as
shown in Figure 9.23. Assume that:
Z
C1
F1 dx + F2 dy = 12,
Z
C2
F1 dx + F2 dy = 10, and
Z
C3
F1 dx + F2 dy = 15.
Figure 9.23: Three circles C1, C2, and C3 and a fish-shaped curve C
(a) Find
R
C F1 dx + F2 dy if C is the oriented fish-shaped curve shown in the figure.
(b) Draw a piecewise smooth oriented closed curve K such that
R
K F1 dx + F2 dy = 1. Your
curve need not be simple. Include in your drawing the direction in which K is traversed.
Section 6 The converse of the mixed partials theorem
6.1. (a) Let C be the curve in R2 − {(0, 0)} parametrized by α(t) = (cos t, −sin t), 0 ≤ t ≤ 2π.
Describe C geometrically, and calculate its winding number using the definition.
(b) More generally, let n be an integer (positive, negative, or zero allowed), and let Cn be
the curve traced out by the path α(t) = (cos nt, sin nt), 0 ≤ t ≤ 2π. Give a geometric",Multivariable_Calculus_Shimamoto.pdf
"the curve traced out by the path α(t) = (cos nt, sin nt), 0 ≤ t ≤ 2π. Give a geometric
description of Cn, and calculate its winding number using the definition.",Multivariable_Calculus_Shimamoto.pdf
"9.7. EXERCISES FOR CHAPTER 9 233
6.2. Let C be a piecewise smooth simple closed curve in R2, oriented counterclockwise, such that
the origin lies in the exterior of C. What are the possible values of the winding number of
C?
6.3. Let F = (F1, F2) be a smooth vector field on the punctured plane U = R2 −{(0, 0)} such that
∂F2
∂x = ∂F1
∂y on U, and let C be a piecewise smooth oriented closed curve in U.
(a) If C does not intersect the x-axis, explain why
R
C F · ds = 0.
(b) Suppose instead that C may intersect the positive x-axis but not the negative x-axis. Is
it necessarily still true that
R
C F · ds = 0? Explain.
6.4. (a) Let c = (c, d) be a point of R2, and let B = B(c, r) be an open ball centered at c. For
the moment, we think of R2 as the uv-plane to avoid having x and y mean too many
different things later on. Let F(u, v) = (F1(u, v), F2(u, v)) be a continuous vector field
on B.
Given any point x = (x, y) in B, let C1 be the curve from ( c, d) to ( x, y) consisting of",Multivariable_Calculus_Shimamoto.pdf
"on B.
Given any point x = (x, y) in B, let C1 be the curve from ( c, d) to ( x, y) consisting of
a vertical line segment followed by a horizontal line segment, as indicated on the left of
Figure 9.24. Define a function f1 : B → R by:
f1(x, y) =
Z
C1
F1 du + F2 dv.
Show that ∂f1
∂x (x, y) = F1(x, y) for all ( x, y) in B. ( Hint: Parametrize C1 and use the
fundamental theorem of calculus, keeping in mind what a partial derivative is.)
Figure 9.24: The curves of integration C1 and C2 for functions f1 and f2, respectively
(b) Similarly, let C2 be the curve from ( c, d) to ( x, y) consisting of a horizontal segment
followed by a vertical segment, as shown on the right of Figure 9.24, and define:
f2(x, y) =
Z
C2
F1 du + F2 dv.
Show that ∂f2
∂y (x, y) = F2(x, y) for all ( x, y) in B.
(c) Now, let U be an open set in R2 such that every pair of points in U can be joined by
a piecewise smooth curve in U. Let a be a point of U. Assume that F = (F1, F2) is a",Multivariable_Calculus_Shimamoto.pdf
"a piecewise smooth curve in U. Let a be a point of U. Assume that F = (F1, F2) is a
smooth vector field on U such that the function f : U → R given in equation (9.14) is
well-defined. That is:
f(x) =
Z x
a
F1 du + F2 dv,",Multivariable_Calculus_Shimamoto.pdf
"234 CHAPTER 9. LINE INTEGRALS
where the notation means
R
C F1 du + F2 dv for any piecewise smooth curve C in U from
a to x.
Show that f is a potential function for F on U, i.e., ∂f
∂x = F1 and ∂f
∂y = F2. In particular,
F is a conservative vector field. ( Hint: Given a point c in U, choose an open ball
B = B(c, r) that is contained in U. Define functions f1 and f2 as in parts (a) and (b).
For x in B, how is f(x) related to f(c) + f1(x) and f(c) + f2(x)?)",Multivariable_Calculus_Shimamoto.pdf
"Chapter 10
Surface integrals
We next study how to integrate vector fields over surfaces. One important caveat: our discussion
applies only to surfaces in R3. For surfaces in Rn with n >3, the expressions that one integrates
are more complicated than vector fields.
It is easy to get lost in the weeds with the details of surface integrals, so, in the first section
of the chapter, we just try to build some intuition about what the integral measures and how to
compute that measurement. The formal definition of the integral appears in the second section.
Calculating surface integrals is not necessarily difficult but it can be messy, so having a range of
options is especially welcome. In addition to the definition and the intuitive approach, sometimes
a surface integral can be converted to a different type of integral altogether. For example, one
of the theorems of the chapter, Stokes’s theorem, relates surface integrals to line integrals, and",Multivariable_Calculus_Shimamoto.pdf
"of the theorems of the chapter, Stokes’s theorem, relates surface integrals to line integrals, and
another, Gauss’s theorem, relates them to triple integrals. It would be a mistake to think of these
results primarily as computational tools, however. Their theoretical consequences are at least as
important, and we try to give a taste of some of the new lines of reasoning about integrals that
they open up. The chapter closes with a couple of sections in which we tie up some loose ends.
This also puts us in position to wrap things up in the next, and final, chapter.
10.1 What the surface integral measures
If S is a surface in R3, we learned in Section 5.5 how to integrate a continuous real-valued function
f : S → R over S. By definition:
ZZ
S
f dS=
ZZ
D
f(σ(s, t))




∂σ
∂s × ∂σ
∂t




ds dt, (10.1)
where σ : D → R3 is a smooth parametrization of S defined on a subset D of the st-plane, as in
Figure 10.1. This is called the integral with respect to surface area. As before, we assume now",Multivariable_Calculus_Shimamoto.pdf
"Figure 10.1. This is called the integral with respect to surface area. As before, we assume now
and in the future that D is the closure of a bounded open set in R2, that is, a bounded open set
together with all its boundary points, such as a closed disk or rectangle (see page 134). As a matter
of full disclosure, we did not show that the value of the integral is independent of the particular
parametrization σ, a point that we address at the end of the chapter.
Now, let F: U → R3 be a vector field on an open set U in R3 that contains S. The idea is
that the integral of F over S measures the amount that F flows through S. This quantity is called
the flux. For example, if F points in a direction normal to the surface, then it is flowing through
effectively, whereas, if it points in a tangent direction, it is not flowing through at all. This stands
235",Multivariable_Calculus_Shimamoto.pdf
"236 CHAPTER 10. SURFACE INTEGRALS
Figure 10.1: A parametrization σ of a surface S in R3
in contrast with the interpretation of the line integral, where we were interested in the degree to
which the vector field flowed along a curve.
To begin, we need to designate a direction of flow through S that is considered to be positive.
Definition. An orientation of a smooth surface S in R3 is a continuous vector field n: S → R3
such that, for every x in S, n(x) is a unit normal vector to S at x. See Figure 10.2. An oriented
surface is a surface S together with a specified orientation n.
Figure 10.2: A surface S with orienting unit normal vector field n
Example 10.1. Suppose that S is a level set defined by f(x, y, z) = c for some smooth real-valued
function f of three variables. If x ∈ S, then, by Proposition 4.17, ∇f(x) is a normal vector to S
at x, so the formula n(x) = 1
∥∇f(x)∥∇f(x) describes an orientation of S, provided that ∇f(x) ̸= 0
for all x in S.",Multivariable_Calculus_Shimamoto.pdf
"at x, so the formula n(x) = 1
∥∇f(x)∥∇f(x) describes an orientation of S, provided that ∇f(x) ̸= 0
for all x in S.
For instance, let S be the sphere x2 + y2 + z2 = a2 of radius a. Here, f(x, y, z) = x2 + y2 + z2,
so ∇f = (2x, 2y, 2z) = 2(x, y, z). As a result, an orientation of the sphere is given by:
n(x, y, z) = 1
2
p
x2 + y2 + z2 · 2(x, y, z) = 1√
a2 (x, y, z) = 1
a(x, y, z)
for all points ( x, y, z) on the sphere. This vector points away from the origin, so we say that it
orients S with the outward normal. See Figure 10.3, left.
We could equally well orient the sphere with the inward unit normal. This would be given by
n(x, y, z) = −1
a(x, y, z).
Or let S be the circular cylinder x2 + y2 = a2. This is a level set of f(x, y, z) = x2 + y2. Then
∇f = (2x, 2y, 0) = 2(x, y,0), and an orientation is:
n(x, y, z) = 1
2
p
x2 + y2 · 2(x, y,0) = 1
a(x, y,0)
for all (x, y, z) on the cylinder. Once again, this is the outward normal (Figure 10.3, right).",Multivariable_Calculus_Shimamoto.pdf
"10.1. WHAT THE SURFACE INTEGRAL MEASURES 237
Figure 10.3: A sphere and cylinder oriented by the outward normal
Figure 10.4: A mammal with orientation 9
In general, let S be an oriented surface with orienting normal n. To measure the flow of a vector
field F = (F1, F2, F3) through S, we find the scalar component Fnorm of F in the normal direction
at every point of S and integrate it over S. This component is given by Fnorm = ∥F∥cos θ, where
θ is the angle between F and n, as shown in Figure 10.5.
Figure 10.5: The component of a vector field in the normal direction at a point x
Hence:
Fnorm = ∥F∥cos θ = ∥F∥∥n∥cos θ = F · n.
The integral of F over S, denoted by
RR
S F · dS, can be expressed as:
ZZ
S
F · dS =
ZZ
S
Fnorm dS =
ZZ
S
F · n dS. (10.2)
9Photo credit: https://pixabay.com/photos/nature-animal-porcupine-mammal-3588682/ by analogicus under
Pixabay license.",Multivariable_Calculus_Shimamoto.pdf
"238 CHAPTER 10. SURFACE INTEGRALS
As the integral of the real-valued function F · n, the expression on the right is an integral with
respect to surface area of the type considered previously in (10.1). We take equation (10.2) as a
tentative definition of the integral of F and use it to work through some examples.
Example 10.2. Let S be the sphere x2 +y2 +z2 = a2 of radius a, oriented by the outward normal.
Find
RR
S F · dS if:
(a) F(x, y, z) = − 1
(x2+y2+z2)3/2 (x, y, z) (i.e., the inverse square field),
(b) F(x, y, z) = (0, 0, z),
(c) F(x, y, z) = (0, 0, −1).
As shown in Example 10.1, the outward unit normal to S is given by n = 1
a(x, y, z).
(a) The inverse square field, illustrated in Figure 10.6, points directly inwards towards the origin,
Figure 10.6: The inverse square field
so F · n is negative at every point of S. In other words, if the positive direction is taken to be
outward, then F is flowing in the negative direction. Thus we expect the integral to be negative.",Multivariable_Calculus_Shimamoto.pdf
"outward, then F is flowing in the negative direction. Thus we expect the integral to be negative.
To calculate it precisely, note that on S, F(x, y, z) = − 1
(a2)3/2 (x, y, z) = − 1
a3 (x, y, z), so:
F · n = − 1
a3 (x, y, z) · 1
a(x, y, z) = − 1
a4

x2 + y2 + z2

= − 1
a4 · a2 = − 1
a2 .
Thus:
ZZ
S
F · dS =
ZZ
S
F · n dS =
ZZ
S
− 1
a2 dS = − 1
a2
ZZ
S
1 dS = − 1
a2 Area (S) = − 1
a2 · 4πa2 = −4π.
This uses the formula that the surface area is 4πa2 from Example 5.19 in Chapter 5. We shall refer
to this surface integral later, so to repeat for emphasis: the integral of the inverse square field over
a sphere of radius a centered at the origin and with the outward orientation is −4π, regardless of
the radius of the sphere.
(b) The vector field F(x, y, z) = (0, 0, z) points straight upward when z >0 and straight downward
when z <0 (Figure 10.7). In either region, the component in the outward direction of S is positive.
Thus we expect the integral to be positive.",Multivariable_Calculus_Shimamoto.pdf
"when z <0 (Figure 10.7). In either region, the component in the outward direction of S is positive.
Thus we expect the integral to be positive.
In fact, F · n = (0, 0, z) · 1
a(x, y, z) = 1
az2, so:
ZZ
S
F · dS = 1
a
ZZ
S
z2 dS.",Multivariable_Calculus_Shimamoto.pdf
"10.1. WHAT THE SURFACE INTEGRAL MEASURES 239
Figure 10.7: The vector field F(x, y, z) = (0, 0, z) flowing through a sphere
This last integral is precisely one of the examples we calculated using a parametrization when we
studied integrals with respect to surface area (Example 5.19 again). The answer is
RR
S z2 dS = 4
3 πa4.
(We also gave a symmetry argument that made use of the facts that
RR
S x2 dS =
RR
S y2 dS =RR
S z2 dS and
RR
S(x2 + y2 + z2) dS =
RR
S a2 dS = a2 · (4πa2) = 4πa4.) Thus:
ZZ
S
F · dS = 1
a · 4
3πa4 = 4
3πa3.
We shall take another look at this example later as well.
(c) Here, F(x, y, z) = (0, 0, −1) is a constant vector field consisting of unit vectors that point straight
down everywhere (Figure 10.8). Roughly speaking, F points inward on the upper hemisphere, i.e.,
the component in the outward direction is negative, and outward on the lower hemisphere. From
the symmetry of the situation, we might expect these two contributions to cancel out exactly, giving",Multivariable_Calculus_Shimamoto.pdf
"the symmetry of the situation, we might expect these two contributions to cancel out exactly, giving
an integral of 0. Informally, the flow in equals the flow out, so the net flow is 0.
Figure 10.8: The vector field F(x, y, z) = (0, 0, −1) flowing through a sphere
To verify this, note that F · n = (0, 0, −1) · 1
a(x, y, z) = −1
az. Thus:
ZZ
S
F · dS =
ZZ
S
−1
az dS= −1
a
ZZ
S
z dS.
By symmetry, this last integral is indeed 0. More precisely, S is symmetric in the xy-plane, i.e.,
(x, y,−z) is in S whenever ( x, y, z) is, and the integrand f(x, y, z) = z satisfies f(x, y,−z) =
−z = −f(x, y, z), so the contributions of the northern and southern hemispheres cancel. ThusRR
S F · dS = 0.",Multivariable_Calculus_Shimamoto.pdf
"240 CHAPTER 10. SURFACE INTEGRALS
In this last example, if we had integrated instead only over the northern hemisphere, the same
line of thought would lead us to expect that the integral is negative. As a test of your intuition
about what the integral represents, you might think about whether this is related to the following
example.
Example 10.3. Let S be the disk of radius a in the xy-plane described by
x2 + y2 ≤ a2, z = 0,
oriented by the upward normal, and let F(x, y, z) = (0 , 0, −1) as in Example 10.2(c) above. See
Figure 10.9. Find
RR
S F · dS.
Figure 10.9: The vector field F(x, y, z) = (0, 0, −1) flowing through a disk in the xy-plane
The upward unit normal to S is n = k = (0, 0, 1), so F·n = (0, 0, −1)·(0, 0, 1) = −1. Therefore:
ZZ
S
F · dS =
ZZ
S
−1 dS = −Area (S) = −πa2.
Continuing with the comments right before the example, the result turned out be negative,
but is the flow of F = (0, 0, −1) through the disk really related to the flow through the northern",Multivariable_Calculus_Shimamoto.pdf
"but is the flow of F = (0, 0, −1) through the disk really related to the flow through the northern
hemisphere? We shall be able to give a precise answer based on general principles once we learn a
little more about surface integrals (see Exercises 4.4 and 5.5 at the end of the chapter).
10.2 The definition of the surface integral
We now present the actual definition of the vector field surface integral
RR
S F · dS. This follows
a strategy we have used before: to define this new type of integral, we parametrize the surface,
substitute for x, y, and z in terms of the two parameters, and then pull back to a double integral
in the parameter plane. Of course, we need to figure out what the pullback should be.
Let S be an oriented surface in R3 with orienting normal n, and let σ : D → R3 be a smooth
parametrization of S with domain D in R2:
σ(s, t) = (x(s, t), y(s, t), z(s, t)), (s, t) ∈ D.
We first describe howσ can be used to give an independent orientation ofS. Recall from Section",Multivariable_Calculus_Shimamoto.pdf
"σ(s, t) = (x(s, t), y(s, t), z(s, t)), (s, t) ∈ D.
We first describe howσ can be used to give an independent orientation ofS. Recall from Section
5.5 that the vectors ∂σ
∂s =
∂x
∂s , ∂y
∂s , ∂z
∂s


and ∂σ
∂t =
∂x
∂t , ∂y
∂t , ∂z
∂t


are “velocity” vectors with respect to
s and t, respectively, and thus are tangent to S. As a result, ∂σ
∂s × ∂σ
∂t is normal to S, and
∂σ
∂s ×∂σ
∂t
∥∂σ
∂s ×∂σ
∂t ∥",Multivariable_Calculus_Shimamoto.pdf
"10.2. THE DEFINITION OF THE SURFACE INTEGRAL 241
is a unit normal, i.e., an orientation of S, provided that ∂σ
∂s × ∂σ
∂t ̸= 0. Since the orientation n that
is given for S is also a unit normal, it must be true that:
n = ±
∂σ
∂s × ∂σ
∂t
∥∂σ
∂s × ∂σ
∂t ∥. (10.3)
We say that σ preserves orientation if the sign is + and that it reverses orientation if −.
Now, to formulate the definition of the surface integral, recall that the idea was that
RR
S F · dS
should equal
RR
S F·n dS. Using (10.3) to substitute for n, this becomes ±
RR
S F·
∂σ
∂s ×∂σ
∂t
∥∂σ
∂s ×∂σ
∂t ∥ dS. Then
the definition of the integral with respect to surface area (10.1) with f = F ·
∂σ
∂s ×∂σ
∂t
∥∂σ
∂s ×∂σ
∂t ∥ gives:
ZZ
S
F · n dS = ±
ZZ
D

F(σ(s, t)) ·
∂σ
∂s × ∂σ
∂t
∥∂σ
∂s × ∂σ
∂t ∥




∂σ
∂s × ∂σ
∂t




ds dt
= ±
ZZ
D
F(σ(s, t)) ·
∂σ
∂s × ∂σ
∂t


ds dt.
This is the thinking behind the following definition.
Definition. Let U be an open set in R3, and let F: U → R3 be a continuous vector field. Let S",Multivariable_Calculus_Shimamoto.pdf
"This is the thinking behind the following definition.
Definition. Let U be an open set in R3, and let F: U → R3 be a continuous vector field. Let S
be an oriented surface contained in U, and let σ : D → R3 be a smooth parametrization of S such
that ∂σ
∂s × ∂σ
∂t is never 0.10 Then the integral of F over S, denoted by
RR
S F · dS, is defined to be:
ZZ
S
F · dS = ±
ZZ
D
F(σ(s, t)) ·
∂σ
∂s × ∂σ
∂t


ds dt, (10.4)
where + is used if σ is orientation-preserving and − if orientation-reversing.
We defer the discussion that the definition is independent of the parametrization σ until the
end of the chapter after we have gotten used to working with the definition and there are fewer
new things to absorb. Also, the definition extends in the obvious way to surfaces that are piecewise
smooth. That is, if a surface S is a union S = S1 ∪ S2 ∪ ··· ∪Sk of oriented smooth surfaces that
intersect at most in pairs along portions of their boundaries, we define:
ZZ
S
F · dS =
ZZ
S1
F · dS +
ZZ
S2",Multivariable_Calculus_Shimamoto.pdf
"intersect at most in pairs along portions of their boundaries, we define:
ZZ
S
F · dS =
ZZ
S1
F · dS +
ZZ
S2
F · dS + ··· +
ZZ
Sk
F · dS.
As a first example, we recalculate an integral that we evaluated before using the tentative
definition in Example 10.2(b).
Example 10.4. Let S be the sphere x2 + y2 + z2 = a2, oriented by the outward normal. FindRR
S F · dS if F(x, y, z) = (0, 0, z).
We apply the definition using the standard parametrization of S with spherical coordinates ϕ
and θ as parameters and ρ = a fixed:
σ(ϕ, θ) = (a sin ϕ cos θ, asin ϕ sin θ, acos ϕ), 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π.
The domain of the parametrization is the rectangle D = [0, π] × [0, 2π] in the ϕθ-plane. See Figure
10.10.
10Actually, it’s allowed for ∂σ
∂s × ∂σ
∂t to be 0 at points ( s, t) on the boundary of D. As noted earlier, as far as
integrals are concerned, bad behavior can be neglected if confined to the boundary (page 130).",Multivariable_Calculus_Shimamoto.pdf
"242 CHAPTER 10. SURFACE INTEGRALS
Figure 10.10: A parametrization of the sphere of radius a
Substituting in terms of the parameters gives F(σ(ϕ, θ)) = (0 , 0, z(ϕ, θ)) = (0 , 0, acos ϕ). In
addition:
∂σ
∂ϕ × ∂σ
∂θ = det


i j k
a cos ϕ cos θ a cos ϕ sin θ −a sin ϕ
−a sin ϕ sin θ a sin ϕ cos θ 0


= ... (we calculated this before in Example 5.19)
= a2 sin ϕ

sin ϕ cos θ, sin ϕ sin θ, cos ϕ


. (10.5)
This vector is guaranteed to be normal to S, but, to determine whether σ is orientation-preserving
or reversing, we must check if it agrees with the given orientation, which is outward. Note that, in
the first octant, where ϕ and θ are both between 0 and π
2 , all components of ∂σ
∂ϕ × ∂σ
∂θ in equation
(10.5) are positive, hence ∂σ
∂ϕ × ∂σ
∂θ points outward. Thus σ is orientation-preserving.
This can also be seen geometrically. At any point x of S, ∂σ
∂ϕ is a tangent vector in the direction
of increasing ϕ. Similarly, ∂σ",Multivariable_Calculus_Shimamoto.pdf
"This can also be seen geometrically. At any point x of S, ∂σ
∂ϕ is a tangent vector in the direction
of increasing ϕ. Similarly, ∂σ
∂θ is tangent in the direction of increasing θ. These vectors are shown
in Figure 10.11. By the right-hand rule, ∂σ
∂ϕ × ∂σ
∂θ points in the outward direction.
Figure 10.11: The tangent vectors ∂σ
∂ϕ and ∂σ
∂θ",Multivariable_Calculus_Shimamoto.pdf
"10.2. THE DEFINITION OF THE SURFACE INTEGRAL 243
To use the definition of the integral (10.4), we integrate the function:
F(σ(ϕ, θ)) ·
∂σ
∂ϕ × ∂σ
∂θ

= (0, 0, acos ϕ) ·

a2 sin ϕ (sin ϕ cos θ, sin ϕ sin θ, cos ϕ)


= a2 sin ϕ

0 + 0 +a cos2 ϕ


= a3 sin ϕ cos2 ϕ.
Thus:
ZZ
S
F · dS = +
ZZ
D
F(σ(ϕ, θ)) ·
∂σ
∂ϕ × ∂σ
∂θ

dϕ dθ
=
ZZ
D
a3 sin ϕ cos2 ϕ dϕ dθ
=
Z 2π
0
Z π
0
a3 sin ϕ cos2 ϕ dϕ

dθ
= a3
Z π
0
sin ϕ cos2 ϕ dϕ
Z 2π
0
1 dθ

= a3

−1
3 cos3 ϕ
π
0

θ
2π
0

= a3 ·

−1
3(−1 − 1)

· 2π
= a3 · 2
3 · 2π
= 4
3πa3.
This agrees with the answer we got earlier in Example 10.2(b), but the calculation here was
at least as messy as the one obtained there by integrating the normal component of F. In other
words, the definition may not always be the most efficient way to go. We’ll return to this example
one more time later and apply a method that’s simplest of all.
Example 10.5. Let F(x, y, z) = (0 , 1, −3) for all ( x, y, z) in R3. This is a constant vector field",Multivariable_Calculus_Shimamoto.pdf
"Example 10.5. Let F(x, y, z) = (0 , 1, −3) for all ( x, y, z) in R3. This is a constant vector field
that we think of as “rain” falling through space. See Figure 10.12, left. Find
RR
S F · dS if:
(a) S is the cone z =
p
x2 + y2, where x2 +y2 ≤ 4, oriented by the upward normal (Figure 10.12,
middle),
(b) S is the disk that caps the cone in part (a), i.e., the disk x2 + y2 ≤ 4 in the plane z = 2,
oriented by the upward normal (Figure 10.12, right).
(a) As we saw in Section 5.4, there are several ways to parametrize the cone, but it’s the graph of
a function of x and y so one way is to use x and y as parameters:
σ(x, y) = (x, y,
p
x2 + y2), x 2 + y2 ≤ 4.
The domain D in the xy-plane is the disk x2 + y2 ≤ 4 of radius 2. Moreover:
∂σ
∂x × ∂σ
∂y = det


i j k
1 0 x√
x2+y2
0 1 y√
x2+y2

 =

− xp
x2 + y2 , − yp
x2 + y2 , 1


.",Multivariable_Calculus_Shimamoto.pdf
"244 CHAPTER 10. SURFACE INTEGRALS
Figure 10.12: Rain (left), the cone z =
p
x2 + y2, x2 + y2 ≤ 4 (middle), and the cap of the cone,
z = 2, x2 + y2 ≤ 4 (right)
The z-component is positive, so ∂σ
∂x × ∂σ
∂y points upward. Thus σ is orientation-preserving. In
addition:
F(σ(x, y)) ·
∂σ
∂x × ∂σ
∂y


= (0, 1, −3) ·

− xp
x2 + y2 , − yp
x2 + y2 , 1


= − yp
x2 + y2 − 3.
Thus, by definition:
ZZ
S
F · dS =
ZZ
D

− yp
x2 + y2 − 3


dx dy
= −
ZZ
D
yp
x2 + y2 dx dy− 3
ZZ
D
1 dx dy.
The first term is 0 using the symmetry of D in the x-axis, and the second is 3 times the area of D.
Therefore
RR
S F · dS = −3 Area (D) = −3 · (π · 22) = −12π.
The negative answer makes sense, since S is oriented so that the upward direction is considered
positive whereas the rain is falling downward.
(b) To integrate over the oriented disk on the right of Figure 10.12, we won’t parametrize at all but
rather go back to integrating the normal component ofF over S, which is what the actual definition",Multivariable_Calculus_Shimamoto.pdf
"rather go back to integrating the normal component ofF over S, which is what the actual definition
formalized. The upward unit normal to S is n = (0, 0, 1), so F·n = (0, 1, −3)·(0, 0, 1) = −3. Hence:
ZZ
S
F · dS =
ZZ
S
F · n dS =
ZZ
S
−3 dS = −3 Area (S) = −3 · (π · 22) = −12π.
We close by picking up the thread with which we ended the previous section, which is to note
that the answers to parts (a) and (b) of the last example are the same. We can interpret this to
say that the flow through the top of the cone equals the flow through the sides. If we could have
been sure that this was true in advance, then we could have calculated the integral over the cone
by replacing it with the much simpler calculation over the capping disk. The next three sections
have something to say about why this would have been valid from a couple of different perspectives.
They also contain the last two big theorems of introductory multivariable calculus.",Multivariable_Calculus_Shimamoto.pdf
"10.3. STOKES’S THEOREM 245
10.3 Stokes’s theorem
There are two types of points on a surface. At some points, one can approach along the surface
from any direction and remain on the surface, at least for a little while, after continuing on through
the point. These are sometimes called interior points . Other points lie on the edge of the
surface in the sense that, after approaching the point from certain directions, one falls off the
surface immediately after continuing through. These are called boundary points. The set of all
boundary points is called the boundary of the surface.
For instance, a hemisphere has a boundary consisting of the equatorial circle, and a cylinder has
a boundary consisting of two pieces, namely, the circles at each end. On the other hand, a sphere
and a torus have no boundary points. See Figure 10.13. In general, the boundary of a surface is
either empty or consists of one or more simple closed curves.",Multivariable_Calculus_Shimamoto.pdf
"either empty or consists of one or more simple closed curves.
Figure 10.13: The boundary of a hemisphere is a circle and that of a cylinder is two circles. The
boundaries of a sphere and torus are empty.
Stokes’s theorem relates a line integral
R
B F · ds around the boundary B of a surface in R3 to a
vector field integral over the surface itself. It can be seen as an extension of Green’s theorem from
double integrals to surface integrals. The theorem is not obvious, and we shall try to show where
it comes from, though in a very special case similar to the one we used to explain Green’s theorem.
Namely, we assume thatS is an oriented surface with a smooth orientation-preserving parametri-
zation σ(s, t) =

x(s, t), y(s, t), z(s, t)


such that:
• the domain of σ is a rectangle R = [a, b] × [c, d] and
• σ is one-to-one and, in particular, transforms the boundary of R precisely onto the boundary
of S.
The situation is shown in Figure 10.14.",Multivariable_Calculus_Shimamoto.pdf
"• σ is one-to-one and, in particular, transforms the boundary of R precisely onto the boundary
of S.
The situation is shown in Figure 10.14.
Figure 10.14: A magic carpet: a surface S with parametrization σ defined on a rectangle R. The
sides of R are oriented so that ∂R = A1 ∪ A2 ∪ (−A3) ∪ (−A4).
Let A denote the boundary of R and B the boundary of S. We orient A counterclockwise and
break it into its four sides A1, A2, A3, A4, where the horizontal sides A1 and A3 are oriented to the",Multivariable_Calculus_Shimamoto.pdf
"246 CHAPTER 10. SURFACE INTEGRALS
right and the vertical sides A2 and A4 are oriented upward. Taking orientation into account, we
have A = A1 ∪ A2 ∪ (−A3) ∪ (−A4). Applying σ then breaks B up into four curves B1, B2, B3, B4,
where Bj = σ(Aj) for j = 1, 2, 3, 4. B inherits the orientation B = B1 ∪ B2 ∪ (−B3) ∪ (−B4).
Let F = (F1, F2, F3) be a smooth vector field defined on an open set in R3 that contains S. We
consider the line integral of F around B, the boundary of S:
Z
B
F · ds =
Z
B1
+
Z
B2
−
Z
B3
−
Z
B4

F · ds.
We use differential form notation:
R
B F · ds =
R
B F1 dx + F2 dy + F3 dz. To cut down on the
amount of notation in any one place, we find
R
B F1 dx,
R
B F2 dy,
R
B F3 dz separately and add the
results at the end.
We can parametrize each of the four pieces of B by applying σ to the corresponding side of A,
fixing one of s or t and using the other as parameter.
For B1: α1(s) = σ(s, c), a ≤ s ≤ b.
For B2: α2(t) = σ(b, t), c ≤ t ≤ d.
For B3: α3(s) = σ(s, d), a ≤ s ≤ b.",Multivariable_Calculus_Shimamoto.pdf
"For B1: α1(s) = σ(s, c), a ≤ s ≤ b.
For B2: α2(t) = σ(b, t), c ≤ t ≤ d.
For B3: α3(s) = σ(s, d), a ≤ s ≤ b.
For B4: α4(t) = σ(a, t), c ≤ t ≤ d.
So for
R
B F1 dx, we integrate over each of B1 through B4 and take the sum with appropri-
ate signs. For example, for B1, the parametrization is α1(s) = σ(s, c) = ( x(s, c), y(s, c), z(s, c)),
where t = c is held fixed. Hence the derivatives with respect to the parameter are deriva-
tives with respect to s, partial derivatives where appropriate. For instance, α′
1(s) = ∂σ
∂s (s, c) =
(∂x
∂s (s, c), ∂y
∂s (s, c), ∂z
∂s (s, c)). As a result, we obtain:
Z
B1
F1 dx =
Z b
a
F1(σ(s, c)) ∂x
∂s (s, c) ds.
Similarly, over B2 with parameter t:
Z
B2
F1 dx =
Z d
c
F1(σ(b, t)) ∂x
∂t (b, t) dt.
Continuing in this way with B3 and B4 and combining the results gives:
Z
B
F1 dx =
Z b
a
F1(σ(s, c)) ∂x
∂s (s, c) ds +
Z d
c
F1(σ(b, t)) ∂x
∂t (b, t) dt
−
Z b
a
F1(σ(s, d)) ∂x
∂s (s, d) ds −
Z d
c
F1(σ(a, t)) ∂x
∂t (a, t) dt. (10.6)",Multivariable_Calculus_Shimamoto.pdf
"Z b
a
F1(σ(s, c)) ∂x
∂s (s, c) ds +
Z d
c
F1(σ(b, t)) ∂x
∂t (b, t) dt
−
Z b
a
F1(σ(s, d)) ∂x
∂s (s, d) ds −
Z d
c
F1(σ(a, t)) ∂x
∂t (a, t) dt. (10.6)
Upon closer examination, each of the terms in this last expression is a line integral over one of
the sides of the boundary of R back in the st-plane. For instance, the first term is
R
A1
(F1 ◦σ) ∂x
∂s ds
and the second is
R
A2
(F1 ◦ σ) ∂x
∂t dt, line integrals over A1 and A2, respectively. In fact, since t is
constant on A1 and s is constant on A2, we can give these terms a more uniform appearance by
writing them as:
Z
A1
(F1 ◦ σ) ∂x
∂s ds + (F1 ◦ σ) ∂x
∂t dt and
Z
A2
(F1 ◦ σ) ∂x
∂s ds + (F1 ◦ σ) ∂x
∂t dt,
respectively. The extra summands are superfluous to the integrals, since the derivatives of the extra
coordinates are zero on the corresponding segment. After converting the remaining two terms to",Multivariable_Calculus_Shimamoto.pdf
"10.3. STOKES’S THEOREM 247
line integrals over A3 and A4, equation (10.6) becomes a line integral over all of A:
Z
B
F1 dx =
Z
A
(F1 ◦ σ) ∂x
∂s ds + (F1 ◦ σ) ∂x
∂t dt.
But now that we are back to a line integral in the plane, Green’s theorem applies, and, as
A = ∂R, we can replace the line integral over A with a double integral over the rectangle R. Thus:
Z
B
F1 dx =
ZZ
R
 ∂
∂s

(F1 ◦ σ) ∂x
∂t


− ∂
∂t

(F1 ◦ σ) ∂x
∂s


ds dt. (10.7)
We work with the integrand of the double integral. Using the product rule to calculate the
partial derivatives with respect to s and t, the integrand becomes:
∂
∂s

F1 ◦ σ) · ∂x
∂t + (F1 ◦ σ) · ∂2x
∂s ∂t− ∂
∂t

F1 ◦ σ) · ∂x
∂s − (F1 ◦ σ) · ∂2x
∂t ∂s.
Thanks to the equality of the mixed partials ∂2x
∂s ∂t = ∂2x
∂t ∂s, the second and fourth terms cancel,
leaving:
∂
∂s

F1 ◦ σ) · ∂x
∂t − ∂
∂t

F1 ◦ σ) · ∂x
∂s .
So after substituting into equation (10.7), at this point, we have:
Z
B
F1 dx =
ZZ
R
 ∂
∂s

F1 ◦ σ) · ∂x
∂t − ∂
∂t

F1 ◦ σ) · ∂x
∂s
",Multivariable_Calculus_Shimamoto.pdf
"∂s .
So after substituting into equation (10.7), at this point, we have:
Z
B
F1 dx =
ZZ
R
 ∂
∂s

F1 ◦ σ) · ∂x
∂t − ∂
∂t

F1 ◦ σ) · ∂x
∂s

ds dt. (10.8)
We compute the partials of F1 ◦ σ using the chain rule. Fans of dependence diagrams will
F1
x y z
s t
Figure 10.15: A dependence diagram for the composition F1 ◦ σ
welcome Figure 10.15. By the chain rule, the integrand in equation (10.8) becomes:
∂F1
∂x
∂x
∂s + ∂F1
∂y
∂y
∂s + ∂F1
∂z
∂z
∂s

 ∂x
∂t −
∂F1
∂x
∂x
∂t + ∂F1
∂y
∂y
∂t + ∂F1
∂z
∂z
∂t

 ∂x
∂s
= ∂F1
∂z
∂z
∂s
∂x
∂t − ∂x
∂s
∂z
∂t


− ∂F1
∂y
∂x
∂s
∂y
∂t − ∂y
∂s
∂x
∂t


,
where ∂F1
∂x , ∂F1
∂y , and ∂F1
∂z are evaluated at σ(s, t) =

x(s, t), y(s, t), z(s, t)


. Substituting this into
(10.8) yields:
Z
B
F1 dx =
ZZ
R
∂F1
∂z
∂z
∂s
∂x
∂t − ∂x
∂s
∂z
∂t


− ∂F1
∂y
∂x
∂s
∂y
∂t − ∂y
∂s
∂x
∂t


ds dt.
The calculations of
R
B F2 dy and
R
B F3 dz are similar and end up giving:
Z
B
F2 dy =
ZZ
R

−∂F2
∂z
∂y
∂s
∂z
∂t − ∂z
∂s
∂y
∂t


+ ∂F2
∂x
∂x
∂s
∂y
∂t − ∂y
∂s
∂x
∂t


ds dt
and",Multivariable_Calculus_Shimamoto.pdf
"R
B F3 dz are similar and end up giving:
Z
B
F2 dy =
ZZ
R

−∂F2
∂z
∂y
∂s
∂z
∂t − ∂z
∂s
∂y
∂t


+ ∂F2
∂x
∂x
∂s
∂y
∂t − ∂y
∂s
∂x
∂t


ds dt
and
Z
B
F3 dz =
ZZ
R
∂F3
∂y
∂y
∂s
∂z
∂t − ∂z
∂s
∂y
∂t


− ∂F3
∂x
∂z
∂s
∂x
∂t − ∂x
∂s
∂z
∂t


ds dt.",Multivariable_Calculus_Shimamoto.pdf
"248 CHAPTER 10. SURFACE INTEGRALS
Adding the three calculations and suitably regrouping the terms results in the formula:
Z
B
F1 dx + F2 dy + F3 dz =
ZZ
R
∂F3
∂y − ∂F2
∂z

∂y
∂s
∂z
∂t − ∂z
∂s
∂y
∂t


+
∂F1
∂z − ∂F3
∂x

∂z
∂s
∂x
∂t − ∂x
∂s
∂z
∂t


+
∂F2
∂x − ∂F1
∂y

∂x
∂s
∂y
∂t − ∂y
∂s
∂x
∂t


ds dt. (10.9)
This looks awful, but actually the ungainly double integral can be expressed much more con-
cisely. The integrand is the dot product of the vectors:
v =
∂F3
∂y − ∂F2
∂z , ∂F1
∂z − ∂F3
∂x , ∂F2
∂x − ∂F1
∂y

and w =
∂y
∂s
∂z
∂t − ∂z
∂s
∂y
∂t , ∂z
∂s
∂x
∂t − ∂x
∂s
∂z
∂t , ∂x
∂s
∂y
∂t − ∂y
∂s
∂x
∂t

.
For v, the components are the differences within the mixed partial pairs ofF, and in fact v = ∇×F,
the curl of F. Likewise, w is also a cross product:
w = det


i j k
∂x
∂s
∂y
∂s
∂z
∂s
∂x
∂t
∂y
∂t
∂z
∂t

 = ∂σ
∂s × ∂σ
∂t .
Thus, after all this work, equation (10.9) becomes:
Z
B
F1 dx + F2 dy + F3 dz =
ZZ
R
(∇ ×F) ·
∂σ
∂s × ∂σ
∂t

ds dt.",Multivariable_Calculus_Shimamoto.pdf
"∂t
∂z
∂t

 = ∂σ
∂s × ∂σ
∂t .
Thus, after all this work, equation (10.9) becomes:
Z
B
F1 dx + F2 dy + F3 dz =
ZZ
R
(∇ ×F) ·
∂σ
∂s × ∂σ
∂t

ds dt.
This last expression is precisely the definition of the surface integral of the vector field ∇ ×F over
S. We have arrived at the final formula:
Z
B
F · ds =
Z
B
F1 dx + F2 dy + F3 dz =
ZZ
S
(∇ ×F) · dS. (10.10)
That was a long calculation. To recap the strategy, to evaluate the line integral over the
boundary B of S, we used a parametrization of S to pull back to a line integral in the plane,
applied Green’s theorem there, and lastly identified the resulting double integral (10.9) as the
pullback of an integral over the surface.
The argument behind (10.10) assumed that the domain of the parametrization is a rectangle, but
the conclusion remains true for surfaces more broadly. In order to formulate a general statement,
we need to say something about the relation between the orientation of a surface and the orientation",Multivariable_Calculus_Shimamoto.pdf
"we need to say something about the relation between the orientation of a surface and the orientation
of its boundary. Let S be a smooth oriented surface in R3 with orienting normal n. We think of n
as defining “up” and imagine walking around S with our feet on the surface and our heads in the
direction of n. Then:
∂S denotes the boundary of S oriented so that, as
you traverse the boundary, S stays on the left.
If S happens to be contained in the plane R2, this is consistent with the way we oriented its
boundary in Green’s theorem, assuming that S is oriented by the upward normal n = (0, 0, 1).",Multivariable_Calculus_Shimamoto.pdf
"10.3. STOKES’S THEOREM 249
Figure 10.16: Three oriented surfaces and their oriented boundaries
For example, let S be a sphere that is oriented by the outward normal. Then the boundary of
the northern hemisphere is the equatorial circle oriented counterclockwise, while the boundary of
the southern hemisphere is the same circle but oriented clockwise. If S is a pair of pants oriented
by the outward normal, then ∂S consists of three closed curves: the waist oriented one way—let’s
call it clockwise—and the bottoms of the two legs oriented counterclockwise. See Figure 10.16.
For a piecewise smooth surface S = S1 ∪ S2 ∪ ··· ∪Sk, we require that the smooth pieces
S1, S2, . . . , Sk are oriented compatibly in the sense that the oriented boundaries of adjacent pieces
are traversed in opposite directions wherever they intersect. These intersections are not considered
to be part of the boundary of the whole surface S. This is illustrated in Figure 10.17.",Multivariable_Calculus_Shimamoto.pdf
"to be part of the boundary of the whole surface S. This is illustrated in Figure 10.17.
Figure 10.17: Compatible orienting normals n1, n2 for adjacent surfaces S1, S2, respectively, with
corresponding orientations of ∂S1, ∂S2
For the extension of (10.10) beyond surfaces whose parameter domains are rectangles, one
approach might be to mimic the strategy we outlined informally for Green’s theorem. Namely,
first verify it in the case where the parameter domain is a union of rectangles that overlap at most
along parts of their boundaries and then argue that the domain of any parametrization can be
approximated arbitrarily well by such a union. We won’t give the details.
With these preliminaries out of the way, we can state the theorem.
Theorem 10.6 (Stokes’s theorem). Let S be a piecewise smooth oriented surface in R3 that is
bounded as a subset of R3, and let F be a smooth vector field defined on an open set containing S.
Then: Z
∂S
F · ds =
ZZ
S
(∇ ×F) · dS.",Multivariable_Calculus_Shimamoto.pdf
"bounded as a subset of R3, and let F be a smooth vector field defined on an open set containing S.
Then: Z
∂S
F · ds =
ZZ
S
(∇ ×F) · dS.
In the form stated, Stokes’s theorem converts a line integral to a surface integral. Since surface
integrals are usually messier than line integrals, this is not the usual way in which the theorem
is applied, at least as far as calculating specific examples goes. (See Exercises 3.3–3.6 for some
exceptions, however.) Sometimes though, the theorem can be used to go the other way, converting
a surface integral to a line integral. This possibility has implications for the general properties of
surface integrals.",Multivariable_Calculus_Shimamoto.pdf
"250 CHAPTER 10. SURFACE INTEGRALS
Example 10.7. Let F be the “falling rain” vector field F(x, y, z) = (0 , 1, −3), and let S be the
cone z =
p
x2 + y2, x2 + y2 ≤ 4, oriented by the upward normal. In Example 10.5, we integrated
F over S using a parametrization. Since we evaluated the integral before, there seems no harm in
giving away a spoiler: the answer is
RR
S F · dS = −12π.
Could this surface integral have been computed using Stokes’s theorem? The answer is yes
provided that there is a vector field G such that F = ∇ ×G, for then Stokes’s theorem givesRR
S F · dS =
RR
S(∇ ×G) · dS =
R
∂S G · ds. In other words, we could find the surface integral of F
over S by computing the line integral of G around the boundary.
We try to find such a G = (G1, G2, G3) by setting F = ∇ ×G and guessing:
F = (0, 1, −3) = det


i j k
∂
∂x
∂
∂y
∂
∂z
G1 G2 G3

.
This leads to the system: 


0 = ∂G3
∂y − ∂G2
∂z
1 = ∂G1
∂z − ∂G3
∂x
−3 = ∂G2
∂x − ∂G1
∂y .",Multivariable_Calculus_Shimamoto.pdf
"

i j k
∂
∂x
∂
∂y
∂
∂z
G1 G2 G3

.
This leads to the system: 


0 = ∂G3
∂y − ∂G2
∂z
1 = ∂G1
∂z − ∂G3
∂x
−3 = ∂G2
∂x − ∂G1
∂y .
From the second equation, we guess G1 = z, from the third G2 = −3x, and from the first G3 = 0.
One can check that these guesses are actually consistent with all three equations. Thus G(x, y, z) =
(z, −3x, 0) works! Our virtuous lifestyle has paid off.
This gives:
ZZ
S
F · dS =
Z
∂S
G · ds =
Z
∂S
z dx− 3x dy+ 0dz =
Z
∂S
z dx− 3x dy.
We calculate the line integral by parametrizing C = ∂S, which is a circle of radius 2 in the plane
z = 2, oriented counterclockwise (see Figure 10.12, middle, or, if you’re willing to peek ahead,
Figure 10.18, left). The circle can be parametrized by α(t) = (2 cost, 2 sint, 2), 0 ≤ t ≤ 2π. Thus:
ZZ
S
F · dS =
Z
C
z dx− 3x dy=
Z 2π
0

2 · (−2 sint) − 6 cost · 2 cost


dt
=
Z 2π
0
(−4 sint − 12 cos2 t) dt
= 4 cost
2π
0 − 12 · 2π
2 (10.11)
= 0 − 12π
= −12π,",Multivariable_Calculus_Shimamoto.pdf
"C
z dx− 3x dy=
Z 2π
0

2 · (−2 sint) − 6 cost · 2 cost


dt
=
Z 2π
0
(−4 sint − 12 cos2 t) dt
= 4 cost
2π
0 − 12 · 2π
2 (10.11)
= 0 − 12π
= −12π,
where, in step (10.11), we used our shortcut to integrate cos 2 t (Exercise 1.1, Chapter 7).
10.4 Curl fields
In the last example, we got the same answer of −12π using Stokes’s theorem as we did in Example
10.5 using a parametrization, but it’s not clear that the Stokes’s theorem approach was any simpler.
Actually, it seemed a little roundabout. The argument has some important consequences, however.
Continuing with the example, let C denote the oriented boundary of the cone S, and suppose
that eS is any piecewise smooth oriented surface having C as its oriented boundary. For instance, eS",Multivariable_Calculus_Shimamoto.pdf
"10.4. CURL FIELDS 251
Figure 10.18: An oriented cone S and two other oriented surfaces eS, all three having the same
oriented boundary
could be the disk in the plane z = 2 that caps off the cone, oriented by the upward normal, which
was also considered in Example 10.5. See Figure 10.18.
If we integrate the falling rain vector field F over the new surface eS, we obtain:
ZZ
eS
F · dS =
ZZ
eS
(∇ ×G) · dS (where G = (z, −3x, 0) as before)
=
Z
∂ eS
G · ds (by Stokes’s theorem)
=
Z
C
G · ds
= −12π (by the calculation in the previous example).
In other words, the integral of F is −12π over all oriented surfaces whose boundary is C.
This idea can be formulated more generally.
Definition. Let U be an open set in R3. A vector field F: U → R3 is called a curl field on U if
there exists a smooth vector field G: U → R3 such that F = ∇ ×G.
For instance, we saw in Example 10.7 that F = (0, 1, −3) is a curl field on R3 with “anti-curl”
G = (z, −3x, 0).",Multivariable_Calculus_Shimamoto.pdf
"For instance, we saw in Example 10.7 that F = (0, 1, −3) is a curl field on R3 with “anti-curl”
G = (z, −3x, 0).
One perspective on curl fields is the following general working principle: curl fields are to surface
integrals as conservative fields are to line integrals. We consider a couple of illustrations of this.
Example 10.8. For line integrals: The integral of a conservative field over an oriented curve
depends only on the endpoints of the curve.
For surface integrals: The integral of a curl field over an oriented surface depends only on the
oriented boundary of the surface. More precisely:
Theorem 10.9. Let S1 and S2 be piecewise smooth oriented surfaces in an open set U in R3 that
have the same oriented boundary, i.e., ∂S1 = ∂S2. If F is a curl field on U, then:
ZZ
S1
F · dS =
ZZ
S2
F · dS.
Proof. Suppose that F = ∇ ×G. Then by Stokes’s theorem, both sides of the equation are equal
to
R
C G · ds, where C = ∂S1 = ∂S2.",Multivariable_Calculus_Shimamoto.pdf
"252 CHAPTER 10. SURFACE INTEGRALS
Example 10.10. For line integrals: The integral of a conservative field over a piecewise smooth
oriented closed curve is 0.
For surface integrals:
Definition. Let S be a piecewise smooth surface in Rn such that every pair of points in S can be
joined by a piecewise smooth curve in S. Then S is called closed if ∂S = ∅ (the empty set).
For example, spheres and tori are closed surfaces, but hemispheres and cylinders are not.
Theorem 10.11. If F is a curl field on an open set U in R3, then
RR
S F·dS = 0 for any piecewise
smooth oriented closed surface S in U.
Proof. Let S be a piecewise smooth oriented closed surface in U. The simplest argument would be
to say that, if F = ∇ ×G, then, by Stokes’s theorem,
RR
S F · dS =
R
∂S G · ds = 0 since ∂S = ∅.
If integrating over the empty set makes you nervous, then perhaps a more aboveboard alternative
would be to cut S into two nonempty piecesS1 and S2 such that S = S1 ∪S2 and S1 and S2 intersect",Multivariable_Calculus_Shimamoto.pdf
"would be to cut S into two nonempty piecesS1 and S2 such that S = S1 ∪S2 and S1 and S2 intersect
only along their common boundary. For instance, S1 could be a small disklike piece of S, and S2
could be what’s left. See Figure 10.19. Then:
ZZ
S
F · dS =
ZZ
S1
F · dS +
ZZ
S2
F · dS
=
Z
∂S1
G · ds +
Z
∂S2
G · ds. (10.12)
Figure 10.19: Dividing S into two pieces, S = S1 ∪ S2
S1 and S2 inherit the orienting unit normal from S, so, in order to keep them on the left, their
boundaries must be traversed in opposite directions. This is consistent with our requirement for
orientations of piecewise smooth surfaces. In other words, ∂S1 and ∂S2 are the same curve but
with opposite orientations. Hence the line integrals in equation (10.12) cancel each other out.
Lastly, we look for an analogue for curl fields of the mixed partials theorem for conservative
fields. In R3, the mixed partials theorem takes the form that, if F is conservative, then ∇× F = 0.",Multivariable_Calculus_Shimamoto.pdf
"fields. In R3, the mixed partials theorem takes the form that, if F is conservative, then ∇× F = 0.
Now, let F be a curl field, say F = ∇ ×G, so (F1, F2, F3) = det


i j k
∂
∂x
∂
∂y
∂
∂z
G1 G2 G3

. Hence:



F1 = ∂G3
∂y − ∂G2
∂z
F2 = ∂G1
∂z − ∂G3
∂x
F3 = ∂G2
∂x − ∂G1
∂y .",Multivariable_Calculus_Shimamoto.pdf
"10.5. GAUSS’S THEOREM 253
Taking certain partial derivatives of F1, F2, F3 introduces the second-order mixed partials of G1,
G2, G3: 


∂F1
∂x = ∂2G3
∂x ∂y− ∂2G2
∂x ∂z
∂F2
∂y = ∂2G1
∂y ∂z− ∂2G3
∂y ∂x
∂F3
∂z = ∂2G2
∂z ∂x− ∂2G1
∂z ∂y.
The terms on the right side are mixed partial pairs that appear with opposite signs. Thus, by the
equality of mixed partials, taking the sum gives:
∂F1
∂x + ∂F2
∂y + ∂F3
∂z = 0. (10.13)
The sum can also be written in terms of the ∇ operator as ∂F1
∂x + ∂F2
∂y + ∂F3
∂z = ( ∂
∂x , ∂
∂y , ∂
∂z ) ·
(F1, F2, F3) = ∇ ·F, where the product is dot product. This quantity has a name.
Definition. If F = (F1, F2, F3) is a smooth vector field on an open set in R3, then:
∇ ·F = ∂F1
∂x + ∂F2
∂y + ∂F3
∂z
is called the divergence of F.
Note that, while the curl of a vector field is again a vector field, the divergence is a real-valued
function.
From (10.13), we have proven the following.",Multivariable_Calculus_Shimamoto.pdf
"function.
From (10.13), we have proven the following.
Theorem 10.12. Let U be an open set in R3, and let F be a smooth vector field on U. If F is a
curl field on U, then:
∇ ·F = 0
at all points of U. Equivalently, given a smooth vector field F on U, if ∇ ·F ̸= 0 at some point,
then F is not a curl field.
Example 10.13. The vector field F(x, y, z) = (2 xy + z2, 2yz + x2, 2xz + y2) is conservative on
R3. It has f(x, y, z) = x2y + y2z + z2x as a potential function. Is F also a curl field on R3?
We could write down the conditions for being a curl field and try to guess a vector field G
such that F = ∇ ×G the way we did before. It is easy, however, to compute the divergence:
∇ ·F = ∂
∂x

2xy + z2

+ ∂
∂y

2yz + x2

+ ∂
∂z

2xz + y2

= 2y + 2z + 2x. Since this is not identically
0, F is not a curl field.
10.5 Gauss’s theorem
Gauss’s theorem relates the integral of a vector field over a closed surface to a triple integral over",Multivariable_Calculus_Shimamoto.pdf
"0, F is not a curl field.
10.5 Gauss’s theorem
Gauss’s theorem relates the integral of a vector field over a closed surface to a triple integral over
the three-dimensional region that the surface encloses. As with Green’s and Stokes’s theorems, we
indicate where the theorem comes from by deriving it in a very special case, then hint at how it
might be extended to more general situations.
Let F = (F1, F2, F3) be a smooth vector field defined on an open set U in R3. We consider the
case of a surface S that is the boundary of a rectangular box W = [a, b] × [c, d] × [e, f] contained
in U. Thus S consists of the six rectangular faces of the box, each parallel to one of the coordinate
planes. We orient each face with the unit normal pointing away from the box, as shown for three
of the faces in Figure 10.20. This makes S into a piecewise smooth oriented closed surface.",Multivariable_Calculus_Shimamoto.pdf
"254 CHAPTER 10. SURFACE INTEGRALS
Figure 10.20: The rectangular box W = [a, b]×[c, d]×[e, f] in R3 and the orientation of its boundary
surface S
We integrate F over S. To do so, we consider each face separately and add the results:
ZZ
S
F · dS =
ZZ
top
+
ZZ
bottom
+
ZZ
front
+
ZZ
back
+
ZZ
left
+
ZZ
right

F · dS. (10.14)
For instance, the top may be parametrized using x and y as parameters with z = f fixed: σ(x, y) =
(x, y, f), a ≤ x ≤ b, c ≤ y ≤ d. Then:
∂σ
∂x × ∂σ
∂y = (1, 0, 0) × (0, 1, 0) = i × j = k = (0, 0, 1).
This points upward, hence away from the box W, so σ is orientation-preserving. Therefore:
ZZ
top
F · dS =
ZZ
[a,b]×[c,d]
F(σ(x, y)) ·
∂σ
∂x × ∂σ
∂y

dx dy
=
ZZ
[a,b]×[c,d]
(F1, F2, F3) · (0, 0, 1) dx dy
=
ZZ
[a,b]×[c,d]
F3(σ(x, y)) dx dy
=
ZZ
[a,b]×[c,d]
F3(x, y, f) dx dy.
For the face on the bottom, the similar parametrization σ(x, y) = (x, y, e) works, only now σ is
orientation-reversing since the orientation of S on the bottom points downward in order to point",Multivariable_Calculus_Shimamoto.pdf
"orientation-reversing since the orientation of S on the bottom points downward in order to point
away from W. This gives:
ZZ
bottom
F · dS = −
ZZ
[a,b]×[c,d]
F3(x, y, e) dx dy.",Multivariable_Calculus_Shimamoto.pdf
"10.5. GAUSS’S THEOREM 255
We add these results and apply the fundamental theorem of calculus:
ZZ
top
F · dS +
ZZ
bottom
F · dS =
ZZ
[a,b]×[c,d]

F3(x, y, f) − F3(x, y, e)


dx dy
=
ZZ
[a,b]×[c,d]

F3(x, y, z)

z=f
z=e

dx dy
=
ZZ
[a,b]×[c,d]
Z f
e
∂F3
∂z (x, y, z) dz

dx dy
=
ZZZ
W
∂F3
∂z (x, y, z) dx dy dz.
The corresponding calculations for the remaining four faces yield:
ZZ
front
F · dS +
ZZ
back
F · dS =
ZZZ
W
∂F1
∂x (x, y, z) dx dy dz
andZZ
left
F · dS +
ZZ
right
F · dS =
ZZZ
W
∂F2
∂y (x, y, z) dx dy dz.
Adding up the results over all the faces as in equation (10.14) then gives:
ZZ
S
F · dS =
ZZZ
W
∂F1
∂x + ∂F2
∂y + ∂F3
∂z


dx dy dz=
ZZZ
W
∇ ·F dx dy dz. (10.15)
One can extend this formula to the case that W is a union of rectangular boxes that intersect
at most along their boundaries by applying equation (10.15) to the individual boxes and adding. A
key observation is that when two neighboring boxes intersect in the interior of W on a portion of a",Multivariable_Calculus_Shimamoto.pdf
"key observation is that when two neighboring boxes intersect in the interior of W on a portion of a
common face, then the orienting normals of each box point in opposite directions in order to point
away from the respective boxes. Hence the normal components F · n are also opposite, so that the
surface integrals over the overlaps cancel out in the sum. This leaves only the surface integral over
the boundary of W. One then claims that equation (10.15) remains true for more general regions
by some sort of limiting argument.
Before stating the theorem, we identify the orientation convention implicit in the argument
given above, namely, if W is a bounded set in R3 whose boundary consists of one or more closed
surfaces, then:
∂W denotes the boundary of W oriented so that
the orienting unit normal points away from W.
An example is shown in Figure 10.21.
The general version of equation (10.15) is then the following.",Multivariable_Calculus_Shimamoto.pdf
"the orienting unit normal points away from W.
An example is shown in Figure 10.21.
The general version of equation (10.15) is then the following.
Theorem 10.14 (Gauss’s theorem). Let U be an open set in R3, and let F be a smooth vector
field on U. If W is a bounded subset of U whose boundary consists of a finite number of piecewise
smooth closed surfaces, then:
ZZ
∂W
F · dS =
ZZZ
W
∇ ·F dx dy dz.",Multivariable_Calculus_Shimamoto.pdf
"256 CHAPTER 10. SURFACE INTEGRALS
Figure 10.21: A solid region W in R3 and its oriented boundary ∂W
In the case that F is a curl field, the left side of Gauss’s theorem is 0. This is because ∂W
consists of closed surfaces and the integral of a curl field over a closed surface is 0 (Theorem 10.11).
The right side is 0, too, because curl fields have divergence 0 (Theorem 10.12). Thus, for curl fields,
Gauss’s theorem reflects properties already known to us.
To illustrate Gauss’s theorem, we begin by taking a new look at two surface integrals we
evaluated earlier.
Example 10.15. Let Sa be the sphere x2 + y2 + z2 = a2 of radius a, oriented by the outward
normal. Find
RR
Sa
F · dS if:
(a) F(x, y, z) = (0, 0, −1),
(b) F(x, y, z) = (0, 0, z).
Figure 10.22: The solid ball Wa of radius a and its oriented boundary, the sphere Sa
(a) This was Example 10.2(c). To answer it using Gauss’s theorem, let Wa be the solid ball
x2 + y2 + z2 ≤ a2 so that Sa = ∂Wa, as in Figure 10.22. Since ∇ ·F = ∂",Multivariable_Calculus_Shimamoto.pdf
"x2 + y2 + z2 ≤ a2 so that Sa = ∂Wa, as in Figure 10.22. Since ∇ ·F = ∂
∂x (0) + ∂
∂y (0) + ∂
∂z (−1) = 0,
Gauss’s theorem says that:
ZZ
Sa
F · dS =
ZZZ
Wa
∇ ·F dx dy dz=
ZZZ
Wa
0 dx dy dz= 0.
(b) This is the third time we have evaluated this integral (see Examples 10.2(b) and 10.4). Here,
∇ ·F = 0 + 0 + 1 = 1, so with Wa as above:
ZZ
Sa
F · dS =
ZZZ
Wa
1 dx dy dz= Vol (Wa) = 4
3πa3,",Multivariable_Calculus_Shimamoto.pdf
"10.6. THE INVERSE SQUARE FIELD 257
where we have used the formula for the volume of a three-dimensional ball from Example 7.8 in
Chapter 7.
It seems odd to say that turning a problem into a triple integral makes it simpler, but, with the
help of Gauss’s theorem, finding the preceding two surface integrals became essentially trivial.
Example 10.16. Let F(x, y, z) = (x, y, z). This vector field radiates directly away from the origin
at every point, and the arrows get longer the further out you go. Let W be a solid region in R3 as
in Gauss’s theorem. Then ∇ ·F = 1 + 1 + 1 = 3, so
RR
∂W F · dS =
RRR
W 3 dx dy dz= 3 Vol (W),
that is:
Vol (W) = 1
3
ZZ
∂W
(x, y, z) · dS.
This is another instance where information about a set can be gleaned from a calculation just
involving its boundary. ( Question: Thinking of the surface integral as the integral of the normal
component and keeping in mind that ∂W is oriented by the outward normal, does it make sense
that
RR",Multivariable_Calculus_Shimamoto.pdf
"component and keeping in mind that ∂W is oriented by the outward normal, does it make sense
that
RR
∂W (x, y, z) · dS is always positive? Note that the origin need not lie in W, so there may
be places where the orienting normal n on the boundary points towards the origin and the normal
component F · n = (x, y, z) · n is negative.)
10.6 The inverse square field
Throughout this section, G denotes the inverse square field G: R3 − {(0, 0, 0)} →R3, given by:
G(x, y, z) =

− x
(x2 + y2 + z2)3/2 , − y
(x2 + y2 + z2)3/2 , − z
(x2 + y2 + z2)3/2

. (10.16)
We study G using Gauss’s theorem, starting with a couple of preliminary facts.
Fact 1. Let Sa be the sphere x2 + y2 + z2 = a2, oriented by the outward normal. Then:
ZZ
Sa
G · dS = −4π, independent of the radius a.
Justification. We calculated and even emphasized this earlier in Example 10.2(a), which was the
first surface integral of a vector field that we found. Note that this implies that G is not a curl",Multivariable_Calculus_Shimamoto.pdf
"first surface integral of a vector field that we found. Note that this implies that G is not a curl
field, since a curl field would integrate to 0 over the closed surface Sa.
Fact 2. ∇ ·G = 0.
Justification. This is a brute force calculation. For instance:
∂G1
∂x = ∂
∂x

− x
(x2 + y2 + z2)3/2

= −(x2 + y2 + z2)3/2 · 1 − x · 3
2 (x2 + y2 + z2)1/2 · 2x
(x2 + y2 + z2)3
= −(x2 + y2 + z2) − 3x2
(x2 + y2 + z2)5/2
= 2x2 − y2 − z2
(x2 + y2 + z2)5/2 .
By symmetric calculations:
∂G2
∂y = 2y2 − x2 − z2
(x2 + y2 + z2)5/2 and ∂G3
∂z = 2z2 − x2 − y2
(x2 + y2 + z2)5/2 .",Multivariable_Calculus_Shimamoto.pdf
"258 CHAPTER 10. SURFACE INTEGRALS
Taking the sum gives ∇ ·G = (2x2−y2−z2)+(2y2−x2−z2)+(2z2−x2−y2)
(x2+y2+z2)5/2 = 0, as claimed.
This fact has an interesting implication, too, for we know that curl fields have divergence 0.
This shows that the converse need not hold, since ∇ ·G = 0 yet G is not a curl field by Fact 1.
Now, let S be a piecewise smooth closed surface in R3 − {(0, 0, 0)}, oriented by the outward
normal. Is that enough for us to be able to say something about
RR
S G·dS? Strikingly, the answer
is yes. There are two cases.
Case 1. Suppose that the origin 0 lies in the exterior of S (Figure 10.23).
Figure 10.23: The origin lies exterior to S.
Let W be the three-dimensional solid inside of S so that S = ∂W . By assumption, 0 /∈ W, so
G is defined throughout W and we may apply Gauss’s theorem. This gives:ZZ
S
G · dS =
ZZZ
W
∇ ·G dx dy dz
=
ZZZ
W
0 dx dy dz (by Fact 2)
= 0.
Case 2. Suppose that 0 lies in the interior of S.",Multivariable_Calculus_Shimamoto.pdf
"S
G · dS =
ZZZ
W
∇ ·G dx dy dz
=
ZZZ
W
0 dx dy dz (by Fact 2)
= 0.
Case 2. Suppose that 0 lies in the interior of S.
The preceding argument is no longer valid, since filling in the interior of S would include the
origin, which is excluded from the domain of G. Instead, we choose a very small sphere Sϵ of radius
ϵ centered at the origin, oriented by the outward normal, and we let W be the region inside S but
outside Sϵ. See Figure 10.24.
Figure 10.24: The origin lies in the interior of S.
Then Gauss’s theorem applies to W, giving:ZZ
∂W
G · dS =
ZZZ
W
∇ ·G dx dy dz= 0, (10.17)",Multivariable_Calculus_Shimamoto.pdf
"10.7. A MORE SUBSTITUTION-FRIENDLY NOTATION FOR SURFACE INTEGRALS 259
as before. This time, however, the boundary of W consists of S and Sϵ, oriented by the normal
pointing away from W. This is the outward normal on S but points inward on Sϵ. In other words,
∂W = S ∪ (−Sϵ), and therefore
RR
∂W G · dS =
RR
S G · dS −
RR
Sϵ
G · dS. By equation (10.17), the
difference is 0, from which it follows that
RR
S G · dS =
RR
Sϵ
G · dS. And, by Fact 1, the integral of
G over any sphere centered at the origin is −4π. Hence:
ZZ
S
G · dS = −4π.
After a slight rearrangement, we can summarize the discussion as follows. For a piecewise
smooth closed surface S that doesn’t pass through the origin and is oriented by the outward
normal:
− 1
4π
ZZ
S
G · dS =
(
0 if 0 lies in the exterior of S,
1 if 0 lies in the interior of S.
As we noted back in Chapter 8, the inverse square field describes the gravitational, or electro-",Multivariable_Calculus_Shimamoto.pdf
"1 if 0 lies in the interior of S.
As we noted back in Chapter 8, the inverse square field describes the gravitational, or electro-
static, force due to an object, or charged particle, located at the origin. If several objects of equal
mass, or equal charge, are placed at the points p1, p2, . . . ,pk in R3, then the total force at a point
x due to these objects is described by the vector field:
F(x) =
kX
i=1
G(x − pi), (10.18)
where G is the same as above (10.16). It is not hard to verify that again ∇· F = 0. A modification
of the argument just given shows that, if S is a piecewise smooth closed surface contained in the
set R3 − {p1, p2, . . . ,pk} and oriented by the outward normal, then:
− 1
4π
ZZ
S
F · dS counts the number of the points p1, p2, . . . ,pk that lie in the interior of S.
This is an instance of what is known in physics asGauss’s law. In effect, the total mass, or charge,
contained inside of S determines, and is determined by, the integral ofF over S. An explicit formula",Multivariable_Calculus_Shimamoto.pdf
"contained inside of S determines, and is determined by, the integral ofF over S. An explicit formula
or parametrization for S is not needed to obtain this information.
10.7 A more substitution-friendly notation for surface integrals
We now fill in a few details that we have been postponing. Some of the arguments are technical,
but they also unveil themes that unify the theory behind what we have been doing.
In the case of line integrals, we have typically replaced the notation
R
C F · ds in favor of the
more suggestive
R
C F1 dx1 + F2 dx2 + ··· + Fn dxn. The latter is useful in that it tells us more or
less what we need to substitute in order to express the integral in terms of the parameter used to
trace out C. We now introduce a corresponding notation for surface integrals.
Let F = (F1, F2, F3) be a continuous vector field on an open set U in R3, and let S be a smooth
oriented surface in U. To integrate F over S, we want to turn the calculation into a double integral",Multivariable_Calculus_Shimamoto.pdf
"oriented surface in U. To integrate F over S, we want to turn the calculation into a double integral
in the two parameters of a parametrization. So let σ : D → R3 be a smooth parametrization of S,
where σ(s, t) = (x(s, t), y(s, t), z(s, t)). For any function of (x, y, z), such as the components F1, F2,
and F3 of F, the parametrization tells us how to substitute for x, y, and z, resulting in a function
of s and t.",Multivariable_Calculus_Shimamoto.pdf
"260 CHAPTER 10. SURFACE INTEGRALS
For simplicity, let’s assume that σ is orientation-preserving. By definition:
ZZ
S
F · dS =
ZZ
D
F(σ(s, t)) ·
∂σ
∂s × ∂σ
∂t

ds dt
=
ZZ
D

F1(σ(s, t)), F2(σ(s, t)), F3(σ(s, t))


· det


i j k
∂x
∂s
∂y
∂s
∂z
∂s
∂x
∂t
∂y
∂t
∂z
∂t

 ds dt
=
ZZ
D

F1(σ(s, t)) · det
∂y
∂s
∂z
∂s∂y
∂t
∂z
∂t

+ F2(σ(s, t)) · det
∂z
∂s
∂x
∂s∂z
∂t
∂x
∂t

+ F3(σ(s, t)) · det
∂x
∂s
∂y
∂s
∂x
∂t
∂y
∂t

ds dt. (10.19)
We adopt the notation:
ZZ
S
F · dS =
ZZ
S
F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy. (10.20)
The “∧” symbol stands for a kind of multiplication known as the wedge product. We develop
some of its properties along the way as we get more practice working with it.
For instance, if F(x, y, z) = ( xyz, x2 + y2 + z2, xy+ yz), then
RR
S F · dS would be writtenRR
S xyz dy∧dz +(x2 +y2 +z2) dz ∧dx+(xy +yz) dx∧dy. The order of the factors in the individual
terms of the notation is important. The pattern is that the subscripts/variables appear in cyclic",Multivariable_Calculus_Shimamoto.pdf
"terms of the notation is important. The pattern is that the subscripts/variables appear in cyclic
order, as indicated in Figure 10.25.
F3
dz
F1
dx
F2
dy
Figure 10.25: The cyclic ordering of coordinates in R3
In comparison with the expanded definition (10.19), the notation in (10.20) says that, as part
of the substitutions for x, y, zin terms of s, t, one should make the substitution, for example:
dy ∧ dz = det
∂y
∂s
∂z
∂s∂y
∂t
∂z
∂t

ds dt.
Actually, this may look more familiar if we take the transpose of the matrix, which does not affect
the value of the determinant:
dy ∧ dz = det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

ds dt. (10.21)
This has the form of a change of variables, i.e., a substitution, for a transformation from the st-
plane to the yz-plane. Similarly, to evaluate an integral presented in the notation of (10.20), one
substitutes:
dz ∧ dx = det
∂z
∂s
∂z
∂t∂x
∂s
∂x
∂t

ds dt
and dx ∧ dy = det
∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

ds dt,
(10.22)",Multivariable_Calculus_Shimamoto.pdf
"10.8. INDEPENDENCE OF PARAMETRIZATION 261
which also have the form of certain changes of variables.
The integrand η = F1 dy∧dz+F2 dz∧dx+F3 dx∧dy of (10.20) is called a differential 2-form.
By comparison, the expressions ω = F1 dx + F2 dy + F3 dz that we integrated over curves in R3 are
called differential 1-forms. To integrate the 2-form η over S, we substitute for x, y, zin terms of
s, tin F1, F2, F3 and replace the wedge products like dy ∧ dz with expressions that come formally
from the change of variables theorem. This converts the integral over S into a double integral over
the parameter domain. The integrand of this double integral is called the pullback of η, denoted
by σ∗(η). From (10.19), it can be written:
σ∗(η) =

(F1 ◦ σ) · det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

+ (F2 ◦ σ) · det
∂z
∂s
∂z
∂t∂x
∂s
∂x
∂t

+ (F3 ◦ σ) · det
∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

ds ∧ dt.
With this notation, the definition of the surface integral takes the form:
ZZ
σ(D)
η =
ZZ
D
σ∗(η), (10.23)",Multivariable_Calculus_Shimamoto.pdf
"∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

ds ∧ dt.
With this notation, the definition of the surface integral takes the form:
ZZ
σ(D)
η =
ZZ
D
σ∗(η), (10.23)
where, on the right, an integral of the form
RR
D f(s, t) ds ∧ dt is defined to be the usual double
integral
RR
D f(s, t) ds dt. Equation (10.23) highlights that the definition has the same form as the
change of variables theorem, continuing an emerging pattern (see (7.3) and (9.6)).
We say more in the next chapter about properties of differential forms, but our discussion so far
can be used to motivate some elementary manipulations. The substitution formulas (10.21) and
(10.22) for dy ∧ dz and the like suggest a connection between wedge products and determinants.
To reinforce this, we adopt the following rules for working with differential forms:
• dz ∧ dy = −dy ∧ dz,
dx ∧ dz = −dz ∧ dx,
dy ∧ dx = −dx ∧ dy,
• dx ∧ dx = 0,
dy ∧ dy = 0,
dz ∧ dz = 0.",Multivariable_Calculus_Shimamoto.pdf
"• dz ∧ dy = −dy ∧ dz,
dx ∧ dz = −dz ∧ dx,
dy ∧ dx = −dx ∧ dy,
• dx ∧ dx = 0,
dy ∧ dy = 0,
dz ∧ dz = 0.
The first collection mirrors the property that interchanging two rows changes the sign of a deter-
minant, and the second mirrors the property that two equal rows imply a determinant of 0. We
postpone until the next chapter the discussion of how these properties may be used.
10.8 Independence of parametrization
Finally, we address the long-standing question of the extent to which the official definition of the
surface integral,
RR
S F·dS = ±
RR
D F(σ(s, t))·
∂σ
∂s × ∂σ
∂t


ds dt, depends on the smooth parametriza-
tion σ : D → R3 of S, where D is a subset of the st-plane and, by assumption, ∂σ
∂s × ∂σ
∂t ̸= 0, except
possibly on the boundary of D. To highlight the role of the parametrization σ, we write
RR
σ F · dS
to denote the quantity
RR
D F(σ(s, t)) ·
∂σ
∂s × ∂σ
∂t


ds dtin the definition.",Multivariable_Calculus_Shimamoto.pdf
"RR
σ F · dS
to denote the quantity
RR
D F(σ(s, t)) ·
∂σ
∂s × ∂σ
∂t


ds dtin the definition.
We continue to assume that σ is orientation-preserving. Let τ : D∗ → R3 be another smooth
parametrization of S. Call its parameters u and v so that τ(u, v) = (x(u, v), y(u, v), z(u, v)). For
simplicity, we assume that σ and τ are one-to-one, except possibly on the boundaries of their
domains. The goal is to compare
RR
σ F · dS and
RR
τ F · dS. We use the formula for the surface",Multivariable_Calculus_Shimamoto.pdf
"262 CHAPTER 10. SURFACE INTEGRALS
integral after it has been expanded out in terms of coordinates as suggested by the differential form
notation, that is:
ZZ
σ
F · dS =
ZZ
D

F1(σ(s, t)) · det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

+ F2(σ(s, t)) · det
∂z
∂s
∂z
∂t∂x
∂s
∂x
∂t

+ F3(σ(s, t)) · det
∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

ds dt, (10.24)
as in equation (10.19), except that the 2 by 2 matrices have been transposed. The expression forRR
τ F · dS is similar, integrating over D∗ and using u’s and v’s instead of s’s and t’s.
If x ∈ S, then x = σ(s, t) for some ( s, t) in D and x = τ(u, v) for some ( u, v) in D∗. We write
this as ( s, t) = T(u, v). This defines a function T : D∗ → D, where, given ( u, v) in D∗, T(u, v) is
the point (s, t) in D such that σ(s, t) = τ(u, v). Then, by construction:
τ(u, v) = σ(s, t) = σ(T(u, v)). (10.25)
See Figure 10.26. (If σ or τ is not one-to-one, which is permitted only on the boundaries of their
Figure 10.26: Two parametrizations σ and τ of the same surface S",Multivariable_Calculus_Shimamoto.pdf
"Figure 10.26: Two parametrizations σ and τ of the same surface S
respective domains, there is some ambiguity on how to define T there, but, as we have observed
before, the integral can tolerate a certain amount of misbehavior on the boundary.)
By equation (10.25) and the chain rule, Dτ(u, v) = Dσ(T(u, v))·DT (u, v) = Dσ(s, t)·DT (u, v),
or: 

∂x
∂u
∂x
∂v∂y
∂u
∂y
∂v∂z
∂u
∂z
∂v

 =


∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t


∂s
∂u
∂s
∂v∂t
∂u
∂t
∂v

. (10.26)
We focus on the 2 by 2 blocks within the first two matrices of this equation. For instance, isolating
the portions of the product involving the second and third rows, as indicated in red above, gives:
∂y
∂u
∂y
∂v∂z
∂u
∂z
∂v

=
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t
 ∂s
∂u
∂s
∂v∂t
∂u
∂t
∂v

=
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

· DT (u, v).
Since the determinant of a product is the product of the determinants—that is, det( AB) =
(det A)(det B)—we obtain:
det
∂y
∂u
∂y
∂v∂z
∂u
∂z
∂v

= det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

· det DT (u, v). (10.27)",Multivariable_Calculus_Shimamoto.pdf
"10.8. INDEPENDENCE OF PARAMETRIZATION 263
Referring to (10.24), the first determinant in the preceding equation appears in the definition ofRR
τ F · dS, while the second appears in the definition of
RR
σ F · dS. By choosing different pairs
of rows in the chain rule (10.26), the same reasoning applies to the other two determinants that
appear in (10.24). In each case, the determinants for τ and σ differ by a factor of det DT (u, v).
In fact, these 2 by 2 determinants are precisely the components of the cross products ∂τ
∂u × ∂τ
∂v
and ∂σ
∂s × ∂σ
∂t , so this calculation shows that:
∂τ
∂u × ∂τ
∂v =
∂σ
∂s × ∂σ
∂t


· det DT (u, v),
where, contrary to standard practice, we have written the scalar factor det DT (u, v) on the right.
The two cross products are normal vectors to S. Since we assumed that σ is orientation-preserving,
we conclude that τ is orientation-preserving if det DT (u, v) is positive and orientation-reversing if
it is negative.",Multivariable_Calculus_Shimamoto.pdf
"we conclude that τ is orientation-preserving if det DT (u, v) is positive and orientation-reversing if
it is negative.
We are ready to compare the integrals obtained using the two parametrizations. Beginning
with the expansion (10.24) with τ as parametrization and then using (10.27) and its analogues to
pull out a factor of det DT (u, v), we obtain:
ZZ
τ
F · dS =
ZZ
D∗

(F1 ◦ τ) · det
∂y
∂u
∂y
∂v∂z
∂u
∂z
∂v

+ (F2 ◦ τ) · det
∂z
∂u
∂z
∂v∂x
∂u
∂x
∂v

+ (F3 ◦ τ) · det
∂x
∂u
∂x
∂v∂y
∂u
∂y
∂v

du dv (10.28)
=
ZZ
D∗

(F1 ◦ τ) · det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

+ (F2 ◦ τ) · det
∂z
∂s
∂z
∂t∂x
∂s
∂x
∂t

+ (F3 ◦ τ) · det
∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

· det DT (u, v) du dv. (10.29)
Next, note that det DT (u, v) = ±|det DT (u, v)|, where the sign depends on whether det DT (u, v)
is positive or negative, or equivalently, as noted above, whether τ is orientation-preserving or
orientation-reversing. Keeping in mind that τ = σ ◦ T, we find:
ZZ
τ
F · dS = ±
ZZ
D∗

(F1 ◦ σ ◦ T) · det
∂y
∂s
∂y",Multivariable_Calculus_Shimamoto.pdf
"orientation-reversing. Keeping in mind that τ = σ ◦ T, we find:
ZZ
τ
F · dS = ±
ZZ
D∗

(F1 ◦ σ ◦ T) · det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

+ (F2 ◦ σ ◦ T) · det
∂z
∂s
∂z
∂t∂x
∂s
∂x
∂t

+ (F3 ◦ σ ◦ T) · det
∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

·
det DT (u, v)
du dv.
This last integral is the result of a change of variables. It is an integral over D∗ that is the
pullback of an integral over D using T as the change of variables transformation from the uv-plane
to the st-plane. The integral over D to which the change of variables is applied is:
±
ZZ
D

(F1 ◦ σ) · det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

+ (F2 ◦ σ) · det
∂z
∂s
∂z
∂t∂x
∂s
∂x
∂t

+ (F3 ◦ σ) · det
∂x
∂s
∂x
∂t∂y
∂s
∂y
∂t

ds dt.
Going back to equation (10.24), this last integral is exactly what would be used to calculate the
surface integral using σ as parametrization. That is, it is ±
RR
σ F · dS. We have attained our
objective.
Proposition 10.17. Let σ : D → R3 be a smooth orientation-preserving parametrization of an",Multivariable_Calculus_Shimamoto.pdf
"RR
σ F · dS. We have attained our
objective.
Proposition 10.17. Let σ : D → R3 be a smooth orientation-preserving parametrization of an
oriented surface S in R3. If τ : D∗ → R3 is another smooth parametrization of S, then:
ZZ
τ
F · dS = ±
ZZ
σ
F · dS,",Multivariable_Calculus_Shimamoto.pdf
"264 CHAPTER 10. SURFACE INTEGRALS
where the sign depends on whether τ is orientation-preserving or orientation-reversing.
In particular, in order to compute
RR
S F·dS =
RR
σ F·dS, it does not matter which orientation-
preserving parametrization σ is used.
For integrals of real-valued functions with respect to surface area, the corresponding assertion
is that
RR
σ f dS=
RR
τ f dSfor any two smooth parametrizations σ, τof S. The proof combines the
ideas just given above with those given in the analogous situation for integrals over curves with
respect to arclength. We won’t go through the details. This may be one of those rare occasions
where it is all right to say that the authorities have weighed in and to let them have their way.
One last remark: We commented earlier that the definition of the vector field surface integral
can be regarded as pulling back the integral of the 2–formη = F1 dy∧dz+F2 dz∧dx+F3 dx∧dy to a",Multivariable_Calculus_Shimamoto.pdf
"can be regarded as pulling back the integral of the 2–formη = F1 dy∧dz+F2 dz∧dx+F3 dx∧dy to a
double integral in the parameter plane (see equation (10.23)). Using that language, our calculation
of the independence of parametrization can be summarized in one line:
ZZ
D∗
τ∗(η) =
Z
D∗
(σ ◦ T)∗(η) =
ZZ
D∗
T∗(σ∗(η)) =
ZZ
D
σ∗(η). (10.30)
In other words, using either σ or τ to pull back η gives the same value.
Don’t panic! The first equation follows since T was defined so that τ = σ ◦ T. The second
equation uses the fact that ( σ ◦ T)∗(η) = T∗(σ∗(η)). This is essentially the chain rule-based
calculation that brought out the factor of det DT (u, v) in going from (10.28) to (10.29), though
we would need to develop the properties of pullbacks more carefully to explain this. (To get a
taste of these properties, see Exercises 4.14–4.18 in Chapter 11.) Finally, the step going from D∗
to D uses the pullback form of the change of variables theorem, which is a refinement that takes",Multivariable_Calculus_Shimamoto.pdf
"to D uses the pullback form of the change of variables theorem, which is a refinement that takes
orientation into account and eliminates the absolute value around the determinant (see Exercise
4.20 in Chapter 11).
Though there are gaps that need to be filled in, the intended lesson here is that differential forms,
properly developed, provide a structure in which the arguments can be presented with remarkable
conciseness. Of course, the proper development may involve lengthy calculations similar to the
ones we presented, but their validity is guided by and rooted in basic principles: the chain rule and
change of variables.
10.9 Exercises for Chapter 10
Section 1 What the surface integral measures
1.1. Let S be the hemisphere x2 + y2 + z2 = a2, z ≥ 0, oriented by the outward normal.
(a) Find
RR
S F · dS if F(x, y, z) = (x, y, z).
(b) Find
RR
S F · dS if F(x, y, z) = (y − z, z− x, x− y).
1.2. Let F(x, y, z) = ( z, x, y), and let S be the triangular surface in R3 with vertices (1 , 0, 0),",Multivariable_Calculus_Shimamoto.pdf
"S F · dS if F(x, y, z) = (y − z, z− x, x− y).
1.2. Let F(x, y, z) = ( z, x, y), and let S be the triangular surface in R3 with vertices (1 , 0, 0),
(0, 1, 0), and (0, 0, 1), oriented by the upward normal. Find
RR
S F · dS.
1.3. Let F be a continuous vector field on R3 such that F(x, y, z) is always a scalar multiple of
(0, 0, 1), i.e., F(x, y, z) =

0, 0, F3(x, y, z)


, and let S be the cylinder x2 + y2 = 1, 0 ≤ z ≤ 5,
oriented by the outward normal. Show that
RR
S F · dS = 0.
1.4. Let U be an open set in R3, and let S be a smooth oriented surface in U with orienting normal
vector field n. Let F be a continuous vector field on U. If F(x) = n(x) for all x in S, prove
that
RR
S F · dS = Area (S).",Multivariable_Calculus_Shimamoto.pdf
"10.9. EXERCISES FOR CHAPTER 10 265
1.5. Let F: R3 → R3 be a vector field such that F(−x) = −F(x) for all x in R3, and let S be
the sphere x2 + y2 + z2 = a2, oriented by the outward normal. Is it necessarily true thatRR
S F · dS = 0? Explain.
Section 2 The definition of the surface integral
2.1. Let S be the portion of the plane x + y + z = 3 for which 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Let
F(x, y, z) = (2y, x, z). If S is oriented by the upward normal, find
RR
S F · dS.
2.2. Let F(x, y, z) = (xz, yz, z2).
(a) Let S be the cylinder x2 + y2 = 4, 1 ≤ z ≤ 3, oriented by the outward normal. FindRR
S F · dS.
(b) Suppose that you “close up” the cylinder S in part (a) by adding the disks x2 + y2 ≤ 4,
z = 3, and x2 + y2 ≤ 4, z = 1, at the top and bottom, respectively, where the disk at
the top is oriented by the upward normal and the one at the bottom is oriented by the
downward normal. Let S1 be the resulting surface: cylinder, top, and bottom. FindRR
S1
F · dS.",Multivariable_Calculus_Shimamoto.pdf
"downward normal. Let S1 be the resulting surface: cylinder, top, and bottom. FindRR
S1
F · dS.
2.3. Let S be the helicoid parametrized by:
σ(r, θ) = (r cos θ, rsin θ, θ), 0 ≤ r ≤ 1, 0 ≤ θ ≤ 4π,
oriented by the upward normal. Find
RR
S F · dS if F(x, y, z) = (y, −x, z2).
2.4. Let F(x, y, z) =

x, −z, 2y(x2 − y2)


. Find
RR
S F · dS if S is the saddle surface z = x2 − y2,
where x2 + y2 ≤ 4, oriented by the upward normal.
2.5. Let W be the solid region in R3 that lies inside the cylinders x2 + z2 = 1 and y2 + z2 = 1 and
above the xy-plane. This solid appeared in Example 5.2 of Chapter 5, shown again in Figure
10.27. Let S be the piecewise smooth surface that is the exposed upper part of W, that is,
the four pieces of the cylinders but not the base, oriented by the upward normal.
Find
RR
S F · dS if F is the vector field F(x, y, z) = (0, 0, z).
Figure 10.27: The surface S consisting of portions of the cylinders x2 + z2 = 1 and y2 + z2 = 1",Multivariable_Calculus_Shimamoto.pdf
"266 CHAPTER 10. SURFACE INTEGRALS
Section 3 Stokes’s theorem
3.1. Consider the vector field F(x, y, z) = (2yz, ysin z, 1 + cosz).
(a) Find a vector field G whose curl is F.
(b) Let S be the half-ellipsoid 4 x2 + 4y2 + z2 = 4, z ≥ 0, oriented by the upward normal.
Use Stokes’s theorem to find
RR
S F · dS.
(c) Find
RR
eS F · dS if eS is the portion of the surface z = 1 − x2 − y2 above the xy-plane,
oriented by the upward normal. ( Hint: Take advantage of what you’ve already done.)
3.2. Consider the vector field F(x, y, z) = (ex+y − xey+z, ey+z − ex+y + yez, −ez).
(a) Is F a conservative vector field? Explain.
(b) Find a vector field G = (G1, G2, G3) such that G2 = 0 and the curl of G is F.
(c) Find
RR
S F · dS if S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0, oriented by the outward
normal.
(d) Find
RR
S F · dS if S is the hemisphere x2 + y2 + z2 = 4, z ≤ 0, oriented by the outward
normal.
(e) Find
RR
S F · dS if S is the cylinder x2 + y2 = 4, 0 ≤ z ≤ 4, oriented by the outward
normal.",Multivariable_Calculus_Shimamoto.pdf
"normal.
(e) Find
RR
S F · dS if S is the cylinder x2 + y2 = 4, 0 ≤ z ≤ 4, oriented by the outward
normal.
Exercises 3.3–3.6 illustrate situations where, by bringing in a surface integral, Stokes’s theorem
can be used to obtain information about line integrals that would be hard to find directly.
3.3. Let F(x, y, z) = (xz + yz, 2xz, 1
2 y2).
(a) Find ∇ ×F.
(b) Let C be the curve of intersection in R3 of the cylinder x2 + y2 = 1 and the saddle
surface z = x2 − y2, oriented counterclockwise as viewed from high above the xy-plane,
looking down. Use Stokes’s theorem to find the line integral
R
C F · ds. ( Hint: Find a
surface whose boundary is C.)
3.4. Let P be a plane in R3 described by the equation ax + by + cz = d. Let C be a piecewise
smooth oriented simple closed curve that lies in P, and let S be the subset of P that is
enclosed by C, i.e., the region of P “inside” C. Show that the area of S is equal to the
absolute value of:
1
2
√
a2 + b2 + c2
Z
C",Multivariable_Calculus_Shimamoto.pdf
"enclosed by C, i.e., the region of P “inside” C. Show that the area of S is equal to the
absolute value of:
1
2
√
a2 + b2 + c2
Z
C
(bz − cy) dx + (cx − az) dy + (ay − bx) dz.
3.5. (a) Find an example of a smooth vector field F = (F1, F2, F3) defined on an open set U of
R3 such that:
• its mixed partials are equal, i.e., ∂F1
∂y = ∂F2
∂x , ∂F1
∂z = ∂F3
∂x , and ∂F2
∂z = ∂F3
∂y , and
• there exists a piecewise smooth oriented simple closed curve C in U such thatR
C F1 dx + F2 dy + F3 dz ̸= 0.
(b) On the other hand, show that, if F is any smooth vector field on an open set U in R3
whose mixed partials are equal, then
R
C F1 dx + F2 dy + F3 dz = 0 for any piecewise
smooth oriented simple closed curve C that is the boundary of an oriented surface S
contained in U.",Multivariable_Calculus_Shimamoto.pdf
"10.9. EXERCISES FOR CHAPTER 10 267
(c) What does part (b) tell you about the curve C you found in part (a)?
3.6. Let C1 and C2 be piecewise smooth simple closed curves contained in the cylinder x2 +y2 = 1
that do not intersect one another, both oriented counterclockwise when viewed from high
above the xy-plane, looking down, as illustrated in Figure 10.28. If F is the vector field
F(x, y, z) = (z − 2y3, 2yz + x5, y2 + x), show that
R
C1
F · ds =
R
C2
F · ds.
Figure 10.28: Two nonintersecting counterclockwise curves C1 and C2 in the cylinder x2 + y2 = 1
Section 4 Curl fields
In Exercises 4.1–4.3, determine whether the given vector field F is a curl field. If it is, find a
vector field G whose curl is F. If not, explain why not.
4.1. F (x, y, z) = (x, y, z)
4.2. F (x, y, z) = (y, z, x)
4.3. F (x, y, z) = (x + 2y + 3z, x4 + y5 + z6, x7y8z9)
4.4. This exercise ties up some loose ends left dangling after Example 10.3 back in Section 10.1.",Multivariable_Calculus_Shimamoto.pdf
"4.4. This exercise ties up some loose ends left dangling after Example 10.3 back in Section 10.1.
Let S1 be the hemisphere x2 + y2 + z2 = a2, z ≥ 0, oriented by the outward normal, and let
S2 be the disk x2 + y2 ≤ a2, z = 0, oriented by the upward normal.
Let F be the vector field F(x, y, z) = (0, 0, −1).
(a) Show that F is a curl field by finding a vector field G whose curl is F.
(b) In Example 10.3, we showed that it is fairly easy to calculate that
RR
S2
F · dS = −πa2.
Without any further calculation, explain why
RR
S1
F · dS = −πa2 as well.
4.5. Let F and G be vector fields on R3.
(a) True or false: If G = F + C for some constant vector C, then F and G have the same
curl. Either prove that the statement is true, or find an example in which it is false.
(b) True or false: If F and G have the same curl, then G = F + C for some constant vector
C. Either prove that the statement is true, or find an example in which it is false.",Multivariable_Calculus_Shimamoto.pdf
"C. Either prove that the statement is true, or find an example in which it is false.
4.6. Does there exist a vector field F on R3, other than the constant vector field F = 0, with the
property that:
•
R
C F · ds = 0 for every piecewise smooth oriented closed curve C in R3 and
•
RR
S F · dS = 0 for every piecewise smooth oriented closed surface S in R3?
If so, find such a vector field, and justify why it works. If no such vector field exists, explain
why not.",Multivariable_Calculus_Shimamoto.pdf
"268 CHAPTER 10. SURFACE INTEGRALS
Section 5 Gauss’s theorem
5.1. Let F(x, y, z) = ( x3, y3, z3). Use Gauss’s theorem to find
RR
S F · dS if S is the sphere x2 +
y2 + z2 = 4, oriented by the outward normal.
5.2. Find
RR
S F · dS if F(x, y, z) = ( x3y3z4, x2y4z4, x2y3z5) and S is the boundary of the cube
W = [0, 1] × [0, 1] × [0, 1], oriented by the normal pointing away from W.
5.3. Let S be the silo-shaped closed surface consisting of:
• the cylinder x2 + y2 = 4, 0 ≤ z ≤ 3,
• a hemispherical cap of radius 2 centered at (0 , 0, 3), i.e., x2 + y2 + (z − 3)2 = 4, z ≥ 3,
and
• the disk x2 + y2 ≤ 4, z = 0, at the base,
all oriented by the normal pointing away from the silo. Find
RR
S F · dS if F(x, y, z) =
(3xz2, 2y, x+ y2 − z3).
5.4. One way to produce a torus is to take the circle C in the xz-plane with center ( a, 0, 0) and
radius b and rotate it about the z-axis. We assume that a > b. The surface that is swept out",Multivariable_Calculus_Shimamoto.pdf
"radius b and rotate it about the z-axis. We assume that a > b. The surface that is swept out
is a torus. The circle C can be parametrized by α(ψ) = (a + b cos ψ, 0, bsin ψ), 0 ≤ ψ ≤ 2π.
The distance of α(ψ) to the z-axis is a+b cos ψ, so rotating α(ψ) counterclockwise by θ about
the z-axis brings it to the point

(a + b cos ψ) cosθ, (a + b cos ψ) sinθ, bsin ψ). This gives the
Figure 10.29: Parametrizing a torus
following parametrization of the torus with ψ and θ as parameters, illustrated in Figure 10.29:
σ(ψ, θ) =

(a + b cos ψ) cosθ, (a + b cos ψ) sinθ, bsin ψ


, 0 ≤ ψ ≤ 2π, 0 ≤ θ ≤ 2π.
Let S denote the resulting torus, and assume that it is oriented by the unit normal vector
that points towards the exterior of the torus.
(a) Show that:
∂σ
∂ψ × ∂σ
∂θ = −b(a + b cos ψ)

cos ψ cos θ, cos ψ sin θ, sin ψ

.",Multivariable_Calculus_Shimamoto.pdf
"10.9. EXERCISES FOR CHAPTER 10 269
(b) Is σ orientation-preserving or orientation-reversing?
(c) Find the surface area of the torus.
(d) Let F(x, y, z) = (x, y, z). Use the parametrization of S and the definition of the surface
integral to find
RR
S F · dS. Then, use your answer and the result of Example 10.16 to
find the volume of the solid torus, that is, the three-dimensional region enclosed by the
torus.
(e) Find the integral of the vector field F(x, y, z) = (0, 0, 1) over S. Use whatever method
seems best. What does your answer say about the flow of F through S?
5.5. We describe an alternative approach to Exercise 4.4. Let F(x, y, z) = (0 , 0, −1). The main
point of the aforementioned exercise is that:
ZZ
S1
F · dS =
ZZ
S2
F · dS, (10.31)
where S1 is the hemisphere x2 + y2 + z2 = a2, z ≥ 0, oriented by the outward normal, and
S2 is the disk x2 + y2 ≤ a2, z = 0, oriented by the upward normal. The argument was based
on Stokes’s theorem.",Multivariable_Calculus_Shimamoto.pdf
"S2 is the disk x2 + y2 ≤ a2, z = 0, oriented by the upward normal. The argument was based
on Stokes’s theorem.
Show that the same conclusion (10.31) can be reached using Gauss’s theorem.
5.6. A bounded solid W in R3 is contained in the region a ≤ z ≤ b, where a and b are constant
and a < b. The boundary of W is the union of three parts, as shown in Figure 10.30, left:
• a top surface T that is a subset of the plane z = b,
• a bottom surface B that is a subset of the plane z = a, and
• a surface S in between that connects the bottom and the top.
Let F(x, y, z) = (0, 0, 1). If the third piece above, S, is oriented by the normal pointing away
from W, express
RR
S F · dS in terms of Area (T) and Area (B).
T
S
B
x2+y2=9,z=2
n–
x² + y² = 4,  z = 8
n
Figure 10.30: A solid region W lying between the planes z = a and z = b whose boundary consists
of a surface S, a top T, and a bottom B (left) and a lumpy shopping bag of volume 75 (right)",Multivariable_Calculus_Shimamoto.pdf
"of a surface S, a top T, and a bottom B (left) and a lumpy shopping bag of volume 75 (right)
5.7. A shopping bag is modeled as a lumpy surface S having an open circular top. Suppose that
the bag is placed in R3 in such a way that the boundary of the opening is the circlex2 +y2 = 4
in the plane z = 8. See Figure 10.30, right. Assume that, when filled exactly to the brim, the
bag has a volume of 75.
Let F(x, y, z) = (x, y, z). If S is oriented by the outward normal, use Gauss’s theorem to findRR
S F · dS. (Note that S is not a closed surface.)",Multivariable_Calculus_Shimamoto.pdf
"270 CHAPTER 10. SURFACE INTEGRALS
Let U be an open set in R3. A smooth real-valued function h: U → R is called harmonic on
U if:
∂2h
∂x2 + ∂2h
∂y2 + ∂2h
∂z2 = 0 at all points of U.
Exercises 5.8–5.10 concern harmonic functions. The results are the analogues of Exercises 4.12–4.15
in Chapter 9, which were about harmonic functions of two variables.
5.8. Show that h(x, y, z) = e−5x sin(3y + 4z) is a harmonic function on R3.
5.9. Let W be a bounded subset of an open set U in R3 whose boundary ∂W consists of a finite
number of piecewise smooth closed surfaces. Assume that every pair of points in W can be
connected by a piecewise smooth curve inW. If h is harmonic on U and if h = 0 at all points of
∂W , prove thath = 0 on all ofW. (Hint: Consider the vector fieldF = h ∇h = h(∂h
∂x , ∂h
∂y , ∂h
∂z ).)
5.10. Let U and W be as in the preceding exercise. Show that, if h1 and h2 are harmonic on U
and if h1 = h2 at all points of ∂W , then h1 = h2 on all of W. Thus a harmonic function is",Multivariable_Calculus_Shimamoto.pdf
"and if h1 = h2 at all points of ∂W , then h1 = h2 on all of W. Thus a harmonic function is
determined on W by its values on the boundary.
Section 6 The inverse square field
6.1. Let Sa denote the sphere x2 + y2 + z2 = a2 of radius a, oriented by the outward normal. If
we did not already know the answer, it might be tempting to calculate the integral of the
inverse square field G(x, y, z) =

− x
(x2+y2+z2)3/2 , − y
(x2+y2+z2)3/2 , − z
(x2+y2+z2)3/2


over Sa by
filling in the sphere with the solid ball Wa given by x2 + y2 + z2 ≤ a2 and using Gauss’s
theorem. Unfortunately, this is not valid (and would give an incorrect answer), because Wa
contains the origin and the origin is not in the domain of G. On Sa, however, G agrees with
the vector field F(x, y, z) =

− x
a3 , − y
a3 , − z
a3


= − 1
a3 (x, y, z), and F is defined on all of R3.
Find
RR
Sa
G · dS by replacing G with F and applying Gauss’s theorem.",Multivariable_Calculus_Shimamoto.pdf
"
− x
a3 , − y
a3 , − z
a3


= − 1
a3 (x, y, z), and F is defined on all of R3.
Find
RR
Sa
G · dS by replacing G with F and applying Gauss’s theorem.
6.2. Let F be a vector field defined on all of R3, except at the two points p = (2 , 0, 0) and
q = (−2, 0, 0). Let S1, S2, and S be the following spheres, centered at (2, 0, 0), (−2, 0, 0), and
(0, 0, 0), respectively, each oriented by the outward normal.
S1: ( x − 2)2 + y2 + z2 = 1
S2: ( x + 2)2 + y2 + z2 = 1
S: x2 + y2 + z2 = 25
Assume that ∇ ·F = 0. If
RR
S1
F · dS = 5 and
RR
S2
F · dS = 6, what is
RR
S F · dS?
6.3. Give the details of the justification of the instance of Gauss’s law cited in the text. That is,
if S is a piecewise smooth closed surface in R3 − {p1, p2, . . . ,pk}, oriented by the outward
normal, and F(x) = Pk
i=1 G(x − pi) is the vector field of (10.18), show that:
− 1
4π
ZZ
S
F · dS equals the number of the points p1, p2, . . . ,pk that lie in the interior of S.",Multivariable_Calculus_Shimamoto.pdf
"− 1
4π
ZZ
S
F · dS equals the number of the points p1, p2, . . . ,pk that lie in the interior of S.
6.4. Suppose that objects of mass m1, m2, . . . , mk, not necessarily equal, are located at the points
p1, p2, . . . ,pk, respectively, in R3. The gravitational force that they exert on another object
whose position is x is described by the vector field:
F(x) =
kX
i=1
mi G(x − pi),",Multivariable_Calculus_Shimamoto.pdf
"10.9. EXERCISES FOR CHAPTER 10 271
where G is our usual inverse square field (10.16). LetS be a piecewise smooth closed surface in
R3−{p1, p2, . . . ,pk}, oriented by the outward normal. What can you say about− 1
4π
RR
S F·dS
in this case?
Section 7 A more substitution-friendly notation for surface integrals
7.1. Let S be the unit sphere x2 + y2 + z2 = 1, oriented by the outward normal. Consider the
surface integral: ZZ
S
xy dy∧ dz + yz dz∧ dx + xz dx∧ dy.
(a) What is the vector field that is being integrated?
(b) Evaluate the integral.
7.2. Suppose that a surface S in R3 is the graph z = f(x, y) of a smooth real-valued function
f : D → R whose domain D is a bounded subset of R2. Let S be oriented by the upward
normal. Show that: ZZ
S
1 dx ∧ dy = Area (D).
In other words, the integral is the area of the projection of S on the xy-plane. Similar
interpretations hold for
RR
S 1 dy ∧ dz and
RR
S 1 dz ∧ dx under analogous hypotheses.
Section 8 Independence of parametrization",Multivariable_Calculus_Shimamoto.pdf
"interpretations hold for
RR
S 1 dy ∧ dz and
RR
S 1 dz ∧ dx under analogous hypotheses.
Section 8 Independence of parametrization
8.1. Let S be the portion of the cone z =
p
x2 + y2 in R3 where 0 ≤ z ≤ 2. In Section 5.4, we
showed that S could be parametrized in more than one way. One possibility is based on using
polar/cylindrical coordinates as parameters (Example 5.14):
σ : [0, 2] × [0, 2π] → R3, σ (s, t) = (s cos t, ssin t, s).
Another is based on spherical coordinates (Example 5.17):
τ : [0, 2
√
2] × [0, 2π] → R3, τ (u, v) =
√
2
2 u cos v,
√
2
2 u sin v,
√
2
2 u


.
Find a formula for a function T : [0, 2
√
2] × [0, 2π] → [0, 2] × [0, 2π] that satisfies τ(u, v) =
σ

T(u, v)


, as in equation (10.25).",Multivariable_Calculus_Shimamoto.pdf
272 CHAPTER 10. SURFACE INTEGRALS,Multivariable_Calculus_Shimamoto.pdf
"Chapter 11
Working with differential forms
This final chapter is a whirlwind survey of how to work with differential forms. We focus particularly
on what differential forms have to say about some of the main theorems we have learned recently.
In what follows, all functions, including vector fields, are assumed to be smooth.
Four theorems, version 1
1. Conservative vector fields. Let C be a piecewise smooth oriented curve in Rn from p
to q. If F = ∇f is a conservative vector field on an open set containing C, then:
Z
C
∇f · ds = f(q) − f(p). (C)
2. Green’s theorem. Let D be a bounded subset of R2. If F = (F1, F2) is a vector field
on an open set containing D, then:
ZZ
D
∂F2
∂x − ∂F1
∂y


dx dy=
Z
∂D
F · ds. (Gr)
3. Stokes’s theorem. Let S be a piecewise smooth oriented surface in R3. If F =
(F1, F2, F3) is a vector field on an open set containing S, then:
ZZ
S
(∇ ×F) · dS =
Z
∂S
F · ds. (S)
4. Gauss’s theorem. Let W be a bounded subset of R3. If F = (F1, F2, F3) is a vector",Multivariable_Calculus_Shimamoto.pdf
"ZZ
S
(∇ ×F) · dS =
Z
∂S
F · ds. (S)
4. Gauss’s theorem. Let W be a bounded subset of R3. If F = (F1, F2, F3) is a vector
field on an open set containing W, then:
ZZZ
W
(∇ ·F) dx dy dz=
ZZ
∂W
F · dS. (Ga)
Our goal is to present just enough of the properties of differential forms to indicate how forms
enter the picture as far as these theorems are concerned. Further information is developed in the
exercises.
273",Multivariable_Calculus_Shimamoto.pdf
"274 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS
11.1 Integrals of differential forms
Roughly speaking, a differential k-form is an expression that can be integrated over an oriented
k-dimensional domain. For instance, in Rn, here are the cases at opposite ends of the range of
dimensions.
• n-forms. In Rn with coordinates x1, x2, . . . , xn, an n-form is an expression:
f dx1 ∧ dx2 ∧ ··· ∧dxn,
where f is a real-valued function of x = (x1, x2, . . . , xn). It could also be written f ∧ dx1 ∧
dx2 ∧ ··· ∧dxn, though f is treated commonly as a scalar factor. If W is an n-dimensional
bounded subset of Rn, which by default we say has the “positive” orientation, then the integralR
W f dx1 ∧ dx2 ∧ ··· ∧dxn is defined to be the usual Riemann integral:
Z
W
f dx1 ∧ dx2 ∧ ··· ∧dxn =
ZZ
···
Z
W
f(x1, x2, . . . , xn) dx1 dx2 ··· dxn. (11.1)
As indicated on the left of equation (11.1), we won’t use multiple integral signs with differential",Multivariable_Calculus_Shimamoto.pdf
"Z
W
f(x1, x2, . . . , xn) dx1 dx2 ··· dxn. (11.1)
As indicated on the left of equation (11.1), we won’t use multiple integral signs with differential
forms. The dimension of the integral can be inferred from the type of form that is being
integrated.
• 0-forms. A 0-form is simply another name for a real-valued functionf. 0-forms are integrated
over zero-dimensional domains, i.e., individual points, by evaluating the function at that
point and where we orient a point by attaching a + or − sign. So
R
+{p} f = f(p) andR
−{p} f = −f(p). If C is an oriented curve from p to q, we define its oriented boundary to
be the oriented points ∂C = (+{q}) ∪ (−{p}).
We restrict our attention here to differential forms on subsets of R3 (or R or R2, depending on
the context), real-valued functions f of three variables (or one or two), and vector fields on subsets
of R3 (or R or R2), since these are the cases of interest in the four big theorems listed above.",Multivariable_Calculus_Shimamoto.pdf
"of R3 (or R or R2), since these are the cases of interest in the four big theorems listed above.
Table 11.1 below gives the notation for 0, 1, 2, and 3-forms and their associated integrals. The
integrals that the notation stands for, shown in the last column, are the kinds of integrals we have
been studying: 1-forms integrate as line integrals, 2-forms as surface integrals, and 3-forms as triple
integrals.
Type of Domain of Notation for integral Meaning
form integration using forms of integral
0-form oriented points
R
+{p} f or
R
−{p} f = f(p) or −f(p)
1-form oriented curves
R
C F1 dx + F2 dy + F3 dz =
R
C F · ds
2-form oriented surfaces
R
S F1 dy ∧ dz + F2 dz ∧ dx
+F3 dx ∧ dy =
RR
S F · dS
3-form 3-dim solids
R
W f dx∧ dy ∧ dz =
RRR
W f dx dy dz
Table 11.1: Integrals of differential forms
In this notation, the four theorems (C), (Gr), (S), and (Ga) translate as integrals of forms as
follows.",Multivariable_Calculus_Shimamoto.pdf
"11.2. DERIVATIVES OF DIFFERENTIAL FORMS 275
Four theorems, version 2
1. Conservative vector fields.
Z
C
∂f
∂x dx + ∂f
∂y dy + ∂f
∂z dz = f(q) − f(p) =
Z
∂C
f (C)
2. Green’s theorem.
Z
D
∂F2
∂x − ∂F1
∂y

dx ∧ dy =
Z
∂D
F1 dx + F2 dy (Gr)
3. Stokes’s theorem.
Z
S
∂F3
∂y − ∂F2
∂z

dy ∧ dz +
∂F1
∂z − ∂F3
∂x

dz ∧ dx +
∂F2
∂x − ∂F1
∂y

dx ∧ dy
=
Z
∂S
F1 dx + F2 dy + F3 dz
(S)
4. Gauss’s theorem.
Z
W
∂F1
∂x + ∂F2
∂y + ∂F3
∂z

dx ∧ dy ∧ dz =
Z
∂W
F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy (Ga)
11.2 Derivatives of differential forms
In this brief foray, we focus on how differential forms behave without worrying about exactly what
they are. Here are some simple rules for working with forms.
1. Arithmetic. These are the rules we encountered earlier that reflect a connection between
differential forms and determinants (see the end of Section 10.7):
(a) dy ∧ dx = −dx ∧ dy, dz ∧ dy = −dy ∧ dz, dx ∧ dz = −dz ∧ dx,
(b) dx ∧ dx = dy ∧ dy = dz ∧ dz = 0.",Multivariable_Calculus_Shimamoto.pdf
"(a) dy ∧ dx = −dx ∧ dy, dz ∧ dy = −dy ∧ dz, dx ∧ dz = −dz ∧ dx,
(b) dx ∧ dx = dy ∧ dy = dz ∧ dz = 0.
These rules have the effect of simplifying what differential forms can look like. For instance, a
2-form should be a generalized double integrand. In R3, such an expression would have the form:
f1 dx ∧ dx + f2 dx ∧ dy + f3 dx ∧ dz + f4 dy ∧ dx + ··· + f9 dz ∧ dz,
where f1, f2, . . . , f9 are real-valued functions of x, y, z. According to the rules, however, three of the
nine terms—the ones involving dx∧dx, dy∧dy, and dz∧dz—are zero and therefore can be omitted.
Similarly, the remaining six terms can be combined in pairs: for instance, f2 dx ∧dy + f4 dy ∧dx =
f2 dx ∧ dy − f4 dx ∧ dy = (f2 − f4) dx ∧ dy. Hence we may assume that any 2-form on a subset of
R3 is a sum of three terms having the form η = F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy.
2. Differentiation. In general, the derivative of a k-form is a ( k + 1)-form. Specifically, for
differential forms on subsets of R3, we define:",Multivariable_Calculus_Shimamoto.pdf
"276 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS
(a) For 0-forms:
d f= ∂f
∂x dx + ∂f
∂y dy + ∂f
∂z dz.
(b) For 1-forms:
d(F1 dx + F2 dy + F3 dz) = dF1 ∧ dx + dF2 ∧ dy + dF3 ∧ dz.
(c) For 2-forms:
d(F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy) = dF1 ∧ dy ∧ dz + dF2 ∧ dz ∧ dx + dF3 ∧ dx ∧ dy.
In (b) and (c), F1, F2, and F3 are 0-forms, so their derivatives are as defined in (a).
Example 11.1. Let x = r cos θ and y = r sin θ. These are real-valued functions of r and θ—in
other words, they are 0-forms—so by definition:
dx = ∂x
∂r dr + ∂x
∂θ dθ = cos θ dr− r sin θ dθ
and dy = ∂y
∂r dr + ∂y
∂θ dθ = sin θ dr+ r cos θ dθ.
Then, by the arithmetic rules:
dx ∧ dy = (cos θ dr− r sin θ dθ) ∧ (sin θ dr+ r cos θ dθ)
= cos θ sin θ dr∧ dr + r cos2 θ dr∧ dθ − r sin2 θ dθ∧ dr − r2 sin θ cos θ dθ∧ dθ
= cos θ sin θ · 0 + r cos2 θ dr∧ dθ − r sin2 θ · (−dr ∧ dθ) − r2 sin θ cos θ · 0
= (r cos2 θ + r sin2 θ) dr ∧ dθ
= r dr∧ dθ.",Multivariable_Calculus_Shimamoto.pdf
"= cos θ sin θ · 0 + r cos2 θ dr∧ dθ − r sin2 θ · (−dr ∧ dθ) − r2 sin θ cos θ · 0
= (r cos2 θ + r sin2 θ) dr ∧ dθ
= r dr∧ dθ.
Example 11.2. Let σ(s, t) = (x(s, t), y(s, t), z(s, t)) be a smooth parametrization of a surface S.
Then x, y, and z are each real-valued functions of s and t, so, for example:
dy = ∂y
∂s ds + ∂y
∂t dt and dz = ∂z
∂s ds + ∂z
∂t dt.
It follows that:
dy ∧ dz = (∂y
∂s ds + ∂y
∂t dt) ∧ (∂z
∂s ds + ∂z
∂t dt)
= 0 + ∂y
∂s
∂z
∂t ds ∧ dt + ∂y
∂t
∂z
∂s dt ∧ ds + 0
=
∂y
∂s
∂z
∂t − ∂y
∂t
∂z
∂s


ds ∧ dt
= det
∂y
∂s
∂y
∂t∂z
∂s
∂z
∂t

ds ∧ dt.
This is the same as the expression for dy ∧dz that we obtained when we discussed the substitutions
used to compute surface integrals, as in equation (10.21) of Chapter 10. Perhaps the calculation
given here is a more organic way of obtaining the substitution. Either way, the formula is consistent
with what the change of variables theorem says to do to pull back an integral with respect to y and
z to one with respect to s and t.",Multivariable_Calculus_Shimamoto.pdf
"11.2. DERIVATIVES OF DIFFERENTIAL FORMS 277
Next, we derive formulas for the derivatives of 1 and 2-forms.
Example 11.3. For a 1-form ω = F1 dx + F2 dy + F3 dz, we find dω by calculating d(F1 dx),
d(F2 dy), and d(F3 dz) separately. For instance, by definition:
d(F1 dx) = dF1 ∧ dx
=
∂F1
∂x dx + ∂F1
∂y dy + ∂F1
∂z dz


∧ dx
= ∂F1
∂x dx ∧ dx + ∂F1
∂y dy ∧ dx + ∂F1
∂z dz ∧ dx
= 0 − ∂F1
∂y dx ∧ dy + ∂F1
∂z dz ∧ dx
= −∂F1
∂y dx ∧ dy + ∂F1
∂z dz ∧ dx.
Similarly:
d(F2 dy) = ∂F2
∂x dx ∧ dy − ∂F2
∂z dy ∧ dz
and d(F3 dz) = −∂F3
∂x dz ∧ dx + ∂F3
∂y dy ∧ dz.
Adding these calculations gives:
d(F1 dx + F2 dy + F3 dz) =
∂F3
∂y − ∂F2
∂z


dy ∧ dz
+
∂F1
∂z − ∂F3
∂x


dz ∧ dx +
∂F2
∂x − ∂F1
∂y


dx ∧ dy,
a formula that is reproduced in row 2(b) of Table 11.2 below.
Example 11.4. For the derivative of a 2-formη = F1 dy∧dz +F2 dz ∧dx+F3 dx∧dy, we compute,
for example:
d(F2 dz ∧ dx) = dF2 ∧ dz ∧ dx
=
∂F2
∂x dx + ∂F2
∂y dy + ∂F2
∂z dz


∧ dz ∧ dx
= ∂F2
∂x dx ∧ dz ∧ dx + ∂F2
∂y dy ∧ dz ∧ dx + ∂F2",Multivariable_Calculus_Shimamoto.pdf
"for example:
d(F2 dz ∧ dx) = dF2 ∧ dz ∧ dx
=
∂F2
∂x dx + ∂F2
∂y dy + ∂F2
∂z dz


∧ dz ∧ dx
= ∂F2
∂x dx ∧ dz ∧ dx + ∂F2
∂y dy ∧ dz ∧ dx + ∂F2
∂z dz ∧ dz ∧ dx
= −∂F2
∂x dx ∧ dx ∧ dz − ∂F2
∂y dy ∧ dx ∧ dz + 0
= 0 + ∂F2
∂y dx ∧ dy ∧ dz
= ∂F2
∂y dx ∧ dy ∧ dz.
Likewise:
d(F1 dy ∧ dz) = ∂F1
∂x dx ∧ dy ∧ dz and d(F3 dx ∧ dy) = ∂F3
∂z dx ∧ dy ∧ dz.",Multivariable_Calculus_Shimamoto.pdf
"278 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS
Summing these calculations gives:
d(F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy) =
∂F1
∂x + ∂F2
∂y + ∂F3
∂z


dx ∧ dy ∧ dz,
as in row 3 of Table 11.2.
We collect the results in a table. Below it, for future reference, we reproduce the most recent
version of our four big theorems (version 2).
1. 0-forms
(a) f = f(x) d f= f′(x) dx
(b) f = f(x, y) d f= ∂f
∂x dx + ∂f
∂y dy
(c) f = f(x, y, z) d f= ∂f
∂x dx + ∂f
∂y dy + ∂f
∂z dz
2. 1-forms
(a) ω = F1(x, y) dx + F2(x, y) dy dω =
∂F2
∂x − ∂F1
∂y


dx ∧ dy
(b) ω = F1(x, y, z) dx + F2(x, y, z) dy dω =
∂F3
∂y − ∂F2
∂z


dy ∧ dz
+F3(x, y, z) dz +
∂F1
∂z − ∂F3
∂x


dz ∧ dx +
∂F2
∂x − ∂F1
∂y


dx ∧ dy
3. 2-forms
η = F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy dη =
∂F1
∂x + ∂F2
∂y + ∂F3
∂z


dx ∧ dy ∧ dz
Table 11.2: Formulas for derivatives
Four theorems, version 2 (again)
1. Conservative vector fields.
Z
C
∂f
∂x dx + ∂f
∂y dy + ∂f
∂z dz = f(q) − f(p) =
Z
∂C
f (C)
2. Green’s theorem.
Z
D
∂F2
∂x − ∂F1
∂y
",Multivariable_Calculus_Shimamoto.pdf
"1. Conservative vector fields.
Z
C
∂f
∂x dx + ∂f
∂y dy + ∂f
∂z dz = f(q) − f(p) =
Z
∂C
f (C)
2. Green’s theorem.
Z
D
∂F2
∂x − ∂F1
∂y

dx ∧ dy =
Z
∂D
F1 dx + F2 dy (Gr)
3. Stokes’s theorem.
Z
S
∂F3
∂y − ∂F2
∂z

dy ∧ dz +
∂F1
∂z − ∂F3
∂x

dz ∧ dx +
∂F2
∂x − ∂F1
∂y

dx ∧ dy
=
Z
∂S
F1 dx + F2 dy + F3 dz
(S)
4. Gauss’s theorem.
Z
W
∂F1
∂x + ∂F2
∂y + ∂F3
∂z

dx ∧ dy ∧ dz =
Z
∂W
F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy (Ga)",Multivariable_Calculus_Shimamoto.pdf
"11.3. A LOOK BACK AT THE THEOREMS OF MULTIVARIABLE CALCULUS 279
11.3 A look back at the theorems of multivariable calculus
Take a moment to compare the formulas in Table 11.2 with the four theorems. It soon becomes
apparent that, in each theorem, a differential form and its derivative appear. In this new framework,
the theorems can be written in spectacularly streamlined form.
Four theorems, final version
1. Conservative vector fields. If f is a real-valued function, then:
Z
C
d f=
Z
∂C
f. (C)
2. Green’s theorem. If ω = F1 dx + F2 dy, then:
Z
D
dω =
Z
∂D
ω. (Gr)
3. Stokes’s theorem. If ω = F1 dx + F2 dy + F3 dz, then:
Z
S
dω =
Z
∂S
ω. (S)
4. Gauss’s theorem. If η = F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy, then:
Z
W
dη =
Z
∂W
η. (Ga)
They’re all the same theorem! But there’s even more to it, for suppose we go back even further,
to first-year calculus and a real-valued function f(x) of one variable. The fundamental theorem of
calculus says that
Rb",Multivariable_Calculus_Shimamoto.pdf
"to first-year calculus and a real-valued function f(x) of one variable. The fundamental theorem of
calculus says that
Rb
a f′(x) dx = f(b) − f(a). Denoting the interval [ a, b] by I and converting to
differential form notation, this becomes:
Z
I
d f=
Z
∂I
f.
In other words, the four theorems (C), (Gr), (S), and (Ga) of multivariable calculus are general-
izations of something that we’ve known for a long time.
The pattern continues in higher dimensions and with more variables. One studies subsets of Rn
that are k-dimensional in the sense that they can be traced out using k independent parameters.
In other words, they are the images of smooth functions of the form ψ : D → Rn, where D ⊂ Rk.
These objects are called k-dimensional manifolds. Differential k-forms can be integrated over
them, or at least over the ones for which there is a notion of orientation.
A typical k-form in Rn is a k-dimensional integrand, that is, a sum of terms of the form",Multivariable_Calculus_Shimamoto.pdf
"A typical k-form in Rn is a k-dimensional integrand, that is, a sum of terms of the form
f dxi1 ∧ dxi2 ∧ ··· ∧dxik , where f is a real-valued function of n variables x1, x2, . . . , xn. By the
arithmetic rules for differential forms, we may assume that the subscripts i1, i2, . . . , ik are distinct
and in fact that they are arranged, say in increasing order i1 < i2 < ··· < ik. If ζ is a k-form
and M is an oriented k-dimensional manifold in Rn with an orientation-preserving parametrization",Multivariable_Calculus_Shimamoto.pdf
"280 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS
ψ : D → Rn as above, then the integral
R
M ζ is defined using substitution and the rules of forms to
pull the integral back to an ordinary Riemann integral over the k-dimensional parameter domain
D in Rk. (See the remarks before Exercise 4.14 at the end of the chapter for more about the
pullback process.) In the end, the relation between the integral and its pullback is summarized by
the formula: Z
M
ζ =
Z
D
ψ∗(ζ). (11.2)
That the value of the integral does not depend on the parametrization rests ultimately on the
change of variables theorem and the chain rule. In fact, you have seen the proof, at least in outline
form. It is (10.30).
Perhaps it is worth reinforcing the point that equation (11.2) agrees with the expressions for
line integrals and surface integrals that we obtained in the previous two chapters. See (9.6) and
(10.23). In particular, line and surface integrals fit in as two special cases of a general type of",Multivariable_Calculus_Shimamoto.pdf
"(10.23). In particular, line and surface integrals fit in as two special cases of a general type of
integral, something that may not have been clear from the original motivations in terms of tangent
and normal components of vector fields.
The general theory has a striking coherence and consistency. The culmination may be a theorem
that describes how information over the entire manifold is related to behavior along the boundary.
The boundary of a k-dimensional manifold M, denoted by ∂M , is (k − 1)-dimensional, as was the
case with curves (k = 1) and surfaces (k = 2). If ω is a differential (k −1)-form, then this theorem,
which is known as Stokes’s theorem, states that:
R
M dω =
R
∂M ω.
It’s the fundamental theorem of calculus. !
11.4 Exercises for Chapter 11
In Exercises 4.1–4.5, calculate the indicated product.
4.1. (x dx+ y dy) ∧ (−y dx+ x dy)
4.2. (x dx+ y dy) ∧ (x dx− y dy)
4.3. (x dx+ y dy+ z dz) ∧ (y dx+ z dy+ x dz)
4.4. (−y dx+ x dy+ z dz) ∧ (yz dy∧ dz + xz dz∧ dx + xy dx∧ dy)",Multivariable_Calculus_Shimamoto.pdf
"4.2. (x dx+ y dy) ∧ (x dx− y dy)
4.3. (x dx+ y dy+ z dz) ∧ (y dx+ z dy+ x dz)
4.4. (−y dx+ x dy+ z dz) ∧ (yz dy∧ dz + xz dz∧ dx + xy dx∧ dy)
4.5. (x2 dy ∧ dz) ∧ (yz dy∧ dz + xz dz∧ dx + xy dx∧ dy)
4.6. (a) If ω = F1 dx1 + F2 dx2 + ··· + Fn dxn is a 1-form on Rn, show that ω ∧ ω = 0.
(b) Find an example of a 2-form η on R4 such that η ∧ η ̸= 0.
In Exercises 4.7–4.11, find the derivative of the given differential form.
4.7. f = x2 + y2
4.8. f = x sin y cos z
4.9. ω = (x2 − y2) dx + 2xy dy",Multivariable_Calculus_Shimamoto.pdf
"11.4. EXERCISES FOR CHAPTER 11 281
4.10. ω = 2xy3z4 dx + 3x2y2z4 dy + 4x2y3z3 dz
4.11. η = (x + 2y + 3z) dy ∧ dz + exyz dz ∧ dx + x4y5 dx ∧ dy
4.12. Let U be an open set in R3.
(a) Let f = f(x, y, z) be a 0-form (= real-valued function) on U. Find d(d f).
(b) Let ω = F1 dx + F2 dy + F3 dz be a 1-form on U. Find d(dω).
(c) To what facts about vector fields do the results of parts (a) and (b) correspond?
4.13. A typical 3-form on R4 has the form:
ζ = F1 dx2 ∧ dx3 ∧ dx4 + F2 dx1 ∧ dx3 ∧ dx4 + F3 dx1 ∧ dx2 ∧ dx4 + F4 dx1 ∧ dx2 ∧ dx3,
where F1, F2, F3, F4 are real-valued functions of x1, x2, x3, x4. The notation is arranged so
that, in the term with leading factor Fi, the differential dxi is omitted.
As a 4-form on R4, dζ has the form dζ = f dx1 ∧dx2 ∧dx3 ∧dx4 for some real-valued function
f. Find a formula for f in terms of F1, F2, F3, and F4.
We lay out more formally some of the rules that govern substitutions and pullbacks. In Rm",Multivariable_Calculus_Shimamoto.pdf
"f. Find a formula for f in terms of F1, F2, F3, and F4.
We lay out more formally some of the rules that govern substitutions and pullbacks. In Rm
with coordinates x1, x2, . . . , xm, a typical differential k-form is a sum of terms of the form:
ζ = f dxi1 ∧ dxi2 ∧ ··· ∧dxik , (11.3)
where f is a real-valued function of x1, x2, . . . , xm and 1 ≤ ij ≤ m for j = 1, 2, . . . , k. Suppose that
each of the coordinates xi can be expressed as a smooth function of n other variables u1, u2, . . . , un,
say xi = γi(u1, u2, . . . , un) for i = 1 , 2, . . . , m. This is equivalent to having a smooth function
γ : Rn → Rm from (u1, u2, . . . , un)-space to ( x1, x2, . . . , xm)-space, where, in terms of component
functions, x = γ(u) =

γ1(u), γ2(u), . . . , γm(u)


.
Using γ to substitute for x1, x2, . . . , xm converts the k-form ζ into a k-form in the variables
u1, u2, . . . , un. This k-form is what we have been calling the pullback of ζ, denoted by γ∗(ζ). For",Multivariable_Calculus_Shimamoto.pdf
"u1, u2, . . . , un. This k-form is what we have been calling the pullback of ζ, denoted by γ∗(ζ). For
instance, substituting x = γ(u) converts f(x) into f(γ(u)) = (f ◦ γ)(u). In other words, f pulls
back to the composition f ◦ γ : Rn → Rm → R. We write this as:
γ∗(f) = f ◦ γ.
Similarly, after substitution, dxi becomes dγi, i.e.:
γ∗(dxi) = dγi,
where the expression on the right is the derivative of the 0-form γi. For instance, if ( x1, x2) =
γ(u1, u2) = ( u1 cos u2, u1 sin u2), then γ∗(dx1) = dγ1 = d(u1 cos u2) = cos u2 du1 − u1 sin u2 du2.
Applying these substitutions en masse, the k-form ζ = f dxi1 ∧ dxi2 ∧ ··· ∧dxik of equation (11.3)
pulls back as:
γ∗(ζ) = (f ◦ γ) dγi1 ∧ dγi2 ∧ ··· ∧dγik .
Perhaps a slicker way of writing this is:
γ∗(f dxi1 ∧ dxi2 ∧ ··· ∧dxik ) = γ∗(f) γ∗(dxi1 ) ∧ γ∗(dxi2 ) ∧ ··· ∧γ∗(dxik ).
If ζ = ζ1 + ζ2 is a sum of k-forms, then carrying out the substitutions in both summands gives the
rule γ∗(ζ1 + ζ2) = γ∗(ζ1) + γ∗(ζ2).",Multivariable_Calculus_Shimamoto.pdf
"If ζ = ζ1 + ζ2 is a sum of k-forms, then carrying out the substitutions in both summands gives the
rule γ∗(ζ1 + ζ2) = γ∗(ζ1) + γ∗(ζ2).
Exercises 4.14–4.18 are intended to provide some practice working with pullbacks.",Multivariable_Calculus_Shimamoto.pdf
"282 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS
4.14. Let α: R → R3 be given by α(t) = (cos t, sin t, t), where we think of R3 as xyz-space.
(a) Find α∗(x2 + y2 + z2).
(b) Show that α∗(dx) = −sin t dt.
(c) Find α∗(dy) and α∗(dz).
(d) Let ω be the 1-form ω = −y dx+ x dy+ z dzon R3. Find α∗(ω).
4.15. Let T : R2 → R2 be the function from the uv-plane to the xy-plane given by T(u, v) =
(u2 − v2, 2uv).
(a) Show that T∗(dx) = 2u du− 2v dv.
(b) Find T∗(dy).
(c) Let ω = −y dx+ x dy. Find T∗(ω).
(d) Let η = xy dx∧ dy. Find T∗(η).
4.16. Let σ : R2 → R3 be the function from the uv-plane to xyz-space given by σ(u, v) = (u cos v,
u sin v, u).
(a) Let ω = x dx+ y dy− z dz. Show that σ∗(ω) = 0.
(b) Let η = z dx∧ dy. Find σ∗(η).
4.17. Let α: R → Rn be a smooth function, where α(t) = (x1(t), x2(t), . . . , xn(t)). If f : Rn → R is
a smooth real-valued function, f = f(x1, x2, . . . , xn), then d fis a 1-form on Rn. Show that:
d(α∗(f)) = α∗(d f).
(Hint: Little Chain Rule.)",Multivariable_Calculus_Shimamoto.pdf
"a smooth real-valued function, f = f(x1, x2, . . . , xn), then d fis a 1-form on Rn. Show that:
d(α∗(f)) = α∗(d f).
(Hint: Little Chain Rule.)
4.18. Let T : R2 → R2 be a smooth function, regarded as a map from the uv-plane to the xy-plane,
so (x, y) = T(u, v) = (T1(u, v), T2(u, v)).
(a) Find T∗(dx) and T∗(dy) in terms of the partial derivatives of T1 and T2.
(b) Let η be a 2-form on the xy-plane: η = f dx∧ dy, where f is a real-valued function of x
and y. Show that:
T∗(η) = (f ◦ T) · det DT du∧ dv.
(c) Let S : R2 → R2 be a smooth function, regarded as a map from the st-plane to the
uv-plane. Then the composition T ◦S : R2 → R2 goes from the st-plane to the xy-plane.
Let η = f dx∧ dy, as in part (b). Show that:
(T ◦ S)∗(η) = S∗(T∗(η)),
i.e., ( T ◦ S)∗ = S∗ ◦ T∗ for 2-forms on R2. (Actually, the same relation holds for
compositions and k-forms in general.)
We next formulate a version of the change of variables theorem for double integrals that is",Multivariable_Calculus_Shimamoto.pdf
"compositions and k-forms in general.)
We next formulate a version of the change of variables theorem for double integrals that is
adapted for differential forms, something we have alluded to before. We are not trying to give
another proof of the theorem. Rather, we take the theorem as known and describe how to restate
it in the language of differential forms.",Multivariable_Calculus_Shimamoto.pdf
"11.4. EXERCISES FOR CHAPTER 11 283
Differential forms are integrated over oriented domains, so, given a bounded subset D of R2, we
first make a choice whether to assign it the positive orientation or the negative orientation. If this
seems too haphazard, think of it as analogous to choosing whether, as a subset of R3, D is oriented
by the upward normal n = k or the downward normal n = −k.
A typical 2-form on D has the form f dx∧ dy, where f is a real-valued function on D. We
define its integral over D by:
Z
D
f dx∧ dy =
( RR
D f(x, y) dx dy if D has the positive orientation,
−
RR
D f(x, y) dx dy if D has the negative orientation,
where, in each case, the integral on the right is the ordinary Riemann double integral. So, for
instance, if D has the negative orientation, then we could also say that, as an integral of a differential
form,
R
D f dy∧ dx is equal to the Riemann integral
RR
D f(x, y) dx dy.",Multivariable_Calculus_Shimamoto.pdf
"form,
R
D f dy∧ dx is equal to the Riemann integral
RR
D f(x, y) dx dy.
Exercises 4.19–4.20 concern how to express the change of variables theorem in terms of integrals
of differential forms.
4.19. We say that a basis ( v, w) of R2 has the positive orientation if det

v w

> 0 and the
negative orientation if det

v w

< 0. Here, we use

v w

to denote the 2 by 2 matrix
whose columns are v and w. In Section 2.5 of Chapter 2, we used the terms “right-handed”
and “left-handed” to describe the same concepts, though there we put the vectors into the
rows of the matrix, i.e., we worked with the transpose, which doesn’t affect the determinant.
Let L: R2 → R2 be a linear transformation represented by a matrix A with respect to the
standard bases. Thus L(x) = Ax. In addition, assume that det A ̸= 0.
(a) Show that

L(e1), L(e2)


has the positive orientation if det A > 0 and the negative
orientation if det A <0.
(b) More generally, let (v, w) be any basis of R2. Show that ( v, w) and
",Multivariable_Calculus_Shimamoto.pdf
"orientation if det A <0.
(b) More generally, let (v, w) be any basis of R2. Show that ( v, w) and

L(v), L(w)


have
the same orientation if det A >0 and opposite orientations if det A <0.
4.20. Let D∗ and D be bounded subsets of R2, and let T : D∗ → D be a smooth function that maps
D∗ onto D and that is one-to-one, except possibly on the boundary of D∗. As in the change of
variables theorem, we think of T as a function from the uv-plane to the xy-plane. Moreover,
assume that det DT (u, v) has the same sign at all points ( u, v) of D∗, except possibly on the
boundary of D∗, where det DT (u, v) = 0 is allowed.
Let D∗ be given an orientation, and letD have the orientation imposed on it byT. That is, let
T(D∗) denote the set D with the same orientation, positive or negative, as D∗ if det DT >0
and the opposite orientation from D∗ if det DT < 0. In the first case, we say that T is
orientation-preserving and, in the second, that it is orientation-reversing. In other words, we",Multivariable_Calculus_Shimamoto.pdf
"orientation-preserving and, in the second, that it is orientation-reversing. In other words, we
categorize how T acts on orientation based on the behavior of its first-order approximation.
Let η = f dx∧ dy be a 2-form on D. Use the change of variables theorem to prove that:
Z
T(D∗)
η =
Z
D∗
T∗(η).
This equation has appeared in various guises throughout the book. You may assume in your
proof that D∗ has the positive orientation. The arguments are similar if it has the negative
orientation. (Hint: See Exercise 4.18.)",Multivariable_Calculus_Shimamoto.pdf
"284 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS
The remaining exercises call for speculation rather than proofs. You are asked to propose
reasonable solutions consistent with patterns that have come before . It is considered a bonus if
what you say is true. The correct answers are known and are covered in more advanced courses that
study manifolds, for instance, advanced real analysis or perhaps differential geometry or differential
topology.
4.21. A surface S in R4 is a two-dimensional subset, something that can be parametrized by a
function σ : D → R4, where D is a subset of R2. Let us call the parameters s and t so
that σ(s, t) = ( x1(s, t), x2(s, t), x3(s, t), x4(s, t)). Assume that it is known how to define an
orientation of S and what it means for a parametrization σ to be orientation-preserving.
(a) Write down the general form of a differential form η in the variables x1, x2, x3, x4 that
could be integrated over S. ( Hint: Six terms.)",Multivariable_Calculus_Shimamoto.pdf
"(a) Write down the general form of a differential form η in the variables x1, x2, x3, x4 that
could be integrated over S. ( Hint: Six terms.)
(b) Let σ : D → R4 be an orientation-preserving parametrization of S. The integral
R
S η
should be defined via the parametrization as a double integral
RR
D f(s, t) ds dtover the
parameter domain D. Propose a formula for the function f.
4.22. Continuing with the preceding problem, assume that it is also known how to orient the
boundary ∂S of an oriented surface S in R4 in a reasonable way. Propose an expression µ
that makes the following statement true:
Z
∂S
F1 dx1 + F2 dx2 + F3 dx3 + F4 dx4 =
Z
S
µ.
4.23. Let W be a bounded four-dimensional region in R4 whose boundary ∂W consists of a finite
number of piecewise smooth three-dimensional subsets. Assuming that it is possible to define
an orientation on ∂W in a reasonable way, propose a 3-form ζ on R4 such that:
Vol (W) =
Z
∂W
ζ,",Multivariable_Calculus_Shimamoto.pdf
"an orientation on ∂W in a reasonable way, propose a 3-form ζ on R4 such that:
Vol (W) =
Z
∂W
ζ,
where the volume on the left refers to four-dimensional volume. More than one answer for ζ
is possible. Try to find one with as much symmetry in the variables x1, x2, x3, x4 as you can.",Multivariable_Calculus_Shimamoto.pdf
"Answers to selected exercises
Section 1.1 Vector arithmetic
1.1. x + y = (5, −3, 9), 2x = (2, 4, 6), 2x − 3y = (−10, 19, −12)
1.3. (−5, 3, −9)
1.5. (b) 1
3 x + 2
3 y
(c) (−1, 1, 2)
Section 1.2 Linear transformations
2.2. T(x + y) = T(x1 + x2, y1 + y2) = (x1 + y1, x2 + y2, 0) = (x1, x2, 0) + (y1, y2, 0) = T(x) +T(y)
T(cx) = T(cx1, cx2) = (cx1, cx2, 0) = c(x1, x2, 0) = c T(x)
Section 1.3 The matrix of a linear transformation
3.1. (a)
1 3
2 4

(b) (23, 34)
(c) (x1 + 3x2, 2x1 + 4x2)
3.3.
0 −1
1 0

3.5.
1 0
0 1

3.9. (a) T(e1) = (1, 2), T(e2) = (2, 4)
(b) all scalar multiples of (1 , 2)
3.11.


1 0
0 1
0 0


3.13.


0 1 0
1 0 0
0 0 −1


285",Multivariable_Calculus_Shimamoto.pdf
"286 ANSWERS TO SELECTED EXERCISES
Section 1.4 Matrix multiplication
4.1. AB =
3 1
1 7

, BA =
6 2
2 4

4.3. AB =


1 0 0
0 1 0
0 0 1

, BA =


1 0 0
0 1 0
0 0 1


4.5. AB =
21 0
0 14

, BA =


13 4 −5
4 5 6
−5 6 17


Section 1.5 The geometry of the dot product
5.1. (a) −3
(b) ∥x∥ =
√
6, ∥y∥ = 3
√
2
(c) arccos(− 1
2
√
3 )
5.3. ± 1√
5 (−2, 1)
Section 1.6 Determinants
6.1. 23
6.3. 0
6.5. 12
6.9. (a) T(x1, x2, x3) = −3x1 + 6x2 − 3x3
(b) T(x + y) = T(x1 + y1, x2 + y2, x3 + y3) = −3(x1 + y1) + 6(x2 + y2) − 3(x3 + y3) =
−3x1 + 6x2 − 3x3 − 3y1 + 6y2 − 3y3 = T(x) + T(y)
T(cx) = T(cx1, cx2, cx3) = −3(cx1) + 6(cx2) − 3(cx3) = c(−3x1 + 6x2 − 3x3) = c T(x)
(c)

−3 6 −3

(d) (−3, 6, −3) · (1, 2, 3) = −3 + 12− 9 = 0
(−3, 6, −3) · (4, 5, 6) = −12 + 30− 18 = 0
Section 2.1 Parametrizations
1.1.
0.5 1 1.5 2
1
2
3
4",Multivariable_Calculus_Shimamoto.pdf
"287
1.3.
-1 0 1 2 3 4 5
1
2
3
4
5
1.5.
1.7.
1.9. α(x) = (x, x2), −1 ≤ x ≤ 1
1.11. α(t) = (a cos t, −a sin t) , 0 ≤ t ≤ 2π
1.13. α(t) = (1 + 4t, 2 + 5t, 3 + 6t)
1.15. α(t) = (1 − t, 0, t)
1.17. (a) v·(p−a)
v·v
(b) a + v·(p−a)
v·v v
1.19. (b) No intersection
(c) Intersection at (2, 1, 3)
Section 2.2 Velocity, acceleration, speed, arclength
2.1. (a) v (t) = (3t2, 6t, 6)
v(t) = 3(t2 + 2)
a(t) = (6t, 6, 0)
(b) 88",Multivariable_Calculus_Shimamoto.pdf
"288 ANSWERS TO SELECTED EXERCISES
2.3. (a) v (t) = (−2 sin 2t, 2 cos 2t, 6t1/2)
v(t) = 2√1 + 9t
a(t) = (−4 cos 2t, −4 sin 2t, 3t−1/2)
(b) 4
27

19
√
19 − 10
√
10


2.5. α(t) = (1 + 1√
14 t, 1 + 2√
14 t, 1 + 3√
14 t)
Section 2.3 Integrals with respect to arclength
3.1. 150
7
3.3. (a) 5
√
2
(b)
√
2
Section 2.4 The geometry of curves: tangent and normal vectors
4.1. T (t) = 1
5

4 cos 4t, −4 sin 4t, 3


N(t) = (−sin 4t, −cos 4t, 0)
T(t) · N(t) = 1
5

−4 cos 4t sin 4t + 4 sin 4t cos 4t + 0


= 0
Section 2.5 The cross product
5.1. (a) (2, −4, 2)
(b) v · (v × w) = 2 − 4 + 2 = 0, w · (v × w) = 2 − 12 + 10 = 0
5.3. (a) (−10, −11, −3)
(b) v · (v × w) = −10 + 22− 12 = 0, w · (v × w) = 20 − 11 − 9 = 0
5.5. (−4, 5, −1)
5.7. (a) (−3, −9, −3)
(b) 3
√
11
(c) u · (v × w) = −12, v · (u × w) = 12, (u × v) · w = −12
(d) 12
5.11. (b) 1
3
√
41
5.15. (a) u · (u × v × w) = det

u u v w

= 0 (two equal rows) and so on
(b) Left-handed
(c) (−4, −4, −4, −4)",Multivariable_Calculus_Shimamoto.pdf
"(d) 12
5.11. (b) 1
3
√
41
5.15. (a) u · (u × v × w) = det

u u v w

= 0 (two equal rows) and so on
(b) Left-handed
(c) (−4, −4, −4, −4)
Section 2.6 The geometry of space curves: Frenet vectors
6.1. (3
5 cos 4t, −3
5 sin 4t, −4
5 )
Section 2.7 Curvature and torsion
7.1. α(t) = (2 cost, 2 sint, 0) is one possibility.",Multivariable_Calculus_Shimamoto.pdf
"289
Section 2.9 The classification of space curves
9.1. (a)
√
2
(b) 1√
2 (−sin t, −cos t, 1)
(c) (−cos t, sin t, 0)
(d) − 1√
2 (sin t, cos t, 1)
(e) 1
2
(f) −1
2
(g) a = 1, b = −1. Both α and β trace out the same helix but in opposite directions. The
rotation F(x, y, z) = (x, −y, −z) about the x-axis by π transforms one parametrization
into the other.
9.3. (a) 2
√
2
(b) 1
2
√
2

1 +
√
3 cost,
√
3 − cos t, −2 sint


(c) 1
2

−
√
3 sint, sin t, −2 cost


(d) 1
2
√
2

1 −
√
3 cost,
√
3 + cost, 2 sint


(e) 1
4
(f) 1
4
(g) It’s congruent to the helix parametrized by β(t) = (2 cost, 2 sint, 2t).
9.7. 1
7
q
19
14
9.9. (a)
√
2
2(1+sin2 t)3/2
(b) τ(t) = 0
9.13. (a) T (t) = −
√
2
2

1, sin 2t, cos 2t


N(t) = (0, −cos 2t, sin 2t)
κ(t) =
√
2
τ(t) =
√
2
(b) α(t) =

−
√
2
2 t, −
√
2
4 +
√
2
4 cos 2t, −
√
2
4 sin 2t

",Multivariable_Calculus_Shimamoto.pdf
"290 ANSWERS TO SELECTED EXERCISES
Section 3.1 Graphs and level sets
1.1. (b)
0
1
2
3
4 5 6
-2 -1 0 1 2
-2
-1
0
1
2
(c)
1.3. (b)
-2
-1
0
1
-2 -1 0 1 2
-2
-1
0
1
2
(c)
1.5. (b)
0
0.5
1
1.5
2
2.5
3
-2 -1 0 1 2
-2
-1
0
1
2",Multivariable_Calculus_Shimamoto.pdf
"291
(c)
1.7. (b)
-1.5
-1
-0.5
0
0.5
1
1.5
-2 -1 0 1 2
-2
-1
0
1
2
(c)
Section 3.2 More surfaces in R3
2.1.
2.3.",Multivariable_Calculus_Shimamoto.pdf
"292 ANSWERS TO SELECTED EXERCISES
2.5. The level set corresponding to c = 0 consists of the single point (0 , 0, 0). Those for c = 1 and
c = 2 are spheres centered at the origin of radius 1 and
√
2, respectively.
2.7. c = −1, 0, 1 (left, center, right)
2.9. c = −1, 0, 1 (left, center, right)
Section 3.3 The equation of a plane
3.1. 4x + 5y + 6z = 32
3.3. For example, p = (9, 0, 0), n = (1, 3, 5)
3.5. Perpendicular
3.7. 6x + 3y + 2z = 6
3.9. 2x + z = 3
3.11. (8
3 , 1
3 , 1
3 )
3.13. (b) It’s the triangular region with vertices (1 , 0, 0), (0, 1, 0), and (0, 0, 1).
3.15. (c) 4√
3
Section 3.4 Open sets
4.1. Let a = (c, d) be a point of U, and let r = min{c, 1 − c, d,1 − d}. Then B(a, r) ⊂ U, so U is
an open set.",Multivariable_Calculus_Shimamoto.pdf
"293
4.3. The point a = (1, 0) is in U, but no open ball centered at a stays within U. Thus U is not
an open set.
Section 3.5 Continuity
5.1. (a) If (x, y) ̸= (0, 0), then, in polar coordinates,
f(x, y) = f(r cos θ, rsin θ) = (r cos θ)3(r sin θ)
r2 = r2 cos3 θ sin θ,
so |f(x, y)| = |r2 cos3 θ sin θ| ≤r2 = ∥(x, y)∥2. The inequality also holds if ( x, y) = (0, 0)
(both sides equal 0). Thus, if ∥(x, y)∥ < a, then |f(x, y)| ≤ ∥(x, y)∥2 < a2.
(b) Let ϵ >0 be given, and let δ = √ϵ. By part (a), if ∥(x, y) − (0, 0)∥ = ∥(x, y)∥ < δ, then
|f(x, y) − f(0, 0)| = |f(x, y)| < δ2 = ϵ. Therefore f is continuous at (0, 0).
Section 3.6 Some properties of continuous functions
6.1. f(x, y) = (1, 2) · (x, y) = x + 2y. We know that the projections x and y are continuous, so, as
an algebraic combination of continuous functions, f is continuous as well.
Section 3.7 The Cauchy-Schwarz and triangle inequalities
7.1. (a) By the triangle inequality,∥v∥ = ∥(v−w)+w∥ ≤ ∥v−w∥+∥w∥, so ∥v∥−∥w∥ ≤ ∥v−w∥.",Multivariable_Calculus_Shimamoto.pdf
"Section 3.7 The Cauchy-Schwarz and triangle inequalities
7.1. (a) By the triangle inequality,∥v∥ = ∥(v−w)+w∥ ≤ ∥v−w∥+∥w∥, so ∥v∥−∥w∥ ≤ ∥v−w∥.
(b) Reversing the roles of v and w gives ∥w∥−∥v∥ ≤ ∥w−v∥ = ∥v−w∥, or −(∥v∥−∥w∥) ≤
∥v − w∥. Hence ±(∥v∥ − ∥w∥) ≤ ∥v − w∥, so
∥v∥ − ∥w∥
 ≤ ∥v − w∥.
7.4. First, note that |cf(x) − cf(a)| = |c||f(x) − f(a)| ≤(|c| + 1)|f(x) − f(a)|. Now, let ϵ >0
be given. Since f is continuous at a, there exists a δ > 0 such that, if ∥x − a∥ < δ,
then |f(x) − f(a)| < ϵ
|c|+1 . Then, for this choice of δ, if ∥x − a∥ < δ, |cf(x) − cf(a)| ≤
(|c| + 1)|f(x) − f(a)| < (|c| + 1) · ϵ
|c|+1 = ϵ. Hence cf is continuous at a.",Multivariable_Calculus_Shimamoto.pdf
"294 ANSWERS TO SELECTED EXERCISES
Section 3.8 Limits
8.1. For sums: Let L = limx→a f(x) and M = limx→a g(x), and consider the functions:
ef(x) =
(
f(x) if x ̸= a,
L if x = a and eg(x) =
(
g(x) if x ̸= a,
M if x = a.
By definition of limit, ef and eg are continuous at a, hence so is their sum. In other words, the
function given by
ef(x) + eg(x) =
(
f(x) + g(x) if x ̸= a,
L + M if x = a
is continuous at a. By definition, this means that lim x→a(f(x) + g(x)) = L + M.
Section 4.1 The first-order approximation
1.1. (a) ∂f
∂x (a) = 4, ∂f
∂y (a) = −2
(b) Df(a) =

4 −2

, ∇f(a) = (4, −2)
(c) ℓ(x, y) = −3 + 4x − 2y
(d) f(2.01, 1.01) = 3.02, ℓ(2.01, 1.01) = 3.02
1.3. (a) ∂f
∂x = 3x2 − 4xy + 3y2
∂f
∂y = −2x2 + 6xy − 12y2
(b)

3x2 − 4xy + 3y2 −2x2 + 6xy − 12y2
1.5. (a) ∂f
∂x = (1 + xy)exy
∂f
∂y = x2exy
(b)

(1 + xy)exy x2exy
1.7. (a) ∂f
∂x = − 2x
(x2+y2)2
∂f
∂y = − 2y
(x2+y2)2
(b)
h
− 2x
(x2+y2)2 − 2y
(x2+y2)2
i
1.9. (a) ∂f
∂x = 1, ∂f
∂y = 2, ∂f
∂z = 3
(b) (1, 2, 3)
1.11. (a) ∂f
∂x =
",Multivariable_Calculus_Shimamoto.pdf
"(x2+y2)2
∂f
∂y = − 2y
(x2+y2)2
(b)
h
− 2x
(x2+y2)2 − 2y
(x2+y2)2
i
1.9. (a) ∂f
∂x = 1, ∂f
∂y = 2, ∂f
∂z = 3
(b) (1, 2, 3)
1.11. (a) ∂f
∂x =

cos(x + 2y) − 2x sin(x + 2y)


e−x2−y2−z2
∂f
∂y = 2

cos(x + 2y) − y sin(x + 2y)


e−x2−y2−z2
∂f
∂z = −2z sin(x + 2y)e−x2−y2−z2
(b) e−x2−y2−z2 
cos(x + 2y) −2x sin(x + 2y) , 2

cos(x + 2y) −y sin(x + 2y)


, −2z sin(x + 2y)


1.13. (a) ∂f
∂x = 2x
x2+y2
∂f
∂y = 2y
x2+y2
∂f
∂z = 0",Multivariable_Calculus_Shimamoto.pdf
"295
(b) ( 2x
x2+y2 , 2y
x2+y2 , 0)
1.17. (a) ∂f
∂x = 2x3
√
x4+y4
∂f
∂y = 2y3
√
x4+y4
(b) ∂f
∂x (0, 0) = 0, ∂f
∂y (0, 0) = 0
Section 4.2 Conditions for differentiability
2.1. lim(x,y)→(0,0) f(x, y) = lim(x,y)→(0,0)(x2 − y2) = 0, while f(0, 0) = π, so f is not continuous at
(0, 0), hence not differentiable there either.
Section 4.4 The C1 test
4.1. For (x, y) ̸= (0, 0), the partial derivatives are given by ∂f
∂x = x√
x2+y2 and ∂f
∂y = y√
x2+y2 . As
algebraic combinations and compositions of continuous functions, both of these are continuous
at all points other than the origin. Thus, by the C1 test, f is differentiable at all points other
than the origin.
4.2. Differentiable at all (x, y) in R2
4.4. Differentiable at all (x, y) where x ̸= 0 and y ̸= 0
4.6. Differentiable at all (w, x, y, z) in R4
Section 4.5 The Little Chain Rule
5.1. (b) (f ◦ α)(t) = f(t2, t3) = t4 + t6, so (f ◦ α)′(t) = 4t3 + 6t5 and (f ◦ α)′(1) = 10.
5.3. ∂f
∂x
dx
dt + ∂f
∂y
dy
dt + ∂f
∂z
dz
dt , where ∂f
∂x , ∂f",Multivariable_Calculus_Shimamoto.pdf
"5.3. ∂f
∂x
dx
dt + ∂f
∂y
dy
dt + ∂f
∂z
dz
dt , where ∂f
∂x , ∂f
∂y , and ∂f
∂z are evaluated at α(t)
Section 4.6 Directional derivatives
6.1. − 5√
17
6.3. 2√
6
6.5. 6√
17 (4, 1)
6.7. r = 5√
2 , s = 5
2
√
2
6.9. (a) 1√
38 (−1, 1, −6)
(b) 1√
38 (1, −1, 6)
(c) 800
√
38 e−42 ◦C/sec
6.11. (a) Direction: u = 1√
13 (3, 2)
Value:
√
13
(b) Direction: u = 1√
13 (−3, −2)
Value: −
√
13",Multivariable_Calculus_Shimamoto.pdf
"296 ANSWERS TO SELECTED EXERCISES
(c) y = 1
3 x2 − 1
3
1 2 3
1
2
3
(d) y = −3
2 ln x
1 2 3
-1
1
2
3
4
5
Section 4.7 ∇f as normal vector
7.1. 2x − 4y − z = −3
7.3. 4y + 3z = 7
7.5. α(t) = (1 + 14t, 2 + 5t, 3 − 8t)
Section 4.8 Higher-order partial derivatives
8.1. ∂2f
∂x2 = 12x2 − 12xy + 6y2
∂2f
∂y ∂x = −6x2 + 12xy − 12y2
∂2f
∂x ∂y = −6x2 + 12xy − 12y2
∂2f
∂y2 = 6x2 − 24xy + 60y2
8.3. ∂2f
∂x2 = (4x2 − 2)e−x2−y2
∂2f
∂y ∂x = 4xye−x2−y2
∂2f
∂x ∂y = 4xye−x2−y2
∂2f
∂y2 = (4y2 − 2)e−x2−y2
8.5. i = 4, j = 3, ∂7f
∂x4 ∂y3 (0, 0) = 144
Section 4.10 Max/min: critical points
10.1. (1, 2) saddle point
10.3. (0, 0) local minimum
10.5. (0, 0) saddle point, (3 , 9) local minimum
10.7. k ̸= 0: (0 , 0) saddle point; ( k
3 , k
3 ) local minimum if k >0, local maximum if k <0
k = 0: (0 , 0) neither local maximum nor minimum (and, in addition, degenerate)",Multivariable_Calculus_Shimamoto.pdf
"297
10.11. y = x + 1
3
-1 1 2 3 4
1
2
3
4
Section 4.11 Classifying nondegenerate critical points
11.1. 1 − 1
2 (x + y)2
11.3. (a) h3 ∂3f
∂x3 + 3h2k ∂3f
∂x2 ∂y + 3hk2 ∂3f
∂x ∂y2 + k3 ∂3f
∂y3
(b) f(a) + ∂f
∂x (a) h + ∂f
∂y (a) k + 1
2
∂2f
∂x2 (a) h2 + 2 ∂2f
∂x ∂y(a) hk + ∂2f
∂y2 (a) k2

+ 1
6
∂3f
∂x3 (a) h3 +
3 ∂3f
∂x2 ∂y (a) h2k + 3 ∂3f
∂x ∂y2 (a) hk2 + ∂3f
∂y3 (a) k3

Section 4.12 Max/min: Lagrange multipliers
12.1. (a) x can be chosen large and positive and y large and negative so that x + 2y = 1. Thus
f(x, y) = xy can be made as large and negative as you like.
(b) Maximum value 1
8 at ( 1
2 , 1
4 )
12.3. Maximum at 1√
29 (2, −3, 4) and minimum at 1√
29 (−2, 3, −4)
12.5. x = 2
3 , y = 1, z = 6
5
12.9. x = 400, y = 150, z = 100
Section 5.1 Volume and iterated integrals
1.1. 49
6
1.3. 7
6
1.5. 26
3 ln 2
1.7. e2 − 2e + 1
1.9. (a)
-1 1 2 3 4
-1
1
2
3
4
(b)
Z 4
3
Z 2
1
f(x, y) dx

dy",Multivariable_Calculus_Shimamoto.pdf
"298 ANSWERS TO SELECTED EXERCISES
1.11. (a)
xy
xy 2
0.25 0.5 0.75 1
0.25
0.5
0.75
1
1.25
(b)
Z 1
0
Z √x
x
f(x, y) dy

dx
1.13. (a)
y= 1-x 2
-1 -0.5 0.5 1
0.25
0.5
0.75
1
1.25
(b)
Z 1
0
Z √
1−y2
−
√
1−y2
f(x, y) dx

dy
1.15. (a)
y=1/x
0 0.25 0.5 0.75 1 1.250
1
2
3
4
(b)
Z 4
1
Z 1
1
y
yexy dx

dy
(c) e4 − 4e
1.17. 2
3
1.19. 2
3
1.21. 1
3
Section 5.2 The double integral
2.1. (a) S = 0 or 1
9
(b) S = 0, 1
16 , 1
8 , 3
16 , or 1
4
(c) If n is odd, S = 0 or 1
n2 . If n is even, S = 0, 1
n2 , 2
n2 , 3
n2 , or 4
n2 .
(e) Let ϵ >0 be given, and let δ =
√ϵ
2 . If R is subdivided into subrectangles of dimensions
△xi by △yj, where △xi < δ and △yj < δ for all i, j, then, by part (d), all Riemann
sums based on the subdivision satisfy |P
i,j f(pij) △xi △yj − 0| < 4δ2 = ϵ. Therefore f
is integrable over R, and
RR
R f(x, y) dA = 0.",Multivariable_Calculus_Shimamoto.pdf
"299
Section 5.3 Interpretations of the double integral
3.1. 18 hundred birds
3.3. (a) 80 miles per hour
(b) All points of R that lie on the circle x2 + y2 = 8
3 , a quarter-circular arc
3.5. (2
3 , 5
3 )
3.7.
q
2
3
Section 5.4 Parametrization of surfaces
4.1. (a) σ : D → R3, σ(x, y) = (x, y,
p
a2 − x2 − y2), where D is the disk x2 + y2 ≤ a2
(b) σ : D → R3, σ(r, θ) = (r cos θ, rsin θ,
√
a2 − r2), where D = [0, a] × [0, 2π]
(c) σ : D → R3, σ(ϕ, θ) = (a sin ϕ cos θ, asin ϕ sin θ, acos ϕ), where D = [0, π
2 ] × [0, 2π]
Section 5.5 Integrals with respect to surface area
5.1. (a) σ : D → R3, σ(x, y) = (x, y,1 − x − y), where D is the triangular region in the xy-plane
with vertices (0, 0), (1, 0), (0, 1)
(b)
√
3
2
(c) 7
√
3
2
5.2. 36π
5.5. (a)
(b) 4(6 +
√
2)π
(c) 2(27 + 10
√
2)π
5.7. (a) 8
(b) 4
Section 5.6 Triple integrals and beyond
6.1. 1
12
6.3. 7
48
6.5. 64
15",Multivariable_Calculus_Shimamoto.pdf
"300 ANSWERS TO SELECTED EXERCISES
6.7. a2
6.8. (a)
(b)
Z √
2
0
Z
q
1−y2
2
0
Z 2
q
1−y2
2 −z2
−2
q
1−y2
2 −z2
f(x, y, z) dx

dz

dy
(c)
Z 2
−2
Z
q
1−x2
4
0
Z
q
2−x2
2 −2z2
0
f(x, y, z) dy

dz

dx
6.10. (a)
(b)
Z 1
0
Z y
0
Z y
z
f(x, y, z) dx

dz

dy
(c)
Z 1
0
Z x
0
Z 1
x
f(x, y, z) dy

dz

dx
Section 6.1 Continuity revisited
1.2. Given any ϵ >0, there exists a δ >0 such that ∥f(x) − L∥ < ϵwhenever ∥x − a∥ < δ, except
possibly when x = a.
Section 6.2 Differentiability revisited
2.1. (a)
2x −2y
2y 2x

(b)
2 −4
4 2

2.3.

1 2 y 9z2
2.5.


1 0
0 1
0 0

",Multivariable_Calculus_Shimamoto.pdf
"301
2.7. (a)


ex+y ex+y
−e−x cos y −e−x sin y
−e−x sin y e −x cos y


(b) All entries of Df(x, y) are continuous on R2, so, by the C1 test, f is differentiable at
every point of R2.
2.9. (a)


a b c
d e f
g h i

 (= A)
Section 6.3 The chain rule: a conceptual approach
3.1. (a) Df(s, t) =


1 −1
2e2s+3t 3e2s+3t
cos t −s sin t

 , Dg(x, y, z) =
yz xz xy
2x 0 3 z2

(b)
2 0
4 −4

(c)


0 0 0
0 0 0
0 0 0


3.3. (a)
1 0
0 1

(b)
1 0
0 1

Section 6.4 The chain rule: a computational approach
4.1. ∂w
∂s = ∂f
∂x + 3∂f
∂y + 5∂f
∂z
∂w
∂t = 2∂f
∂x + 4∂f
∂y + 6∂f
∂z
4.3. (a) ∂w
∂ρ = ∂f
∂x sin ϕ cos θ + ∂f
∂y sin ϕ sin θ + ∂f
∂z cos ϕ
∂w
∂ϕ = ∂f
∂x ρ cos ϕ cos θ + ∂f
∂y ρ cos ϕ sin θ − ∂f
∂z ρ sin ϕ
∂w
∂θ = −∂f
∂x ρ sin ϕ sin θ + ∂f
∂y ρ sin ϕ cos θ
(b) 0
4.5. ∂f
∂x = 5
2 , ∂f
∂y = 3
4
4.9. ∂z
∂x = −1
3 , ∂z
∂y = −2
3
4.11. (b) ∂f
∂x
d2x
dt2 + ∂f
∂y
d2y
dt2 + ∂2f
∂x2
dx
dt

2 + 2 ∂2f
∂x ∂y
dx
dt
dy
dt + ∂2f
∂y2
dy
dt

2
Section 7.1 Change of variables for double integrals",Multivariable_Calculus_Shimamoto.pdf
"∂x
d2x
dt2 + ∂f
∂y
d2y
dt2 + ∂2f
∂x2
dx
dt

2 + 2 ∂2f
∂x ∂y
dx
dt
dy
dt + ∂2f
∂y2
dy
dt

2
Section 7.1 Change of variables for double integrals
1.2. π
8
1.4. (sin 16− sin 4)π",Multivariable_Calculus_Shimamoto.pdf
"302 ANSWERS TO SELECTED EXERCISES
1.6. 2π
1.8. (b) (1 − e−a2
)π
Section 7.3 Examples: linear changes of variables, symmetry
3.1. (a) D is the disk of radius 2 centered at (3 , 2).
(3,2)
D
5
1
2
3
4
5
(b) D∗ is the disk u2 + v2 ≤ 4 of radius 2 centered at the origin.
(c)
ZZ
D∗
(u + v + 5)du dv= 20π
3.4. (a) 150
(b) 100
3.7. (a) T(u, v) =
pu
v , √uv


(b) D∗ = [1, 2] × [1, 4]
(c) 3
2 e(e − 1)
3.9. 8
9
Section 7.4 Change of variables for n-fold integrals
4.1. 48
5 π
4.3. π
6 (1 + 3 ln3
4 )
4.5. (0, 0, 3
8 a)
4.7. 4(2 −
√
2)
4.9. 8π2
15
Section 8.2 Exercises for Chapter 8 (Vector fields)
1.1.
-2 -1 0 1 2
-2
-1
0
1
2",Multivariable_Calculus_Shimamoto.pdf
"303
1.3.
-2 -1 0 1 2
-2
-1
0
1
2
1.5.
-2 -1 0 1 2
-2
-1
0
1
2
1.7. (a) F (x, y) = 1√
4x2+1 (−2x, 1) or its negative
(b) F (x, y) = 1√
4x2+1 (1, 2x) or its negative
1.9. (a) α′(t) =

−2a sin(
√
2t),
√
2a cos(
√
2t)


and F(α(t)) = F
√
2a cos(
√
2t), asin(
√
2t)


=
−2a sin(
√
2t),
√
2a cos(
√
2t)


, so α′(t) = F(α(t)).
(b) x2
2 + y2 = 2a2 cos2(
√
2t)
2 + a2 sin2(
√
2t) = a2 cos2(
√
2t) + a2 sin2(
√
2t) = a2
-2 -1 0 1 2
-2
-1
0
1
2
1.12. (a) α(t) = (x0et, y0et) = et(x0, y0)
(b) The integral curves are rays emanating out from the origin.",Multivariable_Calculus_Shimamoto.pdf
"304 ANSWERS TO SELECTED EXERCISES
Section 9.1 Definitions and examples
1.1. (a)
-2 -1 0 1 2
-2
-1
0
1
2
(b) For instance, any line segment radiating directly away from the origin. Ifα(t) = (t, t), 0 ≤
t ≤ 1, then
R
C x dx+ y dy= 1.
(c) For instance, any circle centered at the origin. If α(t) = (cos t, sin t), 0 ≤ t ≤ 2π, thenR
C x dx+ y dy= 0.
1.3. 1
2 (π + sin2(π2))
1.4. 179
60
1.6. (a) F (x, y) = (x + y, y− x)
(b) 1
(c) 0
(d) 2
1.8. 69
2
1.10. (a)
(b) e4 + 2
Section 9.3 Conservative fields
3.1. (a) It’s conservative with potential function f(x, y) = xy.
(b) 2
3.3. (a) (−6x2y − 8x3z, 6y4z + 6xy2, 12x2z2 − 12y3z2)
(b) Not conservative
3.5. Not conservative
3.7. Conservative with potential function f(x, y, z) = 1
2 x2 + 1
2 y2 + 1
2 z2 + xyz
3.9. (a) q = 1 − 2p",Multivariable_Calculus_Shimamoto.pdf
"305
(b) All positive p
3.13. 3
3.17. The vector field F = ∇(fg) is conservative with potential function fg, so
R
C ∇(fg) · ds = 0
for all piecewise smooth oriented closed curves C. By Exercise 1.19 in Chapter 4, ∇(fg) =
f ∇g + g ∇f, so
R
C(f ∇g + g ∇f) · ds = 0. Hence
R
C(f ∇g) · ds = −
R
C(g ∇f) · ds.
Section 9.4 Green’s theorem
4.1. 2
4.3. 9π3
2
4.6. 4
4.8. 1
2
4.10. 12
Section 9.5 The vector field W
5.1. (a) 27
Section 9.6 The converse of the mixed partials theorem
6.1. (a) C traverses the unit circle once clockwise. Winding number = −1.
(b) Cn goes around the unit circle |n| times, counterclockwise if n >0 and clockwise if n <0.
If n = 0, it stays stuck at the point (1 , 0). Winding number = n.
6.2. Winding number = 0
Section 10.1 What the surface integral measures
1.1. (a) 2πa3
(b) 0
1.2. 1
2
Section 10.2 The definition of the surface integral
2.1. 7
2
2.3. 2π + 32π3
3
2.5. 8
3
Section 10.3 Stokes’s theorem
3.1. (a) For example, G(x, y, z) = (−y cos z, x, y2z)
(b) 2π
(c) 2π",Multivariable_Calculus_Shimamoto.pdf
"2.1. 7
2
2.3. 2π + 32π3
3
2.5. 8
3
Section 10.3 Stokes’s theorem
3.1. (a) For example, G(x, y, z) = (−y cos z, x, y2z)
(b) 2π
(c) 2π
3.3. (a) (y − 2x, x+ y, z)
(b) 3
2 π",Multivariable_Calculus_Shimamoto.pdf
"306 ANSWERS TO SELECTED EXERCISES
Section 10.4 Curl fields
4.1. ∇ ·F = 3, not 0, so F is not a curl field.
4.2. It’s a curl field with G(x, y, z) = (1
2 z2, 1
2 x2, 1
2 y2), for example.
Section 10.5 Gauss’s theorem
5.1. 384
5 π
5.3. 104
3 π
5.7. 225 − 32π
Section 10.6 The inverse square field
6.2. 11
6.4. It’s the total mass contained in the interior of S, i.e., the sum of those mi for which pi is in
the interior of S.
Section 10.7 A more substitution-friendly notation for surface integrals
7.1. (a) F (x, y, z) = (xy, yz, xz)
(b) 0
Section 11.4 Exercises for Chapter 11 (Working with differential forms)
4.1. (x2 + y2) dx ∧ dy
4.3. (xy − z2) dy ∧ dz + (yz − x2) dz ∧ dx + (xz − y2) dx ∧ dy
4.5. 0
4.7. 2x dx+ 2y dy
4.9. 4y dx∧ dy
4.11. (1 + xzexyz) dx ∧ dy ∧ dz
4.13. ∂F1
∂x1
− ∂F2
∂x2
+ ∂F3
∂x3
− ∂F4
∂x4
4.14. (a) 1 + t2
(c) α∗(dy) = cos t dt, α∗(dz) = dt
(d) (1 + t) dt
4.16. (b) u2 du ∧ dv
4.18. (a) T∗(dx) = ∂T1
∂u du + ∂T1
∂v dv
T∗(dy) = ∂T2
∂u du + ∂T2
∂v dv",Multivariable_Calculus_Shimamoto.pdf
"(c) α∗(dy) = cos t dt, α∗(dz) = dt
(d) (1 + t) dt
4.16. (b) u2 du ∧ dv
4.18. (a) T∗(dx) = ∂T1
∂u du + ∂T1
∂v dv
T∗(dy) = ∂T2
∂u du + ∂T2
∂v dv
4.21. (a) F1 dx1 ∧ dx2 + F2 dx1 ∧ dx3 + F3 dx1 ∧ dx4 + F4 dx2 ∧ dx3 + F5 dx2 ∧ dx4 + F6 dx3 ∧ dx4",Multivariable_Calculus_Shimamoto.pdf
"Index
acceleration, 28
add zero, 32, 78
affine function, 84
annulus, 223
arclength, 29, 30
area
as double integral, 131
as line integral, 219
of graph z = f(x, y), 152
of parallelogram, 15, 36, 175
of sphere, 143, 238
of surface, 142
of surface of revolution, 152
average value, 132, 144
basis, 5
binormal, 38
boundary
of curve, 274
of subset of R2, 217
of subset of R3, 255
of surface, 245, 248
bounded
function, 127, 144
set, 127, 144
C1
class, 90
test, 90, 98, 161
C2, 97
C∞, 98
Cartesian product, 127
Cauchy-Schwarz inequality, 71
centroid, 117, 132, 144
chain rule, 161, 211, 262, 264, 280
change of variables, 176, 178, 182, 210, 260,
261, 263, 264, 276, 282, 283
closed ball, 65
closed curve, 212
closed set, 65
closed surface, 252
closure, 134, 235
Cobb-Douglas, 107
complement, 65
composition of functions, 10, 11
conservation of energy, 230
conservative field, 211, 212, 221, 222, 234,
273, 274, 279
continuity, 66, 98, 157
contour map, 57
critical point, 99
cross product, 35
curl, 214",Multivariable_Calculus_Shimamoto.pdf
"conservative field, 211, 212, 221, 222, 234,
273, 274, 279
continuity, 66, 98, 157
contour map, 57
critical point, 99
cross product, 35
curl, 214
curl field, 251
curvature, 40, 227
curve, 25
cylinder, 59, 138, 236
cylindrical coordinates, 137, 182
Darboux formulas, 51
dependence diagram, 163
derivative
matrix, 85, 86, 159
of differential form, 275
of path, 28
determinant, 15
differentiability, 86, 159
differential form
0-form, 274
k-form, 274
n-form, 274
1-form, 204
2-form, 261
directional derivative, 93
divergence, 253
307",Multivariable_Calculus_Shimamoto.pdf
"308 INDEX
dot product, 7, 21
eigenvalue, 120
eigenvector, 120
first-order approximation, 83, 86, 91, 92, 95,
159, 161, 174, 283
flux, 235
Frenet vectors, 39
Frenet-Serret formulas, 42
Fubini’s theorem, 129
fundamental theorem of calculus, 279
Gauss’s law, 259, 270
Gauss’s theorem, 255, 273, 274, 279
global maximum/minimum, 99
gradient, 85, 94
gradient field, 211
graph, 55
Green’s theorem, 217, 273, 274, 279
handedness, 27, 37, 283
harmonic function, 231, 270
helicoid, 151, 152, 265
helix, 27, 39, 40, 41, 55
Hessian, 100
hypar, 74, 80
hyperbolic paraboloid, 58, 74
implicit differentiation, 165, 169
integral
definition, 127, 128, 130, 144
iterated, 123, 129
of density, 131
of differential form, 274
Riemann, 128
with respect to arclength, 30, 203, 208,
227
with respect to surface area, 141, 235,
264
integral curve, 201
integral path, 201, 230
isotherm, 95
Jacobian
determinant, 176
matrix, 86, 159
kinetic energy, 230
Lagrange multipliers, 106
law of cosines, 13
least squares, 117",Multivariable_Calculus_Shimamoto.pdf
"isotherm, 95
Jacobian
determinant, 176
matrix, 86, 159
kinetic energy, 230
Lagrange multipliers, 106
law of cosines, 13
least squares, 117
level set, 57
limit, 72, 158
line integral, 204, 208
linear transformation
definition, 6
matrix of, 8, 9, 175
Little Chain Rule, 92, 93, 94, 103, 112, 163,
165, 211
local maximum/minimum, 99, 100
manifold, 279
marginal productivity, 108
matrix
associativity of multiplication, 20
definition, 8
multiplication, 9, 10
mean value theorem, 89, 112, 116, 168
mixed partials
of real-valued function, 96, 97, 115, 116,
150, 253
of vector field, 214, 221, 223, 252
monkey saddle, 73, 120
nondegenerate critical point, 100
normal vector, 61, 94
octant, 123
one-to-one, 174
open ball, 62
open set, 63
orientation
of basis, 37, 283
of curve, 208
of surface, 236
of surface parametrization, 240
orthogonal, 14
parallelepiped, 36
parametrization
effect on line integral, 207, 208
effect on surface integral, 261, 263
of circle, 26
of curve, 25
of graph, 26
of helix, 27",Multivariable_Calculus_Shimamoto.pdf
"parametrization
effect on line integral, 207, 208
effect on surface integral, 261, 263
of circle, 26
of curve, 25
of graph, 26
of helix, 27
of line, 27",Multivariable_Calculus_Shimamoto.pdf
"INDEX 309
of surface, 134
partial derivative, 85
path, 25
piecewise smooth
curve, 210
surface, 241, 249, 252
plane
as span, 13
equation of, 61
polar coordinates, 69, 136, 176
potential energy, 230
potential function, 211, 212
principal normal, 33
product rules, 32, 41
pullback, 178, 210, 240, 261, 264, 281, 282
related rates, 112
Riemann sum, 30, 127, 130, 131, 141, 174,
180
right-hand rule, 37
rotation matrix, 19
saddle point, 99, 100
saddle surface, 58, 74, 99, 100
sample points, 127
scalar, 3
second derivative test, 100
second-order approximation, 102, 170
separate variables, 149, 179
simple curve, 217
simply connected, 223
smooth, 98, 161
solid torus, 269
span, 13
spectral theorem, 105
speed, 29
sphere, 58, 139, 143, 236, 241
spherical coordinates, 138, 183
standard basis, 5
Stokes’s theorem, 249, 273, 274, 279, 280
substitution, 176, 177, 210, 240, 261, 281
surface integral, 237, 241, 261
symmetric matrix, 105, 119
symmetry
about the origin, 192
in y = x, 192
in y-axis, 180",Multivariable_Calculus_Shimamoto.pdf
"surface integral, 237, 241, 261
symmetric matrix, 105, 119
symmetry
about the origin, 192
in y = x, 192
in y-axis, 180
in plane, 185, 239
torsion, 40, 76
torus, 223, 268
transpose, 16
triangle inequality, 70, 71
unit tangent, 31
unit vector, 14
vector
addition, 3
column, 9
magnitude, 12
norm, 12
scalar multiplication, 3
unit, 14
zero, 3
vector field
circulating, 198, 201, 205, 213, 218
definition, 197
inverse square, 199, 213, 238, 257
winding, 199, 219, 223
velocity, 28
volume
as n-dimensional integral, 144
as double integral, 121, 131
as surface integral, 257
of ball, 183, 187, 193, 257
of parallelepiped, 36
wedge product, 260, 261, 275
winding number, 224
work, 203, 230",Multivariable_Calculus_Shimamoto.pdf
"AC Electrical Circuit Analysis AC Electrical Circuit Analysis 
A Practical ApproachA Practical Approach
James M. FioreJames M. Fiore",ACElectricalCircuitAnalysis.pdf
2,ACElectricalCircuitAnalysis.pdf
"AC Electrical Circuit AnalysisAC Electrical Circuit Analysis
A Practical ApproachA Practical Approach
 
by
 
James M. Fiore
 
 
   
Version 1.1.10, 30 March 2023
3",ACElectricalCircuitAnalysis.pdf
"This AC Electrical Circuit Analysis, by James M. Fiore is copyrighted under the terms of 
a Creative Commons license: 
 
                                                                                                         
This work is freely redistributable for non-commercial use, share-alike with attribution
  Device data sheets and other product information are copyright by their respective owners
and have been obtained through publicly accessible manufacturer's web sites.
Published by James M. Fiore via dissidents 
 
  ISBN13: 979-8605022282
   
For more information or feedback, contact:
James Fiore, Professor
Electrical Engineering Technology
Mohawk Valley Community College
1101 Sherman Drive
Utica, NY 13501                                                
jfiore@mvcc.edu
oer@jimfiore.org
For the latest revisions, related titles, and links to low cost print versions, go to: 
www.mvcc.edu/jfiore or my mirror sites www.dissidents.com and www.jimfiore.org",ACElectricalCircuitAnalysis.pdf
"www.mvcc.edu/jfiore or my mirror sites www.dissidents.com and www.jimfiore.org
YouTube Channel: Electronics with Professor Fiore
Cover art, Chapman's Contribution Redux, by the author 
4",ACElectricalCircuitAnalysis.pdf
"IntroductionIntroduction
Welcome to the AC Electrical Circuit Analysis, an open educational resource (OER). The goal of this text is to 
introduce the theory and practical application of analysis of AC electrical circuits. It assumes familiarity with DC
circuit analysis. If you have not studied DC circuit analysis, it is strongly recommended that you read the 
companion OER text, DC Electrical Circuit Analysis before continuing. Both texts are offered free of charge 
under a Creative Commons non-commercial, share-alike with attribution license. For your convenience, along 
with the free pdf and odt files, print copies are available at a very modest charge. Check my web sites for links.
This text is based on the earlier Workbook for AC Electrical Circuits, which it replaces. The original expository 
text has been greatly expanded and includes many examples along with computer simulations. For the",ACElectricalCircuitAnalysis.pdf
"text has been greatly expanded and includes many examples along with computer simulations. For the 
convenience of those who used the Workbook, many of the problem sets are the same, with some re-ordering 
depending on the chapter.
If you are already familiar with DC Electrical Circuit Analysis, the format of this title is similar. This text picks 
up where the DC text leaves off; beginning with AC concepts such as sinusoidal waveforms, basic Fourier 
decomposition of complex waveforms, complex numbers and the like. Also, reactance and impedance are 
introduced along with phasor diagrams. Chapters on series, parallel and series-parallel RLC circuits commence. 
Following these, network theorems along with nodal and mesh analysis are discussed for the AC case. The text 
completes with chapters on AC power, resonance, and introductions to polyphase systems and magnetic circuits.",ACElectricalCircuitAnalysis.pdf
"completes with chapters on AC power, resonance, and introductions to polyphase systems and magnetic circuits. 
Each chapter begins with a set of learning objectives and concludes with practice exercises that are generally 
divided into four major types: analysis, design, challenge and simulation. Many SPICE-based circuit simulators 
are available, both free and commercial, that can be used with this text. The answers to most odd-numbered 
exercises can be found in the Appendix. A table of standard resistor sizes is also in the Appendix, which is useful
for real-world design problems. If you have any questions regarding this workbook, or are interested in 
contributing to the project, do not hesitate to contact me.
This text is part of a series of OER titles in the areas of electricity, electronics, audio and computer 
programming. It includes three other textbooks covering semiconductor devices, operational amplifiers, and",ACElectricalCircuitAnalysis.pdf
"programming. It includes three other textbooks covering semiconductor devices, operational amplifiers, and 
embedded programming using the C language with the Arduino platform. There is a text covering DC electrical 
circuits similar to this one, and also seven laboratory manuals; one for each of the five texts plus individual titles
covering computer programming using the Python language, and the science of sound. The most recent versions 
of all of my OER texts and manuals may be found at my MVCC   web site as well as my mirror site: 
www.dissidents.com 
This text was created using several free and open software applications including Open Office, Dia, SciDA Vis, 
and XnView.
Special thanks to the following individuals for their efforts in reviewing and proofreading the DC and AC 
Electrical Circuit Analysis texts: Glenn Ballard, John Markham, João Nuno Carvalho, Mark Steffka and Jim 
Noon.
5",ACElectricalCircuitAnalysis.pdf
"For the Have-nots and the Has-beens 

“Well, the people, I would say. There is no patent. Could you patent the sun?”
— Jonas Salk, inventor of the polio vaccine,
when asked who owns the patent to it.
6",ACElectricalCircuitAnalysis.pdf
"Table of ContentsTable of Contents
Chapter 1: Fundamentals . . . . . . . 10
1.0 Chapter Objectives . . . . . . . .  10
1.1 Introduction . . . . . . . . .  10
1.2 Sinusoidal Waveforms . . . . . . .  11
1.3 Basic Fourier Analysis . . . . . . . .  25
1.4 Complex Numbers . . . . . . . .  29
1.5 Reactance and Impedance . . . . . . .  31
1.6 Phasor Diagrams . . . . . . . .  37
Summary . . . . . . . . .  38
Exercises . . . . . . . . .  39
Chapter 2: Series RLC Circuits . . . . . .  44
2.0 Chapter Objectives . . . . . . . .   44
2.1 Introduction . . . . . . . . .   44
2.2 The Series Connection . . . . . . .   44
2.3 Series Impedance . . . . . . . .   45
2.4 Series Circuit Analysis. . . . . . . .   48
2.5 Concerning Practical Inductors . . . . . .   61
Summary . . . . . . . . .   63
Exercises . . . . . . . . .   64
Chapter 3: Parallel RLC Circuits . . . . .  80
3.0 Chapter Objectives . . . . . . . .   80
3.1 Introduction . . . . . . . . .   80
3.2 The Parallel Connection . . . . . .   81",ACElectricalCircuitAnalysis.pdf
"3.0 Chapter Objectives . . . . . . . .   80
3.1 Introduction . . . . . . . . .   80
3.2 The Parallel Connection . . . . . .   81
3.3 Parallel Impedance . . . . . . . .   81
3.4 Parallel Circuit Analysis . . . . . . .   85
Summary . . . . . . . . . 100
Exercises . . . . . . . . . 101
Chapter 4: Series-Parallel RLC Circuits . . . . 110
4.0 Chapter Objectives . . . . . . . .  110
4.1 Introduction . . . . . . . . .  110
4.3 The Series-Parallel Connection . . . . . .  110
4.4 Series-Parallel Impedance . . . . . . .  111
4.5 Series-Parallel Circuit Analysis . . . . . . .  114
Summary . . . . . . . . .  132
Exercises . . . . . . . . .  133
7",ACElectricalCircuitAnalysis.pdf
"Chapter 5: Analysis Theorems and Techniques . . . . 144
5.0 Chapter Objectives . . . . . . . .  144
5.1 Introduction . . . . . . . . .  144
5.2 Source Conversions . . . . . . . .  145
5.3 Superposition Theorem . . . . . . .  153
5.4 Thévenin's and Norton's Theorems . . . . . .  160
5.5 Maximum Power Transfer Theorem . . . . . .  170
5.6 Delta-Y (Pi-T) Conversions . . . . . . .  179
Summary . . . . . . . . .  185
Exercises . . . . . . . . .  187
Chapter 6: Nodal and Mesh Analysis . . . . . 204
6.0 Chapter Objectives . . . . . . . .  204
6.1 Introduction . . . . . . . . .  204
6.2 Nodal Analysis . . . . . . . .  205
6.3 Mesh Analysis . . . . . . . .  222
6.4 Dependent Sources . . . . . . . .  234
Summary . . . . . . . . .  240
Exercises . . . . . . . . .  242
Chapter 7: AC Power . . . . . . . 260
7.0 Chapter Objectives . . . . . . . .   260
7.1 Introduction . . . . . . . . .   260
7.2 Power Waveforms . . . . . . . .   261
7.3 Power Triangle . . . . . . . .   273",ACElectricalCircuitAnalysis.pdf
"7.1 Introduction . . . . . . . . .   260
7.2 Power Waveforms . . . . . . . .   261
7.3 Power Triangle . . . . . . . .   273
7.4 Power Systems . . . . . . . .   280
Summary . . . . . . . . .   286
Exercises . . . . . . . . .   287
Chapter 8: Resonance . . . . . . . 294
8.0 Chapter Objectives . . . . . . . .   294
8.1 Introduction . . . . . . . . .   294
8.2 Series Resonance . . . . . . . .   295
8.3 Parallel Resonance . . . . . . . .   310
Summary . . . . . . . . .   329
Exercises . . . . . . . . .   331
8",ACElectricalCircuitAnalysis.pdf
"Chapter 9: Polyphase Power . . . . . . 338
9.0 Chapter Objectives . . . . . . . .   338
9.1 Introduction . . . . . . . . .   338
9.2 Polyphase Definition . . . . . . . .   339
9.3 Three-Phase Connections . . . . . . .   341
9.4 Power Factor Correction . . . . . . .   354
Summary . . . . . . . . .   364
Exercises . . . . . . . . .   365
Chapter 10: Decibels and Bode Plots . . . . . 372
10.0 Chapter Objectives . . . . . . . .   372
10.1 Introduction . . . . . . . .   372
10.2 The Decibel . . . . . . . . .   372
10.3 Bode Plots . . . . . . . . .   383
10.4 Combining the Elements - Multi-Stage Effects . . . . .   393
Summary . . . . . . . . .   395
Exercises . . . . . . . . .   396
Appendices
A: Standard Component Sizes . . . . . . . 399
B: Methods of Solution of Linear Simultaneous Equations . . . 400
C: Equation Proofs . . . . . . . . 405
D: Answers to Selected Odd-Numbered Problems . . . . 408
E: Questions for Selected Odd Answers . . . . . 422
9",ACElectricalCircuitAnalysis.pdf
"1 1 FundamentalsFundamentals
1.0 Chapter Learning Objectives1.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe sinusoidal waveforms in mathematical terms.
• Express complex waveforms using basic Fourier analysis.
• Perform mathematical operations using complex numbers.
• Draw phasor diagrams for complex numbers.
• Compute and plot capacitive and inductive reactance as a function of frequency.
• Compute and plot complex impedance as a function of frequency.
• Describe the relationships between resistance, reactance, impedance, conductance, susceptance and 
admittance
1.1 Introduction1.1 Introduction
In this chapter we begin our study of alternating current, or AC, electrical circuit analysis. As such, this chapter 
is filled with foundational material that will be used in the remainder of this text. It is critical that the concepts",ACElectricalCircuitAnalysis.pdf
"is filled with foundational material that will be used in the remainder of this text. It is critical that the concepts 
presented here be fully understood in order to achieve success in later chapters. We start with the mathematical 
description of the most simple AC waveform, the sine wave. This includes parameters such as amplitude, 
frequency, period, phase and DC offset. From there we discover how to describe other waveforms in terms of 
combinations of sine waves, and also how to determine the effective, or DC equivalent, value of a sine wave. AC
analysis is practically impossible to perform without the use of complex numbers, and discussion follows with 
their description and proper manipulation. Finally, we introduce the concepts of reactance and impedance, and 
how they relate to resistance. This includes examination using both frequency domain plots and phasor 
diagrams. Phasor diagrams are vector plots and can be used to show the relationships between various voltages",ACElectricalCircuitAnalysis.pdf
"diagrams. Phasor diagrams are vector plots and can be used to show the relationships between various voltages 
in a circuit, as well as between currents or resistive/reactive values.
Many of the topics in this text will echo your studies in DC circuit analysis, such as Ohm's law, Kirchhoff's 
voltage and current laws, series-parallel analysis, nodal analysis, and the like. Thus many concepts will be 
familiar. The major practical difference is that all quantities in DC systems are scalars, that is, they have only 
magnitude. In contrast, AC quantities are vectors, and thus have both magnitude and direction (or more properly,
phase). Consequent;y, mathematical operations such as addition or multiplication have to follow vector rules 
instead of scalar rules. Forgetting to do so is a common error for students new to the subject and one that will 
bring no end of grief. These rules will be reviewed in the section covering complex numbers. Make sure that",ACElectricalCircuitAnalysis.pdf
"bring no end of grief. These rules will be reviewed in the section covering complex numbers. Make sure that 
you have this material mastered before proceeding. 
10",ACElectricalCircuitAnalysis.pdf
"1.2 Sinusoidal Waveforms1.2 Sinusoidal Waveforms
AC, or alternating current, is so named because the current alternates or flips back 
and forth between two polarities. In other words, the current (and consequently the 
voltage) is a function of time. This is fundamentally different from direct current that
is fixed in polarity and generally constant over time. A laboratory DC voltage source,
for example, ideally maintains a set voltage across its terminals and does not vary 
over time. In contrast, as an AC waveform swings back and forth through time, its 
shape can exhibit wide variations ranging from the simple, regular paths of 
laboratory standards such as sine waves, triangle waves and square waves, to the far 
more complex and undulating waveforms produced by musical instruments and the 
human voice. 
The sine wave is the simplest wave that may be created. It represents the motion of a
simple vector rotating at a constant speed, such as the vertical displacement of the",ACElectricalCircuitAnalysis.pdf
"simple vector rotating at a constant speed, such as the vertical displacement of the 
second hand of a clock. An example is shown in Figure 1.1. The horizontal axis plots
time. It increases as we move from left to right (i.e., if point A is to the right of point 
B, then A occurs later in time than does B). The vertical axis is represented here in 
general as a percentage of maximum but would ordinarily be a measurement of 
voltage, current, sound pressure, or the like. 
Note the smooth variation that starts at zero, rises to a positive peak one quarter way 
through, falls back to zero when halfway through, continues to a negative peak three
quarters through, and then rises again to where it started. This process then repeats. 
Each repeat is referred to as a cycle. In Figure 1.1, one complete cycle is shown. 
11
Figure 1.1
A sine wave.
Sine Wave
Percentage of Peak
-100
-80
-60
-40
-20
0
20
40
60
80
100
 
Time (cycles)
0 0.25 0.5 0.75 1",ACElectricalCircuitAnalysis.pdf
"Sine waves exhibit quarter wave symmetry. That is, each quarter (in time) of the 
wave is identical to any other if you simply flip it around the horizontal axis and/or 
rotate it upright about its peak. The time it takes to complete one cycle is called the 
period and is denoted with the symbol T (for Time). The reciprocal of the period is 
the frequency, f. 
 f = 1
T (1.1)
The frequency indicates how many cycles exist in one second. To honor one of the 
19th century researchers in the field, instead of calling the unit “cycles per second”, 
we use Hertz, named after Heinrich Hertz and abbreviated Hz. In Figure 1.2 three 
sine waves are shown with different frequencies; the initial wave (green), a wave at 
twice the frequency (blue), and a third at half the frequency or twice the period (red).
The amplitude (vertical) of the wave can be expressed as a peak quantity, which is 
the change from the center zero line up to the most positive value. Amplitude may",ACElectricalCircuitAnalysis.pdf
"the change from the center zero line up to the most positive value. Amplitude may 
also be expressed as peak-to-peak; the distance from the most negative to the most 
positive. For a sine wave this will always be twice the peak value, although that 
might not be the case for other waves which may be asymmetrical. A series of three 
sine waves with differing amplitudes are shown in Figure 1.3. Along side the initial 
(green) are double amplitude (blue) and half amplitude (red) versions. 
12
Variation in Frequency
Percentage of Peak
-100
-80
-60
-40
-20
0
20
40
60
80
100
 
Time
0 0.2 0.4 0.6 0.8 1
Double period
Initial
Double frequency
Figure 1.2
Sine wave frequency variation.",ACElectricalCircuitAnalysis.pdf
"Combining these parameters, consider the voltage waveform shown in Figure 1.4. 
Here we see two cycles of an AC voltage waveform. 
13
Variation in Amplitude
Percentage of Peak
-200
-150
-100
-50
0
50
100
150
200
 
Time
0 0.2 0.4 0.6 0.8 1
One half amplitude
Initial
Double amplitude
Figure 1.3
Sine wave amplitude variation.
Figure 1.4
Basic sine wave example.
Volts
-5
-4
-3
-2
-1
0
1
2
3
4
5
 
Time (seconds)
0 0.1 0.2 0.3 0.4",ACElectricalCircuitAnalysis.pdf
"The peak value is 4 volts and the peak-to-peak value is 8 volts (typically abbreviated
as “8 V pp”). The period of one cycle is 0.2 seconds, or T = 200 milliseconds. 
Further, the frequency, f = 1/200 milliseconds, or 5 Hz (5 cycles in one second). 
AC waveforms may also be combined with a DC offset. Adding a positive DC level 
shifts the wave up vertically, while a negative DC level shifts the wave down 
vertically. This does not alter the frequency or AC portion of the amplitude (although
the absolute peaks would shift by the DC value). Figure 1.5 shows the effect of 
various DC offsets. Above the initial wave (green) is an otherwise identical wave 
with a positive DC offset equal to 20% of the original peak value (blue). Below the 
original is a third wave (red) that exhibits a negative DC offset equal to half of the 
peak value of the original.
Further, it is possible for a sine wave to be shifted in time compared to some other",ACElectricalCircuitAnalysis.pdf
"peak value of the original.
Further, it is possible for a sine wave to be shifted in time compared to some other 
sine wave or reference. While it is possible to indicate this shift as an absolute time, 
it is more common to do so as a phase shift, that is, the time expressed as a portion 
of the period in degrees. For example, if one sine is ahead of another by one quarter 
of the period, it is said to be leading by 90° (i.e., ¼ of 360°). If it is behind by 
½ of the period, it is said to be lagging by 180° (i.e., later in time by ½ cycle). 
Another way of stating this is that leading waveforms start earlier in time and thus 
are drawn to the left of the reference, while lagging waveforms start later in time and
are drawn to the right.
14
Effect of DC Offset
Percentage of Peak
-150
-100
-50
0
50
100
150
 
Time
0 0.2 0.4 0.6 0.8 1
Negative DC offset, 50%
Initial
Positive DC offset, 20%
Figure 1.5
Sine wave DC offset variation.",ACElectricalCircuitAnalysis.pdf
"Figure 1.6 illustrates the effect of phase shift. Note that in this plot, t = 0 has been 
moved to the center of the horizontal axis. The middle curve is the initial, or 
reference, wave (green). To the left (red) is a wave leading the initial wave by one-
eighth cycle, or 45°. To the right (blue), is a lagging wave of half as much, or 
−22.5°.
Combining the foregoing elements allows us to develop a general format for a sine 
wave (voltage shown):
v(t )=V DC +V P sin(2π f t +θ) (1.2)
Where 
v(t) is the voltage at some time t,
VDC is the DC offset, if any,
VP is the peak value,
f is the frequency,
θ is the phase shift (+ if leading and drawn to the left, − if lagging and drawn
to the right).
For a quick and practical example, the waveform shown in Figure 1.4 has an 
amplitude of 4 volts peak, a frequency of 5 Hz, and no DC offset or phase shift. 
Thus, its expression is v(t )=4sin(2 π5t )
15
Effect of Phase Shift
Percentage of Peak
-100
-80
-60
-40
-20
0
20
40
60
80
100
 
Time",ACElectricalCircuitAnalysis.pdf
"Thus, its expression is v(t )=4sin(2 π5t )
15
Effect of Phase Shift
Percentage of Peak
-100
-80
-60
-40
-20
0
20
40
60
80
100
 
Time
-1 -0.5 0 0.5 1
Leading by 
45 degrees
Initial
Lagging by
22.5 degrees
Figure 1.6
Sine wave phase variation.",ACElectricalCircuitAnalysis.pdf
"To compute a phase shift, first determine the time differential between the waveform
and the reference, which we'll call Δt. The reference may be a fixed point in time 
(e.g., t = 0) or another waveform. Generally, the easy way to do this is to measure 
the difference at the zero-crossings, assuming there is no DC offset. If there is an 
offset, make the measurement where the zero crossing has been shifted to (i.e., at the
DC offset level). Once the difference is found, divide it by the period to represent the
shift as a fraction of a period. As one cycle represents one rotation of the vector, or 
360 degrees, simply multiply the fraction by 360 degrees to find the phase shift in 
degrees. Expressed as a formula:
θ = 360° Δt
T (1.3)
Remember, if the wave is shifted to the left then it is leading and positive, while a 
shift to the right is lagging or delayed in time, and thus negative.
Example 1.1
Write the expression for the waveform shown in Figure 1.7.",ACElectricalCircuitAnalysis.pdf
"shift to the right is lagging or delayed in time, and thus negative.
Example 1.1
Write the expression for the waveform shown in Figure 1.7.
 This waveform superficially may look like the one in Figure 1.4 but don't 
let this fool you. First of all, the time scale is different. For this waveform, 
one cycle completes in 10 milliseconds. Therefore, the frequency is
16
Figure 1.7
Waveform for Example 1.1.
Amps
-3
-2
-1
0
1
2
3
4
5
 
Time (milliseconds)
0 5 10 15 20",ACElectricalCircuitAnalysis.pdf
"f = 1
T
f = 1
10 ms
f = 100 Hz
The second issue is the DC offset. Note that the positive peak occurs at 4 
amps while the negative peak occurs at −2 amps. This indicates a peak-to-
peak value of 6 amps. Without an offset, the positive peak would be at 3 
amps, therefore there is a +1 amp DC offset. The vertical center of the 
waveform is shifted up from 0 amps to +1 amp. This point is at t = 0, 
therefore, there is no phase shift. The resulting expression is:
i(t )=1+3sin(2 π100 t )
Example 1.2
Write the expression for the waveform shown in Figure 1.8. 
First off, the positive peak is 2 volts and the peak-to-peak value is 4 volts. 
Therefore there is no DC offset. The vertical center of the wave does not 
start at t = 0, thus there must be a phase shift. The value at t = 0 is 1.2 volts. 
The wave hits this same amplitude at t = 2 milliseconds and begins to repeat 
another cycle. Consequently the period must be 2 milliseconds. The",ACElectricalCircuitAnalysis.pdf
"The wave hits this same amplitude at t = 2 milliseconds and begins to repeat 
another cycle. Consequently the period must be 2 milliseconds. The 
frequency is the reciprocal of this value, and thus f = 500 Hz. 
17
Figure 1.8
Waveform for Example 1.2.
Volts
-3
-2
-1
0
1
2
3
 
Time (milliseconds)
0 1 2 3 4",ACElectricalCircuitAnalysis.pdf
"The waveform is shifted to the left which indicates a positive or leading 
phase shift. If we examine the second cycle, we see that it hits zero volts at 
1.8 milliseconds. Therefore the shift is 0.2 milliseconds. Expressed in 
degrees this is:
θ = 360° Δt
T
θ = 360° 0.2ms
2 ms
θ = 36°
The final expression is:
 
v(t )=2sin(2π500 t+36 °)
Example 1.3
Draw the waveform corresponding to the following expression.
v(t )=−3+5sin (2π40000 t−72° )
First, note that the −3 volt offset pushes the positive peak down from 5 volts 
to 2 volts, and the negative peak down from −5 volts to −8 volts. The 
frequency of 40 kHz dictates a period of:
T = 1
f
T = 1
40 kHz
T = 25μs
The phase shift of −72° represents 72/360, or 0.2 cycles. This corresponds to
a time delay (shifted right because it's negative) of 0.2 times 25 μs, or 5 μs.
Initially, it is often best to construct the plot via a series of discrete steps 
rather than trying to draw the entire thing in one go. First, draw a sine wave",ACElectricalCircuitAnalysis.pdf
"rather than trying to draw the entire thing in one go. First, draw a sine wave 
with a 5 volt peak amplitude and a period of 25 μs. Now, push the waveform
down 3 volts so that the positive peak is only 2 volts and the negative peak 
is down at −8 volts. Finally, push the newly shifted waveform to the right by
5 μs. The result is shown in Figure 1.9.
18",ACElectricalCircuitAnalysis.pdf
"Laboratory MeasurementsLaboratory Measurements
In the laboratory, a function generator is used to generate sines and other
waveshapes. These devices will allow precise control over both the amplitude and
frequency of the wave along with adding a DC offset, if desired. An example is
shown in Figure 1.10. The corresponding measurement tool is the oscilloscope, or
just scope, for short. 
The oscilloscope is perhaps the most useful and versatile measurement device in
the laboratory. Typically, they feature either two or four input channels, although
more are possible. Each input channel has its own sensitivity adjustment and all
channels share a common time reference. The display draws waveforms in the
same manner as those seen in Figures 1.1 – 1.9. Also, they can plot one voltage
versus another (X – Y mode). Modern oscilloscopes have additional features such as 
the automatic measurement of frequency, amplitude, phase shift, etc., cursor-based",ACElectricalCircuitAnalysis.pdf
"the automatic measurement of frequency, amplitude, phase shift, etc., cursor-based 
measurements, and the ability to save display images as graphics files. An example 
of a four channel digital oscilloscope is shown in Figure 1.11. 
19
Volts
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
 
Time (microseconds)
0 5 10 15 20 25 30 35 40 45 50
Figure 1.9
Waveform for Example 1.3.
Figure 1.10
Laboratory signal generator.",ACElectricalCircuitAnalysis.pdf
"Schematic SymbolsSchematic Symbols
As far as schematics are concerned, the symbols for AC voltage and current sources 
are shown in Figure 1.12. The polarity and direction markings are not absolute; after 
all, these are AC sources whose polarity and directions flip back and forth. The 
markers are instead used to establish a timing reference, especially in circuits 
employing multiple sources. 
It is worth remembering that negating a source is the same as flipping its polarity. 
This was true for DC sources and remains true for AC sources. This is illustrated in 
Figure 1.13. Sometimes flipping or negating source will make analysis a little more 
obvious or easier to visualize. 
20
Figure 1.11
A digital oscilloscope.
Figure 1.12
Schematic symbols for AC 
voltage source (left) and 
current source (right).
Figure 1.13
Polarity/sign equivalence.",ACElectricalCircuitAnalysis.pdf
"Example 1.4
Assume an oscilloscope displays two waves as depicted in Figure 1.14. 
Determine the phase shift of the smaller 20 volt peak (blue) waveform 
relative to the larger 25 volt peak (red) waveform.
First, note that neither wave exhibits a DC offset. If one or both them had an 
offset, the wave(s) would have to be shifted vertically so that their normal 
zero-crossing points would be at the same level. Measuring either wave, the 
period is found to be 1 millisecond. The time shift most easily can be found 
at any of the zero-crossings (there are four locations to choose from). The 
delay is one small deviation, or 0.1 milliseconds, with the smaller wave 
delayed in time, or lagging the larger wave. This indicates a negative phase 
shift.
θ = 360 ° Δt
T
θ = 360° −0.1 ms
1ms
θ =−36 °
21
Figure 1.14
Waveforms for Example 1.4.
Volts
-25
-20
-15
-10
-5
0
5
10
15
20
25
 
Time (milliseconds)
0 0.5 1 1.5 2",ACElectricalCircuitAnalysis.pdf
"Sines and CosinesSines and Cosines
There are a handful of specific phase shifts that are worth a closer look. If a sine 
wave is inverted, that is, flipped upside down, it is indistinguishable from a sine 
wave that has been shifted either +180 or −180 degrees. In other words, such a wave
can be written three different ways: −sin(2πft), sin(2πft − 180°), or sin(2πft + 180°). 
Further, if a sine wave is shifted by +90 degrees (i.e., leading and to the left), it may 
also be referred to as a cosine wave. Thus sin(2πft + 90°) = cos(2πft). Finally, if a 
sine wave is shifted by −90 degrees (i.e., lagging and to the right), it may be referred
to as a negative or inverted cosine wave. Thus sin(2πft − 90°) = −cos(2πft). The 
relationships of these four waves are illustrated in Figure 1.15.
It is also worth noting that the cosine wave represents the first derivative, or slope, 
of the sine wave. As you may recall from other studies, the slope or “steepness” of a",ACElectricalCircuitAnalysis.pdf
"of the sine wave. As you may recall from other studies, the slope or “steepness” of a 
line is the ratio of the vertical change to the horizontal change, sometimes called 
“the rise over the run”. For a voltage, it would be the change in voltage over the 
change in time, or ΔV/Δt. For a smooth, continuously changing curve like a sine 
wave, the slope at a given point is defined properly as the first derivative, or dv/dt in 
this case. To verify that this is true visually, note that the steepest part of the sine 
wave (green) is where it crosses zero amplitude. As it crosses zero while moving 
positive (at t = 0 or t = 1 in Figure 1.15), the cosine (blue) is at its positive peak. As 
the sine cross zero while moving negative (at t = 0.5), the cosine is at its negative 
peak. Further, the sine wave is flat with a slope of zero at its positive and negative 
peaks (at at t = 0.25 and t = 0.75, respectively), and at those times the cosine's 
22
Sines and Cosines
Percentage of Peak
-100
-80",ACElectricalCircuitAnalysis.pdf
"peaks (at at t = 0.25 and t = 0.75, respectively), and at those times the cosine's 
22
Sines and Cosines
Percentage of Peak
-100
-80
-60
-40
-20
0
20
40
60
80
100
 
Time
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-Sine
Sine
Cosine
-Cosine
Figure 1.15
Time relationships between 
sines and cosines.",ACElectricalCircuitAnalysis.pdf
"amplitude is also zero. It is also true that the sine wave is the slope of the negative 
cosine wave, the negative cosine is the slope of the negative sine, and the negative 
sine is the slope of the cosine. Moving in the reverse direction, we can say that the 
anti-derivative (indefinite integral) of a cosine wave is a sine wave, the integral of a 
sine wave is a negative cosine wave, and so forth. These relationships will prove 
most useful when we turn our attention to the response of capacitors and inductors in
AC circuits.
RMS – Root Mean Square MeasurementRMS – Root Mean Square Measurement
 Along with peak and peak-to-peak, amplitude may be given as an RMS (Root Mean 
Square) value. In fact, if peak or peak-to-peak is not specified, the measurement 
is assumed to be RMS. RMS is a special calculation used for finding equivalent DC 
power (very common, for example, with audio power amplifiers). In other words, if",ACElectricalCircuitAnalysis.pdf
"power (very common, for example, with audio power amplifiers). In other words, if 
we are interested in finding the power in a resistor, the calculation must be 
performed using RMS values for voltage or current, not peak or peak-to-peak values.
Failure to do so will result in erroneous powers. This is true regardless of the 
waveshape; be it a sine wave, triangle wave or the complex waves of music signals. 
If a voltage is specified as RMS, it can be treated for power calculations just like an 
equivalently sized DC voltage. For example, a 1 volt RMS sine will produce the 
same power dissipation and heating in a given resistor as will 1 volt DC. For this 
reason, RMS is sometimes referred to as the effective value (i.e., effective DC value).
The name root-mean-square describes the process of determining the effective 
value. First, recall that power is proportional to the square of the voltage or current.",ACElectricalCircuitAnalysis.pdf
"value. First, recall that power is proportional to the square of the voltage or current. 
Thus, our first step will be to square the input waveform. Of course, the waveform is
a function of time and its square will yield some new shape. At this point, we need to
find the average value of this new shape. The reason for this is simple, yet not 
necessarily obvious. Electrical and electronic components have mass, and thus do 
not heat or cool instantly. They exhibit a thermal time constant. Therefore, they 
respond to the average input over time. While we could compute some manner of 
“instantaneous peak power” at some specific instant in time, it does not represent the
equivalent DC power. Once we have obtained the mean value of this squared 
waveform, the corresponding DC value is just the square root of the mean. The result
is a fractional value between zero and one that is used as a scaling factor to turn a",ACElectricalCircuitAnalysis.pdf
"is a fractional value between zero and one that is used as a scaling factor to turn a 
peak value into an RMS value. The value will be unique to the specific waveshape. 
That is, all sines (regardless of phase) have the same factor, all regular triangle 
waves have the same factor, and so on. As we mostly concern ourselves with sines, 
let's take a closer look at determining the RMS factor for them.
We begin with the basic expression for a sine wave without DC offset or phase shift, 
and with an amplitude of one:
v(t )=sin (2π f t )
23",ACElectricalCircuitAnalysis.pdf
"The first step is to square this waveform. A useful trigonometric identity is 
(sin x)2
=1
2 −1
2 cos 2 x
Applying this to our wave yields:
v(t )2
= 1
2 − 1
2 cos(2π2 f t)
This expression describes an inverted cosine wave at twice the original frequency 
and half of the original amplitude, riding on a DC offset equal to its peak value. In 
other words, the negative peak of the cosine is at zero and the positive peak is at 1. 
The next step is to find the average or mean value of this intermediate result. The 
mean is equal to the offset of 0.5. This can be visualized as the area above the offset 
perfectly filling the “dip” below the offset. The final step is to take the square root of
the mean. The square root of 0.5 is equal to one over the square root of two, or 
approximately 0.707. Therefore the RMS value is 0.707 times the peak. Alternately, 
you could divide the peak by square root of two, or approximately 1.414. This 
process is shown graphically in Figure 1.16.",ACElectricalCircuitAnalysis.pdf
"you could divide the peak by square root of two, or approximately 1.414. This 
process is shown graphically in Figure 1.16.
In summation, for sine waves, RMS is always the peak value times 0.707. We could 
also say the RMS value of any sine wave is its peak divided by approximately 1.414. 
Again, these ratios would not necessarily be true of non-sine waves. Details 
regarding other common shapes can be found in Appendix C. Finally, the ratio of the
peak value to the RMS value is called the crest ratio. This is a fixed value for sine 
waves (again, about 1.414), but can be over 10:1 for some kinds of audio signals. 
24
Figure 1.16
Process to find RMS factor for 
sines.
Fraction of Peak
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
 
Time 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Original sine
Sine squared
Mean",ACElectricalCircuitAnalysis.pdf
"WavelengthWavelength
Another item of interest is the speed of propagation of the wave. This varies widely. 
In the case of light in a vacuum (or to a close approximation, an electrical current in 
a wire), the velocity is about 3E8 meters per second (i.e., 300,000 km/s) or about 
186,000 miles per second. 
Given a velocity and a period, we can imagine how far apart the peaks of the wave 
are. This distance is called the wavelength and is denoted by the Greek letter lambda 
λ. Wavelength is equal to the velocity divided by the frequency, λ = v/f. Thus, for a 
loudspeaker producing a 100 Hz sine, as the velocity of sound in air is 344 m/s, then 
λ = 344 m/s / 100 Hz, or 3.44 meters (a little over 11 feet). Notice that the higher the
frequency, the shorter the wavelength. Also, note that the faster the velocity, the 
longer the wavelength. Wavelength calculations are of particular importance in the 
fields of telecommunications and acoustics.",ACElectricalCircuitAnalysis.pdf
"longer the wavelength. Wavelength calculations are of particular importance in the 
fields of telecommunications and acoustics.
1.3 Basic Fourier Analysis1.3 Basic Fourier Analysis
The Fourier theorem, named after the French mathematician Jean-Baptiste Joseph 
Fourier, states that any repetitive waveform can be represented as a collection of sine
and cosine waves of the proper amplitude and frequency. Alternately, it may be 
represented as a series of sine waves each with the proper amplitude, frequency and 
phase. This includes complex signals such as the human voice and musical 
instruments. Consequently, if a system is linear, the response of a system to a 
complex wave may be understood in terms of its response to individual sine waves, 
via superposition. 
In this collection of waves, each component is known as a partial with the lowest 
frequency component known as the fundamental. All other partials are grouped",ACElectricalCircuitAnalysis.pdf
"frequency component known as the fundamental. All other partials are grouped 
together and referred to as overtones. “Regular” waveforms such as square waves 
and triangle waves feature a harmonic overtone sequence meaning that these 
overtones are integer multiples of the fundamental. As a shortcut, they are often 
referred to as just harmonics. 
It might be hard to visualize initially, but like all waves, waves in the shape of a 
square or triangle are made up of a series of sines. The equation for a square wave is:
)2)12sin((12
1)( 1 ftnntv n  
 (1.4)
This says that a square wave of frequency f is made up of an infinite series of sines 
at odd integer multiples of f, with an inverse amplitude characteristic. For example, a
100 Hz square consists of a 100 Hz sine, plus a 300 Hz sine at 1/3 amplitude, plus a 
500 Hz sine at 1/5 amplitude, plus a 700 Hz sine at 1/7 amplitude, and so on. 
25",ACElectricalCircuitAnalysis.pdf
"A triangle wave is similar:
)2)12cos(()12(
1)( 1 2 ftnntv n  
 (1.5)
Thus a triangle wave of frequency f is made up of an infinite series of cosines (sines 
with a 90 degree or one quarter cycle phase shift) at odd integer multiples of f, with 
an inverse square amplitude characteristic. For example, a 100 Hz triangle consists 
of a 100 Hz cosine, plus a 300 Hz cosine at 1/9 amplitude, plus a 500 Hz cosine at 
1/25 amplitude, plus a 700 Hz cosine at 1/49 amplitude, and so on.
A series of graphs showing the construction of a square wave and a triangle wave 
follow. The square wave sequence begins with the fundamental and the first 
harmonic in Figure 1.17. The result is an oddly bumpy wave. The second graph of 
Figure 1.18 adds the next two harmonics. As more harmonics are added, the sides 
get steeper and the top/bottom start to flatten. They flatten because each additional 
harmonic partially cancels some of the peaks and valleys from the previous",ACElectricalCircuitAnalysis.pdf
"harmonic partially cancels some of the peaks and valleys from the previous 
summation. This gives rise to a greater number of undulations with each undulation 
being smaller in amplitude. The sequence finishes with Figure 1.19 showing seven 
harmonics being added with the result approaching a reasonable square wave. As 
ever more harmonics are added, the wave would approach a flat top and bottom with
vertical sides, the idealized square wave. 
26
Building a Square Wave
Voltage
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
 
Time
0 0.2 0.4 0.6 0.8 1
Fundamental plus third harmonic
Fundamental
Third harmonic
Figure 1.17
Building a square wave, part 1.",ACElectricalCircuitAnalysis.pdf
"27
Building a Square Wave
Voltage
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
 
Time
0 0.2 0.4 0.6 0.8 1
Fundamental plus third, 
fifth and seventh harmonics
Fundamental
Third harmonic
Fifth harmonic
Seventh harmonic
Figure 1.18
Building a square wave, part 2.
Figure 1.19
Building a square wave, part 3.
Building a Square Wave
Voltage
-1.0
-0.5
0.0
0.5
1.0
 
Time
0 0.2 0.4 0.6 0.8 1
Fundamental plus 
seven harmonics:
3, 5, 7, 9, 11, 13, 15Fundamental",ACElectricalCircuitAnalysis.pdf
"The triangle sequence begins with a fundamental and the first harmonic as shown in 
Figure 1.20. The resulting combination is already trending away from a simple sine 
shape. The second and final graph, Figure 1.21, shows a total of seven harmonics. 
The result is very close to a triangle, the only obvious deviation is the slight 
rounding at the very peaks. The addition of more harmonics would cause these to 
sharpen further. 
28
Building a Triangle Wave
One Harmonic
Voltage
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
 
Time
0 0.5 1 1.5 2
Fundamental plus 
one harmonic
Fundamental
Harmonic at three
times fundamental
Figure 1.20
Building a triangle wave, 
part 1.
Building a Triangle Wave
Seven Harmonics
Voltage
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
 
Time
0 0.5 1 1.5 2
Fundamental plus 
seven harmonics
Fundamental
Figure 1.21
Building a triangle wave, 
part 2.",ACElectricalCircuitAnalysis.pdf
"1.4 Complex Numbers1.4 Complex Numbers
In AC circuits, parameters such as voltage and current are vectors, that is, they have 
both a magnitude and a phase shift or angle. For example, a voltage might be “12 
volts at an angle of 30 degrees” (or more compactly, 1230º). This is known as 
polar form or magnitude-angle form. Alternately, a vector can be broken into 
rectangular form, that is, its right angle components.
This can be visualized as a right triangle where the magnitude is the hypotenuse, the 
angle is the rotation above or below the horizontal, the horizontal component is the 
side adjacent to the angle and the vertical component is the side opposite of the 
angle. This is shown in Figure 1.22.
Properly, voltage and current vectors are complex numbers that lay on a complex 
plane consisting of a real part and an imaginary part. The horizontal axis is the real 
number axis and the vertical axis is the imaginary number axis. The imaginary axis",ACElectricalCircuitAnalysis.pdf
"number axis and the vertical axis is the imaginary number axis. The imaginary axis 
denotes values times the imaginary operator j (and often referred to as i outside of 
electrical analysis). The j operator is the square root of −1. An example of such a 
value is 3 + j4, in other words, 3 units along the horizontal real axis and 4 units up 
the vertical imaginary axis. This is depicted in Figure 1.23.
29
Horizontal component
Vertical 
component
Magnitude
θ
Figure 1.22
Polar and rectangular forms.
Figure 1.23
Polar and rectangular forms on
the complex plane.
+Real
+Imaginary (+j)
Real component
Imaginary 
component
Magnitude
θ
-Real
-Imaginary (-j)",ACElectricalCircuitAnalysis.pdf
"Converting from one form to another relies on basic trigonometric relations. For 
convenience, the relationships between the magnitude, angle, real component and 
imaginary component are reproduced below:
Magnitude = √Real 2
+Imaginary2
θ = tan−1 Imaginary
Real
Real = Magnitude cosθ
jImaginary = Magnitude sin θ
(1.6a – d)
 
To add or subtract complex quantities, first put them into rectangular form and then 
combine the reals with the reals and the j terms with the j terms as in (3 + j5) +     
(13 − j1) = 16 + j4. These real and imaginary terms must be kept separate. 3 + j5 
does not equal 8 (or even j8). That would be like saying that moving 3 feet to your 
right and 5 feet forward puts you in the same location as moving 8 feet to your right 
(or 8 feet forward). 
The direct way to multiply or divide complex values is to first put them in polar 
form and then multiply or divide the magnitudes. The angles are added together for",ACElectricalCircuitAnalysis.pdf
"form and then multiply or divide the magnitudes. The angles are added together for 
multiplication and subtracted for division. For example, 1230º times 245º is 
2475º while dividing them yields 6−15º. The need for complex numbers will 
become more obvious as we move through the upcoming material. It is imperative 
that you have mastered the manipulation of complex values before moving on to 
subsequent chapters.
Example 1.5
Convert 15 + j20 and 1 k − j2 k into polar form, and 1045º and 0.2−30º 
into rectangular form.
For the first two conversions, use Equations 1.6a and 1.6b.
Magnitude = √152
+202
= 25
θ = tan−1
(
20
15 )= 53.1º
Magnitude = √1 k2+(−2 k)2 = 2.236 k
θ = tan−1
(
−2k
1 k )= −63.4º
The answers for the first part are 2553.1º and 2.236 k−63.4º. 
For the second pair of conversions use Equations 1.6c and 1.6d.
30",ACElectricalCircuitAnalysis.pdf
"Real = 10 cos 45º = 7.07
jImaginary = 10sin 45º = j 7.07
Real = 0.2cos−30º = 0.173
jImaginary = 0.2sin−30º =− j0.1
The answers for the second part are 7.07 + j7.07 and 0.173 − j0.1.
Example 1.6
Perform the following: a) Add 7 + j8 to + j6,   b) Subtract 553.1º from 
10−45º,   c) Divide 2090º by 4−50º,   d) Multiply 90 + j75 by 6 + j10.
a) Add real to real and imaginary to imaginary: 7 + j14.
b) First convert the values into rectangular: 3 + j4 and 7.07 − j7.07. Now 
subtract the first pair from the second pair: 4.07 − j11.07 (or 11.8−69.8º).
c) Divide the magnitudes and subtract the angles: 5140º.
d) First convert the values into polar: 117.1539.8º and 11.6659º. Now 
multiply the magnitudes and add the angles: 1.366 k98.8º.
1.5 Reactance and Impedance1.5 Reactance and Impedance
Unlike a resistor, the voltage and current will not be in phase for an ideal capacitor 
or for an ideal inductor. For the capacitor, the current leads the voltage across the",ACElectricalCircuitAnalysis.pdf
"or for an ideal inductor. For the capacitor, the current leads the voltage across the 
capacitor by 90 degrees. Recall that the voltage across a capacitor cannot change 
instantaneously, i = C dv/dt. For an inductor, the voltage leads the current by 90 
degrees. Similarly, the current through an inductor cannot change instantaneously 
due to v = L di/dt. While ideal capacitors and inductors do not exhibit resistance, the 
voltage does react to the current. Unsurprisingly, we call this characteristic 
reactance and denote it with the letter X. Reactance, like resistance, is a ratio of 
voltage to current. We define capacitive reactance as:
X C = vc
ic
(1.7)
Using a simple sine wave with a peak value of unity, vc = sin 2π f t, ic can be found 
from i = C dv/dt. Recall from our prior work that the derivative or slope of a sine 
wave is a cosine wave, which is in turn equivalent to a sine wave shifted by +90º.
31",ACElectricalCircuitAnalysis.pdf
"ic = C dvc
dt
ic = C d sin(2 π f t )
dt
ic = C sin(2π f t +90º)
Substituting this result into Equation 1.7 yields:
X C = vc
ic
X C = sin(2π f t )
2 π f C sin (2π f t+90º)
X C = 1
2π f C −90º
Or more directly:
X C =− j 1
2π f C (1.8)
In summation, if we divide the capacitor's sinusoidal voltage by its current (Ohm's 
law), we obtain a value with a phase angle of −90°. While the resultant is an ohmic 
value, it cannot be classified as a resistance. Instead, it is referred to as a reactance 
and denoted with the letter X. Thus we can refer to a capacitor's reactance, XC, as 
some number at an angle of −90º, or more conveniently, we simply prepend −j, as in 
XC = −j75 Ω. 
The case for the inductor is similar and left as an exercise. The inductive reactance, 
XL, can be found using:
X L=+j 2π f L (1.9)
An example would be XL = j68 Ω. As stated, while a resistance cannot be added 
directly to a reactance, reactances can be added together so long as we heed the",ACElectricalCircuitAnalysis.pdf
"directly to a reactance, reactances can be added together so long as we heed the 
signs. For example, if we have a capacitive reactance of −j60 Ω in series with an 
inductive reactance of j100 Ω, the result is j40 Ω. This is due to the fact that these 
two items are 180 degrees out of phase with each other, and so they partially cancel. 
Remember, always add or subtract like items: real (resistance) to real, and imaginary
(reactance) to imaginary.
Equations 1.8 and 1.9 are notable because the reactance is not just a function of the 
capacitance or inductance, but also a function of frequency. The reactance of an 
inductor is directly proportional to frequency while the reactance of a capacitor is 
inversely proportional to frequency. The ohmic variations of a 20 Ω resistor, a 
500 μF capacitor and a 500 μH inductor across frequency are shown in Figure 1.24.
32",ACElectricalCircuitAnalysis.pdf
"We can see that the value of resistance does not change with frequency while the 
inductive reactance increases with frequency and the capacitive reactance decreases.
The linear frequency scale makes the capacitor change difficult to see. If this is 
plotted again but using a logarithmic frequency scale as in Figure 1.25, the 
symmetry becomes apparent. 
33
Resistance & Reactance vs. Frequency
Resistance or Reactance (ohms)
0
5
10
15
20
25
30
35
Frequency (Hz)
2,000 4,000 6,000 8,000 1e+04
20 ohm resistor
500 μH Inductor
500 μF capacitor
Figure 1.24
Resistance and reactance 
versus frequency (linear axis).
Figure 1.25
Resistance and reactance 
versus frequency (log axis).
Resistance & Reactance vs. Frequency
Resistance or Reactance (ohms)
0
5
10
15
20
25
30
35
Frequency (Hz)
10 100 1,000 1e+04
20 ohm resistor
500 μH Inductor500 μF capacitor",ACElectricalCircuitAnalysis.pdf
"The effect of both capacitor size and frequency is shown in Figure 1.26 using a log 
frequency axis: the smaller the capacitor, the larger the capacitive reactance at any 
particular frequency.
Similarly, the effect of both inductor size and frequency is shown in Figure 1.27 
using a linear frequency axis: the larger the inductor, the larger the inductive 
reactance at any given frequency.
34
Figure 1.26
Variation of capacitive 
reactance with capacitance and
frequency.
Capacitive Reactance vs. Frequency
Reactance (ohms)
0
5
10
15
20
25
30
35
40
Frequency (Hz)
100 1,000 1e+04 1e+05
10 μF
200 μF
50 μF
1 μF
100 nF
Figure 1.27
Variation of inductive reactance
with inductance and frequency.
Inductive Reactance vs. Frequency
Reactance (ohms)
0
5
10
15
20
25
30
35
40
45
50
Frequency (Hz)
0 2,000 4,000 6,000 8,000 10,000
50 μH
400 μH
3 mH
1 mH",ACElectricalCircuitAnalysis.pdf
"It is worth noting that the plots of Figures 1.26 and 1.27 are for ideal components. In
reality, all components exhibit some resistive, capacitive and inductive effects due to
their construction. For example, eventually, inductive and resistive effects will cause
the capacitive reactance curves of Figure 1.26 to begin to rise at high frequencies. 
Similarly, resistive and capacitive effects will cause the curves Figure 1.27 to flatten 
at very low and very high frequencies.
An interesting observation to remember is that capacitors and inductors are a bit like
a Chimera. That is, they look like different things to different sources, all mashed 
together at once. It would be improper to think of, say, a particular inductor as being 
“so many ohms”. If the source signal is comprised of multiple sine waves, such as a 
square wave or a music waveform, the inductor “looks like” a different ohmic value 
to each of the different frequency components, simultaneously. This is an important",ACElectricalCircuitAnalysis.pdf
"to each of the different frequency components, simultaneously. This is an important 
concept and one that we can take great advantage of, for example, in the design of 
filter circuits.
ImpedanceImpedance
We now arrive at impedance. Impedance is a mixture of resistance and reactance, 
and is denoted by Z. This can be visualized as a series combination of a resistor and 
either a capacitor or an inductor. Examples include Z = 100 − j50 Ω, i.e., 100 ohms 
of resistance in series with 50 ohms of capacitive reactance; and Z = 60045º Ω, 
i.e., a magnitude of 600 ohms that includes resistance and inductive reactance (it 
must be inductive reactance and not capacitive reactance because the sign of the 
angle is positive).
To complete this system, we have susceptance and admittance. Susceptance, B, is 
the reciprocal of reactance. Admittance, Y, is the reciprocal of impedance. These are 
similar to the relation between conductance and resistance, and are convenient for",ACElectricalCircuitAnalysis.pdf
"similar to the relation between conductance and resistance, and are convenient for 
parallel circuit combinations.
Example 1.7
Determine the reactances of a 1 mH inductor and a 2 μF capacitor to a sine 
wave of 2 kHz. Repeat for a frequency of 50 kHz.
Use Equations 1.8 and 1.9. For the capacitor at 2 kHz we have:
X C =− j 1
2π f C
X C =− j 1
2π 2kHz 2μ F
X C =− j 39.8Ω
35",ACElectricalCircuitAnalysis.pdf
"For the second source, 50 kHz is 25 times larger than the original and 
capacitive reactance is inversely proportional to frequency. Therefore XC is 
25 times smaller, or −j1.59 Ω.
For the inductor at 2 kHz,
X L = +j 2π f L
X L = +j 2π 2kHz1 mH
X L = +j 12.57Ω
Inductive reactance is directly proportional to frequency. Thus, increasing f 
by a factor of 25 increases XL by the same factor, resulting in j314.2 Ω.
Example 1.8
Determine the susceptance of an inductor whose reactance is j400 Ω. 
Further, if this inductor is placed in series with a 1000 Ω resistor, determine 
the resulting impedance in polar form, as well as the admittance.
Susceptance is the reciprocal of reactance.
BL = 1
X L
BL = 1
j 400Ω
BL =− j 2.5 millisiemens
The impedance, Z, is the vector sum, or 1000 + j400 Ω. Converting this to 
polar form:
Magnitude = √Real2
+Imaginary2
Magnitude = √10002+4002
Magnitude = 1077
 
θ = tan−1
(
Imaginary
Real )
θ = tan−1
(
400
1000 )
θ = 21.8º",ACElectricalCircuitAnalysis.pdf
"polar form:
Magnitude = √Real2
+Imaginary2
Magnitude = √10002+4002
Magnitude = 1077
 
θ = tan−1
(
Imaginary
Real )
θ = tan−1
(
400
1000 )
θ = 21.8º
The result is Z = 107721.8º Ω. The admittance is the reciprocal, yielding  
Y = 928E-6−21.8º μS.
36",ACElectricalCircuitAnalysis.pdf
"1.6 Phasor Diagrams1.6 Phasor Diagrams
A time domain representation of sine waves as drawn earlier tells us everything we 
need to know about the waves, however it is not the most compact method of 
displaying them, nor are they easy to sketch by hand. Consider the the plot of Figure
1.28. Here, two sine waves (green and red) combine to create a third sine wave 
(blue). While the general idea of the two waves adding is apparent in the graph, it 
takes a moment of inspection to determine the wave magnitudes and the precise 
phase relationships between them. In contrast, phasor diagrams can be used to show
the relationships of multiple sines waves in a simple and easy to read format. They 
can also be used to show how various voltages or currents combine. 
The phasor diagram is based on the complex plane discussed previously where the 
horizontal is the real axis and the vertical is the imaginary (j) axis. The magnitude",ACElectricalCircuitAnalysis.pdf
"horizontal is the real axis and the vertical is the imaginary (j) axis. The magnitude 
and phase of each wave can then be drawn as a vector, and the relationships between
the waves is shown directly. For manual plotting, it is convenient to convert from 
polar form to rectangular form. In the example phasor diagram of Figure 1.29, two 
vectors are shown: 8 + j6 and 5 − j3 (equivalent to 1036.9º and 5.83−31º). 
Phasor diagrams can be used to plot voltages, currents and impedances. We shall 
make considerable use of them in upcoming chapters.
37
Adding Sines
Percentage of Peak
-150
-100
-50
0
50
100
150
 
Time
-1 -0.5 0 0.5 1
Additional wave
Initial
Resultant wave
Figure 1.28
A time domain plot.",ACElectricalCircuitAnalysis.pdf
"1.7 Summary1.7 Summary
The most simple of AC waveforms is the sine wave. It can be thought of as the 
vertical displacement of a vector rotating at a constant speed, like the second hand of
a clock. The length of the second hand represents the height or amplitude of the sine 
wave and the speed of rotation represents its period. As the speed tends to be very 
fast, it is more convenient for us to use frequency, which is just the reciprocal of the 
period. Sines can be displaced vertically, which is also called a DC offset, as well as 
having a horizontal or time shift. When expressed relative to a single cycle, this 
change is referred to as the phase shift. Finally, waveforms such as square waves, 
triangle waves and even more complex waveforms such as voice or music can be 
described in terms of combinations of sounds. Indeed, Fourier analysis tells us that 
any repeating waveform can be described as a series of sines each with the",ACElectricalCircuitAnalysis.pdf
"any repeating waveform can be described as a series of sines each with the 
appropriate frequency, amplitude and phase shift. In order to determine the 
“effective DC value” of a sine wave, that is, the value that produces the same power 
dissipation, RMS values are used. The RMS value of a sine wave is its peak value 
divided by the square root of two (approximately equal to 0.707 of peak).
Complex numbers are used to describe AC voltages and currents, among other 
things. They consist of two parts: a real part and an imaginary part that is plotted 
perpendicular to the real part. All mathematical operations on complex numbers 
must follow vector rules. This includes basic trigonometric operations. 
Reactance can be thought of as the imaginary axis version of resistance. That is, it 
restricts current flow. The difference is that there is a 90 degree phase shift between 
the current and voltage through a reactive element while the two are in phase for",ACElectricalCircuitAnalysis.pdf
"the current and voltage through a reactive element while the two are in phase for 
resistive elements. Capacitive reactance is inversely proportional to frequency while 
inductive reactance is directly proportional to frequency. The combination of 
resistance and reactance is known as impedance. Phasor diagrams may be used to 
plot the components of a complex impedance as well as show the relations between 
voltages or currents in a circuit.
38
Figure 1.29
A phasor diagram.",ACElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. What is a sine wave? Describe its constituent parameters (amplitude, 
frequency, offset, etc.).
2. What is the relationship between the frequency, period and phase of a sine 
wave?
3. What is the mathematical relationship between a sine wave and a cosine 
wave?
4. What is meant by the term RMS (root-mean-square) and what does it have to
do with AC versus DC voltages, currents and powers?
5. Describe the difference between a scalar and a vector.
6. What is a complex number?
7. Detail the relationships between resistance, reactance, impedance, 
conductance, susceptance and admittance.
8. Describe how capacitive reactance varies with frequency.
9. Describe how inductive reactance varies with frequency.
10. Describe a phasor diagram. 
1.8 Exercises 1.8 Exercises 
AnalysisAnalysis
1. Determine the AC peak and RMS voltages, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 10 sin 2π 1000 t",ACElectricalCircuitAnalysis.pdf
"1. Determine the AC peak and RMS voltages, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 10 sin 2π 1000 t
2. Determine the AC peak and RMS voltages, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 0.4 sin 2π 5000 t
3. Determine the peak AC portion voltage, DC offset, frequency, period and 
phase shift for the following expression: v(t) = −3 + 20 sin 2π 50 t
4. Determine the peak AC portion voltage, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 12 + 2 sin 2π 20000 t
5. Determine the AC peak and RMS voltages, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 10 sin (2π 100 t + 45°)
6. Determine the AC peak and RMS voltages, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 5 sin (2π 1000 t − 90°)
7. Determine the peak AC portion voltage, DC offset, frequency, period and",ACElectricalCircuitAnalysis.pdf
"phase shift for the following expression: v(t) = 5 sin (2π 1000 t − 90°)
7. Determine the peak AC portion voltage, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 10 + 1 sin (2π 400 t − 45°)
8. Determine the peak AC portion voltage, DC offset, frequency, period and 
phase shift for the following expression: v(t) = 10 + 10 sin (2π 5000 t + 30°)
39",ACElectricalCircuitAnalysis.pdf
"9. A 1 kHz sine wave has a phase of 72°. Determine the time delay. Repeat for 
a 20 kHz sine wave.
10.  A 2 kHz sine wave has a phase of 18°. Determine the time delay. Repeat for
a 100 kHz sine wave.
11. An oscilloscope measures a time delay of 0.2 milliseconds between a pair of
500 Hz sine waves. Determine the phase shift.
12. An oscilloscope measures a time delay of −10 microseconds between a pair 
of 20 kHz sine waves. Determine the phase shift.
13. Convert the following from rectangular to polar form: a) 10 + j10  b) 5 − j10
c) −100 + j20 d) 3k + j4k 
14. Convert the following from rectangular to polar form: a) 2k + j1.5k            
b) 8 − j8  c) −300 + j300 d) −1k − j1k
15. Convert these from polar to rectangular form: a) 1045°  b) 0.490°          
c) −960° d) 100−45° 
16. Convert these from polar to rectangular form: a) −460°  b) −0.930°         
c) 5120° d) 6−135°
17. Perform the following computations: a) (10 + j10) + (5 + j20)",ACElectricalCircuitAnalysis.pdf
"c) 5120° d) 6−135°
17. Perform the following computations: a) (10 + j10) + (5 + j20)                        
b) (5 + j2) + (−5 + j2)  c) (80 − j2) − (100 + j2)  d) (−65 + j50) − (5 − j200)
18. Perform the following computations: a) (100 + j200) + (75 + j210)                
b) (−35 + j25) + (15 + j8)  c) (500 − j70) − (200 + j30)                                   
d) (−105 + j540) − (5− j200)
19. Perform the following computations: a) (100 + j200) ∙ (75 + j210)                 
b) (−35 + j25) ∙ (15 + j8)  c) (500 − j70) / (200 + j30)                                   
d) (−105 + j540) / (5− j200)
20. Perform the following computations: a) (10 + j10) ∙ (5 + j20)                         
b) (5 + j2) ∙ (−5 + j2)  c) (80 − j2) / (100 + j2)   d) (−65 + j50) / (5− j200)
21. Perform the following computations: a) (100°) ∙ (100°)                            
b) (545°) ∙ (−220°)  c) (20135°) / (40−10°)   d) (80°) / (3245°)",ACElectricalCircuitAnalysis.pdf
"b) (545°) ∙ (−220°)  c) (20135°) / (40−10°)   d) (80°) / (3245°)
22. Perform the following computations: a) (0.30°) ∙ (3180°)                        
b) (5−45°) ∙ (−420°)    c) (0.0595°) / (0.04−20°)                                  
d) (5000°) / (60225°)
23. Perform the following computations: a) (0.30°) + (3180°)                        
b) (5−45°) + (−420°)  c) (0.0595°) − (0.04−20°)                                
d) (5000°) − (60225°)
24. Perform the following computations: a) (100°) + (100°)                           
b) (545°) + (−220°)  c) (20135°) − (40−10°)                                        
d) (80°) − (3245°)
40",ACElectricalCircuitAnalysis.pdf
"25. Determine the capacitive reactance of a 1 μF capacitor at the following 
frequencies: a) 10 Hz  b) 500 Hz    c) 10 kHz  d) 400 kHz  e) 10 MHz
26. Determine the capacitive reactance of a 220 pF capacitor at the following 
frequencies: a) 10 Hz b) 500 Hz    c) 10 kHz  d) 400 kHz  e) 10 MHz
27. Determine the capacitive reactance at 50 Hz for the following capacitors:    
a) 10 pF b) 470 pF  c) 22 nF  d) 33 μF
28. Determine the capacitive reactance at 1 MHz for the following capacitors:   
a) 22 pF b) 560 pF  c) 33 nF d) 4.7 μF
29. Determine the inductive reactance of a 100 mH inductor at the following 
frequencies: a) 10 Hz  b) 500 Hz  c) 10 kHz  d) 400 kHz  e) 10 MHz
30. Determine the inductive reactance of a 100 mH inductor at the following 
frequencies: a) 10 Hz b) 500 Hz  c) 10 kHz  d) 400 kHz  e) 10 MHz
31. Determine the inductive reactance at 1 kHz for the following inductors:      
a) 10 mH  b) 500 mH c) 10 μH  d) 400 μH",ACElectricalCircuitAnalysis.pdf
"31. Determine the inductive reactance at 1 kHz for the following inductors:      
a) 10 mH  b) 500 mH c) 10 μH  d) 400 μH 
32. Determine the inductive reactance at 500 kHz for the following inductors:   
a) 1 mH  b) 40 mH c) 2 μH  d) 50 μH 
33. Draw phasor diagrams for the following: a) 5 + j2  b) −10 −j20  c) 845°     
d) 2−35°
34. Draw phasor diagrams for the following: a) 60j−20  b) −40 + j500                
c) 0.05−45°  d) −1560°
35. The fundamental of a certain square wave is a 5 volt peak, 1 kHz sine. 
Determine the amplitude and frequency of each of the next five harmonics. 
36. The fundamental of a certain triangle wave is a 10 volt peak, 100 Hz sine. 
Determine the amplitude and frequency of each of the next five harmonics.
DesignDesign
37. Determine the capacitance required for the following reactance values at      
1 kHz: a) 560 Ω  b) 330 kΩ  c) 470 kΩ  d) 1.2 kΩ  e) 750 Ω
38. Determine the capacitance required for the following reactance values at",ACElectricalCircuitAnalysis.pdf
"1 kHz: a) 560 Ω  b) 330 kΩ  c) 470 kΩ  d) 1.2 kΩ  e) 750 Ω
38. Determine the capacitance required for the following reactance values at     
20 Hz: a) 56 kΩ  b) 330 kΩ  c) 470 kΩ  d) 1.2 kΩ  e) 750 Ω
39. Determine the inductance required for the following reactance values at    
100 MHz: a) 560 Ω  b) 330 kΩ  c) 470 kΩ  d) 1.2 kΩ  e) 750 Ω
40. Determine the inductance required for the following reactance values at      
25 kHz: a) 56 Ω  b) 33 kΩ c) 470 kΩ  d) 1.2 kΩ  e) 750 Ω
41",ACElectricalCircuitAnalysis.pdf
"41. Which of the following have a reactance of less than 100 Ω for all 
frequencies below 1 kHz? a) 2 mH  b) 99 mH  c) 470 pF  d) 10000 μF
42. Which of the following have a reactance of less than 8 Ω for all frequencies 
above 10 kHz? a) 10 nH b) 5 mH  c) 56 pF  d) 470 μF
43. Which of the following have a reactance of at least 1k Ω for all frequencies 
above 20 kHz? a) 2 mH b) 200 mH  c) 680 pF  d) 33 μF
44. Which of the following have a reactance of at least 75 Ω for all frequencies 
below 5 kHz? a) 680 μH  b) 10 mH  c) 82 pF  d) 33 nF
ChallengeChallenge
45. Determine the negative and positive peak voltages, RMS voltage, DC offset, 
frequency, period and phase shift for the following expression:                      
v(t) = −10 sin (2π 250 t + 180°)
46. Determine the negative and positive peak voltages, DC offset, frequency, 
period and phase shift for the following expression:                                   
v(t) = 1 − 100 sin 2π 50000 t",ACElectricalCircuitAnalysis.pdf
"period and phase shift for the following expression:                                   
v(t) = 1 − 100 sin 2π 50000 t
47. Assume you have a DC coupled oscilloscope set as follows:                           
time base = 100 microseconds/division, vertical sensitivity = 1 volt/division.
Sketch the display of this waveform: v(t) = 2 + 3 sin 2π 2000 t
48. Assume you have a DC coupled oscilloscope set to the following:                 
time base = 20 microseconds/division, vertical sensitivity = 200 
millivolts/division. Sketch the display of this waveform:                            
v(t) = −0.2 + 0.4 sin 2π 10000 t
49. A 200 Ω resistor is in series with a 1 mH inductor. Determine the impedance
of this combination at 200 Hz and at 20 kHz.
50. A 1 kΩ resistor is in series with an inductor. If the combined impedance at 
10 kHz is 1.41 k45°, determine the inductance in mH. 
42",ACElectricalCircuitAnalysis.pdf
"NotesNotes
♫♫♫♫
43",ACElectricalCircuitAnalysis.pdf
"2 2 Series RLC CircuitsSeries RLC Circuits
2.0 Chapter Learning Objectives2.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Utilize Kirchhoff's voltage law, the voltage divider rule and Ohm's law to find node and component 
voltages in series RLC networks that utilize voltage sources or a single current source.
• Compute complex impedance and system current in series RLC circuits.
• Draw phasor diagrams for impedance and component voltages in series RLC circuits.
2.1 Introduction 2.1 Introduction 
In this chapter we introduce series RLC circuits for the AC case. There is much here that will be familiar from 
your prior studies with DC series circuits, however, there will be a few notable changes and perhaps a surprise or
two lurking. The key to most of this is to remember that all computations involve vector quantities. In fact, DC 
can be thought of as a special case of AC; namely, the frequency drops to zero hertz causing XC to approach an",ACElectricalCircuitAnalysis.pdf
"can be thought of as a special case of AC; namely, the frequency drops to zero hertz causing XC to approach an 
open and XL to drop to zero. This leaves us with just resistors and scalar quantities because the phase angles in 
the remainder of the circuit collapse to zero for DC steady state.
Many of the solution techniques from DC analysis will be applicable here. This includes the use of Ohm's law 
and Kirchhoff's voltage law, along with the voltage divider rule. Generally, reactance values will need to be 
computed from capacitor and inductor values before the main analysis may begin. In this chapter, as in the most 
of the remaining chapters, we shall be concerned with determining the circuit response based on a source with a 
single frequency of excitation, in other words, a simple sine wave.
To clarify our analyses, we shall make considerable use of both time domain voltage plots as well as phasor 
diagrams. 
2.2 The Series Connection2.2 The Series Connection",ACElectricalCircuitAnalysis.pdf
"diagrams. 
2.2 The Series Connection2.2 The Series Connection
A series connection is always characterized by a single loop or path for current flow. There are no junctions 
from which current can flow out of or into. Consequently, 
The current is the same everywhere in a series connection. (2.1)
Each component in such a loop will see the same current, regardless of whether it is a resistor, capacitor or 
inductor. Before component voltages can be determined, the capacitive and inductive reactance values must be 
44",ACElectricalCircuitAnalysis.pdf
"computed from the capacitor and inductor values, based on the frequency of the 
driving source. Consequently, if this frequency were to change, the reactances would
change, and this would in turn cause changes in the circulating current and 
component voltages. 
2.3 Series Impedance2.3 Series Impedance
Perhaps the first practical issue we face is determining the effective impedance of an
RLC series loop. For starters, resistors in series simply add. Reactances also add but 
we must be careful of the sign. Inductive reactance and capacitive reactance will 
partially cancel each other. Thus, the impedance in rectangular form is the sum of 
the resistive components for the real portion, plus the sum of the reactances for the 
imaginary (j) portion. We will often find it convenient to express this value in polar 
form.
Example 2.1
What is the impedance of the network shown in Figure 2.1 at a frequency
of 15 kHz?
First we need to find the capacitive reactance value.
X C =− j 1
2π f C",ACElectricalCircuitAnalysis.pdf
"of 15 kHz?
First we need to find the capacitive reactance value.
X C =− j 1
2π f C
X C =− j 1
2π15kHz 910 pF
X C =− j 11.66 k Ω
As there is only one resistor and one capacitor, the result in rectangular form
is 47 k −j11.66 kΩ. In polar form this is:
Magnitude = √Real2
+Imaginary2
Magnitude = √47k2
+(−11.66 k)2
Magnitude = 48.42 k
 
θ = tan−1
(
Imaginary
Real )
θ = tan−1
(
−11.66 k
47 k )
θ =−13.9º
That is, in polar form Z = 48.42E3−13.9° Ω.
45
Figure 2.1
Circuit for Example 2.1.",ACElectricalCircuitAnalysis.pdf
"Example 2.2
Determine the effective impedance of the circuit shown in Figure 2.2 if the
source frequency is 2 kHz. Repeat this for source frequencies of 200 Hz
and 20 kHz. Finally, express the results in both rectangular and polar form.
The first step is to find the reactance values at 2 kHz.
X L = j 2π f L
X L = j 2π2000Hz 15mH
X L = j 188.5Ω
X C =− j 1
2π f C
X C =− j 1
2π2000 Hz 270 nF
X C =− j 294.7 Ω
Combine reals with reals and j terms with j terms.
Z = R+ j X L − j X C
Z = 500+ j 188.5 − j 294.7Ω
Z ≈ 500− j 106.2Ω = 511.2−12° Ω
At 200 Hz XC will ten times larger and XL will be ten times smaller.
XC = −j2947 Ω
XL = j18.85 Ω 
Z = 500 + j18.85 −j2947 Ω
Z ≈ 500 −j2928 Ω = 2970−80.3° Ω
At 20 kHz XC will ten times smaller and XL will be ten times larger.
XC = −j29.47 Ω
XL = j1885 Ω 
Z = 500 + j1885 −j29.47 Ω
Z ≈ 500 + j1856 Ω = 192274.9° Ω
To help visualize this complex impedance, it is useful to construct a phasor plot as",ACElectricalCircuitAnalysis.pdf
"Z = 500 + j1885 −j29.47 Ω
Z ≈ 500 + j1856 Ω = 192274.9° Ω
To help visualize this complex impedance, it is useful to construct a phasor plot as 
shown in Figure 2.3. We will do this for the initial case of 2 kHz. The resistive 
component is the horizontal vector of length 500 (yellow). XL is straight up (blue) at 
188.590°, and XC is straight down (red) at 294.7−90°. By copying the XL vector 
and then shifting it down and next to XC, the difference between the two reactive 
components can be seen (purple component directly above the XL copy). This 
46
Figure 2.2
Circuit for Example 2.2.",ACElectricalCircuitAnalysis.pdf
"reactive sum is then copied and shifted right to join the resistive component to show 
the final result. This is 511.2−12° Ω (green), as expected. 
Example 2.3
Determine the impedance of the network shown in Figure 2.4. If the input
frequency is 1 kHz, determine the capacitor and inductor values.
The reactance values are already given, so we simply add them to determine 
the impedance in rectangular form. Combine reals with reals and j terms 
with j terms, and then convert to polar form.
Z = R+j X L − j X C
Z = 750+j 600 − j 200 Ω
Z = 750+j 400 Ω= 850  28.1° Ω
To find the capacitance and inductance, we simply rearrange the reactance 
formulas and solve. First, the inductor:
X L = j 2π f L
 
L = | X L |
2π f
L = 600Ω
2π1kHz
L ≈ 95.5 mH
47
Figure 2.3
Construction of the impedance 
plot for the network of     
Figure 2.2.
R
jX
-jX-400
300
600
jX
-jX
R
Z
jX
+ jX
L
C
L
C
L-jX
Figure 2.4
Circuit for Example 2.3.",ACElectricalCircuitAnalysis.pdf
"And now for the capacitor:
X C =− j 1
2 π f C
 
C = 1
2π f | X C |
C = 1
2π1 kHz 200 Ω
C = 796 nF
A plot of the impedance vector summation is shown in Figure 2.5. Note how
the three components combine graphically to arrive at Z.
2.4 Series Circuit Analysis2.4 Series Circuit Analysis
The techniques employed for series AC circuit analysis are the same as those used 
for DC. The key item to remember for series circuits, whether AC or DC, as that the 
current will be the same everywhere in a series connection. The major analysis tools 
are Ohm's law, Kirchhoff's voltage law (KVL), and optionally, the voltage divider 
rule. 
48
Figure 2.5
Impedance plot for the network 
of Example 2.3.",ACElectricalCircuitAnalysis.pdf
"KVL states that the sum of voltages around a loop must be zero, or that the sum of 
voltage rises around a loop must equal the sum of voltage drops:
∑ v↑ =  ∑ v↓ (2.2)
The voltage divider rule for AC states that the voltage across any component or 
group of components is proportional to the ratio of the impedance of said component
and the total series impedance. For some component A driven by source e,
vA = e Z A
Z Total
(2.3)
If current is the desired quantity, the equivalent impedance of the network is found 
and then divided into the equivalent source voltage. If the voltages across specific 
components are desired, they can be found using the circulating current and Ohm's 
law. Alternately, they can be computed using the voltage divider rule. If a current 
source is driving the series circuit then the circulating current is the source current. 
Individual component voltages may be found using the source current and Ohm's",ACElectricalCircuitAnalysis.pdf
"Individual component voltages may be found using the source current and Ohm's 
law. The source voltage may be found by summing the component voltages via KVL,
or by determining the equivalent series impedance and multiplying it by the source 
current.
Example 2.4
For the circuit of Figure 2.6, find the circulating current and the voltages
across the capacitor and resistor. 
The first step is to determine XC and then find Ztotal. Then use
Ohm's law to find the circulating current.
X C =− j 1
2π f C
X C =− j 1
2π1000Hz 33nF
X C =− j 4823Ω
ZTotal = R+(− j X C )
ZTotal = 2200 Ω − j 4823Ω
ZTotal = 5301−65.5° Ω
i = v
Z
i = 1V
5301−65.5° Ω
i = 188.6  65.5° μ A
49
Figure 2.6
Circuit for Example 2.4.",ACElectricalCircuitAnalysis.pdf
"Now use Ohm's law to find the component voltages.
vR = i×R
vR = 188.6  65.5° μ A×2200 0° Ω
vR = 0.4149 65.5° V
vC = i× X C
vC = 188.6  65.5° μA×4823−90° Ω
vC = 0.9098−24.5° V
To complete this Example, we shall generate a series of plots: First off, an 
impedance plot showing the vector combination of R and XC leading to ZTotal. 
This is shown in Figure 2.7. 
Next, Figure 2.8 illustrates the phasor diagram of the three voltages. It may 
not be immediately apparent but an examination of this plot shows that it is 
just a counterclockwise rotation of the impedance plot. In fact, it is rotated 
by 65.5 degrees, the phase angle of the current. This should come as no 
surprise as the current is multiplied by each resistance, reactance or 
impedance to develop the voltage, and a vector multiplication simply adds 
the angles. 
Also, note that if we simply summed the magnitudes of the component 
voltages, the result would be considerably larger than the input voltage,",ACElectricalCircuitAnalysis.pdf
"Also, note that if we simply summed the magnitudes of the component 
voltages, the result would be considerably larger than the input voltage, 
seeming to violate KVL. In contrast, once a proper vector summation is 
performed, all is right in paradise.
50
Figure 2.7
Impedance plot for 
Example 2.4.",ACElectricalCircuitAnalysis.pdf
"For further clarity, Figure 2.9 plots the voltages in the time domain. Note 
how the component voltages add graphically resulting in the input voltage.
51
Figure 2.8
Voltage phasor diagram for 
Example 2.4.
Time Domain Graph
Voltage (volts)
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
 
Time (milliseconds)
0 0.25 0.5 0.75 1 1.25 1.5
vc
vin
vR
Figure 2.9
Time domain response for 
Example 2.4.",ACElectricalCircuitAnalysis.pdf
"The time shifts may be computed for verification as follows:
θ = 360° Δt
T
ΔtR = T θR
360 °
Δt R = 1 ms 65.5°
360°
Δt R = 181.9μ s
Δt C = T θC
360 °
ΔtC = 1ms −24.5°
360°
ΔtC =−68μ s
Thus we verify the resistor voltage leading (to the left) by 181.9 μs and the 
capacitor voltage lagging (to the right) by 68 μs, as seen in Figure 2.9.
Computer SimulationComputer Simulation
The circuit of Example 2.4 is captured into a simulator as shown in Figure 2.10. 
Node voltage 1 corresponds to the input voltage and node 2 corresponds to the 
capacitor voltage. The resistor voltage is merely the difference between these two, or
node 1 minus node 2. 
A transient or time domain simulation is performed. The results of the simulation are
shown in Figure 2.11. Note the tight agreement with the computed results as plotted 
in Figure 2.9. 
52
Figure 2.10
The circuit of Example 2.4 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"We will now examine a variety of different series circuits. To ease the computations, 
we will simply state values for the component reactances rather than specify a 
frequency along with inductor and/or capacitor values. 
Example 2.5
Determine the component voltages for the circuit shown in Figure 2.12.
Given the reference direction of the source (which produces a 
counterclockwise reference current), the voltage across the resistor will be 
defined as vb − va. The first step is to find the equivalent series impedance. 
By inspection, this is 180 + j360 Ω, which is equivalent to 402.563.4° Ω. 
The voltages can be found via the voltage divider rule.
vL = e X L
Z
vL = 6 0° V 360  90° Ω
402.5 63.4° Ω
vL = 5.366  26.6° Vp
53
Figure 2.11
Time domain response of 
Example 2.4 from a simulator.
Figure 2.12
Circuit for Example 2.5.",ACElectricalCircuitAnalysis.pdf
"vR = e R
Z
vR = 6 0° V 180  0° Ω
402.5  63.4 ° Ω
vR = 2.683−63.4° Vp
The rectangular versions of vL and vR are 4.8 + j2.4 and 1.2 − j2.4 
respectively, summing up to the source of 60°. This is illustrated in the 
phasor voltage plot of Figure 2.13. 
Compare the plot of Figure 2.13 to that of the RC circuit in Figure 2.8. In 
this circuit, the impedance is inductive. This causes the current to lag the 
voltage, the opposite of the case for the RC circuit. Thus, the resistor voltage
winds up below the real axis instead of above it (remember, the current and 
voltage for a resistor are in phase, and therefore the voltage across the 
resistor will have the same phase angle as the current through it). Once 
again, simply summing the magnitudes of the voltages yields a greater value
than the source, however, a proper vector addition shows that KVL is 
satisfied. 
A time domain plot of the source and component voltages is shown in",ACElectricalCircuitAnalysis.pdf
"satisfied. 
A time domain plot of the source and component voltages is shown in  
Figure 2.14. Note the similarity with the plot of Figure 2.9. Further, note 
how the time positions of the component voltages have swapped.
54
Figure 2.13
Voltage phasor diagram for 
Example 2.5.",ACElectricalCircuitAnalysis.pdf
"Finally, as a source frequency was not specified for this circuit, the time 
scale of Figure 2.14 is calibrated in terms of cycles rather than seconds.
Example 2.6
Find the voltages va and vb in the circuit of Figure 2.15. Assume the current
source is the reference of 0°.
Ohm's law may be used directly as the current is set by the source. Given
the reference direction of the 20 mA source, the reference polarity for the
capacitor voltage will be + to − from bottom to top. Thus, va is negative.
va = vC = −i×X C
va =−20E-3 0° A×1000−90° Ω
va =−20−90° V
Recalling that a negative magnitude is the same as a 180° phase shift, we 
may remove the magnitude's negative sign by adding 180° to the phase 
angle. Therefore, va may also be written as 2090°. 
The node b voltage is found by multiplying the current by the impedance 
seen from node b to ground.
55
Figure 2.14
Time domain response for 
Example 2.5.
Time Domain Graph
Voltage (volts)
-6.0
-4.0
-2.0
0.0
2.0
4.0
6.0
 
Time (cycles)",ACElectricalCircuitAnalysis.pdf
"55
Figure 2.14
Time domain response for 
Example 2.5.
Time Domain Graph
Voltage (volts)
-6.0
-4.0
-2.0
0.0
2.0
4.0
6.0
 
Time (cycles)
0 0.5 1 1.5
vL
vin
vR
Figure 2.15
Circuit for Example 2.6.",ACElectricalCircuitAnalysis.pdf
"vb = i×(R+jX L )
vb = 20  0° mA×(4300+j 150Ω)
vb = 86.05 2° V
Notice the small size of the phase angle. This is expected given that the 
inductive reactance is so much smaller than the resistive component. Finally,
these voltage are RMS as the current is assumed to be RMS (not having been 
labeled as peak or peak-to-peak). 
Example 2.7
Find the voltage vbd in the circuit of Figure 2.16. E1 = 20° volts peak and 
E2 = 560° volts peak. 
The first step is to find the equivalent source voltage. V oltage sources in 
series add, however, note the reference polarity for E2. If we take E1 as the 
system reference, then the combined source is E1 − E2. Using E1 as the 
reference source creates a clockwise reference current and thus a positive 
reference polarity for vbd. If E2 was taken as the reference then vbd would be 
assumed to be negative due to the assumed counterclockwise reference 
direction of the resulting current. Either way will work.
ETotal = E1 −E2
ETotal = 2 0° Vp −5 60 ° Vp",ACElectricalCircuitAnalysis.pdf
"direction of the resulting current. Either way will work.
ETotal = E1 −E2
ETotal = 2 0° Vp −5 60 ° Vp
ETotal = 4.359 −96.6° Vp
We can now use a voltage divider between the impedance seen from node b 
to node d versus the total series impedance. The impedance of the circuit is  
2 k + j7.5 k − j800 Ω, or 2 k + j6.7 k Ω. This is equivalent to 699273.4° Ω.
The impedance between nodes b and d is j7.5 k − j800 Ω, or j6.7 k Ω. In 
polar form this is 670090° Ω.
56
Figure 2.16
Circuit for Example 2.7.",ACElectricalCircuitAnalysis.pdf
"The voltage divider yields:
vbd = e Z bd
Z Total
vbd = 4.359 −96.6° V 6700 90° Ω
6992 73.4 ° Ω
vbd = 4.177−80° Vp
Although we have answered the question, at this point attempting to 
visualize the waveforms in your head to verify the results may be a little 
challenging. No problem! If we also find the voltage across the resistor, we 
can check to see if KVL is verified using a phasor diagram. First, the resistor 
voltage may be found using the voltage divider rule.
vR = e R
Z Total
vR = 4.359 −96.6° V 2000 0° Ω
6992 73.4° Ω
vR = 1.247 −170° Vp
We now plot all four voltages as seen in Figure 2.17 (note the unequal 
horizontal versus vertical scaling for ease of visual analysis).
57
Figure 2.17
Voltage phasor diagram for 
Example 2.7.",ACElectricalCircuitAnalysis.pdf
"KVL tells us that the sum of the voltage rises must equal the sum of the 
voltage drops. We had used E1 as the reference and treated it as a rise. This 
made the reference direction clockwise for the circulating current. As a 
consequence, vR and vbd are voltage drops. E2 is also seen as a drop given its 
reference polarity markings. In other words, E1 should equal the sum of vR, 
vbd and E2. Let's see if this is borne out graphically in the phasor diagram.
Looking at imaginary parts first, E1 is strictly real so the other three vectors 
should sum to zero for their imaginary parts. The imaginary part of E2 is 
about +4.3 while the other two come in around −4.1 and −0.2, so they do 
balance. For the real parts, E2 is +2.5, vbd is about +0.7 and vR is about −1.2. 
These add up to +2, the real component of E1. Everything balances. For 
more exacting results, you could convert each of the polar form results into 
rectangular form and add the reals to the reals, and the imaginaries to the",ACElectricalCircuitAnalysis.pdf
"rectangular form and add the reals to the reals, and the imaginaries to the 
imaginaries, and get the same result. In fact, using the prior recorded values, 
the three drops sum to 1.9970.0013° volts versus precisely 20° volts. 
This small deviation is due to the accumulated rounding and truncation 
errors of the intermediate results.
Sometimes a problem concerns determining resistance or reactance values. The 
solution paths will require use of the analysis rules in reverse. For example, if a 
resistor value is needed to set a specific current, the total required impedance can be 
determined from this current and the given voltage supply. The values of the other 
series components can then be subtracted from the total (using rectangular form),
yielding the required resistor or reactance value. 
Another possibility is determining a specific capacitance or inductance to meet
certain requirements. An example of this would be a simple crossover network for",ACElectricalCircuitAnalysis.pdf
"certain requirements. An example of this would be a simple crossover network for
a loudspeaker system. Generally, a single transducer is incapable of reproducing
music signals of all frequencies at sufficiently loud levels while maintaining low
distortion. Consequently, the frequency range is split into two or more bands with 
each being covered by a transducer optimized to reproduce those frequencies. These 
transducers, along with a few electrical components, are then packaged into an 
appropriate enclosure commonly referred to as a loudspeaker1. In a basic system, the
audible spectrum is split into two parts, as seen in Figure 2.18. The low frequencies 
are handled by a large transducer called a woofer, while the high frequencies are 
handled by a small, light transducer called a tweeter that can follow the rapidly 
changing high frequency content with accuracy. Woofers are not capable of 
reproducing high frequencies, and excessive low frequency content can physically",ACElectricalCircuitAnalysis.pdf
"reproducing high frequencies, and excessive low frequency content can physically 
damage a tweeter. Thus, some means of “steering” the high frequency content to the 
tweeter and the low frequency content to the woofer must be employed. One way to 
do this is to place an inductor or capacitor in series with the transducer. 
1 To add some confusion, the term “loudspeaker” can refer to either the individual 
transducers (engineer-speak) or the entire finished system (audiophile-speak).
58
Figure 2.18
A two-way bookshelf 
loudspeaker system with 
tweeter (top) and woofer 
(bottom).",ACElectricalCircuitAnalysis.pdf
"Figure 2.19 shows a simple way to block low frequencies from a tweeter. In this 
diagram, the loudspeaker is modeled as a resistor. Although this is not perfectly 
accurate, it is sufficient to illustrate the idea. 
To understand this concept, simply think of the circuit as a voltage divider between 
the loudspeaker and the reactance of the capacitor. Capacitive reactance is inversely 
proportional to frequency. The circuit is designed such that XC is much larger than 
than the impedance of the loudspeaker at very low frequencies. This means that very
little of a low frequency signal will reach the loudspeaker as most of that voltage 
will drop across the capacitor. In contrast, at high frequencies XC is much smaller 
than the tweeter's impedance and thus virtually all of the source signal reaches the 
loudspeaker. The transition point between these two regions is called the crossover 
frequency and is usually defined as the point where XC equals the magnitude of the",ACElectricalCircuitAnalysis.pdf
"frequency and is usually defined as the point where XC equals the magnitude of the 
loudspeaker impedance. For a woofer, the capacitor would be replaced by an 
inductor. At low frequencies the inductive reactance would be minimal, thereby 
allowing all of the low frequency content to reach the woofer. The opposite happens 
at high frequencies. The inductive reactance would rise with frequency and 
eventually prevent or “choke off” the signal to the woofer (hence, the slang term 
“choke” for an inductor).
Example 2.8
Referring to the circuit of Figure 2.19, assume the loudspeaker impedance is
80° Ω. Determine a capacitor value for a crossover frequency of 2.5 kHz. 
At this frequency the magnitude of XC should be the same as that of the 
loudspeaker. For a 10 volt RMS input signal, determine the loudspeaker 
voltage at frequencies of 100 Hz, 2.5 kHz and 15 kHz.
The first step is to determine the value of the capacitor such that at 2.5 kHz",ACElectricalCircuitAnalysis.pdf
"voltage at frequencies of 100 Hz, 2.5 kHz and 15 kHz.
The first step is to determine the value of the capacitor such that at 2.5 kHz 
XC equals 8 Ω. To do this, simply rearrange the capacitive reactance formula:
59
Figure 2.19
A basic loudspeaker crossover 
network.",ACElectricalCircuitAnalysis.pdf
"X C =− j 1
2π f C
C = 1
2π f | X C |
C = 1
2π 2500Hz 8Ω
C = 7.96μ F
To find the loudspeaker voltage at the three specified frequencies, first find 
the reactance at that frequency and then use the voltage divider rule. We 
already know that at 2.5 kHz the reactance is −j8 Ω.
vloudspeaker = e Z Load
ZTotal
vloudspeaker = 10  0° Vrms 8Ω
8 − j 8Ω
vloudspeaker = 7.07 45 ° Vrms
At 100 Hz, the frequency has decreased by a factor of 25 so the reactance 
goes up by the same factor yielding −j200 Ω.
vloudspeaker = e Z Load
ZTotal
vloudspeaker = 10  0° Vrms 8Ω
8 − j 200 Ω
vloudspeaker = 0.4 87.7° Vrms
At 15 kHz, the frequency has increased by a factor of 6 so the reactance 
goes down by the same factor yielding −j1.333 Ω.
vloudspeaker = e Z Load
ZTotal
vloudspeaker = 10  0° Vrms 8Ω
8 − j 1.333Ω
vloudspeaker = 9.86 9.5° Vrms
From this tally, it should be obvious that the lowest bass frequencies will 
experience attenuation while the highest frequencies will see little reduction",ACElectricalCircuitAnalysis.pdf
"experience attenuation while the highest frequencies will see little reduction 
in amplitude. This is just what we want. 
The concept of a variation in output voltage with respect to frequency, or 
frequency response, will be examined in much greater detail in upcoming 
chapters.
60",ACElectricalCircuitAnalysis.pdf
"2.5 Concerning Practical Inductors 2.5 Concerning Practical Inductors 
Up to this point, inductors have been treated as ideal components, that is, pure 
inductance. In reality, all inductors have some resistance associated with them due to
the resistance of the wire used to make the coil. This is called ESR, or Equivalent 
Series Resistance. It is also denoted as Rcoil. Ideally, this resistance will be small 
enough to ignore, but ultimately it will place a limit on the performance of any 
circuit that utilizes an inductor. 
While it is possible to measure the resistance of an inductor using a DMM, this will 
not yield an accurate value at all frequencies. In fact, as frequency increases, ESR 
will also increase. This is due to skin effect. At higher frequencies, current is not 
distributed equally throughout the cross-section of a conductor. In fact, it tends to 
“hug” the outer surface or “skin” of the conductor. This reduces the effective cross-",ACElectricalCircuitAnalysis.pdf
"“hug” the outer surface or “skin” of the conductor. This reduces the effective cross-
sectional area and thereby increases the resistance2. In general, as frequency rises, so
does Rcoil. Unfortunately, the situation is further complicated by distributed 
capacitance that will become an issue at still higher frequencies. As a consequence, 
manufacturers will give a “Q plot”, such as the one shown in Figure 2.20. 
2 Recall that R = ρ l / A. A decrease in area A results in increased resistance, R. 
61
Figure 2.20
Inductor Q graph.
Courtesy of TDK",ACElectricalCircuitAnalysis.pdf
"The Q, or quality factor, of an inductor can be defined in terms of the peak energy 
stored in the device versus the energy dissipated per cycle. Keeping time constant, 
we can relate this to the power via the relation i2R. The current necessarily is the 
same for both the reactive and resistive components, therefore the Q of the coil is 
equal to the ratio of its reactance to its resistance at the frequency of interest.
Qcoil = X L
Rcoil
(2.4)
Where
Qcoil is the quality factor of the inductor,
XL is the magnitude of the inductive reactance at the frequency of interest,
Rcoil is the resistance of the inductor at the frequency of interest.
In general, the higher the Qcoil, the better. As can be seen in Figure 2.20, Qcoil is not a 
constant. Indeed, it increases with frequency until a peak is reached, at which point it
begins to fall. 
When dealing with practical inductors, the effective Qcoil can be determined from a",ACElectricalCircuitAnalysis.pdf
"begins to fall. 
When dealing with practical inductors, the effective Qcoil can be determined from a 
graph if the operating frequency is known. Once the Qcoil is found, the effective 
value of Rcoil can be found from Equation 2.4. This value can then be placed in series
with the ideal inductor to create a more accurate result. For analysis, this pair is 
sometimes drawn with a box around it to denote that Rcoil is not a separate physical 
resistor, but is the effective resistance of the inductor.
Example 2.9
Find the voltage across the inductor in the circuit of Figure 2.21. Assume
the source voltage is 200° peak-to-peak at a frequency of 20 kHz. L is
equal to 10 mH, Qcoil is 50, and R1 is 600 Ω.
Remember, the inductor consists of both elements within the dashed
box. First, find the magnitude of the inductive reactance.
| X L | = 2π f L
| X L | = 2π20 kHz 10 mH
| X L | = 1257 Ω
We can now find Rcoil via Equation 2.4:
Qcoil = X L
Rcoil
Therefore,
62
Figure 2.21",ACElectricalCircuitAnalysis.pdf
"| X L | = 2π f L
| X L | = 2π20 kHz 10 mH
| X L | = 1257 Ω
We can now find Rcoil via Equation 2.4:
Qcoil = X L
Rcoil
Therefore,
62
Figure 2.21
Circuit for Example 2.9.",ACElectricalCircuitAnalysis.pdf
"Rcoil = X L
Qcoil
Rcoil = 1257
50
Rcoil = 25.1Ω
The impedance of the inductor is 25.1 + j1257 Ω , or 125788.9° Ω. The 
total impedance is 600 + 25.1 + j1257 Ω , or 140463.6° Ω. We can use a 
voltage divider to find the inductor's voltage.
vind = e Z ind
Z Total
vind = 20  0° Vpp 1257 88.9° Ω
1404 63.6° Ω
vind = 17.9 25.3° Vpp
It is worth noting that, due to the coil resistance, the inductor's current and 
voltage are not precisely 90° out of phase, but rather 88.9°. That's a small 
but measurable deviation.
2.6 Summary2.6 Summary
A series connection is any connection in which the current through one component 
must be identical to the current flowing through any other component in that 
connection. This remains true for AC circuits employing resistors, capacitors and 
inductors. The equivalent impedance of a set of resistors, inductors and capacitors 
placed in series is the vector sum of their resistance and reactance values. Inductive",ACElectricalCircuitAnalysis.pdf
"placed in series is the vector sum of their resistance and reactance values. Inductive 
and capacitive reactance have opposing signs and will partially cancel each other. 
While resistance remains constant across frequency, reactance does not. Capacitive 
reactance decreases with frequency while inductive reactance increases. Therefore, 
the precise impedance value will vary with frequency. If the impedance angle is 
positive, the circuit is said to be inductive. If the impedance angle is negative, the 
circuit is said to be capacitive.
Once inductance and capacitance values have been turned into their corresponding 
reactances at the frequency of interest, Ohm's law can be used to find the voltage 
across any component. The voltage will equal the product of the resistance or 
reactance and the current flowing through it. This is a vector computation. The 
current through a resistor will be in phase with the voltage across it. In contrast, the",ACElectricalCircuitAnalysis.pdf
"current through a resistor will be in phase with the voltage across it. In contrast, the 
inductor voltage will lead the current and the resistor voltage by 90 degrees. Also, 
the capacitor voltage will lag the current and the resistor voltage by 90 degrees. 
From these observations it is apparent that the inductor and capacitor voltages in a 
series connection must be 180 degrees out of phase.
63",ACElectricalCircuitAnalysis.pdf
"Kirchhoff's voltage law (KVL) states that the sum of voltage rises around any series 
loop must equal the sum of voltage drops around that loop. This remains true for AC 
analysis, however, it is imperative that a vector summation is performed and not just 
a simple summation of the magnitudes. If the phase angles are ignored, it is possible 
for the voltage magnitudes to sum to considerably more than the source voltage.
The voltage divider rule (VDR) remains true for AC analysis, but again, it is a vector 
computation. Thus, a rendering using just magnitudes will not produce a correct 
result. For example, a divider between a resistor and inductive reactance of equal 
value will not yield the source voltage splitting in half across each component.
For highest accuracy, the coil resistance, or Rcoil, of inductors should not be ignored. 
It is treated as a small resistance in series with and integral to the inductor. Coil",ACElectricalCircuitAnalysis.pdf
"It is treated as a small resistance in series with and integral to the inductor. Coil 
resistance is computed from the inductor's Q, or quality factor. Q is defined as the 
ratio of XL to Rcoil.
Review QuestionsReview Questions
1. How is the equivalent impedance for a string of series connected resistors, 
capacitors and inductors computed? 
2. Will the impedance of an RLC circuit be the same at all frequencies? 
Why/why not?
3. How are series connected AC voltage sources combined?
4. Does Kirchhoff's voltage law change for AC circuit analysis?
5. Is it possible for the voltage magnitudes of an AC RLC circuit to sum to 
more than the source voltage? Why/why not?
6. How is a phasor impedance plot for a series network related to its phasor 
voltage plot?
2.7 Exercises2.7 Exercises
AnalysisAnalysis
1. Determine the impedance of the circuit of Figure 2.22 for a 1 kHz sine.
2. Determine the impedance of the circuit of Figure 2.22 for a 5 kHz sine.
64
Figure 2.22",ACElectricalCircuitAnalysis.pdf
"3. Determine the impedance of the circuit of Figure 2.23 for a 10 kHz sine.
4. Determine the impedance of the circuit of Figure 2.23 for a 50 kHz sine.
5. Determine the impedance of the circuit of Figure 2.24 for a 1 kHz sine.
6. Determine the impedance of the circuit of Figure 2.24 for a 500 Hz sine.
7. Determine the impedance of the circuit of Figure 2.25.
8. In the circuit of Figure 2.25, if the input frequency is 100 Hz, what is the 
value of the inductor, in mH?
9. In the circuit of Figure 2.25, if the input frequency is 200 Hz, what is the 
value of the capacitor, in μF?
10. Draw the voltage and current waveforms for the circuit of Figure 2.26.
65
Figure 2.23
Figure 2.24
Figure 2.25
Figure 2.26",ACElectricalCircuitAnalysis.pdf
"11. Draw the voltage and current waveforms for the circuit of Figure 2.27.
12. Draw the voltage and current waveforms for the circuit of Figure 2.28 if E is
a one volt peak sine at a frequency of 10 kHz and C = 3.3 nF.
13. Draw the voltage and current waveforms for the circuit of Figure 2.29 if E is
a two volt peak-peak sine at a frequency of 40 Hz and L = 33 mH.
14. Draw the voltage and current waveforms for the circuit of Figure 2.30 if I is 
a 10 μA peak sine at a frequency of 2 kHz and C = 6.8 nF.
66
Figure 2.27
Figure 2.28
Figure 2.29
Figure 2.30",ACElectricalCircuitAnalysis.pdf
"15. Draw the voltage and current waveforms for the circuit of Figure 2.31 if I is 
a two amp peak-peak sine at a frequency of 40 Hz and L = 33 mH.
16. Determine the impedance of the circuit of Figure 2.32.
17. Determine the impedance of the circuit of Figure 2.32 using a frequency of 
10 kHz.
18. For the circuit of Figure 2.32, determine the circulating current and the 
voltages across each component. Draw a phasor diagram of the three 
component voltages. Also find the time delay between the voltages of the 
components.
19. For the circuit of Figure 2.32 using a frequency of 10 kHz, determine the 
circulating current and the voltages across each component. Draw a phasor 
diagram of the three component voltages and determine the time delay 
between the capacitor and resistor voltages.
20. Determine the impedance of the circuit of Figure 2.33.
21. Determine the impedance of the circuit of Figure 2.33 using a frequency of 
10 kHz.
67
Figure 2.31
Figure 2.32
Figure 2.33",ACElectricalCircuitAnalysis.pdf
"22. For the circuit of Figure 2.33, determine the circulating current and the 
voltages across each component. Also find the time delay between the 
voltages of the components.
23. For the circuit of Figure 2.33 with a frequency of 3 kHz, determine the 
circulating current and the voltages across each component. Also find the 
time delay between the voltages of the components.
24. For the circuit of Figure 2.34, determine the circulating current.
25. Determine the impedance of the circuit of Figure 2.34 using a frequency of 
1.5 kHz.
26. For the circuit of Figure 2.34, determine the circulating current and the 
voltages across each component. Also find the time delay between the 
voltages of the components.
27. For the circuit of Figure 2.34 with a frequency of 1.5 kHz, determine the 
circulating current and the voltages across each component. Also find the 
time delay between the voltages of the components.
28. For the circuit of Figure 2.35, determine the circulating current and the",ACElectricalCircuitAnalysis.pdf
"time delay between the voltages of the components.
28. For the circuit of Figure 2.35, determine the circulating current and the 
voltages across each component. Draw a phasor diagram of the three 
component voltages and determine the time delay between the inductor and 
resistor voltages..
68
Figure 2.34
Figure 2.35",ACElectricalCircuitAnalysis.pdf
"29. For the circuit of Figure 2.36, determine the circulating current and the 
voltages across each component.
30. For the circuit of Figure 2.37, determine the circulating current and the 
voltages across each component.
31. For the circuit of Figure 2.38, determine the applied voltage and the voltages
across each component.
32. For the circuit of Figure 2.39, determine the applied voltage and the voltages
across each component.
69
Figure 2.36
Figure 2.37
Figure 2.38
Figure 2.39",ACElectricalCircuitAnalysis.pdf
"33. For the circuit of Figure 2.40, determine the circulating current and the 
voltages across each component.
34. Repeat the previous problem using an input frequency of 10 kHz.
35. For the circuit of Figure 2.41, determine the circulating current and the 
voltages across each component. The source is a 10 volt peak sine at          
20 kHz, R = 200 Ω, C = 100 nF and L = 1 mH.
36. For the circuit of Figure 2.41, find vb and vac.
37. For the circuit of Figure 2.42, find vb and vac. The source is a 50 volt peak-
peak sine at 10 kHz, R = 100 Ω, C = 200 nF and L = 1 mH.
38. For the circuit of preceding problem, determine the circulating current and 
the voltages across each component. 
70
Figure 2.40
Figure 2.41
Figure 2.42",ACElectricalCircuitAnalysis.pdf
"39. For the circuit of Figure 2.43, determine the circulating current and the 
voltages across each component. E is a 1 volt peak 2 kHz sine. Also, draw a 
phasor diagram of the four component voltages.
40. For the circuit of Figure 2.43, find vb and vca. E is a 1 volt peak 2 kHz sine.
41. For the circuit of Figure 2.44, determine vb, vc and vac. E is a 10 volt peak    
15 kHz sine.
42. For the circuit of Figure 2.45, determine the circulating current and the 
voltages across each component. E is a 100 millivolt peak 250 Hz sine. 
Further, draw a phasor diagram of the four component voltages.
71
Figure 2.43
Figure 2.44
Figure 2.45",ACElectricalCircuitAnalysis.pdf
"43. For the circuit of Figure 2.46, determine the circulating current and the 
voltages across each component. E is a 2 volt RMS 1 kHz sine. Also, draw a 
phasor diagram of the four component voltages.
44. For the circuit of Figure 2.47, determine vb, vc and vac. E is a 1 volt peak     
25 kHz sine. 
45. For the circuit of Figure 2.48, determine the voltages across each 
component. The source is a 50 mA peak sine at 15 kHz, R = 200 Ω,             
C = 100 nF and L = 1.5 mH.
72
Figure 2.46
Figure 2.47
Figure 2.48",ACElectricalCircuitAnalysis.pdf
"46. For the circuit of Figure 2.49, determine vac, vb and vc. The source is a 10 mA
peak-peak sine at 50 kHz, R = 2 kΩ, C = 10 nF and L = 800 μH.
47. For the circuit of Figure 2.50, determine the voltages across each 
component. The source is a 2 mA RMS sine at 1 kHz, R = 1.2 kΩ,                
C = 750 nF and L = 6.8 mH.
48. For the circuit of Figure 2.51, determine vac, vb and va. The source is a 2 mA 
peak-peak sine at 300 kHz, R = 560 Ω, C = 6.8 nF and L = 400 μH. 
73
Figure 2.49
Figure 2.50
Figure 2.51",ACElectricalCircuitAnalysis.pdf
"49. For the circuit of Figure 2.52, determine the voltages across each 
component.
50. For the circuit of Figure 2.53, determine the voltages across each 
component. Further, draw a phasor diagram of the four component voltages.
51. For the circuit of Figure 2.54, find vac, vb and vc. The source is 5 mA peak at  
8 kHz.
74
Figure 2.52
Figure 2.53
Figure 2.54",ACElectricalCircuitAnalysis.pdf
"52. For the circuit of Figure 2.55, determine the voltages across each 
component. The source is 20 mA peak at 100 kHz.
53. For the circuit of Figure 2.56, determine the voltages across each 
component.
54. For the circuit of Figure 2.57, determine the voltages vb and vdb. E1 = 20º 
and E2 = 590º.
55. Determine the inductance and capacitance values for the circuit of    
problem 52.
56. For the circuit of Figure 2.57, determine the inductor and capacitor values if 
the source frequency is 12 kHz.
75
Figure 2.55
Figure 2.56
Figure 2.57",ACElectricalCircuitAnalysis.pdf
"57. For the circuit of Figure 2.58, determine the voltages across each 
component. E1 = 10º and E2 = 860º.
DesignDesign
58. Redesign the circuit of Figure 2.32 using a new capacitor such that the 
current magnitude from the source is 100 μA.
59. Redesign the circuit of Figure 2.33 using a new frequency such that the 
current magnitude from the source is 200 μA.
60. For the circuit of Figure 2.32, determine a new capacitor such that |XC| = R.
61. For the circuit of Figure 2.33, determine a new frequency such that |XL| = R.
ChallengeChallenge
62. For the circuit of Figure 2.40, determine a new frequency such that               
|XC| = |XL|.
63. Determine the output voltage across the capacitor of Figure 2.32 at 
frequencies of 100 Hz, 5 kHz and 20 kHz. In light of this, if the input signal 
was a 1 kHz square wave instead of a sine wave as pictured, how would this 
circuit affect the shape of the output waveform (hint: consider 
superposition)?",ACElectricalCircuitAnalysis.pdf
"circuit affect the shape of the output waveform (hint: consider 
superposition)?
64. Assume that you are troubleshooting a circuit like the one shown in      
Figure 2.41. E is a 2 volt peak sine at 2 kHz, R = 390 Ω, C = 100 nF and       
L = 25 mH. The circulating current measures approximately 4 mA with a 
lagging phase angle of just under −40 degrees. What is the likely problem?
65. Given the circuit shown in Figure 2.41, find the values for C and L if the 
source is a 6 volt sine wave at 1 kHz, R = 2 kΩ, vR = 4 V and vL = 5 V .
76
Figure 2.58",ACElectricalCircuitAnalysis.pdf
"66. The circuit of Figure 2.59 can be used as part of a loudspeaker crossover 
network. The goal of this circuit is to steer the low frequency tones to the 
low frequency transducer (labeled here as “Loudspeaker” and often referred 
to as a woofer). A similar network substitutes a capacitor for the inductor to 
steer the high frequency tones to the high frequency transducer (AKA 
tweeter). These networks can be pictured as frequency sensitive voltage 
dividers. At very low frequencies, XC is very large and blocks low frequency
tones from reaching the tweeter. A mirror situation occurs with the 
inductor/woofer variant. The crossover frequency is the frequency at which 
the reactance magnitude equals the resistance. Assuming simple 8 Ω 
resistances for the woofer and tweeter, determine capacitor and inductor 
values that would yield a 1.5 kHz crossover frequency. How might this 
concept be extended to a mid-range loudspeaker that only produces tones in",ACElectricalCircuitAnalysis.pdf
"values that would yield a 1.5 kHz crossover frequency. How might this 
concept be extended to a mid-range loudspeaker that only produces tones in 
the middle of the musical frequency spectrum? (Note, this concept will be 
revisited in the final simulation problem, below, and also in the Simulation 
portion of Chapter 4 which covers series-parallel circuits.) 
SimulationSimulation
67. Simulate the solution of design problem 58 and determine if the values 
produce the required results.
68. Simulate the solution of design problem 59 and determine if the values 
produce the required results.
69. Simulate the solution of design problem 60 and determine if the values 
produce the required results. Hint: if the reactance/resistance magnitudes are
the same, then the voltage magnitudes will be identical.
70. Simulate the solution of design problem 61 and determine if the values 
produce the required results. Hint: if the reactance/resistance magnitudes are",ACElectricalCircuitAnalysis.pdf
"produce the required results. Hint: if the reactance/resistance magnitudes are
the same, then the voltage magnitudes will be identical.
77
Figure 2.59",ACElectricalCircuitAnalysis.pdf
"71. Simulate the solution of challenge problem 62 and determine if the new 
frequency produces the required results. Hint: if the reactance magnitudes 
are the same, then the voltage magnitudes will be identical. Further, their 
phases will cause these voltages to cancel, leaving the resistor voltage equal 
to the source voltage.
72. Using a transient analysis, crosscheck the crossover design of the final 
challenge problem, above. Plot the resistor (loudspeaker) voltage across the 
range of 100 Hz to 20 kHz for both sections.
78",ACElectricalCircuitAnalysis.pdf
"NotesNotes
♫♫♫♫
79",ACElectricalCircuitAnalysis.pdf
"3 3 Parallel RLC CircuitsParallel RLC Circuits
3.0 Chapter Learning Objectives3.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Utilize Kirchhoff's current law, the current divider rule and Ohm's law to find branch currents in parallel
RLC networks that utilize current sources or a single voltage source.
• Compute complex impedance and system voltage in parallel RLC circuits.
• Determine the equivalent value of multiple AC current sources in parallel.
• Draw phasor diagrams for impedance/susceptance and component currents in parallel RLC circuits.
3.1 Introduction 3.1 Introduction 
In this chapter we introduce the analysis of AC RLC circuits configured in parallel. AC parallel circuits echo the 
characteristic of their DC counterpart and many of the solution techniques from DC analysis will be applicable 
here. This includes the use of Ohm's law and Kirchhoff's current law, along with the current divider rule.",ACElectricalCircuitAnalysis.pdf
"here. This includes the use of Ohm's law and Kirchhoff's current law, along with the current divider rule. 
Generally, as with the series circuits presented in the previous chapter, reactance values will need to be 
computed from capacitor and inductor values before the main analysis may begin. Here, as in the most of the 
remaining chapters, we shall be concerned with determining the circuit response based on a source with a single 
frequency of excitation, in other words, a simple sine wave.
Parallel circuits are in many ways the complement of series circuits. The most notable characteristic of a parallel
circuit is that it has only two nodes and each component is connected from one node to the other. There are no 
other connections with which to create a voltage divider. Consequently, all components see the same voltage. 
Currents divide among the components in proportion to their conductance/susceptance (i.e., in inverse 
proportion to their resistance/reactance).",ACElectricalCircuitAnalysis.pdf
"Currents divide among the components in proportion to their conductance/susceptance (i.e., in inverse 
proportion to their resistance/reactance). 
 The key to this is to remember that all computations involve vector quantities. This can lead to some surprising 
results for the uninitiated. For example, due to the 180 degree phase differential between inductors and 
capacitors, it is possible for an individual branch current to be greater than the source current. This does not 
violate Kirchhoff's current law, as we shall see. Indeed, it is reminiscent of a similar situation in AC series 
circuits where an individual component voltage can be greater than the source voltage without violating 
Kirchhoff's voltage law. 
To clarify our analyses, we shall make considerable use of both time domain plots of currents as well as phasor 
diagrams. 
80",ACElectricalCircuitAnalysis.pdf
"3.2 The Parallel Connection3.2 The Parallel Connection
A parallel connection is always characterized by having all components share just 
two nodes. There are no other junctions from which current can flow out of or into, 
or for voltage to split. Consequently, 
The voltage is the same everywhere in a parallel connection. (3.1)
Each component in such a connection will see the same voltage, regardless of 
whether it is a resistor, capacitor or inductor. Before the branch currents can be 
determined, the capacitive and inductive reactance values must be computed. As 
seen from prior work, these are obtained from the capacitor and inductor values, and
based on the frequency of the driving source. Consequently, if this frequency were to
change, the reactances would change, and this would in turn cause changes in the 
various component currents as well as the total current delivered by the source. 
3.3 Parallel Impedance 3.3 Parallel Impedance",ACElectricalCircuitAnalysis.pdf
"various component currents as well as the total current delivered by the source. 
3.3 Parallel Impedance 3.3 Parallel Impedance 
Perhaps the first order of business is to determine equivalent impedance values for 
some collection of parallel components. Recall that the reciprocal of reactance is 
susceptance,
B = 1
X (3.2)
and that the reciprocal of impedance is admittance,
Y = 1
Z (3.3)
The units are siemens for each. It is also worth noting that, due to the division, the 
signs reverse. For example, a capacitive susceptance has an angle of +90 degrees 
and if a complex admittance has a negative angle, then the associated impedance is 
inductive.
The “conductance rule” for parallel combinations studied in the DC case remains 
valid for the AC case, although we generalize it for impedances:
  
Ztotal = 1
1
Z1
+ 1
Z 2
+ 1
Z3
+...+ 1
Z N
(3.4)
81",ACElectricalCircuitAnalysis.pdf
"Each of the individual impedances presented in Equation 3.4 (i.e., Z1, Z2, etc) can 
represent a simple resistance, a pure reactance or a complex impedance. Further, the 
product-sum rule shortcut for two components also remains valid for AC 
components:
Ztotal = Z 1×Z 2
Z 1+Z 2
(3.5)
There is one special case where Equation 3.5 can be “troublesome”, and that's when 
the two impedances consist of a pure capacitive reactance and a pure inductive 
reactance, both of the same magnitude. The two items will effectively cancel each 
other leaving a denominator of zero and an undefined result. While the theoretical 
combination “blows up” and approaches infinity, in reality it is limited by associated
resistances such as Rcoil, and arrives at some finite value This situation is studied in 
great depth in Chapter 8, which covers the concept of resonance.
Example 3.1
Determine the impedance of the network shown in Figure 3.1.
Equation 3.4 would be best here.
Ztotal = 1
1
Z1
+ 1
Z 2
+ 1
Z 3",ACElectricalCircuitAnalysis.pdf
"Example 3.1
Determine the impedance of the network shown in Figure 3.1.
Equation 3.4 would be best here.
Ztotal = 1
1
Z1
+ 1
Z 2
+ 1
Z 3
Z total = 1
1
j 12 kΩ + 1
20 k Ω + 1
− j 48 kΩ
 
Ztotal = 12.49E3 51.3° Ω
This result might be a little surprising to the sharp-eyed. Notice that the 
magnitude of the total is larger than the magnitude of the smallest 
component (the inductor at j12 kΩ). This would never be the case if these 
three components were all resistors: the result would have to be smaller than
the smallest element in the group. 
The reason for this is that the capacitive reactance partially cancels the 
inductive reactance. If the product-sum rule (Equation 3.5) is used with 
these two components, the result is 16E390° or j16 kΩ. Placing that in 
parallel with the 20 kΩ resistor (again using Equation 3.5) leads to the result
computed above.
An admittance diagram is illustrated in Figure 3.2. The vector summation of",ACElectricalCircuitAnalysis.pdf
"computed above.
An admittance diagram is illustrated in Figure 3.2. The vector summation of 
the component conductance and susceptances is verified nicely. 
82
Figure 3.1
Circuit for Example 3.1.",ACElectricalCircuitAnalysis.pdf
"The individual component values are:
BL = 1
j 12 kΩ ≈− j 83.33E-6S
BC = 1
− j 48k Ω ≈ j 20.83E-6S
G = 1
20 kΩ = 50E-6 S
Y total = 1
12.49E3 51.3° Ω ≈ 80.1E-6−51.3° S
In rectangular form Ytotal = 50E−6 − j62.5E−6 S.
83
Figure 3.2
Admittance diagram for the 
network of Figure 3.1.",ACElectricalCircuitAnalysis.pdf
"Example 3.2
Determine the impedance of the network shown in Figure 3.3 at a
frequency of 10 kHz. Repeat this for a frequency of 1 kHz.
First, find the reactances at 10 kHz. For the inductor we find:
X L = j 2π f L
X L = j 2π10 kHz680μ H
X L ≈ j 42.73Ω
And for the capacitor:
X C =− j 1
2 π f C
X C =− j 1
2 π10kHz 470 nF
X C ≈− j 33.86Ω
Now use Equation 3.4 to combine the elements.
Ztotal = 1
1
Z1
+ 1
Z2
+ 1
Z3
Ztotal = 1
1
j 42.73Ω + 1
1.8k Ω + 1
− j 33.86Ω
 
Ztotal = 162.6 −84.8° Ω
In rectangular form this is 14.68 − j161.9 Ω. As the capacitor's reactance is 
the smallest of the three components, it dominates the equivalent impedance 
at this frequency. By working the capacitive reactance formula in reverse, it 
can be shown that the reactive portion of − j161.9 Ω can achieved at this 
frequency by using a capacitance of 98.3 nF. That means that at 10 kHz, this 
parallel network has the same impedance as a 14.68 Ω resistor in series with",ACElectricalCircuitAnalysis.pdf
"parallel network has the same impedance as a 14.68 Ω resistor in series with 
a 98.3 nF capacitor. At any other frequency this will no longer be true, as 
will be illustrated momentarily. 
At 1 kHz, the frequency is reduced by a factor of ten. Therefore, XL will be 
ten times lower, or approximately j4.273 Ω. Further, XC will be ten times 
higher, or about − j338.6 Ω. The inductive reactance will now dominate.
The new impedance is:
84
Figure 3.3
Circuit for Example 3.2.",ACElectricalCircuitAnalysis.pdf
"Ztotal = 1
1
Z1
+ 1
Z 2
+ 1
Z 3
Ztotal = 1
1
j 4.273Ω + 1
1.8k Ω + 1
− j 338.6Ω
 
Ztotal = 4.328  89.9° Ω
In rectangular form this is 10.4E−3 + j4.328 Ω. The result is inductive, the 
opposite of what we saw at 10 kHz. Using the inductive reactance formula, 
it can be shown that at 1 kHz this parallel network has the same impedance 
as a 10.4 milliohm resistor in series with a 689 μH inductor.
3.4 Parallel Circuit Analysis 3.4 Parallel Circuit Analysis 
Kirchhoff's current law (KCL) is the operative rule for parallel circuits. It states that 
the sum of all currents entering and exiting a node must equal zero. Alternately, it 
can be stated as the sum of currents entering a node must equal the sum of currents 
exiting that node. As a pseudo formula: 
∑ I→ = ∑ I← (3.6)
A similar situation occurs in AC parallel circuits to that seen AC series circuits; 
namely that it appears that KCL (like KVL) is “broken”. That is, a simplistic",ACElectricalCircuitAnalysis.pdf
"namely that it appears that KCL (like KVL) is “broken”. That is, a simplistic 
summing of the magnitudes of the currents might not balance. Once again, this is 
because the summation must be a vector summation, paying attention to the phase 
angles of each separate current. 
It is possible to drive a parallel circuit with multiple current sources. These sources 
will add in much the same way that voltage sources in series add, that is, polarity 
and phase must be considered. Ordinarily, voltage sources with differing values are 
not placed in parallel as this violates the basic rule of parallel circuits (voltage being 
the same across all components).
The current divider rule remains valid for AC parallel circuits. Given two 
components, Z1 and Z2, and a current feeding them, IT, the current through one of the 
components will equal the total current times the ratio of the opposite component 
over the sum of the impedance of the pair.
i1 = iT
Z 2
Z1+Z 2
(3.7)
85",ACElectricalCircuitAnalysis.pdf
"This rule is convenient in that the parallel equivalent impedance need not be 
computed, but remember, it is valid only when there are just two components 
involved. 
When analyzing a parallel circuit, if it is being driven by a voltage source, then this 
same voltage must appear across each of the individual components. Ohm's law can 
then be used to determine the individual currents. According to KCL, the total 
current exiting the source must be equal to the sum of these individual currents. For
example, in the circuit shown in Figure 3.4, the voltage E must appear across both 
R and L. Therefore, the currents must be iL = E / XL and iR = E / R, and itotal = iL + iR. 
itotal can also be found be determining the parallel equivalent impedance of R and 
XL, and then dividing this into E. This technique can also be used in reverse in order 
to determine a resistance or reactance value that will produce a given total current:",ACElectricalCircuitAnalysis.pdf
"to determine a resistance or reactance value that will produce a given total current: 
dividing the source by the current yields the equivalent parallel impedance. As one 
of the two is already known, the known component can be used to determine the 
value of the unknown component via Equation 3.4.
Example 3.3
Determine the currents in the circuit shown in Figure 3.4 if the source is 
100º volts peak, XL = j2 kΩ and R = 1 kΩ
The same voltage must appear across all elements in a parallel connection. 
In this case that's 100º volts peak. The two branch currents are found via 
Ohm's law:
iL = v
Z
iL = 10 0° V
j2 kΩ
iL = 5E-3−90° A
iR = v
Z
iR = 10 0° V
1k Ω
iR = 10E-3 0° A
The source current is the sum of the these two currents, or 11.18E−3−26.6º
amps. This can be verified by determining the equivalent parallel impedance
and then applying Ohm's law:
86
Figure 3.4
A simple parallel RL circuit.",ACElectricalCircuitAnalysis.pdf
"ZT = Z 1×Z 2
Z 1+Z 2
ZT = 1k Ω× j 2k Ω
1k Ω+j 2k Ω
ZT = 894  26.6° Ω
As a side note, in rectangular form this is 800 + j400, meaning that this 
parallel combination is equivalent to a series combination of an 800 ohm 
resistance and a 400 ohm inductive reactance. Continuing with Ohm's law, 
we have:
iSource = v
Z T
iSource = 10 0° V
894  26.6° Ω
iSource = 11.18E-3−26.6° A
 A phasor diagram of the currents is shown in Figure 3.5, below.
Current Measurement in the LaboratoryCurrent Measurement in the Laboratory
At this point, a practical question arises; how do we verify these currents in the 
laboratory? After all, the preeminent measurement tool is the oscilloscope, and these
are designed to measure voltage, not current. While it is possible to obtain current 
measurement probes, a simple method to measure current is to use a small current 
sensing resistor with a standard oscilloscope setup. We have seen already that the",ACElectricalCircuitAnalysis.pdf
"sensing resistor with a standard oscilloscope setup. We have seen already that the 
voltage across a resistor must be in phase with the current through it. Therefore, 
87
Figure 3.5
Phasor diagram of currents in 
the circuit of Figure 3.4.",ACElectricalCircuitAnalysis.pdf
"whatever phase angle we see on a resistor's voltage, its current phase angle must be 
the same. The idea is to simply insert a resistor into the branch of which we wish to 
measure the current. As long as the resistance is much smaller than the impedance 
seen in the rest of that branch, it will have little impact on the precise value of the 
current. We can then place a probe across this sensing resistor to read its voltage. 
Ohm's law is then be used to determine the current's magnitude, while the phase 
angle is obtained directly from the oscilloscope (again, because the sensing resistor's
voltage must be in phase with its current, and thus the voltage phase angle is equal to
the current phase angle). This technique will be illustrated in the simulation portion 
of the next example problem.
Example 3.4
Determine the branch currents in the circuit of Figure 3.6.
The first step is to determine the capacitive reactance.
X C =− j 1
2 π f C
X C =− j 1
2 π500 Hz 250 nF
X C ≈− j 1273Ω",ACElectricalCircuitAnalysis.pdf
"The first step is to determine the capacitive reactance.
X C =− j 1
2 π f C
X C =− j 1
2 π500 Hz 250 nF
X C ≈− j 1273Ω
Both the resistor and capacitor will see 20 volts peak from the source. Their 
currents can be determined via Ohm's law:
iC = v
X C
iC = 20 V
1273 − 90° Ω
iC ≈15.71E-3 90° A
iR = v
R
iR = 20 V
220  0° Ω
iR ≈90.91E-3 0° A
The source current is the sum of these two currents, or 92.26E−39.8° 
amps. The source current may also be determined by dividing the source 
voltage by the equivalent parallel impedance, as follows. 
88
Figure 3.6
Circuit for Example 3.4.",ACElectricalCircuitAnalysis.pdf
"ZT = Z 1×Z 2
Z 1+Z2
ZT = 220 Ω×(− j 1273Ω)
220 Ω− j 1273k Ω
ZT = 216.8−9.8° Ω
iSource = v
Z T
iSource = 20  0° V
216.8 −9.8° Ω
iSource = 92.26E-3 9.8° A
A time domain plot of the currents is illustrated in Figure 3.7 along with a 
phasor plot in Figure 3.8. Note that the source current is close in both 
amplitude and phase to the resistor current. By comparison, the capacitor 
current is considerably smaller and with an obvious leading phase shift.
89
Time Domain Graph
Current (milliamps)
-100.0
-50.0
0.0
50.0
100.0
 
Time (milliseconds)
0 0.5 1 1.5 2
isource
ic
ir
Figure 3.7
Current waveforms for the 
circuit of Figure 3.6.
Figure 3.8
Phasor diagram for the circuit
of Figure 3.6.",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
The circuit of Figure 3.6 is captured in a simulator as shown in Figure 3.9. 
Individual 2 ohm resistors are used to sense the currents in the resistor and capacitor 
branches. These sensing resistors are inserted directly above ground for convenience
of measure. In this way a differential measurement is not needed.
The value of a sensing resistor needs to be considerably smaller than the impedance 
of the branch in which it is inserted. Two orders of magnitude (i.e., less than 1 %) 
would be a good place to start. While even smaller values will increase accuracy, 
from a practical standpoint in the lab, such tiny resistor values would produce 
extremely small voltages that would be difficult to measure with great accuracy. 
90
Figure 3.9
Circuit of Figure 3.6 in a 
simulator with added current 
sensing resistors.
Figure 3.10
Simulation results for the 
circuit of Figure 3.9.",ACElectricalCircuitAnalysis.pdf
"A basic transient analysis is then performed as shown in Figure 3.10. The results of 
the simulation are in tight agreement with the plot of Figure 3.7. Note that the 
plotted resistor and capacitor “currents” are, in fact, the voltages across the 
associated sensing resistors (nodes 2 and 3) divided by 2 (ohms), in direct 
application of Ohm's law. Not only does the simulation verify the results computed 
earlier, it also validates the concept of using small current sensing resistors in the 
laboratory.
Example 3.5
Determine the the branch currents in the circuit of Figure 3.11. Use the
source voltage as the reference (0° ).
Although this circuit is drawn a little differently than the prior examples, it 
remains a simple parallel circuit with just two nodes. Therefore, each 
element sees a 2 volt peak potential. Ohm's law will suffice to find the three 
component currents. These are then added to find the source current. Their",ACElectricalCircuitAnalysis.pdf
"component currents. These are then added to find the source current. Their 
reference directions are right-to-left given the source's polarity.
iC = v
X C
iC = 2 0° V
12  −90° Ω
iC ≈0.1667 90° A
iL = v
X L
iL = 2 0° V
8 90 ° Ω
iL = 0.25 −90 ° A
iR = v
R
iR = 2 0° V
10 0° Ω
iR = 0.2 0° A
The sum of these three currents is approximately 0.2167−22.6° amps. This
value can be verified by finding the combined parallel impedance. Using 
Equation 3.4 it can be shown that this impedance is 9.23122.6° Ω. Thus,
91
Figure 3.11
Circuit for Example 3.5.",ACElectricalCircuitAnalysis.pdf
"iSource = v
Z
iSource = 2 0° V
9.231  22.6° Ω
iSource ≈0.2167−22.6° A
A phasor diagram showing the vector summation of the currents is 
illustrated in Figure 3.12. 
Note that iC and iL are in perfect opposition as expected. Subtracting the 
magnitude of iC from iL leaves us with the precise vertical component of the 
source current.
 
92
Figure 3.12
Current phasor diagram for  
the circuit of Figure 3.11.",ACElectricalCircuitAnalysis.pdf
"If a parallel circuit is driven by a current source, as shown in Figure 3.13, there are
two basic methods of solving for the component currents. The first method is to use
the current divider rule. If desired, the component voltage can then be found using
Ohm's law. An alternate method involves finding the parallel equivalent impedance
first and then using Ohm's law to determine the voltage (remember, being a parallel 
circuit, there is only one common voltage). Given the voltage, Ohm's law can be 
used to find the current through one component. To find the current through the 
other, either Ohm's law can be applied a second time or KCL may be used; 
subtracting the current through the first component from the source current. If there 
are more than two components, usually the second method would be the most 
efficient course of action.
It is worth noting that both methods described above will yield the correct answers.",ACElectricalCircuitAnalysis.pdf
"efficient course of action.
It is worth noting that both methods described above will yield the correct answers. 
One is not “more correct” than the other. We can consider each of these as a separate
solution path; that is, a method of arriving at the desired end point. In general, the 
more complex the circuit, the more solution paths there will be. This is good because
one path may be more obvious to you than another. It also allows you a means of 
crosschecking your work.
Example 3.6
Determine the currents in the circuit of Figure 3.13 if the source is 300 
milliamps peak at 10 kHz, R is 750 Ω and L is 15 mH. 
The first step is to determine the inductive reactance.
X L = j 2π f L
X L = j 2π10 kHz15 mH
X L ≈ j 942.4Ω
The current divider rule can be used to find the current through the resistor.
iR = iSource
X L
R+X L
iR = 0.3 0° A 942.4  90 ° Ω
750  0° Ω +942.4  90° Ω
iR = 0.2347 38.5 ° A
KVL can be used to find the remaining current through the inductor as the",ACElectricalCircuitAnalysis.pdf
"750  0° Ω +942.4  90° Ω
iR = 0.2347 38.5 ° A
KVL can be used to find the remaining current through the inductor as the 
source current must equal the sum of the inductor and resistor currents. 
Therefore:
93
Figure 3.13
Parallel network driven by a 
current source.",ACElectricalCircuitAnalysis.pdf
"iL = iSource−i R
iL = 0.3 0° A −0.2347 38.5° A
iL = 0.1868−51.5° A
This value could also have been obtained by applying the current divider 
rule a second time:
iL = iSource
R
R+X L
iL = 0.3 0° A 750 0° Ω
750  0° Ω+942.4  90 ° Ω
iL = 0.1868−51.5° A
A time domain plot of the current waveforms is shown in Figure 3.14. 
From the plot it is apparent that the inductor and resistor currents do indeed 
add up to the source current. Further, the current through the inductor is 
lagging, as expected.
94
Figure 3.14
Current waveforms for the 
circuit of Example 3.6.
Time Domain Graph
Current (amps)
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
 
Time (microseconds)
0 20 40 60 80 100
isource
il
ir",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
The circuit of Example 3.6 is captured in a simulator as shown in Figure 3.15. A pair
of 1 ohm current sensing resistors have been added. By using 1 ohm, the sensing 
voltage will have the same magnitude as the current, with no scaling required.
A transient analysis is performed, plotting the voltages at nodes 2 and 3 along with 
their sum (the source current). The results are shown in Figure 3.16. The plot is 
delayed for a millisecond in order to avoid the initial power-up transient. The results 
are in full agreement with the plot of Figure 3.14.
95
Figure 3.15
Circuit of Example 3.6 in a 
simulator.
Figure 3.16
Results of the transient 
response simulation for the 
circuit of Example 3.6.",ACElectricalCircuitAnalysis.pdf
"Example 3.7
Determine the branch currents in the circuit of Figure 3.17. Use the current 
source as the time reference (i.e., 0°).
Instead of using repeated current dividers here, we can instead find the 
equivalent impedance and then apply Ohm's law to determine the system 
voltage. Each branch current may be obtained by dividing this voltage by the
impedance of that branch; a quick application of Ohm's law.
The system impedance is computed through Equation 3.4 as follows:
Z total = 1
1
Z 1
+ 1
Z 2
+ 1
Z 3
Ztotal = 1
1
− j 12Ω + 1
j 8Ω + 1
10Ω
 
Ztotal = 9.231  22.6° Ω
The system voltage is found using Ohm's law. As the reference direction of 
the current source is flowing down into ground, the branch currents will be 
flowing up. This makes their voltage reference negative with respect to 
ground (i.e., + to − from bottom to top). We can deal with this by describing 
the current source as negative (i.e., flowing out of the top node). The source",ACElectricalCircuitAnalysis.pdf
"the current source as negative (i.e., flowing out of the top node). The source 
can be written as either −10° or as 1180°, it makes no difference.
v = i source×Z total
v =−1 0° A×9.231  22.6°
v ≈−9.231 22.6 ° V
This potential may also be written as 9.231−157.4° volts (remember, an 
inversion is the same as a 180 degree phase shift). 
And now for the branch currents:
96
Figure 3.17
Circuit for Example 3.7.",ACElectricalCircuitAnalysis.pdf
"iC = v
X C
iC = 9.231−157.4° V
12 − 90° Ω
iC ≈0.7692−67.4 ° A
iL = v
X L
iL = 9.231−157.40° V
8 90° Ω
iL = 1.154  112.6° A
iR = v
R
iR = 9.231 −157.4 ° V
10 0° Ω
iR = 0.9231−157.4 ° A
The sum of these three currents is approximately 1180° amps (i.e., 
−10°), the value of the current source, as expected. 
A phasor diagram of the currents is plotted in Figure 3.18.
97
Figure 3.18
Current phasor diagram for  
the circuit of Figure 3.16.",ACElectricalCircuitAnalysis.pdf
"Now for the sneaky bit. In case you didn't notice, the components in this 
circuit are identical to those of Figure 3.11 with the exception of the source. 
As a consequence, the phase angles and magnitudes of the branch currents 
through the three components will remain unchanged relative to each other. 
Due to the change in both magnitude and direction of the source, the entire 
phasor plot has been rotated clockwise by 154.7 degrees and all of the 
magnitudes have been lengthened by the ratio of the voltages (9.231 / 2). 
This is perhaps easiest to see by focusing on the vector for isource in Figure 
3.12. Rotate that vector clockwise, along with every other vector, until isource 
lines up with the negative real axis. The result is the plot of Figure 3.18, 
albeit with newly lengthened magnitudes.
If a parallel circuit contains multiple current sources, they may be added together (a 
vector sum, of course) to generate an equivalent single current source. The analysis",ACElectricalCircuitAnalysis.pdf
"vector sum, of course) to generate an equivalent single current source. The analysis 
then proceeds as before. This is demonstrated in the following example.
Example 3.8
For the circuit of Figure 3.19, determine the currents through the capacitor, 
inductor and resistor, and also determine the system voltage. i1 is 0.50° 
amps and i2 is 345° amps.
The first step here is to determine the effective current driving the circuit. If 
we treat the top node as positive, i2 is entering (positive) while i1 is exiting 
(negative). The combination is:
itotal = i1 +i2
itotal = −0.5 0° A +3 45° A
itotal = 2.67 52.6° A
This current has the same reference direction as source two. A phasor 
diagram of the process is illustrated in Figure 3.20 to help visualize the 
vector addition.
98
Figure 3.19
Circuit for Example 3.8.",ACElectricalCircuitAnalysis.pdf
"The next step is to find the combined impedance so that we can find the 
applied voltage.
Ztotal = 1
1
Z1
+ 1
Z 2
+ 1
Z 3
Ztotal = 1
1
− j 100 Ω + 1
j 50Ω + 1
40 Ω
 
Ztotal = 37.14  21.8° Ω
v = i source×Ztotal
v = 2.67 52.6° A×37.14  21.8°
v ≈99.16  74.4° V
We now apply Ohm's law to each of the components to determine the branch
currents. 
iC = v
X C
iC = 99.16  74.4° V
100 − 90° Ω
iC ≈0.9916 164.4° A
iL = v
X L
iL = 99.16 74.4 ° V
50  90° Ω
iL = 1.983 −15.6° A
99
Figure 3.20
Phasor diagram of the 
combined current sources for 
the circuit for Example 3.8.",ACElectricalCircuitAnalysis.pdf
"iR = v
R
iR = 99.16  74.4° V
40 0° Ω
iR = 2.479  74.4 ° A
The sum of these three currents is approximately 2.6752.6° amps. This 
balances the net entering current from the two sources, verifying KVL. 
3.5 Summary3.5 Summary
In this chapter we have examined parallel circuits using either a single voltage 
source, or one or more AC current sources, along with two or more resistors, 
capacitors and inductors. The defining characteristic of a parallel configuration is 
that all components are connected to just two nodes and that all elements in this 
configuration see the same voltage. If multiple AC current sources are present, they 
may be combined into a single equivalent current source via vector addition. 
Unlike the purely resistive case, the equivalent impedance of a group of parallel RLC
components will not always be smaller than the smallest resistance or reactance in 
the group due to cancellations between the inductance and capacitance. As voltage is",ACElectricalCircuitAnalysis.pdf
"the group due to cancellations between the inductance and capacitance. As voltage is
identical for all components, then the currents through the capacitors must be 180 
degrees out of phase with the currents through the inductors. Thus it is quite possible
that the current through one of the reactance branches could be larger than the 
current supplied by the source. In general, the effective impedance is found by 
summing the individual conductances and susceptances to find the total admittance 
of the group, and then taking the reciprocal of this value. The product-sum rule may 
be used so long as the angles are taken into account (i.e., a vector process). 
Kirchhoff's current law (KCL) states that the sum of currents entering a node must 
equal the sum of currents leaving that node. This remains true for AC circuit 
analysis, however, it must be remembered to always use a vector summation. A 
simple summation of current magnitudes will not achieve proper results.",ACElectricalCircuitAnalysis.pdf
"analysis, however, it must be remembered to always use a vector summation. A 
simple summation of current magnitudes will not achieve proper results.
Individual branch currents in a circuit driven by a voltage source may be determined
by using Ohm's law: simply divide the source voltage by the individual resistance 
and reactance values. These branch currents must sum to the total current delivered 
by the voltage source thanks to KCL. If the parallel network is driven by current 
sources, the individual branch currents can be found by first determining the 
effective parallel impedance and then using Ohm's law to find the system voltage. 
Once the voltage is known, Ohm's law is used again on each component to find the 
associated branch current. Alternately, the current divider rule may be used 
repeatedly on successive pairings of components. 
100",ACElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. How is the equivalent impedance for a group of parallel connected resistors, 
inductors and capacitors computed? 
2. How is the equivalent value for parallel connected AC current sources 
computed?
3. Is the product-sum rule still applicable for AC analysis? Can it be used for 
reactance and/or complex impedance?
4. Define Kirchhoff's current law for AC circuit analysis.
5. Is it possible for a branch current to be larger than the source current in an 
AC parallel circuit? Explain why/why not.
3.6 Exercises3.6 Exercises
AnalysisAnalysis
1. Determine the effective impedance of the network shown in Figure 3.21 at   
10 MHz.
2. Determine the effective impedance of the network shown in Figure 3.22 at 
100 Hz.
3. Determine the effective impedance of the network shown in Figure 3.23 at   
5 kHz.
101
Figure 3.21
Figure 3.22
Figure 3.23",ACElectricalCircuitAnalysis.pdf
"4. Determine the effective impedance of the network shown in Figure 3.24 at   
20 kHz.
5. Determine the effective impedance of the network shown in Figure 3.25.
6. Determine the effective impedance of the network shown in Figure 3.25 if 
the frequency is halved and if the frequency is doubled.
7. For the network shown in Figure 3.21, determine the frequency below which
the impedance is mostly resistive.
8. For the network shown in Figure 3.22, determine the frequency below which
the impedance is mostly inductive.
9. Draw phasor impedance plot for problem 1.
10. Draw phasor impedance plot for problem 2.
11. Determine the three branch currents for the circuit shown in Figure 3.26 and
draw their phasor diagram.
12. Determine the three branch currents for the circuit shown in Figure 3.27 and
draw their phasor diagram.
102
Figure 3.24
Figure 3.25
Figure 3.26
Figure 3.27",ACElectricalCircuitAnalysis.pdf
"13. Determine the four branch currents for the circuit shown in Figure 3.28 and 
draw their phasor diagram.
14. Determine all of the branch currents for the circuit shown in Figure 3.29 
assuming E is a 1 volt RMS sine.
15. Determine all of the branch currents for the circuit shown in Figure 3.30 
given E = 10 volt peak sine, R = 220, XC = −j500. and XL = j1.5 k.
16. Determine all of the branch currents for the circuit shown in Figure 3.31 
given E = 2 volt peak sine, R = 1 k, XC = −j2 k. and XL = j3 k.
103
Figure 3.28
Figure 3.29
Figure 3.30
Figure 3.31",ACElectricalCircuitAnalysis.pdf
"17. Determine the component currents for the circuit shown in Figure 3.32. 
Draw phasor diagram of the source and branch currents.
18. Determine the resistor and capacitor voltages for the circuit shown in  
Figure 3.32.
19. Determine the resistor and inductor voltages for the circuit shown in    
Figure 3.33.
20. Determine the component currents for the circuit shown in Figure 3.33. 
Draw phasor diagram of the source and branch currents.
21. Determine the source voltage for the circuit shown in Figure 3.34. 
22. Determine the component currents for the circuit shown in Figure 3.34. 
Draw the phasor diagram of the source and branch currents.
23. Determine the component currents for the circuit shown in Figure 3.35. I is 
20 mA at 0 degrees.
104
Figure 3.32
Figure 3.33
Figure 3.34
Figure 3.35",ACElectricalCircuitAnalysis.pdf
"24. Determine the source voltage for the circuit shown in Figure 3.35. I is        
20 mA at 0 degrees.
25. Determine the source voltage for the circuit shown in Figure 3.36. Assume 
I1 is 1 mA at 0 degrees and I2 is 2 mA at +90 degrees.
26. Determine the capacitor and inductor currents in the circuit of Figure 3.36. 
Assume I1 is 1 mA at 0 degrees and I2 is 2 mA at +90 degrees.
27. Determine the resistor and capacitor currents in the circuit of Figure 3.37. 
Assume I1 is 20° amps and I2 is 0.545°.
28. Determine the source voltage for the circuit shown in Figure 3.37. Assume 
I1 is 20° A and I2 is 0.545°.
DesignDesign
29. For the network shown in Figure 3.38, determine a value of C such that the 
impedance magnitude of the circuit is 1 kΩ. The source is a 50 Hz sine and 
R is 2.2 kΩ.
105
Figure 3.36
Figure 3.37
Figure 3.38",ACElectricalCircuitAnalysis.pdf
"30. For the network shown in Figure 3.39, determine a value of L such that the 
impedance magnitude of the circuit is 2 kΩ. The source is a 2 MHz sine and 
R is 3.3 kΩ.
31. For the circuit shown in Figure 3.38, determine a value for C such that the 
magnitude of the source current is 1 mA. E is a 2 volt 10 kHz sine and        
R = 8 kΩ.
32. For the network shown in Figure 3.39, determine a value for L such that the 
magnitude of the source current is 10 mA. E is a 25 volt 100 kHz sine and   
R = 4 kΩ.
33. For the network shown in Figure 3.40, determine a value of C such that the 
impedance magnitude of the circuit is 10 k Ω. The source is a 440 Hz sine 
and R is 33 kΩ.
34. For the network shown in Figure 3.41, determine a value of L such that the 
impedance magnitude of the circuit is 200 Ω. The source is a 60 Hz sine and
R is 680 Ω.
35. For the circuit shown in Figure 3.40, determine a value for C such that the",ACElectricalCircuitAnalysis.pdf
"R is 680 Ω.
35. For the circuit shown in Figure 3.40, determine a value for C such that the 
magnitude of the circuit voltage is 200 volts. The source current is a 100 mA
1200 Hz sine and R = 15 kΩ.
36. For the circuit shown in Figure 3.41, determine a value for L such that the 
magnitude of the circuit voltage is 50 volts. The source current is a 2.3 A    
60 Hz sine and R = 330 Ω.
106
Figure 3.39
Figure 3.40
Figure 3.41",ACElectricalCircuitAnalysis.pdf
"37. Given the circuit shown in Figure 3.38, determine a value for C such that the
impedance angle is −45 degrees. The source a 1 volt peak sine at 600 Hz and
R = 680 Ω.
38. Given the circuit shown in Figure 3.39, determine a value for L such that the
impedance angle is 45 degrees. The source a 10 volt peak sine at 100 kHz 
and  R = 1.2 kΩ.
39. Determine a value for C such |XC| = |XL| for the circuit shown in Figure 3.42.
The source frequency is 1 kHz, R = 200 Ω and L = 50 mH. 
40. Determine a value for L such |XC| = |XL| for the circuit shown in Figure 3.43.
The source frequency is 22 kHz, R = 18 kΩ and C = 5 nF. 
41. Add one or more components in parallel with the circuit of Figure 3.22 such 
that the resulting impedance at 20 Hz is 10 Ω with a phase angle of at least 
+30°.
ChallengeChallenge
42. Determine a value for C such that the impedance angle for the circuit shown
in Figure 3.42 is purely resistive (0 degrees). The source frequency is 1 kHz,
R = 200 Ω and L = 50 mH.",ACElectricalCircuitAnalysis.pdf
"in Figure 3.42 is purely resistive (0 degrees). The source frequency is 1 kHz,
R = 200 Ω and L = 50 mH. 
43. Is it possible to change the value of the resistor in Figure 3.34 so that the 
system voltage is 4 volts? If so, what is the value? If not, why not?
44. Is it possible to change the value of the inductor and/or capacitor in Figure 
3.34 so that the system voltage is 4 volts? If so, what is/are the values? If 
not, why not?
107
Figure 3.42
Figure 3.43",ACElectricalCircuitAnalysis.pdf
"45. Assume you are troubleshooting a circuit like the one shown in Figure 3.43. 
I is a 10 mA peak sine at 2 kHz, R = 390 Ω, C = 200 nF and L = 25 mH. The
measured resistor voltage is a little under 2.5 volts. What is the likely 
culprit?
46. Given the circuit shown in Figure 3.43, find the values for C and L if the 
source is a sine wave at 1 kHz, R = 4 kΩ, iSource = 3 mA, iR = 2 mA,               
iL = 5 mA,
SimulationSimulation
47. Using a transient analysis simulation, verify that the source current 
magnitude is 1 mA using the capacitor value determined in design    
problem 31.
48. Using a transient analysis simulation, verify that the source current 
magnitude is 10 mA using the inductor value determined in design    
problem 32.
49. Using a transient analysis simulation, verify that the source voltage 
magnitude is 200 volts using the capacitor value determined in design 
problem 35.
50. Using a transient analysis simulation, verify that the source voltage",ACElectricalCircuitAnalysis.pdf
"problem 35.
50. Using a transient analysis simulation, verify that the source voltage 
magnitude is 50 volts using the inductor value determined in design  
problem 36.
51. Using a transient analysis simulation, verify the design solution for   
problem 39. This can be checked by seeing if the current magnitudes in C 
and L are identical.
52. Using a transient analysis simulation, verify the design solution for   
problem 40. This can be checked by seeing if the current magnitudes in C 
and L are identical.
53. Impedance magnitude as a function of frequency can be investigated by 
driving the circuit with a fixed amplitude current source across a range of 
frequencies. The resulting voltage will be proportional to the effective 
impedance. Investigate this effect by performing an AC analysis on the 
circuits shown in Figures 3.32 and 3.33. Use a frequency range of 10 Hz to 
1 MHz. Before running the simulations, sketch your expected results.",ACElectricalCircuitAnalysis.pdf
"circuits shown in Figures 3.32 and 3.33. Use a frequency range of 10 Hz to 
1 MHz. Before running the simulations, sketch your expected results. 
54. Following the idea presented in the previous problem, investigate the 
impedance as a function of frequency of the circuit shown in Figure 3.43. 
Use R = 1 k, C = 10 nF, and L = 1 mH. Run the simulation from 100 Hz to 
10 MHz. Make sure to sketch your expected results first. 
108",ACElectricalCircuitAnalysis.pdf
"NotesNotes
♫♫♫♫
109",ACElectricalCircuitAnalysis.pdf
"4 4 Series-Parallel RLC CircuitsSeries-Parallel RLC Circuits
4.0 Chapter Learning Objectives4.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Identify series-only and parallel-only sub-groups in series-parallel RLC networks. 
• Compute complex equivalent impedance for series-parallel RLC circuits.
• Simplify an entire RLC network into a simple series or parallel equivalent comprised of complex 
impedances. 
• Utilize KVL, KCL and other techniques to find various voltages and currents in series-parallel RLC 
networks driven by a single effective voltage or current source.
• Interpret phasor diagrams and time domain plots for component voltages and/or currents in series-
parallel RLC circuits.
4.1 Introduction4.1 Introduction
Having completed our examination of strictly series and strictly parallel AC circuits, we turn our attention now 
to somewhat more complex circuits, namely those circuits comprised of components in mixed series-parallel",ACElectricalCircuitAnalysis.pdf
"to somewhat more complex circuits, namely those circuits comprised of components in mixed series-parallel 
arrangements. This chapter deals with a subset of series-parallel RLC circuits, specifically those that are driven 
by a single effective current or voltage source, and which may be simplified using series and parallel component
combinations. The rules and techniques explored for strictly series networks are still applicable to series 
connected subsections of larger circuits. The same is true for the rules and techniques established for strictly 
parallel circuits regarding parallel subsections. Thus, the key to analyzing series-parallel circuits is in 
recognizing those portions of the circuit that form series or parallel sub-circuits, and then applying the series and
parallel analysis rules to those sections. Ohm's law, KVL and KCL may be used in turn to solve portions of the",ACElectricalCircuitAnalysis.pdf
"parallel analysis rules to those sections. Ohm's law, KVL and KCL may be used in turn to solve portions of the 
problem until all currents and voltages are found. As individual voltages and currents are determined, this makes
it easier to apply these rules to determine other values. 
It is often useful to determine the effective impedance of individual sections at the outset in order to facilitate 
circuit analysis. Indeed, continuing the process until the entire network is reduced to a series-only or parallel-
only simplified version is a good starting point. That is, each of the complex impedances that make up the 
series-only or parallel-only simplified equivalent is made up from a sub-circuit which in turn potentially is made
up of other sub-circuits, and so on. The art of examining a complex series-parallel network and being able to 
immediately determine which elements constitute a series connection and which constitute a parallel connection",ACElectricalCircuitAnalysis.pdf
"immediately determine which elements constitute a series connection and which constitute a parallel connection 
is an essential skill and worthy of practice.
110",ACElectricalCircuitAnalysis.pdf
"4.2 The Series-Parallel Connection4.2 The Series-Parallel Connection
Consider the circuit of Figure 4.1. This circuit is neither just a series circuit nor just 
a parallel circuit. If it was a series circuit then the current through all components 
would have to be same, that is, there would no nodes where the current could divide.
This is clearly not the case as the current flowing through the capacitor can divide at 
node b, with one portion flowing down through the resistor and the remainder 
through the inductor. On the other hand, if it was strictly parallel, then all of the 
components would have to exhibit the same voltage and therefore there would be 
only two connection points in the circuit. This is also not the case as there are three 
such points: a, b and ground.
What is true for the circuit of Figure 4.1 is that the resistor and the inductor are in 
parallel. We know this because both components are attached to the same two nodes;",ACElectricalCircuitAnalysis.pdf
"parallel. We know this because both components are attached to the same two nodes;
b and ground, and must exhibit the same voltage, vb. As such, we can find the 
equivalent impedance of this pair and treat the result as a single value, let's call it ZP. 
In this newly simplified circuit, ZP is in series with the capacitor and the source. We 
have simplified the original circuit into a series circuit and thus the series circuit 
analysis rules may be applied. 
4.3 Series-Parallel Impedance4.3 Series-Parallel Impedance
The rules for combining resistors, capacitors and inductors in AC series-parallel 
circuits are similar to those established for combining resistors in DC circuits. 
Obviously, the first item is to determine the reactances of the capacitors and 
inductors. At that point, simple series and parallel combinations can be identified. 
These combinations are each reduced to a complex impedance. Once this is",ACElectricalCircuitAnalysis.pdf
"These combinations are each reduced to a complex impedance. Once this is 
completed, the network is examined again to see if these new complex impedances 
can be identified as parts of new series or parallel sub-circuits, and simplified. This 
process is repeated until we are left with a single complex impedance. Again, it is 
useful to remember that the phase angles of the reactive components can sometimes 
lead to surprising results, such as a series sub-circuit having an impedance 
magnitude smaller than its largest component — something that would never happen
with a network comprised of just resistors. The importance of using vector 
computations cannot be over stressed. 
111
Figure 4.1
A simple series-parallel RLC 
circuit.",ACElectricalCircuitAnalysis.pdf
"Let's begin with a relatively simple series-parallel RLC network where the reactance 
values have already been found.
Example 4.1
Determine the equivalent impedance of the network shown in Figure 4.2.
Looking in from the left side, we note that the inductor and 33 kΩ resistor
are in parallel as they are both tied to the same two nodes. Also, we can
see that the capacitor is in series with the 8.2 kΩ resistor. This series
combination is, in turn, in parallel with the other two parallel components.
Thus, it would make sense to find the series combination first.
Z series = R+(− jX C )
Z series = 8.2 kΩ − j 2k Ω
Z series = 8440  −13.7° Ω
We now place this new complex impedance in parallel with the inductor and
the 33 kΩ resistor.
Z total = 1
1
Z 1
+ 1
Z 2
+ 1
Z 3
Ztotal = 1
1
j 550 Ω + 1
33 kΩ + 1
8440−13.7° Ω
 
Ztotal = 556.8 85.4° Ω
Clearly, the inductor dominates here. The parallel resistor is roughly two 
orders of magnitude larger than the inductive reactance and has minimal",ACElectricalCircuitAnalysis.pdf
"Clearly, the inductor dominates here. The parallel resistor is roughly two 
orders of magnitude larger than the inductive reactance and has minimal 
impact on a parallel combination. Further, the complex impedance derived 
from the capacitor/resistor combination is also considerably larger, and 
given that it has a negative (capacitive) phase angle, it partly cancels the 
inductive reactance. This leaves us with a magnitude a little higher than that 
of the inductive reactance alone, and with a phase angle shifted toward the 
resistive side.  
The series and parallel combinations can be much more complicated than that of the 
prior network. Ladder networks, for example, feature a set of sections that load other
sections, resulting in repeated series and then parallel simplifications. In this 
situation, it is best to start work at the end farthest from the nodes of interest. The 
following example will illustrate this on a modest scale.
112
Figure 4.2
Network for Example 4.1.",ACElectricalCircuitAnalysis.pdf
"Example 4.2
Determine the equivalent impedance of the network shown in Figure 4.3.
Looking in from the right side, we see immediately the 750 Ω resistor. This 
is in series with the sub-circuit comprised of the remaining five components.
This sub-circuit can be seen as the −j800 Ω capacitor in parallel with another
sub-circuit containing the other four components. This four component sub-
circuit consists of the inductor in series with yet another sub-circuit 
consisting of the final two resistors and capacitor. This three element sub-
circuit consists of the 2.2 k Ω resistor in parallel with the series combination 
of the 1 k Ω resistor and the −j400 Ω capacitor. 
The most sensible way to approach this is to start at the left end with the 
simple RC series combination and then work right, toward the nodes of 
interest. We'll number the components from left to right for identification.
Zleft2 = R1+(− jX C1)
Zleft2 = 1k Ω− j 400 Ω
Zleft2 = 1077 −21.8° Ω",ACElectricalCircuitAnalysis.pdf
"interest. We'll number the components from left to right for identification.
Zleft2 = R1+(− jX C1)
Zleft2 = 1k Ω− j 400 Ω
Zleft2 = 1077 −21.8° Ω
We now place this complex impedance in parallel with the 2.2 kΩ resistor. 
This creates a three element sub-circuit which is in series with the inductor.
Z left3 = 1
1
R2
+ 1
Zleft2
Zleft3 = 1
1
2.2 kΩ + 1
1077−21.8° Ω
 
Zleft3 = 734.7 −14.7° Ω
Zleft4 = Z left3 +jX L
Zleft4 = 734.7  −14.7° Ω +j 600Ω
Zleft4 = 822.5  30.2 ° Ω
113
Figure 4.3
Network for Example 4.2.",ACElectricalCircuitAnalysis.pdf
"This group of four is in parallel with the second capacitor of −j800 Ω. 
Finally, we arrive at the equivalent total value by placing the resulting group
of five in series with the 750 Ω resistor.
Z left5 = 1
1
X C2
+ 1
Z left4
Zleft5 = 1
1
− j 800 Ω + 1
822.5  30.2° Ω
 
Zleft5 = 813.4 −31.3° Ω
Ztotal = Z left5 +R3
Ztotal = 813.4  − 31.3° Ω +750 Ω
Ztotal = 1506 −16.3° Ω
In rectangular form this is 1443 −j422.3 Ω, meaning that this network is 
equivalent to a 1443 Ω resistor in series with a capacitive reactance of 
−j422.3 Ω.
 
Series-parallel simplification techniques will not work for all circuits. Some 
networks such as delta or bridge configurations require other techniques that will be 
addressed in later chapters.
4.4 Series-Parallel Analysis4.4 Series-Parallel Analysis
Given the infinite variety of series-parallel configurations, there are myriad ways of 
solving any given circuit for a particular current or voltage. Many solution paths",ACElectricalCircuitAnalysis.pdf
"solving any given circuit for a particular current or voltage. Many solution paths 
exist. This is good, because while you might not see a particular path, there are 
others that will also provide correct results. The only issue is which path is most 
efficient or convenient for you. Suppose we are trying to find vb in the circuit of 
Figure 4.4. How might we approach this problem?
114
Figure 4.4
A series-parallel RLC circuit.",ACElectricalCircuitAnalysis.pdf
"One path would be to find the total impedance seen by the voltage source, Ztotal. 
Dividing the source voltage by this impedance gives us the source current. We could
then perform a current divider between the capacitor and inductor-resistor branches 
to find the inductor current. Once that current is found, it can be multiplied by the 
inductive reactance to find vb. Alternately, having found the total impedance, we 
could compute the voltage divider between the three components on the right and R1 
to find va. Knowing va, a second voltage divider between XL and R2 gives us vb. A 
third possibility would be to find the source current and use that to find va, perhaps 
by finding the drop across R1 and subtracting that from the source, E. Once va is 
found, a voltage divider can be used to find vb. Undoubtedly there are other solution 
paths that will work here. Some are more “computationally expensive” than others,",ACElectricalCircuitAnalysis.pdf
"paths that will work here. Some are more “computationally expensive” than others, 
but as long as you can identify one of them, the answers are within reach. 
Remember, the larger the circuit becomes, the greater the number of possible 
solution paths. Don't fall into the trap of relying on the same “trick” for every 
circuit, though. It is useful to solve these circuits using a variety of techniques as a 
means of cross-checking the results and sharpening your skill set.
Example 4.3
Determine vb for the circuit of Figure 4.5 if the source frequency is 100 Hz.
The first thing to do is to find the capacitive reactance.
X C =− j 1
2 π f C
X C =− j 1
2 π100 Hz 75 nF
X C ≈− j 21.22 kΩ
This reactance is in parallel with the 27 kΩ resistor. Their combination is:
Zrc = R×(− jX C)
R +(− jX C )
Zrc = 27 kΩ×(− j 21.22 kΩ)
27 kΩ − j 21.22 k Ω
Zrc ≈ 16.68E3 −51.8° Ω
This impedance forms a voltage divider with the 47 kΩ resistor to create vb.
vb = esource
Z rc
R+Z rc",ACElectricalCircuitAnalysis.pdf
"27 kΩ − j 21.22 k Ω
Zrc ≈ 16.68E3 −51.8° Ω
This impedance forms a voltage divider with the 47 kΩ resistor to create vb.
vb = esource
Z rc
R+Z rc
vb = 90  0° V 16.68E3 − 51.8Ω
47k Ω +16.68E3 −51.8 ° Ω
vb ≈ 25.5 −38.9° V
115
Figure 4.5
Circuit for Example 4.3.",ACElectricalCircuitAnalysis.pdf
"A time domain plot of vb and the source voltage is shown in Figure 4.6.
Computer SimulationComputer Simulation
To verify the results of the prior example, the circuit of Figure 4.5 is entered into a 
simulator as shown in Figure 4.7.
A time domain or transient analysis is run, examining vb and the source voltage. 
Node 2 corresponds to vb. The results are shown in Figure 4.8.
116
Time Domain Graph
Voltage (volts)
-100
-50
0
50
100
 
Time (milliseconds)
0 2 4 6 8 10
vb
esource
Figure 4.6
Time domain plot of voltages 
for the circuit of Figure 4.5.
Figure 4.7
The circuit of Example 4.3 in 
the simulator.",ACElectricalCircuitAnalysis.pdf
"The plot is delayed one full cycle in order to get past the initial turn-on transient. 
The resulting amplitudes and phase shift line up perfectly with the plot of theoretical
values in Figure 4.6.
Example 4.4
For the circuit of Figure 4.5, determine vab.
This circuit can be analyzed as a pair of voltage dividers. By
definition, vab = va − vb. Numbering the resistors from top to bottom
gives us:
va = esource
R2
R1+R2
va = 100  0° V 40 kΩ
10k Ω + 40 kΩ
va = 80  0° V
117
Figure 4.8
Transient analysis of the circuit
of Example 4.3.
Figure 4.9
Circuit for Example 4.4.",ACElectricalCircuitAnalysis.pdf
"vb = esource
− j X C
− jX C +jX L
vb = 100  0° V − j 800Ω
− j 800 Ω + j 1 kΩ
vb = 400  180° V
This may also be written as −4000°. Now we subtract the two voltages to 
find vab.
vab = va −vb
vab = 80  0° V −400  180° V
vab = 480  0° V
Note that vab is nearly five times larger than the source voltage. This is 
mostly due to the fact that vb itself is four times the source magnitude. Due 
to the fact that XL and XC are relatively close in size, they largely cancel each
other when placed in series. This produces a small net reactance which 
creates a large current. This considerable current then produces large 
voltages across these components. The closer the magnitudes of XL and XC, 
the higher the L and C component voltages. We will examine this effect in 
detail when we discuss resonance in Chapter 8.
To verify this result, we can calculate the voltage across the inductor and 
check to see if KVL is satisfied.
vinductor = esource
j X L
− jX C+jX L
vinductor = 100 0° V j 1k Ω",ACElectricalCircuitAnalysis.pdf
"check to see if KVL is satisfied.
vinductor = esource
j X L
− jX C+jX L
vinductor = 100 0° V j 1k Ω
− j 800 Ω+ j 1k Ω
vinductor = 500 0° V
Adding the vb of −4000° volts to vinductor does indeed yield the source 
voltage of 1000° volts. 
This can be seen graphically in Figure 4.10. First, note that the inductor 
voltage is in phase with the source voltage. This is because the LC branch 
appears to be net inductive, producing a current lagging the source voltage 
by 90 degrees. This same current flows through the inductor, meaning its 
voltage leads this current by 90 degrees, and thus the inductor voltage is in 
phase with the source voltage. The lagging current also flows through the 
capacitor which produces a further 90 degree lag for the capacitor voltage 
(i.e., vb) or 180 degrees total. Combining the large inductor voltage with a 
capacitor voltage that is nearly as large but effectively inverted yields the 
smaller source voltage.
118",ACElectricalCircuitAnalysis.pdf
"Example 4.5
In the circuit of Figure 4.11, determine the current flowing down through the
inductor. Use the source as the reference (0°).
One possible approach for this is to find the equivalent total impedance that 
the source drives in order to find the source current. A current divider can 
then be used between the inductor and the pair of capacitors (all three being 
in parallel). Another option would be to find the impedance of the three 
reactive components and then use the voltage divider rule to find va. 
119
Time Domain Graph
Voltage (volts)
-500
-400
-300
-200
-100
0
100
200
300
400
500
 
Time (cycles)
0 0.2 0.4 0.6 0.8 1
vb
vinductor
esource
Figure 4.10
Time domain plot of the circuit 
of Figure 4.9.
Figure 4.11
Circuit for Example 4.5.",ACElectricalCircuitAnalysis.pdf
"Once va is found, the inductor current can be found using Ohm's law. Each 
of these solution paths requires about as much work as the other so there is 
no clear preference. As we just used the voltage divider rule in the prior 
example, let's use the current divider rule this time.
We are going to need the combined capacitive reactance for the current 
divider, and we will also need it to find the total impedance, so let's do that 
first.
X Ctotal = 1
1
X C1
+ 1
X C2
X Ctotal = 1
1
− j 4 kΩ + 1
− j 8k Ω
 
X Ctotal = 2667−90° Ω
This value is in parallel with the inductive reactance, and that combo is in 
series with the resistor, yielding the total impedance.
Z CL = 1
1
X Ctotal
+ 1
X L
ZCL = 1
1
− j 2667Ω + 1
j 1k Ω
 
ZCL = 1600 90° Ω
Ztotal = R+Z CL
Ztotal = 2 kΩ+1600 90° Ω
Ztotal = 2561 38.7° Ω
The source current is found using Ohm's law.
isource = esource
Z total
isource = 40 0° V
2561 38.7° Ω
 
isource = 15.6E-3−38.7° A",ACElectricalCircuitAnalysis.pdf
"The source current is found using Ohm's law.
isource = esource
Z total
isource = 40 0° V
2561 38.7° Ω
 
isource = 15.6E-3−38.7° A
We now apply a current divider between XCtotal and the inductor.
120",ACElectricalCircuitAnalysis.pdf
"iinductor = isource
X Ctotal
X Ctotal+X L
iinductor = 15.6E-3−38.7° A − j 2667Ω
− j 2667Ω+j 1000Ω
 
iinductor = 24.99E-3 −38.7° A
Once again we see a branch current that is larger in magnitude than the 
source current. This current should produce an inductor voltage of
vinductor = iinductor × X L
vinductor = 24.99E-3 −38.7 ° A×1000 90 ° Ω
vinductor = 24.99  51.3° V
The voltage across the resistor is
vR = i source×R
vR = 15.6E-3 −38.7° A×2000 0° Ω
vR = 31.2−38.7° V
KVL indicates that the sum of vR and vinductor should equal the source of 
400° volts, and it does (within rounding limits). 
 
It is now time for some examples that use current sources.
Example 4.6
Determine va, vb and vab in the circuit of Figure 4.12. Use the source as the
reference angle of 0 degrees.
To find vb we can determine the equivalent impedance of the two
resistors and the inductor and multiply it by the source current. The
rightmost resistor and inductor are in series, yielding 10 + j20 Ω. This",ACElectricalCircuitAnalysis.pdf
"resistors and the inductor and multiply it by the source current. The
rightmost resistor and inductor are in series, yielding 10 + j20 Ω. This
is in parallel with the 18 Ω resistor.
Z b = R1×ZLR
R1 +Z LR
Zb = 18Ω×(10Ω+ j 20Ω)
18Ω+(10Ω+ j 20Ω)
Zb = 11.7 27.9° Ω
121
Figure 4.12
Circuit for Example 4.6.",ACElectricalCircuitAnalysis.pdf
"vb = isource×Zb
vb = 2 0° A×11.7  27.9° Ω
vb = 23.4  27.9° V
The voltage across the capacitor is vab. We can find this through Ohm's law. 
Given the reference direction of the current source, the capacitor's voltage 
reference polarity is + to − from left to right.
vab = i source×X C
vab = 2 0° A×18−90° Ω
vab = 36−90° V
Finally, va is just vab plus vb based on KVL.
va = vab +vb
va = 36 −90° V+23.4  27.9° V
va = 32.48 −50.5° V
A phasor diagram is shown in Figure 4.13. Graphically, it can be seen that 
subtracting vb from va yields vab, as expected. Remember, this is a series-
parallel circuit and therefore we do not see necessarily 0 degree or 90 degree
angles between the various voltages as found in simple series-only circuits. 
122
Figure 4.13
Phasor voltage plot for the 
circuit of Figure 4.12.",ACElectricalCircuitAnalysis.pdf
"Example 4.7
For the circuit of Figure 4.14, determine vb if the 1 amp source is used as the
reference (0°) and the 3 amp source has a 30° lagging phase angle.
The two current sources are in parallel and can be combined together. We 
must be a little careful regarding polarity, though. First of all, a “30° lagging 
phase angle” means that the second source is 3−30° amps. Along with this,
its reference direction is opposite that of the first source. This means that the
second source is negative or inverted by 180 degrees relative to source one. 
Thus, we can treat it as a downward source of −3−30° amps, or 3150° 
amps, whichever we prefer. Now that they're both configured as having a 
downward reference direction, we simply add them together. 
itotal = i1+i2
itotal = 1 0° A+3 150° A
itotal = 2.192 136.8° A
Alternately, we could subtract 3−30° amps from the first source based on 
the reference directions, and note that the resulting direction of the",ACElectricalCircuitAnalysis.pdf
"Alternately, we could subtract 3−30° amps from the first source based on 
the reference directions, and note that the resulting direction of the 
combination is the same as that of the first source. Another option would be 
to reverse the reference direction of the first source. This would yield an 
upward direction with a value of 2.192−43.2° amps.
Having simplified the circuit to a single current source, it should be obvious 
that the inductor is in series with the 22 Ω resistor, and that combination is 
in parallel with both the capacitor and the 33 Ω resistor. Finding that parallel
impedance would allow us to find va. Knowing va, a voltage divider between
the series inductor/resistor combo will yield vb. An important thing to note is
that, given the downward reference direction of the equivalent current 
source, KCL indicates that the current direction through the other 
components must be upward, meaning that both va and vb are negative with 
respect to ground. 
123
Figure 4.14",ACElectricalCircuitAnalysis.pdf
"components must be upward, meaning that both va and vb are negative with 
respect to ground. 
123
Figure 4.14
Circuit for Example 4.7.",ACElectricalCircuitAnalysis.pdf
"Z total = 1
1
X C
+ 1
R1+X L
+ 1
R2
Ztotal = 1
1
− j 80Ω + 1
22 Ω+j 50 Ω+ 1
33Ω
 
Ztotal = 26.38  6.45° Ω
va =−itotal×Ztotal
va =−2.192 136.8° A× 26.38 6.45° Ω
va = 57.8 −36.8° V
If we had reversed the reference direction of the current source, using  
2.192−43.2° amps instead, the leading minus sign would not be required 
and we would arrive at the same result. Continuing,
vb = va(
R1
R1+X L )
vb = 57.8 −36.8° V 22Ω
22Ω+j 50Ω
vb = 23.29  −103° V
Analysis Across the Frequency Domain
For the most part, we have examined the response of a circuit to a single frequency 
of excitation. In many electronic systems, such as in the field of communications, 
numerous frequencies are present simultaneously. Recall from Chapter 1 how 
complex wave shapes such as square waves, triangle waves or music signals can be 
built from a series of sine waves. In such systems, the reactive components behave 
as different values to the various frequencies simultaneously. For example, a",ACElectricalCircuitAnalysis.pdf
"as different values to the various frequencies simultaneously. For example, a 
capacitor may have a reactance of −j400 Ω for a 100 Hz signal while at the same 
time offering a reactance of −j40 Ω for a 1 kHz signal. It is this dynamic quality that 
allows us to design circuits to suppress or block certain frequency components, or to 
select specific frequencies from a large range or spectrum of frequency components. 
We shall introduce this concept by first analyzing the circuit at a couple of specific 
frequencies and then employ a simulator to perform a frequency domain analysis 
(sometimes called an AC analysis) to plot complex response curves of voltage versus
frequency. The concept of frequency domain response will be expanded in upcoming
work, particularly in Chapter 10.
124",ACElectricalCircuitAnalysis.pdf
"Example 4.8
Consider the circuit shown in Figure 4.15. Assume the source is a one volt 
peak sine wave. Determine voltages va, vb and vc if the source frequency is 
10 kHz. Repeat this for an input frequency of 10 Hz.
If we treat E as the input and vc as the final output, this circuit behaves as a 
series of cascading frequency-dependent voltage dividers. Generally 
speaking, at low frequencies the capacitive reactances will be larger than the
associated resistors, and most of the input voltage will make it to node c. At 
high frequencies, the capacitive reactances will be small resulting in 
considerable voltage division at each node. Thus, only a small percentage of 
the input will make it to the final output. In other words, this circuit will 
filter out or remove high frequencies from the input with considerable effect,
much more so than a single RC network.
First, we need to find the three capacitive reactances at 10 kHz. Starting at 
the left, we find
X C =− j 1
2 π f C",ACElectricalCircuitAnalysis.pdf
"First, we need to find the three capacitive reactances at 10 kHz. Starting at 
the left, we find
X C =− j 1
2 π f C
X C =− j 1
2 π10Hz 1μ F
X C ≈− j 15.92 Ω
The other capacitive reactances work out to −j31.83 Ω and −j79.58 Ω. The 
voltage va can be determined by a voltage divider between the 1 kΩ resistor 
and the series-parallel combination of the remaining five components. First, 
the 5 kΩ is in series with the 200 nF. That combination is in parallel with the
500 nF, which is in turn in series with the 2 kΩ resistor. Finally, that group 
of four is in parallel with the 1 μF capacitor. 
The resistors and capacitors are numbered from left to right in the equations 
following. We will need each of the segment impedances for subsequent 
calculations.
125
Figure 4.15
Circuit for Example 4.8.",ACElectricalCircuitAnalysis.pdf
"Z right3 = 1
1
X C2
+ 1
Z right2
Zright3 = 1
1
− j31.83Ω + 1
5000Ω − j 79.58Ω
 
Zright3 = 31.83 −89.6° Ω
Zright4 = R2+Z right3
Zright4 = 2000Ω +31.83−89.6° Ω
 
Zright4 = 2000−0.91° Ω
Zright5 = 1
1
X C1
+ 1
Z right4
Z right5 = 1
1
− j 15.92 Ω + 1
2000−.91° Ω
 
Zright5 = 15.92 −89.5° Ω
At last we come to va:
va = esource
Z right5
R1+Zright5
va = 1 0° V 15.92−89.5° Ω
1000Ω+15.92−89.5° Ω
 
va = 15.92 −88.6° mV
To find vb we perform a voltage divider between the 2 kΩ resistor and Zright3 
using va as the input.
vb = va
Z right3
R2+Z right3
vb = 15.92 −88.6° mV 31.83−89.6° Ω
2000Ω +31.83−89.6° Ω
 
vb = 253.3 −177.3 °μ V
 
126",ACElectricalCircuitAnalysis.pdf
"Finally, to find vc we perform a voltage divider between the 5 kΩ resistor 
and 200 nF capacitor using vb as the input.
vc = vb
X C3
R3 +X C3
vc = 253.3 −177.3° μ V − j 79.58Ω
5000Ω− j 79.58Ω
 
vc = 4.03 93.6° μ V
Obviously, only a tiny percentage of the source signal is found at node c at 
this frequency. 
Repeating this process at 10 Hz yields capacitive reactances of −j15.92 kΩ, 
−j31.83 kΩ and −j79.58 kΩ. At this level, the amount of signal lost through 
each segment is inconsequential. For example, for the final segment the 
voltage divider ratio works out to:
vc
vb
= X C3
R3+X C3
vc
vb
= − j 79.58 k Ω
5000Ω− j 79.58 k Ω
vc
vb
= 0.998 −3.6°
In other words, a mere 0.2% of the signal is lost and there is a modest −3.6° 
phase shift. The results at the other nodes are similar and left as an exercise. 
Thus we see that that the low frequencies are allowed through this network 
while the high frequencies are attenuated.
Computer SimulationComputer Simulation",ACElectricalCircuitAnalysis.pdf
"while the high frequencies are attenuated.
Computer SimulationComputer Simulation
While the results of Example 4.8 should be convincing as to the performance of the 
circuit, it should also be obvious that determining the voltages for any set of source 
frequencies would be a tedious exercise. Fortunately, there are other techniques that 
may be employed, such as those examined in Chapter 10. For now, though, we will 
turn our attention to a simulator. Most simulators offer an AC analysis or frequency 
domain analysis that will create two linked graphs; one for the voltage magnitude 
and another for the phase. We begin by entering the circuit of Figure 4.15 into a 
simulator as shown in Figure 4.16. Even though the schematic shows a 1 kHz source
frequency, the AC analysis will allow us to specify the starting and ending 
frequencies for the plots. In this case, we'll use the 10 Hz and 10 kHz points 
specified in the example. The results are shown in Figure 4.17.
127",ACElectricalCircuitAnalysis.pdf
"The top graph plots the voltages at nodes a, b and c across frequency. It is obvious 
that, as the frequency increases, the voltage at each node decreases. The lower graph
plots the phase shift at each of the nodes and it is apparent that the phase shift 
increases in the negative direction as frequency is increased. This is expected 
because, as the frequency increases, the capacitive reactance decreases, making each
128
Figure 4.16
The circuit of Figure 4.15 in a 
simulator.
Figure 4.17
Frequency domain plots of 
voltage for the circuit of  
Figure 4.15.",ACElectricalCircuitAnalysis.pdf
"parallel combination appear more capacitive, and approaching −90 degrees each. A 
quick check of the voltage magnitudes and phases at 10 Hz indicates that very little 
signal is lost at the three node and that the phase shifts are close to zero. Further, at 
10 kHz, there is considerable signal loss through each section, with each section 
producing nearly −90 degrees, just as calculated. Perhaps the only curious bit here is
the abrupt change in phase shift shown at node c around 300 Hz (red trace). This is 
just an artifact of the plotting software. If an angle goes beyond ±180 degrees, the 
value is rotated back the other way to keep the value within ±180. For instance, 
−185 degrees is the same as +175 degrees.
Combining reactive elements can be a very effective means of selecting out a certain
range of frequencies, as further illustrated in the following example.
Example 4.9
In Chapter 2 we introduced the concept of a loudspeaker crossover network.",ACElectricalCircuitAnalysis.pdf
"Example 4.9
In Chapter 2 we introduced the concept of a loudspeaker crossover network. 
The idea was to “steer” low frequencies to the woofer (low frequency 
transducer) and high frequencies to the tweeter (high frequency transducer). 
An advancement on that simple system is to use a combination of capacitors 
and inductors in place of a simple RC or RL network. One possible 
configuration is illustrated in Figure 4.18. At high frequencies, the capacitive
reactance will be small while the inductive reactance will be large. Thus, 
virtually all of the input signal will reach the loudspeaker. In contrast, at low
frequencies the capacitive reactances will be large and the inductive 
reactance small, resulting in hardly any of the input signal reaching the 
loudspeaker. Somewhere in the middle, a significant portion of the signal 
will make it through. This point is referred to as the crossover frequency. If 
the source voltage is 1 volt peak, determine the voltage developed across an",ACElectricalCircuitAnalysis.pdf
"the source voltage is 1 volt peak, determine the voltage developed across an 
8 Ω loudspeaker at a frequency of 2.6 kHz for this circuit.
First, we need to determine the reactances at the frequency of interest. 
X C1 = − j 1
2π f C
X C1 = − j 1
2π 2.6kHz 5μ F
X C1 ≈− j12.24 Ω
129
Figure 4.18
Circuit for Example 4.9.",ACElectricalCircuitAnalysis.pdf
"The second capacitor is three times as large and therefore its reactance will 
be one-third as much, or −j4.08 Ω. For the inductor,
X L = j 2π f L
X L = j 2π2.6 kHz 360μ H
X L ≈ j 5.88Ω
Now that we have the reactances, the loudspeaker voltage can be computed 
via a pair of voltage dividers. In order find the loudspeaker voltage we'll first
find the voltage developed across the inductor. To find that, we need to find 
the combined impedance of the three components on the right.
Zright3 = 1
1
X L
+ 1
Z right2
Zright3 = 1
1
j 5.88Ω + 1
8Ω − j 4.08Ω
 
Zright3 = 6.44 50.3° Ω
Now for the voltage divider to find vinductor.
vinductor = esource
Z right3
Z right3+X C1
vinductor = 1 0° V 6.44 50.3° Ω
6.44  50.3° Ω− j 12.24 Ω
 
vinductor = 0.77 110.8° V
And now the final voltage divider to find vloudspeaker.
vloudspeaker = vinductor
Zloudspeaker
Z loudspeaker +X C2
vloudspeaker = 0.77 110.8° μ V 8Ω
8Ω− j 4.08Ω
 
vloudspeaker = 0.686  137.9 ° V",ACElectricalCircuitAnalysis.pdf
"vloudspeaker = vinductor
Zloudspeaker
Z loudspeaker +X C2
vloudspeaker = 0.77 110.8° μ V 8Ω
8Ω− j 4.08Ω
 
vloudspeaker = 0.686  137.9 ° V
At this particular frequency the loudspeaker sees about 2/3rds of the source 
voltage. For any higher frequency, the loudspeaker will see a larger share of 
the 1 volt source and for any lower frequency, the loudspeaker see less. At 
very low frequencies, only a few microvolts may get through. 
130",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
In order to get a better sense of the loudspeaker voltage as a function of frequency, 
the circuit of Figure 4.18 is captured in a simulator as shown in Figure 4.19.
An AC analysis simulation is performed with the output shown in Figure 4.20.
131
Figure 4.19
The circuit of Figure 4.18 in a 
simulator.
Figure 4.20
Frequency domain plot of the 
loudspeaker voltage for the 
circuit of  Figure 4.18.",ACElectricalCircuitAnalysis.pdf
"Both the magnitude and phase plots corroborate the calculated loudspeaker voltage 
at 2.6 kHz. The magnitude plot shows that the loudspeaker voltage is very close to 
the input level at frequencies above about 3 kHz. Below this frequency, the 
loudspeaker voltage rolls off considerably. Down at 100 Hz, well into the bass 
region, less than 100 microvolts, or under 0.01% of the input, reaches the 
loudspeaker. This circuit would make for an effective crossover network to a high 
frequency tweeter.
In closing, it is worth noting that a loudspeaker exhibits a complex impedance 
instead of simple resistive value, however, modeling it as an 8 Ω resistor is sufficient
to illustrate the operation of this circuit. We will take a closer look at the impedance 
of loudspeakers and other devices in upcoming chapters.
4.5 Summary4.5 Summary
In this chapter we have determined how to identify basic series-parallel RLC",ACElectricalCircuitAnalysis.pdf
"4.5 Summary4.5 Summary
In this chapter we have determined how to identify basic series-parallel RLC 
networks driven by a single effective voltage or current source. The key to this is to 
identify sub-circuits or subgroups of components that are comprised of either series-
only or parallel-only configurations within themselves. These groupings can then be 
reduced to equivalent impedances using the series and parallel combination 
techniques examined in prior chapters. This process may be repeated until the entire 
circuit is simplified down to either a single series loop or parallel arrangement of 
components driven by a voltage or current source. 
Once a circuit has been simplified, series and parallel analysis techniques, and laws 
such as Ohm's law, Kirchhoff's voltage and current laws, and the voltage and current
divider rules, may be employed to determine various voltages and currents in the 
simplified equivalent. Given these results, the circuit may be expanded back into its",ACElectricalCircuitAnalysis.pdf
"simplified equivalent. Given these results, the circuit may be expanded back into its 
original form in stages, reapplying these rules and techniques to determine voltages 
and currents within the sub-circuits. The process may be iterated until every current 
and voltage in the original circuit is discovered, if desired.
As the impedances of the individual sub-circuits can be anywhere between +90 and 
−90 degrees, phasor diagrams of the various component voltages or currents will no 
longer exhibit the strict right angles seen in series-only and parallel-only circuits. 
What is true is that this perpendicular relationship will still exist among the RLC 
components that comprise a specific series or parallel sub-circuit.
There are infinite varieties of series-parallel RLC configurations and consequently no
single solution technique will work for all of them. In fact, the more complex the 
circuit, the more solution paths that exist for said circuit. Consequently it is prudent",ACElectricalCircuitAnalysis.pdf
"circuit, the more solution paths that exist for said circuit. Consequently it is prudent 
to plan out a solution path instead of just randomly “diving in” as this will lessen the
ultimate effort. 
132",ACElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. In general, describe the process of reducing an AC series-parallel RLC 
network down to a single equivalent impedance.
2. Do Ohm's law, KVL and KCL still apply in AC series-parallel RLC networks?
Why?
3. Is there a finite number of variations of AC series-parallel RLC networks? 
Why/why not?
4. Describe a general procedure to find the voltage between two arbitrary 
points in a series-parallel circuit.
5. In an AC series-parallel RLC circuit, will it always be the case that voltage 
across any resistor is in phase with that resistor's current? Why/why not?
6. In an AC series-parallel RLC circuit, will it always be the case that voltage 
across any inductor leads any resistor's voltage by 90 degrees? Why/why 
not?
4.6 Exercises4.6 Exercises
Analysis Analysis 
Assume the source's angle is 0 degrees unless specified otherwise.
1. Determine the impedance of the circuit of Figure 4.21 at frequencies of      
100 Hz, 10 kHz and 1 MHz.",ACElectricalCircuitAnalysis.pdf
"1. Determine the impedance of the circuit of Figure 4.21 at frequencies of      
100 Hz, 10 kHz and 1 MHz.
2. Determine the impedance of the circuit of Figure 4.22 at frequencies of        
20 Hz, 1 kHz and 20 kHz.
133
Figure 4.21
Figure 4.22",ACElectricalCircuitAnalysis.pdf
"3. Determine the impedance of the circuit of Figure 4.23 at frequencies of      
300 Hz, 30 kHz and 3 MHz.
4. Determine the impedance of the circuit of Figure 4.24 at frequencies of         
1 kHz, 20 kHz and 1 MHz.
5. Determine the impedance of the circuit of Figure 4.25.
6. Determine the impedance of the circuit of Figure 4.26.
134
Figure 4.23
Figure 4.24
Figure 4.25
Figure 4.26",ACElectricalCircuitAnalysis.pdf
"7. For the circuit of Figure 4.27, determine the source current and the current 
through each of the components.
8. For the circuit of Figure 4.27, determine voltages vab and vb.
9. For the circuit of Figure 4.28, determine voltages across R, L and C if the 
source is 7 volts RMS.
10. For the circuit of Figure 4.28, determine the source current and the current 
through each of the three components. Also, draw a phasor diagram of E, vL 
and vR.
11. For the circuit of Figure 4.29, determine the source current and the current 
through each of the components.
12. For the circuit of Figure 4.29, determine voltages vab and vb. Also, draw a 
phasor diagram of E, vab and vb.
135
Figure 4.27
Figure 4.28
Figure 4.29",ACElectricalCircuitAnalysis.pdf
"13. For the circuit of Figure 4.30, determine voltages vab and vb if the source is 
20 volts peak.
14. For the circuit of Figure 4.30, determine the source current and the current 
through each of the four components if the source is 20 volts peak. 
15. For the circuit of Figure 4.31, determine the source current and the current 
through each of the four components. 
16. For the circuit of Figure 4.31, determine voltages vab and vb.
17. For the circuit of Figure 4.32, determine voltages vab and vb if the source is 
100 volts peak.
18. For the circuit of Figure 4.32, determine the currents through the two 
resistors. 
136
Figure 4.30
Figure 4.31
Figure 4.32",ACElectricalCircuitAnalysis.pdf
"19. For the circuit of Figure 4.33, determine the currents each of the three 
components. 
20. For the circuit of Figure 4.33, determine voltages va and vb.
21. For the circuit of Figure 4.34, determine voltages va and vb.
22. For the circuit of Figure 4.34, determine the middle and right branch 
currents and draw a phasor diagram of three circuit currents.
23. For the circuit of Figure 4.35, determine voltages va and vb.
24. For the circuit of Figure 4.35, determine the currents through the two 
resistors. 
137
Figure 4.33
Figure 4.34
Figure 4.35",ACElectricalCircuitAnalysis.pdf
"25. For the circuit of Figure 4.36, determine voltages va and vb. isource = 25 mA.
26. For the circuit of Figure 4.36, determine the currents through the two 
capacitors. 
27. For the circuit of Figure 4.37, determine the current through the capacitor.  
i1 = 10E−30° A and i2 = 3E−390° A.
28. For the circuit of Figure 4.37, determine voltages va and vb.                         
I1 = 10E−30° A and  I2 = 3E−390° A.
29. For the circuit of Figure 4.38, determine voltages va and vb. i1 = 245° A and
i2 = 0.50° A.
30. For the circuit of Figure 4.38, determine the currents through the two 
resistors. i1 = 245° A and i2 = 0.50° A.
138
Figure 4.36
Figure 4.37
Figure 4.38",ACElectricalCircuitAnalysis.pdf
"31. For the bridge circuit of Figure 4.39, determine vab. The source is 50 volts 
peak.
32. For the bridge circuit of Figure 4.40, determine vab. The source is 6 amps 
peak.
DesignDesign
33. Determine a new value for the capacitor in Figure 4.27 such that vb is          
1.5 volts.
34. Determine the required inductive reactance in Figure 4.28 to shift the 
capacitor voltage to half of the source voltage.
35. Determine a new value for the 20 nF capacitor in Figure 4.29 such that the 
resistor current is 2 mA.
139
Figure 4.39
Figure 4.40",ACElectricalCircuitAnalysis.pdf
"36. In the circuit of Figure 4.41, determine a value for L such that the magnitude
of vb equals va/2 if the source frequency is 10 kHz, R = 2.7 kΩ and              
C = 10 nF.
37. Given the circuit of Figure 4.41, determine a value for C such that the 
source current is in phase with the source voltage. The source frequency is   
1 kHz, R = 68 Ω and L = 22 mH.
38. Given the circuit of Figure 4.42, determine a value for L such that vb is         
1 volt. The source is a 6 volt peak sine at 50 kHz, R1 = 510 Ω and                
R2 = 220 Ω.
39. Given the circuit of Figure 4.39, determine a new value for the inductor such
that the magnitude of vb equals the magnitude of va. Assume that the source 
frequency is 20 kHz.
140
Figure 4.41
Figure 4.42",ACElectricalCircuitAnalysis.pdf
"ChallengeChallenge
40. Consider the circuit drawn in Figure 4.43. Using only the available 
components of 1 kΩ, 2.2 kΩ, 1 mH, 5 mH, 10 nF, 75 nF and 560 nF, is it 
possible to configure a circuit such that va is half the magnitude of vb for a 
source frequency of 1 kHz? If so, indicate which values could be used for 
the four components. If not, explain your reasoning.
41. Given the circuit of Figure 4.29, determine the frequency at which vb is half 
of the source voltage.
42. For the circuit of Figure 4.44, determine voltages va, vb , and vc. i1 = 50° A 
and i2 = 390° A.
43. Given the circuit of Figure 4.39, is it possible to change the values of the 
two resistors such that the phase angle of va is the same as that of vb? If so, 
what are the new values, and if not, explain why it is not possible. 
141
Figure 4.43
Figure 4.44",ACElectricalCircuitAnalysis.pdf
"SimulationSimulation
44. Perform a transient analysis to verify the node voltages computed for 
problem 8.
45. Perform a transient analysis to verify the node voltages computed for 
problem 12.
46. Perform a transient analysis to verify the node voltages computed for 
problem 20.
47. Perform a transient analysis to verify the node voltages computed for 
problem 21.
48. Consider the circuit of problem 17. Assuming the source frequency is         
10 kHz, determine values for the capacitors and inductors. Then, use a 
transient analysis to verify the results of problem 17.
49. Perform a transient analysis on the result of problem 33 to verify the 
accuracy of the design.
50. Perform a transient analysis on the result of problem 34 to verify the 
accuracy of the design.
51. Use an AC frequency response simulation to verify the results of       
problem 41.
52. The concept of a loudspeaker crossover network was presented originally in",ACElectricalCircuitAnalysis.pdf
"problem 41.
52. The concept of a loudspeaker crossover network was presented originally in 
Chapter 2,  Series RLC Circuits. In this chapter, we noted that by adding 
more components, it is possible to increase the rate of attenuation. In doing 
so the undesired signals are further reduced in amplitude. The circuits of 
Figure 4.45 and 4.46 (following page) show improved crossovers for a 
woofer and tweeter, respectively. Assuming standard 8 Ω loudspeakers, use 
an AC frequency domain simulation to determine the crossover frequency of
each network. Also, compare the curves at node a to those at node b. Finally,
compare the attenuation slopes to those generated by the simpler crossover 
network presented at the end of Chapter 2. Component values for the 
woofer:  L1 = 760 μH, L2 = 250 μH, C = 10.6 μF. Component values for the 
tweeter: C1 = 5.3 μF, C2 = 16 μF, L = 380 μH.
142",ACElectricalCircuitAnalysis.pdf
"143
Figure 4.45
Figure 4.46",ACElectricalCircuitAnalysis.pdf
"5 5 Analysis Theorems and TechniquesAnalysis Theorems and Techniques
5.0 Chapter Learning Objectives5.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Find the voltage source equivalent of a current source and vice versa.
• Compute voltages and currents in multi-source RLC networks using superposition.
• Simplify RLC networks using Thévenin's and Norton's theorems.
• Determine conditions for maximum power transfer and compute the maximum power.
• Utilize delta-Y and Y-delta conversions for circuit simplification.
5.1 Introduction5.1 Introduction
In this chapter we shall examine a number of theorems and techniques to help us analyze complex circuits and 
address specialized applications. We will begin by examining the concept of source impedance in order to make 
more accurate models of our idealized constant voltage and current sources. This will be a step beyond using a",ACElectricalCircuitAnalysis.pdf
"more accurate models of our idealized constant voltage and current sources. This will be a step beyond using a 
simple resistance as found in the DC case. From there we will investigate how to convert from one type of 
source to another, such as creating a voltage source that is the functional equivalent of a current source. A 
functional equivalent is a source that can be swapped out for another while leaving all of the other circuit 
currents and voltages intact. In other words, all of the circuit's component voltage drops and branch currents will
be identical to those found in the original configuration. This technique is useful in a number of ways, 
particularly in that it can help reduce more complex circuits to simplify analysis. 
The concept of equivalence can be extended beyond just a single source to an entire network. For this we will 
examine Thévenin's and Norton's theorems. Using these theorems, entire circuits utilizing dozens of components",ACElectricalCircuitAnalysis.pdf
"examine Thévenin's and Norton's theorems. Using these theorems, entire circuits utilizing dozens of components
can be modeled as a single source with an associated complex impedance. When coupled with the maximum 
power transfer theorem, these tools will allow us to determine component values that produce the maximum 
amount of load power. 
We will also address a method of analyzing circuits that contain multiple current and/or voltage sources that are 
connected in a non-trivial fashion (i.e., not just series voltage sources or parallel current sources). This is called 
the superposition theorem and it can be applied to any circuit or parameter that meets certain requirements, 
including circuits that have a mix of current sources and voltage sources. Superposition can also be used to 
determine voltages and currents when sources use different frequencies. In fact, one of way of imagining a",ACElectricalCircuitAnalysis.pdf
"determine voltages and currents when sources use different frequencies. In fact, one of way of imagining a 
complex waveshape is to treat it as a series of connected sources, each with a unique frequency, phase and 
amplitude. Superposition we give us a means to handle this new situation.
144",ACElectricalCircuitAnalysis.pdf
"Finally, we will examine how to find equivalent circuits for certain component 
arrangements that use three connecting points, in other words, RLC combinations 
shaped like a triangle or like the letter Y . These are known as delta and Y 
configurations. These configurations are difficult to address with basic series-
parallel simplification techniques. Converting from one configuration to the other 
will help solve that issue.
5.2 Source Conversions5.2 Source Conversions
In DC analysis, we noted that real world sources have practical limits: voltage 
sources cannot produce infinite current and current sources cannot produce infinite 
voltage. A simple way of creating a more accurate model for independent sources is 
to include an internal resistance. For DC voltage sources, a resistance is added in 
series with the source, and for DC current sources a resistance is added in parallel 
with the source. While this works well enough for typical DC sources, the AC",ACElectricalCircuitAnalysis.pdf
"with the source. While this works well enough for typical DC sources, the AC 
situation is a little more complicated.
Models for AC SourcesModels for AC Sources
Just as we added a simple resistance to the DC sources to make improved models, 
we can add a complex impedance to AC sources to do likewise. Once again, it is 
possible to make even more involved models that will be more accurate, but for 
most work, this addition will suffice. Generally, there are wider variations in values 
for the AC case than the DC case.
We can think of AC sources as belonging in one of two broad categories. First, there 
are power generators, that is, systems designed to generate and deliver power for 
other electrical devices. This would include the AC mains system in a residence or a 
portable power generator. At the other end of the spectrum are signal sources such as
transducers and sensors. These devices are generally low power and are not designed",ACElectricalCircuitAnalysis.pdf
"transducers and sensors. These devices are generally low power and are not designed
to produce particularly high currents, quite the opposite of generators.
The model for an AC voltage source adds an impedance in series, as shown in
Figure 5.1. This impedance sets an upper limit on the source's current output. Even
if the output terminals are shorted, the maximum current will be dictated via Ohm's
law to be the source voltage divided by the internal impedance, or E/Zinternal.
Obviously, this internal impedance will create some voltage divider effect with the
attached load. To minimize this effect, the impedance should be as small as
practicably possible. Thus,
The ideal internal impedance of a voltage source is zero ohms (a short). 
It is not always possible to get close to this ideal. In fact, in some situations the AC
source is far removed from the ideal. 
145
Figure 5.1
Practical AC voltage source 
model.",ACElectricalCircuitAnalysis.pdf
"Consider the signal source shown in Figure 5.2. This is a pickup for an electric bass 
guitar. This device is used to translate the motions of guitar strings into a voltage 
that can be fed to an amplifier. It consists of a few thousand turns of very fine 
magnet wire wrapped around a magnet.
When the metal guitar strings vibrate, they alter the surrounding magnetic field. As 
the magnetic field changes, it induces a current, and hence a voltage, in the coil. The 
resulting signal is an electrical analog of the string vibrations. It is, quite literally, an 
AC signal source. But what of its internal impedance? The coil is made of wire 
between AWG 40 and 45, and may be in excess of 1000 feet (300 meters) in length. 
Consulting an AWG table3, we find that this will yield 1000 or more ohms of 
resistance. Of course, a few thousand turns of wire will also create a hefty 
inductance and pickups such as this may exhibit an inductance of a few henries.",ACElectricalCircuitAnalysis.pdf
"inductance and pickups such as this may exhibit an inductance of a few henries. 
Putting these pieces together, the equivalent internal impedance might look 
something like that shown in Figure 5.3. 
Clearly, values like these are a far cry from the fractional ohm values found in
power generation sources, and they present unique challenges. In the circuit of
Figure 5.3, the component labeled Zin represents the input impedance of the
associated amplifier. Typically, it would be highly resistive but the magnitude can
vary quite a bit depending on the application. A guitar or bass amplifier may exhibit 
1 MΩ or more, while an auxiliary input on a home stereo might be 10 kΩ. What 
happens if you plug a guitar into your hi-fi? Obviously, this system creates a voltage 
divider between the internal impedance of the pickup and the input impedance of the
amplifier, but more importantly, the divider is a function of frequency. At low",ACElectricalCircuitAnalysis.pdf
"amplifier, but more importantly, the divider is a function of frequency. At low 
frequencies like 100 Hz, the inductive reactance is small, around j600 Ω, and the 
signal loss to a 10 kΩ input is not that great. On the other hand, at a high frequency 
such as 10 kHz, the inductive reactance will be over j60 kΩ resulting in considerable
signal loss. Thus, the high frequencies are reduced relative to the low frequencies. 
The sonic effect is akin to turning down the treble control — everything will seem 
muffled. If the input impedance is increased considerably, say by a factor of 100, 
then the voltage divider effect at all audible frequencies will be negligible and the 
the instrument will sound true to form. Finally, although this example shows a 
substantial inductance, it is possible for AC sources to have an associated 
capacitance, or even have negligible reactance (i.e., be purely resistive).",ACElectricalCircuitAnalysis.pdf
"capacitance, or even have negligible reactance (i.e., be purely resistive).
3 Such as the one found in Chapter 2 of the companion text, DC Electrical Circuit Analysis,
a companion free OER text by the author.
146
Figure 5.2
An electric bass guitar pickup 
(cover removed).
Figure 5.3
Example of internal 
impedance of guitar pickup.",ACElectricalCircuitAnalysis.pdf
"For a current source, the improved model adds an impedance in parallel, as shown
in Figure 5.4. This impedance sets an upper limit on the source's voltage output. If
the output terminals are opened, the maximum voltage will no longer produce a
huge voltage. Instead, it is dictated by Ohm's law to be the source current times
the internal impedance, or I∙Zinternal. This internal impedance will create some
current divider effect with the attached load. To minimize this effect, the internal
impedance should be as large as practicably possible. Thus,
The ideal internal impedance of a current source is infinite ohms. 
From here on, whenever we deal with practical voltage and current sources, we
understand that these sources have some associated internal resistance, even if
they're not shown explicitly in a schematic diagram. Further, whenever we talk about
ideal sources, we simply use a short for the internal impedance of a voltage source",ACElectricalCircuitAnalysis.pdf
"ideal sources, we simply use a short for the internal impedance of a voltage source 
and an open for the internal impedance of a current source. 
Source EquivalencesSource Equivalences
For any voltage source consisting of an ideal voltage source with a series internal 
impedance, an equivalent current source may be created. Similarly, for any current 
source consisting of an ideal current source with a parallel internal impedance, an 
equivalent voltage source may be created. By “equivalent”, we mean that both 
circuits will produce the same voltage and current to identical loads. Consider the 
simple voltage source on the left side of Figure 5.5. Its equivalent current source is 
shown on the right. 
For reasons that will become apparent under the section on Thévenin's theorem 
following, the internal impedances of these two circuits must be identical if they are 
to behave identically. Knowing that, it is a straightforward process to find the",ACElectricalCircuitAnalysis.pdf
"to behave identically. Knowing that, it is a straightforward process to find the 
required values of the other source. The current/voltage characteristic is linear for 
these circuits, and a straight plot line can be defined by just two points. The two 
obvious points to use are the opened and shorted load cases. In other words, if the 
circuit is equivalent for these two situations, it must work for any load. The shorted 
load case produces a large load current with zero load voltage, and the opened load 
case produces a large load voltage with zero load current. It would not make sense if
the equivalent source could produce a greater current or voltage under the same 
extreme conditions as the original.
147
Figure 5.4
Practical AC current source 
model.
Figure 5.5
A simple AC voltage source 
(left) and corresponding 
current source (right).",ACElectricalCircuitAnalysis.pdf
"For example, given a voltage source, the current that can be developed when the 
load is shorted is E/Zinternal. Under that same load condition, all of the current from 
the current source version must be flowing through the load (otherwise the load isn't 
shorted). Therefore, the value of the equivalent current source must be the current of 
E/Zinternal. Note that the resulting source normally will not have the same phase angle 
as the original source due to the phase angle of the associated impedance. 
To continue, if we look at the open load case for the voltage source, the load current 
would be zero and the load voltage would be the entire source voltage of E. For the 
current source, the load would also see no current and its voltage would be the 
voltage appearing across its internal resistance which is Zinternal times the current 
E/Zinternal, or just E. Thus, the two behave identically at the load limits.",ACElectricalCircuitAnalysis.pdf
"E/Zinternal, or just E. Thus, the two behave identically at the load limits. 
Similarly, if we start with a current source, an open load produces a load voltage of 
I∙Zinternal. Therefore, the equivalent voltage source must have a value of I∙Zinternal. For 
the current source, a shorted load would produce a load current equal to the source 
value, or I. The voltage source version would produce a current of E/Zinternal, where 
the value of E was just found to be equal to I∙Zinternal, and thus the load current would 
be I∙Zinternal/Zinternal, or just I. Once again, the two versions behave identically at the 
load limits. 
Changing the source frequency results in different values for both the reactance and 
the converted source voltage or current, thus the equivalent is valid only for the 
frequency in question.
To summarize the process of source conversion:
• The internal impedance will be the same for both versions.",ACElectricalCircuitAnalysis.pdf
"frequency in question.
To summarize the process of source conversion:
• The internal impedance will be the same for both versions.
• If converting from a voltage source to a current source, the value of the 
current source will be the short circuit current available from the voltage 
source (i.e., shorted load case), and is equal to E/Zinternal.
• If converting from a current source to a voltage source, the value of the 
voltage source will be the open circuit voltage available from the current 
source (i.e., opened load case), and is equal to I∙Zinternal.
• The equivalent is unique to the frequency of the source. 
If a multi-source is being converted (i.e., voltage sources in series or current sources 
in parallel), first combine the sources to arrive at the simplest source and then do the 
conversion. Do not convert the sources first and then try to combine them as you 
will wind up with series-parallel configurations rather than simple sources.",ACElectricalCircuitAnalysis.pdf
"will wind up with series-parallel configurations rather than simple sources. 
Judicious use of source conversions can sometimes simplify multi-source circuits by
allowing converted sources to be combined, resulting in a single source. 
148",ACElectricalCircuitAnalysis.pdf
"Example 5.1
Convert the source of Figure 5.6 into its current source equivalent.            
E = 2 0° volts RMS.
First, the existing impedance of 80 − j60 Ω does not change, it is simply 
moved to a parallel position. To find the value of the current source, 
compute the short circuit current the existing source is capable of  
producing.
I = E
Z
I = 2 0° V
80Ω − j 60Ω
I = 0.02 36.9° A
The result is shown in Figure 5.7. The current source is 20 mA RMS with a 
leading phase angle of 36.9 degrees.
Example 5.2
Convert the source of Figure 5.8 into its voltage source equivalent.             
I = 0.10° amps peak.
Again, the internal impedance of 100 + j20 Ω does not change and we 
simply place it in series. To find the value of the voltage source, find the 
open circuit voltage the existing source is capable of producing.
E = I ×Z
E = 0.1 0° A×(100 +j 20Ω)
E ≈ 10.2  11.3° V
The equivalent is shown in Figure 5.9. The voltage source E is 10.2 volts",ACElectricalCircuitAnalysis.pdf
"E = I ×Z
E = 0.1 0° A×(100 +j 20Ω)
E ≈ 10.2  11.3° V
The equivalent is shown in Figure 5.9. The voltage source E is 10.2 volts 
peak with a leading phase angle of 11.3 degrees. 
149
Figure 5.6
Circuit for Example 5.1.
Figure 5.7
Current source equivalent of 
the source of Figure 5.6.
Figure 5.8
Circuit for Example 5.2.",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
To verify this process, the circuits of Figures 5.8 and 5.9 are entered into a simulator.
The original circuit simply specified an inductive reactance so a convenient 
frequency of 1 kHz was used and then an appropriate inductor was created that 
would yield the desired reactance of j20 Ω. This turned out to 3.183 mH. Further, a 
random load resistor (200 Ω) was picked and applied to both circuits. The result is 
shown in Figure 5.10.
If these two circuits are equivalent, then the voltages seen across the 200 Ω load 
resistors should be identical. These voltages correspond to node voltages 12 and 8. A
quick voltage divider computation shows that the load voltage should be 
approximately 6.7857.5° volts peak. 
A transient analysis is performed, plotting the load voltages. The results are shown 
in Figure 5.11. The plot itself is delayed by one millisecond in order to get past the",ACElectricalCircuitAnalysis.pdf
"in Figure 5.11. The plot itself is delayed by one millisecond in order to get past the 
initial turn-on transient. Further, the current source circuit is shifted by 
approximately 2 microseconds. Without this slight shift in time, the two voltages 
perfectly overlap so that it appears there is only one trace. By looking carefully at 
the graph it can be seen that there are indeed two traces. The amplitudes of these 
waveforms match each other perfectly and also match the computed result. 
150
Figure 5.9
Voltage source equivalent of the
source of Figure 5.8.
Figure 5.10
The circuits of Figures 5.8 and 
5.9 in a simulator.",ACElectricalCircuitAnalysis.pdf
"Now let's turn our attention to using source conversion to simplify and solve a multi-
source circuit.
Example 5.3
For the circuit of Figure 5.12, determine vb. I = 2E−390° amps peak and 
E = 100° volts peak.
One method of solution is to transform the voltage source into a current 
source. By doing so, the entire circuit is reduced to a parallel network. The 
two current sources can then be summed and the remaining three 
components can be combined into one equivalent parallel impedance. At that
point Ohm's law can be used to find vb. 
151
Figure 5.11
Simulation results for the 
source conversion.
Figure 5.12
Circuit for Example 5.3.",ACElectricalCircuitAnalysis.pdf
"The value of the current source is:
I c = E
Z
I c = 10 0° V
27 k Ω
I c = 0.3704E-3 0° A
The converted circuit is shown in Figure 5.13 (in blue). 
Combining the two current sources and using an upward reference direction 
yields:
I total = Ic−I
I total = 0.3704E-3 0° A−2E-3 90° A
I total = 2.034E-3 − 79.5° A
The combined impedance is:
Ztotal = 1
1
X C
+ 1
R1
+ 1
R2
Ztotal = 1
1
− j 10 kΩ + 1
27k Ω + 1
5.6k Ω
 
Ztotal = 4208−24.9 ° Ω
And finally we compute vb:
vb = I total×Ztotal
vb = 2.034E-3  −79.5° A×4208−24.9° Ω
vb = 8.56 −104.4 ° V
We can perform a KCL crosscheck at node b to verify the results in the 
original circuit. Given the reference directions, all currents are flowing out 
of the node except for the current supplied by the voltage source.
152
Figure 5.13
Circuit of Figure 5.12 with 
converted source.",ACElectricalCircuitAnalysis.pdf
"iEsource = E−vb
R1
iEsource = 10 0° V−8.56−104.4 ° V
27 kΩ
iEsource = 0.544E-3  34.4° A
The remaining three currents should add up to this value.
icapacitor = vb
X C
icapacitor = 8.56 −104.4 ° V
− j 10 kΩ
icapacitor = 0.856E-3 −14.4° A
iR2 = vb
R2
iR2 = 8.56 −104.4 ° V
5.6k Ω
iR2 = 1.53E-3−104.4° A
Finally, 0.856E−3−14.4° + 1.53E−3−104.4° + 2E−390° amps does 
indeed equal 0.544E−334.4° amps, within carried rounding error.
5.3 Superposition Theorem5.3 Superposition Theorem
Superposition allows the analysis of multi-source AC series-parallel circuits. 
Superposition can only be applied to networks that are linear and bilateral. 
Fortunately, all of components we have discussed; resistors, capacitors and 
inductors, fall into that category. Further, superposition cannot be used to find values
for non-linear functions, such as power, directly. This is not a limitation though 
because power can be computed from the resulting voltage or current values.",ACElectricalCircuitAnalysis.pdf
"because power can be computed from the resulting voltage or current values. 
The basic idea is to determine the contribution of each source by itself, and then 
combine the results to get the final answer. The contributions are either all voltages 
or all currents, depending on need. We can state the superposition theorem as:
Any voltage or current in a multi-source linear bilateral network may be 
determined by summing the contributions caused by each source acting 
alone, with all other source replaced by their internal impedance.
The process generates a series of new single-source circuits, one for each source. 
These new circuits are then analyzed for the parameter(s) of interest. 
153",ACElectricalCircuitAnalysis.pdf
"Consider the circuit depicted in Figure 5.14.
Here we see two voltage sources, E1 and E2, driving a three element series-parallel 
network. As there are two sources, two derived circuits must be created; one using 
only E1 and the other using only E2. When considering a given source, all other 
sources are replaced by their ideal internal impedance: for a voltage source, that's a 
short; and for a current source, an open. We start by considering E1. In the new 
circuit E2 is replaced with a short. This leaves a fairly simple network where XC and 
XL are in parallel. This combination is in series with R and E1. Using basic series-
parallel techniques, we can solve for desired quantities such as the current flowing 
through R or the voltage vb. It is important to indicate the reference current direction 
and voltage polarity with respect to the source being considered (here, that's left-to-
right and positive, respectively). The process is then repeated for E2, shorting E1",ACElectricalCircuitAnalysis.pdf
"right and positive, respectively). The process is then repeated for E2, shorting E1 
and leaving us with R in parallel with XC, which is in turn in series with XL and E2. 
Note that although in this version Vb is still positive, the reference current direction 
for R is now right-to-left. The numerical results from this version are added to those 
of the E1 version (minding polarities and directions) to achieve the final result. If 
power is needed, it can be computed from these currents or voltages. Note that 
superposition can work with a mix of current sources and voltage sources. The 
practical downside is that for large circuits using many sources, numerous derived 
circuits will need to be analyzed. For example, if there are three voltage sources and 
two current sources, then a total of five derived circuits will be created.
It is also possible to use superposition to find the resulting currents or voltages in a",ACElectricalCircuitAnalysis.pdf
"It is also possible to use superposition to find the resulting currents or voltages in a 
circuit that uses sources with different frequencies. In this instance, the equivalent 
circuits will have different reactance values. In fact, a single non-sinusoidal source 
can be analyzed using this method by treating the source as a series of superimposed
sine waves with each sine source producing a new circuit with its own unique 
reactance values.
To summarize the superposition technique:
• For every voltage or current source in the original circuit, create a new sub-
circuit. The sub-circuits will be identical to the original except that all 
sources other than the one under consideration will be replaced by their 
ideal internal impedance. This means that all remaining voltage sources will 
be shorted and all remaining current sources will be opened.
154
Figure 5.14
A basic multi-source circuit.",ACElectricalCircuitAnalysis.pdf
"• Indicate the reference current directions and voltage polarities on each of the
new sub-circuits, as generated by the source under consideration.
• Solve each of the sub-circuits for the desired voltages and/or currents using 
standard series-parallel analysis techniques. Make sure to note the reference 
voltage polarities and current directions for these items.
• Add all of the contributions from each of the sub-circuits to arrive at the 
final values, being sure to account for current directions and voltage 
polarities in the process.
To illustrate the superposition technique, let's reexamine the dual source circuit 
shown in Figure 5.12 (repeated in Figure 5.15 for ease of reference). We will solve 
this using superposition and compare the results to those of  Example 5.3 which used
source conversion.
Example 5.4
For the circuit of Figure 5.15, determine vb using superposition. 
I = 2E−390° amps peak and E = 100° volts peak.",ACElectricalCircuitAnalysis.pdf
"source conversion.
Example 5.4
For the circuit of Figure 5.15, determine vb using superposition. 
I = 2E−390° amps peak and E = 100° volts peak.
As the circuit has two sources, it will require two sub-circuits. For the 
voltage source, the current source will be replaced with an open. For the 
second circuit utilizing the current source, the voltage source will be 
replaced with a short. First, using the voltage source we find:
155
Figure 5.15
Circuit for Example 5.4.
Figure 5.16
Circuit of Figure 5.15 
considering voltage source.",ACElectricalCircuitAnalysis.pdf
"For this circuit, vb may be determined via a voltage divider. To proceed, we 
need the impedance of the parallel combo on the right.
Z right2 = R× jX C
R − jX C
Zright2 = 5.6 kΩ×(− j 10 kΩ)
5.6 kΩ− j 2k Ω
Zright2 = 4886 − 29.2° Ω
Now for the voltage divider to find the contribution of the first source to vb.
vb1 = E Z right2
Z right2 + R1
vb1 = 10  0° V 4886−29.2° Ω
4886−29.2° Ω +27 k Ω
 
vb1 = 1.558 −24.88° V
We turn our attention to the current source's contribution. We short the 
voltage source and redraw:
This is a simple parallel circuit. We can find vb by placing the resistors and 
capacitors in parallel, and then using Ohm's law. Note that the reference 
direction of the current source is downward, meaning that the component 
current is upward, which makes vb negative (i.e., + to − bottom to top). The 
parallel impedance is:
Ztotal = 1
1
X C
+ 1
R1
+ 1
R2
Ztotal = 1
1
− j 10 kΩ + 1
27k Ω + 1
5.6k Ω
 
Ztotal = 4208−24.9 ° Ω
156
Figure 5.17
Circuit of Figure 5.17",ACElectricalCircuitAnalysis.pdf
"Ztotal = 1
1
X C
+ 1
R1
+ 1
R2
Ztotal = 1
1
− j 10 kΩ + 1
27k Ω + 1
5.6k Ω
 
Ztotal = 4208−24.9 ° Ω
156
Figure 5.17
Circuit of Figure 5.17 
considering current source.",ACElectricalCircuitAnalysis.pdf
"We apply Ohm's law to find this source's contribution to vb.
vb2 = I ×Z total
vb2 =−2E-3 90° A×4208−24.9° Ω
vb2 = 8.416  −114.9° V
The final result is the sum of the two parts:
vb = vb1 +vb2
vb = 1.558 − 24.88° V +8.416 −114.9 ° V
vb = 8.558 −104.4 ° V
This is virtually the same value obtained using the source conversion 
technique in the prior example.
As mentioned previously, superposition can be used to determine the results even 
when the sources use different frequencies. This will be explored in the next 
example.
Example 5.5
For the circuit of Figure 5.18, determine vb.
Using superposition, we derive two new circuits, each with unique reactance
values. The first circuit is shown in Figure 5.19.
157
Figure 5.18
Circuit for Example 5.5.
Figure 5.19
Circuit for Example 5.5,       
first source only.",ACElectricalCircuitAnalysis.pdf
"The reactance values are:
X L = j 2π f L
X L = j 2π1kHz 50 mH
X L ≈ j 314.2Ω
X C =− j 1
2 π f C
X C =− j 1
2 π1kHz 750 nF
X C ≈− j 212.2Ω
A voltage divider can be used to find this portion of vb. The parallel combo 
of the  2 kΩ resistor and capacitor is 211−83.9° Ω.
vb1 = E Z right2
Z right2 + X L
vb1 = 10  0° V 211 −83.9° Ω
211−83.9° Ω +314.2  90 ° Ω
 
vb1 = 19.78 −161.9 ° V
Remember, this waveform is at a frequency of 1 kHz. We can repeat this 
process for the second source which uses 10 kHz. At this new frequency the 
inductive reactance will be ten times larger, or j3142 Ω, and the capacitive 
reactance will be ten times smaller, or −j21.22 Ω. The new circuit is shown 
in Figure 5.20.
Once again, a voltage divider can be used to find this portion of vb. The 
parallel combo of the 2 kΩ resistor and inductor is 168732.5° Ω.
vb2 = E Z left2
Zleft2 + X C
vb2 = 2 0° V 1687 32.5° Ω
1687 32.5° Ω +21.22−90° Ω
 
vb2 = 2.01 0.6° V
158
Figure 5.20
Circuit for Example 5.5,",ACElectricalCircuitAnalysis.pdf
"vb2 = E Z left2
Zleft2 + X C
vb2 = 2 0° V 1687 32.5° Ω
1687 32.5° Ω +21.22−90° Ω
 
vb2 = 2.01 0.6° V
158
Figure 5.20
Circuit for Example 5.5,  
second source only.",ACElectricalCircuitAnalysis.pdf
"This contribution is at a frequency of 10 kHz. Thus, the combination is a 
relatively small 10 kHz sine at about 2 volts peak riding on a 1 kHz sine that
is nearly ten times larger in amplitude. This is shown in Figure 5.21.
Computer SimulationComputer Simulation
In order to verify the two-component waveform of Example 5.5, the circuit of  
Figure 5.18 is captured in a simulator as shown in Figure 5.22.
159
Time Domain Graph
Voltage (volts)
-25
-20
-15
-10
-5
0
5
10
15
20
25
 
Time (msec)
0 0.5 1 1.5 2
vb
vb1
vb2
Figure 5.21
Voltage plot for Example 5.5.
Figure 5.22
Circuit of Example 5.5 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"A transient analysis is performed. The results are illustrated in Figure 5.23.
The results match the computed values nicely. We can see the small amplitude high 
frequency sine wave effectively following the contour of the much larger 1 kHz sine 
wave.
5.4 Thévenin's and Norton's Theorems5.4 Thévenin's and Norton's Theorems
These theorems are related in that they allow complex linear networks to be 
simplified down to a single source with an associated internal impedance. They 
simplify analysis when checking a circuit with multiple possible loads.
Thévenin's TheoremThévenin's Theorem
Thévenin's theorem is named after Léon Charles Thévenin. It states that:
Any single port linear network can be reduced to a simple voltage source, 
Eth, in series with an internal impedance Zth.
160
Figure 5.23
Simulation results for the 
circuit of Example 5.5.",ACElectricalCircuitAnalysis.pdf
"It is important to note that a Thévenin equivalent is valid only at a particular 
frequency. If the system frequency is changed, the reactance and impedance values 
will change and the resulting Eth and Zth values will be altered. Consequently, these 
equivalents are generally not appropriate for a circuit using multiple sources with 
differing frequencies4. A generic example of a Thévenin equivalent is shown in 
Figure 5.24.
The phrase “single port network” means that the original circuit is cut in such a way 
that only two connections exist to the remainder of the circuit. That remainder may 
be a single component or a large multi-component sub-circuit. Eth is the open circuit 
voltage at the port and Zth is the impedance looking back into the port (i.e., the 
equivalent that now drives the remainder). As there are many ways to cut a typical 
circuit, there are many possible Thévenin equivalents.  Consider the circuit shown in
Figure 5.25.",ACElectricalCircuitAnalysis.pdf
"circuit, there are many possible Thévenin equivalents.  Consider the circuit shown in
Figure 5.25.
Suppose we want to find the Thévenin equivalent that drives R2. We cut the circuit 
immediately to the left of R2. That is, The first step is to make the cut, removing the 
remainder of the circuit. In this case the remainder is just R2. We then determine the 
open circuit output voltage at the cut points (i.e., at the open port). This voltage is 
called the Thévenin voltage, Eth. This is shown in Figure 5.26. In a circuit such as 
this, basic series-parallel analysis techniques may be used to find Eth. In this circuit, 
due to the open, no current flows through the inductor, L, and thus no voltage is 
developed across it. Therefore, Eth must equal the voltage developed across the 
capacitor, C.
4 It is possible that an equivalent can be valid across a specified range of frequencies, but it
will not hold for all frequencies.
161
Figure 5.24
Generic Thévenin equivalent 
circuit.",ACElectricalCircuitAnalysis.pdf
"will not hold for all frequencies.
161
Figure 5.24
Generic Thévenin equivalent 
circuit. 
Figure 5.25
Circuit under consideration for 
a Thévenin equivalent.",ACElectricalCircuitAnalysis.pdf
"The second part is finding the Thévenin impedance, Zth. Beginning with the “cut” 
circuit, replace all sources with their ideal internal impedance (thus shorting voltage 
sources and opening current sources). From the perspective of the cut point, look 
back into the circuit and simplify to determine its equivalent impedance. This is 
shown in Figure 5.27. Looking in from where the cut was made (right side), we see 
that R1 and XC are in parallel, and this combination is then in series with XL. Thus, Zth
is equal to jXL + (R1 || −jXC).
As noted earlier, the original circuit could be cut in a number of different ways. We 
might, for example, want to determine the Thévenin equivalent that drives C in the 
original circuit of Figure 5.25. The new port location appears in Figure 5.28.
Clearly, this will result in different values for both Eth and Zth. For example, Zth is 
now R1 || (R2 + jXL). 
A common error is to find Zth from the wrong perspective, namely, finding the",ACElectricalCircuitAnalysis.pdf
"now R1 || (R2 + jXL). 
A common error is to find Zth from the wrong perspective, namely, finding the 
impedance that the source drives. This is flatly incorrect. Remember, Zth is found by 
looking into the port and simplifying whatever is seen from there. One way to 
remember this is that it is possible to create equivalents for multi-source circuits. In 
that instance, there isn't a single driving source, so finding its load impedance is 
nonsensical. 
162
Figure 5.26
Eth,the open circuit output 
voltage.
Figure 5.27
Finding Zth.
Figure 5.28
An alternate port location.",ACElectricalCircuitAnalysis.pdf
"Measuring the Thévenin Equivalent in the LaboratoryMeasuring the Thévenin Equivalent in the Laboratory
In a laboratory situation, the Thévenin equivalent can be found quickly and 
efficiently with the proper tools. First, the circuit is “cut”, leaving just the portion to 
be Thévenized. The voltage at the cut points is measured with an oscilloscope. This 
is Eth. All of the sources are then replaced with their internal impedance, ideally 
shorting voltage sources and opening current sources5. An LCR impedance meter can
then be connected to the port to read Zth. If an impedance meter is not available, then
the source(s) are left in place and an LCR substitution box is placed at the cut points. 
The box is adjusted so that the voltage across it is equal to half of Eth. By the voltage
divider rule, the value of the substitution box must be equal to Zth. In this case, the 
substitution box will yield either an inductance or capacitance value which can then",ACElectricalCircuitAnalysis.pdf
"substitution box will yield either an inductance or capacitance value which can then 
be turned into a reactance given the frequency.
Example 5.6
For the circuit of Figure 5.29, determine the Thévenin equivalent that drives 
the 300 Ω resistor and find vc. Assume the source angle is 0°.
First, let's find Eth, the open circuit output voltage. We cut the circuit so that 
the 300 Ω resistor is removed. Then we determine the voltage at the cut 
points. This circuit is shown in Figure 5.30.
5 A typical laboratory signal generator has a 50 Ω internal impedance, and using this value 
would be more accurate than just replacing the source with a shorting wire.
163
Figure 5.29
Circuit for Example 5.6.
Figure 5.30
Circuit for finding Eth for 
Example 5.6.",ACElectricalCircuitAnalysis.pdf
"There is no current flowing through the inductor due to the open. Therefore, 
the voltage across the inductor is zero. Consequently, Eth is the voltage 
across the capacitor, and that can be found with a voltage divider.
Eth = E X C
X C +R1
Eth = 10  0° V − j 200Ω
− j 200Ω +100 Ω
 
Eth = 8.944 −26.6° V  or 8 − j 4 V
To find Zth, we replace the source with a short and then look back in from the
cut points. The equivalent circuit is shown in Figure 5.31. The inductor is in 
series with the parallel combination of the resistor and capacitor.
Zleft2 = R× jX C
R − jX C
Zleft2 = 100 Ω×(− j 200 Ω)
100 Ω− j 200 Ω
Zleft2 = 89.44  − 26.6° Ω
Zth = Z left2 + X L
Zth = 89.44  −26.6 ° Ω+ j 50Ω
Zth = 80.62  7.12° Ω  or 80 +j 10Ω
The completed equivalent is shown in Figure 5.32.
164
Figure 5.31
Circuit for finding Zth for 
Example 5.6.
Figure 5.32
Completed equivalent for 
Example 5.6.",ACElectricalCircuitAnalysis.pdf
"The voltage across the 300 Ω resistor can be found directly:
vR2 = Eth
R2
Z th + R2
vR2 = 8 − j 4 V 300Ω
80Ω+ j 10 Ω+ 300 Ω
 
vR2 = 7.06 −28.1° V
This value can be verified by following a standard series-parallel 
simplification. For example, the impedance of the three rightmost 
components (181.4−53.97° Ω) forms a voltage divider with the 100 Ω 
resistor and the 10 volt source. This leads to vb (7.156−18.61° volts ). A 
second divider can then be used between vb, the inductor and the 300 Ω 
resistor to find vc, which is 7.06−28.1° volts as expected. The big 
advantage of using the Thévenin equivalent is that we can easily find vc for 
any other value of load because we need only analyze the simpler equivalent
circuit rather than the original.
 
Thévenin's theorem can also be used on multi-source circuits. The technique for 
finding Zth does not change, however, finding Eth is a little more involved, as 
illustrated in the next example.
Example 5.7",ACElectricalCircuitAnalysis.pdf
"finding Zth does not change, however, finding Eth is a little more involved, as 
illustrated in the next example.
Example 5.7
Find vb for the circuit of Figure 5.33 using Thévenin's theorem.
This circuit is similar to the one used in Example 5.5 (Figure 5.18). The 
difference here is that the second source uses the same frequency as the first 
source. The reactance values are:
XL = j314.2 Ω
XC = −j212.2 Ω 
165
Figure 5.33
Circuit for Example 5.7.",ACElectricalCircuitAnalysis.pdf
"The voltage across the 2 kΩ is vb, so we'll treat that resistor as the load in 
order to define the equivalent circuit. This is redrawn in Figure 5.34.
To find Zth, we short the two sources. We're left with the inductor and 
capacitor in parallel. If this is confusing, remember that we are looking from
node b to ground (the cut points), so they are not in series. That is, if a 
sensing current entered at node b, it could split left and right, indicating 
parallel paths, not a series connection.
Z th = − jX C × jX L
− jX C +jX L
Zth = − j 212.2Ω× j 314.2 Ω
− j 212.2 Ω+ j 314.2 Ω
Zth = − j 653.7Ω
We have a few options to find Eth. Superposition could be used, each circuit 
requiring a voltage divider. Alternately, the equivalent is basically a series 
loop as far as finding vb is concerned. Thus, we could find the voltage across
the inductor and subtract that from the left source (assuming a reference 
current direction of clockwise). Finding the inductor voltage requires either",ACElectricalCircuitAnalysis.pdf
"current direction of clockwise). Finding the inductor voltage requires either 
a voltage divider or finding the current. Neither approach is considerably 
less work than the other, and it's probably a good idea to get in a little more 
practice using superposition, so...
Considering the left source, we short the right source and find vb.
 
vbR = E1
X C
X C +X L
vbR = 10  0° V − j 212.2Ω
− j 212.2Ω +j 314.2 Ω
 
vbR = 20.84 180 ° V   or −20.84  0° V
For the right source, we short the left source and find vb. Then we add the 
two contributions to find the final voltage. Note that both sources will 
produce a reference polarity of + to − from top to bottom.
166
Figure 5.34
Circuit for finding equivalents 
for Figure 5.33.",ACElectricalCircuitAnalysis.pdf
"vbL = E2
X C
X C + X L
vbL = 2 0° V j 314.2Ω
− j 212.2 Ω+j 314.2 Ω
 
vbL = 6.16 0° V
The sum of the two is −20.840° + 6.160°, or 14.68180° volts. The 
Thévenin equivalent is a source of 14.68180° volts in series with an 
impedance of −j653.7 Ω. 
To find the voltage across the 2 k Ω resistor, we apply it to the equivalent 
circuit and solve.
vb = Eth
R
R +Z th
vb = 14.68 180 ° V 2 kΩ
2k Ω +(− j 653.7Ω)
 
vb = 13.95 −161.9° V
Computer SimulationComputer Simulation
To verify the results of the preceding example, the circuit of Figure 5.33 is captured 
in a simulator, as shown in Figure 5.35.
Next, a transient analysis is run, plotting the voltage at node 2, which corresponds to
vb in the original circuit. The result is shown in Figure 5.36. The plot is delayed by 
0.1 seconds in order to get past the initial turn-on transient. 
167
Figure 5.35
The circuit of Figure 5.33 
captured in a simulator.",ACElectricalCircuitAnalysis.pdf
"Both the amplitude and phase of the simulation waveform match the computed 
results. 
Norton's TheoremNorton's Theorem
Norton's theorem is named after Edward Lawry Norton. It is the current source 
version of Thévenin's theorem. In other words, complex networks can be reduced to 
a single current source with a parallel internal impedance. Formally, Norton's 
theorem states:
Any single port linear network can be reduced to a simple voltage source, 
In, in parallel with an internal impedance Zn.
The process of finding a Norton equivalent is very similar to finding a Thévenin 
equivalent. First, the Norton impedance is the same as the Thévenin impedance. 
Second, instead of finding the open circuit output voltage, the short circuit output 
current is found. This is the Norton current. Due to the equivalence afforded by 
source conversions, if a Thévenin equivalent for a network can be created, then it 
must be possible to create a Norton equivalent. Indeed, if a Thévenin equivalent is",ACElectricalCircuitAnalysis.pdf
"must be possible to create a Norton equivalent. Indeed, if a Thévenin equivalent is 
found, a source conversion can be performed on it to yield the Norton equivalent. 
168
Figure 5.36
Simulation results for the 
circuit of Figure 5.33.",ACElectricalCircuitAnalysis.pdf
"Example 5.8
Let's reexamine Example 5.6, this time creating a Norton equivalent circuit. 
For convenience, the original circuit of Figure 5.29 is repeated in Figure 
5.37. Once again, the goal will be to determine the equivalent that drives the 
300 Ω resistor and to find vc.
As noted, the Norton impedance, Zn, is the same as Zth. That was 80 + j10 Ω. 
The Norton current, In, is the short-circuit current through the cut points. We 
can think of this as replacing the load resistor with an ammeter. This is the 
same as the current through the inductor. In this situation, the capacitor and 
inductor are in parallel and yield an impedance of j66.67 Ω. Thus, the source
current is:
isource = E
R +Z LC
i source = 10 0°
100Ω +j 66.67Ω
isource = 83.2E-3 −33.7° A
This splits between the capacitor and inductor. Using the current divider rule
we find:
in = iinductor = I source
X C
X C +X L
in = 83.2  − 33.7° A − j 200Ω
− j 200 Ω+j 50Ω
in = 0.1109 −33.7 ° A",ACElectricalCircuitAnalysis.pdf
"we find:
in = iinductor = I source
X C
X C +X L
in = 83.2  − 33.7° A − j 200Ω
− j 200 Ω+j 50Ω
in = 0.1109 −33.7 ° A
A source conversion can be applied to verify this value. The resulting 
voltage source is 8 − j4 volts, precisely the value of the Thévenin equivalent.
 
Ultimately, deciding between using the Thévenin or Norton equivalents is a matter 
of personal taste and convenience. They work equally well.
169
Figure 5.37
Circuit for Example 5.8.",ACElectricalCircuitAnalysis.pdf
"5.5 Maximum Power Transfer Theorem5.5 Maximum Power Transfer Theorem
The concept of maximum power transfer in DC resistive circuits was presented in
earlier work. While maximizing load power is not a goal of all circuit designs, it is
a goal of a portion of them and thus worth a closer look. In short, given an AC
voltage source with internal impedance, as seen in Figure 5.38, a useful question to
ask is “What value of load impedance will yield the maximum amount of power in
the load?” In the DC case, it was discovered that the load resistance must equal the
source resistance in order to achieve maximum load power. In the AC case things
appear to be much more complicated by the possible presence of reactances in both
the source and load.
As a refresher of prior study, consider the basic circuit depicted in Figure 5.38 with
source E, source internal resistance Zi and load impedance Z. For the moment, we 
shall ignore the reactive portions and just describe the load power in terms of the",ACElectricalCircuitAnalysis.pdf
"shall ignore the reactive portions and just describe the load power in terms of the 
load's resistive portion, R. To make the job easier, we may normalize the source 
resistance Ri to 1 Ω.  By doing this, R also becomes a normalized value, that is, it no 
longer represents a simple resistance value but rather represents a ratio in 
comparison to Ri. In this way the analysis will work for any set of source values. 
Note that the value of E will equally scale the power in both Ri and R, so a precise 
value is not needed, and thus, we may as well chose 1 volt for convenience. 
The power in the load can be determined by using I2R where I = E / (Ri+R), yielding
P =(
E
Ri+R )
2
R
Using our normalized values of 1 volt and 1 Ω,
P =(
1
1+R )
2
R    
After expanding we arrive at:  
P = R
R2
+2 R+1
(5.1)
We now have an equation that describes the load power in terms of the load 
resistance. Before we go any further, take a look at what this equation tells you, in",ACElectricalCircuitAnalysis.pdf
"resistance. Before we go any further, take a look at what this equation tells you, in 
general. It is obvious that maximum power will not occur at the extremes. If R = 0 or
R = ∞ (i.e., shorted or opened load) the load power is zero. The maximizing case 
occurs somewhere in the middle. To find the precise value that produces the 
maximum load power, the proof can be divided into two portions. The first involves 
graphing the function and the second requires differential calculus to solve for a 
170
Figure 5.38
Defining maximum power 
transfer.",ACElectricalCircuitAnalysis.pdf
"precise value. We shall proceed with the graphing portion which will lead us to the 
answer. The more rigorous proof of the second portion is detailed in Appendix C. 
The curve of Equation 5.1 is plotted in Figure 5.39. The normalized load resistance 
is set along the horizontal and the normalized power (i.e., for a source of 1 volt) is 
set along the vertical.
An examination of the power curve shows that the peak occurs at R = 1. In other 
words, the load must be equal to the source resistance. Thus, we can say that if no 
reactances are involved, maximum load power occurs when the load resistance 
equals the source resistance. It does not matter if the source is DC or AC.
The graph shown in Figure 5.39 is asymmetric but the concept of resistance ratios is 
key here. This is easier to see if we plot the completed power curve using a 
logarithmic horizontal axis and also scale the vertical axis to 100%, as shown in",ACElectricalCircuitAnalysis.pdf
"logarithmic horizontal axis and also scale the vertical axis to 100%, as shown in 
Figure 5.40. The peak is more apparent and the curve is symmetrical in shape rather 
than lopsided. This reinforces the idea that the ratio of the resistances is what 
matters.
171
Normalized Load Power
0.00
0.05
0.10
0.15
0.20
0.25
0.30
Normalized Load Resistance
0 0.5 1 1.5 2 2.5
R/(R2+2R+1)
Figure 5.39
The resistive portion of the 
power equation plotted.",ACElectricalCircuitAnalysis.pdf
"At this point we may turn our attention to the possible presence of reactances in both
the source and load. It turns out that this is not nearly as complicated as it might 
look. The key is that only resistors dissipate power, not inductors or capacitors6. 
Load power is proportional to iload
2, so our immediate goal is to maximize load 
current for any set of source and load resistances. 
We can modify the original power equation by adding a new term, X, which 
represents the net reactance in the circuit. In other words, X is equal to the sum of 
the reactances in the source impedance and load impedance. The power in the load is
still determined by using I2R, however, we must now include the X term when 
computing the current:
I = E
√((Ri +R)2+X 2)
This leads is to a new load power expression:
P =(
E
√((Ri+R)2
+X 2
))
2
R (5.2)
A cursory look at Equation 5.2 shows that to maximize P, X must be zero. A 
normalized plot of this equation is shown in Figure 5.41 for R = Ri.",ACElectricalCircuitAnalysis.pdf
"A cursory look at Equation 5.2 shows that to maximize P, X must be zero. A 
normalized plot of this equation is shown in Figure 5.41 for R = Ri.
6 Power in AC circuits is examined in great detail in Chapter 7.
172
Load Power as a Percentage of Maximum
0
20
40
60
80
100
  
Normalized Load Resistance
0.01 0.1 1 10 100
R/(R2+2R +1)
Figure 5.40
The load power curve with 
logarithmic axis showing 
symmetry.",ACElectricalCircuitAnalysis.pdf
"A single peak is evident when X is 0. This can be achieved by setting the load
reactance equal in magnitude to the source reactance but with the opposite sign. In
this manner, the reactances will cancel out, leaving a purely resistive circuit with a
minimal value, and thus producing maximal current for that set of resistors. 
173
Figure 5.41
The reactive portion of the 
power equation plotted using 
matched resistance.
Load Power as a Percentage of Maximum
0
20
40
60
80
100
  
Normalized Total Reactance
-10 -5 0 5 10
 
Figure 5.42
The load power surface 
showing variations with both 
load resistance and total 
reactance.",ACElectricalCircuitAnalysis.pdf
"Two variables are involved here, so to further clarify the situation, a 3D surface plot 
of normalized power is shown in Figure 5.42. The vertical axis represents the 
percentage of maximum power while the front and side axes are the normalized total
reactance and load resistance, respectively. A single peak is evident here and 
coincides with X = 0 and R = 1. This is more easily seen by viewing the surface
from the back as shown in Figure 5.43. Note that the highest isocontour encircles
the intersection of X = 0 and R = 1 (i.e., Rload = Ri ).
In sum, we have verified that the resistive portions of the source and load impedance
must be identical and that the reactive portions must be of the same magnitude but of
opposite sign. This configuration is also known as the complex conjugate. Finally, 
we can state:
Maximum load power will be achieved when the load impedance is equal to 
the complex conjugate of the internal impedance of the driving source.",ACElectricalCircuitAnalysis.pdf
"Maximum load power will be achieved when the load impedance is equal to 
the complex conjugate of the internal impedance of the driving source. 
No other value of load impedance will produce a higher load power. We can imagine
two general cases, one with an inductive source impedance and another with a 
capacitive source impedance. These are shown with the proper loads in Figure 5.44.
174
Figure 5.43
An alternate view of the load 
power surface.",ACElectricalCircuitAnalysis.pdf
"To achieve maximum load power in these circuits, Rload = Ri and |jXL| = |−jXC|. Note 
that XL and XC do not have to have the same magnitude as Ri.
While using the complex conjugate produces the maximum load power, it does not 
produce the largest possible load current or load voltage. In fact, this condition 
produces a load voltage and a load current that are half of their maximums. Their 
product, however, is at the maximum. Further, efficiency at maximum load power is 
only 50% (i.e., only half of all generated power goes to the load with the other half 
being wasted internally). Values of R greater than Ri will achieve higher efficiency 
but at reduced load power. Sometimes we favor efficiency over maximal load power.
As any linear single port network can be reduced to something like Figure 5.44 by 
using Thévenin's theorem, combining the two theorems allows us to determine 
maximum power conditions for any impedance in a complex circuit. 
Example 5.9",ACElectricalCircuitAnalysis.pdf
"maximum power conditions for any impedance in a complex circuit. 
Example 5.9
Consider the circuit of Figure 5.45. What is the power generated in the load 
if it is equal to 40 Ω? Further, is that the maximum power that can be 
attained, and if not, what is the maximum load power and what value of load
would be needed?
To find the load power, first find the circulating current, then use power law. 
The total impedance seen by the source is 20 + j10 Ω + 40 Ω, or 60 + j10 Ω.
175
Figure 5.44
Configurations for complex 
conjugate loads.
Figure 5.45
Circuit for Example 5.9.",ACElectricalCircuitAnalysis.pdf
"i = E
Z total
i = 70V
60Ω +j 10Ω
i = 1.151  −9.5° A
As the voltage and current are in phase for a resistor, we can ignore the 
angle for the power calculation.
Pload = i2×Rload
Pload = (1.151 A)2
×40Ω
Pload ≈53 W
This is not the maximum load power that can be achieved because this load 
is not the complex conjugate of the source impedance. The required load for 
maximum load power is shown in Figure 5.46.
We shall repeat the process to find the new load power.
i = E
Ztotal
i = 70 V
40Ω
i = 1.75  0° A
Pload = i2
×Rload
Pload = (1.75 A)2
×20 Ω
Pload = 61.25 W
An alternate method notes that the new circuit's total impedance is purely 
resistive and that the source and load resistances are identical. Therefore the 
voltage source must split evenly across them. In this case that's 35 volts 
RMS each.
176
Figure 5.46
Circuit of Figure 5.45 with 
proper load configuration.",ACElectricalCircuitAnalysis.pdf
"Pload = vR
2
Rload
Pload = (35 V)2
20Ω
Pload = 61.25 W
Example 5.10
For the circuit of Figure 5.47, determine the value of Zload that will achieve 
maximum load power and also determine that power. The current source is 
0.10° amps RMS at 50 kHz.
The first job is determine the inductive reactance at 50 kHz. Recalling that 
XL = j2πfL, this works out to j157 Ω. We now need to find the Thévenin 
equivalent. To find Zth we open the current source and look back in from the 
load. We see the 50 Ω resistor in series with the parallel combination of the 
200 Ω resistor and the inductor. The parallel combination is:
Z = R × jX L
R +jX L
Z = 200 Ω×( j 157 Ω)
200Ω +j 157Ω
Z = 76.3+j 97.2Ω
Therefore, Zth = 126.3 + j97.2 Ω. The complex conjugate is 126.3 − j97.2 Ω. 
The capacitive reactance formula may be used to determine the appropriate 
capacitance value to achieve −j97.2 Ω.
C = 1
2 π f X C
C = 1
2π50 kHz97.2Ω
C = 32.8 nF
 
177
Figure 5.47
Circuit for Example 5.10.",ACElectricalCircuitAnalysis.pdf
"The resulting circuit is shown in Figure 5.48.
To find the load power we need to find Eth. The open circuit output voltage is
the potential appearing across the inductor/resistor pair in Figure 5.47. This 
is because there is no current flowing through the 50 Ω resistor, and 
therefore there is no voltage across it. Eth can be found via Ohm's law as we 
already know the impedance of the parallel branch.
 
Eth = i×Z
Eth = 0.1 0° A×(76.3+j 97.2Ω)
Eth = 12.4 51.9° V
Again, using the complex conjugate, the voltage source splits evenly 
between the resistive components. As the current source was specified as 
RMS, so too will be the equivalent voltage.
Pload = vR
2
Rload
Pload = (6.2 V)2
126.3Ω
Pload = 304.4 mW
This represents the maximum load power that can be achieved in this circuit.
Do not forgot, though, that an equal amount of power is dissipated by the 
source. This produces an efficiency of just 50%.
In summation, we can say that maximum load power is achieved when the load",ACElectricalCircuitAnalysis.pdf
"source. This produces an efficiency of just 50%.
In summation, we can say that maximum load power is achieved when the load 
impedance is equal to the complex conjugate of the internal impedance of the 
circuitry driving the load. Usually, this requires the application of either a Thévenin 
or Norton equivalent. Finally, although maximum power transfer is a desired 
outcome in some situations, it is not desirable in all situations. The reason is one of 
efficiency. At the maximum load power, efficiency is only 50%. In contrast, for load 
impedances that are greater than the source impedance, the load power will 
decrease, however, the efficiency will increase. Increased efficiency is particularly 
important when striving to minimize heat and extend battery life.
178
Figure 5.48
Thévenin equivalent of the 
circuit of Figure 5.47 with 
appropriate load.",ACElectricalCircuitAnalysis.pdf
"5.6 Delta-Y (Pi-T) Conversions5.6 Delta-Y (Pi-T) Conversions
Certain component configurations, such as bridged networks, cannot be reduced to a
single impedance using basic series-parallel conversion techniques. One method for 
simplification involves converting sections into more convenient forms. The 
configurations in question are networks with three external connection points. Due 
to the manner in which they drawn, they are referred to as delta networks and Y 
networks7. These configurations are shown in Figure 5.49. Note that the terminal 
designation of the delta version are upside down compared to those of the Y 
configuration.
These networks can be redrawn without angles. In this form they are known as pi 
(also called “π”) networks and T (also called “tee”) networks. These configurations 
are shown in Figure 5.50.
It is possible to convert back and forth between delta and Y networks in many cases. 
That is, for a given delta network, there may exist a Y network such that the",ACElectricalCircuitAnalysis.pdf
"That is, for a given delta network, there may exist a Y network such that the 
impedances seen between the X, Y and Z terminals are identical, and vice versa. 
Consequently, one configuration can replace another in order to simplify a larger 
circuit. Unlike the DC version, certain AC networks cannot be converted using the 
following technique (see the final Challenge problem for an investigation of this).
 
7 In some sources the capital Greek letter delta (Δ) is used instead of spelling out “delta” 
and the letter Y is spelled out as “wye”. Thus, you may come across discussion of “Δ-Y ”,
“Δ-wye” or “delta-wye” networks. It's all the same stuff. 
179
Figure 5.49
Delta and Y (Δ-Y) networks.
Figure 5.50
Alternate form: Pi and T
(π-T) networks.",ACElectricalCircuitAnalysis.pdf
"Δ-Y ConversionΔ-Y Conversion
A true equivalent circuit would present the same impedance between any two 
terminals as the original circuit. Consider the circuits of Figure 5.49 for the unloaded
case (i.e., just these networks with nothing else connected to them). The equivalent 
impedances seen between each pair of terminals for the delta and the Y respectively 
are:
ZXY = Za || (Zb+Zc) = Zd + Ze (5.3)
ZXZ = Zb || (Za+Zc) = Zd + Zf (5.4)
ZZY = Zc || (Zb+Za) = Ze + Zf (5.5)
Let's assume that we have the delta network and are looking for the Y network 
equivalent. We start by focusing on the final set of terms for each of the three 
expressions (e.g., ZXY = Zd + Ze). Note that we have three equations with three 
unknowns (Zd, Ze and Zf). Thus, they can be solved using a term elimination process. 
If we subtract Equation 5.5 from Equation 5.3, we can eliminate the second 
impedance (Ze) and arrive at a difference between the first and third unknown",ACElectricalCircuitAnalysis.pdf
"impedance (Ze) and arrive at a difference between the first and third unknown 
impedance s (Zd − Zf). This quantity can then be added to Equation 5.4 to eliminate 
the third impedance (Zf), leaving just the first unknown impedance (Zd). 
(Zd + Ze) − (Ze + Zf) = (Zd − Zf) = Za || (Zb+Zc) − Zb || (Za+Zc)
(Zd + Zf) + (Zd − Zf) = 2Zd = 2( Zb || (Za+Zc) + Za || (Zb+Zc) − Zc || (Za+Zb) )
Therefore,
Zd = Zb || (Za+Zc) + Za || (Zb+Zc) − Zc || (Za+Zb) 
which, after simplifying8, is:
Zd = Z a Zb
Z a+Z b+Zc
(5.6)
Similarly, we can show that
Ze = Z a Z c
Z a+Z b+Z c
(5.7)
Z f = Z b Z c
Z a+Z b+Z c
(5.8)
Note that if the magnitudes and angles of three original impedances are identical, the
magnitudes of the Y equivalent impedances will all be one-third of the original  
magnitude, and with the original phase angle. 
8 This process, though not particularly difficult, is somewhat tedious. It is, as they say, “left
as an exercise for the student”.
180",ACElectricalCircuitAnalysis.pdf
"Y-Δ ConversionY-Δ Conversion
For the reverse process of converting from Y to delta, start by noting the similarities 
of the expressions for Zd, Ze and Zf (i.e., Equations 5.6 through 5.8). If two of these 
expressions are divided, a single equation for Za, Zb or Zc will result. For example, 
using Equations 5.6 and 5.7:
Z d
Ze
=
Z a Z b
Z a +Zb +Z c
Z a Z c
Z a +Zb +Z c
 
Z d
Ze
= Z a Zb
Z a Z c
Z d
Z e
= Z b
Z c
Therefore,
Z b
Z c
= Z d
Z e
Z b = Zc Z d
Z e
This process can be repeated for Equations 5.6 and 5.8 to obtain an expression for 
Za. The two expressions for Za and Zb can then be substituted into Equation 5.6 to 
obtain an expression for Zc that utilizes only Zd, Ze and Zf. A similar process is 
followed for Za and Zb resulting in:
Za = Z d Z e+Ze Z f +Z d Z f
Z f
(5.9)
Zb = Z d Ze +Z e Z f +Z d Z f
Ze
(5.10)
Zc = Z d Ze +Z e Z f +Z d Z f
Zd
(5.11)
If the Y network consists of three identical impedances, then the values of the delta",ACElectricalCircuitAnalysis.pdf
"Ze
(5.10)
Zc = Z d Ze +Z e Z f +Z d Z f
Zd
(5.11)
If the Y network consists of three identical impedances, then the values of the delta 
equivalent will all be three times the original magnitude, the inverse of the situation 
when converting from delta to Y .
In summation, equations 5.6, 5.7 and 5.8 can be used to convert a delta network into 
a Y network, and equations 5.9, 5.10 and 5.11 can be used to convert a Y network 
into a delta network. Examples of how to apply this technique to tame up-to-now 
intractable series-parallel networks follow.
181",ACElectricalCircuitAnalysis.pdf
"Example 5.11
Convert the network of Figure 5.51 into its delta configuration equivalent.
Referring back to Figure 5.50, use Equation 5.9 to determine Za.
Za = Z d Z e +Z e Z f +Z d Z f
Z f
Za = (50Ω + j 20 Ω)(10Ω)+(10Ω)(40Ω− j 30Ω)+(50Ω + j 20 Ω)(40 Ω− j 30Ω)
(40Ω − j 30Ω)
Za = 65.6 + j 29.2Ω
Zb and Zc may be determined in similar manner using Equations 5.10 and 
5.11:
Zb = 350 − j80 Ω
Zc = 54.8 − j37.9 Ω
The equivalent is shown in Figure 5.52.
 
182
Figure 5.51
Network for Example 5.11.
Figure 5.52
Equivalent delta network for 
the Y network of Figure 5.51.",ACElectricalCircuitAnalysis.pdf
"Remember, a complex impedance can always be expressed in rectangular form. 
Rectangular form can be expressed directly as a series combination of a resistor and 
either an inductive or capacitive reactance. Even if the original impedances of a 
network are in a parallel form (or even a more complex form), the equivalent can be 
expressed as a series combination.
Example 5.12
Determine va in the circuit of Figure 5.53. Assume the source has a phase 
angle of zero degrees.
This circuit cannot be simplified sufficiently using basic series-parallel 
techniques due to the bridge section. The components between and below 
nodes a and b comprise a delta network, as shown in Figure 5.54. If this 
network is replaced with a Y equivalent, the resulting circuit reduces to a 
simple series-parallel system. 
Before continuing, it would be helpful to determine the impedance of each 
of the parallel sections. For the leftmost pair:
Zleft2 = R× jX L
R + jX L
Z left2 = 10Ω× j100 Ω
10Ω + j 100Ω",ACElectricalCircuitAnalysis.pdf
"of the parallel sections. For the leftmost pair:
Zleft2 = R× jX L
R + jX L
Z left2 = 10Ω× j100 Ω
10Ω + j 100Ω
Zleft2 = 9.9Ω+ j0.99Ω
183
Figure 5.53
Circuit for Example 5.12.
Figure 5.54
Delta network within the  
circuit of Figure 5.53.",ACElectricalCircuitAnalysis.pdf
"In similar manner, the top pair is determined to be 19.8 + j1.98 Ω and the 
rightmost pair is 29.7 + j2.97 Ω. Referring back to Figure 5.50, we can use 
Equation 5.6 to determine Zd. 
Zd = Z a Zb
Za +Z b+Z c
Zd = (19.8Ω+ j 1.98Ω)×(9.9Ω+ j 0.99Ω)
(19.8Ω+ j 1.98Ω)+(9.9Ω+ j 0.99Ω)+(29.7Ω+ j 2.97Ω)
 
Zd = 3.3Ω+ j 0.33Ω
Likewise, we can use Equations 5.7 and 5.8 to determine Ze and Zf.
Ze = 9.9 Ω + j0.99 Ω
Zf = 4.95 Ω + j0.495 Ω
Swapping the equivalent Y network into the original circuit leads us to the 
circuit of Figure 5.55 (Y network shown in blue). This circuit can be 
simplified directly to find va.
In this equivalent circuit, va is simply the source voltage of 10° minus the 
voltage across the 2 Ω resistor. The immediate goal, then, is to find the 
current through that resistor. This can be achieved via a current divider once 
the source current is known. To find the source current, we need to find the 
total impedance of the network. On the upper left side, Zd is in series with",ACElectricalCircuitAnalysis.pdf
"total impedance of the network. On the upper left side, Zd is in series with 
the 2 Ω resistor for a total of 5.3 Ω + j0.33 Ω. This is in parallel with the 
upper right side total of 15.9 Ω + j0.99 Ω. 
184
Figure 5.55
Equivalent circuit for the 
circuit of Figure 5.53.",ACElectricalCircuitAnalysis.pdf
"Zupper = Z upperleft×Z upperright
Z upperleft + Z upperright
Zupper = (5.3Ω + j 0.33Ω)×(15.9Ω + j 0.99Ω)
(5.3Ω + j 0.33Ω) +(15.9Ω+ j0.99Ω)
Zupper = 3.975 Ω + j 0.2475Ω
This is in series with the lower section of 4.95 + j0.495 Ω for a total of 
8.9564.76° Ω. Using Ohm's law, we find the source current:
isource = E
Ztotal
isource = 1 0° V
8.956  4.76° Ω
isource = 0.1117−4.76° A
Now for the current divider and also Ohm's law for the 2 Ω resistor.
i2Ω = isource
Z upperright
Zupperright +Z upperleft
i2Ω = 0.1117−4.76° A 15.9+j 0.99Ω
(15.9+j 0.99Ω)+(5.3+j 0.33Ω)
 
i2Ω = 83.7E-3−4.76° A
v2Ω = i2Ω×R
v2Ω = 83.7E-3−4.76° A×2Ω
v2Ω = 0.1675 −4.76° V
Finally, we subtract that potential from the source to find va.
va = E −v2Ω
va = 1 0° V −0.1675 −4.76° V
va = 0.833  0.95° V
5.7 Summary5.7 Summary
In this chapter we have examined several techniques and theorems to assist with the 
analysis of AC electrical circuits. We began with more practical models for voltage",ACElectricalCircuitAnalysis.pdf
"analysis of AC electrical circuits. We began with more practical models for voltage 
and current sources by adding an internal impedance to set limits on the source's 
maximum output and make it sensitive to output frequency. For a voltage source, 
this impedance is in series, its ideal value being a short, just as it was for the DC 
case. For current sources, the impedance is in parallel, its ideal value being an open. 
185",ACElectricalCircuitAnalysis.pdf
"Source conversions allow us to create an equivalent voltage source for any practical 
current source and vice versa. An equivalent source is one that will create the same 
voltage across (and current into) the remaining circuit as did the original source. In 
some cases, this swap allows multiple sources to be combined into a single source, 
simplifying analysis. If the associated impedance does not have a zero degree phase 
angle, then the converted source will not be in phase with the original, but will 
instead be shifted by the impedance angle.
The superposition theorem states that, for any multi-source linear bilateral network, 
the contributions of each source may be determined independent of all other sources,
the final result being the summation of the contributions. This remains true in the AC
case, however, care must be taken regarding phase shifts when combining the 
various contributions. The original circuit of N sources generates N new circuits, one",ACElectricalCircuitAnalysis.pdf
"various contributions. The original circuit of N sources generates N new circuits, one
for each source under consideration and with all other sources replaced by their ideal
internal impedance.
Thévenin's and Norton's theorems allow the simplification of complex linear single 
port (i.e., two connecting points) networks. The AC Thévenin equivalent consists of 
a voltage source with a series impedance while the AC Norton equivalent consists of 
a current source with a parallel impedance. These impedances can be represented in 
general as a resistance in series with a reactance, and given an operating frequency, 
the reactance can be turned into a capacitance or inductance. These equivalents, 
when replacing the original sub-circuit, will create the same voltage across the 
remainder of the circuit with the same current draw. In other words, the remainder of
the circuit will see no difference between being driven by the original sub-circuit or",ACElectricalCircuitAnalysis.pdf
"the circuit will see no difference between being driven by the original sub-circuit or 
by either the Thévenin or Norton equivalents.
The maximum power transfer theorem states that for a simple voltage source with an
internal impedance driving a simple load, the maximum load power will be achieved
when the load impedance equals the complex conjugate of the internal impedance. 
The complex conjugate has the same real or resistive value, however, the reactive 
portion is of the opposite sign. This results in a cancellation of the reactive 
components, leaving just the resistive portions and maximizing load current. At this 
point, efficiency will be 50%. If the load impedance is higher than the internal 
impedance, the load power will not be as great, however, the system efficiency may 
improve, depending on the phase angle.
Delta-Y conversions allow the generation of equivalent “three connection point” 
impedance networks. RLC networks with three elements in the shape of a triangle or",ACElectricalCircuitAnalysis.pdf
"impedance networks. RLC networks with three elements in the shape of a triangle or 
delta (with one connection point at each corner) may be converted into a three 
element network in the shape of a Y or T, or vice versa. The two versions will 
behave identically to the remainder of the circuit. This allows the simplification of 
some circuits and eases analysis.
186",ACElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. What are the ideal internal impedances of AC voltage and current sources?
2. Outline the process of converting an AC voltage source into an AC current 
source, and vice versa.
3. In general, describe the process of using superposition to analyze a multi-
source circuit.
4. What do Thévenin's and Norton's theorems state? How are they related?
5. What are the conditions to achieve maximum power transfer for AC circuits?
How does this differ from the DC version?
6. What are delta and Y configurations? How are they related?
5.8 Exercises5.8 Exercises
AnalysisAnalysis
1. For the circuit shown in Figure 5.56, use superposition to find vb.
2. For the circuit shown in Figure 5.56, use superposition to find the current 
through the capacitor.
3. Use superposition to find the current through the 82 Ω resistor. For the 
circuit shown in Figure 5.57. 
4. Use superposition to find vb and vcd for the circuit shown in Figure 5.57. 
187
Figure 5.56
Figure 5.57",ACElectricalCircuitAnalysis.pdf
"5. In the circuit of Figure 5.58, use superposition to find vb. Source one is one 
volt peak and source two is two volts peak.
6. In the circuit of Figure 5.58, use superposition to find the currents through 
the two inductors. Source one is two volts peak and source two is three volts
peak.
7. Use superposition to find the current through the 2.2 kΩ resistor for the 
circuit of Figure 5.59. E1 = 10° and E2 = 1090°.
8. Use superposition to find vab for In the circuit of Figure 5.59. E1 = 10° and
E2 = 245°.
9. In the circuit of Figure 5.60, use superposition to find vb and vcd. The sources
are in phase.
188
Figure 5.58
Figure 5.59
Figure 5.60",ACElectricalCircuitAnalysis.pdf
"10. In the circuit of Figure 5.60, use superposition to find the current through 
the capacitor. The sources are in phase.
11. Use superposition to find the two source currents for In the circuit of   
Figure 5.61. Source one is 100 mV peak and source two is 500 mV peak (in 
phase).
12. Use superposition to find vcd for the circuit of Figure 5.61. Source one is    
100 mV peak and source two is 1 V peak (in phase).
13. In the circuit of Figure 5.62, use superposition to find vab.
14. In the circuit of Figure 5.62, use superposition to find the current through 
the 15 kΩ resistor.
15. In the circuit of Figure 5.63, use superposition to find vab.
189
Figure 5.61
Figure 5.62
Figure 5.63",ACElectricalCircuitAnalysis.pdf
"16. In the circuit of Figure 5.63, use superposition to find the current flowing 
through the resistor.
17. For the circuit of Figure 5.64, use superposition to find va and vb. The 
sources are in phase.
18. For the circuit of Figure 5.64, use superposition to find the currents through 
the inductor and capacitor. The sources are in phase.
19. Use superposition in the circuit of Figure 5.65 to find the currents through 
the inductor and capacitor. I1 = 145° and I2 = 245°.
20. Use superposition in the circuit of Figure 5.65 to find vab and vbc. I1 = 10° 
and I2 = 290°.
21. In the circuit of Figure 5.66, Use superposition to find vbc. I1 = 100° and 
I2 = 60°.
22. In the circuit of Figure 5.66, Use superposition to find the current flowing 
through the 2 Ω resistor. I1 = 4120° and I2 = 60°.
190
Figure 5.64
Figure 5.65
Figure 5.66",ACElectricalCircuitAnalysis.pdf
"23. Use superposition to determine the current of source E in the circuit of 
Figure 5.67. E = 40180° and I = 20E−30°. 
24. Use superposition to determine vac in the circuit of Figure 5.67. E = 280° 
and I = 8E−3−180°.
25. Use superposition to determine vb in the circuit of Figure 5.68. I = 3E−30° 
and E = 90°.
26. Use superposition to determine the inductor current in the circuit of     
Figure 5.68. I = 4E−30° and E = 18−45°.
27. For the circuit of Figure 5.69, use superposition to determine the inductor 
current. I = 100E−30° and E = 260°.
28. For the circuit of Figure 5.69, use superposition to determine vab.                  
I = 50E−30° and E = 1890°.
191
Figure 5.67
Figure 5.68
Figure 5.69",ACElectricalCircuitAnalysis.pdf
"29. Use superposition to determine vab in the circuit of Figure 5.70.                     
I = 10E−30° and E = 120°.
30. Use superposition to determine the capacitor current in the circuit of    
Figure 5.70. I = 5E−30° and  E = 18120°.
31. For the circuit of Figure 5.71, determine the Thévenin equivalent that drives 
the 20 nF capacitor.
32. Given the circuit of Figure 5.71, determine the Norton equivalent that drives
the 20 nF capacitor.
33. For the circuit of Figure 5.72, determine the Thévenin and Norton 
equivalents that drive the 600 Ω resistor if the source E = 120°.
34. Given the circuit of Figure 5.72, determine the Thévenin equivalent that 
drives the j1 kΩ inductive reactance if E = 90°.
35. Given the circuit of Figure 5.72, determine the Norton equivalent that drives
the j2.5 kΩ inductive reactance if E = 2445°.
192
Figure 5.70
Figure 5.71
Figure 5.72",ACElectricalCircuitAnalysis.pdf
"36. Use Thévenin's theorem to find vb in the circuit of Figure 5.72 if                  
E = 180°.
37. Use Thévenin's theorem to find vb in the circuit of Figure 5.73.
38. Determine the Thévenin equivalent that drives the 3.9 kΩ + j1 kΩ combo in 
the circuit of Figure 5.73. Does this combo's impedance achieve maximum 
load power? If not, what combo will achieve maximum power and what is 
the resulting power?
39. Determine the Norton equivalent that drives the 500 Ω resistor in the circuit 
of Figure 5.73. Determine the value of component(s) that when placed in 
series with the 500 Ω resistor will achieve maximum load power (i.e., for 
the combo as the load).
40. For the circuit of Figure 5.74, determine the Thévenin and Norton 
equivalents that drive the combo of 36 Ω + j100 Ω. Does this combo achieve
maximum load power? If not, what combo will achieve maximum power 
and what is the resulting power?
193
Figure 5.73
Figure 5.74",ACElectricalCircuitAnalysis.pdf
"41. For the circuit of Figure 5.75, determine the Thévenin and Norton 
equivalents that drive the combo of 300 Ω in parallel with −j1500 Ω. Does 
this combo achieve maximum load power? If not, what combo will achieve 
maximum power and what is the resulting power? E = 1200°.
42. For the circuit of Figure 5.76, determine the Thévenin and Norton 
equivalents that drive the combo of 4.7 kΩ in parallel with j300 Ω. Does this
combo achieve maximum load power? If not, what combo will achieve 
maximum power and what is the resulting power? I = 200E−30°.
43. Determine the equivalent Y (T) network for the circuit of Figure 5.77.        
R1 = R2 = R3 = 10 kΩ and XL1 = XL2 = XL3 = j10 kΩ.
194
Figure 5.75
Figure 5.76
Figure 5.77",ACElectricalCircuitAnalysis.pdf
"44. Determine the equivalent Y (T) network for the circuit of Figure 5.78.
45. Determine the equivalent Y (T) network for the circuit of Figure 5.79.        
R1 = R2 = R3 = 4 kΩ and XC1 = XC2 = XC3 = −j3 kΩ.
46. Determine the equivalent Y (T) network for the circuit of Figure 5.80.
47. Determine the equivalent delta (pi) network for the circuit of Figure 5.81.
195
Figure 5.78
Figure 5.79
Figure 5.80
Figure 5.81",ACElectricalCircuitAnalysis.pdf
"48. Determine the equivalent delta (pi) network for the circuit of Figure 5.82.
49. Determine the equivalent delta (pi) network for the circuit of Figure 5.83.
50. Determine the equivalent delta (pi) network for the circuit of Figure 5.84.
196
Figure 5.82
Figure 5.83
Figure 5.84",ACElectricalCircuitAnalysis.pdf
"51. Find voltage vbc in the circuit of Figure 5.85 through the use of one or more 
delta-Y conversions. E = 100°, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ,               
XC = −j4 kΩ and XL = j8 kΩ.
52. Find voltage vbc in the circuit of Figure 5.86 through the use of one or more 
delta-Y conversions. E = 200°, R1 = 1 kΩ, R2 = 8 kΩ, R3 = 3 kΩ,               
XC = −j4 kΩ and XL = j2 kΩ.
DesignDesign  
53. Design an equivalent current source for Figure 5.87. E = 1290°, R = 1 kΩ 
and XC = − j200 Ω. The source frequency is 10 kHz.
197
Figure 5.85
Figure 5.86
Figure 5. 87",ACElectricalCircuitAnalysis.pdf
"54. Design an equivalent current source for Figure 5.87 if E = 100°,                
R = 2.2 kΩ and C = 100 nF. The source frequency is 1 kHz.
55. Design an equivalent current source for Figure 5.88 if E = 10°, R = 600 Ω 
and L = 2 mH. The source frequency is 20 kHz.
56. Design an equivalent current source for Figure 5.88. E = 290°, R = 10 kΩ 
and XL =  j900 Ω.
57. Design an equivalent voltage source for Figure 5.89. I = 300E−30°,           
R = 4.3 kΩ and XC = −j5 kΩ.
58. Design an equivalent voltage source for Figure 5.89 if I = 100E−3120°,    
R = 75 Ω and L = 1 mH. The source frequency is 10 kHz.
59. Design an equivalent voltage source for Figure 5.90 if I = 10E−30°,          
R = 9.1 kΩ and L = 5 mH. The source frequency is 100 kHz.
60. Design an equivalent voltage source for Figure 5.90 if I = 50E−30°,          
R = 560 Ω and XL = j350 Ω.
198
Figure 5.88
Figure 5.89
Figure 5.90",ACElectricalCircuitAnalysis.pdf
"61. Reconfigure the circuit of Figure 5.91 so that it uses only voltage sources. 
Express all new component and source values in terms of the original labels.
62. Reconfigure the circuit of Figure 5.92 so that it uses only current sources. 
Express all new component and source values in terms of the original labels.
63. Redesign the circuit of Figure 5.56 so that it uses only current sources and 
produces the same node voltages as the original circuit.
64. Consider the 600 Ω resistor to be the load in Figure 5.72. Determine a new 
value for the load in order to achieve maximum load power. Also determine 
the maximum load power.
65. Using Thévenin's theorem with the circuit of Figure 5.73, determine a new 
value of capacitive reactance such that it cancels the Thévenin reactance.
66. Using Norton's theorem with the circuit of Figure 5.74, determine a new 
value of inductive reactance such that the inductor current is 1 mA in 
magnitude.
199
Figure 5.91
Figure 5.92",ACElectricalCircuitAnalysis.pdf
"ChallengeChallenge
67. Redesign the circuit of problem 7 (Figure 5.59) so that it uses only current 
sources and produces the same node voltages as the original circuit.
68. In the circuit of Figure 5.93, use any method or combination of methods to 
determine vab. I1 = 0.050°, I2 = 0.10° and I3 = 0.290°.  
69. In the circuit of Figure 5.94, use any method or combination of methods to 
determine vab. E1 = 50°, E2 = 1090° and E3 = 150°. 
70. In the circuit of Figure 5.94, use any method or combination of methods to 
determine vce. E1 = 150°, E2 = 3090° and E3 = 450°. 
200
Figure 5.93
Figure 5.94",ACElectricalCircuitAnalysis.pdf
"71. Use any method or combination of methods in the circuit of Figure 5.95 to 
determine vad. E1 = 900°, E2 = 1200° and I = 400E−3180°. 
72. Consider the 2.7 kΩ + j4 kΩ combo to be the load in Figure 5.60. Determine
if this value achieves maximum load power. If not, determine a new value 
for the load in order to achieve maximum load power. Also determine the 
maximum load power.
73. In the circuit of Figure 5.96, assume the source E is 120 volts RMS at 60 Hz.
Determine the value for the load, Z, that will produce maximum load power.
Express Z in terms of a resistor and either an inductor or capacitor. Further, 
specify both the series and parallel equivalents for the load.
74. Convert the circuit of Figure 5.97 into the equivalent delta configuration.
201
Figure 5.95
Figure 5.96
Figure 5.97",ACElectricalCircuitAnalysis.pdf
"75. Find the Thévenin equivalent looking into nodes a and b for the circuit of 
Figure 5.98. I = 40°.
76. Find voltage vbc in the circuit of Figure 5.99 through the use of one or more 
delta-Y conversions. E = 1000°, R1 = R2 = 2 kΩ, R3 = 3 kΩ, R4 = 10 kΩ, 
R5 = 5 kΩ, XC1 = XC2 = −j2 kΩ.
77. Given the circuit of Figure 5.100, determine an equivalent circuit using a 
single voltage source. E1 = 1000°, E2 = 60180°, E3 = 4090°,            
E4 = 750°.
202
Figure 5.98
Figure 5.99
Figure 5.100",ACElectricalCircuitAnalysis.pdf
"78. As mentioned earlier, it is possible that certain AC Y and delta networks 
cannot be converted. Consider the circuit of Figure 5.101. Can this circuit be
converted with a practical outcome? Why/why not?
SimulationSimulation
79. Verify the voltage computed for problem 1 by running a transient analysis.
80. Verify the voltage computed for problem 4 by running a transient analysis.
81. Using multiple transient analysis simulations, compare the original circuit of
problem 54 to its converted equivalent. Do this by connecting various 
components to the output terminals, trying several different impedance 
values and checking to see if the two circuits always produce the same 
voltage across this impedance.
82. Using multiple transient analysis simulations, compare the original circuit of
problem 59 to its converted equivalent. Do this by connecting various 
components to the output terminals, trying several different impedance",ACElectricalCircuitAnalysis.pdf
"problem 59 to its converted equivalent. Do this by connecting various 
components to the output terminals, trying several different impedance 
values and checking to see if the two circuits always produce the same 
voltage across this impedance.
83. Run a transient analysis to verify the design of problem 70.
84. Run a transient analysis to verify the design of problem 71.
203
Figure 5.101",ACElectricalCircuitAnalysis.pdf
"6 6 Nodal and Mesh AnalysisNodal and Mesh Analysis
6.0 Chapter Learning Objectives6.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Utilize nodal analysis techniques to solve for voltages in multi-source series-parallel RLC networks.
• Utilize mesh analysis techniques to solve for currents in multi-source series-parallel RLC networks.
• Analyze networks that use dependent voltage and/or current sources.
6.1 Introduction 6.1 Introduction 
This chapter presents methods for the analysis of AC circuits that employ resistors, capacitors and inductors 
along with any number of voltage and/or current sources. The methods of interest are nodal analysis and mesh 
analysis. Nodal analysis is the most general technique and can be applied to virtually any circuit. Mesh analysis 
is nearly as versatile and works well if only voltage sources are present. Both analysis methods generate a",ACElectricalCircuitAnalysis.pdf
"is nearly as versatile and works well if only voltage sources are present. Both analysis methods generate a 
system of simultaneous linear equations that are used to solve the circuit for desired voltages or currents. That is,
the system generates a set of values, either currents or node voltages, rather than individual currents or voltages. 
There are several methods that can be used to solve the simultaneous equations. These include substitution, 
Gauss-Jordan elimination and expansion by minors. These methods are reviewed in Appendix B and are not 
covered in this chapter. Instead, to focus on the circuit analysis aspects with minimal distraction, the 
explanations and examples will simply detail the process of examining the circuit and developing the system of 
equations. The specific technique employed to solve these simultaneous equations depends solely on your 
personal preferences.",ACElectricalCircuitAnalysis.pdf
"equations. The specific technique employed to solve these simultaneous equations depends solely on your 
personal preferences. 
At this point in the study of AC circuits, it is particularly efficient to obtain an advanced scientific calculator that 
can solve the system of equations directly versus working through the solution manually. By doing so, you can 
spend your time more effectively; meaning, mastering the process of circuit analysis and creating the equations. 
Manual solution techniques, though not necessarily difficult, can be tedious, time consuming and error prone. 
Indeed, on larger circuits, there can be a 10:1 differential in time when using a capable calculator versus a 
standard scientific calculator9. If you have not already done so, you should consider obtaining a calculator that 
can solve simultaneous equations with complex coefficients (i.e., the complex real/imaginary quantities we have",ACElectricalCircuitAnalysis.pdf
"can solve simultaneous equations with complex coefficients (i.e., the complex real/imaginary quantities we have
been using). Such calculators can be expensive when purchased new, such as the Texas Instruments TI-89 and 
Nspire models. On the used market, perfectly satisfactory older models such as the TI-85 and TI-86 can be 
found at considerable discount. Another model to consider is the Casio FX-9750GII, although it is not quite as 
powerful as some of the other units mentioned.
9 That's like doing three times as many problems in one-third the time. We don't often get these kinds of opportunities.
204",ACElectricalCircuitAnalysis.pdf
"Along with nodal and mesh, we shall also introduce the concept of dependent AC 
sources. Dependent sources do not exhibit a fixed value, but rather the current or 
voltage is dependent on some other current or voltage in the circuit. What makes this
interesting is that this controlling current or voltage may itself be affected by the 
value produced by the dependent source. Dependent sources are not lab instruments,
like signal or function generators. Instead, they are used to model the behavior of 
active electronic devices such as bipolar and field effect transistors. Mastering the 
analysis of circuits using dependent sources is critical to the understanding of active 
circuitry that use transistors and similar devices10.
6.2 Nodal Analysis6.2 Nodal Analysis
Nodal analysis can be considered a universal solution technique as there are no 
practical circuit configurations that it cannot handle. It does not matter if there are",ACElectricalCircuitAnalysis.pdf
"practical circuit configurations that it cannot handle. It does not matter if there are 
multiple sources or if there are complex configurations that cannot be reduced using 
series-parallel simplification techniques, nodal analysis can handle them all. Further,
nodal analysis tends to “give us what we want”, namely, a set of node voltages for 
the circuit. Once the node voltages are obtained, finding any branch currents or 
component powers becomes an almost trivial exercise. Nodal analysis relies on the 
application of Kirchhoff's current law to create a series of node equations that can be
solved for node voltages. These equations are based on Ohm's law and will be of the 
form i = v/Z, or more generally, i = (1/ZX)∙vA + (1/ZY)∙vB + (1/ZZ)∙vC... 
We will examine two variations on the theme; first, a general version that can be 
used with both voltage and current sources, and a second somewhat quicker version 
that can be used with circuits only driven by current sources.",ACElectricalCircuitAnalysis.pdf
"used with both voltage and current sources, and a second somewhat quicker version 
that can be used with circuits only driven by current sources. 
General MethodGeneral Method
Consider the circuit shown in Figure 6.1. We begin by labeling connection nodes. 
We are interested in identifying current junctions, that is, places where currents can 
combine or split. These are also known as summing nodes and are circled in blue on 
the figure. We do not concern ourselves with points where just two components 
connect without any other connection, such as points a and c. Once the proper nodes
are identified, reference current directions are assigned. The reference current 
directions are chosen arbitrarily and for convenience. They may be the opposite of 
reality. This is not a problem. If we assign directions that are reversed, we'll simply 
wind up with a current version of a double negative, and the computed node voltages
will work out just fine.",ACElectricalCircuitAnalysis.pdf
"wind up with a current version of a double negative, and the computed node voltages
will work out just fine. 
10 For more on transistors and other semiconductors, see Semiconductor Devices: Theory 
and Application. Another free OER text by the author.
205",ACElectricalCircuitAnalysis.pdf
"One node is chosen as the reference. This is the point to which all other node 
voltages are measured against. Typically, the reference node is ground, although it 
does not have to be.
We now write a current summation equation for each summing node, except for the 
reference node. In this circuit there is only one node where currents combine (other 
than ground) and that's node b. Points a and c are places where components connect,
but they are not summing nodes, so we can ignore them for now. Using KCL on 
node b we can say:
   i1 + i2 = i3
Now we'll describe these currents in terms of the source and node voltages, and 
associated components. For example, i3 is the node b voltage divided by −jXC while 
i1 is the voltage across R divided by R. This voltage is va − vb.
va −vb
R + vc −vb
jX L
= vb
− jX C
Noting that va = E1 and vc = E2, with a little algebra this can be reduced to:
(
1
R )E1 +(
1
jX L )E2 = (
1
R + 1
jX L
+ 1
− jX C )vb",ACElectricalCircuitAnalysis.pdf
"= vb
− jX C
Noting that va = E1 and vc = E2, with a little algebra this can be reduced to:
(
1
R )E1 +(
1
jX L )E2 = (
1
R + 1
jX L
+ 1
− jX C )vb
 All quantities are known except for vb and thus it is easily found with a little more 
algebra.  If there had been more nodes, there would have been more equations and 
more unknowns, one for each node. As we shall see, this conductance-voltage 
product format turns out to be a convenient way of writing these equations. Also, 
note that the first two terms on the left reduce to fixed current values.
For current sources, the process is similar but a bit more direct. Consider the circuit 
of Figure 6.2. We start as before, identifying nodes and labeling currents. We then 
write the current summation equations at each node (except for ground). We 
consider currents entering a node as positive and exiting as negative. There are two 
nodes of interest here, and thus, two equations each with two unknowns will be 
generated. 
206
Figure 6.1",ACElectricalCircuitAnalysis.pdf
"nodes of interest here, and thus, two equations each with two unknowns will be 
generated. 
206
Figure 6.1
A simple dual voltage source 
circuit with the currents and 
nodes defined.",ACElectricalCircuitAnalysis.pdf
"Node a: I1 = i3 + i4
Node b: i3 + I2 = i5,  and rearranging in terms of the fixed source,
Node b: I2 = −i3 + i5
The currents are then described by their Ohm's law equivalents:
Node a:  I 1 = va −vb
R2
+ va
R1
Node b:  I 2 = −va −vb
R2
+ vb
jX L
Expanding and collecting terms yields:
Node a:  I 1 = (
1
R1
+ 1
R2 )va −(
1
R2 )vb
Node b:  I 2 = −(
1
R2 )va +(
1
R2
+ 1
jX L )vb
As the impedance values and currents are known, simultaneous equation solution 
techniques may be used to solve for the node voltages. Once again, there are as 
many equations as node voltages. 
One practical point before continuing: It is very important that the coefficients for
the various node voltage terms “line up” when the final system of equations is
written out. That is, there should be a column for the va terms, a column for the vb
terms, and so on. They should not be written out in random order, but rather
following the style shown in Figure 6.3. This format will make it much it easier to",ACElectricalCircuitAnalysis.pdf
"following the style shown in Figure 6.3. This format will make it much it easier to
enter the coefficients into a calculator or solve manually. Further, the set of
coefficients must show diagonal symmetry. That is, if we draw a major diagonal
from upper-left to lower-right (red), whatever coefficients are above-right from the
diagonal should be mirrored below-left of the diagonal (blue, purple, green). If the
set of values does not show diagonal symmetry, an error has been made. You must 
go back and recheck the original node summations. Simple as that. Even the simple 
2x2 of Figure 6.2 shows this symmetry (namely, the coefficient of −1/R2 for vb in the
first equation and va in the second).
207
Figure 6.2
A dual current source circuit 
with the currents and nodes 
defined. 
Figure 6.3
Diagonal symmetry.
(1+j2) a
-(-j2) a
-(1) a
-(-j2) b
(2-j3) b
-(-j3) b
-(1) c
-(-j3) c
(5-j3) c
v vv
v vv
v vv",ACElectricalCircuitAnalysis.pdf
"Example 6.1
Determine the voltage across the inductor in the circuit of Figure 6.4. Source
one is 50° volts RMS and source two 290° volts RMS.
Other than ground, there is only one current summing node in this circuit, 
and that's the junction at the top of the inductor. We will refer to this junction
as node a. Following the outline of Figure 6.1, we define three currents; i1 
entering from the left, i2 entering from the right, and i3 exiting down through 
the inductor. 
   i1 + i2 = i3
Next, these currents are described in terms of the voltages and components. 
We'll number the resistors from left to right.
E1 −va
R1 − jX C
+ E2 −va
R2
= va
jX L
This can be rearranged as:
(
1
R1 − jX C )E1 +(
1
R2 )E2 = (
1
R1 − jX C
+ 1
R2
+ 1
jX L )va
Populate with values:
(
1
500 − j 300Ω)5 0° V +(
1
400 Ω)2 90° V = (
1
500 − j 300Ω + 1
400 Ω + 1
j 200Ω)va
This simplifies to:
8.575E-3 31° A +5E-3 90° A = (5.72E-3−46 ° S)va
Solving for the unknown, we find that va = 2.08798° volts RMS.
208",ACElectricalCircuitAnalysis.pdf
"j 200Ω)va
This simplifies to:
8.575E-3 31° A +5E-3 90° A = (5.72E-3−46 ° S)va
Solving for the unknown, we find that va = 2.08798° volts RMS.
208
Figure 6.4
Circuit for Example 6.1.",ACElectricalCircuitAnalysis.pdf
"Example 6.2
In the circuit of Figure 6.5, determine va and vb. E is 200° volts peak while 
I is 0.10° amps peak. The system frequency is 2 kHz.
There are two nodes of interest here, other than ground. This means we will 
generate two equations with two unknowns (va and vb). Using the standard 
reactance formulas, the inductive and capacitive reactances are found to be 
j125.7 Ω and −j159.2 Ω, respectively. If we assume the reference direction 
for current is from node a to node b, and that the current flow through the 
two center resistors is downward, the equations are:
Node a:  20  0° V −va
100 Ω+ j 125.7Ω = va
250 Ω + va −vb
− j 159.2 Ω
Node b:  va −vb
− j159.2 Ω +0.1 0° A = vb
400Ω
Expanding and collecting terms yields:
Node a:
0.1245 −51.5° A =(
1
250Ω + 1
100Ω+ j 125.7Ω + 1
− j 159.2Ω)va − (
1
− j 159.2Ω)vb
Node b:  
0.1 0° A = − (
1
− j 159.2Ω)va +(
1
400Ω + 1
− j 159.2Ω)vb
These are simplified, ready for manipulation (note diagonal symmetry).",ACElectricalCircuitAnalysis.pdf
"Node b:  
0.1 0° A = − (
1
− j 159.2Ω)va +(
1
400Ω + 1
− j 159.2Ω)vb
These are simplified, ready for manipulation (note diagonal symmetry).
0.1245 −51.5° A = (8E-3 10.1 ° S)va − (6.281E-3  90° S)vb
0.1 0° A = − (6.281E-3  90 ° S)va +(6.76E-3 68.3 ° S)vb
After solving the system of equations, we see that va = 16.240.09° volts 
and vb = 20.99−22.3° volts.
209
Figure 6.5
Circuit for Example 6.2.",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
To verify the results of the preceding example, the circuit of Figure 6.5 is captured in
a simulator as shown in Figure 6.6.
 
A transient analysis is performed on the circuit. Node voltages 1, 3 and 4 are plotted,
corresponding to the voltage source and nodes a and b, respectively, in Figure 6.7. 
 
210
Figure 6.6
The circuit of Example 6.2 
captured in a simulator.
Figure 6.7
Simulation results for the 
circuit of Figure 6.6.",ACElectricalCircuitAnalysis.pdf
"The amplitudes are just as computed. Node voltage a appears to be nearly in phase 
with the voltage source, as expected. Node voltage b lags the source by between 
one-quarter to one-third of a division, or some 30 microseconds. For a 2 kHz source,
this translates to around −22 degrees, verifying the calculated result.
Example 6.3
For the circuit of Figure 6.8, find va and vb. The system frequency is 1 kHz.  
I1 = 2.50° A and I2 = 10° A.
We will generate two equations with two unknowns, va and vb. The reactance
formulas yield j6.28 Ω and −j15.9 Ω for the inductor and capacitor. If we 
assume the reference direction for current is from node a to node b, and that 
the current flow through the capacitor and inductor is from nodes a and b 
downward, the equations are:
Node a:  2.5 0° A = va
− j 15.9Ω + va −vb
10 Ω
Node b:  va −vb
10Ω = 1 0° A + vb
4Ω+ j 6.28Ω
Expanding and collecting terms yields (note diagonal symmetry):
2.5 0° A = (
1
10Ω + 1
− j 15.9Ω)va −(
1
10Ω)vb
−1 0° A =−(
1",ACElectricalCircuitAnalysis.pdf
"4Ω+ j 6.28Ω
Expanding and collecting terms yields (note diagonal symmetry):
2.5 0° A = (
1
10Ω + 1
− j 15.9Ω)va −(
1
10Ω)vb
−1 0° A =−(
1
10Ω)va + (
1
10Ω + 1
4Ω + j 6.28Ω )vb
2.5 0° A = (0.118  32.2 ° S)va − (0.1 0° S) vb
−1 0° A = − (0.1 0° S) va +(0.206 −33.3° S) vb
The results are: va = 30.39-38.7° volts and vb = 11.37−20.8° volts.
211
Figure 6.8
Circuit for Example 6.3.",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
To verify the results of the preceding example, the circuit of Figure 6.8 is captured in
a simulator as shown in Figure 6.9.
 
A transient analysis is performed on the circuit. Node voltages 1 and 2 (i.e., nodes a 
and b, respectively) are plotted in Figure 6.10.
212
Figure 6.9
The circuit of Example 6.3 
captured in a simulator.
Figure 6.10
Simulation results for the 
circuit of Figure 6.9.",ACElectricalCircuitAnalysis.pdf
"The simulation results agree nicely with the computed values in terms of both 
amplitude and phase.
Inspection MethodInspection Method
The system of equations can be obtained directly through inspection if the circuit 
contains current sources and no voltage sources. Let's take another look at the 
equations developed in the preceding example. For convenience, the circuit is 
reproduced in Figure 6.11 with reactance values. I1 = 2.50° A and I2 = 10° A.
2.5 0° A = (
1
10Ω + 1
− j 15.9Ω)va −(
1
10Ω)vb
−1 0° A =−(
1
10Ω)va + (
1
10Ω + 1
4Ω + j 6.28Ω)vb
The top equation was built around a current summation at node a while the bottom 
was built around a summation at node b. The first thing that might be apparent is 
that on the left of the equals signs are the current sources connected to these nodes. 
Positive means the current is entering while negative denotes an exiting current. The 
second thing is that, for the node of interest (node a for the top equation, node b for",ACElectricalCircuitAnalysis.pdf
"second thing is that, for the node of interest (node a for the top equation, node b for 
the bottom), the coefficients represent the items connected to that particular node. 
For example, in the top equation, the components connected to node a are the 10 Ω 
resistor and the − j15.9 Ω reactance. Likewise, in the bottom equation, the 
components connected to node b are the 10 Ω resistor and the 4 + j6.28 Ω 
impedance. The third thing is that the remaining coefficients consist of the 
components that are in common between the node of interest and the other node 
(i.e., 10 Ω connects a to b for the first equation, and also connects b to a for the 
second equation). These other connections always show up as negative. The reason 
for this should be apparent if you examine the structure of the original equations 
from Example 6.3.
If there is no bridging element between a node and the node of interest, then that",ACElectricalCircuitAnalysis.pdf
"from Example 6.3.
If there is no bridging element between a node and the node of interest, then that 
coefficient will be zero. If voltage sources exist in the circuit, source conversions 
can be used to obtain an equivalent circuit that uses only current sources. 
213
Figure 6.11
Circuit of Example 6.3 with 
reactances.",ACElectricalCircuitAnalysis.pdf
"The huge advantage of the inspection method is that it cuts out a time consuming 
and error prone section of the process, namely converting the original KCL 
summations into a set of simplified equations with coefficients for each unknown. 
The inspection method generates the equations directly. To further speed the process,
it can be useful to turn each impedance value into a corresponding admittance value 
before creating the equations. In this way, the reciprocals are computed once for 
each item rather than multiple times in multiple equations. Finally, remember that 
the resulting set of equations must exhibit diagonal symmetry, as shown back in 
Figure 6.3.
The inspection method is summarized as follows:
1. Verify that the circuit uses only current sources and no voltage sources. If 
voltage sources exist, they must be converted to current sources before 
proceeding.
2. Find all of the current summing nodes and number (or letter) them. Also",ACElectricalCircuitAnalysis.pdf
"proceeding.
2. Find all of the current summing nodes and number (or letter) them. Also 
decide on the reference node (usually ground).
3. To generate an equation, locate the first node. This is the node of interest and
the next few steps will be associated with it.
4. Sum the current sources feeding the node of interest. Entering is deemed 
positive while exiting is deemed negative. The sum is placed on one side of 
the equals sign.
5. Next, find all of the impedances connected to the node of interest and write 
them as a sum of admittances on the other side of the equals sign, the group 
being multiplied by this node's voltage (e.g., v1). That makes one term.
6. Now for the other terms. Find all of the admittances that are connected 
between the node of interest and the next node (e.g., node 2). Sum these 
together and multiply the group by this other node's voltage (e.g., v2). 
Subtract that product from the equation built so far. Repeat this process until",ACElectricalCircuitAnalysis.pdf
"Subtract that product from the equation built so far. Repeat this process until
all of the other nodes have been examined (except ground). If there are no 
common impedances between the node of interest and the other node, use 
zero for the coefficient of that node's voltage. Once all other nodes are 
considered, this equation is finished.
7. Go to the next node and treat this as the new node of interest. 
8. Repeat steps 4 through 7 until all nodes have been treated as the node of 
interest. Each iteration creates a new equation. There will be as many 
equations as there are nodes, less the reference node. Check for diagonal 
symmetry and solve.
The inspection method is best observed in action, and is used in the following 
example. 
214",ACElectricalCircuitAnalysis.pdf
"Example 6.4
Write the system of equations for the circuit of Figure 6.12.  I1 = 100° A 
and I2 = 490° A.
We begin at node a, the first node of interest. Find all of the current sources 
connected to this node. All we have is I1. It is exiting, and thus negative.
−10  0° A = ...
Now find all of the items connected to this node and create a sum of 
admittances.
−10  0° A =(
1
4Ω + 1
2Ω + 1
8Ω)va ...
Include the terms that are common between node a and node b. This is 
negative.
−10  0° A =(
1
4Ω + 1
2Ω + 1
8Ω)va −(
1
2Ω)vb ...
And finally, include the terms common between node a and node c. This is 
also negative.
−10  0° A =(
1
4Ω + 1
2Ω + 1
8Ω)va −(
1
2Ω)vb −(
1
8Ω)vc
The first equation is done. We now make node b the node of interest and 
repeat the process.
215
Figure 6.12
Circuit for Example 6.4.",ACElectricalCircuitAnalysis.pdf
"Find all of the current sources connected to this node. There are none.
0 = ...
Find all of the items connected to this node and create a sum of admittances.
0 = ...+(
1
1Ω + 1
2Ω + 1
− j 5Ω)vb ...
Include the terms that are common between node a and node b. This is 
negative and goes into the lead (a before b) to keep everything nicely lined 
up.
0 =−(
1
2Ω)va +(
1
1Ω + 1
2Ω + 1
− j 5Ω)vb ...
Now include the terms common between node b and node c. This is also 
negative and is inserted at the tail (c after b).
0 =−(
1
2Ω)va +(
1
1Ω + 1
2Ω + 1
− j 5Ω)vb −(
1
1Ω)vc
The second equation is finished. We now make node c the node of interest 
and repeat the process for the final time.
Find all of the current sources connected to node c. We have both I1 and I2 
entering.
10 0° A+4  90 ° A = ...
This current is equivalent to 10.7721.8°. Now find all of the items 
connected to this node and create a sum of admittances.
10.77 21.8° A = ...(
1
1Ω + 1
8Ω + 1
j 10Ω)vc",ACElectricalCircuitAnalysis.pdf
"connected to this node and create a sum of admittances.
10.77 21.8° A = ...(
1
1Ω + 1
8Ω + 1
j 10Ω)vc
Include the terms that are common between node c and node a. This is 
negative and goes into the lead (a before c).
10.77 21.8° A =−(
1
8Ω)va ... +(
1
1Ω + 1
8Ω + 1
j 10 Ω)vc
Now include the terms common between node b and node c. This is also 
negative and is inserted in the middle (b before c).
216",ACElectricalCircuitAnalysis.pdf
"10.77 21.8° A =−(
1
8Ω)va −(
1
1Ω)vb +(
1
1Ω + 1
8Ω + 1
j 10 Ω)vc
The third and final equation is finished. The completed set of equations is:
−10 0° A = (
1
4Ω + 1
2Ω + 1
8Ω)va −(
1
2Ω)vb −(
1
8Ω)vc
0 =−(
1
2Ω)va +(
1
1Ω + 1
2Ω + 1
− j 5Ω)vb −(
1
1Ω)vc
10.77  21.8° A =−(
1
8Ω)va −(
1
1Ω)vb +(
1
1Ω + 1
8Ω + 1
j 10Ω)vc
Note that the set exhibits diagonal symmetry and that all coefficient groups 
are negative except for those along the major diagonal. Consequently, the 
coefficient groups may now be simplified to obtain single coefficients for 
the unknowns, and the equations are ready for solution. The results are: 
va = 10.972.8° volts, vb = 23.634.7° volts and vc = 31.237.3° volts. 
These values can be crosschecked by using them to find the currents through
each component, and then verifying KCL for each node.
Although this example may appear to be somewhat long winded, with a little
practice the process will become second nature. At that point, the set of",ACElectricalCircuitAnalysis.pdf
"Although this example may appear to be somewhat long winded, with a little
practice the process will become second nature. At that point, the set of 
equations can be created quickly and with little possibility of error, even for 
large circuits with many nodes. 
Using Source ConversionsUsing Source Conversions
As mentioned previously, given circuits with voltage sources, it may be easier to 
convert them to current sources and then apply the inspection technique rather than 
using the general approach outlined initially. There is one trap to watch out for when
using source conversions: the voltage across or current through a converted 
component will most likely not be the same as the voltage or current in the 
original circuit. This is because the location of the converted component will have 
changed. For example, the circuit of Figure 6.5 (Example 6.2) could be solved using 
the inspection method of nodal analysis by converting the voltage source and its",ACElectricalCircuitAnalysis.pdf
"the inspection method of nodal analysis by converting the voltage source and its 
associated impedance of the 100 Ω resistor in series with the 10 mH inductor into a 
current source. Although the associated impedance still connects to the converted 
source, the other end no longer connects to node a. Rather, it would connect to 
ground. Therefore, the voltage drop across this impedance in the converted circuit is 
not likely to equal the voltage drop seen across it in the original circuit (the only way
they would be equal is if the voltage source E turned out to be 0). 
217",ACElectricalCircuitAnalysis.pdf
"SupernodeSupernode
From time to time you may see a circuit utilizing an ideal voltage source like the
one shown in Figure 6.13. That is, this voltage source does not have a series
impedance associated with it. Without that impedance, it becomes impossible to
create an expression for the current passing through the source using the general
method, and impossible to convert the voltage source into a current source in
order to use the inspection method. There are a few of ways out of this quandary.
The first way is to recognize that all realistic sources have some internal
impedance, so we simply add a very small resistor in series with the source so
that a source conversion is possible. Of course, not just any resistor will do. In
order to maintain accuracy, the newly added resistance has to be much smaller
than any surrounding resistances or reactances. A reduction by two orders of
magnitude generally yields a variation smaller than that produced by component",ACElectricalCircuitAnalysis.pdf
"magnitude generally yields a variation smaller than that produced by component
tolerances in all but high precision circuits and will usually do the trick. Still
smaller values will further increase accuracy. Another way out is to use a 
supernode. 
A supernode is, in effect, the combination of two nodes. It relies on a simple 
observation. If we examine the circuit of Figure 6.13, the path of the voltage source 
produces identical currents flowing into and out of nodes a and b. As a consequence,
if we treat the two nodes as one big node, then when we write a KCL summation, 
these two terms will cancel. To see just how this works, refer to Figure 6.14.
In this version we have replaced the voltage source with its ideal internal impedance;
a short. We have also labeled the two nodes of interest, a and b, and labeled the 
currents, drawn with convenient reference directions. The specific choice of 
direction will not matter, just use whatever scheme seems appropriate.",ACElectricalCircuitAnalysis.pdf
"currents, drawn with convenient reference directions. The specific choice of 
direction will not matter, just use whatever scheme seems appropriate.
Due to the shorted voltage source, nodes a and b are now the same node. Take a look
at the currents entering and exiting this combined “super” node. On the left side 
(formerly node a) we see a constant current Ix entering while i1, i2 and i3 are exiting. 
On the right side (formerly node b) we see the Iy entering along with i1 and i2, and 
218
Figure 6.13
Bridging voltage source 
without an internal 
impedance.
Figure 6.14
Circuit modified for supernode 
analysis.",ACElectricalCircuitAnalysis.pdf
"exiting we see i4. At this point we'll create an expression where all of currents 
entering the super node are on the left side of the equals sign and all of the exiting 
currents are on the right:
Σiin = Σiout
I x+ I y +i1+i2 = i1+i2+i3+i4
This can be simplified to:
I x+ I y = i3 +i4
Writing this in terms of Ohm's law we have:
I x+I y = 1
− j X C
va + 1
j X L
vb
We also know that va − vb = E from the original circuit. We know this because the 
reference polarity of the source is + toward the a node and − toward the b node. 
Therefore it must be  va − vb and not  vb − va. Assuming all sources and components 
are known, that makes two equations with two unknowns, solvable using 
simultaneous equation techniques. This is illustrated in the following example.
Example 6.5
Find va and vb for the circuit of Figure 6.15. E = 160° volts, Ix = 0.10°
amps and Iy = 0.2590° amps.
The circuit is redrawn in Figure 6.16 with nodes and currents labeled.",ACElectricalCircuitAnalysis.pdf
"amps and Iy = 0.2590° amps.
The circuit is redrawn in Figure 6.16 with nodes and currents labeled.
We short the 16 volt source and write a current summation at the a:b
supernode:
Σiin = Σiout
0.1 0° A+0.25 90° A+i1+i2 = i1+i2+i3+i4
This can be simplified to:
0.2693 68.2° A = i3 +i4
Writing this in terms of Ohm's law we have:
0.2693 68.2° A = 1
− j 100 Ωva + 1
j 500Ω vb
0.2693 68.2° A = j 10 mSva − j 2 mSvb
219
Figure 6.15
Circuit for Example 6.5.",ACElectricalCircuitAnalysis.pdf
"We also know that va − vb = 160° volts. Therefore vb = va − 160° volts.
Substituting this into the prior equation yields:
0.2693 68.2° A = j 10 mSva − j2 mS(va −16 0° V)
0.2693 68.2° A = j 10 mSva − j2 mSva +32E-3  90 ° A
0.2394 67.9° A = j 8 mSva
 
va = 29.92 −22.1° V
We know that vb is 160° volts below va, and thus after subtracting,
we find vb = 16.24−43.8° volts. 
To verify, we will perform a KCL summation at each node. For node 
a, assuming i1 exits as drawn: 
i1 = 0.1 0° A − va
− j 100Ω −va−vb
200 Ω
i1 = 0.1 0° A− 29.92 −22.1° V
− j 100 Ω −29.92 −22.1° V−16.24 −43.8° V
200 Ω
i1 = 0.292 −108 ° A
Doing likewise for node b, and assuming i1 enters as drawn:
i1 =−0.25 90° A+ vb
j 500 Ω −va−vb
20 Ω
i1 =−0.25 90° A+16.24 −43.8 ° V
j 500 Ω − 29.92−22.1° V−16.24 −43.8° V
200Ω
i1 = 0.292 −110 ° A
Other than the small deviation due to accumulated rounding, these currents 
match. That means that the current through the voltage source is verified to",ACElectricalCircuitAnalysis.pdf
"match. That means that the current through the voltage source is verified to 
be the same at both terminals, as it must be.
An alternative to the basic supernode technique is to recognize that the two nodes on
either side of the voltage source are effectively locked together by the source 
voltage. That is, if one of the node voltages is found, then the other may be 
determined by adding or subtracting the source voltage to or from the known node 
voltage, depending on the reference polarity. This idea is exploited by simply 
describing one node voltage in terms of the other at the outset. This will reduce the 
total number of unknowns by one and reduce the system of equations by one. The 
technique is illustrated in the example following.
220
Figure 6.16
Circuit modified for supernode
analysis.",ACElectricalCircuitAnalysis.pdf
"Example 6.6
Find node voltages va, vb and vc for the circuit of Figure 6.17. The sources
are: E = 200° volts and I = 245° amps.
Once again we have a situation of a voltage source lacking a series 
impedance which makes a source conversion impossible. Without having to 
short it and thus treating nodes a and c as an explicit supernode, we can take 
an alternate route. We begin by noting that the currents entering and exiting 
the voltage source must be identical. 
The circuit is redrawn in Figure 6.18 with the nodes and convenient current 
directions labeled. The circuit also uses equivalent conductances and  
susceptances in place of the original resistances and reactances in order to 
speed the process of simplifying the equations. Unlike the basic supernode 
technique, this time the voltage source is left in. 
The key observation is that vc = va − 200° V . In other words, vc is locked to 
va and if we find one of them, we can determine the other. Therefore, instead",ACElectricalCircuitAnalysis.pdf
"va and if we find one of them, we can determine the other. Therefore, instead
of writing three equations using three unknowns, we shall instead refer to 
node c in reference to node a. In other words, wherever we need vc we 
instead shall write va − 200° V . Thus, this three node circuit will only need 
two equations.
We begin at node a and apply KCL as usual.
Σiin = Σiout
i1+i3 = i2
This is expanded using Ohm's law and we solve for i1:
i1 = i2 −i3
i1 = j 0.5S va −0.25S(vb −va)
i1 = (0.25 +j 0.5)S va −0.25S vb
On to node b:
I = i3 +i4
2 45 ° A = 0.25S (vb −va ) +0.1S(vb −vc)
2 45 ° A = 0.25S (vb −va ) +0.1S(vb −(va −20  0° V))
2 45 ° A = 0.25S (vb −va ) +0.1S(vb −va +20 0° V)
2 45 ° A = 0.25S (vb −va ) +0.1S(vb −va) +2 0° A
 
1.531 112.5° A =−0.35S va +0.35S vb
221
Figure 6.17
Circuit for Example 6.6.
Figure 6.18
Circuit of Example 6.6 with 
currents labeled and using 
conductances.",ACElectricalCircuitAnalysis.pdf
"And finally node c: 
i4 = i1 +i5
i1 = i4 −i5
i1 = 0.1S(vb −vc)−(− j 0.2S)vc
i1 = 0.1S(vb −(va−20  0° V)) +j 0.2S(va −20  0° V)
i1 = 0.1S(vb −va +20  0° V) +j0.2S(va −20 0° V)
i1 = 0.1S(vb −va) +j 0.2S va +(2 − j 4)A
 
i1 = (−0.1 +j 0.2)S va +0.1S vb +4.472 −63.4° A
The equations for nodes a and c both equal i1, thus they equal each other. 
(0.25+j 0.5)Sva −0.25Svb = (−0.1 +j 0.2)S va +0.1S vb +4.472 −63.4° A
4.472 −63.4 ° A = (0.35 +j 0.3)S va − 0.35S vb
The final equations are:
4.472 −63.4° A = (0.35 +j 0.3)S va − 0.35S vb
1.531  112.5 ° A =−0.35S va +0.35S vb
The solution is va = 9.823 151.3− ° volts and vb = 10.31 176.2− ° volts. As 
vc is 200° volts less than va, then vc = 29 170.6− ° volts. KCL summations 
at each of the three nodes will verify these values.
6.3 Mesh Analysis6.3 Mesh Analysis
Mesh analysis is similar to nodal analysis in that it can handle complex multi-source 
circuits. In some ways it is the mirror image of nodal analysis. While nodal analysis",ACElectricalCircuitAnalysis.pdf
"circuits. In some ways it is the mirror image of nodal analysis. While nodal analysis 
uses Kirchhoff's current law to create a series of current summations at various 
nodes, mesh analysis uses Kirchhoff's voltage law to create a series of loop 
equations that can be solved for mesh currents. The current through any particular 
component may be a mesh current or a combination of mesh currents. Of course, 
once those currents are found, it is a short hop to find any desired voltage. Mesh 
does have one limitation that nodal doesn't: Mesh analysis requires that the circuit  
be planar. That is, the circuit must be able to be drawn on a flat surface without any 
wires crossing each other. Another way of looking at it is that planar circuits can be 
drawn to appear as a series of boxes butting up against each other. To get a visceral 
idea of this notion, grab a piece of paper and place four dots on it. Try to draw a line",ACElectricalCircuitAnalysis.pdf
"idea of this notion, grab a piece of paper and place four dots on it. Try to draw a line 
from each dot to every other dot but without crossing any lines. After a few tries, 
you should be successful. Now try it with five dots. You can't do it unless you “draw 
in the air” and hop over other lines. Obviously, that's possible with real circuits 
because they're 3D. Therefore there are circuits that cannot be solved using mesh. 
222",ACElectricalCircuitAnalysis.pdf
"Consider the circuit of Figure 6.19. This circuit has two voltage sources and
cannot be simplified further, although it can be solved using either superposition
or nodal analysis. For mesh analysis, we begin by designating a set of current
loops. These loops should be minimal in size and together cover all components
at least once. By convention, the loops are drawn with a clockwise reference
direction. There is nothing magical about them being clockwise, it is just a
matter of consistency. The annotated version of the circuit is redrawn in
Figure 6.20. 
Here we have two loop currents, i1 and i2. Note that all components exist in at
least one loop (and sometimes in more than one loop, like capacitor C).
Depending on circuit values, one or more of these loop directions may in fact be 
opposite of reality. This is not a problem. If the true reference direction is opposite, 
then the currents will show up as negative values, and thus we know that the real",ACElectricalCircuitAnalysis.pdf
"then the currents will show up as negative values, and thus we know that the real 
reference direction is counterclockwise. Just remember that a positive result means a
clockwise direction and negative indicates counterclockwise. 
We begin by writing KVL equations for each loop.
Loop 1: E1 = voltage across R + voltage across XC
Loop 2: −E2 = voltage across XC + voltage across XL  
Note that E2 is negative as i2 is drawn flowing out of its negative terminal. Now 
expand the voltage terms using Ohm's law. The resistor and inductor each see a 
single current, i1 and i2, respectively. The capacitor experiences both currents. From 
the perspective of loop 1, i2 is flowing in the opposite direction. Thus, the net current
is i1 − i2. From the perspective of loop 2, i1 is flowing in the opposing direction and 
thus the net current is i2 − i1. The reference voltage polarities reinforce this notion.
Loop 1: E1 = i1 R + (i1 − i2)(−jXC)
Loop 2: −E2 = i2 (jXL) + (i2 − i1)(−jXC)",ACElectricalCircuitAnalysis.pdf
"Loop 1: E1 = i1 R + (i1 − i2)(−jXC)
Loop 2: −E2 = i2 (jXL) + (i2 − i1)(−jXC)
Multiplying out and collecting terms yields:
Loop 1: E1 = (R −jXC) i1 − (−jXC) i2
Loop 2: −E2 = − (−jXC) i1 + (jXL −jXC) i2
223
Figure 6.19
A simple two-source circuit.
Figure 6.20
Circuit with mesh loops and 
voltage polarities drawn.",ACElectricalCircuitAnalysis.pdf
"As the component values and source voltages are known, we have two equations 
with two unknowns. These can be solved for i1 and i2 using the simultaneous 
equation solution techniques of your choice. 
Example 6.7
For the circuit of Figure 6.21, determine vb and vc. The sources are: 
E1 = 90° volts and E2 = 12−90° volts.
 
We begin by labeling our loops, as shown in Figure 6.22. Each loop will 
generate an equation based on a KVL summation around that loop. We will 
number the components from left to right, as usual.
For loop 1:
E1 −E2 = ( jX L1+R1)i1+R2(i1 −i2)
E1 −E2 = (R1+R2+ jX L1)i1 −R2 i2
9 0° V −12 −90 ° V = (20Ω+80Ω+ j 20Ω)i1 −20Ωi2
15 53.1° V = (100Ω+ j 20Ω)i1 −20Ωi2
Repeat for loop 2:
224
Figure 6.21
Circuit for Example 6.7.
Figure 6.22
Circuit of Figure 6.21 with 
current loops drawn.",ACElectricalCircuitAnalysis.pdf
"E2 = (− jX C + jX L2)i2+R2(i2 −i1)
E 2 =−R2 i1+(R2 − jX C + jX L2)i2
12 −90 ° V = −20Ωi1 +(20Ω − j 75+ j 50Ω)i2
12 −90 ° V = −20Ωi1 +(20Ω − j 25Ω)i2
The two loop equations are:
15 53.1° V = (100 Ω+ j 20Ω)i1 −20Ωi2
12 −90 ° V = −20Ωi1 +(20Ω− j 25Ω)i2
The equations show diagonal symmetry. The currents are i1 = 0.178519.9° 
amps and i2 = 0.3529−21.4° amps. To find the voltages vb and vc, we just 
need to apply Ohm's law. The voltage vc is the potential across the j50 Ω 
inductor. 
vc = i2× jX L2
vc = 0.3529  − 21.4° A× j 50Ω
vc=17.64  68.6 ° V
The potential vb is found similarly.
vb = i2×(− jX C +jX L2)
vb = 0.3529  −21.4 °×(− j 75Ω +j 50 Ω)
vb=8.823 −111.4 ° V
For verification, we can also find vb by subtracting the voltage developed 
across the series inductor/resistor pair from the first source.
vb = E1 − i1×(R1+jX L1)
vb = 9 0° V − 0.1785  19.9° A×(80 +j 20 Ω)
vb=8.823 −111.4 ° V
Example 6.8
In the circuit of Figure 6.23, find vb. E = 100° volts peak at a frequency of",ACElectricalCircuitAnalysis.pdf
"vb=8.823 −111.4 ° V
Example 6.8
In the circuit of Figure 6.23, find vb. E = 100° volts peak at a frequency of 
10 kHz.
The circuit is a bridge network. Even though it has only a single voltage 
source, basic series-parallel techniques will not work here. Nodal analysis 
can also work here as can delta-Y conversion, however, mesh is an excellent
choice for this layout. 
225",ACElectricalCircuitAnalysis.pdf
"We start by finding the reactance values. Using the standard reactance 
formulas we find that XL = j628.3 Ω and  XC = −j318.3 Ω. After substituting 
these into the original circuit and defining the loops, we have Figure 6.24.
We have three loops with three unknown currents, and therefore three 
equations. We'll number the resistors from left to right. For loop 1:
E = vC +vR1
E =− jX C (i1 −i2)+R1 (i1 −i3)
E = (R1 − jX C)i1 −(− jX C )i2 −R1 i3
10 0° V = (1k Ω− j 318.3Ω)i1+j318.3 Ωi2 −1k Ωi3
For loop 2:
0 = vC +vL+vR2
0 =− jX C (i2 −i1)+ jX L(i2 −i3)+ jX L i3
0 =−(− jX C)i1+(R2+ X L − jX C )i2 −R2i3
0 =−(− j 318.3Ω)i1+(500 Ω+ j 628.3 − j 318.3Ω)i2 −500Ωi3
0 = j 318.3Ωi1+(500Ω+ j 310Ω)i2 −500Ωi3
226
Figure 6.23
Circuit for Example 6.8.
Figure 6.24
Circuit of Figure 6.23 with 
reactance values and current 
loops drawn.",ACElectricalCircuitAnalysis.pdf
"For loop 3:
0 = vR1+vR2+vR3
0 = R1(i3 −i1)+R2(i3 −i2)+R3i3
0 =−R1i1 −R2 i2+(R1+R2+R3)i3
0 =−1 kΩi1 −500 Ωi2+(1k Ω+500Ω+2k Ω)i3
0 =−1 kΩi1 −500 Ωi2+3.5k Ωi3
The final set of equations is:
10 0° V = (1k Ω − j 318.3Ω)i1+ j 318.3 Ωi2 −1 kΩi3
0 = j 318.3Ω i1+(500Ω+ j 310Ω)i2 −500 Ωi3
0 =−1 kΩi1 −500Ωi2+3.5 k Ωi3
The system of equations has diagonal symmetry. The results are: 
i1 = 10.24E-316° amps, i2 = 6.754E-3−85.7° amps and 
i3 = 2.888E-3−3.13° amps.
For vb, this is just the voltage across the 1 kΩ resistor. Note that a pair of 
meshing currents (i1 and i3) are flowing through that resistor, so we must 
determine the net value of current. If we assume the reference polarity for vb 
is positive, that coincides with the direction of i1, and thus the net current 
must be i1 − i3. The result is inet = 7.57E−323.2° amps. Therefore, 
vb = 7.5723.2° volts.
Computer SimulationComputer Simulation
Figure 6.25 shows the bridge circuit of Example 6.8 captured in a simulator. This",ACElectricalCircuitAnalysis.pdf
"vb = 7.5723.2° volts.
Computer SimulationComputer Simulation
Figure 6.25 shows the bridge circuit of Example 6.8 captured in a simulator. This 
will be used to verify the computed result.
227
Figure 6.25
Circuit of Figure 6.23 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"A transient analysis is run on the circuit, plotting node 2 which corresponds to vb, 
and node 1, the input voltage, which is handy for phase reference. 
Examining the plot, we can see that the node 2 voltage is just above 7.5 volts peak, 
as calculated. Further, this waveform leads the the input waveform by just over a 
quarter of a division. As this plot shows four divisions per cycle, each division is 90 
degrees. This indicates a leading or positive phase shift in the low 20 degree range, 
and that corroborates nicely the computed value of 23.2 degrees.
Inspection MethodInspection Method
Like nodal analysis, it is possible that the system of equations can be obtained 
directly through inspection. This is true only if the circuit contains no current 
sources. Look at the final set of equations derived in Example 6.8 from Figure 6.24. 
A clear pattern will emerge. To generate an equation for a given loop, simply focus",ACElectricalCircuitAnalysis.pdf
"A clear pattern will emerge. To generate an equation for a given loop, simply focus 
on that loop and ask the following questions: What is the total source voltage in this 
loop? This yields the voltage constant on the left side of the equals sign. Next, sum 
the resistance and reactance values in the loop under inspection. This yields the 
coefficient for that current term. For the other current coefficients, sum the 
resistances and reactances that are in common between the loop under inspection 
and the other loops (e.g., for loop 1, XC is in common with loop 2). These values will
always be negative (an exception arises with a “double negative”, as seen with the 
228
Figure 6.26
Transient analysis for the 
circuit of Figure 6.25.",ACElectricalCircuitAnalysis.pdf
"capacitor). As usual, the set of equations produced must exhibit diagonal symmetry. 
While it is possible to extend this technique to include current sources, usually it is  
easier and less error-prone to convert the current sources into voltage sources. Then 
the process can continue with the direct inspection method outlined above. Finally, it
is important to remember that the number of loops determines the number of 
equations to be solved. This method will be illustrated in the example following.
Example 6.9
For the circuit of Figure 6.27, find vb and the current through the 15 Ω 
resistor. E1 = 100° volts peak and E2 = 2090° volts peak.
We identify and label three loops, as shown in Figure 6.28. This circuit 
utilizes only voltage sources and no current sources. Therefore, we can 
apply the inspection method without extra effort.
We start at loop 1 and sum all of the voltage sources. The sole source is E1.
10 0° V = ...",ACElectricalCircuitAnalysis.pdf
"apply the inspection method without extra effort.
We start at loop 1 and sum all of the voltage sources. The sole source is E1.
10 0° V = ...
Now we sum all of the resistances and reactances in this loop. This is the 
coefficient for the first current term.
10 0° V = (20Ω+6Ω+j 8Ω)i1 ...
229
Figure 6.27
Circuit for Example 6.9.
Figure 6.28
Circuit of Figure 6.27 with 
current loops drawn.",ACElectricalCircuitAnalysis.pdf
"We continue the process by determining the components that are in common
between this loop and the next loop. Remember, this coefficient is negative.
10 0° V = (20Ω+6Ω+j 8Ω)i1 −(6Ω+j 8Ω)i2 ...
We repeat the process by determining the common components with the 
next loop. This coefficient is negative. In this situation, no components are 
in common between loops 1 and 3. We shall leave in placeholder with a 
coefficient of zero, just as a reminder that we didn't forget anything and also 
to ensure that the coefficients in the final set of equations line up nicely.
10 0° V = (26Ω+ j 8Ω)i1 −(6Ω+ j 8Ω)i2 −0i3
We now move to loop 2 and repeat the sequence of steps. There is only one 
source, E2, and it shows up negative as mesh current i2 is flowing out of its 
negative terminal. The result is:
−20 90° V =−(6Ω+ j 8Ω)i1+(6Ω+10Ω+ j 8 − j 14Ω)i2 −10Ωi3
−20 90° V =−(6Ω+ j 8Ω)i1+(16Ω − j 6Ω)i2 −10Ωi3
And finally the third loop. Here the second source shows up as positive.",ACElectricalCircuitAnalysis.pdf
"−20 90° V =−(6Ω+ j 8Ω)i1+(16Ω − j 6Ω)i2 −10Ωi3
And finally the third loop. Here the second source shows up as positive.
20  90° V =−0i1−10Ωi2 −(10Ω+15Ω− j 12Ω)i3
20  90° V =−0i1−10Ωi2 −(25Ω − j 12Ω)i3
The completed system of equations is:
 
10 0° V = (26 Ω+ j 8Ω)i1 −(6Ω+ j 8Ω)i2 −0i3
−20 90° V =−(6Ω+ j8Ω)i1+(16Ω − j 6Ω)i2 −10Ωi3
20  90 ° V =−0i1−10Ωi2 −(25Ω − j 12Ω)i3
The system has diagonal symmetry. The resulting currents are:
i1 = 0.6131−15.5° amps, i2 = 0.6687−49.2° amps and 
i3 = 0.561199.3° amps.
The current through the 15 Ω resistor is i3, so that much is done. Regarding 
vb, it can be found using Ohm's law as vb is the series connection of the 6 Ω 
resistor and j8 Ω inductor times the current through them. This current is the 
pair of meshing currents i1 and i2. Assuming the reference polarity for vb is 
positive, that is the direction of i1, and thus the net current must be i1 − i2. 
The result is 0.37565.8° amps. Therefore, vb = 3.75119° volts. To",ACElectricalCircuitAnalysis.pdf
"The result is 0.37565.8° amps. Therefore, vb = 3.75119° volts. To 
crosscheck this, we can subtract the voltage across the 20 Ω resistor from E1.
That's 100° volts minus 20 Ω times 0.6131−15.5° amps, or 3.75119° 
volts, as expected.
230",ACElectricalCircuitAnalysis.pdf
"Supermesh Supermesh 
On occasion you may find a current source which has no associated internal 
impedance, such as the one in the circuit of Figure 6.29. This is similar to the 
situation discussed previously with nodal analysis where a voltage source does not 
have a specified internal impedance. As with nodal, there are two ways of solving 
this predicament. The first technique is to add a very large impedance in parallel 
with the current source and then perform a source conversion on the pair so that the 
inspection method of mesh can be used. The larger the value of this impedance, the 
greater the accuracy. As a general rule it should be at least a couple of orders of 
magnitude larger than any surrounding impedance, and preferably larger. The second
technique is to use supermesh. A supermesh is a larger mesh loop than contains other
mesh loops inside of it.
Refer to the circuit shown in Figure 6.29. In the center we have a current source, Is,",ACElectricalCircuitAnalysis.pdf
"mesh loops inside of it.
Refer to the circuit shown in Figure 6.29. In the center we have a current source, Is, 
which lacks an associated internal impedance. Two traditional mesh loops, i1 and i2, 
are labeled as usual. The problem here is that we cannot use an Ohm's law-based iZ 
voltage drop for vb. We have no way to express this as the voltage across Is is an 
unknown. On the other hand, what we do know is that Is must equal the combination
of the original mesh currents i1 and i2. That is, from the perspective of the first loop, 
Is = i2 − i1. Remember, one or both of the mesh currents could be negative, and thus 
rotating counterclockwise.
At this point we invoke the idea of a supermesh loop. First, we replace the 
problematic current source with its ideal internal impedance, an open. Second, a 
supermesh loop is drawn which encompasses the original two loops. This is shown 
in Figure 6.30. The supermesh loop is shown in red and labeled.
231
Figure 6.29
Circuit for supermesh.",ACElectricalCircuitAnalysis.pdf
"in Figure 6.30. The supermesh loop is shown in red and labeled.
231
Figure 6.29
Circuit for supermesh.
Figure 6.30
Supermesh labeled.",ACElectricalCircuitAnalysis.pdf
"We now perform a KVL summation around the supermesh loop, similar to what we 
have done in prior work. The difference this time around is that we need to 
recognize that components each see one of the original mesh currents; namely i1 or 
i2 here. We do not solve for a supermesh current. Instead, we just use the supermesh 
loop to define the KVL summation. The summation follows:
Σvrises = Σ vdrops
 
E1 = vR+vXL+E2
The voltage drops across the resistor and inductor can be expanded using Ohm's law,
using the original mesh current associated with each component.
E1 −E 2 = i1 R +i2 jX L
 
Also, by inspection, 
I s = i2 −i1   or
i2 = i1 +I s
We now have two equations with two unknowns and can solve for i1 and i2. This 
procedure is illustrated in the following example.
Example 6.10
Find vb for the circuit of Figure 6.31. E1 = 200° volts, E1 = 1890° volts
and IS = 10E−30° amps.
First, we label the loops, as shown in Figure 6.32.
Now we perform a KVL summation around the supermesh loop.",ACElectricalCircuitAnalysis.pdf
"and IS = 10E−30° amps.
First, we label the loops, as shown in Figure 6.32.
Now we perform a KVL summation around the supermesh loop.
Σvrises = Σ vdrops
 
20  0° V = vR +v XL+18 90° V
Expand using Ohm's law and rearrange:
20  0° V −18 90° V = 1kΩi1+j 400 Ωi2
26.9 −42° V = 1kΩi1+j 400 Ωi2
232
Figure 6.31
Circuit for Example 6.10.",ACElectricalCircuitAnalysis.pdf
"By inspection we can see that:
10E-3 0° A = i2 −i1   or
i2 = i1 +10E-3 0° A
We can substitute this expression into the prior supermesh
expression and solve for i1: 
26.9 −42° V = 1kΩi1+ j 400Ωi2
26.9 −42° V = 1kΩi1+ j 400Ω(i1+10E-3 0° A)
26.9 −42° V = 1kΩi1+ j 400Ωi1+4 90° V
29.7 −44.7° V = (1 kΩ+ j 400 Ω)i1
 
i1 ≈ 27.6E-3 −69.5° A
Thus, i2 = 27.6E−3−69.5° amps + 10E−30° amps, or 32.5E−3 52.8− ° 
amps. To determine vb we simply subtract the drop across the 1 kΩ resistor 
from E1:
vb = 20  0° V − i1 1k Ω
vb = 20  0° V − 27.6E-3−69.5° A1 kΩ
vb ≈ 27.85  68.2° V
As a crosscheck, we could also add the voltage across the inductor to E2:
vb = 18 90° V +i2 j 400Ω
vb = 18 90° V +32.5E-3 −52.8° A j400 Ω
vb ≈27.85  68.2° V
Comparison of Nodal and MeshComparison of Nodal and Mesh
Having covered both nodal and mesh in some detail, it is fair to look at the two 
techniques to gauge their strengths and weaknesses. Compared to nodal analysis,",ACElectricalCircuitAnalysis.pdf
"techniques to gauge their strengths and weaknesses. Compared to nodal analysis, 
mesh analysis has the advantage of dealing with impedances rather than admittances
when writing the system of equations. Further, the mesh inspection method works 
with voltage sources, which tends to be convenient for many circuits, while the 
nodal inspection method requires current sources. On the down side, the resulting set
of mesh currents requires further processing in order to find either branch currents or
node voltages. In contrast, nodal analysis produces node voltages directly with no 
further processing. Mesh also has the disadvantage of being limited to planar circuits
while there is no such limit to nodal. Ultimately, instead of thinking in terms of 
which technique is “better” overall, it is more efficient to use the proper tool for the 
job at hand. For example, if a circuit is populated with voltage sources, mesh might",ACElectricalCircuitAnalysis.pdf
"job at hand. For example, if a circuit is populated with voltage sources, mesh might 
be the more efficient route, especially if specific currents are desired. On the other 
233
Figure 6.32
Circuit of Figure 6.31 with 
supermesh labeled.",ACElectricalCircuitAnalysis.pdf
"hand, if you need to find voltages in a circuit that contains numerous current 
sources, nodal would be more effective. 
6.4 Dependent Sources6.4 Dependent Sources
A dependent source is a current or voltage source whose value is not fixed. Instead, 
the value depends on some other circuit current or voltage. The general form for the 
value of a dependent source is N = kM where M and N are currents and/or voltages 
and k is the proportionality constant. For example, the value of a dependent voltage 
source may be a function of a current, so instead of the source being equal to, say, 10
volts, it could be equal to twenty times the current passing through a particular 
resistor, or v = 20i. 
There are four possible dependent sources: They are the voltage-controlled voltage 
source (VCVS), the voltage-controlled current source (VCCS), the current-controlled 
voltage source (CCVS), and the current-controlled current source (CCCS). The",ACElectricalCircuitAnalysis.pdf
"voltage source (CCVS), and the current-controlled current source (CCCS). The 
source and control parameters are the same for both the VCVS and the CCCS so k has
no units, although it may be given as volts/volt and amps/amp, respectively. For the 
VCCS and CCVS, k has units of amps/volt and volts/amp, respectively. These are 
referred to as the transresistance and transconductance of the sources with units of 
ohms and siemens.
The schematic symbols for dependent or controlled sources are usually drawn using 
a diamond instead of a circle. Also, for simulators, there will be a secondary 
connection for the controlling current or voltage. Examples of voltage-controlled 
and current-controlled sources are shown in Figure 6.33. On each of these symbols, 
the control element is shown to the left of the source.
The control portion can be thought of as a connection for a voltmeter or ammeter
which senses the control parameter. These sensing connections are not always",ACElectricalCircuitAnalysis.pdf
"which senses the control parameter. These sensing connections are not always
drawn on a schematic. Instead, the source simply may be labeled as a function, as
in v = 0.02 iX where iX is the controlling current. These simpler controlled sources
are shown in Figure 6.34 and are typical in electronic schematics and texts. In some
cases, these sources are drawn with a circle instead of a diamond. Also, the sine 
wave shape shown here is often omitted from the inside of symbol, however, current
sources are always drawn with an arrow pointing in the reference direction and 
voltages sources always include the reference polarity.
234
Figure 6.33
Dependent source symbols as 
used in common simulators (left
to right): VCVS, CCVS, VCCS, 
and CCCS. 
Figure 6.34
Generic symbols for 
dependent voltage source (left)
and current source (right).",ACElectricalCircuitAnalysis.pdf
"Dependent sources are not “off-the-shelf” items in the same way that a battery or 
signal generator are. Rather, dependent sources are used to model the behavior of 
more complex devices. For example, a bipolar junction transistor commonly is 
modeled as a CCCS while a field effect transistor may be modeled as a VCCS. 
Similarly, many amplifier circuits are modeled as VCVS systems. Solutions for 
circuits using dependent sources follow along the lines of those established for 
independent sources (i.e., the application of Ohm's law, KVL, KCL, etc.), however, 
the sources are now dependent on the remainder of the circuit which tends to 
complicate the analysis. In general, there are two possible circuit configurations for 
dependent sources: isolated and coupled. An example of the isolated form is shown 
in Figure 6.35.
In this example, the dependent source (center, a CCVS) does not interact with the 
sub-circuit on the left driven by the independent source E. Thus it can be analyzed as",ACElectricalCircuitAnalysis.pdf
"sub-circuit on the left driven by the independent source E. Thus it can be analyzed as
two separate circuits as shown in Figure 6.36. 
Solutions for this form are relatively straightforward. The control value for the 
dependent source can be computed directly using standard techniques. Then this 
value is substituted into the dependent source and the analysis continues as normal. 
Sometimes it is convenient if the solution for a particular voltage or current is 
defined in terms of the control parameter rather than as a specific value (e.g., the 
current through a particular component might be 75 i1 instead of just 1 milliamp). 
The second type of circuit (coupled) is somewhat more complex in that the 
dependent source can affect the parameter that controls the dependent source. An 
example is shown in Figure 6.37. 
235
Figure 6.35
Dependent source circuit: 
isolated.
Figure 6.36
Dependent source circuit: 
isolated treated as two circuits.",ACElectricalCircuitAnalysis.pdf
"In this example it should be obvious that the voltage from the dependent source can 
affect the voltage at node a, and it is this very voltage that defines ix, which in turn 
sets up the value of the dependent source. As far as analysis is concerned, either 
mesh or nodal can be used. The dependent source(s) will contribute terms that 
include the controlling parameter(s) so some additional effort will be required. To 
illustrate the technique, consider the circuit of Figure 6.38. We shall use the general 
method of nodal analysis.
We begin by defining current directions. Assume that the currents through R1 and C
are flowing into node a, the current through R2 is flowing out of node a, and the
current through L is flowing out of node b. We shall number the branch currents to
reflect the associated components, from left to right. The resulting KCL
equations are:
Σiin = Σiout
 
Node a: i1 +i3 = i2
Nodeb: k va = i3 +i4
The currents are then described by their Ohm's law equivalents:",ACElectricalCircuitAnalysis.pdf
"equations are:
Σiin = Σiout
 
Node a: i1 +i3 = i2
Nodeb: k va = i3 +i4
The currents are then described by their Ohm's law equivalents:
Node a: E−va
R1
+ vb−va
− jX C
= va
R2
Node b: k va = vb−va
R2
+ vb
jX L
Expanding terms yields:
Node a: E
R1
− va
R1
+ vb
− jX C
− va
− jX C
= va
R2
Nodeb: k va = vb
R2
− va
R2
+ vb
jX L
236
Figure 6.37
Dependent source circuit: 
coupled.
Figure 6.38
Circuit with a voltage-
controlled current source.",ACElectricalCircuitAnalysis.pdf
"Collecting terms and simplifying yields:
Node a : E
R1
= (
1
R1
+ 1
R2
+ 1
− jX C )va − 1
− jX C
vb
Nodeb:   0 =−(k + 1
R2 )va +(
1
R2
+ 1
jX L )vb
At this point, the component values and independent source value would be inserted 
into the equations and the system solved.
Finally, referring back to the prior chapter, it is possible to perform source 
conversions on dependent sources, within limits. The new source will remain a 
dependent source (e.g., VCVS to VCCS). This process is not applicable if the control 
parameter directly involves the internal impedance (i.e., is its voltage or current).
Example 6.11
For the circuit shown in Figure 6.39, determine vc if the source is 10° volt 
peak.
This is an example of the isolated or uncoupled dependent source. The value
of the dependent current source is 30 times the value of the current labeled 
ix, which is the current flowing through the −j2 kΩ capacitive reactance. We",ACElectricalCircuitAnalysis.pdf
"ix, which is the current flowing through the −j2 kΩ capacitive reactance. We 
can find this current first and then determine the resulting value of the 
dependent source. The analysis will require nothing beyond basic series-
parallel techniques. We will number the components from left to right.
The current ix is found via Ohm's law:
 
ix = E
− jX C1
ix = 1 0° V
− j 2 kΩ
ix = 0.5E-3 90 ° A
237
Figure 6.39
Circuit for Example 6.11.",ACElectricalCircuitAnalysis.pdf
"The dependent source is 30 times this value, or 15E−390° amps. Given the
reference direction of this source, the current is flowing upwards through the
65 kΩ resistor and parallel −j50 kΩ capacitor. This establishes vc as 
negative. Multiplying ix by the parallel impedance yields the desired voltage.
Z RC = R3× jX C2
R3 − jX C2
Z RC = 65k Ω×(− j50 k Ω)
65 kΩ− j 50 k Ω
Z RC = 39.6E3 −52.4 ° Ω
vc =−ix ×Z RC
vc =−15E-3 90° Ω×39.6E3 −52.4° Ω
vc = 594 −142.4 ° V
 The next example features a coupled configuration solved using nodal analysis.
Example 6.12
In the circuit of Figure 6.40, determine va. E =200° volts peak at 50 kHz.
In this circuit we have a current controlled voltage source, or CCVS. The 
proper unit for the constant of 1000 is ohms (volts over amps). First, we 
need to determine the reactance value for the inductor. 
X L = 2π f L
X L = 2π50 kHz 4mH
X L = j 1257Ω
There is only one node of interest here (a) so we will only need one KCL",ACElectricalCircuitAnalysis.pdf
"X L = 2π f L
X L = 2π50 kHz 4mH
X L = j 1257Ω
There is only one node of interest here (a) so we will only need one KCL 
equation. The sole unknown is va. Let's assume that the reference directions 
of the currents flowing through the 1 kΩ and 3 kΩ resistors are entering 
node a. We'll call these i1 and i2, respectively. The exiting current is ix.
238
Figure 6.40
Circuit for Example 6.12.",ACElectricalCircuitAnalysis.pdf
"The KCL summation is:
Σiin = Σiout
i1+i2 = i x
This is expanded using Ohm's law (in several steps, for clarity).
ix = i1+i2
va
R2
= E −va
R1+ jX L
+ 1000(Ω)i x−va
R3
va
2k Ω = 20  0° V−va
1k Ω+ j 1257Ω + 1000(Ω)i x−va
3k Ω
20  0° V
1k Ω+ j 1257Ω = va
1 kΩ+ j 1257Ω + va
2 kΩ + va
3k Ω −1000(Ω)i x
3k Ω
12.45−51.5° A = va
1k Ω+ j 1257Ω + va
2k Ω + va
3 kΩ − 1000(Ω)va
2 kΩ×3 kΩ
12.45−51.5° A = va
1k Ω+ j 1257Ω + va
2k Ω + va
3 kΩ − va
6k Ω
12.45−51.5° A = (
1
1 kΩ+ j 1257Ω + 1
2 kΩ + 1
3k Ω − 1
6 kΩ)va
12.45−51.5° A = 1.161E-3−24.8° S va
 
va = 10.7−26.7° V
Computer SimulationComputer Simulation
For verification, the dependent source circuit of Example 6.12 is entered into a 
simulator as shown in Figure 6.41. The results are shown in Figure 6.42. 
239
Figure 6.41
Circuit of Figure 6.40 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"Note the connection to sense the current ix. It is inserted just like an ammeter. As 
mentioned previously, the constant for the dependent source is a transresistance and 
has units of ohms. A transient analysis is run the circuit, plotting the independent 
source, E, as node 1 (blue), and va as node 3 (red). Both the amplitude and lagging 
phase shift line up nicely with the computed result. The voltage of the dependent 
source is also plotted as node 5 (green). Verifying this potential is left as an exercise.
6.5 Summary6.5 Summary
Nodal analysis can be used to solve virtually any complex multi-source AC electrical
circuit. It is based on KCL, writing expressions involving each node in the circuit. A 
system of equations results, there being as many equations as there are nodes in the 
circuit, minus the reference node (typically taken as ground). The set of equations 
will exhibit diagonal symmetry, which can be used as a crosscheck before setting out",ACElectricalCircuitAnalysis.pdf
"will exhibit diagonal symmetry, which can be used as a crosscheck before setting out
to solve them. The solution will be a complete set of node voltages. From these, any 
branch current may be determined as needed. 
There are two different methods of creating the system of equations. The first 
method is deemed the general method and will work for a mix of current sources and
voltage sources. Individual currents are defined based on the node voltages and any 
known current sources. KCL is then applied at each node, followed by simplification 
240
Figure 6.42
Transient analysis for the 
circuit of Figure 6.41.",ACElectricalCircuitAnalysis.pdf
"and combination of terms to arrive at the end equations. The second approach is 
referred to as the inspection method. If the circuit contains only current sources (or if
the voltage sources are converted to equivalent current sources), this method allows 
direct generation of the system of equations without the need for simplification and 
thus is less prone to error. 
Mesh analysis can be used to solve any planar complex multi-source AC electrical 
circuit. In some respects it is the mirror of nodal analysis. It is based on KVL, 
writing expressions involving each closed loop in the circuit. The loops are 
minimally sized and the set of loops must cover every component in the circuit. A 
system of equations results, there being as many equations as there are loops. As 
with nodal analysis, the set of equations will exhibit diagonal symmetry. The 
solution will be a complete set of mesh currents. From these, any branch current and 
node voltage may be determined.",ACElectricalCircuitAnalysis.pdf
"solution will be a complete set of mesh currents. From these, any branch current and 
node voltage may be determined. 
Like nodal, mesh offers two different methods of creating the system of equations. 
The general method will work for a mix of current sources and voltage sources. 
Individual loops are defined based on the meshing currents passing through each 
component. KVL is then applied around each loop, followed by simplification and 
combination of terms to arrive at the end equations. In contrast, if the circuit 
contains only voltage sources (or if the current sources are converted), then the 
inspection method may be used. This method allows direct generation of the system 
of equations and is faster and less error prone. 
Dependent sources are current or voltage sources whose value depends on the 
current or voltage developed in some other part of the circuit. There are four types: 
current controlled current source (CCCS), current controlled voltage source (CCVS),",ACElectricalCircuitAnalysis.pdf
"current controlled current source (CCCS), current controlled voltage source (CCVS), 
voltage controlled current source (VCCS) and voltage controlled voltage source 
(VCVS). These sources are used commonly to model the characteristics of active 
devices such bipolar and field effect transistors. Techniques for solution tend to be a 
bit more involved than when using constant sources, however, nodal analysis in 
particular tends to work well.
Review QuestionsReview Questions
1. Describe the practical differences between nodal analysis and mesh analysis.
2. What is diagonal symmetry? Of what use is it?
3. What are the differences between the general method and the inspection 
method of nodal analysis?
4. What are the differences between the general method and the inspection 
method of mesh analysis?
5. What is a supernode?
6. What is a supermesh?
7. Describe the concept of dependent sources and how they differ from 
independent or constant sources. 
241",ACElectricalCircuitAnalysis.pdf
"6.6 Exercises6.6 Exercises
Analysis Analysis ((All source values are in amps or volts unless specified otherwise)
1. Given the circuit in Figure 6.43, use nodal analysis to determine vc.               
I1 = 30°, I2 = 0.90°.
2. Use nodal analysis to find the current through the 120 Ω resistor in the 
circuit of Figure 6.44. I1 = 0.590°, I2 = 1.60°.
3. Use nodal analysis to find the current through the 43 Ω resistor in the circuit
of Figure 6.44. The sources are in phase.
4. Given the circuit in Figure 6.44, use nodal analysis to determine vb. The 
sources are in phase.
5. Given the circuit in Figure 6.45, determine vc. I1 = 30°, I2 = 20°.
242
Figure 6.43
Figure 6.45
Figure 6.44",ACElectricalCircuitAnalysis.pdf
"6. Use nodal analysis to find the current through the j45 Ω inductor in the 
circuit of Figure 6.45. I1 = 20°, I2 = 1.560°.
7. Use nodal analysis to find the current through the 4 Ω resistor in the circuit 
of Figure 6.46. I1 = 145°, I2 = 245°.
8. Given the circuit in Figure 6.46, use nodal analysis to determine vc.               
I1 = 630°, I2 = 40°.
9. Given the circuit in Figure 6.47, use nodal analysis to determine vac.              
I1 = 100°, I2 = 60°.
10. Use nodal analysis to find the current through the j8 Ω inductor in the circuit
of Figure 6.47. I1 = 30°, I2 = 530°.
11. Use nodal analysis to find the current through the 22 Ω resistor in the circuit
of Figure 6.48. I1 = 800E−30°, I2 = 2.50°, I3 = 220°.
243
Figure 6.46
Figure 6.47
Figure 6.48",ACElectricalCircuitAnalysis.pdf
"12. Given the circuit in Figure 6.48, use nodal analysis to determine vc.               
I1 = 490°, I2 = 10120°, I3 = 50°.
13. Given the circuit in Figure 6.49, use nodal analysis to determine vc.               
I1 = 3E−30°, I2 = 10E−30°, I3 = 2E−30°.
14. Use nodal analysis to find the current through the −j2 kΩ capacitor in the 
circuit of Figure 6.49.  I1 = 1E−30°, I2 = 5E−30°, I3 = 6E−3−90°.
15. Use nodal analysis to find the current through the 3.3 kΩ resistor in the 
circuit of Figure 6.50. E = 360°, I = 4E−3−120°. 
16. Given the circuit in Figure 6.50, write the node equations and determine vc.  
E = 180°, I = 7.5E−3−30°.
17. Given the circuit in Figure 6.51, use nodal analysis to determine vc.               
E = 40180°, I = 20E−30°.
244
Figure 6.49
Figure 6.50
Figure 6.51",ACElectricalCircuitAnalysis.pdf
"18. Use nodal analysis to find the current through the 2.2 kΩ resistor in      
Figure 6.51. E = 2400°, I = 100E−30°.
19. Use nodal analysis to find vbc in the circuit of Figure 6.52. 
20. Use nodal analysis to find the current through the 2.7 kΩ resistor in the 
circuit of Figure 6.53. 
21. Given the circuit in Figure 6.54, use nodal analysis to determine vba.              
E1 = 10°, E2 = 20°.
245
Figure 6.52
Figure 6.53
Figure 6.54",ACElectricalCircuitAnalysis.pdf
"22. Given the circuit in Figure 6.55, use nodal analysis to determine vad.              
E1 = 90°, E2 = 540°.
23. Use nodal analysis to find vcb in the circuit of Figure 6.56. E1 = 10−180°,   
E2 = 250°.
24. Given the circuit in Figure 6.57, use nodal analysis to determine vbc.              
E = 200°,  R1 = 10 kΩ, R2 = 30 kΩ, R3 = 1 kΩ, XC = −j15 kΩ,                      
XL = j20 kΩ.  
246
Figure 6.55
Figure 6.56
Figure 6.57",ACElectricalCircuitAnalysis.pdf
"25. Given the circuit in Figure 6.58, write the mesh loop equations and 
determine vb. 
26. Use mesh analysis to find the current through the 2.7 kΩ resistor in the 
circuit of Figure 6.58. 
27. Use mesh analysis to find the current through the 75 Ω resistor in the circuit 
of Figure 6.52. 
28. Given the circuit in Figure 6.52, write the mesh loop equations and 
determine vc.
29. Given the circuit in Figure 6.53, write the mesh loop equations and 
determine vb.
30. Use mesh analysis to find the current through the 1.8 kΩ resistor in the 
circuit of Figure 6.53. 
31. Use mesh analysis to find the current through the j200 Ω inductor in      
Figure 6.54. E1 = 10°, E2 = 20°.
32. Given the circuit in Figure 6.54, write the mesh loop equations and 
determine vb. Consider using parallel simplification first. E1 = 36−90°,      
E2 = 24−90°.
33. Given the circuit in Figure 6.55, use mesh analysis to determine vcd.              
E1 = 0.10°, E2 = 0.50°.",ACElectricalCircuitAnalysis.pdf
"E2 = 24−90°.
33. Given the circuit in Figure 6.55, use mesh analysis to determine vcd.              
E1 = 0.10°, E2 = 0.50°.
34. Use mesh analysis to find the current through the 600 Ω resistor in the 
circuit of Figure 6.55. E1 = 90°, E2 = 540°.
247
Figure 6.58",ACElectricalCircuitAnalysis.pdf
"35. Use mesh analysis to find the current through the −j200 Ω capacitor in the 
circuit of Figure 6.59. E1 = 180°, E2 = 1290°.
36. Given the circuit in Figure 6.59, use mesh analysis to determine vac.              
E1 = 10°, E2 = 500E−30°.
37. Given the circuit in Figure 6.56, use mesh analysis to determine vc.                
E1 = 10−180°, E2 = 250°.
38. Use mesh analysis to find the current through the 22 kΩ resistor in the 
circuit of Figure 6.56. E1 = 240°, E2 = 360°.
39. Use mesh analysis to find the current through the j300 Ω inductor in    
Figure 6.60. E1 = 10°,  E2 = 1090°.
40. Given the circuit in Figure 6.60, use mesh analysis to determine va.               
E1 = 1000°, E2 = 900°.
41. Given the circuit in Figure 6.57, use mesh analysis to determine vbc.              
E = 100°,  R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j4 kΩ, XL = j8 kΩ.
42. Use mesh analysis to find the current through resistor R3 in the circuit of",ACElectricalCircuitAnalysis.pdf
"42. Use mesh analysis to find the current through resistor R3 in the circuit of 
Figure 6.57. E = 200°,  R1 = 10 kΩ, R2 = 30 kΩ, R3 = 1 kΩ, XC = −j15 kΩ, 
XL = j20 kΩ.
248
Figure 6.59
Figure 6.60",ACElectricalCircuitAnalysis.pdf
"43. Use mesh analysis to find the current through resistor R3 in Figure 6.61.       
E = 600°,  R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j10 kΩ, XL = j20 kΩ.
44. Given the circuit in Figure 6.61, use mesh analysis to determine vbc.             
E = 12090°,  R1 = 100 kΩ, R2 = 20 kΩ, R3 = 10 kΩ, XC = −j5 kΩ,              
XL = j20 kΩ.
45. Given the circuit in Figure 6.62, use mesh analysis to determine vb. Consider
using source conversion. E = 120°, I = 10E−30°.
46. Use mesh analysis to find the current through the 3 Ω resistor in the circuit 
in Figure 6.62. Consider using source conversion. E = 1590°,                      
I = 10E−30°.
47. Use mesh analysis to find the current through the 2.2 kΩ resistor in the 
circuit in Figure 6.63. E = 3.30°, I = 2.1E−30°.
48. Given the circuit in Figure 6.63, use mesh analysis to determine vb.              
E = 100°, I = 30E−390°.
249
Figure 6.61
Figure 6.62
Figure 6.63",ACElectricalCircuitAnalysis.pdf
"49. Given the circuit in Figure 6.64, use nodal analysis to determine vab. 
50. Use nodal analysis to find the current through the 100 mH inductor in the 
circuit of Figure 6.64.
51. Use nodal analysis to find the current through the 330 Ω resistor in the 
circuit of Figure 6.65.
52. Given the circuit in Figure 6.65, write the node equations and determine vb. 
53. Given the circuit in Figure 6.61, use nodal analysis to determine vbc.              
E = 1200°,  R1 = 1 kΩ,  R2 = 2 kΩ, R3 = 3 kΩ, XC = −j10 kΩ,                    
XL = j20 kΩ.
54. Determine the current through the 10 kΩ resistor in the circuit of                
Figure 6.66 if I1 = 10E−3−90°.
55. Determine vb in the circuit of Figure 6.66 if the source I1 = 20E−30°. 
250
Figure 6.64
Figure 6.65
Figure 6.66",ACElectricalCircuitAnalysis.pdf
"56. Determine vc in the circuit of Figure 6.67 if the source E = 3120°.
57. Determine the current through the 5 kΩ resistor in the circuit of Figure 6.67 
if E = 100°.
58. In the circuit of Figure 6.68, determine the capacitor current if the source      
E = 120°.
59. In the circuit of Figure 6.68, determine vc if the source E = 890°.
60. In the circuit of Figure 6.69, determine vb if the source E = 12−90°.
61. In the circuit of Figure 6.69, determine the current flowing into the 1 kΩ 
resistor if the source E = 60°.
251
Figure 6.67
Figure 6.68
Figure 6.69",ACElectricalCircuitAnalysis.pdf
"62. In the circuit of Figure 6.70, determine the current flowing into the 600 Ω 
resistor if I1 = 1E−3180°. 
63. Determine va and vb in the circuit of Figure 6.70 if the source I1 = 2E−30°.
64. Determine va in the circuit of Figure 6.71 if the source E = 20°.
65. Given the circuit in Figure 6.71, determine the current flowing through the  
1 kΩ resistor. Assume that  E = 1545°.
66. Given the circuit in Figure 6.72, determine the current flowing through the  
3 kΩ resistor if the source  E = 2533°.
67. Given the circuit in Figure 6.72, determine vab. Assume the source                 
E = 15−112°.
252
Figure 6.70
Figure 6.71
Figure 6.72",ACElectricalCircuitAnalysis.pdf
"68. In the circuit of Figure 6.73, determine vd.
69. Given the circuit in Figure 6.73, determine the current flowing through the   
1 kΩ resistor.
70. Given the circuit in Figure 6.74, determine the current flowing through the 
100 Ω resistor.
71. Determine vd in the circuit of Figure 6.74.
72. Determine vab in the circuit of Figure 6.75. E = 100°
253
Figure 6.73
Figure 6.74
Figure 6.75",ACElectricalCircuitAnalysis.pdf
"ChallengeChallenge
73. Given the circuit in Figure 6.76, write the node equations. E1 = 500°,         
E2 = 35120°, I = 500E−390°. 
74. Given the circuit in Figure 6.76, use either mesh or nodal analysis to 
determine ved. E1 = 90°, E2 = 120°, I = 50E−30°. 
75. Given the circuit in Figure 6.77, use mesh analysis to determine vfc.               
E1 = 120°, E2 = 480°, E3 = 3670°.
254
Figure 6.76
Figure 6.77",ACElectricalCircuitAnalysis.pdf
"76. Find voltage vbc in the circuit of Figure 6.78 using either mesh or nodal 
analysis. E = 1000°, R1 = R2 = 2 kΩ, R3 = 3 kΩ, R4 = 10 kΩ, R5 = 5 kΩ,  
XC1 = XC2 = −j2 kΩ.
77. Given the circuit in Figure 6.79, use nodal analysis to find vac.                       
I1 = 8E−30°, I2 = 12E−30°, E = 500°.
255
Figure 6.78
Figure 6.79",ACElectricalCircuitAnalysis.pdf
"78. Given the circuit in Figure 6.80, use nodal analysis to determine vad.              
I1 = 0.10°, I2 = 0.20°, I3 = 0.30°.
79. Given the circuit in Figure 6.81, determine vad. E1 = 150°, E2 = 60°,         
I = 100E−30°. 
80. Given the circuit in Figure 6.82, determine vad. E1 = 220°, E2 = −100°,    
I = 2E−30°.
256
Figure 6.80
Figure 6.81
Figure 6.82",ACElectricalCircuitAnalysis.pdf
"81. Given the circuit in Figure 6.83, determine vab. I1 = 1.20°, I2 = 2120°,    
E = 750°.
82. Given the circuit in Figure 6.84, determine vad. I1 = 0.80°, I2 = 0.2180°, 
I3 = 0.10°, E = 150°.
SimulationSimulation
83. Perform a transient analysis simulation on the circuit of problem 25      
(Figure 6.58) to verify the results for vb.
84. Investigate the variation of vb due to frequency in problem 25 (Figure 6.58) 
by performing an AC simulation. Run the simulation from 10 Hz up to     
100 kHz.
85. Investigate the variation of vb due to component tolerance in problem 25 
(Figure 6.58) by performing a Monte Carlo simulation. Apply a 10% 
tolerance to the resistors and capacitor. 
86. Perform a transient analysis simulation on the circuit of problem 28    
(Figure 6.52) to verify the results for vc.
257
Figure 6.83
Figure 6.84",ACElectricalCircuitAnalysis.pdf
"87. Investigate the variation of vb due to frequency in problem 28 (Figure 6.52) 
by performing an AC simulation. Run the simulation from 1 Hz up to         
10 kHz.
88. Investigate the variation of vb due to component tolerance in problem 28 
(Figure 6.52) by performing a Monte Carlo simulation. Apply a 10% 
tolerance to the resistors and capacitors. 
258",ACElectricalCircuitAnalysis.pdf
"NotesNotes
♫♫♫♫
259",ACElectricalCircuitAnalysis.pdf
"7 7 AC PowerAC Power
After completing this chapter, you should be able to: 
• Describe current, voltage and power relationships in AC RLC networks.
• Plot and make use of the power triangle to determine real, apparent and reactive power components in 
an AC power system.
• Compute the power factor of an RLC network. 
• Determine necessary components for basic power factor correction. 
• Perform basic power calculations for systems involving power factor and efficiency.
7.1 Introduction7.1 Introduction
This chapter introduces the concept of power and power waveforms in AC systems. This is an important part of 
AC circuit analysis and turns out to have striking differences compared to the DC counterpart. While it remains 
true that power is the product of current and voltage, a naive application of that definition can lead to erroneous 
conclusions for the AC case. In Chapter One, RMS (i.e., root-mean-square) values were defined and explained.",ACElectricalCircuitAnalysis.pdf
"conclusions for the AC case. In Chapter One, RMS (i.e., root-mean-square) values were defined and explained. 
As a general rule, RMS values are used for power calculations, not peak or peak-to-peak values. Further, while 
complex non-sinusoidal waveshapes are a decided possibility in electronic systems, we shall limit ourselves here
to sinusoids.
One of the tools we shall use is the power triangle. This is a simple trigonometric device designed to illustrate 
the power relations between resistive and reactive components in a complex impedance. One of its parameters is
the power factor, PF. As we shall see, ordinarily we like the power factor to be unity as this implies best use of 
the available current. It turns out that this is not the case in many systems. As a consequence, we shall also 
investigate a simple means of compensating or shifting the power factor back to unity. This is known as power 
factor correction.",ACElectricalCircuitAnalysis.pdf
"investigate a simple means of compensating or shifting the power factor back to unity. This is known as power 
factor correction.
As part of our discussion involving power factor, we shall examine typical applications such as motors. Here we
shall consider the power factor of a motor along with its efficiency. The efficiency is defined as the useful output
power relative to the supplied power and is always less than 100%. Finally, we shall consider basic power factor
correction for this application.
One practical item to remember here is that, while the instantaneous power changes over time due to the 
sinusoidal cycling of the voltage and current, what matters to most electrical and electronic devices is the 
production of internal heat. Devices such as resistors, transistors and so forth, have mass, and thus exhibit a 
thermal time constant. That time constant tends to be much longer than the period of the wave. The effect is an",ACElectricalCircuitAnalysis.pdf
"thermal time constant. That time constant tends to be much longer than the period of the wave. The effect is an 
averaging of the power waveform. In other words, components do not heat up and cool down instantaneously, 
any more than a hot fry pan would drop to room temperature the moment it was removed from its burner. 
260",ACElectricalCircuitAnalysis.pdf
"7.2 Power Waveforms7.2 Power Waveforms
Computation of power in AC systems is somewhat more involved than the DC case 
due to the phase between the current and voltage. It has been stated in prior work 
that power dissipation is characteristic of resistors, and that ideal inductors and 
capacitors do not dissipate power. We shall show precisely why this is the case by 
examining three distinct cases for AC circuits: purely resistive, purely reactive and 
complex impedance. 
Resistive LoadResistive Load
First, consider the case of the purely resistive load, that is, a load with a phase angle 
of 0 degrees. To determine the power, we simply multiply the voltage by the current.
Recall that the basic expression for a sine wave voltage without a DC offset is:
v(t )=V sin (2π f t +θ)
Where 
v(t) is the voltage at some time t,
V is the peak value,
f is the frequency,
θ is the phase shift.
We know that the current and voltage are always in phase for a resistor, and thus θ is",ACElectricalCircuitAnalysis.pdf
"V is the peak value,
f is the frequency,
θ is the phase shift.
We know that the current and voltage are always in phase for a resistor, and thus θ is
zero degrees. Thus, the expression for a sinusoidal current is similar, using I in place
of V for the peak current. We multiply the current and voltage together to arrive at an
expression for power:11
P (t )=v(t)×i (t )
P (t )=V sin(2π ft )×I sin (2π ft )
P (t )=VI(
1
2 − 1
2 cos(2π2 ft ))
P (t )= VI
2 − VI
2 cos(2π2 ft ) (7.1)
The final expression is made of two parts; the first portion which is fixed (not a 
function of time) and the second portion which consists of a negative cosine wave at 
twice the original frequency. This can be visualized as a time shifted sine wave that 
is riding on a DC level which is equal to the peak value of the new sinusoid. This is 
shown in Figure 7.1 using current and voltage peaks normalized to unity. In this 
figure, the current waveform (green) is drawn just slightly above its true value so",ACElectricalCircuitAnalysis.pdf
"figure, the current waveform (green) is drawn just slightly above its true value so 
that it may be seen easily next to the otherwise identical red voltage waveform. 
11 A useful trigonometric identity here is (sin x)2 = ½ – ½ cos 2x
261",ACElectricalCircuitAnalysis.pdf
"The power product is shown in blue. Unless the frequency is ridiculously low, the 
resistor's heating will respond to the average value of this waveform thanks to the 
device's thermal time constant. Due to the fact that sinusoids are symmetrical around
zero, the effective power dissipation averaged over time will be the offset value, or 
VI/2. For example, a one volt peak source delivering a current of one amp peak, as 
shown here, should generate VI/2, or 0.5 watts. This crosschecks nicely with the RMS
calculation of roughly 0.707 volts RMS times 0.707 amps RMS also yielding 0.5 
watts.
Reactive LoadReactive Load
The situation is considerably different if the load is purely reactive. For a load 
consisting of just an inductor, the voltage leads the current by 90 degrees. This is 
equivalent to a cosine wave. Once again, we multiply the voltage by the current to 
arrive at an expression for power:
P (t )=V cos2π ft ×I sin 2π ft
P (t )=VI(
1
2 sin 2π 2 ft)
P (t )= VI
2 sin 2π2 ft (7.2)",ACElectricalCircuitAnalysis.pdf
"arrive at an expression for power:
P (t )=V cos2π ft ×I sin 2π ft
P (t )=VI(
1
2 sin 2π 2 ft)
P (t )= VI
2 sin 2π2 ft (7.2)
262
Figure 7.1
Waveforms for a resistive load 
(the current is shifted slightly 
positive to ease viewing).
AC Power - Resistive
 
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
 
Time
0 0.2 0.4 0.6 0.8 1
Voltage
Current
Power",ACElectricalCircuitAnalysis.pdf
"Note that this expression does not contain a constant term and only contains a time-
varying term. Consequently, without an offset, there is no net power dissipation. The
result is shown in Figure 7.2. Here power is being alternately generated and 
dissipated (i.e., positive values indicate dissipation while negative values indicate 
generation). In this respect, the reactive element can be thought of as alternately 
storing and releasing energy in the manner of an ideal spring being compressed and 
then released. 
Complex Impedance LoadComplex Impedance Load
Finally, we come to the case of a complex load, part resistive and part reactive. 
Given some phase angle, θ, we have:
 
P (t )=V sin 2π ft ×I sin (2π ft +θ)
P (t )=VI(
1
2 cosθ − 1
2 cos(2π 2 ft +θ))
P (t )= VI
2 cosθ − VI
2 cos(2π 2 ft +θ) (7.3)
This expression contains both a constant term and a time varying term, like the case 
for the purely resistive load shown in Equation 7.1. There is, however, an important",ACElectricalCircuitAnalysis.pdf
"for the purely resistive load shown in Equation 7.1. There is, however, an important 
distinction. The constant term is multiplied by the cosine of the impedance angle, a 
263
Figure 7.2
Waveforms for a purely reactive
load.
AC Power - Reactive
 
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
 
Time
0 0.2 0.4 0.6 0.8 1
Voltage
Current
Power",ACElectricalCircuitAnalysis.pdf
"value whose magnitude ranges from 0 up to 1. Therefore, unless θ is zero, the offset 
will not equal the peak value of the sinusoidal portion. This is a particularly 
important point which shall be amplified in a moment. 
Example waveforms using θ = 45° are shown graphically in Figure 7.3. The power 
waveform dips slightly below zero but is not symmetrical around the time axis. 
Consequently, there is some power dissipation but not as much as in the purely 
resistive case. In short,  the long term power average is now a function of the phase 
angle, θ. As cosine θ may range between 0 and 1, the power for the complex 
impedance case will never be more than that of the purely resistive version. Indeed, 
we can see that Equation 7.3 is the general case. If the load is purely resistive then θ 
is zero, and Equation 7.3 reduces to Equation 7.1. Similarly, if the load is purely 
reactive then θ is ±90 degrees, and Equation 7.3 reduces to Equation 7.2.",ACElectricalCircuitAnalysis.pdf
"reactive then θ is ±90 degrees, and Equation 7.3 reduces to Equation 7.2.
While this analysis used an inductive load, the same can be said regarding the 
capacitive case (simply swap the labels for the current and voltage waveforms).
Finally, in the equations above, V and I are peak values. If RMS values are used, 
there is no need to divide VI by 2.
264
AC Power - Complex
 
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
 
Time
0 0.2 0.4 0.6 0.8 1
Voltage
Current
Power
Figure 7.3
Waveforms for a complex 
impedance load.",ACElectricalCircuitAnalysis.pdf
"At this point we can see that resistors dissipate true power but that reactive 
components do not. This raises a practical problem, namely, what to call the current-
voltage product for purely reactive or complex loads. That is, we can't lump together
the current-voltage values for an inductor with those of a resistor any more than we 
would simply add the magnitudes of resistance and reactance. The practical solution 
is that we refer to the “power” in reactive components as reactive power. Reactive 
power uses the symbol Q. Further, the units are not watts, but volt-amps reactive, or 
more commonly, V AR12. Continuing, for a complex impedance we refer to apparent 
power. It uses the symbol S and has units of volt-amps, abbreviated V A. It is called 
apparent power because it appears to be the power if you naively multiply the value 
obtained from a voltmeter by the value obtained from an ammeter. Those devices",ACElectricalCircuitAnalysis.pdf
"obtained from a voltmeter by the value obtained from an ammeter. Those devices 
would not account for the phase angle between the voltage and current, unlike a 
proper power meter, and their product would not be the true power. The various 
power terms are summarized in Figure 7.4.
Quantity Symbol Unit, Abbreviation
Power P watts, W
Apparent Power S volt-amps, V A
Reactive Power Q volt-amps reactive, V AR
A few examples are in order to help solidify these concepts.
Example 7.1
Determine the power dissipated by the resistor in the circuit of Figure 7.5.
Also find the apparent power drawn by the circuit and the reactive power
of the inductor. The source frequency is 1 kHz.
The first item is to find the reactance of the inductor. 
X L = 2π f L
X L = 2π1 kHz1mH
X L = j 6.283 Ω
There are several ways to find power. In a series loop like this, the most 
direct is to use the i2R forms. The source current can be found via Ohm's",ACElectricalCircuitAnalysis.pdf
"direct is to use the i2R forms. The source current can be found via Ohm's 
law. As power calculations utilize RMS values, first find the RMS value of 
the source voltage.
12 For the plural form, some sources use “V ARs” while others use “V AR”. We shall use the 
latter.
265
Figure 7.4
Symbols and units for power 
quantities.
Figure 7.5
Circuit for Example 7.1.",ACElectricalCircuitAnalysis.pdf
"vRMS = v peak
√2
vRMS = 10V
√2
vRMS ≈7.07V
i = v
Z
i = 7.07 V
10+j 6.283Ω
i = 0.5986 − 32.1° A
For the power calculations, we shall only use the magnitudes of the voltage 
and current. Here, the symbol “| |” refers to just the magnitude of the 
reactance or impedance.
P = i2 R
P = (0.5986A)2
10Ω
P ≈3.58 W
Q = i2
| X |
Q = (0.5986 A)2
6.283Ω
Q ≈2.25 VAR, inductive
S = i2 | Z |
S = (0.5986 A)2
|10+j 6.283Ω|
S ≈4.23 VA, inductive
Computer SimulationComputer Simulation
The circuit of Figure 7.5 is captured in a simulator as shown in Figure 7.6. Three 
different transient analysis simulations are run.
 
266
Figure 7.6
The circuit of Figure 7.5 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"The first simulation plots the circulating current (green), the resistor voltage (red) 
and their product (the power, in blue). This is shown in Figure 7.7. We can see that 
current and voltage are perfectly in phase, as expected. Also, the power waveform 
ranges from zero up to about 7 watts. The average of this is approximately half of 
the peak-to-peak, or about 3.5 watts, just as calculated.
The power value can also be computed from voltage times current as a crosscheck. 
The voltage across the resistor can be found via the voltage rule, and its magnitude 
is approximately 5.986 volts RMS. Multiplying this by the RMS current will also 
yield 3.58 watts.
It is instructive to compare these curves to those generated in Figure 7.1 for the 
general resistive case. The current and voltage values in Figure 7.1 were normalized 
to unity so they do not appear to be identical to those of Figure 7.7, however, the",ACElectricalCircuitAnalysis.pdf
"to unity so they do not appear to be identical to those of Figure 7.7, however, the 
important part is that the phase relationships are the same along with the position of 
the power waveform. In both cases the power waveform ranges from a minimum of 
zero up to some maximum value. Consequently, its average value must be half of its 
peak-to-peak value. 
267
Figure 7.7
Transient analysis for the 
resistor of Figure 7.5.",ACElectricalCircuitAnalysis.pdf
"A second set of plots is generated for the inductor. This is shown in Figure 7.8. 
Again, compare this set against the curves seen in Figure 7.2 for the general reactive 
case. We can see that the current (green) is 90 degrees out of phase with the 
inductor's voltage (red) and lagging, as expected. More importantly, we see that the 
power waveform (blue) is centered around zero. The full cycle average of this is 
zero, meaning that no power is dissipated. But how does this square with the 2.25 
V AR reactive power that was calculated for the inductor? A close look at the power 
plot shows that that value corresponds to the maximum value of the reactive power 
waveform (i.e., half of its peak-to-peak value).
Finally, a third set of curves are created for the circuit as a whole. In other words, 
now we're treating the series combination of the inductor and resistor as the load. 
The results are illustrated in Figure 7.9.",ACElectricalCircuitAnalysis.pdf
"now we're treating the series combination of the inductor and resistor as the load. 
The results are illustrated in Figure 7.9.
The computed impedance phase angle was lagging at 32.1 degrees. We can see this 
same shift between the voltage (red) and current (green) waveforms. The interesting 
bit here is the offset and amplitude of the power waveform (blue). The waveform 
has a peak-to-peak value of about 8.5 V A. Once again, the computed value for 
apparent power, S, works out to one-half of the plotted peak-to-peak value. This will 
be the case for P, Q and S. Further, it turns out that if we find the full cycle average 
of this waveform, those small negative peaks would subtract from the total area and 
reduce the value. The result would be the true power of 3.58 watts. We'll look at this 
more closely following another example.
268
Figure 7.8
Transient analysis for the 
inductor of Figure 7.5.",ACElectricalCircuitAnalysis.pdf
"Example 7.2
Determine the power dissipated by the resistor in the circuit of Figure 7.10.
Also find the apparent power drawn by the circuit and the reactive power
of the capacitor. The source frequency is 1 kHz.
The first item is to find the reactance of the capacitor. 
X C = 1
2π f C
X C = 1
2π1kHz 1μ F
X C =− j 15.92Ω
Unlike the the previous example, we shall use the v2/R forms for power as an
alternative. The RMS value of the source voltage is 7.07 volts. First, find vR.
vR = E R
R − jX C
vR = 7.07V 10Ω
10Ω − j 15.92Ω
vR = 3.761 57.9° V
269
Figure 7.9
Transient analysis for the 
resistor and inductor together 
in Figure 7.5.
Figure 7.10
Circuit for Example 7.2.",ACElectricalCircuitAnalysis.pdf
"The capacitor voltage is found via KVL:
vC = E −vR
vC = 7.07 0° −3.761  57.9 ° V
vC = 5.99−32.1° V
For the powers, we just use the magnitude of the voltage.
P = v2
R
P = (3.761 V)2
10Ω
P ≈1.414 W
Q = i2
| X |
Q = (5.99 V)2
15.92Ω
Q ≈2.25 VAR, capacitive
S = E2
| Z |
S = (7.07 V)2
| 10Ω − j 15.92 Ω|
S ≈2.66 VA, capacitive
Computer SimulationComputer Simulation
In order to verify the results, the circuit of Figure 7.10 is captured in a simulator as 
shown in Figure 7.11. Once again, three different transient analysis simulations are 
run; one each for the resistor, the capacitor, and the pair together.
270
Figure 7.11
The circuit of Figure 7.10 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"In Figure 7.12 we see the results of a transient analysis run on the resistor. We can 
see that the voltage and current are in phase. Also, the power waveform swings from
zero up to around 2.8 watts or so. This corresponds to an average value of just under 
1.5 watts, and this agrees nicely with the computed result. 
Figure 7.13 illustrates the results from a transient analysis run on the capacitor. As 
expected, the current is leading the voltage by 90 degrees. We can also see that the 
power waveform is swinging symmetrically around zero, meaning that there is no 
net power dissipation. The peak value is just under 2.5 V AR, aligning nicely with the
calculated value. 
Finally, in Figure 7.14 we have the results of a transient analysis using both the 
capacitor and the resistor as the load. The current waveform is still leading the 
voltage, but by less than 90 degrees. In fact, it leads by about 2/3rds of a division,",ACElectricalCircuitAnalysis.pdf
"voltage, but by less than 90 degrees. In fact, it leads by about 2/3rds of a division, 
some 60° or so, which conforms nicely to the expected impedance angle of  −57.9° 
(i.e. from 10 − j15.92 Ω). The power waveform sits slightly below the horizontal 
axis indicating it is neither true power nor reactive power, but a combination. The 
peak-to-peak value is somewhat over 5 units, indicating an apparent power of just 
over 2.5 V A, again, just as expected.
271
Figure 7.12
Transient analysis for the 
resistor of Figure 7.11.",ACElectricalCircuitAnalysis.pdf
"272
Figure 7.13
Transient analysis for the 
capacitor of Figure 7.11.
Figure 7.14
Transient analysis for the 
resistor and capacitor together 
in Figure 7.11.",ACElectricalCircuitAnalysis.pdf
"7.3 Power Triangle7.3 Power Triangle
The prior section revealed that the phase angle between the current and voltage 
cannot be ignored when computing power. For example, if a 120 volt RMS source 
delivers 2 amps of current, it appears that it delivers 240 watts. This is only true if 
the load is purely resistive. For a complex load, the true power is somewhat less. In 
fact, as we've just seen, if the load is purely reactive, there will be no true load 
power at all. 
Although plotting the current, voltage and power waveforms is instructive, it can be 
somewhat cumbersome. Instead, we use a power triangle as shown in Figure 7.15. 
The horizontal axis represents true power, P, in watts. The vertical axis represents 
reactive power, Q, in V AR. The vector combination of P and Q results in the 
apparent power, S, in V A. Remember, the apparent power is the product of the 
magnitudes of the current and voltage. This is what the power “appears to be” based",ACElectricalCircuitAnalysis.pdf
"magnitudes of the current and voltage. This is what the power “appears to be” based 
on simple current and voltage measurements from a DMM, versus a proper power 
meter. In the resistive case, there is no reactive power and thus S and P are the same.
Consequently, the S vector collapses onto the P vector. In a purely reactive case, 
there is no true power and S and Q are the same; both vectors identical and vertical. 
For the complex case, S is the vector sum of P and Q. This simple graphic nicely 
encapsulates the relationship between the three vectors. Further, given any two of 
the four parts (three vector magnitudes and θ) and with just a little trigonometry, the 
other two parts may be found. 
For example, knowing the real and reactive powers, the apparent power can be 
found via the Pythagorean theorem. Similarly, if the apparent power and angle are 
known, the real and reactive powers may be found using sine and cosine.",ACElectricalCircuitAnalysis.pdf
"known, the real and reactive powers may be found using sine and cosine. 
Remember, apparent power can be found from the product of the RMS voltage and 
current magnitudes for any complex impedance, and θ is the same as the impedance 
angle (i.e., the voltage angle minus the current angle). For your convenience, some 
useful power triangle relationships are listed following.
273
Figure 7.15
The power triangle.
Real 
Power
Reactive 
Power
Real Power, P
Reactive Power, Q
Apparent Power, S
θ",ACElectricalCircuitAnalysis.pdf
"S = vRMS×iRMS (7.4)
S = √P2
+Q2
(7.5)
θ = tan−1 Q
P     (i.e., the impedance angle) (7.6)
P = S cos θ (7.7)
P = √S2
−Q2
(7.8)
Q = S sinθ (7.9)
Q = √S2
−P2
           (7.10)
Power FactorPower Factor
As we are often interested in the true power, it is worth noting that a rearrangement 
of Equation 7.7 shows that the ratio of true power to apparent power is the cosine of 
the impedance angle, P/S  = cos θ. This is known as the power factor and is 
abbreviated PF. Thus, PF = cos θ. Knowing the phase angle and the apparent power, 
true power can be calculated. If PF is positive it is said to be a lagging power factor. 
This is the case for inductive loads where the current is lagging the voltage. In 
contrast, a capacitive load results in a negative or leading PF. Recall that for 
capacitors, current leads the voltage. The sign is only used to indicate leading or 
lagging and will be useful when we examine power factor correction, shortly. For",ACElectricalCircuitAnalysis.pdf
"lagging and will be useful when we examine power factor correction, shortly. For 
example, if a 100 volt RMS source delivers 1 amp for an apparent power of 100 V A 
and the phase angle is −30°, PF is cos(−30°) or 0.866 leading and the true power is  
P = 100 cos(−30°) = 86.6 watts. 
PF = P
S = cosθ    (positive is lagging and inductive) (7.11)
We'll illustrate the use of the power triangle and the power factor in the next thrilling
and action-packed example.13
Example 7.3
Find S, P and Q in the circuit of Figure 7.16. E = 120 volts RMS. The
source frequency is 60 Hz. 
The first step is to determine the inductive reactance.
X L = 2π f L
X L = 2π60 Hz 150 mH
X L = j 56.55Ω
13 Hyperbole seems to work for the movie industry, anyway.
274
Figure 7.16
Circuit for Example 7.3.",ACElectricalCircuitAnalysis.pdf
"From here we can determine the system impedance which will, in turn, 
allow us to determine the source current. 
Z = R× jX L
R + jX L
Z = 160Ω×( j 56.55 Ω)
160 Ω+ j56.55Ω
Z = 53.3 70.5° Ω
isource = esource
Z
i source = 120  0° V
53.3 70.5° Ω
isource = 2.252  − 70.5° A
The apparent power is the product of the magnitudes of circuit voltage and 
current.
S = E×isource
S = 120 V×2.252 A
S = 270.1 VA, inductive
P = S cos θ
P = 270.1 VA cos 70.5°
P = 90 W
Q = S sinθ
Q = 270.1 VA sin 70.5°
Q = 254.7 VAR, inductive
As a crosscheck, the true power can also be determined by squaring the RMS
voltage and then dividing by the 160 Ω resistance. Similarly, dividing the 
squared voltage by XL will generate Q. 
P = vR
2
R
P = 120 V2
160 Ω
P = 90 W
The power triangle for this circuit is shown in Figure 7.17.
 
 
275
Figure 7.17
Power triangle for the circuit  
of Figure 7.16.
90 W
254.7 VAR
270.1 VA
70.5°",ACElectricalCircuitAnalysis.pdf
"Power Factor CorrectionPower Factor Correction
One issue with a reactive load is that the current is higher than it needs to be in order
to achieve a certain true load power. This is wasteful and would require larger 
conductors. To alleviate these issues, an opposite reactance can be added to the load 
such that the resulting load is purely resistive. This can be realized by determining 
the original Q value and then adding a sufficient reactance to produce an additional 
Q of the opposite sign resulting in cancellation. From there, it is short step to 
determine the required impedance. Then, knowing the frequency, the required 
capacitance or inductance can then be found using the appropriate reactance 
formula. This is illustrated in the following example.
Example 7.4
Find circuit PF, S, P and Q for Figure 7.18. E = 200° volts peak at a
frequency of 10 kHz. Also find an appropriate component which when
placed from node a to ground brings PF to unity.",ACElectricalCircuitAnalysis.pdf
"frequency of 10 kHz. Also find an appropriate component which when
placed from node a to ground brings PF to unity.
The inductive reactance of 1 mH at 10 kHz is j62.83 Ω. This is in parallel
with the 100 Ω resistor, which is then in series with the 20 Ω resistor. 
Z = 20Ω + (100 Ω|| j 62.83Ω)
Z = 20Ω + (28.3Ω+ j 45Ω)
Z = 66  43° Ω
This time we shall use peak values to illustrate the difference compared to 
using RMS values as in the previous example. The source current is:
isource = E
Z
isource = 20 0° V
66 43° Ω
isource = 0.303 − 43° A peak
To find S, multiply the magnitudes of source voltage and current. As these 
are peak values, multiply each by 0.707 to arrive at RMS, or just cut the 
answer in half (i.e., 0.7072 is 0.5). This is the apparent power of the whole 
circuit.
S = E×isource
2
S = 20 V×0.303 A
2
S = 3.03 VA, inductive
276
Figure 7.18
Circuit for Example 7.4.",ACElectricalCircuitAnalysis.pdf
"P = S cos θ
P = 3.03VA cos43°
P = 2.22 W
To be clear, 2.22 watts is the combined power dissipation between the two 
resistors.
Q = S sinθ
Q = 3.03 VA sin43 °
Q = 2.07 VAR, inductive
The power triangle for this circuit is shown in Figure 7.19. 
For the second part involving power factor correction, we need to add a 
reactive power equal in magnitude to the existing value but of the opposite 
sign. This means we need to add 2.07 V AR capacitive. The new power 
triangle is shown in Figure 7.20. The vertical components cancel, resulting
in the apparent power equaling the true power with PF = 1.
We can place the correction capacitor where it's convenient in physical
terms, just as long as we add 2.07 V AR capacitive. We don't have a physical
circuit, so that's not a consideration here. A convenient location would be to
place it across the existing resistor-inductor combo. Our goal then is to first 
find the required reactance, and from that, determined the required",ACElectricalCircuitAnalysis.pdf
"place it across the existing resistor-inductor combo. Our goal then is to first 
find the required reactance, and from that, determined the required 
capacitance. We shall do this two different ways. In the first case, we note 
that the capacitor appears across the only other reactive element in the 
circuit. Therefore, in order for them to cancel they must have the same 
reactance magnitude, and thus XC must equal −j62.8 Ω. But what if there 
were multiple reactive elements in the circuit or if it was impractical to 
locate the component there, for example, due to space restrictions? In that 
case, we would simply work the power relation backwards. For comparison, 
suppose we place the capacitor in the same location; from node a to ground. 
A voltage divider can be used to determine the present value of va. This 
works out to 11.4 volts RMS. Using the voltage form of the power rule:
Q = vX
2
X
X = vX
2
Q
X = 11.4V2
2.07VAR
X = 62.8Ω",ACElectricalCircuitAnalysis.pdf
"works out to 11.4 volts RMS. Using the voltage form of the power rule:
Q = vX
2
X
X = vX
2
Q
X = 11.4V2
2.07VAR
X = 62.8Ω
Either way we derive XC, now we just solve the capacitive reactance formula
to find C.
277
Figure 7.19
Power triangle for the circuit  
of Figure 7.18.
2.22 W
2.07 VAR
3.03 VA
43 degrees
Figure 7.20
Power triangle for the circuit  
of Figure 7.18 with power 
factor correction.
2.22 W
2.07 VAR 
Inductive
2.07 VAR 
Capacitive
After, S=P
Before",ACElectricalCircuitAnalysis.pdf
"C = 1
2π f X C
C = 1
2π10 kHz62.8Ω
C = 253.3 nF
The resulting circuit is shown in Figure 7.21.
Computer SimulationComputer Simulation
It is useful to see the reduction in current demand caused by using power factor 
correction. To so so, the circuit of Figure 7.21 is captured in a simulator as shown in 
Figure 7.22. 
We will run two transient analyses to find the source current. The first version will 
be run using the original circuit configuration without the capacitor. The second run 
will include the capacitor. To plot the currents, we will make use of Ohm's law. First 
we obtain the voltages at nodes 1 and 2. Then, using the simulator's post processor, 
we subtract v1 from v2 which yields the voltage across the 20 Ω resistor. This 
quantity is then divided by 20 Ω to arrive at the input current. This is similar to the 
current sense resistor technique used in earlier chapters.   
278
Figure 7.21
The circuit of Figure 7.18 with 
power factor correction.
Figure 7.22",ACElectricalCircuitAnalysis.pdf
"current sense resistor technique used in earlier chapters.   
278
Figure 7.21
The circuit of Figure 7.18 with 
power factor correction.
Figure 7.22
The circuit of Figure 7.21 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"The resulting current for the original circuit is shown in Figure 7.23. Note that the 
peak current is just over 300 milliamps, as calculated in step two of the example.
The second simulation is shown in Figure 7.24, now with power factor correction. 
279
Figure 7.23
Simulation results for the 
circuit of Figure 7.22 without 
power factor correction.
Figure 7.24
Simulation results for the 
circuit of Figure 7.22 with 
power factor correction.",ACElectricalCircuitAnalysis.pdf
"The amplitude here is much reduced, in the range of 160 to 170 milliamps peak. By 
adding the capacitor, the two reactive currents cancel, leaving a parallel impedance 
of just 100 Ω. This is in series with the 20 Ω resistor. Dividing 120 Ω into the 20 volt
peak source yields a peak current of approximately 167 mA, in agreement with the 
simulation. The cancellation of the reactive currents is also shown by the fact that 
the source current is no longer out of phase. In Figure 7.24 the current waveform is 
in phase and starts at zero, as expected. This implies a load that is the equivalent of a
pure resistance. In contrast, the plot of Figure 7.23 shows a current that is lagging by
around one-half division, or about 45 degrees, which unsurprisingly is the 
approximate value of the circuit's impedance angle. Obviously then, this combined 
impedance must be inductive.
To sum up, power factor correction is used to lower current demand while keeping",ACElectricalCircuitAnalysis.pdf
"impedance must be inductive.
To sum up, power factor correction is used to lower current demand while keeping 
load power constant. If the load is fixed, this can be achieved through the use of a 
capacitor (for inductive loads) or an inductor (for capacitive loads). If the load 
demand is dynamic, then a more complex system is required, for example, switching
through a bank of capacitors to get a value close to the precise value needed for that 
particular load. More examples of power factor correction are in the next section.
7.4 7.4 Power SystemsPower Systems
In this section we shall change our focus and consider power systems at a more 
functional or abstract level rather than at the component level. Specifically, we 
would like to consider common loads such as motors, heating elements, lighting 
devices and the like, and how to analyze a system consisting of a variety of different 
loads. For the most part, we shall stick with the basic scheme of multiple loads being",ACElectricalCircuitAnalysis.pdf
"loads. For the most part, we shall stick with the basic scheme of multiple loads being
supplied by a common and ideally constant voltage source with each load 
configured in parallel with the others. Systems do not have to be configured this way
but it is typical of residential and commercial applications. 
Typical LoadsTypical Loads
Three common loads are heating elements, lighting devices and motors or
compressors. The first two are fairly obvious, but the the latter are also quite
common, although often hidden from view inside devices such as refrigerators, air
conditioners and heat pumps. We can classify loads in terms of their typical
impedance. Devices such as baseboard electrical heating units, toasters, coffee 
makers and electric ovens are classified as resistive heating devices. The load they 
present is easily approximated as a simple resistance. The same is true for 
incandescent lights, however, there can be a very large change in resistance between",ACElectricalCircuitAnalysis.pdf
"incandescent lights, however, there can be a very large change in resistance between 
the on and off  states of incandescent lights. Motors, and other devices integrating 
motors such as compressors, present an inductive impedance. This is due to the large
coils of wire, or windings, inside the motor. Any motorized electrical device falls 
into this category, such as a washing machine, drill press or the fan of Figure 7.25.
280
Figure 7.25
A small boxer fan for cooling 
electronic components.",ACElectricalCircuitAnalysis.pdf
"Purely capacitive loads are not nearly as common but some everyday devices exhibit
highly complex impedances that can vary widely in both amplitude and angle. A 
standard moving coil dynamic loudspeaker falls into this category. While you would 
never plug a loudspeaker directly into an AC power system14, the amplifier that 
drives it must be designed to meet the challenges of an impedance whose magnitude 
can change by a factor of ten across the audible range of frequencies. To complicate 
matters, while the phase typically is inductive, at some frequencies it can be purely 
resistive or even capacitive. 
Efficiency  Efficiency  
Efficiency is defined as usable power output divided by applied power input and is 
denoted by the Greek letter eta, η. Normally it is expressed as a percentage and it 
can never be over 100%. 
η = Poutput
Pinput
(7.12)
Some loads, such as typical heating elements, can be thought of as being 100%",ACElectricalCircuitAnalysis.pdf
"can never be over 100%. 
η = Poutput
Pinput
(7.12)
Some loads, such as typical heating elements, can be thought of as being 100% 
efficient, meaning that all of the electrical input is turned into useful output (in this 
case that's heat, although there are more effective ways of using that electrical input 
to generate heat, such as a heat pump15). In contrast, incandescent light bulbs turn 
very little of their input into the desired quantity (light) and thus suffer from low 
efficiency. Just a few percent of the electrical energy fed into an incandescent bulb 
turns into light while the vast majority turns into heat. From that perspective, 
incandescent lights are more efficient at producing heat than light. LED lighting, on 
the other hand, is perhaps an order of magnitude more efficient than incandescent 
lighting, meaning that we can get the same light output for a much smaller electrical 
power input while simultaneously producing less heat.",ACElectricalCircuitAnalysis.pdf
"lighting, meaning that we can get the same light output for a much smaller electrical 
power input while simultaneously producing less heat.
Motors lose power in the form of mechanical losses (e.g., friction) and electrical 
losses (e.g., resistance of windings). Motors are rated in terms of their output power, 
not the power they draw from the source. For example, a motor with a 1 HP rating 
might be said to generate 1 HP (approximately 745.7 W) at the shaft. If the motor is 
90% efficient, the electrical draw would be 745.7 W/0.9 or 829 W. This situation is 
further complicated by the phase angle (i.e., power factor) of the motor due to 
reactive elements, as noted in the previous sections. Compared to motors, home 
14 That is, unless you enjoy hearing a short-lived blast of sound which is followed by a burst
of flames. Plugging a nominal 8 Ω loudspeaker into a 120 V AC outlet would generate 
1800 watts of power, well beyond the design limits of any common loudspeaker.",ACElectricalCircuitAnalysis.pdf
"1800 watts of power, well beyond the design limits of any common loudspeaker. 
15 A heat pump moves heat from one place to another. In a properly designed system it takes
less energy to move heat than it takes to generate it, and thus it's much more effective 
than simple resistive heating when it comes to warming the interior of a building.
281",ACElectricalCircuitAnalysis.pdf
"loudspeakers are particularly inefficient, typically converting only about 1% of the 
electrical input into usable acoustic output. 
Example 7.5
A certain 120 V AC RMS 60 Hz 1.2 HP motor has an efficiency of 90% and a 
lagging power factor of 0.85. Determine the apparent power and the current 
drawn from the system. Also draw the power triangle. 1 HP ≈ 745.7 watts.
The motor output is 1.2 horsepower which is equivalent to:
Pwatts ≈1.2 HP×745.7 W/HP
Pwatts ≈895 W
The input required to achieve this is computed with Equation 7.12.
Pinput = Poutput
η
Pinput = 895 W
0.9
Pinput = 994 W
From Equations 7.10 and 7.11, the remaining powers are:
S = P
PF
S = 994 W
0.85
S = 1170VA
Q = √S2
−P2
Q = √(1170VA)2
−(994 W)2
Q = 617 VAR
The current is found via Equation 7.4, essentially power law:
S = vRMS ×i RMS
iRMS = S
vRMS
iRMS = 1170 VA
120 V
iRMS = 9.75 A
The power triangle is shown in Figure 7.26.
282
Figure 7.26
Power triangle for the motor 
described in Example 7.5.
994 W
617 VAR 
Inductive",ACElectricalCircuitAnalysis.pdf
"The power triangle is shown in Figure 7.26.
282
Figure 7.26
Power triangle for the motor 
described in Example 7.5.
994 W
617 VAR 
Inductive
1170 VA",ACElectricalCircuitAnalysis.pdf
"Example 7.6
For the motor described in Example 7.5, determine an appropriate 
component that when placed in parallel will produce unity power factor. 
Draw the complete power triangle for the system and determine the new 
current draw.
In Example 7.5 the reactive power was determined to be 617 V AR and was 
inductive, so we'll need 617 V AR capacitive to compensate. The new power 
triangle is shown in Figure 7.27. The source voltage was stated to be 120 V . 
We can use a variation on power law to determine the required reactance.
Q = vRMS
2
X C
X C = vRMS
2
Q
X C = 120 V2
617 VAR
X C =− j 23.34 Ω
Now use the capacitive reactance formula to determine the capacitance
value. The line frequency was specified as 60 Hz.
C = 1
2π f X C
C = 1
2π60 Hz23.34Ω
C = 114μ F
After correction, the apparent power and real power are the same.
Thus,
iRMS = S
vRMS
iRMS = 994 VA
120 V
iRMS = 8.28A
The current draw has been reduced by nearly 1.5 amps, a considerable",ACElectricalCircuitAnalysis.pdf
"Thus,
iRMS = S
vRMS
iRMS = 994 VA
120 V
iRMS = 8.28A
The current draw has been reduced by nearly 1.5 amps, a considerable 
savings. The new current is 85% of what it used to be. 
Now let's consider a larger system that contains several devices. 
283
994 W
617 VAR 
Inductive
617 VAR 
Capacitive
After, S=P
Before
Figure 7.27
Power triangle for the 
corrected system of 
Example 7.6.",ACElectricalCircuitAnalysis.pdf
"Example 7.7
The system shown in Figure 7.28 is supplied by a 240 V AC RMS 60 Hz 
source. Load 1 is 1000 watts of resistive heating elements. Load 2 is a 3 HP 
motor with an efficiency of 92% and a lagging power factor of 0.75. Load 3 
is a capacitor bank equivalent to 75 μF. Draw the system power triangle and 
determine whether or not load 3 is appropriate to bring the system power 
factor to unity. 
First, determine the power triangle for the motor. Then we can add the other 
loads to it to create a system power triangle. The motor output is 3 HP which
is equivalent to:
Pwatts ≈3 HP×745.7 W/HP
Pwatts ≈2237W
The required input power is.
Pinput = Poutput
η
Pinput = 2237W
0.92
Pinput = 2432W
The computations for apparent and reactive power follow.
S = P
PF
S = 2432W
0.75
S = 3242VA
Q = √S2
−P2
Q = √(3242VA)2
−(2432W)2
Q = 2144 VAR, inductive
The power triangle for the motor is shown in Figure 7.29.
284
Figure 7.29
Power triangle for the motor 
of Example 7.7.
2432 W
2144 VAR 
Inductive",ACElectricalCircuitAnalysis.pdf
"The power triangle for the motor is shown in Figure 7.29.
284
Figure 7.29
Power triangle for the motor 
of Example 7.7.
2432 W
2144 VAR 
Inductive
3242 VA
Figure 7.28
System block diagram for 
Example 7.7.",ACElectricalCircuitAnalysis.pdf
"Now for the capacitor. First, we need to determine the reactance, then we 
can use the v2/Z form of power law to determine Q.
X C = 1
2π f C
X C = 1
2 π60 Hz 75μ F
X C =− j 35.37Ω
Q = v2
X
Q = (240 V)2
35.37Ω
Q = 1629VAR, capacitive
There are no other reactive elements in the system to consider. We have 515 
fewer capacitive V AR than inductive V AR, so the correction will not be ideal.
We would need to increase the total capacitance by about 24 μF to achieve a 
power factor of unity. The system power diagram with all of the individual 
parts is shown in Figure 7.30. This is then simplified to the final system 
power triangle as shown in Figure 7.31. The system PF is 0.989 lagging.
 
285
2432 W
2144 VAR 
Inductive
1629 VAR 
Capacitive
1000 W
Figure 7.30
Power components for the 
system of Example 7.7.
Figure 7.31
Final power triangle for the 
system of Example 7.7.
3432 W
515 VAR 
Inductive
3470 VA",ACElectricalCircuitAnalysis.pdf
"7.5 Summary7.5 Summary
Power is the product of current and voltage, however, the phase relationship between
the two has a major impact on the result. The most simple and straightforward case 
is when the current and voltage are in phase, meaning the load is a pure resistance. 
In this case, the true power can be computed directly as the product of RMS current 
and voltage. At the other extreme, when the voltage and current are 90 degrees out 
of phase, as in the case of a purely capacitive or inductive load, power is alternately 
generated and dissipated. That is, in a reactive load, true power dissipation is zero. A
mechanical analogy is the storage and release of energy in an ideal spring as it is 
alternately compressed and then allowed to expand. In between these two extremes, 
that is, when the load is a complex impedance, the true power dissipation is 
somewhere between zero and the resistive maximum.",ACElectricalCircuitAnalysis.pdf
"that is, when the load is a complex impedance, the true power dissipation is 
somewhere between zero and the resistive maximum.
The power triangle is used to make visual sense of this situation. It is a right triangle 
comprised of three legs. The horizontal leg represents the true, or resistive, power 
and is denoted by the letter P. It is measured in watts. The vertical leg represents the 
so-called “reactive power”. It is denoted by the letter Q and has units of V AR (volt-
amps-reactive). Q can be either inductive or capacitive. The third leg, the 
hypotenuse, is the apparent power, S. It is measured in V A (volt-amps). It is called 
apparent power because that is what the power appears to be if it is naively 
measured with a voltmeter and an ammeter, ignoring the phase difference between 
them. The angle between the real and apparent powers, θ (theta), is the phase angle 
between the voltage and current. In other words, theta is the impedance angle.",ACElectricalCircuitAnalysis.pdf
"between the voltage and current. In other words, theta is the impedance angle. 
Knowing theta, real power can be determined using right angle trigonometry, 
namely, P = S cos θ. The cosine of theta is also know as the power factor, PF. It 
ranges from 0 (purely reactive) to 1 (purely resistive). Positive or inductive 
impedance angles are said to have lagging power factor while negative or capacitive 
impedance angles produce leading power factor. 
Ideally, loads are purely resistive and have a power factor of unity. If this is not the 
case, then for a given voltage, a higher current is needed to create the same real 
power as in the purely resistive case. This is not advantageous. Power factor 
correction is the process of shifting the power factor back to unity for complex 
loads. This is done by inserting a reactance of the opposite sign to counterbalance 
the reactive portion of the load, for example, adding capacitive reactance to a system",ACElectricalCircuitAnalysis.pdf
"the reactive portion of the load, for example, adding capacitive reactance to a system
that is inductive. The added reactive power must have the same magnitude as the 
original reactive power but be of the opposite sign, resulting in cancellation.
Efficiency is the measure of usable output power to applied power. Ideally, 
electromechanical systems such as motors would be 100% efficient, meaning that 
there is no power loss, but this is not a practical possibility. For example, there will 
always be frictional losses and power losses in wires. 
286",ACElectricalCircuitAnalysis.pdf
"Review QuestionsReview Questions
1. Define apparent power, real power and reactive power.
2. Describe the power triangle.
3. What is power factor, PF? 
4. Describe the difference between leading and lagging power factor.
5. What is power factor correction? 
6. Give examples of resistive loads and inductive loads.
7.6 Exercises7.6 Exercises
AnalysisAnalysis
1. For the circuit shown in Figure 7.32, determine apparent power S, real 
power P, reactive power Q and power factor PF. Also, draw the power 
triangle.
2. For the circuit shown in Figure 7.33, determine apparent power S, real 
power P, reactive power Q and power factor PF. Also, draw the power 
triangle.
287
Figure 7.32
Figure 7.33",ACElectricalCircuitAnalysis.pdf
"3. For the circuit shown in Figure 7.34, determine apparent power S, real 
power P, reactive power Q and power factor PF. Also, draw the power 
triangle. The source is 120 volts.
4. For the circuit shown in Figure 7.35, determine apparent power S, real 
power P, reactive power Q and power factor PF. Also, draw the power 
triangle. The source is 120 volts.
5. For the circuit shown in Figure 7.36, determine apparent power S, real 
power P, reactive power Q and power factor PF. The source is 90 volts,      
XL = j30 Ω, R = 50 Ω.
288
Figure 7.34
Figure 7.35
Figure 7.36",ACElectricalCircuitAnalysis.pdf
"6. For the circuit shown in Figure 7.37, determine apparent power S, real 
power P, reactive power Q and power factor PF. Also, draw the power 
triangle. The source is 240 volts, XC = −j200 Ω, R = 75 Ω.
7. For the circuit shown in Figure 7.38, determine apparent power S, real 
power P, reactive power Q and power factor PF. The source is 120 volts,       
XL = j40 Ω, XC = −j25 Ω, R = 20 Ω.
8. For the circuit shown in Figure 7.38, determine apparent power S, real 
power P, reactive power Q and power factor PF. The source is 120 volts,    
60 Hz. R = 80 Ω, C = 20 μF, L = 400 mH.
9. An audio power amplifier delivers a 30 volt RMS 1 kHz sine to a 
loudspeaker. If the loudspeaker impedance at this frequency is 745°, 
determine the RMS current delivered to the load and the true power.
10. An audio power amplifier delivers an 80 volt peak 35 Hz sine to a 
subwoofer. If the subwoofer impedance at this frequency is 4−30°, 
determine the peak current delivered to the load and the true power.",ACElectricalCircuitAnalysis.pdf
"subwoofer. If the subwoofer impedance at this frequency is 4−30°, 
determine the peak current delivered to the load and the true power.
11. A certain load is specified as drawing 8 kV A with a lagging power factor of 
0.8. Determine the real power P, and the reactive power Q. Further, if the 
source is 120 volts at 60 Hz, determine the effective impedance of the load 
in both polar and rectangular form, and the requisite 
resistance/inductance/capacitance values.
12. A certain load is specified as drawing 20 kV A with a leading power factor of 
0.9. Determine the real power P, the reactive power Q and draw the power 
triangle. If the source is 240 volts at 60 Hz, determine the effective 
impedance of the load in both polar and rectangular form, and the requisite 
resistance/inductance/capacitance values.
289
Figure 7.37
Figure 7.38",ACElectricalCircuitAnalysis.pdf
"13. Consider the system shown in Figure 7.39. E is a standard 120 V input. If 
the three loads are 45 W, 60 W and 75 W incandescent light bulbs, 
respectively, determine the apparent power delivered to the system, the 
source current, the reactive power and the real power.
14. Given the system shown in Figure 7.39, determine the apparent power 
delivered to the system, the source current, the real power, the reactive 
power and the efficiency. E is 120 V . The three loads are resistive heating 
elements of 500 W, 1200 W and 1500 W, respectively. 
15. Consider the system shown in Figure 7.40. E is 240 V . If the three loads are 
200 W, 400 W and 1000 W resistive, respectively, determine the apparent 
power delivered to the system, the real power and the reactive power.
16. Given the system shown in Figure 7.40, determine the apparent power 
delivered to the system, the real power, the reactive power and the 
efficiency. E is 480 V . The three loads are resistive heating elements of",ACElectricalCircuitAnalysis.pdf
"delivered to the system, the real power, the reactive power and the 
efficiency. E is 480 V . The three loads are resistive heating elements of   
1500 W, 2000 W and 3500 W, respectively. 
17. Consider the system shown in Figure 7.39. E is 120 V . Load 1 is 1 kW 
resistive, load 2 is 400 W resistive and load 3 is 600 V AR inductive. 
Determine the apparent power delivered to the system, the source current, 
the reactive power, the real power and the power factor.
18. Consider the system shown in Figure 7.39. E is 240 V . Load 1 is 2 kW 
resistive, load 2 is 800 W resistive and load 3 is 1200 V AR capacitive. 
Determine the apparent power delivered to the system, the source current, 
the real power, the reactive power and the power factor.
19. Given the system shown in Figure 7.39, determine the apparent power 
delivered to the system, the source current, the real power, the reactive 
power and the power factor. E is 120 V . Load 1 is 600 W of incandescent",ACElectricalCircuitAnalysis.pdf
"delivered to the system, the source current, the real power, the reactive 
power and the power factor. E is 120 V . Load 1 is 600 W of incandescent 
lighting, load 2 is 1200 W of heating elements and load 3 is 200 V AR 
capacitive. 
290
Figure 7.40
Figure 7.39",ACElectricalCircuitAnalysis.pdf
"20. Given the system shown in Figure 7.39, determine the apparent power 
delivered to the system, the source current, the real power, the reactive 
power and the power factor. E is 60 V . Load 1 is 90 W of incandescent 
lighting, load 2 is 800 W of heating elements from a dryer and load 3            
200 V AR inductive.
21. A 120 V 3 HP motor draws a real power of 2500 W from the source. 
Determine its efficiency.
22. A 120 V 12 HP motor draws a real power of 10 kW from the source. 
Determine its efficiency.
23. An ideal 120 V 2 HP motor draws an apparent power of 1800 W from the 
source. Determine its power factor.
24. An ideal 120 V 0.3 HP motor draws an apparent power of 270 W from the 
source. Determine its power factor.
25. A 120 V motor is rated at 0.5 HP. It has an efficiency of 78% and a lagging 
power factor of 0.7. Determine the apparent power drawn from the source 
(S), the real power (P), and the reactive power (Q) supplied. Also draw the",ACElectricalCircuitAnalysis.pdf
"power factor of 0.7. Determine the apparent power drawn from the source 
(S), the real power (P), and the reactive power (Q) supplied. Also draw the 
power triangle and find the delivered current.
26. A motor is rated at 10 HP. It has an efficiency of 92% and a lagging power 
factor of 0.8. Determine the apparent power drawn from the source (S), the 
real power (P), and the reactive power (Q) supplied. Also draw the power 
triangle. Finally, determined the current drawn from the 120 V source.
27. Consider the system shown in Figure 7.39. E is 120 V . Load 1 is 1 kW 
resistive, load 2 is 400 W resistive and load 3 is a 1 HP motor that is 80% 
efficient and has a 0.85 lagging power factor. For the system, determine the 
apparent power delivered, the source current, the real power, the reactive 
power and the power factor.
28. Consider the system shown in Figure 7.39. E is 120 V . Load 1 is 2.5 kW 
resistive, load 2 is 500 V AR capacitive and load 3 is a 2 HP motor that is",ACElectricalCircuitAnalysis.pdf
"resistive, load 2 is 500 V AR capacitive and load 3 is a 2 HP motor that is 
85% efficient and has a 0.9 lagging power factor. For the system, determine 
the total power delivered, the source current, the apparent power, the real 
power and the power factor.
291",ACElectricalCircuitAnalysis.pdf
"29. For the system shown in Figure 7.41, E is 240 V . Load 1 is 1.2 kW resistive 
heating, load 2 is 400 W resistive lighting, load 3 is a 0.5 HP motor that is 
80% efficient with a 0.7 lagging power factor, and load 4 is a 1 HP motor 
that is 85% efficient with a 0.8 lagging power factor. For the system, 
determine the apparent power delivered, the source current, the real power, 
the reactive power and the power factor.
DesignDesign
30.  A 120 V 60 Hz source drives a load equivalent to a 75 Ω resistor in parallel 
with a 25 μF capacitor. Determine the appropriate capacitance or inductance 
value to place across this load to produce unity power factor.
31. A 240 V 60 Hz source drives a load equivalent to a 10 Ω resistor in parallel 
with a 50 mH inductor. Determine the appropriate capacitance or inductance
value to place across this load to produce unity power factor.
32. A load of 5030° is driven by a 120 V 60 Hz source. Determine the",ACElectricalCircuitAnalysis.pdf
"value to place across this load to produce unity power factor.
32. A load of 5030° is driven by a 120 V 60 Hz source. Determine the 
appropriate capacitance or inductance value to place across this load to 
produce unity power factor.
33. A load of 50−50° is driven by a 240 V 60 Hz source. Determine the 
appropriate capacitance or inductance value to place across this load to 
produce unity power factor.
34. A certain load is specified as drawing 8 kV A with a lagging power factor of 
0.8. The source is 120 volts at 60 Hz. Determine the appropriate capacitor or
inductor to place in parallel with this load to produce unity power factor.
35.  A certain load is specified as drawing 20 kV A with a leading power factor of
0.9. The source is 240 volts at 60 Hz. Determine the appropriate capacitor or
inductor to place in parallel with this load to produce unity power factor.
36. A 240 V 60 Hz source is connected to a load consisting of heating elements",ACElectricalCircuitAnalysis.pdf
"36. A 240 V 60 Hz source is connected to a load consisting of heating elements 
of 10 kW along with a 15 HP motor with η=90%, PF=0.85. Determine an 
appropriate capacitor or inductor to place in parallel to produce unity power 
factor.
37. A 120 V 60 Hz source is connected to a load consisting of 350 W of resistive
lighting along with a 1.5 HP motor with η=70%, PF=0.75. Determine an 
appropriate component to place in parallel to produce unity power factor.
292
Figure 7.41",ACElectricalCircuitAnalysis.pdf
"ChallengeChallenge
38. A power distribution system for a concert has the following specifications: 
Ten class D audio power amplifiers rated at 2 kW output each with 90% 
efficiency and unity power factor, 10 kW worth of resistive stage lighting to 
illuminate the musicians alongside a troupe of trained dancing kangaroos, a 
3 HP motor used to continuously rotate the drum riser throughout the 
performance (η=80%, PF=0.75) and a 2 HP compressor which inflates and 
deflates a giant rubber T. rex during particularly exciting parts of the show 
(η=85%, PF=0.8). For the system, determine the total power delivered, the 
source current, the apparent power, the real power, and the power factor. 
Finally, make a sketch of this extravaganza with its entertainers in full 
regalia singing their latest tune “Maximum V olume”.
SimulationSimulation
39. Verify the design of problem 28 by performing a transient analysis. The 
design will have been successful if the source current and voltage are in",ACElectricalCircuitAnalysis.pdf
"design will have been successful if the source current and voltage are in 
phase.
40. Verify the design of problem 29 by performing a transient analysis. The 
design will have been successful if the source current and voltage are in 
phase.
293",ACElectricalCircuitAnalysis.pdf
"8 8 RResonanceesonance
8.0 Chapter Learning Objectives8.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Describe series resonance and plot impedance, voltage and current as a function of frequency.
• Describe parallel resonance and plot impedance, voltage and current as a function of frequency.
• Determine system resonant frequency, Q and bandwidth in series and parallel resonant networks.
• Perform series-to-parallel transforms for practical inductors.
8.1 Introduction8.1 Introduction
Resonance can be thought of as a preferred frequency of vibration. In other words, it is a frequency at which a 
system operates with reduced limitation. It is exploited in a variety of areas, for example, a good mechanical 
resonance can be used for the construction of acoustic musical instruments. In this case, we strive to maintain 
the resonant oscillation in order to enhance the instrument's sustain. On the other hand, we might want to control",ACElectricalCircuitAnalysis.pdf
"the resonant oscillation in order to enhance the instrument's sustain. On the other hand, we might want to control
or limit the resonance, as in an automotive suspension system. In electrical systems, resonance tends to produce 
either a maximum or a minimum response to current or voltage. As a result, resonant systems can be used to 
filter out or select specific frequencies across the spectrum. Obvious uses include tuning circuits, oscillators, 
filters and the like. 
There are two basic forms of resonance for electrical circuits: series RLC resonance and parallel RLC resonance. 
Series resonance tends to be the less complicated of the two. Both types share many similarities but in some 
respects they are mirror images of each other. Series and parallel resonant circuits both exhibit wide fluctuations
in impedance magnitude and phase across the frequency spectrum. In the case of series resonance, the",ACElectricalCircuitAnalysis.pdf
"in impedance magnitude and phase across the frequency spectrum. In the case of series resonance, the 
impedance is at a minimum at the resonance frequency. This implies a current maximum if the circuit is driven 
by a voltage source. In contrast, parallel resonance produces an impedance peak at resonance. This implies a 
voltage maximum if the circuit is driven by a current source.
Much of the initial circuit analysis in this chapter concentrates on the response of the circuits across a range of 
frequencies rather than at some random specific frequency. Consequently, we will be making considerable use 
of frequency domain plots; that is, plots of impedance, voltage and current magnitude and phase as they vary 
across the frequency spectrum. Once that is established we can zero in on the area of resonance and determine 
precisely the shape of these curves in the region around the resonant frequency. Along the way we will define",ACElectricalCircuitAnalysis.pdf
"precisely the shape of these curves in the region around the resonant frequency. Along the way we will define 
and utilize new circuit parameters such as Q, bandwidth, half-power frequencies and more.16 
16 “Q” as used in this chapter is not to be confused with the Q used to represent reactive power, although it is related to the
Q, or quality factor, of an inductor. 
294",ACElectricalCircuitAnalysis.pdf
"8.2 Series Resonance8.2 Series Resonance
Let's begin with the simplest RLC circuit; one consisting of a single voltage source 
in series with a single resistor, inductor and capacitor, as shown in Figure 8.1. Of 
particular interest is how the total impedance varies across the frequency spectrum 
and what impact this has on the current and the three component voltages. 
The impedance as seen by the source is simply the sum of the three components, or
Z = R+ jX L − j X C
This can be expanded into
Z= = R+ j 2π f L − j 1
2π f C
The interesting part here is that the first term is not a function of frequency, the 
second term is directly proportional to frequency and the third is inversely 
proportional to frequency. Further, given that the positive and negative reactances 
behave oppositely, it appears that at some frequency they may cancel out, leaving 
just the resistance. 
To refine this, we expect that at low frequencies the capacitor will dominate the",ACElectricalCircuitAnalysis.pdf
"just the resistance. 
To refine this, we expect that at low frequencies the capacitor will dominate the 
impedance. In other words, XC will be the largest of the three ohmic values. This 
means that the overall impedance will tend to mimic both the magnitude and phase 
of the capacitive reactance. On the other hand, at very high frequencies the inductor 
will tend to dominate the impedance. XL will be the largest of the three values. In this
region, the combined impedance will echo that of the inductor. In short, at low 
frequencies the impedance magnitude will be large and the circuit will appear 
capacitive while at high frequencies the impedance magnitude will be large and the 
circuit will appear inductive. In the middle is where things get interesting. 
A plot of the resistance or reactance of the three elements is shown in Figure 8.2. 
The sum of the three is also shown (red). The frequency axis is uses a logarithmic",ACElectricalCircuitAnalysis.pdf
"The sum of the three is also shown (red). The frequency axis is uses a logarithmic 
scale to show the symmetrical nature of the combined impedance curve. 
295
Figure 8.1
A series RLC circuit.",ACElectricalCircuitAnalysis.pdf
"The dip in the center corresponds to an impedance equal to R. At this frequency the 
capacitive and inductive reactances are equal in magnitude and effectively cancel 
each other. All that's left is the resistive component, R. This frequency is known as 
the resonant frequency and is denoted by f0.
The series resonant frequency, f0, is the frequency at which the 
magnitudes of the inductive and capacitive reactances are equal.     (8.1)
This implies that the power factor is unity at resonance. Also, in a real world circuit 
R is the combination of the series resistance plus any resistance from the inductor's 
coil. We can derive a formula for f0 as follows. The definition declares that the 
magnitude of XL must equal the magnitude of XC. Therefore, we can set the 
capacitive and inductive reactance formulas equal to each other and then solve for 
the resulting frequency.
      
X C = 1
2π f C
 
X L = 2π f L
296
Series Resonance Impedance Plot
Impedance Magnitude
0
2
4
6
8
10
0
2
4
6
8",ACElectricalCircuitAnalysis.pdf
"the resulting frequency.
      
X C = 1
2π f C
 
X L = 2π f L
296
Series Resonance Impedance Plot
Impedance Magnitude
0
2
4
6
8
10
0
2
4
6
8
10
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
R+jXL-jXC
XCXL
R
Figure 8.2
Series impedance variation 
across frequency.",ACElectricalCircuitAnalysis.pdf
"X L = X C
2π f 0 L = 1
2π f 0 C
f 0
2
= 1
(2π)2
LC
        f 0 = 1
2π√L C (8.2)
Note that a particular resonant frequency can be obtained through a variety of LC 
pairs. This, along with the value of R, will alter the specific shape of the impedance 
curve in terms of how narrow or broad the dip is. These components will also affect 
how quickly the phase response shifts from fully capacitive (−90°) to fully inductive
(+90°). This shape factor is described by the parameter, Q. The tighter or more 
narrow the curve, the higher the Q. 
Given a constant voltage source, it should be no surprise that a plot of the resulting 
current will be an inversion of the impedance curve. This is shown in Figure 8.3.
If we scale the curves such that they both have a normalized peak of unity, the 
difference in the shapes may be a little easier to see. This is shown in Figure 8.4.
297
Series Resonance Current Plot
Relative Current
0
2
4
6
8
10
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
Low Q
High Q",ACElectricalCircuitAnalysis.pdf
"297
Series Resonance Current Plot
Relative Current
0
2
4
6
8
10
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
Low Q
High Q
Figure 8.3
Series current versus frequency.",ACElectricalCircuitAnalysis.pdf
"At this point we can more precisely define Q. Specifically, the “sharpness” of the 
curve is related to the half-power or “−3 dB” frequencies, f1 and f2.17 These are the 
frequencies at which the current (assuming voltage source drive) falls off to 0.707 of
the maximum value at resonance. Therefore, they represent the frequencies at which 
power will have fallen to one-half of the maximum value seen at resonance (recall 
that power varies as the square of current and that 0.707 squared is approximately 
0.5). f1 is below f0 and f2 is found above. The difference between these two 
frequencies is called the bandwidth, BW. 
BW = f 2 − f 1 (8.3)
Qcircuit = f 0
BW (8.4)
The relationship between these variables is illustrated in Figure 8.5. The vertical axis
is shown as a percentage of maximum. For a series resonant circuit driven by a 
voltage source, this axis is current; however, it can be voltage in the the case of a",ACElectricalCircuitAnalysis.pdf
"voltage source, this axis is current; however, it can be voltage in the the case of a 
parallel resonant circuit, as we shall see. If this plot is compared to the curves in 
Figure 8.4, it should be apparent that for lower Q circuits, f1 and f2 spread apart, 
moving away from the resonant frequency, f0. Thus, for any given f0, a lower Q 
means a wider (larger) bandwidth.
17 Decibels are covered in detail in Chapter 10.
298
Series Resonance Current Plot
Normalized Current
0
0.2
0.4
0.6
0.8
1
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
Low Q
High Q
Figure 8.4
Normalized series current 
versus frequency.",ACElectricalCircuitAnalysis.pdf
"The resonant frequency, f0, in general is not located evenly between f1 and f2. It is, in 
fact, located at their geometric mean. In other words,
f 0 = √ f 1 f 2 (8.5)
From Equation 8.5 we may derive:
f 0
f 1
= f 2
f 0
(8.6)
To find accurate values for f1 and f2 we can define a factor, k0. The derivation of k0 is 
found in Appendix C. 
k0 = 1
2Qcircuit
+
√
1
4Qcircuit
2 +1 (8.7)
f 1 = f 0
k0
(8.8)
f 2 = f 0×k0 (8.9)
299
Figure 8.5
Location of f1 and f2, and 
definition of bandwidth (BW).
Resonance Frequencies
Percent of Maximum
0
10
20
30
40
50
60
70
80
90
100
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
f1 f0 f2
70.7%
BW",ACElectricalCircuitAnalysis.pdf
"For higher Q circuits (Qcircuit ≥ 10), we can approximate symmetry, and thus
f 1 ≈ f 0 − BW
2 (8.10)
f 2 ≈ f 0+ BW
2 (8.11)
As mentioned previously, the Q can be a function of either R or the L/C ratio. In 
Figures 8.6 and 8.7 we have impedance curves for the two cases. The frequency axis
is normalized to f0 (i.e., f0 is unity). In Figure 8.6 we vary the resistance value to see 
how it affects both the magnitude and phase of the impedance across frequency. 
Figure 8.7 is similar except we vary the inductor/capacitor ratio.
Looking first at the phase (blue, left axis), we see in both cases that high Q circuits 
exhibit a quick transition from a negative (capacitive) phase angle to a positive 
(inductive) phase angle. We also notice that the phase shift hits zero at f0, implying 
unity power factor.
The impedance magnitude plots show a slightly different story. While it is true that 
the higher Q plots are sharper, they get that way through different mechanisms. In",ACElectricalCircuitAnalysis.pdf
"the higher Q plots are sharper, they get that way through different mechanisms. In 
the case of the resistor, a lower Q is achieved via a larger resistance. This has the 
effect of blunting the tip of the curve and lowering current flow at f0 when compared
to the high Q case (as seen in Figure 8.3). In contrast, reducing Q by reducing the 
300
Series Resonance Impedance Variation with Q: R Value
Impedance Phase (degrees)
-100
-50
0
50
100
Relative Impedance Magnitude
0
2
4
6
8
10
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
Dashed: Lower Q
via higher R value
Figure 8.6
Magnitude and phase of 
impedance for a variation of 
resistance.",ACElectricalCircuitAnalysis.pdf
"inductor/capacitor ratio broadens the entire curve. The impedance magnitude at the 
dip does not change, and thus the current at f0 does not change. In practical terms, 
the Q for a series circuit, Qseries, may also be defined by the ratio of circuit reactance 
to the total series resistance at resonance.
Qseries = X 0
RT
(8.12)
Where 
Qseries is the Q of the series resonant circuit (i.e., Qcircuit for series),
RT is the total series resistance (Rseries + Rcoil),
X0 is the reactance (either XL or XC) at f0.
From Equation 8.12 we can derive an expression for Qseries in terms of R, L and C as 
follows:
Qseries = X 0
RT
Qseries = √X 0
2
RT
301
Figure 8.7
Magnitude and phase of 
impedance for a variation of 
inductor/capacitor ratio.
Series Resonance Impedance Variation with Q: L/C Ratio
Impedance Phase (degrees)
-100
-50
0
50
100
Relative Impedance Magnitude
0
2
4
6
8
10
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
Dashed: Lower Q
via lower L/C ratio",ACElectricalCircuitAnalysis.pdf
"At resonance XL and XC have the same magnitude, thus we can also say:
Qseries = √X L X C
RT
 
Qseries = 1
RT
√X L X C
 
Qseries = 1
RT √
2 π f L
2π f C
Which simplifies to:
Qseries = 1
RT √
L
C (8.13)
Effect of Q on Component VoltagesEffect of Q on Component Voltages
Q will create a multiplying effect on the inductor and capacitor voltages at 
resonance. At f0, the current through the circuit will equal the source voltage divided 
by R because XC and XL cancel. This current is also flowing through the capacitor 
and inductor. Equation 8.12 shows that their reactances are Q times higher than R, 
and therefore their voltages will be Q times higher than the source voltage. KVL is 
not violated because the voltages across L and C are 180 degrees out of phase and 
cancel each other. As the circuit Q is increased, the voltage multiplying effect 
becomes more pronounced. In extreme cases it is possible to produce inductor and",ACElectricalCircuitAnalysis.pdf
"becomes more pronounced. In extreme cases it is possible to produce inductor and 
capacitor voltages that are more than 100 times larger than the source voltage. As we
move away from the resonant frequency, the multiplying effect decreases. At 
frequencies much lower than f0, almost all of the source voltage will appear across 
the capacitor with little for the resistor and inductor. At much higher frequencies, 
nearly all of the source potential appears across the inductor with nothing seen 
across the capacitor or resistor. This can be seen in Figure 8.8, where the source 
voltage is unity.
As Q decreases, not only do the capacitor and inductor voltages decrease, but 
another effect comes into play. At relatively high Q values, say 10 or more, the 
capacitor and inductor maximum voltages occur at approximately f0. At lower Q 
values the peaks tend to spread apart, with the capacitor's peak below f0 and that of",ACElectricalCircuitAnalysis.pdf
"values the peaks tend to spread apart, with the capacitor's peak below f0 and that of 
the inductor above f0. This is illustrated in Figure 8.9 (again, the source is unity).
302",ACElectricalCircuitAnalysis.pdf
"303
Figure 8.8
Series resonance: component 
voltages for high Q.
Series Resonance Voltage Plot: High Q
Normalized Voltage
0
2
4
6
8
10
12
Normalized Frequency
0.1 1.0 10.0
0.1 1 10
VR
VLVC
Series Resonance Voltage Plot: Low Q
Normalized Voltage
0
0.5
1
1.5
Normalized Frequency
0.0 0.1 1.0 10.0 100.0
0.01 0.1 1 10 100
VR
VLVC
Figure 8.9
Series resonance: component 
voltages for low Q",ACElectricalCircuitAnalysis.pdf
"Example 8.1
Consider the series circuit of Figure 8.10 with the following parameters:
the source is 10 volts peak, L = 1 mH, C = 1 nF and R = 50 Ω. Find the
resonant frequency, the system Q and bandwidth, and  the half-power 
frequencies f1 and f2.
We begin by finding the resonant frequency.
f 0 = 1
2π√ LC
f 0 = 1
2π√1e-3⋅1e-9
f 0 = 159 kHz
We now find the magnitude of the inductive reactance, and from that, the 
system Qseries via Equation 8.12.
X L = 2π f 0 L
X L = 2π159 kHz 1mH
X L = 1000Ω
Qseries = X L
RT
Qseries = 1000Ω
50Ω
Qseries = 20
Knowing the Q, the bandwidth and corner frequencies can be found via 
Equations 8.4, 8.10 and 8.11.
BW = f 0
Q
BW = 159 kHz
20
BW = 7.95 kHz
f 1 = f 0 − BW
2
f 1 = 159 kHz −7.95kHz
2
f 1 ≈155 kHz
304
Figure 8.10
Circuit for Example 8.1.",ACElectricalCircuitAnalysis.pdf
"f 2 = f 0 + BW
2
f 2 = 159 kHz + 7.95kHz
2
f 2 ≈163 kHz
Given the 10 volt peak source, the voltages across the capacitor and inductor
at the resonance frequency of 159 kHz would be Q times greater, or 200 
volts. At higher or lower frequencies, the increased impedance lowers the 
current and also lowers the voltages across the components. At low 
frequencies, most of the source will appear across the capacitor while at high
frequencies the inductor voltage will approach the source voltage.
  
Refining Series QRefining Series Q
As noted in Chapter 2, all inductors have some series resistance associated with 
them, usually called Rcoil. This resistance needs to be included as part of the overall 
circuit resistance, adding to whatever other series resistance exists. While it is 
possible to measure the DC resistance of a coil using a DMM, this will not 
necessarily give an accurate value at high frequencies. Thus, a preferred method is to",ACElectricalCircuitAnalysis.pdf
"necessarily give an accurate value at high frequencies. Thus, a preferred method is to
determine Qcoil at the desired frequency from the inductor's spec sheet, and using the 
calculated reactance at that frequency, determine the value of Rcoil. An example of 
such a curve is shown in Figure 8.11. 
305
Figure 8.11
Inductor Q graph.
Frequency (Hz)
Q
120
100
80
60
40
1 k 10 k 100 k 1 M
B
A
coil",ACElectricalCircuitAnalysis.pdf
"For instance, using curve A, Qcoil at 100 kHz is approximately 90. If XL is 450 Ω at 
this frequency, then Rcoil would be 450 Ω/90, or 5 Ω. 
Effectively, Qcoil sets the ceiling for the Q of the series resonant circuit, Qseries. That 
is, the system Q can never be higher than the coil Q. To do so would require less 
resistance in the loop than Rcoil, which is a practical impossibility. It is also worth 
noting that Rcoil will create a deviation in the inductor voltage compared to the ideal 
case. This is because vL covers the combination of the inductive reactance in series 
with Rcoil, thus the magnitude will be somewhat larger than expected and the angle 
will be less than 90 degrees. These deviations tend to be quite small unless the 
inductor's Q is fairly low and the remaining circuit resistance is not very much larger
than Rcoil.
Example 8.2
For the circuit of Figure 8.11, determine the resonant frequency, the system",ACElectricalCircuitAnalysis.pdf
"than Rcoil.
Example 8.2
For the circuit of Figure 8.11, determine the resonant frequency, the system 
Q, the bandwidth, and the ideal maximum voltage across each of the three 
components. Use curve A from Figure 8.11 for the inductor.
The first item of importance is finding the resonant frequency.
f 0 = 1
2 π√LC
f 0 = 1
2π√22e-3 H50e-9F
f 0 = 4.8kHz
The inductive reactance is:
X L = 2π f 0 L
X L = 2π 4.8kHz 22 mH
X L = 663.3Ω
From the graph, Qcoil is approximately 95, meaning Rcoil is: 
306
Figure 8.12
Circuit for Example 8.2.",ACElectricalCircuitAnalysis.pdf
"Rcoil = X L
Qcoil
Rcoil = 663.3 Ω
95
Rcoil = 7Ω
Combined with the 140 Ω resistor, we are left with 147 Ω, some 5% higher 
than if we had ignored it. The system Q is:
Qseries = X L
RT
Qseries = 663.3Ω
147Ω
Qseries = 4.51
The Q is on the low side but not extremely so. Now for the bandwidth:
BW = f 0
Q
BW = 4.8kHz
4.51
BW = 1.06 kHz
Ideally, at f0 we expect vR will be equal to the source of 1 volt peak while the
inductor and capacitor voltages will be Q times larger, or approximately 4.5 
volts peak. In reality Rcoil will create a voltage divider, reducing the drop 
across the 140 Ω resistor to about 0.95 volts. The change in vL will be 
negligible due to ZL being 663.3489.4° Ω versus the ideal 663.390° Ω. 
The system Q is relatively low (<10), so the vC and vL peaks will shift a little 
from f0, with vC peaking at a slightly lower frequency and vL slightly higher.
Computer SimulationComputer Simulation
Of particular interest in the prior example is the precise shape of the component",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
Of particular interest in the prior example is the precise shape of the component 
responses versus frequency. This can be produced via an AC or frequency domain 
simulation. The circuit of Figure 8.12 is captured in a simulator as shown in Figure 
8.13, and is modified by adding the inductor's coil resistance below the inductor. 
The items of interest are the net resistor voltage which appears between nodes 1 and 
2, the capacitor voltage between nodes 2 and 3, and the inductor voltage which 
appears from node 3 to ground. The analysis is run from 500 Hz to 50 kHz giving us
a factor of 10 in frequency on either side of f0, as seen in Figure 8.14. First, the peaks
are just below 5 kHz, as expected. The resistor voltage (blue) is about 0.95 volts, and
the inductor (red) and capacitor (green) voltages are about 4.5 volts, as calculated.
307",ACElectricalCircuitAnalysis.pdf
"Also, note that there is a slight spread between the peaks of the capacitor and 
inductor voltages, with vC slightly below f0 and vL slightly above, again just as 
expected. At the lowest frequencies, all of the source appears across the capacitor 
while at the highest frequencies all of the source appears across the inductor. Note 
the similarity between these curves and those of Figures 8.8 and 8.9
And now for a change of pace; a design problem.
308
Figure 8.14
Voltage versus frequency for 
each of the three components of
the circuit of Figure 8.13.
Figure 8.13
The circuit of Example 8.2 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"Example 8.3
Design a series resonant circuit with a resonant frequency of 100 kHz and a 
bandwidth of 2 kHz using a 10 mH inductor. Assumes the inductor follows 
curve B in Figure 8.14.
We can find the value for the capacitance by rearranging the resonance 
frequency equation:
f 0 = 1
2π√L C
√L C = 1
2π f 0
C = 1
(2π f 0)2
L
C = 1
(2π100 kHz)210 mH
C = 253.3 pF
Knowing the bandwidth and resonant frequency, we can find the system Q:
Qseries = f 0
BW
Qseries = 100 kHz
2 kHz
Qseries = 50
At resonance, the inductive reactance will be:
X L = 2π f 0 L
X L = 2π100 kHz 10 mH
X L = 6283Ω
The preceding tells us that the total series resistance must be:
Rseries = X L
Qseries
Rseries = 6283Ω
50
Rseries = 125.7Ω
Curve B indicates that Qcoil is approximately 115 at 100 kHz. Thus, Rcoil is: 
309",ACElectricalCircuitAnalysis.pdf
"Rcoil = X L
Qcoil
Rcoil = 6283Ω
115
Rcoil = 54.6Ω
Consequently, we must add 125.7 Ω − 54.6 Ω, or 71.1 Ω, to the series 
network to achieve the desired system Q. Failure to do so will yield a much 
higher Q than specified, resulting in a much reduced bandwidth. The 
completed design is shown in Figure 8.15.
8.3 Parallel Resonance8.3 Parallel Resonance
If the three RLC components are placed in parallel, as in Figure 8.16, a parallel
resonant circuit can result. Typically, it would be driven by a current source as
shown, although this is not a requirement for resonance. Parallel resonance is
slightly more complicated than series resonance due to the fact that the series coil
resistance cannot be lumped in with the remaining circuit resistance as it can with
the series case. In other words, the practical reality is that we have a series-parallel
circuit where the inductor is, in fact, a series combination of the inductance and",ACElectricalCircuitAnalysis.pdf
"circuit where the inductor is, in fact, a series combination of the inductance and
the coil resistance. It turns out that usually this resistance cannot be ignored, even
if it is very small. To alleviate this problem, it is possible to find a parallel
equivalent for the series inductive reactance and associated coil resistance. That is, 
we need a series to parallel transform.
Series to Parallel Inductor TransformSeries to Parallel Inductor Transform
First, let's take a look at what we have in practical terms. A realistic parallel
resonant circuit is illustrated in Figure 8.17. This circuit adds the internal coil
resistance of the inductor to the ideal circuit shown in Figure 8.16. What we
would like to do is derive a means of finding the parallel equivalent of the
inductor with its coil resistance. Certainly, this should be possible to do. After all,
it is a trivial exercise to do the reverse; namely, taking a parallel combination of 
310
Figure 8.16
Ideal parallel resonant circuit.",ACElectricalCircuitAnalysis.pdf
"it is a trivial exercise to do the reverse; namely, taking a parallel combination of 
310
Figure 8.16
Ideal parallel resonant circuit.
Figure 8.15
Completed circuit design for 
Example 8.3.
Figure 8.17
Realistic parallel resonant 
circuit.",ACElectricalCircuitAnalysis.pdf
"an inductor and resistor and finding its series equivalent (i.e., expressing the 
resulting impedance in rectangular form). After completing the process we should 
have an equivalent circuit like that shown in Figure 8.18. In this equivalent circuit, R
and C are the values from the original circuit while L(p) and Rcoil(p) are the parallel 
equivalent transformed values derived from the original inductor. In this version, it 
is easy to combine R in parallel with Rcoil(p) to create a single resistor and thus wind 
up back at our ideal circuit of Figure 8.16. 
For the equivalent transform, refer to Figure 8.19. We start with a practical coil 
consisting of a series combination of resistance and inductive reactance, Rs + jXs.  
We will find the parallel equivalent, Rp || jXp. 
We begin with the reciprocal conductance/resistance rule:
Rs+jX s = 1
1
R p
+ 1
jX p
1
Rs+jX s
= 1
R p
+ 1
jX p
(8.14)
The next step is to isolate the real and imaginary parts of the series version. We can",ACElectricalCircuitAnalysis.pdf
"Rs+jX s = 1
1
R p
+ 1
jX p
1
Rs+jX s
= 1
R p
+ 1
jX p
(8.14)
The next step is to isolate the real and imaginary parts of the series version. We can 
do this by multiplying the left term of Equation 8.10 by the complex conjugate to 
arrive at an equivalent:
1
Rs+jX s
Rs − jX s
Rs − jX s
= Rs
Rs
2
+X s
2 + − jX s
Rs
2
+X s
2
311
Figure 8.18
Transformed (equivalent) 
version of a realistic parallel 
resonant circuit.
Figure 8.19
Series and equivalent parallel 
RL combinations.",ACElectricalCircuitAnalysis.pdf
"Substituting this equivalent back into Equation 8.14 yields,
Rs
Rs
2
+X s
2 + − jX s
Rs
2
+X s
2 = 1
R p
+ 1
jX p
Therefore,
1
Rp
= Rs
Rs
2
+X s
2
1
jX p
= − jX s
Rs
2
+X s
2
Taking the reciprocal results in:
  Rp = Rs
2+X s
2
Rs
(8.15)
jX p = j Rs
2+X s
2
X s
(8.16)
For high Q coils (Qcoil ≥ 10), Xs >> Rs, so we can approximate these as:
  Rp ≈ X s
2
Rs
= Qcoil X s = Qcoil
2
Rs (8.17)
jX p ≈ j X s
2
X s
= jX s (8.18)
Thus, for a high Qcoil, the parallel equivalent reactance is unchanged from the series 
value and the parallel equivalent resistance is the series resistance times the Q of the 
coil squared. Interestingly, Equation 8.17 shows that a smaller RS (which yields a 
proportionally larger Qcoil) results in a larger RP . Thus, the ideal inductor which 
would have no coil resistance results in an Rp of infinity. Due to this resistive 
“inversion” of the series-parallel transform, parallel circuit Q is defined as:
Qparallel = RT
X L
(8.19)
Where",ACElectricalCircuitAnalysis.pdf
"“inversion” of the series-parallel transform, parallel circuit Q is defined as:
Qparallel = RT
X L
(8.19)
Where 
Qparallel is the Q of the parallel resonant circuit (i.e., Qcircuit for parallel),
RT is the total parallel resistance (Rp || R),
XL is the reactance at f0.
312",ACElectricalCircuitAnalysis.pdf
"Based on Equation 8.19 and the development of Equation 8.13, it can be shown that:
Qparallel = RT √
C
L (8.20)
For higher Q circuits (Qparallel ≥ 10), f0 is found as in the series case (repeating):
f 0= 1
2 π√LC (8.2)
For lower Q circuits, f0 will be reduced slightly due to the fact that the transformed 
resistance is frequency dependent. More on this in an upcoming section.
Parallel Resonance ImpedanceParallel Resonance Impedance
A parallel impedance plot is shown in Figure 8.20. The effect is the inverse of the 
series case. At low frequencies the small inductive reactance results in a low 
impedance magnitude with a positive (inductive) phase angle. At high frequencies 
the small capacitive reactance results in a low impedance magnitude with a negative 
(capacitive) phase angle. At resonance the reactive values cancel. This leaves just the
parallel resistive value which produces the characteristic peak in impedance. The",ACElectricalCircuitAnalysis.pdf
"parallel resistive value which produces the characteristic peak in impedance. The 
phase angle is zero, corresponding to a power factor of unity. 
313
Parallel Resonance Impedance Plot
Impedance Phase (degrees)
-100
-50
0
50
100
Normalized Impedance Magnitude
0
0.2
0.4
0.6
0.8
1
Normalized Frequency
0.0 0.1 1.0 10.0 100.0
0.01 0.1 1 10 100
Dashed: Lower Q
Figure 8.20
Impedance plot for parallel 
resonant circuit.",ACElectricalCircuitAnalysis.pdf
"If the parallel resonant circuit is driven by a current source, then the voltage 
produced across the resonant circuit (sometimes referred to as a tank circuit) will 
echo the shape of the impedance magnitude. In other words, it will effectively 
discriminate against high and low frequencies and keep only those signals in the 
vicinity of the resonant frequency. This is one method of making a bandpass filter. 
The lower and upper half-power frequencies, f1 and f2, are found in the same manner 
as in series resonance. 
Repeating for convenience:
BW = f 2 − f 1 (8.3)
Qcircuit = f 0
BW (8.4)
f 0 = √ f 1 f 2 (8.5)
f 0
f 1
= f 2
f 0
(8.6)
 
k0 = 1
2Qcircuit
+
√
1
4Qcircuit
2 +1 (8.7)
f 1 = f 0
k0
(8.8)
f 2 = f 0×k0 (8.9)
For higher Q circuits (Qcircuit ≥ 10), we can approximate symmetry, and thus
f 1 ≈ f 0 − BW
2 (8.10)
f 2 ≈ f 0+ BW
2 (8.11)
Finally, it is worth repeating that for relatively low Q values there will be some",ACElectricalCircuitAnalysis.pdf
"f 1 ≈ f 0 − BW
2 (8.10)
f 2 ≈ f 0+ BW
2 (8.11)
Finally, it is worth repeating that for relatively low Q values there will be some 
shifting of the resonant and half-power frequencies from the equations presented 
above. 
There are some similarities between parallel and series resonance. Like series, as the
parallel Q increases, the impedance curve becomes sharper and the phase change is 
more abrupt. Further, we also see an apparent “Q amplification” effect in parallel 
resonant circuits, however, here it will be the reactive currents that will be increased 
relative to the source current instead of the series component voltages.
314",ACElectricalCircuitAnalysis.pdf
"Note that the parallel resistor can be used to lower the system Q and thus broaden 
the bandwidth, however, the system Q can never be higher than the Q of the inductor
itself. The inductor sets the upper limit on system Q and therefore, how tight the 
bandwidth can be. In other words, Qcircuit ≤ Qcoil. This is the same situation we saw for
series resonance.
Example 8.4
For the circuit of Figure 8.21, determine the resonant frequency, the corner
frequencies of f1 and f2, the bandwidth and the system Q. Also find the
circuit voltage at the resonant frequency. Rcoil = 100 Ω.
First, we'll assume this is a high Q (≥ 10) circuit.
f 0= 1
2π√L C
f 0= 1
2 π√50 mH 910 pF
f 0=23.6 kHz
X L = 2π f L
X L = 2π 23.6kHz 50 mH
X L = 7.41k Ω
Qcoil = X L
Rcoil
Qcoil = 7.41k Ω
100 Ω
Qcoil = 74.1
The parallel equivalent of the coil resistance is:
Rp = Rcoil Qcoil
2
Rp = 100Ω74.12
Rp = 549.5 kΩ
There is no other resistor in parallel with the inductor and capacitor,",ACElectricalCircuitAnalysis.pdf
"Rp = Rcoil Qcoil
2
Rp = 100Ω74.12
Rp = 549.5 kΩ
There is no other resistor in parallel with the inductor and capacitor, 
therefore the equivalent parallel resistance, Rp, is the total resistance of the 
circuit, RT. Consequently, the Q of the circuit must be the same as Qcoil. We 
can verify this as follows:
315
Figure 8.21
Circuit for Example 8.4.",ACElectricalCircuitAnalysis.pdf
"Qparallel = RT
X L
Qparallel = 549.5 k Ω
7.41 kΩ
Qparallel = 74.1
Our initial assumption of high circuit Q is met.
BW = f 0
Qparallel
BW = 23.6 kHz
74.1
BW = 318 Hz
f 1 ≈ f 0 − BW
2
f 1 ≈23.6 kHz − 318Hz
2
f 1 ≈23.44 kHz
f 2 ≈ f 0 + BW
2
f 2 ≈23.6 kHz +318 Hz
2
f 2 ≈23.76 kHz
To find the circuit voltage at f0, simply multiply the resonant impedance of 
549.5 kΩ times the source of 2 mA. This yields approximately 1100 volts. 
Computer SimulationComputer Simulation
The circuit of Example 8.4 is captured in a simulator as shown in Figure 8.22. 
316
Figure 8.22
The circuit of Figure 8.21 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"A frequency domain or AC analysis is run on the circuit, plotting the magnitude of 
the source voltage (node 1) from 2 kHz to 200 kHz. This will give us roughly a 
factor of ten on either side of the resonant frequency. The result is shown in Figure 
8.23. The plot shows a clear and sharp peak in the low 20 kHz region. Note that the 
peak voltage is just over 1000 volts, as predicted. Figure 8.24 shows a magnified 
version of this plot so that we can accurately verify the peak voltage along with f1 
and f2.
Figure 8.24 shows that the peak is indeed approximately 1100volts. The f1 and f2 
frequencies are found at 0.707 times this peak, or some 778 volts. Two measurement
cursors are employed for this task. The Y values are the voltages at the cursor's 
intersection with the curve and the X values are the corresponding frequencies. We 
can see that the results are in tight agreement with the calculations. At levels of",ACElectricalCircuitAnalysis.pdf
"can see that the results are in tight agreement with the calculations. At levels of 
about 777 to 780 volts we obtain f1 and f2 values of approximately 23.44 kHz and 
23.75 kHz, respectively.
317
Figure 8.23
Source voltage of the circuit of 
Figure 8.22.",ACElectricalCircuitAnalysis.pdf
"Example 8.5
For the circuit of Figure 8.25, determine the resonant frequency, the corner 
frequencies of f1 and f2, the bandwidth and the system Q. Also find the 
circuit voltage at the resonant frequency. Rcoil = 100 Ω.
This circuit is identical to the one in the previous example with the 
exception of an added 100 k Ω load resistor. This should lower the system Q
and thus widen the bandwidth. The peak impedance will also be reduced 
which will cause a decrease in the system voltage at resonance. Some 
parameters will not change. They include:
f0 = 23.6 kHz
XL = 7.41 k Ω 
318
Figure 8.24
Magnified peak of the plot of 
Figure 8.23.
Figure 8.25
Circuit for Example 8.5.",ACElectricalCircuitAnalysis.pdf
"Qcoil = 74.1
Rp = 549.5 k Ω 
We'll assume this is a high Q (≥ 10) circuit. 
Rp is in parallel with the load resistance of R yielding an effective parallel 
resistance of 549.5 kΩ || 100 kΩ, or 84.6 kΩ. 
Qparallel = RT
X L
Qparallel = 84.6 k Ω
7.41k Ω
Qparallel = 11.4
The circuit Q is much reduced but our initial assumption of high circuit Q is 
still met. Now we can find the bandwidth and corner frequencies.
BW = f 0
Qparallel
BW = 23.6 kHz
11.4
BW = 2.07kHz
f 1 ≈ f 0 − BW
2
f 1 ≈23.6 kHz − 2.07kHz
2
f 1 ≈22.56 kHz
f 2 ≈ f 0 + BW
2
f 2 ≈23.6 kHz + 2.07kHz
2
f 2 ≈24.64 kHz
The circuit voltage at f0 is reduced to 84.6 kΩ times 2 mA, or 169.2 volts. 
Computer SimulationComputer Simulation
The simulation results follow those of Example 8.4 and are shown in Figure 8.26. 
The results agree with the computed values. The peak voltage has been reduced to 
about 170 volts, and f1 and f2 (found at 0.707 times the peak, or approximately 120 
volts) are about 22.5 kHz and 24.6 kHz, respectively.",ACElectricalCircuitAnalysis.pdf
"about 170 volts, and f1 and f2 (found at 0.707 times the peak, or approximately 120 
volts) are about 22.5 kHz and 24.6 kHz, respectively.
319",ACElectricalCircuitAnalysis.pdf
"Example 8.6
Consider the circuit of Figure 8.27 with the following parameters:          
L=2 mH, C=10 nF, and Qcoil = 25. Determine the resonant frequency
and a value for R such that the system bandwidth is 3 kHz.
As usual, we'll assume this is a high Q (≥ 10) circuit. This is certainly
true of the coil, although we have to determine the resonant frequency
in order to determine the Q of the circuit. 
f 0= 1
2π√LC
f 0= 1
2π √2mH 10 nF
f 0=35.59 kHz
Qparallel = f 0
BW
Qparallel = 35.59 kHz
3 kHz
Qparallel = 11.86
320
Figure 8.26
Simulation results for the 
circuit of Example 8.5.
Figure 8.27
Circuit for Example 8.6.",ACElectricalCircuitAnalysis.pdf
"We have high Q and can continue18. Ultimately, we need to determine the 
total parallel resistance required to achieve this Q. Before we can do that we 
need to determine XL.
X L = 2π f L
X L = 2π35.59 kHz 2mH
X L = 447 Ω
RT = Qparallel×X L
RT = 11.86×447Ω
RT = 5.3k Ω
RT is the parallel combination of R and Rp (the parallel equivalent of Rcoil), so
first we need to find Rcoil. 
Rcoil = X L
Qcoil
Rcoil = 447Ω
25
Rcoil = 17.9Ω
The parallel equivalent resistance of Rcoil is: 
Rp = Rcoil Qcoil
2
Rp = 17.9Ω252
Rp = 11.18 kΩ
Using the conductance rule, we can find the requisite value of R.
R = 1
1
RT
− 1
R p
R = 1
1
5.3k Ω − 1
11.18 k Ω
 
R = 10.08 kΩ
Thus, we need to use a 10.08 kΩ resistor in order to lower the circuit Q 
enough to achieve a 3 kHz bandwidth. Without this resistor, the bandwidth 
will be less than half of what is required.
18 Note that if this value had been greater than 25 we'd be stuck for a different reason;",ACElectricalCircuitAnalysis.pdf
"will be less than half of what is required.
18 Note that if this value had been greater than 25 we'd be stuck for a different reason; 
namely that we'd need to obtain a higher quality inductor because Qcircuit can't be any 
higher than Qcoil.
321",ACElectricalCircuitAnalysis.pdf
"Computer Simulation
Figure 8.28 shows the completed design of the previous example captured in a 
simulator. A 1 mA current source is used for convenience of calculation. 
Given that RT is 5.3 k Ω, the 1 mA current source should produce 5.3 volts at the 
resonance frequency of 35.59 kHz. The results of an AC analysis are shown in 
Figure 8.29.
322
Figure 8.28
Circuit design of Example 8.6 
in a simulator.
Figure 8.29
Frequency response of the 
design from Example 8.6.",ACElectricalCircuitAnalysis.pdf
"First off, the f0 of approximately 35.6 kHz is verified by both the peak in voltage and
the phase angle reaching 0° at this frequency, the latter indicating perfect 
cancellation between the inductor and capacitor (i.e., the circuit impedance is purely 
resistive and achieving unity power factor). The cursors are used to obtain accurate 
values for f1 and f2. These frequencies are reached at 0.707 of the peak of 5.3 volts, 
or about 3.75 volts. The frequencies are approximately 34.15 kHz and 37.15 kHz, 
achieving the desired bandwidth of 3 kHz.
Low Q Parallel Resonance
There are some changes in the computations when Qparallel is low. Generally, this 
means values below 10, although we might think of values between 5 and 10 to be a 
transition region where deviations of two percentage points or less come into play. 
Once the circuit Q falls below 5, the deviations from the high Q equations grow 
rapidly and quickly rise into double digit percentages. The main item of interest here",ACElectricalCircuitAnalysis.pdf
"rapidly and quickly rise into double digit percentages. The main item of interest here
is the shift in f0. 
These deviations are caused by the fact that the approximations used for Equations 
8.17 and 8.18 are no longer true. That is, with low Qcoil values, we can no longer
assume that the transformed XL is the same as the original XL (i.e., Xp and Xs in
Figure 8.19). Given this fact, we can revisit the basic parallel RLC circuit, but this
time using the exact value from the series-to-parallel inductor transform. This is
shown in Figure 8.30. RLoad is the combined resistance of the parallel network while 
Xp is the equivalent value obtained from Equation 8.16 (slightly modified and 
repeated for convenience):
X p = X 2
+R2
X
X and R in this equation are the original series values for the inductor. At f0, the 
magnitudes of the reactances are equal, or XC = Xp, therefore,
X C = X 2
+R2
X
Expanding yields:
1
2π f 0 C = (2π f 0 L)2
+R2
2π f 0 L
Now rearrange and simplify.
2 π f 0 L",ACElectricalCircuitAnalysis.pdf
"X C = X 2
+R2
X
Expanding yields:
1
2π f 0 C = (2π f 0 L)2
+R2
2π f 0 L
Now rearrange and simplify.
2 π f 0 L
2π f 0 C = (2π f 0 L)2
+R2
323
Figure 8.30
Parallel RLC network with 
exact series transform 
equivalent.",ACElectricalCircuitAnalysis.pdf
"L
C = (2π f 0 L)2
+R2
(2π f 0 L)2 = L
C −R2
2π f 0 L = √
L
C −R2
2π f 0 =
√
1
LC − R2
L2
2π f 0 = 1
√LC √1 − C R2
L
And finally we come to:
f 0 = 1
2π√LC √1 − C R2
L (8.21)
If desired, we can treat the first term as the ordinary series resonant frequency and 
the second term as a fractional coefficient, as in:
f 0 = f series k p (8.22)
Where
k p = √1 − C R2
L (8.23)
Using Equation 8.20, kp may also be expressed as:
k p =
√
1 − 1
Q2 (8.24)
Examining Equation 8.23 might lead to some concern, namely, what happens if the 
second term is greater than or equal to 1? Remember, the definition we're using for 
resonance is the frequency at which the reactances cancel, which means the phase 
angle is 0° (unity power factor). If the second term is greater than or equal to 1, the 
phase shift will never reach 0°, and by that definition, we don't really have a 
resonant circuit anymore.
We will explore the reality of this situation by starting with a simple high Q parallel",ACElectricalCircuitAnalysis.pdf
"resonant circuit anymore.
We will explore the reality of this situation by starting with a simple high Q parallel 
circuit and then investigate the changes in the magnitude and phase response as the 
Q is decreased. We begin with the circuit of Figure 8.31.
324",ACElectricalCircuitAnalysis.pdf
"Assuming we have high circuit Q, the resonant frequency is:
f 0= 1
2π√L C
f 0= 1
2 π√1 mH 100 nF
f 0=15.92 kHz
X L = 2π f L
X L = 2π15.92 kHz 1mH
X L = 100Ω
Qcoil = X L
Rcoil
Qcoil = 100 Ω
5Ω
Qcoil = 20
There are no other resistances in the circuit, therefore Qcircuit = Qcoil and our initial 
assumption is correct. The circuit is captured in a simulator and an AC analysis is 
performed. The resulting plots are shown in Figure 8.32.
325
Figure 8.32
Simulation results using a coil 
resistance of 5 Ω (Q of 20).
Figure 8.31
A basic parallel network.",ACElectricalCircuitAnalysis.pdf
"The resonant frequency appears to be just under 16 kHz, as predicted. Cursor-based 
measurement of the frequency at which the phase crosses 0° yields 15.89 kHz. This 
turns out to be even closer than it seems. In spite of the high circuit Q, kp was 
calculated and as expected is very close to unity, namely 0.99875. When multiplied 
by the ideal f0 (i.e., using Equation 8.22), we arrive at 15.90 kHz. Splitting hairs 
perhaps, but it's good to know the deviation is shrinking.
Next, the coil resistance is raised to 50 Ω, yielding a Q of only 2. The simulation is 
run a second time. kp drops to 0.866 with this lowered Q and should produce an f0 of 
approximately 13.77 kHz. The plots are shown in Figure 8.33, and zoomed in for a 
better view. From the lower graph it is obvious that the frequency where the curve 
reaches 0° is just below 14 kHz. Accurate measurement yields 13.78 kHz, right in 
line with the theoretical computation.",ACElectricalCircuitAnalysis.pdf
"reaches 0° is just below 14 kHz. Accurate measurement yields 13.78 kHz, right in 
line with the theoretical computation.
Finally, the coil resistance is increased to 100 Ω. This drops the circuit Q to 1 and 
more importantly, brings kp down to 0. The resulting simulation plots are shown in 
Figure 8.34. At first glance the phase plot looks similar to that of Figure 8.32, 
however, notice that the phase scale has changed with 0° as the maximum. In fact, 
the phase shift never quite reaches 0°. In this regard we can still say that the kp 
equation remains an accurate predictor. 
326
Figure 8.33
Simulation results using a coil 
resistance of 50 Ω (Q of 2).",ACElectricalCircuitAnalysis.pdf
"Alternate Definition for Parallel Resonant Frequency
Instead of defining the parallel resonant frequency as the point where the power 
factor is unity, i.e., where XL and XC have the same magnitude, it can be defined in 
terms of the frequency where the magnitude of the impedance is maximum. For high
Q circuits the two definitions produce essentially the same frequency, however, as 
the circuit Q decreases into the single digits, the frequency of maximum impedance 
begins to deviate from both the high Q idealization and the general unity power 
factor definition. In fact, the frequency of maximum magnitude is situated between 
the two. We shall refer to this frequency as fZ-max to avoid confusion. The formula is:19
f Z-max = f 0
√√
2
Qcircuit
2 +1 − 1
Qcircuit
2 (8.25)
19 For a non-calculus proof, see K. Cartwright, E. Joseph, E. Kaminsky, “Finding the Exact 
Maximum Impedance Resonant Frequency of a Practical Parallel Resonant Circuit Without",ACElectricalCircuitAnalysis.pdf
"Maximum Impedance Resonant Frequency of a Practical Parallel Resonant Circuit Without 
Calculus”, Technology Interface International Journal, vol. 11, no. 1, Fall/Winter 2010. 
[Online Serial]. Available:  http://tiij.org/issues/issues/winter2010/fall_winter_2010.htm 
[Accessed February 15, 2020 ].
327
Figure 8.34
Simulation results using a coil 
resistance of 100 Ω (Q of 1).",ACElectricalCircuitAnalysis.pdf
"This equation will yield a value between the ideal high Q case and the unity power 
factor case. This can be seen in Figure 8.34 where there is still an impedance peak 
(as evidenced by the voltage peak) in spite of the fact that a phase angle of 0° is not 
reached. Furthermore, the frequency of the peak is below that of the high Q case. 
Equation 8.25 predicts a peak at 13.6 kHz which agrees with the value obtained 
from the simulation.
Combination Series and Parallel Resonance
In closing our discussion on resonance we might ask whether or not there are 
practical, everyday examples of systems exhibiting series and parallel resonance in 
series-parallel circuits. The answer is yes. A good example is that of a basic dynamic
moving coil loudspeaker of the type seen in Chapter 2. This is an electro-mechanical
system and thus a proper model has to include the effects of such items as the 
mechanical losses in the system, the mass of the cone, and the like. One possibility",ACElectricalCircuitAnalysis.pdf
"mechanical losses in the system, the mass of the cone, and the like. One possibility 
is shown in Figure 8.35. Lvc and Rvc are the inductance and resistance of the voice 
coil. The remaining components model other aspects of the electro-mechanical 
system. An impedance plot of a typical loudspeaker is shown in Figure 8.36.
328
Figure 8.35
Equivalent electrical network  
of a single dynamic 
loudspeaker.
Adapted from R.H.Small, “Direct-
Radiator Loudspeaker System 
Analysis”, Journal of the Audio 
Engineering Society, June 1972
Figure 8.36
Impedance magnitude and 
phase of a typical dynamic 
loudspeaker.
Courtesy Dayton Audio",ACElectricalCircuitAnalysis.pdf
"The loudspeaker of Figure 8.36 is a medium-size woofer with a nominal impedance 
of 8 Ω. First, note the large variation on both the phase and magnitude of the 
impedance. The parallel items from the model produce an obvious peak in 
impedance just below 30 Hz. This is referred to as the free air resonance and is 
denoted by fs. For this device, the magnitude is over three times the nominal value. 
Also note that the phase angle is 0° at fs, and that the phase is positive (inductive) 
below the resonant frequency and negative (capacitive) above it. This behavior is 
expected from a parallel resonant system. The series elements of the model create 
the rising impedance that is seen following the dip. Note that the phase angle 
continues to increase as frequency rises, indicating the growing dominance of the 
series inductive element. 
8.4 Summary8.4 Summary
Resonance can be described as a preferred mode of vibration, or a frequency at",ACElectricalCircuitAnalysis.pdf
"series inductive element. 
8.4 Summary8.4 Summary
Resonance can be described as a preferred mode of vibration, or a frequency at 
which a system operates particularly well. Resonant systems can be used to filter out
or select specific frequencies across the spectrum. Obvious uses include tuning 
circuits, oscillators, filters and the like. In electrical systems there are two basic 
forms; series RLC resonance and parallel RLC resonance. Series resonance tends to 
be the less complicated of the two. 
For series resonance, the resonant frequency, f0, is defined as the frequency at which 
the magnitude of XL equals the magnitude of XC. In this instance, the reactances 
cancel, leaving the series impedance as R. This creates a U-shaped curve for the 
impedance as it varies across the frequency spectrum. At low frequencies, the 
capacitive reactance dominates and the series impedance is high in magnitude and 
capacitive. At frequencies above the resonant frequency, the inductive reactance",ACElectricalCircuitAnalysis.pdf
"capacitive. At frequencies above the resonant frequency, the inductive reactance 
dominates and the series impedance is again high in magnitude but it is inductive. If 
this circuit is driven with a constant voltage source, the current will be maximum at 
resonance and tail off at lower and higher frequencies. The sharpness of the current 
curve across frequency is a function of the system Q, or quality factor. A high Q 
circuit is one with a very sharp and narrow curve. The “shoulders” of the curve are 
defined as the frequencies at which the power has dropped to one half of the value at
resonance. This corresponds to 0.707 times the current at resonance. The lower 
frequency is f1 and the upper frequency is f2. The difference between the two is 
called the bandwidth, BW. The ratio of resonant frequency to the bandwidth yields 
the circuit Q. Circuit Q can also be found by dividing the magnitude of reactance at",ACElectricalCircuitAnalysis.pdf
"the circuit Q. Circuit Q can also be found by dividing the magnitude of reactance at 
resonance to the total circuit resistance. In high Q circuits it is possible for the 
voltage across the inductor or capacitor to be many times higher than the source 
voltage, higher in fact, by a factor of Q. 
329",ACElectricalCircuitAnalysis.pdf
"Parallel resonance is similar to series resonance but in some ways is like its mirror 
image. In a parallel resonant circuit the inductor will dominate at low frequencies 
and produce a small net impedance. At high frequencies, the capacitor will dominate
and also produce a small net impedance. At resonance, the two effectively will 
cancel and yield a large impedance. In other words, the impedance versus frequency 
curve will appear like an upside down U, producing maximum impedance at 
resonance, and the opposite of the series impedance curve. If this system is driven 
by a constant current source, the resulting voltage will echo the shape of the 
impedance curve, producing maximum voltage at resonance. The upper and lower 
frequencies, along with the bandwidth and system Q, are defined in the same manner
as they are in the series case (with one exception regarding finding Q via resistance 
and reactance).",ACElectricalCircuitAnalysis.pdf
"as they are in the series case (with one exception regarding finding Q via resistance 
and reactance). 
There is one important caveat regarding parallel resonant circuits. Practical inductors
contain a non-trivial series coil resistance. This can play a dominant role in the 
system response. Analysis is generally handled by performing a series to parallel 
transform which creates a parallel resistance out of the inductor's series resistance. 
As a result, system Q can be found as the ratio of effective parallel resistance to 
maximum reactive magnitude, the opposite of the series case. For high Q systems, 
generally taken as 10 or higher, the resonant frequency can use the same equation as 
the series case. For low Q systems, the series to parallel transform creates a shift in 
resonant frequency, making it somewhat lower than the value obtained from the 
basic series equation. Also, the inductor and capacitor currents will be approximately",ACElectricalCircuitAnalysis.pdf
"basic series equation. Also, the inductor and capacitor currents will be approximately
Q times higher than the source current, made possible because they are 180 degrees 
out of phase with each other and effectively cancel. 
In both series and parallel systems, for high Q, f1 and f2 are assumed to lie 
equidistant from f0, splitting BW in half on either side. This is just an approximation 
and errors will grow as the Q decreases. More accurately, the two frequencies lie 
where the ratio of f1/f0 is the same as the ratio of f0/f2. 
Review QuestionsReview Questions
1. Describe the concept of resonance. How is resonance defined in a series 
RLC network?
2. Sketch the impedance versus frequency plot for series resonance.
3. Sketch the impedance versus frequency plot for parallel resonance.
4. Define the terms resonant frequency, bandwidth and Q.
5. How does inductor Q impact system Q in resonant circuits? 
330",ACElectricalCircuitAnalysis.pdf
"8.5 Exercises8.5 Exercises
Inductor Q curves to be used with the exercises below
AnalysisAnalysis
1. A circuit has a resonant frequency of 440 kHz and a system Q of 30. 
Determine the bandwidth and the approximate values for f1 and f2.
2. A circuit has a resonant frequency of 19 kHz and a bandwidth of 500 Hz. 
Determine the system Q and the approximate values for f1 and f2.
3. Find the Qcoil and coil resistance of a 150 μH inductor at 100 kHz using 
device curve A.
4. Find the Qcoil and coil resistance of a 2.2 mH inductor at 50 kHz using 
device curve D.
5. At a certain frequency, an inductor's impedance is 24 + j600 Ω. Determine 
the parallel resistance and reactance that produces the same value. 
331
Frequency (Hz)
Q
50
40
30
20
10
1 k 10 k 100 k 1 M 10 M
A
B
C
D",ACElectricalCircuitAnalysis.pdf
"6. At a certain frequency, an inductor's impedance is 3 + j150 Ω. Determine the
parallel resistance and reactance that produces the same value.
7. A certain 75 μH inductor is described by curve B. Determine the equivalent 
parallel inductor/resistor combination at 1 MHz.
8. A certain 3.3 mH inductor is described by curve C. Determine the equivalent
parallel inductor/resistor combination at 20 kHz.
9. Consider a series circuit consisting of a 2 nF capacitor, an ideal 33 μH 
inductor and a 5 Ω resistor. Determine the resonant frequency, system Q, 
and bandwidth.
10. Consider a series circuit consisting of a 20 nF capacitor, an ideal 100 μH 
inductor and a 2.7 Ω resistor. Determine the resonant frequency, system Q, 
and bandwidth.
11. Consider a series circuit consisting of a 50 nF capacitor, a 20 mH inductor 
with Qcoil of 50 and a 63 Ω resistor. Determine the resonant frequency, 
system Q, and bandwidth.
12. Consider a series circuit consisting of a 200 nF capacitor, a 1 mH inductor",ACElectricalCircuitAnalysis.pdf
"system Q, and bandwidth.
12. Consider a series circuit consisting of a 200 nF capacitor, a 1 mH inductor 
with Qcoil of 65 and a 72 Ω resistor. Determine the resonant frequency, 
system Q, and bandwidth.
13. For the circuit shown in Figure 8.37, determine the resonant frequency, 
system Q and bandwidth. Assume Rcoil = 0 Ω. If the source is 1 volt peak, 
determine the capacitor voltage at resonance.
14. For the circuit shown in Figure 8.38, determine the resonant frequency, 
system Q and bandwidth. Assume Rcoil = 0 Ω. If the source is 10 volts, 
determine the capacitor voltage at resonance.
332
Figure 8.37
Figure 8.38",ACElectricalCircuitAnalysis.pdf
"15. Repeat problem 13 but assume instead that the inductor's Rcoil = 15 Ω.
16.  Repeat problem 12 but assume instead that the inductor follows curve D.
17. For the circuit shown in Figure 8.39, determine the resonant frequency, 
system Q and bandwidth. If the source is 20 mA peak, determine the resistor
and capacitor voltages at resonance.
18. For the circuit shown in Figure 8.40, determine the resonant frequency, 
system Q and bandwidth. If the source is 100 mA, determine the resistor and
capacitor voltages at resonance.
19. For the circuit shown in Figure 8.41, determine the resonant frequency, 
system Q and bandwidth. If the source is 15 volts, determine the inductor 
and capacitor currents at resonance. Assume the inductor's coil resistance is 
3.2 Ω.
333
Figure 8.39
Figure 8.40
Figure 8.41",ACElectricalCircuitAnalysis.pdf
"20. For the circuit shown in Figure 8.42, determine the resonant frequency, 
system Q and bandwidth. If the source is 3 volts, determine the inductor and
capacitor currents at resonance. Assume the inductor's Q is 30.
21. For the circuit shown in Figure 8.43, determine the resonant frequency, 
system Q and bandwidth. If the source is 15 volts, determine the resistor, 
inductor and capacitor currents at resonance.
22. Given the circuit shown in Figure 8.44, determine the resonant frequency, 
system Q and bandwidth. If the source is 2 volts, determine the resistor, 
inductor and capacitor currents at resonance. Assume the inductor's coil 
resistance is 2.5 Ω.
23. For the circuit shown in Figure 8.45, determine the resonant frequency, 
system Q and bandwidth. If the source is 5 volts, determine the resistor, 
inductor and capacitor currents at resonance. Assume the inductor's Q is 40.
334
Figure 8.43
Figure 8.44
Figure 8.42
Figure 8.45",ACElectricalCircuitAnalysis.pdf
"24. Given the circuit shown in Figure 8.46, determine the resonant frequency, 
system Q and bandwidth. If the source is 2.5 mA, determine the resistor 
voltage and the three branch currents at resonance.
25. For the circuit shown in Figure 8.47, determine the resonant frequency, 
system Q and bandwidth. If the source is 500 μA, determine the resistor 
voltage and the three branch currents at resonance. Assume the inductor's Q 
is given by curve C.
26. Given the circuit shown in Figure 8.48, determine the resonant frequency, 
system Q and bandwidth. If the source is 10 mA, determine the resistor 
voltage and the three branch currents at resonance. Assume the inductor's Q 
is given by curve B.
DesignDesign
27. A series resonant circuit has a required f0 of 50 kHz. If a 75 nF capacitor is 
used, determine the required inductance. 
28. A series resonant circuit has a required f0 of 210 kHz. If a 22 μH inductor is 
used, determine the required capacitance. 
335
Figure 8.47
Figure 8.48",ACElectricalCircuitAnalysis.pdf
"used, determine the required capacitance. 
335
Figure 8.47
Figure 8.48
Figure 8.46",ACElectricalCircuitAnalysis.pdf
"29. A parallel resonant circuit consists of a 12 nF capacitor and a 27 μH inductor
with a Qcoil of 55. Determine the required additional parallel resistance to 
achieve a system Q of 40. 
30. A series resonant circuit has a design target of f0=200 kHz with a bandwidth 
of 5 kHz. Which of the inductor curves above (A, B, C, D) represent 
possible candidates, if any, and why/why not?
31. A parallel resonant circuit has a design target of f0=1 MHz with a bandwidth 
of 20 kHz. Which of the inductor curves above (A, B, C, D) represent 
possible candidates, if any, and why/why not?
ChallengeChallenge
32. A parallel resonant circuit has a required f0 of 50 kHz and a bandwidth of      
4 kHz. If a 75 nF capacitor is used and the load impedance is 100 kΩ, 
determine the required inductance and minimum acceptable Qcoil. 
33. A parallel resonant circuit consists of a 150 nF capacitor and a 200 μH 
inductor that has a coil resistance of 1 Ω. The desired bandwidth for the",ACElectricalCircuitAnalysis.pdf
"inductor that has a coil resistance of 1 Ω. The desired bandwidth for the 
network is 2 kHz. Determine the value of resistance to be placed in parallel 
with the network in order to achieve this goal.
34. A resonant circuit consists of a 4 nF capacitor in parallel with a 100 μH coil 
that has a coil resistance of 5 Ω. Determine the resonant frequency and 
bandwidth. Further, assume that this circuit is now loaded by an amplifier 
that has an input impedance equivalent to 10 kΩ resistive in parallel with 
500 pF of input capacitance. Also, the amplifier is connected via 25 feet of 
coaxial cable that exhibits a capacitance of 33 pF per foot. Determine the 
changes in resonant frequency and bandwidth, if any, with this load.
SimulationSimulation
35. Use an AC frequency domain analysis to verify the results of problem 13. 
Plot the resistor voltage from 0.1 f0 to 10 f0.
36. Use an AC frequency domain analysis to verify the results of problem 19.",ACElectricalCircuitAnalysis.pdf
"Plot the resistor voltage from 0.1 f0 to 10 f0.
36. Use an AC frequency domain analysis to verify the results of problem 19. 
Do this by overlapping plots of the resistor, capacitor and inductor voltages 
across a range of 0.1 f0 to 10 f0. 
336",ACElectricalCircuitAnalysis.pdf
"37. Investigate the effects of inductor Q on the system bandwidth of problem 
21. Plot the system voltage from 0.01 f0 to 100 f0 three times, the first using 
the specified coil resistance and then using values ten times larger and ten 
times smaller.
38. Investigate the effects of component tolerance on the system frequency 
response of problem 21. Plot the system voltage from 0.1 f0 to 10 f0 using a 
Monte Carlo variation on the AC frequency domain response. Set a 10% 
tolerance on the capacitor, inductor and resistor but do not alter the coil 
resistance.
39. Use an AC frequency domain analysis to verify the design of problem 27. 
Plot the resistor voltage from 0.1 f0 to 10 f0.
40. Use an AC frequency domain analysis to verify the design of problem 29. 
Plot the system voltage from 0.1 f0 to 10 f0.
41. At high Q values (>10) the capacitor and inductor voltages of series 
resonant circuits will tend to reach maximum very close to the resonant",ACElectricalCircuitAnalysis.pdf
"41. At high Q values (>10) the capacitor and inductor voltages of series 
resonant circuits will tend to reach maximum very close to the resonant 
frequency. At lower Qs, these peaks tend to diverge. A similar situation 
occurs with the currents in parallel resonant circuits. Investigate this effect 
by performing an AC frequency domain analysis on problem 14. Overlay 
plots of vab, vbc and vc for successively larger values of resistance.
42. Investigate the “Q increase” in reactive currents compared to source and 
resistive currents in a parallel resonant circuit. A simple way to verify this is 
by placing AC ammeters in each of the branches of the circuit shown in 
Figure 8.49. Use R = 630 Ω, C = 40 nF, L = 10 μH and i = 1 mA. It is 
worthwhile to compare sets of simulations for different resistor values to see
the current changes relative to the system Q. Slight variations of the source 
frequency may be required to reach the peak.
337
Figure 8.49",ACElectricalCircuitAnalysis.pdf
"9 9 Polyphase PowerPolyphase Power
9.0 Chapter Learning Objectives9.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Define the differences between polyphase and single phase systems and detail their advantages.
• Determine line voltage, line current, phase voltage and phase current in three-phase systems.
• Analyze three-phase systems in both delta (Δ) and Y (wye) configurations.
9.1 Introduction 9.1 Introduction 
In this chapter we shall introduce the concept of polyphase systems. Polyphase systems can be visualized as a 
group of individual sources of the same magnitude that are separated by a certain phase angle such that they are 
evenly divided across a single period. The polyphase load is similarly divided into individual sections or legs. 
By dividing the sources, the application of power can be much more smooth. Further, for the same total load",ACElectricalCircuitAnalysis.pdf
"By dividing the sources, the application of power can be much more smooth. Further, for the same total load 
power, the current delivered by each of the segments is reduced compared to a single phase system. For an 
analogy we could look at a bicycle. A single phase system is like pedaling with only one leg. That is, power is 
applied in a single burst per revolution of the pedal. Having two pedals is like a two-phase system; power is 
delivered twice per revolution, once for the right leg and once for the left. Because there are two pedals, it 
makes sense to separate them physically by 180 degrees or one half of a revolution, otherwise the power 
delivery will not be smooth. It should obvious to anyone who has pedaled a bike that you must pump a single 
pedal much harder using only one leg to achieve the same speed obtained when pumping with both legs. 
Polyphase loads can be balanced or unbalanced. A balanced load means that all legs or sections of the load",ACElectricalCircuitAnalysis.pdf
"Polyphase loads can be balanced or unbalanced. A balanced load means that all legs or sections of the load 
exhibit the same impedance. Consequently, the currents supplied by the sources will be the same except for the 
phase shifts between them. It is possible to create a polyphase system using any number of phases,  however, the
more phases we add, the more complex the construction of the polyphase source and load. Also, the number of 
required conductors between the source and load increases (one per phase). These all increase construction, 
installation and maintenance costs. Polyphase systems using three sections (hereafter simply referred to as 
three-phase systems) are popular because they deliver the benefits of polyphase while limiting the complexity 
and minimizing the costs. Therefore, we shall our limit our discussion to three-phase systems utilizing balanced 
loads.",ACElectricalCircuitAnalysis.pdf
"and minimizing the costs. Therefore, we shall our limit our discussion to three-phase systems utilizing balanced 
loads.
Three-phase systems can be wired in either delta or Y configurations, or a combination. These are reminiscent of
the delta and Y constructs seen in earlier chapters. We shall investigate all of the combinations to determine 
system parameters such as line voltage, line current and load power. We will also investigate power factor 
correction for balanced loads that have a non-negligible phase angle.
338",ACElectricalCircuitAnalysis.pdf
"9.2 Polyphase Definition9.2 Polyphase Definition
A polyphase system uses multiple current-carrying wires with multiple sub-
generators, each with their own unique phase. This allows for considerable delivery 
of power to the load. The most popular scheme is the three-phase configuration. This
can be visualized as three individual sine generators that are interconnected as 
shown in Figure 9.1. To the left is a Y (also known as wye or T) connected generator. 
To the right is a delta (i.e. Δ, and also known as π when drawn upside down) 
connected generator. 
There are multiple ways of reproducing these generators on schematics. Some 
alternate forms for Y -connected generators are shown in Figure 9.2. The lone “tail” 
shown on the version to the right is a connection back to the common center of the 
three sub-generators. This is called the neutral line. It is not always used.
In Figure 9.3 we have some alternate forms for delta-connected generators. Delta",ACElectricalCircuitAnalysis.pdf
"In Figure 9.3 we have some alternate forms for delta-connected generators. Delta 
generators do not have the optional fourth connection as there is no common center 
point. Also, note that the version on the left is drawn upside down (i.e., in the π 
configuration).
339
Figure 9.3
Alternate schematic symbols  
for delta-connected generators.
A
B C
A B
C
Figure 9.1
Three-phase generators:Y-
connected (left) and delta-
connected (right).
Figure 9.2
Alternate schematic symbols  
for Y-connected generators.",ACElectricalCircuitAnalysis.pdf
"Of particular importance is the relative phase of each source. As the load will also 
have three segments or legs (a three-phase load), a consistent delivery of power 
demands that the three sources be spread equally over time. This means that each 
source is one-third of a cycle, or 120 degrees, out of phase with the other legs (i.e., 
leading one and lagging the other). This is shown in Figure 9.4. We shall only 
consider the case of balanced loads, that is, where each leg of the load is identical to
the other legs.
Notice the effectiveness of using the 120 degree stagger. At any time there is always 
either a peak or the shoulders of two adjacent peaks. These peaks can be either 
positive or negative polarity. This makes for a total of six peaks across the 
waveform's period. Therefore, a null is never more than 30 degrees away from the 
nearest peak. Indeed, when it is 30 degrees away, then it's precisely in between two",ACElectricalCircuitAnalysis.pdf
"nearest peak. Indeed, when it is 30 degrees away, then it's precisely in between two 
peaks, and the value from each is the sine of 60 degrees (i.e., 30 degrees off of the 
peak at 90 degrees) or 86.6% of the peak value. Thus, it should be obvious that 
consistent power can be applied by this system. The inevitable car analogy is that we
have gone from a two cylinder engine to a six cylinder engine when moving from 
single phase to three-phase. 
There are several ways to connect three-phase generators to three-phase loads, as we
shall see in the next section.
340
Figure 9.4
Relative phases of the three 
sub-generators.
Three Phase
 
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
 
Time
0 0.5 1 1.5 2
Phase 1 Phase 3Phase 2",ACElectricalCircuitAnalysis.pdf
"9.2 Three-Phase Connections9.2 Three-Phase Connections
It is possible to configure systems using delta- or Y-connected sources with either 
delta- or Y-connected loads. One item to note is that delta-connected systems are 
always three wire systems while Y -connected systems can make use of a fourth 
neutral wire (the common point to which all three sources connect). 
Homogeneous SystemsHomogeneous Systems
The most straightforward systems are delta-to-delta and Y-to-Y . We shall refer to 
these as homogeneous systems as the structures of the generator and load are similar.
Examples are shown in Figures 9.5 and 9.6, respectively.
In these configurations, each leg of the load matches up with a corresponding leg of 
the generator. In the delta-delta configuration of Figure 9.5, it should be obvious just
by inspection that the voltage across any load leg must equal the voltage of the 
corresponding generator leg. For example, the load impedance connected between A'",ACElectricalCircuitAnalysis.pdf
"corresponding generator leg. For example, the load impedance connected between A'
and B' must see the voltage presented by the generator situated between A and B 
because A is directly connected to A' as is B to B'. Similarly, for the Y-Y 
configuration of Figure 9.6, the current through any load leg must equal the current 
flowing through the associated generator leg as there are no other paths for current 
between A and A', B and B', and C and C'.
As the load is balanced and the legs of the generator are identical except for their 
phase, it must be the case that the voltages and currents (and hence the powers) for 
341
Z Z
Z
A
B C
A'
B' C'
+ -
+
-+
-
Figure 9.5
A delta-connected generator 
with a delta-connected load 
(delta-delta).
Z Z
Z
A B
C
A' B'
C'
+
-
+
-
+
-
Figure 9.6
A Y-connected generator with  
a Y-connected load (Y-Y). 
Optional fourth neutral wire 
from center to center shown.",ACElectricalCircuitAnalysis.pdf
"each leg of the load must be the same, with the exception of the phase. This is true 
for both the Y-Y configuration as well as the delta-delta configuration. The tricky bit
here is the difference between a source (or load) current or voltage, and the line 
current or voltage.
Line voltage is the voltage magnitude between any two conductors 
connecting the source to the load, excluding ground or common. 
Line current is the current magnitude flowing in any conductor connecting 
the source to the load, excluding ground or common.
Consider the delta-delta system of Figure 9.5. We have already established that the 
voltage developed by generator A,B must be the same as the voltage across the load 
A',B'. Thus, the voltage measured from the A, A' conductor to the B,B' conductor 
must be the same as source and and load voltages. In other words, in the delta-delta 
configuration, the source, load, and line voltages are all the same.",ACElectricalCircuitAnalysis.pdf
"configuration, the source, load, and line voltages are all the same.
We also found that the source and load currents must be the same for the delta-delta 
configuration, however, this does not imply that the current flowing through the 
wire connecting A to A' must be the same as the current flowing through either the 
generator or the load. After all, two load wires connect to A', not just one. By 
definition, the current flowing through that wire is the line current, and therefore in a
delta-delta configuration, the line current is not the same as the source or load 
currents. To avoid confusion, the voltage or current associated with a single leg is 
referred to as the phase voltage or current versus the line voltage or current.
Turning to the Y-Y configuration of Figure 9.6, we see an opposite situation. The 
source, load, and line currents will all be the same. On the other hand, the line 
voltage comprises two generators, not one (e.g., from A to B or from B to C). Thus,",ACElectricalCircuitAnalysis.pdf
"voltage comprises two generators, not one (e.g., from A to B or from B to C). Thus, 
for a Y-Y configuration the source and load voltages are the same, but they are not 
equal to the line voltage (nor is twice, thanks to the phase shift).
Determining Line Voltage and CurrentDetermining Line Voltage and Current
In order to determine the line voltage for a Y-connected generator (and similarly, the 
line current for a delta connected generator), it is useful to examine a phasor plot of 
the individual generator voltages. This is shown in Figure 9.7. We have three 
voltages of identical amplitude, the only difference between them being their phase. 
Each vector is separated from the others by 120 degrees. Further, each individual 
generator is connected from the common point to one of the external points of A, B 
and C. Line voltage is defined as the potential existing between any two if these 
three points. While it's possible to simply subtract one generator voltage from",ACElectricalCircuitAnalysis.pdf
"three points. While it's possible to simply subtract one generator voltage from 
another to arrive at the difference, there is a nice graphical solution from which we 
can derive a precise formula for the line voltage given the generator voltage.
342",ACElectricalCircuitAnalysis.pdf
"We begin by focusing on quadrants two and three of the phasor plot. This section is 
redrawn in Figure 9.8. In reality, any two vectors can be used for the following 
proof, but this pair turns out to be particularly convenient in its orientation.
For ease of use we shall normalize the magnitude of the generator voltage to unity. 
What we see is that the B and C vectors are split perfectly by the horizontal axis; that
which is above the axis is perfectly mirrored below it. In the upper portion we find a 
right triangle with a hypotenuse of unity (dark red). The angle it makes with the 
horizontal must be half of the angle between it and the C vector. That's half of 120 
343
30°
B
C
1
60°
0.5
X
X
Figure 9.8
Solving for the line voltage of   
a Y-connected generator.
120°
B
A
C
120°
120°
Figure 9.7
Phasor diagram of Y-connected
generator.",ACElectricalCircuitAnalysis.pdf
"degrees, or 60 degrees. As the sum of the interior angles of a triangle must be 180 
degrees, this means that the third angle must be 30 degrees. The horizontal leg of the
triangle (dark yellow or maybe “spicy mustard”) can be determined because we 
know both the hypotenuse and the opposite angle.
opposite = hypotenuse×sin θ
The sine of 30 degrees is exactly 0.5, therefore, the horizontal leg of the triangle 
must be 0.5 times the magnitude of unity, or 0.5. We can use the Pythagorean 
theorem to find the remaining vertical leg (purple).
vertical = √hypotenuse2
−horizontal2
vertical = √12−0.52
vertical = √
3
4
vertical = 1
2 √3
The vertical leg is perfectly mirrored below the horizontal axis. Therefore, the span 
from B to C must be twice this value, or √3. As the voltage developed across each 
leg of the generator is referred to as the generator's phase voltage, we can state:
The line voltage for a Y-connected generator is √3 times its phase voltage.       (9.1)",ACElectricalCircuitAnalysis.pdf
"The line voltage for a Y-connected generator is √3 times its phase voltage.       (9.1)
For example, if the phase voltage of a Y-connected generator is 120 volts, the line 
voltage would be √3 times larger, or approximately 208 volts. 
For a delta-connected generator, the same holds true for the phase and line currents, 
with the proof left as an exercise. That is, 
The line current for a delta-connected generator is √3 times its phase current.    (9.2)
These same relationships hold for the loads as well as the sources, e.g., the current in
a leg of a Y-connected load will be the same as the line current and its phase voltage 
will be √3 times smaller than the line voltage. 
In summation: For delta configurations (generator or load), the phase 
voltage is equal to the line voltage while the line current is larger than the 
phase current by √3. For Y configurations, the phase current is equal to the 
line current while the line voltage is √3 larger than the phase voltage.",ACElectricalCircuitAnalysis.pdf
"line current while the line voltage is √3 larger than the phase voltage.
For homogenous systems, as the generator and load share the same configuration, 
the phase voltages and currents of the load must be identical to those of the 
generator. A useful memory aid is that the power dissipated in the system must equal
the power generated. 
344",ACElectricalCircuitAnalysis.pdf
"Example 9.1
A three-phase delta-connected generator feeds a three-phase delta-connected
load like the system shown in Figure 9.5. Assume the generator phase 
voltage is 120 V AC RMS. The load consists of three identical legs of 50 Ω 
each. Determine the line voltage, load phase voltage, generator phase 
current, line current, load phase current and the total power delivered to the 
load.
As this is a homogenous (delta-delta) system, the load phase voltage and 
current are the same as those of the generator. Therefore, the load phase 
voltage must also be 120 volts. Second, in a delta configuration, the line 
voltage equals the phase voltage, again 120 volts. The load phase current is 
found via Ohm's law and will be an RMS value because the voltage is RMS:
i phase = v phase
Zload
i phase = 120 V
50Ω
i phase = 2.4 A
The generator's phase current must be the same because the generator and 
load have the same configuration. For delta configurations, the line current",ACElectricalCircuitAnalysis.pdf
"load have the same configuration. For delta configurations, the line current 
is √3 times larger than the phase current, thus,
iline = √3×iphase
iline = √3×2.4A
iline ≈ 4.157 A
Finally, total power can be found with a straight application of power law as 
the load is purely resistive and we have RMS values. Remember, this is three
times the power dissipated in one leg.
Ptotal = 3×iphase
2
×R
Ptotal = 3×(2.4 A)2×50Ω
Ptotal = 864 W
This is equivalent to about 1.2 HP. We could have also computed the load 
phase power by using the squared phase voltage divided by the load 
resistance, or by multiplying the phase voltage by the phase current. As this 
is a purely resistive load, there is no phase angle, and thus no power factor 
with which to concern ourselves.
 
345",ACElectricalCircuitAnalysis.pdf
"Example 9.2
A three-phase Y-connected generator feeds a three-phase Y-connected load 
similar to the system shown in Figure 9.6. Assume the generator phase 
voltage is 220 V AC RMS. The load consists of three identical legs of 100 Ω 
each. Determine the line voltage, load phase voltage, generator phase 
current, line current, load phase current and the total power delivered to the 
load.
This is a homogenous (Y -Y) system, therefore the load phase voltage and 
current are the same as those of the generator. Consequently, the load phase 
voltage must be 220 volts. In a Y configuration, the line voltage equals the 
phase voltage times √3. 
vline = √3×vphase
vline = √3×220 V
vline ≈ 381 V
The load phase current is found via Ohm's law and will be an RMS value 
because the voltage is RMS. This is the same as the generator phase current 
and also the line current.
i phase = v phase
Z load
i phase = 220 V
100 Ω
i phase = 2.2A",ACElectricalCircuitAnalysis.pdf
"and also the line current.
i phase = v phase
Z load
i phase = 220 V
100 Ω
i phase = 2.2A
Total power can be found using basic power law as the load is purely 
resistive and we have RMS values. In this case we'll use current times 
voltage for a change of pace.
Ptotal = 3×i phase×vphase
Ptotal = 3×2.2 A×220 V
Ptotal = 1452 W
This is just shy of 2 HP. Once again, this is a purely resistive load and there 
is no phase angle. Thus, the power factor is unity with the real and apparent 
powers being the same.
346",ACElectricalCircuitAnalysis.pdf
"Example 9.3
For the the system shown in Figure 9.9, determine the total apparent and real
power delivered to the load. Also find the line voltage. The phase voltage of 
the source is 240 volts RMS at 60 Hz.
Given the fact that the three load legs are all together at one common point 
(ground), this must be a Y-Y system. Consequently, we know that the line 
voltage must be √3 times the phase voltage of the generator. 
vline = √3×vphase
vline = √3×240 V
vline ≈ 416 V RMS
This is a homogenous system (Y-Y) so we also know that the load voltage is 
equal to the generator voltage, or 240 volts RMS. From that we can find the 
load current (the line current must be the same value because this is a Y-
connected load).
i phase = v phase
Zload
i phase = 240 V
40+j 30Ω
i phase = 4.8 −36.87 ° A
The phase angle is appropriate for the 0° reference generator. The other two 
angles will be off of this by ±120°. The apparent power is simply the 
product of the load current and voltage magnitudes.",ACElectricalCircuitAnalysis.pdf
"angles will be off of this by ±120°. The apparent power is simply the 
product of the load current and voltage magnitudes.
S = 3×iload ×vload
S = 3×4.8A×240 V
S = 3456 VA
347
Figure 9.9
Circuit for Example 9.3.",ACElectricalCircuitAnalysis.pdf
"The real power can be found a few different ways:
P = S×cosθ
P = 3456VA×cos(−36.87 °)
P = 2765W
P = 3×iload
2
×Rload
P = 3×4.8A2
×40Ω
P = 2765W
Computer SimulationComputer Simulation
The circuit of Example 9.3 is worthy of a simulation. The first thing to do is to 
determine an appropriate value of inductance to achieve a reactance of j40 Ω. Given 
the 60 Hz source frequency, this turns out to be approximately 80 mH. The circuit is 
constructed as shown in Figure 9.10. The 240 volt RMS source phase voltage is 
equivalent to approximately 340 volts peak. The positions of the inductor and 
resistor in each leg have been swapped for a reason that will be apparent shortly.
The immediate item of interest is to verify the time shifts and amplitudes of the 
phase voltages. These correspond to nodes 1, 2 and 3. In this configuration, the load 
phase voltage equals the generator phase voltage, thus they should be 340 volts peak
and separated by 120 degrees or 1/3rd of a cycle.",ACElectricalCircuitAnalysis.pdf
"phase voltage equals the generator phase voltage, thus they should be 340 volts peak
and separated by 120 degrees or 1/3rd of a cycle. 
A transient analysis is performed, plotting the node voltages of interest. The result is 
shown in Figure 9.11. The voltages are precisely as expected and the plot compares 
perfectly to the theoretical plot of Figure 9.4.
348
Figure 9.10
The equivalent system of  
Figure 9.9 in a simulator.",ACElectricalCircuitAnalysis.pdf
"Now we check the line voltage. This was calculated to be 416 volts RMS, or 
approximately 588 volts peak. The post processor is used to display the result of 
node voltage 1 minus node voltage 2. This is shown in Figure 9.12. Again, the 
results are as expected with a peak just under 600 volts.
Finally, we will investigate the true load power. Perhaps the easiest way to do this is 
to determine the voltage across the resistive portion of the load. From prior work we 
know that true power is only associated with resistance, not reactance. Thus, all we 
need to do is measure the peak voltage across the resistor. From there, we find its 
RMS equivalent, square it, and divide by the resistor value. This gives us the true 
load power in one leg. For the total power we simply triple the result. Obtaining the 
voltage across the resistor is easy if the resistor is attached to ground. In that case, 
it's just the voltage at the node to which the resistor is connected. This is why the",ACElectricalCircuitAnalysis.pdf
"it's just the voltage at the node to which the resistor is connected. This is why the 
inductor and resistor positions were swapped in the simulation. As they are in series,
it makes no difference to the overall load impedance, however, the new arrangement 
allows us to obtain the resistor voltage directly instead of having to rely on a 
differential voltage obtained through the post processor. 
Another transient analysis is performed, this time plotting the voltage across one of 
the load resistors; namely node 4. The result is shown in Figure 9.13. The peak of 
this waveform is measured to be 271.5 volts, or about 192 volts RMS. Squaring this 
and dividing by 40 Ω yields a little over 921 watts per leg, for a total of about 2765 
watts, as expected.  
349
Figure 9.11
The three load voltages 
simulated from Figure 9.10.",ACElectricalCircuitAnalysis.pdf
"350
Figure 9.12
One of the line voltages 
simulated from Figure 9.10.
Figure 9.13
Simulated voltage across one  
of the load resistors in      
Figure 9.10.",ACElectricalCircuitAnalysis.pdf
"Heterogeneous SystemsHeterogeneous Systems
Systems configured as delta-to-Y and Y-to-delta appear to be a bit more complex 
than homogeneous systems. We shall refer to these as heterogeneous systems as the 
structures of the generator and load are of opposite kind. Examples are shown in 
Figures 9.14 and 9.15, respectively.
These systems are not nearly so difficult as some people think; all you have to do is 
remember statements 9.1 and 9.2. Indeed, the summation is worth repeating here:
For delta configurations (generator or load), the phase voltage is equal to the
line voltage while the line current is larger than the phase current by √3. For 
Y configurations, the phase current is equal to the line current while the line 
voltage is √3 larger than the phase voltage.
You can think of analyzing these systems as a two-step process. First, determine the 
line voltage and current from either the generator or load; and second, transition",ACElectricalCircuitAnalysis.pdf
"line voltage and current from either the generator or load; and second, transition 
from the line to the other side (load or generator). If confusion sets in, remember that
power generated must equal power dissipated or delivered.
In Figure 9.14, the line voltage equals the generator phase voltage. The load is Y-
connected, so each leg sees the line voltage divided by √3. Based on this, each leg of
the load current can be computed. Note that the line current equals the load current. 
The generator phase current will be the line current divided by √3. 
351
A
B C
Z Z
Z
A' B'
C'+ -
+
-+
- Figure 9.14
A delta-connected generator 
with a Y-connected load   
(delta-Y).
Z Z
Z
A'
B' C'
A B
C
+
-
+
-
+
-
Figure 9.15
A Y-connected generator with  
a delta-connected load (Y-
delta).",ACElectricalCircuitAnalysis.pdf
"In Figure 9.15, the line voltage equals √3 times the generator phase voltage. The 
load is delta-connected, so each leg sees the line voltage. Knowing this, each leg of 
the load current can be computed. Also, the line current equals the generator phase 
current, and the load phase current will equal the line current divided by √3. 
Example 9.4
A delta-Y system like the one shown in Figure 9.14 has a generator phase 
voltage of 230 volts RMS at 50 Hz. If the load is 2000° Ω, determine the 
generator phase current, the line voltage, the load phase voltage, the load 
phase current and the total power delivered to the load.  
The generator is delta connected so the line voltage equals the generator 
phase voltage, or 230 volts. The load, being Y-connected, will see a phase 
voltage that is reduced by a factor of √3. 
vload = vline
√3
vload = 230 V
√3
vload ≈ 132.8V RMS
We can use Ohm's law to determine the load phase current.
iload = v phase
Zload
iload = 132.8 V
200  0° Ω",ACElectricalCircuitAnalysis.pdf
"√3
vload = 230 V
√3
vload ≈ 132.8V RMS
We can use Ohm's law to determine the load phase current.
iload = v phase
Zload
iload = 132.8 V
200  0° Ω
iload ≈0.664 A RMS
Being Y-connected, the line current must be the same as the load phase 
current, or 0.664 amps. For delta connections, the line current is √3 times 
larger than the phase current, therefore the generator phase current must be 
√3 times smaller.
igen = iline
√3
igen = 0.664 A
√3
igen ≈ 0.383A RMS
The load is purely resistive and we have RMS values so the total power can 
be found via power law (apparent power equals true power in this case). 
352",ACElectricalCircuitAnalysis.pdf
"Ptotal = 3×iload
2
×R
Ptotal = 3×(0.664 A)2
×200Ω
Ptotal = 264 W
As a crosscheck, the power generated is:
Ptotal = 3×igen×vgen
Ptotal = 3×0.383 A×230 V
Ptotal = 264 W
Power generated equals power dissipated.
Example 9.5
A Y -delta system like the one shown in Figure 9.15 has a generator phase 
voltage of 100 volts RMS at 60 Hz. If the load has a magnitude of 50 Ω with 
a lagging power factor of 0.8, determine the generator phase current, the line
voltage, the load phase voltage, the load phase current and the total true 
power delivered to the load.  
The Y-connected generator creates a line voltage equal to the generator 
phase voltage times √3. This is also the load phase voltage as it is delta-
connected.
vline = √3×vphase
vline = √3×100 V
vline ≈ 173.2 V RMS
The delta-connected load will see a phase voltage that is the same as the line
voltage, or 173.2 volts. From this we can determine the load current. 
 
iload = v phase
Zload
iload = 173.2 V
50 Ω
iload ≈3.464 A RMS",ACElectricalCircuitAnalysis.pdf
"voltage, or 173.2 volts. From this we can determine the load current. 
 
iload = v phase
Zload
iload = 173.2 V
50 Ω
iload ≈3.464 A RMS
As the load is delta-connected, the line current is the load current times √3. 
The generator phase current will be the same as the line current.
iline = √3×iphase
iline = √3×3.464 A
iline = 6 A RMS
353",ACElectricalCircuitAnalysis.pdf
"The true load power can be found several ways. First, we can use the i2 R 
form. To do this we need to find the resistive portion of the load. Recall that 
the power factor is equal to cosine θ. Therefore the impedance angle is:
θ = cos−1
PF
θ = cos−1
0.8
θ ≈ 36.9°
The real part is:
R = Z cosθ
R = 50Ωcos36.9°
R = 40 Ω
Alternately, we could've just multiplied Z by PF to obtain this. Continuing:
 
Ptotal = 3×iload
2×R
Ptotal = 3×(3.464 A)2
×40Ω
Ptotal = 1440 W
We could also find the apparent power and use the power factor.
Ptotal = 3×vload ×iload PF
Ptotal = 3×173.2 V×3.464 A×0.8
Ptotal = 1440 W
As a crosscheck, compare the power dissipated to the power generated.
Ptotal = 3×vgen×i gen×PF
Ptotal = 3×100 V×6A×0.8
Ptotal = 1440 W
9.3 Power Factor Correction9.3 Power Factor Correction
As we saw in earlier work, reactive loads demand higher currents than purely 
resistive loads for a given true load power. The ratio between apparent power, S, and",ACElectricalCircuitAnalysis.pdf
"resistive loads for a given true load power. The ratio between apparent power, S, and
true power, P, is the power factor, PF. Power factor may also be computed as the 
cosine of the load impedance angle. This situation remains for three-phase systems. 
If a balanced three-phase load has a large reactive component, the line current and 
generator phase current will be higher than necessary. The solution to this is power 
factor correction; the introduction of reactive elements that will counterbalance the 
reactive power of the load, essentially providing an opposing current such that the 
reactive currents cancel. 
354",ACElectricalCircuitAnalysis.pdf
"In three-phase systems the situation is potentially complicated by the fact that the 
load is split into three parts and can be either Y-connected or delta-connected. 
The process for three phase is essentially the same as it is for single phase, but with 
a couple slight twists. The first course of action is to determine the reactive power, 
Q, of the load. As we are dealing with balanced loads, it is usually easiest to just 
concentrate on a single leg. There are two basic possibilities. First, if the load 
reactance is known, it is a simple matter to determine the reactive power by finding 
the load phase current, squaring it, and then multiplying by the load reactance. In 
contrast, if the load is described in terms of a power factor, the apparent power can 
be computed from the generator phase voltage and current, and then the power 
factor can be used to find the reactive power (e.g., finding true power and then using",ACElectricalCircuitAnalysis.pdf
"factor can be used to find the reactive power (e.g., finding true power and then using
the Pythagorean theorem). Once the reactive power is known, the required reactance
can be found using power law and either the phase voltage or current. Finally, the 
reactance value is used to determine the component value. As many loads are 
inductive, the compensating component usually will be capacitive. There will three 
units, one for each leg of the load.
For practical purposes, the compensating device is placed across the load terminals 
rather than in series with it. This is true whether the load is Y-connected or delta-
connected. In other words, the compensating devices always will be placed in a delta
configuration. This is true even if the load is Y-connected. We shall look at both 
situations in the next two examples.
Example 9.6
The Y-delta system shown in Figure 9.16 has a generator phase voltage of 
120 volts RMS at 60 Hz. Determine the power factor, the generator phase",ACElectricalCircuitAnalysis.pdf
"The Y-delta system shown in Figure 9.16 has a generator phase voltage of 
120 volts RMS at 60 Hz. Determine the power factor, the generator phase 
current, and the total real and apparent power delivered to the load. Also 
determine components to correct the power factor and the new generator 
phase current.
355
Figure 9.16
Circuit for Example 9.6.",ACElectricalCircuitAnalysis.pdf
"First, the power factor is the cosine of the impedance angle. At 60 Hz, the 
reactance of the 100 mH inductor is −j37.7 Ω. This is in series with the 
resistance for a load impedance of 100 −j37.7 Ω or 106.920.66° per leg. 
The cosine of this angle is 0.9357.
The voltage across each leg of the load will equal the line voltage.
vline = √3×vphase
vline = √3×120 V
vline ≈ 207.8 V RMS
This will produce a load phase current magnitude of:
iload = v phase
Zload
iload = 207.8 V
106.9 Ω
iload ≈1.944 A RMS
Now we can find the load powers.
S = 3×v phase×iphase
S = 3×207.8 V×1.944 A
S ≈ 1212 VA
P = S×PF
P = 1212VA×0.9357
P ≈ 1134 W
Q = S sinθ
Q = 1212VAsin 20.66 °
Q ≈ 427.6 VAR
The load is inductive so the compensation components need to be 
capacitors. Each capacitor needs to create 427.6/3 V AR, or 142.5 V AR. The 
required reactance is:
X C =− j v phase
2
Q
X C =− j (207.8V)2
142.5 VAR
X C ≈− j 303 Ω
And finally, the capacitance value:
356",ACElectricalCircuitAnalysis.pdf
"C = 1
2π f X C
C = 1
2π60 Hz 303Ω
C ≈8.75μ F
These capacitors would be placed directly in parallel with each leg of the 
load and should result in a reduction of the generator and line currents.
Computer SimulationComputer Simulation
To see the effect of power factor correction, the circuit used in Example 9.6 is 
captured in a simulator, as illustrated in Figure 9.17. The goal here is to show the 
reduction in supplied current. To facilitate this, the normal Y-connected three-phase 
source is not used. Instead, three individual sine sources are used, each with a proper
phase shift. A small 1 Ω sensing resistor is inserted in series with one of the sources. 
The voltage across this resistor is easily measured (node 7) and serves as a proxy for
the generator phase current. Compared to the sizes of the other components, this 
extra resistance will have minimal impact on overall circuit behavior, perhaps",ACElectricalCircuitAnalysis.pdf
"extra resistance will have minimal impact on overall circuit behavior, perhaps 
shifting current values around 1% or so. Reducing the resistance to 0.1 Ω will reduce
errors to negligible levels, but 1 Ω is convenient as no scaling is needed and will be 
sufficient to show the effect of power factor correction on source current.
A transient simulation is run on the circuit, plotting the voltage at node 7. This is 
shown in Figure 9.18. Due to the 1 Ω sensing resistance, the voltage value is the 
same as the current value in amps. The peak value of the current is approximately 
4.7 amps. This agrees with the calculated value of 4.76 amps (1.944 amps RMS 
times √2 times √3). Now that a baseline for the current has been established, we turn
our attention to the modified version of the circuit with power factor correction.
357
Figure 9.17
Circuit of Figure 9.16 in a 
simulator.",ACElectricalCircuitAnalysis.pdf
"The three power factor correction capacitors are added in parallel with the existing 
load legs (i.e., from line to line). This is illustrated in Figure 9.19. 
The transient simulation is repeated. The results are shown in Figure 9.20. The peak 
current in this version of the circuit is approximately 4.4 amps. Theoretically, the 
current should be scaled by the power factor, or 0.9357. As previously calculated, 
the value of the original generator phase current is 4.76 amps peak. Multiplying that 
by the power factor yields approximately 4.45 amps peak. The small deviation 
between this result and the simulation is due to the effect of the 1 Ω sensing resistor. 
358
Figure 9.18
Generator phase current for the
original circuit of Figure 9.17.
Figure 9.19
Power factor corrected circuit 
of Figure 9.16 in a simulator.",ACElectricalCircuitAnalysis.pdf
"Example 9.7
The Y-Y system shown in Figure 9.21 has a generator phase voltage of 230 
volts RMS at 50 Hz. The load draws 900 V A with a power factor of 0.85 
lagging. Determine the generator phase current. Also determine components 
to correct the power factor and the new generator phase current once the 
system is corrected.
359
Figure 9.20
Generator phase current for the
power factor corrected circuit 
of Figure 9.19.
Figure 9.21
Circuit for Example 9.7.",ACElectricalCircuitAnalysis.pdf
"We're looking for the generator phase current so let's break this down to a 
single leg, first. The total apparent power, S, is 900 V A. For a single leg that's
300 V A. This is a Y-Y system so the generator phase current and voltage are 
the same as the load phase current and voltage. The current can be found via 
the apparent power.
i = S
v
i = 300 VA
230 V RMS
i ≈ 1.304 A RMS
Given a power factor of 0.85, we can determine the real and reactive 
powers.
P = S×PF
P = 300 VA×0.85
P = 255 W
Q = √S2
−P2
Q = √(300 VA)2−(255 W)2
Q ≈ 158 VAR inductive
For power factor correction, we need 158 V AR capacitive per leg to 
counteract this. These capacitors will be placed across the load terminals in a
delta configuration. As such, they will see the line voltage. For a Y-
connected generator, the line voltage is the phase voltage times √3. The 
result here is 230 volts times √3, or 398.4 volts RMS. From this we may 
determine the required reactance.
X C =− j vphase
2
Q
X C =− j (398.4 V)2",ACElectricalCircuitAnalysis.pdf
"result here is 230 volts times √3, or 398.4 volts RMS. From this we may 
determine the required reactance.
X C =− j vphase
2
Q
X C =− j (398.4 V)2
158 VAR
X C ≈− j 1004.6Ω
The corresponding capacitance value is:
C = 1
2π f X C
C = 1
2π50Hz 1004.6Ω
C ≈3.17μ F
360",ACElectricalCircuitAnalysis.pdf
"Computer SimulationComputer Simulation
In Example 9.7 we computed the generator phase current to be 1.304 amps RMS, 
which is equivalent to 1.844 amps peak. If the corrected circuit is proper, then the 
apparent power should fall to the real power, or 255 watts. The resulting generator 
phase current should be this power divided by the generator phase voltage, 255/230, 
or 1.109 amps RMS. This is equivalent to 1.568 amps peak. (Alternately, we could 
multiply the original current by the power factor because the voltage is constant.)
To verify the effectiveness of the circuit modification, we start by capturing the 
circuit in a simulator, as shown in Figure 9.22.
As in the prior example, the source is built from three discrete sine generators at 
appropriate phases. Beneath one of them, a 1 Ω current sensing resistor is added 
(node 7). This value should produce no more than about 1% deviation as it is a full 
two orders of magnitude smaller than the other circuit resistances.",ACElectricalCircuitAnalysis.pdf
"two orders of magnitude smaller than the other circuit resistances. 
An interesting question for the sharp-eyed observer is how the load resistor and 
inductor values were obtained. This turns out to be not so difficult. We have already 
computed the true and reactive powers for each leg. We also know the load phase 
voltage (it's the same as the generator, 230 volts, as it's a Y-Y connection). 
Therefore, we can find the R and XL values as we have already computed the load 
current and can use this to determine the load impedance, Z. The power factor is 
known, and from this the real and reactive parts can be deduced.
Z = v phase
i
Z = 230 V
1.304 A
Z ≈176Ω
This is the magnitude. For the sake of completeness, the angle is the arccosine of the
361
Figure 9.22
Generator phase current for the
original circuit of Figure 9.21.",ACElectricalCircuitAnalysis.pdf
"power factor, or cos-1(0.85), which is 31.8 degrees. The fastest way to determine R is
to recognize that the real portion is the impedance magnitude times the power factor:
R = Z ×PF
R = 176Ω×0.85
R ≈150Ω
The reactive portion can be found via the Pythagorean theorem or by using the 
power relation. Then we apply the reactance formula to find the inductance.
X L = Q
i2
X L = 158 VAR
(1.304 A RMS)2
X L ≈ 92.9Ω
L = X L
2 π f
L = 92.9Ω
2π50 Hz
L ≈296 mH
The result of a transient analysis is shown in Figure 9.23. The measured peak phase 
current is 1.837 amps. This compares nicely with the expected value of 1.844 amps.
362
Figure 9.23
Generator phase current for the
original circuit of Figure 9.22.",ACElectricalCircuitAnalysis.pdf
"For the comparison, the power factor correction capacitors are added in a delta 
configuration (across the lines) as shown in Figure 9.24. 
Another simulation is run, the result shown in Figure 9.25. The peak current has 
decreased to 1.56 amps. This is just slightly lower than the expected value of 1.568 
amps peak. Again, this small deviation is due to the effect of the sense resistor.
363
Figure 9.24
Power factor corrected circuit 
of Figure 9.21 in a simulator.
Figure 9.25
Generator phase current for the
power factor corrected circuit 
of Figure 9.24.",ACElectricalCircuitAnalysis.pdf
"9.4 Summary9.4 Summary
Polyphase systems can be thought of as a group of individual sources of the same 
magnitude that are synced together and where the load is similarly divided into 
sections or legs. By spreading out the source currents across the waveform's period, 
a smooth application of power to the load can be achieved. Also, for the same line 
current, more power can be delivered to the load than that of a single phase system. 
Loads can be balanced or unbalanced. A balanced load means that all legs of the 
load exhibit an identical impedance. Thus, the currents coming out of the source will
be the same except for the phase shifts spreading them across a single period. While 
any number of phases is possible, three-phase systems are popular as they deliver 
the benefits of polyphase while limiting complexity.
A three-phase source produces currents that are 120 degrees apart. That is, if the first",ACElectricalCircuitAnalysis.pdf
"the benefits of polyphase while limiting complexity.
A three-phase source produces currents that are 120 degrees apart. That is, if the first
signal is taken as the reference, or 0°, then the other two are at 120° and 240°. Both 
the source and the load can be configured in one of two ways: delta or Y . This makes
four possible combinations for the source-load connection; namely Y-Y , delta-delta, 
Y-delta, and delta-Y . In a Y -Y connection the source phase current and load current 
will be the same. The voltage from one line to another will be √3 times larger than 
the phase voltage. In a delta-delta connection the phase and line voltages will be the 
same, but the line current will be √3 times larger than the phase current of the 
generator or load. In delta-Y and Y -delta connections, the source and load no longer 
match configuration so neither the phase voltages nor currents are the same. For the",ACElectricalCircuitAnalysis.pdf
"match configuration so neither the phase voltages nor currents are the same. For the 
portion that is Y-connected, the line current and phase current will be the same. For 
the portion that is delta-connected, the line voltage and phase voltage will be the 
same. For those configurations, the other parameter (voltage or current) will be 
scaled by √3. 
In a system with a balanced load, the system power will simply be three times the 
power of one leg. If the load has a non-negligible phase angle, power factor 
correction can be used to reduce the required line current. The compensating items 
are arranged in a delta configuration, even if the load is Y -connected.
Review QuestionsReview Questions
1. Describe the advantages and disadvantages of polyphase versus single phase
systems.
2. Define the terms delta-connected and Y-connected.
3. How are line and load voltages related for Y-connected loads? 
4. How are line and load currents related for Y-connected loads?",ACElectricalCircuitAnalysis.pdf
"3. How are line and load voltages related for Y-connected loads? 
4. How are line and load currents related for Y-connected loads? 
5. How are line and load voltages related for delta-connected loads?
6. How are line and load currents related for delta-connected loads?
7. Describe a practical connection for power factor correction of a Y-connected
load.
364",ACElectricalCircuitAnalysis.pdf
"9.5 Exercises9.5 Exercises
Unless specified otherwise, assume generator frequencies are 60 Hz for all 
problems.
AnalysisAnalysis
1. As depicted in Figure 9.26, a three-phase delta-connected generator feeds a 
delta-connected load. The generator phase voltage is 120 volts and the load 
consists of three legs of 10 Ω each. Find the voltage across each load leg, 
the line current through the wires connecting the load to to the generator and
the power drawn by the load.
2. Referring to the delta-delta system of Figure 9.26, if the generator phase 
voltage is 230 volts and the load is balanced with each leg at 2 Ω, determine 
the line voltage, line current, generator phase current and load current.
3. The system of Figure 9.27 shows a three-phase Y-connected generator 
feeding a Y-connected load. If the generator phase voltage is 120 volts and 
the load consists of three legs of 20 Ω each, find the line voltage, the line",ACElectricalCircuitAnalysis.pdf
"the load consists of three legs of 20 Ω each, find the line voltage, the line 
current, voltage across each load leg and the total power drawn by the load.
365
Figure 9.26
Figure 9.27",ACElectricalCircuitAnalysis.pdf
"4. Referring to Figure 9.27, if the generator phase voltage is 230 volts and the 
load is balanced with each leg at 12 Ω, determine the line voltage, line 
current, generator phase current, load current, load voltage and total load 
power.
5. As depicted in Figure 9.28, a three-phase delta-connected generator feeds a 
Y-connected load. The generator phase voltage is 120 volts and the load 
consists of balanced legs of 5 Ω each. Find the voltage across each load leg, 
the line current, the line voltage, the generator phase current and the total 
load power.
6. Referring to Figure 9.28, if the generator phase voltage is 400 volts and the 
load is balanced with each leg at 10 Ω, determine the line voltage, line 
current, generator phase current, load current and the voltage across each 
load leg.
7. The system of Figure 9.29 shows a three-phase Y-connected generator 
feeding a delta-connected load. If the generator phase voltage is 120 volts",ACElectricalCircuitAnalysis.pdf
"7. The system of Figure 9.29 shows a three-phase Y-connected generator 
feeding a delta-connected load. If the generator phase voltage is 120 volts 
and the load consists of three legs of 60 Ω each, find the line voltage, the 
line current, voltage across each load leg and the total power drawn by the 
load.
8. Referring to the Y-delta system of Figure 9.28, if the generator phase voltage
is 120 volts and the load is balanced with each leg at 20 Ω, determine the 
line voltage, line current, generator phase current, load current, the voltage 
across each load leg and the total load power.
366
Figure 9.28
Figure 9.29",ACElectricalCircuitAnalysis.pdf
"9. A three-phase delta-connected generator feeds a delta-connected load 
consisting of three legs of 10 Ω in series with j4 Ω of inductive reactance, as
shown in Figure 9.30. If the line voltage is 208 volts, find the voltage across 
each load leg, the current through the wires connecting the load to to the 
generator, and the apparent and real powers drawn by the load.
10. Given the delta-delta system of Figure 9.30, if the generator phase voltage is
120 volts and the load is balanced with each leg at 20 + j10 Ω, determine the
line voltage, line current, generator phase current, load current, the voltage 
across each load leg, and the total real and apparent load powers.
11. A three-phase Y-connected generator feeds a Y-connected load consisting of 
three legs of 10 Ω in series with j4 Ω of inductive reactance, as shown in 
Figure 9.31. If the line voltage is 208 volts, find the voltage across each load
leg, the line current, and the apparent and real powers drawn by the load.",ACElectricalCircuitAnalysis.pdf
"leg, the line current, and the apparent and real powers drawn by the load.
12. Given the Y-Y system of Figure 9.31, if the line voltage is 400 volts and the 
load is balanced with each leg at 100 + j20 Ω, determine the generator phase
voltage, line current, generator phase current, load current, the voltage 
across each load leg, and the total real and apparent load powers.
367
Figure 9.30
Figure 9.31",ACElectricalCircuitAnalysis.pdf
"13. The three-phase system of Figure 9.7 uses a Y-connected generator feeding a
delta-connected load. The load consists of three legs of 40 Ω in series with 
j30 Ω of inductive reactance, as shown in Figure 9.32. If the generator phase
voltage is 230 volts, find the line voltage, the voltage across each load leg, 
the line current, the load current, and the apparent and real powers drawn by
the load.
14. Given the Y-delta system of Figure 9.32, if the line voltage is 400 volts and 
the load is balanced with each leg at 80 + j20 Ω, determine the generator 
phase voltage, line current, generator phase current, load current, the voltage
across each load leg, and the total real and apparent load powers.
15. A 208 three-phase delta-connected generator feeds a Y-connected load 
consisting of three legs of 10 Ω in series with j4 Ω of inductive reactance as 
shown in Figure 9.33. Find the voltage across each load leg, the current",ACElectricalCircuitAnalysis.pdf
"shown in Figure 9.33. Find the voltage across each load leg, the current 
through the wires connecting the load to to the generator, and the apparent 
and real powers drawn by the load.
16. Given the delta-Y system of Figure 9.33, if the line voltage is 400 volts and 
the load is balanced with each leg at 120 + j30 Ω, determine the line current,
generator phase current, load current, the voltage across each load leg, and 
the total real and apparent load powers.
368
Figure 9.32
Figure 9.33",ACElectricalCircuitAnalysis.pdf
"17. A 120 volt three-phase delta-connected generator feeds a delta-connected 
load consisting of three legs of 75 Ω in series with −j10 Ω of capacitive 
reactance as shown in Figure 9.34. Find the voltage across each load leg, the
current through the wires connecting the load to to the generator, and the 
apparent and real powers drawn by the load.
18. A three-phase Y-connected generator feeds a Y-connected load consisting of 
three legs of 150 Ω in series with −j20 Ω of capacitive reactance as shown in
Figure 9.35. If the generator phase voltage is 120 volts, find the line voltage,
the voltage across each load leg, the line current, and the apparent and real 
powers drawn by the load.
DesignDesign
19. Using the delta-delta system of problem 9 and assuming the source 
frequency is 60 Hz, determine appropriate component values to place in 
parallel with each load leg in order to shift the power factor to unity.",ACElectricalCircuitAnalysis.pdf
"frequency is 60 Hz, determine appropriate component values to place in 
parallel with each load leg in order to shift the power factor to unity.  
20. Using the Y-Y system of problem 11 and assuming the source frequency is 
60 Hz, determine appropriate component values to place in parallel with 
each load leg in order to shift the power factor to unity.  
369
Figure 9.34
Figure 9.35",ACElectricalCircuitAnalysis.pdf
"ChallengeChallenge
21. Using the Y-Y system of problem 11 and assuming the source frequency is 
60 Hz, determine appropriate component values to be added to the load in 
order to shift the power factor to unity. These new components should be in 
a delta configuration.
SimulationSimulation
22. Use a transient analysis to verify the phase and line voltage phase 
relationships in problem 1. 
23. Use a transient analysis to verify the results computed for problem 15. 
24. Use a transient analysis to verify the design solution to problem 19. This can
be achieved by ensuring that the voltage and current in each load leg (with 
added correction components) are in phase.
25. Use a transient analysis to verify the design solution to problem 20. This can
be achieved by ensuring that the voltage and current in each load leg (with 
added correction components) are in phase.
370",ACElectricalCircuitAnalysis.pdf
"NotesNotes
♫♫♫♫
371",ACElectricalCircuitAnalysis.pdf
"10 10 Decibels and Bode PlotsDecibels and Bode Plots
10.0 Chapter Learning Objectives10.0 Chapter Learning Objectives
After completing this chapter, you should be able to: 
• Convert between ordinary and decibel based power and voltage gains. 
• Utilize decibel-based voltage and power measurements during circuit analysis. 
• Define and graph a general Bode plot. 
• Detail the differences between lead and lag networks, and graph Bode plots for each. 
• Combine the effects of several lead and lag networks together in order to determine a system Bode plot. 
10.1 Introduction10.1 Introduction
This chapter introduces two related items; the decibel and Bode plots. The decibel measurement scheme is in 
wide use, particularly in the fields of audio and communications. We will be examining its advantages over the 
ordinary system of measurement used up to now and how to convert values of one form into the other. One of",ACElectricalCircuitAnalysis.pdf
"ordinary system of measurement used up to now and how to convert values of one form into the other. One of 
the more important parameters of a circuit is its frequency response, that is, the way in which the circuit 
responds to input signals over a range of frequencies. While we have investigated this to some extent in prior 
work, in this chapter we shall take it to its logical conclusion, namely the Bode plot. A Bode plot is, in fact, a 
pair of plots; one of relative signal magnitude or gain with respect to frequency and a second detailing the phase 
response with respect to frequency. Bode plots are of particular importance in the study of circuits such as 
amplifiers and filters, as well as in systems that make use of negative feedback. The gain magnitude plot makes 
use of a decibel scale and thus it makes sense to begin our study looking at the decibel system: specifically how 
it is defined and its practical use.
10.2 The Decibel10.2 The Decibel",ACElectricalCircuitAnalysis.pdf
"it is defined and its practical use.
10.2 The Decibel10.2 The Decibel
Most people are familiar with the term “decibel” in reference to sound pressure. It’s not uncommon to hear 
someone say something such as “It was 110 decibels at the concert last night and my ears are still ringing.” This 
popular use is somewhat inaccurate, but does show that decibels indicate some sort of quantity or relative level; 
in this case, sound pressure level. 
Decibel Representation of Power and Voltage Gains Decibel Representation of Power and Voltage Gains 
In its simplest form, the decibel is used to measure system gain, such as power or voltage gain, where gain is 
simply the ratio of an output signal to an input signal. For an amplifying circuit, the gain would be greater than 
372",ACElectricalCircuitAnalysis.pdf
"one, but for purely passive systems it will likely be fractional (i.e., the output 
quantity is smaller than the input quantity). For example, a simple voltage divider 
might be said to have a “gain” of 0.2, or some such, meaning that the output signal is
only 20% of the input signal. Unlike the ordinary gain measurements, the decibel 
form is logarithmic. Because of this, it can be very useful for showing ratios of 
change, as well as absolute change. The base unit is the Bel, named after Alexander 
Graham Bell, the noted American scientist and inventor. To convert an ordinary gain
to its Bel counterpart, just take the common log (base 10) of the gain. In equation 
form: 
Bel gain = log10( ordinary gain ) (10.1)
Note that on most hand calculators common log is denoted as “log” while the natural
log is given as “ln”. Unfortunately, some programming languages use “log” to 
indicate natural log and “log10” for common log. More than one student has been",ACElectricalCircuitAnalysis.pdf
"indicate natural log and “log10” for common log. More than one student has been 
bitten by this bug, so be forewarned! As an example, if an amplifier circuit produces 
an output power of 200 milliwatts for an input of 10 milliwatts, we would normally 
say that it has a power gain of: 
G=Pout
Pi n
G=200 mW
10 mW
G=20
For the Bel version, just take the log of this result. 
G '=log10 G
G '=log10 20
G '=1.301
The Bel gain is 1.3 Bels. The term “Bels” is not a unit in the strict sense of the word 
(as in “watts”), but is simply used to indicate that this is not an ordinary gain. In 
contrast, ordinary power and voltage gains are sometimes given units of W/W and 
V/V to distinguish them from Bel gains. Also, note that the symbol for Bel power 
gain is G' and not G. All Bel gains are denoted with the following prime (’) notation 
to avoid confusion. Because Bels tend to be rather large, we typically use one-tenth",ACElectricalCircuitAnalysis.pdf
"to avoid confusion. Because Bels tend to be rather large, we typically use one-tenth 
of a Bel as the norm. The result is the decibel (one-tenth Bel). To convert to decibels,
simply multiply the number of Bels by 10. Our gain of 1.3 Bels is equivalent to 13 
decibels. The units are commonly shortened to dB. Consequently, we may say: 
G '=10 log10G (10.2)
Where the result is in dB.
373",ACElectricalCircuitAnalysis.pdf
"At this point, you may be wondering what the big advantage of the decibel system 
is. To answer this, recall a few log identities. Normal multiplication becomes 
addition in the log system, and division becomes subtraction. Likewise, powers and 
roots become multiplication and division. Because of this, two important things 
show up. First, ratios of change become constant offsets in the decibel system, and 
second, the entire range of values diminishes in size. The result is that a very wide 
range of gains may be represented within a fairly small scope of values, and the 
corresponding calculations can become quicker. 
There are a couple of dB values that are useful to remember, and are illustrated in 
Figure 10.1. With the aid of your calculator, it is very easy to show the following:
Factor
dB Value using
G’= 10 log10  G
1 0 dB 
2 3.01 dB 
4 6.02 dB 
8 9.03 dB 
10 10 dB 
We can also look at fractional factors (i.e., losses instead of gains, Figure 10.2): 
Factor dB Value",ACElectricalCircuitAnalysis.pdf
"2 3.01 dB 
4 6.02 dB 
8 9.03 dB 
10 10 dB 
We can also look at fractional factors (i.e., losses instead of gains, Figure 10.2): 
Factor dB Value
0.5 −3.01 dB 
0.25 −6.02 dB 
0.125  −9.03 dB 
0.1 −10 dB 
If you look carefully, you will notice that a doubling is represented by an increase of 
approximately 3 dB. A factor of 4 is in essence, two doublings. Therefore, it is 
equivalent to 3 dB + 3 dB, or 6 dB. Remember, because we are using logs, 
multiplication turns into simple addition. In a similar manner, a halving is 
represented by approximately −3 dB. The negative sign indicates a reduction. To 
simplify things a bit, think of factors of 2 as ±3 dB, the sign indicating whether you 
are increasing (multiplying), or decreasing (dividing). As you can see, factors of 10 
work out to a very convenient 10 dB. By remembering these two factors, you can 
often estimate a dB conversion without the use of your calculator. For instance, we",ACElectricalCircuitAnalysis.pdf
"often estimate a dB conversion without the use of your calculator. For instance, we 
could rework our initial conversion problem as follows: 
• The amplifier has a gain of 20.
• 20 can be written as 2 times 10.
• The factor of 2 is 3 dB, the factor of 10 is 10 dB.
• The answer must be 3 dB + 10 dB, or 13 dB.
Time for a few examples.
374
Figure 10.1
Positive dB factors.
Figure 10.2
Negative dB factors.",ACElectricalCircuitAnalysis.pdf
"Example 10.1 
An amplifier has a power gain of 800. What is the decibel power gain? 
G '=10 log10 G
G '=10 log10 800
G '=10×2.903
G '=29.03dB
 
We could also use our estimation technique:
• G = 800 = 8∙102
• 8 is equivalent to 3 factors of 2, or 2∙2∙2, and can be expressed as 3 dB +
3 dB + 3 dB, which is, of course, 9 dB 
• 102 is equivalent to 2 factors of 10, or 10 dB + 10 dB = 20 dB. 
Alternately, the power of 2 literally represents 2 Bels, and thus 20 dB. 
• The result is 9 dB + 20 dB, or 29 dB 
Note that if the leading digit is not a power of 2, the estimation will not be 
as precise. For example, if the gain is 850, you know that the decibel gain is 
just a bit over 29 dB. You also know that it must be less than 30 dB 
(1000=103 which is 3 factors of 10, or 30 dB.) As you can see, by using the 
dB form, you tend to concentrate on the magnitude of gain, and not so much
on trailing digits. 
Example 10.2 
An attenuator reduces signal power by a factor of 10,000. What is this loss",ACElectricalCircuitAnalysis.pdf
"on trailing digits. 
Example 10.2 
An attenuator reduces signal power by a factor of 10,000. What is this loss 
expressed in dB? 
G '=10 log10
1
10,000
G '=10×(−4)
G '=−40 dB
    
By using the approximation, we can say, 
1
10,000 =10−4  
The negative exponent tells us we have a loss (negative dB value), and 4 
factors of 10 (i.e., 4 Bels). 
375",ACElectricalCircuitAnalysis.pdf
"G '= −10 dB −10dB −10dB −10 dB 
G '= −40dB
 
Remember, if an increase in signal is produced, the result will be a positive 
dB value. A decrease in signal will always result in a negative dB value. A 
signal that is unchanged indicates a gain of unity, or 0 dB. 
To convert from dB to ordinary form, just invert the steps; that is, divide by ten and 
then take the antilog. 
G=log10
−1 G'
10
 (10.3)
On most hand calculators, base 10 antilog is denoted as 10x. In most computer 
languages, you just raise 10 to the appropriate power, as in G = 10.0**(Gprime / 
10.0) (Python), or use an exponent function, as in pow(10.0, Gprime / 10.0) (C). 
Example 10.3 
An amplifier has a power gain of 23 dB. If the input is 1 mW, what is the 
output? 
In order find the output power, we need to find the ordinary power gain, G. 
G=log10
−1 G '
10
G=log10
−1 23
10
G=199.5
  
Therefore, Pout =199.5 ∙ 1 mW, or 199.5 mW 
You could also use the approximation technique in reverse. To do this, break",ACElectricalCircuitAnalysis.pdf
"G=log10
−1 23
10
G=199.5
  
Therefore, Pout =199.5 ∙ 1 mW, or 199.5 mW 
You could also use the approximation technique in reverse. To do this, break
up the dB gain in 10 dB and 3 dB chunks: 
23dB=3dB+10dB+10dB  
Now replace each chunk with the appropriate factor, and multiply them 
together (remember, when going from log to ordinary form, addition turns 
into multiplication.) 
3dB=2X ,10dB=10X, so,
376",ACElectricalCircuitAnalysis.pdf
"G=2×10×10
G=200
  
While the approximation technique appears to be slower than the calculator, practice
will show otherwise. Being able to quickly estimate dB values can prove to be a 
very handy skill in the electronics field. This is particularly true in larger, multi-stage
designs. 
Example 10.4 
A three-stage amplifier has gains of 10 dB, 16 dB, and 14 dB per section. 
What is the total dB gain? 
Because dB gains are a log form, just add the individual stage gains to arrive
at the system gain. 
G 'total =G'1+G '2+G ' 3
G 'total =10dB+16 dB+14dB
G 'total =40 dB
As you may have noticed, all of the examples up to this point have used power gain 
and not voltage gain. You may be tempted to use the same equations for voltage 
gain. In a word, don’t. If you think back for a moment, you will recall that power 
varies as the square of voltage. In other words, a doubling of voltage will produce a 
quadrupling of power. If you were to use the same dB conversions, a doubling of",ACElectricalCircuitAnalysis.pdf
"quadrupling of power. If you were to use the same dB conversions, a doubling of 
voltage would be 3 dB, yet, because the power has quadrupled, this would indicate a
6 dB rise. Consequently, voltage gain (and current gain as well) are treated in a 
slightly different fashion. We would rather have our doubling of voltage work out to 
6 dB, so that it matches the power calculation. The correction factor is very simple. 
Because power varies as the second power of voltage, the dB form should be twice 
as large for voltage (remember, exponentiation turns into multiplication when using 
logs). Applying this factor to equation 10.2 yields:
A 'v=20 log10 Av
(10.4)
Be careful though, the Bel voltage gain only equals the Bel power gain if the input 
and output impedances of the system are matched (you may recall from your other 
work that it is quite possible to design a circuit with vastly different voltage and 
power gains). If we were to recalculate our earlier table of common factors, we",ACElectricalCircuitAnalysis.pdf
"power gains). If we were to recalculate our earlier table of common factors, we 
would find that a doubling of voltage gain is equivalent to a 6 dB rise, and a ten fold 
increase is equivalent to a 20 dB rise, twice the number of decibels of their power 
gain counterparts. 
377",ACElectricalCircuitAnalysis.pdf
"Note that current gain may be treated in the same manner as voltage gain (although 
this is less commonly done in practice). 
Example 10.5 
A circuit has an output signal of 2 V for an input of 50 mV . What is A'v? 
First find the ordinary gain.
 Av= 2
0.05 =40
Now convert to dB form.
A 'v=20 log10 40
A 'v=20×1.602
A 'v=32.04 dB
  
The approximation technique yields 40=2∙2∙10, or 6 dB + 6 dB + 20 dB, or 
32 dB.
To convert A'v to A, reverse the process.
 
Av=log10
−1 A 'v
20
(10.5)
Example 10.6
An amplifier has a gain of 26 dB. If the input signal is 10 mV , what is the 
output? 
Av=log10
−1 A'v
20
Av=log10
−1 26
20
Av=19.95
V out= Av V i n
V out=19.95×10 mV
 
The final point to note in this section is that, as in the case of power gain, a negative 
decibel value indicates a loss. Therefore, a 2:1 voltage divider would have a gain of 
−6 dB. 
378",ACElectricalCircuitAnalysis.pdf
"Signal Representation in dBW and dBV Signal Representation in dBW and dBV 
As you can see from the preceding section, it is possible to spend considerable time 
converting between decibel gains and ordinary voltages and powers. Because the 
decibel form does offer advantages for gain measurement, it would make sense to 
use a decibel form for power and voltage levels as well. This is a relatively 
straightforward process. There is no reason why we can’t express a power or voltage
in a logarithmic form. Because a dB value just indicates a ratio, all we need to do is 
decide on a reference (i.e., a comparative base for the ratio). For power 
measurements, a likely choice would be 1 watt. In other words, we can describe a 
power as being a certain number of dB above or below 1 watt. Positive values will 
indicate powers greater than 1 watt, while negative values will indicate powers less 
than 1 watt. In general equation form: 
P ' =10log10
P
reference
(10.6)",ACElectricalCircuitAnalysis.pdf
"than 1 watt. In general equation form: 
P ' =10log10
P
reference
(10.6)
The answer will have units of dBW, that is, decibels relative to 1 watt. 
Example 10.7 
A power amplifier has a maximum output of 120 W. What is this power in 
dBW? 
P ' =10log10
P
1 Watt
P ' =10log10
120 W
1W
P ' =20.8 dBW
   
 
There is nothing sacred about the 1 watt reference, short of its convenience. We 
could just as easily choose a different reference. Other common reference points are 
1 milliwatt (dBm) and 1 femtowatt (dBf). Obviously, dBf is used for very low signal
levels, such as those coming from an antenna. dBm is in very wide use in the 
communications industry. To use these other references, just divide the given power 
by the new reference. 
Example 10.8 
A small personal audio music player delivers 200 mW to its headphones. 
What is this output power in dBW, and in dBm? 
For an answer in units of dBW, use the 1 watt reference.
379",ACElectricalCircuitAnalysis.pdf
"P '=10 log10
P
1Watt
P '=10 log10
200 mW
1W
P '=−7dBW
 
For units of dBm, use a 1 milliwatt reference.
P '=10 log10
P
1Watt
P '=10 log10
200 mW
1 mW
P '=23 dBm
200 mW, −7 dBW, and 23 dBm are three ways of saying the same thing. 
Note that the dBW and dBm values are 30 dB apart. This will always be 
true, because the references are a factor of 1000 (30 dB) apart. 
In order to transfer a dBW or similar value into watts, reverse the process. 
P=log10
−1 P '
10 ×reference  (10.7)
Example 10.9 
A studio microphone produces a 12 dBm signal while recording normal 
speech. What is the output power in watts? 
P=log10
−1 P '
10 ×reference
P=log10
−1 12dBm
10 ×1mW
P=15.8mW=0.0158W
 
For voltages, we can use a similar system. A logical reference is 1 V , with the 
resulting units being dBV . As before, these voltage measurements will use a 
multiplier of 20 instead of 10. 
V '=20log10
V
reference
(10.8)
380",ACElectricalCircuitAnalysis.pdf
"Example 10.10 
A test oscillator produces a 2 volt signal. What is this value in dBV? 
V '=20log10
V
reference
V '=20log10
2 V
1V
V '= 6.02dB
 
 
When both circuit gains and signal levels are specified in decibel form, analysis can 
be very quick. Given an input level, simply add the gain to it in order to find the 
output level. Given input and output levels, subtract them in order to find the gain. 
Example 10.11 
A computer hard drive read/write amplifier exhibits a gain of 35 dB. If the 
input signal is −42 dBV , what is the output signal? 
V 'out=V ' i n+A' v
V 'out=−42 dBV+35dB
V 'out=−7dBV
 
Note that the final units are dBV and not dB, thus indicating a voltage and 
not merely a gain. 
Example 10.12 
A guitar power amp needs an input of 20 dBm to achieve an output of 25 
dBW. What is the gain of the amplifier in dB? 
First, it is necessary to convert the power readings so that they share the 
same reference unit. Because dBm represents a reference 30 dB smaller than",ACElectricalCircuitAnalysis.pdf
"same reference unit. Because dBm represents a reference 30 dB smaller than
the dBW reference, just subtract 30 dB to compensate. 
20 dBm=−10 dBW
 
G '=P 'out−P 'i n
G '=25dBW−(−10 dBW)
G '=35 dB
 
381",ACElectricalCircuitAnalysis.pdf
"Note that the units are dB and not dBW. This is very important! Saying that 
the gain is “so many” dBW is the same as saying the gain is “so many” 
watts. Obviously, gains are “pure” numbers and do not carry units such as 
watts or volts. 
The usage of a dB-based system is shown graphically in Figure 10.3. Note how the 
stage gains are added to the input signal to form the output. Even large circuits can 
be quickly analyzed in this form. To make life in the lab even easier, it is possible to 
take measurements directly in dB form. By doing this, you need never convert while
troubleshooting a design. For general-purpose work, voltage measurements are the 
norm, and therefore a dBV scale is often used. 
Items Of Interest In The Laboratory Items Of Interest In The Laboratory 
When using a digital meter on a dBV scale it is possible to “underflow” the meter if 
the signal is too weak. This will happen if you try to measure around zero volts, for",ACElectricalCircuitAnalysis.pdf
"the signal is too weak. This will happen if you try to measure around zero volts, for 
example. If you attempt to calculate the corresponding dBV value, your calculator 
will probably show “error”. The effective value is negative infinite dBV . The meter 
will certainly have a hard time showing this value! Another item of interest revolves 
around the use of dBm measurements. It is common to use a voltmeter to make dBm
measurements, in lieu of a wattmeter. While the connections are considerably 
simpler, a voltmeter cannot measure power. How is this accomplished then? Well, as
long as the circuit impedance is known, power can be derived from a voltage 
measurement. A common impedance in communication systems (such as recording 
studios) is 600 Ω, so a meter can be calibrated to give correct dBm readings by using
Power Law. If this meter is used on a non-600 Ω circuit, the readings will no longer 
reflect accurate dBm values (but will still properly reflect relative changes in dB).",ACElectricalCircuitAnalysis.pdf
"reflect accurate dBm values (but will still properly reflect relative changes in dB). 
Finally, recalling the chapter introduction regarding “110 dB” concert levels, 
properly, that would read “110 dB-SPL”, referring to “Sound Pressure Level”. The 
reference level corresponding to 0 dB-SPL is the quietest sound the average person 
can hear; the threshold of hearing (20 micropascals for young healthy humans). 
Thus, 110 dB-SPL refers to a sound pressure that is 110 dB greater than the threshold
of hearing. Typically, 1 dB represents a “just noticeable difference” in loudness for 
humans, although this depends on the precise frequency and sound pressure.
382
10 dB -6 dB 15 dB
8 dBm 18 dBm 12 dBm 27 dBm
Figure 10.3
Multistage dB application.",ACElectricalCircuitAnalysis.pdf
"10.3 Bode Plots10.3 Bode Plots
The Bode plot is a graphical response prediction technique that is useful for both 
circuit design and analysis. It is named after Hendrik Wade Bode, an American 
engineer known for his work in control systems theory and telecommunications. A 
Bode plot is, in actuality, a pair of plots: One graphs the signal gain or loss of a 
system versus frequency, while the other details the circuit phase versus frequency. 
Both of these items are very important in the design of well-behaved, optimal 
circuits. 
Generally, Bode plots are drawn with logarithmic frequency axes, a decibel gain 
axis, and a phase axis in degrees. First, let’s take a look at the gain plot. A typical 
gain plot is shown Figure 10.4. Remember, “gains” can be fractional, as with a 
voltage divider.
Note how the plot is relatively flat in the middle, or midband, region. The gain value
in this region is known as the midband gain. In purely passive circuits this value",ACElectricalCircuitAnalysis.pdf
"in this region is known as the midband gain. In purely passive circuits this value 
may be fractional (i.e., a negative dB value). At either extreme of the midband 
region, the gain begins to decrease. The gain plot shows two important frequencies, 
f1 and f2. f1 is the lower break frequency while f2 is the upper break frequency. The 
gain at the break frequencies is 3 dB less than the midband gain. These frequencies 
are also known as the half-power points, or corner frequencies. Normally, amplifiers 
are only used for signals between f1 and f2. The exact shape of the rolloff regions will
depend on the design of the circuit. For example, it is possible to design amplifiers 
with no lower break frequency (i.e., a DC amplifier), however, all amplifiers will 
exhibit an upper break. The break points are caused by the presence of circuit 
reactances, typically coupling and stray capacitances. The gain plot is a summation",ACElectricalCircuitAnalysis.pdf
"reactances, typically coupling and stray capacitances. The gain plot is a summation 
of the midband response with the upper and lower frequency limiting networks. 
Let’s take a look at the lower break, f1. 
383
Figure 10.4
Generic gain plot.
f
A'v
A'midband
f1 f2midband",ACElectricalCircuitAnalysis.pdf
"Lead Network Gain Response Lead Network Gain Response 
Reduction in low frequency gain is caused by lead networks. A generic lead network
is shown in Figure 10.5. It gets its name from the fact that the output voltage 
developed across R leads the input. At very high frequencies the circuit will be 
essentially resistive. Conceptually, think of this as a simple voltage divider. The 
divider ratio depends on the reactance of C. As the input frequency drops, Xc 
increases. This makes Vout decrease. At very high frequencies, where Xc<<R, Vout is 
approximately equal to Vin. This can be seen graphically in Figure 10.6, where the 
frequency axis is normalized to fc. The break frequency (i.e., the frequency at which 
the signal has decreased by 3 dB) is found via the standard equation,
f c= 1
2π R C
The response below fc will be a straight line if a decibel gain axis and a logarithmic 
frequency axis are used. This makes for very quick and convenient sketching of",ACElectricalCircuitAnalysis.pdf
"frequency axis are used. This makes for very quick and convenient sketching of 
circuit response. The slope of this line is 6 dB per octave (an octave is a doubling or 
halving of frequency, e.g., 800 Hz is 3 octaves above 100 Hz).20 This range covers a 
factor of two in frequency. This slope may also be expressed as 20 dB per decade, 
where a decade is a factor of 10 in frequency. With reasonable accuracy, this curve 
may be approximated as two line segments, called asymptotes, shown in Figure 10.6
(blue). The shape of this curve is the same for any lead network. Because of this, it is
20 The term octave is borrowed from the field of music. It gets its name from the fact that 
there are eight notes in the standard western scale: do-re-mi-fa-sol-la-ti-do.
384
Figure 10.5
Lead network.
Figure 10.6
Lead network gain plot.
Lead Network Magnitude Response
Amplitude (dB)
-40
-30
-20
-10
0
10
-40
-30
-20
-10
0
10
Normalized frequency
0.0 0.1 1.0 10.0
0.01 0.1 1 10
Exact
Approximate",ACElectricalCircuitAnalysis.pdf
"Amplitude (dB)
-40
-30
-20
-10
0
10
-40
-30
-20
-10
0
10
Normalized frequency
0.0 0.1 1.0 10.0
0.01 0.1 1 10
Exact
Approximate
Slope: 6 dB/octave
or 20 dB/decade",ACElectricalCircuitAnalysis.pdf
"very easy to find the approximate gain at any given frequency as long as fc is known.
It is not necessary to go through reactance and phasor calculations. To create a 
general response equation, start with the voltage divider rule to find the gain: 
V out
V i n
= R
R− j X c
V out
V i n
= R∠0
√R2
+X c
2
∠−arctan X c
R
The magnitude of this is, 
∣Av∣= R
√R2
+X c
2
∣Av∣= 1
√
1+X c
2
R2
(10.9)
Recalling that, 
f c= 1
2 π R C
we may say, 
R= 1
2π f c C
For any frequency of interest, f, 
X c= 1
2 π f C
Equating the two preceding equations yields, 
f c
f = X c
R
(10.10)
Substituting equation 10.10 in equation 10.9 gives, 
Av = 1
√
1+ f c
2
f 2
(10.11)
385",ACElectricalCircuitAnalysis.pdf
"To express Av in dB, substitute equation 10.11 into equation 10.5. 
A 'v=20 log10
1
√
1+ f c
2
f 2
After simplification, the final result is: 
A 'v=−10 log10(1+ f c
2
f 2 )
(10.12)
Where 
fc  is the critical frequency, 
f  is the frequency of interest, 
A'v is the decibel gain at the frequency of interest. 
Example 10.13 
A circuit has a lower break frequency of 40 Hz. How much signal is lost at 
10 Hz? 
A 'v=−10 log10(1+f c
2
f 2 )
A 'v=−10 log10(1+402
102 )
A 'v=−12.3 dB
In other words, the signal level is 12.3 dB lower than it is in the midband. 
Note that 10 Hz is 2 octaves below the break frequency. Because the cutoff 
slope is 6 dB per octave, each octave loses 6 dB. Therefore, the approximate
result is −12 dB, which double-checks the exact result. Without the lead 
network, the gain would stay at 0 dB all the way down to DC (0 Hz.) 
Lead Network Phase Response Lead Network Phase Response 
At very low frequencies, the circuit of Figure 10.5 is largely capacitive. Because of",ACElectricalCircuitAnalysis.pdf
"Lead Network Phase Response Lead Network Phase Response 
At very low frequencies, the circuit of Figure 10.5 is largely capacitive. Because of 
this, the output voltage developed across R leads by 90 degrees. At very high 
frequencies the circuit will be largely resistive. At this point Vout will be in phase with
Vin. At the critical frequency, Vout will lead by 45 degrees. A general phase graph is 
shown in Figure 10.7. As with the gain plot, the phase plot shape is the same for any 
lead network. The general phase equation may be obtained from the voltage divider: 
386",ACElectricalCircuitAnalysis.pdf
"V out
V i n
= R
R− j X c
V out
V i n
= R∠0
√R2
+X c
2
∠−arctan X c
R
The phase portion of this is, 
θ=arctan X c
R
By using equation 10.6, this simplifies to, 
θ=arctan f c
f
(10.13)
Where 
fc is the critical frequency, 
f is the frequency of interest, 
θ is the phase angle at the frequency of interest. 
Often, an approximation, such as the blue line in Figure 10.7, is sufficient. By using 
Equation 10.13, you can show that this approximation is off by no more than 6 
degrees at the corners. 
387
Figure 10.7
Lead network phase response.
Lead Network Phase Response
Phase (degrees)
0
10
20
30
40
50
60
70
80
90
0
10
20
30
40
50
60
70
80
90
Normalized frequency
0.0 0.1 1.0 10.0 100.0
0.01 0.1 1 10 100
Exact
Approximate",ACElectricalCircuitAnalysis.pdf
"Example 10.14 
A telephone amplifier has a lower break frequency of 120 Hz. What is the 
phase response one decade below and one decade above? 
One decade below 120 Hz is 12 Hz, while one decade above is 1.2 kHz. 
θ=arctan f c
f
θ=arctan 120 Hz
12Hz
θ=84.3degrees one decade below f c  (i.e, approaching 90 degrees)
θ=arctan 120 Hz
1.2kHz
θ=5.71degrees one decade above f c  (i.e., approaching 0 degrees)
Remember, if a circuit or amplifier is direct-coupled, and has no lead 
networks, the phase will remain at 0 degrees right back to 0 Hz (DC). 
Lag Network Response Lag Network Response 
Unlike its lead network counterpart, all systems will contain lag networks. In 
essence, it is little more than an inverted lead network. As you can see from Figure 
10.8, it simply transposes the R and C locations. Because of this, the response tends 
to be inverted as well. In terms of gain, Xc is very large at low frequencies, and thus",ACElectricalCircuitAnalysis.pdf
"to be inverted as well. In terms of gain, Xc is very large at low frequencies, and thus 
Vout equals Vin. At high frequencies, Xc decreases, and Vout falls. The break point 
occurs when Xc equals R. The general gain plot is shown in Figure 10.9. Like the 
lead network response, the slope of this curve is −6 dB per octave (or −20 dB per 
decade.) Note that the slope is negative instead of positive. We can derive a general 
gain equation for this circuit in virtually the same manner as we did for the lead 
network. The derivation is left as an exercise. 
A 'v=−10 log10
(1+ f 2
f c
2 )
(10.14)
Where 
fc is the critical frequency, 
f is the frequency of interest, 
A'v is the decibel gain at the frequency of interest. 
388
Figure 10.8
Lag network.",ACElectricalCircuitAnalysis.pdf
"Note that this equation is almost the same as Equation 10.12. The only difference is 
that f and fc have been transposed. 
In a similar vein, we may examine the phase response. At very low frequencies, the 
circuit is basically capacitive. Because the output is taken across C, Vout will be in 
phase with Vin. At very high frequencies, the circuit is essentially resistive. 
Consequently, the output voltage across C will lag by 90 degrees. At the break 
frequency the phase will be −45 degrees. A general phase plot is shown in Figure 
10.10. As with the lead network,we may derive a phase equation. Again, the exact 
steps are very similar, and left as an exercise. 
θ=−90+arctan f c
f
(10.15)
Where 
fc is the critical frequency, 
f is the frequency of interest, 
θ is the phase angle at the frequency of interest. 
389
Lag Network Magnitude Response
Amplitude (dB)
-40
-30
-20
-10
0
10
-40
-30
-20
-10
0
10
Normalized frequency
0.1 1.0 10.0 100.0
0.1 1 10 100
Exact
Approximate",ACElectricalCircuitAnalysis.pdf
"Amplitude (dB)
-40
-30
-20
-10
0
10
-40
-30
-20
-10
0
10
Normalized frequency
0.1 1.0 10.0 100.0
0.1 1 10 100
Exact
Approximate
Slope: -6dB/octave
or -20dB/decade
Figure 10.9
Lag network gain plot.",ACElectricalCircuitAnalysis.pdf
"Example 10.15 
A medical ultra sound transducer feeds a lag network with an upper break 
frequency of 150 kHz. What are the gain and phase values at 1.6 MHz? 
Because this represents a little more than a 1 decade increase, the 
approximate values are −20 dB and −90 degrees, from
Figures 10.7 and 10.8, respectively. The exact values are: 
A 'v=−10 log10(1+f 2
f c
2 )
A 'v=−10 log10(1+1.6 MHz2
150 kHz2 )
A 'v=−20.6dB
θ=−90+arctan f c
f
θ=−90+arctan 150 kHz
1.6MHz
θ=−84.6degrees
The complete Bode plot for this network is shown in Figure
10.11. It is very useful to examine both plots simultaneously.
In this manner you can find the exact phase change for a
given gain quite easily. For example, if you look carefully at
the plots of Figure 10.11, you will note that at the critical
frequency of 150 kHz, the total phase change is −45 degrees.
390
Figure 10.11
Bode plot for 150 kHz lag.
f
A'v
150 kHz
0 dB
-6 dB/octave
Lag Network Phase Response
Phase (degrees)
-90
-80
-70
-60
-50
-40
-30
-20
-10",ACElectricalCircuitAnalysis.pdf
"Bode plot for 150 kHz lag.
f
A'v
150 kHz
0 dB
-6 dB/octave
Lag Network Phase Response
Phase (degrees)
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
Normalized frequency
0.0 0.1 1.0 10.0 100.0
0.01 0.1 1 10 100
Exact
Approximate Figure 10.10
Lag network phase response.
f
θ
0°
-45°
-90°
150 kHz15 kHz 1.5 MHz",ACElectricalCircuitAnalysis.pdf
"Because this circuit involved the use of a single lag network, this is exactly 
what you would expect. 
Rise Time versus Bandwidth Rise Time versus Bandwidth 
For pulse-type signals, the “speed” of a circuit is often expressed in terms of its rise 
time. If a square pulse such as Figure 10.12a is passed into a simple lag network, the 
capacitor charging effect will produce a rounded variation, as seen in Figure 10.12b. 
This effect places an upper limit on the duration of pulses that a given circuit can 
handle without producing excessive distortion. 
By definition, rise time is the amount of time it takes for the signal to traverse from 
10% to 90% of the peak value of the pulse. The shape of this pulse is defined by the 
standard capacitor charge equation examined in earlier course work, and is valid for 
any system with a single clearly dominant lag network. 
V out=V peak (1−ϵ
−t
RC ) (10.16)
In order to find the time internal from the initial starting point to the 10% point, set",ACElectricalCircuitAnalysis.pdf
"V out=V peak (1−ϵ
−t
RC ) (10.16)
In order to find the time internal from the initial starting point to the 10% point, set 
Vout to 0.1Vpeak in Equation 10.16 and solve for t1. 
391
t
Vpeak
t0
Figure 10.12a
Pulse rise time effect:
Input to network.
t
Vpeak
t0 t1 t2
.1Vpeak
.9Vpeak
Figure 10.12b
Pulse rise time effect:
Output of network.",ACElectricalCircuitAnalysis.pdf
"0.1V peak = V peak (1−ϵ
−t1
RC )
0.1V peak = V peak −V peak ϵ
−t1
RC
0.9V peak = V peak ϵ
−t1
RC
0.9 = ϵ
−t1
RC
log 0.9 = −t1
R C
   t1=0.105 RC (10.17)
To find the interval up to the 90% point, follow the same technique using 0.9Vpeak . 
Doing so yields:
t2=2.303 RC (10.18)
The rise time, Tr, is the difference between t1 and t2 
  T r=t 1−t 2
T r=2.303 R C−0.105 RC
T r ≈ 2.2 R C (10.19)
Equation 10.19 ties the rise time to the lag network’s R and C values. These same 
values also set the critical frequency f2. By combining equation 10.15 with the basic 
critical frequency relationship, we can derive an equation relating f2 to Tr. 
f 2= 1
2 π RC
Solving 10.19 in terms of RC, and substituting yields 
f 2= 2.2
2π T r
f 2=0.35
T r
(10.20)
Where 
f2 is the upper critical frequency, 
Tr is the rise time of the output pulse. 
392",ACElectricalCircuitAnalysis.pdf
"Example 10.16 
Determine the rise time for a lag network critical at 100 kHz. 
f 2=0.35
T r
T r=0.35
f 2
T r= 0.35
100 kHz
T r=3.5μs
10.4 Combining the Elements - Multi-Stage Effects 10.4 Combining the Elements - Multi-Stage Effects 
A complete gain or phase plot combines three elements: (1) the midband response, 
(2) the lead response, and (3) the lag response. Normally, a particular design will 
contain multiple lead and lag networks. The complete response is the summation of 
the individual responses. For this reason, it is useful to find the dominant lead and 
lag networks. These are the networks that affect the midband response first. For lead 
networks, the dominant one will be the one with the highest fc . Conversely, the 
dominant lag network will be the one with the lowest fc . It is very common to 
approximate the complete system response by drawing straight-line segments such 
as those given in Figures 10.5 and 10.7. The process goes something like this:",ACElectricalCircuitAnalysis.pdf
"as those given in Figures 10.5 and 10.7. The process goes something like this: 
• Locate all fc s on the frequency axis. 
• Draw a straight line between the dominant lag and lead fc s at the midband 
gain. If the system does not contain any lead networks, continue the 
midband gain line down to DC. 
• Draw a 6 dB per octave slope between the dominant lead and the next lower 
lead network. 
• Because the effects of the networks are cumulative, draw a 12 dB per octave
slope between the second lead fc and the third fc. After the third fc, the slope 
should be 18 dB per octave, after the fourth, 24 dB per octave, and so on. 
• Draw a −6 dB per octave slope between the dominant lag fc and the next 
highest fc. Again, the effects are cumulative, so increase the slope by −6 dB 
at every new fc . 
393",ACElectricalCircuitAnalysis.pdf
"Example 10.17 
Draw the Bode gain plot for the following amplifier: A'v midband = 26 dB, 
one lead network critical at 200 Hz, one lag network critical at 10 kHz, and 
another lag network critical at 30 kHz.
The dominant lag network is 10 kHz. There is only one lead network, so it’s 
dominant by default. 
• Draw a straight line between 200 Hz and 10 kHz at an amplitude of 
26 dB. 
• Draw a 6 dB per octave slope below 200 Hz. To do this, drop down 
one octave (100 Hz) and subtract 6 dB from the present gain (26 dB 
− 6dB = 20 dB.) The line will start at the point 200 Hz/26 dB, and 
pass through the point 100 Hz/20 dB. Because there are no other 
lead networks, this line may be extended to the left edge of the 
graph. 
• Draw a −6 dB per octave slope between 10 kHz and 30 kHz. The 
construction point will be 20 kHz/20 dB. Continue this line to 
30 kHz. The gain at the 30 kHz intersection should be around 
16 dB. The slope above this second fc will be −12 dB per octave.",ACElectricalCircuitAnalysis.pdf
"30 kHz. The gain at the 30 kHz intersection should be around 
16 dB. The slope above this second fc will be −12 dB per octave. 
Therefore, the second construction point should be at 60 kHz/4 dB 
(one octave above 30 kHz, and 12 dB down from the 30 kHz gain). 
Because this is the final lag network, this line may be extended to 
the right edge of the graph. 
A completed graph is shown in Figure 10.13. 
394
Figure 10.13
Gain plot of complete amplifier.
f
A'v
200
26 dB
-12 dB/octave
0 dB
-6 dB/octave
6 dB/octave
10 k 30 k",ACElectricalCircuitAnalysis.pdf
"There is one item that should be noted before we leave this section, and that is the 
concept of narrowing. Narrowing occurs when two or more networks share similar 
critical frequencies, and one of them is a dominant network. The result is that the 
true −3 dB breakpoints may be altered. Here is an extreme example. Assume that a 
circuit has two lag networks, both critical at 1 MHz. A Bode plot would indicate that
the breakpoint is 1 MHz. This is not really true. Remember, the effects of lead and 
lag networks are cumulative. Because each network produces a 3 dB loss at 1 MHz, 
the net loss at this frequency is actually 6 dB. The true −3 dB point will have been 
shifted. The Bode plot only gives you the approximate shape of the response. 
10.6 Summary10.6 Summary
We have seen how to convert gains and signals into a decibel form for both powers 
and voltages. This is convenient because what would require multiplication and",ACElectricalCircuitAnalysis.pdf
"and voltages. This is convenient because what would require multiplication and 
division under the ordinary scheme only requires simple addition and subtraction in 
the dB scheme. Along with this, dB measurement is used almost exclusively for 
Bode gain plots. A Bode plot details a system’s gain magnitude and phase response. 
For gain, the amplitude is measured in dB, while the frequency is normally 
presented in log form. For a phase plot, phase is measured in degrees, and again, the 
frequency axis is logarithmic. The changes in gain and phase at the frequency 
extremes are caused by lead and lag networks. Lead networks cause the low 
frequency gain to roll off. The roll off rate is 6 dB per octave per network. The phase
will change from +90 degrees to 0 degrees per network. Lag networks cause the high
frequency gain to roll off at a rate of −6 dB per octave per network. The phase 
change per lag network is from 0 degrees to −90 degrees. 
Review QuestionsReview Questions",ACElectricalCircuitAnalysis.pdf
"change per lag network is from 0 degrees to −90 degrees. 
Review QuestionsReview Questions
1. What are the advantages of using decibels over the ordinary scheme? 
2. How do decibel power and voltage gain calculations differ? 
3. Define the differences between dB, dBW, dBm, dBV and dBu.
4. Describe a Bode plot.
5. What is a lead network? What general response does it yield? 
6. What is a lag network? What general response does it yield? 
7. What do the terms f1 and f2 indicate about a system’s response? 
8. What are the rolloff slopes for lead and lag networks? 
9. What are the phase changes produced by individual lead and lag networks? 
10. How is rise time related to upper break frequency? 
11. How do multiple lead or lag networks interact to form an overall system 
response? 
12. How does the decibel measurement scheme differ from the ordinary method 
of indicating gains and signal level?
395",ACElectricalCircuitAnalysis.pdf
"10.7 Exercises 10.7 Exercises 
AnalysisAnalysis
 dB emphasis
1. Convert the following power gains into dB form: a) 10  b) 80  c) 500  d) 1  
e) 0.2  f) 0.03. 
2. Convert the following dB power gains into ordinary form: a) 0 dB  b) 12 dB 
c) 33.1 dB  d) 0.2 dB  e) −5.4 dB  f) −20 dB. 
3. An amplifier has an input signal of 1 mW, and produces a 2 W output. What 
is the power gain in dB? 
4. A Hi-Fi power amplifier has a maximum output of 50 W and a power gain 
of 19 dB. What is the maximum input signal power? 
5. An amplifier with a power gain of 27 dB is driven by a 25 mW source. 
Assuming the amplifier doesn’t clip, what is the output signal in watts? 
6. Convert the following voltage gains into dB form: a) 10  b) 40  c) 250  d) 1 
e) 0.5  f) 0.004 
7. Convert the following dB voltage gains into ordinary form: a) 0.5 dB           
b) 0 dB c) 46 dB  d) 10.7 dB  e) −8 dB  f) −14.5 dB 
8. A guitar pre-amp has a gain of 44 dB. If the input signal is 12 mV , what is 
the output signal?",ACElectricalCircuitAnalysis.pdf
"8. A guitar pre-amp has a gain of 44 dB. If the input signal is 12 mV , what is 
the output signal? 
9. A video amplifier has a 140 mV input and a 1.2 V output. What is the 
voltage gain in dB? 
10. The pre-amp in a particular tape deck can output a maximum signal of 4 V . 
If this amplifier has a gain of 18 dB, what is the maximum input signal? 
11. Convert the following powers into dBW: a) 1 W  b) 23 W  c) 6.5 W             
d) 0.2W e) 2.3 mW  f) 1.2 kW  g) 0.045 mW  h) 0.3 μW  i) 5.6E−18 W. 
12. Repeat Problem 11 for units of dBm. 
13. Repeat Problem 11 for units of dBf. 
14. Convert the following voltages into dBV: a) 12.4 V  b) 1 V  c) 0.25 V          
d) 1.414 V  e) 0.1 V  f) 10.6 kV  g) 13 mV  h) 2.78 μV . 
15. A two stage power amplifier has power gains of 12 dB and 16 dB. What is 
the total gain in dB and in ordinary form? 
16. If the amplifier of Problem 15 has an input of −18 dBW, what is the final 
output in dBW? in dBm? in watts?",ACElectricalCircuitAnalysis.pdf
"16. If the amplifier of Problem 15 has an input of −18 dBW, what is the final 
output in dBW? in dBm? in watts? 
17. Referring to Figure 10.3, what are the various stages’ outputs if the input is 
changed to −4 dBm? to −34 dBW? 
396",ACElectricalCircuitAnalysis.pdf
"18. Which amplifier has the greatest power output? a) 50 watts  b) 18 dBW       
c) 50 dBm. 
19. Which amplifier has the greatest power output? a) 200 mW  b) −10 dBW    
c) 22 dBm. 
20. A three stage amplifier has voltage gains of 20 dB, 5 dB, and 12 dB 
respectively. What is the total voltage gain in dB and in ordinary form? 
21. If the circuit of Problem 20 has an input voltage of −16 dBV , what are the 
outputs of the various stages in dBV? In volts? 
22. Repeat Problem 21 for an input of 12 mV . 
23. Which amplifier produces the largest output voltage? a) 15 V  b) 16 dBV
Bode plot emphasis 
24. Given a lead network critical at 3 kHz, what are the gain and phase values at
100 Hz, 3 kHz, and 40 kHz? 
25. Given a lag network tuned to 700 kHz, what are the gain and phase values at
50 kHz, 700 kHz, and 10 MHz? What is the rise time? 
26. A noninverting amplifier has a midband voltage gain of 18 dB and a single 
lag network at 200 kHz. What are the gain and phase values at 30 kHz,",ACElectricalCircuitAnalysis.pdf
"lag network at 200 kHz. What are the gain and phase values at 30 kHz,    
200 kHz, and 1 MHz. What is the rise time? 
27. Repeat Problem 26 for an inverting (−180 degrees) amplifier. 
28. Draw the Bode plot for the circuit of Problem 26. 
29. Draw the Bode plot for the circuit of Problem 27. 
30. An inverting (−180 degrees) amplifier has a midband gain of 32 dB and a 
single lead network critical at 20 Hz (assume the lag network fc is high 
enough to ignore for low frequency calculations). What are the gain and 
phase values at 4 Hz, 20Hz, and 100 Hz? 
31. Repeat Problem 29 with a noninverting amplifier. 
32. Draw the Bode plot for the circuit of Problem 30. 
33. Draw the Bode plot for the circuit of Problem 31. 
34. A noninverting amplifier used for ultrasonic applications has a midband gain
of 41 dB, a lag network critical at 250 kHz, and a lead network critical at   
30 kHz. Draw its gain Bode plot. 
35. Find the gain and phase at 20 kHz, 100 kHz, and 800 kHz for the circuit of",ACElectricalCircuitAnalysis.pdf
"30 kHz. Draw its gain Bode plot. 
35. Find the gain and phase at 20 kHz, 100 kHz, and 800 kHz for the circuit of 
Problem 34. 
36. If the circuit of Problem 34 has a second lag network added at 300 kHz, 
What are the new gain and phase values at 20 kHz, 100 kHz, and 800 kHz? 
37. Draw the gain Bode plot for the circuit of Problem 36. 
397",ACElectricalCircuitAnalysis.pdf
"38. What are the maximum and minimum phase shifts across the entire 
frequency spectrum for the circuit of Problem 36? 
39. A noninverting DC amplifier has a midband gain of 36 dB, and lag networks
at 100 kHz, 750 kHz, and 1.2 MHz. Draw its gain Bode plot. 
40. What are the maximum and minimum phase shifts across the entire 
frequency spectrum for the circuit of Problem 39? 
41. What is the maximum rate of high frequency attenuation for the circuit of 
Problem 39 in dB/Decade? 
42. If an amplifier has two lead networks, what is the maximum rate of low 
frequency attenuation in dB/Octave? 
ChallengeChallenge
43. You would like to use a voltmeter to take dBm readings in a 600 Ω system. 
What voltage should produce 0 dBm?
44. Assuming that it takes about an 8 dB increase in sound pressure level in 
order to produce a sound that is subjectively “twice as loud” to the human 
ear, can a Hi-Fi using a 100 W amplifier sound twice as loud as one with a",ACElectricalCircuitAnalysis.pdf
"ear, can a Hi-Fi using a 100 W amplifier sound twice as loud as one with a 
40 W amplifier (assuming the same loudspeakers)? 
45. Hi-Fi amplifiers are often rated with a “headroom factor” in dB. This 
indicates how much extra power the amplifier can produce for short periods 
of time, over and above its nominal rating. What is the maximum output 
power of a 250 W amplifier with 1.6 dB headroom? 
46. If the amplifier of Problem 34 picks up an extraneous signal that is a −10 
dBV sine wave at 15 kHz, what is the output? 
47. If the amplifier of Problem 39 picks up a high frequency interference signal 
at 30 MHz, how much is it attenuated over a normal signal? If this input 
signal is measured at 2 dBV , what should the output be? 
48. If an amplifier has two lag networks, and both are critical at 2 MHz, is the 
resulting f2 less than, equal to, or greater than 2 MHz? 
49. If an amplifier has two lead networks, and both are critical at 30 Hz, is the",ACElectricalCircuitAnalysis.pdf
"resulting f2 less than, equal to, or greater than 2 MHz? 
49. If an amplifier has two lead networks, and both are critical at 30 Hz, is the 
resulting f1 less than, equal to, or greater than 30 Hz? 
SimulationSimulation
50. Use a simulator to plot the Bode gain response of the circuit in Problem 39. 
51. Use a simulator to plot the Bode phase response of the circuit in Problem 34.
52. Use a simulation program to generate a Bode plot for a lead network 
comprised of a 1 kΩ resistor and a 100 nF capacitor.   
398",ACElectricalCircuitAnalysis.pdf
"Appendix AAppendix A
Standard Component SizesStandard Component Sizes
Passive components (resistors, capacitors and inductors) are available in standard sizes. The tables below are for
resistors. The same digits are used in subsequent decades up to at least 1 Meg ohm (higher decades are not 
shown). Capacitors and inductors are generally not available in as many standard values as are resistors. 
Capacitors below 10 nF (.01 μF) are usually available at the 5% standard digits while larger capacitances tend to
be available at the 20% standards.
5% and 10% standard values, EIA E24 and EIA E125% and 10% standard values, EIA E24 and EIA E12
10% values (EIA E12) are bold
20% values (seldom used) are every fourth value starting from 10 
(i.e., every other 10% value)
10 11 12 13 15 16 18 20 22 24 27 30
33 36 39 43 47 51 56 62 68 75 82 91
1% and 2% standard values, EIA E96 and EIA E481% and 2% standard values, EIA E96 and EIA E48
2% values (EIA E48) are bold",ACElectricalCircuitAnalysis.pdf
"1% and 2% standard values, EIA E96 and EIA E481% and 2% standard values, EIA E96 and EIA E48
2% values (EIA E48) are bold
10.0 10.2 10.5 10.7 11.0 11.3 11.5 11.8 12.1 12.4 12.7 13.0
13.3 13.7 14.0 14.3 14.7 15.0 15.4 15.8 16.2 16.5 16.9 17.4
17.8 18.2 18.7 19.1 19.6 20.0 20.5 21.0 21.5 22.1 22.6 23.2
23.7 24.3 24.9 25.5 26.1 26.7 27.4 28.0 28.7 29.4 30.1 30.9
31.6 32.4 33.2 34.0 34.8 35.7 36.5 37.4 38.3 39.2 40.2 41.2
42.2 43.2 44.2 45.3 46.4 47.5 48.7 49.9 51.1 52.3 53.6 54.9
56.2 57.6 59.0 60.4 61.9 63.4 64.9 66.5 68.1 69.8 71.5 73.2
75.0 76.8 78.7 80.6 82.5 84.5 86.6 88.7 90.9 93.1 95.3 97.6
399",ACElectricalCircuitAnalysis.pdf
"Appendix BAppendix B  
Methods of Solution of Linear Simultaneous EquationsMethods of Solution of Linear Simultaneous Equations
Some circuit analysis methods, such as nodal analysis and mesh analysis, yield a set of linear simultaneous 
equations. There will be as many equations as there are unknowns. For example, a particular circuit might yield 
three equations with three unknown currents (often referred to as a “3 by 3” for the matrix it creates). There are 
several techniques that may be used to solve this system of equations. The methods include graphical, 
substitution, Guass-Jordan elimination and determinants (determinants may be solved via Cramer's Rule/Sarrus’ 
Rule or via expansion by minors). 
GraphicalGraphical
Graphical solutions involve plotting the individual equations on graph paper. The location of where the lines 
cross is the solution to the system (i.e., values that satisfy all of the equations). This technique will not be",ACElectricalCircuitAnalysis.pdf
"cross is the solution to the system (i.e., values that satisfy all of the equations). This technique will not be 
discussed further because it is only practical for two unknowns. It would be very difficult to draw something 
like a four dimensional graph for four equations with four unknowns! 
SubstitutionSubstitution
The idea here is to write one of the equations in terms of one of the unknowns and then substitute this back into 
one of the other equations resulting in a simplified version. This process is iterated for as many unknowns as the
system includes. Take for example the following 2x2:
10 = 20I1 + 8I2
 2 =  8I1 + 4I2
Solve the second equation for I2.
 2 =  8I1 + 4I2
4I2 =   2 − 8I1
 I2 = 0.5 − 2I1
Substitute this back into the first equation and expand/simplify/solve.
10 = 20I1 + 8I2
10 = 20I1 + 8(.5 − 2I1)
10 = 20I1 + 4 − 16I1
10 =  4I1 + 4 
 I1 =  1.5
Finally, substitute this value back into one of the two original equations to determine I2. 
400",ACElectricalCircuitAnalysis.pdf
"2 = 8I1 + 4I2
  2 = 8(1.5) + 4I2
  2 = 12 + 4I2
4I2 = −10 
 I2 = −2.5 
For a 3x3, this process is iterated as follows: Equation 2 would be solved for I3 and this would be substituted 
back into equation 1 yielding a new equation (let’s call it A) with only I1 and I2 terms. Similarly, Equation 3 
would be solved for I3 and this would be substituted back into equation 2 yielding a new equation (let’s call it B)
with only I1 and I2 terms. Equations A and B now make a 2x2 with I1 and I2 as the unknowns and can be solved 
as outlined above. This would yield values for I1 and I2 which could then be substituted into one of the three 
original equations to obtain I3. While the substitution method is perfectly valid for an arbitrarily sized system, it 
proves cumbersome as the system gets larger. 
Gauss-Jordan EliminationGauss-Jordan Elimination
In some respects, Gauss-Jordan is similar to substitution but it tends to involve less overhead for larger systems",ACElectricalCircuitAnalysis.pdf
"In some respects, Gauss-Jordan is similar to substitution but it tends to involve less overhead for larger systems 
and thus is generally preferred. This method involves multiplying one equation by a constant such that when it is
subtracted from another equation, one of the unknown terms disappears. The process is then iterated for as many
unknowns exist in the system. Using the same example from before:
10 = 20I1 + 8I2
 2 =  8I1 + 4I2
Multiply the second equation by the ratio of the coefficients for I2 (8/4 = 2).
 2 =  8I1 + 4I2
 4 = 16I1 + 8I2
Subtract this new equation from the first equation. The I2 terms will cancel leaving just I1.
10 = 20I1 + 8I2
 4 = 16I1 + 8I2
 6 = 4I1
 I1 = 1.5 
Substitute this result back into one of the original equations to obtain I2. For a 3x3, iterate as follows: Using 
equations 1 and 2, multiply equation 2 by the ratio of the coefficients for I3. Subtract this equation from equation",ACElectricalCircuitAnalysis.pdf
"equations 1 and 2, multiply equation 2 by the ratio of the coefficients for I3. Subtract this equation from equation
1 to generate a new equation (let’s call it equation A) that only has I1 and I2 as unknowns. Using equations 2 and 
3, multiply equation 3 by the ratio of the coefficients for I3. Subtract this equation from equation 2 to generate a 
new equation (let’s call it equation B) that only has I1 and I2 as unknowns. Equations A and B now make a 2x2 
with I1 and I2 as the unknowns and can be solved as outlined previously. This would yield values for I1 and I2 
which could then be substituted into one of the three original equations to obtain I3. Like the substitution 
method, Gauss-Jordan grows rapidly as the system size increases. The process tends to be formulaic though, and
thus easier to handle.  
401",ACElectricalCircuitAnalysis.pdf
"DeterminantsDeterminants
Determinants revolve around the concept of a matrix which itself is little more than an ordered collection of 
coefficients and/or constants. It is imperative that the unknowns be in the same order in each equation (i.e., I1 
ascending to IX left to right) A simple coefficient matrix for the original 2x2 example is:
20  8
 8  4
The resultant value (properly referred to as the determinant) for a 2x2 matrix such as this may be solved using 
Sarrus’ Rule: Simply multiply the two values along the upper right-lower left diagonal and then subtract that 
product from the product of the two values found along on the upper left-lower right diagonal. In this example 
that’s: 
20*4 − 8*8 = 16
A solution involves dividing one determinant by another determinant (Cramer's Rule). That is, each matrix is 
solved for its resultant value and then these two values are divided to determine the final answer. One of these",ACElectricalCircuitAnalysis.pdf
"solved for its resultant value and then these two values are divided to determine the final answer. One of these 
matrices will be the coefficient matrix just discussed. This will be placed in the denominator. The numerator 
matrix is a modified version of the basic coefficient matrix. It is created by replacing one column of coefficients 
with the constant values from the original system of equations. For example, the numerator matrix used to find 
I1 would replace the first column (the I1 coefficients 20 and 8) with the constants 10 and 2:
10  8
 2  4
The resultant value is 40 − 16 or 24.
Similarly, the numerator matrix for I2 would replace the I2 coefficients in the second column (8 and 4) with the 
constants 10 and 2:
20 10
 8  2
The resultant value is 40 − 80 or −40. To find any particular unknown, simply divide the modified matrix by the 
basic coefficient matrix.
      10  8
       2  4
I1 = --------
      20  8
       8  4
      24
I1 = -----
      16
402",ACElectricalCircuitAnalysis.pdf
"I1 =  1.5
In like fashion I2 is found:
      20 10
       8  2
I2 = --------
      20  8
       8  4
     −40
I2 = -----
      16
I2 = −2.5
Sarrus’ Rule may also be used with a 3x3 matrix. This is achieved by extending the matrix. Fourth and fifth 
columns are added to the right of the 3x3 matrix by simply making copies of the first two columns. This creates 
three right to left diagonals with three values each and three left to right diagonals with three values each. The 
three values along each diagonal are multiplied together. The three right to left products are then subtracted from
the sum of the three left to right products yielding a single resultant value (the determinant). To create the 
modified numerator matrix, replace the coefficient column of interest with the constant terms and then replicate 
columns one and two. For example, given these three equations:
10 = 20I1 + 8I2 + 3I3
 2 =  8I1 + 4I2 + 5I3
 7 =  3I1 + 5I2 + 6I3",ACElectricalCircuitAnalysis.pdf
"columns one and two. For example, given these three equations:
10 = 20I1 + 8I2 + 3I3
 2 =  8I1 + 4I2 + 5I3
 7 =  3I1 + 5I2 + 6I3
The basic coefficient matrix (i.e., denominator) is:
20  8  3
 8  4  5
 3  5  6
The extended matrix is:
20  8  3  20  8
 8  4  5   8  4
 3  5  6   3  5
The result is: 
20*4*6 + 8*5*3 + 3*8*5 − 3*4*3 − 20*5*5 − 8*8*6
Sarrus’ Rule does not work beyond 3x3. 
403",ACElectricalCircuitAnalysis.pdf
"Expansion by Minors Expansion by Minors 
Expansion by Minors is another method that may be used to generate a determinant solution. This involves 
breaking the matrix into a series of smaller matrices (minors) that are combined using row-column coefficients. 
The position of these coefficients will also indicate whether the determinant of any particular sub-matrix is 
added or subtracted to the total. 
The first step is to establish a single row or column from which to derive the coefficients. This can be any 
horizontal row or vertical column (no diagonals). Each element of the chosen row or column determines the 
associated minor (essentially, that which is left over). Consider the 3x3 used previously:
20  8  3
 8  4  5
 3  5  6
Choosing the top row yields coefficients of 20, 8 and 3. For each of these, blot out its row and column and see 
what is left. This leaves three 2x2 matrices, one for each coefficient. Multiply each coefficient by the",ACElectricalCircuitAnalysis.pdf
"what is left. This leaves three 2x2 matrices, one for each coefficient. Multiply each coefficient by the 
determinant of its 2x2 matrix. To determine whether this result is added or subtracted to the others, the sign may 
be found using the following map for the coefficients:
+  −  +  − etc.
−  +  −  + etc.
+  −  +  − etc.
The origin in the upper left is positive and the signs continually alternate across from it and down from it. The 
result using the top row for the coefficients is found thus (the 2x2 matrices are bold red for clarity):
     4 5       8 5       8 4
20 * 5 6 − 8 * 3 6 + 3 * 3 5
If the second column was used instead (8, 4, 5), the result is found like so:
     8 5      20 3      20 3
−8 * 3 6 + 4 * 3 6 − 5 * 8 5
In closing, whichever method is used, always look for null coefficient terms (that is, places in the equations and 
matrices where the coefficients are zero). Smart use of these can considerably simplify the computations as there",ACElectricalCircuitAnalysis.pdf
"matrices where the coefficients are zero). Smart use of these can considerably simplify the computations as there
are few mathematical operations easier than multiplying by zero. For example, if a particular row of a matrix 
contains a few zeros, that would be a good candidate for the coefficient row when using expansion by minors 
because some 2x2 minors need not be computed (they will just be multiplied by the zero coefficient).
404",ACElectricalCircuitAnalysis.pdf
"Appendix CAppendix C
Equation ProofsEquation Proofs
RMS Equivalents for Non-Sines
For any waveform, the root-mean-square equivalence factor is computed by normalizing the peak value to unity,
squaring the waveform, finding the mean value of that intermediate result, and then taking the square root of the 
mean. This value can never be more than unity. 
  
The RMS value of a square wave is equal to its peak value. This can be proven by observation. Normalize the 
square wave's peak value to unity. Its negative peak will be −1. Squaring these values results in the constant 
value 1 at any time (assuming the rise and fall times are negligible). Obviously, the mean of this is 1 as is its 
square root. Thus the RMS equivalence factor is unity.
For a rectangular pulse that is always positive (i.e., traversing from zero volts to some positive peak and back to 
zero), the RMS value is equal to its peak value times the square root of its duty cycle. As there is no negative",ACElectricalCircuitAnalysis.pdf
"zero), the RMS value is equal to its peak value times the square root of its duty cycle. As there is no negative 
portion to this waveform, we need only examine the positive portion. First, its normalized amplitude is always 
unity, as is its squared value. The mean is simply the time for this positive pulse divided by the period of the 
wave. By definition, that's equivalent to the duty cycle (i.e., the percentage of time the pulse is positive out of a 
full cycle). Thus, the RMS equivalence factor is the square root of the duty cycle.
For a triangle wave, the RMS value is equal to its peak value times one over the square root of three. This can be
proven by first noting that a triangle wave has quarter wave symmetry. Consequently, we need examine only the 
first quarter of a cycle because the other three will produce identical numerical results.
To ease computation, normalize the amplitude and the time for the first quarter cycle to unity. The result is a",ACElectricalCircuitAnalysis.pdf
"To ease computation, normalize the amplitude and the time for the first quarter cycle to unity. The result is a 
straight line that starts at the origin and reaches an amplitude of 1 when time also reaches 1. Written as a 
function of time, the expression for such a voltage is:
v(t )= t
Squaring this gives us t2. To find the mean, we integrate this function:
mean =∫
0
t
t2
dt
mean = 1
3 t3
|t=0
t=1
mean = 1
3
Finally, taking the square root results in a factor of one over the square root of three. QED. 
405",ACElectricalCircuitAnalysis.pdf
"Maximum Power Transfer: Finding the Maximizing Value of P = R/(R2+2R+1)
While the algebraic and graphing technique explored in Chapter 5 leads to a proper answer, it is incomplete. A 
more rigorous treatment using differential calculus follows. We have already determined that the reactive 
portion of the load must have the same magnitude but opposite sign of the internal reactance in order to 
maximize load current, and therefore maximize load power. Thus, we need only examine the resistive portion 
which is described by the equation P = R/(R2+2R+1). This function exhibits a single peak and thus we may find 
the corresponding value by taking the first derivative of the function, setting it to zero (i.e., find the point where 
the slope goes to zero), and solving for R. Chain rule can be used to solve this. Chain rule states: 
dy
du
du
dx
dy
dx 
One way to apply the chain rule is first to rewrite the main equation to remove the numerator R. This effectively",ACElectricalCircuitAnalysis.pdf
"dy
du
du
dx
dy
dx 
One way to apply the chain rule is first to rewrite the main equation to remove the numerator R. This effectively 
removes the issues of having a quotient or product. We divide through by R and arrive at:
RR
P 12
1


or in a somewhat more convenient form:
11 )2(  RRP
Using the chain rule the derivative of this is:
)1()2( 221   RRRdR
dP  
or in “prettier” form:
  21
2
)2(
1



 RR
R
dR
dP
Multiply through by R2
22
2
)12(
1

 RR
R
dR
dP
Which, for a really anal retentive sort of completeness, can be rewritten as:
4
2
)1(
1

 R
R
dR
dP
For dP/dR to be zero, R must equal 1. In other words, it must match the internal resistance. QED.
406",ACElectricalCircuitAnalysis.pdf
"Finding k0 : Determining f1 and f2 in Resonant Circuits
The coefficient k0 was defined such that: f 1 = f 0
k0
 and f 2 = f 0×k0
We start with the definition of circuit Q based on bandwidth and resonant frequency, and expand, solving in 
terms of f2.
Qcircuit = f 0
BW = f 0
f 2 − f 1
f 2 = f 0
Qcircuit
+f 1 (EQ 1)
The resonant frequency is equal to the geometric mean of f1 and f2.
f 0 = √ f 1 f 2
f 1 = f 0
2
f 2
(EQ 2)
Substituting EQ 2 into EQ 1 yields:
f 2 = f 0
Qcircuit
+ f 0
2
f 2
(EQ 3)
We normalize f0, taking it as unity. This means that f2 is now equivalent to k0. Rewriting EQ 3 yields:
k0 = 1
Qcircuit
+ 1
k0
0 = k0
2
− 1
Qcircuit
k0 −1
We can solve this using the quadratic formula where a = 1, b = 1/Qcircuit and c = −1.
−b±√b2
−4ac
2a
Substituting and simplifying results in the equation for k0 :
k0 = 1
2Qcircuit
+
√
1
4Qcircuit
2 +1
407",ACElectricalCircuitAnalysis.pdf
"Appendix DAppendix D  
Answers to Selected Odd-Numbered ProblemsAnswers to Selected Odd-Numbered Problems
1 Fundamentals1 Fundamentals
  1. 10, 7.07, 0, 1 kHz, 1 ms, 0°  3. 20, −3, 50 Hz, 20 ms, 0°
  5. 10, 7.07, 0, 100 Hz, 10 ms, 45°  7. 1, 10, 400 Hz, 2.5 ms, −45°
  9. 200 μs, 10 μs 11. 36°
13. 14.1445°, 11.2−63.4°, 102169°, 5k53.1° 15. 7.07 + j7.07, j0.4, −4.5 + j7.79, 70.7 − j70.7
17. 15 + j30, j4, −20−j4, −70 + j250 19. −34.5k + j36k, −725 + j95, 2.39 −j0.709,       
−2.71 −j0.457
21. 1000°, 10−115°, 0.5145°, 0.25−45° 23. 2.7180°, 4.91−92.7°, 0.076123°, 5444.5°
25. −j15.9 k, −j318, −j15.9, −j0.398, −j15.9E−3 Ω 27. −j318 M, −j6.77 M, −j144.7 k, −j96.5 Ω
29. j6.28, j314, j6.28 k, j251 k, j6.28 M Ω 31. j62.8, j3.14 k, j62.8E−3, j2.51 Ω
33. 35. 1.67 @ 3 kHz, 1 @ 5 kHz, 0.714 @ 7 kHz,      
0.555 @ 9 kHz, 0.455 @ 11 kHz
408",ACElectricalCircuitAnalysis.pdf
"37. 284 nF, 482 pF, 339 pF, 132.6 nF, 212 nF 39. 892 nH, 525 μH, 748 μH, 1.91 μH, 1.19 μH
41. a 43. b
409",ACElectricalCircuitAnalysis.pdf
"2 Series RLC Circuits2 Series RLC Circuits
  1. 2k − j1.94 k Ω  3. 270 + j125.7 Ω
  5. 1 k − j1.278 k Ω  7. 300 − j400
  9. 1.447 μF 11. v(t)=0.1sin2π1000t (i and v are in phase)
13. i is 241 mA p-p and lags by 90° 15. v is 16.6 V p-p and leads by 90°
17. 1 k − j318 Ω  19. i = 953E−617.6° amps, vR = 953E−317.6° 
volts, vC = 303E−3−72.4° volts, delay = 25 μs
21. 1 k + j3.14 kΩ 23. 1 k + j942 Ω
25. 2 k − j33.4 Ω 27. i = 0.4999E−3.957° amps, vR = 0.998.957° 
volts, vC = 16.7E−3−89° volts, delay = 167 μs
29. i = 21.2E−345° amps, vR = 4.2445° volts,       
vC = 4.24−45° volts
31. vS = 60.9−23.2° volts, vR = 560° volts,            
vC = 24−90° volts
33. i = 329E−670.8° amps, vR = 329E−370.8° 
volts, vC = 1.05−19.2° volts, vL = 103E−3160.8° 
volts
35. i = 48.7E−3−13° amps, vR = 9.745−13° volts,  
vC = 3.88−103° volts, vL = 6.12577° volts
37. i = 493E−39.52° amps, vB = 58.241.6° volts, 
vAC = 63−29° volts
39. i = 1E−30° amps, vR = 10° volts,",ACElectricalCircuitAnalysis.pdf
"37. i = 493E−39.52° amps, vB = 58.241.6° volts, 
vAC = 63−29° volts
39. i = 1E−30° amps, vR = 10° volts,                     
vC = 200E−3−90° volts, vL = 200E−390° volts
410",ACElectricalCircuitAnalysis.pdf
"41. i = 2.24E−363.4° amps, vB = 8.94−26.6° volts, 
vC = 2.24153.4° volts, vAC = 12−4.8° volts
43. i = 35.36E−3−45° amps, vC = 2.12−135° volts,
vR = 1.414−45° volts, vL = 3.5445° volts
45. vR = 100° volts, vC = 5.3−90° volts,                 
vL = 7.0790° volts
47. vR = 2.40° volts, vL = 85.4E−390° volts,          
vC = 424E−390° volts
49. vR = 1100° volts, vC = 88−90° volts,                
vL = 22090° volts
51. vAC = 6087.8° volts, vB = 20.1−83.3° volts,      
vC = 20−90° volts
53. vR = 0.32970.8° volts, vL = 0.1034160.8° volts, 
vC = 1.048−19.2° volts
55. L = 79.6 μH, C = 7.24 nF
57. vR = 7.23−130° volts, vL = 3.62−40.1° volts,    
vC = 1.45139.9° volts
59. f = 15.594 kHz
61. f = 3.185 kHz
411",ACElectricalCircuitAnalysis.pdf
"3 Parallel RLC Circuits3 Parallel RLC Circuits
  1. 73.2−12.5° (71.4 −j15.8)  3. 99−8.04° (98 −j13.8)
  5. 182.952.4° (111.5  + j145)  7. f = 45.2 MHz
  9. 11. is = 0.1200010.18°, iR = 0.120°,                        
iC = 377E−690°
13. is = 9.43E−3−58°, iR = 5E−30°,                      iC 
= 2E−390°, iL = 10E−3−90°
15. is = 47.4E−316.3°, iR = 45.5E−30°,                   
iC = 20E−390°, iL = 6.667E−3−90° (all peak)
17. iR = 19.99E−3−1.73°, iC = 603E−688.27° 19. vR = vL = 627E−387° volts
21. vs = 2.82−28.1° volts 23. i2.2k = 5.06E−3−68.2°, i4.7k = 2.37E−3−68.2°,     
iC = 55.7E−321.8°, iL = 37.1E−3−158.2°
25. iC = 670E−6150°, iL = 536E−6−30°,                  27. iR = 1.67E−3−5°, iC = 835E−685°,                    
412",ACElectricalCircuitAnalysis.pdf
"iR = 2.23E−360° iL = 1.044E−3−95°
29. 2.84 μF 31. 7.7 nF
33. 34.5 nF 35. 65.7 nF
37. 390 nF 39. 507 nF
41. 128 mH
4 Series-Parallel RLC Circuits4 Series-Parallel RLC Circuits
  1. Z10k ≈ 1250°,  Z1M ≈ 125−0.8°,                     
Z100M ≈  68.8−24.2° Ω
 3. Z300 ≈ 0.7590°,  Z30k ≈ 77.589.4°,                     
Z3M ≈  9075.4° Ω
  5. Z = 18051.6° Ω  7. is = 9.64E−3−85.6°, iR = 740.7E−60°,                
iC = iL = 9.62E−3−90°,
  9. vR = vC = 3.45−80.5°, vL = 7.2827.9° 11. i50 = 31.4E−690°, i91 = i20 = 12.6E−689.9°,       
is ≈ 44E−690°
13. vb = 9.19−50°, vab = 15.826.5° 15. ij4k = 5E−3−90°, i2.7k = 7.41E−30°,                    
i3.9k = i−j1k = 4.97E−314.4°, is = 12.8E−3−17.1°
17. vb = 1200.7°, vab = 20−175.6° 19. i15k = iL = 887E−6−4°, i12k = 1.12E−33.17°
21. va = 4.814.1°, vb = 6.6214.1° 23. va = 571E−3−26.5°, vb = 537E−3−6.68°
25. va = 19.85127.2°, vb = 23.8127.9° 27. ic = 12.83E−3136.9°
29. va = 7.3261.5°, vb = 7.175.6° 31. vab = −133.30°",ACElectricalCircuitAnalysis.pdf
"25. va = 19.85127.2°, vb = 23.8127.9° 27. ic = 12.83E−3136.9°
29. va = 7.3261.5°, vb = 7.175.6° 31. vab = −133.30°
33. 217 nF 35. 3.24 μF
37. 5.9 μF 39. 19.1 mH
413",ACElectricalCircuitAnalysis.pdf
"5 Analysis Theorems and Techniques5 Analysis Theorems and Techniques
 1.  vb = 2.993.5°  3. i82 = 13.3E−359.8° 
 5. vb = 1.08165°  7. i2.2k = 1.3E−349.2° 
 9. vb = 7.8498.9°, vcd = 10.2−44.2° 11. iS1 = 337E−6−73.6°, iS2 = 401E−6152°
13. vab = 14.3−25.4° 15. vab = 972E−3−166°
17. va = 1.31−174°, vb = 2.58−154° 19. iC = 103E−3101°, iL = 369E−3171°
21. vbc = 19.65−102.5° 23. iE = 14.8E−3177°
25. vb = 10.748.4° 27. iL = 286.7E−340.6°
29. vab = 7.18−115° 31. ZTH = 910°, ETH = 10°
33. ZTH = ZN = j714.3, ETH = 8.5710°,                        
IN = 12E−3−90°
35. ZN = 514.531°, IN = 24E−3−45°
37. vb = 5.69−71.5° 39. ZN = 156417.1°, IN = 19.2E−3−17.1°,               
ZL = 1495 − j461
41. ZTH = ZN = 64.418.8°, ETH = 113.618.8°,           
IN = 1.7650°, Combo needed = 61 − j20.7,                 
P = 52.9 W
43. All three pair = 3.33 k + j3.33 k
45. All three pair = 1.33 k − j1 k 47. Za = Zxy = 3.667 k + j3.667 k,",ACElectricalCircuitAnalysis.pdf
"43. All three pair = 3.33 k + j3.33 k
45. All three pair = 1.33 k − j1 k 47. Za = Zxy = 3.667 k + j3.667 k,                                  
Zb = Zxz = 5.5 k + j5.5 k, Zc = Zyz = 11 k + j11 k
49. Za = Zxy = 1.83 k − j1.83 k,                                      
Zb = Zxz = 2.75 k − j2.75 k, Zc = Zyz = 5.5 k − j5.5 k
51. vbc = 4.554565.4°
53. iS = 11.8E−3101°, in parallel with a series 
combination of 1 kΩ and 79.6 nF
55. iS = 1.54E−3−22.7°, in parallel with a series 
combination of 600 Ω and 2 mH
57. eS = 1.98E3−49.3°, in series with 4.3 kΩ and a 
capacitive reactance of −j5 k
59. eS = 96.319°, in parallel with a series 
combination of 9.1 kΩ and 5 mH
414",ACElectricalCircuitAnalysis.pdf
"61. 63. 
65. −j43.8 Ω
415",ACElectricalCircuitAnalysis.pdf
"6 Nodal and Mesh Analysis6 Nodal and Mesh Analysis
 1. vc = 34.437.7°  3. i43 = 33.2E−344.4°
 5. vc = 180−106°  7. i4Ω = 772E−347.1°
 9. vac = 18.35−80.8° 11. i22 = 3.07−153°
13. vc = 6.09−3.68° 15. i3.3k = 9.86E−318.8°
17. vc = 1838.9° 19. vbc = 1.0959.8°
21. vba = 964E−3127.3° 23. vcb = 15.2149°
25. Loop ordering is left to right
      20° = (4.9 k)i1 − (2.2 k)i2
     −30° = − (2.2 k)i1 + (2.2 k − j106)i2
     vb = 2.99−3.5°
 27. i75 = 8.22E−3−138.5°
29. Loop ordering is left to right
      200° = (6.6 k + j4 k)i1 − (2.7 k + j4 k)i2
     −50° = − (2.7 k + j4 k)i1 + (4.5 k + j3 k)i2
vb = 7.8498.9°
31. ij200 = 4.36E−3−28.5°
33. vcd = 218E−3−153° 35. i−j200 = 49.5E−3−44.8° (up)
37. vc = 14.4133.3° 39. ij300 = 11.7E−3138° (up)
41. vbc = 4.54565.4° 43. iR3 = 9.81E−3175°
45. vb = 9.24−8.7° 47. i2.2k = 1.51E−3−2.04°
49. vab = 14.3−25.4° 51. i330 = 7.76E−314.1°
53. vbc = 58.9175° 55. vb = 2.6842.3°
57. i5k = 49.9E−6−29.9° 59. vc = 106.790.6°",ACElectricalCircuitAnalysis.pdf
"49. vab = 14.3−25.4° 51. i330 = 7.76E−314.1°
53. vbc = 58.9175° 55. vb = 2.6842.3°
57. i5k = 49.9E−6−29.9° 59. vc = 106.790.6°
61. i1k = 17.8E−321.8° 63. va = 9.776°, vb = 0.273−12.3°
65. i1k = 5.24E−332.9° 67. vab = 90.6−112°
69. i1k = 99.95E−60° 71. vd = 16.670.191°
416",ACElectricalCircuitAnalysis.pdf
"7 AC Power7 AC Power
1. S = 6.25 mV A, P = 4.88 mW, Q = 3.9 mV AR (ind), 
PF = 0.781
3. S = 64.4 V A, P = 57.7 W, Q = 28.8 V AR (ind), PF = 
0.894
 5. S = 315 V A, P = 162 W, Q = 270 V AR (ind),         
PF = 0.515
7. S = 751 V A, P = 720 W, Q = 216 V AR (cap),         
PF = 0.958
 9. i = 4.286 A, P = 90.9 W 11. P = 6.4 kW, Q = 4.8 kV AR, L = 2.86 mH
13. S = P = 180 W, Q = 0 V AR, i = 1.5 A 15. S = P = 1600 W, Q = 0 V AR
17. S =  1523 V A, P = 1400 W, Q = 600 V AR,           
PF = 0.919, i = 12.7 A
19. S = 1811 V A, P = 1800 W, Q = 200 V AR,             
PF = 0.994, i = 15.09 A
21. 89.5% 23. 0.829
25. S = 683 V A, P = 478 W, Q = 488 V AR,                   
i = 5.69 A
27. S = 2403 V A, P = 2332 W, PF = 0.971, i = 20 A
29. S = 3154 V A, P = 2943 W, Q = 1135 V AR,           
PF = 0.933
31. 141 μF
33. 102 mH 35. 17.6 mH
37. 260 μF
417",ACElectricalCircuitAnalysis.pdf
"8 Resonance8 Resonance
 1. BW = 14.67 kHz, f1 = 432.7 kHz, f2 = 447.3 kHz  3. Qcoil = 50, Rcoil = 1.88 Ω
 5. 15 kΩ || j600 Ω  7. Rp = 20.36 kΩ, Lp = 75 μH
 9. f0 = 247 kHz, Q = 10.2 11. f0 =  5.03 kHz, Qsys = 8.35, BW = 602 Hz
13. f0 = 10.07 kHz, Qsys = 4.22, BW = 2.39 kHz,          
vC = 4.22
15. f0 = 10.07 kHz, Qsys = 3.51, BW = 2.87 kHz,          
vC = 3.51
17. f0 = 10.07 kHz, Qsys = 4.22, BW = 2.39 Hz,          
vR= 1.5, vC = 6.32
19. f0 = 136.8 kHz, Qsys = 11, BW = 12.4 kHz,              
iL = 424E−3−84.8°, iC = 425.5E−390°
21. f0 = 360.4 kHz, Qsys = 24.1, BW = 14.9 kHz,           
iR = 3.19E−30°, iC = 169.8E−390°,                         
iL = 169.8E−3−88.7° 
23. f0 = 142.4 kHz, Qsys = 14.35, BW = 9.92 kHz,         
iR = 5E−30°, iC = 111.9E−390°,                               
iL = 111.8E−3−88.6°
25. f0 = 225 kHz, Qsys = 25.3, BW = 8.9 kHz,                
iR = 149E−60°, iC = 12.6E−390°,                            
iL = 12.6E−3−88.4°, vR = 1.787
27. 135 μH",ACElectricalCircuitAnalysis.pdf
"iR = 149E−60°, iC = 12.6E−390°,                            
iL = 12.6E−3−88.4°, vR = 1.787
27. 135 μH
29. 6.94 kΩ 31. Qsys = 50, however, none of the inductors exhibit 
Qcoil ≥ 50 at 1 MHz. Therefore there are no viable 
candidates because Qsys can be no larger than Qcoil.
9 Polyphase Power9 Polyphase Power
 1. vLINE = 1200° V (with 120° and 240°, other 
phases not shown from here on), iLINE = 20.8 A,      
iLOAD = 12 A, PLOAD = 4320 W
3. vLINE = 208 V, iLINE = iLOAD = 10.4 A,  PLOAD = 6490 W
 5. vLINE = 120 V, vLOAD = 69.3 V, iLINE = 13.86 A,       
iGEN-PHASE = 8 A, PTOTAL = 2880 W
7. vLINE = vLOAD = 208 V, iLINE = 6 A, PLOAD = 2160 W
 9. vLINE = 208 V, iLINE = 33.4 A, SLOAD = 12 kV A,      
PLOAD = 11.1 kW
11. vLINE = 120 V, iLINE = iLOAD = 11.14 A, SLOAD = 4 kV A,
PLOAD = 3.72 kW
13. vLINE = vLOAD = 398 V, iLINE = 13.8 A,  iLOAD = 7.97 A,
SLOAD = 9.52 kV A, PLOAD = 7.61 kW
15. vLINE = 120 V, iLINE = iLOAD =  11. A, SLOAD = 4 kV A, 
PLOAD = 3.72 kW",ACElectricalCircuitAnalysis.pdf
"SLOAD = 9.52 kV A, PLOAD = 7.61 kW
15. vLINE = 120 V, iLINE = iLOAD =  11. A, SLOAD = 4 kV A, 
PLOAD = 3.72 kW
17. vLINE = 120 V, iLINE =  2.747 A, SLOAD = 571 V A, 
PLOAD = 566 W
19. 91.5 μF
418",ACElectricalCircuitAnalysis.pdf
"10 Decibels and Bode Plots10 Decibels and Bode Plots
1) A) 10 dB  B) 19 dB  C) 26.99 dB  D) 0 dB  E) −6.99 dB  F) −15.23 dB
3) 33 dB 
5) G = 501,   Pout = 12.53 W
7) A) 1.06  B) 1  C) 199.5  D) 3.43 E) 0.398  F) 0.188
9) A = 8.57  A' = 18.66 dB
11) A) 0 dBW B) 13.6 dBW  C) 8.13 dBW  D) −7 dBW  E)  −26.4 dBW  F)30.8 dBW
G) −43.5 dBW  H) −65.2 dBW  I) −172.5 dBW
13) A) 150 dBf B) 163.6 dBf  C) 158.1 dBf  D) 143 dBf  E) 123.6 dBf  F) 180.8 dBf
G) 106.5 dBf  H) 84.8 dBf  I) −22.5 dBf
15) G'total = 28 dB,  G = 631
17) For P'in = 4 dBm: output stage 1 = 6 dBm, stage 2 = 0 dBm, stage 3 = 15 dBm. 
For P'in = −34 dBm: output stage 1 = − 24 dBW, stage 2 = −30 dBW, stage 3 = −15 dBW.
19) a. 200 mW
21) V'out = 21 dBV , 21 dBV = 11.2 V (final output)
For stage 1: 4 dBV = 1.58 V
For stage 2: 9 dBV = 2.82 V
23) a. 15 V
25) At 50 kHz: −0.022 dB, −4.09 degrees
At 700 kHz: −3 dB, −45 degrees
At 10 MHz: −23.1 dB, −86 degrees
Tr = 500 μsec",ACElectricalCircuitAnalysis.pdf
"23) a. 15 V
25) At 50 kHz: −0.022 dB, −4.09 degrees
At 700 kHz: −3 dB, −45 degrees
At 10 MHz: −23.1 dB, −86 degrees
Tr = 500 μsec
27) The amplitude portion does not change. Phases are: At 30 kHz −188.5 degrees, at 200 kHz −225 degrees,      
at 1 MHz −258.7 degrees
419",ACElectricalCircuitAnalysis.pdf
"29)
31) At 4 kHz: 78.7 degrees, at 20 Hz: 45 degrees, at 100 Hz: 11.3 degrees
33)
35) Net gain at 20 kHz = 35.87 dB
Net phase at 20 kHz = 51.7 degrees
At 100 kHz: phase = −5.1 degrees, A'v = 40 dB
At 800 kHz: phase = −70.5 degrees, A'v = 30.5 dB
37)
420
f
A'v
200 kHz
18 dB f
θ
0°
-180°
-270°
2 MHz20 kHz
f
A'v
20 Hz
32 dB
f
θ
0°
90°
200 Hz2 Hz
45°
20 Hz
f
A'v
250 kHz 300 kHz30 kHz",ACElectricalCircuitAnalysis.pdf
"39)
 
41) Each lag network rolls off at 20 dB/decade for a 60 dB/decade total (i.e., above 1.2 MHz).
43) 0.775 V
45) 360 W
47) 71.5 dBV
49) Greater than 30 Hz. 
421
f
A'v
750 kHz
36 dB
1.2 MHz100 kHz",ACElectricalCircuitAnalysis.pdf
"Appendix EAppendix E
A Closing ObservationA Closing Observation
This title is the twelfth in a series of free open educational resources, now at five texts and seven laboratory 
manuals. People occasionally ask why these titles have not gone through the typical route of using a traditional 
publisher (indeed, the Operational Amplifiers & Linear Integrated Circuits text went that route originally and 
then reverted to OER for its third edition). Surely, money is to be made, and after all, this is the culture that 
invented the modern use of the word “monetize”. In a culture that seeks to monetize everything, creating 
something of value and then giving it away is a subversive act. Some people do not see the incessant search for 
profit as necessarily a societal good. 
422",ACElectricalCircuitAnalysis.pdf
"Trinity University Trinity University 
Digital Commons @ Trinity Digital Commons @ Trinity 
Faculty Authored and Edited Books & CDs 
12-2013 
Elementary Differential Equations with Boundary Value Problems Elementary Differential Equations with Boundary Value Problems 
William F. Trench 
Trinity University, wtrench@trinity.edu 
Follow this and additional works at: https://digitalcommons.trinity.edu/mono 
 Part of the Mathematics Commons 
Recommended Citation Recommended Citation 
Trench, William F., ""Elementary Differential Equations with Boundary Value Problems"" (2013). Faculty 
Authored and Edited Books & CDs. 9. 
https://digitalcommons.trinity.edu/mono/9 
This Book is brought to you for free and open access by Digital Commons @ Trinity. It has been accepted for 
inclusion in Faculty Authored and Edited Books & CDs by an authorized administrator of Digital Commons @ Trinity. 
For more information, please contact jcostanz@trinity.edu.",Elementary Differential Equations with Boundary Value Problems.pdf
"ELEMENT ARY
DIFFERENTIAL EQUA TIONS WITH
BOUNDARY V ALUE PROBLEMS
William F . T rench
Andrew G. Cowles Distinguished Professor Emeritus
Department of Mathematics
T rinity University
San Antonio, T exas, USA
wtrench@trinity.edu
This book has been judged to meet the evaluation criteria set by the Edi-
torial Board of the American Institute of Mathematics in con nection with
the Institute’s Open T extbook Initiative . It may be copied, modified, re-
distributed, translated, and built upon subject to the Crea tive Commons
Attribution-NonCommercial-ShareAlike 3.0 Unported Lice nse.
FREE DOWNLOAD: STUDENT SOLUTIONS MANU AL",Elementary Differential Equations with Boundary Value Problems.pdf
"Free Edition 1.01 (December 2013)
This book was published previously by Brooks/Cole Thomson L earning, 2001. This free edition is made
available in the hope that it will be useful as a textbook or re ference. Reproduction is permitted for
any valid noncommercial educational, mathematical, or sci entific purpose. However, charges for profit
beyond reasonable printing costs are prohibited.",Elementary Differential Equations with Boundary Value Problems.pdf
TO BEVERL Y,Elementary Differential Equations with Boundary Value Problems.pdf
"Contents
Chapter 1 Introduction 1
1.1 Applications Leading to Differential Equations
1.2 First Order Equations
5
1.3 Direction Fields for First Order Equations 16
Chapter 2 First Order Equations 30
2.1 Linear First Order Equations 30
2.2 Separable Equations 45
2.3 Existence and Uniqueness of Solutions of Nonlinear Equa tions 55
2.4 T ransformation of Nonlinear Equations into Separable E quations 62
2.5 Exact Equations 73
2.6 Integrating Factors 82
Chapter 3 Numerical Methods
3.1 Euler’s Method
96
3.2 The Improved Euler Method and Related Methods 109
3.3 The Runge-Kutta Method 119
Chapter 4 Applications of First Order Equations1em 130
4.1 Growth and Decay 130
4.2 Cooling and Mixing 140
4.3 Elementary Mechanics 151
4.4 Autonomous Second Order Equations 162
4.5 Applications to Curves 179
Chapter 5 Linear Second Order Equations
5.1 Homogeneous Linear Equations
194
5.2 Constant Coefficient Homogeneous Equations 210
5.3 Nonhomgeneous Linear Equations 221",Elementary Differential Equations with Boundary Value Problems.pdf
"5.1 Homogeneous Linear Equations
194
5.2 Constant Coefficient Homogeneous Equations 210
5.3 Nonhomgeneous Linear Equations 221
5.4 The Method of Undetermined Coefficients I 229
iv",Elementary Differential Equations with Boundary Value Problems.pdf
"5.5 The Method of Undetermined Coefficients II 238
5.6 Reduction of Order 248
5.7 V ariation of Parameters 255
Chapter 6 Applcations of Linear Second Order Equations 268
6.1 Spring Problems I 268
6.2 Spring Problems II 279
6.3 The RLC Circuit 290
6.4 Motion Under a Central Force 296
Chapter 7 Series Solutions of Linear Second Order Equations
7.1 Review of Power Series
306
7.2 Series Solutions Near an Ordinary Point I 319
7.3 Series Solutions Near an Ordinary Point II 334
7.4 Regular Singular Points Euler Equations 342
7.5 The Method of Frobenius I 347
7.6 The Method of Frobenius II 364
7.7 The Method of Frobenius III 378
Chapter 8 Laplace T ransforms
8.1 Introduction to the Laplace T ransform
393
8.2 The Inverse Laplace T ransform 405
8.3 Solution of Initial V alue Problems 413
8.4 The Unit Step Function 419
8.5 Constant Coefficient Equations with Piecewise Continuo us Forcing
Functions 430
8.6 Convolution 440
8.7 Constant Cofficient Equations with Impulses 452",Elementary Differential Equations with Boundary Value Problems.pdf
"Functions 430
8.6 Convolution 440
8.7 Constant Cofficient Equations with Impulses 452
8.8 A Brief T able of Laplace T ransforms
Chapter 9 Linear Higher Order Equations
9.1 Introduction to Linear Higher Order Equations
465
9.2 Higher Order Constant Coefficient Homogeneous Equation s 475
9.3 Undetermined Coefficients for Higher Order Equations 487
9.4 V ariation of Parameters for Higher Order Equations 497
Chapter 10 Linear Systems of Differential Equations
10.1 Introduction to Systems of Differential Equations
507
10.2 Linear Systems of Differential Equations 515
10.3 Basic Theory of Homogeneous Linear Systems 521
10.4 Constant Coefficient Homogeneous Systems I 529",Elementary Differential Equations with Boundary Value Problems.pdf
"vi Contents
10.5 Constant Coefficient Homogeneous Systems II 542
10.6 Constant Coefficient Homogeneous Systems II 556
10.7 V ariation of Parameters for Nonhomogeneous Linear Sys tems 568
Chapter 11 Boundary V alue Problems and Fourier Expansions 580
11.1 Eigenvalue Problems for y′′+ λy = 0 580
11.2 Fourier Series I 586
11.3 Fourier Series II 603
Chapter 12 Fourier Solutions of Partial Differential Equat ions
12.1 The Heat Equation 618
12.2 The W ave Equation 630
12.3 Laplace’s Equation in Rectangular Coordinates 649
12.4 Laplace’s Equation in Polar Coordinates 666
Chapter 13 Boundary V alue Problems for Second Order Linear E quations
13.1 Boundary V alue Problems 676
13.2 Sturm–Liouville Problems 687",Elementary Differential Equations with Boundary Value Problems.pdf
"Preface
Elementary Differential Equations with Boundary V alue Pro blems is written for students in science, en-
gineering, and mathematics who have completed calculus thr ough partial differentiation. If your syllabus
includes Chapter 10 (Linear Systems of Differential Equati ons), your students should have some prepa-
ration in linear algebra.
In writing this book I have been guided by the these principle s:
• An elementary text should be written so the student can read it with comprehension without too
much pain. I have tried to put myself in the student’s place, a nd have chosen to err on the side of
too much detail rather than not enough.
• An elementary text can’t be better than its exercises. This text includes 2041 numbered exercises,
many with several parts. They range in difficulty from routin e to very challenging.
• An elementary text should be written in an informal but math ematically accurate way , illustrated",Elementary Differential Equations with Boundary Value Problems.pdf
"• An elementary text should be written in an informal but math ematically accurate way , illustrated
by appropriate graphics. I have tried to formulate mathemat ical concepts succinctly in language
that students can understand. I have minimized the number of explicitly stated theorems and def-
initions, preferring to deal with concepts in a more convers ational way , copiously illustrated by
299 completely worked out examples. Where appropriate, con cepts and results are depicted in 188
figures.
Although I believe that the computer is an immensely valuabl e tool for learning, doing, and writing
mathematics, the selection and treatment of topics in this t ext reflects my pedagogical orientation along
traditional lines. However, I have incorporated what I beli eve to be the best use of modern technology ,
so you can select the level of technology that you want to incl ude in your course. The text includes 414",Elementary Differential Equations with Boundary Value Problems.pdf
"so you can select the level of technology that you want to incl ude in your course. The text includes 414
exercises – identified by the symbols C and C/G – that call for graphics or computation and graphics.
There are also 79 laboratory exercises – identified by L – that require extensive use of technology . In
addition, several sections include informal advice on the u se of technology . If you prefer not to emphasize
technology , simply ignore these exercises and the advice.
There are two schools of thought on whether techniques and ap plications should be treated together or
separately . I have chosen to separate them; thus, Chapter 2 d eals with techniques for solving first order
equations, and Chapter 4 deals with applications. Similarl y , Chapter 5 deals with techniques for solving
second order equations, and Chapter 6 deals with applicatio ns. However, the exercise sets of the sections
dealing with techniques include some applied problems.",Elementary Differential Equations with Boundary Value Problems.pdf
"dealing with techniques include some applied problems.
Traditionally oriented elementary differential equation s texts are occasionally criticized as being col-
lections of unrelated methods for solving miscellaneous pr oblems. T o some extent this is true; after all,
no single method applies to all situations. Nevertheless, I believe that one idea can go a long way toward
unifying some of the techniques for solving diverse problem s: variation of parameters. I use variation of
parameters at the earliest opportunity in Section 2.1, to so lve the nonhomogeneous linear equation, given
a nontrivial solution of the complementary equation. Y ou ma y find this annoying, since most of us learned
that one should use integrating factors for this task, while perhaps mentioning the variation of parameters
option in an exercise. However, there’s little difference b etween the two approaches, since an integrating",Elementary Differential Equations with Boundary Value Problems.pdf
"option in an exercise. However, there’s little difference b etween the two approaches, since an integrating
factor is nothing more than the reciprocal of a nontrivial so lution of the complementary equation. The
advantage of using variation of parameters here is that it in troduces the concept in its simplest form and
vii",Elementary Differential Equations with Boundary Value Problems.pdf
"viii Preface
focuses the student’s attention on the idea of seeking a solu tion y of a differential equation by writing it
as y = uy1, where y1 is a known solution of related equation and uis a function to be determined. I use
this idea in nonstandard ways, as follows:
• In Section 2.4 to solve nonlinear first order equations, suc h as Bernoulli equations and nonlinear
homogeneous equations.
• In Chapter 3 for numerical solution of semilinear first orde r equations.
• In Section 5.2 to avoid the necessity of introducing comple x exponentials in solving a second or-
der constant coefficient homogeneous equation with charact eristic polynomials that have complex
zeros.
• In Sections 5.4, 5.5, and 9.3 for the method of undetermined coefficients. (If the method of an-
nihilators is your preferred approach to this problem, comp are the labor involved in solving, for
example, y′′+ y′+ y= x4ex by the method of annihilators and the method used in Section 5 .4.)",Elementary Differential Equations with Boundary Value Problems.pdf
"example, y′′+ y′+ y= x4ex by the method of annihilators and the method used in Section 5 .4.)
Introducing variation of parameters as early as possible (S ection 2.1) prepares the student for the con-
cept when it appears again in more complex forms in Section 5. 6, where reduction of order is used not
merely to find a second solution of the complementary equatio n, but also to find the general solution of the
nonhomogeneous equation, and in Sections 5.7, 9.4, and 10.7 , that treat the usual variation of parameters
problem for second and higher order linear equations and for linear systems.
Chapter 11 develops the theory of Fourier series. Section 11 .1 discusses the five main eigenvalue prob-
lems that arise in connection with the method of separation o f variables for the heat and wave equations
and for Laplace’s equation over a rectangular domain:
Problem 1: y′′+ λy = 0 , y (0) = 0 , y (L) = 0
Problem 2: y′′+ λy = 0 , y ′(0) = 0 , y ′(L) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"and for Laplace’s equation over a rectangular domain:
Problem 1: y′′+ λy = 0 , y (0) = 0 , y (L) = 0
Problem 2: y′′+ λy = 0 , y ′(0) = 0 , y ′(L) = 0
Problem 3: y′′+ λy = 0 , y (0) = 0 , y ′(L) = 0
Problem 4: y′′+ λy = 0 , y ′(0) = 0 , y (L) = 0
Problem 5: y′′+ λy = 0 , y (−L) = y(L), y ′(−L) = y′(L)
These problems are handled in a unified way for example, a sing le theorem shows that the eigenvalues
of all five problems are nonnegative.
Section 11.2 presents the Fourier series expansion of funct ions defined on on [−L,L ], interpreting it
as an expansion in terms of the eigenfunctions of Problem 5.
Section 11.3 presents the Fourier sine and cosine expansion s of functions defined on [0,L ], interpreting
them as expansions in terms of the eigenfunctions of Problem s 1 and 2, respectively . In addition, Sec-
tion 11.2 includes what I call the mixed Fourier sine and cosi ne expansions, in terms of the eigenfunctions",Elementary Differential Equations with Boundary Value Problems.pdf
"tion 11.2 includes what I call the mixed Fourier sine and cosi ne expansions, in terms of the eigenfunctions
of Problems 4 and 5, respectively . In all cases, the converge nce properties of these series are deduced
from the convergence properties of the Fourier series discu ssed in Section 11.1.
Chapter 12 consists of four sections devoted to the heat equa tion, the wave equation, and Laplace’s
equation in rectangular and polar coordinates. For all thre e, I consider homogeneous boundary conditions
of the four types occurring in Problems 1-4. I present the met hod of separation of variables as a way of
choosing the appropriate form for the series expansion of th e solution of the given problem, stating—
without belaboring the point—that the expansion may fall sh ort of being an actual solution, and giving
an indication of conditions under which the formal solution is an actual solution. In particular, I found it",Elementary Differential Equations with Boundary Value Problems.pdf
"an indication of conditions under which the formal solution is an actual solution. In particular, I found it
necessary to devote some detail to this question in connecti on with the wave equation in Section 12.2.
In Sections 12.1 (The Heat Equation) and 12.2 (The W ave Equat ion) I devote considerable effort to
devising examples and numerous exercises where the functio ns defining the initial conditions satisfy",Elementary Differential Equations with Boundary Value Problems.pdf
"Preface ix
the homogeneous boundary conditions. Similarly , in most of the examples and exercises Section 12.3
(Laplace’s Equation), the functions defining the boundary c onditions on a given side of the rectangular
domain satisfy homogeneous boundary conditions at the endp oints of the same type (Dirichlet or Neu-
mann) as the boundary conditions imposed on adjacent sides o f the region. Therefore the formal solutions
obtained in many of the examples and exercises are actual sol utions.
Section 13.1 deals with two-point value problems for a secon d order ordinary differential equation.
Conditions for existence and uniqueness of solutions are gi ven, and the construction of Green’s functions
is included.
Section 13.2 presents the elementary aspects of Sturm-Liou ville theory .
Y ou may also find the following to be of interest:
• Section 2.6 deals with integrating factors of the form µ = p(x)q(y), in addition to those of the
form µ = p(x) and µ = q(y) discussed in most texts.",Elementary Differential Equations with Boundary Value Problems.pdf
"form µ = p(x) and µ = q(y) discussed in most texts.
• Section 4.4 makes phase plane analysis of nonlinear second order autonomous equations accessi-
ble to students who have not taken linear algebra, since eige nvalues and eigenvectors do not enter
into the treatment. Phase plane analysis of constant coeffic ient linear systems is included in Sec-
tions 10.4-6.
• Section 4.5 presents an extensive discussion of applicati ons of differential equations to curves.
• Section 6.4 studies motion under a central force, which may be useful to students interested in the
mathematics of satellite orbits.
• Sections 7.5-7 present the method of Frobenius in more deta il than in most texts. The approach
is to systematize the computations in a way that avoids the ne cessity of substituting the unknown
Frobenius series into each equation. This leads to efficienc y in the computation of the coefficients",Elementary Differential Equations with Boundary Value Problems.pdf
"Frobenius series into each equation. This leads to efficienc y in the computation of the coefficients
of the Frobenius solution. It also clarifies the case where th e roots of the indicial equation differ by
an integer (Section 7.7).
• The free Student Solutions Manual contains solutions of mo st of the even-numbered exercises.
• The free Instructor’s Solutions Manual is available by ema il to wtrench@trinity .edu, subject to
verification of the requestor’s faculty status.
The following observations may be helpful as you choose your syllabus:
• Section 2.3 is the only specific prerequisite for Chapter 3. T o accomodate institutions that offer a
separate course in numerical analysis, Chapter 3 is not a pre requisite for any other section in the
text.
• The sections in Chapter 4 are independent of each other, and are not prerequisites for any of the
later chapters. This is also true of the sections in Chapter 6 , except that Section 6.1 is a prerequisite
for Section 6.2.",Elementary Differential Equations with Boundary Value Problems.pdf
"later chapters. This is also true of the sections in Chapter 6 , except that Section 6.1 is a prerequisite
for Section 6.2.
• Chapters 7, 8, and 9 can be covered in any order after the topi cs selected from Chapter 5. For
example, you can proceed directly from Chapter 5 to Chapter 9 .
• The second order Euler equation is discussed in Section 7.4 , where it sets the stage for the method
of Frobenius. As noted at the beginning of Section 7.4, if you want to include Euler equations in
your syllabus while omitting the method of Frobenius, you ca n skip the introductory paragraphs
in Section 7.4 and begin with Definition 7.4.2. Y ou can then co ver Section 7.4 immediately after
Section 5.2.
• Chapters 11, 12, and 13 can be covered at any time after the co mpletion of Chapter 5.
William F . Trench",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 1
Introduction
IN THIS CHAPTER we begin our study of differential equations .
SECTION 1.1 presents examples of applications that lead to d ifferential equations.
SECTION 1.2 introduces basic concepts and definitions conce rning differential equations.
SECTION 1.3 presents a geometric method for dealing with dif ferential equations that has been known
for a very long time, but has become particularly useful and i mportant with the proliferation of readily
available differential equations software.
1",Elementary Differential Equations with Boundary Value Problems.pdf
"2 Chapter 1 Introduction
1.1 APPLICA TIONS LEADING TO DIFFERENTIAL EQUA TIONS
In order to apply mathematical methods to a physical or “real life” problem, we must formulate the prob-
lem in mathematical terms; that is, we must construct a mathematical model for the problem. Many
physical problems concern relationships between changing quantities. Since rates of change are repre-
sented mathematically by derivatives, mathematical model s often involve equations relating an unknown
function and one or more of its derivatives. Such equations a re differential equations . They are the subject
of this book.
Much of calculus is devoted to learning mathematical techni ques that are applied in later courses in
mathematics and the sciences; you wouldn’t have time to lear n much calculus if you insisted on seeing
a specific application of every topic covered in the course. S imilarly , much of this book is devoted to",Elementary Differential Equations with Boundary Value Problems.pdf
"a specific application of every topic covered in the course. S imilarly , much of this book is devoted to
methods that can be applied in later courses. Only a relative ly small part of the book is devoted to
the derivation of specific differential equations from math ematical models, or relating the differential
equations that we study to specific applications. In this sec tion we mention a few such applications.
The mathematical model for an applied problem is almost alwa ys simpler than the actual situation
being studied, since simplifying assumptions are usually r equired to obtain a mathematical problem that
can be solved. For example, in modeling the motion of a fallin g object, we might neglect air resistance
and the gravitational pull of celestial bodies other than Ea rth, or in modeling population growth we might
assume that the population grows continuously rather than i n discrete steps.
A good mathematical model has two important properties:",Elementary Differential Equations with Boundary Value Problems.pdf
"assume that the population grows continuously rather than i n discrete steps.
A good mathematical model has two important properties:
• It’s sufficiently simple so that the mathematical problem c an be solved.
• It represents the actual situation sufficiently well so tha t the solution to the mathematical problem
predicts the outcome of the real problem to within a useful de gree of accuracy . If results predicted
by the model don’t agree with physical observations, the und erlying assumptions of the model must
be revised until satisfactory agreement is obtained.
W e’ll now give examples of mathematical models involving di fferential equations. W e’ll return to these
problems at the appropriate times, as we learn how to solve th e various types of differential equations that
occur in the models.
All the examples in this section deal with functions of time, which we denote by t. If yis a function of
t, y′denotes the derivative of ywith respect to t; thus,
y′= dy
dt.",Elementary Differential Equations with Boundary Value Problems.pdf
"t, y′denotes the derivative of ywith respect to t; thus,
y′= dy
dt.
Population Growth and Decay
Although the number of members of a population (people in a gi ven country , bacteria in a laboratory cul-
ture, wildflowers in a forest, etc.) at any given time tis necessarily an integer, models that use differential
equations to describe the growth and decay of populations us ually rest on the simplifying assumption that
the number of members of the population can be regarded as a di fferentiable function P = P(t). In most
models it is assumed that the differential equation takes th e form
P′= a(P)P, (1.1.1)
where ais a continuous function of P that represents the rate of change of population per unit tim e per
individual. In the Malthusian model , it is assumed that a(P) is a constant, so ( 1.1.1) becomes
P′= aP. (1.1.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.1 Applications Leading to Differential Equations 3
(When you see a name in blue italics, just click on it for infor mation about the person.) This model
assumes that the numbers of births and deaths per unit time ar e both proportional to the population. The
constants of proportionality are the birth rate (births per unit time per individual) and the death rate
(deaths per unit time per individual); ais the birth rate minus the death rate. Y ou learned in calculu s that
if cis any constant then
P = ceat (1.1.3)
satisfies ( 1.1.2), so ( 1.1.2) has infinitely many solutions. T o select the solution of the specific problem
that we’re considering, we must know the population P0 at an initial time, say t = 0 . Setting t = 0 in
(1.1.3) yields c= P(0) = P0, so the applicable solution is
P(t) = P0eat .
This implies that
lim
t→∞
P(t) =
{ ∞ if a> 0,
0 if a< 0;
that is, the population approaches infinity if the birth rate exceeds the death rate, or zero if the death rate",Elementary Differential Equations with Boundary Value Problems.pdf
"lim
t→∞
P(t) =
{ ∞ if a> 0,
0 if a< 0;
that is, the population approaches infinity if the birth rate exceeds the death rate, or zero if the death rate
exceeds the birth rate.
T o see the limitations of the Malthusian model, suppose we’r e modeling the population of a country ,
starting from a time t = 0 when the birth rate exceeds the death rate (so a > 0), and the country’s
resources in terms of space, food supply , and other necessit ies of life can support the existing popula-
tion. Then the prediction P = P0eat may be reasonably accurate as long as it remains within limit s
that the country’s resources can support. However, the mode l must inevitably lose validity when the pre-
diction exceeds these limits. (If nothing else, eventually there won’t be enough space for the predicted
population!)
This flaw in the Malthusian model suggests the need for a model that accounts for limitations of space",Elementary Differential Equations with Boundary Value Problems.pdf
"population!)
This flaw in the Malthusian model suggests the need for a model that accounts for limitations of space
and resources that tend to oppose the rate of population grow th as the population increases. Perhaps the
most famous model of this kind is the V erhulst model, where ( 1.1.2) is replaced by
P′= aP(1 − αP ), (1.1.4)
where α is a positive constant. As long as P is small compared to 1/α , the ratio P′/P is approximately
equal to a. Therefore the growth is approximately exponential; howev er, as P increases, the ratio P′/P
decreases as opposing factors become significant.
Equation ( 1.1.4) is the logistic equation . Y ou will learn how to solve it in Section 1.2. (See Exer-
cise 2.2. 28.) The solution is
P = P0
αP 0 + (1 − αP 0)e−at ,
where P0 = P(0) >0. Therefore limt→∞ P(t) = 1 /α , independent of P0.
Figure 1.1.1 shows typical graphs of P versus tfor various values of P0.
Newton’s Law of Cooling",Elementary Differential Equations with Boundary Value Problems.pdf
"Figure 1.1.1 shows typical graphs of P versus tfor various values of P0.
Newton’s Law of Cooling
According to Newton’s law of cooling , the temperature of a body changes at a rate proportional to t he
difference between the temperature of the body and the tempe rature of the surrounding medium. Thus, if
Tm is the temperature of the medium and T = T(t) is the temperature of the body at time t, then
T′= −k(T − Tm ), (1.1.5)
where kis a positive constant and the minus sign indicates; that the temperature of the body increases with
time if it’s less than the temperature of the medium, or decre ases if it’s greater. W e’ll see in Section 4.2
that if Tm is constant then the solution of ( 1.1.5) is
T = Tm + ( T0 − Tm )e−kt , (1.1.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"4 Chapter 1 Introduction
 P
 t
 1/α
Figure 1.1.1 Solutions of the logistic equation
where T0 is the temperature of the body when t= 0 . Therefore limt→∞ T(t) = Tm , independent of T0.
(Common sense suggests this. Why?)
Figure 1.1.2 shows typical graphs of T versus tfor various values of T0.
Assuming that the medium remains at constant temperature se ems reasonable if we’re considering a
cup of coffee cooling in a room, but not if we’re cooling a huge cauldron of molten metal in the same
room. The difference between the two situations is that the h eat lost by the coffee isn’t likely to raise the
temperature of the room appreciably , but the heat lost by the cooling metal is. In this second situation we
must use a model that accounts for the heat exchanged between the object and the medium. Let T = T(t)
and Tm = Tm (t) be the temperatures of the object and the medium respectivel y , and let T0 and Tm0",Elementary Differential Equations with Boundary Value Problems.pdf
"and Tm = Tm (t) be the temperatures of the object and the medium respectivel y , and let T0 and Tm0
be their initial values. Again, we assume that T and Tm are related by ( 1.1.5). W e also assume that the
change in heat of the object as its temperature changes from T0 to T is a(T − T0) and the change in heat
of the medium as its temperature changes from Tm0 to Tm is am (Tm −Tm0), where aand am are positive
constants depending upon the masses and thermal properties of the object and medium respectively . If
we assume that the total heat of the in the object and the mediu m remains constant (that is, energy is
conserved), then
a(T − T0) + am(Tm − Tm0) = 0 .
Solving this for Tm and substituting the result into ( 1.1.6) yields the differential equation
T′= −k
(
1 + a
am
)
T + k
(
Tm0 + a
am
T0
)
for the temperature of the object. After learning to solve li near first order equations, you’ll be able to
show (Exercise 4.2. 17) that
T = aT0 + amTm0
a+ am
+ am (T0 − Tm0 )
a+ am",Elementary Differential Equations with Boundary Value Problems.pdf
"show (Exercise 4.2. 17) that
T = aT0 + amTm0
a+ am
+ am (T0 − Tm0 )
a+ am
e−k(1+a/a m)t .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.1 Applications Leading to Differential Equations 5
 T
 t t
 Tm
Figure 1.1.2 T emperature according to Newton’s Law of Cooli ng
Glucose Absorption by the Body
Glucose is absorbed by the body at a rate proportional to the a mount of glucose present in the bloodstream.
Let λ denote the (positive) constant of proportionality . Suppos e there are G0 units of glucose in the
bloodstream when t = 0 , and let G = G(t) be the number of units in the bloodstream at time t >0.
Then, since the glucose being absorbed by the body is leaving the bloodstream, Gsatisfies the equation
G′= −λG. (1.1.7)
From calculus you know that if cis any constant then
G= ce−λt (1.1.8)
satisfies ( 1.1.7), so ( 1.1.7) has infinitely many solutions. Setting t = 0 in ( 1.1.8) and requiring that
G(0) = G0 yields c= G0, so
G(t) = G0e−λt .
Now let’s complicate matters by injecting glucose intraven ously at a constant rate of runits of glucose",Elementary Differential Equations with Boundary Value Problems.pdf
"G(0) = G0 yields c= G0, so
G(t) = G0e−λt .
Now let’s complicate matters by injecting glucose intraven ously at a constant rate of runits of glucose
per unit of time. Then the rate of change of the amount of gluco se in the bloodstream per unit time is
G′= −λG + r, (1.1.9)
where the first term on the right is due to the absorption of the glucose by the body and the second term
is due to the injection. After you’ve studied Section 2.1, yo u’ll be able to show (Exercise 2.1. 43) that the
solution of ( 1.1.9) that satisfies G(0) = G0 is
G= r
λ +
(
G0 − r
λ
)
e−λt .",Elementary Differential Equations with Boundary Value Problems.pdf
"6 Chapter 1 Introduction
Graphs of this function are similar to those in Figure 1.1.2. (Why?)
Spread of Epidemics
One model for the spread of epidemics assumes that the number of people infected changes at a rate
proportional to the product of the number of people already i nfected and the number of people who are
susceptible, but not yet infected. Therefore, if S denotes the total population of susceptible people and
I = I(t) denotes the number of infected people at time t, then S − I is the number of people who are
susceptible, but not yet infected. Thus,
I′= rI(S− I),
where ris a positive constant. Assuming that I(0) = I0, the solution of this equation is
I = SI0
I0 + ( S− I0)e−rSt
(Exercise 2.2.29). Graphs of this function are similar to those in Figure 1.1.1. (Why?) Since limt→∞ I(t) =
S, this model predicts that all the susceptible people eventu ally become infected.
Newton’s Second Law of Motion",Elementary Differential Equations with Boundary Value Problems.pdf
"S, this model predicts that all the susceptible people eventu ally become infected.
Newton’s Second Law of Motion
According to Newton’s second law of motion , the instantaneous acceleration aof an object with con-
stant mass mis related to the force F acting on the object by the equation F = ma. For simplicity , let’s
assume that m= 1 and the motion of the object is along a vertical line. Let ybe the displacement of the
object from some reference point on Earth’s surface, measur ed positive upward. In many applications,
there are three kinds of forces that may act on the object:
(a) A force such as gravity that depends only on the position y, which we write as −p(y), where
p(y) >0 if y≥ 0.
(b) A force such as atmospheric resistance that depends on the po sition and velocity of the object, which
we write as −q(y,y ′)y′, where q is a nonnegative function and we’ve put y′“outside” to indicate
that the resistive force is always in the direction opposite to the velocity .",Elementary Differential Equations with Boundary Value Problems.pdf
"that the resistive force is always in the direction opposite to the velocity .
(c) A force f = f(t), exerted from an external source (such as a towline from a hel icopter) that depends
only on t.
In this case, Newton’s second law implies that
y′′= −q(y,y ′)y′− p(y) + f(t),
which is usually rewritten as
y′′+ q(y,y ′)y′+ p(y) = f(t).
Since the second (and no higher) order derivative of y occurs in this equation, we say that it is a second
order differential equation .
Interacting Species: Competition
Let P = P(t) and Q= Q(t) be the populations of two species at time t, and assume that each population
would grow exponentially if the other didn’t exist; that is, in the absence of competition we would have
P′= aP and Q′= bQ, (1.1.10)
where a and b are positive constants. One way to model the effect of compet ition is to assume that
the growth rate per individual of each population is reduced by an amount proportional to the other
population, so ( 1.1.10) is replaced by
P′ = aP − αQ",Elementary Differential Equations with Boundary Value Problems.pdf
"population, so ( 1.1.10) is replaced by
P′ = aP − αQ
Q′ = −βP + bQ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.2 Basic Concepts 7
where α and β are positive constants. (Since negative population doesn’ t make sense, this system works
only while P and Q are both positive.) Now suppose P(0) = P0 > 0 and Q(0) = Q0 > 0. It can
be shown (Exercise 10.4. 42) that there’s a positive constant ρ such that if (P0,Q 0) is above the line L
through the origin with slope ρ, then the species with population P becomes extinct in finite time, but if
(P0,Q 0) is below L, the species with population Qbecomes extinct in finite time. Figure 1.1.3 illustrates
this. The curves shown there are given parametrically by P = P(t),Q = Q(t), t > 0. The arrows
indicate direction along the curves with increasing t.
 P
 Q
 L
Figure 1.1.3 Populations of competing species
1.2 BASIC CONCEPTS
A differential equation is an equation that contains one or more derivatives of an unk nown function.
The order of a differential equation is the order of the highest deriva tive that it contains. A differential",Elementary Differential Equations with Boundary Value Problems.pdf
"The order of a differential equation is the order of the highest deriva tive that it contains. A differential
equation is an ordinary differential equation if it involves an unknown function of only one variable, or a
partial differential equation if it involves partial derivatives of a function of more than one variable. For
now we’ll consider only ordinary differential equations, a nd we’ll just call them differential equations.
Throughout this text, all variables and constants are real u nless it’s stated otherwise. W e’ll usually use
xfor the independent variable unless the independent variab le is time; then we’ll use t.
The simplest differential equations are first order equatio ns of the form
dy
dx = f(x) or, equivalently , y′= f(x),
where f is a known function of x. W e already know from calculus how to find functions that sati sfy this
kind of equation. For example, if
y′= x3,",Elementary Differential Equations with Boundary Value Problems.pdf
"8 Chapter 1 Introduction
then
y =
∫
x3 dx= x4
4 + c,
where cis an arbitrary constant. If n> 1 we can find functions ythat satisfy equations of the form
y(n) = f(x) (1.2.1)
by repeated integration. Again, this is a calculus problem.
Except for illustrative purposes in this section, there’s n o need to consider differential equations like
(1.2.1).W e’ll usually consider differential equations that can b e written as
y(n) = f(x,y,y ′,...,y (n−1)), (1.2.2)
where at least one of the functions y, y′, . . . , y(n−1) actually appears on the right. Here are some exam-
ples:
dy
dx − x2 = 0 (first order) ,
dy
dx + 2 xy2 = −2 (first order) ,
d2y
dx2 + 2 dy
dx + y = 2 x (second order) ,
xy′′′+ y2 = sin x (third order) ,
y(n) + xy′+ 3 y = x (n-th order) .
Although none of these equations is written as in ( 1.2.2), all of them can be written in this form:
y′ = x2,
y′ = −2 − 2xy2,
y′′ = 2 x− 2y′− y,
y′′′= sin x− y2
x ,
y(n) = x− xy′− 3y.
Solutions of Differential Equations",Elementary Differential Equations with Boundary Value Problems.pdf
"y′ = x2,
y′ = −2 − 2xy2,
y′′ = 2 x− 2y′− y,
y′′′= sin x− y2
x ,
y(n) = x− xy′− 3y.
Solutions of Differential Equations
A solution of a differential equation is a function that satisfies the di fferential equation on some open
interval; thus, yis a solution of ( 1.2.2) if yis ntimes differentiable and
y(n) (x) = f(x,y (x),y ′(x),...,y (n−1)(x))
for all xin some open interval (a,b ). In this case, we also say that y is a solution of (1.2.2) on (a,b ).
Functions that satisfy a differential equation at isolated points are not interesting. For example, y = x2
satisfies
xy′+ x2 = 3 x
if and only if x= 0 or x= 1 , but it’s not a solution of this differential equation becau se it does not satisfy
the equation on an open interval.
The graph of a solution of a differential equation is a solution curve . More generally , a curve Cis said
to be an integral curve of a differential equation if every function y = y(x) whose graph is a segment",Elementary Differential Equations with Boundary Value Problems.pdf
"to be an integral curve of a differential equation if every function y = y(x) whose graph is a segment
of C is a solution of the differential equation. Thus, any soluti on curve of a differential equation is an
integral curve, but an integral curve need not be a solution c urve.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.2 Basic Concepts 9
Example 1.2.1 If ais any positive constant, the circle
x2 + y2 = a2 (1.2.3)
is an integral curve of
y′= −x
y. (1.2.4)
T o see this, note that the only functions whose graphs are seg ments of ( 1.2.3) are
y1 =
√
a2 − x2 and y2 = −
√
a2 − x2.
W e leave it to you to verify that these functions both satisfy (1.2.4) on the open interval (−a,a ). However,
(1.2.3) is not a solution curve of ( 1.2.4), since it’s not the graph of a function.
Example 1.2.2 V erify that
y = x2
3 + 1
x (1.2.5)
is a solution of
xy′+ y= x2 (1.2.6)
on (0, ∞ ) and on (−∞ , 0).
Solution Substituting ( 1.2.5) and
y′= 2x
3 − 1
x2
into ( 1.2.6) yields
xy′(x) + y(x) = x
(2x
3 − 1
x2
)
+
(x2
3 + 1
x
)
= x2
for all x̸= 0 . Therefore yis a solution of ( 1.2.6) on (−∞ , 0) and (0, ∞ ). However, yisn’t a solution of
the differential equation on any open interval that contain s x= 0 , since yis not defined at x= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"the differential equation on any open interval that contain s x= 0 , since yis not defined at x= 0 .
Figure 1.2.1 shows the graph of ( 1.2.5). The part of the graph of ( 1.2.5) on (0, ∞ ) is a solution curve
of ( 1.2.6), as is the part of the graph on (−∞ , 0).
Example 1.2.3 Show that if c1 and c2 are constants then
y = ( c1 + c2x)e−x + 2 x− 4 (1.2.7)
is a solution of
y′′+ 2 y′+ y= 2 x (1.2.8)
on (−∞ , ∞ ).
Solution Differentiating ( 1.2.7) twice yields
y′= −(c1 + c2x)e−x + c2e−x + 2
and
y′′= ( c1 + c2x)e−x − 2c2e−x,",Elementary Differential Equations with Boundary Value Problems.pdf
"10 Chapter 1 Introduction
 x
 y
0.5 1.0 1.5 2.0−0.5−1.0−1.5−2.0
2
4
6
8
−2
−4
−6
−8
Figure 1.2.1 y= x2
3 + 1
x
so
y′′+ 2 y′+ y = ( c1 + c2x)e−x − 2c2e−x
+2 [ −(c1 + c2x)e−x + c2e−x + 2 ]
+(c1 + c2x)e−x + 2 x− 4
= (1 − 2 + 1)( c1 + c2x)e−x + ( −2 + 2) c2e−x
+4 + 2 x− 4 = 2 x
for all values of x. Therefore yis a solution of ( 1.2.8) on (−∞ , ∞ ).
Example 1.2.4 Find all solutions of
y(n) = e2x. (1.2.9)
Solution Integrating ( 1.2.9) yields
y(n−1) = e2x
2 + k1,
where k1 is a constant. If n≥ 2, integrating again yields
y(n−2) = e2x
4 + k1x+ k2.
If n≥ 3, repeatedly integrating yields
y= e2x
2n + k1
xn−1
(n− 1)! + k2
xn−2
(n− 2)! + · · ·+ kn, (1.2.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.2 Basic Concepts 11
where k1, k2, . . . , kn are constants. This shows that every solution of ( 1.2.9) has the form ( 1.2.10) for
some choice of the constants k1, k2, . . . , kn. On the other hand, differentiating ( 1.2.10) ntimes shows
that if k1, k2, . . . , kn are arbitrary constants, then the function yin ( 1.2.10) satisfies ( 1.2.9).
Since the constants k1, k2, . . . , kn in ( 1.2.10) are arbitrary , so are the constants
k1
(n− 1)! , k2
(n− 2)! , · · ·, kn.
Therefore Example 1.2.4 actually shows that all solutions of ( 1.2.9) can be written as
y= e2x
2n + c1 + c2x+ · · ·+ cn xn−1,
where we renamed the arbitrary constants in ( 1.2.10) to obtain a simpler formula. As a general rule,
arbitrary constants appearing in solutions of differentia l equations should be simplified if possible. Y ou’ll
see examples of this throughout the text.
Initial V alue Problems
In Example 1.2.4 we saw that the differential equation y(n) = e2x has an infinite family of solutions that",Elementary Differential Equations with Boundary Value Problems.pdf
"Initial V alue Problems
In Example 1.2.4 we saw that the differential equation y(n) = e2x has an infinite family of solutions that
depend upon the narbitrary constants c1, c2, . . . , cn . In the absence of additional conditions, there’s no
reason to prefer one solution of a differential equation ove r another. However, we’ll often be interested
in finding a solution of a differential equation that satisfie s one or more specific conditions. The next
example illustrates this.
Example 1.2.5 Find a solution of
y′= x3
such that y(1) = 2 .
Solution At the beginning of this section we saw that the solutions of y′= x3 are
y= x4
4 + c.
T o determine a value of csuch that y(1) = 2 , we set x= 1 and y= 2 here to obtain
2 = y(1) = 1
4 + c, so c= 7
4 .
Therefore the required solution is
y = x4 + 7
4 .
Figure 1.2.2 shows the graph of this solution. Note that imposing the cond ition y(1) = 2 is equivalent
to requiring the graph of yto pass through the point (1, 2).",Elementary Differential Equations with Boundary Value Problems.pdf
"to requiring the graph of yto pass through the point (1, 2).
W e can rewrite the problem considered in Example 1.2.5 more briefly as
y′= x3, y (1) = 2 .
W e call this an initial value problem . The requirement y(1) = 2 is an initial condition . Initial value
problems can also be posed for higher order differential equ ations. For example,
y′′− 2y′+ 3 y= ex, y (0) = 1 , y ′(0) = 2 (1.2.11)",Elementary Differential Equations with Boundary Value Problems.pdf
"12 Chapter 1 Introduction
is an initial value problem for a second order differential e quation where y and y′are required to have
specified values at x = 0 . In general, an initial value problem for an n-th order differential equation
requires yand its first n− 1 derivatives to have specified values at some point x0. These requirements are
the initial conditions .
1
2
3
4
5
0 1 2−1−2
(1,2)
 x
 y
Figure 1.2.2 y= x2 + 7
4
W e’ll denote an initial value problem for a differential equ ation by writing the initial conditions after
the equation, as in ( 1.2.11). For example, we would write an initial value problem for ( 1.2.2) as
y(n) = f(x,y,y ′,...,y (n−1)), y(x0) = k0, y′(x0) = k1, ..., y (n−1) = kn−1. (1.2.12)
Consistent with our earlier definition of a solution of the di fferential equation in ( 1.2.12), we say that yis
a solution of the initial value problem ( 1.2.12) if yis ntimes differentiable and
y(n) (x) = f(x,y (x),y ′(x),...,y (n−1)(x))",Elementary Differential Equations with Boundary Value Problems.pdf
"a solution of the initial value problem ( 1.2.12) if yis ntimes differentiable and
y(n) (x) = f(x,y (x),y ′(x),...,y (n−1)(x))
for all xin some open interval (a,b ) that contains x0, and ysatisfies the initial conditions in ( 1.2.12). The
largest open interval that contains x0 on which y is defined and satisfies the differential equation is the
interval of validity of y.
Example 1.2.6 In Example 1.2.5 we saw that
y= x4 + 7
4 (1.2.13)
is a solution of the initial value problem
y′= x3, y (1) = 2 .
Since the function in ( 1.2.13) is defined for all x, the interval of validity of this solution is (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.2 Basic Concepts 13
Example 1.2.7 In Example 1.2.2 we verified that
y = x2
3 + 1
x (1.2.14)
is a solution of
xy′+ y= x2
on (0, ∞ ) and on (−∞ , 0). By evaluating ( 1.2.14) at x= ±1, you can see that ( 1.2.14) is a solution of
the initial value problems
xy′+ y = x2, y (1) = 4
3 (1.2.15)
and
xy′+ y = x2, y (−1) = −2
3 . (1.2.16)
The interval of validity of ( 1.2.14) as a solution of ( 1.2.15) is (0, ∞ ), since this is the largest interval that
contains x0 = 1 on which ( 1.2.14) is defined. Similarly , the interval of validity of ( 1.2.14) as a solution of
(1.2.16) is (−∞ , 0), since this is the largest interval that contains x0 = −1 on which ( 1.2.14) is defined.
Free Fall Under Constant Gravity
The term initial value problem originated in problems of motion where the independent vari able is t
(representing elapsed time), and the initial conditions ar e the position and velocity of an object at the
initial (starting) time of an experiment.",Elementary Differential Equations with Boundary Value Problems.pdf
"(representing elapsed time), and the initial conditions ar e the position and velocity of an object at the
initial (starting) time of an experiment.
Example 1.2.8 An object falls under the influence of gravity near Earth’s su rface, where it can be as-
sumed that the magnitude of the acceleration due to gravity i s a constant g.
(a) Construct a mathematical model for the motion of the object i n the form of an initial value problem
for a second order differential equation, assuming that the altitude and velocity of the object at time
t= 0 are known. Assume that gravity is the only force acting on the object.
(b) Solve the initial value problem derived in (a) to obtain the altitude as a function of time.
SO LU TIO N (a) Let y(t) be the altitude of the object at time t. Since the acceleration of the object has
constant magnitude gand is in the downward (negative) direction, ysatisfies the second order equation
y′′= −g,",Elementary Differential Equations with Boundary Value Problems.pdf
"constant magnitude gand is in the downward (negative) direction, ysatisfies the second order equation
y′′= −g,
where the prime now indicates differentiation with respect to t. If y0 and v0 denote the altitude and
velocity when t= 0 , then yis a solution of the initial value problem
y′′= −g, y (0) = y0, y ′(0) = v0. (1.2.17)
SO LU TIO N (b) Integrating ( 1.2.17) twice yields
y′ = −gt+ c1,
y = −gt2
2 + c1t+ c2.
Imposing the initial conditions y(0) = y0 and y′(0) = v0 in these two equations shows that c1 = v0 and
c2 = y0. Therefore the solution of the initial value problem ( 1.2.17) is
y = −gt2
2 + v0t+ y0.",Elementary Differential Equations with Boundary Value Problems.pdf
"14 Chapter 1 Introduction
1.2 Exercises
1. Find the order of the equation.
(a) d2y
dx2 + 2 dy
dx
d3y
dx3 + x= 0 (b) y′′− 3y′+ 2 y= x7
(c) y′− y7 = 0 (d) y′′y− (y′)2 = 2
2. V erify that the function is a solution of the differential eq uation on some interval, for any choice
of the arbitrary constants appearing in the function.
(a) y= ce2x; y′= 2 y
(b) y= x2
3 + c
x; xy′+ y= x2
(c) y= 1
2 + ce−x2
; y′+ 2 xy= x
(d) y= (1 + ce−x2/ 2); (1 − ce−x2/ 2)−1 2y′+ x(y2 − 1) = 0
(e) y= tan
(x3
3 + c
)
; y′= x2(1 + y2)
(f) y= ( c1 + c2x)ex + sin x+ x2; y′′− 2y′+ y = −2 cos x+ x2 − 4x+ 2
(g) y= c1ex + c2x+ 2
x; (1 − x)y′′+ xy′− y= 4(1 − x− x2)x−3
(h) y= x−1/ 2(c1 sin x+ c2 cos x) + 4 x+ 8 ;
x2y′′+ xy′+
(
x2 − 1
4
)
y= 4 x3 + 8 x2 + 3 x− 2
3. Find all solutions of the equation.
(a) y′= −x (b) y′= −xsin x
(c) y′= xln x (d) y′′= xcos x
(e) y′′= 2 xex (f) y′′= 2 x+ sin x+ ex
(g) y′′′= − cos x (h) y′′′= −x2 + ex
(i) y′′′= 7 e4x
4. Solve the initial value problem.
(a) y′= −xex, y (0) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"(g) y′′′= − cos x (h) y′′′= −x2 + ex
(i) y′′′= 7 e4x
4. Solve the initial value problem.
(a) y′= −xex, y (0) = 1
(b) y′= xsin x2, y
(√
π
2
)
= 1
(c) y′= tan x, y (π/ 4) = 3
(d) y′′= x4, y (2) = −1, y ′(2) = −1
(e) y′′= xe2x, y (0) = 7 , y ′(0) = 1
(f) y′′= −xsin x, y (0) = 1 , y ′(0) = −3
(g) y′′′= x2ex, y (0) = 1 , y ′(0) = −2, y ′′(0) = 3
(h) y′′′= 2 + sin 2 x, y (0) = 1 , y ′(0) = −6, y ′′(0) = 3
(i) y′′′= 2 x+ 1 , y (2) = 1 , y ′(2) = −4, y ′′(2) = 7
5. V erify that the function is a solution of the initial value pr oblem.
(a) y= xcos x; y′= cos x− ytan x, y (π/ 4) = π
4
√
2
(b) y= 1 + 2 ln x
x2 + 1
2 ; y′= x2 − 2x2y+ 2
x3 , y (1) = 3
2",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.2 Basic Concepts 15
(c) y= tan
(x2
2
)
; y′= x(1 + y2), y (0) = 0
(d) y= 2
x− 2 ; y′= −y(y+ 1)
x , y (1) = −2
6. V erify that the function is a solution of the initial value pr oblem.
(a) y= x2(1 + ln x); y′′= 3xy′− 4y
x2 , y (e) = 2 e2, y ′(e) = 5 e
(b) y= x2
3 + x− 1; y′′= x2 − xy′+ y+ 1
x2 , y (1) = 1
3 , y ′(1) = 5
3
(c) y= (1 + x2)−1/ 2; y′′= (x2 − 1)y− x(x2 + 1) y′
(x2 + 1) 2 , y (0) = 1 ,
y′(0) = 0
(d) y= x2
1 − x; y′′= 2(x+ y)(xy′− y)
x3 , y (1/ 2) = 1 / 2, y ′(1/ 2) = 3
7. Suppose an object is launched from a point 320 feet above the e arth with an initial velocity of 128
ft/sec upward, and the only force acting on it thereafter is g ravity . T ake g= 32 ft/sec2.
(a) Find the highest altitude attained by the object.
(b) Determine how long it takes for the object to fall to the groun d.
8. Let abe a nonzero real number.
(a) V erify that if cis an arbitrary constant then
y = ( x− c)a (A)
is a solution of
y′= ay(a−1)/a (B)
on (c, ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"8. Let abe a nonzero real number.
(a) V erify that if cis an arbitrary constant then
y = ( x− c)a (A)
is a solution of
y′= ay(a−1)/a (B)
on (c, ∞ ).
(b) Suppose a< 0 or a> 1. Can you think of a solution of (B) that isn’t of the form (A)?
9. V erify that
y=
{ ex − 1, x ≥ 0,
1 − e−x, x< 0,
is a solution of
y′= |y|+ 1
on (−∞ , ∞ ). H IN T : Use the definition of derivative at x= 0 .
10. (a) V erify that if cis any real number then
y= c2 + cx+ 2 c+ 1 (A)
satisfies
y′= −(x+ 2) +
√
x2 + 4 x+ 4 y
2 (B)
on some open interval. Identify the open interval.
(b) V erify that
y1 = −x(x+ 4)
4
also satisfies (B) on some open interval, and identify the ope n interval. (Note that y1 can’t be
obtained by selecting a value of cin (A).)",Elementary Differential Equations with Boundary Value Problems.pdf
"16 Chapter 1 Introduction
1.3 DIRECTION FIELDS FOR FIRST ORDER EQUA TIONS
It’s impossible to find explicit formulas for solutions of so me differential equations. Even if there are
such formulas, they may be so complicated that they’re usele ss. In this case we may resort to graphical
or numerical methods to get some idea of how the solutions of t he given equation behave.
In Section 2.3 we’ll take up the question of existence of solu tions of a first order equation
y′= f(x,y ). (1.3.1)
In this section we’ll simply assume that ( 1.3.1) has solutions and discuss a graphical method for ap-
proximating them. In Chapter 3 we discuss numerical methods for obtaining approximate solutions of
(1.3.1).
Recall that a solution of ( 1.3.1) is a function y= y(x) such that
y′(x) = f(x,y (x))
for all values of xin some interval, and an integral curve is either the graph of a solution or is made up",Elementary Differential Equations with Boundary Value Problems.pdf
"y′(x) = f(x,y (x))
for all values of xin some interval, and an integral curve is either the graph of a solution or is made up
of segments that are graphs of solutions. Therefore, not bei ng able to solve ( 1.3.1) is equivalent to not
knowing the equations of integral curves of ( 1.3.1). However, it’s easy to calculate the slopes of these
curves. T o be specific, the slope of an integral curve of ( 1.3.1) through a given point (x0,y 0) is given by
the number f(x0,y 0). This is the basis of the method of direction fields .
If f is defined on a set R, we can construct a direction field for ( 1.3.1) in Rby drawing a short line
segment through each point (x,y ) in Rwith slope f(x,y ). Of course, as a practical matter, we can’t
actually draw line segments through every point in R; rather, we must select a finite set of points in R.
For example, suppose f is defined on the closed rectangular region
R: {a≤ x≤ b,c ≤ y ≤ d}.
Let
a= x0 <x1 <· · ·<xm = b
be equally spaced points in [a,b ] and",Elementary Differential Equations with Boundary Value Problems.pdf
"R: {a≤ x≤ b,c ≤ y ≤ d}.
Let
a= x0 <x1 <· · ·<xm = b
be equally spaced points in [a,b ] and
c= y0 <y1 <· · ·<yn = d
be equally spaced points in [c,d ]. W e say that the points
(xi,y j ), 0 ≤ i≤ m, 0 ≤ j ≤ n,
form a rectangular grid (Figure 1.3.1). Through each point in the grid we draw a short line segment w ith
slope f(xi ,y j ). The result is an approximation to a direction field for ( 1.3.1) in R. If the grid points are
sufficiently numerous and close together, we can draw approx imate integral curves of ( 1.3.1) by drawing
curves through points in the grid tangent to the line segment s associated with the points in the grid.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.3 Direction Fields for First Order Equations 17
 y
 x
 a  b
 c
 d
Figure 1.3.1 A rectangular grid
Unfortunately , approximating a direction field and graphin g integral curves in this way is too tedious
to be done effectively by hand. However, there is software fo r doing this. As you’ll see, the combina-
tion of direction fields and integral curves gives useful ins ights into the behavior of the solutions of the
differential equation even if we can’t obtain exact solutio ns.
W e’ll study numerical methods for solving a single first orde r equation ( 1.3.1) in Chapter 3. These
methods can be used to plot solution curves of ( 1.3.1) in a rectangular region Rif f is continuous on R.
Figures 1.3.2, 1.3.3, and 1.3.4 show direction fields and solution curves for the differenti al equations
y′= x2 − y2
1 + x2 + y2 , y ′= 1 + xy2 , and y′= x− y
1 + x2 ,
which are all of the form ( 1.3.1) with f continuous for all (x,y ).
−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
4
 y
 x",Elementary Differential Equations with Boundary Value Problems.pdf
"1 + x2 ,
which are all of the form ( 1.3.1) with f continuous for all (x,y ).
−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
4
 y
 x
Figure 1.3.2 A direction field and integral curves
for y= x2 − y2
1 + x2 + y2
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 y
 x
Figure 1.3.3 A direction field and integral curves for
y′= 1 + xy2",Elementary Differential Equations with Boundary Value Problems.pdf
"18 Chapter 1 Introduction
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 y
 x
Figure 1.3.4 A direction and integral curves for y′= x− y
1 + x2
The methods of Chapter 3 won’t work for the equation
y′= −x/y (1.3.2)
if Rcontains part of the x-axis, since f(x,y ) = −x/y is undefined when y = 0 . Similarly , they won’t
work for the equation
y′= x2
1 − x2 − y2 (1.3.3)
if R contains any part of the unit circle x2 + y2 = 1 , because the right side of ( 1.3.3) is undefined if
x2 + y2 = 1 . However, ( 1.3.2) and ( 1.3.3) can written as
y′= A(x,y )
B(x,y ) (1.3.4)
where Aand Bare continuous on any rectangle R. Because of this, some differential equation software
is based on numerically solving pairs of equations of the for m
dx
dt = B(x,y ), dy
dt = A(x,y ) (1.3.5)
where xand yare regarded as functions of a parameter t. If x= x(t) and y= y(t) satisfy these equations,
then
y′= dy
dx = dy
dt
/ dx
dt = A(x,y )
B(x,y ) ,
so y= y(x) satisfies ( 1.3.4).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.3 Direction Fields for First Order Equations 19
Eqns. ( 1.3.2) and ( 1.3.3) can be reformulated as in ( 1.3.4) with
dx
dt = −y, dy
dt = x
and dx
dt = 1 − x2 − y2, dy
dt = x2,
respectively . Even if f is continuous and otherwise “nice” throughout R, your software may require you
to reformulate the equation y′= f(x,y ) as
dx
dt = 1 , dy
dt = f(x,y ),
which is of the form ( 1.3.5) with A(x,y ) = f(x,y ) and B(x,y ) = 1 .
Figure 1.3.5 shows a direction field and some integral curves for ( 1.3.2). As we saw in Example 1.2.1
and will verify again in Section 2.2, the integral curves of ( 1.3.2) are circles centered at the origin.
 x
 y
Figure 1.3.5 A direction field and integral curves for y′= −x
y
Figure 1.3.6 shows a direction field and some integral curves for ( 1.3.3). The integral curves near the
top and bottom are solution curves. However, the integral cu rves near the middle are more complicated.",Elementary Differential Equations with Boundary Value Problems.pdf
"top and bottom are solution curves. However, the integral cu rves near the middle are more complicated.
For example, Figure 1.3.7 shows the integral curve through the origin. The vertices of the dashed rectangle
are on the circle x2 + y2 = 1 (a ≈ .846, b ≈ .533), where all integral curves of ( 1.3.3) have infinite
slope. There are three solution curves of ( 1.3.3) on the integral curve in the figure: the segment above the
level y = bis the graph of a solution on (−∞ ,a ), the segment below the level y = −bis the graph of a
solution on (−a, ∞ ), and the segment between these two levels is the graph of a sol ution on (−a,a ).
USING TECHNOLOGY",Elementary Differential Equations with Boundary Value Problems.pdf
"20 Chapter 1 Introduction
As you study from this book, you’ll often be asked to use compu ter software and graphics. Exercises
with this intent are marked as C (computer or calculator required), C/G (computer and/or graphics
required), or L (laboratory work requiring software and/or graphics). Oft en you may not completely
understand how the software does what it does. This is simila r to the situation most people are in when
they drive automobiles or watch television, and it doesn’t d ecrease the value of using modern technology
as an aid to learning. Just be careful that you use the technol ogy as a supplement to thought rather than a
substitute for it.
 y
 x
Figure 1.3.6 A direction field and integral curves for
y′= x2
1 − x2 − y2
 x
 y
 (a,−b)
 (a,b)(−a,b)
 (−a,−b)
1 2−1−2
1
2
−1
−2
Figure 1.3.7
1.3 Exercises
In Exercises 1–11 a direction field is drawn for the given equation. Sketch some integral curves.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.3 Direction Fields for First Order Equations 21
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
1 A direction field for y′= x
y",Elementary Differential Equations with Boundary Value Problems.pdf
"22 Chapter 1 Introduction
0 0.5 1 1.5 2 2.5 3 3.5 4
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 x
 y
2 A direction field for y′= 2xy2
1 + x2
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
3 A direction field for y′= x2(1 + y2)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.3 Direction Fields for First Order Equations 23
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
4 A direction field for y′= 1
1 + x2 + y2
0 0.5 1 1.5 2 2.5 3
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
5 A direction field for y′= −(2xy2 + y3)",Elementary Differential Equations with Boundary Value Problems.pdf
"24 Chapter 1 Introduction
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
6 A direction field for y′= ( x2 + y2 )1/ 2
0 1 2 3 4 5 6 7
−3
−2
−1
0
1
2
3
 x
 y
7 A direction field for y′= sin xy",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.3 Direction Fields for First Order Equations 25
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
 x
 y
8 A direction field for y′= exy
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
9 A direction field for y′= ( x− y2 )(x2 − y)",Elementary Differential Equations with Boundary Value Problems.pdf
"26 Chapter 1 Introduction
1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 x
 y
10 A direction field for y′= x3y2 + xy3
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.5
1
1.5
2
2.5
3
3.5
4
 x
 y
11 A direction field for y′= sin( x− 2y)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 1.3 Direction Fields for First Order Equations 27
In Exercises 12-22 construct a direction field and plot some integral curves in t he indicated rectangular
region.
12. C/G y′= y(y− 1); {−1 ≤ x≤ 2, −2 ≤ y≤ 2}
13. C/G y′= 2 − 3xy; {−1 ≤ x≤ 4, −4 ≤ y≤ 4}
14. C/G y′= xy(y− 1); {−2 ≤ x≤ 2, −4 ≤ y ≤ 4}
15. C/G y′= 3 x+ y; {−2 ≤ x≤ 2, 0 ≤ y≤ 4}
16. C/G y′= y− x3; {−2 ≤ x≤ 2, −2 ≤ y ≤ 2}
17. C/G y′= 1 − x2 − y2; {−2 ≤ x≤ 2, −2 ≤ y ≤ 2}
18. C/G y′= x(y2 − 1); {−3 ≤ x≤ 3, −3 ≤ y ≤ 2}
19. C/G y′= x
y(y2 − 1) ; {−2 ≤ x≤ 2, −2 ≤ y ≤ 2}
20. C/G y′= xy2
y− 1 ; {−2 ≤ x≤ 2, −1 ≤ y≤ 4}
21. C/G y′= x(y2 − 1)
y ; {−1 ≤ x≤ 1, −2 ≤ y ≤ 2}
22. C/G y′= − x2 + y2
1 − x2 − y2 ; {−2 ≤ x≤ 2, −2 ≤ y≤ 2}
23. L By suitably renaming the constants and dependent variables in the equations
T′= −k(T − Tm ) (A)
and
G′= −λG + r (B)
discussed in Section 1.2 in connection with Newton’s law of c ooling and absorption of glucose in
the body , we can write both as
y′= −ay+ b, (C)",Elementary Differential Equations with Boundary Value Problems.pdf
"discussed in Section 1.2 in connection with Newton’s law of c ooling and absorption of glucose in
the body , we can write both as
y′= −ay+ b, (C)
where a is a positive constant and bis an arbitrary constant. Thus, (A) is of the form (C) with
y = T, a = k, and b = kTm , and (B) is of the form (C) with y = G, a = λ, and b = r. W e’ll
encounter equations of the form (C) in many other applicatio ns in Chapter 2.
Choose a positive aand an arbitrary b. Construct a direction field and plot some integral curves
for (C) in a rectangular region of the form
{0 ≤ t≤ T, c ≤ y ≤ d}
of the ty-plane. V ary T, c, and duntil you discover a common property of all the solutions of ( C).
Repeat this experiment with various choices of aand buntil you can state this property precisely
in terms of aand b.
24. L By suitably renaming the constants and dependent variables in the equations
P′= aP(1 − αP ) (A)
and
I′= rI(S− I) (B)",Elementary Differential Equations with Boundary Value Problems.pdf
"28 Chapter 1 Introduction
discussed in Section 1.1 in connection with V erhulst’s popu lation model and the spread of an
epidemic, we can write both in the form
y′= ay− by2 , (C)
where aand bare positive constants. Thus, (A) is of the form (C) with y = P, a= a, and b= aα,
and (B) is of the form (C) with y= I, a= rS, and b= r. In Chapter 2 we’ll encounter equations
of the form (C) in other applications..
(a) Choose positive numbers aand b. Construct a direction field and plot some integral curves
for (C) in a rectangular region of the form
{0 ≤ t≤ T, 0 ≤ y ≤ d}
of the ty-plane. V ary T and duntil you discover a common property of all solutions of (C)
with y(0) > 0. Repeat this experiment with various choices of aand buntil you can state
this property precisely in terms of aand b.
(b) Choose positive numbers aand b. Construct a direction field and plot some integral curves
for (C) in a rectangular region of the form
{0 ≤ t≤ T, c ≤ y ≤ 0}",Elementary Differential Equations with Boundary Value Problems.pdf
"for (C) in a rectangular region of the form
{0 ≤ t≤ T, c ≤ y ≤ 0}
of the ty-plane. V ary a, b, T and cuntil you discover a common property of all solutions of
(C) with y(0) <0.
Y ou can verify your results later by doing Exercise 2.2. 27.",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 2
First Order Equations
IN THIS CHAPTER we study first order equations for which there are general methods of solution.
SECTION 2.1 deals with linear equations, the simplest kind o f first order equations. In this section we
introduce the method of variation of parameters. The idea un derlying this method will be a unifying
theme for our approach to solving many different kinds of dif ferential equations throughout the book.
SECTION 2.2 deals with separable equations, the simplest no nlinear equations. In this section we intro-
duce the idea of implicit and constant solutions of differen tial equations, and we point out some differ-
ences between the properties of linear and nonlinear equati ons.
SECTION 2.3 discusses existence and uniqueness of solution s of nonlinear equations. Although it may
seem logical to place this section before Section 2.2, we pre sented Section 2.2 first so we could have
illustrative examples in Section 2.3.",Elementary Differential Equations with Boundary Value Problems.pdf
"seem logical to place this section before Section 2.2, we pre sented Section 2.2 first so we could have
illustrative examples in Section 2.3.
SECTION 2.4 deals with nonlinear equations that are not sepa rable, but can be transformed into separable
equations by a procedure similar to variation of parameters .
SECTION 2.5 covers exact differential equations, which are given this name because the method for
solving them uses the idea of an exact differential from calc ulus.
SECTION 2.6 deals with equations that are not exact, but can m ade exact by multiplying them by a
function known called
integrating factor .
29",Elementary Differential Equations with Boundary Value Problems.pdf
"30 Chapter 2 First Order Equations
2.1 LINEAR FIRST ORDER EQUA TIONS
A first order differential equation is said to be linear if it can be written as
y′+ p(x)y = f(x). (2.1.1)
A first order differential equation that can’t be written lik e this is nonlinear. W e say that ( 2.1.1) is
homogeneous if f ≡ 0; otherwise it’s nonhomogeneous. Since y ≡ 0 is obviously a solution of the
homgeneous equation
y′+ p(x)y = 0 ,
we call it the trivial solution . Any other solution is nontrivial.
Example 2.1.1 The first order equations
x2y′+ 3 y = x2,
xy′− 8x2y = sin x,
xy′+ (ln x)y = 0 ,
y′ = x2y− 2,
are not in the form ( 2.1.1), but they are linear, since they can be rewritten as
y′+ 3
x2 y = 1 ,
y′− 8xy = sin x
x ,
y′+ ln x
x y = 0 ,
y′− x2y = −2.
Example 2.1.2 Here are some nonlinear first order equations:
xy′+ 3 y2 = 2 x (because yis squared) ,
yy′ = 3 (because of the product yy′),
y′+ xey = 12 (because of ey ).
General Solution of a Linear First Order Equation",Elementary Differential Equations with Boundary Value Problems.pdf
"yy′ = 3 (because of the product yy′),
y′+ xey = 12 (because of ey ).
General Solution of a Linear First Order Equation
T o motivate a definition that we’ll need, consider the simple linear first order equation
y′= 1
x2 . (2.1.2)
From calculus we know that ysatisfies this equation if and only if
y= − 1
x + c, (2.1.3)
where c is an arbitrary constant. W e call c a parameter and say that ( 2.1.3) defines a one–parameter
family of functions. For each real number c, the function defined by ( 2.1.3) is a solution of ( 2.1.2) on",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 31
(−∞ , 0) and (0, ∞ ); moreover, every solution of ( 2.1.2) on either of these intervals is of the form ( 2.1.3)
for some choice of c. W e say that ( 2.1.3) is the general solution of ( 2.1.2).
W e’ll see that a similar situation occurs in connection with any first order linear equation
y′+ p(x)y = f(x); (2.1.4)
that is, if pand f are continuous on some open interval (a,b ) then there’s a unique formula y = y(x,c )
analogous to ( 2.1.3) that involves xand a parameter cand has the these properties:
• For each fixed value of c, the resulting function of xis a solution of ( 2.1.4) on (a,b ).
• If y is a solution of ( 2.1.4) on (a,b ), then y can be obtained from the formula by choosing c
appropriately .
W e’ll call y= y(x,c ) the general solution of ( 2.1.4).
When this has been established, it will follow that an equati on of the form
P0(x)y′+ P1(x)y= F(x) (2.1.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"When this has been established, it will follow that an equati on of the form
P0(x)y′+ P1(x)y= F(x) (2.1.5)
has a general solution on any open interval (a,b ) on which P0, P1, and F are all continuous and P0 has
no zeros, since in this case we can rewrite ( 2.1.5) in the form ( 2.1.4) with p = P1/P 0 and f = F/P 0,
which are both continuous on (a,b ).
T o avoid awkward wording in examples and exercises, we won’t specify the interval (a,b ) when we
ask for the general solution of a specific linear first order eq uation. Let’s agree that this always means
that we want the general solution on every open interval on wh ich pand f are continuous if the equation
is of the form ( 2.1.4), or on which P0, P1, and F are continuous and P0 has no zeros, if the equation is of
the form ( 2.1.5). W e leave it to you to identify these intervals in specific ex amples and exercises.
For completeness, we point out that if P0, P1, and F are all continuous on an open interval (a,b ), but",Elementary Differential Equations with Boundary Value Problems.pdf
"For completeness, we point out that if P0, P1, and F are all continuous on an open interval (a,b ), but
P0 does have a zero in (a,b ), then ( 2.1.5) may fail to have a general solution on (a,b ) in the sense just
defined. Since this isn’t a major point that needs to be develo ped in depth, we won’t discuss it further;
however, see Exercise 44 for an example.
Homogeneous Linear First Order Equations
W e begin with the problem of finding the general solution of a h omogeneous linear first order equation.
The next example recalls a familiar result from calculus.
Example 2.1.3 Let abe a constant.
(a) Find the general solution of
y′− ay= 0 . (2.1.6)
(b) Solve the initial value problem
y′− ay= 0 , y (x0) = y0.
SO LU TIO N (a) Y ou already know from calculus that if cis any constant, then y = ceax satisfies ( 2.1.6).
However, let’s pretend you’ve forgotten this, and use this p roblem to illustrate a general method for
solving a homogeneous linear first order equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"solving a homogeneous linear first order equation.
W e know that ( 2.1.6) has the trivial solution y ≡ 0. Now suppose yis a nontrivial solution of ( 2.1.6).
Then, since a differentiable function must be continuous, t here must be some open interval I on which y
has no zeros. W e rewrite ( 2.1.6) as
y′
y = a",Elementary Differential Equations with Boundary Value Problems.pdf
"32 Chapter 2 First Order Equations
 x
0.2 0.4 0.6 0.8 1.0
 y
0.5
1.0
1.5
2.0
2.5
3.0
 a = 2
 a = 1.5
 a = 1
 a = −1
 a = −2.5
 a = −4
Figure 2.1.1 Solutions of y′− ay= 0 , y(0) = 1
for xin I. Integrating this shows that
ln |y|= ax+ k, so |y|= ekeax ,
where k is an arbitrary constant. Since eax can never equal zero, y has no zeros, so y is either always
positive or always negative. Therefore we can rewrite yas
y= ceax (2.1.7)
where
c=
{ ek if y> 0,
−ek if y< 0.
This shows that every nontrivial solution of ( 2.1.6) is of the form y = ceax for some nonzero constant c.
Since setting c= 0 yields the trivial solution, all solutions of ( 2.1.6) are of the form ( 2.1.7). Conversely ,
(2.1.7) is a solution of ( 2.1.6) for every choice of c, since differentiating ( 2.1.7) yields y′= aceax = ay.
SO LU TIO N (b) Imposing the initial condition y(x0) = y0 yields y0 = ceax0 , so c= y0e−ax0 and
y= y0e−ax0 eax = y0ea(x−x0).",Elementary Differential Equations with Boundary Value Problems.pdf
"SO LU TIO N (b) Imposing the initial condition y(x0) = y0 yields y0 = ceax0 , so c= y0e−ax0 and
y= y0e−ax0 eax = y0ea(x−x0).
Figure 2.1.1 show the graphs of this function with x0 = 0 , y0 = 1 , and various values of a.
Example 2.1.4 (a) Find the general solution of
xy′+ y = 0 . (2.1.8)
(b) Solve the initial value problem
xy′+ y= 0 , y (1) = 3 . (2.1.9)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 33
SO LU TIO N (a) W e rewrite ( 2.1.8) as
y′+ 1
xy= 0 , (2.1.10)
where xis restricted to either (−∞ , 0) or (0, ∞ ). If yis a nontrivial solution of ( 2.1.10), there must be
some open interval I on which yhas no zeros. W e can rewrite ( 2.1.10) as
y′
y = − 1
x
for xin I. Integrating shows that
ln |y|= − ln |x|+ k, so |y|= ek
|x|.
Since a function that satisfies the last equation can’t chang e sign on either (−∞ , 0) or (0, ∞ ), we can
rewrite this result more simply as
y= c
x (2.1.11)
where
c=
{ ek if y> 0,
−ek if y< 0.
W e’ve now shown that every solution of ( 2.1.10) is given by ( 2.1.11) for some choice of c. (Even though
we assumed that y was nontrivial to derive ( 2.1.11), we can get the trivial solution by setting c = 0 in
(2.1.11).) Conversely , any function of the form ( 2.1.11) is a solution of ( 2.1.10), since differentiating
(2.1.11) yields
y′= − c
x2 ,
and substituting this and ( 2.1.11) into ( 2.1.10) yields
y′+ 1
xy = − c
x2 + 1
x",Elementary Differential Equations with Boundary Value Problems.pdf
"(2.1.11) yields
y′= − c
x2 ,
and substituting this and ( 2.1.11) into ( 2.1.10) yields
y′+ 1
xy = − c
x2 + 1
x
c
x
= − c
x2 + c
x2 = 0 .
Figure 2.1.2 shows the graphs of some solutions corresponding to various values of c
SO LU TIO N (b) Imposing the initial condition y(1) = 3 in ( 2.1.11) yields c = 3 . Therefore the solution
of ( 2.1.9) is
y = 3
x.
The interval of validity of this solution is (0, ∞ ).
The results in Examples 2.1.3(a) and 2.1.4(b) are special cases of the next theorem.
Theorem 2.1.1 If pis continuous on (a,b ), then the general solution of the homogeneous equation
y′+ p(x)y= 0 (2.1.12)
on (a,b ) is
y = ce−P (x) ,
where
P(x) =
∫
p(x) dx (2.1.13)
is any antiderivative of pon (a,b ); that is ,
P′(x) = p(x), a<x<b. (2.1.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"34 Chapter 2 First Order Equations
 x
 y
 c > 0 c < 0
 c > 0  c < 0
Figure 2.1.2 Solutions of xy′+ y= 0 on (0, ∞ ) and (−∞ , 0)
Proof If y = ce−P (x) , differentiating yand using ( 2.1.14) shows that
y′= −P′(x)ce−P (x) = −p(x)ce−P (x) = −p(x)y,
so y′+ p(x)y= 0 ; that is, yis a solution of ( 2.1.12), for any choice of c.
Now we’ll show that any solution of ( 2.1.12) can be written as y = ce−P (x) for some constant c. The
trivial solution can be written this way , with c= 0 . Now suppose yis a nontrivial solution. Then there’s
an open subinterval I of (a,b ) on which yhas no zeros. W e can rewrite ( 2.1.12) as
y′
y = −p(x) (2.1.15)
for xin I. Integrating ( 2.1.15) and recalling ( 2.1.13) yields
ln |y|= −P(x) + k,
where kis a constant. This implies that
|y|= ek e−P (x) .
Since P is defined for all xin (a,b ) and an exponential can never equal zero, we can take I = ( a,b ), so
yhas zeros on (a,b ) ( a,b ), so we can rewrite the last equation as y= ce−P (x) , where
c=
{ ek if y> 0 on (a,b ),",Elementary Differential Equations with Boundary Value Problems.pdf
"yhas zeros on (a,b ) ( a,b ), so we can rewrite the last equation as y= ce−P (x) , where
c=
{ ek if y> 0 on (a,b ),
−ek if y< 0 on (a,b ).
REM A RK : Rewriting a first order differential equation so that one si de depends only on yand y′and the
other depends only on xis called separation of variables . W e did this in Examples 2.1.3 and 2.1.4, and
in rewriting ( 2.1.12) as ( 2.1.15).W e’llapply this method to nonlinear equations in Section 2.2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 35
Linear Nonhomogeneous First Order Equations
W e’ll now solve the nonhomogeneous equation
y′+ p(x)y = f(x). (2.1.16)
When considering this equation we call
y′+ p(x)y= 0
the complementary equation .
W e’ll find solutions of ( 2.1.16) in the form y = uy1, where y1 is a nontrivial solution of the com-
plementary equation and uis to be determined. This method of using a solution of the com plementary
equation to obtain solutions of a nonhomogeneous equation i s a special case of a method called variation
of parameters , which you’ll encounter several times in this book. (Obviou sly , ucan’t be constant, since
if it were, the left side of ( 2.1.16) would be zero. Recognizing this, the early users of this met hod viewed
uas a “parameter” that varies; hence, the name “variation of p arameters. ”)
If
y= uy1, then y′= u′y1 + uy′
1.
Substituting these expressions for yand y′into ( 2.1.16) yields
u′y1 + u(y′
1 + p(x)y1) = f(x),
which reduces to",Elementary Differential Equations with Boundary Value Problems.pdf
"If
y= uy1, then y′= u′y1 + uy′
1.
Substituting these expressions for yand y′into ( 2.1.16) yields
u′y1 + u(y′
1 + p(x)y1) = f(x),
which reduces to
u′y1 = f(x), (2.1.17)
since y1 is a solution of the complementary equation; that is,
y′
1 + p(x)y1 = 0 .
In the proof of Theorem 2.2.1 we saw that y1 has no zeros on an interval where pis continuous. Therefore
we can divide ( 2.1.17) through by y1 to obtain
u′= f(x)/y 1(x).
W e can integrate this (introducing a constant of integratio n), and multiply the result by y1 to get the gen-
eral solution of ( 2.1.16). Before turning to the formal proof of this claim, let’s con sider some examples.
Example 2.1.5 Find the general solution of
y′+ 2 y= x3e−2x. (2.1.18)
By applying (a) of Example 2.1.3 with a = −2, we see that y1 = e−2x is a solution of the com-
plementary equation y′+ 2 y = 0 . Therefore we seek solutions of ( 2.1.18) in the form y = ue−2x, so
that
y′= u′e−2x − 2ue−2x and y′+ 2 y = u′e−2x − 2ue−2x + 2 ue−2x = u′e−2x. (2.1.19)",Elementary Differential Equations with Boundary Value Problems.pdf
"that
y′= u′e−2x − 2ue−2x and y′+ 2 y = u′e−2x − 2ue−2x + 2 ue−2x = u′e−2x. (2.1.19)
Therefore yis a solution of ( 2.1.18) if and only if
u′e−2x = x3e−2x or, equivalently , u′= x3.
Therefore
u= x4
4 + c,",Elementary Differential Equations with Boundary Value Problems.pdf
"36 Chapter 2 First Order Equations
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 x
 y
Figure 2.1.3 A direction field and integral curves for y′+ 2 y= x2e−2x
and
y= ue−2x = e−2x
(x4
4 + c
)
is the general solution of ( 2.1.18).
Figure 2.1.3 shows a direction field and some integral curves for ( 2.1.18).
Example 2.1.6
(a) Find the general solution
y′+ (cot x)y= xcsc x. (2.1.20)
(b) Solve the initial value problem
y′+ (cot x)y = xcsc x, y (π/ 2) = 1 . (2.1.21)
SO LU TIO N (a) Here p(x) = cot xand f(x) = xcsc xare both continuous except at the points x= rπ,
where ris an integer. Therefore we seek solutions of ( 2.1.20) on the intervals (rπ, (r+ 1) π). W e need a
nontrival solution y1 of the complementary equation; thus, y1 must satisfy y′
1 + (cot x)y1 = 0 , which we
rewrite as
y′
1
y1
= − cot x= −cos x
sin x. (2.1.22)
Integrating this yields
ln |y1|= − ln |sin x|,",Elementary Differential Equations with Boundary Value Problems.pdf
"1 + (cot x)y1 = 0 , which we
rewrite as
y′
1
y1
= − cot x= −cos x
sin x. (2.1.22)
Integrating this yields
ln |y1|= − ln |sin x|,
where we take the constant of integration to be zero since we n eed only one function that satisfies ( 2.1.22).
Clearly y1 = 1 / sin xis a suitable choice. Therefore we seek solutions of ( 2.1.20) in the form
y = u
sin x,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 37
so that
y′= u′
sin x − ucos x
sin2 x (2.1.23)
and
y′+ (cot x)y = u′
sin x − ucos x
sin2 x + ucot x
sin x
= u′
sin x − ucos x
sin2 x + ucos x
sin2 x
= u′
sin x.
(2.1.24)
Therefore yis a solution of ( 2.1.20) if and only if
u′/ sin x= xcsc x= x/ sin x or, equivalently , u′= x.
Integrating this yields
u= x2
2 + c, and y= u
sin x = x2
2 sin x + c
sin x. (2.1.25)
is the general solution of ( 2.1.20) on every interval (rπ, (r+ 1) π) (r=integer).
SO LU TIO N (b) Imposing the initial condition y(π/ 2) = 1 in ( 2.1.25) yields
1 = π2
8 + c or c= 1 − π2
8 .
Thus,
y = x2
2 sin x + (1 − π2/ 8)
sin x
is a solution of ( 2.1.21). The interval of validity of this solution is (0,π ); Figure 2.1.4 shows its graph.
1 2 3
− 15
− 10
 − 5
  5
 10
 15
 x
 y
Figure 2.1.4 Solution of y′+ (cot x)y = xcsc x,y (π/ 2) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"38 Chapter 2 First Order Equations
REM A RK : It wasn’t necessary to do the computations ( 2.1.23) and ( 2.1.24) in Example 2.1.6, since we
showed in the discussion preceding Example 2.1.5 that if y = uy1 where y′
1 + p(x)y1 = 0 , then y′+
p(x)y = u′y1. W e did these computations so you would see this happen in thi s specific example. W e
recommend that you include these “unnecesary” computation s in doing exercises, until you’re confident
that you really understand the method. After that, omit them .
W e summarize the method of variation of parameters for solvi ng
y′+ p(x)y= f(x) (2.1.26)
as follows:
(a) Find a function y1 such that
y′
1
y1
= −p(x).
For convenience, take the constant of integration to be zero .
(b) Write
y = uy1 (2.1.27)
to remind yourself of what you’re doing.
(c) Write u′y1 = f and solve for u′; thus, u′= f/y 1 .
(d) Integrate u′to obtain u, with an arbitrary constant of integration.
(e) Substitute uinto ( 2.1.27) to obtain y.
T o solve an equation written as",Elementary Differential Equations with Boundary Value Problems.pdf
"(d) Integrate u′to obtain u, with an arbitrary constant of integration.
(e) Substitute uinto ( 2.1.27) to obtain y.
T o solve an equation written as
P0(x)y′+ P1(x)y = F(x),
we recommend that you divide through by P0(x) to obtain an equation of the form ( 2.1.26) and then
follow this procedure.
Solutions in Integral Form
Sometimes the integrals that arise in solving a linear first o rder equation can’t be evaluated in terms of
elementary functions. In this case the solution must be left in terms of an integral.
Example 2.1.7
(a) Find the general solution of
y′− 2xy= 1 .
(b) Solve the initial value problem
y′− 2xy= 1 , y (0) = y0. (2.1.28)
SO LU TIO N (a) T o apply variation of parameters, we need a nontrivial solut ion y1 of the complementary
equation; thus, y′
1 − 2xy1 = 0 , which we rewrite as
y′
1
y1
= 2 x.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 39
Integrating this and taking the constant of integration to b e zero yields
ln |y1|= x2, so |y1|= ex2
.
W e choose y1 = ex2
and seek solutions of ( 2.1.28) in the form y = uex2
, where
u′ex2
= 1 , so u′= e−x2
.
Therefore
u= c+
∫
e−x2
dx,
but we can’t simplify the integral on the right because there ’s no elementary function with derivative
equal to e−x2
. Therefore the best available form for the general solution of ( 2.1.28) is
y = uex2
= ex2
(
c+
∫
e−x2
dx
)
. (2.1.29)
SO LU TIO N (b) Since the initial condition in ( 2.1.28) is imposed at x0 = 0 , it is convenient to rewrite
(2.1.29) as
y= ex2
(
c+
∫ x
0
e−t2
dt
)
, since
∫ 0
0
e−t2
dt= 0 .
Setting x= 0 and y= y0 here shows that c= y0. Therefore the solution of the initial value problem is
y= ex2
(
y0 +
∫ x
0
e−t2
dt
)
. (2.1.30)
For a given value of y0 and each fixed x, the integral on the right can be evaluated by numerical meth ods.",Elementary Differential Equations with Boundary Value Problems.pdf
"y= ex2
(
y0 +
∫ x
0
e−t2
dt
)
. (2.1.30)
For a given value of y0 and each fixed x, the integral on the right can be evaluated by numerical meth ods.
An alternate procedure is to apply the numerical integratio n procedures discussed in Chapter 3 directly to
the initial value problem ( 2.1.28). Figure 2.1.5 shows graphs of of ( 2.1.30) for several values of y0.
 x
 y
Figure 2.1.5 Solutions of y′− 2xy= 1 , y(0) = y0",Elementary Differential Equations with Boundary Value Problems.pdf
"40 Chapter 2 First Order Equations
An Existence and Uniqueness Theorem
The method of variation of parameters leads to this theorem.
Theorem 2.1.2
Suppose pand f are continuous on an open interval (a,b ), and let y1 be any nontrivial
solution of the complementary equation
y′+ p(x)y= 0
on (a,b ). Then :
(a) The general solution of the nonhomogeneous equation
y′+ p(x)y= f(x) (2.1.31)
on (a,b ) is
y= y1(x)
(
c+
∫
f(x)/y 1(x) dx
)
. (2.1.32)
(b) If x0 is an arbitrary point in (a,b ) and y0 is an arbitrary real number , then the initial value problem
y′+ p(x)y = f(x), y (x0 ) = y0
has the unique solution
y = y1(x)
( y0
y1(x0) +
∫ x
x0
f(t)
y1(t) dt
)
on (a,b ).
Proof (a) T o show that ( 2.1.32) is the general solution of ( 2.1.31) on (a,b ), we must prove that:
(i) If cis any constant, the function yin ( 2.1.32) is a solution of ( 2.1.31) on (a,b ).
(ii) If yis a solution of ( 2.1.31) on (a,b ) then yis of the form ( 2.1.32) for some constant c.",Elementary Differential Equations with Boundary Value Problems.pdf
"(ii) If yis a solution of ( 2.1.31) on (a,b ) then yis of the form ( 2.1.32) for some constant c.
T o prove (i), we first observe that any function of the form ( 2.1.32) is defined on (a,b ), since pand f
are continuous on (a,b ). Differentiating ( 2.1.32) yields
y′= y′
1(x)
(
c+
∫
f(x)/y 1 (x) dx
)
+ f(x).
Since y′
1 = −p(x)y1, this and ( 2.1.32) imply that
y′ = −p(x)y1(x)
(
c+
∫
f(x)/y 1(x) dx
)
+ f(x)
= −p(x)y(x) + f(x),
which implies that yis a solution of ( 2.1.31).
T o prove (ii), suppose yis a solution of ( 2.1.31) on (a,b ). From the proof of Theorem 2.1.1, we know
that y1 has no zeros on (a,b ), so the function u= y/y 1 is defined on (a,b ). Moreover, since
y′= −py+ f and y′
1 = −py1,
u′ = y1y′− y′
1y
y2
1
= y1(−py+ f) − (−py1)y
y2
1
= f
y1
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 41
Integrating u′= f/y 1 yields
u=
(
c+
∫
f(x)/y 1(x) dx
)
,
which implies ( 2.1.32), since y= uy1.
(b) W e’ve proved (a), where ∫ f(x)/y 1(x) dxin ( 2.1.32) is an arbitrary antiderivative of f/y 1 . Now
it’s convenient to choose the antiderivative that equals ze ro when x= x0, and write the general solution
of ( 2.1.31) as
y = y1(x)
(
c+
∫ x
x0
f(t)
y1(t) dt
)
.
Since
y(x0) = y1(x0)
(
c+
∫ x0
x0
f(t)
y1(t) dt
)
= cy1(x0),
we see that y(x0) = y0 if and only if c= y0/y 1(x0).
2.1 Exercises
In Exercises 1– 5 find the general solution.
1. y′+ ay= 0 (a=constant) 2. y′+ 3 x2y = 0
3. xy′+ (ln x)y= 0 4. xy′+ 3 y= 0
5. x2y′+ y = 0
In Exercises 6– 11 solve the initial value problem.
6. y′+
(1 + x
x
)
y= 0 , y (1) = 1
7. xy′+
(
1 + 1
ln x
)
y = 0 , y (e) = 1
8. xy′+ (1 + xcot x)y= 0 , y
(π
2
)
= 2
9. y′−
( 2x
1 + x2
)
y= 0 , y (0) = 2
10. y′+ k
xy = 0 , y (1) = 3 (k= constant)
11. y′+ (tan kx)y= 0 , y (0) = 2 (k= constant)",Elementary Differential Equations with Boundary Value Problems.pdf
"(π
2
)
= 2
9. y′−
( 2x
1 + x2
)
y= 0 , y (0) = 2
10. y′+ k
xy = 0 , y (1) = 3 (k= constant)
11. y′+ (tan kx)y= 0 , y (0) = 2 (k= constant)
In Exercises 12 – 15 find the general solution. Also, plot a direction field and som e integral curves on the
rectangular region {−2 ≤ x≤ 2, −2 ≤ y≤ 2}.
12. C/G y′+ 3 y= 1 13. C/G y′+
(1
x − 1
)
y= − 2
x
14. C/G y′+ 2 xy= xe−x2
15. C/G y′+ 2x
1 + x2 y= e−x
1 + x2",Elementary Differential Equations with Boundary Value Problems.pdf
"42 Chapter 2 First Order Equations
In Exercises 16 – 24 find the general solution.
16. y′+ 1
xy = 7
x2 + 3 17. y′+ 4
x− 1 y = 1
(x− 1)5 + sin x
(x− 1)4
18. xy′+ (1 + 2 x2)y = x3e−x2 19. xy′+ 2 y= 2
x2 + 1
20. y′+ (tan x)y= cos x 21. (1 + x)y′+ 2 y= sin x
1 + x
22. (x− 2)(x− 1)y′− (4x− 3)y= ( x− 2)3
23. y′+ (2 sin xcos x)y= e− sin2 x 24. x2y′+ 3 xy= ex
In Exercises 25– 29 solve the initial value problem and sketch the graph of the so lution.
25. C/G y′+ 7 y= e3x, y (0) = 0
26. C/G (1 + x2)y′+ 4 xy= 2
1 + x2 , y (0) = 1
27. C/G xy′+ 3 y= 2
x(1 + x2) , y (−1) = 0
28. C/G y′+ (cot x)y= cos x, y
(π
2
)
= 1
29. C/G y′+ 1
xy = 2
x2 + 1 , y (−1) = 0
In Exercises 30– 37 solve the initial value problem.
30. (x− 1)y′+ 3 y= 1
(x− 1)3 + sin x
(x− 1)2 , y (0) = 1
31. xy′+ 2 y= 8 x2, y (1) = 3
32. xy′− 2y= −x2, y (1) = 1
33. y′+ 2 xy= x, y (0) = 3
34. (x− 1)y′+ 3 y= 1 + ( x− 1) sec2 x
(x− 1)3 , y (0) = −1
35. (x+ 2) y′+ 4 y= 1 + 2 x2
x(x+ 2) 3 , y (−1) = 2
36. (x2 − 1)y′− 2xy= x(x2 − 1), y (0) = 4",Elementary Differential Equations with Boundary Value Problems.pdf
"(x− 1)3 , y (0) = −1
35. (x+ 2) y′+ 4 y= 1 + 2 x2
x(x+ 2) 3 , y (−1) = 2
36. (x2 − 1)y′− 2xy= x(x2 − 1), y (0) = 4
37. (x2 − 5)y′− 2xy= −2x(x2 − 5), y (2) = 7
In Exercises 38– 42 solve the initial value problem and leave the answer in a form involving a definite
integral. (Y ou can solve these problems numerically by methods discuss ed in Chapter 3. )
38. y′+ 2 xy= x2, y (0) = 3
39. y′+ 1
xy = sin x
x2 , y (1) = 2
40. y′+ y= e−x tan x
x , y (1) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.1 Linear First Order Equations 43
41. y′+ 2x
1 + x2 y= ex
(1 + x2)2 , y (0) = 1
42. xy′+ ( x+ 1) y= ex2
, y (1) = 2
43. Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of
glucose present in the bloodstream. Let λ denote the (positive) constant of proportionality . Now
suppose glucose is injected into a patient’s bloodstream at a constant rate of r units per unit of
time. Let G= G(t) be the number of units in the patient’s bloodstream at time t> 0. Then
G′= −λG + r,
where the first term on the right is due to the absorption of the glucose by the patient’s body and
the second term is due to the injection. Determine Gfor t >0, given that G(0) = G0. Also, find
limt→∞ G(t).
44. (a) L Plot a direction field and some integral curves for
xy′− 2y= −1 (A)
on the rectangular region {−1 ≤ x ≤ 1, −.5 ≤ y ≤ 1.5}. What do all the integral curves
have in common?
(b) Show that the general solution of (A) on (−∞ , 0) and (0, ∞ ) is
y= 1
2 + cx2.",Elementary Differential Equations with Boundary Value Problems.pdf
"have in common?
(b) Show that the general solution of (A) on (−∞ , 0) and (0, ∞ ) is
y= 1
2 + cx2.
(c) Show that yis a solution of (A) on (−∞ , ∞ ) if and only if
y=





1
2 + c1x2, x ≥ 0,
1
2 + c2x2, x< 0,
where c1 and c2 are arbitrary constants.
(d) Conclude from (c) that all solutions of (A) on (−∞ , ∞ ) are solutions of the initial value
problem
xy′− 2y = −1, y (0) = 1
2 .
(e) Use (b) to show that if x0 ̸= 0 and y0 is arbitrary , then the initial value problem
xy′− 2y= −1, y (x0 ) = y0
has infinitely many solutions on ( −∞ , ∞ ). Explain why this does’nt contradict Theorem 2.1.1(b).
45. Suppose f is continuous on an open interval (a,b ) and α is a constant.
(a) Derive a formula for the solution of the initial value proble m
y′+ αy = f(x), y (x0) = y0, (A)
where x0 is in (a,b ) and y0 is an arbitrary real number.
(b) Suppose (a,b ) = ( a, ∞ ), α > 0 and lim
x→∞
f(x) = L. Show that if yis the solution of (A),
then lim
x→∞
y(x) = L/α .",Elementary Differential Equations with Boundary Value Problems.pdf
"44 Chapter 2 First Order Equations
46. Assume that all functions in this exercise are defined on a com mon interval (a,b ).
(a) Prove: If y1 and y2 are solutions of
y′+ p(x)y= f1(x)
and
y′+ p(x)y= f2(x)
respectively , and c1 and c2 are constants, then y= c1y1 + c2y2 is a solution of
y′+ p(x)y= c1f1(x) + c2f2(x).
(This is the principle of superposition. )
(b) Use (a) to show that if y1 and y2 are solutions of the nonhomogeneous equation
y′+ p(x)y= f(x), (A)
then y1 − y2 is a solution of the homogeneous equation
y′+ p(x)y= 0 . (B)
(c) Use (a) to show that if y1 is a solution of (A) and y2 is a solution of (B), then y1 + y2 is a
solution of (A).
47. Some nonlinear equations can be transformed into linear equ ations by changing the dependent
variable. Show that if
g′(y)y′+ p(x)g(y) = f(x)
where y is a function of xand g is a function of y, then the new dependent variable z = g(y)
satisfies the linear equation
z′+ p(x)z = f(x).
48. Solve by the method discussed in Exercise 47.",Elementary Differential Equations with Boundary Value Problems.pdf
"satisfies the linear equation
z′+ p(x)z = f(x).
48. Solve by the method discussed in Exercise 47.
(a) (sec2 y)y′− 3 tan y= −1 (b) ey2
(
2yy′+ 2
x
)
= 1
x2
(c) xy′
y + 2 ln y= 4 x2 (d) y′
(1 + y)2 − 1
x(1 + y) = − 3
x2
49. W e’ve shown that if pand f are continuous on (a,b ) then every solution of
y′+ p(x)y= f(x) (A)
on (a,b ) can be written as y = uy1, where y1 is a nontrivial solution of the complementary equa-
tion for (A) and u′= f/y 1. Now suppose f, f′, . . . , f(m) and p, p′, . . . , p(m−1) are continuous
on (a,b ), where mis a positive integer, and define
f0 = f,
fj = f′
j−1 + pfj−1, 1 ≤ j ≤ m.
Show that
u(j+1) = fj
y1
, 0 ≤ j ≤ m.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.2 Separable Equations 45
2.2 SEP ARABLE EQUA TIONS
A first order differential equation is separable if it can be written as
h(y)y′= g(x), (2.2.1)
where the left side is a product of y′and a function of yand the right side is a function of x. Rewriting
a separable differential equation in this form is called separation of variables. In Section 2.1 we used
separation of variables to solve homogeneous linear equati ons. In this section we’ll apply this method to
nonlinear equations.
T o see how to solve ( 2.2.1), let’s first assume that yis a solution. Let G(x) and H(y) be antiderivatives
of g(x) and h(y); that is,
H′(y) = h(y) and G′(x) = g(x). (2.2.2)
Then, from the chain rule,
d
dxH(y(x)) = H′(y(x))y′(x) = h(y)y′(x).
Therefore ( 2.2.1) is equivalent to
d
dxH(y(x)) = d
dxG(x).
Integrating both sides of this equation and combining the co nstants of integration yields
H(y(x)) = G(x) + c. (2.2.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"d
dxH(y(x)) = d
dxG(x).
Integrating both sides of this equation and combining the co nstants of integration yields
H(y(x)) = G(x) + c. (2.2.3)
Although we derived this equation on the assumption that y is a solution of ( 2.2.1), we can now view it
differently: Any differentiable function ythat satisfies ( 2.2.3) for some constant cis a solution of ( 2.2.1).
T o see this, we differentiate both sides of ( 2.2.3), using the chain rule on the left, to obtain
H′(y(x))y′(x) = G′(x),
which is equivalent to
h(y(x))y′(x) = g(x)
because of ( 2.2.2).
In conclusion, to solve ( 2.2.1) it suffices to find functions G = G(x) and H = H(y) that satisfy
(2.2.2). Then any differentiable function y = y(x) that satisfies ( 2.2.3) is a solution of ( 2.2.1).
Example 2.2.1 Solve the equation
y′= x(1 + y2 ).
Solution Separating variables yields
y′
1 + y2 = x.
Integrating yields
tan−1 y= x2
2 + c
Therefore
y = tan
(x2
2 + c
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"46 Chapter 2 First Order Equations
Example 2.2.2
(a) Solve the equation
y′= −x
y. (2.2.4)
(b) Solve the initial value problem
y′= −x
y, y (1) = 1 . (2.2.5)
(c) Solve the initial value problem
y′= −x
y, y (1) = −2. (2.2.6)
SO LU TIO N (a) Separating variables in ( 2.2.4) yields
yy′= −x.
Integrating yields
y2
2 = −x2
2 + c, or, equivalently , x2 + y2 = 2 c.
The last equation shows that c must be positive if y is to be a solution of ( 2.2.4) on an open interval.
Therefore we let 2c= a2 (with a> 0) and rewrite the last equation as
x2 + y2 = a2. (2.2.7)
This equation has two differentiable solutions for yin terms of x:
y=
√
a2 − x2, −a<x<a, (2.2.8)
and
y= −
√
a2 − x2, −a<x<a. (2.2.9)
The solution curves defined by ( 2.2.8) are semicircles above the x-axis and those defined by ( 2.2.9) are
semicircles below the x-axis (Figure 2.2.1).
SO LU TIO N (b) The solution of ( 2.2.5) is positive when x= 1 ; hence, it is of the form ( 2.2.8). Substituting",Elementary Differential Equations with Boundary Value Problems.pdf
"SO LU TIO N (b) The solution of ( 2.2.5) is positive when x= 1 ; hence, it is of the form ( 2.2.8). Substituting
x= 1 and y= 1 into ( 2.2.7) to satisfy the initial condition yields a2 = 2 ; hence, the solution of ( 2.2.5) is
y=
√
2 − x2, −
√
2 <x<
√
2.
SO LU TIO N (c) The solution of ( 2.2.6) is negative when x = 1 and is therefore of the form ( 2.2.9).
Substituting x = 1 and y = −2 into ( 2.2.7) to satisfy the initial condition yields a2 = 5 . Hence, the
solution of ( 2.2.6) is
y= −
√
5 − x2, −
√
5 <x<
√
5.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.2 Separable Equations 47
 x
 y
1 2−1−2
1
2
−1
−2
(a)
(b)
Figure 2.2.1 (a) y=
√
2 − x2, −
√
2 <x<
√
2; (b) y= −
√
5 − x2, −
√
5 <x<
√
5
Implicit Solutions of Separable Equations
In Examples 2.2.1 and 2.2.2 we were able to solve the equation H(y) = G(x) + c to obtain explicit
formulas for solutions of the given separable differential equations. As we’ll see in the next example,
this isn’t always possible. In this situation we must broade n our definition of a solution of a separable
equation. The next theorem provides the basis for this modifi cation. W e omit the proof, which requires a
result from advanced calculus called as the implicit function theorem .
Theorem 2.2.1 Suppose g = g(x) is continous on (a,b ) and h= h(y) are continuous on (c,d ). Let G
be an antiderivative of g on (a,b ) and let H be an antiderivative of hon (c,d ). Let x0 be an arbitrary
point in (a,b ), let y0 be a point in (c,d ) such that h(y0) ̸= 0 , and define
c= H(y0) − G(x0). (2.2.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"point in (a,b ), let y0 be a point in (c,d ) such that h(y0) ̸= 0 , and define
c= H(y0) − G(x0). (2.2.10)
Then there’s a function y = y(x) defined on some open interval (a1,b 1), where a≤ a1 <x0 <b1 ≤ b,
such that y(x0) = y0 and
H(y) = G(x) + c (2.2.11)
for a1 <x<b 1. Therefore yis a solution of the initial value problem
h(y)y′= g(x), y (x0) = x0. (2.2.12)
It’s convenient to say that ( 2.2.11) with carbitrary is an implicit solution of h(y)y′ = g(x). Curves
defined by ( 2.2.11) are integral curves of h(y)y′= g(x). If csatisfies ( 2.2.10), we’ll say that ( 2.2.11) is
an implicit solution of the initial value problem (2.2.12). However, keep these points in mind:
• For some choices of cthere may not be any differentiable functions ythat satisfy ( 2.2.11).",Elementary Differential Equations with Boundary Value Problems.pdf
"48 Chapter 2 First Order Equations
• The function yin ( 2.2.11) (not ( 2.2.11) itself) is a solution of h(y)y′= g(x).
Example 2.2.3
(a) Find implicit solutions of
y′= 2x+ 1
5y4 + 1 . (2.2.13)
(b) Find an implicit solution of
y′= 2x+ 1
5y4 + 1 , y (2) = 1 . (2.2.14)
SO LU TIO N (a) Separating variables yields
(5y4 + 1) y′= 2 x+ 1 .
Integrating yields the implicit solution
y5 + y= x2 + x+ c. (2.2.15)
of ( 2.2.13).
SO LU TIO N (b) Imposing the initial condition y(2) = 1 in ( 2.2.15) yields 1 + 1 = 4 + 2 + c, so c= −4.
Therefore
y5 + y = x2 + x− 4
is an implicit solution of the initial value problem ( 2.2.14). Although more than one differentiable func-
tion y = y(x) satisfies 2.2.13) near x = 1 , it can be shown that there’s only one such function that
satisfies the initial condition y(1) = 2 .
Figure 2.2.2 shows a direction field and some integral curves for ( 2.2.13).
Constant Solutions of Separable Equations
An equation of the form
y′= g(x)p(y)
is separable, since it can be rewritten as
1",Elementary Differential Equations with Boundary Value Problems.pdf
"Constant Solutions of Separable Equations
An equation of the form
y′= g(x)p(y)
is separable, since it can be rewritten as
1
p(y) y′= g(x).
However, the division by p(y) is not legitimate if p(y) = 0 for some values of y. The next two examples
show how to deal with this problem.
Example 2.2.4 Find all solutions of
y′= 2 xy2. (2.2.16)
Solution Here we must divide by p(y) = y2 to separate variables. This isn’t legitimate if yis a solution
of (
2.2.16) that equals zero for some value of x. One such solution can be found by inspection: y ≡ 0.
Now suppose yis a solution of ( 2.2.16) that isn’t identically zero. Since yis continuous there must be an
interval on which yis never zero. Since division by y2 is legitimate for xin this interval, we can separate
variables in ( 2.2.16) to obtain
y′
y2 = 2 x.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.2 Separable Equations 49
1 1.5 2 2.5 3 3.5 4
−1
−0.5
0
0.5
1
1.5
2
 x
 y
Figure 2.2.2 A direction field and integral curves for y′= 2x+ 1
5y4 + 1
Integrating this yields
− 1
y = x2 + c,
which is equivalent to
y = − 1
x2 + c. (2.2.17)
W e’ve now shown that if y is a solution of ( 2.2.16) that is not identically zero, then y must be of the
form ( 2.2.17). By substituting ( 2.2.17) into ( 2.2.16), you can verify that ( 2.2.17) is a solution of ( 2.2.16).
Thus, solutions of ( 2.2.16) are y ≡ 0 and the functions of the form ( 2.2.17). Note that the solution y ≡ 0
isn’t of the form ( 2.2.17) for any value of c.
Figure 2.2.3 shows a direction field and some integral curves for ( 2.2.16)
Example 2.2.5 Find all solutions of
y′= 1
2 x(1 − y2). (2.2.18)
Solution Here we must divide by p(y) = 1 − y2 to separate variables. This isn’t legitimate if y is a
solution of ( 2.2.18) that equals ±1 for some value of x. T wo such solutions can be found by inspection:",Elementary Differential Equations with Boundary Value Problems.pdf
"solution of ( 2.2.18) that equals ±1 for some value of x. T wo such solutions can be found by inspection:
y ≡ 1 and y ≡ − 1. Now suppose yis a solution of ( 2.2.18) such that 1 − y2 isn’t identically zero. Since
1 − y2 is continuous there must be an interval on which 1 − y2 is never zero. Since division by 1 − y2 is
legitimate for xin this interval, we can separate variables in ( 2.2.18) to obtain
2y′
y2 − 1 = −x.",Elementary Differential Equations with Boundary Value Problems.pdf
"50 Chapter 2 First Order Equations
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 y
 x
Figure 2.2.3 A direction field and integral curves for y′= 2 xy2
A partial fraction expansion on the left yields
[ 1
y− 1 − 1
y+ 1
]
y′= −x,
and integrating yields
ln
⏐
⏐
⏐
⏐
y− 1
y+ 1
⏐
⏐
⏐
⏐= −x2
2 + k;
hence, ⏐
⏐
⏐
⏐
y− 1
y+ 1
⏐
⏐
⏐
⏐= ek e−x2/ 2.
Since y(x) ̸= ±1 for xon the interval under discussion, the quantity (y− 1)/ (y+ 1) can’t change sign
in this interval. Therefore we can rewrite the last equation as
y− 1
y+ 1 = ce−x2/ 2,
where c= ±ek , depending upon the sign of (y− 1)/ (y+ 1) on the interval. Solving for yyields
y = 1 + ce−x2/ 2
1 − ce−x2/ 2 . (2.2.19)
W e’ve now shown that if yis a solution of ( 2.2.18) that is not identically equal to ±1, then ymust be
as in ( 2.2.19). By substituting ( 2.2.19) into ( 2.2.18) you can verify that ( 2.2.19) is a solution of ( 2.2.18).
Thus, the solutions of ( 2.2.18) are y ≡ 1, y ≡ − 1 and the functions of the form ( 2.2.19). Note that the",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.2 Separable Equations 51
constant solution y ≡ 1 can be obtained from this formula by taking c = 0 ; however, the other constant
solution, y≡ − 1, can’t be obtained in this way .
Figure 2.2.4 shows a direction field and some integrals for ( 2.2.18).
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−3
−2
−1
0
1
2
3
 x
 y
Figure 2.2.4 A direction field and integral curves for y′= x(1 − y2)
2
Differences Between Linear and Nonlinear Equations
Theorem 2.1.2 states that if pand f are continuous on (a,b ) then every solution of
y′+ p(x)y= f(x)
on (a,b ) can be obtained by choosing a value for the constant cin the general solution, and if x0 is any
point in (a,b ) and y0 is arbitrary , then the initial value problem
y′+ p(x)y= f(x), y (x0) = y0
has a solution on (a,b ).
The not true for nonlinear equations. First, we saw in Exampl es 2.2.4 and 2.2.5 that a nonlinear
equation may have solutions that can’t be obtained by choosi ng a specific value of a constant appearing",Elementary Differential Equations with Boundary Value Problems.pdf
"equation may have solutions that can’t be obtained by choosi ng a specific value of a constant appearing
in a one-parameter family of solutions. Second, it is in gene ral impossible to determine the interval
of validity of a solution to an initial value problem for a non linear equation by simply examining the
equation, since the interval of validity may depend on the in itial condition. For instance, in Example 2.2.2
we saw that the solution of dy
dx = −x
y, y (x0 ) = y0
is valid on (−a,a ), where a=
√
x2
0 + y2
0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"52 Chapter 2 First Order Equations
Example 2.2.6 Solve the initial value problem
y′= 2 xy2, y (0) = y0
and determine the interval of validity of the solution.
Solution First suppose y0 ̸= 0 . From Example
2.2.4, we know that ymust be of the form
y = − 1
x2 + c. (2.2.20)
Imposing the initial condition shows that c = −1/y 0. Substituting this into ( 2.2.20) and rearranging
terms yields the solution
y= y0
1 − y0x2 .
This is also the solution if y0 = 0 . If y0 < 0, the denominator isn’t zero for any value of x, so the the
solution is valid on (−∞ , ∞ ). If y0 >0, the solution is valid only on (−1/ √y0, 1/ √y0).
2.2 Exercises
In Exercises 1– 6 find all solutions.
1. y′= 3x2 + 2 x+ 1
y− 2 2. (sin x)(sin y) + (cos y)y′= 0
3. xy′+ y2 + y= 0 4. y′ln |y|+ x2y= 0
5. (3y3 + 3 ycos y+ 1) y′+ (2x+ 1) y
1 + x2 = 0
6. x2yy′= ( y2 − 1)3/ 2
In Exercises 7– 10 find all solutions. Also, plot a direction field and some integ ral curves on the indicated
rectangular region.",Elementary Differential Equations with Boundary Value Problems.pdf
"In Exercises 7– 10 find all solutions. Also, plot a direction field and some integ ral curves on the indicated
rectangular region.
7. C/G y′= x2(1 + y2 ); {−1 ≤ x≤ 1, −1 ≤ y≤ 1}
8. C/G y′(1 + x2) + xy= 0; {−2 ≤ x≤ 2, −1 ≤ y ≤ 1}
9. C/G y′= ( x− 1)(y− 1)(y− 2); {−2 ≤ x≤ 2, −3 ≤ y≤ 3}
10. C/G (y− 1)2y′= 2 x+ 3; {−2 ≤ x≤ 2, −2 ≤ y≤ 5}
In Exercises 11 and 12 solve the initial value problem.
11. y′= x2 + 3 x+ 2
y− 2 , y (1) = 4
12. y′+ x(y2 + y) = 0 , y (2) = 1
In Exercises 13-16 solve the initial value problem and graph the solution.
13. C/G (3y2 + 4 y)y′+ 2 x+ cos x= 0 , y (0) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.2 Separable Equations 53
14. C/G y′+ (y+ 1)( y− 1)(y− 2)
x+ 1 = 0 , y (1) = 0
15. C/G y′+ 2 x(y+ 1) = 0 , y (0) = 2
16. C/G y′= 2 xy(1 + y2 ), y (0) = 1
In Exercises 17– 23 solve the initial value problem and find the interval of valid ity of the solution.
17. y′(x2 + 2) + 4 x(y2 + 2 y+ 1) = 0 , y (1) = −1
18. y′= −2x(y2 − 3y+ 2) , y (0) = 3
19. y′= 2x
1 + 2 y, y (2) = 0 20. y′= 2 y− y2 , y (0) = 1
21. x+ yy′= 0 , y (3) = −4
22. y′+ x2(y+ 1)( y− 2)2 = 0 , y (4) = 2
23. (x+ 1)( x− 2)y′+ y= 0 , y (1) = −3
24. Solve y′= (1 + y2 )
(1 + x2) explicitly . H IN T : Use the identity tan(A+ B) = tan A+ tan B
1 − tan Atan B.
25. Solve y′√
1 − x2 +
√
1 − y2 = 0 explicitly . H IN T : Use the identity sin(A− B) = sin Acos B−
cos Asin B.
26. Solve y′= cos x
sin y, y (π) = π
2 explicitly . H IN T : Use the identity cos(x+ π/ 2) = − sin xand the
periodicity of the cosine.
27. Solve the initial value problem
y′= ay− by2, y (0) = y0.
Discuss the behavior of the solution if (a) y0 ≥ 0; (b) y0 <0.",Elementary Differential Equations with Boundary Value Problems.pdf
"27. Solve the initial value problem
y′= ay− by2, y (0) = y0.
Discuss the behavior of the solution if (a) y0 ≥ 0; (b) y0 <0.
28. The population P = P(t) of a species satisfies the logistic equation
P′= aP(1 − αP )
and P(0) = P0 >0. Find P for t> 0, and find limt→∞ P(t).
29. An epidemic spreads through a population at a rate proportio nal to the product of the number of
people already infected and the number of people susceptibl e, but not yet infected. Therefore, if
Sdenotes the total population of susceptible people and I = I(t) denotes the number of infected
people at time t, then
I′= rI(S− I),
where r is a positive constant. Assuming that I(0) = I0, find I(t) for t >0, and show that
limt→∞ I(t) = S.
30. L The result of Exercise 29 is discouraging: if any susceptible member of the group is in itially
infected, then in the long run all susceptible members are in fected! On a more hopeful note,",Elementary Differential Equations with Boundary Value Problems.pdf
"infected, then in the long run all susceptible members are in fected! On a more hopeful note,
suppose the disease spreads according to the model of Exerci se 29, but there’s a medication that
cures the infected population at a rate proportional to the n umber of infected individuals. Now the
equation for the number of infected individuals becomes
I′= rI(S− I) − qI (A)
where qis a positive constant.",Elementary Differential Equations with Boundary Value Problems.pdf
"54 Chapter 2 First Order Equations
(a) Choose rand Spositive. By plotting direction fields and solutions of (A) o n suitable rectan-
gular grids
R= {0 ≤ t≤ T, 0 ≤ I ≤ d}
in the (t,I )-plane, verify that if Iis any solution of (A) such that I(0) >0, then limt→∞ I(t) =
S− q/r if q <rSand limt→∞ I(t) = 0 if q≥ rS.
(b) T o verify the experimental results of (a), use separation of variables to solve (A) with initial
condition I(0) = I0 > 0, and find limt→∞ I(t). H IN T : There are three cases to consider:
(i) q <rS; (ii) q>rS ; (iii) q= rS.
31. L Consider the differential equation
y′= ay− by2 − q, (A)
where a, bare positive constants, and qis an arbitrary constant. Suppose ydenotes a solution of
this equation that satisfies the initial condition y(0) = y0.
(a) Choose aand bpositive and q <a2/ 4b. By plotting direction fields and solutions of (A) on
suitable rectangular grids
R= {0 ≤ t≤ T, c ≤ y≤ d} (B)
in the (t,y )-plane, discover that there are numbers y1 and y2 with y1 < y2 such that if",Elementary Differential Equations with Boundary Value Problems.pdf
"suitable rectangular grids
R= {0 ≤ t≤ T, c ≤ y≤ d} (B)
in the (t,y )-plane, discover that there are numbers y1 and y2 with y1 < y2 such that if
y0 >y1 then limt→∞ y(t) = y2, and if y0 <y1 then y(t) = −∞ for some finite value of t.
(What happens if y0 = y1?)
(b) Choose a and b positive and q = a2/ 4b. By plotting direction fields and solutions of (A)
on suitable rectangular grids of the form (B), discover that there’s a number y1 such that if
y0 ≥ y1 then limt→∞ y(t) = y1, while if y0 <y1 then y(t) = −∞ for some finite value of
t.
(c) Choose positive a, b and q > a2/ 4b. By plotting direction fields and solutions of (A) on
suitable rectangular grids of the form (B), discover that no matter what y0 is, y(t) = −∞ for
some finite value of t.
(d) V erify your results experiments analytically . Start by sep arating variables in (A) to obtain
y′
ay− by2 − q = 1 .
T o decide what to do next you’ll have to use the quadratic form ula. This should lead you to",Elementary Differential Equations with Boundary Value Problems.pdf
"y′
ay− by2 − q = 1 .
T o decide what to do next you’ll have to use the quadratic form ula. This should lead you to
see why there are three cases. T ake it from there!
Because of its role in the transition between these three cas es, q0 = a2/ 4bis called a
bifur-
cation value of q. In general, if qis a parameter in any differential equation, q0 is said to be a
bifurcation value of qif the nature of the solutions of the equation with q <q0 is qualitatively
different from the nature of the solutions with q >q0.
32. L By plotting direction fields and solutions of
y′= qy− y3 ,
convince yourself that q0 = 0 is a bifurcation value of q for this equation. Explain what makes
you draw this conclusion.
33. Suppose a disease spreads according to the model of Exercise 29, but there’s a medication that
cures the infected population at a constant rate of qindividuals per unit time, where q >0. Then
the equation for the number of infected individuals becomes
I′= rI(S− I) − q.",Elementary Differential Equations with Boundary Value Problems.pdf
"the equation for the number of infected individuals becomes
I′= rI(S− I) − q.
Assuming that I(0) = I0 >0, use the results of Exercise 31 to describe what happens as t→ ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equatio ns 55
34. Assuming that p̸≡0, state conditions under which the linear equation
y′+ p(x)y= f(x)
is separable. If the equation satisfies these conditions, so lve it by separation of variables and by
the method developed in Section 2.1.
Solve the equations in Exercises 35– 38 using variation of parameters followed by separation of var iables.
35. y′+ y= 2xe−x
1 + yex
36. xy′− 2y= x6
y+ x2
37. y′− y= (x+ 1) e4x
(y+ ex)2 38. y′− 2y= xe2x
1 − ye−2x
39. Use variation of parameters to show that the solutions of the following equations are of the form
y= uy1, where usatisfies a separable equation u′= g(x)p(u). Find y1 and gfor each equation.
(a) xy′+ y = h(x)p(xy) (b) xy′− y= h(x)p
(y
x
)
(c) y′+ y = h(x)p(ex y) (d) xy′+ ry = h(x)p(xr y)
(e) y′+ v′(x)
v(x) y = h(x)p(v(x)y)
2.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUA TIO NS",Elementary Differential Equations with Boundary Value Problems.pdf
"(e) y′+ v′(x)
v(x) y = h(x)p(v(x)y)
2.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUA TIO NS
Although there are methods for solving some nonlinear equat ions, it’s impossible to find useful formulas
for the solutions of most. Whether we’re looking for exact so lutions or numerical approximations, it’s
useful to know conditions that imply the existence and uniqu eness of solutions of initial value problems
for nonlinear equations. In this section we state such a cond ition and illustrate it with examples.
 y
 x
 a  b
 c
 d
Figure 2.3.1 An open rectangle",Elementary Differential Equations with Boundary Value Problems.pdf
"56 Chapter 2 First Order Equations
Some terminology: an open rectangle Ris a set of points (x,y ) such that
a<x<b and c<y <d
(Figure 2.3.1). W e’ll denote this set by R : {a < x < b,c < y < d}. “Open” means that the boundary
rectangle (indicated by the dashed lines in Figure 2.3.1) isn’t included in R.
The next theorem gives sufficient conditions for existence a nd uniqueness of solutions of initial value
problems for first order nonlinear differential equations. W e omit the proof, which is beyond the scope of
this book.
Theorem 2.3.1
(a) If f is continuous on an open rectangle
R: {a<x<b,c<y<d }
that contains (x0,y 0) then the initial value problem
y′= f(x,y ), y (x0) = y0 (2.3.1)
has at least one solution on some open subinterval of (a,b ) that contains x0.
(b) If both f and fy are continuous on Rthen (2.3.1) has a unique solution on some open subinterval
of (a,b ) that contains x0.
It’s important to understand exactly what Theorem 2.3.1 says.",Elementary Differential Equations with Boundary Value Problems.pdf
"of (a,b ) that contains x0.
It’s important to understand exactly what Theorem 2.3.1 says.
• (a) is an existence theorem . It guarantees that a solution exists on some open interval t hat contains
x0, but provides no information on how to find the solution, or to determine the open interval on
which it exists. Moreover, (a) provides no information on the number of solutions that ( 2.3.1) may
have. It leaves open the possibility that ( 2.3.1) may have two or more solutions that differ for values
of xarbitrarily close to x0. W e will see in Example 2.3.6 that this can happen.
• (b) is a uniqueness theorem . It guarantees that ( 2.3.1) has a unique solution on some open interval
(a,b) that contains x0. However, if (a,b ) ̸= ( −∞ , ∞ ), ( 2.3.1) may have more than one solution on
a larger interval that contains (a,b ). For example, it may happen that b< ∞ and all solutions have
the same values on (a,b ), but two solutions y1 and y2 are defined on some interval (a,b 1) with",Elementary Differential Equations with Boundary Value Problems.pdf
"the same values on (a,b ), but two solutions y1 and y2 are defined on some interval (a,b 1) with
b1 >b, and have different values for b<x<b 1; thus, the graphs of the y1 and y2 “branch off” in
different directions at x= b. (See Example 2.3.7 and Figure 2.3.3). In this case, continuity implies
that y1(b) = y2(b) (call their common value y), and y1 and y2 are both solutions of the initial value
problem
y′= f(x,y ), y (b) = y (2.3.2)
that differ on every open interval that contains b. Therefore f or fy must have a discontinuity at
some point in each open rectangle that contains (b, y), since if this were not so, ( 2.3.2) would have
a unique solution on some open interval that contains b. W e leave it to you to give a similar analysis
of the case where a> −∞ .
Example 2.3.1 Consider the initial value problem
y′= x2 − y2
1 + x2 + y2 , y (x0 ) = y0. (2.3.3)
Since
f(x,y ) = x2 − y2
1 + x2 + y2 and fy (x,y ) = − 2y(1 + 2 x2)
(1 + x2 + y2)2",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equatio ns 57
are continuous for all (x,y ), Theorem 2.3.1 implies that if (x0,y 0) is arbitrary , then ( 2.3.3) has a unique
solution on some open interval that contains x0.
Example 2.3.2 Consider the initial value problem
y′= x2 − y2
x2 + y2 , y (x0 ) = y0. (2.3.4)
Here
f(x,y ) = x2 − y2
x2 + y2 and fy (x,y ) = − 4x2y
(x2 + y2 )2
are continuous everywhere except at (0, 0). If (x0,y 0) ̸= (0 , 0), there’s an open rectangle Rthat contains
(x0,y 0) that does not contain (0, 0). Since f and fy are continuous on R, Theorem 2.3.1 implies that if
(x0,y 0) ̸= (0 , 0) then ( 2.3.4) has a unique solution on some open interval that contains x0.
Example 2.3.3 Consider the initial value problem
y′= x+ y
x− y, y (x0) = y0. (2.3.5)
Here
f(x,y ) = x+ y
x− y and fy (x,y ) = 2x
(x− y)2
are continuous everywhere except on the line y= x. If y0 ̸= x0, there’s an open rectangle Rthat contains",Elementary Differential Equations with Boundary Value Problems.pdf
"x− y and fy (x,y ) = 2x
(x− y)2
are continuous everywhere except on the line y= x. If y0 ̸= x0, there’s an open rectangle Rthat contains
(x0,y 0) that does not intersect the line y= x. Since fand fy are continuous on R, Theorem 2.3.1 implies
that if y0 ̸= x0, ( 2.3.5) has a unique solution on some open interval that contains x0.
Example 2.3.4 In Example 2.2.4 we saw that the solutions of
y′= 2 xy2 (2.3.6)
are
y≡ 0 and y= − 1
x2 + c,
where cis an arbitrary constant. In particular, this implies that n o solution of ( 2.3.6) other than y ≡ 0 can
equal zero for any value of x. Show that Theorem 2.3.1(b) implies this.
Solution W e’ll obtain a contradiction by assuming that ( 2.3.6) has a solution y1 that equals zero for some
value of x, but isn’t identically zero. If y1 has this property , there’s a point x0 such that y1(x0) = 0 , but
y1(x) ̸= 0 for some value of xin every open interval that contains x0. This means that the initial value
problem
y′= 2 xy2, y (x0) = 0 (2.3.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"y1(x) ̸= 0 for some value of xin every open interval that contains x0. This means that the initial value
problem
y′= 2 xy2, y (x0) = 0 (2.3.7)
has two solutions y ≡ 0 and y = y1 that differ for some value of xon every open interval that contains
x0. This contradicts Theorem 2.3.1(b), since in ( 2.3.6) the functions
f(x,y ) = 2 xy2 and fy (x,y ) = 4 xy.
are both continuous for all (x,y ), which implies that ( 2.3.7) has a unique solution on some open interval
that contains x0.",Elementary Differential Equations with Boundary Value Problems.pdf
"58 Chapter 2 First Order Equations
Example 2.3.5 Consider the initial value problem
y′= 10
3 xy2/ 5, y (x0 ) = y0. (2.3.8)
(a) For what points (x0,y 0) does Theorem 2.3.1(a) imply that ( 2.3.8) has a solution?
(b) For what points (x0,y 0) does Theorem 2.3.1(b) imply that ( 2.3.8) has a unique solution on some
open interval that contains x0?
SO LU TIO N (a) Since
f(x,y ) = 10
3 xy2/ 5
is continuous for all (x,y ), Theorem 2.3.1 implies that ( 2.3.8) has a solution for every (x0,y 0).
SO LU TIO N (b) Here
fy (x,y ) = 4
3 xy−3/ 5
is continuous for all (x,y ) with y ̸= 0 . Therefore, if y0 ̸= 0 there’s an open rectangle on which both f
and fy are continuous, and Theorem 2.3.1 implies that ( 2.3.8) has a unique solution on some open interval
that contains x0.
If y= 0 then fy (x,y ) is undefined, and therefore discontinuous; hence, Theorem 2.3.1 does not apply
to ( 2.3.8) if y0 = 0 .
Example 2.3.6 Example 2.3.5 leaves open the possibility that the initial value problem
y′= 10",Elementary Differential Equations with Boundary Value Problems.pdf
"to ( 2.3.8) if y0 = 0 .
Example 2.3.6 Example 2.3.5 leaves open the possibility that the initial value problem
y′= 10
3 xy2/ 5, y (0) = 0 (2.3.9)
has more than one solution on every open interval that contai ns x0 = 0 . Show that this is true.
Solution By inspection, y≡ 0 is a solution of the differential equation
y′= 10
3 xy2/ 5. (2.3.10)
Since y≡ 0 satisfies the initial condition y(0) = 0 , it’s a solution of ( 2.3.9).
Now suppose y is a solution of ( 2.3.10) that isn’t identically zero. Separating variables in ( 2.3.10)
yields
y−2/ 5y′= 10
3 x
on any open interval where yhas no zeros. Integrating this and rewriting the arbitrary c onstant as 5c/ 3
yields
5
3 y3/ 5 = 5
3 (x2 + c).
Therefore
y = ( x2 + c)5/ 3. (2.3.11)
Since we divided by yto separate variables in ( 2.3.10), our derivation of ( 2.3.11) is legitimate only on
open intervals where y has no zeros. However, ( 2.3.11) actually defines y for all x, and differentiating
(2.3.11) shows that
y′= 10
3 x(x2 + c)2/ 3 = 10",Elementary Differential Equations with Boundary Value Problems.pdf
"(2.3.11) shows that
y′= 10
3 x(x2 + c)2/ 3 = 10
3 xy2/ 5, −∞ <x< ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equatio ns 59
 x
 y
Figure 2.3.2 T wo solutions ( y= 0 and y = x1/ 2) of ( 2.3.9) that differ on every interval containing
x0 = 0
Therefore ( 2.3.11) satisfies ( 2.3.10) on (−∞ , ∞ ) even if c ≤ 0, so that y(
√
|c|) = y(−
√
|c|) = 0 . In
particular, taking c= 0 in ( 2.3.11) yields
y= x10/ 3
as a second solution of ( 2.3.9). Both solutions are defined on (−∞ , ∞ ), and they differ on every open
interval that contains x0 = 0 (see Figure 2.3.2.) In fact, there are four distinct solutions of ( 2.3.9) defined
on (−∞ , ∞ ) that differ from each other on every open interval that conta ins x0 = 0 . Can you identify
the other two?
Example 2.3.7 From Example
2.3.5, the initial value problem
y′= 10
3 xy2/ 5, y (0) = −1 (2.3.12)
has a unique solution on some open interval that contains x0 = 0 . Find a solution and determine the
largest open interval (a,b ) on which it’s unique.",Elementary Differential Equations with Boundary Value Problems.pdf
"largest open interval (a,b ) on which it’s unique.
Solution Let ybe any solution of ( 2.3.12). Because of the initial condition y(0) = −1 and the continuity
of y, there’s an open interval I that contains x0 = 0 on which yhas no zeros, and is consequently of the
form ( 2.3.11). Setting x= 0 and y= −1 in ( 2.3.11) yields c= −1, so
y= ( x2 − 1)5/ 3 (2.3.13)
for xin I. Therefore every solution of ( 2.3.12) differs from zero and is given by ( 2.3.13) on (−1, 1);
that is, ( 2.3.13) is the unique solution of ( 2.3.12) on (−1, 1). This is the largest open interval on which
(2.3.12) has a unique solution. T o see this, note that ( 2.3.13) is a solution of ( 2.3.12) on (−∞ , ∞ ). From",Elementary Differential Equations with Boundary Value Problems.pdf
"60 Chapter 2 First Order Equations
Exercise 2.2. 15, there are infinitely many other solutions of ( 2.3.12) that differ from ( 2.3.13) on every
open interval larger than (−1, 1). One such solution is
y =
{
(x2 − 1)5/ 3, −1 ≤ x≤ 1,
0, |x|>1.
(Figure 2.3.3).
1−1  x
 y
(0, −1)
Figure 2.3.3 T wo solutions of ( 2.3.12) on (−∞ , ∞ )
that coincide on (−1, 1), but on no larger open
interval
 x
 y
(0,1)
Figure 2.3.4 The unique solution of ( 2.3.14)
Example 2.3.8 From Example 2.3.5, the initial value problem
y′= 10
3 xy2/ 5, y (0) = 1 (2.3.14)
has a unique solution on some open interval that contains x0 = 0 . Find the solution and determine the
largest open interval on which it’s unique.
Solution Let ybe any solution of (
2.3.14). Because of the initial condition y(0) = 1 and the continuity
of y, there’s an open interval I that contains x0 = 0 on which yhas no zeros, and is consequently of the
form ( 2.3.11). Setting x= 0 and y= 1 in ( 2.3.11) yields c= 1 , so
y= ( x2 + 1) 5/ 3 (2.3.15)",Elementary Differential Equations with Boundary Value Problems.pdf
"form ( 2.3.11). Setting x= 0 and y= 1 in ( 2.3.11) yields c= 1 , so
y= ( x2 + 1) 5/ 3 (2.3.15)
for xin I. Therefore every solution of ( 2.3.14) differs from zero and is given by ( 2.3.15) on (−∞ , ∞ );
that is, ( 2.3.15) is the unique solution of ( 2.3.14) on (−∞ , ∞ ). Figure 2.3.4 shows the graph of this
solution.
2.3 Exercises
In Exercises 1-13 find all (x0,y 0) for which Theorem 2.3.1 implies that the initial value problem y′=
f(x,y ), y(x0 ) = y0 has (a) a solution (b) a unique solution on some open interval that contains x0.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equatio ns 61
1. y′= x2 + y2
sin x
2. y′= ex + y
x2 + y2
3. y′= tan xy 4. y′= x2 + y2
ln xy
5. y′= ( x2 + y2)y1/ 3
6. y′= 2 xy
7. y′= ln(1 + x2 + y2) 8. y′= 2x+ 3 y
x− 4y
9. y′= ( x2 + y2)1/ 2 10. y′= x(y2 − 1)2/ 3
11. y′= ( x2 + y2)2 12. y′= ( x+ y)1/ 2
13. y′= tan y
x− 1
14. Apply Theorem 2.3.1 to the initial value problem
y′+ p(x)y = q(x), y (x0 ) = y0
for a linear equation, and compare the conclusions that can b e drawn from it to those that follow
from Theorem 2.1.2.
15. (a) V erify that the function
y=
{
(x2 − 1)5/ 3, −1 <x< 1,
0, |x| ≥1,
is a solution of the initial value problem
y′= 10
3 xy2/ 5, y (0) = −1
on (−∞ , ∞ ). H IN T : Y ou’ll need the definition
y′(x) = lim
x→ x
y(x) − y(x)
x− x
to verify that ysatisfies the differential equation at x= ±1.
(b) V erify that if ϵi = 0 or 1 for i= 1 , 2 and a, b> 1, then the function
y=



















ϵ1(x2 − a2)5/ 3, −∞ <x< −a,
0, −a≤ x≤ − 1,",Elementary Differential Equations with Boundary Value Problems.pdf
"y=



















ϵ1(x2 − a2)5/ 3, −∞ <x< −a,
0, −a≤ x≤ − 1,
(x2 − 1)5/ 3, −1 <x< 1,
0, 1 ≤ x≤ b,
ϵ2(x2 − b2)5/ 3, b<x< ∞ ,
is a solution of the initial value problem of (a) on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"62 Chapter 2 First Order Equations
16. Use the ideas developed in Exercise 15 to find infinitely many solutions of the initial value problem
y′= y2/ 5 , y (0) = 1
on (−∞ , ∞ ).
17. Consider the initial value problem
y′= 3 x(y− 1)1/ 3, y (x0) = y0. (A)
(a) For what points (x0,y 0) does Theorem 2.3.1 imply that (A) has a solution?
(b) For what points (x0,y 0) does Theorem 2.3.1 imply that (A) has a unique solution on some
open interval that contains x0?
18. Find nine solutions of the initial value problem
y′= 3 x(y− 1)1/ 3, y (0) = 1
that are all defined on (−∞ , ∞ ) and differ from each other for values of xin every open interval
that contains x0 = 0 .
19. From Theorem 2.3.1, the initial value problem
y′= 3 x(y− 1)1/ 3, y (0) = 9
has a unique solution on an open interval that contains x0 = 0 . Find the solution and determine
the largest open interval on which it’s unique.
20. (a) From Theorem 2.3.1, the initial value problem
y′= 3 x(y− 1)1/ 3, y (3) = −7 (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"the largest open interval on which it’s unique.
20. (a) From Theorem 2.3.1, the initial value problem
y′= 3 x(y− 1)1/ 3, y (3) = −7 (A)
has a unique solution on some open interval that contains x0 = 3 . Determine the largest such
open interval, and find the solution on this interval.
(b) Find infinitely many solutions of (A), all defined on (−∞ , ∞ ).
21. Prove:
(a) If
f(x,y 0) = 0 , a<x<b, (A)
and x0 is in (a,b ), then y≡ y0 is a solution of
y′= f(x,y ), y (x0) = y0
on (a,b ).
(b) If f and fy are continuous on an open rectangle that contains (x0,y 0) and (A) holds, no
solution of y′= f(x,y ) other than y≡ y0 can equal y0 at any point in (a,b ).
2.4 TRANSFORMA TION OF NONLINEAR EQUA TIONS INTO SEP ARABLE EQUA TIONS
In Section 2.1 we found that the solutions of a linear nonhomo geneous equation
y′+ p(x)y= f(x)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.4 Transformation of Nonlinear Equations into Separable Equa tions 63
are of the form y= uy1, where y1 is a nontrivial solution of the complementary equation
y′+ p(x)y= 0 (2.4.1)
and uis a solution of
u′y1(x) = f(x).
Note that this last equation is separable, since it can be rew ritten as
u′= f(x)
y1(x) .
In this section we’ll consider nonlinear differential equa tions that are not separable to begin with, but can
be solved in a similar fashion by writing their solutions in t he form y = uy1, where y1 is a suitably chosen
known function and usatisfies a separable equation. W e’llsay in this case that we transformed the given
equation into a separable equation.
Bernoulli Equations
A Bernoulli equation is an equation of the form
y′+ p(x)y = f(x)yr , (2.4.2)
where r can be any real number other than 0 or 1. (Note that ( 2.4.2) is linear if and only if r = 0 or
r= 1 .) W e can transform ( 2.4.2) into a separable equation by variation of parameters: if y1 is a nontrivial",Elementary Differential Equations with Boundary Value Problems.pdf
"r= 1 .) W e can transform ( 2.4.2) into a separable equation by variation of parameters: if y1 is a nontrivial
solution of ( 2.4.1), substituting y= uy1 into ( 2.4.2) yields
u′y1 + u(y′
1 + p(x)y1) = f(x)(uy1)r ,
which is equivalent to the separable equation
u′y1(x) = f(x) ( y1(x))r ur or u′
ur = f(x) ( y1(x))r−1 ,
since y′
1 + p(x)y1 = 0 .
Example 2.4.1 Solve the Bernoulli equation
y′− y= xy2. (2.4.3)
Solution Since y1 = ex is a solution of y′− y = 0 , we look for solutions of ( 2.4.3) in the form y= uex,
where
u′ex = xu2e2x or, equivalently , u′= xu2ex.
Separating variables yields
u′
u2 = xex,
and integrating yields
− 1
u = ( x− 1)ex + c.
Hence,
u= − 1
(x− 1)ex + c",Elementary Differential Equations with Boundary Value Problems.pdf
"64 Chapter 2 First Order Equations
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 x
 y
Figure 2.4.1 A direction field and integral curves for y′− y= xy2
and
y= − 1
x− 1 + ce−x .
Figure 2.4.1 shows direction field and some integral curves of ( 2.4.3).
Other Nonlinear Equations That Can be Transformed Into Sepa rable Equations
W e’ve seen that the nonlinear Bernoulli equation can be tran sformed into a separable equation by the
substitution y = uy1 if y1 is suitably chosen. Now let’s discover a sufficient conditio n for a nonlinear
first order differential equation
y′= f(x,y ) (2.4.4)
to be transformable into a separable equation in the same way . Substituting y= uy1 into ( 2.4.4) yields
u′y1(x) + uy′
1(x) = f(x,uy 1(x)),
which is equivalent to
u′y1(x) = f(x,uy 1(x)) − uy′
1(x). (2.4.5)
If
f(x,uy 1(x)) = q(u)y′
1(x)
for some function q, then ( 2.4.5) becomes
u′y1(x) = ( q(u) − u)y′
1(x), (2.4.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"1(x). (2.4.5)
If
f(x,uy 1(x)) = q(u)y′
1(x)
for some function q, then ( 2.4.5) becomes
u′y1(x) = ( q(u) − u)y′
1(x), (2.4.6)
which is separable. After checking for constant solutions u≡ u0 such that q(u0) = u0, we can separate
variables to obtain
u′
q(u) − u = y′
1(x)
y1(x) .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.4 Transformation of Nonlinear Equations into Separable Equa tions 65
Homogeneous Nonlinear Equations
In the text we’ll consider only the most widely studied class of equations for which the method of the
preceding paragraph works. Other types of equations appear in Exercises 44– 51.
The differential equation ( 2.4.4) is said to be homogeneous if xand y occur in f in such a way that
f(x,y ) depends only on the ratio y/x ; that is, ( 2.4.4) can be written as
y′= q(y/x ), (2.4.7)
where q= q(u) is a function of a single variable. For example,
y′= y+ xe−y/x
x = y
x + e−y/x
and
y′= y2 + xy− x2
x2 =
(y
x
) 2
+ y
x − 1
are of the form ( 2.4.7), with
q(u) = u+ e−u and q(u) = u2 + u− 1,
respectively . The general method discussed above can be app lied to ( 2.4.7) with y1 = x(and therefore
y′
1 = 1) . Thus, substituting y= uxin ( 2.4.7) yields
u′x+ u= q(u),
and separation of variables (after checking for constant so lutions u≡ u0 such that q(u0) = u0) yields
u′
q(u) − u = 1
x.",Elementary Differential Equations with Boundary Value Problems.pdf
"u′x+ u= q(u),
and separation of variables (after checking for constant so lutions u≡ u0 such that q(u0) = u0) yields
u′
q(u) − u = 1
x.
Before turning to examples, we point out something that you m ay’ve have already noticed: the defini-
tion of homogeneous equation given here isn’t the same as the definition given in Section 2. 1, where we
said that a linear equation of the form
y′+ p(x)y= 0
is homogeneous. W e make no apology for this inconsistency , s ince we didn’t create it historically , homo-
geneous has been used in these two inconsistent ways. The one having t o do with linear equations is the
most important. This is the only section of the book where the meaning defined here will apply .
Since y/x is in general undefined if x = 0 , we’ll consider solutions of nonhomogeneous equations
only on open intervals that do not contain the point x= 0 .
Example 2.4.2 Solve
y′= y+ xe−y/x
x . (2.4.8)
Solution Substituting y= uxinto ( 2.4.8) yields
u′x+ u= ux+ xe−ux/x
x = u+ e−u.",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 2.4.2 Solve
y′= y+ xe−y/x
x . (2.4.8)
Solution Substituting y= uxinto ( 2.4.8) yields
u′x+ u= ux+ xe−ux/x
x = u+ e−u.
Simplifying and separating variables yields
euu′= 1
x.",Elementary Differential Equations with Boundary Value Problems.pdf
"66 Chapter 2 First Order Equations
Integrating yields eu = ln |x|+ c. Therefore u= ln(ln |x|+ c) and y= ux= xln(ln |x|+ c).
Figure 2.4.2 shows a direction field and integral curves for ( 2.4.8).
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 x
 y
Figure 2.4.2 A direction field and some integral curves for y′= y+ xe−y/x
x
Example 2.4.3
(a) Solve
x2y′= y2 + xy− x2. (2.4.9)
(b) Solve the initial value problem
x2y′= y2 + xy− x2, y (1) = 2 . (2.4.10)
SO LU TIO N (a) W e first find solutions of ( 2.4.9) on open intervals that don’t contain x = 0 . W e can
rewrite ( 2.4.9) as
y′= y2 + xy− x2
x2
for xin any such interval. Substituting y = uxyields
u′x+ u= (ux)2 + x(ux) − x2
x2 = u2 + u− 1,
so
u′x= u2 − 1. (2.4.11)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.4 Transformation of Nonlinear Equations into Separable Equa tions 67
By inspection this equation has the constant solutions u≡ 1 and u≡ − 1. Therefore y = xand y = −x
are solutions of ( 2.4.9). If uis a solution of ( 2.4.11) that doesn’t assume the values ±1 on some interval,
separating variables yields
u′
u2 − 1 = 1
x,
or, after a partial fraction expansion,
1
2
[ 1
u− 1 − 1
u+ 1
]
u′= 1
x.
Multiplying by 2 and integrating yields
ln
⏐
⏐
⏐
⏐
u− 1
u+ 1
⏐
⏐
⏐
⏐= 2 ln |x|+ k,
or ⏐
⏐
⏐
⏐
u− 1
u+ 1
⏐
⏐
⏐
⏐= ek x2,
which holds if u− 1
u+ 1 = cx2 (2.4.12)
where cis an arbitrary constant. Solving for uyields
u= 1 + cx2
1 − cx2 .
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 y
 x
Figure 2.4.3 A direction field and integral curves for
x2y′= y2 + xy− x2
 x
 y
1
2
Figure 2.4.4 Solutions of x2y′= y2 + xy− x2,
y(1) = 2
Therefore
y= ux= x(1 + cx2)
1 − cx2 (2.4.13)
is a solution of ( 2.4.10) for any choice of the constant c. Setting c = 0 in ( 2.4.13) yields the solution",Elementary Differential Equations with Boundary Value Problems.pdf
"y= ux= x(1 + cx2)
1 − cx2 (2.4.13)
is a solution of ( 2.4.10) for any choice of the constant c. Setting c = 0 in ( 2.4.13) yields the solution
y = x. However, the solution y = −xcan’t be obtained from ( 2.4.13). Thus, the solutions of ( 2.4.9) on
intervals that don’t contain x= 0 are y = −xand functions of the form ( 2.4.13).",Elementary Differential Equations with Boundary Value Problems.pdf
"68 Chapter 2 First Order Equations
The situation is more complicated if x= 0 is the open interval. First, note that y = −xsatisfies ( 2.4.9)
on (−∞ , ∞ ). If c1 and c2 are arbitrary constants, the function
y=







x(1 + c1x2)
1 − c1x2 , a<x< 0,
x(1 + c2x2)
1 − c2x2 , 0 ≤ x<b,
(2.4.14)
is a solution of ( 2.4.9) on (a,b ), where
a=



− 1
√c1
if c1 >0,
−∞ if c1 ≤ 0,
and b=



1
√c2
if c2 >0,
∞ if c2 ≤ 0.
W e leave it to you to verify this. T o do so, note that if yis any function of the form ( 2.4.13) then y(0) = 0
and y′(0) = 1 .
Figure 2.4.3 shows a direction field and some integral curves for ( 2.4.9).
SO LU TIO N (b) W e could obtain cby imposing the initial condition y(1) = 2 in ( 2.4.13), and then solving
for c. However, it’s easier to use ( 2.4.12). Since u = y/x , the initial condition y(1) = 2 implies that
u(1) = 2 . Substituting this into ( 2.4.12) yields c= 1 / 3. Hence, the solution of ( 2.4.10) is
y = x(1 + x2/ 3)
1 − x2/ 3 .",Elementary Differential Equations with Boundary Value Problems.pdf
"u(1) = 2 . Substituting this into ( 2.4.12) yields c= 1 / 3. Hence, the solution of ( 2.4.10) is
y = x(1 + x2/ 3)
1 − x2/ 3 .
The interval of validity of this solution is (−
√
3,
√
3). However, the largest interval on which ( 2.4.10)
has a unique solution is (0,
√
3). T o see this, note from ( 2.4.14) that any function of the form
y=







x(1 + cx2)
1 − cx2 , a<x ≤ 0,
x(1 + x2/ 3)
1 − x2/ 3 , 0 ≤ x<
√
3,
(2.4.15)
is a solution of ( 2.4.10) on (a,
√
3), where a= −1/ √cif c> 0 or a= −∞ if c≤ 0. (Why doesn’t this
contradict Theorem 2.3.1?)
Figure 2.4.4 shows several solutions of the initial value problem ( 2.4.10). Note that these solutions
coincide on (0,
√
3).
In the last two examples we were able to solve the given equati ons explicitly . However, this isn’t always
possible, as you’ll see in the exercises.
2.4 Exercises
In Exercises 1– 4 solve the given Bernoulli equation.
1. y′+ y= y2 2. 7xy′− 2y= −x2
y6
3. x2y′+ 2 y= 2 e1/x y1/ 2 4. (1 + x2)y′+ 2 xy= 1
(1 + x2)y",Elementary Differential Equations with Boundary Value Problems.pdf
"In Exercises 1– 4 solve the given Bernoulli equation.
1. y′+ y= y2 2. 7xy′− 2y= −x2
y6
3. x2y′+ 2 y= 2 e1/x y1/ 2 4. (1 + x2)y′+ 2 xy= 1
(1 + x2)y
In Exercises 5 and 6 find all solutions. Also, plot a direction field and some integ ral curves on the
indicated rectangular region.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.4 Transformation of Nonlinear Equations into Separable Equa tions 69
5. C/G y′− xy= x3y3; {−3 ≤ x≤ 3, 2 ≤ y≥ 2}
6. C/G y′− 1 + x
3x y= y4; {−2 ≤ x≤ 2, −2 ≤ y≤ 2}
In Exercises 7– 11 solve the initial value problem.
7. y′− 2y= xy3, y (0) = 2
√
2
8. y′− xy= xy3/ 2, y (1) = 4
9. xy′+ y= x4y4 , y (1) = 1 / 2
10. y′− 2y= 2 y1/ 2, y (0) = 1
11. y′− 4y= 48x
y2 , y (0) = 1
In Exercises 12 and 13 solve the initial value problem and graph the solution.
12. C/G x2y′+ 2 xy= y3, y (1) = 1 /
√
2
13. C/G y′− y= xy1/ 2, y (0) = 4
14. Y ou may have noticed that the logistic equation
P′= aP(1 − αP )
from V erhulst’s model for population growth can be written i n Bernoulli form as
P′− aP = −aαP 2.
This isn’t particularly interesting, since the logistic eq uation is separable, and therefore solvable
by the method studied in Section 2.2. So let’s consider a more complicated model, where a is
a positive constant and α is a positive continuous function of ton [0, ∞ ). The equation for this
model is",Elementary Differential Equations with Boundary Value Problems.pdf
"a positive constant and α is a positive continuous function of ton [0, ∞ ). The equation for this
model is
P′− aP = −aα(t)P2,
a non-separable Bernoulli equation.
(a) Assuming that P(0) = P0 >0, find P for t >0. H IN T : Express your result in terms of the
integral ∫t
0 α(τ)eaτ dτ.
(b) V erify that your result reduces to the known results for the M althusian model where α = 0 ,
and the V erhulst model where α is a nonzero constant.
(c) Assuming that
lim
t→∞
e−at
∫ t
0
α(τ)eaτ dτ = L
exists (finite or infinite), find limt→∞ P(t).
In Exercises 15– 18 solve the equation explicitly.
15. y′= y+ x
x
16. y′= y2 + 2 xy
x2
17. xy3y′= y4 + x4
18. y′= y
x + sec y
x",Elementary Differential Equations with Boundary Value Problems.pdf
"70 Chapter 2 First Order Equations
In Exercises 19-21 solve the equation explicitly. Also, plot a direction field a nd some integral curves on
the indicated rectangular region.
19. C/G x2y′= xy+ x2 + y2; {−8 ≤ x≤ 8, −8 ≤ y≤ 8}
20. C/G xyy′= x2 + 2 y2; {−4 ≤ x≤ 4, −4 ≤ y≤ 4}
21. C/G y′= 2y2 + x2e−(y/x )2
2xy ; {−8 ≤ x≤ 8, −8 ≤ y ≤ 8}
In Exercises 22– 27 solve the initial value problem.
22. y′= xy+ y2
x2 , y (−1) = 2
23. y′= x3 + y3
xy2 , y (1) = 3
24. xyy′+ x2 + y2 = 0 , y (1) = 2
25. y′= y2 − 3xy− 5x2
x2 , y (1) = −1
26. x2y′= 2 x2 + y2 + 4 xy, y (1) = 1
27. xyy′= 3 x2 + 4 y2, y (1) =
√
3
In Exercises 28– 34 solve the given homogeneous equation implicitly.
28. y′= x+ y
x− y
29. (y′x− y)(ln |y| −ln |x|) = x
30. y′= y3 + 2 xy2 + x2y+ x3
x(y+ x)2
31. y′= x+ 2 y
2x+ y
32. y′= y
y− 2x 33. y′= xy2 + 2 y3
x3 + x2y+ xy2
34. y′= x3 + x2y+ 3 y3
x3 + 3 xy2
35. L
(a) Find a solution of the initial value problem
x2y′= y2 + xy− 4x2, y (−1) = 0 (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"x3 + x2y+ xy2
34. y′= x3 + x2y+ 3 y3
x3 + 3 xy2
35. L
(a) Find a solution of the initial value problem
x2y′= y2 + xy− 4x2, y (−1) = 0 (A)
on the interval (−∞ , 0). V erify that this solution is actually valid on (−∞ , ∞ ).
(b) Use Theorem 2.3.1 to show that (A) has a unique solution on (−∞ , 0).
(c) Plot a direction field for the differential equation in (A) on a square
{−r≤ x≤ r, −r≤ y≤ r},
where ris any positive number. Graph the solution you obtained in (a) on this field.
(d) Graph other solutions of (A) that are defined on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.4 Transformation of Nonlinear Equations into Separable Equa tions 71
(e) Graph other solutions of (A) that are defined only on interval s of the form (−∞ ,a ), where
is a finite positive number.
36. L
(a) Solve the equation
xyy′= x2 − xy+ y2 (A)
implicitly .
(b) Plot a direction field for (A) on a square
{0 ≤ x≤ r, 0 ≤ y ≤ r}
where ris any positive number.
(c) Let K be a positive integer. (Y ou may have to try several choices fo r K.) Graph solutions of
the initial value problems
xyy′= x2 − xy+ y2 , y (r/ 2) = kr
K,
for k= 1 , 2, . . . , K. Based on your observations, find conditions on the positive numbers x0
and y0 such that the initial value problem
xyy′= x2 − xy+ y2, y (x0) = y0, (B)
has a unique solution (i) on (0, ∞ ) or (ii) only on an interval (a, ∞ ), where a> 0?
(d) What can you say about the graph of the solution of (B) as x → ∞ ? (Again, assume that
x0 >0 and y0 >0.)
37. L
(a) Solve the equation
y′= 2y2 − xy+ 2 x2
xy+ 2 x2 (A)
implicitly .",Elementary Differential Equations with Boundary Value Problems.pdf
"x0 >0 and y0 >0.)
37. L
(a) Solve the equation
y′= 2y2 − xy+ 2 x2
xy+ 2 x2 (A)
implicitly .
(b) Plot a direction field for (A) on a square
{−r≤ x≤ r, −r≤ y≤ r}
where r is any positive number. By graphing solutions of (A), determ ine necessary and
sufficient conditions on (x0,y 0) such that (A) has a solution on (i) (−∞ , 0) or (ii) (0, ∞ )
such that y(x0) = y0.
38. L Follow the instructions of Exercise 37 for the equation
y′= xy+ x2 + y2
xy .
39. L Pick any nonlinear homogeneous equation y′= q(y/x ) you like, and plot direction fields on
the square {−r≤ x≤ r, −r≤ y ≤ r}, where r >0. What happens to the direction field as you
vary r? Why?",Elementary Differential Equations with Boundary Value Problems.pdf
"72 Chapter 2 First Order Equations
40. Prove: If ad− bc̸= 0 , the equation
y′= ax+ by+ α
cx+ dy+ β
can be transformed into the homogeneous nonlinear equation
dY
dX = aX+ bY
cX+ dY
by the substitution x= X− X0, y = Y − Y0, where X0 and Y0 are suitably chosen constants.
In Exercises 41-43 use a method suggested by Exercise 40 to solve the given equation implicitly.
41. y′= −6x+ y− 3
2x− y− 1 42. y′= 2x+ y+ 1
x+ 2 y− 4
43. y′= −x+ 3 y− 14
x+ y− 2
In Exercises 44– 51 find a function y1 such that the substitution y = uy1 transforms the given equation
into a separable equation of the form (2.4.6). Then solve the given equation explicitly.
44. 3xy2y′= y3 + x 45. xyy′= 3 x6 + 6 y2
46. x3y′= 2( y2 + x2y− x4) 47. y′= y2e−x + 4 y+ 2 ex
48. y′= y2 + ytan x+ tan 2 x
sin2 x
49. x(ln x)2y′= −4(ln x)2 + yln x+ y2
50. 2x(y+ 2 √x)y′= ( y+ √x)2 51. (y+ ex2
)y′= 2 x(y2 + yex2
+ e2x2
)
52. Solve the initial value problem
y′+ 2
xy= 3x2y2 + 6 xy+ 2
x2(2xy+ 3) , y (2) = 2 .
53. Solve the initial value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"+ e2x2
)
52. Solve the initial value problem
y′+ 2
xy= 3x2y2 + 6 xy+ 2
x2(2xy+ 3) , y (2) = 2 .
53. Solve the initial value problem
y′+ 3
xy= 3x4y2 + 10 x2y+ 6
x3(2x2y+ 5) , y (1) = 1 .
54. Prove: If yis a solution of a homogeneous nonlinear equation y′= q(y/x ), so is y1 = y(ax)/a ,
where ais any nonzero constant.
55. A generalized Riccati equation is of the form
y′= P(x) + Q(x)y+ R(x)y2 . (A)
(If R≡ − 1, (A) is a Riccati equation .) Let y1 be a known solution and yan arbitrary solution of
(A). Let z= y− y1. Show that zis a solution of a Bernoulli equation with n= 2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.5 Exact Equations 73
In Exercises 56– 59, given that y1 is a solution of the given equation, use the method suggested by Exercise
55 to find other solutions.
56. y′= 1 + x− (1 + 2 x)y+ xy2; y1 = 1
57. y′= e2x + (1 − 2ex)y+ y2 ; y1 = ex
58. xy′= 2 − x+ (2 x− 2)y− xy2; y1 = 1
59. xy′= x3 + (1 − 2x2)y+ xy2; y1 = x
2.5 EXACT EQUA TIONS
In this section it’s convenient to write first order differen tial equations in the form
M(x,y ) dx+ N(x,y ) dy= 0 . (2.5.1)
This equation can be interpreted as
M(x,y ) + N(x,y ) dy
dx = 0 , (2.5.2)
where xis the independent variable and yis the dependent variable, or as
M(x,y ) dx
dy + N(x,y ) = 0 , (2.5.3)
where y is the independent variable and xis the dependent variable. Since the solutions of ( 2.5.2) and
(2.5.3) will often have to be left in implicit, form we’ll say that F(x,y ) = cis an implicit solution of
(2.5.1) if every differentiable function y = y(x) that satisfies F(x,y ) = c is a solution of ( 2.5.2) and",Elementary Differential Equations with Boundary Value Problems.pdf
"(2.5.1) if every differentiable function y = y(x) that satisfies F(x,y ) = c is a solution of ( 2.5.2) and
every differentiable function x= x(y) that satisfies F(x,y ) = cis a solution of ( 2.5.3).
Here are some examples:
Equation ( 2.5.1) Equation ( 2.5.2) Equation ( 2.5.3)
3x2y2 dx+ 2 x3ydy = 0 3x2y2 + 2 x3y dy
dx = 0 3x2y2 dx
dy + 2 x3y= 0
(x2 + y2) dx+ 2 xydy = 0 (x2 + y2 ) + 2 xy dy
dx = 0 (x2 + y2 ) dx
dy + 2 xy= 0
3ysin xdx − 2xycos xdy = 0 3ysin x− 2xycos xdy
dx = 0 3ysin xdx
dy − 2xycos x= 0
Note that a separable equation can be written as ( 2.5.1) as
M(x) dx+ N(y) dy = 0 .
W e’ll develop a method for solving ( 2.5.1) under appropriate assumptions on M and N. This method
is an extension of the method of separation of variables (Exe rcise 41). Before stating it we consider an
example.",Elementary Differential Equations with Boundary Value Problems.pdf
"74 Chapter 2 First Order Equations
Example 2.5.1 Show that
x4y3 + x2y5 + 2 xy= c (2.5.4)
is an implicit solution of
(4x3y3 + 2 xy5 + 2 y) dx+ (3 x4y2 + 5 x2y4 + 2 x) dy= 0 . (2.5.5)
Solution Regarding yas a function of xand differentiating ( 2.5.4) implicitly with respect to xyields
(4x3y3 + 2 xy5 + 2 y) + (3 x4y2 + 5 x2y4 + 2 x) dy
dx = 0 .
Similarly , regarding xas a function of yand differentiating ( 2.5.4) implicitly with respect to yyields
(4x3y3 + 2 xy5 + 2 y) dx
dy + (3 x4y2 + 5 x2y4 + 2 x) = 0 .
Therefore ( 2.5.4) is an implicit solution of ( 2.5.5) in either of its two possible interpretations.
Y ou may think this example is pointless, since concocting a d ifferential equation that has a given
implicit solution isn’t particularly interesting. Howeve r, it illustrates the next important theorem, which
we’ll prove by using implicit differentiation, as in Exampl e 2.5.1.
Theorem 2.5.1 If F = F(x,y ) has continuous partial derivatives Fx and Fy , then",Elementary Differential Equations with Boundary Value Problems.pdf
"we’ll prove by using implicit differentiation, as in Exampl e 2.5.1.
Theorem 2.5.1 If F = F(x,y ) has continuous partial derivatives Fx and Fy , then
F(x,y ) = c (c=constant), (2.5.6)
is an implicit solution of the differential equation
Fx(x,y ) dx+ Fy (x,y ) dy= 0 . (2.5.7)
Proof Regarding yas a function of xand differentiating ( 2.5.6) implicitly with respect to xyields
Fx(x,y ) + Fy (x,y ) dy
dx = 0 .
On the other hand, regarding xas a function of y and differentiating ( 2.5.6) implicitly with respect to y
yields
Fx(x,y ) dx
dy + Fy(x,y ) = 0 .
Thus, ( 2.5.6) is an implicit solution of ( 2.5.7) in either of its two possible interpretations.
W e’ll say that the equation
M(x,y ) dx+ N(x,y ) dy= 0 (2.5.8)
is exact on an an open rectangle Rif there’s a function F = F(x,y ) such Fx and Fy are continuous, and
Fx(x,y ) = M(x,y ) and Fy (x,y ) = N(x,y ) (2.5.9)
for all (x,y ) in R. This usage of “exact” is related to its usage in calculus, wh ere the expression
Fx(x,y ) dx+ Fy (x,y ) dy",Elementary Differential Equations with Boundary Value Problems.pdf
"for all (x,y ) in R. This usage of “exact” is related to its usage in calculus, wh ere the expression
Fx(x,y ) dx+ Fy (x,y ) dy
(obtained by substituting ( 2.5.9) into the left side of ( 2.5.8)) is the exact differential of F.
Example 2.5.1 shows that it’s easy to solve ( 2.5.8) if it’s exact and we know a function F that satisfies
(2.5.9). The important questions are:",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.5 Exact Equations 75
QU ESTIO N 1. Given an equation ( 2.5.8), how can we determine whether it’s exact?
QU ESTIO N 2. If ( 2.5.8) is exact, how do we find a function F satisfying ( 2.5.9)?
T o discover the answer to Question 1, assume that there’s a fu nction F that satisfies ( 2.5.9) on some
open rectangle R, and in addition that F has continuous mixed partial derivatives Fxy and Fyx. Then a
theorem from calculus implies that
Fxy = Fyx . (2.5.10)
If Fx = M and Fy = N, differentiating the first of these equations with respect t o yand the second with
respect to xyields
Fxy = My and Fyx = Nx. (2.5.11)
From ( 2.5.10) and ( 2.5.11), we conclude that a necessary condition for exactness is th at My = Nx . This
motivates the next theorem, which we state without proof.
Theorem 2.5.2
[The Exactness Condition ] Suppose M and N are continuous and have continuous par-
tial derivatives My and Nx on an open rectangle R. Then
M(x,y ) dx+ N(x,y ) dy= 0
is exact on Rif and only if",Elementary Differential Equations with Boundary Value Problems.pdf
"tial derivatives My and Nx on an open rectangle R. Then
M(x,y ) dx+ N(x,y ) dy= 0
is exact on Rif and only if
My (x,y ) = Nx(x,y ) (2.5.12)
for all (x,y ) in R..
T o help you remember the exactness condition, observe that t he coefficients of dxand dy are differ-
entiated in ( 2.5.12) with respect to the “opposite” variables; that is, the coef ficient of dxis differentiated
with respect to y, while the coefficient of dyis differentiated with respect to x.
Example 2.5.2 Show that the equation
3x2ydx + 4 x3 dy= 0
is not exact on any open rectangle.
Solution Here
M(x,y ) = 3 x2y and N(x,y ) = 4 x3
so
My (x,y ) = 3 x2 and Nx(x,y ) = 12 x2.
Therefore My = Nx on the line x= 0 , but not on any open rectangle, so there’s no function F such that
Fx(x,y ) = M(x,y ) and Fy (x,y ) = N(x,y ) for all (x,y ) on any open rectangle.
The next example illustrates two possible methods for findin g a function F that satisfies the condition
Fx = M and Fy = N if Mdx + Ndy = 0 is exact.
Example 2.5.3 Solve",Elementary Differential Equations with Boundary Value Problems.pdf
"Fx = M and Fy = N if Mdx + Ndy = 0 is exact.
Example 2.5.3 Solve
(4x3y3 + 3 x2) dx+ (3 x4y2 + 6 y2 ) dy= 0 . (2.5.13)
Solution (Method 1) Here
M(x,y ) = 4 x3y3 + 3 x2, N (x,y ) = 3 x4y2 + 6 y2,",Elementary Differential Equations with Boundary Value Problems.pdf
"76 Chapter 2 First Order Equations
and
My (x,y ) = Nx (x,y ) = 12 x3y2
for all (x,y ). Therefore Theorem 2.5.2 implies that there’s a function F such that
Fx(x,y ) = M(x,y ) = 4 x3y3 + 3 x2 (2.5.14)
and
Fy(x,y ) = N(x,y ) = 3 x4y2 + 6 y2 (2.5.15)
for all (x,y ). T o find F, we integrate ( 2.5.14) with respect to xto obtain
F(x,y ) = x4y3 + x3 + φ(y), (2.5.16)
where φ(y) is the “constant” of integration. (Here φ is “constant” in that it’s independent of x, the variable
of integration.) If φ is any differentiable function of ythen F satisfies ( 2.5.14). T o determine φ so that F
also satisfies ( 2.5.15), assume that φ is differentiable and differentiate F with respect to y. This yields
Fy (x,y ) = 3 x4y2 + φ′(y).
Comparing this with ( 2.5.15) shows that
φ′(y) = 6 y2.
W e integrate this with respect to yand take the constant of integration to be zero because we’re interested
only in finding some F that satisfies ( 2.5.14) and ( 2.5.15). This yields
φ(y) = 2 y3.",Elementary Differential Equations with Boundary Value Problems.pdf
"only in finding some F that satisfies ( 2.5.14) and ( 2.5.15). This yields
φ(y) = 2 y3.
Substituting this into ( 2.5.16) yields
F(x,y ) = x4y3 + x3 + 2 y3 . (2.5.17)
Now Theorem 2.5.1 implies that
x4y3 + x3 + 2 y3 = c
is an implicit solution of ( 2.5.13). Solving this for yyields the explicit solution
y=
(c− x3
2 + x4
) 1/ 3
.
Solution (Method 2) Instead of first integrating ( 2.5.14) with respect to x, we could begin by integrating
(2.5.15) with respect to yto obtain
F(x,y ) = x4y3 + 2 y3 + ψ (x), (2.5.18)
where ψ is an arbitrary function of x. T o determine ψ , we assume that ψ is differentiable and differentiate
F with respect to x, which yields
Fx(x,y ) = 4 x3y3 + ψ ′(x).
Comparing this with ( 2.5.14) shows that
ψ ′(x) = 3 x2.
Integrating this and again taking the constant of integrati on to be zero yields
ψ (x) = x3.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.5 Exact Equations 77
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 y
 x
Figure 2.5.1 A direction field and integral curves for (4x3y3 + 3 x2) dx+ (3 x4y2 + 6 y2 ) dy= 0
Substituting this into ( 2.5.18) yields ( 2.5.17).
Figure 2.5.1 shows a direction field and some integral curves of ( 2.5.13),
Here’s a summary of the procedure used in Method 1 of this exam ple. Y ou should summarize procedure
used in Method 2.
Procedure For Solving An Exact Equation
Step 1. Check that the equation
M(x,y ) dx+ N(x,y ) dy= 0 (2.5.19)
satisfies the exactness condition My = Nx. If not, don’t go further with this procedure.
Step 2. Integrate
∂F (x,y )
∂x = M(x,y )
with respect to xto obtain
F(x,y ) = G(x,y ) + φ(y), (2.5.20)
where Gis an antiderivative of M with respect to x, and φ is an unknown function of y.
Step 3. Differentiate ( 2.5.20) with respect to yto obtain
∂F (x,y )
∂y = ∂G(x,y )
∂y + φ′(y).",Elementary Differential Equations with Boundary Value Problems.pdf
"Step 3. Differentiate ( 2.5.20) with respect to yto obtain
∂F (x,y )
∂y = ∂G(x,y )
∂y + φ′(y).
Step 4. Equate the right side of this equation to N and solve for φ′; thus,
∂G(x,y )
∂y + φ′(y) = N(x,y ), so φ′(y) = N(x,y ) − ∂G(x,y )
∂y .",Elementary Differential Equations with Boundary Value Problems.pdf
"78 Chapter 2 First Order Equations
Step 5. Integrate φ′with respect to y, taking the constant of integration to be zero, and substitu te the
result in ( 2.5.20) to obtain F(x,y ).
Step 6. Set F(x,y ) = cto obtain an implicit solution of ( 2.5.19). If possible, solve for yexplicitly as a
function of x.
It’s a common mistake to omit Step 6. However, it’s important to include this step, since F isn’t itself
a solution of ( 2.5.19).
Many equations can be conveniently solved by either of the tw o methods used in Example 2.5.3. How-
ever, sometimes the integration required in one approach is more difficult than in the other. In such cases
we choose the approach that requires the easier integration .
Example 2.5.4 Solve the equation
(yexy tan x+ exy sec2 x) dx+ xexy tan xdy = 0 . (2.5.21)
Solution W e leave it to you to check that My = Nx on any open rectangle where tan xand sec xare
defined. Here we must find a function F such that
Fx(x,y ) = yexy tan x+ exy sec2 x (2.5.22)
and",Elementary Differential Equations with Boundary Value Problems.pdf
"defined. Here we must find a function F such that
Fx(x,y ) = yexy tan x+ exy sec2 x (2.5.22)
and
Fy (x,y ) = xexy tan x. (2.5.23)
It’s difficult to integrate ( 2.5.22) with respect to x, but easy to integrate ( 2.5.23) with respect to y. This
yields
F(x,y ) = exy tan x+ ψ (x). (2.5.24)
Differentiating this with respect to xyields
Fx(x,y ) = yexy tan x+ exy sec2 x+ ψ ′(x).
Comparing this with ( 2.5.22) shows that ψ ′(x) = 0 . Hence, ψ is a constant, which we can take to be zero
in ( 2.5.24), and
exy tan x= c
is an implicit solution of ( 2.5.21).
Attempting to apply our procedure to an equation that isn’t e xact will lead to failure in Step 4, since
the function
N − ∂G
∂y
won’t be independent of xif My ̸= Nx (Exercise 31), and therefore can’t be the derivative of a function
of yalone. Here’s an example that illustrates this.
Example 2.5.5 V erify that the equation
3x2y2 dx+ 6 x3ydy = 0 (2.5.25)",Elementary Differential Equations with Boundary Value Problems.pdf
"of yalone. Here’s an example that illustrates this.
Example 2.5.5 V erify that the equation
3x2y2 dx+ 6 x3ydy = 0 (2.5.25)
is not exact, and show that the procedure for solving exact eq uations fails when applied to ( 2.5.25).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.5 Exact Equations 79
Solution Here
My (x,y ) = 6 x2y and Nx(x,y ) = 18 x2y,
so ( 2.5.25) isn’t exact. Nevertheless, let’s try to find a function F such that
Fx(x,y ) = 3 x2y2 (2.5.26)
and
Fy (x,y ) = 6 x3y. (2.5.27)
Integrating ( 2.5.26) with respect to xyields
F(x,y ) = x3y2 + φ(y),
and differentiating this with respect to yyields
Fy (x,y ) = 2 x3y+ φ′(y).
For this equation to be consistent with ( 2.5.27),
6x3y= 2 x3y+ φ′(y),
or
φ′(y) = 4 x3y.
This is a contradiction, since φ′must be independent of x. Therefore the procedure fails.
2.5 Exercises
In Exercises 1– 17 determine which equations are exact and solve them.
1. 6x2y2 dx+ 4 x3ydy = 0
2. (3ycos x+ 4 xex + 2 x2ex) dx+ (3 sin x+ 3) dy= 0
3. 14x2y3 dx+ 21 x2y2 dy= 0
4. (2x− 2y2) dx+ (12 y2 − 4xy) dy= 0
5. (x+ y)2 dx+ ( x+ y)2 dy= 0 6. (4x+ 7 y) dx+ (3 x+ 4 y) dy = 0
7. (−2y2 sin x+ 3 y3 − 2x) dx+ (4 ycos x+ 9 xy2) dy= 0
8. (2x+ y) dx+ (2 y+ 2 x) dy= 0
9. (3x2 + 2 xy+ 4 y2) dx+ ( x2 + 8 xy+ 18 y) dy = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"7. (−2y2 sin x+ 3 y3 − 2x) dx+ (4 ycos x+ 9 xy2) dy= 0
8. (2x+ y) dx+ (2 y+ 2 x) dy= 0
9. (3x2 + 2 xy+ 4 y2) dx+ ( x2 + 8 xy+ 18 y) dy = 0
10. (2x2 + 8 xy+ y2) dx+ (2 x2 + xy3/ 3) dy= 0
11.
(1
x + 2 x
)
dx+
(1
y + 2 y
)
dy= 0
12. (ysin xy+ xy2 cos xy) dx+ ( xsin xy+ xy2 cos xy) dy= 0
13. xdx
(x2 + y2)3/ 2 + ydy
(x2 + y2 )3/ 2 = 0
14.
(
ex(x2y2 + 2 xy2) + 6 x
)
dx+ (2 x2yex + 2) dy= 0
15.
(
x2ex2+y (2x2 + 3) + 4 x
)
dx+ ( x3ex2+y − 12y2) dy= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"80 Chapter 2 First Order Equations
16. (exy (x4y+ 4 x3) + 3 y) dx+ ( x5exy + 3 x) dy= 0
17. (3x2 cos xy− x3ysin xy+ 4 x) dx+ (8 y− x4 sin xy) dy = 0
In Exercises 18– 22 solve the initial value problem.
18. (4x3y2 − 6x2y− 2x− 3) dx+ (2 x4y− 2x3) dy= 0 , y (1) = 3
19. (−4ycos x+ 4 sin xcos x+ sec 2 x) dx+ (4 y− 4 sin x) dy= 0 , y (π/ 4) = 0
20. (y3 − 1)ex dx+ 3 y2(ex + 1) dy= 0 , y (0) = 0
21. (sin x− ysin x− 2 cos x) dx+ cos xdy = 0 , y (0) = 1
22. (2x− 1)(y− 1) dx+ ( x+ 2)( x− 3) dy= 0 , y (1) = −1
23. C/G Solve the exact equation
(7x+ 4 y) dx+ (4 x+ 3 y) dy = 0 .
Plot a direction field and some integral curves for this equat ion on the rectangle
{−1 ≤ x≤ 1, −1 ≤ y ≤ 1}.
24. C/G Solve the exact equation
ex(x4y2 + 4 x3y2 + 1) dx+ (2 x4yex + 2 y) dy = 0 .
Plot a direction field and some integral curves for this equat ion on the rectangle
{−2 ≤ x≤ 2, −1 ≤ y ≤ 1}.
25. C/G Plot a direction field and some integral curves for the exact e quation
(x3y4 + x) dx+ ( x4y3 + y) dy = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"{−2 ≤ x≤ 2, −1 ≤ y ≤ 1}.
25. C/G Plot a direction field and some integral curves for the exact e quation
(x3y4 + x) dx+ ( x4y3 + y) dy = 0
on the rectangle {−1 ≤ x≤ 1, −1 ≤ y≤ 1}. (See Exercise 37(a)).
26. C/G Plot a direction field and some integral curves for the exact e quation
(3x2 + 2 y) dx+ (2 y+ 2 x) dy= 0
on the rectangle {−2 ≤ x≤ 2, −2 ≤ y≤ 2}. (See Exercise 37(b)).
27. L
(a) Solve the exact equation
(x3y4 + 2 x) dx+ ( x4y3 + 3 y) dy = 0 (A)
implicitly .
(b) For what choices of (x0,y 0) does Theorem 2.3.1 imply that the initial value problem
(x3y4 + 2 x) dx+ ( x4y3 + 3 y) dy = 0 , y (x0) = y0, (B)
has a unique solution on an open interval (a,b ) that contains x0?",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.5 Exact Equations 81
(c) Plot a direction field and some integral curves for (A) on a rec tangular region centered at the
origin. What is the interval of validity of the solution of (B )?
28. L
(a) Solve the exact equation
(x2 + y2) dx+ 2 xydy = 0 (A)
implicitly .
(b) For what choices of (x0,y 0) does Theorem 2.3.1 imply that the initial value problem
(x2 + y2) dx+ 2 xydy = 0 , y (x0) = y0, (B)
has a unique solution y= y(x) on some open interval (a,b ) that contains x0?
(c) Plot a direction field and some integral curves for (A). From t he plot determine, the interval
(a,b ) of (b), the monotonicity properties (if any) of the solution of (B) , and limx→ a+ y(x)
and limx→ b− y(x). H IN T : Y our answers will depend upon which quadrant contains (x0,y 0).
29. Find all functions M such that the equation is exact.
(a) M(x,y ) dx+ ( x2 − y2 ) dy= 0
(b) M(x,y ) dx+ 2 xysin xcos ydy = 0
(c) M(x,y ) dx+ ( ex − ey sin x) dy= 0
30. Find all functions N such that the equation is exact.",Elementary Differential Equations with Boundary Value Problems.pdf
"(b) M(x,y ) dx+ 2 xysin xcos ydy = 0
(c) M(x,y ) dx+ ( ex − ey sin x) dy= 0
30. Find all functions N such that the equation is exact.
(a) (x3y2 + 2 xy+ 3 y2) dx+ N(x,y ) dy= 0
(b) (ln xy+ 2 ysin x) dx+ N(x,y ) dy= 0
(c) (xsin x+ ysin y) dx+ N(x,y ) dy= 0
31. Suppose M,N, and their partial derivatives are continuous on an open rect angle R, and Gis an
antiderivative of M with respect to x; that is,
∂G
∂x = M.
Show that if My ̸= Nx in Rthen the function
N − ∂G
∂y
is not independent of x.
32. Prove: If the equations M1 dx+N1 dy= 0 and M2 dx+N2 dy= 0 are exact on an open rectangle
R, so is the equation
(M1 + M2) dx+ ( N1 + N2) dy= 0 .
33. Find conditions on the constants A, B, C, and Dsuch that the equation
(Ax+ By) dx+ ( Cx+ Dy) dy = 0
is exact.
34. Find conditions on the constants A, B, C, D, E, and F such that the equation
(Ax2 + Bxy+ Cy2) dx+ ( Dx2 + Exy+ Fy2) dy= 0
is exact.",Elementary Differential Equations with Boundary Value Problems.pdf
"82 Chapter 2 First Order Equations
35. Suppose M and N are continuous and have continuous partial derivatives My and Nx that satisfy
the exactness condition My = Nx on an open rectangle R. Show that if (x,y ) is in Rand
F(x,y ) =
∫ x
x0
M(s,y 0) ds+
∫ y
y0
N(x,t ) dt,
then Fx = M and Fy = N.
36. Under the assumptions of Exercise 35, show that
F(x,y ) =
∫ y
y0
N(x0,s ) ds+
∫ x
x0
M(t,y ) dt.
37. Use the method suggested by Exercise 35, with (x0,y 0) = (0 , 0), to solve the these exact equa-
tions:
(a) (x3y4 + x) dx+ ( x4y3 + y) dy = 0
(b) (x2 + y2 ) dx+ 2 xydy = 0
(c) (3x2 + 2 y) dx+ (2 y+ 2 x) dy= 0
38. Solve the initial value problem
y′+ 2
xy= − 2xy
x2 + 2 x2y+ 1 , y (1) = −2.
39. Solve the initial value problem
y′− 3
xy= 2x4(4x3 − 3y)
3x5 + 3 x3 + 2 y, y (1) = 1 .
40. Solve the initial value problem
y′+ 2 xy= −e−x2
(
3x+ 2 yex2
2x+ 3 yex2
)
, y (0) = −1.
41. Rewrite the separable equation
h(y)y′= g(x) (A)
as an exact equation
M(x,y ) dx+ N(x,y ) dy= 0 . (B)",Elementary Differential Equations with Boundary Value Problems.pdf
"(
3x+ 2 yex2
2x+ 3 yex2
)
, y (0) = −1.
41. Rewrite the separable equation
h(y)y′= g(x) (A)
as an exact equation
M(x,y ) dx+ N(x,y ) dy= 0 . (B)
Show that applying the method of this section to (B) yields th e same solutions that would be
obtained by applying the method of separation of variables t o (A)
42. Suppose all second partial derivatives of M = M(x,y ) and N = N(x,y ) are continuous and
Mdx + Ndy = 0 and −Ndx + Mdy = 0 are exact on an open rectangle R. Show that
Mxx + Myy = Nxx + Nyy = 0 on R.
43. Suppose all second partial derivatives of F = F(x,y ) are continuous and Fxx + Fyy = 0 on an
open rectangle R. (A function with these properties is said to be harmonic; see also Exercise 42.)
Show that −Fy dx + Fx dy = 0 is exact on R, and therefore there’s a function G such that
Gx = −Fy and Gy = Fx in R. (A function G with this property is said to be a harmonic
conjugate of F.)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.6 Exact Equations 83
44. V erify that the following functions are harmonic, and find al l their harmonic conjugates. (See
Exercise 43.)
(a) x2 − y2 (b) ex cos y (c) x3 − 3xy2
(d) cos xcosh y (e) sin xcosh y
2.6 INTEGRA TING F ACTORS
In Section 2.5 we saw that if M, N, My and Nx are continuous and My = Nx on an open rectangle R
then
M(x,y ) dx+ N(x,y ) dy= 0 (2.6.1)
is exact on R. Sometimes an equation that isn’t exact can be made exact by m ultiplying it by an appro-
priate function. For example,
(3x+ 2 y2) dx+ 2 xydy = 0 (2.6.2)
is not exact, since My (x,y ) = 4 y̸= Nx(x,y ) = 2 yin ( 2.6.2). However, multiplying ( 2.6.2) by xyields
(3x2 + 2 xy2) dx+ 2 x2ydy = 0 , (2.6.3)
which is exact, since My (x,y ) = Nx(x,y ) = 4 xyin ( 2.6.3). Solving ( 2.6.3) by the procedure given in
Section 2.5 yields the implicit solution
x3 + x2y2 = c.
A function µ = µ(x,y ) is an integrating factor for ( 2.6.1) if
µ(x,y )M(x,y ) dx+ µ(x,y )N(x,y ) dy = 0 (2.6.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"x3 + x2y2 = c.
A function µ = µ(x,y ) is an integrating factor for ( 2.6.1) if
µ(x,y )M(x,y ) dx+ µ(x,y )N(x,y ) dy = 0 (2.6.4)
is exact. If we know an integrating factor µ for ( 2.6.1), we can solve the exact equation ( 2.6.4) by the
method of Section 2.5. It would be nice if we could say that ( 2.6.1) and ( 2.6.4) always have the same
solutions, but this isn’t so. For example, a solution y = y(x) of ( 2.6.4) such that µ(x,y (x)) = 0 on
some interval a < x < bcould fail to be a solution of ( 2.6.1) (Exercise 1), while ( 2.6.1) may have a
solution y = y(x) such that µ(x,y (x)) isn’t even defined (Exercise 2). Similar comments apply if y is
the independent variable and x is the dependent variable in ( 2.6.1) and ( 2.6.4). However, if µ(x,y ) is
defined and nonzero for all (x,y ), ( 2.6.1) and ( 2.6.4) are equivalent; that is, they have the same solutions.
Finding Integrating Factors
By applying Theorem 2.5.2 (with M and N replaced by µM and µN ), we see that ( 2.6.4) is exact on an",Elementary Differential Equations with Boundary Value Problems.pdf
"Finding Integrating Factors
By applying Theorem 2.5.2 (with M and N replaced by µM and µN ), we see that ( 2.6.4) is exact on an
open rectangle Rif µM , µN , (µM )y , and (µN )x are continuous and
∂
∂y (µM ) = ∂
∂x (µN ) or, equivalently , µy M + µMy = µxN + µNx
on R. It’s better to rewrite the last equation as
µ(My − Nx) = µxN − µy M, (2.6.5)
which reduces to the known result for exact equations; that i s, if My = Nx then ( 2.6.5) holds with µ = 1 ,
so ( 2.6.1) is exact.
Y ou may think ( 2.6.5) is of little value, since it involves partial derivatives of the unknown integrating
factor µ, and we haven’t studied methods for solving such equations. However, we’ll now show that
(2.6.5) is useful if we restrict our search to integrating factors t hat are products of a function of xand a",Elementary Differential Equations with Boundary Value Problems.pdf
"84 Chapter 2 Integrating Factors
function of y; that is, µ(x,y ) = P(x)Q(y). W e’re not saying that every equation Mdx + Ndy = 0
has an integrating factor of this form; rather, we’re saying that some equations have such integrating
factors.W e’llnow develop a way to determine whether a given equation has such an integrating factor,
and a method for finding the integrating factor in this case.
If µ(x,y ) = P(x)Q(y), then µx(x,y ) = P′(x)Q(y) and µy (x,y ) = P(x)Q′(y), so ( 2.6.5) becomes
P(x)Q(y)(My − Nx ) = P′(x)Q(y)N − P(x)Q′(y)M, (2.6.6)
or, after dividing through by P(x)Q(y),
My − Nx = P′(x)
P(x) N − Q′(y)
Q(y) M. (2.6.7)
Now let
p(x) = P′(x)
P(x) and q(y) = Q′(y)
Q(y) ,
so ( 2.6.7) becomes
My − Nx = p(x)N − q(y)M. (2.6.8)
W e obtained ( 2.6.8) by assuming that Mdx +Ndy = 0 has an integrating factor µ(x,y ) = P(x)Q(y).
However, we can now view ( 2.6.7) differently: If there are functions p= p(x) and q = q(y) that satisfy
(2.6.8) and we define
P(x) = ±e
∫
p(x) dx and Q(y) = ±e
∫",Elementary Differential Equations with Boundary Value Problems.pdf
"(2.6.8) and we define
P(x) = ±e
∫
p(x) dx and Q(y) = ±e
∫
q(y) dy , (2.6.9)
then reversing the steps that led from ( 2.6.6) to ( 2.6.8) shows that µ(x,y ) = P(x)Q(y) is an integrating
factor for Mdx + Ndy = 0 . In using this result, we take the constants of integration i n ( 2.6.9) to be zero
and choose the signs conveniently so the integrating factor has the simplest form.
There’s no simple general method for ascertaining whether f unctions p= p(x) and q= q(y) satisfying
(2.6.8) exist. However, the next theorem gives simple sufficient co nditions for the given equation to have
an integrating factor that depends on only one of the indepen dent variables xand y, and for finding an
integrating factor in this case.
Theorem 2.6.1
Let M,N,M y , and Nx be continuous on an open rectangle R. Then:
(a) If (My − Nx)/N is independent of yon Rand we define
p(x) = My − Nx
N
then
µ(x) = ±e
∫
p(x) dx (2.6.10)
is an integrating factor for
M(x,y ) dx+ N(x,y ) dy= 0 (2.6.11)
on R.",Elementary Differential Equations with Boundary Value Problems.pdf
"p(x) = My − Nx
N
then
µ(x) = ±e
∫
p(x) dx (2.6.10)
is an integrating factor for
M(x,y ) dx+ N(x,y ) dy= 0 (2.6.11)
on R.
(b) If (Nx − My )/M is independent of xon Rand we define
q(y) = Nx − My
M ,
then
µ(y) = ±e
∫
q(y) dy (2.6.12)
is an integrating factor for (2.6.11) on R.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.6 Exact Equations 85
Proof (a) If (My − Nx)/N is independent of y, then ( 2.6.8) holds with p= ( My − Nx)/N and q≡ 0.
Therefore
P(x) = ±e
∫
p(x) dx and Q(y) = ±e
∫
q(y) dy = ±e0 = ±1,
so ( 2.6.10) is an integrating factor for ( 2.6.11) on R.
(b) If (Nx − My )/M is independent of xthen eqrefeq:2.6.8 holds with p≡ 0 and q= ( Nx − My )/M ,
and a similar argument shows that ( 2.6.12) is an integrating factor for ( 2.6.11) on R.
The next two examples show how to apply Theorem 2.6.1.
Example 2.6.1 Find an integrating factor for the equation
(2xy3 − 2x3y3 − 4xy2 + 2 x) dx+ (3 x2y2 + 4 y) dy= 0 (2.6.13)
and solve the equation.
Solution In (
2.6.13)
M = 2 xy3 − 2x3y3 − 4xy2 + 2 x, N = 3 x2y2 + 4 y,
and
My − Nx = (6 xy2 − 6x3y2 − 8xy) − 6xy2 = −6x3y2 − 8xy,
so ( 2.6.13) isn’t exact. However,
My − Nx
N = −6x3y2 + 8 xy
3x2y2 + 4 y = −2x
is independent of y, so Theorem 2.6.1(a) applies with p(x) = −2x. Since
∫
p(x) dx= −
∫
2xdx = −x2,
µ(x) = e−x2",Elementary Differential Equations with Boundary Value Problems.pdf
"N = −6x3y2 + 8 xy
3x2y2 + 4 y = −2x
is independent of y, so Theorem 2.6.1(a) applies with p(x) = −2x. Since
∫
p(x) dx= −
∫
2xdx = −x2,
µ(x) = e−x2
is an integrating factor. Multiplying ( 2.6.13) by µ yields the exact equation
e−x2
(2xy3 − 2x3y3 − 4xy2 + 2 x) dx+ e−x2
(3x2y2 + 4 y) dy = 0 . (2.6.14)
T o solve this equation, we must find a function F such that
Fx(x,y ) = e−x2
(2xy3 − 2x3y3 − 4xy2 + 2 x) (2.6.15)
and
Fy(x,y ) = e−x2
(3x2y2 + 4 y). (2.6.16)
Integrating ( 2.6.16) with respect to yyields
F(x,y ) = e−x2
(x2y3 + 2 y2 ) + ψ (x). (2.6.17)
Differentiating this with respect to xyields
Fx(x,y ) = e−x2
(2xy3 − 2x3y3 − 4xy2) + ψ ′(x).
Comparing this with ( 2.6.15) shows that ψ ′(x) = 2 xe−x2
; therefore, we can let ψ (x) = −e−x2
in
(2.6.17) and conclude that
e−x2 (
y2 (x2y+ 2) − 1
)
= c
is an implicit solution of ( 2.6.14). It is also an implicit solution of ( 2.6.13).
Figure 2.6.1 shows a direction field and some integal curves for ( 2.6.13)",Elementary Differential Equations with Boundary Value Problems.pdf
"86 Chapter 2 Integrating Factors
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−4
−3
−2
−1
0
1
2
3
4
 y
 x
Figure 2.6.1 A direction field and integral curves for
(2xy3 − 2x3y3 − 4xy2 + 2 x) dx+ (3 x2y2 + 4 y) dy= 0
Example 2.6.2 Find an integrating factor for
2xy3 dx+ (3 x2y2 + x2y3 + 1) dy= 0 (2.6.18)
and solve the equation.
Solution In (
2.6.18),
M = 2 xy3, N = 3 x2y2 + x2y3 + 1 ,
and
My − Nx = 6 xy2 − (6xy2 + 2 xy3) = −2xy3 ,
so ( 2.6.18) isn’t exact. Moreover,
My − Nx
N = − 2xy3
3x2y2 + x2y2 + 1
is not independent of y, so Theorem 2.6.1(a) does not apply . However, Theorem 2.6.1(b) does apply ,
since
Nx − My
M = 2xy3
2xy3 = 1
is independent of x, so we can take q(y) = 1 . Since
∫
q(y) dy =
∫
dy= y,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.6 Exact Equations 87
µ(y) = ey is an integrating factor. Multiplying ( 2.6.18) by µ yields the exact equation
2xy3ey dx+ (3 x2y2 + x2y3 + 1) ey dy= 0 . (2.6.19)
T o solve this equation, we must find a function F such that
Fx(x,y ) = 2 xy3ey (2.6.20)
and
Fy (x,y ) = (3 x2y2 + x2y3 + 1) ey . (2.6.21)
Integrating ( 2.6.20) with respect to xyields
F(x,y ) = x2y3 ey + φ(y). (2.6.22)
Differentiating this with respect to yyields
Fy = (3 x2y2 + x2y3)ey + φ′(y),
and comparing this with ( 2.6.21) shows that φ′(y) = ey . Therefore we set φ(y) = ey in ( 2.6.22) and
conclude that
(x2y3 + 1) ey = c
is an implicit solution of ( 2.6.19). It is also an implicit solution of ( 2.6.18). Figure 2.6.2 shows a direction
field and some integral curves for ( 2.6.18).
−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
4
 y
 x
Figure 2.6.2 A direction field and integral curves for 2xy3ey dx+ (3 x2y2 + x2y3 + 1) ey dy= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
4
 y
 x
Figure 2.6.2 A direction field and integral curves for 2xy3ey dx+ (3 x2y2 + x2y3 + 1) ey dy= 0
Theorem 2.6.1 does not apply in the next example, but the more general argum ent that led to Theo-
rem 2.6.1 provides an integrating factor.",Elementary Differential Equations with Boundary Value Problems.pdf
"88 Chapter 2 Integrating Factors
Example 2.6.3 Find an integrating factor for
(3xy+ 6 y2 ) dx+ (2 x2 + 9 xy) dy= 0 (2.6.23)
and solve the equation.
Solution In (
2.6.23)
M = 3 xy+ 6 y2, N = 2 x2 + 9 xy,
and
My − Nx = (3 x+ 12 y) − (4x+ 9 y) = −x+ 3 y.
Therefore My − Nx
M = −x+ 3 y
3xy+ 6 y2 and Nx − My
N = x− 3y
2x2 + 9 xy,
so Theorem 2.6.1 does not apply . Following the more general argument that led to Theorem 2.6.1, we
look for functions p= p(x) and q= q(y) such that
My − Nx = p(x)N − q(y)M;
that is,
−x+ 3 y= p(x)(2x2 + 9 xy) − q(y)(3xy+ 6 y2).
Since the left side contains only first degree terms in xand y, we rewrite this equation as
xp(x)(2x+ 9 y) − yq(y)(3x+ 6 y) = −x+ 3 y.
This will be an identity if
xp(x) = A and yq(y) = B, (2.6.24)
where Aand Bare constants such that
−x+ 3 y= A(2x+ 9 y) − B(3x+ 6 y),
or, equivalently ,
−x+ 3 y= (2 A− 3B)x+ (9 A− 6B)y.
Equating the coefficients of xand yon both sides shows that the last equation holds for all (x,y ) if
2A− 3B = −1
9A− 6B = 3 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Equating the coefficients of xand yon both sides shows that the last equation holds for all (x,y ) if
2A− 3B = −1
9A− 6B = 3 ,
which has the solution A= 1 , B= 1 . Therefore ( 2.6.24) implies that
p(x) = 1
x and q(y) = 1
y.
Since ∫
p(x) dx= ln |x| and
∫
q(y) dy = ln |y|,
we can let P(x) = xand Q(y) = y; hence, µ(x,y ) = xy is an integrating factor. Multiplying ( 2.6.23)
by µ yields the exact equation
(3x2y2 + 6 xy3) dx+ (2 x3y+ 9 x2y2) dy= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.6 Exact Equations 89
−4 −3 −2 −1 0 1 2 3 4
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
 y
 x
Figure 2.6.3 A direction field and integral curves for (3xy+ 6 y2) dx+ (2 x2 + 9 xy) dy= 0
W e leave it to you to use the method of Section 2.5 to show that t his equation has the implicit solution
x3y2 + 3 x2y3 = c. (2.6.25)
This is also an implicit solution of ( 2.6.23). Since x≡ 0 and y ≡ 0 satisfy ( 2.6.25), you should check to
see that x≡ 0 and y≡ 0 are also solutions of ( 2.6.23). (Why is it necesary to check this?)
Figure 2.6.3 shows a direction field and integral curves for ( 2.6.23).
See Exercise 28 for a general discussion of equations like ( 2.6.23).
Example 2.6.4 The separable equation
− ydx + ( x+ x6) dy= 0 (2.6.26)
can be converted to the exact equation
− dx
x+ x6 + dy
y = 0 (2.6.27)
by multiplying through by the integrating factor
µ(x,y ) = 1
y(x+ x6) .
However, to solve ( 2.6.27) by the method of Section 2.5 we would have to evaluate the nas ty integral
∫ dx
x+ x6 .",Elementary Differential Equations with Boundary Value Problems.pdf
"µ(x,y ) = 1
y(x+ x6) .
However, to solve ( 2.6.27) by the method of Section 2.5 we would have to evaluate the nas ty integral
∫ dx
x+ x6 .
Instead, we solve ( 2.6.26) explicitly for yby finding an integrating factor of the form µ(x,y ) = xayb.",Elementary Differential Equations with Boundary Value Problems.pdf
"90 Chapter 2 Integrating Factors
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 y
 x
Figure 2.6.4 A direction field and integral curves for −ydx + ( x+ x6) dy= 0
Solution In ( 2.6.26)
M = −y, N = x+ x6,
and
My − Nx = −1 − (1 + 6 x5) = −2 − 6x5.
W e look for functions p= p(x) and q= q(y) such that
My − Nx = p(x)N − q(y)M;
that is,
− 2 − 6x5 = p(x)(x+ x6) + q(y)y. (2.6.28)
The right side will contain the term −6x5 if p(x) = −6/x . Then ( 2.6.28) becomes
−2 − 6x5 = −6 − 6x5 + q(y)y,
so q(y) = 4 /y . Since ∫
p(x) dx= −
∫ 6
xdx= −6 ln |x|= ln 1
x6 ,
and ∫
q(y) dy =
∫ 4
ydy= 4 ln |y|= ln y4 ,
we can take P(x) = x−6 and Q(y) = y4 , which yields the integrating factor µ(x,y ) = x−6y4. Multi-
plying ( 2.6.26) by µ yields the exact equation
− y5
x6 dx+
(y4
x5 + y4
)
dy= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.6 Exact Equations 91
W e leave it to you to use the method of the Section 2.5 to show th at this equation has the implicit solution
(y
x
) 5
+ y5 = k.
Solving for yyields
y= k1/ 5x(1 + x5)−1/ 5,
which we rewrite as
y= cx(1 + x5)−1/ 5
by renaming the arbitrary constant. This is also a solution o f ( 2.6.26).
Figure 2.6.4 shows a direction field and some integral curves for ( 2.6.26).
2.6 Exercises
1. (a) V erify that µ(x,y ) = yis an integrating factor for
ydx +
(
2x+ 1
y
)
dy= 0 (A)
on any open rectangle that does not intersect the xaxis or, equivalently , that
y2 dx+ (2 xy+ 1) dy= 0 (B)
is exact on any such rectangle.
(b) V erify that y≡ 0 is a solution of (B), but not of (A).
(c) Show that
y(xy+ 1) = c (C)
is an implicit solution of (B), and explain why every differe ntiable function y = y(x) other
than y≡ 0 that satisfies (C) is also a solution of (A).
2. (a) V erify that µ(x,y ) = 1 / (x− y)2 is an integrating factor for
−y2 dx+ x2 dy= 0 (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"than y≡ 0 that satisfies (C) is also a solution of (A).
2. (a) V erify that µ(x,y ) = 1 / (x− y)2 is an integrating factor for
−y2 dx+ x2 dy= 0 (A)
on any open rectangle that does not intersect the line y = xor, equivalently , that
− y2
(x− y)2 dx+ x2
(x− y)2 dy= 0 (B)
is exact on any such rectangle.
(b) Use Theorem 2.2.1 to show that xy
(x− y) = c (C)
is an implicit solution of (B), and explain why it’s also an im plicit solution of (A)
(c) V erify that y= xis a solution of (A), even though it can’t be obtained from (C) .
In Exercises 3– 16 find an integrating factor; that is a function of only one vari able, and solve the given
equation.
3. ydx − xdy = 0 4. 3x2ydx + 2 x3 dy= 0
5. 2y3 dx+ 3 y2 dy= 0 6. (5xy+ 2 y+ 5) dx+ 2 xdy = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"92 Chapter 2 Integrating Factors
7. (xy+ x+ 2 y+ 1) dx+ ( x+ 1) dy= 0
8. (27xy2 + 8 y3) dx+ (18 x2y+ 12 xy2) dy= 0
9. (6xy2 + 2 y) dx+ (12 x2y+ 6 x+ 3) dy= 0
10. y2 dx+
(
xy2 + 3 xy+ 1
y
)
dy= 0
11. (12x3y+ 24 x2y2) dx+ (9 x4 + 32 x3y+ 4 y) dy = 0
12. (x2y+ 4 xy+ 2 y) dx+ ( x2 + x) dy= 0
13. −ydx + ( x4 − x) dy= 0
14. cos xcos ydx + (sin xcos y− sin xsin y+ y) dy = 0
15. (2xy+ y2) dx+ (2 xy+ x2 − 2x2y2 − 2xy3) dy= 0
16. ysin ydx + x(sin y− ycos y) dy = 0
In Exercises 17– 23 find an integrating factor of the form µ(x,y ) = P(x)Q(y) and solve the given
equation.
17. y(1 + 5 ln |x|) dx+ 4 xln |x|dy= 0
18. (αy + γxy) dx+ ( βx + δxy) dy= 0
19. (3x2y3 − y2 + y) dx+ ( −xy+ 2 x) dy= 0
20. 2ydx + 3( x2 + x2y3 ) dy= 0
21. (acos xy− ysin xy) dx+ ( bcos xy− xsin xy) dy = 0
22. x4y4 dx+ x5y3 dy= 0
23. y(xcos x+ 2 sin x) dx+ x(y+ 1) sin xdy = 0
In Exercises 24– 27 find an integrating factor and solve the equation. Plot a dire ction field and some",Elementary Differential Equations with Boundary Value Problems.pdf
"23. y(xcos x+ 2 sin x) dx+ x(y+ 1) sin xdy = 0
In Exercises 24– 27 find an integrating factor and solve the equation. Plot a dire ction field and some
integral curves for the equation in the indicated rectangul ar region.
24. C/G (x4y3 + y) dx+ ( x5y2 − x) dy= 0; {−1 ≤ x≤ 1, −1 ≤ y≤ 1}
25. C/G (3xy+ 2 y2 + y) dx+ ( x2 + 2 xy+ x+ 2 y) dy = 0; {−2 ≤ x≤ 2, −2 ≤ y ≤ 2}
26. C/G (12xy+ 6 y3) dx+ (9 x2 + 10 xy2) dy= 0; {−2 ≤ x≤ 2, −2 ≤ y≤ 2}
27. C/G (3x2y2 + 2 y) dx+ 2 xdy = 0; {−4 ≤ x≤ 4, −4 ≤ y ≤ 4}
28. Suppose a, b, c, and dare constants such that ad− bc ̸= 0 , and let m and n be arbitrary real
numbers. Show that
(axmy+ byn+1) dx+ ( cxm+1 + dxyn ) dy= 0
has an integrating factor µ(x,y ) = xα yβ .
29. Suppose M, N, Mx , and Ny are continuous for all (x,y ), and µ = µ(x,y ) is an integrating factor
for
M(x,y ) dx+ N(x,y ) dy= 0 . (A)
Assume that µx and µy are continuous for all (x,y ), and suppose y = y(x) is a differentiable",Elementary Differential Equations with Boundary Value Problems.pdf
"for
M(x,y ) dx+ N(x,y ) dy= 0 . (A)
Assume that µx and µy are continuous for all (x,y ), and suppose y = y(x) is a differentiable
function such that µ(x,y (x)) = 0 and µx(x,y (x)) ̸= 0 for all xin some interval I. Show that yis
a solution of (A) on I.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 2.6 Exact Equations 93
30. According to Theorem 2.1.2, the general solution of the linear nonhomogeneous equatio n
y′+ p(x)y= f(x) (A)
is
y = y1(x)
(
c+
∫
f(x)/y 1 (x) dx
)
, (B)
where y1 is any nontrivial solution of the complementary equation y′+ p(x)y = 0 . In this exercise
we obtain this conclusion in a different way . Y ou may find it in structive to apply the method
suggested here to solve some of the exercises in Section 2.1.
(a) Rewrite (A) as
[p(x)y− f(x)] dx+ dy= 0 , (C)
and show that µ = ±e
∫
p(x) dx is an integrating factor for (C).
(b) Multiply (A) through by µ = ±e
∫
p(x) dx and verify that the resulting equation can be rewrit-
ten as
(µ(x)y)′= µ(x)f(x).
Then integrate both sides of this equation and solve for yto show that the general solution of
(A) is
y = 1
µ(x)
(
c+
∫
f(x)µ(x) dx
)
.
Why is this form of the general solution equivalent to (B)?",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 3
Numerical Methods
In this chapter we study numerical methods for solving a first order differential equation
y′= f(x,y ).
SECTION 3.1 deals with Euler’s method , which is really too crude to be of much use in practical appli -
cations. However, its simplicity allows for an introductio n to the ideas required to understand the better
methods discussed in the other two sections.
SECTION 3.2 discusses improvements on Euler’s method.
SECTION 3.3 deals with the
Runge-Kutta method, perhaps the most widely used method for numerical
solution of differential equations.
95",Elementary Differential Equations with Boundary Value Problems.pdf
"96 Chapter 3 Numerical Methods
3.1 EULER’S METHOD
If an initial value problem
y′= f(x,y ), y (x0) = y0 (3.1.1)
can’t be solved analytically , it’s necessary to resort to nu merical methods to obtain useful approximations
to a solution of ( 3.1.1). W e’ll consider such methods in this chapter.
W e’re interested in computing approximate values of the sol ution of ( 3.1.1) at equally spaced points
x0, x1, . . . , xn = bin an interval [x0,b ]. Thus,
xi = x0 + ih, i = 0 , 1,...,n,
where
h= b− x0
n .
W e’ll denote the approximate values of the solution at these points by y0, y1, . . . , yn ; thus, yi is an
approximation to y(xi). W e’ll call
ei = y(xi ) − yi
the error at the ith step. Because of the initial condition y(x0) = y0, we’ll always have e0 = 0 . However,
in general ei ̸= 0 if i> 0.
W e encounter two sources of error in applying a numerical met hod to solve an initial value problem:
• The formulas defining the method are based on some sort of app roximation. Errors due to the",Elementary Differential Equations with Boundary Value Problems.pdf
"• The formulas defining the method are based on some sort of app roximation. Errors due to the
inaccuracy of the approximation are called truncation errors .
• Computers do arithmetic with a fixed number of digits, and th erefore make errors in evaluating
the formulas defining the numerical methods. Errors due to th e computer’s inability to do exact
arithmetic are called roundoff errors.
Since a careful analysis of roundoff error is beyond the scop e of this book, we’ll consider only trunca-
tion errors.
Euler’s Method
The simplest numerical method for solving ( 3.1.1) is Euler’s method . This method is so crude that it is
seldom used in practice; however, its simplicity makes it us eful for illustrative purposes.
Euler’s method is based on the assumption that the tangent li ne to the integral curve of ( 3.1.1) at
(xi,y (xi)) approximates the integral curve over the interval [xi,x i+1]. Since the slope of the integral",Elementary Differential Equations with Boundary Value Problems.pdf
"(xi,y (xi)) approximates the integral curve over the interval [xi,x i+1]. Since the slope of the integral
curve of ( 3.1.1) at (xi,y (xi)) is y′(xi) = f(xi ,y (xi)), the equation of the tangent line to the integral
curve at (xi,y (xi)) is
y = y(xi) + f(xi,y (xi))(x− xi). (3.1.2)
Setting x= xi+1 = xi + hin ( 3.1.2) yields
yi+1 = y(xi) + hf(xi,y (xi)) (3.1.3)
as an approximation to y(xi+1 ). Since y(x0) = y0 is known, we can use ( 3.1.3) with i= 0 to compute
y1 = y0 + hf(x0 ,y 0).
However, setting i= 1 in ( 3.1.3) yields
y2 = y(x1) + hf(x1 ,y (x1)),",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.1 Euler’s Method 97
which isn’t useful, since we don’t know y(x1). Therefore we replace y(x1) by its approximate value y1
and redefine
y2 = y1 + hf(x1 ,y 1).
Having computed y2, we can compute
y3 = y2 + hf(x2 ,y 2).
In general, Euler’s method starts with the known value y(x0 ) = y0 and computes y1, y2, . . . , yn succes-
sively by with the formula
yi+1 = yi + hf(xi ,y i), 0 ≤ i≤ n− 1. (3.1.4)
The next example illustrates the computational procedure i ndicated in Euler’s method.
Example 3.1.1 Use Euler’s method with h= 0 .1 to find approximate values for the solution of the initial
value problem
y′+ 2 y= x3e−2x, y (0) = 1 (3.1.5)
at x= 0 .1, 0.2, 0.3.
Solution W e rewrite ( 3.1.5) as
y′= −2y+ x3e−2x, y (0) = 1 ,
which is of the form ( 3.1.1), with
f(x,y ) = −2y+ x3e−2x, x0 = 0 , and y0 = 1 .
Euler’s method yields
y1 = y0 + hf(x0,y 0)
= 1 + ( .1)f(0, 1) = 1 + ( .1)(−2) = .8,
y2 = y1 + hf(x1,y 1)
= .8 + ( .1)f(.1,. 8) = .8 + ( .1) (−2(.8) + ( .1)3e−. 2) = .640081873,",Elementary Differential Equations with Boundary Value Problems.pdf
"= 1 + ( .1)f(0, 1) = 1 + ( .1)(−2) = .8,
y2 = y1 + hf(x1,y 1)
= .8 + ( .1)f(.1,. 8) = .8 + ( .1) (−2(.8) + ( .1)3e−. 2) = .640081873,
y3 = y2 + hf(x2,y 2)
= .640081873 + ( .1)
(
−2(.640081873) + ( .2)3e−. 4)
= .512601754.
W e’ve written the details of these computations to ensure th at you understand the procedure. However,
in the rest of the examples as well as the exercises in this cha pter, we’ll assume that you can use a
programmable calculator or a computer to carry out the neces sary computations.
Examples Illustrating The Error in Euler’s Method
Example 3.1.2 Use Euler’s method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find approxi-
mate values of the solution of the initial value problem
y′+ 2 y= x3e−2x, y (0) = 1
at x= 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct solution
y= e−2x
4 (x4 + 4) , (3.1.6)
which can be obtained by the method of Section 2.1. (V erify .)",Elementary Differential Equations with Boundary Value Problems.pdf
"98 Chapter 3 Numerical Methods
Solution T able 3.1.1 shows the values of the exact solution ( 3.1.6) at the specified points, and the ap-
proximate values of the solution at these points obtained by Euler’s method with step sizes h = 0 .1,
h = 0 .05, and h= 0 .025. In examining this table, keep in mind that the approximate v alues in the col-
umn corresponding to h= .05 are actually the results of 20 steps with Euler’s method. W e h aven’t listed
the estimates of the solution obtained for x= 0 .05, 0.15, . . . , since there’s nothing to compare them with
in the column corresponding to h= 0 .1. Similarly , the approximate values in the column correspon ding
to h= 0 .025 are actually the results of 40 steps with Euler’s method.
T able 3.1.1. Numerical solution of y′+ 2 y= x3e−2x, y(0) = 1 , by Euler’s method.
x h= 0 .1 h= 0 .05 h= 0 .025 Exact
0.0 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.810005655 0.814518349 0.818751221",Elementary Differential Equations with Boundary Value Problems.pdf
"x h= 0 .1 h= 0 .05 h= 0 .025 Exact
0.0 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.810005655 0.814518349 0.818751221
0.2 0.640081873 0.656266437 0.663635953 0.670588174
0.3 0.512601754 0.532290981 0.541339495 0.549922980
0.4 0.411563195 0.432887056 0.442774766 0.452204669
0.5 0.332126261 0.353785015 0.363915597 0.373627557
0.6 0.270299502 0.291404256 0.301359885 0.310952904
0.7 0.222745397 0.242707257 0.252202935 0.261398947
0.8 0.186654593 0.205105754 0.213956311 0.222570721
0.9 0.159660776 0.176396883 0.184492463 0.192412038
1.0 0.139778910 0.154715925 0.162003293 0.169169104
Y ou can see from T able 3.1.1 that decreasing the step size improves the accuracy of Euler ’s method.
For example,
yexact (1) − yapprox (1) ≈



.0293 with h= 0 .1,
.0144 with h= 0 .05,
.0071 with h= 0 .025.
Based on this scanty evidence, you might guess that the error in approximating the exact solution at a
fixed",Elementary Differential Equations with Boundary Value Problems.pdf
".0071 with h= 0 .025.
Based on this scanty evidence, you might guess that the error in approximating the exact solution at a
fixed
value of xby Euler’s method is roughly halved when the step size is halv ed. Y ou can find more evidence
to support this conjecture by examining T able 3.1.2, which lists the approximate values of yexact − yapprox at
x= 0 .1, 0.2, . . . , 1.0.
T able 3.1.2. Errors in approximate solutions of y′+ 2 y = x3e−2x, y(0) = 1 , obtained by
Euler’s method.
x h= 0 .1 h= 0 .05 h= 0 .025
0.1 0.0187 0.0087 0.0042
0.2 0.0305 0.0143 0.0069
0.3 0.0373 0.0176 0.0085
0.4 0.0406 0.0193 0.0094
0.5 0.0415 0.0198 0.0097
0.6 0.0406 0.0195 0.0095
0.7 0.0386 0.0186 0.0091
0.8 0.0359 0.0174 0.0086
0.9 0.0327 0.0160 0.0079
1.0 0.0293 0.0144 0.0071
Example 3.1.3 T ables 3.1.3 and 3.1.4 show analogous results for the nonlinear initial value prob lem
y′= −2y2 + xy+ x2, y(0) = 1 , (3.1.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.1 Euler’s Method 99
except in this case we can’t solve ( 3.1.7) exactly . The results in the “Exact” column were obtained by
using a more accurate numerical method known as the Runge-Kutta method with a small step size. They
are exact to eight decimal places.
Since we think it’s important in evaluating the accuracy of t he numerical methods that we’ll be studying
in this chapter, we often include a column listing values of t he exact solution of the initial value problem,
even if the directions in the example or exercise don’t speci fically call for it. If quotation marks are
included in the heading, the values were obtained by applyin g the Runge-Kutta method in a way that’s
explained in Section 3.3. If quotation marks are not include d, the values were obtained from a known
formula for the solution. In either case, the values are exac t to eight places to the right of the decimal
point.
T able 3.1.3. Numerical solution of y′= −2y2 + xy+ x2, y(0) = 1 , by Euler’s method.",Elementary Differential Equations with Boundary Value Problems.pdf
"point.
T able 3.1.3. Numerical solution of y′= −2y2 + xy+ x2, y(0) = 1 , by Euler’s method.
x h= 0 .1 h= 0 .05 h= 0 .025 “Exact”
0.0 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.821375000 0.829977007 0.837584494
0.2 0.681000000 0.707795377 0.719226253 0.729641890
0.3 0.605867800 0.633776590 0.646115227 0.657580377
0.4 0.559628676 0.587454526 0.600045701 0.611901791
0.5 0.535376972 0.562906169 0.575556391 0.587575491
0.6 0.529820120 0.557143535 0.569824171 0.581942225
0.7 0.541467455 0.568716935 0.581435423 0.593629526
0.8 0.569732776 0.596951988 0.609684903 0.621907458
0.9 0.614392311 0.641457729 0.654110862 0.666250842
1.0 0.675192037 0.701764495 0.714151626 0.726015790
T able 3.1.4. Errors in approximate solutions of y′= −2y2 + xy+ x2, y(0) = 1 , obtained
by Euler’s method.
x h= 0 .1 h= 0 .05 h= 0 .025
0.1 0.0376 0.0162 0.0076
0.2 0.0486 0.0218 0.0104
0.3 0.0517 0.0238 0.0115
0.4 0.0523 0.0244 0.0119
0.5 0.0522 0.0247 0.0121
0.6 0.0521 0.0248 0.0121",Elementary Differential Equations with Boundary Value Problems.pdf
"0.1 0.0376 0.0162 0.0076
0.2 0.0486 0.0218 0.0104
0.3 0.0517 0.0238 0.0115
0.4 0.0523 0.0244 0.0119
0.5 0.0522 0.0247 0.0121
0.6 0.0521 0.0248 0.0121
0.7 0.0522 0.0249 0.0122
0.8 0.0522 0.0250 0.0122
0.9 0.0519 0.0248 0.0121
1.0 0.0508 0.0243 0.0119
Truncation Error in Euler’s Method
Consistent with the results indicated in T ables 3.1.1– 3.1.4, we’ll now show that under reasonable as-
sumptions on f there’s a constant K such that the error in approximating the solution of the init ial value
problem
y′= f(x,y ), y (x0 ) = y0,
at a given point b>x 0 by Euler’s method with step size h= ( b− x0)/n satisfies the inequality
|y(b) − yn | ≤Kh,",Elementary Differential Equations with Boundary Value Problems.pdf
"100 Chapter 3 Numerical Methods
where K is a constant independent of n.
There are two sources of error (not counting roundoff) in Eul er’s method:
1. The error committed in approximating the integral curve by t he tangent line ( 3.1.2) over the interval
[xi,x i+1].
2. The error committed in replacing y(xi ) by yi in ( 3.1.2) and using ( 3.1.4) rather than ( 3.1.2) to
compute yi+1.
Euler’s method assumes that yi+1 defined in ( 3.1.2) is an approximation to y(xi+1 ). W e call the error
in this approximation the local truncation error at the ith step , and denote it by Ti; thus,
Ti = y(xi+1) − y(xi ) − hf(xi ,y (xi)). (3.1.8)
W e’ll now use T aylor’s theorem to estimate Ti, assuming for simplicity that f, fx, and fy are continuous
and bounded for all (x,y ). Then y′′exists and is bounded on [x0,b ]. T o see this, we differentiate
y′(x) = f(x,y (x))
to obtain
y′′(x) = fx(x,y (x)) + fy (x,y (x))y′(x)
= fx(x,y (x)) + fy (x,y (x))f(x,y (x)).",Elementary Differential Equations with Boundary Value Problems.pdf
"y′(x) = f(x,y (x))
to obtain
y′′(x) = fx(x,y (x)) + fy (x,y (x))y′(x)
= fx(x,y (x)) + fy (x,y (x))f(x,y (x)).
Since we assumed that f, fx and fy are bounded, there’s a constant M such that
|fx(x,y (x)) + fy (x,y (x))f(x,y (x))| ≤M, x 0 <x<b,
which implies that
|y′′(x)| ≤M, x 0 <x<b. (3.1.9)
Since xi+1 = xi + h, T aylor’s theorem implies that
y(xi+1 ) = y(xi ) + hy′(xi) + h2
2 y′′(˜xi),
where ˜xi is some number between xi and xi+1. Since y′(xi) = f(xi ,y (xi)) this can be written as
y(xi+1) = y(xi) + hf(xi,y (xi)) + h2
2 y′′(˜xi),
or, equivalently ,
y(xi+1) − y(xi ) − hf(xi ,y (xi)) = h2
2 y′′(˜xi).
Comparing this with ( 3.1.8) shows that
Ti = h2
2 y′′(˜xi).
Recalling ( 3.1.9), we can establish the bound
|Ti| ≤Mh2
2 , 1 ≤ i≤ n. (3.1.10)
Although it may be difficult to determine the constant M, what is important is that there’s an M such that
(3.1.10) holds. W e say that the local truncation error of Euler’s met hod is of order h2, which we write as
O(h2).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.1 Euler’s Method 101
Note that the magnitude of the local truncation error in Eule r’s method is determined by the second
derivative y′′of the solution of the initial value problem. Therefore the l ocal truncation error will be
larger where |y′′|is large, or smaller where |y′′|is small.
Since the local truncation error for Euler’s method is O(h2), it’s reasonable to expect that halving h
reduces the local truncation error by a factor of 4. This is tr ue, but halving the step size also requires twice
as many steps to approximate the solution at a given point. T o analyze the overall effect of truncation
error in Euler’s method, it’s useful to derive an equation re lating the errors
ei+1 = y(xi+1 ) − yi+1 and ei = y(xi) − yi.
T o this end, recall that
y(xi+1 ) = y(xi ) + hf(xi ,y (xi)) + Ti (3.1.11)
and
yi+1 = yi + hf(xi ,y i). (3.1.12)
Subtracting ( 3.1.12) from ( 3.1.11) yields
ei+1 = ei + h[f(xi,y (xi)) − f(xi,y i)] + Ti. (3.1.13)",Elementary Differential Equations with Boundary Value Problems.pdf
"and
yi+1 = yi + hf(xi ,y i). (3.1.12)
Subtracting ( 3.1.12) from ( 3.1.11) yields
ei+1 = ei + h[f(xi,y (xi)) − f(xi,y i)] + Ti. (3.1.13)
The last term on the right is the local truncation error at the ith step. The other terms reflect the way errors
made at previous steps affect ei+1 . Since |Ti| ≤Mh2/ 2, we see from ( 3.1.13) that
|ei+1| ≤ |ei|+ h|f(xi,y (xi)) − f(xi,y i)|+ Mh2
2 . (3.1.14)
Since we assumed that fy is continuous and bounded, the mean value theorem implies th at
f(xi ,y (xi)) − f(xi,y i) = fy (xi,y ∗
i )(y(xi ) − yi) = fy (xi,y ∗
i )ei,
where y∗
i is between yi and y(xi ). Therefore
|f(xi,y (xi)) − f(xi,y i)| ≤R|ei|
for some constant R. From this and ( 3.1.14),
|ei+1| ≤(1 + Rh)|ei|+ Mh2
2 , 0 ≤ i≤ n− 1. (3.1.15)
For convenience, let C = 1 + Rh. Since e0 = y(x0) − y0 = 0 , applying ( 3.1.15) repeatedly yields
|e1| ≤ Mh2
2
|e2| ≤ C|e1|+ Mh2
2 ≤ (1 + C) Mh2
2
|e3| ≤ C|e2|+ Mh2
2 ≤ (1 + C+ C2) Mh2
2
.
.
.
|en | ≤ C|en−1|+ Mh2
2 ≤ (1 + C+ · · ·+ Cn−1) Mh2
2 . (3.1.16)",Elementary Differential Equations with Boundary Value Problems.pdf
"2
|e2| ≤ C|e1|+ Mh2
2 ≤ (1 + C) Mh2
2
|e3| ≤ C|e2|+ Mh2
2 ≤ (1 + C+ C2) Mh2
2
.
.
.
|en | ≤ C|en−1|+ Mh2
2 ≤ (1 + C+ · · ·+ Cn−1) Mh2
2 . (3.1.16)
Recalling the formula for the sum of a geometric series, we se e that
1 + C+ · · ·+ Cn−1 = 1 − Cn
1 − C = (1 + Rh)n − 1
Rh",Elementary Differential Equations with Boundary Value Problems.pdf
"102 Chapter 3 Numerical Methods
(since C = 1 + Rh). From this and ( 3.1.16),
|y(b) − yn |= |en | ≤(1 + Rh)n − 1
R
Mh
2 . (3.1.17)
Since T aylor’s theorem implies that
1 + Rh<e Rh
(verify),
(1 + Rh)n <enRh = eR(b−x0) (since nh= b− x0).
This and ( 3.1.17) imply that
|y(b) − yn | ≤Kh, (3.1.18)
with
K = MeR(b−x0) − 1
2R .
Because of ( 3.1.18) we say that the global truncation error of Euler’s method is of order h, which we
write as O(h).
Semilinear Equations and V ariation of Parameters
An equation that can be written in the form
y′+ p(x)y = h(x,y ) (3.1.19)
with p ̸≡0 is said to be semilinear. (Of course, ( 3.1.19) is linear if his independent of y.) One way to
apply Euler’s method to an initial value problem
y′+ p(x)y= h(x,y ), y (x0 ) = y0 (3.1.20)
for ( 3.1.19) is to think of it as
y′= f(x,y ), y (x0 ) = y0,
where
f(x,y ) = −p(x)y+ h(x,y ).
However, we can also start by applying variation of paramete rs to ( 3.1.20), as in Sections 2.1 and 2.4;",Elementary Differential Equations with Boundary Value Problems.pdf
"where
f(x,y ) = −p(x)y+ h(x,y ).
However, we can also start by applying variation of paramete rs to ( 3.1.20), as in Sections 2.1 and 2.4;
thus, we write the solution of ( 3.1.20) as y= uy1, where y1 is a nontrivial solution of the complementary
equation y′+ p(x)y = 0 . Then y= uy1 is a solution of ( 3.1.20) if and only if uis a solution of the initial
value problem
u′= h(x,uy 1(x))/y 1(x), u (x0) = y(x0)/y 1(x0). (3.1.21)
W e can apply Euler’s method to obtain approximate values u0, u1, . . . , un of this initial value problem,
and then take
yi = uiy1(xi)
as approximate values of the solution of ( 3.1.20). W e’ll call this procedure the Euler semilinear method .
The next two examples show that the Euler and Euler semilinea r methods may yield drastically different
results.
Example 3.1.4 In Example
2.1.7 we had to leave the solution of the initial value problem
y′− 2xy= 1 , y (0) = 3 (3.1.22)
in the form
y= ex2
(
3 +
∫ x
0
e−t2
dt
)
(3.1.23)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.1 Euler’s Method 103
because it was impossible to evaluate this integral exactly in terms of elementary functions. Use step
sizes h= 0 .2, h= 0 .1, and h= 0 .05 to find approximate values of the solution of ( 3.1.22) at x= 0 , 0.2,
0.4, 0.6, . . . , 2.0 by (a) Euler’s method; (b) the Euler semilinear method.
SO LU TIO N (a) Rewriting ( 3.1.22) as
y′= 1 + 2 xy, y (0) = 3 (3.1.24)
and applying Euler’s method with f(x,y ) = 1 + 2 xyyields the results shown in T able 3.1.5. Because of
the large differences between the estimates obtained for th e three values of h, it would be clear that these
results are useless even if the “exact” values were not inclu ded in the table.
T able 3.1.5. Numerical solution of y′− 2xy= 1 , y(0) = 3 , with Euler’s method.
x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.200000000 3.262000000 3.294348537 3.327851973
0.4 3.656000000 3.802028800 3.881421103 3.966059348",Elementary Differential Equations with Boundary Value Problems.pdf
"0.2 3.200000000 3.262000000 3.294348537 3.327851973
0.4 3.656000000 3.802028800 3.881421103 3.966059348
0.6 4.440960000 4.726810214 4.888870783 5.067039535
0.8 5.706790400 6.249191282 6.570796235 6.936700945
1.0 7.732963328 8.771893026 9.419105620 10.184923955
1.2 11.026148659 13.064051391 14.405772067 16.067111677
1.4 16.518700016 20.637273893 23.522935872 27.289392347
1.6 25.969172024 34.570423758 41.033441257 50.000377775
1.8 42.789442120 61.382165543 76.491018246 98.982969504
2.0 73.797840446 115.440048291 152.363866569 211.954462214
It’s easy to see why Euler’s method yields such poor results. Recall that the constant M in ( 3.1.10) –
which plays an important role in determining the local trunc ation error in Euler’s method – must be an
upper bound for the values of the second derivative y′′of the solution of the initial value problem ( 3.1.22)
on (0, 2). The problem is that y′′assumes very large values on this interval. T o see this, we di fferentiate
(3.1.24) to obtain",Elementary Differential Equations with Boundary Value Problems.pdf
"on (0, 2). The problem is that y′′assumes very large values on this interval. T o see this, we di fferentiate
(3.1.24) to obtain
y′′(x) = 2 y(x) + 2 xy′(x) = 2 y(x) + 2 x(1 + 2 xy(x)) = 2(1 + 2 x2)y(x) + 2 x,
where the second equality follows again from ( 3.1.24). Since ( 3.1.23) implies that y(x) >3ex2
if x> 0,
y′′(x) >6(1 + 2 x2)ex2
+ 2 x, x> 0.
For example, letting x= 2 shows that y′′(2) >2952.
SO LU TIO N (b) Since y1 = ex2
is a solution of the complementary equation y′− 2xy= 0 , we can apply
the Euler semilinear method to ( 3.1.22), with
y = uex2
and u′= e−x2
, u (0) = 3 .
The results listed in T able 3.1.6 are clearly better than those obtained by Euler’s method.
T able 3.1.6. Numerical solution of y′− 2xy= 1 , y(0) = 3 , by the Euler semilinear method.",Elementary Differential Equations with Boundary Value Problems.pdf
"104 Chapter 3 Numerical Methods
x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.330594477 3.329558853 3.328788889 3.327851973
0.4 3.980734157 3.974067628 3.970230415 3.966059348
0.6 5.106360231 5.087705244 5.077622723 5.067039535
0.8 7.021003417 6.980190891 6.958779586 6.936700945
1.0 10.350076600 10.269170824 10.227464299 10.184923955
1.2 16.381180092 16.226146390 16.147129067 16.067111677
1.4 27.890003380 27.592026085 27.441292235 27.289392347
1.6 51.183323262 50.594503863 50.298106659 50.000377775
1.8 101.424397595 100.206659076 99.595562766 98.982969504
2.0 217.301032800 214.631041938 213.293582978 211.954462214
W e can’t give a general procedure for determining in advance whether Euler’s method or the semilinear
Euler method will produce better results for a given semilin ear initial value problem ( 3.1.19). As a rule of
thumb, the Euler semilinear method will yield better result s than Euler’s method if |u′′|is small on [x0,b ],",Elementary Differential Equations with Boundary Value Problems.pdf
"thumb, the Euler semilinear method will yield better result s than Euler’s method if |u′′|is small on [x0,b ],
while Euler’s method yields better results if |u′′|is large on [x0,b ]. In many cases the results obtained by
the two methods don’t differ appreciably . However, we propo se the an intuitive way to decide which is
the better method: Try both methods with multiple step sizes , as we did in Example 3.1.4, and accept the
results obtained by the method for which the approximations change less as the step size decreases.
Example 3.1.5 Applying Euler’s method with step sizes h= 0 .1, h= 0 .05, and h= 0 .025 to the initial
value problem
y′− 2y= x
1 + y2 , y (1) = 7 (3.1.25)
on [1, 2] yields the results in T able 3.1.7. Applying the Euler semilinear method with
y= ue2x and u′= xe−2x
1 + u2e4x , u (1) = 7 e−2
yields the results in T able 3.1.8. Since the latter are clearly less dependent on step size tha n the former,",Elementary Differential Equations with Boundary Value Problems.pdf
"1 + u2e4x , u (1) = 7 e−2
yields the results in T able 3.1.8. Since the latter are clearly less dependent on step size tha n the former,
we conclude that the Euler semilinear method is better than E uler’s method for ( 3.1.25). This conclusion
is supported by comparing the approximate results obtained by the two methods with the “exact” values
of the solution.
T able 3.1.7. Numerical solution of y′− 2y= x/ (1 + y2), y(1) = 7 , by Euler’s method.
x h= 0 .1 h= 0 .05 h= 0 .025 “Exact”
1.0 7.000000000 7.000000000 7.000000000 7.000000000
1.1 8.402000000 8.471970569 8.510493955 8.551744786
1.2 10.083936450 10.252570169 10.346014101 10.446546230
1.3 12.101892354 12.406719381 12.576720827 12.760480158
1.4 14.523152445 15.012952416 15.287872104 15.586440425
1.5 17.428443554 18.166277405 18.583079406 19.037865752
1.6 20.914624471 21.981638487 22.588266217 23.253292359
1.7 25.097914310 26.598105180 27.456479695 28.401914416
1.8 30.117766627 32.183941340 33.373738944 34.690375086",Elementary Differential Equations with Boundary Value Problems.pdf
"1.7 25.097914310 26.598105180 27.456479695 28.401914416
1.8 30.117766627 32.183941340 33.373738944 34.690375086
1.9 36.141518172 38.942738252 40.566143158 42.371060528
2.0 43.369967155 47.120835251 49.308511126 51.752229656",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.1 Euler’s Method 105
T able 3.1.8. Numerical solution of y′− 2y= x/ (1 + y2), y(1) = 7 , by the Euler semilinear
method.
x h= 0 .1 h= 0 .05 h= 0 .025 “Exact”
1.0 7.000000000 7.000000000 7.000000000 7.000000000
1.1 8.552262113 8.551993978 8.551867007 8.551744786
1.2 10.447568674 10.447038547 10.446787646 10.446546230
1.3 12.762019799 12.761221313 12.760843543 12.760480158
1.4 15.588535141 15.587448600 15.586934680 15.586440425
1.5 19.040580614 19.039172241 19.038506211 19.037865752
1.6 23.256721636 23.254942517 23.254101253 23.253292359
1.7 28.406184597 28.403969107 28.402921581 28.401914416
1.8 34.695649222 34.692912768 34.691618979 34.690375086
1.9 42.377544138 42.374180090 42.372589624 42.371060528
2.0 51.760178446 51.756054133 51.754104262 51.752229656
Example 3.1.6 Applying Euler’s method with step sizes h= 0 .1, h= 0 .05, and h= 0 .025 to the initial
value problem
y′+ 3 x2y= 1 + y2, y (2) = 2 (3.1.26)",Elementary Differential Equations with Boundary Value Problems.pdf
"value problem
y′+ 3 x2y= 1 + y2, y (2) = 2 (3.1.26)
on [2, 3] yields the results in T able 3.1.9. Applying the Euler semilinear method with
y= ue−x3
and u′= ex3
(1 + u2e−2x3
), u (2) = 2 e8
yields the results in T able 3.1.10. Noting the close agreement among the three columns of T able 3.1.9
(at least for larger values of x) and the lack of any such agreement among the columns of T able 3.1.10,
we conclude that Euler’s method is better than the Euler semi linear method for ( 3.1.26). Comparing the
results with the exact values supports this conclusion.
T able 3.1.9. Numerical solution of y′+ 3 x2y= 1 + y2, y (2) = 2 , by Euler’s method.
x h= 0 .1 h= 0 .05 h= 0 .025 “Exact”
2.0 2.000000000 2.000000000 2.000000000 2.000000000
2.1 0.100000000 0.493231250 0.609611171 0.701162906
2.2 0.068700000 0.122879586 0.180113445 0.236986800
2.3 0.069419569 0.070670890 0.083934459 0.103815729
2.4 0.059732621 0.061338956 0.063337561 0.068390786
2.5 0.056871451 0.056002363 0.056249670 0.057281091",Elementary Differential Equations with Boundary Value Problems.pdf
"2.4 0.059732621 0.061338956 0.063337561 0.068390786
2.5 0.056871451 0.056002363 0.056249670 0.057281091
2.6 0.050560917 0.051465256 0.051517501 0.051711676
2.7 0.048279018 0.047484716 0.047514202 0.047564141
2.8 0.042925892 0.043967002 0.043989239 0.044014438
2.9 0.042148458 0.040839683 0.040857109 0.040875333
3.0 0.035985548 0.038044692 0.038058536 0.038072838
T able 3.1.10. Numerical solution of y′+ 3x2y= 1 + y2, y (2) = 2 , by the Euler semilinear
method.",Elementary Differential Equations with Boundary Value Problems.pdf
"106 Chapter 3 Numerical Methods
x h= 0 .1 h= 0 .05 h= 0 .025 “Exact”
x h= 0 .1 h= 0 .05 h= 0 .025 h= .0125
2.0 2.000000000 2.000000000 2.000000000 2.000000000
2.1 0.708426286 0.702568171 0.701214274 0.701162906
2.2 0.214501852 0.222599468 0.228942240 0.236986800
2.3 0.069861436 0.083620494 0.092852806 0.103815729
2.4 0.032487396 0.047079261 0.056825805 0.068390786
2.5 0.021895559 0.036030018 0.045683801 0.057281091
2.6 0.017332058 0.030750181 0.040189920 0.051711676
2.7 0.014271492 0.026931911 0.036134674 0.047564141
2.8 0.011819555 0.023720670 0.032679767 0.044014438
2.9 0.009776792 0.020925522 0.029636506 0.040875333
3.0 0.008065020 0.018472302 0.026931099 0.038072838
In the next two sections we’ll study other numerical methods for solving initial value problems, called
the improved Euler method , the midpoint method , Heun’s method and the Runge-Kutta method . If the
initial value problem is semilinear as in ( 3.1.19), we also have the option of using variation of parameters",Elementary Differential Equations with Boundary Value Problems.pdf
"initial value problem is semilinear as in ( 3.1.19), we also have the option of using variation of parameters
and then applying the given numerical method to the initial v alue problem ( 3.1.21) for u. By analogy
with the terminology used here, we’ll call the resulting pro cedure the improved Euler semilinear method ,
the midpoint semilinear method , Heun’s semilinear method or the Runge -Kutta semilinear method , as the
case may be.
3.1 Exercises
Y ou may want to save the results of these exercises, sincewe’ ll revisit in the next two sections. In Exer-
cises 1– 5 use Euler’s method to find approximate values of the solution of the given initial value problem
at the points xi = x0 + ih, where x0 is the point wher the initial condition is imposed and i = 1 , 2, 3.
The purpose of these exercises is to familiarize you with the computational procedure of Euler’s method.
1. C y′= 2 x2 + 3 y2 − 2, y (2) = 1; h= 0 .05
2. C y′= y+
√
x2 + y2, y (0) = 1; h= 0 .1",Elementary Differential Equations with Boundary Value Problems.pdf
"1. C y′= 2 x2 + 3 y2 − 2, y (2) = 1; h= 0 .05
2. C y′= y+
√
x2 + y2, y (0) = 1; h= 0 .1
3. C y′+ 3 y= x2 − 3xy+ y2 , y (0) = 2; h= 0 .05
4. C y′= 1 + x
1 − y2 , y (2) = 3; h= 0 .1
5. C y′+ x2y = sin xy, y (1) = π; h= 0 .2
6. C Use Euler’s method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find approximate
values of the solution of the initial value problem
y′+ 3 y = 7 e4x, y (0) = 2
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct
solution y= e4x + e−3x, which can be obtained by the method of Section 2.1. Present y our results
in a table like T able 3.1.1.
7. C Use Euler’s method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find approximate
values of the solution of the initial value problem
y′+ 2
xy= 3
x3 + 1 , y (1) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.1 Euler’s Method 107
at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0. Compare these approximate values with the values of the exa ct
solution
y= 1
3x2 (9 ln x+ x3 + 2) ,
which can be obtained by the method of Section 2.1. Present yo ur results in a table like T able 3.1.1.
8. C Use Euler’s method with step sizes h= 0 .05, h= 0 .025, and h= 0 .0125 to find approximate
values of the solution of the initial value problem
y′= y2 + xy− x2
x2 , y (1) = 2
at x = 1 .0, 1.05, 1.10, 1.15, . . . , 1.5. Compare these approximate values with the values of the
exact solution
y = x(1 + x2/ 3)
1 − x2/ 3
obtained in Example 2.4.3. Present your results in a table like T able 3.1.1.
9. C In Example 2.2.3 it was shown that
y5 + y = x2 + x− 4
is an implicit solution of the initial value problem
y′= 2x+ 1
5y4 + 1 , y (2) = 1 . (A)
Use Euler’s method with step sizes h= 0 .1, h= 0 .05, and h= 0 .025 to find approximate values",Elementary Differential Equations with Boundary Value Problems.pdf
"y′= 2x+ 1
5y4 + 1 , y (2) = 1 . (A)
Use Euler’s method with step sizes h= 0 .1, h= 0 .05, and h= 0 .025 to find approximate values
of the solution of (A) at x = 2 .0, 2.1, 2.2, 2.3, . . . , 3.0. Present your results in tabular form. T o
check the error in these approximate values, construct anot her table of values of the residual
R(x,y ) = y5 + y− x2 − x+ 4
for each value of (x,y ) appearing in the first table.
10. C Y ou can see from Example 2.5.1 that
x4y3 + x2y5 + 2 xy= 4
is an implicit solution of the initial value problem
y′= − 4x3y3 + 2 xy5 + 2 y
3x4y2 + 5 x2y4 + 2 x, y (1) = 1 . (A)
Use Euler’s method with step sizes h= 0 .1, h= 0 .05, and h= 0 .025 to find approximate values
of the solution of (A) at x = 1 .0, 1.1, 1.2, 1.3, . . . , 2.0. Present your results in tabular form. T o
check the error in these approximate values, construct anot her table of values of the residual
R(x,y ) = x4y3 + x2y5 + 2 xy− 4
for each value of (x,y ) appearing in the first table.",Elementary Differential Equations with Boundary Value Problems.pdf
"R(x,y ) = x4y3 + x2y5 + 2 xy− 4
for each value of (x,y ) appearing in the first table.
11. C Use Euler’s method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find approximate
values of the solution of the initial value problem
(3y2 + 4 y)y′+ 2 x+ cos x= 0 , y (0) = 1; (Exercise 2.2. 13)
at x= 0 , 0.1, 0.2, 0.3, . . . , 1.0.",Elementary Differential Equations with Boundary Value Problems.pdf
"108 Chapter 3 Numerical Methods
12. C Use Euler’s method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find approximate
values of the solution of the initial value problem
y′+ (y+ 1)( y− 1)(y− 2)
x+ 1 = 0 , y (1) = 0 (Exercise 2.2. 14)
at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0.
13. C Use Euler’s method and the Euler semilinear method with step sizes h= 0 .1, h= 0 .05, and
h= 0 .025 to find approximate values of the solution of the initial valu e problem
y′+ 3 y = 7 e−3x, y (0) = 6
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct
solution y = e−3x(7x+ 6) , which can be obtained by the method of Section 2.1. Do you not ice
anything special about the results? Explain.
The linear initial value problems in Exercises 14– 19 can’t be solved exactly in terms of known elementary
functions. In each exercise, use Euler’s method and the Eule r semilinear methods with the indicated step",Elementary Differential Equations with Boundary Value Problems.pdf
"functions. In each exercise, use Euler’s method and the Eule r semilinear methods with the indicated step
sizes to find approximate values of the solution of the given i nitial value problem at 11 equally spaced
points (including the endpoints) in the interval.
14. C y′− 2y= 1
1 + x2 , y (2) = 2 ; h= 0 .1, 0.05, 0.025 on [2, 3]
15. C y′+ 2 xy= x2, y (0) = 3 (Exercise 2.1. 38); h= 0 .2, 0.1, 0.05 on [0, 2]
16. C y′+ 1
xy= sin x
x2 , y (1) = 2 ; (Exercise 2.1. 39); h= 0 .2, 0.1, 0.05 on [1, 3]
17. C y′+ y= e−x tan x
x , y (1) = 0 ; (Exercise 2.1. 40); h= 0 .05, 0.025, 0.0125 on [1, 1.5]
18. C y′+ 2x
1 + x2 y= ex
(1 + x2)2 , y (0) = 1 ; (Exercise 2.1. 41); h= 0 .2, 0.1, 0.05 on [0, 2]
19. C xy′+ ( x+ 1) y = ex2
, y (1) = 2 ; (Exercise 2.1. 42); h= 0 .05, 0.025, 0.0125 on [1, 1.5]
In Exercises 20– 22, use Euler’s method and the Euler semilinear method with the indicated step sizes
to find approximate values of the solution of the given initia l value problem at 11 equally spaced points",Elementary Differential Equations with Boundary Value Problems.pdf
"to find approximate values of the solution of the given initia l value problem at 11 equally spaced points
(including the endpoints) in the interval.
20. C y′+ 3 y= xy2(y+ 1) , y (0) = 1 ; h= 0 .1, 0.05, 0.025 on [0, 1]
21. C y′− 4y= x
y2(y+ 1) , y (0) = 1 ; h= 0 .1, 0.05, 0.025 on [0, 1]
22. C y′+ 2 y= x2
1 + y2 , y (2) = 1 ; h= 0 .1, 0.05, 0.025 on [2, 3]
23. NU M ERICA L QUA D RA TU RE . The fundamental theorem of calculus says that if f is continuous
on a closed interval [a,b ] then it has an antiderivative F such that F′(x) = f(x) on [a,b ] and
∫ b
a
f(x) dx= F(b) − F(a). (A)
This solves the problem of evaluating a definite integral if t he integrand f has an antiderivative that
can be found and evaluated easily . However, if f doesn’t have this property , (A) doesn’t provide",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.2 The Improved Euler Method and Related Methods 109
a useful way to evaluate the definite integral. In this case we must resort to approximate methods.
There’s a class of such methods called numerical quadrature , where the approximation takes the
form ∫ b
a
f(x) dx≈
n∑
i=0
cif(xi), (B)
where a = x0 < x1 < · · ·< xn = bare suitably chosen points and c0, c1, . . . , cn are suitably
chosen constants. W e call (B) a quadrature formula .
(a) Derive the quadrature formula
∫ b
a
f(x) dx≈ h
n−1∑
i=0
f(a+ ih) (where h= ( b− a)/n ) (C)
by applying Euler’s method to the initial value problem
y′= f(x), y (a) = 0 .
(b) The quadrature formula (C) is sometimes called the left rectangle rule . Draw a figure that
justifies this terminology .
(c) L For several choices of a, b, and A, apply (C) to f(x) = Awith n= 10 , 20, 40, 80, 160, 320.
Compare your results with the exact answers and explain what you find.",Elementary Differential Equations with Boundary Value Problems.pdf
"Compare your results with the exact answers and explain what you find.
(d) L For several choices of a, b, A, and B, apply (C) to f(x) = A+ Bxwith n= 10 , 20, 40,
80, 160, 320. Compare your results with the exact answers and explain wha t you find.
3.2 THE IMPROVED EULER METHOD AND RELA TED METHODS
In Section 3.1 we saw that the global truncation error of Eule r’s method is O(h), which would seem to
imply that we can achieve arbitrarily accurate results with Euler’s method by simply choosing the step size
sufficiently small. However, this isn’t a good idea, for two r easons. First, after a certain point decreasing
the step size will increase roundoff errors to the point wher e the accuracy will deteriorate rather than
improve. The second and more important reason is that in most applications of numerical methods to an
initial value problem
y′= f(x,y ), y (x0 ) = y0, (3.2.1)
the expensive part of the computation is the evaluation of f. Therefore we want methods that give good",Elementary Differential Equations with Boundary Value Problems.pdf
"y′= f(x,y ), y (x0 ) = y0, (3.2.1)
the expensive part of the computation is the evaluation of f. Therefore we want methods that give good
results for a given number of such evaluations. This is what m otivates us to look for numerical methods
better than Euler’s.
T o clarify this point, suppose we want to approximate the val ue of eby applying Euler’s method to the
initial value problem
y′= y, y (0) = 1 , (with solution y= ex)
on [0, 1], with h = 1 / 12, 1/ 24, and 1/ 48, respectively . Since each step in Euler’s method requires
one evaluation of f, the number of evaluations of f in each of these attempts is n = 12 , 24, and 48,
respectively . In each case we accept yn as an approximation to e. The second column of T able 3.2.1
shows the results. The first column of the table indicates the number of evaluations of f required to
obtain the approximation, and the last column contains the v alue of erounded to ten significant figures.",Elementary Differential Equations with Boundary Value Problems.pdf
"obtain the approximation, and the last column contains the v alue of erounded to ten significant figures.
In this section we’ll study the improved Euler method , which requires two evaluations of f at each
step. W e’ve used this method with h = 1 / 6, 1/ 12, and 1/ 24. The required number of evaluations of f",Elementary Differential Equations with Boundary Value Problems.pdf
"110 Chapter 3 Numerical Methods
were 12, 24, and 48, as in the three applications of Euler’s method; however, yo u can see from the third
column of T able 3.2.1 that the approximation to eobtained by the improved Euler method with only 12
evaluations of f is better than the approximation obtained by Euler’s method with 48 evaluations.
In Section 3.1 we’ll study the Runge-Kutta method , which requires four evaluations of f at each step.
W e’ve used this method with h = 1 / 3, 1/ 6, and 1/ 12. The required number of evaluations of f were
again 12, 24, and 48, as in the three applications of Euler’s method and the impro ved Euler method;
however, you can see from the fourth column of T able 3.2.1 that the approximation to e obtained by
the Runge-Kutta method with only 12 evaluations of f is better than the approximation obtained by the
improved Euler method with 48 evaluations.
T able 3.2.1. Approximations to eobtained by three numerical methods.
n Euler Improved Euler Runge-Kutta Exact",Elementary Differential Equations with Boundary Value Problems.pdf
"T able 3.2.1. Approximations to eobtained by three numerical methods.
n Euler Improved Euler Runge-Kutta Exact
12 2.613035290 2.707188994 2.718069764 2.718281828
24 2.663731258 2.715327371 2.718266612 2.718281828
48 2.690496599 2.717519565 2.718280809 2.718281828
The Improved Euler Method
The improved Euler method for solving the initial value problem ( 3.2.1) is based on approximating the
integral curve of ( 3.2.1) at (xi,y (xi)) by the line through (xi,y (xi)) with slope
mi = f(xi ,y (xi)) + f(xi+1,y (xi+1))
2 ;
that is, mi is the average of the slopes of the tangents to the integral cu rve at the endpoints of [xi,x i+1].
The equation of the approximating line is therefore
y= y(xi) + f(xi ,y (xi)) + f(xi+1,y (xi+1))
2 (x− xi). (3.2.2)
Setting x= xi+1 = xi + hin ( 3.2.2) yields
yi+1 = y(xi ) + h
2 (f(xi,y (xi)) + f(xi+1 ,y (xi+1))) (3.2.3)
as an approximation to y(xi+1). As in our derivation of Euler’s method, we replace y(xi) (unknown if",Elementary Differential Equations with Boundary Value Problems.pdf
"2 (f(xi,y (xi)) + f(xi+1 ,y (xi+1))) (3.2.3)
as an approximation to y(xi+1). As in our derivation of Euler’s method, we replace y(xi) (unknown if
i> 0) by its approximate value yi; then ( 3.2.3) becomes
yi+1 = yi + h
2 (f(xi,y i) + f(xi+1,y (xi+1)) .
However, this still won’t work, because we don’t know y(xi+1), which appears on the right. W e overcome
this by replacing y(xi+1) by yi + hf(xi ,y i), the value that the Euler method would assign to yi+1.
Thus, the improved Euler method starts with the known value y(x0) = y0 and computes y1, y2, . . . , yn
successively with the formula
yi+1 = yi + h
2 (f(xi ,y i) + f(xi+1 ,y i + hf(xi ,y i))) . (3.2.4)
The computation indicated here can be conveniently organiz ed as follows: given yi, compute
k1i = f(xi,y i),
k2i = f(xi + h,y i + hk1i) ,
yi+1 = yi + h
2 (k1i + k2i).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.2 The Improved Euler Method and Related Methods 111
The improved Euler method requires two evaluations of f(x,y ) per step, while Euler’s method requires
only one. However, we’ll see at the end of this section that if fsatisfies appropriate assumptions, the local
truncation error with the improved Euler method is O(h3), rather than O(h2) as with Euler’s method.
Therefore the global truncation error with the improved Eul er method is O(h2); however, we won’t prove
this.
W e note that the magnitude of the local truncation error in th e improved Euler method and other
methods discussed in this section is determined by the third derivative y′′′of the solution of the initial
value problem. Therefore the local truncation error will be larger where |y′′′|is large, or smaller where
|y′′′|is small.
The next example, which deals with the initial value problem considered in Example 3.1.1, illustrates
the computational procedure indicated in the improved Eule r method.",Elementary Differential Equations with Boundary Value Problems.pdf
"the computational procedure indicated in the improved Eule r method.
Example 3.2.1 Use the improved Euler method with h= 0 .1 to find approximate values of the solution
of the initial value problem
y′+ 2 y= x3e−2x, y (0) = 1 (3.2.5)
at x= 0 .1, 0.2, 0.3.
Solution As in Example 3.1.1, we rewrite ( 3.2.5) as
y′= −2y+ x3e−2x, y (0) = 1 ,
which is of the form ( 3.2.1), with
f(x,y ) = −2y+ x3e−2x, x0 = 0 , and y0 = 1 .
The improved Euler method yields
k10 = f(x0 ,y 0) = f(0, 1) = −2,
k20 = f(x1 ,y 0 + hk10) = f(.1, 1 + ( .1)(−2))
= f(.1,. 8) = −2(.8) + ( .1)3e−. 2 = −1.599181269,
y1 = y0 + h
2 (k10 + k20),
= 1 + ( .05)(−2 − 1.599181269) = .820040937,
k11 = f(x1 ,y 1) = f(.1,. 820040937) = −2(.820040937) + ( .1)3e−. 2 = −1.639263142,
k21 = f(x2 ,y 1 + hk11) = f(.2,. 820040937 + .1(−1.639263142)),
= f(.2,. 656114622) = −2(.656114622) + ( .2)3e−. 4 = −1.306866684,
y2 = y1 + h
2 (k11 + k21),
= .820040937 + ( .05)(−1.639263142 − 1.306866684) = .672734445,",Elementary Differential Equations with Boundary Value Problems.pdf
"y2 = y1 + h
2 (k11 + k21),
= .820040937 + ( .05)(−1.639263142 − 1.306866684) = .672734445,
k12 = f(x2 ,y 2) = f(.2,. 672734445) = −2(.672734445) + ( .2)3e−. 4 = −1.340106330,
k22 = f(x3 ,y 2 + hk12) = f(.3,. 672734445 + .1(−1.340106330)),
= f(.3,. 538723812) = −2(.538723812) + ( .3)3e−. 6 = −1.062629710,
y3 = y2 + h
2 (k12 + k22)
= .672734445 + ( .05)(−1.340106330 − 1.062629710) = .552597643.
Example 3.2.2 T able 3.2.2 shows results of using the improved Euler method with step si zes h = 0 .1
and h= 0 .05 to find approximate values of the solution of the initial valu e problem
y′+ 2 y= x3e−2x, y (0) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"112 Chapter 3 Numerical Methods
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. For comparison, it also shows the corresponding approxima te values
obtained with Euler’s method in 3.1.2, and the values of the exact solution
y= e−2x
4 (x4 + 4) .
The results obtained by the improved Euler method with h= 0 .1 are better than those obtained by Euler’s
method with h= 0 .05.
T able 3.2.2. Numerical solution of y′+ 2 y= x3e−2x, y(0) = 1 , by Euler’s method and the
improved Euler method.
x h= 0 .1 h= 0 .05 h= 0 .1 h= 0 .05 Exact
0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.810005655 0.820040937 0.819050572 0.818751221
0.2 0.640081873 0.656266437 0.672734445 0.671086455 0.670588174
0.3 0.512601754 0.532290981 0.552597643 0.550543878 0.549922980
0.4 0.411563195 0.432887056 0.455160637 0.452890616 0.452204669
0.5 0.332126261 0.353785015 0.376681251 0.374335747 0.373627557
0.6 0.270299502 0.291404256 0.313970920 0.311652239 0.310952904",Elementary Differential Equations with Boundary Value Problems.pdf
"0.5 0.332126261 0.353785015 0.376681251 0.374335747 0.373627557
0.6 0.270299502 0.291404256 0.313970920 0.311652239 0.310952904
0.7 0.222745397 0.242707257 0.264287611 0.262067624 0.261398947
0.8 0.186654593 0.205105754 0.225267702 0.223194281 0.222570721
0.9 0.159660776 0.176396883 0.194879501 0.192981757 0.192412038
1.0 0.139778910 0.154715925 0.171388070 0.169680673 0.169169104
Euler Improved Euler Exact
Example 3.2.3 T able 3.2.3 shows analogous results for the nonlinear initial value pro blem
y′= −2y2 + xy+ x2, y(0) = 1 .
W e applied Euler’s method to this problem in Example 3.1.3.
T able 3.2.3. Numerical solution of y′= −2y2 + xy+ x2, y(0) = 1 , by Euler’s method and
the improved Euler method.
x h= 0 .1 h= 0 .05 h= 0 .1 h= 0 .05 “Exact”
0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.821375000 0.840500000 0.838288371 0.837584494
0.2 0.681000000 0.707795377 0.733430846 0.730556677 0.729641890",Elementary Differential Equations with Boundary Value Problems.pdf
"0.1 0.800000000 0.821375000 0.840500000 0.838288371 0.837584494
0.2 0.681000000 0.707795377 0.733430846 0.730556677 0.729641890
0.3 0.605867800 0.633776590 0.661600806 0.658552190 0.657580377
0.4 0.559628676 0.587454526 0.615961841 0.612884493 0.611901791
0.5 0.535376972 0.562906169 0.591634742 0.588558952 0.587575491
0.6 0.529820120 0.557143535 0.586006935 0.582927224 0.581942225
0.7 0.541467455 0.568716935 0.597712120 0.594618012 0.593629526
0.8 0.569732776 0.596951988 0.626008824 0.622898279 0.621907458
0.9 0.614392311 0.641457729 0.670351225 0.667237617 0.666250842
1.0 0.675192037 0.701764495 0.730069610 0.726985837 0.726015790
Euler Improved Euler “Exact”
Example 3.2.4 Use step sizes h= 0 .2, h= 0 .1, and h= 0 .05 to find approximate values of the solution
of
y′− 2xy= 1 , y (0) = 3 (3.2.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.2 The Improved Euler Method and Related Methods 113
at x = 0 , 0.2, 0.4, 0.6, . . . , 2.0 by (a) the improved Euler method; (b) the improved Euler semilinear
method. (W e used Euler’s method and the Euler semilinear met hod on this problem in 3.1.4.)
SO LU TIO N (a) Rewriting ( 3.2.6) as
y′= 1 + 2 xy, y (0) = 3
and applying the improved Euler method with f(x,y ) = 1 + 2 xyyields the results shown in T able 3.2.4.
SO LU TIO N (b) Since y1 = ex2
is a solution of the complementary equation y′− 2xy= 0 , we can apply
the improved Euler semilinear method to ( 3.2.6), with
y = uex2
and u′= e−x2
, u (0) = 3 .
The results listed in T able 3.2.5 are clearly better than those obtained by the improved Euler method.
T able 3.2.4. Numerical solution of y′− 2xy= 1 , y(0) = 3 , by the improved Euler method.
x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.328000000 3.328182400 3.327973600 3.327851973",Elementary Differential Equations with Boundary Value Problems.pdf
"x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.328000000 3.328182400 3.327973600 3.327851973
0.4 3.964659200 3.966340117 3.966216690 3.966059348
0.6 5.057712497 5.065700515 5.066848381 5.067039535
0.8 6.900088156 6.928648973 6.934862367 6.936700945
1.0 10.065725534 10.154872547 10.177430736 10.184923955
1.2 15.708954420 15.970033261 16.041904862 16.067111677
1.4 26.244894192 26.991620960 27.210001715 27.289392347
1.6 46.958915746 49.096125524 49.754131060 50.000377775
1.8 89.982312641 96.200506218 98.210577385 98.982969504
2.0 184.563776288 203.151922739 209.464744495 211.954462214
T able 3.2.5. Numerical solution of y′−2xy= 1 , y(0) = 3 , by the improved Euler semilinear
method.
x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.326513400 3.327518315 3.327768620 3.327851973
0.4 3.963383070 3.965392084 3.965892644 3.966059348
0.6 5.063027290 5.066038774 5.066789487 5.067039535",Elementary Differential Equations with Boundary Value Problems.pdf
"0.4 3.963383070 3.965392084 3.965892644 3.966059348
0.6 5.063027290 5.066038774 5.066789487 5.067039535
0.8 6.931355329 6.935366847 6.936367564 6.936700945
1.0 10.178248417 10.183256733 10.184507253 10.184923955
1.2 16.059110511 16.065111599 16.066611672 16.067111677
1.4 27.280070674 27.287059732 27.288809058 27.289392347
1.6 49.989741531 49.997712997 49.999711226 50.000377775
1.8 98.971025420 98.979972988 98.982219722 98.982969504
2.0 211.941217796 211.951134436 211.953629228 211.954462214",Elementary Differential Equations with Boundary Value Problems.pdf
"114 Chapter 3 Numerical Methods
A Family of Methods with O(h3) Local Truncation Error
W e’ll now derive a class of methods with O(h3) local truncation error for solving ( 3.2.1). For simplicity ,
we assume that f, fx, fy , fxx, fyy , and fxy are continuous and bounded for all (x,y ). This implies that
if yis the solution of ( 3.2.1 then y′′and y′′′are bounded (Exercise 31).
W e begin by approximating the integral curve of ( 3.2.1) at (xi,y (xi)) by the line through (xi,y (xi))
with slope
mi = σy′(xi) + ρy′(xi + θh),
where σ, ρ, and θ are constants that we’ll soon specify; however, we insist at the outset that 0 < θ≤ 1,
so that
xi <xi + θh ≤ xi+1.
The equation of the approximating line is
y = y(xi ) + mi(x− xi)
= y(xi ) + [ σy′(xi) + ρy′(xi + θh)] ( x− xi). (3.2.7)
Setting x= xi+1 = xi + hin ( 3.2.7) yields
ˆyi+1 = y(xi ) + h[σy′(xi) + ρy′(xi + θh)]
as an approximation to y(xi+1 ).
T o determine σ, ρ, and θ so that the error
Ei = y(xi+1 ) − ˆyi+1",Elementary Differential Equations with Boundary Value Problems.pdf
"ˆyi+1 = y(xi ) + h[σy′(xi) + ρy′(xi + θh)]
as an approximation to y(xi+1 ).
T o determine σ, ρ, and θ so that the error
Ei = y(xi+1 ) − ˆyi+1
= y(xi+1 ) − y(xi) − h[σy′(xi) + ρy′(xi + θh)] (3.2.8)
in this approximation is O(h3), we begin by recalling from T aylor’s theorem that
y(xi+1 ) = y(xi ) + hy′(xi) + h2
2 y′′(xi) + h3
6 y′′′(ˆxi),
where ˆxi is in (xi,x i+1). Since y′′′is bounded this implies that
y(xi+1 ) − y(xi) − hy′(xi) − h2
2 y′′(xi) = O(h3).
Comparing this with ( 3.2.8) shows that Ei = O(h3) if
σy′(xi) + ρy′(xi + θh) = y′(xi) + h
2 y′′(xi) + O(h2). (3.2.9)
However, applying T aylor’s theorem to y′shows that
y′(xi + θh) = y′(xi) + θhy′′(xi) + (θh)2
2 y′′′(xi),
where xi is in (xi,x i + θh). Since y′′′is bounded, this implies that
y′(xi + θh) = y′(xi) + θhy′′(xi) + O(h2).
Substituting this into ( 3.2.9) and noting that the sum of two O(h2) terms is again O(h2) shows that
Ei = O(h3) if
(σ + ρ)y′(xi) + ρθhy ′′(xi) = y′(xi) + h
2 y′′(xi),",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.2 The Improved Euler Method and Related Methods 115
which is true if
σ + ρ = 1 and ρθ = 1
2 . (3.2.10)
Since y′= f(x,y ), we can now conclude from ( 3.2.8) that
y(xi+1) = y(xi) + h[σf (xi ,y i) + ρf(xi + θh,y (xi + θh))] + O(h3) (3.2.11)
if σ, ρ, and θ satisfy ( 3.2.10). However, this formula would not be useful even if we knew y(xi) exactly
(as we would for i= 0 ), since we still wouldn’t know y(xi + θh) exactly . T o overcome this difficulty , we
again use T aylor’s theorem to write
y(xi + θh) = y(xi) + θhy′(xi) + h2
2 y′′(˜xi),
where ˜xi is in (xi,x i + θh). Since y′(xi) = f(xi ,y (xi)) and y′′is bounded, this implies that
|y(xi + θh) − y(xi ) − θhf(xi ,y (xi))| ≤Kh2 (3.2.12)
for some constant K. Since fy is bounded, the mean value theorem implies that
|f(xi + θh,u ) − f(xi + θh,v )| ≤M|u− v|
for some constant M. Letting
u= y(xi + θh) and v= y(xi) + θhf(xi ,y (xi))
and recalling ( 3.2.12) shows that
f(xi + θh,y (xi + θh)) = f(xi + θh,y (xi ) + θhf(xi ,y (xi))) + O(h2).",Elementary Differential Equations with Boundary Value Problems.pdf
"and recalling ( 3.2.12) shows that
f(xi + θh,y (xi + θh)) = f(xi + θh,y (xi ) + θhf(xi ,y (xi))) + O(h2).
Substituting this into ( 3.2.11) yields
y(xi+1 ) = y(xi ) + h[σf (xi,y (xi))+
ρf(xi + θh,y (xi) + θhf(xi ,y (xi)))] + O(h3).
This implies that the formula
yi+1 = yi + h[σf (xi ,y i) + ρf(xi + θh,y i + θhf(xi ,y i))]
has O(h3) local truncation error if σ, ρ, and θ satisfy ( 3.2.10). Substituting σ = 1 − ρ and θ = 1 / 2ρ here
yields
yi+1 = yi + h
[
(1 − ρ)f(xi ,y i) + ρf
(
xi + h
2ρ,y i + h
2ρf(xi ,y i)
)]
. (3.2.13)
The computation indicated here can be conveniently organiz ed as follows: given yi, compute
k1i = f(xi ,y i),
k2i = f
(
xi + h
2ρ,y i + h
2ρk1i
)
,
yi+1 = yi + h[(1 − ρ)k1i + ρk2i].
Consistent with our requirement that 0 <θ < 1, we require that ρ ≥ 1/ 2. Letting ρ = 1 / 2 in ( 3.2.13)
yields the improved Euler method ( 3.2.4). Letting ρ = 3 / 4 yields Heun’s method ,
yi+1 = yi + h
[ 1
4 f(xi ,y i) + 3
4 f
(
xi + 2
3 h,y i + 2
3 hf(xi ,y i)
)]
,",Elementary Differential Equations with Boundary Value Problems.pdf
"116 Chapter 3 Numerical Methods
which can be organized as
k1i = f(xi,y i),
k2i = f
(
xi + 2h
3 ,y i + 2h
3 k1i
)
,
yi+1 = yi + h
4 (k1i + 3 k2i).
Letting ρ = 1 yields the midpoint method ,
yi+1 = yi + hf
(
xi + h
2 ,y i + h
2 f(xi ,y i)
)
,
which can be organized as
k1i = f(xi ,y i),
k2i = f
(
xi + h
2 ,y i + h
2 k1i
)
,
yi+1 = yi + hk2i.
Examples involving the midpoint method and Heun’s method ar e given in Exercises 23-30.
3.2 Exercises
Most of the following numerical exercises involve initial v alue problems considered in the exercises in
Section 3.1. Y ou’ll find it instructive to compare the result s that you obtain here with the corresponding
results that you obtained in Section 3.1.
In Exercises 1– 5 use the improved Euler method to find approximate values of th e solution of the given
initial value problem at the points xi = x0 + ih, where x0 is the point where the initial condition is
imposed and i= 1 , 2, 3.
1. C y′= 2 x2 + 3 y2 − 2, y (2) = 1; h= 0 .05
2. C y′= y+
√",Elementary Differential Equations with Boundary Value Problems.pdf
"imposed and i= 1 , 2, 3.
1. C y′= 2 x2 + 3 y2 − 2, y (2) = 1; h= 0 .05
2. C y′= y+
√
x2 + y2, y (0) = 1; h= 0 .1
3. C y′+ 3 y= x2 − 3xy+ y2 , y (0) = 2; h= 0 .05
4. C y′= 1 + x
1 − y2 , y (2) = 3; h= 0 .1
5. C y′+ x2y = sin xy, y (1) = π; h= 0 .2
6. C Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
y′+ 3 y = 7 e4x, y (0) = 2
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct
solution y= e4x + e−3x, which can be obtained by the method of Section 2.1. Present y our results
in a table like T able 3.2.2.
7. C Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
y′+ 2
xy= 3
x3 + 1 , y (1) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.2 The Improved Euler Method and Related Methods 117
at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0. Compare these approximate values with the values of the exa ct
solution
y= 1
3x2 (9 ln x+ x3 + 2)
which can be obtained by the method of Section 2.1. Present yo ur results in a table like T able 3.2.2.
8. C Use the improved Euler method with step sizes h= 0 .05, h= 0 .025, and h= 0 .0125 to find
approximate values of the solution of the initial value prob lem
y′= y2 + xy− x2
x2 , y (1) = 2 ,
at x = 1 .0, 1.05, 1.10, 1.15, . . . , 1.5. Compare these approximate values with the values of the
exact solution
y = x(1 + x2/ 3)
1 − x2/ 3
obtained in Example 2.4.3. Present your results in a table like T able 3.2.2.
9. C In Example 3.2.2 it was shown that
y5 + y = x2 + x− 4
is an implicit solution of the initial value problem
y′= 2x+ 1
5y4 + 1 , y (2) = 1 . (A)
Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find",Elementary Differential Equations with Boundary Value Problems.pdf
"y′= 2x+ 1
5y4 + 1 , y (2) = 1 . (A)
Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of (A) at x= 2 .0, 2.1, 2.2, 2.3, . . . , 3.0. Present your results in
tabular form. T o check the error in these approximate values , construct another table of values of
the residual
R(x,y ) = y5 + y− x2 − x+ 4
for each value of (x,y ) appearing in the first table.
10. C Y ou can see from Example 2.5.1 that
x4y3 + x2y5 + 2 xy= 4
is an implicit solution of the initial value problem
y′= − 4x3y3 + 2 xy5 + 2 y
3x4y2 + 5 x2y4 + 2 x, y (1) = 1 . (A)
Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of (A) at x= 1 .0, 1.14, 1.2, 1.3, . . . , 2.0. Present your results
in tabular form. T o check the error in these approximate valu es, construct another table of values
of the residual
R(x,y ) = x4y3 + x2y5 + 2 xy− 4",Elementary Differential Equations with Boundary Value Problems.pdf
"in tabular form. T o check the error in these approximate valu es, construct another table of values
of the residual
R(x,y ) = x4y3 + x2y5 + 2 xy− 4
for each value of (x,y ) appearing in the first table.
11. C Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
(3y2 + 4 y)y′+ 2 x+ cos x= 0 , y (0) = 1 (Exercise 2.2. 13)
at x= 0 , 0.1, 0.2, 0.3, . . . , 1.0.",Elementary Differential Equations with Boundary Value Problems.pdf
"118 Chapter 3 Numerical Methods
12. C Use the improved Euler method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
y′+ (y+ 1)( y− 1)(y− 2)
x+ 1 = 0 , y (1) = 0 (Exercise 2.2. 14)
at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0.
13. C Use the improved Euler method and the improved Euler semilin ear method with step sizes
h = 0 .1, h = 0 .05, and h = 0 .025 to find approximate values of the solution of the initial valu e
problem
y′+ 3 y= e−3x(1 − 2x), y (0) = 2 ,
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct
solution y = e−3x(2 + x− x2), which can be obtained by the method of Section 2.1. Do you
notice anything special about the results? Explain.
The linear initial value problems in Exercises 14– 19 can’t be solved exactly in terms of known elementary
functions. In each exercise use the improved Euler and impro ved Euler semilinear methods with the",Elementary Differential Equations with Boundary Value Problems.pdf
"functions. In each exercise use the improved Euler and impro ved Euler semilinear methods with the
indicated step sizes to find approximate values of the soluti on of the given initial value problem at 11
equally spaced points (including the endpoints) in the inte rval.
14. C y′− 2y= 1
1 + x2 , y (2) = 2 ; h= 0 .1, 0.05, 0.025 on [2, 3]
15. C y′+ 2 xy= x2, y (0) = 3 ; h= 0 .2, 0.1, 0.05 on [0, 2] (Exercise 2.1. 38)
16. C y′+ 1
xy= sin x
x2 , y (1) = 2 , h= 0 .2, 0.1, 0.05 on [1, 3] (Exercise 2.1.39)
17. C y′+ y= e−x tan x
x , y (1) = 0 ; h= 0 .05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1. 40),
18. C y′+ 2x
1 + x2 y= ex
(1 + x2)2 , y (0) = 1 ; h= 0 .2, 0.1, 0.05 on [0, 2] (Exercise 2.1. 41)
19. C xy′+ ( x+ 1) y = ex2
, y (1) = 2 ; h= 0 .05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1. 42)
In Exercises 20– 22 use the improved Euler method and the improved Euler semilin ear method with the
indicated step sizes to find approximate values of the soluti on of the given initial value problem at 11",Elementary Differential Equations with Boundary Value Problems.pdf
"indicated step sizes to find approximate values of the soluti on of the given initial value problem at 11
equally spaced points (including the endpoints) in the inte rval.
20. C y′+ 3 y= xy2(y+ 1) , y (0) = 1 ; h= 0 .1, 0.05, 0.025 on [0, 1]
21. C y′− 4y= x
y2(y+ 1) , y (0) = 1 ; h= 0 .1, 0.05, 0.025 on [0, 1]
22. C y′+ 2 y= x2
1 + y2 , y (2) = 1 ; h= 0 .1, 0.05, 0.025 on [2, 3]
23. C Do Exercise 7 with “improved Euler method” replaced by “midpoint method. ”
24. C Do Exercise 7 with “improved Euler method” replaced by “Heun’s method. ”
25. C Do Exercise 8 with “improved Euler method” replaced by “midpoint method. ”
26. C Do Exercise 8 with “improved Euler method” replaced by “Heun’s method. ”
27. C Do Exercise 11 with “improved Euler method” replaced by “midpoint method. ”",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.3 The Runge-Kutta Method 119
28. C Do Exercise 11 with “improved Euler method” replaced by “Heun’s method. ”
29. C Do Exercise 12 with “improved Euler method” replaced by “midpoint method. ”
30. C Do Exercise 12 with “improved Euler method” replaced by “Heun’s method. ”
31. Show that if f, fx, fy , fxx , fyy , and fxy are continuous and bounded for all (x,y ) and y is the
solution of the initial value problem
y′= f(x,y ), y (x0 ) = y0,
then y′′and y′′′are bounded.
32. NU M ERICA L QUA D RA TU RE (see Exercise 3.1. 23).
(a) Derive the quadrature formula
∫ b
a
f(x) dx≈ .5h(f(a) + f(b)) + h
n−1∑
i=1
f(a+ ih) (where h= ( b− a)/n ) (A)
by applying the improved Euler method to the initial value pr oblem
y′= f(x), y (a) = 0 .
(b) The quadrature formula (A) is called the trapezoid rule . Draw a figure that justifies this
terminology .
(c) L For several choices of a, b, A, and B, apply (A) to f(x) = A + Bx, with n =",Elementary Differential Equations with Boundary Value Problems.pdf
"terminology .
(c) L For several choices of a, b, A, and B, apply (A) to f(x) = A + Bx, with n =
10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain wha t you
find.
(d) L For several choices of a, b, A, B, and C, apply (A) to f(x) = A+ Bx + Cx2, with
n= 10 , 20, 40, 80, 160, 320. Compare your results with the exact answers and explain wha t
you find.
3.3 THE RUNGE-KUTT A METHOD
In general, if kis any positive integer and fsatisfies appropriate assumptions, there are numerical met hods
with local truncation error O(hk+1) for solving an initial value problem
y′= f(x,y ), y (x0 ) = y0. (3.3.1)
Moreover, it can be shown that a method with local truncation error O(hk+1) has global truncation error
O(hk ). In Sections 3.1 and 3.2 we studied numerical methods where k = 1 and k = 2 . W e’ll skip
methods for which k = 3 and proceed to the Runge-Kutta method, the most widely used method, for",Elementary Differential Equations with Boundary Value Problems.pdf
"methods for which k = 3 and proceed to the Runge-Kutta method, the most widely used method, for
which k = 4 . The magnitude of the local truncation error is determined b y the fifth derivative y(5) of
the solution of the initial value problem. Therefore the loc al truncation error will be larger where |y(5)|
is large, or smaller where |y(5)|is small. The Runge-Kutta method computes approximate valu es y1, y2,
. . . , yn of the solution of ( 3.3.1) at x0, x0 + h, . . . , x0 + nhas follows: Given yi, compute
k1i = f(xi ,y i),
k2i = f
(
xi + h
2 ,y i + h
2 k1i
)
,
k3i = f
(
xi + h
2 ,y i + h
2 k2i
)
,
k4i = f(xi + h,y i + hk3i),
and
yi+1 = yi + h
6 (k1i + 2 k2i + 2 k3i + k4i).
The next example, which deals with the initial value problem considered in Examples 3.1.1 and 3.2.1,
illustrates the computational procedure indicated in the R unge-Kutta method.",Elementary Differential Equations with Boundary Value Problems.pdf
"120 Chapter 3 Numerical Methods
Example 3.3.1 Use the Runge-Kutta method with h= 0 .1 to find approximate values for the solution of
the initial value problem
y′+ 2 y= x3e−2x, y (0) = 1 , (3.3.2)
at x= 0 .1, 0.2.
Solution Again we rewrite ( 3.3.2) as
y′= −2y+ x3e−2x, y (0) = 1 ,
which is of the form ( 3.3.1), with
f(x,y ) = −2y+ x3e−2x, x0 = 0 , and y0 = 1 .
The Runge-Kutta method yields
k10 = f(x0,y 0) = f(0, 1) = −2,
k20 = f(x0 + h/ 2,y 0 + hk10/ 2) = f(.05, 1 + ( .05)(−2))
= f(.05,. 9) = −2(.9) + ( .05)3e−. 1 = −1.799886895,
k30 = f(x0 + h/ 2,y 0 + hk20/ 2) = f(.05, 1 + ( .05)(−1.799886895))
= f(.05,. 910005655) = −2(.910005655) + ( .05)3e−. 1 = −1.819898206,
k40 = f(x0 + h,y 0 + hk30) = f(.1, 1 + ( .1)(−1.819898206))
= f(.1,. 818010179) = −2(.818010179) + ( .1)3e−. 2 = −1.635201628,
y1 = y0 + h
6 (k10 + 2 k20 + 2 k30 + k40),
= 1 + .1
6 (−2 + 2( −1.799886895) + 2( −1.819898206) − 1.635201628) = .818753803,",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 = y0 + h
6 (k10 + 2 k20 + 2 k30 + k40),
= 1 + .1
6 (−2 + 2( −1.799886895) + 2( −1.819898206) − 1.635201628) = .818753803,
k11 = f(x1,y 1) = f(.1,. 818753803) = −2(.818753803)) + ( .1)3e−. 2 = −1.636688875,
k21 = f(x1 + h/ 2,y 1 + hk11/ 2) = f(.15,. 818753803 + ( .05)(−1.636688875))
= f(.15,. 736919359) = −2(.736919359) + ( .15)3e−. 3 = −1.471338457,
k31 = f(x1 + h/ 2,y 1 + hk21/ 2) = f(.15,. 818753803 + ( .05)(−1.471338457))
= f(.15,. 745186880) = −2(.745186880) + ( .15)3e−. 3 = −1.487873498,
k41 = f(x1 + h,y 1 + hk31) = f(.2,. 818753803 + ( .1)(−1.487873498))
= f(.2,. 669966453) = −2(.669966453) + ( .2)3e−. 4 = −1.334570346,
y2 = y1 + h
6 (k11 + 2 k21 + 2 k31 + k41),
= .818753803 + .1
6 (−1.636688875 + 2( −1.471338457) + 2( −1.487873498) − 1.334570346)
= .670592417.
The Runge-Kutta method is sufficiently accurate for most app lications.
Example 3.3.2 T able 3.3.1 shows results of using the Runge-Kutta method with step size s h= 0 .1 and",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 3.3.2 T able 3.3.1 shows results of using the Runge-Kutta method with step size s h= 0 .1 and
h= 0 .05 to find approximate values of the solution of the initial valu e problem
y′+ 2 y= x3e−2x, y (0) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.3 The Runge-Kutta Method 121
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. For comparison, it also shows the corresponding approxima te values
obtained with the improved Euler method in Example 3.2.2, and the values of the exact solution
y= e−2x
4 (x4 + 4) .
The results obtained by the Runge-Kutta method are clearly b etter than those obtained by the improved
Euler method in fact; the results obtained by the Runge-Kutt a method with h= 0 .1 are better than those
obtained by the improved Euler method with h= 0 .05.
T able 3.3.1. Numerical solution of y′+ 2 y = x3e−2x, y(0) = 1 , by the Runge-Kuttta
method and the improved Euler method.
x h= 0 .1 h= 0 .05 h= 0 .1 h= 0 .05 Exact
0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.820040937 0.819050572 0.818753803 0.818751370 0.818751221
0.2 0.672734445 0.671086455 0.670592417 0.670588418 0.670588174
0.3 0.552597643 0.550543878 0.549928221 0.549923281 0.549922980",Elementary Differential Equations with Boundary Value Problems.pdf
"0.2 0.672734445 0.671086455 0.670592417 0.670588418 0.670588174
0.3 0.552597643 0.550543878 0.549928221 0.549923281 0.549922980
0.4 0.455160637 0.452890616 0.452210430 0.452205001 0.452204669
0.5 0.376681251 0.374335747 0.373633492 0.373627899 0.373627557
0.6 0.313970920 0.311652239 0.310958768 0.310953242 0.310952904
0.7 0.264287611 0.262067624 0.261404568 0.261399270 0.261398947
0.8 0.225267702 0.223194281 0.222575989 0.222571024 0.222570721
0.9 0.194879501 0.192981757 0.192416882 0.192412317 0.192412038
1.0 0.171388070 0.169680673 0.169173489 0.169169356 0.169169104
Improved Euler Runge-Kutta Exact
Example 3.3.3 T able 3.3.2 shows analogous results for the nonlinear initial value pro blem
y′= −2y2 + xy+ x2, y(0) = 1 .
W e applied the improved Euler method to this problem in Examp le 3.
T able 3.3.2. Numerical solution of y′= −2y2 + xy+ x2, y(0) = 1 , by the Runge-Kuttta
method and the improved Euler method.
x h= 0 .1 h= 0 .05 h= 0 .1 h= 0 .05 “Exact”",Elementary Differential Equations with Boundary Value Problems.pdf
"method and the improved Euler method.
x h= 0 .1 h= 0 .05 h= 0 .1 h= 0 .05 “Exact”
0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.840500000 0.838288371 0.837587192 0.837584759 0.837584494
0.2 0.733430846 0.730556677 0.729644487 0.729642155 0.729641890
0.3 0.661600806 0.658552190 0.657582449 0.657580598 0.657580377
0.4 0.615961841 0.612884493 0.611903380 0.611901969 0.611901791
0.5 0.591634742 0.588558952 0.587576716 0.587575635 0.587575491
0.6 0.586006935 0.582927224 0.581943210 0.581942342 0.581942225
0.7 0.597712120 0.594618012 0.593630403 0.593629627 0.593629526
0.8 0.626008824 0.622898279 0.621908378 0.621907553 0.621907458
0.9 0.670351225 0.667237617 0.666251988 0.666250942 0.666250842
1.0 0.730069610 0.726985837 0.726017378 0.726015908 0.726015790
Improved Euler Runge-Kutta “Exact”",Elementary Differential Equations with Boundary Value Problems.pdf
"122 Chapter 3 Numerical Methods
Example 3.3.4 T ables 3.3.3 and 3.3.4 show results obtained by applying the Runge-Kutta and Runge -
Kutta semilinear methods to to the initial value problem
y′− 2xy= 1 , y(0) = 3 ,
which we considered in Examples 3.1.4 and 3.2.4.
T able 3.3.3. Numerical solution of y′− 2xy= 1 , y(0) = 3 , by the Runge-Kutta method.
x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.327846400 3.327851633 3.327851952 3.327851973
0.4 3.966044973 3.966058535 3.966059300 3.966059348
0.6 5.066996754 5.067037123 5.067039396 5.067039535
0.8 6.936534178 6.936690679 6.936700320 6.936700945
1.0 10.184232252 10.184877733 10.184920997 10.184923955
1.2 16.064344805 16.066915583 16.067098699 16.067111677
1.4 27.278771833 27.288605217 27.289338955 27.289392347
1.6 49.960553660 49.997313966 50.000165744 50.000377775
1.8 98.834337815 98.971146146 98.982136702 98.982969504
2.0 211.393800152 211.908445283 211.951167637 211.954462214",Elementary Differential Equations with Boundary Value Problems.pdf
"1.8 98.834337815 98.971146146 98.982136702 98.982969504
2.0 211.393800152 211.908445283 211.951167637 211.954462214
T able 3.3.4. Numerical solution of y′− 2xy = 1 , y(0) = 3 , by the Runge-Kutta semilinear
method.
x h= 0 .2 h= 0 .1 h= 0 .05 “Exact”
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.327853286 3.327852055 3.327851978 3.327851973
0.4 3.966061755 3.966059497 3.966059357 3.966059348
0.6 5.067042602 5.067039725 5.067039547 5.067039535
0.8 6.936704019 6.936701137 6.936700957 6.936700945
1.0 10.184926171 10.184924093 10.184923963 10.184923955
1.2 16.067111961 16.067111696 16.067111678 16.067111677
1.4 27.289389418 27.289392167 27.289392335 27.289392347
1.6 50.000370152 50.000377302 50.000377745 50.000377775
1.8 98.982955511 98.982968633 98.982969450 98.982969504
2.0 211.954439983 211.954460825 211.954462127 211.954462214
The Case Where x0 Isn’t The Left Endpoint
So far in this chapter we’ve considered numerical methods fo r solving an initial value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"The Case Where x0 Isn’t The Left Endpoint
So far in this chapter we’ve considered numerical methods fo r solving an initial value problem
y′= f(x,y ), y (x0) = y0 (3.3.3)
on an interval [x0,b ], for which x0 is the left endpoint. W e haven’t discussed numerical method s for
solving ( 3.3.3) on an interval [a,x 0], for which x0 is the right endpoint. T o be specific, how can we
obtain approximate values y−1, y−2, . . . , y−n of the solution of ( 3.3.3) at x0 − h,...,x 0 − nh, where
h= ( x0 − a)/n ? Here’s the answer to this question:
Consider the initial value problem
z′= −f(−x,z ), z (−x0) = y0, (3.3.4)
on the interval [−x0, −a], for which −x0 is the left endpoint. Use a numerical method to obtain approx i-
mate values z1, z2, . . . , zn of the solution of ( 3.3.4) at −x0 + h, −x0 + 2 h, . . . , −x0 + nh= −a. Then",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.3 The Runge-Kutta Method 123
y−1 = z1, y−2 = z2, ... , y−n = zn are approximate values of the solution of ( 3.3.3) at x0 − h, x0 − 2h,
. . . , x0 − nh= a.
The justification for this answer is sketched in Exercise 23. Note how easy it is to make the change the
given problem ( 3.3.3) to the modified problem ( 3.3.4): first replace f by −f and then replace x, x0, and
yby −x, −x0, and z, respectively .
Example 3.3.5 Use the Runge-Kutta method with step size h = 0 .1 to find approximate values of the
solution of
(y− 1)2y′= 2 x+ 3 , y (1) = 4 (3.3.5)
at x= 0 , 0.1, 0.2, . . . , 1.
Solution W e first rewrite ( 3.3.5) in the form ( 3.3.3) as
y′= 2x+ 3
(y− 1)2 , y (1) = 4 . (3.3.6)
Since the initial condition y(1) = 4 is imposed at the right endpoint of the interval [0, 1], we apply the
Runge-Kutta method to the initial value problem
z′= 2x− 3
(z− 1)2 , z (−1) = 4 (3.3.7)
on the interval [−1, 0]. (Y ou should verify that ( 3.3.7) is related to ( 3.3.6) as ( 3.3.4) is related to ( 3.3.3).)",Elementary Differential Equations with Boundary Value Problems.pdf
"(z− 1)2 , z (−1) = 4 (3.3.7)
on the interval [−1, 0]. (Y ou should verify that ( 3.3.7) is related to ( 3.3.6) as ( 3.3.4) is related to ( 3.3.3).)
T able 3.3.5 shows the results. Reversing the order of the rows in T able 3.3.5 and changing the signs of
the values of xyields the first two columns of T able 3.3.6. The last column of T able 3.3.6 shows the exact
values of y, which are given by
y= 1 + (3 x2 + 9 x+ 15) 1/ 3.
(Since the differential equation in ( 3.3.6) is separable, this formula can be obtained by the method of
Section 2.2.)
T able 3.3.5. Numerical solution of z′= 2x− 3
(z− 1)2 , z(−1) = 4 , on [−1, 0].
x z
-1.0 4.000000000
-0.9 3.944536474
-0.8 3.889298649
-0.7 3.834355648
-0.6 3.779786399
-0.5 3.725680888
-0.4 3.672141529
-0.3 3.619284615
-0.2 3.567241862
-0.1 3.516161955
0.0 3.466212070
T able 3.3.6. Numerical solution of (y− 1)2y′= 2 x+ 3 , y(1) = 4 , on [0, 1].",Elementary Differential Equations with Boundary Value Problems.pdf
"124 Chapter 3 Numerical Methods
x y Exact
0.00 3.466212070 3.466212074
0.10 3.516161955 3.516161958
0.20 3.567241862 3.567241864
0.30 3.619284615 3.619284617
0.40 3.672141529 3.672141530
0.50 3.725680888 3.725680889
0.60 3.779786399 3.779786399
0.70 3.834355648 3.834355648
0.80 3.889298649 3.889298649
0.90 3.944536474 3.944536474
1.00 4.000000000 4.000000000
W e leave it to you to develop a procedure for handling the nume rical solution of ( 3.3.3) on an interval
[a,b ] such that a<x 0 <b (Exercises 26 and 27).
3.3 Exercises
Most of the following numerical exercises involve initial v alue problems considered in the exercises in
Sections 3.2. Y ou’ll find it instructive to compare the resul ts that you obtain here with the corresponding
results that you obtained in those sections.
In Exercises
1– 5 use the Runge-Kutta method to find approximate values of the s olution of the given initial",Elementary Differential Equations with Boundary Value Problems.pdf
"In Exercises
1– 5 use the Runge-Kutta method to find approximate values of the s olution of the given initial
value problem at the points xi = x0 + ih, where x0 is the point where the initial condition is imposed
and i= 1 , 2.
1. C y′= 2 x2 + 3 y2 − 2, y (2) = 1; h= 0 .05
2. C y′= y+
√
x2 + y2, y (0) = 1; h= 0 .1
3. C y′+ 3 y= x2 − 3xy+ y2 , y (0) = 2; h= 0 .05
4. C y′= 1 + x
1 − y2 , y (2) = 3; h= 0 .1
5. C y′+ x2y = sin xy, y (1) = π; h= 0 .2
6. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
y′+ 3 y= 7 e4x, y (0) = 2 ,
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct
solution y= e4x + e−3x, which can be obtained by the method of Section 2.1. Present y our results
in a table like T able 3.3.1.
7. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find",Elementary Differential Equations with Boundary Value Problems.pdf
"in a table like T able 3.3.1.
7. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
y′+ 2
xy= 3
x3 + 1 , y (1) = 1
at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0. Compare these approximate values with the values of the exa ct
solution
y= 1
3x2 (9 ln x+ x3 + 2) ,
which can be obtained by the method of Section 2.1. Present yo ur results in a table like T able 3.3.1.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.3 The Runge-Kutta Method 125
8. C Use the Runge-Kutta method with step sizes h = 0 .05, h = 0 .025, and h = 0 .0125 to find
approximate values of the solution of the initial value prob lem
y′= y2 + xy− x2
x2 , y (1) = 2
at x = 1 .0, 1.05, 1.10, 1.15 . . . , 1.5. Compare these approximate values with the values of the
exact solution
y= x(1 + x2/ 3)
1 − x2/ 3 ,
which was obtained in Example 2.2.3. Present your results in a table like T able 3.3.1.
9. C In Example 2.2.3 it was shown that
y5 + y = x2 + x− 4
is an implicit solution of the initial value problem
y′= 2x+ 1
5y4 + 1 , y (2) = 1 . (A)
Use the Runge-Kutta method with step sizes h= 0 .1, h = 0 .05, and h = 0 .025 to find approxi-
mate values of the solution of (A) at x= 2 .0, 2.1, 2.2, 2.3, . . . , 3.0. Present your results in tabular
form. T o check the error in these approximate values, constr uct another table of values of the
residual
R(x,y ) = y5 + y− x2 − x+ 4
for each value of (x,y ) appearing in the first table.",Elementary Differential Equations with Boundary Value Problems.pdf
"residual
R(x,y ) = y5 + y− x2 − x+ 4
for each value of (x,y ) appearing in the first table.
10. C Y ou can see from Example 2.5.1 that
x4y3 + x2y5 + 2 xy= 4
is an implicit solution of the initial value problem
y′= − 4x3y3 + 2 xy5 + 2 y
3x4y2 + 5 x2y4 + 2 x, y (1) = 1 . (A)
Use the Runge-Kutta method with step sizes h= 0 .1, h = 0 .05, and h = 0 .025 to find approxi-
mate values of the solution of (A) at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0. Present your results in tabular
form. T o check the error in these approximate values, constr uct another table of values of the
residual
R(x,y ) = x4y3 + x2y5 + 2 xy− 4
for each value of (x,y ) appearing in the first table.
11. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
(3y2 + 4 y)y′+ 2 x+ cos x= 0 , y (0) = 1 (Exercise 2.2. 13),
at x= 0 , 0.1, 0.2, 0.3, . . . , 1.0.",Elementary Differential Equations with Boundary Value Problems.pdf
"126 Chapter 3 Numerical Methods
12. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of the initial value prob lem
y′+ (y+ 1)( y− 1)(y− 2)
x+ 1 = 0 , y (1) = 0 (Exercise 2.2. 14),
at x= 1 .0, 1.1, 1.2, 1.3, . . . , 2.0.
13. C Use the Runge-Kutta method and the Runge-Kutta semilinear m ethod with step sizes h= 0 .1,
h= 0 .05, and h= 0 .025 to find approximate values of the solution of the initial valu e problem
y′+ 3 y= e−3x(1 − 4x+ 3 x2 − 4x3), y (0) = −3
at x = 0 , 0.1, 0.2, 0.3, . . . , 1.0. Compare these approximate values with the values of the exa ct
solution y= −e−3x(3 − x+ 2x2 − x3 + x4), which can be obtained by the method of Section 2.1.
Do you notice anything special about the results? Explain.
The linear initial value problems in Exercises 14– 19 can’t be solved exactly in terms of known elementary
functions. In each exercise use the Runge-Kutta and the Rung e-Kutta semilinear methods with the indi-",Elementary Differential Equations with Boundary Value Problems.pdf
"functions. In each exercise use the Runge-Kutta and the Rung e-Kutta semilinear methods with the indi-
cated step sizes to find approximate values of the solution of the given initial value problem at 11 equally
spaced points (including the endpoints) in the interval.
14. C y′− 2y= 1
1 + x2 , y (2) = 2 ; h= 0 .1, 0.05, 0.025 on [2, 3]
15. C y′+ 2 xy= x2, y (0) = 3 ; h= 0 .2, 0.1, 0.05 on [0, 2] (Exercise 2.1. 38)
16. C y′+ 1
xy= sin x
x2 , y (1) = 2; h= 0 .2, 0.1, 0.05 on [1, 3] (Exercise 2.1. 39)
17. C y′+ y= e−x tan x
x , y (1) = 0; h= 0 .05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1. 40)
18. C y′+ 2x
1 + x2 y= ex
(1 + x2)2 , y (0) = 1; h= 0 .2, 0.1, 0.05 on [0, 2] (Exercise 2.1, 41)
19. C xy′+ ( x+ 1) y = ex2
, y (1) = 2 ; h= 0 .05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1. 42)
In Exercises 20– 22 use the Runge-Kutta method and the Runge-Kutta semilinear m ethod with the indi-
cated step sizes to find approximate values of the solution of the given initial value problem at 11 equally",Elementary Differential Equations with Boundary Value Problems.pdf
"cated step sizes to find approximate values of the solution of the given initial value problem at 11 equally
spaced points (including the endpoints) in the interval.
20. C y′+ 3 y= xy2(y+ 1) , y (0) = 1 ; h= 0 .1, 0.05, 0.025 on [0, 1]
21. C y′− 4y= x
y2(y+ 1) , y (0) = 1 ; h= 0 .1, 0.05, 0.025 on [0, 1]
22. C y′+ 2 y= x2
1 + y2 , y (2) = 1 ; h= 0 .1, 0.05, 0.025 on [2, 3]
23. C Suppose a<x 0, so that −x0 <−a. Use the chain rule to show that if zis a solution of
z′= −f(−x,z ), z (−x0) = y0,
on [−x0, −a], then y = z(−x) is a solution of
y′= f(x,y ), y (x0 ) = y0,
on [a,x 0].",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 3.3 The Runge-Kutta Method 127
24. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of
y′= y2 + xy− x2
x2 , y (2) = −1
at x = 1 .1, 1.2, 1.3, . . . 2.0. Compare these approximate values with the values of the exa ct
solution
y= x(4 − 3x2)
4 + 3 x2 ,
which can be obtained by referring to Example 2.4.3.
25. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of
y′= −x2y− xy2, y (1) = 1
at x= 0 , 0.1, 0.2, . . . , 1.
26. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of
y′+ 1
xy= 7
x2 + 3 , y (1) = 3
2
at x= 0 .5, 0.6,. . . , 1.5. Compare these approximate values with the values of the exa ct solution
y = 7 ln x
x + 3x
2 ,
which can be obtained by the method discussed in Section 2.1.",Elementary Differential Equations with Boundary Value Problems.pdf
"y = 7 ln x
x + 3x
2 ,
which can be obtained by the method discussed in Section 2.1.
27. C Use the Runge-Kutta method with step sizes h = 0 .1, h = 0 .05, and h = 0 .025 to find
approximate values of the solution of
xy′+ 2 y= 8 x2, y (2) = 5
at x = 1 .0, 1.1, 1.2, . . . , 3.0. Compare these approximate values with the values of the exa ct
solution
y = 2 x2 − 12
x2 ,
which can be obtained by the method discussed in Section 2.1.
28. NU M ERICA L QUA D RA TU RE (see Exercise 3.1. 23).
(a) Derive the quadrature formula
∫ b
a
f(x) dx≈ h
6 (f(a) + f(b)) + h
3
n−1∑
i=1
f(a+ ih) + 2h
3
n∑
i=1
f(a+ (2 i− 1)h/ 2) (A)
(where h= ( b− a)/n ) by applying the Runge-Kutta method to the initial value prob lem
y′= f(x), y (a) = 0 .
This quadrature formula is called Simpson’s Rule .",Elementary Differential Equations with Boundary Value Problems.pdf
"128 Chapter 3 Numerical Methods
(b) L For several choices of a, b, A, B, C, and Dapply (A) to f(x) = A+ Bx+ Cx+ Dx3,
with n= 10 , 20, 40, 80, 160, 320. Compare your results with the exact answers and explain
what you find.
(c) L For several choices of a, b, A, B, C, D, and Eapply (A) to f(x) = A+ Bx+ Cx2 +
Dx3 + Ex4, with n= 10 , 20, 40, 80, 160, 320. Compare your results with the exact answers
and explain what you find.",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 4
Applications of First Order Equations
IN THIS CHAPTER we consider applications of first order diffe rential equations.
SECTION 4.1 begins with a discussion of exponential growth a nd decay , which you have probably al-
ready seen in calculus. W e consider applications to radioac tive decay , carbon dating, and compound
interest. W e also consider more complicated problems where the rate of change of a quantity is in part
proportional to the magnitude of the quantity , but is also in fluenced by other other factors for example, a
radioactive susbstance is manufactured at a certain rate, b ut decays at a rate proportional to its mass, or a
saver makes regular deposits in a savings account that draws compound interest.
SECTION 4.2 deals with applications of Newton’s law of cooli ng and with mixing problems.
SECTION 4.3 discusses applications to elementary mechanic s involving Newton’s second law of mo-",Elementary Differential Equations with Boundary Value Problems.pdf
"SECTION 4.3 discusses applications to elementary mechanic s involving Newton’s second law of mo-
tion. The problems considered include motion under the influ ence of gravity in a resistive medium, and
determining the initial velocity required to launch a satel lite.
SECTION 4.4 deals with methods for dealing with a type of seco nd order equation that often arises in
applications of Newton’s second law of motion, by reformula ting it as first order equation with a different
independent variable. Although the method doesn’t usually lead to an explicit solution of the given
equation, it does provide valuable insights into the behavi or of the solutions.
SECTION 4.5 deals with applications of differential equati ons to curves.
129",Elementary Differential Equations with Boundary Value Problems.pdf
"130 Chapter 4 Applications of First Order Equations
4.1 GROWTH AND DECA Y
Since the applications in this section deal with functions o f time, we’ll denote the independent variable
by t. If Qis a function of t, Q′will denote the derivative of Qwith respect to t; thus,
Q′= dQ
dt .
Exponential Growth and Decay
One of the most common mathematical models for a physical pro cess is the exponential model , where
it’s assumed that the rate of change of a quantity Qis proportional to Q; thus
Q′= aQ, (4.1.1)
where ais the constant of proportionality .
From Example 3, the general solution of ( 4.1.1) is
Q= ceat
and the solution of the initial value problem
Q′= aQ, Q (t0) = Q0
is
Q= Q0ea(t−t0). (4.1.2)
Since the solutions of Q′ = aQ are exponential functions, we say that a quantity Q that satisfies this
equation grows exponentially if a> 0, or decays exponentially if a< 0 (Figure 4.1.1).
Radioactive Decay",Elementary Differential Equations with Boundary Value Problems.pdf
"equation grows exponentially if a> 0, or decays exponentially if a< 0 (Figure 4.1.1).
Radioactive Decay
Experimental evidence shows that radioactive material dec ays at a rate proportional to the mass of the
material present. According to this model the mass Q(t) of a radioactive material present at time t
satisfies ( 4.1.1), where ais a negative constant whose value for any given material mus t be determined
by experimental observation. For simplicity , we’ll replac e the negative constant aby −k, where k is a
positive number that we’ll call the decay constant of the material. Thus, ( 4.1.1) becomes
Q′= −kQ.
If the mass of the material present at t= t0 is Q0, the mass present at time tis the solution of
Q′= −kQ, Q (t0) = Q0.
From ( 4.1.2) with a= −k, the solution of this initial value problem is
Q= Q0e−k(t−t0). (4.1.3)
The half–life τ of a radioactive material is defined to be the time required fo r half of its mass to decay;
that is, if Q(t0) = Q0, then
Q(τ + t0) = Q0
2 . (4.1.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"that is, if Q(t0) = Q0, then
Q(τ + t0) = Q0
2 . (4.1.4)
From ( 4.1.3) with t= τ + t0, ( 4.1.4) is equivalent to
Q0e−kτ = Q0
2 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.1 Growth and Decay 131
 Q
 t
 a > 0
 a < 0
 Q0
Figure 4.1.1 Exponential growth and decay
so
e−kτ = 1
2 .
T aking logarithms yields
−kτ = ln 1
2 = − ln 2 ,
so the half-life is
τ = 1
kln 2. (4.1.5)
(Figure 4.1.2). The half-life is independent of t0 and Q0, since it’s determined by the properties of
material, not by the amount of the material present at any par ticular time.
Example 4.1.1 A radioactive substance has a half-life of 1620 years.
(a) If its mass is now 4 g (grams), how much will be left 810 years fr om now?
(b) Find the time t1 when 1.5 g of the substance remain.
SO LU TIO N (a) From ( 4.1.3) with t0 = 0 and Q0 = 4 ,
Q= 4 e−kt , (4.1.6)
where we determine kfrom ( 4.1.5), with τ= 1620 years:
k= ln 2
τ = ln 2
1620 .
Substituting this in ( 4.1.6) yields
Q= 4 e−(t ln 2) / 1620. (4.1.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"132 Chapter 4 Applications of First Order Equations
 Q
 t τ
 Q0
 .5Q0
Figure 4.1.2 Half-life of a radioactive substance
Therefore the mass left after 810 years will be
Q(810) = 4 e−(810 ln 2) / 1620 = 4 e−(ln 2) / 2
= 2
√
2 g.
SO LU TIO N (b) Setting t= t1 in ( 4.1.7) and requiring that Q(t1) = 1 .5 yields
3
2 = 4 e(−t1 ln 2) / 1620.
Dividing by 4 and taking logarithms yields
ln 3
8 = −t1 ln 2
1620 .
Since ln 3/ 8 = − ln 8 / 3,
t1 = 1620 ln 8/ 3
ln 2 ≈ 2292.4 years.
Interest Compounded Continuously
Suppose we deposit an amount of money Q0 in an interest-bearing account and make no further deposits
or withdrawals for tyears, during which the account bears interest at a constant annual rate r. T o calculate
the value of the account at the end of tyears, we need one more piece of information: how the interes t
is added to the account, or—as the bankers say—how it is compounded. If the interest is compounded",Elementary Differential Equations with Boundary Value Problems.pdf
"is added to the account, or—as the bankers say—how it is compounded. If the interest is compounded
annually , the value of the account is multiplied by 1 + rat the end of each year. This means that after t
years the value of the account is
Q(t) = Q0(1 + r)t.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.1 Growth and Decay 133
If interest is compounded semiannually , the value of the acc ount is multiplied by (1 + r/ 2) every 6
months. Since this occurs twice annually , the value of the ac count after tyears is
Q(t) = Q0
(
1 + r
2
) 2t
.
In general, if interest is compounded ntimes per year, the value of the account is multiplied ntimes per
year by (1 + r/n ); therefore, the value of the account after tyears is
Q(t) = Q0
(
1 + r
n
) nt
. (4.1.8)
Thus, increasing the frequency of compounding increases th e value of the account after a fixed period
of time. T able 4.1.7 shows the effect of increasing the number of compoundings ov er t= 5 years on an
initial deposit of Q0 = 100 (dollars), at an annual interest rate of 6%.
T able 4.1.7. T able The effect of compound interest
n $100
(
1 + .06
n
) 5n
(number of compoundings (value in dollars
per year) after 5 years)
1 $133.82
2 $134.39
4 $134.68
8 $134.83
364 $134.98",Elementary Differential Equations with Boundary Value Problems.pdf
"n $100
(
1 + .06
n
) 5n
(number of compoundings (value in dollars
per year) after 5 years)
1 $133.82
2 $134.39
4 $134.68
8 $134.83
364 $134.98
Y ou can see from T able 4.1.7 that the value of the account after 5 years is an increasing fu nction of
n. Now suppose the maximum allowable rate of interest on savin gs accounts is restricted by law , but
the time intervals between successive compoundings isn’t ; then competing banks can attract savers by
compounding often. The ultimate step in this direction is to compound continuously , by which we mean
that n→ ∞ in ( 4.1.8). Since we know from calculus that
lim
n→∞
(
1 + r
n
) n
= er ,
this yields
Q(t) = lim
n→∞
Q0
(
1 + r
n
) nt
= Q0
[
lim
n→∞
(
1 + r
n
) n] t
= Q0ert .
Observe that Q= Q0ert is the solution of the initial value problem
Q′= rQ, Q (0) = Q0;
that is, with continuous compounding the value of the accoun t grows exponentially .",Elementary Differential Equations with Boundary Value Problems.pdf
"134 Chapter 4 Applications of First Order Equations
Example 4.1.2 If $150 is deposited in a bank that pays 5 1
2 % annual interest compounded continuously ,
the value of the account after tyears is
Q(t) = 150 e. 055t
dollars. (Note that it’s necessary to write the interest rat e as a decimal; thus, r = .055.) Therefore, after
t= 10 years the value of the account is
Q(10) = 150 e. 55 ≈ $259.99.
Example 4.1.3 W e wish to accumulate $10,000 in 10 years by making a single de posit in a savings
account bearing 5 1
2 % annual interest compounded continuously . How much must we deposit in the
account?
Solution The value of the account at time tis
Q(t) = Q0e. 055t. (4.1.9)
Since we want Q(10) to be $10,000, the initial deposit Q0 must satisfy the equation
10000 = Q0e. 55, (4.1.10)
obtained by setting t= 10 and Q(10) = 10000 in (
4.1.9). Solving ( 4.1.10) for Q0 yields
Q0 = 10000 e−. 55 ≈ $5769.50.
Mixed Growth and Decay",Elementary Differential Equations with Boundary Value Problems.pdf
"obtained by setting t= 10 and Q(10) = 10000 in (
4.1.9). Solving ( 4.1.10) for Q0 yields
Q0 = 10000 e−. 55 ≈ $5769.50.
Mixed Growth and Decay
Example 4.1.4 A radioactive substance with decay constant kis produced at a constant rate of aunits of
mass per unit time.
(a) Assuming that Q(0) = Q0, find the mass Q(t) of the substance present at time t.
(b) Find limt→∞ Q(t).
SO LU TIO N (a) Here
Q′= rate of increase of Q− rate of decrease of Q.
The rate of increase is the constant a. Since Qis radioactive with decay constant k, the rate of decrease
is kQ. Therefore
Q′= a− kQ.
This is a linear first order differential equation. Rewritin g it and imposing the initial condition shows that
Qis the solution of the initial value problem
Q′+ kQ = a, Q (0) = Q0. (4.1.11)
Since e−kt is a solution of the complementary equation, the solutions o f ( 4.1.11) are of the form Q =
ue−kt , where u′e−kt = a, so u′= aekt . Hence,
u= a
kekt + c",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.1 Growth and Decay 135
 Q
 t
 a/k
Figure 4.1.3 Q(t) approaches the steady state value a
k as t→ ∞
and
Q= ue−kt = a
k + ce−kt .
Since Q(0) = Q0, setting t= 0 here yields
Q0 = a
k + c or c= Q0 − a
k.
Therefore
Q= a
k +
(
Q0 − a
k
)
e−kt . (4.1.12)
SO LU TIO N (b) Since k> 0, limt→∞ e−kt = 0 , so from ( 4.1.12)
lim
t→∞
Q(t) = a
k.
This limit depends only on aand k, and not on Q0. W e say that a/k is the steady state value of Q. From
(4.1.12) we also see that Qapproaches its steady state value from above if Q0 > a/k, or from below if
Q0 <a/k . If Q0 = a/k , then Qremains constant (Figure 4.1.3).
Carbon Dating
The fact that Q approaches a steady state value in the situation discussed i n Example 4 underlies the
method of carbon dating , devised by the American chemist and Nobel Prize Winner W .S. Libby.
Carbon 12 is stable, but carbon-14, which is produced by cosm ic bombardment of nitrogen in the upper",Elementary Differential Equations with Boundary Value Problems.pdf
"Carbon 12 is stable, but carbon-14, which is produced by cosm ic bombardment of nitrogen in the upper
atmosphere, is radioactive with a half-life of about 5570 ye ars. Libby assumed that the quantity of carbon-
12 in the atmosphere has been constant throughout time, and t hat the quantity of radioactive carbon-14",Elementary Differential Equations with Boundary Value Problems.pdf
"136 Chapter 4 Applications of First Order Equations
achieved its steady state value long ago as a result of its cre ation and decomposition over millions of
years. These assumptions led Libby to conclude that the rati o of carbon-14 to carbon-12 has been nearly
constant for a long time. This constant, which we denote by R, has been determined experimentally .
Living cells absorb both carbon-12 and carbon-14 in the prop ortion in which they are present in the
environment. Therefore the ratio of carbon-14 to carbon-12 in a living cell is always R. However, when
the cell dies it ceases to absorb carbon, and the ratio of carb on-14 to carbon-12 decreases exponentially
as the radioactive carbon-14 decays. This is the basis for th e method of carbon dating, as illustrated in
the next example.
Example 4.1.5 An archaeologist investigating the site of an ancient villa ge finds a burial ground where",Elementary Differential Equations with Boundary Value Problems.pdf
"the next example.
Example 4.1.5 An archaeologist investigating the site of an ancient villa ge finds a burial ground where
the amount of carbon-14 present in individual remains is bet ween 42 and 44% of the amount present in
live individuals. Estimate the age of the village and the len gth of time for which it survived.
Solution Let Q= Q(t) be the quantity of carbon-14 in an individual set of remains tyears after death,
and let Q0 be the quantity that would be present in live individuals. Si nce carbon-14 decays exponentially
with half-life 5570 years, its decay constant is
k= ln 2
5570 .
Therefore
Q= Q0e−t(ln 2) / 5570
if we choose our time scale so that t0 = 0 is the time of death. If we know the present value of Qwe can
solve this equation for t, the number of years since death occurred. This yields
t= −5570 ln Q/Q 0
ln 2 .
It is given that Q= .42Q0 in the remains of individuals who died first. Therefore these deaths occurred
about
t1 = −5570 ln .42
ln 2 ≈ 6971",Elementary Differential Equations with Boundary Value Problems.pdf
"ln 2 .
It is given that Q= .42Q0 in the remains of individuals who died first. Therefore these deaths occurred
about
t1 = −5570 ln .42
ln 2 ≈ 6971
years ago. For the most recent deaths, Q= .44Q0; hence, these deaths occurred about
t2 = −5570 ln .44
ln 2 ≈ 6597
years ago. Therefore it’s reasonable to conclude that the vi llage was founded about 7000 years ago, and
lasted for about 400 years.
A Savings Program
Example 4.1.6 A person opens a savings account with an initial deposit of $1 000 and subsequently
deposits $50 per week. Find the value Q(t) of the account at time t> 0, assuming that the bank pays 6%
interest compounded continuously .
Solution Observe that Q isn’t continuous, since there are 52 discrete deposits per y ear of $50 each.
T o construct a mathematical model for this problem in the for m of a differential equation, we make
the simplifying assumption that the deposits are made conti nuously at a rate of $2600 per year. This",Elementary Differential Equations with Boundary Value Problems.pdf
"the simplifying assumption that the deposits are made conti nuously at a rate of $2600 per year. This
is essential, since solutions of differential equations ar e continuous functions. With this assumption, Q
increases continuously at the rate
Q′= 2600 + .06Q",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.1 Growth and Decay 137
and therefore Qsatisfies the differential equation
Q′− .06Q= 2600 . (4.1.13)
(Of course, we must recognize that the solution of this equat ion is an approximation to the true value of
Qat any given time. W e’ll discuss this further below .) Since e. 06t is a solution of the complementary
equation, the solutions of ( 4.1.13) are of the form Q = ue. 06t, where u′e. 06t = 2600 . Hence, u′ =
2600e−. 06t,
u= −2600
.06 e−. 06t + c
and
Q= ue. 06t = −2600
.06 + ce. 06t. (4.1.14)
Setting t= 0 and Q= 1000 here yields
c= 1000 + 2600
.06 ,
and substituting this into ( 4.1.14) yields
Q= 1000 e. 06t + 2600
.06 (e. 06t − 1), (4.1.15)
where the first term is the value due to the initial deposit and the second is due to the subsequent weekly
deposits.
Mathematical models must be tested for validity by comparin g predictions based on them with the
actual outcome of experiments. Example 6 is unusual in that w e can compute the exact value of the",Elementary Differential Equations with Boundary Value Problems.pdf
"actual outcome of experiments. Example 6 is unusual in that w e can compute the exact value of the
account at any specified time and compare it with the approxim ate value predicted by ( 4.1.15) (See
Exercise 21.). The follwing table gives a comparison for a ten year perio d. Each exact answer corresponds
to the time of the year-end deposit, and each year is assumed t o have exactly 52 weeks.
Y ear Approximate V alue of Q Exact V alue of P Error Percentage Error
(Example 4.1.6) (Exercise 21) Q− P (Q− P)/P
1 $ 3741.42 $ 3739.87 $ 1.55 .0413%
2 6652.36 6649.17 3.19 .0479
3 9743.30 9738.37 4.93 .0506
4 13,025.38 13,018.60 6.78 .0521
5 16,510.41 16,501.66 8.75 .0530
6 20,210.94 20,200.11 10.83 .0536
7 24,140.30 24,127.25 13.05 .0541
8 28,312.63 28,297.23 15.40 .0544
9 32,742.97 32,725.07 17.90 .0547
10 37,447.27 37,426.72 20.55 .0549",Elementary Differential Equations with Boundary Value Problems.pdf
"138 Chapter 4 Applications of First Order Equations
4.1 Exercises
1. The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left
at time t> 0 if Q(0) = 20 g.
2. The half-life of a radioactive substance is 2 days. Find the t ime required for a given amount of the
material to decay to 1/10 of its original mass.
3. A radioactive material loses 25% of its mass in 10 minutes. Wh at is its half-life?
4. A tree contains a known percentage p0 of a radioactive substance with half-life τ. When the tree
dies the substance decays and isn’t replaced. If the percent age of the substance in the fossilized
remains of such a tree is found to be p1, how long has the tree been dead?
5. If tp and tq are the times required for a radioactive material to decay to 1/p and 1/q times its
original mass (respectively), how are tp and tq related?
6. Find the decay constant kfor a radioactive substance, given that the mass of the subst ance is Q1",Elementary Differential Equations with Boundary Value Problems.pdf
"6. Find the decay constant kfor a radioactive substance, given that the mass of the subst ance is Q1
at time t1 and Q2 at time t2.
7. A process creates a radioactive substance at the rate of 2 g/h r and the substance decays at a rate
proportional to its mass, with constant of proportionality k= .1(hr)−1. If Q(t) is the mass of the
substance at time t, find limt→∞ Q(t).
8. A bank pays interest continuously at the rate of 6%. How long d oes it take for a deposit of Q0 to
grow in value to 2Q0?
9. At what rate of interest, compounded continuously , will a ba nk deposit double in value in 8 years?
10. A savings account pays 5% per annum interest compounded cont inuously . The initial deposit is
Q0 dollars. Assume that there are no subsequent withdrawals or deposits.
(a) How long will it take for the value of the account to triple?
(b) What is Q0 if the value of the account after 10 years is $100,000 dollars ?",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) How long will it take for the value of the account to triple?
(b) What is Q0 if the value of the account after 10 years is $100,000 dollars ?
11. A candymaker makes 500 pounds of candy per week, while his lar ge family eats the candy at a
rate equal to Q(t)/ 10 pounds per week, where Q(t) is the amount of candy present at time t.
(a) Find Q(t) for t> 0 if the candymaker has 250 pounds of candy at t= 0 .
(b) Find limt→∞ Q(t).
12. Suppose a substance decays at a yearly rate equal to half the s quare of the mass of the substance
present. If we start with 50 g of the substance, how long will i t be until only 25 g remain?
13. A super bread dough increases in volume at a rate proportiona l to the volume V present. If V
increases by a factor of 10 in 2 hours and V(0) = V0, find V at any time t. How long will it take
for V to increase to 100V0?
14. A radioactive substance decays at a rate proportional to the amount present, and half the original",Elementary Differential Equations with Boundary Value Problems.pdf
"for V to increase to 100V0?
14. A radioactive substance decays at a rate proportional to the amount present, and half the original
quantity Q0 is left after 1500 years. In how many years would the original amount be reduced to
3Q0/ 4? How much will be left after 2000 years?
15. A wizard creates gold continuously at the rate of 1 ounce per h our, but an assistant steals it con-
tinuously at the rate of 5% of however much is there per hour. L et W(t) be the number of ounces
that the wizard has at time t. Find W(t) and limt→∞ W(t) if W(0) = 1 .
16. A process creates a radioactive substance at the rate of 1 g/h r, and the substance decays at an hourly
rate equal to 1/10 of the mass present (expressed in grams). A ssuming that there are initially 20 g,
find the mass S(t) of the substance present at time t, and find limt→∞ S(t).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.1 Growth and Decay 139
17. A tank is empty at t = 0 . W ater is added to the tank at the rate of 10 gal/min, but it lea ks out
at a rate (in gallons per minute) equal to the number of gallon s in the tank. What is the smallest
capacity the tank can have if this process is to continue fore ver?
18. A person deposits $25,000 in a bank that pays 5% per year inter est, compounded continuously .
The person continuously withdraws from the account at the ra te of $750 per year. Find V(t), the
value of the account at time tafter the initial deposit.
19. A person has a fortune that grows at rate proportional to the s quare root of its worth. Find the
worth W of the fortune as a function of tif it was $1 million 6 months ago and is $4 million today .
20. Let p= p(t) be the quantity of a product present at time t. The product is manufactured continu-
ously at a rate proportional to p, with proportionality constant 1/2, and it’s consumed cont inuously",Elementary Differential Equations with Boundary Value Problems.pdf
"ously at a rate proportional to p, with proportionality constant 1/2, and it’s consumed cont inuously
at a rate proportional to p2, with proportionality constant 1/8. Find p(t) if p(0) = 100 .
21. (a) In the situation of Example 4.1.6 find the exact value P(t) of the person’s account after t
years, where tis an integer. Assume that each year has exactly 52 weeks, and include the
year-end deposit in the computation.
HIN T : At time t the initial $1000 has been on deposit for t years. There have been 52t
deposits of $50 each. The first $50 has been on deposit for t− 1/ 52 years, the second for
t− 2/ 52 years · · ·in general, the jth $50 has been on deposit for t− j/ 52 years (1 ≤
j ≤ 52t). Find the present value of each $50 deposit assuming 6% interest compounded
continuously, and use the formula
1 + x+ x2 + · · ·+ xn = 1 − xn+1
1 − x (x̸= 1)
to find their total value.
(b) Let
p(t) = Q(t) − P(t)
P(t)
be the relative error after tyears. Find
p(∞ ) = lim
t→∞
p(t).",Elementary Differential Equations with Boundary Value Problems.pdf
"1 − x (x̸= 1)
to find their total value.
(b) Let
p(t) = Q(t) − P(t)
P(t)
be the relative error after tyears. Find
p(∞ ) = lim
t→∞
p(t).
22. A homebuyer borrows P0 dollars at an annual interest rate r, agreeing to repay the loan with equal
monthly payments of M dollars per month over N years.
(a) Derive a differential equation for the loan principal (amou nt that the homebuyer owes) P(t)
at time t >0, making the simplifying assumption that the homebuyer repa ys the loan con-
tinuously rather than in discrete steps. (See Example 4.1.6 .)
(b) Solve the equation derived in (a).
(c) Use the result of (b) to determine an approximate value for M assuming that each year has
exactly 12 months of equal length.
(d) It can be shown that the exact value of M is given by
M = rP0
12
(1 − (1 + r/ 12)−12N ) −1
.
Compare the value of M obtained from the answer in (c) to the exact value if (i) P0 =
$50, 000, r= 7 1
2 %, N = 20 (ii) P0 = $150 , 000, r= 9 .0%, N = 30 .",Elementary Differential Equations with Boundary Value Problems.pdf
"$50, 000, r= 7 1
2 %, N = 20 (ii) P0 = $150 , 000, r= 9 .0%, N = 30 .
23. Assume that the homebuyer of Exercise 22 elects to repay the loan continuously at the rate of αM
dollars per month, where α is a constant greater than 1. (This is called accelerated payment .)",Elementary Differential Equations with Boundary Value Problems.pdf
"140 Chapter 4 Applications of First Order Equations
(a) Determine the time T(α) when the loan will be paid off and the amount S(α) that the home-
buyer will save.
(b) Suppose P0 = $50 , 000, r= 8 %, and N = 15 . Compute the savings realized by accelerated
payments with α = 1 .05, 1.10, and 1.15.
24. A benefactor wishes to establish a trust fund to pay a researc her’s salary for T years. The salary
is to start at S0 dollars per year and increase at a fractional rate of aper year. Find the amount
of money P0 that the benefactor must deposit in a trust fund paying inter est at a rate rper year.
Assume that the researcher’s salary is paid continuously , t he interest is compounded continuously ,
and the salary increases are granted continuously .
25. L A radioactive substance with decay constant kis produced at the rate of
at
1 + btQ(t)
units of mass per unit time, where a and b are positive constants and Q(t) is the mass of the",Elementary Differential Equations with Boundary Value Problems.pdf
"at
1 + btQ(t)
units of mass per unit time, where a and b are positive constants and Q(t) is the mass of the
substance present at time t; thus, the rate of production is small at the start and tends t o slow when
Qis large.
(a) Set up a differential equation for Q.
(b) Choose your own positive values for a, b, k, and Q0 = Q(0). Use a numerical method to
discover what happens to Q(t) as t→ ∞ . (Be precise, expressing your conclusions in terms
of a, b, k. However, no proof is required.)
26. L Follow the instructions of Exercise 25, assuming that the substance is produced at the rate of
at/ (1 + bt(Q(t))2) units of mass per unit of time.
27. L Follow the instructions of Exercise 25, assuming that the substance is produced at the rate of
at/ (1 + bt) units of mass per unit of time.
4.2 COOLING AND MIXING
Newton’s Law of Cooling
Newton’s law of cooling states that if an object with tempera ture T(t) at time t is in a medium with",Elementary Differential Equations with Boundary Value Problems.pdf
"4.2 COOLING AND MIXING
Newton’s Law of Cooling
Newton’s law of cooling states that if an object with tempera ture T(t) at time t is in a medium with
temperature Tm (t), the rate of change of T at time tis proportional to T(t) − Tm (t); thus, T satisfies a
differential equation of the form
T′= −k(T − Tm ). (4.2.1)
Here k >0, since the temperature of the object must decrease if T >Tm , or increase if T <Tm . W e’ll
call kthe temperature decay constant of the medium .
For simplicity , in this section we’ll assume that the medium is maintained at a constant temperature Tm .
This is another example of building a simple mathematical mo del for a physical phenomenon. Like most
mathematical models it has its limitations. For example, it ’s reasonable to assume that the temperature of
a room remains approximately constant if the cooling object is a cup of coffee, but perhaps not if it’s a
huge cauldron of molten metal. (For more on this see Exercise 17.)
T o solve ( 4.2.1), we rewrite it as",Elementary Differential Equations with Boundary Value Problems.pdf
"huge cauldron of molten metal. (For more on this see Exercise 17.)
T o solve ( 4.2.1), we rewrite it as
T′+ kT = kTm.
Since e−kt is a solution of the complementary equation, the solutions o f this equation are of the form
T = ue−kt , where u′e−kt = kTm , so u′= kTmekt . Hence,
u= Tm ekt + c,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.2 Cooling and Mixing 141
so
T = ue−kt = Tm + ce−kt .
If T(0) = T0, setting t= 0 here yields c= T0 − Tm , so
T = Tm + ( T0 − Tm )e−kt . (4.2.2)
Note that T − Tm decays exponentially , with decay constant k.
Example 4.2.1 A ceramic insulator is baked at 400◦C and cooled in a room in which the temperature is
25◦C. After 4 minutes the temperature of the insulator is 200◦C. What is its temperature after 8 minutes?
Solution Here T0 = 400 and Tm = 25 , so ( 4.2.2) becomes
T = 25 + 375 e−kt . (4.2.3)
W e determine kfrom the stated condition that T(4) = 200 ; that is,
200 = 25 + 375 e−4k;
hence,
e−4k = 175
375 = 7
15 .
T aking logarithms and solving for kyields
k= −1
4 ln 7
15 = 1
4 ln 15
7 .
Substituting this into ( 4.2.3) yields
T = 25 + 375 e− t
4 ln 15
7
(Figure 4.2.1). Therefore the temperature of the insulator after 8 minute s is
T(8) = 25 + 375 e−2 ln 15
7
= 25 + 375
(7
15
) 2
≈ 107◦C.",Elementary Differential Equations with Boundary Value Problems.pdf
"4 ln 15
7
(Figure 4.2.1). Therefore the temperature of the insulator after 8 minute s is
T(8) = 25 + 375 e−2 ln 15
7
= 25 + 375
(7
15
) 2
≈ 107◦C.
Example 4.2.2 An object with temperature 72◦F is placed outside, where the temperature is −20◦F . At
11:05 the temperature of the object is 60◦F and at 11:07 its temperature is 50◦F . At what time was the
object placed outside?
Solution Let T(t) be the temperature of the object at time t. For convenience, we choose the origin
t0 = 0 of the time scale to be 11:05 so that T0 = 60 . W e must determine the time τ when T(τ) = 72 .
Substituting T0 = 60 and Tm = −20 into (
4.2.2) yields
T = −20 + (60 − (−20)) e−kt
or
T = −20 + 80 e−kt . (4.2.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"142 Chapter 4 Applications of First Order Equations
 t
 T
5 10 15 20 25 30
100
150
200
250
300
350
400
50
Figure 4.2.1 T = 25 + 375 e−(t/ 4) ln 15 / 7
W e obtain kfrom the stated condition that the temperature of the object is 50 ◦F at 11:07. Since 11:07 is
t= 2 on our time scale, we can determine kby substituting T = 50 and t= 2 into ( 4.2.4) to obtain
50 = −20 + 80 e−2k
(Figure 4.2.2); hence,
e−2k = 70
80 = 7
8 .
T aking logarithms and solving for kyields
k= −1
2 ln 7
8 = 1
2 ln 8
7 .
Substituting this into ( 4.2.4) yields
T = −20 + 80 e− t
2 ln 8
7 ,
and the condition T(τ) = 72 implies that
72 = −20 + 80 e− τ
2 ln 8
7 ;
hence,
e− τ
2 ln 8
7 = 92
80 = 23
20 .
T aking logarithms and solving for τ yields
τ = −2 ln 23
20
ln 8
7
≈ − 2.09 min.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.2 Cooling and Mixing 143
 T
 t
−5 5 10 15 20 25 30 35 40
20
40
60
80
−20
100
 T=72
Figure 4.2.2 T = −20 + 80 e− t
2 ln 8
7
Therefore the object was placed outside about 2 minutes and 5 seconds before 11:05; that is, at 11:02:55.
Mixing Problems
In the next two examples a saltwater solution with a given con centration (weight of salt per unit volume
of solution) is added at a specified rate to a tank that initial ly contains saltwater with a different concentra-
tion. The problem is to determine the quantity of salt in the t ank as a function of time. This is an example
of a mixing problem . T o construct a tractable mathematical model for mixing pro blems we assume in
our examples (and most exercises) that the mixture is stirre d instantly so that the salt is always uniformly
distributed throughout the mixture. Exercises 22 and 23 deal with situations where this isn’t so, but the
distribution of salt becomes approximately uniform as t→ ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"distribution of salt becomes approximately uniform as t→ ∞ .
Example 4.2.3 A tank initially contains 40 pounds of salt dissolved in 600 g allons of water. Starting at
t0 = 0 , water that contains 1/2 pound of salt per gallon is poured in to the tank at the rate of 4 gal/min and
the mixture is drained from the tank at the same rate (Figure 4.2.3).
(a) Find a differential equation for the quantity Q(t) of salt in the tank at time t >0, and solve the
equation to determine Q(t).
(b) Find limt→∞ Q(t).
SO LU TIO N (a) T o find a differential equation for Q, we must use the given information to derive an
expression for Q′. But Q′is the rate of change of the quantity of salt in the tank change s with respect to
time; thus, if rate in denotes the rate at which salt enters the tank and rate out denotes the rate by which
it leaves, then
Q′= rate in − rate out . (4.2.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"144 Chapter 4 Applications of First Order Equations
600 gal
4 gal/min; .5 lb/gal
4 gal/min
Figure 4.2.3 A mixing problem
The rate in is (1
2 lb/gal
)
× (4 gal/min) = 2 lb/min.
Determining the rate out requires a little more thought. W e’ re removing 4 gallons of the mixture per
minute, and there are always 600 gallons in the tank; that is, we’re removing 1/ 150 of the mixture per
minute. Since the salt is evenly distributed in the mixture, we are also removing 1/ 150 of the salt per
minute. Therefore, if there are Q(t) pounds of salt in the tank at time t, the rate out at any time t is
Q(t)/ 150. Alternatively , we can arrive at this conclusion by arguing that
rate out = ( concentration) × (rate of flow out )
= ( lb/gal) × (gal/min)
= Q(t)
600 × 4 = Q(t)
150 .
W e can now write ( 4.2.5) as
Q′= 2 − Q
150 .
This first order equation can be rewritten as
Q′+ Q
150 = 2 .
Since e−t/ 150 is a solution of the complementary equation, the solutions o f this equation are of the form",Elementary Differential Equations with Boundary Value Problems.pdf
"Q′+ Q
150 = 2 .
Since e−t/ 150 is a solution of the complementary equation, the solutions o f this equation are of the form
Q= ue−t/ 150, where u′e−t/ 150 = 2 , so u′= 2 et/ 150. Hence,
u= 300 et/ 150 + c,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.2 Cooling and Mixing 145
100 200 300 400 500 600 700 800 900
50
100
150
200
250
300
 t
 Q
Figure 4.2.4 Q= 300 − 260e−t/ 150
so
Q= ue−t/ 150 = 300 + ce−t/ 150 (4.2.6)
(Figure 4.2.4). Since Q(0) = 40 , c= −260; therefore,
Q= 300 − 260e−t/ 150.
SO LU TIO N (b) From ( 4.2.6), we see that that limt→∞ Q(t) = 300 for any value of Q(0). This is intu-
itively reasonable, since the incoming solution contains 1 /2 pound of salt per gallon and there are always
600 gallons of water in the tank.
Example 4.2.4 A 500-liter tank initially contains 10 g of salt dissolved in 200 liters of water. Starting
at t0 = 0 , water that contains 1/4 g of salt per liter is poured into the tank at the rate of 4 liters/min and
the mixture is drained from the tank at the rate of 2 liters/mi n (Figure
4.2.5). Find a differential equation
for the quantity Q(t) of salt in the tank at time tprior to the time when the tank overflows and find the
concentration K(t) (g/liter ) of salt in the tank at any such time.",Elementary Differential Equations with Boundary Value Problems.pdf
"concentration K(t) (g/liter ) of salt in the tank at any such time.
Solution W e first determine the amount W(t) of solution in the tank at any time t prior to overflow .
Since W(0) = 200 and we’re adding 4 liters/min while removing only 2 liters/m in, there’s a net gain of
2 liters/min in the tank; therefore,
W(t) = 2 t+ 200 .
Since W(150) = 500 liters (capacity of the tank), this formula is valid for 0 ≤ t≤ 150.
Now let Q(t) be the number of grams of salt in the tank at time t, where 0 ≤ t ≤ 150. As in
Example 4.2.3,
Q′= rate in − rate out . (4.2.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"146 Chapter 4 Applications of First Order Equations
2t+200 liters
4 liters/min; .25 g/liter
Figure 4.2.5 Another mixing problem
The rate in is (1
4 g/liter
)
× (4 liters/min ) = 1 g/min. (4.2.8)
T o determine the rate out, we observe that since the mixture i s being removed from the tank at the constant
rate of 2 liters/min and there are 2t+ 200 liters in the tank at time t, the fraction of the mixture being
removed per minute at time tis
2
2t+ 200 = 1
t+ 100 .
W e’re removing this same fraction of the salt per minute. The refore, since there are Q(t) grams of salt in
the tank at time t,
rate out = Q(t)
t+ 100 . (4.2.9)
Alternatively , we can arrive at this conclusion by arguing t hat
rate out = ( concentration) × (rate of flow out ) = ( g/liter) × (liters/min)
= Q(t)
2t+ 200 × 2 = Q(t)
t+ 100 .
Substituting ( 4.2.8) and ( 4.2.9) into ( 4.2.7) yields
Q′= 1 − Q
t+ 100 , so Q′+ 1
t+ 100 Q= 1 . (4.2.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"= Q(t)
2t+ 200 × 2 = Q(t)
t+ 100 .
Substituting ( 4.2.8) and ( 4.2.9) into ( 4.2.7) yields
Q′= 1 − Q
t+ 100 , so Q′+ 1
t+ 100 Q= 1 . (4.2.10)
By separation of variables, 1/ (t+ 100) is a solution of the complementary equation, so the solution s of
(4.2.10) are of the form
Q= u
t+ 100 , where u′
t+ 100 = 1 , so u′= t+ 100 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.2 Cooling and Mixing 147
Hence,
u= (t+ 100) 2
2 + c. (4.2.11)
Since Q(0) = 10 and u= ( t+ 100) Q, ( 4.2.11) implies that
(100)(10) = (100)2
2 + c,
so
c= 100(10) − (100)2
2 = −4000
and therefore
u= (t+ 100) 2
2 − 4000.
Hence,
Q= u
t+ 200 = t+ 100
2 − 4000
t+ 100 .
Now let K(t) be the concentration of salt at time t. Then
K(t) = 1
4 − 2000
(t+ 100) 2
(Figure 4.2.6).
200 400 600 800 1000
 t
.05
.15
.25
.10
.20
 K
Figure 4.2.6 K(t) = 1
4 − 2000
(t+ 100) 2",Elementary Differential Equations with Boundary Value Problems.pdf
"148 Chapter 4 Applications of First Order Equations
4.2 Exercises
1. A thermometer is moved from a room where the temperature is 70◦F to a freezer where the tem-
perature is 12◦F. After 30 seconds the thermometer reads 40◦F . What does it read after 2 minutes?
2. A fluid initially at 100◦C is placed outside on a day when the temperature is −10◦C, and the
temperature of the fluid drops 20◦C in one minute. Find the temperature T(t) of the fluid for
t> 0.
3. At 12:00 PM a thermometer reading 10◦F is placed in a room where the temperature is 70◦F . It
reads 56◦ when it’s placed outside, where the temperature is 5◦F , at 12:03. What does it read at
12:05 PM ?
4. A thermometer initially reading 212◦F is placed in a room where the temperature is 70◦F . After 2
minutes the thermometer reads 125◦F .
(a) What does the thermometer read after 4 minutes?
(b) When will the thermometer read 72◦F?
(c) When will the thermometer read 69◦F?",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) What does the thermometer read after 4 minutes?
(b) When will the thermometer read 72◦F?
(c) When will the thermometer read 69◦F?
5. An object with initial temperature 150◦C is placed outside, where the temperature is 35◦C. Its
temperatures at 12:15 and 12:20 are 120◦C and 90◦C, respectively .
(a) At what time was the object placed outside?
(b) When will its temperature be 40◦C?
6. An object is placed in a room where the temperature is 20◦C. The temperature of the object drops
by 5◦C in 4 minutes and by 7◦C in 8 minutes. What was the temperature of the object when it w as
initially placed in the room?
7. A cup of boiling water is placed outside at 1:00 PM . One minute later the temperature of the water
is 152◦F . After another minute its temperature is 112◦F . What is the outside temperature?
8. A tank initially contains 40 gallons of pure water. A solutio n with 1 gram of salt per gallon of",Elementary Differential Equations with Boundary Value Problems.pdf
"8. A tank initially contains 40 gallons of pure water. A solutio n with 1 gram of salt per gallon of
water is added to the tank at 3 gal/min, and the resulting solu tion dranes out at the same rate. Find
the quantity Q(t) of salt in the tank at time t> 0.
9. A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. W ater with 1/2
pound of salt per gallon is added to the tank at 6 gal/min, and t he resulting solution leaves at the
same rate. Find the quantity Q(t) of salt in the tank at time t> 0.
10. A tank initially contains 100 liters of a salt solution with a concentration of .1 g/liter. A solution
with a salt concentration of .3 g/liter is added to the tank at 5 liters/min, and the resulting mixture
is drained out at the same rate. Find the concentration K(t) of salt in the tank as a function of t.
11. A 200 gallon tank initially contains 100 gallons of water wit h 20 pounds of salt. A salt solution",Elementary Differential Equations with Boundary Value Problems.pdf
"11. A 200 gallon tank initially contains 100 gallons of water wit h 20 pounds of salt. A salt solution
with 1/4 pound of salt per gallon is added to the tank at 4 gal/m in, and the resulting mixture is
drained out at 2 gal/min. Find the quantity of salt in the tank as it’s about to overflow .
12. Suppose water is added to a tank at 10 gal/min, but leaks out at the rate of 1/5 gal/min for each
gallon in the tank. What is the smallest capacity the tank can have if the process is to continue
indefinitely?
13. A chemical reaction in a laboratory with volume V (in ft 3) produces q1 ft3/min of a noxious gas as
a byproduct. The gas is dangerous at concentrations greater than c, but harmless at concentrations
≤ c. Intake fans at one end of the laboratory pull in fresh air at t he rate of q2 ft3/min and exhaust
fans at the other end exhaust the mixture of gas and air from th e laboratory at the same rate.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.2 Cooling and Mixing 149
Assuming that the gas is always uniformly distributed in the room and its initial concentration c0
is at a safe level, find the smallest value of q2 required to maintain safe conditions in the laboratory
for all time.
14. A 1200-gallon tank initially contains 40 pounds of salt diss olved in 600 gallons of water. Starting
at t0 = 0 , water that contains 1/2 pound of salt per gallon is added to t he tank at the rate of 6
gal/min and the resulting mixture is drained from the tank at 4 gal/min. Find the quantity Q(t) of
salt in the tank at any time t> 0 prior to overflow .
15. T ank T1 initially contain 50 gallons of pure water. Starting at t0 = 0 , water that contains 1 pound
of salt per gallon is poured into T1 at the rate of 2 gal/min. The mixture is drained from T1 at the
same rate into a second tank T2, which initially contains 50 gallons of pure water. Also sta rting at",Elementary Differential Equations with Boundary Value Problems.pdf
"same rate into a second tank T2, which initially contains 50 gallons of pure water. Also sta rting at
t0 = 0 , a mixture from another source that contains 2 pounds of salt per gallon is poured into T2
at the rate of 2 gal/min. The mixture is drained from T2 at the rate of 4 gal/min.
(a) Find a differential equation for the quantity Q(t) of salt in tank T2 at time t> 0.
(b) Solve the equation derived in (a) to determine Q(t).
(c) Find limt→∞ Q(t).
16. Suppose an object with initial temperature T0 is placed in a sealed container, which is in turn placed
in a medium with temperature Tm . Let the initial temperature of the container be S0 . Assume that
the temperature of the object does not affect the temperatur e of the container, which in turn does
not affect the temperature of the medium. (These assumption s are reasonable, for example, if the
object is a cup of coffee, the container is a house, and the med ium is the atmosphere.)",Elementary Differential Equations with Boundary Value Problems.pdf
"object is a cup of coffee, the container is a house, and the med ium is the atmosphere.)
(a) Assuming that the container and the medium have distinct tem perature decay constants kand
km respectively , use Newton’s law of cooling to find the tempera tures S(t) and T(t) of the
container and object at time t.
(b) Assuming that the container and the medium have the same temp erature decay constant k,
use Newton’s law of cooling to find the temperatures S(t) and T(t) of the container and
object at time t.
(c) Find lim .t→∞ S(t) and limt→∞ T(t) .
17. In our previous examples and exercises concerning Newton’s law of cooling we assumed that the
temperature of the medium remains constant. This model is ad equate if the heat lost or gained by
the object is insignificant compared to the heat required to c ause an appreciable change in the tem-
perature of the medium. If this isn’t so, we must use a model th at accounts for the heat exchanged",Elementary Differential Equations with Boundary Value Problems.pdf
"perature of the medium. If this isn’t so, we must use a model th at accounts for the heat exchanged
between the object and the medium. Let T = T(t) and Tm = Tm (t) be the temperatures of the
object and the medium, respectively , and let T0 and Tm0 be their initial values. Again, we assume
that T and Tm are related by Newton’s law of cooling,
T′= −k(T − Tm ). (A)
W e also assume that the change in heat of the object as its temp erature changes from T0 to T is
a(T − T0) and that the change in heat of the medium as its temperature ch anges from Tm0 to Tm
is am (Tm − Tm0), where aand am are positive constants depending upon the masses and therma l
properties of the object and medium, respectively . If we ass ume that the total heat of the system
consisting of the object and the medium remains constant (th at is, energy is conserved), then
a(T − T0) + am(Tm − Tm0) = 0 . (B)
(a) Equation (A) involves two unknown functions T and Tm . Use (A) and (B) to derive a differ-",Elementary Differential Equations with Boundary Value Problems.pdf
"a(T − T0) + am(Tm − Tm0) = 0 . (B)
(a) Equation (A) involves two unknown functions T and Tm . Use (A) and (B) to derive a differ-
ential equation involving only T.",Elementary Differential Equations with Boundary Value Problems.pdf
"150 Chapter 4 Applications of First Order Equations
(b) Find T(t) and Tm (t) for t> 0.
(c) Find limt→∞ T(t) and limt→∞ Tm (t).
18. Control mechanisms allow fluid to flow into a tank at a rate prop ortional to the volume V of fluid
in the tank, and to flow out at a rate proportional to V2. Suppose V(0) = V0 and the constants of
proportionality are aand b, respectively . Find V(t) for t> 0 and find limt→∞ V(t).
19. Identical tanks T1 and T2 initially contain W gallons each of pure water. Starting at t0 = 0 , a
salt solution with constant concentration cis pumped into T1 at r gal/min and drained from T1
into T2 at the same rate. The resulting mixture in T2 is also drained at the same rate. Find the
concentrations c1(t) and c2(t) in tanks T1 and T2 for t> 0.
20. An infinite sequence of identical tanks T1, T2, . . . , Tn , . . . , initially contain W gallons each of
pure water. They are hooked together so that fluid drains from Tn into Tn+1 (n= 1 , 2, · · ·). A salt",Elementary Differential Equations with Boundary Value Problems.pdf
"pure water. They are hooked together so that fluid drains from Tn into Tn+1 (n= 1 , 2, · · ·). A salt
solution is circulated through the tanks so that it enters an d leaves each tank at the constant rate of
rgal/min. The solution has a concentration of cpounds of salt per gallon when it enters T1.
(a) Find the concentration cn(t) in tank Tn for t> 0.
(b) Find limt→∞ cn(t) for each n.
21. T anks T1 and T2 have capacities W1 and W2 liters, respectively . Initially they are both full of dye
solutions with concentrations c1 and c2 grams per liter. Starting at t0 = 0 , the solution from T1 is
pumped into T2 at a rate of rliters per minute, and the solution from T2 is pumped into T1 at the
same rate.
(a) Find the concentrations c1(t) and c2(t) of the dye in T1 and T2 for t> 0.
(b) Find limt→∞ c1(t) and limt→∞ c2(t).
22. L Consider the mixing problem of Example 4.2.3, but without the assumption that the mixture",Elementary Differential Equations with Boundary Value Problems.pdf
"(b) Find limt→∞ c1(t) and limt→∞ c2(t).
22. L Consider the mixing problem of Example 4.2.3, but without the assumption that the mixture
is stirred instantly so that the salt is always uniformly dis tributed throughout the mixture. Assume
instead that the distribution approaches uniformity as t→ ∞ . In this case the differential equation
for Qis of the form
Q′+ a(t)
150 Q= 2
where limt→∞ a(t) = 1 .
(a) Assuming that Q(0) = Q0, can you guess the value of limt→∞ Q(t)?.
(b) Use numerical methods to confirm your guess in the these cases :
(i) a(t) = t/ (1 + t) (ii) a(t) = 1 − e−t2
(iii) a(t) = 1 − sin(e−t ).
23. L Consider the mixing problem of Example 4.2.4 in a tank with infinite capacity , but without
the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed
throughout the mixture. Assume instead that the distributi on approaches uniformity as t→ ∞ . In
this case the differential equation for Qis of the form
Q′+ a(t)
t+ 100 Q= 1",Elementary Differential Equations with Boundary Value Problems.pdf
"this case the differential equation for Qis of the form
Q′+ a(t)
t+ 100 Q= 1
where limt→∞ a(t) = 1 .
(a) Let K(t) be the concentration of salt at time t. Assuming that Q(0) = Q0, can you guess
the value of limt→∞ K(t)?
(b) Use numerical methods to confirm your guess in the these cases :
(i) a(t) = t/ (1 + t) (ii) a(t) = 1 − e−t2
(iii) a(t) = 1 + sin( e−t ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.3 Elementary Mechanics 151
4.3 ELEMENT ARY MECHANICS
Newton’s Second Law of Motion
In this section we consider an object with constant mass m moving along a line under a force F. Let
y = y(t) be the displacement of the object from a reference point on th e line at time t, and let v = v(t)
and a = a(t) be the velocity and acceleration of the object at time t. Thus, v = y′and a = v′= y′′,
where the prime denotes differentiation with respect to t. Newton’s second law of motion asserts that the
force F and the acceleration aare related by the equation
F = ma. (4.3.1)
Units
In applications there are three main sets of units in use for l ength, mass, force, and time: the cgs, mks, and
British systems. All three use the second as the unit of time. T able 1 shows the other units. Consistent
with ( 4.3.1), the unit of force in each system is defined to be the force req uired to impart an acceleration
of (one unit of length) /s 2 to one unit of mass.
Length Force Mass",Elementary Differential Equations with Boundary Value Problems.pdf
"of (one unit of length) /s 2 to one unit of mass.
Length Force Mass
cgs centimeter (cm) dyne (d) gram (g)
mks meter (m) newton (N) kilogram (kg)
British foot (ft) pound (lb) slug (sl)
T able 1.
If we assume that Earth is a perfect sphere with constant mass density , Newton’s law of gravitation
(discussed later in this section) asserts that the force exe rted on an object by Earth’s gravitational field
is proportional to the mass of the object and inversely propo rtional to the square of its distance from the
center of Earth. However, if the object remains sufficiently close to Earth’s surface, we may assume that
the gravitational force is constant and equal to its value at the surface. The magnitude of this force is
mg, where g is called the acceleration due to gravity . (T o be completely accurate, g should be called
the magnitude of the acceleration due to gravity at Earth’s surf ace.) This quantity has been determined
experimentally . Approximate values of gare
g = 980 cm/s2 (cgs)",Elementary Differential Equations with Boundary Value Problems.pdf
"experimentally . Approximate values of gare
g = 980 cm/s2 (cgs)
g = 9 .8 m/s2 (mks)
g = 32 ft/s2 (British).
In general, the force F in ( 4.3.1) may depend upon t, y, and y′. Since a= y′′, ( 4.3.1) can be written in
the form
my′′= F(t,y,y ′), (4.3.2)
which is a second order equation. W e’ll consider this equati on with restrictions on F later; however, since
Chapter 2 dealt only with first order equations, we consider h ere only problems in which ( 4.3.2) can be
recast as a first order equation. This is possible if F does not depend on y, so ( 4.3.2) is of the form
my′′= F(t,y ′).
Letting v= y′and v′= y′′yields a first order equation for v:
mv′= F(t,v ). (4.3.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"152 Chapter 4 Applications of First Order Equations
Solving this equation yields vas a function of t. If we know y(t0 ) for some time t0, we can integrate vto
obtain yas a function of t.
Equations of the form ( 4.3.3) occur in problems involving motion through a resisting med ium.
Motion Through a Resisting Medium Under Constant Gravitati onal Force
Now we consider an object moving vertically in some medium. W e assume that the only forces acting on
the object are gravity and resistance from the medium. W e als o assume that the motion takes place close
to Earth’s surface and take the upward direction to be positi ve, so the gravitational force can be assumed
to have the constant value −mg. W e’ll see that, under reasonable assumptions on the resist ing force, the
velocity approaches a limit as t→ ∞ . W e call this limit the terminal velocity .
Example 4.3.1 An object with mass mmoves under constant gravitational force through a medium t hat",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 4.3.1 An object with mass mmoves under constant gravitational force through a medium t hat
exerts a resistance with magnitude proportional to the spee d of the object. (Recall that the speed of an
object is |v|, the absolute value of its velocity v.) Find the velocity of the object as a function of t, and
find the terminal velocity . Assume that the initial velocity is v0.
Solution The total force acting on the object is
F = −mg+ F1, (4.3.4)
where −mgis the force due to gravity and F1 is the resisting force of the medium, which has magnitude
k|v|, where k is a positive constant. If the object is moving downward ( v ≤ 0), the resisting force is
upward (Figure 4.3.1(a)), so
F1 = k|v|= k(−v) = −kv.
On the other hand, if the object is moving upward ( v ≥ 0), the resisting force is downward (Fig-
ure 4.3.1(b)), so
F1 = −k|v|= −kv.
Thus, ( 4.3.4) can be written as
F = −mg− kv, (4.3.5)
regardless of the sign of the velocity .
From Newton’s second law of motion,
F = ma= mv′,",Elementary Differential Equations with Boundary Value Problems.pdf
"Thus, ( 4.3.4) can be written as
F = −mg− kv, (4.3.5)
regardless of the sign of the velocity .
From Newton’s second law of motion,
F = ma= mv′,
so ( 4.3.5) yields
mv′= −mg− kv,
or
v′+ k
mv= −g. (4.3.6)
Since e−kt/m is a solution of the complementary equation, the solutions o f ( 4.3.6) are of the form v =
ue−kt/m , where u′e−kt/m = −g, so u′= −gekt/m . Hence,
u= −mg
k ekt/m + c,
so
v= ue−kt/m = −mg
k + ce−kt/m . (4.3.7)
Since v(0) = v0,
v0 = −mg
k + c,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.3 Elementary Mechanics 153
 v
  v
 F1 = − kv
 F1 = − kv
(a) (b)
Figure 4.3.1 Resistive forces
so
c= v0 + mg
k
and ( 4.3.7) becomes
v= −mg
k +
(
v0 + mg
k
)
e−kt/m .
Letting t→ ∞ here shows that the terminal velocity is
lim
t→∞
v(t) = −mg
k ,
which is independent of the initial velocity v0 (Figure 4.3.2).
Example 4.3.2 A 960-lb object is given an initial upward velocity of 60 ft/s near the surface of Earth.
The atmosphere resists the motion with a force of 3 lb for each ft/s of speed. Assuming that the only other
force acting on the object is constant gravity , find its veloc ity v as a function of t, and find its terminal
velocity .
Solution Since mg = 960 and g = 32 , m= 960 / 32 = 30 . The atmospheric resistance is −3vlb if vis
expressed in feet per second. Therefore
30v′= −960 − 3v,
which we rewrite as
v′+ 1
10 v= −32.",Elementary Differential Equations with Boundary Value Problems.pdf
"154 Chapter 4 Applications of First Order Equations
 − mg/k
 t
 v
Figure 4.3.2 Solutions of mv′= −mg− kv
Since e−t/ 10 is a solution of the complementary equation, the solutions o f this equation are of the form
v= ue−t/ 10, where u′e−t/ 10 = −32, so u′= −32et/ 10. Hence,
u= −320et/ 10 + c,
so
v= ue−t/ 10 = −320 + ce−t/ 10. (4.3.8)
The initial velocity is 60 ft/s in the upward (positive) dire ction; hence, v0 = 60 . Substituting t = 0 and
v= 60 in ( 4.3.8) yields
60 = −320 + c,
so c= 380 , and ( 4.3.8) becomes
v = −320 + 380 e−t/ 10 ft/s
The terminal velocity is
lim
t→∞
v(t) = −320 ft/s.
Example 4.3.3 A 10 kg mass is given an initial velocity v0 ≤ 0 near Earth’s surface. The only forces
acting on it are gravity and atmospheric resistance proport ional to the square of the speed. Assuming that
the resistance is 8 N if the speed is 2 m/s, find the velocity of t he object as a function of t, and find the
terminal velocity .",Elementary Differential Equations with Boundary Value Problems.pdf
"the resistance is 8 N if the speed is 2 m/s, find the velocity of t he object as a function of t, and find the
terminal velocity .
Solution Since the object is falling, the resistance is in the upward ( positive) direction. Hence,
mv′= −mg+ kv2, (4.3.9)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.3 Elementary Mechanics 155
where kis a constant. Since the magnitude of the resistance is 8 N whe n v= 2 m/s,
k(22) = 8 ,
so k= 2 N-s2/ m2. Since m= 10 and g= 9 .8, ( 4.3.9) becomes
10v′= −98 + 2 v2 = 2( v2 − 49). (4.3.10)
If v0 = −7, then v≡ − 7 for all t≥ 0. If v0 ̸= −7, we separate variables to obtain
1
v2 − 49 v′= 1
5 , (4.3.11)
which is convenient for the required partial fraction expan sion
1
v2 − 49 = 1
(v− 7)(v+ 7) = 1
14
[ 1
v− 7 − 1
v+ 7
]
. (4.3.12)
Substituting ( 4.3.12) into ( 4.3.11) yields
1
14
[ 1
v− 7 − 1
v+ 7
]
v′= 1
5 ,
so [ 1
v− 7 − 1
v+ 7
]
v′= 14
5 .
Integrating this yields
ln |v− 7| −ln |v+ 7 |= 14 t/ 5 + k.
Therefore ⏐
⏐
⏐
⏐
v− 7
v+ 7
⏐
⏐
⏐
⏐= eke14t/ 5.
Since Theorem
2.3.1 implies that (v − 7)/ (v + 7) can’t change sign (why?), we can rewrite the last
equation as
v− 7
v+ 7 = ce14t/ 5, (4.3.13)
which is an implicit solution of ( 4.3.10). Solving this for vyields
v= −7 c+ e−14t/ 5
c− e−14t/ 5 . (4.3.14)
Since v(0) = v0, it ( 4.3.13) implies that",Elementary Differential Equations with Boundary Value Problems.pdf
"v= −7 c+ e−14t/ 5
c− e−14t/ 5 . (4.3.14)
Since v(0) = v0, it ( 4.3.13) implies that
c= v0 − 7
v0 + 7 .
Substituting this into ( 4.3.14) and simplifying yields
v= −7 v0(1 + e−14t/ 5) − 7(1 − e−14t/ 5)
v0(1 − e−14t/ 5) − 7(1 + e−14t/ 5 .
Since v0 ≤ 0, vis defined and negative for all t> 0. The terminal velocity is
lim
t→∞
v(t) = −7 m/s,
independent of v0. More generally , it can be shown (Exercise 11) that if vis any solution of ( 4.3.9) such
that v0 ≤ 0 then
lim
t→∞
v(t) = −
√ mg
k
(Figure 4.3.3).",Elementary Differential Equations with Boundary Value Problems.pdf
"156 Chapter 4 Applications of First Order Equations
 t
 v
v = − (mg/k) 1/2
Figure 4.3.3 Solutions of mv′= −mg+ kv2, v(0) = v0 ≤ 0
Example 4.3.4 A 10-kg mass is launched vertically upward from Earth’s surf ace with an initial velocity
of v0 m/s. The only forces acting on the mass are gravity and atmosp heric resistance proportional to the
square of the speed. Assuming that the atmospheric resistan ce is 8 N if the speed is 2 m/s, find the time
T required for the mass to reach maximum altitude.
Solution The mass will climb while v >0 and reach its maximum altitude when v = 0 . Therefore v >0
for 0 ≤ t<T and v(T) = 0 . Although the mass of the object and our assumptions concern ing the forces
acting on it are the same as those in Example 3, ( 4.3.10) does not apply here, since the resisting force is
negative if v >0; therefore, we replace ( 4.3.10) by
10v′= −98 − 2v2. (4.3.15)
Separating variables yields
5
v2 + 49 v′= −1,
and integrating this yields
5
7 tan−1 v
7 = −t+ c.",Elementary Differential Equations with Boundary Value Problems.pdf
"10v′= −98 − 2v2. (4.3.15)
Separating variables yields
5
v2 + 49 v′= −1,
and integrating this yields
5
7 tan−1 v
7 = −t+ c.
(Recall that tan−1 uis the number θ such that −π/ 2 <θ<π/ 2 and tan θ = u.) Since v(0) = v0,
c= 5
7 tan−1 v0
7 ,
so vis defined implicitly by
5
7 tan−1 v
7 = −t+ 5
7 tan−1 v0
7 , 0 ≤ t≤ T. (4.3.16)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.3 Elementary Mechanics 157
0.2 0.4 0.6 0.8 1
10
20
30
40
50
 t
 v
Figure 4.3.4 Solutions of ( 4.3.15) for various v0 >0
Solving this for vyields
v = 7 tan
(
−7t
5 + tan −1 v0
7
)
. (4.3.17)
Using the identity
tan(A− B) = tan A− tan B
1 + tan Atan B
with A= tan −1(v0/ 7) and B= 7 t/ 5, and noting that tan(tan−1 θ) = θ, we can simplify ( 4.3.17) to
v= 7 v0 − 7 tan(7t/ 5)
7 + v0 tan(7t/ 5) .
Since v(T) = 0 and tan−1(0) = 0 , ( 4.3.16) implies that
−T + 5
7 tan−1 v0
7 = 0 .
Therefore
T = 5
7 tan−1 v0
7 .
Since tan−1(v0/ 7) <π/ 2 for all v0, the time required for the mass to reach its maximum altitude is less
than 5π
14 ≈ 1.122 s
regardless of the initial velocity . Figure 4.3.4 shows graphs of vover [0,T ] for various values of v0.",Elementary Differential Equations with Boundary Value Problems.pdf
"158 Chapter 4 Applications of First Order Equations
 y = − R
 y = 0
 y = h
 y
Figure 4.3.5 Escape velocity
Escape V elocity
Suppose a space vehicle is launched vertically and its fuel i s exhausted when the vehicle reaches an
altitude habove Earth, where his sufficiently large so that resistance due to Earth’s atmos phere can be
neglected. Let t= 0 be the time when burnout occurs. Assuming that the gravitati onal forces of all other
celestial bodies can be neglected, the motion of the vehicle for t >0 is that of an object with constant
mass munder the influence of Earth’s gravitational force, which we now assume to vary inversely with
the square of the distance from Earth’s center; thus, if we ta ke the upward direction to be positive then
gravitational force on the vehicle at an altitude yabove Earth is
F = − K
(y+ R)2 , (4.3.18)
where Ris Earth’s radius (Figure 4.3.5).
Since F = −mgwhen y= 0 , setting y= 0 in ( 4.3.18) yields
−mg= − K
R2 ;",Elementary Differential Equations with Boundary Value Problems.pdf
"F = − K
(y+ R)2 , (4.3.18)
where Ris Earth’s radius (Figure 4.3.5).
Since F = −mgwhen y= 0 , setting y= 0 in ( 4.3.18) yields
−mg= − K
R2 ;
therefore K = mgR2 and ( 4.3.18) can be written more specifically as
F = − mgR2
(y+ R)2 . (4.3.19)
From Newton’s second law of motion,
F = md2y
dt2 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.3 Elementary Mechanics 159
so ( 4.3.19) implies that
d2y
dt2 = − gR2
(y+ R)2 . (4.3.20)
W e’ll show that there’s a number ve, called the escape velocity , with these properties:
1. If v0 ≥ ve then v(t) > 0 for all t >0, and the vehicle continues to climb for all t >0; that is,
it “escapes” Earth. (Is it really so obvious that limt→∞ y(t) = ∞ in this case? For a proof, see
Exercise 20.)
2. If v0 < ve then v(t) decreases to zero and becomes negative. Therefore the vehic le attains a
maximum altitude ym and falls back to Earth.
Since ( 4.3.20) is second order, we can’t solve it by methods discussed so fa r. However, we’re concerned
with vrather than y, and vis easier to find. Since v= y′the chain rule implies that
d2y
dt2 = dv
dt = dv
dy
dy
dt = vdv
dy.
Substituting this into ( 4.3.20) yields the first order separable equation
vdv
dy = − gR2
(y+ R)2 . (4.3.21)
When t = 0 , the velocity is v0 and the altitude is h. Therefore we can obtain v as a function of y by",Elementary Differential Equations with Boundary Value Problems.pdf
"vdv
dy = − gR2
(y+ R)2 . (4.3.21)
When t = 0 , the velocity is v0 and the altitude is h. Therefore we can obtain v as a function of y by
solving the initial value problem
vdv
dy = − gR2
(y+ R)2 , v (h) = v0.
Integrating ( 4.3.21) with respect to yyields
v2
2 = gR2
y+ R + c. (4.3.22)
Since v(h) = v0,
c= v2
0
2 − gR2
h+ R,
so ( 4.3.22) becomes
v2
2 = gR2
y+ R +
(v2
0
2 − gR2
h+ R
)
. (4.3.23)
If
v0 ≥
(2gR2
h+ R
) 1/ 2
,
the parenthetical expression in ( 4.3.23) is nonnegative, so v(y) >0 for y >h. This proves that there’s an
escape velocity ve. W e’ll now prove that
ve =
(2gR2
h+ R
) 1/ 2
by showing that the vehicle falls back to Earth if
v0 <
(2gR2
h+ R
) 1/ 2
. (4.3.24)",Elementary Differential Equations with Boundary Value Problems.pdf
"160 Chapter 4 Applications of First Order Equations
If ( 4.3.24) holds then the parenthetical expression in ( 4.3.23) is negative and the vehicle will attain a
maximum altitude ym >h that satisfies the equation
0 = gR2
ym + R +
(v2
0
2 − gR2
h+ R
)
.
The velocity will be zero at the maximum altitude, and the obj ect will then fall to Earth under the influence
of gravity .
4.3 Exercises
Except where directed otherwise, assume that the magnitude of the gravitational force on an object with
mass mis constant and equal to mg. In exercises involving vertical motion take the upward dir ection to
be positive.
1. A firefighter who weighs 192 lb slides down an infinitely long fir e pole that exerts a frictional
resistive force with magnitude proportional to his speed, w ith k = 2 .5 lb-s/ft. Assuming that he
starts from rest, find his velocity as a function of time and fin d his terminal velocity .
2. A firefighter who weighs 192 lb slides down an infinitely long fir e pole that exerts a frictional",Elementary Differential Equations with Boundary Value Problems.pdf
"2. A firefighter who weighs 192 lb slides down an infinitely long fir e pole that exerts a frictional
resistive force with magnitude proportional to her speed, w ith constant of proportionality k. Find
k, given that her terminal velocity is -16 ft/s, and then find he r velocity vas a function of t. Assume
that she starts from rest.
3. A boat weighs 64,000 lb. Its propellor produces a constant th rust of 50,000 lb and the water exerts
a resistive force with magnitude proportional to the speed, with k = 2000 lb-s/ft. Assuming that
the boat starts from rest, find its velocity as a function of ti me, and find its terminal velocity .
4. A constant horizontal force of 10 N pushes a 20 kg-mass throug h a medium that resists its motion
with .5 N for every m/s of speed. The initial velocity of the ma ss is 7 m/s in the direction opposite
to the direction of the applied force. Find the velocity of th e mass for t> 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"to the direction of the applied force. Find the velocity of th e mass for t> 0.
5. A stone weighing 1/2 lb is thrown upward from an initial heigh t of 5 ft with an initial speed of 32
ft/s. Air resistance is proportional to speed, with k = 1 / 128 lb-s/ft. Find the maximum height
attained by the stone.
6. A 3200-lb car is moving at 64 ft/s down a 30-degree grade when i t runs out of fuel. Find its
velocity after that if friction exerts a resistive force wit h magnitude proportional to the square of
the speed, with k= 1 lb-s2/ ft2. Also find its terminal velocity .
7. A 96 lb weight is dropped from rest in a medium that exerts a res istive force with magnitude
proportional to the speed. Find its velocity as a function of time if its terminal velocity is -128 ft/s.
8. An object with mass mmoves vertically through a medium that exerts a resistive fo rce with magni-
tude proportional to the speed. Let y = y(t) be the altitude of the object at time t, with y(0) = y0.",Elementary Differential Equations with Boundary Value Problems.pdf
"tude proportional to the speed. Let y = y(t) be the altitude of the object at time t, with y(0) = y0.
Use the results of Example 4.3.1 to show that
y(t) = y0 + m
k(v0 − v− gt).
9. An object with mass mis launched vertically upward with initial velocity v0 from Earth’s surface
(y0 = 0 ) in a medium that exerts a resistive force with magnitude pro portional to the speed. Find
the time T when the object attains its maximum altitude ym . Then use the result of Exercise 8 to
find ym.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.3 Elementary Mechanics 161
10. An object weighing 256 lb is dropped from rest in a medium that exerts a resistive force with
magnitude proportional to the square of the speed. The magni tude of the resisting force is 1 lb
when |v|= 4 ft/s. Find vfor t> 0, and find its terminal velocity .
11. An object with mass mis given an initial velocity v0 ≤ 0 in a medium that exerts a resistive force
with magnitude proportional to the square of the speed. Find the velocity of the object for t >0,
and find its terminal velocity .
12. An object with mass m is launched vertically upward with initial velocity v0 in a medium that
exerts a resistive force with magnitude proportional to the square of the speed.
(a) Find the time T when the object reaches its maximum altitude.
(b) Use the result of Exercise 11 to find the velocity of the object for t>T .
13. L An object with mass mis given an initial velocity v0 ≤ 0 in a medium that exerts a resistive",Elementary Differential Equations with Boundary Value Problems.pdf
"13. L An object with mass mis given an initial velocity v0 ≤ 0 in a medium that exerts a resistive
force of the form a|v|/ (1 + |v|), where ais positive constant.
(a) Set up a differential equation for the speed of the object.
(b) Use your favorite numerical method to solve the equation you found in (a), to convince your-
self that there’s a unique number a0 such that limt→∞ s(t) = ∞ if a≤ a0 and limt→∞ s(t)
exists (finite) if a > a0. (W e say that a0 is the bifurcation value of a.) Try to find a0 and
limt→∞ s(t) in the case where a>a 0. H IN T : See Exercise 14.
14. An object of mass m falls in a medium that exerts a resistive force f = f(s), where s = |v|is
the speed of the object. Assume that f(0) = 0 and f is strictly increasing and differentiable on
(0, ∞ ).
(a) Write a differential equation for the speed s = s(t) of the object. T ake it as given that all
solutions of this equation with s(0) ≥ 0 are defined for all t >0 (which makes good sense
on physical grounds).",Elementary Differential Equations with Boundary Value Problems.pdf
"solutions of this equation with s(0) ≥ 0 are defined for all t >0 (which makes good sense
on physical grounds).
(b) Show that if lims→∞ f(s) ≤ mgthen limt→∞ s(t) = ∞ .
(c) Show that if lims→∞ f(s) >mg then limt→∞ s(t) = sT (terminal speed), where f(sT ) =
mg. H IN T : Use Theorem 2.3.1.
15. A 100-g mass with initial velocity v0 ≤ 0 falls in a medium that exerts a resistive force proportional
to the fourth power of the speed. The resistance is .1 N if the speed is 3 m/s.
(a) Set up the initial value problem for the velocity vof the mass for t> 0.
(b) Use Exercise 14(c) to determine the terminal velocity of the object.
(c) C T o confirm your answer to (b), use one of the numerical methods studied in Chapter 3
to compute approximate solutions on [0, 1] (seconds) of the initial value problem of (a), with
initial values v0 = 0 , −2, −4, . . . , −12. Present your results in graphical form similar to
Figure 4.3.3.",Elementary Differential Equations with Boundary Value Problems.pdf
"initial values v0 = 0 , −2, −4, . . . , −12. Present your results in graphical form similar to
Figure 4.3.3.
16. A 64-lb object with initial velocity v0 ≤ 0 falls through a dense fluid that exerts a resistive force
proportional to the square root of the speed. The resistance is 64 lb if the speed is 16 ft/s.
(a) Set up the initial value problem for the velocity vof the mass for t> 0.
(b) Use Exercise 14(c) to determine the terminal velocity of the object.
(c) C T o confirm your answer to (b), use one of the numerical methods studied in Chapter 3
to compute approximate solutions on [0, 4] (seconds) of the initial value problem of (a), with
initial values v0 = 0 , −5, −10, . . . , −30. Present your results in graphical form similar to
Figure 4.3.3.
In Exercises 17-20, assume that the force due to gravity is given by Newton’s law of gravitation. T ake the
upward direction to be positive.",Elementary Differential Equations with Boundary Value Problems.pdf
"162 Chapter 4 Applications of First Order Equations
17. A space probe is to be launched from a space station 200 miles a bove Earth. Determine its escape
velocity in miles/s. T ake Earth’s radius to be 3960 miles.
18. A space vehicle is to be launched from the moon, which has a rad ius of about 1080 miles. The
acceleration due to gravity at the surface of the moon is abou t 5.31 ft/s2. Find the escape velocity
in miles/s.
19. (a) Show that Eqn. ( 4.3.23) can be rewritten as
v2 = h− y
y+ Rv2
e + v2
0 .
(b) Show that if v0 = ρve with 0 ≤ ρ< 1, then the maximum altitude ym attained by the space
vehicle is
ym = h+ Rρ2
1 − ρ2 .
(c) By requiring that v(ym ) = 0 , use Eqn. ( 4.3.22) to deduce that if v0 <ve then
|v|= ve
[ (1 − ρ2)(ym − y)
y+ R
] 1/ 2
,
where ym and ρ are as defined in (b) and y≥ h.
(d) Deduce from (c) that if v <ve, the vehicle takes equal times to climb from y = hto y = ym
and to fall back from y= ym to y = h.",Elementary Differential Equations with Boundary Value Problems.pdf
"(d) Deduce from (c) that if v <ve, the vehicle takes equal times to climb from y = hto y = ym
and to fall back from y= ym to y = h.
20. In the situation considered in the discussion of escape velo city , show that limt→∞ y(t) = ∞ if
v(t) >0 for all t> 0.
HIN T : Use a proof by contradiction. Assume that there’s a number ym such that y(t) ≤ ym for all
t> 0. Deduce from this that there’s positive number α such that y′′(t) ≤ − α for all t≥ 0. Show
that this contradicts the assumption that v(t) >0 for all t> 0.
4.4 AUTONOMOUS SECOND ORDER EQUA TIONS
A second order differential equation that can be written as
y′′= F(y,y ′) (4.4.1)
where F is independent of t, is said to be autonomous. An autonomous second order equation can be
converted into a first order equation relating v= y′and y. If we let v= y′, ( 4.4.1) becomes
v′= F(y,v ). (4.4.2)
Since
v′= dv
dt = dv
dy
dy
dt = vdv
dy, (4.4.3)
(4.4.2) can be rewritten as
vdv
dy = F(y,v ). (4.4.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"v′= F(y,v ). (4.4.2)
Since
v′= dv
dt = dv
dy
dy
dt = vdv
dy, (4.4.3)
(4.4.2) can be rewritten as
vdv
dy = F(y,v ). (4.4.4)
The integral curves of ( 4.4.4) can be plotted in the (y,v ) plane, which is called the P oincaré phase plane
of ( 4.4.1). If y is a solution of ( 4.4.1) then y = y(t),v = y′(t) is a parametric equation for an integral",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 163
curve of ( 4.4.4). W e’ll call these integral curves trajectories of ( 4.4.1), and we’ll call ( 4.4.4) the phase
plane equivalent of ( 4.4.1).
In this section we’ll consider autonomous equations that ca n be written as
y′′+ q(y,y ′)y′+ p(y) = 0 . (4.4.5)
Equations of this form often arise in applications of Newton ’s second law of motion. For example,
suppose y is the displacement of a moving object with mass m. It’s reasonable to think of two types
of time-independent forces acting on the object. One type - s uch as gravity - depends only on position.
W e could write such a force as −mp(y). The second type - such as atmospheric resistance or frictio n -
may depend on position and velocity . (Forces that depend on v elocity are called damping forces.) W e
write this force as −mq(y,y ′)y′, where q(y,y ′) is usually a positive function and we’ve put the factor y′",Elementary Differential Equations with Boundary Value Problems.pdf
"write this force as −mq(y,y ′)y′, where q(y,y ′) is usually a positive function and we’ve put the factor y′
outside to make it explicit that the force is in the direction opposing the motion. In this case Newton’s,
second law of motion leads to ( 4.4.5).
The phase plane equivalent of ( 4.4.5) is
vdv
dy + q(y,v )v+ p(y) = 0 . (4.4.6)
Some statements that we’ll be making about the properties of (4.4.5) and ( 4.4.6) are intuitively reasonable,
but difficult to prove. Therefore our presentation in this se ction will be informal: we’ll just say things
without proof, all of which are true if we assume that p= p(y) is continuously differentiable for all yand
q= q(y,v ) is continuously differentiable for all (y,v ). W e begin with the following statements:
• Statement 1. If y0 and v0 are arbitrary real numbers then ( 4.4.5) has a unique solution on (−∞ , ∞ )
such that y(0) = y0 and y′(0) = v0.",Elementary Differential Equations with Boundary Value Problems.pdf
"• Statement 1. If y0 and v0 are arbitrary real numbers then ( 4.4.5) has a unique solution on (−∞ , ∞ )
such that y(0) = y0 and y′(0) = v0.
• Statement 2.) If y = y(t) is a solution of ( 4.4.5) and τ is any constant then y1 = y(t− τ) is also a
solution of ( 4.4.5), and yand y1 have the same trajectory .
• Statement 3. If two solutions y and y1 of ( 4.4.5) have the same trajectory then y1(t) = y(t− τ)
for some constant τ.
• Statement 4. Distinct trajectories of ( 4.4.5) can’t intersect; that is, if two trajectories of ( 4.4.5)
intersect, they are identical.
• Statement 5. If the trajectory of a solution of ( 4.4.5) is a closed curve then (y(t),v (t)) traverses
the trajectory in a finite time T, and the solution is periodic with period T; that is, y(t+ T) = y(t)
for all tin (−∞ , ∞ ).
If y is a constant such that p(y) = 0 then y ≡ y is a constant solution of ( 4.4.5). W e say that y is an",Elementary Differential Equations with Boundary Value Problems.pdf
"for all tin (−∞ , ∞ ).
If y is a constant such that p(y) = 0 then y ≡ y is a constant solution of ( 4.4.5). W e say that y is an
equilibrium of ( 4.4.5) and (y, 0) is a critical point of the phase plane equivalent equation ( 4.4.6). W e say
that the equilibrium and the critical point are stable if, for any given ϵ> 0 no matter how small , there’s a
δ >0, sufficiently small , such that if
√
(y0 − y)2 + v2
0 <δ
then the solution of the initial value problem
y′′+ q(y,y ′)y′+ p(y) = 0 , y (0) = y0, y ′(0) = v0
satisfies the inequality √
(y(t) − y)2 + ( v(t))2 <ϵ",Elementary Differential Equations with Boundary Value Problems.pdf
"164 Chapter 4 Applications of First Order Equations
for all t> 0. Figure 4.4.1 illustrates the geometrical interpretation of this definit ion in the Poincaré phase
plane: if (y0,v 0) is in the smaller shaded circle (with radius δ), then (y(t),v (t)) must be in in the larger
circle (with radius ϵ) for all t> 0.
 y  y
 v
 ε
 δ
Figure 4.4.1 Stability: if (y0,v 0) is in the smaller circle then (y(t),v (t)) is in the larger circle for all
t> 0
If an equilibrium and the associated critical point are not s table, we say they are unstable. T o see if
you really understand what stable means, try to give a direct definition of unstable (Exercise 22). W e’ll
illustrate both definitions in the following examples.
The Undamped Case
W e’ll begin with the case where q≡ 0, so ( 4.4.5) reduces to
y′′+ p(y) = 0 . (4.4.7)
W e say that this equation - as well as any physical situation t hat it may model - is undamped. The phase
plane equivalent of ( 4.4.7) is the separable equation
vdv
dy + p(y) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"plane equivalent of ( 4.4.7) is the separable equation
vdv
dy + p(y) = 0 .
Integrating this yields
v2
2 + P(y) = c, (4.4.8)
where cis a constant of integration and P(y) = ∫ p(y) dyis an antiderivative of p.
If ( 4.4.7) is the equation of motion of an object of mass m, then mv2 / 2 is the kinetic energy and
mP(y) is the potential energy of the object; thus, ( 4.4.8) says that the total energy of the object remains",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 165
constant, or is conserved. In particular, if a trajectory passes through a given point (y0,v 0) then
c= v2
0
2 + P(y0).
Example 4.4.1 [The Undamped Spring - Mass System ] Consider an object with mass msuspended from
a spring and moving vertically . Let ybe the displacement of the object from the position it occupi es when
suspended at rest from the spring (Figure 4.4.2).
 y
(a)
 0
(b) (c)
Figure 4.4.2 (a) y> 0 (b) y= 0 (c) y< 0
Assume that if the length of the spring is changed by an amount ∆ L(positive or negative), then the
spring exerts an opposing force with magnitude k|∆ L|, where k is a positive constant. In Section 6.1 it
will be shown that if the mass of the spring is negligible comp ared to mand no other forces act on the
object then Newton’s second law of motion implies that
my′′= −ky, (4.4.9)
which can be written in the form ( 4.4.7) with p(y) = ky/m . This equation can be solved easily by a",Elementary Differential Equations with Boundary Value Problems.pdf
"my′′= −ky, (4.4.9)
which can be written in the form ( 4.4.7) with p(y) = ky/m . This equation can be solved easily by a
method that we’ll study in Section 5.2, but that method isn’t available here. Instead, we’ll consider the
phase plane equivalent of ( 4.4.9).
From ( 4.4.3), we can rewrite ( 4.4.9) as the separable equation
mvdv
dy = −ky.
Integrating this yields
mv2
2 = −ky2
2 + c,",Elementary Differential Equations with Boundary Value Problems.pdf
"166 Chapter 4 Applications of First Order Equations
 y
 v
Figure 4.4.3 Trajectories of my′′+ ky = 0
which implies that
mv2 + ky2 = ρ (4.4.10)
(ρ = 2 c). This defines an ellipse in the Poincaré phase plane (Figure 4.4.3).
W e can identify ρ by setting t= 0 in ( 4.4.10); thus, ρ = mv2
0 + ky2
0 , where y0 = y(0) and v0 = v(0).
T o determine the maximum and minimum values of ywe set v= 0 in ( 4.4.10); thus,
ymax = R and ymin = −R, with R=
√ ρ
k. (4.4.11)
Equation ( 4.4.9) has exactly one equilibrium, y = 0 , and it’s stable. Y ou can see intuitively why this is
so: if the object is displaced in either direction from equil ibrium, the spring tries to bring it back.
In this case we can find yexplicitly as a function of t. (Don’t expect this to happen in more complicated
problems!) If v >0 on an interval I, ( 4.4.10) implies that
dy
dt = v=
√
ρ − ky2
m
on I. This is equivalent to
√
k√
ρ − ky2
dy
dt = ω0, where ω0 =
√
k
m. (4.4.12)
Since ∫ √
kdy√
ρ − ky2
= sin −1
(√
k
ρy
)
+ c= sin −1
(y
R
)",Elementary Differential Equations with Boundary Value Problems.pdf
"m
on I. This is equivalent to
√
k√
ρ − ky2
dy
dt = ω0, where ω0 =
√
k
m. (4.4.12)
Since ∫ √
kdy√
ρ − ky2
= sin −1
(√
k
ρy
)
+ c= sin −1
(y
R
)
+ c",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 167
 y = R
 y = − R
 t
 y
Figure 4.4.4 y = Rsin(ω0t+ φ)
(see ( 4.4.11)), ( 4.4.12) implies that that there’s a constant φ such that
sin−1
(y
R
)
= ω0t+ φ
or
y = Rsin(ω0t+ φ)
for all tin I. Although we obtained this function by assuming that v >0, you can easily verify that y
satisfies ( 4.4.9) for all values of t. Thus, the displacement varies periodically between −Rand R, with
period T = 2 π/ω 0 (Figure 4.4.4). (If you’ve taken a course in elementary mechanics you may r ecognize
this as simple harmonic motion .)
Example 4.4.2 [The Undamped P endulum ] Now we consider the motion of a pendulum with mass m,
attached to the end of a weightless rod with length Lthat rotates on a frictionless axle (Figure 4.4.5). W e
assume that there’s no air resistance.
Let ybe the angle measured from the rest position (vertically dow nward) of the pendulum, as shown in",Elementary Differential Equations with Boundary Value Problems.pdf
"assume that there’s no air resistance.
Let ybe the angle measured from the rest position (vertically dow nward) of the pendulum, as shown in
Figure 4.4.5. Newton’s second law of motion says that the product of mand the tangential acceleration
equals the tangential component of the gravitational force ; therefore, from Figure 4.4.5,
mLy′′= −mgsin y,
or
y′′= − g
Lsin y. (4.4.13)
Since sin nπ = 0 if nis any integer, ( 4.4.13) has infinitely many equilibria yn = nπ. If nis even, the
mass is directly below the axle (Figure 4.4.6 (a)) and gravity opposes any deviation from the equilibrium.",Elementary Differential Equations with Boundary Value Problems.pdf
"168 Chapter 4 Applications of First Order Equations
 m
 y  L
Figure 4.4.5 The undamped pendulum
(a) Stable equilibrium (b) Unstable equilibrium
Figure 4.4.6 (a) Stable equilibrium (b) Unstable equilibrium
However, if n is odd, the mass is directly above the axle (Figure 4.4.6 (b)) and gravity increases any
deviation from the equilibrium. Therefore we conclude on ph ysical grounds that y2m = 2 mπ is stable
and y2m+1 = (2 m+ 1) π is unstable.
The phase plane equivalent of ( 4.4.13) is
vdv
dy = − g
Lsin y,
where v= y′is the angular velocity of the pendulum. Integrating this yi elds
v2
2 = g
Lcos y+ c. (4.4.14)
If v = v0 when y= 0 , then
c= v2
0
2 − g
L,
so ( 4.4.14) becomes
v2
2 = v2
0
2 − g
L(1 − cos y) = v2
0
2 − 2g
L sin2 y
2 ,
which is equivalent to
v2 = v2
0 − v2
c sin2 y
2 , (4.4.15)
where
vc = 2
√ g
L.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 169
The curves defined by ( 4.4.15) are the trajectories of ( 4.4.13). They are periodic with period 2π in y,
which isn’t surprising, since if y = y(t) is a solution of ( 4.4.13) then so is yn = y(t) + 2 nπ for any
integer n. Figure 4.4.7 shows trajectories over the interval [−π,π ]. From ( 4.4.15), you can see that if
|v0|>vc then vis nonzero for all t, which means that the object whirls in the same direction for ever, as in
Figure 4.4.8. The trajectories associated with this whirling motion are above the upper dashed curve and
below the lower dashed curve in Figure 4.4.7. Y ou can also see from ( 4.4.15) that if 0 < |v0|< vc,then
v= 0 when y= ±ymax, where
ymax = 2 sin −1(|v0|/v c).
In this case the pendulum oscillates periodically between −ymax and ymax, as shown in Figure 4.4.9. The
trajectories associated with this kind of motion are the ova ls between the dashed curves in Figure 4.4.7.",Elementary Differential Equations with Boundary Value Problems.pdf
"trajectories associated with this kind of motion are the ova ls between the dashed curves in Figure 4.4.7.
It can be shown (see Exercise 21 for a partial proof) that the period of the oscillation is
T = 8
∫ π/ 2
0
dθ√
v2c − v2
0 sin2 θ
. (4.4.16)
Although this integral can’t be evaluated in terms of famili ar elementary functions, you can see that it’s
finite if |v0|<vc.
The dashed curves in Figure 4.4.7 contain four trajectories. The critical points (π, 0) and (−π, 0) are
the trajectories of the unstable equilibrium solutions y = ±π. The upper dashed curve connecting (but
not including) them is obtained from initial conditions of t he form y(t0 ) = 0 , v(t0) = vc. If y is any
solution with this trajectory then
lim
t→∞
y(t) = π and lim
t→−∞
y(t) = −π.
The lower dashed curve connecting (but not including) them i s obtained from initial conditions of the
form y(t0 ) = 0 , v(t0) = −vc. If yis any solution with this trajectory then
lim
t→∞
y(t) = −π and lim
t→−∞
y(t) = π.",Elementary Differential Equations with Boundary Value Problems.pdf
"form y(t0 ) = 0 , v(t0) = −vc. If yis any solution with this trajectory then
lim
t→∞
y(t) = −π and lim
t→−∞
y(t) = π.
Consistent with this, the integral ( 4.4.16) diverges to ∞ if v0 = ±vc . (Exercise 21) .
Since the dashed curves separate trajectories of whirling s olutions from trajectories of oscillating solu-
tions, each of these curves is called a separatrix.
In general, if ( 4.4.7) has both stable and unstable equilibria then the separatri ces are the curves given
by ( 4.4.8) that pass through unstable critical points. Thus, if (y, 0) is an unstable critical point, then
v2
2 + P(y) = P(y) (4.4.17)
defines a separatrix passing through (y, 0).",Elementary Differential Equations with Boundary Value Problems.pdf
"170 Chapter 4 Applications of First Order Equations
 π − π
 x
 y
Figure 4.4.7 Trajectories of the undamped pendulum
Figure 4.4.8 The whirling undamped pendulum
 ymax − y max
Figure 4.4.9 The oscillating undamped pendulum
Stability and Instability Conditions for y′′+ p(y) = 0
It can be shown (Exercise 23) that an equilibrium yof an undamped equation
y′′+ p(y) = 0 (4.4.18)
is stable if there’s an open interval (a,b ) containing ysuch that
p(y) <0 if a<y< y and p(y) >0 if y<y<b. (4.4.19)
If we regard p(y) as a force acting on a unit mass, ( 4.4.19) means that the force resists all sufficiently
small displacements from y.
W e’ve already seen examples illustrating this principle. T he equation ( 4.4.9) for the undamped spring-
mass system is of the form ( 4.4.18) with p(y) = ky/m , which has only the stable equilibrium y = 0 . In
this case ( 4.4.19) holds with a= −∞ and b= ∞ . The equation ( 4.4.13) for the undamped pendulum is",Elementary Differential Equations with Boundary Value Problems.pdf
"this case ( 4.4.19) holds with a= −∞ and b= ∞ . The equation ( 4.4.13) for the undamped pendulum is
of the form ( 4.4.18) with p(y) = ( g/L ) sin y. W e’ve seen that y = 2 mπ is a stable equilibrium if mis an
integer. In this case
p(y) = sin y< 0 if (2m− 1)π<y< 2mπ
and
p(y) >0 if 2mπ <y< (2m+ 1) π.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 171
It can also be shown (Exercise 24) that yis unstable if there’s a b> ysuch that
p(y) <0 if y <y<b (4.4.20)
or an a< ysuch that
p(y) >0 if a<y< y. (4.4.21)
If we regard p(y) as a force acting on a unit mass, ( 4.4.20) means that the force tends to increase all
sufficiently small positive displacements from y, while ( 4.4.21) means that the force tends to increase the
magnitude of all sufficiently small negative displacements from y.
The undamped pendulum also illustrates this principle. W e’ ve seen that y= (2 m+ 1) π is an unstable
equilibrium if mis an integer. In this case
sin y< 0 if (2m+ 1) π<y < (2m+ 2) π,
so ( 4.4.20) holds with b= (2 m+ 2) π, and
sin y> 0 if 2mπ <y <(2m+ 1) π,
so ( 4.4.21) holds with a= 2 mπ.
Example 4.4.3 The equation
y′′+ y(y− 1) = 0 (4.4.22)
is of the form ( 4.4.18) with p(y) = y(y− 1). Therefore y = 0 and y = 1 are the equilibria of ( 4.4.22).
Since
y(y− 1) >0 if y <0 or y> 1,
<0 if 0 <y< 1,",Elementary Differential Equations with Boundary Value Problems.pdf
"Since
y(y− 1) >0 if y <0 or y> 1,
<0 if 0 <y< 1,
y = 0 is unstable and y= 1 is stable.
The phase plane equivalent of ( 4.4.22) is the separable equation
vdv
dy + y(y− 1) = 0 .
Integrating yields
v2
2 + y3
3 − y2
2 = C,
which we rewrite as
v2 + 1
3 y2 (2y− 3) = c (4.4.23)
after renaming the constant of integration. These are the tr ajectories of ( 4.4.22). If y is any solution of
(4.4.22), the point (y(t),v (t)) moves along the trajectory of yin the direction of increasing yin the upper
half plane ( v= y′>0), or in the direction of decreasing yin the lower half plane ( v= y′<0).
Figure 4.4.10 shows typical trajectories. The dashed curve through the cr itical point (0, 0), obtained by
setting c = 0 in ( 4.4.23), separates the y-v plane into regions that contain different kinds of trajecto ries;
again, we call this curve a separatrix. Trajectories in the region bounded by the closed loop (b) are closed",Elementary Differential Equations with Boundary Value Problems.pdf
"again, we call this curve a separatrix. Trajectories in the region bounded by the closed loop (b) are closed
curves, so solutions associated with them are periodic. Sol utions associated with other trajectories are not
periodic. If yis any such solution with trajectory not on the separatrix, t hen
lim
t→∞
y(t) = −∞ , lim
t→−∞
y(t) = −∞ ,
lim
t→∞
v(t) = −∞ , lim
t→−∞
v(t) = ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"172 Chapter 4 Applications of First Order Equations
1  y
 v
(a) (b)
(c) (b)
Figure 4.4.10 Trajectories of y′′+ y(y− 1) = 0
The separatrix contains four trajectories of ( 4.4.22). One is the point (0, 0), the trajectory of the equi-
librium y = 0 . Since distinct trajectories can’t intersect, the segment s of the separatrix marked (a), (b),
and (c) – which don’t include (0, 0) – are distinct trajectories, none of which can be traversed i n finite
time. Solutions with these trajectories have the following asymptotic behavior:
lim
t→∞
y(t) = 0 , lim
t→−∞
y(t) = −∞ ,
lim
t→∞
v(t) = 0 , lim
t→−∞
v(t) = ∞ (on (a))
lim
t→∞
y(t) = 0 , lim
t→−∞
y(t) = 0 ,
lim
t→∞
v(t) = 0 , lim
t→−∞
v(t) = 0 (on (b))
lim
t→∞
y(t) = −∞ , lim
t→−∞
y(t) = 0 ,
lim
t→∞
v(t) = −∞ , lim
t→−∞
v(t) = 0 (on (c)).
.
The Damped Case
The phase plane equivalent of the damped autonomous equatio n
y′′+ q(y,y ′)y′+ p(y) = 0 (4.4.24)
is
vdv
dy + q(y,v )v+ p(y) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
".
The Damped Case
The phase plane equivalent of the damped autonomous equatio n
y′′+ q(y,y ′)y′+ p(y) = 0 (4.4.24)
is
vdv
dy + q(y,v )v+ p(y) = 0 .
This equation isn’t separable, so we can’t solve it for v in terms of y, as we did in the undamped case,
and conservation of energy doesn’t hold. (For example, ener gy expended in overcoming friction is lost.)
However, we can study the qualitative behavior of its soluti ons by rewriting it as
dv
dy = −q(y,v ) − p(y)
v (4.4.25)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 173
and considering the direction fields for this equation. In th e following examples we’ll also be showing
computer generated trajectories of this equation, obtaine d by numerical methods. The exercises call for
similar computations. The methods discussed in Chapter 3 ar e not suitable for this task, since p(y)/v in
(4.4.25) is undefined on the yaxis of the Poincaré phase plane. Therefore we’re forced to a pply numerical
methods briefly discussed in Section 10.1 to the system
y′ = v
v′ = −q(y,v )v− p(y),
which is equivalent to ( 4.4.24) in the sense defined in Section 10.1. Fortunately , most diff erential equation
software packages enable you to do this painlessly .
In the text we’ll confine ourselves to the case where qis constant, so ( 4.4.24) and ( 4.4.25) reduce to
y′′+ cy′+ p(y) = 0 (4.4.26)
and
dv
dy = −c− p(y)
v .
(W e’ll consider more general equations in the exercises.) T he constant cis called the damping constant .",Elementary Differential Equations with Boundary Value Problems.pdf
"and
dv
dy = −c− p(y)
v .
(W e’ll consider more general equations in the exercises.) T he constant cis called the damping constant .
In situations where ( 4.4.26) is the equation of motion of an object, c is positive; however, there are
situations where cmay be negative.
The Damped Spring-Mass System
Earlier we considered the spring - mass system under the assu mption that the only forces acting on the
object were gravity and the spring’s resistance to changes i n its length. Now we’ll assume that some
mechanism (for example, friction in the spring or atmospher ic resistance) opposes the motion of the
object with a force proportional to its velocity . In Section 6.1 it will be shown that in this case Newton’s
second law of motion implies that
my′′+ cy′+ ky = 0 , (4.4.27)
where c > 0 is the damping constant . Again, this equation can be solved easily by a method that
we’ll study in Section 5.2, but that method isn’t available h ere. Instead, we’ll consider its phase plane",Elementary Differential Equations with Boundary Value Problems.pdf
"we’ll study in Section 5.2, but that method isn’t available h ere. Instead, we’ll consider its phase plane
equivalent, which can be written in the form ( 4.4.25) as
dv
dy = − c
m − ky
mv. (4.4.28)
(A minor note: the c in ( 4.4.26) actually corresponds to c/m in this equation.) Figure 4.4.11 shows a
typical direction field for an equation of this form. Recalli ng that motion along a trajectory must be in the
direction of increasing yin the upper half plane ( v >0) and in the direction of decreasing yin the lower
half plane ( v <0), you can infer that all trajectories approach the origin in clockwise fashion. T o confirm
this, Figure 4.4.12 shows the same direction field with some trajectories filled i n. All the trajectories
shown there correspond to solutions of the initial value pro blem
my′′+ cy′+ ky = 0 , y (0) = y0, y ′(0) = v0,
where
mv2
0 + ky2
0 = ρ (a positive constant );
thus, if there were no damping ( c = 0 ), all the solutions would have the same dashed elliptic traj ectory ,",Elementary Differential Equations with Boundary Value Problems.pdf
"0 + ky2
0 = ρ (a positive constant );
thus, if there were no damping ( c = 0 ), all the solutions would have the same dashed elliptic traj ectory ,
shown in Figure 4.4.14.",Elementary Differential Equations with Boundary Value Problems.pdf
"174 Chapter 4 Applications of First Order Equations
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 v
 y
Figure 4.4.11 A typical direction field for
my′′+ cy′+ ky = 0 with 0 <c<c 1
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 v
 y
Figure 4.4.12 Figure 4.4.11 with some trajectories
added
Solutions corresponding to the trajectories in Figure 4.4.12 cross the y-axis infinitely many times. The
corresponding solutions are said to be oscillatory (Figure 4.4.13) It is shown in Section 6.2 that there’s
a number c1 such that if 0 ≤ c < c1 then all solutions of ( 4.4.27) are oscillatory , while if c ≥ c1, no
solutions of ( 4.4.27) have this property . (In fact, no solution not identically z ero can have more than two
zeros in this case.) Figure 4.4.14 shows a direction field and some integral curves for ( 4.4.28) in this case.
 t
 y
Figure 4.4.13 An oscillatory solution of my′′+ cy′+ ky = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"t
 y
Figure 4.4.13 An oscillatory solution of my′′+ cy′+ ky = 0
Example 4.4.4 (The Damped Pendulum) Now we return to the pendulum. If we assume that some
mechanism (for example, friction in the axle or atmospheric resistance) opposes the motion of the pen-
dulum with a force proportional to its angular velocity , New ton’s second law of motion implies that
mLy′′= −cy′− mgsin y, (4.4.29)
where c >0 is the damping constant. (Again, a minor note: the c in ( 4.4.26) actually corresponds to",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 175
c/mL in this equation.) T o plot a direction field for ( 4.4.29) we write its phase plane equivalent as
dv
dy = − c
mL − g
Lvsin y.
Figure 4.4.15 shows trajectories of four solutions of ( 4.4.29), all satisfying y(0) = 0 . For each m= 0 , 1,
2, 3, imparting the initial velocity v(0) = vm causes the pendulum to make mcomplete revolutions and
then settle into decaying oscillation about the stable equi librium y= 2 mπ.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
 v
 y
Figure 4.4.14 A typical direction field for my′′+ cy′+ ky = 0 with c>c 1
 y
 v
 v3
 v2
 v1
 v0
 y = 2π  y = 4π  y = 6π
Figure 4.4.15 Four trajectories of the damped pendulum",Elementary Differential Equations with Boundary Value Problems.pdf
"176 Chapter 4 Applications of First Order Equations
4.4 Exercises
In Exercises 1– 4 find the equations of the trajectories of the given undamped e quation. Identify the
equilibrium solutions, determine whether they are stable o r unstable, and plot some trajectories. H IN T :
Use Eqn. (4.4.8) to obtain the equations of the trajectories.
1. C/G y′′+ y3 = 0 2. C/G y′′+ y2 = 0
3. C/G y′′+ y|y|= 0 4. C/G y′′+ ye−y = 0
In Exercises 5– 8 find the equations of the trajectories of the given undamped e quation. Identify the
equilibrium solutions, determine whether they are stable o r unstable, and find the equations of the sepa-
ratrices (that is, the curves through the unstable equilibr ia). Plot the separatrices and some trajectories
in each of the regions of P oincaré plane determined by them. HIN T : Use Eqn. (4.4.17) to determine the
separatrices.
5. C/G y′′− y3 + 4 y= 0 6. C/G y′′+ y3 − 4y= 0
7. C/G y′′+ y(y2 − 1)(y2 − 4) = 0 8. C/G y′′+ y(y− 2)(y− 1)(y+ 2) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"separatrices.
5. C/G y′′− y3 + 4 y= 0 6. C/G y′′+ y3 − 4y= 0
7. C/G y′′+ y(y2 − 1)(y2 − 4) = 0 8. C/G y′′+ y(y− 2)(y− 1)(y+ 2) = 0
In Exercises 9– 12 plot some trajectories of the given equation for various val ues (positive, negative, zero)
of the parameter a. Find the equilibria of the equation and classify them as sta ble or unstable. Explain
why the phase plane plots corresponding to positive and nega tive values of adiffer so markedly. Can you
think of a reason why zero deserves to be called the critical value of a?
9. L y′′+ y2 − a= 0 10. L y′′+ y3 − ay= 0
11. L y′′− y3 + ay = 0 12. L y′′+ y− ay3 = 0
In Exercises 13-18 plot trajectories of the given equation for c = 0 and small nonzero (positive and
negative) values of cto observe the effects of damping.
13. L y′′+ cy′+ y3 = 0 14. L y′′+ cy′− y= 0
15. L y′′+ cy′+ y3 = 0 16. L y′′+ cy′+ y2 = 0
17. L y′′+ cy′+ y|y|= 0 18. L y′′+ y(y− 1) + cy= 0
19. L The van der P ol equation
y′′− µ(1 − y2)y′+ y = 0 , (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"17. L y′′+ cy′+ y|y|= 0 18. L y′′+ y(y− 1) + cy= 0
19. L The van der P ol equation
y′′− µ(1 − y2)y′+ y = 0 , (A)
where µ is a positive constant and y is electrical current (Section 6.3), arises in the study of a n
electrical circuit whose resistive properties depend upon the current. The damping term
−µ(1 − y2 )y′works to reduce |y|if |y|< 1 or to increase |y|if |y|> 1. It can be shown that
van der Pol’s equation has exactly one closed trajectory , wh ich is called a limit cycle . Trajectories",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.4 Autonomous Second Order Equations 177
inside the limit cycle spiral outward to it, while trajector ies outside the limit cycle spiral inward to it
(Figure 4.4.16). Use your favorite differential equations software to ver ify this for µ = .5, 1.1.5, 2.
Use a grid with −4 <y <4 and −4 <v <4.
 y
 v
Figure 4.4.16 Trajectories of van der Pol’s equation
20. L Rayleigh’s equation ,
y′′− µ(1 − (y′)2/ 3)y′+ y= 0
also has a limit cycle. Follow the directions of Exercise 19 for this equation.
21. In connection with Eqn ( 4.4.15), suppose y(0) = 0 and y′(0) = v0, where 0 <v0 <vc.
(a) Let T1 be the time required for yto increase from zero to ymax = 2 sin −1(v0/v c). Show that
dy
dt =
√
v2
0 − v2c sin2 y/ 2, 0 ≤ t<T 1. (A)
(b) Separate variables in (A) and show that
T1 =
∫ ymax
0
du√
v2
0 − v2c sin2 u/ 2
(B)
(c) Substitute sin u/ 2 = ( v0/v c) sin θ in (B) to obtain
T1 = 2
∫ π/ 2
0
dθ√
v2c − v2
0 sin2 θ
. (C)",Elementary Differential Equations with Boundary Value Problems.pdf
"178 Chapter 4 Applications of First Order Equations
(d) Conclude from symmetry that the time required for (y(t),v (t)) to traverse the trajectory
v2 = v2
0 − v2
c sin2 y/ 2
is T = 4 T1, and that consequently y(t + T) = y(t) and v(t + T) = v(t); that is, the
oscillation is periodic with period T.
(e) Show that if v0 = vc, the integral in (C) is improper and diverges to ∞ . Conclude from this
that y(t) <π for all tand limt→∞ y(t) = π.
22. Give a direct definition of an unstable equilibrium of y′′+ p(y) = 0 .
23. Let pbe continuous for all yand p(0) = 0 . Suppose there’s a positive number ρ such that p(y) >0
if 0 <y ≤ ρ and p(y) <0 if −ρ ≤ y< 0. For 0 <r ≤ ρ let
α(r) = min
{ ∫ r
0
p(x) dx,
∫ 0
−r
|p(x)|dx
}
and β(r) = max
{ ∫ r
0
p(x) dx,
∫ 0
−r
|p(x)|dx
}
.
Let ybe the solution of the initial value problem
y′′+ p(y) = 0 , y (0) = v0, y ′(0) = v0,
and define c(y0,v 0) = v2
0 + 2
∫y0
0 p(x) dx.
(a) Show that
0 <c(y0,v 0) <v2
0 + 2 β(|y0|) if 0 <|y0| ≤ρ.
(b) Show that
v2 + 2
∫ y
0",Elementary Differential Equations with Boundary Value Problems.pdf
"and define c(y0,v 0) = v2
0 + 2
∫y0
0 p(x) dx.
(a) Show that
0 <c(y0,v 0) <v2
0 + 2 β(|y0|) if 0 <|y0| ≤ρ.
(b) Show that
v2 + 2
∫ y
0
p(x) dx= c(y0 ,v 0), t> 0.
(c) Conclude from (b) that if c(y0,v 0) <2α(r) then |y|<r, t> 0.
(d) Given ϵ> 0, let δ> 0 be chosen so that
δ2 + 2 β(δ) <max
{
ϵ2/ 2, 2α(ϵ/
√
2)
}
.
Show that if
√
y2
0 + v2
0 < δthen
√
y2 + v2 < ϵfor t >0, which implies that y = 0 is a
stable equilibrium of y′′+ p(y) = 0 .
(e) Now let pbe continuous for all y and p(y) = 0 , where y is not necessarily zero. Suppose
there’s a positive number ρ such that p(y) > 0 if y < y≤ y+ ρ and p(y) < 0 if y− ρ ≤
y< y. Show that yis a stable equilibrium of y′′+ p(y) = 0 .
24. Let pbe continuous for all y.
(a) Suppose p(0) = 0 and there’s a positive number ρ such that p(y) <0 if 0 <y ≤ ρ. Let ϵbe
any number such that 0 <ϵ<ρ . Show that if yis the solution of the initial value problem
y′′+ p(y) = 0 , y (0) = y0, y ′(0) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"any number such that 0 <ϵ<ρ . Show that if yis the solution of the initial value problem
y′′+ p(y) = 0 , y (0) = y0, y ′(0) = 0
with 0 < y0 < ϵ, then y(t) ≥ ϵ for some t > 0. Conclude that y = 0 is an unstable
equilibrium of y′′+ p(y) = 0 . H IN T : Let k= min y0≤x≤ϵ (−p(x)), which is positive. Show
that if y(t) <ϵ for 0 ≤ t<T then kT2 <2(ϵ− y0).
(b) Now let p(y) = 0 , where yisn’t necessarily zero. Suppose there’s a positive number ρ such
that p(y) <0 if y<y ≤ y+ ρ. Show that yis an unstable equilibrium of y′′+ p(y) = 0 .
(c) Modify your proofs of (a) and (b) to show that if there’s a positive number ρ such that
p(y) >0 if y− ρ ≤ y< y, then yis an unstable equilibrium of y′′+ p(y) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 179
4.5 APPLICA TIONS TO CURVES
One-Parameter Families of Curves
W e begin with two examples of families of curves generated by varying a parameter over a set of real
numbers.
Example 4.5.1 For each value of the parameter c, the equation
y− cx2 = 0 (4.5.1)
defines a curve in the xy-plane. If c ̸= 0 , the curve is a parabola through the origin, opening upward i f
c> 0 or downward if c< 0. If c= 0 , the curve is the xaxis (Figure
4.5.1).
 x
 y
Figure 4.5.1 A family of curves defined by y− cx2 = 0
Example 4.5.2 For each value of the parameter cthe equation
y= x+ c (4.5.2)
defines a line with slope 1(Figure 4.5.2).
Definition 4.5.1 An equation that can be written in the form
H(x,y,c ) = 0 (4.5.3)
is said to define a one-parameter family of curves if, for each value of cin in some nonempty set of real
numbers, the set of points (x,y ) that satisfy ( 4.5.3) forms a curve in the xy-plane.",Elementary Differential Equations with Boundary Value Problems.pdf
"180 Chapter 4 Applications of First Order Equations
 x
 y
Figure 4.5.2 A family of lines defined by y = x+ c
 x
 y
Figure 4.5.3 A family of circles defined by
x2 + y2 − c2 = 0
Equations ( 4.5.1) and ( 4.5.2) define one–parameter families of curves. (Although ( 4.5.2) isn’t in the
form ( 4.5.3), it can be written in this form as y− x− c= 0 .)
Example 4.5.3 If c> 0, the graph of the equation
x2 + y2 − c= 0 (4.5.4)
is a circle with center at (0, 0) and radius √c. If c = 0 , the graph is the single point (0, 0). (W e don’t
regard a single point as a curve.) If c <0, the equation has no graph. Hence, ( 4.5.4) defines a one–
parameter family of curves for positive values of c. This family consists of all circles centered at (0, 0)
(Figure 4.5.3).
Example 4.5.4 The equation
x2 + y2 + c2 = 0
does not define a one-parameter family of curves, since no (x,y ) satisfies the equation if c̸= 0 , and only
the single point (0, 0) satisfies it if c= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"the single point (0, 0) satisfies it if c= 0 .
Recall from Section 1.2 that the graph of a solution of a diffe rential equation is called an integral curve
of the equation. Solving a first order differential equation usually produces a one–parameter family of
integral curves of the equation. Here we are interested in th e converse problem: given a one–parameter
family of curves, is there a first order differential equatio n for which every member of the family is an
integral curve. This suggests the next definition.
Definition 4.5.2
If every curve in a one-parameter family defined by the equati on
H(x,y,c ) = 0 (4.5.5)
is an integral curve of the first order differential equation
F(x,y,y ′) = 0 , (4.5.6)
then ( 4.5.6) is said to be a differential equation for the family.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 181
T o find a differential equation for a one–parameter family we differentiate its defining equation ( 4.5.5)
implicitly with respect to x, to obtain
Hx(x,y,c ) + Hy(x,y,c )y′= 0 . (4.5.7)
If this equation doesn’t, then it’s a differential equation for the family . If it does contain c, it may be
possible to obtain a differential equation for the family by eliminating cbetween ( 4.5.5) and ( 4.5.7).
Example 4.5.5 Find a differential equation for the family of curves defined by
y= cx2. (4.5.8)
Solution Differentiating ( 4.5.8) with respect to xyields
y′= 2 cx.
Therefore c= y′/ 2x, and substituting this into ( 4.5.8) yields
y = xy′
2
as a differential equation for the family of curves defined by (4.5.8). The graph of any function of the
form y = cx2 is an integral curve of this equation.
The next example shows that members of a given family of curve s may be obtained by joining integral
curves for more than one differential equation.
Example 4.5.6",Elementary Differential Equations with Boundary Value Problems.pdf
"curves for more than one differential equation.
Example 4.5.6
(a) Try to find a differential equation for the family of lines tan gent to the parabola y= x2.
(b) Find two tangent lines to the parabola y = x2 that pass through (2, 3), and find the points of
tangency .
SO LU TIO N (a) The equation of the line through a given point (x0,y 0) with slope mis
y = y0 + m(x− x0). (4.5.9)
If (x0,y 0) is on the parabola, then y0 = x2
0 and the slope of the tangent line through ( x0,x 2
0) is m= 2 x0;
hence, (
4.5.9) becomes
y = x2
0 + 2 x0(x− x0),
or, equivalently ,
y= −x2
0 + 2 x0x. (4.5.10)
Here x0 plays the role of the constant cin Definition
4.5.1; that is, varying x0 over (−∞ , ∞ ) produces
the family of tangent lines to the parabola y = x2.
Differentiating ( 4.5.10) with respect to xyields y′= 2 x0.. W e can express x0 in terms of xand yby
rewriting ( 4.5.10) as
x2
0 − 2x0x+ y= 0
and using the quadratic formula to obtain
x0 = x±
√
x2 − y. (4.5.11)",Elementary Differential Equations with Boundary Value Problems.pdf
"182 Chapter 4 Applications of First Order Equations
W e must choose the plus sign in ( 4.5.11) if x<x 0 and the minus sign if x>x 0; thus,
x0 =
(
x+
√
x2 − y
)
if x<x 0
and
x0 =
(
x−
√
x2 − y
)
if x>x 0.
Since y′= 2 x0, this implies that
y′= 2
(
x+
√
x2 − y
)
, if x<x 0 (4.5.12)
and
y′= 2
(
x−
√
x2 − y
)
, if x>x 0. (4.5.13)
Neither ( 4.5.12) nor ( 4.5.13) is a differential equation for the family of tangent lines t o the parabola
y = x2. However, if each tangent line is regarded as consisting of t wo tangent half lines joined at the
point of tangency , ( 4.5.12) is a differential equation for the family of tangent half li nes on which xis less
than the abscissa of the point of tangency (Figure 4.5.4(a)), while ( 4.5.13) is a differential equation for
the family of tangent half lines on which xis greater than this abscissa (Figure 4.5.4(b)). The parabola
y = x2 is also an integral curve of both ( 4.5.12) and ( 4.5.13).
 y  y
 x x
(a) (b)
Figure 4.5.4",Elementary Differential Equations with Boundary Value Problems.pdf
"y = x2 is also an integral curve of both ( 4.5.12) and ( 4.5.13).
 y  y
 x x
(a) (b)
Figure 4.5.4
SO LU TIO N (b) From ( 4.5.10) the point (x,y ) = (2 , 3) is on the tangent line through (x0,x 2
0) if and only
if
3 = −x2
0 + 4 x0,
which is equivalent to
x2
0 − 4x0 + 3 = ( x0 − 3)(x0 − 1) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 183
Letting x0 = 3 in ( 4.5.10) shows that (2, 3) is on the line
y= −9 + 6 x,
which is tangent to the parabola at (x0,x 2
0) = (3 , 9), as shown in Figure
4.5.5
Letting x0 = 1 in ( 4.5.10) shows that (2, 3) is on the line
y= −1 + 2 x,
which is tangent to the parabola at (x0,x 2
0) = (1 , 1), as shown in Figure
4.5.5.
 y
1
2
3
4
5
6
7
8
9
10
11
 x
1 2 3
 y = x2
Figure 4.5.5
Geometric Problems
W e now consider some geometric problems that can be solved by means of differential equations.
Example 4.5.7 Find curves y= y(x) such that every point (x0,y (x0)) on the curve is the midpoint of the
line segment with endpoints on the coordinate axes and tange nt to the curve at (x0,y (x0)) (Figure 4.5.6).
Solution The equation of the line tangent to the curve at P = ( x0,y (x0) is
y= y(x0) + y′(x0)(x− x0).
If we denote the xand yintercepts of the tangent line by xI and yI (Figure 4.5.6), then
0 = y(x0) + y′(x0)(xI − x0) (4.5.14)
and",Elementary Differential Equations with Boundary Value Problems.pdf
"If we denote the xand yintercepts of the tangent line by xI and yI (Figure 4.5.6), then
0 = y(x0) + y′(x0)(xI − x0) (4.5.14)
and
yI = y(x0 ) − y′(x0)x0. (4.5.15)",Elementary Differential Equations with Boundary Value Problems.pdf
"184 Chapter 4 Applications of First Order Equations
From Figure 4.5.6, P is the midpoint of the line segment connecting (xI , 0) and (0,y I ) if and only if
xI = 2 x0 and yI = 2 y(x0). Substituting the first of these conditions into ( 4.5.14) or the second into
(4.5.15) yields
y(x0) + y′(x0)x0 = 0 .
Since x0 is arbitrary we drop the subscript and conclude that y= y(x) satisfies
y+ xy′= 0 ,
which can be rewritten as
(xy)′= 0 .
Integrating yields xy= c, or
y = c
x.
If c = 0 this curve is the line y = 0 , which does not satisfy the geometric requirements imposed by the
problem; thus, c̸= 0 , and the solutions define a family of hyperbolas (Figure 4.5.7).
 x
 y
 xI .5 xI
 yI
.5 yI
Figure 4.5.6
 x
 y
Figure 4.5.7
Example 4.5.8 Find curves y = y(x) such that the tangent line to the curve at any point (x0,y (x0))
intersects the x-axis at (x2
0, 0). Figure
4.5.8 illustrates the situation in the case where the curve is in th e
first quadrant and 0 <x< 1.",Elementary Differential Equations with Boundary Value Problems.pdf
"intersects the x-axis at (x2
0, 0). Figure
4.5.8 illustrates the situation in the case where the curve is in th e
first quadrant and 0 <x< 1.
Solution The equation of the line tangent to the curve at (x0,y (x0)) is
y= y(x0) + y′(x0)(x− x0).
Since (x2
0, 0) is on the tangent line,
0 = y(x0 ) + y′(x0)(x2
0 − x0).
Since x0 is arbitrary we drop the subscript and conclude that y= y(x) satisfies
y+ y′(x2 − x) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 185
 x0x0
2  x
 y
Figure 4.5.8
 x = 1
 x
 y
Figure 4.5.9
Therefor
y′
y = − 1
x2 − x = − 1
x(x− 1) = 1
x − 1
x− 1 ,
so
ln |y|= ln |x| −ln |x− 1|+ k= ln
⏐
⏐
⏐
⏐
x
x− 1
⏐
⏐
⏐
⏐+ k,
and
y = cx
x− 1 .
If c= 0 , the graph of this function is the x-axis. If c̸= 0 , it’s a hyperbola with vertical asymptote x= 1
and horizontal asymptote y = c. Figure 4.5.9 shows the graphs for c̸= 0 .
Orthogonal Trajectories
T wo curves C1 and C2 are said to be orthogonal at a point of intersection (x0,y 0) if they have perpen-
dicular tangents at (x0,y 0). (Figure 4.5.10). A curve is said to be an orthogonal trajectory of a given
family of curves if it’s orthogonal to every curve in the fami ly . For example, every line through the origin
is an orthogonal trajectory of the family of circles centere d at the origin. Conversely , any such circle is
an orthogonal trajectory of the family of lines through the o rigin (Figure 4.5.11).",Elementary Differential Equations with Boundary Value Problems.pdf
"an orthogonal trajectory of the family of lines through the o rigin (Figure 4.5.11).
Orthogonal trajectories occur in many physical applicatio ns. For example, if u = u(x,y ) is the tem-
perature at a point (x,y ), the curves defined by
u(x,y ) = c (4.5.16)
are called isothermal curves. The orthogonal trajectories of this family are call ed heat-flow lines, because
at any given point the direction of maximum heat flow is perpen dicular to the isothermal through the
point. If urepresents the potential energy of an object moving under a f orce that depends upon (x,y ), the
curves ( 4.5.16) are called equipotentials , and the orthogonal trajectories are called lines of force .
From analytic geometry we know that two nonvertical lines L1 and L2 with slopes m1 and m2, re-
spectively , are perpendicular if and only if m2 = −1/m 1; therefore, the integral curves of the differential
equation
y′= − 1
f(x,y )",Elementary Differential Equations with Boundary Value Problems.pdf
"186 Chapter 4 Applications of First Order Equations
 x
 y
Figure 4.5.10 Curves orthogonal at a point of
intersection
 x
 y
Figure 4.5.11 Orthogonal families of circles and
lines
are orthogonal trajectories of the integral curves of the di fferential equation
y′= f(x,y ),
because at any point (x0,y 0) where curves from the two families intersect the slopes of th e respective
tangent lines are
m1 = f(x0,y 0) and m2 = − 1
f(x0,y 0) .
This suggests a method for finding orthogonal trajectories o f a family of integral curves of a first order
equation.
Finding Orthogonal T rajectories
Step 1. Find a differential equation
y′= f(x,y )
for the given family .
Step 2. Solve the differential equation
y′= − 1
f(x,y )
to find the orthogonal trajectories.
Example 4.5.9 Find the orthogonal trajectories of the family of circles
x2 + y2 = c2 (c> 0). (4.5.17)
Solution T o find a differential equation for the family of circles we di fferentiate (
4.5.17) implicitly with
respect to xto obtain
2x+ 2 yy′= 0 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 187
or
y′= −x
y.
Therefore the integral curves of
y′= y
x
are orthogonal trajectories of the given family . W e leave it to you to verify that the general solution of
this equation is
y= kx,
where kis an arbitrary constant. This is the equation of a nonvertic al line through (0, 0). The y axis is
also an orthogonal trajectory of the given family . Therefor e every line through the origin is an orthogonal
trajectory of the given family ( 4.5.17) (Figure 4.5.11). This is consistent with the theorem of plane
geometry which states that a diameter of a circle and a tangen t line to the circle at the end of the diameter
are perpendicular.
Example 4.5.10 Find the orthogonal trajectories of the family of hyperbola s
xy= c (c̸= 0) (4.5.18)
(Figure
4.5.7).
Solution Differentiating ( 4.5.18) implicitly with respect to xyields
y+ xy′= 0 ,
or
y′= − y
x;
thus, the integral curves of
y′= x
y",Elementary Differential Equations with Boundary Value Problems.pdf
"(Figure
4.5.7).
Solution Differentiating ( 4.5.18) implicitly with respect to xyields
y+ xy′= 0 ,
or
y′= − y
x;
thus, the integral curves of
y′= x
y
are orthogonal trajectories of the given family . Separatin g variables yields
y′y= x
and integrating yields
y2 − x2 = k,
which is the equation of a hyperbola if k̸= 0 , or of the lines y= xand y = −xif k= 0 (Figure 4.5.12).
Example 4.5.11 Find the orthogonal trajectories of the family of circles de fined by
(x− c)2 + y2 = c2 (c̸= 0) . (4.5.19)
These circles are centered on the x-axis and tangent to the y-axis (Figure 4.5.13(a)).
Solution Multiplying out the left side of ( 4.5.19) yields
x2 − 2cx+ y2 = 0 , (4.5.20)",Elementary Differential Equations with Boundary Value Problems.pdf
"188 Chapter 4 Applications of First Order Equations
 x
 y
Figure 4.5.12 Orthogonal trajectories of the hyperbolas xy= c
and differentiating this implicitly with respect to xyields
2(x− c) + 2 yy′= 0 . (4.5.21)
From ( 4.5.20),
c= x2 + y2
2x ,
so
x− c= x− x2 + y2
2x = x2 − y2
2x .
Substituting this into ( 4.5.21) and solving for y′yields
y′= y2 − x2
2xy . (4.5.22)
The curves defined by ( 4.5.19) are integral curves of ( 4.5.22), and the integral curves of
y′= 2xy
x2 − y2
are orthogonal trajectories of the family ( 4.5.19). This is a homogeneous nonlinear equation, which we
studied in Section 2.4. Substituting y = uxyields
u′x+ u= 2x(ux)
x2 − (ux)2 = 2u
1 − u2 ,
so
u′x= 2u
1 − u2 − u= u(u2 + 1)
1 − u2 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 189
Separating variables yields
1 − u2
u(u2 + 1) u′= 1
x,
or, equivalently , [ 1
u − 2u
u2 + 1
]
u′= 1
x.
Therefore
ln |u| −ln(u2 + 1) = ln |x|+ k.
By substituting u= y/x , we see that
ln |y| −ln |x| −ln(x2 + y2) + ln( x2) = ln |x|+ k,
which, since ln(x2) = 2 ln |x|, is equivalent to
ln |y| −ln(x2 + y2) = k,
or
|y|= ek (x2 + y2).
T o see what these curves are we rewrite this equation as
x2 + |y|2 − e−k|y|= 0
and complete the square to obtain
x2 + ( |y| −e−k / 2)2 = ( e−k / 2)2.
This can be rewritten as
x2 + ( y− h)2 = h2,
where
h=





e−k
2 if y≥ 0,
−e−k
2 if y≤ 0.
Thus, the orthogonal trajectories are circles centered on t he y axis and tangent to the x axis (Fig-
ure 4.5.13(b)). The circles for which h > 0 are above the x-axis, while those for which h < 0 are
below .
 y  y
 x x
(a) (b)
Figure 4.5.13 (a) The circles (x− c)2 + y2 = c2 (b) The circles x2 + ( y− h)2 = h2",Elementary Differential Equations with Boundary Value Problems.pdf
"190 Chapter 4 Applications of First Order Equations
4.5 Exercises
In Exercises 1– 8 find a first order differential equation for the given family o f curves.
1. y(x2 + y2) = c 2. exy = cy
3. ln |xy|= c(x2 + y2 ) 4. y= x1/ 2 + cx
5. y= ex2
+ ce−x2
6. y= x3 + c
x
7. y= sin x+ cex 8. y= ex + c(1 + x2)
9. Show that the family of circles
(x− x0)2 + y2 = 1 , −∞ <x0 <∞ ,
can be obtained by joining integral curves of two first order d ifferential equations. More specifi-
cally , find differential equations for the families of semic ircles
(x− x0)2 + y2 = 1 , x0 <x<x 0 + 1 , −∞ <x0 <∞ ,
(x− x0)2 + y2 = 1 , x0 − 1 <x<x 0, −∞ <x0 <∞ .
10. Suppose f and gare differentiable for all x. Find a differential equation for the family of functions
y= f + cg(c=constant).
In Exercises 11– 13 find a first order differential equation for the given family o f curves.
11. Lines through a given point (x0,y 0).
12. Circles through (−1, 0) and (1, 0).
13. Circles through (0, 0) and (0, 2).",Elementary Differential Equations with Boundary Value Problems.pdf
"11. Lines through a given point (x0,y 0).
12. Circles through (−1, 0) and (1, 0).
13. Circles through (0, 0) and (0, 2).
14. Use the method Example 4.5.6(a) to find the equations of lines through the given points tangen t to
the parabola y= x2. Also, find the points of tangency .
(a) (5, 9) (b) (6, 11) (c) (−6, 20) (d) (−3, 5)
15. (a) Show that the equation of the line tangent to the circle
x2 + y2 = 1 (A)
at a point (x0,y 0) on the circle is
y = 1 − x0x
y0
if x0 ̸= ±1. (B)
(b) Show that if y′is the slope of a nonvertical tangent line to the circle (A) an d (x,y ) is a point
on the tangent line then
(y′)2(x2 − 1) − 2xyy′+ y2 − 1 = 0 . (C)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 4.5 Applications to Curves 191
(c) Show that the segment of the tangent line (B) on which (x− x0)/y 0 >0 is an integral curve
of the differential equation
y′= xy−
√
x2 + y2 − 1
x2 − 1 , (D)
while the segment on which (x− x0)/y 0 <0 is an integral curve of the differential equation
y′= xy+
√
x2 + y2 − 1
x2 − 1 . (E)
HIN T : Use the quadratic formula to solve (C) for y′. Then substitute (B) for y and choose
the ± sign in the quadratic formula so that the resulting expressi on for y′reduces to the
known slope y′= −x0/y 0.
(d) Show that the upper and lower semicircles of (A) are also inte gral curves of (D) and (E).
(e) Find the equations of two lines through (5,5) tangent to the c ircle (A), and find the points of
tangency .
16. (a) Show that the equation of the line tangent to the parabola
x= y2 (A)
at a point (x0,y 0) ̸= (0 , 0) on the parabola is
y = y0
2 + x
2y0
. (B)
(b) Show that if y′is the slope of a nonvertical tangent line to the parabola (A) and (x,y ) is a",Elementary Differential Equations with Boundary Value Problems.pdf
"y = y0
2 + x
2y0
. (B)
(b) Show that if y′is the slope of a nonvertical tangent line to the parabola (A) and (x,y ) is a
point on the tangent line then
4x2(y′)2 − 4xyy′+ x= 0 . (C)
(c) Show that the segment of the tangent line defined in (a) on which x>x 0 is an integral curve
of the differential equation
y′= y+
√
y2 − x
2x , (D)
while the segment on which x<x 0 is an integral curve of the differential equation
y′= y−
√
y2 − x
2x , (E)
HIN T : Use the quadratic formula to solve (C) for y′. Then substitute (B) for y and choose
the ± sign in the quadratic formula so that the resulting expressi on for y′reduces to the
known slope y′= 1
2y0
.
(d) Show that the upper and lower halves of the parabola (A), give n by y = √xand y = −√x
for x> 0, are also integral curves of (D) and (E).
17. Use the results of Exercise 16 to find the equations of two lines tangent to the parabola x= y2 and
passing through the given point. Also find the points of tange ncy .",Elementary Differential Equations with Boundary Value Problems.pdf
"passing through the given point. Also find the points of tange ncy .
(a) (−5, 2) (b) (−4, 0) (c) (7, 4) (d) (5, −3)
18. Find a curve y = y(x) through (1,2) such that the tangent to the curve at any point (x0,y (x0))
intersects the xaxis at xI = x0/ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"192 Chapter 4 Applications of First Order Equations
19. Find all curves y = y(x) such that the tangent to the curve at any point (x0,y (x0)) intersects the
xaxis at xI = x3
0.
20. Find all curves y = y(x) such that the tangent to the curve at any point passes through a given
point (x1,y 1).
21. Find a curve y = y(x) through (1, −1) such that the tangent to the curve at any point (x0,y (x0))
intersects the yaxis at yI = x3
0.
22. Find all curves y = y(x) such that the tangent to the curve at any point (x0,y (x0)) intersects the
yaxis at yI = x0.
23. Find a curve y = y(x) through (0, 2) such that the normal to the curve at any point (x0,y (x0))
intersects the xaxis at xI = x0 + 1 .
24. Find a curve y = y(x) through (2, 1) such that the normal to the curve at any point (x0,y (x0))
intersects the yaxis at yI = 2 y(x0).
In Exercises
25– 29 find the orthogonal trajectories of the given family of curve s.
25. x2 + 2 y2 = c2 26. x2 + 4 xy+ y2 = c
27. y= ce2x 28. xyex2
= c
29. y= cex
x",Elementary Differential Equations with Boundary Value Problems.pdf
"25. x2 + 2 y2 = c2 26. x2 + 4 xy+ y2 = c
27. y= ce2x 28. xyex2
= c
29. y= cex
x
30. Find a curve through (−1, 3) orthogonal to every parabola of the form
y= 1 + cx2
that it intersects. Which of these parabolas does the desire d curve intersect?
31. Show that the orthogonal trajectories of
x2 + 2 axy+ y2 = c
satisfy
|y− x|a+1|y+ x|a−1 = k.
32. If lines Land L1 intersect at (x0,y 0) and α is the smallest angle through which Lmust be rotated
counterclockwise about (x0,y 0) to bring it into coincidence with L1, we say that α is the angle
from Lto L1; thus, 0 ≤ α < π. If Land L1 are tangents to curves C and C1, respectively , that
intersect at (x0,y 0), we say that C1 intersects Cat the angle α. Use the identity
tan(A+ B) = tan A+ tan B
1 − tan Atan B
to show that if Cand C1 are intersecting integral curves of
y′= f(x,y ) and y′= f(x,y ) + tan α
1 − f(x,y ) tan α
(
α ̸= π
2
)
,
respectively , then C1 intersects Cat the angle α.",Elementary Differential Equations with Boundary Value Problems.pdf
"y′= f(x,y ) and y′= f(x,y ) + tan α
1 − f(x,y ) tan α
(
α ̸= π
2
)
,
respectively , then C1 intersects Cat the angle α.
33. Use the result of Exercise 32 to find a family of curves that intersect every nonvertical li ne through
the origin at the angle α = π/ 4.
34. Use the result of Exercise 32 to find a family of curves that intersect every circle centere d at the
origin at a given angle α ̸= π/ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 5
Linear Second Order Equations
IN THIS CHAPTER we study a particularly important class of se cond order equations. Because of
their many applications in science and engineering, second order differential equation have historically
been the most thoroughly studied class of differential equa tions. Research on the theory of second order
differential equations continues to the present day . This c hapter is devoted to second order equations that
can be written in the form
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x).
Such equations are said to be linear. As in the case of first order linear equations, (A) is said to b e
homogeneous if F ≡ 0, or nonhomogeneous if F ̸≡0.
SECTION 5.1 is devoted to the theory of homogeneous linear eq uations.
SECTION 5.2 deals with homogeneous equations of the special form
ay′′+ by′+ cy= 0 ,
where a, b, and care constant ( a̸= 0 ). When you’ve completed this section you’ll know everythin g there
is to know about solving such equations.",Elementary Differential Equations with Boundary Value Problems.pdf
"where a, b, and care constant ( a̸= 0 ). When you’ve completed this section you’ll know everythin g there
is to know about solving such equations.
SECTION 5.3 presents the theory of nonhomogeneous linear eq uations.
SECTIONS 5.4 AND 5.5 present the
method of undetermined coefficients , which can be used to solve
nonhomogeneous equations of the form
ay′′+ by′+ cy= F(x),
where a, b, and care constants and F has a special form that is still sufficiently general to occur in many
applications. In this section we make extensive use of the id ea of variation of parameters introduced in
Chapter 2.
SECTION 5.6 deals with
reduction of order , a technique based on the idea of variation of parameters,
which enables us to find the general solution of a nonhomogene ous linear second order equation provided
that we know one nontrivial (not identically zero) solution of the associated homogeneous equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"that we know one nontrivial (not identically zero) solution of the associated homogeneous equation.
SECTION 5.6 deals with the method traditionally called variation of parameters , which enables us to
find the general solution of a nonhomogeneous linear second o rder equation provided that we know two
nontrivial solutions (with nonconstant ratio) of the assoc iated homogeneous equation.
193",Elementary Differential Equations with Boundary Value Problems.pdf
"194 Chapter 5 Linear Second Order Equations
5.1 HOMOGENEOUS LINEAR EQUA TIONS
A second order differential equation is said to be linear if it can be written as
y′′+ p(x)y′+ q(x)y = f(x). (5.1.1)
W e call the function f on the right a forcing function , since in physical applications it’s often related to
a force acting on some system modeled by the differential equ ation. W e say that ( 5.1.1) is homogeneous
if f ≡ 0 or nonhomogeneous if f ̸≡0. Since these definitions are like the corresponding definiti ons in
Section 2.1 for the linear first order equation
y′+ p(x)y = f(x), (5.1.2)
it’s natural to expect similarities between methods of solv ing ( 5.1.1) and ( 5.1.2). However, solving ( 5.1.1)
is more difficult than solving ( 5.1.2). For example, while Theorem 2.1.1 gives a formula for the general
solution of ( 5.1.2) in the case where f ≡ 0 and Theorem 2.1.2 gives a formula for the case where f ̸≡0,",Elementary Differential Equations with Boundary Value Problems.pdf
"solution of ( 5.1.2) in the case where f ≡ 0 and Theorem 2.1.2 gives a formula for the case where f ̸≡0,
there are no formulas for the general solution of ( 5.1.1) in either case. Therefore we must be content to
solve linear second order equations of special forms.
In Section 2.1 we considered the homogeneous equation y′+p(x)y = 0 first, and then used a nontrivial
solution of this equation to find the general solution of the n onhomogeneous equation y′+ p(x)y = f(x).
Although the progression from the homogeneous to the nonhom ogeneous case isn’t that simple for the
linear second order equation, it’s still necessary to solve the homogeneous equation
y′′+ p(x)y′+ q(x)y= 0 (5.1.3)
in order to solve the nonhomogeneous equation ( 5.1.1). This section is devoted to ( 5.1.3).
The next theorem gives sufficient conditions for existence a nd uniqueness of solutions of initial value
problems for ( 5.1.3). W e omit the proof.",Elementary Differential Equations with Boundary Value Problems.pdf
"The next theorem gives sufficient conditions for existence a nd uniqueness of solutions of initial value
problems for ( 5.1.3). W e omit the proof.
Theorem 5.1.1 Suppose pand qare continuous on an open interval (a,b ), let x0 be any point in (a,b ),
and let k0 and k1 be arbitrary real numbers . Then the initial value problem
y′′+ p(x)y′+ q(x)y = 0 , y(x0) = k0, y′(x0) = k1
has a unique solution on (a,b ).
Since y ≡ 0 is obviously a solution of ( 5.1.3) we call it the trivial solution. Any other solution is
nontrivial. Under the assumptions of Theorem 5.1.1, the only solution of the initial value problem
y′′+ p(x)y′+ q(x)y= 0 , y(x0 ) = 0 , y′(x0) = 0
on (a,b ) is the trivial solution (Exercise 24).
The next three examples illustrate concepts that we’ll deve lop later in this section. Y ou shouldn’t be
concerned with how to find the given solutions of the equations in these examples. This will be explained
in later sections.
Example 5.1.1 The coefficients of y′and yin
y′′− y = 0 (5.1.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"in later sections.
Example 5.1.1 The coefficients of y′and yin
y′′− y = 0 (5.1.4)
are the constant functions p ≡ 0 and q ≡ − 1, which are continuous on (−∞ , ∞ ). Therefore Theo-
rem
5.1.1 implies that every initial value problem for ( 5.1.4) has a unique solution on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 195
(a) V erify that y1 = ex and y2 = e−x are solutions of ( 5.1.4) on (−∞ , ∞ ).
(b) V erify that if c1 and c2 are arbitrary constants, y = c1ex +c2e−x is a solution of ( 5.1.4) on (−∞ , ∞ ).
(c) Solve the initial value problem
y′′− y = 0 , y (0) = 1 , y ′(0) = 3 . (5.1.5)
SO LU TIO N (a) If y1 = ex then y′
1 = ex and y′′
1 = ex = y1, so y′′
1 −y1 = 0 . If y2 = e−x, then y′
2 = −e−x
and y′′
2 = e−x = y2, so y′′
2 − y2 = 0 .
SO LU TIO N (b) If
y= c1ex + c2e−x (5.1.6)
then
y′= c1ex − c2e−x (5.1.7)
and
y′′= c1ex + c2e−x,
so
y′′− y = ( c1ex + c2e−x) − (c1ex + c2e−x)
= c1(ex − ex) + c2(e−x − e−x) = 0
for all x. Therefore y= c1ex + c2e−x is a solution of ( 5.1.4) on (−∞ , ∞ ).
SO LU TIO N (c) W e can solve ( 5.1.5) by choosing c1 and c2 in ( 5.1.6) so that y(0) = 1 and y′(0) = 3 .
Setting x= 0 in ( 5.1.6) and ( 5.1.7) shows that this is equivalent to
c1 + c2 = 1
c1 − c2 = 3 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Setting x= 0 in ( 5.1.6) and ( 5.1.7) shows that this is equivalent to
c1 + c2 = 1
c1 − c2 = 3 .
Solving these equations yields c1 = 2 and c2 = −1. Therefore y = 2 ex − e−x is the unique solution of
(5.1.5) on (−∞ , ∞ ).
Example 5.1.2 Let ω be a positive constant. The coefficients of y′and yin
y′′+ ω2y = 0 (5.1.8)
are the constant functions p ≡ 0 and q ≡ ω2, which are continuous on (−∞ , ∞ ). Therefore Theo-
rem 5.1.1 implies that every initial value problem for ( 5.1.8) has a unique solution on (−∞ , ∞ ).
(a) V erify that y1 = cos ωx and y2 = sin ωx are solutions of ( 5.1.8) on (−∞ , ∞ ).
(b) V erify that if c1 and c2 are arbitrary constants then y= c1 cos ωx + c2 sin ωx is a solution of ( 5.1.8)
on (−∞ , ∞ ).
(c) Solve the initial value problem
y′′+ ω2y = 0 , y (0) = 1 , y ′(0) = 3 . (5.1.9)
SO LU TIO N (a) If y1 = cos ωx then y′
1 = −ω sin ωx and y′′
1 = −ω2 cos ωx = −ω2y1, so y′′
1 + ω2 y1 = 0 .
If y2 = sin ωx then, y′
2 = ω cos ωx and y′′
2 = −ω2 sin ωx = −ω2y2, so y′′",Elementary Differential Equations with Boundary Value Problems.pdf
"1 = −ω sin ωx and y′′
1 = −ω2 cos ωx = −ω2y1, so y′′
1 + ω2 y1 = 0 .
If y2 = sin ωx then, y′
2 = ω cos ωx and y′′
2 = −ω2 sin ωx = −ω2y2, so y′′
2 + ω2y2 = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"196 Chapter 5 Linear Second Order Equations
SO LU TIO N (b) If
y= c1 cos ωx + c2 sin ωx (5.1.10)
then
y′= ω(−c1 sin ωx + c2 cos ωx) (5.1.11)
and
y′′= −ω2(c1 cos ωx + c2 sin ωx),
so
y′′+ ω2y = −ω2(c1 cos ωx + c2 sin ωx) + ω2(c1 cos ωx + c2 sin ωx)
= c1ω2(− cos ωx + cos ωx) + c2ω2(− sin ωx + sin ωx) = 0
for all x. Therefore y= c1 cos ωx + c2 sin ωx is a solution of ( 5.1.8) on (−∞ , ∞ ).
SO LU TIO N (c) T o solve ( 5.1.9), we must choosing c1 and c2 in ( 5.1.10) so that y(0) = 1 and y′(0) = 3 .
Setting x= 0 in ( 5.1.10) and ( 5.1.11) shows that c1 = 1 and c2 = 3 /ω . Therefore
y = cos ωx + 3
ω sin ωx
is the unique solution of ( 5.1.9) on (−∞ , ∞ ).
Theorem 5.1.1 implies that if k0 and k1 are arbitrary real numbers then the initial value problem
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 , y (x0 ) = k0, y ′(x0) = k1 (5.1.12)
has a unique solution on an interval (a,b ) that contains x0, provided that P0, P1, and P2 are continuous",Elementary Differential Equations with Boundary Value Problems.pdf
"has a unique solution on an interval (a,b ) that contains x0, provided that P0, P1, and P2 are continuous
and P0 has no zeros on (a,b ). T o see this, we rewrite the differential equation in ( 5.1.12) as
y′′+ P1(x)
P0(x) y′+ P2(x)
P0(x) y= 0
and apply Theorem 5.1.1 with p= P1/P 0 and q= P2/P 0.
Example 5.1.3 The equation
x2y′′+ xy′− 4y= 0 (5.1.13)
has the form of the differential equation in ( 5.1.12), with P0(x) = x2, P1(x) = x, and P2(x) = −4,
which are are all continuous on (−∞ , ∞ ). However, since P(0) = 0 we must consider solutions of
(5.1.13) on (−∞ , 0) and (0, ∞ ). Since P0 has no zeros on these intervals, Theorem 5.1.1 implies that the
initial value problem
x2y′′+ xy′− 4y= 0 , y (x0) = k0, y ′(x0) = k1
has a unique solution on (0, ∞ ) if x0 >0, or on (−∞ , 0) if x0 <0.
(a) V erify that y1 = x2 is a solution of ( 5.1.13) on (−∞ , ∞ ) and y2 = 1 /x 2 is a solution of ( 5.1.13)
on (−∞ , 0) and (0, ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) V erify that y1 = x2 is a solution of ( 5.1.13) on (−∞ , ∞ ) and y2 = 1 /x 2 is a solution of ( 5.1.13)
on (−∞ , 0) and (0, ∞ ).
(b) V erify that if c1 and c2 are any constants then y= c1x2 + c2/x 2 is a solution of ( 5.1.13) on (−∞ , 0)
and (0, ∞ ).
(c) Solve the initial value problem
x2y′′+ xy′− 4y= 0 , y (1) = 2 , y ′(1) = 0 . (5.1.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 197
(d) Solve the initial value problem
x2y′′+ xy′− 4y= 0 , y (−1) = 2 , y ′(−1) = 0 . (5.1.15)
SO LU TIO N (a) If y1 = x2 then y′
1 = 2 xand y′′
1 = 2 , so
x2y′′
1 + xy′
1 − 4y1 = x2(2) + x(2x) − 4x2 = 0
for xin (−∞ , ∞ ). If y2 = 1 /x 2, then y′
2 = −2/x 3 and y′′
2 = 6 /x 4, so
x2y′′
2 + xy′
2 − 4y2 = x2
(6
x4
)
− x
(2
x3
)
− 4
x2 = 0
for xin (−∞ , 0) or (0, ∞ ).
SO LU TIO N (b) If
y= c1x2 + c2
x2 (5.1.16)
then
y′= 2 c1x− 2c2
x3 (5.1.17)
and
y′′= 2 c1 + 6c2
x4 ,
so
x2y′′+ xy′− 4y = x2
(
2c1 + 6c2
x4
)
+ x
(
2c1x− 2c2
x3
)
− 4
(
c1x2 + c2
x2
)
= c1(2x2 + 2 x2 − 4x2) + c2
(6
x2 − 2
x2 − 4
x2
)
= c1 ·0 + c2 ·0 = 0
for xin (−∞ , 0) or (0, ∞ ).
SO LU TIO N (c) T o solve ( 5.1.14), we choose c1 and c2 in ( 5.1.16) so that y(1) = 2 and y′(1) = 0 . Setting
x= 1 in ( 5.1.16) and ( 5.1.17) shows that this is equivalent to
c1 + c2 = 2
2c1 − 2c2 = 0 .
Solving these equations yields c1 = 1 and c2 = 1 . Therefore y = x2 + 1 /x 2 is the unique solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"c1 + c2 = 2
2c1 − 2c2 = 0 .
Solving these equations yields c1 = 1 and c2 = 1 . Therefore y = x2 + 1 /x 2 is the unique solution of
(5.1.14) on (0, ∞ ).
SO LU TIO N (d) W e can solve ( 5.1.15) by choosing c1 and c2 in ( 5.1.16) so that y(−1) = 2 and y′(−1) =
0. Setting x= −1 in ( 5.1.16) and ( 5.1.17) shows that this is equivalent to
c1 + c2 = 2
−2c1 + 2 c2 = 0 .
Solving these equations yields c1 = 1 and c2 = 1 . Therefore y = x2 + 1 /x 2 is the unique solution of
(5.1.15) on (−∞ , 0).",Elementary Differential Equations with Boundary Value Problems.pdf
"198 Chapter 5 Linear Second Order Equations
Although the formulas for the solutions of ( 5.1.14) and ( 5.1.15) are both y = x2 + 1 /x 2, you should
not conclude that these two initial value problems have the s ame solution. Remember that a solution of
an initial value problem is defined on an interval that contains the initial point ; therefore, the solution
of ( 5.1.14) is y = x2 + 1 /x 2 on the interval (0, ∞ ), which contains the initial point x0 = 1 , while the
solution of ( 5.1.15) is y= x2 + 1 /x 2 on the interval (−∞ , 0), which contains the initial point x0 = −1.
The General Solution of a Homogeneous Linear Second Order Eq uation
If y1 and y2 are defined on an interval (a,b ) and c1 and c2 are constants, then
y= c1y1 + c2y2
is a linear combination of y1 and y2. For example, y = 2 cos x+ 7 sin x is a linear combination of
y1 = cos xand y2 = sin x, with c1 = 2 and c2 = 7 .
The next theorem states a fact that we’ve already verified in E xamples 5.1.1, 5.1.2, and 5.1.3.",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 = cos xand y2 = sin x, with c1 = 2 and c2 = 7 .
The next theorem states a fact that we’ve already verified in E xamples 5.1.1, 5.1.2, and 5.1.3.
Theorem 5.1.2 If y1 and y2 are solutions of the homogeneous equation
y′′+ p(x)y′+ q(x)y= 0 (5.1.18)
on (a,b ), then any linear combination
y= c1y1 + c2y2 (5.1.19)
of y1 and y2 is also a solution of (5.1.18) on (a,b ).
Proof If
y= c1y1 + c2y2
then
y′= c1y′
1 + c2y′
2 and y′′= c1y′′
1 + c2y′′
2 .
Therefore
y′′+ p(x)y′+ q(x)y = ( c1y′′
1 + c2y′′
2 ) + p(x)(c1y′
1 + c2y′
2) + q(x)(c1y1 + c2y2)
= c1 (y′′
1 + p(x)y′
1 + q(x)y1) + c2 (y′′
2 + p(x)y′
2 + q(x)y2)
= c1 ·0 + c2 ·0 = 0 ,
since y1 and y2 are solutions of ( 5.1.18).
W e say that {y1,y 2}is a fundamental set of solutions of (5.1.18) on (a,b ) if every solution of ( 5.1.18)
on (a,b ) can be written as a linear combination of y1 and y2 as in ( 5.1.19). In this case we say that
(5.1.19) is general solution of (5.1.18) on (a,b ).
Linear Independence",Elementary Differential Equations with Boundary Value Problems.pdf
"(5.1.19) is general solution of (5.1.18) on (a,b ).
Linear Independence
W e need a way to determine whether a given set {y1,y 2}of solutions of ( 5.1.18) is a fundamental set.
The next definition will enable us to state necessary and suffi cient conditions for this.
W e say that two functions y1 and y2 defined on an interval (a,b ) are linearly independent on (a,b ) if
neither is a constant multiple of the other on (a,b ). (In particular, this means that neither can be the trivial
solution of ( 5.1.18), since, for example, if y1 ≡ 0 we could write y1 = 0 y2.) W e’ll also say that the set
{y1,y 2}is linearly independent on (a,b ).
Theorem 5.1.3 Suppose pand qare continuous on (a,b ). Then a set {y1,y 2}of solutions of
y′′+ p(x)y′+ q(x)y= 0 (5.1.20)
on (a,b ) is a fundamental set if and only if {y1,y 2}is linearly independent on (a,b ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 199
W e’ll present the proof of Theorem 5.1.3 in steps worth regarding as theorems in their own right.
However, let’s first interpret Theorem 5.1.3 in terms of Examples 5.1.1, 5.1.2, and 5.1.3.
Example 5.1.4
(a) Since ex/e −x = e2x is nonconstant, Theorem 5.1.3 implies that y = c1ex + c2e−x is the general
solution of y′′− y= 0 on (−∞ , ∞ ).
(b) Since cos ωx/ sin ωx = cot ωx is nonconstant, Theorem 5.1.3 implies that y = c1 cos ωx +
c2 sin ωx is the general solution of y′′+ ω2y= 0 on (−∞ , ∞ ).
(c) Since x2/x −2 = x4 is nonconstant, Theorem 5.1.3 implies that y = c1x2 + c2/x 2 is the general
solution of x2y′′+ xy′− 4y= 0 on (−∞ , 0) and (0, ∞ ).
The Wronskian and Abel’s Formula
T o motivate a result that we need in order to prove Theorem 5.1.3, let’s see what is required to prove that
{y1,y 2}is a fundamental set of solutions of ( 5.1.20) on (a,b ). Let x0 be an arbitrary point in (a,b ), and",Elementary Differential Equations with Boundary Value Problems.pdf
"{y1,y 2}is a fundamental set of solutions of ( 5.1.20) on (a,b ). Let x0 be an arbitrary point in (a,b ), and
suppose yis an arbitrary solution of ( 5.1.20) on (a,b ). Then yis the unique solution of the initial value
problem
y′′+ p(x)y′+ q(x)y = 0 , y (x0 ) = k0, y ′(x0) = k1; (5.1.21)
that is, k0 and k1 are the numbers obtained by evaluating y and y′at x0. Moreover, k0 and k1 can be
any real numbers, since Theorem 5.1.1 implies that ( 5.1.21) has a solution no matter how k0 and k1 are
chosen. Therefore {y1,y 2}is a fundamental set of solutions of ( 5.1.20) on (a,b ) if and only if it’s possible
to write the solution of an arbitrary initial value problem ( 5.1.21) as y = c1y1 + c2y2. This is equivalent
to requiring that the system
c1y1(x0) + c2y2(x0) = k0
c1y′
1(x0) + c2y′
2(x0) = k1
(5.1.22)
has a solution (c1,c 2) for every choice of (k0,k 1). Let’s try to solve ( 5.1.22).
Multiplying the first equation in ( 5.1.22) by y′
2(x0) and the second by y2(x0) yields
c1y1(x0)y′",Elementary Differential Equations with Boundary Value Problems.pdf
"Multiplying the first equation in ( 5.1.22) by y′
2(x0) and the second by y2(x0) yields
c1y1(x0)y′
2(x0) + c2y2(x0)y′
2(x0) = y′
2(x0)k0
c1y′
1(x0)y2(x0) + c2y′
2(x0)y2(x0) = y2(x0)k1,
and subtracting the second equation here from the first yield s
(y1(x0)y′
2 (x0) − y′
1(x0)y2(x0)) c1 = y′
2(x0)k0 − y2(x0)k1. (5.1.23)
Multiplying the first equation in ( 5.1.22) by y′
1(x0) and the second by y1(x0) yields
c1y1(x0)y′
1(x0) + c2y2(x0)y′
1(x0) = y′
1(x0)k0
c1y′
1(x0)y1(x0) + c2y′
2(x0)y1(x0) = y1(x0)k1,
and subtracting the first equation here from the second yield s
(y1(x0)y′
2 (x0) − y′
1(x0)y2(x0)) c2 = y1(x0)k1 − y′
1(x0)k0. (5.1.24)
If
y1(x0)y′
2(x0) − y′
1(x0)y2 (x0) = 0 ,
it’s impossible to satisfy ( 5.1.23) and ( 5.1.24) (and therefore ( 5.1.22)) unless k0 and k1 happen to satisfy
y1(x0)k1 − y′
1(x0)k0 = 0
y′
2(x0)k0 − y2(x0)k1 = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"200 Chapter 5 Linear Second Order Equations
On the other hand, if
y1(x0)y′
2(x0) − y′
1(x0)y2(x0) ̸= 0 (5.1.25)
we can divide ( 5.1.23) and ( 5.1.24) through by the quantity on the left to obtain
c1 = y′
2(x0)k0 − y2(x0)k1
y1(x0)y′
2(x0) − y′
1(x0)y2 (x0)
c2 = y1(x0)k1 − y′
1(x0)k0
y1(x0)y′
2(x0) − y′
1(x0)y2 (x0) ,
(5.1.26)
no matter how k0 and k1 are chosen. This motivates us to consider conditions on y1 and y2 that imply
(5.1.25).
Theorem 5.1.4 Suppose pand qare continuous on (a,b ), let y1 and y2 be solutions of
y′′+ p(x)y′+ q(x)y= 0 (5.1.27)
on (a,b ), and define
W = y1y′
2 − y′
1y2. (5.1.28)
Let x0 be any point in (a,b ). Then
W(x) = W(x0)e
−
∫x
x0
p(t) dt
, a<x<b. (5.1.29)
Therefore either W has no zeros in (a,b ) or W ≡ 0 on (a,b ).
Proof Differentiating ( 5.1.28) yields
W′= y′
1y′
2 + y1y′′
2 − y′
1y′
2 − y′′
1 y2 = y1y′′
2 − y′′
1 y2. (5.1.30)
Since y1 and y2 both satisfy ( 5.1.27),
y′′
1 = −py′
1 − qy1 and y′′
2 = −py′
2 − qy2.
Substituting these into ( 5.1.30) yields",Elementary Differential Equations with Boundary Value Problems.pdf
"1 y2. (5.1.30)
Since y1 and y2 both satisfy ( 5.1.27),
y′′
1 = −py′
1 − qy1 and y′′
2 = −py′
2 − qy2.
Substituting these into ( 5.1.30) yields
W′ = −y1
(
py′
2 + qy2
)
+ y2
(
py′
1 + qy1
)
= −p(y1y′
2 − y2y′
1) − q(y1y2 − y2y1)
= −p(y1y′
2 − y2y′
1) = −pW.
Therefore W′+ p(x)W = 0 ; that is, W is the solution of the initial value problem
y′+ p(x)y= 0 , y (x0 ) = W(x0).
W e leave it to you to verify by separation of variables that th is implies ( 5.1.29). If W(x0) ̸= 0 , ( 5.1.29)
implies that W has no zeros in (a,b ), since an exponential is never zero. On the other hand, if W(x0) = 0 ,
(5.1.29) implies that W(x) = 0 for all xin (a,b ).
The function W defined in ( 5.1.28) is the Wronskian of {y1,y 2}. Formula ( 5.1.29) is Abel’s formula .
The Wronskian of {y1,y 2}is usually written as the determinant
W =
⏐
⏐
⏐
⏐
⏐
y1 y2
y′
1 y′
2
⏐
⏐
⏐
⏐
⏐.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 201
The expressions in ( 5.1.26) for c1 and c2 can be written in terms of determinants as
c1 = 1
W(x0)
⏐
⏐
⏐
⏐
⏐
k0 y2(x0)
k1 y′
2(x0)
⏐
⏐
⏐
⏐
⏐ and c2 = 1
W(x0)
⏐
⏐
⏐
⏐
⏐
y1(x0) k0
y′
1(x0) k1
⏐
⏐
⏐
⏐
⏐.
If you’ve taken linear algebra you may recognize this as
Cramer’s rule .
Example 5.1.5 V erify Abel’s formula for the following differential equat ions and the corresponding so-
lutions, from Examples 5.1.1, 5.1.2, and 5.1.3:
(a) y′′− y = 0; y1 = ex, y2 = e−x
(b) y′′+ ω2y = 0; y1 = cos ωx, y 2 = sin ωx
(c) x2y′′+ xy′− 4y= 0; y1 = x2, y2 = 1 /x 2
SO LU TIO N (a) Since p≡ 0, we can verify Abel’s formula by showing that W is constant, which is true,
since
W(x) =
⏐
⏐
⏐
⏐
⏐
ex e−x
ex −e−x
⏐
⏐
⏐
⏐
⏐= ex(−e−x ) − ex e−x = −2
for all x.
SO LU TIO N (b) Again, since p≡ 0, we can verify Abel’s formula by showing that W is constant, which
is true, since
W(x) =
⏐
⏐
⏐
⏐
⏐
cos ωx sin ωx
−ω sin ωx ω cos ωx
⏐
⏐
⏐
⏐
⏐
= cos ωx(ω cos ωx) − (−ω sin ωx) sin ωx",Elementary Differential Equations with Boundary Value Problems.pdf
"is true, since
W(x) =
⏐
⏐
⏐
⏐
⏐
cos ωx sin ωx
−ω sin ωx ω cos ωx
⏐
⏐
⏐
⏐
⏐
= cos ωx(ω cos ωx) − (−ω sin ωx) sin ωx
= ω(cos2 ωx + sin 2 ωx) = ω
for all x.
SO LU TIO N (c) Computing the Wronskian of y1 = x2 and y2 = 1 /x 2 directly yields
W =
⏐
⏐
⏐
⏐
⏐
x2 1/x 2
2x −2/x 3
⏐
⏐
⏐
⏐
⏐= x2
(
− 2
x3
)
− 2x
(1
x2
)
= − 4
x. (5.1.31)
T o verify Abel’s formula we rewrite the differential equati on as
y′′+ 1
xy′− 4
x2 y = 0
to see that p(x) = 1 /x . If x0 and xare either both in (−∞ , 0) or both in (0, ∞ ) then
∫ x
x0
p(t) dt=
∫ x
x0
dt
t = ln
(x
x0
)
,
so Abel’s formula becomes
W(x) = W(x0)e− ln(x/x 0) = W(x0) x0
x
= −
(4
x0
) (x0
x
)
from ( 5.1.31)
= − 4
x,",Elementary Differential Equations with Boundary Value Problems.pdf
"202 Chapter 5 Linear Second Order Equations
which is consistent with ( 5.1.31).
The next theorem will enable us to complete the proof of Theor em 5.1.3.
Theorem 5.1.5 Suppose pand qare continuous on an open interval (a,b ), let y1 and y2 be solutions of
y′′+ p(x)y′+ q(x)y= 0 (5.1.32)
on (a,b ), and let W = y1y′
2 − y′
1y2. Then y1 and y2 are linearly independent on (a,b ) if and only if W
has no zeros on (a,b ).
Proof W e first show that if W(x0) = 0 for some x0 in (a,b ), then y1 and y2 are linearly dependent on
(a,b ). Let Ibe a subinterval of (a,b ) on which y1 has no zeros. (If there’s no such subinterval, y1 ≡ 0 on
(a,b ), so y1 and y2 are linearly independent, and we’re finished with this part o f the proof.) Then y2/y 1
is defined on I, and (y2
y1
) ′
= y1y′
2 − y′
1y2
y2
1
= W
y2
1
. (5.1.33)
However, if W(x0) = 0 , Theorem 5.1.4 implies that W ≡ 0 on (a,b ). Therefore ( 5.1.33) implies that",Elementary Differential Equations with Boundary Value Problems.pdf
") ′
= y1y′
2 − y′
1y2
y2
1
= W
y2
1
. (5.1.33)
However, if W(x0) = 0 , Theorem 5.1.4 implies that W ≡ 0 on (a,b ). Therefore ( 5.1.33) implies that
(y2/y 1)′≡ 0, so y2/y 1 = c(constant) on I. This shows that y2(x) = cy1(x) for all xin I. However, we
want to show that y2 = cy1(x) for all xin (a,b ). Let Y = y2 − cy1. Then Y is a solution of ( 5.1.32)
on (a,b ) such that Y ≡ 0 on I, and therefore Y′≡ 0 on I. Consequently , if x0 is chosen arbitrarily in I
then Y is a solution of the initial value problem
y′′+ p(x)y′+ q(x)y= 0 , y (x0) = 0 , y ′(x0) = 0 ,
which implies that Y ≡ 0 on (a,b ), by the paragraph following Theorem 5.1.1. (See also Exercise 24).
Hence, y2 − cy1 ≡ 0 on (a,b ), which implies that y1 and y2 are not linearly independent on (a,b ).
Now suppose W has no zeros on (a,b ). Then y1 can’t be identically zero on (a,b ) (why not?), and
therefore there is a subinterval I of (a,b ) on which y1 has no zeros. Since ( 5.1.33) implies that y2/y 1 is",Elementary Differential Equations with Boundary Value Problems.pdf
"therefore there is a subinterval I of (a,b ) on which y1 has no zeros. Since ( 5.1.33) implies that y2/y 1 is
nonconstant on I, y2 isn’t a constant multiple of y1 on (a,b ). A similar argument shows that y1 isn’t a
constant multiple of y2 on (a,b ), since
(y1
y2
) ′
= y′
1y2 − y1y′
2
y2
2
= −W
y2
2
on any subinterval of (a,b ) where y2 has no zeros.
W e can now complete the proof of Theorem 5.1.3. From Theorem 5.1.5, two solutions y1 and y2 of
(5.1.32) are linearly independent on (a,b ) if and only if W has no zeros on (a,b ). From Theorem 5.1.4
and the motivating comments preceding it, {y1,y 2}is a fundamental set of solutions of ( 5.1.32) if and
only if W has no zeros on (a,b ). Therefore {y1,y 2}is a fundamental set for ( 5.1.32) on (a,b ) if and only
if {y1,y 2}is linearly independent on (a,b ).
The next theorem summarizes the relationships among the con cepts discussed in this section.",Elementary Differential Equations with Boundary Value Problems.pdf
"if {y1,y 2}is linearly independent on (a,b ).
The next theorem summarizes the relationships among the con cepts discussed in this section.
Theorem 5.1.6 Suppose pand qare continuous on an open interval (a,b ) and let y1 and y2 be solutions
of
y′′+ p(x)y′+ q(x)y= 0 (5.1.34)
on (a,b ). Then the following statements are equivalent ; that is , they are either all true or all false .
(a) The general solution of (5.1.34) on (a,b ) is y= c1y1 + c2y2.
(b) {y1,y 2}is a fundamental set of solutions of (5.1.34) on (a,b ).
(c) {y1,y 2}is linearly independent on (a,b ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 203
(d) The Wronskian of {y1,y 2}is nonzero at some point in (a,b ).
(e) The Wronskian of {y1,y 2}is nonzero at all points in (a,b ).
W e can apply this theorem to an equation written as
P0(x)y′′+ P1(x)y′+ P2(x)y = 0
on an interval (a,b ) where P0, P1, and P2 are continuous and P0 has no zeros.
Theorem 5.1.7 Suppose cis in (a,b ) and α and β are real numbers, not both zero. Under the assump-
tions of Theorem 5.1.7, suppose y1 and y2 are solutions of (5.1.34) such that
αy1(c) + βy′
1(c) = 0 and αy2(c) + βy′
2 (c) = 0 . (5.1.35)
Then {y1,y 2}isn’t linearly independent on (a,b ).
Proof Since α and β are not both zero, ( 5.1.35) implies that
⏐
⏐
⏐
⏐
y1(c) y′
1(c)
y2(c) y′
2(c)
⏐
⏐
⏐
⏐= 0 , so
⏐
⏐
⏐
⏐
y1(c) y2(c)
y′
1(c) y′
2(c)
⏐
⏐
⏐
⏐= 0
and Theorem
5.1.6 implies the stated conclusion.
5.1 Exercises
1. (a) V erify that y1 = e2x and y2 = e5x are solutions of
y′′− 7y′+ 10 y= 0 (A)
on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"5.1.6 implies the stated conclusion.
5.1 Exercises
1. (a) V erify that y1 = e2x and y2 = e5x are solutions of
y′′− 7y′+ 10 y= 0 (A)
on (−∞ , ∞ ).
(b) V erify that if c1 and c2 are arbitrary constants then y= c1e2x + c2e5x is a solution of (A) on
(−∞ , ∞ ).
(c) Solve the initial value problem
y′′− 7y′+ 10 y= 0 , y (0) = −1, y ′(0) = 1 .
(d) Solve the initial value problem
y′′− 7y′+ 10 y= 0 , y (0) = k0, y ′(0) = k1.
2. (a) V erify that y1 = ex cos xand y2 = ex sin xare solutions of
y′′− 2y′+ 2 y= 0 (A)
on (−∞ , ∞ ).
(b) V erify that if c1 and c2 are arbitrary constants then y = c1ex cos x+ c2ex sin xis a solution
of (A) on (−∞ , ∞ ).
(c) Solve the initial value problem
y′′− 2y′+ 2 y= 0 , y (0) = 3 , y ′(0) = −2.",Elementary Differential Equations with Boundary Value Problems.pdf
"204 Chapter 5 Linear Second Order Equations
(d) Solve the initial value problem
y′′− 2y′+ 2 y= 0 , y (0) = k0, y ′(0) = k1.
3. (a) V erify that y1 = ex and y2 = xex are solutions of
y′′− 2y′+ y= 0 (A)
on (−∞ , ∞ ).
(b) V erify that if c1 and c2 are arbitrary constants then y = ex(c1 + c2x) is a solution of (A) on
(−∞ , ∞ ).
(c) Solve the initial value problem
y′′− 2y′+ y= 0 , y (0) = 7 , y ′(0) = 4 .
(d) Solve the initial value problem
y′′− 2y′+ y= 0 , y (0) = k0, y ′(0) = k1.
4. (a) V erify that y1 = 1 / (x− 1) and y2 = 1 / (x+ 1) are solutions of
(x2 − 1)y′′+ 4 xy′+ 2 y= 0 (A)
on (−∞ , −1), (−1, 1), and (1, ∞ ). What is the general solution of (A) on each of these
intervals?
(b) Solve the initial value problem
(x2 − 1)y′′+ 4 xy′+ 2 y= 0 , y (0) = −5, y ′(0) = 1 .
What is the interval of validity of the solution?
(c) C/G Graph the solution of the initial value problem.
(d) V erify Abel’s formula for y1 and y2, with x0 = 0 .
5. Compute the Wronskians of the given sets of functions.",Elementary Differential Equations with Boundary Value Problems.pdf
"(d) V erify Abel’s formula for y1 and y2, with x0 = 0 .
5. Compute the Wronskians of the given sets of functions.
(a) {1,e x} (b) {ex,e x sin x}
(c) {x+ 1 ,x 2 + 2 } (d) {x1/ 2,x −1/ 3}
(e) {sin x
x , cos x
x } (f) {xln |x|,x 2 ln |x|}
(g) {ex cos √x,e x sin √x}
6. Find the Wronskian of a given set {y1,y 2}of solutions of
y′′+ 3( x2 + 1) y′− 2y= 0 ,
given that W(π) = 0 .
7. Find the Wronskian of a given set {y1,y 2}of solutions of
(1 − x2)y′′− 2xy′+ α(α + 1) y= 0 ,
given that W(0) = 1 . (This is Legendre’s equation .)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 205
8. Find the Wronskian of a given set {y1,y 2}of solutions of
x2y′′+ xy′+ ( x2 − ν2)y = 0 ,
given that W(1) = 1 . (This is Bessel’s equation .)
9. (This exercise shows that if you know one nontrivial solutio n of y′′+ p(x)y′+ q(x)y = 0 , you
can use Abel’s formula to find another.)
Suppose pand qare continuous and y1 is a solution of
y′′+ p(x)y′+ q(x)y= 0 (A)
that has no zeros on (a,b ). Let P(x) =
∫
p(x) dxbe any antiderivative of pon (a,b ).
(a) Show that if K is an arbitrary nonzero constant and y2 satisfies
y1y′
2 − y′
1y2 = Ke−P (x) (B)
on (a,b ), then y2 also satisfies (A) on (a,b ), and {y1,y 2}is a fundamental set of solutions
on (A) on (a,b ).
(b) Conclude from (a) that if y2 = uy1 where u′= Ke−P (x)
y2
1 (x) , then {y1,y 2}is a fundamental
set of solutions of (A) on (a,b ).
In Exercises 10– 23 use the method suggested by Exercise 9 to find a second solution y2 that isn’t a",Elementary Differential Equations with Boundary Value Problems.pdf
"set of solutions of (A) on (a,b ).
In Exercises 10– 23 use the method suggested by Exercise 9 to find a second solution y2 that isn’t a
constant multiple of the solution y1. Choose K conveniently to simplify y2.
10. y′′− 2y′− 3y= 0 ; y1 = e3x
11. y′′− 6y′+ 9 y= 0 ; y1 = e3x
12. y′′− 2ay′+ a2y= 0 (a= constant); y1 = eax
13. x2y′′+ xy′− y= 0 ; y1 = x
14. x2y′′− xy′+ y= 0 ; y1 = x
15. x2y′′− (2a− 1)xy′+ a2y= 0 (a= nonzero constant); x> 0; y1 = xa
16. 4x2y′′− 4xy′+ (3 − 16x2)y= 0 ; y1 = x1/ 2e2x
17. (x− 1)y′′− xy′+ y= 0 ; y1 = ex
18. x2y′′− 2xy′+ ( x2 + 2) y= 0 ; y1 = xcos x
19. 4x2(sin x)y′′− 4x(xcos x+ sin x)y′+ (2 xcos x+ 3 sin x)y= 0 ; y1 = x1/ 2
20. (3x− 1)y′′− (3x+ 2) y′− (6x− 8)y = 0 ; y1 = e2x
21. (x2 − 4)y′′+ 4 xy′+ 2 y= 0 ; y1 = 1
x− 2
22. (2x+ 1) xy′′− 2(2x2 − 1)y′− 4(x+ 1) y= 0 ; y1 = 1
x
23. (x2 − 2x)y′′+ (2 − x2)y′+ (2 x− 2)y= 0 ; y1 = ex
24. Suppose pand qare continuous on an open interval (a,b ) and let x0 be in (a,b ). Use Theorem 5.1.1",Elementary Differential Equations with Boundary Value Problems.pdf
"24. Suppose pand qare continuous on an open interval (a,b ) and let x0 be in (a,b ). Use Theorem 5.1.1
to show that the only solution of the initial value problem
y′′+ p(x)y′+ q(x)y= 0 , y (x0) = 0 , y ′(x0) = 0
on (a,b ) is the trivial solution y ≡ 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"206 Chapter 5 Linear Second Order Equations
25. Suppose P0, P1, and P2 are continuous on (a,b ) and let x0 be in (a,b ). Show that if either of the
following statements is true then P0(x) = 0 for some xin (a,b ).
(a) The initial value problem
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 , y (x0 ) = k0, y ′(x0) = k1
has more than one solution on (a,b ).
(b) The initial value problem
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 , y (x0) = 0 , y ′(x0) = 0
has a nontrivial solution on (a,b ).
26. Suppose pand qare continuous on (a,b ) and y1 and y2 are solutions of
y′′+ p(x)y′+ q(x)y= 0 (A)
on (a,b ). Let
z1 = αy1 + βy2 and z2 = γy1 + δy2,
where α, β, γ, and δ are constants. Show that if {z1,z 2}is a fundamental set of solutions of (A)
on (a,b ) then so is {y1,y 2}.
27. Suppose pand qare continuous on (a,b ) and {y1,y 2}is a fundamental set of solutions of
y′′+ p(x)y′+ q(x)y= 0 (A)
on (a,b ). Let
z1 = αy1 + βy2 and z2 = γy1 + δy2,",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′+ p(x)y′+ q(x)y= 0 (A)
on (a,b ). Let
z1 = αy1 + βy2 and z2 = γy1 + δy2,
where α,β,γ , and δ are constants. Show that {z1,z 2}is a fundamental set of solutions of (A) on
(a,b ) if and only if αγ − βδ ̸= 0 .
28. Suppose y1 is differentiable on an interval (a,b ) and y2 = ky1, where kis a constant. Show that
the Wronskian of {y1,y 2}is identically zero on (a,b ).
29. Let
y1 = x3 and y2 =
{ x3, x ≥ 0,
−x3, x< 0.
(a) Show that the Wronskian of {y1,y 2}is defined and identically zero on (−∞ , ∞ ).
(b) Suppose a< 0 <b. Show that {y1,y 2}is linearly independent on (a,b ).
(c) Use Exercise 25(b) to show that these results don’t contradict Theorem 5.1.5, because neither
y1 nor y2 can be a solution of an equation
y′′+ p(x)y′+ q(x)y= 0
on (a,b ) if pand qare continuous on (a,b ).
30. Suppose pand qare continuous on (a,b ) and {y1,y 2}is a set of solutions of
y′′+ p(x)y′+ q(x)y= 0
on (a,b ) such that either y1(x0) = y2(x0) = 0 or y′
1(x0) = y′
2(x0) = 0 for some x0 in (a,b ).",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′+ p(x)y′+ q(x)y= 0
on (a,b ) such that either y1(x0) = y2(x0) = 0 or y′
1(x0) = y′
2(x0) = 0 for some x0 in (a,b ).
Show that {y1,y 2}is linearly dependent on (a,b ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 207
31. Suppose pand qare continuous on (a,b ) and {y1,y 2}is a fundamental set of solutions of
y′′+ p(x)y′+ q(x)y= 0
on (a,b ). Show that if y1(x1) = y1(x2) = 0 , where a<x 1 <x2 <b, then y2(x) = 0 for some x
in (x1,x 2). H IN T : Show that if y2 has no zeros in (x1,x 2), then y1/y 2 is either strictly increasing
or strictly decreasing on (x1,x 2), and deduce a contradiction.
32. Suppose pand qare continuous on (a,b ) and every solution of
y′′+ p(x)y′+ q(x)y= 0 (A)
on (a,b ) can be written as a linear combination of the twice different iable functions {y1,y 2}. Use
Theorem 5.1.1 to show that y1 and y2 are themselves solutions of (A) on (a,b ).
33. Suppose p1, p2, q1, and q2 are continuous on (a,b ) and the equations
y′′+ p1(x)y′+ q1(x)y= 0 and y′′+ p2(x)y′+ q2(x)y= 0
have the same solutions on (a,b ). Show that p1 = p2 and q1 = q2 on (a,b ). H IN T : Use Abel’s
formula.",Elementary Differential Equations with Boundary Value Problems.pdf
"have the same solutions on (a,b ). Show that p1 = p2 and q1 = q2 on (a,b ). H IN T : Use Abel’s
formula.
34. (For this exercise you have to know about 3 × 3 determinants.) Show that if y1 and y2 are twice
continuously differentiable on (a,b ) and the Wronskian W of {y1,y 2}has no zeros in (a,b ) then
the equation
1
W
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y y 1 y2
y′ y′
1 y′
2
y′′ y′′
1 y′′
2
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 0
can be written as
y′′+ p(x)y′+ q(x)y = 0 , (A)
where pand q are continuous on (a,b ) and {y1,y 2}is a fundamental set of solutions of (A) on
(a,b ). H IN T : Expand the determinant by cofactors of its first column.
35. Use the method suggested by Exercise
34 to find a linear homogeneous equation for which the
given functions form a fundamental set of solutions on some i nterval.
(a) ex cos 2 x, e x sin 2x (b) x, e 2x
(c) x, x ln x (d) cos(ln x), sin(ln x)
(e) cosh x, sinh x (f) x2 − 1, x 2 + 1
36. Suppose pand qare continuous on (a,b ) and {y1,y 2}is a fundamental set of solutions of",Elementary Differential Equations with Boundary Value Problems.pdf
"(e) cosh x, sinh x (f) x2 − 1, x 2 + 1
36. Suppose pand qare continuous on (a,b ) and {y1,y 2}is a fundamental set of solutions of
y′′+ p(x)y′+ q(x)y= 0 (A)
on (a,b ). Show that if yis a solution of (A) on (a,b ), there’s exactly one way to choose c1 and c2
so that y = c1y1 + c2y2 on (a,b ).
37. Suppose pand qare continuous on (a,b ) and x0 is in (a,b ). Let y1 and y2 be the solutions of
y′′+ p(x)y′+ q(x)y= 0 (A)
such that
y1(x0) = 1 , y ′
1(x0) = 0 and y2(x0) = 0 , y′
2(x0) = 1 .
(Theorem 5.1.1 implies that each of these initial value problems has a uniqu e solution on (a,b ).)",Elementary Differential Equations with Boundary Value Problems.pdf
"208 Chapter 5 Linear Second Order Equations
(a) Show that {y1,y 2}is linearly independent on (a,b ).
(b) Show that an arbitrary solution yof (A) on (a,b ) can be written as y = y(x0 )y1 + y′(x0)y2.
(c) Express the solution of the initial value problem
y′′+ p(x)y′+ q(x)y= 0 , y (x0) = k0, y ′(x0) = k1
as a linear combination of y1 and y2.
38. Find solutions y1 and y2 of the equation y′′= 0 that satisfy the initial conditions
y1(x0) = 1 , y ′
1(x0) = 0 and y2(x0) = 0 , y ′
2(x0) = 1 .
Then use Exercise 37 (c) to write the solution of the initial value problem
y′′= 0 , y (0) = k0, y ′(0) = k1
as a linear combination of y1 and y2.
39. Let x0 be an arbitrary real number. Given (Example 5.1.1) that ex and e−x are solutions of y′′−y =
0, find solutions y1 and y2 of y′′− y = 0 such that
y1(x0) = 1 , y ′
1(x0) = 0 and y2(x0) = 0 , y′
2(x0) = 1 .
Then use Exercise 37 (c) to write the solution of the initial value problem
y′′− y= 0 , y (x0) = k0, y ′(x0) = k1
as a linear combination of y1 and y2.",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′− y= 0 , y (x0) = k0, y ′(x0) = k1
as a linear combination of y1 and y2.
40. Let x0 be an arbitrary real number. Given (Example 5.1.2) that cos ωx and sin ωx are solutions of
y′′+ ω2y = 0 , find solutions of y′′+ ω2y= 0 such that
y1(x0) = 1 , y ′
1(x0) = 0 and y2(x0) = 0 , y′
2(x0) = 1 .
Then use Exercise 37 (c) to write the solution of the initial value problem
y′′+ ω2y= 0 , y (x0) = k0, y ′(x0) = k1
as a linear combination of y1 and y2. Use the identities
cos(A+ B) = cos Acos B− sin Asin B
sin(A+ B) = sin Acos B+ cos Asin B
to simplify your expressions for y1, y2, and y.
41. Recall from Exercise 4 that 1/ (x− 1) and 1/ (x+ 1) are solutions of
(x2 − 1)y′′+ 4 xy′+ 2 y= 0 (A)
on (−1, 1). Find solutions of (A) such that
y1(0) = 1 , y ′
1(0) = 0 and y2(0) = 0 , y′
2(0) = 1 .
Then use Exercise 37 (c) to write the solution of initial value problem
(x2 − 1)y′′+ 4 xy′+ 2 y= 0 , y (0) = k0, y ′(0) = k1
as a linear combination of y1 and y2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.1 Homogeneous Linear Equations 209
42. (a) V erify that y1 = x2 and y2 = x3 satisfy
x2y′′− 4xy′+ 6 y= 0 (A)
on (−∞ , ∞ ) and that {y1,y 2}is a fundamental set of solutions of (A) on (−∞ , 0) and
(0, ∞ ).
(b) Let a1, a2, b1, and b2 be constants. Show that
y =
{ a1x2 + a2x3, x ≥ 0,
b1x2 + b2x3, x< 0
is a solution of (A) on (−∞ , ∞ ) if and only if a1 = b1. From this, justify the statement that
yis a solution of (A) on (−∞ , ∞ ) if and only if
y =
{ c1x2 + c2x3, x ≥ 0,
c1x2 + c3x3, x< 0,
where c1, c2, and c3 are arbitrary constants.
(c) For what values of k0 and k1 does the initial value problem
x2y′′− 4xy′+ 6 y= 0 , y (0) = k0, y ′(0) = k1
have a solution? What are the solutions?
(d) Show that if x0 ̸= 0 and k0,k 1 are arbitrary constants, the initial value problem
x2y′′− 4xy′+ 6 y = 0 , y (x0) = k0, y ′(x0) = k1 (B)
has infinitely many solutions on (−∞ , ∞ ). On what interval does (B) have a unique solution?
43. (a) V erify that y1 = xand y2 = x2 satisfy
x2y′′− 2xy′+ 2 y= 0 (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"43. (a) V erify that y1 = xand y2 = x2 satisfy
x2y′′− 2xy′+ 2 y= 0 (A)
on (−∞ , ∞ ) and that {y1,y 2}is a fundamental set of solutions of (A) on (−∞ , 0) and
(0, ∞ ).
(b) Let a1, a2, b1, and b2 be constants. Show that
y=
{ a1x+ a2x2, x ≥ 0,
b1x+ b2x2, x< 0
is a solution of (A) on (−∞ , ∞ ) if and only if a1 = b1 and a2 = b2. From this, justify the
statement that the general solution of (A) on (−∞ , ∞ ) is y = c1x+ c2x2, where c1 and c2
are arbitrary constants.
(c) For what values of k0 and k1 does the initial value problem
x2y′′− 2xy′+ 2 y= 0 , y (0) = k0, y ′(0) = k1
have a solution? What are the solutions?
(d) Show that if x0 ̸= 0 and k0,k 1 are arbitrary constants then the initial value problem
x2y′′− 2xy′+ 2 y = 0 , y (x0) = k0, y ′(x0) = k1
has a unique solution on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"210 Chapter 5 Linear Second Order Equations
44. (a) V erify that y1 = x3 and y2 = x4 satisfy
x2y′′− 6xy′+ 12 y= 0 (A)
on (−∞ , ∞ ), and that {y1,y 2}is a fundamental set of solutions of (A) on (−∞ , 0) and
(0, ∞ ).
(b) Show that yis a solution of (A) on (−∞ , ∞ ) if and only if
y =
{ a1x3 + a2x4, x ≥ 0,
b1x3 + b2x4, x< 0,
where a1, a2, b1, and b2 are arbitrary constants.
(c) For what values of k0 and k1 does the initial value problem
x2y′′− 6xy′+ 12 y= 0 , y (0) = k0, y ′(0) = k1
have a solution? What are the solutions?
(d) Show that if x0 ̸= 0 and k0,k 1 are arbitrary constants then the initial value problem
x2y′′− 6xy′+ 12 y= 0 , y (x0 ) = k0, y ′(x0) = k1 (B)
has infinitely many solutions on (−∞ , ∞ ). On what interval does (B) have a unique solution?
5.2 CONST ANT COEFFICIENT HOMOGENEOUS EQUA TIONS
If a,b , and care real constants and a̸= 0 , then
ay′′+ by′+ cy= F(x)
is said to be a constant coefficient equation . In this section we consider the homogeneous constant coef-",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy= F(x)
is said to be a constant coefficient equation . In this section we consider the homogeneous constant coef-
ficient equation
ay′′+ by′+ cy= 0 . (5.2.1)
As we’ll see, all solutions of ( 5.2.1) are defined on (−∞ , ∞ ). This being the case, we’ll omit references
to the interval on which solutions are defined, or on which a gi ven set of solutions is a fundamental set,
etc., since the interval will always be (−∞ , ∞ ).
The key to solving ( 5.2.1) is that if y = erx where r is a constant then the left side of ( 5.2.1) is a
multiple of erx ; thus, if y = erx then y′= rerx and y′′= r2erx , so
ay′′+ by′+ cy= ar2erx + brerx + cerx = ( ar2 + br+ c)erx . (5.2.2)
The quadratic polynomial
p(r) = ar2 + br+ c
is the characteristic polynomial of ( 5.2.1), and p(r) = 0 is the characteristic equation . From ( 5.2.2) we
can see that y = erx is a solution of ( 5.2.1) if and only if p(r) = 0 .
The roots of the characteristic equation are given by the qua dratic formula
r= −b±
√
b2 − 4ac",Elementary Differential Equations with Boundary Value Problems.pdf
"The roots of the characteristic equation are given by the qua dratic formula
r= −b±
√
b2 − 4ac
2a . (5.2.3)
W e consider three cases:",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.2 Constant Coefficient Homogeneous Equations 211
CA SE 1. b2 − 4ac> 0, so the characteristic equation has two distinct real roots .
CA SE 2. b2 − 4ac= 0 , so the characteristic equation has a repeated real root.
CA SE 3. b2 − 4ac< 0, so the characteristic equation has complex roots.
In each case we’ll start with an example.
Case 1: Distinct Real Roots
Example 5.2.1
(a) Find the general solution of
y′′+ 6 y′+ 5 y = 0 . (5.2.4)
(b) Solve the initial value problem
y′′+ 6 y′+ 5 y = 0 , y (0) = 3 , y′(0) = −1. (5.2.5)
SO LU TIO N (a) The characteristic polynomial of ( 5.2.4) is
p(r) = r2 + 6 r+ 5 = ( r+ 1)( r+ 5) .
Since p(−1) = p(−5) = 0 , y1 = e−x and y2 = e−5x are solutions of ( 5.2.4). Since y2/y 1 = e−4x is
nonconstant, 5.1.6 implies that the general solution of ( 5.2.4) is
y= c1e−x + c2e−5x. (5.2.6)
SO LU TIO N (b) W e must determine c1 and c2 in ( 5.2.6) so that ysatisfies the initial conditions in ( 5.2.5).
Differentiating ( 5.2.6) yields
y′= −c1e−x − 5c2e−5x. (5.2.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"Differentiating ( 5.2.6) yields
y′= −c1e−x − 5c2e−5x. (5.2.7)
Imposing the initial conditions y(0) = 3 , y′(0) = −1 in ( 5.2.6) and ( 5.2.7) yields
c1 + c2 = 3
−c1 − 5c2 = −1.
The solution of this system is c1 = 7 / 2,c 2 = −1/ 2. Therefore the solution of ( 5.2.5) is
y= 7
2 e−x − 1
2 e−5x.
Figure 5.2.1 is a graph of this solution.
If the characteristic equation has arbitrary distinct real roots r1 and r2, then y1 = er1 x and y2 = er2 x
are solutions of ay′′+ by′+ cy= 0 . Since y2/y 1 = e(r2 −r1)x is nonconstant, Theorem 5.1.6 implies that
{y1,y 2}is a fundamental set of solutions of ay′′+ by′+ cy= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"212 Chapter 5 Linear Second Order Equations
1
2
3
4321 5
 x
 y
Figure 5.2.1 y = 7
2 e−x − 1
2 e−5x
Case 2: A Repeated Real Root
Example 5.2.2
(a) Find the general solution of
y′′+ 6 y′+ 9 y = 0 . (5.2.8)
(b) Solve the initial value problem
y′′+ 6 y′+ 9 y = 0 , y (0) = 3 , y′(0) = −1. (5.2.9)
SO LU TIO N (a) The characteristic polynomial of ( 5.2.8) is
p(r) = r2 + 6 r+ 9 = ( r+ 3) 2,
so the characteristic equation has the repeated real root r1 = −3. Therefore y1 = e−3x is a solution
of ( 5.2.8). Since the characteristic equation has no other roots, ( 5.2.8) has no other solutions of the
form erx . W e look for solutions of the form y = uy1 = ue−3x, where uis a function that we’ll now
determine. (This should remind you of the method of variatio n of parameters used in Section 2.1 to
solve the nonhomogeneous equation y′+ p(x)y = f(x), given a solution y1 of the complementary
equation y′+ p(x)y = 0 . It’s also a special case of a method called reduction of order that we’ll study",Elementary Differential Equations with Boundary Value Problems.pdf
"equation y′+ p(x)y = 0 . It’s also a special case of a method called reduction of order that we’ll study
in Section 5.6. For other ways to obtain a second solution of ( 5.2.8) that’s not a multiple of e−3x, see
Exercises 5.1. 9, 5.1. 12, and 33.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.2 Constant Coefficient Homogeneous Equations 213
If y= ue−3x, then
y′= u′e−3x − 3ue−3x and y′′= u′′e−3x − 6u′e−3x + 9 ue−3x,
so
y′′+ 6 y′+ 9 y = e−3x [(u′′− 6u′+ 9 u) + 6( u′− 3u) + 9 u]
= e−3x [u′′− (6 − 6)u′+ (9 − 18 + 9) u] = u′′e−3x.
Therefore y = ue−3x is a solution of ( 5.2.8) if and only if u′′= 0 , which is equivalent to u= c1 + c2x,
where c1 and c2 are constants. Therefore any function of the form
y= e−3x(c1 + c2x) (5.2.10)
is a solution of ( 5.2.8). Letting c1 = 1 and c2 = 0 yields the solution y1 = e−3x that we already knew .
Letting c1 = 0 and c2 = 1 yields the second solution y2 = xe−3x. Since y2/y 1 = xis nonconstant, 5.1.6
implies that {y1,y 2}is fundamental set of solutions of ( 5.2.8), and ( 5.2.10) is the general solution.
SO LU TIO N (b) Differentiating ( 5.2.10) yields
y′= −3e−3x(c1 + c2x) + c2e−3x. (5.2.11)
Imposing the initial conditions y(0) = 3 , y′(0) = −1 in ( 5.2.10) and ( 5.2.11) yields c1 = 3 and −3c1 +",Elementary Differential Equations with Boundary Value Problems.pdf
"y′= −3e−3x(c1 + c2x) + c2e−3x. (5.2.11)
Imposing the initial conditions y(0) = 3 , y′(0) = −1 in ( 5.2.10) and ( 5.2.11) yields c1 = 3 and −3c1 +
c2 = −1, so c2 = 8 . Therefore the solution of ( 5.2.9) is
y= e−3x(3 + 8 x).
Figure 5.2.2 is a graph of this solution.
1
2
3
1 2 3
 x
 y
Figure 5.2.2 y = e−3x(3 + 8 x)",Elementary Differential Equations with Boundary Value Problems.pdf
"214 Chapter 5 Linear Second Order Equations
If the characteristic equation of ay′′+ by′+ cy = 0 has an arbitrary repeated root r1, the characteristic
polynomial must be
p(r) = a(r− r1)2 = a(r2 − 2r1r+ r2
1).
Therefore
ar2 + br+ c= ar2 − (2ar1)r+ ar2
1,
which implies that b = −2ar1 and c = ar2
1. Therefore ay′′+ by′+ cy = 0 can be written as a(y′′−
2r1y′+ r2
1y) = 0 . Since a̸= 0 this equation has the same solutions as
y′′− 2r1y′+ r2
1y= 0 . (5.2.12)
Since p(r1) = 0 , t y1 = er1 x is a solution of ay′′+ by′+ cy = 0 , and therefore of ( 5.2.12). Proceeding
as in Example 5.2.2, we look for other solutions of ( 5.2.12) of the form y = uer1x; then
y′= u′er1 x + ruer1x and y′′= u′′er1 x + 2 r1u′er1 x + r2
1 uer1 x,
so
y′′− 2r1y′+ r2
1y = erx [
(u′′+ 2 r1u′+ r2
1 u) − 2r1(u′+ r1u) + r2
1u
]
= er1 x [
u′′+ (2 r1 − 2r1)u′+ ( r2
1 − 2r2
1 + r2
1)u
]
= u′′er1 x.
Therefore y = uer1 x is a solution of ( 5.2.12) if and only if u′′= 0 , which is equivalent to u= c1 + c2x,",Elementary Differential Equations with Boundary Value Problems.pdf
"1 − 2r2
1 + r2
1)u
]
= u′′er1 x.
Therefore y = uer1 x is a solution of ( 5.2.12) if and only if u′′= 0 , which is equivalent to u= c1 + c2x,
where c1 and c2 are constants. Hence, any function of the form
y= er1 x(c1 + c2x) (5.2.13)
is a solution of ( 5.2.12). Letting c1 = 1 and c2 = 0 here yields the solution y1 = er1 x that we already
knew . Letting c1 = 0 and c2 = 1 yields the second solution y2 = xer1 x. Since y2/y 1 = xis noncon-
stant, 5.1.6 implies that {y1,y 2}is a fundamental set of solutions of ( 5.2.12), and ( 5.2.13) is the general
solution.
Case 3: Complex Conjugate Roots
Example 5.2.3
(a) Find the general solution of
y′′+ 4 y′+ 13 y= 0 . (5.2.14)
(b) Solve the initial value problem
y′′+ 4 y′+ 13 y= 0 , y (0) = 2 , y′(0) = −3. (5.2.15)
SO LU TIO N (a) The characteristic polynomial of ( 5.2.14) is
p(r) = r2 + 4 r+ 13 = r2 + 4 r+ 4 + 9 = ( r+ 2) 2 + 9 .
The roots of the characteristic equation are r1 = −2 + 3 iand r2 = −2 − 3i. By analogy with Case 1, it’s",Elementary Differential Equations with Boundary Value Problems.pdf
"The roots of the characteristic equation are r1 = −2 + 3 iand r2 = −2 − 3i. By analogy with Case 1, it’s
reasonable to expect that e(−2+3i)x and e(−2−3i)x are solutions of ( 5.2.14). This is true (see Exercise 34);
however, there are difficulties here, since you are probably not familiar with exponential functions with
complex arguments, and even if you are, it’s inconvenient to work with them, since they are complex–
valued. W e’ll take a simpler approach, which we motivate as f ollows: the exponential notation suggests
that
e(−2+3i)x = e−2xe3ix and e(−2−3i)x = e−2xe−3ix ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.2 Constant Coefficient Homogeneous Equations 215
so even though we haven’t defined e3ix and e−3ix, it’s reasonable to expect that every linear combination
of e(−2+3i)x and e(−2−3i)x can be written as y = ue−2x, where udepends upon x. T o determine u, we
note that if y= ue−2x then
y′= u′e−2x − 2ue−2x and y′′= u′′e−2x − 4u′e−2x + 4 ue−2x,
so
y′′+ 4 y′+ 13 y = e−2x [(u′′− 4u′+ 4 u) + 4( u′− 2u) + 13 u]
= e−2x [u′′− (4 − 4)u′+ (4 − 8 + 13) u] = e−2x(u′′+ 9 u).
Therefore y = ue−2x is a solution of ( 5.2.14) if and only if
u′′+ 9 u= 0 .
From Example 5.1.2, the general solution of this equation is
u= c1 cos 3 x+ c2 sin 3 x.
Therefore any function of the form
y= e−2x(c1 cos 3 x+ c2 sin 3x) (5.2.16)
is a solution of ( 5.2.14). Letting c1 = 1 and c2 = 0 yields the solution y1 = e−2x cos 3x. Letting c1 = 0
and c2 = 1 yields the second solution y2 = e−2x sin 3 x. Since y2/y 1 = tan 3 xis nonconstant, 5.1.6",Elementary Differential Equations with Boundary Value Problems.pdf
"and c2 = 1 yields the second solution y2 = e−2x sin 3 x. Since y2/y 1 = tan 3 xis nonconstant, 5.1.6
implies that {y1,y 2}is a fundamental set of solutions of ( 5.2.14), and ( 5.2.16) is the general solution.
SO LU TIO N (b) Imposing the condition y(0) = 2 in ( 5.2.16) shows that c1 = 2 . Differentiating ( 5.2.16)
yields
y′= −2e−2x(c1 cos 3 x+ c2 sin 3 x) + 3 e−2x(−c1 sin 3 x+ c2 cos 3 x),
and imposing the initial condition y′(0) = −3 here yields −3 = −2c1 + 3 c2 = −4 + 3 c2, so c2 = 1 / 3.
Therefore the solution of ( 5.2.15) is
y= e−2x(2 cos 3 x+ 1
3 sin 3 x).
Figure 5.2.3 is a graph of this function.
Now suppose the characteristic equation of ay′′+ by′+ cy = 0 has arbitrary complex roots; thus,
b2 − 4ac< 0 and, from ( 5.2.3), the roots are
r1 = −b+ i
√
4ac− b2
2a , r 2 = −b− i
√
4ac− b2
2a ,
which we rewrite as
r1 = λ + iω, r 2 = λ − iω, (5.2.17)
with
λ = − b
2a, ω =
√
4ac− b2
2a .
Don’t memorize these formulas. Just remember that r1 and r2 are of the form ( 5.2.17), where λ is an",Elementary Differential Equations with Boundary Value Problems.pdf
"with
λ = − b
2a, ω =
√
4ac− b2
2a .
Don’t memorize these formulas. Just remember that r1 and r2 are of the form ( 5.2.17), where λ is an
arbitrary real number and ω is positive; λ and ω are the real and imaginary parts , respectively , of r1.
Similarly , λ and −ω are the real and imaginary parts of r2. W e say that r1 and r2 are complex conjugates ,",Elementary Differential Equations with Boundary Value Problems.pdf
"216 Chapter 5 Linear Second Order Equations
1
2
1 2
 x
 y
Figure 5.2.3 y= e−2x(2 cos 3 x+ 1
3 sin 3 x)
which means that they have the same real part and their imagin ary parts have the same absolute values,
but opposite signs.
As in Example 5.2.3, it’s reasonable to to expect that the solutions of ay′′+ by′+ cy = 0 are linear
combinations of e(λ +iω )x and e(λ −iω )x. Again, the exponential notation suggests that
e(λ +iω )x = eλx eiωx and e(λ −iω )x = eλx e−iωx ,
so even though we haven’t defined eiωx and e−iωx , it’s reasonable to expect that every linear combination
of e(λ +iω )x and e(λ −iω )x can be written as y = ueλx , where udepends upon x. T o determine uwe first
observe that since r1 = λ + iω and r2 = λ − iω are the roots of the characteristic equation, pmust be of
the form
p(r) = a(r− r1)(r− r2)
= a(r− λ − iω)(r− λ + iω)
= a
[
(r− λ)2 + ω2]
= a(r2 − 2λr + λ2 + ω2).
Therefore ay′′+ by′+ cy= 0 can be written as
a[ y′′− 2λy′+ ( λ2 + ω2)y] = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"= a(r− λ − iω)(r− λ + iω)
= a
[
(r− λ)2 + ω2]
= a(r2 − 2λr + λ2 + ω2).
Therefore ay′′+ by′+ cy= 0 can be written as
a[ y′′− 2λy′+ ( λ2 + ω2)y] = 0 .
Since a̸= 0 this equation has the same solutions as
y′′− 2λy′+ ( λ2 + ω2)y= 0 . (5.2.18)
T o determine uwe note that if y= ueλx then
y′= u′eλx + λueλx and y′′= u′′eλx + 2 λu′eλx + λ2ueλx .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.2 Constant Coefficient Homogeneous Equations 217
Substituting these expressions into ( 5.2.18) and dropping the common factor eλx yields
(u′′+ 2 λu′+ λ2u) − 2λ(u′+ λu) + ( λ2 + ω2)u= 0 ,
which simplifies to
u′′+ ω2u= 0 .
From Example 5.1.2, the general solution of this equation is
u= c1 cos ωx + c2 sin ωx.
Therefore any function of the form
y= eλx (c1 cos ωx + c2 sin ωx) (5.2.19)
is a solution of ( 5.2.18). Letting c1 = 1 and c2 = 0 here yields the solution y1 = eλx cos ωx. Letting
c1 = 0 and c2 = 1 yields a second solution y2 = eλx sin ωx. Since y2/y 1 = tan ωx is nonconstant,
so Theorem 5.1.6 implies that {y1,y 2}is a fundamental set of solutions of ( 5.2.18), and ( 5.2.19) is the
general solution.
Summary
The next theorem summarizes the results of this section.
Theorem 5.2.1
Let p(r) = ar2 + br+ cbe the characteristic polynomial of
ay′′+ by′+ cy= 0 . (5.2.20)
Then:
(a) If p(r) = 0 has distinct real roots r1 and r2, then the general solution of (5.2.20) is",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy= 0 . (5.2.20)
Then:
(a) If p(r) = 0 has distinct real roots r1 and r2, then the general solution of (5.2.20) is
y= c1er1 x + c2er2 x.
(b) If p(r) = 0 has a repeated root r1, then the general solution of (5.2.20) is
y= er1 x(c1 + c2x).
(c) If p(r) = 0 has complex conjugate roots r1 = λ + iω and r2 = λ − iω (where ω > 0), then the
general solution of (5.2.20) is
y= eλx (c1 cos ωx + c2 sin ωx).
5.2 Exercises
In Exercises 1– 12 find the general solution.
1. y′′+ 5 y′− 6y= 0 2. y′′− 4y′+ 5 y= 0
3. y′′+ 8 y′+ 7 y= 0 4. y′′− 4y′+ 4 y= 0
5. y′′+ 2 y′+ 10 y= 0 6. y′′+ 6 y′+ 10 y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"218 Chapter 5 Linear Second Order Equations
7. y′′− 8y′+ 16 y= 0 8. y′′+ y′= 0
9. y′′− 2y′+ 3 y= 0 10. y′′+ 6 y′+ 13 y= 0
11. 4y′′+ 4 y′+ 10 y= 0 12. 10y′′− 3y′− y = 0
In Exercises 13– 17 solve the initial value problem.
13. y′′+ 14 y′+ 50 y= 0 , y (0) = 2 , y ′(0) = −17
14. 6y′′− y′− y = 0 , y (0) = 10 , y ′(0) = 0
15. 6y′′+ y′− y = 0 , y (0) = −1, y ′(0) = 3
16. 4y′′− 4y′− 3y= 0 , y (0) = 13
12 , y ′(0) = 23
24
17. 4y′′− 12y′+ 9 y= 0 , y (0) = 3 , y ′(0) = 5
2
In Exercises 18– 21 solve the initial value problem and graph the solution.
18. C/G y′′+ 7 y′+ 12 y= 0 , y (0) = −1, y ′(0) = 0
19. C/G y′′− 6y′+ 9 y= 0 , y (0) = 0 , y ′(0) = 2
20. C/G 36y′′− 12y′+ y = 0 , y (0) = 3 , y ′(0) = 5
2
21. C/G y′′+ 4 y′+ 10 y= 0 , y (0) = 3 , y ′(0) = −2
22. (a) Suppose yis a solution of the constant coefficient homogeneous equati on
ay′′+ by′+ cy = 0 . (A)
Let z(x) = y(x− x0), where x0 is an arbitrary real number. Show that
az′′+ bz′+ cz = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy = 0 . (A)
Let z(x) = y(x− x0), where x0 is an arbitrary real number. Show that
az′′+ bz′+ cz = 0 .
(b) Let z1(x) = y1(x− x0) and z2(x) = y2(x− x0), where {y1,y 2}is a fundamental set of
solutions of (A). Show that {z1,z 2}is also a fundamental set of solutions of (A).
(c) The statement of Theorem 5.2.1 is convenient for solving an initial value problem
ay′′+ by′+ cy= 0 , y (0) = k0, y ′(0) = k1,
where the initial conditions are imposed at x0 = 0 . However, if the initial value problem is
ay′′+ by′+ cy= 0 , y (x0 ) = k0, y ′(x0) = k1, (B)
where x0 ̸= 0 , then determining the constants in
y= c1er1 x + c2er2 x, y = er1 x(c1 + c2x), or y = eλx (c1 cos ωx + c2 sin ωx)
(whichever is applicable) is more complicated. Use (b) to restate Theorem 5.2.1 in a form
more convenient for solving (B).
In Exercises 23– 28 use a method suggested by Exercise 22 to solve the initial value problem.
23. y′′+ 3 y′+ 2 y= 0 , y (1) = −1, y ′(1) = 4",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.2 Constant Coefficient Homogeneous Equations 219
24. y′′− 6y′− 7y= 0 , y (2) = −1
3 , y ′(2) = −5
25. y′′− 14y′+ 49 y= 0 , y (1) = 2 , y ′(1) = 11
26. 9y′′+ 6 y′+ y= 0 , y (2) = 2 , y ′(2) = −14
3
27. 9y′′+ 4 y= 0 , y (π/ 4) = 2 , y ′(π/ 4) = −2
28. y′′+ 3 y= 0 , y (π/ 3) = 2 , y ′(π/ 3) = −1
29. Prove: If the characteristic equation of
ay′′+ by′+ cy = 0 (A)
has a repeated negative root or two roots with negative real p arts, then every solution of (A) ap-
proaches zero as x→ ∞ .
30. Suppose the characteristic polynomial of ay′′+ by′+ cy= 0 has distinct real roots r1 and r2. Use
a method suggested by Exercise 22 to find a formula for the solution of
ay′′+ by′+ cy= 0 , y (x0) = k0, y ′(x0) = k1.
31. Suppose the characteristic polynomial of ay′′+ by′+ cy = 0 has a repeated real root r1. Use a
method suggested by Exercise 22 to find a formula for the solution of
ay′′+ by′+ cy= 0 , y (x0) = k0, y ′(x0) = k1.",Elementary Differential Equations with Boundary Value Problems.pdf
"method suggested by Exercise 22 to find a formula for the solution of
ay′′+ by′+ cy= 0 , y (x0) = k0, y ′(x0) = k1.
32. Suppose the characteristic polynomial of ay′′+ by′+ cy= 0 has complex conjugate roots λ ± iω.
Use a method suggested by Exercise 22 to find a formula for the solution of
ay′′+ by′+ cy= 0 , y (x0) = k0, y ′(x0) = k1.
33. Suppose the characteristic equation of
ay′′+ by′+ cy = 0 (A)
has a repeated real root r1. T emporarily , think of erx as a function of two real variables xand r.
(a) Show that
a ∂2
∂2x(erx ) + b ∂
∂x (erx ) + cerx = a(r− r1)2erx . (B)
(b) Differentiate (B) with respect to rto obtain
a ∂
∂r
(∂2
∂2x(erx )
)
+ b ∂
∂r
(∂
∂x (erx )
)
+ c(xerx ) = [2 + ( r− r1)x]a(r− r1)erx . (C)
(c) Reverse the orders of the partial differentiations in the fir st two terms on the left side of (C)
to obtain
a ∂2
∂x2 (xerx ) + b ∂
∂x (xerx ) + c(xerx ) = [2 + ( r− r1)x]a(r− r1)erx . (D)
(d) Set r= r1 in (B) and (D) to see that y1 = er1 x and y2 = xer1 x are solutions of (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"220 Chapter 5 Linear Second Order Equations
34. In calculus you learned that eu, cos u, and sin ucan be represented by the infinite series
eu =
∞∑
n=0
un
n! = 1 + u
1! + u2
2! + u3
3! + · · ·+ un
n! + · · · (A)
cos u=
∞∑
n=0
(−1)n u2n
(2n)! = 1 − u2
2! + u4
4! + · · ·+ ( −1)n u2n
(2n)! + · · ·, (B)
and
sin u=
∞∑
n=0
(−1)n u2n+1
(2n+ 1)! = u− u3
3! + u5
5! + · · ·+ ( −1)n u2n+1
(2n+ 1)! + · · · (C)
for all real values of u. Even though you have previously considered (A) only for rea l values of u,
we can set u= iθ, where θ is real, to obtain
eiθ =
∞∑
n=0
(iθ)n
n! . (D)
Given the proper background in the theory of infinite series w ith complex terms, it can be shown
that the series in (D) converges for all real θ.
(a) Recalling that i2 = −1, write enough terms of the sequence {in}to convince yourself that
the sequence is repetitive:
1,i, −1, −i, 1,i, −1, −i, 1,i, −1, −i, 1,i, −1, −i, · ··.
Use this to group the terms in (D) as
eiθ =
(
1 − θ2
2 + θ4
4 + · · ·
)
+ i
(
θ − θ3
3! + θ5",Elementary Differential Equations with Boundary Value Problems.pdf
"Use this to group the terms in (D) as
eiθ =
(
1 − θ2
2 + θ4
4 + · · ·
)
+ i
(
θ − θ3
3! + θ5
5! + · · ·
)
=
∞∑
n=0
(−1)n θ2n
(2n)! + i
∞∑
n=0
(−1)n θ2n+1
(2n+ 1)! .
By comparing this result with (B) and (C), conclude that
eiθ = cos θ + isin θ. (E)
This is Euler’s identity .
(b) Starting from
eiθ 1 eiθ 2 = (cos θ1 + isin θ1)(cos θ2 + isin θ2),
collect the real part (the terms not multiplied by i) and the imaginary part (the terms multi-
plied by i) on the right, and use the trigonometric identities
cos(θ1 + θ2) = cos θ1 cos θ2 − sin θ1 sin θ2
sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2
to verify that
ei(θ 1+θ 2 ) = eiθ 1 eiθ 2 ,
as you would expect from the use of the exponential notation eiθ .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.3 Nonhomogeneous Linear Equations 221
(c) If α and β are real numbers, define
eα +iβ = eα eiβ = eα (cos β + isin β). (F)
Show that if z1 = α1 + iβ1 and z2 = α2 + iβ2 then
ez1+z2 = ez1 ez2 .
(d) Let a, b, and cbe real numbers, with a ̸= 0 . Let z = u+ iv where uand v are real-valued
functions of x. Then we say that zis a solution of
ay′′+ by′+ cy= 0 (G)
if uand vare both solutions of (G). Use Theorem 5.2.1(c) to verify that if the characteristic
equation of (G) has complex conjugate roots λ ± iω then z1 = e(λ +iω )x and z2 = e(λ −iω )x
are both solutions of (G).
5.3 NONHOMOGENEOUS LINEAR EQUA TIONS
W e’ll now consider the nonhomogeneous linear second order e quation
y′′+ p(x)y′+ q(x)y = f(x), (5.3.1)
where the forcing function f isn’t identically zero. The next theorem, an extension of Th eorem 5.1.1,
gives sufficient conditions for existence and uniqueness of solutions of initial value problems for ( 5.3.1).
W e omit the proof, which is beyond the scope of this book.",Elementary Differential Equations with Boundary Value Problems.pdf
"W e omit the proof, which is beyond the scope of this book.
Theorem 5.3.1
Suppose p, ,q and f are continuous on an open interval (a,b ), let x0 be any point in
(a,b ), and let k0 and k1 be arbitrary real numbers . Then the initial value problem
y′′+ p(x)y′+ q(x)y= f(x), y (x0) = k0, y ′(x0) = k1
has a unique solution on (a,b ).
T o find the general solution of ( 5.3.1) on an interval (a,b ) where p, q, and f are continuous, it’s
necessary to find the general solution of the associated homo geneous equation
y′′+ p(x)y′+ q(x)y= 0 (5.3.2)
on (a,b ). W e call ( 5.3.2) the complementary equation for ( 5.3.1).
The next theorem shows how to find the general solution of ( 5.3.1) if we know one solution yp of
(5.3.1) and a fundamental set of solutions of ( 5.3.2). W e call yp a particular solution of ( 5.3.1); it can be
any solution that we can find, one way or another.
Theorem 5.3.2
Suppose p,q, and f are continuous on (a,b ). Let yp be a particular solution of
y′′+ p(x)y′+ q(x)y= f(x) (5.3.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"Theorem 5.3.2
Suppose p,q, and f are continuous on (a,b ). Let yp be a particular solution of
y′′+ p(x)y′+ q(x)y= f(x) (5.3.3)
on (a,b ), and let {y1,y 2}be a fundamental set of solutions of the complementary equat ion
y′′+ p(x)y′+ q(x)y= 0 (5.3.4)
on (a,b ). Then yis a solution of (5.3.3) on (a,b ) if and only if
y= yp + c1y1 + c2y2, (5.3.5)
where c1 and c2 are constants.",Elementary Differential Equations with Boundary Value Problems.pdf
"222 Chapter 5 Linear Second Order Equations
Proof W e first show that y in ( 5.3.5) is a solution of ( 5.3.3) for any choice of the constants c1 and c2.
Differentiating ( 5.3.5) twice yields
y′= y′
p + c1y′
1 + c2y′
2 and y′′= y′′
p + c1y′′
1 + c2y′′
2 ,
so
y′′+ p(x)y′+ q(x)y = ( y′′
p + c1y′′
1 + c2y′′
2 ) + p(x)(y′
p + c1y′
1 + c2y′
2)
+q(x)(yp + c1y1 + c2y2)
= ( y′′
p + p(x)y′
p + q(x)yp) + c1(y′′
1 + p(x)y′
1 + q(x)y1)
+c2(y′′
2 + p(x)y′
2 + q(x)y2)
= f + c1 ·0 + c2 ·0 = f,
since yp satisfies ( 5.3.3) and y1 and y2 satisfy ( 5.3.4).
Now we’ll show that every solution of ( 5.3.3) has the form ( 5.3.5) for some choice of the constants c1
and c2. Suppose yis a solution of ( 5.3.3). W e’ll show that y− yp is a solution of ( 5.3.4), and therefore of
the form y− yp = c1y1 + c2y2, which implies ( 5.3.5). T o see this, we compute
(y− yp)′′+ p(x)(y− yp )′+ q(x)(y− yp) = ( y′′− y′′
p ) + p(x)(y′− y′
p )
+q(x)(y− yp )
= ( y′′+ p(x)y′+ q(x)y)
−(y′′
p + p(x)y′
p + q(x)yp )
= f(x) − f(x) = 0 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"p ) + p(x)(y′− y′
p )
+q(x)(y− yp )
= ( y′′+ p(x)y′+ q(x)y)
−(y′′
p + p(x)y′
p + q(x)yp )
= f(x) − f(x) = 0 ,
since yand yp both satisfy ( 5.3.3).
W e say that ( 5.3.5) is the general solution of (5.3.3) on (a,b ).
If P0, P1, and F are continuous and P0 has no zeros on (a,b ), then Theorem 5.3.2 implies that the
general solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x) (5.3.6)
on (a,b ) is y = yp + c1y1 + c2y2, where yp is a particular solution of ( 5.3.6) on (a,b ) and {y1,y 2}is a
fundamental set of solutions of
P0(x)y′′+ P1(x)y′+ P2(x)y = 0
on (a,b ). T o see this, we rewrite ( 5.3.6) as
y′′+ P1(x)
P0(x) y′+ P2(x)
P0(x) y = F(x)
P0(x)
and apply Theorem 5.3.2 with p= P1/P 0, q= P2/P 0, and f = F/P 0.
T o avoid awkward wording in examples and exercises, we won’t specify the interval (a,b ) when we ask
for the general solution of a specific linear second order equ ation, or for a fundamental set of solutions of",Elementary Differential Equations with Boundary Value Problems.pdf
"for the general solution of a specific linear second order equ ation, or for a fundamental set of solutions of
a homogeneous linear second order equation. Let’s agree tha t this always means that we want the general
solution (or a fundamental set of solutions, as the case may b e) on every open interval on which p, q, and
f are continuous if the equation is of the form ( 5.3.3), or on which P0, P1, P2, and F are continuous and
P0 has no zeros, if the equation is of the form ( 5.3.6). W e leave it to you to identify these intervals in
specific examples and exercises.
For completeness, we point out that if P0, P1, P2, and F are all continuous on an open interval (a,b ),
but P0 does have a zero in (a,b ), then ( 5.3.6) may fail to have a general solution on (a,b ) in the sense
just defined. Exercises 42– 44 illustrate this point for a homogeneous equation.
In this section we to limit ourselves to applications of Theo rem 5.3.2 where we can guess at the form
of the particular solution.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.3 Nonhomogeneous Linear Equations 223
Example 5.3.1
(a) Find the general solution of
y′′+ y = 1 . (5.3.7)
(b) Solve the initial value problem
y′′+ y = 1 , y (0) = 2 , y ′(0) = 7 . (5.3.8)
SO LU TIO N (a) W e can apply Theorem 5.3.2 with (a,b ) = ( −∞ , ∞ ), since the functions p ≡ 0, q ≡ 1,
and f ≡ 1 in ( 5.3.7) are continuous on (−∞ , ∞ ). By inspection we see that yp ≡ 1 is a particular solution
of ( 5.3.7). Since y1 = cos xand y2 = sin xform a fundamental set of solutions of the complementary
equation y′′+ y = 0 , the general solution of ( 5.3.7) is
y= 1 + c1 cos x+ c2 sin x. (5.3.9)
SO LU TIO N (b) Imposing the initial condition y(0) = 2 in ( 5.3.9) yields 2 = 1 + c1, so c1 = 1 . Differen-
tiating ( 5.3.9) yields
y′= −c1 sin x+ c2 cos x.
Imposing the initial condition y′(0) = 7 here yields c2 = 7 , so the solution of ( 5.3.8) is
y = 1 + cos x+ 7 sin x.
Figure 5.3.1 is a graph of this function.
Example 5.3.2
(a) Find the general solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"y = 1 + cos x+ 7 sin x.
Figure 5.3.1 is a graph of this function.
Example 5.3.2
(a) Find the general solution of
y′′− 2y′+ y= −3 − x+ x2. (5.3.10)
(b) Solve the initial value problem
y′′− 2y′+ y= −3 − x+ x2, y (0) = −2, y ′(0) = 1 . (5.3.11)
SO LU TIO N (a) The characteristic polynomial of the complementary equati on
y′′− 2y′+ y= 0
is r2 − 2r + 1 = ( r − 1)2, so y1 = ex and y2 = xex form a fundamental set of solutions of the
complementary equation. T o guess a form for a particular sol ution of ( 5.3.10), we note that substituting a
second degree polynomial yp = A+ Bx+ Cx2 into the left side of ( 5.3.10) will produce another second
degree polynomial with coefficients that depend upon A, B, and C. The trick is to choose A, B, and C
so the polynomials on the two sides of ( 5.3.10) have the same coefficients; thus, if
yp = A+ Bx+ Cx2 then y′
p = B+ 2 Cx and y′′
p = 2 C,
so
y′′
p − 2y′
p + yp = 2 C− 2(B+ 2 Cx) + ( A+ Bx+ Cx2)
= (2 C− 2B+ A) + ( −4C+ B)x+ Cx2 = −3 − x+ x2.",Elementary Differential Equations with Boundary Value Problems.pdf
"224 Chapter 5 Linear Second Order Equations
1 2 3 4 5 6
2
4
6
8
− 2
− 4
− 6
− 8
 x
 y
Figure 5.3.1 y = 1 + cos x+ 7 sin x
Equating coefficients of like powers of xon the two sides of the last equality yields
C = 1
B− 4C = −1
A− 2B+ 2 C = −3,
so C = 1 , B = −1 + 4 C= 3 , and A= −3 − 2C+ 2 B= 1 . Therefore yp = 1 + 3 x+ x2 is a particular
solution of ( 5.3.10) and Theorem 5.3.2 implies that
y= 1 + 3 x+ x2 + ex(c1 + c2x) (5.3.12)
is the general solution of ( 5.3.10).
SO LU TIO N (b) Imposing the initial condition y(0) = −2 in ( 5.3.12) yields −2 = 1 + c1, so c1 = −3.
Differentiating ( 5.3.12) yields
y′= 3 + 2 x+ ex(c1 + c2x) + c2ex,
and imposing the initial condition y′(0) = 1 here yields 1 = 3 + c1 + c2, so c2 = 1 . Therefore the
solution of ( 5.3.11) is
y = 1 + 3 x+ x2 − ex (3 − x).
Figure 5.3.2 is a graph of this solution.
Example 5.3.3 Find the general solution of
x2y′′+ xy′− 4y= 2 x4 (5.3.13)
on (−∞ , 0) and (0, ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.3 Nonhomogeneous Linear Equations 225
0.5 1.0 1.5 2.0
2
4
6
8
10
12
14
16
 x
 y
Figure 5.3.2 y = 1 + 3 x+ x2 − ex (3 − x)
Solution In Example 5.1.3, we verified that y1 = x2 and y2 = 1 /x 2 form a fundamental set of solutions
of the complementary equation
x2y′′+ xy′− 4y= 0
on (−∞ , 0) and (0, ∞ ). T o find a particular solution of ( 5.3.13), we note that if yp = Ax4, where Ais a
constant then both sides of ( 5.3.13) will be constant multiples of x4 and we may be able to choose Aso
the two sides are equal. This is true in this example, since if yp = Ax4 then
x2y′′
p + xy′
p − 4yp = x2(12Ax2) + x(4Ax3) − 4Ax4 = 12 Ax4 = 2 x4
if A= 1 / 6; therefore, yp = x4/ 6 is a particular solution of ( 5.3.13) on (−∞ , ∞ ). Theorem 5.3.2 implies
that the general solution of ( 5.3.13) on (−∞ , 0) and (0, ∞ ) is
y= x4
6 + c1x2 + c2
x2 .
The Principle of Superposition
The next theorem enables us to break a nonhomogeous equation into simpler parts, find a particular",Elementary Differential Equations with Boundary Value Problems.pdf
"6 + c1x2 + c2
x2 .
The Principle of Superposition
The next theorem enables us to break a nonhomogeous equation into simpler parts, find a particular
solution for each part, and then combine their solutions to o btain a particular solution of the original
problem.
Theorem 5.3.3
[The Principle of Superposition ] Suppose yp1 is a particular solution of
y′′+ p(x)y′+ q(x)y= f1(x)
on (a,b ) and yp2 is a particular solution of
y′′+ p(x)y′+ q(x)y= f2(x)",Elementary Differential Equations with Boundary Value Problems.pdf
"226 Chapter 5 Linear Second Order Equations
on (a,b ). Then
yp = yp1 + yp2
is a particular solution of
y′′+ p(x)y′+ q(x)y= f1(x) + f2(x)
on (a,b ).
Proof If yp = yp1 + yp2 then
y′′
p + p(x)y′
p + q(x)yp = ( yp1 + yp2 )′′+ p(x)(yp1 + yp2 )′+ q(x)(yp1 + yp2 )
= (y′′
p1 + p(x)y′
p1 + q(x)yp1
) + (y′′
p2 + p(x)y′
p2 + q(x)yp2
)
= f1(x) + f2(x).
It’s easy to generalize Theorem 5.3.3 to the equation
y′′+ p(x)y′+ q(x)y= f(x) (5.3.14)
where
f = f1 + f2 + · · ·+ fk;
thus, if ypi is a particular solution of
y′′+ p(x)y′+ q(x)y = fi (x)
on (a,b ) for i = 1 , 2, . . . , k, then yp1 + yp2 + · · ·+ ypk is a particular solution of ( 5.3.14) on (a,b ).
Moreover, by a proof similar to the proof of Theorem 5.3.3 we can formulate the principle of superposition
in terms of a linear equation written in the form
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x)
(Exercise 39); that is, if yp1 is a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F1(x)
on (a,b ) and yp2 is a particular solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"(Exercise 39); that is, if yp1 is a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F1(x)
on (a,b ) and yp2 is a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F2(x)
on (a,b ), then yp1 + yp2 is a solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F1(x) + F2(x)
on (a,b ).
Example 5.3.4 The function yp1 = x4/ 15 is a particular solution of
x2y′′+ 4 xy′+ 2 y= 2 x4 (5.3.15)
on (−∞ , ∞ ) and yp2 = x2/ 3 is a particular solution of
x2y′′+ 4 xy′+ 2 y= 4 x2 (5.3.16)
on (−∞ , ∞ ). Use the principle of superposition to find a particular solu tion of
x2y′′+ 4 xy′+ 2 y = 2 x4 + 4 x2 (5.3.17)
on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.3 Nonhomogeneous Linear Equations 227
Solution The right side F(x) = 2 x4 + 4 x2 in ( 5.3.17) is the sum of the right sides
F1(x) = 2 x4 and F2(x) = 4 x2.
in ( 5.3.15) and ( 5.3.16). Therefore the principle of superposition implies that
yp = yp1 + yp2 = x4
15 + x2
3
is a particular solution of ( 5.3.17).
5.3 Exercises
In Exercises 1– 6 find a particular solution by the method used in Example 5.3.2. Then find the general
solution and, where indicated, solve the initial value prob lem and graph the solution.
1. y′′+ 5 y′− 6y= 22 + 18 x− 18x2
2. y′′− 4y′+ 5 y= 1 + 5 x
3. y′′+ 8 y′+ 7 y= −8 − x+ 24 x2 + 7 x3
4. y′′− 4y′+ 4 y= 2 + 8 x− 4x2
5. C/G y′′+ 2 y′+ 10 y= 4 + 26 x+ 6 x2 + 10 x3, y (0) = 2 , y ′(0) = 9
6. C/G y′′+ 6 y′+ 10 y= 22 + 20 x, y (0) = 2 , y′(0) = −2
7. Show that the method used in Example 5.3.2 won’t yield a particular solution of
y′′+ y′= 1 + 2 x+ x2; (A)
that is, (A) does’nt have a particular solution of the form yp = A+ Bx+ Cx2, where A, B, and
Care constants.",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′+ y′= 1 + 2 x+ x2; (A)
that is, (A) does’nt have a particular solution of the form yp = A+ Bx+ Cx2, where A, B, and
Care constants.
In Exercises 8– 13 find a particular solution by the method used in Example 5.3.3.
8. x2y′′+ 7 xy′+ 8 y = 6
x
9. x2y′′− 7xy′+ 7 y= 13 x1/ 2
10. x2y′′− xy′+ y= 2 x3
11. x2y′′+ 5 xy′+ 4 y= 1
x3
12. x2y′′+ xy′+ y= 10 x1/ 3 13. x2y′′− 3xy′+ 13 y= 2 x4
14. Show that the method suggested for finding a particular solut ion in Exercises 8-13 won’t yield a
particular solution of
x2y′′+ 3 xy′− 3y= 1
x3 ; (A)
that is, (A) doesn’t have a particular solution of the form yp = A/x 3.
15. Prove: If a, b, c, α, and M are constants and M ̸= 0 then
ax2y′′+ bxy′+ cy= Mxα
has a particular solution yp = Axα (A= constant) if and only if aα(α − 1) + bα + c̸= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"228 Chapter 5 Linear Second Order Equations
If a, b, c, and α are constants, then
a(eαx )′′+ b(eαx )′+ ceαx = ( aα2 + bα + c)eαx .
Use this in Exercises 16– 21 to find a particular solution . Then find the general solution a nd, where
indicated, solve the initial value problem and graph the sol ution.
16. y′′+ 5 y′− 6y= 6 e3x 17. y′′− 4y′+ 5 y= e2x
18. C/G y′′+ 8 y′+ 7 y= 10 e−2x, y (0) = −2, y′(0) = 10
19. C/G y′′− 4y′+ 4 y= ex, y (0) = 2 , y ′(0) = 0
20. y′′+ 2 y′+ 10 y= ex/ 2 21. y′′+ 6 y′+ 10 y= e−3x
22. Show that the method suggested for finding a particular solut ion in Exercises 16-21 won’t yield a
particular solution of
y′′− 7y′+ 12 y= 5 e4x; (A)
that is, (A) doesn’t have a particular solution of the form yp = Ae4x.
23. Prove: If α and M are constants and M ̸= 0 then constant coefficient equation
ay′′+ by′+ cy= Meαx
has a particular solution yp = Aeαx (A = constant) if and only if eαx isn’t a solution of the
complementary equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy= Meαx
has a particular solution yp = Aeαx (A = constant) if and only if eαx isn’t a solution of the
complementary equation.
If ω is a constant, differentiating a linear combination of cos ωx and sin ωx with respect to x yields
another linear combination of cos ωx and sin ωx. In Exercises 24– 29 use this to find a particular solution
of the equation. Then find the general solution and, where ind icated, solve the initial value problem and
graph the solution.
24. y′′− 8y′+ 16 y= 23 cos x− 7 sin x
25. y′′+ y′= −8 cos 2 x+ 6 sin 2 x
26. y′′− 2y′+ 3 y= −6 cos 3 x+ 6 sin 3 x
27. y′′+ 6 y′+ 13 y= 18 cos x+ 6 sin x
28. C/G y′′+ 7 y′+ 12 y= −2 cos 2 x+ 36 sin 2 x, y (0) = −3, y ′(0) = 3
29. C/G y′′− 6y′+ 9 y= 18 cos 3 x+ 18 sin 3 x, y (0) = 2 , y ′(0) = 2
30. Find the general solution of
y′′+ ω2
0 y = Mcos ωx + Nsin ωx,
where M and N are constants and ω and ω0 are distinct positive numbers.",Elementary Differential Equations with Boundary Value Problems.pdf
"30. Find the general solution of
y′′+ ω2
0 y = Mcos ωx + Nsin ωx,
where M and N are constants and ω and ω0 are distinct positive numbers.
31. Show that the method suggested for finding a particular solut ion in Exercises 24-29 won’t yield a
particular solution of
y′′+ y= cos x+ sin x; (A)
that is, (A) does not have a particular solution of the form yp = Acos x+ Bsin x.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.4 The Method of Undetermined Coefficients 229
32. Prove: If M, N are constants (not both zero) and ω >0, the constant coefficient equation
ay′′+ by′+ cy= Mcos ωx + Nsin ωx (A)
has a particular solution that’s a linear combination of cos ωx and sin ωx if and only if the left side
of (A) is not of the form a(y′′+ω2y), so that cos ωx and sin ωx are solutions of the complementary
equation.
In Exercises 33– 38 refer to the cited exercises and use the principal of superpo sition to find a particular
solution. Then find the general solution.
33. y′′+ 5 y′− 6y= 22 + 18 x− 18x2 + 6 e3x (See Exercises 1 and 16.)
34. y′′− 4y′+ 5 y= 1 + 5 x+ e2x (See Exercises 2 and 17.)
35. y′′+ 8 y′+ 7 y= −8 − x+ 24 x2 + 7 x3 + 10 e−2x (See Exercises 3 and 18.)
36. y′′− 4y′+ 4 y= 2 + 8 x− 4x2 + ex (See Exercises 4 and 19.)
37. y′′+ 2 y′+ 10 y= 4 + 26 x+ 6 x2 + 10 x3 + ex/ 2 (See Exercises 5 and 20.)
38. y′′+ 6 y′+ 10 y= 22 + 20 x+ e−3x (See Exercises 6 and 21.)
39. Prove: If yp1 is a particular solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"38. y′′+ 6 y′+ 10 y= 22 + 20 x+ e−3x (See Exercises 6 and 21.)
39. Prove: If yp1 is a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y = F1(x)
on (a,b ) and yp2 is a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y = F2(x)
on (a,b ), then yp = yp1 + yp2 is a solution of
P0(x)y′′+ P1(x)y′+ P2(x)y = F1(x) + F2(x)
on (a,b ).
40. Suppose p, q, and f are continuous on (a,b ). Let y1, y2, and yp be twice differentiable on (a,b ),
such that y= c1y1 + c2y2 + yp is a solution of
y′′+ p(x)y′+ q(x)y= f
on (a,b ) for every choice of the constants c1,c 2. Show that y1 and y2 are solutions of the comple-
mentary equation on (a,b ).
5.4 THE METHOD OF UNDETERMINED COEFFICIENTS I
In this section we consider the constant coefficient equatio n
ay′′+ by′+ cy= eαx G(x), (5.4.1)
where α is a constant and Gis a polynomial.
From Theorem 5.3.2, the general solution of ( 5.4.1) is y = yp + c1y1 + c2y2, where yp is a particular",Elementary Differential Equations with Boundary Value Problems.pdf
"where α is a constant and Gis a polynomial.
From Theorem 5.3.2, the general solution of ( 5.4.1) is y = yp + c1y1 + c2y2, where yp is a particular
solution of ( 5.4.1) and {y1,y 2}is a fundamental set of solutions of the complementary equat ion
ay′′+ by′+ cy= 0 .
In Section 5.2 we showed how to find {y1,y 2}. In this section we’ll show how to find yp. The procedure
that we’ll use is called the method of undetermined coefficients .
Our first example is similar to Exercises 16– 21.",Elementary Differential Equations with Boundary Value Problems.pdf
"230 Chapter 5 Linear Second Order Equations I
Example 5.4.1 Find a particular solution of
y′′− 7y′+ 12 y= 4 e2x. (5.4.2)
Then find the general solution.
Solution Substituting yp = Ae2x for yin (
5.4.2) will produce a constant multiple of Ae2x on the left side
of ( 5.4.2), so it may be possible to choose Aso that yp is a solution of ( 5.4.2). Let’s try it; if yp = Ae2x
then
y′′
p − 7y′
p + 12 yp = 4 Ae2x − 14Ae2x + 12 Ae2x = 2 Ae2x = 4 e2x
if A = 2 . Therefore yp = 2 e2x is a particular solution of ( 5.4.2). T o find the general solution, we note
that the characteristic polynomial of the complementary eq uation
y′′− 7y′+ 12 y= 0 (5.4.3)
is p(r) = r2 − 7r+ 12 = ( r− 3)(r− 4), so {e3x,e 4x}is a fundamental set of solutions of ( 5.4.3).
Therefore the general solution of ( 5.4.2) is
y = 2 e2x + c1e3x + c2e4x.
Example 5.4.2 Find a particular solution of
y′′− 7y′+ 12 y= 5 e4x. (5.4.4)
Then find the general solution.
Solution Fresh from our success in finding a particular solution of (",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′− 7y′+ 12 y= 5 e4x. (5.4.4)
Then find the general solution.
Solution Fresh from our success in finding a particular solution of (
5.4.2) — where we chose yp = Ae2x
because the right side of ( 5.4.2) is a constant multiple of e2x — it may seem reasonable to try yp = Ae4x
as a particular solution of ( 5.4.4). However, this won’t work, since we saw in Example 5.4.1 that e4x is
a solution of the complementary equation ( 5.4.3), so substituting yp = Ae4x into the left side of ( 5.4.4)
produces zero on the left, no matter how we choose A. T o discover a suitable form for yp , we use the same
approach that we used in Section 5.2 to find a second solution o f
ay′′+ by′+ cy = 0
in the case where the characteristic equation has a repeated real root: we look for solutions of ( 5.4.4) in
the form y= ue4x, where uis a function to be determined. Substituting
y= ue4x, y ′= u′e4x + 4 ue4x, and y′′= u′′e4x + 8 u′e4x + 16 ue4x (5.4.5)
into ( 5.4.4) and canceling the common factor e4x yields",Elementary Differential Equations with Boundary Value Problems.pdf
"y= ue4x, y ′= u′e4x + 4 ue4x, and y′′= u′′e4x + 8 u′e4x + 16 ue4x (5.4.5)
into ( 5.4.4) and canceling the common factor e4x yields
(u′′+ 8 u′+ 16 u) − 7(u′+ 4 u) + 12 u= 5 ,
or
u′′+ u′= 5 .
By inspection we see that up = 5 xis a particular solution of this equation, so yp = 5 xe4x is a particular
solution of ( 5.4.4). Therefore
y = 5 xe4x + c1e3x + c2e4x
is the general solution.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.4 The Method of Undetermined Coefficients 231
Example 5.4.3 Find a particular solution of
y′′− 8y′+ 16 y= 2 e4x. (5.4.6)
Solution Since the characteristic polynomial of the complementary e quation
y′′− 8y′+ 16 y= 0 (5.4.7)
is p(r) = r2 − 8r+ 16 = ( r− 4)2, both y1 = e4x and y2 = xe4x are solutions of ( 5.4.7). Therefore
(5.4.6) does not have a solution of the form yp = Ae4x or yp = Axe4x. As in Example 5.4.2, we look
for solutions of ( 5.4.6) in the form y = ue4x, where uis a function to be determined. Substituting from
(5.4.5) into ( 5.4.6) and canceling the common factor e4x yields
(u′′+ 8 u′+ 16 u) − 8(u′+ 4 u) + 16 u= 2 ,
or
u′′= 2 .
Integrating twice and taking the constants of integration t o be zero shows that up = x2 is a particular
solution of this equation, so yp = x2e4x is a particular solution of ( 5.4.4). Therefore
y= e4x(x2 + c1 + c2x)
is the general solution.
The preceding examples illustrate the following facts conc erning the form of a particular solution yp",Elementary Differential Equations with Boundary Value Problems.pdf
"y= e4x(x2 + c1 + c2x)
is the general solution.
The preceding examples illustrate the following facts conc erning the form of a particular solution yp
of a constant coefficent equation
ay′′+ by′+ cy= keαx ,
where kis a nonzero constant:
(a) If eαx isn’t a solution of the complementary equation
ay′′+ by′+ cy = 0 , (5.4.8)
then yp = Aeαx , where Ais a constant. (See Example 5.4.1).
(b) If eαx is a solution of ( 5.4.8) but xeαx is not, then yp = Axeαx , where A is a constant. (See
Example 5.4.2.)
(c) If both eαx and xeαx are solutions of ( 5.4.8), then yp = Ax2eαx , where A is a constant. (See
Example 5.4.3.)
See Exercise 30 for the proofs of these facts.
In all three cases you can just substitute the appropriate fo rm for yp and its derivatives directly into
ay′′
p + by′
p + cyp = keαx ,
and solve for the constant A, as we did in Example 5.4.1. (See Exercises 31– 33.) However, if the equation
is
ay′′+ by′+ cy = keαx G(x),",Elementary Differential Equations with Boundary Value Problems.pdf
"and solve for the constant A, as we did in Example 5.4.1. (See Exercises 31– 33.) However, if the equation
is
ay′′+ by′+ cy = keαx G(x),
where Gis a polynomial of degree greater than zero, we recommend tha t you use the substitution y =
ueαx as we did in Examples 5.4.2 and 5.4.3. The equation for uwill turn out to be
au′′+ p′(α)u′+ p(α)u= G(x), (5.4.9)",Elementary Differential Equations with Boundary Value Problems.pdf
"232 Chapter 5 Linear Second Order Equations I
where p(r) = ar2 + br+ cis the characteristic polynomial of the complementary equa tion and p′(r) =
2ar+ b(Exercise 30); however, you shouldn’t memorize this since it’s easy to de rive the equation for
u in any particular case. Note, however, that if eαx is a solution of the complementary equation then
p(α) = 0 , so ( 5.4.9) reduces to
au′′+ p′(α)u′= G(x),
while if both eαx and xeαx are solutions of the complementary equation then p(r) = a(r− α)2 and
p′(r) = 2 a(r− α), so p(α) = p′(α) = 0 and ( 5.4.9) reduces to
au′′= G(x).
Example 5.4.4 Find a particular solution of
y′′− 3y′+ 2 y= e3x(−1 + 2 x+ x2). (5.4.10)
Solution Substituting
y= ue3x, y ′= u′e3x + 3 ue3x, and y′′= u′′e3x + 6 u′e3x + 9 ue3x
into ( 5.4.10) and canceling e3x yields
(u′′+ 6 u′+ 9 u) − 3(u′+ 3 u) + 2 u= −1 + 2 x+ x2,
or
u′′+ 3 u′+ 2 u= −1 + 2 x+ x2. (5.4.11)
As in Example 2, in order to guess a form for a particular solution of ( 5.4.11), we note that substituting a",Elementary Differential Equations with Boundary Value Problems.pdf
"u′′+ 3 u′+ 2 u= −1 + 2 x+ x2. (5.4.11)
As in Example 2, in order to guess a form for a particular solution of ( 5.4.11), we note that substituting a
second degree polynomial up = A+ Bx+ Cx2 for uin the left side of ( 5.4.11) produces another second
degree polynomial with coefficients that depend upon A, B, and C; thus,
if up = A+ Bx+ Cx2 then u′
p = B+ 2 Cx and u′′
p = 2 C.
If up is to satisfy (
5.4.11), we must have
u′′
p + 3 u′
p + 2 up = 2 C+ 3( B+ 2 Cx) + 2( A+ Bx+ Cx2)
= (2 C+ 3 B+ 2 A) + (6 C+ 2 B)x+ 2 Cx2 = −1 + 2 x+ x2.
Equating coefficients of like powers of xon the two sides of the last equality yields
2C = 1
2B+ 6 C = 2
2A+ 3 B+ 2 C = −1.
Solving these equations for C, B, and A (in that order) yields C = 1 / 2,B = −1/ 2,A = −1/ 4.
Therefore
up = −1
4 (1 + 2 x− 2x2)
is a particular solution of ( 5.4.11), and
yp = upe3x = −e3x
4 (1 + 2 x− 2x2)
is a particular solution of ( 5.4.10).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.4 The Method of Undetermined Coefficients 233
Example 5.4.5 Find a particular solution of
y′′− 4y′+ 3 y= e3x(6 + 8 x+ 12 x2). (5.4.12)
Solution Substituting
y= ue3x, y ′= u′e3x + 3 ue3x, and y′′= u′′e3x + 6 u′e3x + 9 ue3x
into ( 5.4.12) and canceling e3x yields
(u′′+ 6 u′+ 9 u) − 4(u′+ 3 u) + 3 u= 6 + 8 x+ 12 x2,
or
u′′+ 2 u′= 6 + 8 x+ 12 x2. (5.4.13)
There’s no uterm in this equation, since e3x is a solution of the complementary equation for ( 5.4.12).
(See Exercise 30.) Therefore ( 5.4.13) does not have a particular solution of the form up = A+ Bx+ Cx2
that we used successfully in Example 5.4.4, since with this choice of up,
u′′
p + 2 u′
p = 2 C+ ( B+ 2 Cx)
can’t contain the last term ( 12x2) on the right side of (
5.4.13). Instead, let’s try up = Ax+ Bx2 + Cx3
on the grounds that
u′
p = A+ 2 Bx+ 3 Cx2 and u′′
p = 2 B+ 6 Cx
together contain all the powers of xthat appear on the right side of (
5.4.13).
Substituting these expressions in place of u′and u′′in ( 5.4.13) yields",Elementary Differential Equations with Boundary Value Problems.pdf
"5.4.13).
Substituting these expressions in place of u′and u′′in ( 5.4.13) yields
(2B+ 6 Cx) + 2( A+ 2 Bx+ 3 Cx2) = (2 B+ 2 A) + (6 C+ 4 B)x+ 6 Cx2 = 6 + 8 x+ 12 x2.
Comparing coefficients of like powers of xon the two sides of the last equality shows that up satisfies
(5.4.13) if
6C = 12
4B+ 6 C = 8
2A+ 2 B = 6 .
Solving these equations successively yields C = 2 , B= −1, and A= 4 . Therefore
up = x(4 − x+ 2 x2)
is a particular solution of ( 5.4.13), and
yp = upe3x = xe3x(4 − x+ 2 x2)
is a particular solution of ( 5.4.12).
Example 5.4.6 Find a particular solution of
4y′′+ 4 y′+ y= e−x/ 2(−8 + 48 x+ 144 x2). (5.4.14)
Solution Substituting
y= ue−x/ 2, y ′= u′e−x/ 2 − 1
2 ue−x/ 2, and y′′= u′′e−x/ 2 − u′e−x/ 2 + 1
4 ue−x/ 2",Elementary Differential Equations with Boundary Value Problems.pdf
"234 Chapter 5 Linear Second Order Equations I
into ( 5.4.14) and canceling e−x/ 2 yields
4
(
u′′− u′+ u
4
)
+ 4
(
u′− u
2
)
+ u= 4 u′′= −8 + 48 x+ 144 x2,
or
u′′= −2 + 12 x+ 36 x2, (5.4.15)
which does not contain u or u′because e−x/ 2 and xe−x/ 2 are both solutions of the complementary
equation. (See Exercise 30.) T o obtain a particular solution of ( 5.4.15) we integrate twice, taking the
constants of integration to be zero; thus,
u′
p = −2x+ 6 x2 + 12 x3 and up = −x2 + 2 x3 + 3 x4 = x2(−1 + 2 x+ 3 x2).
Therefore
yp = upe−x/ 2 = x2e−x/ 2(−1 + 2 x+ 3 x2)
is a particular solution of (
5.4.14).
Summary
The preceding examples illustrate the following facts conc erning particular solutions of a constant coef-
ficent equation of the form
ay′′+ by′+ cy= eαx G(x),
where Gis a polynomial (see Exercise 30):
(a) If eαx isn’t a solution of the complementary equation
ay′′+ by′+ cy = 0 , (5.4.16)
then yp = eαx Q(x), where Qis a polynomial of the same degree as G. (See Example 5.4.4).",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy = 0 , (5.4.16)
then yp = eαx Q(x), where Qis a polynomial of the same degree as G. (See Example 5.4.4).
(b) If eαx is a solution of ( 5.4.16) but xeαx is not, then yp = xeαx Q(x), where Qis a polynomial of
the same degree as G. (See Example 5.4.5.)
(c) If both eαx and xeαx are solutions of ( 5.4.16), then yp = x2eαx Q(x), where Qis a polynomial of
the same degree as G. (See Example 5.4.6.)
In all three cases, you can just substitute the appropriate f orm for yp and its derivatives directly into
ay′′
p + by′
p + cyp = eαx G(x),
and solve for the coefficients of the polynomial Q. However, if you try this you will see that the compu-
tations are more tedious than those that you encounter by mak ing the substitution y = ueαx and finding
a particular solution of the resulting equation for u. (See Exercises 34-36.) In Case (a) the equation for u
will be of the form
au′′+ p′(α)u′+ p(α)u= G(x),",Elementary Differential Equations with Boundary Value Problems.pdf
"will be of the form
au′′+ p′(α)u′+ p(α)u= G(x),
with a particular solution of the form up = Q(x), a polynomial of the same degree as G, whose coeffi-
cients can be found by the method used in Example 5.4.4. In Case (b) the equation for uwill be of the
form
au′′+ p′(α)u′= G(x)
(no uterm on the left), with a particular solution of the form up = xQ(x), where Qis a polynomial of
the same degree as Gwhose coefficents can be found by the method used in Example 5.4.5. In Case (c)
the equation for uwill be of the form
au′′= G(x)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.4 The Method of Undetermined Coefficients 235
with a particular solution of the form up = x2Q(x) that can be obtained by integrating G(x)/a twice and
taking the constants of integration to be zero, as in Example 5.4.6.
Using the Principle of Superposition
The next example shows how to combine the method of undetermi ned coefficients and Theorem 5.3.3,
the principle of superposition.
Example 5.4.7 Find a particular solution of
y′′− 7y′+ 12 y= 4 e2x + 5 e4x. (5.4.17)
Solution In Example
5.4.1 we found that yp1 = 2 e2x is a particular solution of
y′′− 7y′+ 12 y= 4 e2x,
and in Example 5.4.2 we found that yp2 = 5 xe4x is a particular solution of
y′′− 7y′+ 12 y= 5 e4x.
Therefore the principle of superposition implies that yp = 2 e2x +5xe4x is a particular solution of ( 5.4.17).
5.4 Exercises
In Exercises 1– 14 find a particular solution.
1. y′′− 3y′+ 2 y= e3x(1 + x) 2. y′′− 6y′+ 5 y= e−3x(35 − 8x)
3. y′′− 2y′− 3y= ex (−8 + 3 x) 4. y′′+ 2 y′+ y= e2x(−7 − 15x+ 9 x2)",Elementary Differential Equations with Boundary Value Problems.pdf
"1. y′′− 3y′+ 2 y= e3x(1 + x) 2. y′′− 6y′+ 5 y= e−3x(35 − 8x)
3. y′′− 2y′− 3y= ex (−8 + 3 x) 4. y′′+ 2 y′+ y= e2x(−7 − 15x+ 9 x2)
5. y′′+ 4 y= e−x(7 − 4x+ 5 x2) 6. y′′− y′− 2y= ex(9 + 2 x− 4x2)
7. y′′− 4y′− 5y= −6xe−x 8. y′′− 3y′+ 2 y= ex(3 − 4x)
9. y′′+ y′− 12y= e3x(−6 + 7 x) 10. 2y′′− 3y′− 2y = e2x(−6 + 10 x)
11. y′′+ 2 y′+ y = e−x(2 + 3 x) 12. y′′− 2y′+ y= ex(1 − 6x)
13. y′′− 4y′+ 4 y= e2x(1 − 3x+ 6 x2)
14. 9y′′+ 6 y′+ y= e−x/ 3(2 − 4x+ 4 x2)
In Exercises 15– 19 find the general solution.
15. y′′− 3y′+ 2 y= e3x(1 + x) 16. y′′− 6y′+ 8 y= ex(11 − 6x)
17. y′′+ 6 y′+ 9 y= e2x(3 − 5x) 18. y′′+ 2 y′− 3y= −16xex
19. y′′− 2y′+ y = ex(2 − 12x)",Elementary Differential Equations with Boundary Value Problems.pdf
"236 Chapter 5 Linear Second Order Equations I
In Exercises 20– 23 solve the initial value problem and plot the solution.
20. C/G y′′− 4y′− 5y= 9 e2x(1 + x), y (0) = 0 , y ′(0) = −10
21. C/G y′′+ 3 y′− 4y= e2x(7 + 6 x), y (0) = 2 , y ′(0) = 8
22. C/G y′′+ 4 y′+ 3 y= −e−x(2 + 8 x), y (0) = 1 , y ′(0) = 2
23. C/G y′′− 3y′− 10y= 7 e−2x, y (0) = 1 , y ′(0) = −17
In Exercises 24– 29 use the principle of superposition to find a particular solut ion.
24. y′′+ y′+ y= xex + e−x (1 + 2 x)
25. y′′− 7y′+ 12 y= −ex (17 − 42x) − e3x
26. y′′− 8y′+ 16 y= 6 xe4x + 2 + 16 x+ 16 x2
27. y′′− 3y′+ 2 y= −e2x(3 + 4 x) − ex
28. y′′− 2y′+ 2 y= ex (1 + x) + e−x(2 − 8x+ 5 x2)
29. y′′+ y = e−x(2 − 4x+ 2 x2) + e3x(8 − 12x− 10x2)
30. (a) Prove that yis a solution of the constant coefficient equation
ay′′+ by′+ cy = eαx G(x) (A)
if and only if y= ueαx , where usatisfies
au′′+ p′(α)u′+ p(α)u= G(x) (B)
and p(r) = ar2 + br+ cis the characteristic polynomial of the complementary equa tion
ay′′+ by′+ cy = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"au′′+ p′(α)u′+ p(α)u= G(x) (B)
and p(r) = ar2 + br+ cis the characteristic polynomial of the complementary equa tion
ay′′+ by′+ cy = 0 .
For the rest of this exercise, let Gbe a polynomial. Give the requested proofs for the case
where
G(x) = g0 + g1x+ g2x2 + g3x3.
(b) Prove that if eαx isn’t a solution of the complementary equation then (B) has a particular
solution of the form up = A(x), where A is a polynomial of the same degree as G, as in
Example 5.4.4. Conclude that (A) has a particular solution of the form yp = eαx A(x).
(c) Show that if eαx is a solution of the complementary equation and xeαx isn’t , then (B)
has a particular solution of the form up = xA(x), where A is a polynomial of the same
degree as G, as in Example 5.4.5. Conclude that (A) has a particular solution of the form
yp = xeαx A(x).
(d) Show that if eαx and xeαx are both solutions of the complementary equation then (B) ha s a",Elementary Differential Equations with Boundary Value Problems.pdf
"yp = xeαx A(x).
(d) Show that if eαx and xeαx are both solutions of the complementary equation then (B) ha s a
particular solution of the form up = x2A(x), where Ais a polynomial of the same degree as
G, and x2A(x) can be obtained by integrating G/a twice, taking the constants of integration
to be zero, as in Example 5.4.6. Conclude that (A) has a particular solution of the form
yp = x2eαx A(x).
Exercises 31– 36 treat the equations considered in Examples 5.4.1– 5.4.6. Substitute the suggested form
of yp into the equation and equate the resulting coefficients of li ke functions on the two sides of the
resulting equation to derive a set of simultaneous equation s for the coefficients in yp . Then solve for the
coefficients to obtain yp. Compare the work you’ve done with the work required to obtai n the same results
in Examples 5.4.1– 5.4.6.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.4 The Method of Undetermined Coefficients 237
31. Compare with Example 5.4.1:
y′′− 7y′+ 12 y= 4 e2x; yp = Ae2x
32. Compare with Example 5.4.2:
y′′− 7y′+ 12 y= 5 e4x; yp = Axe4x
33. Compare with Example 5.4.3.
y′′− 8y′+ 16 y= 2 e4x; yp = Ax2e4x
34. Compare with Example 5.4.4:
y′′− 3y′+ 2 y = e3x(−1 + 2 x+ x2), y p = e3x(A+ Bx+ Cx2)
35. Compare with Example 5.4.5:
y′′− 4y′+ 3 y= e3x(6 + 8 x+ 12 x2), y p = e3x(Ax+ Bx2 + Cx3)
36. Compare with Example 5.4.6:
4y′′+ 4 y′+ y = e−x/ 2(−8 + 48 x+ 144 x2), y p = e−x/ 2(Ax2 + Bx3 + Cx4)
37. Write y = ueαx to find the general solution.
(a) y′′+ 2 y′+ y= e−x
√x (b) y′′+ 6 y′+ 9 y= e−3x ln x
(c) y′′− 4y′+ 4 y= e2x
1 + x (d) 4y′′+ 4 y′+ y= 4 e−x/ 2
(1
x + x
)
38. Suppose α ̸= 0 and kis a positive integer. In most calculus books integrals like ∫ xk eαx dxare
evaluated by integrating by parts ktimes. This exercise presents another method. Let
y=
∫
eαx P(x) dx
with
P(x) = p0 + p1x+ · · ·+ pk xk, (where pk ̸= 0 ).
(a) Show that y = eαx u, where",Elementary Differential Equations with Boundary Value Problems.pdf
"y=
∫
eαx P(x) dx
with
P(x) = p0 + p1x+ · · ·+ pk xk, (where pk ̸= 0 ).
(a) Show that y = eαx u, where
u′+ αu = P(x). (A)
(b) Show that (A) has a particular solution of the form
up = A0 + A1x+ · · ·+ Akxk,
where Ak , Ak−1, . . . , A0 can be computed successively by equating coefficients of xk ,x k−1,..., 1
on both sides of the equation
u′
p + αup = P(x).
(c) Conclude that
∫
eαx P(x) dx= (A0 + A1x+ · · ·+ Ak xk) eαx + c,
where cis a constant of integration.",Elementary Differential Equations with Boundary Value Problems.pdf
"238 Chapter 5 Linear Second Order Equations
39. Use the method of Exercise 38 to evaluate the integral.
(a)
∫
ex(4 + x) dx (b)
∫
e−x(−1 + x2) dx
(c) ∫ x3e−2x dx (d) ∫ ex(1 + x)2 dx
(e) ∫ e3x(−14 + 30 x+ 27 x2) dx (f) ∫ e−x (1 + 6 x2 − 14x3 + 3 x4) dx
40. Use the method suggested in Exercise 38 to evaluate ∫ xkeαx dx, where kis an arbitrary positive
integer and α ̸= 0 .
5.5 THE METHOD OF UNDETERMINED COEFFICIENTS II
In this section we consider the constant coefficient equatio n
ay′′+ by′+ cy= eλx (P(x) cos ωx + Q(x) sin ωx) (5.5.1)
where λ and ω are real numbers, ω ̸= 0 , and P and Qare polynomials. W e want to find a particular
solution of ( 5.5.1). As in Section 5.4, the procedure that we will use is called the method of undetermined
coefficients .
Forcing Functions Without Exponential Factors
W e begin with the case where λ = 0 in ( 5.5.1); thus, we we want to find a particular solution of
ay′′+ by′+ cy= P(x) cos ωx + Q(x) sin ωx, (5.5.2)
where P and Qare polynomials.",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy= P(x) cos ωx + Q(x) sin ωx, (5.5.2)
where P and Qare polynomials.
Differentiating xr cos ωx and xr sin ωx yields
d
dxxr cos ωx = −ωxr sin ωx + rxr−1 cos ωx
and d
dxxr sin ωx = ωxr cos ωx + rxr−1 sin ωx.
This implies that if
yp = A(x) cos ωx + B(x) sin ωx
where Aand Bare polynomials, then
ay′′
p + by′
p + cyp = F(x) cos ωx + G(x) sin ωx,
where F and Gare polynomials with coefficients that can be expressed in te rms of the coefficients of A
and B. This suggests that we try to choose Aand Bso that F = P and G = Q, respectively . Then yp
will be a particular solution of ( 5.5.2). The next theorem tells us how to choose the proper form for yp.
For the proof see Exercise 37.
Theorem 5.5.1 Suppose ω is a positive number and P and Qare polynomials . Let kbe the larger of the
degrees of P and Q. Then the equation
ay′′+ by′+ cy= P(x) cos ωx + Q(x) sin ωx
has a particular solution
yp = A(x) cos ωx + B(x) sin ωx, (5.5.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.5 The Method of Undetermined Coefficients II 239
where
A(x) = A0 + A1x+ · · ·+ Ak xk and B(x) = B0 + B1x+ · · ·+ Bk xk,
provided that cos ωx and sin ωx are not solutions of the complementary equation . The solutions of
a(y′′+ ω2y) = P(x) cos ωx + Q(x) sin ωx
(for which cos ωx and sin ωx are solutions of the complementary equation ) are of the form (5.5.3), where
A(x) = A0x+ A1x2 + · · ·+ Ak xk+1 and B(x) = B0x+ B1x2 + · · ·+ Bk xk+1.
For an analog of this theorem that’s applicable to ( 5.5.1), see Exercise 38.
Example 5.5.1 Find a particular solution of
y′′− 2y′+ y = 5 cos 2 x+ 10 sin 2 x. (5.5.4)
Solution In ( 5.5.4) the coefficients of cos 2xand sin 2xare both zero degree polynomials (constants).
Therefore Theorem 5.5.1 implies that ( 5.5.4) has a particular solution
yp = Acos 2x+ Bsin 2 x.
Since
y′
p = −2Asin 2 x+ 2 Bcos 2x and y′′
p = −4(Acos 2 x+ Bsin 2 x),
replacing yby yp in ( 5.5.4) yields
y′′
p − 2y′
p + yp = −4(Acos 2 x+ Bsin 2 x) − 4(−Asin 2 x+ Bcos 2 x)",Elementary Differential Equations with Boundary Value Problems.pdf
"p = −4(Acos 2 x+ Bsin 2 x),
replacing yby yp in ( 5.5.4) yields
y′′
p − 2y′
p + yp = −4(Acos 2 x+ Bsin 2 x) − 4(−Asin 2 x+ Bcos 2 x)
+(Acos 2 x+ Bsin 2 x)
= ( −3A− 4B) cos 2 x+ (4 A− 3B) sin 2 x.
Equating the coefficients of cos 2xand sin 2 xhere with the corresponding coefficients on the right side
of ( 5.5.4) shows that yp is a solution of ( 5.5.4) if
−3A− 4B = 5
4A− 3B = 10 .
Solving these equations yields A= 1 , B= −2. Therefore
yp = cos 2 x− 2 sin 2 x
is a particular solution of ( 5.5.4).
Example 5.5.2 Find a particular solution of
y′′+ 4 y= 8 cos 2 x+ 12 sin 2 x. (5.5.5)
Solution The procedure used in Example 5.5.1 doesn’t work here; substituting yp = Acos 2x+ Bsin 2 x
for yin ( 5.5.5) yields
y′′
p + 4 yp = −4(Acos 2 x+ Bsin 2 x) + 4( Acos 2x+ Bsin 2 x) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"240 Chapter 5 Linear Second Order Equations
for any choice of A and B, since cos 2 xand sin 2 xare both solutions of the complementary equation
for ( 5.5.5). W e’re dealing with the second case mentioned in Theorem 5.5.1, and should therefore try a
particular solution of the form
yp = x(Acos 2x+ Bsin 2 x). (5.5.6)
Then
y′
p = Acos 2 x+ Bsin 2 x+ 2 x(−Asin 2 x+ Bcos 2 x)
and y′′
p = −4Asin 2 x+ 4 Bcos 2x− 4x(Acos 2x+ Bsin 2 x)
= −4Asin 2 x+ 4 Bcos 2x− 4yp (see ( 5.5.6)),
so
y′′
p + 4 yp = −4Asin 2 x+ 4 Bcos 2x.
Therefore yp is a solution of ( 5.5.5) if
−4Asin 2 x+ 4 Bcos 2 x= 8 cos 2 x+ 12 sin 2 x,
which holds if A= −3 and B = 2 . Therefore
yp = −x(3 cos 2 x− 2 sin 2 x)
is a particular solution of ( 5.5.5).
Example 5.5.3 Find a particular solution of
y′′+ 3 y′+ 2 y= (16 + 20 x) cos x+ 10 sin x. (5.5.7)
Solution The coefficients of cos xand sin xin ( 5.5.7) are polynomials of degree one and zero, respec-",Elementary Differential Equations with Boundary Value Problems.pdf
"Solution The coefficients of cos xand sin xin ( 5.5.7) are polynomials of degree one and zero, respec-
tively . Therefore Theorem 5.5.1 tells us to look for a particular solution of ( 5.5.7) of the form
yp = ( A0 + A1x) cos x+ ( B0 + B1 x) sin x. (5.5.8)
Then
y′
p = ( A1 + B0 + B1x) cos x+ ( B1 − A0 − A1x) sin x (5.5.9)
and
y′′
p = (2 B1 − A0 − A1x) cos x− (2A1 + B0 + B1x) sin x, (5.5.10)
so
y′′
p + 3 y′
p + 2 yp = [ A0 + 3 A1 + 3 B0 + 2 B1 + ( A1 + 3 B1)x] cos x
+ [B0 + 3 B1 − 3A0 − 2A1 + ( B1 − 3A1)x] sin x. (5.5.11)
Comparing the coefficients of xcos x, xsin x, cos x, and sin xhere with the corresponding coefficients
in ( 5.5.7) shows that yp is a solution of ( 5.5.7) if
A1 + 3 B1 = 20
−3A1 + B1 = 0
A0 + 3 B0 + 3 A1 + 2 B1 = 16
−3A0 + B0 − 2A1 + 3 B1 = 10 .
Solving the first two equations yields A1 = 2 , B1 = 6 . Substituting these into the last two equations
yields
A0 + 3 B0 = 16 − 3A1 − 2B1 = −2
−3A0 + B0 = 10 + 2 A1 − 3B1 = −4.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.5 The Method of Undetermined Coefficients II 241
Solving these equations yields A0 = 1 , B0 = −1. Substituting A0 = 1 , A1 = 2 , B0 = −1, B1 = 6 into
(5.5.8) shows that
yp = (1 + 2 x) cos x− (1 − 6x) sin x
is a particular solution of ( 5.5.7).
A Useful Observation
In ( 5.5.9), ( 5.5.10), and ( 5.5.11) the polynomials multiplying sin xcan be obtained by replacing A0,A 1,B 0,
and B1 by B0, B1 , −A0, and −A1, respectively , in the polynomials mutiplying cos x. An analogous re-
sult applies in general, as follows (Exercise 36).
Theorem 5.5.2 If
yp = A(x) cos ωx + B(x) sin ωx,
where A(x) and B(x) are polynomials with coefficients A0 . . . , Ak and B0, . . . , Bk , then the polynomials
multiplying sin ωx in
y′
p , y ′′
p , ay ′′
p + by′
p + cyp and y′′
p + ω2yp
can be obtained by replacing A0, . . . , Ak by B0, . . . , Bk and B0, . . . , Bk by −A0, . . . , −Ak in the
corresponding polynomials multiplying cos ωx.",Elementary Differential Equations with Boundary Value Problems.pdf
"corresponding polynomials multiplying cos ωx.
W e won’t use this theorem in our examples, but we recommend th at you use it to check your manipu-
lations when you work the exercises.
Example 5.5.4 Find a particular solution of
y′′+ y= (8 − 4x) cos x− (8 + 8 x) sin x. (5.5.12)
Solution According to Theorem
5.5.1, we should look for a particular solution of the form
yp = ( A0x+ A1x2) cos x+ ( B0 x+ B1x2) sin x, (5.5.13)
since cos xand sin xare solutions of the complementary equation. However, let’ s try
yp = ( A0 + A1x) cos x+ ( B0 + B1x) sin x (5.5.14)
first, so you can see why it doesn’t work. From ( 5.5.10),
y′′
p = (2 B1 − A0 − A1x) cos x− (2A1 + B0 + B1x) sin x,
which together with ( 5.5.14) implies that
y′′
p + yp = 2 B1 cos x− 2A1 sin x.
Since the right side of this equation does not contain xcos xor xsin x, ( 5.5.14) can’t satisfy ( 5.5.12) no
matter how we choose A0, A1, B0, and B1.
Now let yp be as in ( 5.5.13). Then
y′
p = [ A0 + (2 A1 + B0)x+ B1x2] cos x
+
[",Elementary Differential Equations with Boundary Value Problems.pdf
"matter how we choose A0, A1, B0, and B1.
Now let yp be as in ( 5.5.13). Then
y′
p = [ A0 + (2 A1 + B0)x+ B1x2] cos x
+
[
B0 + (2 B1 − A0)x− A1x2]
sin x
and y′′
p =
[
2A1 + 2 B0 − (A0 − 4B1)x− A1x2]
cos x
+ [ 2B1 − 2A0 − (B0 + 4 A1)x− B1 x2] sin x,",Elementary Differential Equations with Boundary Value Problems.pdf
"242 Chapter 5 Linear Second Order Equations
so
y′′
p + yp = (2 A1 + 2 B0 + 4 B1x) cos x+ (2 B1 − 2A0 − 4A1x) sin x.
Comparing the coefficients of cos xand sin xhere with the corresponding coefficients in ( 5.5.12) shows
that yp is a solution of ( 5.5.12) if
4B1 = −4
−4A1 = −8
2B0 + 2 A1 = 8
−2A0 + 2 B1 = −8.
The solution of this system is A1 = 2 , B1 = −1, A0 = 3 , B0 = 2 . Therefore
yp = x[(3 + 2 x) cos x+ (2 − x) sin x]
is a particular solution of ( 5.5.12).
Forcing Functions with Exponential Factors
T o find a particular solution of
ay′′+ by′+ cy= eλx (P(x) cos ωx + Q(x) sin ωx) (5.5.15)
when λ ̸= 0 , we recall from Section 5.4 that substituting y = ueλx into ( 5.5.15) will produce a constant
coefficient equation for uwith the forcing function P(x) cos ωx + Q(x) sin ωx. W e can find a particular
solution up of this equation by the procedure that we used in Examples 5.5.1– 5.5.4. Then yp = upeλx is
a particular solution of ( 5.5.15).
Example 5.5.5 Find a particular solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"a particular solution of ( 5.5.15).
Example 5.5.5 Find a particular solution of
y′′− 3y′+ 2 y= e−2x [2 cos 3 x− (34 − 150x) sin 3 x] . (5.5.16)
Solution Let y = ue−2x. Then
y′′− 3y′+ 2 y = e−2x [(u′′− 4u′+ 4 u) − 3(u′− 2u) + 2 u]
= e−2x(u′′− 7u′+ 12 u)
= e−2x [2 cos 3 x− (34 − 150x) sin 3 x]
if
u′′− 7u′+ 12 u= 2 cos 3 x− (34 − 150x) sin 3 x. (5.5.17)
Since cos 3xand sin 3 xaren’t solutions of the complementary equation
u′′− 7u′+ 12 u= 0 ,
Theorem 5.5.1 tells us to look for a particular solution of ( 5.5.17) of the form
up = ( A0 + A1x) cos 3 x+ ( B0 + B1x) sin 3 x. (5.5.18)
Then
u′
p = ( A1 + 3 B0 + 3 B1x) cos 3 x+ ( B1 − 3A0 − 3A1x) sin 3 x
and u′′
p = ( −9A0 + 6 B1 − 9A1x) cos 3 x− (9B0 + 6 A1 + 9 B1x) sin 3 x,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.5 The Method of Undetermined Coefficients II 243
so
u′′
p − 7u′
p + 12 up = [3 A0 − 21B0 − 7A1 + 6 B1 + (3 A1 − 21B1)x] cos 3 x
+ [21 A0 + 3 B0 − 6A1 − 7B1 + (21 A1 + 3 B1)x] sin 3 x.
Comparing the coefficients of xcos 3x, xsin 3 x, cos 3 x, and sin 3 xhere with the corresponding coeffi-
cients on the right side of (
5.5.17) shows that up is a solution of ( 5.5.17) if
3A1 − 21B1 = 0
21A1 + 3 B1 = 150
3A0 − 21B0 − 7A1 + 6 B1 = 2
21A0 + 3 B0 − 6A1 − 7B1 = −34.
(5.5.19)
Solving the first two equations yields A1 = 7 , B1 = 1 . Substituting these values into the last two
equations of ( 5.5.19) yields
3A0 − 21B0 = 2 + 7 A1 − 6B1 = 45
21A0 + 3 B0 = −34 + 6 A1 + 7 B1 = 15 .
Solving this system yields A0 = 1 , B0 = −2. Substituting A0 = 1 , A1 = 7 , B0 = −2, and B1 = 1 into
(5.5.18) shows that
up = (1 + 7 x) cos 3 x− (2 − x) sin 3 x
is a particular solution of ( 5.5.17). Therefore
yp = e−2x [(1 + 7 x) cos 3 x− (2 − x) sin 3 x]
is a particular solution of ( 5.5.16).",Elementary Differential Equations with Boundary Value Problems.pdf
"is a particular solution of ( 5.5.17). Therefore
yp = e−2x [(1 + 7 x) cos 3 x− (2 − x) sin 3 x]
is a particular solution of ( 5.5.16).
Example 5.5.6 Find a particular solution of
y′′+ 2 y′+ 5 y= e−x [(6 − 16x) cos 2 x− (8 + 8 x) sin 2 x] . (5.5.20)
Solution Let y = ue−x. Then
y′′+ 2 y′+ 5 y = e−x [(u′′− 2u′+ u) + 2( u′− u) + 5 u]
= e−x(u′′+ 4 u)
= e−x [(6 − 16x) cos 2 x− (8 + 8 x) sin 2 x]
if
u′′+ 4 u= (6 − 16x) cos 2 x− (8 + 8 x) sin 2 x. (5.5.21)
Since cos 2xand sin 2 xare solutions of the complementary equation
u′′+ 4 u= 0 ,
Theorem 5.5.1 tells us to look for a particular solution of ( 5.5.21) of the form
up = ( A0x+ A1x2) cos 2 x+ ( B0x+ B1x2) sin 2 x.",Elementary Differential Equations with Boundary Value Problems.pdf
"244 Chapter 5 Linear Second Order Equations
Then
u′
p =
[
A0 + (2 A1 + 2 B0)x+ 2 B1x2]
cos 2x
+
[
B0 + (2 B1 − 2A0)x− 2A1x2]
sin 2 x
and u′′
p = [ 2A1 + 4 B0 − (4A0 − 8B1)x− 4A1x2] cos 2x
+
[
2B1 − 4A0 − (4B0 + 8 A1)x− 4B1x2]
sin 2 x,
so
u′′
p + 4 up = (2 A1 + 4 B0 + 8 B1x) cos 2 x+ (2 B1 − 4A0 − 8A1x) sin 2 x.
Equating the coefficients of xcos 2x, xsin 2 x, cos 2x, and sin 2 xhere with the corresponding coefficients
on the right side of (
5.5.21) shows that up is a solution of ( 5.5.21) if
8B1 = −16
−8A1 = − 8
4B0 + 2 A1 = 6
−4A0 + 2 B1 = −8.
(5.5.22)
The solution of this system is A1 = 1 , B1 = −2, B0 = 1 , A0 = 1 . Therefore
up = x[(1 + x) cos 2 x+ (1 − 2x) sin 2 x]
is a particular solution of ( 5.5.21), and
yp = xe−x [(1 + x) cos 2 x+ (1 − 2x) sin 2 x]
is a particular solution of ( 5.5.20).
Y ou can also find a particular solution of ( 5.5.20) by substituting
yp = xe−x [(A0 + A1x) cos 2 x+ ( B0 + B1x) sin 2 x]",Elementary Differential Equations with Boundary Value Problems.pdf
"Y ou can also find a particular solution of ( 5.5.20) by substituting
yp = xe−x [(A0 + A1x) cos 2 x+ ( B0 + B1x) sin 2 x]
for yin ( 5.5.20) and equating the coefficients of xe−x cos 2x, xe−x sin 2 x, e−x cos 2 x, and e−x sin 2 xin
the resulting expression for
y′′
p + 2 y′
p + 5 yp
with the corresponding coefficients on the right side of ( 5.5.20). (See Exercise 38). This leads to the same
system ( 5.5.22) of equations for A0, A1, B0, and B1 that we obtained in Example 5.5.6. However, if you
try this approach you’ll see that deriving ( 5.5.22) this way is much more tedious than the way we did it
in Example 5.5.6.
5.5 Exercises
In Exercises 1– 17 find a particular solution.
1. y′′+ 3 y′+ 2 y= 7 cos x− sin x
2. y′′+ 3 y′+ y = (2 − 6x) cos x− 9 sin x
3. y′′+ 2 y′+ y = ex(6 cos x+ 17 sin x)
4. y′′+ 3 y′− 2y= −e2x(5 cos 2 x+ 9 sin 2 x)
5. y′′− y′+ y= ex(2 + x) sin x
6. y′′+ 3 y′− 2y= e−2x [(4 + 20 x) cos 3 x+ (26 − 32x) sin 3 x]",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.5 The Method of Undetermined Coefficients II 245
7. y′′+ 4 y= −12 cos 2 x− 4 sin 2 x
8. y′′+ y = ( −4 + 8 x) cos x+ (8 − 4x) sin x
9. 4y′′+ y= −4 cos x/ 2 − 8xsin x/ 2
10. y′′+ 2 y′+ 2 y= e−x(8 cos x− 6 sin x)
11. y′′− 2y′+ 5 y= ex [(6 + 8 x) cos 2 x+ (6 − 8x) sin 2 x]
12. y′′+ 2 y′+ y = 8 x2 cos x− 4xsin x
13. y′′+ 3 y′+ 2 y= (12 + 20 x+ 10 x2) cos x+ 8 xsin x
14. y′′+ 3 y′+ 2 y= (1 − x− 4x2) cos 2 x− (1 + 7 x+ 2 x2) sin 2 x
15. y′′− 5y′+ 6 y= −ex [
(4 + 6 x− x2) cos x− (2 − 4x+ 3 x2) sin x
]
16. y′′− 2y′+ y = −ex [ (3 + 4 x− x2) cos x+ (3 − 4x− x2) sin x]
17. y′′− 2y′+ 2 y= ex [
(2 − 2x− 6x2) cos x+ (2 − 10x+ 6 x2) sin x
]
In Exercises 1– 17 find a particular solution and graph it.
18. C/G y′′+ 2 y′+ y= e−x [(5 − 2x) cos x− (3 + 3 x) sin x]
19. C/G y′′+ 9 y= −6 cos 3 x− 12 sin 3 x
20. C/G y′′+ 3 y′+ 2 y= (1 − x− 4x2) cos 2 x− (1 + 7 x+ 2 x2) sin 2 x
21. C/G y′′+ 4 y′+ 3 y= e−x [ (2 + x+ x2) cos x+ (5 + 4 x+ 2 x2) sin x]
In Exercises 22– 26 solve the initial value problem.",Elementary Differential Equations with Boundary Value Problems.pdf
"21. C/G y′′+ 4 y′+ 3 y= e−x [ (2 + x+ x2) cos x+ (5 + 4 x+ 2 x2) sin x]
In Exercises 22– 26 solve the initial value problem.
22. y′′− 7y′+ 6 y= −ex (17 cos x− 7 sin x), y (0) = 4 , y′(0) = 2
23. y′′− 2y′+ 2 y= −ex (6 cos x+ 4 sin x), y (0) = 1 , y′(0) = 4
24. y′′+ 6 y′+ 10 y= −40ex sin x, y (0) = 2 , y ′(0) = −3
25. y′′− 6y′+ 10 y= −e3x(6 cos x+ 4 sin x), y (0) = 2 , y ′(0) = 7
26. y′′− 3y′+ 2 y= e3x [21 cos x− (11 + 10 x) sin x] , y(0) = 0 , y ′(0) = 6
In Exercises 27– 32 use the principle of superposition to find a particular solut ion. Where indicated, solve
the initial value problem.
27. y′′− 2y′− 3y= 4 e3x + ex(cos x− 2 sin x)
28. y′′+ y = 4 cos x− 2 sin x+ xex + e−x
29. y′′− 3y′+ 2 y= xex + 2 e2x + sin x
30. y′′− 2y′+ 2 y= 4 xex cos x+ xe−x + 1 + x2
31. y′′− 4y′+ 4 y= e2x(1 + x) + e2x(cos x− sin x) + 3 e3x + 1 + x
32. y′′− 4y′+ 4 y= 6 e2x + 25 sin x, y (0) = 5 , y′(0) = 3
In Exercises 33– 35 solve the initial value problem and graph the solution.",Elementary Differential Equations with Boundary Value Problems.pdf
"32. y′′− 4y′+ 4 y= 6 e2x + 25 sin x, y (0) = 5 , y′(0) = 3
In Exercises 33– 35 solve the initial value problem and graph the solution.
33. C/G y′′+ 4 y= −e−2x [(4 − 7x) cos x+ (2 − 4x) sin x] , y(0) = 3 , y ′(0) = 1
34. C/G y′′+ 4 y′+ 4 y= 2 cos 2 x+ 3 sin 2 x+ e−x, y (0) = −1, y′(0) = 2",Elementary Differential Equations with Boundary Value Problems.pdf
"246 Chapter 5 Linear Second Order Equations
35. C/G y′′+ 4 y= ex(11 + 15 x) + 8 cos 2 x− 12 sin 2 x, y (0) = 3 , y′(0) = 5
36. (a) V erify that if
yp = A(x) cos ωx + B(x) sin ωx
where Aand Bare twice differentiable, then
y′
p = ( A′+ ωB ) cos ωx + ( B′− ωA) sin ωx and
y′′
p = ( A′′+ 2 ωB ′− ω2A) cos ωx + ( B′′− 2ωA′− ω2B) sin ωx.
(b) Use the results of (a) to verify that
ay′′
p + by′
p + cyp = [ (c− aω2)A+ bωB + 2 aωB ′+ bA′+ aA′′] cos ωx +
[
−bωA + ( c− aω2)B− 2aωA′+ bB′+ aB′′]
sin ωx.
(c) Use the results of (a) to verify that
y′′
p + ω2yp = ( A′′+ 2 ωB ′) cos ωx + ( B′′− 2ωA′) sin ωx.
(d) Prove Theorem 5.5.2.
37. Let a, b, c, and ω be constants, with a̸= 0 and ω >0, and let
P(x) = p0 + p1x+ · · ·+ pkxk and Q(x) = q0 + q1x+ · · ·+ qkxk,
where at least one of the coefficients pk, qk is nonzero, so kis the larger of the degrees of P and Q.
(a) Show that if cos ωx and sin ωx are not solutions of the complementary equation
ay′′+ by′+ cy = 0 ,
then there are polynomials",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) Show that if cos ωx and sin ωx are not solutions of the complementary equation
ay′′+ by′+ cy = 0 ,
then there are polynomials
A(x) = A0 + A1x+ · · ·+ Ak xk and B(x) = B0 + B1x+ · · ·+ Bk xk (A)
such that
(c− aω2)A+ bωB + 2 aωB ′+ bA′+ aA′′ = P
−bωA + ( c− aω2)B− 2aωA′+ bB′+ aB′′ = Q,
where (Ak ,B k), (Ak−1,B k−1), . . . , (A0 ,B 0) can be computed successively by solving the
systems
(c− aω2)Ak + bωBk = pk
−bωAk + ( c− aω2)Bk = qk,
and, if 1 ≤ r≤ k,
(c− aω2)Ak−r + bωBk−r = pk−r + · · ·
−bωAk−r + ( c− aω2)Bk−r = qk−r + · · ·,
where the terms indicated by “ · · ·” depend upon the previously computed coefficients with
subscripts greater than k− r. Conclude from this and Exercise 36(b) that
yp = A(x) cos ωx + B(x) sin ωx (B)
is a particular solution of
ay′′+ by′+ cy= P(x) cos ωx + Q(x) sin ωx.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.5 The Method of Undetermined Coefficients II 247
(b) Conclude from Exercise 36(c) that the equation
a(y′′+ ω2y) = P(x) cos ωx + Q(x) sin ωx (C)
does not have a solution of the form (B) with Aand B as in (A). Then show that there are
polynomials
A(x) = A0x+ A1x2 + · · ·+ Ak xk+1 and B(x) = B0x+ B1x2 + · · ·+ Bk xk+1
such that
a(A′′+ 2 ωB ′) = P
a(B′′− 2ωA′) = Q,
where the pairs (Ak ,B k), (Ak−1,B k−1), . . . , (A0,B 0) can be computed successively as
follows:
Ak = − qk
2aω(k+ 1)
Bk = pk
2aω(k+ 1) ,
and, if k≥ 1,
Ak−j = − 1
2ω
[ qk−j
a(k− j+ 1) − (k− j+ 2) Bk−j+1
]
Bk−j = 1
2ω
[ pk−j
a(k− j+ 1) − (k− j+ 2) Ak−j+1
]
for 1 ≤ j ≤ k. Conclude that (B) with this choice of the polynomials Aand Bis a particular
solution of (C).
38. Show that Theorem 5.5.1 implies the next theorem: Suppose ω is a positive number and P and Q
are polynomials. Let kbe the larger of the degrees of P and Q. Then the equation
ay′′+ by′+ cy= eλx (P(x) cos ωx + Q(x) sin ωx)
has a particular solution",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy= eλx (P(x) cos ωx + Q(x) sin ωx)
has a particular solution
yp = eλx (A(x) cos ωx + B(x) sin ωx) , (A)
where
A(x) = A0 + A1x+ · · ·+ Akxk and B(x) = B0 + B1x+ · · ·+ Bk xk,
provided that eλx cos ωx and eλx sin ωx are not solutions of the complementary equation. The
equation
a
[
y′′− 2λy′+ ( λ2 + ω2)y
]
= eλx (P(x) cos ωx + Q(x) sin ωx)
(for which eλx cos ωx and eλx sin ωx are solutions of the complementary equation ) has a partic-
ular solution of the form (A), where
A(x) = A0x+ A1x2 + · · ·+ Ak xk+1 and B(x) = B0x+ B1x2 + · · ·+ Bk xk+1.",Elementary Differential Equations with Boundary Value Problems.pdf
"248 Chapter 5 Linear Second Order Equations
39. This exercise presents a method for evaluating the integral
y =
∫
eλx (P(x) cos ωx + Q(x) sin ωx) dx
where ω ̸= 0 and
P(x) = p0 + p1x+ · · ·+ pkxk, Q (x) = q0 + q1x+ · · ·+ qkxk.
(a) Show that y = eλx u, where
u′+ λu = P(x) cos ωx + Q(x) sin ωx. (A)
(b) Show that (A) has a particular solution of the form
up = A(x) cos ωx + B(x) sin ωx,
where
A(x) = A0 + A1x+ · · ·+ Ak xk, B (x) = B0 + B1 x+ · · ·+ Bk xk,
and the pairs of coefficients (Ak,B k ), (Ak−1,B k−1), . . . , (A0 ,B 0) can be computed succes-
sively as the solutions of pairs of equations obtained by equ ating the coefficients of xr cos ωx
and xr sin ωx for r= k, k− 1, . . . , 0.
(c) Conclude that
∫
eλx (P(x) cos ωx + Q(x) sin ωx) dx= eλx (A(x) cos ωx + B(x) sin ωx) + c,
where cis a constant of integration.
40. Use the method of Exercise 39 to evaluate the integral.
(a)
∫
x2 cos xdx (b)
∫
x2ex cos xdx
(c) ∫ xe−x sin 2 xdx (d) ∫ x2e−x sin xdx",Elementary Differential Equations with Boundary Value Problems.pdf
"40. Use the method of Exercise 39 to evaluate the integral.
(a)
∫
x2 cos xdx (b)
∫
x2ex cos xdx
(c) ∫ xe−x sin 2 xdx (d) ∫ x2e−x sin xdx
(e) ∫ x3ex sin xdx (f) ∫ ex [xcos x− (1 + 3 x) sin x] dx
(g) ∫ e−x [ (1 + x2) cos x+ (1 − x2) sin x] dx
5.6 REDUCTION OF ORDER
In this section we give a method for finding the general soluti on of
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x) (5.6.1)
if we know a nontrivial solution y1 of the complementary equation
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 . (5.6.2)
The method is called reduction of order because it reduces the task of solving ( 5.6.1) to solving a first
order equation. Unlike the method of undetermined coefficie nts, it does not require P0, P1, and P2 to be
constants, or F to be of any special form.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.6 Reduction of Order 249
By now you shoudn’t be surprised that we look for solutions of (5.6.1) in the form
y = uy1 (5.6.3)
where uis to be determined so that ysatisfies ( 5.6.1). Substituting ( 5.6.3) and
y′ = u′y1 + uy′
1
y′′ = u′′y1 + 2 u′y′
1 + uy′′
1
into ( 5.6.1) yields
P0(x)(u′′y1 + 2 u′y′
1 + uy′′
1 ) + P1(x)(u′y1 + uy′
1) + P2(x)uy1 = F(x).
Collecting the coefficients of u, u′, and u′′yields
(P0y1)u′′+ (2 P0y′
1 + P1y1)u′+ ( P0y′′
1 + P1y′
1 + P2y1)u= F. (5.6.4)
However, the coefficient of uis zero, since y1 satisfies ( 5.6.2). Therefore ( 5.6.4) reduces to
Q0(x)u′′+ Q1(x)u′= F, (5.6.5)
with
Q0 = P0y1 and Q1 = 2 P0y′
1 + P1y1.
(It isn’t worthwhile to memorize the formulas for Q0 and Q1!) Since ( 5.6.5) is a linear first order equation
in u′, we can solve it for u′by variation of parameters as in Section 1.2, integrate the s olution to obtain
u, and then obtain yfrom ( 5.6.3).
Example 5.6.1
(a) Find the general solution of
xy′′− (2x+ 1) y′+ ( x+ 1) y = x2, (5.6.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"u, and then obtain yfrom ( 5.6.3).
Example 5.6.1
(a) Find the general solution of
xy′′− (2x+ 1) y′+ ( x+ 1) y = x2, (5.6.6)
given that y1 = ex is a solution of the complementary equation
xy′′− (2x+ 1) y′+ ( x+ 1) y= 0 . (5.6.7)
(b) As a byproduct of (a), find a fundamental set of solutions of ( 5.6.7).
SO LU TIO N (a) If y= uex, then y′= u′ex + uex and y′′= u′′ex + 2 u′ex + uex, so
xy′′− (2x+ 1) y′+ ( x+ 1) y = x(u′′ex + 2 u′ex + uex)
−(2x+ 1)( u′ex + uex) + ( x+ 1) uex
= ( xu′′− u′)ex.
Therefore y = uex is a solution of ( 5.6.6) if and only if
(xu′′− u′)ex = x2,
which is a first order equation in u′. W e rewrite it as
u′′− u′
x = xe−x. (5.6.8)",Elementary Differential Equations with Boundary Value Problems.pdf
"250 Chapter 5 Linear Second Order Equations
T o focus on how we apply variation of parameters to this equat ion, we temporarily write z = u′, so that
(5.6.8) becomes
z′− z
x = xe−x. (5.6.9)
W e leave it to you to show (by separation of variables) that z1 = xis a solution of the complementary
equation
z′− z
x = 0
for ( 5.6.9). By applying variation of parameters as in Section 1.2, we c an now see that every solution of
(5.6.9) is of the form
z= vx where v′x= xe−x, so v′= e−x and v= −e−x + C1.
Since u′= z= vx, uis a solution of ( 5.6.8) if and only if
u′= vx= −xe−x + C1x.
Integrating this yields
u= ( x+ 1) e−x + C1
2 x2 + C2.
Therefore the general solution of ( 5.6.6) is
y = uex = x+ 1 + C1
2 x2ex + C2ex. (5.6.10)
SO LU TIO N (b) By letting C1 = C2 = 0 in ( 5.6.10), we see that yp1 = x+ 1 is a solution of ( 5.6.6). By
letting C1 = 2 and C2 = 0 , we see that yp2 = x+1+x2ex is also a solution of ( 5.6.6). Since the difference",Elementary Differential Equations with Boundary Value Problems.pdf
"letting C1 = 2 and C2 = 0 , we see that yp2 = x+1+x2ex is also a solution of ( 5.6.6). Since the difference
of two solutions of ( 5.6.6) is a solution of ( 5.6.7), y2 = yp1 − yp2 = x2ex is a solution of ( 5.6.7). Since
y2/y 1 is nonconstant and we already know that y1 = ex is a solution of ( 5.6.6), Theorem 5.1.6 implies
that {ex,x 2ex}is a fundamental set of solutions of ( 5.6.7).
Although ( 5.6.10) is a correct form for the general solution of ( 5.6.6), it’s silly to leave the arbitrary co-
efficient of x2ex as C1/ 2 where C1 is an arbitrary constant. Moreover, it’s sensible to make th e subscripts
of the coefficients of y1 = ex and y2 = x2ex consistent with the subscripts of the functions themselves .
Therefore we rewrite ( 5.6.10) as
y= x+ 1 + c1ex + c2x2ex
by simply renaming the arbitrary constants. W e’ll also do th is in the next two examples, and in the
answers to the exercises.
Example 5.6.2
(a) Find the general solution of
x2y′′+ xy′− y= x2 + 1 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"answers to the exercises.
Example 5.6.2
(a) Find the general solution of
x2y′′+ xy′− y= x2 + 1 ,
given that y1 = xis a solution of the complementary equation
x2y′′+ xy′− y = 0 . (5.6.11)
As a byproduct of this result, find a fundamental set of soluti ons of (
5.6.11).
(b) Solve the initial value problem
x2y′′+ xy′− y= x2 + 1 , y (1) = 2 , y′(1) = −3. (5.6.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.6 Reduction of Order 251
SO LU TIO N (a) If y= ux, then y′= u′x+ uand y′′= u′′x+ 2 u′, so
x2y′′+ xy′− y = x2(u′′x+ 2 u′) + x(u′x+ u) − ux
= x3u′′+ 3 x2u′.
Therefore y = uxis a solution of ( 5.6.12) if and only if
x3u′′+ 3 x2u′= x2 + 1 ,
which is a first order equation in u′. W e rewrite it as
u′′+ 3
xu′= 1
x + 1
x3 . (5.6.13)
T o focus on how we apply variation of parameters to this equat ion, we temporarily write z = u′, so that
(5.6.13) becomes
z′+ 3
xz= 1
x + 1
x3 . (5.6.14)
W e leave it to you to show by separation of variables that z1 = 1 /x 3 is a solution of the complementary
equation
z′+ 3
xz = 0
for ( 5.6.14). By variation of parameters, every solution of ( 5.6.14) is of the form
z = v
x3 where v′
x3 = 1
x + 1
x3 , so v′= x2 + 1 and v= x3
3 + x+ C1.
Since u′= z= v/x 3, uis a solution of ( 5.6.14) if and only if
u′= v
x3 = 1
3 + 1
x2 + C1
x3 .
Integrating this yields
u= x
3 − 1
x − C1
2x2 + C2.
Therefore the general solution of ( 5.6.12) is
y= ux= x2
3 − 1 − C1",Elementary Differential Equations with Boundary Value Problems.pdf
"x3 = 1
3 + 1
x2 + C1
x3 .
Integrating this yields
u= x
3 − 1
x − C1
2x2 + C2.
Therefore the general solution of ( 5.6.12) is
y= ux= x2
3 − 1 − C1
2x + C2x. (5.6.15)
Reasoning as in the solution of Example 5.6.1(a), we conclude that y1 = x and y2 = 1 /x form a
fundamental set of solutions for ( 5.6.11).
As we explained above, we rename the constants in ( 5.6.15) and rewrite it as
y= x2
3 − 1 + c1x+ c2
x. (5.6.16)
SO LU TIO N (b) Differentiating ( 5.6.16) yields
y′= 2x
3 + c1 − c2
x2 . (5.6.17)",Elementary Differential Equations with Boundary Value Problems.pdf
"252 Chapter 5 Linear Second Order Equations
Setting x = 1 in ( 5.6.16) and ( 5.6.17) and imposing the initial conditions y(1) = 2 and y′(1) = −3
yields
c1 + c2 = 8
3
c1 − c2 = −11
3 .
Solving these equations yields c1 = −1/ 2, c2 = 19 / 6. Therefore the solution of ( 5.6.12) is
y = x2
3 − 1 − x
2 + 19
6x.
Using reduction of order to find the general solution of a homo geneous linear second order equation
leads to a homogeneous linear first order equation in u′that can be solved by separation of variables. The
next example illustrates this.
Example 5.6.3 Find the general solution and a fundamental set of solutions of
x2y′′− 3xy′+ 3 y= 0 , (5.6.18)
given that y1 = xis a solution.
Solution If y= uxthen y′= u′x+ uand y′′= u′′x+ 2 u′, so
x2y′′− 3xy′+ 3 y = x2(u′′x+ 2 u′) − 3x(u′x+ u) + 3 ux
= x3u′′− x2u′.
Therefore y = uxis a solution of (
5.6.18) if and only if
x3u′′− x2u′= 0 .
Separating the variables u′and xyields
u′′
u′ = 1
x,
so
ln |u′|= ln |x|+ k, or, equivalently , u′= C1x.
Therefore",Elementary Differential Equations with Boundary Value Problems.pdf
"x3u′′− x2u′= 0 .
Separating the variables u′and xyields
u′′
u′ = 1
x,
so
ln |u′|= ln |x|+ k, or, equivalently , u′= C1x.
Therefore
u= C1
2 x2 + C2,
so the general solution of ( 5.6.18) is
y= ux= C1
2 x3 + C2x,
which we rewrite as
y= c1x+ c2x3.
Therefore {x,x 3}is a fundamental set of solutions of ( 5.6.18).
5.6 Exercises
In Exercises 1– 17 find the general solution, given that y1 satisfies the complementary equation. As a
byproduct, find a fundamental set of solutions of the complem entary equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.6 Reduction of Order 253
1. (2x+ 1) y′′− 2y′− (2x+ 3) y= (2 x+ 1) 2; y1 = e−x
2. x2y′′+ xy′− y= 4
x2 ; y1 = x
3. x2y′′− xy′+ y= x; y1 = x
4. y′′− 3y′+ 2 y= 1
1 + e−x ; y1 = e2x
5. y′′− 2y′+ y = 7 x3/ 2ex; y1 = ex
6. 4x2y′′+ (4 x− 8x2)y′+ (4 x2 − 4x− 1)y = 4 x1/ 2ex(1 + 4 x); y1 = x1/ 2ex
7. y′′− 2y′+ 2 y= ex sec x; y1 = ex cos x
8. y′′+ 4 xy′+ (4 x2 + 2) y= 8 e−x(x+2); y1 = e−x2
9. x2y′′+ xy′− 4y= −6x− 4; y1 = x2
10. x2y′′+ 2 x(x− 1)y′+ ( x2 − 2x+ 2) y = x3e2x; y1 = xe−x
11. x2y′′− x(2x− 1)y′+ ( x2 − x− 1)y= x2ex; y1 = xex
12. (1 − 2x)y′′+ 2 y′+ (2 x− 3)y= (1 − 4x+ 4 x2)ex; y1 = ex
13. x2y′′− 3xy′+ 4 y = 4 x4; y1 = x2
14. 2xy′′+ (4 x+ 1) y′+ (2 x+ 1) y= 3 x1/ 2e−x ; y1 = e−x
15. xy′′− (2x+ 1) y′+ ( x+ 1) y= −ex ; y1 = ex
16. 4x2y′′− 4x(x+ 1) y′+ (2 x+ 3) y= 4 x5/ 2e2x; y1 = x1/ 2
17. x2y′′− 5xy′+ 8 y = 4 x2; y1 = x2
In Exercises 18– 30 find a fundamental set of solutions, given that y1 is a solution.
18. xy′′+ (2 − 2x)y′+ ( x− 2)y= 0; y1 = ex
19. x2y′′− 4xy′+ 6 y = 0; y1 = x2",Elementary Differential Equations with Boundary Value Problems.pdf
"18. xy′′+ (2 − 2x)y′+ ( x− 2)y= 0; y1 = ex
19. x2y′′− 4xy′+ 6 y = 0; y1 = x2
20. x2(ln |x|)2y′′− (2xln |x|)y′+ (2 + ln |x|)y= 0; y1 = ln |x|
21. 4xy′′+ 2 y′+ y= 0; y1 = sin √x
22. xy′′− (2x+ 2) y′+ ( x+ 2) y= 0; y1 = ex
23. x2y′′− (2a− 1)xy′+ a2y= 0; y1 = xa
24. x2y′′− 2xy′+ ( x2 + 2) y= 0; y1 = xsin x
25. xy′′− (4x+ 1) y′+ (4 x+ 2) y= 0; y1 = e2x
26. 4x2(sin x)y′′− 4x(xcos x+ sin x)y′+ (2 xcos x+ 3 sin x)y= 0; y1 = x1/ 2
27. 4x2y′′− 4xy′+ (3 − 16x2)y= 0; y1 = x1/ 2e2x
28. (2x+ 1) xy′′− 2(2x2 − 1)y′− 4(x+ 1) y= 0; y1 = 1 /x
29. (x2 − 2x)y′′+ (2 − x2)y′+ (2 x− 2)y= 0; y1 = ex
30. xy′′− (4x+ 1) y′+ (4 x+ 2) y= 0; y1 = e2x
In Exercises 31– 33 solve the initial value problem, given that y1 satisfies the complementary equation.
31. x2y′′− 3xy′+ 4 y = 4 x4, y (−1) = 7 , y ′(−1) = −8; y1 = x2
32. (3x− 1)y′′− (3x+ 2) y′− (6x− 8)y = 0 , y (0) = 2 , y′(0) = 3; y1 = e2x",Elementary Differential Equations with Boundary Value Problems.pdf
"254 Chapter 5 Linear Second Order Equations
33. (x+ 1) 2y′′− 2(x+ 1) y′− (x2 + 2 x− 1)y = ( x+ 1) 3ex, y (0) = 1 , y ′(0) = − 1;
y1 = ( x+ 1) ex
In Exercises 34 and 35 solve the initial value problem and graph the solution, give n that y1 satisfies the
complementary equation.
34. C/G x2y′′+ 2 xy′− 2y= x2, y (1) = 5
4 , y′(1) = 3
2 ; y1 = x
35. C/G (x2 − 4)y′′+ 4 xy′+ 2 y= x+ 2 , y (0) = −1
3 , y ′(0) = −1; y1 = 1
x− 2
36. Suppose p1 and p2 are continuous on (a,b ). Let y1 be a solution of
y′′+ p1(x)y′+ p2(x)y= 0 (A)
that has no zeros on (a,b ), and let x0 be in (a,b ). Use reduction of order to show that y1 and
y2(x) = y1(x)
∫ x
x0
1
y2
1 (t) exp
(
−
∫ t
x0
p1(s) ds
)
dt
form a fundamental set of solutions of (A) on (a,b ). (N OTE : This exercise is related to Exercise 9.)
37. The nonlinear first order equation
y′+ y2 + p(x)y+ q(x) = 0 (A)
is a Riccati equation . (See Exercise 2.4. 55.) Assume that pand qare continuous.
(a) Show that yis a solution of (A) if and only if y= z′/z , where",Elementary Differential Equations with Boundary Value Problems.pdf
"is a Riccati equation . (See Exercise 2.4. 55.) Assume that pand qare continuous.
(a) Show that yis a solution of (A) if and only if y= z′/z , where
z′′+ p(x)z′+ q(x)z= 0 . (B)
(b) Show that the general solution of (A) is
y= c1z′
1 + c2z′
2
c1z1 + c2z2
, (C)
where {z1,z 2}is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants.
(c) Does the formula (C) imply that the first order equation (A) ha s a two–parameter family of
solutions? Explain your answer.
38. Use a method suggested by Exercise 37 to find all solutions. of the equation.
(a) y′+ y2 + k2 = 0 (b) y′+ y2 − 3y+ 2 = 0
(c) y′+ y2 + 5 y− 6 = 0 (d) y′+ y2 + 8 y+ 7 = 0
(e) y′+ y2 + 14 y+ 50 = 0 (f) 6y′+ 6 y2 − y− 1 = 0
(g) 36y′+ 36 y2 − 12y+ 1 = 0
39. Use a method suggested by Exercise 37 and reduction of order to find all solutions of the equation,
given that y1 is a solution.
(a) x2(y′+ y2) − x(x+ 2) y+ x+ 2 = 0; y1 = 1 /x
(b) y′+ y2 + 4 xy+ 4 x2 + 2 = 0; y1 = −2x",Elementary Differential Equations with Boundary Value Problems.pdf
"given that y1 is a solution.
(a) x2(y′+ y2) − x(x+ 2) y+ x+ 2 = 0; y1 = 1 /x
(b) y′+ y2 + 4 xy+ 4 x2 + 2 = 0; y1 = −2x
(c) (2x+ 1)( y′+ y2) − 2y− (2x+ 3) = 0; y1 = −1
(d) (3x− 1)(y′+ y2) − (3x+ 2) y− 6x+ 8 = 0; y1 = 2",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.7 V ariation of Parameters 255
(e) x2(y′+ y2) + xy+ x2 − 1
4 = 0; y1 = − tan x− 1
2x
(f) x2(y′+ y2) − 7xy+ 7 = 0; y1 = 1 /x
40. The nonlinear first order equation
y′+ r(x)y2 + p(x)y+ q(x) = 0 (A)
is the generalized Riccati equation . (See Exercise 2.4. 55.) Assume that pand q are continuous
and ris differentiable.
(a) Show that yis a solution of (A) if and only if y= z′/rz , where
z′′+
[
p(x) − r′(x)
r(x)
]
z′+ r(x)q(x)z = 0 . (B)
(b) Show that the general solution of (A) is
y= c1z′
1 + c2z′
2
r(c1z1 + c2z2) ,
where {z1,z 2}is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants.
5.7 V ARIA TION OF P ARAMETERS
In this section we give a method called variation of parameters for finding a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x) (5.7.1)
if we know a fundamental set {y1,y 2}of solutions of the complementary equation
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 . (5.7.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"if we know a fundamental set {y1,y 2}of solutions of the complementary equation
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 . (5.7.2)
Having found a particular solution yp by this method, we can write the general solution of ( 5.7.1) as
y= yp + c1y1 + c2y2.
Since we need only one nontrivial solution of ( 5.7.2) to find the general solution of ( 5.7.1) by reduction
of order, it’s natural to ask why we’re interested in variati on of parameters, which requires two linearly
independent solutions of ( 5.7.2) to achieve the same goal. Here’s the answer:
• If we already know two linearly independent solutions of ( 5.7.2) then variation of parameters will
probably be simpler than reduction of order.
• V ariation of parameters generalizes naturally to a method for finding particular solutions of higher
order linear equations (Section 9.4) and linear systems of e quations (Section 10.7), while reduction
of order doesn’t.",Elementary Differential Equations with Boundary Value Problems.pdf
"order linear equations (Section 9.4) and linear systems of e quations (Section 10.7), while reduction
of order doesn’t.
• V ariation of parameters is a powerful theoretical tool use d by researchers in differential equations.
Although a detailed discussion of this is beyond the scope of this book, you can get an idea of what
it means from Exercises 37– 39.",Elementary Differential Equations with Boundary Value Problems.pdf
"256 Chapter 5 Linear Second Order Equations
W e’ll now derive the method. As usual, we consider solutions of ( 5.7.1) and ( 5.7.2) on an interval (a,b )
where P0, P1, P2, and F are continuous and P0 has no zeros. Suppose that {y1,y 2}is a fundamental
set of solutions of the complementary equation ( 5.7.2). W e look for a particular solution of ( 5.7.1) in the
form
yp = u1y1 + u2y2 (5.7.3)
where u1 and u2 are functions to be determined so that yp satisfies ( 5.7.1). Y ou may not think this is a
good idea, since there are now two unknown functions to be det ermined, rather than one. However, since
u1 and u2 have to satisfy only one condition (that yp is a solution of ( 5.7.1)), we can impose a second
condition that produces a convenient simplification, as fol lows.
Differentiating ( 5.7.3) yields
y′
p = u1y′
1 + u2y′
2 + u′
1y1 + u′
2y2. (5.7.4)
As our second condition on u1 and u2 we require that
u′
1y1 + u′
2y2 = 0 . (5.7.5)
Then (
5.7.4) becomes
y′
p = u1y′
1 + u2y′
2; (5.7.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"2y2. (5.7.4)
As our second condition on u1 and u2 we require that
u′
1y1 + u′
2y2 = 0 . (5.7.5)
Then (
5.7.4) becomes
y′
p = u1y′
1 + u2y′
2; (5.7.6)
that is, ( 5.7.5) permits us to differentiate yp (once!) as if u1 and u2 are constants. Differentiating ( 5.7.4)
yields
y′′
p = u1y′′
1 + u2y′′
2 + u′
1y′
1 + u′
2y′
2. (5.7.7)
(There are no terms involving u′′
1 and u′′
2 here, as there would be if we hadn’t required (
5.7.5).) Substitut-
ing ( 5.7.3), ( 5.7.6), and ( 5.7.7) into ( 5.7.1) and collecting the coefficients of u1 and u2 yields
u1(P0y′′
1 + P1y′
1 + P2y1) + u2(P0y′′
2 + P1y′
2 + P2y2) + P0(u′
1y′
1 + u′
2y′
2) = F.
As in the derivation of the method of reduction of order, the c oefficients of u1 and u2 here are both zero
because y1 and y2 satisfy the complementary equation. Hence, we can rewrite t he last equation as
P0(u′
1y′
1 + u′
2y′
2) = F. (5.7.8)
Therefore yp in (
5.7.3) satisfies ( 5.7.1) if
u′
1y1 + u′
2y2 = 0
u′
1y′
1 + u′
2y′
2 = F
P0
, (5.7.9)",Elementary Differential Equations with Boundary Value Problems.pdf
"P0(u′
1y′
1 + u′
2y′
2) = F. (5.7.8)
Therefore yp in (
5.7.3) satisfies ( 5.7.1) if
u′
1y1 + u′
2y2 = 0
u′
1y′
1 + u′
2y′
2 = F
P0
, (5.7.9)
where the first equation is the same as ( 5.7.5) and the second is from ( 5.7.8).
W e’ll now show that you can always solve ( 5.7.9) for u′
1 and u′
2. (The method that we use here will
always work, but simpler methods usually work when you’re de aling with specific equations.) T o obtain
u′
1, multiply the first equation in (
5.7.9) by y′
2 and the second equation by y2. This yields
u′
1y1y′
2 + u′
2y2y′
2 = 0
u′
1y′
1y2 + u′
2y′
2y2 = Fy2
P0
.
Subtracting the second equation from the first yields
u′
1(y1y′
2 − y′
1y2) = −Fy2
P0
. (5.7.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.7 V ariation of Parameters 257
Since {y1,y 2}is a fundamental set of solutions of ( 5.7.2) on (a,b ), Theorem 5.1.6 implies that the
Wronskian y1y′
2 − y′
1y2 has no zeros on (a,b ). Therefore we can solve ( 5.7.10) for u′
1, to obtain
u′
1 = − Fy2
P0(y1y′
2 − y′
1y2) . (5.7.11)
W e leave it to you to start from ( 5.7.9) and show by a similar argument that
u′
2 = Fy1
P0(y1y′
2 − y′
1y2) . (5.7.12)
W e can now obtain u1 and u2 by integrating u′
1 and u′
2. The constants of integration can be taken to be
zero, since any choice of u1 and u2 in (
5.7.3) will suffice.
Y ou should not memorize ( 5.7.11) and ( 5.7.12). On the other hand, you don’t want to rederive the
whole procedure for every specific problem. W e recommend the a compromise:
(a) Write
yp = u1y1 + u2y2 (5.7.13)
to remind yourself of what you’re doing.
(b) Write the system
u′
1y1 + u′
2y2 = 0
u′
1y′
1 + u′
2y′
2 = F
P0
(5.7.14)
for the specific problem you’re trying to solve.
(c) Solve ( 5.7.14) for u′
1 and u′",Elementary Differential Equations with Boundary Value Problems.pdf
"u′
1y1 + u′
2y2 = 0
u′
1y′
1 + u′
2y′
2 = F
P0
(5.7.14)
for the specific problem you’re trying to solve.
(c) Solve ( 5.7.14) for u′
1 and u′
2 by any convenient method.
(d) Obtain u1 and u2 by integrating u′
1 and u′
2, taking the constants of integration to be zero.
(e) Substitute u1 and u2 into (
5.7.13) to obtain yp.
Example 5.7.1 Find a particular solution yp of
x2y′′− 2xy′+ 2 y= x9/ 2, (5.7.15)
given that y1 = xand y2 = x2 are solutions of the complementary equation
x2y′′− 2xy′+ 2 y= 0 .
Then find the general solution of ( 5.7.15).
Solution W e set
yp = u1x+ u2x2,
where
u′
1x+ u′
2x2 = 0
u′
1 + 2 u′
2x = x9/ 2
x2 = x5/ 2.
From the first equation, u′
1 = −u′
2x. Substituting this into the second equation yields u′
2x = x5/ 2, so
u′
2 = x3/ 2 and therefore u′
1 = −u′
2x= −x5/ 2. Integrating and taking the constants of integration to be
zero yields
u1 = −2
7 x7/ 2 and u2 = 2
5 x5/ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"258 Chapter 5 Linear Second Order Equations
Therefore
yp = u1x+ u2x2 = −2
7 x7/ 2x+ 2
5 x5/ 2x2 = 4
35 x9/ 2,
and the general solution of ( 5.7.15) is
y= 4
35 x9/ 2 + c1x+ c2x2.
Example 5.7.2 Find a particular solution yp of
(x− 1)y′′− xy′+ y= ( x− 1)2, (5.7.16)
given that y1 = xand y2 = ex are solutions of the complementary equation
(x− 1)y′′− xy′+ y = 0 .
Then find the general solution of ( 5.7.16).
Solution W e set
yp = u1x+ u2ex,
where
u′
1x+ u′
2ex = 0
u′
1 + u′
2ex = (x− 1)2
x− 1 = x− 1.
Subtracting the first equation from the second yields −u′
1(x− 1) = x− 1, so u′
1 = −1. From this and
the first equation, u′
2 = −xe−x u′
1 = xe−x. Integrating and taking the constants of integration to be z ero
yields
u1 = −x and u2 = −(x+ 1) e−x.
Therefore
yp = u1x+ u2ex = ( −x)x+ ( −(x+ 1) e−x)ex = −x2 − x− 1,
so the general solution of (
5.7.16) is
y= yp + c1x+ c2ex = −x2 − x− 1 + c1x+ c2ex = −x2 − 1 + ( c1 − 1)x+ c2ex. (5.7.17)",Elementary Differential Equations with Boundary Value Problems.pdf
"so the general solution of (
5.7.16) is
y= yp + c1x+ c2ex = −x2 − x− 1 + c1x+ c2ex = −x2 − 1 + ( c1 − 1)x+ c2ex. (5.7.17)
However, since c1 is an arbitrary constant, so is c1 −1; therefore, we improve the appearance of this result
by renaming the constant and writing the general solution as
y= −x2 − 1 + c1x+ c2ex. (5.7.18)
There’s nothing wrong with leaving the general solution of ( 5.7.16) in the form ( 5.7.17); however, we
think you’ll agree that ( 5.7.18) is preferable. W e can also view the transition from ( 5.7.17) to ( 5.7.18)
differently . In this example the particular solution yp = −x2−x−1 contained the term −x, which satisfies
the complementary equation. W e can drop this term and redefin e yp = −x2 − 1, since −x2 − x− 1 is a
solution of ( 5.7.16) and xis a solution of the complementary equation; hence, −x2 −1 = ( −x2−x−1)+x
is also a solution of ( 5.7.16). In general, it’s always legitimate to drop linear combina tions of {y1,y 2}",Elementary Differential Equations with Boundary Value Problems.pdf
"is also a solution of ( 5.7.16). In general, it’s always legitimate to drop linear combina tions of {y1,y 2}
from particular solutions obtained by variation of paramet ers. (See Exercise 36 for a general discussion
of this question.) W e’ll do this in the following examples an d in the answers to exercises that ask for a
particular solution. Therefore, don’t be concerned if your answer to such an exercise differs from ours
only by a solution of the complementary equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.7 V ariation of Parameters 259
Example 5.7.3 Find a particular solution of
y′′+ 3 y′+ 2 y= 1
1 + ex . (5.7.19)
Then find the general solution.
Solution
The characteristic polynomial of the complementary equati on
y′′+ 3 y′+ 2 y = 0 (5.7.20)
is p(r) = r2 + 3 r+ 2 = ( r+ 1)( r+ 2) , so y1 = e−x and y2 = e−2x form a fundamental set of solutions
of (
5.7.20). W e look for a particular solution of ( 5.7.19) in the form
yp = u1e−x + u2e−2x,
where
u′
1e−x + u′
2e−2x = 0
−u′
1e−x − 2u′
2e−2x = 1
1 + ex .
Adding these two equations yields
−u′
2e−2x = 1
1 + ex , so u′
2 = − e2x
1 + ex .
From the first equation,
u′
1 = −u′
2e−x = ex
1 + ex .
Integrating by means of the substitution v= ex and taking the constants of integration to be zero yields
u1 =
∫ ex
1 + ex dx=
∫ dv
1 + v = ln(1 + v) = ln(1 + ex)
and
u2 = −
∫ e2x
1 + ex dx= −
∫ v
1 + vdv=
∫ [ 1
1 + v − 1
]
dv
= ln(1 + v) − v= ln(1 + ex) − ex.
Therefore
yp = u1e−x + u2e−2x
= [ln(1 + ex )]e−x + [ln(1 + ex) − ex] e−2x,
so",Elementary Differential Equations with Boundary Value Problems.pdf
"∫ v
1 + vdv=
∫ [ 1
1 + v − 1
]
dv
= ln(1 + v) − v= ln(1 + ex) − ex.
Therefore
yp = u1e−x + u2e−2x
= [ln(1 + ex )]e−x + [ln(1 + ex) − ex] e−2x,
so
yp = (e−x + e−2x) ln(1 + ex) − e−x.
Since the last term on the right satisfies the complementary e quation, we drop it and redefine
yp =
(
e−x + e−2x)
ln(1 + ex ).
The general solution of ( 5.7.19) is
y= yp + c1e−x + c2e−2x =
(
e−x + e−2x)
ln(1 + ex) + c1e−x + c2e−2x.",Elementary Differential Equations with Boundary Value Problems.pdf
"260 Chapter 5 Linear Second Order Equations
Example 5.7.4 Solve the initial value problem
(x2 − 1)y′′+ 4 xy′+ 2 y = 2
x+ 1 , y (0) = −1, y ′(0) = −5, (5.7.21)
given that
y1 = 1
x− 1 and y2 = 1
x+ 1
are solutions of the complementary equation
(x2 − 1)y′′+ 4 xy′+ 2 y= 0 .
Solution W e first use variation of parameters to find a particular solut ion of
(x2 − 1)y′′+ 4 xy′+ 2 y= 2
x+ 1
on (−1, 1) in the form
yp = u1
x− 1 + u2
x+ 1 ,
where
u′
1
x− 1 + u′
2
x+ 1 = 0 (5.7.22)
− u′
1
(x− 1)2 − u′
2
(x+ 1) 2 = 2
(x+ 1)( x2 − 1) .
Multiplying the first equation by 1/ (x− 1) and adding the result to the second equation yields
[ 1
x2 − 1 − 1
(x+ 1) 2
]
u′
2 = 2
(x+ 1)( x2 − 1) . (5.7.23)
Since [ 1
x2 − 1 − 1
(x+ 1) 2
]
= (x+ 1) − (x− 1)
(x+ 1)( x2 − 1) = 2
(x+ 1)( x2 − 1) ,
(5.7.23) implies that u′
2 = 1 . From (
5.7.22),
u′
1 = −x− 1
x+ 1 u′
2 = −x− 1
x+ 1 .
Integrating and taking the constants of integration to be ze ro yields
u1 = −
∫ x− 1
x+ 1 dx= −
∫ x+ 1 − 2
x+ 1 dx
=
∫ [ 2
x+ 1 − 1
]",Elementary Differential Equations with Boundary Value Problems.pdf
"x+ 1 .
Integrating and taking the constants of integration to be ze ro yields
u1 = −
∫ x− 1
x+ 1 dx= −
∫ x+ 1 − 2
x+ 1 dx
=
∫ [ 2
x+ 1 − 1
]
dx= 2 ln( x+ 1) − x
and
u2 =
∫
dx= x.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.7 V ariation of Parameters 261
Therefore
yp = u1
x− 1 + u2
x+ 1 = [2 ln( x+ 1) − x] 1
x− 1 + x 1
x+ 1
= 2 ln(x+ 1)
x− 1 + x
[ 1
x+ 1 − 1
x− 1
]
= 2 ln(x+ 1)
x− 1 − 2x
(x+ 1)( x− 1) .
However, since
2x
(x+ 1)( x− 1) =
[ 1
x+ 1 + 1
x− 1
]
is a solution of the complementary equation, we redefine
yp = 2 ln( x+ 1)
x− 1 .
Therefore the general solution of ( 5.7.24) is
y = 2 ln( x+ 1)
x− 1 + c1
x− 1 + c2
x+ 1 . (5.7.24)
Differentiating this yields
y′= 2
x2 − 1 − 2 ln(x+ 1)
(x− 1)2 − c1
(x− 1)2 − c2
(x+ 1) 2 .
Setting x = 0 in the last two equations and imposing the initial condition s y(0) = −1 and y′(0) = −5
yields the system
−c1 + c2 = −1
−2 − c1 − c2 = −5.
The solution of this system is c1 = 2 , c2 = 1 . Substituting these into ( 5.7.24) yields
y = 2 ln(x+ 1)
x− 1 + 2
x− 1 + 1
x+ 1
= 2 ln(x+ 1)
x− 1 + 3x+ 1
x2 − 1
as the solution of ( 5.7.21). Figure 5.7.1 is a graph of the solution.
Comparison of Methods",Elementary Differential Equations with Boundary Value Problems.pdf
"x− 1 + 2
x− 1 + 1
x+ 1
= 2 ln(x+ 1)
x− 1 + 3x+ 1
x2 − 1
as the solution of ( 5.7.21). Figure 5.7.1 is a graph of the solution.
Comparison of Methods
W e’ve now considered three methods for solving nonhomogene ous linear equations: undetermined co-
efficients, reduction of order, and variation of parameters . It’s natural to ask which method is best for a
given problem. The method of undetermined coefficients shou ld be used for constant coefficient equa-
tions with forcing functions that are linear combinations o f polynomials multiplied by functions of the
form eαx , eλx cos ωx, or eλx sin ωx. Although the other two methods can be used to solve such prob lems,
they will be more difficult except in the most trivial cases, b ecause of the integrations involved.
If the equation isn’t a constant coefficient equation or the f orcing function isn’t of the form just spec-
ified, the method of undetermined coefficients does not apply and the choice is necessarily between the",Elementary Differential Equations with Boundary Value Problems.pdf
"ified, the method of undetermined coefficients does not apply and the choice is necessarily between the
other two methods. The case could be made that reduction of or der is better because it requires only
one solution of the complementary equation while variation of parameters requires two. However, vari-
ation of parameters will probably be easier if you already kn ow a fundamental set of solutions of the
complementary equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"262 Chapter 5 Linear Second Order Equations
1.0−1.0 0.50.5−0.5
10
20
30
40
50
−10
−20
−30
−40
−50
 x
 y
Figure 5.7.1 y= 2 ln(x+ 1)
x− 1 + 3x+ 1
x2 − 1
5.7 Exercises
In Exercises 1– 6 use variation of parameters to find a particular solution.
1. y′′+ 9 y= tan 3 x 2. y′′+ 4 y= sin 2 xsec2 2x
3. y′′− 3y′+ 2 y= 4
1 + e−x
4. y′′− 2y′+ 2 y= 3 ex sec x
5. y′′− 2y′+ y = 14 x3/ 2ex 6. y′′− y= 4e−x
1 − e−2x",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.7 V ariation of Parameters 263
In Exercises 7– 29 use variation of parameters to find a particular solution, gi ven the solutions y1, y2 of
the complementary equation.
7. x2y′′+ xy′− y= 2 x2 + 2; y1 = x, y 2 = 1
x
8. xy′′+ (2 − 2x)y′+ ( x− 2)y= e2x; y1 = ex , y 2 = ex
x
9. 4x2y′′+ (4 x− 8x2)y′+ (4 x2 − 4x− 1)y = 4 x1/ 2ex, x> 0;
y1 = x1/ 2ex, y2 = x−1/ 2ex
10. y′′+ 4 xy′+ (4 x2 + 2) y= 4 e−x(x+2); y1 = e−x2
, y 2 = xe−x2
11. x2y′′− 4xy′+ 6 y = x5/ 2, x> 0; y1 = x2, y2 = x3
12. x2y′′− 3xy′+ 3 y = 2 x4 sin x; y1 = x, y2 = x3
13. (2x+ 1) y′′− 2y′− (2x+ 3) y= (2 x+ 1) 2e−x; y1 = e−x , y 2 = xex
14. 4xy′′+ 2 y′+ y= sin √x; y1 = cos √x, y 2 = sin √x
15. xy′′− (2x+ 2) y′+ ( x+ 2) y= 6 x3ex ; y1 = ex, y 2 = x3ex
16. x2y′′− (2a− 1)xy′+ a2y= xa+1; y1 = xa, y 2 = xa ln x
17. x2y′′− 2xy′+ ( x2 + 2) y= x3 cos x; y1 = xcos x, y 2 = xsin x
18. xy′′− y′− 4x3y = 8 x5; y1 = ex2
, y2 = e−x2
19. (sin x)y′′+ (2 sin x− cos x)y′+ (sin x− cos x)y= e−x; y1 = e−x, y 2 = e−x cos x",Elementary Differential Equations with Boundary Value Problems.pdf
"18. xy′′− y′− 4x3y = 8 x5; y1 = ex2
, y2 = e−x2
19. (sin x)y′′+ (2 sin x− cos x)y′+ (sin x− cos x)y= e−x; y1 = e−x, y 2 = e−x cos x
20. 4x2y′′− 4xy′+ (3 − 16x2)y= 8 x5/ 2; y1 = √xe2x, y2 = √xe−2x
21. 4x2y′′− 4xy′+ (4 x2 + 3) y= x7/ 2; y1 = √xsin x, y2 = √xcos x
22. x2y′′− 2xy′− (x2 − 2)y= 3 x4; y1 = xex, y2 = xe−x
23. x2y′′− 2x(x+ 1) y′+ ( x2 + 2 x+ 2) y = x3ex; y1 = xex, y 2 = x2ex
24. x2y′′− xy′− 3y= x3/ 2; y1 = 1 /x, y 2 = x3
25. x2y′′− x(x+ 4) y′+ 2( x+ 3) y= x4ex; y1 = x2, y 2 = x2ex
26. x2y′′− 2x(x+ 2) y′+ ( x2 + 4 x+ 6) y = 2 xex; y1 = x2ex, y 2 = x3ex
27. x2y′′− 4xy′+ ( x2 + 6) y= x4; y1 = x2 cos x, y 2 = x2 sin x
28. (x− 1)y′′− xy′+ y= 2( x− 1)2ex; y1 = x, y 2 = ex
29. 4x2y′′− 4x(x+ 1) y′+ (2 x+ 3) y= x5/ 2ex; y1 = √x, y 2 = √xex
In Exercises 30– 32 use variation of parameters to solve the initial value probl em, given y1,y 2 are solu-
tions of the complementary equation.
30. (3x− 1)y′′− (3x+ 2) y′− (6x− 8)y = (3 x− 1)2e2x, y (0) = 1 , y′(0) = 2 ;
y1 = e2x, y2 = xe−x",Elementary Differential Equations with Boundary Value Problems.pdf
"tions of the complementary equation.
30. (3x− 1)y′′− (3x+ 2) y′− (6x− 8)y = (3 x− 1)2e2x, y (0) = 1 , y′(0) = 2 ;
y1 = e2x, y2 = xe−x
31. (x− 1)2y′′− 2(x− 1)y′+ 2 y= ( x− 1)2, y (0) = 3 , y ′(0) = −6;
y1 = x− 1, y2 = x2 − 1
32. (x− 1)2y′′− (x2 − 1)y′+ ( x+ 1) y = ( x− 1)3ex, y (0) = 4 , y ′(0) = −6;
y1 = ( x− 1)ex , y 2 = x− 1",Elementary Differential Equations with Boundary Value Problems.pdf
"264 Chapter 5 Linear Second Order Equations
In Exercises 33– 35 use variation of parameters to solve the initial value probl em and graph the solution,
given that y1,y 2 are solutions of the complementary equation.
33. C/G (x2 − 1)y′′+ 4 xy′+ 2 y= 2 x, y (0) = 0 , y′(0) = −2; y1 = 1
x− 1 , y2 = 1
x+ 1
34. C/G x2y′′+ 2 xy′− 2y= −2x2, y (1) = 1 , y′(1) = −1; y1 = x, y2 = 1
x2
35. C/G (x+ 1)(2 x+ 3) y′′+ 2( x+ 2) y′− 2y= (2 x+ 3) 2, y (0) = 0 , y ′(0) = 0 ;
y1 = x+ 2 , y 2 = 1
x+ 1
36. Suppose
yp = y+ a1y1 + a2y2
is a particular solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x), (A)
where y1 and y2 are solutions of the complementary equation
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 .
Show that yis also a solution of (A).
37. Suppose p, q, and f are continuous on (a,b ) and let x0 be in (a,b ). Let y1 and y2 be the solutions
of
y′′+ p(x)y′+ q(x)y= 0
such that
y1(x0) = 1 , y ′
1(x0) = 0 , y 2(x0) = 0 , y ′
2(x0) = 1 .
Use variation of parameters to show that the solution of the i nitial value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"y1(x0) = 1 , y ′
1(x0) = 0 , y 2(x0) = 0 , y ′
2(x0) = 1 .
Use variation of parameters to show that the solution of the i nitial value problem
y′′+ p(x)y′+ q(x)y = f(x), y (x0 ) = k0, y′(x0) = k1,
is
y(x) = k0y1(x) + k1y2(x)
+
∫ x
x0
(y1(t)y2(x) − y1(x)y2(t)) f(t) exp
(∫ t
x0
p(s) ds
)
dt.
HIN T : Use Abel’s formula for the Wronskian of {y1,y 2}, and integrate u′
1 and u′
2 from x0 to x.
Show also that
y′(x) = k0y′
1(x) + k1y′
2(x)
+
∫ x
x0
(y1(t)y′
2(x) − y′
1(x)y2(t)) f(t) exp
(∫ t
x0
p(s) ds
)
dt.
38. Suppose f is continuous on an open interval that contains x0 = 0 . Use variation of parameters to
find a formula for the solution of the initial value problem
y′′− y = f(x), y (0) = k0, y ′(0) = k1.
39. Suppose f is continuous on (a, ∞ ), where a< 0, so x0 = 0 is in (a, ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 5.7 V ariation of Parameters 265
(a) Use variation of parameters to find a formula for the solution of the initial value problem
y′′+ y= f(x), y (0) = k0, y ′(0) = k1.
HIN T : Y ou will need the addition formulas for the sine and cosine:
sin(A+ B) = sin Acos B+ cos Asin B
cos(A+ B) = cos Acos B− sin Asin B.
For the rest of this exercise assume that the improper integr al ∫∞
0 f(t) dtis absolutely convergent.
(b) Show that if yis a solution of
y′′+ y= f(x) (A)
on (a, ∞ ), then
lim
x→∞
(y(x) − A0 cos x− A1 sin x) = 0 (B)
and
lim
x→∞
(y′(x) + A0 sin x− A1 cos x) = 0 , (C)
where
A0 = k0 −
∫ ∞
0
f(t) sin tdt and A1 = k1 +
∫ ∞
0
f(t) cos tdt.
HIN T : Recall from calculus that if ∫∞
0 f(t) dtconverges absolutely, then limx→∞
∫∞
x |f(t)|dt = 0 .
(c) Show that if A0 and A1 are arbitrary constants, then there’s a unique solution of y′′+ y =
f(x) on (a, ∞ ) that satisfies (B) and (C).",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 6
Applications of Linear Second Order
Equations
IN THIS CHAPTER we study applications of linear second order equations.
SECTIONS 6.1 AND 6.2 is about spring–mass systems.
SECTION 6.2 is about RLC circuits, the electrical analogs of spring–mass systems.
SECTION 6.3 is about motion of an object under a central force , which is particularly relevant in the
space age, since, for example, a satellite moving in orbit su bject only to Earth’s gravity is experiencing
motion under a central force.
267",Elementary Differential Equations with Boundary Value Problems.pdf
"268 Chapter 6 Applications of Linear Second Order Equations
6.1 SPRING PROBLEMS I
W e consider the motion of an object of mass m, suspended from a spring of negligible mass. W e say that
the spring–mass system is in equilibrium when the object is at rest and the forces acting on it sum to zer o.
The position of the object in this case is the equilibrium position . W e define yto be the displacement of
the object from its equilibrium position (Figure 6.1.1), measured positive upward.
 y
(a)
0
(b) (c)
Figure 6.1.1 (a) y> 0 (b) y = 0 , (c) y< 0 Figure 6.1.2 A spring – mass system with damping
Our model accounts for the following kinds of forces acting o n the object:
• The force −mg, due to gravity .
• A force Fs exerted by the spring resisting change in its length. The natural length of the spring
is its length with no mass attached. W e assume that the spring obeys Hooke’s law : If the length
of the spring is changed by an amount ∆ Lfrom its natural length, then the spring exerts a force",Elementary Differential Equations with Boundary Value Problems.pdf
"of the spring is changed by an amount ∆ Lfrom its natural length, then the spring exerts a force
Fs = k∆ L, where kis a positive number called the spring constant . If the spring is stretched then
∆ L> 0 and Fs >0, so the spring force is upward, while if the spring is compres sed then ∆ L< 0
and Fs <0, so the spring force is downward.
• A damping force Fd = −cy′that resists the motion with a force proportional to the velo city of
the object. It may be due to air resistance or friction in the s pring. However, a convenient way to
visualize a damping force is to assume that the object is rigi dly attached to a piston with negligible
mass immersed in a cylinder (called a dashpot) filled with a viscous liquid (Figure 6.1.2). As the
piston moves, the liquid exerts a damping force. W e say that t he motion is undamped if c = 0 , or
damped if c> 0.
• An external force F, other than the force due to gravity , that may vary with t, but is independent of",Elementary Differential Equations with Boundary Value Problems.pdf
"damped if c> 0.
• An external force F, other than the force due to gravity , that may vary with t, but is independent of
displacement and velocity . W e say that the motion is free if F ≡ 0, or forced if F ̸≡0.
From Newton’s second law of motion,
my′′= −mg+ Fd + Fs + F = −mg− cy′+ Fs + F. (6.1.1)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.1 Spring Problems I 269
 y
(a)
0
(b)
 L
 ∆  L
Figure 6.1.3 (a) Natural length of spring (b) Spring stretched by mass
W e must now relate Fs to y. In the absence of external forces the object stretches the s pring by an amount
∆ lto assume its equilibrium position (Figure 6.1.3). Since the sum of the forces acting on the object is
then zero, Hooke’s Law implies that mg = k∆ l. If the object is displaced y units from its equilibrium
position, the total change in the length of the spring is ∆ L= ∆ l− y, so Hooke’s law implies that
Fs = k∆ L= k∆ l− ky.
Substituting this into ( 6.1.1) yields
my′′= −mg− cy′+ k∆ L− ky+ F.
Since mg= k∆ lthis can be written as
my′′+ cy′+ ky = F. (6.1.2)
W e call this the equation of motion .
Simple Harmonic Motion
Throughout the rest of this section we’ll consider spring–m ass systems without damping; that is, c = 0 .
W e’ll consider systems with damping in the next section.",Elementary Differential Equations with Boundary Value Problems.pdf
"W e’ll consider systems with damping in the next section.
W e first consider the case where the motion is also free; that i s, F=0. W e begin with an example.
Example 6.1.1 An object stretches a spring 6 inches in equilibrium.
(a) Set up the equation of motion and find its general solution.
(b) Find the displacement of the object for t >0 if it’s initially displaced 18 inches above equilibrium
and given a downward velocity of 3 ft/s.",Elementary Differential Equations with Boundary Value Problems.pdf
"270 Chapter 6 Applications of Linear Second Order Equations
SO LU TIO N (a) Setting c= 0 and F = 0 in ( 6.1.2) yields the equation of motion
my′′+ ky = 0 ,
which we rewrite as
y′′+ k
my= 0 . (6.1.3)
Although we would need the weight of the object to obtain kfrom the equation mg = k∆ lwe can obtain
k/m from ∆ l alone; thus, k/m = g/ ∆ l. Consistent with the units used in the problem statement, we
take g = 32 ft/s2. Although ∆ l is stated in inches, we must convert it to feet to be consisten t with this
choice of g; that is, ∆ l= 1 / 2 ft. Therefore
k
m = 32
1/ 2 = 64
and ( 6.1.3) becomes
y′′+ 64 y= 0 . (6.1.4)
The characteristic equation of ( 6.1.4) is
r2 + 64 = 0 ,
which has the zeros r= ±8i. Therefore the general solution of ( 6.1.4) is
y = c1 cos 8t+ c2 sin 8 t. (6.1.5)
SO LU TIO N (b) The initial upward displacement of 18 inches is positive and must be expressed in feet.
The initial downward velocity is negative; thus,
y(0) = 3
2 and y′(0) = −3.
Differentiating ( 6.1.5) yields",Elementary Differential Equations with Boundary Value Problems.pdf
"The initial downward velocity is negative; thus,
y(0) = 3
2 and y′(0) = −3.
Differentiating ( 6.1.5) yields
y′= −8c1 sin 8 t+ 8 c2 cos 8t. (6.1.6)
Setting t = 0 in ( 6.1.5) and ( 6.1.6) and imposing the initial conditions shows that c1 = 3 / 2 and c2 =
−3/ 8. Therefore
y = 3
2 cos 8t− 3
8 sin 8t,
where yis in feet (Figure 6.1.4).
W e’ll now consider the equation
my′′+ ky = 0
where mand kare arbitrary positive numbers. Dividing through by mand defining ω0 =
√
k/m yields
y′′+ ω2
0 y= 0 .
The general solution of this equation is
y= c1 cos ω0t+ c2 sin ω0t. (6.1.7)
W e can rewrite this in a more useful form by defining
R=
√
c2
1 + c2
2, (6.1.8)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.1 Spring Problems I 271
0.5 1.5 2.5 1.0 2.0 3.0
0.5
1.5
1.0
2.0
−0.5
−1.5
−1.0
−2.0
 x
 y
Figure 6.1.4 y= 3
2 cos 8t− 3
8 sin 8 t
and
c1 = Rcos φ and c2 = Rsin φ. (6.1.9)
Substituting from ( 6.1.9) into ( 6.1.7) and applying the identity
cos ω0tcos φ + sin ω0tsin φ = cos( ω0t− φ)
yields
y = Rcos(ω0t− φ). (6.1.10)
From ( 6.1.8) and ( 6.1.9) we see that the Rand φ can be interpreted as polar coordinates of the point
with rectangular coordinates (c1,c 2) (Figure 6.1.5). Given c1 and c2, we can compute Rfrom ( 6.1.8).
From ( 6.1.8) and ( 6.1.9), we see that φ is related to c1 and c2 by
cos φ = c1√
c2
1 + c2
2
and sin φ = c2
√
c2
1 + c2
2
.
There are infinitely many angles φ, differing by integer multiples of 2π, that satisfy these equations. W e
will always choose φ so that −π ≤ φ<π .
The motion described by (
6.1.7) or ( 6.1.10) is simple harmonic motion . W e see from either of these
equations that the motion is periodic, with period
T = 2 π/ω 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"6.1.7) or ( 6.1.10) is simple harmonic motion . W e see from either of these
equations that the motion is periodic, with period
T = 2 π/ω 0.
This is the time required for the object to complete one full c ycle of oscillation (for example, to move from
its highest position to its lowest position and back to its hi ghest position). Since the highest and lowest
positions of the object are y = Rand y = −R, we say that Ris the amplitude of the oscillation. The",Elementary Differential Equations with Boundary Value Problems.pdf
"272 Chapter 6 Applications of Linear Second Order Equations
 θ
 c1
 c2
 R
Figure 6.1.5 R=
√
c2
1 + c2
2; c1 = Rcos φ; c2 = Rsin φ
angle φ in ( 6.1.10) is the phase angle . It’s measured in radians. Equation ( 6.1.10) is the amplitude–phase
form of the displacement. If tis in seconds then ω0 is in radians per second (rad/s); it’s the frequency of
the motion. It is also called the natural frequency of the spring–mass system without damping.
Example 6.1.2 W e found the displacement of the object in Example 6.1.1 to be
y = 3
2 cos 8t− 3
8 sin 8t.
Find the frequency , period, amplitude, and phase angle of th e motion.
Solution The frequency is ω0 = 8 rad/s, and the period is T = 2 π/ω 0 = π/ 4 s. Since c1 = 3 / 2 and
c2 = −3/ 8, the amplitude is
R=
√
c2
1 + c2
2 =
√
(3
2
) 2
+
(3
8
) 2
= 3
8
√
17.
The phase angle is determined by
cos φ =
3
2
3
8
√
17 = 4√
17 (6.1.11)
and
sin φ = −3
8
3
8
√
17 = − 1√
17 . (6.1.12)
Using a calculator, we see from ( 6.1.11) that
φ ≈ ± .245 rad.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.1 Spring Problems I 273
Since sin φ< 0 (see ( 6.1.12)), the minus sign applies here; that is,
φ ≈ − .245 rad.
Example 6.1.3 The natural length of a spring is 1 m. An object is attached to i t and the length of the
spring increases to 102 cm when the object is in equilibrium. Then the object is initially displaced
downward 1 cm and given an upward velocity of 14 cm/s. Find the displacement for t >0. Also, find
the natural frequency , period, amplitude, and phase angle o f the resulting motion. Express the answers in
cgs units.
Solution In cgs units g= 980 cm/s2. Since ∆ l= 2 cm, ω2
0 = g/ ∆ l= 490 . Therefore
y′′+ 490 y= 0 , y (0) = −1, y ′(0) = 14 .
The general solution of the differential equation is
y = c1 cos 7
√
10t+ c2 sin 7
√
10t,
so
y′= 7
√
10
(
−c1 sin 7
√
10t+ c2 cos 7
√
10t
)
.
Substituting the initial conditions into the last two equat ions yields c1 = −1 and c2 = 2 /
√
10. Hence,
y= − cos 7
√
10t+ 2√
10 sin 7
√
10t.
The frequency is 7
√",Elementary Differential Equations with Boundary Value Problems.pdf
"√
10. Hence,
y= − cos 7
√
10t+ 2√
10 sin 7
√
10t.
The frequency is 7
√
10 rad/s, and the period is T = 2 π/ (7
√
10) s. The amplitude is
R=
√
c2
1 + c2
2 =
√
(−1)2 +
( 2√
10
) 2
=
√
7
5 cm.
The phase angle is determined by
cos φ = c1
R = −
√
5
7 and sin φ = c2
R =
√
2
7 .
Therefore φ is in the second quadrant and
φ = cos −1
(
−
√
5
7
)
≈ 2.58 rad.
Undamped Forced Oscillation
In many mechanical problems a device is subjected to periodi c external forces. For example, soldiers
marching in cadence on a bridge cause periodic disturbances in the bridge, and the engines of a propeller
driven aircraft cause periodic disturbances in its wings. I n the absence of sufficient damping forces, such
disturbances – even if small in magnitude – can cause structu ral breakdown if they are at certain critical
frequencies. T o illustrate, this we’ll consider the motion of an object in a spring–mass system without
damping, subject to an external force
F(t) = F0 cos ωt",Elementary Differential Equations with Boundary Value Problems.pdf
"274 Chapter 6 Applications of Linear Second Order Equations
where F0 is a constant. In this case the equation of motion ( 6.1.2) is
my′′+ ky = F0 cos ωt,
which we rewrite as
y′′+ ω2
0 y = F0
m cos ωt (6.1.13)
with ω0 =
√
k/m . W e’ll see from the next two examples that the solutions of ( 6.1.13) with ω ̸= ω0
behave very differently from the solutions with ω = ω0.
Example 6.1.4 Solve the initial value problem
y′′+ ω2
0 y= F0
m cos ωt, y (0) = 0 , y ′(0) = 0 , (6.1.14)
given that ω ̸= ω0.
Solution W e first obtain a particular solution of ( 6.1.13) by the method of undetermined coefficients.
Since ω ̸= ω0, cos ωt isn’t a solution of the complementary equation
y′′+ ω2
0 y= 0 .
Therefore ( 6.1.13) has a particular solution of the form
yp = Acos ωt + Bsin ωt.
Since
y′′
p = −ω2 (Acos ωt + Bsin ωt),
y′′
p + ω2
0 yp = F0
m cos ωt
if and only if
(ω2
0 − ω2) (Acos ωt + Bsin ωt) = F0
m cos ωt.
This holds if and only if
A= F0
m(ω2
0 − ω2) and B = 0 ,
so
yp = F0
m(ω2
0 − ω2) cos ωt.",Elementary Differential Equations with Boundary Value Problems.pdf
"(ω2
0 − ω2) (Acos ωt + Bsin ωt) = F0
m cos ωt.
This holds if and only if
A= F0
m(ω2
0 − ω2) and B = 0 ,
so
yp = F0
m(ω2
0 − ω2) cos ωt.
The general solution of ( 6.1.13) is
y= F0
m(ω2
0 − ω2) cos ωt + c1 cos ω0t+ c2 sin ω0t, (6.1.15)
so
y′= −ωF0
m(ω2
0 − ω2) sin ωt + ω0(−c1 sin ω0t+ c2 cos ω0t).
The initial conditions y(0) = 0 and y′(0) = 0 in ( 6.1.14) imply that
c1 = − F0
m(ω2
0 − ω2) and c2 = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.1 Spring Problems I 275
Substituting these into ( 6.1.15) yields
y= F0
m(ω2
0 − ω2) (cos ωt − cos ω0t). (6.1.16)
It is revealing to write this in a different form. W e start wit h the trigonometric identities
cos(α − β) = cos α cos β + sin α sin β
cos(α + β) = cos α cos β − sin α sin β.
Subtracting the second identity from the first yields
cos(α − β) − cos(α + β) = 2 sin α sin β (6.1.17)
Now let
α − β = ωt and α + β = ω0t, (6.1.18)
so that
α = (ω0 + ω)t
2 and β = (ω0 − ω)t
2 . (6.1.19)
Substituting ( 6.1.18) and ( 6.1.19) into ( 6.1.17) yields
cos ωt − cos ω0t= 2 sin (ω0 − ω)t
2 sin (ω0 + ω)t
2 ,
and substituting this into ( 6.1.16) yields
y = R(t) sin (ω0 + ω)t
2 , (6.1.20)
where
R(t) = 2F0
m(ω2
0 − ω2) sin (ω0 − ω)t
2 . (6.1.21)
From ( 6.1.20) we can regard yas a sinusoidal variation with frequency (ω0 + ω)/ 2 and variable ampli-
tude |R(t)|. In Figure 6.1.6 the dashed curve above the taxis is y= |R(t)|, the dashed curve below the t",Elementary Differential Equations with Boundary Value Problems.pdf
"tude |R(t)|. In Figure 6.1.6 the dashed curve above the taxis is y= |R(t)|, the dashed curve below the t
axis is y = −|R(t)|, and the displacement yappears as an oscillation bounded by them. The oscillation
of yfor ton an interval between successive zeros of R(t) is called a beat.
Y ou can see from ( 6.1.20) and ( 6.1.21) that
|y(t)| ≤ 2|F0|
m|ω2
0 − ω2|;
moreover, if ω + ω0 is sufficiently large compared with ω − ω0, then |y|assumes values close to (perhaps
equal to) this upper bound during each beat. However, the osc illation remains bounded for all t. (This
assumes that the spring can withstand deflections of this siz e and continue to obey Hooke’s law .) The
next example shows that this isn’t so if ω = ω0.
Example 6.1.5 Find the general solution of
y′′+ ω2
0 y= F0
m cos ω0t. (6.1.22)",Elementary Differential Equations with Boundary Value Problems.pdf
"276 Chapter 6 Applications of Linear Second Order Equations
 t
 y
Figure 6.1.6 Undamped oscillation with beats
Solution W e first obtain a particular solution yp of ( 6.1.22). Since cos ω0tis a solution of the comple-
mentary equation, the form for yp is
yp = t(Acos ω0t+ Bsin ω0t). (6.1.23)
Then
y′
p = Acos ω0t+ Bsin ω0t+ ω0t(−Asin ω0t+ Bcos ω0t)
and
y′′
p = 2 ω0(−Asin ω0t+ Bcos ω0t) − ω2
0t(Acos ω0t+ Bsin ω0t). (6.1.24)
From ( 6.1.23) and ( 6.1.24), we see that yp satisfies ( 6.1.22) if
−2Aω0 sin ω0t+ 2 Bω 0 cos ω0t= F0
m cos ω0t;
that is, if
A= 0 and B= F0
2mω0
.
Therefore
yp = F0t
2mω0
sin ω0t
is a particular solution of ( 6.1.22). The general solution of ( 6.1.22) is
y = F0t
2mω0
sin ω0t+ c1 cos ω0t+ c2 sin ω0t.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.1 Spring Problems I 277
 t
 y
 y = F0 t / 2mω 0
 y = − F 0 t / 2mω 0
Figure 6.1.7 Unbounded displacement due to resonance
The graph of yp is shown in Figure 6.1.7, where it can be seen that yp oscillates between the dashed lines
y= F0t
2mω0
and y= − F0t
2mω0
with increasing amplitude that approaches ∞ as t → ∞ . Of course, this means that the spring must
eventually fail to obey Hooke’s law or break.
This phenomenon of unbounded displacements of a spring–mas s system in response to a periodic
forcing function at its natural frequency is called resonance. More complicated mechanical structures
can also exhibit resonance–like phenomena. For example, rh ythmic oscillations of a suspension bridge
by wind forces or of an airplane wing by periodic vibrations o f reciprocating engines can cause damage
or even failure if the frequencies of the disturbances are cl ose to critical frequencies determined by the
parameters of the mechanical system in question.
6.1 Exercises",Elementary Differential Equations with Boundary Value Problems.pdf
"parameters of the mechanical system in question.
6.1 Exercises
In the following exercises assume that there’s no damping.
1. C/G An object stretches a spring 4 inches in equilibrium. Find an d graph its displacement for
t >0 if it’s initially displaced 36 inches above equilibrium and given a downward velocity of 2
ft/s.
2. An object stretches a string 1.2 inches in equilibrium. Find its displacement for t > 0 if it’s
initially displaced 3 inches below equilibrium and given a d ownward velocity of 2 ft/s.
3. A spring with natural length .5 m has length 50.5 cm with a mass of 2 gm suspended from it.
The mass is initially displaced 1.5 cm below equilibrium and released with zero velocity . Find its
displacement for t> 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"278 Chapter 6 Applications of Linear Second Order Equations
4. An object stretches a spring 6 inches in equilibrium. Find it s displacement for t> 0 if it’s initially
displaced 3 inches above equilibrium and given a downward ve locity of 6 inches/s. Find the
frequency , period, amplitude and phase angle of the motion.
5. C/G An object stretches a spring 5 cm in equilibrium. It is initia lly displaced 10 cm above
equilibrium and given an upward velocity of .25 m/s. Find and graph its displacement for t >0.
Find the frequency , period, amplitude, and phase angle of th e motion.
6. A 10 kg mass stretches a spring 70 cm in equilibrium. Suppose a 2 kg mass is attached to the
spring, initially displaced 25 cm below equilibrium, and gi ven an upward velocity of 2 m/s. Find
its displacement for t> 0. Find the frequency , period, amplitude, and phase angle of t he motion.
7. A weight stretches a spring 1.5 inches in equilibrium. The we ight is initially displaced 8 inches",Elementary Differential Equations with Boundary Value Problems.pdf
"7. A weight stretches a spring 1.5 inches in equilibrium. The we ight is initially displaced 8 inches
above equilibrium and given a downward velocity of 4 ft/s. Fi nd its displacement for t> 0.
8. A weight stretches a spring 6 inches in equilibrium. The weig ht is initially displaced 6 inches
above equilibrium and given a downward velocity of 3 ft/s. Fi nd its displacement for t> 0.
9. A spring–mass system has natural frequency 7
√
10 rad/s. The natural length of the spring is .7 m.
What is the length of the spring when the mass is in equilibriu m?
10. A 64 lb weight is attached to a spring with constant k= 8 lb/ft and subjected to an external force
F(t) = 2 sin t. The weight is initially displaced 6 inches above equilibri um and given an upward
velocity of 2 ft/s. Find its displacement for t> 0.
11. A unit mass hangs in equilibrium from a spring with constant k= 1 / 16. Starting at t= 0 , a force
F(t) = 3 sin tis applied to the mass. Find its displacement for t> 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"F(t) = 3 sin tis applied to the mass. Find its displacement for t> 0.
12. C/G A 4 lb weight stretches a spring 1 ft in equilibrium. An extern al force F(t) = .25 sin 8 t
lb is applied to the weight, which is initially displaced 4 in ches above equilibrium and given a
downward velocity of 1 ft/s. Find and graph its displacement for t> 0.
13. A 2 lb weight stretches a spring 6 inches in equilibrium. An ex ternal force F(t) = sin 8 tlb is ap-
plied to the weight, which is released from rest 2 inches belo w equilibrium. Find its displacement
for t> 0.
14. A 10 gm mass suspended on a spring moves in simple harmonic mot ion with period 4 s. Find the
period of the simple harmonic motion of a 20 gm mass suspended from the same spring.
15. A 6 lb weight stretches a spring 6 inches in equilibrium. Supp ose an external force F(t) =
3
16 sin ωt + 3
8 cos ωt lb is applied to the weight. For what value of ω will the displacement",Elementary Differential Equations with Boundary Value Problems.pdf
"3
16 sin ωt + 3
8 cos ωt lb is applied to the weight. For what value of ω will the displacement
be unbounded? Find the displacement if ω has this value. Assume that the motion starts from
equilibrium with zero initial velocity .
16. C/G A 6 lb weight stretches a spring 4 inches in equilibrium. Supp ose an external force F(t) =
4 sin ωt − 6 cos ωt lb is applied to the weight. For what value of ω will the displacement be
unbounded? Find and graph the displacement if ω has this value. Assume that the motion starts
from equilibrium with zero initial velocity .
17. A mass of one kg is attached to a spring with constant k = 4 N/m. An external force F(t) =
− cos ωt − 2 sin ωt n is applied to the mass. Find the displacement y for t >0 if ω equals the
natural frequency of the spring–mass system. Assume that th e mass is initially displaced 3 m
above equilibrium and given an upward velocity of 450 cm/s.",Elementary Differential Equations with Boundary Value Problems.pdf
"above equilibrium and given an upward velocity of 450 cm/s.
18. An object is in simple harmonic motion with frequency ω0, with y(0) = y0 and y′(0) = v0. Find
its displacement for t >0. Also, find the amplitude of the oscillation and give formula s for the
sine and cosine of the initial phase angle.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Spring Problems II 279
19. T wo objects suspended from identical springs are set into mo tion. The period of one object is
twice the period of the other. How are the weights of the two ob jects related?
20. T wo objects suspended from identical springs are set into mo tion. The weight of one object is
twice the weight of the other. How are the periods of the resul ting motions related?
21. T wo identical objects suspended from different springs are set into motion. The period of one
motion is 3 times the period of the other. How are the two sprin g constants related?
6.2 SPRING PROBLEMS II
Free Vibrations With Damping
In this section we consider the motion of an object in a spring –mass system with damping. W e start with
unforced motion, so the equation of motion is
my′′+ cy′+ ky = 0 . (6.2.1)
Now suppose the object is displaced from equilibrium and giv en an initial velocity . Intuition suggests that",Elementary Differential Equations with Boundary Value Problems.pdf
"my′′+ cy′+ ky = 0 . (6.2.1)
Now suppose the object is displaced from equilibrium and giv en an initial velocity . Intuition suggests that
if the damping force is sufficiently weak the resulting motio n will be oscillatory , as in the undamped case
considered in the previous section, while if it’s sufficient ly strong the object may just move slowly toward
the equilibrium position without ever reaching it. W e’ll no w confirm these intuitive ideas mathematically .
The characteristic equation of ( 6.2.1) is
mr2 + cr+ k= 0 .
The roots of this equation are
r1 = −c−
√
c2 − 4mk
2m and r2 = −c+
√
c2 − 4mk
2m . (6.2.2)
In Section 5.2 we saw that the form of the solution of ( 6.2.1) depends upon whether c2 − 4mkis positive,
negative, or zero. W e’ll now consider these three cases.
Underdamped Motion
W e say the motion is underdamped if c<
√
4mk. In this case r1 and r2 in ( 6.2.2) are complex conjugates,
which we write as
r1 = − c
2m − iω1 and r2 = − c
2m + iω1,
where
ω1 =
√
4mk− c2
2m .",Elementary Differential Equations with Boundary Value Problems.pdf
"4mk. In this case r1 and r2 in ( 6.2.2) are complex conjugates,
which we write as
r1 = − c
2m − iω1 and r2 = − c
2m + iω1,
where
ω1 =
√
4mk− c2
2m .
The general solution of ( 6.2.1) in this case is
y= e−ct/ 2m(c1 cos ω1t+ c2 sin ω1t).
By the method used in Section 6.1 to derive the amplitude–pha se form of the displacement of an object
in simple harmonic motion, we can rewrite this equation as
y = Re−ct/ 2m cos(ω1t− φ), (6.2.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"280 Chapter 6 Applications of Linear Second Order Equations
 x
 y
 y = Re−ct / 2m
 y = −− Re −ct / 2m
Figure 6.2.1 Underdamped motion
where
R=
√
c2
1 + c2
2, R cos φ = c1, and Rsin φ = c2.
The factor Re−ct/ 2m in (
6.2.3) is called the time–varying amplitude of the motion, the quantity ω1 is
called the frequency, and T = 2 π/ω 1 (which is the period of the cosine function in ( 6.2.3) is called the
quasi–period . A typical graph of ( 6.2.3) is shown in Figure 6.2.1. As illustrated in that figure, the graph
of yoscillates between the dashed exponential curves y = ±Re−ct/ 2m.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Spring Problems II 281
Overdamped Motion
W e say the motion is overdamped if c >
√
4mk. In this case the zeros r1 and r2 of the characteristic
polynomial are real, with r1 <r2 <0 (see ( 6.2.2)), and the general solution of ( 6.2.1) is
y= c1er1 t + c2er2 t .
Again limt→∞ y(t) = 0 as in the underdamped case, but the motion isn’t oscillatory , since ycan’t equal
zero for more than one value of tunless c1 = c2 = 0 . (Exercise 23.)
Critically Damped Motion
W e say the motion is critically damped if c =
√
4mk. In this case r1 = r2 = −c/ 2mand the general
solution of ( 6.2.1) is
y= e−ct/ 2m(c1 + c2t).
Again limt→∞ y(t) = 0 and the motion is nonoscillatory , since y can’t equal zero for more than one
value of tunless c1 = c2 = 0 . (Exercise 22).
Example 6.2.1 Suppose a 64 lb weight stretches a spring 6 inches in equilibr ium and a dashpot provides
a damping force of clb for each ft/sec of velocity .",Elementary Differential Equations with Boundary Value Problems.pdf
"a damping force of clb for each ft/sec of velocity .
(a) Write the equation of motion of the object and determine the v alue of c for which the motion is
critically damped.
(b) Find the displacement y for t >0 if the motion is critically damped and the initial condition s are
y(0) = 1 and y′(0) = 20 .
(c) Find the displacement y for t >0 if the motion is critically damped and the initial condition s are
y(0) = 1 and y′(0) = −20.
SO LU TIO N (a) Here m= 2 slugs and k= 64 /. 5 = 128 lb/ft. Therefore the equation of motion ( 6.2.1) is
2y′′+ cy′+ 128 y= 0 . (6.2.4)
The characteristic equation is
2r2 + cr+ 128 = 0 ,
which has roots
r= −c±
√
c2 − 8 ·128
4 .
Therefore the damping is critical if
c=
√
8 ·128 = 32 lb–sec/ft .
SO LU TIO N (b) Setting c= 32 in ( 6.2.4) and cancelling the common factor 2 yields
y′′+ 16 y+ 64 y= 0 .
The characteristic equation is
r2 + 16 r+ 64 y= ( r+ 8) 2 = 0 .
Hence, the general solution is
y = e−8t (c1 + c2t). (6.2.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"282 Chapter 6 Applications of Linear Second Order Equations
0.2 0.4 0.6 0.8 1.0
0.5
1.5
1.0
2.0
−0.5
(a)
(b)
 x
 y
Figure 6.2.2 (a) y= e−8t (1 + 28 t) (b) y = e−8t(1 − 12t)
Differentiating this yields
y′= −8y+ c2e−8t. (6.2.6)
Imposing the initial conditions y(0) = 1 and y′(0) = 20 in the last two equations shows that 1 = c1 and
20 = −8 + c2. Hence, the solution of the initial value problem is
y= e−8t(1 + 28 t).
Therefore the object approaches equilibrium from above as t→ ∞ . There’s no oscillation.
SO LU TIO N (c) Imposing the initial conditions y(0) = 1 and y′(0) = −20 in ( 6.2.5) and ( 6.2.6) yields
1 = c1 and −20 = −8 + c2. Hence, the solution of this initial value problem is
y= e−8t(1 − 12t).
Therefore the object moves downward through equilibrium ju st once, and then approaches equilibrium
from below as t → ∞ . Again, there’s no oscillation. The solutions of these two i nitial value problems
are graphed in Figure 6.2.2.",Elementary Differential Equations with Boundary Value Problems.pdf
"from below as t → ∞ . Again, there’s no oscillation. The solutions of these two i nitial value problems
are graphed in Figure 6.2.2.
Example 6.2.2 Find the displacement of the object in Example 6.2.1 if the damping constant is c = 4
lb–sec/ft and the initial conditions are y(0) = 1 .5 ft and y′(0) = −3 ft/sec.
Solution With c= 4 , the equation of motion ( 6.2.4) becomes
y′′+ 2 y′+ 64 y= 0 (6.2.7)
after cancelling the common factor 2. The characteristic eq uation
r2 + 2 r+ 64 = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Spring Problems II 283
has complex conjugate roots
r= −2 ± √4 − 4 ·64
2 = −1 ± 3
√
7i.
Therefore the motion is underdamped and the general solutio n of ( 6.2.7) is
y= e−t (c1 cos 3
√
7t+ c2 sin 3
√
7t).
Differentiating this yields
y′= −y+ 3
√
7e−t (−c1 sin 3
√
7t+ c2 cos 3
√
7t).
Imposing the initial conditions y(0) = 1 .5 and y′(0) = −3 in the last two equations yields 1.5 = c1 and
−3 = −1.5 + 3
√
7c2. Hence, the solution of the initial value problem is
y= e−t
(3
2 cos 3
√
7t− 1
2
√
7 sin 3
√
7t
)
. (6.2.8)
The amplitude of the function in parentheses is
R=
√ (3
2
) 2
+
( 1
2
√
7
) 2
=
√
9
4 + 1
4 ·7 =
√
64
4 ·7 = 4√
7 .
Therefore we can rewrite ( 6.2.8) as
y= 4√
7 e−t cos(3
√
7t− φ),
where
cos φ = 3
2R = 3
√
7
8 and sin φ = − 1
2
√
7R = −1
8 .
Therefore φ ∼
= −.125 radians.
Example 6.2.3 Let the damping constant in Example 1 be c= 40 lb–sec/ft. Find the displacement yfor
t> 0 if y(0) = 1 and y′(0) = 1 .
Solution With c= 40 , the equation of motion (
6.2.4) reduces to",Elementary Differential Equations with Boundary Value Problems.pdf
"t> 0 if y(0) = 1 and y′(0) = 1 .
Solution With c= 40 , the equation of motion (
6.2.4) reduces to
y′′+ 20 y′+ 64 y= 0 (6.2.9)
after cancelling the common factor 2. The characteristic eq uation
r2 + 20 r+ 64 = ( r+ 16)( r+ 4) = 0
has the roots r1 = −4 and r2 = −16. Therefore the general solution of ( 6.2.9) is
y = c1e−4t + c2e−16t. (6.2.10)
Differentiating this yields
y′= −4e−4t − 16c2e−16t.",Elementary Differential Equations with Boundary Value Problems.pdf
"284 Chapter 6 Applications of Linear Second Order Equations
0.2 0.4 0.6 0.8 1.0 1.2
0.2
0.4
0.6
0.8
1.0
 x
 y
Figure 6.2.3 y = 17
12 e−4t − 5
12 e−16t
The last two equations and the initial conditions y(0) = 1 and y′(0) = 1 imply that
c1 + c2 = 1
−4c1 − 16c2 = 1 .
The solution of this system is c1 = 17 / 12, c2 = −5/ 12. Substituting these into ( 6.2.10) yields
y = 17
12 e−4t − 5
12 e−16t
as the solution of the given initial value problem (Figure 6.2.3).
Forced Vibrations With Damping
Now we consider the motion of an object in a spring-mass syste m with damping, under the influence of a
periodic forcing function F(t) = F0 cos ωt, so that the equation of motion is
my′′+ cy′+ ky = F0 cos ωt. (6.2.11)
In Section 6.1 we considered this equation with c= 0 and found that the resulting displacement yassumed
arbitrarily large values in the case of resonance (that is, w hen ω = ω0 =
√
k/m ). Here we’ll see that in",Elementary Differential Equations with Boundary Value Problems.pdf
"arbitrarily large values in the case of resonance (that is, w hen ω = ω0 =
√
k/m ). Here we’ll see that in
the presence of damping the displacement remains bounded fo r all t, and the initial conditions have little
effect on the motion as t→ ∞ . In fact, we’ll see that for large tthe displacement is closely approximated
by a function of the form
y= Rcos(ωt − φ), (6.2.12)
where the amplitude Rdepends upon m, c, k, F0, and ω. W e’re interested in the following question:",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Spring Problems II 285
QU ESTIO N :Assuming that m, c, k, and F0 are held constant, what value of ω produces the largest
amplitude Rin (6.2.12), and what is this largest amplitude?
T o answer this question, we must solve ( 6.2.11) and determine Rin terms of F0,ω 0,ω , and c. W e can
obtain a particular solution of ( 6.2.11) by the method of undetermined coefficients. Since cos ωt does not
satisfy the complementary equation
my′′+ cy′+ ky = 0 ,
we can obtain a particular solution of ( 6.2.11) in the form
yp = Acos ωt + Bsin ωt. (6.2.13)
Differentiating this yields
y′
p = ω(−Asin ωt + Bcos ωt)
and
y′′
p = −ω2 (Acos ωt + Bsin ωt).
From the last three equations,
my′′
p + cy′
p + kyp = ( −mω2 A+ cωB + kA) cos ωt + ( −mω2 B− cωA + kB) sin ωt,
so yp satisfies ( 6.2.11) if
(k− mω2)A+ cωB = F0
−cωA + ( k− mω2 )B = 0 .
Solving for Aand Band substituting the results into ( 6.2.13) yields
yp = F0
(k− mω2)2 + c2ω2
[
(k− mω2 ) cos ωt + cω sin ωt
]
,",Elementary Differential Equations with Boundary Value Problems.pdf
"Solving for Aand Band substituting the results into ( 6.2.13) yields
yp = F0
(k− mω2)2 + c2ω2
[
(k− mω2 ) cos ωt + cω sin ωt
]
,
which can be written in amplitude–phase form as
yp = F0√
(k− mω2 )2 + c2ω2 cos(ωt − φ), (6.2.14)
where
cos φ = k− mω2
√
(k− mω2 )2 + c2ω2 and sin φ = cω√
(k− mω2 )2 + c2ω2 . (6.2.15)
T o compare this with the undamped forced vibration that we co nsidered in Section 6.1 it’s useful to
write
k− mω2 = m
(k
m − ω2
)
= m(ω2
0 − ω2), (6.2.16)
where ω0 =
√
k/m is the natural angular frequency of the undamped simple harm onic motion of an
object with mass mon a spring with constant k. Substituting ( 6.2.16) into ( 6.2.14) yields
yp = F0√
m2(ω2
0 − ω2)2 + c2ω2 cos(ωt − φ). (6.2.17)
The solution of an initial value problem
my′′+ cy′+ ky = F0 cos ωt, y (0) = y0, y ′(0) = v0,",Elementary Differential Equations with Boundary Value Problems.pdf
"286 Chapter 6 Applications of Linear Second Order Equations
is of the form y= yc + yp, where yc has one of the three forms
yc = e−ct/ 2m(c1 cos ω1t+ c2 sin ω1t),
yc = e−ct/ 2m(c1 + c2t),
yc = c1er1 t + c2er2 t (r1,r 2 <0).
In all three cases limt→∞ yc(t) = 0 for any choice of c1 and c2. For this reason we say that yc is the
transient component of the solution y. The behavior of yfor large tis determined by yp, which we call the
steady state component of y. Thus, for large tthe motion is like simple harmonic motion at the frequency
of the external force.
The amplitude of yp in ( 6.2.17) is
R= F0√
m2(ω2
0 − ω2)2 + c2ω2 , (6.2.18)
which is finite for all ω; that is, the presence of damping precludes the phenomenon o f resonance that we
encountered in studying undamped vibrations under a period ic forcing function. W e’ll now find the value
ωmax of ω for which Ris maximized. This is the value of ω for which the function
ρ(ω) = m2(ω2
0 − ω2)2 + c2ω2",Elementary Differential Equations with Boundary Value Problems.pdf
"ωmax of ω for which Ris maximized. This is the value of ω for which the function
ρ(ω) = m2(ω2
0 − ω2)2 + c2ω2
in the denominator of ( 6.2.18) attains its minimum value. By rewriting this as
ρ(ω) = m2(ω4
0 + ω4) + ( c2 − 2m2ω2
0 )ω2, (6.2.19)
you can see that ρ is a strictly increasing function of ω2 if
c≥
√
2m2ω2
0 =
√
2mk.
(Recall that ω2
0 = k/m ). Therefore ωmax = 0 if this inequality holds. From ( 6.2.15), you can see that
φ = 0 if ω = 0 . In this case, ( 6.2.14) reduces to
yp = F0√
m2ω4
0
= F0
k ,
which is consistent with Hooke’s law: if the mass is subjecte d to a constant force F0, its displacement
should approach a constant yp such that kyp = F0. Now suppose c<
√
2mk. Then, from ( 6.2.19),
ρ′(ω) = 2 ω(2m2ω2 + c2 − 2m2ω2
0 ),
and ωmax is the value of ω for which the expression in parentheses equals zero; that is ,
ωmax =
√
ω2
0 − c2
2m2 =
√
k
m
(
1 − c2
2km
)
.
(T o see that ρ(ωmax) is the minimum value of ρ(ω), note that ρ′(ω) < 0 if ω < ωmax and ρ′(ω) > 0",Elementary Differential Equations with Boundary Value Problems.pdf
"ωmax =
√
ω2
0 − c2
2m2 =
√
k
m
(
1 − c2
2km
)
.
(T o see that ρ(ωmax) is the minimum value of ρ(ω), note that ρ′(ω) < 0 if ω < ωmax and ρ′(ω) > 0
if ω > ωmax.) Substituting ω = ωmax in ( 6.2.18) and simplifying shows that the maximum amplitude
Rmax is
Rmax = 2mF0
c
√
4mk− c2 if c<
√
2mk.
W e summarize our results as follows.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Spring Problems II 287
Theorem 6.2.1 Suppose we consider the amplitude Rof the steady state component of the solution of
my′′+ cy′+ ky = F0 cos ωt
as a function of ω.
(a) If c≥
√
2mk, the maximum amplitude is Rmax = F0/k and it’s attained when ω = ωmax = 0 .
(b) If c<
√
2mk, the maximum amplitude is
Rmax = 2mF0
c
√
4mk− c2 , (6.2.20)
and it’s attained when
ω = ωmax =
√
k
m
(
1 − c2
2km
)
. (6.2.21)
Note that Rmax and ωmax are continuous functions of c, for c≥ 0, since ( 6.2.20) and ( 6.2.21) reduce to
Rmax = F0/k and ωmax = 0 if c=
√
2km.",Elementary Differential Equations with Boundary Value Problems.pdf
"288 Chapter 6 Applications of Linear Second Order Equations
6.2 Exercises
1. A 64 lb object stretches a spring 4 ft in equilibrium. It is att ached to a dashpot with damping
constant c= 8 lb-sec/ft. The object is initially displaced 18 inches abov e equilibrium and given a
downward velocity of 4 ft/sec. Find its displacement and tim e–varying amplitude for t> 0.
2. C/G A 16 lb weight is attached to a spring with natural length 5 ft. With the weight attached,
the spring measures 8.2 ft. The weight is initially displace d 3 ft below equilibrium and given an
upward velocity of 2 ft/sec. Find and graph its displacement for t >0 if the medium resists the
motion with a force of one lb for each ft/sec of velocity . Also , find its time–varying amplitude.
3. C/G An 8 lb weight stretches a spring 1.5 inches. It is attached to a dashpot with damping
constant c=8 lb-sec/ft. The weight is initially displaced 3 inches abo ve equilibrium and given an",Elementary Differential Equations with Boundary Value Problems.pdf
"constant c=8 lb-sec/ft. The weight is initially displaced 3 inches abo ve equilibrium and given an
upward velocity of 6 ft/sec. Find and graph its displacement for t> 0.
4. A 96 lb weight stretches a spring 3.2 ft in equilibrium. It is a ttached to a dashpot with damping
constant c=18 lb-sec/ft. The weight is initially displaced 15 inches b elow equilibrium and given a
downward velocity of 12 ft/sec. Find its displacement for t> 0.
5. A 16 lb weight stretches a spring 6 inches in equilibrium. It i s attached to a damping mechanism
with constant c. Find all values of csuch that the free vibration of the weight has infinitely many
oscillations.
6. An 8 lb weight stretches a spring .32 ft. The weight is initial ly displaced 6 inches above equilibrium
and given an upward velocity of 4 ft/sec. Find its displaceme nt for t >0 if the medium exerts a
damping force of 1.5 lb for each ft/sec of velocity .",Elementary Differential Equations with Boundary Value Problems.pdf
"damping force of 1.5 lb for each ft/sec of velocity .
7. A 32 lb weight stretches a spring 2 ft in equilibrium. It is att ached to a dashpot with constant c= 8
lb-sec/ft. The weight is initially displaced 8 inches below equilibrium and released from rest. Find
its displacement for t> 0.
8. A mass of 20 gm stretches a spring 5 cm. The spring is attached t o a dashpot with damping
constant 400 dyne sec/cm. Determine the displacement for t> 0 if the mass is initially displaced
9 cm above equilibrium and released from rest.
9. A 64 lb weight is suspended from a spring with constant k = 25 lb/ft. It is initially displaced 18
inches above equilibrium and released from rest. Find its di splacement for t >0 if the medium
resists the motion with 6 lb of force for each ft/sec of veloci ty .
10. A 32 lb weight stretches a spring 1 ft in equilibrium. The weig ht is initially displaced 6 inches
above equilibrium and given a downward velocity of 3 ft/sec. Find its displacement for t >0 if",Elementary Differential Equations with Boundary Value Problems.pdf
"above equilibrium and given a downward velocity of 3 ft/sec. Find its displacement for t >0 if
the medium resists the motion with a force equal to 3 times the speed in ft/sec.
11. An 8 lb weight stretches a spring 2 inches. It is attached to a d ashpot with damping constant
c=4 lb-sec/ft. The weight is initially displaced 3 inches abo ve equilibrium and given a downward
velocity of 4 ft/sec. Find its displacement for t> 0.
12. C/G A 2 lb weight stretches a spring .32 ft. The weight is initiall y displaced 4 inches below
equilibrium and given an upward velocity of 5 ft/sec. The med ium provides damping with constant
c= 1 / 8 lb-sec/ft. Find and graph the displacement for t> 0.
13. An 8 lb weight stretches a spring 8 inches in equilibrium. It i s attached to a dashpot with damping
constant c = .5 lb-sec/ft and subjected to an external force F(t) = 4 cos 2 t lb. Determine the
steady state component of the displacement for t> 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Spring Problems II 289
14. A 32 lb weight stretches a spring 1 ft in equilibrium. It is att ached to a dashpot with constant
c = 12 lb-sec/ft. The weight is initially displaced 8 inches above equilibrium and released from
rest. Find its displacement for t> 0.
15. A mass of one kg stretches a spring 49 cm in equilibrium. A dash pot attached to the spring
supplies a damping force of 4 N for each m/sec of speed. The mas s is initially displaced 10 cm
above equilibrium and given a downward velocity of 1 m/sec. F ind its displacement for t> 0.
16. A mass of 100 grams stretches a spring 98 cm in equilibrium. A d ashpot attached to the spring
supplies a damping force of 600 dynes for each cm/sec of speed . The mass is initially displaced 10
cm above equilibrium and given a downward velocity of 1 m/sec . Find its displacement for t> 0.
17. A 192 lb weight is suspended from a spring with constant k= 6 lb/ft and subjected to an external",Elementary Differential Equations with Boundary Value Problems.pdf
"17. A 192 lb weight is suspended from a spring with constant k= 6 lb/ft and subjected to an external
force F(t) = 8 cos 3 tlb. Find the steady state component of the displacement for t >0 if the
medium resists the motion with a force equal to 8 times the spe ed in ft/sec.
18. A 2 gm mass is attached to a spring with constant 20 dyne/cm. Fi nd the steady state component of
the displacement if the mass is subjected to an external forc e F(t) = 3 cos 4 t− 5 sin 4 tdynes and
a dashpot supplies 4 dynes of damping for each cm/sec of veloc ity .
19. C/G A 96 lb weight is attached to a spring with constant 12 lb/ft. F ind and graph the steady state
component of the displacement if the mass is subjected to an e xternal force F(t) = 18 cos t−9 sin t
lb and a dashpot supplies 24 lb of damping for each ft/sec of ve locity .
20. A mass of one kg stretches a spring 49 cm in equilibrium. It is a ttached to a dashpot that supplies a",Elementary Differential Equations with Boundary Value Problems.pdf
"20. A mass of one kg stretches a spring 49 cm in equilibrium. It is a ttached to a dashpot that supplies a
damping force of 4 N for each m/sec of speed. Find the steady st ate component of its displacement
if it’s subjected to an external force F(t) = 8 sin 2 t− 6 cos 2 tN.
21. A mass mis suspended from a spring with constant kand subjected to an external force F(t) =
α cos ω0t+ β sin ω0t, where ω0 is the natural frequency of the spring–mass system without d amp-
ing. Find the steady state component of the displacement if a dashpot with constant csupplies
damping.
22. Show that if c1 and c2 are not both zero then
y= er1 t(c1 + c2t)
can’t equal zero for more than one value of t.
23. Show that if c1 and c2 are not both zero then
y = c1er1 t + c2er2 t
can’t equal zero for more than one value of t.
24. Find the solution of the initial value problem
my′′+ cy′+ ky = 0 , y (0) = y0, y′(0) = v0,
given that the motion is underdamped, so the general solutio n of the equation is",Elementary Differential Equations with Boundary Value Problems.pdf
"my′′+ cy′+ ky = 0 , y (0) = y0, y′(0) = v0,
given that the motion is underdamped, so the general solutio n of the equation is
y= e−ct/ 2m(c1 cos ω1t+ c2 sin ω1t).
25. Find the solution of the initial value problem
my′′+ cy′+ ky = 0 , y (0) = y0, y′(0) = v0,
given that the motion is overdamped, so the general solution of the equation is
y = c1er1 t + c2er2 t (r1,r 2 <0).",Elementary Differential Equations with Boundary Value Problems.pdf
"290 Chapter 6 Applications of Linear Second Order Equations
26. Find the solution of the initial value problem
my′′+ cy′+ ky = 0 , y (0) = y0, y′(0) = v0,
given that the motion is critically damped, so that the gener al solution of the equation is of the
form
y= er1 t(c1 + c2t) (r1 <0).
6.3 THE RLC CIRCUIT
In this section we consider the RLCcircuit, shown schematically in Figure 6.3.1. As we’ll see, the RLC
circuit is an electrical analog of a spring-mass system with damping.
Nothing happens while the switch is open (dashed line). When the switch is closed (solid line) we say
that the circuit is closed . Differences in electrical potential in a closed circuit ca use current to flow in
the circuit. The battery or generator in Figure 6.3.1 creates a difference in electrical potential E = E(t)
between its two terminals, which we’ve marked arbitrarily a s positive and negative. (W e could just as",Elementary Differential Equations with Boundary Value Problems.pdf
"between its two terminals, which we’ve marked arbitrarily a s positive and negative. (W e could just as
well interchange the markings.) W e’ll say that E(t) >0 if the potential at the positive terminal is greater
than the potential at the negative terminal, E(t) < 0 if the potential at the positive terminal is less than
the potential at the negative terminal, and E(t) = 0 if the potential is the same at the two terminals. W e
call Ethe impressed voltage .
Induction Coil(Inductance  L)
+ −−
Capacitor(Capacitance  C)
+
−−
Resistor(Resistance  R)
+
−−
Battery or Generator
(Impressed Voltage  E=E(t))
+ −−
Switch
Figure 6.3.1 An RLC circuit
At any time t, the same current flows in all points of the circuit. W e denote current by I = I(t). W e
say that I(t) >0 if the direction of flow is around the circuit from the positiv e terminal of the battery or
generator back to the negative terminal, as indicated by the arrows in Figure 6.3.1 I(t) <0 if the flow is",Elementary Differential Equations with Boundary Value Problems.pdf
"generator back to the negative terminal, as indicated by the arrows in Figure 6.3.1 I(t) <0 if the flow is
in the opposite direction, and I(t) = 0 if no current flows at time t.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.3 The RLC Circuit 291
Differences in potential occur at the resistor, induction c oil, and capacitor in Figure 6.3.1. Note that the
two sides of each of these components are also identified as po sitive and negative. The voltage drop across
each component is defined to be the potential on the positive s ide of the component minus the potential
on the negative side. This terminology is somewhat misleadi ng, since “drop” suggests a decrease even
though changes in potential are signed quantities and there fore may be increases. Nevertheless, we’ll go
along with tradition and call them voltage drops. The voltag e drop across the resistor in Figure 6.3.1 is
given by
VR = IR, (6.3.1)
where I is current and Ris a positive constant, the resistance of the resistor. The voltage drop across the
induction coil is given by
VI = LdI
dt = LI′, (6.3.2)
where Lis a positive constant, the inductance of the coil.",Elementary Differential Equations with Boundary Value Problems.pdf
"induction coil is given by
VI = LdI
dt = LI′, (6.3.2)
where Lis a positive constant, the inductance of the coil.
A capacitor stores electrical charge Q = Q(t), which is related to the current in the circuit by the
equation
Q(t) = Q0 +
∫ t
0
I(τ) dτ, (6.3.3)
where Q0 is the charge on the capacitor at t= 0 . The voltage drop across a capacitor is given by
VC = Q
C, (6.3.4)
where Cis a positive constant, the capacitance of the capacitor.
T able 6.3.8 names the units for the quantities that we’ve discussed. The units are defined so that
1 volt = 1 ampere ·1 ohm
= 1 henry ·1 ampere/ second
= 1 coulomb/ farad
and
1 ampere = 1 coulomb/ second.
T able 6.3.8. Electrical Units
Symbol Name Unit
E Impressed V oltage volt
I Current ampere
Q Charge coulomb
R Resistance ohm
L Inductance henry
C Capacitance farad
According to Kirchoff ’s law , the sum of the voltage drops in a closed RLCcircuit equals the impressed
voltage. Therefore, from ( 6.3.1), ( 6.3.2), and ( 6.3.4),
LI′+ RI+ 1
CQ= E(t). (6.3.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"292 Chapter 6 Applications of Linear Second Order Equations
This equation contains two unknowns, the current I in the circuit and the charge Q on the capacitor.
However, ( 6.3.3) implies that Q′= I, so ( 6.3.5) can be converted into the second order equation
LQ′′+ RQ′+ 1
CQ= E(t) (6.3.6)
in Q. T o find the current flowing in an RLC circuit, we solve ( 6.3.6) for Qand then differentiate the
solution to obtain I.
In Sections 6.1 and 6.2 we encountered the equation
my′′+ cy′+ ky = F(t) (6.3.7)
in connection with spring-mass systems. Except for notatio n this equation is the same as ( 6.3.6). The
correspondence between electrical and mechanical quantit ies connected with ( 6.3.6) and ( 6.3.7) is shown
in T able 6.3.9.
T able 6.3.9. Electrical and Mechanical Units
Electrical Mechanical
charge Q displacement y
curent I velocity y′
impressed voltage E(t) external force F(t)
inductance L Mass m
resistance R damping c
1/capacitance 1/C cpring constant k",Elementary Differential Equations with Boundary Value Problems.pdf
"curent I velocity y′
impressed voltage E(t) external force F(t)
inductance L Mass m
resistance R damping c
1/capacitance 1/C cpring constant k
The equivalence between ( 6.3.6) and ( 6.3.7) is an example of how mathematics unifies fundamental
similarities in diverse physical phenomena. Since we’ve al ready studied the properties of solutions of
(6.3.7) in Sections 6.1 and 6.2, we can obtain results concerning so lutions of ( 6.3.6) by simpling changing
notation, according to T able 6.3.8.
Free Oscillations
W e say that an RLC circuit is in free oscillation if E(t) = 0 for t> 0, so that ( 6.3.6) becomes
LQ′′+ RQ′+ 1
CQ= 0 . (6.3.8)
The characteristic equation of ( 6.3.8) is
Lr2 + Rr+ 1
C = 0 ,
with roots
r1 = −R−
√
R2 − 4L/C
2L and r2 = −R+
√
R2 − 4L/C
2L . (6.3.9)
There are three cases to consider, all analogous to the cases considered in Section 6.2 for free vibrations
of a damped spring-mass system.
CA SE 1. The oscillation is underdamped if R<
√",Elementary Differential Equations with Boundary Value Problems.pdf
"of a damped spring-mass system.
CA SE 1. The oscillation is underdamped if R<
√
4L/C . In this case, r1 and r2 in ( 6.3.9) are complex
conjugates, which we write as
r1 = − R
2L + iω1 and r2 = − R
2L − iω1,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.3 The RLC Circuit 293
where
ω1 =
√
4L/C − R2
2L .
The general solution of ( 6.3.8) is
Q= e−Rt/ 2L(c1 cos ω1t+ c2 sin ω1t),
which we can write as
Q= Ae−Rt/ 2L cos(ω1t− φ), (6.3.10)
where
A=
√
c2
1 + c2
2, A cos φ = c1, and Asin φ = c2.
In the idealized case where R= 0 , the solution (
6.3.10) reduces to
Q= Acos
( t√
LC
− φ
)
,
which is analogous to the simple harmonic motion of an undamp ed spring-mass system in free vibration.
Actual RLCcircuits are usually underdamped, so the case we’ve just con sidered is the most important.
However, for completeness we’ll consider the other two poss ibilities.
CA SE 2. The oscillation is overdamped if R >
√
4L/C . In this case, the zeros r1 and r2 of the
characteristic polynomial are real, with r1 <r2 <0 (see ( 6.3.9)), and the general solution of ( 6.3.8) is
Q= c1er1 t + c2er2 t. (6.3.11)
CA SE 3. The oscillation is critically damped if R =
√
4L/C . In this case, r1 = r2 = −R/ 2Land
the general solution of ( 6.3.8) is",Elementary Differential Equations with Boundary Value Problems.pdf
"CA SE 3. The oscillation is critically damped if R =
√
4L/C . In this case, r1 = r2 = −R/ 2Land
the general solution of ( 6.3.8) is
Q= e−Rt/ 2L(c1 + c2t). (6.3.12)
If R ̸= 0 , the exponentials in ( 6.3.10), ( 6.3.11), and ( 6.3.12) are negative, so the solution of any
homogeneous initial value problem
LQ′′+ RQ′+ 1
CQ= 0 , Q (0) = Q0, Q ′(0) = I0,
approaches zero exponentially as t → ∞ . Thus, all such solutions are transient, in the sense defined
Section 6.2 in the discussion of forced vibrations of a sprin g-mass system with damping.
Example 6.3.1 At t= 0 a current of 2 amperes flows in an RLC circuit with resistance R= 40 ohms,
inductance L = .2 henrys, and capacitance C = 10 −5 farads. Find the current flowing in the circuit at
t> 0 if the initial charge on the capacitor is 1 coulomb. Assume th at E(t) = 0 for t> 0.
Solution The equation for the charge Qis
1
5 Q′′+ 40 Q′+ 10000 Q= 0 ,
or
Q′′+ 200 Q′+ 50000 Q= 0 . (6.3.13)
Therefore we must solve the initial value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"1
5 Q′′+ 40 Q′+ 10000 Q= 0 ,
or
Q′′+ 200 Q′+ 50000 Q= 0 . (6.3.13)
Therefore we must solve the initial value problem
Q′′+ 200 Q′+ 50000 Q= 0 , Q (0) = 1 , Q ′(0) = 2 . (6.3.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"294 Chapter 6 Applications of Linear Second Order Equations
The desired current is the derivative of the solution of this initial value problem.
The characteristic equation of ( 6.3.13) is
r2 + 200 r+ 50000 = 0 ,
which has complex zeros r= −100 ± 200i. Therefore the general solution of ( 6.3.13) is
Q= e−100t(c1 cos 200 t+ c2 sin 200 t). (6.3.15)
Differentiating this and collecting like terms yields
Q′= −e−100t [(100c1 − 200c2) cos 200 t+ (100 c2 + 200 c1) sin 200 t] . (6.3.16)
T o find the solution of the initial value problem ( 6.3.14), we set t= 0 in ( 6.3.15) and ( 6.3.16) to obtain
c1 = Q(0) = 1 and − 100c1 + 200 c2 = Q′(0) = 2;
therefore, c1 = 1 and c2 = 51 / 100, so
Q= e−100t
(
cos 200 t+ 51
100 sin 200 t
)
is the solution of ( 6.3.14). Differentiating this yields
I = e−100t(2 cos 200 t− 251 sin 200 t).
Forced Oscillations With Damping
An initial value problem for ( 6.3.6) has the form
LQ′′+ RQ′+ 1
CQ= E(t), Q (0) = Q0, Q ′(0) = I0, (6.3.17)",Elementary Differential Equations with Boundary Value Problems.pdf
"Forced Oscillations With Damping
An initial value problem for ( 6.3.6) has the form
LQ′′+ RQ′+ 1
CQ= E(t), Q (0) = Q0, Q ′(0) = I0, (6.3.17)
where Q0 is the initial charge on the capacitor and I0 is the initial current in the circuit. W e’ve already
seen that if E ≡ 0 then all solutions of ( 6.3.17) are transient. If E ̸≡0, we know that the solution of
(6.3.17) has the form Q = Qc + Qp, where Qc satisfies the complementary equation, and approaches
zero exponentially as t→ ∞ for any initial conditions , while Qp depends only on Eand is independent
of the initial conditions. As in the case of forced oscillati ons of a spring-mass system with damping, we
call Qp the steady state charge on the capacitor of the RLC circuit. Since I = Q′= Q′
c + Q′
p and Q′
c
also tends to zero exponentially as t→ ∞ , we say that Ic = Q′
c is the
transient current and Ip = Q′
p is
the
steady state current. In most applications we’re interested only in the s teady state charge and current.",Elementary Differential Equations with Boundary Value Problems.pdf
"transient current and Ip = Q′
p is
the
steady state current. In most applications we’re interested only in the s teady state charge and current.
Example 6.3.2 Find the amplitude-phase form of the steady state current in the RLC circuit in Fig-
ure 6.3.1 if the impressed voltage, provided by an alternating curren t generator, is E(t) = E0 cos ωt.
Solution W e’ll first find the steady state charge on the capacitor as a pa rticular solution of
LQ′′+ RQ′+ 1
CQ= E0 cos ωt.
T o do, this we’ll simply reinterpret a result obtained in Sec tion 6.2, where we found that the steady state
solution of
my′′+ cy′+ ky = F0 cos ωt",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.3 The RLC Circuit 295
is
yp = F0√
(k− mω2 )2 + c2ω2 cos(ωt − φ),
where
cos φ = k− mω2
√
(k− mω2 )2 + c2ω2 and sin φ = cω√
(k− mω2 )2 + c2ω2 .
(See Equations ( 6.2.14) and ( 6.2.15).) By making the appropriate changes in the symbols (accord ing to
T able 2) yields the steady state charge
Qp = E0√
(1/C − Lω2)2 + R2ω2 cos(ωt − φ),
where
cos φ = 1/C − Lω2
√
(1/C − Lω2)2 + R2ω2 and sin φ = Rω√
(1/C − Lω2)2 + R2ω2 .
Therefore the steady state current in the circuit is
Ip = Q′
p = − ωE 0
√
(1/C − Lω2)2 + R2ω2 sin(ωt − φ).
6.3 Exercises
In Exercises 1-5 find the current in the RLC circuit, assuming that E(t) = 0 for t> 0.
1. R= 3 ohms; L= .1 henrys; C = .01 farads; Q0 = 0 coulombs; I0 = 2 amperes.
2. R= 2 ohms; L= .05 henrys; C = .01 farads’; Q0 = 2 coulombs; I0 = −2 amperes.
3. R= 2 ohms; L= .1 henrys; C = .01 farads; Q0 = 2 coulombs; I0 = 0 amperes.
4. R= 6 ohms; L= .1 henrys; C = .004 farads’; Q0 = 3 coulombs; I0 = −10 amperes.",Elementary Differential Equations with Boundary Value Problems.pdf
"4. R= 6 ohms; L= .1 henrys; C = .004 farads’; Q0 = 3 coulombs; I0 = −10 amperes.
5. R= 4 ohms; L= .05 henrys; C = .008 farads; Q0 = −1 coulombs; I0 = 2 amperes.
In Exercises 6-10 find the steady state current in the circuit described by the e quation.
6. 1
10 Q′′+ 3 Q′+ 100 Q= 5 cos 10 t− 5 sin 10 t
7. 1
20 Q′′+ 2 Q′+ 100 Q= 10 cos 25 t− 5 sin 25 t
8. 1
10 Q′′+ 2 Q′+ 100 Q= 3 cos 50 t− 6 sin 50 t
9. 1
10 Q′′+ 6 Q′+ 250 Q= 10 cos 100 t+ 30 sin 100 t
10. 1
20 Q′′+ 4 Q′+ 125 Q= 15 cos 30 t− 30 sin 30 t
11. Show that if E(t) = Ucos ωt + V sin ωt where Uand V are constants then the steady state current
in the RLC circuit shown in Figure 6.3.1 is
Ip = ω2RE(t) + (1 /C − Lω2)E′(t)
∆ ,
where
∆ = (1 /C − Lω2)2 + R2ω2.",Elementary Differential Equations with Boundary Value Problems.pdf
"296 Chapter 6 Applications of Linear Second Order Equations
12. Find the amplitude of the steady state current Ip in the RLC circuit shown in Figure 6.3.1 if
E(t) = Ucos ωt + V sin ωt, where U and V are constants. Then find the value ω0 of ω maximizes
the amplitude, and find the maximum amplitude.
In Exercises 13-17 plot the amplitude of the steady state current against ω. Estimate the value of ω that
maximizes the amplitude of the steady state current, and est imate this maximum amplitude. HIN T : Y ou
can confirm your results by doing Exercise 12.
13. L 1
10 Q′′+ 3 Q′+ 100 Q= Ucos ωt + V sin ωt
14. L 1
20 Q′′+ 2 Q′+ 100 Q= Ucos ωt + V sin ωt
15. L 1
10 Q′′+ 2 Q′+ 100 Q= Ucos ωt + V sin ωt
16. L 1
10 Q′′+ 6 Q′+ 250 Q= Ucos ωt + V sin ωt
17. L 1
20 Q′′+ 4 Q′+ 125 Q= Ucos ωt + V sin ωt
6.4 MOTION UNDER A CENTRAL FORCE
W e’ll now study the motion of a object moving under the influen ce of a central force ; that is, a force",Elementary Differential Equations with Boundary Value Problems.pdf
"6.4 MOTION UNDER A CENTRAL FORCE
W e’ll now study the motion of a object moving under the influen ce of a central force ; that is, a force
whose magnitude at any point P other than the origin depends only on the distance from P to the origin,
and whose direction at P is parallel to the line connecting P and the origin, as indicated in Figure 6.4.1
for the case where the direction of the force at every point is toward the origin. Gravitational forces
are central forces; for example, as mentioned in Section 4.3 , if we assume that Earth is a perfect sphere
with constant mass density then Newton’s law of gravitation asserts that the force exerted on an object
by Earth’s gravitational field is proportional to the mass of the object and inversely proportional to the
square of its distance from the center of Earth, which we take to be the origin.
If the initial position and velocity vectors of an object mov ing under a central force are parallel, then",Elementary Differential Equations with Boundary Value Problems.pdf
"If the initial position and velocity vectors of an object mov ing under a central force are parallel, then
the subsequent motion is along the line from the origin to the initial position. Here we’ll assume that the
initial position and velocity vectors are not parallel; in t his case the subsequent motion is in the plane
determined by them. For convenience we take this to be the xy-plane. W e’ll consider the problem of
determining the curve traversed by the object. W e call this c urve the orbit.
W e can represent a central force in terms of polar coordinate s
x= rcos θ, y = rsin θ
as
F(r,θ ) = f(r)(cos θi + sin θ j).
W e assume that f is continuous for all r >0. The magnitude of F at (x,y ) = ( rcos θ,r sin θ) is |f(r)|,
so it depends only on the distance rfrom the point to the origin the direction of F is from the point to the
origin if f(r) <0, or from the origin to the point if f(r) >0. W e’ll show that the orbit of an object with
mass mmoving under this force is given by
r(θ) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"mass mmoving under this force is given by
r(θ) = 1
u(θ) ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.4 Motion Under a Central Force 297
 x
 y
Figure 6.4.1
where uis solution of the differential equation
d2u
dθ2 + u= − 1
mh2u2 f(1/u ), (6.4.1)
and his a constant defined below .
Newton’s second law of motion ( F = ma) says that the polar coordinates r= r(t) and θ = θ(t) of the
particle satisfy the vector differential equation
m(rcos θ i + rsin θj)′′= f(r)(cos θ i + sin θj). (6.4.2)
T o deal with this equation we introduce the unit vectors
e1 = cos θi + sin θ j and e2 = − sin θi + cos θj.
Note that e1 points in the direction of increasing r and e2 points in the direction of increasing θ (Fig-
ure 6.4.2); moreover,
de1
dθ = e2, de2
dθ = −e1, (6.4.3)
and
e1 ·e2 = cos θ(− sin θ) + sin θ cos θ = 0 ,
so e1 and e2 are perpendicular. Recalling that the single prime (′) stands for differentiation with respect
to t, we see from ( 6.4.3) and the chain rule that
e′
1 = θ′e2 and e′
2 = −θ′e1. (6.4.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"298 Chapter 6 Applications of Linear Second Order Equations
 e1
 e2
 x
 y
Figure 6.4.2
Now we can write ( 6.4.2) as
m(re1)′′= f(r)e1. (6.4.5)
But
(re1)′= r′e1 + re′
1 = r′e1 + rθ′e2
(from (
6.4.4)), and
(re1)′′ = ( r′e1 + rθ′e2)′
= r′′e1 + r′e′
1 + ( rθ′′+ r′θ′)e2 + rθ′e′
2
= r′′e1 + r′θ′e2 + ( rθ′′+ r′θ′)e2 − r(θ′)2e1 (from (
6.4.4))
=
(
r′′− r(θ′)2)
e1 + ( rθ′′+ 2 r′θ′)e2.
Substituting this into ( 6.4.5) yields
m(r′′− r(θ′)2) e1 + m(rθ′′+ 2 r′θ′)e2 = f(r)e1.
By equating the coefficients of e1 and e2 on the two sides of this equation we see that
m
(
r′′− r(θ′)2)
= f(r) (6.4.6)
and
rθ′′+ 2 r′θ′= 0 .
Multiplying the last equation by ryields
r2θ′′+ 2 rr′θ′= ( r2θ′)′= 0 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.4 Motion Under a Central Force 299
so
r2θ′= h, (6.4.7)
where his a constant that we can write in terms of the initial conditi ons as
h= r2(0)θ′(0).
Since the initial position and velocity vectors are
r(0)e1(0) and r′(0)e1(0) + r(0)θ′(0)e2(0),
our assumption that these two vectors are not parallel impli es that θ′(0) ̸= 0 , so h̸= 0 .
Now let u= 1 /r . Then u2 = θ′/h (from ( 6.4.7)) and
r′= − u′
u2 = −h
(u′
θ′
)
,
which implies that
r′= −hdu
dθ , (6.4.8)
since
u′
θ′ = du
dt
/ dθ
dt = du
dθ .
Differentiating ( 6.4.8) with respect to tyields
r′′= −hd
dt
(du
dθ
)
= −hd2u
dθ2 θ′,
which implies that
r′′= −h2u2 d2u
dθ2 since θ′= hu2.
Substituting from these equalities into ( 6.4.6) and recalling that r= 1 /u yields
−m
(
h2u2 d2u
dθ2 + 1
uh2u4
)
= f(1/u ),
and dividing through by −mh2u2 yields ( 6.4.1).
Eqn. ( 6.4.7) has the following geometrical interpretation, which is kn own as Kepler’s Second Law .",Elementary Differential Equations with Boundary Value Problems.pdf
"and dividing through by −mh2u2 yields ( 6.4.1).
Eqn. ( 6.4.7) has the following geometrical interpretation, which is kn own as Kepler’s Second Law .
Theorem 6.4.1 The position vector of an object moving under a central force sweeps out equal areas in
equal times; more precisely , if θ(t1) ≤ θ(t2 ) then the (signed) area of the sector
{(x,y ) = ( rcos θ,r sin θ) : 0 ≤ r≤ r(θ), θ(t1) ≤ θ(t2)}
(Figure 6.4.3) is given by
A= h(t2 − t1)
2 ,
where h= r2θ′, which we have shown to be constant.",Elementary Differential Equations with Boundary Value Problems.pdf
"300 Chapter 6 Applications of Linear Second Order Equations
 θ  = θ  ( t2 )
 θ  = θ  ( t1 )
 x
 y
Figure 6.4.3
Proof Recall from calculus that the area of the shaded sector in Fig ure 6.4.3 is
A= 1
2
∫ θ (t2)
θ (t1)
r2(θ) dθ,
where r= r(θ) is the polar representation of the orbit. Making the change o f variable θ = θ(t) yields
A= 1
2
∫ t2
t1
r2(θ(t))θ′(t) dt. (6.4.9)
But ( 6.4.7) and ( 6.4.9) imply that
A= 1
2
∫ t2
t1
hdt = h(t2 − t1)
2 ,
which completes the proof.
Motion Under an Inverse Square Law Force
In the special case where f(r) = −mk/r 2 = −mku2, so F can be interpreted as a gravitational force,
(6.4.1) becomes
d2u
dθ2 + u= k
h2 . (6.4.10)
The general solution of the complementary equation
d2u
dθ2 + u= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.4 Motion Under a Central Force 301
can be written in amplitude–phase form as
u= Acos(θ − φ),
where A ≥ 0 and φ is a phase angle. Since up = k/h 2 is a particular solution of ( 6.4.10), the general
solution of ( 6.4.10) is
u= Acos(θ − φ) + k
h2 ;
hence, the orbit is given by
r=
(
Acos(θ − φ) + k
h2
) −1
,
which we rewrite as
r= ρ
1 + ecos(θ − φ) , (6.4.11)
where
ρ = h2
k and e= Aρ.
A curve satisfying ( 6.4.11) is a conic section with a focus at the origin (Exercise 1). The nonnegative
constant eis the eccentricity of the orbit, which is an ellipse if e< 1 ellipse (a circle if e= 0 ), a parabola
if e= 1 , or a hyperbola if e> 1.
 A
 P
 x
 y
Figure 6.4.4
If the orbit is an ellipse, then the minimum and maximum value s of rare
rmin = ρ
1 + e (the perihelion distance , attained when θ = φ)
rmax = ρ
1 − e (the aphelion distance , attained when θ = φ + π).
Figure 6.4.4 shows a typical elliptic orbit. The point P on the orbit where r = rmin is the perigee and",Elementary Differential Equations with Boundary Value Problems.pdf
"Figure 6.4.4 shows a typical elliptic orbit. The point P on the orbit where r = rmin is the perigee and
the point Awhere r= rmax is the apogee.",Elementary Differential Equations with Boundary Value Problems.pdf
"302 Chapter 6 Applications of Linear Second Order Equations
For example, Earth’s orbit around the Sun is approximately a n ellipse with e≈ .017, rmin ≈ 91 × 106
miles, and rmax ≈ 95 × 106 miles. Halley’s comet has a very elongated approximately el liptical orbit
around the sun, with e ≈ .967, rmin ≈ 55 × 106 miles, and rmax ≈ 33 × 108 miles. Some comets (the
nonrecurring type) have parabolic or hyperbolic orbits.
6.4 Exercises
1. Find the equation of the curve
r= ρ
1 + ecos(θ − φ) (A)
in terms of (X,Y ) = ( rcos(θ − φ),r sin(θ − φ)), which are rectangular coordinates with respect
to the axes shown in Figure 6.4.5. Use your results to verify that (A) is the equation of an elli pse
if 0 <e< 1, a parabola if e= 1 , or a hyperbola if e> 1. If e< 1, leave your answer in the form
(X− X0)2
a2 + (Y − Y0)2
b2 = 1 ,
and show that the area of the ellipse is
A= πρ 2
(1 − e2)3/ 2 .
Then use Theorem 6.4.1 to show that the time required for the object to traverse the e ntire orbit is
T = 2πρ 2",Elementary Differential Equations with Boundary Value Problems.pdf
"A= πρ 2
(1 − e2)3/ 2 .
Then use Theorem 6.4.1 to show that the time required for the object to traverse the e ntire orbit is
T = 2πρ 2
h(1 − e2)3/ 2 .
(This is Kepler’s third law ; T is called the period of the orbit.)
 x
 y
 X Y
 φ
Figure 6.4.5",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.4 Motion Under a Central Force 303
2. Suppose an object with mass mmoves in the xy-plane under the central force
F(r,θ ) = −mk
r2 (cos θ i + sin θj),
where kis a positive constant. As we shown, the orbit of the object is given by
r= ρ
1 + ecos(θ − φ) .
Determine ρ, e, and φ in terms of the initial conditions
r(0) = r0, r ′(0) = r′
0, and θ(0) = θ0 , θ ′(0) = θ′
0.
Assume that the initial position and velocity vectors are no t collinear.
3. Suppose we wish to put a satellite with mass minto an elliptical orbit around Earth. Assume that
the only force acting on the object is Earth’s gravity , given by
F(r,θ ) = −mg
(R2
r2
)
(cos θ i + sin θj),
where Ris Earth’s radius, g is the acceleration due to gravity at Earth’s surface, and rand θ are
polar coordinates in the plane of the orbit, with the origin a t Earth’s center.
(a) Find the eccentricity required to make the aphelion and peri helion distances equal to Rγ1
and Rγ2, respectively , where 1 <γ 1 <γ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) Find the eccentricity required to make the aphelion and peri helion distances equal to Rγ1
and Rγ2, respectively , where 1 <γ 1 <γ 2.
(b) Find the initial conditions
r(0) = r0, r ′(0) = r′
0, and θ(0) = θ0, θ ′(0) = θ′
0
required to make the initial point the perigee, and the motio n along the orbit in the direction
of increasing θ. H IN T : Use the results of Exercise 2.
4. An object with mass mmoves in a spiral orbit r= cθ2 under a central force
F(r,θ ) = f(r)(cos θi + sin θ j).
Find f.
5. An object with mass mmoves in the orbit r= r0eγθ under a central force
F(r,θ ) = f(r)(cos θi + sin θ j).
Find f.
6. Suppose an object with mass mmoves under the central force
F(r,θ ) = −mk
r3 (cos θ i + sin θj),
with
r(0) = r0, r ′(0) = r′
0, and θ(0) = θ0 , θ ′(0) = θ′
0,
where h= r2
0θ′
0 ̸= 0 .
(a) Set up a second order initial value problem for u= 1 /r as a function of θ.
(b) Determine r= r(θ) if (i) h2 <k; (ii) h2 = k; (iii) h2 >k.",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 7
Series Solutions of Linear SecondEquations
IN THIS CHAPTER we study a class of second order differential equations that occur in many applica-
tions, but can’t be solved in closed form in terms of elementa ry functions. Here are some examples:
(1) Bessel’s equation
x2y′′+ xy′+ ( x2 − ν2)y = 0 ,
which occurs in problems displaying cylindrical symmetry , such as diffraction of light through a circular
aperture, propagation of electromagnetic radiation throu gh a coaxial cable, and vibrations of a circular
drum head.
(2)
Airy’s equation ,
y′′− xy= 0 ,
which occurs in astronomy and quantum physics.
(3) Legendre’s equation
(1 − x2)y′′− 2xy′+ α(α + 1) y= 0 ,
which occurs in problems displaying spherical symmetry , pa rticularly in electromagnetism.
These equations and others considered in this chapter can be written in the form
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 , (A)
where P0, P1, and P2 are polynomials with no common factor. For most equations th at occur in appli-",Elementary Differential Equations with Boundary Value Problems.pdf
"P0(x)y′′+ P1(x)y′+ P2(x)y = 0 , (A)
where P0, P1, and P2 are polynomials with no common factor. For most equations th at occur in appli-
cations, these polynomials are of degree two or less. W e’ll i mpose this restriction, although the methods
that we’ll develop can be extended to the case where the coeffi cient functions are polynomials of arbitrary
degree, or even power series that converge in some circle aro und the origin in the complex plane.
Since (A) does not in general have closed form solutions, we s eek series representations for solutions.
W e’ll see that if P0(0) ̸= 0 then solutions of (A) can be written as power series
y=
∞∑
n=0
an xn
that converge in an open interval centered at x= 0 .
305",Elementary Differential Equations with Boundary Value Problems.pdf
"306 Chapter 6 Applications of Linear Second Order Equations
SECTION 7.1 reviews the properties of power series.
SECTIONS 7.2 AND 7.3 are devoted to finding power series solut ions of (A) in the case where P0(0) ̸= 0 .
The situation is more complicated if P0(0) = 0 ; however, if P1 and P2 satisfy assumptions that apply to
most equations of interest, then we’re able to use a modified s eries method to obtain solutions of (A).
SECTION 7.4 introduces the appropriate assumptions on P1 and P2 in the case where P0(0) = 0 , and
deals with
Euler’s equation
ax2y′′+ bxy′+ cy= 0 ,
where a, b, and care constants. This is the simplest equation that satisfies t hese assumptions.
SECTIONS 7.5 –7.7 deal with three distinct cases satisfying the assumptions introduced in Section 7.4.
In all three cases, (A) has at least one solution of the form
y1 = xr
∞∑
n=0
anxn ,
where rneed not be an integer. The problem is that there are three pos sibilities – each requiring a different",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 = xr
∞∑
n=0
anxn ,
where rneed not be an integer. The problem is that there are three pos sibilities – each requiring a different
approach – for the form of a second solution y2 such that {y1,y 2}is a fundamental pair of solutions of (A).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 307
7.1 REVIEW OF POWER SERIES
Many applications give rise to differential equations with solutions that can’t be expressed in terms of
elementary functions such as polynomials, rational functi ons, exponential and logarithmic functions, and
trigonometric functions. The solutions of some of the most i mportant of these equations can be expressed
in terms of power series. W e’ll study such equations in this c hapter. In this section we review relevant
properties of power series. W e’ll omit proofs, which can be f ound in any standard calculus text.
Definition 7.1.1 An infinite series of the form
∞∑
n=0
an(x− x0)n , (7.1.1)
where x0 and a0, a1, . . . , an, . . . are constants , is called a power series in x− x0. W e say that the power
series ( 7.1.1) converges for a given xif the limit
lim
N →∞
N∑
n=0
an (x− x0)n
exists; otherwise, we say that the power series diverges for the given x.",Elementary Differential Equations with Boundary Value Problems.pdf
"lim
N →∞
N∑
n=0
an (x− x0)n
exists; otherwise, we say that the power series diverges for the given x.
A power series in x− x0 must converge if x = x0, since the positive powers of x− x0 are all zero
in this case. This may be the only value of xfor which the power series converges. However, the next
theorem shows that if the power series converges for some x̸= x0 then the set of all values of xfor which
it converges forms an interval.
Theorem 7.1.2
F or any power series
∞∑
n=0
an (x− x0)n ,
exactly one of the these statements is true :
(i) The power series converges only for x= x0.
(ii) The power series converges for all values of x.
(iii) There’s a positive number Rsuch that the power series converges if |x− x0|< Rand diverges
if |x− x0|>R.
In case (iii) we say that Ris the radius of convergence of the power series. For convenience, we include
the other two cases in this definition by defining R= 0 in case (i) and R= ∞ in case (ii). W e define the
open interval of convergence of ∑ ∞",Elementary Differential Equations with Boundary Value Problems.pdf
"the other two cases in this definition by defining R= 0 in case (i) and R= ∞ in case (ii). W e define the
open interval of convergence of ∑ ∞
n=0 an (x− x0)n to be
(x0 − R,x 0 + R) if 0 <R< ∞ , or (−∞ , ∞ ) if R= ∞ .
If Ris finite, no general statement can be made concerning conver gence at the endpoints x= x0 ± Rof
the open interval of convergence; the series may converge at one or both points, or diverge at both.
Recall from calculus that a series of constants ∑ ∞
n=0 αn is said to
converge absolutely if the series
of absolute values ∑ ∞
n=0 |αn|converges. It can be shown that a power series ∑ ∞
n=0 an (x− x0)n with
a positive radius of convergence Rconverges absolutely in its open interval of convergence; t hat is, the
series ∞∑
n=0
|an||x− x0|n",Elementary Differential Equations with Boundary Value Problems.pdf
"308 Chapter 7 Series Solutions of Linear Second Equations
of absolute values converges if |x− x0|< R. However, if R <∞ , the series may fail to converge
absolutely at an endpoint x0 ± R, even if it converges there.
The next theorem provides a useful method for determining th e radius of convergence of a power
series. It’s derived in calculus by applying the ratio test t o the corresponding series of absolute values.
For related theorems see Exercises 2 and 4.
Theorem 7.1.3 Suppose there’s an integer N such that an ̸= 0 if n≥ N and
lim
n→∞
⏐
⏐
⏐
⏐
an+1
an
⏐
⏐
⏐
⏐= L,
where 0 ≤ L≤ ∞ . Then the radius of convergence of ∑ ∞
n=0 an(x− x0)n is R= 1 /L, which should be
interpreted to mean that R= 0 if L= ∞ , or R= ∞ if L= 0 .
Example 7.1.1 Find the radius of convergence of the series:
(a)
∞∑
n=0
n!xn (b)
∞∑
n=10
(−1)n xn
n! (c)
∞∑
n=0
2nn2(x− 1)n .
SO LU TIO N (a) Here an = n!, so
lim
n→∞
⏐
⏐
⏐
⏐
an+1
an
⏐
⏐
⏐
⏐= lim
n→∞
(n+ 1)!
n! = lim
n→∞
(n+ 1) = ∞ .
Hence, R= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"n! (c)
∞∑
n=0
2nn2(x− 1)n .
SO LU TIO N (a) Here an = n!, so
lim
n→∞
⏐
⏐
⏐
⏐
an+1
an
⏐
⏐
⏐
⏐= lim
n→∞
(n+ 1)!
n! = lim
n→∞
(n+ 1) = ∞ .
Hence, R= 0 .
SO LU TIO N (b) Here an = (1) n /n ! for n≥ N = 10 , so
lim
n→∞
⏐
⏐
⏐
⏐
an+1
an
⏐
⏐
⏐
⏐= lim
n→∞
n!
(n+ 1)! = lim
n→∞
1
n+ 1 = 0 .
Hence, R= ∞ .
SO LU TIO N (c) Here an = 2 n n2, so
lim
n→∞
⏐
⏐
⏐
⏐
an+1
an
⏐
⏐
⏐
⏐= lim
n→∞
2n+1(n+ 1) 2
2n n2 = 2 lim
n→∞
(
1 + 1
n
) 2
= 2 .
Hence, R= 1 / 2.
T aylor Series
If a function f has derivatives of all orders at a point x = x0, then the T aylor series of f about x0 is
defined by
∞∑
n=0
f(n)(x0)
n! (x− x0)n .
In the special case where x0 = 0 , this series is also called the Maclaurin series of f.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 309
T aylor series for most of the common elementary functions co nverge to the functions on their open
intervals of convergence. For example, you are probably fam iliar with the following Maclaurin series:
ex =
∞∑
n=0
xn
n! , −∞ <x< ∞ , (7.1.2)
sin x =
∞∑
n=0
(−1)n x2n+1
(2n+ 1)! , −∞ <x< ∞ , (7.1.3)
cos x =
∞∑
n=0
(−1)n x2n
(2n)! , −∞ <x< ∞ , (7.1.4)
1
1 − x =
∞∑
n=0
xn , −1 <x< 1. (7.1.5)
Differentiation of Power Series
A power series with a positive radius of convergence defines a function
f(x) =
∞∑
n=0
an (x− x0)n
on its open interval of convergence. W e say that the series represents f on the open interval of conver-
gence. A function f represented by a power series may be a familiar elementary fu nction as in ( 7.1.2)–
(7.1.5); however, it often happens that f isn’t a familiar function, so the series actually defines f.
The next theorem shows that a function represented by a power series has derivatives of all orders on",Elementary Differential Equations with Boundary Value Problems.pdf
"The next theorem shows that a function represented by a power series has derivatives of all orders on
the open interval of convergence of the power series, and pro vides power series representations of the
derivatives.
Theorem 7.1.4
A power series
f(x) =
∞∑
n=0
an (x− x0)n
with positive radius of convergence Rhas derivatives of all orders in its open interval of converg ence,
and successive derivatives can be obtained by repeatedly di fferentiating term by term ; that is,
f′(x) =
∞∑
n=1
nan (x− x0)n−1, (7.1.6)
f′′(x) =
∞∑
n=2
n(n− 1)an (x− x0)n−2, (7.1.7)
.
.
.
f(k) (x) =
∞∑
n=k
n(n− 1) · · ·(n− k+ 1) an(x− x0)n−k.
(7.1.8)
Moreover, all of these series have the same radius of convergence R.",Elementary Differential Equations with Boundary Value Problems.pdf
"310 Chapter 7 Series Solutions of Linear Second Equations
Example 7.1.2 Let f(x) = sin x. From ( 7.1.3),
f(x) =
∞∑
n=0
(−1)n x2n+1
(2n+ 1)! .
From ( 7.1.6),
f′(x) =
∞∑
n=0
(−1)n d
dx
[ x2n+1
(2n+ 1)!
]
=
∞∑
n=0
(−1)n x2n
(2n)! ,
which is the series ( 7.1.4) for cos x.
Uniqueness of Power Series
The next theorem shows that if f is defined by a power series in x− x0 with a positive radius of conver-
gence, then the power series is the T aylor series of f about x0.
Theorem 7.1.5 If the power series
f(x) =
∞∑
n=0
an (x− x0)n
has a positive radius of convergence, then
an = f(n)(x0)
n! ; (7.1.9)
that is, ∑ ∞
n=0 an (x− x0)n is the T aylor series of f about x0.
This result can be obtained by setting x= x0 in ( 7.1.8), which yields
f(k)(x0) = k(k− 1) · · ·1 ·ak = k!ak.
This implies that
ak = f(k)(x0)
k! .
Except for notation, this is the same as ( 7.1.9).
The next theorem lists two important properties of power ser ies that follow from Theorem 7.1.5.
Theorem 7.1.6
(a) If
∞∑
n=0
an (x− x0)n =
∞∑",Elementary Differential Equations with Boundary Value Problems.pdf
"The next theorem lists two important properties of power ser ies that follow from Theorem 7.1.5.
Theorem 7.1.6
(a) If
∞∑
n=0
an (x− x0)n =
∞∑
n=0
bn (x− x0)n
for all xin an open interval that contains x0, then an = bn for n= 0 , 1, 2, . . . .
(b) If
∞∑
n=0
an(x− x0)n = 0
for all xin an open interval that contains x0, then an = 0 for n= 0 , 1, 2, . . . .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 311
T o obtain (a) we observe that the two series represent the same function f on the open interval; hence,
Theorem 7.1.5 implies that
an = bn = f(n)(x0)
n! , n = 0 , 1, 2,....
(b) can be obtained from (a) by taking bn = 0 for n= 0 , 1, 2, . . . .
T aylor Polynomials
If f has N derivatives at a point x0, we say that
TN (x) =
N∑
n=0
f(n)(x0)
n! (x− x0)n
is the N-th T aylor polynomial of f about x0. This definition and Theorem 7.1.5 imply that if
f(x) =
∞∑
n=0
an (x− x0)n ,
where the power series has a positive radius of convergence , then the T aylor polynomials of f about x0
are given by
TN (x) =
N∑
n=0
an(x− x0)n .
In numerical applications, we use the T aylor polynomials to approximate f on subintervals of the open
interval of convergence of the power series. For example, ( 7.1.2) implies that the T aylor polynomial TN
of f(x) = ex is
TN (x) =
N∑
n=0
xn
n! .",Elementary Differential Equations with Boundary Value Problems.pdf
"interval of convergence of the power series. For example, ( 7.1.2) implies that the T aylor polynomial TN
of f(x) = ex is
TN (x) =
N∑
n=0
xn
n! .
The solid curve in Figure 7.1.1 is the graph of y = ex on the interval [0, 5]. The dotted curves in
Figure 7.1.1 are the graphs of the T aylor polynomials T1, . . . , T6 of y = ex about x0 = 0 . From this
figure, we conclude that the accuracy of the approximation of y = ex by its T aylor polynomial TN
improves as N increases.
Shifting the Summation Index
In Definition 7.1.1 of a power series in x− x0, the n-th term is a constant multiple of (x− x0)n. This
isn’t true in ( 7.1.6), ( 7.1.7), and ( 7.1.8), where the general terms are constant multiples of (x− x0)n−1,
(x− x0)n−2, and (x− x0)n−k, respectively . However, these series can all be rewritten s o that their n-th
terms are constant multiples of (x− x0)n. For example, letting n= k+ 1 in the series in ( 7.1.6) yields
f′(x) =
∞∑
k=0
(k+ 1) ak+1(x− x0)k, (7.1.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"terms are constant multiples of (x− x0)n. For example, letting n= k+ 1 in the series in ( 7.1.6) yields
f′(x) =
∞∑
k=0
(k+ 1) ak+1(x− x0)k, (7.1.10)
where we start the new summation index kfrom zero so that the first term in (
7.1.10) (obtained by setting
k= 0 ) is the same as the first term in ( 7.1.6) (obtained by setting n= 1 ). However, the sum of a series is
independent of the symbol used to denote the summation index , just as the value of a definite integral is
independent of the symbol used to denote the variable of inte gration. Therefore we can replace kby nin
(7.1.10) to obtain
f′(x) =
∞∑
n=0
(n+ 1) an+1(x− x0)n, (7.1.11)",Elementary Differential Equations with Boundary Value Problems.pdf
"312 Chapter 7 Series Solutions of Linear Second Equations
 x
 y
1 2 3 4 5
 N = 1
 N = 2
 N = 3
 N = 4
 N = 5
 N = 6
Figure 7.1.1 Approximation of y = ex by T aylor polynomials about x= 0
where the general term is a constant multiple of (x− x0)n.
It isn’t really necessary to introduce the intermediate sum mation index k. W e can obtain ( 7.1.11)
directly from ( 7.1.6) by replacing nby n+ 1 in the general term of ( 7.1.6) and subtracting 1 from the
lower limit of ( 7.1.6). More generally , we use the following procedure for shifti ng indices.
Shifting the Summation Index in a Power Series
For any integer k, the power series
∞∑
n=n0
bn (x− x0)n−k
can be rewritten as ∞∑
n=n0−k
bn+k(x− x0)n ;
that is, replacing n by n+ k in the general term and subtracting k from the lower limit of summation
leaves the series unchanged.
Example 7.1.3 Rewrite the following power series from ( 7.1.7) and ( 7.1.8) so that the general term in",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 313
each is a constant multiple of (x− x0)n :
(a)
∞∑
n=2
n(n− 1)an(x− x0)n−2 (b)
∞∑
n=k
n(n− 1) · · ·(n− k+ 1) an(x− x0)n−k.
SO LU TIO N (a) Replacing n by n+ 2 in the general term and subtracting 2 from the lower limit of
summation yields
∞∑
n=2
n(n− 1)an (x− x0)n−2 =
∞∑
n=0
(n+ 2)( n+ 1) an+2(x− x0)n.
SO LU TIO N (b) Replacing n by n + k in the general term and subtracting k from the lower limit of
summation yields
∞∑
n=k
n(n− 1) · · ·(n− k+ 1) an (x− x0)n−k =
∞∑
n=0
(n+ k)(n+ k− 1) · · ·(n+ 1) an+k(x− x0)n .
Example 7.1.4 Given that
f(x) =
∞∑
n=0
an xn,
write the function xf′′as a power series in which the general term is a constant multi ple of xn.
Solution From Theorem 7.1.4 with x0 = 0 ,
f′′(x) =
∞∑
n=2
n(n− 1)an xn−2.
Therefore
xf′′(x) =
∞∑
n=2
n(n− 1)an xn−1.
Replacing nby n+ 1 in the general term and subtracting 1 from the lower limit of summation yields
xf′′(x) =
∞∑
n=1
(n+ 1) nan+1xn .
W e can also write this as
xf′′(x) =
∞∑
n=0",Elementary Differential Equations with Boundary Value Problems.pdf
"xf′′(x) =
∞∑
n=1
(n+ 1) nan+1xn .
W e can also write this as
xf′′(x) =
∞∑
n=0
(n+ 1) nan+1xn ,
since the first term in this last series is zero. (W e’ll see lat er that sometimes it’s useful to include zero
terms at the beginning of a series.)
Linear Combinations of Power Series
If a power series is multiplied by a constant, then the consta nt can be placed inside the summation; that
is,
c
∞∑
n=0
an (x− x0)n =
∞∑
n=0
can (x− x0)n.",Elementary Differential Equations with Boundary Value Problems.pdf
"314 Chapter 7 Series Solutions of Linear Second Equations
T wo power series
f(x) =
∞∑
n=0
an (x− x0)n and g(x) =
∞∑
n=0
bn (x− x0)n
with positive radii of convergence can be added term by term a t points common to their open intervals of
convergence; thus, if the first series converges for |x− x0|<R1 and the second converges for |x− x0|<
R2, then
f(x) + g(x) =
∞∑
n=0
(an + bn )(x− x0)n
for |x− x0|< R, where Ris the smaller of R1 and R2. More generally , linear combinations of power
series can be formed term by term; for example,
c1f(x) + c2f(x) =
∞∑
n=0
(c1an + c2bn)(x− x0)n .
Example 7.1.5 Find the Maclaurin series for cosh xas a linear combination of the Maclaurin series for
ex and e−x.
Solution By definition,
cosh x= 1
2 ex + 1
2 e−x.
Since
ex =
∞∑
n=0
xn
n! and e−x =
∞∑
n=0
(−1)n xn
n! ,
it follows that
cosh x=
∞∑
n=0
1
2 [1 + ( −1)n ] xn
n! . (7.1.12)
Since
1
2 [1 + ( −1)n ] =
{
1 if n= 2 m, an even integer ,
0 if n= 2 m+ 1 , an odd integer ,
we can rewrite ( 7.1.12) more simply as",Elementary Differential Equations with Boundary Value Problems.pdf
"n! . (7.1.12)
Since
1
2 [1 + ( −1)n ] =
{
1 if n= 2 m, an even integer ,
0 if n= 2 m+ 1 , an odd integer ,
we can rewrite ( 7.1.12) more simply as
cosh x=
∞∑
m=0
x2m
(2m)! .
This result is valid on (−∞ , ∞ ), since this is the open interval of convergence of the Maclau rin series for
ex and e−x.
Example 7.1.6 Suppose
y=
∞∑
n=0
an xn
on an open interval I that contains the origin.
(a) Express
(2 − x)y′′+ 2 y
as a power series in xon I.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 315
(b) Use the result of (a) to find necessary and sufficient conditions on the coefficient s {an }for yto be
a solution of the homogeneous equation
(2 − x)y′′+ 2 y = 0 (7.1.13)
on I.
SO LU TIO N (a) From ( 7.1.7) with x0 = 0 ,
y′′=
∞∑
n=2
n(n− 1)anxn−2.
Therefore
(2 − x)y′′+ 2 y = 2 y′′− xy′+ 2 y
=
∞∑
n=2
2n(n− 1)an xn−2 −
∞∑
n=2
n(n− 1)anxn−1 +
∞∑
n=0
2anxn . (7.1.14)
T o combine the three series we shift indices in the first two to make their general terms constant multiples
of xn ; thus,
∞∑
n=2
2n(n− 1)an xn−2 =
∞∑
n=0
2(n+ 2)( n+ 1) an+2xn (7.1.15)
and ∞∑
n=2
n(n− 1)anxn−1 =
∞∑
n=1
(n+ 1) nan+1xn =
∞∑
n=0
(n+ 1) nan+1xn , (7.1.16)
where we added a zero term in the last series so that when we sub stitute from ( 7.1.15) and ( 7.1.16) into
(7.1.14) all three series will start with n= 0 ; thus,
(2 − x)y′′+ 2 y=
∞∑
n=0
[2(n+ 2)( n+ 1) an+2 − (n+ 1) nan+1 + 2 an ]xn. (7.1.17)
SO LU TIO N (b) From ( 7.1.17) we see that ysatisfies ( 7.1.13) on I if",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
n=0
[2(n+ 2)( n+ 1) an+2 − (n+ 1) nan+1 + 2 an ]xn. (7.1.17)
SO LU TIO N (b) From ( 7.1.17) we see that ysatisfies ( 7.1.13) on I if
2(n+ 2)( n+ 1) an+2 − (n+ 1) nan+1 + 2 an = 0 , n = 0 , 1, 2,.... (7.1.18)
Conversely , Theorem 7.1.6 (b) implies that if y= ∑ ∞
n=0 anxn satisfies (
7.1.13) on I, then ( 7.1.18) holds.
Example 7.1.7 Suppose
y=
∞∑
n=0
an (x− 1)n
on an open interval I that contains x0 = 1 . Express the function
(1 + x)y′′+ 2( x− 1)2y′+ 3 y (7.1.19)
as a power series in x− 1 on I.",Elementary Differential Equations with Boundary Value Problems.pdf
"316 Chapter 7 Series Solutions of Linear Second Equations
Solution Since we want a power series in x− 1, we rewrite the coefficient of y′′in ( 7.1.19) as 1 + x=
2 + ( x− 1), so ( 7.1.19) becomes
2y′′+ ( x− 1)y′′+ 2( x− 1)2y′+ 3 y.
From ( 7.1.6) and ( 7.1.7) with x0 = 1 ,
y′=
∞∑
n=1
nan(x− 1)n−1 and y′′=
∞∑
n=2
n(n− 1)an(x− 1)n−2.
Therefore
2y′′ =
∞∑
n=2
2n(n− 1)an(x− 1)n−2,
(x− 1)y′′ =
∞∑
n=2
n(n− 1)an(x− 1)n−1,
2(x− 1)2y′ =
∞∑
n=1
2nan(x− 1)n+1,
3y =
∞∑
n=0
3an(x− 1)n.
Before adding these four series we shift indices in the first t hree so that their general terms become
constant multiples of (x− 1)n. This yields
2y′′ =
∞∑
n=0
2(n+ 2)( n+ 1) an+2(x− 1)n, (7.1.20)
(x− 1)y′′ =
∞∑
n=0
(n+ 1) nan+1(x− 1)n , (7.1.21)
2(x− 1)2y′ =
∞∑
n=1
2(n− 1)an−1(x− 1)n , (7.1.22)
3y =
∞∑
n=0
3an (x− 1)n , (7.1.23)
where we added initial zero terms to the series in ( 7.1.21) and ( 7.1.22). Adding ( 7.1.20)–( 7.1.23) yields
(1 + x)y′′+ 2( x− 1)2y′+ 3 y = 2 y′′+ ( x− 1)y′′+ 2( x− 1)2y′+ 3 y
=
∞∑
n=0",Elementary Differential Equations with Boundary Value Problems.pdf
"(1 + x)y′′+ 2( x− 1)2y′+ 3 y = 2 y′′+ ( x− 1)y′′+ 2( x− 1)2y′+ 3 y
=
∞∑
n=0
bn (x− 1)n ,
where
b0 = 4 a2 + 3 a0, (7.1.24)
bn = 2( n+ 2)( n+ 1) an+2 + ( n+ 1) nan+1 + 2( n− 1)an−1 + 3 an , n≥ 1. (7.1.25)
The formula ( 7.1.24) for b0 can’t be obtained by setting n= 0 in ( 7.1.25), since the summation in ( 7.1.22)
begins with n= 1 , while those in ( 7.1.20), ( 7.1.21), and ( 7.1.23) begin with n= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 317
7.1 Exercises
1. For each power series use Theorem 7.1.3 to find the radius of convergence R. If R >0, find the
open interval of convergence.
(a)
∞∑
n=0
(−1)n
2nn (x− 1)n (b)
∞∑
n=0
2nn(x− 2)n
(c)
∞∑
n=0
n!
9n xn (d)
∞∑
n=0
n(n+ 1)
16n (x− 2)n
(e)
∞∑
n=0
(−1)n 7n
n! xn (f)
∞∑
n=0
3n
4n+1(n+ 1) 2 (x+ 7) n
2. Suppose there’s an integer M such that bm ̸= 0 for m≥ M, and
lim
m→∞
⏐
⏐
⏐
⏐
bm+1
bm
⏐
⏐
⏐
⏐= L,
where 0 ≤ L≤ ∞ . Show that the radius of convergence of
∞∑
m=0
bm(x− x0)2m
is R = 1 /
√
L, which is interpreted to mean that R = 0 if L = ∞ or R = ∞ if L = 0 . H IN T :
Apply Theorem 7.1.3 to the series ∑ ∞
m=0 bmzm and then let z= ( x− x0)2.
3. For each power series, use the result of Exercise
2 to find the radius of convergence R. If R> 0,
find the open interval of convergence.
(a)
∞∑
m=0
(−1)m (3m+ 1)( x− 1)2m+1 (b)
∞∑
m=0
(−1)m m(2m+ 1)
2m (x+ 2) 2m
(c)
∞∑
m=0
m!
(2m)! (x− 1)2m (d)
∞∑
m=0
(−1)m m!
9m (x+ 8) 2m
(e)
∞∑
m=0
(−1)m (2m− 1)
3m x2m+1 (f)",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
m=0
(−1)m m(2m+ 1)
2m (x+ 2) 2m
(c)
∞∑
m=0
m!
(2m)! (x− 1)2m (d)
∞∑
m=0
(−1)m m!
9m (x+ 8) 2m
(e)
∞∑
m=0
(−1)m (2m− 1)
3m x2m+1 (f)
∞∑
m=0
(x− 1)2m
4. Suppose there’s an integer M such that bm ̸= 0 for m≥ M, and
lim
m→∞
⏐
⏐
⏐
⏐
bm+1
bm
⏐
⏐
⏐
⏐= L,
where 0 ≤ L≤ ∞ . Let kbe a positive integer. Show that the radius of convergence of
∞∑
m=0
bm(x− x0)km
is R = 1 /
k√
L, which is interpreted to mean that R = 0 if L = ∞ or R = ∞ if L = 0 . H IN T :
Apply Theorem 7.1.3 to the series ∑ ∞
m=0 bmzm and then let z= ( x− x0)k .
5. For each power series use the result of Exercise
4 to find the radius of convergence R. If R >0,
find the open interval of convergence.",Elementary Differential Equations with Boundary Value Problems.pdf
"318 Chapter 7 Series Solutions of Linear Second Equations
(a)
∞∑
m=0
(−1)m
(27)m (x− 3)3m+2 (b)
∞∑
m=0
x7m+6
m
(c)
∞∑
m=0
9m (m+ 1)
(m+ 2) (x− 3)4m+2 (d)
∞∑
m=0
(−1)m 2m
m! x4m+3
(e)
∞∑
m=0
m!
(26)m (x+ 1) 4m+3 (f)
∞∑
m=0
(−1)m
8mm(m+ 1) (x− 1)3m+1
6. L Graph y= sin xand the T aylor polynomial
T2M +1(x) =
M∑
n=0
(−1)n x2n+1
(2n+ 1)!
on the interval (−2π, 2π) for M = 1 , 2, 3, . . . , until you find a value of M for which there’s no
perceptible difference between the two graphs.
7. L Graph y= cos xand the T aylor polynomial
T2M (x) =
M∑
n=0
(−1)n x2n
(2n)!
on the interval (−2π, 2π) for M = 1 , 2, 3, . . . , until you find a value of M for which there’s no
perceptible difference between the two graphs.
8. L Graph y= 1 / (1 − x) and the T aylor polynomial
TN (x) =
N∑
n=0
xn
on the interval [0,. 95] for N = 1 , 2, 3, . . . , until you find a value of N for which there’s no
perceptible difference between the two graphs. Choose the s cale on the y-axis so that 0 ≤ y ≤ 20.",Elementary Differential Equations with Boundary Value Problems.pdf
"perceptible difference between the two graphs. Choose the s cale on the y-axis so that 0 ≤ y ≤ 20.
9. L Graph y= cosh xand the T aylor polynomial
T2M (x) =
M∑
n=0
x2n
(2n)!
on the interval (−5, 5) for M = 1 , 2, 3, . . . , until you find a value of M for which there’s no
perceptible difference between the two graphs. Choose the s cale on the y-axis so that 0 ≤ y ≤ 75.
10. L Graph y= sinh xand the T aylor polynomial
T2M +1(x) =
M∑
n=0
x2n+1
(2n+ 1)!
on the interval (−5, 5) for M = 0 , 1, 2, . . . , until you find a value of M for which there’s no per-
ceptible difference between the two graphs. Choose the scal e on the y-axis so that −75 ≤ y≤ 75.
In Exercises 11– 15 find a power series solution y(x) = ∑ ∞
n=0 an xn.
11. (2 + x)y′′+ xy′+ 3 y 12. (1 + 3 x2)y′′+ 3 x2y′− 2y
13. (1 + 2 x2)y′′+ (2 − 3x)y′+ 4 y 14. (1 + x2)y′′+ (2 − x)y′+ 3 y",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.1 Review of Power Series 319
15. (1 + 3 x2)y′′− 2xy′+ 4 y
16. Suppose y(x) = ∑ ∞
n=0 an(x+ 1) n on an open interval that contains x0 = − 1. Find a power
series in x+ 1 for
xy′′+ (4 + 2 x)y′+ (2 + x)y.
17. Suppose y(x) = ∑ ∞
n=0 an (x− 2)n on an open interval that contains x0 = 2 . Find a power series
in x− 2 for
x2y′′+ 2 xy′− 3xy.
18.
L Do the following experiment for various choices of real numb ers a0 and a1.
(a) Use differential equations software to solve the initial va lue problem
(2 − x)y′′+ 2 y= 0 , y (0) = a0, y ′(0) = a1,
numerically on (−1.95, 1.95). Choose the most accurate method your software package
provides. (See Section 10.1 for a brief discussion of one suc h method.)
(b) For N = 2 , 3, 4, . . . , compute a2, . . . , aN from Eqn.( 7.1.18) and graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on the same axes. Continue increasing N until it’s obvious
that there’s no point in continuing. (This sounds vague, but you’ll know when to stop.)",Elementary Differential Equations with Boundary Value Problems.pdf
"that there’s no point in continuing. (This sounds vague, but you’ll know when to stop.)
19. L Follow the directions of Exercise 18 for the initial value problem
(1 + x)y′′+ 2( x− 1)2y′+ 3 y= 0 , y (1) = a0, y ′(1) = a1,
on the interval (0, 2). Use Eqns. ( 7.1.24) and ( 7.1.25) to compute {an}.
20. Suppose the series ∑ ∞
n=0 an xn converges on an open interval (−R,R ), let rbe an arbitrary real
number, and define
y(x) = xr
∞∑
n=0
an xn =
∞∑
n=0
anxn+r
on (0,R ). Use Theorem
7.1.4 and the rule for differentiating the product of two function s to show
that
y′(x) =
∞∑
n=0
(n+ r)an xn+r−1,
y′′(x) =
∞∑
n=0
(n+ r)(n+ r− 1)anxn+r−2,
.
.
.
y(k)(x) =
∞∑
n=0
(n+ r)(n+ r− 1) · · ·(n+ r− k)anxn+r−k
on (0,R )",Elementary Differential Equations with Boundary Value Problems.pdf
"320 Chapter 7 Series Solutions of Linear Second Order Equations
In Exercises 21– 26 let ybe as defined in Exercise 20, and write the given expression in the form xr ∑ ∞
n=0 bnxn.
21. x2(1 − x)y′′+ x(4 + x)y′+ (2 − x)y
22. x2(1 + x)y′′+ x(1 + 2 x)y′− (4 + 6 x)y
23. x2(1 + x)y′′− x(1 − 6x− x2)y′+ (1 + 6 x+ x2)y
24. x2(1 + 3 x)y′′+ x(2 + 12 x+ x2)y′+ 2 x(3 + x)y
25. x2(1 + 2 x2)y′′+ x(4 + 2 x2)y′+ 2(1 − x2)y
26. x2(2 + x2)y′′+ 2 x(5 + x2)y′+ 2(3 − x2)y7.2 SERIES SOLUTIONS NEAR AN ORDINARY POINT I
Many physical applications give rise to second order homoge neous linear differential equations of the
form
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 , (7.2.1)
where P0, P1, and P2 are polynomials. Usually the solutions of these equations c an’t be expressed in
terms of familiar elementary functions. Therefore we’ll co nsider the problem of representing solutions of
(7.2.1) with series.
W e assume throughout that P0, P1 and P2 have no common factors. Then we say that x0 is an ordinary",Elementary Differential Equations with Boundary Value Problems.pdf
"(7.2.1) with series.
W e assume throughout that P0, P1 and P2 have no common factors. Then we say that x0 is an ordinary
point of ( 7.2.1) if P0(x0) ̸= 0 , or a singular point if P0(x0) = 0 . For Legendre’s equation,
(1 − x2)y′′− 2xy′+ α(α + 1) y= 0 , (7.2.2)
x0 = 1 and x0 = −1 are singular points and all other points are ordinary points . For Bessel’s equation,
x2y′′+ xy′+ ( x2 − ν2)y = 0 ,
x0 = 0 is a singular point and all other points are ordinary points. If P0 is a nonzero constant as in Airy’s
equation,
y′′− xy= 0 , (7.2.3)
then every point is an ordinary point.
Since polynomials are continuous everywhere, P1/P 0 and P2/P 0 are continuous at any point x0 that
isn’t a zero of P0. Therefore, if x0 is an ordinary point of ( 7.2.1) and a0 and a1 are arbitrary real numbers,
then the initial value problem
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 , y (x0) = a0, y ′(x0) = a1 (7.2.4)
has a unique solution on the largest open interval that conta ins x0 and does not contain any zeros of P0.",Elementary Differential Equations with Boundary Value Problems.pdf
"has a unique solution on the largest open interval that conta ins x0 and does not contain any zeros of P0.
T o see this, we rewrite the differential equation in ( 7.2.4) as
y′′+ P1(x)
P0(x) y′+ P2(x)
P0(x) y= 0
and apply Theorem 5.1.1 with p = P1/P 0 and q = P2/P 0. In this section and the next we consider the
problem of representing solutions of ( 7.2.1) by power series that converge for values of xnear an ordinary
point x0.
W e state the next theorem without proof.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 321
Theorem 7.2.1 Suppose P0, P1, and P2 are polynomials with no common factor and P0 isn’t identically
zero. Let x0 be a point such that P0(x0) ̸= 0 , and let ρ be the distance from x0 to the nearest zero of P0
in the complex plane. (If P0 is constant, then ρ = ∞ .) Then every solution of
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 (7.2.5)
can be represented by a power series
y=
∞∑
n=0
an(x− x0)n (7.2.6)
that converges at least on the open interval (x0 −ρ,x 0 +ρ). ( If P0 is nonconstant , so that ρ is necessarily
finite , then the open interval of convergence of (7.2.6) may be larger than (x0 −ρ,x 0 +ρ). If P0 is constant
then ρ = ∞ and (x0 − ρ,x 0 + ρ) = ( −∞ , ∞ ).)
W e call ( 7.2.6) a power series solution in x− x0 of ( 7.2.5). W e’ll now develop a method for finding
power series solutions of ( 7.2.5). For this purpose we write ( 7.2.5) as Ly= 0 , where
Ly= P0y′′+ P1y′+ P2y. (7.2.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"power series solutions of ( 7.2.5). For this purpose we write ( 7.2.5) as Ly= 0 , where
Ly= P0y′′+ P1y′+ P2y. (7.2.7)
Theorem 7.2.1 implies that every solution of Ly= 0 on (x0 − ρ,x 0 + ρ) can be written as
y =
∞∑
n=0
an(x− x0)n .
Setting x= x0 in this series and in the series
y′=
∞∑
n=1
nan(x− x0)n−1
shows that y(x0) = a0 and y′(x0) = a1. Since every initial value problem ( 7.2.4) has a unique solution,
this means that a0 and a1 can be chosen arbitrarily , and a2, a3, . . . are uniquely determined by them.
T o find a2, a3, . . . , we write P0, P1, and P2 in powers of x− x0, substitute
y =
∞∑
n=0
an(x− x0)n ,
y′=
∞∑
n=1
nan(x− x0)n−1,
y′′=
∞∑
n=2
n(n− 1)an(x− x0)n−2
into ( 7.2.7), and collect the coefficients of like powers of x− x0. This yields
Ly=
∞∑
n=0
bn(x− x0)n , (7.2.8)
where {b0,b 1,...,b n,... }are expressed in terms of {a0,a 1,...,a n,... }and the coefficients of P0, P1,
and P2, written in powers of x− x0. Since ( 7.2.8) and (a) of Theorem 7.1.6 imply that Ly = 0 if and",Elementary Differential Equations with Boundary Value Problems.pdf
"and P2, written in powers of x− x0. Since ( 7.2.8) and (a) of Theorem 7.1.6 imply that Ly = 0 if and
only if bn = 0 for n ≥ 0, all power series solutions in x− x0 of Ly = 0 can be obtained by choosing
a0 and a1 arbitrarily and computing a2, a3, . . . , successively so that bn = 0 for n ≥ 0. For simplicity ,
we call the power series obtained this way the power series in x− x0 for the general solution of Ly= 0 ,
without explicitly identifying the open interval of conver gence of the series.",Elementary Differential Equations with Boundary Value Problems.pdf
"322 Chapter 7 Series Solutions of Linear Second Order Equations
Example 7.2.1 Let x0 be an arbitrary real number. Find the power series in x−x0 for the general solution
of
y′′+ y= 0 . (7.2.9)
Solution Here
Ly= y′′+ y.
If
y =
∞∑
n=0
an(x− x0)n ,
then
y′′=
∞∑
n=2
n(n− 1)an (x− x0)n−2,
so
Ly=
∞∑
n=2
n(n− 1)an (x− x0)n−2 +
∞∑
n=0
an (x− x0)n .
T o collect coefficients of like powers of x− x0, we shift the summation index in the first sum. This yields
Ly=
∞∑
n=0
(n+ 2)( n+ 1) an+2(x− x0)n +
∞∑
n=0
an (x− x0)n =
∞∑
n=0
bn(x− x0)n ,
with
bn = ( n+ 2)( n+ 1) an+2 + an.
Therefore Ly= 0 if and only if
an+2 = −an
(n+ 2)( n+ 1) , n ≥ 0, (7.2.10)
where a0 and a1 are arbitrary . Since the indices on the left and right sides o f ( 7.2.10) differ by two, we
write ( 7.2.10) separately for neven (n= 2 m) and nodd (n= 2 m+ 1) . This yields
a2m+2 = −a2m
(2m+ 2)(2 m+ 1) , m ≥ 0, (7.2.11)
and
a2m+3 = −a2m+1
(2m+ 3)(2 m+ 2) , m ≥ 0. (7.2.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"a2m+2 = −a2m
(2m+ 2)(2 m+ 1) , m ≥ 0, (7.2.11)
and
a2m+3 = −a2m+1
(2m+ 3)(2 m+ 2) , m ≥ 0. (7.2.12)
Computing the coefficients of the even powers of x− x0 from ( 7.2.11) yields
a2 = − a0
2 ·1
a4 = − a2
4 ·3 = − 1
4 ·3
(
− a0
2 ·1
)
= a0
4 ·3 ·2 ·1 ,
a6 = − a4
6 ·5 = − 1
6 ·5
( a0
4 ·3 ·2 ·1
)
= − a0
6 ·5 ·4 ·3 ·2 ·1 ,
and, in general,
a2m = ( −1)m a0
(2m)! , m ≥ 0. (7.2.13)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 323
Computing the coefficients of the odd powers of x− x0 from ( 7.2.12) yields
a3 = − a1
3 ·2
a5 = − a3
5 ·4 = − 1
5 ·4
(
− a1
3 ·2
)
= a1
5 ·4 ·3 ·2 ,
a7 = − a5
7 ·6 = − 1
7 ·6
( a1
5 ·4 ·3 ·2
)
= − a1
7 ·6 ·5 ·4 ·3 ·2 ,
and, in general,
a2m+1 = (−1)ma1
(2m+ 1)! m≥ 0. (7.2.14)
Thus, the general solution of ( 7.2.9) can be written as
y=
∞∑
m=0
a2m(x− x0)2m +
∞∑
m=0
a2m+1(x− x0)2m+1,
or, from ( 7.2.13) and ( 7.2.14), as
y= a0
∞∑
m=0
(−1)m (x− x0)2m
(2m)! + a1
∞∑
m=0
(−1)m (x− x0)2m+1
(2m+ 1)! . (7.2.15)
If we recall from calculus that
∞∑
m=0
(−1)m (x− x0)2m
(2m)! = cos( x− x0) and
∞∑
m=0
(−1)m (x− x0)2m+1
(2m+ 1)! = sin( x− x0),
then ( 7.2.15) becomes
y= a0 cos(x− x0) + a1 sin(x− x0),
which should look familiar.
Equations like ( 7.2.10), ( 7.2.11), and ( 7.2.12), which define a given coefficient in the sequence {an}
in terms of one or more coefficients with lesser indices are ca lled recurrence relations . When we use a",Elementary Differential Equations with Boundary Value Problems.pdf
"in terms of one or more coefficients with lesser indices are ca lled recurrence relations . When we use a
recurrence relation to compute terms of a sequence we’re com puting recursively.
In the remainder of this section we consider the problem of fin ding power series solutions in x− x0
for equations of the form
(1 + α(x− x0)2) y′′+ β(x− x0)y′+ γy = 0 . (7.2.16)
Many important equations that arise in applications are of t his form with x0 = 0 , including Legendre’s
equation ( 7.2.2), Airy’s equation ( 7.2.3), Chebyshev’s equation ,
(1 − x2)y′′− xy′+ α2y = 0 ,
and Hermite’s equation ,
y′′− 2xy′+ 2 αy = 0 .
Since
P0(x) = 1 + α(x− x0)2
in ( 7.2.16), the point x0 is an ordinary point of ( 7.2.16), and Theorem 7.2.1 implies that the solutions of
(7.2.16) can be written as power series in x−x0 that converge on the interval (x0 −1/
√
|α|,x 0+1/
√
|α|)",Elementary Differential Equations with Boundary Value Problems.pdf
"324 Chapter 7 Series Solutions of Linear Second Order Equations
if α ̸= 0 , or on (−∞ , ∞ ) if α = 0 . W e’ll see that the coefficients in these power series can be o btained
by methods similar to the one used in Example 7.2.1.
T o simplify finding the coefficients, we introduce some notat ion for products:
s∏
j=r
bj = br br+1 · · ·bs if s≥ r.
Thus,
7∏
j=2
bj = b2b3b4b5b6b7,
4∏
j=0
(2j+ 1) = (1)(3)(5)(7)(9) = 945 ,
and
2∏
j=2
j2 = 2 2 = 4 .
W e define s∏
j=r
bj = 1 if s<r,
no matter what the form of bj .
Example 7.2.2 Find the power series in xfor the general solution of
(1 + 2 x2)y′′+ 6 xy′+ 2 y= 0 . (7.2.17)
Solution Here
Ly= (1 + 2 x2)y′′+ 6 xy′+ 2 y.
If
y=
∞∑
n=0
an xn
then
y′=
∞∑
n=1
nan xn−1 and y′′=
∞∑
n=2
n(n− 1)anxn−2,
so
Ly = (1 + 2 x2)
∞∑
n=2
n(n− 1)anxn−2 + 6 x
∞∑
n=1
nanxn−1 + 2
∞∑
n=0
anxn
=
∞∑
n=2
n(n− 1)anxn−2 +
∞∑
n=0
[2n(n− 1) + 6 n+ 2] anxn
=
∞∑
n=2
n(n− 1)anxn−2 + 2
∞∑
n=0
(n+ 1) 2an xn.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 325
T o collect coefficients of xn , we shift the summation index in the first sum. This yields
Ly=
∞∑
n=0
(n+ 2)( n+ 1) an+2xn + 2
∞∑
n=0
(n+ 1) 2anxn =
∞∑
n=0
bn xn,
with
bn = ( n+ 2)( n+ 1) an+2 + 2( n+ 1) 2an , n ≥ 0.
T o obtain solutions of ( 7.2.17), we set bn = 0 for n≥ 0. This is equivalent to the recurrence relation
an+2 = −2 n+ 1
n+ 2 an, n ≥ 0. (7.2.18)
Since the indices on the left and right differ by two, we write (7.2.18) separately for n = 2 m and
n= 2 m+ 1 , as in Example 7.2.1. This yields
a2m+2 = −2 2m+ 1
2m+ 2 a2m = −2m+ 1
m+ 1 a2m, m ≥ 0, (7.2.19)
and
a2m+3 = −2 2m+ 2
2m+ 3 a2m+1 = −4 m+ 1
2m+ 3 a2m+1 , m ≥ 0. (7.2.20)
Computing the coefficients of even powers of xfrom ( 7.2.19) yields
a2 = −1
1 a0,
a4 = −3
2 a2 =
(
−3
2
)(
−1
1
)
a0 = 1 ·3
1 ·2 a0,
a6 = −5
3 a4 = −5
3
(1 ·3
1 ·2
)
a0 = −1 ·3 ·5
1 ·2 ·3 a0,
a8 = −7
4 a6 = −7
4
(
−1 ·3 ·5
1 ·2 ·3
)
a0 = 1 ·3 ·5 ·7
1 ·2 ·3 ·4 a0.
In general,
a2m = ( −1)m
∏ m",Elementary Differential Equations with Boundary Value Problems.pdf
"3
(1 ·3
1 ·2
)
a0 = −1 ·3 ·5
1 ·2 ·3 a0,
a8 = −7
4 a6 = −7
4
(
−1 ·3 ·5
1 ·2 ·3
)
a0 = 1 ·3 ·5 ·7
1 ·2 ·3 ·4 a0.
In general,
a2m = ( −1)m
∏ m
j=1(2j− 1)
m! a0, m ≥ 0. (7.2.21)
(Note that ( 7.2.21) is correct for m= 0 because we defined ∏ 0
j=1 bj = 1 for any bj .)
Computing the coefficients of odd powers of xfrom (
7.2.20) yields
a3 = −4 1
3 a1,
a5 = −4 2
5 a3 = −4 2
5
(
−4 1
3
)
a1 = 4 2 1 ·2
3 ·5 a1,
a7 = −4 3
7 a5 = −4 3
7
(
42 1 ·2
3 ·5
)
a1 = −43 1 ·2 ·3
3 ·5 ·7 a1,
a9 = −4 4
9 a7 = −4 4
9
(
43 1 ·2 ·3
3 ·5 ·7
)
a1 = 4 4 1 ·2 ·3 ·4
3 ·5 ·7 ·9 a1.",Elementary Differential Equations with Boundary Value Problems.pdf
"326 Chapter 7 Series Solutions of Linear Second Order Equations
In general,
a2m+1 = (−1)m 4mm!∏ m
j=1(2j+ 1) a1, m ≥ 0. (7.2.22)
From (
7.2.21) and ( 7.2.22),
y = a0
∞∑
m=0
(−1)m
∏ m
j=1(2j− 1)
m! x2m + a1
∞∑
m=0
(−1)m 4mm!∏ m
j=1(2j+ 1) x2m+1.
is the power series in xfor the general solution of (
7.2.17). Since P0(x) = 1+2 x2 has no real zeros, Theo-
rem 5.1.1 implies that every solution of ( 7.2.17) is defined on (−∞ , ∞ ). However, since P0(±i/
√
2) = 0 ,
Theorem 7.2.1 implies only that the power series converges in (−1/
√
2, 1/
√
2) for any choice of a0 and
a1.
The results in Examples 7.2.1 and 7.2.2 are consequences of the following general theorem.
Theorem 7.2.2 The coefficients {an}in any solution y= ∑ ∞
n=0 an (x− x0)n of
(1 + α(x− x0)2) y′′+ β(x− x0)y′+ γy = 0 (7.2.23)
satisfy the recurrence relation
an+2 = − p(n)
(n+ 2)( n+ 1) an, n ≥ 0, (7.2.24)
where
p(n) = αn(n− 1) + βn + γ. (7.2.25)
Moreover, the coefficients of the even and odd powers of x− x0 can be computed separately as",Elementary Differential Equations with Boundary Value Problems.pdf
"where
p(n) = αn(n− 1) + βn + γ. (7.2.25)
Moreover, the coefficients of the even and odd powers of x− x0 can be computed separately as
a2m+2 = − p(2m)
(2m+ 2)(2 m+ 1) a2m, m ≥ 0 (7.2.26)
and
a2m+3 = − p(2m+ 1)
(2m+ 3)(2 m+ 2) a2m+1, m ≥ 0, (7.2.27)
where a0 and a1 are arbitrary.
Proof Here
Ly= (1 + α(x− x0)2)y′′+ β(x− x0)y′+ γy.
If
y =
∞∑
n=0
an(x− x0)n ,
then
y′=
∞∑
n=1
nan (x− x0)n−1 and y′′=
∞∑
n=2
n(n− 1)an(x− x0)n−2.
Hence,
Ly =
∞∑
n=2
n(n− 1)an (x− x0)n−2 +
∞∑
n=0
[αn(n− 1) + βn + γ] an (x− x0)n
=
∞∑
n=2
n(n− 1)an (x− x0)n−2 +
∞∑
n=0
p(n)an (x− x0)n,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 327
from ( 7.2.25). T o collect coefficients of powers of x− x0, we shift the summation index in the first sum.
This yields
Ly=
∞∑
n=0
[(n+ 2)( n+ 1) an+2 + p(n)an] ( x− x0)n .
Thus, Ly= 0 if and only if
(n+ 2)( n+ 1) an+2 + p(n)an = 0 , n ≥ 0,
which is equivalent to ( 7.2.24). Writing ( 7.2.24) separately for the cases where n= 2 mand n= 2 m+ 1
yields ( 7.2.26) and ( 7.2.27).
Example 7.2.3 Find the power series in x− 1 for the general solution of
(2 + 4 x− 2x2)y′′− 12(x− 1)y′− 12y= 0 . (7.2.28)
Solution W e must first write the coefficient P0(x) = 2 + 4 x− x2 in powers of x− 1. T o do this, we
write x= ( x− 1) + 1 in P0(x) and then expand the terms, collecting powers of x− 1; thus,
2 + 4 x− 2x2 = 2 + 4[( x− 1) + 1] − 2[(x− 1) + 1] 2
= 4 − 2(x− 1)2.
Therefore we can rewrite ( 7.2.28) as
(4 − 2(x− 1)2) y′′− 12(x− 1)y′− 12y= 0 ,
or, equivalently , (
1 − 1
2 (x− 1)2
)
y′′− 3(x− 1)y′− 3y= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Therefore we can rewrite ( 7.2.28) as
(4 − 2(x− 1)2) y′′− 12(x− 1)y′− 12y= 0 ,
or, equivalently , (
1 − 1
2 (x− 1)2
)
y′′− 3(x− 1)y′− 3y= 0 .
This is of the form ( 7.2.23) with α = −1/ 2, β = −3, and γ = −3. Therefore, from ( 7.2.25)
p(n) = −n(n− 1)
2 − 3n− 3 = −(n+ 2)( n+ 3)
2 .
Hence, Theorem 7.2.2 implies that
a2m+2 = − p(2m)
(2m+ 2)(2 m+ 1) a2m
= (2m+ 2)(2 m+ 3)
2(2m+ 2)(2 m+ 1) a2m = 2m+ 3
2(2m+ 1) a2m, m ≥ 0
and
a2m+3 = − p(2m+ 1)
(2m+ 3)(2 m+ 2) a2m+1
= (2m+ 3)(2 m+ 4)
2(2m+ 3)(2 m+ 2) a2m+1 = m+ 2
2(m+ 1) a2m+1, m ≥ 0.
W e leave it to you to show that
a2m = 2m+ 1
2m a0 and a2m+1 = m+ 1
2m a1, m ≥ 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"328 Chapter 7 Series Solutions of Linear Second Order Equations
which implies that the power series in x− 1 for the general solution of ( 7.2.28) is
y= a0
∞∑
m=0
2m+ 1
2m (x− 1)2m + a1
∞∑
m=0
m+ 1
2m (x− 1)2m+1.
In the examples considered so far we were able to obtain close d formulas for coefficients in the power
series solutions. In some cases this is impossible, and we mu st settle for computing a finite number of
terms in the series. The next example illustrates this with a n initial value problem.
Example 7.2.4 Compute a0, a1, . . . , a7 in the series solution y = ∑ ∞
n=0 an xn of the initial value prob-
lem
(1 + 2 x2)y′′+ 10 xy′+ 8 y = 0 , y (0) = 2 , y ′(0) = −3. (7.2.29)
Solution Since α = 2 , β = 10 , and γ = 8 in (
7.2.29),
p(n) = 2 n(n− 1) + 10 n+ 8 = 2( n+ 2) 2.
Therefore
an+2 = −2 (n+ 2) 2
(n+ 2)( n+ 1) an = −2 n+ 2
n+ 1 an, n ≥ 0.
Writing this equation separately for n= 2 mand n= 2 m+ 1 yields
a2m+2 = −2 (2m+ 2)
2m+ 1 a2m = −4 m+ 1
2m+ 1 a2m, m ≥ 0 (7.2.30)
and",Elementary Differential Equations with Boundary Value Problems.pdf
"n+ 1 an, n ≥ 0.
Writing this equation separately for n= 2 mand n= 2 m+ 1 yields
a2m+2 = −2 (2m+ 2)
2m+ 1 a2m = −4 m+ 1
2m+ 1 a2m, m ≥ 0 (7.2.30)
and
a2m+3 = −2 2m+ 3
2m+ 2 a2m+1 = −2m+ 3
m+ 1 a2m+1, m ≥ 0. (7.2.31)
Starting with a0 = y(0) = 2 , we compute a2,a 4, and a6 from ( 7.2.30):
a2 = −4 1
1 2 = −8,
a4 = −4 2
3 (−8) = 64
3 ,
a6 = −4 3
5
(64
3
)
= −256
5 .
Starting with a1 = y′(0) = −3, we compute a3,a 5 and a7 from ( 7.2.31):
a3 = −3
1 (−3) = 9 ,
a5 = −5
2 9 = −45
2 ,
a7 = −7
3
(
−45
2
)
= 105
2 .
Therefore the solution of ( 7.2.29) is
y= 2 − 3x− 8x2 + 9 x3 + 64
3 x4 − 45
2 x5 − 256
5 x6 + 105
2 x7 + · · ·.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 329
USING TECHNOLOGY
Computing coefficients recursively as in Example 7.2.4 is tedious. W e recommend that you do this
kind of computation by writing a short program to implement t he appropriate recurrence relation on a
calculator or computer. Y ou may wish to do this in verifying e xamples and doing exercises (identified by
the symbol C ) in this chapter that call for numerical computation of the c oefficients in series solutions.
W e obtained the answers to these exercises by using software that can produce answers in the form of
rational numbers. However, it’s perfectly acceptable - and more practical - to get your answers in decimal
form. Y ou can always check them by converting our fractions t o decimals.
If you’re interested in actually using series to compute num erical approximations to solutions of a
differential equation, then whether or not there’s a simple closed form for the coefficents is essentially",Elementary Differential Equations with Boundary Value Problems.pdf
"differential equation, then whether or not there’s a simple closed form for the coefficents is essentially
irrelevant. For computational purposes it’s usually more e fficient to start with the given coefficients
a0 = y(x0) and a1 = y′(x0), compute a2, . . . , aN recursively , and then compute approximate values of
the solution from the T aylor polynomial
TN (x) =
N∑
n=0
an(x− x0)n .
The trick is to decide how to choose N so the approximation y(x) ≈ TN (x) is sufficiently accurate on
the subinterval of the interval of convergence that you’re i nterested in. In the computational exercises
in this and the next two sections, you will often be asked to ob tain the solution of a given problem by
numerical integration with software of your choice (see Sec tion 10.1 for a brief discussion of one such
method), and to compare the solution obtained in this way wit h the approximations obtained with TN for",Elementary Differential Equations with Boundary Value Problems.pdf
"method), and to compare the solution obtained in this way wit h the approximations obtained with TN for
various values of N. This is a typical textbook kind of exercise, designed to giv e you insight into how
the accuracy of the approximation y(x) ≈ TN (x) behaves as a function of N and the interval that you’re
working on. In real life, you would choose one or the other of t he two methods (numerical integration or
series solution). If you choose the method of series solutio n, then a practical procedure for determining
a suitable value of N is to continue increasing N until the maximum of |TN − TN −1|on the interval of
interest is within the margin of error that you’re willing to accept.
In doing computational problems that call for numerical sol ution of differential equations you should
choose the most accurate numerical integration procedure y our software supports, and experiment with",Elementary Differential Equations with Boundary Value Problems.pdf
"choose the most accurate numerical integration procedure y our software supports, and experiment with
the step size until you’re confident that the numerical resul ts are sufficiently accurate for the problem at
hand.
7.2 Exercises
In Exercises 1 – 8 find the power series in xfor the general solution.
1. (1 + x2)y′′+ 6 xy′+ 6 y= 0 2. (1 + x2)y′′+ 2 xy′− 2y= 0
3. (1 + x2)y′′− 8xy′+ 20 y= 0 4. (1 − x2)y′′− 8xy′− 12y= 0
5. (1 + 2 x2)y′′+ 7 xy′+ 2 y= 0 6. (1 + x2)y′′+ 2 xy′+ 1
4 y = 0
7. (1 − x2)y′′− 5xy′− 4y= 0 8. (1 + x2)y′′− 10xy′+ 28 y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"330 Chapter 7 Series Solutions of Linear Second Order Equations
9. L
(a) Find the power series in xfor the general solution of y′′+ xy′+ 2 y= 0 .
(b) For several choices of a0 and a1, use differential equations software to solve the initial v alue
problem
y′′+ xy′+ 2 y = 0 , y (0) = a0, y ′(0) = a1, (A)
numerically on (−5, 5).
(c) For fixed rin {1, 2, 3, 4, 5}graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on (−r,r ). Continue increasing Nuntil there’s no perceptible
difference between the two graphs.
10. L Follow the directions of Exercise 9 for the differential equation
y′′+ 2 xy′+ 3 y= 0 .
In Exercises 11 – 13 find a0, . . . , aN for N at least 7 in the power series solution y = ∑ ∞
n=0 anxn of the
initial value problem.
11.
C (1 + x2)y′′+ xy′+ y= 0 , y (0) = 2 , y ′(0) = −1
12. C (1 + 2 x2)y′′− 9xy′− 6y= 0 , y (0) = 1 , y ′(0) = −1
13. C (1 + 8 x2)y′′+ 2 y = 0 , y (0) = 2 , y ′(0) = −1",Elementary Differential Equations with Boundary Value Problems.pdf
"12. C (1 + 2 x2)y′′− 9xy′− 6y= 0 , y (0) = 1 , y ′(0) = −1
13. C (1 + 8 x2)y′′+ 2 y = 0 , y (0) = 2 , y ′(0) = −1
14. L Do the next experiment for various choices of real numbers a0, a1, and r, with 0 <r <1/
√
2.
(a) Use differential equations software to solve the initial va lue problem
(1 − 2x2)y′′− xy′+ 3 y= 0 , y (0) = a0, y ′(0) = a1, (A)
numerically on (−r,r ).
(b) For N = 2 , 3, 4, . . . , compute a2, . . . , aN in the power series solution y = ∑ ∞
n=0 anxn of
(A), and graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on (−r,r ). Continue increasing Nuntil there’s no perceptible
difference between the two graphs.
15.
L Do (a) and (b) for several values of rin (0, 1):
(a) Use differential equations software to solve the initial va lue problem
(1 + x2)y′′+ 10 xy′+ 14 y= 0 , y (0) = 5 , y ′(0) = 1 , (A)
numerically on (−r,r ).
(b) For N = 2 , 3, 4, . . . , compute a2, . . . , aN in the power series solution y = ∑ ∞
n=0 anxn of
(A) , and graph
TN (x) =
N∑
n=0
an xn",Elementary Differential Equations with Boundary Value Problems.pdf
"(b) For N = 2 , 3, 4, . . . , compute a2, . . . , aN in the power series solution y = ∑ ∞
n=0 anxn of
(A) , and graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on (−r,r ). Continue increasing N until there’s no percep-
tible difference between the two graphs. What happens to the required N as r→ 1?
(c) Try (a) and (b) with r= 1 .2. Explain your results.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 331
In Exercises 16 – 20 find the power series in x− x0 for the general solution.
16. y′′− y = 0; x0 = 3 17. y′′−(x−3)y′−y= 0; x0 = 3
18. (1 − 4x+ 2 x2)y′′+ 10( x− 1)y′+ 6 y= 0; x0 = 1
19. (11 − 8x+ 2 x2)y′′− 16(x− 2)y′+ 36 y= 0; x0 = 2
20. (5 + 6 x+ 3 x2)y′′+ 9( x+ 1) y′+ 3 y= 0; x0 = −1
In Exercises 21 – 26 find a0, . . . , aN for N at least 7 in the power series y = ∑ ∞
n=0 an (x− x0)n for the
solution of the initial value problem. T ake x0 to be the point where the initial conditions are imposed.
21.
C (x2 − 4)y′′− xy′− 3y= 0 , y (0) = −1, y ′(0) = 2
22. C y′′+ ( x− 3)y′+ 3 y= 0 , y (3) = −2, y ′(3) = 3
23. C (5 − 6x+ 3 x2)y′′+ ( x− 1)y′+ 12 y= 0 , y (1) = −1, y ′(1) = 1
24. C (4x2 − 24x+ 37) y′′+ y = 0 , y (3) = 4 , y ′(3) = −6
25. C (x2 − 8x+ 14) y′′− 8(x− 4)y′+ 20 y= 0 , y (4) = 3 , y ′(4) = −4
26. C (2x2 + 4 x+ 5) y′′− 20(x+ 1) y′+ 60 y= 0 , y (−1) = 3 , y ′(−1) = −3
27. (a) Find a power series in xfor the general solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"26. C (2x2 + 4 x+ 5) y′′− 20(x+ 1) y′+ 60 y= 0 , y (−1) = 3 , y ′(−1) = −3
27. (a) Find a power series in xfor the general solution of
(1 + x2)y′′+ 4 xy′+ 2 y = 0 . (A)
(b) Use (a) and the formula
1
1 − r = 1 + r+ r2 + · · ·+ rn + · · ·(−1 <r< 1)
for the sum of a geometric series to find a closed form expressi on for the general solution of
(A) on (−1, 1).
(c) Show that the expression obtained in (b) is actually the general solution of of (A) on (−∞ , ∞ ).
28. Use Theorem 7.2.2 to show that the power series in xfor the general solution of
(1 + αx2)y′′+ βxy′+ γy = 0
is
y= a0
∞∑
m=0
(−1)m


m−1∏
j=0
p(2j)

 x2m
(2m)! + a1
∞∑
m=0
(−1)m


m−1∏
j=0
p(2j+ 1)

 x2m+1
(2m+ 1)! .
29. Use Exercise 28 to show that all solutions of
(1 + αx2)y′′+ βxy′+ γy = 0
are polynomials if and only if
αn(n− 1) + βn + γ = α(n− 2r)(n− 2s− 1),
where rand sare nonnegative integers.",Elementary Differential Equations with Boundary Value Problems.pdf
"332 Chapter 7 Series Solutions of Linear Second Order Equations
30. (a) Use Exercise 28 to show that the power series in xfor the general solution of
(1 − x2)y′′− 2bxy′+ α(α + 2 b− 1)y= 0
is y= a0y1 + a1y2, where
y1 =
∞∑
m=0


m−1∏
j=0
(2j− α)(2j+ α + 2 b− 1)

 x2m
(2m)!
and
y2 =
∞∑
m=0


m−1∏
j=0
(2j+ 1 − α)(2j+ α + 2 b)

 x2m+1
(2m+ 1)! .
(b) Suppose 2bisn’t a negative odd integer and k is a nonnegative integer. Show that y1 is a
polynomial of degree 2ksuch that y1(−x) = y1(x) if α = 2 k, while y2 is a polynomial of
degree 2k+1 such that y2(−x) = −y2(−x) if α = 2 k+1. Conclude that if nis a nonnegative
integer, then there’s a polynomial Pn of degree nsuch that Pn(−x) = ( −1)n Pn(x) and
(1 − x2)P′′
n − 2bxP′
n + n(n+ 2 b− 1)Pn = 0 . (A)
(c) Show that (A) implies that
[(1 − x2)bP′
n]′= −n(n+ 2 b− 1)(1 − x2)b−1Pn,
and use this to show that if mand nare nonnegative integers, then
[(1 − x2)bP′
n]′Pm − [(1 − x2)bP′
m]′Pn =
[m(m+ 2 b− 1) − n(n+ 2 b− 1)] (1 − x2)b−1Pm Pn.
(B)",Elementary Differential Equations with Boundary Value Problems.pdf
"[(1 − x2)bP′
n]′Pm − [(1 − x2)bP′
m]′Pn =
[m(m+ 2 b− 1) − n(n+ 2 b− 1)] (1 − x2)b−1Pm Pn.
(B)
(d) Now suppose b> 0. Use (B) and integration by parts to show that if m̸= n, then
∫ 1
−1
(1 − x2)b−1Pm(x)Pn(x) dx= 0 .
(W e say that Pm and Pn are orthogonal on (−1, 1) with respect to the weighting function
(1 − x2)b−1.)
31. (a) Use Exercise 28 to show that the power series in x for the general solution of Hermite’s
equation
y′′− 2xy′+ 2 αy = 0
is y= a0y1 + a1y1, where
y1 =
∞∑
m=0


m−1∏
j=0
(2j− α)

 2mx2m
(2m)!
and
y2 =
∞∑
m=0


m−1∏
j=0
(2j+ 1 − α)

 2mx2m+1
(2m+ 1)! .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.2 Series Solutions Near an Ordinary Point I 333
(b) Suppose k is a nonnegative integer. Show that y1 is a polynomial of degree 2k such that
y1(−x) = y1(x) if α = 2 k, while y2 is a polynomial of degree 2k+ 1 such that y2(−x) =
−y2(−x) if α = 2 k+1. Conclude that if nis a nonnegative integer then there’s a polynomial
Pn of degree nsuch that Pn (−x) = ( −1)n Pn(x) and
P′′
n − 2xP′
n + 2 nPn = 0 . (A)
(c) Show that (A) implies that
[e−x2
P′
n]′= −2ne−x2
Pn,
and use this to show that if mand nare nonnegative integers, then
[e−x2
P′
n ]′Pm − [e−x2
P′
m]′Pn = 2( m− n)e−x2
PmPn. (B)
(d) Use (B) and integration by parts to show that if m̸= n, then
∫ ∞
−∞
e−x2
Pm(x)Pn (x) dx= 0 .
(W e say that Pm and Pn are orthogonal on (−∞ , ∞ ) with respect to the weighting function
e−x2
.)
32. Consider the equation (
1 + αx3)
y′′+ βx2y′+ γxy = 0 , (A)
and let p(n) = αn(n− 1) + βn + γ. (The special case y′′− xy= 0 of (A) is Airy’s equation.)",Elementary Differential Equations with Boundary Value Problems.pdf
"1 + αx3)
y′′+ βx2y′+ γxy = 0 , (A)
and let p(n) = αn(n− 1) + βn + γ. (The special case y′′− xy= 0 of (A) is Airy’s equation.)
(a) Modify the argument used to prove Theorem 7.2.2 to show that
y =
∞∑
n=0
an xn
is a solution of (A) if and only if a2 = 0 and
an+3 = − p(n)
(n+ 3)( n+ 2) an , n ≥ 0.
(b) Show from (a) that an = 0 unless n= 3 mor n= 3 m+ 1 for some nonnegative integer m,
and that
a3m+3 = − p(3m)
(3m+ 3)(3 m+ 2) a3m, m ≥ 0,
and
a3m+4 = − p(3m+ 1)
(3m+ 4)(3 m+ 3) a3m+1, m ≥ 0,
where a0 and a1 may be specified arbitrarily .
(c) Conclude from (b) that the power series in xfor the general solution of (A) is
y= a0
∞∑
m=0
(−1)m


m−1∏
j=0
p(3j)
3j+ 2

 x3m
3m m!
+a1
∞∑
m=0
(−1)m


m−1∏
j=0
p(3j+ 1)
3j+ 4

 x3m+1
3m m! .",Elementary Differential Equations with Boundary Value Problems.pdf
"334 Chapter 7 Series Solutions of Linear Second Order Equations
In Exercises 33 – 37 use the method of Exercise 32 to find the power series in xfor the general solution.
33. y′′− xy= 0 34. (1 − 2x3)y′′− 10x2y′− 8xy= 0
35. (1 + x3)y′′+ 7 x2y′+ 9 xy= 0 36. (1 − 2x3)y′′+ 6 x2y′+ 24 xy= 0
37. (1 − x3)y′′+ 15 x2y′− 63xy= 0
38. Consider the equation (
1 + αxk+2)
y′′+ βxk+1y′+ γxk y = 0 , (A)
where kis a positive integer, and let p(n) = αn(n− 1) + βn + γ.
(a) Modify the argument used to prove Theorem 7.2.2 to show that
y =
∞∑
n=0
an xn
is a solution of (A) if and only if an = 0 for 2 ≤ n≤ k+ 1 and
an+k+2 = − p(n)
(n+ k+ 2)( n+ k+ 1) an , n ≥ 0.
(b) Show from (a) that an = 0 unless n= ( k+ 2)mor n= ( k+ 2)m+ 1 for some nonnegative
integer m, and that
a(k+2)(m+1) = − p((k+ 2) m)
(k+ 2)( m+ 1)[( k+ 2)( m+ 1) − 1] a(k+2)m, m ≥ 0,
and
a(k+2)(m+1)+1 = − p((k+ 2) m+ 1)
[(k+ 2)( m+ 1) + 1]( k+ 2)( m+ 1) a(k+2)m+1, m ≥ 0,
where a0 and a1 may be specified arbitrarily .",Elementary Differential Equations with Boundary Value Problems.pdf
"and
a(k+2)(m+1)+1 = − p((k+ 2) m+ 1)
[(k+ 2)( m+ 1) + 1]( k+ 2)( m+ 1) a(k+2)m+1, m ≥ 0,
where a0 and a1 may be specified arbitrarily .
(c) Conclude from (b) that the power series in xfor the general solution of (A) is
y= a0
∞∑
m=0
(−1)m


m−1∏
j=0
p((k+ 2) j)
(k+ 2)( j+ 1) − 1

 x(k+2)m
(k+ 2) mm!
+a1
∞∑
m=0
(−1)m


m−1∏
j=0
p((k+ 2) j+ 1)
(k+ 2)( j+ 1) + 1

 x(k+2)m+1
(k+ 2) mm! .
In Exercises 39 – 44 use the method of Exercise 38 to find the power series in xfor the general solution.
39. (1 + 2 x5)y′′+ 14 x4y′+ 10 x3y = 0
40. y′′+ x2y= 0 41. y′′+ x6y′+ 7 x5y = 0
42. (1 + x8)y′′− 16x7y′+ 72 x6y = 0
43. (1 − x6)y′′− 12x5y′− 30x4y = 0
44. y′′+ x5y′+ 6 x4y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.3 Series Solutions Near an Ordinary Point II 335
7.3 SERIES SOLUTIONS NEAR AN ORDINARY POINT II
In this section we continue to find series solutions
y=
∞∑
n=0
an(x− x0)n
of initial value problems
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 , y (x0 ) = a0, y ′(x0) = a1, (7.3.1)
where P0,P 1, and P2 are polynomials and P0(x0) ̸= 0 , so x0 is an ordinary point of ( 7.3.1). However,
here we consider cases where the differential equation in ( 7.3.1) is not of the form
(1 + α(x− x0)2) y′′+ β(x− x0)y′+ γy = 0 ,
so Theorem 7.2.2 does not apply , and the computation of the coefficients {an }is more complicated. For
the equations considered here it’s difficult or impossible t o obtain an explicit formula for an in terms of
n. Nevertheless, we can calculate as many coefficients as we wi sh. The next three examples illustrate
this.
Example 7.3.1 Find the coefficients a0, . . . , a7 in the series solution y= ∑ ∞
n=0 an xn of the initial value
problem",Elementary Differential Equations with Boundary Value Problems.pdf
"this.
Example 7.3.1 Find the coefficients a0, . . . , a7 in the series solution y= ∑ ∞
n=0 an xn of the initial value
problem
(1 + x+ 2 x2)y′′+ (1 + 7 x)y′+ 2 y= 0 , y (0) = −1, y ′(0) = −2. (7.3.2)
Solution Here
Ly= (1 + x+ 2 x2)y′′+ (1 + 7 x)y′+ 2 y.
The zeros (−1 ± i
√
7)/ 4 of P0(x) = 1 + x+ 2 x2 have absolute value 1/
√
2, so Theorem 7.2.2 implies
that the series solution converges to the solution of ( 7.3.2) on (−1/
√
2, 1/
√
2). Since
y =
∞∑
n=0
an xn, y ′=
∞∑
n=1
nan xn−1 and y′′=
∞∑
n=2
n(n− 1)anxn−2,
Ly =
∞∑
n=2
n(n− 1)an xn−2 +
∞∑
n=2
n(n− 1)anxn−1 + 2
∞∑
n=2
n(n− 1)an xn
+
∞∑
n=1
nanxn−1 + 7
∞∑
n=1
nanxn + 2
∞∑
n=0
anxn .
Shifting indices so the general term in each series is a const ant multiple of xn yields
Ly =
∞∑
n=0
(n+ 2)( n+ 1) an+2xn +
∞∑
n=0
(n+ 1) nan+1xn + 2
∞∑
n=0
n(n− 1)anxn
+
∞∑
n=0
(n+ 1) an+1xn + 7
∞∑
n=0
nanxn + 2
∞∑
n=0
anxn =
∞∑
n=0
bn xn,",Elementary Differential Equations with Boundary Value Problems.pdf
"336 Chapter 7 Series Solutions of Linear Second Order Equations
where
bn = ( n+ 2)( n+ 1) an+2 + ( n+ 1) 2an+1 + ( n+ 2)(2 n+ 1) an .
Therefore y = ∑ ∞
n=0 anxn is a solution of Ly= 0 if and only if
an+2 = −n+ 1
n+ 2 an+1 − 2n+ 1
n+ 1 an, n≥ 0. (7.3.3)
From the initial conditions in ( 7.3.2), a0 = y(0) = −1 and a1 = y′(0) = −2. Setting n = 0 in ( 7.3.3)
yields
a2 = −1
2 a1 − a0 = −1
2 (−2) − (−1) = 2 .
Setting n= 1 in ( 7.3.3) yields
a3 = −2
3 a2 − 3
2 a1 = −2
3 (2) − 3
2 (−2) = 5
3 .
W e leave it to you to compute a4,a 5,a 6,a 7 from ( 7.3.3) and show that
y= −1 − 2x+ 2 x2 + 5
3 x3 − 55
12 x4 + 3
4 x5 + 61
8 x6 − 443
56 x7 + · · ·.
W e also leave it to you (Exercise 13) to verify numerically that the T aylor polynomials TN (x) = ∑ N
n=0 an xn
converge to the solution of (
7.3.2) on (−1/
√
2, 1/
√
2).
Example 7.3.2 Find the coefficients a0, . . . , a5 in the series solution
y=
∞∑
n=0
an (x+ 1) n
of the initial value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"√
2, 1/
√
2).
Example 7.3.2 Find the coefficients a0, . . . , a5 in the series solution
y=
∞∑
n=0
an (x+ 1) n
of the initial value problem
(3 + x)y′′+ (1 + 2 x)y′− (2 − x)y = 0 , y (−1) = 2 , y ′(−1) = −3. (7.3.4)
Solution Since the desired series is in powers of x+ 1 we rewrite the differential equation in ( 7.3.4) as
Ly= 0 , with
Ly= (2 + ( x+ 1)) y′′− (1 − 2(x+ 1)) y′− (3 − (x+ 1)) y.
Since
y=
∞∑
n=0
an (x+ 1) n, y ′=
∞∑
n=1
nan (x+ 1) n−1 and y′′=
∞∑
n=2
n(n− 1)an(x+ 1) n−2,
Ly = 2
∞∑
n=2
n(n− 1)an(x+ 1) n−2 +
∞∑
n=2
n(n− 1)an(x+ 1) n−1
−
∞∑
n=1
nan (x+ 1) n−1 + 2
∞∑
n=1
nan (x+ 1) n
−3
∞∑
n=0
an (x+ 1) n +
∞∑
n=0
an (x+ 1) n+1.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.3 Series Solutions Near an Ordinary Point II 337
Shifting indices so that the general term in each series is a c onstant multiple of (x+ 1) n yields
Ly = 2
∞∑
n=0
(n+ 2)( n+ 1) an+2(x+ 1) n +
∞∑
n=0
(n+ 1) nan+1(x+ 1) n
−
∞∑
n=0
(n+ 1) an+1(x+ 1) n +
∞∑
n=0
(2n− 3)an (x+ 1) n +
∞∑
n=1
an−1(x+ 1) n
=
∞∑
n=0
bn(x+ 1) n,
where
b0 = 4 a2 − a1 − 3a0
and
bn = 2( n+ 2)( n+ 1) an+2 + ( n2 − 1)an+1 + (2 n− 3)an + an−1, n ≥ 1.
Therefore y = ∑ ∞
n=0 an(x+ 1) n is a solution of Ly= 0 if and only if
a2 = 1
4 (a1 + 3 a0) (7.3.5)
and
an+2 = − 1
2(n+ 2)( n+ 1)
[
(n2 − 1)an+1 + (2 n− 3)an + an−1
]
, n ≥ 1. (7.3.6)
From the initial conditions in ( 7.3.4), a0 = y(−1) = 2 and a1 = y′(−1) = −3. W e leave it to you to
compute a2, . . . , a5 with ( 7.3.5) and ( 7.3.6) and show that the solution of ( 7.3.4) is
y = −2 − 3(x+ 1) + 3
4 (x+ 1) 2 − 5
12 (x+ 1) 3 + 7
48 (x+ 1) 4 − 1
60 (x+ 1) 5 + · · ·.
W e also leave it to you (Exercise 14) to verify numerically that the T aylor polynomials TN (x) = ∑ N",Elementary Differential Equations with Boundary Value Problems.pdf
"48 (x+ 1) 4 − 1
60 (x+ 1) 5 + · · ·.
W e also leave it to you (Exercise 14) to verify numerically that the T aylor polynomials TN (x) = ∑ N
n=0 an xn
converge to the solution of (
7.3.4) on the interval of convergence of the power series solution .
Example 7.3.3 Find the coefficients a0, . . . , a5 in the series solution y= ∑ ∞
n=0 an xn of the initial value
problem
y′′+ 3 xy′+ (4 + 2 x2)y = 0 , y (0) = 2 , y ′(0) = −3. (7.3.7)
Solution Here
Ly= y′′+ 3 xy′+ (4 + 2 x2)y.
Since
y =
∞∑
n=0
anxn , y ′=
∞∑
n=1
nanxn−1, and y′′=
∞∑
n=2
n(n− 1)anxn−2,
Ly =
∞∑
n=2
n(n− 1)anxn−2 + 3
∞∑
n=1
nanxn + 4
∞∑
n=0
anxn + 2
∞∑
n=0
an xn+2.
Shifting indices so that the general term in each series is a c onstant multiple of xn yields
Ly=
∞∑
n=0
(n+ 2)( n+ 1) an+2xn +
∞∑
n=0
(3n+ 4) an xn + 2
∞∑
n=2
an−2xn =
∞∑
n=0
bnxn",Elementary Differential Equations with Boundary Value Problems.pdf
"338 Chapter 7 Series Solutions of Linear Second Order Equations
where
b0 = 2 a2 + 4 a0, b 1 = 6 a3 + 7 a1,
and
bn = ( n+ 2)( n+ 1) an+2 + (3 n+ 4) an + 2 an−2, n ≥ 2.
Therefore y = ∑ ∞
n=0 anxn is a solution of Ly= 0 if and only if
a2 = −2a0, a 3 = −7
6 a1, (7.3.8)
and
an+2 = − 1
(n+ 2)( n+ 1) [(3n+ 4) an + 2 an−2] , n ≥ 2. (7.3.9)
From the initial conditions in ( 7.3.7), a0 = y(0) = 2 and a1 = y′(0) = −3. W e leave it to you to
compute a2, . . . , a5 with ( 7.3.8) and ( 7.3.9) and show that the solution of ( 7.3.7) is
y= 2 − 3x− 4x2 + 7
2 x3 + 3 x4 − 79
40 x5 + · · ·.
W e also leave it to you (Exercise 15) to verify numerically that the T aylor polynomials TN (x) = ∑ N
n=0 an xn
converge to the solution of (
7.3.9) on the interval of convergence of the power series solution .
7.3 Exercises
In Exercises 1– 12 find the coefficients a0,. . . , aN for N at least 7 in the series solution y = ∑ ∞
n=0 anxn
of the initial value problem.
1.
C (1 + 3 x)y′′+ xy′+ 2 y= 0 , y (0) = 2 , y ′(0) = −3",Elementary Differential Equations with Boundary Value Problems.pdf
"n=0 anxn
of the initial value problem.
1.
C (1 + 3 x)y′′+ xy′+ 2 y= 0 , y (0) = 2 , y ′(0) = −3
2. C (1 + x+ 2 x2)y′′+ (2 + 8 x)y′+ 4 y = 0 , y (0) = −1, y ′(0) = 2
3. C (1 − 2x2)y′′+ (2 − 6x)y′− 2y= 0 , y (0) = 1 , y ′(0) = 0
4. C (1 + x+ 3 x2)y′′+ (2 + 15 x)y′+ 12 y= 0 , y (0) = 0 , y ′(0) = 1
5. C (2 + x)y′′+ (1 + x)y′+ 3 y= 0 , y (0) = 4 , y ′(0) = 3
6. C (3 + 3 x+ x2)y′′+ (6 + 4 x)y′+ 2 y = 0 , y (0) = 7 , y ′(0) = 3
7. C (4 + x)y′′+ (2 + x)y′+ 2 y= 0 , y (0) = 2 , y ′(0) = 5
8. C (2 − 3x+ 2 x2)y′′− (4 − 6x)y′+ 2 y= 0 , y (1) = 1 , y ′(1) = −1
9. C (3x+ 2 x2)y′′+ 10(1 + x)y′+ 8 y = 0 , y (−1) = 1 , y ′(−1) = −1
10. C (1 − x+ x2)y′′− (1 − 4x)y′+ 2 y= 0 , y (1) = 2 , y ′(1) = −1
11. C (2 + x)y′′+ (2 + x)y′+ y= 0 , y (−1) = −2, y ′(−1) = 3
12. C x2y′′− (6 − 7x)y′+ 8 y = 0 , y (1) = 1 , y ′(1) = −2
13. L Do the following experiment for various choices of real numb ers a0, a1, and r, with 0 <r <
1/
√
2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.3 Series Solutions Near an Ordinary Point II 339
(a) Use differential equations software to solve the initial va lue problem
(1 + x+ 2 x2)y′′+ (1 + 7 x)y′+ 2 y= 0 , y (0) = a0, y ′(0) = a1, (A)
numerically on (−r,r ). (See Example 7.3.1.)
(b) For N = 2 , 3, 4, . . . , compute a2, . . . , aN in the power series solution y = ∑ ∞
n=0 anxn of
(A), and graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on (−r,r ). Continue increasing Nuntil there’s no perceptible
difference between the two graphs.
14.
L Do the following experiment for various choices of real numb ers a0, a1, and r, with 0 <r <2.
(a) Use differential equations software to solve the initial va lue problem
(3 + x)y′′+ (1 + 2 x)y′− (2 − x)y = 0 , y (−1) = a0, y ′(−1) = a1, (A)
numerically on (−1 − r, −1 + r). (See Example 7.3.2. Why this interval?)
(b) For N = 2 , 3, 4, . . . , compute a2,...,a N in the power series solution
y=
∞∑
n=0
an (x+ 1) n
of (A), and graph
TN (x) =
N∑
n=0
an (x+ 1) n",Elementary Differential Equations with Boundary Value Problems.pdf
"(b) For N = 2 , 3, 4, . . . , compute a2,...,a N in the power series solution
y=
∞∑
n=0
an (x+ 1) n
of (A), and graph
TN (x) =
N∑
n=0
an (x+ 1) n
and the solution obtained in (a) on (−1 − r, −1 + r). Continue increasing N until there’s no
perceptible difference between the two graphs.
15. L Do the following experiment for several choices of a0, a1, and r, with r> 0.
(a) Use differential equations software to solve the initial va lue problem
y′′+ 3 xy′+ (4 + 2 x2)y = 0 , y (0) = a0, y ′(0) = a1, (A)
numerically on (−r,r ). (See Example 7.3.3.)
(b) Find the coefficients a0, a1, . . . , aN in the power series solution y = ∑ ∞
n=0 anxn of (A),
and graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on (−r,r ). Continue increasing Nuntil there’s no perceptible
difference between the two graphs.",Elementary Differential Equations with Boundary Value Problems.pdf
"340 Chapter 7 Series Solutions of Linear Second Order Equations
16. L Do the following experiment for several choices of a0 and a1.
(a) Use differential equations software to solve the initial va lue problem
(1 − x)y′′− (2 − x)y′+ y = 0 , y (0) = a0, y ′(0) = a1, (A)
numerically on (−r,r ).
(b) Find the coefficients a0, a1, . . . , aN in the power series solution y = ∑ N
n=0 anxn of (A),
and graph
TN (x) =
N∑
n=0
an xn
and the solution obtained in (a) on (−r,r ). Continue increasing Nuntil there’s no perceptible
difference between the two graphs. What happens as you let r→ 1?
17.
L Follow the directions of Exercise 16 for the initial value problem
(1 + x)y′′+ 3 y′+ 32 y= 0 , y (0) = a0, y ′(0) = a1.
18. L Follow the directions of Exercise 16 for the initial value problem
(1 + x2)y′′+ y′+ 2 y= 0 , y (0) = a0, y ′(0) = a1.
In Exercises 19– 28 find the coefficients a0, . . . , aN for N at least 7 in the series solution
y=
∞∑
n=0
an(x− x0)n",Elementary Differential Equations with Boundary Value Problems.pdf
"In Exercises 19– 28 find the coefficients a0, . . . , aN for N at least 7 in the series solution
y=
∞∑
n=0
an(x− x0)n
of the initial value problem. T ake x0 to be the point where the initial conditions are imposed.
19. C (2 + 4 x)y′′− 4y′− (6 + 4 x)y= 0 , y (0) = 2 , y ′(0) = −7
20. C (1 + 2 x)y′′− (1 − 2x)y′− (3 − 2x)y= 0 , y (1) = 1 , y ′(1) = −2
21. C (5 + 2 x)y′′− y′+ (5 + x)y= 0 , y (−2) = 2 , y ′(−2) = −1
22. C (4 + x)y′′− (4 + 2 x)y′+ (6 + x)y= 0 , y (−3) = 2 , y ′(−3) = −2
23. C (2 + 3 x)y′′− xy′+ 2 xy= 0 , y (0) = −1, y ′(0) = 2
24. C (3 + 2 x)y′′+ 3 y′− xy= 0 , y (−1) = 2 , y ′(−1) = −3
25. C (3 + 2 x)y′′− 3y′− (2 + x)y = 0 , y (−2) = −2, y ′(−2) = 3
26. C (10 − 2x)y′′+ (1 + x)y= 0 , y (2) = 2 , y ′(2) = −4
27. C (7 + x)y′′+ (8 + 2 x)y′+ (5 + x)y= 0 , y (−4) = 1 , y ′(−4) = 2
28. C (6 + 4 x)y′′+ (1 + 2 x)y= 0 , y (−1) = −1, y ′(−1) = 2
29. Show that the coefficients in the power series in xfor the general solution of
(1 + αx + βx2)y′′+ ( γ + δx)y′+ ϵy = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"29. Show that the coefficients in the power series in xfor the general solution of
(1 + αx + βx2)y′′+ ( γ + δx)y′+ ϵy = 0
satisfy the recurrrence relation
an+2 = −γ + αn
n+ 2 an+1 − βn(n− 1) + δn + ϵ
(n+ 2)( n+ 1) an.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.3 Series Solutions Near an Ordinary Point II 341
30. (a) Let α and β be constants, with β ̸= 0 . Show that y= ∑ ∞
n=0 an xn is a solution of
(1 + αx + βx2)y′′+ (2 α + 4 βx)y′+ 2 βy = 0 (A)
if and only if
an+2 + αan+1 + βan = 0 , n ≥ 0. (B)
An equation of this form is called a
second order homogeneous linear difference equation .
The polynomial p(r) = r2 + αr + β is called the characteristic polynomial of (B). If r1 and
r2 are the zeros of p, then 1/r 1 and 1/r 2 are the zeros of
P0(x) = 1 + αx + βx2.
(b) Suppose p(r) = ( r− r1)(r− r2) where r1 and r2 are real and distinct, and let ρ be the
smaller of the two numbers {1/ |r1|, 1/ |r2|}. Show that if c1 and c2 are constants then the
sequence
an = c1rn
1 + c2rn
2 , n ≥ 0
satisfies (B). Conclude from this that any function of the for m
y=
∞∑
n=0
(c1rn
1 + c2rn
2 )xn
is a solution of (A) on (−ρ,ρ ).
(c) Use (b) and the formula for the sum of a geometric series to show that t he functions
y1 = 1
1 − r1x and y2 = 1
1 − r2x",Elementary Differential Equations with Boundary Value Problems.pdf
"(c) Use (b) and the formula for the sum of a geometric series to show that t he functions
y1 = 1
1 − r1x and y2 = 1
1 − r2x
form a fundamental set of solutions of (A) on (−ρ,ρ ).
(d) Show that {y1,y 2}is a fundamental set of solutions of (A) on any interval that d oes’nt
contain either 1/r 1 or 1/r 2.
(e) Suppose p(r) = ( r− r1)2, and let ρ = 1 / |r1|. Show that if c1 and c2 are constants then the
sequence
an = ( c1 + c2n)rn
1 , n ≥ 0
satisfies (B). Conclude from this that any function of the for m
y =
∞∑
n=0
(c1 + c2n)rn
1 xn
is a solution of (A) on (−ρ,ρ ).
(f) Use (e) and the formula for the sum of a geometric series to sho w that the functions
y1 = 1
1 − r1x and y2 = x
(1 − r1x)2
form a fundamental set of solutions of (A) on (−ρ,ρ ).
(g) Show that {y1,y 2}is a fundamental set of solutions of (A) on any interval that d oes not
contain 1/r 1.",Elementary Differential Equations with Boundary Value Problems.pdf
"342 Chapter 7 Series Solutions of Linear Second Order Equations
31. Use the results of Exercise 30 to find the general solution of the given equation on any inter val on
which polynomial multiplying y′′has no zeros.
(a) (1 + 3 x+ 2 x2)y′′+ (6 + 8 x)y′+ 4 y= 0
(b) (1 − 5x+ 6 x2)y′′− (10 − 24x)y′+ 12 y= 0
(c) (1 − 4x+ 4 x2)y′′− (8 − 16x)y′+ 8 y= 0
(d) (4 + 4 x+ x2)y′′+ (8 + 4 x)y′+ 2 y= 0
(e) (4 + 8 x+ 3 x2)y′′+ (16 + 12 x)y′+ 6 y= 0
In Exercises 32– 38 find the coefficients a0, . . . , aN for N at least 7 in the series solution y = ∑ ∞
n=0 anxn
of the initial value problem.
32.
C y′′+ 2 xy′+ (3 + 2 x2)y = 0 , y (0) = 1 , y ′(0) = −2
33. C y′′− 3xy′+ (5 + 2 x2)y = 0 , y (0) = 1 , y ′(0) = −2
34. C y′′+ 5 xy′− (3 − x2)y = 0 , y (0) = 6 , y ′(0) = −2
35. C y′′− 2xy′− (2 + 3 x2)y = 0 , y (0) = 2 , y ′(0) = −5
36. C y′′− 3xy′+ (2 + 4 x2)y = 0 , y (0) = 3 , y ′(0) = 6
37. C 2y′′+ 5 xy′+ (4 + 2 x2)y= 0 , y (0) = 3 , y ′(0) = −2
38. C 3y′′+ 2 xy′+ (4 − x2)y = 0 , y (0) = −2, y ′(0) = 3",Elementary Differential Equations with Boundary Value Problems.pdf
"37. C 2y′′+ 5 xy′+ (4 + 2 x2)y= 0 , y (0) = 3 , y ′(0) = −2
38. C 3y′′+ 2 xy′+ (4 − x2)y = 0 , y (0) = −2, y ′(0) = 3
39. Find power series in xfor the solutions y1 and y2 of
y′′+ 4 xy′+ (2 + 4 x2)y = 0
such that y1(0) = 1 , y′
1(0) = 0 , y2(0) = 0 , y′
2(0) = 1 , and identify y1 and y2 in terms of
familiar elementary functions.
In Exercises 40– 49 find the coefficients a0, . . . , aN for N at least 7 in the series solution
y=
∞∑
n=0
an(x− x0)n
of the initial value problem. T ake x0 to be the point where the initial conditions are imposed.
40. C (1 + x)y′′+ x2y′+ (1 + 2 x)y= 0 , y (0) − 2, y ′(0) = 3
41. C y′′+ (1 + 2 x+ x2)y′+ 2 y= 0 , y (0) = 2 , y ′(0) = 3
42. C (1 + x2)y′′+ (2 + x2)y′+ xy= 0 , y (0) = −3, y ′(0) = 5
43. C (1 + x)y′′+ (1 − 3x+ 2 x2)y′− (x− 4)y= 0 , y (1) = −2, y ′(1) = 3
44. C y′′+ (13 + 12 x+ 3 x2)y′+ (5 + 2 x), y (−2) = 2 , y ′(−2) = −3
45. C (1 + 2 x+ 3 x2)y′′+ (2 − x2)y′+ (1 + x)y= 0 , y (0) = 1 , y ′(0) = −2",Elementary Differential Equations with Boundary Value Problems.pdf
"44. C y′′+ (13 + 12 x+ 3 x2)y′+ (5 + 2 x), y (−2) = 2 , y ′(−2) = −3
45. C (1 + 2 x+ 3 x2)y′′+ (2 − x2)y′+ (1 + x)y= 0 , y (0) = 1 , y ′(0) = −2
46. C (3 + 4 x+ x2)y′′− (5 + 4 x− x2)y′− (2 + x)y= 0 , y (−2) = 2 , y ′(−2) = −1
47. C (1 + 2 x+ x2)y′′+ (1 − x)y= 0 , y (0) = 2 , y ′(0) = −1
48. C (x− 2x2)y′′+ (1 + 3 x− x2)y′+ (2 + x)y= 0 , y (1) = 1 , y ′(1) = 0
49. C (16 − 11x+ 2 x2)y′′+ (10 − 6x+ x2)y′− (2 − x)y, y (3) = 1 , y ′(3) = −2",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.4 Regular Singular Points: Euler Equations 343
7.4 REGULAR SINGULAR POINTS EULER EQUA TIONS
This section sets the stage for Sections 1.5, 1.6, and 1.7. If you’re not interested in those sections, but wish
to learn about Euler equations, omit the introductory parag raphs and start reading at Definition 7.4.2.
In the next three sections we’ll continue to study equations of the form
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 (7.4.1)
where P0, P1, and P2 are polynomials, but the emphasis will be different from tha t of Sections 7.2 and
7.3, where we obtained solutions of ( 7.4.1) near an ordinary point x0 in the form of power series in
x− x0. If x0 is a singular point of ( 7.4.1) (that is, if P(x0) = 0 ), the solutions can’t in general be
represented by power series in x− x0. Nevertheless, it’s often necessary in physical applicati ons to study
the behavior of solutions of ( 7.4.1) near a singular point. Although this can be difficult in the a bsence of",Elementary Differential Equations with Boundary Value Problems.pdf
"the behavior of solutions of ( 7.4.1) near a singular point. Although this can be difficult in the a bsence of
some sort of assumption on the nature of the singular point, e quations that satisfy the requirements of the
next definition can be solved by series methods discussed in t he next three sections. Fortunately , many
equations arising in applications satisfy these requireme nts.
Definition 7.4.1 Let P0, P1, and P2 be polynomials with no common factor and suppose P0(x0) = 0 .
Then x0 is a regular singular point of the equation
P0(x)y′′+ P1(x)y′+ P2(x)y = 0 (7.4.2)
if ( 7.4.2) can be written as
(x− x0)2A(x)y′′+ ( x− x0)B(x)y′+ C(x)y= 0 (7.4.3)
where A, B, and C are polynomials and A(x0) ̸= 0 ; otherwise, x0 is an irregular singular point of
(7.4.2).
Example 7.4.1 Bessel’s equation,
x2y′′+ xy′+ ( x2 − ν2)y = 0 , (7.4.4)
has the singular point x0 = 0 . Since this equation is of the form ( 7.4.3) with x0 = 0 , A(x) = 1 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"x2y′′+ xy′+ ( x2 − ν2)y = 0 , (7.4.4)
has the singular point x0 = 0 . Since this equation is of the form ( 7.4.3) with x0 = 0 , A(x) = 1 ,
B(x) = 1 , and C(x) = x2 − ν2, it follows that x0 = 0 is a regular singular point of ( 7.4.4).
Example 7.4.2 Legendre’s equation,
(1 − x2)y′′− 2xy′+ α(α + 1) y= 0 , (7.4.5)
has the singular points x0 = ±1. Mutiplying through by 1 − xyields
(x− 1)2(x+ 1) y′′+ 2 x(x− 1)y′− α(α + 1)( x− 1)y= 0 ,
which is of the form ( 7.4.3) with x0 = 1 , A(x) = x+ 1 , B(x) = 2 x, and C(x) = −α(α + 1)( x− 1).
Therefore x0 = 1 is a regular singular point of ( 7.4.5). W e leave it to you to show that x0 = −1 is also a
regular singular point of ( 7.4.5).",Elementary Differential Equations with Boundary Value Problems.pdf
"344 Chapter 7 Series Solutions of Linear Second Order Equations
Example 7.4.3 The equation
x3y′′+ xy′+ y= 0
has an irregular singular point at x0 = 0 . (V erify .)
For convenience we restrict our attention to the case where x0 = 0 is a regular singular point of ( 7.4.2).
This isn’t really a restriction, since if x0 ̸= 0 is a regular singular point of ( 7.4.2) then introducing the
new independent variable t = x− x0 and the new unknown Y(t) = y(t+ x0) leads to a differential
equation with polynomial coefficients that has a regular sin gular point at t0 = 0 . This is illustrated in
Exercise 22 for Legendre’s equation, and in Exercise 23 for the general case.
Euler Equations
The simplest kind of equation with a regular singular point a t x0 = 0 is the Euler equation, defined as
follows.
Definition 7.4.2
An Euler equation is an equation that can be written in the form
ax2y′′+ bxy′+ cy= 0 , (7.4.6)
where a,b , and care real constants and a̸= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"An Euler equation is an equation that can be written in the form
ax2y′′+ bxy′+ cy= 0 , (7.4.6)
where a,b , and care real constants and a̸= 0 .
Theorem 5.1.1 implies that ( 7.4.6) has solutions defined on (0, ∞ ) and (−∞ , 0), since ( 7.4.6) can be
rewritten as
ay′′+ b
xy′+ c
x2 y= 0 .
For convenience we’ll restrict our attention to the interva l (0, ∞ ). (Exercise 19 deals with solutions of
(7.4.6) on (−∞ , 0).) The key to finding solutions on (0, ∞ ) is that if x >0 then xr is defined as a
real-valued function on (0, ∞ ) for all values of r, and substituting y= xr into ( 7.4.6) produces
ax2(xr )′′+ bx(xr )′+ cxr = ax2r(r− 1)xr−2 + bxrxr−1 + cxr
= [ ar(r− 1) + br+ c]xr. (7.4.7)
The polynomial
p(r) = ar(r− 1) + br+ c
is called the indicial polynomial of ( 7.4.6), and p(r) = 0 is its indicial equation . From ( 7.4.7) we can see
that y = xr is a solution of ( 7.4.6) on (0, ∞ ) if and only if p(r) = 0 . Therefore, if the indicial equation",Elementary Differential Equations with Boundary Value Problems.pdf
"that y = xr is a solution of ( 7.4.6) on (0, ∞ ) if and only if p(r) = 0 . Therefore, if the indicial equation
has distinct real roots r1 and r2 then y1 = xr1 and y2 = xr2 form a fundamental set of solutions of ( 7.4.6)
on (0, ∞ ), since y2/y 1 = xr2−r1 is nonconstant. In this case
y = c1xr1 + c2xr2
is the general solution of ( 7.4.6) on (0, ∞ ).
Example 7.4.4 Find the general solution of
x2y′′− xy′− 8y= 0 (7.4.8)
on (0, ∞ ).
Solution The indicial polynomial of ( 7.4.8) is
p(r) = r(r− 1) − r− 8 = ( r− 4)(r+ 2) .
Therefore y1 = x4 and y2 = x−2 are solutions of ( 7.4.8) on (0, ∞ ), and its general solution on (0, ∞ ) is
y= c1x4 + c2
x2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.4 Regular Singular Points: Euler Equations 345
Example 7.4.5 Find the general solution of
6x2y′′+ 5 xy′− y = 0 (7.4.9)
on (0, ∞ ).
Solution The indicial polynomial of ( 7.4.9) is
p(r) = 6 r(r− 1) + 5 r− 1 = (2 r− 1)(3r+ 1) .
Therefore the general solution of ( 7.4.9) on (0, ∞ ) is
y = c1x1/ 2 + c2x−1/ 3.
If the indicial equation has a repeated root r1, then y1 = xr1 is a solution of
ax2y′′+ bxy′+ cy= 0 , (7.4.10)
on (0, ∞ ), but ( 7.4.10) has no other solution of the form y = xr . If the indicial equation has complex
conjugate zeros then ( 7.4.10) has no real–valued solutions of the form y = xr . Fortunately we can use
the results of Section 5.2 for constant coefficient equation s to solve ( 7.4.10) in any case.
Theorem 7.4.3 Suppose the roots of the indicial equation
ar(r− 1) + br+ c= 0 (7.4.11)
are r1 and r2. Then the general solution of the Euler equation
ax2y′′+ bxy′+ cy = 0 (7.4.12)
on (0, ∞ ) is
y = c1xr1 + c2xr2 if r1 and r2 are distinct real numbers ;",Elementary Differential Equations with Boundary Value Problems.pdf
"ax2y′′+ bxy′+ cy = 0 (7.4.12)
on (0, ∞ ) is
y = c1xr1 + c2xr2 if r1 and r2 are distinct real numbers ;
y = xr1 (c1 + c2 ln x) if r1 = r2 ;
y = xλ [c1 cos ( ω ln x) + c2 sin (ω ln x)] if r1,r 2 = λ ± iω with ω >0.
Proof W e first show that y = y(x) satisfies ( 7.4.12) on (0, ∞ ) if and only if Y(t) = y(et ) satisfies the
constant coefficient equation
ad2Y
dt2 + ( b− a) dY
dt + cY = 0 (7.4.13)
on (−∞ , ∞ ). T o do this, it’s convenient to write x = et , or, equivalently , t = ln x; thus, Y(t) = y(x),
where x= et . From the chain rule,
dY
dt = dy
dx
dx
dt
and, since
dx
dt = et = x,
it follows that dY
dt = xdy
dx. (7.4.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"346 Chapter 7 Series Solutions of Linear Second Order Equations
Differentiating this with respect to tand using the chain rule again yields
d2Y
dt2 = d
dt
(dY
dt
)
= d
dt
(
xdy
dx
)
= dx
dt
dy
dx + xd2y
dx2
dx
dt
= xdy
dx + x2 d2y
dx2
(
since dx
dt = x
)
.
From this and ( 7.4.14),
x2 d2y
dx2 = d2Y
dt2 − dY
dt .
Substituting this and ( 7.4.14) into ( 7.4.12) yields ( 7.4.13). Since ( 7.4.11) is the characteristic equation of
(7.4.13), Theorem 5.2.1 implies that the general solution of ( 7.4.13) on (−∞ , ∞ ) is
Y(t) = c1er1 t + c2er2 t if r1 and r2 are distinct real numbers;
Y(t) = er1 t(c1 + c2t) if r1 = r2;
Y(t) = eλt (c1 cos ωt + c2 sin ωt) if r1,r 2 = λ ± iω with ω ̸= 0 .
Since Y(t) = y(et ), substituting t = ln xin the last three equations shows that the general solution o f
(7.4.12) on (0, ∞ ) has the form stated in the theorem.
Example 7.4.6 Find the general solution of
x2y′′− 5xy′+ 9 y= 0 (7.4.15)
on (0, ∞ ).
Solution The indicial polynomial of ( 7.4.15) is",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 7.4.6 Find the general solution of
x2y′′− 5xy′+ 9 y= 0 (7.4.15)
on (0, ∞ ).
Solution The indicial polynomial of ( 7.4.15) is
p(r) = r(r− 1) − 5r+ 9 = ( r− 3)2.
Therefore the general solution of ( 7.4.15) on (0, ∞ ) is
y= x3(c1 + c2 ln x).
Example 7.4.7 Find the general solution of
x2y′′+ 3 xy′+ 2 y= 0 (7.4.16)
on (0, ∞ ).
Solution The indicial polynomial of ( 7.4.16) is
p(r) = r(r− 1) + 3 r+ 2 = ( r+ 1) 2 + 1 .
The roots of the indicial equation are r= −1 ± iand the general solution of ( 7.4.16) on (0, ∞ ) is
y = 1
x[c1 cos(ln x) + c2 sin(ln x)] .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.4 Regular Singular Points: Euler Equations 347
7.4 Exercises
In Exercises 1– 18 find the general solution of the given Euler equation on (0, ∞ ).
1. x2y′′+ 7 xy′+ 8 y = 0 2. x2y′′− 7xy′+ 7 y= 0
3. x2y′′− xy′+ y= 0 4. x2y′′+ 5 xy′+ 4 y= 0
5. x2y′′+ xy′+ y= 0 6. x2y′′− 3xy′+ 13 y= 0
7. x2y′′+ 3 xy′− 3y = 0 8. 12x2y′′− 5xy′′+ 6 y= 0
9. 4x2y′′+ 8 xy′+ y = 0 10. 3x2y′′− xy′+ y= 0
11. 2x2y′′− 3xy′+ 2 y= 0 12. x2y′′+ 3 xy′+ 5 y= 0
13. 9x2y′′+ 15 xy′+ y= 0 14. x2y′′− xy′+ 10 y= 0
15. x2y′′− 6y= 0 16. 2x2y′′+ 3 xy′− y= 0
17. x2y′′− 3xy′+ 4 y = 0 18. 2x2y′′+ 10 xy′+ 9 y= 0
19. (a) Adapt the proof of Theorem 7.4.3 to show that y= y(x) satisfies the Euler equation
ax2y′′+ bxy′+ cy = 0 (7.4.1)
on (−∞ , 0) if and only if Y(t) = y(−et )
ad2Y
dt2 + ( b− a) dY
dt + cY = 0 .
on (−∞ , ∞ ).
(b) Use (a) to show that the general solution of ( 7.4.1) on (−∞ , 0) is
y = c1|x|r1 + c2|x|r2 if r1 and r2 are distinct real numbers;
y = |x|r1 (c1 + c2 ln |x|) if r1 = r2;",Elementary Differential Equations with Boundary Value Problems.pdf
"y = c1|x|r1 + c2|x|r2 if r1 and r2 are distinct real numbers;
y = |x|r1 (c1 + c2 ln |x|) if r1 = r2;
y = |x|λ [c1 cos ( ω ln |x|) + c2 sin (ω ln |x|)] if r1,r 2 = λ ± iω with ω >0.
20. Use reduction of order to show that if
ar(r− 1) + br+ c= 0
has a repeated root r1 then y= xr1 (c1 + c2 ln x) is the general solution of
ax2y′′+ bxy′+ cy= 0
on (0, ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"348 Chapter 7 Series Solutions of Linear Second Order Equations
21. A nontrivial solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= 0
is said to be oscillatory on an interval (a,b ) if it has infinitely many zeros on (a,b ). Otherwise y
is said to be nonoscillatory on (a,b ). Show that the equation
x2y′′+ ky = 0 ( k= constant)
has oscillatory solutions on (0, ∞ ) if and only if k >1/ 4.
22. In Example 7.4.2 we saw that x0 = 1 and x0 = −1 are regular singular points of Legendre’s
equation
(1 − x2)y′′− 2xy′+ α(α + 1) y= 0 . (A)
(a) Introduce the new variables t= x− 1 and Y(t) = y(t+ 1) , and show that yis a solution of
(A) if and only if Y is a solution of
t(2 + t) d2Y
dt2 + 2(1 + t) dY
dt − α(α + 1) Y = 0 ,
which has a regular singular point at t0 = 0 .
(b) Introduce the new variables t= x+ 1 and Y(t) = y(t− 1), and show that yis a solution of
(A) if and only if Y is a solution of
t(2 − t) d2Y
dt2 + 2(1 − t) dY
dt + α(α + 1) Y = 0 ,
which has a regular singular point at t0 = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"(A) if and only if Y is a solution of
t(2 − t) d2Y
dt2 + 2(1 − t) dY
dt + α(α + 1) Y = 0 ,
which has a regular singular point at t0 = 0 .
23. Let P0,P 1, and P2 be polynomials with no common factor, and suppose x0 ̸= 0 is a singular point
of
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 . (A)
Let t= x− x0 and Y(t) = y(t+ x0).
(a) Show that yis a solution of (A) if and only if Y is a solution of
R0(t) d2Y
dt2 + R1(t) dY
dt + R2(t)Y = 0 . (B)
where
Ri(t) = Pi(t+ x0), i = 0 , 1, 2.
(b) Show that R0, R1, and R2 are polynomials in twith no common factors, and R0(0) = 0 ;
thus, t0 = 0 is a singular point of (B).
7.5 THE METHOD OF FROBENIUS I
In this section we begin to study series solutions of a homoge neous linear second order differential equa-
tion with a regular singular point at x0 = 0 , so it can be written as
x2A(x)y′′+ xB(x)y′+ C(x)y= 0 , (7.5.1)
where A, B, Care polynomials and A(0) ̸= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 349
W e’ll see that ( 7.5.1) always has at least one solution of the form
y = xr
∞∑
n=0
anxn
where a0 ̸= 0 and ris a suitably chosen number. The method we will use to find solu tions of this form
and other forms that we’ll encounter in the next two sections is called the method of Frobenius , and we’ll
call them Frobenius solutions .
It can be shown that the power series ∑ ∞
n=0 anxn in a Frobenius solution of (
7.5.1) converges on some
open interval (−ρ,ρ ), where 0 <ρ ≤ ∞ . However, since xr may be complex for negative xor undefined
if x= 0 , we’ll consider solutions defined for positive values of x. Easy modifications of our results yield
solutions defined for negative values of x. (Exercise 54).
W e’ll restrict our attention to the case where A, B, and C are polynomials of degree not greater than
two, so ( 7.5.1) becomes
x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2x2)y = 0 , (7.5.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"two, so ( 7.5.1) becomes
x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2x2)y = 0 , (7.5.2)
where αi, βi, and γi are real constants and α0 ̸= 0 . Most equations that arise in applications can be
written this way . Some examples are
αx2y′′+ βxy′+ γy = 0 (Euler’s equation) ,
x2y′′+ xy′+ ( x2 − ν2)y = 0 (Bessel’s equation) ,
and
xy′′+ (1 − x)y′+ λy = 0 , (Laguerre’s equation ),
where we would multiply the last equation through by x to put it in the form ( 7.5.2). However, the
method of Frobenius can be extended to the case where A, B, and Care functions that can be represented
by power series in xon some interval that contains zero, and A0(0) ̸= 0 (Exercises 57 and 58).
The next two theorems will enable us to develop systematic me thods for finding Frobenius solutions
of ( 7.5.2).
Theorem 7.5.1 Let
Ly= x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2x2)y,
and define
p0(r) = α0r(r− 1) + β0r+ γ0,
p1(r) = α1r(r− 1) + β1r+ γ1,
p2(r) = α2r(r− 1) + β2r+ γ2.
Suppose the series",Elementary Differential Equations with Boundary Value Problems.pdf
"and define
p0(r) = α0r(r− 1) + β0r+ γ0,
p1(r) = α1r(r− 1) + β1r+ γ1,
p2(r) = α2r(r− 1) + β2r+ γ2.
Suppose the series
y=
∞∑
n=0
an xn+r (7.5.3)
converges on (0,ρ ). Then
Ly=
∞∑
n=0
bnxn+r (7.5.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"350 Chapter 7 Series Solutions of Linear Second Order Equations
on (0,ρ ), where
b0 = p0(r)a0,
b1 = p0(r+ 1) a1 + p1(r)a0,
bn = p0(n+ r)an + p1(n+ r− 1)an−1 + p2(n+ r− 2)an−2, n ≥ 2.
(7.5.5)
Proof W e begin by showing that if yis given by ( 7.5.3) and α, β, and γ are constants, then
αx2y′′+ βxy′+ γy =
∞∑
n=0
p(n+ r)an xn+r , (7.5.6)
where
p(r) = αr (r− 1) + βr + γ.
Differentiating (3) twice yields
y′=
∞∑
n=0
(n+ r)an xn+r−1 (7.5.7)
and
y′′=
∞∑
n=0
(n+ r)(n+ r− 1)an xn+r−2. (7.5.8)
Multiplying ( 7.5.7) by xand ( 7.5.8) by x2 yields
xy′=
∞∑
n=0
(n+ r)anxn+r
and
x2y′′=
∞∑
n=0
(n+ r)(n+ r− 1)anxn+r .
Therefore
αx2y′′+ βxy′+ γy =
∞∑
n=0
[α(n+ r)(n+ r− 1) + β(n+ r) + γ] anxn+r
=
∞∑
n=0
p(n+ r)an xn+r ,
which proves ( 7.5.6).
Multiplying ( 7.5.6) by xyields
x(αx2y′′+ βxy′+ γy) =
∞∑
n=0
p(n+ r)anxn+r+1 =
∞∑
n=1
p(n+ r− 1)an−1xn+r . (7.5.9)
Multiplying ( 7.5.6) by x2 yields
x2(αx2y′′+ βxy′+ γy) =
∞∑
n=0
p(n+ r)anxn+r+2 =
∞∑
n=2
p(n+ r− 2)an−2xn+r . (7.5.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 351
T o use these results, we rewrite
Ly= x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2x2)y
as
Ly = (α0x2y′′+ β0 xy′+ γ0 y) + x(α1x2y′′+ β1xy′+ γ1y)
+ x2 (α2x2y′′+ β2xy′+ γ2 y) .
(7.5.11)
From ( 7.5.6) with p= p0,
α0x2y′′+ β0xy′+ γ0y=
∞∑
n=0
p0(n+ r)an xn+r .
From ( 7.5.9) with p= p1,
x
(
α1x2y′′+ β1xy′+ γ1y
)
=
∞∑
n=1
p1(n+ r− 1)an−1xn+r .
From ( 7.5.10) with p= p2,
x2 (
α2x2y′′+ β2xy′+ γ2y
)
=
∞∑
n=2
p2(n+ r− 2)an−2xn+r .
Therefore we can rewrite ( 7.5.11) as
Ly=
∞∑
n=0
p0(n+ r)anxn+r +
∞∑
n=1
p1(n+ r− 1)an−1xn+r
+
∞∑
n=2
p2(n+ r− 2)an−2xn+r ,
or
Ly = p0(r)a0xr + [ p0(r+ 1) a1 + p1(r)a2] xr+1
+
∞∑
n=2
[p0(n+ r)an + p1(n+ r− 1)an−1 + p2(n+ r− 2)an−2] xn+r ,
which implies ( 7.5.4) with {bn}defined as in ( 7.5.5).
Theorem 7.5.2 Let
Ly= x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2x2)y,
where α0 ̸= 0 , and define
p0(r) = α0r(r− 1) + β0r+ γ0,
p1(r) = α1r(r− 1) + β1r+ γ1,
p2(r) = α2r(r− 1) + β2r+ γ2.",Elementary Differential Equations with Boundary Value Problems.pdf
"352 Chapter 7 Series Solutions of Linear Second Order Equations
Suppose ris a real number such that p0(n+ r) is nonzero for all positive integers n. Define
a0(r) = 1 ,
a1(r) = − p1(r)
p0(r+ 1) ,
an (r) = −p1(n+ r− 1)an−1(r) + p2(n+ r− 2)an−2(r)
p0(n+ r) , n ≥ 2.
(7.5.12)
Then the Frobenius series
y(x,r ) = xr
∞∑
n=0
an(r)xn (7.5.13)
converges and satisfies
Ly(x,r ) = p0(r)xr (7.5.14)
on the interval (0,ρ ), where ρ is the distance from the origin to the nearest zero of A(x) = α0 + α1x+
α2x2 in the complex plane . (If Ais constant, then ρ = ∞ .)
If {an(r)}is determined by the recurrence relation ( 7.5.12) then substituting an = an(r) into ( 7.5.5)
yields b0 = p0(r) and bn = 0 for n≥ 1, so ( 7.5.4) reduces to ( 7.5.14). W e omit the proof that the series
(7.5.13) converges on (0,ρ ).
If αi = βi = γi = 0 for i= 1 , 2, then Ly= 0 reduces to the Euler equation
α0x2y′′+ β0xy′+ γ0y = 0 .
Theorem 7.4.3 shows that the solutions of this equation are determined by t he zeros of the indicial poly-",Elementary Differential Equations with Boundary Value Problems.pdf
"α0x2y′′+ β0xy′+ γ0y = 0 .
Theorem 7.4.3 shows that the solutions of this equation are determined by t he zeros of the indicial poly-
nomial
p0(r) = α0r(r− 1) + β0r+ γ0.
Since ( 7.5.14) implies that this is also true for the solutions of Ly= 0 , we’ll also say that p0 is the indicial
polynomial of ( 7.5.2), and that p0(r) = 0 is the indicial equation of Ly = 0 . W e’ll consider only cases
where the indicial equation has real roots r1 and r2, with r1 ≥ r2.
Theorem 7.5.3 Let Land {an(r)}be as in Theorem 7.5.2, and suppose the indicial equation p0(r) = 0
of Ly= 0 has real roots r1 and r2, where r1 ≥ r2. Then
y1(x) = y(x,r 1) = xr1
∞∑
n=0
an(r1)xn
is a Frobenius solution of Ly= 0 . Moreover , if r1 − r2 isn’t an integer then
y2(x) = y(x,r 2) = xr2
∞∑
n=0
an(r2)xn
is also a Frobenius solution of Ly= 0 , and {y1,y 2}is a fundamental set of solutions.
Proof Since r1 and r2 are roots of p0(r) = 0 , the indicial polynomial can be factored as
p0(r) = α0(r− r1)(r− r2). (7.5.15)
Therefore",Elementary Differential Equations with Boundary Value Problems.pdf
"Proof Since r1 and r2 are roots of p0(r) = 0 , the indicial polynomial can be factored as
p0(r) = α0(r− r1)(r− r2). (7.5.15)
Therefore
p0(n+ r1) = nα0(n+ r1 − r2),",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 353
which is nonzero if n >0, since r1 − r2 ≥ 0. Therefore the assumptions of Theorem 7.5.2 hold with
r= r1, and ( 7.5.14) implies that Ly1 = p0(r1)xr1 = 0 .
Now suppose r1 − r2 isn’t an integer. From ( 7.5.15),
p0(n+ r2) = nα0(n− r1 + r2) ̸= 0 if n= 1 , 2, · · ·.
Hence, the assumptions of Theorem 7.5.2 hold with r= r2, and ( 7.5.14) implies that Ly2 = p0(r2)xr2 =
0. W e leave the proof that {y1,y 2}is a fundamental set of solutions as an exercise (Exercise 52).
It isn’t always possible to obtain explicit formulas for the coefficients in Frobenius solutions. However,
we can always set up the recurrence relations and use them to c ompute as many coefficients as we want.
The next example illustrates this.
Example 7.5.1 Find a fundamental set of Frobenius solutions of
2x2(1 + x+ x2)y′′+ x(9 + 11 x+ 11 x2)y′+ (6 + 10 x+ 7 x2)y= 0 . (7.5.16)
Compute just the first six coefficients a0,. . . , a5 in each solution.",Elementary Differential Equations with Boundary Value Problems.pdf
"2x2(1 + x+ x2)y′′+ x(9 + 11 x+ 11 x2)y′+ (6 + 10 x+ 7 x2)y= 0 . (7.5.16)
Compute just the first six coefficients a0,. . . , a5 in each solution.
Solution For the given equation, the polynomials defined in Theorem
7.5.2 are
p0(r) = 2 r(r− 1) + 9 r+ 6 = (2 r+ 3)( r+ 2) ,
p1(r) = 2 r(r− 1) + 11 r+ 10 = (2 r+ 5)( r+ 2) ,
p2(r) = 2 r(r− 1) + 11 r+ 7 = (2 r+ 7)( r+ 1) .
The zeros of the indicial polynomial p0 are r1 = −3/ 2 and r2 = −2, so r1 − r2 = 1 / 2. Therefore
Theorem 7.5.3 implies that
y1 = x−3/ 2
∞∑
n=0
an (−3/ 2)xn and y2 = x−2
∞∑
n=0
an(−2)xn (7.5.17)
form a fundamental set of Frobenius solutions of ( 7.5.16). T o find the coefficients in these series, we use
the recurrence relation of Theorem 7.5.2; thus,
a0(r) = 1 ,
a1(r) = − p1(r)
p0(r+ 1) = −(2r+ 5)( r+ 2)
(2r+ 5)( r+ 3) = −r+ 2
r+ 3 ,
an(r) = −p1(n+ r− 1)an−1 + p2(n+ r− 2)an−2
p0(n+ r)
= −(n+ r+ 1)(2 n+ 2 r+ 3) an−1(r) + ( n+ r− 1)(2n+ 2 r+ 3) an−2(r)
(n+ r+ 2)(2 n+ 2 r+ 3)
= −(n+ r+ 1) an−1(r) + ( n+ r− 1)an−2(r)
n+ r+ 2 , n ≥ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"= −(n+ r+ 1)(2 n+ 2 r+ 3) an−1(r) + ( n+ r− 1)(2n+ 2 r+ 3) an−2(r)
(n+ r+ 2)(2 n+ 2 r+ 3)
= −(n+ r+ 1) an−1(r) + ( n+ r− 1)an−2(r)
n+ r+ 2 , n ≥ 2.
Setting r= −3/ 2 in these equations yields
a0(−3/ 2) = 1 ,
a1(−3/ 2) = −1/ 3,
an (−3/ 2) = −(2n− 1)an−1(−3/ 2) + (2 n− 5)an−2(−3/ 2)
2n+ 1 , n ≥ 2,
(7.5.18)",Elementary Differential Equations with Boundary Value Problems.pdf
"354 Chapter 7 Series Solutions of Linear Second Order Equations
and setting r= −2 yields
a0(−2) = 1 ,
a1(−2) = 0 ,
an (−2) = −(n− 1)an−1(−2) + ( n− 3)an−2(−2)
n , n ≥ 2.
(7.5.19)
Calculating with ( 7.5.18) and ( 7.5.19) and substituting the results into ( 7.5.17) yields the fundamental set
of Frobenius solutions
y1 = x−3/ 2
(
1 − 1
3 x+ 2
5 x2 − 5
21 x3 + 7
135 x4 + 76
1155 x5 + · · ·
)
,
y2 = x−2
(
1 + 1
2 x2 − 1
3 x3 + 1
8 x4 + 1
30 x5 + · · ·
)
.
Special Cases With T wo T erm Recurrence Relations
For n ≥ 2, the recurrence relation ( 7.5.12) of Theorem 7.5.2 involves the three coefficients an (r),
an−1(r), and an−2(r). W e’ll now consider some special cases where ( 7.5.12) reduces to a two term
recurrence relation; that is, a relation involving only an(r) and an−1(r) or only an (r) and an−2(r).
This simplification often makes it possible to obtain explic it formulas for the coefficents of Frobenius
solutions.
W e first consider equations of the form",Elementary Differential Equations with Boundary Value Problems.pdf
"solutions.
W e first consider equations of the form
x2(α0 + α1x)y′′+ x(β0 + β1x)y′+ ( γ0 + γ1x)y= 0
with α0 ̸= 0 . For this equation, α2 = β2 = γ2 = 0 , so p2 ≡ 0 and the recurrence relations in
Theorem 7.5.2 simplify to
a0(r) = 1 ,
an (r) = −p1(n+ r− 1)
p0(n+ r) an−1(r), n ≥ 1. (7.5.20)
Example 7.5.2 Find a fundamental set of Frobenius solutions of
x2(3 + x)y′′+ 5 x(1 + x)y′− (1 − 4x)y= 0 . (7.5.21)
Give explicit formulas for the coefficients in the solutions .
Solution For this equation, the polynomials defined in Theorem 7.5.2 are
p0(r) = 3 r(r− 1) + 5 r− 1 = (3 r− 1)(r+ 1) ,
p1(r) = r(r− 1) + 5 r+ 4 = ( r+ 2) 2,
p2(r) = 0 .
The zeros of the indicial polynomial p0 are r1 = 1 / 3 and r2 = −1, so r1 − r2 = 4 / 3. Therefore
Theorem 7.5.3 implies that
y1 = x1/ 3
∞∑
n=0
an(1/ 3)xn and y2 = x−1
∞∑
n=0
an(−1)xn",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 355
form a fundamental set of Frobenius solutions of ( 7.5.21). T o find the coefficients in these series, we use
the recurrence relationss ( 7.5.20); thus,
a0(r) = 1 ,
an(r) = −p1(n+ r− 1)
p0(n+ r) an−1(r)
= − (n+ r+ 1) 2
(3n+ 3 r− 1)(n+ r+ 1) an−1(r)
= − n+ r+ 1
3n+ 3 r− 1 an−1(r), n ≥ 1.
(7.5.22)
Setting r= 1 / 3 in ( 7.5.22) yields
a0(1/ 3) = 1 ,
an (1/ 3) = −3n+ 4
9n an−1(1/ 3), n ≥ 1.
By using the product notation introduced in Section 7.2 and p roceeding as we did in the examples in that
section yields
an (1/ 3) =
(−1)n ∏ n
j=1(3j+ 4)
9nn! , n ≥ 0.
Therefore
y1 = x1/ 3
∞∑
n=0
(−1)n ∏ n
j=1(3j+ 4)
9nn! xn
is a Frobenius solution of ( 7.5.21).
Setting r= −1 in ( 7.5.22) yields
a0(−1) = 1 ,
an(−1) = − n
3n− 4 an−1(−1), n ≥ 1,
so
an(−1) = (−1)n n!∏ n
j=1(3j− 4) .
Therefore
y2 = x−1
∞∑
n=0
(−1)n n!
∏ n
j=1(3j− 4) xn
is a Frobenius solution of (
7.5.21), and {y1,y 2}is a fundamental set of solutions.
W e now consider equations of the form",Elementary Differential Equations with Boundary Value Problems.pdf
"∏ n
j=1(3j− 4) xn
is a Frobenius solution of (
7.5.21), and {y1,y 2}is a fundamental set of solutions.
W e now consider equations of the form
x2(α0 + α2x2)y′′+ x(β0 + β2x2)y′+ ( γ0 + γ2x2)y= 0 (7.5.23)
with α0 ̸= 0 . For this equation, α1 = β1 = γ1 = 0 , so p1 ≡ 0 and the recurrence relations in
Theorem 7.5.2 simplify to
a0(r) = 1 ,
a1(r) = 0 ,
an (r) = −p2(n+ r− 2)
p0(n+ r) an−2(r), n ≥ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"356 Chapter 7 Series Solutions of Linear Second Order Equations
Since a1(r) = 0 , the last equation implies that an (r) = 0 if nis odd, so the Frobenius solutions are of
the form
y(x,r ) = xr
∞∑
m=0
a2m(r)x2m,
where
a0(r) = 1 ,
a2m(r) = −p2(2m+ r− 2)
p0(2m+ r) a2m−2(r), m ≥ 1. (7.5.24)
Example 7.5.3 Find a fundamental set of Frobenius solutions of
x2(2 − x2)y′′− x(3 + 4 x2)y′+ (2 − 2x2)y= 0 . (7.5.25)
Give explicit formulas for the coefficients in the solutions .
Solution For this equation, the polynomials defined in Theorem 7.5.2 are
p0(r) = 2 r(r− 1) − 3r+ 2 = ( r− 2)(2r− 1),
p1(r) = 0
p2(r) = − [r(r− 1) + 4 r+ 2] = −(r+ 1)( r+ 2) .
The zeros of the indicial polynomial p0 are r1 = 2 and r2 = 1 / 2, so r1 − r2 = 3 / 2. Therefore
Theorem 7.5.3 implies that
y1 = x2
∞∑
m=0
a2m(1/ 3)x2m and y2 = x1/ 2
∞∑
m=0
a2m(1/ 2)x2m
form a fundamental set of Frobenius solutions of ( 7.5.25). T o find the coefficients in these series, we use
the recurrence relation ( 7.5.24); thus,
a0(r) = 1 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"the recurrence relation ( 7.5.24); thus,
a0(r) = 1 ,
a2m(r) = −p2(2m+ r− 2)
p0(2m+ r) a2m−2(r)
= (2m+ r)(2m+ r− 1)
(2m+ r− 2)(4m+ 2 r− 1) a2m−2(r), m ≥ 1.
(7.5.26)
Setting r= 2 in ( 7.5.26) yields
a0(2) = 1 ,
a2m(2) = (m+ 1)(2 m+ 1)
m(4m+ 3) a2m−2(2), m ≥ 1,
so
a2m(2) = ( m+ 1)
m∏
j=1
2j+ 1
4j+ 3 .
Therefore
y1 = x2
∞∑
m=0
(m+ 1)


m∏
j=1
2j+ 1
4j+ 3

 x2m
is a Frobenius solution of (
7.5.25).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 357
Setting r= 1 / 2 in ( 7.5.26) yields
a0(1/ 2) = 1 ,
a2m(1/ 2) = (4m− 1)(4m+ 1)
8m(4m− 3) a2m−2(1/ 2), m ≥ 1,
so
a2m(1/ 2) = 1
8mm!
m∏
j=1
(4j− 1)(4j+ 1)
4j− 3 .
Therefore
y2 = x1/ 2
∞∑
m=0
1
8mm!


m∏
j=1
(4j− 1)(4j+ 1)
4j− 3

 x2m
is a Frobenius solution of (
7.5.25) and {y1,y 2}is a fundamental set of solutions.
REM A RK : Thus far, we considered only the case where the indicial equ ation has real roots that don’t
differ by an integer, which allows us to apply Theorem 7.5.3. However, for equations of the form ( 7.5.23),
the sequence {a2m(r)}in ( 7.5.24) is defined for r= r2 if r1 − r2 isn’t an even integer. It can be shown
(Exercise 56) that in this case
y1 = xr1
∞∑
m=0
a2m(r1)x2m and y2 = xr2
∞∑
m=0
a2m(r2)x2m
form a fundamental set Frobenius solutions of ( 7.5.23).
USING TECHNOLOGY
As we said at the end of Section 7.2, if you’re interested in ac tually using series to compute numerical",Elementary Differential Equations with Boundary Value Problems.pdf
"USING TECHNOLOGY
As we said at the end of Section 7.2, if you’re interested in ac tually using series to compute numerical
approximations to solutions of a differential equation, th en whether or not there’s a simple closed form
for the coefficents is essentially irrelevant; recursive co mputation is usually more efficient. Since it’s also
laborious, we encourage you to write short programs to imple ment recurrence relations on a calculator or
computer, even in exercises where this is not specifically re quired.
In practical use of the method of Frobenius when x0 = 0 is a regular singular point, we’re interested
in how well the functions
yN (x,r i) = xri
N∑
n=0
an(ri)xn, i = 1 , 2,
approximate solutions to a given equation when ri is a zero of the indicial polynomial. In dealing with
the corresponding problem for the case where x0 = 0 is an ordinary point, we used numerical integration",Elementary Differential Equations with Boundary Value Problems.pdf
"the corresponding problem for the case where x0 = 0 is an ordinary point, we used numerical integration
to solve the differential equation subject to initial condi tions y(0) = a0, y ′(0) = a1, and compared the
result with values of the T aylor polynomial
TN (x) =
N∑
n=0
anxn.
W e can’t do that here, since in general we can’t prescribe arb itrary initial values for solutions of a dif-
ferential equation at a singular point. Therefore, motivat ed by Theorem 7.5.2 (specifically , ( 7.5.14)), we
suggest the following procedure.",Elementary Differential Equations with Boundary Value Problems.pdf
"358 Chapter 7 Series Solutions of Linear Second Order Equations
V erification Procedure
Let Land Yn(x; ri) be defined by
Ly= x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2x2)y
and
yN (x; ri) = xri
N∑
n=0
an (ri)xn,
where the coefficients {an(ri)}N
n=0 are computed as in (
7.5.12), Theorem 7.5.2. Compute the error
EN (x; ri) = x−ri LyN (x; ri)/α 0 (7.5.27)
for various values of N and various values of xin the interval (0,ρ ), with ρ as defined in Theorem 7.5.2.
The multiplier x−ri /α 0 on the right of ( 7.5.27) eliminates the effects of small or large values of xri
near x = 0 , and of multiplication by an arbitrary constant. In some exe rcises you will be asked to
estimate the maximum value of EN (x; ri) on an interval (0,δ ] by computing EN (xm; ri) at the M points
xm = mδ/M, m = 1 , 2, . . . , M, and finding the maximum of the absolute values:
σN (δ) = max {|EN (xm ; ri)|, m = 1 , 2,...,M }. (7.5.28)",Elementary Differential Equations with Boundary Value Problems.pdf
"xm = mδ/M, m = 1 , 2, . . . , M, and finding the maximum of the absolute values:
σN (δ) = max {|EN (xm ; ri)|, m = 1 , 2,...,M }. (7.5.28)
(For simplicity , this notation ignores the dependence of th e right side of the equation on iand M.)
T o implement this procedure, you’ll have to write a computer program to calculate {an(ri)}from the
applicable recurrence relation, and to evaluate EN (x; ri).
The next exercise set contains five exercises specifically id entified by L that ask you to implement the
verification procedure. These particular exercises were ch osen arbitrarily you can just as well formulate
such laboratory problems for any of the equations in any of th e Exercises 1– 10, 14-25, and 28– 51
7.5 Exercises
This set contains exercises specifically identified by L that ask you to implement the verification pro-
cedure. These particular exercises were chosen arbitraril y you can just as well formulate such laboratory",Elementary Differential Equations with Boundary Value Problems.pdf
"cedure. These particular exercises were chosen arbitraril y you can just as well formulate such laboratory
problems for any of the equations in Exercises 1– 10, 14-25, and 28– 51.
In Exercises 1– 10 find a fundamental set of Frobenius solutions. Compute a0, a1 . . . , aN for N at least 7
in each solution.
1. C 2x2(1 + x+ x2)y′′+ x(3 + 3 x+ 5 x2)y′− y = 0
2. C 3x2y′′+ 2 x(1 + x− 2x2)y′+ (2 x− 8x2)y = 0
3. C x2(3 + 3 x+ x2)y′′+ x(5 + 8 x+ 7 x2)y′− (1 − 2x− 9x2)y = 0
4. C 4x2y′′+ x(7 + 2 x+ 4 x2)y′− (1 − 4x− 7x2)y= 0
5. C 12x2(1 + x)y′′+ x(11 + 35 x+ 3 x2)y′− (1 − 10x− 5x2)y = 0
6. C x2(5 + x+ 10 x2)y′′+ x(4 + 3 x+ 48 x2)y′+ ( x+ 36 x2)y = 0
7. C 8x2y′′− 2x(3 − 4x− x2)y′+ (3 + 6 x+ x2)y= 0
8. C 18x2(1 + x)y′′+ 3 x(5 + 11 x+ x2)y′− (1 − 2x− 5x2)y= 0
9. C x(3 + x+ x2)y′′+ (4 + x− x2)y′+ xy= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 359
10. C 10x2(1 + x+ 2 x2)y′′+ x(13 + 13 x+ 66 x2)y′− (1 + 4 x+ 10 x2)y= 0
11. L The Frobenius solutions of
2x2(1 + x+ x2)y′′+ x(9 + 11 x+ 11 x2)y′+ (6 + 10 x+ 7 x2)y = 0
obtained in Example 7.5.1 are defined on (0,ρ ), where ρ is defined in Theorem 7.5.2. Find ρ.
Then do the following experiments for each Frobenius soluti on, with M = 20 and δ = .5ρ, .7ρ,
and .9ρ in the verification procedure described at the end of this sec tion.
(a) Compute σN (δ) (see Eqn. ( 7.5.28)) for N = 5 , 10, 15,. . . , 50.
(b) Find N such that σN (δ) <10−5.
(c) Find N such that σN (δ) <10−10.
12. L By Theorem 7.5.2 the Frobenius solutions of the equation in Exercise 4 are defined on (0, ∞ ).
Do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with M = 20 and
δ = 1 , 2, and 3 in the verification procedure described at the end of this sec tion.
13. L The Frobenius solutions of the equation in Exercise 6 are defined on (0,ρ ), where ρ is defined",Elementary Differential Equations with Boundary Value Problems.pdf
"13. L The Frobenius solutions of the equation in Exercise 6 are defined on (0,ρ ), where ρ is defined
in Theorem 7.5.2. Find ρ and do experiments (a), (b), and (c) of Exercise 11 for each Frobenius
solution, with M = 20 and δ = .3ρ, .4ρ, and .5ρ, in the verification procedure described at the
end of this section.
In Exercises 14– 25 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi-
cients in each solution.
14. 2x2y′′+ x(3 + 2 x)y′− (1 − x)y = 0
15. x2(3 + x)y′′+ x(5 + 4 x)y′− (1 − 2x)y= 0
16. 2x2y′′+ x(5 + x)y′− (2 − 3x)y= 0
17. 3x2y′′+ x(1 + x)y′− y= 0
18. 2x2y′′− xy′+ (1 − 2x)y= 0
19. 9x2y′′+ 9 xy′− (1 + 3 x)y= 0
20. 3x2y′′+ x(1 + x)y′− (1 + 3 x)y= 0
21. 2x2(3 + x)y′′+ x(1 + 5 x)y′+ (1 + x)y= 0
22. x2(4 + x)y′′− x(1 − 3x)y′+ y = 0
23. 2x2y′′+ 5 xy′+ (1 + x)y= 0
24. x2(3 + 4 x)y′′+ x(5 + 18 x)y′− (1 − 12x)y= 0
25. 6x2y′′+ x(10 − x)y′− (2 + x)y = 0
26. L By Theorem 7.5.2 the Frobenius solutions of the equation in Exercise 17 are defined on (0, ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"25. 6x2y′′+ x(10 − x)y′− (2 + x)y = 0
26. L By Theorem 7.5.2 the Frobenius solutions of the equation in Exercise 17 are defined on (0, ∞ ).
Do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with M = 20 and
δ = 3 , 6, 9, and 12 in the verification procedure described at the end of this sec tion.
27. L The Frobenius solutions of the equation in Exercise 22 are defined on (0,ρ ), where ρ is defined
in Theorem 7.5.2. Find ρ and do experiments (a), (b), and (c) of Exercise 11 for each Frobenius
solution, with M = 20 and δ = .25ρ, .5ρ, and .75ρ in the verification procedure described at the
end of this section.",Elementary Differential Equations with Boundary Value Problems.pdf
"360 Chapter 7 Series Solutions of Linear Second Order Equations
In Exercises 28– 32 find a fundamental set of Frobenius solutions. Compute coeffi cients a0, . . . , aN for N
at least 7 in each solution.
28. C x2(8 + x)y′′+ x(2 + 3 x)y′+ (1 + x)y = 0
29. C x2(3 + 4 x)y′′+ x(11 + 4 x)y′− (3 + 4 x)y= 0
30. C 2x2(2 + 3 x)y′′+ x(4 + 11 x)y′− (1 − x)y= 0
31. C x2(2 + x)y′′+ 5 x(1 − x)y′− (2 − 8x)y
32. C x2(6 + x)y′′+ x(11 + 4 x)y′+ (1 + 2 x)y= 0
In Exercises 33– 46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi-
cients in each solution.
33. 8x2y′′+ x(2 + x2)y′+ y= 0
34. 8x2(1 − x2)y′′+ 2 x(1 − 13x2)y′+ (1 − 9x2)y= 0
35. x2(1 + x2)y′′− 2x(2 − x2)y′+ 4 y= 0
36. x(3 + x2)y′′+ (2 − x2)y′− 8xy= 0
37. 4x2(1 − x2)y′′+ x(7 − 19x2)y′− (1 + 14 x2)y= 0
38. 3x2(2 − x2)y′′+ x(1 − 11x2)y′+ (1 − 5x2)y= 0
39. 2x2(2 + x2)y′′− x(12 − 7x2)y′+ (7 + 3 x2)y= 0
40. 2x2(2 + x2)y′′+ x(4 + 7 x2)y′− (1 − 3x2)y = 0
41. 2x2(1 + 2 x2)y′′+ 5 x(1 + 6 x2)y′− (2 − 40x2)y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"40. 2x2(2 + x2)y′′+ x(4 + 7 x2)y′− (1 − 3x2)y = 0
41. 2x2(1 + 2 x2)y′′+ 5 x(1 + 6 x2)y′− (2 − 40x2)y = 0
42. 3x2(1 + x2)y′′+ 5 x(1 + x2)y′− (1 + 5 x2)y = 0
43. x(1 + x2)y′′+ (4 + 7 x2)y′+ 8 xy= 0
44. x2(2 + x2)y′′+ x(3 + x2)y′− y= 0
45. 2x2(1 + x2)y′′+ x(3 + 8 x2)y′− (3 − 4x2)y = 0
46. 9x2y′′+ 3 x(3 + x2)y′− (1 − 5x2)y = 0
In Exercises 47– 51 find a fundamental set of Frobenius solutions. Compute the co efficients a0, . . . , a2M
for M at least 7 in each solution.
47. C 6x2y′′+ x(1 + 6 x2)y′+ (1 + 9 x2)y = 0
48. C x2(8 + x2)y′′+ 7 x(2 + x2)y′− (2 − 9x2)y= 0
49. C 9x2(1 + x2)y′′+ 3 x(3 + 13 x2)y′− (1 − 25x2)y= 0
50. C 4x2(1 + x2)y′′+ 4 x(1 + 6 x2)y′− (1 − 25x2)y= 0
51. C 8x2(1 + 2 x2)y′′+ 2 x(5 + 34 x2)y′− (1 − 30x2)y = 0
52. Suppose r1 >r2, a0 = b0 = 1 , and the Frobenius series
y1 = xr1
∞∑
n=0
anxn and y2 = xr2
∞∑
n=0
bnxn
both converge on an interval (0,ρ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 361
(a) Show that y1 and y2 are linearly independent on (0,ρ ). H IN T : Show that if c1 and c2 are
constants such that c1y1 + c2y2 ≡ 0 on (0,ρ ), then
c1xr1−r2
∞∑
n=0
an xn + c2
∞∑
n=0
bnxn = 0 , 0 <x<ρ.
Then let x→ 0+ to conclude that c2 = 0 .
(b) Use the result of (b) to complete the proof of Theorem 7.5.3.
53. The equation
x2y′′+ xy′+ ( x2 − ν2)y = 0 (7.5.1)
is Bessel’s equation of order ν. (Here ν is a parameter, and this use of “order” should not be con-
fused with its usual use as in “the order of the equation. ”) Th e solutions of ( 7.5.1) are Bessel functions of order
ν.
(a) Assuming that ν isn’t an integer, find a fundamental set of Frobenius solutio ns of ( 7.5.1).
(b) If ν = 1 / 2, the solutions of ( 7.5.1) reduce to familiar elementary functions. Identify these
functions.
54. (a) V erify that
d
dx(|x|rxn) = ( n+ r)|x|rxn−1 and d2
dx2 (|x|rxn) = ( n+ r)(n+ r− 1)|x|rxn−2
if x̸= 0 .
(b) Let",Elementary Differential Equations with Boundary Value Problems.pdf
"functions.
54. (a) V erify that
d
dx(|x|rxn) = ( n+ r)|x|rxn−1 and d2
dx2 (|x|rxn) = ( n+ r)(n+ r− 1)|x|rxn−2
if x̸= 0 .
(b) Let
Ly= x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2x2)y′+ ( γ0 + γ1x+ γ2 x2)y= 0 .
Show that if xr ∑ ∞
n=0 an xn is a solution of Ly = 0 on (0,ρ ) then |x|r ∑ ∞
n=0 an xn is a
solution on (−ρ, 0) and (0,ρ ).
55. (a) Deduce from Eqn. (
7.5.20) that
an (r) = ( −1)n
n∏
j=1
p1(j+ r− 1)
p0(j+ r) .
(b) Conclude that if p0(r) = α0(r− r1)(r− r2) where r1 − r2 is not an integer, then
y1 = xr1
∞∑
n=0
an(r1)xn and y2 = xr2
∞∑
n=0
an (r2)xn
form a fundamental set of Frobenius solutions of
x2(α0 + α1x)y′′+ x(β0 + β1x)y′+ ( γ0 + γ1x)y= 0 .
(c) Show that if p0 satisfies the hypotheses of (b) then
y1 = xr1
∞∑
n=0
(−1)n
n! ∏ n
j=1(j+ r1 − r2)
(γ1
α0
) n
xn
and
y2 = xr2
∞∑
n=0
(−1)n
n! ∏ n
j=1(j+ r2 − r1)
(γ1
α0
) n
xn
form a fundamental set of Frobenius solutions of
α0x2y′′+ β0xy′+ ( γ0 + γ1x)y= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"362 Chapter 7 Series Solutions of Linear Second Order Equations
56. Let
Ly= x2(α0 + α2x2)y′′+ x(β0 + β2x2)y′+ ( γ0 + γ2x2)y= 0
and define
p0(r) = α0r(r− 1) + β0r+ γ0 and p2(r) = α2r(r− 1) + β2r+ γ2.
(a) Use Theorem 7.5.2 to show that if
a0(r) = 1 ,
p0(2m+ r)a2m(r) + p2(2m+ r− 2)a2m−2(r) = 0 , m ≥ 1, (7.5.1)
then the Frobenius series y(x,r ) = xr ∑ ∞
m=0 a2mx2m satisfies Ly(x,r ) = p0(r)xr .
(b) Deduce from (
7.5.1) that if p0(2m+ r) is nonzero for every positive integer mthen
a2m(r) = ( −1)m
m∏
j=1
p2(2j+ r− 2)
p0(2j+ r) .
(c) Conclude that if p0(r) = α0(r− r1)(r− r2) where r1 − r2 is not an even integer, then
y1 = xr1
∞∑
m=0
a2m(r1)x2m and y2 = xr2
∞∑
m=0
a2m(r2)x2m
form a fundamental set of Frobenius solutions of Ly= 0 .
(d) Show that if p0 satisfies the hypotheses of (c) then
y1 = xr1
∞∑
m=0
(−1)m
2mm! ∏ m
j=1(2j+ r1 − r2)
(γ2
α0
) m
x2m
and
y2 = xr2
∞∑
m=0
(−1)m
2mm! ∏ m
j=1(2j+ r2 − r1)
(γ2
α0
) m
x2m
form a fundamental set of Frobenius solutions of
α0x2y′′+ β0xy′+ ( γ0 + γ2x2)y= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"y2 = xr2
∞∑
m=0
(−1)m
2mm! ∏ m
j=1(2j+ r2 − r1)
(γ2
α0
) m
x2m
form a fundamental set of Frobenius solutions of
α0x2y′′+ β0xy′+ ( γ0 + γ2x2)y= 0 .
57. Let
Ly= x2q0(x)y′′+ xq1(x)y′+ q2(x)y,
where
q0(x) =
∞∑
j=0
αj xj , q 1(x) =
∞∑
j=0
βj xj , q 2(x) =
∞∑
j=0
γj xj ,
and define
pj (r) = αj r(r− 1) + βj r+ γj , j = 0 , 1,....
Let y = xr ∑ ∞
n=0 anxn . Show that
Ly= xr
∞∑
n=0
bn xn,
where
bn =
n∑
j=0
pj (n+ r− j)an−j .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.5 The Method of Frobenius I 363
58. (a) Let Lbe as in Exercise 57. Show that if
y(x,r ) = xr
∞∑
n=0
an(r)xn
where
a0(r) = 1 ,
an (r) = − 1
p0(n+ r)
n∑
j=1
pj (n+ r− j)an−j (r), n ≥ 1,
then
Ly(x,r ) = p0(r)xr .
(b) Conclude that if
p0(r) = α0(r− r1)(r− r2)
where r1 − r2 isn’t an integer then y1 = y(x,r 1) and y2 = y(x,r 2) are solutions of Ly= 0 .
59. Let
Ly= x2(α0 + αq xq )y′′+ x(β0 + βq xq )y′+ ( γ0 + γq xq )y
where qis a positive integer, and define
p0(r) = α0r(r− 1) + β0r+ γ0 and pq (r) = αq r(r− 1) + βq r+ γq .
(a) Show that if
y(x,r ) = xr
∞∑
m=0
aqm (r)xqm
where
a0(r) = 1 ,
aqm(r) = −pq (q(m− 1) + r)
p0(qm+ r) aq(m−1) (r), m ≥ 1, (7.5.1)
then
Ly(x,r ) = p0(r)xr .
(b) Deduce from ( 7.5.1) that
aqm(r) = ( −1)m
m∏
j=1
pq (q(j− 1) + r)
p0(qj+ r) .
(c) Conclude that if p0(r) = α0(r− r1)(r− r2) where r1 − r2 is not an integer multiple of q,
then
y1 = xr1
∞∑
m=0
aqm (r1)xqm and y2 = xr2
∞∑
m=0
aqm (r2)xqm
form a fundamental set of Frobenius solutions of Ly= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"364 Chapter 7 Series Solutions of Linear Second Order Equations
(d) Show that if p0 satisfies the hypotheses of (c) then
y1 = xr1
∞∑
m=0
(−1)m
qmm! ∏ m
j=1(qj+ r1 − r2)
(γq
α0
) m
xqm
and
y2 = xr2
∞∑
m=0
(−1)m
qmm! ∏ m
j=1(qj+ r2 − r1)
(γq
α0
) m
xqm
form a fundamental set of Frobenius solutions of
α0x2y′′+ β0xy′+ ( γ0 + γq xq )y= 0 .
60. (a) Suppose α0,α 1, and α2 are real numbers with α0 ̸= 0 , and {an}∞
n=0 is defined by
α0a1 + α1a0 = 0
and
α0an + α1an−1 + α2an−2 = 0 , n ≥ 2.
Show that
(α0 + α1x+ α2x2)
∞∑
n=0
an xn = α0a0,
and infer that ∞∑
n=0
anxn = α0a0
α0 + α1x+ α2x2 .
(b) With α0,α 1, and α2 as in (a), consider the equation
x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1 x+ β2x2)y′+ ( γ0 + γ1x+ γ2 x2)y= 0 , (7.5.1)
and define
pj (r) = αj r(r− 1) + βj r+ γj , j = 0 , 1, 2.
Suppose
p1(r− 1)
p0(r) = α1
α0
, p2(r− 2)
p0(r) = α2
α0
,
and
p0(r) = α0(r− r1)(r− r2),
where r1 >r2. Show that
y1 = xr1
α0 + α1x+ α2x2 and y2 = xr2
α0 + α1x+ α2x2",Elementary Differential Equations with Boundary Value Problems.pdf
"p0(r) = α1
α0
, p2(r− 2)
p0(r) = α2
α0
,
and
p0(r) = α0(r− r1)(r− r2),
where r1 >r2. Show that
y1 = xr1
α0 + α1x+ α2x2 and y2 = xr2
α0 + α1x+ α2x2
form a fundamental set of Frobenius solutions of ( 7.5.1) on any interval (0,ρ ) on which
α0 + α1x+ α2x2 has no zeros.
In Exercises 61– 68 use the method suggested by Exercise 60 to find the general solution on some interval
(0,ρ ).
61. 2x2(1 + x)y′′− x(1 − 3x)y′+ y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 365
62. 6x2(1 + 2 x2)y′′+ x(1 + 50 x2)y′+ (1 + 30 x2)y = 0
63. 28x2(1 − 3x)y′′− 7x(5 + 9 x)y′+ 7(2 + 9 x)y= 0
64. 9x2(5 + x)y′′+ 9 x(5 + 3 x)y′− (5 − 8x)y= 0
65. 8x2(2 − x2)y′′+ 2 x(10 − 21x2)y′− (2 + 35 x2)y = 0
66. 4x2(1 + 3 x+ x2)y′′− 4x(1 − 3x− 3x2)y′+ 3(1 − x+ x2)y= 0
67. 3x2(1 + x)2y′′− x(1 − 10x− 11x2)y′+ (1 + 5 x2)y= 0
68. 4x2(3 + 2 x+ x2)y′′− x(3 − 14x− 15x2)y′+ (3 + 7 x2)y = 0
7.6 THE METHOD OF FROBENIUS II
In this section we discuss a method for finding two linearly in dependent Frobenius solutions of a homo-
geneous linear second order equation near a regular singula r point in the case where the indicial equation
has a repeated real root. As in the preceding section, we cons ider equations that can be written as
x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1 x+ β2x2)y′+ ( γ0 + γ1x+ γ2 x2)y= 0 (7.6.1)
where α0 ̸= 0 . W e assume that the indicial equation p0(r) = 0 has a repeated real root r1. In this case",Elementary Differential Equations with Boundary Value Problems.pdf
"where α0 ̸= 0 . W e assume that the indicial equation p0(r) = 0 has a repeated real root r1. In this case
Theorem 7.5.3 implies that ( 7.6.1) has one solution of the form
y1 = xr1
∞∑
n=0
an xn,
but does not provide a second solution y2 such that {y1,y 2}is a fundamental set of solutions. The
following extension of Theorem 7.5.2 provides a way to find a second solution.
Theorem 7.6.1 Let
Ly= x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1 x+ β2x2)y′+ ( γ0 + γ1x+ γ2 x2)y, (7.6.2)
where α0 ̸= 0 and define
p0(r) = α0r(r− 1) + β0r+ γ0,
p1(r) = α1r(r− 1) + β1r+ γ1,
p2(r) = α2r(r− 1) + β2r+ γ2.
Suppose ris a real number such that p0(n+ r) is nonzero for all positive integers n, and define
a0(r) = 1 ,
a1(r) = − p1(r)
p0(r+ 1) ,
an(r) = −p1(n+ r− 1)an−1(r) + p2(n+ r− 2)an−2(r)
p0(n+ r) , n ≥ 2.
Then the Frobenius series
y(x,r ) = xr
∞∑
n=0
an(r)xn (7.6.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"366 Chapter 7 Series Solutions of Linear Second Order Equations
satisfies
Ly(x,r ) = p0(r)xr . (7.6.4)
Moreover,
∂y
∂r (x,r ) = y(x,r ) ln x+ xr
∞∑
n=1
a′
n(r)xn , (7.6.5)
and
L
(∂y
∂r (x,r )
)
= p′
0(r)xr + xr p0(r) ln x. (7.6.6)
Proof Theorem 7.5.2 implies ( 7.6.4). Differentiating formally with respect to rin ( 7.6.3) yields
∂y
∂r (x,r ) = ∂
∂r (xr )
∞∑
n=0
an (r)xn + xr
∞∑
n=1
a′
n (r)xn
= xr ln x
∞∑
n=0
an(r)xn + xr
∞∑
n=1
a′
n (r)xn
= y(x,r ) ln x+ xr
∞∑
n=1
a′
n(r)xn ,
which proves (
7.6.5).
T o prove that ∂y(x,r )/∂r satisfies ( 7.6.6), we view y in ( 7.6.2) as a function y = y(x,r ) of two
variables, where the prime indicates partial differentiat ion with respect to x; thus,
y′= y′(x,r ) = ∂y
∂x (x,r ) and y′′= y′′(x,r ) = ∂2y
∂x2 (x,r ).
With this notation we can use ( 7.6.2) to rewrite ( 7.6.4) as
x2q0(x) ∂2y
∂x2 (x,r ) + xq1(x) ∂y
∂x (x,r ) + q2(x)y(x,r ) = p0(r)xr , (7.6.7)
where
q0(x) = α0 + α1x+ α2x2,
q1(x) = β0 + β1x+ β2x2,
q2(x) = γ0 + γ1x+ γ2x2.",Elementary Differential Equations with Boundary Value Problems.pdf
"∂x2 (x,r ) + xq1(x) ∂y
∂x (x,r ) + q2(x)y(x,r ) = p0(r)xr , (7.6.7)
where
q0(x) = α0 + α1x+ α2x2,
q1(x) = β0 + β1x+ β2x2,
q2(x) = γ0 + γ1x+ γ2x2.
Differentiating both sides of ( 7.6.7) with respect to ryields
x2q0(x) ∂3y
∂r∂x 2 (x,r ) + xq1(x) ∂2y
∂r∂x (x,r ) + q2(x) ∂y
∂r (x,r ) = p′
0(r)xr + p0(r)xr ln x.
By changing the order of differentiation in the first two term s on the left we can rewrite this as
x2q0(x) ∂3y
∂x2∂r (x,r ) + xq1(x) ∂2y
∂x∂r (x,r ) + q2(x) ∂y
∂r (x,r ) = p′
0(r)xr + p0(r)xr ln x,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 367
or
x2q0(x) ∂2
∂x2
(∂y
∂r (x,r )
)
+ xq1(x) ∂
∂r
(∂y
∂x (x,r )
)
+ q2(x) ∂y
∂r (x,r ) = p′
0(r)xr + p0(r)xr ln x,
which is equivalent to (
7.6.6).
Theorem 7.6.2 Let Lbe as in Theorem 7.6.1 and suppose the indicial equation p0(r) = 0 has a repeated
real root r1. Then
y1(x) = y(x,r 1) = xr1
∞∑
n=0
an(r1)xn
and
y2(x) = ∂y
∂r (x,r 1) = y1(x) ln x+ xr1
∞∑
n=1
a′
n(r1)xn (7.6.8)
form a fundamental set of solutions of Ly= 0 .
Proof Since r1 is a repeated root of p0(r) = 0 , the indicial polynomial can be factored as
p0(r) = α0(r− r1)2,
so
p0(n+ r1) = α0n2,
which is nonzero if n >0. Therefore the assumptions of Theorem 7.6.1 hold with r = r1, and ( 7.6.4)
implies that Ly1 = p0(r1)xr1 = 0 . Since
p′
0(r) = 2 α(r− r1)
it follows that p′
0(r1) = 0 , so (
7.6.6) implies that
Ly2 = p′
0(r1)xr1 + xr1 p0(r1) ln x= 0 .
This proves that y1 and y2 are both solutions of Ly= 0 . W e leave the proof that {y1,y 2}is a fundamental
set as an exercise (Exercise
53).",Elementary Differential Equations with Boundary Value Problems.pdf
"This proves that y1 and y2 are both solutions of Ly= 0 . W e leave the proof that {y1,y 2}is a fundamental
set as an exercise (Exercise
53).
Example 7.6.1 Find a fundamental set of solutions of
x2(1 − 2x+ x2)y′′− x(3 + x)y′+ (4 + x)y= 0 . (7.6.9)
Compute just the terms involving xn+r1 , where 0 ≤ n≤ 4 and r1 is the root of the indicial equation.
Solution For the given equation, the polynomials defined in Theorem 7.6.1 are
p0(r) = r(r− 1) − 3r+ 4 = ( r− 2)2,
p1(r) = −2r(r− 1) − r+ 1 = −(r− 1)(2r+ 1) ,
p2(r) = r(r− 1).
Since r1 = 2 is a repeated root of the indicial polynomial p0, Theorem 7.6.2 implies that
y1 = x2
∞∑
n=0
an(2)xn and y2 = y1 ln x+ x2
∞∑
n=1
a′
n (2)xn (7.6.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"368 Chapter 7 Series Solutions of Linear Second Order Equations
form a fundamental set of Frobenius solutions of ( 7.6.9). T o find the coefficients in these series, we use
the recurrence formulas from Theorem 7.6.1:
a0(r) = 1 ,
a1(r) = − p1(r)
p0(r+ 1) = −(r− 1)(2r+ 1)
(r− 1)2 = 2r+ 1
r− 1 ,
an (r) = −p1(n+ r− 1)an−1(r) + p2(n+ r− 2)an−2(r)
p0(n+ r)
= (n+ r− 2) [(2n+ 2 r− 1)an−1(r) − (n+ r− 3)an−2(r)]
(n+ r− 2)2
= (2n+ 2 r− 1)
(n+ r− 2) an−1(r) − (n+ r− 3)
(n+ r− 2) an−2(r),n ≥ 2.
(7.6.11)
Differentiating yields
a′
1(r) = − 3
(r− 1)2 ,
a′
n(r) = 2n+ 2 r− 1
n+ r− 2 a′
n−1(r) − n+ r− 3
n+ r− 2 a′
n−2(r)
− 3
(n+ r− 2)2 an−1(r) − 1
(n+ r− 2)2 an−2(r), n ≥ 2.
(7.6.12)
Setting r= 2 in ( 7.6.11) and ( 7.6.12) yields
a0(2) = 1 ,
a1(2) = 5 ,
an (2) = (2n+ 3)
n an−1(2) − (n− 1)
n an−2(2), n ≥ 2
(7.6.13)
and
a′
1(2) = −3,
a′
n(2) = 2n+ 3
n a′
n−1(2) − n− 1
n a′
n−2(2) − 3
n2 an−1(2) − 1
n2 an−2(2), n ≥ 2.
(7.6.14)
Computing recursively with ( 7.6.13) and ( 7.6.14) yields",Elementary Differential Equations with Boundary Value Problems.pdf
"n(2) = 2n+ 3
n a′
n−1(2) − n− 1
n a′
n−2(2) − 3
n2 an−1(2) − 1
n2 an−2(2), n ≥ 2.
(7.6.14)
Computing recursively with ( 7.6.13) and ( 7.6.14) yields
a0(2) = 1 , a1(2) = 5 , a2(2) = 17 , a3(2) = 143
3 , a4(2) = 355
3 ,
and
a′
1(2) = −3, a′
2(2) = −29
2 , a′
3(2) = −859
18 , a′
4(2) = −4693
36 .
Substituting these coefficients into ( 7.6.10) yields
y1 = x2
(
1 + 5 x+ 17 x2 + 143
3 x3 + 355
3 x4 + · · ·
)
and
y2 = y1 ln x− x3
(
3 + 29
2 x+ 859
18 x2 + 4693
36 x3 + · · ·
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 369
Since the recurrence formula ( 7.6.11) involves three terms, it’s not possible to obtain a simple e xplicit
formula for the coefficients in the Frobenius solutions of ( 7.6.9). However, as we saw in the preceding
sections, the recurrrence formula for {an(r)}involves only two terms if either α1 = β1 = γ1 = 0 or
α2 = β2 = γ2 = 0 in ( 7.6.1). In this case, it’s often possible to find explicit formulas for the coefficients.
The next two examples illustrate this.
Example 7.6.2 Find a fundamental set of Frobenius solutions of
2x2(2 + x)y′′+ 5 x2y′+ (1 + x)y = 0 . (7.6.15)
Give explicit formulas for the coefficients in the solutions .
Solution For the given equation, the polynomials defined in Theorem
7.6.1 are
p0(r) = 4 r(r− 1) + 1 = (2 r− 1)2,
p1(r) = 2 r(r− 1) + 5 r+ 1 = ( r+ 1)(2 r+ 1) ,
p2(r) = 0 .
Since r1 = 1 / 2 is a repeated zero of the indicial polynomial p0, Theorem 7.6.2 implies that
y1 = x1/ 2
∞∑
n=0
an (1/ 2)xn (7.6.16)
and",Elementary Differential Equations with Boundary Value Problems.pdf
"p2(r) = 0 .
Since r1 = 1 / 2 is a repeated zero of the indicial polynomial p0, Theorem 7.6.2 implies that
y1 = x1/ 2
∞∑
n=0
an (1/ 2)xn (7.6.16)
and
y2 = y1 ln x+ x1/ 2
∞∑
n=1
a′
n (1/ 2)xn (7.6.17)
form a fundamental set of Frobenius solutions of (
7.6.15). Since p2 ≡ 0, the recurrence formulas in
Theorem 7.6.1 reduce to
a0(r) = 1 ,
an (r) = −p1(n+ r− 1)
p0(n+ r) an−1(r),
= −(n+ r)(2n+ 2 r− 1)
(2n+ 2 r− 1)2 an−1(r),
= − n+ r
2n+ 2 r− 1 an−1(r), n ≥ 0.
W e leave it to you to show that
an (r) = ( −1)n
n∏
j=1
j+ r
2j+ 2 r− 1 , n ≥ 0. (7.6.18)
Setting r= 1 / 2 yields
an(1/ 2) = ( −1)n
n∏
j=1
j+ 1 / 2
2j = ( −1)n
n∏
j=1
2j+ 1
4j ,
=
(−1)n ∏ n
j=1(2j+ 1)
4nn! , n ≥ 0.
(7.6.19)",Elementary Differential Equations with Boundary Value Problems.pdf
"370 Chapter 7 Series Solutions of Linear Second Order Equations
Substituting this into ( 7.6.16) yields
y1 = x1/ 2
∞∑
n=0
(−1)n ∏ n
j=1(2j+ 1)
4nn! xn.
T o obtain y2 in ( 7.6.17), we must compute a′
n (1/ 2) for n = 1 , 2,. . . . W e’ll do this by logarithmic
differentiation. From (
7.6.18),
|an(r)|=
n∏
j=1
|j+ r|
|2j+ 2 r− 1|, n ≥ 1.
Therefore
ln |an(r)|=
n∑
j=1
(ln |j+ r| −ln |2j+ 2 r− 1|) .
Differentiating with respect to ryields
a′
n (r)
an (r) =
n∑
j=1
( 1
j+ r − 2
2j+ 2 r− 1
)
.
Therefore
a′
n (r) = an(r)
n∑
j=1
( 1
j+ r − 2
2j+ 2 r− 1
)
.
Setting r= 1 / 2 here and recalling ( 7.6.19) yields
a′
n(1/ 2) =
(−1)n ∏ n
j=1(2j+ 1)
4nn!


n∑
j=1
1
j+ 1 / 2 −
n∑
j=1
1
j

 . (7.6.20)
Since
1
j+ 1 / 2 − 1
j = j− j− 1/ 2
j(j+ 1 / 2) = − 1
j(2j+ 1) ,
(7.6.20) can be rewritten as
a′
n(1/ 2) = −
(−1)n ∏ n
j=1(2j+ 1)
4nn!
n∑
j=1
1
j(2j+ 1) .
Therefore, from ( 7.6.17),
y2 = y1 ln x− x1/ 2
∞∑
n=1
(−1)n ∏ n
j=1(2j+ 1)
4n n!


n∑
j=1
1
j(2j+ 1)

 xn.",Elementary Differential Equations with Boundary Value Problems.pdf
"j=1(2j+ 1)
4nn!
n∑
j=1
1
j(2j+ 1) .
Therefore, from ( 7.6.17),
y2 = y1 ln x− x1/ 2
∞∑
n=1
(−1)n ∏ n
j=1(2j+ 1)
4n n!


n∑
j=1
1
j(2j+ 1)

 xn.
Example 7.6.3 Find a fundamental set of Frobenius solutions of
x2(2 − x2)y′′− 2x(1 + 2 x2)y′+ (2 − 2x2)y= 0 . (7.6.21)
Give explicit formulas for the coefficients in the solutions .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 371
Solution For ( 7.6.21), the polynomials defined in Theorem 7.6.1 are
p0(r) = 2 r(r− 1) − 2r+ 2 = 2( r− 1)2,
p1(r) = 0 ,
p2(r) = −r(r− 1) − 4r− 2 = −(r+ 1)( r+ 2) .
As in Section 7.5, since p1 ≡ 0, the recurrence formulas of Theorem 7.6.1 imply that an(r) = 0 if nis
odd, and
a0(r) = 1 ,
a2m(r) = −p2(2m+ r− 2)
p0(2m+ r) a2m−2(r)
= (2m+ r− 1)(2m+ r)
2(2m+ r− 1)2 a2m−2(r)
= 2m+ r
2(2m+ r− 1) a2m−2(r), m ≥ 1.
Since r1 = 1 is a repeated root of the indicial polynomial p0, Theorem 7.6.2 implies that
y1 = x
∞∑
m=0
a2m(1)x2m (7.6.22)
and
y2 = y1 ln x+ x
∞∑
m=1
a′
2m(1)x2m (7.6.23)
form a fundamental set of Frobenius solutions of (
7.6.21). W e leave it to you to show that
a2m(r) = 1
2m
m∏
j=1
2j+ r
2j+ r− 1 . (7.6.24)
Setting r= 1 yields
a2m(1) = 1
2m
m∏
j=1
2j+ 1
2j =
∏ m
j=1(2j+ 1)
4mm! , (7.6.25)
and substituting this into ( 7.6.22) yields
y1 = x
∞∑
m=0
∏ m
j=1(2j+ 1)
4m m! x2m.
T o obtain y2 in ( 7.6.23), we must compute a′",Elementary Differential Equations with Boundary Value Problems.pdf
"4mm! , (7.6.25)
and substituting this into ( 7.6.22) yields
y1 = x
∞∑
m=0
∏ m
j=1(2j+ 1)
4m m! x2m.
T o obtain y2 in ( 7.6.23), we must compute a′
2m(1) for m = 1 , 2, . . . . Again we use logarithmic
differentiation. From (
7.6.24),
|a2m(r)|= 1
2m
m∏
j=1
|2j+ r|
|2j+ r− 1|.
T aking logarithms yields
ln |a2m(r)|= −mln 2 +
m∑
j=1
(ln |2j+ r| −ln |2j+ r− 1|) .",Elementary Differential Equations with Boundary Value Problems.pdf
"372 Chapter 7 Series Solutions of Linear Second Order Equations
Differentiating with respect to ryields
a′
2m(r)
a2m(r) =
m∑
j=1
( 1
2j+ r − 1
2j+ r− 1
)
.
Therefore
a′
2m(r) = a2m(r)
m∑
j=1
( 1
2j+ r − 1
2j+ r− 1
)
.
Setting r= 1 and recalling ( 7.6.25) yields
a′
2m(1) =
∏ m
j=1(2j+ 1)
4mm!
m∑
j=1
( 1
2j+ 1 − 1
2j
)
. (7.6.26)
Since 1
2j+ 1 − 1
2j = − 1
2j(2j+ 1) ,
(7.6.26) can be rewritten as
a′
2m(1) = −
∏ m
j=1(2j+ 1)
2 ·4mm!
m∑
j=1
1
j(2j+ 1) .
Substituting this into ( 7.6.23) yields
y2 = y1 ln x− x
2
∞∑
m=1
∏ m
j=1(2j+ 1)
4mm!


m∑
j=1
1
j(2j+ 1)

 x2m.
If the solution y1 = y(x,r 1) of Ly = 0 reduces to a finite sum, then there’s a difficulty in using
logarithmic differentiation to obtain the coefficients {a′
n(r1)}in the second solution. The next example
illustrates this difficulty and shows how to overcome it.
Example 7.6.4 Find a fundamental set of Frobenius solutions of
x2y′′− x(5 − x)y′+ (9 − 4x)y= 0 . (7.6.27)
Give explicit formulas for the coefficients in the solutions .",Elementary Differential Equations with Boundary Value Problems.pdf
"x2y′′− x(5 − x)y′+ (9 − 4x)y= 0 . (7.6.27)
Give explicit formulas for the coefficients in the solutions .
Solution For (
7.6.27) the polynomials defined in Theorem 7.6.1 are
p0(r) = r(r− 1) − 5r+ 9 = ( r− 3)2,
p1(r) = r− 4,
p2(r) = 0 .
Since r1 = 3 is a repeated zero of the indicial polynomial p0, Theorem 7.6.2 implies that
y1 = x3
∞∑
n=0
an (3)xn (7.6.28)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 373
and
y2 = y1 ln x+ x3
∞∑
n=1
a′
n(3)xn (7.6.29)
are linearly independent Frobenius solutions of (
7.6.27). T o find the coefficients in ( 7.6.28) we use the
recurrence formulas
a0(r) = 1 ,
an (r) = −p1(n+ r− 1)
p0(n+ r) an−1(r)
= − n+ r− 5
(n+ r− 3)2 an−1(r), n ≥ 1.
W e leave it to you to show that
an(r) = ( −1)n
n∏
j=1
j+ r− 5
(j+ r− 3)2 . (7.6.30)
Setting r= 3 here yields
an (3) = ( −1)n
n∏
j=1
j− 2
j2 ,
so a1(3) = 1 and an (3) = 0 if n≥ 2. Substituting these coefficients into ( 7.6.28) yields
y1 = x3(1 + x).
T o obtain y2 in ( 7.6.29) we must compute a′
n (3) for n= 1 , 2, . . . . Let’s first try logarithmic differenti-
ation. From (
7.6.30),
|an(r)|=
n∏
j=1
|j+ r− 5|
|j+ r− 3|2 , n ≥ 1,
so
ln |an (r)|=
n∑
j=1
(ln |j+ r− 5| −2 ln |j+ r− 3|) .
Differentiating with respect to ryields
a′
n (r)
an (r) =
n∑
j=1
( 1
j+ r− 5 − 2
j+ r− 3
)
.
Therefore
a′
n (r) = an(r)
n∑
j=1
( 1
j+ r− 5 − 2
j+ r− 3
)
. (7.6.31)",Elementary Differential Equations with Boundary Value Problems.pdf
"a′
n (r)
an (r) =
n∑
j=1
( 1
j+ r− 5 − 2
j+ r− 3
)
.
Therefore
a′
n (r) = an(r)
n∑
j=1
( 1
j+ r− 5 − 2
j+ r− 3
)
. (7.6.31)
However, we can’t simply set r= 3 here if n≥ 2, since the bracketed expression in the sum correspond-
ing to j = 2 contains the term 1/ (r− 3). In fact, since an (3) = 0 for n ≥ 2, the formula ( 7.6.31) for
a′
n (r) is actually an indeterminate form at r= 3 .
W e overcome this difficulty as follows. From (
7.6.30) with n= 1 ,
a1(r) = − r− 4
(r− 2)2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"374 Chapter 7 Series Solutions of Linear Second Order Equations
Therefore
a′
1(r) = r− 6
(r− 2)3 ,
so
a′
1(3) = −3. (7.6.32)
From (
7.6.30) with n≥ 2,
an (r) = ( −1)n (r− 4)(r− 3)
∏ n
j=3(j+ r− 5)
∏ n
j=1(j+ r− 3)2 = ( r− 3)cn(r),
where
cn (r) = ( −1)n (r− 4)
∏ n
j=3(j+ r− 5)
∏ n
j=1(j+ r− 3)2 , n ≥ 2.
Therefore
a′
n (r) = cn (r) + ( r− 3)c′
n(r), n ≥ 2,
which implies that a′
n (3) = cn (3) if n≥ 3. W e leave it to you to verify that
a′
n (3) = cn (3) = (−1)n+1
n(n− 1)n! , n ≥ 2.
Substituting this and ( 7.6.32) into ( 7.6.29) yields
y2 = x3(1 + x) ln x− 3x4 − x3
∞∑
n=2
(−1)n
n(n− 1)n! xn.
7.6 Exercises
In Exercises 1– 11 find a fundamental set of Frobenius solutions. Compute the te rms involving xn+r1 ,
where 0 ≤ n≤ N (N at least 7) and r1 is the root of the indicial equation. Optionally , write a com puter
program to implement the applicable recurrence formulas an d take N >7.
1. C x2y′′− x(1 − x)y′+ (1 − x2)y = 0
2. C x2(1 + x+ 2 x2)y′+ x(3 + 6 x+ 7 x2)y′+ (1 + 6 x− 3x2)y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"1. C x2y′′− x(1 − x)y′+ (1 − x2)y = 0
2. C x2(1 + x+ 2 x2)y′+ x(3 + 6 x+ 7 x2)y′+ (1 + 6 x− 3x2)y= 0
3. C x2(1 + 2 x+ x2)y′′+ x(1 + 3 x+ 4 x2)y′− x(1 − 2x)y= 0
4. C 4x2(1 + x+ x2)y′′+ 12 x2(1 + x)y′+ (1 + 3 x+ 3 x2)y= 0
5. C x2(1 + x+ x2)y′′− x(1 − 4x− 2x2)y′+ y= 0
6. C 9x2y′′+ 3 x(5 + 3 x− 2x2)y′+ (1 + 12 x− 14x2)y= 0
7. C x2y′′+ x(1 + x+ x2)y′+ x(2 − x)y= 0
8. C x2(1 + 2 x)y′′+ x(5 + 14 x+ 3 x2)y′+ (4 + 18 x+ 12 x2)y = 0
9. C 4x2y′′+ 2 x(4 + x+ x2)y′+ (1 + 5 x+ 3 x2)y= 0
10. C 16x2y′′+ 4 x(6 + x+ 2 x2)y′+ (1 + 5 x+ 18 x2)y = 0
11. C 9x2(1 + x)y′′+ 3 x(5 + 11 x− x2)y′+ (1 + 16 x− 7x2)y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 375
In Exercises 12– 22 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi-
cients.
12. 4x2y′′+ (1 + 4 x)y= 0
13. 36x2(1 − 2x)y′′+ 24 x(1 − 9x)y′+ (1 − 70x)y= 0
14. x2(1 + x)y′′− x(3 − x)y′+ 4 y= 0
15. x2(1 − 2x)y′′− x(5 − 4x)y′+ (9 − 4x)y= 0
16. 25x2y′′+ x(15 + x)y′+ (1 + x)y= 0
17. 2x2(2 + x)y′′+ x2y′+ (1 − x)y= 0
18. x2(9 + 4 x)y′′+ 3 xy′+ (1 + x)y = 0
19. x2y′′− x(3 − 2x)y′+ (4 + 3 x)y= 0
20. x2(1 − 4x)y′′+ 3 x(1 − 6x)y′+ (1 − 12x)y= 0
21. x2(1 + 2 x)y′′+ x(3 + 5 x)y′+ (1 − 2x)y= 0
22. 2x2(1 + x)y′′− x(6 − x)y′+ (8 − x)y= 0
In Exercises 23– 27 find a fundamental set of Frobenius solutions. Compute the te rms involving xn+r1 ,
where 0 ≤ n≤ N (N at least 7) and r1 is the root of the indicial equation. Optionally, write a com puter
program to implement the applicable recurrence formulas an d take N >7.
23. C x2(1 + 2 x)y′′+ x(5 + 9 x)y′+ (4 + 3 x)y= 0
24. C x2(1 − 2x)y′′− x(5 + 4 x)y′+ (9 + 4 x)y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"23. C x2(1 + 2 x)y′′+ x(5 + 9 x)y′+ (4 + 3 x)y= 0
24. C x2(1 − 2x)y′′− x(5 + 4 x)y′+ (9 + 4 x)y= 0
25. C x2(1 + 4 x)y′′− x(1 − 4x)y′+ (1 + x)y= 0
26. C x2(1 + x)y′′+ x(1 + 2 x)y′+ xy= 0
27. C x2(1 − x)y′′+ x(7 + x)y′+ (9 − x)y= 0
In Exercises 28– 38 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi-
cients.
28. x2y′′− x(1 − x2)y′+ (1 + x2)y= 0
29. x2(1 + x2)y′′− 3x(1 − x2)y′+ 4 y= 0
30. 4x2y′′+ 2 x3y′+ (1 + 3 x2)y= 0
31. x2(1 + x2)y′′− x(1 − 2x2)y′+ y= 0
32. 2x2(2 + x2)y′′+ 7 x3y′+ (1 + 3 x2)y= 0
33. x2(1 + x2)y′′− x(1 − 4x2)y′+ (1 + 2 x2)y = 0
34. 4x2(4 + x2)y′′+ 3 x(8 + 3 x2)y′+ (1 − 9x2)y= 0
35. 3x2(3 + x2)y′′+ x(3 + 11 x2)y′+ (1 + 5 x2)y= 0
36. 4x2(1 + 4 x2)y′′+ 32 x3y′+ y = 0
37. 9x2y′′− 3x(7 − 2x2)y′+ (25 + 2 x2)y= 0
38. x2(1 + 2 x2)y′′+ x(3 + 7 x2)y′+ (1 − 3x2)y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"376 Chapter 7 Series Solutions of Linear Second Order Equations
In Exercises 39– 43 find a fundamental set of Frobenius solutions. Compute the te rms involving x2m+r1 ,
where 0 ≤ m≤ M(M at least 3) and r1 is the root of the indicial equation. Optionally, write a com puter
program to implement the applicable recurrence formulas an d take M >3.
39. C x2(1 + x2)y′′+ x(3 + 8 x2)y′+ (1 + 12 x2)y
40. C x2y′′− x(1 − x2)y′+ (1 + x2)y= 0
41. C x2(1 − 2x2)y′′+ x(5 − 9x2)y′+ (4 − 3x2)y = 0
42. C x2(2 + x2)y′′+ x(14 − x2)y′+ 2(9 + x2)y= 0
43. C x2(1 + x2)y′′+ x(3 + 7 x2)y′+ (1 + 8 x2)y= 0
In Exercises 44– 52 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi-
cients.
44. x2(1 − 2x)y′′+ 3 xy′+ (1 + 4 x)y= 0
45. x(1 + x)y′′+ (1 − x)y′+ y = 0
46. x2(1 − x)y′′+ x(3 − 2x)y′+ (1 + 2 x)y= 0
47. 4x2(1 + x)y′′− 4x2y′+ (1 − 5x)y= 0
48. x2(1 − x)y′′− x(3 − 5x)y′+ (4 − 5x)y= 0
49. x2(1 + x2)y′′− x(1 + 9 x2)y′+ (1 + 25 x2)y = 0
50. 9x2y′′+ 3 x(1 − x2)y′+ (1 + 7 x2)y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"48. x2(1 − x)y′′− x(3 − 5x)y′+ (4 − 5x)y= 0
49. x2(1 + x2)y′′− x(1 + 9 x2)y′+ (1 + 25 x2)y = 0
50. 9x2y′′+ 3 x(1 − x2)y′+ (1 + 7 x2)y = 0
51. x(1 + x2)y′′+ (1 − x2)y′− 8xy= 0
52. 4x2y′′+ 2 x(4 − x2)y′+ (1 + 7 x2)y = 0
53. Under the assumptions of Theorem 7.6.2, suppose the power series
∞∑
n=0
an(r1)xn and
∞∑
n=1
a′
n (r1)xn
converge on (−ρ,ρ ).
(a) Show that
y1 = xr1
∞∑
n=0
an(r1)xn and y2 = y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
are linearly independent on (0,ρ ). H IN T : Show that if c1 and c2 are constants such that
c1y1 + c2y2 ≡ 0 on (0,ρ ), then
(c1 + c2 ln x)
∞∑
n=0
an (r1)xn + c2
∞∑
n=1
a′
n (r1)xn = 0 , 0 <x<ρ.
Then let x→ 0+ to conclude that c2 = 0 .
(b) Use the result of (a) to complete the proof of Theorem
7.6.2.
54. Let
Ly= x2(α0 + α1x)y′′+ x(β0 + β1x)y′+ ( γ0 + γ1x)y",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.6 The Method of Frobenius II 377
and define
p0(r) = α0r(r− 1) + β0r+ γ0 and p1(r) = α1r(r− 1) + β1r+ γ1.
Theorem 7.6.1 and Exercise 7.5. 55(a) imply that if
y(x,r ) = xr
∞∑
n=0
an (r)xn
where
an(r) = ( −1)n
n∏
j=1
p1(j+ r− 1)
p0(j+ r) ,
then
Ly(x,r ) = p0(r)xr .
Now suppose p0(r) = α0(r− r1)2 and p1(k+ r1) ̸= 0 if kis a nonnegative integer.
(a) Show that Ly= 0 has the solution
y1 = xr1
∞∑
n=0
an (r1)xn ,
where
an (r1) = (−1)n
αn
0 (n!)2
n∏
j=1
p1(j+ r1 − 1).
(b) Show that Ly= 0 has the second solution
y2 = y1 ln x+ xr1
∞∑
n=1
an (r1)Jn xn ,
where
Jn =
n∑
j=1
p′
1(j+ r1 − 1)
p1(j+ r1 − 1) − 2
n∑
j=1
1
j.
(c) Conclude from (a) and (b) that if γ1 ̸= 0 then
y1 = xr1
∞∑
n=0
(−1)n
(n!)2
(γ1
α0
) n
xn
and
y2 = y1 ln x− 2xr1
∞∑
n=1
(−1)n
(n!)2
(γ1
α0
) n


n∑
j=1
1
j

 xn
are solutions of
α0x2y′′+ β0xy′+ ( γ0 + γ1x)y= 0 .
(The conclusion is also valid if γ1 = 0 . Why?)",Elementary Differential Equations with Boundary Value Problems.pdf
"378 Chapter 7 Series Solutions of Linear Second Order Equations
55. Let
Ly= x2(α0 + αq xq )y′′+ x(β0 + βq xq )y′+ ( γ0 + γq xq )y
where qis a positive integer, and define
p0(r) = α0r(r− 1) + β0r+ γ0 and pq (r) = αq r(r− 1) + βq r+ γq .
Suppose
p0(r) = α0(r− r1)2 and pq (r) ̸≡0.
(a) Recall from Exercise 7.5. 59 that Ly = 0 has the solution
y1 = xr1
∞∑
m=0
aqm(r1)xqm,
where
aqm (r1) = (−1)m
(q2α0)m(m!)2
m∏
j=1
pq (q(j− 1) + r1) .
(b) Show that Ly= 0 has the second solution
y2 = y1 ln x+ xr1
∞∑
m=1
a′
qm (r1)Jm xqm ,
where
Jm =
m∑
j=1
p′
q (q(j− 1) + r1)
pq (q(j− 1) + r1) − 2
q
m∑
j=1
1
j.
(c) Conclude from (a) and (b) that if γq ̸= 0 then
y1 = xr1
∞∑
m=0
(−1)m
(m!)2
( γq
q2α0
) m
xqm
and
y2 = y1 ln x− 2
qxr1
∞∑
m=1
(−1)m
(m!)2
( γq
q2α0
) m


m∑
j=1
1
j

 xqm
are solutions of
α0x2y′′+ β0xy′+ ( γ0 + γq xq )y= 0 .
56. The equation
xy′′+ y′+ xy= 0
is
Bessel’s equation of order zero. (See Exercise 53.) Find two linearly independent Frobenius
solutions of this equation.",Elementary Differential Equations with Boundary Value Problems.pdf
"xy′′+ y′+ xy= 0
is
Bessel’s equation of order zero. (See Exercise 53.) Find two linearly independent Frobenius
solutions of this equation.
57. Suppose the assumptions of Exercise 7.5. 53 hold, except that
p0(r) = α0(r− r1)2.
Show that
y1 = xr1
α0 + α1x+ α2x2 and y2 = xr1 ln x
α0 + α1x+ α2x2
are linearly independent Frobenius solutions of
x2(α0 + α1x+ α2x2)y′′+ x(β0 + β1x+ β2 x2)y′+ ( γ0 + γ1x+ γ2x2)y = 0
on any interval (0,ρ ) on which α0 + α1x+ α2x2 has no zeros.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 379
In Exercises 58– 65 use the method suggested by Exercise 57 to find the general solution on some interval
(0,ρ ).
58. 4x2(1 + x)y′′+ 8 x2y′+ (1 + x)y= 0
59. 9x2(3 + x)y′′+ 3 x(3 + 7 x)y′+ (3 + 4 x)y= 0
60. x2(2 − x2)y′′− x(2 + 3 x2)y′+ (2 − x2)y= 0
61. 16x2(1 + x2)y′′+ 8 x(1 + 9 x2)y′+ (1 + 49 x2)y = 0
62. x2(4 + 3 x)y′′− x(4 − 3x)y′+ 4 y= 0
63. 4x2(1 + 3 x+ x2)y′′+ 8 x2(3 + 2 x)y′+ (1 + 3 x+ 9 x2)y = 0
64. x2(1 − x)2y′′− x(1 + 2 x− 3x2)y′+ (1 + x2)y= 0
65. 9x2(1 + x+ x2)y′′+ 3 x(1 + 7 x+ 13 x2)y′+ (1 + 4 x+ 25 x2)y = 0
66. (a) Let Land y(x,r ) be as in Exercises 57 and 58. Extend Theorem 7.6.1 by showing that
L
(∂y
∂r (x,r )
)
= p′
0(r)xr + xr p0(r) ln x.
(b) Show that if
p0(r) = α0(r− r1)2
then
y1 = y(x,r 1) and y2 = ∂y
∂r (x,r 1)
are solutions of Ly= 0 .
7.7 THE METHOD OF FROBENIUS III
In Sections 7.5 and 7.6 we discussed methods for finding Frobe nius solutions of a homogeneous linear",Elementary Differential Equations with Boundary Value Problems.pdf
"7.7 THE METHOD OF FROBENIUS III
In Sections 7.5 and 7.6 we discussed methods for finding Frobe nius solutions of a homogeneous linear
second order equation near a regular singular point in the ca se where the indicial equation has a repeated
root or distinct real roots that don’t differ by an integer. I n this section we consider the case where the
indicial equation has distinct real roots that differ by an i nteger. W e’ll limit our discussion to equations
that can be written as
x2(α0 + α1x)y′′+ x(β0 + β1x)y′+ ( γ0 + γ1x)y= 0 (7.7.1)
or
x2(α0 + α2x2)y′′+ x(β0 + β2x2)y′+ ( γ0 + γ2x2)y = 0 ,
where the roots of the indicial equation differ by a positive integer.
W e begin with a theorem that provides a fundamental set of sol utions of equations of the form ( 7.7.1).
Theorem 7.7.1 Let
Ly= x2(α0 + α1x)y′′+ x(β0 + β1x)y′+ ( γ0 + γ1x)y,
where α0 ̸= 0 , and define
p0(r) = α0r(r− 1) + β0r+ γ0,
p1(r) = α1r(r− 1) + β1r+ γ1.",Elementary Differential Equations with Boundary Value Problems.pdf
"380 Chapter 7 Series Solutions of Linear Second Order Equations
Suppose ris a real number such that p0(n+ r) is nonzero for all positive integers n, and define
a0(r) = 1 ,
an (r) = −p1(n+ r− 1)
p0(n+ r) an−1(r), n ≥ 1. (7.7.2)
Let r1 and r2 be the roots of the indicial equation p0(r) = 0 , and suppose r1 = r2 + k, where k is a
positive integer . Then
y1 = xr1
∞∑
n=0
an (r1)xn
is a Frobenius solution of Ly= 0 . Moreover , if we define
a0(r2) = 1 ,
an (r2) = −p1(n+ r2 − 1)
p0(n+ r2) an−1(r2), 1 ≤ n≤ k− 1, (7.7.3)
and
C = −p1(r1 − 1)
kα 0
ak−1(r2), (7.7.4)
then
y2 = xr2
k−1∑
n=0
an (r2)xn + C
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
(7.7.5)
is also a solution of Ly= 0 , and {y1,y 2}is a fundamental set of solutions.
Proof Theorem 7.5.3 implies that Ly1 = 0 . W e’ll now show that Ly2 = 0 . Since Lis a linear operator,
this is equivalent to showing that
L
(
xr2
k−1∑
n=0
an(r2)xn
)
+ CL
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
= 0 . (7.7.6)
T o verify this, we’ll show that
L
(
xr2
k−1∑
n=0
an(r2)xn
)",Elementary Differential Equations with Boundary Value Problems.pdf
"L
(
xr2
k−1∑
n=0
an(r2)xn
)
+ CL
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
= 0 . (7.7.6)
T o verify this, we’ll show that
L
(
xr2
k−1∑
n=0
an(r2)xn
)
= p1(r1 − 1)ak−1(r2)xr1 (7.7.7)
and
L
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
= kα 0xr1 . (7.7.8)
This will imply that Ly2 = 0 , since substituting (
7.7.7) and ( 7.7.8) into ( 7.7.6) and using ( 7.7.4) yields
Ly2 = [ p1(r1 − 1)ak−1(r2) + Ckα 0] xr1
= [ p1(r1 − 1)ak−1(r2) − p1(r1 − 1)ak−1(r2)] xr1 = 0 .
W e’ll prove ( 7.7.8) first. From Theorem 7.6.1,
L
(
y(x,r ) ln x+ xr
∞∑
n=1
a′
n(r)xn
)
= p′
0(r)xr + xr p0(r) ln x.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 381
Setting r= r1 and recalling that p0(r1) = 0 and y1 = y(x,r 1) yields
L
(
y1 ln x+ xr1
∞∑
n=1
a′
n(r1)xn
)
= p′
0(r1)xr1 . (7.7.9)
Since r1 and r2 are the roots of the indicial equation, the indicial polynom ial can be written as
p0(r) = α0(r− r1)(r− r2) = α0
[
r2 − (r1 + r2)r+ r1r2
]
.
Differentiating this yields
p′
0(r) = α0(2r− r1 − r2).
Therefore p′
0(r1) = α0(r1 − r2) = kα 0, so (
7.7.9) implies ( 7.7.8).
Before proving ( 7.7.7), we first note an (r2) is well defined by ( 7.7.3) for 1 ≤ n ≤ k − 1, since
p0(n+ r2) ̸= 0 for these values of n. However, we can’t define an (r2) for n ≥ k with ( 7.7.3), since
p0(k+ r2) = p0(r1) = 0 . For convenience, we define an (r2) = 0 for n≥ k. Then, from Theorem 7.5.1,
L
(
xr2
k−1∑
n=0
an(r2)xn
)
= L
(
xr2
∞∑
n=0
an (r2)xn
)
= xr2
∞∑
n=0
bnxn, (7.7.10)
where b0 = p0(r2) = 0 and
bn = p0(n+ r2)an (r2) + p1(n+ r2 − 1)an−1(r2), n ≥ 1.",Elementary Differential Equations with Boundary Value Problems.pdf
"an(r2)xn
)
= L
(
xr2
∞∑
n=0
an (r2)xn
)
= xr2
∞∑
n=0
bnxn, (7.7.10)
where b0 = p0(r2) = 0 and
bn = p0(n+ r2)an (r2) + p1(n+ r2 − 1)an−1(r2), n ≥ 1.
If 1 ≤ n ≤ k− 1, then ( 7.7.3) implies that bn = 0 . If n ≥ k+ 1 , then bn = 0 because an−1(r2) =
an (r2) = 0 . Therefore ( 7.7.10) reduces to
L
(
xr2
k−1∑
n=0
an (r2)xn
)
= [ p0(k+ r2)ak (r2) + p1(k+ r2 − 1)ak−1(r2)] xk+r2 .
Since ak(r2) = 0 and k+ r2 = r1, this implies ( 7.7.7).
W e leave the proof that {y1,y 2}is a fundamental set as an exercise (Exercise 41).
Example 7.7.1 Find a fundamental set of Frobenius solutions of
2x2(2 + x)y′′− x(4 − 7x)y′− (5 − 3x)y= 0 .
Give explicit formulas for the coefficients in the solutions .
Solution For the given equation, the polynomials defined in Theorem 7.7.1 are
p0(r) = 4 r(r− 1) − 4r− 5 = (2 r+ 1)(2 r− 5),
p1(r) = 2 r(r− 1) + 7 r+ 3 = ( r+ 1)(2 r+ 3) .
The roots of the indicial equation are r1 = 5 / 2 and r2 = −1/ 2, so k = r1 − r2 = 3 . Therefore
Theorem 7.7.1 implies that
y1 = x5/ 2
∞∑
n=0",Elementary Differential Equations with Boundary Value Problems.pdf
"The roots of the indicial equation are r1 = 5 / 2 and r2 = −1/ 2, so k = r1 − r2 = 3 . Therefore
Theorem 7.7.1 implies that
y1 = x5/ 2
∞∑
n=0
an (5/ 2)xn (7.7.11)
and
y2 = x−1/ 2
2∑
n=0
an (−1/ 2) + C
(
y1 ln x+ x5/ 2
∞∑
n=1
a′
n(5/ 2)xn
)
(7.7.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"382 Chapter 7 Series Solutions of Linear Second Order Equations
(with Cas in ( 7.7.4)) form a fundamental set of solutions of Ly= 0 . The recurrence formula ( 7.7.2) is
a0(r) = 1 ,
an (r) = −p1(n+ r− 1)
p0(n+ r) an−1(r)
= − (n+ r)(2n+ 2 r+ 1)
(2n+ 2 r+ 1)(2 n+ 2 r− 5) an−1(r),
= − n+ r
2n+ 2 r− 5 an−1(r), n≥ 1,
(7.7.13)
which implies that
an (r) = ( −1)n
n∏
j=1
j+ r
2j+ 2 r− 5 , n≥ 0. (7.7.14)
Therefore
an(5/ 2) =
(−1)n ∏ n
j=1(2j+ 5)
4nn! . (7.7.15)
Substituting this into ( 7.7.11) yields
y1 = x5/ 2
∞∑
n=0
(−1)n ∏ n
j=1(2j+ 5)
4nn! xn.
T o compute the coefficients a0(−1/ 2),a 1(−1/ 2) and a2(−1/ 2) in y2, we set r = −1/ 2 in ( 7.7.13)
and apply the resulting recurrence formula for n= 1 , 2; thus,
a0(−1/ 2) = 1 ,
an(−1/ 2) = − 2n− 1
4(n− 3) an−1(−1/ 2), n= 1 , 2.
The last formula yields
a1(−1/ 2) = 1 / 8 and a2(−1/ 2) = 3 / 32.
Substituting r1 = 5 / 2,r 2 = −1/ 2,k = 3 , and α0 = 4 into ( 7.7.4) yields C = −15/ 128. Therefore,
from ( 7.7.12),
y2 = x−1/ 2
(
1 + 1
8 x+ 3
32 x2
)
− 15
128",Elementary Differential Equations with Boundary Value Problems.pdf
"from ( 7.7.12),
y2 = x−1/ 2
(
1 + 1
8 x+ 3
32 x2
)
− 15
128
(
y1 ln x+ x5/ 2
∞∑
n=1
a′
n (5/ 2)xn
)
. (7.7.16)
W e use logarithmic differentiation to obtain obtain a′
n(r). From (
7.7.14),
|an(r)|=
n∏
j=1
|j+ r|
|2j+ 2 r− 5|, n≥ 1.
Therefore
ln |an(r)|=
n∑
j=1
(ln |j+ r| −ln |2j+ 2 r− 5|) .
Differentiating with respect to ryields
a′
n (r)
an (r) =
n∑
j=1
( 1
j+ r − 2
2j+ 2 r− 5
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 383
Therefore
a′
n (r) = an(r)
n∑
j=1
( 1
j+ r − 2
2j+ 2 r− 5
)
.
Setting r= 5 / 2 here and recalling ( 7.7.15) yields
a′
n(5/ 2) =
(−1)n ∏ n
j=1(2j+ 5)
4nn!
n∑
j=1
( 1
j+ 5 / 2 − 1
j
)
. (7.7.17)
Since 1
j+ 5 / 2 − 1
j = − 5
j(2j+ 5) ,
we can rewrite ( 7.7.17) as
a′
n(5/ 2) = −5
(−1)n ∏ n
j=1(2j+ 5)
4nn!


n∑
j=1
1
j(2j+ 5)

 .
Substituting this into (
7.7.16) yields
y2 = x−1/ 2
(
1 + 1
8 x+ 3
32 x2
)
− 15
128 y1 ln x
+ 75
128 x5/ 2
∞∑
n=1
(−1)n ∏ n
j=1(2j+ 5)
4n n!


n∑
j=1
1
j(2j+ 5)

 xn.
If C = 0 in ( 7.7.4), there’s no need to compute
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
in the formula (
7.7.5) for y2. Therefore it’s best to compute C before computing {a′
n(r1)}∞
n=1. This is
illustrated in the next example. (See also Exercises
44 and 45.)
Example 7.7.2 Find a fundamental set of Frobenius solutions of
x2(1 − 2x)y′′+ x(8 − 9x)y′+ (6 − 3x)y= 0 .
Give explicit formulas for the coefficients in the solutions .",Elementary Differential Equations with Boundary Value Problems.pdf
"x2(1 − 2x)y′′+ x(8 − 9x)y′+ (6 − 3x)y= 0 .
Give explicit formulas for the coefficients in the solutions .
Solution For the given equation, the polynomials defined in Theorem 7.7.1 are
p0(r) = r(r− 1) + 8 r+ 6 = ( r+ 1)( r+ 6)
p1(r) = −2r(r− 1) − 9r− 3 = −(r+ 3)(2 r+ 1) .
The roots of the indicial equation are r1 = −1 and r2 = −6, so k= r1−r2 = 5 . Therefore Theorem 7.7.1
implies that
y1 = x−1
∞∑
n=0
an(−1)xn (7.7.18)",Elementary Differential Equations with Boundary Value Problems.pdf
"384 Chapter 7 Series Solutions of Linear Second Order Equations
and
y2 = x−6
4∑
n=0
an(−6) + C
(
y1 ln x+ x−1
∞∑
n=1
a′
n (−1)xn
)
(7.7.19)
(with Cas in (
7.7.4)) form a fundamental set of solutions of Ly= 0 . The recurrence formula ( 7.7.2) is
a0(r) = 1 ,
an(r) = −p1(n+ r− 1)
p0(n+ r) an−1(r)
= (n+ r+ 2)(2 n+ 2 r− 1)
(n+ r+ 1)( n+ r+ 6) an−1(r), n≥ 1,
(7.7.20)
which implies that
an(r) =
n∏
j=1
(j+ r+ 2)(2 j+ 2 r− 1)
(j+ r+ 1)( j+ r+ 6)
=


n∏
j=1
j+ r+ 2
j+ r+ 1




n∏
j=1
2j+ 2 r− 1
j+ r+ 6

 .
(7.7.21)
Since n∏
j=1
j+ r+ 2
j+ r+ 1 = (r+ 3)( r+ 4) · · ·(n+ r+ 2)
(r+ 2)( r+ 3) · · ·(n+ r+ 1) = n+ r+ 2
r+ 2
because of cancellations, ( 7.7.21) simplifies to
an (r) = n+ r+ 2
r+ 2
n∏
j=1
2j+ 2 r− 1
j+ r+ 6 .
Therefore
an (−1) = ( n+ 1)
n∏
j=1
2j− 3
j+ 5 .
Substituting this into ( 7.7.18) yields
y1 = x−1
∞∑
n=0
(n+ 1)


n∏
j=1
2j− 3
j+ 5

 xn.
T o compute the coefficients a0(−6),...,a 4(−6) in y2, we set r = −6 in (
7.7.20) and apply the",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 = x−1
∞∑
n=0
(n+ 1)


n∏
j=1
2j− 3
j+ 5

 xn.
T o compute the coefficients a0(−6),...,a 4(−6) in y2, we set r = −6 in (
7.7.20) and apply the
resulting recurrence formula for n= 1 , 2, 3, 4; thus,
a0(−6) = 1 ,
an (−6) = (n− 4)(2n− 13)
n(n− 5) an−1(−6), n= 1 , 2, 3, 4.
The last formula yields
a1(−6) = −33
4 , a2(−6) = 99
4 , a3(−6) = −231
8 , a4(−6) = 0 .
Since a4(−6) = 0 , ( 7.7.4) implies that the constant Cin ( 7.7.19) is zero. Therefore ( 7.7.19) reduces to
y2 = x−6
(
1 − 33
4 x+ 99
4 x2 − 231
8 x3
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 385
W e now consider equations of the form
x2(α0 + α2x2)y′′+ x(β0 + β2x2)y′+ ( γ0 + γ2x2)y = 0 ,
where the roots of the indicial equation are real and differ b y an even integer. The case where the roots
are real and differ by an odd integer can be handled by the meth od discussed in 56.
The proof of the next theorem is similar to the proof of Theore m 7.7.1 (Exercise 43).
Theorem 7.7.2 Let
Ly= x2(α0 + α2x2)y′′+ x(β0 + β2x2)y′+ ( γ0 + γ2x2)y,
where α0 ̸= 0 , and define
p0(r) = α0r(r− 1) + β0r+ γ0,
p2(r) = α2r(r− 1) + β2r+ γ2.
Suppose ris a real number such that p0(2m+ r) is nonzero for all positive integers m, and define
a0(r) = 1 ,
a2m(r) = −p2(2m+ r− 2)
p0(2m+ r) a2m−2(r), m ≥ 1. (7.7.22)
Let r1 and r2 be the roots of the indicial equation p0(r) = 0 , and suppose r1 = r2 + 2 k, where kis a
positive integer . Then
y1 = xr1
∞∑
m=0
a2m(r1)x2m
is a Frobenius solution of Ly= 0 . Moreover , if we define
a0(r2) = 1 ,
a2m(r2) = −p2(2m+ r2 − 2)",Elementary Differential Equations with Boundary Value Problems.pdf
"positive integer . Then
y1 = xr1
∞∑
m=0
a2m(r1)x2m
is a Frobenius solution of Ly= 0 . Moreover , if we define
a0(r2) = 1 ,
a2m(r2) = −p2(2m+ r2 − 2)
p0(2m+ r2) a2m−2(r2), 1 ≤ m≤ k− 1
and
C = −p2(r1 − 2)
2kα 0
a2k−2(r2), (7.7.23)
then
y2 = xr2
k−1∑
m=0
a2m(r2)x2m + C
(
y1 ln x+ xr1
∞∑
m=1
a′
2m(r1)x2m
)
(7.7.24)
is also a solution of Ly= 0 , and {y1,y 2}is a fundamental set of solutions.Example 7.7.3 Find a fundamental set of Frobenius solutions of
x2(1 + x2)y′′+ x(3 + 10 x2)y′− (15 − 14x2)y= 0 .
Give explicit formulas for the coefficients in the solutions .",Elementary Differential Equations with Boundary Value Problems.pdf
"386 Chapter 7 Series Solutions of Linear Second Order Equations
Solution For the given equation, the polynomials defined in Theorem 7.7.2 are
p0(r) = r(r− 1) + 3 r− 15 = ( r− 3)(r+ 5)
p2(r) = r(r− 1) + 10 r+ 14 = ( r+ 2)( r+ 7) .
The roots of the indicial equation are r1 = 3 and r2 = −5, so k = ( r1 − r2)/ 2 = 4 . Therefore
Theorem 7.7.2 implies that
y1 = x3
∞∑
m=0
a2m(3)x2m (7.7.25)
and
y2 = x−5
3∑
m=0
a2m(−5)x2m + C
(
y1 ln x+ x3
∞∑
m=1
a′
2m(3)x2m
)
(with Cas in (
7.7.23)) form a fundamental set of solutions of Ly= 0 . The recurrence formula ( 7.7.22) is
a0(r) = 1 ,
a2m(r) = −p2(2m+ r− 2)
p0(2m+ r) a2m−2(r)
= − (2m+ r)(2m+ r+ 5)
(2m+ r− 3)(2m+ r+ 5) a2m−2(r)
= − 2m+ r
2m+ r− 3 a2m−2(r), m≥ 1,
(7.7.26)
which implies that
a2m(r) = ( −1)m
m∏
j=1
2j+ r
2j+ r− 3 , m≥ 0. (7.7.27)
Therefore
a2m(3) =
(−1)m ∏ m
j=1(2j+ 3)
2mm! . (7.7.28)
Substituting this into ( 7.7.25) yields
y1 = x3
∞∑
m=0
(−1)m ∏ m
j=1(2j+ 3)
2mm! x2m.",Elementary Differential Equations with Boundary Value Problems.pdf
"Therefore
a2m(3) =
(−1)m ∏ m
j=1(2j+ 3)
2mm! . (7.7.28)
Substituting this into ( 7.7.25) yields
y1 = x3
∞∑
m=0
(−1)m ∏ m
j=1(2j+ 3)
2mm! x2m.
T o compute the coefficients a2(−5), a4(−5), and a6(−5) in y2, we set r = −5 in ( 7.7.26) and apply
the resulting recurrence formula for m= 1 , 2, 3; thus,
a2m(−5) = − 2m− 5
2(m− 4) a2m−2(−5), m= 1 , 2, 3.
This yields
a2(−5) = −1
2 , a4(−5) = 1
8 , a6(−5) = 1
16 .
Substituting r1 = 3 , r2 = −5, k = 4 , and α0 = 1 into ( 7.7.23) yields C = −3/ 16. Therefore, from
(7.7.24),
y2 = x−5
(
1 − 1
2 x2 + 1
8 x4 + 1
16 x6
)
− 3
16
(
y1 ln x+ x3
∞∑
m=1
a′
2m(3)x2m
)
. (7.7.29)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 387
T o obtain a′
2m(r) we use logarithmic differentiation. From (
7.7.27),
|a2m(r)|=
m∏
j=1
|2j+ r|
|2j+ r− 3|, m≥ 1.
Therefore
ln |a2m(r)|=
n∑
j=1
(ln |2j+ r| −ln |2j+ r− 3|) .
Differentiating with respect to ryields
a′
2m(r)
a2m(r) =
m∑
j=1
( 1
2j+ r − 1
2j+ r− 3
)
.
Therefore
a′
2m(r) = a2m(r)
n∑
j=1
( 1
2j+ r − 1
2j+ r− 3
)
.
Setting r= 3 here and recalling ( 7.7.28) yields
a′
2m(3) =
(−1)m ∏ m
j=1(2j+ 3)
2mm!
m∑
j=1
( 1
2j+ 3 − 1
2j
)
. (7.7.30)
Since 1
2j+ 3 − 1
2j = − 3
2j(2j+ 3) ,
we can rewrite ( 7.7.30) as
a′
2m(3) = −3
2
(−1)n ∏ m
j=1(2j+ 3)
2mm!


n∑
j=1
1
j(2j+ 3)

 .
Substituting this into (
7.7.29) yields
y2 = x−5
(
1 − 1
2 x2 + 1
8 x4 + 1
16 x6
)
− 3
16 y1 ln x
+ 9
32 x3
∞∑
m=1
(−1)m ∏ m
j=1(2j+ 3)
2mm!


m∑
j=1
1
j(2j+ 3)

 x2m.
Example 7.7.4 Find a fundamental set of Frobenius solutions of
x2(1 − 2x2)y′′+ x(7 − 13x2)y′− 14x2y= 0 .
Give explicit formulas for the coefficients in the solutions .",Elementary Differential Equations with Boundary Value Problems.pdf
"x2(1 − 2x2)y′′+ x(7 − 13x2)y′− 14x2y= 0 .
Give explicit formulas for the coefficients in the solutions .
Solution For the given equation, the polynomials defined in Theorem
7.7.2 are
p0(r) = r(r− 1) + 7 r = r(r+ 6) ,
p2(r) = −2r(r− 1) − 13r− 14 = −(r+ 2)(2 r+ 7) .",Elementary Differential Equations with Boundary Value Problems.pdf
"388 Chapter 7 Series Solutions of Linear Second Order Equations
The roots of the indicial equation are r1 = 0 and r2 = −6, so k = ( r1 − r2)/ 2 = 3 . Therefore
Theorem 7.7.2 implies that
y1 =
∞∑
m=0
a2m(0)x2m, (7.7.31)
and
y2 = x−6
2∑
m=0
a2m(−6)x2m + C
(
y1 ln x+
∞∑
m=1
a′
2m(0)x2m
)
(7.7.32)
(with Cas in (
7.7.23)) form a fundamental set of solutions of Ly= 0 . The recurrence formulas ( 7.7.22)
are
a0(r) = 1 ,
a2m(r) = −p2(2m+ r− 2)
p0(2m+ r) a2m−2(r)
= (2m+ r)(4m+ 2 r+ 3)
(2m+ r)(2m+ r+ 6) a2m−2(r)
= 4m+ 2 r+ 3
2m+ r+ 6 a2m−2(r), m≥ 1,
(7.7.33)
which implies that
a2m(r) =
m∏
j=1
4j+ 2 r+ 3
2j+ r+ 6 .
Setting r= 0 yields
a2m(0) = 6
∏ m
j=1(4j+ 3)
2m (m+ 3)! .
Substituting this into ( 7.7.31) yields
y1 = 6
∞∑
m=0
∏ m
j=1(4j+ 3)
2m(m+ 3)! x2m.
T o compute the coefficients a0(−6), a2(−6), and a4(−6) in y2, we set r = −6 in ( 7.7.33) and apply
the resulting recurrence formula for m= 1 , 2; thus,
a0(−6) = 1 ,
a2m(−6) = 4m− 9
2m a2m−2(−6), m= 1 , 2.
The last formula yields
a2(−6) = −5",Elementary Differential Equations with Boundary Value Problems.pdf
"the resulting recurrence formula for m= 1 , 2; thus,
a0(−6) = 1 ,
a2m(−6) = 4m− 9
2m a2m−2(−6), m= 1 , 2.
The last formula yields
a2(−6) = −5
2 and a4(−6) = 5
8 .
Since p2(−2) = 0 , the constant C in ( 7.7.23) is zero. Therefore ( 7.7.32) reduces to
y2 = x−6
(
1 − 5
2 x2 + 5
8 x4
)
.
7.7 Exercises
In Exercises 1– 40 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi-
cients.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 389
1. x2y′′− 3xy′+ (3 + 4 x)y= 0
2. xy′′+ y= 0
3. 4x2(1 + x)y′′+ 4 x(1 + 2 x)y′− (1 + 3 x)y= 0
4. xy′′+ xy′+ y= 0
5. 2x2(2 + 3 x)y′′+ x(4 + 21 x)y′− (1 − 9x)y= 0
6. x2y′′+ x(2 + x)y′− (2 − 3x)y= 0
7. 4x2y′′+ 4 xy′− (9 − x)y= 0
8. x2y′′+ 10 xy′+ (14 + x)y= 0
9. 4x2(1 + x)y′′+ 4 x(3 + 8 x)y′− (5 − 49x)y= 0
10. x2(1 + x)y′′− x(3 + 10 x)y′+ 30 xy= 0
11. x2y′′+ x(1 + x)y′− 3(3 + x)y= 0
12. x2y′′+ x(1 − 2x)y′− (4 + x)y= 0
13. x(1 + x)y′′− 4y′− 2y= 0
14. x2(1 + 2 x)y′′+ x(9 + 13 x)y′+ (7 + 5 x)y= 0
15. 4x2y′′− 2x(4 − x)y′− (7 + 5 x)y= 0
16. 3x2(3 + x)y′′− x(15 + x)y′− 20y= 0
17. x2(1 + x)y′′+ x(1 − 10x)y′− (9 − 10x)y= 0
18. x2(1 + x)y′′+ 3 x2y′− (6 − x)y= 0
19. x2(1 + 2 x)y′′− 2x(3 + 14 x)y′+ (6 + 100 x)y= 0
20. x2(1 + x)y′′− x(6 + 11 x)y′+ (6 + 32 x)y= 0
21. 4x2(1 + x)y′′+ 4 x(1 + 4 x)y′− (49 + 27 x)y= 0
22. x2(1 + 2 x)y′′− x(9 + 8 x)y′− 12xy= 0
23. x2(1 + x2)y′′− x(7 − 2x2)y′+ 12 y= 0
24. x2y′′− x(7 − x2)y′+ 12 y= 0
25. xy′′− 5y′+ xy= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"22. x2(1 + 2 x)y′′− x(9 + 8 x)y′− 12xy= 0
23. x2(1 + x2)y′′− x(7 − 2x2)y′+ 12 y= 0
24. x2y′′− x(7 − x2)y′+ 12 y= 0
25. xy′′− 5y′+ xy= 0
26. x2y′′+ x(1 + 2 x2)y′− (1 − 10x2)y = 0
27. x2y′′− xy′− (3 − x2)y= 0
28. 4x2y′′+ 2 x(8 + x2)y′+ (5 + 3 x2)y = 0
29. x2y′′+ x(1 + x2)y′− (1 − 3x2)y= 0
30. x2y′′+ x(1 − 2x2)y′− 4(1 + 2 x2)y = 0
31. 4x2y′′+ 8 xy′− (35 − x2)y= 0
32. 9x2y′′− 3x(11 + 2 x2)y′+ (13 + 10 x2)y = 0
33. x2y′′+ x(1 − 2x2)y′− 4(1 − x2)y = 0
34. x2y′′+ x(1 − 3x2)y′− 4(1 − 3x2)y = 0
35. x2(1 + x2)y′′+ x(5 + 11 x2)y′+ 24 x2y = 0
36. 4x2(1 + x2)y′′+ 8 xy′− (35 − x2)y = 0
37. x2(1 + x2)y′′− x(5 − x2)y′− (7 + 25 x2)y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"390 Chapter 7 Series Solutions of Linear Second Order Equations
38. x2(1 + x2)y′′+ x(5 + 2 x2)y′− 21y= 0
39. x2(1 + 2 x2)y′′− x(3 + x2)y′− 2x2y= 0
40. 4x2(1 + x2)y′′+ 4 x(2 + x2)y′− (15 + x2)y = 0
41. (a) Under the assumptions of Theorem 7.7.1, show that
y1 = xr1
∞∑
n=0
an (r1)xn
and
y2 = xr2
k−1∑
n=0
an(r2)xn + C
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
are linearly independent. H IN T : Show that if c1 and c2 are constants such that c1y1 + c2y2 ≡
0 on an interval (0,ρ ), then
x−r2 (c1y1(x) + c2y2(x)) = 0 , 0 <x<ρ.
Then let x→ 0+ to conclude that c2=0.
(b) Use the result of (a) to complete the proof of Theorem
7.7.1.
42. Find a fundamental set of Frobenius solutions of Bessel’s eq uation
x2y′′+ xy′+ ( x2 − ν2)y = 0
in the case where ν is a positive integer.
43. Prove Theorem 7.7.2.
44. Under the assumptions of Theorem 7.7.1, show that C = 0 if and only if p1(r2 + ) = 0 for some
integer in {0, 1,...,k − 1}.",Elementary Differential Equations with Boundary Value Problems.pdf
"44. Under the assumptions of Theorem 7.7.1, show that C = 0 if and only if p1(r2 + ) = 0 for some
integer in {0, 1,...,k − 1}.
45. Under the assumptions of Theorem 7.7.2, show that C = 0 if and only if p2(r2 + 2) = 0 for some
integer ℓin {0, 1,...,k − 1}.
46. Let
Ly= α0x2y′′+ β0xy′+ ( γ0 + γ1x)y
and define
p0(r) = α0r(r− 1) + β0r+ γ0.
Show that if
p0(r) = α0(r− r1)(r− r2)
where r1 − r2 = k, a positive integer, then Ly= 0 has the solutions
y1 = xr1
∞∑
n=0
(−1)n
n! ∏ n
j=1(j+ k)
(γ1
α0
) n
xn
and
y2 = xr2
k−1∑
n=0
(−1)n
n! ∏ n
j=1(j− k)
(γ1
α0
) n
xn
− 1
k!(k− 1)!
(γ1
α0
) k

 y1 ln x− xr1
∞∑
n=1
(−1)n
n! ∏ n
j=1(j+ k)
(γ1
α0
) n


n∑
j=1
2j+ k
j(j+ k)

 xn

 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 7.7 The Method of Frobenius III 391
47. Let
Ly= α0x2y′′+ β0xy′+ ( γ0 + γ2x2)y
and define
p0(r) = α0r(r− 1) + β0r+ γ0.
Show that if
p0(r) = α0(r− r1)(r− r2)
where r1 − r2 = 2 k, an even positive integer, then Ly= 0 has the solutions
y1 = xr1
∞∑
m=0
(−1)m
4mm! ∏ m
j=1(j+ k)
(γ2
α0
) m
x2m
and
y2 = xr2
k−1∑
m=0
(−1)m
4mm! ∏ m
j=1(j− k)
(γ2
α0
) m
x2m
− 2
4kk!(k− 1)!
(γ2
α0
) k

 y1 ln x− xr1
2
∞∑
m=1
(−1)m
4m m! ∏ m
j=1(j+ k)
(γ2
α0
) m


m∑
j=1
2j+ k
j(j+ k)

 x2m

 .
48. Let Lbe as in Exercises 7.5.
57 and 7.5. 58, and suppose the indicial polynomial of Ly= 0 is
p0(r) = α0(r− r1)(r− r2),
with k = r1 − r2, where kis a positive integer. Define a0(r) = 1 for all r. If ris a real number
such that p0(n+ r) is nonzero for all positive integers n, define
an(r) = − 1
p0(n+ r)
n∑
j=1
pj (n+ r− j)an−j (r), n≥ 1,
and let
y1 = xr1
∞∑
n=0
an(r1)xn.
Define
an (r2) = − 1
p0(n+ r2)
n∑
j=1
pj (n+ r2 − j)an−j (r2) if n≥ 1 and n̸= k,
and let ak(r2) be arbitrary .",Elementary Differential Equations with Boundary Value Problems.pdf
"and let
y1 = xr1
∞∑
n=0
an(r1)xn.
Define
an (r2) = − 1
p0(n+ r2)
n∑
j=1
pj (n+ r2 − j)an−j (r2) if n≥ 1 and n̸= k,
and let ak(r2) be arbitrary .
(a) Conclude from Exercise 7.6.. 66 that
L
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
= kα 0xr1 .
(b) Conclude from Exercise 7.5..
57 that
L
(
xr2
∞∑
n=0
an(r2)xn
)
= Axr1 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"392 Chapter 7 Series Solutions of Linear Second Order Equations
where
A=
k∑
j=1
pj (r1 − j)ak−j (r2).
(c) Show that y1 and
y2 = xr2
∞∑
n=0
an (r2)xn − A
kα 0
(
y1 ln x+ xr1
∞∑
n=1
a′
n (r1)xn
)
form a fundamental set of Frobenius solutions of Ly= 0 .
(d) Show that choosing the arbitrary quantity ak(r2) to be nonzero merely adds a multiple of y1
to y2. Conclude that we may as well take ak (r2) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 8
Laplace T ransforms
IN THIS CHAPTER we study the method of Laplace transforms , which illustrates one of the basic prob-
lem solving techniques in mathematics: transform a difficul t problem into an easier one, solve the lat-
ter, and then use its solution to obtain a solution of the orig inal problem. The method discussed here
transforms an initial value problem for a constant coefficie nt equation into an algebraic equation whose
solution can then be used to solve the initial value problem. In some cases this method is merely an
alternative procedure for solving problems that can be solv ed equally well by methods that we considered
previously; however, in other cases the method of Laplace tr ansforms is more efficient than the methods
previously discussed. This is especially true in physical p roblems dealing with discontinuous forcing
functions.
SECTION 8.1 defines the Laplace transform and developes its p roperties.",Elementary Differential Equations with Boundary Value Problems.pdf
"functions.
SECTION 8.1 defines the Laplace transform and developes its p roperties.
SECTION 8.2 deals with the problem of finding a function that h as a given Laplace transform.
SECTION 8.3 applies the Laplace transform to solve initial v alue problems for constant coefficient second
order differential equations on (0, ∞ ).
SECTION 8.4 introduces the unit step function.
SECTION 8.5 uses the unit step function to solve constant coe fficient equations with piecewise continu-
ous forcing functions.
SECTION 8.6 deals with the convolution theorem, an importan t theoretical property of the Laplace trans-
form.
SECTION 8.7 introduces the idea of impulsive force, and trea ts constant coefficient equations with im-
pulsive forcing functions.
SECTION 8.8 is a brief table of Laplace transforms.
393",Elementary Differential Equations with Boundary Value Problems.pdf
"394 Chapter 8 Laplace Transforms
8.1 INTRODUCTION TO THE LAPLACE TRANSFORM
Definition of the Laplace Transform
T o define the Laplace transform, we first recall the definition of an improper integral. If g is integrable
over the interval [a,T ] for every T >a, then the improper integral of gover [a, ∞ ) is defined as
∫ ∞
a
g(t) dt= lim
T →∞
∫ T
a
g(t) dt. (8.1.1)
W e say that the improper integral converges if the limit in ( 8.1.1) exists; otherwise, we say that the
improper integral diverges or does not exist . Here’s the definition of the Laplace transform of a function
f.
Definition 8.1.1 Let f be defined for t≥ 0 and let sbe a real number . Then the Laplace transform of f
is the function F defined by
F(s) =
∫ ∞
0
e−stf(t) dt, (8.1.2)
for those values of sfor which the improper integral converges .
It is important to keep in mind that the variable of integrati on in ( 8.1.2) is t, while s is a parameter",Elementary Differential Equations with Boundary Value Problems.pdf
"It is important to keep in mind that the variable of integrati on in ( 8.1.2) is t, while s is a parameter
independent of t. W e use tas the independent variable for f because in applications the Laplace transform
is usually applied to functions of time.
The Laplace transform can be viewed as an operator L that transforms the function f = f(t) into the
function F = F(s). Thus, ( 8.1.2) can be expressed as
F = L(f).
The functions f and F form a transform pair , which we’ll sometimes denote by
f(t) ↔ F(s).
It can be shown that if F(s) is defined for s= s0 then it’s defined for all s>s 0 (Exercise 14(b)).
Computation of Some Simple Laplace Transforms
Example 8.1.1 Find the Laplace transform of f(t) = 1 .
Solution From ( 8.1.2) with f(t) = 1 ,
F(s) =
∫ ∞
0
e−st dt= lim
T →∞
∫ T
0
e−st dt.
If s̸= 0 then ∫ T
0
e−st dt= −1
se−st
⏐
⏐
⏐
T
0
= 1 − e−sT
s . (8.1.3)
Therefore
lim
T →∞
∫ T
0
e−st dt=
{ 1
s, s> 0,
∞ , s< 0.
(8.1.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.1 Introduction to the Laplace Transform 395
If s= 0 the integrand reduces to the constant 1, and
lim
T →∞
∫ T
0
1 dt= lim
T →∞
∫ T
0
1 dt= lim
T →∞
T = ∞ .
Therefore F(0) is undefined, and
F(s) =
∫ ∞
0
e−st dt= 1
s, s> 0.
This result can be written in operator notation as
L(1) = 1
s, s> 0,
or as the transform pair
1 ↔ 1
s, s> 0.
REM A RK : It is convenient to combine the steps of integrating from 0 to T and letting T → ∞ . Therefore,
instead of writing ( 8.1.3) and ( 8.1.4) as separate steps we write
∫ ∞
0
e−st dt= −1
se−st
⏐
⏐
⏐
∞
0
=
{ 1
s, s> 0,
∞ , s< 0.
W e’ll follow this practice throughout this chapter.
Example 8.1.2 Find the Laplace transform of f(t) = t.
Solution From ( 8.1.2) with f(t) = t,
F(s) =
∫ ∞
0
e−st tdt. (8.1.5)
If s̸= 0 , integrating by parts yields
∫ ∞
0
e−st tdt = −te−st
s
⏐
⏐
⏐
⏐
∞
0
+ 1
s
∫ ∞
0
e−st dt= −
[ t
s + 1
s2
]
e−st
⏐
⏐
⏐
⏐
∞
0
=
{ 1
s2 , s> 0,
∞ , s< 0.
If s= 0 , the integral in ( 8.1.5) becomes
∫ ∞
0
tdt = t2
2
⏐
⏐
⏐
⏐
∞
0
= ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"0
e−st dt= −
[ t
s + 1
s2
]
e−st
⏐
⏐
⏐
⏐
∞
0
=
{ 1
s2 , s> 0,
∞ , s< 0.
If s= 0 , the integral in ( 8.1.5) becomes
∫ ∞
0
tdt = t2
2
⏐
⏐
⏐
⏐
∞
0
= ∞ .
Therefore F(0) is undefined and
F(s) = 1
s2 , s> 0.
This result can also be written as
L(t) = 1
s2 , s> 0,
or as the transform pair
t↔ 1
s2 , s> 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"396 Chapter 8 Laplace Transforms
Example 8.1.3 Find the Laplace transform of f(t) = eat , where ais a constant.
Solution From ( 8.1.2) with f(t) = eat ,
F(s) =
∫ ∞
0
e−steat dt.
Combining the exponentials yields
F(s) =
∫ ∞
0
e−(s−a)t dt.
However, we know from Example 8.1.1 that
∫ ∞
0
e−st dt= 1
s, s> 0.
Replacing sby s− ahere shows that
F(s) = 1
s− a, s>a.
This can also be written as
L(eat ) = 1
s− a, s>a, or eat ↔ 1
s− a, s>a.
Example 8.1.4 Find the Laplace transforms of f(t) = sin ωt and g(t) = cos ωt, where ω is a constant.
Solution Define
F(s) =
∫ ∞
0
e−st sin ωtdt (8.1.6)
and
G(s) =
∫ ∞
0
e−st cos ωtdt. (8.1.7)
If s> 0, integrating ( 8.1.6) by parts yields
F(s) = −e−st
s sin ωt
⏐
⏐
⏐
∞
0
+ ω
s
∫ ∞
0
e−st cos ωtdt,
so
F(s) = ω
sG(s). (8.1.8)
If s> 0, integrating ( 8.1.7) by parts yields
G(s) = −e−st cos ωt
s
⏐
⏐
⏐
∞
0
− ω
s
∫ ∞
0
e−st sin ωtdt,
so
G(s) = 1
s − ω
sF(s).
Now substitute from ( 8.1.8) into this to obtain
G(s) = 1
s − ω2
s2 G(s).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.1 Introduction to the Laplace Transform 397
Solving this for G(s) yields
G(s) = s
s2 + ω2 , s> 0.
This and ( 8.1.8) imply that
F(s) = ω
s2 + ω2 , s> 0.
T ables of Laplace transforms
Extensive tables of Laplace transforms have been compiled a nd are commonly used in applications. The
brief table of Laplace transforms in the Appendix will be ade quate for our purposes.
Example 8.1.5 Use the table of Laplace transforms to find L(t3e4t ).
Solution The table includes the transform pair
tn eat ↔ n!
(s− a)n+1 .
Setting n= 3 and a= 4 here yields
L(t3e4t) = 3!
(s− 4)4 = 6
(s− 4)4 .
W e’ll sometimes write Laplace transforms of specific functi ons without explicitly stating how they are
obtained. In such cases you should refer to the table of Lapla ce transforms.
Linearity of the Laplace Transform
The next theorem presents an important property of the Lapla ce transform.
Theorem 8.1.2 [Linearity Property ] Suppose L(fi) is defined for s > si, 1 ≤ i ≤ n). Let s0 be the",Elementary Differential Equations with Boundary Value Problems.pdf
"Theorem 8.1.2 [Linearity Property ] Suppose L(fi) is defined for s > si, 1 ≤ i ≤ n). Let s0 be the
largest of the numbers s1, s2, . . . , sn , and let c1, c2,. . . , cn be constants . Then
L(c1f1 + c2f2 + · · ·+ cn fn ) = c1L(f1) + c2L(f2) + · · ·+ cnL(fn ) for s>s 0.
Proof W e give the proof for the case where n= 2 . If s>s 0 then
L(c1f1 + c2f2) =
∫ ∞
0
e−st (c1f1(t) + c2f2(t))) dt
= c1
∫ ∞
0
e−stf1(t) dt+ c2
∫ ∞
0
e−stf2(t) dt
= c1L(f1) + c2L(f2).
Example 8.1.6 Use Theorem 8.1.2 and the known Laplace transform
L(eat ) = 1
s− a
to find L(cosh bt) (b̸= 0) .",Elementary Differential Equations with Boundary Value Problems.pdf
"398 Chapter 8 Laplace Transforms
Solution By definition,
cosh bt= ebt + e−bt
2 .
Therefore
L(cosh bt) = L
(1
2 ebt + 1
2 e−bt
)
= 1
2 L(ebt) + 1
2 L(e−bt) (linearity property)
= 1
2
1
s− b + 1
2
1
s+ b,
(8.1.9)
where the first transform on the right is defined for s > band the second for s >−b; hence, both are
defined for s> |b|. Simplifying the last expression in ( 8.1.9) yields
L(cosh bt) = s
s2 − b2 , s> |b|.
The First Shifting Theorem
The next theorem enables us to start with known transform pai rs and derive others. (For other results of
this kind, see Exercises 6 and 13.)
Theorem 8.1.3 [First Shifting Theorem ] If
F(s) =
∫ ∞
0
e−st f(t) dt (8.1.10)
is the Laplace transform of f(t) for s > s0, then F(s− a) is the Laplace transform of eat f(t) for
s>s 0 + a.
PRO O F. Replacing sby s− ain ( 8.1.10) yields
F(s− a) =
∫ ∞
0
e−(s−a)tf(t) dt (8.1.11)
if s− a>s 0; that is, if s>s 0 + a. However, ( 8.1.11) can be rewritten as
F(s− a) =
∫ ∞
0
e−st (
eat f(t)
)
dt,
which implies the conclusion.",Elementary Differential Equations with Boundary Value Problems.pdf
"if s− a>s 0; that is, if s>s 0 + a. However, ( 8.1.11) can be rewritten as
F(s− a) =
∫ ∞
0
e−st (
eat f(t)
)
dt,
which implies the conclusion.
Example 8.1.7 Use Theorem
8.1.3 and the known Laplace transforms of 1, t, cos ωt, and sin ωt to find
L(eat ), L(teat ), L(eλt sin ωt), and L(eλt cos ωt).
Solution In the following table the known transform pairs are listed o n the left and the required transform
pairs listed on the right are obtained by applying Theorem 8.1.3.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.1 Introduction to the Laplace Transform 399
f(t) ↔ F(s) eat f(t) ↔ F(s− a)
1 ↔ 1
s, s> 0 eat ↔ 1
(s− a) , s>a
t↔ 1
s2 , s> 0 teat ↔ 1
(s− a)2 , s>a
sin ωt ↔ ω
s2 + ω2 , s> 0 eλt sin ωt ↔ ω
(s− λ)2 + ω2 , s>λ
cos ωt ↔ s
s2 + ω2 , s> 0 eλt sin ωt ↔ s− λ
(s− λ)2 + ω2 , s>λ
Existence of Laplace Transforms
Not every function has a Laplace transform. For example, it c an be shown (Exercise 3) that
∫ ∞
0
e−stet2
dt= ∞
for every real number s. Hence, the function f(t) = et2
does not have a Laplace transform.
Our next objective is to establish conditions that ensure th e existence of the Laplace transform of a
function. W e first review some relevant definitions from calc ulus.
Recall that a limit
lim
t→ t0
f(t)
exists if and only if the one-sided limits
lim
t→ t0−
f(t) and lim
t→ t0+
f(t)
both exist and are equal; in this case,
lim
t→ t0
f(t) = lim
t→ t0−
f(t) = lim
t→ t0+
f(t).
Recall also that f is continuous at a point t0 in an open interval (a,b ) if and only if
lim
t→ t0",Elementary Differential Equations with Boundary Value Problems.pdf
"t→ t0
f(t) = lim
t→ t0−
f(t) = lim
t→ t0+
f(t).
Recall also that f is continuous at a point t0 in an open interval (a,b ) if and only if
lim
t→ t0
f(t) = f(t0),
which is equivalent to
lim
t→ t0+
f(t) = lim
t→ t0−
f(t) = f(t0 ). (8.1.12)
For simplicity , we define
f(t0+) = lim
t→ t0+
f(t) and f(t0 −) = lim
t→ t0−
f(t),
so ( 8.1.12) can be expressed as
f(t0 +) = f(t0 −) = f(t0).
If f(t0 +) and f(t0 −) have finite but distinct values, we say that f has a jump discontinuity at t0, and
f(t0 +) − f(t0−)
is called the jump in f at t0 (Figure 8.1.1).",Elementary Differential Equations with Boundary Value Problems.pdf
"400 Chapter 8 Laplace Transforms
 t0
 x
 y
 f (t0+)
 f (t0−)
Figure 8.1.1 A jump discontinuity
If f(t0+) and f(t0−) are finite and equal, but either f isn’t defined at t0 or it’s defined but
f(t0 ) ̸= f(t0 +) = f(t0 −),
we say that f has a removable discontinuity at t0 (Figure 8.1.2). This terminolgy is appropriate since a
function f with a removable discontinuity at t0 can be made continuous at t0 by defining (or redefining)
f(t0 ) = f(t0 +) = f(t0 −).
REM A RK : W e know from calculus that a definite integral isn’t affecte d by changing the values of its
integrand at isolated points. Therefore, redefining a funct ion f to make it continuous at removable dis-
continuities does not change L(f).
Definition 8.1.4
(i) A function f is said to be piecewise continuous on a finite closed interval [0,T ] if f(0+) and
f(T−) are finite and f is continuous on the open interval (0,T ) except possibly at finitely many
points, where f may have jump discontinuities or removable discontinuitie s.",Elementary Differential Equations with Boundary Value Problems.pdf
"points, where f may have jump discontinuities or removable discontinuitie s.
(ii) A function f is said to be piecewise continuous on the infinite interval [0, ∞ ) if it’s piecewise
continuous on [0,T ] for every T >0.
Figure 8.1.3 shows the graph of a typical piecewise continuous function.
It is shown in calculus that if a function is piecewise contin uous on a finite closed interval then it’s
integrable on that interval. But if f is piecewise continuous on [0, ∞ ), then so is e−stf(t), and therefore
∫ T
0
e−st f(t) dt",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.1 Introduction to the Laplace Transform 401
 t0
 x
 f(t0)
 f(t0−) = f(t 0+)
Figure 8.1.2
 a  b
 x
 y
Figure 8.1.3 A piecewise continuous function on
[a,b ]
exists for every T >0. However, piecewise continuity alone does not guarantee th at the improper integral
∫ ∞
0
e−st f(t) dt= lim
T →∞
∫ T
0
e−st f(t) dt (8.1.13)
converges for s in some interval (s0, ∞ ). For example, we noted earlier that ( 8.1.13) diverges for all
s if f(t) = et2
. Stated informally , this occurs because et2
increases too rapidly as t → ∞ . The next
definition provides a constraint on the growth of a function t hat guarantees convergence of its Laplace
transform for sin some interval (s0, ∞ ) .
Definition 8.1.5 A function f is said to be of exponential order s0 if there are constants M and t0 such
that
|f(t)| ≤Mes0t, t ≥ t0. (8.1.14)
In situations where the specific value of s0 is irrelevant we say simply that f is of exponential order .",Elementary Differential Equations with Boundary Value Problems.pdf
"that
|f(t)| ≤Mes0t, t ≥ t0. (8.1.14)
In situations where the specific value of s0 is irrelevant we say simply that f is of exponential order .
The next theorem gives useful sufficient conditions for a fun ction f to have a Laplace transform. The
proof is sketched in Exercise 10.
Theorem 8.1.6 If f is piecewise continuous on [0, ∞ ) and of exponential order s0, then L(f) is defined
for s>s 0.
REM A RK : W e emphasize that the conditions of Theorem 8.1.6 are sufficient, but not necessary , for f to
have a Laplace transform. For example, Exercise 14(c) shows that f may have a Laplace transform even
though f isn’t of exponential order.
Example 8.1.8 If f is bounded on some interval [t0, ∞ ), say
|f(t)| ≤M, t ≥ t0,
then ( 8.1.14) holds with s0 = 0 , so f is of exponential order zero. Thus, for example, sin ωt and cos ωt
are of exponential order zero, and Theorem 8.1.6 implies that L(sin ωt) and L(cos ωt) exist for s >0.
This is consistent with the conclusion of Example 8.1.4.",Elementary Differential Equations with Boundary Value Problems.pdf
"402 Chapter 8 Laplace Transforms
Example 8.1.9 It can be shown that if limt→∞ e−s0t f(t) exists and is finite then f is of exponential
order s0 (Exercise 9). If α is any real number and s0 >0 then f(t) = tα is of exponential order s0, since
lim
t→∞
e−s0t tα = 0 ,
by L ’Hôpital’s rule. If α ≥ 0, f is also continuous on [0, ∞ ). Therefore Exercise 9 and Theorem 8.1.6
imply that L(tα ) exists for s≥ s0. However, since s0 is an arbitrary positive number, this really implies
that L(tα ) exists for all s> 0. This is consistent with the results of Example 8.1.2 and Exercises 6 and 8.
Example 8.1.10 Find the Laplace transform of the piecewise continuous func tion
f(t) =
{
1, 0 ≤ t< 1,
−3e−t , t ≥ 1.
Solution Since f is defined by different formulas on [0, 1) and [1, ∞ ), we write
F(s) =
∫ ∞
0
e−st f(t) dt=
∫ 1
0
e−st(1) dt+
∫ ∞
1
e−st (−3e−t ) dt.
Since
∫ 1
0
e−st dt=



1 − e−s
s , s ̸= 0 ,
1, s = 0 ,
and ∫ ∞
1
e−st(−3e−t ) dt= −3
∫ ∞
1
e−(s+1)t dt= −3e−(s+1)
s+ 1 , s> −1,
it follows that",Elementary Differential Equations with Boundary Value Problems.pdf
"Since
∫ 1
0
e−st dt=



1 − e−s
s , s ̸= 0 ,
1, s = 0 ,
and ∫ ∞
1
e−st(−3e−t ) dt= −3
∫ ∞
1
e−(s+1)t dt= −3e−(s+1)
s+ 1 , s> −1,
it follows that
F(s) =





1 − e−s
s − 3 e−(s+1)
s+ 1 , s> −1, s̸= 0 ,
1 − 3
e , s = 0 .
This is consistent with Theorem 8.1.6, since
|f(t)| ≤3e−t , t ≥ 1,
and therefore f is of exponential order s0 = −1.
REM A RK : In Section 8.4 we’ll develop a more efficient method for findi ng Laplace transforms of piece-
wise continuous functions.
Example 8.1.11 W e stated earlier that
∫ ∞
0
e−stet2
dt= ∞
for all s, so Theorem 8.1.6 implies that f(t) = et2
is not of exponential order, since
lim
t→∞
et2
Mes0t = lim
t→∞
1
Met2−s0t = ∞ ,
so
et2
>Me s0t
for sufficiently large values of t, for any choice of M and s0 (Exercise 3).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.1 Introduction to the Laplace Transform 403
8.1 Exercises
1. Find the Laplace transforms of the following functions by ev aluating the integral F(s) =
∫∞
0 e−stf(t) dt.
(a) t (b) te−t (c) sinh bt
(d) e2t − 3et (e) t2
2. Use the table of Laplace transforms to find the Laplace transf orms of the following functions.
(a) cosh tsin t (b) sin2 t (c) cos2 2t
(d) cosh2 t (e) tsinh 2 t (f) sin tcos t
(g) sin
(
t+ π
4
)
(h) cos 2t− cos 3t (i) sin 2 t+ cos 4 t
3. Show that ∫ ∞
0
e−stet2
dt= ∞
for every real number s.
4. Graph the following piecewise continuous functions and eva luate f(t+), f(t−), and f(t) at each
point of discontinuity .
(a) f(t) =



−t, 0 ≤ t< 2,
t− 4, 2 ≤ t< 3,
1, t ≥ 3.
(b) f(t) =



t2 + 2 , 0 ≤ t< 1,
4, t = 1 ,
t, t> 1.
(c) f(t) =



sin t, 0 ≤ t<π/ 2,
2 sin t, π/ 2 ≤ t<π,
cos t, t ≥ π.
(d) f(t) =











t, 0 ≤ t< 1,
2, t = 1 ,
2 − t, 1 ≤ t< 2,
3, t = 2 ,
6, t> 2.
5. Find the Laplace transform:
(a) f(t) =
{ e−t , 0 ≤ t< 1,",Elementary Differential Equations with Boundary Value Problems.pdf
"










t, 0 ≤ t< 1,
2, t = 1 ,
2 − t, 1 ≤ t< 2,
3, t = 2 ,
6, t> 2.
5. Find the Laplace transform:
(a) f(t) =
{ e−t , 0 ≤ t< 1,
e−2t, t ≥ 1. (b) f(t) =
{ 1, 0 ≤ t< 4,
t, t ≥ 4.
(c) f(t) =
{ t, 0 ≤ t< 1,
1, t ≥ 1. (d) f(t) =
{ tet , 0 ≤ t< 1,
et , t ≥ 1.
6. Prove that if f(t) ↔ F(s) then tkf(t) ↔ (−1)k F(k)(s). H IN T : Assume that it’s permissible to
differentiate the integral ∫∞
0 e−st f(t) dtwith respect to sunder the integral sign.
7. Use the known Laplace transforms
L(eλt sin ωt) = ω
(s− λ)2 + ω2 and L(eλt cos ωt) = s− λ
(s− λ)2 + ω2
and the result of Exercise 6 to find L(teλt cos ωt) and L(teλt sin ωt).
8. Use the known Laplace transform L(1) = 1 /s and the result of Exercise 6 to show that
L(tn ) = n!
sn+1 , n = integer.
9. (a) Show that if limt→∞ e−s0tf(t) exists and is finite then f is of exponential order s0.
(b) Show that if f is of exponential order s0 then limt→∞ e−st f(t) = 0 for all s>s 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"404 Chapter 8 Laplace Transforms
(c) Show that if f is of exponential order s0 and g(t) = f(t+ τ) where τ >0, then gis also of
exponential order s0.
10. Recall the next theorem from calculus.
TH EO REM A. Let gbe integrable on [0,T ] for every T >0. Suppose there’s a function wdefined
on some interval [τ, ∞ ) (with τ ≥ 0) such that |g(t)| ≤w(t) for t≥ τ and ∫∞
τ w(t) dtconverges.
Then ∫∞
0 g(t) dtconverges.
Use Theorem A to show that if f is piecewise continuous on [0, ∞ ) and of exponential order s0,
then f has a Laplace transform F(s) defined for s>s 0.
11. Prove: If f is piecewise continuous and of exponential order then lims→∞ F(s) = 0 .
12. Prove: If f is continuous on [0, ∞ ) and of exponential order s0 >0, then
L
(∫ t
0
f(τ) dτ
)
= 1
sL(f), s>s 0.
HIN T : Use integration by parts to evaluate the transform on the lef t.
13. Suppose fis piecewise continuous and of exponential order, and that limt→ 0+ f(t)/t exists. Show
that
L
(f(t)
t
)
=
∫ ∞
s
F(r) dr.",Elementary Differential Equations with Boundary Value Problems.pdf
"13. Suppose fis piecewise continuous and of exponential order, and that limt→ 0+ f(t)/t exists. Show
that
L
(f(t)
t
)
=
∫ ∞
s
F(r) dr.
HIN T : Use the results of Exercises 6 and 11.
14. Suppose f is piecewise continuous on [0, ∞ ).
(a) Prove: If the integral g(t) = ∫t
0 e−s0τ f(τ) dτ satisfies the inequality |g(t)| ≤M (t ≥ 0),
then f has a Laplace transform F(s) defined for s > s0. H IN T : Use integration by parts to
show that
∫ T
0
e−st f(t) dt= e−(s−s0)T g(T) + ( s− s0)
∫ T
0
e−(s−s0)tg(t) dt.
(b) Show that if L(f) exists for s= s0 then it exists for s>s 0. Show that the function
f(t) = tet2
cos(et2
)
has a Laplace transform defined for s> 0, even though f isn’t of exponential order.
(c) Show that the function
f(t) = tet2
cos(et2
)
has a Laplace transform defined for s> 0, even though f isn’t of exponential order.
15. Use the table of Laplace transforms and the result of Exercis e 13 to find the Laplace transforms of
the following functions.
(a) sin ωt
t (ω >0) (b) cos ωt − 1",Elementary Differential Equations with Boundary Value Problems.pdf
"the following functions.
(a) sin ωt
t (ω >0) (b) cos ωt − 1
t (ω >0) (c) eat − ebt
t
(d) cosh t− 1
t (e) sinh2 t
t
16. The gamma function is defined by
Γ( α) =
∫ ∞
0
xα −1e−x dx,
which can be shown to converge if α> 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.2 The Inverse Laplace Transform 405
(a) Use integration by parts to show that
Γ( α + 1) = αΓ( α), α> 0.
(b) Show that Γ( n+ 1) = n! if n= 1 , 2, 3,. . . .
(c) From (b) and the table of Laplace transforms,
L(tα ) = Γ( α + 1)
sα +1 , s> 0,
if α is a nonnegative integer. Show that this formula is valid for any α> −1. H IN T : Change
the variable of integration in the integral for Γ( α + 1) .
17. Suppose f is continuous on [0,T ] and f(t+ T) = f(t) for all t≥ 0. (W e say in this case that f
is periodic with period T.)
(a) Conclude from Theorem 8.1.6 that the Laplace transform of f is defined for s >0. H IN T :
Since f is continuous on [0,T ] and periodic with period T, it’s bounded on [0, ∞ ).
(b) (b) Show that
F(s) = 1
1 − e−sT
∫ T
0
e−stf(t) dt, s> 0.
HIN T : Write
F(s) =
∞∑
n=0
∫ (n+1)T
nT
e−stf(t) dt.
Then show that ∫ (n+1)T
nT
e−stf(t) dt= e−nsT
∫ T
0
e−st f(t) dt,
and recall the formula for the sum of a geometric series.",Elementary Differential Equations with Boundary Value Problems.pdf
"∫ (n+1)T
nT
e−stf(t) dt.
Then show that ∫ (n+1)T
nT
e−stf(t) dt= e−nsT
∫ T
0
e−st f(t) dt,
and recall the formula for the sum of a geometric series.
18. Use the formula given in Exercise 17(b) to find the Laplace transforms of the given periodic
functions:
(a) f(t) =
{ t, 0 ≤ t< 1,
2 − t, 1 ≤ t< 2, f(t+ 2) = f(t), t ≥ 0
(b) f(t) =
{
1, 0 ≤ t< 1
2 ,
−1, 1
2 ≤ t< 1, f(t+ 1) = f(t), t ≥ 0
(c) f(t) = |sin t|
(d) f(t) =
{
sin t, 0 ≤ t<π,
0, π ≤ t< 2π, f(t+ 2 π) = f(t)
8.2 THE INVERSE LAPLACE TRANSFORM
Definition of the Inverse Laplace Transform
In Section 8.1 we defined the Laplace transform of f by
F(s) = L(f) =
∫ ∞
0
e−st f(t) dt.
W e’ll also say that f is an inverse Laplace Transform of F, and write
f = L−1(F).",Elementary Differential Equations with Boundary Value Problems.pdf
"406 Chapter 8 Laplace Transforms
T o solve differential equations with the Laplace transform , we must be able to obtain f from its transform
F. There’s a formula for doing this, but we can’t use it because it requires the theory of functions of a
complex variable. Fortunately , we can use the table of Lapla ce transforms to find inverse transforms that
we’ll need.
Example 8.2.1 Use the table of Laplace transforms to find
(a) L−1
( 1
s2 − 1
)
and (b) L−1
( s
s2 + 9
)
.
SO LU TIO N (a) Setting b= 1 in the transform pair
sinh bt↔ b
s2 − b2
shows that
L−1
( 1
s2 − 1
)
= sinh t.
SO LU TIO N (b) Setting ω = 3 in the transform pair
cos ωt ↔ s
s2 + ω2
shows that
L−1
( s
s2 + 9
)
= cos 3 t.
The next theorem enables us to find inverse transforms of line ar combinations of transforms in the
table. W e omit the proof.
Theorem 8.2.1
[Linearity Property ] If F1, F2, . . . , Fn are Laplace transforms and c1, c2, . . . , cn are
constants, then",Elementary Differential Equations with Boundary Value Problems.pdf
"Theorem 8.2.1
[Linearity Property ] If F1, F2, . . . , Fn are Laplace transforms and c1, c2, . . . , cn are
constants, then
L−1(c1F1 + c2F2 + · · ·+ cnFn ) = c1L−1(F1) + c2L−1(F2) + · · ·+ cn L−1Fn.
Example 8.2.2 Find
L−1
( 8
s+ 5 + 7
s2 + 3
)
.
Solution From the table of Laplace transforms in Section 8.8„
eat ↔ 1
s− a and sin ωt ↔ ω
s2 + ω2 .
Theorem 8.2.1 with a= −5 and ω =
√
3 yields
L−1
( 8
s+ 5 + 7
s2 + 3
)
= 8 L−1
( 1
s+ 5
)
+ 7 L−1
( 1
s2 + 3
)
= 8 L−1
( 1
s+ 5
)
+ 7√
3 L−1
( √
3
s2 + 3
)
= 8 e−5t + 7√
3 sin
√
3t.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.2 The Inverse Laplace Transform 407
Example 8.2.3 Find
L−1
( 3s+ 8
s2 + 2 s+ 5
)
.
Solution Completing the square in the denominator yields
3s+ 8
s2 + 2 s+ 5 = 3s+ 8
(s+ 1) 2 + 4 .
Because of the form of the denominator, we consider the trans form pairs
e−t cos 2 t↔ s+ 1
(s+ 1) 2 + 4 and e−t sin 2 t↔ 2
(s+ 1) 2 + 4 ,
and write
L−1
( 3s+ 8
(s+ 1) 2 + 4
)
= L−1
( 3s+ 3
(s+ 1) 2 + 4
)
+ L−1
( 5
(s+ 1) 2 + 4
)
= 3 L−1
( s+ 1
(s+ 1) 2 + 4
)
+ 5
2 L−1
( 2
(s+ 1) 2 + 4
)
= e−t (3 cos 2 t+ 5
2 sin 2t).
REM A RK : W e’ll often write inverse Laplace transforms of specific fu nctions without explicitly stating
how they are obtained. In such cases you should refer to the ta ble of Laplace transforms in Section 8.8.
Inverse Laplace Transforms of Rational Functions
Using the Laplace transform to solve differential equation s often requires finding the inverse transform
of a rational function
F(s) = P(s)
Q(s) ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Using the Laplace transform to solve differential equation s often requires finding the inverse transform
of a rational function
F(s) = P(s)
Q(s) ,
where P and Qare polynomials in swith no common factors. Since it can be shown that lims→∞ F(s) =
0 if F is a Laplace transform, we need only consider the case where d egree(P) <degree(Q). T o obtain
L−1(F), we find the partial fraction expansion of F, obtain inverse transforms of the individual terms in
the expansion from the table of Laplace transforms, and use t he linearity property of the inverse transform.
The next two examples illustrate this.
Example 8.2.4 Find the inverse Laplace transform of
F(s) = 3s+ 2
s2 − 3s+ 2 . (8.2.1)
Solution (M ETH O D 1) Factoring the denominator in ( 8.2.1) yields
F(s) = 3s+ 2
(s− 1)(s− 2) . (8.2.2)
The form for the partial fraction expansion is
3s+ 2
(s− 1)(s− 2) = A
s− 1 + B
s− 2 . (8.2.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"408 Chapter 8 Laplace Transforms
Multiplying this by (s− 1)(s− 2) yields
3s+ 2 = ( s− 2)A+ ( s− 1)B.
Setting s= 2 yields B= 8 and setting s= 1 yields A= −5. Therefore
F(s) = − 5
s− 1 + 8
s− 2
and
L−1(F) = −5L−1
( 1
s− 1
)
+ 8 L−1
( 1
s− 2
)
= −5et + 8 e2t.
Solution (M ETH O D 2) W e don’t really have to multiply ( 8.2.3) by (s− 1)(s− 2) to compute A and
B. W e can obtain Aby simply ignoring the factor s− 1 in the denominator of ( 8.2.2) and setting s = 1
elsewhere; thus,
A= 3s+ 2
s− 2
⏐
⏐
⏐
⏐
s=1
= 3 ·1 + 2
1 − 2 = −5. (8.2.4)
Similarly , we can obtain B by ignoring the factor s− 2 in the denominator of ( 8.2.2) and setting s = 2
elsewhere; thus,
B= 3s+ 2
s− 1
⏐
⏐
⏐
⏐
s=2
= 3 ·2 + 2
2 − 1 = 8 . (8.2.5)
T o justify this, we observe that multiplying ( 8.2.3) by s− 1 yields
3s+ 2
s− 2 = A+ ( s− 1) B
s− 2 ,
and setting s= 1 leads to ( 8.2.4). Similarly , multiplying ( 8.2.3) by s− 2 yields
3s+ 2
s− 1 = ( s− 2) A
s− 2 + B",Elementary Differential Equations with Boundary Value Problems.pdf
"3s+ 2
s− 2 = A+ ( s− 1) B
s− 2 ,
and setting s= 1 leads to ( 8.2.4). Similarly , multiplying ( 8.2.3) by s− 2 yields
3s+ 2
s− 1 = ( s− 2) A
s− 2 + B
and setting s = 2 leads to ( 8.2.5). (It isn’t necesary to write the last two equations. W e wrot e them only
to justify the shortcut procedure indicated in ( 8.2.4) and ( 8.2.5).)
The shortcut employed in the second solution of Example 8.2.4 is Heaviside’s method . The next theo-
rem states this method formally . For a proof and an extension of this theorem, see Exercise 10.
Theorem 8.2.2 Suppose
F(s) = P(s)
(s− s1)(s− s2) · · ·(s− sn ) , (8.2.6)
where s1, s2 , . . . ,s n are distinct and P is a polynomial of degree less than n. Then
F(s) = A1
s− s1
+ A2
s− s2
+ · · ·+ An
s− sn
,
where Ai can be computed from (8.2.6) by ignoring the factor s− si and setting s= si elsewhere.
Example 8.2.5 Find the inverse Laplace transform of
F(s) = 6 + ( s+ 1)( s2 − 5s+ 11)
s(s− 1)(s− 2)(s+ 1) . (8.2.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.2 The Inverse Laplace Transform 409
Solution The partial fraction expansion of ( 8.2.7) is of the form
F(s) = A
s + B
s− 1 + C
s− 2 + D
s+ 1 . (8.2.8)
T o find A, we ignore the factor sin the denominator of ( 8.2.7) and set s= 0 elsewhere. This yields
A= 6 + (1)(11)
(−1)(−2)(1) = 17
2 .
Similarly , the other coefficients are given by
B= 6 + (2)(7)
(1)(−1)(2) = −10,
C = 6 + 3(5)
2(1)(3) = 7
2 ,
and
D= 6
(−1)(−2)(−3) = −1.
Therefore
F(s) = 17
2
1
s − 10
s− 1 + 7
2
1
s− 2 − 1
s+ 1
and
L−1(F) = 17
2 L−1
(1
s
)
− 10L−1
( 1
s− 1
)
+ 7
2 L−1
( 1
s− 2
)
− L−1
( 1
s+ 1
)
= 17
2 − 10et + 7
2 e2t − e−t .
REM A RK : W e didn’t “multiply out” the numerator in ( 8.2.7) before computing the coefficients in ( 8.2.8),
since it wouldn’t simplify the computations.
Example 8.2.6 Find the inverse Laplace transform of
F(s) = 8 − (s+ 2)(4 s+ 10)
(s+ 1)( s+ 2) 2 . (8.2.9)
Solution The form for the partial fraction expansion is
F(s) = A
s+ 1 + B
s+ 2 + C
(s+ 2) 2 . (8.2.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"(s+ 1)( s+ 2) 2 . (8.2.9)
Solution The form for the partial fraction expansion is
F(s) = A
s+ 1 + B
s+ 2 + C
(s+ 2) 2 . (8.2.10)
Because of the repeated factor (s+ 2) 2 in ( 8.2.9), Heaviside’s method doesn’t work. Instead, we find a
common denominator in ( 8.2.10). This yields
F(s) = A(s+ 2) 2 + B(s+ 1)( s+ 2) + C(s+ 1)
(s+ 1)( s+ 2) 2 . (8.2.11)
If ( 8.2.9) and ( 8.2.11) are to be equivalent, then
A(s+ 2) 2 + B(s+ 1)( s+ 2) + C(s+ 1) = 8 − (s+ 2)(4 s+ 10) . (8.2.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"410 Chapter 8 Laplace Transforms
The two sides of this equation are polynomials of degree two. From a theorem of algebra, they will be
equal for all s if they are equal for any three distinct values of s. W e may determine A, B and C by
choosing convenient values of s.
The left side of ( 8.2.12) suggests that we take s= −2 to obtain C = −8, and s= −1 to obtain A= 2 .
W e can now choose any third value of s to determine B. T aking s = 0 yields 4A+ 2 B+ C = −12.
Since A= 2 and C = −8 this implies that B= −6. Therefore
F(s) = 2
s+ 1 − 6
s+ 2 − 8
(s+ 2) 2
and
L−1(F) = 2 L−1
( 1
s+ 1
)
− 6L−1
( 1
s+ 2
)
− 8L−1
( 1
(s+ 2) 2
)
= 2 e−t − 6e−2t − 8te−2t .
Example 8.2.7 Find the inverse Laplace transform of
F(s) = s2 − 5s+ 7
(s+ 2) 3 .
Solution The form for the partial fraction expansion is
F(s) = A
s+ 2 + B
(s+ 2) 2 + C
(s+ 2) 3 .
The easiest way to obtain A, B, and Cis to expand the numerator in powers of s+ 2 . This yields",Elementary Differential Equations with Boundary Value Problems.pdf
"F(s) = A
s+ 2 + B
(s+ 2) 2 + C
(s+ 2) 3 .
The easiest way to obtain A, B, and Cis to expand the numerator in powers of s+ 2 . This yields
s2 − 5s+ 7 = [( s+ 2) − 2]2 − 5[(s+ 2) − 2] + 7 = ( s+ 2) 2 − 9(s+ 2) + 21 .
Therefore
F(s) = (s+ 2) 2 − 9(s+ 2) + 21
(s+ 2) 3
= 1
s+ 2 − 9
(s+ 2) 2 + 21
(s+ 2) 3
and
L−1(F) = L−1
( 1
s+ 2
)
− 9L−1
( 1
(s+ 2) 2
)
+ 21
2 L−1
( 2
(s+ 2) 3
)
= e−2t
(
1 − 9t+ 21
2 t2
)
.
Example 8.2.8 Find the inverse Laplace transform of
F(s) = 1 − s(5 + 3 s)
s[(s+ 1) 2 + 1] . (8.2.13)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.2 The Inverse Laplace Transform 411
Solution One form for the partial fraction expansion of F is
F(s) = A
s + Bs+ C
(s+ 1) 2 + 1 . (8.2.14)
However, we see from the table of Laplace transforms that the inverse transform of the second fraction
on the right of ( 8.2.14) will be a linear combination of the inverse transforms
e−t cos t and e−t sin t
of s+ 1
(s+ 1) 2 + 1 and 1
(s+ 1) 2 + 1
respectively . Therefore, instead of ( 8.2.14) we write
F(s) = A
s + B(s+ 1) + C
(s+ 1) 2 + 1 . (8.2.15)
Finding a common denominator yields
F(s) = A
[
(s+ 1) 2 + 1
]
+ B(s+ 1) s+ Cs
s[(s+ 1) 2 + 1] . (8.2.16)
If ( 8.2.13) and ( 8.2.16) are to be equivalent, then
A[ (s+ 1) 2 + 1 ] + B(s+ 1) s+ Cs = 1 − s(5 + 3 s).
This is true for all sif it’s true for three distinct values of s. Choosing s= 0 , −1, and 1 yields the system
2A = 1
A− C = 3
5A+ 2 B+ C = −7.
Solving this system yields
A= 1
2 , B = −7
2 , C = −5
2 .
Hence, from ( 8.2.15),
F(s) = 1
2s − 7
2
s+ 1
(s+ 1) 2 + 1 − 5
2
1
(s+ 1) 2 + 1 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Solving this system yields
A= 1
2 , B = −7
2 , C = −5
2 .
Hence, from ( 8.2.15),
F(s) = 1
2s − 7
2
s+ 1
(s+ 1) 2 + 1 − 5
2
1
(s+ 1) 2 + 1 .
Therefore
L−1(F) = 1
2 L−1
(1
s
)
− 7
2 L−1
( s+ 1
(s+ 1) 2 + 1
)
− 5
2 L−1
( 1
(s+ 1) 2 + 1
)
= 1
2 − 7
2 e−t cos t− 5
2 e−t sin t.
Example 8.2.9 Find the inverse Laplace transform of
F(s) = 8 + 3 s
(s2 + 1)( s2 + 4) . (8.2.17)",Elementary Differential Equations with Boundary Value Problems.pdf
"412 Chapter 8 Laplace Transforms
Solution The form for the partial fraction expansion is
F(s) = A+ Bs
s2 + 1 + C+ Ds
s2 + 4 .
The coefficients A, B, C and D can be obtained by finding a common denominator and equating t he
resulting numerator to the numerator in ( 8.2.17). However, since there’s no first power of sin the denom-
inator of ( 8.2.17), there’s an easier way: the expansion of
F1(s) = 1
(s2 + 1)( s2 + 4)
can be obtained quickly by using Heaviside’s method to expan d
1
(x+ 1)( x+ 4) = 1
3
( 1
x+ 1 − 1
x+ 4
)
and then setting x= s2 to obtain
1
(s2 + 1)( s2 + 4) = 1
3
( 1
s2 + 1 − 1
s2 + 4
)
.
Multiplying this by 8 + 3 syields
F(s) = 8 + 3 s
(s2 + 1)( s2 + 4) = 1
3
(8 + 3 s
s2 + 1 − 8 + 3 s
s2 + 4
)
.
Therefore
L−1(F) = 8
3 sin t+ cos t− 4
3 sin 2t− cos 2t.
USING TECHNOLOGY
Some software packages that do symbolic algebra can find part ial fraction expansions very easily . W e
recommend that you use such a package if one is available to yo u, but only after you’ve done enough",Elementary Differential Equations with Boundary Value Problems.pdf
"recommend that you use such a package if one is available to yo u, but only after you’ve done enough
partial fraction expansions on your own to master the techni que.
8.2 Exercises
1. Use the table of Laplace transforms to find the inverse Laplac e transform.
(a) 3
(s− 7)4 (b) 2s− 4
s2 − 4s+ 13 (c) 1
s2 + 4 s+ 20
(d) 2
s2 + 9 (e) s2 − 1
(s2 + 1) 2 (f) 1
(s− 2)2 − 4
(g) 12s− 24
(s2 − 4s+ 85) 2 (h) 2
(s− 3)2 − 9 (i) s2 − 4s+ 3
(s2 − 4s+ 5) 2
2. Use Theorem 8.2.1 and the table of Laplace transforms to find the inverse Laplac e transform.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.2 The Inverse Laplace Transform 413
(a) 2s+ 3
(s− 7)4 (b) s2 − 1
(s− 2)6 (c) s+ 5
s2 + 6 s+ 18
(d) 2s+ 1
s2 + 9 (e) s
s2 + 2 s+ 1 (f) s+ 1
s2 − 9
(g) s3 + 2 s2 − s− 3
(s+ 1) 4 (h) 2s+ 3
(s− 1)2 + 4 (i) 1
s − s
s2 + 1
(j) 3s+ 4
s2 − 1 (k) 3
s− 1 + 4s+ 1
s2 + 9 (l) 3
(s+ 2) 2 − 2s+ 6
s2 + 4
3. Use Heaviside’s method to find the inverse Laplace transform .
(a) 3 − (s+ 1)( s− 2)
(s+ 1)( s+ 2)( s− 2) (b) 7 + ( s+ 4)(18 − 3s)
(s− 3)(s− 1)(s+ 4)
(c) 2 + ( s− 2)(3 − 2s)
(s− 2)(s+ 2)( s− 3) (d) 3 − (s− 1)(s+ 1)
(s+ 4)( s− 2)(s− 1)
(e) 3 + ( s− 2)(10 − 2s− s2)
(s− 2)(s+ 2)( s− 1)(s+ 3) (f) 3 + ( s− 3)(2s2 + s− 21)
(s− 3)(s− 1)(s+ 4)( s− 2)
4. Find the inverse Laplace transform.
(a) 2 + 3 s
(s2 + 1)( s+ 2)( s+ 1) (b) 3s2 + 2 s+ 1
(s2 + 1)( s2 + 2 s+ 2)
(c) 3s+ 2
(s− 2)(s2 + 2 s+ 5) (d) 3s2 + 2 s+ 1
(s− 1)2(s+ 2)( s+ 3)
(e) 2s2 + s+ 3
(s− 1)2(s+ 2) 2 (f) 3s+ 2
(s2 + 1)( s− 1)2
5. Use the method of Example 8.2.9 to find the inverse Laplace transform.
(a) 3s+ 2
(s2 + 4)( s2 + 9) (b) −4s+ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"(s2 + 1)( s− 1)2
5. Use the method of Example 8.2.9 to find the inverse Laplace transform.
(a) 3s+ 2
(s2 + 4)( s2 + 9) (b) −4s+ 1
(s2 + 1)( s2 + 16) (c) 5s+ 3
(s2 + 1)( s2 + 4)
(d) −s+ 1
(4s2 + 1)( s2 + 1) (e) 17s− 34
(s2 + 16)(16 s2 + 1) (f) 2s− 1
(4s2 + 1)(9 s2 + 1)
6. Find the inverse Laplace transform.
(a) 17s− 15
(s2 − 2s+ 5)( s2 + 2 s+ 10) (b) 8s+ 56
(s2 − 6s+ 13)( s2 + 2 s+ 5)
(c) s+ 9
(s2 + 4 s+ 5)( s2 − 4s+ 13) (d) 3s− 2
(s2 − 4s+ 5)( s2 − 6s+ 13)
(e) 3s− 1
(s2 − 2s+ 2)( s2 + 2 s+ 5) (f) 20s+ 40
(4s2 − 4s+ 5)(4 s2 + 4 s+ 5)
7. Find the inverse Laplace transform.
(a) 1
s(s2 + 1) (b) 1
(s− 1)(s2 − 2s+ 17)
(c) 3s+ 2
(s− 2)(s2 + 2 s+ 10) (d) 34 − 17s
(2s− 1)(s2 − 2s+ 5)
(e) s+ 2
(s− 3)(s2 + 2 s+ 5) (f) 2s− 2
(s− 2)(s2 + 2 s+ 10)
8. Find the inverse Laplace transform.",Elementary Differential Equations with Boundary Value Problems.pdf
"414 Chapter 8 Laplace Transforms
(a) 2s+ 1
(s2 + 1)( s− 1)(s− 3) (b) s+ 2
(s2 + 2 s+ 2)( s2 − 1)
(c) 2s− 1
(s2 − 2s+ 2)( s+ 1)( s− 2) (d) s− 6
(s2 − 1)(s2 + 4)
(e) 2s− 3
s(s− 2)(s2 − 2s+ 5) (f) 5s− 15
(s2 − 4s+ 13)( s− 2)(s− 1)
9. Given that f(t) ↔ F(s), find the inverse Laplace transform of F(as− b), where a> 0.
10. (a) If s1, s2, . . . , sn are distinct and P is a polynomial of degree less than n, then
P(s)
(s− s1)(s− s2) · · ·(s− sn ) = A1
s− s1
+ A2
s− s2
+ · · ·+ An
s− sn
.
Multiply through by s− si to show that Ai can be obtained by ignoring the factor s− si on
the left and setting s= si elsewhere.
(b) Suppose P and Q1 are polynomials such that degree (P) ≤ degree(Q1) and Q1(s1) ̸= 0 .
Show that the coefficient of 1/ (s− s1) in the partial fraction expansion of
F(s) = P(s)
(s− s1 )Q1(s)
is P(s1)/Q 1(s1).
(c) Explain how the results of (a) and (b) are related.
8.3 SOLUTION OF INITIAL V ALUE PROBLEMS
Laplace Transforms of Derivatives",Elementary Differential Equations with Boundary Value Problems.pdf
"is P(s1)/Q 1(s1).
(c) Explain how the results of (a) and (b) are related.
8.3 SOLUTION OF INITIAL V ALUE PROBLEMS
Laplace Transforms of Derivatives
In the rest of this chapter we’ll use the Laplace transform to solve initial value problems for constant
coefficient second order equations. T o do this, we must know h ow the Laplace transform of f′is related
to the Laplace transform of f. The next theorem answers this question.
Theorem 8.3.1 Suppose f is continuous on [0, ∞ ) and of exponential order s0, and f′is piecewise
continuous on [0, ∞ ). Then f and f′have Laplace transforms for s>s 0, and
L(f′) = sL(f) − f(0). (8.3.1)
Proof
W e know from Theorem 8.1.6 that L(f) is defined for s > s0. W e first consider the case where f′is
continuous on [0, ∞ ). Integration by parts yields
∫ T
0
e−stf′(t) dt = e−st f(t)
⏐
⏐
⏐
T
0
+ s
∫ T
0
e−st f(t) dt
= e−sT f(T) − f(0) + s
∫ T
0
e−st f(t) dt
(8.3.2)
for any T >0. Since f is of exponential order s0, limT →∞ e−sT f(T) = 0 and the last integral in (",Elementary Differential Equations with Boundary Value Problems.pdf
"= e−sT f(T) − f(0) + s
∫ T
0
e−st f(t) dt
(8.3.2)
for any T >0. Since f is of exponential order s0, limT →∞ e−sT f(T) = 0 and the last integral in (
8.3.2)
converges as T → ∞ if s>s 0. Therefore
∫ ∞
0
e−st f′(t) dt = −f(0) + s
∫ ∞
0
e−st f(t) dt
= −f(0) + sL(f),",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.3 Solution of Initial V alue Problems 415
which proves ( 8.3.1). Now suppose T >0 and f′is only piecewise continuous on [0,T ], with discon-
tinuities at t1 < t2 < · · ·< tn−1. For convenience, let t0 = 0 and tn = T. Integrating by parts
yields
∫ ti
ti−1
e−st f′(t) dt = e−st f(t)
⏐
⏐
⏐
ti
ti−1
+ s
∫ ti
ti−1
e−st f(t) dt
= e−sti f(ti ) − e−sti−1 f(ti−1) + s
∫ ti
ti−1
e−stf(t) dt.
Summing both sides of this equation from i= 1 to nand noting that
(
e−st1 f(t1) − e−st0 f(t0)
)
+
(
e−st2 f(t2) − e−st1 f(t1 )
)
+ · · ·+
(
e−stN f(tN ) − e−stN−1 f(tN −1 )
)
= e−stN f(tN ) − e−st0 f(t0 ) = e−sT f(T) − f(0)
yields (
8.3.2), so ( 8.3.1) follows as before.
Example 8.3.1 In Example 8.1.4 we saw that
L(cos ωt) = s
s2 + ω2 .
Applying ( 8.3.1) with f(t) = cos ωt shows that
L(−ω sin ωt) = s s
s2 + ω2 − 1 = − ω2
s2 + ω2 .
Therefore
L(sin ωt) = ω
s2 + ω2 ,
which agrees with the corresponding result obtained in 8.1.4.
In Section 2.1 we showed that the solution of the initial valu e problem",Elementary Differential Equations with Boundary Value Problems.pdf
"s2 + ω2 ,
which agrees with the corresponding result obtained in 8.1.4.
In Section 2.1 we showed that the solution of the initial valu e problem
y′= ay, y (0) = y0, (8.3.3)
is y = y0eat . W e’ll now obtain this result by using the Laplace transform .
Let Y(s) = L(y) be the Laplace transform of the unknown solution of ( 8.3.3). T aking Laplace trans-
forms of both sides of ( 8.3.3) yields
L(y′) = L(ay),
which, by Theorem 8.3.1, can be rewritten as
sL(y) − y(0) = aL(y),
or
sY(s) − y0 = aY(s).
Solving for Y(s) yields
Y(s) = y0
s− a,
so
y= L−1(Y(s)) = L−1
( y0
s− a
)
= y0L−1
( 1
s− a
)
= y0eat ,
which agrees with the known result.
W e need the next theorem to solve second order differential e quations using the Laplace transform.",Elementary Differential Equations with Boundary Value Problems.pdf
"416 Chapter 8 Laplace Transforms
Theorem 8.3.2 Suppose f and f′are continuous on [0, ∞ ) and of exponential order s0, and that f′′is
piecewise continuous on [0, ∞ ). Then f, f′, and f′′have Laplace transforms for s>s 0,
L(f′) = sL(f) − f(0), (8.3.4)
and
L(f′′) = s2L(f) − f′(0) − sf(0). (8.3.5)
Proof Theorem 8.3.1 implies that L(f′) exists and satisfies ( 8.3.4) for s > s0. T o prove that L(f′′)
exists and satisfies ( 8.3.5) for s > s0, we first apply Theorem 8.3.1 to g = f′. Since g satisfies the
hypotheses of Theorem 8.3.1, we conclude that L(g′) is defined and satisfies
L(g′) = sL(g) − g(0)
for s>s 0. However, since g′= f′′, this can be rewritten as
L(f′′) = sL(f′) − f′(0).
Substituting ( 8.3.4) into this yields ( 8.3.5).
Solving Second Order Equations with the Laplace Transform
W e’ll now use the Laplace transform to solve initial value pr oblems for second order equations.
Example 8.3.2 Use the Laplace transform to solve the initial value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 8.3.2 Use the Laplace transform to solve the initial value problem
y′′− 6y′+ 5 y= 3 e2t, y (0) = 2 , y ′(0) = 3 . (8.3.6)
Solution T aking Laplace transforms of both sides of the differential equation in ( 8.3.6) yields
L(y′′− 6y′+ 5 y) = L(3e2t) = 3
s− 2 ,
which we rewrite as
L(y′′) − 6L(y′) + 5 L(y) = 3
s− 2 . (8.3.7)
Now denote L(y) = Y(s). Theorem 8.3.2 and the initial conditions in ( 8.3.6) imply that
L(y′) = sY(s) − y(0) = sY(s) − 2
and
L(y′′) = s2Y(s) − y′(0) − sy(0) = s2Y(s) − 3 − 2s.
Substituting from the last two equations into ( 8.3.7) yields
(s2Y(s) − 3 − 2s) − 6 ( sY(s) − 2) + 5 Y(s) = 3
s− 2 .
Therefore
(s2 − 6s+ 5) Y(s) = 3
s− 2 + (3 + 2 s) + 6( −2), (8.3.8)
so
(s− 5)(s− 1)Y(s) = 3 + ( s− 2)(2s− 9)
s− 2 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.3 Solution of Initial V alue Problems 417
and
Y(s) = 3 + ( s− 2)(2s− 9)
(s− 2)(s− 5)(s− 1) .
Heaviside’s method yields the partial fraction expansion
Y(s) = − 1
s− 2 + 1
2
1
s− 5 + 5
2
1
s− 1 ,
and taking the inverse transform of this yields
y= −e2t + 1
2 e5t + 5
2 et
as the solution of ( 8.3.6).
It isn’t necessary to write all the steps that we used to obtai n ( 8.3.8). T o see how to avoid this, let’s
apply the method of Example 8.3.2 to the general initial value problem
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1. (8.3.9)
T aking Laplace transforms of both sides of the differential equation in ( 8.3.9) yields
aL(y′′) + bL(y′) + cL(y) = F(s). (8.3.10)
Now let Y(s) = L(y). Theorem 8.3.2 and the initial conditions in ( 8.3.9) imply that
L(y′) = sY(s) − k0 and L(y′′) = s2Y(s) − k1 − k0s.
Substituting these into ( 8.3.10) yields
a
(
s2Y(s) − k1 − k0s
)
+ b(sY(s) − k0) + cY(s) = F(s). (8.3.11)
The coefficient of Y(s) on the left is the characteristic polynomial",Elementary Differential Equations with Boundary Value Problems.pdf
"a
(
s2Y(s) − k1 − k0s
)
+ b(sY(s) − k0) + cY(s) = F(s). (8.3.11)
The coefficient of Y(s) on the left is the characteristic polynomial
p(s) = as2 + bs+ c
of the complementary equation for ( 8.3.9). Using this and moving the terms involving k0 and k1 to the
right side of ( 8.3.11) yields
p(s)Y(s) = F(s) + a(k1 + k0s) + bk0. (8.3.12)
This equation corresponds to ( 8.3.8) of Example 8.3.2. Having established the form of this equation in
the general case, it is preferable to go directly from the ini tial value problem to this equation. Y ou may
find it easier to remember ( 8.3.12) rewritten as
p(s)Y(s) = F(s) + a(y′(0) + sy(0)) + by(0). (8.3.13)
Example 8.3.3 Use the Laplace transform to solve the initial value problem
2y′′+ 3 y′+ y = 8 e−2t, y (0) = −4, y′(0) = 2 . (8.3.14)
Solution The characteristic polynomial is
p(s) = 2 s2 + 3 s+ 1 = (2 s+ 1)( s+ 1)",Elementary Differential Equations with Boundary Value Problems.pdf
"418 Chapter 8 Laplace Transforms
and
F(s) = L(8e−2t) = 8
s+ 2 ,
so ( 8.3.13) becomes
(2s+ 1)( s+ 1) Y(s) = 8
s+ 2 + 2(2 − 4s) + 3( −4).
Solving for Y(s) yields
Y(s) = 4 (1 − (s+ 2)( s+ 1))
(s+ 1 / 2)(s+ 1)( s+ 2) .
Heaviside’s method yields the partial fraction expansion
Y(s) = 4
3
1
s+ 1 / 2 − 8
s+ 1 + 8
3
1
s+ 2 ,
so the solution of ( 8.3.14) is
y = L−1(Y(s)) = 4
3 e−t/ 2 − 8e−t + 8
3 e−2t
(Figure 8.3.1).
1 2 3 4 5 6 7
−1
−2
−3
−4
 t
 y
Figure 8.3.1 y = 4
3 e−t/ 2 − 8e−t + 8
3 e−2t
1 2 3 4 5 6 7
1
−1
−2
−3
−4
.5
 t
 y
Figure 8.3.2 y = 1
2 − 7
2 e−t cos t− 5
2 e−t sin t
Example 8.3.4 Solve the initial value problem
y′′+ 2 y′+ 2 y= 1 , y (0) = −3, y′(0) = 1 . (8.3.15)
Solution The characteristic polynomial is
p(s) = s2 + 2 s+ 2 = ( s+ 1) 2 + 1
and
F(s) = L(1) = 1
s,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.3 Solution of Initial V alue Problems 419
so ( 8.3.13) becomes [
(s+ 1) 2 + 1
]
Y(s) = 1
s + 1 ·(1 − 3s) + 2( −3).
Solving for Y(s) yields
Y(s) = 1 − s(5 + 3 s)
s[(s+ 1) 2 + 1] .
In Example 8.2.8 we found the inverse transform of this function to be
y= 1
2 − 7
2 e−t cos t− 5
2 e−t sin t
(Figure 8.3.2), which is therefore the solution of ( 8.3.15).
REM A RK : In our examples we applied Theorems 8.3.1 and 8.3.2 without verifying that the unknown
function y satisfies their hypotheses. This is characteristic of the fo rmal manipulative way in which the
Laplace transform is used to solve differential equations. Any doubts about the validity of the method for
solving a given equation can be resolved by verifying that th e resulting function y is the solution of the
given problem.
8.3 Exercises
In Exercises 1– 31 use the Laplace transform to solve the initial value problem .
1. y′′+ 3 y′+ 2 y= et , y (0) = 1 , y ′(0) = −6
2. y′′− y′− 6y= 2 , y (0) = 1 , y ′(0) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"1. y′′+ 3 y′+ 2 y= et , y (0) = 1 , y ′(0) = −6
2. y′′− y′− 6y= 2 , y (0) = 1 , y ′(0) = 0
3. y′′+ y′− 2y= 2 e3t, y (0) = −1, y ′(0) = 4
4. y′′− 4y= 2 e3t , y (0) = 1 , y ′(0) = −1
5. y′′+ y′− 2y= e3t, y (0) = 1 , y ′(0) = −1
6. y′′+ 3 y′+ 2 y= 6 et , y (0) = 1 , y ′(0) = −1
7. y′′+ y = sin 2 t, y (0) = 0 , y ′(0) = 1
8. y′′− 3y′+ 2 y= 2 e3t , y (0) = 1 , y ′(0) = −1
9. y′′− 3y′+ 2 y= e4t , y (0) = 1 , y ′(0) = −2
10. y′′− 3y′+ 2 y= e3t , y (0) = −1, y ′(0) = −4
11. y′′+ 3 y′+ 2 y= 2 et , y (0) = 0 , y ′(0) = −1
12. y′′+ y′− 2y= −4, y (0) = 2 , y ′(0) = 3
13. y′′+ 4 y= 4 , y (0) = 0 , y ′(0) = 1
14. y′′− y′− 6y= 2 , y (0) = 1 , y ′(0) = 0
15. y′′+ 3 y′+ 2 y= et , y (0) = 0 , y ′(0) = 1
16. y′′− y = 1 , y (0) = 1 , y ′(0) = 0
17. y′′+ 4 y= 3 sin t, y (0) = 1 , y ′(0) = −1
18. y′′+ y′= 2 e3t, y (0) = −1, y ′(0) = 4
19. y′′+ y = 1 , y (0) = 2 , y ′(0) = 0
20. y′′+ y = t, y (0) = 0 , y ′(0) = 2",Elementary Differential Equations with Boundary Value Problems.pdf
"420 Chapter 8 Laplace Transforms
21. y′′+ y = t− 3 sin 2 t, y (0) = 1 , y ′(0) = −3
22. y′′+ 5 y′+ 6 y= 2 e−t , y (0) = 1 , y ′(0) = 3
23. y′′+ 2 y′+ y = 6 sin t− 4 cos t, y (0) = −1, y′(0) = 1
24. y′′− 2y′− 3y= 10 cos t, y (0) = 2 , y ′(0) = 7
25. y′′+ y = 4 sin t+ 6 cos t, y (0) = −6, y′(0) = 2
26. y′′+ 4 y= 8 sin 2 t+ 9 cos t, y (0) = 1 , y ′(0) = 0
27. y′′− 5y′+ 6 y= 10 et cos t, y (0) = 2 , y ′(0) = 1
28. y′′+ 2 y′+ 2 y= 2 t, y (0) = 2 , y ′(0) = −7
29. y′′− 2y′+ 2 y= 5 sin t+ 10 cos t, y (0) = 1 , y′(0) = 2
30. y′′+ 4 y′+ 13 y= 10 e−t − 36et , y (0) = 0 , y′(0) = −16
31. y′′+ 4 y′+ 5 y= e−t (cos t+ 3 sin t), y (0) = 0 , y ′(0) = 4
32. 2y′′− 3y′− 2y= 4 et , y (0) = 1 , y′(0) = −2
33. 6y′′− y′− y = 3 e2t, y (0) = 0 , y′(0) = 0
34. 2y′′+ 2 y′+ y= 2 t, y (0) = 1 , y′(0) = −1
35. 4y′′− 4y′+ 5 y= 4 sin t− 4 cos t, y (0) = 0 , y′(0) = 11 / 17
36. 4y′′+ 4 y′+ y= 3 sin t+ cos t, y (0) = 2 , y′(0) = −1
37. 9y′′+ 6 y′+ y= 3 e3t , y (0) = 0 , y′(0) = −3",Elementary Differential Equations with Boundary Value Problems.pdf
"36. 4y′′+ 4 y′+ y= 3 sin t+ cos t, y (0) = 2 , y′(0) = −1
37. 9y′′+ 6 y′+ y= 3 e3t , y (0) = 0 , y′(0) = −3
38. Suppose a,b , and care constants and a̸= 0 . Let
y1 = L−1
( as+ b
as2 + bs+ c
)
and y2 = L−1
( a
as2 + bs+ c
)
.
Show that
y1(0) = 1 , y ′
1(0) = 0 and y2(0) = 0 , y ′
2(0) = 1 .
HIN T : Use the Laplace transform to solve the initial value problem s
ay′′+ by′+ cy = 0 , y (0) = 1 , y ′(0) = 0
ay′′+ by′+ cy = 0 , y (0) = 0 , y ′(0) = 1 .
8.4 THE UNIT STEP FUNCTION
In the next section we’ll consider initial value problems
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1,
where a, b, and care constants and f is piecewise continuous. In this section we’ll develop proc edures
for using the table of Laplace transforms to find Laplace tran sforms of piecewise continuous functions,
and to find the piecewise continuous inverses of Laplace tran sforms.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.4 The Unit Step Function 421
Example 8.4.1 Use the table of Laplace transforms to find the Laplace transf orm of
f(t) =
{
2t+ 1 , 0 ≤ t< 2,
3t, t ≥ 2 (8.4.1)
(Figure 8.4.1).
Solution Since the formula for f changes at t= 2 , we write
L(f) =
∫ ∞
0
e−st f(t) dt
=
∫ 2
0
e−st(2t+ 1) dt+
∫ ∞
2
e−st(3t) dt.
(8.4.2)
T o relate the first term to a Laplace transform, we add and subt ract
∫ ∞
2
e−st (2t+ 1) dt
in ( 8.4.2) to obtain
L(f) =
∫ ∞
0
e−st(2t+ 1) dt+
∫ ∞
2
e−st(3t− 2t− 1) dt
=
∫ ∞
0
e−st(2t+ 1) dt+
∫ ∞
2
e−st(t− 1) dt
= L(2t+ 1) +
∫ ∞
2
e−st(t− 1) dt.
(8.4.3)
T o relate the last integral to a Laplace transform, we make th e change of variable x= t− 2 and rewrite
the integral as
∫ ∞
2
e−st (t− 1) dt =
∫ ∞
0
e−s(x+2)(x+ 1) dx
= e−2s
∫ ∞
0
e−sx (x+ 1) dx.
Since the symbol used for the variable of integration has no e ffect on the value of a definite integral, we
can now replace xby the more standard tand write
∫ ∞
2
e−st(t− 1) dt= e−2s
∫ ∞
0
e−st (t+ 1) dt= e−2sL(t+ 1) .",Elementary Differential Equations with Boundary Value Problems.pdf
"can now replace xby the more standard tand write
∫ ∞
2
e−st(t− 1) dt= e−2s
∫ ∞
0
e−st (t+ 1) dt= e−2sL(t+ 1) .
This and ( 8.4.3) imply that
L(f) = L(2t+ 1) + e−2sL(t+ 1) .
Now we can use the table of Laplace transforms to find that
L(f) = 2
s2 + 1
s + e−2s
(1
s2 + 1
s
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"422 Chapter 8 Laplace Transforms
1 2 3 4
1
2
3
4
5
6
7
8
9
10
11
12
 y
 t
Figure 8.4.1 The piecewise continuous function
(8.4.1)
1
 τ  t
 y
Figure 8.4.2 y= u(t− τ)
Laplace Transforms of Piecewise Continuous Functions
W e’ll now develop the method of Example 8.4.1 into a systematic way to find the Laplace transform of a
piecewise continuous function. It is convenient to introdu ce the unit step function , defined as
u(t) =
{ 0, t< 0
1, t ≥ 0. (8.4.4)
Thus, u(t) “steps” from the constant value 0 to the constant value 1 at t= 0 . If we replace tby t− τ in
(8.4.4), then
u(t− τ) =
{ 0, t<τ,
1, t ≥ τ ;
that is, the step now occurs at t= τ (Figure 8.4.2).
The step function enables us to represent piecewise continu ous functions conveniently . For example,
consider the function
f(t) =
{
f0(t), 0 ≤ t<t 1,
f1(t), t ≥ t1, (8.4.5)
where we assume that f0 and f1 are defined on [0, ∞ ), even though they equal f only on the indicated
intervals. This assumption enables us to rewrite ( 8.4.5) as",Elementary Differential Equations with Boundary Value Problems.pdf
"intervals. This assumption enables us to rewrite ( 8.4.5) as
f(t) = f0(t) + u(t− t1) (f1(t) − f0(t)) . (8.4.6)
T o verify this, note that if t<t 1 then u(t− t1) = 0 and ( 8.4.6) becomes
f(t) = f0(t) + (0) ( f1(t) − f0(t)) = f0(t).
If t≥ t1 then u(t− t1) = 1 and ( 8.4.6) becomes
f(t) = f0(t) + (1) ( f1(t) − f0(t)) = f1(t).
W e need the next theorem to show how ( 8.4.6) can be used to find L(f).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.4 The Unit Step Function 423
Theorem 8.4.1 Let g be defined on [0, ∞ ). Suppose τ ≥ 0 and L(g(t+ τ)) exists for s > s0. Then
L(u(t− τ)g(t)) exists for s>s 0, and
L(u(t− τ)g(t)) = e−sτ L(g(t+ τ)) .
Proof By definition,
L(u(t− τ)g(t)) =
∫ ∞
0
e−stu(t− τ)g(t) dt.
From this and the definition of u(t− τ),
L(u(t− τ)g(t)) =
∫ τ
0
e−st (0) dt+
∫ ∞
τ
e−stg(t) dt.
The first integral on the right equals zero. Introducing the n ew variable of integration x = t− τ in the
second integral yields
L(u(t− τ)g(t)) =
∫ ∞
0
e−s(x+τ ) g(x+ τ) dx= e−sτ
∫ ∞
0
e−sx g(x+ τ) dx.
Changing the name of the variable of integration in the last i ntegral from xto tyields
L(u(t− τ)g(t)) = e−sτ
∫ ∞
0
e−st g(t+ τ) dt= e−sτ L(g(t+ τ)).
Example 8.4.2 Find
L(u(t− 1)(t2 + 1) ) .
Solution Here τ = 1 and g(t) = t2 + 1 , so
g(t+ 1) = ( t+ 1) 2 + 1 = t2 + 2 t+ 2 .
Since
L(g(t+ 1)) = 2
s3 + 2
s2 + 2
s,
Theorem 8.4.1 implies that
L
(
u(t− 1)(t2 + 1)
)
= e−s
(2
s3 + 2
s2 + 2
s
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Since
L(g(t+ 1)) = 2
s3 + 2
s2 + 2
s,
Theorem 8.4.1 implies that
L
(
u(t− 1)(t2 + 1)
)
= e−s
(2
s3 + 2
s2 + 2
s
)
.
Example 8.4.3 Use Theorem 8.4.1 to find the Laplace transform of the function
f(t) =
{
2t+ 1 , 0 ≤ t< 2,
3t, t ≥ 2,
from Example 8.4.1.
Solution W e first write f in the form ( 8.4.6) as
f(t) = 2 t+ 1 + u(t− 2)(t− 1).",Elementary Differential Equations with Boundary Value Problems.pdf
"424 Chapter 8 Laplace Transforms
Therefore
L(f) = L(2t+ 1) + L(u(t− 2)(t− 1))
= L(2t+ 1) + e−2sL(t+ 1) (from Theorem 8.4.1)
= 2
s2 + 1
s + e−2s
(1
s2 + 1
s
)
,
which is the result obtained in Example 8.4.1.
Formula ( 8.4.6) can be extended to more general piecewise continuous funct ions. For example, we can
write
f(t) =





f0(t), 0 ≤ t<t 1,
f1(t), t 1 ≤ t<t 2,
f2(t), t ≥ t2,
as
f(t) = f0(t) + u(t− t1) (f1(t) − f0(t)) + u(t− t2) (f2(t) − f1(t))
if f0, f1, and f2 are all defined on [0, ∞ ).
Example 8.4.4 Find the Laplace transform of
f(t) =









1, 0 ≤ t< 2,
−2t+ 1 , 2 ≤ t< 3,
3t, 3 ≤ t< 5,
t− 1, t ≥ 5
(8.4.7)
(Figure
8.4.3).
Solution In terms of step functions,
f(t) = 1 + u(t− 2)(−2t+ 1 − 1) + u(t− 3)(3t+ 2 t− 1)
+u(t− 5)(t− 1 − 3t),
or
f(t) = 1 − 2u(t− 2)t+ u(t− 3)(5t− 1) − u(t− 5)(2t+ 1) .
Now Theorem 8.4.1 implies that
L(f) = L(1) − 2e−2sL(t+ 2) + e−3sL(5(t+ 3) − 1) − e−5sL(2(t+ 5) + 1)
= L(1) − 2e−2sL(t+ 2) + e−3sL(5t+ 14) − e−5sL(2t+ 11)
= 1
s − 2e−2s
(1
s2 + 2
s",Elementary Differential Equations with Boundary Value Problems.pdf
"L(f) = L(1) − 2e−2sL(t+ 2) + e−3sL(5(t+ 3) − 1) − e−5sL(2(t+ 5) + 1)
= L(1) − 2e−2sL(t+ 2) + e−3sL(5t+ 14) − e−5sL(2t+ 11)
= 1
s − 2e−2s
(1
s2 + 2
s
)
+ e−3s
(5
s2 + 14
s
)
− e−5s
(2
s2 + 11
s
)
.
The trigonometric identities
sin(A+ B) = sin Acos B+ cos Asin B (8.4.8)
cos(A+ B) = cos Acos B− sin Asin B (8.4.9)
are useful in problems that involve shifting the arguments o f trigonometric functions. W e’ll use these
identities in the next example.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.4 The Unit Step Function 425
1 2 3 4 5 6
2
4
6
8
10
12
14
16
−6
−4
−2
 t
 y
Figure 8.4.3 The piecewise contnuous function ( 8.4.7)
Example 8.4.5 Find the Laplace transform of
f(t) =









sin t, 0 ≤ t< π
2 ,
cos t− 3 sin t, π
2 ≤ t<π,
3 cos t, t ≥ π
(8.4.10)
(Figure 8.4.4).
Solution In terms of step functions,
f(t) = sin t+ u(t− π/ 2)(cos t− 4 sin t) + u(t− π)(2 cos t+ 3 sin t).
Now Theorem 8.4.1 implies that
L(f) = L(sin t) + e− π
2 sL
(
cos
(
t+ π
2
)
− 4 sin
(
t+ π
2
))
+e−πs L(2 cos(t+ π) + 3 sin( t+ π)) . (8.4.11)
Since
cos
(
t+ π
2
)
− 4 sin
(
t+ π
2
)
= − sin t− 4 cos t
and
2 cos(t+ π) + 3 sin( t+ π) = −2 cos t− 3 sin t,
we see from ( 8.4.11) that
L(f) = L(sin t) − e−πs/ 2L(sin t+ 4 cos t) − e−πs L(2 cos t+ 3 sin t)
= 1
s2 + 1 − e− π
2 s
(1 + 4 s
s2 + 1
)
− e−πs
(3 + 2 s
s2 + 1
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"426 Chapter 8 Laplace Transforms
1 2 3 4 5 6
1
−1
2
−2
3
−3
 t
 y
Figure 8.4.4 The piecewise continuous function ( 8.4.10)
The Second Shifting Theorem
Replacing g(t) by g(t− τ) in Theorem 8.4.1 yields the next theorem.
Theorem 8.4.2 [Second Shifting Theorem ] If τ ≥ 0 and L(g) exists for s>s 0 then L(u(t− τ)g(t− τ))
exists for s>s 0 and
L(u(t− τ)g(t− τ)) = e−sτ L(g(t)),
or , equivalently,
if g(t) ↔ G(s), then u(t− τ)g(t− τ) ↔ e−sτ G(s). (8.4.12)
REM A RK : Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by eat
corresponds to shifting the argument of its transform by aunits. Theorem 8.4.2 states that multiplying a
Laplace transform by the exponential e−τ s corresponds to shifting the argument of the inverse transfo rm
by τ units.
Example 8.4.6 Use ( 8.4.12) to find
L−1
(e−2s
s2
)
.
Solution T o apply ( 8.4.12) we let τ = 2 and G(s) = 1 /s 2. Then g(t) = tand ( 8.4.12) implies that
L−1
(e−2s
s2
)
= u(t− 2)(t− 2).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.4 The Unit Step Function 427
Example 8.4.7 Find the inverse Laplace transform hof
H(s) = 1
s2 − e−s
(1
s2 + 2
s
)
+ e−4s
(4
s3 + 1
s
)
,
and find distinct formulas for hon appropriate intervals.
Solution Let
G0(s) = 1
s2 , G 1(s) = 1
s2 + 2
s, G 2(s) = 4
s3 + 1
s.
Then
g0(t) = t, g1(t) = t+ 2 , g2(t) = 2 t2 + 1 .
Hence, ( 8.4.12) and the linearity of L−1 imply that
h(t) = L−1 (G0(s)) − L−1 (
e−sG1(s)
)
+ L−1 (
e−4sG2(s)
)
= t− u(t− 1) [(t− 1) + 2] + u(t− 4)
[
2(t− 4)2 + 1
]
= t− u(t− 1)(t+ 1) + u(t− 4)(2t2 − 16t+ 33) ,
which can also be written as
h(t) =





t, 0 ≤ t< 1,
−1, 1 ≤ t< 4,
2t2 − 16t+ 32 , t ≥ 4.
Example 8.4.8 Find the inverse transform of
H(s) = 2s
s2 + 4 − e− π
2 s 3s+ 1
s2 + 9 + e−πs s+ 1
s2 + 6 s+ 10 .
Solution Let
G0(s) = 2s
s2 + 4 , G 1(s) = −(3s+ 1)
s2 + 9 ,
and
G2(s) = s+ 1
s2 + 6 s+ 10 = (s+ 3) − 2
(s+ 3) 2 + 1 .
Then
g0(t) = 2 cos 2 t, g 1(t) = −3 cos 3 t− 1
3 sin 3 t,
and
g2(t) = e−3t (cos t− 2 sin t).",Elementary Differential Equations with Boundary Value Problems.pdf
"G2(s) = s+ 1
s2 + 6 s+ 10 = (s+ 3) − 2
(s+ 3) 2 + 1 .
Then
g0(t) = 2 cos 2 t, g 1(t) = −3 cos 3 t− 1
3 sin 3 t,
and
g2(t) = e−3t (cos t− 2 sin t).
Therefore ( 8.4.12) and the linearity of L−1 imply that
h(t) = 2 cos 2 t− u(t− π/ 2)
[
3 cos 3( t− π/ 2) + 1
3 sin 3
(
t− π
2
) ]
+u(t− π)e−3(t−π ) [cos(t− π) − 2 sin( t− π)] .",Elementary Differential Equations with Boundary Value Problems.pdf
"428 Chapter 8 Laplace Transforms
Using the trigonometric identities ( 8.4.8) and ( 8.4.9), we can rewrite this as
h(t) = 2 cos 2 t+ u(t− π/ 2) (3 sin 3 t− 1
3 cos 3 t)
−u(t− π)e−3(t−π )(cos t− 2 sin t)
(8.4.13)
(Figure 8.4.5).
1 2 3 4 5 6
1
2
3
4
5
−1
−2
−3
−4
−5
−6
−7
 t
 y
Figure 8.4.5 The piecewise continouous function ( 8.4.13)
8.4 Exercises
In Exercises 1– 6 find the Laplace transform by the method of Example 8.4.1. Then express the given
function f in terms of unit step functions as in Eqn. ( 8.4.6), and use Theorem 8.4.1 to find L(f). Where
indicated by C/G , graph f.
1. f(t) =
{ 1, 0 ≤ t< 4,
t, t ≥ 4.
2. f(t) =
{
t, 0 ≤ t< 1,
1, t ≥ 1.
3. C/G f(t) =
{ 2t− 1, 0 ≤ t< 2,
t, t ≥ 2.
4. C/G f(t) =
{ 1, 0 ≤ t< 1,
t+ 2 , t ≥ 1.
5. f(t) =
{ t− 1, 0 ≤ t< 2,
4, t ≥ 2.
6. f(t) =
{ t2, 0 ≤ t< 1,
0, t ≥ 1.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.4 The Unit Step Function 429
In Exercises 7– 18 express the given function f in terms of unit step functions and use Theorem 8.4.1 to
find L(f). Where indicated by C/G , graph f.
7. f(t) =
{ 0, 0 ≤ t< 2,
t2 + 3 t, t ≥ 2.
8. f(t) =
{
t2 + 2 , 0 ≤ t< 1,
t, t ≥ 1.
9. f(t) =
{ tet , 0 ≤ t< 1,
et , t ≥ 1.
10. f(t) =
{ e −t , 0 ≤ t< 1,
e−2t, t ≥ 1.
11. f(t) =







−t, 0 ≤ t< 2,
t− 4, 2 ≤ t< 3,
1, t ≥ 3.
12. f(t) =







0, 0 ≤ t< 1,
t, 1 ≤ t< 2,
0, t ≥ 2.
13. f(t) =







t, 0 ≤ t< 1,
t2, 1 ≤ t< 2,
0, t ≥ 2.
14. f(t) =







t, 0 ≤ t< 1,
2 − t, 1 ≤ t< 2,
6, t> 2.
15.
C/G f(t) =









sin t, 0 ≤ t< π
2 ,
2 sin t, π
2 ≤ t<π,
cos t, t ≥ π.
16. C/G f(t) =







2, 0 ≤ t< 1,
−2t+ 2 , 1 ≤ t< 3,
3t, t ≥ 3.
17.
C/G f(t) =







3, 0 ≤ t< 2,
3t+ 2 , 2 ≤ t< 4,
4t, t ≥ 4.
18.
C/G f(t) =
{ (t+ 1) 2, 0 ≤ t< 1,
(t+ 2) 2, t ≥ 1.
In Exercises 19– 28 use Theorem 8.4.2 to express the inverse transforms in terms of step functions , and",Elementary Differential Equations with Boundary Value Problems.pdf
"{ (t+ 1) 2, 0 ≤ t< 1,
(t+ 2) 2, t ≥ 1.
In Exercises 19– 28 use Theorem 8.4.2 to express the inverse transforms in terms of step functions , and
then find distinct formulas the for inverse transforms on the appropriate intervals, as in Example 8.4.7.
Where indicated by C/G , graph the inverse transform.
19. H(s) = e−2s
s− 2
20. H(s) = e−s
s(s+ 1)
21. C/G H(s) = e−s
s3 + e−2s
s2
22. C/G H(s) =
(2
s + 1
s2
)
+ e−s
(3
s − 1
s2
)
+ e−3s
(1
s + 1
s2
)",Elementary Differential Equations with Boundary Value Problems.pdf
"430 Chapter 8 Laplace Transforms
23. H(s) =
(5
s − 1
s2
)
+ e−3s
(6
s + 7
s2
)
+ 3e−6s
s3
24. H(s) = e−πs (1 − 2s)
s2 + 4 s+ 5
25. C/G H(s) =
(1
s − s
s2 + 1
)
+ e− π
2 s
(3s− 1
s2 + 1
)
26. H(s) = e−2s
[ 3(s− 3)
(s+ 1)( s− 2) − s+ 1
(s− 1)(s− 2)
]
27. H(s) = 1
s + 1
s2 + e−s
(3
s + 2
s2
)
+ e−3s
(4
s + 3
s2
)
28. H(s) = 1
s − 2
s3 + e−2s
(3
s − 1
s3
)
+ e−4s
s2
29. Find L(u(t− τ)).
30. Let {tm }∞
m=0 be a sequence of points such that t0 = 0 , tm+1 > tm, and limm→∞ tm = ∞ . For
each nonnegative integer m, let fm be continuous on [tm, ∞ ), and let f be defined on [0, ∞ ) by
f(t) = fm (t), tm ≤ t<t m+1 (m= 0 , 1,... ).
Show that f is piecewise continuous on [0, ∞ ) and that it has the step function representation
f(t) = f0(t) +
∞∑
m=1
u(t− tm ) (fm (t) − fm−1(t)) , 0 ≤ t< ∞ .
How do we know that the series on the right converges for all tin [0, ∞ )?
31. In addition to the assumptions of Exercise
30, assume that
|fm (t)| ≤Mes0t, t≥ tm , m= 0 , 1,..., (A)
and that the series ∞∑
m=0",Elementary Differential Equations with Boundary Value Problems.pdf
"31. In addition to the assumptions of Exercise
30, assume that
|fm (t)| ≤Mes0t, t≥ tm , m= 0 , 1,..., (A)
and that the series ∞∑
m=0
e−ρt m (B)
converges for some ρ> 0. Using the steps listed below , show that L(f) is defined for s>s 0 and
L(f) = L(f0) +
∞∑
m=1
e−stm L(gm) (C)
for s>s 0 + ρ, where
gm(t) = fm (t+ tm ) − fm−1(t+ tm ).
(a) Use (A) and Theorem 8.1.6 to show that
L(f) =
∞∑
m=0
∫ tm+1
tm
e−stfm (t) dt (D)
is defined for s>s 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 431
(b) Show that (D) can be rewritten as
L(f) =
∞∑
m=0
(∫ ∞
tm
e−st fm (t) dt−
∫ ∞
tm+1
e−stfm (t) dt
)
. (E)
(c) Use (A), the assumed convergence of (B), and the comparison t est to show that the series
∞∑
m=0
∫ ∞
tm
e−stfm (t) dt and
∞∑
m=0
∫ ∞
tm+1
e−st fm (t) dt
both converge (absolutely) if s>s 0 + ρ.
(d) Show that (E) can be rewritten as
L(f) = L(f0) +
∞∑
m=1
∫ ∞
tm
e−st (fm (t) − fm−1 (t)) dt
if s>s 0 + ρ.
(e) Complete the proof of (C).
32. Suppose {tm}∞
m=0 and {fm}∞
m=0 satisfy the assumptions of Exercises
30 and 31, and there’s a
positive constant K such that tm ≥ Km for m sufficiently large. Show that the series (B) of
Exercise 31 converges for any ρ > 0, and conclude from this that (C) of Exercise 31 holds for
s>s 0.
In Exercises 33– 36 find the step function representation of f and use the result of Exercise 32 to find",Elementary Differential Equations with Boundary Value Problems.pdf
"s>s 0.
In Exercises 33– 36 find the step function representation of f and use the result of Exercise 32 to find
L(f). HIN T : Y ou will need formulas related to the formula for the sum of a geometric series.
33. f(t) = m+ 1 , m≤ t<m + 1 ( m= 0 , 1, 2,... )
34. f(t) = ( −1)m , m≤ t<m + 1 ( m= 0 , 1, 2,... )
35. f(t) = ( m+ 1) 2, m≤ t<m + 1 ( m= 0 , 1, 2,... )
36. f(t) = ( −1)m m, m≤ t<m + 1 ( m= 0 , 1, 2,... )
8.5 CONST ANT COEEFFICIENT EQUA TIONS WITH PIECEWISE CONTINUOU S FORCING FUNC-
TIONS
W e’ll now consider initial value problems of the form
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1, (8.5.1)
where a, b, and care constants ( a ̸= 0 ) and f is piecewise continuous on [0, ∞ ). Problems of this kind
occur in situations where the input to a physical system unde rgoes instantaneous changes, as when a
switch is turned on or off or the forces acting on the system ch ange abruptly .
It can be shown (Exercises 23 and 24) that the differential equation in ( 8.5.1) has no solutions on an",Elementary Differential Equations with Boundary Value Problems.pdf
"It can be shown (Exercises 23 and 24) that the differential equation in ( 8.5.1) has no solutions on an
open interval that contains a jump discontinuity of f. Therefore we must define what we mean by a
solution of ( 8.5.1) on [0, ∞ ) in the case where f has jump discontinuities. The next theorem motivates
our definition. W e omit the proof.",Elementary Differential Equations with Boundary Value Problems.pdf
"432 Chapter 8 Laplace Transforms
Theorem 8.5.1 Suppose a,b , and care constants (a̸= 0) , and f is piecewise continuous on [0, ∞ ). with
jump discontinuities at t1, . . . , tn , where
0 <t1 <· · ·<tn .
Let k0 and k1 be arbitrary real numbers. Then there is a unique function ydefined on [0, ∞ ) with these
properties:
(a) y(0) = k0 and y′(0) = k1.
(b) yand y′are continuous on [0, ∞ ).
(c) y′′is defined on every open subinterval of [0, ∞ ) that does not contain any of the points t1, . . . , tn,
and
ay′′+ by′+ cy = f(t)
on every such subinterval.
(d) y′′has limits from the right and left at t1, . . . ,t n .
W e define the function yof Theorem 8.5.1 to be the solution of the initial value problem ( 8.5.1).
W e begin by considering initial value problems of the form
ay′′+ by′+ cy=
{
f0(t), 0 ≤ t<t 1,
f1(t), t ≥ t1, y(0) = k0, y ′(0) = k1, (8.5.2)
where the forcing function has a single jump discontinuity a t t1.
W e can solve ( 8.5.2) by the these steps:",Elementary Differential Equations with Boundary Value Problems.pdf
"where the forcing function has a single jump discontinuity a t t1.
W e can solve ( 8.5.2) by the these steps:
Step 1. Find the solution y0 of the initial value problem
ay′′+ by′+ cy= f0(t), y (0) = k0, y ′(0) = k1.
Step 2. Compute c0 = y0(t1) and c1 = y′
0(t1).
Step 3. Find the solution y1 of the initial value problem
ay′′+ by′+ cy = f1(t), y (t1) = c0, y ′(t1) = c1.
Step 4. Obtain the solution yof ( 8.5.2) as
y=
{
y0(t), 0 ≤ t<t 1
y1(t), t ≥ t1.
It is shown in Exercise 23 that y′exists and is continuous at t1. The next example illustrates this
procedure.
Example 8.5.1 Solve the initial value problem
y′′+ y= f(t), y (0) = 2 , y′(0) = −1, (8.5.3)
where
f(t) =



1, 0 ≤ t< π
2 ,
−1, t ≥ π
2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 433
1 2 3 4 5 6
1
2
−1
−2
 t
 y
Figure 8.5.1 Graph of ( 8.5.4)
Solution The initial value problem in Step 1 is
y′′+ y= 1 , y (0) = 2 , y ′(0) = −1.
W e leave it to you to verify that its solution is
y0 = 1 + cos t− sin t.
Doing Step 2 yields y0(π/ 2) = 0 and y′
0(π/ 2) = −1, so the second initial value problem is
y′′+ y= −1, y
(π
2
)
= 0 , y′
(π
2
)
= −1.
W e leave it to you to verify that the solution of this problem i s
y1 = −1 + cos t+ sin t.
Hence, the solution of ( 8.5.3) is
y=



1 + cos t− sin t, 0 ≤ t< π
2 ,
−1 + cos t+ sin t, t ≥ π
2
(8.5.4)
(Figure:8.5.1).
If f0 and f1 are defined on [0, ∞ ), we can rewrite ( 8.5.2) as
ay′′+ by′+ cy= f0(t) + u(t− t1) (f1(t) − f0(t)) , y (0) = k0, y ′(0) = k1,
and apply the method of Laplace transforms. W e’ll now solve t he problem considered in Example 8.5.1
by this method.",Elementary Differential Equations with Boundary Value Problems.pdf
"434 Chapter 8 Laplace Transforms
Example 8.5.2 Use the Laplace transform to solve the initial value problem
y′′+ y= f(t), y (0) = 2 , y′(0) = −1, (8.5.5)
where
f(t) =



1, 0 ≤ t< π
2 ,
−1, t ≥ π
2 .
Solution Here
f(t) = 1 − 2u
(
t− π
2
)
,
so Theorem 8.4.1 (with g(t) = 1 ) implies that
L(f) = 1 − 2e−πs/ 2
s .
Therefore, transforming ( 8.5.5) yields
(s2 + 1) Y(s) = 1 − 2e−πs/ 2
s − 1 + 2 s,
so
Y(s) = (1 − 2e−πs/ 2)G(s) + 2s− 1
s2 + 1 , (8.5.6)
with
G(s) = 1
s(s2 + 1) .
The form for the partial fraction expansion of Gis
1
s(s2 + 1) = A
s + Bs+ C
s2 + 1 . (8.5.7)
Multiplying through by s(s2 + 1) yields
A(s2 + 1) + ( Bs+ C)s= 1 ,
or
(A+ B)s2 + Cs+ A= 1 .
Equating coefficients of like powers of son the two sides of this equation shows that A= 1 , B = −A=
−1 and C = 0 . Hence, from ( 8.5.7),
G(s) = 1
s − s
s2 + 1 .
Therefore
g(t) = 1 − cos t.
From this, ( 8.5.6), and Theorem 8.4.2,
y= 1 − cos t− 2u
(
t− π
2
)(
1 − cos
(
t− π
2
))
+ 2 cos t− sin t.",Elementary Differential Equations with Boundary Value Problems.pdf
"s − s
s2 + 1 .
Therefore
g(t) = 1 − cos t.
From this, ( 8.5.6), and Theorem 8.4.2,
y= 1 − cos t− 2u
(
t− π
2
)(
1 − cos
(
t− π
2
))
+ 2 cos t− sin t.
Simplifying this (recalling that cos(t− π/ 2) = sin t) yields
y= 1 + cos t− sin t− 2u
(
t− π
2
)
(1 − sin t),",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 435
or
y=



1 + cos t− sin t, 0 ≤ t< π
2 ,
−1 + cos t+ sin t, t ≥ π
2 ,
which is the result obtained in Example 8.5.1.
REM A RK : It isn’t obvious that using the Laplace transform to solve ( 8.5.2) as we did in Example 8.5.2
yields a function ywith the properties stated in Theorem 8.5.1; that is, such that yand y′are continuous
on [0, ∞ ) and y′′has limits from the right and left at t1. However, this is true if f0 and f1 are continuous
and of exponential order on [0, ∞ ). A proof is sketched in Exercises 8.6. 11–8.6.13.
Example 8.5.3 Solve the initial value problem
y′′− y= f(t), y (0) = −1, y′(0) = 2 , (8.5.8)
where
f(t) =
{ t, 0 ≤ t< 1,
1, t ≥ 1.
Solution Here
f(t) = t− u(t− 1)(t− 1),
so
L(f) = L(t) − L (u(t− 1)(t− 1))
= L(t) − e−sL(t) (from Theorem 8.4.1)
= 1
s2 − e−s
s2 .
Since transforming ( 8.5.8) yields
(s2 − 1)Y(s) = L(f) + 2 − s,
we see that
Y(s) = (1 − e−s)H(s) + 2 − s
s2 − 1 , (8.5.9)",Elementary Differential Equations with Boundary Value Problems.pdf
"= 1
s2 − e−s
s2 .
Since transforming ( 8.5.8) yields
(s2 − 1)Y(s) = L(f) + 2 − s,
we see that
Y(s) = (1 − e−s)H(s) + 2 − s
s2 − 1 , (8.5.9)
where
H(s) = 1
s2(s2 − 1) = 1
s2 − 1 − 1
s2 ;
therefore
h(t) = sinh t− t. (8.5.10)
Since
L−1
(2 − s
s2 − 1
)
= 2 sinh t− cosh t,
we conclude from ( 8.5.9), ( 8.5.10), and Theorem 8.4.1 that
y = sinh t− t− u(t− 1) (sinh(t− 1) − t+ 1) + 2 sinh t− cosh t,
or
y = 3 sinh t− cosh t− t− u(t− 1) (sinh(t− 1) − t+ 1) (8.5.11)
W e leave it to you to verify that yand y′are continuous and y′′has limits from the right and left at t1 = 1 .",Elementary Differential Equations with Boundary Value Problems.pdf
"436 Chapter 8 Laplace Transforms
Example 8.5.4 Solve the initial value problem
y′′+ y = f(t), y (0) = 0 , y′(0) = 0 , (8.5.12)
where
f(t) =









0, 0 ≤ t< π
4 ,
cos 2t, π
4 ≤ t<π,
0, t ≥ π.
Solution Here
f(t) = u(t− π/ 4) cos 2 t− u(t− π) cos 2 t,
so
L(f) = L(u(t− π/ 4) cos 2 t) − L (u(t− π) cos 2 t)
= e−πs/ 4L(cos 2(t+ π/ 4)) − e−πs L(cos 2(t+ π))
= −e−πs/ 4L(sin 2 t) − e−πs L(cos 2t)
= −2e−πs/ 4
s2 + 4 − se−πs
s2 + 4 .
Since transforming ( 8.5.12) yields
(s2 + 1) Y(s) = L(f),
we see that
Y(s) = e−πs/ 4H1(s) + e−πs H2(s), (8.5.13)
where
H1(s) = − 2
(s2 + 1)( s2 + 4) and H2(s) = − s
(s2 + 1)( s2 + 4) . (8.5.14)
T o simplify the required partial fraction expansions, we fir st write
1
(x+ 1)( x+ 4) = 1
3
[ 1
x+ 1 − 1
x+ 4
]
.
Setting x= s2 and substituting the result in ( 8.5.14) yields
H1(s) = −2
3
[ 1
s2 + 1 − 1
s2 + 4
]
and H2(s) = −1
3
[ s
s2 + 1 − s
s2 + 4
]
.
The inverse transforms are
h1(t) = −2
3 sin t+ 1
3 sin 2t and h2(t) = −1
3 cos t+ 1
3 cos 2t.",Elementary Differential Equations with Boundary Value Problems.pdf
"s2 + 4
]
and H2(s) = −1
3
[ s
s2 + 1 − s
s2 + 4
]
.
The inverse transforms are
h1(t) = −2
3 sin t+ 1
3 sin 2t and h2(t) = −1
3 cos t+ 1
3 cos 2t.
From ( 8.5.13) and Theorem 8.4.2,
y = u
(
t− π
4
)
h1
(
t− π
4
)
+ u(t− π)h2(t− π). (8.5.15)
Since
h1
(
t− π
4
)
= −2
3 sin
(
t− π
4
)
+ 1
3 sin 2
(
t− π
4
)
= −
√
2
3 (sin t− cos t) − 1
3 cos 2t",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 437
1.0
0.5
−0.5
−1.0
1 2 3 4 5 6  t
 y
Figure 8.5.2 Graph of ( 8.5.16)
and
h2(t− π) = −1
3 cos(t− π) + 1
3 cos 2( t− π)
= 1
3 cos t+ 1
3 cos 2t,
(8.5.15) can be rewritten as
y = −1
3 u
(
t− π
4
)( √
2(sin t− cos t) + cos 2 t
)
+ 1
3 u(t− π)(cos t+ cos 2 t)
or
y=













0, 0 ≤ t< π
4 ,
−
√
2
3 (sin t− cos t) − 1
3 cos 2 t, π
4 ≤ t<π,
−
√
2
3 sin t+ 1 +
√
2
3 cos t, t ≥ π.
(8.5.16)
W e leave it to you to verify that y and y′are continuous and y′′has limits from the right and left at
t1 = π/ 4 and t2 = π (Figure 8.5.2).
8.5 Exercises
In Exercises 1– 20 use the Laplace transform to solve the initial value problem . Where indicated by
C/G , graph the solution.",Elementary Differential Equations with Boundary Value Problems.pdf
"438 Chapter 8 Laplace Transforms
1. y′′+ y =
{ 3, 0 ≤ t<π,
0, t ≥ π,
y(0) = 0 , y ′(0) = 0
2. y′′+ y =
{ 3, 0 ≤ t< 4,
; 2 t− 5, t> 4, y(0) = 1 , y ′(0) = 0
3. y′′− 2y′=
{ 4, 0 ≤ t< 1,
6, t ≥ 1,
y(0) = −6, y ′(0) = 1
4. y′′− y =
{ e2t, 0 ≤ t< 2,
1, t ≥ 2,
y(0) = 3 , y ′(0) = −1
5. y′′− 3y′+ 2 y=







0, 0 ≤ t< 1,
1, 1 ≤ t< 2,
−1, t ≥ 2,
y(0) = −3, y ′(0) = 1
6.
C/G y′′+ 4 y=
{ |sin t|, 0 ≤ t< 2π,
0, t ≥ 2π,
y(0) = −3, y ′(0) = 1
7. y′′− 5y′+ 4 y=







1, 0 ≤ t< 1
−1, 1 ≤ t< 2,
0, t ≥ 2,
y(0) = 3 , y ′(0) = −5
8. y′′+ 9 y=





cos t, 0 ≤ t< 3π
2 ,
sin t, t ≥ 3π
2 ,
y(0) = 0 , y′(0) = 0
9. C/G y′′+ 4 y=



t, 0 ≤ t< π
2 ,
π, t ≥ π
2 ,
y(0) = 0 , y ′(0) = 0
10. y′′+ y =
{ t, 0 ≤ t<π,
−t, t ≥ π,
y(0) = 0 , y′(0) = 0
11. y′′− 3y′+ 2 y=
{ 0, 0 ≤ t< 2,
2t− 4, t ≥ 2, , y (0) = 0 , y ′(0) = 0
12. y′′+ y =
{ t, 0 ≤ t< 2π,
−2t, t ≥ 2π, y(0) = 1 , y ′(0) = 2
13. C/G y′′+ 3 y′+ 2 y=
{ 1, 0 ≤ t< 2,
−1, t ≥ 2, y(0) = 0 , y′(0) = 0
14. y′′− 4y′+ 3 y=
{
−1, 0 ≤ t< 1,",Elementary Differential Equations with Boundary Value Problems.pdf
"−2t, t ≥ 2π, y(0) = 1 , y ′(0) = 2
13. C/G y′′+ 3 y′+ 2 y=
{ 1, 0 ≤ t< 2,
−1, t ≥ 2, y(0) = 0 , y′(0) = 0
14. y′′− 4y′+ 3 y=
{
−1, 0 ≤ t< 1,
1, t ≥ 1, y(0) = 0 , y′(0) = 0
15. y′′+ 2 y′+ y =
{ et , 0 ≤ t< 1,
et − 1, t ≥ 1, y(0) = 3 , y′(0) = −1
16. y′′+ 2 y′+ y =
{ 4et , 0 ≤ t< 1,
0, t ≥ 1, y(0) = 0 , y′(0) = 0
17. y′′+ 3 y′+ 2 y=
{ e−t , 0 ≤ t< 1,
0, t ≥ 1, y(0) = 1 , y′(0) = −1",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 439
18. y′′− 4y′+ 4 y=
{ e2t , 0 ≤ t< 2,
−e2t , t ≥ 2, y(0) = 0 , y′(0) = −1
19. C/G y′′=



t2, 0 ≤ t< 1,
−t, 1 ≤ t< 2,
t+ 1 , t ≥ 2,
y(0) = 1 , y′(0) = 0
20. y′′+ 2 y′+ 2 y=



1, 0 ≤ t< 2π,
t, 2π ≤ t< 3π,
−1, t ≥ 3π,
y(0) = 2 , y ′(0) = −1
21. Solve the initial value problem
y′′= f(t), y (0) = 0 , y ′(0) = 0 ,
where
f(t) = m+ 1 , m ≤ t<m + 1 , m = 0 , 1, 2,....
22. Solve the given initial value problem and find a formula that d oes not involve step functions and
represents yon each interval of continuity of f.
(a) y′′+ y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = m+ 1 , mπ ≤ t< (m+ 1) π, m = 0 , 1, 2,... .
(b) y′′+ y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = ( m+ 1) t, 2mπ ≤ t <2(m+ 1) π, m = 0 , 1, 2,... HIN T : Y ou’ll need the
formula
1 + 2 + · · ·+ m= m(m+ 1)
2 .
(c) y′′+ y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = ( −1)m , mπ ≤ t< (m+ 1) π, m = 0 , 1, 2,....",Elementary Differential Equations with Boundary Value Problems.pdf
"formula
1 + 2 + · · ·+ m= m(m+ 1)
2 .
(c) y′′+ y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = ( −1)m , mπ ≤ t< (m+ 1) π, m = 0 , 1, 2,....
(d) y′′− y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = m+ 1 , m ≤ t< (m+ 1) , m = 0 , 1, 2,....
HIN T : Y ou will need the formula
1 + r+ · · ·+ rm = 1 − rm+1
1 − r (r̸= 1) .
(e) y′′+ 2 y′+ 2 y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = ( m+ 1)(sin t+ 2 cos t), 2mπ ≤ t< 2(m+ 1) π, m = 0 , 1, 2,....
(See the hint in (d).)
(f) y′′− 3y′+ 2 y= f(t), y (0) = 0 , y ′(0) = 0 ;
f(t) = m+ 1 , m ≤ t<m + 1 , m = 0 , 1, 2,....
(See the hints in (b) and (d).)
23. (a) Let g be continuous on (α,β ) and differentiable on the (α,t 0) and (t0,β ). Suppose A =
limt→ t0− g′(t) and B = lim t→ t0+ g′(t) both exist. Use the mean value theorem to show
that
lim
t→ t0−
g(t) − g(t0)
t− t0
= A and lim
t→ t0+
g(t) − g(t0)
t− t0
= B.
(b) Conclude from (a) that g′(t0) exists and g′is continuous at t0 if A= B.",Elementary Differential Equations with Boundary Value Problems.pdf
"440 Chapter 8 Laplace Transforms
(c) Conclude from (a) that if gis differentiable on (α,β ) then g′can’t have a jump discontinuity
on (α,β ).
24. (a) Let a, b, and cbe constants, with a̸= 0 . Let f be piecewise continuous on an interval (α,β ),
with a single jump discontinuity at a point t0 in (α,β ). Suppose yand y′are continuous on
(α,β ) and y′′on (α,t 0) and (t0,β ). Suppose also that
ay′′+ by′+ cy= f(t) (A)
on (α,t 0) and (t0,β ). Show that
y′′(t0+) − y′′(t0−) = f(t0 +) − f(t0 −)
a ̸= 0 .
(b) Use (a) and Exercise 23(c) to show that (A) does not have solutions on any interval (α,β )
that contains a jump discontinuity of f.
25. Suppose P0,P 1, and P2 are continuous and P0 has no zeros on an open interval (a,b ), and that F
has a jump discontinuity at a point t0 in (a,b ). Show that the differential equation
P0(t)y′′+ P1(t)y′+ P2(t)y= F(t)
has no solutions on (a,b ).H IN T : Generalize the result of Exercise 24 and use Exercise 23(c).",Elementary Differential Equations with Boundary Value Problems.pdf
"P0(t)y′′+ P1(t)y′+ P2(t)y= F(t)
has no solutions on (a,b ).H IN T : Generalize the result of Exercise 24 and use Exercise 23(c).
26. Let 0 = t0 <t1 <· · ·<tn . Suppose fm is continuous on [tm, ∞ ) for m= 1 ,...,n . Let
f(t) =
{ fm(t), t m ≤ t<t m+1, m = 1 ,...,n − 1,
fn (t), t ≥ tn .
Show that the solution of
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1,
as defined following Theorem 8.5.1, is given by
y=















z0(t), 0 ≤ t<t 1,
z0(t) + z1(t), t 1 ≤ t<t 2,
.
.
.
z0 + · · ·+ zn−1(t), t n−1 ≤ t<t n,
z0 + · · ·+ zn (t), t ≥ tn ,
where z0 is the solution of
az′′+ bz′+ cz= f0(t), z (0) = k0, z ′(0) = k1
and zm is the solution of
az′′+ bz′+ cz = fm (t) − fm−1 (t), z (tm ) = 0 , z ′(tm) = 0
for m= 1 ,...,n .
8.6 CONVOLUTION",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.6 Convolution 441
In this section we consider the problem of finding the inverse Laplace transform of a product H(s) =
F(s)G(s), where F and G are the Laplace transforms of known functions f and g. T o motivate our
interest in this problem, consider the initial value proble m
ay′′+ by′+ cy = f(t), y (0) = 0 , y ′(0) = 0 .
T aking Laplace transforms yields
(as2 + bs+ c)Y(s) = F(s),
so
Y(s) = F(s)G(s), (8.6.1)
where
G(s) = 1
as2 + bs+ c.
Until now wen’t been interested in the factorization indica ted in ( 8.6.1), since we dealt only with differ-
ential equations with specific forcing functions. Hence, we could simply do the indicated multiplication
in ( 8.6.1) and use the table of Laplace transforms to find y = L−1(Y). However, this isn’t possible if we
want a formula for yin terms of f, which may be unspecified.
T o motivate the formula for L−1(FG), consider the initial value problem
y′− ay= f(t), y (0) = 0 , (8.6.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"T o motivate the formula for L−1(FG), consider the initial value problem
y′− ay= f(t), y (0) = 0 , (8.6.2)
which we first solve without using the Laplace transform. The solution of the differential equation in
(8.6.2) is of the form y= ueat where
u′= e−at f(t).
Integrating this from 0 to tand imposing the initial condition u(0) = y(0) = 0 yields
u=
∫ t
0
e−aτ f(τ) dτ.
Therefore
y(t) = eat
∫ t
0
e−aτ f(τ) dτ =
∫ t
0
ea(t−τ )f(τ) dτ. (8.6.3)
Now we’ll use the Laplace transform to solve ( 8.6.2) and compare the result to ( 8.6.3). T aking Laplace
transforms in ( 8.6.2) yields
(s− a)Y(s) = F(s),
so
Y(s) = F(s) 1
s− a,
which implies that
y(t) = L−1
(
F(s) 1
s− a
)
. (8.6.4)
If we now let g(t) = eat , so that
G(s) = 1
s− a,
then ( 8.6.3) and ( 8.6.4) can be written as
y(t) =
∫ t
0
f(τ)g(t− τ) dτ",Elementary Differential Equations with Boundary Value Problems.pdf
"442 Chapter 8 Laplace Transforms
and
y= L−1(FG),
respectively . Therefore
L−1(FG) =
∫ t
0
f(τ)g(t− τ) dτ (8.6.5)
in this case.
This motivates the next definition.
Definition 8.6.1 The convolution f ∗gof two functions f and gis defined by
(f ∗g)(t) =
∫ t
0
f(τ)g(t− τ) dτ.
It can be shown (Exercise 6) that f ∗g= g∗f; that is,
∫ t
0
f(t− τ)g(τ) dτ =
∫ t
0
f(τ)g(t− τ) dτ.
Eqn. ( 8.6.5) shows that L−1(FG) = f ∗g in the special case where g(t) = eat . This next theorem
states that this is true in general.
Theorem 8.6.2
[The Convolution Theorem ] If L(f) = F and L(g) = G, then
L(f ∗g) = FG.
A complete proof of the convolution theorem is beyond the sco pe of this book. However, we’ll assume
that f ∗g has a Laplace transform and verify the conclusion of the theo rem in a purely computational
way . By the definition of the Laplace transform,
L(f ∗g) =
∫ ∞
0
e−st(f ∗g)(t) dt=
∫ ∞
0
e−st
∫ t
0
f(τ)g(t− τ) dτdt.",Elementary Differential Equations with Boundary Value Problems.pdf
"way . By the definition of the Laplace transform,
L(f ∗g) =
∫ ∞
0
e−st(f ∗g)(t) dt=
∫ ∞
0
e−st
∫ t
0
f(τ)g(t− τ) dτdt.
This iterated integral equals a double integral over the reg ion shown in Figure 8.6.1. Reversing the order
of integration yields
L(f ∗g) =
∫ ∞
0
f(τ)
∫ ∞
τ
e−st g(t− τ) dtdτ. (8.6.6)
However, the substitution x= t− τ shows that
∫ ∞
τ
e−stg(t− τ) dt =
∫ ∞
0
e−s(x+τ ) g(x) dx
= e−sτ
∫ ∞
0
e−sx g(x) dx= e−sτ G(s).
Substituting this into ( 8.6.6) and noting that G(s) is independent of τ yields
L(f ∗g) =
∫ ∞
0
e−sτ f(τ)G(s) dτ
= G(s)
∫ ∞
0
e−st f(τ) dτ = F(s)G(s).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.6 Convolution 443
 t = τ
 t
 τ
Figure 8.6.1
Example 8.6.1 Let
f(t) = eat and g(t) = ebt (a̸= b).
V erify that L(f ∗g) = L(f)L(g), as implied by the convolution theorem.
Solution W e first compute
(f ∗g)(t) =
∫ t
0
eaτ eb(t−τ ) dτ = ebt
∫ t
0
e(a−b)τ dτ
= ebt e(a−b)τ
a− b
⏐
⏐
⏐
⏐
t
0
= ebt [ e(a−b)t − 1]
a− b
= eat − ebt
a− b .
Since
eat ↔ 1
s− a and ebt ↔ 1
s− b,
it follows that
L(f ∗g) = 1
a− b
[ 1
s− a − 1
s− b
]
= 1
(s− a)(s− b)
= L(eat)L(ebt ) = L(f)L(g).",Elementary Differential Equations with Boundary Value Problems.pdf
"444 Chapter 8 Laplace Transforms
A Formula for the Solution of an Initial V alue Problem
The convolution theorem provides a formula for the solution of an initial value problem for a linear
constant coefficient second order equation with an unspecifi ed. The next three examples illustrate this.
Example 8.6.2 Find a formula for the solution of the initial value problem
y′′− 2y′+ y = f(t), y (0) = k0, y ′(0) = k1. (8.6.7)
Solution T aking Laplace transforms in ( 8.6.7) yields
(s2 − 2s+ 1) Y(s) = F(s) + ( k1 + k0s) − 2k0.
Therefore
Y(s) = 1
(s− 1)2 F(s) + k1 + k0s− 2k0
(s− 1)2
= 1
(s− 1)2 F(s) + k0
s− 1 + k1 − k0
(s− 1)2 .
From the table of Laplace transforms,
L−1
( k0
s− 1 + k1 − k0
(s− 1)2
)
= et (k0 + ( k1 − k0)t) .
Since 1
(s− 1)2 ↔ tet and F(s) ↔ f(t),
the convolution theorem implies that
L−1
( 1
(s− 1)2 F(s)
)
=
∫ t
0
τeτ f(t− τ) dτ.
Therefore the solution of ( 8.6.7) is
y(t) = et (k0 + ( k1 − k0)t) +
∫ t
0
τeτ f(t− τ) dτ.",Elementary Differential Equations with Boundary Value Problems.pdf
"L−1
( 1
(s− 1)2 F(s)
)
=
∫ t
0
τeτ f(t− τ) dτ.
Therefore the solution of ( 8.6.7) is
y(t) = et (k0 + ( k1 − k0)t) +
∫ t
0
τeτ f(t− τ) dτ.
Example 8.6.3 Find a formula for the solution of the initial value problem
y′′+ 4 y= f(t), y (0) = k0, y ′(0) = k1. (8.6.8)
Solution T aking Laplace transforms in ( 8.6.8) yields
(s2 + 4) Y(s) = F(s) + k1 + k0s.
Therefore
Y(s) = 1
(s2 + 4) F(s) + k1 + k0s
s2 + 4 .
From the table of Laplace transforms,
L−1
(k1 + k0s
s2 + 4
)
= k0 cos 2t+ k1
2 sin 2 t.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.6 Convolution 445
Since 1
(s2 + 4) ↔ 1
2 sin 2 t and F(s) ↔ f(t),
the convolution theorem implies that
L−1
( 1
(s2 + 4) F(s)
)
= 1
2
∫ t
0
f(t− τ) sin 2 τdτ.
Therefore the solution of ( 8.6.8) is
y(t) = k0 cos 2 t+ k1
2 sin 2 t+ 1
2
∫ t
0
f(t− τ) sin 2 τdτ.
Example 8.6.4 Find a formula for the solution of the initial value problem
y′′+ 2 y′+ 2 y= f(t), y (0) = k0, y ′(0) = k1. (8.6.9)
Solution T aking Laplace transforms in ( 8.6.9) yields
(s2 + 2 s+ 2) Y(s) = F(s) + k1 + k0s+ 2 k0.
Therefore
Y(s) = 1
(s+ 1) 2 + 1 F(s) + k1 + k0s+ 2 k0
(s+ 1) 2 + 1
= 1
(s+ 1) 2 + 1 F(s) + (k1 + k0) + k0(s+ 1)
(s+ 1) 2 + 1 .
From the table of Laplace transforms,
L−1
((k1 + k0) + k0(s+ 1)
(s+ 1) 2 + 1
)
= e−t ((k1 + k0) sin t+ k0 cos t) .
Since 1
(s+ 1) 2 + 1 ↔ e−t sin t and F(s) ↔ f(t),
the convolution theorem implies that
L−1
( 1
(s+ 1) 2 + 1 F(s)
)
=
∫ t
0
f(t− τ)e−τ sin τdτ.
Therefore the solution of ( 8.6.9) is
y(t) = e−t ((k1 + k0) sin t+ k0 cos t) +
∫ t
0
f(t− τ)e−τ sin τdτ. (8.6.10)",Elementary Differential Equations with Boundary Value Problems.pdf
")
=
∫ t
0
f(t− τ)e−τ sin τdτ.
Therefore the solution of ( 8.6.9) is
y(t) = e−t ((k1 + k0) sin t+ k0 cos t) +
∫ t
0
f(t− τ)e−τ sin τdτ. (8.6.10)
Evaluating Convolution Integrals
W e’ll say that an integral of the form ∫t
0 u(τ)v(t− τ) dτ is a convolution integral . The convolution
theorem provides a convenient way to evaluate convolution i ntegrals.",Elementary Differential Equations with Boundary Value Problems.pdf
"446 Chapter 8 Laplace Transforms
Example 8.6.5 Evaluate the convolution integral
h(t) =
∫ t
0
(t− τ)5τ7dτ.
Solution W e could evaluate this integral by expanding (t− τ)5 in powers of τ and then integrating.
However, the convolution theorem provides an easier way . Th e integral is the convolution of f(t) = t5
and g(t) = t7. Since
t5 ↔ 5!
s6 and t7 ↔ 7!
s8 ,
the convolution theorem implies that
h(t) ↔ 5!7!
s14 = 5!7!
13!
13!
s14 ,
where we have written the second equality because
13!
s14 ↔ t13.
Hence,
h(t) = 5!7!
13! t13.
Example 8.6.6 Use the convolution theorem and a partial fraction expansio n to evaluate the convolution
integral
h(t) =
∫ t
0
sin a(t− τ) cos bτdτ (|a| ̸= |b|).
Solution Since
sin at↔ a
s2 + a2 and cos bt↔ s
s2 + b2 ,
the convolution theorem implies that
H(s) = a
s2 + a2
s
s2 + b2 .
Expanding this in a partial fraction expansion yields
H(s) = a
b2 − a2
[ s
s2 + a2 − s
s2 + b2
]
.
Therefore
h(t) = a
b2 − a2 (cos at− cos bt) .
V olterra Integral Equations",Elementary Differential Equations with Boundary Value Problems.pdf
"H(s) = a
b2 − a2
[ s
s2 + a2 − s
s2 + b2
]
.
Therefore
h(t) = a
b2 − a2 (cos at− cos bt) .
V olterra Integral Equations
An equation of the form
y(t) = f(t) +
∫ t
0
k(t− τ)y(τ) dτ (8.6.11)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.6 Convolution 447
is a V olterra integral equation . Here f and k are given functions and y is unknown. Since the integral
on the right is a convolution integral, the convolution theo rem provides a convenient formula for solving
(8.6.11). T aking Laplace transforms in ( 8.6.11) yields
Y(s) = F(s) + K(s)Y(s),
and solving this for Y(s) yields
Y(s) = F(s)
1 − K(s) .
W e then obtain the solution of ( 8.6.11) as y= L−1(Y).
Example 8.6.7 Solve the integral equation
y(t) = 1 + 2
∫ t
0
e−2(t−τ )y(τ) dτ. (8.6.12)
Solution T aking Laplace transforms in ( 8.6.12) yields
Y(s) = 1
s + 2
s+ 2 Y(s),
and solving this for Y(s) yields
Y(s) = 1
s + 2
s2 .
Hence,
y(t) = 1 + 2 t.
Transfer Functions
The next theorem presents a formula for the solution of the ge neral initial value problem
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1,
where we assume for simplicity that f is continuous on [0, ∞ ) and that L(f) exists. In Exercises 11– 14",Elementary Differential Equations with Boundary Value Problems.pdf
"where we assume for simplicity that f is continuous on [0, ∞ ) and that L(f) exists. In Exercises 11– 14
it’s shown that the formula is valid under much weaker condit ions on f.
Theorem 8.6.3 Suppose f is continuous on [0, ∞ ) and has a Laplace transform . Then the solution of the
initial value problem
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1, (8.6.13)
is
y(t) = k0y1(t) + k1y2(t) +
∫ t
0
w(τ)f(t− τ) dτ, (8.6.14)
where y1 and y2 satisfy
ay′′
1 + by′
1 + cy1 = 0 , y 1(0) = 1 , y ′
1(0) = 0 , (8.6.15)
and
ay′′
2 + by′
2 + cy2 = 0 , y 2(0) = 0 , y ′
2(0) = 1 , (8.6.16)
and
w(t) = 1
ay2(t). (8.6.17)",Elementary Differential Equations with Boundary Value Problems.pdf
"448 Chapter 8 Laplace Transforms
Proof T aking Laplace transforms in ( 8.6.13) yields
p(s)Y(s) = F(s) + a(k1 + k0s) + bk0,
where
p(s) = as2 + bs+ c.
Hence,
Y(s) = W(s)F(s) + V(s) (8.6.18)
with
W(s) = 1
p(s) (8.6.19)
and
V(s) = a(k1 + k0s) + bk0
p(s) . (8.6.20)
T aking Laplace transforms in ( 8.6.15) and ( 8.6.16) shows that
p(s)Y1(s) = as+ b and p(s)Y2 (s) = a.
Therefore
Y1(s) = as+ b
p(s)
and
Y2(s) = a
p(s) . (8.6.21)
Hence, ( 8.6.20) can be rewritten as
V(s) = k0Y1(s) + k1Y2(s).
Substituting this into ( 8.6.18) yields
Y(s) = k0Y1(s) + k1Y2(s) + 1
aY2(s)F(s).
T aking inverse transforms and invoking the convolution the orem yields ( 8.6.14). Finally , ( 8.6.19) and
(8.6.21) imply ( 8.6.17).
It is useful to note from ( 8.6.14) that yis of the form
y= v+ h,
where
v(t) = k0y1(t) + k1y2(t)
depends on the initial conditions and is independent of the f orcing function, while
h(t) =
∫ t
0
w(τ)f(t− τ) dτ",Elementary Differential Equations with Boundary Value Problems.pdf
"where
v(t) = k0y1(t) + k1y2(t)
depends on the initial conditions and is independent of the f orcing function, while
h(t) =
∫ t
0
w(τ)f(t− τ) dτ
depends on the forcing function and is independent of the ini tial conditions. If the zeros of the character-
istic polynomial
p(s) = as2 + bs+ c
of the complementary equation have negative real parts, the n y1 and y2 both approach zero as t → ∞ ,
so limt→∞ v(t) = 0 for any choice of initial conditions. Moreover, the value of h(t) is essentially",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.6 Convolution 449
independent of the values of f(t− τ) for large τ, since limτ →∞ w(τ) = 0 . In this case we say that vand
hare transient and steady state components , respectively , of the solution yof ( 8.6.13). These definitions
apply to the initial value problem of Example 8.6.4, where the zeros of
p(s) = s2 + 2 s+ 2 = ( s+ 1) 2 + 1
are −1 ± i. From ( 8.6.10), we see that the solution of the general initial value probl em of Example 8.6.4
is y = v+ h, where
v(t) = e−t ((k1 + k0) sin t+ k0 cos t)
is the transient component of the solution and
h(t) =
∫ t
0
f(t− τ)e−τ sin τdτ
is the steady state component. The definitions don’t apply to the initial value problems considered in
Examples 8.6.2 and 8.6.3, since the zeros of the characteristic polynomials in these two examples don’t
have negative real parts.
In physical applications where the input f and the output y of a device are related by ( 8.6.13), the",Elementary Differential Equations with Boundary Value Problems.pdf
"have negative real parts.
In physical applications where the input f and the output y of a device are related by ( 8.6.13), the
zeros of the characteristic polynomial usually do have nega tive real parts. Then W = L(w) is called the
transfer function of the device. Since
H(s) = W(s)F(s),
we see that
W(s) = H(s)
F(s)
is the ratio of the transform of the steady state output to the transform of the input.
Because of the form of
h(t) =
∫ t
0
w(τ)f(t− τ) dτ,
wis sometimes called the weighting function of the device, since it assigns weights to past values of the
input f. It is also called the impulse response of the device, for reasons discussed in the next section.
Formula ( 8.6.14) is given in more detail in Exercises 8– 10 for the three possible cases where the zeros
of p(s) are real and distinct, real and repeated, or complex conjuga tes, respectively .
8.6 Exercises
1. Express the inverse transform as an integral.
(a) 1
s2(s2 + 4) (b) s
(s+ 2)( s2 + 9)
(c) s
(s2 + 4)( s2 + 9) (d) s",Elementary Differential Equations with Boundary Value Problems.pdf
"8.6 Exercises
1. Express the inverse transform as an integral.
(a) 1
s2(s2 + 4) (b) s
(s+ 2)( s2 + 9)
(c) s
(s2 + 4)( s2 + 9) (d) s
(s2 + 1) 2
(e) 1
s(s− a) (f) 1
(s+ 1)( s2 + 2 s+ 2)
(g) 1
(s+ 1) 2(s2 + 4 s+ 5) (h) 1
(s− 1)3(s+ 2) 2",Elementary Differential Equations with Boundary Value Problems.pdf
"450 Chapter 8 Laplace Transforms
(i) s− 1
s2(s2 − 2s+ 2) (j) s(s+ 3)
(s2 + 4)( s2 + 6 s+ 10)
(k) 1
(s− 3)5s6 (l) 1
(s− 1)3(s2 + 4)
(m) 1
s2(s− 2)3 (n) 1
s7(s− 2)6
2. Find the Laplace transform.
(a)
∫ t
0
sin aτ cos b(t− τ) dτ (b)
∫ t
0
eτ sin a(t− τ) dτ
(c)
∫ t
0
sinh aτ cosh a(t− τ) dτ (d)
∫ t
0
τ(t− τ) sin ωτ cos ω(t− τ) dτ
(e) et
∫ t
0
sin ωτ cos ω(t− τ) dτ (f) et
∫ t
0
τ2(t− τ)eτ dτ
(g) e−t
∫ t
0
e−τ τ cos ω(t− τ) dτ (h) et
∫ t
0
e2τ sinh(t− τ) dτ
(i)
∫ t
0
τe2τ sin 2( t− τ) dτ (j)
∫ t
0
(t− τ)3eτ dτ
(k)
∫ t
0
τ6e−(t−τ ) sin 3( t− τ) dτ (l)
∫ t
0
τ2(t− τ)3 dτ
(m)
∫ t
0
(t− τ)7e−τ sin 2τdτ (n)
∫ t
0
(t− τ)4 sin 2 τdτ
3. Find a formula for the solution of the initial value problem.
(a) y′′+ 3 y′+ y= f(t), y (0) = 0 , y ′(0) = 0
(b) y′′+ 4 y= f(t), y (0) = 0 , y ′(0) = 0
(c) y′′+ 2 y′+ y= f(t), y (0) = 0 , y ′(0) = 0
(d) y′′+ k2y= f(t), y (0) = 1 , y ′(0) = −1
(e) y′′+ 6 y′+ 9 y= f(t), y (0) = 0 , y ′(0) = −2
(f) y′′− 4y= f(t), y (0) = 0 , y ′(0) = 3",Elementary Differential Equations with Boundary Value Problems.pdf
"(d) y′′+ k2y= f(t), y (0) = 1 , y ′(0) = −1
(e) y′′+ 6 y′+ 9 y= f(t), y (0) = 0 , y ′(0) = −2
(f) y′′− 4y= f(t), y (0) = 0 , y ′(0) = 3
(g) y′′− 5y′+ 6 y= f(t), y (0) = 1 , y ′(0) = 3
(h) y′′+ ω2y = f(t), y (0) = k0, y ′(0) = k1
4. Solve the integral equation.
(a) y(t) = t−
∫ t
0
(t− τ)y(τ) dτ
(b) y(t) = sin t− 2
∫ t
0
cos(t− τ)y(τ) dτ
(c) y(t) = 1 + 2
∫ t
0
y(τ) cos( t− τ) dτ (d) y(t) = t+
∫ t
0
y(τ)e−(t−τ ) dτ
(e) y′(t) = t+
∫ t
0
y(τ) cos( t− τ) dτ, y(0) = 4",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.6 Convolution 451
(f) y(t) = cos t− sin t+
∫ t
0
y(τ) sin(t− τ) dτ
5. Use the convolution theorem to evaluate the integral.
(a)
∫ t
0
(t− τ)7τ8 dτ (b)
∫ t
0
(t− τ)13τ7 dτ
(c)
∫ t
0
(t− τ)6τ7 dτ (d)
∫ t
0
e−τ sin(t− τ) dτ
(e)
∫ t
0
sin τ cos 2(t− τ) dτ
6. Show that ∫ t
0
f(t− τ)g(τ) dτ =
∫ t
0
f(τ)g(t− τ) dτ
by introducing the new variable of integration x= t− τ in the first integral.
7. Use the convolution theorem to show that if f(t) ↔ F(s) then
∫ t
0
f(τ) dτ ↔ F(s)
s .
8. Show that if p(s) = as2 + bs+ chas distinct real zeros r1 and r2 then the solution of
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1
is
y(t) = k0
r2er1 t − r1er2 t
r2 − r1
+ k1
er2 t − er1t
r2 − r1
+ 1
a(r2 − r1)
∫ t
0
(er2 τ − er1τ )f(t− τ) dτ.
9. Show that if p(s) = as2 + bs+ chas a repeated real zero r1 then the solution of
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1
is
y(t) = k0(1 − r1t)er1 t + k1ter1 t + 1
a
∫ t
0
τer1 τ f(t− τ) dτ.",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1
is
y(t) = k0(1 − r1t)er1 t + k1ter1 t + 1
a
∫ t
0
τer1 τ f(t− τ) dτ.
10. Show that if p(s) = as2 + bs+ chas complex conjugate zeros λ ± iω then the solution of
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1
is
y(t) = eλt
[
k0(cos ωt − λ
ω sin ωt) + k1
ω sin ωt
]
+ 1
aω
∫ t
0
eλt f(t− τ) sin ωτdτ.",Elementary Differential Equations with Boundary Value Problems.pdf
"452 Chapter 8 Laplace Transforms
11. Let
w= L−1
( 1
as2 + bs+ c
)
,
where a,b , and care constants and a̸= 0 .
(a) Show that wis the solution of
aw′′+ bw′+ cw= 0 , w (0) = 0 , w ′(0) = 1
a.
(b) Let f be continuous on [0, ∞ ) and define
h(t) =
∫ t
0
w(t− τ)f(τ) dτ.
Use Leibniz’s rule for differentiating an integral with respect to a parameter to show that his
the solution of
ah′′+ bh′+ ch= f, h (0) = 0 , h ′(0) = 0 .
(c) Show that the function y in Eqn. ( 8.6.14) is the solution of Eqn. ( 8.6.13) provided that f is
continuous on [0, ∞ ); thus, it’s not necessary to assume that f has a Laplace transform.
12. Consider the initial value problem
ay′′+ by′+ cy = f(t), y (0) = 0 , y ′(0) = 0 , (A)
where a,b , and care constants, a̸= 0 , and
f(t) =
{
f0(t), 0 ≤ t<t 1,
f1(t), t ≥ t1.
Assume that f0 is continuous and of exponential order on [0, ∞ ) and f1 is continuous and of
exponential order on [t1, ∞ ). Let
p(s) = as2 + bs+ c.
(a) Show that the Laplace transform of the solution of (A) is",Elementary Differential Equations with Boundary Value Problems.pdf
"exponential order on [t1, ∞ ). Let
p(s) = as2 + bs+ c.
(a) Show that the Laplace transform of the solution of (A) is
Y(s) = F0(s) + e−st1 G(s)
p(s)
where g(t) = f1(t+ t1) − f0(t+ t1).
(b) Let wbe as in Exercise 11. Use Theorem 8.4.2 and the convolution theorem to show that the
solution of (A) is
y(t) =
∫ t
0
w(t− τ)f0(τ) dτ + u(t− t1)
∫ t−t1
0
w(t− t1 − τ)g(τ) dτ
for t> 0.
(c) Henceforth, assume only that f0 is continuous on [0, ∞ ) and f1 is continuous on [t1, ∞ ).
Use Exercise 11 (a) and (b) to show that
y′(t) =
∫ t
0
w′(t− τ)f0(τ) dτ + u(t− t1)
∫ t−t1
0
w′(t− t1 − τ)g(τ) dτ",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.7 Constant Coefficient Equations with Impulses 453
for t> 0, and
y′′(t) = f(t)
a +
∫ t
0
w′′(t− τ)f0(τ) dτ + u(t− t1)
∫ t−t1
0
w′′(t− t1 − τ)g(τ) dτ
for 0 < t < t1 and t > t1. Also, show y satisfies the differential equation in (A) on (0,t 1)
and (t1, ∞ ).
(d) Show that yand y′are continuous on [0, ∞ ).
13. Suppose
f(t) =















f0(t), 0 ≤ t<t 1,
f1(t), t 1 ≤ t<t 2,
.
.
.
fk−1(t), t k−1 ≤ t<t k ,
fk (t), t ≥ tk,
where fm is continuous on [tm, ∞ ) for m= 0 ,...,k (let t0 = 0 ), and define
gm (t) = fm (t+ tm) − fm−1 (t+ tm), m= 1 ,...,k.
Extend the results of Exercise
12 to show that the solution of
ay′′+ by′+ cy= f(t), y (0) = 0 , y ′(0) = 0
is
y(t) =
∫ t
0
w(t− τ)f0(τ) dτ +
k∑
m=1
u(t− tm)
∫ t−tm
0
w(t− tm − τ)gm(τ) dτ.
14. Let {tm }∞
m=0 be a sequence of points such that t0 = 0 , tm+1 > tm, and limm→∞ tm = ∞ . For
each nonegative integer mlet fm be continuous on [tm, ∞ ), and let f be defined on [0, ∞ ) by
f(t) = fm (t), t m ≤ t<t m+1 m= 0 , 1, 2 ....
Let",Elementary Differential Equations with Boundary Value Problems.pdf
"each nonegative integer mlet fm be continuous on [tm, ∞ ), and let f be defined on [0, ∞ ) by
f(t) = fm (t), t m ≤ t<t m+1 m= 0 , 1, 2 ....
Let
gm(t) = fm (t+ tm) − fm−1(t+ tm ), m = 1 ,...,k.
Extend the results of Exercise
13 to show that the solution of
ay′′+ by′+ cy= f(t), y (0) = 0 , y ′(0) = 0
is
y(t) =
∫ t
0
w(t− τ)f0(τ) dτ +
∞∑
m=1
u(t− tm)
∫ t−tm
0
w(t− tm − τ)gm(τ) dτ.
HIN T : See Exercise 30.
8.7 CONST ANT COEFFICIENT EQUA TIONS WITH IMPULSES
So far in this chapter, we’ve considered initial value probl ems for the constant coefficient equation
ay′′+ by′+ cy= f(t),",Elementary Differential Equations with Boundary Value Problems.pdf
"454 Chapter 8 Laplace Transforms
where f is continuous or piecewise continuous on [0, ∞ ). In this section we consider initial value prob-
lems where f represents a force that’s very large for a short time and zero otherwise. W e say that such
forces are impulsive. Impulsive forces occur, for example, when two objects coll ide. Since it isn’t feasible
to represent such forces as continuous or piecewise continu ous functions, we must construct a different
mathematical model to deal with them.
If f is an integrable function and f(t) = 0 for toutside of the interval [t0,t 0 + h], then
∫t0+h
t0
f(t) dt
is called the total impulse of f. W e’re interested in the idealized situation where his so small that the
total impulse can be assumed to be applied instantaneously a t t = t0. W e say in this case that f is an
impulse function . In particular, we denote by δ(t− t0) the impulse function with total impulse equal to",Elementary Differential Equations with Boundary Value Problems.pdf
"impulse function . In particular, we denote by δ(t− t0) the impulse function with total impulse equal to
one, applied at t= t0. (The impulse function δ(t) obtained by setting t0 = 0 is the Dirac δ function.) It
must be understood, however, that δ(t− t0) isn’t a function in the standard sense, since our “definition ”
implies that δ(t− t0) = 0 if t̸= t0, while
∫ t0
t0
δ(t− t0) dt= 1 .
From calculus we know that no function can have these propert ies; nevertheless, there’s a branch of
mathematics known as the theory of distributions where the definition can be made rigorous. Since the
theory of distributions is beyond the scope of this book, we’ ll take an intuitive approach to impulse
functions.
Our first task is to define what we mean by the solution of the ini tial value problem
ay′′+ by′+ cy = δ(t− t0), y (0) = 0 , y ′(0) = 0 ,
where t0 is a fixed nonnegative number. The next theorem will motivate our definition.",Elementary Differential Equations with Boundary Value Problems.pdf
"ay′′+ by′+ cy = δ(t− t0), y (0) = 0 , y ′(0) = 0 ,
where t0 is a fixed nonnegative number. The next theorem will motivate our definition.
Theorem 8.7.1 Suppose t0 ≥ 0. F or each positive number h, let yh be the solution of the initial value
problem
ay′′
h + by′
h + cyh = fh (t), y h (0) = 0 , y ′
h (0) = 0 , (8.7.1)
where
fh (t) =







0, 0 ≤ t<t 0,
1/h, t 0 ≤ t<t 0 + h,
0, t ≥ t0 + h,
(8.7.2)
so fh has unit total impulse equal to the area of the shaded rectang le in Figure
8.7.1. Then
lim
h→ 0+
yh (t) = u(t− t0)w(t− t0), (8.7.3)
where
w= L−1
( 1
as2 + bs+ c
)
.
Proof T aking Laplace transforms in ( 8.7.1) yields
(as2 + bs+ c)Yh(s) = Fh(s),
so
Yh (s) = Fh(s)
as2 + bs+ c.
The convolution theorem implies that
yh (t) =
∫ t
0
w(t− τ)fh (τ) dτ.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.7 Constant Coefficient Equations with Impulses 455
 1/h
 t0  t0+h  t
 y
Figure 8.7.1 y= fh (t)
Therefore, ( 8.7.2) implies that
yh (t) =













0, 0 ≤ t<t 0,
1
h
∫ t
t0
w(t− τ) dτ, t 0 ≤ t≤ t0 + h,
1
h
∫ t0+h
t0
w(t− τ) dτ, t>t 0 + h.
(8.7.4)
Since yh (t) = 0 for all hif 0 ≤ t≤ t0, it follows that
lim
h→ 0+
yh (t) = 0 if 0 ≤ t≤ t0. (8.7.5)
W e’ll now show that
lim
h→ 0+
yh (t) = w(t− t0) if t>t 0. (8.7.6)
Suppose tis fixed and t>t 0. From ( 8.7.4),
yh (t) = 1
h
∫ t0+h
t0
w(t− τ)dτ if h<t − t0. (8.7.7)
Since
1
h
∫ t0+h
t0
dτ = 1 , (8.7.8)
we can write
w(t− t0) = 1
hw(t− t0)
∫ t0+h
t0
dτ = 1
h
∫ t0+h
t0
w(t− t0) dτ.",Elementary Differential Equations with Boundary Value Problems.pdf
"456 Chapter 8 Laplace Transforms
From this and ( 8.7.7),
yh (t) − w(t− t0) = 1
h
∫ t0+h
t0
(w(t− τ) − w(t− t0)) dτ.
Therefore
|yh (t) − w(t− t0)| ≤1
h
∫ t0+h
t0
|w(t− τ) − w(t− t0)|dτ. (8.7.9)
Now let Mh be the maximum value of |w(t− τ) − w(t− t0)|as τ varies over the interval [t0,t 0 + h].
(Remember that tand t0 are fixed.) Then ( 8.7.8) and ( 8.7.9) imply that
|yh(t) − w(t− t0)| ≤1
hMh
∫ t0+h
t0
dτ = Mh . (8.7.10)
But limh→ 0+ Mh = 0 , since wis continuous. Therefore ( 8.7.10) implies ( 8.7.6). This and ( 8.7.5) imply
(8.7.3).
Theorem 8.7.1 motivates the next definition.
Definition 8.7.2 If t0 >0, then the solution of the initial value problem
ay′′+ by′+ cy = δ(t− t0), y (0) = 0 , y ′(0) = 0 , (8.7.11)
is defined to be
y = u(t− t0)w(t− t0),
where
w= L−1
( 1
as2 + bs+ c
)
.
In physical applications where the input f and the output y of a device are related by the differential
equation
ay′′+ by′+ cy= f(t),",Elementary Differential Equations with Boundary Value Problems.pdf
"as2 + bs+ c
)
.
In physical applications where the input f and the output y of a device are related by the differential
equation
ay′′+ by′+ cy= f(t),
wis called the impulse response of the device. Note that wis the solution of the initial value problem
aw′′+ bw′+ cw= 0 , w (0) = 0 , w ′(0) = 1 /a, (8.7.12)
as can be seen by using the Laplace transform to solve this pro blem. (V erify .) On the other hand, we can
solve ( 8.7.12) by the methods of Section 5.2 and show that wis defined on (−∞ , ∞ ) by
w= er2 t − er1t
a(r2 − r1) , w = 1
ater1 t , or w= 1
aω eλt sin ωt, (8.7.13)
depending upon whether the polynomial p(r) = ar2 + br+ chas distinct real zeros r1 and r2, a repeated
zero r1, or complex conjugate zeros λ ± iω. (In most physical applications, the zeros of the character istic
polynomial have negative real parts, so limt→∞ w(t) = 0 .) This means that y = u(t− t0)w(t− t0) is
defined on (−∞ , ∞ ) and has the following properties:
y(t) = 0 , t<t 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"defined on (−∞ , ∞ ) and has the following properties:
y(t) = 0 , t<t 0,
ay′′+ by′+ cy = 0 on (−∞ ,t 0) and (t0, ∞ ),
and
y′
−(t0) = 0 , y ′
+(t0) = 1 /a (8.7.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.7 Constant Coefficient Equations with Impulses 457
 t0
 y
 t
Figure 8.7.2 An illustration of Theorem 8.7.1
(remember that y′
−(t0) and y′
+(t0) are derivatives from the right and left, respectively) and y′(t0) does
not exist. Thus, even though we defined y = u(t−t0 )w(t−t0 ) to be the solution of ( 8.7.11), this function
doesn’t satisfy the differential equation in ( 8.7.11) at t0, since it isn’t differentiable there; in fact ( 8.7.14)
indicates that an impulse causes a jump discontinuity in vel ocity . (T o see that this is reasonable, think of
what happens when you hit a ball with a bat.) This means that th e initial value problem ( 8.7.11) doesn’t
make sense if t0 = 0 , since y′(0) doesn’t exist in this case. However y = u(t)w(t) can be defined to be
the solution of the modified initial value problem
ay′′+ by′+ cy= δ(t), y (0) = 0 , y ′
− (0) = 0 ,
where the condition on the derivative at t= 0 has been replaced by a condition on the derivative from the
left.",Elementary Differential Equations with Boundary Value Problems.pdf
"− (0) = 0 ,
where the condition on the derivative at t= 0 has been replaced by a condition on the derivative from the
left.
Figure 8.7.2 illustrates Theorem 8.7.1 for the case where the impulse response wis the first expression
in ( 8.7.13) and r1 and r2 are distinct and both negative. The solid curve in the figure i s the graph of w.
The dashed curves are solutions of ( 8.7.1) for various values of h. As hdecreases the graph of yh moves
to the left toward the graph of w.
Example 8.7.1 Find the solution of the initial value problem
y′′− 2y′+ y= δ(t− t0), y (0) = 0 , y ′(0) = 0 , (8.7.15)
where t0 >0. Then interpret the solution for the case where t0 = 0 .
Solution Here
w= L−1
( 1
s2 − 2s+ 1
)
= L−1
( 1
(s− 1)2
)
= te−t ,",Elementary Differential Equations with Boundary Value Problems.pdf
"458 Chapter 8 Laplace Transforms
0.1
0.2
0.3
0.4
 t0  t0 + 1  t 0 + 2  t0 + 3  t 0 + 4  t 0 + 5  t 0 + 6  t 0 + 7  t
 y
Figure 8.7.3 y = u(t− t0)(t− t0)e−(t−t0)
so Definition 8.7.2 yields
y= u(t− t0)(t− t0)e−(t−t0)
as the solution of ( 8.7.15) if t0 > 0. If t0 = 0 , then ( 8.7.15) doesn’t have a solution; however, y =
u(t)te−t (which we would usually write simply as y = te−t ) is the solution of the modified initial value
problem
y′′− 2y′+ y= δ(t), y (0) = 0 , y ′
−(0) = 0 .
The graph of y= u(t− t0)(t− t0)e−(t−t0) is shown in Figure 8.7.3
Definition 8.7.2 and the principle of superposition motivate the next definit ion.
Definition 8.7.3 Suppose α is a nonzero constant and f is piecewise continuous on [0, ∞ ). If t0 > 0,
then the solution of the initial value problem
ay′′+ by′+ cy= f(t) + αδ (t− t0), y (0) = k0, y ′(0) = k1
is defined to be
y(t) = ˆ y(t) + αu(t− t0)w(t− t0),
where ˆyis the solution of
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1.",Elementary Differential Equations with Boundary Value Problems.pdf
"is defined to be
y(t) = ˆ y(t) + αu(t− t0)w(t− t0),
where ˆyis the solution of
ay′′+ by′+ cy= f(t), y (0) = k0, y ′(0) = k1.
This definition also applies if t0 = 0 , provided that the initial condition y′(0) = k1 is replaced by
y′
− (0) = k1.
Example 8.7.2 Solve the initial value problem
y′′+ 6 y′+ 5 y = 3 e−2t + 2 δ(t− 1), y (0) = −3, y ′(0) = 2 . (8.7.16)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.7 Constant Coefficient Equations with Impulses 459
Solution W e leave it to you to show that the solution of
y′′+ 6 y′+ 5 y= 3 e−2t, y (0) = −3, y′(0) = 2
is
ˆy= −e−2t + 1
2 e−5t − 5
2 e−t .
Since
w(t) = L−1
( 1
s2 + 6 s+ 5
)
= L−1
( 1
(s+ 1)( s+ 5)
)
= 1
4 L−1
( 1
s+ 1 − 1
s+ 5
)
= e−t − e−5t
4 ,
the solution of ( 8.7.16) is
y = −e−2t + 1
2 e−5t − 5
2 e−t + u(t− 1) e−(t−1) − e−5(t−1)
2 (8.7.17)
(Figure 8.7.4) .
 t
 y
1 2 3 4
−1
−2
−3
 t = 1
Figure 8.7.4 Graph of ( 8.7.17)
1 2 3 4 5 6 7 8 9 10 11 12 13
1
2
3
4
5
−1
−2
−3
 t = π  t = 2π
 t
 y
Figure 8.7.5 Graph of ( 8.7.19)
Definition 8.7.3 can be extended in the obvious way to cover the case where the f orcing function
contains more than one impulse.
Example 8.7.3 Solve the initial value problem
y′′+ y= 1 + 2 δ(t− π) − 3δ(t− 2π), y (0) = −1, y′(0) = 2 . (8.7.18)
Solution W e leave it to you to show that
ˆy= 1 − 2 cos t+ 2 sin t
is the solution of
y′′+ y= 1 , y (0) = −1, y ′(0) = 2 .
Since
w= L−1
( 1
s2 + 1
)
= sin t,",Elementary Differential Equations with Boundary Value Problems.pdf
"460 Chapter 8 Laplace Transforms
the solution of ( 8.7.18) is
y = 1 − 2 cos t+ 2 sin t+ 2 u(t− π) sin(t− π) − 3u(t− 2π) sin(t− 2π)
= 1 − 2 cos t+ 2 sin t− 2u(t− π) sin t− 3u(t− 2π) sin t,
or
y=





1 − 2 cos t+ 2 sin t, 0 ≤ t<π,
1 − 2 cos t, π ≤ t< 2π,
1 − 2 cos t− 3 sin t, t ≥ 2π
(8.7.19)
(Figure
8.7.5).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.7 Constant Coefficient Equations with Impulses 461
8.7 Exercises
In Exercises 1– 20 solve the initial value problem. Where indicated by C/G , graph the solution.
1. y′′+ 3 y′+ 2 y= 6 e2t + 2 δ(t− 1), y (0) = 2 , y ′(0) = −6
2. C/G y′′+ y′− 2y= −10e−t + 5 δ(t− 1), y (0) = 7 , y ′(0) = −9
3. y′′− 4y= 2 e−t + 5 δ(t− 1), y (0) = −1, y ′(0) = 2
4. C/G y′′+ y= sin 3 t+ 2 δ(t− π/ 2), y (0) = 1 , y ′(0) = −1
5. y′′+ 4 y= 4 + δ(t− 3π), y (0) = 0 , y ′(0) = 1
6. y′′− y = 8 + 2 δ(t− 2), y (0) = −1, y ′(0) = 1
7. y′′+ y′= et + 3 δ(t− 6), y (0) = −1, y ′(0) = 4
8. y′′+ 4 y= 8 e2t + δ(t− π/ 2), y (0) = 8 , y ′(0) = 0
9. C/G y′′+ 3 y′+ 2 y= 1 + δ(t− 1), y (0) = 1 , y ′(0) = −1
10. y′′+ 2 y′+ y = et + 2 δ(t− 2), y (0) = −1, y ′(0) = 2
11. C/G y′′+ 4 y= sin t+ δ(t− π/ 2), y (0) = 0 , y ′(0) = 2
12. y′′+ 2 y′+ 2 y= δ(t− π) − 3δ(t− 2π), y (0) = −1, y ′(0) = 2
13. y′′+ 4 y′+ 13 y= δ(t− π/ 6) + 2 δ(t− π/ 3), y (0) = 1 , y ′(0) = 2
14. 2y′′− 3y′− 2y= 1 + δ(t− 2), y (0) = −1, y ′(0) = 2",Elementary Differential Equations with Boundary Value Problems.pdf
"13. y′′+ 4 y′+ 13 y= δ(t− π/ 6) + 2 δ(t− π/ 3), y (0) = 1 , y ′(0) = 2
14. 2y′′− 3y′− 2y= 1 + δ(t− 2), y (0) = −1, y ′(0) = 2
15. 4y′′− 4y′+ 5 y= 4 sin t− 4 cos t+ δ(t− π/ 2) − δ(t− π), y (0) = 1 , y ′(0) = 1
16. y′′+ y = cos 2 t+ 2 δ(t− π/ 2) − 3δ(t− π), y (0) = 0 , y ′(0) = −1
17. C/G y′′− y= 4 e−t − 5δ(t− 1) + 3 δ(t− 2), y (0) = 0 , y ′(0) = 0
18. y′′+ 2 y′+ y = et − δ(t− 1) + 2 δ(t− 2), y (0) = 0 , y ′(0) = −1
19. y′′+ y = f(t) + δ(t− 2π), y (0) = 0 , y ′(0) = 1 , and
f(t) =
{
sin 2 t, 0 ≤ t<π,
0, t ≥ π.
20. y′′+ 4 y= f(t) + δ(t− π) − 3δ(t− 3π/ 2), y (0) = 1 , y ′(0) = −1, and
f(t) =
{
1, 0 ≤ t<π/ 2,
2, t ≥ π/ 2
21. y′′+ y = δ(t), y (0) = 1 , y ′
−(0) = −2
22. y′′− 4y= 3 δ(t), y (0) = −1, y ′
−(0) = 7
23. y′′+ 3 y′+ 2 y= −5δ(t), y (0) = 0 , y ′
−(0) = 0
24. y′′+ 4 y′+ 4 y= −δ(t), y (0) = 1 , y ′
−(0) = 5
25. 4y′′+ 4 y′+ y= 3 δ(t), y (0) = 1 , y ′
− (0) = −6
In Exercises 26-28, solve the initial value problem
ay′′
h + by′
h + cyh =







0, 0 ≤ t<t 0,
1/h, t 0 ≤ t<t 0 + h,",Elementary Differential Equations with Boundary Value Problems.pdf
"− (0) = −6
In Exercises 26-28, solve the initial value problem
ay′′
h + by′
h + cyh =







0, 0 ≤ t<t 0,
1/h, t 0 ≤ t<t 0 + h,
0, t ≥ t0 + h,
yh (0) = 0 , y ′
h (0) = 0 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"462 Chapter 8 Laplace Transforms
where t0 >0 and h> 0. Then find
w= L−1
( 1
as2 + bs+ c
)
and verify Theorem 8.7.1 by graphing wand yh on the same axes, for small positive values of h.
26. L y′′+ 2 y′+ 2 y= fh (t), y (0) = 0 , y ′(0) = 0
27. L y′′+ 2 y′+ y = fh (t), y (0) = 0 , y ′(0) = 0
28. L y′′+ 3 y′+ 2 y= fh (t), y (0) = 0 , y ′(0) = 0
29. Recall from Section 6.2 that the displacement of an object of mass min a spring–mass system in
free damped oscillation is
my′′+ cy′+ ky = 0 , y (0) = y0, y ′(0) = v0,
and that ycan be written as
y = Re−ct/ 2m cos(ω1t− φ)
if the motion is underdamped. Suppose y(τ) = 0 . Find the impulse that would have to be applied
to the object at t= τ to put it in equilibrium.
30. Solve the initial value problem. Find a formula that does not involve step functions and represents
yon each subinterval of [0, ∞ ) on which the forcing function is zero.
(a) y′′− y =
∞∑
k=1
δ(t− k), y (0) = 0 , y ′(0) = 1
(b) y′′+ y =
∞∑
k=1
δ(t− 2kπ), y (0) = 0 , y ′(0) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) y′′− y =
∞∑
k=1
δ(t− k), y (0) = 0 , y ′(0) = 1
(b) y′′+ y =
∞∑
k=1
δ(t− 2kπ), y (0) = 0 , y ′(0) = 1
(c) y′′− 3y′+ 2 y=
∞∑
k=1
δ(t− k), y (0) = 0 , y ′(0) = 1
(d) y′′+ y =
∞∑
k=1
δ(t− kπ), y (0) = 0 , y ′(0) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 8.8 A Brief T able of Laplace Transforms 463
8.8 A BRIEF T ABLE OF LAPLACE TRANSFORMS
f(t) F(s)
1 1
s (s> 0)
tn n!
sn+1 (s> 0)
(n= integer >0)
tp, p> −1 Γ( p+ 1)
s(p+1) (s> 0)
eat 1
s− a (s>a )
tneat n!
(s− a)n+1 (s> 0)
(n= integer >0)
cos ωt s
s2 + ω2 (s> 0)
sin ωt ω
s2 + ω2 (s> 0)
eλt cos ωt s− λ
(s− λ)2 + ω2 (s>λ )
eλt sin ωt ω
(s− λ)2 + ω2 (s>λ )
cosh bt s
s2 − b2 (s> |b|)
sinh bt b
s2 − b2 (s> |b|)
tcos ωt s2 − ω2
(s2 + ω2)2 (s> 0)",Elementary Differential Equations with Boundary Value Problems.pdf
"464 Chapter 8 Laplace Transforms
tsin ωt 2ωs
(s2 + ω2)2 (s> 0)
sin ωt − ωt cos ωt 2ω3
(s2 + ω2)2 (s> 0)
ωt − sin ωt ω3
s2(s2 + ω2)2 (s> 0)
1
t sin ωt arctan
(ω
s
)
(s> 0)
eat f(t) F(s− a)
tkf(t) ( −1)kF(k)(s)
f(ωt) 1
ω F
(s
ω
)
, ω > 0
u(t− τ) e−τ s
s (s> 0)
u(t− τ)f(t− τ) (τ >0) e−τ s F(s)
∫ t
o
f(τ)g(t− τ) dτ F (s) ·G(s)
δ(t− a) e−as (s> 0)",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 9
Linear Higher Order Equations
IN THIS CHAPTER we extend the results obtained in Chapter 5 fo r linear second order equations to
linear higher order equations.
SECTION 9.1 presents a theoretical introduction to linear h igher order equations.
SECTION 9.2 discusses higher order constant coefficient hom ogeneous equations.
SECTION 9.3 presents the method of undetermined coefficient s for higher order equations.
SECTION 9.4 extends the method of variation of parameters to higher order equations.
465",Elementary Differential Equations with Boundary Value Problems.pdf
"466 Chapter 9 Linear Higher Order Equations
9.1 INTRODUCTION TO LINEAR HIGHER ORDER EQUA TIONS
An nth order differential equation is said to be linear if it can be written in the form
y(n) + p1(x)y(n−1) + · · ·+ pn(x)y = f(x). (9.1.1)
W e considered equations of this form with n = 1 in Section 2.1 and with n = 2 in Chapter 5. In this
chapter nis an arbitrary positive integer.
In this section we sketch the general theory of linear nth order equations. Since this theory has already
been discussed for n= 2 in Sections 5.1 and 5.3, we’ll omit proofs.
For convenience, we consider linear differential equation s written as
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn (x)y= F(x), (9.1.2)
which can be rewritten as ( 9.1.1) on any interval on which P0 has no zeros, with p1 = P1/P 0, . . . ,
pn = Pn/P 0 and f = F/P 0. For simplicity , throughout this chapter we’ll abbreviate the left side of
(9.1.2) by Ly; that is,
Ly= P0y(n) + P1y(n−1) + · · ·+ Pny.",Elementary Differential Equations with Boundary Value Problems.pdf
"(9.1.2) by Ly; that is,
Ly= P0y(n) + P1y(n−1) + · · ·+ Pny.
W e say that the equation Ly = F is normal on (a,b ) if P0, P1, . . . , Pn and F are continuous on (a,b )
and P0 has no zeros on (a,b ). If this is so then Ly = F can be written as ( 9.1.1) with p1, . . . , pn and f
continuous on (a,b ).
The next theorem is analogous to Theorem 5.3.1.
Theorem 9.1.1 Suppose Ly= F is normal on (a,b ), let x0 be a point in (a,b ), and let k0, k1, . . . , kn−1
be arbitrary real numbers . Then the initial value problem
Ly= F, y (x0) = k0, y ′(x0) = k1,..., y (n−1)(x0) = kn−1
has a unique solution on (a,b ).
Homogeneous Equations
Eqn. ( 9.1.2) is said to be homogeneous if F ≡ 0 and nonhomogeneous otherwise. Since y ≡ 0 is
obviously a solution of Ly= 0 , we call it the trivial solution. Any other solution is nontrivial.
If y1, y2, . . . , yn are defined on (a,b ) and c1, c2, . . . , cn are constants, then
y = c1y1 + c2y2 + · · ·+ cn yn (9.1.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"If y1, y2, . . . , yn are defined on (a,b ) and c1, c2, . . . , cn are constants, then
y = c1y1 + c2y2 + · · ·+ cn yn (9.1.3)
is a linear combination of {y1,y 2 ...,y n}. It’s easy to show that if y1, y2, . . . , yn are solutions of Ly= 0
on (a,b ), then so is any linear combination of {y1,y 2,...,y n}. (See the proof of Theorem 5.1.2.) W e
say that {y1,y 2,...,y n}is a fundamental set of solutions of Ly= 0 on (a,b ) if every solution of Ly= 0
on (a,b ) can be written as a linear combination of {y1,y 2,...,y n}, as in ( 9.1.3). In this case we say that
(9.1.3) is the general solution of Ly= 0 on (a,b ).
It can be shown (Exercises 14 and 15) that if the equation Ly = 0 is normal on (a,b ) then it has in-
finitely many fundamental sets of solutions on (a,b ). The next definition will help to identify fundamental
sets of solutions of Ly= 0 .
W e say that {y1,y 2,...,y n}is linearly independent on (a,b ) if the only constants c1, c2, . . . , cn such
that",Elementary Differential Equations with Boundary Value Problems.pdf
"sets of solutions of Ly= 0 .
W e say that {y1,y 2,...,y n}is linearly independent on (a,b ) if the only constants c1, c2, . . . , cn such
that
c1y1(x) + c2y2(x) + · · ·+ cnyn (x) = 0 , a<x<b, (9.1.4)
are c1 = c2 = · · ·= cn = 0 . If ( 9.1.4) holds for some set of constants c1, c2, . . . , cn that are not all zero,
then {y1,y 2,...,y n }is linearly dependent on (a,b )
The next theorem is analogous to Theorem 5.1.3.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.1 Introduction to Linear Higher Order Equations 467
Theorem 9.1.2 If Ly = 0 is normal on (a,b ), then a set {y1,y 2,...,y n }of nsolutions of Ly = 0 on
(a,b ) is a fundamental set if and only if it’s linearly independent on (a,b ).
Example 9.1.1 The equation
x3y′′′− x2y′′− 2xy′+ 6 y = 0 (9.1.5)
is normal and has the solutions y1 = x2, y2 = x3, and y3 = 1 /x on (−∞ , 0) and (0, ∞ ). Show that
{y1,y 2,y 3}is linearly independent on (−∞ , 0) and (0, ∞ ). Then find the general solution of ( 9.1.5) on
(−∞ , 0) and (0, ∞ ).
Solution Suppose
c1x2 + c2x3 + c3
x = 0 (9.1.6)
on (0, ∞ ). W e must show that c1 = c2 = c3 = 0 . Differentiating ( 9.1.6) twice yields the system
c1x2 + c2x3 + c3
x = 0
2c1x+ 3 c2x2 − c3
x2 = 0
2c1 + 6 c2x+ 2c3
x3 = 0 .
(9.1.7)
If ( 9.1.7) holds for all xin (0, ∞ ), then it certainly holds at x= 1 ; therefore,
c1 + c2 + c3 = 0
2c1 + 3 c2 − c3 = 0
2c1 + 6 c2 + 2 c3 = 0 .
(9.1.8)",Elementary Differential Equations with Boundary Value Problems.pdf
"c1 + c2 + c3 = 0
2c1 + 3 c2 − c3 = 0
2c1 + 6 c2 + 2 c3 = 0 .
(9.1.8)
By solving this system directly , you can verify that it has on ly the trivial solution c1 = c2 = c3 =
0; however, for our purposes it’s more useful to recall from li near algebra that a homogeneous linear
system of nequations in nunknowns has only the trivial solution if its determinant is nonzero. Since the
determinant of ( 9.1.8) is ⏐
⏐
⏐
⏐
⏐
⏐
1 1 1
2 3 −1
2 6 2
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
1 0 0
2 1 −3
2 4 0
⏐
⏐
⏐
⏐
⏐
⏐
= 12 ,
it follows that (
9.1.8) has only the trivial solution, so {y1,y 2,y 3}is linearly independent on (0, ∞ ). Now
Theorem 9.1.2 implies that
y= c1x2 + c2x3 + c3
x
is the general solution of ( 9.1.5) on (0, ∞ ). T o see that this is also true on (−∞ , 0), assume that ( 9.1.6)
holds on (−∞ , 0). Setting x= −1 in ( 9.1.7) yields
c1 − c2 − c3 = 0
−2c1 + 3 c2 − c3 = 0
2c1 − 6c2 − 2c3 = 0 .
Since the determinant of this system is
⏐
⏐
⏐
⏐
⏐
⏐
1 −1 −1
−2 3 −1
2 −6 −2
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
1 0 0",Elementary Differential Equations with Boundary Value Problems.pdf
"2c1 − 6c2 − 2c3 = 0 .
Since the determinant of this system is
⏐
⏐
⏐
⏐
⏐
⏐
1 −1 −1
−2 3 −1
2 −6 −2
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
1 0 0
−2 1 −3
2 −4 0
⏐
⏐
⏐
⏐
⏐
⏐
= −12,
it follows that c1 = c2 = c3 = 0 ; that is, {y1,y 2,y 3}is linearly independent on (−∞ , 0).",Elementary Differential Equations with Boundary Value Problems.pdf
"468 Chapter 9 Linear Higher Order Equations
Example 9.1.2 The equation
y(4) + y′′′− 7y′′− y′+ 6 y= 0 (9.1.9)
is normal and has the solutions y1 = ex, y2 = e−x, y3 = e2x, and y4 = e−3x on (−∞ , ∞ ). (V erify .)
Show that {y1,y 2,y 3,y 4}is linearly independent on (−∞ , ∞ ). Then find the general solution of ( 9.1.9).
Solution Suppose c1, c2, c3, and c4 are constants such that
c1ex + c2e−x + c3e2x + c4e−3x = 0 (9.1.10)
for all x. W e must show that c1 = c2 = c3 = c4 = 0 . Differentiating ( 9.1.10) three times yields the
system
c1ex + c2e−x + c3e2x + c4e−3x = 0
c1ex − c2e−x + 2 c3e2x − 3c4e−3x = 0
c1ex + c2e−x + 4 c3e2x + 9 c4e−3x = 0
c1ex − c2e−x + 8 c3e2x − 27c4e−3x = 0 .
(9.1.11)
If ( 9.1.11) holds for all x, then it certainly holds for x= 0 . Therefore
c1 + c2 + c3 + c4 = 0
c1 − c2 + 2 c3 − 3c4 = 0
c1 + c2 + 4 c3 + 9 c4 = 0
c1 − c2 + 8 c3 − 27c4 = 0 .
The determinant of this system is
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 1
1 −1 2 −3
1 1 4 9
1 −1 8 −27
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 1",Elementary Differential Equations with Boundary Value Problems.pdf
"The determinant of this system is
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 1
1 −1 2 −3
1 1 4 9
1 −1 8 −27
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 1
0 −2 1 −4
0 0 3 8
0 −2 7 −28
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
−2 1 −4
0 3 8
−2 7 −28
⏐
⏐
⏐
⏐
⏐
⏐
=
⏐
⏐
⏐
⏐
⏐
⏐
−2 1 −4
0 3 8
0 6 −24
⏐
⏐
⏐
⏐
⏐
⏐
= −2
⏐
⏐
⏐
⏐
3 8
6 −24
⏐
⏐
⏐
⏐= 240 ,
(9.1.12)
so the system has only the trivial solution c1 = c2 = c3 = c4 = 0 . Now Theorem
9.1.2 implies that
y= c1ex + c2e−x + c3e2x + c4e−3x
is the general solution of ( 9.1.9).
The Wronskian
W e can use the method used in Examples 9.1.1 and 9.1.2 to test nsolutions {y1,y 2,...,y n}of any nth
order equation Ly = 0 for linear independence on an interval (a,b ) on which the equation is normal.
Thus, if c1, c2 ,. . . , cn are constants such that
c1y1 + c2y2 + · · ·+ cnyn = 0 , a<x<b,
then differentiating n− 1 times leads to the n× nsystem of equations
c1y1(x) + c2y2(x)+ · · ·+cn yn (x) = 0
c1y′
1(x) + c2y′
2(x)+ · · ·+cn y′
n (x) = 0
.
.
.
c1y(n−1)
1 (x) + c2y(n−1)",Elementary Differential Equations with Boundary Value Problems.pdf
"c1y1(x) + c2y2(x)+ · · ·+cn yn (x) = 0
c1y′
1(x) + c2y′
2(x)+ · · ·+cn y′
n (x) = 0
.
.
.
c1y(n−1)
1 (x) + c2y(n−1)
2 (x)+ · · ·+cn y(n−1)
n (x) = 0
(9.1.13)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.1 Introduction to Linear Higher Order Equations 469
for c1, c2, . . . , cn . For a fixed x, the determinant of this system is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1(x) y2(x) · · · yn (x)
y′
1(x) y′
2(x) · · · y′
n (x)
.
.
. .
.
. . . . .
.
.
y(n−1)
1 (x) y(n−1)
2 (x) · · ·y(n−1)
n (x)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
W e call this determinant the
Wronskian of {y1,y 2,...,y n }. If W(x) ̸= 0 for some xin (a,b ) then the
system ( 9.1.13) has only the trivial solution c1 = c2 = · · ·= cn = 0 , and Theorem 9.1.2 implies that
y = c1y1 + c2y2 + · · ·+ cn yn
is the general solution of Ly= 0 on (a,b ).
The next theorem generalizes Theorem 5.1.4. The proof is sketched in (Exercises 17– 20).
Theorem 9.1.3 Suppose the homogeneous linear nth order equation
P0(x)y(n) + P1(x)yn−1 + · · ·+ Pn(x)y = 0 (9.1.14)
is normal on (a,b ), let y1,y 2, . . . , yn be solutions of (9.1.14) on (a,b ), and let x0 be in (a,b ). Then the
Wronskian of {y1,y 2,...,y n }is given by
W(x) = W(x0) exp
{
−
∫ x
x0
P1(t)",Elementary Differential Equations with Boundary Value Problems.pdf
"Wronskian of {y1,y 2,...,y n }is given by
W(x) = W(x0) exp
{
−
∫ x
x0
P1(t)
P0(t) dt
}
, a<x<b. (9.1.15)
Therefore, either W has no zeros in (a,b ) or W ≡ 0 on (a,b ).
Formula ( 9.1.15) is Abel’s formula .
The next theorem is analogous to Theorem 5.1.6..
Theorem 9.1.4 Suppose Ly = 0 is normal on (a,b ) and let y1, y2, . . . , yn be nsolutions of Ly = 0 on
(a,b ). Then the following statements are equivalent ; that is , they are either all true or all false :
(a) The general solution of Ly= 0 on (a,b ) is y= c1y1 + c2y2 + · · ·+ cnyn .
(b) {y1,y 2,...,y n}is a fundamental set of solutions of Ly= 0 on (a,b ).
(c) {y1,y 2,...,y n}is linearly independent on (a,b ).
(d) The Wronskian of {y1,y 2,...,y n }is nonzero at some point in (a,b ).
(e) The Wronskian of {y1,y 2,...,y n }is nonzero at all points in (a,b ).
Example 9.1.3 In Example 9.1.1 we saw that the solutions y1 = x2, y2 = x3, and y3 = 1 /x of
x3y′′′− x2y′′− 2xy′+ 6 y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 9.1.3 In Example 9.1.1 we saw that the solutions y1 = x2, y2 = x3, and y3 = 1 /x of
x3y′′′− x2y′′− 2xy′+ 6 y = 0
are linearly independent on (−∞ , 0) and (0, ∞ ). Calculate the Wronskian of {y1,y 2,y 3}.
Solution If x̸= 0 , then
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x2 x3 1
x
2x 3x2 − 1
x2
2 6 x 2
x3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 2 x3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 x 1
x3
2 3 x − 1
x3
1 3 x 1
x3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,",Elementary Differential Equations with Boundary Value Problems.pdf
"470 Chapter 9 Linear Higher Order Equations
where we factored x2, x, and 2 out of the first, second, and third rows of W(x), respectively . Adding the
second row of the last determinant to the first and third rows y ields
W(x) = 2 x3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
3 4 x 0
2 3 x − 1
x3
3 6 x 0
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 2 x3
(1
x3
) ⏐
⏐
⏐
⏐
3 4 x
3 6 x
⏐
⏐
⏐
⏐= 12 x.
Therefore W(x) ̸= 0 on (−∞ , 0) and (0, ∞ ).
Example 9.1.4 In Example
9.1.2 we saw that the solutions y1 = ex , y2 = e−x, y3 = e2x, and y4 = e−3x
of
y(4) + y′′′− 7y′′− y′+ 6 y= 0
are linearly independent on every open interval. Calculate the Wronskian of {y1,y 2,y 3,y 4}.
Solution For all x,
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
ex e−x e2x e−3x
ex −e−x 2e2x −3e−3x
ex e−x 4e2x 9e−3x
ex −e−x 8e2x −27e−3x
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
Factoring the exponential common factor from each row yield s
W(x) = e−x
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 1
1 −1 2 −3
1 1 4 9
1 −1 8 −27
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 240 e−x,
from (
9.1.12).",Elementary Differential Equations with Boundary Value Problems.pdf
"W(x) = e−x
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 1
1 −1 2 −3
1 1 4 9
1 −1 8 −27
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 240 e−x,
from (
9.1.12).
REM A RK : Under the assumptions of Theorem 9.1.4, it isn’t necessary to obtain a formula for W(x). Just
evaluate W(x) at a convenient point in (a,b ), as we did in Examples 9.1.1 and 9.1.2.
Theorem 9.1.5 Suppose cis in (a,b ) and α1,α 2, . . . , are real numbers, not all zero. Under the assump-
tions of Theorem 10.3.3, suppose y1 and y2 are solutions of (5.1.35) such that
αyi(c) + y′
i(c) + · · ·+ y(n−1)
i (c) = 0 , 1 ≤ i≤ n. (9.1.16)
Then {y1,y 2,...y n }isn’t linearly independent on (a,b ).
Proof Since α1, α2, . . . , αn are not all zero, ( 9.1.14) implies that
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1(c) y′
1(c) · · ·y(n−1)
1 (c)
y2(c) y′
2(c) · · ·y(n−1)
2 (c)
.
.
. .
.
. . . . .
.
.
yn (c) y′
n (c) · · ·y(n−1)
n (c)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 0 ,
so ⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1(c) y2(c) · · · yn(c)
y′
1(c) y′
2(c) · · · y′
n(c)
.
.
. .
.
. . . . .
.
.
y(n−1)
1 (c) y(n−1)
2 (c)(c) · · ·y(n−1)
n (c)(c)
⏐
⏐",Elementary Differential Equations with Boundary Value Problems.pdf
"⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1(c) y2(c) · · · yn(c)
y′
1(c) y′
2(c) · · · y′
n(c)
.
.
. .
.
. . . . .
.
.
y(n−1)
1 (c) y(n−1)
2 (c)(c) · · ·y(n−1)
n (c)(c)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 0
and Theorem
9.1.4 implies the stated conclusion.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.1 Introduction to Linear Higher Order Equations 471
General Solution of a Nonhomogeneous Equation
The next theorem is analogous to Theorem 5.3.2. It shows how to find the general solution of Ly = F
if we know a particular solution of Ly = F and a fundamental set of solutions of the complementary
equation Ly= 0 .
Theorem 9.1.6 Suppose Ly = F is normal on (a,b ). Let yp be a particular solution of Ly = F on
(a,b ), and let {y1,y 2,...,y n }be a fundamental set of solutions of the complementary equat ion Ly= 0
on (a,b ). Then yis a solution of Ly= F on (a,b ) if and only if
y = yp + c1y1 + c2y2 + · · ·+ cnyn ,
where c1,c 2,...,c n are constants.
The next theorem is analogous to Theorem 5.3.2.
Theorem 9.1.7 [The Principle of Superposition ] Suppose for each i = 1 , 2, . . . , r, the function ypi is a
particular solution of Ly= Fi on (a,b ). Then
yp = yp1 + yp2 + · · ·+ ypr
is a particular solution of
Ly= F1(x) + F2(x) + · · ·+ Fr (x)
on (a,b ).",Elementary Differential Equations with Boundary Value Problems.pdf
"particular solution of Ly= Fi on (a,b ). Then
yp = yp1 + yp2 + · · ·+ ypr
is a particular solution of
Ly= F1(x) + F2(x) + · · ·+ Fr (x)
on (a,b ).
W e’ll apply Theorems 9.1.6 and 9.1.7 throughout the rest of this chapter.
9.1 Exercises
1. V erify that the given function is the solution of the initial value problem.
(a) x3y′′′− 3x2y′′+ 6 xy′− 6y= −24
x, y (−1) = 0 , y′(−1) = 0 , y ′′(−1) = 0 ;
y= −6x− 8x2 − 3x3 + 1
x
(b) y′′′− 1
xy′′− y′+ 1
xy = x2 − 4
x4 , y (1) = 3
2 , y ′(1) = 1
2 , y′′(1) = 1 ;
y = x+ 1
2x
(c) xy′′′− y′′− xy′+ y= x2, y (1) = 2 , y ′(1) = 5 , y ′′(1) = −1;
y = −x2 − 2 + 2 e(x−1) − e−(x−1) + 4 x
(d) 4x3y′′′+ 4 x2y′′− 5xy′+ 2 y= 30 x2, y (1) = 5 , y ′(1) = 17
2 ;
y′′(1) = 63
4 ; y= 2 x2 ln x− x1/ 2 + 2 x−1/ 2 + 4 x2
(e) x4y(4) − 4x3y′′′+ 12 x2y′′− 24xy′+ 24 y= 6 x4, y (1) = −2,
y′(1) = −9, y ′′(1) = −27, y ′′′(1) = −52;
y = x4 ln x+ x− 2x2 + 3 x3 − 4x4
(f) xy(4) − y′′′− 4xy′′+ 4 y′= 96 x2, y (1) = −5, y ′(1) = −24
y′′(1) = −36; y′′′(1) = −48; y = 9 − 12x+ 6 x2 − 8x3",Elementary Differential Equations with Boundary Value Problems.pdf
"472 Chapter 9 Linear Higher Order Equations
2. Solve the initial value problem
x3y′′′− x2y′′− 2xy′+ 6 y= 0 , y (−1) = −4, y ′(−1) = −14, y ′′(−1) = −20.
HIN T : See Example 9.1.1.
3. Solve the initial value problem
y(4) + y′′′− 7y′′− y′+ 6 y= 0 , y (0) = 5 , y ′(0) = −6, y ′′(0) = 10 , y ′′′(0) − 36.
HIN T : See Example 9.1.2.
4. Find solutions y1, y2, . . . , yn of the equation y(n) = 0 that satisfy the initial conditions
y(j)
i (x0) =
{
0, j ̸= i− 1,
1, j = i− 1, 1 ≤ i≤ n.
5. (a) V erify that the function
y= c1x3 + c2x2 + c3
x
satisfies
x3y′′′− x2y′′− 2xy′+ 6 y= 0 (A)
if c1, c2, and c3 are constants.
(b) Use (a) to find solutions y1, y2, and y3 of (A) such that
y1(1) = 1 , y ′
1(1) = 0 , y ′′
1 (1) = 0
y2(1) = 0 , y ′
2(1) = 1 , y ′′
2 (1) = 0
y3(1) = 0 , y ′
3(1) = 0 , y ′′
3 (1) = 1 .
(c) Use (b) to find the solution of (A) such that
y(1) = k0, y ′(1) = k1, y ′′(1) = k2.
6. V erify that the given functions are solutions of the given eq uation, and show that they form a",Elementary Differential Equations with Boundary Value Problems.pdf
"y(1) = k0, y ′(1) = k1, y ′′(1) = k2.
6. V erify that the given functions are solutions of the given eq uation, and show that they form a
fundamental set of solutions of the equation on any interval on which the equation is normal.
(a) y′′′+ y′′− y′− y= 0; {ex, e−x, xe−x}
(b) y′′′− 3y′′+ 7 y′− 5y= 0; {ex, ex cos 2x, ex sin 2 x}.
(c) xy′′′− y′′− xy′+ y= 0; {ex, e−x, x}
(d) x2y′′′+ 2 xy′′− (x2 + 2) y= 0; {ex/x, e −x/x, 1}
(e) (x2 − 2x+ 2) y′′′− x2y′′+ 2 xy′− 2y= 0; {x, x2, ex}
(f) (2x− 1)y(4) − 4xy′′′+ (5 − 2x)y′′+ 4 xy′− 4y = 0; {x, ex, e−x,e 2x}
(g) xy(4) − y′′′− 4xy′+ 4 y′= 0; {1,x 2, e2x, e−2x}
7. Find the Wronskian W of a set of three solutions of
y′′′+ 2 xy′′+ ex y′− y= 0 ,
given that W(0) = 2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.1 Introduction to Linear Higher Order Equations 473
8. Find the Wronskian W of a set of four solutions of
y(4) + (tan x)y′′′+ x2y′′+ 2 xy= 0 ,
given that W(π/ 4) = K.
9. (a) Evaluate the Wronskian W {ex, xex, x2ex }. Evaluate W(0).
(b) V erify that y1, y2, and y3 satisfy
y′′′− 3y′′+ 3 y′− y= 0 . (A)
(c) Use W(0) from (a) and Abel’s formula to calculate W(x).
(d) What is the general solution of (A)?
10. Compute the Wronskian of the given set of functions.
(a) {1, ex, e−x} (b) {ex, ex sin x, ex cos x}
(c) {2, x+ 1 , x2 + 2 } (d) x, xln x, 1/x }
(e) {1, x, x2
2! , x3
3! , · · ·, xn
n! } (f) {ex, e−x, x}
(g) {ex/x, e −x/x, 1} (h) {x, x2, ex}
(i) {x, x3, 1/x, 1/x 2} (j) {ex, e−x, x, e2x}
(k) {e2x, e−2x, 1, x2}
11. Suppose Ly = 0 is normal on (a,b ) and x0 is in (a,b ). Use Theorem 9.1.1 to show that y ≡ 0 is
the only solution of the initial value problem
Ly= 0 , y (x0) = 0 , y ′(x0) = 0 ,...,y (n−1)(x0) = 0 ,
on (a,b ).",Elementary Differential Equations with Boundary Value Problems.pdf
"the only solution of the initial value problem
Ly= 0 , y (x0) = 0 , y ′(x0) = 0 ,...,y (n−1)(x0) = 0 ,
on (a,b ).
12. Prove: If y1, y2, . . . , yn are solutions of Ly= 0 and the functions
zi =
n∑
j=1
aij yj , 1 ≤ i≤ n,
form a fundamental set of solutions of Ly= 0 , then so do y1, y2, . . . , yn .
13. Prove: If
y = c1y1 + c2y2 + · · ·+ ck yk + yp
is a solution of a linear equation Ly = F for every choice of the constants c1, c2 ,. . . , ck , then
Lyi = 0 for 1 ≤ i≤ k.
14. Suppose Ly = 0 is normal on (a,b ) and let x0 be in (a,b ). For 1 ≤ i ≤ n, let yi be the solution
of the initial value problem
Lyi = 0 , y (j)
i (x0) =
{
0, j ̸= i− 1,
1, j = i− 1, 1 ≤ i≤ n,
where x0 is an arbitrary point in (a,b ). Show that any solution of Ly= 0 on (a,b ), can be written
as
y= c1y1 + c2y2 + · · ·+ cnyn ,
with cj = y(j−1)(x0).",Elementary Differential Equations with Boundary Value Problems.pdf
"474 Chapter 9 Linear Higher Order Equations
15. Suppose {y1,y 2,...,y n}is a fundamental set of solutions of
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn(x)y = 0
on (a,b ), and let
z1 = a11y1 + a12y2 + · · ·+ a1nyn
z2 = a21y1 + a22y2 + · · ·+ a2nyn
.
.
.
.
.
.
.
.
.
.
.
.
zn = an1y1 + an2y2 + · · ·+ annyn ,
where the {aij }are constants. Show that {z1,z 2,...,z n }is a fundamental set of solutions of (A)
if and only if the determinant ⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
a11 a12 · · ·a1n
a21 a22 · · ·a2n
.
.
. .
.
. . . . .
.
.
an1 an2 · · ·ann
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
is nonzero.H IN T : The determinant of a product of n× nmatrices equals the product of the deter-
minants.
16. Show that {y1,y 2,...,y n }is linearly dependent on (a,b ) if and only if at least one of the functions
y1, y2, . . . , yn can be written as a linear combination of the others on (a,b ).
T ake the following as a hint in Exercises
17– 19:
By the definition of determinant,
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
a11 a12 · · ·a1n
a21 a22 · · ·a2n
.
.
. .
.
. . . . .",Elementary Differential Equations with Boundary Value Problems.pdf
"17– 19:
By the definition of determinant,
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
a11 a12 · · ·a1n
a21 a22 · · ·a2n
.
.
. .
.
. . . . .
.
.
an1 an2 · · ·ann
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
=
∑
±a1i1 a2i2 ,...,a nin ,
where the sum is over all permutations (i1,i 2,...,i n) of (1, 2,...,n ) and the choice of + or − in each
term depends only on the permutation associated with that te rm.
17. Prove: If
A(u1,u 2,...,u n) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
a11 a12 · · · a1n
a21 a22 · · · a2n
.
.
. .
.
. . . . .
.
.
an−1, 1 an−1, 2 · · ·an−1,n
u1 u2 · · · un
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,
then
A(u1 + v1,u 2 + v2,...,u n + vn ) = A(u1,u 2,...,u n) + A(v1,v 2,...,v n).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.1 Introduction to Linear Higher Order Equations 475
18. Let
F =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
f11 f12 · · ·f1n
f21 f22 · · ·f2n
.
.
. .
.
. . . . .
.
.
fn1 fn2 · · ·fnn
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,
where fij (1 ≤ i, j ≤ n) is differentiable. Show that
F′= F1 + F2 + · · ·+ Fn,
where Fi is the determinant obtained by differentiating the ith row of F.
19. Use Exercise
18 to show that if W is the Wronskian of the n-times differentiable functions y1, y2,
. . . , yn , then
W′=
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 · · · yn
y′
1 y′
2 · · · y′
n
.
.
.
.
.
. . . . .
.
.
y(n−2)
1 y(n−2)
2 · · ·y(n−2)
n
y(n)
1 y(n)
2 · · ·y(n)
n
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
20. Use Exercises
17 and 19 to show that if W is the Wronskian of solutions {y1,y 2,...,y n}of the
normal equation
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn (x)y= 0 , (A)
then W′= −P1W/P 0. Derive Abel’s formula (Eqn. ( 9.1.15)) from this. H IN T : Use (A) to write
y(n) in terms of y,y ′,...,y (n−1).
21. Prove Theorem 9.1.6.",Elementary Differential Equations with Boundary Value Problems.pdf
"y(n) in terms of y,y ′,...,y (n−1).
21. Prove Theorem 9.1.6.
22. Prove Theorem 9.1.7.
23. Show that if the Wronskian of the n-times continuously differentiable functions {y1,y 2,...,y n }
has no zeros in (a,b ), then the differential equation obtained by expanding the d eterminant
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y y 1 y2 · · ·yn
y′ y′
1 y′
2 · · ·y′
n
.
.
. .
.
. .
.
. . . . .
.
.
y(n) y(n)
1 y(n)
2 · · ·y(n)
n
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 0 ,
in cofactors of its first column is normal and has {y1,y 2,...,y n}as a fundamental set of solutions
on (a,b ).
24. Use the method suggested by Exercise
23 to find a linear homogeneous equation such that the
given set of functions is a fundamental set of solutions on in tervals on which the Wronskian of the
set has no zeros.
(a) {x, x2 − 1, x2 + 1 } (b) {ex, e−x , x}
(c) {ex, xe−x, 1} (d) {x, x2, ex}",Elementary Differential Equations with Boundary Value Problems.pdf
"476 Chapter 9 Linear Higher Order Equations
(e) {x, x2, 1/x } (f) {x+ 1 , ex, e3x}
(g) {x, x3, 1/x, 1/x 2} (h) {x, xln x, 1/x, x 2}
(i) {ex, e−x , x, e2x} (j) {e2x, e−2x, 1, x2}
9.2 HIGHER ORDER CONST ANT COEFFICIENT HOMOGENEOUS EQUA TIONS
If a0, a1, . . . , an are constants and a0 ̸= 0 , then
a0y(n) + a1y(n−1) + · · ·+ any = F(x)
is said to be a constant coefficient equation . In this section we consider the homogeneous constant coef-
ficient equation
a0y(n) + a1y(n−1) + · · ·+ any = 0 . (9.2.1)
Since ( 9.2.1) is normal on (−∞ , ∞ ), the theorems in Section 9.1 all apply with (a,b ) = ( −∞ , ∞ ).
As in Section 5.2, we call
p(r) = a0rn + a1rn−1 + · · ·+ an (9.2.2)
the characteristic polynomial of ( 9.2.1). W e saw in Section 5.2 that when n= 2 the solutions of ( 9.2.1)
are determined by the zeros of the characteristic polynomia l. This is also true when n > 2, but the
situation is more complicated in this case. Consequently , w e take a different approach here than in
Section 5.2.",Elementary Differential Equations with Boundary Value Problems.pdf
"situation is more complicated in this case. Consequently , w e take a different approach here than in
Section 5.2.
If kis a positive integer, let Dk stand for the k-th derivative operator; that is
Dk y= y(k).
If
q(r) = b0rm + b1rm−1 + · · ·+ bm
is an arbitrary polynomial, define the operator
q(D) = b0Dm + b1Dm−1 + · · ·+ bm
such that
q(D)y = ( b0Dm + b1Dm−1 + · · ·+ bm)y = b0y(m) + b1y(m−1) + · · ·+ bmy
whenever yis a function with mderivatives. W e call q(D) a polynomial operator.
With pas in ( 9.2.2),
p(D) = a0Dn + a1Dn−1 + · · ·+ an,
so ( 9.2.1) can be written as p(D)y = 0 . If ris a constant then
p(D)erx = (a0Dn erx + a1Dn−1erx + · · ·+ an erx )
= ( a0rn + a1rn−1 + · · ·+ an )erx ;
that is
p(D)(erx ) = p(r)erx .
This shows that y = erx is a solution of ( 9.2.1) if p(r) = 0 . In the simplest case, where phas ndistinct
real zeros r1, r2,. . . , rn , this argument yields nsolutions
y1 = er1 x, y 2 = er2 x,..., y n = ern x.",Elementary Differential Equations with Boundary Value Problems.pdf
"real zeros r1, r2,. . . , rn , this argument yields nsolutions
y1 = er1 x, y 2 = er2 x,..., y n = ern x.
It can be shown (Exercise 39) that the Wronskian of {er1x,e r2 x,...,e rn x}is nonzero if r1, r2, . . . , rn
are distinct; hence, {er1 x,e r2x,...,e rnx}is a fundamental set of solutions of p(D)y = 0 in this case.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 477
Example 9.2.1
(a) Find the general solution of
y′′′− 6y′′+ 11 y′− 6y= 0 . (9.2.3)
(b) Solve the initial value problem
y′′′− 6y′′+ 11 y′− 6y= 0 , y (0) = 4 , y ′(0) = 5 , y ′′(0) = 9 . (9.2.4)
Solution The characteristic polynomial of ( 9.2.3) is
p(r) = r3 − 6r2 + 11 r− 6 = ( r− 1)(r− 2)(r− 3).
Therefore {ex,e 2x,e 3x}is a set of solutions of ( 9.2.3). It is a fundamental set, since its Wronskian is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
⏐
⏐
⏐
⏐
⏐
⏐
= e6x
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1
1 2 3
1 4 9
⏐
⏐
⏐
⏐
⏐
⏐
= 2 e6x ̸= 0 .
Therefore the general solution of (
9.2.3) is
y= c1ex + c2e2x + c3e3x. (9.2.5)
SO LU TIO N (b) W e must determine c1, c2 and c3 in ( 9.2.5) so that y satisfies the initial conditions in
(9.2.4). Differentiating ( 9.2.5) twice yields
y′ = c1ex + 2 c2e2x + 3 c3e3x
y′′ = c1ex + 4 c2e2x + 9 c3e3x. (9.2.6)
Setting x= 0 in ( 9.2.5) and ( 9.2.6) and imposing the initial conditions yields",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′ = c1ex + 4 c2e2x + 9 c3e3x. (9.2.6)
Setting x= 0 in ( 9.2.5) and ( 9.2.6) and imposing the initial conditions yields
c1 + c2 + c3 = 4
c1 + 2 c2 + 3 c3 = 5
c1 + 4 c2 + 9 c3 = 9 .
The solution of this system is c1 = 4 , c2 = −1, c3 = 1 . Therefore the solution of ( 9.2.4) is
y= 4 ex − e2x + e3x
(Figure 9.2.1).
Now we consider the case where the characteristic polynomia l ( 9.2.2) does not have n distinct real
zeros. For this purpose it is useful to define what we mean by a f actorization of a polynomial operator.
W e begin with an example.
Example 9.2.2 Consider the polynomial
p(r) = r3 − r2 + r− 1
and the associated polynomial operator
p(D) = D3 − D2 + D− 1.",Elementary Differential Equations with Boundary Value Problems.pdf
"478 Chapter 9 Linear Higher Order Equations
0.5 1 1.5 2.0 .0
100
200
300
400
500
600
700
 x
 y
Figure 9.2.1 y= 4 ex − e2x + e3x
Since p(r) can be factored as
p(r) = ( r− 1)(r2 + 1) = ( r2 + 1)( r− 1),
it’s reasonable to expect that p(D) can be factored as
p(D) = ( D− 1)(D2 + 1) = ( D2 + 1)( D− 1). (9.2.7)
However, before we can make this assertion we must define what we mean by saying that two operators
are equal, and what we mean by the products of operators in ( 9.2.7). W e say that two operators are equal
if they apply to the same functions and always produce the sam e result. The definitions of the products in
(9.2.7) is this: if yis any three-times differentiable function then
(a) (D− 1)(D2 + 1) yis the function obtained by first applying D2 + 1 to yand then applying D− 1
to the resulting function
(b) (D2 + 1)( D− 1)yis the function obtained by first applying D− 1 to yand then applying D2 + 1
to the resulting function.
From (a),
(D− 1)(D2 + 1) y = ( D− 1)[(D2 + 1) y]",Elementary Differential Equations with Boundary Value Problems.pdf
"to the resulting function.
From (a),
(D− 1)(D2 + 1) y = ( D− 1)[(D2 + 1) y]
= ( D− 1)(y′′+ y) = D(y′′+ y) − (y′′+ y)
= ( y′′′+ y′) − (y′′+ y)
= y′′′− y′′+ y′− y= ( D3 − D2 + D− 1)y.
(9.2.8)
This implies that
(D− 1)(D2 + 1) = ( D3 − D2 + D− 1).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 479
From (b),
(D2 + 1)( D− 1)y = ( D2 + 1)[( D− 1)y]
= ( D2 + 1)( y′− y) = D2(y′− y) + ( y′− y)
= ( y′′′− y′′) + ( y′− y)
= y′′′− y′′+ y′− y= ( D3 − D2 + D− 1)y,
(9.2.9)
(D2 + 1)( D− 1) = ( D3 − D2 + D− 1),
which completes the justification of ( 9.2.7).
Example 9.2.3 Use the result of Example 9.2.2 to find the general solution of
y′′′− y′′+ y′− y = 0 . (9.2.10)
Solution From ( 9.2.8), we can rewrite ( 9.2.10) as
(D− 1)(D2 + 1) y = 0 ,
which implies that any solution of (D2 + 1) y = 0 is a solution of ( 9.2.10). Therefore y1 = cos xand
y2 = sin xare solutions of ( 9.2.10).
From ( 9.2.9), we can rewrite ( 9.2.10) as
(D2 + 1)( D− 1)y = 0 ,
which implies that any solution of (D− 1)y = 0 is a solution of ( 9.2.10). Therefore y3 = ex is solution
of ( 9.2.10).
The Wronskian of {ex, cos x, sin x}is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
cos x sin x e x
− sin x cos x e x
− cos x − sin x e x
⏐
⏐
⏐
⏐
⏐
⏐
.
Since
W(0) =
⏐
⏐
⏐
⏐
⏐
⏐
1 0 1
0 1 1
−1 0 1
⏐",Elementary Differential Equations with Boundary Value Problems.pdf
"W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
cos x sin x e x
− sin x cos x e x
− cos x − sin x e x
⏐
⏐
⏐
⏐
⏐
⏐
.
Since
W(0) =
⏐
⏐
⏐
⏐
⏐
⏐
1 0 1
0 1 1
−1 0 1
⏐
⏐
⏐
⏐
⏐
⏐
= 2 ,
{cos x, sin x,e x}is linearly independent and
y = c1 cos x+ c2 sin x+ c3ex
is the general solution of (
9.2.10).
Example 9.2.4 Find the general solution of
y(4) − 16y= 0 . (9.2.11)
Solution The characteristic polynomial of ( 9.2.11) is
p(r) = r4 − 16 = ( r2 − 4)(r2 + 4) = ( r− 2)(r+ 2)( r2 + 4) .
By arguments similar to those used in Examples 9.2.2 and 9.2.3, it can be shown that ( 9.2.11) can be
written as
(D2 + 4)( D+ 2)( D− 2)y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"480 Chapter 9 Linear Higher Order Equations
or
(D2 + 4)( D− 2)(D+ 2) y= 0
or
(D− 2)(D+ 2)( D2 + 4) y= 0 .
Therefore yis a solution of ( 9.2.11) if it’s a solution of any of the three equations
(D− 2)y= 0 , (D+ 2) y= 0 , (D2 + 4) y= 0 .
Hence, {e2x,e −2x, cos 2x, sin 2x}is a set of solutions of ( 9.2.11). The Wronskian of this set is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
e2x e−2x cos 2x sin 2 x
2e2x −2e−2x −2 sin 2 x 2 cos 2 x
4e2x 4e−2x −4 cos 2 x −4 sin 2 x
8e2x −8e−2x 8 sin 2 x −8 cos 2 x
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
Since
W(0) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 1 0
2 −2 0 2
4 4 −4 0
8 −8 0 −8
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= −512,
{e2x,e −2x, cos 2x, sin 2x}is linearly independent, and
y1 = c1e2x + c2e−2x + c3 cos 2x+ c4 sin 2 x
is the general solution of (
9.2.11).
It is known from algebra that every polynomial
p(r) = a0rn + a1rn−1 + · · ·+ an
with real coefficients can be factored as
p(r) = a0p1(r)p2(r) · · ·pk (r),
where no pair of the polynomials p1, p2, . . . , pk has a commom factor and each is either of the form",Elementary Differential Equations with Boundary Value Problems.pdf
"p(r) = a0p1(r)p2(r) · · ·pk (r),
where no pair of the polynomials p1, p2, . . . , pk has a commom factor and each is either of the form
pj (r) = ( r− rj )mj , (9.2.12)
where rj is real and mj is a positive integer, or
pj (r) =
[
(r− λj )2 + ω2
j
] mj
, (9.2.13)
where λj and ωj are real, ωj ̸= 0 , and mj is a positive integer. If ( 9.2.12) holds then rj is a real zero of
p, while if ( 9.2.13) holds then λ + iω and λ − iω are complex conjugate zeros of p. In either case, mj is
the multiplicity of the zero(s).
By arguments similar to those used in our examples, it can be s hown that
p(D) = a0p1(D)p2 (D) · · ·pk(D) (9.2.14)
and that the order of the factors on the right can be chosen arb itrarily . Therefore, if pj (D)y = 0 for some
j then p(D)y = 0 . T o see this, we simply rewrite ( 9.2.14) so that pj (D) is applied first. Therefore the",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 481
problem of finding solutions of p(D)y = 0 with pas in ( 9.2.14) reduces to finding solutions of each of
these equations
pj (D)y = 0 , 1 ≤ j ≤ k,
where pj is a power of a first degree term or of an irreducible quadratic . T o find a fundamental set of
solutions {y1,y 2,...,y n }of p(D)y = 0 , we find fundamental set of solutions of each of the equations
and take {y1,y 2,...,y n }to be the set of all functions in these separate fundamental s ets. In Exercise 40
we sketch the proof that {y1,y 2,...,y n}is linearly independent, and therefore a fundamental set of
solutions of p(D)y = 0 .
T o apply this procedure to general homogeneous constant coe fficient equations, we must be able to
find fundamental sets of solutions of equations of the form
(D− a)my = 0
and [
(D− λ)2 + ω2] m
y = 0 ,
where mis an arbitrary positive integer. The next two theorems show how to do this.
Theorem 9.2.1 If mis a positive integer , then",Elementary Differential Equations with Boundary Value Problems.pdf
"y = 0 ,
where mis an arbitrary positive integer. The next two theorems show how to do this.
Theorem 9.2.1 If mis a positive integer , then
{eax,xe ax,...,x m−1eax} (9.2.15)
is a fundamental set of solutions of
(D− a)m y= 0 . (9.2.16)
Proof W e’ll show that if
f(x) = c1 + c2x+ · · ·+ cmxm−1
is an arbitrary polynomial of degree ≤ m− 1, then y = eaxf is a solution of ( 9.2.16). First note that if g
is any differentiable function then
(D− a)eax g= Deax g− aeaxg= aeaxg+ eaxg′− aeaxg,
so
(D− a)eaxg= eaxg′. (9.2.17)
Therefore
(D− a)eax f = eax f′ (from ( 9.2.17) with g= f)
(D− a)2eaxf = ( D− a)eaxf′= eaxf′′ (from ( 9.2.17) with g= f′)
(D− a)3eaxf = ( D− a)eaxf′′= eax f′′′ (from ( 9.2.17) with g= f′′)
.
.
.
(D− a)m eaxf = ( D− a)eaxf(m−1) = eaxf(m) (from (
9.2.17) with g= f(m−1) ).
Since f(m) = 0 , the last equation implies that y = eaxf is a solution of ( 9.2.16) if f is any polynomial",Elementary Differential Equations with Boundary Value Problems.pdf
"9.2.17) with g= f(m−1) ).
Since f(m) = 0 , the last equation implies that y = eaxf is a solution of ( 9.2.16) if f is any polynomial
of degree ≤ m− 1. In particular, each function in ( 9.2.15) is a solution of ( 9.2.16). T o see that ( 9.2.15)
is linearly independent (and therefore a fundamental set of solutions of ( 9.2.16)), note that if
c1eax + c2xeax + c· · ·+ cm−1xm−1eax = 0
for all xin some interval (a,b ), then
c1 + c2x+ c· · ·+ cm−1xm−1 = 0
for all xin (a,b ). However, we know from algebra that if this polynomial has mo re than m−1 zeros then
c1 = c2 = · · ·= cn = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"482 Chapter 9 Linear Higher Order Equations
Example 9.2.5 Find the general solution of
y′′′+ 3 y′′+ 3 y′+ y = 0 . (9.2.18)
Solution The characteristic polynomial of ( 9.2.18) is
p(r) = r3 + 3 r2 + 3 r+ 1 = ( r+ 1) 3.
Therefore ( 9.2.18) can be written as
(D+ 1) 3y= 0 ,
so Theorem 9.2.1 implies that the general solution of ( 9.2.18) is
y = e−x(c1 + c2x+ c3x2).
The proof of the next theorem is sketched in Exercise 41.
Theorem 9.2.2 If ω ̸= 0 and mis a positive integer , then
{eλx cos ωx,xe λx cos ωx, ...,x m−1eλx cos ωx,
eλx sin ωx,xe λx sin ωx, ...,x m−1eλx sin ωx}
is a fundamental set of solutions of
[(D− λ)2 + ω2]my= 0 .
Example 9.2.6 Find the general solution of
(D2 + 4 D+ 13) 3y= 0 . (9.2.19)
Solution The characteristic polynomial of ( 9.2.19) is
p(r) = ( r2 + 4 r+ 13) 3 = ((r+ 2) 2 + 9 ) 3
.
Therefore ( 9.2.19) can be be written as
[(D+ 2) 2 + 9] 3y = 0 ,
so Theorem 9.2.2 implies that the general solution of ( 9.2.19) is",Elementary Differential Equations with Boundary Value Problems.pdf
".
Therefore ( 9.2.19) can be be written as
[(D+ 2) 2 + 9] 3y = 0 ,
so Theorem 9.2.2 implies that the general solution of ( 9.2.19) is
y= ( a1 + a2x+ a3x2)e−2x cos 3 x+ ( b1 + b2x+ b3x2)e−2x sin 3 x.
Example 9.2.7 Find the general solution of
y(4) + 4 y′′′+ 6 y′′+ 4 y′= 0 . (9.2.20)
Solution The characteristic polynomial of ( 9.2.20) is
p(r) = r4 + 4 r3 + 6 r2 + 4 r
= r(r3 + 4 r2 + 6 r+ 4)
= r(r+ 2)( r2 + 2 r+ 2)
= r(r+ 2)[( r+ 1) 2 + 1] .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 483
Therefore ( 9.2.20) can be written as
[(D+ 1) 2 + 1]( D+ 2) Dy = 0 .
Fundamental sets of solutions of
[ (D+ 1) 2 + 1 ] y= 0 , (D+ 2) y = 0 , and Dy = 0 .
are given by
{e−x cos x,e −x sin x}, {e−2x}, and {1},
respectively . Therefore the general solution of ( 9.2.20) is
y = e−x(c1 cos x+ c2 sin x) + c3e−2x + c4.
Example 9.2.8 Find a fundamental set of solutions of
[(D+ 1) 2 + 1] 2(D− 1)3(D+ 1) D2y= 0 . (9.2.21)
Solution A fundamental set of solutions of ( 9.2.21) can be obtained by combining fundamental sets of
solutions of [
(D+ 1) 2 + 1
] 2
y= 0 , (D− 1)3y = 0 ,
(D+ 1) y = 0 , and D2y = 0 .
Fundamental sets of solutions of these equations are given b y
{e−x cos x,xe −x cos x,e −x sin x,xe −x sin x}, {ex,xe x,x 2ex },
{e−x}, and {1,x },
respectively . These ten functions form a fundamental set of solutions of ( 9.2.21).
9.2 Exercises
In Exercises 1– 14 find the general solution.",Elementary Differential Equations with Boundary Value Problems.pdf
"respectively . These ten functions form a fundamental set of solutions of ( 9.2.21).
9.2 Exercises
In Exercises 1– 14 find the general solution.
1. y′′′− 3y′′+ 3 y′− y = 0 2. y(4) + 8 y′′− 9y= 0
3. y′′′− y′′+ 16 y′− 16y= 0 4. 2y′′′+ 3 y′′− 2y′− 3y= 0
5. y′′′+ 5 y′′+ 9 y′+ 5 y = 0 6. 4y′′′− 8y′′+ 5 y′− y= 0
7. 27y′′′+ 27 y′′+ 9 y′+ y= 0 8. y(4) + y′′= 0
9. y(4) − 16y= 0 10. y(4) + 12 y′′+ 36 y= 0
11. 16y(4) − 72y′′+ 81 y= 0 12. 6y(4) + 5 y′′′+ 7 y′′+ 5 y′+ y= 0
13. 4y(4) + 12 y′′′+ 3 y′′− 13y′− 6y= 0
14. y(4) − 4y′′′+ 7 y′′− 6y′+ 2 y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"484 Chapter 9 Linear Higher Order Equations
In Exercises 15– 27 solve the initial value problem. Where indicated by C/G , graph the solution.
15. y′′′− 2y′′+ 4 y′− 8y = 0 , y (0) = 2 , y ′(0) = −2, y′′(0) = 0
16. y′′′+ 3 y′′− y′− 3y= 0 , y (0) = 0 , y ′(0) = 14 , y ′′(0) = −40
17. C/G y′′′− y′′− y′+ y= 0 , y (0) = −2, y ′(0) = 9 , y ′′(0) = 4
18. C/G y′′′− 2y′− 4y= 0 , y (0) = 6 , y ′(0) = 3 , y ′′(0) = 22
19. C/G
3y′′′− y′′− 7y′+ 5 y = 0 , y (0) = 14
5 , y ′(0) = 0 , y ′′(0) = 10
20. y′′′− 6y′′+ 12 y′− 8y= 0 , y (0) = 1 , y ′(0) = −1, y ′′(0) = −4
21. 2y′′′− 11y′′+ 12 y′+ 9 y= 0 , y (0) = 6 , y ′(0) = 3 , y ′′(0) = 13
22. 8y′′′− 4y′′− 2y′+ y = 0 , y (0) = 4 , y ′(0) = −3, y ′′(0) = −1
23. y(4) − 16y= 0 , y (0) = 2 , y′(0) = 2 , y′′(0) = −2, y′′′(0) = 0
24. y(4) − 6y′′′+ 7 y′′+ 6 y′− 8y= 0 , y (0) = −2, y ′(0) = −8, y ′′(0) = −14,
y′′′(0) = −62
25. 4y(4) − 13y′′+ 9 y= 0 , y (0) = 1 , y ′(0) = 3 , y ′′(0) = 1 , y ′′′(0) = 3",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′′(0) = −62
25. 4y(4) − 13y′′+ 9 y= 0 , y (0) = 1 , y ′(0) = 3 , y ′′(0) = 1 , y ′′′(0) = 3
26. y(4) + 2 y′′′− 2y′′− 8y′− 8y= 0 , y (0) = 5 , y ′(0) = −2, y ′′(0) = 6 , y ′′′(0) = 8
27. C/G 4y(4) + 8 y′′′+ 19 y′′+ 32 y′+ 12 y = 0 , y (0) = 3 , y ′(0) = −3, y ′′(0) = −7
2 ,
y′′′(0) = 31
4
28. Find a fundamental set of solutions of the given equation, an d verify that it’s a fundamental set by
evaluating its Wronskian at x= 0 .
(a) (D− 1)2(D− 2)y= 0 (b) (D2 + 4)( D− 3)y = 0
(c) (D2 + 2 D+ 2)( D− 1)y = 0 (d) D3(D− 1)y= 0
(e) (D2 − 1)(D2 + 1) y= 0 (f) (D2 − 2D+ 2)( D2 + 1) y= 0
In Exercises 29– 38 find a fundamental set of solutions.
29. (D2 + 6 D+ 13)( D− 2)2D3y = 0
30. (D− 1)2(2D− 1)3(D2 + 1) y= 0
31. (D2 + 9) 3D2y = 0 32. (D− 2)3(D+ 1) 2Dy = 0
33. (D2 + 1)( D2 + 9) 2(D− 2)y = 0 34. (D4 − 16)2y = 0
35. (4D2 + 4 D+ 9) 3y= 0 36. D3(D− 2)2(D2 + 4) 2y= 0
37. (4D2 + 1) 2(9D2 + 4) 3y = 0 38.
[
(D− 1)4 − 16
]
y= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 485
39. It can be shown that
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
1 1 · · · 1
a1 a2 · · ·an
a2
1 a2
2 · · ·a2
n
.
.
. .
.
. . . . .
.
.
an−1
1 an−1
2 · · ·an−1
n
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
=
∏
1≤i<j ≤n
(aj − ai), (A)
where the left side is the
V andermonde determinant and the right side is the product of all factors
of the form (aj − ai) with iand jbetween 1 and nand i<j .
(a) V erify (A) for n= 2 and n= 3 .
(b) Find the Wronskian of {ea1x, e a2 x,...,e an x}.
40. A theorem from algebra says that if P1 and P2 are polynomials with no common factors then there
are polynomials Q1 and Q2 such that
Q1P1 + Q2P2 = 1 .
This implies that
Q1(D)P1(D)y+ Q2(D)P2(D)y = y
for every function ywith enough derivatives for the left side to be defined.
(a) Use this to show that if P1 and P2 have no common factors and
P1(D)y = P2(D)y = 0
then y= 0 .
(b) Suppose P1 and P2 are polynomials with no common factors. Let u1, . . . , ur be linearly",Elementary Differential Equations with Boundary Value Problems.pdf
"P1(D)y = P2(D)y = 0
then y= 0 .
(b) Suppose P1 and P2 are polynomials with no common factors. Let u1, . . . , ur be linearly
independent solutions of P1(D)y = 0 and let v1, . . . , vs be linearly independent solutions of
P2(D)y = 0 . Use (a) to show that {u1,...,u r ,v 1,...,v s}is a linearly independent set.
(c) Suppose the characteristic polynomial of the constant coef ficient equation
a0y(n) + a1y(n−1) + · · ·+ an y= 0 (A)
has the factorization
p(r) = a0p1(r)p2(r) · · ·pk(r),
where each pj is of the form
pj (r) = ( r− rj )nj or pj (r) = [( r− λj )2 + w2
j ]mj (ωj >0)
and no two of the polynomials p1, p2, . . . , pk have a common factor. Show that we can
find a fundamental set of solutions {y1,y 2,...,y n}of (A) by finding a fundamental set of
solutions of each of the equations
pj (D)y = 0 , 1 ≤ j ≤ k,
and taking {y1,y 2,...,y n }to be the set of all functions in these separate fundamental s ets.",Elementary Differential Equations with Boundary Value Problems.pdf
"486 Chapter 9 Linear Higher Order Equations
41. (a) Show that if
z = p(x) cos ωx + q(x) sin ωx, (A)
where pand qare polynomials of degree ≤ k, then
(D2 + ω2)z = p1(x) cos ωx + q1(x) sin ωx,
where p1 and q1 are polynomials of degree ≤ k− 1.
(b) Apply (a) mtimes to show that if zis of the form (A) where pand qare polynomial of degree
≤ m− 1, then
(D2 + ω2)m z = 0 . (B)
(c) Use Eqn. ( 9.2.17) to show that if y = eλx zthen
[(D− λ)2 + ω2]my = eλx (D2 + ω2)m z.
(d) Conclude from (b) and (c) that if pand qare arbitrary polynomials of degree ≤ m− 1 then
y= eλx (p(x) cos ωx + q(x) sin ωx)
is a solution of
[(D− λ)2 + ω2]my= 0 . (C)
(e) Conclude from (d) that the functions
eλx cos ωx,xe λx cos ωx, ...,x m−1eλx cos ωx,
eλx sin ωx,xe λx sin ωx, ...,x m−1eλx sin ωx (D)
are all solutions of (C).
(f) Complete the proof of Theorem 9.2.2 by showing that the functions in (D) are linearly inde-
pendent.
42. (a) Use the trigonometric identities
cos(A+ B) = cos Acos B− sin Asin B",Elementary Differential Equations with Boundary Value Problems.pdf
"pendent.
42. (a) Use the trigonometric identities
cos(A+ B) = cos Acos B− sin Asin B
sin(A+ B) = cos Asin B+ sin Acos B
to show that
(cos A+ isin A)(cos B+ isin B) = cos( A+ B) + isin(A+ B).
(b) Apply (a) repeatedly to show that if nis a positive integer then
n∏
k=1
(cos Ak + isin Ak) = cos( A1 + A2 + · · ·+ An) + isin(A1 + A2 + · · ·+ An).
(c) Infer from (b) that if nis a positive integer then
(cos θ + isin θ)n = cos nθ + isin nθ. (A)
(d) Show that (A) also holds if n = 0 or a negative integer. H IN T : V erify by direct calculation
that
(cos θ + isin θ)−1 = (cos θ − isin θ).
Then replace θ by −θ in (A).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 487
(e) Now suppose nis a positive integer. Infer from (A) that if
zk = cos
(2kπ
n
)
+ isin
(2kπ
n
)
, k = 0 , 1,...,n − 1,
and
ζk = cos
((2k+ 1) π
n
)
+ isin
((2k+ 1) π
n
)
, k = 0 , 1,...,n − 1,
then
zn
k = 1 and ζn
k = −1, k = 0 , 1,...,n − 1.
(Why don’t we also consider other integer values for k?)
(f) Let ρ be a positive number. Use (e) to show that
zn − ρ = ( z− ρ1/n z0)(z− ρ1/n z1) · · ·(z− ρ1/n zn−1)
and
zn + ρ = ( z− ρ1/n ζ0)(z− ρ1/n ζ1) · · ·(z− ρ1/n ζn−1).
43. Use (e) of Exercise 42 to find a fundamental set of solutions of the given equation.
(a) y′′′− y= 0 (b) y′′′+ y = 0
(c) y(4) + 64 y= 0 (d) y(6) − y = 0
(e) y(6) + 64 y= 0 (f)
[
(D− 1)6 − 1
]
y= 0
(g) y(5) + y(4) + y′′′+ y′′+ y′+ y= 0
44. An equation of the form
a0xny(n) + a1xn−1y(n−1) + · · ·+ an−1xy′+ an y= 0 , x> 0, (A)
where a0, a1, . . . , an are constants, is an Euler or equidimensional equation.
Show that if
x= et and Y(t) = y(x(t)), (B)
then
xdy",Elementary Differential Equations with Boundary Value Problems.pdf
"where a0, a1, . . . , an are constants, is an Euler or equidimensional equation.
Show that if
x= et and Y(t) = y(x(t)), (B)
then
xdy
dx = dY
dt
x2 d2y
dx2 = d2Y
dt2 − dY
dt
x3 d3y
dx3 = d3Y
dt3 − 3 d2Y
dt2 + 2 dY
dt .
In general, it can be shown that if ris any integer ≥ 2 then
xr dr y
dxr = dr Y
dtr + A1r
dr−1Y
dtr−1 + · · ·+ Ar−1,r
dY
dt
where A1r , . . . , Ar−1,r are integers. Use these results to show that the substitutio n (B) transforms
(A) into a constant coefficient equation for Y as a function of t.",Elementary Differential Equations with Boundary Value Problems.pdf
"488 Chapter 9 Linear Higher Order Equations
45. Use Exercise 44 to show that a function y= y(x) satisfies the equation
a0x3y′′′+ a1x2y′′+ a2xy′+ a3y= 0 , (A)
on (0, ∞ ) if and only if the function Y(t) = y(et ) satisfies
a0
d3Y
dt3 + ( a1 − 3a0) d2Y
dt2 + ( a2 − a1 + 2 a0) dY
dt + a3Y = 0 .
Assuming that a0, a1, a2, a3 are real and a0 ̸= 0 , find the possible forms for the general solution
of (A).
9.3 UNDETERMINED COEFFICIENTS FOR HIGHER ORDER EQUA TIONS
In this section we consider the constant coefficient equatio n
a0y(n) + a1y(n−1) + · · ·+ an y= F(x), (9.3.1)
where n≥ 3 and F is a linear combination of functions of the form
eαx (
p0 + p1x+ · · ·+ pkxk)
or
eλx [(p0 + p1x+ · · ·+ pkxk ) cos ωx + (q0 + q1x+ · · ·+ qkxk ) sin ωx] .
From Theorem 9.1.5, the general solution of ( 9.3.1) is y= yp + yc, where yp is a particular solution of
(9.3.1) and yc is the general solution of the complementary equation
a0y(n) + a1y(n−1) + · · ·+ any = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"(9.3.1) and yc is the general solution of the complementary equation
a0y(n) + a1y(n−1) + · · ·+ any = 0 .
In Section 9.2 we learned how to find yc . Here we will learn how to find yp when the forcing function
has the form stated above. The procedure that we use is a gener alization of the method that we used in
Sections 5.4 and 5.5, and is again called method of undetermined coefficients . Since the underlying ideas
are the same as those in Sections 5.4 and 5.5, we’ll give an inf ormal presentation based on examples.
Forcing Functions of the Form eαx (
p0 + p1x+ · · ·+ pk xk)
W e first consider equations of the form
a0y(n) + a1y(n−1) + · · ·+ an y= eαx (
p0 + p1x+ · · ·+ pkxk)
.
Example 9.3.1 Find a particular solution of
y′′′+ 3 y′′+ 2 y′− y = ex(21 + 24 x+ 28 x2 + 5 x3). (9.3.2)
Solution Substituting
y = uex,
y′ = ex(u′+ u),
y′′ = ex(u′′+ 2 u′+ u),
y′′′= ex(u′′′+ 3 u′′+ 3 u′+ u)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.3 Undetermined Coefficients for Higher Order Equations 489
into ( 9.3.2) and canceling ex yields
(u′′′+ 3 u′′+ 3 u′+ u) + 3( u′′+ 2 u′+ u) + 2( u′+ u) − u= 21 + 24 x+ 28 x2 + 5 x3,
or
u′′′+ 6 u′′+ 11 u′+ 5 u= 21 + 24 x+ 28 x2 + 5 x3. (9.3.3)
Since the unknown uappears on the left, we can see that ( 9.3.3) has a particular solution of the form
up = A+ Bx+ Cx2 + Dx3.
Then
u′
p = B+ 2 Cx+ 3 Dx2
u′′
p = 2 C+ 6 Dx
u′′′
p = 6 D.
Substituting from the last four equations into the left side of (
9.3.3) yields
u′′′
p + 6 u′′
p + 11 u′
p + 5 up = 6 D+ 6(2 C+ 6 Dx) + 11( B+ 2 Cx+ 3 Dx2)
+5(A+ Bx+ Cx2 + Dx3)
= (5 A+ 11 B+ 12 C+ 6 D) + (5 B+ 22 C+ 36 D)x
+(5C+ 33 D)x2 + 5 Dx3.
Comparing coefficients of like powers of xon the right sides of this equation and (
9.3.3) shows that up
satisfies ( 9.3.3) if
5D = 5
5C+ 33 D = 28
5B+ 22 C+ 36 D = 24
5A+ 11 B+ 12 C+ 6 D = 21 .
Solving these equations successively yields D= 1 , C = −1, B= 2 , A= 1 . Therefore
up = 1 + 2 x− x2 + x3",Elementary Differential Equations with Boundary Value Problems.pdf
"5A+ 11 B+ 12 C+ 6 D = 21 .
Solving these equations successively yields D= 1 , C = −1, B= 2 , A= 1 . Therefore
up = 1 + 2 x− x2 + x3
is a particular solution of ( 9.3.3), so
yp = ex up = ex(1 + 2 x− x2 + x3)
is a particular solution of ( 9.3.2) (Figure 9.3.1).
Example 9.3.2 Find a particular solution of
y(4) − y′′′− 6y′′+ 4 y′+ 8 y= e2x(4 + 19 x+ 6 x2). (9.3.4)
Solution Substituting
y = ue2x,
y′ = e2x(u′+ 2 u),
y′′ = e2x(u′′+ 4 u′+ 4 u),
y′′′= e2x(u′′′+ 6 u′′+ 12 u′+ 8 u),
y(4) = e2x(u(4) + 8 u′′′+ 24 u′′+ 32 u′+ 16 u)",Elementary Differential Equations with Boundary Value Problems.pdf
"490 Chapter 9 Linear Higher Order Equations
1 2
10
20
30
40
50
 x
 y
Figure 9.3.1 yp = ex(1 + 2 x− x2 + x3)
into ( 9.3.4) and canceling e2x yields
(u(4) + 8 u′′′+ 24 u′′+ 32 u′+ 16 u) − (u′′′+ 6 u′′+ 12 u′+ 8 u)
−6(u′′+ 4 u′+ 4 u) + 4( u′+ 2 u) + 8 u= 4 + 19 x+ 6 x2,
or
u(4) + 7 u′′′+ 12 u′′= 4 + 19 x+ 6 x2. (9.3.5)
Since neither unor u′appear on the left, we can see that ( 9.3.5) has a particular solution of the form
up = Ax2 + Bx3 + Cx4. (9.3.6)
Then
u′
p = 2 Ax+ 3 Bx2 + 4 Cx3
u′′
p = 2 A+ 6 Bx+ 12 Cx2
u′′′
p = 6 B+ 24 Cx
u(4)
p = 24 C.
Substituting u′′
p , u′′′
p , and u(4)
p into the left side of (
9.3.5) yields
u(4)
p + 7 u′′′
p + 12 u′′
p = 24 C+ 7(6 B+ 24 Cx) + 12(2 A+ 6 Bx+ 12 Cx2)
= (24 A+ 42 B+ 24 C) + (72 B+ 168 C)x+ 144 Cx2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.3 Undetermined Coefficients for Higher Order Equations 491
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
.05
.10
.15
.20
.25
.30
 x
 y
Figure 9.3.2 yp = x2e2x
24 (−4 + 4 x+ x2)
Comparing coefficients of like powers of xon the right sides of this equation and ( 9.3.5) shows that up
satisfies ( 9.3.5) if
144C = 6
72B+ 168 C = 19
24A+ 42 B+ 24 C = 4 .
Solving these equations successively yields C = 1 / 24, B = 1 / 6, A = −1/ 6. Substituting these into
(9.3.6) shows that
up = x2
24 (−4 + 4 x+ x2)
is a particular solution of ( 9.3.5), so
yp = e2xup = x2e2x
24 (−4 + 4 x+ x2)
is a particular solution of ( 9.3.4). (Figure 9.3.2).
Forcing Functions of the Form eλx (P(x) cos ωx + Q(x) sin ωx)
W e now consider equations of the form
a0y(n) + a1y(n−1) + · · ·+ an y= eλx (P(x) cos ωx + Q(x) sin ωx) ,
where P and Qare polynomials.
Example 9.3.3 Find a particular solution of
y′′′+ y′′− 4y′− 4y= ex[(5 − 5x) cos x+ (2 + 5 x) sin x]. (9.3.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"492 Chapter 9 Linear Higher Order Equations
Solution Substituting
y = uex,
y′ = ex(u′+ u),
y′′ = ex(u′′+ 2 u′+ u),
y′′′= ex(u′′′+ 3 u′′+ 3 u′+ u)
into ( 9.3.7) and canceling ex yields
(u′′′+ 3 u′′+ 3 u′+ u) + ( u′′+ 2 u′+ u) − 4(u′+ u) − 4u= (5 − 5x) cos x+ (2 + 5 x) sin x,
or
u′′′+ 4 u′′+ u′− 6u= (5 − 5x) cos x+ (2 + 5 x) sin x. (9.3.8)
Since cos xand sin xare not solutions of the complementary equation
u′′′+ 4 u′′+ u′− 6u= 0 ,
a theorem analogous to Theorem 5.5.1 implies that ( 9.3.8) has a particular solution of the form
up = ( A0 + A1x) cos x+ ( B0 + B1x) sin x. (9.3.9)
Then
u′
p = ( A1 + B0 + B1x) cos x+ ( B1 − A0 − A1x) sin x,
u′′
p = (2 B1 − A0 − A1x) cos x− (2A1 + B0 + B1x) sin x,
u′′′
p = −(3A1 + B0 + B1x) cos x− (3B1 − A0 − A1x) sin x,
so
u′′′
p + 4 u′′
p + u′
p − 6up = − [10A0 + 2 A1 − 8B1 + 10 A1x] cos x
− [10B0 + 2 B1 + 8 A1 + 10 B1x] sin x.
Comparing the coefficients of xcos x, xsin x, cos x, and sin xhere with the corresponding coefficients
in (",Elementary Differential Equations with Boundary Value Problems.pdf
"− [10B0 + 2 B1 + 8 A1 + 10 B1x] sin x.
Comparing the coefficients of xcos x, xsin x, cos x, and sin xhere with the corresponding coefficients
in (
9.3.8) shows that up is a solution of ( 9.3.8) if
−10A1 = −5
−10B1 = 5
−10A0 − 2A1 + 8 B1 = 5
−10B0 − 2B1 − 8A1 = 2 .
Solving the first two equations yields A1 = 1 / 2, B1 = −1/ 2. Substituting these into the last two
equations yields
−10A0 = 5 + 2 A1 − 8B1 = 10
−10B0 = 2 + 2 B1 + 8 A1 = 5 ,
so A0 = −1, B0 = −1/ 2. Substituting A0 = −1, A1 = 1 / 2, B0 = −1/ 2, B1 = −1/ 2 into ( 9.3.9)
shows that
up = −1
2 [(2 − x) cos x+ (1 + x) sin x]
is a particular solution of ( 9.3.8), so
yp = exup = −ex
2 [(2 − x) cos x+ (1 + x) sin x]
is a particular solution of ( 9.3.7) (Figure 9.3.3).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.3 Undetermined Coefficients for Higher Order Equations 493
1 2 3 4
20
40
60
80
100
−20
 x
 y
Figure 9.3.3 yp = exup = −ex
2 [(2 − x) cos x+ (1 + x) sin x]
Example 9.3.4 Find a particular solution of
y′′′+ 4 y′′+ 6 y′+ 4 y= e−x [(1 − 6x) cos x− (3 + 2 x) sin x] . (9.3.10)
Solution Substituting
y = ue−x,
y′ = e−x(u′− u),
y′′ = e−x(u′′− 2u′+ u),
y′′′= e−x(u′′′− 3u′′+ 3 u′− u)
into ( 9.3.10) and canceling e−x yields
(u′′′− 3u′′+ 3 u′− u) + 4( u′′− 2u′+ u) + 6( u′− u) + 4 u= (1 − 6x) cos x− (3 + 2 x) sin x,
or
u′′′+ u′′+ u′+ u= (1 − 6x) cos x− (3 + 2 x) sin x. (9.3.11)
Since cos xand sin xare solutions of the complementary equation
u′′′+ u′′+ u′+ u= 0 ,
a theorem analogous to Theorem 5.5.1 implies that ( 9.3.11) has a particular solution of the form
up = ( A0x+ A1x2) cos x+ ( B0x+ B1x2) sin x. (9.3.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"494 Chapter 9 Linear Higher Order Equations
Then
u′
p = [ A0 + (2 A1 + B0)x+ B1x2] cos x+ [ B0 + (2 B1 − A0)x− A1x2] sin x,
u′′
p = [2 A1 + 2 B0 − (A0 − 4B1)x− A1x2] cos x
+[2B1 − 2A0 − (B0 + 4 A1)x− B1x2] sin x,
u′′′
p = −[3A0 − 6B1 + (6 A1 + B0)x+ B1x2] cos x
−[3B0 + 6 A1 + (6 B1 − A0)x− A1x2] sin x,
so
u′′′
p + u′′
p + u′
p + up = −[2A0 − 2B0 − 2A1 − 6B1 + (4 A1 − 4B1)x] cos x
−[2B0 + 2 A0 − 2B1 + 6 A1 + (4 B1 + 4 A1)x] sin x.
Comparing the coefficients of xcos x, xsin x, cos x, and sin xhere with the corresponding coefficients
in (
9.3.11) shows that up is a solution of ( 9.3.11) if
−4A1 + 4 B1 = −6
−4A1 − 4B1 = −2
−2A0 + 2 B0 + 2 A1 + 6 B1 = 1
−2A0 − 2B0 − 6A1 + 2 B1 = −3.
Solving the first two equations yields A1 = 1 , B1 = −1/ 2. Substituting these into the last two equations
yields
−2A0 + 2 B0 = 1 − 2A1 − 6B1 = 2
−2A0 − 2B0 = −3 + 6 A1 − 2B1 = 4 ,
1 2 3 4 5 6 7 8 9
0.1
0.2
−0.1
−0.2
−0.3
−0.4
−0.5
 x
 y
Figure 9.3.4 yp = −xe−x
2 [(3 − 2x) cos x+ (1 + x) sin x]",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.3 Undetermined Coefficients for Higher Order Equations 495
so A0 = −3/ 2 and B0 = −1/ 2. Substituting A0 = −3/ 2, A1 = 1 , B0 = −1/ 2, B1 = −1/ 2 into
(9.3.12) shows that
up = −x
2 [(3 − 2x) cos x+ (1 + x) sin x]
is a particular solution of ( 9.3.11), so
yp = e−xup = −xe−x
2 [(3 − 2x) cos x+ (1 + x) sin x]
(Figure 9.3.4) is a particular solution of ( 9.3.10).
9.3 Exercises
In Exercises 1– 59 find a particular solution.
1. y′′′− 6y′′+ 11 y′− 6y= −e−x (4 + 76 x− 24x2)
2. y′′′− 2y′′− 5y′+ 6 y = e−3x(32 − 23x+ 6 x2)
3. 4y′′′+ 8 y′′− y′− 2y = −ex (4 + 45 x+ 9 x2)
4. y′′′+ 3 y′′− y′− 3y= e−2x(2 − 17x+ 3 x2)
5. y′′′+ 3 y′′− y′− 3y= ex(−1 + 2 x+ 24 x2 + 16 x3)
6. y′′′+ y′′− 2y= ex(14 + 34 x+ 15 x2)
7. 4y′′′+ 8 y′′− y′− 2y = −e−2x(1 − 15x)
8. y′′′− y′′− y′+ y= ex(7 + 6 x)
9. 2y′′′− 7y′′+ 4 y′+ 4 y= e2x(17 + 30 x)
10. y′′′− 5y′′+ 3 y′+ 9 y = 2 e3x(11 − 24x2)
11. y′′′− 7y′′+ 8 y′+ 16 y= 2 e4x(13 + 15 x)
12. 8y′′′− 12y′′+ 6 y′− y= ex/ 2(1 + 4 x)",Elementary Differential Equations with Boundary Value Problems.pdf
"10. y′′′− 5y′′+ 3 y′+ 9 y = 2 e3x(11 − 24x2)
11. y′′′− 7y′′+ 8 y′+ 16 y= 2 e4x(13 + 15 x)
12. 8y′′′− 12y′′+ 6 y′− y= ex/ 2(1 + 4 x)
13. y(4) + 3 y′′′− 3y′′− 7y′+ 6 y= −e−x (12 + 8 x− 8x2)
14. y(4) + 3 y′′′+ y′′− 3y′− 2y= −3e2x(11 + 12 x)
15. y(4) + 8 y′′′+ 24 y′′+ 32 y′= −16e−2x(1 + x+ x2 − x3)
16. 4y(4) − 11y′′− 9y′− 2y= −ex(1 − 6x)
17. y(4) − 2y′′′+ 3 y′− y= ex(3 + 4 x+ x2)
18. y(4) − 4y′′′+ 6 y′′− 4y′+ 2 y= e2x(24 + x+ x4)
19. 2y(4) + 5 y′′′− 5y′− 2y= 18 ex(5 + 2 x)
20. y(4) + y′′′− 2y′′− 6y′− 4y= −e2x(4 + 28 x+ 15 x2)
21. 2y(4) + y′′′− 2y′− y= 3 e−x/ 2(1 − 6x)
22. y(4) − 5y′′+ 4 y= ex(3 + x− 3x2)
23. y(4) − 2y′′′− 3y′′+ 4 y′+ 4 y= e2x(13 + 33 x+ 18 x2)
24. y(4) − 3y′′′+ 4 y′= e2x(15 + 26 x+ 12 x2)
25. y(4) − 2y′′′+ 2 y′− y= ex(1 + x)
26. 2y(4) − 5y′′′+ 3 y′′+ y′− y = ex(11 + 12 x)",Elementary Differential Equations with Boundary Value Problems.pdf
"496 Chapter 9 Linear Higher Order Equations
27. y(4) + 3 y′′′+ 3 y′′+ y′= e−x(5 − 24x+ 10 x2)
28. y(4) − 7y′′′+ 18 y′′− 20y′+ 8 y= e2x(3 − 8x− 5x2)
29. y′′′− y′′− 4y′+ 4 y= e−x [(16 + 10 x) cos x+ (30 − 10x) sin x]
30. y′′′+ y′′− 4y′− 4y= e−x [(1 − 22x) cos 2 x− (1 + 6 x) sin 2 x]
31. y′′′− y′′+ 2 y′− 2y= e2x[(27 + 5 x− x2) cos x+ (2 + 13 x+ 9 x2) sin x]
32. y′′′− 2y′′+ y′− 2y= −ex[(9 − 5x+ 4 x2) cos 2 x− (6 − 5x− 3x2) sin 2 x]
33. y′′′+ 3 y′′+ 4 y′+ 12 y= 8 cos 2 x− 16 sin 2 x
34. y′′′− y′′+ 2 y= ex[(20 + 4 x) cos x− (12 + 12 x) sin x]
35. y′′′− 7y′′+ 20 y′− 24y= −e2x[(13 − 8x) cos 2 x− (8 − 4x) sin 2 x]
36. y′′′− 6y′′+ 18 y′= −e3x[(2 − 3x) cos 3 x− (3 + 3 x) sin 3 x]
37. y(4) + 2 y′′′− 2y′′− 8y′− 8y= ex (8 cos x+ 16 sin x)
38. y(4) − 3y′′′+ 2 y′′+ 2 y′− 4y= ex (2 cos 2 x− sin 2 x)
39. y(4) − 8y′′′+ 24 y′′− 32y′+ 15 y= e2x(15xcos 2x+ 32 sin 2 x)
40. y(4) + 6 y′′′+ 13 y′′+ 12 y′+ 4 y= e−x[(4 − x) cos x− (5 + x) sin x]
41. y(4) + 3 y′′′+ 2 y′′− 2y′− 4y= −e−x (cos x− sin x)",Elementary Differential Equations with Boundary Value Problems.pdf
"40. y(4) + 6 y′′′+ 13 y′′+ 12 y′+ 4 y= e−x[(4 − x) cos x− (5 + x) sin x]
41. y(4) + 3 y′′′+ 2 y′′− 2y′− 4y= −e−x (cos x− sin x)
42. y(4) − 5y′′′+ 13 y′′− 19y′+ 10 y= ex(cos 2x+ sin 2 x)
43. y(4) + 8 y′′′+ 32 y′′+ 64 y′+ 39 y= e−2x[(4 − 15x) cos 3 x− (4 + 15 x) sin 3 x]
44. y(4) − 5y′′′+ 13 y′′− 19y′+ 10 y= ex[(7 + 8 x) cos 2 x+ (8 − 4x) sin 2 x]
45. y(4) + 4 y′′′+ 8 y′′+ 8 y′+ 4 y= −2e−x (cos x− 2 sin x)
46. y(4) − 8y′′′+ 32 y′′− 64y′+ 64 y= e2x(cos 2 x− sin 2 x)
47. y(4) − 8y′′′+ 26 y′′− 40y′+ 25 y= e2x[3 cos x− (1 + 3 x) sin x]
48. y′′′− 4y′′+ 5 y′− 2y = e2x − 4ex − 2 cos x+ 4 sin x
49. y′′′− y′′+ y′− y= 5 e2x + 2 ex − 4 cos x+ 4 sin x
50. y′′′− y′= −2(1 + x) + 4 ex − 6e−x + 96 e3x
51. y′′′− 4y′′+ 9 y′− 10y= 10 e2x + 20 ex sin 2 x− 10
52. y′′′+ 3 y′′+ 3 y′+ y = 12 e−x + 9 cos 2 x− 13 sin 2 x
53. y′′′+ y′′− y′− y= 4 e−x(1 − 6x) − 2xcos x+ 2(1 + x) sin x
54. y(4) − 5y′′+ 4 y= −12ex + 6 e−x + 10 cos x
55. y(4) − 4y′′′+ 11 y′′− 14y′+ 10 y= −ex(sin x+ 2 cos 2 x)",Elementary Differential Equations with Boundary Value Problems.pdf
"54. y(4) − 5y′′+ 4 y= −12ex + 6 e−x + 10 cos x
55. y(4) − 4y′′′+ 11 y′′− 14y′+ 10 y= −ex(sin x+ 2 cos 2 x)
56. y(4) + 2 y′′′− 3y′′− 4y′+ 4 y= 2 ex(1 + x) + e−2x
57. y(4) + 4 y= sinh xcos x− cosh xsin x
58. y(4) + 5 y′′′+ 9 y′′+ 7 y′+ 2 y= e−x(30 + 24 x) − e−2x
59. y(4) − 4y′′′+ 7 y′′− 6y′+ 2 y= ex (12x− 2 cos x+ 2 sin x)
In Exercises 60– 68 find the general solution.
60. y′′′− y′′− y′+ y= e2x(10 + 3 x)
61. y′′′+ y′′− 2y= −e3x(9 + 67 x+ 17 x2)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.3 Undetermined Coefficients for Higher Order Equations 497
62. y′′′− 6y′′+ 11 y′− 6y= e2x(5 − 4x− 3x2)
63. y′′′+ 2 y′′+ y′= −2e−x (7 − 18x+ 6 x2)
64. y′′′− 3y′′+ 3 y′− y = ex(1 + x)
65. y(4) − 2y′′+ y = −e−x (4 − 9x+ 3 x2)
66. y′′′+ 2 y′′− y′− 2y= e−2x [(23 − 2x) cos x+ (8 − 9x) sin x]
67. y(4) − 3y′′′+ 4 y′′− 2y′= ex [(28 + 6 x) cos 2 x+ (11 − 12x) sin 2 x]
68. y(4) − 4y′′′+ 14 y′′− 20y′+ 25 y= ex [(2 + 6 x) cos 2 x+ 3 sin 2 x]
In Exercises 69– 74 solve the initial value problem and graph the solution.
69. C/G y′′′− 2y′′− 5y′+ 6 y= 2 ex(1 − 6x), y (0) = 2 , y ′(0) = 7 , y ′′(0) = 9
70. C/G y′′′− y′′− y′+ y= −e−x (4 − 8x), y (0) = 2 , y ′(0) = 0 , y ′′(0) = 0
71. C/G 4y′′′− 3y′− y= e−x/ 2(2 − 3x), y (0) = −1, y ′(0) = 15 , y ′′(0) = −17
72. C/G y(4) + 2 y′′′+ 2 y′′+ 2 y′+ y = e−x(20 − 12x), y(0) = 3 , y′(0) = −4, y′′(0) =
7, y′′′(0) = −22
73. C/G y′′′+ 2 y′′+ y′+ 2 y= 30 cos x− 10 sin x, y (0) = 3 , y ′(0) = −4, y ′′(0) = 16",Elementary Differential Equations with Boundary Value Problems.pdf
"7, y′′′(0) = −22
73. C/G y′′′+ 2 y′′+ y′+ 2 y= 30 cos x− 10 sin x, y (0) = 3 , y ′(0) = −4, y ′′(0) = 16
74. C/G y(4) − 3y′′′+ 5 y′′− 2y′= −2ex(cos x− sin x), y(0) = 2 , y′(0) = 0 , y′′(0) = − 1,
y′′′(0) = −5
75. Prove: A function yis a solution of the constant coefficient nonhomogeneous equ ation
a0y(n) + a1y(n−1) + · · ·+ an y= eαx G(x) (A)
if and only if y= ueαx , where usatisfies the differential equation
a0u(n) + p(n−1)(α)
(n− 1)! u(n−1) + p(n−2)(α)
(n− 2)! u(n−2) + · · ·+ p(α)u= G(x) (B)
and
p(r) = a0rn + a1rn−1 + · · ·+ an
is the characteristic polynomial of the complementary equa tion
a0y(n) + a1y(n−1) + · · ·+ an y= 0 .
76. Prove:
(a) The equation
a0u(n) + p(n−1)(α)
(n− 1)! u(n−1) + p(n−2)(α)
(n− 2)! u(n−2) + · · ·+ p(α)u
= (p0 + p1x+ · · ·+ pkxk) cos ωx
+ (q0 + q1x+ · · ·+ qkxk) sin ωx
(A)
has a particular solution of the form
up = xm (u0 + u1x+ · · ·+ ukxk ) cos ωx + (v0 + v1x+ · · ·+ vkxk) sin ωx.",Elementary Differential Equations with Boundary Value Problems.pdf
"498 Chapter 9 Linear Higher Order Equations
(b) If λ + iω is a zero of pwith multiplicity m≥ 1, then (A) can be written as
a(u′′+ ω2u) =
(
p0 + p1x+ · · ·+ pk xk)
cos ωx +
(
q0 + q1x+ · · ·+ qkxk)
sin ωx,
which has a particular solution of the form
up = U(x) cos ωx + V(x) sin ωx,
where
U(x) = u0x+ u1x2 + · · ·+ uk xk+1, V(x) = v0x+ v1x2 + · · ·+ vk xk+1
and
a(U′′(x) + 2 ωV ′(x)) = p0 + p1x+ · · ·+ pkxk
a(V′′(x) − 2ωU ′(x)) = q0 + q1x+ · · ·+ qkxk.
9.4 V ARIA TION OF P ARAMETERS FOR HIGHER ORDER EQUA TIONS
Derivation of the method
W e assume throughout this section that the nonhomogeneous l inear equation
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn(x)y = F(x) (9.4.1)
is normal on an interval (a,b ). W e’ll abbreviate this equation as Ly= F, where
Ly= P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn(x)y.
When we speak of solutions of this equation and its complemen tary equation Ly= 0 , we mean solutions
on (a,b ). W e’ll show how to use the method of variation of parameters t o find a particular solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"on (a,b ). W e’ll show how to use the method of variation of parameters t o find a particular solution of
Ly= F, provided that we know a fundamental set of solutions {y1,y 2,...,y n}of Ly= 0 .
W e seek a particular solution of Ly= F in the form
yp = u1y1 + u2y2 + · · ·+ unyn (9.4.2)
where {y1,y 2,...,y n}is a known fundamental set of solutions of the complementary equation
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn(x)y = 0
and u1, u2, . . . , un are functions to be determined. W e begin by imposing the foll owing n− 1 conditions
on u1,u 2,...,u n:
u′
1y1 + u′
2y2+ · · ·+u′
n yn = 0
u′
1y′
1 + u′
2y′
2+ · · ·+u′
n y′
n = 0
.
.
.
u′
1y(n−2)
1 + u′
2y(n−2)
2 + · · ·+u′
n y(n−2)
n = 0 .
(9.4.3)
These conditions lead to simple formulas for the first n− 1 derivatives of yp:
y(r)
p = u1y(r)
1 + u2y(r)
2 · · ·+ uny(r)
n , 0 ≤ r≤ n− 1. (9.4.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.4 V ariation of Parameters for Higher Order Equations 499
These formulas are easy to remember, since they look as thoug h we obtained them by differentiating
(9.4.2) n− 1 times while treating u1, u2, . . . , un as constants. T o see that ( 9.4.3) implies ( 9.4.4), we first
differentiate ( 9.4.2) to obtain
y′
p = u1y′
1 + u2y′
2 + · · ·+ un y′
n + u′
1y1 + u′
2y2 + · · ·+ u′
nyn ,
which reduces to
y′
p = u1y′
1 + u2y′
2 + · · ·+ uny′
n
because of the first equation in (
9.4.3). Differentiating this yields
y′′
p = u1y′′
1 + u2y′′
2 + · · ·+ un y′′
n + u′
1y′
1 + u′
2y′
2 + · · ·+ u′
ny′
n ,
which reduces to
y′′
p = u1y′′
1 + u2y′′
2 + · · ·+ uny′′
n
because of the second equation in (
9.4.3). Continuing in this way yields ( 9.4.4).
The last equation in ( 9.4.4) is
y(n−1)
p = u1y(n−1)
1 + u2y(n−1)
2 + · · ·+ uny(n−1)
n .
Differentiating this yields
y(n)
p = u1y(n)
1 + u2y(n)
2 + · · ·+ uny(n)
n + u′
1y(n−1)
1 + u′
2y(n−1)
2 + · · ·+ u′
n y(n−1)
n .
Substituting this and (",Elementary Differential Equations with Boundary Value Problems.pdf
"y(n)
p = u1y(n)
1 + u2y(n)
2 + · · ·+ uny(n)
n + u′
1y(n−1)
1 + u′
2y(n−1)
2 + · · ·+ u′
n y(n−1)
n .
Substituting this and (
9.4.4) into ( 9.4.1) yields
u1Ly1 + u2Ly2 + · · ·+ un Lyn + P0(x)
(
u′
1y(n−1)
1 + u′
2y(n−1)
2 + · · ·+ u′
ny(n−1)
n
)
= F(x).
Since Lyi = 0 (1 ≤ i≤ n), this reduces to
u′
1y(n−1)
1 + u′
2y(n−1)
2 + · · ·+ u′
n y(n−1)
n = F(x)
P0(x) .
Combining this equation with ( 9.4.3) shows that
yp = u1y1 + u2y2 + · · ·+ unyn
is a solution of ( 9.4.1) if
u′
1y1 + u′
2y2+ · · ·+u′
nyn = 0
u′
1y′
1 + u′
2y′
2+ · · ·+u′
ny′
n = 0
.
.
.
u′
1y(n−2)
1 + u′
2y(n−2)
2 + · · ·+u′
ny(n−2)
n = 0
u′
1y(n−1)
1 + u′
2y(n−1)
2 + · · ·+u′
ny(n−1)
n = F/P 0,
which can be written in matrix form as











y1 y2 · · · yn
y′
1 y′
2 · · · y′
n
.
.
. .
.
. . . . .
.
.
y(n−2)
1 y(n−2)
2 · · ·y(n−2)
n
y(n−1)
1 y(n−1)
2 · · ·y(n−1)
n


















u′
1
u′
2
.
.
.
u′
n−1
u′
n







=







0
0
.
.
.
0
F/P 0







. (9.4.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"500 Chapter 9 Linear Higher Order Equations
The determinant of this system is the Wronskian W of the fundamental set of solutions {y1,y 2,...,y n},
which has no zeros on (a,b ), by Theorem 9.1.4. Solving ( 9.4.5) by Cramer’s rule yields
u′
j = ( −1)n−j FWj
P0W, 1 ≤ j ≤ n, (9.4.6)
where Wj is the Wronskian of the set of functions obtained by deleting yj from {y1,y 2,...,y n }and
keeping the remaining functions in the same order. Equivale ntly , Wj is the determinant obtained by
deleting the last row and j-th column of W.
Having obtained u′
1, u′
2,...,u ′
n, we can integrate to obtain u1, u2,...,u n. As in Section 5.7, we take
the constants of integration to be zero, and we drop any linea r combination of {y1,y 2,...,y n}that may
appear in yp .
REM A RK : For efficiency , it’s best to compute W1, W2, . . . , Wn first, and then compute W by expanding
in cofactors of the last row; thus,
W =
n∑
j=1
(−1)n−j y(n−1)
j Wj .
Third Order Equations
If n= 3 , then
W =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 y3
y′",Elementary Differential Equations with Boundary Value Problems.pdf
"in cofactors of the last row; thus,
W =
n∑
j=1
(−1)n−j y(n−1)
j Wj .
Third Order Equations
If n= 3 , then
W =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 y3
y′
1 y′
2 y′
3
y′′
1 y′′
2 y′′
3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
Therefore
W1 =
⏐
⏐
⏐
⏐
⏐
y2 y3
y′
2 y′
3
⏐
⏐
⏐
⏐
⏐, W 2 =
⏐
⏐
⏐
⏐
⏐
y1 y3
y′
1 y′
3
⏐
⏐
⏐
⏐
⏐, W 3 =
⏐
⏐
⏐
⏐
⏐
y1 y2
y′
1 y′
2
⏐
⏐
⏐
⏐
⏐,
and (
9.4.6) becomes
u′
1 = FW1
P0W, u ′
2 = −FW2
P0W, u ′
3 = FW3
P0W. (9.4.7)
Example 9.4.1 Find a particular solution of
xy′′′− y′′− xy′+ y= 8 x2ex, (9.4.8)
given that y1 = x, y2 = ex , and y3 = e−x form a fundamental set of solutions of the complementary
equation. Then find the general solution of ( 9.4.8).
Solution W e seek a particular solution of ( 9.4.8) of the form
yp = u1x+ u2ex + u3e−x .
The Wronskian of {y1,y 2,y 3}is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
x e x e−x
1 ex −e−x
0 ex e−x
⏐
⏐
⏐
⏐
⏐
⏐
,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.4 V ariation of Parameters for Higher Order Equations 501
so
W1 =
⏐
⏐
⏐
⏐
ex e−x
ex −e−x
⏐
⏐
⏐
⏐= −2,
W2 =
⏐
⏐
⏐
⏐
x e −x
1 −e−x
⏐
⏐
⏐
⏐= −e−x(x+ 1) ,
W3 =
⏐
⏐
⏐
⏐
x e x
1 ex
⏐
⏐
⏐
⏐= ex(x− 1).
Expanding W by cofactors of the last row yields
W = 0 W1 − exW2 + e−xW3 = 0( −2) − ex (−e−x (x+ 1) ) + e−xex(x− 1) = 2 x.
Since F(x) = 8 x2ex and P0(x) = x,
F
P0W = 8x2ex
x·2x = 4 ex.
Therefore, from ( 9.4.7)
u′
1 = 4 exW1 = 4 ex(−2) = −8ex ,
u′
2 = −4exW2 = −4ex (
−e−x(x+ 1)
)
= 4( x+ 1) ,
u′
3 = 4 exW3 = 4 ex (ex(x− 1)) = 4 e2x(x− 1).
Integrating and taking the constants of integration to be ze ro yields
u1 = −8ex, u 2 = 2( x+ 1) 2,u 3 = e2x(2x− 3).
Hence,
yp = u1y1 + u2y2 + u3y3
= ( −8ex )x+ ex(2(x+ 1) 2) + e−x (
e2x(2x− 3)
)
= ex(2x2 − 2x− 1).
Since −ex is a solution of the complementary equation, we redefine
yp = 2 xex(x− 1).
Therefore the general solution of (
9.4.8) is
y= 2 xex(x− 1) + c1x+ c2ex + c3e−x.
Fourth Order Equations
If n= 4 , then
W =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 y3 y4",Elementary Differential Equations with Boundary Value Problems.pdf
"9.4.8) is
y= 2 xex(x− 1) + c1x+ c2ex + c3e−x.
Fourth Order Equations
If n= 4 , then
W =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 y3 y4
y′
1 y′
2 y′
3 y′
4
y′′
1 y′′
2 y′′
3 y′′
4
y′′′
1 y′′′
2 y′′′
3 y′′′
4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,",Elementary Differential Equations with Boundary Value Problems.pdf
"502 Chapter 9 Linear Higher Order Equations
Therefore
W1 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y2 y3 y4
y′
2 y′
3 y′
4
y′′
2 y′′
3 y′′
4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
, W 2 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y3 y4
y′
1 y′
3 y′
4
y′′
1 y′′
3 y′′
4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,
W3 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 y4
y′
1 y′
2 y′
4
y′′
1 y′′
2 y′′
4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
, W 4 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 y3
y′
1 y′
2 y′
3
y′′
1 y′′
2 y′′
3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,
and (
9.4.6) becomes
u′
1 = −FW1
P0W, u ′
2 = FW2
P0W, u ′
3 = −FW3
P0W, u ′
4 = FW4
P0W. (9.4.9)
Example 9.4.2 Find a particular solution of
x4y(4) + 6 x3y′′′+ 2 x2y′′− 4xy′+ 4 y= 12 x2, (9.4.10)
given that y1 = x, y2 = x2, y3 = 1 /x and y4 = 1 /x 2 form a fundamental set of solutions of the
complementary equation. Then find the general solution of ( 9.4.10) on (−∞ , 0) and (0, ∞ ).
Solution W e seek a particular solution of ( 9.4.10) of the form
yp = u1x+ u2x2 + u3
x + u4
x2 .
The Wronskian of {y1,y 2,y 3,y 4}is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x x 2 1/x −1/x 2
1 2 x −1/x 2 −2/x 3
0 2 2 /x 3 6/x 4
0 0 −6/x 4 −24/x 5
⏐",Elementary Differential Equations with Boundary Value Problems.pdf
"x + u4
x2 .
The Wronskian of {y1,y 2,y 3,y 4}is
W(x) =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x x 2 1/x −1/x 2
1 2 x −1/x 2 −2/x 3
0 2 2 /x 3 6/x 4
0 0 −6/x 4 −24/x 5
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,
so
W1 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x2 1/x 1/x 2
2x −1/x 2 −2/x 3
2 2 /x 3 6/x 4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= −12
x4 ,
W2 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x 1/x 1/x 2
1 −1/x 2 −2/x 3
0 2 /x 3 6/x 4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= − 6
x5 ,
W3 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x x 2 1/x 2
1 2 x −2/x 3
0 2 6 /x 4
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 12
x2 ,
W4 =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x x 2 1/x
1 2 x −1/x 2
0 2 2 /x 3
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= 6
x.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.4 V ariation of Parameters for Higher Order Equations 503
Expanding W by cofactors of the last row yields
W = −0W1 + 0 W2 −
(
− 6
x4
)
W3 +
(
−24
x5
)
W4
= 6
x4
12
x2 − 24
x5
6
x = −72
x6 .
Since F(x) = 12 x2 and P0(x) = x4,
F
P0W = 12x2
x4
(
−x6
72
)
= −x4
6 .
Therefore, from ( 9.4.9),
u′
1 = −
(
−x4
6
)
W1 = x4
6
(
−12
x4
)
= −2,
u′
2 = − x4
6 W2 = −x4
6
(
− 6
x5
)
= 1
x,
u′
3 = −
(
−x4
6
)
W3 = x4
6
12
x2 = 2 x2,
u′
4 = − x4
6 W4 = −x4
6
6
x = −x3.
Integrating these and taking the constants of integration t o be zero yields
u1 = −2x, u 2 = ln |x|, u 3 = 2x3
3 ,u 4 = −x4
4 .
Hence,
yp = u1y1 + u2y2 + u3y3 + u4y4
= ( −2x)x+ (ln |x|)x2 + 2x3
3
1
x +
(
−x4
4
) 1
x2
= x2 ln |x| −19x2
12 .
Since −19x2/ 12 is a solution of the complementary equation, we redefine
yp = x2 ln |x|.
Therefore
y= x2 ln |x|+ c1x+ c2x2 + c3
x + c4
x2
is the general solution of ( 9.4.10) on (−∞ , 0) and (0, ∞ ).
9.4 Exercises",Elementary Differential Equations with Boundary Value Problems.pdf
"yp = x2 ln |x|.
Therefore
y= x2 ln |x|+ c1x+ c2x2 + c3
x + c4
x2
is the general solution of ( 9.4.10) on (−∞ , 0) and (0, ∞ ).
9.4 Exercises
In Exercises 1– 21 find a particular solution, given the fundamental set of solu tions of the complementary
equation.
1. x3y′′′− x2(x+ 3) y′′+ 2 x(x+ 3) y′− 2(x+ 3) y = −4x4; {x, x2, xex}",Elementary Differential Equations with Boundary Value Problems.pdf
"504 Chapter 9 Linear Higher Order Equations
2. y′′′+ 6 xy′′+ (6 + 12 x2)y′+ (12 x+ 8 x3)y = x1/ 2e−x2
; {e−x2
, xe−x2
, x2e−x2
}
3. x3y′′′− 3x2y′′+ 6 xy′− 6y= 2 x; {x,x 2,x 3}
4. x2y′′′+ 2 xy′′− (x2 + 2) y′= 2 x2; {1, ex/x, e −x/x }
5. x3y′′′− 3x2(x+ 1) y′′+ 3 x(x2 + 2 x+ 2) y′− (x3 + 3 x2 + 6 x+ 6) y= x4e−3x;
{xex, x2ex, x3ex}
6. x(x2 − 2)y′′′+ ( x2 − 6)y′′+ x(2 − x2)y′+ (6 − x2)y= 2( x2 − 2)2; {ex, e−x , 1/x }
7. xy′′′− (x− 3)y′′− (x+ 2) y′+ ( x− 1)y = −4e−x ; {ex, ex /x,e −x /x }
8. 4x3y′′′+ 4 x2y′′− 5xy′+ 2 y= 30 x2; {√x, 1/ √x, x2}
9. x(x2 − 1)y′′′+ (5 x2 + 1) y′′+ 2 xy′− 2y= 12 x2; {x, 1/ (x− 1), 1/ (x+ 1) }
10. x(1 − x)y′′′+ ( x2 − 3x+ 3) y′′+ xy′− y = 2( x− 1)2; {x, 1/x,e x/x }
11. x3y′′′+ x2y′′− 2xy′+ 2 y= x2; {x, x2, 1/x }
12. xy′′′− y′′− xy′+ y = x2; {x, ex, e−x}
13. xy(4) + 4 y′′′= 6 ln |x|; {1,x, x 2, 1/x }
14. 16x4y(4) + 96 x3y′′′+ 72 x2y′′− 24xy′+ 9 y= 96 x5/ 2; {√x, 1/ √x, x3/ 2, x−3/ 2}
15. x(x2 − 6)y(4) + 2( x2 − 12)y′′′+ x(6 − x2)y′′+ 2(12 − x2)y′= 2( x2 − 6)2;",Elementary Differential Equations with Boundary Value Problems.pdf
"15. x(x2 − 6)y(4) + 2( x2 − 12)y′′′+ x(6 − x2)y′′+ 2(12 − x2)y′= 2( x2 − 6)2;
{1, 1/x, e x, e−x}
16. x4y(4) − 4x3y′′′+ 12 x2y′′− 24xy′+ 24 y= x4; {x, x2, x3, x4}
17. x4y(4) − 4x3y′′′+ 2 x2(6 − x2)y′′+ 4 x(x2 − 6)y′+ ( x4 − 4x2 + 24) y= 4 x5ex ;
{xex, x2ex, xe−x, x2e−x}
18. x4y(4) + 6 x3y′′′+ 2 x2y′′− 4xy′+ 4 y= 12 x2; {x,x 2, 1/x, 1/x 2}
19. xy(4) + 4 y′′′− 2xy′′− 4y′+ xy= 4 ex; {ex, e−x , ex/x, e −x/x }
20. xy(4) +(4 −6x)y′′′+(13x−18)y′′+(26 −12x)y′+(4x−12)y= 3 ex; {ex, e2x, ex /x, e 2x/x }
21. x4y(4) − 4x3y′′′+ x2(12 − x2)y′′+ 2 x(x2 − 12)y′+ 2(12 − x2)y= 2 x5; {x, x2, xex, xe−x}
In Exercises 22– 33 solve the initial value problem, given the fundamental set o f solutions of the comple-
mentary equation. Where indicated by C/G , graph the solution.
22. C/G x3y′′′−2x2y′′+3xy′−3y= 4 x, y (1) = 4 , y ′(1) = 4 , y ′′(1) = 2 ; {x, x3, xln x}
23. x3y′′′− 5x2y′′+ 14 xy′− 18y= x3, y (1) = 0 , y ′(1) = 1 , y ′′(1) = 7 ; {x2, x3, x3 ln x}",Elementary Differential Equations with Boundary Value Problems.pdf
"23. x3y′′′− 5x2y′′+ 14 xy′− 18y= x3, y (1) = 0 , y ′(1) = 1 , y ′′(1) = 7 ; {x2, x3, x3 ln x}
24. (5 − 6x)y′′′+ (12 x− 4)y′′+ (6 x− 23)y′+ (22 − 12x)y= −(6x− 5)2ex
y(0) = −4, y ′(0) = −3
2 , y ′′(0) = −19; {ex, e2x, xe−x}
25. x3y′′′− 6x2y′′+ 16 xy′− 16y= 9 x4, y (1) = 2 , y ′(1) = 1 , y ′′(1) = 5 ;
{x,x 4, x4 ln |x|}
26. C/G (x2 − 2x+ 2) y′′′− x2y′′+ 2 xy′− 2y= ( x2 − 2x+ 2) 2, y (0) = 0 , y ′(0) = 5 ,
y′′(0) = 0 ; {x, x2, ex }
27. x3y′′′+ x2y′′− 2xy′+ 2 y= x(x+ 1) , y (−1) = −6, y ′(−1) = 43
6 , y ′′(−1) = −5
2 ;
{x,x 2, 1/x }",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 9.4 V ariation of Parameters for Higher Order Equations 505
28. (3x− 1)y′′′− (12x− 1)y′′+ 9( x+ 1) y′− 9y = 2 ex(3x− 1)2, y (0) = 3
4 ,
y′(0) = 5
4 , y ′′(0) = 1
4 ; {x+ 1 , ex, e3x}
29. C/G (x2 − 2)y′′′− 2xy′′+ (2 − x2)y′+ 2 xy= 2( x2 − 2)2, y (0) = 1 , y ′(0) = −5,
y′′(0) = 5 ; {x2, ex , e−x}
30. C/G x4y(4) + 3x3y′′′−x2y′′+ 2xy′−2y = 9 x2, y (1) = −7, y ′(1) = −11, y ′′(1) = −5,
y′′′(1) = 6; {x, x2, 1/x,x ln x}
31. (2x− 1)y(4) − 4xy′′′+ (5 − 2x)y′′+ 4 xy′− 4y= 6(2 x− 1)2, y (0) = 55
4 , y ′(0) = 0 ,
y′′(0) = 13 , y ′′′(0) = 1 ; {x, ex, e−x, e2x}
32. 4x4y(4) + 24x3y′′′+ 23x2y′′− xy′+ y = 6 x, y(1) = 2 , y ′(1) = 0 , y ′′(1) = 4 , y ′′′(1) =
−37
4 ; {x, √x, 1/x, 1/ √x}
33. x4y(4) + 5 x3y′′′− 3x2y′′− 6xy′+ 6 y= 40 x3, y (−1) = −1, y′(−1) = −7,
y′′(−1) = −1, y ′′′(−1) = −31; {x, x3, 1/x, 1/x 2}
34. Suppose the equation
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn (x)y= F(x) (A)
is normal on an interval (a,b ). Let {y1,y 2,...,y n }be a fundamental set of solutions of its com-",Elementary Differential Equations with Boundary Value Problems.pdf
"is normal on an interval (a,b ). Let {y1,y 2,...,y n }be a fundamental set of solutions of its com-
plementary equation on (a,b ), let W be the Wronskian of {y1,y 2,...,y n }, and let Wj be the
determinant obtained by deleting the last row and the j-th column of W. Suppose x0 is in (a,b ),
let
uj (x) = ( −1)(n−j)
∫ x
x0
F(t)Wj (t)
P0(t)W(t) dt, 1 ≤ j ≤ n,
and define
yp = u1y1 + u2y2 + · · ·+ un yn .
(a) Show that yp is a solution of (A) and that
y(r)
p = u1y(r)
1 + u2y(r)
2 · · ·+ uny(r)
n , 1 ≤ r≤ n− 1,
and
y(n)
p = u1y(n)
1 + u2y(n)
2 + · · ·+ uny(n)
n + F
P0
.
HIN T : See the derivation of the method of variation of parameters a t the beginning of the
section.
(b) Show that yp is the solution of the initial value problem
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn(x)y = F(x),
y(x0) = 0 , y′(x0) = 0 ,..., y (n−1)(x0) = 0 .
(c) Show that yp can be written as
yp(x) =
∫ x
x0
G(x,t )F(t) dt,",Elementary Differential Equations with Boundary Value Problems.pdf
"506 Chapter 9 Linear Higher Order Equations
where
G(x,t ) = 1
P0(t)W(t)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1(t) y2(t) · · · yn (t)
y′
1(t) y′
2(t) · · · y′
n (t)
.
.
. .
.
. . . . .
.
.
y(n−2)
1 (t) y(n−2)
2 (t) · · ·y(n−2)
n (t)
y1(x) y2(x) · · · yn (x)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
,
which is called the
Green’s function for (A).
(d) Show that
∂j G(x,t )
∂xj = 1
P0(t)W(t)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1(t) y2(t) · · · yn (t)
y′
1(t) y′
2(t) · · · y′
n (t)
.
.
.
.
.
. . . . .
.
.
y(n−2)
1 (t) y(n−2)
2 (t) · · ·y(n−2)
n (t)
y(j)
1 (x) y(j)
2 (x) · · ·y(j)
n (x)
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
, 0 ≤ j ≤ n.
(e) Show that if a<t<b then
∂j G(x,t )
∂xj
⏐
⏐
⏐
⏐
x=t
=



0, 1 ≤ j ≤ n− 2,
1
P0(t) , j = n− 1.
(f) Show that
y(j)
p (x) =









∫ x
x0
∂j G(x,t )
∂xj F(t) dt, 1 ≤ j ≤ n− 1,
F(x)
P0(x) +
∫ x
x0
∂(n)G(x,t )
∂xn F(t) dt, j = n.
In Exercises 35– 42 use the method suggested by Exercise 34 to find a particular solution in the form
yp =
∫x
x0",Elementary Differential Equations with Boundary Value Problems.pdf
"x0
∂(n)G(x,t )
∂xn F(t) dt, j = n.
In Exercises 35– 42 use the method suggested by Exercise 34 to find a particular solution in the form
yp =
∫x
x0
G(x,t )F(t) dt, given the indicated fundamental set of solutions. Assume t hat xand x0 are in
an interval on which the equation is normal.
35. y′′′+ 2 y′− y′− 2y= F(x); {ex, e−x,e −2x}
36. x3y′′′+ x2y′′− 2xy′+ 2 y= F(x); {x,x 2, 1/x }
37. x3y′′′− x2(x+ 3) y′′+ 2 x(x+ 3) y′− 2(x+ 3) y = F(x); {x,x 2,xe x}
38. x(1 − x)y′′′+ ( x2 − 3x+ 3) y′′+ xy′− y = F(x); {x, 1/x,e x /x }
39. y(4) − 5y′′+ 4 y= F(x); {ex, e−x, e2x, e−2x}
40. xy(4) + 4 y′′′= F(x); {1,x, x 2, 1/x }
41. x4y(4) + 6 x3y′′′+ 2 x2y′′− 4xy′+ 4 y= F(x); {x,x 2, 1/x, 1/x 2}
42. xy(4) − y′′′− 4xy′+ 4 y′= F(x); {1,x 2, e2x,e −2x}",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 10
Linear Systems of Differential
Equations
IN THIS CHAPTER we consider systems of differential equatio ns involving more than one unknown
function. Such systems arise in many physical applications .
SECTION 10.1 presents examples of physical situations that lead to systems of differential equations.
SECTION 10.2 discusses linear systems of differential equa tions.
SECTION 10.3 deals with the basic theory of homogeneous line ar systems.
SECTIONS 10.4, 10.5, AND 10.6 present the theory of constant coefficient homogeneous systems.
SECTION 10.7 presents the method of variation of parameters for nonhomogeneous linear systems.
507",Elementary Differential Equations with Boundary Value Problems.pdf
"508 Chapter 10 Linear Systems of Differential Equations
10.1 INTRODUCTION TO SYSTEMS OF DIFFERENTIAL EQUA TIONS
Many physical situations are modelled by systems of n differential equations in nunknown functions,
where n ≥ 2. The next three examples illustrate physical problems that lead to systems of differential
equations. In these examples and throughout this chapter we ’ll denote the independent variable by t.
Example 10.1.1 T anks T1 and T2 contain 100 gallons and 300 gallons of salt solutions, respe ctively . Salt
solutions are simultaneously added to both tanks from exter nal sources, pumped from each tank to the
other, and drained from both tanks (Figure 10.1.1). A solution with 1 pound of salt per gallon is pumped
into T1 from an external source at 5 gal/min, and a solution with 2 pounds of salt per gallon is pumped
into T2 from an external source at 4 gal/min. The solution from T1 is pumped into T2 at 2 gal/min, and the",Elementary Differential Equations with Boundary Value Problems.pdf
"into T2 from an external source at 4 gal/min. The solution from T1 is pumped into T2 at 2 gal/min, and the
solution from T2 is pumped into T1 at 3 gal/min. T1 is drained at 6 gal/min and T2 is drained at 3 gal/min.
Let Q1(t) and Q2(t) be the number of pounds of salt in T1 and T2, respectively , at time t >0. Derive a
system of differential equations for Q1 and Q2. Assume that both mixtures are well stirred.
6 gal/min 3 gal/min
2 gal/min
3 gal/min
 T
1  T2
5 gal/min; 1 lb/gal 4 gal/min; 2 lb/gal
300 gal
100 gal
Figure 10.1.1
Solution As in Section 4.2, let rate in and rate out denote the rates (lb/min) at which salt enters and
leaves a tank; thus,
Q′
1 = ( rate in )1 − (rate out )1,
Q′
2 = ( rate in )2 − (rate out )2.
Note that the volumes of the solutions in T1 and T2 remain constant at 100 gallons and 300 gallons,
respectively .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.1 Introduction to Systems of Differential Equations 509
T1 receives salt from the external source at the rate of
(1 lb/gal) × (5 gal/min) = 5 lb/min ,
and from T2 at the rate of
(lb/gal in T2) × (3 gal/min) = 1
300 Q2 × 3 = 1
100 Q2 lb/min.
Therefore
(rate in) 1 = 5 + 1
100 Q2. (10.1.1)
Solution leaves T1 at the rate of 8 gal/min, since 6 gal/min are drained and 2 gal/ min are pumped to T2;
hence,
(rate out )1 = ( lb/gal in T 1) × (8 gal/min) = 1
100 Q1 × 8 = 2
25 Q1. (10.1.2)
Eqns. ( 10.1.1) and ( 10.1.2) imply that
Q′
1 = 5 + 1
100 Q2 − 2
25 Q1. (10.1.3)
T2 receives salt from the external source at the rate of
(2 lb/gal) × (4 gal/min) = 8 lb/min ,
and from T1 at the rate of
(lb/gal in T1) × (2 gal/min) = 1
100 Q1 × 2 = 1
50 Q1 lb/min.
Therefore
(rate in) 2 = 8 + 1
50 Q1. (10.1.4)
Solution leaves T2 at the rate of 6 gal/min, since 3 gal/min are drained and 3 gal/min are pumped to T1;
hence,
(rate out )2 = ( lb/gal in T 2) × (6 gal/min) = 1
300 Q2 × 6 = 1
50 Q2. (10.1.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"hence,
(rate out )2 = ( lb/gal in T 2) × (6 gal/min) = 1
300 Q2 × 6 = 1
50 Q2. (10.1.5)
Eqns. ( 10.1.4) and ( 10.1.5) imply that
Q′
2 = 8 + 1
50 Q1 − 1
50 Q2. (10.1.6)
W e say that ( 10.1.3) and ( 10.1.6) form a system of two first order equations in two unknowns , and write
them together as
Q′
1 = 5 − 2
25 Q1 + 1
100 Q2
Q′
2 = 8 + 1
50 Q1 − 1
50 Q2.
Example 10.1.2 A mass m1 is suspended from a rigid support on a spring S1 and a second mass m2
is suspended from the first on a spring S2 (Figure 10.1.2). The springs obey Hooke’s law , with spring
constants k1 and k2. Internal friction causes the springs to exert damping forc es proportional to the rates
of change of their lengths, with damping constants c1 and c2. Let y1 = y1(t) and y2 = y2(t) be the
displacements of the two masses from their equilibrium posi tions at time t, measured positive upward.
Derive a system of differential equations for y1 and y2, assuming that the masses of the springs are",Elementary Differential Equations with Boundary Value Problems.pdf
"Derive a system of differential equations for y1 and y2, assuming that the masses of the springs are
negligible and that vertical external forces F1 and F2 also act on the objects.",Elementary Differential Equations with Boundary Value Problems.pdf
"510 Chapter 10 Linear Systems of Differential Equations
Mass  m1
Mass  m2
 y1
 y2
Spring  S1
Spring  S2
Figure 10.1.2
Solution In equilibrium, S1 supports both m1 and m2 and S2 supports only m2. Therefore, if ∆ ℓ1 and
∆ ℓ2 are the elongations of the springs in equilibrium then
(m1 + m2)g= k1∆ ℓ1 and m2g= k2∆ ℓ2. (10.1.7)
Let H1 be the Hooke’s law force acting on m1, and let D1 be the damping force on m1. Similarly , let
H2 and D2 be the Hooke’s law and damping forces acting on m2. According to Newton’s second law of
motion,
m1y′′
1 = −m1 g+ H1 + D1 + F1,
m2y′′
2 = −m2 g+ H2 + D2 + F2. (10.1.8)
When the displacements are y1 and y2, the change in length of S1 is −y1 + ∆ ℓ1 and the change in length
of S2 is −y2 + y1 + ∆ ℓ2. Both springs exert Hooke’s law forces on m1, while only S2 exerts a Hooke’s
law force on m2. These forces are in directions that tend to restore the spri ngs to their natural lengths.
Therefore",Elementary Differential Equations with Boundary Value Problems.pdf
"law force on m2. These forces are in directions that tend to restore the spri ngs to their natural lengths.
Therefore
H1 = k1(−y1 + ∆ ℓ1) − k2(−y2 + y1 + ∆ ℓ2) and H2 = k2(−y2 + y1 + ∆ ℓ2). (10.1.9)
When the velocities are y′
1 and y′
2, S1 and S2 are changing length at the rates −y′
1 and −y′
2 + y′
1, respec-
tively . Both springs exert damping forces on m1, while only S2 exerts a damping force on m2. Since the
force due to damping exerted by a spring is proportional to th e rate of change of length of the spring and
in a direction that opposes the change, it follows that
D1 = −c1y′
1 + c2(y′
2 − y′
1) and D2 = −c2(y′
2 − y′
1). (10.1.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.1 Introduction to Systems of Differential Equations 511
From ( 10.1.8), ( 10.1.9), and ( 10.1.10),
m1y′′
1 = −m1 g+ k1(−y1 + ∆ ℓ1) − k2(−y2 + y1 + ∆ ℓ2)
−c1y′
1 + c2(y′
2 − y′
1) + F1
= −(m1 g− k1∆ ℓ1 + k2∆ ℓ2) − k1y1 + k2(y2 − y1)
−c1y′
1 + c2(y′
2 − y′
1) + F1
(10.1.11)
and
m2y′′
2 = −m2g+ k2(−y2 + y1 + ∆ ℓ2) − c2(y′
2 − y′
1) + F2
= −(m2g− k2∆ ℓ2) − k2(y2 − y1) − c2(y′
2 − y′
1) + F2. (10.1.12)
From ( 10.1.7),
m1g− k1∆ ℓ1 + k2∆ ℓ2 = −m2g+ k2∆ ℓ2 = 0 .
Therefore we can rewrite ( 10.1.11) and ( 10.1.12) as
m1y′′
1 = −(c1 + c2)y′
1 + c2y′
2 − (k1 + k2)y1 + k2y2 + F1
m2y′′
2 = c2y′
1 − c2y′
2 + k2y1 − k2y2 + F2.
Example 10.1.3 Let X = X(t) = x(t) i + y(t) j + z(t) k be the position vector at time tof an object
with mass m, relative to a rectangular coordinate system with origin at Earth’s center (Figure 10.1.3).
According to Newton’s law of gravitation, Earth’s gravitat ional force F = F(x,y,z ) on the object is",Elementary Differential Equations with Boundary Value Problems.pdf
"According to Newton’s law of gravitation, Earth’s gravitat ional force F = F(x,y,z ) on the object is
inversely proportional to the square of the distance of the o bject from Earth’s center, and directed toward
the center; thus,
F = K
∥X∥2
(
− X
∥X∥
)
= −K xi + yj + zk
(x2 + y2 + z2)3/ 2 , (10.1.13)
where K is a constant. T o determine K, we observe that the magnitude of F is
∥F∥= K ∥X∥
∥X∥3 = K
∥X∥2 = K
(x2 + y2 + z2) .
Let Rbe Earth’s radius. Since ∥F∥= mgwhen the object is at Earth’s surface,
mg = K
R2 , so K = mgR2.
Therefore we can rewrite ( 10.1.13) as
F = −mgR2 xi + yj + zk
(x2 + y2 + z2)3/ 2 .
Now suppose F is the only force acting on the object. According to Newton’s second law of motion,
F = mX′′; that is,
m(x′′i + y′′j + z′′k) = −mgR2 xi + yj + zk
(x2 + y2 + z2)3/ 2 .
Cancelling the common factor mand equating components on the two sides of this equation yie lds the
system
x′′ = − gR2x
(x2 + y2 + z2)3/ 2
y′′ = − gR2y
(x2 + y2 + z2)3/ 2
z′′ = − gR2z
(x2 + y2 + z2)3/ 2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"system
x′′ = − gR2x
(x2 + y2 + z2)3/ 2
y′′ = − gR2y
(x2 + y2 + z2)3/ 2
z′′ = − gR2z
(x2 + y2 + z2)3/ 2 .
(10.1.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"512 Chapter 10 Linear Systems of Differential Equations
 x
 y
 z
 X(t)
Figure 10.1.3
Rewriting Higher Order Systems as First Order Systems
A system of the form
y′
1 = g1(t,y 1,y 2,...,y n)
y′
2 = g2(t,y 1,y 2,...,y n)
.
.
.
y′
n = gn(t,y 1,y 2,...,y n)
(10.1.15)
is called a
first order system , since the only derivatives occurring in it are first derivat ives. The derivative
of each of the unknowns may depend upon the independent varia ble and all the unknowns, but not on
the derivatives of other unknowns. When we wish to emphasize the number of unknown functions in
(10.1.15) we will say that ( 10.1.15) is an n× nsystem.
Systems involving higher order derivatives can often be ref ormulated as first order systems by intro-
ducing additional unknowns. The next two examples illustra te this.
Example 10.1.4 Rewrite the system
m1y′′
1 = −(c1 + c2)y′
1 + c2y′
2 − (k1 + k2)y1 + k2y2 + F1
m2y′′
2 = c2y′
1 − c2y′
2 + k2y1 − k2y2 + F2. (10.1.16)",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 10.1.4 Rewrite the system
m1y′′
1 = −(c1 + c2)y′
1 + c2y′
2 − (k1 + k2)y1 + k2y2 + F1
m2y′′
2 = c2y′
1 − c2y′
2 + k2y1 − k2y2 + F2. (10.1.16)
derived in Example 10.1.2 as a system of first order equations.
Solution If we define v1 = y′
1 and v2 = y′
2, then v′
1 = y′′
1 and v′
2 = y′′
2 , so ( 10.1.16) becomes
m1v′
1 = −(c1 + c2)v1 + c2v2 − (k1 + k2)y1 + k2y2 + F1
m2v′
2 = c2v1 − c2v2 + k2y1 − k2y2 + F2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.1 Introduction to Systems of Differential Equations 513
Therefore {y1,y 2,v 1,v 2}satisfies the 4 × 4 first order system
y′
1 = v1
y′
2 = v2
v′
1 = 1
m1
[−(c1 + c2)v1 + c2v2 − (k1 + k2)y1 + k2y2 + F1]
v′
2 = 1
m2
[c2v1 − c2v2 + k2y1 − k2y2 + F2] .
(10.1.17)
REM A RK : The difference in form between ( 10.1.15) and ( 10.1.17), due to the way in which the unknowns
are denoted in the two systems, isn’t important; ( 10.1.17) is a first order system, in that each equation in
(10.1.17) expresses the first derivative of one of the unknown functio ns in a way that does not involve
derivatives of any of the other unknowns.
Example 10.1.5 Rewrite the system
x′′ = f(t,x,x ′,y,y ′,y ′′)
y′′′= g(t,x,x ′,y,y ′y′′)
as a first order system.
Solution W e regard x, x′, y, y′, and y′′as unknown functions, and rename them
x= x1, x′= x2, y = y1, y ′= y2, y ′′= y3.
These unknowns satisfy the system
x′
1 = x2
x′
2 = f(t,x 1,x 2,y 1,y 2,y 3)
y′
1 = y2
y′
2 = y3
y′
3 = g(t,x 1,x 2,y 1,y 2,y 3).",Elementary Differential Equations with Boundary Value Problems.pdf
"These unknowns satisfy the system
x′
1 = x2
x′
2 = f(t,x 1,x 2,y 1,y 2,y 3)
y′
1 = y2
y′
2 = y3
y′
3 = g(t,x 1,x 2,y 1,y 2,y 3).
Rewriting Scalar Differential Equations as Systems
In this chapter we’ll refer to differential equations invol ving only one unknown function as scalar differ-
ential equations. Scalar differential equations can be rew ritten as systems of first order equations by the
method illustrated in the next two examples.
Example 10.1.6 Rewrite the equation
y(4) + 4 y′′′+ 6 y′′+ 4 y′+ y= 0 (10.1.18)
as a 4 × 4 first order system.
Solution W e regard y, y′, y′′, and y′′′as unknowns and rename them
y= y1, y ′= y2, y ′′= y3, and y′′′= y4.
Then y(4) = y′
4, so (
10.1.18) can be written as
y′
4 + 4 y4 + 6 y3 + 4 y2 + y1 = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"514 Chapter 10 Linear Systems of Differential Equations
Therefore {y1,y 2,y 3,y 4}satisfies the system
y′
1 = y2
y′
2 = y3
y′
3 = y4
y′
4 = −4y4 − 6y3 − 4y2 − y1.
Example 10.1.7 Rewrite
x′′′= f(t,x,x ′,x ′′)
as a system of first order equations.
Solution W e regard x, x′, and x′′as unknowns and rename them
x= y1, x ′= y2, and x′′= y3.
Then
y′
1 = x′= y2, y ′
2 = x′′= y3, and y′
3 = x′′′.
Therefore {y1,y 2,y 3}satisfies the first order system
y′
1 = y2
y′
2 = y3
y′
3 = f(t,y 1,y 2,y 3).
Since systems of differential equations involving higher d erivatives can be rewritten as first order sys-
tems by the method used in Examples
10.1.5 – 10.1.7 , we’ll consider only first order systems.
Numerical Solution of Systems
The numerical methods that we studied in Chapter 3 can be exte nded to systems, and most differential
equation software packages include programs to solve syste ms of equations. W e won’t go into detail on",Elementary Differential Equations with Boundary Value Problems.pdf
"equation software packages include programs to solve syste ms of equations. W e won’t go into detail on
numerical methods for systems; however, for illustrative p urposes we’ll describe the Runge-Kutta method
for the numerical solution of the initial value problem
y′
1 = g1(t,y 1,y 2), y 1(t0) = y10,
y′
2 = g2(t,y 1,y 2), y 2(t0) = y20
at equally spaced points t0, t1, . . . , tn = bin an interval [t0,b ]. Thus,
ti = t0 + ih, i = 0 , 1,...,n,
where
h= b− t0
n .
W e’ll denote the approximate values of y1 and y2 at these points by y10 ,y 11,...,y 1n and y20 ,y 21,...,y 2n.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.1 Introduction to Systems of Differential Equations 515
The Runge-Kutta method computes these approximate values a s follows: given y1i and y2i, compute
I1i = g1(ti,y 1i,y 2i),
J1i = g2(ti,y 1i,y 2i),
I2i = g1
(
ti + h
2 ,y 1i + h
2 I1i,y 2i + h
2 J1i
)
,
J2i = g2
(
ti + h
2 ,y 1i + h
2 I1i,y 2i + h
2 J1i
)
,
I3i = g1
(
ti + h
2 ,y 1i + h
2 I2i,y 2i + h
2 J2i
)
,
J3i = g2
(
ti + h
2 ,y 1i + h
2 I2i,y 2i + h
2 J2i
)
,
I4i = g1(ti + h,y 1i + hI3i,y 2i + hJ3i),
J4i = g2(ti + h,y 1i + hI3i,y 2i + hJ3i),
and
y1,i +1 = y1i + h
6 (I1i + 2 I2i + 2 I3i + I4i),
y2,i +1 = y2i + h
6 (J1i + 2 J2i + 2 J3i + J4i)
for i= 0 , . . . , n−1. Under appropriate conditions on g1 and g2, it can be shown that the global truncation
error for the Runge-Kutta method is O(h4), as in the scalar case considered in Section 3.3.
10.1 Exercises
1. T anks T1 and T2 contain 50 gallons and 100 gallons of salt solutions, respec tively . A solution",Elementary Differential Equations with Boundary Value Problems.pdf
"10.1 Exercises
1. T anks T1 and T2 contain 50 gallons and 100 gallons of salt solutions, respec tively . A solution
with 2 pounds of salt per gallon is pumped into T1 from an external source at 1 gal/min, and a
solution with 3 pounds of salt per gallon is pumped into T2 from an external source at 2 gal/min.
The solution from T1 is pumped into T2 at 3 gal/min, and the solution from T2 is pumped into T1
at 4 gal/min. T1 is drained at 2 gal/min and T2 is drained at 1 gal/min. Let Q1(t) and Q2(t) be the
number of pounds of salt in T1 and T2, respectively , at time t> 0. Derive a system of differential
equations for Q1 and Q2. Assume that both mixtures are well stirred.
2. T wo 500 gallon tanks T1 and T2 initially contain 100 gallons each of salt solution. A solut ion
with 2 pounds of salt per gallon is pumped into T1 from an external source at 6 gal/min, and a
solution with 1 pound of salt per gallon is pumped into T2 from an external source at 5 gal/min.",Elementary Differential Equations with Boundary Value Problems.pdf
"solution with 1 pound of salt per gallon is pumped into T2 from an external source at 5 gal/min.
The solution from T1 is pumped into T2 at 2 gal/min, and the solution from T2 is pumped into T1
at 1 gal/min. Both tanks are drained at 3 gal/min. Let Q1(t) and Q2(t) be the number of pounds
of salt in T1 and T2, respectively , at time t >0. Derive a system of differential equations for Q1
and Q2 that’s valid until a tank is about to overflow . Assume that bot h mixtures are well stirred.
3. A mass m1 is suspended from a rigid support on a spring S1 with spring constant k1 and damping
constant c1. A second mass m2 is suspended from the first on a spring S2 with spring constant k2
and damping constant c2, and a third mass m3 is suspended from the second on a spring S3 with
spring constant k3 and damping constant c3. Let y1 = y1(t), y2 = y2(t), and y3 = y3(t) be the
displacements of the three masses from their equilibrium po sitions at time t, measured positive",Elementary Differential Equations with Boundary Value Problems.pdf
"displacements of the three masses from their equilibrium po sitions at time t, measured positive
upward. Derive a system of differential equations for y1, y2 and y3, assuming that the masses of
the springs are negligible and that vertical external force s F1, F2, and F3 also act on the masses.",Elementary Differential Equations with Boundary Value Problems.pdf
"516 Chapter 10 Linear Systems of Differential Equations
4. Let X = xi + yj + zk be the position vector of an object with mass m, expressed in terms of a
rectangular coordinate system with origin at Earth’s cente r (Figure 10.1.3). Derive a system of dif-
ferential equations for x, y, and z, assuming that the object moves under Earth’s gravitationa l force
(given by Newton’s law of gravitation, as in Example 10.1.3 ) and a resistive force proportional to
the speed of the object. Let α be the constant of proportionality .
5. Rewrite the given system as a first order system.
(a) x′′′= f(t,x,y,y ′)
y′′= g(t,y,y ′) (b)
u′= f(t,u,v,v ′,w ′)
v′′= g(t,u,v,v ′,w )
w′′= h(t,u,v,v ′,w,w ′)
(c) y′′′= f(t,y,y ′,y ′′) (d) y(4) = f(t,y )
(e) x′′= f(t,x,y )
y′′= g(t,x,y )
6. Rewrite the system (
10.1.14) of differential equations derived in Example 10.1.3 as a first order
system.
7. Formulate a version of Euler’s method (Section 3.1) for the n umerical solution of the initial value
problem
y′",Elementary Differential Equations with Boundary Value Problems.pdf
"system.
7. Formulate a version of Euler’s method (Section 3.1) for the n umerical solution of the initial value
problem
y′
1 = g1(t,y 1,y 2), y 1(t0) = y10,
y′
2 = g2(t,y 1,y 2), y 2(t0) = y20,
on an interval [t0,b ].
8. Formulate a version of the improved Euler method (Section 3. 2) for the numerical solution of the
initial value problem
y′
1 = g1(t,y 1,y 2), y 1(t0) = y10,
y′
2 = g2(t,y 1,y 2), y 2(t0) = y20,
on an interval [t0,b ].
10.2 LINEAR SYSTEMS OF DIFFERENTIAL EQUA TIONS
A first order system of differential equations that can be wri tten in the form
y′
1 = a11(t)y1 + a12(t)y2 + · · ·+ a1n(t)yn + f1(t)
y′
2 = a21(t)y1 + a22(t)y2 + · · ·+ a2n(t)yn + f2(t)
.
.
.
y′
n = an1(t)y1 + an2(t)y2 + · · ·+ ann(t)yn + fn (t)
(10.2.1)
is called a
linear system .
The linear system ( 10.2.1) can be written in matrix form as





y′
1
y′
2
.
.
.
y′
n




 =





a11(t) a12 (t) · · ·a1n(t)
a21(t) a22 (t) · · ·a2n(t)
.
.
. .
.
. . . . .
.
.
an1(t) an2(t) · · ·ann(t)






",Elementary Differential Equations with Boundary Value Problems.pdf
".
.
y′
n




 =





a11(t) a12 (t) · · ·a1n(t)
a21(t) a22 (t) · · ·a2n(t)
.
.
. .
.
. . . . .
.
.
an1(t) an2(t) · · ·ann(t)










y1
y2
.
.
.
yn




 +





f1(t)
f2(t)
.
.
.
fn (t)




 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.2 Linear Systems of Differential Equations 517
or more briefly as
y′= A(t)y + f (t), (10.2.2)
where
y =





y1
y2
.
.
.
yn




 , A (t) =





a11(t) a12(t) · · ·a1n(t)
a21(t) a22(t) · · ·a2n(t)
.
.
. .
.
. . . . .
.
.
an1(t) an2(t) · · ·ann(t)




 , and f (t) =





f1(t)
f2(t)
.
.
.
fn (t)




 .
W e call Athe
coefficient matrix of ( 10.2.2) and f the forcing function . W e’ll say that Aand f are con-
tinuous if their entries are continuous. If f = 0, then ( 10.2.2) is homogeneous; otherwise, ( 10.2.2) is
nonhomogeneous.
An initial value problem for ( 10.2.2) consists of finding a solution of ( 10.2.2) that equals a given
constant vector
k =





k1
k2
.
.
.
kn




 .
at some initial point t0. W e write this initial value problem as
y′= A(t)y + f (t), y(t0) = k.
The next theorem gives sufficient conditions for the existen ce of solutions of initial value problems for
(
10.2.2). W e omit the proof.",Elementary Differential Equations with Boundary Value Problems.pdf
"The next theorem gives sufficient conditions for the existen ce of solutions of initial value problems for
(
10.2.2). W e omit the proof.
Theorem 10.2.1 Suppose the coefficient matrix Aand the forcing function f are continuous on (a,b ), let
t0 be in (a,b ), and let k be an arbitrary constant n-vector . Then the initial value problem
y′= A(t)y + f (t), y(t0) = k
has a unique solution on (a,b ).
Example 10.2.1
(a) Write the system
y′
1 = y1 + 2 y2 + 2 e4t
y′
2 = 2 y1 + y2 + e4t (10.2.3)
in matrix form and conclude from Theorem 10.2.1 that every initial value problem for ( 10.2.3) has
a unique solution on (−∞ , ∞ ).
(b) V erify that
y = 1
5
[ 8
7
]
e4t + c1
[ 1
1
]
e3t + c2
[ 1
−1
]
e−t (10.2.4)
is a solution of (
10.2.3) for all values of the constants c1 and c2.
(c) Find the solution of the initial value problem
y′=
[
1 2
2 1
]
y +
[
2
1
]
e4t , y(0) = 1
5
[
3
22
]
. (10.2.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"518 Chapter 10 Linear Systems of Differential Equations
SO LU TIO N (a) The system ( 10.2.3) can be written in matrix form as
y′=
[ 1 2
2 1
]
y +
[ 2
1
]
e4t .
An initial value problem for (
10.2.3) can be written as
y′=
[ 1 2
2 1
]
y +
[ 2
1
]
e4t, y (t0 ) =
[ k1
k2
]
.
Since the coefficient matrix and the forcing function are bot h continuous on (−∞ , ∞ ), Theorem
10.2.1
implies that this problem has a unique solution on (−∞ , ∞ ).
SO LU TIO N (b) If y is given by ( 10.2.4), then
Ay + f = 1
5
[ 1 2
2 1
][ 8
7
]
e4t + c1
[ 1 2
2 1
][ 1
1
]
e3t
+c2
[ 1 2
2 1
][ 1
−1
]
e−t +
[ 2
1
]
e4t
= 1
5
[
22
23
]
e4t + c1
[
3
3
]
e3t + c2
[
−1
1
]
e−t +
[
2
1
]
e4t
= 1
5
[
32
28
]
e4t + 3 c1
[
1
1
]
e3t − c2
[
1
−1
]
e−t = y′.
SO LU TIO N (c) W e must choose c1 and c2 in ( 10.2.4) so that
1
5
[
8
7
]
+ c1
[
1
1
]
+ c2
[
1
−1
]
= 1
5
[
3
22
]
,
which is equivalent to [ 1 1
1 −1
][ c1
c2
]
=
[ −1
3
]
.
Solving this system yields c1 = 1 , c2 = −2, so
y = 1
5
[
8
7
]
e4t +
[
1
1
]
e3t − 2
[
1
−1
]
e−t",Elementary Differential Equations with Boundary Value Problems.pdf
"1 −1
][ c1
c2
]
=
[ −1
3
]
.
Solving this system yields c1 = 1 , c2 = −2, so
y = 1
5
[
8
7
]
e4t +
[
1
1
]
e3t − 2
[
1
−1
]
e−t
is the solution of (
10.2.5).
REM A RK : The theory of n× nlinear systems of differential equations is analogous to th e theory of the
scalar n-th order equation
P0(t)y(n) + P1(t)y(n−1) + · · ·+ Pn (t)y= F(t), (10.2.6)
as developed in Sections 9.1. For example, by rewriting ( 10.2.6) as an equivalent linear system it can be
shown that Theorem 10.2.1 implies Theorem 9.1.1 (Exercise 12).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.2 Linear Systems of Differential Equations 519
10.2 Exercises
1. Rewrite the system in matrix form and verify that the given ve ctor function satisfies the system for
any choice of the constants c1 and c2.
(a) y′
1 = 2 y1 + 4 y2
y′
2 = 4 y1 + 2 y2; y = c1
[ 1
1
]
e6t + c2
[ 1
−1
]
e−2t
(b) y′
1 = −2y1 − 2y2
y′
2 = −5y1 + y2; y = c1
[
1
1
]
e−4t + c2
[
−2
5
]
e3t
(c) y′
1 = −4y1 − 10y2
y′
2 = 3 y1 + 7 y2; y = c1
[ −5
3
]
e2t + c2
[ 2
−1
]
et
(d) y′
1 = 2 y1 + y2
y′
2 = y1 + 2 y2; y = c1
[ 1
1
]
e3t + c2
[ 1
−1
]
et
2. Rewrite the system in matrix form and verify that the given ve ctor function satisfies the system for
any choice of the constants c1, c2, and c3.
(a)
y′
1 = −y1 + 2 y2 + 3 y3
y′
2 = y2 + 6 y3
y′
3 = −2y3;
y = c1


1
1
0

 et + c2


1
0
0

 e−t + c3


1
−2
1

 e−2t
(b)
y′
1 = 2 y2 + 2 y3
y′
2 = 2 y1 + 2 y3
y′
3 = 2 y1 + 2 y2;
y = c1


−1
0
1

 e−2t + c2


0
−1
1

 e−2t + c3


1
1
1

 e4t
(c)
y′
1 = −y1 + 2 y2 + 2 y3
y′
2 = 2 y1 − y2 + 2 y3",Elementary Differential Equations with Boundary Value Problems.pdf
"y′
3 = 2 y1 + 2 y2;
y = c1


−1
0
1

 e−2t + c2


0
−1
1

 e−2t + c3


1
1
1

 e4t
(c)
y′
1 = −y1 + 2 y2 + 2 y3
y′
2 = 2 y1 − y2 + 2 y3
y′
3 = 2 y1 + 2 y2 − y3;
y = c1


−1
0
1

 e−3t + c2


0
−1
1

 e−3t + c3


1
1
1

 e3t
(d)
y′
1 = 3 y1 − y2 − y3
y′
2 = −2y1 + 3 y2 + 2 y3
y′
3 = 4 y1 − y2 − 2y3;
y = c1


1
0
1

 e2t + c2


1
−1
1

 e3t + c3


1
−3
7

 e−t
3. Rewrite the initial value problem in matrix form and verify t hat the given vector function is a
solution.
(a) y′
1 = y1 + y2
y′
2 = −2y1 + 4 y2,
y1(0) = 1
y2(0) = 0; y = 2
[ 1
1
]
e2t −
[ 1
2
]
e3t
(b) y′
1 = 5 y1 + 3 y2
y′
2 = −y1 + y2,
y1(0) = 12
y2(0) = −6; y = 3
[
1
−1
]
e2t + 3
[
3
−1
]
e4t",Elementary Differential Equations with Boundary Value Problems.pdf
"520 Chapter 10 Linear Systems of Differential Equations
4. Rewrite the initial value problem in matrix form and verify t hat the given vector function is a
solution.
(a)
y′
1 = 6 y1 + 4 y2 + 4 y3
y′
2 = −7y1 − 2y2 − y3,
y′
3 = 7 y1 + 4 y2 + 3 y3
,
y1(0) = 3
y2(0) = −6
y3(0) = 4
y =


1
−1
1

 e6t + 2


1
−2
1

 e2t +


0
−1
1

 e−t
(b)
y′
1 = 8 y1 + 7 y2 + 7 y3
y′
2 = −5y1 − 6y2 − 9y3,
y′
3 = 5 y1 + 7 y2 + 10 y3,
y1(0) = 2
y2(0) = −4
y3(0) = 3
y =


1
−1
1

 e8t +


0
−1
1

 e3t +


1
−2
1

 et
5. Rewrite the system in matrix form and verify that the given ve ctor function satisfies the system for
any choice of the constants c1 and c2.
(a) y′
1 = −3y1 + 2 y2 + 3 − 2t
y′
2 = −5y1 + 3 y2 + 6 − 3t
y = c1
[ 2 cos t
3 cos t− sin t
]
+ c2
[ 2 sin t
3 sin t+ cos t
]
+
[ 1
t
]
(b) y′
1 = 3 y1 + y2 − 5et
y′
2 = −y1 + y2 + et
y = c1
[
−1
1
]
e2t + c2
[
1 + t
−t
]
e2t +
[
1
3
]
et
(c) y′
1 = −y1 − 4y2 + 4 et + 8 tet
y′
2 = −y1 − y2 + e3t + (4 t+ 2) et
y = c1
[ 2
1
]",Elementary Differential Equations with Boundary Value Problems.pdf
"y = c1
[
−1
1
]
e2t + c2
[
1 + t
−t
]
e2t +
[
1
3
]
et
(c) y′
1 = −y1 − 4y2 + 4 et + 8 tet
y′
2 = −y1 − y2 + e3t + (4 t+ 2) et
y = c1
[ 2
1
]
e−3t + c2
[ −2
1
]
et +
[ e3t
2tet
]
(d) y′
1 = −6y1 − 3y2 + 14 e2t + 12 et
y′
2 = y1 − 2y2 + 7 e2t − 12et
y = c1
[ −3
1
]
e−5t + c2
[ −1
1
]
e−3t +
[ e2t + 3 et
2e2t − 3et
]
6. Convert the linear scalar equation
P0(t)y(n) + P1(t)y(n−1) + · · ·+ Pn(t)y(t) = F(t) (A)
into an equivalent n× nsystem
y′= A(t)y + f (t),
and show that Aand f are continuous on an interval (a,b ) if and only if (A) is normal on (a,b ).
7. A matrix function
Q(t) =





q11(t) q12(t) · · ·q1s(t)
q21(t) q22(t) · · ·q2s(t)
.
.
. .
.
. . . . .
.
.
qr1(t) qr2(t) · · ·qrs(t)




",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.2 Linear Systems of Differential Equations 521
is said to be differentiable if its entries {qij }are differentiable. Then the derivative Q′is defined
by
Q′(t) =





q′
11(t) q′
12(t) · · ·q′
1s(t)
q′
21(t) q′
22(t) · · ·q′
2s(t)
.
.
. .
.
. . . . .
.
.
q′
r1(t) q′
r2(t) · · ·q′
rs (t)




 .
(a) Prove: If P and Qare differentiable matrices such that P + Qis defined and if c1 and c2 are
constants, then
(c1P + c2Q)′= c1P′+ c2Q′.
(b) Prove: If P and Qare differentiable matrices such that PQ is defined, then
(PQ)′= P′Q+ PQ′.
8. V erify that Y′= AY.
(a) Y =
[ e6t e−2t
e6t −e−2t
]
, A =
[ 2 4
4 2
]
(b) Y =
[ e−4t −2e3t
e−4t 5e3t
]
, A =
[ −2 −2
−5 1
]
(c) Y =
[ −5e2t 2et
3e2t −et
]
, A =
[ −4 −10
3 7
]
(d) Y =
[ e3t et
e3t −et
]
, A =
[ 2 1
1 2
]
(e) Y =


et e−t e−2t
et 0 −2e−2t
0 0 e−2t

 , A =


−1 2 3
0 1 6
0 0 −2


(f) Y =


−e−2t −e−2t e4t
0 e−2t e4t
e−2t 0 e4t

 , A =


0 2 2
2 0 2
2 2 0


(g) Y =


e3t e−3t 0
e3t 0 −e−3t
e3t e−3t e−3t

 , A =
",Elementary Differential Equations with Boundary Value Problems.pdf
"
(f) Y =


−e−2t −e−2t e4t
0 e−2t e4t
e−2t 0 e4t

 , A =


0 2 2
2 0 2
2 2 0


(g) Y =


e3t e−3t 0
e3t 0 −e−3t
e3t e−3t e−3t

 , A =


−9 6 6
−6 3 6
−6 6 3


(h) Y =


e2t e3t e−t
0 −e3t −3e−t
e2t e3t 7e−t

 , A =


3 −1 −1
−2 3 2
4 −1 −2


9. Suppose
y1 =
[ y11
y21
]
and y2 =
[ y12
y22
]
are solutions of the homogeneous system
y′= A(t)y, (A)
and define
Y =
[ y11 y12
y21 y22
]
.
(a) Show that Y′= AY.
(b) Show that if c is a constant vector then y = Yc is a solution of (A).
(c) State generalizations of (a) and (b) for n× nsystems.",Elementary Differential Equations with Boundary Value Problems.pdf
"522 Chapter 10 Linear Systems of Differential Equations
10. Suppose Y is a differentiable square matrix.
(a) Find a formula for the derivative of Y2.
(b) Find a formula for the derivative of Yn, where nis any positive integer.
(c) State how the results obtained in (a) and (b) are analogous to results from calculus concerning
scalar functions.
11. It can be shown that if Y is a differentiable and invertible square matrix function, then Y−1 is
differentiable.
(a) Show that ( Y−1)′= −Y−1Y′Y−1. (Hint: Differentiate the identity Y−1Y = I.)
(b) Find the derivative of Y−n = (Y−1) n
, where nis a positive integer.
(c) State how the results obtained in (a) and (b) are analogous to results from calculus concerning
scalar functions.
12. Show that Theorem 10.2.1 implies Theorem 9.1.1. H IN T : Write the scalar equation
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn (x)y= F(x)
as an n× nsystem of linear equations.
13. Suppose y is a solution of the n× nsystem y′= A(t)y on (a,b ), and that the n× nmatrix P is",Elementary Differential Equations with Boundary Value Problems.pdf
"as an n× nsystem of linear equations.
13. Suppose y is a solution of the n× nsystem y′= A(t)y on (a,b ), and that the n× nmatrix P is
invertible and differentiable on (a,b ). Find a matrix Bsuch that the function x = Py is a solution
of x′= Bx on (a,b ).
10.3 BASIC THEORY OF HOMOGENEOUS LINEAR SYSTEMS
In this section we consider homogeneous linear systems y′= A(t)y, where A = A(t) is a continuous
n× n matrix function on an interval (a,b ). The theory of linear homogeneous systems has much in
common with the theory of linear homogeneous scalar equatio ns, which we considered in Sections 2.1,
5.1, and 9.1.
Whenever we refer to solutions of y′= A(t)y we’ll mean solutions on (a,b ). Since y ≡ 0 is obviously
a solution of y′= A(t)y, we call it the trivial solution. Any other solution is nontrivial.
If y1, y2, . . . , yn are vector functions defined on an interval (a,b ) and c1, c2, . . . , cn are constants,
then
y = c1y1 + c2y2 + · · ·+ cnyn (10.3.1)",Elementary Differential Equations with Boundary Value Problems.pdf
"then
y = c1y1 + c2y2 + · · ·+ cnyn (10.3.1)
is a linear combination of y1, y2, . . . , yn . It’s easy show that if y1, y2, . . . , yn are solutions of y′= A(t)y
on (a,b ), then so is any linear combination of y1, y2, . . . , yn (Exercise 1). W e say that {y1, y2,..., yn}
is a fundamental set of solutions of y′= A(t)y on (a,b ) on if every solution of y′= A(t)y on (a,b ) can
be written as a linear combination of y1, y2, . . . , yn , as in ( 10.3.1). In this case we say that ( 10.3.1) is
the general solution of y′= A(t)y on (a,b ).
It can be shown that if Ais continuous on (a,b ) then y′= A(t)y has infinitely many fundamental sets
of solutions on (a,b ) (Exercises 15 and 16). The next definition will help to characterize fundamental
sets of solutions of y′= A(t)y.
W e say that a set {y1, y2,..., yn}of n-vector functions is linearly independent on (a,b ) if the only
constants c1, c2, . . . , cn such that
c1y1(t) + c2y2(t) + · · ·+ cn yn(t) = 0 , a<t<b, (10.3.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"constants c1, c2, . . . , cn such that
c1y1(t) + c2y2(t) + · · ·+ cn yn(t) = 0 , a<t<b, (10.3.2)
are c1 = c2 = · · ·= cn = 0 . If ( 10.3.2) holds for some set of constants c1, c2, . . . , cn that are not all
zero, then {y1, y2,..., yn}is linearly dependent on (a,b )
The next theorem is analogous to Theorems 5.1.3 and 9.1.2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.3 Basic Theory of Homogeneous Linear System 523
Theorem 10.3.1 Suppose the n×nmatrix A= A(t) is continuous on (a,b ). Then a set {y1, y2,..., yn}
of nsolutions of y′= A(t)y on (a,b ) is a fundamental set if and only if it’s linearly independent on (a,b ).
Example 10.3.1 Show that the vector functions
y1 =


et
0
e−t

 , y2 =


0
e3t
1

 , and y3 =


e2t
e3t
0


are linearly independent on every interval (a,b ).
Solution Suppose
c1


et
0
e−t

 + c2


0
e3t
1

 + c3


e2t
e3t
0

 =


0
0
0

 , a<t<b.
W e must show that c1 = c2 = c3 = 0 . Rewriting this equation in matrix form yields


et 0 e2t
0 e3t e3t
e−t 1 0




c1
c2
c3

 =


0
0
0

 , a<t<b.
Expanding the determinant of this system in cofactors of the entries of the first row yields
⏐
⏐
⏐
⏐
⏐
⏐
et 0 e2t
0 e3t e3t
e−t 1 0
⏐
⏐
⏐
⏐
⏐
⏐
= et
⏐
⏐
⏐
⏐
e3t e3t
1 0
⏐
⏐
⏐
⏐− 0
⏐
⏐
⏐
⏐
0 e3t
e−t 0
⏐
⏐
⏐
⏐+ e2t
⏐
⏐
⏐
⏐
0 e3t
e−t 1
⏐
⏐
⏐
⏐
= et (−e3t) + e2t (−e2t ) = −2e4t .",Elementary Differential Equations with Boundary Value Problems.pdf
"⏐
⏐
⏐
⏐
⏐
⏐
= et
⏐
⏐
⏐
⏐
e3t e3t
1 0
⏐
⏐
⏐
⏐− 0
⏐
⏐
⏐
⏐
0 e3t
e−t 0
⏐
⏐
⏐
⏐+ e2t
⏐
⏐
⏐
⏐
0 e3t
e−t 1
⏐
⏐
⏐
⏐
= et (−e3t) + e2t (−e2t ) = −2e4t .
Since this determinant is never zero, c1 = c2 = c3 = 0 .
W e can use the method in Example 10.3.1 to test nsolutions {y1, y2,..., yn}of any n× nsystem
y′= A(t)y for linear independence on an interval (a,b ) on which Ais continuous. T o explain this (and
for other purposes later), it’s useful to write a linear comb ination of y1, y2, . . . , yn in a different way . W e
first write the vector functions in terms of their components as
y1 =





y11
y21
.
.
.
yn1




 , y2 =





y12
y22
.
.
.
yn2




 ,..., yn =





y1n
y2n
.
.
.
ynn




 .
If
y = c1y1 + c2y2 + · · ·+ cnyn
then
y = c1





y11
y21
.
.
.
yn1




 + c2





y12
y22
.
.
.
yn2




 + · · ·+ cn





y1n
y2n
.
.
.
ynn





=





y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
. .
.
. . . . .
.
.
yn1 yn2 · · ·ynn








",Elementary Differential Equations with Boundary Value Problems.pdf
"




y1n
y2n
.
.
.
ynn





=





y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
. .
.
. . . . .
.
.
yn1 yn2 · · ·ynn










c1
c2
.
.
.
cn




 .",Elementary Differential Equations with Boundary Value Problems.pdf
"524 Chapter 10 Linear Systems of Differential Equations
This shows that
c1y1 + c2y2 + · · ·+ cn yn = Yc, (10.3.3)
where
c =





c1
c2
.
.
.
cn





and
Y = [ y1 y2 · · ·yn ] =





y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
. .
.
. . . . .
.
.
yn1 yn2 · · ·ynn




 ; (10.3.4)
that is, the columns of Y are the vector functions y1, y2,..., yn.
For reference below , note that
Y′ = [ y′
1 y′
2 · · ·y′
n]
= [ Ay1 Ay2 · · ·Ayn ]
= A[y1 y2 · · ·yn ] = AY;
that is, Y satisfies the matrix differential equation
Y′= AY.
The determinant of Y,
W =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
.
.
.
. . . . .
.
.
yn1 yn2 · · ·ynn
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
(10.3.5)
is called the
Wronskian of {y1, y2,..., yn }. It can be shown (Exercises 2 and 3) that this definition is
analogous to definitions of the Wronskian of scalar function s given in Sections 5.1 and 9.1. The next
theorem is analogous to Theorems 5.1.4 and 9.1.3. The proof is sketched in Exercise 4 for n= 2 and in",Elementary Differential Equations with Boundary Value Problems.pdf
"theorem is analogous to Theorems 5.1.4 and 9.1.3. The proof is sketched in Exercise 4 for n= 2 and in
Exercise 5 for general n.
Theorem 10.3.2 [Abel’s F ormula ] Suppose the n× n matrix A = A(t) is continuous on (a,b ), let
y1, y2, . . . , yn be solutions of y′ = A(t)y on (a,b ), and let t0 be in (a,b ). Then the Wronskian of
{y1, y2,..., yn}is given by
W(t) = W(t0) exp
(∫ t
t0
[
a11(s) + a22(s) + · · ·+ ann(s)] ds
)
, a<t<b. (10.3.6)
Therefore, either W has no zeros in (a,b ) or W ≡ 0 on (a,b ).
REM A RK : The sum of the diagonal entries of a square matrix A is called the trace of A, denoted by
tr(A). Thus, for an n× nmatrix A,
tr(A) = a11 + a22 + · · ·+ ann,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.3 Basic Theory of Homogeneous Linear System 525
and ( 10.3.6) can be written as
W(t) = W(t0) exp
(∫ t
t0
tr(A(s)) ds
)
, a<t<b.
The next theorem is analogous to Theorems 5.1.6 and 9.1.4.
Theorem 10.3.3 Suppose the n× nmatrix A = A(t) is continuous on (a,b ) and let y1, y2, . . . ,yn be
solutions of y′= A(t)y on (a,b ). Then the following statements are equivalent; that is, the y are either
all true or all false:
(a) The general solution of y′= A(t)y on (a,b ) is y = c1y1 + c2y2 + · · ·+ cnyn , where c1, c2, . . . ,
cn are arbitrary constants.
(b) {y1, y2,..., yn }is a fundamental set of solutions of y′= A(t)y on (a,b ).
(c) {y1, y2,..., yn }is linearly independent on (a,b ).
(d) The Wronskian of {y1, y2,..., yn}is nonzero at some point in (a,b ).
(e) The Wronskian of {y1, y2,..., yn}is nonzero at all points in (a,b ).
W e say that Y in ( 10.3.4) is a fundamental matrix for y′ = A(t)y if any (and therefore all) of the",Elementary Differential Equations with Boundary Value Problems.pdf
"W e say that Y in ( 10.3.4) is a fundamental matrix for y′ = A(t)y if any (and therefore all) of the
statements ( a)-(e) of Theorem 10.3.2 are true for the columns of Y. In this case, ( 10.3.3) implies that the
general solution of y′= A(t)y can be written as y = Yc, where c is an arbitrary constant n-vector.
Example 10.3.2 The vector functions
y1 =
[ −e2t
2e2t
]
and y2 =
[ −e−t
e−t
]
are solutions of the constant coefficient system
y′=
[ −4 −3
6 5
]
y (10.3.7)
on (−∞ , ∞ ). (V erify .)
(a) Compute the Wronskian of {y1, y2}directly from the definition ( 10.3.5)
(b) V erify Abel’s formula ( 10.3.6) for the Wronskian of {y1, y2}.
(c) Find the general solution of ( 10.3.7).
(d) Solve the initial value problem
y′=
[ −4 −3
6 5
]
y, y(0) =
[ 4
−5
]
. (10.3.8)
SO LU TIO N (a) From ( 10.3.5)
W(t) =
⏐
⏐
⏐
⏐
−e2t −e−t
2e2t e−t
⏐
⏐
⏐
⏐= e2t e−t
[ −1 −1
2 1
]
= et . (10.3.9)
SO LU TIO N (b) Here
A=
[ −4 −3
6 5
]
,",Elementary Differential Equations with Boundary Value Problems.pdf
"526 Chapter 10 Linear Systems of Differential Equations
so tr (A) = −4 + 5 = 1 . If t0 is an arbitrary real number then ( 10.3.6) implies that
W(t) = W(t0) exp
(∫ t
t0
1 ds
)
=
⏐
⏐
⏐
⏐
−e2t0 −e−t0
2e2t0 e−t0
⏐
⏐
⏐
⏐e(t−t0) = et0 et−t0 = et ,
which is consistent with (
10.3.9).
SO LU TIO N (c) Since W(t) ̸= 0 , Theorem 10.3.3 implies that {y1, y2}is a fundamental set of solutions
of ( 10.3.7) and
Y =
[ −e2t −e−t
2e2t e−t
]
is a fundamental matrix for ( 10.3.7). Therefore the general solution of ( 10.3.7) is
y = c1y1 + c2y2 = c1
[ −e2t
2e2t
]
+ c2
[ −e−t
e−t
]
=
[ −e2t −e−t
2e2t e−t
][ c1
c2
]
. (10.3.10)
SO LU TIO N (d) Setting t= 0 in ( 10.3.10) and imposing the initial condition in ( 10.3.8) yields
c1
[
−1
2
]
+ c2
[
−1
1
]
=
[
4
−5
]
.
Thus,
−c1 − c2 = 4
2c1 + c2 = −5.
The solution of this system is c1 = −1, c2 = −3. Substituting these values into (
10.3.10) yields
y = −
[
−e2t
2e2t
]
− 3
[
−e−t
e−t
]
=
[
e2t + 3 e−t
−2e2t − 3e−t
]
as the solution of ( 10.3.8).
10.3 Exercises",Elementary Differential Equations with Boundary Value Problems.pdf
"10.3.10) yields
y = −
[
−e2t
2e2t
]
− 3
[
−e−t
e−t
]
=
[
e2t + 3 e−t
−2e2t − 3e−t
]
as the solution of ( 10.3.8).
10.3 Exercises
1. Prove: If y1, y2, . . . , yn are solutions of y′= A(t)y on (a,b ), then any linear combination of y1,
y2, . . . , yn is also a solution of y′= A(t)y on (a,b ).
2. In Section 5.1 the Wronskian of two solutions y1 and y2 of the scalar second order equation
P0(x)y′′+ P1(x)y′+ P2(x)y= 0 (A)
was defined to be
W =
⏐
⏐
⏐
⏐
y1 y2
y′
1 y′
2
⏐
⏐
⏐
⏐.
(a) Rewrite (A) as a system of first order equations and show that W is the Wronskian (as defined
in this section) of two solutions of this system.
(b) Apply Eqn. (
10.3.6) to the system derived in (a), and show that
W(x) = W(x0) exp
{
−
∫ x
x0
P1(s)
P0(s) ds
}
,
which is the form of Abel’s formula given in Theorem 9.1.3.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.3 Basic Theory of Homogeneous Linear System 527
3. In Section 9.1 the Wronskian of nsolutions y1, y2, . . . , yn of the n−th order equation
P0(x)y(n) + P1(x)y(n−1) + · · ·+ Pn(x)y = 0 (A)
was defined to be
W =
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
y1 y2 · · · yn
y′
1 y′
2 · · · y′
n
.
.
.
.
.
. . . . .
.
.
y(n−1)
1 y(n−1)
2 · · ·y(n−1)
n
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
(a) Rewrite (A) as a system of first order equations and show that W is the Wronskian (as defined
in this section) of nsolutions of this system.
(b) Apply Eqn. (
10.3.6) to the system derived in (a), and show that
W(x) = W(x0) exp
{
−
∫ x
x0
P1(s)
P0(s) ds
}
,
which is the form of Abel’s formula given in Theorem 9.1.3.
4. Suppose
y1 =
[ y11
y21
]
and y2 =
[ y12
y22
]
are solutions of the 2 × 2 system y′= Ay on (a,b ), and let
Y =
[ y11 y12
y21 y22
]
and W =
⏐
⏐
⏐
⏐
y11 y12
y21 y22
⏐
⏐
⏐
⏐;
thus, W is the Wronskian of {y1, y2}.
(a) Deduce from the definition of determinant that
W′=
⏐
⏐
⏐
⏐
y′
11 y′
12
y21 y22
⏐
⏐
⏐
⏐+
⏐
⏐
⏐
⏐
y11 y12",Elementary Differential Equations with Boundary Value Problems.pdf
"⏐
⏐;
thus, W is the Wronskian of {y1, y2}.
(a) Deduce from the definition of determinant that
W′=
⏐
⏐
⏐
⏐
y′
11 y′
12
y21 y22
⏐
⏐
⏐
⏐+
⏐
⏐
⏐
⏐
y11 y12
y′
21 y′
22
⏐
⏐
⏐
⏐.
(b) Use the equation Y′= A(t)Y and the definition of matrix multiplication to show that
[y′
11 y′
12 ] = a11[y11 y12 ] + a12[y21 y22]
and
[y′
21 y′
22 ] = a21[y11 y12] + a22[y21 y22 ].
(c) Use properties of determinants to deduce from (a) and (a) that
⏐
⏐
⏐
⏐
y′
11 y′
12
y21 y22
⏐
⏐
⏐
⏐= a11W and
⏐
⏐
⏐
⏐
y11 y12
y′
21 y′
22
⏐
⏐
⏐
⏐= a22W.
(d) Conclude from (c) that
W′= ( a11 + a22)W,
and use this to show that if a<t 0 <b then
W(t) = W(t0) exp
(∫ t
t0
[a11(s) + a22(s)] ds
)
a<t<b.",Elementary Differential Equations with Boundary Value Problems.pdf
"528 Chapter 10 Linear Systems of Differential Equations
5. Suppose the n× nmatrix A= A(t) is continuous on (a,b ). Let
Y =





y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
.
.
.
. . . . .
.
.
yn1 yn2 · · ·ynn




 ,
where the columns of Y are solutions of y′= A(t)y. Let
ri = [ yi1 yi2 ...y in ]
be the ith row of Y, and let W be the determinant of Y.
(a) Deduce from the definition of determinant that
W′= W1 + W2 + · · ·+ Wn,
where, for 1 ≤ m≤ n, the ith row of Wm is ri if i̸= m, and r′
m if i= m.
(b) Use the equation Y′= AY and the definition of matrix multiplication to show that
r′
m = am1r1 + am2r2 + · · ·+ amnrn .
(c) Use properties of determinants to deduce from (b) that
det(Wm) = ammW.
(d) Conclude from (a) and (c) that
W′= ( a11 + a22 + · · ·+ ann)W,
and use this to show that if a<t 0 <b then
W(t) = W(t0) exp
(∫ t
t0
[
a11(s) + a22(s) + · · ·+ ann(s)] ds
)
, a<t<b.
6. Suppose the n× nmatrix Ais continuous on (a,b ) and t0 is a point in (a,b ). Let Y be a funda-",Elementary Differential Equations with Boundary Value Problems.pdf
"t0
[
a11(s) + a22(s) + · · ·+ ann(s)] ds
)
, a<t<b.
6. Suppose the n× nmatrix Ais continuous on (a,b ) and t0 is a point in (a,b ). Let Y be a funda-
mental matrix for y′= A(t)y on (a,b ).
(a) Show that Y(t0) is invertible.
(b) Show that if k is an arbitrary n-vector then the solution of the initial value problem
y′= A(t)y, y(t0) = k
is
y = Y(t)Y−1(t0)k.
7. Let
A=
[ 2 4
4 2
]
, y1 =
[ e6t
e6t
]
, y2 =
[ e−2t
−e−2t
]
, k =
[ −3
9
]
.
(a) V erify that {y1, y2}is a fundamental set of solutions for y′= Ay.
(b) Solve the initial value problem
y′= Ay, y(0) = k. (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.3 Basic Theory of Homogeneous Linear System 529
(c) Use the result of Exercise 6(b) to find a formula for the solution of (A) for an arbitrary initi al
vector k.
8. Repeat Exercise 7 with
A=
[ −2 −2
−5 1
]
, y1 =
[ e−4t
e−4t
]
, y2 =
[ −2e3t
5e3t
]
, k =
[ 10
−4
]
.
9. Repeat Exercise 7 with
A=
[ −4 −10
3 7
]
, y1 =
[ −5e2t
3e2t
]
, y2 =
[ 2et
−et
]
, k =
[ −19
11
]
.
10. Repeat Exercise 7 with
A=
[
2 1
1 2
]
, y1 =
[
e3t
e3t
]
, y2 =
[
et
−et
]
, k =
[
2
8
]
.
11. Let
A =


3 −1 −1
−2 3 2
4 −1 −2

 ,
y1 =


e2t
0
e2t

 , y2 =


e3t
−e3t
e3t

 , y3 =


e−t
−3e−t
7e−t

 , k =


2
−7
20

 .
(a) V erify that {y1, y2, y3}is a fundamental set of solutions for y′= Ay.
(b) Solve the initial value problem
y′= Ay, y(0) = k. (A)
(c) Use the result of Exercise
6(b) to find a formula for the solution of (A) for an arbitrary initi al
vector k.
12. Repeat Exercise 11 with
A =


0 2 2
2 0 2
2 2 0

 ,
y1 =


−e−2t
0
e−2t

 , y2 =


−e−2t
e−2t
0

 , y3 =


e4t",Elementary Differential Equations with Boundary Value Problems.pdf
"vector k.
12. Repeat Exercise 11 with
A =


0 2 2
2 0 2
2 2 0

 ,
y1 =


−e−2t
0
e−2t

 , y2 =


−e−2t
e−2t
0

 , y3 =


e4t
e4t
e4t

 , k =


0
−9
12

 .
13. Repeat Exercise
11 with
A =


−1 2 3
0 1 6
0 0 −2

 ,
y1 =


et
et
0

 , y2 =


e−t
0
0

 , y3 =


e−2t
−2e−2t
e−2t

 , k =


5
5
−1

 .",Elementary Differential Equations with Boundary Value Problems.pdf
"530 Chapter 10 Linear Systems of Differential Equations
14. Suppose Y and Z are fundamental matrices for the n× nsystem y′= A(t)y. Then some of the
four matrices YZ−1, Y−1Z, Z−1Y, ZY−1 are necessarily constant. Identify them and prove that
they are constant.
15. Suppose the columns of an n× nmatrix Y are solutions of the n× nsystem y′= Ay and C is
an n× nconstant matrix.
(a) Show that the matrix Z = YC satisfies the differential equation Z′= AZ.
(b) Show that Zis a fundamental matrix for y′= A(t)y if and only if Cis invertible and Y is a
fundamental matrix for y′= A(t)y.
16. Suppose the n× nmatrix A = A(t) is continuous on (a,b ) and t0 is in (a,b ). For i = 1 , 2, . . . ,
n, let yi be the solution of the initial value problem y′
i = A(t)yi, yi(t0) = ei, where
e1 =





1
0
.
.
.
0




 , e2 =





0
1
.
.
.
0




 , · · ·en =





0
0
.
.
.
1




 ;
that is, the jth component of ei is 1 if j = i, or 0 if j ̸= i.",Elementary Differential Equations with Boundary Value Problems.pdf
"




0
1
.
.
.
0




 , · · ·en =





0
0
.
.
.
1




 ;
that is, the jth component of ei is 1 if j = i, or 0 if j ̸= i.
(a) Show that {y1, y2,..., yn}is a fundamental set of solutions of y′= A(t)y on (a,b ).
(b) Conclude from (a) and Exercise
15 that y′= A(t)y has infinitely many fundamental sets of
solutions on (a,b ).
17. Show that Y is a fundamental matrix for the system y′= A(t)y if and only if Y−1 is a funda-
mental matrix for y′ = −AT (t)y, where AT denotes the transpose of A. H IN T : See Exercise
11.
18. Let Zbe the fundamental matrix for the constant coefficient syste m y′= Ay such that Z(0) = I.
(a) Show that Z(t)Z(s) = Z(t+ s) for all s and t. H IN T : F or fixed slet Γ 1(t) = Z(t)Z(s)
and Γ 2(t) = Z(t+ s). Show that Γ 1 and Γ 2 are both solutions of the matrix initial value
problem Γ ′= AΓ , Γ(0) = Z(s). Then conclude from Theorem 10.2.1 that Γ 1 = Γ 2.
(b) Show that (Z(t))−1 = Z(−t).",Elementary Differential Equations with Boundary Value Problems.pdf
"problem Γ ′= AΓ , Γ(0) = Z(s). Then conclude from Theorem 10.2.1 that Γ 1 = Γ 2.
(b) Show that (Z(t))−1 = Z(−t).
(c) The matrix Z defined above is sometimes denoted by etA . Discuss the motivation for this
notation.
10.4 CONST ANT COEFFICIENT HOMOGENEOUS SYSTEMS I
W e’ll now begin our study of the homogeneous system
y′= Ay, (10.4.1)
where Ais an n× nconstant matrix. Since Ais continuous on (−∞ , ∞ ), Theorem 10.2.1 implies that
all solutions of ( 10.4.1) are defined on (−∞ , ∞ ). Therefore, when we speak of solutions of y′= Ay,
we’ll mean solutions on (−∞ , ∞ ).
In this section we assume that all the eigenvalues of A are real and that A has a set of n linearly
independent eigenvectors. In the next two sections we consi der the cases where some of the eigenvalues
of Aare complex, or where Adoes not have nlinearly independent eigenvectors.
In Example 10.3.2 we showed that the vector functions
y1 =
[ −e2t
2e2t
]
and y2 =
[ −e−t
e−t
]
form a fundamental set of solutions of the system
y′=",Elementary Differential Equations with Boundary Value Problems.pdf
"In Example 10.3.2 we showed that the vector functions
y1 =
[ −e2t
2e2t
]
and y2 =
[ −e−t
e−t
]
form a fundamental set of solutions of the system
y′=
[ −4 −3
6 5
]
y, (10.4.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.4 Constant Coefficient Homogeneous Systems I 531
but we did not show how we obtained y1 and y2 in the first place. T o see how these solutions can be
obtained we write ( 10.4.2) as
y′
1 = −4y1 − 3y2
y′
2 = 6 y1 + 5 y2
(10.4.3)
and look for solutions of the form
y1 = x1eλt and y2 = x2eλt , (10.4.4)
where x1, x2, and λ are constants to be determined. Differentiating ( 10.4.4) yields
y′
1 = λx1eλt and y′
2 = λx2eλt .
Substituting this and ( 10.4.4) into ( 10.4.3) and canceling the common factor eλt yields
−4x1 − 3x2 = λx1
6x1 + 5 x2 = λx2.
For a given λ, this is a homogeneous algebraic system, since it can be rewr itten as
(−4 − λ)x1 − 3x2 = 0
6x1 + (5 − λ)x2 = 0 . (10.4.5)
The trivial solution x1 = x2 = 0 of this system isn’t useful, since it corresponds to the triv ial solution
y1 ≡ y2 ≡ 0 of ( 10.4.3), which can’t be part of a fundamental set of solutions of ( 10.4.2). Therefore we",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 ≡ y2 ≡ 0 of ( 10.4.3), which can’t be part of a fundamental set of solutions of ( 10.4.2). Therefore we
consider only those values of λ for which ( 10.4.5) has nontrivial solutions. These are the values of λ for
which the determinant of ( 10.4.5) is zero; that is,
⏐
⏐
⏐
⏐
−4 − λ −3
6 5 − λ
⏐
⏐
⏐
⏐ = ( −4 − λ)(5 − λ) + 18
= λ2 − λ − 2
= ( λ − 2)(λ + 1) = 0 ,
which has the solutions λ1 = 2 and λ2 = −1.
T aking λ = 2 in (
10.4.5) yields
−6x1 − 3x2 = 0
6x1 + 3 x2 = 0 ,
which implies that x1 = −x2/ 2, where x2 can be chosen arbitrarily . Choosing x2 = 2 yields the solution
y1 = −e2t, y2 = 2 e2t of ( 10.4.3). W e can write this solution in vector form as
y1 =
[ −1
2
]
e2t. (10.4.6)
T aking λ = −1 in (
10.4.5) yields the system
−3x1 − 3x2 = 0
6x1 + 6 x2 = 0 ,
so x1 = −x2. T aking x2 = 1 here yields the solution y1 = −e−t , y2 = e−t of ( 10.4.3). W e can write
this solution in vector form as
y2 =
[ −1
1
]
e−t . (10.4.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"this solution in vector form as
y2 =
[ −1
1
]
e−t . (10.4.7)
In ( 10.4.6) and ( 10.4.7) the constant coefficients in the arguments of the exponenti al functions are the
eigenvalues of the coefficient matrix in ( 10.4.2), and the vector coefficients of the exponential functions
are associated eigenvectors. This illustrates the next the orem.",Elementary Differential Equations with Boundary Value Problems.pdf
"532 Chapter 10 Linear Systems of Differential Equations
Theorem 10.4.1 Suppose the n×nconstant matrix Ahas nreal eigenvalues λ1,λ 2,...,λ n (which need
not be distinct ) with associated linearly independent eigenvectors x1, x2,..., xn. Then the functions
y1 = x1eλ 1 t, y2 = x2eλ 2t, ..., yn = xneλ n t
form a fundamental set of solutions of y′= Ay; that is , the general solution of this system is
y = c1x1eλ 1t + c2x2eλ 2t + · · ·+ cnxn eλ nt .
Proof Differentiating yi = xieλ i t and recalling that Axi = λixi yields
y′
i = λixieλ i t = Axieλ i t = Ayi.
This shows that yi is a solution of y′= Ay.
The Wronskian of {y1, y2,..., yn}is
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x11eλ 1t x12eλ 2t · · ·x1neλ n t
x21eλ 1t x22eλ 2t · · ·x2neλ n t
.
.
. .
.
. . . . .
.
.
xn1eλ 1t xn2eλ 2t · · ·xnneλx nt
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
= eλ 1t eλ 2t · · ·eλ nt
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x11 x12 · · ·x1n
x21 x22 · · ·x2n
.
.
. .
.
. . . . .
.
.
xn1 xn2 · · ·xnn
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.",Elementary Differential Equations with Boundary Value Problems.pdf
"= eλ 1t eλ 2t · · ·eλ nt
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
x11 x12 · · ·x1n
x21 x22 · · ·x2n
.
.
. .
.
. . . . .
.
.
xn1 xn2 · · ·xnn
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
⏐
.
Since the columns of the determinant on the right are x1, x2, . . . , xn , which are assumed to be linearly
independent, the determinant is nonzero. Therefore Theore m
10.3.3 implies that {y1, y2,..., yn}is a
fundamental set of solutions of y′= Ay.
Example 10.4.1
(a) Find the general solution of
y′=
[ 2 4
4 2
]
y. (10.4.8)
(b) Solve the initial value problem
y′=
[ 2 4
4 2
]
y, y(0) =
[ 5
−1
]
. (10.4.9)
SO LU TIO N (a) The characteristic polynomial of the coefficient matrix Ain ( 10.4.8) is
⏐
⏐
⏐
⏐
2 − λ 4
4 2 − λ
⏐
⏐
⏐
⏐ = ( λ − 2)2 − 16
= ( λ − 2 − 4)(λ − 2 + 4)
= ( λ − 6)(λ + 2) .
Hence, λ1 = 6 and λ2 = −2 are eigenvalues of A. T o obtain the eigenvectors, we must solve the system
[ 2 − λ 4
4 2 − λ
][ x1
x2
]
=
[ 0
0
]
(10.4.10)
with λ = 6 and λ = −2. Setting λ = 6 in (
10.4.10) yields
[ −4 4
4 −4
][ x1
x2
]
=
[ 0
0
]
,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.4 Constant Coefficient Homogeneous Systems I 533
which implies that x1 = x2. T aking x2 = 1 yields the eigenvector
x1 =
[ 1
1
]
,
so
y1 =
[ 1
1
]
e6t
is a solution of (
10.4.8). Setting λ = −2 in ( 10.4.10) yields
[ 4 4
4 4
][ x1
x2
]
=
[ 0
0
]
,
which implies that x1 = −x2. T aking x2 = 1 yields the eigenvector
x2 =
[ −1
1
]
,
so
y2 =
[ −1
1
]
e−2t
is a solution of (
10.4.8). From Theorem 10.4.1, the general solution of ( 10.4.8) is
y = c1y1 + c2y2 = c1
[ 1
1
]
e6t + c2
[ −1
1
]
e−2t. (10.4.11)
SO LU TIO N (b) T o satisfy the initial condition in ( 10.4.9), we must choose c1 and c2 in ( 10.4.11) so that
c1
[
1
1
]
+ c2
[
−1
1
]
=
[
5
−1
]
.
This is equivalent to the system
c1 − c2 = 5
c1 + c2 = −1,
so c1 = 2 ,c 2 = −3. Therefore the solution of (
10.4.9) is
y = 2
[
1
1
]
e6t − 3
[
−1
1
]
e−2t,
or, in terms of components,
y1 = 2 e6t + 3 e−2t, y 2 = 2 e6t − 3e−2t.
Example 10.4.2
(a) Find the general solution of
y′=


3 −1 −1
−2 3 2
4 −1 −2

 y. (10.4.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"534 Chapter 10 Linear Systems of Differential Equations
(b) Solve the initial value problem
y′=


3 −1 −1
−2 3 2
4 −1 −2

 y, y(0) =


2
−1
8

 . (10.4.13)
SO LU TIO N (a) The characteristic polynomial of the coefficient matrix Ain ( 10.4.12) is
⏐
⏐
⏐
⏐
⏐
⏐
3 − λ −1 −1
−2 3 − λ 2
4 −1 −2 − λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 2)(λ − 3)(λ + 1) .
Hence, the eigenvalues of Aare λ1 = 2 , λ2 = 3 , and λ3 = −1. T o find the eigenvectors, we must solve
the system 

3 − λ −1 −1
−2 3 − λ 2
4 −1 −2 − λ




x1
x2
x3

 =


0
0
0

 (10.4.14)
with λ = 2 , 3, −1. With λ = 2 , the augmented matrix of (
10.4.14) is





1 −1 −1
.
.
. 0
−2 1 2 .
.
. 0
4 −1 −4 .
.
. 0




 ,
which is row equivalent to 




1 0 −1 .
.
. 0
0 1 0 .
.
. 0
0 0 0 .
.
. 0




 .
Hence, x1 = x3 and x2 = 0 . T aking x3 = 1 yields
y1 =


1
0
1

 e2t
as a solution of (
10.4.12). With λ = 3 , the augmented matrix of ( 10.4.14) is





0 −1 −1 .
.
. 0
−2 0 2
.
.
. 0
4 −1 −5 .
.
. 0




 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"as a solution of (
10.4.12). With λ = 3 , the augmented matrix of ( 10.4.14) is





0 −1 −1 .
.
. 0
−2 0 2
.
.
. 0
4 −1 −5 .
.
. 0




 ,
which is row equivalent to 




1 0 −1 .
.
. 0
0 1 1 .
.
. 0
0 0 0
.
.
. 0




 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.4 Constant Coefficient Homogeneous Systems I 535
Hence, x1 = x3 and x2 = −x3. T aking x3 = 1 yields
y2 =


1
−1
1

 e3t
as a solution of (
10.4.12). With λ = −1, the augmented matrix of ( 10.4.14) is





4 −1 −1 .
.
. 0
−2 4 2 .
.
. 0
4 −1 −1 .
.
. 0




 ,
which is row equivalent to 




1 0 −1
7
.
.
. 0
0 1 3
7
.
.
. 0
0 0 0
.
.
. 0




 .
Hence, x1 = x3/ 7 and x2 = −3x3/ 7. T aking x3 = 7 yields
y3 =


1
−3
7

 e−t
as a solution of (
10.4.12). By Theorem 10.4.1, the general solution of ( 10.4.12) is
y = c1


1
0
1

 e2t + c2


1
−1
1

 e3t + c3


1
−3
7

 e−t ,
which can also be written as
y =


e2t e3t e−t
0 −e3t −3e−t
e2t e3t 7e−t




c1
c2
c3

 . (10.4.15)
SO LU TIO N (b) T o satisfy the initial condition in ( 10.4.13) we must choose c1, c2, c3 in ( 10.4.15) so that


1 1 1
0 −1 −3
1 1 7




c1
c2
c3

 =


2
−1
8

 .
Solving this system yields c1 = 3 , c2 = −2, c3 = 1 . Hence, the solution of (
10.4.13) is
y =",Elementary Differential Equations with Boundary Value Problems.pdf
"0 −1 −3
1 1 7




c1
c2
c3

 =


2
−1
8

 .
Solving this system yields c1 = 3 , c2 = −2, c3 = 1 . Hence, the solution of (
10.4.13) is
y =


e2t e3t e−t
0 −e3t −3e−t
e2t e3t 7e−t




3
−2
1


= 3


1
0
1

 e2t − 2


1
−1
1

 e3t +


1
−3
7

 e−t .",Elementary Differential Equations with Boundary Value Problems.pdf
"536 Chapter 10 Linear Systems of Differential Equations
Example 10.4.3 Find the general solution of
y′=


−3 2 2
2 −3 2
2 2 −3

 y. (10.4.16)
Solution The characteristic polynomial of the coefficient matrix Ain (
10.4.16) is
⏐
⏐
⏐
⏐
⏐
⏐
−3 − λ 2 2
2 −3 − λ 2
2 2 −3 − λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 1)(λ + 5) 2.
Hence, λ1 = 1 is an eigenvalue of multiplicity 1, while λ2 = −5 is an eigenvalue of multiplicity 2.
Eigenvectors associated with λ1 = 1 are solutions of the system with augmented matrix





−4 2 2
.
.
. 0
2 −4 2 .
.
. 0
2 2 −4 .
.
. 0




 ,
which is row equivalent to 




1 0 −1 .
.
. 0
0 1 −1
.
.
. 0
0 0 0 .
.
. 0




 .
Hence, x1 = x2 = x3, and we choose x3 = 1 to obtain the solution
y1 =


1
1
1

 et (10.4.17)
of (
10.4.16). Eigenvectors associated with λ2 = −5 are solutions of the system with augmented matrix





2 2 2 .
.
. 0
2 2 2 .
.
. 0
2 2 2 .
.
. 0




 .
Hence, the components of these eigenvectors need only satis fy the single condition",Elementary Differential Equations with Boundary Value Problems.pdf
"




2 2 2 .
.
. 0
2 2 2 .
.
. 0
2 2 2 .
.
. 0




 .
Hence, the components of these eigenvectors need only satis fy the single condition
x1 + x2 + x3 = 0 .
Since there’s only one equation here, we can choose x2 and x3 arbitrarily . W e obtain one eigenvector by
choosing x2 = 0 and x3 = 1 , and another by choosing x2 = 1 and x3 = 0 . In both cases x1 = −1.
Therefore 

−1
0
1

 and


−1
1
0

",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.4 Constant Coefficient Homogeneous Systems I 537
are linearly independent eigenvectors associated with λ2 = −5, and the corresponding solutions of
(10.4.16) are
y2 =


−1
0
1

 e−5t and y3 =


−1
1
0

 e−5t.
Because of this and (
10.4.17), Theorem 10.4.1 implies that the general solution of ( 10.4.16) is
y = c1


1
1
1

 et + c2


−1
0
1

 e−5t + c3


−1
1
0

 e−5t .
Geometric Properties of Solutions when n= 2
W e’ll now consider the geometric properties of solutions of a 2 × 2 constant coefficient system
[ y′
1
y′
2
]
=
[ a11 a12
a21 a22
][ y1
y2
]
. (10.4.18)
It is convenient to think of a “ y1-y2 plane,"" where a point is identified by rectangular coordinat es (y1 ,y 2).
If y =
[ y1
y2
]
is a non-constant solution of ( 10.4.18), then the point (y1(t),y 2(t)) moves along a curve
C in the y1-y2 plane as t varies from −∞ to ∞ . W e call C the trajectory of y. (W e also say that",Elementary Differential Equations with Boundary Value Problems.pdf
"C in the y1-y2 plane as t varies from −∞ to ∞ . W e call C the trajectory of y. (W e also say that
C is a trajectory of the system ( 10.4.18).) I’s important to note that C is the trajectory of infinitely
many solutions of ( 10.4.18), since if τ is any real number, then y(t − τ) is a solution of ( 10.4.18)
(Exercise 28(b)), and (y1(t− τ),y 2(t− τ)) also moves along C as tvaries from −∞ to ∞ . Moreover,
Exercise 28(c) implies that distinct trajectories of ( 10.4.18) can’t intersect, and that two solutions y1 and
y2 of ( 10.4.18) have the same trajectory if and only if y2(t) = y1(t− τ) for some τ.
From Exercise 28(a), a trajectory of a nontrivial solution of ( 10.4.18) can’t contain (0, 0), which we
define to be the trajectory of the trivial solution y ≡ 0. More generally , if y =
[ k1
k2
]
̸= 0 is a constant
solution of ( 10.4.18) (which could occur if zero is an eigenvalue of the matrix of ( 10.4.18)), we define the
trajectory of y to be the single point (k1,k 2).",Elementary Differential Equations with Boundary Value Problems.pdf
"trajectory of y to be the single point (k1,k 2).
T o be specific, this is the question: What do the trajectories look like, and how are they traversed? In
this section we’ll answer this question, assuming that the m atrix
A=
[ a11 a12
a21 a22
]
of ( 10.4.18) has real eigenvalues λ1 and λ2 with associated linearly independent eigenvectors x1 and x2.
Then the general solution of ( 10.4.18) is
y = c1x1eλ 1t + c2x2eλ 2t . (10.4.19)
W e’ll consider other situations in the next two sections.
W e leave it to you (Exercise 35) to classify the trajectories of ( 10.4.18) if zero is an eigenvalue of A.
W e’ll confine our attention here to the case where both eigenv alues are nonzero. In this case the simplest
situation is where λ1 = λ2 ̸= 0 , so ( 10.4.19) becomes
y = ( c1x1 + c2x2)eλ 1 t.
Since x1 and x2 are linearly independent, an arbitrary vector x can be written as x = c1x1 + c2x2.
Therefore the general solution of ( 10.4.18) can be written as y = xeλ 1t where x is an arbitrary 2-",Elementary Differential Equations with Boundary Value Problems.pdf
"Therefore the general solution of ( 10.4.18) can be written as y = xeλ 1t where x is an arbitrary 2-
vector, and the trajectories of nontrivial solutions of ( 10.4.18) are half-lines through (but not including)",Elementary Differential Equations with Boundary Value Problems.pdf
"538 Chapter 10 Linear Systems of Differential Equations
the origin. The direction of motion is away from the origin if λ1 >0 (Figure 10.4.1), toward it if λ1 <0
(Figure 10.4.2). (In these and the next figures an arrow through a point indic ates the direction of motion
along the trajectory through the point.)
 y1
 y2
Figure 10.4.1 Trajectories of a 2 × 2 system with a
repeated positive eigenvalue
 y1
 y2
Figure 10.4.2 Trajectories of a 2 × 2 system with a
repeated negative eigenvalue
Now suppose λ2 > λ1, and let L1 and L2 denote lines through the origin parallel to x1 and x2,
respectively . By a half-line of L1 (or L2), we mean either of the rays obtained by removing the origin
from L1 (or L2).
Letting c2 = 0 in ( 10.4.19) yields y = c1x1eλ 1t . If c1 ̸= 0 , the trajectory defined by this solution is a
half-line of L1. The direction of motion is away from the origin if λ1 > 0, toward the origin if λ1 < 0.
Similarly , the trajectory of y = c2x2eλ 2t with c2 ̸= 0 is a half-line of L2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Similarly , the trajectory of y = c2x2eλ 2t with c2 ̸= 0 is a half-line of L2.
Henceforth, we assume that c1 and c2 in ( 10.4.19) are both nonzero. In this case, the trajectory of
(10.4.19) can’t intersect L1 or L2, since every point on these lines is on the trajectory of a sol ution for
which either c1 = 0 or c2 = 0 . (Remember: distinct trajectories can’t intersect!). The refore the trajectory
of ( 10.4.19) must lie entirely in one of the four open sectors bounded by L1 and L2, but do not any point
on L1 or L2. Since the initial point (y1(0),y 2(0)) defined by
y(0) = c1x1 + c2x2
is on the trajectory , we can determine which sector contains the trajectory from the signs of c1 and c2, as
shown in Figure 10.4.3.
The direction of y(t) in ( 10.4.19) is the same as that of
e−λ 2t y(t) = c1x1e−(λ 2 −λ 1)t + c2x2 (10.4.20)
and of
e−λ 1t y(t) = c1x1 + c2x2e(λ 2−λ 1)t. (10.4.21)
Since the right side of ( 10.4.20) approaches c2x2 as t→ ∞ , the trajectory is asymptotically parallel to L2",Elementary Differential Equations with Boundary Value Problems.pdf
"Since the right side of ( 10.4.20) approaches c2x2 as t→ ∞ , the trajectory is asymptotically parallel to L2
as t→ ∞ . Since the right side of ( 10.4.21) approaches c1x1 as t→ −∞ , the trajectory is asymptotically
parallel to L1 as t→ −∞ .
The shape and direction of traversal of the trajectory of ( 10.4.19) depend upon whether λ1 and λ2 are
both positive, both negative, or of opposite signs. W e’ll no w analyze these three cases.
Henceforth ∥u∥denote the length of the vector u.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.4 Constant Coefficient Homogeneous Systems I 539
 x2
 x1
 c1 > 0, c2 < 0
 c1 > 0,  c2 > 0
 c1 < 0, c2 > 0
 c1 < 0,  c2 < 0
 L1
 L2
Figure 10.4.3 Four open sectors bounded by L1 and
L2
 y1
 y2
 L1
 L2
Figure 10.4.4 T wo positive eigenvalues; motion
away from origin
Case 1: λ2 >λ 1 >0
Figure 10.4.4 shows some typical trajectories. In this case, limt→−∞ ∥y(t)∥= 0 , so the trajectory is not
only asymptotically parallel to L1 as t→ −∞ , but is actually asymptotically tangent to L1 at the origin.
On the other hand, limt→∞ ∥y(t)∥= ∞ and
lim
t→∞

y(t) − c2x2eλ 2t 
 = lim
t→∞
∥c1x1eλ 1t ∥= ∞ ,
so, although the trajectory is asymptotically parallel to L2 as t → ∞ , it’s not asymptotically tangent to
L2. The direction of motion along each trajectory is away from t he origin.
Case 2: 0 >λ 2 >λ 1
Figure 10.4.5 shows some typical trajectories. In this case, limt→∞ ∥y(t)∥ = 0 , so the trajectory is",Elementary Differential Equations with Boundary Value Problems.pdf
"Case 2: 0 >λ 2 >λ 1
Figure 10.4.5 shows some typical trajectories. In this case, limt→∞ ∥y(t)∥ = 0 , so the trajectory is
asymptotically tangent to L2 at the origin as t→ ∞ . On the other hand, limt→−∞ ∥y(t)∥= ∞ and
lim
t→−∞

y(t) − c1x1eλ 1t 
 = lim
t→−∞
∥c2x2eλ 2t ∥= ∞ ,
so, although the trajectory is asymptotically parallel to L1 as t→ −∞ , it’s not asymptotically tangent to
it. The direction of motion along each trajectory is toward t he origin.
Case 3: λ2 >0 >λ 1
Figure 10.4.6 shows some typical trajectories. In this case,
lim
t→∞
∥y(t)∥= ∞ and lim
t→∞

y(t) − c2x2eλ 2t 
 = lim
t→∞
∥c1x1eλ 1t ∥= 0 ,
so the trajectory is asymptotically tangent to L2 as t→ ∞ . Similarly ,
lim
t→−∞
∥y(t)∥= ∞ and lim
t→−∞

y(t) − c1x1eλ 1t 
 = lim
t→−∞
∥c2x2eλ 2t ∥= 0 ,
so the trajectory is asymptotically tangent to L1 as t→ −∞ . The direction of motion is toward the origin
on L1 and away from the origin on L2. The direction of motion along any other trajectory is away f rom",Elementary Differential Equations with Boundary Value Problems.pdf
"on L1 and away from the origin on L2. The direction of motion along any other trajectory is away f rom
L1, toward L2.",Elementary Differential Equations with Boundary Value Problems.pdf
"540 Chapter 10 Linear Systems of Differential Equations
 y1
 y2
 L1
 L2
Figure 10.4.5 T wo negative eigenvalues; motion
toward the origin
 y1
 y2
 L1
 L2
Figure 10.4.6 Eigenvalues of different signs
10.4 Exercises
In Exercises 1– 15 find the general solution.
1. y′=
[ 1 2
2 1
]
y
2. y′= 1
4
[
−5 3
3 −5
]
y
3. y′= 1
5
[ −4 3
−2 −11
]
y 4. y′=
[ −1 −4
−1 −1
]
y
5. y′=
[ 2 −4
−1 −1
]
y 6. y′=
[ 4 −3
2 −1
]
y
7. y′=
[ −6 −3
1 −2
]
y 8. y′=


1 −1 −2
1 −2 −3
−4 1 −1

 y
9. y′=


−6 −4 −8
−4 0 −4
−8 −4 −6

 y 10. y′=


3 5 8
1 −1 −2
−1 −1 −1

 y
11. y′=


1 −1 2
12 −4 10
−6 1 −7

 y 12. y′=


4 −1 −4
4 −3 −2
1 −1 −1

 y
13. y′=


−2 2 −6
2 6 2
−2 −2 2

 y 14. y′=


3 2 −2
−2 7 −2
−10 10 −5

 y",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.4 Constant Coefficient Homogeneous Systems I 541
15. y′=


3 1 −1
3 5 1
−6 2 4

 y
In Exercises
16– 27 solve the initial value problem.
16. y′=
[ −7 4
−6 7
]
y, y(0) =
[ 2
−4
]
17. y′= 1
6
[
7 2
−2 2
]
y, y(0) =
[
0
−3
]
18. y′=
[ 21 −12
24 −15
]
y, y(0) =
[ 5
3
]
19. y′=
[ −7 4
−6 7
]
y, y(0) =
[ −1
7
]
20. y′= 1
6


1 2 0
4 −1 0
0 0 3

 y, y(0) =


4
7
1


21. y′= 1
3


2 −2 3
−4 4 3
2 1 0

 y, y(0) =


1
1
5


22. y′=


6 −3 −8
2 1 −2
3 −3 −5

 y, y(0) =


0
−1
−1


23. y′= 1
3


2 4 −7
1 5 −5
−4 4 −1

 y, y(0) =


4
1
3


24. y′=


3 0 1
11 −2 7
1 0 3

 y, y(0) =


2
7
6


25. y′=


−2 −5 −1
−4 −1 1
4 5 3

 y, y(0) =


8
−10
−4


26. y′=


3 −1 0
4 −2 0
4 −4 2

 y, y(0) =


7
10
2


27. y′=


−2 2 6
2 6 2
−2 −2 2

 y, y(0) =


6
−10
7


28. Let Abe an n× nconstant matrix. Then Theorem
10.2.1 implies that the solutions of
y′= Ay (A)
are all defined on (−∞ , ∞ ).",Elementary Differential Equations with Boundary Value Problems.pdf
"

6
−10
7


28. Let Abe an n× nconstant matrix. Then Theorem
10.2.1 implies that the solutions of
y′= Ay (A)
are all defined on (−∞ , ∞ ).
(a) Use Theorem 10.2.1 to show that the only solution of (A) that can ever equal the ze ro vector
is y ≡ 0.",Elementary Differential Equations with Boundary Value Problems.pdf
"542 Chapter 10 Linear Systems of Differential Equations
(b) Suppose y1 is a solution of (A) and y2 is defined by y2(t) = y1(t− τ), where τ is an
arbitrary real number. Show that y2 is also a solution of (A).
(c) Suppose y1 and y2 are solutions of (A) and there are real numbers t1 and t2 such that
y1(t1) = y2(t2). Show that y2(t) = y1(t− τ) for all t, where τ = t2 − t1. H IN T :
Show that y1(t− τ) and y2(t) are solutions of the same initial value problem for (A), and
apply the uniqueness assertion of Theorem 10.2.1.
In Exercises 29- 34 describe and graph trajectories of the given system.
29. C/G y′=
[ 1 1
1 −1
]
y
30.
C/G y′=
[ −4 3
−2 −11
]
y
31. C/G y′=
[ 9 −3
−1 11
]
y 32. C/G y′=
[ −1 −10
−5 4
]
y
33. C/G y′=
[ 5 −4
1 10
]
y 34. C/G y′=
[ −7 1
3 −5
]
y
35. Suppose the eigenvalues of the 2 × 2 matrix Aare λ = 0 and µ ̸= 0 , with corresponding eigen-
vectors x1 and x2. Let L1 be the line through the origin parallel to x1.",Elementary Differential Equations with Boundary Value Problems.pdf
"vectors x1 and x2. Let L1 be the line through the origin parallel to x1.
(a) Show that every point on L1 is the trajectory of a constant solution of y′= Ay.
(b) Show that the trajectories of nonconstant solutions of y′= Ay are half-lines parallel to x2
and on either side of L1, and that the direction of motion along these trajectories i s away
from L1 if µ> 0, or toward L1 if µ< 0.
The matrices of the systems in Exercises
36-41 are singular . Describe and graph the trajectories of
nonconstant solutions of the given systems.
36. C/G y′=
[
−1 1
1 −1
]
y
37.
C/G y′=
[ −1 −3
2 6
]
y
38. C/G y′=
[ 1 −3
−1 3
]
y 39. C/G y′=
[ 1 −2
−1 2
]
y
40. C/G y′=
[ −4 −4
1 1
]
y 41. C/G y′=
[ 3 −1
−3 1
]
y
42. L Let P = P(t) and Q= Q(t) be the populations of two species at time t, and assume that each
population would grow exponentially if the other didn’t exi st; that is, in the absence of competition,
P′= aP and Q′= bQ, (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"population would grow exponentially if the other didn’t exi st; that is, in the absence of competition,
P′= aP and Q′= bQ, (A)
where aand bare positive constants. One way to model the effect of compet ition is to assume
that the growth rate per individual of each population is red uced by an amount proportional to the
other population, so (A) is replaced by
P′ = aP − αQ
Q′ = −βP + bQ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 543
where α and β are positive constants. (Since negative population doesn’ t make sense, this system
holds only while P and Qare both positive.) Now suppose P(0) = P0 >0 and Q(0) = Q0 >0.
(a) For several choices of a, b, α, and β, verify experimentally (by graphing trajectories of (A)
in the P-Qplane) that there’s a constant ρ > 0 (depending upon a, b, α, and β) with the
following properties:
(i) If Q0 >ρP 0, then P decreases monotonically to zero in finite time, during which Q
remains positive.
(ii) If Q0 <ρP 0, then Qdecreases monotonically to zero in finite time, during which P
remains positive.
(b) Conclude from (a) that exactly one of the species becomes extinct in finite time if Q0 ̸= ρP0.
Determine experimentally what happens if Q0 = ρP0.
(c) Confirm your experimental results and determine γ by expressing the eigenvalues and asso-
ciated eigenvectors of
A=
[ a −α
−β b
]",Elementary Differential Equations with Boundary Value Problems.pdf
"(c) Confirm your experimental results and determine γ by expressing the eigenvalues and asso-
ciated eigenvectors of
A=
[ a −α
−β b
]
in terms of a, b, α, and β, and applying the geometric arguments developed at the end o f this
section.
10.5 CONST ANT COEFFICIENT HOMOGENEOUS SYSTEMS II
W e saw in Section 10.4 that if an n× nconstant matrix Ahas nreal eigenvalues λ1, λ2, . . . , λn (which
need not be distinct) with associated linearly independent eigenvectors x1, x2, . . . , xn, then the general
solution of y′= Ay is
y = c1x1eλ 1t + c2x2eλ 2t + · · ·+ cnxn eλ nt .
In this section we consider the case where Ahas nreal eigenvalues, but does not have nlinearly indepen-
dent eigenvectors. It is shown in linear algebra that this oc curs if and only if Ahas at least one eigenvalue
of multiplicity r >1 such that the associated eigenspace has dimension less than r. In this case A is
said to be defective. Since it’s beyond the scope of this book to give a complete an alysis of systems with",Elementary Differential Equations with Boundary Value Problems.pdf
"said to be defective. Since it’s beyond the scope of this book to give a complete an alysis of systems with
defective coefficient matrices, we will restrict our attent ion to some commonly occurring special cases.
Example 10.5.1 Show that the system
y′=
[ 11 −25
4 −9
]
y (10.5.1)
does not have a fundamental set of solutions of the form {x1eλ 1t , x2eλ 2 t}, where λ1 and λ2 are eigenval-
ues of the coefficient matrix Aof ( 10.5.1) and x1, and x2 are associated linearly independent eigenvectors.
Solution The characteristic polynomial of Ais
⏐
⏐
⏐
⏐
11 − λ −25
4 −9 − λ
⏐
⏐
⏐
⏐ = ( λ − 11)(λ + 9) + 100
= λ2 − 2λ + 1 = ( λ − 1)2.
Hence, λ = 1 is the only eigenvalue of A. The augmented matrix of the system (A− I)x = 0 is

 10 −25 .
.
. 0
4 −10 .
.
. 0

 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"544 Chapter 10 Linear Systems of Differential Equations
which is row equivalent to 



1 −5
2
.
.
. 0
0 0 .
.
. 0



 .
Hence, x1 = 5 x2/ 2 where x2 is arbitrary . Therefore all eigenvectors of Aare scalar multiples of x1 =[ 5
2
]
, so Adoes not have a set of two linearly independent eigenvectors .
From Example 10.5.1, we know that all scalar multiples of y1 =
[ 5
2
]
et are solutions of (
10.5.1);
however, to find the general solution we must find a second solu tion y2 such that {y1, y2}is linearly
independent. Based on your recollection of the procedure fo r solving a constant coefficient scalar equation
ay′′+ by′+ cy = 0
in the case where the characteristic polynomial has a repeat ed root, you might expect to obtain a second
solution of ( 10.5.1) by multiplying the first solution by t. However, this yields y2 =
[ 5
2
]
tet , which
doesn’t work, since
y′
2 =
[ 5
2
]
(tet + et ), while
[ 11 −25
4 −9
]
y2 =
[ 5
2
]
tet .
The next theorem shows what to do in this situation.",Elementary Differential Equations with Boundary Value Problems.pdf
"doesn’t work, since
y′
2 =
[ 5
2
]
(tet + et ), while
[ 11 −25
4 −9
]
y2 =
[ 5
2
]
tet .
The next theorem shows what to do in this situation.
Theorem 10.5.1
Suppose the n×nmatrix Ahas an eigenvalue λ1 of multiplicity ≥ 2 and the associated
eigenspace has dimension 1; that is , all λ1-eigenvectors of Aare scalar multiples of an eigenvector x.
Then there are infinitely many vectors u such that
(A− λ1I)u = x. (10.5.2)
Moreover, if u is any such vector then
y1 = xeλ 1t and y2 = ueλ 1t + xteλ 1 t (10.5.3)
are linearly independent solutions of y′= Ay.
A complete proof of this theorem is beyond the scope of this bo ok. The difficulty is in proving that
there’s a vector u satisfying ( 10.5.2), since det(A− λ1I) = 0 . W e’ll take this without proof and verify
the other assertions of the theorem.
W e already know that y1 in ( 10.5.3) is a solution of y′= Ay. T o see that y2 is also a solution, we
compute
y′
2 − Ay2 = λ1ueλ 1t + xeλ 1 t + λ1xteλ 1t − Aueλ 1t − Axteλ 1 t",Elementary Differential Equations with Boundary Value Problems.pdf
"compute
y′
2 − Ay2 = λ1ueλ 1t + xeλ 1 t + λ1xteλ 1t − Aueλ 1t − Axteλ 1 t
= ( λ1u + x − Au)eλ 1t + ( λ1x − Ax)teλ 1 t.
Since Ax = λ1x, this can be written as
y′
2 − Ay2 = − ((A− λ1I)u − x) eλ 1t ,
and now ( 10.5.2) implies that y′
2 = Ay2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 545
T o see that y1 and y2 are linearly independent, suppose c1 and c2 are constants such that
c1y1 + c2y2 = c1xeλ 1t + c2(ueλ 1t + xteλ 1t ) = 0. (10.5.4)
W e must show that c1 = c2 = 0 . Multiplying ( 10.5.4) by e−λ 1t shows that
c1x + c2(u + xt) = 0. (10.5.5)
By differentiating this with respect to t, we see that c2x = 0, which implies c2 = 0 , because x ̸= 0.
Substituting c2 = 0 into ( 10.5.5) yields c1x = 0, which implies that c1 = 0 , again because x ̸= 0
Example 10.5.2 Use Theorem 10.5.1 to find the general solution of the system
y′=
[ 11 −25
4 −9
]
y (10.5.6)
considered in Example 10.5.1.
Solution In Example 10.5.1 we saw that λ1 = 1 is an eigenvalue of multiplicity 2 of the coefficient
matrix Ain ( 10.5.6), and that all of the eigenvectors of Aare multiples of
x =
[ 5
2
]
.
Therefore
y1 =
[ 5
2
]
et
is a solution of (
10.5.6). From Theorem 10.5.1, a second solution is given by y2 = uet + xtet , where",Elementary Differential Equations with Boundary Value Problems.pdf
"x =
[ 5
2
]
.
Therefore
y1 =
[ 5
2
]
et
is a solution of (
10.5.6). From Theorem 10.5.1, a second solution is given by y2 = uet + xtet , where
(A− I)u = x. The augmented matrix of this system is

 10 −25 .
.
. 5
4 −10 .
.
. 2

 ,
which is row equivalent to 
 1 −5
2
.
.
. 1
2
0 0 .
.
. 0

.
Therefore the components of u must satisfy
u1 − 5
2 u2 = 1
2 ,
where u2 is arbitrary . W e choose u2 = 0 , so that u1 = 1 / 2 and
u =
[ 1
2
0
]
.
Thus,
y2 =
[ 1
0
] et
2 +
[ 5
2
]
tet .",Elementary Differential Equations with Boundary Value Problems.pdf
"546 Chapter 10 Linear Systems of Differential Equations
Since y1 and y2 are linearly independent by Theorem 10.5.1, they form a fundamental set of solutions of
(10.5.6). Therefore the general solution of ( 10.5.6) is
y = c1
[
5
2
]
et + c2
([
1
0
] et
2 +
[
5
2
]
tet
)
.
Note that choosing the arbitrary constant u2 to be nonzero is equivalent to adding a scalar multiple of
y1 to the second solution y2 (Exercise 33).
Example 10.5.3 Find the general solution of
y′=


3 4 −10
2 1 −2
2 2 −5

 y. (10.5.7)
Solution The characteristic polynomial of the coefficient matrix Ain (
10.5.7) is
⏐
⏐
⏐
⏐
⏐
⏐
3 − λ 4 −10
2 1 − λ −2
2 2 −5 − λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 1)(λ + 1) 2.
Hence, the eigenvalues are λ1 = 1 with multiplicity 1 and λ2 = −1 with multiplicity 2.
Eigenvectors associated with λ1 = 1 must satisfy (A− I)x = 0. The augmented matrix of this system
is 




2 4 −10
.
.
. 0
2 0 −2 .
.
. 0
2 2 −6 .
.
. 0




 ,
which is row equivalent to 




1 0 −1 .
.
. 0
0 1 −2
.
.
. 0
0 0 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"is 




2 4 −10
.
.
. 0
2 0 −2 .
.
. 0
2 2 −6 .
.
. 0




 ,
which is row equivalent to 




1 0 −1 .
.
. 0
0 1 −2
.
.
. 0
0 0 0 .
.
. 0




 .
Hence, x1 = x3 and x2 = 2 x3, where x3 is arbitrary . Choosing x3 = 1 yields the eigenvector
x1 =


1
2
1

 .
Therefore
y1 =


1
2
1

 et
is a solution of (
10.5.7).
Eigenvectors associated with λ2 = −1 satisfy (A+ I)x = 0. The augmented matrix of this system is





4 4 −10 .
.
. 0
2 2 −2 .
.
. 0
2 2 −4 .
.
. 0




 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 547
which is row equivalent to 




1 1 0 .
.
. 0
0 0 1 .
.
. 0
0 0 0 .
.
. 0




 .
Hence, x3 = 0 and x1 = −x2, where x2 is arbitrary . Choosing x2 = 1 yields the eigenvector
x2 =


−1
1
0

 ,
so
y2 =


−1
1
0

 e−t
is a solution of (
10.5.7).
Since all the eigenvectors of Aassociated with λ2 = −1 are multiples of x2, we must now use Theo-
rem 10.5.1 to find a third solution of ( 10.5.7) in the form
y3 = ue−t +


−1
1
0

 te−t , (10.5.8)
where u is a solution of (A+ I)u = x2. The augmented matrix of this system is





4 4 −10 .
.
. −1
2 2 −2 .
.
. 1
2 2 −4 .
.
. 0




 ,
which is row equivalent to 




1 1 0
.
.
. 1
0 0 1 .
.
. 1
2
0 0 0 .
.
. 0




 .
Hence, u3 = 1 / 2 and u1 = 1 − u2, where u2 is arbitrary . Choosing u2 = 0 yields
u =


1
0
1
2

 ,
and substituting this into (
10.5.8) yields the solution
y3 =


2
0
1

 e−t
2 +


−1
1
0

 te−t
of (
10.5.7).",Elementary Differential Equations with Boundary Value Problems.pdf
"548 Chapter 10 Linear Systems of Differential Equations
Since the Wronskian of {y1, y2, y3}at t= 0 is
⏐
⏐
⏐
⏐
⏐
⏐
1 −1 1
2 1 0
1 0 1
2
⏐
⏐
⏐
⏐
⏐
⏐
= 1
2 ,
{y1, y2, y3}is a fundamental set of solutions of ( 10.5.7). Therefore the general solution of ( 10.5.7) is
y = c1


1
2
1

 et + c2


−1
1
0

 e−t + c3




2
0
1

 e−t
2 +


−1
1
0

 te−t

 .
Theorem 10.5.2
Suppose the n×nmatrix Ahas an eigenvalue λ1 of multiplicity ≥ 3 and the associated
eigenspace is one–dimensional ; that is , all eigenvectors associated with λ1 are scalar multiples of the
eigenvector x. Then there are infinitely many vectors u such that
(A− λ1I)u = x, (10.5.9)
and, if u is any such vector , there are infinitely many vectors v such that
(A− λ1I)v = u. (10.5.10)
If u satisfies (10.5.9) and v satisfies (10.5.10), then
y1 = xeλ 1 t,
y2 = ueλ 1t + xteλ 1t , and
y3 = veλ 1 t + uteλ 1t + x t2eλ 1t
2
are linearly independent solutions of y′= Ay.",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 = xeλ 1 t,
y2 = ueλ 1t + xteλ 1t , and
y3 = veλ 1 t + uteλ 1t + x t2eλ 1t
2
are linearly independent solutions of y′= Ay.
Again, it’s beyond the scope of this book to prove that there a re vectors u and v that satisfy ( 10.5.9)
and ( 10.5.10). Theorem 10.5.1 implies that y1 and y2 are solutions of y′= Ay. W e leave the rest of the
proof to you (Exercise 34).
Example 10.5.4 Use Theorem 10.5.2 to find the general solution of
y′=


1 1 1
1 3 −1
0 2 2

 y. (10.5.11)
Solution The characteristic polynomial of the coefficient matrix Ain (
10.5.11) is
⏐
⏐
⏐
⏐
⏐
⏐
1 − λ 1 1
1 3 − λ −1
0 2 2 − λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 2)3.
Hence, λ1 = 2 is an eigenvalue of multiplicity 3. The associated eigenvectors satisfy (A− 2I)x = 0.
The augmented matrix of this system is





−1 1 1 .
.
. 0
1 1 −1 .
.
. 0
0 2 0 .
.
. 0




 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 549
which is row equivalent to 




1 0 −1 .
.
. 0
0 1 0 .
.
. 0
0 0 0 .
.
. 0




 .
Hence, x1 = x3 and x2 = 0 , so the eigenvectors are all scalar multiples of
x1 =


1
0
1

 .
Therefore
y1 =


1
0
1

 e2t
is a solution of (
10.5.11).
W e now find a second solution of ( 10.5.11) in the form
y2 = ue2t +


1
0
1

 te2t,
where u satisfies (A− 2I)u = x1. The augmented matrix of this system is





−1 1 1 .
.
. 1
1 1 −1
.
.
. 0
0 2 0 .
.
. 1




 ,
which is row equivalent to 




1 0 −1 .
.
. −1
2
0 1 0 .
.
. 1
2
0 0 0 .
.
. 0




 .
Letting u3 = 0 yields u1 = −1/ 2 and u2 = 1 / 2; hence,
u = 1
2


−1
1
0


and
y2 =


−1
1
0

 e2t
2 +


1
0
1

 te2t
is a solution of (
10.5.11).
W e now find a third solution of ( 10.5.11) in the form
y3 = ve2t +


−1
1
0

 te2t
2 +


1
0
1

 t2e2t
2",Elementary Differential Equations with Boundary Value Problems.pdf
"550 Chapter 10 Linear Systems of Differential Equations
where v satisfies (A− 2I)v = u. The augmented matrix of this system is





−1 1 1 .
.
. −1
2
1 1 −1 .
.
. 1
2
0 2 0 .
.
. 0




 ,
which is row equivalent to 




1 0 −1
.
.
. 1
2
0 1 0 .
.
. 0
0 0 0 .
.
. 0




 .
Letting v3 = 0 yields v1 = 1 / 2 and v2 = 0 ; hence,
v = 1
2


1
0
0

 .
Therefore
y3 =


1
0
0

 e2t
2 +


−1
1
0

 te2t
2 +


1
0
1

 t2e2t
2
is a solution of ( 10.5.11). Since y1, y2, and y3 are linearly independent by Theorem 10.5.2, they form a
fundamental set of solutions of ( 10.5.11). Therefore the general solution of ( 10.5.11) is
y = c1


1
0
1

 e2t + c2




−1
1
0

 e2t
2 +


1
0
1

 te2t


+c3




1
0
0

 e2t
2 +


−1
1
0

 te2t
2 +


1
0
1

 t2e2t
2

 .
Theorem 10.5.3
Suppose the n×nmatrix Ahas an eigenvalue λ1 of multiplicity ≥ 3 and the associated",Elementary Differential Equations with Boundary Value Problems.pdf
"

−1
1
0

 te2t
2 +


1
0
1

 t2e2t
2

 .
Theorem 10.5.3
Suppose the n×nmatrix Ahas an eigenvalue λ1 of multiplicity ≥ 3 and the associated
eigenspace is two–dimensional; that is, all eigenvectors o f Aassociated with λ1 are linear combinations
of two linearly independent eigenvectors x1 and x2. Then there are constants α and β (not both zero )
such that if
x3 = αx1 + βx2 , (10.5.12)
then there are infinitely many vectors u such that
(A− λ1I)u = x3. (10.5.13)
If u satisfies (10.5.13), then
y1 = x1eλ 1t ,
y2 = x2eλ 1t , and
y3 = ueλ 1t + x3teλ 1t , (10.5.14)
are linearly independent solutions of y′= Ay.
W e omit the proof of this theorem.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 551
Example 10.5.5 Use Theorem 10.5.3 to find the general solution of
y′=


0 0 1
−1 1 1
−1 0 2

 y. (10.5.15)
Solution The characteristic polynomial of the coefficient matrix Ain (
10.5.15) is
⏐
⏐
⏐
⏐
⏐
⏐
−λ 0 1
−1 1 − λ 1
−1 0 2 − λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 1)3.
Hence, λ1 = 1 is an eigenvalue of multiplicity 3. The associated eigenvectors satisfy (A− I)x = 0. The
augmented matrix of this system is 




−1 0 1 .
.
. 0
−1 0 1 .
.
. 0
−1 0 1 .
.
. 0




 ,
which is row equivalent to 




1 0 −1 .
.
. 0
0 0 0 .
.
. 0
0 0 0
.
.
. 0




 .
Hence, x1 = x3 and x2 is arbitrary , so the eigenvectors are of the form
x1 =


x3
x2
x3

 = x3


1
0
1

 + x2


0
1
0

 .
Therefore the vectors
x1 =


1
0
1

 and x2 =


0
1
0

 (10.5.16)
form a basis for the eigenspace, and
y1 =


1
0
1

 et and y2 =


0
1
0

 et
are linearly independent solutions of (
10.5.15).",Elementary Differential Equations with Boundary Value Problems.pdf
"
 (10.5.16)
form a basis for the eigenspace, and
y1 =


1
0
1

 et and y2 =


0
1
0

 et
are linearly independent solutions of (
10.5.15).
T o find a third linearly independent solution of ( 10.5.15), we must find constants α and β (not both
zero) such that the system
(A− I)u = αx1 + βx2 (10.5.17)
has a solution u. The augmented matrix of this system is





−1 0 1 .
.
. α
−1 0 1 .
.
. β
−1 0 1 .
.
. α




 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"552 Chapter 10 Linear Systems of Differential Equations
which is row equivalent to 




1 0 −1 .
.
. −α
0 0 0 .
.
. β − α
0 0 0 .
.
. 0




 . (10.5.18)
Therefore (
10.5.17) has a solution if and only if β = α, where α is arbitrary . If α = β = 1 then ( 10.5.12)
and ( 10.5.16) yield
x3 = x1 + x2 =


1
0
1

 +


0
1
0

 =


1
1
1

 ,
and the augmented matrix (
10.5.18) becomes





1 0 −1 .
.
. −1
0 0 0 .
.
. 0
0 0 0 .
.
. 0




 .
This implies that u1 = −1 + u3, while u2 and u3 are arbitrary . Choosing u2 = u3 = 0 yields
u =


−1
0
0

 .
Therefore (
10.5.14) implies that
y3 = uet + x3tet =


−1
0
0

 et +


1
1
1

 tet
is a solution of (
10.5.15). Since y1, y2, and y3 are linearly independent by Theorem 10.5.3, they form a
fundamental set of solutions for ( 10.5.15). Therefore the general solution of ( 10.5.15) is
y = c1


1
0
1

 et + c2


0
1
0

 et + c3




−1
0
0

 et +


1
1
1

 tet

 .",Elementary Differential Equations with Boundary Value Problems.pdf
"y = c1


1
0
1

 et + c2


0
1
0

 et + c3




−1
0
0

 et +


1
1
1

 tet

 .
Geometric Properties of Solutions when n= 2
W e’ll now consider the geometric properties of solutions of a 2 × 2 constant coefficient system
[ y′
1
y′
2
]
=
[ a11 a12
a21 a22
][ y1
y2
]
(10.5.19)
under the assumptions of this section; that is, when the matr ix
A=
[ a11 a12
a21 a22
]
has a repeated eigenvalue λ1 and the associated eigenspace is one-dimensional. In this c ase we know
from Theorem 10.5.1 that the general solution of ( 10.5.19) is
y = c1xeλ 1t + c2(ueλ 1t + xteλ 1 t), (10.5.20)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 553
where x is an eigenvector of Aand u is any one of the infinitely many solutions of
(A− λ1I)u = x. (10.5.21)
W e assume that λ1 ̸= 0 .
x
 u
 c2 > 0 c2 < 0
L
Positive Half−Plane
Negative Half−Plane
Figure 10.5.1 Positive and negative half-planes
Let Ldenote the line through the origin parallel to x. By a half-line of Lwe mean either of the rays
obtained by removing the origin from L. Eqn. ( 10.5.20) is a parametric equation of the half-line of Lin
the direction of x if c1 > 0, or of the half-line of Lin the direction of −x if c1 < 0. The origin is the
trajectory of the trivial solution y ≡ 0.
Henceforth, we assume that c2 ̸= 0 . In this case, the trajectory of ( 10.5.20) can’t intersect L, since
every point of Lis on a trajectory obtained by setting c2 = 0 . Therefore the trajectory of ( 10.5.20) must
lie entirely in one of the open half-planes bounded by L, but does not contain any point on L. Since the",Elementary Differential Equations with Boundary Value Problems.pdf
"lie entirely in one of the open half-planes bounded by L, but does not contain any point on L. Since the
initial point (y1(0),y 2(0)) defined by y(0) = c1x1 + c2u is on the trajectory , we can determine which
half-plane contains the trajectory from the sign of c2, as shown in Figure 552. For convenience we’ll call
the half-plane where c2 >0 the positive half-plane . Similarly , the-half plane where c2 <0 is the negative
half-plane. Y ou should convince yourself (Exercise 35) that even though there are infinitely many vectors
u that satisfy ( 10.5.21), they all define the same positive and negative half-planes . In the figures simply
regard u as an arrow pointing to the positive half-plane, since wen’t attempted to give u its proper length
or direction in comparison with x. For our purposes here, only the relative orientation of x and u is
important; that is, whether the positive half-plane is to th e right of an observer facing the direction of x",Elementary Differential Equations with Boundary Value Problems.pdf
"important; that is, whether the positive half-plane is to th e right of an observer facing the direction of x
(as in Figures 10.5.2 and 10.5.5), or to the left of the observer (as in Figures 10.5.3 and 10.5.4).
Multiplying ( 10.5.20) by e−λ 1t yields
e−λ 1t y(t) = c1x + c2u + c2tx.
Since the last term on the right is dominant when |t|is large, this provides the following information on
the direction of y(t):
(a) Along trajectories in the positive half-plane ( c2 >0), the direction of y(t) approaches the direction
of x as t→ ∞ and the direction of −x as t→ −∞ .
(b) Along trajectories in the negative half-plane ( c2 <0), the direction of y(t) approaches the direction
of −x as t→ ∞ and the direction of x as t→ −∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"554 Chapter 10 Linear Systems of Differential Equations
Since
lim
t→∞
∥y(t)∥= ∞ and lim
t→−∞
y(t) = 0 if λ1 >0,
or
lim
t−→∞
∥y(t)∥= ∞ and lim
t→∞
y(t) = 0 if λ1 <0,
there are four possible patterns for the trajectories of ( 10.5.19), depending upon the signs of c2 and λ1.
Figures 10.5.2-10.5.5 illustrate these patterns, and reveal the following princi ple:
If λ1 and c2 have the same sign then the direction of the traectory approa ches the direction of −x as
∥y∥ → 0 and the direction of x as ∥y∥ → ∞ . If λ1 and c2 have opposite signs then the direction of the
trajectory approaches the direction of x as ∥y∥ → 0 and the direction of −x as ∥y∥ → ∞ .
 y1
 y2
 u
 x
 L
Figure 10.5.2 Positive eigenvalue; motion away
from the origin
 y1
 y2
 u
 x
 L
Figure 10.5.3 Positive eigenvalue; motion away
from the origin
 y1
 y2
 u
 x
 L
Figure 10.5.4 Negative eigenvalue; motion toward
the origin
 y1
 y2
 x
 L
 u
Figure 10.5.5 Negative eigenvalue; motion toward
the origin",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.5 Constant Coefficient Homogeneous Systems II 555
10.5 Exercises
In Exercises 1– 12 find the general solution.
1. y′=
[
3 4
−1 7
]
y
2. y′=
[ 0 −1
1 −2
]
y
3. y′=
[ −7 4
−1 −11
]
y 4. y′=
[ 3 1
−1 1
]
y
5. y′=
[ 4 12
−3 −8
]
y 6. y′=
[ −10 9
−4 2
]
y
7. y′=
[ −13 16
−9 11
]
y 8. y′=


0 2 1
−4 6 1
0 4 2

 y
9. y′= 1
3


1 1 −3
−4 −4 3
−2 1 0

 y 10. y′=


−1 1 −1
−2 0 2
−1 3 −1

 y
11. y′=


4 −2 −2
−2 3 −1
2 −1 3

 y 12. y′=


6 −5 3
2 −1 3
2 1 1

 y
In Exercises
13– 23 solve the initial value problem.
13. y′=
[ −11 8
−2 −3
]
y, y(0) =
[ 6
2
]
14. y′=
[ 15 −9
16 −9
]
y, y(0) =
[ 5
8
]
15. y′=
[
−3 −4
1 −7
]
y, y(0) =
[
2
3
]
16. y′=
[
−7 24
−6 17
]
y, y(0) =
[
3
1
]
17. y′=
[ −7 3
−3 −1
]
y, y(0) =
[ 0
2
]
18. y′=


−1 1 0
1 −1 −2
−1 −1 −1

 y, y(0) =


6
5
−7


19. y′=


−2 2 1
−2 2 1
−3 3 2

 y, y(0) =


−6
−2
0


20. y′=


−7 −4 4
−1 0 1
−9 −5 6

 y, y(0) =


−6
9
−1

",Elementary Differential Equations with Boundary Value Problems.pdf
"556 Chapter 10 Linear Systems of Differential Equations
21. y′=


−1 −4 −1
3 6 1
−3 −2 3

 y, y(0) =


−2
1
3


22. y′=


4 −8 −4
−3 −1 −3
1 −1 9

 y, y(0) =


−4
1
−3


23. y′=


−5 −1 11
−7 1 13
−4 0 8

 y, y(0) =


0
2
2


The coefficient matrices in Exercises
24– 32 have eigenvalues of multiplicity 3. Find the general solution.
24. y′=


5 −1 1
−1 9 −3
−2 2 4

 y
25. y′=


1 10 −12
2 2 3
2 −1 6

 y
26. y′=


−6 −4 −4
2 −1 1
2 3 1

 y 27. y′=


0 2 −2
−1 5 −3
1 1 1

 y
28. y′=


−2 −12 10
2 −24 11
2 −24 8

 y 29. y′=


−1 −12 8
1 −9 4
1 −6 1

 y
30. y′=


−4 0 −1
−1 −3 −1
1 0 −2

 y 31. y′=


−3 −3 4
4 5 −8
2 3 −5

 y
32. y′=


−3 −1 0
1 −1 0
−1 −1 −2

y
33. Under the assumptions of Theorem
10.5.1, suppose u and ˆu are vectors such that
(A− λ1I)u = x and (A− λ1I)ˆu = x,
and let
y2 = ueλ 1t + xteλ 1 t and ˆy2 = ˆueλ 1t + xteλ 1t .
Show that y2 − ˆy2 is a scalar multiple of y1 = xeλ 1t .",Elementary Differential Equations with Boundary Value Problems.pdf
"and let
y2 = ueλ 1t + xteλ 1 t and ˆy2 = ˆueλ 1t + xteλ 1t .
Show that y2 − ˆy2 is a scalar multiple of y1 = xeλ 1t .
34. Under the assumptions of Theorem 10.5.2, let
y1 = xeλ 1t ,
y2 = ueλ 1t + xteλ 1t , and
y3 = veλ 1t + uteλ 1t + x t2eλ 1t
2 .
Complete the proof of Theorem 10.5.2 by showing that y3 is a solution of y′ = Ay and that
{y1, y2, y3}is linearly independent.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.6 Constant Coefficient Homogeneous Systems III 557
35. Suppose the matrix
A=
[ a11 a12
a21 a22
]
has a repeated eigenvalue λ1 and the associated eigenspace is one-dimensional. Let x be a λ1-
eigenvector of A. Show that if (A− λ1I)u1 = x and (A− λ1I)u2 = x, then u2 − u1 is parallel
to x. Conclude from this that all vectors u such that (A− λ1 I)u = x define the same positive and
negative half-planes with respect to the line Lthrough the origin parallel to x.
In Exercises 36- 45 plot trajectories of the given system.
36. C/G y′=
[ −3 −1
4 1
]
y
37. C/G y′=
[ 2 −1
1 0
]
y
38. C/G y′=
[ −1 −3
3 5
]
y 39. C/G y′=
[ −5 3
−3 1
]
y
40. C/G y′=
[ −2 −3
3 4
]
y 41. C/G y′=
[ −4 −3
3 2
]
y
42. C/G y′=
[
0 −1
1 −2
]
y 43. C/G y′=
[
0 1
−1 2
]
y
44. C/G y′=
[ −2 1
−1 0
]
y 45. C/G y′=
[ 0 −4
1 −4
]
y
10.6 CONST ANT COEFFICIENT HOMOGENEOUS SYSTEMS III
W e now consider the system y′ = Ay, where A has a complex eigenvalue λ = α + iβ with β ̸= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"]
y
10.6 CONST ANT COEFFICIENT HOMOGENEOUS SYSTEMS III
W e now consider the system y′ = Ay, where A has a complex eigenvalue λ = α + iβ with β ̸= 0 .
W e continue to assume that Ahas real entries, so the characteristic polynomial of Ahas real coefficients.
This implies that λ = α − iβ is also an eigenvalue of A.
An eigenvector x of Aassociated with λ = α + iβ will have complex entries, so we’ll write
x = u + iv
where u and v have real entries; that is, u and v are the real and imaginary parts of x. Since Ax = λx,
A(u + iv) = ( α + iβ)(u + iv). (10.6.1)
T aking complex conjugates here and recalling that Ahas real entries yields
A(u − iv) = ( α − iβ)(u − iv),
which shows that x = u − iv is an eigenvector associated with λ = α − iβ. The complex conjugate
eigenvalues λ and λ can be separately associated with linearly independent sol utions y′= Ay; however,
we won’t pursue this approach, since solutions obtained in t his way turn out to be complex–valued.",Elementary Differential Equations with Boundary Value Problems.pdf
"we won’t pursue this approach, since solutions obtained in t his way turn out to be complex–valued.
Instead, we’ll obtain solutions of y′= Ay in the form
y = f1u + f2v (10.6.2)
where f1 and f2 are real–valued scalar functions. The next theorem shows ho w to do this.",Elementary Differential Equations with Boundary Value Problems.pdf
"558 Chapter 10 Linear Systems of Differential Equations
Theorem 10.6.1 Let A be an n× nmatrix with real entries . Let λ = α + iβ (β ̸= 0 ) be a complex
eigenvalue of Aand let x = u + iv be an associated eigenvector , where u and v have real components .
Then u and v are both nonzero and
y1 = eαt (u cos βt − v sin βt) and y2 = eαt (u sin βt + v cos βt),
which are the real and imaginary parts of
eαt (cos βt + isin βt)(u + iv), (10.6.3)
are linearly independent solutions of y′= Ay.
Proof A function of the form ( 10.6.2) is a solution of y′= Ay if and only if
f′
1u + f′
2v = f1Au + f2Av. (10.6.4)
Carrying out the multiplication indicated on the right side of ( 10.6.1) and collecting the real and imaginary
parts of the result yields
A(u + iv) = ( αu − βv) + i(αv + βu).
Equating real and imaginary parts on the two sides of this equ ation yields
Au = αu − βv
Av = αv + βu.
W e leave it to you (Exercise 25) to show from this that u and v are both nonzero. Substituting from these",Elementary Differential Equations with Boundary Value Problems.pdf
"Au = αu − βv
Av = αv + βu.
W e leave it to you (Exercise 25) to show from this that u and v are both nonzero. Substituting from these
equations into ( 10.6.4) yields
f′
1u + f′
2v = f1(αu − βv) + f2(αv + βu)
= ( αf1 + βf2 )u + ( −βf1 + αf2)v.
This is true if
f′
1 = αf1 + βf2
f′
2 = −βf1 + αf2, or, equivalently , f′
1 − αf1 = βf2
f′
2 − αf2 = −βf1 .
If we let f1 = g1eαt and f2 = g2eαt , where g1 and g2 are to be determined, then the last two equations
become
g′
1 = βg2
g′
2 = −βg1 ,
which implies that
g′′
1 = βg′
2 = −β2g1,
so
g′′
1 + β2g1 = 0 .
The general solution of this equation is
g1 = c1 cos βt + c2 sin βt.
Moreover, since g2 = g′
1/β ,
g2 = −c1 sin βt + c2 cos βt.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.6 Constant Coefficient Homogeneous Systems III 559
Multiplying g1 and g2 by eαt shows that
f1 = eαt ( c1 cos βt + c2 sin βt),
f2 = eαt (−c1 sin βt + c2 cos βt).
Substituting these into ( 10.6.2) shows that
y = eαt [(c1 cos βt + c2 sin βt)u + ( −c1 sin βt + c2 cos βt)v]
= c1eαt (u cos βt − v sin βt) + c2eαt (u sin βt + v cos βt) (10.6.5)
is a solution of y′= Ay for any choice of the constants c1 and c2. In particular, by first taking c1 = 1
and c2 = 0 and then taking c1 = 0 and c2 = 1 , we see that y1 and y2 are solutions of y′= Ay. W e leave
it to you to verify that they are, respectively , the real and i maginary parts of ( 10.6.3) (Exercise 26), and
that they are linearly independent (Exercise 27).
Example 10.6.1 Find the general solution of
y′=
[ 4 −5
5 −2
]
y. (10.6.6)
Solution The characteristic polynomial of the coefficient matrix Ain ( 10.6.6) is
⏐
⏐
⏐
⏐
4 − λ −5
5 −2 − λ
⏐
⏐
⏐
⏐= ( λ − 1)2 + 16 .",Elementary Differential Equations with Boundary Value Problems.pdf
"]
y. (10.6.6)
Solution The characteristic polynomial of the coefficient matrix Ain ( 10.6.6) is
⏐
⏐
⏐
⏐
4 − λ −5
5 −2 − λ
⏐
⏐
⏐
⏐= ( λ − 1)2 + 16 .
Hence, λ = 1 + 4 iis an eigenvalue of A. The associated eigenvectors satisfy (A− (1 + 4 i) I) x = 0.
The augmented matrix of this system is

 3 − 4i −5 .
.
. 0
5 −3 − 4i
.
.
. 0

 ,
which is row equivalent to 
 1 −3+4i
5
.
.
. 0
0 0
.
.
. 0

 .
Therefore x1 = (3 + 4 i)x2/ 5. T aking x2 = 5 yields x1 = 3 + 4 i, so
x =
[
3 + 4 i
5
]
is an eigenvector. The real and imaginary parts of
et (cos 4 t+ isin 4 t)
[ 3 + 4 i
5
]
are
y1 = et
[ 3 cos 4 t− 4 sin 4 t
5 cos 4 t
]
and y2 = et
[ 3 sin 4 t+ 4 cos 4 t
5 sin 4 t
]
,
which are linearly independent solutions of (
10.6.6). The general solution of ( 10.6.6) is
y = c1et
[
3 cos 4 t− 4 sin 4 t
5 cos 4 t
]
+ c2et
[
3 sin 4 t+ 4 cos 4 t
5 sin 4 t
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"560 Chapter 10 Linear Systems of Differential Equations
Example 10.6.2 Find the general solution of
y′=
[ −14 39
−6 16
]
y. (10.6.7)
Solution The characteristic polynomial of the coefficient matrix Ain ( 10.6.7) is
⏐
⏐
⏐
⏐
−14 − λ 39
−6 16 − λ
⏐
⏐
⏐
⏐= ( λ − 1)2 + 9 .
Hence, λ = 1 + 3 iis an eigenvalue of A. The associated eigenvectors satisfy (A− (1 + 3 i)I) x = 0.
The augmented augmented matrix of this system is

 −15 − 3i 39 .
.
. 0
−6 15 − 3i
.
.
. 0

 ,
which is row equivalent to 
 1 −5+i
2
.
.
. 0
0 0 .
.
. 0

 .
Therefore x1 = (5 − i)/ 2. T aking x2 = 2 yields x1 = 5 − i, so
x =
[ 5 − i
2
]
is an eigenvector. The real and imaginary parts of
et (cos 3 t+ isin 3 t)
[ 5 − i
2
]
are
y1 = et
[ sin 3 t+ 5 cos 3 t
2 cos 3 t
]
and y2 = et
[ − cos 3t+ 5 sin 3 t
2 sin 3 t
]
,
which are linearly independent solutions of (
10.6.7). The general solution of ( 10.6.7) is
y = c1et
[
sin 3 t+ 5 cos 3 t
2 cos 3 t
]
+ c2et
[
− cos 3t+ 5 sin 3 t
2 sin 3 t
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"10.6.7). The general solution of ( 10.6.7) is
y = c1et
[
sin 3 t+ 5 cos 3 t
2 cos 3 t
]
+ c2et
[
− cos 3t+ 5 sin 3 t
2 sin 3 t
]
.
Example 10.6.3 Find the general solution of
y′=


−5 5 4
−8 7 6
1 0 0

 y. (10.6.8)
Solution The characteristic polynomial of the coefficient matrix Ain (
10.6.8) is
⏐
⏐
⏐
⏐
⏐
⏐
−5 − λ 5 4
−8 7 − λ 6
1 0 −λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 2)(λ2 + 1) .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.6 Constant Coefficient Homogeneous Systems III 561
Hence, the eigenvalues of Aare λ1 = 2 , λ2 = i, and λ3 = −i. The augmented matrix of (A− 2I)x = 0
is 




−7 5 4 .
.
. 0
−8 5 6 .
.
. 0
1 0 −2
.
.
. 0




 ,
which is row equivalent to 




1 0 −2 .
.
. 0
0 1 −2 .
.
. 0
0 0 0 .
.
. 0




 .
Therefore x1 = x2 = 2 x3. T aking x3 = 1 yields
x1 =


2
2
1

 ,
so
y1 =


2
2
1

 e2t
is a solution of (
10.6.8).
The augmented matrix of (A− iI)x = 0 is





−5 − i 5 4 .
.
. 0
−8 7 − i 6 .
.
. 0
1 0 −i .
.
. 0




 ,
which is row equivalent to 




1 0 −i .
.
. 0
0 1 1 − i .
.
. 0
0 0 0 .
.
. 0




 .
Therefore x1 = ix3 and x2 = −(1 − i)x3. T aking x3 = 1 yields the eigenvector
x2 =


i
−1 + i
1

 .
The real and imaginary parts of
(cos t+ isin t)


i
−1 + i
1


are
y2 =


− sin t
− cos t− sin t
cos t

 and y3 =


cos t
cos t− sin t
sin t

 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"562 Chapter 10 Linear Systems of Differential Equations
which are solutions of ( 10.6.8). Since the Wronskian of {y1, y2, y3}at t= 0 is
⏐
⏐
⏐
⏐
⏐
⏐
2 0 1
2 −1 1
1 1 0
⏐
⏐
⏐
⏐
⏐
⏐
= 1 ,
{y1, y2, y3}is a fundamental set of solutions of (
10.6.8). The general solution of ( 10.6.8) is
y = c1


2
2
1

 e2t + c2


− sin t
− cos t− sin t
cos t

 + c3


cos t
cos t− sin t
sin t

 .
Example 10.6.4 Find the general solution of
y′=


1 −1 −2
1 3 2
1 −1 2

 y. (10.6.9)
Solution The characteristic polynomial of the coefficient matrix Ain (
10.6.9) is
⏐
⏐
⏐
⏐
⏐
⏐
1 − λ −1 −2
1 3 − λ 2
1 −1 2 − λ
⏐
⏐
⏐
⏐
⏐
⏐
= −(λ − 2) ((λ − 2)2 + 4 ) .
Hence, the eigenvalues of A are λ1 = 2 , λ2 = 2 + 2 i, and λ3 = 2 − 2i. The augmented matrix of
(A− 2I)x = 0 is 




−1 −1 −2 .
.
. 0
1 1 2
.
.
. 0
1 −1 0 .
.
. 0




 ,
which is row equivalent to 




1 0 1 .
.
. 0
0 1 1 .
.
. 0
0 0 0
.
.
. 0




 .
Therefore x1 = x2 = −x3. T aking x3 = 1 yields
x1 =


−1
−1
1

 ,
so
y1 =


−1",Elementary Differential Equations with Boundary Value Problems.pdf
"



1 0 1 .
.
. 0
0 1 1 .
.
. 0
0 0 0
.
.
. 0




 .
Therefore x1 = x2 = −x3. T aking x3 = 1 yields
x1 =


−1
−1
1

 ,
so
y1 =


−1
−1
1

 e2t
is a solution of (
10.6.9).
The augmented matrix of (A− (2 + 2 i)I) x = 0 is





−1 − 2i −1 −2 .
.
. 0
1 1 − 2i 2 .
.
. 0
1 −1 −2i .
.
. 0




 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.6 Constant Coefficient Homogeneous Systems III 563
which is row equivalent to 




1 0 −i .
.
. 0
0 1 i .
.
. 0
0 0 0 .
.
. 0




 .
Therefore x1 = ix3 and x2 = −ix3. T aking x3 = 1 yields the eigenvector
x2 =


i
−i
1


The real and imaginary parts of
e2t(cos 2 t+ isin 2 t)


i
−i
1


are
y2 = e2t


− sin 2 t
sin 2t
cos 2t

 and y2 = e2t


cos 2 t
− cos 2 t
sin 2 t

 ,
which are solutions of (
10.6.9). Since the Wronskian of {y1, y2, y3}at t= 0 is
⏐
⏐
⏐
⏐
⏐
⏐
−1 0 1
−1 0 −1
1 1 0
⏐
⏐
⏐
⏐
⏐
⏐
= −2,
{y1, y2, y3}is a fundamental set of solutions of (
10.6.9). The general solution of ( 10.6.9) is
y = c1


−1
−1
1

 e2t + c2e2t


− sin 2 t
sin 2 t
cos 2 t

 + c3e2t


cos 2 t
− cos 2 t
sin 2 t

 .
Geometric Properties of Solutions when n= 2
W e’ll now consider the geometric properties of solutions of a 2 × 2 constant coefficient system
[ y′
1
y′
2
]
=
[ a11 a12
a21 a22
][ y1
y2
]
(10.6.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"[ y′
1
y′
2
]
=
[ a11 a12
a21 a22
][ y1
y2
]
(10.6.10)
under the assumptions of this section; that is, when the matr ix
A=
[ a11 a12
a21 a22
]
has a complex eigenvalue λ = α + iβ (β ̸= 0 ) and x = u + iv is an associated eigenvector, where
u and v have real components. T o describe the trajectories accurat ely it’s necessary to introduce a new
rectangular coordinate system in the y1-y2 plane. This raises a point that hasn’t come up before: It is
always possible to choose x so that (u, v) = 0 . A special effort is required to do this, since not every
eigenvector has this property . However, if we know an eigenv ector that doesn’t, we can multiply it by a
suitable complex constant to obtain one that does. T o see thi s, note that if x is a λ-eigenvector of Aand
kis an arbitrary real number, then
x1 = (1 + ik)x = (1 + ik)(u + iv) = ( u − kv) + i(v + ku)",Elementary Differential Equations with Boundary Value Problems.pdf
"564 Chapter 10 Linear Systems of Differential Equations
is also a λ-eigenvector of A, since
Ax1 = A((1 + ik)x) = (1 + ik)Ax = (1 + ik)λx = λ((1 + ik)x) = λx1.
The real and imaginary parts of x1 are
u1 = u − kv and v1 = v + ku, (10.6.11)
so
(u1, v1) = ( u − kv, v + ku) = −
[
(u, v)k2 + ( ∥v∥2 − ∥u∥2)k− (u, v)
]
.
Therefore (u1, v1) = 0 if
(u, v)k2 + ( ∥v∥2 − ∥u∥2)k− (u, v) = 0 . (10.6.12)
If (u, v) ̸= 0 we can use the quadratic formula to find two real values of k such that (u1, v1) = 0
(Exercise 28).
Example 10.6.5 In Example 10.6.1 we found the eigenvector
x =
[ 3 + 4 i
5
]
=
[ 3
5
]
+ i
[ 4
0
]
for the matrix of the system (
10.6.6). Here u =
[ 3
5
]
and v =
[ 4
0
]
are not orthogonal, since
(u, v) = 12 . Since ∥v∥2 − ∥u∥2 = −18, (
10.6.12) is equivalent to
2k2 − 3k− 2 = 0 .
The zeros of this equation are k1 = 2 and k2 = −1/ 2. Letting k= 2 in ( 10.6.11) yields
u1 = u − 2v =
[ −5
5
]
and v1 = v + 2 u =
[ 10
10
]
,
and (u1, v1) = 0 . Letting k= −1/ 2 in (
10.6.11) yields
u1 = u + v
2 =",Elementary Differential Equations with Boundary Value Problems.pdf
"u1 = u − 2v =
[ −5
5
]
and v1 = v + 2 u =
[ 10
10
]
,
and (u1, v1) = 0 . Letting k= −1/ 2 in (
10.6.11) yields
u1 = u + v
2 =
[ 5
5
]
and v1 = v − u
2 = 1
2
[ −5
5
]
,
and again (u1, v1) = 0 .
(The numbers don’t always work out as nicely as in this exampl e. Y ou’ll need a calculator or computer
to do Exercises 29-40.)
Henceforth, we’ll assume that (u, v) = 0 . Let U and V be unit vectors in the directions of u and v,
respectively; that is, U = u/ ∥u∥and V = v/ ∥v∥. The new rectangular coordinate system will have the
same origin as the y1-y2 system. The coordinates of a point in this system will be deno ted by (z1,z 2),
where z1 and z2 are the displacements in the directions of U and V, respectively .
From ( 10.6.5), the solutions of ( 10.6.10) are given by
y = eαt [(c1 cos βt + c2 sin βt)u + ( −c1 sin βt + c2 cos βt)v] . (10.6.13)
For convenience, let’s call the curve traversed by e−αt y(t) a shadow trajectory of ( 10.6.10). Multiplying
(10.6.13) by e−αt yields",Elementary Differential Equations with Boundary Value Problems.pdf
"For convenience, let’s call the curve traversed by e−αt y(t) a shadow trajectory of ( 10.6.10). Multiplying
(10.6.13) by e−αt yields
e−αt y(t) = z1(t)U + z2(t)V,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.6 Constant Coefficient Homogeneous Systems III 565
where
z1(t) = ∥u∥(c1 cos βt + c2 sin βt)
z2(t) = ∥v∥(−c1 sin βt + c2 cos βt).
Therefore
(z1(t))2
∥u∥2 + (z2(t))2
∥v∥2 = c2
1 + c2
2
(verify!), which means that the shadow trajectories of (
10.6.10) are ellipses centered at the origin, with
axes of symmetry parallel to U and V. Since
z′
1 = β∥u∥
∥v∥ z2 and z′
2 = −β∥v∥
∥u∥z1,
the vector from the origin to a point on the shadow ellipse rot ates in the same direction that V would have
to be rotated by π/ 2 radians to bring it into coincidence with U (Figures 10.6.1 and 10.6.2).
 y1
 y2
 V
 U
Figure 10.6.1 Shadow trajectories traversed
clockwise
 y1
 y2
 U
 V
Figure 10.6.2 Shadow trajectories traversed
counterclockwise
If α = 0 , then any trajectory of ( 10.6.10) is a shadow trajectory of ( 10.6.10); therefore, if λ is
purely imaginary , then the trajectories of ( 10.6.10) are ellipses traversed periodically as indicated in Fig-
ures 10.6.1 and 10.6.2.
If α> 0, then
lim
t→∞",Elementary Differential Equations with Boundary Value Problems.pdf
"ures 10.6.1 and 10.6.2.
If α> 0, then
lim
t→∞
∥y(t)∥= ∞ and lim
t→−∞
y(t) = 0 ,
so the trajectory spirals away from the origin as t varies from −∞ to ∞ . The direction of the spiral
depends upon the relative orientation of U and V, as shown in Figures 10.6.3 and 10.6.4.
If α< 0, then
lim
t→−∞
∥y(t)∥= ∞ and lim
t→∞
y(t) = 0 ,
so the trajectory spirals toward the origin as tvaries from −∞ to ∞ . Again, the direction of the spiral
depends upon the relative orientation of U and V, as shown in Figures 10.6.5 and 10.6.6.",Elementary Differential Equations with Boundary Value Problems.pdf
"566 Chapter 10 Linear Systems of Differential Equations
 y1
 y2
 V
 U
Figure 10.6.3 α> 0; shadow trajectory spiraling
outward
 y1
 y2
 U
 V
Figure 10.6.4 α> 0; shadow trajectory spiraling
outward
 y1
 y2
 V
 U
Figure 10.6.5 α< 0; shadow trajectory spiraling
inward
 y1
 y2
 U
 V
Figure 10.6.6 α< 0; shadow trajectory spiraling
inward
10.6 Exercises
In Exercises 1– 16 find the general solution.
1. y′=
[ −1 2
−5 5
]
y
2. y′=
[ −11 4
−26 9
]
y
3. y′=
[
1 2
−4 5
]
y 4. y′=
[
5 −6
3 −1
]
y",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.6 Constant Coefficient Homogeneous Systems III 567
5. y′=


3 −3 1
0 2 2
5 1 1

 y 6. y′=


−3 3 1
1 −5 −3
−3 7 3

 y
7. y′=


2 1 −1
0 1 1
1 0 1

 y 8. y′=


−3 1 −3
4 −1 2
4 −2 3

 y
9. y′=
[ 5 −4
10 1
]
y 10. y′= 1
3
[ 7 −5
2 5
]
y
11. y′=
[ 3 2
−5 1
]
y 12. y′=
[ 34 52
−20 −30
]
y
13. y′=


1 1 2
1 0 −1
−1 −2 −1

 y 14. y′=


3 −4 −2
−5 7 −8
−10 13 −8

 y
15. y′=


6 0 −3
−3 3 3
1 −2 6

 y′ 16. y′=


1 2 −2
0 2 −1
1 0 0

 y′
In Exercises
17– 24 solve the initial value problem.
17. y′=
[ 4 −6
3 −2
]
y, y(0) =
[ 5
2
]
18. y′=
[ 7 15
−3 1
]
y, y(0) =
[ 5
1
]
19. y′=
[ 7 −15
3 −5
]
y, y(0) =
[ 17
7
]
20. y′= 1
6
[
4 −2
5 2
]
y, y(0) =
[
1
−1
]
21. y′=


5 2 −1
−3 2 2
1 3 2

 y, y(0) =


4
0
6


22. y′=


4 4 0
8 10 −20
2 3 −2

 y, y(0) =


8
6
5


23. y′=


1 15 −15
−6 18 −22
−3 11 −15

 y, y(0) =


15
17
10


24. y′=


4 −4 4
−10 3 15
2 −3 1

 y, y(0) =


16
14
6

",Elementary Differential Equations with Boundary Value Problems.pdf
"568 Chapter 10 Linear Systems of Differential Equations
25. Suppose an n× nmatrix Awith real entries has a complex eigenvalue λ = α + iβ (β ̸= 0 ) with
associated eigenvector x = u + iv, where u and v have real components. Show that u and v are
both nonzero.
26. V erify that
y1 = eαt (u cos βt − v sin βt) and y2 = eαt (u sin βt + v cos βt),
are the real and imaginary parts of
eαt (cos βt + isin βt)(u + iv).
27. Show that if the vectors u and v are not both 0 and β ̸= 0 then the vector functions
y1 = eαt (u cos βt − v sin βt) and y2 = eαt (u sin βt + v cos βt)
are linearly independent on every interval. H IN T : There are two cases to consider: (i) {u, v}
linearly independent, and (ii) {u, v}linearly dependent. In either case, exploit the the linear
independence of {cos βt, sin βt}on every interval.
28. Suppose u =
[ u1
u2
]
and v =
[ v1
v2
]
are not orthogonal; that is, (u, v) ̸= 0 .
(a) Show that the quadratic equation
(u, v)k2 + ( ∥v∥2 − ∥u∥2)k− (u, v) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"[ u1
u2
]
and v =
[ v1
v2
]
are not orthogonal; that is, (u, v) ̸= 0 .
(a) Show that the quadratic equation
(u, v)k2 + ( ∥v∥2 − ∥u∥2)k− (u, v) = 0
has a positive root k1 and a negative root k2 = −1/k 1.
(b) Let u(1)
1 = u − k1v, v(1)
1 = v + k1u, u(2)
1 = u − k2v, and v(2)
1 = v + k2u, so that
(u(1)
1 , v(1)
1 ) = ( u(2)
1 , v(2)
1 ) = 0 , from the discussion given above. Show that
u(2)
1 = v(1)
1
k1
and v(2)
1 = −u(1)
1
k1
.
(c) Let U1, V1, U2, and V2 be unit vectors in the directions of u(1)
1 , v(1)
1 , u(2)
1 , and v(2)
1 ,
respectively . Conclude from (a) that U2 = V1 and V2 = −U1, and that therefore the
counterclockwise angles from U1 to V1 and from U2 to V2 are both π/ 2 or both −π/ 2.
In Exercises
29-32 find vectors U and V parallel to the axes of symmetry of the trajectories, and plo t
some typical trajectories.
29. C/G y′=
[ 3 −5
5 −3
]
y
30. C/G y′=
[ −15 10
−25 15
]
y
31. C/G y′=
[ −4 8
−4 4
]
y 32. C/G y′=
[ −3 −15
3 3
]
y",Elementary Differential Equations with Boundary Value Problems.pdf
"some typical trajectories.
29. C/G y′=
[ 3 −5
5 −3
]
y
30. C/G y′=
[ −15 10
−25 15
]
y
31. C/G y′=
[ −4 8
−4 4
]
y 32. C/G y′=
[ −3 −15
3 3
]
y
In Exercises 33-40 find vectors U and V parallel to the axes of symmetry of the shadow trajectories, and
plot a typical trajectory.
33. C/G y′=
[ −5 6
−12 7
]
y
34. C/G y′=
[ 5 −12
6 −7
]
y",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.7 V ariation of Parameters for Nonhomogeneous Linear Systems 569
35. C/G y′=
[ 4 −5
9 −2
]
y 36. C/G y′=
[ −4 9
−5 2
]
y
37. C/G y′=
[
−1 10
−10 −1
]
y 38. C/G y′=
[
−1 −5
20 −1
]
y
39. C/G y′=
[ −7 10
−10 9
]
y 40. C/G y′=
[ −7 6
−12 5
]
y
10.7 V ARIA TION OF P ARAMETERS FOR NONHOMOGENEOUS LINEAR SYSTEMS
W e now consider the nonhomogeneous linear system
y′= A(t)y + f (t),
where Ais an n× nmatrix function and f is an n-vector forcing function. Associated with this system is
the complementary system y′= A(t)y.
The next theorem is analogous to Theorems 5.3.2 and 9.1.5. It shows how to find the general solution
of y′= A(t)y + f (t) if we know a particular solution of y′= A(t)y + f (t) and a fundamental set of
solutions of the complementary system. W e leave the proof as an exercise (Exercise 21).
Theorem 10.7.1 Suppose the n× n matrix function A and the n-vector function f are continuous on",Elementary Differential Equations with Boundary Value Problems.pdf
"Theorem 10.7.1 Suppose the n× n matrix function A and the n-vector function f are continuous on
(a,b ). Let yp be a particular solution of y′ = A(t)y + f (t) on (a,b ), and let {y1, y2,..., yn}be a
fundamental set of solutions of the complementary equation y′= A(t)y on (a,b ). Then y is a solution
of y′= A(t)y + f (t) on (a,b ) if and only if
y = yp + c1y1 + c2y2 + · · ·+ cn yn ,
where c1,c 2, . . . , cn are constants.
Finding a Particular Solution of a Nonhomogeneous System
W e now discuss an extension of the method of variation of para meters to linear nonhomogeneous systems.
This method will produce a particular solution of a nonhomog enous system y′= A(t)y + f (t) provided
that we know a fundamental matrix for the complementary syst em. T o derive the method, suppose Y is a
fundamental matrix for the complementary system; that is,
Y =





y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
. .
.
. . . . .
.
.
yn1 yn2 · · ·ynn




 ,
where
y1 =





y11
y21
.
.
.
yn1



",Elementary Differential Equations with Boundary Value Problems.pdf
"Y =





y11 y12 · · ·y1n
y21 y22 · · ·y2n
.
.
. .
.
. . . . .
.
.
yn1 yn2 · · ·ynn




 ,
where
y1 =





y11
y21
.
.
.
yn1




 , y2 =





y12
y22
.
.
.
yn2




 , · · ·, yn =





y1n
y2n
.
.
.
ynn





is a fundamental set of solutions of the complementary syste m. In Section 10.3 we saw that Y′= A(t)Y.
W e seek a particular solution of
y′= A(t)y + f (t) (10.7.1)
of the form
yp = Yu, (10.7.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"570 Chapter 10 Linear Systems of Differential Equations
where u is to be determined. Differentiating ( 10.7.2) yields
y′
p = Y′u + Yu′
= AYu + Yu′(since Y′= AY)
= Ayp + Yu′(since Yu = yp ).
Comparing this with ( 10.7.1) shows that yp = Yu is a solution of ( 10.7.1) if and only if
Yu′= f .
Thus, we can find a particular solution yp by solving this equation for u′, integrating to obtain u, and
computing Yu. W e can take all constants of integration to be zero, since an y particular solution will
suffice.
Exercise 22 sketches a proof that this method is analogous to the method o f variation of parameters
discussed in Sections 5.7 and 9.4 for scalar linear equation s.
Example 10.7.1
(a) Find a particular solution of the system
y′=
[ 1 2
2 1
]
y +
[ 2e4t
e4t
]
, (10.7.3)
which we considered in Example
10.2.1.
(b) Find the general solution of ( 10.7.3).
SO LU TIO N (a) The complementary system is
y′=
[ 1 2
2 1
]
y. (10.7.4)
The characteristic polynomial of the coefficient matrix is
⏐
⏐
⏐
⏐",Elementary Differential Equations with Boundary Value Problems.pdf
"SO LU TIO N (a) The complementary system is
y′=
[ 1 2
2 1
]
y. (10.7.4)
The characteristic polynomial of the coefficient matrix is
⏐
⏐
⏐
⏐
1 − λ 2
2 1 − λ
⏐
⏐
⏐
⏐= ( λ + 1)( λ − 3).
Using the method of Section 10.4, we find that
y1 =
[
e3t
e3t
]
and y2 =
[
e−t
−e−t
]
are linearly independent solutions of (
10.7.4). Therefore
Y =
[
e3t e−t
e3t −e−t
]
is a fundamental matrix for ( 10.7.4). W e seek a particular solution yp = Yu of ( 10.7.3), where Yu′= f ;
that is, [ e3t e−t
e3t −e−t
][ u′
1
u′
2
]
=
[ 2e4t
e4t
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.7 V ariation of Parameters for Nonhomogeneous Linear Systems 571
The determinant of Y is the Wronskian
⏐
⏐
⏐
⏐
e3t e−t
e3t −e−t
⏐
⏐
⏐
⏐= −2e2t .
By Cramer’s rule,
u′
1 = − 1
2e2t
⏐
⏐
⏐
⏐
2e4t e−t
e4t −e−t
⏐
⏐
⏐
⏐ = 3e3t
2e2t = 3
2 et ,
u′
2 = − 1
2e2t
⏐
⏐
⏐
⏐
e3t 2e4t
e3t e4t
⏐
⏐
⏐
⏐ = e7t
2e2t = 1
2 e5t.
Therefore
u′= 1
2
[ 3et
e5t
]
.
Integrating and taking the constants of integration to be ze ro yields
u = 1
10
[
15et
e5t
]
,
so
yp = Yu = 1
10
[ e3t e−t
e3t −e−t
] [ 15et
e5t
]
= 1
5
[ 8e4t
7e4t
]
is a particular solution of ( 10.7.3).
SO LU TIO N (b) From Theorem 10.7.1, the general solution of ( 10.7.3) is
y = yp + c1y1 + c2y2 = 1
5
[ 8e4t
7e4t
]
+ c1
[ e3t
e3t
]
+ c2
[ e−t
−e−t
]
, (10.7.5)
which can also be written as
y = 1
5
[ 8e4t
7e4t
]
+
[ e3t e−t
e3t −e−t
]
c,
where c is an arbitrary constant vector.
Writing ( 10.7.5) in terms of coordinates yields
y1 = 8
5 e4t + c1e3t + c2e−t
y2 = 7
5 e4t + c1e3t − c2e−t ,
so our result is consistent with Example 10.2.1. .",Elementary Differential Equations with Boundary Value Problems.pdf
"y1 = 8
5 e4t + c1e3t + c2e−t
y2 = 7
5 e4t + c1e3t − c2e−t ,
so our result is consistent with Example 10.2.1. .
If Aisn’t a constant matrix, it’s usually difficult to find a funda mental set of solutions for the system
y′ = A(t)y. It is beyond the scope of this text to discuss methods for doi ng this. Therefore, in the
following examples and in the exercises involving systems w ith variable coefficient matrices we’ll provide
fundamental matrices for the complementary systems withou t explaining how they were obtained.
Example 10.7.2 Find a particular solution of
y′=
[ 2 2 e−2t
2e2t 4
]
y +
[ 1
1
]
, (10.7.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"572 Chapter 10 Linear Systems of Differential Equations
given that
Y =
[ e4t −1
e6t e2t
]
is a fundamental matrix for the complementary system.
Solution W e seek a particular solution yp = Yu of (
10.7.6) where Yu′= f ; that is,
[
e4t −1
e6t e2t
][
u′
1
u′
2
]
=
[
1
1
]
.
The determinant of Y is the Wronskian
⏐
⏐
⏐
⏐
e4t −1
e6t e2t
⏐
⏐
⏐
⏐= 2 e6t.
By Cramer’s rule,
u′
1 = 1
2e6t
⏐
⏐
⏐
⏐
1 −1
1 e2t
⏐
⏐
⏐
⏐ = e2t + 1
2e6t = e−4t + e−6t
2
u′
2 = 1
2e6t
⏐
⏐
⏐
⏐
e4t 1
e6t 1
⏐
⏐
⏐
⏐ = e4t − e6t
2e6t = e−2t − 1
2 .
Therefore
u′= 1
2
[ e−4t + e−6t
e−2t − 1
]
.
Integrating and taking the constants of integration to be ze ro yields
u = − 1
24
[ 3e−4t + 2 e−6t
6e−2t + 12 t
]
,
so
yp = Yu = − 1
24
[ e4t −1
e6t e2t
][ 3e−4t + 2 e−6t
6e−2t + 12 t
]
= 1
24
[ 4e−2t + 12 t− 3
−3e2t(4t+ 1) − 8
]
is a particular solution of ( 10.7.6).
Example 10.7.3 Find a particular solution of
y′= − 2
t2
[ t −3t2
1 −2t
]
y + t2
[ 1
1
]
, (10.7.7)
given that
Y =
[
2t 3t2
1 2 t
]",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 10.7.3 Find a particular solution of
y′= − 2
t2
[ t −3t2
1 −2t
]
y + t2
[ 1
1
]
, (10.7.7)
given that
Y =
[
2t 3t2
1 2 t
]
is a fundamental matrix for the complementary system on (−∞ , 0) and (0, ∞ ).
Solution W e seek a particular solution yp = Yu of (
10.7.7) where Yu′= f ; that is,
[ 2t 3t2
1 2 t
][ u′
1
u′
2
]
=
[ t2
t2
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.7 V ariation of Parameters for Nonhomogeneous Linear Systems 573
The determinant of Y is the Wronskian
⏐
⏐
⏐
⏐
2t 3t2
1 2 t
⏐
⏐
⏐
⏐= t2.
By Cramer’s rule,
u′
1 = 1
t2
⏐
⏐
⏐
⏐
t2 3t2
t2 2t
⏐
⏐
⏐
⏐ = 2t3 − 3t4
t2 = 2 t− 3t2,
u′
2 = 1
t2
⏐
⏐
⏐
⏐
2t t 2
1 t2
⏐
⏐
⏐
⏐ = 2t3 − t2
t2 = 2 t− 1.
Therefore
u′=
[ 2t− 3t2
2t− 1
]
.
Integrating and taking the constants of integration to be ze ro yields
u =
[ t2 − t3
t2 − t
]
,
so
yp = Yu =
[ 2t 3t2
1 2 t
][ t2 − t3
t2 − t
]
=
[ t3(t− 1)
t2(t− 1)
]
is a particular solution of ( 10.7.7).
Example 10.7.4
(a) Find a particular solution of
y′=


2 −1 −1
1 0 −1
1 −1 0

 y +


et
0
e−t

 . (10.7.8)
(b) Find the general solution of (
10.7.8).
SO LU TIO N (a) The complementary system for ( 10.7.8) is
y′=


2 −1 −1
1 0 −1
1 −1 0

 y. (10.7.9)
The characteristic polynomial of the coefficient matrix is
⏐
⏐
⏐
⏐
⏐
⏐
2 − λ −1 −1
1 −λ −1
1 −1 −λ
⏐
⏐
⏐
⏐
⏐
⏐
= −λ(λ − 1)2.
Using the method of Section 10.4, we find that
y1 =


1
1
1

 , y2 =
",Elementary Differential Equations with Boundary Value Problems.pdf
"⏐
⏐
⏐
⏐
⏐
⏐
2 − λ −1 −1
1 −λ −1
1 −1 −λ
⏐
⏐
⏐
⏐
⏐
⏐
= −λ(λ − 1)2.
Using the method of Section 10.4, we find that
y1 =


1
1
1

 , y2 =


et
et
0

 , and y3 =


et
0
et

",Elementary Differential Equations with Boundary Value Problems.pdf
"574 Chapter 10 Linear Systems of Differential Equations
are linearly independent solutions of ( 10.7.9). Therefore
Y =


1 et et
1 et 0
1 0 et


is a fundamental matrix for (
10.7.9). W e seek a particular solution yp = Yu of ( 10.7.8), where Yu′= f ;
that is, 

1 et et
1 et 0
1 0 et




u′
1
u′
2
u′
3

 =


et
0
e−t

 .
The determinant of Y is the Wronskian
⏐
⏐
⏐
⏐
⏐
⏐
1 et et
1 et 0
1 0 et
⏐
⏐
⏐
⏐
⏐
⏐
= −e2t .
Thus, by Cramer’s rule,
u′
1 = − 1
e2t
⏐
⏐
⏐
⏐
⏐
⏐
et et et
0 et 0
e−t 0 et
⏐
⏐
⏐
⏐
⏐
⏐
= −e3t − et
e2t = e−t − et
u′
2 = − 1
e2t
⏐
⏐
⏐
⏐
⏐
⏐
1 et et
1 0 0
1 e−t et
⏐
⏐
⏐
⏐
⏐
⏐
= −1 − e2t
e2t = 1 − e−2t
u′
3 = − 1
e2t
⏐
⏐
⏐
⏐
⏐
⏐
1 et et
1 et 0
1 0 e−t
⏐
⏐
⏐
⏐
⏐
⏐
= e2t
e2t = 1 .
Therefore
u′=


e−t − et
1 − e−2t
1

 .
Integrating and taking the constants of integration to be ze ro yields
u =



−et − e−t
e−2t
2 + t
t


 ,
so
yp = Yu =


1 et et
1 et 0
1 0 et





−et − e−t
e−2t
2 + t
t


 =





et (2t− 1) − e−t
2
et (t− 1) − e−t
2",Elementary Differential Equations with Boundary Value Problems.pdf
"e−2t
2 + t
t


 ,
so
yp = Yu =


1 et et
1 et 0
1 0 et





−et − e−t
e−2t
2 + t
t


 =





et (2t− 1) − e−t
2
et (t− 1) − e−t
2
et (t− 1) − e−t





is a particular solution of (
10.7.8).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.7 V ariation of Parameters for Nonhomogeneous Linear Systems 575
SO LU TIO N (a) From Theorem 10.7.1 the general solution of ( 10.7.8) is
y = yp + c1y1 + c2y2 + c3y3 =





et (2t− 1) − e−t
2
et (t− 1) − e−t
2
et (t− 1) − e−t




 + c1


1
1
1

 + c2


et
et
0

 + c3


et
0
et

 ,
which can be written as
y = yp + Yc =





et (2t− 1) − e−t
2
et (t− 1) − e−t
2
et (t− 1) − e−t




 +


1 et et
1 et 0
1 0 et

 c
where c is an arbitrary constant vector.
Example 10.7.5 Find a particular solution of
y′= 1
2


3 e−t −e2t
0 6 0
−e−2t e−3t −1

 y +


1
et
e−t

 , (10.7.10)
given that
Y =


et 0 e2t
0 e3t e3t
e−t 1 0


is a fundamental matrix for the complementary system.
Solution W e seek a particular solution of (
10.7.10) in the form yp = Yu, where Yu′= f ; that is,


et 0 e2t
0 e3t e3t
e−t 1 0




u′
1
u′
2
u′
3

 =


1
et
e−t

 .
The determinant of Y is the Wronskian
⏐
⏐
⏐
⏐
⏐
⏐
et 0 e2t
0 e3t e3t
e−t 1 0
⏐
⏐
⏐
⏐
⏐
⏐",Elementary Differential Equations with Boundary Value Problems.pdf
"e−t 1 0




u′
1
u′
2
u′
3

 =


1
et
e−t

 .
The determinant of Y is the Wronskian
⏐
⏐
⏐
⏐
⏐
⏐
et 0 e2t
0 e3t e3t
e−t 1 0
⏐
⏐
⏐
⏐
⏐
⏐
= −2e4t .
By Cramer’s rule,
u′
1 = − 1
2e4t
⏐
⏐
⏐
⏐
⏐
⏐
1 0 e2t
et e3t e3t
e−t 1 0
⏐
⏐
⏐
⏐
⏐
⏐
= e4t
2e4t = 1
2
u′
2 = − 1
2e4t
⏐
⏐
⏐
⏐
⏐
⏐
et 1 e2t
0 et e3t
e−t e−t 0
⏐
⏐
⏐
⏐
⏐
⏐
= e3t
2e4t = 1
2 e−t
u′
3 = − 1
2e4t
⏐
⏐
⏐
⏐
⏐
⏐
et 0 1
0 e3t et
e−t 1 e−t
⏐
⏐
⏐
⏐
⏐
⏐
= −e3t − 2e2t
2e4t = 2e−2t − e−t
2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"576 Chapter 10 Linear Systems of Differential Equations
Therefore
u′= 1
2


1
e−t
2e−2t − e−t

 .
Integrating and taking the constants of integration to be ze ro yields
u = 1
2


t
−e−t
e−t − e−2t

 ,
so
yp = Yu = 1
2


et 0 e2t
0 e3t e3t
e−t 1 0




t
−e−t
e−t − e−2t

 = 1
2


et (t+ 1) − 1
−et
e−t (t− 1)


is a particular solution of (
10.7.10).
10.7 Exercises
In Exercises 1– 10 find a particular solution.
1. y′=
[
−1 −4
−1 −1
]
y +
[
21e4t
8e−3t
]
2. y′= 1
5
[
−4 3
−2 −11
]
y +
[
50e3t
10e−3t
]
3. y′=
[ 1 2
2 1
]
y +
[ 1
t
]
4. y′=
[ −4 −3
6 5
]
y +
[ 2
−2et
]
5. y′=
[ −6 −3
1 −2
]
y +
[ 4e−3t
4e−5t
]
6. y′=
[ 0 1
−1 0
]
y +
[ 1
t
]
7. y′=


3 1 −1
3 5 1
−6 2 4

 y +


3
6
3

 8. y′=


3 −1 −1
−2 3 2
4 −1 −2

 y +


1
et
et


9. y′=


−3 2 2
2 −3 2
2 2 −3

 y +


et
e−5t
et


10. y′= 1
3


1 1 −3
−4 −4 3
−2 1 0

 y +


et
et
et


In Exercises",Elementary Differential Equations with Boundary Value Problems.pdf
"
1
et
et


9. y′=


−3 2 2
2 −3 2
2 2 −3

 y +


et
e−5t
et


10. y′= 1
3


1 1 −3
−4 −4 3
−2 1 0

 y +


et
et
et


In Exercises
11– 20 find a particular solution, given that Y is a fundamental matrix for the complementary
system.
11. y′= 1
t
[ 1 t
−t 1
]
y + t
[ cos t
sin t
]
; Y = t
[ cos t sin t
− sin t cos t
]
12. y′= 1
t
[
1 t
t 1
]
y +
[
t
t2
]
; Y = t
[
et e−t
et −e−t
]",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 10.7 V ariation of Parameters for Nonhomogeneous Linear Systems 577
13. y′= 1
t2 − 1
[ t −1
−1 t
]
y + t
[ 1
−1
]
; Y =
[ t 1
1 t
]
14. y′= 1
3
[ 1 −2e−t
2et −1
]
y +
[ e2t
e−2t
]
; Y =
[ 2 e−t
et 2
]
15. y′= 1
2t4
[
3t3 t6
1 −3t3
]
y + 1
t
[
t2
1
]
; Y = 1
t2
[
t3 t4
−1 t
]
16. y′=




1
t− 1 − e−t
t− 1
et
t+ 1
1
t+ 1



 y +
[ t2 − 1
t2 − 1
]
; Y =
[ t e −t
et t
]
17. y′= 1
t


1 1 0
0 2 1
−2 2 2

 y +


1
2
1

 Y =


t2 t3 1
t2 2t3 −1
0 2 t3 2


18. y′=


3 et e2t
e−t 2 et
e−2t e−t 1

 y +


e3t
0
0

 ; Y =


e5t e2t 0
e4t 0 et
e3t −1 −1


19. y′= 1
t


1 t 0
0 1 t
0 −t 1

 y +


t
t
t

 ; Y = t


1 cos t sin t
0 − sin t cos t
0 − cos t − sin t


20. y′= −1
t


e−t −t 1 − e−t
e−t 1 −t− e−t
e−t −t 1 − e−t

 y + 1
t


et
0
et

 ; Y = 1
t


et e−t t
et −e−t e−t
et e−t 0


21. Prove Theorem
10.7.1.
22. (a) Convert the scalar equation
P0(t)y(n) + P1(t)y(n−1) + · · ·+ Pn(t)y = F(t) (A)
into an equivalent n× nsystem",Elementary Differential Equations with Boundary Value Problems.pdf
"

21. Prove Theorem
10.7.1.
22. (a) Convert the scalar equation
P0(t)y(n) + P1(t)y(n−1) + · · ·+ Pn(t)y = F(t) (A)
into an equivalent n× nsystem
y′= A(t)y + f (t). (B)
(b) Suppose (A) is normal on an interval (a,b ) and {y1,y 2,...,y n }is a fundamental set of
solutions of
P0(t)y(n) + P1(t)y(n−1) + · · ·+ Pn (t)y= 0 (C)
on (a,b ). Find a corresponding fundamental matrix Y for
y′= A(t)y (D)
on (a,b ) such that
y= c1y1 + c2y2 + · · ·+ cn yn
is a solution of (C) if and only if y = Yc with
c =





c1
c2
.
.
.
cn





is a solution of (D).",Elementary Differential Equations with Boundary Value Problems.pdf
"578 Chapter 10 Linear Systems of Differential Equations
(c) Let yp = u1y1 + u1y2 + · · ·+ unyn be a particular solution of (A), obtained by the method
of variation of parameters for scalar equations as given in S ection 9.4, and define
u =





u1
u2
.
.
.
un




 .
Show that yp = Yu is a solution of (B).
(d) Let yp = Yu be a particular solution of (B), obtained by the method of var iation of param-
eters for systems as given in this section. Show that yp = u1y1 + u1y2 + · · ·+ unyn is a
solution of (A).
23. Suppose the n× nmatrix function Aand the n–vector function f are continuous on (a,b ). Let
t0 be in (a,b ), let k be an arbitrary constant vector, and let Y be a fundamental matrix for the
homogeneous system y′= A(t)y. Use variation of parameters to show that the solution of the
initial value problem
y′= A(t)y + f (t), y(t0) = k
is
y(t) = Y(t)
(
Y−1(t0)k +
∫ t
t0
Y−1(s)f (s) ds
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 11
Boundary V alue Problems and Fourier
Expansions
IN THIS CHAPTER we develop series representations of functi ons that will be used to solve partial
differential equations in Chapter 12.
SECTION 11.1 deals with five boundary value problems for the d ifferential equation
y′′+ λy = 0 .
They are related to problems in partial differential equati ons that will be discussed in Chapter 12. W e
define what is meant by eigenvalues and eigenfunctions of the boundary value problems, and show that
the eigenfunctions have a property called
orthogonality.
SECTION 11.2 introduces Fourier series , which are expansions of given functions in term of sines and
cosines.
SECTION 11.3 deals with expansions of functions in terms of t he eigenfunctions of four of the eigenvalue
problems discussed in Section 11.1. They are all related to t he Fourier series discussed in Section 11.2.
581",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.1 Eigenvalue Problems for y′′+ λy = 0 581
11.1 EIGENV ALUE PROBLEMS FOR y′′+ λy = 0
In Chapter 12 we’ll study partial differential equations th at arise in problems of heat conduction, wave
propagation, and potential theory . The purpose of this chap ter is to develop tools required to solve these
equations. In this section we consider the following proble ms, where λ is a real number and L> 0:
Problem 1: y′′+ λy = 0 , y (0) = 0 , y (L) = 0
Problem 2: y′′+ λy = 0 , y ′(0) = 0 , y ′(L) = 0
Problem 3: y′′+ λy = 0 , y (0) = 0 , y ′(L) = 0
Problem 4: y′′+ λy = 0 , y ′(0) = 0 , y (L) = 0
Problem 5: y′′+ λy = 0 , y (−L) = y(L), y ′(−L) = y′(L)
In each problem the conditions following the differential e quation are called boundary conditions . Note
that the boundary conditions in Problem 5, unlike those in Pr oblems 1-4, don’t require that yor y′be zero
at the boundary points, but only that yhave the same value at x= ±L, and that y′have the same value",Elementary Differential Equations with Boundary Value Problems.pdf
"at the boundary points, but only that yhave the same value at x= ±L, and that y′have the same value
at x= ±L. W e say that the boundary conditions in Problem 5 are periodic.
Obviously , y ≡ 0 (the trivial solution) is a solution of Problems 1-5 for any v alue of λ. For most values
of λ, there are no other solutions. The interesting question is t his:
F or what values of λ does the problem have nontrivial solutions, and what are the y?
A value of λ for which the problem has a nontrivial solution is an eigenvalue of the problem, and
the nontrivial solutions are λ-eigenfunctions, or eigenfunctions associated with λ. Note that a nonzero
constant multiple of a λ-eigenfunction is again a λ-eigenfunction.
Problems 1-5 are called eigenvalue problems . Solving an eigenvalue problem means finding all its
eigenvalues and associated eigenfunctions. W e’ll take it a s given here that all the eigenvalues of Prob-
lems 1-5 are real numbers. This is proved in a more general set ting in Section 13.2.",Elementary Differential Equations with Boundary Value Problems.pdf
"lems 1-5 are real numbers. This is proved in a more general set ting in Section 13.2.
Theorem 11.1.1 Problems 1– 5 have no negative eigenvalues. Moreover , λ = 0 is an eigenvalue of
Problems 2 and 5, with associated eigenfunction y0 = 1 , but λ = 0 isn’t an eigenvalue of Problems 1, 3,
or 4.
Proof W e consider Problems 1-4, and leave Problem 5 to you (Exercis e 1). If y′′+ λy = 0 , then
y(y′′+ λy) = 0 , so ∫ L
0
y(x)(y′′(x) + λy(x)) dx= 0;
therefore,
λ
∫ L
0
y2(x) dx= −
∫ L
0
y(x)y′′(x) dx. (11.1.1)
Integration by parts yields
∫ L
0
y(x)y′′(x) dx = y(x)y′(x)
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
(y′(x))2 dx
= y(L)y′(L) − y(0)y′(0) −
∫ L
0
(y′(x))2 dx.
(11.1.2)
However, if ysatisfies any of the boundary conditions of Problems 1-4, the n
y(L)y′(L) − y(0)y′(0) = 0;",Elementary Differential Equations with Boundary Value Problems.pdf
"582 Chapter 11 Boundary V alue Problems and Fourier Expansions
hence, ( 11.1.1) and ( 11.1.2) imply that
λ
∫ L
0
y2 (x) dx=
∫ L
0
(y′(x))2 dx.
If y ̸≡0, then ∫L
0 y2 (x) dx> 0. Therefore λ ≥ 0 and, if λ = 0 , then y′(x) = 0 for all xin (0,L ) (why?),
and y is constant on (0,L ). Any constant function satisfies the boundary conditions of Problem 2, so
λ = 0 is an eigenvalue of Problem 2 and any nonzero constant functi on is an associated eigenfunction.
However, the only constant function that satisfies the bound ary conditions of Problems 1, 3, or 4 is y ≡ 0.
Therefore λ = 0 isn’t an eigenvalue of any of these problems.
Example 11.1.1 (Problem 1) Solve the eigenvalue problem
y′′+ λy = 0 , y (0) = 0 , y (L) = 0 . (11.1.3)
Solution From Theorem 11.1.1, any eigenvalues of ( 11.1.3) must be positive. If ysatisfies ( 11.1.3) with
λ> 0, then
y= c1 cos
√
λx + c2 sin
√
λx,
where c1 and c2 are constants. The boundary condition y(0) = 0 implies that c1 = 0 . Therefore
y = c2 sin
√",Elementary Differential Equations with Boundary Value Problems.pdf
"λ> 0, then
y= c1 cos
√
λx + c2 sin
√
λx,
where c1 and c2 are constants. The boundary condition y(0) = 0 implies that c1 = 0 . Therefore
y = c2 sin
√
λx . Now the boundary condition y(L) = 0 implies that c2 sin
√
λL = 0 . T o make
c2 sin
√
λL = 0 with c2 ̸= 0 , we must choose
√
λ = nπ/L , where nis a positive integer. Therefore
λn = n2π2/L 2 is an eigenvalue and
yn = sin nπx
L
is an associated eigenfunction.
For future reference, we state the result of Example 11.1.1 as a theorem.
Theorem 11.1.2 The eigenvalue problem
y′′+ λy = 0 , y (0) = 0 , y (L) = 0
has infinitely many positive eigenvalues λn = n2π2/L 2, with associated eigenfunctions
yn = sin nπx
L , n = 1 , 2, 3,....
There are no other eigenvalues.
W e leave it to you to prove the next theorem about Problem 2 by a n argument like that of Exam-
ple 11.1.1 (Exercise 17).
Theorem 11.1.3 The eigenvalue problem
y′′+ λy = 0 , y ′(0) = 0 , y ′(L) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"ple 11.1.1 (Exercise 17).
Theorem 11.1.3 The eigenvalue problem
y′′+ λy = 0 , y ′(0) = 0 , y ′(L) = 0
has the eigenvalue λ0 = 0 , with associated eigenfunction y0 = 1 , and infinitely many positive eigenvalues
λn = n2π2/L 2, with associated eigenfunctions
yn = cos nπx
L ,n = 1 , 2, 3 ....
There are no other eigenvalues.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.1 Eigenvalue Problems for y′′+ λy = 0 583
Example 11.1.2 (Problem 3) Solve the eigenvalue problem
y′′+ λy = 0 , y (0) = 0 , y ′(L) = 0 . (11.1.4)
Solution From Theorem 11.1.1, any eigenvalues of ( 11.1.4) must be positive. If ysatisfies ( 11.1.4) with
λ> 0, then
y= c1 cos
√
λx + c2 sin
√
λx,
where c1 and c2 are constants. The boundary condition y(0) = 0 implies that c1 = 0 . Therefore
y = c2 sin
√
λx . Hence, y′ = c2
√
λ cos
√
λx and the boundary condition y′(L) = 0 implies that
c2 cos
√
λL = 0 . T o make c2 cos
√
λL = 0 with c2 ̸= 0 we must choose
√
λ = (2n− 1)π
2L ,
where nis a positive integer. Then λn = (2 n− 1)2π2/ 4L2 is an eigenvalue and
yn = sin (2n− 1)πx
2L
is an associated eigenfunction.
For future reference, we state the result of Example 11.1.2 as a theorem.
Theorem 11.1.4 The eigenvalue problem
y′′+ λy = 0 , y (0) = 0 , y ′(L) = 0
has infinitely many positive eigenvalues λn = (2 n− 1)2π2/ 4L2, with associated eigenfunctions
yn = sin (2n− 1)πx",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′+ λy = 0 , y (0) = 0 , y ′(L) = 0
has infinitely many positive eigenvalues λn = (2 n− 1)2π2/ 4L2, with associated eigenfunctions
yn = sin (2n− 1)πx
2L , n = 1 , 2, 3,....
There are no other eigenvalues.
W e leave it to you to prove the next theorem about Problem 4 by a n argument like that of Exam-
ple 11.1.2 (Exercise 18).
Theorem 11.1.5 The eigenvalue problem
y′′+ λy = 0 , y ′(0) = 0 , y (L) = 0
has infinitely many positive eigenvalues λn = (2 n− 1)2π2/ 4L2, with associated eigenfunctions
yn = cos (2n− 1)πx
2L , n = 1 , 2, 3,....
There are no other eigenvalues.
Example 11.1.3 (Problem 5) Solve the eigenvalue problem
y′′+ λy = 0 , y (−L) = y(L), y ′(−L) = y′(L). (11.1.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"584 Chapter 11 Boundary V alue Problems and Fourier Expansions
Solution From Theorem 11.1.1, λ = 0 is an eigenvalue of ( 11.1.5) with associated eigenfunction y0 = 1 ,
and any other eigenvalues must be positive. If ysatisfies ( 11.1.5) with λ> 0, then
y= c1 cos
√
λx + c2 sin
√
λx, (11.1.6)
where c1 and c2 are constants. The boundary condition y(−L) = y(L) implies that
c1 cos(−
√
λL ) + c2 sin(−
√
λL ) = c1 cos
√
λL + c2 sin
√
λL. (11.1.7)
Since
cos(−
√
λL ) = cos
√
λL and sin(−
√
λL ) = − sin
√
λL, (11.1.8)
(11.1.7) implies that
c2 sin
√
λL = 0 . (11.1.9)
Differentiating ( 11.1.6) yields
y′=
√
λ
(
−c1 sin
√
λx + c2 cos
√
λx
)
.
The boundary condition y′(−L) = y′(L) implies that
−c1 sin(−
√
λL ) + c2 cos(−
√
λL ) = −c1 sin
√
λL + c2 cos
√
λL,
and ( 11.1.8) implies that
c1 sin
√
λL = 0 . (11.1.10)
Eqns. ( 11.1.9) and ( 11.1.10) imply that c1 = c2 = 0 unless
√
λ = nπ/L , where nis a positive integer.",Elementary Differential Equations with Boundary Value Problems.pdf
"c1 sin
√
λL = 0 . (11.1.10)
Eqns. ( 11.1.9) and ( 11.1.10) imply that c1 = c2 = 0 unless
√
λ = nπ/L , where nis a positive integer.
In this case ( 11.1.9) and ( 11.1.10) both hold for arbitrary c1 and c2. The eigenvalue determined in this
way is λn = n2π2/L 2, and each such eigenvalue has the linearly independent asso ciated eigenfunctions
cos nπx
L and sin nπx
L .
For future reference we state the result of Example 11.1.3 as a theorem.
Theorem 11.1.6 The eigenvalue problem
y′′+ λy = 0 , y (−L) = y(L), y ′(−L) = y′(L),
has the eigenvalue λ0 = 0 , with associated eigenfunction y0 = 1 and infinitely many positive eigenvalues
λn = n2π2/L 2, with associated eigenfunctions
y1n = cos nπx
L and y2n = sin nπx
L , n = 1 , 2, 3,....
There are no other eigenvalues.
Orthogonality
W e say that two integrable functions f and gare orthogonal on an interval [a,b ] if
∫ b
a
f(x)g(x) dx= 0 .
More generally , we say that the functions φ1, φ2, . . . , φn, . . . (finitely or infinitely many) are orthogonal",Elementary Differential Equations with Boundary Value Problems.pdf
"∫ b
a
f(x)g(x) dx= 0 .
More generally , we say that the functions φ1, φ2, . . . , φn, . . . (finitely or infinitely many) are orthogonal
on [a,b ] if ∫ b
a
φi(x)φj (x) dx= 0 whenever i̸= j.
The importance of orthogonality will become clear when we st udy Fourier series in the next two sections.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.1 Eigenvalue Problems for y′′+ λy = 0 585
Example 11.1.4 Show that the eigenfunctions
1, cos πx
L , sin πx
L , cos 2πx
L , sin 2πx
L ,..., cos nπx
L , sin nπx
L ,... (11.1.11)
of Problem 5 are orthogonal on [−L,L ].
Solution W e must show that ∫ L
−L
f(x)g(x) dx= 0 (11.1.12)
whenever f and gare distinct functions from ( 11.1.11). If ris any nonzero integer, then
∫ L
−L
cos rπx
L dx= L
rπ sin rπx
L
⏐
⏐
⏐
⏐
L
−L
= 0 . (11.1.13)
and ∫ L
−L
sin rπx
L dx= − L
rπ cos rπx
L
⏐
⏐
⏐
⏐
L
−L
= 0 .
Therefore (
11.1.12) holds if f ≡ 1 and gis any other function in ( 11.1.11).
If f(x) = cos mπx/L and g(x) = cos nπx/L where mand nare distinct positive integers, then
∫ L
−L
f(x)g(x) dx=
∫ L
−L
cos mπx
L cos nπx
L dx. (11.1.14)
T o evaluate this integral, we use the identity
cos Acos B= 1
2 [cos(A− B) + cos( A+ B)]
with A= mπx/L and B= nπx/L . Then ( 11.1.14) becomes
∫ L
−L
f(x)g(x) dx= 1
2
[ ∫ L
−L
cos (m− n)πx
L dx+
∫ L
−L
cos (m+ n)πx
L dx
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"with A= mπx/L and B= nπx/L . Then ( 11.1.14) becomes
∫ L
−L
f(x)g(x) dx= 1
2
[ ∫ L
−L
cos (m− n)πx
L dx+
∫ L
−L
cos (m+ n)πx
L dx
]
.
Since m− nand m+ nare both nonzero integers, ( 11.1.13) implies that the integrals on the right are
both zero. Therefore ( 11.1.12) is true in this case.
If f(x) = sin mπx/L and g(x) = sin nπx/L where mand nare distinct positive integers, then
∫ L
−L
f(x)g(x) dx=
∫ L
−L
sin mπx
L sin nπx
L dx. (11.1.15)
T o evaluate this integral, we use the identity
sin Asin B= 1
2 [cos(A− B) − cos(A+ B)]
with A= mπx/L and B= nπx/L . Then ( 11.1.15) becomes
∫ L
−L
f(x)g(x) dx= 1
2
[ ∫ L
−L
cos (m− n)πx
L dx−
∫ L
−L
cos (m+ n)πx
L dx
]
= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"586 Chapter 11 Boundary V alue Problems and Fourier Expansions
If f(x) = sin mπx/L and g(x) = cos nπx/L where mand nare positive integers (not necessarily
distinct), then ∫ L
−L
f(x)g(x) dx=
∫ L
−L
sin mπx
L cos nπx
L dx= 0
because the integrand is an odd function and the limits are sy mmetric about x= 0 .
Exercises 19-22 ask you to verify that the eigenfunctions of Problems 1-4 are orthogonal on [0,L ].
However, this also follows from a general theorem that we’ll prove in Chapter 13.
11.1 Exercises
1. Prove that λ = 0 is an eigenvalue of Problem 5 with associated eigenfunction y0 = 1 , and that any
other eigenvalues must be positive. H IN T : See the proof of Theorem 11.1.1.
In Exercises 2-16 solve the eigenvalue problem.
2. y′′+ λy = 0 , y (0) = 0 , y (π) = 0
3. y′′+ λy = 0 , y ′(0) = 0 , y ′(π) = 0
4. y′′+ λy = 0 , y(0) = 0 , y′(π) = 0
5. y′′+ λy = 0 , y′(0) = 0 , y(π) = 0
6. y′′+ λy = 0 , y (−π) = y(π), y ′(−π) = y′(π)
7. y′′+ λy = 0 , y ′(0) = 0 , y ′(1) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"5. y′′+ λy = 0 , y′(0) = 0 , y(π) = 0
6. y′′+ λy = 0 , y (−π) = y(π), y ′(−π) = y′(π)
7. y′′+ λy = 0 , y ′(0) = 0 , y ′(1) = 0
8. y′′+ λy = 0 , y′(0) = 0 , y(1) = 0
9. y′′+ λy = 0 , y (0) = 0 , y (1) = 0
10. y′′+ λy = 0 , y (−1) = y(1), y ′(−1) = y′(1)
11. y′′+ λy = 0 , y(0) = 0 , y′(1) = 0
12. y′′+ λy = 0 , y (−2) = y(2), y ′(−2) = y′(2)
13. y′′+ λy = 0 , y (0) = 0 , y (2) = 0
14. y′′+ λy = 0 , y′(0) = 0 , y(3) = 0
15. y′′+ λy = 0 , y(0) = 0 , y′(1/ 2) = 0
16. y′′+ λy = 0 , y ′(0) = 0 , y ′(5) = 0
17. Prove Theorem 11.1.3.
18. Prove Theorem 11.1.5.
19. V erify that the eigenfunctions
sin πx
L , sin 2πx
L ,..., sin nπx
L ,...
of Problem 1 are orthogonal on [0,L ].
20. V erify that the eigenfunctions
1, cos πx
L , cos 2πx
L ,..., cos nπx
L ,...
of Problem 2 are orthogonal on [0,L ].",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 587
21. V erify that the eigenfunctions
sin πx
2L, sin 3πx
2L ,..., sin (2n− 1)πx
2L ,...
of Problem 3 are orthogonal on [0,L ].
22. V erify that the eigenfunctions
cos πx
2L, cos 3πx
2L ,..., cos (2n− 1)πx
2L ,...
of Problem 4 are orthogonal on [0,L ].
In Exercises 23-26 solve the eigenvalue problem.
23. y′′+ λy = 0 , y (0) = 0 ,
∫ L
0
y(x) dx= 0
24. y′′+ λy = 0 , y ′(0) = 0 ,
∫ L
0
y(x) dx= 0
25. y′′+ λy = 0 , y (L) = 0 ,
∫ L
0
y(x) dx= 0
26. y′′+ λy = 0 , y ′(L) = 0 ,
∫ L
0
y(x) dx= 0
11.2 FOURIER EXP ANSIONS I
In Example 11.1.4 and Exercises 11.1. 4–11.1. 22 we saw that the eigenfunctions of Problem 5 are orthog-
onal on [−L,L ] and the eigenfunctions of Problems 1–4 are orthogonal on [0,L ]. In this section and the
next we introduce some series expansions in terms of these ei genfunctions. W e’ll use these expansions to
solve partial differential equations in Chapter 12.
Theorem 11.2.1",Elementary Differential Equations with Boundary Value Problems.pdf
"solve partial differential equations in Chapter 12.
Theorem 11.2.1
Suppose the functions φ1,φ 2,φ 3, . . . , are orthogonal on [a,b ] and
∫ b
a
φ2
n (x) dx̸= 0 , n = 1 , 2, 3,.... (11.2.1)
Let c1,c 2,c 3,. . . be constants such that the partial sums fN (x) = ∑ N
m=1 cmφm(x) satisfy the inequalities
|fN (x)| ≤M, a ≤ x≤ b, N = 1 , 2, 3,...
for some constant M <∞ . Suppose also that the series
f(x) =
∞∑
m=1
cmφm (x) (11.2.2)
converges and is integrable on [a,b ]. Then
cn =
∫ b
a
f(x)φn (x) dx
∫ b
a
φ2
n(x) dx
, n = 1 , 2, 3,.... (11.2.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"588 Chapter 11 Boundary V alue Problems and Fourier Expansions
Proof Multiplying ( 11.2.2) by φn and integrating yields
∫ b
a
f(x)φn (x) dx=
∫ b
a
φn (x)
( ∞∑
m=1
cmφm (x)
)
dx. (11.2.4)
It can be shown that the boundedness of the partial sums {fN }∞
N =1 and the integrability of f allow us to
interchange the operations of integration and summation on the right of (
11.2.4), and rewrite ( 11.2.4) as
∫ b
a
f(x)φn (x) dx=
∞∑
m=1
cm
∫ b
a
φn(x)φm(x) dx. (11.2.5)
(This isn’t easy to prove.) Since
∫ b
a
φn(x)φm(x) dx= 0 if m̸= n,
(11.2.5) reduces to ∫ b
a
f(x)φn (x) dx= cn
∫ b
a
φ2
n(x) dx.
Now (
11.2.1) implies ( 11.2.3).
Theorem 11.2.1 motivates the next definition.
Definition 11.2.2 Suppose φ1,φ 2, . . . , φn,. . . are orthogonal on [a,b ] and ∫b
a φ2
n(x) dx̸= 0 , n= 1 , 2, 3,
. . . . Let f be integrable on [a,b ], and define
cn =
∫ b
a
f(x)φn (x) dx
∫ b
a
φ2
n(x) dx
, n = 1 , 2, 3,.... (11.2.6)
Then the infinite series ∑ ∞",Elementary Differential Equations with Boundary Value Problems.pdf
". . . . Let f be integrable on [a,b ], and define
cn =
∫ b
a
f(x)φn (x) dx
∫ b
a
φ2
n(x) dx
, n = 1 , 2, 3,.... (11.2.6)
Then the infinite series ∑ ∞
n=1 cn φn(x) is called the F ourier expansion of f in terms of the orthogonal
set {φn}∞
n=1, and c1, c2, . . . , cn, . . . are called the F ourier coefficients of f with respect to {φn}∞
n=1. W e
indicate the relationship between f and its Fourier expansion by
f(x) ∼
∞∑
n=1
cn φn(x), a ≤ x≤ b. (11.2.7)
Y ou may wonder why we don’t write
f(x) =
∞∑
n=1
cn φn(x), a ≤ x≤ b,
rather than ( 11.2.7). Unfortunately , this isn’t always true. The series on the r ight may diverge for some or
all values of xin [a,b ], or it may converge to f(x) for some values of xand not for others. So, for now ,
we’ll just think of the series as being associated with f because of the definition of the coefficients {cn},
and we’ll indicate this association informally as in ( 11.2.7).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 589
Fourier Series
W e’ll now study Fourier expansions in terms of the eigenfunc tions
1, cos πx
L , sin πx
L , cos 2πx
L , sin 2πx
L ,..., cos nπx
L , sin nπx
L ,....
of Problem 5. If f is integrable on [−L,L ], its Fourier expansion in terms of these functions is called the
F ourier series of f on [−L,L ]. Since ∫ L
−L
12 dx= 2 L,
∫ L
−L
cos2 nπx
L dx= 1
2
∫ L
−L
(
1 + cos 2nπx
L
)
dx= 1
2
(
x+ L
2nπ sin 2nπx
L
) ⏐
⏐
⏐
⏐
L
−L
= L,
and
∫ L
−L
sin2 nπx
L dx= 1
2
∫ L
−L
(
1 − cos 2nπx
L
)
dx= 1
2
(
x− L
2nπ sin 2nπx
L
)
,
⏐
⏐
⏐
⏐
L
−L
= L,
we see from (
11.2.6) that the Fourier series of f on [−L,L ] is
a0 +
∞∑
n=1
(
an cos nπx
L + bn sin nπx
L
)
,
where
a0 = 1
2L
∫ L
−L
f(x) dx,
an = 1
L
∫ L
−L
f(x) cos nπx
L dx, and bn = 1
L
∫ L
−L
f(x) sin nπx
L dx,n ≥ 1.
Note that a0 is the average value of f on [−L,L ], while an and bn (for n ≥ 1) are twice the average
values of
f(x) cos nπx
L and f(x) sin nπx
L
on [−L,L ], respectively .",Elementary Differential Equations with Boundary Value Problems.pdf
"values of
f(x) cos nπx
L and f(x) sin nπx
L
on [−L,L ], respectively .
Convergence of Fourier Series
The question of convergence of Fourier series for arbitrary integrable functions is beyond the scope of
this book. However, we can state a theorem that settles this q uestion for most functions that arise in
applications.
Definition 11.2.3
A function f is said to be piecewise smooth on [a,b ] if:
(a) f has at most finitely many points of discontinuity in (a,b );
(b) f′exists and is continuous except possibly at finitely many poi nts in (a,b );
(c) f(x0+) = lim x→ x0+ f(x) and f′(x0+) = lim x→ x0+ f′(x) exist if a≤ x0 <b;
(d) f(x0−) = lim x→ x0− f(x) and f′(x0−) = lim x→ x0− f′(x) exist if a<x 0 ≤ b.
Since f and f′are required to be continuous at all but finitely many points i n [a,b ], f(x0+) = f(x0−)
and f′(x0+) = f′(x0−) for all but finitely many values of x0 in (a,b ). Recall from Section 8.1 that f is
said to have a jump discontinuity at x0 if f(x0 +) ̸= f(x0−).",Elementary Differential Equations with Boundary Value Problems.pdf
"said to have a jump discontinuity at x0 if f(x0 +) ̸= f(x0−).
The next theorem gives sufficient conditions for convergenc e of a Fourier series. The proof is beyond
the scope of this book.",Elementary Differential Equations with Boundary Value Problems.pdf
"590 Chapter 11 Boundary V alue Problems and Fourier Expansions
Theorem 11.2.4 If f is piecewise smooth on [−L,L ], then the F ourier series
F(x) = a0 +
∞∑
n=1
(
an cos nπx
L + bn sin nπx
L
)
(11.2.8)
of f on [−L,L ] converges for all xin [−L,L ]; moreover,
F(x) =











f(x) if −L<x<L and f is continuous at x
f(x−) + f(x+)
2 if −L<x<L and f is discontinuous at x
f(−L+) + f(L−)
2 if x= Lor x= −L.
Since f(x+) = f(x−) if f is continuous at x, we can also say that
F(x) =





f(x+) + f(x−)
2 if − L<x<L,
f(L−) + f(−L+)
2 if x= ±L.
Note that F is itself piecewise smooth on [−L,L ], and F(x) = f(x) at all points in the open interval
(−L,L ) where f is continuous. Since the series in ( 11.2.8) converges to F(x) for all xin [−L,L ], you
may be tempted to infer that the error
EN (x) =
⏐
⏐
⏐
⏐
⏐F(x) − a0 −
N∑
n=1
(
an cos nπx
L + bn sin nπx
L
) ⏐
⏐
⏐
⏐
⏐
can be made as small as we please for all xin [−L,L ] by choosing N sufficiently large. However, this",Elementary Differential Equations with Boundary Value Problems.pdf
"N∑
n=1
(
an cos nπx
L + bn sin nπx
L
) ⏐
⏐
⏐
⏐
⏐
can be made as small as we please for all xin [−L,L ] by choosing N sufficiently large. However, this
isn’t true if f has a discontinuity somewhere in (−L,L ), or if f(−L+) ̸= f(L−). Here’s the situation in
this case.
If f has a jump discontinuity at a point α in (−L,L ), there will be sequences of points {uN }and {vN }
in (−L,α ) and (α,L ), respectively, such that
lim
N →∞
uN = lim
N →∞
vN = α
and
EN (uN ) ≈ .09|f(α−) − f(α+)| and EN (vN ) ≈ .09|f(α−) − f(α+)|.
Thus, the maximum value of the error EN (x) near α does not approach zero as N → ∞ , but just occurs
closer and closer to (and on both sides of ) α, and is essentially independent of N.
If f(−L+) ̸= f(L−), then there will be sequences of points {uN }and {vN }in (−L,L ) such that
lim
N →∞
uN = −L, lim
N →∞
vN = L,
EN (uN ) ≈ .09|f(−L+) − f(L−)| and EN (vN ) ≈ .09|f(−L+) − f(L−)|.
This is the Gibbs phenomenon . Having been alerted to it, you may see it in Figures 11.2.2– 11.2.4,",Elementary Differential Equations with Boundary Value Problems.pdf
"This is the Gibbs phenomenon . Having been alerted to it, you may see it in Figures 11.2.2– 11.2.4,
below; however, we’ll give a specific example at the end of thi s section.
Example 11.2.1 Find the Fourier series of the piecewise smooth function
f(x) =
{ −x, −2 <x< 0,
1
2 , 0 <x< 2
on [−2, 2] (Figure 11.2.1). Determine the sum of the Fourier series for −2 ≤ x≤ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 591
1
2
1 2−1−2
 x
 y
Figure 11.2.1
Solution Note that wen’t bothered to define f(−2), f(0), and f(2). No matter how they may be defined,
f is piecewise smooth on [−2, 2], and the coefficients in the Fourier series
F(x) = a0 +
∞∑
n=1
(
an cos nπx
2 + bn sin nπx
2
)
are not affected by them. In any case, Theorem 11.2.4 implies that F(x) = f(x) in (−2, 0) and (0, 2),
where f is continuous, while
F(−2) = F(2) = f(−2+) + f(2−)
2 = 1
2
(
2 + 1
2
)
= 5
4
and
F(0) = f(0−) + f(0+)
2 = 1
2
(
0 + 1
2
)
= 1
4 .
T o summarize,
F(x) =



















5
4 , x = −2
−x, −2 <x< 0,
1
4 , x = 0 ,
1
2 , 0 <x< 2,
5
4 , x = 2 .
W e compute the Fourier coefficients as follows:
a0 = 1
4
∫ 2
−2
f(x) dx= 1
4
[ ∫ 0
−2
(−x) dx+
∫ 2
0
1
2 dx
]
= 3
4 .",Elementary Differential Equations with Boundary Value Problems.pdf
"592 Chapter 11 Boundary V alue Problems and Fourier Expansions
If n≥ 1, then
an = 1
2
∫ 2
−2
f(x) cos nπx
2 dx= 1
2
[ ∫ 0
−2
(−x) cos nπx
2 dx+
∫ 2
0
1
2 cos nπx
2 dx
]
= 2
n2π2 (cos nπ − 1),
and
bn = 1
2
∫ 2
−2
f(x) sin nπx
2 dx= 1
2
[ ∫ 0
−2
(−x) sin nπx
2 dx+
∫ 2
0
1
2 sin nπx
2 dx
]
= 1
2nπ (1 + 3 cos nπ).
Therefore
F(x) = 3
4 + 2
π2
∞∑
n=1
cos nπ − 1
n2 cos nπx
2 + 1
2π
∞∑
n=1
1 + 3 cos nπ
n sin nπx
2 .
Figure 11.2.2 shows how the partial sum
Fm(x) = 3
4 + 2
π2
m∑
n=1
cos nπ − 1
n2 cos nπx
2 + 1
2π
m∑
n=1
1 + 3 cos nπ
n sin nπx
2
approximates f(x) for m= 5 (dotted curve), m= 10 (dashed curve), and m= 15 (solid curve).
1
2
1 2−1−2
 x
 y
Figure 11.2.2",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 593
Even and Odd Functions
Computing the Fourier coefficients of a function f can be tedious; however, the computation can often be
simplified by exploiting symmetries in f or some of its terms. T o focus on this, we recall some concepts
that you studied in calculus. Let uand vbe defined on [−L,L ] and suppose that
u(−x) = u(x) and v(−x) = −v(x), −L≤ x≤ L.
Then we say that uis an even function and vis an odd function . Note that:
• The product of two even functions is even.
• The product of two odd functions is even.
• The product of an even function and an odd function is odd.
Example 11.2.2 The functions u(x) = cos ωx and u(x) = x2 are even, while v(x) = sin ωx and
v(x) = x3 are odd. The function w(x) = ex is neither even nor odd.
Y ou learned parts (a) and (b) of the next theorem in calculus, and the other parts follow fr om them
(Exercise
1).
Theorem 11.2.5 Suppose uis even and vis odd on [−L,L ]. Then:
(a)
∫ L
−L
u(x) dx= 2
∫ L
0
u(x) dx, (b)
∫ L
−L",Elementary Differential Equations with Boundary Value Problems.pdf
"(Exercise
1).
Theorem 11.2.5 Suppose uis even and vis odd on [−L,L ]. Then:
(a)
∫ L
−L
u(x) dx= 2
∫ L
0
u(x) dx, (b)
∫ L
−L
v(x) dx= 0 ,
(c)
∫ L
−L
u(x) cos nπx
L dx= 2
∫ L
0
u(x) cos nπx
L dx,
(d)
∫ L
−L
v(x) sin nπx
L dx= 2
∫ L
0
v(x) sin nπx
L dx,
(e)
∫ L
−L
u(x) sin nπx
L dx= 0 and (f)
∫ L
−L
v(x) cos nπx
L dx= 0 .
Example 11.2.3 Find the Fourier series of f(x) = x2 − xon [−2, 2], and determine its sum for −2 ≤
x≤ 2.
Solution Since L= 2 ,
F(x) = a0 +
∞∑
n=1
(
an cos nπx
2 + bn sin nπx
2
)
where
a0 = 1
4
∫ 2
−2
(x2 − x) dx, (11.2.9)
an = 1
2
∫ 2
−2
(x2 − x) cos nπx
2 dx, n = 1 , 2, 3,..., (11.2.10)
and
bn = 1
2
∫ 2
−2
(x2 − x) sin nπx
2 dx, n = 1 , 2, 3,.... (11.2.11)",Elementary Differential Equations with Boundary Value Problems.pdf
"594 Chapter 11 Boundary V alue Problems and Fourier Expansions
W e simplify the evaluation of these integrals by using Theor em 11.2.5 with u(x) = x2 and v(x) = x;
thus, from ( 11.2.9),
a0 = 1
2
∫ 2
0
x2 dx= x3
6
⏐
⏐
⏐
⏐
2
0
= 4
3 .
From ( 11.2.10),
an =
∫ 2
0
x2 cos nπx
2 dx= 2
nπ
[
x2 sin nπx
2
⏐
⏐
⏐
⏐
2
0
− 2
∫ 2
0
xsin nπx
2 dx
]
= 8
n2π2
[
xcos nπx
2
⏐
⏐
⏐
⏐
2
0
−
∫ 2
0
cos nπx
2 dx
]
= 8
n2π2
[
2 cos nπ − 2
nπ sin nπx
2
⏐
⏐
⏐
⏐
2
0
]
= ( −1)n 16
n2π2 .
From ( 11.2.11),
bn = −
∫ 2
0
xsin nπx
2 dx= 2
nπ
[
xcos nπx
2
⏐
⏐
⏐
⏐
2
0
−
∫ 2
0
cos nπx
2 dx
]
= 2
nπ
[
2 cos nπ − 2
nπ sin nπx
2
⏐
⏐
⏐
⏐
2
0
]
= ( −1)n 4
nπ .
Therefore
F(x) = 4
3 + 16
π2
∞∑
n=1
(−1)n
n2 cos nπx
2 + 4
π
∞∑
n=1
(−1)n
n sin nπx
2 .
Theorem 11.2.4 implies that
F(x) =



4, x = −2,
x2 − x, −2 <x< 2,
4, x = 2 .
Figure
11.2.3 shows how the partial sum
Fm(x) = 4
3 + 16
π2
m∑
n=1
(−1)n
n2 cos nπx
2 + 4
π
m∑
n=1
(−1)n
n sin nπx
2",Elementary Differential Equations with Boundary Value Problems.pdf
"x2 − x, −2 <x< 2,
4, x = 2 .
Figure
11.2.3 shows how the partial sum
Fm(x) = 4
3 + 16
π2
m∑
n=1
(−1)n
n2 cos nπx
2 + 4
π
m∑
n=1
(−1)n
n sin nπx
2
approximates f(x) for m= 5 (dotted curve), m= 10 (dashed curve), and m= 15 (solid curve).
Theorem 11.2.5 ilmplies the next theorem follows.
Theorem 11.2.6 Suppose f is integrable on [−L,L ].
(a) If f is even , the F ourier series of f on [−L,L ] is
F(x) = a0 +
∞∑
n=1
an cos nπx
L ,
where
a0 = 1
L
∫ L
0
f(x) dx and an = 2
L
∫ L
0
f(x) cos nπx
L dx, n ≥ 1.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 595
1−1 2−2
1
2
3
4
5
6
 x
 y
Figure 11.2.3 Approximation of f(x) = x2 − xby partial sums of its Fourier series on [−2, 2]
(b) If f is odd , the F ourier series of f on [−L,L ] is
F(x) =
∞∑
n=1
bn sin nπx
L ,
where
bn = 2
L
∫ L
0
f(x) sin nπx
L dx.
Example 11.2.4 Find the Fourier series of f(x) = xon [−π,π ], and determine its sum for −π ≤ x≤ π.
Solution Since f is odd and L= π,
F(x) =
∞∑
n=1
bn sin nx
where
bn = 2
π
∫ π
0
xsin nxdx = − 2
nπ
[
xcos nx
⏐
⏐
⏐
⏐
π
0
−
∫ π
0
cos nxdx
]
= − 2
ncos nπ + 2
n2π sin nx
⏐
⏐
⏐
⏐
π
0
= ( −1)n+1 2
n.
Therefore
F(x) = −2
∞∑
n=1
(−1)n
n sin nx.",Elementary Differential Equations with Boundary Value Problems.pdf
"596 Chapter 11 Boundary V alue Problems and Fourier Expansions
1 2 3−1−2−3
1
2
3
−1
−2
−3
 x
 y
Figure 11.2.4 Approximation of f(x) = xby partial sums of its Fourier series on [−π,π ]
Theorem 11.2.4 implies that
F(x) =



0, x = −π,
x, −π <x<π,
0, x = π.
Figure
11.2.4 shows how the partial sum
Fm (x) = −2
m∑
n=1
(−1)n
n sin nx
approximates f(x) for m= 5 (dotted curve), m= 10 (dashed curve), and m= 15 (solid curve).
Example 11.2.5 Find the Fourier series of f(x) = |x|on [−π,π ] and determine its sum for
−π ≤ x≤ π.
Solution Since f is even and L= π,
F(x) = a0 +
∞∑
n=1
an cos nx.
Since f(x) = xif x≥ 0,
a0 = 1
π
∫ π
0
xdx = x2
2π
⏐
⏐
⏐
⏐
π
0
= π
2",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 597
and, if n≥ 1,
an = 2
π
∫ π
0
xcos nxdx = 2
nπ
[
xsin nx
⏐
⏐
⏐
⏐
π
0
−
∫ π
0
sin nxdx
]
= 2
n2π cos nx
⏐
⏐
⏐
⏐
π
0
= 2
n2π (cos nπ − 1) = 2
n2π [(−1)n − 1].
Therefore
F(x) = π
2 + 2
π
∑
n=0
(−1)n − 1
n2 cos nx. (11.2.12)
However, since
(−1)n − 1 =
{ 0 if n= 2 m,
−2 if n= 2 m+ 1 ,
the terms in ( 11.2.12) for which n= 2 mare all zeros. Therefore we only to include the terms for whic h
n= 2 m+ 1 ; that is, we can rewrite ( 11.2.12) as
F(x) = π
2 − 4
π
∞∑
m=0
1
(2m+ 1) 2 cos(2m+ 1) x.
However, since the name of the index of summation doesn’t mat ter, we prefer to replace m by n, and
write
F(x) = π
2 − 4
π
∞∑
n=0
1
(2n+ 1) 2 cos(2n+ 1) x.
Since |x|is continuous for all xand | −π|= |π|, Theorem 11.2.4 implies that F(x) = |x|for all xin
[−π,π ].
Example 11.2.6 Find the Fourier series of f(x) = x(x2 − L2) on [−L,L ], and determine its sum for
−L≤ x≤ L.
Solution Since f is odd,
F(x) =
∞∑
n=1
bn sin nπx
L ,
where
bn = 2
L
∫ L
0
x(x2 − L2) sin nπx
L dx
= − 2",Elementary Differential Equations with Boundary Value Problems.pdf
"−L≤ x≤ L.
Solution Since f is odd,
F(x) =
∞∑
n=1
bn sin nπx
L ,
where
bn = 2
L
∫ L
0
x(x2 − L2) sin nπx
L dx
= − 2
nπ
[
x(x2 − L2) cos nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
(3x2 − L2) cos nπx
L dx
]
= 2L
n2π2
[
(3x2 − L2) sin nπx
L
⏐
⏐
⏐
⏐
L
0
− 6
∫ L
0
xsin nπx
L dx
]
= 12L2
n3π3
[
xcos nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
cos nπx
L dx
]
= ( −1)n 12L3
n3π3 .
Therefore
F(x) = 12L3
π3
∞∑
n=1
(−1)n
n3 sin nπx
L .
Theorem 11.2.4 implies that F(x) = x(x2 − L2) for all xin [−L,L ].",Elementary Differential Equations with Boundary Value Problems.pdf
"598 Chapter 11 Boundary V alue Problems and Fourier Expansions
Example 11.2.7 (Gibbs Phenomenon) The Fourier series of
f(x) =





0, −1 <x< −1
2 ,
1, −1
2 <x< 1
2 ,
0, 1
2 <x< 1
on [−1, 1] is
F(x) = 1
2 + 2
π
∞∑
n=1
(−1)n−1
2n− 1 cos(2n− 1)πx.
(V erify .) According to Theorem 11.2.4,
F(x) =











0, −1 ≤ x< −1
2 ,
1
2 , x = −1
2 ,
1, −1
2 <x< 1
2 ,
1
2 , x = 1
2 ,
0, 1
2 <x ≤ 1;
thus, F (as well as f) has unit jump discontinuities at x= ±1
2 . Figures 11.2.6-11.2.7 show the graphs of
y = f(x) and
y = F2N −1(x) = 1
2 + 2
π
N∑
n=1
(−1)n−1
2n− 1 cos(2n− 1)πx
for N = 10 , 20, and 30. Y ou can see that although F2N −1 approximates F (and therefore f) well on
larger intervals as N increases, the maximum absolute values of the errors remain approximately equal
to .09, but occur closer to the discontinuities x= ±1
2 as N increases.
1−1
 y = 1.00
 y = 1.09
 y = − 0.09
 x
 y
Figure 11.2.5 The Gibbs Phenomenon:
Example 11.2.7, N = 10
1−1
 y = 1.00
 y = 1.09
 y = − 0.09
 x
 y",Elementary Differential Equations with Boundary Value Problems.pdf
"1−1
 y = 1.00
 y = 1.09
 y = − 0.09
 x
 y
Figure 11.2.5 The Gibbs Phenomenon:
Example 11.2.7, N = 10
1−1
 y = 1.00
 y = 1.09
 y = − 0.09
 x
 y
Figure 11.2.6 The Gibbs Phenomenon:
Example 11.2.7, N = 20",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 599
1−1
 y = 1.00
 y = 1.09
 y = − .09
 x
 y
Figure 11.2.7 The Gibbs Phenomenon: Example 11.2.7, N = 30
USING TECHNOLOGY
The computation of Fourier coefficients will be tedious in ma ny of the exercises in this chapter and
the next. T o learn the technique, we recommend that you do som e exercises in each section “by hand, ”
perhaps using the table of integrals at the front of the book. However, we encourage you to use your
favorite symbolic computation software in the more difficul t problems.
11.2 Exercises
1. Prove Theorem 11.1.5.
In Exercises 2-16 find the F ourier series of f on [−L,L ] and determine its sum for −L≤ x≤ L. Where
indicated by C , graph f and
Fm(x) = a0 +
m∑
n=1
(
an cos nπx
L + bn sin nπx
L
)
on the same axes for various values of m.
2. C L= 1 ; f(x) = 2 − x
3. L= π; f(x) = 2 x− 3x2
4. L= 1 ; f(x) = 1 − 3x2
5. L= π; f(x) = |sin x|",Elementary Differential Equations with Boundary Value Problems.pdf
"600 Chapter 11 Boundary V alue Problems and Fourier Expansions
6. C L= π; f(x) = xcos x
7. L= π; f(x) = |x|cos x
8. C L= π; f(x) = xsin x
9. L= π; f(x) = |x|sin x
10. L= 1 ; f(x) =







0, −1 <x< 1
2 ,
cos πx, −1
2 <x< 1
2 ,
0, 1
2 <x< 1
11. L= 1 ; f(x) =







0, −1 <x< 1
2 ,
xcos πx, −1
2 <x< 1
2 ,
0, 1
2 <x< 1
12. L= 1 ; f(x) =







0, −1 <x< 1
2 ,
sin πx, −1
2 <x< 1
2 ,
0, 1
2 <x< 1
13. L= 1 ; f(x) =







0, −1 <x< 1
2 ,
|sin πx|, −1
2 <x< 1
2 ,
0, 1
2 <x< 1
14. L= 1 ; f(x) =







0, −1 <x< 1
2 ,
xsin πx, −1
2 <x< 1
2 ,
0, 1
2 <x< 1
15. C L= 4 ; f(x) =
{ 0, −4 <x< 0,
x, 0 <x< 4
16. C L= 1 ; f(x) =
{ x2, −1 <x< 0,
1 − x2, 0 <x< 1
17. L V erify the Gibbs phenomenon for f(x) =



2, −2 <x< −1,
1, −1 <x< 1,
−1, 1 <x< 2.
18.
L V erify the Gibbs phenomenon for f(x) =



2, −3 <x< −2,
3, −2 <x< 2,
1, 2 <x< 3.
19. Deduce from Example
11.2.5 that
∞∑
n=0
1
(2n+ 1) 2 = π2
8 .
20. (a) Find the Fourier series of f(x) = ex on [−π,π ].",Elementary Differential Equations with Boundary Value Problems.pdf
"3, −2 <x< 2,
1, 2 <x< 3.
19. Deduce from Example
11.2.5 that
∞∑
n=0
1
(2n+ 1) 2 = π2
8 .
20. (a) Find the Fourier series of f(x) = ex on [−π,π ].
(b) Deduce from (a) that
∞∑
n=0
1
n2 + 1 = π coth π − 1
2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.2 Fourier Expansions I 601
21. Find the Fourier series of f(x) = ( x− π) cos xon [−π,π ].
22. Find the Fourier series of f(x) = ( x− π) sin xon [−π,π ].
23. Find the Fourier series of f(x) = sin kx(k̸= integer) on [−π,π ].
24. Find the Fourier series of f(x) = cos kx(k̸= integer) on [−π,π ].
25. (a) Suppose g′is continuous on [a,b ] and ω ̸= 0 . Use integration by parts to show that there’s a
constant M such that
⏐
⏐
⏐
⏐
⏐
∫ b
a
g(x) cos ωxdx
⏐
⏐
⏐
⏐
⏐≤ M
ω and
⏐
⏐
⏐
⏐
⏐
∫ b
a
g(x) sin ωxdx
⏐
⏐
⏐
⏐
⏐≤ M
ω , ω > 0.
(b) Show that the conclusion of (a) also holds if gis piecewise smooth on [a,b ]. (This is a special
case of Riemann’s Lemma .
(c) W e say that a sequence {αn}∞
n=1 is
of order n−k and write αn = O(1/n k) if there’s a
constant M such that
|αn|< M
nk , n = 1 , 2, 3,....
Let {an}∞
n=1 and {bn }∞
n=1 be the Fourier coefficients of a piecewise smooth function. C on-
clude from (b) that an = O(1/n ) and bn = O(1/n ).",Elementary Differential Equations with Boundary Value Problems.pdf
"Let {an}∞
n=1 and {bn }∞
n=1 be the Fourier coefficients of a piecewise smooth function. C on-
clude from (b) that an = O(1/n ) and bn = O(1/n ).
26. (a) Suppose f(−L) = f(L), f′(−L) = f′(L), f′is continuous, and f′′is piecewise continuous
on [−L,L ]. Use Theorem
11.2.4 and integration by parts to show that
f(x) = a0 +
∞∑
n=1
(
an cos nπx
L + bn sin nπx
L
)
, −L≤ x≤ L,
with
a0 = 1
2L
∫ L
−L
f(x) dx,
an = − L
n2π2
∫ L
−L
f′′(x) cos nπx
L dx, and bn = − L
n2π2
∫ L
−L
f′′(x) sin nπx
L dx,n ≥ 1.
(b) Show that if, in addition to the assumptions in (a), f′′is continuous and f′′′is piecewise
continuous on [−L,L ], then
an = L2
n3π3
∫ L
−L
f′′′(x) sin nπx
L dx.
27. Show that if f is integrable on [−L,L ] and
f(x+ L) = f(x), −L<x< 0
(Figure 11.2.8), then the Fourier series of f on [−L,L ] has the form
A0 +
∞∑
n=1
(
An cos 2nπ
L + Bn sin 2nπ
L
)
where
A0 = 1
L
∫ L
0
f(x) dx,",Elementary Differential Equations with Boundary Value Problems.pdf
"602 Chapter 11 Boundary V alue Problems and Fourier Expansions
and
An = 2
L
∫ L
0
f(x) cos 2nπx
L dx, B n = 2
L
∫ L
0
f(x) sin 2nπx
L dx, n = 1 , 2, 3,....
 L − L  x
 y
Figure 11.2.8 y= f(x), where f(x+ L) = f(x),
−L<x< 0
 − L  L  x
 y
Figure 11.2.9 y= f(x), where f(x+ L) = −f(x),
−L<x< 0
28. Show that if f is integrable on [−L,L ] and
f(x+ L) = −f(x), −L<x< 0
(Figure 11.2.9), then the Fourier series of f on [−L,L ] has the form
∞∑
n=1
(
An cos (2n− 1)πx
L + Bn sin (2n− 1)πx
L
)
,
where
An = 2
L
∫ L
0
f(x) cos (2n− 1)πx
L dx and Bn = 2
L
∫ L
0
f(x) sin (2n− 1)πx
L dx, n = 1 , 2, 3,....
29. Suppose φ1, φ2, . . . , φm are orthogonal on [a,b ] and
∫ b
a
φ2
n(x) dx̸= 0 , n = 1 , 2,...,m.
If a1, a2, . . . , am are arbitrary real numbers, define
Pm = a1φ1 + a2φ2 + · · ·+ amφm .
Let
Fm = c1φ1 + c2φ2 + · · ·+ cmφm ,
where
cn =
∫b
a f(x)φn (x) dx
∫b
a φ2
n(x) dx
;
that is, c1, c2, . . . , cm are Fourier coefficients of f.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 603
(a) Show that ∫ b
a
(f(x) − Fm(x))φn (x) dx= 0 , n = 1 , 2,...,m.
(b) Show that ∫ b
a
(f(x) − Fm(x))2 dx≤
∫ b
a
(f(x) − Pm(x))2 dx,
with equality if and only if an = cn , n= 1 , 2,...,m .
(c) Show that ∫ b
a
(f(x) − Fm(x))2 dx=
∫ b
a
f2 (x) dx−
m∑
n=1
c2
n
∫ b
a
φ2
n dx.
(d) Conclude from (c) that
m∑
n=1
c2
n
∫ b
a
φ2
n(x) dx≤
∫ b
a
f2(x) dx.
30. If A0, A1, . . . , Am and B1, B2 , . . . , Bm are arbitrary constants we say that
Pm(x) = A0 +
m∑
n=1
(
An cos nπx
L + Bn sin nπx
L
)
is a trigonometric polynomial of degree ≤ m.
Now let
a0 +
∞∑
n=1
(
an cos nπx
L + bn sin nπx
L
)
be the Fourier series of an integrable function f on [−L,L ], and let
Fm(x) = a0 +
m∑
n=1
(
an cos nπx
L + bn sin nπx
L
)
.
(a) Conclude from Exercise 29(b) that
∫ L
−L
(f(x) − Fm(x))2 dx≤
∫ L
−L
(f(x) − Pm(x))2 dx,
with equality if and only if An = an , n= 0 , 1, . . . , m, and Bn = bn , n= 1 , 2, . . . , m.
(b) Conclude from Exercise 29(d) that
2a2
0 +
m∑
n=1
(a2
n + b2",Elementary Differential Equations with Boundary Value Problems.pdf
"(b) Conclude from Exercise 29(d) that
2a2
0 +
m∑
n=1
(a2
n + b2
n ) ≤ 1
L
∫ L
−L
f2 (x) dx
for every m≥ 0.
(c) Conclude from (b) that limn→∞ an = lim n→∞ bn = 0 .
11.3 FOURIER EXP ANSIONS II",Elementary Differential Equations with Boundary Value Problems.pdf
"604 Chapter 11 Boundary V alue Problems and Fourier Expansions
In this section we discuss Fourier expansions in terms of the eigenfunctions of Problems 1-4 for Sec-
tion 11.1.
Fourier Cosine Series
From Exercise 11.1. 20, the eigenfunctions
1, cos πx
L , cos 2πx
L ,..., cos nπx
L ,...
of the boundary value problem
y′′+ λy = 0 , y ′(0) = 0 , y ′(L) = 0 (11.3.1)
(Problem 2) are orthogonal on [0,L ]. If f is integrable on [0,L ] then the Fourier expansion of f in terms
of these functions is called the F ourier cosine series of f on [0,L ]. This series is
a0 +
∞∑
n=1
an cos nπx
L ,
where
a0 =
∫ L
0
f(x) dx
∫ L
0
dx
= 1
L
∫ L
0
f(x) dx
and
an =
∫ L
0
f(x) cos nπx
L dx
∫ L
0
cos2 nπx
L dx
= 2
L
∫ L
0
f(x) cos nπx
L dx, n = 1 , 2, 3,....
Comparing this definition with Theorem 6(a) shows that the Fourier cosine series of f on [0,L ] is the
Fourier series of the function
f1(x) =
{
f(−x), −L<x< 0,
f(x), 0 ≤ x≤ L,
obtained by extending f over [−L,L ] as an even function (Figure 11.3.1).",Elementary Differential Equations with Boundary Value Problems.pdf
"Fourier series of the function
f1(x) =
{
f(−x), −L<x< 0,
f(x), 0 ≤ x≤ L,
obtained by extending f over [−L,L ] as an even function (Figure 11.3.1).
Applying Theorem 11.2.4 to f1 yields the next theorem.
Theorem 11.3.1 If f is piecewise smooth on [0,L ], then the F ourier cosine series
C(x) = a0 +
∞∑
n=1
an cos nπx
L
of f on [0,L ], with
a0 = 1
L
∫ L
0
f(x) dx and an = 2
L
∫ L
0
f(x) cos nπx
L dx, n = 1 , 2, 3,...,
converges for all xin [0,L ]; moreover,
C(x) =













f(0+) if x= 0
f(x) if 0 <x<L and f is continuous at x
f(x−) + f(x+)
2 if 0 <x<L and f is discontinuous at x
f(L−) if x= L.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 605
 L − L
 y = f(x) y = f(−x)
 x
 y
Figure 11.3.1
Example 11.3.1 Find the Fourier cosine series of f(x) = xon [0,L ].
Solution The coefficients are
a0 = 1
L
∫ L
0
xdx = 1
L
x2
2
⏐
⏐
⏐
⏐
L
0
= L
2
and, if n≥ 1
an = 2
L
∫ L
0
xcos nπx
L dx= 2
nπ
[
xsin nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
sin nπx
L dx
]
= − 2
nπ
∫ L
0
sin nπx
L dx= 2L
n2π2 cos nπx
L
⏐
⏐
⏐
⏐
L
0
= 2L
n2π2 [(−1)n − 1]
=



− 4L
(2m− 1)2π2 if n= 2 m− 1,
0 if n= 2 m.
Therefore
C(x) = L
2 − 4L
π2
∞∑
n=1
1
(2n− 1)2 cos (2n− 1)πx
L .
Theorem 11.3.1 implies that
C(x) = x, 0 ≤ x≤ L.
Fourier Sine Series",Elementary Differential Equations with Boundary Value Problems.pdf
"606 Chapter 11 Boundary V alue Problems and Fourier Expansions
From Exercise 11.1. 19, the eigenfunctions
sin πx
L , sin 2πx
L ,..., sin nπx
L ,...
of the boundary value problem
y′′+ λy = 0 , y (0) = 0 , y (L) = 0
(Problem 1) are orthogonal on [0,L ]. If f is integrable on [0,L ] then the Fourier expansion of f in terms
of these functions is called the F ourier sine series of f on [0,L ]. This series is
∞∑
n=1
bn sin nπx
L ,
where
bn =
∫ L
0
f(x) sin nπx
L dx
∫ L
0
sin2 nπx
L dx
= 2
L
∫ L
0
f(x) sin nπx
L dx, n = 1 , 2, 3,....
Comparing this definition with Theorem 6(b) shows that the Fourier sine series of f on [0,L ] is the
Fourier series of the function
f2(x) =
{
−f(−x), −L<x< 0,
f(x), 0 ≤ x≤ L,
obtained by extending f over [−L,L ] as an odd function (Figure 11.3.2).
Applying Theorem 11.2.4 to f2 yields the next theorem.
Theorem 11.3.2 If f is piecewise smooth on [0,L ], then the F ourier sine series
S(x) =
∞∑
n=1
bn sin nπx
L
of f on [0,L ], with
bn = 2
L
∫ L
0
f(x) sin nπx
L dx,",Elementary Differential Equations with Boundary Value Problems.pdf
"S(x) =
∞∑
n=1
bn sin nπx
L
of f on [0,L ], with
bn = 2
L
∫ L
0
f(x) sin nπx
L dx,
converges for all xin [0,L ]; moreover,
S(x) =













0 if x= 0
f(x) if 0 <x<L and f is continuous at x
f(x−) + f(x+)
2 if 0 <x<L and f is discontinuous at x
0 if x= L.
Example 11.3.2 Find the Fourier sine series of f(x) = xon [0,L ].",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 607
 L − L
 y = f(x)
 y = − f(−x)
 x
 y
Figure 11.3.2
Solution The coefficients are
bn = 2
L
∫ L
0
xsin nπx
L dx= − 2
nπ
[
xcos nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
cos nπx
L dx
]
= ( −1)n+1 2L
nπ + 2L
n2π2 sin nπx
L
⏐
⏐
⏐
⏐
L
0
= ( −1)n+1 2L
nπ .
Therefore
S(x) = −2L
π
∞∑
n=1
(−1)n
n sin nπx
L .
Theorem 11.3.2 implies that
S(x) =
{ x, 0 ≤ x<L,
0, x = L.
Mixed Fourier Cosine Series
From Exercise 11.1. 22, the eigenfunctions
cos πx
2L, cos 3πx
2L ,..., cos (2n− 1)πx
2L ,...
of the boundary value problem
y′′+ λy = 0 , y ′(0) = 0 , y (L) = 0 (11.3.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"608 Chapter 11 Boundary V alue Problems and Fourier Expansions
(Problem 4) are orthogonal on [0,L ]. If f is integrable on [0,L ] then the Fourier expansion of f in terms
of these functions is ∞∑
n=1
cn cos (2n− 1)πx
2L ,
where
cn =
∫ L
0
f(x) cos (2n− 1)πx
2L dx
∫ L
0
cos2 (2n− 1)πx
L dx
= 2
L
∫ L
0
f(x) cos (2n− 1)πx
2L dx.
W e’ll call this expansion the mixed F ourier cosine series of f on [0,L ], because the boundary condi-
tions of ( 11.3.2) are “mixed” in that they require yto be zero at one boundary point and y′to be zero at the
other. By contrast, the “ordinary” Fourier cosine series is associated with ( 11.3.1), where the boundary
conditions require that y′be zero at both endpoints.
It can be shown (Exercise 57) that the mixed Fourier cosine series of f on [0,L ] is simply the restriction
to [0,L ] of the Fourier cosine series of
f3(x) =
{ f(x), 0 ≤ x≤ L,
−f(2L− x), L<x ≤ 2L
on [0, 2L] (Figure 11.3.3).
 y = f(x)
 L  2L
 y = − f(2L−x)
 x
 y
Figure 11.3.3",Elementary Differential Equations with Boundary Value Problems.pdf
"f3(x) =
{ f(x), 0 ≤ x≤ L,
−f(2L− x), L<x ≤ 2L
on [0, 2L] (Figure 11.3.3).
 y = f(x)
 L  2L
 y = − f(2L−x)
 x
 y
Figure 11.3.3
Applying Theorem 11.3.1 with f replaced by f3 and Lreplaced by 2Lyields the next theorem.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 609
Theorem 11.3.3 If f is piecewise smooth on [0,L ], then the mixed F ourier cosine series
CM (x) =
∞∑
n=1
cn cos (2n− 1)πx
2L
of f on [0,L ], with
cn = 2
L
∫ L
0
f(x) cos (2n− 1)πx
2L dx,
converges for all xin [0,L ]; moreover,
CM (x) =













f(0+) if x= 0
f(x) if 0 <x<L and f is continuous at x
f(x−) + f(x+)
2 if 0 <x<L and f is discontinuous at x
0 if x= L.
Example 11.3.3 Find the mixed Fourier cosine series of f(x) = x− Lon [0,L ].
Solution The coefficients are
cn = 2
L
∫ L
0
(x− L) cos (2n− 1)πx
2L dx
= 4
(2n− 1)π
[
(x− L) sin (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
sin (2n− 1)πx
2L dx
]
= 8L
(2n− 1)2π2 cos (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
= − 8L
(2n− 1)2π2 .
Therefore
CM (x) = −8L
π2
∞∑
n=1
1
(2n− 1)2 cos (2n− 1)πx
2L .
Theorem 11.3.3 implies that
CM (x) = x− L, 0 ≤ x≤ L.
Mixed Fourier Sine Series
From Exercise 11.1. 21, the eigenfunctions
sin πx
2L, sin 3πx
2L ,..., sin (2n− 1)πx
2L ,...
of the boundary value problem",Elementary Differential Equations with Boundary Value Problems.pdf
"Mixed Fourier Sine Series
From Exercise 11.1. 21, the eigenfunctions
sin πx
2L, sin 3πx
2L ,..., sin (2n− 1)πx
2L ,...
of the boundary value problem
y′′+ λy = 0 , y (0) = 0 , y ′(L) = 0
(Problem 3) are orthogonal on [0,L ]. If f is integrable on [0,L ], then the Fourier expansion of f in terms
of these functions is ∞∑
n=1
dn sin (2n− 1)πx
2L ,",Elementary Differential Equations with Boundary Value Problems.pdf
"610 Chapter 11 Boundary V alue Problems and Fourier Expansions
where
dn =
∫ L
0
f(x) sin (2n− 1)πx
2L dx
∫ L
0
sin2 (2n− 1)πx
2L dx
= 2
L
∫ L
0
f(x) sin (2n− 1)πx
2L dx.
W e’ll call this expansion the mixed F ourier sine series of f on [0,L ].
It can be shown (Exercise 58) that the mixed Fourier sine series of f on [0,L ] is simply the restriction
to [0,L ] of the Fourier sine series of
f4(x) =
{ f(x), 0 ≤ x≤ L,
f(2L− x), L<x ≤ 2L,
on [0, 2L] (Figure 11.3.4).
 L  2L
 y = f(x)  y = f(2L−x)
 x
 y
Figure 11.3.4
Applying Theorem 11.3.2 with f replaced by f4 and Lreplaced by 2Lyields the next theorem.
Theorem 11.3.4 If f is piecewise smooth on [0,L ], then the mixed F ourier sine series
SM (x) =
∞∑
n=1
dn sin (2n− 1)πx
2L
of f on [0,L ], with
dn = 2
L
∫ L
0
f(x) sin (2n− 1)πx
2L dx,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 611
converges for all xin [0,L ]; moreover,
SM (x) =













0 if x= 0
f(x) if 0 <x<L and f is continuous at x
f(x−) + f(x+)
2 if 0 <x<L and f is discontinuous at x
f(L−) if x= L.
Example 11.3.4 Find the mixed Fourier sine series of f(x) = xon [0,L ].
Solution The coefficients are
dn = 2
L
∫ L
0
xsin (2n− 1)πx
2L dx
= − 4
(2n− 1)π
[
xcos (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
cos (2n− 1)πx
2L dx
]
= 4
(2n− 1)π
∫ L
0
cos (2n− 1)πx
2L dx
= 8L
(2n− 1)2π2 sin (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
= ( −1)n+1 8L
(2n− 1)2π2 .
Therefore
SM (x) = −8L
π2
∞∑
n=1
(−1)n
(2n− 1)2 sin (2n− 1)πx
2L .
Theorem 11.3.4 implies that
SM (x) = x, 0 ≤ x≤ L.
A Useful Observation
In applications involving expansions in terms of the eigenf unctions of Problems 1-4, the functions being
expanded are often polynomials that satisfy the boundary co nditions of the problem under consideration.",Elementary Differential Equations with Boundary Value Problems.pdf
"expanded are often polynomials that satisfy the boundary co nditions of the problem under consideration.
In this case the next theorem presents an efficient way to obta in the coefficients in the expansion.
Theorem 11.3.5
(a) If f′(0) = f′(L) = 0 , f′′is continuous , and f′′′is piecewise continuous on [0,L ], then
f(x) = a0 +
∞∑
n=1
an cos nπx
L , 0 ≤ x≤ L, (11.3.3)
with
a0 = 1
L
∫ L
0
f(x) dx and an = 2L2
n3π3
∫ L
0
f′′′(x) sin nπx
L dx, n ≥ 1. (11.3.4)
Now suppose f′is continuous and f′′is piecewise continuous on [0,L ].
(b) If f(0) = f(L) = 0 , then
f(x) =
∞∑
n=1
bn sin nπx
L , 0 ≤ x≤ L,",Elementary Differential Equations with Boundary Value Problems.pdf
"612 Chapter 11 Boundary V alue Problems and Fourier Expansions
with
bn = − 2L
n2π2
∫ L
0
f′′(x) sin nπx
L dx. (11.3.5)
(c) If f′(0) = f(L) = 0 , then
f(x) =
∞∑
n=1
cn cos (2n− 1)πx
2L , 0 ≤ x≤ L,
with
cn = − 8L
(2n− 1)2π2
∫ L
0
f′′(x) cos (2n− 1)πx
2L dx. (11.3.6)
(d) If f(0) = f′(L) = 0 , then
f(x) =
∞∑
n=1
dn sin (2n− 1)πx
2L , 0 ≤ x≤ L,
with
dn = − 8L
(2n− 1)2π2
∫ L
0
f′′(x) sin (2n− 1)πx
2L dx. (11.3.7)
Proof W e’ll prove (a) and leave the rest to you (Exercises 35, 42, and 50). Since f is continuous on
[0,L ], Theorem 11.3.1 implies ( 11.3.3) with a0, a1, a2,... as defined in Theorem 11.3.1. W e already know
that a0 is as in ( 11.3.4). If n≥ 1, integrating twice by parts yields
an = 2
L
∫ L
0
f(x) cos nπx
L dx
= 2
nπ
[
f(x) sin nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
f′(x) sin nπx
L dx
]
= − 2
nπ
∫ L
0
f′(x) sin nπx
L dx(since sin 0 = sin nπ = 0 )
= 2L
n2π2
[
f′(x) cos nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
f′′(x) cos nπx
L
]
dx
= − 2L
n2π2
∫ L
0
f′′(x) cos nπx
L dx(since f′(0) = f′(L) = 0 )
= − 2L2
n3π3
[",Elementary Differential Equations with Boundary Value Problems.pdf
"= 2L
n2π2
[
f′(x) cos nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
f′′(x) cos nπx
L
]
dx
= − 2L
n2π2
∫ L
0
f′′(x) cos nπx
L dx(since f′(0) = f′(L) = 0 )
= − 2L2
n3π3
[
f′′(x) sin nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
f′′′(x) sin nπx
L dx
]
= 2L2
n3π3
∫ L
0
f′′′(x) sin nπx
L dx(since sin 0 = sin nπ = 0 ).
(By an argument similar to one used in the proof of Theorem 8.3 .1, the last integration by parts is le-
gitimate in the case where f′′′is undefined at finitely many points in [0,L ], so long as it’s piecewise
continuous on [0,L ].) This completes the proof.
Example 11.3.5 Find the Fourier cosine expansion of f(x) = x2(3L− 2x) on [0,L ].",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 613
Solution Here
a0 = 1
L
∫ L
0
(3Lx2 − 2x3) dx= 1
L
(
Lx3 − x4
2
) ⏐
⏐
⏐
⏐
L
0
= L3
2
and
an = 2
L
∫ L
0
(3Lx2 − 2x3) cos nπx
L dx, n ≥ 1.
Evaluating this integral directly is laborious. However, s ince f′(x) = 6 Lx− 6x2, we see that f′(0) =
f′(L) = 0 . Since f′′′(x) = −12, we see from ( 11.3.4) that if n≥ 1 then
an = −24L2
n3π3
∫ L
0
sin nπx
L dx= 24L3
n4π4 cos nπx
L
⏐
⏐
⏐
⏐
L
0
= 24L3
n4π4 [(−1)n − 1]
=



− 48L3
(2m− 1)4π4 if n= 2 m− 1,
0 if n= 2 m.
Therefore
C(x) = L3
2 − 48L3
π4
∞∑
n=1
1
(2n− 1)4 cos (2n− 1)πx
L .
Example 11.3.6 Find the Fourier sine expansion of f(x) = x(x2 − 3Lx+ 2 L2) on [0,L ].
Solution Since f(0) = f(L) = 0 and f′′(x) = 6( x− L), we see from ( 11.3.5) that
bn = − 12L
n2π2
∫ L
0
(x− L) sin nπx
L dx
= 12L2
n3π3
[
(x− L) cos nπx
L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
cos nπx
L dx
]
= 12L2
n3π3
[
L− L
nπ sin nπx
L
⏐
⏐
⏐
⏐
L
0
]
= 12L3
n3π3 .
Therefore
S(x) = 12L3
π3
∞∑
n=1
1
n3 sin nπx
L .",Elementary Differential Equations with Boundary Value Problems.pdf
"L
⏐
⏐
⏐
⏐
L
0
−
∫ L
0
cos nπx
L dx
]
= 12L2
n3π3
[
L− L
nπ sin nπx
L
⏐
⏐
⏐
⏐
L
0
]
= 12L3
n3π3 .
Therefore
S(x) = 12L3
π3
∞∑
n=1
1
n3 sin nπx
L .
Example 11.3.7 Find the mixed Fourier cosine expansion of f(x) = 3 x3 − 4Lx2 + L3 on [0,L ].
Solution Since f′(0) = f(L) = 0 and f′′(x) = 2(9 x− 4L), we see from ( 11.3.6) that
cn = − 16L
(2n− 1)2π2
∫ L
0
(9x− 4L) cos (2n− 1)πx
2L dx
= − 32L2
(2n− 1)3π3
[
(9x− 4L) sin (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
− 9
∫ L
0
sin (2n− 1)πx
2L
]
dx
= − 32L2
(2n− 1)3π3
[
(−1)n+15L+ 18L
(2n− 1)π cos (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
]
= 32L3
(2n− 1)3π3
[
(−1)n 5 + 18
(2n− 1)π
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"614 Chapter 11 Boundary V alue Problems and Fourier Expansions
Therefore
CM (x) = 32L3
π3
∞∑
n=1
1
(2n− 1)3
[
(−1)n 5 + 18
(2n− 1)π
]
cos (2n− 1)πx
2L .
Example 11.3.8 Find the mixed Fourier sine expansion of
f(x) = x(2x2 − 9Lx+ 12 L2)
on [0,L ].
Solution Since f(0) = f′(L) = 0 , and f′′(x) = 6(2 x− 3L), we see from ( 11.3.7) that
dn = − 48L
(2n− 1)2π2
∫ L
0
(2x− 3L) sin (2n− 1)πx
2L dx
= 96L2
(2n− 1)3π3
[
(2x− 3L) cos (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
− 2
∫ L
0
cos (2n− 1)πx
2L dx
]
= 96L2
(2n− 1)3π3
[
3L− 4L
(2n− 1)π sin (2n− 1)πx
2L
⏐
⏐
⏐
⏐
L
0
]
= 96L3
(2n− 1)3π3
[
3 + ( −1)n 4
(2n− 1)π
]
.
Therefore
SM (x) = 96L3
π3
∞∑
n=1
1
(2n− 1)3
[
3 + ( −1)n 4
(2n− 1)π
]
sin (2n− 1)πx
2L .
11.3 Exercises
In exercises marked by C graph f and some partial sums of the required series. If the interval is [0,L ],
choose a specific value of Lfor the graph.
In Exercises 1-10 find the F ourier cosine series.
1. f(x) = x2; [0,L ]
2. C f(x) = 1 − x; [0, 1]
3. C f(x) = x2 − 2Lx; [0,L ]",Elementary Differential Equations with Boundary Value Problems.pdf
"In Exercises 1-10 find the F ourier cosine series.
1. f(x) = x2; [0,L ]
2. C f(x) = 1 − x; [0, 1]
3. C f(x) = x2 − 2Lx; [0,L ]
4. f(x) = sin kx (k̸= integer); [0,π ]
5. C f(x) =
{ 1, 0 ≤ x≤ L
2
0, L
2 <x<L ; [0,L ]
6. f(x) = x2 − L2; [0,L ]
7. f(x) = ( x− 1)2; [0, 1]
8. f(x) = ex; [0,π ]
9. C f(x) = x(L− x); [0,L ]
10. C f(x) = x(x− 2L); [0,L ]",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 615
In Exercises 11-17 find the F ourier sine series.
11. C f(x) = 1 ; [0,L ]
12. C f(x) = 1 − x; [0, 1]
13. f(x) = cos kx (k̸= integer); [0,π ]
14. C f(x) =
{ 1, 0 ≤ x≤ L
2
0, L
2 <x<L ; [0,L ]
15. C f(x) =
{ x, 0 ≤ x≤ L
2 ,
L− x, L
2 ≤ x≤ L; [0,L ].
16. C f(x) = xsin x; [0,π ]
17. f(x) = ex; [0,π ]
In Exercises 18-24 find the mixed F ourier cosine series.
18. C f(x) = 1 ; [0,L ]
19. f(x) = x2; [0,L ]
20. C f(x) = x; [0, 1]
21. C f(x) =
{ 1, 0 ≤ x≤ L
2
0, L
2 <x<L ; [0,L ]
22. f(x) = cos x; [0,π ]
23. f(x) = sin x; [0,π ]
24. C f(x) = x(L− x); [0,L ]
In Exercises 25-30 find the mixed F ourier sine series.
25. C f(x) = 1 ; [0,L ]
26. f(x) = x2; [0,L ]
27. C f(x) =
{ 1, 0 ≤ x≤ L
2
0, L
2 <x<L ; [0,L ]
28. f(x) = cos x; [0,π ]
29. f(x) = sin x; [0,π ]
30. C f(x) = x(L− x); [0,L ].
In Exercises 31-34 use Theorem 11.3.5(a) to find the F ourier cosine series of f on [0,L ].
31. f(x) = 3 x2(x2 − 2L2)
32. f(x) = x3(3x− 4L)
33. f(x) = x2(3x2 − 8Lx+ 6 L2)",Elementary Differential Equations with Boundary Value Problems.pdf
"31. f(x) = 3 x2(x2 − 2L2)
32. f(x) = x3(3x− 4L)
33. f(x) = x2(3x2 − 8Lx+ 6 L2)
34. f(x) = x2(x− L)2
35. (a) Prove Theorem 11.3.5(b).",Elementary Differential Equations with Boundary Value Problems.pdf
"616 Chapter 11 Boundary V alue Problems and Fourier Expansions
(b) In addition to the assumptions of Theorem 11.3.5(b), suppose f′′(0) = f′′(L) = 0 , f′′′is
continuous, and f(4) is piecewise continuous on [0,L ]. Show that
bn = 2L3
n4π4
∫ L
0
f(4) (x) sin nπx
L dx, n ≥ 1.
In Exercises 36-41 use Theorem 11.3.5(b) or , where applicable, Exercise 11.1.35(b), to find the F ourier
sine series of f on [0,L ].
36. C f(x) = x(L− x)
37. C f(x) = x2(L− x)
38. f(x) = x(L2 − x2)
39. f(x) = x(x3 − 2Lx2 + L3)
40. f(x) = x(3x4 − 10L2x2 + 7 L4)
41. f(x) = x(3x4 − 5Lx3 + 2 L4)
42. (a) Prove Theorem 11.3.5(c).
(b) In addition to the assumptions of Theorem 11.3.5(c), suppose f′′(L) = 0 , f′′is continuous,
and f′′′is piecewise continuous on [0,L ]. Show that
cn = 16L2
(2n− 1)3π3
∫ L
0
f′′′(x) sin (2n− 1)πx
2L dx, n ≥ 1.
In Exercises 43-49 use Theorem 11.3.5(c) or , where applicable, Exercise 11.1.42(b), to find the mixed
F ourier cosine series of f on [0,L ].
43. C f(x) = x2(L− x)
44. f(x) = L2 − x2",Elementary Differential Equations with Boundary Value Problems.pdf
"F ourier cosine series of f on [0,L ].
43. C f(x) = x2(L− x)
44. f(x) = L2 − x2
45. f(x) = L3 − x3
46. f(x) = 2 x3 + 3 Lx2 − 5L3
47. f(x) = 4 x3 + 3 Lx2 − 7L3
48. f(x) = x4 − 2Lx3 + L4
49. f(x) = x4 − 4Lx3 + 6 L2x2 − 3L4
50. (a) Prove Theorem 11.3.5(d).
(b) In addition to the assumptions of Theorem 11.3.5(d), suppose f′′(0) = 0 , f′′is continuous,
and f′′′is piecewise continuous on [0,L ]. Show that
dn = − 16L2
(2n− 1)3π3
∫ L
0
f′′′(x) cos (2n− 1)πx
2L dx, n ≥ 1.
In Exercises 51-56 use Theorem 11.3.5(d) or , where applicable, Exercise 50(b), to find the mixed F ourier
sine series of the f on [0,L ].
51. f(x) = x(2L− x)
52. f(x) = x2(3L− 2x)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 11.3 Fourier Expansions II 617
53. f(x) = ( x− L)3 + L3
54. f(x) = x(x2 − 3L2)
55. f(x) = x3(3x− 4L)
56. f(x) = x(x3 − 2Lx2 + 2 L3)
57. Show that the mixed Fourier cosine series of f on [0,L ] is the restriction to [0,L ] of the Fourier
cosine series of
f3(x) =
{ f(x), 0 ≤ x≤ L,
−f(2L− x), L<x ≤ 2L
on [0, 2L]. Use this to prove Theorem 11.3.3.
58. Show that the mixed Fourier sine series of f on [0,L ] is the restriction to [0,L ] of the Fourier sine
series of
f4(x) =
{ f(x), 0 ≤ x≤ L,
f(2L− x), L<x ≤ 2L
on [0, 2L]. Use this to prove Theorem 11.3.4.
59. Show that the Fourier sine series of f on [0,L ] is the restriction to [0,L ] of the Fourier sine series
of
f3(x) =
{
f(x), 0 ≤ x≤ L,
−f(2L− x), L<x ≤ 2L
on [0, 2L].
60. Show that the Fourier cosine series of f on [0,L ] is the restriction to [0,L ] of the Fourier cosine
series of
f4(x) =
{ f(x), 0 ≤ x≤ L,
f(2L− x), L<x ≤ 2L
on [0, 2L].",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 12
Fourier Solutions of Partial Differential
IN THIS CHAPTER we use the series discussed in Chapter 11 to so lve partial differential equations that
arise in problems of mathematical physics.
SECTION 12.1 deals with the partial differential equation
ut = a2uxx,
which arises in problems of conduction of heat.
SECTION 12.2 deals with the partial differential equation
utt = a2uxx,
which arises in the problem of the vibrating string.
SECTION 12.3 deals with the partial differential equation
uxx + uyy = 0 ,
which arises in steady state problems of heat conduction and potential theory .
SECTION 12.4 deals with the partial differential equation
urr + 1
rur + 1
r2 uθθ = 0 ,
which is the equivalent to the equation studied in Section 1. 3 when the independent variables are polar
coordinates.
619",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.1 The Heat Equation 619
12.1 THE HEA T EQUA TION
W e begin the study of partial differential equations with th e problem of heat flow in a uniform bar of
length L, situated on the xaxis with one end at the origin and the other at x= L(Figure 12.1.1).
W e assume that the bar is perfectly insulated except possibl y at its endpoints, and that the temperature
is constant on each cross section and therefore depends only on xand t. W e also assume that the thermal
properties of the bar are independent of x and t. In this case, it can be shown that the temperature
u= u(x,t ) at time tat a point xunits from the origin satisfies the partial differential equ ation
ut = a2uxx, 0 <x<L, t> 0,
where ais a positive constant determined by the thermal properties . This is the heat equation .
 x = 0  x = L
 x
Figure 12.1.1 A uniform bar of length L
T o determine u, we must specify the temperature at every point in the bar whe n t= 0 , say
u(x, 0) = f(x), 0 ≤ x≤ L.",Elementary Differential Equations with Boundary Value Problems.pdf
"T o determine u, we must specify the temperature at every point in the bar whe n t= 0 , say
u(x, 0) = f(x), 0 ≤ x≤ L.
W e call this the initial condition. W e must also specify boundary conditions that umust satisfy at the
ends of the bar for all t> 0. W e’ll call this problem an initial-boundary value problem.
W e begin with the boundary conditions u(0,t ) = u(L,t ) = 0 , and write the initial-boundary value
problem as
ut = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) = f(x), 0 ≤ x≤ L.
(12.1.1)
Our method of solving this problem is called separation of variables (not to be confused with method
of separation of variables used in Section 2.2 for solving or dinary differential equations). W e begin by
looking for functions of the form
v(x,t ) = X(x)T(t)
that are not identically zero and satisfy
vt = a2vxx, v (0,t ) = 0 , v (L,t ) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"620 Chapter 12 Fourier Solutions of Partial Differential
for all (x,t ). Since
vt = XT′ and vxx = X′′T,
vt = a2vxx if and only if
XT′= a2X′′T,
which we rewrite as
T′
a2T = X′′
X .
Since the expression on the left is independent of xwhile the one on the right is independent of t, this
equation can hold for all (x,t ) only if the two sides equal the same constant, which we call a separation
constant, and write it as −λ; thus,
X′′
X = T′
a2T = −λ.
This is equivalent to
X′′+ λX = 0
and
T′= −a2λT. (12.1.2)
Since v(0,t ) = X(0)T(t) = 0 and v(L,t ) = X(L)T(t) = 0 and we don’t want T to be identically zero,
X(0) = 0 and X(L) = 0 . Therefore λ must be an eigenvalue of the boundary value problem
X′′+ λX = 0 , X (0) = 0 , X (L) = 0 , (12.1.3)
and X must be a λ-eigenfunction. From Theorem 11.1.2, the eigenvalues of ( 12.1.3) are λn = n2π2/L 2,
with associated eigenfunctions
Xn = sin nπx
L , n = 1 , 2, 3,....
Substituting λ = n2π2/L 2 into ( 12.1.2) yields
T′= −(n2π2a2/L 2)T,
which has the solution",Elementary Differential Equations with Boundary Value Problems.pdf
"Xn = sin nπx
L , n = 1 , 2, 3,....
Substituting λ = n2π2/L 2 into ( 12.1.2) yields
T′= −(n2π2a2/L 2)T,
which has the solution
Tn = e−n2π 2 a2t/L 2
.
Now let
vn (x,t ) = Xn (x)Tn (t) = e−n2π 2a2 t/L 2
sin nπx
L , n = 1 , 2, 3,...
Since
vn (x, 0) = sin nπx
L ,
vn satisfies ( 12.1.1) with f(x) = sin nπx/L . More generally , if α1,...,α m are constants and
um (x,t ) =
m∑
n=1
αne−n2π 2 a2t/L 2
sin nπx
L ,
then um satisfies ( 12.1.1) with
f(x) =
m∑
n=1
αn sin nπx
L .
This motivates the next definition.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.1 The Heat Equation 621
Definition 12.1.1 The formal solution of the initial-boundary value problem
ut = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) = f(x), 0 ≤ x≤ L
(12.1.4)
is
u(x,t ) =
∞∑
n=1
αne−n2π 2 a2t/L 2
sin nπx
L , (12.1.5)
where
S(x) =
∞∑
n=1
αn sin nπx
L
is the Fourier sine series of f on [0,L ]; that is,
αn = 2
L
∫ L
0
f(x) sin nπx
L dx.
W e use the term “formal solution” in this definition because i t’s not in general true that the infinite
series in ( 12.1.5) actually satisfies all the requirements of the initial-bou ndary value problem ( 12.1.4)
when it does, we say that it’s an actual solution of ( 12.1.4).
Because of the negative exponentials in ( 12.1.5), uconverges for all (x,t ) with t >0 (Exercise 54).
Since each term in ( 12.1.5) satisfies the heat equation and the boundary conditions in ( 12.1.4), ualso has
these properties if ut and uxx can be obtained by differentiating the series in ( 12.1.5) term by term once",Elementary Differential Equations with Boundary Value Problems.pdf
"these properties if ut and uxx can be obtained by differentiating the series in ( 12.1.5) term by term once
with respect to tand twice with respect to x, for t> 0. However, it’s not always legitimate to differentiate
an infinite series term by term. The next theorem gives a usefu l sufficient condition for legitimacy of term
by term differentiation of an infinite series. W e omit the pro of.
Theorem 12.1.2 A convergent infinite series
W(z) =
∞∑
n=1
wn(z)
can be differentiated term by term on a closed interval [z1,z 2] to obtain
W′(z) =
∞∑
n=1
w′
n(z)
(where the derivatives at z = z1 and z = z2 are one-sided ) provided that w′
n is continuous on [z1,z 2]
and
|w′
n(z)| ≤Mn , z 1 ≤ z ≤ z2, n = 1 , 2, 3,...,
where M1,M 2, . . . , Mn , . . . , are constants such that the series ∑ ∞
n=1 Mn converges.
Theorem 12.1.2, applied twice with z = xand once with z = t, shows that uxx and ut can be obtained",Elementary Differential Equations with Boundary Value Problems.pdf
"n=1 Mn converges.
Theorem 12.1.2, applied twice with z = xand once with z = t, shows that uxx and ut can be obtained
by differentiating uterm by term if t >0 (Exercise 54). Therefore usatisfies the heat equation and the
boundary conditions in ( 12.1.4) for t> 0. Therefore, since u(x, 0) = S(x) for 0 ≤ x≤ L, uis an actual
solution of ( 12.1.4) if and only if S(x) = f(x) for 0 ≤ x≤ L. From Theorem 11.3.2, this is true if f is
continuous and piecewise smooth on [0,L ], and f(0) = f(L) = 0 .
In this chapter we’ll define formal solutions of several kind s of problems. When we ask you to solve
such problems, we always mean that you should find a formal sol ution.",Elementary Differential Equations with Boundary Value Problems.pdf
"622 Chapter 12 Fourier Solutions of Partial Differential
Example 12.1.1 Solve ( 12.1.4) with f(x) = x(x2 − 3Lx+ 2 L2).
Solution From Example 11.3.6, the Fourier sine series of f on [0,L ] is
S(x) = 12L3
π3
∞∑
n=1
1
n3 sin nπx
L .
Therefore
u(x,t ) = 12L3
π3
∞∑
n=1
1
n3 e−n2π 2 a2t/L 2
sin nπx
L .
If both ends of bar are insulated so that no heat can pass throu gh them, then the boundary conditions
are
ux(0,t ) = 0 , u x(L,t ) = 0 , t> 0.
W e leave it to you (Exercise 1) to use the method of separation of variables and Theorem 11.1.3 to
motivate the next definition.
Definition 12.1.3
The formal solution of the initial-boundary value problem
ut = a2uxx, 0 <x<L, t> 0,
ux(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) = f(x), 0 ≤ x≤ L
(12.1.6)
is
u(x,t ) = α0 +
∞∑
n=1
αne−n2π 2 a2t/L 2
cos nπx
L ,
where
C(x) = α0 +
∞∑
n=1
αn cos nπx
L
is the Fourier cosine series of f on [0,L ]; that is ,
α0 = 1
L
∫ L
0
f(x) dx and αn = 2
L
∫ L
0
f(x) cos nπx
L dx, n = 1 , 2, 3,....",Elementary Differential Equations with Boundary Value Problems.pdf
"αn cos nπx
L
is the Fourier cosine series of f on [0,L ]; that is ,
α0 = 1
L
∫ L
0
f(x) dx and αn = 2
L
∫ L
0
f(x) cos nπx
L dx, n = 1 , 2, 3,....
Example 12.1.2 Solve ( 12.1.6) with f(x) = x.
Solution From Example 11.3.1, the Fourier cosine series of f on [0,L ] is
C(x) = L
2 − 4L
π2
∞∑
n=1
1
(2n− 1)2 cos (2n− 1)πx
L .
Therefore
u(x,t ) = L
2 − 4L
π2
∞∑
n=1
1
(2n− 1)2 e−(2n−1)2π 2 a2t/L 2
cos (2n− 1)πx
L .
W e leave it to you (Exercise 2) to use the method of separation of variables and Theorem 11. 1.4 to
motivate the next definition.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.1 The Heat Equation 623
Definition 12.1.4 The formal solution of the initial-boundary value problem
ut = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) = f(x), 0 ≤ x≤ L
(12.1.7)
is
u(x,t ) =
∞∑
n=1
αne−(2n−1)2π 2 a2t/ 4L2
sin (2n− 1)πx
2L ,
where
SM (x) =
∞∑
n=1
αn sin (2n− 1)πx
2L
is the mixed Fourier sine series of f on [0,L ]; that is ,
αn = 2
L
∫ L
0
f(x) sin (2n− 1)πx
2L dx.
Example 12.1.3 Solve ( 12.1.7) with f(x) = x.
Solution From Example 11.3.4, the mixed Fourier sine series of f on [0,L ] is
SM (x) = −8L
π2
∞∑
n=1
(−1)n
(2n− 1)2 sin (2n− 1)πx
2L .
Therefore
u(x,t ) = −8L
π2
∞∑
n=1
(−1)n
(2n− 1)2 e−(2n−1)2π 2a2 t/ 4L2
sin (2n− 1)πx
2L .
Figure 12.1.2 shows a graph of u = u(x,t ) plotted with respect to xfor various values of t. The line
y = xcorresponds to t = 0 . The other curves correspond to positive values of t. As tincreases, the
graphs approach the line u= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"y = xcorresponds to t = 0 . The other curves correspond to positive values of t. As tincreases, the
graphs approach the line u= 0 .
W e leave it to you (Exercise 3) to use the method of separation of variables and Theorem 11.1.5 to
motivate the next definition.
Definition 12.1.5
The formal solution of the initial-boundary value problem
ut = a2uxx, 0 <x<L, t> 0,
ux(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) = f(x), 0 ≤ x≤ L
(12.1.8)
is
u(x,t ) =
∞∑
n=1
αne−(2n−1)2π 2 a2t/ 4L2
cos (2n− 1)πx
2L ,
where
CM (x) =
∞∑
n=1
αn cos (2n− 1)πx
2L
is the mixed Fourier cosine series of f on [0,L ]; that is,
αn = 2
L
∫ L
0
f(x) cos (2n− 1)πx
2L dx.",Elementary Differential Equations with Boundary Value Problems.pdf
"624 Chapter 12 Fourier Solutions of Partial Differential
 L
 x
 u
 y = x
Figure 12.1.2
Example 12.1.4 Solve ( 12.1.8) with f(x) = x− L.
Solution From Example 11.3.3, the mixed Fourier cosine series of f on [0,L ] is
CM (x) = −8L
π2
∞∑
n=1
1
(2n− 1)2 cos (2n− 1)πx
2L .
Therefore
u(x,t ) = −8L
π2
∞∑
n=1
1
(2n− 1)2 e−(2n−1)2π 2 a2t/ 4L2
cos (2n− 1)πx
2L .
Nonhomogeneous Problems
A problem of the form
ut = a2uxx + h(x), 0 <x<L, t> 0,
u(0,t ) = u0, u (L,t ) = uL, t> 0,
u(x, 0) = f(x), 0 ≤ x≤ L
(12.1.9)
can be transformed to a problem that can be solved by separati on of variables. W e write
u(x,t ) = v(x,t ) + q(x), (12.1.10)
where qis to be determined. Then
ut = vt and uxx = vxx + q′′",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.1 The Heat Equation 625
so usatisfies ( 12.1.9) if vsatisfies
vt = a2vxx + a2q′′(x) + h(x), 0 <x<L, t> 0,
v(0,t ) = u0 − q(0), v (L,t ) = uL − q(L), t> 0,
v(x, 0) = f(x) − q(x), 0 ≤ x≤ L.
This reduces to
vt = a2vxx, 0 <x<L, t> 0,
v(0,t ) = 0 , v (L,t ) = 0 , t> 0,
v(x, 0) = f(x) − q(x), 0 ≤ x≤ L
(12.1.11)
if
a2q′′+ h(x) = 0 , q (0) = u0, q (L) = uL.
W e can obtain q by integrating q′′= −h/a 2 twice and choosing the constants of integration so that
q(0) = u0 and q(L) = uL. Then we can solve ( 12.1.11) for vby separation of variables, and ( 12.1.10) is
the solution of ( 12.1.9).
Example 12.1.5 Solve
ut = uxx − 2, 0 <x< 1, t> 0,
u(0,t ) = −1, u (1,t ) = 1 , t> 0,
u(x, 0) = x3 − 2x2 + 3 x− 1, 0 ≤ x≤ 1.
Solution W e leave it to you to show that
q(x) = x2 + x− 1
satisfies
q′′− 2 = 0 , q (0) = −1, q (1) = 1 .
Therefore
u(x,t ) = v(x,t ) + x2 + x− 1,
where
vt = vxx, 0 <x< 1, t> 0,
v(0,t ) = 0 , v (1,t ) = 0 , t> 0,
and
v(x, 0) = x3 − 2x2 + 3 x− 1 − x2 − x+ 1 = x(x2 − 3x+ 2) .",Elementary Differential Equations with Boundary Value Problems.pdf
"where
vt = vxx, 0 <x< 1, t> 0,
v(0,t ) = 0 , v (1,t ) = 0 , t> 0,
and
v(x, 0) = x3 − 2x2 + 3 x− 1 − x2 − x+ 1 = x(x2 − 3x+ 2) .
From Example 12.1.1 with a= 1 and L= 1 ,
v(x,t ) = 12
π3
∞∑
n=1
1
n3 e−n2π 2t sin nπx.
Therefore
u(x,t ) = x2 + x− 1 + 12
π3
∞∑
n=1
1
n3 e−n2π 2 t sin nπx.
A similar procedure works if the boundary conditions in ( 12.1.11) are replaced by mixed boundary
conditions
ux(0,t ) = u0, u (L,t ) = uL, t> 0",Elementary Differential Equations with Boundary Value Problems.pdf
"626 Chapter 12 Fourier Solutions of Partial Differential
or
u(0,t ) = u0, u x(L,t ) = uL, t> 0;
however, this isn’t true in general for the boundary conditi ons
ux(0,t ) = u0, u x(L,t ) = uL, t> 0.
(See Exercise 47.)
USING TECHNOLOGY
Numerical experiments can enhance your understanding of th e solutions of initial-boundary value prob-
lems. T o be specific, consider the formal solution
u(x,t ) =
∞∑
n=1
αne−n2π 2 a2t/L 2
sin nπx
L ,
of ( 12.1.4), where
S(x) =
∞∑
n=1
αn sin nπx
L
is the Fourier sine series of f on [0,L ]. Consider the m-th partial sum
um (x,t ) =
m∑
n=1
αne−n2π 2 a2t/L 2
sin nπx
L . (12.1.12)
For several fixed values of t(including t= 0 ), graph um(x,t ) versus t. In some cases it may be useful to
graph the curves corresponding to the various values of ton the same axes in other cases you may want to
graph the various curves sucessively (for increasing value s of t), and create a primitive motion picture on",Elementary Differential Equations with Boundary Value Problems.pdf
"graph the various curves sucessively (for increasing value s of t), and create a primitive motion picture on
your monitor. Repeat this experiment for several values of m, to compare how the results depend upon
mfor small and large values of t. However, keep in mind that the meanings of “small” and “larg e” in this
case depend upon the constants a2 and L2. A good way to handle this is to rewrite ( 12.1.12) as
um(x,t ) =
m∑
n=1
αne−n2τ sin nπx
L ,
where
τ = π2a2t
L2 , (12.1.13)
and graph um versus xfor selected values of τ.
These comments also apply to the situations considered in De finitions 12.1.3-12.1.5, except that ( 12.1.13)
should be replaced by
τ = π2a2t
4L2 ,
in Definitions 12.1.4 and 12.1.5.
In some of the exercises we say “perform numerical experimen ts. ” This means that you should perform
the computations just described with the formal solution ob tained in the exercise.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.1 The Heat Equation 627
12.1 Exercises
1. Explain Definition 12.1.3.
2. Explain Definition 12.1.4.
3. Explain Definition 12.1.5.
4. C Perform numerical experiments with the formal solution obt ained in Example 12.1.1.
5. C Perform numerical experiments with the formal solution obt ained in Example 12.1.2.
6. C Perform numerical experiments with the formal solution obt ained in Example 12.1.3.
7. C Perform numerical experiments with the formal solution obt ained in Example 12.1.4.
In Exercises 8-42 solve the initial-boundary value problem. Where indicated by C , perform numerical
experiments. T o simplify the computation of coefficients in some of these problems, check first to see if
u(x, 0) is a polynomial that satisfies the boundary conditions. If it does, apply Theorem 11.3.5; also, see
Exercises 11.3.35(b), 11.3.42(b), and 11.3.50(b).
8. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(1 − x), 0 ≤ x≤ 1
9. ut = 9 uxx, 0 <x< 4, t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"8. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(1 − x), 0 ≤ x≤ 1
9. ut = 9 uxx, 0 <x< 4, t> 0,
u(0,t ) = 0 , u (4,t ) = 0 , t> 0,
u(x, 0) = 1 , 0 ≤ x≤ 4
10. ut = 3 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = xsin x, 0 ≤ x≤ π
11. C ut = 9 uxx, 0 <x< 2, t> 0,
u(0,t ) = 0 , u (2,t ) = 0 , t> 0,
u(x, 0) = x2(2 − x), 0 ≤ x≤ 2
12. ut = 4 uxx, 0 <x< 3, t> 0,
u(0,t ) = 0 , u (3,t ) = 0 , t> 0,
u(x, 0) = x(9 − x2), 0 ≤ x≤ 3
13. ut = 4 uxx, 0 <x< 2, t> 0,
u(0,t ) = 0 , u (2,t ) = 0 , t> 0,
u(x, 0) =
{ x, 0 ≤ x≤ 1,
2 − x, 1 ≤ x≤ 2.
14. ut = 7 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(3x4 − 10x2 + 7) , 0 ≤ x≤ 1
15. ut = 5 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(x3 − 2x2 + 1) , 0 ≤ x≤ 1
16. ut = 2 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(3x4 − 5x3 + 2) , 0 ≤ x≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"628 Chapter 12 Fourier Solutions of Partial Differential
17. C ut = 9 uxx, 0 <x< 4, t> 0,
ux(0,t ) = 0 , u x(4,t ) = 0 , t> 0,
u(x, 0) = x2, 0 ≤ x≤ 4
18. ut = 4 uxx, 0 <x< 2, t> 0,
ux(0,t ) = 0 , u x(2,t ) = 0 , t> 0,
u(x, 0) = x(x− 4), 0 ≤ x≤ 2
19. C ut = 9 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x(1 − x), 0 ≤ x≤ 1
20. ut = 3 uxx, 0 <x< 2, t> 0,
ux(0,t ) = 0 , u x(2,t ) = 0 , t> 0,
u(x, 0) = 2 x2(3 − x), 0 ≤ x≤ 2
21. ut = 5 uxx, 0 <x<
√
2, t> 0,
ux(0,t ) = 0 , u x(
√
2,t ) = 0 , t> 0,
u(x, 0) = 3 x2(x2 − 4), 0 ≤ x≤
√
2
22. C ut = 3 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x3(3x− 4), 0 ≤ x≤ 1
23. ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x2(3x2 − 8x+ 6) , 0 ≤ x≤ 1
24. ut = uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = x2(x− π)2, 0 ≤ x≤ π
25. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = sin πx, 0 ≤ x≤ 1
26. C ut = 3 uxx, 0 <x<π, t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"25. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = sin πx, 0 ≤ x≤ 1
26. C ut = 3 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = x(π − x), 0 ≤ x≤ π
27. ut = 5 uxx, 0 <x< 2, t> 0,
u(0,t ) = 0 , u x(2,t ) = 0 , t> 0,
u(x, 0) = x(4 − x), 0 ≤ x≤ 2
28. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x2(3 − 2x), 0 ≤ x≤ 1
29. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = ( x− 1)3 + 1 , 0 ≤ x≤ 1
30. C ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x(x2 − 3), 0 ≤ x≤ 1
31. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x3(3x− 4), 0 ≤ x≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.1 The Heat Equation 629
32. ut = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x(x3 − 2x2 + 2) , 0 ≤ x≤ 1
33. ut = 3 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = x2(π − x), 0 ≤ x≤ π
34. ut = 16 uxx, 0 <x< 2π, t> 0,
ux(0,t ) = 0 , u (2π,t ) = 0 , t> 0,
u(x, 0) = 4 , 0 ≤ x≤ 2π
35. ut = 9 uxx, 0 <x< 4, t> 0,
ux(0,t ) = 0 , u (4,t ) = 0 , t> 0,
u(x, 0) = x2, 0 ≤ x≤ 4
36. C ut = 3 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 1 − x, 0 ≤ x≤ 1
37. ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 1 − x3, 0 ≤ x≤ 1
38. ut = 7 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = π2 − x2, 0 ≤ x≤ π
39. ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 4 x3 + 3 x2 − 7, 0 ≤ x≤ 1
40. ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 2 x3 + 3 x2 − 5, 0 ≤ x≤ 1
41. C ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 2 x3 + 3 x2 − 5, 0 ≤ x≤ 1
41. C ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x4 − 4x3 + 6 x2 − 3, 0 ≤ x≤ 1
42. ut = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x4 − 2x3 + 1 , 0 ≤ x≤ 1
In Exercises 43-46 solve the initial-boundary value problem. P erform numeric al experiments for specific
values of Land a.
43. C ut = a2uxx, 0 <x<L, t> 0,
ux(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) =
{
1, 0 ≤ x≤ L
2 ,
0, L
2 <x<L
44. C ut = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) =
{ 1, 0 ≤ x≤ L
2 ,
0, L
2 <x<L",Elementary Differential Equations with Boundary Value Problems.pdf
"630 Chapter 12 Fourier Solutions of Partial Differential
45. C ut = a2uxx, 0 <x<L, t> 0,
ux(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) =
{ 1, 0 ≤ x≤ L
2 ,
0, L
2 <x<L
46. C ut = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) =
{ 1, 0 ≤ x≤ L
2 ,
0, L
2 <x<L
47. Let hbe continuous on [0,L ] and let u0, uL, and abe constants, with a> 0. Show that it’s always
possible to find a function qthat satisfies (a), (b), or (c), but that this isn’t so for (d).
(a) a2q′′+ h= 0 , q (0) = u0, q (L) = uL
(b) a2q′′+ h= 0 , q ′(0) = u0, q (L) = uL
(c) a2q′′+ h= 0 , q (0) = u0, q ′(L) = uL
(d) a2q′′+ h= 0 , q ′(0) = u0, q ′(L) = uL
In Exercises 48-53 solve the nonhomogeneous initial-boundary value problem
48. ut = 9 uxx − 54x, 0 <x< 4, t> 0,
u(0,t ) = 1 , u (4,t ) = 61 , t> 0,
u(x, 0) = 2 − x+ x3, 0 ≤ x≤ 4
49. ut = uxx − 2, 0 <x< 1, t> 0,
u(0,t ) = 1 , u (1,t ) = 3 , t> 0,
u(x, 0) = 2 x2 + 1 , 0 ≤ x≤ 1
50. ut = 3 uxx − 18x, 0 <x< 1, t> 0,
ux(0,t ) = −1, u (1,t ) = −1, t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"u(0,t ) = 1 , u (1,t ) = 3 , t> 0,
u(x, 0) = 2 x2 + 1 , 0 ≤ x≤ 1
50. ut = 3 uxx − 18x, 0 <x< 1, t> 0,
ux(0,t ) = −1, u (1,t ) = −1, t> 0,
u(x, 0) = x3 − 2x, 0 ≤ x≤ 1
51. ut = 9 uxx − 18, 0 <x< 4, t> 0,
ux(0,t ) = −1, u (4,t ) = 10 , t> 0,
u(x, 0) = 2 x2 − x− 2, 0 ≤ x≤ 4
52. ut = uxx + π2 sin πx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = −π, t> 0,
u(x, 0) = 2 sin πx, 0 ≤ x≤ 1
53. ut = uxx − 6x, 0 <x<L, t> 0,
u(0,t ) = 3 , u x(1,t ) = 2 , t> 0,
u(x, 0) = x3 − x2 + x+ 3 , 0 ≤ x≤ 1
54. In this exercise take it as given that the infinite series ∑ ∞
n=1 npe−qn2
converges for all pif q >0,
and, where appropriate, use the comparison test for absolut e convergence of an infinite series.
Let
u(x,t ) =
∞∑
n=1
αne−n2π 2 a2t/L 2
sin nπx
L
where
αn = 2
L
∫ L
0
f(x) sin nπx
L dx
and f is piecewise smooth on [0,L ].",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 631
(a) Show that uis defined for (x,t ) such that t> 0.
(b) For fixed t> 0, use Theorem 12.1.2 with z = xto show that
ux(x,t ) = π
L
∞∑
n=1
nαne−n2π 2 a2t/L 2
cos nπx
L , −∞ <x< ∞ .
(c) Starting from the result of (a), use Theorem 12.1.2 with z = xto show that, for a fixed t> 0,
uxx(x,t ) = − π2
L2
∞∑
n=1
n2αne−n2π 2 a2t/L 2
sin nπx
L , −∞ <x< ∞ .
(d) For fixed but arbitrary x, use Theorem 12.1.2 with z= tto show that
ut(x,t ) = −π2a2
L2
∞∑
n=1
n2αne−n2π 2 a2t/L 2
sin nπx
L ,
if t>t 0 >0, where t0 is an arbitrary positive number. Then argue that since t0 is arbitrary ,
the conclusion holds for all t> 0.
(e) Conclude from (c) and (d) that
ut = a2uxx, −∞ <x< ∞ , t> 0.
By repeatedly applying the arguments in (a) and (c), it can be shown that ucan be differentiated
term by term any number of times with respect to xand/or tif t> 0.
12.2 THE W A VE EQUA TION
In this section we consider initial-boundary value problem s of the form
utt = a2uxx, 0 <x<L, t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"12.2 THE W A VE EQUA TION
In this section we consider initial-boundary value problem s of the form
utt = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) = f(x), u t (x, 0) = g(x), 0 ≤ x≤ L,
(12.2.1)
where ais a constant and f and gare given functions of x.
The partial differential equation utt = a2uxx is called the wave equation . It is necessary to specify
both f and gbecause the wave equation is a second order equation in tfor each fixed x.
This equation and its generalizations
utt = a2(uxx + uyy ) and utt = a2(uxx + uyy + uzz )
to two and three space dimensions have important applicatio ns to the the propagation of electromagnetic,
sonic, and water waves.
The Vibrating String
W e motivate the study of the wave equation by considering its application to the vibrations of a string –
such as a violin string – tightly stretched in equilibrium al ong the x-axis in the xu-plane and tied to the
points (0, 0) and (L, 0) (Figure 12.2.1).",Elementary Differential Equations with Boundary Value Problems.pdf
"points (0, 0) and (L, 0) (Figure 12.2.1).
If the string is plucked in the vertical direction and releas ed at time t = 0 , it will oscillate in the xu-
plane. Let u(x,t ) denote the displacement of the point on the string above (or b elow) the abscissa xat
time t.
W e’ll show that it’s reasonable to assume that u satisfies the wave equation under the following as-
sumptions:",Elementary Differential Equations with Boundary Value Problems.pdf
"632 Chapter 12 Fourier Solutions of Partial Differential Equations
 L  x
 u
Figure 12.2.1 A stretched string
1. The mass density (mass per unit length) ρ of the string is constant throughout the string.
2. The tension T induced by tightly stretching the string along the x-axis is so great that all other
forces, such as gravity and air resistance, can be neglected .
3. The tension at any point on the string acts along the tangent t o the string at that point, and the
magnitude of its horizontal component is always equal to T, the tension in the string in equilibrium.
4. The slope of the string at every point remains sufficiently sm all so that we can make the approxima-
tion √
1 + u2
x ≈ 1. (12.2.2)
Figure
12.2.2 shows a segment of the displaced string at a time t >0. (Don’t think that the figure is
necessarily inconsistent with Assumption 4; we exaggerate d the slope for clarity .)
The vectors T1 and T2 are the forces due to tension, acting along the tangents to th e segment at its",Elementary Differential Equations with Boundary Value Problems.pdf
"The vectors T1 and T2 are the forces due to tension, acting along the tangents to th e segment at its
endpoints. From Newton’s second law of motion, T1 − T2 is equal to the mass times the acceleration of
the center of mass of the segment. The horizontal and vertica l components of T1 − T2 are
|T2|cos θ2 − |T1|cos θ1 and |T2|sin θ2 − |T1|sin θ1,
respectively . Since
|T2|cos θ2 = |T1|cos θ1 = T (12.2.3)
by assumption, the net horizontal force is zero, so there’s n o horizontal acceleration. Since the initial
horizontal velocity is zero, there’s no horizontal motion.
Applying Newton’s second law of motion in the vertical direc tion yields
|T2|sin θ2 − |T1|sin θ1 = ρ∆ sutt(x,t ), (12.2.4)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 633
 T1
 θ 2
 T2
 x
 u
 θ 1
Figure 12.2.2 A segment of the displaced string
where ∆ sis the length of the segment and xis the abscissa of the center of mass; hence,
x< x<x + ∆ x.
From calculus, we know that
∆ s=
∫ x+∆ x
x
√
1 + u2
x(σ,t ) dσ;
however, because of (
12.2.2), we make the approximation
∆ s≈
∫ x+∆ x
x
1 dσ = ∆ x,
so ( 12.2.4) becomes
|T2|sin θ2 − |T1|sin θ1 = ρ∆ xutt(x,t ).
Therefore
|T2|sin θ2 − |T1|sin θ1
∆ x = ρutt(x,t ).
Recalling ( 12.2.3), we divide by T to obtain
tan θ2 − tan θ1
∆ x = ρ
Tutt(x,t ). (12.2.5)
Since tan θ1 = ux(x,t ) and tan θ2 = ux(x+ ∆ x,t ), ( 12.2.5) is equivalent to
ux(x+ ∆ x) − ux(x,t )
∆ x = ρ
Tutt(x,t ).",Elementary Differential Equations with Boundary Value Problems.pdf
"634 Chapter 12 Fourier Solutions of Partial Differential Equations
Letting ∆ x→ 0 yields
uxx(x,t ) = ρ
Tutt(x,t ),
which we rewrite as utt = a2uxx, with a2 = T/ρ .
The Formal Solution
As in Section 12.1, we use separation of variables to obtain a suitable definition for the formal solution of
(12.2.1). W e begin by looking for functions of the form v(x,t ) = X(x)T(t) that are not identically zero
and satisfy
vtt = a2vxx, v (0,t ) = 0 , v (L,t ) = 0
for all (x,t ). Since
vtt = XT′′ and vxx = X′′T,
vtt = a2vxx if and only if
XT′′= a2X′′T,
which we rewrite as
T′′
a2T = X′′
X .
For this to hold for all (x,t ), the two sides must equal the same constant; thus,
X′′
X = T′′
a2T = −λ,
which is equivalent to
X′′+ λX = 0
and
T′′+ a2λT = 0 . (12.2.6)
Since v(0,t ) = X(0)T(t) = 0 and v(L,t ) = X(L)T(t) = 0 and we don’t want T to be identically zero,
X(0) = 0 and X(L) = 0 . Therefore λ must be an eigenvalue of
X′′+ λX = 0 , X (0) = 0 , X (L) = 0 , (12.2.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"X(0) = 0 and X(L) = 0 . Therefore λ must be an eigenvalue of
X′′+ λX = 0 , X (0) = 0 , X (L) = 0 , (12.2.7)
and X must be a λ-eigenfunction. From Theorem 11.1.2, the eigenvalues of ( 12.2.7) are λn = n2π2/L 2,
with associated eigenfunctions
Xn = sin nπx
L , n = 1 , 2, 3,....
Substituting λ = n2π2/L 2 into ( 12.2.6) yields
T′′+ ( n2π2a2/L 2)T = 0 ,
which has the general solution
Tn = αn cos nπat
L + βn L
nπa sin nπat
L ,
where αn and βn are constants. Now let
vn(x,t ) = Xn (x)Tn (t) =
(
αn cos nπat
L + βn L
nπa sin nπat
L
)
sin nπx
L .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 635
Then
∂vn
∂t (x,t ) =
(
−nπa
L αn sin nπat
L + βn cos nπat
L
)
sin nπx
L ,
so
vn (x, 0) = αn sin nπx
L and ∂vn
∂t (x, 0) = βn sin nπx
L .
Therefore vn satisfies ( 12.2.1) with f(x) = αn sin nπx/L and g(x) = βn cos nπx/L . More generally , if
α1, α2, . . . , αm and β1, β2 ,. . . , βm are constants and
um(x,t ) =
m∑
n=1
(
αn cos nπat
L + βn L
nπa sin nπat
L
)
sin nπx
L ,
then um satisfies ( 12.2.1) with
f(x) =
m∑
n=1
αn sin nπx
L and g(x) =
m∑
n=1
βn sin nπx
L .
This motivates the next definition.
Definition 12.2.1
If f and gare piecewise smooth of [0,L ], then the formal solution of ( 12.2.1) is
u(x,t ) =
∞∑
n=1
(
αn cos nπat
L + βn L
nπa sin nπat
L
)
sin nπx
L , (12.2.8)
where
Sf (x) =
∞∑
n=1
αn sin nπx
L and Sg (x) =
∞∑
n=1
βn sin nπx
L
are the Fourier sine series of f and gon [0,L ]; that is,
αn = 2
L
∫ L
0
f(x) sin nπx
L dx and βn = 2
L
∫ L
0
g(x) sin nπx
L dx.",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
n=1
βn sin nπx
L
are the Fourier sine series of f and gon [0,L ]; that is,
αn = 2
L
∫ L
0
f(x) sin nπx
L dx and βn = 2
L
∫ L
0
g(x) sin nπx
L dx.
Since there are no convergence-producing factors in ( 12.2.8) like the negative exponentials in tthat
appear in formal solutions of initial-boundary value probl ems for the heat equation, it isn’t obvious that
(12.2.8) even converges for any values of xand t, let alone that it can be differentiated term by term to
show that utt = a2uxx. However, the next theorem guarantees that the series conve rges not only for
0 ≤ x≤ Land t≥ 0, but for −∞ <x< ∞ and −∞ <t< ∞ .
Theorem 12.2.2 If f and g are pieceswise smooth on [0,L ], then uin (12.2.1) converges for all (x,t ),
and can be written as
u(x,t ) = 1
2 [Sf (x+ at) + Sf (x− at)] + 1
2a
∫ x+at
x−at
Sg (τ) dτ. (12.2.9)
Proof Setting A= nπx/L and B= nπat/L in the identities
sin Acos B= 1
2 [sin(A+ B) + sin( A− B)]
and
sin Asin B= −1
2 [cos(A+ B) − cos(A− B)]",Elementary Differential Equations with Boundary Value Problems.pdf
"636 Chapter 12 Fourier Solutions of Partial Differential Equations
yields
cos nπat
L sin nπx
L = 1
2
[
sin nπ(x+ at)
L + sin nπ(x− at)
L
]
(12.2.10)
and
sin nπat
L sin nπx
L = −1
2
[
cos nπ(x+ at)
L − cos nπ(x− at)
L
]
= nπ
2L
∫ x+at
x−at
sin nπτ
L dτ.
(12.2.11)
From ( 12.2.10),
∞∑
n=1
αn cos nπat
L sin nπx
L = 1
2
∞∑
n=1
αn
(
sin nπ(x+ at)
L + sin nπ(x− at)
L
)
= 1
2 [Sf (x+ at) + Sf (x− at)].
(12.2.12)
Since it can be shown that a Fourier sine series can be integra ted term by term between any two limits,
(12.2.11) implies that
∞∑
n=1
βn L
nπa sin nπat
L sin nπx
L = 1
2a
∞∑
n=1
βn
∫ x+at
x−at
sin nπτ
L dτ
= 1
2a
∫ x+at
x−at
(∞∑
n=1
βn sin nπτ
L
)
dτ
= 1
2a
∫ x+at
x−at
Sg (τ) dτ.
This and ( 12.2.12) imply ( 12.2.9), which completes the proof.
As we’ll see below , if Sg is differentiable and Sf is twice differentiable on (−∞ , ∞ ), then ( 12.2.9)
satisfies utt = a2uxx for all (x,t ). W e need the next theorem to formulate conditions on f and gsuch",Elementary Differential Equations with Boundary Value Problems.pdf
"satisfies utt = a2uxx for all (x,t ). W e need the next theorem to formulate conditions on f and gsuch
that Sf and Sg to have these properties.
Theorem 12.2.3 Suppose h is differentiable on [0,L ]; that is, h′(x) exists for 0 < x < L,and the
one-sided derivatives
h′
+(0) = lim
x→ 0+
h(x) − h(0)
x and h′
−(L) = lim
x→ L−
h(x) − h(L)
x− L
both exist.
(a) Let pbe the odd periodic extension of hto (−∞ , ∞ ); that is ,
p(x) =
{ h(x), 0 ≤ x≤ L,
−h(−x), −L<x< 0, and p(x+ 2 L) = p(x), −∞ <x< ∞ .
Then pis differentiable on (−∞ , ∞ ) if and only if
h(0) = h(L) = 0 . (12.2.13)
(b) Let qbe the even periodic extension of hto (−∞ , ∞ ); that is ,
q(x) =
{ h(x), 0 ≤ x≤ L,
h(−x), −L<x< 0, and q(x+ 2 L) = q(x), −∞ <x< ∞ .
Then qis differentiable on (−∞ , ∞ ) if and only if
h′
+(0) = h′
−(L) = 0 . (12.2.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 637
Proof Throughout this proof, kdenotes an integer. Since f is differentiable on the open interval (0,L ),
both pand qare differentiable on every open interval ((k− 1)L,kL ). Thus, we need only to determine
whether pand qare differentiable at x= kLfor every k.
(a) From Figure 12.2.3, pis discontinuous at x= 2 kLif h(0) ̸= 0 and discontinuous at x= (2 k−1)L
if h(L) ̸= 0 . Therefore pis not differentiable on (−∞ , ∞ ) unless h(0) = h(L) = 0 . From Figure 12.2.4,
if h(0) = h(L) = 0 , then
p′(2kL) = h′
+(0) and p′((2k− 1)L) = h′
−(L)
for every k; therefore, pis differentiable on (−∞ , ∞ ).
 x = 0  x = L  x = 2L x = − L x = − 2L x = − 3L  x =  3L
 x
 y
Figure 12.2.3 The odd extension of a function that
does not satisfy ( 12.2.13)
 x = 0  x = L  x = 2L x = − L x = − 2L x = − 3L  x =  3L
 x
 y
Figure 12.2.4 The odd extension of a function that
satisfies ( 12.2.13)
(b) From Figure 12.2.5,
q′
−(2kL) = −h′
+(0) and q′
+(2kL) = h′
+(0),",Elementary Differential Equations with Boundary Value Problems.pdf
"x
 y
Figure 12.2.4 The odd extension of a function that
satisfies ( 12.2.13)
(b) From Figure 12.2.5,
q′
−(2kL) = −h′
+(0) and q′
+(2kL) = h′
+(0),
so qis differentiable at x= 2 kLif and only if h′
+(0) = 0 . Also,
q′
−((2k− 1)L) = h′
−(L) and q′
+((2k− 1)L) = −h′
−(L),
so q is differentiable at x = (2 k − 1)L if and only if h′
−(L) = 0 . Therefore q is differentiable on
(−∞ , ∞ ) if and only if h′
+(0) = h′
−(L) = 0 , as in Figure
12.2.6. This completes the proof.
 x = 0  x = L  x = 2L x = − L x = − 2L x = − 3L  x =  3L
 x
 y
Figure 12.2.5 The even extension of a function that
doesn’t satisfy ( 12.2.14)
 x = 0  x = L  x = 2L x = − L x = − 2L x = − 3L  x =  3L
 x
 y
Figure 12.2.6 The even extension of a function that
satisfies ( 12.2.14)",Elementary Differential Equations with Boundary Value Problems.pdf
"638 Chapter 12 Fourier Solutions of Partial Differential Equations
Theorem 12.2.4 The formal solution of (12.2.1) is an actual solution if gis differentiable on [0,L ] and
g(0) = g(L) = 0 , (12.2.15)
while f is twice differentiable on [0,L ] and
f(0) = f(L) = 0 (12.2.16)
and
f′′
+(0) = f′′
− (L) = 0 . (12.2.17)
Proof W e first show that Sg is differentiable and Sf is twice differentiable on (−∞ , ∞ ). W e’ll then
differentiate ( 12.2.9) twice with respect to xand tand verify that ( 12.2.9) is an actual solution of ( 12.2.1).
Since f and gare continuous on (0,L ), Theorem 11.3.2 implies that Sf (x) = f(x) and Sg (x) = g(x)
on [0,L ]. Therefore Sf and Sg are the odd periodic extensions of f and g. Since f and gare differen-
tiable on [0,L ], ( 12.2.15), ( 12.2.16), and Theorem 12.2.3(a) imply that Sf and Sg are differentiable on
(−∞ , ∞ ).
Since S′
f (x) = f′(x) on [0,L ] (one-sided derivatives at the endpoints), and S′
f is even (the derivative
of an odd function is even), S′",Elementary Differential Equations with Boundary Value Problems.pdf
"Since S′
f (x) = f′(x) on [0,L ] (one-sided derivatives at the endpoints), and S′
f is even (the derivative
of an odd function is even), S′
f is the even periodic extension of f′. By assumption, f′is differentiable
on [0,L ]. Because of ( 12.2.17), Theorem 12.2.3(b) with h = f′and q = S′
f implies that S′′
f exists on
(−∞ , ∞ ).
Now we can differentiate ( 12.2.9) twice with respect to xand t:
ux(x,t ) = 1
2 [S′
f (x+ at) + S′
f (x− at)] + 1
2a[Sg (x+ at) − Sg (x− at)],
uxx(x,t ) = 1
2 [S′′
f (x+ at) + S′′
f (x− at)] + 1
2a[S′
g (x+ at) − S′
g (x− at)], (12.2.18)
ut (x,t ) = a
2 [S′
f (x+ at) − S′
f (x− at)] + 1
2 [Sg (x+ at) + Sg (x− at)], (12.2.19)
and
utt(x,t ) = a2
2 [S′′
f (x+ at) − S′′
f (x− at)] + a
2 [S′
g (x+ at) − S′
g (x− at)]. (12.2.20)
Comparing ( 12.2.18) and ( 12.2.20) shows that utt(x,t ) = a2uxx(x,t ) for all (x,t ).
From ( 12.2.8), u(0,t ) = u(L,t ) = 0 for all t. From ( 12.2.9), u(x, 0) = Sf (x) for all x, and therefore,
in particular,
u(x, 0) = f(x), 0 ≤ x<L.",Elementary Differential Equations with Boundary Value Problems.pdf
"From ( 12.2.8), u(0,t ) = u(L,t ) = 0 for all t. From ( 12.2.9), u(x, 0) = Sf (x) for all x, and therefore,
in particular,
u(x, 0) = f(x), 0 ≤ x<L.
From ( 12.2.19), ut(x, 0) = Sg (x) for all x, and therefore, in particular,
ut(x, 0) = g(x), 0 ≤ x<L.
Therefore uis an actual solution of ( 12.2.1). This completes the proof.
Eqn ( 12.2.9) is called d’Alembert’s solution of ( 12.2.1). Although d’Alembert’s solution was useful
for proving Theorem 12.2.4 and is very useful in a slightly different context (Exercise s 63-68), ( 12.2.8)
is preferable for computational purposes.
Example 12.2.1 Solve (
12.2.1) with
f(x) = x(x3 − 2Lx2 + L2) and g(x) = x(L− x).",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 639
Solution W e leave it to you to verify that f and gsatisfy the assumptions of Theorem 12.2.4.
From Exercise 11.3. 39,
Sf (x) = 96L4
π5
∞∑
n=1
1
(2n− 1)5 sin (2n− 1)πx
L .
From Exercise 11.3. 36,
Sg (x) = 8L2
π3
∞∑
n=1
1
(2n− 1)3 sin (2n− 1)πx
L .
From ( 12.2.8),
u(x,t ) = 96L4
π5
∞∑
n=1
1
(2n− 1)5 cos (2n− 1)πat
L sin (2n− 1)πx
L
+ 8L3
aπ4
∞∑
n=1
1
(2n− 1)4 sin (2n− 1)πat
L sin (2n− 1)πx
L .
Theorem 12.1.2 implies that uxx and utt can be obtained by term by term differentiation, for all (x,t ),
so utt = a2uxx for all (x,t ) (Exercise 62). Moreover, Theorem 11.3.2 implies that Sf (x) = f(x) and
Sg (x) = g(x) if 0 ≤ x≤ L. Therefore u(x, 0) = f(x) and ut (x, 0) = g(x) if 0 ≤ x≤ L. Hence, uis
an actual solution of the initial-boundary value problem.
REM A RK : In solving a specific initial-boundary value problem ( 12.2.1), it’s convenient to solve the prob-",Elementary Differential Equations with Boundary Value Problems.pdf
"REM A RK : In solving a specific initial-boundary value problem ( 12.2.1), it’s convenient to solve the prob-
lem with g ≡ 0, then with f ≡ 0, and add the solutions to obtain the solution of the given pro blem.
Because of this, either f ≡ 0 or g≡ 0 in all the specific initial-boundary value problems in the ex ercises.
The Plucked String
If f and g don’t satisfy the assumptions of Theorem 12.2.4, then ( 12.2.8) isn’t an actual solution of
(12.2.1) in fact, it can be shown that ( 12.2.1) doesn’t have an actual solution in this case. Nevertheless , u
is defined for all (x,t ), and we can see from ( 12.2.18) and ( 12.2.20) that utt(x,t ) = a2uxx(x,t ) for all
(x,t ) such that S′′
f (x± at) and S′
g (x± at) exist. Moreover, umay still provide a useful approximation
to the vibration of the string; a laboratory experiment can c onfirm or deny this.
W e’ll now consider the initial-boundary value problem ( 12.2.1) with
f(x) =
{
x, 0 ≤ x≤ L
2 ,
L− x, L
2 ≤ x≤ L (12.2.21)",Elementary Differential Equations with Boundary Value Problems.pdf
"W e’ll now consider the initial-boundary value problem ( 12.2.1) with
f(x) =
{
x, 0 ≤ x≤ L
2 ,
L− x, L
2 ≤ x≤ L (12.2.21)
and g≡ 0. Since f isn’t differentiable at x= L/ 2, it does’nt satisfy the assumptions of Theorem 12.2.4,
so the formal solution of ( 12.2.1) can’t be an actual solution. Nevertheless, it’s instructi ve to investigate
the properties of the formal solution.
The graph of f is shown in Figure 12.2.7. Intuitively , we are plucking the string by half its length a t the
middle. Y ou’re right if you think this is an extraordinarily large displacement; however, we could remove
this objection by multiplying the function in Figure 12.2.7 by a small constant. Since this would just
multiply the formal solution by the same constant, we’ll lea ve f as we’ve defined it. Similar comments
apply to the exercises.
From Exercise 11.3. 15, the Fourier sine series of f on [0,L ] is
Sf (x) = 4L
π2
∞∑
n=1
(−1)n+1
(2n− 1)2 sin (2n− 1)πx
L ,",Elementary Differential Equations with Boundary Value Problems.pdf
"640 Chapter 12 Fourier Solutions of Partial Differential Equations
 L .5 L
 .5 L
 x
 y
Figure 12.2.7 Graph of ( 12.2.21)
which converges to f for all xin [0,L ], by Theorem 11.3.2. Therefore
u(x,t ) = 4L
π2
∞∑
n=1
(−1)n+1
(2n− 1)2 cos (2n− 1)πat
L sin (2n− 1)πx
L . (12.2.22)
This series converges absolutely for all (x,t ) by the comparison test, since the series
∞∑
n=1
1
(2n− 1)2
converges. Moreover, ( 12.2.22) satisfies the boundary conditions
u(0,t ) = u(L,t ) = 0 , t> 0,
and the initial condition
u(x, 0) = f(x), 0 ≤ x≤ L.
However, we can’t justify differentiating ( 12.2.22) term by term even once, and formally differentiating it
twice term by term produces a series that diverges for all (x,t ). (V erify .). Therefore we use d’Alembert’s
form
u(x,t ) = 1
2 [Sf (x+ at) + Sf (x− at)] (12.2.23)
for uto study its derivatives. Figure 12.2.8 shows the graph of Sf , which is the odd periodic extension
of f. Y ou can see from the graph that Sf is differentiable at x(and S′",Elementary Differential Equations with Boundary Value Problems.pdf
"of f. Y ou can see from the graph that Sf is differentiable at x(and S′
f (x) = ±1) if and only if xisn’t an
odd multiple of L/ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 641
 x = 0  x = L  x = 2L x = − L x = − 2L x = − 3L  x =  3L
 x
 y
 − .5L
  .5L
Figure 12.2.8 The odd periodic extension of
(12.2.21)
 x = 0  x = L  x = 2L x = − L x = − 2L x = − 3L  x =  3L
 x
 y
 .5L
 − .5L
Figure 12.2.9 Graphs of y= Sf (x− at) (dashed)
and y = Sf (x− at) (solid), with f as in ( 12.2.21)
In Figure 12.2.9 the dashed and solid curves are the graphs of y = Sf (x− at) and y = Sf (x+ at)
respectively , for a fixed value of t. As t increases the dashed curve moves to the right and the solid
curve moves to the left. For this reason, we say that the funct ions u1(x,t ) = Sf (x+ at) and u2(x,t ) =
Sf (x− at) are traveling waves . Note that u1 satisfies the wave equation at (x,t ) if x+ at isn’t an
odd multiple of L/ 2 and u2 satisfies the wave equation at (x,t ) if x− atisn’t an odd multiple of L/ 2.
Therefore ( 12.2.23) (or, equivalently , ( 12.2.22)) satisfies utt(x,t ) = a2uxx(x,t ) = 0 for all (x,t ) such",Elementary Differential Equations with Boundary Value Problems.pdf
"Therefore ( 12.2.23) (or, equivalently , ( 12.2.22)) satisfies utt(x,t ) = a2uxx(x,t ) = 0 for all (x,t ) such
that neither x− atnor x+ atis an odd multiple of L/ 2.
W e conclude by finding an explicit formula for u(x,t ) under the assumption that
0 ≤ x≤ L and 0 ≤ t≤ L/ 2a. (12.2.24)
T o see how this formula can be used to compute u(x,t ) for 0 ≤ x ≤ Land arbitrary t, we refer you to
Exercise 16.
From Figure 12.2.10,
Sf (x− at) =
{ x− at, 0 ≤ x≤ L
2 + at,
L− x+ at, L
2 + at≤ x≤ L
and
Sf (x+ at) =
{ x+ at, 0 ≤ x≤ L
2 − at,
L− x− at, L
2 − at≤ x≤ L
if (x,t ) satisfies ( 12.2.24).
Therefore, from ( 12.2.23),
u(x,t ) =







x, 0 ≤ x≤ L
2 − at,
L
2 − at, L
2 − at≤ x≤ L
2 + at,
L− x, L
2 − at≤ x≤ L
if (x,t ) satisfies ( 12.2.24). Figure 12.2.11 is the graph of this function on [0,L ] for a fixed tin (0,L/ 2a).",Elementary Differential Equations with Boundary Value Problems.pdf
"642 Chapter 12 Fourier Solutions of Partial Differential Equations
 L at  .5L
 .5L
 x =.5L − at  x =.5L + at
 x
 y
Figure 12.2.10 The part of the graph from
Figure 12.2.9 on [0,L ]
 L
 .5L
 y = .5L − at
 x =.5L + at x = .5L − at
 x
 y
Figure 12.2.11 The graph of ( 12.2.23) on [0,L ] for
a fixed tin (0,L/ 2a)
USING TECHNOLOGY
Although the formal solution
u(x,t ) =
∞∑
n=1
(
αn cos nπat
L + βn L
nπa sin nπat
L
)
sin nπx
L
of ( 12.2.1) is defined for all (x,t ), we’re mainly interested in its behavior for 0 ≤ x≤ Land t ≥ 0. In
fact, it’s sufficient to consider only values of tin the interval 0 ≤ t< 2L/a , since
u(x,t + 2 kL/a ) = u(x,t )
for all (x,t ) if kis an integer. (V erify .)
Y ou can create an animation of the motion of the string by perf orming the following numerical experi-
ment.
Let mand kbe positive integers. Let
tj = 2Lj
ka , j = 0 , 1,...k ;
thus, t0, t1, . . . tk are equally spaced points in [0, 2L/a ]. For each j = 0 , 1 ,2, . . . k, graph the partial sum
um (x,t j ) =
m∑
n=1",Elementary Differential Equations with Boundary Value Problems.pdf
"thus, t0, t1, . . . tk are equally spaced points in [0, 2L/a ]. For each j = 0 , 1 ,2, . . . k, graph the partial sum
um (x,t j ) =
m∑
n=1
(
αn cos nπatj
L + βn L
nπa sin nπatj
L
)
sin nπx
L
on [0,L ] as a function of x. Write your program so that each graph remains displayed on t he monitor for
a short time, and is then deleted and replaced by the next. Rep eat this procedure for various values of m
and k.
W e suggest that you perform experiments of this kind in the ex ercises marked C , without other
specific instructions. (These exercises were chosen arbitr arily; the experiment is worthwhile in all the
exercises dealing with specific initial-boundary value pro blems.) In some of the exercises the formal
solutions have other forms, defined in Exercises 17, 34, and 49; however, the idea of the experiment is
the same.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 643
12.2 Exercises
In Exercises 1-15 solve the initial-boundary value problem. In some of these e xercises, Theorem 11.3.5(b)
or Exercise 11.3.35 will simplify the computation of the coefficients in the F our ier sine series.
1. utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) =
{
x, 0 ≤ x≤ 1
2 ,
1 − x, 1
2 ≤ x≤ 1 , 0 ≤ x≤ 1
2. utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(1 − x), u t(x, 0) = 0 , 0 ≤ x≤ 1
3. utt = 7 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x2(1 − x), u t(x, 0) = 0 , 0 ≤ x≤ 1
4. C utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(1 − x), 0 ≤ x≤ 1
5. utt = 7 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 ut (x, 0) = x2(1 − x), 0 ≤ x≤ 1
6. utt = 64 uxx, 0 <x< 3, t> 0,
u(0,t ) = 0 , u (3,t ) = 0 , t> 0,
u(x, 0) = x(x2 − 9), u t(x, 0) = 0 , 0 ≤ x≤ 3
7. utt = 4 uxx, 0 <x< 1, t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"6. utt = 64 uxx, 0 <x< 3, t> 0,
u(0,t ) = 0 , u (3,t ) = 0 , t> 0,
u(x, 0) = x(x2 − 9), u t(x, 0) = 0 , 0 ≤ x≤ 3
7. utt = 4 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(x3 − 2x2 + 1) , u t(x, 0) = 0 , 0 ≤ x≤ 1
8. C utt = 64 uxx, 0 <x< 3, t> 0,
u(0,t ) = 0 , u (3,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(x2 − 9), 0 ≤ x≤ 3
9. utt = 4 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(x3 − 2x2 + 1) , 0 ≤ x≤ 1
10. utt = 5 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = xsin x, u t(x, 0) = 0 , 0 ≤ x≤ π
11. utt = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(3x4 − 5x3 + 2) , u t (x, 0) = 0 , 0 ≤ x≤ 1
12. C utt = 5 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = xsin x, 0 ≤ x≤ π
13. utt = uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(3x4 − 5x3 + 2) , 0 ≤ x≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"644 Chapter 12 Fourier Solutions of Partial Differential Equations
14. utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x(3x4 − 10x2 + 7) , u t(x, 0) = 0 , 0 ≤ x≤ 1
15. C utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 ut (x, 0) = x(3x4 − 10x2 + 7) , 0 ≤ x≤ 1
16. W e saw that the displacement of the plucked string is, on the o ne hand,
u(x,t ) = 4L
π2
∞∑
n=1
(−1)n+1
(2n− 1)2 cos (2n− 1)πat
L sin (2n− 1)πx
L , 0 ≤ x≤ L, t ≥ 0, (A)
and, on the other hand,
u(x,τ ) =







x, 0 ≤ x≤ L
2 − aτ,
L
2 − aτ, L
2 − aτ ≤ x≤ L
2 + aτ,
L− x, L
2 − aτ ≤ x≤ L.
(B)
if 0 ≤ τ ≤ L/ 2a. The first objective of this exercise is to show that (B) can be used to compute
u(x,t ) for 0 ≤ x≤ Land all t> 0.
(a) Show that if t> 0, there’s a nonnegative integer msuch that either
(i) t= mL
a + τ or (ii) t= (m+ 1) L
a − τ,
where 0 ≤ τ ≤ L/ 2a.
(b) Use (A) to show that u(x,t ) = ( −1)m u(x,τ ) if (i) holds, while u(x,t ) = ( −1)m+1u(x,τ )
if (ii) holds.",Elementary Differential Equations with Boundary Value Problems.pdf
"a − τ,
where 0 ≤ τ ≤ L/ 2a.
(b) Use (A) to show that u(x,t ) = ( −1)m u(x,τ ) if (i) holds, while u(x,t ) = ( −1)m+1u(x,τ )
if (ii) holds.
(c) L Perform the following experiment for specific values of Land aand various values of
mand k: Let
tj = Lj
2ka, j = 0 , 1,...k ;
thus, t0, t1, . . . , tk are equally spaced points in [0,L/ 2a]. For each j = 0 , 1 , 2,. . . , k,
graph the mth partial sum of (A) and u(x,t j ) computed from (B) on the same axis. Create
an animation, as described in the remarks on using technolog y at the end of the section.
17. If a string vibrates with the end at x= 0 free to move in a frictionless vertical track and the end at
x= Lfixed, then the initial-boundary value problem for its displ acement takes the form
utt = a2uxx, 0 <x<L, t> 0,
ux(0,t ) = 0 , u (L,t ) = 0 , t> 0,
u(x, 0) = f(x), u t(x, 0) = g(x), 0 ≤ x≤ L.
(A)
Justify defining the formal solution of (A) to be
u(x,t ) =
∞∑
n=1
(
αn cos (2n− 1)πat
2L + 2Lβn
(2n− 1)πa sin (2n− 1)πat
2L
)
cos (2n− 1)πx
2L ,",Elementary Differential Equations with Boundary Value Problems.pdf
"(A)
Justify defining the formal solution of (A) to be
u(x,t ) =
∞∑
n=1
(
αn cos (2n− 1)πat
2L + 2Lβn
(2n− 1)πa sin (2n− 1)πat
2L
)
cos (2n− 1)πx
2L ,
where
CM f(x) =
∞∑
n=1
αn cos (2n− 1)πx
2L and CM g(x) =
∞∑
n=1
βn cos (2n− 1)πx
2L",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 645
are the mixed Fourier cosine series of f and gon [0,L ]; that is,
αn = 2
L
∫ L
0
f(x) cos (2n− 1)πx
2L dx and βn = 2
L
∫ L
0
g(x) cos (2n− 1)πx
2L dx.
In Exercises 18-31, use Exercise 17 to solve the initial-boundary value problem. In some of thes e exercises
Theorem 11.3.5(c) or Exercise 11.3.42(b) will simplify the computation of the coefficients in the mixe d
F ourier cosine series.
18. utt = 9 uxx, 0 <x< 2, t> 0,
ux(0,t ) = 0 , u (2,t ) = 0 , t> 0,
u(x, 0) = 4 − x2, u t(x, 0) = 0 , 0 ≤ x≤ 2
19. utt = 4 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x2(1 − x), u t(x, 0) = 0 , 0 ≤ x≤ 1
20. utt = 9 uxx, 0 <x< 2, t> 0,
ux(0,t ) = 0 , u (2,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = 4 − x2, 0 ≤ x≤ 2
21. utt = 4 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x2(1 − x), 0 ≤ x≤ 1
22. C utt = 5 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 2 x3 + 3 x2 − 5, u t (x, 0) = 0 , 0 ≤ x≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"22. C utt = 5 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 2 x3 + 3 x2 − 5, u t (x, 0) = 0 , 0 ≤ x≤ 1
23. utt = 3 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = π3 − x3, u t(x, 0) = 0 , 0 ≤ x≤ π
24. utt = 5 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = 2 x3 + 3 x2 − 5, 0 ≤ x≤ 1
25. C utt = 3 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u (π,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = π3 − x3, 0 ≤ x≤ π
26. utt = 9 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x4 − 2x3 + 1 , u t(x, 0) = 0 , 0 ≤ x≤ 1
27. utt = 7 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 4 x3 + 3 x2 − 7, u t (x, 0) = 0 , 0 ≤ x≤ 1
28. utt = 9 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x4 − 2x3 + 1 , 0 ≤ x≤ 1
29. C utt = 7 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = 4 x3 + 3 x2 − 7, 0 ≤ x≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"646 Chapter 12 Fourier Solutions of Partial Differential Equations
30. utt = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = x4 − 4x3 + 6 x2 − 3, u t (x, 0) = 0 , 0 ≤ x≤ 1
31. utt = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u (1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x4 − 4x3 + 6 x2 − 3, 0 ≤ x≤ 1
32. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value
problem in Exercise 17.
33. Use the result of Exercise 32 to show that the formal solution of the initial-boundary val ue problem
in Exercise 17 is an actual solution if gis differentiable and f is twice differentiable on [0,L ] and
g′
+(0) = g(L) = f′
+(0) = f(L) = f′′
− (L) = 0 .
HIN T : See Exercise 11.3.57, and apply Theorem 12.2.3 with Lreplaced by 2L.
34. Justify defining the formal solution of the initial-boundar y value problem
utt = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) = f(x), u t (x, 0) = g(x), 0 ≤ x≤ L
to be
u(x,t ) =
∞∑
n=1
(
αn cos (2n− 1)πat",Elementary Differential Equations with Boundary Value Problems.pdf
"utt = a2uxx, 0 <x<L, t> 0,
u(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) = f(x), u t (x, 0) = g(x), 0 ≤ x≤ L
to be
u(x,t ) =
∞∑
n=1
(
αn cos (2n− 1)πat
2L + 2Lβn
(2n− 1)πa sin (2n− 1)πat
2L
)
sin (2n− 1)πx
2L ,
where
SM f(x) =
∞∑
n=1
αn sin (2n− 1)πx
2L and SM g(x) =
∞∑
n=1
βn sin (2n− 1)πx
2L
are the mixed Fourier sine series of f and gon [0,L ]; that is,
αn = 2
L
∫ L
0
f(x) sin (2n− 1)πx
2L dx and βn = 2
L
∫ L
0
g(x) sin (2n− 1)πx
2L dx.
In Exercises 35-46 use Exercise 34 to solve the initial-boundary value problem. In some of thes e exercises
Theorem 11.3.5(d) or Exercise 11.3.50(b) will simplify the computation of the coefficients in the mixe d
F ourier sine series.
35. utt = 64 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = x(2π − x), u t (x, 0) = 0 , 0 ≤ x≤ π
36. utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x2(3 − 2x), u t (x, 0) = 0 , 0 ≤ x≤ 1
37. utt = 64 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u x(π,t ) = 0 , t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"u(x, 0) = x2(3 − 2x), u t (x, 0) = 0 , 0 ≤ x≤ 1
37. utt = 64 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(2π − x), 0 ≤ x≤ π",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 647
38. C utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x2(3 − 2x), 0 ≤ x≤ 1
39. utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = ( x− 1)3 + 1 , u t (x, 0) = 0 , 0 ≤ x≤ 1
40. utt = 3 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = x(x2 − 3π2), u t (x, 0) = 0 , 0 ≤ x≤ π
41. utt = 9 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = ( x− 1)3 + 1 , 0 ≤ x≤ 1
42. utt = 3 uxx, 0 <x<π, t> 0,
u(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(x2 − 3π2), 0 ≤ x≤ π
43. utt = 5 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x3(3x− 4), u t (x, 0) = 0 , 0 ≤ x≤ 1
44. C utt = 16 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x(x3 − 2x2 + 2) , u t(x, 0) = 0 , 0 ≤ x≤ 1
45. utt = 5 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x3(3x− 4), 0 ≤ x≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"45. utt = 5 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x3(3x− 4), 0 ≤ x≤ 1
46. C utt = 16 uxx, 0 <x< 1, t> 0,
u(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x(x3 − 2x2 + 2) , 0 ≤ x≤ 1
47. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value
problem in Exercise 34.
48. Use the result of Exercise 47 to show that the formal solution of the initial-boundary val ue problem
in Exercise 34 is an actual solution if gis differentiable and f is twice differentiable on [0,L ] and
f(0) = f′
−(L) = g(0) = g′
−(L) = f′′
+(0) = 0 .
HIN T : See Exercise 11.3.58 and apply Theorem 12.2.3 with Lreplaced by 2L.
49. Justify defining the formal solution of the initial-boundar y value problem
utt = a2uxx, 0 <x<L, t> 0,
ux(0,t ) = 0 , u x(L,t ) = 0 , t> 0,
u(x, 0) = f(x), u t(x, 0) = g(x), 0 ≤ x≤ L.
to be
u(x,t ) = α0 + β0t+
∞∑
n=1
(
αn cos nπat
L + Lβn
nπa sin nπat
L
)
cos nπx
L ,
where
Cf (x) = α0 +
∞∑
n=1",Elementary Differential Equations with Boundary Value Problems.pdf
"to be
u(x,t ) = α0 + β0t+
∞∑
n=1
(
αn cos nπat
L + Lβn
nπa sin nπat
L
)
cos nπx
L ,
where
Cf (x) = α0 +
∞∑
n=1
αn cos nπx
L and Cg (x) = β0 +
∞∑
n=1
βn cos nπx
L",Elementary Differential Equations with Boundary Value Problems.pdf
"648 Chapter 12 Fourier Solutions of Partial Differential Equations
are the Fourier cosine series of f and gon [0,L ]; that is,
α0 = 1
L
∫ L
0
f(x) dx, β 0 = 1
L
∫ L
0
g(x) dx,
αn = 2
L
∫ L
0
f(x) cos nπx
L dx, and βn = 2
L
∫ L
0
g(x) cos nπx
L dx, n = 1 , 2, 3,....
In Exercises 50-59 use Exercise 49 to solve the initial-boundary value problem. In some of thes e exercises
Theorem 11.3.5(a) will simplify the computation of the coefficients in the F our ier cosine series.
50. utt = 5 uxx, 0 <x< 2, t> 0,
ux(0,t ) = 0 , u x(2,t ) = 0 , t> 0,
u(x, 0) = 2 x2(3 − x), u t (x, 0) = 0 , 0 ≤ x≤ 2
51. utt = 5 uxx, 0 <x< 2, t> 0,
ux(0,t ) = 0 , u x(2,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = 2 x2(3 − x), 0 ≤ x≤ 2
52. utt = 4 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = x3(3x− 4π), u t(x, 0) = 0 , 0 ≤ x≤ π
53. utt = 7 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 3 x2(x2 − 2), u t(x, 0) = 0 , 0 ≤ x≤ 1
54. C utt = 4 uxx, 0 <x<π, t> 0,",Elementary Differential Equations with Boundary Value Problems.pdf
"ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 3 x2(x2 − 2), u t(x, 0) = 0 , 0 ≤ x≤ 1
54. C utt = 4 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x3(3x− 4π), 0 ≤ x≤ π
55. utt = 7 uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = 3 x2(x2 − 2), 0 ≤ x≤ 1
56. utt = 16 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = x2(x− π)2, u t(x, 0) = 0 , 0 ≤ x≤ π
57. C utt = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = x2(3x2 − 8x+ 6) , u t (x, 0) = 0 , 0 ≤ x≤ 1
58. utt = 16 uxx, 0 <x<π, t> 0,
ux(0,t ) = 0 , u x(π,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x2(x− π)2, 0 ≤ x≤ π
59. C utt = uxx, 0 <x< 1, t> 0,
ux(0,t ) = 0 , u x(1,t ) = 0 , t> 0,
u(x, 0) = 0 , u t(x, 0) = x2(3x2 − 8x+ 6) , 0 ≤ x≤ 1
60. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value
problem in Exercise 49.",Elementary Differential Equations with Boundary Value Problems.pdf
"60. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value
problem in Exercise 49.
61. Use the result of Exercise 60 to show that the formal solution of the initial-boundary val ue problem
in Exercise 49 is an actual solution if gis differentiable and f is twice differentiable on [0,L ] and
f′
+(0) = f′
−(L) = g′
+(0) = g′
−(L) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.2 The W ave Equation 649
62. Suppose λ and µ are constants and either pn (x) = cos nλx or pn (x) = sin nλx, while either
qn(t) = cos nµt or qn(t) = sin nµt for n= 1 , 2, 3, . . . . Let
u(x,t ) =
∞∑
n=1
knpn(x)qn (t), (A)
where {kn}∞
n=1 are constants.
(a) Show that if ∑ ∞
n=1 |kn|converges then u(x,t ) converges for all (x,t ).
(b) Use Theorem 12.1.2 to show that if ∑ ∞
n=1 n|kn|converges then (A) can be differentiated
term by term with respect to xand tfor all (x,t ); that is,
ux(x,t ) =
∞∑
n=1
knp′
n(x)qn(t)
and
ut(x,t ) =
∞∑
n=1
knpn(x)q′
n (t).
(c) Suppose ∑ ∞
n=1 n2|kn|converges. Show that
uxx(x,y ) =
∞∑
n=1
kn p′′
n(x)qn(t)
and
utt(x,y ) =
∞∑
n=1
knpn (x)q′′
n(t)
(d) Suppose ∑ ∞
n=1 n2|αn|and ∑ ∞
n=1 n|βn |both converge. Show that the formal solution
u(x,t ) =
∞∑
n=1
(
αn cos nπat
L + βn L
nπa sin nπat
L
)
sin nπx
L
of Equation 12.2.1 satisfies utt = a2uxx for all (x,t ).
This conclusion also applies to the formal solutions defined in Exercises 17, 34, and 49.",Elementary Differential Equations with Boundary Value Problems.pdf
"L
of Equation 12.2.1 satisfies utt = a2uxx for all (x,t ).
This conclusion also applies to the formal solutions defined in Exercises 17, 34, and 49.
63. Suppose gis differentiable and f is twice differentiable on (−∞ , ∞ ), and let
u0(x,t ) = f(x+ at) + f(x− at)
2 and u1(x,t ) = 1
2a
∫ x+at
x−at
g(u) du.
(a) Show that
∂2u0
∂t2 = a2 ∂2u0
∂x2 , −∞ <x< ∞ , t> 0,
and
u0(x, 0) = f(x), ∂u0
∂t (x, 0) = 0 , −∞ <x< ∞ .
(b) Show that
∂2u1
∂t2 = a2 ∂2u1
∂x2 , −∞ <x< ∞ , t> 0,
and
u1(x, 0) = 0 , ∂u1
∂t (x, 0) = g(x), −∞ <x< ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"650 Chapter 12 Fourier Solutions of Partial Differential Equations
(c) Solve
utt = a2uxx, −∞ <t< ∞ , t> 0,
u(x, 0) = f(x), u t(x, 0) = g(x), −∞ <x< ∞ .
In Exercises 64-68 use the result of Exercise 63 to find a solution of
utt = a2uxx, −∞ <x< ∞
that satisfies the given initial conditions.
64. u(x, 0) = x, ut(x, 0) = 4 ax, −∞ <x< ∞
65. u(x, 0) = x2, ut (x, 0) = 1 , −∞ <x< ∞
66. u(x, 0) = sin x, ut (x, 0) = acos x, −∞ <x< ∞
67. u(x, 0) = x3, ut (x, 0) = 6 x2, −∞ <x< ∞
68. u(x, 0) = xsin x, ut(x, 0) = sin x, −∞ <x< ∞
12.3 LAPLACE’S EQUA TION IN RECT ANGULAR COORDINA TES
The temperature u= u(x,y,t ) in a two-dimensional plate satisfies the two-dimensional he at equation
ut = a2(uxx + uyy ), (12.3.1)
where (x,y ) varies over the interior of the plate and t> 0. T o find a solution of ( 12.3.1), it’s necessary to
specify the initial temperature u(x,y, 0) and conditions that must be satisfied on the boundary . Howeve r,
as t→ ∞ , the influence of the initial condition decays, so
lim
t→∞",Elementary Differential Equations with Boundary Value Problems.pdf
"as t→ ∞ , the influence of the initial condition decays, so
lim
t→∞
ut (x,y,t ) = 0
and the temperature approaches a steady state distribution u= u(x,y ) that satisfies
uxx + uyy = 0 . (12.3.2)
This is Laplace’s equation . This equation also arises in applications to fluid mechanic s and potential
theory; in fact, it is also called the potential equation . W e seek solutions of ( 12.3.2) in a region Rthat
satisfy specified conditions – called boundary conditions – on the boundary of R. For example, we may
require u to assume prescribed values on the boundary . This is called a Dirichlet condition , and the
problem is called a Dirichlet problem . Or, we may require the normal derivative of uat each point (x,y )
on the boundary to assume prescribed values. This is called a Neumann condition , and the problem is
called a Neumann problem . In some problems we impose Dirichlet conditions on part of t he boundary",Elementary Differential Equations with Boundary Value Problems.pdf
"called a Neumann problem . In some problems we impose Dirichlet conditions on part of t he boundary
and Neumann conditions on the rest. Then we say that the bound ary conditions and the problem are
mixed.
Solving boundary value problems for ( 12.3.2) over general regions is beyond the scope of this book,
so we consider only very simple regions. W e begin by consider ing the rectangular region shown in
Figure 12.3.1.
The possible boundary conditions for this region can be writ ten as
(1 − α)u(x, 0) + αuy(x, 0) = f0(x), 0 ≤ x≤ a,
(1 − β)u(x,b ) + βuy (x,b ) = f1(x), 0 ≤ x≤ a,
(1 − γ)u(0,y ) + γux(0,y ) = g0(y), 0 ≤ y≤ b,
(1 − δ)u(a,y ) + δux(a,y ) = g1(y), 0 ≤ y≤ b,",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 651
 y
 x a
 b
Figure 12.3.1 A rectangular region and its boundary
where α, β, γ, and δ can each be either 0 or 1; thus, there are 16 possibilities. Let BVP (α,β,γ,δ )(f0,f 1,g 0,g 1)
denote the problem of finding a solution of ( 12.3.2) that satisfies these conditions. This is a Dirichlet
problem if
α = β = γ = δ = 0
(Figure 12.3.2), or a Neumann problem if
α = β = γ = δ = 1
(Figure 12.3.3). The other 14 problems are mixed.
 y
 x a
 b
 uxx + uyy = 0
 u(x,0) = f0(x)
 u(x,b) = f1(x)
 u(0,y) = g0(y)  u(a,y) = g1(y)
Figure 12.3.2 A Dirichlet problem
 y
 x a
 b
 uxx + uyy = 0
 uy(x,0) = f0(x)
 uy(x,b) = f1(x)
 ux(0,y) = g0(y)  ux(a,y) = g1(y)
Figure 12.3.3 A Neumann problem
For given (α,β,γ,δ ), the sum of solutions of
BVP(α,β,γ,δ )(f0, 0, 0, 0), BVP(α,β,γ,δ )(0,f 1, 0, 0),",Elementary Differential Equations with Boundary Value Problems.pdf
"652 Chapter 12 Fourier Solutions of Partial Differential Equations
BVP(α,β,γ,δ )(0, 0,g 0, 0), and BVP (α,β,γ,δ )(0, 0, 0,g 1)
is a solution of
BVP(α,β,γ,δ )(f0,f 1,g 0,g 1).
Therefore we concentrate on problems where only one of the fu nctions f0, f1, g0, g2 isn’t identically
zero. There are 64 (count them!) problems of this form. Each h as homogeneous boundary conditions on
three sides of the rectangle, and a nonhomogeneous boundary condition on the fourth. W e use separation
of variables to find infinitely many functions that satisfy La place’s equation and the three homogeneous
boundary conditions in the open rectangle. W e then use these solutions as building blocks to construct a
formal solution of Laplace’s equation that also satisfies th e nonhomogeneous boundary condition. Since
it’s not feasible to consider all 64 cases, we’ll restrict ou r attention in the text to just four. Others are
discussed in the exercises.
If v(x,y ) = X(x)Y(y) then
vxx + vyy = X′′Y + XY′′= 0",Elementary Differential Equations with Boundary Value Problems.pdf
"discussed in the exercises.
If v(x,y ) = X(x)Y(y) then
vxx + vyy = X′′Y + XY′′= 0
for all (x,y ) if and only if
X′′
X = −Y′′
Y = k
for all (x,y ), where kis a separation constant. This equation is equivalent to
X′′− kX = 0 , Y ′′+ kY = 0 . (12.3.3)
From here, the strategy depends upon the boundary condition s. W e illustrate this by examples.
Example 12.3.1 Define the formal solution of
uxx + uyy = 0 , 0 <x<a, 0 <y<b,
u(x, 0) = f(x), u (x,b ) = 0 , 0 ≤ x≤ a,
u(0,y ) = 0 , u (a,y ) = 0 , 0 ≤ y≤ b
(12.3.4)
(Figure 12.3.4).
Solution The boundary conditions in ( 12.3.4) require products v(x,y ) = X(x)Y(y) such that X(0) =
X(a) = Y(b) = 0 ; hence, we let k= −λ in ( 12.3.3). Thus, X and Y must satisfy
X′′+ λX = 0 , X (0) = 0 , X (a) = 0 (12.3.5)
and
Y′′− λY = 0 , Y (b) = 0 . (12.3.6)
From Theorem 11.1.2, the eigenvalues of ( 12.3.5) are λn = n2π2/a 2, with associated eigenfunctions
Xn = sin nπx
a , n = 1 , 2, 3,....
Substituting λ = n2π2/a 2 into ( 12.3.6) yields",Elementary Differential Equations with Boundary Value Problems.pdf
"Xn = sin nπx
a , n = 1 , 2, 3,....
Substituting λ = n2π2/a 2 into ( 12.3.6) yields
Y′′− (n2π2/a 2)Y = 0 , Y (b) = 0 ,
so we could take
Yn = sinh nπ(b− y)
a ; (12.3.7)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 653
 y
 x a
 b
 uxx + uyy = 0
 u(x,0) = f(x)
 u(x,b) = 0
 u(0,y) = 0  u(a,y) = 0
Figure 12.3.4 The boundary value problem ( 12.3.4)
however, because of the nonhomogeneous Dirichlet conditio n at y= 0 , it’s better to require that Yn (0) =
1, which can be achieved by dividing the right side of ( 12.3.7) by its value at y= 0 ; thus, we take
Yn = sinh nπ(b− y)/a
sinh nπb/a .
Then
vn (x,y ) = Xn (x)Yn (y) = sinh nπ(b− y)/a
sinh nπb/a sin nπx
a ,
so vn (x, 0) = sin nπx/a and vn satisfies ( 12.3.4) with f(x) = sin nπx/a . More generally , if α1, . . . ,
αm are arbitrary constants then
um (x,y ) =
m∑
n=1
αn
sinh nπ(b− y)/a
sinh nπb/a sin nπx
a
satisfies ( 12.3.4) with
f(x) =
m∑
n=1
αn sin nπx
L .
Therefore, if f is an arbitrary piecewise smooth function on [0,a ], we define the formal solution of
(12.3.4) to be
u(x,y ) =
∞∑
n=1
αn
sinh nπ(b− y)/a
sinh nπb/a sin nπx
a , (12.3.8)
where
S(x) =
∞∑
n=1
αn sin nπx
a",Elementary Differential Equations with Boundary Value Problems.pdf
"654 Chapter 12 Fourier Solutions of Partial Differential Equations
is the Fourier sine series of f on [0,a ]; that is,
αn = 2
a
∫ a
0
f(x) sin nπx
a dx, n = 1 , 2, 3,....
If y<b then
sinh nπ(b− y)/a
sinh nπb/a ≈ e−nπy/a (12.3.9)
for large n, so the series in ( 12.3.8) converges if 0 <y<b ; moreover, since also
cosh nπ(b− y)/a
sinh nπb/a ≈ e−nπy/a
for large n, Theorem 12.1.2 applied twice with z = xand twice with z = t, shows that uxx and uyy can
be obtained by differentiating uterm by term if 0 <y<b . (Exercise 37). Therefore usatisfies Laplace’s
equation in the interior of the rectangle in Figure 12.3.4. Moreover, the series in ( 12.3.8) also converges
on the boundary of the rectangle, and satisfies the three homo geneous boundary conditions conditions in
(12.3.4). Therefore, since u(x, 0) = S(x) for 0 ≤ x≤ L, uis an actual solution of ( 12.3.5) if and only if
S(x) = f(x) for 0 ≤ x≤ a. From Theorem 11.3.2, this is true if f is continuous and piecewise smooth",Elementary Differential Equations with Boundary Value Problems.pdf
"S(x) = f(x) for 0 ≤ x≤ a. From Theorem 11.3.2, this is true if f is continuous and piecewise smooth
on [0,L ], and f(0) = f(L) = 0 .
Example 12.3.2 Solve ( 12.3.4) with f(x) = x(x2 − 3ax+ 2 a2).
Solution From Example 11.3.6,
S(x) = 12a3
π3
∞∑
n=1
1
n3 sin nπx
a .
Therefore
u(x,y ) = 12a3
π3
∞∑
n=1
sinh nπ(b− y)/a
n3 sinh nπb/a sin nπx
a . (12.3.10)
T o compute approximate values of u(x,y ), we must use partial sums of the form
um(x,y ) = 12a3
π3
m∑
n=1
sinh nπ(b− y)/a
n3 sinh nπb/a sin nπx
a .
Because of ( 12.3.9), small values of mprovide sufficient accuracy for most applications if 0 < y < b.
Moreover, the n3 in the denominator in ( 12.3.10) ensures that this is also true for y = 0 . For graphing
purposes, we chose a= 2 , b= 1 , and m= 10 . Figure 12.3.5 shows the surface
u= u(x,y ), 0 ≤ x≤ 2, 0 ≤ y ≤ 1,
while Figure 12.3.6 shows the curves
u= u(x, 0.1k), 0 ≤ x≤ 2, k = 0 , 1,..., 10.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 655
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
0
0.5
1
1.5
2
2.5
3
Figure 12.3.5
1 2
1
2
3
 x
 y
Figure 12.3.6
Example 12.3.3 Define the formal solution of
uxx + uyy = 0 , 0 <x<a, 0 <y<b,
u(x, 0) = 0 , u y(x,b ) = f(x), 0 ≤ x≤ a,
ux(0,y ) = 0 , u x(a,y ) = 0 , 0 ≤ y≤ b
(12.3.11)
(Figure 12.3.7).
 y
 x a
 b
 uxx + uyy = 0
 u(x,0) = 0
 uy(x,b) = f(x)
 ux(0,y) = 0  ux(a,y) = 0
Figure 12.3.7 The boundary value problem ( 12.3.11)
Solution The boundary conditions in ( 12.3.11) require products v(x,y ) = X(x)Y(y) such that X′(0) =",Elementary Differential Equations with Boundary Value Problems.pdf
"656 Chapter 12 Fourier Solutions of Partial Differential Equations
X′(a) = Y(0) = 0 ; hence, we let k= −λ in ( 12.3.3). Thus, X and Y must satisfy
X′′+ λX = 0 , X ′(0) = 0 , X ′(a) = 0 (12.3.12)
and
Y′′− λY = 0 , Y (0) = 0 . (12.3.13)
From Theorem 11.1.3, the eigenvalues of ( 12.3.12) are λ = 0 , with associated eigenfunction X0 = 1 ,
and λn = n2π2/a 2, with associated eigenfunctions
Xn = cos nπx
a , n = 1 , 2, 3,....
Since Y0 = y satisfies ( 12.3.13) with λ = 0 , we take v0(x,y ) = X0(x)Y0(y) = y. Substituting
λ = n2π2/a 2 into ( 12.3.13) yields
Y′′− (n2π2/a 2)Y = 0 , Y (0) = 0 ,
so we could take
Yn = sinh nπy
a . (12.3.14)
However, because of the nonhomogeneous Neumann condition a t y= b, it’s better to require that Y′
n (b) =
1, which can be achieved by dividing the right side of ( 12.3.14) by the value of its derivative at y = b;
thus,
Yn = asinh nπy/a
nπ cosh nπb/a .
Then
vn (x,y ) = Xn (x)Yn (y) = asinh nπy/a
nπ cosh nπb/a cos nπx
a ,
so
∂vn
∂y (x,b ) = cos nπx
a .",Elementary Differential Equations with Boundary Value Problems.pdf
"thus,
Yn = asinh nπy/a
nπ cosh nπb/a .
Then
vn (x,y ) = Xn (x)Yn (y) = asinh nπy/a
nπ cosh nπb/a cos nπx
a ,
so
∂vn
∂y (x,b ) = cos nπx
a .
Therefore vn satisfies ( 12.3.11) with f(x) = cos nπx/a . More generally , if α0, . . . , αm are arbitrary
constants then
um (x,y ) = α0y+ a
π
m∑
n=1
αn
sinh nπy/a
ncosh nπb/a cos nπx
a
satisfies ( 12.3.11) with
f(x) = α0 +
m∑
n=1
αn cos nπx
L .
Therefore, if f is an arbitrary piecewise smooth function on [0,a ] we define the formal solution of
(12.3.11) to be
u(x,y ) = α0y+ a
π
∞∑
n=1
αn
sinh nπy/a
ncosh nπb/a cos nπx
a ,
where
C(x) = α0 +
∞∑
n=1
αn cos nπx
a
is the Fourier cosine series of f on [0,a ]; that is,
α0 = 1
a
∫ a
0
f(x) dx and αn = 2
a
∫ a
0
f(x) cos nπx
a dx, n = 1 , 2, 3,....",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 657
Example 12.3.4 Solve ( 12.3.11) with f(x) = x.
Solution From Example 11.3.1,
C(x) = a
2 − 4a
π2
∞∑
n=1
1
(2n− 1)2 cos (2n− 1)πx
a .
Therefore
u(x,y ) = ay
2 − 4a2
π3
∞∑
n=1
sinh(2n− 1)πy/a
(2n− 1)3 cosh(2n− 1)πb/a cos (2n− 1)πx
a . (12.3.15)
For graphing purposes, we chose a = 2 , b = 1 , and retained the terms through n = 10 in ( 12.3.15).
Figure 12.3.8 shows the surface
u= u(x,y ), 0 ≤ x≤ 2, 0 ≤ y ≤ 1,
while Figure 12.3.9 shows the curves
u= u(x,. 1k), 0 ≤ x≤ 2, k = 0 , 1,..., 10.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
0
0.5
1
1.5
Figure 12.3.8
1 2
1
 x
 y
Figure 12.3.9
Example 12.3.5 Define the formal solution of
uxx + uyy = 0 , 0 <x<a, 0 <y<b,
u(x, 0) = 0 , u y(x,b ) = 0 , 0 ≤ x≤ a,
u(0,y ) = g(y), u x(a,y ) = 0 , 0 ≤ y≤ b
(12.3.16)
(Figure 12.3.10).
Solution The boundary conditions in ( 12.3.16) require products v(x,y ) = X(x)Y(y) such that Y(0) =",Elementary Differential Equations with Boundary Value Problems.pdf
"(12.3.16)
(Figure 12.3.10).
Solution The boundary conditions in ( 12.3.16) require products v(x,y ) = X(x)Y(y) such that Y(0) =
Y′(b) = X′(a) = 0 ; hence, we let k= λ in ( 12.3.3). Thus, X and Y must satisfy
X′′− λX = 0 , X ′(a) = 0 (12.3.17)",Elementary Differential Equations with Boundary Value Problems.pdf
"658 Chapter 12 Fourier Solutions of Partial Differential Equations
 y
 x a
 b
 uxx + uyy = 0
 u(x,0) = 0
 uy(x,b) = 0
 u(0,y) = g(y)  ux(a,y) = 0
Figure 12.3.10 The boundary value problem ( 12.3.16)
and
Y′′+ λY = 0 , Y (0) = 0 , Y ′(b) = 0 . (12.3.18)
From Theorem 11.1.4, the eigenvalues of ( 12.3.18) are λn = (2 n− 1)2π2/ 4b2, with associated eigen-
functions
Yn = sin (2n− 1)πy
2b , n = 1 , 2, 3,....
Substituting λ = (2 n− 1)2π2/ 4b2 into ( 12.3.17) yields
X′′− ((2n− 1)2π2/ 4b2)X = 0 , X ′(a) = 0 ,
so we could take
Xn = cosh (2n− 1)π(x− a)
2b . (12.3.19)
However, because of the nonhomogeneous Dirichlet conditio n at x= 0 , it’s better to require that Xn (0) =
1, which can be achieved by dividing the right side of ( 12.3.19) by its value at x= 0 ; thus,
Xn = cosh(2n− 1)π(x− a)/ 2b
cosh(2n− 1)πa/ 2b .
Then
vn(x,y ) = Xn (x)Yn (y) = cosh(2n− 1)π(x− a)/ 2b
cosh(2n− 1)πa/ 2b sin (2n− 1)πy
2b ,
so
vn (0,y ) = sin (2n− 1)πy
2b .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 659
Therefore vn satisfies ( 12.3.16) with g(y) = sin(2 n− 1)πy/ 2b. More generally , if α1, . . . , αm are
arbitrary constants then
um (x,y ) =
m∑
n=1
αn
cosh(2n− 1)π(x− a)/ 2b
cosh(2n− 1)πa/ 2b sin (2n− 1)πy
2b
satisfies ( 12.3.16) with
g(y) =
m∑
n=1
αn sin (2n− 1)πy
2b .
Thus, if gis an arbitrary piecewise smooth function on [0,b ], we define the formal solution of ( 12.3.16)
to be
u(x,y ) =
∞∑
n=1
αn
cosh(2n− 1)π(x− a)/ 2b
cosh(2n− 1)πa/ 2b sin (2n− 1)πy
2b ,
where
SM (x) =
∞∑
n=1
αn sin (2n− 1)πy
2b
is the mixed Fourier sine series of gon [0,b ]; that is,
αn = 2
b
∫ b
0
g(y) sin (2n− 1)πy
2b dy.
Example 12.3.6 Solve ( 12.3.16) with g(y) = y(2y2 − 9by+ 12 b2).
Solution From Example 11.3.8,
SM (y) = 96b3
π3
∞∑
n=1
1
(2n− 1)3
[
3 + ( −1)n 4
(2n− 1)π
]
sin (2n− 1)πy
2b .
Therefore
u(x,y ) = 96b3
π3
∞∑
n=1
cosh(2n− 1)π(x− a)/ 2b
(2n− 1)3 cosh(2n− 1)πa/ 2b
[
3 + ( −1)n 4
(2n− 1)π
]
sin (2n− 1)πy
2b .",Elementary Differential Equations with Boundary Value Problems.pdf
"sin (2n− 1)πy
2b .
Therefore
u(x,y ) = 96b3
π3
∞∑
n=1
cosh(2n− 1)π(x− a)/ 2b
(2n− 1)3 cosh(2n− 1)πa/ 2b
[
3 + ( −1)n 4
(2n− 1)π
]
sin (2n− 1)πy
2b .
Example 12.3.7 Define the formal solution of
uxx + uyy = 0 , 0 <x<a, 0 <y<b,
uy(x, 0) = 0 , u (x,b ) = 0 , 0 ≤ x≤ a,
ux(0,y ) = 0 , u x(a,y ) = g(y), 0 ≤ y≤ b
(12.3.20)
(Figure 12.3.11).
Solution The boundary conditions in ( 12.3.20) require products v(x,y ) = X(x)Y(y) such that Y′(0) =
Y(b) = X′(0) = 0 ; hence, we let k= λ in ( 12.3.3). Thus, X and Y must satisfy
X′′− λX = 0 , X ′(0) = 0 (12.3.21)
and
Y′′+ λY = 0 , Y ′(0) = 0 , Y (b) = 0 . (12.3.22)",Elementary Differential Equations with Boundary Value Problems.pdf
"660 Chapter 12 Fourier Solutions of Partial Differential Equations
 y
 x a
 b
 uxx + uyy = 0
 uy(x,0) = 0
 u(x,b) = 0
 ux(0,y) = 0  ux(a,y) = g(y)
Figure 12.3.11 The boundary value problem ( 12.3.20)
From Theorem 11.1.4, the eigenvalues of ( 12.3.22) are λn = (2 n− 1)2π2/ 4b2, with associated eigen-
functions
Yn = cos (2n− 1)πy
2b , n = 1 , 2, 3,....
Substituting λ = (2 n− 1)2π2/ 4b2 into ( 12.3.21) yields
X′′− ((2n− 1)2π2/ 4b2)X = 0 , X ′(0) = 0 ,
so we could take
Xn = cosh (2n− 1)πx
2b . (12.3.23)
However, because of the nonhomogeneous Neumann condition a t x = a, it’s better to require that
X′
n (a) = 1 , which can be achieved by dividing the right side of ( 12.3.23) by the value of its deriva-
tive at x= a; thus,
Xn = 2bcosh(2n− 1)πx/ 2b
(2n− 1)π sinh(2n− 1)πa/ 2b.
Then
vn (x,y ) = Xn (x)Yn(y) = 2bcosh(2n− 1)πx/ 2b
(2n− 1)π sinh(2n− 1)πa/ 2bcos (2n− 1)πy
2b ,
so
∂vn
∂x (a,y ) = cos (2n− 1)πy
2b .",Elementary Differential Equations with Boundary Value Problems.pdf
"Then
vn (x,y ) = Xn (x)Yn(y) = 2bcosh(2n− 1)πx/ 2b
(2n− 1)π sinh(2n− 1)πa/ 2bcos (2n− 1)πy
2b ,
so
∂vn
∂x (a,y ) = cos (2n− 1)πy
2b .
Therefore vn satisfies ( 12.3.20) with g(y) = cos(2 n− 1)πy/ 2b. More generally , if α1, . . . , αm are
arbitrary constants then
um(x,y ) = 2b
π
m∑
n=1
αn
cosh(2n− 1)πx/ 2b
(2n− 1) sinh(2 n− 1)πa/ 2bcos (2n− 1)πy
2b",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 661
satisfies ( 12.3.20) with
g(y) =
∞∑
n=1
αn cos (2n− 1)πy
2b .
Therefore, if g is an arbitrary piecewise smooth function on [0,b ], we define the formal solution of
(12.3.20) to be
u(x,y ) = 2b
π
∞∑
n=1
αn
cosh(2n− 1)πx/ 2b
(2n− 1) sinh(2 n− 1)πa/ 2bcos (2n− 1)πy
2b ,
where
CM (y) =
∞∑
n=1
αn cos (2n− 1)πy
2b
is the mixed Fourier cosine series of gon [0,b ]; that is,
αn = 2
b
∫ b
0
g(y) cos (2n− 1)πy
2b dy.
Example 12.3.8 Solve ( 12.3.20) with g(y) = y− b.
Solution From Example 11.3.3,
CM (y) = − 8b
π2
∞∑
n=1
1
(2n− 1)2 cos (2n− 1)πy
2b .
Therefore
u(x,y ) = −16b2
π3
∞∑
n=1
cosh(2n− 1)πx/ 2b
(2n− 1)3 sinh(2n− 1)πa/ 2bcos (2n− 1)πy
2b .
Laplace’s Equation for a Semi-Infinite Strip
W e now seek solutions of Laplace’s equation on the semi-infin ite strip
S : {0 <x<a, y > 0}
(Figure 12.3.12) that satisfy homogeneous boundary conditions at x= 0 and x= a, and a nonhomoge-",Elementary Differential Equations with Boundary Value Problems.pdf
"S : {0 <x<a, y > 0}
(Figure 12.3.12) that satisfy homogeneous boundary conditions at x= 0 and x= a, and a nonhomoge-
neous Dirichlet or Neumann condition at y= 0 . An example of such a problem is
uxx + uyy = 0 , 0 <x<a, y > 0,
u(x, 0) = f(x), 0 ≤ x≤ a,
u(0,y ) = 0 , u (a,y ) = 0 , y > 0,
(12.3.24)
The boundary conditions in this problem are not sufficient to determine u, for if u0 = u0(x,y ) is a
solution and Kis a constant then
u1(x,y ) = u0(x,y ) + Ksin πx
a sinh πy
a .
is also a solution. (V erify .) However, if we also require — on physical grounds — that the solution remain
bounded for all (x,y ) in Sthen K = 0 and this difficulty is eliminated.",Elementary Differential Equations with Boundary Value Problems.pdf
"662 Chapter 12 Fourier Solutions of Partial Differential Equations
 y
 x a
 uxx + uyy = 0
 u(x,0) = f(x)
 u(0,y) = 0  u(a,y) = 0
Figure 12.3.12 A boundary value problem on a semi-infinite st rip
Example 12.3.9 Define the bounded formal solution of ( 12.3.24).
Solution Proceeding as in the solution of Example 12.3.1, we find that the building block functions are
of the form
vn (x,y ) = Yn (y) sin nπx
a ,
where
Y′′
n − (n2π2/a 2)Yn = 0 .
Therefore
Yn = c1enπy/a + c2e−nπy/a
where c1 and c2 are constants. Although the boundary conditions in ( 12.3.24) don’t restrict c1, and c2,
we must set c1 = 0 to ensure that Yn is bounded. Letting c2 = 1 yields
vn(x,y ) = e−nπy/a sin nπx
a ,
and we define the bounded formal solution of ( 12.3.24) to be
u(x,y ) =
∞∑
n=1
bne−nπy/a sin nπx
a ,
where
S(x) =
∞∑
n=1
bn sin nπx
a
is the Fourier sine series of f on [0,a ].",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 663
See Exercises 29-34 for other boundary value problems on a semi-infinite strip.
12.3 Exercises
In Exercises 1-16 apply the definition developed in Example 1 to solve the bound ary value problem. (Use
Theorem 11.3.5 where it applies.) Where indicated by C , graph the surface u= u(x,y ), 0 ≤ x≤ a,
0 ≤ y ≤ b.
1. uxx + uyy = 0 , 0 <x< 1, 0 <y <1,
u(x, 0) = x(1 − x), u (x, 1) = 0 , 0 ≤ x≤ 1,
u(0,y ) = 0 , u (1,y ) = 0 , 0 ≤ y≤ 1
2. uxx + uyy = 0 , 0 <x< 2, 0 <y <3,
u(x, 0) = x2(2 − x), u (x, 3) = 0 , 0 ≤ x≤ 2,
u(0,y ) = 0 , u (2,y ) = 0 , 0 ≤ y≤ 3
3. C uxx + uyy = 0 , 0 <x< 2, 0 <y< 2,
u(x, 0) =
{ x, 0 ≤ x≤ 1,
2 − x, 1 ≤ x≤ 2, u(x, 2) = 0 , 0 ≤ x≤ 2,
u(0,y ) = 0 , u (2,y ) = 0 , 0 ≤ y≤ 2
4. uxx + uyy = 0 , 0 <x<π, 0 <y< 1,
u(x, 0) = xsin x, u (x,π ) = 0 , 0 ≤ x≤ π,
u(0,y ) = 0 , u (π,y ) = 0 , 0 ≤ y ≤ 1
5. uxx + uyy = 0 , 0 <x< 3, 0 <y <2,
u(x, 0) = 0 , u y(x, 2) = x2, 0 ≤ x≤ 3,
ux(0,y ) = 0 , u x(3,y ) = 0 , 0 ≤ y≤ 2",Elementary Differential Equations with Boundary Value Problems.pdf
"5. uxx + uyy = 0 , 0 <x< 3, 0 <y <2,
u(x, 0) = 0 , u y(x, 2) = x2, 0 ≤ x≤ 3,
ux(0,y ) = 0 , u x(3,y ) = 0 , 0 ≤ y≤ 2
6. uxx + uyy = 0 , 0 <x< 1, 0 <y <2,
u(x, 0) = 0 , u y(x, 2) = 1 − x, 0 ≤ x≤ 1,
ux(0,y ) = 0 , u x(1,y ) = 0 , 0 ≤ y≤ 2
7. uxx + uyy = 0 , 0 <x< 2, 0 <y <2,
u(x, 0) = 0 , u y(x, 2) = x2 − 4, 0 ≤ x≤ 2,
ux(0,y ) = 0 , u x(2,y ) = 0 , 0 ≤ y≤ 2
8. uxx + uyy = 0 , 0 <x< 1, 0 <y <1,
u(x, 0) = 0 , u y(x, 1) = ( x− 1)2, 0 ≤ x≤ 1,
ux(0,y ) = 0 , u x(1,y ) = 0 , 0 ≤ y≤ 1
9. C uxx + uyy = 0 , 0 <x< 3, 0 <y< 2,
u(x, 0) = 0 , u y(x, 2) = 0 , 0 ≤ x≤ 3,
u(0,y ) = y(4 − y), u x(3,y ) = 0 , 0 ≤ y≤ 2
10. uxx + uyy = 0 , 0 <x< 2, 0 <y <1,
u(x, 0) = 0 , u y(x, 1) = 0 , 0 ≤ x≤ 2,
u(0,y ) = y2(3 − 2y), u x(2,y ) = 0 , 0 ≤ y≤ 1
11. uxx + uyy = 0 , 0 <x< 2, 0 <y <2,
u(x, 0) = 0 , u y(x, 2) = 0 , 0 ≤ x≤ 2,
u(0,y ) = ( y− 2)3 + 8 , u x(2,y ) = 0 , 0 ≤ y≤ 2
12. uxx + uyy = 0 , 0 <x< 3, 0 <y <1,
u(x, 0) = 0 , u y(x, 1) = 0 , 0 ≤ x≤ 3,
u(0,y ) = y(2y2 − 9y+ 12) , u x(3,y ) = 0 , 0 ≤ y≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"664 Chapter 12 Fourier Solutions of Partial Differential Equations
13. C uxx + uyy = 0 , 0 <x< 1, 0 <y<π,
uy(x, 0) = 0 , u (x,π ) = 0 , 0 ≤ x≤ 1,
ux(0,y ) = 0 , u x(1,y ) = sin y, 0 ≤ y≤ π
14. uxx + uyy = 0 , 0 <x< 2, 0 <y <3,
uy(x, 0) = 0 , u (x, 3) = 0 , 0 ≤ x≤ 2,
ux(0,y ) = 0 , u x(2,y ) = y(3 − y), 0 ≤ y ≤ 3
15. uxx + uyy = 0 , 0 <x< 1, 0 <y <π,
uy(x, 0) = 0 , u (x,π ) = 0 , 0 ≤ x≤ 1,
ux(0,y ) = 0 , u x(1,y ) = π2 − y2 , 0 ≤ y≤ π
16. uxx + uyy = 0 , 0 <x< 1, 0 <y <1,
uy(x, 0) = 0 , u (x, 1) = 0 , 0 ≤ x≤ 1,
ux(0,y ) = 0 , u x(1,y ) = 1 − y3 , 0 ≤ y≤ 1
In Exercises 17-28 define the formal solution of
uxx + uyy = 0 , 0 <x<a, 0 <y <b
that satisfies the given boundary conditions for general a, b, and f or g. Then solve the boundary value
problem for the specified a, b, and f or g. (Use Theorem 11.3.5 where it applies.) Where indicated by C
, graph the surface u= u(x,y ), 0 ≤ x≤ a, 0 ≤ y≤ b.
17. C u(x, 0) = 0 , u (x,b ) = f(x), 0 <x<a,
u(0,y ) = 0 , u (a,y ) = 0 , 0 <y <b",Elementary Differential Equations with Boundary Value Problems.pdf
", graph the surface u= u(x,y ), 0 ≤ x≤ a, 0 ≤ y≤ b.
17. C u(x, 0) = 0 , u (x,b ) = f(x), 0 <x<a,
u(0,y ) = 0 , u (a,y ) = 0 , 0 <y <b
a= 3 , b= 2 , f(x) = x(3 − x)
18. u(x, 0) = f(x), u (x,b ) = 0 , 0 <x<a,
ux(0,y ) = 0 , u x(a,y ) = 0 , 0 <y<b
a= 2 , b= 1 , f(x) = x2(x− 2)2
19. u(x, 0) = f(x), u (x,b ) = 0 , 0 <x<a,
ux(0,y ) = 0 , u (a,y ) = 0 , 0 <y<b
a= 1 , b= 2 , f(x) = 3 x3 − 4x2 + 1
20. u(x, 0) = f(x), u (x,b ) = 0 , 0 <x<a,
u(0,y ) = 0 , u x(a,y ) = 0 , 0 <y<b
a= 3 , b= 2 , f(x) = x(6 − x)
21. u(x, 0) = f(x), u y(x,b ) = 0 , 0 <x<a,
u(0,y ) = 0 , u (a,y ) = 0 , 0 <y <b
a= π, b= 2 , f(x) = x(π2 − x2)
22. uy(x, 0) = 0 , u (x,b ) = f(x), 0 <x<a,
ux(0,y ) = 0 , u x(a,y ) = 0 , 0 <y<b
a= π, b= 1 , f(x) = x2(x− π)2
23. C uy (x, 0) = f(x), u (x,b ) = 0 , 0 <x<a,
u(0,y ) = 0 , u (a,y ) = 0 , 0 <y <b
a= π, b= 1 , f(x) =
{ x, 0 ≤ x≤ π
2 ,
π − x, π
2 ≤ x≤ π
24. u(x, 0) = 0 , u (x,b ) = 0 , 0 <x<a,
ux(0,y ) = 0 , u (a,y ) = g(y), 0 <y<b
a= 1 , b= 1 , g(y) = y(y3 − 2y2 + 1)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.3 Laplace’s Equation in Rectangular Coordinates 665
25. C uy (x, 0) = 0 , u (x,b ) = 0 , 0 <x<a,
ux(0,y ) = 0 , u (a,y ) = g(y), 0 <y<b
a= 2 , b= 2 , g(y) = 4 − y2
26. u(x, 0) = 0 , u (x,b ) = 0 , 0 <x<a,
ux(0,y ) = 0 , u x(a,y ) = g(y), 0 <y<b
a= 1 , b= 4 , g(y) =
{ y, 0 ≤ y ≤ 2,
4 − y, 2 ≤ y ≤ 4
27. u(x, 0) = 0 , u y(x,b ) = 0 , 0 <x<a,
ux(0,y ) = g(y), u x(a,y ) = 0 , 0 <y<b
a= 1 , b= π, g(y) = y2 (3π − 2y)
28. uy(x, 0) = 0 , u y(x,b ) = 0 , 0 <x<a,
ux(0,y ) = g(y), u (a,y ) = 0 , 0 <y<b
a= 2 , b= π, g(y) = y
In Exercises 29-34 define the bounded formal solution of
uxx + uyy = 0 , 0 <x<a, y> 0
that satisfies the given boundary conditions for general aand f. Then solve the boundary value problem
for the specified aand f.
29. u(x, 0) = f(x), 0 <x<a ,
ux(0,y ) = 0 , u x(a,y ) = 0 , y> 0
a= πf (x) = x2(3π − 2x)
30. u(x, 0) = f(x), 0 <x<a ,
ux(0,y ) = 0 , u (a,y ) = 0 , y> 0
a= 3 , f(x) = 9 − x2
31. u(x, 0) = f(x), 0 <x<a ,
u(0,y ) = 0 , u x(a,y ) = 0 , y> 0",Elementary Differential Equations with Boundary Value Problems.pdf
"30. u(x, 0) = f(x), 0 <x<a ,
ux(0,y ) = 0 , u (a,y ) = 0 , y> 0
a= 3 , f(x) = 9 − x2
31. u(x, 0) = f(x), 0 <x<a ,
u(0,y ) = 0 , u x(a,y ) = 0 , y> 0
a= π, f(x) = x(2π − x)
32. uy(x, 0) = f(x), 0 <x<a ,
u(0,y ) = 0 , u (a,y ) = 0 , y> 0
a= π, f(x) = x2(π − x)
33. uy(x, 0) = f(x), 0 <x<a ,
ux(0,y ) = 0 , u (a,y ) = 0 , y> 0
a= 7 , f(x) = x(7 − x)
34. uy(x, 0) = f(x), 0 <x<a ,
u(0,y ) = 0 , u x(a,y ) = 0 , y> 0
a= 5 , f(x) = x(5 − x)
35. Define the formal solution of the Dirichlet problem
uxx + uyy = 0 , 0 <x<a, 0 <y<b,
u(x, 0) = f0(x), u (x,b ) = f1(x), 0 ≤ x≤ a,
u(0,y ) = g0(y), u (a,y ) = g1(y), 0 ≤ y≤ b
36. Show that the Neumann Problem
uxx + uyy = 0 , 0 <x<a, 0 <y<b,
uy(x, 0) = f0(x), u y(x,b ) = f1(x), 0 ≤ x≤ a,
ux(0,y ) = g0(y), u x(a,y ) = g1(y), 0 ≤ y≤ b",Elementary Differential Equations with Boundary Value Problems.pdf
"666 Chapter 12 Fourier Solutions of Partial Differential Equations
has no solution unless
∫ a
0
f0(x) dx=
∫ a
0
f1(x) dx=
∫ b
0
g0(y) dy =
∫ b
0
g1(y) dy = 0 .
In this case it has infinitely many formal solutions. Find the m.
37. In this exercise take it as given that the infinite series ∑ ∞
n=1 np e−qn converges for all pif q >0,
and, where appropriate, use the comparison test for absolut e convergence of an infinite series.
Let
u(x,y ) =
∞∑
n=1
αn
sinh nπ(b− y)/a
sinh nπb/a sin nπx
a ,
where
αn = 2
a
∫ a
0
f(x) sin nπx
a dx
and f is piecewise smooth on [0,a ].
(a) V erify the approximations
sinh nπ(b− y)/a
sinh nπb/a ≈ e−nπy/a , y <b, (A)
and
cosh nπ(b− y)/a
sinh nπb/a ≈ e−nπy/a , y <b (B)
for large n.
(b) Use (A) to show that uis defined for (x,y ) such that 0 <y<b .
(c) For fixed yin (0,b ), use (A) and Theorem 12.1.2 with z= xto show that
ux(x,y ) = π
a
∞∑
n=1
nαn
sinh nπ(b− y)/a
sinh nπb/a cos nπx
a , −∞ <x< ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"(c) For fixed yin (0,b ), use (A) and Theorem 12.1.2 with z= xto show that
ux(x,y ) = π
a
∞∑
n=1
nαn
sinh nπ(b− y)/a
sinh nπb/a cos nπx
a , −∞ <x< ∞ .
(d) Starting from the result of (b), use (A) and Theorem 12.1.2 with z = xto show that, for a
fixed yin (0,b ),
uxx(x,y ) = −π2
a2
∞∑
n=1
n2αn
sinh nπ(b− y)/a
sinh nπb/a sin nπx
a , −∞ <x< ∞ .
(e) For fixed but arbitrary x, use (B) and Theorem 12.1.2 with z= yto show that
uy (x,y ) = −π
a
∞∑
n=1
nαn
cosh nπ(b− y)/a
sinh nπb/a sin nπx
a
if 0 < y0 < y < b, where y0 is an arbitrary number in (0,b ). Then argue that since y0 can
be chosen arbitrarily small, the conclusion holds for all yin (0,b ).
(f) Starting from the result of (e), use (A) and Theorem 12.1.2 to show that
uyy (x,y ) = π2
a2
∞∑
n=1
n2αn
sinh nπ(b− y)/a
sinh nπb/a sin nπx
a , 0 <y <b.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.4 Laplace’s Equation in Polar Coordinates 667
(g) Conclude that usatisfies Laplace’s equation for all (x,y ) such that 0 <y <b.
By repeatedly applying the arguments in (c)– (f), it can be shown that ucan be differentiated term
by term any number of times with respect to xand/or yif 0 <y <b.
12.4 LAPLACE’S EQUA TION IN POLAR COORDINA TES
In Section 12.3 we solved boundary value problems for Laplac e’s equation over a rectangle with sides
parallel to the x,y -axes. Now we’ll consider boundary value problems for Lapla ce’s equation over regions
with boundaries best described in terms of polar coordinate s. In this case it’s appropriate to regard uas
function of (r,θ ) and write Laplace’s equation in polar form as
urr + 1
rur + 1
r2 uθθ = 0 , (12.4.1)
where
r=
√
x2 + y2 and θ = cos −1 x
r = sin −1 x
r.
W e begin with the case where the region is a circular disk with radius ρ, centered at the origin; that is,
we want to define a formal solution of the boundary value probl em",Elementary Differential Equations with Boundary Value Problems.pdf
"we want to define a formal solution of the boundary value probl em
urr + 1
rur + 1
r2 uθθ = 0 , 0 <r<ρ, −π ≤ θ<π,
u(ρ,θ ) = f(θ), −π ≤ θ<π
(12.4.2)
(Figure 12.4.1). Note that ( 12.4.2) imposes no restriction on u(r,θ ) when r = 0 . W e’ll address this
question at the appropriate time.
 ur r + r−1  ur + r−2  uθ  θ  = 0 
 u(ρ ,θ ) = f(θ )
 x
 y
Figure 12.4.1 The boundary value problem ( 12.4.2)",Elementary Differential Equations with Boundary Value Problems.pdf
"668 Chapter 12 Fourier Solutions of Partial Differential Equations
W e first look for products v(r,θ ) = R(r)Θ( θ) that satisfy ( 12.4.1). For this function,
vrr + 1
rvr + 1
r2 vθθ = R′′Θ + 1
rR′Θ + 1
r2 RΘ ′′= 0
for all (r,θ ) with r̸= 0 if
r2R′′+ rR′
R = −Θ ′′
Θ = λ,
where λ is a separation constant. (V erify .) This equation is equiva lent to
Θ ′′+ λΘ = 0
and
r2R′′+ rR′− λR = 0 . (12.4.3)
Since (r,π ) and (r, −π) are the polar coordinates of the same point, we impose period ic boundary con-
ditions on Θ ; that is,
Θ ′′+ λΘ = 0 , Θ( −π) = Θ( π), Θ ′(−π) = Θ ′(π). (12.4.4)
Since we don’t want RΘ to be identically zero, λ must be an eigenvalue of ( 12.4.4) and Θ must be an
associated eigenfunction. From Theorem 11.1.6, the eigenvalues of ( 12.4.4) are λ0 = 0 with associated
eigenfunctions Θ 0 = 1 and, for n = 1 , 2, 3,..., λ n = n2, with associated eigenfunction cos nθ and
sin nθ therefore,
Θ n = αn cos nθ + βn sin nθ
where αn and βn are constants.",Elementary Differential Equations with Boundary Value Problems.pdf
"sin nθ therefore,
Θ n = αn cos nθ + βn sin nθ
where αn and βn are constants.
Substituting λ = 0 into ( 12.4.3) yields the
r2R′′+ rR′= 0 ,
so
R′′
0
R′
0
= −1
r,
R′
0 = c1
r ,
and
R0 = c2 + c1 ln r. (12.4.5)
If c1 ̸= 0 then
lim
r→ 0+
|R0(r)|= ∞ ,
which doesn’t make sense if we interpret u0(r,θ ) = R0(r)Θ 0(θ) = R0(r) as the steady state temperature
distribution in a disk whose boundary is maintained at the co nstant temperature R0(ρ). Therefore we now
require R0 to be bounded as r → 0+. This implies that c1 = 0 , and we take c2 = 1 . Thus, R0 = 1 and
v0(r,θ ) = R0(r)Θ 0(θ) = 1 . Note that v0 satisfies ( 12.4.2) with f(θ) = 1 .
Substituting λ = n2 into ( 12.4.3) yields the Euler equation
r2R′′
n + rR′
n − n2Rn = 0 (12.4.6)
for Rn. The indicial polynomial of this equation is
s(s− 1) + s− n2 = ( s− n)(s+ n),",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.4 Laplace’s Equation in Polar Coordinates 669
so the general solution of ( 12.4.6) is
Rn = c1rn + c2r−n , (12.4.7)
by Theorem 7.4.3. Consistent with our previous assumption on R0, we now require Rn to be bounded as
r→ 0+. This implies that c2 = 0 , and we choose c1 = ρ−n . Then Rn(r) = rn /ρ n , so
vn (r,θ ) = Rn(r)Θ n (θ) = rn
ρn (αn cos nθ + sin nθ).
Now vn satisfies ( 12.4.2) with
f(θ) = αn cos nθ + βn sin nθ.
More generally , if α0, α1,. . . , αm and β1, β2, . . . , βm are arbitrary constants then
um(r,θ ) = α0 +
m∑
n=1
rn
ρn (αn cos nθ + βn sin nθ)
satisfies ( 12.4.2) with
f(θ) = α0 +
m∑
n=1
(αn cos nθ + βn sin nθ).
This motivates the next definition.
Definition 12.4.1
The bounded formal solution of the boundary value problem ( 12.4.2) is
u(r,θ ) = α0 +
∞∑
n=1
rn
ρn (αn cos nθ + βn sin nθ), (12.4.8)
where
F(θ) = α0 +
∞∑
n=1
(αn cos nθ + βn sin nθ)
is the Fourier series of f on [−π,π ]; that is,
α0 = 1
2π
∫ π
−π
f(θ) dθ,
and
αn = 1
π
∫ π
−π
f(θ) cos nθdθ and βn = 1
π
∫ π
−π",Elementary Differential Equations with Boundary Value Problems.pdf
"is the Fourier series of f on [−π,π ]; that is,
α0 = 1
2π
∫ π
−π
f(θ) dθ,
and
αn = 1
π
∫ π
−π
f(θ) cos nθdθ and βn = 1
π
∫ π
−π
f(θ) sin nθdθ, n = 1 , 2, 3,....
Since ∑ ∞
n=0 nk(r/ρ )n converges for every kif 0 < r < ρ, Theorem
12.1.2 can be used to show that
if 0 <r<ρ then ( 12.4.8) can be differentiated term by term any number of times with r espect to both r
and θ. Since the terms in ( 12.4.8) satisfy Laplace’s equation if r> 0, ( 12.4.8) satisfies Laplace’s equation
if 0 <r <ρ . Therefore, since u(ρ,θ ) = F(θ), uis an actual solution of ( 12.4.2) if and only if
F(θ) = f(θ), −π ≤ θ<π.
From Theorem 11.2.4, this is true if fis continuous and piecewise smooth on [−π,π ] and f(−π) = f(π).
Example 12.4.1 Find the bounded formal solution of ( 12.4.2) with f(θ) = θ(π2 − θ2).",Elementary Differential Equations with Boundary Value Problems.pdf
"670 Chapter 12 Fourier Solutions of Partial Differential Equations
Solution From Example 11.2.6,
θ(π2 − θ2) = 12
∞∑
n=1
(−1)n
n3 sin nθ, −π ≤ θ ≤ π,
so
u(r,θ ) = 12
∞∑
n=1
rn
ρn
(−1)n
n3 sin nθ, 0 ≤ r≤ ρ, −π ≤ θ ≤ π.
Example 12.4.2 Define the formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , ρ 0 <r <ρ, −π ≤ θ<π,
u(ρ0,θ ) = 0 , u (ρ,θ ) = f(θ), −π ≤ θ<π,
(12.4.9)
where 0 <ρ 0 <ρ (Figure 12.4.2).
 ur r + r−1  ur + r−2  uθ  θ  = 0 
 u(ρ ,θ ) = f(θ )
 u(ρ 0,θ ) = 0
 x
 y
Figure 12.4.2 The boundary value problem ( 12.4.9)
Solution W e use separation of variables exactly as before, except tha t now we choose the constants in
(12.4.5) and ( 12.4.7) so that Rn (ρ0) = 0 for n = 0 , 1, 2,. . . . In view of the nonhomogeneous Dirichlet
condition on the boundary r = ρ, it’s also convenient to require that Rn(ρ) = 1 for n= 0 , 1, 2,. . . . W e
leave it to you to verify that
R0(r) = ln r/ρ 0
ln ρ/ρ 0
and Rn = ρ−n
0 rn − ρn
0 r−n
ρ−n
0 ρn − ρn
0 ρ−n , n = 1 , 2, 3,...",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.4 Laplace’s Equation in Polar Coordinates 671
satisfy these requirements. Therefore
v0(ρ,θ ) = ln r/ρ 0
ln ρ/ρ 0
and
vn (r,θ ) = ρ−n
0 rn − ρn
0 r−n
ρ−n
0 ρn − ρn
0 ρ−n (αn cos nθ + βn sin nθ), n = 1 , 2, 3,...,
where αn and βn are arbitrary constants.
If α0, α1,. . . , αm and β1, β2, . . . , βm are arbitrary constants then
um(r,θ ) = α0
ln r/ρ 0
ln ρ/ρ 0
+
m∑
n=1
ρ−n
0 rn − ρn
0 r−n
ρ−n
0 ρn − ρn
0 ρ−n (αn cos nθ + βn sin nθ)
satisfies (
12.4.9), with
f(θ) = α0 +
m∑
n=1
(αn cos nθ + βn sin nθ).
This motivates us to define the formal solution of ( 12.4.9) for general f to be
u(r,θ ) = α0
ln r/ρ 0
ln ρ/ρ 0
+
∞∑
n=1
ρ−n
0 rn − ρn
0 r−n
ρ−n
0 ρn − ρn
0 ρ−n (αn cos nθ + βn sin nθ),
where
F(θ) = α0 +
∞∑
n=1
(αn cos nθ + βn sin nθ)
is the Fourier series of f on [−π,π ].
Example 12.4.3 Define the bounded formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , 0 <r<ρ, 0 <θ<γ,
u(ρ,θ ) = f(θ), 0 ≤ θ ≤ γ,
u(r, 0) = 0 , u (r,γ ) = 0 , 0 <r<ρ,
(12.4.10)
where 0 <γ < 2π (Figure 12.4.3).",Elementary Differential Equations with Boundary Value Problems.pdf
"rur + 1
r2 uθθ = 0 , 0 <r<ρ, 0 <θ<γ,
u(ρ,θ ) = f(θ), 0 ≤ θ ≤ γ,
u(r, 0) = 0 , u (r,γ ) = 0 , 0 <r<ρ,
(12.4.10)
where 0 <γ < 2π (Figure 12.4.3).
Solution Now v(r,θ ) = R(r)Θ( θ), where
r2R′′+ rR′− λR = 0 (12.4.11)
and
Θ ′′+ λΘ = 0 , Θ(0) = 0 , Θ( γ) = 0 . (12.4.12)
From Theorem 11.1.2, the eigenvalues of ( 12.4.12) are λn = n2π2/γ 2, with associated eigenfunction
Θ n = sin nπθ/γ , n= 1 , 2, 3,. . . . Substituting λ = n2π2/γ 2 into ( 12.4.11) yields the Euler equation
r2R′′+ rR′
n − n2π2
γ2 R= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"672 Chapter 12 Fourier Solutions of Partial Differential Equations
 ur r + r−1  ur + r−2  uθ  θ  = 0 
 u(ρ ,θ ) = f(θ ) u(r,γ ) = 0
 γ  x
 y
Figure 12.4.3 The boundary value problem ( 12.4.10)
The indicial polynomial of this equation is
s(s− 1) + s− n2π2
γ2 =
(
s− nπ
γ
)(
s+ nπ
γ
)
,
so
Rn = c1rnπ/γ + c2r−nπ/γ ,
by Theorem 7.4.3. T o obtain a solution that remains bounded as r → 0+ we let c2 = 0 . Because of the
Dirichlet condition at r= ρ, it’s convenient to have r(ρ) = 1 ; therefore we take c1 = ρ−nπ/γ , so
Rn(r) = rnπ/γ
ρnπ/γ .
Now
vn (r,θ ) = Rn(r)Θ n (θ) = rnπ/γ
ρnπ/γ sin nπθ
γ
satisfies ( 12.4.10) with
f(θ) = sin nπθ
γ .
More generally , if α1, α2, . . . , αm and are arbitrary constants then
um (r,θ ) =
m∑
n=1
αn
rnπ/γ
ρnπ/γ sin nπθ
γ
satisfies ( 12.4.10) with
f(θ) =
m∑
n=1
αn sin nπθ
γ .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 12.4 Laplace’s Equation in Polar Coordinates 673
This motivates us to define the bounded formal solution of ( 12.4.10) to be
um (r,θ ) =
∞∑
n=1
αn
rnπ/γ
ρnπ/γ sin nπθ
γ ,
where
S(θ) =
∞∑
n=1
αn sin nπθ
γ
is the Fourier sine expansion of f on [0,γ ]; that is,
αn = 2
γ
∫ γ
0
f(θ) sin nπθ
γ dθ.
12.4 Exercises
1. Define the formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , ρ 0 <r<ρ, −π ≤ θ<π,
u(ρ0,θ ) = f(θ), u (ρ,θ ) = 0 , −π ≤ θ<π,
where 0 <ρ 0 <ρ .
2. Define the formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , ρ 0 <r<ρ, 0 <θ<γ,
u(ρ0,θ ) = 0 , u (ρ,θ ) = f(θ), 0 ≤ θ ≤ γ,
u(r, 0) = 0 , u (r,γ ) = 0 , ρ 0 <r<ρ,
where 0 <γ < 2π and 0 <ρ 0 <ρ .
3. Define the formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , ρ 0 <r<ρ, 0 <θ<γ,
u(ρ0,θ ) = 0 , u r (ρ,θ ) = g(θ), 0 ≤ θ ≤ γ,
uθ (r, 0) = 0 , u θ (r,γ ) = 0 , ρ 0 <r <ρ,
where 0 <γ < 2π and 0 <ρ 0 <ρ .
4. Define the bounded formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , 0 <r<ρ, 0 <θ<γ,
u(ρ,θ ) = f(θ), 0 ≤ θ ≤ γ,
uθ (r, 0) = 0 , u (r,γ ) = 0 , 0 <r <ρ,",Elementary Differential Equations with Boundary Value Problems.pdf
"urr + 1
rur + 1
r2 uθθ = 0 , 0 <r<ρ, 0 <θ<γ,
u(ρ,θ ) = f(θ), 0 ≤ θ ≤ γ,
uθ (r, 0) = 0 , u (r,γ ) = 0 , 0 <r <ρ,
where 0 <γ < 2π.",Elementary Differential Equations with Boundary Value Problems.pdf
"674 Chapter 12 Fourier Solutions of Partial Differential Equations
5. Define the formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , ρ 0 <r<ρ, 0 <θ<γ,
ur (ρ0,θ ) = g(θ), u r (ρ,θ ) = 0 , 0 ≤ θ ≤ γ,
u(r, 0) = 0 , u θ (r,γ ) = 0 , ρ 0 <r<ρ,
where 0 <γ < 2π and 0 <ρ 0 <ρ .
6. Define the bounded formal solution of
urr + 1
rur + 1
r2 uθθ = 0 , 0 <r<ρ, 0 <θ<γ,
u(ρ,θ ) = f(θ), 0 ≤ θ ≤ γ,
uθ (r, 0) = 0 , u θ (r,γ ) = 0 , 0 <r<ρ,
where 0 <γ < 2π.
7. Show that the Neumann problem
urr + 1
rur + 1
r2 uθθ = 0 , 0 <r<ρ, −π ≤ θ<π,
ur (ρ,θ ) = f(θ), −π ≤ θ <π
has no bounded formal solution unless ∫π
−π f(θ) dθ = 0 . In this case it has infinitely many solu-
tions. Find those solutions.",Elementary Differential Equations with Boundary Value Problems.pdf
"CHAPTER 13
Boundary V alue Problems for Second
Order Ordinary Differential Equations
IN THIS CHAPTER we discuss boundary value problems and eigen value problems for linear second
order ordinary differential equations.
Section 13.1 discusses point two-point boundary value prob lems for linear second order ordinary differ-
ential equations.
Section 13.2 deals with generalizations of the eigenvalue p roblems considered in Section 11.1
677",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.1 T wo-Point Boundary V alue Problems 677
13.1 TWO-POINT BOUNDARY V ALUE PROBLEMS
In Section 5.3 we considered initial value problems for the l inear second order equation
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x). (13.1.1)
Suppose P0, P1, P2, and F are continuous and P0 has no zeros on an open interval (a,b ). From Theo-
rem 5.3.1, if x0 is in (a,b ) and k1 and k2 are arbitrary real numbers then ( 13.1.1) has a unique solution
on (a,b ) such that y(x0 ) = k1 and y′(x0) = k2. Now we consider a different problem for ( 13.1.1).
PROBLEM Suppose P0, P1, P2, and F are continous and P0 has no zeros on a closed interval [a,b ].
Let α, β, ρ, and δ be real numbers such that
α2 + β2 ̸= 0 and ρ2 + δ2 ̸= 0 , (13.1.2)
and let k1 and k2 be arbitrary real numbers. Find a solution of
P0(x)y′′+ P1(x)y′+ P2(x)y= F(x) (13.1.3)
on the closed interval [a,b ] such that
αy (a) + βy′(a) = k1 (13.1.4)
and
ρy(b) + δy′(b) = k2. (13.1.5)",Elementary Differential Equations with Boundary Value Problems.pdf
"P0(x)y′′+ P1(x)y′+ P2(x)y= F(x) (13.1.3)
on the closed interval [a,b ] such that
αy (a) + βy′(a) = k1 (13.1.4)
and
ρy(b) + δy′(b) = k2. (13.1.5)
The assumptions stated in this problem apply throughout thi s section and won’t be repeated. Note
that we imposed conditions on P0, P1, P2, and F on the closed interval [a,b ], and we are interested in
solutions of ( 13.1.3) on the closed interval. This is different from the situatio n considered in Chapter 5,
where we imposed conditions on P0, P1, P2, and F on the open interval (a,b ) and we were interested in
solutions on the open interval. There is really no problem he re; we can always extend P0, P1, P2, and
F to an open interval (c,d ) (for example, by defining them to be constant on (c,d ] and [b,d )) so that
they are continuous and P0 has no zeros on [c,d ]. Then we can apply the theorems from Chapter 5 to the
equation
y′′+ P1(x)
P0(x) y′+ P2(x)
P0(x) y = F(x)
P0(x)
on (c,d ) to draw conclusions about solutions of ( 13.1.3) on [a,b ].",Elementary Differential Equations with Boundary Value Problems.pdf
"equation
y′′+ P1(x)
P0(x) y′+ P2(x)
P0(x) y = F(x)
P0(x)
on (c,d ) to draw conclusions about solutions of ( 13.1.3) on [a,b ].
W e call aand bboundary points. The conditions ( 13.1.4) and ( 13.1.5) are boundary conditions , and the
problem is a two-point boundary value problem or, for simplicity , a boundary value problem . (W e used
similar terminology in Chapter 12 with a different meaning; both meanings are in common usage.) W e
require ( 13.1.2) to insure that we’re imposing a sensible condition at each b oundary point. For example,
if α2 + β2 = 0 then α = β = 0 , so αy (a) + βy′(a) = 0 for all choices of y(a) and y′(a). Therefore
(13.1.4) is an impossible condition if k1 ̸= 0 , or no condition at all if k1 = 0 .
W e abbreviate ( 13.1.1) as Ly= F, where
Ly= P0(x)y′′+ P1(x)y′+ P0(x)y,
and we denote
B1(y) = αy (a) + βy′(a) and B2(y) = ρy(b) + δy′(b).
W e combine ( 13.1.3), ( 13.1.4), and ( 13.1.5) as
Ly= F, B 1(y) = k1, B 2(y) = k2. (13.1.6)",Elementary Differential Equations with Boundary Value Problems.pdf
"678 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
This boundary value problem is homogeneous if F = 0 and k1 = k2 = 0 ; otherwise it’s nonhomoge-
neous.
W e leave it to you (Exercise 1) to verify that B1 and B2 are linear operators; that is, if c1 and c2 are
constants then
Bi(c1y1 + c2y2) = c1Bi (y1) + c2Bi (y2), i = 1 , 2. (13.1.7)
The next three examples show that the question of existence a nd uniqueness for solutions of boundary
value problems is more complicated than for initial value pr oblems.
Example 13.1.1 Consider the boundary value problem
y′′+ y= 1 , y (0) = 0 , y (π/ 2) = 0 .
The general solution of y′′+ y= 1 is
y= 1 + c1 sin x+ c2 cos x,
so y(0) = 0 if and only if c2 = −1 and y(π/ 2) = 0 if and only if c1 = −1. Therefore
y= 1 − sin x− cos x
is the unique solution of the boundary value problem.
Example 13.1.2 Consider the boundary value problem
y′′+ y = 1 , y (0) = 0 , y (π) = 0 .
Again, the general solution of y′′+ y= 1 is",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 13.1.2 Consider the boundary value problem
y′′+ y = 1 , y (0) = 0 , y (π) = 0 .
Again, the general solution of y′′+ y= 1 is
y= 1 + c1 sin x+ c2 cos x,
so y(0) = 0 if and only if c2 = −1, but y(π) = 0 if and only if c2 = 1 . Therefore the boundary value
problem has no solution.
Example 13.1.3 Consider the boundary value problem
y′′+ y= sin 2 x, y (0) = 0 , y (π) = 0 .
Y ou can use the method of undetermined coefficients (Section 5.5) to find that the general solution of
y′′+ y = sin 2 xis
y= −sin 2 x
3 + c1 sin x+ c2 cos x.
The boundary conditions y(0) = 0 and y(π) = 0 both require that c2 = 0 , but they don’t restrict c1.
Therefore the boundary value problem has infinitely many sol utions
y= −sin 2 x
3 + c1 sin x,
where c1 is arbitrary .
Theorem 13.1.1 If z1 and z2 are solutions of Ly= 0 such that either B1(z1) = B1(z2) = 0 or B2 (z1) =
B2(z2 ) = 0 , then {z1,z 2}is linearly dependent. Equivalently , if {z1,z 2}is linearly independent, then
B2
1 (z1) + B2
1 (z2) ̸= 0 and B2",Elementary Differential Equations with Boundary Value Problems.pdf
"B2(z2 ) = 0 , then {z1,z 2}is linearly dependent. Equivalently , if {z1,z 2}is linearly independent, then
B2
1 (z1) + B2
1 (z2) ̸= 0 and B2
2 (z1) + B2
2 (z2) ̸= 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.1 T wo-Point Boundary V alue Problems 679
Proof Recall that B1(z) = αz (a) + βz′(a) and α2 + β2 ̸= 0 . Therefore, if B1(z1) = B1(z2) = 0 then
(α,β ) is a nontrivial solution of the system
αz1(a) + βz′
1(a) = 0
αz2(a) + βz′(a) = 0 .
This implies that
z1(a)z′
2(a) − z′
1(a)z2(a) = 0 ,
so {z1,z 2}is linearly dependent, by Theorem 5.1.6. W e leave it to you to show that {z1,z 2}is linearly
dependent if B2(z1) = B2(z2) = 0 .
Theorem 13.1.2 The following statements are equivalent ; that is , they are either all true or all false .
(a) There’s a fundamental set {z1,z 2}of solutions of Ly= 0 such that
B1(z1)B2 (z2) − B1(z2)B2(z1) ̸= 0 . (13.1.8)
(b) If {y1,y 2}is a fundamental set of solutions of Ly= 0 then
B1(y1)B2 (y2) − B1(y2)B2(y1) ̸= 0 . (13.1.9)
(c) F or each continuous F and pair of constants (k1,k 2), the boundary value problem
Ly= F, B 1(y) = k1, B 2(y) = k2
has a unique solution .
(d) The homogeneous boundary value problem
Ly= 0 , B 1(y) = 0 , B 2(y) = 0 (13.1.10)",Elementary Differential Equations with Boundary Value Problems.pdf
"Ly= F, B 1(y) = k1, B 2(y) = k2
has a unique solution .
(d) The homogeneous boundary value problem
Ly= 0 , B 1(y) = 0 , B 2(y) = 0 (13.1.10)
has only the trivial solution y= 0 .
(e) The homogeneous equation Ly= 0 has linearly independent solutions z1 and z2 such that B1(z1) =
0 and B2 (z2) = 0 .
Proof W e’ll show that
(a) = ⇒ (b) = ⇒ (c) = ⇒ (d) = ⇒ (e) = ⇒ (a).
(a) =⇒ (b): Since {z1,z 2}is a fundamental set of solutions for Ly = 0 , there are constants a1, a2,
b1, and b2 such that
y1 = a1z1 + a2z2
y2 = b1z1 + b2z2. (13.1.11)
Moreover, ⏐
⏐
⏐
⏐
a1 a2
b1 b2
⏐
⏐
⏐
⏐̸= 0 . (13.1.12)
because if this determinant were zero, its rows would be line arly dependent and therefore {y1,y 2}would
be linearly dependent, contrary to our assumption that {y1,y 2}is a fundamental set of solutions of Ly=
0. From (
13.1.7) and ( 13.1.11),
[ B1(y1) B2(y1)
B1(y2) B2(y2)
]
=
[ a1 a2
b1 b2
][ B1(z1) B2(z1)
B1(z2) B2(z2)
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"680 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
Since the determinant of a product of matrices is the product of the determinants of the matrices, ( 13.1.8)
and ( 13.1.12) imply ( 13.1.9).
(b) =⇒ (c): Since {y1,y 2}is a fundamental set of solutions of Ly = 0 , the general solution of
Ly= F is
y= yp + c1y1 + c2y2,
where c1 and c2 are arbitrary constants and yp is a particular solution of Ly= F. T o satisfy the boundary
conditions, we must choose c1 and c2 so that
k1 = B1(yp ) + c1B1(y1) + c2B1(y2)
k2 = B2(yp ) + c1B2(y1) + c2B2(y2),
(recall ( 13.1.7)), which is equivalent to
c1B1 (y1) + c2B1(y2) = k1 − B1(yp )
c1B2(y1) + c2B2(22) = k2 − B2(yp ).
From ( 13.1.9), this system always has a unique solution (c1,c 2).
(c) =⇒ (d): Obviously , y = 0 is a solution of ( 13.1.10). From (c) with F = 0 and k1 = k2 = 0 , it’s
the only solution.
(d) =⇒ (e): Let {y1,y 2}be a fundamental system for Ly= 0 and let",Elementary Differential Equations with Boundary Value Problems.pdf
"the only solution.
(d) =⇒ (e): Let {y1,y 2}be a fundamental system for Ly= 0 and let
z1 = B1 (y2)y1 − B1(y1)y2 and z2 = B2(y2)y1 − B2(y1)y2.
Then B1(z1) = 0 and B2(z2) = 0 . T o see that z1 and z2 are linearly independent, note that
a1z1 + a2z2 = a1[B1(y2 )y1 − B1(y1 )y2] + a2[B2(y2)y1 − B2(y1)y2]
= [ B1(y2)a1 + B2(y2 )a2]y1 − [B1(y1)a1 + B2(y1)a2]y2.
Therefore, since y1 and y2 are linearly independent, a1z1 + a2z2 = 0 if and only if
[ B1(y1) B2(y1)
B1(y2) B2(y2)
][ a1
a2
]
=
[ 0
0
]
.
If this system has a nontrivial solution then so does the syst em
[ B1(y1) B1(y2)
B2(y1) B2(y2)
][ c1
c2
]
=
[ 0
0
]
.
This and (
13.1.7) imply that y= c1z1 + c2z2 is a nontrivial solution of ( 13.1.10), which contradicts (d).
(e) =⇒ (a). Theorem 13.1.1 implies that if B1(z1) = 0 and B2(z2) = 0 then B1(z2) ̸= 0 and
B2(z1 ) ̸= 0 . This implies ( 13.1.8), which completes the proof.
Example 13.1.4 Solve the boundary value problem
x2y′′− 2xy′+ 2 y− 2x3 = 0 , y (1) = 4 , y ′(2) = 3 , (13.1.13)",Elementary Differential Equations with Boundary Value Problems.pdf
"Example 13.1.4 Solve the boundary value problem
x2y′′− 2xy′+ 2 y− 2x3 = 0 , y (1) = 4 , y ′(2) = 3 , (13.1.13)
given that {x,x 2}is a fundamental set of solutions of the complementary equat ion
Solution Using variation of parameters (Section 5.7), you can show th at yp = x3 is a solution of the
complementary equation
x2y′′− 2xy′+ 2 y= 2 x3 = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.1 T wo-Point Boundary V alue Problems 681
Therefore the solution of ( 13.1.13) can be written as
y= x3 + c1x+ c2x2.
Then
y′= 3 x2 + c1 + 2 c2x,
and imposing the boundary conditions yields the system
c1 + c2 = 3
c1 + 4 c2 = −9,
so c1 = 7 and c2 = −4. Therefore
y = x3 + 7 x− 4x2
is the unique solution of ( 13.1.13)
Example 13.1.5 Solve the boundary value problem
y′′− 7y′+ 12 y= 4 e2x, y (0) = 3 , y (1) = 5 e2.
Solution From Example 5.4.1, yp = 2 e2x is a particular solution of
y′′− 7y′+ 12 y= 4 e2x. (13.1.14)
Since {e3x,e 4x}is a fundamental set for the complementary equation, we coul d write the solution of
(13.1.13) as
y= 2 e2x + c1e3x + c2e4x
and determine c1 and c2 by imposing the boundary conditions. However, this would le ad to some tedious
algebra, and the form of the solution would be very unappeali ng. (Try it!) In this case it’s convenient to
use the fundamental system {z1,z 2}mentioned in Theorem 13.1.2(e); that is, we choose {z1,z 2}so that",Elementary Differential Equations with Boundary Value Problems.pdf
"use the fundamental system {z1,z 2}mentioned in Theorem 13.1.2(e); that is, we choose {z1,z 2}so that
B1(z1 ) = z1(0) = 0 and B2(z2) = z2(1) = 0 . It is easy to see that
z1 = e3x − e4x and z2 = e3(x−1) − e4(x−1)
satisfy these requirements. Now we write the solution of ( 13.1.14) as
y= 2 e2x + c1
(e3x − e4x) + c2
(
e3(x−1) − e4(x−1)
)
.
Imposing the boundary conditions y(0) = 3 and y(1) = 5 e2 yields
3 = 2 + c2e−4(e− 1) and 5e2 = 2 e2 + c1e3(1 − e).
Therefore
c1 = 3
e(1 − e) , c 2 = e4
e− 1 ,
and
y= 2 e2x + 3
e(1 − e) (e3x − e4x) + e4
e− 1 (e3(x−1) − e4(x−1)).
Sometimes it’s useful to have a formula for the solution of a g eneral boundary problem. Our next
theorem addresses this question.",Elementary Differential Equations with Boundary Value Problems.pdf
"682 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
Theorem 13.1.3 Suppose the homogeneous boundary value problem
Ly= 0 , B 1(y) = 0 , B 2(y) = 0 (13.1.15)
has only the trivial solution . Let y1 and y2 be linearly independent solutions of Ly = 0 such that
B1(y1 ) = 0 and B2(y2) = 0 , and let
W = y1y′
2 − y′
1y2.
Then the unique solution of
Ly= F, B 1(y) = 0 , B 2(y) = 0 (13.1.16)
is
y(x) = y1(x)
∫ b
x
F(t)y2(t)
P0(t)W(t) dt+ y2(x)
∫ x
a
F(t)y1(t)
P0(t)W(t) dt. (13.1.17)
Proof In Section 5.7 we saw that if
y= u1y1 + u2y2 (13.1.18)
where
u′
1y1 + u′
2y2 = 0
u′
1y′
1 + u′
2y′
2 = F,
then Ly= F. Solving for u′
1 and u′
2 yields
u′
1 = − Fy2
P0W and u′
2 = Fy1
P0W,
which hold if
u1(x) =
∫ b
x
F(t)y2(t)
P0(t)W(t) dt and u2(x) =
∫ x
a
F(t)y1(t)
P0(t)W(t) dt.
This and ( 13.1.18) show that ( 13.1.17) is a solution of Ly= F. Differentiating ( 13.1.17) yields
y′(x) = y′
1(x)
∫ b
x
F(t)y2(t)
P0(t)W(t) dt+ y′
2(x)
∫ x
a
F(t)y1(t)
P0(t)W(t) dt. (13.1.19)",Elementary Differential Equations with Boundary Value Problems.pdf
"y′(x) = y′
1(x)
∫ b
x
F(t)y2(t)
P0(t)W(t) dt+ y′
2(x)
∫ x
a
F(t)y1(t)
P0(t)W(t) dt. (13.1.19)
(V erify .) From ( 13.1.17) and ( 13.1.19),
B1(y) = B1(y1 )
∫ b
a
F(t)y2(t)
P0(t)W(t) dt= 0
because B1(y1) = 0 , and
B2(y) = B2(y2 )
∫ b
a
F(t)y1(t)
P0(t)W(t) dt= 0
because B2(y2) = 0 . Hence, ysatisfies ( 13.1.16). This completes the proof.
W e can rewrite ( 13.1.17) as
y=
∫ b
a
G(x,t )F(t) dt, (13.1.20)
where
G(x,t ) =







y1(t)y2(x)
P0(t)W(t) , a ≤ t≤ x,
y1(x)y2(t)
P0(t)W(t) , x ≤ t≤ b.
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.1 T wo-Point Boundary V alue Problems 683
This is the Green’s function for the boundary value problem ( 13.1.16). The Green’s function is related
to the boundary value problem ( 13.1.16) in much the same way that the inverse of a square matrix A is
related to the linear algebraic system y = Ax; just as we substitute the given vector y into the formula
x = A−1y to solve y = Ax, we substitute the given function F into the formula ( 13.1.20) to obtain the
solution of ( 13.1.16). The analogy goes further: just as A−1 exists if and only if Ax = 0 has only the
trivial solution, the boundary value problem ( 13.1.16) has a Green’s function if and only the homogeneous
boundary value problem ( 13.1.15) has only the trivial solution.
W e leave it to you (Exercise 32) to show that the assumptions of Theorem 13.1.3 imply that the unique
solution of the boundary value problem
Ly= F, B 1(y) = k1, B 2(y) = k2
is
y(x) =
∫ b
a
G(x,t )F(t) dt+ k2
B2(y1) y1 + k1
B1(y2) y2.",Elementary Differential Equations with Boundary Value Problems.pdf
"solution of the boundary value problem
Ly= F, B 1(y) = k1, B 2(y) = k2
is
y(x) =
∫ b
a
G(x,t )F(t) dt+ k2
B2(y1) y1 + k1
B1(y2) y2.
Example 13.1.6 Solve the boundary value problem
y′′+ y= F(x). y (0) + y′(0) = 0 , y (π) − y′(π) = 0 , (13.1.21)
and find the Green’s function for this problem.
Solution Here
B1(y) = y(0) + y′(0) and B2(y) = y(π) − y′(π).
Let {z1,z 2}= {cos x, sin x}, which is a fundamental set of solutions of y′′+ y= 0 . Then
B1(z1) = (cos x− sin x)
⏐
⏐
x=0 = 1
B2(z1) = (cos x+ sin x)
⏐
⏐
x=π = −1
and
B1(z2) = (sin x+ cos x)
⏐
⏐
x=0 = 1
B2(z2) = (sin x− cos x)
⏐
⏐
x=π = 1 .
Therefore
B1(z1)B2(z2) − B1(z2)B2(z1 ) = 2 ,
so Theorem
13.1.2 implies that ( 13.1.21) has a unique solution. Let
y1 = B1(z2)z1 − B1(z1)z2 = cos x− sin x
and
y2 = B2(z2)z1 − B2(z1)z2 = cos x+ sin x.
Then B1(y1) = 0 , B2(y2) = 0 , and the Wronskian of {y1,y 2}is
W(x) =
⏐
⏐
⏐
⏐
cos x− sin x cos x+ sin x
− sin x− cos x − sin x+ cos x
⏐
⏐
⏐
⏐= 2 .
Since P0 = 1 , (
13.1.17) yields the solution",Elementary Differential Equations with Boundary Value Problems.pdf
"W(x) =
⏐
⏐
⏐
⏐
cos x− sin x cos x+ sin x
− sin x− cos x − sin x+ cos x
⏐
⏐
⏐
⏐= 2 .
Since P0 = 1 , (
13.1.17) yields the solution
y(x) = cos x− sin x
2
∫ π
x
F(t)(cos t+ sin t) dt
+ cos x+ sin x
2
∫ x
0
F(t)(cos t− sin t) dt.",Elementary Differential Equations with Boundary Value Problems.pdf
"684 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
The Green’s function is
G(x,t ) =







(cos t− sin t)(cos x+ sin x)
2 , 0 ≤ t≤ x,
(cos x− sin x)(cos t+ sin t)
2 , x ≤ t≤ π.
W e’ll now consider the situation not covered by Theorem 13.1.3.
Theorem 13.1.4 Suppose the homogeneous boundary value problem
Ly= 0 , B 1(y) = 0 , B 2(y) = 0 (13.1.22)
has a nontrivial solution y1, and let y2 be any solution of Ly= 0 that isn’t a constant multiple of y1. Let
W = y1y′
2 − y′
1y2. If ∫ b
a
F(t)y1(t)
P0(t)W(t) dt= 0 , (13.1.23)
then the homogeneous boundary value problem
Ly= F, B 1(y) = 0 , B 2(y) = 0 (13.1.24)
has infinitely many solutions , all of the form y= yp + c1y1, where
yp = y1(x)
∫ b
x
F(t)y2(t)
P0(t)W(t) dt+ y2(x)
∫ x
a
F(t)y1(t)
P0(t)W(t) dt
and c1 is a constant . If ∫ b
a
F(t)y1(t)
P0(t)W(t) dt̸= 0 ,
then (13.1.24) has no solution .
Proof From the proof of Theorem 13.1.3, yp is a particular solution of Ly= F, and
y′
p (x) = y′
1(x)
∫ b
x",Elementary Differential Equations with Boundary Value Problems.pdf
"then (13.1.24) has no solution .
Proof From the proof of Theorem 13.1.3, yp is a particular solution of Ly= F, and
y′
p (x) = y′
1(x)
∫ b
x
F(t)y2(t)
P0(t)W(t) dt+ y′
2(x)
∫ x
a
F(t)y1(t)
P0(t)W(t) dt.
Therefore the general solution of ( 13.1.22) is of the form
y= yp + c1y1 + c2y2,
where c1 and c2 are constants. Then
B1(y) = B1(yp + c1y1 + c2y2) = B1(yp) + c1B1(y1) + c2B1y2
= B1(y1)
∫ b
a
F(t)y2(t)
P0(t)W(t) dt+ c1B1(y1) + c2B1 (y2)
= c2B1(y2 )
Since B1(y1) = 0 , Theorem 13.1.1 implies that B1(y2) ̸= 0 ; hence, B1(y) = 0 if and only if c2 = 0 .
Therefore y = yp + c1y1 and
B2(y) = B2(yp + c1y1) = B2(y2)
∫ b
a
F(t)y1(t)
P0(t)W(t) dt+ c1B2(y1)
= B2(y2)
∫ b
a
F(t)y1(t)
P0(t)W(t) dt,
since B2(y1 ) = 0 . From Theorem 13.1.1, B2(y2) ̸= 0 (since B2(y1 = 0 ). Therefore Ly= 0 if and only
if ( 13.1.23) holds. This completes the proof.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.1 T wo-Point Boundary V alue Problems 685
Example 13.1.7 Applying Theorem 13.1.4 to the boundary value problem
y′′+ y = F(x), y (0) = 0 , y (π) = 0 (13.1.25)
explains the Examples 13.1.2 and 13.1.3. The complementary equation y′′+ y = 0 has the linear inde-
pendent solutions y1 = sin xand y2 = cos x, and y1 satisfies both boundary conditions. Since P0 = 1
and
W =
⏐
⏐
⏐
⏐
sin x cos x
cos x − sin x
⏐
⏐
⏐
⏐= −1,
(
13.1.23) reduces to ∫ π
0
F(x) sin xdx = 0 .
From Example 13.1.2, F(x) = 1 and
∫ π
0
F(x) sin xdx =
∫ π
0
sin xdx = 2 ,
so Theorem 13.1.3 implies that ( 13.1.25) has no solution. In Example 13.1.3,
F(x) = sin 2 x= 2 sin xcos x
and ∫ π
0
F(x) sin xdx = 2
∫ π
0
sin2 xcos xdx = 2
3 sin3 x
⏐
⏐
⏐
⏐
π
0
= 0 ,
so Theorem
13.1.3 implies that ( 13.1.25) has infinitely many solutions, differing by constant multi ples of
y1(x) = sin x.
13.1 Exercises
1. V erify that B1 and B2 are linear operators; that is, if c1 and c2 are constants then",Elementary Differential Equations with Boundary Value Problems.pdf
"y1(x) = sin x.
13.1 Exercises
1. V erify that B1 and B2 are linear operators; that is, if c1 and c2 are constants then
Bi(c1y1 + c2y2) = c1Bi (y1) + c2Bi (y2), i = 1 , 2.
In Exercises 2– 7 solve the boundary value problem.
2. y′′− y = x, y(0) = −2, y(1) = 1
3. y′′= 2 − 3x, y(0) = 0 , y(1) − y′(1) = 0
4. y′′− y = x, y(0) + y′(0) = 3 , y(1) − y′(1) = 2
5. y′′+ 4 y= 1 , y(0) = 3 , y(π/ 2) + y′(π/ 2) = −7
6. y′′− 2y′+ y = 2 ex, y(0) − 2y′(0) = 3 , y(1) + y′(1) = 6 e
7. y′′− 7y′+ 12 y= 4 e2x, y(0) + y′(0) = 8 , y(1) = −7e2 (see Example 13.1.5)
8. State a condition on F such that the boundary value problem
y′′= F(x), y (0) = 0 , y (1) − y′(1) = 0
has a solution, and find all solutions.",Elementary Differential Equations with Boundary Value Problems.pdf
"686 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
9. (a) State a condition on aand bsuch that the boundary value problem
y′′+ y = F(x), y (a) = 0 , y (b) = 0 (A)
has a unique solution for every continuous F, and find the solution by the method used to
prove Theorem 13.1.3
(b) In the case where aand bdon’t satisfy the condition you gave for (a), state necessary and
sufficient on F such that (A) has a solution, and find all solutions by the meth od used to
prove Theorem 13.1.4.
10. Follow the instructions in Exercise 9 for the boundary value problem
y′′+ y= F(x), y (a) = 0 , y ′(b) = 0 .
11. Follow the instructions in Exercise 9 for the boundary value problem
y′′+ y = F(x), y ′(a) = 0 , y ′(b) = 0 .
In Exercises 12– 15 find a formula for the solution of the boundary problem by the m ethod used to prove
Theorem 13.1.3. Assume that a<b .
12. y′′− y = F(x), y(a) = 0 , y(b) = 0
13. y′′− y = F(x), y(a) = 0 , y′(b) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Theorem 13.1.3. Assume that a<b .
12. y′′− y = F(x), y(a) = 0 , y(b) = 0
13. y′′− y = F(x), y(a) = 0 , y′(b) = 0
14. y′′− y = F(x), y′(a) = 0 , y′(b) = 0
15. y′′− y = F(x), y(a) − y′(a) = 0 , y(b) + y′(b) = 0
In Exercises 16– 19 find all values of ω such that boundary problem has a unique solution, and find the
solution by the method used to prove Theorem 13.1.3. F or other values of ω, find conditions on F such
that the problem has a solution, and find all solutions by the m ethod used to prove Theorem 13.1.4.
16. y′′+ ω2y = F(x), y(0) = 0 , y(π) = 0
17. y′′+ ω2y = F(x), y(0) = 0 , y′(π) = 0
18. y′′+ ω2y = F(x), y′(0) = 0 , y(π) = 0
19. y′′+ ω2y = F(x), y′(0) = 0 , y′(π) = 0
20. Let {z1,z 2}be a fundamental set of solutions of Ly = 0 . Given that the homogeneous boundary
value problem
Ly= 0 , B 1(y) = 0 , B 2(y) = 0
has a nontrivial solution, express it explicity in terms of z1 and z2.
21. If the boundary value problem has a solution for every contin uous F, then find the Green’s function",Elementary Differential Equations with Boundary Value Problems.pdf
"21. If the boundary value problem has a solution for every contin uous F, then find the Green’s function
for the problem and use it to write an explicit formula for the solution. Otherwise, if the boundary
value problem does not have a solution for every continuous F, find a necessary and sufficient
condition on F for the problem to have a solution, and find all solutions. Ass ume that a<b .
(a) y′′= F(x), y(a) = 0 , y(b) = 0
(b) y′′= F(x), y(a) = 0 , y′(b) = 0
(c) y′′= F(x), y′(a) = 0 , y(b) = 0
(d) y′′= F(x), y′(a) = 0 , y′(b) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.1 T wo-Point Boundary V alue Problems 687
22. Find the Green’s function for the boundary value problem
y′′= F(x), y (0) − 2y′(0) = 0 , y (1) + 2 y′(1) = 0 . (A)
Then use the Green’s function to solve (A) with (a) F(x) = 1 , (b) F(x) = x, and (c) F(x) = x2.
23. Find the Green’s function for the boundary value problem
x2y′′+ xy′+ ( x2 − 1/ 4)y= F(x), y (π/ 2) = 0 , y (π) = 0 , (A)
given that
y1(x) = cos x√x and y2(x) = sin x√x
are solutions of the complementary equation. Then use the Gr een’s function to solve (A) with (a)
F(x) = x3/ 2 and (b) F(x) = x5/ 2.
24. Find the Green’s function for the boundary value problem
x2y′′− 2xy′+ 2 y= F(x), y (1) = 0 , y (2) = 0 , (A)
given that {x,x 2}is a fundamental set of solutions of the complementary equat ion. Then use the
Green’s function to solve (A) with (a) F(x) = 2 x3 and (b) F(x) = 6 x4.
25. Find the Green’s function for the boundary value problem
x2y′′+ xy′− y= F(x), y (1) − 2y′(1) = 0 , y ′(2) = 0 , (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"25. Find the Green’s function for the boundary value problem
x2y′′+ xy′− y= F(x), y (1) − 2y′(1) = 0 , y ′(2) = 0 , (A)
given that {x, 1/x }is a fundamental set of solutions of the complementary equat ion. Then use the
Green’s function to solve (A) with (a) F(x) = 1 , (b) F(x) = x2, and (c) F(x) = x3.
In Exercises 26– 30 find necessary and sufficient conditions on α, β, ρ, and δ for the boundary value
problem to have a unique solution for every continuous F, and find the Green’s function.
26. y′′= F(x), αy (0) + βy′(0) = 0 , ρy(1) + δy′(1) = 0
27. y′′+ y = F(x), αy (0) + βy′(0) = 0 , ρy(π) + δy′(π) = 0
28. y′′+ y = F(x), αy (0) + βy′(0) = 0 , ρy(π/ 2) + δy′(π/ 2) = 0
29. y′′− 2y′+ 2 y= F(x), αy (0) + βy′(0) = 0 , ρy(π) + δy′(π) = 0
30. y′′− 2y′+ 2 y= F(x), αy (0) + βy′(0) = 0 , ρy(π/ 2) + δy′(π/ 2) = 0
31. Find necessary and sufficient conditions on α, β, ρ, and δ for the boundary value problem
y′′− y = F(x), αy (a) + βy′(a) = 0 , ρy (b) + δy′(b) = 0 (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′− y = F(x), αy (a) + βy′(a) = 0 , ρy (b) + δy′(b) = 0 (A)
to have a unique solution for every continuous F, and find the Green’s function for (A). Assume
that a<b .
32. Show that the assumptions of Theorem 13.1.3 imply that the unique solution of
Ly= F, B 1(y) = k1, B 2(y) = f2
is
y=
∫ b
a
G(x,t )F(t) dt+ k2
B2
(y1)y1 + k1
B1(y2 ) y2.",Elementary Differential Equations with Boundary Value Problems.pdf
"688 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
13.2 STURM-LIOUVILLE PROBLEMS
In this section we consider eigenvalue problems of the form
P0(x)y′′+ P1(x)y′+ P2(x)y+ λR(x)y= 0 , B 1(y) = 0 , B 2(y) = 0 , (13.2.1)
where
B1(y) = αy (a) + βy′(a) and B2(y) = ρy(b) + δy′(b).
As in Section 13.1, α, β, ρ, and δ are real numbers, with
α2 + β2 >0 and ρ2 + δ2 >0,
P0, P1, P2, and Rare continuous, and P0 and Rare positive on [a,b ].
W e say that λ is an eigenvalue of ( 13.2.1) if ( 13.2.1) has a nontrivial solution y. In this case, y is an
eigenfunction associated with λ, or a λ-eigenfunction. Solving the eigenvalue problem means finding all
eigenvalues and associated eigenfunctions of ( 13.2.1).
Example 13.2.1 Solve the eigenvalue problem
y′+ 3 y′+ 2 y+ λy = 0 , y (0) = 0 , y (1) = 0 . (13.2.2)
Solution The characteristic equation of ( 13.2.2) is
r2 + 3 r+ 2 + λ = 0 ,
with zeros
r1 = −3 + √1 − 4λ
2 and r2 = −3 − √1 − 4λ
2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Solution The characteristic equation of ( 13.2.2) is
r2 + 3 r+ 2 + λ = 0 ,
with zeros
r1 = −3 + √1 − 4λ
2 and r2 = −3 − √1 − 4λ
2 .
If λ < 1/ 4 then r1 and r2 are real and distinct, so the general solution of the differe ntial equation in
(13.2.2) is
y= c1er1 t + c2er2 t .
The boundary conditions require that
c1 + c2 = 0
c1er1 + c2er2 = 0 .
Since the determinant of this system is er2 − er1 ̸= 0 , the system has only the trivial solution. Therefore
λ isn’t an eigenvalue of ( 13.2.2).
If λ = 1 / 4 then r1 = r2 = −3/ 2, so the general solution of ( 13.2.2) is
y= e−3x/ 2(c1 + c2x).
The boundary condition y(0) = 0 requires that c1 = 0 , so y = c2xe−3x/ 2 and the boundary condition
y(0) requires that c2 = 0 . Therefore λ = 1 / 4 isn’t an eigenvalue of ( 13.2.2).
If λ> 1/ 4 then
r1 = −3
2 + iω and r2 = −3
2 − iω,
with
ω =
√
4λ − 1
2 or, equivalently , λ = 1 + 4 ω2
4 . (13.2.3)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 689
In this case the general solution of the differential equati on in ( 13.2.2) is
y= e−3x/ 2(c1 cos ωx + c2 sin ωx).
The boundary condition y(0) = 0 requires that c1 = 0 , so y = c2e−3x/ 2 sin ωx, which holds with c2 ̸= 0
if and only if ω = nπ, where nis an integer. W e may assume that nis a positive integer. (Why?). From
(13.2.3), the eigenvalues are λn = (1 + 4 n2π2)/ 4, with associated eigenfunctions
yn = e−3x/ 2 sin nπx, n = 1 , 2, 3,....
Example 13.2.2 Solve the eigenvalue problem
x2y′′+ xy′+ λy = 0 , y (1) = 0 , y (2) = 0 . (13.2.4)
Solution If λ = 0 , the differential equation in ( 13.2.4) reduces to x(xy′)′= 0 , so xy′= c1,
y′= c1
x, and y = c1 ln x+ c2.
The boundary condition y(1) = 0 requires that c2 = 0 , so y= c1 ln x. The boundary condition y(2) = 0
requires that c1 ln 2 = 0 , so c1 = 0 . Therefore zero isn’t an eigenvalue of ( 13.2.4).
If λ< 0, we write λ = −k2 with k >0, so ( 13.2.4) becomes
x2y′′+ xy′− k2y= 0 ,",Elementary Differential Equations with Boundary Value Problems.pdf
"If λ< 0, we write λ = −k2 with k >0, so ( 13.2.4) becomes
x2y′′+ xy′− k2y= 0 ,
an Euler equation (Section 7.4) with indicial equation
r2 − k2 = ( r− k)(r+ k) = 0 .
Therefore
y= c1xk + c2x−k.
The boundary conditions require that
c1 + c2 = 0
2kc1 + 2 −kc2 = 0 .
Since the determinant of this system is 2−k − 2k ̸= 0 , c1 = c2 = 0 . Therefore ( 13.2.4) has no negative
eigenvalues.
If λ> 0 we write λ = k2 with k> 0. Then ( 13.2.4) becomes
x2y′′+ xy′+ k2y= 0 ,
an Euler equation with indicial equation
r2 + k2 = ( r− ik)(r+ ik) = 0 ,
so
y= c1 cos(kln x) + c2 sin(kln x).
The boundary condition y(1) = 0 requires that c1 = 0 . Therefore y = c2 sin(kln x). This holds with
c2 ̸= 0 if and only if k = nπ/ ln 2, where nis a positive integer. Hence, the eigenvalues of ( 13.2.4) are
λn = ( nπ/ ln 2)2, with associated eigenfunctions
yn = sin
(nπ
ln 2 ln x
)
, n = 1 , 2, 3,....
For theoretical purposes, it’s useful to rewrite the differ ential equation in ( 13.2.1) in a different form,",Elementary Differential Equations with Boundary Value Problems.pdf
"(nπ
ln 2 ln x
)
, n = 1 , 2, 3,....
For theoretical purposes, it’s useful to rewrite the differ ential equation in ( 13.2.1) in a different form,
provided by the next theorem.",Elementary Differential Equations with Boundary Value Problems.pdf
"690 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
Theorem 13.2.1 If P0, P1, P2, and R are continuous and P0 and R are positive on a closed interval
[a,b ], then the equation
P0(x)y′′+ P1(x)y′+ P2(x)y+ λR(x)y= 0 (13.2.5)
can be rewritten as
(p(x)y′)′+ q(x)y+ λr(x)y = 0 , (13.2.6)
where p, p′, qand rare continuous and pand rare positive on [a,b ].
Proof W e begin by rewriting ( 13.2.5) as
y′′+ u(x)y′+ v(x)y+ λR1(x)y= 0 , (13.2.7)
with u= P1/P 0, v= P2/P 0, and R1 = R/P 0. (Note that R1 is positive on [a,b ].) Now let p(x) = eU (x),
where U is any antiderivative of u. Then pis positive on [a,b ] and, since U′= u,
p′(x) = p(x)u(x) (13.2.8)
is continuous on [a,b ]. Multiplying ( 13.2.7) by p(x) yields
p(x)y′′+ p(x)u(x)y′+ p(x)v(x)y+ λp(x)R1(x)y= 0 . (13.2.9)
Since pis positive on [a,b ], this equation has the same solutions as ( 13.2.5). From ( 13.2.8),
(p(x)y′)′= p(x)y′′+ p′(x)y′= p(x)y′′+ p(x)u(x)y′,",Elementary Differential Equations with Boundary Value Problems.pdf
"Since pis positive on [a,b ], this equation has the same solutions as ( 13.2.5). From ( 13.2.8),
(p(x)y′)′= p(x)y′′+ p′(x)y′= p(x)y′′+ p(x)u(x)y′,
so ( 13.2.9) can be rewritten as in ( 13.2.6), with q(x) = p(x)v(x) and r(x) = p(x)R1(x). This completes
the proof.
It is to be understood throughout the rest of this section tha t p, q, and rhave the properties stated in
Theorem 13.2.1. Moreover, whenever we write Lyin a general statement, we mean
Ly= ( p(x)y′)′+ q(x)y.
The differential equation ( 13.2.6) is called a Sturm– Liouville equation , and the eigenvalue problem
(p(x)y′)′+ q(x)y+ λr(x)y = 0 , B 1(y) = 0 , B 2 (y) = 0 , (13.2.10)
which is equivalent to ( 13.2.1), is called a Sturm-Liouville problem .
Example 13.2.3 Rewrite the eigenvalue problem
y′′+ 3 y′+ (2 + λ)y= 0 , y (0) = 0 , y (1) = 0 (13.2.11)
of Example 13.2.1 as a Sturm-Liouville problem.
Solution Comparing ( 13.2.11) to ( 13.2.7) shows that u(x) = 3 , so we take U(x) = 3 xand p(x) = e3x.",Elementary Differential Equations with Boundary Value Problems.pdf
"of Example 13.2.1 as a Sturm-Liouville problem.
Solution Comparing ( 13.2.11) to ( 13.2.7) shows that u(x) = 3 , so we take U(x) = 3 xand p(x) = e3x.
Multiplying the differential equation in ( 13.2.11) by e3x yields
e3x(y′′+ 3 y′) + 2 e3xy+ λe3xy = 0 .
Since
e3x(y′′+ 3 y′) = ( e3xy′)′,
(13.2.11) is equivalent to the Sturm–Liouville problem
(e3xy′)′+ 2 e3xy+ λe3xy= 0 , y (0) = 0 , y (1) = 0 . (13.2.12)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 691
Example 13.2.4 Rewrite the eigenvalue problem
x2y′′+ xy′+ λy = 0 , y (1) = 0 , y (2) = 0 (13.2.13)
of Example 13.2.2 as a Sturm-Liouville problem.
Solution Dividing the differential equation in ( 13.2.13) by x2 yields
y′′+ 1
xy′+ λ
x2 y= 0 .
Comparing this to ( 13.2.7) shows that u(x) = 1 /x , so we take U(x) = ln x and p(x) = eln x = x.
Multiplying the differntial equation by xyields
xy′′+ y′+ λ
xy= 0 .
Since
xy′′+ y′= ( xy′)′,
(13.2.13) is equivalent to the Sturm–Liouville problem
(xy′)′+ λ
xy= 0 , y (1) = 0 , y (2) = 0 . (13.2.14)
Problems 1–4 of Section 11.1 are Sturm–Liouville problems. (Problem 5 isn’t , although some authors
use a definition of Sturm-Liouville problem that does include it.) W e were able to find the eigenvalues
of Problems 1-4 explicitly because in each problem the coeffi cients in the boundary conditions satisfy
αβ = 0 and ρδ = 0 ; that is, each boundary condition involves either y or y′, but not both. If this isn’t",Elementary Differential Equations with Boundary Value Problems.pdf
"αβ = 0 and ρδ = 0 ; that is, each boundary condition involves either y or y′, but not both. If this isn’t
true then the eigenvalues can’t in general be expressed exac tly by simple formulas; rather, approximate
values must be obtained by numerical solution of equations d erived by requiring the determinants of
certain 2 × 2 systems of homogeneous equations to be zero. T o apply the num erical methods effectively ,
graphical methods must be used to determine approximate loc ations of the zeros of these determinants.
Then the zeros can be computed accurately by numerical metho ds.
Example 13.2.5 Solve the Sturm–Liouville problem
y′′+ λy = 0 , y (0) + y′(0) = 0 , y (1) + 3 y′(1) = 0 . (13.2.15)
Solution If λ = 0 , the differential equation in ( 13.2.15) reduces to y′′= 0 , with general solution
y = c1 + c2x. The boundary conditions require that
c1 + c2 = 0
c1 + 4 c2 = 0 ,
so c1 = c2 = 0 . Therefore zero isn’t an eigenvalue of ( 13.2.15).",Elementary Differential Equations with Boundary Value Problems.pdf
"y = c1 + c2x. The boundary conditions require that
c1 + c2 = 0
c1 + 4 c2 = 0 ,
so c1 = c2 = 0 . Therefore zero isn’t an eigenvalue of ( 13.2.15).
If λ < 0, we write λ = −k2 where k >0, and the differential equation in ( 13.2.15) becomes y′′−
k2y= 0 , with general solution
y = c1 cosh kx+ c2 sinh kx, (13.2.16)
so
y′= k(c1 sinh kx+ c2 cosh kx).",Elementary Differential Equations with Boundary Value Problems.pdf
"692 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
1
2
3
4
1 2 3 4
 k
 u
Figure 13.2.1 u= tanh kand u= −2k/ (1 − 3k2)
The boundary conditions require that
c1 + kc2 = 0
(cosh k+ 3 ksinh k)c1 + (sinh k+ 3 kcosh k)c2 = 0 . (13.2.17)
The determinant of this system is
DN (k) =
⏐
⏐
⏐
⏐
1 k
cosh k+ 3 ksinh k sinh k+ 3 kcosh k
⏐
⏐
⏐
⏐
= (1 − 3k2) sinh k+ 2 kcosh k.
Therefore the system (
13.2.17) has a nontrivial solution if and only if DN (k) = 0 or, equivalently ,
tanh k= − 2k
1 − 3k2 . (13.2.18)
The graph of the right side (Figure 13.2.1) has a vertical asymptote at k= 1 /
√
3. Since the two sides have
different signs if k <1/
√
3, this equation has no solution in (0, 1/
√
3). Figure 13.2.1 shows the graphs
of the two sides of ( 13.2.18) on an interval to the right of the vertical asymptote, which is indicated by
the dashed line. Y ou can see that the two curves intersect nea r k0 = 1 .2, Given this estmate, you can use",Elementary Differential Equations with Boundary Value Problems.pdf
"the dashed line. Y ou can see that the two curves intersect nea r k0 = 1 .2, Given this estmate, you can use
Newton’s to compute k0 more accurately . W e computed k0 ≈ 1.1219395. Therefore −k2
0 ≈ − 1.2587483
is an eigenvalue of ( 13.2.15). From ( 13.2.16) and the first equation in ( 13.2.17),
y0 = k0 cosh k0x− sinh k0x.
If λ> 0 we write λ = k2 where k >0, and differential equation in ( 13.2.15) becomes y′′+ k2y = 0 ,
with general solution
y = cos kx+ c2 sin kx, (13.2.19)",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 693
so
y′= k(−c1 sin kx+ c2 cos kx).
The boundary conditions require that
c1 + kc2 = 0
(cos k− 3ksin k)c1 + (sin k+ 3 kcos k)c2 = 0 . (13.2.20)
The determinant of this system is
DP (k) =
⏐
⏐
⏐
⏐
1 k
cos k− 3ksin k sin k+ 3 kcos k
⏐
⏐
⏐
⏐
= (1 + 3 k2) sin k+ 2 kcos k.
The system (
13.2.20) has a nontrivial solution if and only if DP (k) = 0 or, equivalently ,
tan k= − 2k
1 + 3 k2 .
Figure 13.2.2 shows the graphs of the two sides of this equation. Y ou can see from the figure that the
graphs intersect at infinitely many points kn ≈ nπ (n= 1 , 2, 3,. . . ), where the error in this approximation
approaches zero as n → ∞ . Given this estimate, you can use Newton’s method to compute kn more
accurately . W e computed
k1 ≈ 2.9256856,
k2 ≈ 6.1765914,
k3 ≈ 9.3538959,
k4 ≈ 12.5132570.
The estimates of the corresponding eigenvalues λn = k2
n are
λ1 ≈ 8.5596361,
λ2 ≈ 38.1502809,
λ3 ≈ 87.4953676,
λ4 ≈ 156.5815998.",Elementary Differential Equations with Boundary Value Problems.pdf
"k4 ≈ 12.5132570.
The estimates of the corresponding eigenvalues λn = k2
n are
λ1 ≈ 8.5596361,
λ2 ≈ 38.1502809,
λ3 ≈ 87.4953676,
λ4 ≈ 156.5815998.
From ( 13.2.19) and the first equation in ( 13.2.20),
yn = kn cos knx− sin knx
is an eigenfunction associated with λn
Since the differential equations in ( 13.2.12) and ( 13.2.14) are more complicated than those in ( 13.2.11)
and ( 13.2.13) respectively , what is the point of Theorem 13.2.1? The point is this: to solve a specific
problem, it may be better to deal with it directly , as we did in Examples 13.2.1 and 13.2.2; however, we’ll
see that transforming the general eigenvalue problem ( 13.2.1) to the Sturm–Liouville problem ( 13.2.10)
leads to results applicable to all eigenvalue problems of the form ( 13.2.1).
Theorem 13.2.2 If
Ly= ( p(x)y′)′+ q(x)y
and uand v are twice continuously functions on [a,b ] that satisfy the boundary conditions B1(y) = 0
and B2(y) = 0 , then ∫ b
a
[u(x)Lv(x) − v(x)Lu(x)] dx= 0 . (13.2.21)",Elementary Differential Equations with Boundary Value Problems.pdf
"694 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
 k = π  k = 2π  k = 3π  k = 4π
 k
 u
1
− 1
Figure 13.2.2 u= tan kand u= −2k/ (1 + k)
Proof Integration by parts yields
∫ b
a
[u(x)Lv(x) − v(x)Lu(x)] dx =
∫ b
a
[u(x)(p(x)v′(x))′− v(x)(p(x)u′(x))′] dx
= p(x)[u(x)v′(x) − u′(x)v(x)]
⏐
⏐
⏐
⏐
b
a
−
∫ b
a
p(x)[u′(x)v′(x) − u′(x)v′(x)] dx.
Since the last integral equals zero,
∫ b
a
[u(x)Lv(x) − v(x)Lu(x)] dx= p(x)[u(x)v′(x) − u′(x)v′(x)]
⏐
⏐
⏐
⏐
b
a
. (13.2.22)
By assumption, B1(u) = B1(v) = 0 and B2(u) = B2(v) = 0 . Therefore
αu(a) + βu′(a) = 0
αv (a) + βv′(a) = 0 and ρu(b) + δu′(b) = 0
ρv(b) + δv′(b) = 0 .
Since α2 + β2 >0 and ρ2 + δ2 >0, the determinants of these two systems must both be zero; tha t is,
u(a)v′(a) − u′(a)v(a) = u(b)v′(b) − u′(b)v(b) = 0 .
This and (
13.2.22) imply ( 13.2.21), which completes the proof.
The next theorem shows that a Sturm–Liouville problem has no complex eigenvalues.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 695
Theorem 13.2.3 If λ = p+ qiwith q̸= 0 then the boundary value problem
Ly+ λr(x)y= 0 , B 1(y) = 0 , B 2(y) = 0
has only the trivial solution.
Proof For this theorem to make sense, we must consider complex-val ued solutions of
Ly+ ( p+ iq)r(x,y )y = 0 . (13.2.23)
If y = u+ iv where u and v are real-valued and twice differentiable, we define y′ = u′+ iv′and
y′′= u′′+ iv′′. W e say that yis a solution of ( 13.2.23) if the real and imaginary parts of the left side of
(13.2.23) are both zero. Since Ly= ( p(x)′y)′+ q(x)yand p, q, and rare real-valued,
Ly+ λr(x)y = L(u+ iv) + ( p+ iq)r(x)(u+ iv)
= Lu+ r(x)(pu− qv) + i[Lv+ r(x)(pu+ qv)],
so Ly+ λr(x)y = 0 if and only if
Lu+ r(x)(pu− qv) = 0
Lv+ r(x)(qu+ pv) = 0 .
Multiplying the first equation by vand the second by uyields
vLu+ r(x)(puv− qv2) = 0
uLv+ r(x)(qu2 + puv) = 0 .
Subtracting the first equation from the second yields
uLv− vLu+ qr(x)(u2 + v2 ) = 0 ,
so ∫ b
a
[u(x)Lv(x) − v(x)Lu(x)] dx+
∫ b
a",Elementary Differential Equations with Boundary Value Problems.pdf
"Subtracting the first equation from the second yields
uLv− vLu+ qr(x)(u2 + v2 ) = 0 ,
so ∫ b
a
[u(x)Lv(x) − v(x)Lu(x)] dx+
∫ b
a
r(x)[u2(x) + v2(x)] dx= 0 . (13.2.24)
Since
B1(y) = B1(u+ iv) = B1(u) + iB1(v)
and
B2(y) = B2(u+ iv) = B2(u) + iB2(v),
B1(y) = 0 and B2(y) = 0 implies that
B1 (u) = B2(u) = B1 (v) = B2(v) = 0 .
Therefore Theorem 13.2.2 implies that first integral in ( 13.2.24) equals zero, so ( 13.2.24) reduces to
q
∫ b
a
r(x)[u2(x) + v2(x)] dx= 0 .
Since r is positive on [a,b ] and q ̸= 0 by assumption, this implies that u ≡ 0 and v ≡ 0 on [a,b ].
Therefore y ≡ 0 on [a,b ], which completes the proof.",Elementary Differential Equations with Boundary Value Problems.pdf
"696 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
Theorem 13.2.4 If λ1 and λ2 are distinct eigenvalues of the Sturm–Liouville problem
Ly+ λr(x)y= 0 , B 1(y) = 0 , B 2(y) = 0 (13.2.25)
with associated eigenfunctions uand vrespectively, then
∫ b
a
r(x)u(x)v(x) dx= 0 . (13.2.26)
Proof Since uand vsatisfy the boundary conditions in ( 13.2.25), Theorem 13.2.2 implies that
∫ b
a
[u(x)Lv(x) − v(x)Lu(x)] dx= 0 .
Since Lu= −λ1ruand Lv= −λ2rv, this implies that
(λ1 − λ2)
∫ b
a
r(x)u(x)v(x) dx= 0 .
Since λ1 ̸= λ2, this implies ( 13.2.26), which completes the proof.
If uand vare any integrable functions on [a,b ] and
∫ b
a
r(x)u(x)v(x) dx= 0 ,
we say that uand vorthogonal on [a,b ] with respect to r= r(x).
Theorem 13.1.1 implies the next theorem.
Theorem 13.2.5 If u̸≡0 and vboth satisfy
Ly+ λr(x)y= 0 , B 1(y) = 0 , B 2(y) = 0 ,
then v= cufor some constant c.
W e’ve now proved parts of the next theorem. A complete proof i s beyond the scope of this book.",Elementary Differential Equations with Boundary Value Problems.pdf
"then v= cufor some constant c.
W e’ve now proved parts of the next theorem. A complete proof i s beyond the scope of this book.
Theorem 13.2.6 The set of all eigenvalues of the Sturm–Liouville problem
Ly+ λr(x)y= 0 , B 1(y) = 0 , B 2(y) = 0
can be ordered as
λ1 <λ 2 <· · ·<λ n <· · ·,
and
lim
n→∞
λn = ∞ .
F or each n, if yn is an arbitrary λn -eigenfunction, then every λn -eigenfunction is a constant multiple of
yn . If m̸= n,y m and yn are orthogonal [a,b ] with respect to r= r(x); that is ,
∫ b
a
r(x)ym (x)yn (x) dx= 0 . (13.2.27)
Y ou may want to verify ( 13.2.27) for the eigenfunctions obtained in Examples 13.2.1 and 13.2.2.
In conclusion, we mention the next theorem. The proof is beyo nd the scope of this book.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 697
Theorem 13.2.7 Let λ1 <λ 2 <· · ·<λ n <· · ·be the eigenvalues of the Sturm–Liouville problem
Ly+ λr(x)y= 0 , B 1(y) = 0 , B 2(y) = 0 ,
with associated eigenvectors y1,y 2, . . . , yn , . . . . Suppose f is piecewise smooth (Definition 11.2.3) on
[a,b ]. F or each n, let
cn =
∫ b
a
r(x)f(x)yn (x) dx
∫ b
a
r(x)y2
n (x) dx
.
Then
f(x−) + f(x+)
2 =
∞∑
n=1
cnyn (x)
for all xin the open interval (a,b ).
13.2 Exercises
In Exercises 1– 7 rewrite the equation in Sturm–Liouville form (with λ = 0) . Assume that b,c,α, and
ν are contants .
1. y′′+ by′+ cy = 0
2. x2y′′+ xy′+ ( x2 − ν2)y= 0 (Bessel’s equation)
3. (1 − x2)y′′− xy′+ α2y= 0 (Chebyshev’s equation)
4. x2y′′+ bxy′+ cy = 0 (Euler’s equation)
5. y′′− 2xy′+ 2 αy = 0 (Hermite’s equation)
6. xy′′+ (1 − x)y′+ αy = 0 (Laguerre’s equation)
7. (1 − x2)y′′− 2xy′+ α(α + 1) y = 0 (Legendre’s equation)
8. In Example 13.2.4 we found that the eigenvalue problem
x2y′′+ xy′+ λy = 0 , y (1) = 0 , y (2) = 0 (A)",Elementary Differential Equations with Boundary Value Problems.pdf
"8. In Example 13.2.4 we found that the eigenvalue problem
x2y′′+ xy′+ λy = 0 , y (1) = 0 , y (2) = 0 (A)
is equivalent to the Sturm-Liouville problem
(xy′)′+ λ
xy= 0 , y (1) = 0 , y (2) = 0 . (B)
Multiply the differential equation in (B) by yand integrate to show that
λ
∫ 2
1
y2 (x)
x dx=
∫ 2
1
x(y′(x))2 dx.
Conclude from this that the eigenvalues of (A) are all positi ve.
9. Solve the eigenvalue problem
y′′+ 2 y′+ y+ λy = 0 , y (0) = 0 , y (1) = 0 .",Elementary Differential Equations with Boundary Value Problems.pdf
"698 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
10. Solve the eigenvalue problem
y′′+ 2 y′+ y+ λy = 0 , y ′(0) = 0 , y ′(1) = 0 .
In Exercises 11– 20 : (a) Determine whether λ = 0 is an eigenvalue. If it is, find an associated eigenfunc-
tion.
(b) Compute the negative eigenvalues with errors not greater th an 5 × 10−8. State the form of the
associated eigenfunctions.
(c) Compute the first four positive eigenvalues with errors not g reater than 5 × 10−8. State the form of
the associated eigenfunctions.
11. C y′′+ λy = 0 , y(0) + 2 y′(0) = 0 , y(2) = 0
12. C y′′+ λy = 0 , y′(0) = 0 , y(1) − 2y′(1) = 0
13. C y′′+ λy = 0 , y(0) − y′(0) = 0 , y′(π) = 0
14. C y′′+ λy = 0 , y(0) + 2 y′(0) = 0 , y(π) = 0
15. C y′′+ λy = 0 , y′(0) = 0 , y(2) − y′(2) = 0
16. C y′′+ λy = 0 , y(0) + y′(0) = 0 , y(2) + 2 y′(2) = 0
17. C y′′+ λy = 0 , y(0) + 2 y′(0) = 0 , y(3) − 2y′(3) = 0
18. C y′′+ λy = 0 , 3y(0) + y′(0) = 0 , 3y(2) − 2y′(2) = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"17. C y′′+ λy = 0 , y(0) + 2 y′(0) = 0 , y(3) − 2y′(3) = 0
18. C y′′+ λy = 0 , 3y(0) + y′(0) = 0 , 3y(2) − 2y′(2) = 0
19. C y′′+ λy = 0 , y(0) + 2 y′(0) = 0 , y(3) − y′(3) = 0
20. C y′′+ λy = 0 , 5y(0) + 2 y′(0) = 0 , 5y(1) − 2y′(1) = 0
21. Find the first five eigenvalues of the boundary value problem
y′′+ 2 y′+ y+ λy = 0 , y (0) = 0 , y ′(1) = 0
with errors not greater than 5 × 10−8. State the form of the associated eigenfunctions.
In Exercises 22– 24 take it as given that {xekx ,xe −kx }and {xcos kx,x sin kx}are fundamental sets of
solutions of
x2y′′− 2xy′+ 2 y− k2x2y= 0
and
x2y′′− 2xy′+ 2 y+ k2x2y= 0 ,
respectively.
22. Solve the eigenvalue problem for
x2y′′− 2xy′+ 2 y+ λx2y = 0 , y (1) = 0 , y (2) = 0 .
23. C Find the first five eigenvalues of
x2y′′− 2xy′+ 2 y+ λx2y = 0 , y ′(1) = 0 , y (2) = 0
with errors no greater than 5 × 10−8. State the form of the associated eienfunctions.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 699
24. C Find the first five eigenvalues of
x2y′′− 2xy′+ 2 y+ λx2y = 0 , y (1) = 0 , y ′(2) = 0
with errors no greater than 5 × 10−8. State the form of the associated eienfunctions.
25. Consider the Sturm-Liouville problem
y′′+ λy = 0 , y (0) = 0 , y (L) + δy′(L) = 0 . (A)
(a) Show that (A) can’t have more than one negative eigenvalue, a nd find the values of δ for which
it has one.
(b) Find all values of δ such that λ = 0 is an eigenvalue of (A).
(c) Show that λ = k2 with k >0 is an eigenvalue of (A) if and only if
tan kL= −δk. (B)
(d) For n= 1 , 2, . . . , let yn be an eigenfunction associated with λn = k2
n . From Theorem
13.2.4,
ym and yn are orthogonal over [0,L ] if m̸= n. V erify this directly . H IN T : Integrate by parts twice
and use (B).
26. Solve the Sturm-Liouville problem
y′′+ λy = 0 , y (0) + αy ′(0) = 0 , y (π) + αy ′(π) = 0 ,
where α ̸= 0 .
27. L Consider the Sturm-Liouville problem",Elementary Differential Equations with Boundary Value Problems.pdf
"y′′+ λy = 0 , y (0) + αy ′(0) = 0 , y (π) + αy ′(π) = 0 ,
where α ̸= 0 .
27. L Consider the Sturm-Liouville problem
y′′+ λy = 0 , y (0) + αy ′(0) = 0 , y (1) + ( α − 1)y′(1) = 0 , (A)
where 0 <α< 1.
(a) Show that λ = 0 is an eigenvalue of (A), and find an associated eigenfunction .
(b) Show that (A) has a negative eigenvalue, and find the form of an associated eigenfunction.
(c) Give a graphical argument to show that (A) has infinitely many positive eigenvalues λ1 <λ 2 <
· · ·<λ n <· · ·, and state the form of the associated eigenfunctions.
Exercises 28– 30 deal with the Sturm–Liouville problem
y′′+ λy = 0 , αy (0) + βy′(0), ρy (L) + δy′(L) = 0 , (SL)
where α2 + β2 >0 and ρ2 + δ2 >0.
28. Show that λ = 0 is an eigenvalue of (SL) if and only if
α(ρL + δ) − βρ = 0 .
29. L The point of this exercise is that (SL) can’t have more than tw o negative eigenvalues.
(a) Show that λ is a negative eigenvalue of (SL) if and only if λ = −k2, where k is a positive
solution of",Elementary Differential Equations with Boundary Value Problems.pdf
"(a) Show that λ is a negative eigenvalue of (SL) if and only if λ = −k2, where k is a positive
solution of
(αρ − βδk 2) sinh kL+ k(αδ − βρ ) cosh kL.",Elementary Differential Equations with Boundary Value Problems.pdf
"700 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
(b) Suppose αδ − βρ = 0 . Show that (SL) has a negative eigenvalue if and only if αρ and βδ are
both nonzero. Find the negative eigenvalue and an associate d eigenfunction. H IN T : Show that in
this case ρ = pα and s= qβ, where q̸= 0 .
(c) Suppose βρ − αδ ̸= 0 . W e know from Section 11.1 that (SL) has no negative eigenval ues if
αρ = 0 and βδ = 0 . Assume that either αρ ̸= 0 or βδ ̸= 0 . Then we can rewrite (A) as
tanh kL= k(βρ − αδ )
αρ − βδk 2 .
By graphing both sides of this equation on the same axes (ther e are several possibilities for the right
side), show that it has at most two positive solutions, so (SL ) has at most two negative eigenvalues.
30. L The point of this exercise is that (SL) has infinitely many pos itive eigenvalues λ1 < λ2 <
· · ·<λ n <· · ·, and that limn→∞ λn = ∞ .
(a) Show that λ is a positive eigenvalue of (SL) if and only if λ = k2, where kis a positive solution
of",Elementary Differential Equations with Boundary Value Problems.pdf
"· · ·<λ n <· · ·, and that limn→∞ λn = ∞ .
(a) Show that λ is a positive eigenvalue of (SL) if and only if λ = k2, where kis a positive solution
of
(αρ + βδk 2) sin kL+ k(αδ − βρ ) cos kL= 0 . (A)
(b) Suppose αδ − βρ = 0 . Show that the positive eigenvalues of (SL) are λn = ( nπ/L )2, n= 1 ,
2, 3, . . . . H IN T : Recall the hint in Exercise 29(b).
Now suppose αδ − βρ ̸= 0 . From Section 11.1, if αρ = 0 and βδ = 0 , then (SL) has the
eigenvalues
λn = [(2 n− 1)π/ 2L]2, n = 1 , 2, 3,...
(why?), so let’s suppose in addition that at least one of the p roducts αρ and βδ is nonzero. Then
we can rewrite (A) as
tan kL= k(βρ − αδ )
αρ − βδk 2 . (B)
By graphing both sides of this equation on the same axes (ther e are several possibilities for the
right side), convince yourself of the following:
(c) If βδ = 0 , there’s a positive integer N such that (B) has one solution kn in each of the intervals
((2n− 1)π/L, (2n+ 1) π/L )) , n = N,N + 1 ,N + 2 ,..., (C)
and either
lim
n→∞
(
kn − (2n− 1)π
2L
)",Elementary Differential Equations with Boundary Value Problems.pdf
"((2n− 1)π/L, (2n+ 1) π/L )) , n = N,N + 1 ,N + 2 ,..., (C)
and either
lim
n→∞
(
kn − (2n− 1)π
2L
)
= 0 or lim
n→∞
(
kn − (2n+ 1) π
2L
)
= 0 .
(d) If βδ ̸= 0 , there’s a positive integer N such that (B) has one solution kn in each of the intervals
(C) and
lim
n→∞
(
kn − nπ
N
)
= 0 .
31. The following Sturm–Liouville problems are genera1izatio ns of Problems 1–4 of Section 11.1.
Problem 1: (p(x)y′)′+ λr(x)y= 0 , y(a) = 0 , y(b) = 0
Problem 2: (p(x)y′)′+ λr(x)y= 0 , y′(a) = 0 , y′(b) = 0
Problem 3: (p(x)y′)′+ λr(x)y= 0 , y(a) = 0 , y′(b) = 0
Problem 4: (p(x)y′)′+ λr(x)y= 0 , y′(a) = 0 , y(b) = 0
Prove: Problems 1–4 have no negative eigenvalues. Moreover , λ = 0 is an eigenvalue of Problem 2
with associated eigenfunction y0 = 1 , but λ = 0 isn’t an eigenvalue of Problems 1, 3, and 4. H IN T :
See the proof of Theorem 11.1.1.",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 701
32. Show that the eigenvalues of the Sturm–Liouville problem
(p(x)y′)′+ λr(x)y= 0 , αy (a) + βy′(a) = 0 , ρy (b) + δy′(b)
are all positive if αβ ≤ 0, ρδ ≥ 0, and (αβ )2 + ( ρδ)2 >0.",Elementary Differential Equations with Boundary Value Problems.pdf
"702 Chapter 13 Boundary V alue Problems for Second Order Ordinary Differen tial Equations
A BRIEF T ABLE OF INTEGRALS
∫
uα du= uα +1
α + 1 + c, α ̸= −1
∫ du
u = ln |u|+ c
∫
cos udu = sin u+ c
∫
sin udu = − cos u+ c
∫
tan udu = − ln |cos u|+ c
∫
cot udu = ln |sin u|+ c
∫
sec2 udu = tan u+ c
∫
csc2 udu = − cot u+ c
∫
sec udu = ln |sec u+ tan u|+ c
∫
cos2 udu = u
2 + 1
4 sin 2 u+ c
∫
sin2 udu = u
2 − 1
4 sin 2 u+ c
∫ du
1 + u2 du= tan −1 u+ c
∫ du√
1 − u2 du= sin −1 u+ c
∫ 1
u2 − 1 du= 1
2 ln
⏐
⏐
⏐
⏐
u− 1
u+ 1
⏐
⏐
⏐
⏐+ c
∫
cosh udu = sinh u+ c
∫
sinh udu = cosh u+ c
∫
udv = uv−
∫
v du
∫
ucos udu = usin u+ cos u+ c",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 13.2 Sturm-Liouville Problems 703
∫
usin udu = −ucos u+ sin u+ c
∫
ueu du= ueu − eu + c
∫
eλu cos ωudu = eλu (λ cos ωu + ω sin ωu)
λ2 + ω2 + c
∫
eλu sin ωudu = eλu (λ sin ωu − ω cos ωu)
λ2 + ω2 + c
∫
ln |u|du= uln |u| −u+ c
∫
uln |u|du= u2 ln |u|
2 − u2
4 + c
∫
cos ω1ucos ω2udu = sin(ω1 + ω2)u
2(ω1 + ω2) + sin(ω1 − ω2)u
2(ω1 − ω2) + c (ω1 ̸= ±ω2)
∫
sin ω1usin ω2udu = −sin(ω1 + ω2)u
2(ω1 + ω2) + sin(ω1 − ω2)u
2(ω1 − ω2) + c (ω1 ̸= ±ω2)
∫
sin ω1ucos ω2udu = −cos(ω1 + ω2)u
2(ω1 + ω2) − cos(ω1 − ω2)u
2(ω1 − ω2) + c (ω1 ̸= ±ω2)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected
Exercises
Section 1.2 Answers, pp. 14– 15
1.2.1 (p. 14) (a) 3 (b) 2 (c) 1 (d) 2
1.2.3 (p. 14) (a) y= −x2
2 + c (b) y = xcos x− sin x+ c
(c) y = x2
2 ln x− x2
4 + c(d) y= −xcos x+ 2 sin x+ c1 + c2x
(e) y= (2 x− 4)ex + c1 + c2x (f) y = x3
3 − sin x+ ex + c1 + c2x
(g) y = sin x+ c1 + c2x+ c3x2 (h) y= −x5
60 + ex + c1 + c2x+ c3x2
(i) y = 7
64 e4x + c1 + c2x+ c3x2
1.2.4 (p. 14) (a) y= −(x− 1)ex (b) y= 1 − 1
2 cos x2 (c) y= 3 − ln(
√
2 cos x)
(d) y= −47
15 − 37
5 (x− 2) + x5
30 (e) y = 1
4 xe2x − 1
4 e2x + 29
4
(f) y= xsin x+ 2 cos x− 3x− 1 (g) y = ( x2 − 6x+ 12) ex + x2
2 − 8x− 11
(h) y= x3
3 + cos 2x
6 + 7
4 x2 − 6x+ 7
8 (i) y = x4
12 + x3
6 + 1
2 (x− 2)2 − 26
3 (x− 2) − 5
3
1.2.7 (p. 15) (a) 576 ft (b) 10 s 1.2.8 (p. 15) (b) y= 0 1.2.10 (p. 15) (a) (−2c−2, ∞ ) ( −∞ , ∞ )
705",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 705
Section 2.1 Answers, pp. 41– 44
2.1.1 (p. 41) y = e−ax 2.1.2 (p. 41) y= ce−x3
2.1.3 (p. 41) y= ce−(ln x)2/ 2
2.1.4 (p. 41) y = c
x3 2.1.5 (p. 41) y = ce1/x 2.1.6 (p. 41) y = e−(x−1)
x 2.1.7 (p. 41) y= e
xln x
2.1.8 (p. 41) y = π
xsin x 2.1.9 (p. 41) y= 2(1 + x2) 2.1.10 (p. 41) y = 3 x−k
2.1.11 (p. 41) y = c(cos kx)1/k 2.1.12 (p. 41) y = 1
3 + ce−3x 2.1.13 (p. 41) y = 2
x + c
xex
2.1.14 (p. 41) y= e−x2
(x2
2 + c
)
2.1.15 (p. 41) y = −e−x + c
1 + x2 2.1.16 (p. 42) y = 7 ln |x|
x + 3
2 x+ c
x
2.1.17 (p. 42) y = ( x− 1)−4(ln |x− 1| −cos x+ c) 2.1.18 (p. 42) y= e−x2
(x3
4 + c
x
)
2.1.19 (p. 42) y = 2 ln |x|
x2 + 1
2 + c
x2 2.1.20 (p. 42) y = ( x+ c) cos x 2.1.21 (p. 42) y = c− cos x
(1 + x)2
2.1.22 (p. 42) y = −1
2
(x− 2)3
(x− 1) + c(x− 2)5
(x− 1) 2.1.23 (p. 42) y= ( x+ c)e− sin2 x
2.1.24 (p. 42) y = ex
x2 − ex
x3 + c
x2 . y= e3x − e−7x
10 2.1.26 (p. 42) 2x+ 1
(1 + x2)2
2.1.27 (p. 42) y= 1
x2 ln
(1 + x2
2
)
2.1.29 (p. 42) y = 2 ln |x|
x + x
2 − 1",Elementary Differential Equations with Boundary Value Problems.pdf
"x2 − ex
x3 + c
x2 . y= e3x − e−7x
10 2.1.26 (p. 42) 2x+ 1
(1 + x2)2
2.1.27 (p. 42) y= 1
x2 ln
(1 + x2
2
)
2.1.29 (p. 42) y = 2 ln |x|
x + x
2 − 1
2x 2.1.28 (p. 42) y= 1
2 (sin x+ csc x)
2.1.29 (p. 42) y = 2 ln |x|
x + x
2 − 1
2x 2.1.30 (p. 42) y= ( x− 1)−3 [ln(1 − x) − cos x]
2.1.31 (p. 42) y= 2 x2 + 1
x2 (0, ∞ ) 2.1.32 (p. 42) y = x2(1−ln x) 2.1.33 (p. 42) y = 1
2 + 5
2 e−x2
2.1.34 (p. 42) y = ln |x− 1|+ tan x+ 1
(x− 1)3 2.1.35 (p. 42) y= ln |x|+ x2 + 1
(x+ 2) 4
2.1.36 (p. 42) y = ( x2 − 1)
(1
2 ln |x2 − 1| −4
)
2.1.37 (p. 42) y = −(x2 − 5)
(
7 + ln |x2 − 5|
)
2.1.38 (p. 42) y = e−x2
(
3 +
∫ x
0
t2et2
dt
)
2.1.39 (p. 42) y = 1
x
(
2 +
∫ x
1
sin t
t dt
)
2.1.40 (p. 42) y= e−x
∫ x
1
tan t
t dt
2.1.41 (p. 43) y= 1
1 + x2
(
1 +
∫ x
0
et
1 + t2 dt
)
2.1.42 (p. 43) y= 1
x
(
2e−(x−1) + e−x
∫ x
1
et et2
dt
)",Elementary Differential Equations with Boundary Value Problems.pdf
"706 Answers to Selected Exercises
2.1.43 (p. 43) G= r
λ +
(
G0 − r
λ
)
e−λt limt→∞ G(t) = r
λ 2.1.45 (p. 43) (a) y= y0e−a(x−x0) + e−ax
∫ x
x0
eatf(t) dt
2.1.48 (p. 44) (a) y= tan −1
(1
3 + ce3x
)
(b) y= ±
[
ln
(1
x + c
x2
)] 1/ 2
(c) y= exp
(
x2 + c
x2
)
(d) y = −1 + x
c+ 3 ln |x|
Section 2.2 Answers, pp. 52– 55
2.2.1 (p. 52) y = 2 ±
√
2(x3 + x2 + x+ c)
2.2.2 (p. 52) ln(|sin y|) = cos x+ c; y ≡ kπ, k= integer
2.2.3 (p. 52) y = c
x− c y ≡ − 1 2.2.4 (p. 52) (ln y)2
2 = −x3
3 + c
2.2.5 (p. 52) y3 + 3 sin y+ ln |y|+ ln(1 + x2) + tan −1 x= c; y≡ 0
2.2.6 (p. 52) y = ±
(
1 +
( x
1 + cx
) 2) 1/ 2
; y≡ ± 1
2.2.7 (p. 52) y= tan
(x3
3 + c
)
2.2.8 (p. 52) y= c√
1 + x2 2.2.9 (p. 52) y = 2 − ce(x−1)2/ 2
1 − ce(x−1)2/ 2 ; y ≡ 1
2.2.10 (p. 52) y = 1 +
(
3x2 + 9 x+ c)1/ 3
2.2.11 (p. 52) y = 2 +
√
2
3 x3 + 3 x2 + 4 x− 11
3 2.2.12 (p. 52) y = e−(x2 −4)/ 2
2 − e−(x2−4)/ 2
2.2.13 (p. 52) y3 + 2y2 + x2 + sin x= 3 2.2.14 (p. 53) (y+ 1)( y− 1)−3(y− 2)2 = −256(x+ 1) −6
2.2.15 (p. 53) y= −1+3e−x2",Elementary Differential Equations with Boundary Value Problems.pdf
"2 − e−(x2−4)/ 2
2.2.13 (p. 52) y3 + 2y2 + x2 + sin x= 3 2.2.14 (p. 53) (y+ 1)( y− 1)−3(y− 2)2 = −256(x+ 1) −6
2.2.15 (p. 53) y= −1+3e−x2
2.2.16 (p. 53) y= 1√
2e−2x2
− 1
2.2.17 (p. 53) y≡ − 1; ( −∞ , ∞ )
2.2.18 (p. 53) y = 4 − e−x2
2 − e−x2 ; ( −∞ , ∞ ) 2.2.19 (p. 53) y= −1 +
√
4x2 − 15
2 ;
(√
15
2 , ∞
)
2.2.20 (p. 53) y = 2
1 + e−2x (−∞ , ∞ ) 2.2.21 (p. 53) y = −
√
25 − x2; (−5, 5)
2.2.22 (p. 53) y ≡ 2, (−∞ , ∞ ) 2.2.23 (p. 53) y= 3
(x+ 1
2x− 4
) 1/ 3
; (−∞ , 2)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 707
2.2.24 (p. 53) y = x+ c
1 − cx 2.2.25 (p. 53) y = −xcos c+
√
1 − x2 sin c; y≡ 1; y≡ − 1
2.2.26 (p. 53) y = −x+ 3 π/ 2 2.2.28 (p. 53) P = P0
αP 0 + (1 − αP 0)e−at ; limt→∞ P(t) = 1 /α
2.2.29 (p. 53) I = SI0
I0 + ( S− I0)e−rSt
2.2.30 (p. 53) If q = rS then I = I0
1 + rI0t and limt→∞ I(t) = 0 . If q ̸= Rs, then I =
αI0
I0 + ( α − I0)e−rαt . If q<rs , then limt→∞ I(t) = α = S− q
r
if q >rS, then limt→∞ I(t) = 0 2.2.34 (p. 55) f = ap, where a=constant
2.2.35 (p. 55) y = e−x (
−1 ±
√
2x2 + c
)
2.2.36 (p. 55) y = x2 (
−1 +
√
x2 + c
)
2.2.37 (p. 55) y = ex (
−1 + (3 xex + c)1/ 3)
2.2.38 (p. 55) y = e2x(1 ±
√
c− x2) 2.2.39 (p. 55) (a) y1 = 1 /x ; g(x) = h(x)
(b) y1 = x; g(x) = h(x)/x 2 (c) y1 = e−x; g(x) = ex h(x)
(d) y1 = x−r ; g(x) = xr−1h(x) (e) y1 = 1 /v (x); g(x) = v(x)h(x)
Section 2.3 Answers, pp. 60– 62
2.3.1 (p. 60) (a) , (b) x0 ̸= kπ (k= integer) 2.3.2 (p. 60) (a) , (b) (x0,y 0) ̸= (0 , 0)
2.3.3 (p. 61) (a) , (b) x0y0 ̸= (2 k+ 1) π",Elementary Differential Equations with Boundary Value Problems.pdf
"2.3.1 (p. 60) (a) , (b) x0 ̸= kπ (k= integer) 2.3.2 (p. 60) (a) , (b) (x0,y 0) ̸= (0 , 0)
2.3.3 (p. 61) (a) , (b) x0y0 ̸= (2 k+ 1) π
2 (k= integer) 2.3.4 (p. 61) (a) , (b) x0y0 >0 and x0y0 ̸= 1
2.3.5 (p. 61) (a) all (x0,y 0) (b) (x0,y 0) with y0 ̸= 0 2.3.6 (p. 61) (a) , (b) all (x0,y 0)
2.3.7 (p. 61) (a) , (b) all (x0,y 0) 2.3.8 (p. 61) (a) , (b) (x0,y 0) such that x0 ̸= 4 y0
2.3.9 (p. 61) (a) all (x0,y 0) (b) all (x0,y 0) ̸= (0 , 0) 2.3.10 (p. 61) (a) all (x0,y 0)
(b) all (x0,y 0) with y0 ̸= ±1 2.3.11 (p. 61) (a) , (b) all (x0,y 0)",Elementary Differential Equations with Boundary Value Problems.pdf
"708 Answers to Selected Exercises
2.3.12 (p. 61) (a) , (b) all (x0,y 0) such that x0 + y0 >0
2.3.13 (p. 61) (a) , (b) all (x0,y 0) with x0 ̸= 1 , y 0 ̸= (2 k+ 1) π
2 (k= integer)
2.3.16 (p. 61) y =
(3
5 x+ 1
) 5/ 3
, −∞ <x< ∞ , is a solution.
Also,
y =
{ 0, −∞ <x ≤ − 5
3
(3
5 x+ 1
) 5/ 3
, −5
3 <x< ∞
is a solution, For every a≥ 5
3 , the following function is also a solution:
y =







(3
5 (x+ a)
) 5/ 3
, −∞ <x< −a,
0, −a≤ x≤ − 5
3
(3
5 x+ 1 ) 5/ 3
, −5
3 <x< ∞ .
2.3.17 (p. 62) (a) all (x0,y 0) (b) all (x0,y 0) with y0 ̸= 1
2.3.18 (p. 62) y1 ≡ 1; y2 = 1 + |x|3; y3 = 1 − |x|3; y4 = 1 + x3; y5 = 1 − x3
y6 =
{ 1 + x3, x ≥ 0,
1, x< 0 ; y7 =
{ 1 − x3, x ≥ 0,
1, x< 0 ;
y8 =
{ 1, x ≥ 0,
1 + x3, x< 0 ; y9 =
{ 1, x ≥ 0,
1 − x3, x< 0
2.3.19 (p. 62) y = 1 + ( x2 + 4) 3/ 2, −∞ <x< ∞
2.3.20 (p. 62) (a) The solution is unique on (0, ∞ ). It is given by
y=
{
1, 0 <x ≤
√
5,
1 − (x2 − 5)3/ 2,
√
5 <x< ∞
(b)
y=
{ 1, −∞ <x ≤
√
5,
1 − (x2 − 5)3/ 2,
√
5 <x< ∞",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 709
is a solution of (A) on (−∞ , ∞ ). If α ≥ 0, then
y=



1 + ( x2 − α2)3/ 2, −∞ <x< −α,
1, −α ≤ x≤
√
5,
1 − (x2 − 5)3/ 2,
√
5 <x< ∞ ,
and
y=



1 − (x2 − α2)3/ 2, −∞ <x< −α,
1, −α ≤ x≤
√
5,
1 − (x2 − 5)3/ 2,
√
5 <x< ∞ ,
are also solutions of (A) on (−∞ , ∞ ).
Section 2.4 Answers, pp. 68– 72
2.4.1 (p. 68) y = 1
1 − cex 2.4.2 (p. 68) y= x2/ 7(c− ln |x|)1/ 7 2.4.3 (p. 68) y = e2/x (c− 1/x )2
2.4.4 (p. 68) y = ±
√2x+ c
1 + x2 2.4.5 (p. 68) y = ±(1 − x2 + ce−x2
)−1/ 2
2.4.6 (p. 68) y=
[ x
3(1 − x) + ce−x
] 1/ 3
2.4.7 (p. 69) y= 2
√
2√1 − 4x 2.4.8 (p. 69) y=
[
1 − 3
2 e−(x2−1)/ 4
] −2
2.4.9 (p. 69) y = 1
x(11 − 3x)1/ 3 2.4.10 (p. 69) y= (2 ex − 1)2
2.4.11 (p. 69) y = (2 e12x − 1 − 12x)1/ 3 2.4.12 (p. 69) y=
[ 5x
2(1 + 4 x5)
] 1/ 2
2.4.13 (p. 69) y = (4 ex/ 2 − x− 2)2
2.4.14 (p. 69) P = P0eat
1 + aP0
∫t
0 α(τ)eaτ dτ
; limt→∞ P(t) =



∞ if L= 0 ,
0 if L= ∞ ,
1/aL if 0 <L< ∞ .
2.4.15 (p.
69) y = x(ln |x|+ c) 2.4.16 (p. 69) y= cx2
1 − cx y= −x",Elementary Differential Equations with Boundary Value Problems.pdf
"∫t
0 α(τ)eaτ dτ
; limt→∞ P(t) =



∞ if L= 0 ,
0 if L= ∞ ,
1/aL if 0 <L< ∞ .
2.4.15 (p.
69) y = x(ln |x|+ c) 2.4.16 (p. 69) y= cx2
1 − cx y= −x
2.4.17 (p. 69) y = ±x(4 ln |x|+ c)1/ 4 2.4.18 (p. 69) y = xsin−1(ln |x|+ c)
2.4.19 (p. 69) y = xtan(ln |x|+ c) 2.4.20 (p. 69) y= ±x
√
cx2 − 1
2.4.21 (p. 70) y = ±xln(ln |x|+ c) 2.4.22 (p. 70) y= − 2x
2 ln |x|+ 1
2.4.23 (p. 70) y = x(3 ln x+ 27) 1/ 3 2.4.24 (p. 70) y = 1
x
(9 − x4
2
) 1/ 2
2.4.25 (p. 70) y= −x",Elementary Differential Equations with Boundary Value Problems.pdf
"710 Answers to Selected Exercises
2.4.26 (p. 70) y= −x(4x− 3)
(2x− 3) 2.4.27 (p. 70) y= x
√
4x6 − 1 2.4.28 (p. 70) tan−1 y
x − 1
2 ln(x2 + y2) = c
2.4.29 (p. 70) (x+ y) ln |x|+ y(1 − ln |y|) + cx= 0 2.4.30 (p. 70) (y+ x)3 = 3 x3(ln |x|+ c)
2.4.31 (p. 70) (y+ x) = c(y− x)3; y = x; y= −x
2.4.32 (p. 70) y2 (y− 3x) = c; y ≡ 0; y = 3 x
2.4.33 (p. 70) (x−y)3 (x+y) = cy2 x4; y= 0; y= x; y= −x 2.4.34 (p. 70) y
x + y3
x3 = ln |x|+c
2.4.40 (p. 71) Choose X0 and Y0 so that
aX0 + bY0 = α
cX0 + dY0 = β.
2.4.41 (p. 72) (y+ 2 x+ 1) 4(2y− 6x− 3) = c; y = 3 x+ 3 / 2; y= −2x− 1
2.4.42 (p. 72) (y+ x− 1)(y− x− 5)3 = c; y = x+ 5; y= −x+ 1
2.4.43 (p. 72) ln |y − x− 6| − 2(x+ 2)
y− x− 6 = c; y = x+ 6 2.4.44 (p. 72) (y1 = x1/ 3) y =
x1/ 3(ln |x|+ c)1/ 3
2.4.45 (p. 72) y1 = x3; y= ±x3√
cx6 − 1 2.4.46 (p. 72) y1 = x2; y = x2(1 + cx4)
1 − cx4 y = −x2
2.4.47 (p. 72) y1 = ex ; y= −ex(1 − 2cex)
1 − cex ; y= −2ex
2.4.48 (p. 72) y1 = tan x; y= tan x tan(ln |tan x|+ c)",Elementary Differential Equations with Boundary Value Problems.pdf
"1 − cx4 y = −x2
2.4.47 (p. 72) y1 = ex ; y= −ex(1 − 2cex)
1 − cex ; y= −2ex
2.4.48 (p. 72) y1 = tan x; y= tan x tan(ln |tan x|+ c)
2.4.49 (p. 72) y1 = ln x; y= 2 ln x (1 + c(ln x)4)
1 − c(ln x)4 ; y= −2 ln x
2.4.50 (p. 72) y1 = x1/ 2; y = x1/ 2(−2 ±
√
ln |x|+ c)
2.4.51 (p. 72) y1 = ex2
; y= ex2
(−1 ±
√
2x2 + c) 2.4.52 (p. 72) y= −3 + √1 + 60 x
2x
2.4.53 (p. 72) y = −5 + √1 + 48 x
2x2 2.4.56 (p. 72) y= 1 + 1
x+ 1 + cex",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 711
2.4.57 (p. 72) y = ex − 1
1 + ce−x 2.4.58 (p. 72) y = 1 − 1
x(1 − cx) 2.4.59 (p. 72) y = x− 2x
x2 + c",Elementary Differential Equations with Boundary Value Problems.pdf
"712 Answers to Selected Exercises
Section 2.5 Answers, pp. 79– 82
2.5.1 (p. 79) 2x3y2 = c 2.5.2 (p. 79) 3ysin x+ 2 x2ex + 3 y= c 2.5.3 (p. 79) Not exact
2.5.4 (p. 79) x2 − 2xy2 + 4 y3 = c 2.5.5 (p. 79) x+ y= c 2.5.6 (p. 79) Not exact
2.5.7 (p. 79) 2y2 cos x+ 3 xy3 − x2 = c 2.5.8 (p. 79) Not exact
2.5.9 (p. 79) x3 +x2y+4xy2 +9y2 = c 2.5.10 (p. 79) Not exact 2.5.11 (p. 79) ln |xy|+x2 +y2 = c
2.5.12 (p. 79) Not exact 2.5.13 (p. 79) x2 + y2 = c 2.5.14 (p. 79) x2y2 ex + 2 y+ 3 x2 = c
2.5.15 (p. 79) x3ex2+y − 4y3 + 2 x2 = c 2.5.16 (p. 79) x4exy + 3 xy= c
2.5.17 (p. 79) x3 cos xy+ 4 y2 + 2 x2 = c 2.5.18 (p. 79) y = x+
√
2x2 + 3 x− 1
x2
2.5.19 (p. 79) y = sin x−
√
1 − tan x
2 2.5.20 (p. 79) y=
(ex − 1
ex + 1
) 1/ 3
2.5.21 (p. 79) y = 1 + 2 tan x 2.5.22 (p. 79) y= x2 − x+ 6
(x+ 2)( x− 3)
2.5.23 (p. 80) 7x2
2 + 4 xy+ 3y2
2 = c 2.5.24 (p. 80) (x4y2 + 1) ex + y2 = c
2.5.29 (p. 81) (a) M(x,y ) = 2 xy+ f(x) (b) M(x,y ) = 2(sin x+ xcos x)(ysin y+ cos y) + f(x)
(c) M(x,y ) = yex − ey cos x+ f(x)",Elementary Differential Equations with Boundary Value Problems.pdf
"2.5.29 (p. 81) (a) M(x,y ) = 2 xy+ f(x) (b) M(x,y ) = 2(sin x+ xcos x)(ysin y+ cos y) + f(x)
(c) M(x,y ) = yex − ey cos x+ f(x)
2.5.30 (p. 81) (a) N(x,y ) = x4y
2 + x2 + 6 xy+ g(y) (b) N(x,y ) = x
y + 2 ysin x+ g(y)
(c) N(x,y ) = x(sin y+ ycos y) + g(y)
2.5.33 (p. 81) B= C 2.5.34 (p. 81) B = 2 D, E = 2 C
2.5.37 (p. 82) (a) 2x2 + x4y4 + y2 = c(b) x3 + 3 xy2 = c(c) x3 + y2 + 2 xy= c
2.5.38 (p. 82) y = −1 − 1
x2 2.5.39 (p. 82) y = x3
(
−3(x2 + 1) +
√
9x4 + 34 x2 + 21
2
)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 713
2.5.40 (p. 82) y = −e−x2
(
2x+
√
9 − 5x2
3
)
.
2.5.44 (p. 82) (a) G(x,y ) = 2 xy+ c(b) G(x,y ) = ex sin y+ c
(c) G(x,y ) = 3 x2y− y3 + c(d) G(x,y ) = − sin xsinh y+ c
(e) G(x,y ) = cos xsinh y+ c
Section 2.6 Answers, pp. 91– 93
2.6.3 (p. 91) µ(x) = 1 /x 2; y= cxand µ(y) = 1 /y 2; x= cy
2.6.4 (p. 91) µ(x) = x−3/ 2; x3/ 2y= c 2.6.5 (p. 91) µ(y) = 1 /y 3; y3e2x = c
2.6.6 (p. 91) µ(x) = e5x/ 2; e5x/ 2(xy+ 1) = c 2.6.7 (p. 92) µ(x) = ex; ex (xy+ y+ x) = c
2.6.8 (p. 92) µ(x) = x; x2y2 (9x+ 4y) = c 2.6.9 (p. 92) µ(y) = y2 ; y3 (3x2y+ 2x+ 1) = c 2.6.10
(p. 92) µ(y) = yey ; ey (xy3 + 1) = c 2.6.11 (p. 92) µ(y) = y2 ; y3 (3x4 + 8 x3y+ y) = c
2.6.12 (p. 92) µ(x) = xex; x2y(x+ 1) ex = c
2.6.13 (p. 92) µ(x) = ( x3 − 1)−4/ 3; xy(x3 − 1)−1/ 3 = cand x≡ 1
2.6.14 (p. 92) µ(y) = ey ; ey (sin xcos y+y−1) = c 2.6.15 (p. 92) µ(y) = e−y2
; xye−y2
(x+y) = c
2.6.16 (p. 92) xy
sin y = cand y = kπ (k = integer) 2.6.17 (p. 92) µ(x,y ) = x4y3; x5y4 ln x = c",Elementary Differential Equations with Boundary Value Problems.pdf
"; xye−y2
(x+y) = c
2.6.16 (p. 92) xy
sin y = cand y = kπ (k = integer) 2.6.17 (p. 92) µ(x,y ) = x4y3; x5y4 ln x = c
2.6.18 (p. 92) µ(x,y ) = 1 /xy ; |x|α |y|β eγx eδy = cand x≡ 0, y≡ 0
2.6.19 (p. 92) µ(x,y ) = x−2y−3 ; 3 x2y2 + y= 1 + cxy2 and x≡ 0, y≡ 0
2.6.20 (p. 92) µ(x,y ) = x−2y−1 ; − 2
x + y3 + 3 ln |y|= cand x≡ 0, y≡ 0
2.6.21 (p. 92) µ(x,y ) = eaxeby ; eaxeby cos xy = c",Elementary Differential Equations with Boundary Value Problems.pdf
"714 Answers to Selected Exercises
2.6.22 (p. 92) µ(x,y ) = x−4y−3 (and others) xy= c 2.6.23 (p. 92) µ(x,y ) = xey ; x2yey sin x= c
2.6.24 (p. 92) µ(x) = 1 /x 2; x3y3
3 − y
x = c 2.6.25 (p. 92) µ(x) = x+ 1 ; y(x+ 1) 2(x+ y) = c
2.6.26 (p. 92) µ(x,y ) = x2y2 ; x3y3(3x+ 2 y2) = c
2.6.27 (p. 92) µ(x,y ) = x−2y−2 ; 3x2y= cxy+ 2 and x≡ 0, y≡ 0
Section 3.1 Answers, pp. 106– 108
3.1.1 (p. 106) y1 = 1 .450000000, y2 = 2 .085625000, y3 = 3 .079099746
3.1.2 (p. 106) y1 = 1 .200000000, y2 = 1 .440415946, y3 = 1 .729880994
3.1.3 (p. 106) y1 = 1 .900000000, y2 = 1 .781375000, y3 = 1 .646612970
3.1.4 (p. 106) y1 = 2 .962500000, y2 = 2 .922635828, y3 = 2 .880205639
3.1.5 (p. 106) y1 = 2 .513274123, y2 = 1 .814517822, y3 = 1 .216364496
3.1.6 (p. 106) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 48.298147362 51.492825643 53.076673685 54.647937102
3.1.7 (p. 106) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
2.0 1.390242009 1.370996758 1.361921132 1.353193719",Elementary Differential Equations with Boundary Value Problems.pdf
"3.1.7 (p. 106) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
2.0 1.390242009 1.370996758 1.361921132 1.353193719
3.1.8 (p. 107) x h = 0 . 05 h = 0 . 025 h = 0 . 0125 Exact
1.50 7.886170437 8.852463793 9.548039907 10.500000000
3.1.9 (p. 107)
x h = 0 . 1 h = 0 . 05 h = 0 . 025 h = 0 . 1 h = 0 . 05 h = 0 . 025
3.0 1.469458241 1.462514486 1.459217010 0.3210 0.1537 0.0753
Approximate Solutions Residuals
3.1.10 (p. 107)
x h = 0 . 1 h = 0 . 05 h = 0 . 025 h = 0 . 1 h = 0 . 05 h = 0 . 025
2.0 0.473456737 0.483227470 0.487986391 -0.3129 -0.1563 -0.0781
Approximate Solutions Residuals
3.1.11 (p. 107) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
1.0 0.691066797 0.676269516 0.668327471 0.659957689",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 715
3.1.12 (p. 108) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
2.0 -0.772381768 -0.761510960 -0.756179726 -0.750912371
3.1.13 (p. 108)
Euler’ s method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 0.538871178 0.593002325 0.620131525 0.647231889
Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 0.647231889 0.647231889 0.647231889 0.647231889
Applying variation of parameters to the given initial value problem yields
y = ue−3x, where (A) u′= 7 , u (0) = 6 . Since u′′= 0 , Euler’s method yields the exact
solution of (A). Therefore the Euler semilinear method prod uces the exact solution of the
given problem
.
3.1.14 (p.
108)
Euler’ s method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
3.0 12.804226135 13.912944662 14.559623055 15.282004826
Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
3.0 15.354122287 15.317257705 15.299429421 15.282004826
3.1.15 (p. 108)
Euler’ s method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”",Elementary Differential Equations with Boundary Value Problems.pdf
"3.0 15.354122287 15.317257705 15.299429421 15.282004826
3.1.15 (p. 108)
Euler’ s method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”
2.0 0.867565004 0.885719263 0.895024772 0.904276722
Euler semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”
2.0 0.569670789 0.720861858 0.808438261 0.904276722
3.1.16 (p. 108)
Euler’ s method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”
3.0 0.922094379 0.945604800 0.956752868 0.967523153
Euler semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”
3.0 0.993954754 0.980751307 0.974140320 0.967523153",Elementary Differential Equations with Boundary Value Problems.pdf
"716 Answers to Selected Exercises
3.1.17 (p. 108)
Euler’ s method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact”
1.50 0.319892131 0.330797109 0.337020123 0.343780513
Euler semilinear method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact”
1.50 0.305596953 0.323340268 0.333204519 0.343780513
3. 1. 18 (p. 108)
Euler’ s method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”
2.0 0.754572560 0.743869878 0.738303914 0.732638628
Euler semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact”
2.0 0.722610454 0.727742966 0.730220211 0.732638628
3. 1. 19 (p. 108)
Euler’ s method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact”
1.50 2.175959970 2.210259554 2.227207500 2.244023982
Euler semilinear method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact”
1.50 2.117953342 2.179844585 2.211647904 2.244023982
3. 1. 20 (p. 108)
Euler’ s method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
1.0 0.032105117 0.043997045 0.050159310 0.056415515
Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”",Elementary Differential Equations with Boundary Value Problems.pdf
"1.0 0.032105117 0.043997045 0.050159310 0.056415515
Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
1.0 0.056020154 0.056243980 0.056336491 0.056415515
3. 1. 21 (p. 108)
Euler’ s method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
1.0 28.987816656 38.426957516 45.367269688 54.729594761
Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
1.0 54.709134946 54.724150485 54.728228015 54.729594761
3. 1. 22 (p. 108)
Euler’ s method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
3.0 1.361427907 1.361320824 1.361332589 1.361383810
Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact”
3.0 1.291345518 1.326535737 1.344004102 1.361383810
Section 3.2 Answers, pp. 116– 108",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 717
3. 2. 1 (p. 116) y1 = 1 . 542812500, y 2 = 2 . 421622101, y 3 = 4 . 208020541
3. 2. 2 (p. 116) y1 = 1 . 220207973, y 2 = 1 . 489578775 y3 = 1 . 819337186
3. 2. 3 (p. 116) y1 = 1 . 890687500, y 2 = 1 . 763784003, y 3 = 1 . 622698378
3. 2. 4 (p. 116) y1 = 2 . 961317914 y2 = 2 . 920132727 y3 = 2 . 876213748.
3. 2. 5 (p. 116) y1 = 2 . 478055238, y 2 = 1 . 844042564, y 3 = 1 . 313882333
3. 2. 6 (p. 116) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 56.134480009 55.003390448 54.734674836 54.647937102
3. 2. 7 (p. 116) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
2.0 1.353501839 1.353288493 1.353219485 1.353193719
3. 2. 8 (p. 117) x h = 0 . 05 h = 0 . 025 h = 0 . 0125 Exact
1.50 10.141969585 10.396770409 10.472502111 10.500000000
3. 2. 9 (p. 117)
x h = 0 . 1 h = 0 . 05 h = 0 . 025 h = 0 . 1 h = 0 . 05 h = 0 . 025
3.0 1.455674816 1.455935127 1.456001289 -0.00818 -0.00207 -0.000518
Approximate Solutions Residuals
3. 2. 10 (p. 117)",Elementary Differential Equations with Boundary Value Problems.pdf
"3.0 1.455674816 1.455935127 1.456001289 -0.00818 -0.00207 -0.000518
Approximate Solutions Residuals
3. 2. 10 (p. 117)
x h = 0 . 1 h = 0 . 05 h = 0 . 025 h = 0 . 1 h = 0 . 05 h = 0 . 025
2.0 0.492862999 0.492709931 0.492674855 0.00335 0.000777 0.000187
Approximate Solutions Residuals
3. 2. 11 (p. 117) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.660268159 0.660028505 0.659974464 0.659957689
3. 2. 12 (p. 118) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
2.0 -0.749751364 -0.750637632 -0.750845571 -0.750912371
3. 2. 13 (p. 118) Applying variation of parameters to the given initial value problem
y = ue−3x , where (A) u′= 1 − 2x, u (0) = 2 . Since u′′′= 0 , the improved Euler method yields
the exact solution of (A). Therefore the improved Euler semi linear method produces the exact solution
of the given problem.
Improved Euler method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 0.105660401 0.100924399 0.099893685 0.099574137
Improved Euler semilinear method",Elementary Differential Equations with Boundary Value Problems.pdf
"Improved Euler method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 0.105660401 0.100924399 0.099893685 0.099574137
Improved Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 0.099574137 0.099574137 0.099574137 0.099574137
3. 2. 14 (p. 118)
Improved Euler method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 15.107600968 15.234856000 15.269755072 15.282004826
Improved Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 15.285231726 15.282812424 15.282206780 15.282004826",Elementary Differential Equations with Boundary Value Problems.pdf
"718 Answers to Selected Exercises
3. 2. 15 (p. 118)
Improved Euler method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.924335375 0.907866081 0.905058201 0.904276722
Improved Euler semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.969670789 0.920861858 0.908438261 0.904276722
3. 2. 16 (p. 118)
Improved Euler method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
3.0 0.967473721 0.967510790 0.967520062 0.967523153
Improved Euler semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
3.0 0.967473721 0.967510790 0.967520062 0.967523153
3. 2. 17 (p. 118)
Improved Euler method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 0.349176060 0.345171664 0.344131282 0.343780513
Improved Euler semilinear method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 0.349350206 0.345216894 0.344142832 0.343780513
3. 2. 18 (p. 118)
Improved Euler method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.732679223 0.732721613 0.732667905 0.732638628
Improved Euler semilinear method",Elementary Differential Equations with Boundary Value Problems.pdf
"Improved Euler method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.732679223 0.732721613 0.732667905 0.732638628
Improved Euler semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.732166678 0.732521078 0.732609267 0.732638628
3. 2. 19 (p. 118)
Improved Euler method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 2.247880315 2.244975181 2.244260143 2.244023982
Improved Euler semilinear method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 2.248603585 2.245169707 2.244310465 2.244023982
3. 2. 20 (p. 118)
Improved Euler method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.059071894 0.056999028 0.056553023 0.056415515
Improved Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.056295914 0.056385765 0.056408124 0.056415515",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 719
3. 2. 21 (p. 118)
Improved Euler method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 50.534556346 53.483947013 54.391544440 54.729594761
Improved Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 54.709041434 54.724083572 54.728191366 54.729594761
3. 2. 22 (p. 118)
Improved Euler method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 1.361395309 1.361379259 1.361382239 1.361383810
Improved Euler semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 1.375699933 1.364730937 1.362193997 1.361383810
3. 2. 23 (p. 118) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
2.0 1.349489056 1.352345900 1.352990822 1.353193719
3. 2. 24 (p. 118) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
2.0 1.350890736 1.352667599 1.353067951 1.353193719
3. 2. 25 (p. 118) x h = 0 . 05 h = 0 . 025 h = 0 . 0125 Exact
1.50 10.133021311 10.391655098 10.470731411 10.500000000
3. 2. 26 (p. 118) x h = 0 . 05 h = 0 . 025 h = 0 . 0125 Exact",Elementary Differential Equations with Boundary Value Problems.pdf
"1.50 10.133021311 10.391655098 10.470731411 10.500000000
3. 2. 26 (p. 118) x h = 0 . 05 h = 0 . 025 h = 0 . 0125 Exact
1.50 10.136329642 10.393419681 10.470731411 10.500000000
3. 2. 27 (p. 118) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.660846835 0.660189749 0.660016904 0.659957689
3. 2. 28 (p. 119) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.660658411 0.660136630 0.660002840 0.659957689
3. 2. 29 (p. 119) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
2.0 -0.750626284 -0.750844513 -0.750895864 -0.751331499
3. 2. 30 (p. 119) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
2.0 -0.750335016 -0.750775571 -0.750879100 -0.751331499
Section 3.3 Answers, pp. 124– 127
3. 3. 1 (p. 124) y1 = 1 . 550598190, y 2 = 2 . 469649729 3. 3. 2 (p. 124) y1 = 1 . 221551366, y 2 = 1 . 492920208
3. 3. 3 (p. 124) y1 = 1 . 890339767, y 2 = 1 . 763094323 3. 3. 4 (p. 124) y1 = 2 . 961316248 y2 = 2 . 920128958.
3. 3. 5 (p. 124) y1 = 2 . 475605264, y 2 = 1 . 825992433",Elementary Differential Equations with Boundary Value Problems.pdf
"3. 3. 5 (p. 124) y1 = 2 . 475605264, y 2 = 1 . 825992433
3. 3. 6 (p. 124) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
1.0 54.654509699 54.648344019 54.647962328 54.647937102",Elementary Differential Equations with Boundary Value Problems.pdf
"720 Answers to Selected Exercises
3. 3. 7 (p. 124) x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
2.0 1.353191745 1.353193606 1.353193712 1.353193719
3. 3. 8 (p. 125) x h = 0 . 05 h = 0 . 025 h = 0 . 0125 Exact
1.50 10.498658198 10.499906266 10.499993820 10.500000000
3. 3. 9 (p. 125)
x h = 0 . 1 h = 0 . 05 h = 0 . 025 h = 0 . 1 h = 0 . 05 h = 0 . 025
3.0 1.456023907 1.456023403 1.456023379 0.0000124 0.000000611 0.0000000333
Approximate Solutions Residuals
3. 3. 10 (p. 125)
x h = 0 . 1 h = 0 . 05 h = 0 . 025 h = 0 . 1 h = 0 . 05 h = 0 . 025
2.0 0.492663789 0.492663738 0.492663736 0.000000902 0.0000000508 0.00000000302
Approximate Solutions Residuals
3. 3. 11 (p. 125) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.659957046 0.659957646 0.659957686 0.659957689
3. 3. 12 (p. 126) x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
2.0 -0.750911103 -0.750912294 -0.750912367 -0.750912371
3. 3. 13 (p. 126) Applying variation of parameters to the given initial value problem yields",Elementary Differential Equations with Boundary Value Problems.pdf
"2.0 -0.750911103 -0.750912294 -0.750912367 -0.750912371
3. 3. 13 (p. 126) Applying variation of parameters to the given initial value problem yields
y = ue−3x , where (A) u′= 1 − 4x + 3 x2 − 4x3, u (0) = −3. Since u(5) = 0 , the Runge-Kutta
method yields the exact solution of (A). Therefore the Euler semilinear method produces the exact
solution of the given problem.
Runge-Kutta method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
0.0 -3.000000000 -3.000000000 -3.000000000 -3.000000000
0.1 -2.162598011 -2.162526572 -2.162522707 -2.162522468
0.2 -1.577172164 -1.577070939 -1.577065457 -1.577065117
0.3 -1.163350794 -1.163242678 -1.163236817 -1.163236453
0.4 -0.868030294 -0.867927182 -0.867921588 -0.867921241
0.5 -0.655542739 -0.655450183 -0.655445157 -0.655444845
0.6 -0.501535352 -0.501455325 -0.501450977 -0.501450707
0.7 -0.389127673 -0.389060213 -0.389056546 -0.389056318
0.8 -0.306468018 -0.306412184 -0.306409148 -0.306408959
0.9 -0.245153433 -0.245107859 -0.245105379 -0.245105226",Elementary Differential Equations with Boundary Value Problems.pdf
"0.8 -0.306468018 -0.306412184 -0.306409148 -0.306408959
0.9 -0.245153433 -0.245107859 -0.245105379 -0.245105226
1.0 -0.199187198 -0.199150401 -0.199148398 -0.199148273
Runge-Kutta semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 Exact
0.0 -3.000000000 -3.000000000 -3.000000000 -3.000000000
0.1 -2.162522468 -2.162522468 -2.162522468 -2.162522468
0.2 -1.577065117 -1.577065117 -1.577065117 -1.577065117
0.3 -1.163236453 -1.163236453 -1.163236453 -1.163236453
0.4 -0.867921241 -0.867921241 -0.867921241 -0.867921241
0.5 -0.655444845 -0.655444845 -0.655444845 -0.655444845
0.6 -0.501450707 -0.501450707 -0.501450707 -0.501450707
0.7 -0.389056318 -0.389056318 -0.389056318 -0.389056318
0.8 -0.306408959 -0.306408959 -0.306408959 -0.306408959
0.9 -0.245105226 -0.245105226 -0.245105226 -0.245105226
1.0 -0.199148273 -0.199148273 -0.199148273 -0.199148273",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 721
3. 3. 14 (p. 126)
Runge-Kutta method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 15.281660036 15.281981407 15.282003300 15.282004826
Runge-Kutta semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 15.282005990 15.282004899 15.282004831 15.282004826
3. 3. 15 (p. 126)
Runge-Kutta method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.904678156 0.904295772 0.904277759 0.904276722
Runge-Kutta semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.904592215 0.904297062 0.904278004 0.904276722
3. 3. 16 (p. 126)
Runge-Kutta method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
3.0 0.967523147 0.967523152 0.967523153 0.967523153
Runge-Kutta semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
3.0 0.967523147 0.967523152 0.967523153 0.967523153
3. 3. 17 (p. 126)
Runge-Kutta method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 0.343839158 0.343784814 0.343780796 0.343780513
Runge-Kutta semilinear method",Elementary Differential Equations with Boundary Value Problems.pdf
"x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 0.343839158 0.343784814 0.343780796 0.343780513
Runge-Kutta semilinear method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.00 0.000000000 0.000000000 0.000000000 0.000000000
1.05 0.028121022 0.028121010 0.028121010 0.028121010
1.10 0.055393494 0.055393466 0.055393465 0.055393464
1.15 0.082164048 0.082163994 0.082163990 0.082163990
1.20 0.108862698 0.108862597 0.108862591 0.108862590
1.25 0.136058715 0.136058528 0.136058517 0.136058516
1.30 0.164564862 0.164564496 0.164564473 0.164564471
1.35 0.195651074 0.195650271 0.195650219 0.195650216
1.40 0.231542288 0.231540164 0.231540027 0.231540017
1.45 0.276818775 0.276811011 0.276810491 0.276810456
1.50 0.343839124 0.343784811 0.343780796 0.343780513
3. 3. 18 (p. 126)
Runge-Kutta method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.732633229 0.732638318 0.732638609 0.732638628
Runge-Kutta semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""",Elementary Differential Equations with Boundary Value Problems.pdf
"2.0 0.732633229 0.732638318 0.732638609 0.732638628
Runge-Kutta semilinear method
x h = 0 . 2 h = 0 . 1 h = 0 . 05 “Exact""
2.0 0.732639212 0.732638663 0.732638630 0.732638628",Elementary Differential Equations with Boundary Value Problems.pdf
"722 Answers to Selected Exercises
3. 3. 19 (p. 126)
Runge-Kutta method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 2.244025683 2.244024088 2.244023989 2.244023982
Runge-Kutta semilinear method
x h = 0 . 0500 h = 0 . 0250 h = 0 . 0125 “Exact""
1.50 2.244025081 2.244024051 2.244023987 2.244023982
3. 3. 20 (p. 126)
Runge-Kutta method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.056426886 0.056416137 0.056415552 0.056415515
Runge-Kutta semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 0.056415185 0.056415495 0.056415514 0.056415515
3. 3. 21 (p. 126)
Runge-Kutta method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 54.695901186 54.727111858 54.729426250 54.729594761
Runge-Kutta semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
1.0 54.729099966 54.729561720 54.729592658 54.729594761
3. 3. 22 (p. 126)
Runge-Kutta method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 1.361384082 1.361383812 1.361383809 1.361383810
Runge-Kutta semilinear method",Elementary Differential Equations with Boundary Value Problems.pdf
"Runge-Kutta method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 1.361384082 1.361383812 1.361383809 1.361383810
Runge-Kutta semilinear method
x h = 0 . 1 h = 0 . 05 h = 0 . 025 “Exact""
3.0 1.361456502 1.361388196 1.361384079 1.361383810
3. 3. 24 (p. 127) x h = . 1 h = . 05 h = . 025 Exact
2.00 -1.000000000 -1.000000000 -1.000000000 -1.000000000
3. 3. 25 (p. 127) x h = . 1 h = . 05 h = . 025 “Exact""
1.00 1.000000000 1.000000000 1.000000000 1.000000000
3. 3. 26 (p. 127) x h = . 1 h = . 05 h = . 025 Exact
1.50 4.142171279 4.142170553 4.142170508 4.142170505
3. 3. 27 (p. 127) x h = . 1 h = . 05 h = . 025 Exact
3.0 16.666666988 16.666666687 16.666666668 16.666666667
Section 4.1 Answers, pp. 138– 140
4. 1. 1 (p. 138) Q = 20 e−(t ln 2) / 3200 g 4. 1. 2 (p. 138) 2 ln 10
ln 2 days 4. 1. 3 (p. 138) τ = 10 ln 2
ln 4 / 3 minutes",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 723
4. 1. 4 (p. 138) τ ln(p0/p 1 )
ln 2 4. 1. 5 (p. 138) tp
tq
= ln p
ln q 4. 1. 6 (p. 138) k = 1
t2 − t1
ln Q1
Q2
4. 1. 7 (p. 138) 20 g
4. 1. 8 (p. 138) 50 ln 2
3 yrs 4. 1. 9 (p. 138) 25
2 ln 2 %
4. 1. 10 (p. 138) (a) = 20 ln 3 yr (b). Q0 = 100000 e−. 5 4. 1. 11 (p. 138) (a) Q(t) = 5000 − 4750e−t/ 10 (b) 5000
lbs
4. 1. 12 (p.
138) 1
25 yrs; 4. 1. 13 (p. 138) V = V0 et ln 10/ 2 4 hours
4. 1. 14 (p. 138) 1500 ln 4
3
ln 2 yrs; 2 −4/ 3 Q0 4. 1. 15 (p. 138) W (t) = 20 − 19e−t/ 20; limt→∞ W (t) = 20 ounces
4. 1. 16 (p. 138) S(t) = 10(1 + e−t/ 10); lim t→∞ S(t) = 10 g 4. 1. 17 (p. 139) 10 gallons
4. 1. 18 (p. 139) V (t) = 15000 + 10000 et/ 20 4. 1. 19 (p. 139) W (t) = 4 × 106 (t + 1) 2 dollars t years from now
4. 1. 20 (p. 139) p = 100
25 − 24e−t/ 2 4. 1. 21 (p. 139) (a) P (t) = 1000 e. 06t + 50 e. 06t − 1
e. 06/ 52 − 1 (b) 5. 64 × 10−4
4. 1. 22 (p. 139) (a) P ′= rP − 12M (b) P = 12M
r (1 − ert ) + P0ert (c) M ≈ rP0
12(1 − e−rN )",Elementary Differential Equations with Boundary Value Problems.pdf
"e. 06/ 52 − 1 (b) 5. 64 × 10−4
4. 1. 22 (p. 139) (a) P ′= rP − 12M (b) P = 12M
r (1 − ert ) + P0ert (c) M ≈ rP0
12(1 − e−rN )
(d) For (i) approximate M = $402 . 25, exact M = $402 . 80
for (ii) approximate M = $1206 . 05, exact M = $1206 . 93.
4. 1. 23 (p. 139) (a) T (α ) = − 1
r ln
(
1 −
(
1 − e−rN )/α
))
years
S(α ) = P0
(1 − e−rN )
[
rN + α ln
(
1 − (1 − e−rN )/α
)]
(b) T (1. 05) = 13 . 69 yrs, S(1. 05) = $3579 . 94 T (1. 10) = 12 . 61 yrs,
S(1. 10) = $6476 . 63 T (1. 15) = 11 . 70 yrs, S(1. 15) = $8874 . 98.
4. 1. 24 (p. 140) P0 =



S0 (1 − e(a−r)T )
r − a if a ̸= r,
S0 T if a = r.
Section 4.2 Answers, pp. 148– 150
4. 2. 1 (p. 148) ≈ 15. 15◦F 4. 2. 2 (p. 148) T = −10 + 110 e−t ln 11
9 4. 2. 3 (p. 148) ≈ 24. 33◦F
4. 2. 4 (p. 148) (a) 91. 30◦F (b) 8.99 minutes after being placed outside (c) never
4. 2. 5 (p. 148) (a) 12:11:32 (b) 12:47:33 4. 2. 6 (p. 148) (85/ 3)◦C 4. 2. 7 (p. 148) 32◦F 4. 2. 8 (p. 148) Q(t) = 40(1 − e−3t/ 40)",Elementary Differential Equations with Boundary Value Problems.pdf
"4. 2. 5 (p. 148) (a) 12:11:32 (b) 12:47:33 4. 2. 6 (p. 148) (85/ 3)◦C 4. 2. 7 (p. 148) 32◦F 4. 2. 8 (p. 148) Q(t) = 40(1 − e−3t/ 40)
4. 2. 9 (p. 148) Q(t) = 30 − 20e−t/ 10 4. 2. 10 (p. 148) K (t) = . 3 − . 2e−t/ 20 4. 2. 11 (p. 148) Q(50) = 47 . 5
(pounds)
4. 2. 12 (p.
148) 50 gallons 4. 2. 13 (p. 148) min q2 = q1/ c 4. 2. 14 (p. 149) Q = t + 300 − 234 × 105
(t + 300) 2 , 0 ≤ t ≤ 300
4. 2. 15 (p. 149) (a) Q′+ 2
25 Q = 6 − 2e−t/ 25 (b) Q = 75 − 50e−t/ 25 − 25e−2t/ 25 (c) 75
4. 2. 16 (p. 149) (a) T = Tm + ( T0 − Tm)e−kt + k(S0 − Tm)
(k − km )
(
e−kmt − e−kt )
(b) T = Tm + k(S0 − Tm )te−kt + ( T0 − Tm )e−kt (c) limt→∞ T (t) = lim t→∞ S(t) = Tm
4. 2. 17 (p. 149) (a) T ′= −k
(
1 + a
am
)
T + k
(
Tm0 + a
am
T0
)
(b) T = aT0 + amTm0
a + am
+ am (T0 − Tm0 )
a + am
e−k(1+a/a m )t ,
Tm = aT0 + am Tm0
a + am
+ a(Tm0 − T0)
a + am
e−k(1+a/a m )t ; (c) limt→∞ T (t) = lim t→∞ Tm (t) = aT0 + am Tm0
a + am
4. 2. 18 (p. 150) V = a
b
V0
V0 − (V0 − a/b ) e−at , limt→∞ V (t) = a/b
4. 2. 19 (p. 150) c1 = c",Elementary Differential Equations with Boundary Value Problems.pdf
"a + am
4. 2. 18 (p. 150) V = a
b
V0
V0 − (V0 − a/b ) e−at , limt→∞ V (t) = a/b
4. 2. 19 (p. 150) c1 = c
(
1 − e−rt/W )
, c2 = c
(
1 − e−rt/W − r
W te−rt/W
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"724 Answers to Selected Exercises
4. 2. 20 (p. 150) (a) cn = c
(
1 − e−rt/W
n−1∑
j=0
1
j!
(rt
W
) j
)
(b) c (c) 0
4. 2. 21 (p. 150) Let c∞ = c1W1 + c2W2
W1 + W2
, α = c2W 2
2 − c1W 2
1
W1 + W2
, and β = W1 + W2
W1 W2
. Then:
(a) c1(t) = c∞ + α
W1
e−rβt , c2(t) = c∞ − α
W2
e−rβt
(b) limt→∞ c1(t) = lim t→∞ c2(t) = c∞
Section 4.3 Answers, pp. 160– 162
4. 3. 1 (p. 160) v = − 384
5
(
1 − e−5t/ 12)
; − 384
5 ft/s 4. 3. 2 (p. 160) k = 12; v = −16(1 − e−2t )
4. 3. 3 (p. 160) v = 25(1 − e−t );25 ft/s 4. 3. 4 (p. 160) v = 20 − 27e−t/ 40 4. 3. 5 (p. 160) ≈ 17. 10 ft
4. 3. 6 (p. 160) v = − 40(13 + 3 e−4t/ 5)
13 − 3e−4t/ 5 ; -40 ft/s 4. 3. 7 (p. 160) v = −128(1 − e−t/ 4)
4. 3. 9 (p. 160) T = m
k ln
(
1 + v0k
mg
)
; ym = y0 + m
k
[
v0 − mg
k ln
(
1 + v0k
mg
)]
4. 3. 10 (p. 161) v = − 64(1 − e−t )
1 + e−t ; -64 ft/s
4. 3. 11 (p. 161) v = α v0(1 + e−βt ) − α (1 − e−βt )
α (1 + e−βt ) − v0(1 − e−βt ) ; −α , where α =
√ mg
k and β = 2
√
kg
m .
4. 3. 12 (p. 161) T =
√
m
kg tan−1
(
v0
√
k
mg
)
v = −
√ mg",Elementary Differential Equations with Boundary Value Problems.pdf
"α (1 + e−βt ) − v0(1 − e−βt ) ; −α , where α =
√ mg
k and β = 2
√
kg
m .
4. 3. 12 (p. 161) T =
√
m
kg tan−1
(
v0
√
k
mg
)
v = −
√ mg
k ; 1 − e−2
√
gk
m (t−T )
1 + e−2
√
gk
m (t−T )
4. 3. 13 (p. 161) s′= mg − as
s + 1 ; a0 = mg. 4. 3. 14 (p. 161) (a) ms′= mg − f (s)
4. 3. 15 (p. 161) (a) v′= −9. 8 + v4/ 81 (b) vT ≈ − 5. 308 m/s
4. 3. 16 (p. 161) (a) v′= −32 + 8
√
|v|; vT = −16 ft/s (b) From Exercise 4.3. 14(c), vT is the negative
number such that −32 + 8
√
|vT |= 0 ; thus, vT = −16 ft/s.
4. 3. 17 (p. 162) ≈ 6. 76 miles/s 4. 3. 18 (p. 162) ≈ 1. 47 miles/s 4. 3. 20 (p. 162) α = gR2
(ym + R)2
Section 4.4 Answers, pp. 176– 177
4. 4. 1 (p. 176) y = 0 is a stable equilibrium; trajectories are v2 + y4
4 = c
4. 4. 2 (p. 176) y = 0 is an unstable equilibrium; trajectories are v2 + 2y3
3 = c
4. 4. 3 (p. 176) y = 0 is a stable equilibrium; trajectories are v2 + 2|y|3
3 = c
4. 4. 4 (p. 176) y = 0 is a stable equilibrium; trajectories are v2 − e−y (y + 1) = c",Elementary Differential Equations with Boundary Value Problems.pdf
"3 = c
4. 4. 4 (p. 176) y = 0 is a stable equilibrium; trajectories are v2 − e−y (y + 1) = c
4. 4. 5 (p. 176) equilibria: 0 (stable) and −2, 2 (unstable); trajectories: 2v2 − y4 + 8 y2 = c;
separatrix: 2v2 − y4 + 8 y2 = 16
4. 4. 6 (p. 176) equilibria: 0 (unstable) and −2, 2 (stable); trajectories: 2v2 + y4 − 8y2 = c;
separatrix: 2v2 + y4 − 8y2 = 0
4. 4. 7 (p. 176) equilibria: 0, −2, 2 (stable), −1, 1 (unstable); trajectories:
6v2 + y2 (2y4 − 15y2 + 24) = c; separatrix: 6v2 + y2(2y4 − 15y2 + 24) = 11
4. 4. 8 (p. 176) equilibria: 0, 2 (stable) and −2, 1 (unstable);
trajectories: 30v2 + y2(12y3 − 15y2 − 80y + 120) = c;
separatrices: 30v2 + y2 (12y3 − 15y2 − 80y + 120) = 496 and",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 725
30v2 + y2 (12y3 − 15y2 − 80y + 120) = 37
4. 4. 9 (p. 176) No equilibria if a < 0; 0 is unstable if a = 0 ; √a is stable and
−√a is unstable if a > 0.
* 4. 4. 10 (p. 176) 0 is a stable equilibrium if a ≤ 0; −√a and √a are stable and 0 is unstable if a > 0.
4. 4. 11 (p. 176) 0 is unstable if a ≤ 0; −√a and √a are unstable and 0 is stable if a > 0.
4. 4. 12 (p. 176) 0 is stable if a ≤ 0; 0 is stable and −√a and √a are unstable if a ≤ 0.
4. 4. 22 (p. 178) An equilibrium solution y of y′′+ p(y) = 0 is unstable if there’ s an ϵ > 0
such that, for every δ > 0, there’ s a solution of (A) with
√
(y(0) − y)2 + v2 (0) < δ , but
√
(y(t) − y)2 + v2(t) ≥
ϵ for some t > 0.
Section 4.5 Answers, pp. 190– 192
4. 5. 1 (p. 190) y′= − 2xy
x2 + 3 y2 4. 5. 2 (p. 190) y′= − y2
(xy − 1) 4. 5. 3 (p. 190) y′= − y(x2 + y2 − 2x2 ln |xy|)
x(x2 + y2 − 2y2 ln |xy|) .
4. 5. 4 (p. 190) xy′− y = − x1/ 2
2 4. 5. 5 (p. 190) y′+ 2 xy = 4 xex2
4. 5. 6 (p. 190) xy′+ y = 4 x3",Elementary Differential Equations with Boundary Value Problems.pdf
"x(x2 + y2 − 2y2 ln |xy|) .
4. 5. 4 (p. 190) xy′− y = − x1/ 2
2 4. 5. 5 (p. 190) y′+ 2 xy = 4 xex2
4. 5. 6 (p. 190) xy′+ y = 4 x3
4. 5. 7 (p. 190) y′− y = cos x − sin x 4. 5. 8 (p. 190) (1 + x2)y′− 2xy = (1 − x)2ex
4. 5. 10 (p. 190) y′g − yg′= f ′g − f g′. 4. 5. 11 (p. 190) (x − x0)y′= y − y0 4. 5. 12 (p. 190) y′(y2 − x2 +
1) + 2 xy = 0 4. 5. 13 (p. 190) 2x(y − 1)y′− y2 + x2 + 2 y = 0
4. 5. 14 (p. 190) (a) y = −81 + 18 x, (9, 81) y = −1 + 2 x, (1, 1)
(b) y = −121 + 22 x, (11, 121) y = −1 + 2 x, (1, 1)
(c) y = −100 − 20x, (−10, 100) y = −4 − 4x, (−2, 4)
(d) y = −25 − 10x, (−5, 25) y = −1 − 2x, (−1, 1)
4. 5. 15 (p. 190) (e) y = 5 + 3 x
4 , (−3/ 5, 4/ 5) y = − 5 − 4x
3 , (4/ 5, −3/ 5)
4. 5. 17 (p. 191) (a) y = − 1
2 (1 + x), (1, −1); y = 5
2 + x
10 , (25, 5)
(b) y = 1
4 (4 + x), (4, 2) y = − 1
4 (4 + x), (4, −2);
(c) y = 1
2 (1 + x), (1, 1) y = 7
2 + x
14 , (49, 7)
(d) y = − 1
2 (1 + x), (1, −1) y = − 5
2 − x
10 , (25, −5)
4. 5. 18 (p. 191) y = 2 x2 4. 5. 19 (p. 192) y = cx√
|x2 − 1|",Elementary Differential Equations with Boundary Value Problems.pdf
"2 + x
14 , (49, 7)
(d) y = − 1
2 (1 + x), (1, −1) y = − 5
2 − x
10 , (25, −5)
4. 5. 18 (p. 191) y = 2 x2 4. 5. 19 (p. 192) y = cx√
|x2 − 1|
4. 5. 20 (p. 192) y = y1 + c(x − x1)
4. 5. 21 (p. 192) y = − x3
2 − x
2 4. 5. 22 (p. 192) y = −x ln |x|+ cx 4. 5. 23 (p. 192) y =
√
2x + 4
4. 5. 24 (p. 192) y =
√
x2 − 3 4. 5. 25 (p. 192) y = kx2 4. 5. 26 (p. 192) (y − x)3(y + x) = k
4. 5. 27 (p. 192) y2 = −x + k 4. 5. 28 (p. 192) y2 = − 1
2 ln(1 + 2 x2) + k
4. 5. 29 (p. 192) y2 = −2x − ln(x − 1)2 + k 4. 5. 30 (p. 192) y = 1 +
√
9 − x2
2 ; those with c > 0
4. 5. 33 (p. 192) tan−1 y
x − 1
2 ln(x2 + y2) = k 4. 5. 34 (p. 192) 1
2 ln(x2 + y2) + (tan α ) tan −1 y
x = k
Section 5.1 Answers, pp. 203– 210
5. 1. 1 (p. 203) (c) y = −2e2x + e5x (d) y = (5 k0 − k1) e2x
3 + ( k1 − 2k0 ) e5x
3 .
5. 1. 2 (p. 203) (c) y = ex (3 cos x − 5 sin x) (d) y = ex (k0 cos x + ( k1 − k0 ) sin x)
5. 1. 3 (p. 204) (c) y = ex (7 − 3x) (d) y = ex (k0 + ( k1 − k0 )x)",Elementary Differential Equations with Boundary Value Problems.pdf
"726 Answers to Selected Exercises
5. 1. 4 (p. 204) (a) y = c1
x − 1 + c2
x + 1 (b) y = 2
x − 1 − 3
x + 1 ; ( −1, 1)
5. 1. 5 (p. 204) (a) ex (b) e2x cos x (c) x2 + 2 x − 2 (d) − 5
6 x−5/ 6 (e) − 1
x2 (f) (x ln |x|)2 (g) e2x
2√x
5. 1. 6 (p. 204) 0 5. 1. 7 (p. 204) W (x) = (1 − x2)−1 5. 1. 8 (p. 205) W (x) = 1
x 5. 1. 10 (p. 205) y2 = e−x
5. 1. 11 (p. 205) y2 = xe3x 5. 1. 12 (p. 205) y2 = xeax 5. 1. 13 (p. 205) y2 = 1
x 5. 1. 14 (p. 205) y2 = x ln x
5. 1. 15 (p. 205) y2 = xa ln x 5. 1. 16 (p. 205) y2 = x1/ 2e−2x 5. 1. 17 (p. 205) y2 = x 5. 1. 18 (p. 205)
y2 = x sin x 5. 1. 19 (p. 205) y2 = x1/ 2 cos x 5. 1. 20 (p. 205) y2 = xe−x 5. 1. 21 (p. 205) y2 = 1
x2 − 4
5. 1. 22 (p. 205) y2 = e2x
5. 1. 23 (p. 205) y2 = x2 5. 1. 35 (p. 207) (a) y′′− 2y′+ 5 y = 0 (b) (2x − 1)y′′− 4xy′+ 4 y = 0 (c)
x2y′′− xy′+ y = 0
(d) x2y′′+ xy′+ y = 0 (e) y′′− y = 0 (f) xy′′− y′= 0
5. 1. 37 (p. 207) (c) y = k0 y1 + k1y2 5. 1. 38 (p. 208) y1 = 1 , y2 = x − x0; y = k0 + k1(x − x0)",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 1. 37 (p. 207) (c) y = k0 y1 + k1y2 5. 1. 38 (p. 208) y1 = 1 , y2 = x − x0; y = k0 + k1(x − x0)
5. 1. 39 (p. 208) y1 = cosh( x − x0), y2 = sinh( x − x0 ); y = k0 cosh(x − x0) + k1 sinh(x − x0)
5. 1. 40 (p. 208) y1 = cos ω (x − x0), y2 = 1
ω sin ω (x − x0) y = k0 cos ω (x − x0) + k1
ω sin ω (x − x0)
5. 1. 41 (p. 208) y1 = 1
1 − x2 , y2 = x
1 − x2 y = k0 + k1x
1 − x2
5. 1. 42 (p. 209) (c) k0 = k1 = 0 ; y =
{
c1x2 + c2x3, x ≥ 0,
c1x2 + c3x3, x < 0
(d) (0, ∞ ) if x0 > 0, (−∞ , 0) if x0 < 0
5. 1. 43 (p. 209) (c) k0 = 0 , k1 arbitrary y = k1x + c2x2
5. 1. 44 (p. 210) (c) k0 = k1 = 0 y =
{
a1 x3 + a2x4, x ≥ 0,
b1x3 + b2 x4, x < 0
(d) (0, ∞ ) if x0 > 0, (−∞ , 0) if x0 < 0
Section 5.2 Answers, pp. 217– 220
5. 2. 1 (p. 217) y = c1e−6x +c2ex 5. 2. 2 (p. 217) y = e2x (c1 cos x+c2 sin x) 5. 2. 3 (p. 217) y = c1e−7x+c2e−x
5. 2. 4 (p. 217) y = e2x(c1 + c2x) 5. 2. 5 (p. 217) y = e−x(c1 cos 3x + c2 sin 3x)",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 2. 4 (p. 217) y = e2x(c1 + c2x) 5. 2. 5 (p. 217) y = e−x(c1 cos 3x + c2 sin 3x)
5. 2. 6 (p. 217) y = e−3x(c1 cos x + c2 sin x) 5. 2. 7 (p. 217) y = e4x (c1 + c2x) 5. 2. 8 (p. 217) y = c1 + c2e−x
5. 2. 9 (p. 217) y = ex(c1 cos
√
2x + c2 sin
√
2x) 5. 2. 10 (p. 217) y = e−3x (c1 cos 2 x + c2 sin 2 x)
5. 2. 11 (p. 217) y = e−x/ 2
(
c1 cos 3x
2 + c2 sin 3x
2
)
5. 2. 12 (p. 217) y = c1e−x/ 5 + c2ex/ 2
5. 2. 13 (p. 218) y = e−7x(2 cos x − 3 sin x) 5. 2. 14 (p. 218) y = 4 ex/ 2 + 6 e−x/ 3 5. 2. 15 (p. 218) y =
3ex/ 3 − 4e−x/ 2
5. 2. 16 (p. 218) y = e−x/ 2
3 + 3e3x/ 2
4 5. 2. 17 (p. 218) y = e3x/ 2 (3 − 2x) 5. 2. 18 (p. 218) y = 3 e−4x − 4e−3x
5. 2. 19 (p. 218) y = 2 xe3x 5. 2. 20 (p. 218) y = ex/ 6 (3+2x) 5. 2. 21 (p. 218) y = e−2x
(
3 cos
√
6x + 2
√
6
3 sin
√
6x
)
5. 2. 23 (p. 218) y = 2 e−(x−1) − 3e−2(x−1) 5. 2. 24 (p. 219) y = 1
3 e−(x−2) − 2
3 e7(x−2)
5. 2. 25 (p. 219) y = e7(x−1) (2 − 3(x − 1)) 5. 2. 26 (p. 219) y = e−(x−2)/ 3 (2 − 4(x − 2))
5. 2. 27 (p. 219) y = 2 cos 2
3
(
x − π
4
)",Elementary Differential Equations with Boundary Value Problems.pdf
"3 e7(x−2)
5. 2. 25 (p. 219) y = e7(x−1) (2 − 3(x − 1)) 5. 2. 26 (p. 219) y = e−(x−2)/ 3 (2 − 4(x − 2))
5. 2. 27 (p. 219) y = 2 cos 2
3
(
x − π
4
)
− 3 sin 2
3
(
x − π
4
)
5. 2. 28 (p. 219) y = 2 cos
√
3
(
x − π
3
)
− 1√
3
sin
√
3
(
x − π
3
)
5. 2. 30 (p. 219) y = k0
r2 − r1
(
r2er1 (x−x0 ) − r1er2 (x−x0 ))
+ k1
r2 − r1
(
er2 (x−x0 ) − er1(x−x0 ))
5. 2. 31 (p. 219) y = er1(x−x0 ) [k0 + ( k1 − r1 k0)(x − x0)]",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 727
5. 2. 32 (p. 219) y = eλ (x−x0 )
[
k0 cos ω (x − x0) +
(k1 − λk 0
ω
)
sin ω (x − x0)
]
Section 5.3 Answers, pp. 227– 229
5. 3. 1 (p. 227) yp = −1 + 2 x + 3 x2 ; y = −1 + 2 x + 3 x2 + c1e−6x + c2ex
5. 3. 2 (p. 227) yp = 1 + x; y = 1 + x + e2x (c1 cos x + c2 sin x)
5. 3. 3 (p. 227) yp = −x + x3; y = −x + x3 + c1e−7x + c2e−x
5. 3. 4 (p. 227) yp = 1 − x2; y = 1 − x2 + e2x (c1 + c2x)
5. 3. 5 (p. 227) yp = 2 x + x3 ; y = 2 x + x3 + e−x(c1 cos 3 x + c2 sin 3 x);
y = 2 x + x3 + e−x (2 cos 3 x + 3 sin 3 x)
5. 3. 6 (p. 227) yp = 1 + 2 x; y = 1 + 2 x + e−3x(c1 cos x + c2 sin x); y = 1 + 2 x + e−3x(cos x − sin x)
5. 3. 8 (p. 227) yp = 2
x 5. 3. 9 (p. 227) yp = 4 x1/ 2 5. 3. 10 (p. 227) yp = x3
2 5. 3. 11 (p. 227) yp = 1
x3
5. 3. 12 (p. 227) yp = 9 x1/ 3 5. 3. 13 (p. 227) yp = 2x4
13 5. 3. 16 (p. 228) yp = e3x
3 ; y = e3x
3 + c1e−6x + c2ex
5. 3. 17 (p. 228) yp = e2x; y = e2x (1 + c1 cos x + c2 sin x)",Elementary Differential Equations with Boundary Value Problems.pdf
"13 5. 3. 16 (p. 228) yp = e3x
3 ; y = e3x
3 + c1e−6x + c2ex
5. 3. 17 (p. 228) yp = e2x; y = e2x (1 + c1 cos x + c2 sin x)
5. 3. 18 (p. 228) y = −2e−2x ; y = −2e−2x + c1e−7x + c2e−x; y = −2e−2x − e−7x + e−x
5. 3. 19 (p. 228) yp = ex; y = ex + e2x (c1 + c2x); y = ex + e2x(1 − 3x)
5. 3. 20 (p. 228) yp = 4
45 ex/ 2 ; y = 4
45 ex/ 2 + e−x (c1 cos 3 x + c2 sin 3 x)
5. 3. 21 (p. 228) yp = e−3x; y = e−3x (1 + c1 cos x + c2 sin x)
5. 3. 24 (p. 228) yp = cos x − sin x; y = cos x − sin x + e4x (c1 + c2x)
5. 3. 25 (p. 228) yp = cos 2 x − 2 sin 2 x; y = cos 2 x − 2 sin 2 x + c1 + c2e−x
5. 3. 26 (p. 228) yp = cos 3 x; y = cos 3 x + ex (c1 cos
√
2x + c2 sin
√
2x)
5. 3. 27 (p. 228) yp = cos x + sin x; y = cos x + sin x + e−3x (c1 cos 2 x + c2 sin 2 x)
5. 3. 28 (p. 228) yp = −2 cos 2 x + sin 2 x; y = −2 cos 2 x + sin 2 x + c1e−4x + c2e−3x
y = −2 cos 2 x + sin 2 x + 2 e−4x − 3e−3x
5. 3. 29 (p. 228) yp = cos 3 x − sin 3 x; y = cos 3 x − sin 3x + e3x (c1 + c2x)
y = cos 3 x − sin 3 x + e3x(1 + 2 x)",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 3. 29 (p. 228) yp = cos 3 x − sin 3 x; y = cos 3 x − sin 3x + e3x (c1 + c2x)
y = cos 3 x − sin 3 x + e3x(1 + 2 x)
5. 3. 30 (p. 228) y = 1
ω 2
0 − ω 2 (M cos ωx + N sin ωx ) + c1 cos ω 0x + c2 sin ω 0x
5. 3. 33 (p. 229) yp = −1 + 2 x + 3 x2 + e3x
3 ; y = −1 + 2 x + 3 x2 + e3x
3 + c1e−6x + c2ex
5. 3. 34 (p. 229) yp = 1 + x + e2x; y = 1 + x + e2x (1 + c1 cos x + c2 sin x)
5. 3. 35 (p. 229) yp = −x + x3 − 2e−2x ; y = −x + x3 − 2e−2x + c1e−7x + c2e−x
5. 3. 36 (p. 229) yp = 1 − x2 + ex; y = 1 − x2 + ex + e2x(c1 + c2x)
5. 3. 37 (p. 229) yp = 2 x + x3 + 4
45 ex/ 2; y = 2 x + x3 + 4
45 ex/ 2 + e−x(c1 cos 3 x + c2 sin 3x)
5. 3. 38 (p. 229) yp = 1 + 2 x + e−3x; y = 1 + 2 x + e−3x (1 + c1 cos x + c2 sin x)
Section 5.4 Answers, pp. 235– 238
5. 4. 1 (p. 235) yp = e3x
(
− 1
4 + x
2
)
5. 4. 2 (p. 235) yp = e−3x
(
1 − x
4
)
5. 4. 3 (p. 235) yp = ex
(
2 − 3x
4
)
5. 4. 4 (p. 235) yp = e2x (1 − 3x + x2 ) 5. 4. 5 (p. 235) yp = e−x(1 + x2 ) 5. 4. 6 (p. 235) yp = ex(−2 + x + 2x2 )",Elementary Differential Equations with Boundary Value Problems.pdf
"(
2 − 3x
4
)
5. 4. 4 (p. 235) yp = e2x (1 − 3x + x2 ) 5. 4. 5 (p. 235) yp = e−x(1 + x2 ) 5. 4. 6 (p. 235) yp = ex(−2 + x + 2x2 )
5. 4. 7 (p. 235) yp = xe−x
(1
6 + x
2
)
5. 4. 8 (p. 235) yp = xex (1 + 2 x) 5. 4. 9 (p. 235) yp = xe3x
(
−1 + x
2
)
5. 4. 10 (p. 235) yp = xe2x(−2+x) 5. 4. 11 (p. 235) yp = x2e−x
(
1 + x
2
)
5. 4. 12 (p. 235) yp = x2ex
(1
2 − x
)
5. 4. 13 (p. 235) yp = x2e2x
2 (1 − x + x2) 5. 4. 14 (p. 235) yp = x2e−x/ 3
27 (3 − 2x + x2 )",Elementary Differential Equations with Boundary Value Problems.pdf
"728 Answers to Selected Exercises
5. 4. 15 (p. 235) y = e3x
4 (−1 + 2 x) + c1ex + c2e2x 5. 4. 16 (p. 235) y = ex(1 − 2x) + c1e2x + c2e4x
5. 4. 17 (p. 235) y = e2x
5 (1 − x) + e−3x (c1 + c2x) 5. 4. 18 (p. 235) y = xex(1 − 2x) + c1ex + c2e−3x
5. 4. 19 (p. 235) y = ex [
x2(1 − 2x) + c1 + c2x
]
5. 4. 20 (p. 236) y = −e2x(1 + x) + 2 e−x − e5x
5. 4. 21 (p. 236) y = xe2x + 3 ex − e−4x 5. 4. 22 (p. 236) y = e−x (2 + x − 2x2) − e−3x
5. 4. 23 (p. 236) y = e−2x(3 − x) − 2e5x 5. 4. 24 (p. 236) yp = − ex
3 (1 − x) + e−x(3 + 2 x)
5. 4. 25 (p. 236) yp = ex(3 + 7 x) + xe3x 5. 4. 26 (p. 236) yp = x3e4x + 1 + 2 x + x2
5. 4. 27 (p. 236) yp = xe2x(1 − 2x) + xex 5. 4. 28 (p. 236) yp = ex(1 + x) + x2e−x
5. 4. 29 (p. 236) yp = x2e−x + e3x (1 − x2) 5. 4. 31 (p. 237) yp = 2 e2x 5. 4. 32 (p. 237) yp = 5 xe4x
5. 4. 33 (p. 237) yp = x2e4x 5. 4. 34 (p. 237) yp = − e3x
4 (1+2 x−2x2 ) 5. 4. 35 (p. 237) yp = xe3x(4−x+2x2 )
5. 4. 36 (p. 237) yp = x2e−x/ 2(−1 + 2 x + 3 x2)
5. 4. 37 (p. 237) (a) y = e−x
(4",Elementary Differential Equations with Boundary Value Problems.pdf
"4 (1+2 x−2x2 ) 5. 4. 35 (p. 237) yp = xe3x(4−x+2x2 )
5. 4. 36 (p. 237) yp = x2e−x/ 2(−1 + 2 x + 3 x2)
5. 4. 37 (p. 237) (a) y = e−x
(4
3 x3/ 2 + c1x + c2
)
(b) y = e−3x
[
x2
4 (2 ln x − 3) + c1x + c2
]
(c) y = e2x [(x + 1) ln |x + 1 |+ c1x + c2] (d) y = e−x/ 2
(
x ln |x|+ x3
6 + c1x + c2
)
5. 4. 39 (p. 238) (a) ex (3 + x) + c (b) −e−x (1 + x)2 + c (c) − e−2x
8 (3 + 6 x + 6 x2 + 4 x3) + c
(d) ex(1 + x2 ) + c (e) e3x (−6 + 4 x + 9 x2) + c (f) −e−x(1 − 2x3 + 3 x4) + c
5. 4. 40 (p. 238) (−1)kk!eαx
α k+1
k∑
r=0
(−αx )r
r! + c
Section 5.5 Answers, pp. 244– 248
5. 5. 1 (p. 244) yp = cos x + 2 sin x 5. 5. 2 (p. 244) yp = cos x + (2 − 2x) sin x
5. 5. 3 (p. 244) yp = ex(−2 cos x + 3 sin x)
5. 5. 4 (p. 244) yp = e2x
2 (cos 2 x − sin 2x) 5. 5. 5 (p. 244) yp = −ex(x cos x − sin x)
5. 5. 6 (p. 244) yp = e−2x(1 − 2x)(cos 3 x − sin 3x) 5. 5. 7 (p. 245) yp = x(cos 2 x − 3 sin 2 x)
5. 5. 8 (p. 245) yp = −x [(2 − x) cos x + (3 − 2x) sin x] 5. 5. 9 (p. 245) yp = x
[
x cos
(x
2
)
− 3 sin
(x
2
)]",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 5. 8 (p. 245) yp = −x [(2 − x) cos x + (3 − 2x) sin x] 5. 5. 9 (p. 245) yp = x
[
x cos
(x
2
)
− 3 sin
(x
2
)]
5. 5. 10 (p. 245) yp = xe−x(3 cos x + 4 sin x) 5. 5. 11 (p. 245) yp = xex [(−1 + x) cos 2 x + (1 + x) sin 2 x]
5. 5. 12 (p. 245) yp = −(14 − 10x) cos x − (2 + 8 x − 4x2) sin x.
5. 5. 13 (p. 245) yp = (1 + 2 x + x2) cos x + (1 + 3 x2) sin x 5. 5. 14 (p. 245) yp = x2
2 (cos 2x − sin 2 x)
5. 5. 15 (p. 245) yp = ex(x2 cos x + 2 sin x) 5. 5. 16 (p. 245) yp = ex(1 − x2)(cos x + sin x)
5. 5. 17 (p. 245) yp = ex(x2 − x3)(cos x + sin x) 5. 5. 18 (p. 245) yp = e−x [(1 + 2 x) cos x − (1 − 3x) sin x]
5. 5. 19 (p. 245) yp = x(2 cos 3 x − sin 3x) 5. 5. 20 (p. 245) yp = −x3 cos x + ( x + 2 x2) sin x
5. 5. 21 (p. 245) yp = −e−x [
(x + x2) cos x − (1 + 2 x) sin x
]
5. 5. 22 (p. 245) y = ex(2 cos x + 3 sin x) + 3 ex − e6x 5. 5. 23 (p. 245) y = ex [(1 + 2 x) cos x + (1 − 3x) sin x]",Elementary Differential Equations with Boundary Value Problems.pdf
"]
5. 5. 22 (p. 245) y = ex(2 cos x + 3 sin x) + 3 ex − e6x 5. 5. 23 (p. 245) y = ex [(1 + 2 x) cos x + (1 − 3x) sin x]
5. 5. 24 (p. 245) y = ex (cos x−2 sin x)+e−3x(cos x+sin x) 5. 5. 25 (p. 245) y = e3x [(2 + 2 x) cos x − (1 + 3 x) sin x]
5. 5. 26 (p. 245) y = e3x [(2 + 3 x) cos x + (4 − x) sin x]+3ex−5e2x 5. 5. 27 (p. 245) yp = xe3x − ex
5 (cos x − 2 sin x)
5. 5. 28 (p. 245) yp = x(cos x + 2 sin x) − ex
2 (1 − x) + e−x
2
5. 5. 29 (p. 245) yp = − xex
2 (2 + x) + 2 xe2x + 1
10 (3 cos x + sin x)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 729
5. 5. 30 (p. 245) yp = xex(cos x + x sin x) + e−x
25 (4 + 5 x) + 1 + x + x2
2
5. 5. 31 (p. 245) yp = x2e2x
6 (3 + x) − e2x(cos x − sin x) + 3 e3x + 1
4 (2 + x)
5. 5. 32 (p. 245) y = (1 − 2x + 3 x2)e2x + 4 cos x + 3 sin x 5. 5. 33 (p. 245) y = xe−2x cos x + 3 cos 2 x
5. 5. 34 (p. 245) y = − 3
8 cos 2 x + 1
4 sin 2x + e−x − 13
8 e−2x − 3
4 xe−2x
5. 5. 40 (p. 248) (a) 2x cos x − (2 − x2) sin x + c (b) − ex
2
[
(1 − x2) cos x − (1 − x)2 sin x
]
+ c
(c) − e−x
25 [(4 + 10 x) cos 2 x − (3 − 5x) sin 2 x] + c
(d) − e−x
2
[
(1 + x)2 cos x − (1 − x2) sin x
]
+ c
(e) − ex
2
[
x(3 − 3x + x2 ) cos x − (3 − 3x + x3) sin x
]
+ c
(f) −ex [(1 − 2x) cos x + (1 + x) sin x] + c (g) e−x [x cos x + x(1 + x) sin x] + c
Section 5.6 Answers, pp. 253– 255
5. 6. 1 (p. 253) y = 1 − 2x + c1e−x + c2xex; {e−x , xe x} 5. 6. 2 (p. 253) y = 4
3x2 + c1x + c2
x ; {x, 1/x }
5. 6. 3 (p. 253) y = x(ln |x|)2
2 + c1x + c2x ln |x|; {x, x ln |x|}",Elementary Differential Equations with Boundary Value Problems.pdf
"3x2 + c1x + c2
x ; {x, 1/x }
5. 6. 3 (p. 253) y = x(ln |x|)2
2 + c1x + c2x ln |x|; {x, x ln |x|}
5. 6. 4 (p. 253) y = ( e2x + ex ) ln(1 + e−x) + c1e2x + c2ex ; {e2x , e x}
5. 6. 5 (p. 253) y = ex
(4
5 x7/ 2 + c1 + c2x
)
; {ex, xe x}
5. 6. 6 (p. 253) y = ex(2x3/ 2 + x1/ 2 ln x + c1x1/ 2 + c2x−1/ 2); {x1/ 2 ex, x −1/ 2e−x }
5. 6. 7 (p. 253) y = ex(x sin x + cos x ln |cos x|+ c1 cos x + c2 sin x); {ex cos x, e x sin x}
5. 6. 8 (p. 253) y = e−x2
(2e−2x + c1 + c2x); {e−x2
, xe −x2
}
5. 6. 9 (p. 253) y = 2 x + 1 + c1x2 + c2
x2 ; {x2, 1/x 2 }
5. 6. 10 (p. 253) y = xe2x
9 + xe−x(c1 + c2x); {xe−x, x 2e−x}
5. 6. 11 (p. 253) y = xex
(x
3 + c1 + c2
x2
)
; {xex , e x/x }
5. 6. 12 (p. 253) y = − (2x − 1)2ex
8 + c1ex + c2xe−x; {ex, xe −x}
5. 6. 13 (p. 253) y = x4 + c1x2 + c2x2 ln |x|; {x2, x 2 ln |x|}
5. 6. 14 (p. 253) y = e−x(x3/ 2 + c1 + c2x1/ 2); {e−x, x 1/ 2e−x }
5. 6. 15 (p. 253) y = ex (x+c1+c2x2); {ex , x 2ex} 5. 6. 16 (p. 253) y = x1/ 2
(
e2x
2 + c1 + c2ex
)
; {x1/ 2 , x 1/ 2ex}",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 6. 15 (p. 253) y = ex (x+c1+c2x2); {ex , x 2ex} 5. 6. 16 (p. 253) y = x1/ 2
(
e2x
2 + c1 + c2ex
)
; {x1/ 2 , x 1/ 2ex}
5. 6. 17 (p. 253) y = −2x2 ln x + c1x2 + c2x4; {x2, x 4} 5. 6. 18 (p. 253) {ex , e x/x } 5. 6. 19 (p. 253) {x2, x 3}
5. 6. 20 (p. 253) {ln |x|, x ln |x|}5. 6. 21 (p. 253) {sin √x, cos √x} 5. 6. 22 (p. 253) {ex, x 3ex } 5. 6. 23 (p. 253)
{xa , x a ln x}
5. 6. 24 (p. 253) {x sin x, x cos x} 5. 6. 25 (p. 253) {e2x , x 2e2x} 5. 6. 26 (p. 253) {x1/ 2, x 1/ 2 cos x}
5. 6. 27 (p. 253) {x1/ 2e2x , x 1/ 2e−2x } 5. 6. 28 (p. 253) {1/x, e 2x } 5. 6. 29 (p. 253) {ex , x 2} 5. 6. 30 (p. 253)
{e2x , x 2e2x } 5. 6. 31 (p. 253) y = x4 + 6 x2 − 8x2 ln |x|
5. 6. 32 (p. 253) y = 2 e2x − xe−x 5. 6. 33 (p. 254) y = (x + 1)
4
[
−ex (3 − 2x) + 7 e−x ]
5. 6. 34 (p. 254) y = x2
4 + x 5. 6. 35 (p. 254) y = (x + 2) 2
6(x − 2) + 2x
x2 − 4",Elementary Differential Equations with Boundary Value Problems.pdf
"730 Answers to Selected Exercises
5. 6. 38 (p. 254) (a) y = −kc1 sin kx + kc2 cos kx
c1 cos kx + c2 sin kx (b) y = c1 + 2 c2ex
c1 + c2ex
(c) y = −6c1 + c2e7x
c1 + c2e7x (d) y = − 7c1 + c2e6x
c1 + c2e6x
(e) y = − (7c1 − c2) cos x + ( c1 + 7 c2) sin x
c1 cos x + c2 sin x
(f) y = −2c1 + 3 c2e5x/ 6
6(c1 + c2e5x/ 6) (g) y = c1 + c2(x + 6)
6(c1 + c2x)
5. 6. 39 (p. 254) (a) y = c1 + c2ex(1 + x)
x(c1 + c2ex ) (b) y = −2c1x + c2(1 − 2x2 )
c1 + c2x
(c) y = −c1 + c2e2x(x + 1)
c1 + c2xe2x (d) y = 2c1 + c2e−3x (1 − x)
c1 + c2xe−3x
(e) y = (2c2x − c1) cos x − (2c1x + c2) sin x
2x(c1 cos x + c2 sin x) (f) y = c1 + 7 c2x6
x(c1 + c2x6)
Section 5.7 Answers, pp. 262– 264
5. 7. 1 (p. 262) yp = − cos 3 x ln |sec 3 x + tan 3 x|
9 5. 7. 2 (p. 262) yp = − sin 2x ln |cos 2 x|
4 + x cos 2 x
2
5. 7. 3 (p. 262) yp = 4 ex(1 + ex) ln(1 + e−x) 5. 7. 4 (p. 262) yp = 3 ex (cos x ln |cos x|+ x sin x)
5. 7. 5 (p. 262) yp = 8
5 x7/ 2ex 5. 7. 6 (p. 262) yp = ex ln(1 − e−2x ) − e−x ln(e2x − 1) 5. 7. 7 (p. 263) yp =",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 7. 5 (p. 262) yp = 8
5 x7/ 2ex 5. 7. 6 (p. 262) yp = ex ln(1 − e−2x ) − e−x ln(e2x − 1) 5. 7. 7 (p. 263) yp =
2(x2 − 3)
3
5. 7. 8 (p. 263) yp = e2x
x 5. 7. 9 (p. 263) yp = x1/ 2ex ln x 5. 7. 10 (p. 263) yp = e−x(x+2)
5. 7. 11 (p. 263) yp = −4x5/ 2 5. 7. 12 (p. 263) yp = −2x2 sin x−2x cos x 5. 7. 13 (p. 263) yp = − xe−x(x + 1)
2
5. 7. 14 (p. 263) yp = −
√x cos √x
2 5. 7. 15 (p. 263) yp = 3x4 ex
2 5. 7. 16 (p. 263) yp = xa+1
5. 7. 17 (p. 263) yp = x2 sin x
2 5. 7. 18 (p. 263) yp = −2x2 5. 7. 19 (p. 263) yp = −e−x sin x
5. 7. 20 (p. 263) yp = −
√x
2 5. 7. 21 (p. 263) yp = x3/ 2
4 5. 7. 22 (p. 263) yp = −3x2
5. 7. 23 (p. 263) yp = x3ex
2 5. 7. 24 (p. 263) yp = − 4x3/ 2
15 5. 7. 25 (p. 263) yp = x3ex 5. 7. 26 (p. 263)
yp = xex
5. 7. 27 (p. 263) yp = x2 5. 7. 28 (p. 263) yp = xex(x − 2) 5. 7. 29 (p. 263) yp = √xex(x − 1)/ 4
5. 7. 30 (p. 263) y = e2x(3x2 − 2x + 6)
6 + xe−x
3 5. 7. 31 (p. 263) y = ( x − 1)2 ln(1 − x) + 2 x2 − 5x + 3",Elementary Differential Equations with Boundary Value Problems.pdf
"5. 7. 30 (p. 263) y = e2x(3x2 − 2x + 6)
6 + xe−x
3 5. 7. 31 (p. 263) y = ( x − 1)2 ln(1 − x) + 2 x2 − 5x + 3
5. 7. 32 (p. 263) y = ( x2−1)ex −5(x−1) 5. 7. 33 (p. 264) y = x(x2 + 6)
3(x2 − 1) 5. 7. 34 (p. 264) y = − x2
2 + x + 1
2x2
5. 7. 35 (p. 264) y = x2(4x + 9)
6(x + 1)
5. 7. 38 (p. 264) (a) y = k0 cosh x + k1 sinh x +
∫ x
0
sinh(x − t)f (t) dt
(b) y′= k0 sinh x + k1 cosh x +
∫ x
0
cosh(x − t)f (t) dt
5. 7. 39 (p. 264) (a) y(x) = k0 cos x + k1 sin x +
∫ x
0
sin(x − t)f (t) dt
(b) y′(x) = −k0 sin x + k1 cos x +
∫x
0 cos(x − t)f (t) dt",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 731
Section 6.1 Answers, pp. 277– 278
6. 1. 1 (p. 277) y = 3 cos 4
√
6t − 1
2
√
6
sin 4
√
6t ft 6. 1. 2 (p. 277) y = − 1
4 cos 8
√
5t − 1
4
√
5
sin 8
√
5t ft
6. 1. 3 (p. 277) y = 1 . 5 cos 14
√
10t cm
6. 1. 4 (p. 277) y = 1
4 cos 8 t − 1
16 sin 8 t ft; R =
√
17
16 ft; ω 0 = 8 rad/s; T = π/ 4 s;
φ ≈ − . 245 rad ≈ − 14. 04◦;
6. 1. 5 (p. 277) y = 10 cos 14 t + 25
14 sin 14 t cm; R = 5
14
√
809 cm; ω 0 = 14 rad/s; T = π/ 7 s;
φ ≈ . 177 rad ≈ 10. 12◦
6. 1. 6 (p. 277) y = − 1
4 cos
√
70 t + 2√
70
sin
√
70 t m; R = 1
4
√
67
35 m ω 0 =
√
70 rad/s;
T = 2 π/
√
70 s; φ ≈ 2. 38 rad ≈ 136. 28◦
6. 1. 7 (p. 277) y = 2
3 cos 16t − 1
4 sin 16 t ft 6. 1. 8 (p. 278) y = 1
2 cos 8 t − 3
8 sin 8 t ft 6. 1. 9 (p. 278) . 72 m
6. 1. 10 (p. 278) y = 1
3 sin t + 1
2 cos 2 t + 5
6 sin 2t ft 6. 1. 11 (p. 278) y = 16
5
(
4 sin t
4 − sin t
)
6. 1. 12 (p. 278) y = − 1
16 sin 8t + 1
3 cos 4
√
2t − 1
8
√
2
sin 4
√
2t
6. 1. 13 (p. 278) y = −t cos 8 t − 1
6 cos 8 t + 1",Elementary Differential Equations with Boundary Value Problems.pdf
"5
(
4 sin t
4 − sin t
)
6. 1. 12 (p. 278) y = − 1
16 sin 8t + 1
3 cos 4
√
2t − 1
8
√
2
sin 4
√
2t
6. 1. 13 (p. 278) y = −t cos 8 t − 1
6 cos 8 t + 1
8 sin 8t ft 6. 1. 14 (p. 278) T = 4
√
2 s
6. 1. 15 (p. 278) ω = 8 rad/s y = − t
16 (− cos 8 t + 2 sin 8 t) + 1
128 sin 8 t ft
6. 1. 16 (p. 278) ω = 4
√
6 rad/s; y = − t√
6
[ 8
3 cos 4
√
6t + 4 sin 4
√
6t
]
+ 1
9 sin 4
√
6t ft
6. 1. 17 (p. 278) y = t
2 cos 2t − t
4 sin 2t + 3 cos 2 t + 2 sin 2 t m
6. 1. 18 (p. 278) y = y0 cos ω 0 t + v0
ω 0
sin ω 0 t; R = 1
ω 0
√
(ω 0 y0)2 + ( v0)2;
cos φ = y0 ω 0
√
(ω 0 y0)2 + ( v0)2
; sin φ = v0
√
(ω 0y0 )2 + ( v0)2
6. 1. 19 (p. 278) The object with the longer period weighs four times as much as the other .
6. 1. 20 (p. 278) T2 =
√
2T1, where T1 is the period of the smaller object.
6. 1. 21 (p. 278) k1 = 9 k2, where k1 is the spring constant of the system with the shorter period.
Section 6.2 Answers, pp. 287– 289
6. 2. 1 (p. 287) y = e−2t
2 (3 cos 2 t − sin 2 t) ft;
√
5
2 e−2t ft",Elementary Differential Equations with Boundary Value Problems.pdf
"Section 6.2 Answers, pp. 287– 289
6. 2. 1 (p. 287) y = e−2t
2 (3 cos 2 t − sin 2 t) ft;
√
5
2 e−2t ft
6. 2. 2 (p. 287) y = −e−t
(
3 cos 3 t + 1
3 sin 3t
)
ft
√
82
3 e−t ft
6. 2. 3 (p. 287) y = e−16t
(1
4 + 10 t
)
ft 6. 2. 4 (p. 287) y = − e−3t
4 (5 cos t + 63 sin t) ft
6. 2. 5 (p. 287) 0 ≤ c < 8 lb-sec/ft 6. 2. 6 (p. 287) y = 1
2 e−3t
(
cos
√
91t + 11√
91
sin
√
91t
)
ft
6. 2. 7 (p. 287) y = − e−4t
3 (2 + 8 t) ft 6. 2. 8 (p. 287) y = e−10t
(
9 cos 4
√
6t + 45
2
√
6
sin 4
√
6t
)
cm
6. 2. 9 (p. 287) y = e−3t/ 2
(
3
2 cos
√
41
2 t + 9
2
√
41
sin
√
41
2 t
)
ft",Elementary Differential Equations with Boundary Value Problems.pdf
"732 Answers to Selected Exercises
6. 2. 10 (p. 287) y = e− 3
2 t
(
1
2 cos
√
119
2 t − 9
2
√
119
sin
√
119
2 t
)
ft
6. 2. 11 (p. 287) y = e−8t
(
1
4 cos 8
√
2t − 1
4
√
2
sin 8
√
2t
)
ft
6. 2. 12 (p. 287) y = e−t
(
− 1
3 cos 3
√
11t + 14
9
√
11
sin 3
√
11t
)
ft
6. 2. 13 (p. 287) yp = 22
61 cos 2 t + 2
61 sin 2 t ft 6. 2. 14 (p. 288) y = − 2
3 (e−8t − 2e−4t )
6. 2. 15 (p. 288) y = e−2t
(1
10 cos 4 t − 1
5 sin 4 t
)
m 6. 2. 16 (p. 288) y = e−3t (10 cos t − 70 sin t) cm
6. 2. 17 (p. 288) yp = − 2
15 cos 3 t + 1
15 sin 3t ft
6. 2. 18 (p. 288) yp = 11
100 cos 4t + 27
100 sin 4 t cm 6. 2. 19 (p. 288) yp = 42
73 cos t + 39
73 sin t ft
6. 2. 20 (p. 288) y = − 1
2 cos 2 t + 1
4 sin 2t m 6. 2. 21 (p. 288) yp = 1
cω 0
(−β cos ω 0t + α sin ω 0t)
6. 2. 24 (p. 288) y = e−ct/ 2m
(
y0 cos ω 1t + 1
ω 1
(v0 + cy0
2m ) sin ω 1 t
)
6. 2. 25 (p. 288) y = r2 y0 − v0
r2 − r1
er1t + v0 − r1y0
r2 − r1
er2t 6. 2. 26 (p. 289) y = er1 t (y0 + ( v0 − r1y0 )t)
Section 6.3 Answers, pp. 294– 295",Elementary Differential Equations with Boundary Value Problems.pdf
"r2 − r1
er1t + v0 − r1y0
r2 − r1
er2t 6. 2. 26 (p. 289) y = er1 t (y0 + ( v0 − r1y0 )t)
Section 6.3 Answers, pp. 294– 295
6. 3. 1 (p. 294) I = e−15t
(
2 cos 5
√
15t − 6√
31
sin 5
√
31t
)
6. 3. 2 (p. 294) I = e−20t (2 cos 40 t − 101 sin 40 t) 6. 3. 3 (p. 294) I = − 200
3 e−10t sin 30t
6. 3. 4 (p. 294) I = −10e−30t (cos 40t + 18 sin 40 t) 6. 3. 5 (p. 294) I = −e−40t (2 cos 30 t − 86 sin 30 t)
6. 3. 6 (p. 294) Ip = − 1
3 (cos 10t + 2 sin 10 t) 6. 3. 7 (p. 294) Ip = 20
37 (cos 25t − 6 sin 25 t)
6. 3. 8 (p. 294) Ip = 3
13 (8 cos 50 t − sin 50 t) 6. 3. 9 (p. 294) Ip = 20
123 (17 sin 100 t − 11 cos 100 t)
6. 3. 10 (p. 294) Ip = − 45
52 (cos 30t + 8 sin 30 t)
6. 3. 12 (p. 295) ω 0 = 1 /
√
LC maximum amplitude =
√
U 2 + V 2/R
Section 6.4 Answers, pp. 301– 302
6. 4. 1 (p. 301) If e = 1 , then Y 2 = ρ (ρ − 2X ); if e ̸= 1
(
X + eρ
1 − e2
) 2
+ Y 2
1 − e2 = ρ 2
(1 − e2)2 if ;
e < 1 let X0 = − eρ
1 − e2 , a = ρ
1 − e2 , b = ρ√
1 − e2 .
6. 4. 2 (p. 302) Let h = r2
0 θ ′
0; then ρ = h2
k , e =
[ (ρ",Elementary Differential Equations with Boundary Value Problems.pdf
"1 − e2 = ρ 2
(1 − e2)2 if ;
e < 1 let X0 = − eρ
1 − e2 , a = ρ
1 − e2 , b = ρ√
1 − e2 .
6. 4. 2 (p. 302) Let h = r2
0 θ ′
0; then ρ = h2
k , e =
[ (ρ
r0
− 1
) 2
+
(
ρr ′
0
h
) 2
] 1/ 2
. If e = 0 , then
θ 0 is undefined, but also irrelevant if e ̸= 0 then φ = θ 0 − α , where −π ≤ α < π , cos α =
1
e
(ρ
r0
− 1
)
and sin α = ρr ′
0
eh .
6. 4. 3 (p. 302) (a) e = γ 2 − γ 1
γ 1 + γ 2
(b) r0 = Rγ 1, r′
0 = 0 , θ 0 arbitrary, θ ′
0 =
[
2gγ 2
Rγ 3
1 (γ 1 + γ 2)
] 1/ 2
6. 4. 4 (p. 302) f (r) = −mh2
(6c
r4 + 1
r3
)
6. 4. 5 (p. 302) f (r) = − mh2(γ 2 + 1)
r3
6. 4. 6 (p. 302) (a) d2u
dθ 2 +
(
1 − k
h2
)
u = 0 , u (θ 0 ) = 1
r0
, du(θ 0)
dθ = − r′
0
h . (b) with γ =",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 733
⏐
⏐
⏐1 − k
h2
⏐
⏐
⏐
1/ 2
: (i) r = r0
(
cosh γ (θ − θ 0) − r0 r′
0
γh sinh γ (θ − θ 0)
) −1
(ii) r = r0
(
1 − r0r′
0
h (θ − θ 0)
) −1
;
(iii) r = r0
(
cos γ (θ − θ 0 ) − r0 r′
0
γh sin γ (θ − θ 0 )
) −1
Section 7.1 Answers, pp. 316– 319
7. 1. 1 (p. 316) (a) R = 2 ; I = ( −1, 3); (b) R = 1 / 2; I = (3 / 2, 5/ 2) (c) R = 0 ; (d) R = 16 ;
I = ( −14, 18) (e) R = ∞ ; I = ( −∞ , ∞ ) (f) R = 4 / 3; I = ( −25/ 3, −17/ 3)
7. 1. 3 (p. 316) (a) R = 1 ; I = (0 , 2) (b) R =
√
2; I = ( −2 −
√
2, −2 +
√
2); (c) R = ∞ ;
I = ( −∞ , ∞ ) (d) R = 0 (e) R =
√
3; I = ( −
√
3,
√
3) (f) R = 1 I = (0 , 2)
7. 1. 5 (p. 316) (a) R = 3 ; I = (0 , 6) (b) R = 1 ; I = ( −1, 1) (c) R = 1 /
√
3
I = (3 − 1/
√
3, 3 + 1 /
√
3) (d) R = ∞ ; I = ( −∞ , ∞ ) (e) R = 0 (f) R = 2 ;
I = ( −1, 3)
7. 1. 11 (p. 317) bn = 2( n + 2)( n + 1) an+2 + ( n + 1) nan+1 + ( n + 3) an
7. 1. 12 (p. 317) b0 = 2 a2 − 2a0 bn = ( n + 2)( n + 1) an+2 + [3 n(n − 1) − 2]an + 3( n − 1)an−1 , n ≥ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"7. 1. 12 (p. 317) b0 = 2 a2 − 2a0 bn = ( n + 2)( n + 1) an+2 + [3 n(n − 1) − 2]an + 3( n − 1)an−1 , n ≥ 1
7. 1. 13 (p. 317) bn = ( n + 2)( n + 1) an+2 + 2( n + 1) an+1 + (2 n2 − 5n + 4) an
7. 1. 14 (p. 317) bn = ( n + 2)( n + 1) an+2 + 2( n + 1) an+1 + ( n2 − 2n + 3) an
7. 1. 15 (p. 318) bn = ( n + 2)( n + 1) an+2 + (3 n2 − 5n + 4) an
7. 1. 16 (p. 318) b0 = −2a2 + 2 a1 + a0 ,
bn = −(n + 2)( n + 1) an+2 + ( n + 1)( n + 2) an+1 + (2 n + 1) an + an−1 , n ≥ 2
7. 1. 17 (p. 318) b0 = 8 a2 + 4 a1 − 6a0 ,
bn = 4( n + 2)( n + 1) an+2 + 4( n + 1) 2an+1 + ( n2 + n − 6)an − 3an−1 , n ≥ 1
7. 1. 21 (p. 319) b0 = ( r + 1)( r + 2) a0 ,
bn = ( n + r + 1)( n + r + 2) an − (n + r − 2)2an−1 , n ≥ 1.
7. 1. 22 (p. 319) b0 = ( r − 2)(r + 2) a0 ,
bn = ( n + r − 2)(n + r + 2) an + ( n + r + 2)( n + r − 3)an−1 , n ≥ 14
7. 1. 23 (p. 319) b0 = ( r − 1)2a0 , b1 = r2a1 + ( r + 2)( r + 3) a0,
bn = ( n + r − 1)2 an + ( n + r + 1)( n + r + 2) an−1 + ( n + r − 1)an−2 , n ≥ 2",Elementary Differential Equations with Boundary Value Problems.pdf
"bn = ( n + r − 1)2 an + ( n + r + 1)( n + r + 2) an−1 + ( n + r − 1)an−2 , n ≥ 2
7. 1. 24 (p. 319) b0 = r(r + 1) a0 , b1 = ( r + 1)( r + 2) a1 + 3( r + 1)( r + 2) a0 ,
bn = ( n + r)(n + r + 1) an + 3( n + r)(n + r + 1) an−1 + ( n + r)an−2 , n ≥ 2
7. 1. 25 (p. 319) b0 = ( r + 2)( r + 1) a0 b1 = ( r + 3)( r + 2) a1 ,
bn = ( n + r + 2)( n + r + 1) an + 2( n + r − 1)(n + r − 3)an−2, n ≥ 2
7. 1. 26 (p. 319) b0 = 2( r + 1)( r + 3) a0 , b1 = 2( r + 2)( r + 4) a1 ,
bn = 2( n + r + 1)( n + r + 3) an + ( n + r − 3)(n + r)an−2 , n ≥ 2
Section 7.2 Answers, pp. 328– 333
7. 2. 1 (p. 328) y = a0
∞∑
m=0
(−1)m(2m + 1) x2m + a1
∞∑
m=0
(−1)m(m + 1) x2m+1
7. 2. 2 (p. 328) y = a0
∞∑
m=0
(−1)m+1 x2m
2m − 1 + a1 x
7. 2. 3 (p. 328) y = a0(1 − 10x2 + 5 x4) + a1
(
x − 2x3 + 1
5 x5
)
7. 2. 4 (p. 328) y = a0
∞∑
m=0
(m + 1)(2 m + 1) x2m + a1
3
∞∑
m=0
(m + 1)(2 m + 3) x2m+1
7. 2. 5 (p. 328) y = a0
∞∑
m=0
(−1)m
[
m−1∏
j=0
4j + 1
2j + 1
]
x2m + a1
∞∑
m=0
(−1)m
[
m−1∏
j=0
(4j + 3)
]
x2m+1
2mm!",Elementary Differential Equations with Boundary Value Problems.pdf
"734 Answers to Selected Exercises
7. 2. 6 (p. 328) y = a0
∞∑
m=0
(−1)m
[ m−1∏
j=0
(4j + 1) 2
2j + 1
]
x2m
8m m! + a1
∞∑
m=0
(−1)m
[ m−1∏
j=0
(4j + 3) 2
2j + 3
]
x2m+1
8m m!
7. 2. 7 (p. 328) y = a0
∞∑
m=0
2mm!∏ m−1
j=0 (2j + 1)
x2m + a1
∞∑
m=0
∏ m−1
j=0 (2j + 3)
2m m! x2m+1
7. 2. 8 (p. 328) y = a0
(
1 − 14x2 + 35
3 x4
)
+ a1
(
x − 3x3 + 3
5 x5 + 1
35 x7
)
7. 2. 9 (p. 329) (a) y = a0
∞∑
m=0
(−1)m x2m
∏ m−1
j=0 (2j + 1)
+ a1
∞∑
m=0
(−1)m x2m+1
2m m!
7. 2. 10 (p. 329) (a) y = a0
∞∑
m=0
(−1)m
[ m−1∏
j=0
4j + 3
2j + 1
]
x2m
2m m! + a1
∞∑
m=0
(−1)m
[ m−1∏
j=0
4j + 5
2j + 3
]
x2m+1
2m m!
7. 2. 11 (p. 329) y = 2 − x − x2 + 1
3 x3 + 5
12 x4 − 1
6 x5 − 17
72 x6 + 13
126 x7 + · · ·
7. 2. 12 (p. 329) y = 1 − x + 3 x2 − 5
2 x3 + 5 x4 − 21
8 x5 + 3 x6 − 11
16 x7 + · · ·
7. 2. 13 (p. 329) y = 2 − x − 2x2 + 1
3 x3 + 3 x4 − 5
6 x5 − 49
5 x6 + 45
14 x7 + · · ·
7. 2. 16 (p. 330) y = a0
∞∑
m=0
(x − 3)2m
(2m)! + a1
∞∑
m=0
(x − 3)2m+1
(2m + 1)!
7. 2. 17 (p. 330) y = a0
∞∑
m=0
(x − 3)2m
2mm! + a1
∞∑
m=0",Elementary Differential Equations with Boundary Value Problems.pdf
"7. 2. 16 (p. 330) y = a0
∞∑
m=0
(x − 3)2m
(2m)! + a1
∞∑
m=0
(x − 3)2m+1
(2m + 1)!
7. 2. 17 (p. 330) y = a0
∞∑
m=0
(x − 3)2m
2mm! + a1
∞∑
m=0
(x − 3)2m+1
∏ m−1
j=0 (2j + 3)
7. 2. 18 (p. 330) y = a0
∞∑
m=0
[ m−1∏
j=0
(2j + 3)
]
(x − 1)2m
m! + a1
∞∑
m=0
4m(m + 1)!∏ m−1
j=0 (2j + 3)
(x − 1)2m+1
7. 2. 19 (p. 330) y = a0
(
1 − 6(x − 2)2 + 4
3 (x − 2)4 + 8
135 (x − 2)6
)
+ a1
(
(x − 2) − 10
9 (x − 2)3
)
7. 2. 20 (p. 330) y = a0
∞∑
m=0
(−1)m
[ m−1∏
j=0
(2j + 1)
]
3m
4mm! (x + 1) 2m + a1
∞∑
m=0
(−1)m 3m m!∏ m−1
j=0 (2j + 3)
(x + 1) 2m+1
7. 2. 21 (p. 330) y = −1 + 2 x + 3
8 x2 − 1
3 x3 − 3
128 x4 − 1
1024 x6 + · · ·
7. 2. 22 (p. 330) y = −2 + 3( x − 3) + 3( x − 3)2 − 2(x − 3)3 − 5
4 (x − 3)4 + 3
5 (x − 3)5 + 7
24 (x − 3)6 − 4
35 (x − 3)7 + · · ·
7. 2. 23 (p. 330) y = −1 + ( x − 1) + 3( x − 1)2 − 5
2 (x − 1)3 − 27
4 (x − 1)4 + 21
4 (x − 1)5 + 27
2 (x − 1)6 − 81
8 (x − 1)7 + · · ·
7. 2. 24 (p. 330) y = 4 − 6(x − 3) − 2(x − 3)2 + ( x − 3)3 + 3
2 (x − 3)4 − 5
4 (x − 3)5 − 49
20 (x − 3)6 + 135",Elementary Differential Equations with Boundary Value Problems.pdf
"2 (x − 1)6 − 81
8 (x − 1)7 + · · ·
7. 2. 24 (p. 330) y = 4 − 6(x − 3) − 2(x − 3)2 + ( x − 3)3 + 3
2 (x − 3)4 − 5
4 (x − 3)5 − 49
20 (x − 3)6 + 135
56 (x − 3)7 + · · ·
7. 2. 25 (p. 330) y = 3 − 4(x − 4) + 15( x − 4)2 − 4(x − 4)3 + 15
4 (x − 4)4 − 1
5 (x − 4)5
7. 2. 26 (p. 330) y = 3 − 3(x + 1) − 30(x + 1) 2 + 20
3 (x + 1) 3 + 20( x + 1) 4 − 4
3 (x + 1) 5 − 8
9 (x + 1) 6
7. 2. 27 (p. 330) (a) y = a0
∞∑
m=0
(−1)m x2m + a1
∞∑
m=0
(−1)mx2m+1 (b)y = a0 + a1x
1 + x2
7. 2. 33 (p. 333) y = a0
∞∑
m=0
x3m
3mm!
∏ m−1
j=0 (3j + 2)
+ a1
∞∑
m=0
x3m+1
3m m!
∏ m−1
j=0 (3j + 4)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 735
7. 2. 34 (p. 333) y = a0
∞∑
m=0
(2
3
) m
[ m−1∏
j=0
(3j + 2)
]
x3m
m! + a1
∞∑
m=0
6m m!∏ m−1
j=0 (3j + 4)
x3m+1
7. 2. 35 (p. 333) y = a0
∞∑
m=0
(−1)m 3mm!∏ m−1
j=0 (3j + 2)
x3m + a1
∞∑
m=0
(−1)m
[ m−1∏
j=0
(3j + 4)
]
x3m+1
3mm!
7. 2. 36 (p. 333) y = a0 (1 − 4x3 + 4 x6) + a1
∞∑
m=0
2m
[
m−1∏
j=0
3j − 5
3j + 4
]
x3m+1
7. 2. 37 (p. 333) y = a0
(
1 + 21
2 x3 + 42
5 x6 + 7
20 x9
)
+ a1
(
x + 4 x4 + 10
7 x7
)
7. 2. 39 (p. 333) y = a0
∞∑
m=0
(−2)m
[ m−1∏
j=0
5j + 1
5j + 4
]
x5m + a1
∞∑
m=0
(
− 2
5
) m
[ m−1∏
j=0
(5j + 2)
]
x5m+1
m!
7. 2. 40 (p. 333) y = a0
∞∑
m=0
(−1)m x4m
4mm!
∏ m−1
j=0 (4j + 3)
+ a1
∞∑
m=0
(−1)m x4m+1
4m m!
∏ m−1
j=0 (4j + 5)
7. 2. 41 (p. 333) y = a0
∞∑
m=0
(−1)m x7m
∏ m−1
j=0 (7j + 6)
+ a1
∞∑
m=0
(−1)m x7m+1
7mm!
7. 2. 42 (p. 333) y = a0
(
1 − 9
7 x8
)
+ a1
(
x − 7
9 x9
)
7. 2. 43 (p. 333) y = a0
∞∑
m=0
x6m + a1
∞∑
m=0
x6m+1
7. 2. 44 (p. 333) y = a0
∞∑
m=0
(−1)m x6m
∏ m−1
j=0 (6j + 5)
+ a1
∞∑
m=0
(−1)m x6m+1
6mm!",Elementary Differential Equations with Boundary Value Problems.pdf
")
7. 2. 43 (p. 333) y = a0
∞∑
m=0
x6m + a1
∞∑
m=0
x6m+1
7. 2. 44 (p. 333) y = a0
∞∑
m=0
(−1)m x6m
∏ m−1
j=0 (6j + 5)
+ a1
∞∑
m=0
(−1)m x6m+1
6mm!
Section 7.3 Answers, pp. 337– 341
7. 3. 1 (p. 337) y = 2 − 3x − 2x2 + 7
2 x3 − 55
12 x4 + 59
8 x5 − 83
6 x6 + 9547
336 x7 + · · ·
7. 3. 2 (p. 337) y = −1 + 2 x − 4x3 + 4 x4 + 4 x5 − 12x6 + 4 x7 + · · ·
7. 3. 3 (p. 337) y = 1 + x2 − 2
3 x3 + 11
6 x4 − 9
5 x5 + 329
90 x6 − 1301
315 x7 + · · ·
7. 3. 4 (p. 337) y = x − x2 − 7
2 x3 + 15
2 x4 + 45
8 x5 − 261
8 x6 + 207
16 x7 + · · ·
7. 3. 5 (p. 337) y = 4 + 3 x − 15
4 x2 + 1
4 x3 + 11
16 x4 − 5
16 x5 + 1
20 x6 + 1
120 x7 + · · ·
7. 3. 6 (p. 337) y = 7 + 3 x − 16
3 x2 + 13
3 x3 − 23
9 x4 + 10
9 x5 − 7
27 x6 − 1
9 x7 + · · ·
7. 3. 7 (p. 337) y = 2 + 5 x − 7
4 x2 − 3
16 x3 + 37
192 x4 − 7
192 x5 − 1
1920 x6 + 19
11520 x7 + · · ·
7. 3. 8 (p. 337) y = 1 − (x − 1) + 4
3 (x − 1)3 − 4
3 (x − 1)4 − 4
5 (x − 1)5 + 136
45 (x − 1)6 − 104
63 (x − 1)7 + · · ·
7. 3. 9 (p. 337) y = 1 − (x + 1) + 4( x + 1) 2 − 13",Elementary Differential Equations with Boundary Value Problems.pdf
"3 (x − 1)3 − 4
3 (x − 1)4 − 4
5 (x − 1)5 + 136
45 (x − 1)6 − 104
63 (x − 1)7 + · · ·
7. 3. 9 (p. 337) y = 1 − (x + 1) + 4( x + 1) 2 − 13
3 (x + 1) 3 + 77
6 (x + 1) 4 − 278
15 (x + 1) 5 + 1942
45 (x + 1) 6 − 23332
315 (x + 1) 7 + · · ·
7. 3. 10 (p. 337) y = 2 − (x − 1) − 1
2 (x − 1)2 + 5
3 (x − 1)3 − 19
12 (x − 1)4 + 7
30 (x − 1)5 + 59
45 (x − 1)6 − 1091
630 (x − 1)7 + · · ·
7. 3. 11 (p. 337) y = −2 + 3( x + 1) − 1
2 (x + 1) 2 − 2
3 (x + 1) 3 + 5
8 (x + 1) 4 − 11
30 (x + 1) 5 + 29
144 (x + 1) 6 − 101
840 (x + 1) 7 + · · ·
7. 3. 12 (p. 337) y = 1 − 2(x − 1) − 3(x − 1)2 + 8( x − 1)3 − 4(x − 1)4 − 42
5 (x − 1)5 + 19( x − 1)6 − 604
35 (x − 1)7 + · · ·
7. 3. 19 (p. 339) y = 2 − 7x − 4x2 − 17
6 x3 − 3
4 x4 − 9
40 x5 + · · ·",Elementary Differential Equations with Boundary Value Problems.pdf
"736 Answers to Selected Exercises
7. 3. 20 (p. 339) y = 1 − 2(x − 1) + 1
2 (x − 1)2 − 1
6 (x − 1)3 + 5
36 (x − 1)4 − 73
1080 (x − 1)5 + · · ·
7. 3. 21 (p. 339) y = 2 − (x + 2) − 7
2 (x + 2) 2 + 4
3 (x + 2) 3 − 1
24 (x + 2) 4 + 1
60 (x + 2) 5 + · · ·
7. 3. 22 (p. 339) y = 2 − 2(x + 3) − (x + 3) 2 + ( x + 3) 3 − 11
12 (x + 3) 4 + 67
60 (x + 3) 5 + · · ·
7. 3. 23 (p. 339) y = −1 + 2 x + 1
3 x3 − 5
12 x4 + 2
5 x5 + · · ·
7. 3. 24 (p. 339) y = 2 − 3(x + 1) + 7
2 (x + 1) 2 − 5(x + 1) 3 + 197
24 (x + 1) 4 − 287
20 (x + 1) 5 + · · ·
7. 3. 25 (p. 339) y = −2 + 3( x + 2) − 9
2 (x + 2) 2 + 11
6 (x + 2) 3 + 5
24 (x + 2) 4 + 7
20 (x + 2) 5 + · · ·
7. 3. 26 (p. 339) y = 2 − 4(x − 2) − 1
2 (x − 2)2 + 2
9 (x − 2)3 + 49
432 (x − 2)4 + 23
1080 (x − 2)5 + · · ·
7. 3. 27 (p. 339) y = 1 + 2( x + 4) − 1
6 (x + 4) 2 − 10
27 (x + 4) 3 + 19
648 (x + 4) 4 + 13
324 (x + 4) 5 + · · ·
7. 3. 28 (p. 339) y = −1 + 2( x + 1) − 1
4 (x + 1) 2 + 1
2 (x + 1) 3 − 65
96 (x + 1) 4 + 67
80 (x + 1) 5 + · · ·",Elementary Differential Equations with Boundary Value Problems.pdf
"324 (x + 4) 5 + · · ·
7. 3. 28 (p. 339) y = −1 + 2( x + 1) − 1
4 (x + 1) 2 + 1
2 (x + 1) 3 − 65
96 (x + 1) 4 + 67
80 (x + 1) 5 + · · ·
7. 3. 31 (p. 341) (a) y = c1
1 + x + c2
1 + 2 x (b) y = c1
1 − 2x + c2
1 − 3x (c) y = c1
1 − 2x + c2x
(1 − 2x)2
(d) y = c1
2 + x + c2x
(2 + x)2 (e) y = c1
2 + x + c2
2 + 3 x
7. 3. 32 (p. 341) y = 1 − 2x − 3
2 x2 + 5
3 x3 + 17
24 x4 − 11
20 x5 + · · ·
7. 3. 33 (p. 341) y = 1 − 2x − 5
2 x2 + 2
3 x3 − 3
8 x4 + 1
3 x5 + · · ·
7. 3. 34 (p. 341) y = 6 − 2x + 9 x2 + 2
3 x3 − 23
4 x4 − 3
10 x5 + · · ·
7. 3. 35 (p. 341) y = 2 − 5x + 2 x2 − 10
3 x3 + 3
2 x4 − 25
12 x5 + · · ·
7. 3. 36 (p. 341) y = 3 + 6 x − 3x2 + x3 − 2x4 − 17
20 x5 + · · ·
7. 3. 37 (p. 341) y = 3 − 2x − 3x2 + 3
2 x3 + 3
2 x4 − 49
80 x5 + · · ·
7. 3. 38 (p. 341) y = −2 + 3 x + 4
3 x2 − x3 − 19
54 x4 + 13
60 x5 + · · ·
7. 3. 39 (p. 341) y1 =
∞∑
m=0
(−1)mx2m
m! = e−x2
, y 2 =
∞∑
m=0
(−1)mx2m+1
m! = xe−x2
7. 3. 40 (p. 341) y = −2 + 3 x + x2 − 1
6 x3 − 3
4 x4 + 31
120 x5 + · · ·",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
m=0
(−1)mx2m
m! = e−x2
, y 2 =
∞∑
m=0
(−1)mx2m+1
m! = xe−x2
7. 3. 40 (p. 341) y = −2 + 3 x + x2 − 1
6 x3 − 3
4 x4 + 31
120 x5 + · · ·
7. 3. 41 (p. 341) y = 2 + 3 x − 7
2 x2 − 5
6 x3 + 41
24 x4 + 41
120 x5 + · · ·
7. 3. 42 (p. 341) y = −3 + 5 x − 5x2 + 23
6 x3 − 23
12 x4 + 11
30 x5 + · · ·
7. 3. 43 (p. 341) y = −2 + 3( x − 1) + 3
2 (x − 1)2 − 17
12 (x − 1)3 − 1
12 (x − 1)4 + 1
8 (x − 1)5 + · · ·
7. 3. 44 (p. 341) y = 2 − 3(x + 2) + 1
2 (x + 2) 2 − 1
3 (x + 2) 3 + 31
24 (x + 2) 4 − 53
120 (x + 2) 5 + · · ·
7. 3. 45 (p. 341) y = 1 − 2x + 3
2 x2 − 11
6 x3 + 15
8 x4 − 71
60 x5 + · · ·
7. 3. 46 (p. 341) y = 2 − (x + 2) − 7
2 (x + 2) 2 − 43
6 (x + 2) 3 − 203
24 (x + 2) 4 − 167
30 (x + 2) 5 + · · ·
7. 3. 47 (p. 341) y = 2 − x − x2 + 7
6 x3 − x4 + 89
120 x5 + · · ·",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 737
7. 3. 48 (p. 341) y = 1 + 3
2 (x − 1)2 + 1
6 (x − 1)3 − 1
8 (x − 1)5 + · · ·
7. 3. 49 (p. 341) y = 1 − 2(x − 3) + 1
2 (x − 3)2 − 1
6 (x − 3)3 + 1
4 (x − 3)4 − 1
6 (x − 3)5 + · · ·
Section 7.4 Answers, pp. 346– 347
7. 4. 1 (p. 346) y = c1x−4 + c2x−2 7. 4. 2 (p. 346) y = c1x + c2x7
7. 4. 3 (p. 346) y = x(c1 + c2 ln x) 7. 4. 4 (p. 346) y = x−2(c1 + c2 ln x)
7. 4. 5 (p. 346) y = c1 cos(ln x) + c2 sin(ln x) 7. 4. 6 (p. 346) y = x2[c1 cos(3 ln x) + c2 sin(3 ln x)]
7. 4. 7 (p. 346) y = c1x + c2
x3 7. 4. 8 (p. 346) y = c1x2/ 3 + c2x3/ 4 7. 4. 9 (p. 346) y = x−1/ 2(c1 + c2 ln x)
7. 4. 10 (p. 346) y = c1x+c2x1/ 3 7. 4. 11 (p. 346) y = c1x2+c2x1/ 2 7. 4. 12 (p. 346) y = 1
x [c1 cos(2 ln x) + c2 sin(2 ln x]
7. 4. 13 (p. 346) y = x−1/ 3(c1 + c2 ln x) 7. 4. 14 (p. 346) y = x [c1 cos(3 ln x) + c2 sin(3 ln x)]
7. 4. 15 (p. 346) y = c1x3 + c2
x2 7. 4. 16 (p. 346) y = c1
x + c2x1/ 2 7. 4. 17 (p. 346) y = x2(c1 + c2 ln x)
7. 4. 18 (p. 346) y = 1
x2
[
c1 cos
(
1√
2
ln x
)",Elementary Differential Equations with Boundary Value Problems.pdf
"x2 7. 4. 16 (p. 346) y = c1
x + c2x1/ 2 7. 4. 17 (p. 346) y = x2(c1 + c2 ln x)
7. 4. 18 (p. 346) y = 1
x2
[
c1 cos
(
1√
2
ln x
)
+ c2 sin
(
1√
2
ln x
)]
Section 7.5 Answers, pp. 357– 364
7. 5. 1 (p. 357) y1 = x1/ 2
(
1 − 1
5 x − 2
35 x2 + 31
315 x3 + · · ·
)
y2 = x−1
(
1 + x + 1
2 x2 − 1
6 x3 + · · ·
)
;
7. 5. 2 (p. 357) y1 = x1/ 3
(
1 − 2
3 x + 8
9 x2 − 40
81 x3 + · · ·
)
; y2 = 1 − x + 6
5 x2 − 4
5 x3 + · · ·
7. 5. 3 (p. 357) y1 = x1/ 3
(
1 − 4
7 x − 7
45 x2 + 970
2457 x3 + · · ·
)
; y2 = x−1
(
1 − x2 + 2
3 x3 + · · ·
)
7. 5. 4 (p. 357) y1 = x1/ 4
(
1 − 1
2 x − 19
104 x2 + 1571
10608 x3 + · · ·
)
; y2 = x−1
(
1 + 2 x − 11
6 x2 − 1
7 x3 + · · ·
)
7. 5. 5 (p. 357) y1 = x1/ 3
(
1 − x + 28
31 x2 − 1111
1333 x3 + · · ·
)
; y2 = x−1/ 4
(
1 − x + 7
8 x2 − 19
24 x3 + · · ·
)
;
7. 5. 6 (p. 357) y1 = x1/ 5
(
1 − 6
25 x − 1217
625 x2 + 41972
46875 x3 + · · ·
)
; y2 = x − 1
4 x2 − 35
18 x3 + 11
12 x4 + · · ·
7. 5. 7 (p. 357) y1 = x3/ 2
(
1 − x + 11
26 x2 − 109
1326 x3 + · · ·
)
; y2 = x1/ 4
(",Elementary Differential Equations with Boundary Value Problems.pdf
")
; y2 = x − 1
4 x2 − 35
18 x3 + 11
12 x4 + · · ·
7. 5. 7 (p. 357) y1 = x3/ 2
(
1 − x + 11
26 x2 − 109
1326 x3 + · · ·
)
; y2 = x1/ 4
(
1 + 4 x − 131
24 x2 + 39
14 x3 + · · ·
)
7. 5. 8 (p. 357) y1 = x1/ 3
(
1 − 1
3 x + 2
15 x2 − 5
63 x3 + · · ·
)
; y2 = x−1/ 6
(
1 − 1
12 x2 + 1
18 x3 + · · ·
)
7. 5. 9 (p. 357) y1 = 1 − 1
14 x2 + 1
105 x3 + · · ·; y2 = x−1/ 3
(
1 − 1
18 x − 71
405 x2 + 719
34992 x3 + · · ·
)
7. 5. 10 (p. 358) y1 = x1/ 5
(
1 + 3
17 x − 7
153 x2 − 547
5661 x3 + · · ·
)
; y2 = x−1/ 2
(
1 + x + 14
13 x2 − 556
897 x3 + · · ·
)
7. 5. 14 (p. 358) y1 = x1/ 2
∞∑
n=0
(−2)n
∏ n
j=1(2j + 3) xn; y2 = x−1
∞∑
n=0
(−1)n
n! xn
7. 5. 15 (p. 358) y1 = x1/ 3
∞∑
n=0
(−1)n ∏ n
j=1(3j + 1)
9nn! xn; x−1
7. 5. 16 (p. 358) y1 = x1/ 2
∞∑
n=0
(−1)n
2nn! xn; y2 = 1
x2
∞∑
n=0
(−1)n
∏ n
j=1(2j − 5) xn
7. 5. 17 (p.
358) y1 = x
∞∑
n=0
(−1)n
∏ n
j=1(3j + 4) xn; y2 = x−1/ 3
∞∑
n=0
(−1)n
3n n! xn
7. 5. 18 (p. 358) y1 = x
∞∑
n=0
2n
n!
∏ n
j=1(2j + 1) xn; y2 = x1/ 2
∞∑
n=0
2n
n!
∏ n
j=1(2j − 1) xn",Elementary Differential Equations with Boundary Value Problems.pdf
"738 Answers to Selected Exercises
7. 5. 19 (p. 358) y1 = x1/ 3
∞∑
n=0
1
n!
∏ n
j=1(3j + 2) xn; y2 = x−1/ 3
∞∑
n=0
1
n!
∏ n
j=1(3j − 2) xn
7. 5. 20 (p.
358) y1 = x
(
1 + 2
7 x + 1
70 x2
)
; y2 = x−1/ 3
∞∑
n=0
(−1)n
3n n!
( n∏
j=1
3j − 13
3j − 4
)
xn
7. 5. 21 (p. 358) y1 = x1/ 2
∞∑
n=0
(−1)n
( n∏
j=1
2j + 1
6j + 1
)
; xn y2 = x1/ 3
∞∑
n=0
(−1)n
9n n!
( n∏
j=1
(3j + 1)
)
xn
7. 5. 22 (p.
358) y1 = x
∞∑
n=0
(−1)n(n + 2)!
2
∏ n
j=1(4j + 3) ; xn y2 = x1/ 4
∞∑
n=0
(−1)n
16n n!
n∏
j=1
(4j + 5) xn
7. 5. 23 (p.
358) y1 = x−1/ 2
∞∑
n=0
(−1)n
n!
∏ n
j=1(2j + 1) xn; y2 = x−1
∞∑
n=0
(−1)n
n!
∏ n
j=1(2j − 1) xn
7. 5. 24 (p.
358) y1 = x1/ 3
∞∑
n=0
(−1)n
n!
(2
9
) n
( n∏
j=1
(6j + 5)
)
xn; y2 = x−1
∞∑
n=0
(−1)n2n
(n∏
j=1
2j − 1
3j − 4
)
xn
7. 5. 25 (p. 358) y1 = 4 x1/ 3
∞∑
n=0
1
6nn!(3n + 4) xn; x−1
7. 5. 28 (p. 359) y1 = x1/ 2
(
1 − 9
40 x + 5
128 x2 − 245
39936 x3 + · · ·
)
; y2 = x1/ 4
(
1 − 25
96 x + 675
14336 x2 − 38025
5046272 x3 + · · ·
)
7. 5. 29 (p. 359) y1 = x1/ 3
(
1 + 32
117 x − 28",Elementary Differential Equations with Boundary Value Problems.pdf
"39936 x3 + · · ·
)
; y2 = x1/ 4
(
1 − 25
96 x + 675
14336 x2 − 38025
5046272 x3 + · · ·
)
7. 5. 29 (p. 359) y1 = x1/ 3
(
1 + 32
117 x − 28
1053 x2 + 4480
540189 x3 + · · ·
)
; y2 = x−3
(
1 + 32
7 x + 48
7 x2
)
7. 5. 30 (p. 359) y1 = x1/ 2
(
1 − 5
8 x + 55
96 x2 − 935
1536 x3 + · · ·
)
; y2 = x−1/ 2
(
1 + 1
4 x − 5
32 x2 − 55
384 x3 + · · ·
)
.
7. 5. 31 (p. 359) y1 = x1/ 2
(
1 − 3
4 x + 5
96 x2 + 5
4224 x3 + · · ·
)
; y2 = x−2 (
1 + 8 x + 60 x2 − 160x3 + · · ·
)
7. 5. 32 (p. 359) y1 = x−1/ 3
(
1 − 10
63 x + 200
7371 x2 − 17600
3781323 x3; + · · ·
)
; y2 = x−1/ 2
(
1 − 3
20 x + 9
352 x2 − 105
23936 x3 + · · ·
)
7. 5. 33 (p. 359) y1 = x1/ 2
∞∑
m=0
(−1)m
8m m!
(m∏
j=1
4j − 3
8j + 1
)
x2m; y2 = x1/ 4
∞∑
m=0
(−1)m
16m m!
(m∏
j=1
8j − 7
8j − 1
)
x2m
7. 5. 34 (p. 359) y1 = x1/ 2
∞∑
m=0
(m∏
j=1
8j − 3
8j + 1
)
x2m ; y2 = x1/ 4
∞∑
m=0
1
2mm!
(m∏
j=1
(2j − 1)
)
x2m
7. 5. 35 (p.
359) y1 = x4
∞∑
m=0
(−1)m(m + 1) x2m ; y2 = −x
∞∑
m=0
(−1)m (2m − 1)x2m
7. 5. 36 (p. 359) y1 = x1/ 3
∞∑
m=0
(−1)m
18mm!",Elementary Differential Equations with Boundary Value Problems.pdf
"(2j − 1)
)
x2m
7. 5. 35 (p.
359) y1 = x4
∞∑
m=0
(−1)m(m + 1) x2m ; y2 = −x
∞∑
m=0
(−1)m (2m − 1)x2m
7. 5. 36 (p. 359) y1 = x1/ 3
∞∑
m=0
(−1)m
18mm!
(m∏
j=1
(6j − 17)
)
x2m; y2 = 1 + 4
5 x2 + 8
55 x4
7. 5. 37 (p. 359) y1 = x1/ 4
∞∑
m=0
(m∏
j=1
8j + 1
8j + 5
)
x2m ; y2 = x−1
∞∑
m=0
∏ m
j=1(2j − 1)
2m m! x2m
7. 5. 38 (p. 359) y1 = x1/ 2
∞∑
m=0
1
8mm!
(m∏
j=1
(4j − 1)
)
x2m ; y2 = x1/ 3
∞∑
m=0
2m
(m∏
j=1
3j − 1
12j − 1
)
x2m
7. 5. 39 (p. 359) y1 = x7/ 2
∞∑
m=0
(−1)m
∏ m
j=1(4j + 5)
8m m! x2m ; y2 = x1/ 2
∞∑
m=0
(−1)m
4m
(
m∏
j=1
4j − 1
2j − 3
)
x2m",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 739
7. 5. 40 (p. 359) y1 = x1/ 2
∞∑
m=0
(−1)m
4m
(m∏
j=1
4j − 1
2j + 1
)
x2m; y2 = x−1/ 2
∞∑
m=0
(−1)m
8mm!
(m∏
j=1
(4j − 3)
)
x2m
7. 5. 41 (p.
359) y1 = x1/ 2
∞∑
m=0
(−1)m
m!
(m∏
j=1
(2j + 1)
)
x2m ; y2 = 1
x2
∞∑
m=0
(−2)m
(m∏
j=1
4j − 3
4j − 5
)
x2m
7. 5. 42 (p. 359) y1 = x1/ 3
∞∑
m=0
(−1)m
(
m∏
j=1
3j − 4
3j + 2
)
x2m; y2 = x−1(1 + x2)
7. 5. 43 (p. 359) y1 =
∞∑
m=0
(−1)m 2m(m + 1)!∏ m
j=1(2j + 3) x2m; y2 = 1
x3
∞∑
m=0
(−1)m
∏ m
j=1(2j − 1)
2m m! x2m
7. 5. 44 (p. 359) y1 = x1/ 2
∞∑
m=0
(−1)m
8m m!
(
m∏
j=1
(4j − 3)2
4j + 3
)
x2m; y2 = x−1
∞∑
m=0
(−1)m
2mm!
(
m∏
j=1
(2j − 3)2
4j − 3
)
x2m
7. 5. 45 (p. 359) y1 = x
∞∑
m=0
(−2)m
(m∏
j=1
2j + 1
4j + 5
)
x2m; y2 = x−3/ 2
∞∑
m=0
(−1)m
4m m!
(m∏
j=1
(4j − 3)
)
x2m
7. 5. 46 (p.
359) y1 = x1/ 3
∞∑
m=0
(−1)m
2m
∏ m
j=1(3j + 1) x2m; y2 = x−1/ 3
∞∑
m=0
(−1)m
6m m! x2m
7. 5. 47 (p. 359) y1 = x1/ 2
(
1 − 6
13 x2 + 36
325 x4 − 216
12025 x6 + · · ·
)
; y2 = x1/ 3
(
1 − 1
2 x2 + 1
8 x4 − 1
48 x6 + · · ·
)",Elementary Differential Equations with Boundary Value Problems.pdf
"6m m! x2m
7. 5. 47 (p. 359) y1 = x1/ 2
(
1 − 6
13 x2 + 36
325 x4 − 216
12025 x6 + · · ·
)
; y2 = x1/ 3
(
1 − 1
2 x2 + 1
8 x4 − 1
48 x6 + · · ·
)
7. 5. 48 (p. 359) y1 = x1/ 4
(
1 − 13
64 x2 + 273
8192 x4 − 2639
524288 x6 + · · ·
)
; y2 = x−1
(
1 − 1
3 x2 + 2
33 x4 − 2
209 x6 + · · ·
)
7. 5. 49 (p. 359) y1 = x1/ 3
(
1 − 3
4 x2 + 9
14 x4 − 81
140 x6 + · · ·
)
; y2 = x−1/ 3
(
1 − 2
3 x2 + 5
9 x4 − 40
81 x6 + · · ·
)
7. 5. 50 (p. 359) y1 = x1/ 2
(
1 − 3
2 x2 + 15
8 x4 − 35
16 x6 + · · ·
)
; y2 = x−1/ 2
(
1 − 2x2 + 8
3 x4 − 16
5 x6 + · · ·
)
7. 5. 51 (p. 359) y1 = x1/ 4
(
1 − x2 + 3
2 x4 − 5
2 x6 + · · ·
)
; y2 = x−1/ 2
(
1 − 2
5 x2 + 36
65 x4 − 408
455 x6 + · · ·
)
7. 5. 53 (p. 360) (a) y1 = xν
∞∑
m=0
(−1)m
4m m!
∏ m
j=1(j + ν ) x2m; y2 = x−ν
∞∑
m=0
(−1)m
4mm!
∏ m
j=1(j − ν ) x2m
y1 = sin x
√x ; y2 = cos x√x
7. 5. 61 (p. 363) y1 = x1/ 2
1 + x ; y2 = x
1 + x 7. 5. 62 (p. 364) y1 = x1/ 3
1 + 2 x2 ; y2 = x1/ 2
1 + 2 x2
7. 5. 63 (p. 364) y1 = x1/ 4
1 − 3x ; y2 = x2",Elementary Differential Equations with Boundary Value Problems.pdf
"1 + x ; y2 = x
1 + x 7. 5. 62 (p. 364) y1 = x1/ 3
1 + 2 x2 ; y2 = x1/ 2
1 + 2 x2
7. 5. 63 (p. 364) y1 = x1/ 4
1 − 3x ; y2 = x2
1 − 3x 7. 5. 64 (p. 364) y1 = x1/ 3
5 + x ; y2 = x−1/ 3
5 + x
7. 5. 65 (p. 364) y1 = x1/ 4
2 − x2 ; y2 = x−1/ 2
2 − x2 7. 5. 66 (p. 364) y1 = x1/ 2
1 + 3 x + x2 ; y2 = x3/ 2
1 + 3 x + x2
7. 5. 67 (p. 364) y1 = x
(1 + x)2 ; y2 = x1/ 3
(1 + x)2 7. 5. 68 (p. 364) y1 = x
3 + 2 x + x2 ; y2 = x1/ 4
3 + 2 x + x2
Section 7.6 Answers, pp. 373– 378
7. 6. 1 (p. 373) y1 = x
(
1 − x + 3
4 x2 − 13
36 x3 + · · ·
)
; y2 = y1 ln x + x2
(
1 − x + 65
108 x2 + · · ·
)
7. 6. 2 (p. 373) y1 = x−1
(
1 − 2x + 9
2 x2 − 20
3 x3 + · · ·
)
; y2 = y1 ln x + 1 − 15
4 x + 133
18 x2 + · · ·
7. 6. 3 (p. 373) y1 = 1 + x − x2 + 1
3 x3 + · · ·; y2 = y1 ln x − x
(
3 − 1
2 x − 31
18 x2 + · · ·
)",Elementary Differential Equations with Boundary Value Problems.pdf
"740 Answers to Selected Exercises
7. 6. 4 (p. 373) y1 = x1/ 2
(
1 − 2x + 5
2 x2 − 2x3 + · · ·
)
; y2 = y1 ln x + x3/ 2
(
1 − 9
4 x + 17
6 x2 + · · ·
)
7. 6. 5 (p. 373) y1 = x
(
1 − 4x + 19
2 x2 − 49
3 x3 + · · ·
)
; y2 = y1 ln x + x2
(
3 − 43
4 x + 208
9 x2 + · · ·
)
7. 6. 6 (p. 373) y1 = x−1/ 3
(
1 − x + 5
6 x2 − 1
2 x3 + · · ·
)
; y2 = y1 ln x + x2/ 3
(
1 − 11
12 x + 25
36 x2 + · · ·
)
7. 6. 7 (p. 373) y1 = 1 − 2x + 7
4 x2 − 7
9 x3 + · · ·; y2 = y1 ln x + x
(
3 − 15
4 x + 239
108 x2 + · · ·
)
7. 6. 8 (p. 373) y1 = x−2
(
1 − 2x + 5
2 x2 − 3x3 + · · ·
)
; y2 = y1 ln x + 3
4 − 13
6 x + · · ·
7. 6. 9 (p. 373) y1 = x−1/ 2
(
1 − x + 1
4 x2 + 1
18 x3 + · · ·
)
; y2 = y1 ln x + x1/ 2
(3
2 − 13
16 x + 1
54 x2 + · · ·
)
7. 6. 10 (p. 373) y1 = x−1/ 4
(
1 − 1
4 x − 7
32 x2 + 23
384 x3 + · · ·
)
; y2 = y1 ln x + x3/ 4
(1
4 + 5
64 x − 157
2304 x2 + · · ·
)
7. 6. 11 (p. 373) y1 = x−1/ 3
(
1 − x + 7
6 x2 − 23
18 x3 + · · ·
)
; y2 = y1 ln x − x5/ 3
(1
12 − 13
108 x · · ·
)",Elementary Differential Equations with Boundary Value Problems.pdf
"64 x − 157
2304 x2 + · · ·
)
7. 6. 11 (p. 373) y1 = x−1/ 3
(
1 − x + 7
6 x2 − 23
18 x3 + · · ·
)
; y2 = y1 ln x − x5/ 3
(1
12 − 13
108 x · · ·
)
7. 6. 12 (p. 373) y1 = x1/ 2
∞∑
n=0
(−1)n
(n!)2 xn; y2 = y1 ln x − 2x1/ 2
∞∑
n=1
(−1)n
(n!)2
( n∑
j=1
1
j
)
xn;
7. 6. 13 (p. 374) y1 = x1/ 6
∞∑
n=0
(2
3
) n
∏ n
j=1(3j + 1)
n! xn;
y2 = y1 ln x − x1/ 6
∞∑
n=1
(2
3
) n
∏ n
j=1(3j + 1)
n!
( n∑
j=1
1
j(3j + 1)
)
xn
7. 6. 14 (p. 374) y1 = x2
∞∑
n=0
(−1)n (n + 1) 2xn; y2 = y1 ln x − 2x2
∞∑
n=1
(−1)nn(n + 1) xn
7. 6. 15 (p. 374) y1 = x3
∞∑
n=0
2n(n + 1) xn; y2 = y1 ln x − x3
∞∑
n=1
2n nxn
7. 6. 16 (p. 374) y1 = x1/ 5
∞∑
n=0
(−1)n ∏ n
j=1(5j + 1)
125n (n!)2 xn;
y2 = y1 ln x − x1/ 5
∞∑
n=1
(−1)n ∏ n
j=1(5j + 1)
125n (n!)2
( n∑
j=1
5j + 2
j(5j + 1)
)
xn
7. 6. 17 (p. 374) y1 = x1/ 2
∞∑
n=0
(−1)n ∏ n
j=1(2j − 3)
4nn! xn;
y2 = y1 ln x + 3 x1/ 2
∞∑
n=1
(−1)n ∏ n
j=1(2j − 3)
4nn!
( n∑
j=1
1
j(2j − 3)
)
xn
7. 6. 18 (p. 374) y1 = x1/ 3
∞∑
n=0
(−1)n ∏ n
j=1(6j − 7)2
81n (n!)2 xn;
y2 = y1 ln x + 14 x1/ 3
∞∑
n=1",Elementary Differential Equations with Boundary Value Problems.pdf
"j=1(2j − 3)
4nn!
( n∑
j=1
1
j(2j − 3)
)
xn
7. 6. 18 (p. 374) y1 = x1/ 3
∞∑
n=0
(−1)n ∏ n
j=1(6j − 7)2
81n (n!)2 xn;
y2 = y1 ln x + 14 x1/ 3
∞∑
n=1
(−1)n ∏ n
j=1(6j − 7)2
81n (n!)2
( n∑
j=1
1
j(6j − 7) )
)
xn
7. 6. 19 (p. 374) y1 = x2
∞∑
n=0
(−1)n ∏ n
j=1(2j + 5)
(n!)2 xn;
y2 = y1 ln x − 2x2
∞∑
n=1
(−1)n ∏ n
j=1(2j + 5)
(n!)2
( n∑
j=1
(j + 5)
j(2j + 5)
)
xn",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 741
7. 6. 20 (p. 374) y1 = 1
x
∞∑
n=0
2n ∏ n
j=1(2j − 1)
n! xn;
y2 = y1 ln x + 1
x
∞∑
n=1
2n ∏ n
j=1(2j − 1)
n!
(
n∑
j=1
1
j(2j − 1)
)
xn
7. 6. 21 (p. 374) y1 = 1
x
∞∑
n=0
(−1)n ∏ n
j=1(2j − 5)
n! xn;
y2 = y1 ln x + 5
x
∞∑
n=1
(−1)n ∏ n
j=1(2j − 5)
n!
( n∑
j=1
1
j(2j − 5)
)
xn
7. 6. 22 (p. 374) y1 = x2
∞∑
n=0
(−1)n ∏ n
j=1(2j + 3)
2n n! xn;
y2 = y1 ln x − 3x2
∞∑
n=0
(−1)n ∏ n
j=1(2j + 3)
2nn!
( n∑
j=1
1
j(2j + 3)
)
xn
7. 6. 23 (p. 374) y1 = x−2
(
1 + 3 x + 3
2 x2 − 1
2 x3 + · · ·
)
; y2 = y1 ln x − 5x−1
(
1 + 5
4 x − 1
4 x2 + · · ·
)
7. 6. 24 (p. 374) y1 = x3(1 + 20 x + 180 x2 + 1120 x3 + · · ·; y2 = y1 ln x − x4
(
26 + 324 x + 6968
3 x2 + · · ·
)
7. 6. 25 (p. 374) y1 = x
(
1 − 5x + 85
4 x2 − 3145
36 x3 + · · ·
)
; y2 = y1 ln x + x2
(
2 − 39
4 x + 4499
108 x2 + · · ·
)
7. 6. 26 (p. 374) y1 = 1 − x + 3
4 x2 − 7
12 x3 + · · ·; y2 = y1 ln x + x
(
1 − 3
4 x + 5
9 x2 + · · ·
)
7. 6. 27 (p. 374) y1 = x−3(1 + 16 x + 36 x2 + 16 x3 + · · ·); y2 = y1 ln x − x−2
(",Elementary Differential Equations with Boundary Value Problems.pdf
"12 x3 + · · ·; y2 = y1 ln x + x
(
1 − 3
4 x + 5
9 x2 + · · ·
)
7. 6. 27 (p. 374) y1 = x−3(1 + 16 x + 36 x2 + 16 x3 + · · ·); y2 = y1 ln x − x−2
(
40 + 150 x + 280
3 x2 + · · ·
)
7. 6. 28 (p. 374) y1 = x
∞∑
m=0
(−1)m
2mm! x2m ; y2 = y1 ln x − x
2
∞∑
m=1
(−1)m
2mm!
(m∑
j=1
1
j
)
x2m
7. 6. 29 (p. 374) y1 = x2
∞∑
m=0
(−1)m(m + 1) x2m ; y2 = y1 ln x − x2
2
∞∑
m=1
(−1)mmx2m
7. 6. 30 (p. 374) y1 = x1/ 2
∞∑
m=0
(−1)m
4m m! x2m; y2 = y1 ln x − x1/ 2
2
∞∑
m=1
(−1)m
4mm!
(m∑
j=1
1
j
)
x2m
7. 6. 31 (p. 374) y1 = x
∞∑
m=0
(−1)m ∏ m
j=1(2j − 1)
2mm! x2m;
y2 = y1 ln x + x
2
∞∑
m=1
(−1)m ∏ m
j=1(2j − 1)
2mm!
(
m∑
j=1
1
j(2j − 1)
)
x2m
7. 6. 32 (p. 374) y1 = x1/ 2
∞∑
m=0
(−1)m ∏ m
j=1(4j − 1)
8mm! x2m ;
y2 = y1 ln x + x1/ 2
2
∞∑
m=1
(−1)m ∏ m
j=1(4j − 1)
8m m!
(m∑
j=1
1
j(4j − 1)
)
x2m
7. 6. 33 (p. 374) y1 = x
∞∑
m=0
(−1)m ∏ m
j=1(2j + 1)
2mm! x2m;
y2 = y1 ln x − x
2
∞∑
m=1
(−1)m ∏ m
j=1(2j + 1)
2mm!
(m∑
j=1
1
j(2j + 1)
)
x2m",Elementary Differential Equations with Boundary Value Problems.pdf
"742 Answers to Selected Exercises
7. 6. 34 (p. 374) y1 = x−1/ 4
∞∑
m=0
(−1)m ∏ m
j=1(8j − 13)
(32)m m! x2m ;
y2 = y1 ln x + 13
2 x−1/ 4
∞∑
m=1
(−1)m ∏ m
j=1(8j − 13)
(32)m m!
(
m∑
j=1
1
j(8j − 13)
)
x2m
7. 6. 35 (p. 374) y1 = x1/ 3
∞∑
m=0
(−1)m ∏ m
j=1(3j − 1)
9mm! x2m ;
y2 = y1 ln x + x1/ 3
2
∞∑
m=1
(−1)m ∏ m
j=1(3j − 1)
9m m!
(m∑
j=1
1
j(3j − 1)
)
x2m
7. 6. 36 (p. 374) y1 = x1/ 2
∞∑
m=0
(−1)m ∏ m
j=1(4j − 3)(4j − 1)
4m(m!)2 x2m ;
y2 = y1 ln x + x1/ 2
∞∑
m=1
(−1)m ∏ m
j=1(4j − 3)(4j − 1)
4m(m!)2
(m∑
j=1
8j − 3
j(4j − 3)(4j − 1)
)
x2m
7. 6. 37 (p. 374) y1 = x5/ 3
∞∑
m=0
(−1)m
3m m! x2m; y2 = y21 ln x − x5/ 3
2
∞∑
m=1
(−1)m
3mm!
(m∑
j=1
1
j
)
x2m
7. 6. 38 (p. 374) y1 = 1
x
∞∑
m=0
(−1)m ∏ m
j=1(4j − 7)
2mm! x2m;
y2 = y1 ln x + 7
2x
∞∑
m=1
(−1)m ∏ m
j=1(4j − 7)
2mm!
(
m∑
j=1
1
j(4j − 7)
)
x2m
7. 6. 39 (p. 374) y1 = x−1
(
1 − 3
2 x2 + 15
8 x4 − 35
16 x6 + · · ·
)
; y2 = y1 ln x + x
(1
4 − 13
32 x2 + 101
192 x4 + · · ·
)
7. 6. 40 (p. 375) y1 = x
(
1 − 1
2 x2 + 1
8 x4 − 1
48 x6 + · · ·
)",Elementary Differential Equations with Boundary Value Problems.pdf
"16 x6 + · · ·
)
; y2 = y1 ln x + x
(1
4 − 13
32 x2 + 101
192 x4 + · · ·
)
7. 6. 40 (p. 375) y1 = x
(
1 − 1
2 x2 + 1
8 x4 − 1
48 x6 + · · ·
)
; y2 = y1 ln x + x3
(1
4 − 3
32 x2 + 11
576 x4 + · · ·
)
7. 6. 41 (p. 375) y1 = x−2
(
1 − 3
4 x2 − 9
64 x4 − 25
256 x6 + · · ·
)
; y2 = y1 ln x + 1
2 − 21
128 x2 − 215
1536 x4 + · · ·
7. 6. 42 (p. 375) y1 = x−3
(
1 − 17
8 x2 + 85
256 x4 − 85
18432 x6 + · · ·
)
; y2 = y1 ln x + x−1
(25
8 − 471
512 x2 + 1583
110592 x4 + · · ·
)
7. 6. 43 (p. 375) y1 = x−1
(
1 − 3
4 x2 + 45
64 x4 − 175
256 x6 + · · ·
)
; y2 = y1 ln x − x
(1
4 − 33
128 x2 + 395
1536 x4 + · · ·
)
7. 6. 44 (p. 375) y1 = 1
x ; y2 = y1 ln x − 6 + 6 x − 8
3 x2
7. 6. 45 (p. 375) y1 = 1 − x; y2 = y1 ln x + 4 x
7. 6. 46 (p. 375) y1 = (x − 1)2
x ; y2 = y1 ln x + 3 − 3x + 2
∞∑
n=2
1
n(n2 − 1) xn
7. 6. 47 (p. 375) y1 = x1/ 2(x + 1) 2; y2 = y1 ln x − x3/ 2
(
3 + 3 x + 2
∞∑
n=2
(−1)n
n(n2 − 1) xn
)
7. 6. 48 (p. 375) y1 = x2(1 − x)3; y2 = y1 ln x + x3
(
4 − 7x + 11
3 x2 − 6
∞∑
n=3
1",Elementary Differential Equations with Boundary Value Problems.pdf
"(
3 + 3 x + 2
∞∑
n=2
(−1)n
n(n2 − 1) xn
)
7. 6. 48 (p. 375) y1 = x2(1 − x)3; y2 = y1 ln x + x3
(
4 − 7x + 11
3 x2 − 6
∞∑
n=3
1
n(n − 2)(n2 − 1) xn
)
7. 6. 49 (p. 375) y1 = x − 4x3 + x5; y2 = y1 ln x + 6 x3 − 3x5
7. 6. 50 (p. 375) y1 = x1/ 3
(
1 − 1
6 x2
)
; y2 = y1 ln x + x7/ 3
(
1
4 − 1
12
∞∑
m=1
1
6m m(m + 1)( m + 1)! x2m
)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 743
7. 6. 51 (p. 375) y1 = (1 + x2)2; y2 = y1 ln x − 3
2 x2 − 3
2 x4 +
∞∑
m=3
(−1)m
m(m − 1)(m − 2) x2m
7. 6. 52 (p. 375) y1 = x−1/ 2
(
1 − 1
2 x2 + 1
32 x4
)
; y2 = y1 ln x + x3/ 2
(
5
8 − 9
128 x2 +
∞∑
m=2
1
4m+1(m − 1)m(m + 1)( m + 1)! x2m
)
.
7. 6. 56 (p. 377) y1 =
∞∑
m=0
(−1)m
4m (m!)2 x2m ; y2 = y1 ln x −
∞∑
m=1
(−1)m
4m(m!)2
( m∑
j=1
1
j
)
x2m
7. 6. 58 (p. 378) x1/ 2
1 + x ; x1/ 2 ln x
1 + x 7. 6. 59 (p. 378) x1/ 3
3 + x ; x1/ 3 ln x
3 + x
7. 6. 60 (p. 378) x
2 − x2 ; x ln x
2 − x2 7. 6. 61 (p. 378) x1/ 4
1 + x2 ; x1/ 4 ln x
1 + x2
7. 6. 62 (p. 378) x
4 + 3 x ; x ln x
4 + 3 x 7. 6. 63 (p. 378) x1/ 2
1 + 3 x + x2 ; x1/ 2 ln x
1 + 3 x + x2
7. 6. 64 (p. 378) x
(1 − x)2 ; x ln x
(1 − x)2 7. 6. 65 (p. 378) x1/ 3
1 + x + x2 ; x1/ 3 ln x
1 + x + x2",Elementary Differential Equations with Boundary Value Problems.pdf
"744 Answers to Selected Exercises
Section 7.7 Answers, pp. 388– 390
7. 7. 1 (p. 388) y1 = 2 x3
∞∑
n=0
(−4)n
n!(n + 2)! xn; y2 = x + 4 x2 − 8
(
y1 ln x − 4
∞∑
n=1
(−4)n
n!(n + 2)!
( n∑
j=1
j + 1
j(j + 2)
)
xn
)
7. 7. 2 (p. 388) y1 = x
∞∑
n=0
(−1)n
n!(n + 1)! xn; y2 = 1 − y1 ln x + x
∞∑
n=1
(−1)n
n!(n + 1)!
( n∑
j=1
2j + 1
j(j + 1)
)
xn
7. 7. 3 (p. 388) y1 = x1/ 2; y2 = x−1/ 2 + y1 ln x + x1/ 2
∞∑
n=1
(−1)n
n xn
7. 7. 4 (p. 388) y1 = x
∞∑
n=0
(−1)n
n! xn = xe−x; y2 = 1 − y1 ln x + x
∞∑
n=1
(−1)n
n!
( n∑
j=1
1
j
)
xn
7. 7. 5 (p. 388) y1 = x1/ 2
∞∑
n=0
(
− 3
4
) n
∏ n
j=1(2j + 1)
n! xn;
y2 = x−1/ 2 − 3
4
(
y1 ln x − x1/ 2
∞∑
n=1
(
− 3
4
) n
∏ n
j=1(2j + 1)
n!
( n∑
j=1
1
j(2j + 1)
)
xn
)
7. 7. 6 (p. 388) y1 = x
∞∑
n=0
(−1)n
n! xn = xe−x; y2 = x−2
(
1 + 1
2 x + 1
2 x2
)
− 1
2
(
y1 ln x − x
∞∑
n=1
(−1)n
n!
( n∑
j=1
1
j
)
xn
)
7. 7. 7 (p. 388) y1 = 6 x3/ 2
∞∑
n=0
(−1)n
4nn!(n + 3)! xn;
y2 = x−3/ 2
(
1 + 1
8 x + 1
64 x2
)
− 1
768
(
y1 ln x − 6x3/ 2
∞∑
n=1
(−1)n
4nn!(n + 3)!
(
n∑
j=1
2j + 3",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
n=0
(−1)n
4nn!(n + 3)! xn;
y2 = x−3/ 2
(
1 + 1
8 x + 1
64 x2
)
− 1
768
(
y1 ln x − 6x3/ 2
∞∑
n=1
(−1)n
4nn!(n + 3)!
(
n∑
j=1
2j + 3
j(j + 3)
)
xn
)
7. 7. 8 (p. 388) y1 = 120
x2
∞∑
n=0
(−1)n
n!(n + 5)! xn;
y2 = x−7
(
1 + 1
4 x + 1
24 x2 + 1
144 x3 + 1
576 x4
)
− 1
2880
(
y1 ln x − 120
x2
∞∑
n=1
(−1)n
n!(n + 5)!
( n∑
j=1
2j + 5
j(j + 5)
)
xn
)
7. 7. 9 (p. 388) y1 = x1/ 2
6
∞∑
n=0
(−1)n (n + 1)( n + 2)( n + 3) xn;
y2 = x−5/ 2
(
1 + 1
2 x + x2
)
− 3y1 ln x + 3
2 x1/ 2
∞∑
n=1
(−1)n(n + 1)( n + 2)( n + 3)
( n∑
j=1
1
j(j + 3)
)
xn
7. 7. 10 (p. 388) y1 = x4
(
1 − 2
5 x
)
y2 = 1 + 10 x + 50 x2 + 200 x3 − 300
(
y1 ln x + 27
25 x5 − 1
30 x6
)
7. 7. 11 (p. 388) y1 = x3; y2 = x−3
(
1 − 6
5 x + 3
4 x2 − 1
3 x3 + 1
8 x4 − 1
20 x5
)
− 1
120
(
y1 ln x + x3
∞∑
n=1
(−1)n6!
n(n + 6)! xn
)
7. 7. 12 (p. 388) y1 = x2
∞∑
n=0
1
n!
( n∏
j=1
2j + 3
j + 4
)
xn;
y2 = x−2
(
1 + x + 1
4 x2 − 1
12 x3
)
− 1
16 y1 ln x + x2
8
∞∑
n=1
1
n!
( n∏
j=1
2j + 3
j + 4
)( n∑
j=1
(j2 + 3 j + 6)
j(j + 4)(2 j + 3)
)
xn",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 745
7. 7. 13 (p. 388) y1 = x5
∞∑
n=0
(−1)n (n + 1)( n + 2) xn; y2 = 1 − x
2 + x2
6
7. 7. 14 (p. 388) y1 = 1
x
∞∑
n=0
(−1)n
n!
( n∏
j=1
(j + 3)(2 j − 3)
j + 6
)
xn; y2 = x−7
(
1 + 26
5 x + 143
20 x2
)
7. 7. 15 (p. 388) y1 = x7/ 2
∞∑
n=0
(−1)n
2n(n + 4)! xn; y2 = x−1/ 2
(
1 − 1
2 x + 1
8 x2 − 1
48 x3
)
7. 7. 16 (p. 388) y1 = x10/ 3
∞∑
n=0
(−1)n(n + 1)
9n
( n∏
j=1
3j + 7
j + 4
)
xn; y2 = x−2/ 3
(
1 + 4
27 x − 1
243 x2
)
7. 7. 17 (p. 388) y1 = x3
7∑
n=0
(−1)n(n + 1)
( n∏
j=1
j − 8
j + 6
)
xn; y2 = x−3
(
1 + 52
5 x + 234
5 x2 + 572
5 x3 + 143 x4
)
7. 7. 18 (p. 388) y1 = x3
∞∑
n=0
(−1)n
n!
( n∏
j=1
(j + 3) 2
j + 5
)
xn; y2 = x−2
(
1 + 1
4 x
)
7. 7. 19 (p. 388) y1 = x6
4∑
n=0
(−1)n 2n
( n∏
j=1
j − 5
j + 5
)
xn; y2 = x(1 + 18 x + 144 x2 + 672 x3 + 2016 x4)
7. 7. 20 (p. 388) y1 = x6
(
1 + 2
3 x + 1
7 x2
)
; y2 = x
(
1 + 21
4 x + 21
2 x2 + 35
4 x3
)
7. 7. 21 (p. 388) y1 = x7/ 2
∞∑
n=0
(−1)n (n + 1) xn; y2 = x−7/ 2
(
1 − 5
6 x + 2
3 x2 − 1
2 x3 + 1
3 x4 − 1
6 x5
)",Elementary Differential Equations with Boundary Value Problems.pdf
"1 + 21
4 x + 21
2 x2 + 35
4 x3
)
7. 7. 21 (p. 388) y1 = x7/ 2
∞∑
n=0
(−1)n (n + 1) xn; y2 = x−7/ 2
(
1 − 5
6 x + 2
3 x2 − 1
2 x3 + 1
3 x4 − 1
6 x5
)
7. 7. 22 (p. 388) y1 = x10
6
∞∑
n=0
(−1)n2n(n + 1)( n + 2)( n + 3) xn ;
y2 =
(
1 − 4
3 x + 5
3 x2 − 40
21 x3 + 40
21 x4 − 32
21 x5 + 16
21 x6
)
7. 7. 23 (p. 388) y1 = x6
∞∑
m=0
(−1)m ∏ m
j=1(2j + 5)
2mm! x2m;
y2 = x2
(
1 + 3
2 x2
)
− 15
2 y1 ln x + 75
2 x6
∞∑
m=1
(−1)m ∏ m
j=1(2j + 5)
2m+1m!
(m∑
j=1
1
j(2j + 5)
)
x2m
7. 7. 24 (p. 388) y1 = x6
∞∑
m=0
(−1)m
2m m! x2m = x6e−x2 / 2;
y2 = x2
(
1 + 1
2 x2
)
− 1
2 y1 ln x + x6
4
∞∑
m=1
(−1)m
2mm!
(m∑
j=1
1
j
)
x2m
7. 7. 25 (p. 388) y1 = 6 x6
∞∑
m=0
(−1)m
4mm!(m + 3)! x2m;
y2 = 1 + 1
8 x2 + 1
64 x4 − 1
384
(
y1 ln x − 3x6
∞∑
m=1
(−1)m
4mm!(m + 3)!
(m∑
j=1
2j + 3
j(j + 3)
)
x2m
)
7. 7. 26 (p. 388) y1 = x
2
∞∑
m=0
(−1)m(m + 2)
m! x2m ;
y2 = x−1 − 4y1 ln x + x
∞∑
m=1
(−1)m(m + 2)
m!
(m∑
j=1
j2 + 4 j + 2
j(j + 1)( j + 2)
)
x2m",Elementary Differential Equations with Boundary Value Problems.pdf
"746 Answers to Selected Exercises
7. 7. 27 (p. 388) y1 = 2 x3
∞∑
m=0
(−1)m
4mm!(m + 2)! x2m;
y2 = x−1
(
1 + 1
4 x2
)
− 1
16
(
y1 ln x − 2x3
∞∑
m=1
(−1)m
4mm!(m + 2)!
(
m∑
j=1
j + 1
j(j + 2)
)
x2m
)
7. 7. 28 (p. 388) y1 = x−1/ 2
∞∑
m=0
(−1)m ∏ m
j=1(2j − 1)
8mm!(m + 1)! x2m ;
y2 = x−5/ 2 + 1
4 y1 ln x − x−1/ 2
∞∑
m=1
(−1)m ∏ m
j=1(2j − 1)
8m+1m!(m + 1)!
(m∑
j=1
2j2 − 2j − 1
j(j + 1)(2 j − 1)
)
x2m
7. 7. 29 (p. 388) y1 = x
∞∑
m=0
(−1)m
2mm! x2m = xe−x2/ 2; y2 = x−1 − y1 ln x + x
2
∞∑
m=1
(−1)m
2m m!
( m∑
j=1
1
j
)
x2m
7. 7. 30 (p. 388) y1 = x2
∞∑
m=0
1
m! x2m = x2 ex2
; y2 = x−2(1 − x2) − 2y1 ln x + x2
∞∑
m=1
1
m!
(m∑
j=1
1
j
)
x2m
7. 7. 31 (p. 388) y1 = 6 x5/ 2
∞∑
m=0
(−1)m
16m m!(m + 3)! x2m;
y2 = x−7/ 2
(
1 + 1
32 x2 + 1
1024 x4
)
− 1
24576
(
y1 ln x − 3x5/ 2
∞∑
m=1
(−1)m
16m m!(m + 3)!
( m∑
j=1
2j + 3
j(j + 3)
)
x2m
)
7. 7. 32 (p. 388) y1 = 2 x13/ 3
∞∑
m=0
∏ m
j=1(3j + 1)
9m m!(m + 2)! x2m;
y2 = x1/ 3
(
1 + 2
9 x2
)
+ 2
81
(
y1 ln x − x13/ 3
∞∑
m=0
∏ m
j=1(3j + 1)
9m m!(m + 2)!
(m∑",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
m=0
∏ m
j=1(3j + 1)
9m m!(m + 2)! x2m;
y2 = x1/ 3
(
1 + 2
9 x2
)
+ 2
81
(
y1 ln x − x13/ 3
∞∑
m=0
∏ m
j=1(3j + 1)
9m m!(m + 2)!
(m∑
j=1
3j2 + 2 j + 2
j(j + 2)(3 j + 1)
)
x2m
)
7. 7. 33 (p. 388) y1 = x2; y2 = x−2(1 + 2 x2) − 2
(
y1 ln x + x2
∞∑
m=1
1
m(m + 2)! x2m
)
7. 7. 34 (p. 388) y1 = x2
(
1 − 1
2 x2
)
; y2 = x−2
(
1 + 9
2 x2
)
− 27
2
(
y1 ln x + 7
12 x4 − x2
∞∑
m=2
(3
2
) m
m(m − 1)(m + 2)! x2m
)
7. 7. 35 (p. 388) y1 =
∞∑
m=0
(−1)m(m + 1) x2m; y2 = x−4
7. 7. 36 (p. 388) y1 = x5/ 2
∞∑
m=0
(−1)m
(m + 1)( m + 2)( m + 3) x2m; y2 = x−7/ 2(1 + x2)2
7. 7. 37 (p. 388) y1 = x7
5
∞∑
m=0
(−1)m(m + 5) x2m; y2 = x−1 (
1 − 2x2 + 3 x4 − 4x6)
7. 7. 38 (p. 389) y1 = x3
∞∑
m=0
(−1)m m + 1
2m
(m∏
j=1
2j + 1
j + 5
)
x2m ; y2 = x−7
(
1 + 21
8 x2 + 35
16 x4 + 35
64 x6
)
7. 7. 39 (p. 389) y1 = 2 x4
∞∑
m=0
(−1)m
∏ m
j=1(4j + 5)
2m(m + 2)! x2m; y2 = 1 − 1
2 x2
7. 7. 40 (p. 389) y1 = x3/ 2
∞∑
m=0
(−1)m ∏ m
j=1(2j − 1)
2m−1 (m + 2)! x2m ; y2 = x−5/ 2
(
1 + 3
2 x2
)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 747
7. 7. 42 (p. 389) y1 = xν
∞∑
m=0
(−1)m
4m m!
∏ m
j=1(j + ν ) x2m;
y2 = x−ν
ν −1∑
m=0
(−1)m
4mm!
∏ m
j=1(j − ν ) x2m − 2
4ν ν !(ν − 1)!
(
y1 ln x − xν
2
∞∑
m=1
(−1)m
4m m!
∏ m
j=1(j + ν )
(
m∑
j=1
2j + ν
j(j + ν )
)
x2m
)
Section 8.1 Answers, pp. 402– 404
8. 1. 1 (p. 402) (a) 1
s2 (b) 1
(s + 1) 2 (c) b
s2 − b2 (d) −2s + 5
(s − 1)(s − 2) (e) 2
s3
8. 1. 2 (p. 402) (a) s2 + 2
[(s − 1)2 + 1] [( s + 1) 2 + 1] (b) 2
s(s2 + 4) (c) s2 + 8
s(s2 + 16) (d) s2 − 2
s(s2 − 4)
(e) 4s
(s2 − 4)2 (f) 1
s2 + 4 (g) 1√
2
s + 1
s2 + 1 (h) 5s
(s2 + 4)( s2 + 9) (i) s3 + 2 s2 + 4 s + 32
(s2 + 4)( s2 + 16)
8. 1. 4 (p. 402) (a) f (3−) = −1, f (3) = f (3+) = 1 (b) f (1−) = 3 , f (1) = 4 , f (1+) = 1
(c) f
(π
2 −
)
= 1 , f
(π
2
)
= f
(π
2 +
)
= 2 , f (π −) = 0 , f (π ) = f (π +) = −1
(d) f (1−) = 1 , f (1) = 2 , f (1+) = 1 , f (2−) = 0 , f (2) = 3 , f (2+) = 6
8. 1. 5 (p. 402) (a) 1 − e−(s+1)
s + 1 + e−(s+2)
s + 2 (b) 1
s + e−4s
(1
s2 + 3
s
)
(c) 1 − e−s
s2 (d) 1 − e−(s−1)
(s − 1)2",Elementary Differential Equations with Boundary Value Problems.pdf
"8. 1. 5 (p. 402) (a) 1 − e−(s+1)
s + 1 + e−(s+2)
s + 2 (b) 1
s + e−4s
(1
s2 + 3
s
)
(c) 1 − e−s
s2 (d) 1 − e−(s−1)
(s − 1)2
8. 1. 7 (p. 402) L(eλt cos ωt ) = (s − λ )2 − ω 2
((s − λ )2 + ω 2)2 L(eλt sin ωt ) = 2ω (s − λ )
((s − λ )2 + ω 2)2
8. 1. 15 (p. 403) (a) tan−1 ω
s , s > 0 (b) 1
2 ln s2
s2 + ω 2 , s > 0 (c) ln s − b
s − a , s > max(a, b )
(d) 1
2 ln s2
s2 − 1 , s > 1 (e) 1
4 ln s2
s2 − 4 , s > 2
8. 1. 18 (p. 404) (a) 1
s2 tanh s
2 (b) 1
s tanh s
4 (c) 1
s2 + 1 coth πs
2 (d) 1
(s2 + 1)(1 − e−πs )
Section 8.2 Answers, pp. 411– 413
8. 2. 1 (p. 411) (a) t3e7t
2 (b) 2e2t cos 3 t (c) e−2t
4 sin 4t (d) 2
3 sin 3 t (e) t cos t
(f) e2t
2 sinh 2t (g) 2te2t
3 sin 9 t (h) 2e3t
3 sinh 3t (i) e2t t cos t
8. 2. 2 (p. 411) (a) t2e7t + 17
6 t3e7t (b) e2t
(1
6 t3 + 1
6 t4 + 1
40 t5
)
(c) e−3t
(
cos 3 t + 2
3 sin 3t
)
(d) 2 cos 3 t + 1
3 sin 3 t (e) (1 − t)e−t (f) cosh 3t + 1
3 sinh 3t (g)
(
1 − t − t2 − 1
6 t3
)
e−t
(h) et
(
2 cos 2 t + 5
2 sin 2 t
)",Elementary Differential Equations with Boundary Value Problems.pdf
"3 sin 3t
)
(d) 2 cos 3 t + 1
3 sin 3 t (e) (1 − t)e−t (f) cosh 3t + 1
3 sinh 3t (g)
(
1 − t − t2 − 1
6 t3
)
e−t
(h) et
(
2 cos 2 t + 5
2 sin 2 t
)
(i) 1 − cos t (j) 3 cosh t + 4 sinh t (k) 3et + 4 cos 3 t + 1
3 sin 3t
(l) 3te−2t − 2 cos 2 t − 3 sin 2 t
8. 2. 3 (p. 412) (a) 1
4 e2t − 1
4 e−2t − e−t (b) 1
5 e−4t − 41
5 et + 5 e3t (c) − 1
2 e2t − 13
10 e−2t − 1
5 e3t
(d) − 2
5 e−4t − 3
5 et (e) 3
20 e2t − 37
12 e−2t + 1
3 et + 8
5 e−3t (f) 39
10 et + 3
14 e3t + 23
105 e−4t − 7
3 e2t
8. 2. 4 (p. 412) (a) 4
5 e−2t − 1
2 e−t − 3
10 cos t + 11
10 sin t (b) 2
5 sin t + 6
5 cos t + 7
5 e−t sin t − 6
5 e−t cos t
(c) 8
13 e2t − 8
13 e−t cos 2 t + 15
26 e−t sin 2 t (d) 1
2 tet + 3
8 et + e−2t − 11
8 e−3t
(e) 2
3 tet + 1
9 et + te−2t − 1
9 e−2t (f) −et + 5
2 tet + cos t − 3
2 sin t
8. 2. 5 (p. 412) (a) 3
5 cos 2 t + 1
5 sin 2 t − 3
5 cos 3 t − 2
15 sin 3t (b) − 4
15 cos t + 1
15 sin t + 4
15 cos 4 t − 1
60 sin 4t",Elementary Differential Equations with Boundary Value Problems.pdf
"748 Answers to Selected Exercises
(c) 5
3 cos t + sin t − 5
3 cos 2 t − 1
2 sin 2t (d) − 1
3 cos t
2 + 2
3 sin t
2 + 1
3 cos t − 1
3 sin t
(e) 1
15 cos t
4 − 8
15 sin t
4 − 1
15 cos 4 t + 1
30 sin 4t (f) 2
5 cos t
3 − 3
5 sin t
3 − 2
5 cos t
2 + 2
5 sin t
2
8. 2. 6 (p. 412) (a) et (cos 2t + sin 2 t) − e−t
(
cos 3 t + 4
3 sin 3 t
)
(b) e3t
(
− cos 2 t + 3
2 sin 2t
)
+ e−t
(
cos 2 t + 1
2 sin 2 t
)
(c) e−2t
(1
8 cos t + 1
4 sin t
)
− e2t
(1
8 cos 3t − 1
12 sin 3 t
)
(d) e2t
(
cos t + 1
2 sin t
)
− e3t
(
cos 2 t − 1
4 sin 2t
)
(e) et
(1
5 cos t + 2
5 sin t
)
− e−t
(1
5 cos 2 t + 2
5 sin 2 t
)
(f) et/ 2
(
− cos t + 9
8 sin t
)
+ e−t/ 2
(
cos t − 1
8 sin t
)
8. 2. 7 (p. 412) (a) 1−cos t (b) et
16 (1 − cos 4 t) (c) 4
9 e2t + 5
9 e−t sin 3 t − 4
9 e−t cos 3 t (d) 3et/ 2 − 7
2 et sin 2 t − 3et cos 2 t
(e) 1
4 e3t − 1
4 e−t cos 2t (f) 1
9 e2t − 1
9 e−t cos 3 t + 5
9 e−t sin 3t
8. 2. 8 (p. 412) (a) − 3
10 sin t + 2
5 cos t − 3
4 et + 7
20 e3t (b) − 3
5 e−t sin t + 1
5 e−t cos t − 1
2 e−t + 3",Elementary Differential Equations with Boundary Value Problems.pdf
"9 e−t cos 3 t + 5
9 e−t sin 3t
8. 2. 8 (p. 412) (a) − 3
10 sin t + 2
5 cos t − 3
4 et + 7
20 e3t (b) − 3
5 e−t sin t + 1
5 e−t cos t − 1
2 e−t + 3
10 et
(c) − 1
10 et sin t − 7
10 et cos t + 1
5 e−t + 1
2 e2t (d) − 1
2 et + 7
10 e−t − 1
5 cos 2 t + 3
5 sin 2t
(e) 3
10 + 1
10 e2t + 1
10 et sin 2 t − 2
5 et cos 2 t (f) − 4
9 e2t cos 3 t + 1
3 e2t sin 3 t − 5
9 e2t + et
8. 2. 9 (p. 413) 1
a e
b
a t f
(t
a
)
Section 8.3 Answers, pp. 418– 419
8. 3. 1 (p. 418) y = 1
6 et − 9
2 e−t + 16
3 e−2t 8. 3. 2 (p. 418) y = − 1
3 + 8
15 e3t + 4
5 e−2t
8. 3. 3 (p. 418) y = − 23
15 e−2t + 1
3 et + 1
5 e3t 8. 3. 4 (p. 418) y = − 1
4 e2t + 17
20 e−2t + 2
5 e3t
8. 3. 5 (p. 418) y = 11
15 e−2t + 1
6 et + 1
10 e3t 8. 3. 6 (p. 418) y = et + 2 e−2t − 2e−t
8. 3. 7 (p. 418) y = 5
3 sin t − 1
3 sin 2 t 8. 3. 8 (p. 418) y = 4 et − 4e2t + e3t
8. 3. 9 (p. 418) y = − 7
2 e2t + 13
3 et + 1
6 e4t 8. 3. 10 (p. 418) y = 5
2 et − 4e2t + 1
2 e3t
8. 3. 11 (p. 418) y = 1
3 et − 2e−t + 5",Elementary Differential Equations with Boundary Value Problems.pdf
"8. 3. 9 (p. 418) y = − 7
2 e2t + 13
3 et + 1
6 e4t 8. 3. 10 (p. 418) y = 5
2 et − 4e2t + 1
2 e3t
8. 3. 11 (p. 418) y = 1
3 et − 2e−t + 5
3 e−2t 8. 3. 12 (p. 418) y = 2 − e−2t + et
8. 3. 13 (p. 418) y = 1 − cos 2 t + 1
2 sin 2 t 8. 3. 14 (p. 418) y = − 1
3 + 8
15 e3t + 4
5 e−2t
8. 3. 15 (p. 418) y = 1
6 et − 2
3 e−2t + 1
2 e−t 8. 3. 16 (p. 418) y = −1 + et + e−t
8. 3. 17 (p. 418) y = cos 2 t − sin 2t + sin t 8. 3. 18 (p. 418) y = 7
3 − 7
2 e−t + 1
6 e3t
8. 3. 19 (p. 418) y = 1 + cos t 8. 3. 20 (p. 418) y = t + sin t 8. 3. 21 (p. 418) y = t − 6 sin t + cos t + sin 2 t
8. 3. 22 (p. 418) y = e−t + 4 e−2t − 4e−3t 8. 3. 23 (p. 418) y = −3 cos t − 2 sin t + e−t (2 + 5 t)
8. 3. 24 (p. 419) y = − sin t − 2 cos t + 3 e3t + e−t 8. 3. 25 (p. 419) y = (3 t + 4) sin t − (2t + 6) cos t
8. 3. 26 (p. 419) y = −(2t + 2) cos 2 t + sin 2 t + 3 cos t 8. 3. 27 (p. 419) y = et (cos t − 3 sin t) + e3t",Elementary Differential Equations with Boundary Value Problems.pdf
"8. 3. 26 (p. 419) y = −(2t + 2) cos 2 t + sin 2 t + 3 cos t 8. 3. 27 (p. 419) y = et (cos t − 3 sin t) + e3t
8. 3. 28 (p. 419) y = −1 + t + e−t (3 cos t − 5 sin t) 8. 3. 29 (p. 419) y = 4 cos t − 3 sin t − et (3 cos t − 8 sin t)
8. 3. 30 (p. 419) y = e−t − 2et + e−2t (cos 3t − 11/ 3 sin 3 t)
8. 3. 31 (p. 419) y = e−t (sin t − cos t) + e−2t (cos t + 4 sin t)
8. 3. 32 (p. 419) y = 1
5 e2t − 4
3 et + 32
15 e−t/ 2 8. 3. 33 (p. 419) y = 1
7 e2t − 2
5 et/ 2 + 9
35 e−t/ 3
8. 3. 34 (p. 419) y = e−t/ 2(5 cos( t/ 2) − sin(t/ 2)) + 2 t − 4
8. 3. 35 (p. 419) y = 1
17
(
12 cos t + 20 sin t − 3et/ 2(4 cos t + sin t)
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 749
8. 3. 36 (p. 419) y = e−t/ 2
10 (5t + 26) − 1
5 (3 cos t + sin t) 8. 3. 37 (p. 419) y = 1
100
(
3e3t − et/ 3(3 + 310 t)
)
Section 8.4 Answers, pp. 427– 430
8. 4. 1 (p. 427) 1 + u(t − 4)(t − 1); 1
s + e−4s
(1
s2 + 3
s
)
8. 4. 2 (p. 427) t + u(t − 1)(1 − t); 1 − e−s
s2
8. 4. 3 (p. 427) 2t − 1 − u(t − 2)(t − 1);
(2
s2 − 1
s
)
− e−2s
(1
s2 + 1
s
)
8. 4. 4 (p. 427) 1 + u(t − 1)(t + 1) ; 1
s + e−s
(1
s2 + 2
s
)
8. 4. 5 (p. 427) t − 1 + u(t − 2)(5 − t); 1
s2 − 1
s − e−2s
(1
s2 − 3
s
)
8. 4. 6 (p. 427) t2 (1 − u(t − 1)); 2
s3 − e−s
(2
s3 + 2
s2 + 1
s
)
8. 4. 7 (p. 428) u(t − 2)(t2 + 3 t); e−2s
(2
s3 + 7
s2 + 10
s
)
8. 4. 8 (p. 428) t2 + 2 + u(t − 1)(t − t2 − 2); 2
s3 + 2
s − e−s
(2
s3 + 1
s2 + 2
s
)
8. 4. 9 (p. 428) tet + u(t − 1)(et − tet); 1 − e−(s−1)
(s − 1)2
8. 4. 10 (p. 428) e−t + u(t − 1)(e−2t − e−t ) ; 1 − e−(s+1)
s + 1 + e−(s+2)
s + 2
8. 4. 11 (p. 428) −t + 2 u(t − 2)(t − 2) − u(t − 3)(t − 5); − 1
s2 + 2e−2s
s2 + e−3s
(2
s − 1
s2
)",Elementary Differential Equations with Boundary Value Problems.pdf
"s + 1 + e−(s+2)
s + 2
8. 4. 11 (p. 428) −t + 2 u(t − 2)(t − 2) − u(t − 3)(t − 5); − 1
s2 + 2e−2s
s2 + e−3s
(2
s − 1
s2
)
8. 4. 12 (p. 428) [u(t − 1) − u(t − 2)] t ; e−s
(1
s2 + 1
s
)
− e−2s
(1
s2 + 2
s
)
8. 4. 13 (p. 428) t + u(t − 1)(t2 − t) − u(t − 2)t2; 1
s2 + e−s
(2
s3 + 1
s2
)
− e−2s
(2
s3 + 4
s2 + 4
s
)
8. 4. 14 (p. 428) t + u(t − 1)(2 − 2t) + u(t − 2)(4 + t); 1
s2 − 2 e−s
s2 + e−2s
(1
s2 + 6
s
)
8. 4. 15 (p. 428) sin t + u(t − π/ 2) sin t + u(t − π )(cos t − 2 sin t); 1 + e− π
2 ss − e−πs (s − 2)
s2 + 1
8. 4. 16 (p. 428) 2 − 2u(t − 1)t + u(t − 3)(5t − 2); 2
s − e−s
(2
s2 + 2
s
)
+ e−3s
(5
s2 + 13
s
)
8. 4. 17 (p. 428) 3 + u(t − 2)(3t − 1) + u(t − 4)(t − 2); 3
s + e−2s
(3
s2 + 5
s
)
+ e−4s
(1
s2 + 2
s
)
8. 4. 18 (p. 428) (t + 1) 2 + u(t − 1)(2t + 3) ; 2
s3 + 2
s2 + 1
s + e−s
(2
s2 + 5
s
)
8. 4. 19 (p. 428) u(t − 2)e2(t−2) =
{
0, 0 ≤ t < 2,
e2(t−2), t ≥ 2.
8. 4. 20 (p. 428) u(t − 1)
(
1 − e−(t−1))
=
{
0, 0 ≤ t < 1,
1 − e−(t−1), t ≥ 1.
8. 4. 21 (p. 428) u(t − 1) (t − 1)2",Elementary Differential Equations with Boundary Value Problems.pdf
"{
0, 0 ≤ t < 2,
e2(t−2), t ≥ 2.
8. 4. 20 (p. 428) u(t − 1)
(
1 − e−(t−1))
=
{
0, 0 ≤ t < 1,
1 − e−(t−1), t ≥ 1.
8. 4. 21 (p. 428) u(t − 1) (t − 1)2
2 + u(t − 2)(t − 2) =











0, 0 ≤ t < 1,
(t − 1)2
2 , 1 ≤ t < 2,
t2 − 3
2 , t ≥ 2.",Elementary Differential Equations with Boundary Value Problems.pdf
"750 Answers to Selected Exercises
8. 4. 22 (p. 428) 2 + t + u(t − 1)(4 − t) + u(t − 3)(t − 2) =





2 + t, 0 ≤ t < 1,
6, 1 ≤ t < 3,
t + 4 , t ≥ 3.
8. 4. 23 (p.
429) 5 − t + u(t − 3)(7t − 15) + 3
2 u(t − 6)(t − 6)2 =





5 − t, 0 ≤ t < 3,
6t − 10, 3 ≤ t < 6,
44 − 12t + 3
2 t2 , t ≥ 6.
8. 4. 24 (p. 429) u(t − π )e−2(t−π ) (2 cos t − 5 sin t) =
{
0, 0 ≤ t < π,
e−2(t−π )(2 cos t − 5 sin t), t ≥ π.
8. 4. 25 (p. 429) 1 − cos t + u(t − π/ 2)(3 sin t + cos t) =



1 − cos t, 0 ≤ t < π
2 ,
1 + 3 sin t, t ≥ π
2 .
8. 4. 26 (p. 429) u(t − 2)
(
4e−(t−2) − 4e2(t−2) + 2 e(t−2) )
=
{
0, 0 ≤ t < 2,
4e−(t−2) − 4e2(t−2) + 2 e(t−2) , t ≥ 2.
8. 4. 27 (p. 429) 1 + t + u(t − 1)(2t + 1) + u(t − 3)(3t − 5) =





t + 1 , 0 ≤ t < 1,
3t + 2 , 1 ≤ t < 3,
6t − 3, t ≥ 3.
8. 4. 28 (p.
429) 1 − t2 + u(t − 2)
(
− t2
2 + 2 t + 1
)
+ u(t − 4)(t − 4) =











1 − t2, 0 ≤ t < 2
− 3t2
2 + 2 t + 2 , 2 ≤ t < 4,
− 3t2
2 + 3 t − 2, t ≥ 4.
8. 4. 29 (p. 429) e−τ s",Elementary Differential Equations with Boundary Value Problems.pdf
")
+ u(t − 4)(t − 4) =











1 − t2, 0 ≤ t < 2
− 3t2
2 + 2 t + 2 , 2 ≤ t < 4,
− 3t2
2 + 3 t − 2, t ≥ 4.
8. 4. 29 (p. 429) e−τ s
s 8. 4. 30 (p. 429) For each t only finitely many terms are nonzero.
8. 4. 33 (p. 430) 1 +
∞∑
m=1
u(t − m); 1
s(1 − e−s ) 8. 4. 34 (p. 430) 1 + 2
∞∑
m=1
(−1)mu(t − m); 1
s ; 1 − e−s
1 + e−s
8. 4. 35 (p. 430) 1 +
∞∑
m=1
(2m + 1) u(t − m); e−s(1 + e−s)
s(1 − e−s)2 8. 4. 36 (p. 430)
∞∑
m=1
(−1)m(2m − 1)u(t − m); 1
s
(1 − es )
(1 + es )2
Section 8.5 Answers, pp. 437– 439
8. 5. 1 (p. 437) y = 3(1 − cos t) − 3u(t − π )(1 + cos t)
8. 5. 2 (p. 437) y = 3 −2 cos t+2u(t−4) (t − 4 − sin(t − 4)) 8. 5. 3 (p. 437) y = − 15
2 + 3
2 e2t − 2t + u(t − 1)
2 (e2(t−1) − 2t + 1)
8. 5. 4 (p. 437) y = 1
2 et + 13
6 e−t + 1
3 e2t + u(t − 2)
(
−1 + 1
2 et−2 + 1
2 e−(t−2) + 1
2 et+2 − 1
6 e−(t−6) − 1
3 e2t
)
8. 5. 5 (p. 437) y = −7et + 4 e2t + u(t − 1)
(1
2 − et−1 + 1
2 e2(t−1)
)
− 2u(t − 2)
(1
2 − et−2 + 1
2 e2(t−2)
)
8. 5. 6 (p. 437) y = 1
3 sin 2t − 3 cos 2 t + 1",Elementary Differential Equations with Boundary Value Problems.pdf
"(1
2 − et−1 + 1
2 e2(t−1)
)
− 2u(t − 2)
(1
2 − et−2 + 1
2 e2(t−2)
)
8. 5. 6 (p. 437) y = 1
3 sin 2t − 3 cos 2 t + 1
3 sin t − 2u(t − π )
(1
3 sin t + 1
6 sin 2 t
)
+ u(t − 2π )
(1
3 sin t − 1
6 sin 2t
)
8. 5. 7 (p. 437) y = 1
4 − 31
12 e4t + 16
3 et + u(t − 1)
(2
3 et−1 − 1
6 e4(t−1) − 1
2
)
+ u(t − 2)
(1
4 + 1
12 e4(t−2) − 1
3 et−2
)
8. 5. 8 (p. 437) y = 1
8 (cos t − cos 3 t) − 1
8 u
(
t − 3π
2
)(
sin t − cos t + sin 3 t − 1
3 cos 3 t
)
8. 5. 9 (p. 437) y = t
4 − 1
8 sin 2t + 1
8 u
(
t − π
2
)
(π cos 2 t − sin 2 t + 2 π − 2t)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 751
8. 5. 10 (p. 437) y = t − sin t − 2u(t − π )(t + sin t + π cos t)
8. 5. 11 (p. 437) y = u(t − 2)
(
t − 1
2 + e2(t−2)
2 − 2et−2
)
8. 5. 12 (p. 437) y = t + sin t + cos t − u(t − 2π )(3t − 3 sin t − 6π cos t)
8. 5. 13 (p. 437) y = 1
2 + 1
2 e−2t − e−t + u(t − 2)
(
2e−(t−2) − e−2(t−2) − 1
)
8. 5. 14 (p. 437) y = − 1
3 − 1
6 e3t + 1
2 et + u(t − 1)
(2
3 + 1
3 e3(t−1) − et−1
)
8. 5. 15 (p. 437) y = 1
4
(
et + e−t (11 + 6 t)
)
+ u(t − 1)(te−(t−1) − 1)
8. 5. 16 (p. 437) y = et − e−t − 2te−t − u(t − 1)
(
et − e−(t−2) − 2(t − 1)e−(t−2))
8. 5. 17 (p. 437) y = te−t + e−2t + u(t − 1)
(
e−t (2 − t) − e−(2t−1) )
8. 5. 18 (p. 438) y = y = t2 e2t
2 − te2t − u(t − 2)(t − 2)2e2t
8. 5. 19 (p. 438) y = t4
12 + 1 − 1
12 u(t − 1)(t4 + 2 t3 − 10t + 7) + 1
6 u(t − 2)(2t3 + 3 t2 − 36t + 44)
8. 5. 20 (p. 438) y = 1
2 e−t (3 cos t + sin t) + 1
2
−u(t − 2π )
(
e−(t−2π )
(
(π − 1) cos t + 2π − 1
2 sin t
)
+ 1 − t
2
)
− 1
2 u(t − 3π )
(
e−(t−3π )(3π cos t + (3 π + 1) sin t) + t
)",Elementary Differential Equations with Boundary Value Problems.pdf
"2
−u(t − 2π )
(
e−(t−2π )
(
(π − 1) cos t + 2π − 1
2 sin t
)
+ 1 − t
2
)
− 1
2 u(t − 3π )
(
e−(t−3π )(3π cos t + (3 π + 1) sin t) + t
)
8. 5. 21 (p. 438) y = t2
2 +
∞∑
m=1
u(t − m) (t − m)2
2
8. 5. 22 (p. 438) (a) y =
{ 2m + 1 − cos t, 2mπ ≤ t < (2m + 1) π (m = 0 , 1, . . . )
2m, (2m − 1)π ≤ t < 2mπ (m = 1 , 2, . . . )
(b) y = ( m + 1)( t − sin t − mπ cos t), 2mπ ≤ t < (2m + 2) π (m = 0 , 1, . . . )
(c) y = ( −1)m − (2m + 1) cos t, mπ ≤ t < (m + 1) π (m = 0 , 1, . . . )
(d) y = em+1 − 1
2(e − 1) (et−m + e−t) − m − 1, m ≤ t < m + 1 ( m = 0 , 1 . . . )
(e) y =
(
m + 1 −
(
e2(m+1)π − 1
e2π − 1
)
e−t
)
sin t 2mπ ≤ t < 2(m + 1) π (m = 0 , 1, . . . )
(f) y = m + 1
2 − et−m em+1 − 1
e − 1 + 1
2 e2(t−m) e2m+2 − 1
e2 − 1 , m ≤ t < m + 1 ( m = 0 , 1, . . . )",Elementary Differential Equations with Boundary Value Problems.pdf
"752 Answers to Selected Exercises
Section 8.6 Answers, pp. 448– 452
8. 6. 1 (p. 448) (a) 1
2
∫ t
0
τ sin 2( t − τ ) dτ (b)
∫ t
0
e−2τ cos 3( t − τ ) dτ
(c) 1
2
∫ t
0
sin 2τ cos 3( t − τ ) dτ or 1
3
∫ t
0
sin 3τ cos 2( t − τ ) dτ (d)
∫ t
0
cos τ sin(t − τ ) dτ
(e)
∫ t
0
eaτ dτ (f) e−t
∫ t
0
sin(t − τ ) dτ (g) e−2t
∫ t
0
τ e τ sin(t − τ ) dτ
(h) e−2t
2
∫ t
0
τ 2(t − τ )e3τ dτ (i)
∫ t
0
(t − τ )eτ cos τ dτ (j)
∫ t
0
e−3τ cos τ cos 2(t − τ ) dτ
(k) 1
4!5!
∫ t
0
τ 4 (t − τ )5e3τ dτ (l) 1
4
∫ t
0
τ 2 eτ sin 2(t − τ ) dτ
(m) 1
2
∫ t
0
τ (t − τ )2 e2(t−τ ) dτ (n) 1
5!6!
∫ t
0
(t − τ )5e2(t−τ ) τ 6 dτ
8. 6. 2 (p. 449) (a) as
(s2 + a2 )(s2 + b2 ) (b) a
(s − 1)(s2 + a2 ) (c) as
(s2 − a2 )2 (d) 2ωs (s2 − ω 2 )
(s2 + ω 2 )4
(e) (s − 1)ω
((s − 1)2 + ω 2)2 (f) 2
(s − 2)3(s − 1)2 (g) s + 1
(s + 2) 2 [(s + 1) 2 + ω 2 ]
(h) 1
(s − 3) ((s − 1)2 − 1) (i) 2
(s − 2)2 (s2 + 4) (j) 6
s4 (s − 1) (k) 3 ·6!
s7 [(s + 1) 2 + 9]
(l) 12
s7 (m) 2 ·7!
s8 [(s + 1) 2 + 4] (n) 48
s5(s2 + 4)
8. 6. 3 (p. 449) (a) y = 2√
5",Elementary Differential Equations with Boundary Value Problems.pdf
"s4 (s − 1) (k) 3 ·6!
s7 [(s + 1) 2 + 9]
(l) 12
s7 (m) 2 ·7!
s8 [(s + 1) 2 + 4] (n) 48
s5(s2 + 4)
8. 6. 3 (p. 449) (a) y = 2√
5
∫ t
0
f (t − τ )e−3τ / 2 sinh
√
5τ
2 dτ (b) y = 1
2
∫ t
0
f (t − τ ) sin 2 τ dτ
(c) y =
∫ t
0
τ e −τ f (t − τ ) dτ (d) y(t) = − 1
k sin kt + cos kt + 1
k
∫ t
0
f (t − τ ) sin kτ dτ
(e) y = −2te−3t +
∫ t
0
τ e −3τ f (t − τ ) dτ (f) y = 3
2 sinh 2t + 1
2
∫ t
0
f (t − τ ) sinh 2 τ dτ
(g) y = e3t +
∫ t
0
(e3τ − e2τ )f (t − τ ) dτ (h) y = k1
ω sin ωt + k0 cos ωt + 1
ω
∫ t
0
f (t − τ ) sin ωτ dτ
8. 6. 4 (p. 449) (a) y = sin t(b) y = te−t (c) y = 1 + 2 tet (d) y = t + t2
2
(e) y = 4 + 5
2 t2 + 1
24 t4 (f) y = 1 − t
8. 6. 5 (p. 450) (a) 7!8!
16! t16 (b) 13!7!
21! t21 (c) 6!7!
14! t14 (d) 1
2 (e−t + sin t − cos t) (e) 1
3 (cos t − cos 2 t)
Section 8.7 Answers, pp. 460– 461
8. 7. 1 (p. 460) y = 1
2 e2t − 4e−t + 11
2 e−2t + 2 u(t − 1)(e−(t−1) − e−2(t−1))
8. 7. 2 (p. 460) y = 2 e−2t + 5 e−t + 5
3 u(t − 1)(e(t−1) − e−2(t−1))
8. 7. 3 (p. 460) y = 1
6 e2t − 2
3 e−t − 1",Elementary Differential Equations with Boundary Value Problems.pdf
"8. 7. 2 (p. 460) y = 2 e−2t + 5 e−t + 5
3 u(t − 1)(e(t−1) − e−2(t−1))
8. 7. 3 (p. 460) y = 1
6 e2t − 2
3 e−t − 1
2 e−2t + 5
2 u(t − 1) sinh 2( t − 1)
8. 7. 4 (p. 460) y = 1
8 (8 cos t − 5 sin t − sin 3 t) − 2u(t − π/ 2) cos t
8. 7. 5 (p. 460) y = 1 − cos 2 t + 1
2 sin 2 t + 1
2 u(t − 3π ) sin 2 t
8. 7. 6 (p. 460) y = 4 et + 3 e−t − 8 + 2 u(t − 2) sinh( t − 2)
8. 7. 7 (p. 460) y = 1
2 et − 7
2 e−t + 2 + 3 u(t − 6)(1 − e−(t−6))",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 753
8. 7. 8 (p. 460) y = e2t + 7 cos 2 t − sin 2 t − 1
2 u(t − π/ 2) sin 2 t
8. 7. 9 (p. 460) y = 1
2 (1 + e−2t ) + u(t − 1)(e−(t−1) − e−2(t−1))
8. 7. 10 (p. 460) y = 1
4 et + 1
4 e−t (2t − 5) + 2 u(t − 2)(t − 2)e−(t−2)
8. 7. 11 (p. 460) y = 1
6 (2 sin t + 5 sin 2 t) − 1
2 u(t − π/ 2) sin 2 t
8. 7. 12 (p. 460) y = e−t (sin t − cos t) − e−(t−π ) sin t − 3u(t − 2π )e−(t−2π ) sin t
8. 7. 13 (p. 460) y = e−2t
(
cos 3 t + 4
3 sin 3t
)
− 1
3 u(t − π/ 6)e−2(t−π/ 6) cos 3 t − 2
3 u(t − π/ 3)e−2(t−π/ 3) sin 3 t
8. 7. 14 (p. 460) y = 7
10 e2t − 6
5 e−t/ 2 − 1
2 + 1
5 u(t − 2)(e2(t−2) − e−(t−2)/ 2)
8. 7. 15 (p. 460) y = 1
17 (12 cos t + 20 sin t) + 1
34 et/ 2(10 cos t − 11 sin t) − u(t − π/ 2)e(2t−π )/ 4 cos t
+u(t − π )e(t−π )/ 2 sin t
8. 7. 16 (p. 460) y = 1
3 (cos t − cos 2t − 3 sin t) − 2u(t − π/ 2) cos t + 3 u(t − π ) sin t
8. 7. 17 (p. 460) y = et − e−t(1 + 2 t) − 5u(t − 1) sinh( t − 1) + 3 u(t − 2) sinh( t − 2)
8. 7. 18 (p. 460) y = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"8. 7. 17 (p. 460) y = et − e−t(1 + 2 t) − 5u(t − 1) sinh( t − 1) + 3 u(t − 2) sinh( t − 2)
8. 7. 18 (p. 460) y = 1
4 (et − e−t (1 + 6 t)) − u(t − 1)e−(t−1) + 2 u(t − 2)e−(t−2))
8. 7. 19 (p. 460) y = 5
3 sin t − 1
3 sin 2 t + 1
3 u(t − π )(sin 2 t + 2 sin t) + u(t − 2π ) sin t
8. 7. 20 (p. 460) y = 3
4 cos 2 t − 1
2 sin 2 t + 1
4 + 1
4 u(t − π/ 2)(1 + cos 2 t) + 1
2 u(t − π ) sin 2 t + 3
2 u(t − 3π/ 2) sin 2 t
8. 7. 21 (p. 460) y = cos t − sin t 8. 7. 22 (p. 460) y = 1
4 (8e3t − 12e−2t )
8. 7. 23 (p. 460) y = 5( e−2t − e−t ) 8. 7. 24 (p. 460) y = e−2t (1 + 6 t)
8. 7. 25 (p. 460) y = 1
4 e−t/ 2(4 − 19t)
8. 7. 29 (p. 461) y = ( −1)k mω 1Re−cτ / 2m δ (t − τ ) if ω 1τ − φ = (2 k + 1) π/ 2(k = integer)
8. 7. 30 (p. 461) (a) y = (em+1 − 1)(et−m − e−t )
2(e − 1) , m ≤ t < m + 1 , ( m = 0 , 1, . . . )
(b) y = ( m + 1) sin t, 2mπ ≤ t < 2(m + 1) π , ( m = 0 , 1, . . . )
(c) y = e2(t−m) e2m+2 − 1
e2 − 1 − e(t−m) em+1 − 1
e − 1 , m ≤ t < m + 1 (m = 0 , 1, . . . )
(d) y =
{
0, 2mπ ≤ t < (2m + 1) π,",Elementary Differential Equations with Boundary Value Problems.pdf
"(c) y = e2(t−m) e2m+2 − 1
e2 − 1 − e(t−m) em+1 − 1
e − 1 , m ≤ t < m + 1 (m = 0 , 1, . . . )
(d) y =
{
0, 2mπ ≤ t < (2m + 1) π,
− sin t, (2m + 1) π ≤ t < (2m + 2) π,
(m = 0 , 1,. . . )
Section 9.1 Answers, pp. 470– 474
9. 1. 2 (p. 471) y = 2 x2 − 3x3 + 1
x 9. 1. 3 (p. 471) y = 2 ex+3e−x −e2x+e−3x 9. 1. 4 (p. 471) yi = (x − x0)i−1
(i − 1)! , 1 ≤ i ≤ n
9. 1. 5 (p. 471) (b) y1 = − 1
2 x3 + x2 + 1
2x , y 2 = 1
3 x2 − 1
3x , y 3 = 1
4 x3 − 1
3 x2 + 1
12x
(c) y = k0y1 + k1y2 + k2 y3
9. 1. 7 (p. 471) 2e−x2
9. 1. 8 (p. 472)
√
2K cos x 9. 1. 9 (p. 472) (a) W (x) = 2 e3x (d) y = ex (c1 +c2x +c3x2)
9. 1. 10 (p. 472) (a) 2 (b) −e3x (c) 4 (d) 4/x 2 (e) 1 (f) 2x (g) 2/x 2 (h) ex(x2 − 2x + 2)
(i) −240/x 5 (j) 6e2x (2x − 1)(l) −128x
9. 1. 24 (p. 474) (a) y′′′= 0 (b) xy′′′− y′′− xy′+ y = 0 (c) (2x − 3)y′′′− 2y′′− (2x − 5)y′= 0
(d) (x2 − 2x + 2) y′′′− x2y′′+ 2 xy′− 2y = 0 (e) x3y′′′+ x2y′′− 2xy′+ 2 y = 0
(f) (3x − 1)y′′′− (12x − 1)y′′+ 9( x + 1) y′− 9y = 0
(g) x4y(4) + 5 x3 y′′′− 3x2 y′′− 6xy′+ 6 y = 0",Elementary Differential Equations with Boundary Value Problems.pdf
"754 Answers to Selected Exercises
(h) x4y(4) + 3 x2 y′′′− x2y′′+ 2 xy′− 2y = 0
(i) (2x − 1)y(4) − 4xy′′′+ (5 − 2x)y′′+ 4 xy′− 4y = 0
(j) xy(4) − y′′′− 4xy′′+ 4 y′= 0
Section 9.2 Answers, pp. 482– 487
9. 2. 1 (p. 482) y = ex(c1 + c2x + c3x2 ) 9. 2. 2 (p. 482) y = c1ex + c2e−x + c3 cos 3 x + c4 sin 3x
9. 2. 3 (p. 482) y = c1ex + c2 cos 4 x + c3 sin 4 x 9. 2. 4 (p. 482) y = c1ex + c2e−x + c3e−3x/ 2
9. 2. 5 (p. 482) y = c1e−x + e−2x (c1 cos x + c2 sin x) 9. 2. 6 (p. 482) y = c1ex + ex/ 2(c2 + c3x)
9. 2. 7 (p. 482) y = e−x/ 3(c1 + c2x + c3x2) 9. 2. 8 (p. 482) y = c1 + c2x + c3 cos x + c4 sin x
9. 2. 9 (p. 482) y = c1e2x + c2e−2x + c3 cos 2 x + c4 sin 2x
9. 2. 10 (p. 482) y = ( c1 + c2x) cos
√
6x + ( c3 + c4x) sin
√
6x
9. 2. 11 (p. 482) y = e3x/ 2 (c1 + c2x) + e−3x/ 2 (c3 + c4x)
9. 2. 12 (p. 482) y = c1e−x/ 2 + c2e−x/ 3 + c3 cos x + c4 sin x
9. 2. 13 (p. 482) y = c1ex +c2e−2x +c3e−x/ 2 +c4e−3x/ 2 9. 2. 14 (p. 482) y = ex(c1 +c2x +c3 cos x +c4 sin x)",Elementary Differential Equations with Boundary Value Problems.pdf
"9. 2. 13 (p. 482) y = c1ex +c2e−2x +c3e−x/ 2 +c4e−3x/ 2 9. 2. 14 (p. 482) y = ex(c1 +c2x +c3 cos x +c4 sin x)
9. 2. 15 (p. 483) y = cos 2 x − 2 sin 2 x + e2x 9. 2. 16 (p. 483) y = 2 ex + 3 e−x − 5e−3x
9. 2. 17 (p. 483) y = 2 ex + 3 xex − 4e−x
9. 2. 18 (p. 483) y = 2 e−x cos x − 3e−x sin x + 4 e2x 9. 2. 19 (p. 483) y = 9
5 e−5x/ 3 + ex (1 + 2 x)
9. 2. 20 (p. 483) y = e2x(1 − 3x + 2 x2 ) 9. 2. 21 (p. 483) y = e3x (2 − x) + 4 e−x/ 2
9. 2. 22 (p. 483) y = ex/ 2(1 − 2x) + 3 e−x/ 2 9. 2. 23 (p. 483) y = 1
8 (5e2x + e−2x + 10 cos 2 x + 4 sin 2 x)
9. 2. 24 (p. 483) y = −4ex + e2x − e4x + 2 e−x 9. 2. 25 (p. 483) y = 2 ex − e−x
9. 2. 26 (p. 483) y = e2x + e−2x + e−x(3 cos x + sin x) 9. 2. 27 (p. 483) y = 2 e−x/ 2 + cos 2 x − sin 2 x
9. 2. 28 (p. 483) (a) {ex , xe x, e 2x }: 1 (b) {cos 2 x, sin 2x, e 3x}: 26
(c) {e−x cos x, e −x sin x, e x }: 5 (d) {1, x, x 2, e x}2ex
(e) {ex, e −x, cos x, sin x}8 (f) {cos x, sin x, e x cos x, e x sin x}: 5",Elementary Differential Equations with Boundary Value Problems.pdf
"(c) {e−x cos x, e −x sin x, e x }: 5 (d) {1, x, x 2, e x}2ex
(e) {ex, e −x, cos x, sin x}8 (f) {cos x, sin x, e x cos x, e x sin x}: 5
9. 2. 29 (p. 483) {e−3x cos 2x, e −3x sin 2x, e 2x, xe 2x, 1, x, x 2}
9. 2. 30 (p. 483) {ex, xe x , e x/ 2, xe x/ 2, x 2ex/ 2, cos x, sin x}
9. 2. 31 (p. 483) {cos 3 x, x cos 3 x, x 2 cos 3 x, sin 3 x, x sin 3 x, x 2 sin 3x, 1, x }
9. 2. 32 (p. 483) {e2x, xe 2x, x 2 e2x, e −x, xe −x, 1}
9. 2. 33 (p. 483) {cos x, sin x, cos 3x, x cos 3 x, sin 3 x, x sin 3 x, e 2x }
9. 2. 34 (p. 483) {e2x, xe 2x, e −2x , xe −2x , cos 2 x, x cos 2x, sin 2x, x sin 2 x}
9. 2. 35 (p. 483) {e−x/ 2 cos 2x, xe −x/ 2 cos 2 x, x 2e−x/ 2 cos 2 x, e −x/ 2 sin 2x, xe −x/ 2 sin 2 x,
x2e−x/ 2 sin 2x}
9. 2. 36 (p. 483) {1, x, x 2, e 2x , xe 2x, cos 2 x, x cos 2 x, sin 2x, x sin 2x}
9. 2. 37 (p. 483) {cos(x/ 2), x cos(x/ 2), sin(x/ 2), x sin(x/ 2), cos 2 x/ 3 x cos(2x/ 3),
x2 cos(2x/ 3), sin(2x/ 3), x sin(2x/ 3), x 2 sin(2x/ 3)}",Elementary Differential Equations with Boundary Value Problems.pdf
"9. 2. 37 (p. 483) {cos(x/ 2), x cos(x/ 2), sin(x/ 2), x sin(x/ 2), cos 2 x/ 3 x cos(2x/ 3),
x2 cos(2x/ 3), sin(2x/ 3), x sin(2x/ 3), x 2 sin(2x/ 3)}
9. 2. 38 (p. 483) {e−x, e 3x , e x cos 2 x, e x sin 2 x} 9. 2. 39 (p. 484) (b) e(a1+a2+···+an)x
∏
1≤i<j≤n
(aj − ai )
9. 2. 43 (p. 486) (a)
{
ex, e −x/ 2 cos
(√
3
2 x
)
, e −x/ 2 sin
(√
3
2 x
)}
(b)
{
e−x , e x/ 2 cos
(√
3
2 x
)
, e x/ 2 sin
(√
3
2 x
)}
(c) {e2x cos 2 x, e 2x sin 2 x, e −2x cos 2 x, e −2x sin 2 x}
(d)
{
ex, e −x, e x/ 2 cos
(√
3
2 x
)
, e x/ 2 sin
(√
3
2 x
)
, e −x/ 2 cos
(√
3
2 x
)
, e −x/ 2 sin
(√
3
2 x
)}
(e) {cos 2 x, sin 2 x, e −
√
3x cos x, e −
√
3x sin x, e
√
3x cos x, e
√
3x sin x}
(f)
{
1, e 2x, e 3x/ 2 cos
(√
3
2 x
)
, e 3x/ 2 sin
(√
3
2 x
)
, e x/ 2 cos
(√
3
2 x
)
, e x/ 2 sin
(√
3
2 x
)}",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 755
(g)
{
e−x, e x/ 2 cos
(√
3
2 x
)
, e x/ 2 sin
(√
3
2 x
)
, e −x/ 2 cos
(√
3
2 x
)
, e −x/ 2 sin
(√
3
2 x
)}
9. 2. 45 (p. 487) y = c1xr1 + c2xr2 + c3xr3 (r1, r 2 , r 3 distinct); y = c1xr1 + ( c2 + c3 ln x)xr2 (r1, r 2
distinct); y = [ c1 + c2 ln x + c3(ln x)2]xr1 ; y = c1xr1 + xλ [c2 cos(ω ln x) + c3 sin(ω ln x)]
Section 9.3 Answers, pp. 494– 496
9. 3. 1 (p. 494) yp = e−x(2+x−x2 ) 9. 3. 2 (p. 494) yp = − e−3x
4 (3−x+x2 ) 9. 3. 3 (p. 494) yp = ex(1+x−x2 )
9. 3. 4 (p. 494) yp = e−2x(1 − 5x + x2). 9. 3. 5 (p. 494) yp = − xex
2 (1 − x + x2 − x3)
9. 3. 6 (p. 494) yp = x2ex(1 + x) 9. 3. 7 (p. 494) yp = xe−2x
2 (2 + x) 9. 3. 8 (p. 494) yp = x2ex
2 (2 + x)
9. 3. 9 (p. 494) yp = x2e2x
2 (1+2 x) 9. 3. 10 (p. 494) yp = x2e3x(2+ x−x2 ) 9. 3. 11 (p. 494) yp = x2e4x (2+ x)
9. 3. 12 (p. 494) yp = x3ex/ 2
48 (1 + x) 9. 3. 13 (p. 494) yp = e−x(1 − 2x + x2 ) 9. 3. 14 (p. 494) yp = e2x(1 − x)
9. 3. 15 (p. 494) yp = e−2x(1 + x + x2 − x3) 9. 3. 16 (p. 494) yp = ex",Elementary Differential Equations with Boundary Value Problems.pdf
"9. 3. 15 (p. 494) yp = e−2x(1 + x + x2 − x3) 9. 3. 16 (p. 494) yp = ex
3 (1 − x) 9. 3. 17 (p. 494) yp = ex(1 + x)2
9. 3. 18 (p. 494) yp = xex(1 + x3) 9. 3. 19 (p. 494) yp = xex(2 + x) 9. 3. 20 (p. 494) yp = xe2x
6 (1 − x2)
9. 3. 21 (p. 494) yp = 4 xe−x/ 2(1 + x) 9. 3. 22 (p. 494) yp = xex
6 (1 + x2)
9. 3. 23 (p. 494) yp = x2e2x
6 (1 + x + x2) 9. 3. 24 (p. 494) yp = x2e2x
6 (3 + x + x2) 9. 3. 25 (p. 494) yp =
x3ex
48 (2 + x)
9. 3. 26 (p. 494) yp = x3ex
6 (1 + x) 9. 3. 27 (p. 495) yp = − x3e−x
6 (1 − x + x2) 9. 3. 28 (p. 495) yp =
x3e2x
12 (2 + x − x2)
9. 3. 29 (p. 495) yp = e−x [(1 + x) cos x + (2 − x) sin x] 9. 3. 30 (p. 495) yp = e−x [(1 − x) cos 2 x + (1 + x) sin 2 x]
9. 3. 31 (p. 495) yp = e2x[(1 + x − x2) cos x + (1 + 2 x) sin x]
9. 3. 32 (p. 495) yp = ex
2 [(1 + x) cos 2 x + (1 − x + x2) sin 2 x] 9. 3. 33 (p. 495) yp = x
13 (8 cos 2 x + 14 sin 2 x)
9. 3. 34 (p. 495) yp = xex[(1 + x) cos x + (3 + x) sin x] 9. 3. 35 (p. 495) yp = xe2x
2 [(3 − x) cos 2 x + sin 2 x]",Elementary Differential Equations with Boundary Value Problems.pdf
"13 (8 cos 2 x + 14 sin 2 x)
9. 3. 34 (p. 495) yp = xex[(1 + x) cos x + (3 + x) sin x] 9. 3. 35 (p. 495) yp = xe2x
2 [(3 − x) cos 2 x + sin 2 x]
9. 3. 36 (p. 495) yp = − xe3x
12 (x cos 3 x + sin 3 x) 9. 3. 37 (p. 495) yp = − ex
10 (cos x + 7 sin x)
9. 3. 38 (p. 495) yp = ex
12 (cos 2 x − sin 2x) 9. 3. 39 (p. 495) yp = xe2x cos 2 x
9. 3. 40 (p. 495) yp = − e−x
2 [(1 + x) cos x + (2 − x) sin x] 9. 3. 41 (p. 495) yp = xe−x
10 (cos x + 2 sin x)
9. 3. 42 (p. 495) yp = xex
40 (3 cos 2 x − sin 2 x) 9. 3. 43 (p. 495) yp = xe−2x
8 [(1 − x) cos 3 x + (1 + x) sin 3 x]
9. 3. 44 (p. 495) yp = − xex
4 (1 + x) sin 2 x 9. 3. 45 (p. 495) yp = x2e−x
4 (cos x − 2 sin x)
9. 3. 46 (p. 495) yp = − x2e2x
32 (cos 2 x − sin 2x) 9. 3. 47 (p. 495) yp = x2e2x
8 (1 + x) sin x
9. 3. 48 (p. 495) yp = 2 x2ex + xe2x − cos x 9. 3. 49 (p. 495) yp = e2x + xex + 2 x cos x
9. 3. 50 (p. 495) yp = 2 x + x2 + 2 xex − 3xe−x + 4 e3x",Elementary Differential Equations with Boundary Value Problems.pdf
"9. 3. 48 (p. 495) yp = 2 x2ex + xe2x − cos x 9. 3. 49 (p. 495) yp = e2x + xex + 2 x cos x
9. 3. 50 (p. 495) yp = 2 x + x2 + 2 xex − 3xe−x + 4 e3x
9. 3. 51 (p. 495) yp = xex(cos 2 x −2 sin 2 x) + 2xe2x + 1 9. 3. 52 (p. 495) yp = x2e−2x (1+ 2x)−cos 2x + sin 2x",Elementary Differential Equations with Boundary Value Problems.pdf
"756 Answers to Selected Exercises
9. 3. 53 (p. 495) yp = 2 x2(1 + x)e−x + x cos x − 2 sin x 9. 3. 54 (p. 495) yp = 2 xex + xe−x + cos x
9. 3. 55 (p. 495) yp = xex
6 (cos x + sin 2 x) 9. 3. 56 (p. 495) yp = x2
54 [(2 + 2 x)ex + 3 e−2x]
9. 3. 57 (p. 495) yp = x
8 sinh x sin x 9. 3. 58 (p. 495) yp = x3(1 + x)e−x + xe−2x
9. 3. 59 (p. 495) yp = xex(2x2 + cos x + sin x) 9. 3. 60 (p. 495) y = e2x (1 + x) + c1e−x + ex (c2 + c3x)
9. 3. 61 (p. 495) y = e3x
(
1 − x − x2
2
)
+ c1ex + e−x(c2 cos x + c3 sin x)
9. 3. 62 (p. 496) y = xe2x(1 + x)2 + c1ex + c2e2x + c3e3x
9. 3. 63 (p. 496) y = x2e−x(1 − x)2 + c1 + e−x(c2 + c3x)
9. 3. 64 (p. 496) y = x3ex
24 (4 + x) + ex(c1 + c2x + c3x2)
9. 3. 65 (p. 496) y = x2e−x
16 (1 + 2 x − x2) + ex(c1 + c2x) + e−x(c3 + c4x)
9. 3. 66 (p. 496) y = e−2x
[(
1 + x
2
)
cos x +
(3
2 − 2x
)
sin x
]
+ c1ex + c2e−x + c3e−2x
9. 3. 67 (p. 496) y = −xex sin 2 x + c1 + c2ex + ex(c3 cos x + c4 sin x)
9. 3. 68 (p. 496) y = − x2ex",Elementary Differential Equations with Boundary Value Problems.pdf
"(3
2 − 2x
)
sin x
]
+ c1ex + c2e−x + c3e−2x
9. 3. 67 (p. 496) y = −xex sin 2 x + c1 + c2ex + ex(c3 cos x + c4 sin x)
9. 3. 68 (p. 496) y = − x2ex
16 (1 + x) cos 2 x + ex [(c1 + c2x) cos 2 x + ( c3 + c4x) sin 2 x]
9. 3. 69 (p. 496) y = ( x2 + 2) ex − e−2x + e3x 9. 3. 70 (p. 496) y = e−x(1 + x + x2) + (1 − x)ex
9. 3. 71 (p. 496) y =
(
x2
12 + 16
)
xe−x/ 2 − ex 9. 3. 72 (p. 496) y = (2 − x)(x2 + 1) e−x + cos x − sin x
9. 3. 73 (p. 496) y = (2 − x) cos x − (1 − 7x) sin x + e−2x 9. 3. 74 (p. 496) 2 + ex [(1 + x) cos x − sin x − 1]
Section 9.4 Answers, pp. 502– 505
9. 4. 1 (p. 502) yp = 2 x3 9. 4. 2 (p. 503) yp = 8
105 x7/ 2 e−x2
9. 4. 3 (p. 503) yp = x ln |x|
9. 4. 4 (p. 503) yp = − 2(x2 + 2)
x 9. 4. 5 (p. 503) yp = − xe−3x
64 9. 4. 6 (p. 503) yp = − 2x2
3
9. 4. 7 (p. 503) yp = − e−x(x + 1)
x 9. 4. 8 (p. 503) yp = 2 x2 ln |x|9. 4. 9 (p. 503) yp = x2 + 1
9. 4. 10 (p. 503) yp = 2x2 + 6
3 9. 4. 11 (p. 503) yp = x2 ln |x|
3 9. 4. 12 (p. 503) yp = −x2 − 2
9. 4. 13 (p. 503) 1
4 x3 ln |x| −25",Elementary Differential Equations with Boundary Value Problems.pdf
"9. 4. 10 (p. 503) yp = 2x2 + 6
3 9. 4. 11 (p. 503) yp = x2 ln |x|
3 9. 4. 12 (p. 503) yp = −x2 − 2
9. 4. 13 (p. 503) 1
4 x3 ln |x| −25
48 x3 9. 4. 14 (p. 503) yp = x5/ 2
4 9. 4. 15 (p. 503) yp = x(12 − x2)
6
9. 4. 16 (p. 503) yp = x4 ln |x|
6 9. 4. 17 (p. 503) yp = x3ex
2 9. 4. 18 (p. 503) yp = x2 ln |x|
9. 4. 19 (p. 503) yp = xex
2 9. 4. 20 (p. 503) yp = 3xex
2 9. 4. 21 (p. 503) yp = −x3
9. 4. 22 (p. 503) y = −x(ln x)2 + 3 x + x3 − 2x ln x 9. 4. 23 (p. 503) y = x3
2 (ln |x|)2 + x2 − x3 + 2 x3 ln |x|
9. 4. 24 (p. 503) y = − 1
2 (3x + 1) xex − 3ex − e2x + 4 xe−x 9. 4. 25 (p. 503) y = 3
2 x4(ln x)2 + 3 x − x4 + 2 x4 ln x
9. 4. 26 (p. 503) y = − x4 + 12
6 + 3 x − x2 + 2 ex 9. 4. 27 (p. 503) y =
(
x2
3 − x
2
)
ln |x|+ 4 x − 2x2
9. 4. 28 (p. 504) y = − xex(1 + 3 x)
2 + x + 1
2 − ex
4 + e3x
2 9. 4. 29 (p. 504) y = −8x + 2x2 − 2x3 + 2ex − e−x
9. 4. 30 (p. 504) y = 3 x2 ln x − 7x2 9. 4. 31 (p. 504) y = 3(4x2 + 9)
2 + x
2 − ex
2 + e−x
2 + e2x
4",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 757
9. 4. 32 (p. 504) y = x ln x + x − √x + 1
x + 1√x . 9. 4. 33 (p. 504) y = x3 ln |x|+ x − 2x3 + 1
x − 1
x2
9. 4. 35 (p. 505) yp =
∫ x
x0
e(x−t) − 3e−(x−t) + 2 e−2(x−t)
6 F (t) dt 9. 4. 36 (p. 505) yp =
∫ x
x0
(x − t)2(2x + t)
6xt3 F (t) dt
9. 4. 37 (p. 505) yp =
∫ x
x0
xe(x−t) − x2 + x(t − 1)
t4 F (t) dt 9. 4. 38 (p. 505) yp =
∫ x
x0
x2 − t(t − 2) − 2te(x−t)
2x(t − 1)2 F (t) dt
9. 4. 39 (p. 505) yp =
∫ x
x0
e2(x−t) − 2e(x−t) + 2 e−(x−t) − e−2(x−t)
12 F (t) dt
9. 4. 40 (p. 505) yp =
∫ x
x0
(x − t)3
6x F (t) dt
9. 4. 41 (p. 505) yp =
∫ x
x0
(x + t)(x − t)3
12x2t3 F (t) dt
9. 4. 42 (p. 505) yp =
∫ x
x0
e2(x−t)(1 + 2 t) + e−2(x−t) (1 − 2t) − 4x2 + 4 t2 − 2
32t2 F (t) dt
Section 10.1 Answers, pp. 514– 515
10. 1. 1 (p. 514)
Q′
1 = 2 − 1
10 Q1 + 1
25 Q2
Q′
2 = 6 + 3
50 Q1 − 1
20 Q2 .
10. 1. 2 (p. 514)
Q′
1 = 12 − 5
100 + 2 t Q1 + 1
100 + 3 t Q2
Q′
2 = 5 + 1
50 + t Q1 − 4
100 + 3 t Q2.
10. 1. 3 (p. 514) m1y′′
1 = −(c1 + c2)y′
1 + c2y′",Elementary Differential Equations with Boundary Value Problems.pdf
"Q′
1 = 12 − 5
100 + 2 t Q1 + 1
100 + 3 t Q2
Q′
2 = 5 + 1
50 + t Q1 − 4
100 + 3 t Q2.
10. 1. 3 (p. 514) m1y′′
1 = −(c1 + c2)y′
1 + c2y′
2 − (k1 + k2 )y1 + k2y2 + F1
m2y′′
2 = ( c2 − c3)y′
1 − (c2 + c3)y′
2 + c3y′
3 + ( k2 − k3)y1 − (k2 + k3 )y2 + k3y3 + F2
m3 y′′
3 = c3y′
1 + c3y′
2 − c3y′
3 + k3 y1 + k3y2 − k3 y3 + F3
10. 1. 4 (p. 515) x′′= − α
m x′+ gR2 x
(x2 + y2 + z2)3/ 2 y′′= − α
m y′+ gR2 y
(x2 + y2 + z2)3/ 2
z′′= − α
m z′+ gR2 z
(x2 + y2 + z2)3/ 2
10. 1. 5 (p. 515) (a)
x′
1 = x2
x′
2 = x3
x′
3 = f (t, x 1, y 1, y 2)
y′
1 = y2
y′
2 = g(t, y 1, y 2)
(b)
u′
1 = f (t, u 1 , v 1, v 2, w 2)
; v′
1 = v2
v′
2 = g(t, u 1, v 1, v 2, w 1)
w′
1 = w2
w′
2 = h(t, u 1, v 1, v 2, w 1, w 2)
(c)
y′
1 = y2
y′
2 = y3
y′
3 = f (t, y 1, y 2, y 3)
(d)
y′
1 = y2
y′
2 = y3
y′
3 = y4
y′
4 = f (t, y 1)
(e)
x′
1 = x2
x′
2 = f (t, x 1, y 1)
y′
1 = y2
y′
2 = g(t, x 1, y 1)
10. 1. 6 (p.
515)
x′ = x1
y′ = y1
z′ = z1
x′
1 = − gR2 x
(x2 + y2 + z2)3/ 2
y′
1 = − gR2 y
(x2 + y2 + z2)3/ 2
z′
1 = − gR2 z",Elementary Differential Equations with Boundary Value Problems.pdf
"y′
2 = g(t, x 1, y 1)
10. 1. 6 (p.
515)
x′ = x1
y′ = y1
z′ = z1
x′
1 = − gR2 x
(x2 + y2 + z2)3/ 2
y′
1 = − gR2 y
(x2 + y2 + z2)3/ 2
z′
1 = − gR2 z
(x2 + y2 + z2)3/ 2
Section 10.2 Answers, pp. 518– 521",Elementary Differential Equations with Boundary Value Problems.pdf
"758 Answers to Selected Exercises
10. 2. 1 (p. 518) (a) y′=
[
2 4
4 2
]
y (b) y′=
[
−2 −2
−5 1
]
y
(c) y′=
[
−4 −10
3 7
]
y (d) y′=
[
2 1
1 2
]
y
10. 2. 2 (p.
518) (a) y′=
[ −1 2 3
0 1 6
0 0 −2
]
y (b) y′=
[ 0 2 2
2 0 2
2 2 0
]
y
(c) y′=
[ −1 2 2
2 −1 2
2 2 −1
]
y (d) y′=
[ 3 −1 −1
−2 3 2
4 −1 −2
]
y
10. 2. 3 (p.
518) (a) y′=
[
1 1
−2 4
]
y, y(0) =
[
1
0
]
(b) y′=
[
5 3
−1 1
]
y, y(0) =
[
9
−5
]
10. 2. 4 (p.
519) (a) y′=
[ 6 4 4
−7 −2 −1
7 4 3
]
y, y(0) =
[ 3
−6
4
]
(b) y′=
[ 8 7 7
−5 −6 −9
5 7 10
]
y, y(0) =
[ 2
−4
3
]
10. 2. 5 (p.
519) (a) y′=
[
−3 2
−5 3
]
+
[
3 − 2t
6 − 3t
]
(b) y′=
[
3 1
−1 1
]
y +
[
−5et
et
]
10. 2. 10 (p. 521) (a) d
dt Y 2 = Y ′Y + Y Y ′
(b) d
dt Y n = Y ′Y n−1 + Y Y ′Y n−2 + Y 2Y ′Y n−3 + · · ·+ Y n−1Y ′=
n−1∑
r=0
Y r Y ′Y n−r−1
10. 2. 13 (p. 521) B = ( P ′+ P A)P −1.
Section 10.3 Answers, pp. 525– 529
10. 3. 2 (p. 525) y′=


0 1
− P2(x)
P0(x) − P1(x)
P0(x)

 y 10 . 3. 3 (p.
526) y′=






0 1 · · · 0
.
.
.
.
.
. . . . .
.
.
0 0 · · · 1
− Pn(x)",Elementary Differential Equations with Boundary Value Problems.pdf
"

0 1
− P2(x)
P0(x) − P1(x)
P0(x)

 y 10 . 3. 3 (p.
526) y′=






0 1 · · · 0
.
.
.
.
.
. . . . .
.
.
0 0 · · · 1
− Pn(x)
P0(x) − Pn−1(x)
P0(x) · · · −P1(x)
P0(x)






y
10. 3. 7 (p.
527) (b) y =
[
3e6t − 6e−2t
3e6t + 6 e−2t
]
(c) y = 1
2
[
e6t + e−2t e6t − e−2t
e6t − e−2t e6t + e−2t
]
k
10. 3. 8 (p. 528) (b) y =
[
6e−4t + 4 e3t
6e−4t − 10e3t
]
(c) y = 1
7
[
5e−4t + 2 e3t 2e−4t − 2e3t
5e−4t − 5e3t 2e−4t + 5 e3t
]
k
10. 3. 9 (p. 528) (b) y =
[
−15e2t − 4et
9e2t + 2 et
]
(c) y =
[
−5e2t + 6 et −10e2t + 10 et
3e2t − 3et 6e2t − 5et
]
k
10. 3. 10 (p. 528) (b) y =
[
5e3t − 3et
5e3t + 3 et
]
(c) y = 1
2
[
e3t + et e3t − et
e3t − et e3t + et
]
k
10. 3. 11 (p. 528) (b) y =
[ e2t − 2e3t + 3 e−t
2e3t − 9e−t
e2t − 2e3t + 21 e−t
]
(c) y = 1
6
[ 4e2t + 3 e3t − e−t 6e2t − 6e3t 2e2t − 3e3t + e−t
−3e3t + 3 e−t 6e3t 3e3t − 3e−t
4e2t + 3 e3t − 7e−t 6e2t − 6e3t 2e2t − 3e3t + 7 e−t
]
k",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 759
10. 3. 12 (p. 528) (b) y = 1
3
[ −e−2t + e4t
−10e−2t + e4t
11e−2t + e4t
]
(c) y = 1
3
[ 2e−2t + e4t −e−2t + e4t −e−2t + e4t
−e−2t + e4t 2e−2t + e4t −e−2t + e4t
−e−2t + e4t −e−2t + e4t 2e−2t + e4t
]
k
10. 3. 13 (p. 528) (b) y =
[ 3et + 3 e−t − e−2t
3et + 2 e−2t
−e−2t
]
(c) y =
[ e−t et − e−t 2et − 3e−t + e−2t
0 et 2et − 2e−2t
0 0 e−2t
]
k
10. 3. 14 (p. 529) Y Z −1 and ZY −1
Section 10.4 Answers, pp. 539– 541
10. 4. 1 (p. 539) y = c1
[
1
1
]
e3t + c2
[
1
−1
]
e−t 10. 4. 2 (p.
539) y = c1
[
1
1
]
e−t/ 2 + c2
[
−1
1
]
e−2t
10. 4. 3 (p.
539) y = c1
[
−3
1
]
e−t + c2
[
−1
2
]
e−2t 10. 4. 4 (p.
539) y = c1
[
2
1
]
e−3t + c2
[
−2
1
]
et
10. 4. 5 (p.
539) y = c1
[
1
1
]
e−2t + c1
[
−4
1
]
e3t 10. 4. 6 (p.
539) y = c1
[
3
2
]
e2t + c2
[
1
1
]
et
10. 4. 7 (p.
539) y = c1
[
−3
1
]
e−5t + c2
[
−1
1
]
e−3t
10. 4. 8 (p.
539) y = c1
[ 1
2
1
]
e−3t + c2
[ −1
−4
1
]
e−t + c3
[ −1
−1
1
]
e2t
10. 4. 9 (p.
539) y = c1
[ 2
1
2
]
e−16t + c2
[ −1
2
0
]
e2t + c3
[ −1
0
1
]
e2t",Elementary Differential Equations with Boundary Value Problems.pdf
"539) y = c1
[ 1
2
1
]
e−3t + c2
[ −1
−4
1
]
e−t + c3
[ −1
−1
1
]
e2t
10. 4. 9 (p.
539) y = c1
[ 2
1
2
]
e−16t + c2
[ −1
2
0
]
e2t + c3
[ −1
0
1
]
e2t
10. 4. 10 (p.
539) y = c1
[ −2
−4
3
]
et + c2
[ −1
1
0
]
e−2t + c3
[ −7
−5
4
]
e2t
10. 4. 11 (p.
539) y = c1
[ −1
−1
1
]
e−2t + c2
[ −1
−2
1
]
e−3t + c3
[ −2
−6
3
]
e−5t
10. 4. 12 (p.
539) y = c1
[ 11
7
1
]
e3t + c2
[ 1
2
1
]
e−2t + c3
[ 1
1
1
]
e−t
10. 4. 13 (p.
539) y = c1
[ 4
−1
1
]
e−4t + c2
[ −1
−1
1
]
e6t + c3
[ −1
0
1
]
e4t
10. 4. 14 (p.
539) y = c1
[ 1
1
5
]
e−5t + c2
[ −1
0
1
]
e5t + c3
[ 1
1
0
]
e5t
10. 4. 15 (p.
539) y = c1
[ 1
−1
2
]
+ c2
[ −1
0
3
]
e6t + c3
[ 1
3
0
]
e6t
10. 4. 16 (p.
540) y = −
[
2
6
]
e5t +
[
4
2
]
e−5t 10. 4. 17 (p.
540) y =
[
2
−4
]
et/ 2 +
[
−2
1
]
et
10. 4. 18 (p.
540) y =
[
7
7
]
e9t −
[
2
4
]
e−3t 10. 4. 19 (p.
540) y =
[
3
9
]
e5t −
[
4
2
]
e−5t",Elementary Differential Equations with Boundary Value Problems.pdf
"760 Answers to Selected Exercises
10. 4. 20 (p. 540) y =
[ 5
5
0
]
et/ 2 +
[ 0
0
1
]
et/ 2 +
[ −1
2
0
]
e−t/ 2 10. 4. 21 (p.
540) y =
[ 3
3
3
]
et +
[ −2
−2
2
]
e−t
10. 4. 22 (p.
540) y =
[ 2
−2
2
]
et −
[ 3
0
3
]
e−2t +
[ 1
1
0
]
e3t
10. 4. 23 (p.
540) y = −
[ 1
2
1
]
et +
[ 4
2
4
]
e−t +
[ 1
1
0
]
e2t
10. 4. 24 (p.
540) y =
[ −2
−2
2
]
e2t −
[ 0
3
0
]
e−2t +
[ 4
12
4
]
e4t
10. 4. 25 (p.
540) y =
[ −1
−1
1
]
e−6t +
[ 2
−2
2
]
e2t +
[ 7
−7
−7
]
e4t
10. 4. 26 (p.
540) y =
[ 1
4
4
]
e−t +
[ 6
6
−2
]
e2t 10. 4. 27 (p.
540) y =
[ 4
−2
2
]
+
[ 3
−9
6
]
e4t +
[ −1
1
−1
]
e2t
10. 4. 29 (p.
541) Half lines of L1 : y2 = y1 and L2 : y2 = −y1 are trajectories other trajectories
are asymptotically tangent to L1 as t → −∞ and asymptotically tangent to L2 as t → ∞ .
10. 4. 30 (p. 541) Half lines of L1 : y2 = −2y1 and L2 : y2 = −y1/ 3 are trajectories
other trajectories are asymptotically parallel to L1 as t → −∞ and asymptotically tangent to L2 as
t → ∞ .",Elementary Differential Equations with Boundary Value Problems.pdf
"other trajectories are asymptotically parallel to L1 as t → −∞ and asymptotically tangent to L2 as
t → ∞ .
10. 4. 31 (p. 541) Half lines of L1 : y2 = y1/ 3 and L2 : y2 = −y1 are trajectories other trajectories
are asymptotically tangent to L1 as t → −∞ and asymptotically parallel to L2 as t → ∞ .
10. 4. 32 (p. 541) Half lines of L1 : y2 = y1/ 2 and L2 : y2 = −y1 are trajectories other trajectories
are asymptotically tangent to L1 as t → −∞ and asymptotically tangent to L2 as t → ∞ .
10. 4. 33 (p. 541) Half lines of L1 : y2 = −y1/ 4 and L2 : y2 = −y1 are trajectories other trajectories
are asymptotically tangent to L1 as t → −∞ and asymptotically parallel to L2 as t → ∞ .
10. 4. 34 (p. 541) Half lines of L1 : y2 = −y1 and L2 : y2 = 3 y1 are trajectories other trajectories
are asymptotically parallel to L1 as t → −∞ and asymptotically tangent to L2 as t → ∞ .
10. 4. 36 (p. 541) Points on L2 : y2 = y1 are trajectories of constant solutions. The trajectories",Elementary Differential Equations with Boundary Value Problems.pdf
"10. 4. 36 (p. 541) Points on L2 : y2 = y1 are trajectories of constant solutions. The trajectories
of nonconstant solutions are half-lines on either side of L1 , parallel to
[
1
−1
]
, traversed toward L1.
10. 4. 37 (p. 541) Points on L1 : y2 = −y1/ 3 are trajectories of constant solutions. The trajectories
of nonconstant solutions are half-lines on either side of L1, parallel to
[
−1
2
]
, traversed away from
L1 .
10. 4. 38 (p.
541) Points on L1 : y2 = y1/ 3 are trajectories of constant solutions. The trajectories
of nonconstant solutions are half-lines on either side of L1, parallel to
[
1
−1
]
,
[
1
]
−1, traversed
away from L1 .",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 761
10. 4. 39 (p. 541) Points on L1 : y2 = y1/ 2 are trajectories of constant solutions. The trajectories
of nonconstant solutions are half-lines on either side of L1 , parallel to
[
1
−1
]
, L1 .
10. 4. 40 (p. 541) Points on L2 : y2 = −y1 are trajectories of constant solutions. The trajectories
of nonconstant solutions are half-lines on either side of L2 , parallel to
[
−4
1
]
, traversed toward L1.
10. 4. 41 (p.
541) Points on L1 : y2 = 3 y1 are trajectories of constant solutions. The trajectories
of nonconstant solutions are half-lines on either side of L1, parallel to
[
1
−1
]
, traversed away from
L1 .
Section 10.5 Answers, pp. 554– 556
10. 5. 1 (p. 554) y = c1
[
2
1
]
e5t + c2
([
−1
0
]
e5t +
[
2
1
]
te5t
)
.
10. 5. 2 (p.
554) y = c1
[
1
1
]
e−t + c2
([
1
0
]
e−t +
[
1
1
]
te−t
)
10. 5. 3 (p.
554) y = c1
[
−2
1
]
e−9t + c2
([
−1
0
]
e−9t +
[
−2
1
]
te−9t
)
10. 5. 4 (p.
554) y = c1
[
−1
1
]
e2t + c2
([
−1
0
]
e2t +
[
−1
1
]
te2t
)
10. 5. 5 (p.
554) c1",Elementary Differential Equations with Boundary Value Problems.pdf
"[
−2
1
]
e−9t + c2
([
−1
0
]
e−9t +
[
−2
1
]
te−9t
)
10. 5. 4 (p.
554) y = c1
[
−1
1
]
e2t + c2
([
−1
0
]
e2t +
[
−1
1
]
te2t
)
10. 5. 5 (p.
554) c1
[
−2
1
]
+ c2
([
−1
0
]
e−2t
3 +
[
−2
1
]
te−2t
)
10. 5. 6 (p.
554) y = c1
[
3
2
]
e−4t + c2
([
−1
0
]
e−4t
2 +
[
3
2
]
te−4t
)
10. 5. 7 (p.
554) y = c1
[
4
3
]
e−t + c2
([
−1
0
]
e−t
3 +
[
4
3
]
te−t
)
10. 5. 8 (p.
554) y = c1
[ −1
−1
2
]
+ c2
[ 1
1
2
]
e4t + c3
([ 0
1
0
]
e4t
2 +
[ 1
1
2
]
te4t
)
10. 5. 9 (p.
554) y = c1
[ −1
1
1
]
et + c2
[ 1
−1
1
]
e−t + c3
([ 0
3
0
]
e−t +
[ 1
−1
1
]
te−t
)
.
10. 5. 10 (p.
554) y = c1
[ 0
1
1
]
e2t + c2
[ 1
0
1
]
e−2t + c3
([ 1
1
0
]
e−2t
2 +
[ 1
0
1
]
te−2t
)
10. 5. 11 (p.
554) y = c1
[ −2
−3
1
]
e2t + c2
[ 0
−1
1
]
e4t + c3
([ 1
0
0
]
e4t
2 +
[ 0
−1
1
]
te4t
)
10. 5. 12 (p.
554) y = c1
[ −1
−1
1
]
e−2t + c2
[ 1
1
1
]
e4t + c3
([ 1
0
0
]
e4t
2 +
[ 1
1
1
]
te4t
)
.
10. 5. 13 (p.
554) y =
[
6
2
]
e−7t −
[
8
4
]
te−7t 10. 5. 14 (p.
554) y =
[
5
8
]
e3t −
[
12
16
]
te3t",Elementary Differential Equations with Boundary Value Problems.pdf
"762 Answers to Selected Exercises
10. 5. 15 (p. 554) y =
[
2
3
]
e−5t −
[
8
4
]
te−5t 10. 5. 16 (p.
554) y =
[
3
1
]
e5t −
[
12
6
]
te5t
10. 5. 17 (p.
554) y =
[
0
2
]
e−4t +
[
6
6
]
te−4t
10. 5. 18 (p.
554) y =
[ 4
8
−6
]
et +
[ 2
−3
−1
]
e−2t +
[ −1
1
0
]
te−2t
10. 5. 19 (p.
554) y =
[ 3
3
6
]
e2t −
[ 9
5
6
]
+
[ 2
2
0
]
t
10. 5. 20 (p.
554) y = −
[ 2
0
2
]
e−3t +
[ −4
9
1
]
et −
[ 0
4
4
]
tet
10. 5. 21 (p.
555) y =
[ −2
2
2
]
e4t +
[ 0
−1
1
]
e2t +
[ 3
−3
3
]
te2t
10. 5. 22 (p.
555) y = −
[ 1
1
0
]
e−4t +
[ −3
2
−3
]
e8t +
[ 8
0
−8
]
te8t
10. 5. 23 (p.
555) y =
[ 3
6
3
]
e4t −
[ 3
4
1
]
+
[ 8
4
4
]
t
10. 5. 24 (p.
555) y = c1
[ 0
1
1
]
e6t + c2
([ −1
1
0
]
e6t
4 +
[ 0
1
1
]
te6t
)
+c3
([ 1
1
0
]
e6t
8 +
[ −1
1
0
]
te6t
4 +
[ 0
1
1
]
t2e6t
2
)
10. 5. 25 (p. 555) y = c1
[ −1
1
1
]
e3t + c2
([ 1
0
0
]
e3t
2 +
[ −1
1
1
]
te3t
)
+c3
([ 1
2
0
]
e3t
36 +
[ 1
0
0
]
te3t
2 +
[ −1
1
1
]
t2e3t
2
)
10. 5. 26 (p. 555) y = c1
[ 0
−1
1
]
e−2t + c2
([ −1
1
0
]
e−2t +
[ 0
−1
1
]
te−2t
)
+c3
([ 3
−2",Elementary Differential Equations with Boundary Value Problems.pdf
"0
]
e3t
36 +
[ 1
0
0
]
te3t
2 +
[ −1
1
1
]
t2e3t
2
)
10. 5. 26 (p. 555) y = c1
[ 0
−1
1
]
e−2t + c2
([ −1
1
0
]
e−2t +
[ 0
−1
1
]
te−2t
)
+c3
([ 3
−2
0
]
e−2t
4 +
[ −1
1
0
]
te−2t +
[ 0
−1
1
]
t2 e−2t
2
)",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 763
10. 5. 27 (p. 555) y = c1
[ 0
1
1
]
e2t + c2
([ 1
1
0
]
e2t
2 +
[ 0
1
1
]
te2t
)
+c3
([ −1
1
0
]
e2t
8 +
[ 1
1
0
]
te2t
2 +
[ 0
1
1
]
t2e2t
2
)
10. 5. 28 (p. 555) y = c1
[ −2
1
2
]
e−6t + c2
(
−
[ 6
1
0
]
e−6t
6 +
[ −2
1
2
]
te−6t
)
+c3
(
−
[ 12
1
0
]
e−6t
36 −
[ 6
1
0
]
te−6t
6 +
[ −2
1
2
]
t2e−6t
2
)
.
10. 5. 29 (p. 555) y = c1
[ −4
0
1
]
e−3t + c2
[ 6
1
0
]
e−3t + c3
([ 1
0
0
]
e−3t +
[ 2
1
1
]
te−3t
)
10. 5. 30 (p.
555) y = c1
[ −1
0
1
]
e−3t + c2
[ 0
1
0
]
e−3t + c3
([ 1
0
0
]
e−3t +
[ −1
−1
1
]
te−3t
)
10. 5. 31 (p.
555) y = c1
[ 2
0
1
]
e−t + c2
[ −3
2
0
]
e−t + c3
([ 1
0
0
]
e−t
2 +
[ −1
2
1
]
te−t
)
10. 5. 32 (p.
555) y = c1
[ −1
1
0
]
e−2t + c2
[ 0
0
1
]
e−2t + c3
([ −1
0
0
]
e−2t +
[ 1
−1
1
]
te−2t
)
Section 10.6 Answers, pp.
565– 568
10. 6. 1 (p. 565) y = c1e2t
[
3 cos t + sin t
5 cos t
]
+ c2e2t
[
3 sin t − cos t
5 sin t
]
.
10. 6. 2 (p. 565) y = c1e−t
[
5 cos 2 t + sin 2 t
13 cos 2 t
]
+ c2e−t
[
5 sin 2 t − cos 2 t
13 sin 2 t
]
.",Elementary Differential Equations with Boundary Value Problems.pdf
"]
+ c2e2t
[
3 sin t − cos t
5 sin t
]
.
10. 6. 2 (p. 565) y = c1e−t
[
5 cos 2 t + sin 2 t
13 cos 2 t
]
+ c2e−t
[
5 sin 2 t − cos 2 t
13 sin 2 t
]
.
10. 6. 3 (p. 565) y = c1e3t
[
cos 2t + sin 2 t
2 cos 2 t
]
+ c2e3t
[
sin 2t − cos 2 t
2 sin 2 t
]
.
10. 6. 4 (p. 565) y = c1e2t
[
cos 3t − sin 3 t
cos 3 t
]
+ c2e2t
[
sin 3t + cos 3 t
sin 3t
]
.
10. 6. 5 (p. 566) y = c1
[ −1
−1
2
]
e−2t + c2e4t
[ cos 2t − sin 2 t
cos 2t + sin 2 t
2 cos 2 t
]
+ c3e4t
[ sin 2 t + cos 2 t
sin 2 t − cos 2 t
2 sin 2 t
]
.
10. 6. 6 (p.
566) y = c1
[ −1
−1
1
]
e−t + c2e−2t
[ cos 2t − sin 2 t
− cos 2t − sin 2 t
2 cos 2 t
]
+ c3e−2t
[ sin 2 t + cos 2 t
− sin 2t + cos 2 t
2 sin 2 t
]
10. 6. 7 (p.
566) y = c1
[ 1
1
1
]
e2t + c2et
[ − sin t
sin t
cos t
]
+ c3et
[ cos t
− cos t
sin t
]",Elementary Differential Equations with Boundary Value Problems.pdf
"764 Answers to Selected Exercises
10. 6. 8 (p. 566) y = c1
[ −1
1
1
]
et + c2e−t
[ − sin 2 t − cos 2 t
2 cos 2 t
2 cos 2 t
]
+ c3e−t
[ cos 2 t − sin 2t
2 sin 2 t
2 sin 2 t
]
10. 6. 9 (p.
566) y = c1e3t
[
cos 6t − 3 sin 6 t
5 cos 6 t
]
+ c2e3t
[
sin 6 t + 3 cos 6 t
5 sin 6 t
]
10. 6. 10 (p. 566) y = c1e2t
[
cos t − 3 sin t
2 cos t
]
+ c2e2t
[
sin t + 3 cos t
2 sin t
]
10. 6. 11 (p. 566) y = c1e2t
[
3 sin 3 t − cos 3 t
5 cos 3 t
]
+ c2e2t
[
−3 cos 3 t − sin 3t
5 sin 3 t
]
10. 6. 12 (p. 566) y = c1e2t
[
sin 4t − 8 cos 4 t
5 cos 4 t
]
+ c2e2t
[
− cos 4t − 8 sin 4 t
5 sin 4 t
]
10. 6. 13 (p. 566) y = c1
[ −1
1
1
]
e−2t + c2et
[ sin t
− cos t
cos t
]
+ c3et
[ − cos t
− sin t
sin t
]
10. 6. 14 (p.
566) y = c1
[ 2
2
1
]
e−2t + c2e2t
[ − cos 3 t − sin 3t
− sin 3 t
cos 3 t
]
+ c3e2t
[ − sin 3 t + cos 3 t
cos 3t
sin 3t
]
10. 6. 15 (p.
566) y = c1
[ 1
2
1
]
e3t + c2e6t
[ − sin 3 t
sin 3 t
cos 3 t
]
+ c3e6t
[ cos 3 t
− cos 3 t
sin 3t
]
10. 6. 16 (p.
566) y = c1
[ 1
1
1
]
et + c2et",Elementary Differential Equations with Boundary Value Problems.pdf
"566) y = c1
[ 1
2
1
]
e3t + c2e6t
[ − sin 3 t
sin 3 t
cos 3 t
]
+ c3e6t
[ cos 3 t
− cos 3 t
sin 3t
]
10. 6. 16 (p.
566) y = c1
[ 1
1
1
]
et + c2et
[ 2 cos t − 2 sin t
cos t − sin t
2 cos t
]
+ c3et
[ 2 sin t + 2 cos t
cos t + sin t
2 sin t
]
10. 6. 17 (p.
566) y = et
[
5 cos 3 t + sin 3 t
2 cos 3 t + 3 sin 3 t
]
10. 6. 18 (p. 566) y = e4t
[
5 cos 6 t + 5 sin 6 t
cos 6 t − 3 sin 6 t
]
10. 6. 19 (p. 566) y = et
[
17 cos 3 t − sin 3 t
7 cos 3 t + 3 sin 3 t
]
10. 6. 20 (p. 566) y = et/ 2
[
cos(t/ 2) + sin( t/ 2)
− cos(t/ 2) + 2 sin( t/ 2)
]
10. 6. 21 (p. 566) y =
[ 1
−1
2
]
et + e4t
[ 3 cos t + sin t
cos t − 3 sin t
4 cos t − 2 sin t
]
10. 6. 22 (p.
566) y =
[ 4
4
2
]
e8t + e2t
[ 4 cos 2 t + 8 sin 2 t
−6 sin 2 t + 2 cos 2 t
3 cos 2 t + sin 2 t
]
10. 6. 23 (p.
566) y =
[ 0
3
3
]
e−4t + e4t
[ 15 cos 6 t + 10 sin 6 t
14 cos 6 t − 8 sin 6 t
7 cos 6 t − 4 sin 6 t
]
10. 6. 24 (p.
566) y =
[ 6
−3
3
]
e8t +
[ 10 cos 4 t − 4 sin 4 t
17 cos 4 t − sin 4 t
3 cos 4 t − 7 sin 4 t
]
10. 6. 29 (p.",Elementary Differential Equations with Boundary Value Problems.pdf
"7 cos 6 t − 4 sin 6 t
]
10. 6. 24 (p.
566) y =
[ 6
−3
3
]
e8t +
[ 10 cos 4 t − 4 sin 4 t
17 cos 4 t − sin 4 t
3 cos 4 t − 7 sin 4 t
]
10. 6. 29 (p.
567) U = 1√
2
[
−1
1
]
, V = 1√
2
[
1
1
]
10. 6. 30 (p.
567) U ≈
[
. 5257
. 8507
]
, V ≈
[
−. 8507
. 5257
]",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 765
10. 6. 31 (p. 567) U ≈
[
. 8507
. 5257
]
,
V ≈
[
−. 5257
. 8507
]
10. 6. 32 (p. 567) U ≈
[
−. 9732
. 2298
]
, V ≈
[
. 2298
. 9732
]
10. 6. 33 (p. 567) U ≈
[
. 5257
. 8507
]
, V ≈
[
−. 8507
. 5257
]
10. 6. 34 (p. 567) U ≈
[
−. 5257
. 8507
]
, V ≈
[
. 8507
. 5257
]
10. 6. 35 (p. 568) U ≈
[
−. 8817
. 4719
]
, V ≈
[
. 4719
. 8817
]
10. 6. 36 (p. 568) U ≈
[
. 8817
. 4719
]
, V ≈
[
−. 4719
. 8817
]
10. 6. 37 (p. 568) U =
[
0
1
]
, V =
[
−1
0
]
10. 6. 38 (p.
568) U =
[
0
1
]
, V =
[
1
0
]
10. 6. 39 (p.
568) U = 1√
2
[
1
1
]
, V = 1
√
2
[
−1
1
]
10. 6. 40 (p.
568) U ≈
[
. 5257
. 8507
]
, V ≈
[
−. 8507
. 5257
]
Section 10.7 Answers, pp. 575– 577
10. 7. 1 (p. 575)
[
5e4t + e−3t (2 + 8 t)
−e4t − e−3t (1 − 4t)
]
10. 7. 2 (p. 575)
[
13e3t + 3 e−3t
−e3t − 11e−3t
]
10. 7. 3 (p. 575) 1
9
[
7 − 6t
−11 + 3 t
]
10. 7. 4 (p. 575)
[
5 − 3et
−6 + 5 et
]
10. 7. 5 (p. 575)
[
e−5t (3 + 6 t) + e−3t (3 − 2t)
−e−5t (3 + 2 t) − e−3t (1 − 2t)
]
10. 7. 6 (p. 575)
[
t
0
]",Elementary Differential Equations with Boundary Value Problems.pdf
"[
5 − 3et
−6 + 5 et
]
10. 7. 5 (p. 575)
[
e−5t (3 + 6 t) + e−3t (3 − 2t)
−e−5t (3 + 2 t) − e−3t (1 − 2t)
]
10. 7. 6 (p. 575)
[
t
0
]
10. 7. 7 (p. 575) − 1
6
[ 2 − 6t
7 + 6 t
1 − 12t
]
10. 7. 8 (p. 575) − 1
6
[ 3et + 4
6et − 4
10
]
10. 7. 9 (p. 575) 1
18
[ et(1 + 12 t) − e−5t (1 + 6 t)
−2et (1 − 6t) − e−5t (1 − 12t)
et(1 + 12 t) − e−5t (1 + 6 t)
]
10. 7. 10 (p. 575) 1
3
[ 2et
et
2et
]
10. 7. 11 (p. 575)
[
t sin t
0
]
10. 7. 12 (p. 575) −
[
t2
2t
]
10. 7. 13 (p. 575) (t − 1) (ln |t − 1|+ t)
[
1
−1
]
10. 7. 14 (p. 575) 1
9
[
5e2t − e−3t
e3t − 5e−2t
]
10. 7. 15 (p. 576) 1
4t
[
2t3 ln |t|+ t3(t + 2)
2 ln |t|+ 3 t − 2
]
10. 7. 16 (p. 576) 1
2
[
te−t(t + 2) + ( t3 − 2)
tet(t − 2) + ( t3 + 2)
]
10. 7. 17 (p. 576) −
[ t
t
t
]
10. 7. 18 (p.
576) 1
4
[ −3et
1
e−t
]
10. 7. 19 (p. 576)
[ 2t2 + t
t
−t
]
10. 7. 20 (p. 576) et
4t
[ 2t + 1
2t − 1
2t + 1
]
10. 7. 22 (p. 576) (a) y′=






0 1 · · · 0
0 0 · · · 0
.
.
.
.
.
. . . . .
.
.
0 0 · · · 1
−Pn(t)/P 0(t) −Pn−1/P 0(t) · · · −P1(t)/P 0(t)
",Elementary Differential Equations with Boundary Value Problems.pdf
"]
10. 7. 22 (p. 576) (a) y′=






0 1 · · · 0
0 0 · · · 0
.
.
.
.
.
. . . . .
.
.
0 0 · · · 1
−Pn(t)/P 0(t) −Pn−1/P 0(t) · · · −P1(t)/P 0(t)






y +




0
0
.
.
.
F (t)/P 0(t)



 .",Elementary Differential Equations with Boundary Value Problems.pdf
"766 Answers to Selected Exercises
(b)




y1 y2 · · · yn
y′
1 y′
2 · · · y′
n
.
.
.
.
.
. . . . .
.
.
y(n−1)
1 y(n−1)
2 · · ·y(n−1)
n




Section 11.1 Answers, pp.
585– 586
11. 1. 2 (p. 585) λ n = n2 , yn = sin nx, n = 1 , 2, 3, . . .
11. 1. 3 (p. 585) λ 0 = 0 , y0 = 1 ; λ n = n2 , yn = cos nx, n = 1 , 2, 3, . . .
11. 1. 4 (p. 585) λ n = (2n − 1)2
4 , yn = sin (2n − 1)x
2 , n = 1 , 2, 3, . . . ,
11. 1. 5 (p. 585) λ n = (2n − 1)2
4 , yn = cos (2n − 1)x
2 , n = 1 , 2, 3, . . .
11. 1. 6 (p. 585) λ 0 = 0 , y0 = 1 , λ n = n2 , y1n = cos nx, y2n = sin nx, n = 1 , 2, 3, . . .
11. 1. 7 (p. 585) λ n = n2 π 2, yn = cos nπx , n = 1 , 2, 3, . . .
11. 1. 8 (p. 585) λ n = (2n − 1)2π 2
4 , yn = cos (2n − 1)πx
2 , n = 1 , 2, 3, . . .
11. 1. 9 (p. 585) λ n = n2 π 2, yn = sin nπx , n = 1 , 2, 3, . . .
11. 1. 10 (p. 585) λ 0 = 0 , y0 = 1 , λ n = n2π 2 , y1n = cos nπx , y2n = sin nπx , n = 1 , 2, 3, . . .
11. 1. 11 (p. 585) λ n = (2n − 1)2π 2
4 , yn = sin (2n − 1)πx
2 , n = 1 , 2, 3, . . .",Elementary Differential Equations with Boundary Value Problems.pdf
"11. 1. 11 (p. 585) λ n = (2n − 1)2π 2
4 , yn = sin (2n − 1)πx
2 , n = 1 , 2, 3, . . .
11. 1. 12 (p. 585) λ 0 = 0 , y0 = 1 , λ n = n2 π 2
4 , y1n = cos nπx
2 , y2n = sin nπx
2 , n = 1 , 2, 3, . . .
11. 1. 13 (p. 585) λ n = n2 π 2
4 , yn = sin nπx
2 , n = 1 , 2, 3, . . .
11. 1. 14 (p. 585) λ n = (2n − 1)2π 2
36 , yn = cos (2n − 1)πx
6 , n = 1 , 2, 3, . . .
11. 1. 15 (p. 585) λ n = (2 n − 1)2π 2 , yn = sin(2 n − 1)πx , n = 1 , 2, 3, . . .
11. 1. 16 (p. 585) λ n = n2 π 2
25 , yn = cos nπx
5 , n = 1 , 2, 3, . . .
11. 1. 23 (p. 586) λ n = 4 n2 π 2 /L 2 yn = sin 2nπx
L , n = 1 , 2, 3, . . .
11. 1. 24 (p. 586) λ n = n2 π 2 /L 2 yn = cos nπx
L , n = 1 , 2, 3, . . .
11. 1. 25 (p. 586) λ n = 4 n2 π 2 /L 2 yn = sin 2nπx
L , n = 1 , 2, 3, . . .
11. 1. 26 (p. 586) λ n = n2 π 2 /L 2 yn = cos nπx
L , n = 1 , 2, 3, . . . .
Section 11.2 Answers, pp. 598– 602
11. 2. 2 (p. 598) F (x) = 2 + 2
π
∞∑
n=1
(−1)n
n sin nπx ; F (x) =
{ 2, x = −1,
2 − x, −1 < x < 1,
2, x = 1
11. 2. 3 (p. 598) F (x) = −π 2 − 12
∞∑",Elementary Differential Equations with Boundary Value Problems.pdf
"11. 2. 2 (p. 598) F (x) = 2 + 2
π
∞∑
n=1
(−1)n
n sin nπx ; F (x) =
{ 2, x = −1,
2 − x, −1 < x < 1,
2, x = 1
11. 2. 3 (p. 598) F (x) = −π 2 − 12
∞∑
n=1
(−1)n
n2 cos nx − 4
∞∑
n=1
(−1)n
n sin nx;
F (x) =
{ −3π 2, x = −π,
2x − 3x2 , −π < x < π,
−3π 2, x = π
11. 2. 4 (p. 598) F (x) = − 12
π 2
∞∑
n=1
(−1)n cos nπx
n2 ; F (x) = 1 − 3x2 −1 ≤ x ≤ 1",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 767
11. 2. 5 (p. 598) F (x) = 2
π − 4
π
∞∑
n=1
1
4n2 − 1 cos 2nx; F (x) = |sin x|, −π ≤ x ≤ π
11. 2. 6 (p. 599) F (x) = − 1
2 sin x + 2
∞∑
n=2
(−1)n n
n2 − 1 sin nx;; F (x) = x cos x, −π ≤ x ≤ π
11. 2. 7 (p. 599) F (x) = − 2
π + π
2 cos x − 4
π
∞∑
n=1
4n2 + 1
(4n2 − 1)2 cos 2 nx;
F (x) = |x|cos x, −π ≤ x ≤ π
11. 2. 8 (p. 599) F (x) = 1 − 1
2 cos x − 2
∞∑
n=2
(−1)n
n2 − 1 cos nx; F (x) = x sin x, −π ≤ x ≤ π
11. 2. 9 (p. 599) F (x) = π
2 sin x − 16
π
∞∑
n=1
n
(4n2 − 1)2 sin 2nx; F (x) = |x|sin x, −π ≤ x ≤ π
11. 2. 10 (p. 599) F (x) = 1
π + 1
2 cos πx − 2
π
∞∑
n=1
(−1)n
4n2 − 1 cos 2 nπx ; F (x) = f (x), −1 ≤ x ≤ 1
11. 2. 11 (p. 599) F (x) = 1
4π sin πx − 8
π 2
∞∑
n=1
(−1)n n
(4n2 − 1)2 sin 2 nπx ;
− 1
4π
∞∑
n=1
(−1)n
n(n + 1) sin(2n + 1) πx F (x) = f (x), −1 ≤ x ≤ 1
11. 2. 12 (p. 599) F (x) = 1
2 sin πx − 4
π
∞∑
n=1
(−1)n n
4n2 − 1 sin 2 nπx ; F (x) =









0, −1 ≤ x < 1
2 ,
− 1
2 , x = − 1
2 ,
sin πx, − 1
2 < x < 1
2 ,
1
2 , x = 1
2 ,
0, 1",Elementary Differential Equations with Boundary Value Problems.pdf
"π
∞∑
n=1
(−1)n n
4n2 − 1 sin 2 nπx ; F (x) =









0, −1 ≤ x < 1
2 ,
− 1
2 , x = − 1
2 ,
sin πx, − 1
2 < x < 1
2 ,
1
2 , x = 1
2 ,
0, 1
2 < x ≤ 1
11. 2. 13 (p. 599) F (x) = 1
π + 1
π cos πx − 2
π
∞∑
n=2
1
n2 − 1
(
1 − n sin nπ
2
)
cos nπx ;
F (x) =









0, −1 ≤ x < 1
2 ,
1
2 , x = −1,
|sin πx |, − 1
2 < x < 1
2 ,
1
2 , x = 1 ,
0, 1
2 < x ≤ 1
11. 2. 14 (p. 599) F (x) = 1
π 2 + 1
4π cos πx + 2
π 2
∞∑
n=1
(−1)n 4n2 + 1
(4n2 − 1)2 cos 2 nπx
+ 1
4π
∞∑
n=1
(−1)n 2n + 1
n(n + 1) cos(2n + 1) πx ;
F (x) =









0, −1 ≤ x < 1
2 ,
1
4 , x = − 1
2 ,
x sin πx, − 1
2 < x < 1
2 ,
1
4 , x = 1
2 ,
0, 1
2 < x ≤ 1,
11. 2. 15 (p. 599) F (x) = 1 − 8
π 2
∞∑
n=0
1
(2n + 1) 2 cos (2n + 1) πx
4 − 4
π
∞∑
n=1
(−1)n
n sin nπx
4 ;",Elementary Differential Equations with Boundary Value Problems.pdf
"768 Answers to Selected Exercises
F (x) =





2, x = −4,
0, −4 < x < 0,
x, 0 ≤ x < 4,
2, x = 4
11. 2. 16 (p.
599) F (x) = 1
2 + 1
π
∞∑
n=1
1
n sin 2nπx + 8
π 3
∞∑
n=0
1
(2n + 1) 3 sin(2n + 1) πx ;
F (x) =









1
2 , x = −1,
x2 , −1 < x < 0,
1
2 , x = 0 ,
1 − x2, 0 < x < 1,
1
2 , x = 1
11. 2. 17 (p. 599) F (x) = 3
4 + 1
π
∞∑
n=1
1
n sin nπ
2 cos nπx
2 + 3
π
∞∑
n=1
1
n
(
cos nπ − cos nπ
2
)
sin nπx
2
11. 2. 18 (p. 599) F (x) = 5
2 + 3
π
∞∑
n=1
1
n sin 2nπ
3 cos nπx
3 + 1
π
∞∑
n=1
1
n
(
cos nπ − cos 2nπ
3
)
sin nπx
3
11. 2. 20 (p. 599) F (x) = sinh π
π
(
1 + 2
∞∑
n=1
(−1)n
n2 + 1 cos nx − 2
∞∑
n=1
(−1)n n
n2 + 1 sin nx
)
11. 2. 21 (p. 599) F (x) = −π cos x − 1
2 sin x + 2
∞∑
n=2
(−1)n n
n2 − 1 sin nx
11. 2. 22 (p. 600) F (x) = 1 − 1
2 cos x − π sin x − 2
∞∑
n=2
(−1)n
n2 − 1 cos nx
11. 2. 23 (p. 600) F (x) = − 2 sin kπ
π
∞∑
n=1
(−1)n n
n2 − k2 sin nx
11. 2. 24 (p. 600) F (x) = sin kπ
π
[
1
k − 2k
∞∑
n=1
(−1)n
n2 − k2 cos nx
]
Section 11.3 Answers, pp. 613– 616",Elementary Differential Equations with Boundary Value Problems.pdf
"π
∞∑
n=1
(−1)n n
n2 − k2 sin nx
11. 2. 24 (p. 600) F (x) = sin kπ
π
[
1
k − 2k
∞∑
n=1
(−1)n
n2 − k2 cos nx
]
Section 11.3 Answers, pp. 613– 616
11. 3. 1 (p. 613) C(x) = L2
3 + 4L2
π 2
∞∑
n=1
(−1)n
n2 cos nπx
L
11. 3. 2 (p. 613) C(x) = 1
2 + 4
π 2
∞∑
n=1
1
(2n − 1)2 cos(2n − 1)πx
11. 3. 3 (p. 613) C(x) = − 2L2
3 + 4L2
π 2
∞∑
n=1
1
n2 cos nπx
L
11. 3. 4 (p. 613) C(x) = 1 − cos kπ
kπ − 2k
π
∞∑
n=1
[1 − (−1)n cos kπ ]
n2 − k2 cos nx.
11. 3. 5 (p. 613) C(x) = 1
2 − 2
π
∞∑
n=1
(−1)n
2n − 1 cos (2n − 1)πx
L
11. 3. 6 (p. 613) C(x) = − 2L2
3 + 4L2
π 2
∞∑
n=1
(−1)n
n2 cos nπx
L",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 769
11. 3. 7 (p. 613) C(x) = 1
3 + 4
π 2
∞∑
n=1
1
n2 cos nπx
11. 3. 8 (p. 613) C(x) = eπ − 1
π + 2
π
∞∑
n=1
[(−1)neπ − 1]
(n2 + 1) cos nx
11. 3. 9 (p. 613) C(x) = L2
6 − L2
π 2
∞∑
n=1
1
n2 cos 2nπx
L
11. 3. 10 (p. 613) C(x) = − 2L2
3 + 4L2
π 2
∞∑
n=1
1
n2 cos nπx
L
11. 3. 11 (p. 614) S(x) = 4
π
∞∑
n=1
1
(2n − 1) sin (2n − 1)πx
L
11. 3. 12 (p. 614) S(x) = 2
π
∞∑
n=1
1
n sin nπx
11. 3. 13 (p. 614) S(x) = 2
π
∞∑
n=1
[1 − (−1)n cos kπ ] n
n2 − k2 sin nx
11. 3. 14 (p. 614) S(x) = 2
π
∞∑
n=1
1
n
[
1 − cos nπ
2
]
sin nπx
L
11. 3. 15 (p. 614) S(x) = 4L
π 2
∞∑
n=1
(−1)n+1
(2n − 1)2 sin (2n − 1)πx
L
11. 3. 16 (p. 614) S(x) = π
2 sin x − 16
π
∞∑
n=1
n
(4n2 − 1)2 sin 2 nx
11. 3. 17 (p. 614) S(x) = − 2
π
∞∑
n=1
n[(−1)neπ − 1]
(n2 + 1) sin nx
11. 3. 18 (p. 614) CM (x) = − 4
π
∞∑
n=1
(−1)n
2n − 1 cos (2n − 1)πx
2L
11. 3. 19 (p. 614) CM (x) = − 4L2
π
∞∑
n=1
(−1)n
2n − 1
[
1 − 8
(2n − 1)2π 2
]
cos (2n − 1)πx
2L
11. 3. 20 (p. 614) CM (x) = − 4
π
∞∑
n=1
[
(−1)n + 2
(2n − 1)π",Elementary Differential Equations with Boundary Value Problems.pdf
"π
∞∑
n=1
(−1)n
2n − 1
[
1 − 8
(2n − 1)2π 2
]
cos (2n − 1)πx
2L
11. 3. 20 (p. 614) CM (x) = − 4
π
∞∑
n=1
[
(−1)n + 2
(2n − 1)π
]
cos (2n − 1)πx
2 .
11. 3. 21 (p. 614) CM (x) = − 4
π
∞∑
n=1
1
2n − 1 cos (2n + 1) π
4 cos (2n − 1)πx
2L
11. 3. 22 (p. 614) CM (x) = 4
π
∞∑
n=1
(−1)n 2n − 1
(2n − 3)(2n + 1) cos (2n − 1)x
2
11. 3. 23 (p. 614) CM (x) = − 8
π
∞∑
n=1
1
(2n − 3)(2n + 1) cos (2n − 1)x
2
11. 3. 24 (p. 614) CM (x) = − 8L2
π 2
∞∑
n=1
1
(2n − 1)2
[
1 + 4(−1)n
(2n − 1)π
]
cos (2n − 1)πx
2L",Elementary Differential Equations with Boundary Value Problems.pdf
"770 Answers to Selected Exercises
11. 3. 25 (p. 614) SM (x) = 4
π
∞∑
n=1
1
(2n − 1) sin (2n − 1)πx
2L
11. 3. 26 (p. 614) SM (x) = − 16L2
π 2
∞∑
n=1
1
(2n − 1)2
[
(−1)n + 2
(2n − 1)π
]
sin (2n − 1)πx
2L
11. 3. 27 (p. 614) SM (x) = 4
π
∞∑
n=1
1
2n − 1
[
1 − cos (2n − 1)π )
4
]
sin (2n − 1)πx
2L
11. 3. 28 (p. 614) SM (x) = 4
π
∞∑
n=1
2n − 1
(2n − 3)(2n + 1) sin (2n − 1)x
2
11. 3. 29 (p. 614) SM (x) = 8
π
∞∑
n=1
(−1)n
(2n − 3)(2n + 1) sin (2n − 1)x
2
11. 3. 30 (p. 614) SM (x) = 8L2
π 2
∞∑
n=1
1
(2n − 1)2
[
(−1)n + 4
(2n − 1)π
]
sin (2n − 1)πx
2L
11. 3. 31 (p. 614) C(x) = − 7L4
5 − 144L4
π 4
∞∑
n=1
(−1)n
n4 cos nπx
L
11. 3. 32 (p. 614) C(x) = − 2L4
5 − 48L4
π 4
∞∑
n=1
1 + ( −1)n2
n4 cos nπx
L
11. 3. 33 (p. 614) C(x) = 3L4
5 − 48L4
π 4
∞∑
n=1
2 + ( −1)n
n4 cos nπx
L
11. 3. 34 (p. 614) C(x) = L4
30 − 3L4
π 4
∞∑
n=1
1
n4 cos 2nπx
L
11. 3. 36 (p. 615) S(x) = 8L2
π 3
∞∑
n=1
1
(2n − 1)3 sin (2n − 1)πx
L
11. 3. 37 (p. 615) S(x) = − 4L3
π 3
∞∑
n=1
(1 + ( −1)n2)
n3 sin nπx
L",Elementary Differential Equations with Boundary Value Problems.pdf
"L
11. 3. 36 (p. 615) S(x) = 8L2
π 3
∞∑
n=1
1
(2n − 1)3 sin (2n − 1)πx
L
11. 3. 37 (p. 615) S(x) = − 4L3
π 3
∞∑
n=1
(1 + ( −1)n2)
n3 sin nπx
L
11. 3. 38 (p. 615) S(x) = − 12L3
π 3
∞∑
n=1
(−1)n
n3 sin nπx
L
11. 3. 39 (p. 615) S(x) = 96L4
π 5
∞∑
n=1
1
(2n − 1)5 sin (2n − 1)πx
L
11. 3. 40 (p. 615) S(x) = − 720L5
π 5
∞∑
n=1
(−1)n
n5 sin nπx
L
11. 3. 41 (p. 615) S(x) = − 240L5
π 5
∞∑
n=1
1 + ( −1)n2
n5 sin nπx
L
11. 3. 43 (p. 615) CM (x) = − 64L3
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n + 3
(2n − 1)π
]
cos (2n − 1)πx
2L
11. 3. 44 (p. 615) CM (x) = − 32L2
π 3
∞∑
n=1
(−1)n
(2n − 1)3 cos (2n − 1)πx
2L",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 771
11. 3. 45 (p. 615) CM (x) = − 96L3
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n + 2
(2n − 1)π
]
cos (2n − 1)πx
2L
11. 3. 46 (p. 615) CM (x) = 96L3
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n3 + 4
(2n − 1)π
]
cos (2n − 1)πx
2L
11. 3. 47 (p. 615) CM (x) = 96L3
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n5 + 8
(2n − 1)π
]
cos (2n − 1)πx
2L
11. 3. 48 (p. 615) CM (x) = − 384L4
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n 4
(2n − 1)π
]
cos (2n − 1)πx
2L
11. 3. 49 (p. 615) CM (x) = − 768L4
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n 2
(2n − 1)π
]
cos (2n − 1)πx
2L
11. 3. 51 (p. 615) SM (x) = 32L2
π 3
∞∑
n=1
1
(2n − 1)3 sin (2n − 1)πx
2L
11. 3. 52 (p. 615) SM (x) = − 96L3
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 4
(2n − 1)π
]
sin (2n − 1)πx
2L
11. 3. 53 (p. 616) SM (x) = 96L3
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 2
(2n − 1)π
]
sin (2n − 1)πx
2L
11. 3. 54 (p. 616) SM (x) = 192L3
π 4
∞∑
n=1
(−1)n
(2n − 1)4 sin (2n − 1)πx
2L
11. 3. 55 (p. 616) SM (x) = 1536L4
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 3
(2n − 1)π
]
sin (2n − 1)πx
2L",Elementary Differential Equations with Boundary Value Problems.pdf
"π 4
∞∑
n=1
(−1)n
(2n − 1)4 sin (2n − 1)πx
2L
11. 3. 55 (p. 616) SM (x) = 1536L4
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 3
(2n − 1)π
]
sin (2n − 1)πx
2L
11. 3. 56 (p. 616) SM (x) = 384L4
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 4
(2n − 1)π
]
sin (2n − 1)πx
2L
Section 12.1 Answers, pp. 626– 629
12. 1. 8 (p. 626) u(x, t ) = 8
π 3
∞∑
n=1
1
(2n − 1)3 e−(2n−1)2 π 2t sin(2n − 1)πx
12. 1. 9 (p. 626) u(x, t ) = 4
π
∞∑
n=1
1
(2n − 1) e−9(2n−1)2 π 2t/ 16 sin (2n − 1)πx
4
12. 1. 10 (p. 626) u(x, t ) = π
2 e−3t sin x − 16
π
∞∑
n=1
n
(4n2 − 1)2 e−12n2 t sin 2 nx
12. 1. 11 (p. 626) u(x, t ) = − 32
π 3
∞∑
n=1
(1 + ( −1)n2)
n3 e−9n2π 2 t/ 4 sin nπx
2
12. 1. 12 (p. 626) u(x, t ) = − 324
π 3
∞∑
n=1
(−1)n
n3 e−4n2π 2t/ 9 sin nπx
3
12. 1. 13 (p. 626) u(x, t ) = 8
π 2
∞∑
n=1
(−1)n+1
(2n − 1)2 e−(2n−1)2 π 2t sin (2n − 1)πx
2
12. 1. 14 (p. 626) u(x, t ) = − 720
π 5
∞∑
n=1
(−1)n
n5 e−7n2π 2t sin nπx",Elementary Differential Equations with Boundary Value Problems.pdf
"772 Answers to Selected Exercises
12. 1. 15 (p. 626) u(x, t ) = 96
π 5
∞∑
n=1
1
(2n − 1)5 e−5(2n−1)2 π 2t sin(2n − 1)πx
12. 1. 16 (p. 626) u(x, t ) = − 240
π 5
∞∑
n=1
1 + ( −1)n2
n5 e−2n2 π 2t sin nπx .
12. 1. 17 (p. 627) u(x, t ) = 16
3 + 64
π 2
∞∑
n=1
(−1)n
n2 e−9π 2n2 t/ 16 cos nπx
4
12. 1. 18 (p. 627) u(x, t ) = − 8
3 + 16
π 2
∞∑
n=1
1
n2 e−n2π 2 t cos nπx
2
12. 1. 19 (p. 627) u(x, t ) = 1
6 − 1
π 2
∞∑
n=1
1
n2 e−36n2 π 2t cos 2nπx
12. 1. 20 (p. 627) u(x, t ) = 4 − 384
π 4
∞∑
n=1
1
(2n − 1)4 e−3(2n−1)2π 2 t/ 4 cos (2n − 1)πx
2
12. 1. 21 (p. 627) u(x, y ) = − 28
5 − 576
π 4
∞∑
n=1
(−1)n
n4 e−5n2 π 2t/ 2 cos nπx√
2
12. 1. 22 (p. 627) u(x, t ) = − 2
5 − 48
π 4
∞∑
n=1
1 + ( −1)n2
n4 e−3n2 π 2t cos nπx
12. 1. 23 (p. 627) u(x, t ) = 3
5 − 48
π 4
∞∑
n=1
2 + ( −1)n
n4 e−n2π 2 t cos nπx
12. 1. 24 (p. 627) u(x, t ) = π 4
30 − 3
∞∑
n=1
1
n4 e−4n2 t cos 2 nx
12. 1. 25 (p. 627) u(x, t ) = 8
π
∞∑
n=1
(−1)n
(2n + 1)(2 n − 3) e−(2n−1)2 π 2t/ 4 sin (2n − 1)πx
2
12. 1. 26 (p. 627) u(x, t ) = 8
∞∑",Elementary Differential Equations with Boundary Value Problems.pdf
"12. 1. 25 (p. 627) u(x, t ) = 8
π
∞∑
n=1
(−1)n
(2n + 1)(2 n − 3) e−(2n−1)2 π 2t/ 4 sin (2n − 1)πx
2
12. 1. 26 (p. 627) u(x, t ) = 8
∞∑
n=1
1
(2n − 1)2
[
(−1)n + 4
(2n − 1)π
]
e−3(2n−1)2 t/ 4 sin (2n − 1)x
2
12. 1. 27 (p. 627) u(x, t ) = 128
π 3
∞∑
n=1
1
(2n − 1)3 e−5(2n−1)2 t/ 16 sin (2n − 1)πx
4
12. 1. 28 (p. 627) u(x, t ) = − 96
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 4
(2n − 1)π
]
e−(2n−1)2π 2 t/ 4 sin (2n − 1)πx
2
12. 1. 29 (p. 627) u(x, t ) = 96
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 2
(2n − 1)π
]
e−(2n−1)2 π 2t/ 4 sin (2n − 1)πx
2
12. 1. 30 (p. 627) u(x, t ) = 192
π 4
∞∑
n=1
(−1)n
(2n − 1)4 e−(2n−1)2 π 2t/ 4 sin (2n − 1)πx
2
12. 1. 31 (p. 627) u(x, t ) = 1536
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 3
(2n − 1)π
]
e−(2n−1)2 π 2t/ 4 sin (2n − 1)πx
2
12. 1. 32 (p. 628) u(x, t ) = 384
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 4
(2n − 1)π
]
e−(2n−1)2π 2 t/ 4 sin (2n − 1)πx
2",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 773
12. 1. 33 (p. 628) u(x, t ) = −64
∞∑
n=1
e−3(2n−1)2 t/ 4
(2n − 1)3
[
(−1)n + 3
(2n − 1)π
]
cos (2n − 1)x
2
12. 1. 34 (p. 628) u(x, t ) = − 16
π
∞∑
n=1
(−1)n
2n − 1 e−(2n−1)2 t cos (2n − 1)x
4
12. 1. 35 (p. 628) u(x, t ) = − 64
π
∞∑
n=1
(−1)n
2n − 1
[
1 − 8
(2n − 1)2π 2
]
e−9(2n−1)2 π 2t/ 64 cos (2n − 1)πx
8
12. 1. 36 (p. 628) u(x, t ) = 8
π 2
∞∑
n=1
1
(2n − 1)2 e−3(2n−1)2 π 2t/ 4 cos (2n − 1)πx
2
12. 1. 37 (p. 628) u(x, t ) = − 96
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n + 2
(2n − 1)π
]
e−(2n−1)2π 2t/ 4 cos (2n − 1)πx
2
12. 1. 38 (p. 628) u(x, t ) = − 32
π
∞∑
n=1
(−1)n
(2n − 1)3 e−7(2n−1)2 t/ 4 cos (2n − 1)x
2
12. 1. 39 (p. 628) u(x, t ) = 96
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n5 + 8
(2n − 1)π
]
e−(2n−1)2 π 2t/ 4 cos (2n − 1)πx
2
12. 1. 40 (p. 628) u(x, t ) = 96
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n3 + 4
(2n − 1)π
]
e−(2n−1)2 π 2t/ 4 cos (2n − 1)πx
2
12. 1. 41 (p. 628) u(x, t ) = − 768
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n2
(2n − 1)π
]
e−(2n−1)2π 2t/ 4 cos (2n − 1)πx
2",Elementary Differential Equations with Boundary Value Problems.pdf
"2
12. 1. 41 (p. 628) u(x, t ) = − 768
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n2
(2n − 1)π
]
e−(2n−1)2π 2t/ 4 cos (2n − 1)πx
2
12. 1. 42 (p. 628) u(x, t ) = − 384
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n4
(2n − 1)π
]
e−(2n−1)2π 2t/ 4 cos (2n − 1)πx
2
12. 1. 43 (p. 628) u(x, t ) = 1
2 − 2
π
∞∑
n=1
(−1)n
2n − 1 e−(2n−1)2 π 2a2t/L 2
cos (2n − 1)πx
L
12. 1. 44 (p. 628) u(x, t ) = 2
π
∞∑
n=1
1
n
[
1 − cos nπ
2
]
e−n2π 2a2t/L 2
sin nπx
L
12. 1. 45 (p. 629) u(x, t ) = 4
π
∞∑
n=1
1
2n − 1 sin (2n − 1)π
4 e−(2n−1)2π 2a2t/ 4L2
cos (2n − 1)πx
2L
12. 1. 46 (p. 629) u(x, t ) = 4
π
∞∑
n=1
1
2n − 1
[
1 − cos (2n − 1)π )
4
]
e−(2n−1)2 π 2a2t/ 4L2
sin (2n − 1)πx
2L
12. 1. 48 (p. 629) u(x, t ) = 1 − x + x3 + 4
π
∞∑
n=1
e−9π 2 (2n−1)2t/ 16
(2n − 1) sin (2n − 1)πx
4
12. 1. 49 (p. 629) u(x, t ) = 1 + x + x2 − 8
π 3
∞∑
n=1
e−(2n−1)2π 2t
(2n − 1)3 sin(2n − 1)πx
12. 1. 50 (p. 629) u(x, t ) = −1 − x + x3 + 8
π 2
∞∑
n=1
1
(2n − 1)2 e−3(2n−1)2 π 2t/ 4 cos (2n − 1)πx
2
12. 1. 51 (p. 629) u(x, t ) = x2 − x − 2 − 64
π
∞∑
n=1",Elementary Differential Equations with Boundary Value Problems.pdf
"π 2
∞∑
n=1
1
(2n − 1)2 e−3(2n−1)2 π 2t/ 4 cos (2n − 1)πx
2
12. 1. 51 (p. 629) u(x, t ) = x2 − x − 2 − 64
π
∞∑
n=1
(−1)n
2n − 1
[
1 − 8
(2n − 1)2 π 2
]
e−9(2n−1)2π 2 t/ 64 cos (2n − 1)πx
8",Elementary Differential Equations with Boundary Value Problems.pdf
"774 Answers to Selected Exercises
12. 1. 52 (p. 629) u(x, t ) = sin πx + 8
π
∞∑
n=1
(−1)n
(2n + 1)(2 n − 3) e−(2n−1)2π 2 t/ 4 sin (2n − 1)πx
2
12. 1. 53 (p. 629) u(x, t ) = x3 − x + 3 + 32
π 3
∞∑
n=1
e−(2n−1)2π 2t/ 4
(2n − 1)3 sin (2n − 1)πx
2
Section 12.2 Answers, pp. 642– 649
12. 2. 1 (p. 642) u(x, t ) = 4
3π 3
∞∑
n=1
(−1)n+1
(2n − 1)3 sin 3(2n − 1)πt sin(2n − 1)πx
12. 2. 2 (p. 642) u(x, t ) = 8
π 3
∞∑
n=1
1
(2n − 1)3 cos 3(2n − 1)πt sin(2n − 1)πx
12. 2. 3 (p. 642) u(x, t ) = − 4
π 3
∞∑
n=1
(1 + ( −1)n2)
n3 cos n
√
7 πt sin nπx
12. 2. 4 (p. 642) u(x, t ) = 8
3π 4
∞∑
n=1
1
(2n − 1)4 sin 3(2n − 1)πt sin(2n − 1)πx
12. 2. 5 (p. 642) u(x, t ) = − 4√
7 π 4
∞∑
n=1
(1 + ( −1)n2)
n4 sin n
√
7 πt sin nπx
12. 2. 6 (p. 642) u(x, t ) = 324
π 3
∞∑
n=1
(−1)n
n3 cos 8nπt
3 sin nπx
3
12. 2. 7 (p. 642) u(x, t ) = 96
π 5
∞∑
n=1
1
(2n − 1)5 cos 2(2n − 1)πt sin(2n − 1)πx
12. 2. 8 (p. 642) u(x, t ) = 243
2π 4
∞∑
n=1
(−1)n
n4 sin 8nπt
3 sin nπx
3
12. 2. 9 (p. 642) u(x, t ) = 48
π 6
∞∑
n=1
1",Elementary Differential Equations with Boundary Value Problems.pdf
"12. 2. 8 (p. 642) u(x, t ) = 243
2π 4
∞∑
n=1
(−1)n
n4 sin 8nπt
3 sin nπx
3
12. 2. 9 (p. 642) u(x, t ) = 48
π 6
∞∑
n=1
1
(2n − 1)6 sin 2(2n − 1)πt sin(2n − 1)πx .
12. 2. 10 (p. 642) u(x, t ) = π
2 cos
√
5 t sin x − 16
π
∞∑
n=1
n
(4n2 − 1)2 cos 2 n
√
5 t sin 2nx
12. 2. 11 (p. 642) u(x, t ) = − 240
π 5
∞∑
n=1
1 + ( −1)n2
n5 cos nπt sin nπx
12. 2. 12 (p. 642) u(x, t ) = π
2
√
5
sin
√
5 t sin x − 8
π
√
5
∞∑
n=1
1
(4n2 − 1)2 sin 2 n
√
5 t sin 2 nx
12. 2. 13 (p. 642) u(x, t ) = − 240
π 6
∞∑
n=1
1 + ( −1)n2
n6 sin nπt sin nπx
12. 2. 14 (p. 643) u(x, t ) = − 720
π 5
∞∑
n=1
(−1)n
n5 cos 3 nπt sin nπx
12. 2. 15 (p. 643) u(x, t ) = − 240
π 6
∞∑
n=1
(−1)n
n6 sin 3 nπt sin nπx
12. 2. 18 (p. 644) u(x, t ) = − 128
π 3
∞∑
n=1
(−1)n
(2n − 1)3 cos 3(2n − 1)πt
4 cos (2n − 1)πx
4",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 775
12. 2. 19 (p. 644) u(x, t ) = − 64
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n + 3
(2n − 1)π
]
cos(2n − 1)πt cos (2n − 1)πx
2
12. 2. 20 (p. 644) u(x, t ) = − 512
3π 4
∞∑
n=1
(−1)n
(2n − 1)4 sin 3(2n − 1)πt
4 cos (2n − 1)πx
4
12. 2. 21 (p. 644) u(x, t ) = − 64
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 3
(2n − 1)π
]
sin(2n − 1)πt cos (2n − 1)πx
2
12. 2. 22 (p. 644) u(x, t ) = 96
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n3 + 4
(2n − 1)π
]
cos (2n − 1)
√
5 πt
2 cos (2n − 1)πx
2
12. 2. 23 (p. 644) u(x, t ) = −96
∞∑
n=1
1
(2n − 1)3
[
(−1)n + 2
(2n − 1)π
]
cos (2n − 1)
√
3 t
2 cos (2n − 1)x
2
12. 2. 24 (p. 644) u(x, t ) = 192
π 4
√
5
∞∑
n=1
1
(2n − 1)4
[
(−1)n3 + 4
(2n − 1)π
]
sin (2n − 1)
√
5 πt
2 cos (2n − 1)πx
2
12. 2. 25 (p. 644) u(x, t ) = − 192√
3
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 2
(2n − 1)π
]
sin (2n − 1)
√
3t
2 sin (2n − 1)x
2
12. 2. 26 (p. 644) u(x, t ) = − 384
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n4
(2n − 1)π
]
cos 3(2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 27 (p. 644) u(x, t ) = 96
π 3
∞∑
n=1
1",Elementary Differential Equations with Boundary Value Problems.pdf
"π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n4
(2n − 1)π
]
cos 3(2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 27 (p. 644) u(x, t ) = 96
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n5 + 8
(2n − 1)π
]
cos (2n − 1)
√
7 πt
2 cos (2n − 1)πx
2
12. 2. 28 (p. 644) u(x, t ) = − 768
3π 5
∞∑
n=1
1
(2n − 1)5
[
1 + (−1)n4
(2n − 1)π
]
sin 3(2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 29 (p. 644) u(x, t ) = 192
π 4
√
7
∞∑
n=1
1
(2n − 1)4
[
(−1)n5 + 8
(2n − 1)π
]
sin (2n − 1)
√
7 πt
2 cos (2n − 1)πx
2
12. 2. 30 (p. 645) u(x, t ) = − 768
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + (−1)n2
(2n − 1)π
]
cos (2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 31 (p. 645) u(x, t ) = − 1536
π 5
∞∑
n=1
1
(2n − 1)5
[
1 + (−1)n2
(2n − 1)π
]
sin (2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 32 (p. 645) u(x, t ) = 1
2 [CM f(x + at) + CM f(x − at)] + 1
2a
∫ x+at
x−at
CM g(τ ) dτ
12. 2. 35 (p. 645) u(x, t ) = 32
π
∞∑
n=1
1
(2n − 1)3 cos 4(2n − 1)t sin (2n − 1)x
2
12. 2. 36 (p. 645) u(x, t ) = − 96
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 4
(2n − 1)π
]
cos 3(2n − 1)πt
2 sin (2n − 1)πx
2",Elementary Differential Equations with Boundary Value Problems.pdf
"2
12. 2. 36 (p. 645) u(x, t ) = − 96
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 4
(2n − 1)π
]
cos 3(2n − 1)πt
2 sin (2n − 1)πx
2
12. 2. 37 (p. 645) u(x, t ) = 8
π
∞∑
n=1
1
(2n − 1)4 sin 4(2n − 1)t sin (2n − 1)x
2
12. 2. 38 (p. 646) u(x, t ) = − 64
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + ( −1)n 4
(2n − 1)π
]
sin 3(2n − 1)πt
2 sin (2n − 1)πx
2",Elementary Differential Equations with Boundary Value Problems.pdf
"776 Answers to Selected Exercises
12. 2. 39 (p. 646) u(x, t ) = 96
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + ( −1)n 2
(2n − 1)π
]
cos 3(2n − 1)πt
2 sin (2n − 1)πx
2
12. 2. 40 (p. 646) u(x, t ) = 192
π
∞∑
n=1
(−1)n
(2n − 1)4 cos (2n − 1)
√
3 t
2 sin (2n − 1)x
2
12. 2. 41 (p. 646) u(x, t ) = 64
π 4
∞∑
n=1
1
(2n − 1)4
[
1 + ( −1)n 2
(2n − 1)π
]
sin 3(2n − 1)πt
2 sin (2n − 1)πx
2
12. 2. 42 (p. 646) u(x, t ) = 384√
3 π
∞∑
n=1
(−1)n
(2n − 1)5 sin (2n − 1)
√
3 t
2 sin (2n − 1)x
2
12. 2. 43 (p. 646) u(x, t ) = 1536
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 3
(2n − 1)π
]
cos (2n − 1)
√
5 πt
2 sin (2n − 1)πx
2
12. 2. 44 (p. 646) u(x, t ) = 384
π 4
∞∑
n=1
1
(2n − 1)4
[
(−1)n + 4
(2n − 1)π
]
cos(2n − 1)πt sin (2n − 1)πx
2
12. 2. 45 (p. 646) u(x, t ) = 3072√
5 π 5
∞∑
n=1
1
(2n − 1)5
[
(−1)n + 3
(2n − 1)π
]
sin (2n − 1)
√
5 πt
2 sin (2n − 1)πx
2
12. 2. 46 (p. 646) u(x, t ) = 384
π 5
∞∑
n=1
1
(2n − 1)5
[
(−1)n + 4
(2n − 1)π
]
sin(2n − 1)πt sin (2n − 1)πx
2
12. 2. 47 (p. 646) u(x, t ) = 1",Elementary Differential Equations with Boundary Value Problems.pdf
"2
12. 2. 46 (p. 646) u(x, t ) = 384
π 5
∞∑
n=1
1
(2n − 1)5
[
(−1)n + 4
(2n − 1)π
]
sin(2n − 1)πt sin (2n − 1)πx
2
12. 2. 47 (p. 646) u(x, t ) = 1
2 [SM f(x + at) + SM f(x − at)] + 1
2a
∫ x+at
x−at
SM g(τ ) dτ
12. 2. 50 (p. 647) u(x, t ) = 4 − 768
π 4
∞∑
n=1
1
(2n − 1)4 cos
√
5(2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 51 (p. 647) u(x, t ) = 4 t − 1536√
5 π 5
∞∑
n=1
1
(2n − 1)5 sin
√
5(2n − 1)πt
2 cos (2n − 1)πx
2
12. 2. 52 (p. 647) u(x, t ) = − 2π 4
5 − 48
∞∑
n=1
1 + ( −1)n2
n4 cos 2 nt cos nx
12. 2. 53 (p. 647) u(x, t ) = − 7
5 − 144
π 4
∞∑
n=1
(−1)n
n4 cos n
√
7 πt cos nπx
12. 2. 54 (p. 647) u(x, t ) = − 2π 4 t
5 − 24
∞∑
n=1
1 + ( −1)n2
n5 sin 2nt cos nx
12. 2. 55 (p. 647) u(x, t ) = − 7t
5 − 144
π 5
√
7
∞∑
n=1
(−1)n
n5 sin n
√
7 πt cos nπx
12. 2. 56 (p. 647) u(x, t ) = π 4
30 − 3
∞∑
n=1
1
n4 cos 8 nt cos 2 nx
12. 2. 57 (p. 647) u(x, t ) = 3
5 − 48
π 4
∞∑
n=1
2 + ( −1)n
n4 cos nπt cos nπx
12. 2. 58 (p. 647) u(x, t ) = π 4 t
30 − 3
8
∞∑
n=1
1
n5 sin 8nt cos 2 nx",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 777
12. 2. 59 (p. 647) u(x, t ) = 3t
5 − 48
π 5
∞∑
n=1
2 + ( −1)n
n5 sin nπt cos nπx
12. 2. 60 (p. 647) u(x, t ) = 1
2 [Cf (x + at) + Cf (x − at)] + 1
2a
∫x+at
x−at Cg (τ ) dτ
12. 2. 63 (p. 648) (c) u(x, t ) = f (x + at) + f (x − at)
2 + 1
2a
∫ x+at
x−at
g(u) du
12. 2. 64 (p. 649) u(x, t ) = x(1 + 4 at) 12. 2. 65 (p. 649) u(x, t ) = x2 + a2 t2 + t
12. 2. 66 (p. 649) u(x, t ) = sin( x + at) 12. 2. 67 (p. 649) u(x, t ) = x3 + 6 tx2 + 3 a2 t2x + 2 a2 t3
12. 2. 68 (p. 649) u(x, t ) = x sin x cos at + at cos x sin at + sin x sin at
a
Section 12.3 Answers, pp. 662– 665
12. 3. 1 (p. 662) u(x, y ) = 8
π 3
∞∑
n=1
sinh(2n − 1)π (1 − y)
(2n − 1)3 sinh(2n − 1)π sin(2n − 1)πx
12. 3. 2 (p. 662) u(x, y ) = − 32
π 3
∞∑
n=1
(1 + ( −1)n2) sinh nπ (3 − y)/ 2
n3 sinh 3nπ/ 2 sin nπx
2
12. 3. 3 (p. 662) u(x, y ) = 8
π 2
∞∑
n=1
(−1)n+1 sinh(2n − 1)π (1 − y/ 2)
(2n − 1)2 sinh(2n − 1)π sin (2n − 1)πx
2
12. 3. 4 (p. 662) u(x, y ) = π
2
sinh(1 − y)
sinh 1 sin x − 16
π
∞∑
n=1",Elementary Differential Equations with Boundary Value Problems.pdf
"n=1
(−1)n+1 sinh(2n − 1)π (1 − y/ 2)
(2n − 1)2 sinh(2n − 1)π sin (2n − 1)πx
2
12. 3. 4 (p. 662) u(x, y ) = π
2
sinh(1 − y)
sinh 1 sin x − 16
π
∞∑
n=1
n sinh 2n(1 − y)
(4n2 − 1)2 sinh 2n sin 2 nx
12. 3. 5 (p. 662) u(x, y ) = 3 y + 108
π 3
∞∑
n=1
(−1)n sinh nπy/ 3
n3 cosh 2nπ/ 3 cos nπx
3
12. 3. 6 (p. 662) u(x, y ) = y
2 + 4
π 3
∞∑
n=1
sinh(2n − 1)πy
(2n − 1)3 cosh 2(2n − 1)π cos(2n − 1)πx
12. 3. 7 (p. 662) u(x, y ) = − 8y
3 + 32
π 3
∞∑
n=1
(−1)n sinh nπy/ 2
n3 cosh nπ cos nπx
2
12. 3. 8 (p. 662) u(x, y ) = y
3 + 4
π 3
∞∑
n=1
sinh nπy
n3 cosh nπ cos nπx
12. 3. 9 (p. 662) u(x, y ) = 128
π 3
∞∑
n=1
cosh(2n − 1)π (x − 3)/ 4
(2n − 1)3 cosh 3(2n − 1)π/ 4 sin (2n − 1)πy
4
12. 3. 10 (p. 662) u(x, y ) = − 96
π 3
∞∑
n=1
[
1 + ( −1)n 4
(2n − 1)π
]
cosh(2n − 1)π (x − 2)/ 2
(2n − 1)3 cosh(2n − 1)π sin (2n − 1)πy
2
12. 3. 11 (p. 662) u(x, y ) = 768
π 3
∞∑
n=1
[
1 + ( −1)n 2
(2n − 1)π
]
cosh(2n − 1)π (x − 2)/ 4
(2n − 1)3 cosh(2n − 1)π/ 2 sin (2n − 1)πy
4
12. 3. 12 (p. 662) u(x, y ) = 96
π 3
∞∑
n=1
[",Elementary Differential Equations with Boundary Value Problems.pdf
"∞∑
n=1
[
1 + ( −1)n 2
(2n − 1)π
]
cosh(2n − 1)π (x − 2)/ 4
(2n − 1)3 cosh(2n − 1)π/ 2 sin (2n − 1)πy
4
12. 3. 12 (p. 662) u(x, y ) = 96
π 3
∞∑
n=1
[
3 + ( −1)n 4
(2n − 1)π
]
cosh(2n − 1)π (x − 3)/ 2
(2n − 1)3 cosh 3(2 n − 1)π/ 2 sin (2n − 1)πy
2
12. 3. 13 (p. 663) u(x, y ) = − 16
π
∞∑
n=1
cosh(2n − 1)x/ 2
(2n − 3)(2n + 1)(2 n − 1) sinh(2n − 1)/ 2 cos (2n − 1)y
2
12. 3. 14 (p. 663) u(x, y ) = − 432
π 3
∞∑
n=1
[
1 + 4(−1)n
(2n − 1)π
]
cosh(2n − 1)πx/ 6
(2n − 1)3 sinh(2n − 1)π/ 3 cos (2n − 1)πy
6",Elementary Differential Equations with Boundary Value Problems.pdf
"778 Answers to Selected Exercises
12. 3. 15 (p. 663) u(x, y ) = − 64
π
∞∑
n=1
(−1)n cosh(2n − 1)x/ 2
(2n − 1)4 sinh(2n − 1)/ 2 cos (2n − 1)y
2 .
12. 3. 16 (p. 663) u(x, y ) = − 192
π 4
∞∑
n=1
cosh(2n − 1)πx/ 2
(2n − 1)4 sinh(2n − 1)π/ 2
[
(−1)n + 2
(2n − 1)π
]
cos (2n − 1)πy
2
12. 3. 17 (p. 663) u(x, y ) =
∞∑
n=1
α n
sinh nπy/a
sinh nπb/a sin nπx
a , α n = 2
a
∫ a
0
f (x) sin nπx
a dx
u(x, y ) = 72
π 3
∞∑
n=1
sinh(2n − 1)πy/ 3
(2n − 1)3 sinh 2(2n − 1)π/ 3 sin (2n − 1)πx
3
12. 3. 18 (p. 663) u(x, y ) = α 0(1 − y/b ) +
∞∑
n=1
α n
sinh nπ (b − y)/a
sinh nπb/a cos nπx
a , α 0 = 1
a
∫ a
0
f (x) dx,
α n = 2
a
∫ a
0
f (x) cos nπx
a dx, n ≥ 1
u(x, y ) = 8(1 − y)
15 − 48
π 4
∞∑
n=1
1
n4
sinh nπ (1 − y)
sinh nπ cos nπx
12. 3. 19 (p. 663) u(x, y ) =
∞∑
n=1
α n
sinh(2n − 1)π (b − y)/ 2a
sinh(2n − 1)πb/ 2a cos (2n − 1)πx
2a ,
α n = 2
a
∫ a
0
f (x) cos (2n − 1)πx
2a dx
u(x, y ) = 288
π 3
∞∑
n=1
sinh(2n − 1)π (2 − y)/ 6
(2n − 1)3 sinh(2n − 1)π/ 3 sin (2n − 1)πx
6
12. 3. 20 (p. 663) u(x, y ) =
∞∑
n=1",Elementary Differential Equations with Boundary Value Problems.pdf
"2a dx
u(x, y ) = 288
π 3
∞∑
n=1
sinh(2n − 1)π (2 − y)/ 6
(2n − 1)3 sinh(2n − 1)π/ 3 sin (2n − 1)πx
6
12. 3. 20 (p. 663) u(x, y ) =
∞∑
n=1
α n
sinh(2n − 1)π (b − y)/ 2a
sinh(2n − 1)πb/ 2a sin (2n − 1)πx
2a ,
α n = 2
a
∫ a
0
f (x) sin (2n − 1)πx
2a dx
u(x, y ) = 32
π 3
∞∑
n=1
[
(−1)n5 + 18
(2n − 1)π
]
sinh(2n − 1)π (2 − y)/ 2
(2n − 1)3 sinh(2n − 1)π cos (2n − 1)πx
2 .
12. 3. 21 (p. 663) u(x, y ) =
∞∑
n=1
α n
cosh nπ (y − b)/a
cosh nπb/a sin nπx
a , α n = 2
a
∫ a
0
f (x) sin nπx
a dx
u(x, y ) = −12
∞∑
n=1
(−1)n cosh n(y − 2)
n3 cosh 2n sin nx
12. 3. 22 (p. 663) u(x, y ) = α 0 +
∞∑
n=1
α n
cosh nπy/a
cosh nπb/a cos nπx
a , α 0 = 1
a
∫ a
0
f (x) dx,
α n = 2
a
∫ a
0
f (x) cos nπx
a dx, n ≥ 1
u(x, y ) = π 4
30 − 3
∞∑
n=1
1
n4
cosh 2ny
cos 2 n cos 2 nx
12. 3. 23 (p. 663) u(x, y ) = a
π
∞∑
n=1
α n
sinh nπ (y − b)/a
n cosh nπb/a sin nπx
a , α n = 2
a
∫ a
0
f (x) sin nπx
a dx
u(x, y ) = 4
π
∞∑
n=1
(−1)n+1 sinh(2n − 1)(y − 1)
(2n − 1)3 cosh(2n − 1) sin(2n − 1)x",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 779
12. 3. 24 (p. 663) u(x, y ) =
∞∑
n=1
α n
cosh nπx/b
cosh nπa/b sin nπy
b , α n = 2
b
∫ b
0
g(y) sin nπy
b dy
u(x, y ) = 96
π 5
∞∑
n=1
cosh(2n − 1)πx
(2n − 1)5 cosh(2n − 1)π sin(2n − 1)πy .
12. 3. 25 (p. 664) u(x, y ) =
∞∑
n=1
α n
cosh(2n − 1)πx/ 2b
cosh(2n − 1)πa/ 2b cos (2n − 1)πy
2b ,
α n = 2
b
∫ b
0
g(y) cos (2n − 1)πy
2b dy
u(x, y ) = − 128
π 3
∞∑
n=1
(−1)n cosh(2n − 1)πx/ 4
(2n − 1)3 cosh(2n − 1)π/ 2 cos (2n − 1)πy
4 .
12. 3. 26 (p. 664) u(x, y ) = b
π
∞∑
n=1
α n
cosh nπx/b
n sinh nπa/b sin nπy
b , α n = 2
b
∫ b
0
g(y) sin nπy
b dy
u(x, y ) = 64
π 3
∞∑
n=1
(−1)n+1 cosh(2n − 1)πx/ 4
(2n − 1)3 sinh(2n − 1)π/ 4 sin (2n − 1)πy
4
12. 3. 27 (p. 664) u(x, y ) = − 2b
π
∞∑
n=1
α n
cosh(2n − 1)π (x − a)/ 2b
(2n − 1) sinh(2n − 1)πa/ 2b sin (2n − 1)y
2b ,
α n = 2
b
∫ b
0
g(y) sin (2n − 1)πy
2b dy
u(x, y ) = 192
∞∑
n=1
[
1 + ( −1)n 4
(2n − 1)π
]
cosh(2n − 1)(x − 1)/ 2
(2n − 1)4 sinh(2n − 1)/ 2 sin (2n − 1)y
2 .
12. 3. 28 (p. 664) u(x, y ) = α 0(x − a) + b
π
∞∑",Elementary Differential Equations with Boundary Value Problems.pdf
"n=1
[
1 + ( −1)n 4
(2n − 1)π
]
cosh(2n − 1)(x − 1)/ 2
(2n − 1)4 sinh(2n − 1)/ 2 sin (2n − 1)y
2 .
12. 3. 28 (p. 664) u(x, y ) = α 0(x − a) + b
π
∞∑
n=1
α n
sinh nπ (x − a)/b
n cosh nπa/b cos nπy
b , α 0 = 1
b
∫ b
0
g(y) cos nπy
b dy,
α n = 2
b
∫ b
0
g(y) cos nπy
b dy
u(x, y ) = π (x − 2)
2 − 4
π
∞∑
n=1
sinh(2n − 1)(x − 2)
(2n − 1)3 cosh 2(2n − 1) cos(2n − 1)y.
12. 3. 29 (p. 664) u(x, y ) = α 0 +
∞∑
n=1
α ne−nπy/a cos nπx
a , α 0 = 1
a
∫ a
0
f (x) dx,
α n = 2
a
∫ a
0
f (x) cos nπx
a dx, n ≥ 1
u(x, y ) = π 3
2 − 48
π
∑ ∞
n=1
1
(2n−1)4 e−(2n−1)y cos(2n − 1)x
12. 3. 30 (p. 664) u(x, y ) =
∞∑
n=1
α ne−(2n−1)πy/ 2a cos (2n − 1)πx
2a , α n = 2
a
∫ a
0
f (x) cos (2n − 1)πx
2a dx
u(x, y ) = − 288
π 3
∞∑
n=1
(−1)n
(2n − 1)3 e−(2n−1)πy/ 6 cos (2n − 1)πx
6
12. 3. 31 (p. 664) u(x, y ) =
∞∑
n=1
α ne−(2n−1)πy/ 2a sin (2n − 1)πx
2a , α n = 2
a
∫ a
0
f (x) sin (2n − 1)πx
2a dx
u(x, y ) = 32
π
∞∑
n=1
1
(2n − 1)3 e−(2n−1)y/ 2 sin (2n − 1)x
2 .",Elementary Differential Equations with Boundary Value Problems.pdf
"780 Answers to Selected Exercises
12. 3. 32 (p. 664) u(x, y ) = − a
π
∞∑
n=1
α n
n e−nπy/a sin nπx
a , α n = 2
a
∫ a
0
f (x) sin nπx
a dx
u(x) = 4
∞∑
n=1
(1 + ( −1)n2)
n4 e−ny sin nx
12. 3. 33 (p. 664) u(x, y ) = − 2a
π
∞∑
n=1
α n
2n − 1 e−(2n−1)πy/ 2a cos (2n − 1)πx
2a , α n = 2
a
∫ a
0
f (x) cos (2n − 1)πx
2a dx
u(x, y ) = 5488
π 3
∞∑
n=1
1
(2n − 1)3
[
1 + 4(−1)n
(2n − 1)π
]
e−(2n−1)πy/ 14 cos (2n − 1)πx
14
12. 3. 34 (p. 664) u(x, y ) = − 2a
π
∞∑
n=1
α n
2n − 1 e−(2n−1)πy/ 2a sin (2n − 1)πx
2a , α n = 2
a
∫ a
0
f (x) sin (2n − 1)πx
2a dx
u(x, y ) = − 2000
π 3
∞∑
n=1
1
(2n − 1)3
[
(−1)n + 4
(2n − 1)π
]
e−(2n−1)πy/ 10 sin (2n − 1)πx
10
12. 3. 35 (p. 664) u(x, y ) =
∞∑
n=1
An sinh nπ (b − y)/a + Bn sinh nπy/a
sinh nπb/a sin nπx
a
+
∞∑
n=1
Cn sinh nπ (a − x)/b + Dn sinh nπx/b
sinh nπa/b sin nπy
b
12. 3. 36 (p. 664) u(x, y ) = C + a
π
∞∑
n=1
Bn cosh nπy/a − An cosh nπ (y − b)/a
n sinh nπb/a cos nπx
a
+ b
π
∞∑
n=1
Dn cosh nπx/b − Cn cosh nπ (x − a)/b
n sinh nπa/b cos nπy
b",Elementary Differential Equations with Boundary Value Problems.pdf
"π
∞∑
n=1
Bn cosh nπy/a − An cosh nπ (y − b)/a
n sinh nπb/a cos nπx
a
+ b
π
∞∑
n=1
Dn cosh nπx/b − Cn cosh nπ (x − a)/b
n sinh nπa/b cos nπy
b
Section 12.4 Answers, pp. 672– 673
12. 4. 1 (p. 672) u(r, θ ) = α 0
ln r/ρ
ln ρ 0/ρ +
∞∑
n=1
rn ρ −n − ρ nr−n
ρ n
0 ρ −n − ρ nρ −n
0
(α n cos nθ + β n sin nθ ) α 0 = 1
2π
∫ π
−π
f (θ ) dθ ,
and α n = 1
π
∫ π
−π
f (θ ) cos nθ dθ , β n = 1
π
∫ π
−π
f (θ ) sin nθ dθ , n = 1 , 2, 3, . . .
12. 4. 2 (p. 672) u(r, θ ) =
∞∑
n=1
α n
ρ −nπ/γ
0 rnπ/γ − ρ nπ/γ
0 r−nπ/γ
ρ −nπ/γ
0 ρ nπ/γ − ρ nπ/γ
0 ρ −nπ/γ
sin nπθ
γ
α n = 1
γ
∫ γ
0
f (θ ) sin nπθ
γ dθ , n = 1 , 2, 3,. . .
12. 4. 3 (p. 672) u(r, θ ) = ρα 0 ln r
ρ 0
+ ργ
π
∞∑
n=1
α n
n
ρ
−nπ/γ
0 rnπ/γ − ρ
nπ/γ
0 r−nπ/γ
ρ −nπ/γ
0 ρ nπ/γ + ρ nπ/γ
0 ρ −nπ/γ
cos nπθ
γ
α 0 = 1
γ
∫ γ
0
f (θ ) dθ , α n = 2
γ
∫ γ
0
f (θ ) cos nπθ
γ dθ , n = 1 , 2, 3,. . .
12. 4. 4 (p. 672) u(r, θ ) =
∞∑
n=1
α n
r(2n−1)π/ 2γ
ρ (2n−1)π/ 2γ cos (2n − 1)πθ
2γ
α n = 2
γ
∫ γ
0
f (θ ) cos (2n − 1)πθ
2γ dθ , n = 1 , 2, 3,. . .",Elementary Differential Equations with Boundary Value Problems.pdf
"12. 4. 4 (p. 672) u(r, θ ) =
∞∑
n=1
α n
r(2n−1)π/ 2γ
ρ (2n−1)π/ 2γ cos (2n − 1)πθ
2γ
α n = 2
γ
∫ γ
0
f (θ ) cos (2n − 1)πθ
2γ dθ , n = 1 , 2, 3,. . .
12. 4. 5 (p. 673) u(r, θ ) = 2γρ 0
π
∞∑
n=1
α n
2n − 1
ρ −(2n−1)π/ 2γ r(2n−1)π/ 2γ + ρ (2n−1)π/ 2γ r−(2n−1)π/ 2γ
ρ −(2n−1)π/ 2γ ρ (2n−1)π/ 2γ
0 − ρ (2n−1)π/ 2γ ρ −(2n−1)π/ 2γ
0
sin (2n − 1)πθ
2γ ,",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 781
α n = 2
γ
∫ γ
0
g(θ ) sin (2n − 1)πθ
2γ dθ , n = 1 , 2, 3,. . .
12. 4. 6 (p. 673) u(r, θ ) = α 0 +
∞∑
n=1
α n
rnπ/γ
ρ nπ/γ cos nπθ
γ α 0 = 1
γ
∫ γ
0
f (θ ) dθ ,
α n = 2
γ
∫ γ
0
f (θ ) cos nπθ
γ dθ , n = 1 , 2, 3,. . .
12. 4. 7 (p. 673) vn(r, θ ) = rn
nρ n−1 (α n cos nθ + sin nθ )
u(r, θ ) = c +
∞∑
n=1
rn
nρ n−1 (α n cos nθ + β n sin nθ ) α n = 1
π
∫ π
−π
f (θ ) cos nθ dθ ,
β n = 1
π
∫ π
−π
f (θ ) sin nθ dθ , n = 1 , 2, 3,. . .
Section 13.1 Answers, pp. 684– 686
13. 1. 2 (p. 684) y = −x + 2
e − 1
(
ex − e(x−1) )
13. 1. 3 (p. 684) y = x2 − x3
3 + cx with c arbitrary
13. 1. 4 (p. 684) y = −x + 2 ex + e−(x−1) 13. 1. 5 (p. 684) y = 1
4 + 11
4 cos 2x + 9
4 sin 2x
13. 1. 6 (p. 684) y = ( x2 +13−8x)ex 13. 1. 7 (p. 684) y = 2 e2x + 3(5e3x − 4e4x )
e(15 − 16e) + 2e4(4e3(x−1) − 3e4(x−1) )
16e − 15
13. 1. 8 (p. 684)
∫ b
a
tF (t) dt = 0 y = −x
∫ 1
x
F (t) dt −
∫ x
0
tF (t) dt + c1x with c1 arbitrary
13. 1. 9 (p. 685) (a) b − a ̸= kπ (k = integer)
y = sin(x − a)",Elementary Differential Equations with Boundary Value Problems.pdf
"∫ b
a
tF (t) dt = 0 y = −x
∫ 1
x
F (t) dt −
∫ x
0
tF (t) dt + c1x with c1 arbitrary
13. 1. 9 (p. 685) (a) b − a ̸= kπ (k = integer)
y = sin(x − a)
sin(b − a)
∫ b
x
F (t) sin(t − b) dt + sin(x − b)
sin(b − a)
∫ x
a
F (t) sin(t − a) dt
(b)
∫ b
a
F (t) sin(t − a) dt = 0
y = −sin(x − a)
∫ b
x
F (t) cos( t − a) dt − cos(x − a)
∫ x
a
F (t) sin(t − a) dt
+c1 sin(x − a) with c1 arbitrary
13. 1. 10 (p. 685) (a) b − a ̸= ( k + 1 / 2)π (k = integer)
y = − sin(x − a)
cos(b − a)
∫ b
x
F (t) cos( t − b) dt − cos(x − b)
cos(b − a)
∫ b
x
F (t) sin( t − a) dt
(b)
∫ b
a
F (t) sin(t − a) dt = 0
y = −sin(x − a)
∫ b
x
F (t) cos( t − a) dt − cos(x − a)
∫ x
a
F (t) sin(t − a) dt
+c1 sin(x − a) with c1 arbitrary
13. 1. 11 (p. 685) (a) b − a ̸= kπ (k = integer)
y = cos(x − a)
sin(b − a)
∫ b
x
F (t) cos(t − b) dt + cos(x − b)
sin(b − a)
∫ x
a
F (t) cos(t − a) dt
(b)
∫ b
a
F (t) cos(t − a) dt = 0
y = cos( x − a)
∫ b
x
F (t) sin(t − a) dt + sin( x − a)
∫ x
a
F (t) cos(t − a) dt
+c1 cos(x − a) with c1 arbitrary",Elementary Differential Equations with Boundary Value Problems.pdf
"782 Answers to Selected Exercises
13. 1. 12 (p. 685) y = sinh(x − a)
sinh(b − a)
∫ b
x
F (t) sinh( t − b) dt + sinh(x − b)
sinh(b − a)
∫ x
a
F (t) sinh( t − a) dt
13. 1. 13 (p. 685) y = − sinh(x − a)
cosh(b − a)
∫ b
x
F (t) cosh(t − b) dt − cosh(x − b)
cosh(b − a)
∫ x
a
F (t) sinh(t − a) dt
13. 1. 14 (p. 685) y = − cosh(x − a)
sinh(b − a)
∫ b
x
F (t) cosh(t − b) dt − cosh(x − b)
sinh(b − a)
∫ x
a
F (t) cosh(t − a) dt
13. 1. 15 (p. 685) y = − 1
2
(
ex
∫ b
x
e−t F (t) dt + e−x
∫ x
a
et F (t) dt
)
13. 1. 16 (p. 685) If ω isn’t a positive integer, then
y = 1
ω sin ωπ
(
sin ωx
∫ π
x
F (t) sin ω (t − π ) dt + sin ω (x − π )
∫ x
0
F (t) sin ωt dt
)
.
If ω = n (positive integer), then
∫ π
0
F (t) sin nt dt = 0 is necessary for existence
of a solution. In this case,
y = − 1
n
(
sin nx
∫ π
x
F (t) cos nt dt + cos nx
∫ x
0
F (t) sin nt dt
)
+ c1 sin nx
with c1 arbitrary.
13. 1. 17 (p. 685) If ω ̸= n + 1 / 2 (n = integer), then
y = − sin ωx
ω cos ωπ
∫ π
x
F (t) cos ω (t − π ) dt − cos ω (x − π )",Elementary Differential Equations with Boundary Value Problems.pdf
"with c1 arbitrary.
13. 1. 17 (p. 685) If ω ̸= n + 1 / 2 (n = integer), then
y = − sin ωx
ω cos ωπ
∫ π
x
F (t) cos ω (t − π ) dt − cos ω (x − π )
ω cos ωπ
∫ x
0
F (t) sin ωt dt.
If ω = n + 1 / 2 (n = integer), then
∫ π
0
F (t) sin(n + 1 / 2)t dt = 0 is necessary
for existence of a solution. In this case,
y = − sin(n + 1 / 2)x
n + 1 / 2
∫ π
x
F (t) cos(n + 1 / 2)t dt
− cos(n + 1 / 2)x
n + / 2
∫ x
0
F (t) sin(n + 1 / 2)t dt + c1 sin(n + 1 / 2)x
with c1 arbitrary,
13. 1. 18 (p. 685) If ω ̸= n + 1 / 2 (n = integer), then
y = cos ωx
ω cos ωπ
∫ π
x
F (t) sin ω (t − π ) dt + sin ω (x − π )
ω cos ωπ
∫ x
0
F (t) cos ωt dt.
If ω = n + 1 / 2 (n = integer), then
∫ π
0
F (t) cos(n + 1 / 2)t dt = 0 is necessary
for existence of a solution. In this case,
y = cos(n + 1 / 2)x
n + 1 / 2
∫ π
x
F (t) sin(n + 1 / 2)t dt
+ sin(n + 1 / 2)x
n + / 2
∫ x
0
F (t) cos( n + 1 / 2)t dt + c1 cos(n + 1 / 2)x
with c1 arbitrary.
13. 1. 19 (p. 685) If ω isn’t a positive integer, then
y = 1
ω sin ωπ
(
cos ωx
∫ π
x",Elementary Differential Equations with Boundary Value Problems.pdf
"with c1 arbitrary.
13. 1. 19 (p. 685) If ω isn’t a positive integer, then
y = 1
ω sin ωπ
(
cos ωx
∫ π
x
F (t) cos ω (t − π ) dt + cos ω (x − π )
∫ x
0
F (t) cos ωt dt
)
.",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 783
If ω = n (positive integer), then
∫ π
0
F (t) cos nt dt = 0 is necessary for existence
of a solution. In this case,
y = − 1
n
(
cos nx
∫ π
x
F (t) sin nt dt + sin nx
∫ x
0
F (t) cos nt dt
)
+ c1 cos nx
with c1 arbitrary.
13. 1. 20 (p. 685) y1 = B1(z2)z1 − B1 (z1)z2
13. 1. 21 (p. 685) (a) G(x, t ) =





(t − a)(x − b)
b − a a ≤ t ≤ x,
(x − a)(t − b)
b − a) x ≤ t ≤ b
y = 1
b − a
(
(x − a)
∫ b
x
(t − b)F (t) dt + ( x − b)
∫ x
a
(t − a)F (t) dt
)
(b) G(x, t ) =
{
a − t a ≤ t ≤ x
a − x x ≤ t ≤ b y = ( a − x)
∫ b
x
F (t) dt +
∫ x
a
(a − t)F (t) dt
(c) G(x, t ) =
{
x − b a ≤ t ≤ x
t − b x ≤ t ≤ b y =
∫ b
x
(t − b)F (t) dt + ( x − b)
∫ x
a
F (t) dt
(d)
∫ b
a
F (t) dt = 0 is a necessary condition for existence of a solution. Then
y =
∫ b
x
tF (t) dt + x
∫ x
a
F (t) dt + c1 with c1 arbitrary.
13. 1. 22 (p. 686) G(x, t ) =





− (2 + t)(3 − x)
5 , 0 ≤ t ≤ x,
− (2 + x)(3 − t)
5 , x ≤ t ≤ 1
(a) y = x2 − x − 2
2 (b) y = 5x2 − 7x − 14
30",Elementary Differential Equations with Boundary Value Problems.pdf
"13. 1. 22 (p. 686) G(x, t ) =





− (2 + t)(3 − x)
5 , 0 ≤ t ≤ x,
− (2 + x)(3 − t)
5 , x ≤ t ≤ 1
(a) y = x2 − x − 2
2 (b) y = 5x2 − 7x − 14
30
(c) y = 5x4 − 9x − 18
60
13. 1. 23 (p. 686) G(x, t ) =





cos t sin x
t3/ 2√x , π
2 ≤ t ≤ x,
cos x sin t
t3/ 2√x , x ≤ t ≤ π
(a) y = 1 + cos x − sin x√x (b) y = x + π cos x − π/ 2 sin x√x
13. 1. 24 (p. 686) G(x, t ) =





(t − 1)x(x − 2)
t3 , 1 ≤ t ≤ x,
x(x − 1)(t − 2)
t3 , x ≤ t ≤ 2
(a) y = x(x − 1)(x − 2) (b) y = x(x − 1)(x − 2)(x + 3)
13. 1. 25 (p. 686) G(x, t ) =





− 1
22
(
3 + 1
t2
)(
x + 4
x
)
, 1 ≤ x ≤ t,
− 1
22
(
3x + 1
x
)(
1 + 4
t2
)
, x ≤ t ≤ 2
(a) y = x2 − 11x + 4
11x (b) y = 11x3 − 45x2 − 4
33x (c) y = 11x4 − 139x2 − 28
88x",Elementary Differential Equations with Boundary Value Problems.pdf
"784 Answers to Selected Exercises
13. 1. 26 (p. 686) α (ρ + δ ) − βρ ̸= 0 G(x, t ) =





(β − αt )(ρ + δ − ρx )
α (ρ + δ ) − βρ , 0 ≤ t ≤ x,
(β − αx )(ρ + δ − ρt )
α (ρ + δ ) − βρ , x ≤ t ≤ 1
13. 1. 27 (p. 686) αδ − βρ ̸= 0 G(x, t ) =





(β cos t − α sin t)(δ cos x − ρ sin x)
αδ − βρ , 0 ≤ t ≤ x,
(β cos x − α sin x)(δ cos t − ρ sin t)
αδ − βρ , x ≤ t ≤ π
13. 1. 28 (p. 686) αρ + βδ ̸= 0 G(x, t ) =





(β cos t − α sin t)(ρ cos x + δ sin x)
αρ + βδ x ≤ t ≤ π
(β cos x − α sin x)(ρ cos t + δ sin t)
αρ + βδ 0 ≤ t ≤ x
13. 1. 29 (p. 686) αδ −βρ ̸= 0 G(x, t ) =







ex−t)(β cos t − (α + β ) sin t)(δ cos x − (ρ + δ ) sin x)
αδ − βρ 0 ≤ t ≤ x,
ex−t (β cos x − (α + β ) sin x)(δ cos t − (ρ + δ ) sin t)
αδ − βρ , x ≤ t ≤ π
13. 1. 30 (p. 686) βδ +(α +β )(ρ +δ ) ̸= 0 G(x, t ) =







ex−t (β cos t − (α + β ) sin t)((ρ + δ ) cos x + δ sin x)
βδ + ( α + β )(ρ + δ , 0 ≤ t ≤ x,
ex−t (β cos x − (α + β ) sin x)((ρ + δ ) cos t + δ sin t)",Elementary Differential Equations with Boundary Value Problems.pdf
"ex−t (β cos t − (α + β ) sin t)((ρ + δ ) cos x + δ sin x)
βδ + ( α + β )(ρ + δ , 0 ≤ t ≤ x,
ex−t (β cos x − (α + β ) sin x)((ρ + δ ) cos t + δ sin t)
βδ + ( α + β )(ρ + δ , x ≤ t ≤ π/ 2
13. 1. 31 (p. 686) (ρ + δ )(α − β )e(b−a) − (ρ − δ )(α + β )e(a−b) ̸= 0
G(x, t ) =







((α − β )e(t−a) − (α + β )e−(t−a))((ρ − δ )e(x−b) − (ρ + δ )e−(x−b))
2[(ρ + δ )(α − β )e(b−a) − (ρ − δ )(α + β )e(a−b)] , 0 ≤ t ≤ x,
((α − β )e(x−a) − (α + β )e−(x−a))((ρ − δ )e(t−b) − (ρ + δ )e−(t−b))
2[(ρ + δ )(α − β )e(b−a) − (ρ − δ )(α + β )e(a−b)] x ≤ t ≤ π
Section 13.2 Answers, pp. 696– 700
13. 2. 1 (p. 696) (ebxy′)′+cebxy = 0 13. 2. 2 (p. 696) (xy′)′+
(
x − ν 2
x
)
y = 0 13. 2. 3 (p. 696) (
√
1 − x2y′)′+ α 2
√
1 − x2 y = 0
13. 2. 4 (p. 696) (xby′)′+cxb−2y = 0 13. 2. 5 (p. 696) (e−x2
y′)′+ 2αe −x2
y = 0 13. 2. 6 (p. 696) (xe−xy′)′+
αe −xy = 0
13. 2. 7 (p. 696) ((1 − x2)y′)′+ α (α + 1) y = 0
13. 2. 9 (p. 696) λ n = n2 π 2, yn = e−x sin nπx (n = positive integer)",Elementary Differential Equations with Boundary Value Problems.pdf
"αe −xy = 0
13. 2. 7 (p. 696) ((1 − x2)y′)′+ α (α + 1) y = 0
13. 2. 9 (p. 696) λ n = n2 π 2, yn = e−x sin nπx (n = positive integer)
13. 2. 10 (p. 697) λ 0 = −1, y0 = 1 λ n = n2π 2 , yn = e−x(nπ cos nπx + sin nπx ) (n = positive
integer)
13. 2. 11 (p. 697) (a) λ = 0 is an eigenvalue y0 = 2 − x (b) none (c) 5. 0476821, 14. 9198790,
29. 7249673, 49. 4644528 y = 2
√
λ cos
√
λ x − sin
√
λx
13. 2. 12 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −0. 5955245 y = cosh
√
−λ x (c) 8. 8511386,
38. 4741053, 87. 8245457, 156. 9126094 y = cos
√
λx
13. 2. 13 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) none (c) 0. 1470328, 1. 4852833, 4. 5761411,
9. 6059439 y =
√
λ cos
√
λx + sin
√
λx
13. 2. 14 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −0. 1945921
y = 2
√
−λ cosh
√
−λ x − sinh
√
−λ x (c) 1. 9323619, 5. 9318981, 11. 9317920,
19. 9317507 y = 2
√
λ cos
√
λ x − sin
√
λ x",Elementary Differential Equations with Boundary Value Problems.pdf
"Answers to Selected Exercises 785
13. 2. 15 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −1. 0664054 y = cosh
√
−λ x (c) 1. 5113188,
8. 8785880, 21. 2104662, 38. 4805610 y = cos
√
λ x
13. 2. 16 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −1. 0239346
y =
√
−λ cosh
√
−λ x − sinh
√
−λ x (c) 2. 0565705, 9. 3927144, 21. 7169130,
38. 9842177 y =
√
λ cos
√
λ x − sin
√
λ x
13. 2. 17 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −0. 4357577,
y = 2
√
−λ cosh
√
−λ x − sinh
√
−λ x (c) 0. 3171423, 3. 7055350, 9. 1970150,
16. 8760401 y = 2
√
λ cos
√
λ x − sin
√
λ x
13. 2. 18 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −2. 1790546, −9. 0006633
y =
√
−λ cosh
√
−λ x − 3 sinh
√
−λ x
(c) 5. 8453181, 17. 9260967, 35. 1038567, 57. 2659330 y =
√
λ cos
√
λ x − 3 sin
√
λ x
13. 2. 19 (p. 697) (a) λ = 0 is an eigenvalue y0 = 2 − x (b) −1. 0273046
y = 2
√
−λ cosh
√
−λ x − sinh
√
−λ x (c) 8. 8694608, 16. 5459202, 26. 4155505,
38. 4784094 y = 2
√
λ cos
√
λ x − sin
√
λ x",Elementary Differential Equations with Boundary Value Problems.pdf
"y = 2
√
−λ cosh
√
−λ x − sinh
√
−λ x (c) 8. 8694608, 16. 5459202, 26. 4155505,
38. 4784094 y = 2
√
λ cos
√
λ x − sin
√
λ x
13. 2. 20 (p. 697) (a) λ = 0 isn’t an eigenvalue (b) −7. 9394171, −3. 1542806
y = 2
√
−λ cosh
√
−λ x − 5 sinh
√
−λ x (c) 29. 3617465, 78. 777456, 147. 8866417,
236. 7229622 y = 2
√
λ cos
√
λ x − 5 sin
√
λ x
13. 2. 21 (p. 697) λ = 0 , y = xe−x 20. 1907286, 118. 8998692, 296. 5544121, 553. 1646458
y = e−x sin
√
λ x
13. 2. 22 (p. 697) λ n = n2 π 2 , yn = x sin nπ (x − 2) (n = positive integer)
13. 2. 23 (p. 697) λ = 0 , y = x(2 − x) 20 . 1907286, 118. 8998692, 296. 5544121
553. 1646458, y = x sin
√
λ (x − 2)
13. 2. 24 (p. 698) 3. 3730893, 23. 1923372, 62. 6797232, 121. 8999231, 200. 8578309
y = x sin
√
λ (x − 1)
13. 2. 25 (p. 698) (a) −L < δ < 0 (b) δ = −L
13. 2. 26 (p. 698) λ 0 = −1/α 2 y0 = e−x/α λ n = n2 , yn = nα cos nx − sin nx, n = 1 , 2, . . .
13. 2. 27 (p. 698) (a) y = x − α (b) y = αk cosh kx − sin kx (c) y = αk cos kx − sin kx",Elementary Differential Equations with Boundary Value Problems.pdf
"13. 2. 27 (p. 698) (a) y = x − α (b) y = αk cosh kx − sin kx (c) y = αk cos kx − sin kx
13. 2. 29 (p. 698) (b) λ = −α 2/β 2 y = e−αx/β",Elementary Differential Equations with Boundary Value Problems.pdf
"Index
A
Abel’ s formula,
199– 202, 468
Accelerated payment, 139
Acceleration due to gravity, 151
Airy’ s equation, 319
Amplitude,
of oscillation, 271
time-varying, 279
Amplitude–phase form, 272
Aphelion distance, 300
Apogee, 300
Applications,
of first order equations, 130– 192
autonomous second order equations, 162– 179
cooling problems, 140– 141
curves, 179– 192
elementary mechanics, 151– 177
growth and decay, 130– 140
mixing problems, 143– 150
of linear second order equations, 268– 302
motion under a central force, 295– 302
motion under inverse square law force, 299–
301
RLC circuit, 289– 295
spring–mass systems, 268– 289
Autonomous second order equations, 162– 183
conversion to first order equations, 162
damped 173– 178
pendulum 174
spring–mass system, 173
Newton’ s second law of motion and, 163
undamped 164– 172
pendulum 173– 169
spring–mass system, 164– 173
stability and instability conditions for, 170– 178
B
Beat,
275
Bernoulli’ s equation, 63– 64
Bessel functions of order ν , 360",Elementary Differential Equations with Boundary Value Problems.pdf
"stability and instability conditions for, 170– 178
B
Beat,
275
Bernoulli’ s equation, 63– 64
Bessel functions of order ν , 360
Bessel’ s equation, 205 287, 348
of order ν , 360
of order zero, 377
ordinary point of, 319
singular point of, 319, 342
Bifurcation value, 54, 176
Birth rate, 2
Boundary conditions, 580
in heat equation, 618
Laplace equation, 649– 651
periodic, 580
separated, 676
for two-point boundary value problems, 676
Boundary points, 676
Boundary value problems, 618
initial-, 618
mixed, 649
two-point, 676– 686
assumptions, 676
boundary conditions for, 676
defined, 676
Green’ s functions for, 681
homogeneous and nonhomogeneous, 676
orthogonality and, 583
Capacitance, 290
Capacitor, 290
Carbon dating, 136
Central force,
motion under a, 295– 302
in terms of polar coordinates,
Characteristic equation, 210
with complex conjugate roots, 214– 217
with disinct real roots, 211– 217
with repeated real root, 212, 217
Characteristic polynomial, 210, 340, 475
Charge, 290",Elementary Differential Equations with Boundary Value Problems.pdf
"with disinct real roots, 211– 217
with repeated real root, 212, 217
Characteristic polynomial, 210, 340, 475
Charge, 290
steady state, 293
Chebyshev polynomials, 322
Chebshev’ s equation, 322
787",Elementary Differential Equations with Boundary Value Problems.pdf
"788 Index
Circuit, RLC. See RLC circuit
Closed Circuit, 289
Coefficient(s) See also Constant coefficient equations
computing recursively, 322
Fourier, 587
in Frobenius solutions, 352– 358
undetermined, method of, 229– 248, 475– 496
principle of superposition and, 235
Coefficient matrix, 516, 516
Competition, species, 6, 541
Complementary equation, 35, 469
Complementary system, 568
Compound interest, continuous, 131, 134
Constant,
damping, 173
decay, 130
separation, 619
spring, 268
temperature decay, 140
Constant coefficient equations, 210, 475
homogeneous, 210– 221
with complex conjugate roots, 214– 217
with distinct real roots, 211, 217
higher order . See Higher order constant coef-
ficient homogeneous equations
with repeated real roots, 212, 217
with impulses, 452– 460
nonhomogeneous, 229– 248
with piecewise continuous forcing functions, 430–
439
Constant coefficient homogeneous linear systems of
differential equations, 529– 568
geometric properties of solutions,",Elementary Differential Equations with Boundary Value Problems.pdf
"439
Constant coefficient homogeneous linear systems of
differential equations, 529– 568
geometric properties of solutions,
when n = 2 , 536– 539, 551– 554, 562– 565
with complex eigenvalue of constant matrix, 556–
565
with defective constant matrix, 558– 556
with linearly independent eigenvetors, 529– 541
Constant solutions of separable first order equations,
48– 53
Converge absolutely, 307
Convergence,
of improper integral, 393
open interval of, 306
radius of, 306
Convergent infinite series, 620
Convergent power series, 306
Convolution, 440– 452
convolution integral, 445– 450
defined, 441
theorem, 441
transfer functions, 446– 448
V olterra integral equation, 445
Cooling, Newton’ s law of, 3, 140
Cooling problems, 140– 141, 148– 148
Cosine series, Fourier, 603– 604
mixed, 606– 608
Critically damped motion, 280– 281
oscillation, 291– 294
Critical point, 163
Current, 289
steady state, 293
transient, 293
Curves, 179– 192
equipotential, 185
geometric problems, 183
isothermal, 185",Elementary Differential Equations with Boundary Value Problems.pdf
"Critical point, 163
Current, 289
steady state, 293
transient, 293
Curves, 179– 192
equipotential, 185
geometric problems, 183
isothermal, 185
one-parameter famlies of, 179– 183 subsubitem
defined, 179
differential equation for, 180
orthogonal trajectories, 190– 190, 192
finding, 186– 190
D
D’Alembert’ s solution,
637
Damped autonomous second order equations, 172–
179
for pendulum, 174
for spring-mass system, 173
Damped motion, 268
Damping,
RLC circuit in forced oscllation with, 293
spring-mass systems with, 173, 268, 279– 288
critically damped motion, 280– 283
forced vibrations, 283– 287
free vibrations, 279– 283
overdamped motion, 279
underdamped motion, 279
spring-mass systems without, 269– 277
forced oscillation, 273– 277
Damping constant, 173
Damping forces, 163, 268
Dashpot, 268
Dating, carbon, 135– 136
Death rate 3
Decay, See Exponential growth and decay,
Decay constant, 130
Derivatives, Laplace transform of, 413– 415
Differential equations,
defined, 8
order of, 8
ordinary, 8",Elementary Differential Equations with Boundary Value Problems.pdf
"Decay constant, 130
Derivatives, Laplace transform of, 413– 415
Differential equations,
defined, 8
order of, 8
ordinary, 8
partial, 8
solutions of, 9– 11
Differentiation of power series, 308
Dirac, Paul A. M., 452",Elementary Differential Equations with Boundary Value Problems.pdf
"Index 789
Dirac delta function, 452
Direction fields for first order equations, 16– 27
Dirichlet, Peter G. L., 662
Dirichlet condition, 662
Dirichlet problem, 662
Discontinuity,
jump, 398
removable, 408
Distributions, theory of, 453
Divergence of improper integral, 393
Divergent power series, 306
E
Eccentricity of orbit,
300
Eigenfunction associated with λ , 580, 689
Eigenvalue, 580, 689
Eigenvalue problems, See also Boundary value prob-
lems
Sturm-Liouville problems, 686– 700
defined, 689
orthogonality in, 695
solving, 688
Elliptic orbit, 300
Epidemics 5– 53
Equidimensional equation, 486
Equilibrium, 163
spring-mass system, 268
Equilibrium position, 268
Equipotentials, 185
Error(s),
in applying numerical methods, 96
in Euler’ s method, 97– 102
at the i-th step, 96
truncation, 96
global, 102, 111, 119
local, 100
local, numerical methods with O(h3 ), 114–
116
Escape velocity, 159
Euler’ s equation, 343– 346, 229
Euler’ s identity, 96– 108
Euler’ s method, 96– 108
error in, 97– 100",Elementary Differential Equations with Boundary Value Problems.pdf
"116
Escape velocity, 159
Euler’ s equation, 343– 346, 229
Euler’ s identity, 96– 108
Euler’ s method, 96– 108
error in, 97– 100
truncation, 100– 102
improved, 109– 114
semilinear, 102– 106
step size and accuracy of, 97
Even functions, 592
Exact first order equations, 73– 82
implicit solutions of, 73– 74
procedurs for solving, 76
Exactness condition, 75
Existence of solutions of nonlinear first order equa-
tions, 56– 62
Existence theorem, 40, 56
Exponential growth and decay, 130– 140
carbon dating, 136
interest compounded continuously, 134
mixed growth and decay, 134
radioactive decay, 130
savings program, 136
Exponential order, function of, 400
F
First order equations,
31– 93
applications of See under Applications.
autonomous second order equation converted to,
162
direction fields for, 16– 20
exact, 73– 82
implicit solution of, 73
procedurs for solving, 76
linear, 31– 44
homogeneous, 106– 35
nonhomogeneous, 35– 41
solutions of, 30
nonlinear, 41, 52, 56– 72",Elementary Differential Equations with Boundary Value Problems.pdf
"procedurs for solving, 76
linear, 31– 44
homogeneous, 106– 35
nonhomogeneous, 35– 41
solutions of, 30
nonlinear, 41, 52, 56– 72
existence and uniqueness of solutions of, 56–
62
transformation into separables, 62– 72
numerical methods for solving. See Numerical
method
separable, 45– 55, 68– 72
constant solutions of, 48– 50
implicit solutions of, 47– 48
First order systems of equations,
higher order systems written as, 511
scalar differential equations written as, 512
First shifting theorem, 397
Force(s)
damping, 163, 268
gravitational, 151, 158
impulsive, 453
lines of, 185
motion under central, 296– 302
motion under inverse square law, 299– 302
Forced motion, 268
oscillation
damped, 293– 294
undamped, 273– 277
vibrations, 283– 287
Forcing function, 194
without exponential factors, 516, 516 487 – 494
with exponential factors, 244– 244",Elementary Differential Equations with Boundary Value Problems.pdf
"790 Index
piecewise continuous constant equations with,
430– 439
Fourier coefficients, 587
Fourier series, 586– 616
convergence of, 589
cosine series, 603– 604
convergence of, 607
mixed, 606
defined 588
even and odd functions, 592– 589
sine series, 605
convergence of, 605
mixed, 609
Fourier solutions of partial differential equations, 618–
673
heat equation, 618– 629
boundary conditions in, 618
defined, 618
formal and actual solutions of, 620
initial-boundary value problem, 618– 625
initial condition, 618
nonhomogeneous problems, 623– 625
separation of variables to solve, 618– 620
Laplace’ s equation, 649– 673
boundary conditions, 649– 662
defined, 649
formal solutions of, 651– 662
in polar coordinates, 666– 673
for semiinfinite strip, 660
Free fall under constant gravity, 13
Free motion, 268
oscillation, RLC circuit in, 291– 292
vibrations, 279– 283
Frequency, 279
of simple harmonic motion, 292
Frobenius solutions, 347– 390
indicial equation with distinct real roots differ-",Elementary Differential Equations with Boundary Value Problems.pdf
"vibrations, 279– 283
Frequency, 279
of simple harmonic motion, 292
Frobenius solutions, 347– 390
indicial equation with distinct real roots differ-
ing by an integer, 378– 390
indicial equation with distinct real roots not dif-
fering by an integer, 351– 364
indicial equation with repeated root, 364– 378
power series in, 348
recurrence relationship in, 350
two term, 353– 355
verifying, 357
Function(s)
even and odd, 592
piecewise smooth, 588
Fundamental matrix, 524
Fundamental set of solutions, of higher order constant
coefficient homogeneous equations, 480–
482
of homogeneous linear second order equations,
198, 202
of homogeneous linear systems of differential
equations, 522, 524
of linear higher order equations, 466
G
Gamma function,
403
Generalized Riccati equation, 72, 255
General solution
of higher order constant coefficient homogeneous
equations, 475– 480
of homogeneous linear second order equations,
198
of homogeneous linear systems of differential
equations, 521, 524",Elementary Differential Equations with Boundary Value Problems.pdf
"equations, 475– 480
of homogeneous linear second order equations,
198
of homogeneous linear systems of differential
equations, 521, 524
of linear higher order equations, 465, 469
of nonhomogeneous linear first order equations,
30, 40
of nonhomogeneous linear second order equa-
tions, 221, 248– 255
Geometric problems, 235– 184
Gibbs phenomenon, 589, 597
Global truncation error in Euler’ s method, 102
Glucose absorption by the body, 5
Gravitation, Newton’ s law of, 151, 176, 295, 510, 515
Gravity, acceleration due to, 151
Green’ s function, 504, 681, 682, 685– 686
Grid, rectangular, 17
Growth and decay,
carbon dating, 136
exponential, 130– 140
interest compounded continuously, 133– 134
mixed growth and decay, 134
radioactive decay, 130
savings program, 136
H
Half-life,
130
Half-line, 536
Half-plane, 552
Harmonic conjugate function, 82
Harmonic function, 82
Harmonic motion, simple, 166, 271 292 , 292
amplitude of oscillation, 272
natural frequancy of, 272
phase angle of, 272",Elementary Differential Equations with Boundary Value Problems.pdf
"Harmonic function, 82
Harmonic motion, simple, 166, 271 292 , 292
amplitude of oscillation, 272
natural frequancy of, 272
phase angle of, 272
Heat equation, 618– 629
boundary conditions in, 618
defined, 618
formal and actual solutions of, 620
initial-boundary value problems, 618, 625
initial condition, 618
nonhomogeneous problems, 623– 625",Elementary Differential Equations with Boundary Value Problems.pdf
"Index 791
separation of variables to solve, 618
Heat flow lines, 185
Heaviside’ s method, 407, 412
Hermite’ s equation, 322
Heun’ s method, 116
Higher order constant coefficient homogeneous equa-
tions, 475– 487
characteristic polynomial of, 482– 478
fundamental sets of solutions of, 480
general solution of, 476– 482
Homogeneous linear first order equations, 30– 34
general solutions of, 33
separation of variables in, 35
Homogeneous linear higher order equations, 465
Homogeneous linear second order equations, 194– 221
constant coefficient, 210– 221
with complex conjugate roots, 214– 217
with distinct real roots, 210– 217
with repeated real roots, 210– 214, 217
solutions of, 194, 198
the Wronskian and Abel’ s formula, 199– 202
Homogeneous linear systems of differential equations,
516
basic theory of, 521– 528
constant coefficient, 529– 568
with complex eigenvalues of coefficient ma-
trix, 556– 568
with defective coefficient matrix, 542– 551
geometric properties of solutions when n =",Elementary Differential Equations with Boundary Value Problems.pdf
"with complex eigenvalues of coefficient ma-
trix, 556– 568
with defective coefficient matrix, 542– 551
geometric properties of solutions when n =
2, 529– 539, 551– 554, 562– 565
with linearly independent eigenvectors, 529–
539 subitem fundamental set of solutions
of, 521, 524
general solution of, 521, 524
trivial and nontrivial solution of, 521
Wronskian of solution set of, 523
Homogeneous nonlinear equations
defined, 65
transformation into separable equations, 65– 68
Hooke’ s law, 268– 269
I
Imaginary part,
215
Implicit function theorem, 47
Implicit solution(s) 73– 74
of exact first order equations, 73– 74
of initial value problems, 47
of separable first order equations, 47– 49
Impressed voltage, 53
Improper integral, 393
Improved Euler method, 108– 112 120 – 122
semilinear, 112– 114
Impulse function, 452
Impulse response, 448, 455
Impulses, constant coefficient equations with, 452–
461
Independence, linear
of n function, 466
of two functions, 198
of vector functions, 525",Elementary Differential Equations with Boundary Value Problems.pdf
"Impulses, constant coefficient equations with, 452–
461
Independence, linear
of n function, 466
of two functions, 198
of vector functions, 525
Indicial equation, 343, 351
with distinct real roots differing by an integer,
378– 390
with distinct real roots not differing by an inte-
ger, 351– 364
with repeated root, 364– 378
Indicial polynomial, 343, 351
Inductance, 290
Infinite series, convergent, 620
Initial-boundary value problem, 618
heat equation, 618– 625
wave equation, 630– 649
Initial conditions, 11
Initial value problems, 11– 14
implicit solution of, 47
Laplace transforms to solve, 413– 419
formula for, 443– 444
second order equations, 415– 419
Integral curves, 9– 9, 413– 27,
Integrals,
convolution, 445– 445
improper, 393
Integrating factors, 82– 93
finding, 83– 93
Interest compounded continuously, 131– 134
Interval of validity, 12
Inverse Laplace transforms, 404– 413
defined, 404
linearity property of, 405
of rational functions, 406– 413",Elementary Differential Equations with Boundary Value Problems.pdf
"Interval of validity, 12
Inverse Laplace transforms, 404– 413
defined, 404
linearity property of, 405
of rational functions, 406– 413
Inverse square law force, motion under, 299– 301
Irregular singular point, 342
Isothermal curves, 185
J
Jump discontinuity,
398
K
Kepler’ s second law,
296
Kepler’ s third law, 301
Kirchoff’ s Law, 290
L
Laguerre’ s equation,
348
Lambda-eigenfunctions, 580, 687
Laplace’ s equation, 649– 673",Elementary Differential Equations with Boundary Value Problems.pdf
"792 Index
boundary conditions, 649– 651
defined, 649
formal solutions of, 651– 662
in polar coordinates, 666– 673
for semi-infinte strip, 660
Laplace transforms, 393– 461
computation of simple, 393– 396
of constant coefficient equations
with impulses 452– 461
with piecewise continuous forcing functions,
430– 439
convolution, 440– 452
convolution integral, 445
defined, 441
theorem, 441
transfer functions, 446– 448
definition of, 393
existence of, 398
First shifting theorem, 397
inverse, 404
defined, 403
linearity property of, 405
of rational functons, 406– 411
linearity of, 396
of piecewise continuous functions, 421– 430
unit step function and, 419– 430
Second shifting theorem, 425
to solve initial value problems, 413– 419
derivatives , 413– 415
formula for, 443– 444
second order equations, 415
tables of, 396
Legendre’ s equation, 204, 319
ordinary points of, 319
singular points of, 319, 347
Limit, 398
Limit cycle, 176
Linear combination(s), 198, 465, 521
of power series, 313– 316",Elementary Differential Equations with Boundary Value Problems.pdf
"ordinary points of, 319
singular points of, 319, 347
Limit, 398
Limit cycle, 176
Linear combination(s), 198, 465, 521
of power series, 313– 316
Linear difference equations, second order homoge-
neous, 340
Linear first order equations, 30– 44
homogeneous, 30– 35
general solution of, 33
separation of variables, 35
nonhomogeneous, 30, 35– 41
general solution of, 35– 41
solutions in integral form, 38– 39
variation of parameters to solve, 35, 38
solutions of, 30– 31
Linear higher order equations, 465– 505
fundamental set of solutions of, 465, 466
general solution of, 465, 469
higher order constant coefficient homogeneous
equations, 475– 487 characteristic polyomial
of 475– 480
fundamental sets of solutions of, 479– 482
general solution of, 476– 478
homogeneous, 465
nonhomogeneous, 465, 469
trivial and nontrivial solutions of, 465
undetermined coefficients for, 487– 496
variation of parameters for, 497– 505
derivation of method, 497– 499
fourth order equations, 501– 497",Elementary Differential Equations with Boundary Value Problems.pdf
"undetermined coefficients for, 487– 496
variation of parameters for, 497– 505
derivation of method, 497– 499
fourth order equations, 501– 497
third order equations, 499
Wronskian of solutions of 467– 469
Linear independence 198
of n functions, 466
of two functions, 198
of vector functions, 521– 523
Linearity,
of inverse Laplace transform, 405
of Laplace transform, 396
Linear second order equations, 194– 264
applications of. See under Applications
defined, 194
homogeneous, 194– 221
constant coefficient, 210– 201
solutions of, 194– 198
the Wronskian and Abel’ s formula, 199– 202
nonhomnogeneous, 194, 221– 264, 465, 469
comparison of methods for solving, 203
complementary equation for, 221
constant coefficient, 229– 248
general solution of, 221– 225
particular solution of, 221, 225– 227
reduction of order to find general solution of,
248– 255
superposition principle and, 225– 227
undetermined coefficients method for, 229–
248
variation of parameters to find particular so-
lution of, 255– 264",Elementary Differential Equations with Boundary Value Problems.pdf
"superposition principle and, 225– 227
undetermined coefficients method for, 229–
248
variation of parameters to find particular so-
lution of, 255– 264
series solutions of, 306– 390
Euler’ s equation, 343– 347
Frobenius solutions, 347– 390
near an ordinary point, 319– 339
with regular singular points, 342– 347
Linear systems of differential equations, 515– 577
defined, 515
homogeneous, 515
basic theory of, 521– 529
constant coefficient, 529– 568
fundamental set of solutions of, 521– 524
general solution of, 521, 524",Elementary Differential Equations with Boundary Value Problems.pdf
"Index 793
linear indeopendence of, 521, 524
trivial and nontrivial solution of, 521
Wronskian of solution set of, 523
nonhomogeneous, 516
variation of parameters for, 568– 576
solutions to initial value problem, 515– 517
Lines of force, 185
Liouville, Joseph, 689
local truncation error, 100– 102
numerical methods with O(h3), 114-116
Logistic equation 3
M
Maclaurin series,
308
Magnitude of acceleration due to gravity at Earth’ s
surface, 151
Malthusian model, 2
Mathematical models, 2
validity of, 137, 140, 149
Matrix/matrices, 516– 518
coefficient,
complex eigenvalue of, 557– 568
defective, 542
fundamental, 524
Mechanics, elementary, 151– 178
escape velocity, 158– 159 162 , 162
motion through resisting medium under constant
gravitational force, 151– 157
Newton’ s second law of motion, 151– 151
pendulum motion
damped, 174– 176
undamped, 173– 169
spring-mass system
damped, 173– 174, 269, 279– 289
undamped, 164– 173,268
units used in, 151
Midpoint method, 109",Elementary Differential Equations with Boundary Value Problems.pdf
"damped, 174– 176
undamped, 173– 169
spring-mass system
damped, 173– 174, 269, 279– 289
undamped, 164– 173,268
units used in, 151
Midpoint method, 109
Mixed boundary value problems, 649
Mixed Fourier cosine series, 606– 608
Mixed Fourier sine series, 609
Mixed growth and decay, 134
Mixing problems, 143– 148
Models, mathematical, 2– 3
validity of, 137, 140, 149
Motion,
damped, 268
critically, 280
overdamped, 279– 280
underdamped, 279
elementary, See Mechanics, elementary
equation of, 269
forced, 270
free, 270
Newton’ s second law of, 6, 151– 151, 163, 165,
173, 173– 174, 268, 296, 510
autonomous second order equations and, 163
simple harmonic, 166, 269– 273
amplitude of oscillation, 271
frequency of, 272
phase angle of, 271
through resisting medium under constant gravi-
tational force, 152– 157
under a central force, 295– 302
under inverse square law force, 299– 301
undamped, 268
Multiplicity, 479
N
Natural frequency,
272
Natural length of spring, 268
Negative half plane, 552",Elementary Differential Equations with Boundary Value Problems.pdf
"undamped, 268
Multiplicity, 479
N
Natural frequency,
272
Natural length of spring, 268
Negative half plane, 552
Neumann condition, 649
Neumann problem, 649
Newton’ s law of cooling, 3, 140– 141, 148– 150
Newton’ s law of gravitation, 151, 176, 295, 510, 519
Newton’ s second law of motion, 151– 151, 163, 166,
173, 176, 268, 296, 509, 510
autonomous second order equations and, 163
Nonhomogeneous linear second order equations, 30,
35, 41
general solution of, 35– 38 40 – 41
solutions in integral form, 38
variation of parameters to solve, 35, 38
Nonhomogeneous linear second order equations, 194,
221– 264
comparison of methods for solving, 262
complementary equation for, 221, 221
constant coefficient, 229– 255
general solution of, 221– 225
particular solution of, 221, 221– 226, 229– 235,
255– 262
reduction of order to find general solution of,
248– 255
superposition principle and, 225– 223
undetermined coefficients method for, 229– 255
forcing functions with exponential factors, 242–
244",Elementary Differential Equations with Boundary Value Problems.pdf
"248– 255
superposition principle and, 225– 223
undetermined coefficients method for, 229– 255
forcing functions with exponential factors, 242–
244
forcing functions without exponential factors,
238– 241
superposition principle and, 235
variation of parameters to find particular so-
lution of, 255– 264
Nonhomogeneous linear systems of differential equa-
tions, 515
variation of parameters for, 568– 577
Nonlinear first order equations, 52 56 – 72",Elementary Differential Equations with Boundary Value Problems.pdf
"794 Index
existence and uniqueness of solutions of, 56– 72
transformation into separable equations, 62– 72
Nonoscillatory solution, 358
Nontrivial solutions
of homogeneous linear first order equations, 30
of homogeneous linear higher order equations,
465
of homogeneous linear second order equations,
194
of homogeneous linear systems of differential
equations, 521
Numerical methods, 96– 127, 514
with O(h3 ) local truncation, 114– 116
error in, 96
Euler’ s method, 96– 108
error in, 97– 102
semilinear, 102– 106
step size and accuracy of, 97
truncation error in, 99– 102
Heun’ s method, 115
semilinear, 106
improved Euler method, 106, 109– 112
semilinear, 112
midpoint, 116
Runge-Kutta method, 98, 106 119 – 127, 513–
514
for cases where x0 isn’t the left endpoint, 122–
124
semilinear, 106, 122
for systems of differential equations, 514
Numerical quadrature, 119, 127
O
Odd functions,
592
One-parameter families of curves, 179– 183
defined, 179
differential equation for, 180",Elementary Differential Equations with Boundary Value Problems.pdf
"Numerical quadrature, 119, 127
O
Odd functions,
592
One-parameter families of curves, 179– 183
defined, 179
differential equation for, 180
One-parameter families of functions, 30
Open interval of convergence, 306
Open rectangle, 56
Orbit, 301
eccentricity of, 300
elliptic, 300
period of, 301
Order of differential equation, 8
Ordinary differential equation,
defined, 8
Ordinary point, series solutions of linear second order
equations near, 319– 341
Orthogonality, 583– 581
in Sturm-Liouville problems, 695
Orthogonal trajectories, 186– 190,
finding, 186
Orthogonal with respect to a weighting function, 331,
331
Oscillation
amplitude of, 271
critically damped, 292
overdamped, 292
RLC circuit in forced, with damping, 293– 293
RLC circuit in free, 291– 293
undamped forced, 273– 277
underdamped, 291
Oscillatory solutions, 232– 176, 347
Overdamped motion, 279– 279
P
Partial differential equations
defined,
8
Fourier solutions of, 618– 673
heat equation, 618– 630
Laplace’ s equation, 649– 673",Elementary Differential Equations with Boundary Value Problems.pdf
"P
Partial differential equations
defined,
8
Fourier solutions of, 618– 673
heat equation, 618– 630
Laplace’ s equation, 649– 673
wave equation 630– 649
Partial fraction expansions, software packages to find,
411
Particular solutions of nonhomogeneous higher equa-
tions, 469, 487– 505
Particular solutions of nonhomogeneous linear sec-
ond order equations, 221, 225– 226, 229–
235, 255– 261
Particular solutions of nonhomogeneous linear sys-
tems equations, 568– 577
Pendulum
damped, 174– 176
undamped, 173– 169
Perigee, 300
Perihelion distance, 300
Periodic functions, 404
Period of orbit, 296
Phase angle of simple harmonic motion, 271– 272
Phase plane equivalent, 162
Piecewise continuous functions, 399
forcing, constant coeffocient equations with, 430–
439
Piecewise smooth function, 588
Laplace transforms of 398– 401, 421– 430
unit step functions and, 419– 430
Plucked string, wave equation applied to, 638– 642
Poinccaré, Henri, 162
Polar coordinates
central force in terms of, 296– 298",Elementary Differential Equations with Boundary Value Problems.pdf
"Plucked string, wave equation applied to, 638– 642
Poinccaré, Henri, 162
Polar coordinates
central force in terms of, 296– 298
in amplitude-phase form, 271
Laplace’ s equation in, 666– 673
Polynomial(s)
characteristic, 210, 340, 482",Elementary Differential Equations with Boundary Value Problems.pdf
"Index 795
of higher order constant coefficient homoge-
neous equations, 475– 478
Chebyshev, 322
indicial, 343, 351
T aylor, 309
trigonometric, 602
Polynomial operator, 475
Population growth and decay, 2
Positive half-plane, 552
Potential equation, 649
Power series, 306– 319
convergent, 306– 307
defined, 306
differentiation of, 308– 309
divergent, 306
linear combinations of, 313– 316
radius of convergence of, 306, 307
shifting summation index in, 310– 312
solutions of linear second order equations, rep-
resented by, 319– 341
T aylor polynomials, 309
T aylor series, 308
uniqueness of 309– 309
Q
Quasi-period,
279
R
Radioactive decay,
130– 131
Radius of convergence of power series, 306, 307
Rational functions, inverse Laplace transforms of, 406–
413
Rayleigh, Lord, 171
Rayleigh’ s equation, 177
Real part, 215
Rectangle, open, 56
Rectangular grid, 17
Recurrence relations, 322
in Frobenius solutions, 351
two term, 353– 355
Reduction of order, 213, 248– 255
Regular singular points, 342– 347",Elementary Differential Equations with Boundary Value Problems.pdf
"Recurrence relations, 322
in Frobenius solutions, 351
two term, 353– 355
Reduction of order, 213, 248– 255
Regular singular points, 342– 347
at x0 = 0 , 347– 364
Removable discontinuity, 398
Resistance, 290
Resistor, 290
Resonance, 277
Ricatti, Jacopo Francesco, 72
Ricatti equation, 72
RLC circuit, 289– 294
closed, 289
in forced oscillation with danping, 293
in free oscillation, 291– 293
Roundoff errors, 96
Runge-Kutta method, 98, 119– 127, 514
for cases where x0 isn’t the left endpoint, 122
for linear systems of differential equations, 514
semilinear, 106, 122
S
Savings program, growth of,
136
Scalar differential equations, 512
Second order differential equation, 6
autonomous, 162– 178
conversion to first order equation, 162
damped, 173– 178
Newton’ s second law of mation and, 163
undamped, 164– 169
Laplace transform to solve, 415– 418
linear, See linear second equations
two-point boundary value problems for, 676–
686
assumptions, 676
boundary conditions for, 676
defined, 676",Elementary Differential Equations with Boundary Value Problems.pdf
"linear, See linear second equations
two-point boundary value problems for, 676–
686
assumptions, 676
boundary conditions for, 676
defined, 676
Green’ s function for, 682
homogeneous and nonhomogeneous, 676
procedure for solving, 677
Second order homogeneous linear difference equa-
tion, 340
Second shifting Theorem, 425– 427
Semilinear Euler method, 102
Semilinear improved Euler method, 106, 112
Semilinear Runge-Kutta method, 108, 124
Separable first order equations, 45– 55
constant solutions of, 48– 50
implicit solutions, 47
transfomations of nonlinear equations to, 62– 63
Bernoulli’ s equation, 62– 68
homogeneous nonlinear equations, 65– 68
other equations, 64
Separated boundary conditions, 676
Separation constant, 619
Separation of variables, 35, 45
to solve heat equation, 618
to solve Laplace’ s equation, 651– 662, 667– 673
to solve wave equation, 632
Separatrix, 171, 170
Series, power . See Power series
Series solution of linear second order equations, 306-
390",Elementary Differential Equations with Boundary Value Problems.pdf
"to solve wave equation, 632
Separatrix, 171, 170
Series, power . See Power series
Series solution of linear second order equations, 306-
390
Frobenius solutions, 347– 390
near an ordinary point, 319
Shadow trajectory, 564– 565
Shifting theorem
first, 397",Elementary Differential Equations with Boundary Value Problems.pdf
"796 Index
second, 425– 427
Simple harmonic motion, 269– 273
amplitude of oscillation, 271
natural frequency of, 272
phase angle of, 272
Simpson’ s rule, 127
Singular point, 319
irregular, 342
regular, 342– 347
Solution(s), 9– 10 See also Frobenius solutions Non-
trivial solutions Series solutions of linear
second order equations Trivial solution
nonoscillatory,
358
oscillatory, 358
Solution curve, 9– 9
Species, interacting, 6, 540
Spring, natural length of, 268, 269
Spring constant, 268
Spring-mass systems, 268– 289
damped, 172, 269, 287– 289
critically damped motion, 287– 283
forced vibrations, 283– 287
free vibrations, 283– 285
overdamped motion, 279
underdamped motion, 279
in equilibrium, 268
simple harmonic motion, 269– 273
amplitude of oscillation, 272
natural frequency of, 272
phase angle of, 272
undamped, 164– 166, 269– 277
forced oscillation, 273– 287
Stability of equilibrium and critical point, 163– 164
Steady state, 135
Steady state charge, 293",Elementary Differential Equations with Boundary Value Problems.pdf
"forced oscillation, 273– 287
Stability of equilibrium and critical point, 163– 164
Steady state, 135
Steady state charge, 293
Steady state component, 285, 447
Steady state current, 293
String motion, wave equation applied to, 630– 638
plucked, 638– 642
vibrating, 630– 638
Sturm-Liouville equation, 689
Sturm-Liouville expansion, 696
Sturm-Liouville problems, 687– 700
defined, 687
orthogonality in, 695
solving, 687
Summation index in power series, 310– 312
Superposition, principle of, 44, 225, 235, 470
method of undetermine coefficients and, 235
Systems of differential equations, 507– 518 See also
Linear systems of differential equations
first order
higher order systems rewritten as, 322– 512
scalar differential equations rewritten as, 512
numerical solutions of, 514
two first order equations in two unknowns, 507–
510
T
T angent lines,
181
T aylor polynomials, 309
T aylor Series, 308
T emperature, Newton’ s law of cooling, 3 140 – 141,
148– 149
T emperature decay constant of the medium, 140",Elementary Differential Equations with Boundary Value Problems.pdf
"T aylor Series, 308
T emperature, Newton’ s law of cooling, 3 140 – 141,
148– 149
T emperature decay constant of the medium, 140
T erminal velocity, 152
Time-varying amplitude, 279
T otal impulse, 452
Trajectory(ies),
of autonomous second order equations, 162
orthogonal, 186– 190
finding, 186– 244
shadow, 564
of 2 × 2 systems, 536– 539, 551– 554, 562– 565
Transfer functions, 446
Transformation of nonlinear equations to separable
first order, equations, 62– 81
Bernoulli’ s equation, 63
homogeneous nonlinear equations, 65– 68
other equations, 64
Transform pair, 393
Transient current, 293
Transient components, 285, 447
Transient solutions, 292
Trapezoid rule, 119
Trivial solution,
of homogeneous linear first order equations, 30
of homogeneous linear second order equations,
194
of homogeneous linear systems of differential
equations, 521
of linear higher order differential equations, 465
Truncation error(s), 96
in Euler’ s method, 100
global, 102, 109
local, 100",Elementary Differential Equations with Boundary Value Problems.pdf
"equations, 521
of linear higher order differential equations, 465
Truncation error(s), 96
in Euler’ s method, 100
global, 102, 109
local, 100
numerical methods with O(h3 ), 114– 116
T wo-point boundary value problems, 676– 686
assumptions, 676
boundary conditions for, 676
defined, 676
Green’ s function for, 681
homogeneous and nonhomogeneous, 677
U",Elementary Differential Equations with Boundary Value Problems.pdf
"Index 797
Undamped autonomous second order equations, 164–
171
pendulum, 173– 169
spring-mass system, 164– 166
stability and instabilty conditions for, 170– 171
Undamped motion, 268
Underdamped motion, 279
Underdamped oscillation, 291
Undetermined coefficients
for linear higher order equations, 487– 496
forcing functions, 487– 494
for linear second order equations, 229– 248
principle of superposition, 235
Uniqueness of solutions of nonlinear first equations,
56– 62
Uniqueness theorem, 40, 56, 194, 465, 516
Unit step function, 422– 430
V
V alidity, interval of,
12
V andermonde, 484
V andermonde determinant, 484
van der Pol’ s equation, 176
V ariables, separation of, 35, 45
V ariation of parameters
for linear first order equations, 35
for linear higher order equations, 497– 505
derivation of method, 497– 498
fourth order equations, 501– 502
third order equations, 499
for linear higher second order equations, 255
for nonhomogeneous linear systems of differen-
tial equations, 568– 577",Elementary Differential Equations with Boundary Value Problems.pdf
"third order equations, 499
for linear higher second order equations, 255
for nonhomogeneous linear systems of differen-
tial equations, 568– 577
V elocity
escape, 158– 151
terminal, 152– 156
V erhulst, Pierre, 3
V erhulst model, 3, 27, 69
Vibrating strings, wave equation applied to, 630
Vibrations
forced, 283– 287
free, 279– 283
V oltage, impressed, 289
V oltage drop, 290
V olterra, Vito 445
V olterra integral equation, 445
W
W ave equation,
630– 649
defined, 630
plucked string, 637– 642
vibrating string, 630– 637
assumptions, 630
formal solution, 632– 637
W ave, traveling, 639
orthogonal with respect to, 587 331
Wronskian
of solutions of homogeneous linear systems of
differential equations, 523
of solutions of homogeneous second differential
equations, 199– 201
of solutions of homogeneous linear higher order
differential equations, 467– 469",Elementary Differential Equations with Boundary Value Problems.pdf